Numerical Assignment 2

7/30/2019 Numerical Assignment 2

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Numerical Methods

Professor Dennis Giannacopoulos

Assignment 2

Nassir Abou Ziki

Monday Nov 12 2012-10-15


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Problem 1In this problem we consider two first-order triangular finite elements used to solve the Laplace equation

for electrostatic potential. We first find a local S-matrix for each triangle, and then the global S-matrix

for the mesh which consists of just these two triangles. Figure 1 shows the local (disjoint) and global(conjoint) node-numberings respectively along with the (x, y)-coordinates of the element vertices in

meters.

First we will derive the equations that will allow us to calculate the local S-matrix for each triangle. Since

we are assuming the potential Uis linear overxandy then we can write:

For a given triangle (assume vertices are number 1 through 3), then at vertex 1 we have:

We get similar expressions at vertices 2 and 3 and we can write:

Using Cramers rule we can write an expression ofaas

Figure 1: This figure shows the position of the nodes in meters as well as the

numbering of the nodes for the disjoint (a) elements and the conjoint (b) elements.


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The determinant in the denominator is just twice the area of the triangle denoted2A, then

Similarly we get an expression for band c:

Now we can write a, b and c in terms ofU1, U2, and U3 by substituting the previous expressions intothe

following equation,

And thus we can now write

Where we have,

Notice that equation (1.6) for the total potential is now given in terms of the three unknownsU1, U2,

and U3 only since we eliminated a, b and c by substitution.

Now that we have an expression of the potential of one triangle (e) we look at the contribution to the

energy from that triangle, which is given by,

|| This can be written in matrix form:


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Such that,

( ) Using equation 1.13 and equations 1.7-1.9 we can find the expressions of for i= 1,2,3 and j=1,2,3

Not that is in fact symmetric thus equations 1.14 through 1.19 are sufficient to find .Now we can find the local matrices, let the lower triangle be triangle 1 and let the upper triangle be

triangle 2, then we have:


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[

]

The global S-matrix is given by:

Where the matrix Cis:

[

]

Thus the global S-matrix is:

These calculations were done by hand.


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Problem 2In this problem we use a program Simple2D to solve for the electrostatic potential for one-quarter of

the cross-section of a rectangular coaxial cable shown in figure 2. The Simple2D uses the finite element

method to solve for the static potential.

We used the two-element mesh shown in Figure 1(b) as a building block to construct a finite element

mesh for one-quarter of the cross section of the coaxial cable, the mesh is shown in Figure 3. We then

used that mesh to write the input file for Simple2D. The first part of the input file defines the position of

the nodes. Then the elements are entered in the input file where they were defined by going through

the nodes in a counter-clockwise fashion. Finally, the last part of the input file defines the Dirchelet

Boundary Conditions. The input file is named inputfile.txt it can be found in the assignment directory.

Also I inserted the input file in Appendix 2a for reference.

Figure 2 A sketch of the coaxial cable with the

boundary conditions


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Once we wrote the input file, I created a directory and placed in it the input file, an empty output file

labeled outputfile.txt along with the Simple2D executable file. Then I ran Simple2D using the following

command in command prompt:

C:\path to directory\ simple2D outputfile.txt

The program solves for the potential at the nodes of the mesh and saves the result into the output file,

which can also be found in the assignment folder as well as in Appendix 2b. To determine the potential

at (x,y) = (0.06, 0.04) we just need to find the node associated with these coordinates in the output file.

Notice from Figure 3 that the point (x,y) = (0.06, 0.04), is in fact node number 16. Therefore the

potential at (x,y) = (0.06, 0.04) is given by:

V=3.68423 volts

Figure 3 This figure shows the finite element mesh configuration with the node

numberings and the boundary conditions. Note that the width of the mesh is 0.1

meters and the length is 0.1 meters. Hence we are looking at the lower left

quarter of the initial problem


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Problem 3In this problem we are going to solve the same problem as described in question number 2. However,

we will use different methods to get the solution. We will be using the Conjugate Gradient method as

well as Cholesky decomposition. In order, to solve this problem we define a mesh using a finite-difference node spacing, h=0.02m in x and y directions for the same one-quarter cross-section of the

system shown in Figure 2. The mesh is shown in Figure 4.

We used a five-point difference formula to write the problem in the form of:

A is 19 by 19 coefficient matrix, the column vectorx represents the unknown potentials at the 19 nodesand the column vector b is given by the problem. Note that most of the entries of the vector b are zero

(except at 9, 10, 13, 16 and 19 where these entries take the value of -10). I wrote theAmatrix by hand

and entered it to Matlab. The data corresponding to this problem is saved in the file data.mat.

Figure 4 This figure shows the finite element mesh with the boundary

conditions as well as the numbering of the nodes at which we need to solve

for the electric potential.


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I wrote a program implementing the conjugate gradient method (un-preconditioned), the code is

included under the function conjugateGrad in Appendix 3. Also the code needed to solve the

problem using Cholesky Decomposition from Assignemnt 1 is included in Appendix 3 (functions

cholDecomposition, solveLinear and solve_helper). The matrix operation functions

are not included in the Appendix for simplicity also because these functions are trivial, they are found in

the assignment directory.

a. We test the matrix A using the Cholesky Decompostion program. We basically run in Matlab thefollowing command

cholDecomposition(A)

The output is a 19 by 19 lower triangular matrix with complex entries. Therefore, therefore A is not

positive definite. We would like to have a system in the form of where A is symmetric positivedefinite. To achieve this purpose we left multiply the equation by

to get:

Hence now we have a modified system:

, where and Amodand bmodare saved in the data.mat file. The modified matrix is now guaranteed to be symmetric

and positive definite.

b. Solving using Cholesky matlab code and output:>> load('data.mat')

>> solveLinear(Amod,bmod)

ans =

0.6381

1.3111

2.0175

2.6394

2.8350

1.24112.5890

4.1194

5.7050

6.0612

1.7373

3.6842


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6.1661

2.0240

4.2445

6.8608

2.1143

4.4090

7.0327

Solving using conjugate gradient method, matlab code and output:

>> [x,iter,res]=conjugateGrad(Amod,bmod,0)

x =

0.6381

1.3111

2.0175

2.6394

2.8350

1.2411

2.5890

4.1194

5.7050

6.0612

1.7373

3.6842

6.1661

2.0240

4.2445

6.8608

2.1143

4.4090

7.0327

iter =

20

res =


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50.4665 29.6248

31.1892 15.0240

21.5185 9.4059

17.0146 8.1961

14.4802 6.1396

10.9324 5.8752

10.0132 7.5790

11.9813 5.3248

10.3042 6.1069

8.4821 4.5521

7.2796 2.5955

6.3425 2.9610

3.0684 1.3835

1.8087 0.8348

2.0684 1.0711

1.6836 0.7185

0.5139 0.2248

0.0140 0.0060

0.0000 0.0000

0.0000 0.0000

Note that the solution from the Cholesky Decomposition method is almost identical to the solution

given by the conjugate gradient method (vector x in the output).

c. The plots of the norms

0

10

20

30

40

50

60

0 5 10 15 20 25

norm

Number of Iterations

The variation of the 2-norm and infinite norm

as a function of the number of iterations

2-norm

infinitenorm

Plot 1 This plot shows the variation of the 2-norm and the infinite nor as a function of the number of

iterations of the conjugate gradient method. Note that both go to zero and thus we have convergence


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Appendix

App en dix 2a

The input file:

0.000 0.000

0.020 0.000

0.040 0.000

0.060 0.000

0.080 0.000

0.100 0.000

0.000 0.020

0.020 0.020

0.040 0.020

0.060 0.020

0.080 0.020

0.100 0.020

0.000 0.040

0.020 0.040

0.040 0.040

0.060 0.040

0.080 0.040

0.100 0.040

0.000 0.060

0.020 0.060

0.040 0.060

0.060 0.060


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0.080 0.060

0.100 0.060

0.000 0.080

0.020 0.080

0.040 0.080

0.060 0.080

0.080 0.080

0.100 0.080

0.000 0.100

0.020 0.100

0.040 0.100

0.060 0.100

/

2 7 1 0.

2 8 7 0.

3 8 2 0.

3 9 8 0.

4 9 3 0.

4 10 9 0.

5 10 4 0.

5 11 10 0.

6 11 5 0.

6 12 11 0.

8 13 7 0.

8 14 13 0.


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9 14 8 0.

9 15 14 0.

10 15 9 0.

10 16 15 0.

11 16 10 0.

11 17 16 0.

12 17 11 0.

12 18 17 0.

14 19 13 0.

14 20 19 0.

15 20 14 0.

15 21 20 0.

16 21 15 0.

16 22 21 0.

17 22 16 0.

17 23 22 0.

18 23 17 0.

18 24 23 0.

20 25 19 0.

20 26 25 0.

21 26 20 0.

21 27 26 0.

22 27 21 0.

22 28 27 0.

23 28 22 0.


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23 29 28 0.

24 29 23 0.

24 30 29 0.

26 31 25 0.

26 32 31 0.

27 32 26 0.

27 33 32 0.

28 33 27 0.

28 34 33 0.

/

1 0.000

2 0.000

3 0.000

4 0.000

5 0.000

6 0.000

7 0.000

13 0.000

19 0.000

25 0.000

31 0.000

28 10.00

29 10.00

30 10.00

34 10.00


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App en dix 2b

The output file:

Input node list

n x y

1 0.00000 0.00000

2 0.02000 0.00000

3 0.04000 0.00000

4 0.06000 0.00000

5 0.08000 0.00000

6 0.10000 0.00000

7 0.00000 0.02000

8 0.02000 0.02000

9 0.04000 0.02000

10 0.06000 0.02000

11 0.08000 0.02000

12 0.10000 0.02000

13 0.00000 0.04000

14 0.02000 0.04000

15 0.04000 0.04000

16 0.06000 0.04000

17 0.08000 0.04000

18 0.10000 0.04000

19 0.00000 0.06000

20 0.02000 0.06000


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21 0.04000 0.06000

22 0.06000 0.06000

23 0.08000 0.06000

24 0.10000 0.06000

25 0.00000 0.08000

26 0.02000 0.08000

27 0.04000 0.08000

28 0.06000 0.08000

29 0.08000 0.08000

30 0.10000 0.08000

31 0.00000 0.10000

32 0.02000 0.10000

33 0.04000 0.10000

34 0.06000 0.10000

Input element list

i j k Source

2 7 1 .00000E+00

2 8 7 .00000E+00

3 8 2 .00000E+00

3 9 8 .00000E+00

4 9 3 .00000E+00

4 10 9 .00000E+00


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20 26 25 .00000E+00

21 26 20 .00000E+00

21 27 26 .00000E+00

22 27 21 .00000E+00

22 28 27 .00000E+00

23 28 22 .00000E+00

23 29 28 .00000E+00

24 29 23 .00000E+00

24 30 29 .00000E+00

26 31 25 .00000E+00

26 32 31 .00000E+00

27 32 26 .00000E+00

27 33 32 .00000E+00

28 33 27 .00000E+00

28 34 33 .00000E+00

Input fixed potentials

node value

1 0.00000

2 0.00000

3 0.00000

4 0.00000

5 0.00000


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10 0.06000 0.02000 1.73734

11 0.08000 0.02000 2.02403

12 0.10000 0.02000 2.11426

13 0.00000 0.04000 0.00000

14 0.02000 0.04000 1.31112

15 0.04000 0.04000 2.58895

16 0.06000 0.04000 3.68423

17 0.08000 0.04000 4.24452

18 0.10000 0.04000 4.40899

19 0.00000 0.06000 0.00000

20 0.02000 0.06000 2.01747

21 0.04000 0.06000 4.11938

22 0.06000 0.06000 6.16611

23 0.08000 0.06000 6.86082

24 0.10000 0.06000 7.03266

25 0.00000 0.08000 0.00000

26 0.02000 0.08000 2.63936

27 0.04000 0.08000 5.70500

28 0.06000 0.08000 10.00000

29 0.08000 0.08000 10.00000

30 0.10000 0.08000 10.00000

31 0.00000 0.10000 0.00000

32 0.02000 0.10000 2.83499

33 0.04000 0.10000 6.06125

34 0.06000 0.10000 10.00000


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App en dix 3

% This function is an implentation of the non pre-conditioned conjugate% gradient method to solve a system of equations Ax= b.function [x,iter,res]=conjugateGrad(A,b,es)

%get the size of the square matrix An=size(A,1);

%set iteration to zeroiter=0;

%start with zero as an initial guessx=zeros(n,1);

%find the residual vectorr=b-(mat_times(A,x));

%set the initial direction vector to be equal to the first residualp=r;

%loop until we are donewhile(1)

%find alpha using p, r and A

alpha=mat_times(mat_transpose(p),r)/mat_times(mat_times(mat_transpose(p),A),p);

%the new minimum point is given byx=x+(alpha*p);

%find the new residualr=b-mat_times(A,x);

%set column one of res to be the 2-norm at each iter and column two to%be the infinite normres(iter+1,1)=two_norm(r);res(iter+1,2)=infiniteNorm(r);

%check if we are doneiter=iter+1;if iter>n || max(r)


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------------------------------------------------------------------------------------------------------------

function L=cholDecomposition(X)%Runs the Cholesky Decomposition Algorithm on Matrix X and saves the lower% matrix in L.

% Check that the matrix is a square matrix[m,n]=size(X);if m~=n

error ('The matrix is not a square matrix!!');end

%Check if the matrix is symmetricif X ~= mat_transpose(X)

error ('The matrix is not symmetric!!');end

% The cholesky algortithm to find L

L= zeros(n,n);for j=1:n

L(j,j)=+sqrt(X(j,j));for i = j+1:n

L(i,j)=X(i,j)/L(j,j);for k = j+1:i

X(i,k)=X(i,k)-(L(i,j)*L(k,j));end

endend

end

------------------------------------------------------------------------------------------------------------------------------------

function x=solve_helper(L,b)%This function solves the system Ax=b by taking as an argument the matrix L%which is the lower matrix resulting from the cholesky decomposition of A%and the second argument is the vector b. The function returns the vector x

L_trans=mat_transpose(L);

[m,n]=size(L);y=zeros(n,1);

y(1)=b(1)/L(1,1);for i=2:n

tmp=0;for j=1:i-1

tmp=tmp+(L(i,j)*y(j));endy(i)=(b(i)-tmp)/L(i,i);

end

% Finding from y=L'x


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x=zeros(n,1);x(n)=y(n)/L_trans(n,n);for i=n-1:-1:1

tmp=0;for j=i+1:n

tmp=tmp+(L_trans(i,j)*(x(j)));endx(i)=(y(i)-tmp)/L_trans(i,i);

end

end-------------------------------------------------------------------------------------------------------------------------------------

function x = solveLinear(A,b)%this function takes a matrix and a vector and solves the system Ax=b using%a cholesky deomposition

L=cholDecomposition(A);x=solve_helper(L,b);

end--------------------------------------------------------------------------------------------------------------------------------------

function m= two_norm(V)% This function takes a column vector V and returns the 2-norm of the% vector which is just the euclidean norm.

n=size(V,1);s=0;for i=1:n

s=s+((V(i)^2));endm=sqrt(s);

end--------------------------------------------------------------------------------------------------------------------------------------

function m=infiniteNorm(V)%this functions computes the infinite norm of a column vector Vn=size(V,1);m=0;for i=1:n

y=abs(V(i));if y > m

m=y;end

end

end

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Numerical Assignment 2