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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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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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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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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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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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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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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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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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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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