cs305/mm355 maple practical sessions

7/23/2019 CS305/MM355 Maple Practical Sessions

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CS305/MM355 Maple Practical Sessions.

Work Sheet Five.

Take this work sheet with you to the practical class. You are to read through it in

the practical, reproducing for yourself all of the Maple output on the sheet. Only

then should you attempt the exercises on the sheet at the back. If you have any

difficulties, seek assistance from the tutor. Practical work will account for up to

30% of the final mark.

You can save your Maple session at any time by clicking on the 'Save As....' option

in the 'File' tab; I suggest that you save the current sheet as 'session5.mws'.

WHEN YOU HAVE COMPLETED THE SOLUTIONS ON EXERCISE SHEET 5, SAVE THEM IN A FILE

CALLED 'solutions2.mws'. AT THE BEGINNING OF THE FILE, SPECIFY YOUR NAME,

STUDENT NUMBER AND CLASS. PRINT OUT THE FILE 'solutions2.mws' AND GIVE IT TO

THE THE LECTURER EITHER IN CLASS, IN A LABORATORY SESSION, OR BY PUSHING IT

UNDER HIS OFFICE DOOR (MARTIN MEERE, ROOM C120, ARUS DE BRUN). THE DEADLINE FOR

SUBMITTING YOUR WORK FOR PROBLEM SHEET FIVE IS THE 7th OF MARCH. COPYING WILL

BE SEVERELY PUNISHED; IF YOU SHARE YOUR WORK, YOU WILL ALSO BE SHARING YOUR

MARKS.

The Rayleigh-Ritz Method

We illustrate how the Rayleigh-Ritz procedure can be implemented in Maple via an example. We

begin with a bit of revision. Consider the following boundary value problem on 0 Int({Diff(w(x),x)*Diff(u(x),x)+w(x)*u(x)-w(x)*sin(x)},x=0..1)=0;


2/7

=d

0

1

{ }+

d

d

x( )w x

d

d

x( )u x ( )w x ( )u x ( )w x ( )sin x x 0

or

=( )B ,u w ( )l wwhere

=( )B ,u w d

0

1

{ }+

d

d

x( )w x

d

d

x( )u x ( )w x ( )u x x

=( )l w d0

1

( )w x ( )sin x x

We must make a choice for our approximation functions. We will consider the particular case N=3.

Recall that our first approximation function phi0(x) must satisfy the essential boundary condition,while phi1(x), phi2(x), phi3(x) must satisfy the homogeneous form of the essential boundary

condition. With this in mind, we make the following choice

> restart:for i from 0 to 3 do phi[i](x):=x^i end do;

:=( )0x 1

:=( )1 x x

:=( )2 x x2

:=( )3 x x3

In the Rayleigh Ritz method, we must solve the following equation for (c1,c2,c3):

=

B11 B12 B13

B21 B22 B23

B31 B32 B33

c1

c2

c3

F1

F2

F3

where

=Bij ( )B ,i j=Fi ( )l i ( )B ,i 0

> for i to 3 do

F[i]:=int(phi[i](x)*sin(x),x=0..1)-int(diff(phi[0](x),x)*diff(ph

i[i](x),x)+phi[0](x)*phi[i](x),x=0..1)end do:

> for i to 3 do

for j to 3 do

B[i,j]:=int(diff(phi[i](x),x)*diff(phi[j](x),x)+phi[i](x)*phi[j]

(x),x=0..1);

end do:

end do:

> BC:=matrix([[B[1,1],B[1,2],B[1,3]],[B[2,1],B[2,2],B[2,3]],[B[3,1

],B[3,2],B[3,3]]]);


3/7

:=BC

4

3

5

4

6

5

5

4

23

15

5

3

6

5

5

3

68

35> FC:=vector([F[1],F[2],F[3]]);

:=FC

, , ( )sin 1 ( )cos 1

1

2+ ( )cos 1 2 ( )sin 1

7

3 5 ( )cos 1 3 ( )sin 1

1

4

To find the c's, we simply multiply (inverse ofBC)*F:

>with(linalg):c:=evalf(multiply(inverse(BC),FC));

:=c [ ], ,-0.42916838 0.45483397 -0.16262447

Our approximation is now given by

> U3:=phi[0](x)+c[1]*phi[1](x)+c[2]*phi[2](x)+c[3]*phi[3](x);

:=U3 + 1 0.42916838x 0.45483397x2 0.16262447x3It is useful for comparative purposes to calculate the exact solution. We can calculate this as

follows:

> deq:=-diff(u(x),x,x)+u(x)=sin(x);

:=deq = +

d

d2

x2

( )u x ( )u x ( )sin x

>bcs:=u(0)=1,D(u)(1)=0;

:=bcs ,=( )u 0 1 =( )( )D u 1 0> exact:=dsolve({deq,bcs},u(x));

exact ( )u x1

4

ex

( ) + +2 ( )cosh 1 2 ( )sinh 1 ( )cos 1( )cosh 1

=:=

1

4

e( )x

( )+ +2 ( )cosh 1 2 ( )sinh 1 ( )cos 1( )cosh 1

1

2( )sin x+ +

We now plot the exact solution and our approximation on the same graph for comparative

purposes:

>plot([U3,rhs(exact)],x=0..1);

It is also worth plotting the error, that is, rhs(exact)-U3:

>plot(rhs(exact)-U3,x=0..1);


4/7

The Galerkin Method

We now obtain an approximation to the same problem using the Galerkin method. In the Galerkin

method, we begin by writing our differential equation in the general form

=( )A u fwhereAis a differential operator. For our particular probem, we have

> restart:A(u)=-diff(u(x),x,x)+u(x),f=sin(x);

,=( )A u +

d

d2

x2

( )u x ( )u x =f ( )sin x

We will make an approximation withN=3again. We now have to choose our approximation

functions such that both the essential and natural boundary conditions are satisified. We make the

following choice:

>phi[0](x):=1; for i from 1 to 3 do

phi[i](x):=x^(i+1)-(i+1)*x^i/i end do;

:=( )0 x 1

:=( )1 x x2

2x

:=( )2 x x3

3

2x

2

:=( )3 x x4

4

3x

3

In the Galerkin method, we must solve the following equations for the coefficients c1,c2,c3:

=

A11 A12 A13A

21 A

22 A

23

A31 A32 A32

c1c

2

c3

F1F

2

F3

where

:=Fi d

0

1

i ( )f ( )A 0 x

:=Aij d

0

1

i ( )A j x

We now denote byA_phi[i] the operatorAacting onphi[i].


5/7

> for i from 0 to 3 do

A_phi[i]:=-diff(phi[i](x),x,x)+phi[i](x)

end do:


F[i]:=int(phi[i](x)*(sin(x)-A_phi[0]),x=0..1)

end do:

> for i from 1 to 3 dofor j from 1 to 3 do

A[i,j]:=int(phi[i](x)*A_phi[j],x=0..1)

end do:

end do:

>AC:=matrix([[A[1,1],A[1,2],A[1,3]],[A[2,1],A[2,2],A[2,3]],[A[3,1

],A[3,2],A[3,3]]]);

:=AC

28

15

43

60

122

315

43

60

11

28

71

280

122

315

71

280

58

315

> F:=vector([F[1],F[2],F[3]]);

:=F

, , +

4

33 ( )cos 1 +

13

4

7

2( )cos 1 6 ( )sin 1

362

15

59

3( )cos 1 16 ( )sin 1

>with(linalg):c:=evalf(multiply(inverse(AC),F));

:=c [ ], ,0.2182873 -0.1871590 0.0367546


:=U3 + +1. 0.4990258000x2 0.4365746x 0.2361651333x3 0.0367546x4

The Collocation Method

We illustrate the method with the example discussed above. We choose the same approximation

functions as those for the Galerkin method, that is,

> restart:with(linalg):phi[0](x):=1:for i from 1 to 3 do

phi[i](x):=x^(i+1)-(i+1)*x^i/i end do:

Warning, the protected names norm and trace have been redefined andunprotected


:=U3 + + +1 c1 ( )x2

2x c2

x3

3

2x

2c3

x4

4

3x

3

The residual is given by:

> R:=-diff(U3,x,x)+U3-sin(x);

R 2c1

c2

( )6x 3 c3 ( )12x2

8x 1 c1

( )x2 2x c2

x3

3

2x

2c

3

x4

4

3x

3 + + + +:=

( )sin x


6/7

We choose the collocation pointsx=1/4,x=1/2,x=3/4:

> eqnset:={subs(x=1/4,R)=0,subs(x=1/2,R)=0,subs(x=3/4,R)=0};

eqnset = + + + 39

16c

1

91

64c

2

947

768c

31

sin

1

40 = + +

11

4c

11

43

48c

3

1

4c

2

sin

1

20, ,{:=

= + 47

16c

1

123

64c

2

255

256c

3

1

sin3

40}

> sols:=evalf(solve(eqnset,{c[1],c[2],c[3]}));

:=sols { }, ,=c1 0.2182606603 =c3 0.0367218111 =c2 -0.1869829286> U3:=phi[0](x)+.2182606603*phi[1](x)-.1869829286*phi[2](x)+.36721

8111e-1*phi[3](x);

:=U3 + +1. 0.4987350532x2 0.4365213206x 0.2359453434x3 0.0367218111x4

> eqn:=-diff(u(x),x,x)+u(x)=sin(x):bcs:=u(0)=1,D(u)(1)=0:exact:=ds

olve({eqn,bcs},u(x)):

>plot(rhs(exact)-U3,x=0..1);

The Least-Squared Method

We solve the same problem as before, but now using the least-squared method. We again have:

,=( )A u +

d

d2

x2

( )u x ( )u x =f ( )sin x

We will make an approximation withN=3again. We use the same approximaton functions as

those for the Galerkin method:

> restart:phi[0](x):=1:for i from 1 to 3 do

phi[i](x):=x^(i+1)-(i+1)*x^i/i end do:

In the least-squared method, we must solve the following equations for the coefficients c1 ,c2, c3:

=

A11 A12 A13

A21 A22 A23

A31 A32 A32

c1

c2

c3

F1

F2

F3

where

=Aij d

0

1

( )A i ( )A j x


7/7

:=Fi d

0

1

( )f ( )A 0 ( )A i x


A_phi[i]:=-diff(phi[i](x),x,x)+phi[i](x)

end do;

:=A_phi0 1

:=A_phi1 + 2 x2

2x

:=A_phi2 + + 6x 3 x3

3

2x

2

:=A_phi3 + + 12x2

8x x4

4

3x

3


F[i]:=int(A_phi[i]*(sin(x)-A_phi[0]),x=0..1)

end do;

:=F1 5 ( )cos 14

3

:=F2 +

25

4

13

2( )cos 1 12 ( )sin 1

:=F3

722

15

119

3( )cos 1 32 ( )sin 1


for j from 1 to 3 doA[i,j]:=int(A_phi[i]*A_phi[j],x=0..1)

end do:

end do:

>AC:=matrix([[A[1,1],A[1,2],A[1,3]],[A[2,1],A[2,2],A[2,3]],[A[3,1

],A[3,2],A[3,3]]]);

:=AC

36

5

73

60

206

315

73

60

517

140

687

280

206

315

687

280

778

315

> F:=vector([F[1],F[2],F[3]]);

:=F

, ,5 ( )cos 1

4

3+

25

4

13

2( )cos 1 12 ( )sin 1

722

15

119

3( )cos 1 32 ( )sin 1

>with(linalg):c:=evalf(multiply(inverse(AC),F));

:=c [ ], ,0.2183328560 -0.18737550 0.03693630


:=U3 + +1. 0.4993961060x2

0.4366657120x 0.2366239000x3

0.03693630x4

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