COMPUTER ORIENTED NEUMARICAL METHODS

8/9/2019 COMPUTER ORIENTED NEUMARICAL METHODS

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CHAPTER 5

Ordinary Differential Equations

function , its derivatives, and the variable upon which they depend. The most generalform of an ordinary differential equation is given by

(6.1)

Where m represents the highest order derivative, and y and its derivatives are functions

0),.....,,,( )(1 =mn y y y yt φ

of t. The order of the differential equation is the order of its highest derivative and its

degree is the degree of the derivative of the highest order after the equation has been

rationalized in derivatives. If no roduct of the de endent ariable t with itself or

any one of its derivatives occur, then the equation is said to be linear, otherwise it is

nonlinear. A linear differential equation of order m can be expressed in the form

(6.2))()()( )(0

t r t yt pm

p p =∑= φ


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t . .

of order m can be written as (6.3)

Then the equation (6.3) is called a canonical representation of the differential equation

),.....,,,( )1(1)( −= mnm y y y yt F y

(6.1). In a such form ,the highest order derivative is expressed in terms of the lower

order derivatives and he independent variable.

A general solution of an ordinary differential equation such as(6.1) is a relation between y,

t and m arbitrary constants , which satisfies the equation, but which contains no

derivatives. The solution may be an implicit relation of the form

(6.4)0),...,,,( 21 =mccc yt wr an exp c t unct on o t o t e orm .

The m arbitrary constants can be determined by prescribing m conditions of

the form 6.6

.,...,, 21 mcccw y =

mccc ,..., 21

1,.....,2,1,0,)( −== mvt v


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At on point t = , which are called initial conditions. The point is called the initial

point. The differential equation (6.1) together with the initial conditions (6.6) is called an

0t 0t

m th order initial value problem.

If the m conditions are prescribed at more than one point to determine the m arbitrary

.

the boundary value problem .

3


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EULER METHOD:

we write as (6.58)

where This is called the Euler method. Applying the method at the mesh

1,......,2,1,0,1 −=+=+ N jhf uu j j j).,( j j j ut f f =

, jt ),(1 j j j j ut f huu =−+

points j=0,1,2,….,N-1 , we obtain the numerical solution of (6.8) as

001 += hf uu

11

112

....................

−− +=

+=

N N N hf uu

hf uu

Choosing a suitable value of h and the initial condition, is obtained from the first

equation , is obtained from the second equation and so on. The method (6.58) is called

1u

2u

an explicit method , since using and we can calculate directly. The

Euler method is the simplest method to solve (6.8).

hu j , j f 1+ ju


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The truncation error in the method is given by)( 111 ut uT j j j −= +++

)(2

)],()([)( ''2

1 ξ uh

ut hf t ut u j j j j =+−= +

Where If we denote.1+


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The local truncation error is of O ( h .

Example 6.9:

)2h 0→

Use the Euler method to solve numerically the initial value problem.

With h=0.2,0.1 and 0.05 on the interval [0.1].Neglecting the round off errors, determine the bound for

1)0(,2 2' =−= utuu

the error. Applying Richardson’s extrapolation to improve the counted value u(1.0 .We have

With h=0.2 . The initial condition gives4,3,2,1,0;2

21 =−=+ juht uu j j j j

.10 =uFor j=0:U(0.2)

For j=1:

1,0 00 == ut .0.12 20001 =−=≈ uht uu

1,2.0 11 == ut

For j=2 :.92.0)1)(2.0)(2.0(21

2)4.0(2

21112

=−=−=≈ uht uuu

92.0,4.0 22 == ut

Similarly , we get

.78458.0)92.0)(4.0)(2.0(292.0

2)6.0(2

22223

=−=−=≈ uht uuu

.50706.0)1(,63684.0)8.0( 54 =≈=≈ uuuu

en = . , we getFor j=0 :

0.12)1.0(

1,020001

10

=−=≈==

uht uuu

ut

6


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22.0

1,1.02

11

−=≈==

uht uuu

ut For j=1 :

.98.0)1)(1.0)(1.0(21 2 =−=For j=2 :

23.0

98.0,2.02

22

−=≈==

uht uuu

ut

Similarly, we get.94158.0)98.0)(2.0)(1.0(298.0 2 =−=

.82525.0)5.0(,88839.0)4.0( 54 =≈=≈ uuuu

.50364.0)0.1(

,56011.0)9.0(,62202.0)8.0(

,68835.0)7.0(,75715.)6.0(

10

98

76

=≈

=≈=≈=≈=≈

uu

uuuu

uuouu

or = . , we ge

,92906.0)3.0(,9517.0)25.0(

,97054.0)2.0(,9851.0)15.0(

,995.0)1.0(,0.1)05.0(

≈≈

≈≈≈≈

≈≈

uu

uu

uu

,67928.0)7.0(,71225.0)65.0(

,74561.0)6.0(,77899.0)55.0(

,81195.0)5.0(,84401.0)45.0(,..,..

≈≈≈≈≈≈

uu

uu

uu

.50179.0)0.1(,52831.0)95.0(

,55615.0)9.0(,58527.0)85.0(,61559.0)8.0(,64698.0)75.0(

≈≈≈≈≈≈

uu

uuuu

7


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The truncation error in the Euler method is given by

''2

uh

TE =

.)(2

)(2

2

max10

22

t uh

uh

TE nt

n

≤≤≤= ξ

2 .u =

232

2

10

2

2)1(

)31(22 max ht t hTE t ≤+−≤ ≤≤

As the absolute maximum of in[0,1] is 2.

The error in Euler method is of the form

)31( 2t −

Richardson’s extrapolation gives....)()( 321 +++=− hchchchut u j j

2/2)1()1(

− −− k k k

huhu12

)( −=

k u

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Backward Euler Method:

−The equation at the mesh point may be written as 6.60

where . This is called the Backward Euler Method . The solution values

are determined from the following equations

11 ++ += j j j uu

1321 ,....,, + juuuu

),( 111 +++ = j j j ut f f ),( j j ut f

h

=−1+= jt t

hf uu

hf uu

+=

+=

212

101

The method (6.60) involves the unknown value on both sides of the equation and hence N N N hf uu += −1

...................

1+ ju 1+ juimp icit y. Suc a met o is ca e an mp c t met o .Since f(t,u) is nonlinear,(6.60)produces a nonlinear algebraic equation for the solution of . For thesolution of this nonlinear equation, we can use the Newton ‐RAPHSON METHOD . Applying the Newton ‐Raphson method to (6.60). We get

1+ ju

Or after rearrangement as ∂

−++= ++

++++

+u

uuut f huu s j s

j s

j j j s

j )(),()(1

)1(1

)(11

)1(1

.),()(1 11111 s

j j

s

j j

s

j

s

j ut hf uuuuuh +++++

+ +−=− ∂−

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Where is the sth iterate of and is evaluated at ( ) . We iterate (6.61) until

(6.62)

)(1

s ju + 1+ ju

u f

∂∂ )(

11 , s j j ut ++

TOLuu s s ≤−+ )()1(

Where TOL is a specified absolute error tolerance. Since the solution u(t) ofVaries continuously, we may take the initial approximation as The local truncation error at thepoint in the method(6.60) is given by

00 )(),,( η == t uut f dt du

j j uu =+0

1=1+ j

))](,()([)(

)(

2

111

111

+++

+++

+−=−=

j j j j

j j j

t ut hf t ut u

ut uT

(6.63)

It may be noted that the truncation errors of Euler and backward Euler methods are equal but are ofopposite signs.

.),(''2 1

+


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We have 8.01)(')(

111 += +++ ut u F

s

j j j

Taking , we obtain from (6.64), for

)64.6,.....(2,1,0)(' )( 111

1 =−=

+

+

+

+

+ su F uu s j j j

0 j j uu =+

)0(1

,

,93070332.0,14896552.1)(',00038050.0)(,93103448.0,16.1)(',08.0)(

,,.

3272

)2(1

)1(1

)1(1

)1(

1

)0(

1

)0(

1

011

======

===

−

uu F u F uu f u F

uu

Therefore , u(0.2) 93070331.01 =≈ u,.,.,. 111 === uuu

,93070331.0,4.0:1 1)0(

22 ==== uut j

,82247016.0,26319054.1)(',00000034.0)(

,82247043.0,26365260.1)(',00182463.0)(

,82391436.0,29782506.01)(',13859338.0)(

)3(2

)2(2

)2(2

)2(2

)1(2

)1(2

)1(2

)0(2

)0(2

======

===

uu F u F

uu F u F

uu F u F

Therefore , u(0.4) = =0.82247016.The exact values are u(0.2)=0.96153846,u(0.4)=0.86206897.

2u

..,.'. 22,

2 ==−= − uuu

11


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MID – POINT METHOD:

The equation may be written as (6.65)

This is called the mid ‐point method. The solution values are given by

),(2

11 j j

j j ut f huu =− −+ .1...,2,1,0,211 −=+= −+ N jhf uu j j j

213

102

....................2

2

+=

+=

hf uu

hf uu

(6.66)The value is known from the initial condition. The value is unknown and has to be determined

by some other method after which (6.66) is used to determine successively .

11 2 −− += N N N hf uu1u 1u

N uuu ,...., 32

The local truncation error is given by

3

11

111

))](,(2[)(

)(

−+

+++

+−=−=

j j j j

j j j

h

t ut hf ut u

ut uT

.

The bound on the truncation error is given by (6.68)

.11,3 +−


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Example 6.13: Solve the initial value problem

1)0(,2' 2 =−= utuu

, . , , .

compute u(0,2). Determine the percentage relative error at t=1.The mid ‐point method gives

(6.69)4,3,2,1,8.04 212

11 =−=−= −−+ jut uuht uu j j j j j j j

We calculate from the Taylor series method of second order. We have1u

)0(''2

)0(')0()(2

1 uh

huuhuu ++=≈

''' .

We obtain

,, −===

.96.0)2(2

04.001)2.0(1 =−++=≈ uu

, . , ,.,,, 5432

,4.0,852544.0,96.0:2

.852544.0)96.0)(2.0)(8.0(1

.8.0)4.0(

,.,.,:

221

2

21102

110

=====−=

−=≈====

t uu j

ut uuu

uu

8.0)8.0(

6.0,7274139.0,858544.0:3

.7274139.0)852544.0)(4.0)(8.0(96.0

...

23324

332

2

2213

−=≈====

=−=−=≈

ut uuu

t uu j

ut uuu

134981176.0)5985611.0)(8.0)(8.0(7274139.0

8.0)0.1(

,8.0,5985611.0,7274139.0:4

.5985611.0)7274139.0)(6.0)(2.0(4852544.0

2

24435

443

=−=−=≈

=====−

ut uuu

t uu j


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The exact solution is Therefore , u(1)=0.5.).1/(1)( 2t t u +=

*The percentage error is defined as

Where u is the exact solution and is the approximate solution. Therefore:

100*. ue p

−

=

*u

%.38.0100*5.0

4981176.05.0. =

−=e p

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6.4 SINGLESTEP METHODS:

(6.70)Can be classified mainly into two types. They are (1) single step methods and (2) multistep methods.

],[,)(),,(' 000 bt t t uut f u == η

, .Thus, a general single step method can be written as

(6.71)),,,,( 111 huut t huu j j j j j j +++ += φ huut t .

. This function is called the increment function . If can be obtained simply by evaluatingthe right hand side of (6.71) , then the method is called an explicit method . In this case , the method is

of the form

),,( hut φ φ 1+ ju

(6.72)

If the right hand side of (6.71) depends on also, then it is called an implicit method. The general

).,,(1 hut huu j j j j φ +=+

1+ ju

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Taylor Series Method:

The fundamental numerical method for the solution of 6.70 is the Ta lor series method.

We assume that the function u(t) can be expanded in Taylor series about any point ,that is jt

11

....)('')(!2

1)(')()()( 2 t ut t t ut t t ut u j j j j j +−+−+=

(6.75)

This expansion holds for and

( ).

!1!t ut t

pt ut t

p j j j j +−

++−+

],[ 0 bt t .10


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Denote by , the value obtained from by using an approximate valueIn place of the exact value . Neglecting the error term, we have the method

)),(,( ht ut h j jφ ),,( hut h j jφ ju)( jt u

.

to approximate . The error or the truncation error of the method is given by

....,,,,,,1 −==+ uuu j j j j

)( 1+ jt u

)(1 11

1 ht uhT p p θ += +++ .

The method (6.67) is called the Taylor series method of the order p . Substituting p=1 , in (6.76) we get p +

),('1 j j j j j ut hf uhuuu +=+=+ . ,

order 1 .To apply (6.76) , it is necessary to know If and are known, then

the derivatives can be calculated as follows :

).(),.....('),( )( j p

j j t ut ut u jt )( jt u

t u,Next , the differential equation u’=f(t,u) is differentiated to obtain expressions for the

higher order derivatives of u(t) .Thus , we have

j)).(,()(' j j j t ut f t u =

),(' ut f u =

..........................................................

)(2''' 2 ut uuututt

ut

ff f f f f ff f u

u

++++=+=

17

Where represents the partial derivatives of f with respect to t and u and so on. The values....., ut f f


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can be computed by substituting Therefore, if and are known exactly ,then (6.76) can be compute with an error

. jt t = jt )( jt uu

The number of terms to be included in (6.76) is fixed by the permissible error. If this error is and the

( ))(

!1)1(1 ht u

ph

j p p θ ++

++

ε series is truncated at the term then

Or (6.78)

)()( j p t u

ε θ ε θ

)!1()()!1()(

)(1

)1(1

+


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Example 6.15 :

Given the initial value ro ram u 0 =0' 22

Determine the first three non ‐zero terms in the Taylor series for u(t) and hence obtain the value for u(1) .Also determine t when the error in u(t) obtained from the first two non ‐zero terms is to be less than

after roundin .

,

6−

We have u(0)=0,u’(0)=0u’’=2t+2uu’,u’’(0)=0u’’’=2+2 uu’’+ u’’’ 0 =22)'(u

0)0(),'''''10'5(2

0)0(],)''(3''''4[2

0)0(),'''3'''(2

)6()4()5()6(

)5(2)4()5(

)4()4(

=++=

=++==+=

uuuuuuuu

uuuuuuu

uuuuuu

'''120''45'10[2

0)0()0()0(

80)0(),)'''(10''15'6(2

)11(2)5()6()4(

)7()8()9()10()11(

)10()9()8(

+++====

=+++=

uuuuuuuuu

uuu

uuuuuuuuu

Thus , the Taylor series for u(t) becomes . The approximate value of u(1) is givenby . If only the first two terms are used , then the value of t is obtained from

,

1172

20792

631

31

)( t t t t u ++=350168.0

20792

631

31

)1( ≈++=u

19

.solving we get .711 10*5.02079

2 −

COMPUTER ORIENTED NEUMARICAL METHODS

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