Research ArticleStable Numerical Evaluation of Finite HankelTransforms and Their Application
Manoj P Tripathi12 B P Singh3 and Om P Singh1
1 Department of Mathematical Sciences Indian Institute of Technology Banaras Hindu University Varanasi 221005 India2Department of Mathematics Udai Pratap Autonomous College Varanasi 221002 India3 Department of Mathematics IEC College of Engineering amp Technology Greater Noida 201306 India
Correspondence should be addressed to Om P Singh singhomgmailcom
Received 1 June 2014 Revised 21 September 2014 Accepted 23 September 2014 Published 13 November 2014
Academic Editor Baruch Cahlon
Copyright copy 2014 Manoj P Tripathi et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
A new stable algorithm based on hat functions for numerical evaluation of Hankel transform of order ] gt minus1 is proposed inthis paper The hat basis functions are used as a basis to expand a part of the integrand 119903119891(119903) appearing in the Hankel transformintegralThis leads to a very simple efficient and stable algorithm for the numerical evaluation of Hankel transformThe novelty ofour paper is that we give error and stability analysis of the algorithm and corroborate our theoretical findings by various numericalexperiments Finally an application of the proposed algorithm is given for solving the heat equation in an infinite cylinder with aradiation condition
1 Introduction
Classically the Hankel transform 119865] of order ] of a function119891(119903) is defined by
119865] (119901) = intinfin
0
119903119891 (119903) 119869] (119901119903) 119889119903 119901 gt 0 (1)
As the Hankel transform is self-reciprocal its inverse is givenby
119891 (119903) = int
infin
0
119901119865] (119901) 119869] (119901119903) 119889119901 (2)
where 119869] is the ]th order Bessel function of first kindThis form of Hankel transform (HT) has the advantage
of reducing to the Fourier sine or cosine transform when ] =plusmn12TheHankel transform arises naturally in the discussionof problems posed in cylindrical coordinates and hence as aresult of separation of variables involving Bessel functionsSo it has found wide range of applications related to theproblems inmathematical physics possessing axial symmetry[1] But analytical evaluations of 119865](119901) are rare so numericalmethods are important
The usual classical methods like Trapezoidal rule Cotesrule and so forth connected with replacing the integrandby sequence of polynomials have high accuracy if integrandis smooth But 119903119891(119903)119869](119901119903) and 119901119865](119901)119869](119901119903) are rapidlyoscillating functions for large 119903 and 119901 respectively
To overcome these difficulties two different techniquesare available in the literature The first is the fast Hankeltransform as proposed by Siegman [2] Here by substitutionand scaling the problem is transformed in the space of thelogarithmic coordinates and the fast Fourier transform inthat space But it involves the conventional errors arisingwhen a nonperiodic function is replaced by its periodicextension Moreover it is sensitive to the smoothness offunction in that space The second method is based on theuse of Filon quadrature philosophy [3] In Filon quadraturephilosophy the integrand is separated into the productof an (assumed) slowly varying component and a rapidlyoscillating component In the context of Hankel transformthe former is 119903119891(119903) and the latter is 119869](119901119903) But the errorassociated with Filon quadrature philosophy is appreciablefor 119901 lt 1There are several extrapolationmethods developedin the eighties In particular the papers by Levin and Sidi[4ndash7] are relevant Levin and Sidi in 1981 [4] developed very
Hindawi Publishing CorporationInternational Journal of AnalysisVolume 2014 Article ID 670562 11 pageshttpdxdoiorg1011552014670562
2 International Journal of Analysis
general class of extrapolation methods for various types ofinfinite integrals whether oscillatory or not Sidi in [5ndash7]emphasized oscillatory integral involving Fourier andHankelTransforms among other more general cases In additionthe subject is dealt with in great detail in a book by Sidi [8Chapter 5]
Later in 2004 Guizar-Sicairos and Gutierrez-Vega [9]obtained a powerful scheme to calculate the HT of order] ge 0 by extending the zero order HT algorithm of Yu etal [10] to higher orders Their algorithm is based on theorthogonality properties of Bessel functions Postnikov [11]proposed for the first time a novel and powerful method forcomputing zero and first order HT by using Haar waveletsRefining the idea of Postnikov [11] Singh et al [12 13]obtained three efficient algorithms for numerical evaluationof HT of order ] gt minus1 using linear Legendre multi-wavelets Legendre wavelets and rationalized Haar waveletswhich were shown to be superior to the other mentionedalgorithms Recently Singh et al [14] and Pandey et al[15] tested their proposed algorithms on some examplesfor their stability with respect to noise 120572120579
119894where 120579
119894is a
uniform random variable in [minus1 1] added to the input signal119891(119903) But none of the above mentioned algorithms wereanalyzed theoretically for stability and no error analysis wasprovided
The lack of error and stability analysis in the papers citedabove motivated us for the present work In the presentpaper we use hat basis functions for approximation of119903119891(119903) and replace it by its approximation in (1) therebygetting an efficient and stable algorithm for the numericalevaluation of the HT of order ] gt minus1 Error and stabilityanalysis of the algorithm are given in Sections 4 and 5respectively and further corroborated by the numericalexperiments performed on the test functions in Section 6Test functions with known analytic HT are used with ran-dom noise term 120572120579
119894added to the input field 119891(119903) where
120579119894is a uniform random variable with values in [minus1 1]
to illustrate the stability and efficiency of the proposedalgorithm In Section 7 the algorithm is applied for solvingthe heat equation in an infinite cylinder with a radiationcondition
2 Hat Functions and TheirAssociated Properties
Hat functions are defined on the domain [0 1] These arecontinuous functions with shape of hats when plotted ontwo-dimensional planes The interval [0 1] is divided into 119899subintervals [119894ℎ (119894+1)ℎ] 119894 = 0 1 2 119899minus1 of equal lengthsℎ where ℎ = 1119899 The hat functionrsquos family of first (119899 + 1) hatfunctions is defined as follows [16]
1205950(119905) =
ℎ minus 119905
ℎ
0 le 119905 lt ℎ
0 otherwise
120595119894(119905)
=
119905 minus (119894 minus 1) ℎ
ℎ
(119894 minus 1) ℎ le 119905 lt 119894ℎ
(119894 + 1) ℎ minus 119905
ℎ
119894ℎ le 119905 lt (119894 + 1) ℎ 119894 = 1 2 119899 minus 1
0 otherwise
120595119899(119905) =
119905 minus (1 minus ℎ)
ℎ
1 minus ℎ le 119905 le 1
0 otherwise(3)
From the definition of hat functions it is obvious that
120595119894(119896ℎ) =
1 119894 = 119896
0 119894 = 119896
(4)
The hat functions 120595119895(119905) are continuous linearly independent
and are in 1198712[0 1]A function 119891 isin 1198712[0 1]may be approximated as
119891 (119905) cong
119894=119899
sum
119894=0
119891119894120595119894(119905) = 119891
01205950(119905) + 119891
11205951(119905)
+ 11989121205952(119905) + sdot sdot sdot + 119891
119899120595119899(119905)
(5)
The important aspect of using extended hat functions inthe approximation of function 119891(119905) lies in the fact that thecoefficients 119891
119894in (5) are given by
119891119894= 119891 (119894ℎ) for 119894 = 0 1 2 119899 where ℎ = 1
119899
(6)
3 Method of Solution
To derive the algorithm we first assume that the domainspace of input signal 119891(119903) extends over a limited region 0 le119903 le 119877 From physical point of view this assumption isreasonable due to the fact that the input signal 119891(119903) whichrepresents the physical field either is zero or has an infinitelylong decaying tail outside a disc of finite radius 119877 Thereforeinmany practical applications either the input signal119891(119903) hasa compact support or for a given 120576 gt 0 there exists a positivereal 119877 such that | intinfin
119877
119903119891(119903)119869](119901119903)119889119903| lt 120576 which is the case if119891(119903) = 119900(119903
120582
) where 120582 lt minus32 as 119903 rarr infin Hence in eithercase from (1) we have
119865] (119901) = int119877
0
119903119891 (119903) 119869] (119901119903) 119889119903 (7)
By scaling (7) may be written as
119865] (119901) = int1
0
119903119891 (119903) 119869] (119901119903) 119889119903 (8)
which is known as finite Hankel transform (FHT) [14 15 17]The FHT is a good approximation of the HT given by (1)Writing 119903119891(119903) = 119892(119903) in (8) we get
119865] (119901) = int1
0
119892 (119903) 119869] (119901119903) 119889119903 (9)
International Journal of Analysis 3
Using (5) and (6) we may approximate 119892(119903) as
119892 (119903) cong
119899
sum
119894=0
119892 (119894ℎ) 120595119894(119903) (10)
From (10) (9) may be written as
119865] (119901) cong int1
0
(
119899
sum
119894=0
119892 (119894ℎ) 120595119894(119903)) 119869] (119901119903) 119889119903
=
119899
sum
119894=0
119892 (119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
(11)
Using (3) (11) may be expressed as
119865] (119901) cong 119892 (0) intℎ
0
(
ℎ minus 119903
ℎ
) 119869] (119901119903) 119889119903
+
119899minus1
sum
119894=1
119892 (119894ℎ) [int
119894ℎ
(119894minus1)ℎ
(
119903 minus (119894 minus 1) ℎ
ℎ
) 119869] (119901119903) 119889119903
+int
(119894+1)ℎ
119894ℎ
((119894 + 1) ℎ minus 119903
ℎ
) 119869] (119901119903) 119889119903]
+ 119892 (119899ℎ) int
1
1minusℎ
(
119903 minus (1 minus ℎ)
ℎ
) 119869] (119901119903) 119889119903
(12)
It is noteworthy here that the integrals involved in (12) areevaluated by using the following formulae
int
119886
0
119869] (119905) 119889119905 = 2 lim119873rarrinfin
119873
sum
119899=0
119869]+2119899+1 (119886) Re (]) gt minus1 (13)
(see [18 p 333])
int
119886
0
1199051minus]119869] (119905) 119889119905 =
1
2]minus1 ]
minus 1198861minus]119869]minus1 (119886) (14)
(see [18 p 333])
int
119886
0
119905120583
119869] (119905) 119889119905
= 119886120583(] + 120583 + 1) 2
(] minus 120583 + 1) 2
times lim119873rarrinfin
119873
sum
119899=0
(] + 2119899 + 1) Γ (((] minus 120583 + 1) 2) + 119899)Γ (((] + 120583 + 3) 2) + 119899)
times 119869]+2119899+1 (119886) Re (] + 120583 + 1) gt 0
(15)
(see [19 p 480])
4 Error Analysis
Let the RHS of (10) be denoted by 119892119899(119903) that is
119892119899(119903) =
119899
sum
119894=0
119892 (119894ℎ) 120595119894(119903) (16)
Now replacing 119892(119903) by 119892119899(119903) in (9) we define an 119899th approxi-
mate 119865]119899(119901) of the FHT 119865](119901) as follows
Definition 1 An 119899th approximate finite Hankel transform of119891(119903) denoted by 119865]119899(119901) is defined as
119865]119899 (119901) = int1
0
119892119899(119903) 119869] (119901119903) 119889119903
=
119899
sum
119894=0
119892 (119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
(17)
Let 120576119899(119901) denote the absolute error between the FHT 119865](119901)
and its 119899th approximate 119865]119899(119901) then we have the following
Theorem2 If119892(119903) is approximated by the family of first (119899+1)hat functions as given in (10) then
(i) |119892(119895ℎ) minus 119892119899(119895ℎ)| = 0 for 119895 = 0 1 2 119899
(ii) |119892(119903) minus 119892119899(119903)| le (12119899
2
)|11989210158401015840
(119895ℎ)| + 119874(11198993
) for 119895ℎ lt119903 lt (119895 + 1)ℎ 119895 = 0 1 2 119899 minus 1
(iii) 120576119899(119901) = |119865](119901) minus 119865]119899(119901)| le 1198722119899
2
+ 119874(11198993
) where|11989210158401015840
(119895ℎ)| le 119872
Proof (i) From (16) and (4) the value of 119892119899(119903) at 119895th nodal
point 119903 = 119895ℎ 119895 = 0 1 2 119899 is given by 119892119899(119895ℎ) =
sum119899
119894=0119892(119894ℎ)120595
119894(119895ℎ) = 119892(119895ℎ)
So |119892(119895ℎ) minus 119892119899(119895ℎ)| = 0 for 119895 = 0 1 2 119899
(ii) If 119903 lies between two consecutive integer multiples ofℎ that is 119895ℎ lt 119903 lt (119895 + 1)ℎ 119895 = 0 1 2 119899 minus 1 then from(16) we have
119892119899(119903) =
119899
sum
119894=0
119892 (119894ℎ) 120595119894(119903)
= 119892 (119895ℎ) 120595119895(119903) + 119892 ((119895 + 1) ℎ) 120595
119895+1(119903)
= 119892 (119895ℎ) (
(119895 + 1) ℎ minus 119903
ℎ
)
+ 119892 (119895ℎ + ℎ) (
119903 minus 119895ℎ
ℎ
) (Using (3))
= 119892 (119895ℎ) minus 119895ℎ(
119892 (119895ℎ + ℎ) minus 119892 (119895ℎ)
ℎ
)
+ 119903(
119892 (119895ℎ + ℎ) minus 119892 (119895ℎ)
ℎ
)
= 119892 (119895ℎ) + (119903 minus 119895ℎ)(
119892 (119895ℎ + ℎ) minus 119892 (119895ℎ)
ℎ
)
(18)
As ℎ rarr 0 from (18) we obtain
119892119899(119903) cong 119892 (119895ℎ) + (119903 minus 119895ℎ) 119892
1015840
(119895ℎ) (19)By expanding 119892
119899(119903) in the form of Taylorrsquos series in the
powers of (119903 minus 119895ℎ) we have
119892 (119903) =
infin
sum
119896=0
(119903 minus 119895ℎ)119896
119896
119892(119896)
(119895ℎ) (20)
4 International Journal of Analysis
where 119892(119896) denotes the 119896th order derivative of 119892(119903) Using (19)and (20) the error between exact and approximate values of119892(119903) is given by
119892 (119903) minus 119892119899(119903) =
infin
sum
119896=2
(119903 minus 119895ℎ)119896
119896
119892(119896)
(119895ℎ)
=
(119903 minus 119895ℎ)2
2
11989210158401015840
(119895ℎ) + 119874 (119903 minus 119895ℎ)3
(21)
Since (119903 minus 119895ℎ) lt ℎ and 119899ℎ = 1 from (21) we get
1003816100381610038161003816119892 (119903) minus 119892
119899(119903)1003816100381610038161003816le
1
21198992
1003816100381610038161003816100381611989210158401015840
(119895ℎ)
10038161003816100381610038161003816+ 119874(
1
1198993) (22)
(iii) The absolute error 120576119899(119901) between exact FHT 119865](119901)
and 119899th approximate FHT 119865]119899(119901) is given by
120576119899(119901) =
10038161003816100381610038161003816119865] (119901) minus 119865]119899 (119901)
10038161003816100381610038161003816
=
100381610038161003816100381610038161003816100381610038161003816
int
1
0
119892 (119903) 119869] (119901119903) 119889119903 minus int1
0
119892119899(119903) 119869] (119901119903) 119889119903
100381610038161003816100381610038161003816100381610038161003816
=
10038161003816100381610038161003816100381610038161003816100381610038161003816
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
(119892 (119903) minus 119892119899(119903)) 119869] (119901119903) 119889119903
10038161003816100381610038161003816100381610038161003816100381610038161003816
(23)
As |119869](119901119903)| le 1 we have
120576119899(119901) le
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
1003816100381610038161003816119892 (119903) minus 119892
119899(119903)1003816100381610038161003816119889119903
le
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
(
1
21198992
1003816100381610038161003816100381611989210158401015840
(119895ℎ)
10038161003816100381610038161003816+ 119874(
1
1198993))119889119903
(Follows from (22))
(24)
If |11989210158401015840(119895ℎ)| le 119872 then it follows that
120576119899(119901) le
119872
21198992
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
119889119903
+ 119874(
1
1198993)
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
119889119903
=
119872
21198992+ 119874(
1
1198993) as 119899ℎ = 1
(25)
5 Stability Analysis
In this section the stability of the proposed algorithm isanalyzed under the influence of noise In what follows theexact data function is denoted by 119891(119903) and the noisy datafunction119891120572(119903) is obtained by adding a randomnoise120572 to119891(119903)such that 119891120572(119903
119894) = 119891(119903
119894) + 120572120579
119894 where 119903
119894= 119894ℎ 119894 = 0 1 2 119899
119899ℎ = 30 and 120579119894is a uniform random variable with values in
[minus1 1] such that max |119891120572(119903119894) minus 119891(119903
119894)| le 120572 0 le 119894 le 119899 If 119903119891120572(119903)
is denoted by 119892120572(119903) and the approximate HT of the perturbedfunction is denoted by 119865120572]119899(119901) then from (17)
119865120572
]119899 (119901) =
119899
sum
119894=0
119892120572
(119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
=
119899
sum
119894=0
119894ℎ (119891 (119894ℎ) + 120572120579119894) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
(26)
From (26) and (17) we have
10038161003816100381610038161003816119865120572
]119899 (119901) minus 119865]119899 (119901)10038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816100381610038161003816
119899
sum
119894=0
119892120572
(119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
minus
119899
sum
119894=0
119892 (119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
1003816100381610038161003816100381610038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816100381610038161003816
119899
sum
119894=0
119894ℎ (120572120579119894) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
1003816100381610038161003816100381610038161003816100381610038161003816
le
119899
sum
119894=0
119894ℎ1003816100381610038161003816120572120579119894
1003816100381610038161003816int
1
0
1003816100381610038161003816120595119894(119903)1003816100381610038161003816
1003816100381610038161003816119869] (119901119903)
1003816100381610038161003816119889119903
le
119899
sum
119894=0
119894ℎ1003816100381610038161003816120572120579119894
1003816100381610038161003816int
1
0
1003816100381610038161003816120595119894(119903)1003816100381610038161003816119889119903 (As 100381610038161003816
1003816119869] (119901119903)
1003816100381610038161003816le 1)
(27)
Let 120579119872= max|120579
119894| then substituting 120595
119894(119903) from (3) we get
10038161003816100381610038161003816119865120572
]119899 (119901) minus 119865]119899 (119901)10038161003816100381610038161003816
le 120579119872ℎ |120572|
times [
119899minus1
sum
119894=1
119894 (int
119894ℎ
(119894minus1)ℎ
(119903 minus (119894 minus 1) ℎ)
ℎ
119889119903
+int
(119894+1)ℎ
119894ℎ
((119894 + 1) ℎ minus 119903)
ℎ
119889119903)
+ 119899int
1
1minusℎ
(119903 minus (1 minus ℎ))
ℎ
119889119903]
= 120579119872|120572| [ℎ
2
119899minus1
sum
119894=1
119894 + 119899
ℎ2
2
] = 120579119872|120572| (
1198992
ℎ2
2
)
le|120572|
2
(As 120579119872le 1 119899ℎ = 1)
(28)
Thus we have proved that
Theorem 3 When the input signal 119891(119903) is corrupted withnoise 120572 the proposed algorithm reduces the noise at least bya factor of 12 in the output data 119865]119899(119901)
International Journal of Analysis 5
Table 1 Least square errors 119864119895(119901)2
in different examples for 119899 = 100
119864119895(119901)2
Example 1 Example 2 Example 3 Example 4 Example 510038171003817100381710038171198640(119901)
10038171003817100381710038172
72696119890 minus 005 11078119890 minus 016 49160119890 minus 006 69741119890 minus 005 66819119890 minus 006
10038171003817100381710038171198641(119901)
10038171003817100381710038172
69980119890 minus 005 13724119890 minus 004 15965119890 minus 004 73362119890 minus 005 60667119890 minus 005
10038171003817100381710038171198642(119901)
10038171003817100381710038172
22536119890 minus 004 33312119890 minus 004 34473119890 minus 004 17687119890 minus 004 13147119890 minus 004
6 Numerical Illustrations
The test problems included in this paper are solved withand without random perturbations (noises) to illustrate theefficiency and stability of proposed algorithm by choosingthree different values of noise 120572 as 120572
0= 0 120572
1= 0002 and
1205722= 0005 (as discussed in the beginning of Section 5) In the
following examples the errors119864119895(119901) 119895 = 0 1 2 are computed
and their graphs are sketched for 119899 = 100 where 119864119895(119901) =
(Exact FHT 119865](119901)minus approximate FHT 119865120572119895]119899(119901) obtained from(26) with random noise 120572
119895)
Further the parameter 119901 ranges between 0 and 30 in stepsof 02 Figures 3 and 6 depict the graph of |119865120572119895]119899(119901) minus 119865]119899(119901)|119895 = 1 2 for different test functions in Examples 1 and 2respectively These graphs are in conformity withTheorem 3For all the illustrations the computations are done in MAT-LAB 701 For different examples the least square errors119864119895(119901)2involved in computations of approximate FHT with
noises 120572119895 119895 = 0 1 2 are given in Table 1These are calculated
using the formula
10038171003817100381710038171003817119864119895(119901)
100381710038171003817100381710038172=radicsum119899
119894=0(119864119895(119901119894))
2
119899 + 1
(29)
where 119901119894is taken in steps of 02 in the range [0 30]
Example 1 Consider the function 119891(119903) = radic(1 minus 1199032) 0 le 119903 le1 given in [14 20] for which
1198651(119901) =
120587
1198692
1(1199012)
2119901
0 lt 119901 lt infin
0 119901 = 0
(30)
Numerical evaluation of 1198651(119901) has been achieved by Barakat
and Sandler [20] by using Filon quadrature philosophy butagain the associated error is appreciable for 119901 lt 1 whereasour method gives almost zero error in that range Equation(12) is used to obtain the 119899th approximate HT 119865
1119899(119901) The
comparison between exact HT 1198651(119901) and approximate HT
1198651119899(119901) is shown in Figure 1 In Figure 2 the errors 119864
0(119901)
1198641(119901) and119864
2(119901) for 119899 = 100 are plotted In Figure 3 |119865120572
1119899(119901)minus
1198651119899(119901)| is shown for noises 120572 = 0002 0005 This figure is in
conformity withTheorem 3
Example 2 (Sombrero function) A very important and oftenused function is the Circ function that can be defined as
Circ( 119903119886
) =
1 119903 le 119886
0 119903 gt 119886
(31)
0 5 10 15 20 25 30
0
002
004
006
008
01
012
014
016
018
p
Exac
t and
appr
oxim
ate F
HT
minus002
Figure 1 Exact HT 1198651(119901) (solid line) and approximate HT 119865
1119899(119901)
(dotted line) for 119899 = 100 Example 1
The zeroth order HT of Circ(119903119886) is the Sombrero function[13 21] that will be written as 119878
0(119901) with the following
analytical expression
1198780(119901) = 119886
21198691(119886119901)
119886119901
(32)
We use (12) to obtain the approximation for the FHT 1198650119899(119901)
In Figure 4 we sketch the error 1198640(119901) between exact HT
1198780(119901) and approximate FHT119865
0119899(119901) of Circ function (without
noise) for 119886 = 1 and 119899 = 10 It is evident that even forsuch a small value of 119899 the error is appreciably small Figure 5compares the errors 119864
1(119901) and 119864
2(119901) for 119899 = 100 Figure 6
shows the plot of |1198651205720119899(119901) minus 119865
0119899(119901)| for 120572 = 0002 and 120572 =
0005 with 119899 = 100
Example 3 Consider the function 119891(119903) = (2120587)[arccos(119903) minus119903radic(1 minus 119903
2)] 0 le 119903 le 1 given in [14 22] forwhich zeroth order
exact HT is
1198650(119901) = 2
1198692
1(1199012)
1199012
0 le 119901 lt infin (33)
It is a well-known pair which arises in optical diffractiontheory [23] The function 119891(119903) is known as optical transferfunction of an aberration-free optical system with a circularaperture and 119865
0(119901) is the corresponding spread function
Barakat and Parshall [22] evaluated1198650(119901) using Filon quadra-
ture philosophy but the associated error is again appreciablefor 119901 lt 1 whereas our method gives significantly small error
6 International Journal of Analysis
0 5 10 15 20 25 30
0
1
2
3
p
minus5
minus4
minus3
minus2
minus1
times10minus4
E0(p)E1(p)
andE2(p)
E0(p)
E1(p)
E2(p)
Figure 2 1198640(119901) 119864
1(119901) 119864
2(119901) for 119899 = 100 Example 1
0 5 10 15 20 25 300
1
2
3
4
5
6
7
8
p
times10minus4
For noise 120572 = 0005
For noise 120572 = 0002
Figure 3 Plot of |1198651205721119899(119901)minus119865
1119899(119901)| for 120572 = 0002 and 120572 = 0005with
119899 = 100 Example 1
in that range In Figure 7 the error 1198640(119901) is shown Figure 8
is the comparison between 1198641(119901) and 119864
2(119901) for 119899 = 100
Example 4 Consider the function [24] 119891(119903) = 119903](1198862
minus 1199032
)120583
119867(119886 minus 119903) Re(]) gt minus1Re(120583) gt minus1 0 lt 119903 lt 119886 where119867(119903) is a unit step function
defined by
119867(119903) =
1 119903 ge 0
0 119903 lt 0
(34)
The ]th order HT of 119891(119903) is given by
119865] (119901) = 2]119886120583+]+1
119901minus120583minus1
120583 + 1 119869]+120583+1 (119886119901) (35)
Using the method of solution developed in Section 3 theapproximate HT 119865]119899(119901) has been calculated for ] = 32
0 5 10 15 20 25 30p
0
05
1
15
2
25
3
minus05
E0(p)
times10minus11
Figure 4 1198640(119901) for 119899 = 10 Example 2
0 5 10 15 20 25 30p
0
2
minus14
minus12
minus10
minus8
minus6
minus4
minus2
times10minus4
E1(p)
andE2(p)
E1(p)
E2(p)
Figure 5 1198641(119901) and 119864
2(119901) for 119899 = 100 Example 2
0 5 10 15 20 25 30p
0
02
04
06
08
1
12times10minus3
For noise 120572 = 0005
For noise 120572 = 0002
Figure 6 Plot of |1198651205720119899(119901)minus119865
0119899(119901)| for 120572 = 0002 and 120572 = 0005with
119899 = 100 Example 2
International Journal of Analysis 7
0
01
02
03
04
05
06
07
08
09
1
0 5 10 15 20 25 30p
times10minus5
E0(p)
Figure 7 1198640(119901) for 119899 = 100 in Example 3
0
2
0 5 10 15 20 25 30p
times10minus4
E1(p)
andE2(p)
E1(p)
E2(p)
minus12
minus10
minus8
minus6
minus4
minus2
Figure 8 1198641(119901) and 119864
2(119901) for 119899 = 100 in Example 3
0
001
002
003
004
005
006
007
008
Exac
t and
appr
oxim
ate F
HT
0 5 10 15 20 25 30p
minus002
minus001
Figure 9 Exact HT 11986532(119901) (solid line) and approximate HT
11986532119899
(119901) (dotted line) for 120583 = 12 and 119899 = 100 Example 4
0
1
2
0 5 10 15 20 25 30p
minus4
minus3
minus2
minus1
times10minus4
E0(p)
E1(p)
E2(p)
E0(p)E1(p)
andE2(p)
Figure 10 1198640(119901) 119864
1(119901) 119864
2(119901) for 119899 = 100 in Example 4
0
05
1
0 5 10 15 20 25 30p
Erro
r (w
ithou
t noi
se )E0(p)
minus2
minus15
minus1
minus05
times10minus5
Figure 11 1198640(119901) for 119899 = 100 in Example 5
120583 = 12 and 119886 = 1 The comparison between 11986532119899
(119901) andexact FHT 119865
32(119901) is shown in Figure 9 Further in Figure 10
a comparison between the different errors 1198640(119901) 119864
1(119901) and
1198642(119901) is shown for 119899 = 100
Example 5 In this example we choose as a test function thegeneralized version of the top-hat function given as 119891(119903) =119903][119867(119903) minus119867(119903 minus 119886)] 119886 gt 0 and119867(119903) is the unit step function
defined by (34) Then
119865] (119901) =119869]+1 (119901)
119901
(36)
In [9] authors took 119886 = 1 and ] = 4 for numericalcalculations We take 119886 = 1 ] = 5 and observe that theassociated errors with and without random noises are quitesmall The error 119864
0(119901) (without noise) is shown in Figure 11
8 International Journal of Analysis
0
05
0 5 10 15 20 25 30p
minus3
minus25
minus2
minus15
minus1
minus05
times10minus4
2E1(p)
andE2(p)
2E1(p)
E2(p)
Figure 12 Comparison between 1198641(119901) and 119864
2(119901) for 119899 = 100 in
Example 5
0 02 04 06 08 1
0
02
04
06
08
1
12
r
minus02
f(r)
ua(r0)
andf(r)
ua(r 0)
Figure 13 The initial condition function 119891(119903) (solid red line) and119906119886(119903 0) (= lim
119905rarr0+119906119886(119903 119905)) (lowastblue)
0 01 02 03 04 05 06 07 08 09
0
1
2
3
4
r
minus3
minus2
minus1
times10minus3
E(r)
Figure 14 Error between 119891(119903) and 119906119886(119903 0)
0 02 04 06 08 1
0
02
04
06
08
1
12
r
Solu
tion
for d
iffer
ent t
minus02
u0 for t = 0
u1 for t = 001
u2 for t = 002
u3 for t = 004
Figure 15The various profiles of the solutions 119906119886(119903 119905) at fixed times
0 01 02 03 04 05 06 07 08 09 10
1
2
r
Abso
lute
erro
rtimes10minus4
For t = 0
For t = 001
For t = 002
For t = 004
Figure 16 The absolute error between 119906119886(119903 119905) and 119906(119903 119905) for
different 119905
0
05
1
0
05
1
0
02
04
06
08
1
rt
minus02
u(rt)
Figure 17 The solution 119906(119903 119905) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
International Journal of Analysis 9
0
05
1
0
05
10
02
04
06
08
1
rt
ua(rt)
Figure 18 The approximate solution 119906119886(119903 119905) using 119899th approximate
FHT 1198655119899(1205735119904) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
for 119899 = 100 In Figure 12 errors 1198641(119901) and 119864
2(119901) for 119899 = 100
are shown
7 Applications
In this section we solve heat equation in cylindrical coor-dinates inside an infinitely long cylinder of radius unity byusing the theory of finite Hankel transform (FHT) developedin the preceding pages The initial temperature is givenby 119891(119903) and radiation takes place at the surface into thesurrounding medium maintained at zero temperature Weseek a function 119906(119903 119905) where 119903 is radius and 119905 is time (119906 doesnot depend upon 120579 and 119905) in the following application Themathematical model of this problem is the diffusion equationin polar coordinates given by
1205972
119906 (119903 119905)
1205972119903
+
1
119903
120597119906 (119903 119905)
120597119903
minus
]2
1199032119906 (119903 119905)
=
120597119906 (119903 119905)
120597119905
(0 lt 119903 lt 1 0 lt 119905 lt infin)
(37)
with the following initial and boundary conditions
(i) as 119905 rarr 0+ 119906(119903 119905) rarr 119891(119903) = 119903
][119867(119903) minus 119867(119903 minus 1)]
where 0 le 119903 le 1 and119867(119903) is unit step function givenby (34)
(ii) as 119903 rarr 1minus (120597119906120597119903) + 119866119906 rarr 0 for each fixed 119905 gt 0
and 119866 gt 0
When 119906(119903 119905) denotes the temperature within the cylinder119866 gt 0means that heat is being radiated away from the surfaceof the cylinder
LetΩ]119903 denote the Bessel differential operator (1205972
1205972
119903) +
(1119903)(120597120597119903) minus (]21199032) Then differential equation (37) can bewritten as
Ω]119903119906 =120597119906
120597119905
(38)
Applying the ]th order finite Hankel transform operator toboth sides of the differential equation (38) and using (8) weget
int
1
0
119903Ω]119903119906119869] (120573]119904119903) 119889119903 = int1
0
119903 (
120597119906
120597119905
) 119869] (120573]119904119903) 119889119903 (39)
where 120573]119904 is the 119904th positive root of equation [24]
119866119869] (119911) + 1199111198691015840
] (119911) = 0 ] ge minus1
2
(40)
The LHS of (39) may be written as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= int
1
0
(
1205972
119906
1205972119903
) 119903119869] (120573]119904119903) 119889119903
+ int
1
0
(
120597119906
120597119903
) 119869] (120573]119904119903) 119889119903 minus ]2 int1
0
(
119906
119903
) 119869] (120573]119904119903) 119889119903
= lim119903rarr1minus
120597119906 (119903 119905)
120597119903
119869] (120573]119904) minus 120573]119904 int1
0
119903
120597119906
120597119903
1198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
= minus119866119869] (120573]119904) lim119903rarr1minus119906 (119903 119905) minus 120573]119904 int
1
0
120597119906
120597119903
1199031198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903 (As 119903 997888rarr 1minus
120597119906
120597119903
+ 119866119906 997888rarr 0)
(41)
Since 120573]119904 is a root of 119866119869](119911) + 1199111198691015840
](119911) = 0 we have 119866119869](120573]119904) =minus120573]119904119869
1015840
](120573]119904) and thus
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
minus 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
+ 120573]119904 int1
0
119906 (1198691015840
] (120573]119904119903) + 120573]11990411990311986910158401015840
] (120573]119904119903)) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
(42)
Using the fact that 119869](119909) is a solution of 1199092(11988921199101198891199092) +119909(119889119910119889119909) + (119909
2
minus ]2)119910 = 0 (42) may be simplified as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= minus1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
(43)
10 International Journal of Analysis
From (43) and (39) we have
minus 1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
=
120597
120597119905
(int
1
0
119903119906 (119903 119905) 119869] (120573]119904119903) 119889119903)
(44)
If FHT int10
119906(119903 119905)119903119869](120573]119904119903) 119889119903 of 119906(119903 119905) is denoted by119880(120573]119904 119905)then (44) may be written as
120597119880 (120573]119904 119905)
120597119905
= minus1205732
]119904119880(120573]119904 119905) (45)
whose solution is given by
119880(120573]119904 119905) = 119860 (120573]119904) 119890minus1205732]119904119905 (46)
The constant of integration 119860(120573]119904) in (46) is determined byinitial condition Thus
119860 (120573]119904) = 119865] (120573]119904) = int1
0
119891 (119903) 119903119869] (120573]119904119903) 119889119903 (47)
Hence from (47) (46) becomes
119880 (120573]119904 119905) = 119865] (120573]119904) 119890minus1205732]119904119905 (48)
Therefore by inversion theorem [25] we have
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 1205732]11990411986910158402] (120573]119904)
= lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 11986621198692] (120573]119904)
(As 119866119869] (120573]119904) = minus120573]1199041198691015840
] (120573]119904))
(49)
So
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
119869] (120573]119904119903)
1198692
] (120573]119904) (50)
We want to prove that 119906(119903 119905) given by (50) is truly a solutionof (37) that satisfies the given initial and boundary conditionsTo achieve this we need the following well known estimates[17]
119865] (120573]119904) = 119874(1
12057332
]119904)
120573]119904 asymp 120587(119904 +1
4
)
1198692
] (120573]119904) asymp2
120587120573]119904
as 119904 997888rarr infin
(51)
Using the above estimates we see that series (50) and theseries obtained by applyingΩ]119903 and 120597120597119905 separately under thesummation sign of (50) converge uniformly on 0 lt 119903 lt 1 and119905 gt 0 Hence by applyingΩ]119903 minus 120597120597119905 on (50) we get
(Ω]119903 minus120597
120597119905
) 119906 (119903 119905)
= ( lim119873rarrinfin
119873
sum
119904=1
21205734
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
sdot [
[
(1205732
]1199041199032
11986910158401015840
] (120573]119904119903) + 120573]1199041199031198691015840
] (120573]119904119903)
+ (1205732
]1199041199032
minus ]2) 119869] (120573]119904119903))
times (1205732
]1199041199032
1198692
] (120573]119904))minus1
]
]
) = 0
(52)
Hence we see that (50) satisfies the differential equation (37)Let us verify the boundary condition (ii) we have
lim119903rarr1minus[
120597119906 (119903 119905)
120597119903
+ 119866119906 (119903 119905)]
= lim119903rarr1minus
[
[
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662) 1198692] (120573]119904)
times (120573]1199041198691015840
] (120573]119904119903) + 119866119869] (120573]119904119903))]
]
(53)
and since the convergence is uniform we can take thelim119903rarr1minus inside the summation sign and arrive at the conclu-
sion as 120573]119904rsquos are the roots of the equation119866119869](119911)+1199111198691015840
](119911) = 0The initial condition (i) is already taken care of as we
evaluated the constant 119860(120573]119904) by using it Through Figures13ndash18 we establish the accuracy of the proposed methodFor numerical solution of differential equation (37) we havetaken ] = 5 and 119866 = 1 We have used MATLAB routineldquofzerordquo to find zeros of (40) and we have drawn Figures 13ndash18 by truncating series (50) at 119873 = 30 While evaluatingthe solution 119906(119903 119905) from (50) we have evaluated 119865
5(1205735119904) first
from its analytical expression given by (36) and denoted thesolution thus obtained by 119906(119903 119905) in Figures 13ndash18 and thenevaluating 119899th approximates FHT 119865
5119899(1205735119904) for 119899 = 100
using our proposed algorithm for evaluation of the FHT asgiven by (17) This solution is denoted by 119906
119886(119903 119905) in the above
mentioned figuresFigure 13 compares the given initial condition 119891(119903) with
119906119886(119903 119905) as 119905 rarr 0
+ and Figure 14 shows the correspondingerror 119864(119903) = 119906
119886(119903 0) minus 119891(119903) Figure 15 depicts the various
profiles of 119906119886(119903 119905) at fixed times 119905
119895= 0 1100 150 125
where the various profiles are denoted by 1199060 1199061 1199062 and
1199063 The absolute errors between 119906
119886(119903 119905) and 119906(119903 119905) at fixed
times 119905119895= 0 1100 150 125 are shown in Figure 16 As
the maximum possible error occurs in the neighborhood of
International Journal of Analysis 11
0 and 0001 we have restricted 119905 in (0 1] in Figures 17 and18 representing 119906(119903 119905) and 119906
119886(119903 119905) respectively and note that
they are in good agreement in the range
8 Conclusions
A new stable and efficient algorithm based on hat functionsfor the numerical evaluation of Hankel transform is proposedand analyzed Choosing hat basis functions to expand theinput signal 119903119891(119903) makes our algorithm attractive in theirapplication in the applied physical problems as they eliminatethe problems connected with the Gibbs phenomenon takingplace in [9 11] We have given for the first time (as per ourinformation) error and stability analysis which was one ofthe main drawbacks of the earlier algorithms and by variousnumerical experiments we have corroborated our theoreticalfindings Stability with respect to the data is restored andexcellent accuracy is obtained even for small sample intervaland high noise levels in the data From the various figuresit is obvious that the algorithm is consistent and does notdepend on the particular choice of the input signal Theaccuracy and simplicity of the algorithm provide an edgeover the many other algorithms Finally an application of theproposed algorithm is given for solving the heat equation inan infinite cylinder with a radiation condition
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] I N Sneddon The Use of Integral Transforms McGraw-Hill1972
[2] A E Siegman ldquoQuasi fastHankel transformrdquoOptics Letters vol1 no 1 pp 13ndash15 1977
[3] L Filon ldquoOn quadrature formula for trigonometric integralsrdquoProceedings of the Royal Society of Edinburgh vol 49 pp 38ndash471928-1929
[4] D Levin andA Sidi ldquoTwo new classes of nonlinear transforma-tions for accelerating the convergence of infinite integrals andseriesrdquo Applied Mathematics and Computation vol 9 no 3 pp175ndash215 1981
[5] A Sidi ldquoExtrapolation methods for oscillatory infinite inte-gralsrdquo Journal of the Institute of Mathematics and its Applica-tions vol 26 no 1 pp 1ndash20 1980
[6] A Sidi ldquoA user-friendly extrapolation method for oscillatoryinfinite integralsrdquoMathematics of Computation vol 51 no 183pp 249ndash266 1988
[7] A Sidi ldquoComputation of infinite integrals involving Besselfunctions of arbitrary order by the 119863-transformationrdquo Journalof Computational and Applied Mathematics vol 78 no 1 pp125ndash130 1997
[8] A Sidi Practical Extrapolation Methods Theory and Applica-tions vol 10 of Cambridge Monographs on Applied and Compu-tational Mathematics Cambridge University Press CambridgeUK 2003
[9] M Guizar-Sicairos and J C Gutierrez-Vega ldquoComputation ofquasi-discrete Hankel transforms of integer order for propagat-ing optical wave fieldsrdquo Journal of the Optical Society of AmericaA Optics Image Science and Vision vol 21 no 1 pp 53ndash582004
[10] L Yu M Huang M Chen W Chen W Huang and Z ZhuldquoQuasi-discrete Hankel transformrdquoOptics Letters vol 23 no 6pp 409ndash411 1998
[11] E B Postnikov ldquoAbout calculation of the Hankel transformusing preliminary wavelet transformrdquo Journal of Applied Math-ematics no 6 pp 319ndash325 2003
[12] V K Singh O P Singh and R K Pandey ldquoNumerical evalu-ation of the Hankel transform by using linear Legendre multi-waveletsrdquoComputer Physics Communications vol 179 no 6 pp424ndash429 2008
[13] V K Singh O P Singh and R K Pandey ldquoEfficient algorithmsto compute Hankel transforms using waveletsrdquo ComputerPhysics Communications vol 179 no 11 pp 812ndash818 2008
[14] O P Singh V K Singh and R K Pandey ldquoAn efficient and sta-ble algorithm for numerical evaluation of Hankel transformsrdquoJournal of Applied Mathematics amp Informatics vol 28 no 5-6pp 1055ndash1071 2010
[15] R K Pandey O P Singh and V K Singh ldquoA stable algorithmfor numerical evaluation of Hankel transforms using Haarwaveletsrdquo Numerical Algorithms vol 53 no 4 pp 451ndash4662010
[16] E Babolian and M Mordad ldquoA numerical method for solvingsystems of linear and nonlinear integral equations of the secondkind by hat basis functionsrdquo Computers amp Mathematics withApplications vol 62 no 1 pp 187ndash198 2011
[17] R S Pathak and O P Singh ldquoFinite Hankel transforms ofdistributionsrdquo Pacific Journal of Mathematics vol 99 no 2 pp439ndash458 1982
[18] A Erdelyi EdTables of Integral TransformsMcGraw-Hill NewYork NY USA 1954
[19] M Abramowitz and I A Stegun Handbook of MathematicalFunctions with Formulas Graphs and Mathematical TablesDover New York NY USA 1965
[20] R Barakat and B H Sandler ldquoEvaluation of first-order Hankeltransforms using Filon quadrature philosophyrdquo Applied Mathe-matics Letters vol 11 no 1 pp 127ndash131 1998
[21] J D Secada ldquoNumerical evaluation of the Hankel transformrdquoComputer Physics Communications vol 116 no 2-3 pp 278ndash294 1999
[22] R Barakat and E Parshall ldquoNumerical evaluation of the zero-order Hankel transform using Filon quadrature philosophyrdquoApplied Mathematics Letters vol 9 no 5 pp 21ndash26 1996
[23] J Gaskell Linear Systems Fourier Transforms and Opticschapter 11 John Wiley amp Sons New York NY USA 1978
[24] A D Poularika ldquoThe Hankel transformrdquo in The Handbook ofFormulas and Tables for Signal Processing CRC Press LLC BocaRaton Fla USA 1999
[25] G NWatsonTheory of Bessel Functions Cambridge UniversityPress Cambridge UK 2nd edition 1958
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 International Journal of Analysis
general class of extrapolation methods for various types ofinfinite integrals whether oscillatory or not Sidi in [5ndash7]emphasized oscillatory integral involving Fourier andHankelTransforms among other more general cases In additionthe subject is dealt with in great detail in a book by Sidi [8Chapter 5]
Later in 2004 Guizar-Sicairos and Gutierrez-Vega [9]obtained a powerful scheme to calculate the HT of order] ge 0 by extending the zero order HT algorithm of Yu etal [10] to higher orders Their algorithm is based on theorthogonality properties of Bessel functions Postnikov [11]proposed for the first time a novel and powerful method forcomputing zero and first order HT by using Haar waveletsRefining the idea of Postnikov [11] Singh et al [12 13]obtained three efficient algorithms for numerical evaluationof HT of order ] gt minus1 using linear Legendre multi-wavelets Legendre wavelets and rationalized Haar waveletswhich were shown to be superior to the other mentionedalgorithms Recently Singh et al [14] and Pandey et al[15] tested their proposed algorithms on some examplesfor their stability with respect to noise 120572120579
119894where 120579
119894is a
uniform random variable in [minus1 1] added to the input signal119891(119903) But none of the above mentioned algorithms wereanalyzed theoretically for stability and no error analysis wasprovided
The lack of error and stability analysis in the papers citedabove motivated us for the present work In the presentpaper we use hat basis functions for approximation of119903119891(119903) and replace it by its approximation in (1) therebygetting an efficient and stable algorithm for the numericalevaluation of the HT of order ] gt minus1 Error and stabilityanalysis of the algorithm are given in Sections 4 and 5respectively and further corroborated by the numericalexperiments performed on the test functions in Section 6Test functions with known analytic HT are used with ran-dom noise term 120572120579
119894added to the input field 119891(119903) where
120579119894is a uniform random variable with values in [minus1 1]
to illustrate the stability and efficiency of the proposedalgorithm In Section 7 the algorithm is applied for solvingthe heat equation in an infinite cylinder with a radiationcondition
2 Hat Functions and TheirAssociated Properties
Hat functions are defined on the domain [0 1] These arecontinuous functions with shape of hats when plotted ontwo-dimensional planes The interval [0 1] is divided into 119899subintervals [119894ℎ (119894+1)ℎ] 119894 = 0 1 2 119899minus1 of equal lengthsℎ where ℎ = 1119899 The hat functionrsquos family of first (119899 + 1) hatfunctions is defined as follows [16]
1205950(119905) =
ℎ minus 119905
ℎ
0 le 119905 lt ℎ
0 otherwise
120595119894(119905)
=
119905 minus (119894 minus 1) ℎ
ℎ
(119894 minus 1) ℎ le 119905 lt 119894ℎ
(119894 + 1) ℎ minus 119905
ℎ
119894ℎ le 119905 lt (119894 + 1) ℎ 119894 = 1 2 119899 minus 1
0 otherwise
120595119899(119905) =
119905 minus (1 minus ℎ)
ℎ
1 minus ℎ le 119905 le 1
0 otherwise(3)
From the definition of hat functions it is obvious that
120595119894(119896ℎ) =
1 119894 = 119896
0 119894 = 119896
(4)
The hat functions 120595119895(119905) are continuous linearly independent
and are in 1198712[0 1]A function 119891 isin 1198712[0 1]may be approximated as
119891 (119905) cong
119894=119899
sum
119894=0
119891119894120595119894(119905) = 119891
01205950(119905) + 119891
11205951(119905)
+ 11989121205952(119905) + sdot sdot sdot + 119891
119899120595119899(119905)
(5)
The important aspect of using extended hat functions inthe approximation of function 119891(119905) lies in the fact that thecoefficients 119891
119894in (5) are given by
119891119894= 119891 (119894ℎ) for 119894 = 0 1 2 119899 where ℎ = 1
119899
(6)
3 Method of Solution
To derive the algorithm we first assume that the domainspace of input signal 119891(119903) extends over a limited region 0 le119903 le 119877 From physical point of view this assumption isreasonable due to the fact that the input signal 119891(119903) whichrepresents the physical field either is zero or has an infinitelylong decaying tail outside a disc of finite radius 119877 Thereforeinmany practical applications either the input signal119891(119903) hasa compact support or for a given 120576 gt 0 there exists a positivereal 119877 such that | intinfin
119877
119903119891(119903)119869](119901119903)119889119903| lt 120576 which is the case if119891(119903) = 119900(119903
120582
) where 120582 lt minus32 as 119903 rarr infin Hence in eithercase from (1) we have
119865] (119901) = int119877
0
119903119891 (119903) 119869] (119901119903) 119889119903 (7)
By scaling (7) may be written as
119865] (119901) = int1
0
119903119891 (119903) 119869] (119901119903) 119889119903 (8)
which is known as finite Hankel transform (FHT) [14 15 17]The FHT is a good approximation of the HT given by (1)Writing 119903119891(119903) = 119892(119903) in (8) we get
119865] (119901) = int1
0
119892 (119903) 119869] (119901119903) 119889119903 (9)
International Journal of Analysis 3
Using (5) and (6) we may approximate 119892(119903) as
119892 (119903) cong
119899
sum
119894=0
119892 (119894ℎ) 120595119894(119903) (10)
From (10) (9) may be written as
119865] (119901) cong int1
0
(
119899
sum
119894=0
119892 (119894ℎ) 120595119894(119903)) 119869] (119901119903) 119889119903
=
119899
sum
119894=0
119892 (119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
(11)
Using (3) (11) may be expressed as
119865] (119901) cong 119892 (0) intℎ
0
(
ℎ minus 119903
ℎ
) 119869] (119901119903) 119889119903
+
119899minus1
sum
119894=1
119892 (119894ℎ) [int
119894ℎ
(119894minus1)ℎ
(
119903 minus (119894 minus 1) ℎ
ℎ
) 119869] (119901119903) 119889119903
+int
(119894+1)ℎ
119894ℎ
((119894 + 1) ℎ minus 119903
ℎ
) 119869] (119901119903) 119889119903]
+ 119892 (119899ℎ) int
1
1minusℎ
(
119903 minus (1 minus ℎ)
ℎ
) 119869] (119901119903) 119889119903
(12)
It is noteworthy here that the integrals involved in (12) areevaluated by using the following formulae
int
119886
0
119869] (119905) 119889119905 = 2 lim119873rarrinfin
119873
sum
119899=0
119869]+2119899+1 (119886) Re (]) gt minus1 (13)
(see [18 p 333])
int
119886
0
1199051minus]119869] (119905) 119889119905 =
1
2]minus1 ]
minus 1198861minus]119869]minus1 (119886) (14)
(see [18 p 333])
int
119886
0
119905120583
119869] (119905) 119889119905
= 119886120583(] + 120583 + 1) 2
(] minus 120583 + 1) 2
times lim119873rarrinfin
119873
sum
119899=0
(] + 2119899 + 1) Γ (((] minus 120583 + 1) 2) + 119899)Γ (((] + 120583 + 3) 2) + 119899)
times 119869]+2119899+1 (119886) Re (] + 120583 + 1) gt 0
(15)
(see [19 p 480])
4 Error Analysis
Let the RHS of (10) be denoted by 119892119899(119903) that is
119892119899(119903) =
119899
sum
119894=0
119892 (119894ℎ) 120595119894(119903) (16)
Now replacing 119892(119903) by 119892119899(119903) in (9) we define an 119899th approxi-
mate 119865]119899(119901) of the FHT 119865](119901) as follows
Definition 1 An 119899th approximate finite Hankel transform of119891(119903) denoted by 119865]119899(119901) is defined as
119865]119899 (119901) = int1
0
119892119899(119903) 119869] (119901119903) 119889119903
=
119899
sum
119894=0
119892 (119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
(17)
Let 120576119899(119901) denote the absolute error between the FHT 119865](119901)
and its 119899th approximate 119865]119899(119901) then we have the following
Theorem2 If119892(119903) is approximated by the family of first (119899+1)hat functions as given in (10) then
(i) |119892(119895ℎ) minus 119892119899(119895ℎ)| = 0 for 119895 = 0 1 2 119899
(ii) |119892(119903) minus 119892119899(119903)| le (12119899
2
)|11989210158401015840
(119895ℎ)| + 119874(11198993
) for 119895ℎ lt119903 lt (119895 + 1)ℎ 119895 = 0 1 2 119899 minus 1
(iii) 120576119899(119901) = |119865](119901) minus 119865]119899(119901)| le 1198722119899
2
+ 119874(11198993
) where|11989210158401015840
(119895ℎ)| le 119872
Proof (i) From (16) and (4) the value of 119892119899(119903) at 119895th nodal
point 119903 = 119895ℎ 119895 = 0 1 2 119899 is given by 119892119899(119895ℎ) =
sum119899
119894=0119892(119894ℎ)120595
119894(119895ℎ) = 119892(119895ℎ)
So |119892(119895ℎ) minus 119892119899(119895ℎ)| = 0 for 119895 = 0 1 2 119899
(ii) If 119903 lies between two consecutive integer multiples ofℎ that is 119895ℎ lt 119903 lt (119895 + 1)ℎ 119895 = 0 1 2 119899 minus 1 then from(16) we have
119892119899(119903) =
119899
sum
119894=0
119892 (119894ℎ) 120595119894(119903)
= 119892 (119895ℎ) 120595119895(119903) + 119892 ((119895 + 1) ℎ) 120595
119895+1(119903)
= 119892 (119895ℎ) (
(119895 + 1) ℎ minus 119903
ℎ
)
+ 119892 (119895ℎ + ℎ) (
119903 minus 119895ℎ
ℎ
) (Using (3))
= 119892 (119895ℎ) minus 119895ℎ(
119892 (119895ℎ + ℎ) minus 119892 (119895ℎ)
ℎ
)
+ 119903(
119892 (119895ℎ + ℎ) minus 119892 (119895ℎ)
ℎ
)
= 119892 (119895ℎ) + (119903 minus 119895ℎ)(
119892 (119895ℎ + ℎ) minus 119892 (119895ℎ)
ℎ
)
(18)
As ℎ rarr 0 from (18) we obtain
119892119899(119903) cong 119892 (119895ℎ) + (119903 minus 119895ℎ) 119892
1015840
(119895ℎ) (19)By expanding 119892
119899(119903) in the form of Taylorrsquos series in the
powers of (119903 minus 119895ℎ) we have
119892 (119903) =
infin
sum
119896=0
(119903 minus 119895ℎ)119896
119896
119892(119896)
(119895ℎ) (20)
4 International Journal of Analysis
where 119892(119896) denotes the 119896th order derivative of 119892(119903) Using (19)and (20) the error between exact and approximate values of119892(119903) is given by
119892 (119903) minus 119892119899(119903) =
infin
sum
119896=2
(119903 minus 119895ℎ)119896
119896
119892(119896)
(119895ℎ)
=
(119903 minus 119895ℎ)2
2
11989210158401015840
(119895ℎ) + 119874 (119903 minus 119895ℎ)3
(21)
Since (119903 minus 119895ℎ) lt ℎ and 119899ℎ = 1 from (21) we get
1003816100381610038161003816119892 (119903) minus 119892
119899(119903)1003816100381610038161003816le
1
21198992
1003816100381610038161003816100381611989210158401015840
(119895ℎ)
10038161003816100381610038161003816+ 119874(
1
1198993) (22)
(iii) The absolute error 120576119899(119901) between exact FHT 119865](119901)
and 119899th approximate FHT 119865]119899(119901) is given by
120576119899(119901) =
10038161003816100381610038161003816119865] (119901) minus 119865]119899 (119901)
10038161003816100381610038161003816
=
100381610038161003816100381610038161003816100381610038161003816
int
1
0
119892 (119903) 119869] (119901119903) 119889119903 minus int1
0
119892119899(119903) 119869] (119901119903) 119889119903
100381610038161003816100381610038161003816100381610038161003816
=
10038161003816100381610038161003816100381610038161003816100381610038161003816
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
(119892 (119903) minus 119892119899(119903)) 119869] (119901119903) 119889119903
10038161003816100381610038161003816100381610038161003816100381610038161003816
(23)
As |119869](119901119903)| le 1 we have
120576119899(119901) le
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
1003816100381610038161003816119892 (119903) minus 119892
119899(119903)1003816100381610038161003816119889119903
le
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
(
1
21198992
1003816100381610038161003816100381611989210158401015840
(119895ℎ)
10038161003816100381610038161003816+ 119874(
1
1198993))119889119903
(Follows from (22))
(24)
If |11989210158401015840(119895ℎ)| le 119872 then it follows that
120576119899(119901) le
119872
21198992
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
119889119903
+ 119874(
1
1198993)
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
119889119903
=
119872
21198992+ 119874(
1
1198993) as 119899ℎ = 1
(25)
5 Stability Analysis
In this section the stability of the proposed algorithm isanalyzed under the influence of noise In what follows theexact data function is denoted by 119891(119903) and the noisy datafunction119891120572(119903) is obtained by adding a randomnoise120572 to119891(119903)such that 119891120572(119903
119894) = 119891(119903
119894) + 120572120579
119894 where 119903
119894= 119894ℎ 119894 = 0 1 2 119899
119899ℎ = 30 and 120579119894is a uniform random variable with values in
[minus1 1] such that max |119891120572(119903119894) minus 119891(119903
119894)| le 120572 0 le 119894 le 119899 If 119903119891120572(119903)
is denoted by 119892120572(119903) and the approximate HT of the perturbedfunction is denoted by 119865120572]119899(119901) then from (17)
119865120572
]119899 (119901) =
119899
sum
119894=0
119892120572
(119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
=
119899
sum
119894=0
119894ℎ (119891 (119894ℎ) + 120572120579119894) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
(26)
From (26) and (17) we have
10038161003816100381610038161003816119865120572
]119899 (119901) minus 119865]119899 (119901)10038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816100381610038161003816
119899
sum
119894=0
119892120572
(119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
minus
119899
sum
119894=0
119892 (119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
1003816100381610038161003816100381610038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816100381610038161003816
119899
sum
119894=0
119894ℎ (120572120579119894) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
1003816100381610038161003816100381610038161003816100381610038161003816
le
119899
sum
119894=0
119894ℎ1003816100381610038161003816120572120579119894
1003816100381610038161003816int
1
0
1003816100381610038161003816120595119894(119903)1003816100381610038161003816
1003816100381610038161003816119869] (119901119903)
1003816100381610038161003816119889119903
le
119899
sum
119894=0
119894ℎ1003816100381610038161003816120572120579119894
1003816100381610038161003816int
1
0
1003816100381610038161003816120595119894(119903)1003816100381610038161003816119889119903 (As 100381610038161003816
1003816119869] (119901119903)
1003816100381610038161003816le 1)
(27)
Let 120579119872= max|120579
119894| then substituting 120595
119894(119903) from (3) we get
10038161003816100381610038161003816119865120572
]119899 (119901) minus 119865]119899 (119901)10038161003816100381610038161003816
le 120579119872ℎ |120572|
times [
119899minus1
sum
119894=1
119894 (int
119894ℎ
(119894minus1)ℎ
(119903 minus (119894 minus 1) ℎ)
ℎ
119889119903
+int
(119894+1)ℎ
119894ℎ
((119894 + 1) ℎ minus 119903)
ℎ
119889119903)
+ 119899int
1
1minusℎ
(119903 minus (1 minus ℎ))
ℎ
119889119903]
= 120579119872|120572| [ℎ
2
119899minus1
sum
119894=1
119894 + 119899
ℎ2
2
] = 120579119872|120572| (
1198992
ℎ2
2
)
le|120572|
2
(As 120579119872le 1 119899ℎ = 1)
(28)
Thus we have proved that
Theorem 3 When the input signal 119891(119903) is corrupted withnoise 120572 the proposed algorithm reduces the noise at least bya factor of 12 in the output data 119865]119899(119901)
International Journal of Analysis 5
Table 1 Least square errors 119864119895(119901)2
in different examples for 119899 = 100
119864119895(119901)2
Example 1 Example 2 Example 3 Example 4 Example 510038171003817100381710038171198640(119901)
10038171003817100381710038172
72696119890 minus 005 11078119890 minus 016 49160119890 minus 006 69741119890 minus 005 66819119890 minus 006
10038171003817100381710038171198641(119901)
10038171003817100381710038172
69980119890 minus 005 13724119890 minus 004 15965119890 minus 004 73362119890 minus 005 60667119890 minus 005
10038171003817100381710038171198642(119901)
10038171003817100381710038172
22536119890 minus 004 33312119890 minus 004 34473119890 minus 004 17687119890 minus 004 13147119890 minus 004
6 Numerical Illustrations
The test problems included in this paper are solved withand without random perturbations (noises) to illustrate theefficiency and stability of proposed algorithm by choosingthree different values of noise 120572 as 120572
0= 0 120572
1= 0002 and
1205722= 0005 (as discussed in the beginning of Section 5) In the
following examples the errors119864119895(119901) 119895 = 0 1 2 are computed
and their graphs are sketched for 119899 = 100 where 119864119895(119901) =
(Exact FHT 119865](119901)minus approximate FHT 119865120572119895]119899(119901) obtained from(26) with random noise 120572
119895)
Further the parameter 119901 ranges between 0 and 30 in stepsof 02 Figures 3 and 6 depict the graph of |119865120572119895]119899(119901) minus 119865]119899(119901)|119895 = 1 2 for different test functions in Examples 1 and 2respectively These graphs are in conformity withTheorem 3For all the illustrations the computations are done in MAT-LAB 701 For different examples the least square errors119864119895(119901)2involved in computations of approximate FHT with
noises 120572119895 119895 = 0 1 2 are given in Table 1These are calculated
using the formula
10038171003817100381710038171003817119864119895(119901)
100381710038171003817100381710038172=radicsum119899
119894=0(119864119895(119901119894))
2
119899 + 1
(29)
where 119901119894is taken in steps of 02 in the range [0 30]
Example 1 Consider the function 119891(119903) = radic(1 minus 1199032) 0 le 119903 le1 given in [14 20] for which
1198651(119901) =
120587
1198692
1(1199012)
2119901
0 lt 119901 lt infin
0 119901 = 0
(30)
Numerical evaluation of 1198651(119901) has been achieved by Barakat
and Sandler [20] by using Filon quadrature philosophy butagain the associated error is appreciable for 119901 lt 1 whereasour method gives almost zero error in that range Equation(12) is used to obtain the 119899th approximate HT 119865
1119899(119901) The
comparison between exact HT 1198651(119901) and approximate HT
1198651119899(119901) is shown in Figure 1 In Figure 2 the errors 119864
0(119901)
1198641(119901) and119864
2(119901) for 119899 = 100 are plotted In Figure 3 |119865120572
1119899(119901)minus
1198651119899(119901)| is shown for noises 120572 = 0002 0005 This figure is in
conformity withTheorem 3
Example 2 (Sombrero function) A very important and oftenused function is the Circ function that can be defined as
Circ( 119903119886
) =
1 119903 le 119886
0 119903 gt 119886
(31)
0 5 10 15 20 25 30
0
002
004
006
008
01
012
014
016
018
p
Exac
t and
appr
oxim
ate F
HT
minus002
Figure 1 Exact HT 1198651(119901) (solid line) and approximate HT 119865
1119899(119901)
(dotted line) for 119899 = 100 Example 1
The zeroth order HT of Circ(119903119886) is the Sombrero function[13 21] that will be written as 119878
0(119901) with the following
analytical expression
1198780(119901) = 119886
21198691(119886119901)
119886119901
(32)
We use (12) to obtain the approximation for the FHT 1198650119899(119901)
In Figure 4 we sketch the error 1198640(119901) between exact HT
1198780(119901) and approximate FHT119865
0119899(119901) of Circ function (without
noise) for 119886 = 1 and 119899 = 10 It is evident that even forsuch a small value of 119899 the error is appreciably small Figure 5compares the errors 119864
1(119901) and 119864
2(119901) for 119899 = 100 Figure 6
shows the plot of |1198651205720119899(119901) minus 119865
0119899(119901)| for 120572 = 0002 and 120572 =
0005 with 119899 = 100
Example 3 Consider the function 119891(119903) = (2120587)[arccos(119903) minus119903radic(1 minus 119903
2)] 0 le 119903 le 1 given in [14 22] forwhich zeroth order
exact HT is
1198650(119901) = 2
1198692
1(1199012)
1199012
0 le 119901 lt infin (33)
It is a well-known pair which arises in optical diffractiontheory [23] The function 119891(119903) is known as optical transferfunction of an aberration-free optical system with a circularaperture and 119865
0(119901) is the corresponding spread function
Barakat and Parshall [22] evaluated1198650(119901) using Filon quadra-
ture philosophy but the associated error is again appreciablefor 119901 lt 1 whereas our method gives significantly small error
6 International Journal of Analysis
0 5 10 15 20 25 30
0
1
2
3
p
minus5
minus4
minus3
minus2
minus1
times10minus4
E0(p)E1(p)
andE2(p)
E0(p)
E1(p)
E2(p)
Figure 2 1198640(119901) 119864
1(119901) 119864
2(119901) for 119899 = 100 Example 1
0 5 10 15 20 25 300
1
2
3
4
5
6
7
8
p
times10minus4
For noise 120572 = 0005
For noise 120572 = 0002
Figure 3 Plot of |1198651205721119899(119901)minus119865
1119899(119901)| for 120572 = 0002 and 120572 = 0005with
119899 = 100 Example 1
in that range In Figure 7 the error 1198640(119901) is shown Figure 8
is the comparison between 1198641(119901) and 119864
2(119901) for 119899 = 100
Example 4 Consider the function [24] 119891(119903) = 119903](1198862
minus 1199032
)120583
119867(119886 minus 119903) Re(]) gt minus1Re(120583) gt minus1 0 lt 119903 lt 119886 where119867(119903) is a unit step function
defined by
119867(119903) =
1 119903 ge 0
0 119903 lt 0
(34)
The ]th order HT of 119891(119903) is given by
119865] (119901) = 2]119886120583+]+1
119901minus120583minus1
120583 + 1 119869]+120583+1 (119886119901) (35)
Using the method of solution developed in Section 3 theapproximate HT 119865]119899(119901) has been calculated for ] = 32
0 5 10 15 20 25 30p
0
05
1
15
2
25
3
minus05
E0(p)
times10minus11
Figure 4 1198640(119901) for 119899 = 10 Example 2
0 5 10 15 20 25 30p
0
2
minus14
minus12
minus10
minus8
minus6
minus4
minus2
times10minus4
E1(p)
andE2(p)
E1(p)
E2(p)
Figure 5 1198641(119901) and 119864
2(119901) for 119899 = 100 Example 2
0 5 10 15 20 25 30p
0
02
04
06
08
1
12times10minus3
For noise 120572 = 0005
For noise 120572 = 0002
Figure 6 Plot of |1198651205720119899(119901)minus119865
0119899(119901)| for 120572 = 0002 and 120572 = 0005with
119899 = 100 Example 2
International Journal of Analysis 7
0
01
02
03
04
05
06
07
08
09
1
0 5 10 15 20 25 30p
times10minus5
E0(p)
Figure 7 1198640(119901) for 119899 = 100 in Example 3
0
2
0 5 10 15 20 25 30p
times10minus4
E1(p)
andE2(p)
E1(p)
E2(p)
minus12
minus10
minus8
minus6
minus4
minus2
Figure 8 1198641(119901) and 119864
2(119901) for 119899 = 100 in Example 3
0
001
002
003
004
005
006
007
008
Exac
t and
appr
oxim
ate F
HT
0 5 10 15 20 25 30p
minus002
minus001
Figure 9 Exact HT 11986532(119901) (solid line) and approximate HT
11986532119899
(119901) (dotted line) for 120583 = 12 and 119899 = 100 Example 4
0
1
2
0 5 10 15 20 25 30p
minus4
minus3
minus2
minus1
times10minus4
E0(p)
E1(p)
E2(p)
E0(p)E1(p)
andE2(p)
Figure 10 1198640(119901) 119864
1(119901) 119864
2(119901) for 119899 = 100 in Example 4
0
05
1
0 5 10 15 20 25 30p
Erro
r (w
ithou
t noi
se )E0(p)
minus2
minus15
minus1
minus05
times10minus5
Figure 11 1198640(119901) for 119899 = 100 in Example 5
120583 = 12 and 119886 = 1 The comparison between 11986532119899
(119901) andexact FHT 119865
32(119901) is shown in Figure 9 Further in Figure 10
a comparison between the different errors 1198640(119901) 119864
1(119901) and
1198642(119901) is shown for 119899 = 100
Example 5 In this example we choose as a test function thegeneralized version of the top-hat function given as 119891(119903) =119903][119867(119903) minus119867(119903 minus 119886)] 119886 gt 0 and119867(119903) is the unit step function
defined by (34) Then
119865] (119901) =119869]+1 (119901)
119901
(36)
In [9] authors took 119886 = 1 and ] = 4 for numericalcalculations We take 119886 = 1 ] = 5 and observe that theassociated errors with and without random noises are quitesmall The error 119864
0(119901) (without noise) is shown in Figure 11
8 International Journal of Analysis
0
05
0 5 10 15 20 25 30p
minus3
minus25
minus2
minus15
minus1
minus05
times10minus4
2E1(p)
andE2(p)
2E1(p)
E2(p)
Figure 12 Comparison between 1198641(119901) and 119864
2(119901) for 119899 = 100 in
Example 5
0 02 04 06 08 1
0
02
04
06
08
1
12
r
minus02
f(r)
ua(r0)
andf(r)
ua(r 0)
Figure 13 The initial condition function 119891(119903) (solid red line) and119906119886(119903 0) (= lim
119905rarr0+119906119886(119903 119905)) (lowastblue)
0 01 02 03 04 05 06 07 08 09
0
1
2
3
4
r
minus3
minus2
minus1
times10minus3
E(r)
Figure 14 Error between 119891(119903) and 119906119886(119903 0)
0 02 04 06 08 1
0
02
04
06
08
1
12
r
Solu
tion
for d
iffer
ent t
minus02
u0 for t = 0
u1 for t = 001
u2 for t = 002
u3 for t = 004
Figure 15The various profiles of the solutions 119906119886(119903 119905) at fixed times
0 01 02 03 04 05 06 07 08 09 10
1
2
r
Abso
lute
erro
rtimes10minus4
For t = 0
For t = 001
For t = 002
For t = 004
Figure 16 The absolute error between 119906119886(119903 119905) and 119906(119903 119905) for
different 119905
0
05
1
0
05
1
0
02
04
06
08
1
rt
minus02
u(rt)
Figure 17 The solution 119906(119903 119905) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
International Journal of Analysis 9
0
05
1
0
05
10
02
04
06
08
1
rt
ua(rt)
Figure 18 The approximate solution 119906119886(119903 119905) using 119899th approximate
FHT 1198655119899(1205735119904) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
for 119899 = 100 In Figure 12 errors 1198641(119901) and 119864
2(119901) for 119899 = 100
are shown
7 Applications
In this section we solve heat equation in cylindrical coor-dinates inside an infinitely long cylinder of radius unity byusing the theory of finite Hankel transform (FHT) developedin the preceding pages The initial temperature is givenby 119891(119903) and radiation takes place at the surface into thesurrounding medium maintained at zero temperature Weseek a function 119906(119903 119905) where 119903 is radius and 119905 is time (119906 doesnot depend upon 120579 and 119905) in the following application Themathematical model of this problem is the diffusion equationin polar coordinates given by
1205972
119906 (119903 119905)
1205972119903
+
1
119903
120597119906 (119903 119905)
120597119903
minus
]2
1199032119906 (119903 119905)
=
120597119906 (119903 119905)
120597119905
(0 lt 119903 lt 1 0 lt 119905 lt infin)
(37)
with the following initial and boundary conditions
(i) as 119905 rarr 0+ 119906(119903 119905) rarr 119891(119903) = 119903
][119867(119903) minus 119867(119903 minus 1)]
where 0 le 119903 le 1 and119867(119903) is unit step function givenby (34)
(ii) as 119903 rarr 1minus (120597119906120597119903) + 119866119906 rarr 0 for each fixed 119905 gt 0
and 119866 gt 0
When 119906(119903 119905) denotes the temperature within the cylinder119866 gt 0means that heat is being radiated away from the surfaceof the cylinder
LetΩ]119903 denote the Bessel differential operator (1205972
1205972
119903) +
(1119903)(120597120597119903) minus (]21199032) Then differential equation (37) can bewritten as
Ω]119903119906 =120597119906
120597119905
(38)
Applying the ]th order finite Hankel transform operator toboth sides of the differential equation (38) and using (8) weget
int
1
0
119903Ω]119903119906119869] (120573]119904119903) 119889119903 = int1
0
119903 (
120597119906
120597119905
) 119869] (120573]119904119903) 119889119903 (39)
where 120573]119904 is the 119904th positive root of equation [24]
119866119869] (119911) + 1199111198691015840
] (119911) = 0 ] ge minus1
2
(40)
The LHS of (39) may be written as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= int
1
0
(
1205972
119906
1205972119903
) 119903119869] (120573]119904119903) 119889119903
+ int
1
0
(
120597119906
120597119903
) 119869] (120573]119904119903) 119889119903 minus ]2 int1
0
(
119906
119903
) 119869] (120573]119904119903) 119889119903
= lim119903rarr1minus
120597119906 (119903 119905)
120597119903
119869] (120573]119904) minus 120573]119904 int1
0
119903
120597119906
120597119903
1198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
= minus119866119869] (120573]119904) lim119903rarr1minus119906 (119903 119905) minus 120573]119904 int
1
0
120597119906
120597119903
1199031198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903 (As 119903 997888rarr 1minus
120597119906
120597119903
+ 119866119906 997888rarr 0)
(41)
Since 120573]119904 is a root of 119866119869](119911) + 1199111198691015840
](119911) = 0 we have 119866119869](120573]119904) =minus120573]119904119869
1015840
](120573]119904) and thus
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
minus 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
+ 120573]119904 int1
0
119906 (1198691015840
] (120573]119904119903) + 120573]11990411990311986910158401015840
] (120573]119904119903)) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
(42)
Using the fact that 119869](119909) is a solution of 1199092(11988921199101198891199092) +119909(119889119910119889119909) + (119909
2
minus ]2)119910 = 0 (42) may be simplified as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= minus1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
(43)
10 International Journal of Analysis
From (43) and (39) we have
minus 1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
=
120597
120597119905
(int
1
0
119903119906 (119903 119905) 119869] (120573]119904119903) 119889119903)
(44)
If FHT int10
119906(119903 119905)119903119869](120573]119904119903) 119889119903 of 119906(119903 119905) is denoted by119880(120573]119904 119905)then (44) may be written as
120597119880 (120573]119904 119905)
120597119905
= minus1205732
]119904119880(120573]119904 119905) (45)
whose solution is given by
119880(120573]119904 119905) = 119860 (120573]119904) 119890minus1205732]119904119905 (46)
The constant of integration 119860(120573]119904) in (46) is determined byinitial condition Thus
119860 (120573]119904) = 119865] (120573]119904) = int1
0
119891 (119903) 119903119869] (120573]119904119903) 119889119903 (47)
Hence from (47) (46) becomes
119880 (120573]119904 119905) = 119865] (120573]119904) 119890minus1205732]119904119905 (48)
Therefore by inversion theorem [25] we have
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 1205732]11990411986910158402] (120573]119904)
= lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 11986621198692] (120573]119904)
(As 119866119869] (120573]119904) = minus120573]1199041198691015840
] (120573]119904))
(49)
So
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
119869] (120573]119904119903)
1198692
] (120573]119904) (50)
We want to prove that 119906(119903 119905) given by (50) is truly a solutionof (37) that satisfies the given initial and boundary conditionsTo achieve this we need the following well known estimates[17]
119865] (120573]119904) = 119874(1
12057332
]119904)
120573]119904 asymp 120587(119904 +1
4
)
1198692
] (120573]119904) asymp2
120587120573]119904
as 119904 997888rarr infin
(51)
Using the above estimates we see that series (50) and theseries obtained by applyingΩ]119903 and 120597120597119905 separately under thesummation sign of (50) converge uniformly on 0 lt 119903 lt 1 and119905 gt 0 Hence by applyingΩ]119903 minus 120597120597119905 on (50) we get
(Ω]119903 minus120597
120597119905
) 119906 (119903 119905)
= ( lim119873rarrinfin
119873
sum
119904=1
21205734
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
sdot [
[
(1205732
]1199041199032
11986910158401015840
] (120573]119904119903) + 120573]1199041199031198691015840
] (120573]119904119903)
+ (1205732
]1199041199032
minus ]2) 119869] (120573]119904119903))
times (1205732
]1199041199032
1198692
] (120573]119904))minus1
]
]
) = 0
(52)
Hence we see that (50) satisfies the differential equation (37)Let us verify the boundary condition (ii) we have
lim119903rarr1minus[
120597119906 (119903 119905)
120597119903
+ 119866119906 (119903 119905)]
= lim119903rarr1minus
[
[
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662) 1198692] (120573]119904)
times (120573]1199041198691015840
] (120573]119904119903) + 119866119869] (120573]119904119903))]
]
(53)
and since the convergence is uniform we can take thelim119903rarr1minus inside the summation sign and arrive at the conclu-
sion as 120573]119904rsquos are the roots of the equation119866119869](119911)+1199111198691015840
](119911) = 0The initial condition (i) is already taken care of as we
evaluated the constant 119860(120573]119904) by using it Through Figures13ndash18 we establish the accuracy of the proposed methodFor numerical solution of differential equation (37) we havetaken ] = 5 and 119866 = 1 We have used MATLAB routineldquofzerordquo to find zeros of (40) and we have drawn Figures 13ndash18 by truncating series (50) at 119873 = 30 While evaluatingthe solution 119906(119903 119905) from (50) we have evaluated 119865
5(1205735119904) first
from its analytical expression given by (36) and denoted thesolution thus obtained by 119906(119903 119905) in Figures 13ndash18 and thenevaluating 119899th approximates FHT 119865
5119899(1205735119904) for 119899 = 100
using our proposed algorithm for evaluation of the FHT asgiven by (17) This solution is denoted by 119906
119886(119903 119905) in the above
mentioned figuresFigure 13 compares the given initial condition 119891(119903) with
119906119886(119903 119905) as 119905 rarr 0
+ and Figure 14 shows the correspondingerror 119864(119903) = 119906
119886(119903 0) minus 119891(119903) Figure 15 depicts the various
profiles of 119906119886(119903 119905) at fixed times 119905
119895= 0 1100 150 125
where the various profiles are denoted by 1199060 1199061 1199062 and
1199063 The absolute errors between 119906
119886(119903 119905) and 119906(119903 119905) at fixed
times 119905119895= 0 1100 150 125 are shown in Figure 16 As
the maximum possible error occurs in the neighborhood of
International Journal of Analysis 11
0 and 0001 we have restricted 119905 in (0 1] in Figures 17 and18 representing 119906(119903 119905) and 119906
119886(119903 119905) respectively and note that
they are in good agreement in the range
8 Conclusions
A new stable and efficient algorithm based on hat functionsfor the numerical evaluation of Hankel transform is proposedand analyzed Choosing hat basis functions to expand theinput signal 119903119891(119903) makes our algorithm attractive in theirapplication in the applied physical problems as they eliminatethe problems connected with the Gibbs phenomenon takingplace in [9 11] We have given for the first time (as per ourinformation) error and stability analysis which was one ofthe main drawbacks of the earlier algorithms and by variousnumerical experiments we have corroborated our theoreticalfindings Stability with respect to the data is restored andexcellent accuracy is obtained even for small sample intervaland high noise levels in the data From the various figuresit is obvious that the algorithm is consistent and does notdepend on the particular choice of the input signal Theaccuracy and simplicity of the algorithm provide an edgeover the many other algorithms Finally an application of theproposed algorithm is given for solving the heat equation inan infinite cylinder with a radiation condition
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] I N Sneddon The Use of Integral Transforms McGraw-Hill1972
[2] A E Siegman ldquoQuasi fastHankel transformrdquoOptics Letters vol1 no 1 pp 13ndash15 1977
[3] L Filon ldquoOn quadrature formula for trigonometric integralsrdquoProceedings of the Royal Society of Edinburgh vol 49 pp 38ndash471928-1929
[4] D Levin andA Sidi ldquoTwo new classes of nonlinear transforma-tions for accelerating the convergence of infinite integrals andseriesrdquo Applied Mathematics and Computation vol 9 no 3 pp175ndash215 1981
[5] A Sidi ldquoExtrapolation methods for oscillatory infinite inte-gralsrdquo Journal of the Institute of Mathematics and its Applica-tions vol 26 no 1 pp 1ndash20 1980
[6] A Sidi ldquoA user-friendly extrapolation method for oscillatoryinfinite integralsrdquoMathematics of Computation vol 51 no 183pp 249ndash266 1988
[7] A Sidi ldquoComputation of infinite integrals involving Besselfunctions of arbitrary order by the 119863-transformationrdquo Journalof Computational and Applied Mathematics vol 78 no 1 pp125ndash130 1997
[8] A Sidi Practical Extrapolation Methods Theory and Applica-tions vol 10 of Cambridge Monographs on Applied and Compu-tational Mathematics Cambridge University Press CambridgeUK 2003
[9] M Guizar-Sicairos and J C Gutierrez-Vega ldquoComputation ofquasi-discrete Hankel transforms of integer order for propagat-ing optical wave fieldsrdquo Journal of the Optical Society of AmericaA Optics Image Science and Vision vol 21 no 1 pp 53ndash582004
[10] L Yu M Huang M Chen W Chen W Huang and Z ZhuldquoQuasi-discrete Hankel transformrdquoOptics Letters vol 23 no 6pp 409ndash411 1998
[11] E B Postnikov ldquoAbout calculation of the Hankel transformusing preliminary wavelet transformrdquo Journal of Applied Math-ematics no 6 pp 319ndash325 2003
[12] V K Singh O P Singh and R K Pandey ldquoNumerical evalu-ation of the Hankel transform by using linear Legendre multi-waveletsrdquoComputer Physics Communications vol 179 no 6 pp424ndash429 2008
[13] V K Singh O P Singh and R K Pandey ldquoEfficient algorithmsto compute Hankel transforms using waveletsrdquo ComputerPhysics Communications vol 179 no 11 pp 812ndash818 2008
[14] O P Singh V K Singh and R K Pandey ldquoAn efficient and sta-ble algorithm for numerical evaluation of Hankel transformsrdquoJournal of Applied Mathematics amp Informatics vol 28 no 5-6pp 1055ndash1071 2010
[15] R K Pandey O P Singh and V K Singh ldquoA stable algorithmfor numerical evaluation of Hankel transforms using Haarwaveletsrdquo Numerical Algorithms vol 53 no 4 pp 451ndash4662010
[16] E Babolian and M Mordad ldquoA numerical method for solvingsystems of linear and nonlinear integral equations of the secondkind by hat basis functionsrdquo Computers amp Mathematics withApplications vol 62 no 1 pp 187ndash198 2011
[17] R S Pathak and O P Singh ldquoFinite Hankel transforms ofdistributionsrdquo Pacific Journal of Mathematics vol 99 no 2 pp439ndash458 1982
[18] A Erdelyi EdTables of Integral TransformsMcGraw-Hill NewYork NY USA 1954
[19] M Abramowitz and I A Stegun Handbook of MathematicalFunctions with Formulas Graphs and Mathematical TablesDover New York NY USA 1965
[20] R Barakat and B H Sandler ldquoEvaluation of first-order Hankeltransforms using Filon quadrature philosophyrdquo Applied Mathe-matics Letters vol 11 no 1 pp 127ndash131 1998
[21] J D Secada ldquoNumerical evaluation of the Hankel transformrdquoComputer Physics Communications vol 116 no 2-3 pp 278ndash294 1999
[22] R Barakat and E Parshall ldquoNumerical evaluation of the zero-order Hankel transform using Filon quadrature philosophyrdquoApplied Mathematics Letters vol 9 no 5 pp 21ndash26 1996
[23] J Gaskell Linear Systems Fourier Transforms and Opticschapter 11 John Wiley amp Sons New York NY USA 1978
[24] A D Poularika ldquoThe Hankel transformrdquo in The Handbook ofFormulas and Tables for Signal Processing CRC Press LLC BocaRaton Fla USA 1999
[25] G NWatsonTheory of Bessel Functions Cambridge UniversityPress Cambridge UK 2nd edition 1958
Submit your manuscripts athttpwwwhindawicom
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Analysis 3
Using (5) and (6) we may approximate 119892(119903) as
119892 (119903) cong
119899
sum
119894=0
119892 (119894ℎ) 120595119894(119903) (10)
From (10) (9) may be written as
119865] (119901) cong int1
0
(
119899
sum
119894=0
119892 (119894ℎ) 120595119894(119903)) 119869] (119901119903) 119889119903
=
119899
sum
119894=0
119892 (119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
(11)
Using (3) (11) may be expressed as
119865] (119901) cong 119892 (0) intℎ
0
(
ℎ minus 119903
ℎ
) 119869] (119901119903) 119889119903
+
119899minus1
sum
119894=1
119892 (119894ℎ) [int
119894ℎ
(119894minus1)ℎ
(
119903 minus (119894 minus 1) ℎ
ℎ
) 119869] (119901119903) 119889119903
+int
(119894+1)ℎ
119894ℎ
((119894 + 1) ℎ minus 119903
ℎ
) 119869] (119901119903) 119889119903]
+ 119892 (119899ℎ) int
1
1minusℎ
(
119903 minus (1 minus ℎ)
ℎ
) 119869] (119901119903) 119889119903
(12)
It is noteworthy here that the integrals involved in (12) areevaluated by using the following formulae
int
119886
0
119869] (119905) 119889119905 = 2 lim119873rarrinfin
119873
sum
119899=0
119869]+2119899+1 (119886) Re (]) gt minus1 (13)
(see [18 p 333])
int
119886
0
1199051minus]119869] (119905) 119889119905 =
1
2]minus1 ]
minus 1198861minus]119869]minus1 (119886) (14)
(see [18 p 333])
int
119886
0
119905120583
119869] (119905) 119889119905
= 119886120583(] + 120583 + 1) 2
(] minus 120583 + 1) 2
times lim119873rarrinfin
119873
sum
119899=0
(] + 2119899 + 1) Γ (((] minus 120583 + 1) 2) + 119899)Γ (((] + 120583 + 3) 2) + 119899)
times 119869]+2119899+1 (119886) Re (] + 120583 + 1) gt 0
(15)
(see [19 p 480])
4 Error Analysis
Let the RHS of (10) be denoted by 119892119899(119903) that is
119892119899(119903) =
119899
sum
119894=0
119892 (119894ℎ) 120595119894(119903) (16)
Now replacing 119892(119903) by 119892119899(119903) in (9) we define an 119899th approxi-
mate 119865]119899(119901) of the FHT 119865](119901) as follows
Definition 1 An 119899th approximate finite Hankel transform of119891(119903) denoted by 119865]119899(119901) is defined as
119865]119899 (119901) = int1
0
119892119899(119903) 119869] (119901119903) 119889119903
=
119899
sum
119894=0
119892 (119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
(17)
Let 120576119899(119901) denote the absolute error between the FHT 119865](119901)
and its 119899th approximate 119865]119899(119901) then we have the following
Theorem2 If119892(119903) is approximated by the family of first (119899+1)hat functions as given in (10) then
(i) |119892(119895ℎ) minus 119892119899(119895ℎ)| = 0 for 119895 = 0 1 2 119899
(ii) |119892(119903) minus 119892119899(119903)| le (12119899
2
)|11989210158401015840
(119895ℎ)| + 119874(11198993
) for 119895ℎ lt119903 lt (119895 + 1)ℎ 119895 = 0 1 2 119899 minus 1
(iii) 120576119899(119901) = |119865](119901) minus 119865]119899(119901)| le 1198722119899
2
+ 119874(11198993
) where|11989210158401015840
(119895ℎ)| le 119872
Proof (i) From (16) and (4) the value of 119892119899(119903) at 119895th nodal
point 119903 = 119895ℎ 119895 = 0 1 2 119899 is given by 119892119899(119895ℎ) =
sum119899
119894=0119892(119894ℎ)120595
119894(119895ℎ) = 119892(119895ℎ)
So |119892(119895ℎ) minus 119892119899(119895ℎ)| = 0 for 119895 = 0 1 2 119899
(ii) If 119903 lies between two consecutive integer multiples ofℎ that is 119895ℎ lt 119903 lt (119895 + 1)ℎ 119895 = 0 1 2 119899 minus 1 then from(16) we have
119892119899(119903) =
119899
sum
119894=0
119892 (119894ℎ) 120595119894(119903)
= 119892 (119895ℎ) 120595119895(119903) + 119892 ((119895 + 1) ℎ) 120595
119895+1(119903)
= 119892 (119895ℎ) (
(119895 + 1) ℎ minus 119903
ℎ
)
+ 119892 (119895ℎ + ℎ) (
119903 minus 119895ℎ
ℎ
) (Using (3))
= 119892 (119895ℎ) minus 119895ℎ(
119892 (119895ℎ + ℎ) minus 119892 (119895ℎ)
ℎ
)
+ 119903(
119892 (119895ℎ + ℎ) minus 119892 (119895ℎ)
ℎ
)
= 119892 (119895ℎ) + (119903 minus 119895ℎ)(
119892 (119895ℎ + ℎ) minus 119892 (119895ℎ)
ℎ
)
(18)
As ℎ rarr 0 from (18) we obtain
119892119899(119903) cong 119892 (119895ℎ) + (119903 minus 119895ℎ) 119892
1015840
(119895ℎ) (19)By expanding 119892
119899(119903) in the form of Taylorrsquos series in the
powers of (119903 minus 119895ℎ) we have
119892 (119903) =
infin
sum
119896=0
(119903 minus 119895ℎ)119896
119896
119892(119896)
(119895ℎ) (20)
4 International Journal of Analysis
where 119892(119896) denotes the 119896th order derivative of 119892(119903) Using (19)and (20) the error between exact and approximate values of119892(119903) is given by
119892 (119903) minus 119892119899(119903) =
infin
sum
119896=2
(119903 minus 119895ℎ)119896
119896
119892(119896)
(119895ℎ)
=
(119903 minus 119895ℎ)2
2
11989210158401015840
(119895ℎ) + 119874 (119903 minus 119895ℎ)3
(21)
Since (119903 minus 119895ℎ) lt ℎ and 119899ℎ = 1 from (21) we get
1003816100381610038161003816119892 (119903) minus 119892
119899(119903)1003816100381610038161003816le
1
21198992
1003816100381610038161003816100381611989210158401015840
(119895ℎ)
10038161003816100381610038161003816+ 119874(
1
1198993) (22)
(iii) The absolute error 120576119899(119901) between exact FHT 119865](119901)
and 119899th approximate FHT 119865]119899(119901) is given by
120576119899(119901) =
10038161003816100381610038161003816119865] (119901) minus 119865]119899 (119901)
10038161003816100381610038161003816
=
100381610038161003816100381610038161003816100381610038161003816
int
1
0
119892 (119903) 119869] (119901119903) 119889119903 minus int1
0
119892119899(119903) 119869] (119901119903) 119889119903
100381610038161003816100381610038161003816100381610038161003816
=
10038161003816100381610038161003816100381610038161003816100381610038161003816
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
(119892 (119903) minus 119892119899(119903)) 119869] (119901119903) 119889119903
10038161003816100381610038161003816100381610038161003816100381610038161003816
(23)
As |119869](119901119903)| le 1 we have
120576119899(119901) le
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
1003816100381610038161003816119892 (119903) minus 119892
119899(119903)1003816100381610038161003816119889119903
le
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
(
1
21198992
1003816100381610038161003816100381611989210158401015840
(119895ℎ)
10038161003816100381610038161003816+ 119874(
1
1198993))119889119903
(Follows from (22))
(24)
If |11989210158401015840(119895ℎ)| le 119872 then it follows that
120576119899(119901) le
119872
21198992
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
119889119903
+ 119874(
1
1198993)
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
119889119903
=
119872
21198992+ 119874(
1
1198993) as 119899ℎ = 1
(25)
5 Stability Analysis
In this section the stability of the proposed algorithm isanalyzed under the influence of noise In what follows theexact data function is denoted by 119891(119903) and the noisy datafunction119891120572(119903) is obtained by adding a randomnoise120572 to119891(119903)such that 119891120572(119903
119894) = 119891(119903
119894) + 120572120579
119894 where 119903
119894= 119894ℎ 119894 = 0 1 2 119899
119899ℎ = 30 and 120579119894is a uniform random variable with values in
[minus1 1] such that max |119891120572(119903119894) minus 119891(119903
119894)| le 120572 0 le 119894 le 119899 If 119903119891120572(119903)
is denoted by 119892120572(119903) and the approximate HT of the perturbedfunction is denoted by 119865120572]119899(119901) then from (17)
119865120572
]119899 (119901) =
119899
sum
119894=0
119892120572
(119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
=
119899
sum
119894=0
119894ℎ (119891 (119894ℎ) + 120572120579119894) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
(26)
From (26) and (17) we have
10038161003816100381610038161003816119865120572
]119899 (119901) minus 119865]119899 (119901)10038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816100381610038161003816
119899
sum
119894=0
119892120572
(119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
minus
119899
sum
119894=0
119892 (119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
1003816100381610038161003816100381610038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816100381610038161003816
119899
sum
119894=0
119894ℎ (120572120579119894) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
1003816100381610038161003816100381610038161003816100381610038161003816
le
119899
sum
119894=0
119894ℎ1003816100381610038161003816120572120579119894
1003816100381610038161003816int
1
0
1003816100381610038161003816120595119894(119903)1003816100381610038161003816
1003816100381610038161003816119869] (119901119903)
1003816100381610038161003816119889119903
le
119899
sum
119894=0
119894ℎ1003816100381610038161003816120572120579119894
1003816100381610038161003816int
1
0
1003816100381610038161003816120595119894(119903)1003816100381610038161003816119889119903 (As 100381610038161003816
1003816119869] (119901119903)
1003816100381610038161003816le 1)
(27)
Let 120579119872= max|120579
119894| then substituting 120595
119894(119903) from (3) we get
10038161003816100381610038161003816119865120572
]119899 (119901) minus 119865]119899 (119901)10038161003816100381610038161003816
le 120579119872ℎ |120572|
times [
119899minus1
sum
119894=1
119894 (int
119894ℎ
(119894minus1)ℎ
(119903 minus (119894 minus 1) ℎ)
ℎ
119889119903
+int
(119894+1)ℎ
119894ℎ
((119894 + 1) ℎ minus 119903)
ℎ
119889119903)
+ 119899int
1
1minusℎ
(119903 minus (1 minus ℎ))
ℎ
119889119903]
= 120579119872|120572| [ℎ
2
119899minus1
sum
119894=1
119894 + 119899
ℎ2
2
] = 120579119872|120572| (
1198992
ℎ2
2
)
le|120572|
2
(As 120579119872le 1 119899ℎ = 1)
(28)
Thus we have proved that
Theorem 3 When the input signal 119891(119903) is corrupted withnoise 120572 the proposed algorithm reduces the noise at least bya factor of 12 in the output data 119865]119899(119901)
International Journal of Analysis 5
Table 1 Least square errors 119864119895(119901)2
in different examples for 119899 = 100
119864119895(119901)2
Example 1 Example 2 Example 3 Example 4 Example 510038171003817100381710038171198640(119901)
10038171003817100381710038172
72696119890 minus 005 11078119890 minus 016 49160119890 minus 006 69741119890 minus 005 66819119890 minus 006
10038171003817100381710038171198641(119901)
10038171003817100381710038172
69980119890 minus 005 13724119890 minus 004 15965119890 minus 004 73362119890 minus 005 60667119890 minus 005
10038171003817100381710038171198642(119901)
10038171003817100381710038172
22536119890 minus 004 33312119890 minus 004 34473119890 minus 004 17687119890 minus 004 13147119890 minus 004
6 Numerical Illustrations
The test problems included in this paper are solved withand without random perturbations (noises) to illustrate theefficiency and stability of proposed algorithm by choosingthree different values of noise 120572 as 120572
0= 0 120572
1= 0002 and
1205722= 0005 (as discussed in the beginning of Section 5) In the
following examples the errors119864119895(119901) 119895 = 0 1 2 are computed
and their graphs are sketched for 119899 = 100 where 119864119895(119901) =
(Exact FHT 119865](119901)minus approximate FHT 119865120572119895]119899(119901) obtained from(26) with random noise 120572
119895)
Further the parameter 119901 ranges between 0 and 30 in stepsof 02 Figures 3 and 6 depict the graph of |119865120572119895]119899(119901) minus 119865]119899(119901)|119895 = 1 2 for different test functions in Examples 1 and 2respectively These graphs are in conformity withTheorem 3For all the illustrations the computations are done in MAT-LAB 701 For different examples the least square errors119864119895(119901)2involved in computations of approximate FHT with
noises 120572119895 119895 = 0 1 2 are given in Table 1These are calculated
using the formula
10038171003817100381710038171003817119864119895(119901)
100381710038171003817100381710038172=radicsum119899
119894=0(119864119895(119901119894))
2
119899 + 1
(29)
where 119901119894is taken in steps of 02 in the range [0 30]
Example 1 Consider the function 119891(119903) = radic(1 minus 1199032) 0 le 119903 le1 given in [14 20] for which
1198651(119901) =
120587
1198692
1(1199012)
2119901
0 lt 119901 lt infin
0 119901 = 0
(30)
Numerical evaluation of 1198651(119901) has been achieved by Barakat
and Sandler [20] by using Filon quadrature philosophy butagain the associated error is appreciable for 119901 lt 1 whereasour method gives almost zero error in that range Equation(12) is used to obtain the 119899th approximate HT 119865
1119899(119901) The
comparison between exact HT 1198651(119901) and approximate HT
1198651119899(119901) is shown in Figure 1 In Figure 2 the errors 119864
0(119901)
1198641(119901) and119864
2(119901) for 119899 = 100 are plotted In Figure 3 |119865120572
1119899(119901)minus
1198651119899(119901)| is shown for noises 120572 = 0002 0005 This figure is in
conformity withTheorem 3
Example 2 (Sombrero function) A very important and oftenused function is the Circ function that can be defined as
Circ( 119903119886
) =
1 119903 le 119886
0 119903 gt 119886
(31)
0 5 10 15 20 25 30
0
002
004
006
008
01
012
014
016
018
p
Exac
t and
appr
oxim
ate F
HT
minus002
Figure 1 Exact HT 1198651(119901) (solid line) and approximate HT 119865
1119899(119901)
(dotted line) for 119899 = 100 Example 1
The zeroth order HT of Circ(119903119886) is the Sombrero function[13 21] that will be written as 119878
0(119901) with the following
analytical expression
1198780(119901) = 119886
21198691(119886119901)
119886119901
(32)
We use (12) to obtain the approximation for the FHT 1198650119899(119901)
In Figure 4 we sketch the error 1198640(119901) between exact HT
1198780(119901) and approximate FHT119865
0119899(119901) of Circ function (without
noise) for 119886 = 1 and 119899 = 10 It is evident that even forsuch a small value of 119899 the error is appreciably small Figure 5compares the errors 119864
1(119901) and 119864
2(119901) for 119899 = 100 Figure 6
shows the plot of |1198651205720119899(119901) minus 119865
0119899(119901)| for 120572 = 0002 and 120572 =
0005 with 119899 = 100
Example 3 Consider the function 119891(119903) = (2120587)[arccos(119903) minus119903radic(1 minus 119903
2)] 0 le 119903 le 1 given in [14 22] forwhich zeroth order
exact HT is
1198650(119901) = 2
1198692
1(1199012)
1199012
0 le 119901 lt infin (33)
It is a well-known pair which arises in optical diffractiontheory [23] The function 119891(119903) is known as optical transferfunction of an aberration-free optical system with a circularaperture and 119865
0(119901) is the corresponding spread function
Barakat and Parshall [22] evaluated1198650(119901) using Filon quadra-
ture philosophy but the associated error is again appreciablefor 119901 lt 1 whereas our method gives significantly small error
6 International Journal of Analysis
0 5 10 15 20 25 30
0
1
2
3
p
minus5
minus4
minus3
minus2
minus1
times10minus4
E0(p)E1(p)
andE2(p)
E0(p)
E1(p)
E2(p)
Figure 2 1198640(119901) 119864
1(119901) 119864
2(119901) for 119899 = 100 Example 1
0 5 10 15 20 25 300
1
2
3
4
5
6
7
8
p
times10minus4
For noise 120572 = 0005
For noise 120572 = 0002
Figure 3 Plot of |1198651205721119899(119901)minus119865
1119899(119901)| for 120572 = 0002 and 120572 = 0005with
119899 = 100 Example 1
in that range In Figure 7 the error 1198640(119901) is shown Figure 8
is the comparison between 1198641(119901) and 119864
2(119901) for 119899 = 100
Example 4 Consider the function [24] 119891(119903) = 119903](1198862
minus 1199032
)120583
119867(119886 minus 119903) Re(]) gt minus1Re(120583) gt minus1 0 lt 119903 lt 119886 where119867(119903) is a unit step function
defined by
119867(119903) =
1 119903 ge 0
0 119903 lt 0
(34)
The ]th order HT of 119891(119903) is given by
119865] (119901) = 2]119886120583+]+1
119901minus120583minus1
120583 + 1 119869]+120583+1 (119886119901) (35)
Using the method of solution developed in Section 3 theapproximate HT 119865]119899(119901) has been calculated for ] = 32
0 5 10 15 20 25 30p
0
05
1
15
2
25
3
minus05
E0(p)
times10minus11
Figure 4 1198640(119901) for 119899 = 10 Example 2
0 5 10 15 20 25 30p
0
2
minus14
minus12
minus10
minus8
minus6
minus4
minus2
times10minus4
E1(p)
andE2(p)
E1(p)
E2(p)
Figure 5 1198641(119901) and 119864
2(119901) for 119899 = 100 Example 2
0 5 10 15 20 25 30p
0
02
04
06
08
1
12times10minus3
For noise 120572 = 0005
For noise 120572 = 0002
Figure 6 Plot of |1198651205720119899(119901)minus119865
0119899(119901)| for 120572 = 0002 and 120572 = 0005with
119899 = 100 Example 2
International Journal of Analysis 7
0
01
02
03
04
05
06
07
08
09
1
0 5 10 15 20 25 30p
times10minus5
E0(p)
Figure 7 1198640(119901) for 119899 = 100 in Example 3
0
2
0 5 10 15 20 25 30p
times10minus4
E1(p)
andE2(p)
E1(p)
E2(p)
minus12
minus10
minus8
minus6
minus4
minus2
Figure 8 1198641(119901) and 119864
2(119901) for 119899 = 100 in Example 3
0
001
002
003
004
005
006
007
008
Exac
t and
appr
oxim
ate F
HT
0 5 10 15 20 25 30p
minus002
minus001
Figure 9 Exact HT 11986532(119901) (solid line) and approximate HT
11986532119899
(119901) (dotted line) for 120583 = 12 and 119899 = 100 Example 4
0
1
2
0 5 10 15 20 25 30p
minus4
minus3
minus2
minus1
times10minus4
E0(p)
E1(p)
E2(p)
E0(p)E1(p)
andE2(p)
Figure 10 1198640(119901) 119864
1(119901) 119864
2(119901) for 119899 = 100 in Example 4
0
05
1
0 5 10 15 20 25 30p
Erro
r (w
ithou
t noi
se )E0(p)
minus2
minus15
minus1
minus05
times10minus5
Figure 11 1198640(119901) for 119899 = 100 in Example 5
120583 = 12 and 119886 = 1 The comparison between 11986532119899
(119901) andexact FHT 119865
32(119901) is shown in Figure 9 Further in Figure 10
a comparison between the different errors 1198640(119901) 119864
1(119901) and
1198642(119901) is shown for 119899 = 100
Example 5 In this example we choose as a test function thegeneralized version of the top-hat function given as 119891(119903) =119903][119867(119903) minus119867(119903 minus 119886)] 119886 gt 0 and119867(119903) is the unit step function
defined by (34) Then
119865] (119901) =119869]+1 (119901)
119901
(36)
In [9] authors took 119886 = 1 and ] = 4 for numericalcalculations We take 119886 = 1 ] = 5 and observe that theassociated errors with and without random noises are quitesmall The error 119864
0(119901) (without noise) is shown in Figure 11
8 International Journal of Analysis
0
05
0 5 10 15 20 25 30p
minus3
minus25
minus2
minus15
minus1
minus05
times10minus4
2E1(p)
andE2(p)
2E1(p)
E2(p)
Figure 12 Comparison between 1198641(119901) and 119864
2(119901) for 119899 = 100 in
Example 5
0 02 04 06 08 1
0
02
04
06
08
1
12
r
minus02
f(r)
ua(r0)
andf(r)
ua(r 0)
Figure 13 The initial condition function 119891(119903) (solid red line) and119906119886(119903 0) (= lim
119905rarr0+119906119886(119903 119905)) (lowastblue)
0 01 02 03 04 05 06 07 08 09
0
1
2
3
4
r
minus3
minus2
minus1
times10minus3
E(r)
Figure 14 Error between 119891(119903) and 119906119886(119903 0)
0 02 04 06 08 1
0
02
04
06
08
1
12
r
Solu
tion
for d
iffer
ent t
minus02
u0 for t = 0
u1 for t = 001
u2 for t = 002
u3 for t = 004
Figure 15The various profiles of the solutions 119906119886(119903 119905) at fixed times
0 01 02 03 04 05 06 07 08 09 10
1
2
r
Abso
lute
erro
rtimes10minus4
For t = 0
For t = 001
For t = 002
For t = 004
Figure 16 The absolute error between 119906119886(119903 119905) and 119906(119903 119905) for
different 119905
0
05
1
0
05
1
0
02
04
06
08
1
rt
minus02
u(rt)
Figure 17 The solution 119906(119903 119905) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
International Journal of Analysis 9
0
05
1
0
05
10
02
04
06
08
1
rt
ua(rt)
Figure 18 The approximate solution 119906119886(119903 119905) using 119899th approximate
FHT 1198655119899(1205735119904) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
for 119899 = 100 In Figure 12 errors 1198641(119901) and 119864
2(119901) for 119899 = 100
are shown
7 Applications
In this section we solve heat equation in cylindrical coor-dinates inside an infinitely long cylinder of radius unity byusing the theory of finite Hankel transform (FHT) developedin the preceding pages The initial temperature is givenby 119891(119903) and radiation takes place at the surface into thesurrounding medium maintained at zero temperature Weseek a function 119906(119903 119905) where 119903 is radius and 119905 is time (119906 doesnot depend upon 120579 and 119905) in the following application Themathematical model of this problem is the diffusion equationin polar coordinates given by
1205972
119906 (119903 119905)
1205972119903
+
1
119903
120597119906 (119903 119905)
120597119903
minus
]2
1199032119906 (119903 119905)
=
120597119906 (119903 119905)
120597119905
(0 lt 119903 lt 1 0 lt 119905 lt infin)
(37)
with the following initial and boundary conditions
(i) as 119905 rarr 0+ 119906(119903 119905) rarr 119891(119903) = 119903
][119867(119903) minus 119867(119903 minus 1)]
where 0 le 119903 le 1 and119867(119903) is unit step function givenby (34)
(ii) as 119903 rarr 1minus (120597119906120597119903) + 119866119906 rarr 0 for each fixed 119905 gt 0
and 119866 gt 0
When 119906(119903 119905) denotes the temperature within the cylinder119866 gt 0means that heat is being radiated away from the surfaceof the cylinder
LetΩ]119903 denote the Bessel differential operator (1205972
1205972
119903) +
(1119903)(120597120597119903) minus (]21199032) Then differential equation (37) can bewritten as
Ω]119903119906 =120597119906
120597119905
(38)
Applying the ]th order finite Hankel transform operator toboth sides of the differential equation (38) and using (8) weget
int
1
0
119903Ω]119903119906119869] (120573]119904119903) 119889119903 = int1
0
119903 (
120597119906
120597119905
) 119869] (120573]119904119903) 119889119903 (39)
where 120573]119904 is the 119904th positive root of equation [24]
119866119869] (119911) + 1199111198691015840
] (119911) = 0 ] ge minus1
2
(40)
The LHS of (39) may be written as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= int
1
0
(
1205972
119906
1205972119903
) 119903119869] (120573]119904119903) 119889119903
+ int
1
0
(
120597119906
120597119903
) 119869] (120573]119904119903) 119889119903 minus ]2 int1
0
(
119906
119903
) 119869] (120573]119904119903) 119889119903
= lim119903rarr1minus
120597119906 (119903 119905)
120597119903
119869] (120573]119904) minus 120573]119904 int1
0
119903
120597119906
120597119903
1198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
= minus119866119869] (120573]119904) lim119903rarr1minus119906 (119903 119905) minus 120573]119904 int
1
0
120597119906
120597119903
1199031198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903 (As 119903 997888rarr 1minus
120597119906
120597119903
+ 119866119906 997888rarr 0)
(41)
Since 120573]119904 is a root of 119866119869](119911) + 1199111198691015840
](119911) = 0 we have 119866119869](120573]119904) =minus120573]119904119869
1015840
](120573]119904) and thus
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
minus 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
+ 120573]119904 int1
0
119906 (1198691015840
] (120573]119904119903) + 120573]11990411990311986910158401015840
] (120573]119904119903)) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
(42)
Using the fact that 119869](119909) is a solution of 1199092(11988921199101198891199092) +119909(119889119910119889119909) + (119909
2
minus ]2)119910 = 0 (42) may be simplified as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= minus1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
(43)
10 International Journal of Analysis
From (43) and (39) we have
minus 1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
=
120597
120597119905
(int
1
0
119903119906 (119903 119905) 119869] (120573]119904119903) 119889119903)
(44)
If FHT int10
119906(119903 119905)119903119869](120573]119904119903) 119889119903 of 119906(119903 119905) is denoted by119880(120573]119904 119905)then (44) may be written as
120597119880 (120573]119904 119905)
120597119905
= minus1205732
]119904119880(120573]119904 119905) (45)
whose solution is given by
119880(120573]119904 119905) = 119860 (120573]119904) 119890minus1205732]119904119905 (46)
The constant of integration 119860(120573]119904) in (46) is determined byinitial condition Thus
119860 (120573]119904) = 119865] (120573]119904) = int1
0
119891 (119903) 119903119869] (120573]119904119903) 119889119903 (47)
Hence from (47) (46) becomes
119880 (120573]119904 119905) = 119865] (120573]119904) 119890minus1205732]119904119905 (48)
Therefore by inversion theorem [25] we have
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 1205732]11990411986910158402] (120573]119904)
= lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 11986621198692] (120573]119904)
(As 119866119869] (120573]119904) = minus120573]1199041198691015840
] (120573]119904))
(49)
So
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
119869] (120573]119904119903)
1198692
] (120573]119904) (50)
We want to prove that 119906(119903 119905) given by (50) is truly a solutionof (37) that satisfies the given initial and boundary conditionsTo achieve this we need the following well known estimates[17]
119865] (120573]119904) = 119874(1
12057332
]119904)
120573]119904 asymp 120587(119904 +1
4
)
1198692
] (120573]119904) asymp2
120587120573]119904
as 119904 997888rarr infin
(51)
Using the above estimates we see that series (50) and theseries obtained by applyingΩ]119903 and 120597120597119905 separately under thesummation sign of (50) converge uniformly on 0 lt 119903 lt 1 and119905 gt 0 Hence by applyingΩ]119903 minus 120597120597119905 on (50) we get
(Ω]119903 minus120597
120597119905
) 119906 (119903 119905)
= ( lim119873rarrinfin
119873
sum
119904=1
21205734
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
sdot [
[
(1205732
]1199041199032
11986910158401015840
] (120573]119904119903) + 120573]1199041199031198691015840
] (120573]119904119903)
+ (1205732
]1199041199032
minus ]2) 119869] (120573]119904119903))
times (1205732
]1199041199032
1198692
] (120573]119904))minus1
]
]
) = 0
(52)
Hence we see that (50) satisfies the differential equation (37)Let us verify the boundary condition (ii) we have
lim119903rarr1minus[
120597119906 (119903 119905)
120597119903
+ 119866119906 (119903 119905)]
= lim119903rarr1minus
[
[
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662) 1198692] (120573]119904)
times (120573]1199041198691015840
] (120573]119904119903) + 119866119869] (120573]119904119903))]
]
(53)
and since the convergence is uniform we can take thelim119903rarr1minus inside the summation sign and arrive at the conclu-
sion as 120573]119904rsquos are the roots of the equation119866119869](119911)+1199111198691015840
](119911) = 0The initial condition (i) is already taken care of as we
evaluated the constant 119860(120573]119904) by using it Through Figures13ndash18 we establish the accuracy of the proposed methodFor numerical solution of differential equation (37) we havetaken ] = 5 and 119866 = 1 We have used MATLAB routineldquofzerordquo to find zeros of (40) and we have drawn Figures 13ndash18 by truncating series (50) at 119873 = 30 While evaluatingthe solution 119906(119903 119905) from (50) we have evaluated 119865
5(1205735119904) first
from its analytical expression given by (36) and denoted thesolution thus obtained by 119906(119903 119905) in Figures 13ndash18 and thenevaluating 119899th approximates FHT 119865
5119899(1205735119904) for 119899 = 100
using our proposed algorithm for evaluation of the FHT asgiven by (17) This solution is denoted by 119906
119886(119903 119905) in the above
mentioned figuresFigure 13 compares the given initial condition 119891(119903) with
119906119886(119903 119905) as 119905 rarr 0
+ and Figure 14 shows the correspondingerror 119864(119903) = 119906
119886(119903 0) minus 119891(119903) Figure 15 depicts the various
profiles of 119906119886(119903 119905) at fixed times 119905
119895= 0 1100 150 125
where the various profiles are denoted by 1199060 1199061 1199062 and
1199063 The absolute errors between 119906
119886(119903 119905) and 119906(119903 119905) at fixed
times 119905119895= 0 1100 150 125 are shown in Figure 16 As
the maximum possible error occurs in the neighborhood of
International Journal of Analysis 11
0 and 0001 we have restricted 119905 in (0 1] in Figures 17 and18 representing 119906(119903 119905) and 119906
119886(119903 119905) respectively and note that
they are in good agreement in the range
8 Conclusions
A new stable and efficient algorithm based on hat functionsfor the numerical evaluation of Hankel transform is proposedand analyzed Choosing hat basis functions to expand theinput signal 119903119891(119903) makes our algorithm attractive in theirapplication in the applied physical problems as they eliminatethe problems connected with the Gibbs phenomenon takingplace in [9 11] We have given for the first time (as per ourinformation) error and stability analysis which was one ofthe main drawbacks of the earlier algorithms and by variousnumerical experiments we have corroborated our theoreticalfindings Stability with respect to the data is restored andexcellent accuracy is obtained even for small sample intervaland high noise levels in the data From the various figuresit is obvious that the algorithm is consistent and does notdepend on the particular choice of the input signal Theaccuracy and simplicity of the algorithm provide an edgeover the many other algorithms Finally an application of theproposed algorithm is given for solving the heat equation inan infinite cylinder with a radiation condition
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] I N Sneddon The Use of Integral Transforms McGraw-Hill1972
[2] A E Siegman ldquoQuasi fastHankel transformrdquoOptics Letters vol1 no 1 pp 13ndash15 1977
[3] L Filon ldquoOn quadrature formula for trigonometric integralsrdquoProceedings of the Royal Society of Edinburgh vol 49 pp 38ndash471928-1929
[4] D Levin andA Sidi ldquoTwo new classes of nonlinear transforma-tions for accelerating the convergence of infinite integrals andseriesrdquo Applied Mathematics and Computation vol 9 no 3 pp175ndash215 1981
[5] A Sidi ldquoExtrapolation methods for oscillatory infinite inte-gralsrdquo Journal of the Institute of Mathematics and its Applica-tions vol 26 no 1 pp 1ndash20 1980
[6] A Sidi ldquoA user-friendly extrapolation method for oscillatoryinfinite integralsrdquoMathematics of Computation vol 51 no 183pp 249ndash266 1988
[7] A Sidi ldquoComputation of infinite integrals involving Besselfunctions of arbitrary order by the 119863-transformationrdquo Journalof Computational and Applied Mathematics vol 78 no 1 pp125ndash130 1997
[8] A Sidi Practical Extrapolation Methods Theory and Applica-tions vol 10 of Cambridge Monographs on Applied and Compu-tational Mathematics Cambridge University Press CambridgeUK 2003
[9] M Guizar-Sicairos and J C Gutierrez-Vega ldquoComputation ofquasi-discrete Hankel transforms of integer order for propagat-ing optical wave fieldsrdquo Journal of the Optical Society of AmericaA Optics Image Science and Vision vol 21 no 1 pp 53ndash582004
[10] L Yu M Huang M Chen W Chen W Huang and Z ZhuldquoQuasi-discrete Hankel transformrdquoOptics Letters vol 23 no 6pp 409ndash411 1998
[11] E B Postnikov ldquoAbout calculation of the Hankel transformusing preliminary wavelet transformrdquo Journal of Applied Math-ematics no 6 pp 319ndash325 2003
[12] V K Singh O P Singh and R K Pandey ldquoNumerical evalu-ation of the Hankel transform by using linear Legendre multi-waveletsrdquoComputer Physics Communications vol 179 no 6 pp424ndash429 2008
[13] V K Singh O P Singh and R K Pandey ldquoEfficient algorithmsto compute Hankel transforms using waveletsrdquo ComputerPhysics Communications vol 179 no 11 pp 812ndash818 2008
[14] O P Singh V K Singh and R K Pandey ldquoAn efficient and sta-ble algorithm for numerical evaluation of Hankel transformsrdquoJournal of Applied Mathematics amp Informatics vol 28 no 5-6pp 1055ndash1071 2010
[15] R K Pandey O P Singh and V K Singh ldquoA stable algorithmfor numerical evaluation of Hankel transforms using Haarwaveletsrdquo Numerical Algorithms vol 53 no 4 pp 451ndash4662010
[16] E Babolian and M Mordad ldquoA numerical method for solvingsystems of linear and nonlinear integral equations of the secondkind by hat basis functionsrdquo Computers amp Mathematics withApplications vol 62 no 1 pp 187ndash198 2011
[17] R S Pathak and O P Singh ldquoFinite Hankel transforms ofdistributionsrdquo Pacific Journal of Mathematics vol 99 no 2 pp439ndash458 1982
[18] A Erdelyi EdTables of Integral TransformsMcGraw-Hill NewYork NY USA 1954
[19] M Abramowitz and I A Stegun Handbook of MathematicalFunctions with Formulas Graphs and Mathematical TablesDover New York NY USA 1965
[20] R Barakat and B H Sandler ldquoEvaluation of first-order Hankeltransforms using Filon quadrature philosophyrdquo Applied Mathe-matics Letters vol 11 no 1 pp 127ndash131 1998
[21] J D Secada ldquoNumerical evaluation of the Hankel transformrdquoComputer Physics Communications vol 116 no 2-3 pp 278ndash294 1999
[22] R Barakat and E Parshall ldquoNumerical evaluation of the zero-order Hankel transform using Filon quadrature philosophyrdquoApplied Mathematics Letters vol 9 no 5 pp 21ndash26 1996
[23] J Gaskell Linear Systems Fourier Transforms and Opticschapter 11 John Wiley amp Sons New York NY USA 1978
[24] A D Poularika ldquoThe Hankel transformrdquo in The Handbook ofFormulas and Tables for Signal Processing CRC Press LLC BocaRaton Fla USA 1999
[25] G NWatsonTheory of Bessel Functions Cambridge UniversityPress Cambridge UK 2nd edition 1958
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 International Journal of Analysis
where 119892(119896) denotes the 119896th order derivative of 119892(119903) Using (19)and (20) the error between exact and approximate values of119892(119903) is given by
119892 (119903) minus 119892119899(119903) =
infin
sum
119896=2
(119903 minus 119895ℎ)119896
119896
119892(119896)
(119895ℎ)
=
(119903 minus 119895ℎ)2
2
11989210158401015840
(119895ℎ) + 119874 (119903 minus 119895ℎ)3
(21)
Since (119903 minus 119895ℎ) lt ℎ and 119899ℎ = 1 from (21) we get
1003816100381610038161003816119892 (119903) minus 119892
119899(119903)1003816100381610038161003816le
1
21198992
1003816100381610038161003816100381611989210158401015840
(119895ℎ)
10038161003816100381610038161003816+ 119874(
1
1198993) (22)
(iii) The absolute error 120576119899(119901) between exact FHT 119865](119901)
and 119899th approximate FHT 119865]119899(119901) is given by
120576119899(119901) =
10038161003816100381610038161003816119865] (119901) minus 119865]119899 (119901)
10038161003816100381610038161003816
=
100381610038161003816100381610038161003816100381610038161003816
int
1
0
119892 (119903) 119869] (119901119903) 119889119903 minus int1
0
119892119899(119903) 119869] (119901119903) 119889119903
100381610038161003816100381610038161003816100381610038161003816
=
10038161003816100381610038161003816100381610038161003816100381610038161003816
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
(119892 (119903) minus 119892119899(119903)) 119869] (119901119903) 119889119903
10038161003816100381610038161003816100381610038161003816100381610038161003816
(23)
As |119869](119901119903)| le 1 we have
120576119899(119901) le
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
1003816100381610038161003816119892 (119903) minus 119892
119899(119903)1003816100381610038161003816119889119903
le
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
(
1
21198992
1003816100381610038161003816100381611989210158401015840
(119895ℎ)
10038161003816100381610038161003816+ 119874(
1
1198993))119889119903
(Follows from (22))
(24)
If |11989210158401015840(119895ℎ)| le 119872 then it follows that
120576119899(119901) le
119872
21198992
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
119889119903
+ 119874(
1
1198993)
119899minus1
sum
119895=0
int
(119895+1)ℎ
119895ℎ
119889119903
=
119872
21198992+ 119874(
1
1198993) as 119899ℎ = 1
(25)
5 Stability Analysis
In this section the stability of the proposed algorithm isanalyzed under the influence of noise In what follows theexact data function is denoted by 119891(119903) and the noisy datafunction119891120572(119903) is obtained by adding a randomnoise120572 to119891(119903)such that 119891120572(119903
119894) = 119891(119903
119894) + 120572120579
119894 where 119903
119894= 119894ℎ 119894 = 0 1 2 119899
119899ℎ = 30 and 120579119894is a uniform random variable with values in
[minus1 1] such that max |119891120572(119903119894) minus 119891(119903
119894)| le 120572 0 le 119894 le 119899 If 119903119891120572(119903)
is denoted by 119892120572(119903) and the approximate HT of the perturbedfunction is denoted by 119865120572]119899(119901) then from (17)
119865120572
]119899 (119901) =
119899
sum
119894=0
119892120572
(119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
=
119899
sum
119894=0
119894ℎ (119891 (119894ℎ) + 120572120579119894) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
(26)
From (26) and (17) we have
10038161003816100381610038161003816119865120572
]119899 (119901) minus 119865]119899 (119901)10038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816100381610038161003816
119899
sum
119894=0
119892120572
(119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
minus
119899
sum
119894=0
119892 (119894ℎ) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
1003816100381610038161003816100381610038161003816100381610038161003816
=
1003816100381610038161003816100381610038161003816100381610038161003816
119899
sum
119894=0
119894ℎ (120572120579119894) int
1
0
120595119894(119903) 119869] (119901119903) 119889119903
1003816100381610038161003816100381610038161003816100381610038161003816
le
119899
sum
119894=0
119894ℎ1003816100381610038161003816120572120579119894
1003816100381610038161003816int
1
0
1003816100381610038161003816120595119894(119903)1003816100381610038161003816
1003816100381610038161003816119869] (119901119903)
1003816100381610038161003816119889119903
le
119899
sum
119894=0
119894ℎ1003816100381610038161003816120572120579119894
1003816100381610038161003816int
1
0
1003816100381610038161003816120595119894(119903)1003816100381610038161003816119889119903 (As 100381610038161003816
1003816119869] (119901119903)
1003816100381610038161003816le 1)
(27)
Let 120579119872= max|120579
119894| then substituting 120595
119894(119903) from (3) we get
10038161003816100381610038161003816119865120572
]119899 (119901) minus 119865]119899 (119901)10038161003816100381610038161003816
le 120579119872ℎ |120572|
times [
119899minus1
sum
119894=1
119894 (int
119894ℎ
(119894minus1)ℎ
(119903 minus (119894 minus 1) ℎ)
ℎ
119889119903
+int
(119894+1)ℎ
119894ℎ
((119894 + 1) ℎ minus 119903)
ℎ
119889119903)
+ 119899int
1
1minusℎ
(119903 minus (1 minus ℎ))
ℎ
119889119903]
= 120579119872|120572| [ℎ
2
119899minus1
sum
119894=1
119894 + 119899
ℎ2
2
] = 120579119872|120572| (
1198992
ℎ2
2
)
le|120572|
2
(As 120579119872le 1 119899ℎ = 1)
(28)
Thus we have proved that
Theorem 3 When the input signal 119891(119903) is corrupted withnoise 120572 the proposed algorithm reduces the noise at least bya factor of 12 in the output data 119865]119899(119901)
International Journal of Analysis 5
Table 1 Least square errors 119864119895(119901)2
in different examples for 119899 = 100
119864119895(119901)2
Example 1 Example 2 Example 3 Example 4 Example 510038171003817100381710038171198640(119901)
10038171003817100381710038172
72696119890 minus 005 11078119890 minus 016 49160119890 minus 006 69741119890 minus 005 66819119890 minus 006
10038171003817100381710038171198641(119901)
10038171003817100381710038172
69980119890 minus 005 13724119890 minus 004 15965119890 minus 004 73362119890 minus 005 60667119890 minus 005
10038171003817100381710038171198642(119901)
10038171003817100381710038172
22536119890 minus 004 33312119890 minus 004 34473119890 minus 004 17687119890 minus 004 13147119890 minus 004
6 Numerical Illustrations
The test problems included in this paper are solved withand without random perturbations (noises) to illustrate theefficiency and stability of proposed algorithm by choosingthree different values of noise 120572 as 120572
0= 0 120572
1= 0002 and
1205722= 0005 (as discussed in the beginning of Section 5) In the
following examples the errors119864119895(119901) 119895 = 0 1 2 are computed
and their graphs are sketched for 119899 = 100 where 119864119895(119901) =
(Exact FHT 119865](119901)minus approximate FHT 119865120572119895]119899(119901) obtained from(26) with random noise 120572
119895)
Further the parameter 119901 ranges between 0 and 30 in stepsof 02 Figures 3 and 6 depict the graph of |119865120572119895]119899(119901) minus 119865]119899(119901)|119895 = 1 2 for different test functions in Examples 1 and 2respectively These graphs are in conformity withTheorem 3For all the illustrations the computations are done in MAT-LAB 701 For different examples the least square errors119864119895(119901)2involved in computations of approximate FHT with
noises 120572119895 119895 = 0 1 2 are given in Table 1These are calculated
using the formula
10038171003817100381710038171003817119864119895(119901)
100381710038171003817100381710038172=radicsum119899
119894=0(119864119895(119901119894))
2
119899 + 1
(29)
where 119901119894is taken in steps of 02 in the range [0 30]
Example 1 Consider the function 119891(119903) = radic(1 minus 1199032) 0 le 119903 le1 given in [14 20] for which
1198651(119901) =
120587
1198692
1(1199012)
2119901
0 lt 119901 lt infin
0 119901 = 0
(30)
Numerical evaluation of 1198651(119901) has been achieved by Barakat
and Sandler [20] by using Filon quadrature philosophy butagain the associated error is appreciable for 119901 lt 1 whereasour method gives almost zero error in that range Equation(12) is used to obtain the 119899th approximate HT 119865
1119899(119901) The
comparison between exact HT 1198651(119901) and approximate HT
1198651119899(119901) is shown in Figure 1 In Figure 2 the errors 119864
0(119901)
1198641(119901) and119864
2(119901) for 119899 = 100 are plotted In Figure 3 |119865120572
1119899(119901)minus
1198651119899(119901)| is shown for noises 120572 = 0002 0005 This figure is in
conformity withTheorem 3
Example 2 (Sombrero function) A very important and oftenused function is the Circ function that can be defined as
Circ( 119903119886
) =
1 119903 le 119886
0 119903 gt 119886
(31)
0 5 10 15 20 25 30
0
002
004
006
008
01
012
014
016
018
p
Exac
t and
appr
oxim
ate F
HT
minus002
Figure 1 Exact HT 1198651(119901) (solid line) and approximate HT 119865
1119899(119901)
(dotted line) for 119899 = 100 Example 1
The zeroth order HT of Circ(119903119886) is the Sombrero function[13 21] that will be written as 119878
0(119901) with the following
analytical expression
1198780(119901) = 119886
21198691(119886119901)
119886119901
(32)
We use (12) to obtain the approximation for the FHT 1198650119899(119901)
In Figure 4 we sketch the error 1198640(119901) between exact HT
1198780(119901) and approximate FHT119865
0119899(119901) of Circ function (without
noise) for 119886 = 1 and 119899 = 10 It is evident that even forsuch a small value of 119899 the error is appreciably small Figure 5compares the errors 119864
1(119901) and 119864
2(119901) for 119899 = 100 Figure 6
shows the plot of |1198651205720119899(119901) minus 119865
0119899(119901)| for 120572 = 0002 and 120572 =
0005 with 119899 = 100
Example 3 Consider the function 119891(119903) = (2120587)[arccos(119903) minus119903radic(1 minus 119903
2)] 0 le 119903 le 1 given in [14 22] forwhich zeroth order
exact HT is
1198650(119901) = 2
1198692
1(1199012)
1199012
0 le 119901 lt infin (33)
It is a well-known pair which arises in optical diffractiontheory [23] The function 119891(119903) is known as optical transferfunction of an aberration-free optical system with a circularaperture and 119865
0(119901) is the corresponding spread function
Barakat and Parshall [22] evaluated1198650(119901) using Filon quadra-
ture philosophy but the associated error is again appreciablefor 119901 lt 1 whereas our method gives significantly small error
6 International Journal of Analysis
0 5 10 15 20 25 30
0
1
2
3
p
minus5
minus4
minus3
minus2
minus1
times10minus4
E0(p)E1(p)
andE2(p)
E0(p)
E1(p)
E2(p)
Figure 2 1198640(119901) 119864
1(119901) 119864
2(119901) for 119899 = 100 Example 1
0 5 10 15 20 25 300
1
2
3
4
5
6
7
8
p
times10minus4
For noise 120572 = 0005
For noise 120572 = 0002
Figure 3 Plot of |1198651205721119899(119901)minus119865
1119899(119901)| for 120572 = 0002 and 120572 = 0005with
119899 = 100 Example 1
in that range In Figure 7 the error 1198640(119901) is shown Figure 8
is the comparison between 1198641(119901) and 119864
2(119901) for 119899 = 100
Example 4 Consider the function [24] 119891(119903) = 119903](1198862
minus 1199032
)120583
119867(119886 minus 119903) Re(]) gt minus1Re(120583) gt minus1 0 lt 119903 lt 119886 where119867(119903) is a unit step function
defined by
119867(119903) =
1 119903 ge 0
0 119903 lt 0
(34)
The ]th order HT of 119891(119903) is given by
119865] (119901) = 2]119886120583+]+1
119901minus120583minus1
120583 + 1 119869]+120583+1 (119886119901) (35)
Using the method of solution developed in Section 3 theapproximate HT 119865]119899(119901) has been calculated for ] = 32
0 5 10 15 20 25 30p
0
05
1
15
2
25
3
minus05
E0(p)
times10minus11
Figure 4 1198640(119901) for 119899 = 10 Example 2
0 5 10 15 20 25 30p
0
2
minus14
minus12
minus10
minus8
minus6
minus4
minus2
times10minus4
E1(p)
andE2(p)
E1(p)
E2(p)
Figure 5 1198641(119901) and 119864
2(119901) for 119899 = 100 Example 2
0 5 10 15 20 25 30p
0
02
04
06
08
1
12times10minus3
For noise 120572 = 0005
For noise 120572 = 0002
Figure 6 Plot of |1198651205720119899(119901)minus119865
0119899(119901)| for 120572 = 0002 and 120572 = 0005with
119899 = 100 Example 2
International Journal of Analysis 7
0
01
02
03
04
05
06
07
08
09
1
0 5 10 15 20 25 30p
times10minus5
E0(p)
Figure 7 1198640(119901) for 119899 = 100 in Example 3
0
2
0 5 10 15 20 25 30p
times10minus4
E1(p)
andE2(p)
E1(p)
E2(p)
minus12
minus10
minus8
minus6
minus4
minus2
Figure 8 1198641(119901) and 119864
2(119901) for 119899 = 100 in Example 3
0
001
002
003
004
005
006
007
008
Exac
t and
appr
oxim
ate F
HT
0 5 10 15 20 25 30p
minus002
minus001
Figure 9 Exact HT 11986532(119901) (solid line) and approximate HT
11986532119899
(119901) (dotted line) for 120583 = 12 and 119899 = 100 Example 4
0
1
2
0 5 10 15 20 25 30p
minus4
minus3
minus2
minus1
times10minus4
E0(p)
E1(p)
E2(p)
E0(p)E1(p)
andE2(p)
Figure 10 1198640(119901) 119864
1(119901) 119864
2(119901) for 119899 = 100 in Example 4
0
05
1
0 5 10 15 20 25 30p
Erro
r (w
ithou
t noi
se )E0(p)
minus2
minus15
minus1
minus05
times10minus5
Figure 11 1198640(119901) for 119899 = 100 in Example 5
120583 = 12 and 119886 = 1 The comparison between 11986532119899
(119901) andexact FHT 119865
32(119901) is shown in Figure 9 Further in Figure 10
a comparison between the different errors 1198640(119901) 119864
1(119901) and
1198642(119901) is shown for 119899 = 100
Example 5 In this example we choose as a test function thegeneralized version of the top-hat function given as 119891(119903) =119903][119867(119903) minus119867(119903 minus 119886)] 119886 gt 0 and119867(119903) is the unit step function
defined by (34) Then
119865] (119901) =119869]+1 (119901)
119901
(36)
In [9] authors took 119886 = 1 and ] = 4 for numericalcalculations We take 119886 = 1 ] = 5 and observe that theassociated errors with and without random noises are quitesmall The error 119864
0(119901) (without noise) is shown in Figure 11
8 International Journal of Analysis
0
05
0 5 10 15 20 25 30p
minus3
minus25
minus2
minus15
minus1
minus05
times10minus4
2E1(p)
andE2(p)
2E1(p)
E2(p)
Figure 12 Comparison between 1198641(119901) and 119864
2(119901) for 119899 = 100 in
Example 5
0 02 04 06 08 1
0
02
04
06
08
1
12
r
minus02
f(r)
ua(r0)
andf(r)
ua(r 0)
Figure 13 The initial condition function 119891(119903) (solid red line) and119906119886(119903 0) (= lim
119905rarr0+119906119886(119903 119905)) (lowastblue)
0 01 02 03 04 05 06 07 08 09
0
1
2
3
4
r
minus3
minus2
minus1
times10minus3
E(r)
Figure 14 Error between 119891(119903) and 119906119886(119903 0)
0 02 04 06 08 1
0
02
04
06
08
1
12
r
Solu
tion
for d
iffer
ent t
minus02
u0 for t = 0
u1 for t = 001
u2 for t = 002
u3 for t = 004
Figure 15The various profiles of the solutions 119906119886(119903 119905) at fixed times
0 01 02 03 04 05 06 07 08 09 10
1
2
r
Abso
lute
erro
rtimes10minus4
For t = 0
For t = 001
For t = 002
For t = 004
Figure 16 The absolute error between 119906119886(119903 119905) and 119906(119903 119905) for
different 119905
0
05
1
0
05
1
0
02
04
06
08
1
rt
minus02
u(rt)
Figure 17 The solution 119906(119903 119905) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
International Journal of Analysis 9
0
05
1
0
05
10
02
04
06
08
1
rt
ua(rt)
Figure 18 The approximate solution 119906119886(119903 119905) using 119899th approximate
FHT 1198655119899(1205735119904) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
for 119899 = 100 In Figure 12 errors 1198641(119901) and 119864
2(119901) for 119899 = 100
are shown
7 Applications
In this section we solve heat equation in cylindrical coor-dinates inside an infinitely long cylinder of radius unity byusing the theory of finite Hankel transform (FHT) developedin the preceding pages The initial temperature is givenby 119891(119903) and radiation takes place at the surface into thesurrounding medium maintained at zero temperature Weseek a function 119906(119903 119905) where 119903 is radius and 119905 is time (119906 doesnot depend upon 120579 and 119905) in the following application Themathematical model of this problem is the diffusion equationin polar coordinates given by
1205972
119906 (119903 119905)
1205972119903
+
1
119903
120597119906 (119903 119905)
120597119903
minus
]2
1199032119906 (119903 119905)
=
120597119906 (119903 119905)
120597119905
(0 lt 119903 lt 1 0 lt 119905 lt infin)
(37)
with the following initial and boundary conditions
(i) as 119905 rarr 0+ 119906(119903 119905) rarr 119891(119903) = 119903
][119867(119903) minus 119867(119903 minus 1)]
where 0 le 119903 le 1 and119867(119903) is unit step function givenby (34)
(ii) as 119903 rarr 1minus (120597119906120597119903) + 119866119906 rarr 0 for each fixed 119905 gt 0
and 119866 gt 0
When 119906(119903 119905) denotes the temperature within the cylinder119866 gt 0means that heat is being radiated away from the surfaceof the cylinder
LetΩ]119903 denote the Bessel differential operator (1205972
1205972
119903) +
(1119903)(120597120597119903) minus (]21199032) Then differential equation (37) can bewritten as
Ω]119903119906 =120597119906
120597119905
(38)
Applying the ]th order finite Hankel transform operator toboth sides of the differential equation (38) and using (8) weget
int
1
0
119903Ω]119903119906119869] (120573]119904119903) 119889119903 = int1
0
119903 (
120597119906
120597119905
) 119869] (120573]119904119903) 119889119903 (39)
where 120573]119904 is the 119904th positive root of equation [24]
119866119869] (119911) + 1199111198691015840
] (119911) = 0 ] ge minus1
2
(40)
The LHS of (39) may be written as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= int
1
0
(
1205972
119906
1205972119903
) 119903119869] (120573]119904119903) 119889119903
+ int
1
0
(
120597119906
120597119903
) 119869] (120573]119904119903) 119889119903 minus ]2 int1
0
(
119906
119903
) 119869] (120573]119904119903) 119889119903
= lim119903rarr1minus
120597119906 (119903 119905)
120597119903
119869] (120573]119904) minus 120573]119904 int1
0
119903
120597119906
120597119903
1198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
= minus119866119869] (120573]119904) lim119903rarr1minus119906 (119903 119905) minus 120573]119904 int
1
0
120597119906
120597119903
1199031198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903 (As 119903 997888rarr 1minus
120597119906
120597119903
+ 119866119906 997888rarr 0)
(41)
Since 120573]119904 is a root of 119866119869](119911) + 1199111198691015840
](119911) = 0 we have 119866119869](120573]119904) =minus120573]119904119869
1015840
](120573]119904) and thus
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
minus 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
+ 120573]119904 int1
0
119906 (1198691015840
] (120573]119904119903) + 120573]11990411990311986910158401015840
] (120573]119904119903)) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
(42)
Using the fact that 119869](119909) is a solution of 1199092(11988921199101198891199092) +119909(119889119910119889119909) + (119909
2
minus ]2)119910 = 0 (42) may be simplified as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= minus1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
(43)
10 International Journal of Analysis
From (43) and (39) we have
minus 1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
=
120597
120597119905
(int
1
0
119903119906 (119903 119905) 119869] (120573]119904119903) 119889119903)
(44)
If FHT int10
119906(119903 119905)119903119869](120573]119904119903) 119889119903 of 119906(119903 119905) is denoted by119880(120573]119904 119905)then (44) may be written as
120597119880 (120573]119904 119905)
120597119905
= minus1205732
]119904119880(120573]119904 119905) (45)
whose solution is given by
119880(120573]119904 119905) = 119860 (120573]119904) 119890minus1205732]119904119905 (46)
The constant of integration 119860(120573]119904) in (46) is determined byinitial condition Thus
119860 (120573]119904) = 119865] (120573]119904) = int1
0
119891 (119903) 119903119869] (120573]119904119903) 119889119903 (47)
Hence from (47) (46) becomes
119880 (120573]119904 119905) = 119865] (120573]119904) 119890minus1205732]119904119905 (48)
Therefore by inversion theorem [25] we have
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 1205732]11990411986910158402] (120573]119904)
= lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 11986621198692] (120573]119904)
(As 119866119869] (120573]119904) = minus120573]1199041198691015840
] (120573]119904))
(49)
So
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
119869] (120573]119904119903)
1198692
] (120573]119904) (50)
We want to prove that 119906(119903 119905) given by (50) is truly a solutionof (37) that satisfies the given initial and boundary conditionsTo achieve this we need the following well known estimates[17]
119865] (120573]119904) = 119874(1
12057332
]119904)
120573]119904 asymp 120587(119904 +1
4
)
1198692
] (120573]119904) asymp2
120587120573]119904
as 119904 997888rarr infin
(51)
Using the above estimates we see that series (50) and theseries obtained by applyingΩ]119903 and 120597120597119905 separately under thesummation sign of (50) converge uniformly on 0 lt 119903 lt 1 and119905 gt 0 Hence by applyingΩ]119903 minus 120597120597119905 on (50) we get
(Ω]119903 minus120597
120597119905
) 119906 (119903 119905)
= ( lim119873rarrinfin
119873
sum
119904=1
21205734
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
sdot [
[
(1205732
]1199041199032
11986910158401015840
] (120573]119904119903) + 120573]1199041199031198691015840
] (120573]119904119903)
+ (1205732
]1199041199032
minus ]2) 119869] (120573]119904119903))
times (1205732
]1199041199032
1198692
] (120573]119904))minus1
]
]
) = 0
(52)
Hence we see that (50) satisfies the differential equation (37)Let us verify the boundary condition (ii) we have
lim119903rarr1minus[
120597119906 (119903 119905)
120597119903
+ 119866119906 (119903 119905)]
= lim119903rarr1minus
[
[
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662) 1198692] (120573]119904)
times (120573]1199041198691015840
] (120573]119904119903) + 119866119869] (120573]119904119903))]
]
(53)
and since the convergence is uniform we can take thelim119903rarr1minus inside the summation sign and arrive at the conclu-
sion as 120573]119904rsquos are the roots of the equation119866119869](119911)+1199111198691015840
](119911) = 0The initial condition (i) is already taken care of as we
evaluated the constant 119860(120573]119904) by using it Through Figures13ndash18 we establish the accuracy of the proposed methodFor numerical solution of differential equation (37) we havetaken ] = 5 and 119866 = 1 We have used MATLAB routineldquofzerordquo to find zeros of (40) and we have drawn Figures 13ndash18 by truncating series (50) at 119873 = 30 While evaluatingthe solution 119906(119903 119905) from (50) we have evaluated 119865
5(1205735119904) first
from its analytical expression given by (36) and denoted thesolution thus obtained by 119906(119903 119905) in Figures 13ndash18 and thenevaluating 119899th approximates FHT 119865
5119899(1205735119904) for 119899 = 100
using our proposed algorithm for evaluation of the FHT asgiven by (17) This solution is denoted by 119906
119886(119903 119905) in the above
mentioned figuresFigure 13 compares the given initial condition 119891(119903) with
119906119886(119903 119905) as 119905 rarr 0
+ and Figure 14 shows the correspondingerror 119864(119903) = 119906
119886(119903 0) minus 119891(119903) Figure 15 depicts the various
profiles of 119906119886(119903 119905) at fixed times 119905
119895= 0 1100 150 125
where the various profiles are denoted by 1199060 1199061 1199062 and
1199063 The absolute errors between 119906
119886(119903 119905) and 119906(119903 119905) at fixed
times 119905119895= 0 1100 150 125 are shown in Figure 16 As
the maximum possible error occurs in the neighborhood of
International Journal of Analysis 11
0 and 0001 we have restricted 119905 in (0 1] in Figures 17 and18 representing 119906(119903 119905) and 119906
119886(119903 119905) respectively and note that
they are in good agreement in the range
8 Conclusions
A new stable and efficient algorithm based on hat functionsfor the numerical evaluation of Hankel transform is proposedand analyzed Choosing hat basis functions to expand theinput signal 119903119891(119903) makes our algorithm attractive in theirapplication in the applied physical problems as they eliminatethe problems connected with the Gibbs phenomenon takingplace in [9 11] We have given for the first time (as per ourinformation) error and stability analysis which was one ofthe main drawbacks of the earlier algorithms and by variousnumerical experiments we have corroborated our theoreticalfindings Stability with respect to the data is restored andexcellent accuracy is obtained even for small sample intervaland high noise levels in the data From the various figuresit is obvious that the algorithm is consistent and does notdepend on the particular choice of the input signal Theaccuracy and simplicity of the algorithm provide an edgeover the many other algorithms Finally an application of theproposed algorithm is given for solving the heat equation inan infinite cylinder with a radiation condition
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] I N Sneddon The Use of Integral Transforms McGraw-Hill1972
[2] A E Siegman ldquoQuasi fastHankel transformrdquoOptics Letters vol1 no 1 pp 13ndash15 1977
[3] L Filon ldquoOn quadrature formula for trigonometric integralsrdquoProceedings of the Royal Society of Edinburgh vol 49 pp 38ndash471928-1929
[4] D Levin andA Sidi ldquoTwo new classes of nonlinear transforma-tions for accelerating the convergence of infinite integrals andseriesrdquo Applied Mathematics and Computation vol 9 no 3 pp175ndash215 1981
[5] A Sidi ldquoExtrapolation methods for oscillatory infinite inte-gralsrdquo Journal of the Institute of Mathematics and its Applica-tions vol 26 no 1 pp 1ndash20 1980
[6] A Sidi ldquoA user-friendly extrapolation method for oscillatoryinfinite integralsrdquoMathematics of Computation vol 51 no 183pp 249ndash266 1988
[7] A Sidi ldquoComputation of infinite integrals involving Besselfunctions of arbitrary order by the 119863-transformationrdquo Journalof Computational and Applied Mathematics vol 78 no 1 pp125ndash130 1997
[8] A Sidi Practical Extrapolation Methods Theory and Applica-tions vol 10 of Cambridge Monographs on Applied and Compu-tational Mathematics Cambridge University Press CambridgeUK 2003
[9] M Guizar-Sicairos and J C Gutierrez-Vega ldquoComputation ofquasi-discrete Hankel transforms of integer order for propagat-ing optical wave fieldsrdquo Journal of the Optical Society of AmericaA Optics Image Science and Vision vol 21 no 1 pp 53ndash582004
[10] L Yu M Huang M Chen W Chen W Huang and Z ZhuldquoQuasi-discrete Hankel transformrdquoOptics Letters vol 23 no 6pp 409ndash411 1998
[11] E B Postnikov ldquoAbout calculation of the Hankel transformusing preliminary wavelet transformrdquo Journal of Applied Math-ematics no 6 pp 319ndash325 2003
[12] V K Singh O P Singh and R K Pandey ldquoNumerical evalu-ation of the Hankel transform by using linear Legendre multi-waveletsrdquoComputer Physics Communications vol 179 no 6 pp424ndash429 2008
[13] V K Singh O P Singh and R K Pandey ldquoEfficient algorithmsto compute Hankel transforms using waveletsrdquo ComputerPhysics Communications vol 179 no 11 pp 812ndash818 2008
[14] O P Singh V K Singh and R K Pandey ldquoAn efficient and sta-ble algorithm for numerical evaluation of Hankel transformsrdquoJournal of Applied Mathematics amp Informatics vol 28 no 5-6pp 1055ndash1071 2010
[15] R K Pandey O P Singh and V K Singh ldquoA stable algorithmfor numerical evaluation of Hankel transforms using Haarwaveletsrdquo Numerical Algorithms vol 53 no 4 pp 451ndash4662010
[16] E Babolian and M Mordad ldquoA numerical method for solvingsystems of linear and nonlinear integral equations of the secondkind by hat basis functionsrdquo Computers amp Mathematics withApplications vol 62 no 1 pp 187ndash198 2011
[17] R S Pathak and O P Singh ldquoFinite Hankel transforms ofdistributionsrdquo Pacific Journal of Mathematics vol 99 no 2 pp439ndash458 1982
[18] A Erdelyi EdTables of Integral TransformsMcGraw-Hill NewYork NY USA 1954
[19] M Abramowitz and I A Stegun Handbook of MathematicalFunctions with Formulas Graphs and Mathematical TablesDover New York NY USA 1965
[20] R Barakat and B H Sandler ldquoEvaluation of first-order Hankeltransforms using Filon quadrature philosophyrdquo Applied Mathe-matics Letters vol 11 no 1 pp 127ndash131 1998
[21] J D Secada ldquoNumerical evaluation of the Hankel transformrdquoComputer Physics Communications vol 116 no 2-3 pp 278ndash294 1999
[22] R Barakat and E Parshall ldquoNumerical evaluation of the zero-order Hankel transform using Filon quadrature philosophyrdquoApplied Mathematics Letters vol 9 no 5 pp 21ndash26 1996
[23] J Gaskell Linear Systems Fourier Transforms and Opticschapter 11 John Wiley amp Sons New York NY USA 1978
[24] A D Poularika ldquoThe Hankel transformrdquo in The Handbook ofFormulas and Tables for Signal Processing CRC Press LLC BocaRaton Fla USA 1999
[25] G NWatsonTheory of Bessel Functions Cambridge UniversityPress Cambridge UK 2nd edition 1958
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Analysis 5
Table 1 Least square errors 119864119895(119901)2
in different examples for 119899 = 100
119864119895(119901)2
Example 1 Example 2 Example 3 Example 4 Example 510038171003817100381710038171198640(119901)
10038171003817100381710038172
72696119890 minus 005 11078119890 minus 016 49160119890 minus 006 69741119890 minus 005 66819119890 minus 006
10038171003817100381710038171198641(119901)
10038171003817100381710038172
69980119890 minus 005 13724119890 minus 004 15965119890 minus 004 73362119890 minus 005 60667119890 minus 005
10038171003817100381710038171198642(119901)
10038171003817100381710038172
22536119890 minus 004 33312119890 minus 004 34473119890 minus 004 17687119890 minus 004 13147119890 minus 004
6 Numerical Illustrations
The test problems included in this paper are solved withand without random perturbations (noises) to illustrate theefficiency and stability of proposed algorithm by choosingthree different values of noise 120572 as 120572
0= 0 120572
1= 0002 and
1205722= 0005 (as discussed in the beginning of Section 5) In the
following examples the errors119864119895(119901) 119895 = 0 1 2 are computed
and their graphs are sketched for 119899 = 100 where 119864119895(119901) =
(Exact FHT 119865](119901)minus approximate FHT 119865120572119895]119899(119901) obtained from(26) with random noise 120572
119895)
Further the parameter 119901 ranges between 0 and 30 in stepsof 02 Figures 3 and 6 depict the graph of |119865120572119895]119899(119901) minus 119865]119899(119901)|119895 = 1 2 for different test functions in Examples 1 and 2respectively These graphs are in conformity withTheorem 3For all the illustrations the computations are done in MAT-LAB 701 For different examples the least square errors119864119895(119901)2involved in computations of approximate FHT with
noises 120572119895 119895 = 0 1 2 are given in Table 1These are calculated
using the formula
10038171003817100381710038171003817119864119895(119901)
100381710038171003817100381710038172=radicsum119899
119894=0(119864119895(119901119894))
2
119899 + 1
(29)
where 119901119894is taken in steps of 02 in the range [0 30]
Example 1 Consider the function 119891(119903) = radic(1 minus 1199032) 0 le 119903 le1 given in [14 20] for which
1198651(119901) =
120587
1198692
1(1199012)
2119901
0 lt 119901 lt infin
0 119901 = 0
(30)
Numerical evaluation of 1198651(119901) has been achieved by Barakat
and Sandler [20] by using Filon quadrature philosophy butagain the associated error is appreciable for 119901 lt 1 whereasour method gives almost zero error in that range Equation(12) is used to obtain the 119899th approximate HT 119865
1119899(119901) The
comparison between exact HT 1198651(119901) and approximate HT
1198651119899(119901) is shown in Figure 1 In Figure 2 the errors 119864
0(119901)
1198641(119901) and119864
2(119901) for 119899 = 100 are plotted In Figure 3 |119865120572
1119899(119901)minus
1198651119899(119901)| is shown for noises 120572 = 0002 0005 This figure is in
conformity withTheorem 3
Example 2 (Sombrero function) A very important and oftenused function is the Circ function that can be defined as
Circ( 119903119886
) =
1 119903 le 119886
0 119903 gt 119886
(31)
0 5 10 15 20 25 30
0
002
004
006
008
01
012
014
016
018
p
Exac
t and
appr
oxim
ate F
HT
minus002
Figure 1 Exact HT 1198651(119901) (solid line) and approximate HT 119865
1119899(119901)
(dotted line) for 119899 = 100 Example 1
The zeroth order HT of Circ(119903119886) is the Sombrero function[13 21] that will be written as 119878
0(119901) with the following
analytical expression
1198780(119901) = 119886
21198691(119886119901)
119886119901
(32)
We use (12) to obtain the approximation for the FHT 1198650119899(119901)
In Figure 4 we sketch the error 1198640(119901) between exact HT
1198780(119901) and approximate FHT119865
0119899(119901) of Circ function (without
noise) for 119886 = 1 and 119899 = 10 It is evident that even forsuch a small value of 119899 the error is appreciably small Figure 5compares the errors 119864
1(119901) and 119864
2(119901) for 119899 = 100 Figure 6
shows the plot of |1198651205720119899(119901) minus 119865
0119899(119901)| for 120572 = 0002 and 120572 =
0005 with 119899 = 100
Example 3 Consider the function 119891(119903) = (2120587)[arccos(119903) minus119903radic(1 minus 119903
2)] 0 le 119903 le 1 given in [14 22] forwhich zeroth order
exact HT is
1198650(119901) = 2
1198692
1(1199012)
1199012
0 le 119901 lt infin (33)
It is a well-known pair which arises in optical diffractiontheory [23] The function 119891(119903) is known as optical transferfunction of an aberration-free optical system with a circularaperture and 119865
0(119901) is the corresponding spread function
Barakat and Parshall [22] evaluated1198650(119901) using Filon quadra-
ture philosophy but the associated error is again appreciablefor 119901 lt 1 whereas our method gives significantly small error
6 International Journal of Analysis
0 5 10 15 20 25 30
0
1
2
3
p
minus5
minus4
minus3
minus2
minus1
times10minus4
E0(p)E1(p)
andE2(p)
E0(p)
E1(p)
E2(p)
Figure 2 1198640(119901) 119864
1(119901) 119864
2(119901) for 119899 = 100 Example 1
0 5 10 15 20 25 300
1
2
3
4
5
6
7
8
p
times10minus4
For noise 120572 = 0005
For noise 120572 = 0002
Figure 3 Plot of |1198651205721119899(119901)minus119865
1119899(119901)| for 120572 = 0002 and 120572 = 0005with
119899 = 100 Example 1
in that range In Figure 7 the error 1198640(119901) is shown Figure 8
is the comparison between 1198641(119901) and 119864
2(119901) for 119899 = 100
Example 4 Consider the function [24] 119891(119903) = 119903](1198862
minus 1199032
)120583
119867(119886 minus 119903) Re(]) gt minus1Re(120583) gt minus1 0 lt 119903 lt 119886 where119867(119903) is a unit step function
defined by
119867(119903) =
1 119903 ge 0
0 119903 lt 0
(34)
The ]th order HT of 119891(119903) is given by
119865] (119901) = 2]119886120583+]+1
119901minus120583minus1
120583 + 1 119869]+120583+1 (119886119901) (35)
Using the method of solution developed in Section 3 theapproximate HT 119865]119899(119901) has been calculated for ] = 32
0 5 10 15 20 25 30p
0
05
1
15
2
25
3
minus05
E0(p)
times10minus11
Figure 4 1198640(119901) for 119899 = 10 Example 2
0 5 10 15 20 25 30p
0
2
minus14
minus12
minus10
minus8
minus6
minus4
minus2
times10minus4
E1(p)
andE2(p)
E1(p)
E2(p)
Figure 5 1198641(119901) and 119864
2(119901) for 119899 = 100 Example 2
0 5 10 15 20 25 30p
0
02
04
06
08
1
12times10minus3
For noise 120572 = 0005
For noise 120572 = 0002
Figure 6 Plot of |1198651205720119899(119901)minus119865
0119899(119901)| for 120572 = 0002 and 120572 = 0005with
119899 = 100 Example 2
International Journal of Analysis 7
0
01
02
03
04
05
06
07
08
09
1
0 5 10 15 20 25 30p
times10minus5
E0(p)
Figure 7 1198640(119901) for 119899 = 100 in Example 3
0
2
0 5 10 15 20 25 30p
times10minus4
E1(p)
andE2(p)
E1(p)
E2(p)
minus12
minus10
minus8
minus6
minus4
minus2
Figure 8 1198641(119901) and 119864
2(119901) for 119899 = 100 in Example 3
0
001
002
003
004
005
006
007
008
Exac
t and
appr
oxim
ate F
HT
0 5 10 15 20 25 30p
minus002
minus001
Figure 9 Exact HT 11986532(119901) (solid line) and approximate HT
11986532119899
(119901) (dotted line) for 120583 = 12 and 119899 = 100 Example 4
0
1
2
0 5 10 15 20 25 30p
minus4
minus3
minus2
minus1
times10minus4
E0(p)
E1(p)
E2(p)
E0(p)E1(p)
andE2(p)
Figure 10 1198640(119901) 119864
1(119901) 119864
2(119901) for 119899 = 100 in Example 4
0
05
1
0 5 10 15 20 25 30p
Erro
r (w
ithou
t noi
se )E0(p)
minus2
minus15
minus1
minus05
times10minus5
Figure 11 1198640(119901) for 119899 = 100 in Example 5
120583 = 12 and 119886 = 1 The comparison between 11986532119899
(119901) andexact FHT 119865
32(119901) is shown in Figure 9 Further in Figure 10
a comparison between the different errors 1198640(119901) 119864
1(119901) and
1198642(119901) is shown for 119899 = 100
Example 5 In this example we choose as a test function thegeneralized version of the top-hat function given as 119891(119903) =119903][119867(119903) minus119867(119903 minus 119886)] 119886 gt 0 and119867(119903) is the unit step function
defined by (34) Then
119865] (119901) =119869]+1 (119901)
119901
(36)
In [9] authors took 119886 = 1 and ] = 4 for numericalcalculations We take 119886 = 1 ] = 5 and observe that theassociated errors with and without random noises are quitesmall The error 119864
0(119901) (without noise) is shown in Figure 11
8 International Journal of Analysis
0
05
0 5 10 15 20 25 30p
minus3
minus25
minus2
minus15
minus1
minus05
times10minus4
2E1(p)
andE2(p)
2E1(p)
E2(p)
Figure 12 Comparison between 1198641(119901) and 119864
2(119901) for 119899 = 100 in
Example 5
0 02 04 06 08 1
0
02
04
06
08
1
12
r
minus02
f(r)
ua(r0)
andf(r)
ua(r 0)
Figure 13 The initial condition function 119891(119903) (solid red line) and119906119886(119903 0) (= lim
119905rarr0+119906119886(119903 119905)) (lowastblue)
0 01 02 03 04 05 06 07 08 09
0
1
2
3
4
r
minus3
minus2
minus1
times10minus3
E(r)
Figure 14 Error between 119891(119903) and 119906119886(119903 0)
0 02 04 06 08 1
0
02
04
06
08
1
12
r
Solu
tion
for d
iffer
ent t
minus02
u0 for t = 0
u1 for t = 001
u2 for t = 002
u3 for t = 004
Figure 15The various profiles of the solutions 119906119886(119903 119905) at fixed times
0 01 02 03 04 05 06 07 08 09 10
1
2
r
Abso
lute
erro
rtimes10minus4
For t = 0
For t = 001
For t = 002
For t = 004
Figure 16 The absolute error between 119906119886(119903 119905) and 119906(119903 119905) for
different 119905
0
05
1
0
05
1
0
02
04
06
08
1
rt
minus02
u(rt)
Figure 17 The solution 119906(119903 119905) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
International Journal of Analysis 9
0
05
1
0
05
10
02
04
06
08
1
rt
ua(rt)
Figure 18 The approximate solution 119906119886(119903 119905) using 119899th approximate
FHT 1198655119899(1205735119904) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
for 119899 = 100 In Figure 12 errors 1198641(119901) and 119864
2(119901) for 119899 = 100
are shown
7 Applications
In this section we solve heat equation in cylindrical coor-dinates inside an infinitely long cylinder of radius unity byusing the theory of finite Hankel transform (FHT) developedin the preceding pages The initial temperature is givenby 119891(119903) and radiation takes place at the surface into thesurrounding medium maintained at zero temperature Weseek a function 119906(119903 119905) where 119903 is radius and 119905 is time (119906 doesnot depend upon 120579 and 119905) in the following application Themathematical model of this problem is the diffusion equationin polar coordinates given by
1205972
119906 (119903 119905)
1205972119903
+
1
119903
120597119906 (119903 119905)
120597119903
minus
]2
1199032119906 (119903 119905)
=
120597119906 (119903 119905)
120597119905
(0 lt 119903 lt 1 0 lt 119905 lt infin)
(37)
with the following initial and boundary conditions
(i) as 119905 rarr 0+ 119906(119903 119905) rarr 119891(119903) = 119903
][119867(119903) minus 119867(119903 minus 1)]
where 0 le 119903 le 1 and119867(119903) is unit step function givenby (34)
(ii) as 119903 rarr 1minus (120597119906120597119903) + 119866119906 rarr 0 for each fixed 119905 gt 0
and 119866 gt 0
When 119906(119903 119905) denotes the temperature within the cylinder119866 gt 0means that heat is being radiated away from the surfaceof the cylinder
LetΩ]119903 denote the Bessel differential operator (1205972
1205972
119903) +
(1119903)(120597120597119903) minus (]21199032) Then differential equation (37) can bewritten as
Ω]119903119906 =120597119906
120597119905
(38)
Applying the ]th order finite Hankel transform operator toboth sides of the differential equation (38) and using (8) weget
int
1
0
119903Ω]119903119906119869] (120573]119904119903) 119889119903 = int1
0
119903 (
120597119906
120597119905
) 119869] (120573]119904119903) 119889119903 (39)
where 120573]119904 is the 119904th positive root of equation [24]
119866119869] (119911) + 1199111198691015840
] (119911) = 0 ] ge minus1
2
(40)
The LHS of (39) may be written as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= int
1
0
(
1205972
119906
1205972119903
) 119903119869] (120573]119904119903) 119889119903
+ int
1
0
(
120597119906
120597119903
) 119869] (120573]119904119903) 119889119903 minus ]2 int1
0
(
119906
119903
) 119869] (120573]119904119903) 119889119903
= lim119903rarr1minus
120597119906 (119903 119905)
120597119903
119869] (120573]119904) minus 120573]119904 int1
0
119903
120597119906
120597119903
1198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
= minus119866119869] (120573]119904) lim119903rarr1minus119906 (119903 119905) minus 120573]119904 int
1
0
120597119906
120597119903
1199031198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903 (As 119903 997888rarr 1minus
120597119906
120597119903
+ 119866119906 997888rarr 0)
(41)
Since 120573]119904 is a root of 119866119869](119911) + 1199111198691015840
](119911) = 0 we have 119866119869](120573]119904) =minus120573]119904119869
1015840
](120573]119904) and thus
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
minus 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
+ 120573]119904 int1
0
119906 (1198691015840
] (120573]119904119903) + 120573]11990411990311986910158401015840
] (120573]119904119903)) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
(42)
Using the fact that 119869](119909) is a solution of 1199092(11988921199101198891199092) +119909(119889119910119889119909) + (119909
2
minus ]2)119910 = 0 (42) may be simplified as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= minus1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
(43)
10 International Journal of Analysis
From (43) and (39) we have
minus 1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
=
120597
120597119905
(int
1
0
119903119906 (119903 119905) 119869] (120573]119904119903) 119889119903)
(44)
If FHT int10
119906(119903 119905)119903119869](120573]119904119903) 119889119903 of 119906(119903 119905) is denoted by119880(120573]119904 119905)then (44) may be written as
120597119880 (120573]119904 119905)
120597119905
= minus1205732
]119904119880(120573]119904 119905) (45)
whose solution is given by
119880(120573]119904 119905) = 119860 (120573]119904) 119890minus1205732]119904119905 (46)
The constant of integration 119860(120573]119904) in (46) is determined byinitial condition Thus
119860 (120573]119904) = 119865] (120573]119904) = int1
0
119891 (119903) 119903119869] (120573]119904119903) 119889119903 (47)
Hence from (47) (46) becomes
119880 (120573]119904 119905) = 119865] (120573]119904) 119890minus1205732]119904119905 (48)
Therefore by inversion theorem [25] we have
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 1205732]11990411986910158402] (120573]119904)
= lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 11986621198692] (120573]119904)
(As 119866119869] (120573]119904) = minus120573]1199041198691015840
] (120573]119904))
(49)
So
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
119869] (120573]119904119903)
1198692
] (120573]119904) (50)
We want to prove that 119906(119903 119905) given by (50) is truly a solutionof (37) that satisfies the given initial and boundary conditionsTo achieve this we need the following well known estimates[17]
119865] (120573]119904) = 119874(1
12057332
]119904)
120573]119904 asymp 120587(119904 +1
4
)
1198692
] (120573]119904) asymp2
120587120573]119904
as 119904 997888rarr infin
(51)
Using the above estimates we see that series (50) and theseries obtained by applyingΩ]119903 and 120597120597119905 separately under thesummation sign of (50) converge uniformly on 0 lt 119903 lt 1 and119905 gt 0 Hence by applyingΩ]119903 minus 120597120597119905 on (50) we get
(Ω]119903 minus120597
120597119905
) 119906 (119903 119905)
= ( lim119873rarrinfin
119873
sum
119904=1
21205734
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
sdot [
[
(1205732
]1199041199032
11986910158401015840
] (120573]119904119903) + 120573]1199041199031198691015840
] (120573]119904119903)
+ (1205732
]1199041199032
minus ]2) 119869] (120573]119904119903))
times (1205732
]1199041199032
1198692
] (120573]119904))minus1
]
]
) = 0
(52)
Hence we see that (50) satisfies the differential equation (37)Let us verify the boundary condition (ii) we have
lim119903rarr1minus[
120597119906 (119903 119905)
120597119903
+ 119866119906 (119903 119905)]
= lim119903rarr1minus
[
[
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662) 1198692] (120573]119904)
times (120573]1199041198691015840
] (120573]119904119903) + 119866119869] (120573]119904119903))]
]
(53)
and since the convergence is uniform we can take thelim119903rarr1minus inside the summation sign and arrive at the conclu-
sion as 120573]119904rsquos are the roots of the equation119866119869](119911)+1199111198691015840
](119911) = 0The initial condition (i) is already taken care of as we
evaluated the constant 119860(120573]119904) by using it Through Figures13ndash18 we establish the accuracy of the proposed methodFor numerical solution of differential equation (37) we havetaken ] = 5 and 119866 = 1 We have used MATLAB routineldquofzerordquo to find zeros of (40) and we have drawn Figures 13ndash18 by truncating series (50) at 119873 = 30 While evaluatingthe solution 119906(119903 119905) from (50) we have evaluated 119865
5(1205735119904) first
from its analytical expression given by (36) and denoted thesolution thus obtained by 119906(119903 119905) in Figures 13ndash18 and thenevaluating 119899th approximates FHT 119865
5119899(1205735119904) for 119899 = 100
using our proposed algorithm for evaluation of the FHT asgiven by (17) This solution is denoted by 119906
119886(119903 119905) in the above
mentioned figuresFigure 13 compares the given initial condition 119891(119903) with
119906119886(119903 119905) as 119905 rarr 0
+ and Figure 14 shows the correspondingerror 119864(119903) = 119906
119886(119903 0) minus 119891(119903) Figure 15 depicts the various
profiles of 119906119886(119903 119905) at fixed times 119905
119895= 0 1100 150 125
where the various profiles are denoted by 1199060 1199061 1199062 and
1199063 The absolute errors between 119906
119886(119903 119905) and 119906(119903 119905) at fixed
times 119905119895= 0 1100 150 125 are shown in Figure 16 As
the maximum possible error occurs in the neighborhood of
International Journal of Analysis 11
0 and 0001 we have restricted 119905 in (0 1] in Figures 17 and18 representing 119906(119903 119905) and 119906
119886(119903 119905) respectively and note that
they are in good agreement in the range
8 Conclusions
A new stable and efficient algorithm based on hat functionsfor the numerical evaluation of Hankel transform is proposedand analyzed Choosing hat basis functions to expand theinput signal 119903119891(119903) makes our algorithm attractive in theirapplication in the applied physical problems as they eliminatethe problems connected with the Gibbs phenomenon takingplace in [9 11] We have given for the first time (as per ourinformation) error and stability analysis which was one ofthe main drawbacks of the earlier algorithms and by variousnumerical experiments we have corroborated our theoreticalfindings Stability with respect to the data is restored andexcellent accuracy is obtained even for small sample intervaland high noise levels in the data From the various figuresit is obvious that the algorithm is consistent and does notdepend on the particular choice of the input signal Theaccuracy and simplicity of the algorithm provide an edgeover the many other algorithms Finally an application of theproposed algorithm is given for solving the heat equation inan infinite cylinder with a radiation condition
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] I N Sneddon The Use of Integral Transforms McGraw-Hill1972
[2] A E Siegman ldquoQuasi fastHankel transformrdquoOptics Letters vol1 no 1 pp 13ndash15 1977
[3] L Filon ldquoOn quadrature formula for trigonometric integralsrdquoProceedings of the Royal Society of Edinburgh vol 49 pp 38ndash471928-1929
[4] D Levin andA Sidi ldquoTwo new classes of nonlinear transforma-tions for accelerating the convergence of infinite integrals andseriesrdquo Applied Mathematics and Computation vol 9 no 3 pp175ndash215 1981
[5] A Sidi ldquoExtrapolation methods for oscillatory infinite inte-gralsrdquo Journal of the Institute of Mathematics and its Applica-tions vol 26 no 1 pp 1ndash20 1980
[6] A Sidi ldquoA user-friendly extrapolation method for oscillatoryinfinite integralsrdquoMathematics of Computation vol 51 no 183pp 249ndash266 1988
[7] A Sidi ldquoComputation of infinite integrals involving Besselfunctions of arbitrary order by the 119863-transformationrdquo Journalof Computational and Applied Mathematics vol 78 no 1 pp125ndash130 1997
[8] A Sidi Practical Extrapolation Methods Theory and Applica-tions vol 10 of Cambridge Monographs on Applied and Compu-tational Mathematics Cambridge University Press CambridgeUK 2003
[9] M Guizar-Sicairos and J C Gutierrez-Vega ldquoComputation ofquasi-discrete Hankel transforms of integer order for propagat-ing optical wave fieldsrdquo Journal of the Optical Society of AmericaA Optics Image Science and Vision vol 21 no 1 pp 53ndash582004
[10] L Yu M Huang M Chen W Chen W Huang and Z ZhuldquoQuasi-discrete Hankel transformrdquoOptics Letters vol 23 no 6pp 409ndash411 1998
[11] E B Postnikov ldquoAbout calculation of the Hankel transformusing preliminary wavelet transformrdquo Journal of Applied Math-ematics no 6 pp 319ndash325 2003
[12] V K Singh O P Singh and R K Pandey ldquoNumerical evalu-ation of the Hankel transform by using linear Legendre multi-waveletsrdquoComputer Physics Communications vol 179 no 6 pp424ndash429 2008
[13] V K Singh O P Singh and R K Pandey ldquoEfficient algorithmsto compute Hankel transforms using waveletsrdquo ComputerPhysics Communications vol 179 no 11 pp 812ndash818 2008
[14] O P Singh V K Singh and R K Pandey ldquoAn efficient and sta-ble algorithm for numerical evaluation of Hankel transformsrdquoJournal of Applied Mathematics amp Informatics vol 28 no 5-6pp 1055ndash1071 2010
[15] R K Pandey O P Singh and V K Singh ldquoA stable algorithmfor numerical evaluation of Hankel transforms using Haarwaveletsrdquo Numerical Algorithms vol 53 no 4 pp 451ndash4662010
[16] E Babolian and M Mordad ldquoA numerical method for solvingsystems of linear and nonlinear integral equations of the secondkind by hat basis functionsrdquo Computers amp Mathematics withApplications vol 62 no 1 pp 187ndash198 2011
[17] R S Pathak and O P Singh ldquoFinite Hankel transforms ofdistributionsrdquo Pacific Journal of Mathematics vol 99 no 2 pp439ndash458 1982
[18] A Erdelyi EdTables of Integral TransformsMcGraw-Hill NewYork NY USA 1954
[19] M Abramowitz and I A Stegun Handbook of MathematicalFunctions with Formulas Graphs and Mathematical TablesDover New York NY USA 1965
[20] R Barakat and B H Sandler ldquoEvaluation of first-order Hankeltransforms using Filon quadrature philosophyrdquo Applied Mathe-matics Letters vol 11 no 1 pp 127ndash131 1998
[21] J D Secada ldquoNumerical evaluation of the Hankel transformrdquoComputer Physics Communications vol 116 no 2-3 pp 278ndash294 1999
[22] R Barakat and E Parshall ldquoNumerical evaluation of the zero-order Hankel transform using Filon quadrature philosophyrdquoApplied Mathematics Letters vol 9 no 5 pp 21ndash26 1996
[23] J Gaskell Linear Systems Fourier Transforms and Opticschapter 11 John Wiley amp Sons New York NY USA 1978
[24] A D Poularika ldquoThe Hankel transformrdquo in The Handbook ofFormulas and Tables for Signal Processing CRC Press LLC BocaRaton Fla USA 1999
[25] G NWatsonTheory of Bessel Functions Cambridge UniversityPress Cambridge UK 2nd edition 1958
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 International Journal of Analysis
0 5 10 15 20 25 30
0
1
2
3
p
minus5
minus4
minus3
minus2
minus1
times10minus4
E0(p)E1(p)
andE2(p)
E0(p)
E1(p)
E2(p)
Figure 2 1198640(119901) 119864
1(119901) 119864
2(119901) for 119899 = 100 Example 1
0 5 10 15 20 25 300
1
2
3
4
5
6
7
8
p
times10minus4
For noise 120572 = 0005
For noise 120572 = 0002
Figure 3 Plot of |1198651205721119899(119901)minus119865
1119899(119901)| for 120572 = 0002 and 120572 = 0005with
119899 = 100 Example 1
in that range In Figure 7 the error 1198640(119901) is shown Figure 8
is the comparison between 1198641(119901) and 119864
2(119901) for 119899 = 100
Example 4 Consider the function [24] 119891(119903) = 119903](1198862
minus 1199032
)120583
119867(119886 minus 119903) Re(]) gt minus1Re(120583) gt minus1 0 lt 119903 lt 119886 where119867(119903) is a unit step function
defined by
119867(119903) =
1 119903 ge 0
0 119903 lt 0
(34)
The ]th order HT of 119891(119903) is given by
119865] (119901) = 2]119886120583+]+1
119901minus120583minus1
120583 + 1 119869]+120583+1 (119886119901) (35)
Using the method of solution developed in Section 3 theapproximate HT 119865]119899(119901) has been calculated for ] = 32
0 5 10 15 20 25 30p
0
05
1
15
2
25
3
minus05
E0(p)
times10minus11
Figure 4 1198640(119901) for 119899 = 10 Example 2
0 5 10 15 20 25 30p
0
2
minus14
minus12
minus10
minus8
minus6
minus4
minus2
times10minus4
E1(p)
andE2(p)
E1(p)
E2(p)
Figure 5 1198641(119901) and 119864
2(119901) for 119899 = 100 Example 2
0 5 10 15 20 25 30p
0
02
04
06
08
1
12times10minus3
For noise 120572 = 0005
For noise 120572 = 0002
Figure 6 Plot of |1198651205720119899(119901)minus119865
0119899(119901)| for 120572 = 0002 and 120572 = 0005with
119899 = 100 Example 2
International Journal of Analysis 7
0
01
02
03
04
05
06
07
08
09
1
0 5 10 15 20 25 30p
times10minus5
E0(p)
Figure 7 1198640(119901) for 119899 = 100 in Example 3
0
2
0 5 10 15 20 25 30p
times10minus4
E1(p)
andE2(p)
E1(p)
E2(p)
minus12
minus10
minus8
minus6
minus4
minus2
Figure 8 1198641(119901) and 119864
2(119901) for 119899 = 100 in Example 3
0
001
002
003
004
005
006
007
008
Exac
t and
appr
oxim
ate F
HT
0 5 10 15 20 25 30p
minus002
minus001
Figure 9 Exact HT 11986532(119901) (solid line) and approximate HT
11986532119899
(119901) (dotted line) for 120583 = 12 and 119899 = 100 Example 4
0
1
2
0 5 10 15 20 25 30p
minus4
minus3
minus2
minus1
times10minus4
E0(p)
E1(p)
E2(p)
E0(p)E1(p)
andE2(p)
Figure 10 1198640(119901) 119864
1(119901) 119864
2(119901) for 119899 = 100 in Example 4
0
05
1
0 5 10 15 20 25 30p
Erro
r (w
ithou
t noi
se )E0(p)
minus2
minus15
minus1
minus05
times10minus5
Figure 11 1198640(119901) for 119899 = 100 in Example 5
120583 = 12 and 119886 = 1 The comparison between 11986532119899
(119901) andexact FHT 119865
32(119901) is shown in Figure 9 Further in Figure 10
a comparison between the different errors 1198640(119901) 119864
1(119901) and
1198642(119901) is shown for 119899 = 100
Example 5 In this example we choose as a test function thegeneralized version of the top-hat function given as 119891(119903) =119903][119867(119903) minus119867(119903 minus 119886)] 119886 gt 0 and119867(119903) is the unit step function
defined by (34) Then
119865] (119901) =119869]+1 (119901)
119901
(36)
In [9] authors took 119886 = 1 and ] = 4 for numericalcalculations We take 119886 = 1 ] = 5 and observe that theassociated errors with and without random noises are quitesmall The error 119864
0(119901) (without noise) is shown in Figure 11
8 International Journal of Analysis
0
05
0 5 10 15 20 25 30p
minus3
minus25
minus2
minus15
minus1
minus05
times10minus4
2E1(p)
andE2(p)
2E1(p)
E2(p)
Figure 12 Comparison between 1198641(119901) and 119864
2(119901) for 119899 = 100 in
Example 5
0 02 04 06 08 1
0
02
04
06
08
1
12
r
minus02
f(r)
ua(r0)
andf(r)
ua(r 0)
Figure 13 The initial condition function 119891(119903) (solid red line) and119906119886(119903 0) (= lim
119905rarr0+119906119886(119903 119905)) (lowastblue)
0 01 02 03 04 05 06 07 08 09
0
1
2
3
4
r
minus3
minus2
minus1
times10minus3
E(r)
Figure 14 Error between 119891(119903) and 119906119886(119903 0)
0 02 04 06 08 1
0
02
04
06
08
1
12
r
Solu
tion
for d
iffer
ent t
minus02
u0 for t = 0
u1 for t = 001
u2 for t = 002
u3 for t = 004
Figure 15The various profiles of the solutions 119906119886(119903 119905) at fixed times
0 01 02 03 04 05 06 07 08 09 10
1
2
r
Abso
lute
erro
rtimes10minus4
For t = 0
For t = 001
For t = 002
For t = 004
Figure 16 The absolute error between 119906119886(119903 119905) and 119906(119903 119905) for
different 119905
0
05
1
0
05
1
0
02
04
06
08
1
rt
minus02
u(rt)
Figure 17 The solution 119906(119903 119905) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
International Journal of Analysis 9
0
05
1
0
05
10
02
04
06
08
1
rt
ua(rt)
Figure 18 The approximate solution 119906119886(119903 119905) using 119899th approximate
FHT 1198655119899(1205735119904) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
for 119899 = 100 In Figure 12 errors 1198641(119901) and 119864
2(119901) for 119899 = 100
are shown
7 Applications
In this section we solve heat equation in cylindrical coor-dinates inside an infinitely long cylinder of radius unity byusing the theory of finite Hankel transform (FHT) developedin the preceding pages The initial temperature is givenby 119891(119903) and radiation takes place at the surface into thesurrounding medium maintained at zero temperature Weseek a function 119906(119903 119905) where 119903 is radius and 119905 is time (119906 doesnot depend upon 120579 and 119905) in the following application Themathematical model of this problem is the diffusion equationin polar coordinates given by
1205972
119906 (119903 119905)
1205972119903
+
1
119903
120597119906 (119903 119905)
120597119903
minus
]2
1199032119906 (119903 119905)
=
120597119906 (119903 119905)
120597119905
(0 lt 119903 lt 1 0 lt 119905 lt infin)
(37)
with the following initial and boundary conditions
(i) as 119905 rarr 0+ 119906(119903 119905) rarr 119891(119903) = 119903
][119867(119903) minus 119867(119903 minus 1)]
where 0 le 119903 le 1 and119867(119903) is unit step function givenby (34)
(ii) as 119903 rarr 1minus (120597119906120597119903) + 119866119906 rarr 0 for each fixed 119905 gt 0
and 119866 gt 0
When 119906(119903 119905) denotes the temperature within the cylinder119866 gt 0means that heat is being radiated away from the surfaceof the cylinder
LetΩ]119903 denote the Bessel differential operator (1205972
1205972
119903) +
(1119903)(120597120597119903) minus (]21199032) Then differential equation (37) can bewritten as
Ω]119903119906 =120597119906
120597119905
(38)
Applying the ]th order finite Hankel transform operator toboth sides of the differential equation (38) and using (8) weget
int
1
0
119903Ω]119903119906119869] (120573]119904119903) 119889119903 = int1
0
119903 (
120597119906
120597119905
) 119869] (120573]119904119903) 119889119903 (39)
where 120573]119904 is the 119904th positive root of equation [24]
119866119869] (119911) + 1199111198691015840
] (119911) = 0 ] ge minus1
2
(40)
The LHS of (39) may be written as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= int
1
0
(
1205972
119906
1205972119903
) 119903119869] (120573]119904119903) 119889119903
+ int
1
0
(
120597119906
120597119903
) 119869] (120573]119904119903) 119889119903 minus ]2 int1
0
(
119906
119903
) 119869] (120573]119904119903) 119889119903
= lim119903rarr1minus
120597119906 (119903 119905)
120597119903
119869] (120573]119904) minus 120573]119904 int1
0
119903
120597119906
120597119903
1198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
= minus119866119869] (120573]119904) lim119903rarr1minus119906 (119903 119905) minus 120573]119904 int
1
0
120597119906
120597119903
1199031198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903 (As 119903 997888rarr 1minus
120597119906
120597119903
+ 119866119906 997888rarr 0)
(41)
Since 120573]119904 is a root of 119866119869](119911) + 1199111198691015840
](119911) = 0 we have 119866119869](120573]119904) =minus120573]119904119869
1015840
](120573]119904) and thus
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
minus 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
+ 120573]119904 int1
0
119906 (1198691015840
] (120573]119904119903) + 120573]11990411990311986910158401015840
] (120573]119904119903)) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
(42)
Using the fact that 119869](119909) is a solution of 1199092(11988921199101198891199092) +119909(119889119910119889119909) + (119909
2
minus ]2)119910 = 0 (42) may be simplified as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= minus1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
(43)
10 International Journal of Analysis
From (43) and (39) we have
minus 1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
=
120597
120597119905
(int
1
0
119903119906 (119903 119905) 119869] (120573]119904119903) 119889119903)
(44)
If FHT int10
119906(119903 119905)119903119869](120573]119904119903) 119889119903 of 119906(119903 119905) is denoted by119880(120573]119904 119905)then (44) may be written as
120597119880 (120573]119904 119905)
120597119905
= minus1205732
]119904119880(120573]119904 119905) (45)
whose solution is given by
119880(120573]119904 119905) = 119860 (120573]119904) 119890minus1205732]119904119905 (46)
The constant of integration 119860(120573]119904) in (46) is determined byinitial condition Thus
119860 (120573]119904) = 119865] (120573]119904) = int1
0
119891 (119903) 119903119869] (120573]119904119903) 119889119903 (47)
Hence from (47) (46) becomes
119880 (120573]119904 119905) = 119865] (120573]119904) 119890minus1205732]119904119905 (48)
Therefore by inversion theorem [25] we have
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 1205732]11990411986910158402] (120573]119904)
= lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 11986621198692] (120573]119904)
(As 119866119869] (120573]119904) = minus120573]1199041198691015840
] (120573]119904))
(49)
So
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
119869] (120573]119904119903)
1198692
] (120573]119904) (50)
We want to prove that 119906(119903 119905) given by (50) is truly a solutionof (37) that satisfies the given initial and boundary conditionsTo achieve this we need the following well known estimates[17]
119865] (120573]119904) = 119874(1
12057332
]119904)
120573]119904 asymp 120587(119904 +1
4
)
1198692
] (120573]119904) asymp2
120587120573]119904
as 119904 997888rarr infin
(51)
Using the above estimates we see that series (50) and theseries obtained by applyingΩ]119903 and 120597120597119905 separately under thesummation sign of (50) converge uniformly on 0 lt 119903 lt 1 and119905 gt 0 Hence by applyingΩ]119903 minus 120597120597119905 on (50) we get
(Ω]119903 minus120597
120597119905
) 119906 (119903 119905)
= ( lim119873rarrinfin
119873
sum
119904=1
21205734
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
sdot [
[
(1205732
]1199041199032
11986910158401015840
] (120573]119904119903) + 120573]1199041199031198691015840
] (120573]119904119903)
+ (1205732
]1199041199032
minus ]2) 119869] (120573]119904119903))
times (1205732
]1199041199032
1198692
] (120573]119904))minus1
]
]
) = 0
(52)
Hence we see that (50) satisfies the differential equation (37)Let us verify the boundary condition (ii) we have
lim119903rarr1minus[
120597119906 (119903 119905)
120597119903
+ 119866119906 (119903 119905)]
= lim119903rarr1minus
[
[
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662) 1198692] (120573]119904)
times (120573]1199041198691015840
] (120573]119904119903) + 119866119869] (120573]119904119903))]
]
(53)
and since the convergence is uniform we can take thelim119903rarr1minus inside the summation sign and arrive at the conclu-
sion as 120573]119904rsquos are the roots of the equation119866119869](119911)+1199111198691015840
](119911) = 0The initial condition (i) is already taken care of as we
evaluated the constant 119860(120573]119904) by using it Through Figures13ndash18 we establish the accuracy of the proposed methodFor numerical solution of differential equation (37) we havetaken ] = 5 and 119866 = 1 We have used MATLAB routineldquofzerordquo to find zeros of (40) and we have drawn Figures 13ndash18 by truncating series (50) at 119873 = 30 While evaluatingthe solution 119906(119903 119905) from (50) we have evaluated 119865
5(1205735119904) first
from its analytical expression given by (36) and denoted thesolution thus obtained by 119906(119903 119905) in Figures 13ndash18 and thenevaluating 119899th approximates FHT 119865
5119899(1205735119904) for 119899 = 100
using our proposed algorithm for evaluation of the FHT asgiven by (17) This solution is denoted by 119906
119886(119903 119905) in the above
mentioned figuresFigure 13 compares the given initial condition 119891(119903) with
119906119886(119903 119905) as 119905 rarr 0
+ and Figure 14 shows the correspondingerror 119864(119903) = 119906
119886(119903 0) minus 119891(119903) Figure 15 depicts the various
profiles of 119906119886(119903 119905) at fixed times 119905
119895= 0 1100 150 125
where the various profiles are denoted by 1199060 1199061 1199062 and
1199063 The absolute errors between 119906
119886(119903 119905) and 119906(119903 119905) at fixed
times 119905119895= 0 1100 150 125 are shown in Figure 16 As
the maximum possible error occurs in the neighborhood of
International Journal of Analysis 11
0 and 0001 we have restricted 119905 in (0 1] in Figures 17 and18 representing 119906(119903 119905) and 119906
119886(119903 119905) respectively and note that
they are in good agreement in the range
8 Conclusions
A new stable and efficient algorithm based on hat functionsfor the numerical evaluation of Hankel transform is proposedand analyzed Choosing hat basis functions to expand theinput signal 119903119891(119903) makes our algorithm attractive in theirapplication in the applied physical problems as they eliminatethe problems connected with the Gibbs phenomenon takingplace in [9 11] We have given for the first time (as per ourinformation) error and stability analysis which was one ofthe main drawbacks of the earlier algorithms and by variousnumerical experiments we have corroborated our theoreticalfindings Stability with respect to the data is restored andexcellent accuracy is obtained even for small sample intervaland high noise levels in the data From the various figuresit is obvious that the algorithm is consistent and does notdepend on the particular choice of the input signal Theaccuracy and simplicity of the algorithm provide an edgeover the many other algorithms Finally an application of theproposed algorithm is given for solving the heat equation inan infinite cylinder with a radiation condition
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] I N Sneddon The Use of Integral Transforms McGraw-Hill1972
[2] A E Siegman ldquoQuasi fastHankel transformrdquoOptics Letters vol1 no 1 pp 13ndash15 1977
[3] L Filon ldquoOn quadrature formula for trigonometric integralsrdquoProceedings of the Royal Society of Edinburgh vol 49 pp 38ndash471928-1929
[4] D Levin andA Sidi ldquoTwo new classes of nonlinear transforma-tions for accelerating the convergence of infinite integrals andseriesrdquo Applied Mathematics and Computation vol 9 no 3 pp175ndash215 1981
[5] A Sidi ldquoExtrapolation methods for oscillatory infinite inte-gralsrdquo Journal of the Institute of Mathematics and its Applica-tions vol 26 no 1 pp 1ndash20 1980
[6] A Sidi ldquoA user-friendly extrapolation method for oscillatoryinfinite integralsrdquoMathematics of Computation vol 51 no 183pp 249ndash266 1988
[7] A Sidi ldquoComputation of infinite integrals involving Besselfunctions of arbitrary order by the 119863-transformationrdquo Journalof Computational and Applied Mathematics vol 78 no 1 pp125ndash130 1997
[8] A Sidi Practical Extrapolation Methods Theory and Applica-tions vol 10 of Cambridge Monographs on Applied and Compu-tational Mathematics Cambridge University Press CambridgeUK 2003
[9] M Guizar-Sicairos and J C Gutierrez-Vega ldquoComputation ofquasi-discrete Hankel transforms of integer order for propagat-ing optical wave fieldsrdquo Journal of the Optical Society of AmericaA Optics Image Science and Vision vol 21 no 1 pp 53ndash582004
[10] L Yu M Huang M Chen W Chen W Huang and Z ZhuldquoQuasi-discrete Hankel transformrdquoOptics Letters vol 23 no 6pp 409ndash411 1998
[11] E B Postnikov ldquoAbout calculation of the Hankel transformusing preliminary wavelet transformrdquo Journal of Applied Math-ematics no 6 pp 319ndash325 2003
[12] V K Singh O P Singh and R K Pandey ldquoNumerical evalu-ation of the Hankel transform by using linear Legendre multi-waveletsrdquoComputer Physics Communications vol 179 no 6 pp424ndash429 2008
[13] V K Singh O P Singh and R K Pandey ldquoEfficient algorithmsto compute Hankel transforms using waveletsrdquo ComputerPhysics Communications vol 179 no 11 pp 812ndash818 2008
[14] O P Singh V K Singh and R K Pandey ldquoAn efficient and sta-ble algorithm for numerical evaluation of Hankel transformsrdquoJournal of Applied Mathematics amp Informatics vol 28 no 5-6pp 1055ndash1071 2010
[15] R K Pandey O P Singh and V K Singh ldquoA stable algorithmfor numerical evaluation of Hankel transforms using Haarwaveletsrdquo Numerical Algorithms vol 53 no 4 pp 451ndash4662010
[16] E Babolian and M Mordad ldquoA numerical method for solvingsystems of linear and nonlinear integral equations of the secondkind by hat basis functionsrdquo Computers amp Mathematics withApplications vol 62 no 1 pp 187ndash198 2011
[17] R S Pathak and O P Singh ldquoFinite Hankel transforms ofdistributionsrdquo Pacific Journal of Mathematics vol 99 no 2 pp439ndash458 1982
[18] A Erdelyi EdTables of Integral TransformsMcGraw-Hill NewYork NY USA 1954
[19] M Abramowitz and I A Stegun Handbook of MathematicalFunctions with Formulas Graphs and Mathematical TablesDover New York NY USA 1965
[20] R Barakat and B H Sandler ldquoEvaluation of first-order Hankeltransforms using Filon quadrature philosophyrdquo Applied Mathe-matics Letters vol 11 no 1 pp 127ndash131 1998
[21] J D Secada ldquoNumerical evaluation of the Hankel transformrdquoComputer Physics Communications vol 116 no 2-3 pp 278ndash294 1999
[22] R Barakat and E Parshall ldquoNumerical evaluation of the zero-order Hankel transform using Filon quadrature philosophyrdquoApplied Mathematics Letters vol 9 no 5 pp 21ndash26 1996
[23] J Gaskell Linear Systems Fourier Transforms and Opticschapter 11 John Wiley amp Sons New York NY USA 1978
[24] A D Poularika ldquoThe Hankel transformrdquo in The Handbook ofFormulas and Tables for Signal Processing CRC Press LLC BocaRaton Fla USA 1999
[25] G NWatsonTheory of Bessel Functions Cambridge UniversityPress Cambridge UK 2nd edition 1958
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Analysis 7
0
01
02
03
04
05
06
07
08
09
1
0 5 10 15 20 25 30p
times10minus5
E0(p)
Figure 7 1198640(119901) for 119899 = 100 in Example 3
0
2
0 5 10 15 20 25 30p
times10minus4
E1(p)
andE2(p)
E1(p)
E2(p)
minus12
minus10
minus8
minus6
minus4
minus2
Figure 8 1198641(119901) and 119864
2(119901) for 119899 = 100 in Example 3
0
001
002
003
004
005
006
007
008
Exac
t and
appr
oxim
ate F
HT
0 5 10 15 20 25 30p
minus002
minus001
Figure 9 Exact HT 11986532(119901) (solid line) and approximate HT
11986532119899
(119901) (dotted line) for 120583 = 12 and 119899 = 100 Example 4
0
1
2
0 5 10 15 20 25 30p
minus4
minus3
minus2
minus1
times10minus4
E0(p)
E1(p)
E2(p)
E0(p)E1(p)
andE2(p)
Figure 10 1198640(119901) 119864
1(119901) 119864
2(119901) for 119899 = 100 in Example 4
0
05
1
0 5 10 15 20 25 30p
Erro
r (w
ithou
t noi
se )E0(p)
minus2
minus15
minus1
minus05
times10minus5
Figure 11 1198640(119901) for 119899 = 100 in Example 5
120583 = 12 and 119886 = 1 The comparison between 11986532119899
(119901) andexact FHT 119865
32(119901) is shown in Figure 9 Further in Figure 10
a comparison between the different errors 1198640(119901) 119864
1(119901) and
1198642(119901) is shown for 119899 = 100
Example 5 In this example we choose as a test function thegeneralized version of the top-hat function given as 119891(119903) =119903][119867(119903) minus119867(119903 minus 119886)] 119886 gt 0 and119867(119903) is the unit step function
defined by (34) Then
119865] (119901) =119869]+1 (119901)
119901
(36)
In [9] authors took 119886 = 1 and ] = 4 for numericalcalculations We take 119886 = 1 ] = 5 and observe that theassociated errors with and without random noises are quitesmall The error 119864
0(119901) (without noise) is shown in Figure 11
8 International Journal of Analysis
0
05
0 5 10 15 20 25 30p
minus3
minus25
minus2
minus15
minus1
minus05
times10minus4
2E1(p)
andE2(p)
2E1(p)
E2(p)
Figure 12 Comparison between 1198641(119901) and 119864
2(119901) for 119899 = 100 in
Example 5
0 02 04 06 08 1
0
02
04
06
08
1
12
r
minus02
f(r)
ua(r0)
andf(r)
ua(r 0)
Figure 13 The initial condition function 119891(119903) (solid red line) and119906119886(119903 0) (= lim
119905rarr0+119906119886(119903 119905)) (lowastblue)
0 01 02 03 04 05 06 07 08 09
0
1
2
3
4
r
minus3
minus2
minus1
times10minus3
E(r)
Figure 14 Error between 119891(119903) and 119906119886(119903 0)
0 02 04 06 08 1
0
02
04
06
08
1
12
r
Solu
tion
for d
iffer
ent t
minus02
u0 for t = 0
u1 for t = 001
u2 for t = 002
u3 for t = 004
Figure 15The various profiles of the solutions 119906119886(119903 119905) at fixed times
0 01 02 03 04 05 06 07 08 09 10
1
2
r
Abso
lute
erro
rtimes10minus4
For t = 0
For t = 001
For t = 002
For t = 004
Figure 16 The absolute error between 119906119886(119903 119905) and 119906(119903 119905) for
different 119905
0
05
1
0
05
1
0
02
04
06
08
1
rt
minus02
u(rt)
Figure 17 The solution 119906(119903 119905) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
International Journal of Analysis 9
0
05
1
0
05
10
02
04
06
08
1
rt
ua(rt)
Figure 18 The approximate solution 119906119886(119903 119905) using 119899th approximate
FHT 1198655119899(1205735119904) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
for 119899 = 100 In Figure 12 errors 1198641(119901) and 119864
2(119901) for 119899 = 100
are shown
7 Applications
In this section we solve heat equation in cylindrical coor-dinates inside an infinitely long cylinder of radius unity byusing the theory of finite Hankel transform (FHT) developedin the preceding pages The initial temperature is givenby 119891(119903) and radiation takes place at the surface into thesurrounding medium maintained at zero temperature Weseek a function 119906(119903 119905) where 119903 is radius and 119905 is time (119906 doesnot depend upon 120579 and 119905) in the following application Themathematical model of this problem is the diffusion equationin polar coordinates given by
1205972
119906 (119903 119905)
1205972119903
+
1
119903
120597119906 (119903 119905)
120597119903
minus
]2
1199032119906 (119903 119905)
=
120597119906 (119903 119905)
120597119905
(0 lt 119903 lt 1 0 lt 119905 lt infin)
(37)
with the following initial and boundary conditions
(i) as 119905 rarr 0+ 119906(119903 119905) rarr 119891(119903) = 119903
][119867(119903) minus 119867(119903 minus 1)]
where 0 le 119903 le 1 and119867(119903) is unit step function givenby (34)
(ii) as 119903 rarr 1minus (120597119906120597119903) + 119866119906 rarr 0 for each fixed 119905 gt 0
and 119866 gt 0
When 119906(119903 119905) denotes the temperature within the cylinder119866 gt 0means that heat is being radiated away from the surfaceof the cylinder
LetΩ]119903 denote the Bessel differential operator (1205972
1205972
119903) +
(1119903)(120597120597119903) minus (]21199032) Then differential equation (37) can bewritten as
Ω]119903119906 =120597119906
120597119905
(38)
Applying the ]th order finite Hankel transform operator toboth sides of the differential equation (38) and using (8) weget
int
1
0
119903Ω]119903119906119869] (120573]119904119903) 119889119903 = int1
0
119903 (
120597119906
120597119905
) 119869] (120573]119904119903) 119889119903 (39)
where 120573]119904 is the 119904th positive root of equation [24]
119866119869] (119911) + 1199111198691015840
] (119911) = 0 ] ge minus1
2
(40)
The LHS of (39) may be written as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= int
1
0
(
1205972
119906
1205972119903
) 119903119869] (120573]119904119903) 119889119903
+ int
1
0
(
120597119906
120597119903
) 119869] (120573]119904119903) 119889119903 minus ]2 int1
0
(
119906
119903
) 119869] (120573]119904119903) 119889119903
= lim119903rarr1minus
120597119906 (119903 119905)
120597119903
119869] (120573]119904) minus 120573]119904 int1
0
119903
120597119906
120597119903
1198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
= minus119866119869] (120573]119904) lim119903rarr1minus119906 (119903 119905) minus 120573]119904 int
1
0
120597119906
120597119903
1199031198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903 (As 119903 997888rarr 1minus
120597119906
120597119903
+ 119866119906 997888rarr 0)
(41)
Since 120573]119904 is a root of 119866119869](119911) + 1199111198691015840
](119911) = 0 we have 119866119869](120573]119904) =minus120573]119904119869
1015840
](120573]119904) and thus
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
minus 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
+ 120573]119904 int1
0
119906 (1198691015840
] (120573]119904119903) + 120573]11990411990311986910158401015840
] (120573]119904119903)) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
(42)
Using the fact that 119869](119909) is a solution of 1199092(11988921199101198891199092) +119909(119889119910119889119909) + (119909
2
minus ]2)119910 = 0 (42) may be simplified as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= minus1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
(43)
10 International Journal of Analysis
From (43) and (39) we have
minus 1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
=
120597
120597119905
(int
1
0
119903119906 (119903 119905) 119869] (120573]119904119903) 119889119903)
(44)
If FHT int10
119906(119903 119905)119903119869](120573]119904119903) 119889119903 of 119906(119903 119905) is denoted by119880(120573]119904 119905)then (44) may be written as
120597119880 (120573]119904 119905)
120597119905
= minus1205732
]119904119880(120573]119904 119905) (45)
whose solution is given by
119880(120573]119904 119905) = 119860 (120573]119904) 119890minus1205732]119904119905 (46)
The constant of integration 119860(120573]119904) in (46) is determined byinitial condition Thus
119860 (120573]119904) = 119865] (120573]119904) = int1
0
119891 (119903) 119903119869] (120573]119904119903) 119889119903 (47)
Hence from (47) (46) becomes
119880 (120573]119904 119905) = 119865] (120573]119904) 119890minus1205732]119904119905 (48)
Therefore by inversion theorem [25] we have
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 1205732]11990411986910158402] (120573]119904)
= lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 11986621198692] (120573]119904)
(As 119866119869] (120573]119904) = minus120573]1199041198691015840
] (120573]119904))
(49)
So
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
119869] (120573]119904119903)
1198692
] (120573]119904) (50)
We want to prove that 119906(119903 119905) given by (50) is truly a solutionof (37) that satisfies the given initial and boundary conditionsTo achieve this we need the following well known estimates[17]
119865] (120573]119904) = 119874(1
12057332
]119904)
120573]119904 asymp 120587(119904 +1
4
)
1198692
] (120573]119904) asymp2
120587120573]119904
as 119904 997888rarr infin
(51)
Using the above estimates we see that series (50) and theseries obtained by applyingΩ]119903 and 120597120597119905 separately under thesummation sign of (50) converge uniformly on 0 lt 119903 lt 1 and119905 gt 0 Hence by applyingΩ]119903 minus 120597120597119905 on (50) we get
(Ω]119903 minus120597
120597119905
) 119906 (119903 119905)
= ( lim119873rarrinfin
119873
sum
119904=1
21205734
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
sdot [
[
(1205732
]1199041199032
11986910158401015840
] (120573]119904119903) + 120573]1199041199031198691015840
] (120573]119904119903)
+ (1205732
]1199041199032
minus ]2) 119869] (120573]119904119903))
times (1205732
]1199041199032
1198692
] (120573]119904))minus1
]
]
) = 0
(52)
Hence we see that (50) satisfies the differential equation (37)Let us verify the boundary condition (ii) we have
lim119903rarr1minus[
120597119906 (119903 119905)
120597119903
+ 119866119906 (119903 119905)]
= lim119903rarr1minus
[
[
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662) 1198692] (120573]119904)
times (120573]1199041198691015840
] (120573]119904119903) + 119866119869] (120573]119904119903))]
]
(53)
and since the convergence is uniform we can take thelim119903rarr1minus inside the summation sign and arrive at the conclu-
sion as 120573]119904rsquos are the roots of the equation119866119869](119911)+1199111198691015840
](119911) = 0The initial condition (i) is already taken care of as we
evaluated the constant 119860(120573]119904) by using it Through Figures13ndash18 we establish the accuracy of the proposed methodFor numerical solution of differential equation (37) we havetaken ] = 5 and 119866 = 1 We have used MATLAB routineldquofzerordquo to find zeros of (40) and we have drawn Figures 13ndash18 by truncating series (50) at 119873 = 30 While evaluatingthe solution 119906(119903 119905) from (50) we have evaluated 119865
5(1205735119904) first
from its analytical expression given by (36) and denoted thesolution thus obtained by 119906(119903 119905) in Figures 13ndash18 and thenevaluating 119899th approximates FHT 119865
5119899(1205735119904) for 119899 = 100
using our proposed algorithm for evaluation of the FHT asgiven by (17) This solution is denoted by 119906
119886(119903 119905) in the above
mentioned figuresFigure 13 compares the given initial condition 119891(119903) with
119906119886(119903 119905) as 119905 rarr 0
+ and Figure 14 shows the correspondingerror 119864(119903) = 119906
119886(119903 0) minus 119891(119903) Figure 15 depicts the various
profiles of 119906119886(119903 119905) at fixed times 119905
119895= 0 1100 150 125
where the various profiles are denoted by 1199060 1199061 1199062 and
1199063 The absolute errors between 119906
119886(119903 119905) and 119906(119903 119905) at fixed
times 119905119895= 0 1100 150 125 are shown in Figure 16 As
the maximum possible error occurs in the neighborhood of
International Journal of Analysis 11
0 and 0001 we have restricted 119905 in (0 1] in Figures 17 and18 representing 119906(119903 119905) and 119906
119886(119903 119905) respectively and note that
they are in good agreement in the range
8 Conclusions
A new stable and efficient algorithm based on hat functionsfor the numerical evaluation of Hankel transform is proposedand analyzed Choosing hat basis functions to expand theinput signal 119903119891(119903) makes our algorithm attractive in theirapplication in the applied physical problems as they eliminatethe problems connected with the Gibbs phenomenon takingplace in [9 11] We have given for the first time (as per ourinformation) error and stability analysis which was one ofthe main drawbacks of the earlier algorithms and by variousnumerical experiments we have corroborated our theoreticalfindings Stability with respect to the data is restored andexcellent accuracy is obtained even for small sample intervaland high noise levels in the data From the various figuresit is obvious that the algorithm is consistent and does notdepend on the particular choice of the input signal Theaccuracy and simplicity of the algorithm provide an edgeover the many other algorithms Finally an application of theproposed algorithm is given for solving the heat equation inan infinite cylinder with a radiation condition
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] I N Sneddon The Use of Integral Transforms McGraw-Hill1972
[2] A E Siegman ldquoQuasi fastHankel transformrdquoOptics Letters vol1 no 1 pp 13ndash15 1977
[3] L Filon ldquoOn quadrature formula for trigonometric integralsrdquoProceedings of the Royal Society of Edinburgh vol 49 pp 38ndash471928-1929
[4] D Levin andA Sidi ldquoTwo new classes of nonlinear transforma-tions for accelerating the convergence of infinite integrals andseriesrdquo Applied Mathematics and Computation vol 9 no 3 pp175ndash215 1981
[5] A Sidi ldquoExtrapolation methods for oscillatory infinite inte-gralsrdquo Journal of the Institute of Mathematics and its Applica-tions vol 26 no 1 pp 1ndash20 1980
[6] A Sidi ldquoA user-friendly extrapolation method for oscillatoryinfinite integralsrdquoMathematics of Computation vol 51 no 183pp 249ndash266 1988
[7] A Sidi ldquoComputation of infinite integrals involving Besselfunctions of arbitrary order by the 119863-transformationrdquo Journalof Computational and Applied Mathematics vol 78 no 1 pp125ndash130 1997
[8] A Sidi Practical Extrapolation Methods Theory and Applica-tions vol 10 of Cambridge Monographs on Applied and Compu-tational Mathematics Cambridge University Press CambridgeUK 2003
[9] M Guizar-Sicairos and J C Gutierrez-Vega ldquoComputation ofquasi-discrete Hankel transforms of integer order for propagat-ing optical wave fieldsrdquo Journal of the Optical Society of AmericaA Optics Image Science and Vision vol 21 no 1 pp 53ndash582004
[10] L Yu M Huang M Chen W Chen W Huang and Z ZhuldquoQuasi-discrete Hankel transformrdquoOptics Letters vol 23 no 6pp 409ndash411 1998
[11] E B Postnikov ldquoAbout calculation of the Hankel transformusing preliminary wavelet transformrdquo Journal of Applied Math-ematics no 6 pp 319ndash325 2003
[12] V K Singh O P Singh and R K Pandey ldquoNumerical evalu-ation of the Hankel transform by using linear Legendre multi-waveletsrdquoComputer Physics Communications vol 179 no 6 pp424ndash429 2008
[13] V K Singh O P Singh and R K Pandey ldquoEfficient algorithmsto compute Hankel transforms using waveletsrdquo ComputerPhysics Communications vol 179 no 11 pp 812ndash818 2008
[14] O P Singh V K Singh and R K Pandey ldquoAn efficient and sta-ble algorithm for numerical evaluation of Hankel transformsrdquoJournal of Applied Mathematics amp Informatics vol 28 no 5-6pp 1055ndash1071 2010
[15] R K Pandey O P Singh and V K Singh ldquoA stable algorithmfor numerical evaluation of Hankel transforms using Haarwaveletsrdquo Numerical Algorithms vol 53 no 4 pp 451ndash4662010
[16] E Babolian and M Mordad ldquoA numerical method for solvingsystems of linear and nonlinear integral equations of the secondkind by hat basis functionsrdquo Computers amp Mathematics withApplications vol 62 no 1 pp 187ndash198 2011
[17] R S Pathak and O P Singh ldquoFinite Hankel transforms ofdistributionsrdquo Pacific Journal of Mathematics vol 99 no 2 pp439ndash458 1982
[18] A Erdelyi EdTables of Integral TransformsMcGraw-Hill NewYork NY USA 1954
[19] M Abramowitz and I A Stegun Handbook of MathematicalFunctions with Formulas Graphs and Mathematical TablesDover New York NY USA 1965
[20] R Barakat and B H Sandler ldquoEvaluation of first-order Hankeltransforms using Filon quadrature philosophyrdquo Applied Mathe-matics Letters vol 11 no 1 pp 127ndash131 1998
[21] J D Secada ldquoNumerical evaluation of the Hankel transformrdquoComputer Physics Communications vol 116 no 2-3 pp 278ndash294 1999
[22] R Barakat and E Parshall ldquoNumerical evaluation of the zero-order Hankel transform using Filon quadrature philosophyrdquoApplied Mathematics Letters vol 9 no 5 pp 21ndash26 1996
[23] J Gaskell Linear Systems Fourier Transforms and Opticschapter 11 John Wiley amp Sons New York NY USA 1978
[24] A D Poularika ldquoThe Hankel transformrdquo in The Handbook ofFormulas and Tables for Signal Processing CRC Press LLC BocaRaton Fla USA 1999
[25] G NWatsonTheory of Bessel Functions Cambridge UniversityPress Cambridge UK 2nd edition 1958
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 International Journal of Analysis
0
05
0 5 10 15 20 25 30p
minus3
minus25
minus2
minus15
minus1
minus05
times10minus4
2E1(p)
andE2(p)
2E1(p)
E2(p)
Figure 12 Comparison between 1198641(119901) and 119864
2(119901) for 119899 = 100 in
Example 5
0 02 04 06 08 1
0
02
04
06
08
1
12
r
minus02
f(r)
ua(r0)
andf(r)
ua(r 0)
Figure 13 The initial condition function 119891(119903) (solid red line) and119906119886(119903 0) (= lim
119905rarr0+119906119886(119903 119905)) (lowastblue)
0 01 02 03 04 05 06 07 08 09
0
1
2
3
4
r
minus3
minus2
minus1
times10minus3
E(r)
Figure 14 Error between 119891(119903) and 119906119886(119903 0)
0 02 04 06 08 1
0
02
04
06
08
1
12
r
Solu
tion
for d
iffer
ent t
minus02
u0 for t = 0
u1 for t = 001
u2 for t = 002
u3 for t = 004
Figure 15The various profiles of the solutions 119906119886(119903 119905) at fixed times
0 01 02 03 04 05 06 07 08 09 10
1
2
r
Abso
lute
erro
rtimes10minus4
For t = 0
For t = 001
For t = 002
For t = 004
Figure 16 The absolute error between 119906119886(119903 119905) and 119906(119903 119905) for
different 119905
0
05
1
0
05
1
0
02
04
06
08
1
rt
minus02
u(rt)
Figure 17 The solution 119906(119903 119905) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
International Journal of Analysis 9
0
05
1
0
05
10
02
04
06
08
1
rt
ua(rt)
Figure 18 The approximate solution 119906119886(119903 119905) using 119899th approximate
FHT 1198655119899(1205735119904) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
for 119899 = 100 In Figure 12 errors 1198641(119901) and 119864
2(119901) for 119899 = 100
are shown
7 Applications
In this section we solve heat equation in cylindrical coor-dinates inside an infinitely long cylinder of radius unity byusing the theory of finite Hankel transform (FHT) developedin the preceding pages The initial temperature is givenby 119891(119903) and radiation takes place at the surface into thesurrounding medium maintained at zero temperature Weseek a function 119906(119903 119905) where 119903 is radius and 119905 is time (119906 doesnot depend upon 120579 and 119905) in the following application Themathematical model of this problem is the diffusion equationin polar coordinates given by
1205972
119906 (119903 119905)
1205972119903
+
1
119903
120597119906 (119903 119905)
120597119903
minus
]2
1199032119906 (119903 119905)
=
120597119906 (119903 119905)
120597119905
(0 lt 119903 lt 1 0 lt 119905 lt infin)
(37)
with the following initial and boundary conditions
(i) as 119905 rarr 0+ 119906(119903 119905) rarr 119891(119903) = 119903
][119867(119903) minus 119867(119903 minus 1)]
where 0 le 119903 le 1 and119867(119903) is unit step function givenby (34)
(ii) as 119903 rarr 1minus (120597119906120597119903) + 119866119906 rarr 0 for each fixed 119905 gt 0
and 119866 gt 0
When 119906(119903 119905) denotes the temperature within the cylinder119866 gt 0means that heat is being radiated away from the surfaceof the cylinder
LetΩ]119903 denote the Bessel differential operator (1205972
1205972
119903) +
(1119903)(120597120597119903) minus (]21199032) Then differential equation (37) can bewritten as
Ω]119903119906 =120597119906
120597119905
(38)
Applying the ]th order finite Hankel transform operator toboth sides of the differential equation (38) and using (8) weget
int
1
0
119903Ω]119903119906119869] (120573]119904119903) 119889119903 = int1
0
119903 (
120597119906
120597119905
) 119869] (120573]119904119903) 119889119903 (39)
where 120573]119904 is the 119904th positive root of equation [24]
119866119869] (119911) + 1199111198691015840
] (119911) = 0 ] ge minus1
2
(40)
The LHS of (39) may be written as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= int
1
0
(
1205972
119906
1205972119903
) 119903119869] (120573]119904119903) 119889119903
+ int
1
0
(
120597119906
120597119903
) 119869] (120573]119904119903) 119889119903 minus ]2 int1
0
(
119906
119903
) 119869] (120573]119904119903) 119889119903
= lim119903rarr1minus
120597119906 (119903 119905)
120597119903
119869] (120573]119904) minus 120573]119904 int1
0
119903
120597119906
120597119903
1198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
= minus119866119869] (120573]119904) lim119903rarr1minus119906 (119903 119905) minus 120573]119904 int
1
0
120597119906
120597119903
1199031198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903 (As 119903 997888rarr 1minus
120597119906
120597119903
+ 119866119906 997888rarr 0)
(41)
Since 120573]119904 is a root of 119866119869](119911) + 1199111198691015840
](119911) = 0 we have 119866119869](120573]119904) =minus120573]119904119869
1015840
](120573]119904) and thus
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
minus 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
+ 120573]119904 int1
0
119906 (1198691015840
] (120573]119904119903) + 120573]11990411990311986910158401015840
] (120573]119904119903)) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
(42)
Using the fact that 119869](119909) is a solution of 1199092(11988921199101198891199092) +119909(119889119910119889119909) + (119909
2
minus ]2)119910 = 0 (42) may be simplified as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= minus1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
(43)
10 International Journal of Analysis
From (43) and (39) we have
minus 1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
=
120597
120597119905
(int
1
0
119903119906 (119903 119905) 119869] (120573]119904119903) 119889119903)
(44)
If FHT int10
119906(119903 119905)119903119869](120573]119904119903) 119889119903 of 119906(119903 119905) is denoted by119880(120573]119904 119905)then (44) may be written as
120597119880 (120573]119904 119905)
120597119905
= minus1205732
]119904119880(120573]119904 119905) (45)
whose solution is given by
119880(120573]119904 119905) = 119860 (120573]119904) 119890minus1205732]119904119905 (46)
The constant of integration 119860(120573]119904) in (46) is determined byinitial condition Thus
119860 (120573]119904) = 119865] (120573]119904) = int1
0
119891 (119903) 119903119869] (120573]119904119903) 119889119903 (47)
Hence from (47) (46) becomes
119880 (120573]119904 119905) = 119865] (120573]119904) 119890minus1205732]119904119905 (48)
Therefore by inversion theorem [25] we have
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 1205732]11990411986910158402] (120573]119904)
= lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 11986621198692] (120573]119904)
(As 119866119869] (120573]119904) = minus120573]1199041198691015840
] (120573]119904))
(49)
So
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
119869] (120573]119904119903)
1198692
] (120573]119904) (50)
We want to prove that 119906(119903 119905) given by (50) is truly a solutionof (37) that satisfies the given initial and boundary conditionsTo achieve this we need the following well known estimates[17]
119865] (120573]119904) = 119874(1
12057332
]119904)
120573]119904 asymp 120587(119904 +1
4
)
1198692
] (120573]119904) asymp2
120587120573]119904
as 119904 997888rarr infin
(51)
Using the above estimates we see that series (50) and theseries obtained by applyingΩ]119903 and 120597120597119905 separately under thesummation sign of (50) converge uniformly on 0 lt 119903 lt 1 and119905 gt 0 Hence by applyingΩ]119903 minus 120597120597119905 on (50) we get
(Ω]119903 minus120597
120597119905
) 119906 (119903 119905)
= ( lim119873rarrinfin
119873
sum
119904=1
21205734
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
sdot [
[
(1205732
]1199041199032
11986910158401015840
] (120573]119904119903) + 120573]1199041199031198691015840
] (120573]119904119903)
+ (1205732
]1199041199032
minus ]2) 119869] (120573]119904119903))
times (1205732
]1199041199032
1198692
] (120573]119904))minus1
]
]
) = 0
(52)
Hence we see that (50) satisfies the differential equation (37)Let us verify the boundary condition (ii) we have
lim119903rarr1minus[
120597119906 (119903 119905)
120597119903
+ 119866119906 (119903 119905)]
= lim119903rarr1minus
[
[
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662) 1198692] (120573]119904)
times (120573]1199041198691015840
] (120573]119904119903) + 119866119869] (120573]119904119903))]
]
(53)
and since the convergence is uniform we can take thelim119903rarr1minus inside the summation sign and arrive at the conclu-
sion as 120573]119904rsquos are the roots of the equation119866119869](119911)+1199111198691015840
](119911) = 0The initial condition (i) is already taken care of as we
evaluated the constant 119860(120573]119904) by using it Through Figures13ndash18 we establish the accuracy of the proposed methodFor numerical solution of differential equation (37) we havetaken ] = 5 and 119866 = 1 We have used MATLAB routineldquofzerordquo to find zeros of (40) and we have drawn Figures 13ndash18 by truncating series (50) at 119873 = 30 While evaluatingthe solution 119906(119903 119905) from (50) we have evaluated 119865
5(1205735119904) first
from its analytical expression given by (36) and denoted thesolution thus obtained by 119906(119903 119905) in Figures 13ndash18 and thenevaluating 119899th approximates FHT 119865
5119899(1205735119904) for 119899 = 100
using our proposed algorithm for evaluation of the FHT asgiven by (17) This solution is denoted by 119906
119886(119903 119905) in the above
mentioned figuresFigure 13 compares the given initial condition 119891(119903) with
119906119886(119903 119905) as 119905 rarr 0
+ and Figure 14 shows the correspondingerror 119864(119903) = 119906
119886(119903 0) minus 119891(119903) Figure 15 depicts the various
profiles of 119906119886(119903 119905) at fixed times 119905
119895= 0 1100 150 125
where the various profiles are denoted by 1199060 1199061 1199062 and
1199063 The absolute errors between 119906
119886(119903 119905) and 119906(119903 119905) at fixed
times 119905119895= 0 1100 150 125 are shown in Figure 16 As
the maximum possible error occurs in the neighborhood of
International Journal of Analysis 11
0 and 0001 we have restricted 119905 in (0 1] in Figures 17 and18 representing 119906(119903 119905) and 119906
119886(119903 119905) respectively and note that
they are in good agreement in the range
8 Conclusions
A new stable and efficient algorithm based on hat functionsfor the numerical evaluation of Hankel transform is proposedand analyzed Choosing hat basis functions to expand theinput signal 119903119891(119903) makes our algorithm attractive in theirapplication in the applied physical problems as they eliminatethe problems connected with the Gibbs phenomenon takingplace in [9 11] We have given for the first time (as per ourinformation) error and stability analysis which was one ofthe main drawbacks of the earlier algorithms and by variousnumerical experiments we have corroborated our theoreticalfindings Stability with respect to the data is restored andexcellent accuracy is obtained even for small sample intervaland high noise levels in the data From the various figuresit is obvious that the algorithm is consistent and does notdepend on the particular choice of the input signal Theaccuracy and simplicity of the algorithm provide an edgeover the many other algorithms Finally an application of theproposed algorithm is given for solving the heat equation inan infinite cylinder with a radiation condition
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] I N Sneddon The Use of Integral Transforms McGraw-Hill1972
[2] A E Siegman ldquoQuasi fastHankel transformrdquoOptics Letters vol1 no 1 pp 13ndash15 1977
[3] L Filon ldquoOn quadrature formula for trigonometric integralsrdquoProceedings of the Royal Society of Edinburgh vol 49 pp 38ndash471928-1929
[4] D Levin andA Sidi ldquoTwo new classes of nonlinear transforma-tions for accelerating the convergence of infinite integrals andseriesrdquo Applied Mathematics and Computation vol 9 no 3 pp175ndash215 1981
[5] A Sidi ldquoExtrapolation methods for oscillatory infinite inte-gralsrdquo Journal of the Institute of Mathematics and its Applica-tions vol 26 no 1 pp 1ndash20 1980
[6] A Sidi ldquoA user-friendly extrapolation method for oscillatoryinfinite integralsrdquoMathematics of Computation vol 51 no 183pp 249ndash266 1988
[7] A Sidi ldquoComputation of infinite integrals involving Besselfunctions of arbitrary order by the 119863-transformationrdquo Journalof Computational and Applied Mathematics vol 78 no 1 pp125ndash130 1997
[8] A Sidi Practical Extrapolation Methods Theory and Applica-tions vol 10 of Cambridge Monographs on Applied and Compu-tational Mathematics Cambridge University Press CambridgeUK 2003
[9] M Guizar-Sicairos and J C Gutierrez-Vega ldquoComputation ofquasi-discrete Hankel transforms of integer order for propagat-ing optical wave fieldsrdquo Journal of the Optical Society of AmericaA Optics Image Science and Vision vol 21 no 1 pp 53ndash582004
[10] L Yu M Huang M Chen W Chen W Huang and Z ZhuldquoQuasi-discrete Hankel transformrdquoOptics Letters vol 23 no 6pp 409ndash411 1998
[11] E B Postnikov ldquoAbout calculation of the Hankel transformusing preliminary wavelet transformrdquo Journal of Applied Math-ematics no 6 pp 319ndash325 2003
[12] V K Singh O P Singh and R K Pandey ldquoNumerical evalu-ation of the Hankel transform by using linear Legendre multi-waveletsrdquoComputer Physics Communications vol 179 no 6 pp424ndash429 2008
[13] V K Singh O P Singh and R K Pandey ldquoEfficient algorithmsto compute Hankel transforms using waveletsrdquo ComputerPhysics Communications vol 179 no 11 pp 812ndash818 2008
[14] O P Singh V K Singh and R K Pandey ldquoAn efficient and sta-ble algorithm for numerical evaluation of Hankel transformsrdquoJournal of Applied Mathematics amp Informatics vol 28 no 5-6pp 1055ndash1071 2010
[15] R K Pandey O P Singh and V K Singh ldquoA stable algorithmfor numerical evaluation of Hankel transforms using Haarwaveletsrdquo Numerical Algorithms vol 53 no 4 pp 451ndash4662010
[16] E Babolian and M Mordad ldquoA numerical method for solvingsystems of linear and nonlinear integral equations of the secondkind by hat basis functionsrdquo Computers amp Mathematics withApplications vol 62 no 1 pp 187ndash198 2011
[17] R S Pathak and O P Singh ldquoFinite Hankel transforms ofdistributionsrdquo Pacific Journal of Mathematics vol 99 no 2 pp439ndash458 1982
[18] A Erdelyi EdTables of Integral TransformsMcGraw-Hill NewYork NY USA 1954
[19] M Abramowitz and I A Stegun Handbook of MathematicalFunctions with Formulas Graphs and Mathematical TablesDover New York NY USA 1965
[20] R Barakat and B H Sandler ldquoEvaluation of first-order Hankeltransforms using Filon quadrature philosophyrdquo Applied Mathe-matics Letters vol 11 no 1 pp 127ndash131 1998
[21] J D Secada ldquoNumerical evaluation of the Hankel transformrdquoComputer Physics Communications vol 116 no 2-3 pp 278ndash294 1999
[22] R Barakat and E Parshall ldquoNumerical evaluation of the zero-order Hankel transform using Filon quadrature philosophyrdquoApplied Mathematics Letters vol 9 no 5 pp 21ndash26 1996
[23] J Gaskell Linear Systems Fourier Transforms and Opticschapter 11 John Wiley amp Sons New York NY USA 1978
[24] A D Poularika ldquoThe Hankel transformrdquo in The Handbook ofFormulas and Tables for Signal Processing CRC Press LLC BocaRaton Fla USA 1999
[25] G NWatsonTheory of Bessel Functions Cambridge UniversityPress Cambridge UK 2nd edition 1958
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Analysis 9
0
05
1
0
05
10
02
04
06
08
1
rt
ua(rt)
Figure 18 The approximate solution 119906119886(119903 119905) using 119899th approximate
FHT 1198655119899(1205735119904) for 0 lt 119905 lt 1 and 0 lt 119903 lt 1
for 119899 = 100 In Figure 12 errors 1198641(119901) and 119864
2(119901) for 119899 = 100
are shown
7 Applications
In this section we solve heat equation in cylindrical coor-dinates inside an infinitely long cylinder of radius unity byusing the theory of finite Hankel transform (FHT) developedin the preceding pages The initial temperature is givenby 119891(119903) and radiation takes place at the surface into thesurrounding medium maintained at zero temperature Weseek a function 119906(119903 119905) where 119903 is radius and 119905 is time (119906 doesnot depend upon 120579 and 119905) in the following application Themathematical model of this problem is the diffusion equationin polar coordinates given by
1205972
119906 (119903 119905)
1205972119903
+
1
119903
120597119906 (119903 119905)
120597119903
minus
]2
1199032119906 (119903 119905)
=
120597119906 (119903 119905)
120597119905
(0 lt 119903 lt 1 0 lt 119905 lt infin)
(37)
with the following initial and boundary conditions
(i) as 119905 rarr 0+ 119906(119903 119905) rarr 119891(119903) = 119903
][119867(119903) minus 119867(119903 minus 1)]
where 0 le 119903 le 1 and119867(119903) is unit step function givenby (34)
(ii) as 119903 rarr 1minus (120597119906120597119903) + 119866119906 rarr 0 for each fixed 119905 gt 0
and 119866 gt 0
When 119906(119903 119905) denotes the temperature within the cylinder119866 gt 0means that heat is being radiated away from the surfaceof the cylinder
LetΩ]119903 denote the Bessel differential operator (1205972
1205972
119903) +
(1119903)(120597120597119903) minus (]21199032) Then differential equation (37) can bewritten as
Ω]119903119906 =120597119906
120597119905
(38)
Applying the ]th order finite Hankel transform operator toboth sides of the differential equation (38) and using (8) weget
int
1
0
119903Ω]119903119906119869] (120573]119904119903) 119889119903 = int1
0
119903 (
120597119906
120597119905
) 119869] (120573]119904119903) 119889119903 (39)
where 120573]119904 is the 119904th positive root of equation [24]
119866119869] (119911) + 1199111198691015840
] (119911) = 0 ] ge minus1
2
(40)
The LHS of (39) may be written as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= int
1
0
(
1205972
119906
1205972119903
) 119903119869] (120573]119904119903) 119889119903
+ int
1
0
(
120597119906
120597119903
) 119869] (120573]119904119903) 119889119903 minus ]2 int1
0
(
119906
119903
) 119869] (120573]119904119903) 119889119903
= lim119903rarr1minus
120597119906 (119903 119905)
120597119903
119869] (120573]119904) minus 120573]119904 int1
0
119903
120597119906
120597119903
1198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
= minus119866119869] (120573]119904) lim119903rarr1minus119906 (119903 119905) minus 120573]119904 int
1
0
120597119906
120597119903
1199031198691015840
] (120573]119904119903) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903 (As 119903 997888rarr 1minus
120597119906
120597119903
+ 119866119906 997888rarr 0)
(41)
Since 120573]119904 is a root of 119866119869](119911) + 1199111198691015840
](119911) = 0 we have 119866119869](120573]119904) =minus120573]119904119869
1015840
](120573]119904) and thus
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
minus 120573]1199041198691015840
] (120573]119904) lim119903rarr1minus119906 (119903 119905)
+ 120573]119904 int1
0
119906 (1198691015840
] (120573]119904119903) + 120573]11990411990311986910158401015840
] (120573]119904119903)) 119889119903
minus ]2 int1
0
119906
119903
119869] (120573]119904119903) 119889119903
(42)
Using the fact that 119869](119909) is a solution of 1199092(11988921199101198891199092) +119909(119889119910119889119909) + (119909
2
minus ]2)119910 = 0 (42) may be simplified as
int
1
0
119903 (
1205972
119906
1205972119903
+
1
119903
120597119906
120597119903
minus
]2
1199032119906) 119869] (120573]119904119903) 119889119903
= minus1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
(43)
10 International Journal of Analysis
From (43) and (39) we have
minus 1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
=
120597
120597119905
(int
1
0
119903119906 (119903 119905) 119869] (120573]119904119903) 119889119903)
(44)
If FHT int10
119906(119903 119905)119903119869](120573]119904119903) 119889119903 of 119906(119903 119905) is denoted by119880(120573]119904 119905)then (44) may be written as
120597119880 (120573]119904 119905)
120597119905
= minus1205732
]119904119880(120573]119904 119905) (45)
whose solution is given by
119880(120573]119904 119905) = 119860 (120573]119904) 119890minus1205732]119904119905 (46)
The constant of integration 119860(120573]119904) in (46) is determined byinitial condition Thus
119860 (120573]119904) = 119865] (120573]119904) = int1
0
119891 (119903) 119903119869] (120573]119904119903) 119889119903 (47)
Hence from (47) (46) becomes
119880 (120573]119904 119905) = 119865] (120573]119904) 119890minus1205732]119904119905 (48)
Therefore by inversion theorem [25] we have
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 1205732]11990411986910158402] (120573]119904)
= lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 11986621198692] (120573]119904)
(As 119866119869] (120573]119904) = minus120573]1199041198691015840
] (120573]119904))
(49)
So
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
119869] (120573]119904119903)
1198692
] (120573]119904) (50)
We want to prove that 119906(119903 119905) given by (50) is truly a solutionof (37) that satisfies the given initial and boundary conditionsTo achieve this we need the following well known estimates[17]
119865] (120573]119904) = 119874(1
12057332
]119904)
120573]119904 asymp 120587(119904 +1
4
)
1198692
] (120573]119904) asymp2
120587120573]119904
as 119904 997888rarr infin
(51)
Using the above estimates we see that series (50) and theseries obtained by applyingΩ]119903 and 120597120597119905 separately under thesummation sign of (50) converge uniformly on 0 lt 119903 lt 1 and119905 gt 0 Hence by applyingΩ]119903 minus 120597120597119905 on (50) we get
(Ω]119903 minus120597
120597119905
) 119906 (119903 119905)
= ( lim119873rarrinfin
119873
sum
119904=1
21205734
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
sdot [
[
(1205732
]1199041199032
11986910158401015840
] (120573]119904119903) + 120573]1199041199031198691015840
] (120573]119904119903)
+ (1205732
]1199041199032
minus ]2) 119869] (120573]119904119903))
times (1205732
]1199041199032
1198692
] (120573]119904))minus1
]
]
) = 0
(52)
Hence we see that (50) satisfies the differential equation (37)Let us verify the boundary condition (ii) we have
lim119903rarr1minus[
120597119906 (119903 119905)
120597119903
+ 119866119906 (119903 119905)]
= lim119903rarr1minus
[
[
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662) 1198692] (120573]119904)
times (120573]1199041198691015840
] (120573]119904119903) + 119866119869] (120573]119904119903))]
]
(53)
and since the convergence is uniform we can take thelim119903rarr1minus inside the summation sign and arrive at the conclu-
sion as 120573]119904rsquos are the roots of the equation119866119869](119911)+1199111198691015840
](119911) = 0The initial condition (i) is already taken care of as we
evaluated the constant 119860(120573]119904) by using it Through Figures13ndash18 we establish the accuracy of the proposed methodFor numerical solution of differential equation (37) we havetaken ] = 5 and 119866 = 1 We have used MATLAB routineldquofzerordquo to find zeros of (40) and we have drawn Figures 13ndash18 by truncating series (50) at 119873 = 30 While evaluatingthe solution 119906(119903 119905) from (50) we have evaluated 119865
5(1205735119904) first
from its analytical expression given by (36) and denoted thesolution thus obtained by 119906(119903 119905) in Figures 13ndash18 and thenevaluating 119899th approximates FHT 119865
5119899(1205735119904) for 119899 = 100
using our proposed algorithm for evaluation of the FHT asgiven by (17) This solution is denoted by 119906
119886(119903 119905) in the above
mentioned figuresFigure 13 compares the given initial condition 119891(119903) with
119906119886(119903 119905) as 119905 rarr 0
+ and Figure 14 shows the correspondingerror 119864(119903) = 119906
119886(119903 0) minus 119891(119903) Figure 15 depicts the various
profiles of 119906119886(119903 119905) at fixed times 119905
119895= 0 1100 150 125
where the various profiles are denoted by 1199060 1199061 1199062 and
1199063 The absolute errors between 119906
119886(119903 119905) and 119906(119903 119905) at fixed
times 119905119895= 0 1100 150 125 are shown in Figure 16 As
the maximum possible error occurs in the neighborhood of
International Journal of Analysis 11
0 and 0001 we have restricted 119905 in (0 1] in Figures 17 and18 representing 119906(119903 119905) and 119906
119886(119903 119905) respectively and note that
they are in good agreement in the range
8 Conclusions
A new stable and efficient algorithm based on hat functionsfor the numerical evaluation of Hankel transform is proposedand analyzed Choosing hat basis functions to expand theinput signal 119903119891(119903) makes our algorithm attractive in theirapplication in the applied physical problems as they eliminatethe problems connected with the Gibbs phenomenon takingplace in [9 11] We have given for the first time (as per ourinformation) error and stability analysis which was one ofthe main drawbacks of the earlier algorithms and by variousnumerical experiments we have corroborated our theoreticalfindings Stability with respect to the data is restored andexcellent accuracy is obtained even for small sample intervaland high noise levels in the data From the various figuresit is obvious that the algorithm is consistent and does notdepend on the particular choice of the input signal Theaccuracy and simplicity of the algorithm provide an edgeover the many other algorithms Finally an application of theproposed algorithm is given for solving the heat equation inan infinite cylinder with a radiation condition
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] I N Sneddon The Use of Integral Transforms McGraw-Hill1972
[2] A E Siegman ldquoQuasi fastHankel transformrdquoOptics Letters vol1 no 1 pp 13ndash15 1977
[3] L Filon ldquoOn quadrature formula for trigonometric integralsrdquoProceedings of the Royal Society of Edinburgh vol 49 pp 38ndash471928-1929
[4] D Levin andA Sidi ldquoTwo new classes of nonlinear transforma-tions for accelerating the convergence of infinite integrals andseriesrdquo Applied Mathematics and Computation vol 9 no 3 pp175ndash215 1981
[5] A Sidi ldquoExtrapolation methods for oscillatory infinite inte-gralsrdquo Journal of the Institute of Mathematics and its Applica-tions vol 26 no 1 pp 1ndash20 1980
[6] A Sidi ldquoA user-friendly extrapolation method for oscillatoryinfinite integralsrdquoMathematics of Computation vol 51 no 183pp 249ndash266 1988
[7] A Sidi ldquoComputation of infinite integrals involving Besselfunctions of arbitrary order by the 119863-transformationrdquo Journalof Computational and Applied Mathematics vol 78 no 1 pp125ndash130 1997
[8] A Sidi Practical Extrapolation Methods Theory and Applica-tions vol 10 of Cambridge Monographs on Applied and Compu-tational Mathematics Cambridge University Press CambridgeUK 2003
[9] M Guizar-Sicairos and J C Gutierrez-Vega ldquoComputation ofquasi-discrete Hankel transforms of integer order for propagat-ing optical wave fieldsrdquo Journal of the Optical Society of AmericaA Optics Image Science and Vision vol 21 no 1 pp 53ndash582004
[10] L Yu M Huang M Chen W Chen W Huang and Z ZhuldquoQuasi-discrete Hankel transformrdquoOptics Letters vol 23 no 6pp 409ndash411 1998
[11] E B Postnikov ldquoAbout calculation of the Hankel transformusing preliminary wavelet transformrdquo Journal of Applied Math-ematics no 6 pp 319ndash325 2003
[12] V K Singh O P Singh and R K Pandey ldquoNumerical evalu-ation of the Hankel transform by using linear Legendre multi-waveletsrdquoComputer Physics Communications vol 179 no 6 pp424ndash429 2008
[13] V K Singh O P Singh and R K Pandey ldquoEfficient algorithmsto compute Hankel transforms using waveletsrdquo ComputerPhysics Communications vol 179 no 11 pp 812ndash818 2008
[14] O P Singh V K Singh and R K Pandey ldquoAn efficient and sta-ble algorithm for numerical evaluation of Hankel transformsrdquoJournal of Applied Mathematics amp Informatics vol 28 no 5-6pp 1055ndash1071 2010
[15] R K Pandey O P Singh and V K Singh ldquoA stable algorithmfor numerical evaluation of Hankel transforms using Haarwaveletsrdquo Numerical Algorithms vol 53 no 4 pp 451ndash4662010
[16] E Babolian and M Mordad ldquoA numerical method for solvingsystems of linear and nonlinear integral equations of the secondkind by hat basis functionsrdquo Computers amp Mathematics withApplications vol 62 no 1 pp 187ndash198 2011
[17] R S Pathak and O P Singh ldquoFinite Hankel transforms ofdistributionsrdquo Pacific Journal of Mathematics vol 99 no 2 pp439ndash458 1982
[18] A Erdelyi EdTables of Integral TransformsMcGraw-Hill NewYork NY USA 1954
[19] M Abramowitz and I A Stegun Handbook of MathematicalFunctions with Formulas Graphs and Mathematical TablesDover New York NY USA 1965
[20] R Barakat and B H Sandler ldquoEvaluation of first-order Hankeltransforms using Filon quadrature philosophyrdquo Applied Mathe-matics Letters vol 11 no 1 pp 127ndash131 1998
[21] J D Secada ldquoNumerical evaluation of the Hankel transformrdquoComputer Physics Communications vol 116 no 2-3 pp 278ndash294 1999
[22] R Barakat and E Parshall ldquoNumerical evaluation of the zero-order Hankel transform using Filon quadrature philosophyrdquoApplied Mathematics Letters vol 9 no 5 pp 21ndash26 1996
[23] J Gaskell Linear Systems Fourier Transforms and Opticschapter 11 John Wiley amp Sons New York NY USA 1978
[24] A D Poularika ldquoThe Hankel transformrdquo in The Handbook ofFormulas and Tables for Signal Processing CRC Press LLC BocaRaton Fla USA 1999
[25] G NWatsonTheory of Bessel Functions Cambridge UniversityPress Cambridge UK 2nd edition 1958
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
10 International Journal of Analysis
From (43) and (39) we have
minus 1205732
]119904 int1
0
119906 (119903 119905) 119903119869] (120573]119904119903) 119889119903
=
120597
120597119905
(int
1
0
119903119906 (119903 119905) 119869] (120573]119904119903) 119889119903)
(44)
If FHT int10
119906(119903 119905)119903119869](120573]119904119903) 119889119903 of 119906(119903 119905) is denoted by119880(120573]119904 119905)then (44) may be written as
120597119880 (120573]119904 119905)
120597119905
= minus1205732
]119904119880(120573]119904 119905) (45)
whose solution is given by
119880(120573]119904 119905) = 119860 (120573]119904) 119890minus1205732]119904119905 (46)
The constant of integration 119860(120573]119904) in (46) is determined byinitial condition Thus
119860 (120573]119904) = 119865] (120573]119904) = int1
0
119891 (119903) 119903119869] (120573]119904119903) 119889119903 (47)
Hence from (47) (46) becomes
119880 (120573]119904 119905) = 119865] (120573]119904) 119890minus1205732]119904119905 (48)
Therefore by inversion theorem [25] we have
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 1205732]11990411986910158402] (120573]119904)
= lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905119869] (120573]119904119903)
(1205732
]119904 minus ]2) 1198692] (120573]119904) + 11986621198692] (120573]119904)
(As 119866119869] (120573]119904) = minus120573]1199041198691015840
] (120573]119904))
(49)
So
119906 (119903 119905) = lim119873rarrinfin
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
119869] (120573]119904119903)
1198692
] (120573]119904) (50)
We want to prove that 119906(119903 119905) given by (50) is truly a solutionof (37) that satisfies the given initial and boundary conditionsTo achieve this we need the following well known estimates[17]
119865] (120573]119904) = 119874(1
12057332
]119904)
120573]119904 asymp 120587(119904 +1
4
)
1198692
] (120573]119904) asymp2
120587120573]119904
as 119904 997888rarr infin
(51)
Using the above estimates we see that series (50) and theseries obtained by applyingΩ]119903 and 120597120597119905 separately under thesummation sign of (50) converge uniformly on 0 lt 119903 lt 1 and119905 gt 0 Hence by applyingΩ]119903 minus 120597120597119905 on (50) we get
(Ω]119903 minus120597
120597119905
) 119906 (119903 119905)
= ( lim119873rarrinfin
119873
sum
119904=1
21205734
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662)
sdot [
[
(1205732
]1199041199032
11986910158401015840
] (120573]119904119903) + 120573]1199041199031198691015840
] (120573]119904119903)
+ (1205732
]1199041199032
minus ]2) 119869] (120573]119904119903))
times (1205732
]1199041199032
1198692
] (120573]119904))minus1
]
]
) = 0
(52)
Hence we see that (50) satisfies the differential equation (37)Let us verify the boundary condition (ii) we have
lim119903rarr1minus[
120597119906 (119903 119905)
120597119903
+ 119866119906 (119903 119905)]
= lim119903rarr1minus
[
[
119873
sum
119904=1
21205732
]119904119865] (120573]119904) 119890minus1205732]119904119905
(1205732
]119904 minus ]2 + 1198662) 1198692] (120573]119904)
times (120573]1199041198691015840
] (120573]119904119903) + 119866119869] (120573]119904119903))]
]
(53)
and since the convergence is uniform we can take thelim119903rarr1minus inside the summation sign and arrive at the conclu-
sion as 120573]119904rsquos are the roots of the equation119866119869](119911)+1199111198691015840
](119911) = 0The initial condition (i) is already taken care of as we
evaluated the constant 119860(120573]119904) by using it Through Figures13ndash18 we establish the accuracy of the proposed methodFor numerical solution of differential equation (37) we havetaken ] = 5 and 119866 = 1 We have used MATLAB routineldquofzerordquo to find zeros of (40) and we have drawn Figures 13ndash18 by truncating series (50) at 119873 = 30 While evaluatingthe solution 119906(119903 119905) from (50) we have evaluated 119865
5(1205735119904) first
from its analytical expression given by (36) and denoted thesolution thus obtained by 119906(119903 119905) in Figures 13ndash18 and thenevaluating 119899th approximates FHT 119865
5119899(1205735119904) for 119899 = 100
using our proposed algorithm for evaluation of the FHT asgiven by (17) This solution is denoted by 119906
119886(119903 119905) in the above
mentioned figuresFigure 13 compares the given initial condition 119891(119903) with
119906119886(119903 119905) as 119905 rarr 0
+ and Figure 14 shows the correspondingerror 119864(119903) = 119906
119886(119903 0) minus 119891(119903) Figure 15 depicts the various
profiles of 119906119886(119903 119905) at fixed times 119905
119895= 0 1100 150 125
where the various profiles are denoted by 1199060 1199061 1199062 and
1199063 The absolute errors between 119906
119886(119903 119905) and 119906(119903 119905) at fixed
times 119905119895= 0 1100 150 125 are shown in Figure 16 As
the maximum possible error occurs in the neighborhood of
International Journal of Analysis 11
0 and 0001 we have restricted 119905 in (0 1] in Figures 17 and18 representing 119906(119903 119905) and 119906
119886(119903 119905) respectively and note that
they are in good agreement in the range
8 Conclusions
A new stable and efficient algorithm based on hat functionsfor the numerical evaluation of Hankel transform is proposedand analyzed Choosing hat basis functions to expand theinput signal 119903119891(119903) makes our algorithm attractive in theirapplication in the applied physical problems as they eliminatethe problems connected with the Gibbs phenomenon takingplace in [9 11] We have given for the first time (as per ourinformation) error and stability analysis which was one ofthe main drawbacks of the earlier algorithms and by variousnumerical experiments we have corroborated our theoreticalfindings Stability with respect to the data is restored andexcellent accuracy is obtained even for small sample intervaland high noise levels in the data From the various figuresit is obvious that the algorithm is consistent and does notdepend on the particular choice of the input signal Theaccuracy and simplicity of the algorithm provide an edgeover the many other algorithms Finally an application of theproposed algorithm is given for solving the heat equation inan infinite cylinder with a radiation condition
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] I N Sneddon The Use of Integral Transforms McGraw-Hill1972
[2] A E Siegman ldquoQuasi fastHankel transformrdquoOptics Letters vol1 no 1 pp 13ndash15 1977
[3] L Filon ldquoOn quadrature formula for trigonometric integralsrdquoProceedings of the Royal Society of Edinburgh vol 49 pp 38ndash471928-1929
[4] D Levin andA Sidi ldquoTwo new classes of nonlinear transforma-tions for accelerating the convergence of infinite integrals andseriesrdquo Applied Mathematics and Computation vol 9 no 3 pp175ndash215 1981
[5] A Sidi ldquoExtrapolation methods for oscillatory infinite inte-gralsrdquo Journal of the Institute of Mathematics and its Applica-tions vol 26 no 1 pp 1ndash20 1980
[6] A Sidi ldquoA user-friendly extrapolation method for oscillatoryinfinite integralsrdquoMathematics of Computation vol 51 no 183pp 249ndash266 1988
[7] A Sidi ldquoComputation of infinite integrals involving Besselfunctions of arbitrary order by the 119863-transformationrdquo Journalof Computational and Applied Mathematics vol 78 no 1 pp125ndash130 1997
[8] A Sidi Practical Extrapolation Methods Theory and Applica-tions vol 10 of Cambridge Monographs on Applied and Compu-tational Mathematics Cambridge University Press CambridgeUK 2003
[9] M Guizar-Sicairos and J C Gutierrez-Vega ldquoComputation ofquasi-discrete Hankel transforms of integer order for propagat-ing optical wave fieldsrdquo Journal of the Optical Society of AmericaA Optics Image Science and Vision vol 21 no 1 pp 53ndash582004
[10] L Yu M Huang M Chen W Chen W Huang and Z ZhuldquoQuasi-discrete Hankel transformrdquoOptics Letters vol 23 no 6pp 409ndash411 1998
[11] E B Postnikov ldquoAbout calculation of the Hankel transformusing preliminary wavelet transformrdquo Journal of Applied Math-ematics no 6 pp 319ndash325 2003
[12] V K Singh O P Singh and R K Pandey ldquoNumerical evalu-ation of the Hankel transform by using linear Legendre multi-waveletsrdquoComputer Physics Communications vol 179 no 6 pp424ndash429 2008
[13] V K Singh O P Singh and R K Pandey ldquoEfficient algorithmsto compute Hankel transforms using waveletsrdquo ComputerPhysics Communications vol 179 no 11 pp 812ndash818 2008
[14] O P Singh V K Singh and R K Pandey ldquoAn efficient and sta-ble algorithm for numerical evaluation of Hankel transformsrdquoJournal of Applied Mathematics amp Informatics vol 28 no 5-6pp 1055ndash1071 2010
[15] R K Pandey O P Singh and V K Singh ldquoA stable algorithmfor numerical evaluation of Hankel transforms using Haarwaveletsrdquo Numerical Algorithms vol 53 no 4 pp 451ndash4662010
[16] E Babolian and M Mordad ldquoA numerical method for solvingsystems of linear and nonlinear integral equations of the secondkind by hat basis functionsrdquo Computers amp Mathematics withApplications vol 62 no 1 pp 187ndash198 2011
[17] R S Pathak and O P Singh ldquoFinite Hankel transforms ofdistributionsrdquo Pacific Journal of Mathematics vol 99 no 2 pp439ndash458 1982
[18] A Erdelyi EdTables of Integral TransformsMcGraw-Hill NewYork NY USA 1954
[19] M Abramowitz and I A Stegun Handbook of MathematicalFunctions with Formulas Graphs and Mathematical TablesDover New York NY USA 1965
[20] R Barakat and B H Sandler ldquoEvaluation of first-order Hankeltransforms using Filon quadrature philosophyrdquo Applied Mathe-matics Letters vol 11 no 1 pp 127ndash131 1998
[21] J D Secada ldquoNumerical evaluation of the Hankel transformrdquoComputer Physics Communications vol 116 no 2-3 pp 278ndash294 1999
[22] R Barakat and E Parshall ldquoNumerical evaluation of the zero-order Hankel transform using Filon quadrature philosophyrdquoApplied Mathematics Letters vol 9 no 5 pp 21ndash26 1996
[23] J Gaskell Linear Systems Fourier Transforms and Opticschapter 11 John Wiley amp Sons New York NY USA 1978
[24] A D Poularika ldquoThe Hankel transformrdquo in The Handbook ofFormulas and Tables for Signal Processing CRC Press LLC BocaRaton Fla USA 1999
[25] G NWatsonTheory of Bessel Functions Cambridge UniversityPress Cambridge UK 2nd edition 1958
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Analysis 11
0 and 0001 we have restricted 119905 in (0 1] in Figures 17 and18 representing 119906(119903 119905) and 119906
119886(119903 119905) respectively and note that
they are in good agreement in the range
8 Conclusions
A new stable and efficient algorithm based on hat functionsfor the numerical evaluation of Hankel transform is proposedand analyzed Choosing hat basis functions to expand theinput signal 119903119891(119903) makes our algorithm attractive in theirapplication in the applied physical problems as they eliminatethe problems connected with the Gibbs phenomenon takingplace in [9 11] We have given for the first time (as per ourinformation) error and stability analysis which was one ofthe main drawbacks of the earlier algorithms and by variousnumerical experiments we have corroborated our theoreticalfindings Stability with respect to the data is restored andexcellent accuracy is obtained even for small sample intervaland high noise levels in the data From the various figuresit is obvious that the algorithm is consistent and does notdepend on the particular choice of the input signal Theaccuracy and simplicity of the algorithm provide an edgeover the many other algorithms Finally an application of theproposed algorithm is given for solving the heat equation inan infinite cylinder with a radiation condition
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] I N Sneddon The Use of Integral Transforms McGraw-Hill1972
[2] A E Siegman ldquoQuasi fastHankel transformrdquoOptics Letters vol1 no 1 pp 13ndash15 1977
[3] L Filon ldquoOn quadrature formula for trigonometric integralsrdquoProceedings of the Royal Society of Edinburgh vol 49 pp 38ndash471928-1929
[4] D Levin andA Sidi ldquoTwo new classes of nonlinear transforma-tions for accelerating the convergence of infinite integrals andseriesrdquo Applied Mathematics and Computation vol 9 no 3 pp175ndash215 1981
[5] A Sidi ldquoExtrapolation methods for oscillatory infinite inte-gralsrdquo Journal of the Institute of Mathematics and its Applica-tions vol 26 no 1 pp 1ndash20 1980
[6] A Sidi ldquoA user-friendly extrapolation method for oscillatoryinfinite integralsrdquoMathematics of Computation vol 51 no 183pp 249ndash266 1988
[7] A Sidi ldquoComputation of infinite integrals involving Besselfunctions of arbitrary order by the 119863-transformationrdquo Journalof Computational and Applied Mathematics vol 78 no 1 pp125ndash130 1997
[8] A Sidi Practical Extrapolation Methods Theory and Applica-tions vol 10 of Cambridge Monographs on Applied and Compu-tational Mathematics Cambridge University Press CambridgeUK 2003
[9] M Guizar-Sicairos and J C Gutierrez-Vega ldquoComputation ofquasi-discrete Hankel transforms of integer order for propagat-ing optical wave fieldsrdquo Journal of the Optical Society of AmericaA Optics Image Science and Vision vol 21 no 1 pp 53ndash582004
[10] L Yu M Huang M Chen W Chen W Huang and Z ZhuldquoQuasi-discrete Hankel transformrdquoOptics Letters vol 23 no 6pp 409ndash411 1998
[11] E B Postnikov ldquoAbout calculation of the Hankel transformusing preliminary wavelet transformrdquo Journal of Applied Math-ematics no 6 pp 319ndash325 2003
[12] V K Singh O P Singh and R K Pandey ldquoNumerical evalu-ation of the Hankel transform by using linear Legendre multi-waveletsrdquoComputer Physics Communications vol 179 no 6 pp424ndash429 2008
[13] V K Singh O P Singh and R K Pandey ldquoEfficient algorithmsto compute Hankel transforms using waveletsrdquo ComputerPhysics Communications vol 179 no 11 pp 812ndash818 2008
[14] O P Singh V K Singh and R K Pandey ldquoAn efficient and sta-ble algorithm for numerical evaluation of Hankel transformsrdquoJournal of Applied Mathematics amp Informatics vol 28 no 5-6pp 1055ndash1071 2010
[15] R K Pandey O P Singh and V K Singh ldquoA stable algorithmfor numerical evaluation of Hankel transforms using Haarwaveletsrdquo Numerical Algorithms vol 53 no 4 pp 451ndash4662010
[16] E Babolian and M Mordad ldquoA numerical method for solvingsystems of linear and nonlinear integral equations of the secondkind by hat basis functionsrdquo Computers amp Mathematics withApplications vol 62 no 1 pp 187ndash198 2011
[17] R S Pathak and O P Singh ldquoFinite Hankel transforms ofdistributionsrdquo Pacific Journal of Mathematics vol 99 no 2 pp439ndash458 1982
[18] A Erdelyi EdTables of Integral TransformsMcGraw-Hill NewYork NY USA 1954
[19] M Abramowitz and I A Stegun Handbook of MathematicalFunctions with Formulas Graphs and Mathematical TablesDover New York NY USA 1965
[20] R Barakat and B H Sandler ldquoEvaluation of first-order Hankeltransforms using Filon quadrature philosophyrdquo Applied Mathe-matics Letters vol 11 no 1 pp 127ndash131 1998
[21] J D Secada ldquoNumerical evaluation of the Hankel transformrdquoComputer Physics Communications vol 116 no 2-3 pp 278ndash294 1999
[22] R Barakat and E Parshall ldquoNumerical evaluation of the zero-order Hankel transform using Filon quadrature philosophyrdquoApplied Mathematics Letters vol 9 no 5 pp 21ndash26 1996
[23] J Gaskell Linear Systems Fourier Transforms and Opticschapter 11 John Wiley amp Sons New York NY USA 1978
[24] A D Poularika ldquoThe Hankel transformrdquo in The Handbook ofFormulas and Tables for Signal Processing CRC Press LLC BocaRaton Fla USA 1999
[25] G NWatsonTheory of Bessel Functions Cambridge UniversityPress Cambridge UK 2nd edition 1958
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
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