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page 39 E763(part 2) Numerical Methods
)()(')(')()(' xqxtxqxtxg +=
)()('')(')('2)('')()('' xqxtxqxtxqxtxg ++=
L (4.71)
=
=i
j
jiji xQxt
jij
ixg
0
)()()( )()(
)!(!
!)(
and since we are interested on the values of the derivatives atx = 0, we have:
=
=
i
j
jiji Qtjij
ig
0
)()()( )0()0()!(!
!)0( (4.72)
The first derivative of a polynomial, say, )(xQn is 121 2 bxbxnb nn +++
L , so the second
derivative will be: 232 232)1( bxbxnbn nn +++
L and so on. Then these derivatives
evaluated atx = 0 are successively: jbjbbbb !,,,)432(,)32(,2, 4321 L
Then, we can write (4.72) as: =
=
=
=i
j
jij
i
j
jiji bcibjicj
jij
ig
00
)( !)!(!)!(!
!)0( and equating
this to the ith derivative of )(xPm evaluated atx = 0, that is, iai! gives:
=
=i
j
jiji bca0
, but since 10 =b we can finally write:
i
i
j
jiji cbca =
=
1
0
for i = 1, , k, (k= m + n). (4.73)
where we take the coefficients 0=ia for mi > and 0=ib for ni > .
Example
Consider the functionx
xf
=1
1)( . This function has a pole at x = 1 and polynomial
approximations will not perform well.
The Taylor coefficients of this function are given by: !2
)12(531
j
jc jj
=
L
So the first five are: 10 =c ,2
11 =c ,
8
32 =c ,
16
53 =c ,
128
354 =c which gives a polynomial of
order 4.
From the equations above we can calculate the Pad coefficients. We choose: m = n =2, so
k= m + n = 4.
Taking 10 =b , 000 abc = gives 10 =a
The other equations (from asking the derivatives to fit) are:
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E763 (part 2) Numerical Methods page 40
i
i
j
jiji cbca =
=
1
0
for i = 1, , k, (k= m + n)
and in detail:
4132231404
31221303
211202
1101
cbcbcbcbca
cbcbcbca
cbcbca
cbca
=
==
=
but form = n =2, we have 04343 ==== bbaa and the system can be written as:
41322
31220
211202
1101
000
000
cbcbc
cbcbc
cbcbca
cbca
=
=
=
=
and we can see that the second set of equations can be solved for the coefficients bi. Re-writing
this system:
42213
32012
cbcbc
cbcbc
=+
=+
In general, when 2kmn == as in this case the matrix that define the system will be of the
form:
+++
0321
3012
2101
1210
rrrr
rrrr
rrrr
rrrr
nnn
n
n
n
L
LLLLL
L
L
L
This is a special kind of matrix called the Toeplitz matrix that has the same element along each
diagonal so it is defined by a total of 2n+1 numbers. There are methods for solving systems with
this matrix and in particular,
Solving the system will give:
]161,43,1[ =a and ]165,45,1[ =b
giving:
2
2
2210
22102
20.31251.251
0.06250.751)(
xx
xx
xbxbb
xaxaaxR
+
+=
++
++=
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page 41 E763(part 2) Numerical Methods
Fig. 4.12 shows the function
xxf
=
1
1)( (blue line) and the Pad
approximant (red line)
2
222
0.31251.251
0.06250.751)(
x
xxxR
+
+=
The Taylor polynomial up to 4th order:4
43
32
210)( xcxcxcxccxP ++++=
with a green line.0 1
0
2
4
6
8
Fig. 4.12
It is clear that the Pad approximant gives a better fit, particularly closer to the pole.
Increasing the order, the Pad approximation
gets closer to the singularity.
Fig. 4.13 shows the approximations for
m = n = 1, 2, 3 and 4.
The poles closer to 1 of this functions are:
333.1:)(11 xR
106.1:)(22 xR
052.1:)(33 xR
031.1:)(44 xR
0 0.5 10
2
4
6
8
Fig. 4.13
We can see that even )(11 xR , a ratio of linear functions ofx, gives a better approximation than
the 4th order Taylor polynomial.
Exercise 4.7
Using the Taylor (McLaurin) expansion of xxf cos)( = , truncated to order 4:
2421)(
424
0
xxxcxt
k
kk +==
=
find the Pad approximant:
012
2
012
2
2
222
)(
)()(
bxbxb
axaxa
xQ
xPxR
++
++==
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E763 (part 2) Numerical Methods page 68
n subintervals.
Change of Variable
The above procedure was developed for definite integrals over the interval [1, 1]. Integralsover other intervals can be calculated after a change of variables. For example if the integral tocalculate is:
b
a
dttf )(
To change from the interval [a, b] to [1, 1] the following change of variable can be made:
2)(
2)(
ab
batx
+ or
22
bax
abt
++
so dx
abdt
2
=
Then, the integral can be approximated by:
=
++
n
i
ni
ni
b
a
bax
abfw
abdttf
1222
)( (6.43)
Exercise 6.11
Use Gaussian quadrature to calculate: +25.1
25.0
)5.0(sin dxx
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page 87 E763(part 2) Numerical Methods
Example
For the problem of the square coaxial (or square capacitor) seen earlier, the BV problem is
defined by the Laplace equation: 2= 0 with some boundary conditions (L = 2 , u = ,s = 0).
An appropriate functional (variational expression) for this case is:
J() = ( )2 d (given) (8.21)
Using Rayleigh-Ritz: J() = djbj x,y( )j =1
N
2
d
= dj bj (x,y)j=1
N
2
d
J() = didjbi (x,y) bj (x,y)j =1
N
i =1
N
d (8.22)
This can be written as the matrix equation: J(d) = dTAd,
where d = { dj} and aij = bi bj
d
Now, find stationary value:J
di= 0 for all i: i = 1, ... ,N
so, applying it to (8.22):J
di= aij dj
j=1
N
= 0 , for all i: i = 1, ... ,N
And the problem reduces to the matrix equation: A d = 0 (8.23)
Solving the system of equations (8.23) we obtain the coefficients dj and the unknown
function can be obtained as:
u(x,y) = dj bj (x,y)j=1
N
We can see that both methods, the weighted residuals and the variational method transform
the BV problem into an algebraic, matrix problem.
One of the first steps in the implementation of either method is the choice of appropriate
expansion functions to use in the Rayleigh-Ritz procedure: basis or trial functions and weighting
functions.
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E763 (part 2) Numerical Methods page 88
The finite element method provides a simple form to construct these functions and
implementing these methods.
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page A1 E763(part 2) Numerical Methods Appendix
APPENDIX
1. Taylor theorem
For a continuous function we have )()()(' afxfdttf
x
a
= , then, we can write:
+=x
a
dttfafxf )(')()( or )()()( 0 xRafxf += (A1.1)
where the reminder is =x
a
dttfxR )(')(0 .
We can now integrateR0 by parts using:)(
)()(' )2(
txvdtdv
dttfdutfu
==
==giving:
+==x
a
x
a
x
a
dttftxtftxdttfxR )()()(')()(')( )2(0 which gives after solving and
substituting in (A1.1):
++=x
a
dttftxafaxafxf )()()(')()()( )2( or
)())((')()( 1 xRaxafafxf ++= (A1.2)
We can also integrateR1 by parts using this time:
2
)()(
)()(2
)3()2(
txvdttxdv
dttfdutfu
==
==which gives:
+==x
a
x
a
x
a
dttftxtxtf
dttftxxR )()()(2
)()()()( )3(22
)2()2(
1 and again, after
substituting in (A1.2) gives:
+++=x
a
dttftxaxafaxafafxf )()()(2
)())((')()( )3(22)2(
Proceeding in this way we get the expansion:
nn
n
Raxn
afax
afax
afaxafafxf ++++++= )(
!
)()(
!3
)()(
2
)())((')()(
)(3
)3(2
)2(
L (A1.3)
where the reminder can be written as: +
=x
a
nn
n dttfn
txxR )(
!
)()( )1( (A1.4)
To find a more useful for for the reminder we need to invoke some general mathematicaltheorems:
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E763 (part 2) Numerical Methods Appendix page A2
First Theorem for the Mean Value of Integrals
If the function )(tg is continuous and integrable in the interval [a, x], then there exists a point
between a and x such that: ))(()( axgdttg
x
a
= .
This says simply that the integral can be represented by an average value of the function )(g
times the length of the interval. Because this average value must be between the minimum and
the maximum values and the function is continuous, there will be a point for which the
function has this value.
And in a more complex form:
Second Theorem for the Mean Value of Integrals
Now if the functionsgand h are continuous and integrable in the interval and h does not changesign in the interval, there exists a point between a andx such that:
=x
a
x
a
dtthgdtthtg )()()()(
If we now use this second theorem on expression (A1.4), with: )()( )1( tftg n+= and
!
)()(
n
txth
n= , we get:
= +
x
a
nn
n dtn
txfxR
!
)()()( )1( which can be integrated, giving:
1)1(
)()!1(
)()( +
+
+
= nn
n txn
fxR
(A1.5)
for the reminder of the Taylor expansion, where is appoint between a andx.
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page A9 E763(part 2) Numerical Methods Appendix
4. A variational formulation for the wave equation
Equation (8.20) in the example on the use of the variational method shows a variational
formulation for the one-dimensional wave equation. In the following we will prove that indeed
both problems are equivalent.
Equation (8.20) stated:
k2 =s.v.
dy
dx
2
dx
a
b
y2 dx
a
b
(A4.1)
Lets examine a variation ofk2 produced by a small perturbation y on the solutiony:
k2(y + y) = k2 + k2 =
d(y + y)dx
2
dx
a
b
(y +y)2 dxa
b
And re-writing:
k2 +k2( ) (y +y)2 dx
a
b
=d(y + y)
dx
2
dx
a
b
(A4.2)
Expanding and neglecting higher order variations:
k2 +k2( ) (y2 + 2yy) dx
a
b
=dy
dx
2
+ 2dy
dx
dy
dx
dx
a
b
or
k2
y2
dx
a
b
+ 2k2 yy dxa
b
+ k2 y2 dxa
b
=dy
dx
2
dx
a
b
+ 2dy
dx
dy
dxdx
a
b
(A4.3)
and now using (A4.1):
2k2
yy dx
a
b
+k2 y2 dxa
b
= 2 dydxdydx
dx
a
b
(A4.4)
Now, since we want k2 to be stationary about the solution functiony, we make k2 = 0, and we
examine what conditions this imposes on the functiony:
k2
yy dx
a
b
=dy
dx
dy
dxdx
a
b
(A4.5)
Integrating the RHS by parts:
k2
yy dx
a
b
= dydxy ab y d
2y
dx2dx
a
b
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E763 (part 2) Numerical Methods Appendix page A10
Or re-arranging:
yd2y
dx2+ k2y
dx
a
b
=dy
dxy
a
b
(A4.6)
Since y is arbitrary, (A4.6) can only be valid if both sides are zero. That means thaty shouldsatisfy the differential equation:
d2y
dx 2+ k2y = 0 (A4.7)
and any of the boundary conditions:
dy
dx= 0 at a and b or y = 0 at a and b (fixed values ofy at the ends). (A4.8)
Summarizing, we can see that imposing the condition of stationarity of (A4.1) with respect
to small variations of the function y leads toy satisfying the differential equation (A4.7), which
is the wave equation, and any of the boundary conditions (A4.8); that is, either fixed values ofy
at the ends (Dirichlet B.C.), or zero normal derivative (Neumann B.C.).
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E763 (part 2) Numerical Methods Appendix page A12
6. Shape Functions (Interpolation Functions)
The function u is approximated in each triangle by a first order function (a plane). This
will be given by an expression of the form: p + qx + ry which can also be written as a vector
product (equation 9.14).
u(x,y) = p + qx + ry = (1 x y)
p
q
r
(A6.1)
Evaluating this expression at each node of a triangle (with nodes numbered 1, 2 and 3):
u1 = p+ qx1 + ry1u2 = p + qx2 + ry2
u3 = p+ qx3 + ry3
u1
u2
u3
=
1 x1 y1
1 x2 y2
1 x3 y3
p
q
r
(A6.2)
And from here we can calculate the value of the constants p, q and rin terms of the nodal values
and the coordinates of the nodes:
p
q
r
=
1 x1 y1
1 x2 y2
1 x3 y3
1u1
u2
u3
(A6.3)
Replacing (A6. 3) in (A6.1):
u(x,y) = (1 x y)
1 x1 y1
1 x2 y2
1 x3 y3
1u1
u2
u3
(A6.4)
and comparing with (9.15), written as a vector product:
u(x,y) u(x,y) = u1N1(x,y) + u2N2 (x,y) + u3N3(x,y) = (N1 N2 N3 )
u1
u2
u3
(A6.5)
we have finally:
(N1 N2 N3)= (1 x y)
1 x1 y1
1 x2 y2
1 x3 y3
1
(A6.6)
Solving the right hand side (inverting the matrix and multiplying), gives the expression for
each shape function (9.16):
Ni (x,y) =1
2Aai + bix + ciy( ) (A6.7)
whereA is the area of the triangle and
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page A13 E763(part 2) Numerical Methods Appendix
a1 =x2y3 x3y2 b1 =y2 y3 c1 =x3 x2
a2 =x3y1 x1y3 b2 =y3 y1 c2 =x1 x3 (A6.8)
a3 =x1y2 x2y1 b3 =y1 y2 c3 =x2 x1
Note that from (A5.1), the area of the triangle can be written as:
A = 12 a1 + a2 + a3( ) (A6.9)
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