matachi / mve255-matematisk-analys-i-flera-variabler

MATLAB Exercises from MVE255 at Chalmers, http://www.math.chalmers.se/Math/Grundutb/CTH/mve255/1213/

MVE255, Matematisk analys i flera variabler M, 2012/13

Daniel Jonsson, IT, Chalmers

Course site

Exercise 2

>> jacobi(@funk2, [1;2])

ans =

   -4.0000         0
   -2.0000   -1.0000

>> newton(@funk2, [1;1], 0.1)

ans =

    1.0000
    2.0000

>> newton(@funk2, [-2;-1], 10^-6)

ans =

   -1.0000
   -2.0000

Dugga 1

Question 1

>> jacobi(@funk3, [1;2;3])

ans =

    0.7539    2.2617    1.5078
    1.6765    0.8382    0.5588

Question 4

>> newton(@funk4, [-1;-1;-5], 10^-6)

ans =

   -0.4343
   -0.4343
    5.3028

>> newton(@funk4, [-1;-1;-1], 10^-6)

ans =

    1.0000
    1.0000
    1.0000

Question 5

Plot f(x,y) = (2x + y - 3)^2 + (x*y - 1)^2 in Matlab:

>> x = -5:.5:5;
>> y = x;
>> [X, Y] = meshgrid(x, y);
>> Z = (2.*X + Y - 3).^2 + (X.*Y - 1).^2;
>> surfc(X, Y, Z);

meshgrid documentation plot3 documentation

Other way to set x and y:

>> x = linspace(-10,10);
>> y = x;

Exercise 3

The PDF hand-in document: exercise3/documentation.pdf. This contains all found extreme points, eigenvalues and a bunch of images.
The Google Docs version can be accessed here. Anyone has access to view the document and write comments in it.

Part 1

extremeFinder finds an extreme point close to [0;0]:

>> [p, e] = extremeFinder(@ex3funk1, [0;0], 10e-5)
p =
   -0.4214
    0.1330
e =
    2.3243
    8.9020

p = Coordinate for the point, (-0.4214, 0.1330).
e = Eigenvalues for the Hessian matrix. Both are positive, meaning it's positive definite and it's a local minimum point.

Part 2

>> [p, e] = extremeFinder(@ex3funk2, [0;0;0], 10e-5)
p =
    1.0000
    2.0000
         0
e =
   -2.0000
    2.0000
    6.0000

The Hessian is indefinite, because there are both positive and negative eigenvalues. Therefore it's a saddle point.

Exercise 4

FEM1 1.4

n=101;
p=linspace(0,1,n);
t=[1:n-1; 2:n; ones(1,n-1)];
e=[1 n; 1 2];
[U, A, b] = MyPoissonSolver(p, t, e, @EqData14, @BdryData14);
U2 = p./2.*(1.-p); % real solution
plot(p,U,p,U2)

FEM1 1.5

[U, A, b] = MyPoissonSolver(p, t, e, @EqData15, @BdryData15);
U2 = -p.^2./2 + p + 1; % real solution
plot(p,U,p,U2)

Problem 3

[U, A, b] = MyPoissonSolver(p, t, e, @EqData3, @BdryData3);
plot(p,U)

Explanation: a(x) = The thermal conductivity coefficient. Half-way through the plate there is another material with 10 times higher thermal conductivity as the material to the left. So it will lead heat 10 times better and get 1/10th as hot.

Problem 4

Analogy between thermal conductivity and mechanic (In Firefox the link will go to the correct page and position):
a ↔ EA
j ↔ N
g ↔ P
f ↔ KxA

[U, A, b] = MyPoissonSolver(p, t, e, @EqData4, @BdryData4);
plot(p,U)

FEM1 Intro - Stången/The Rod

Stången: Introduktion till MyPoisson och Dugga 2

Example 1

L=1;    % the length of the rod
n=101;  % the number of points, n-1 is the number of intervals
p=linspace(0,L,n);
t=[1:n-1; 2:n; ones(1,n-1)];
e=[1 n; 1 2];
[U,A,b]=MyPoissonSolver(p,t,e,@EqDataStang11, @BdryDataStang11);
plot(p,U)
dL=U(end) % the extension of the rod

Example 2

Radius as a parameter:

L=1;    % the length of the rod
n=101;  % the number of points, n-1 is the number of intervals
p=linspace(0,L,n);
t=[1:n-1; 2:n; ones(1,n-1)];
e=[1 n; 1 2];

P=10;   % N force
r=0.01; % m radius
EqData=@(x,tag) EqDataStang12(x,tag,r);
BdryData=@(x,tag) BdryDataStang12(x,tag,P);
[U,A,b]=MyPoissonSolver(p,t,e,EqData,BdryData);
dL=U(end) % the extension of the rod
plot(p,U)

Plot a table with the two columns radius of rod and extension:

P=10;   % N
BdryData=@(x,tag)BdryDataStang12(x,tag,P);
rr=[]; dL=[];
for r=0.01:.001:0.03
    EqData=@(x,tag)EqDataStang12(x,tag,r);
    [U,A,b]=MyPoissonSolver(p,t,e,EqData,BdryData);
    dL=[dL;U(end)];     % save the extension in a column matrix
    rr=[rr;r];          % save the radius in a column matrix
end
[rr,dL]                 % display a table

If we want the rod to extend 0.001 units, we can use the function stangfunk and solve it using newton to get the most optimal radius of the rod:

R=newton(@stangfunk,.01,1e-6)

Dugga 2

Question 5

Determine the second degree taylor series to the function (x^2 - y - 1)^(9/4) around the coordinate (-1, -1).

Calculate the partial derivates with Mathematica:

Clear[f];
f[x_, y_] := ((x^2 - y - 1)^(9/4));
D[D[f[x, y], y], y] /. x -> -1 /. y -> -1
D[f[x, y], x] /. x -> -1 /. y -> -1

Answer: 1 - 9/2*(x+1) - 9/4*(y+1) + 63/8*(x+1)^2 + 45/8*(x+1)(y+1) + 45/32*(y+1)^2

Question 10

The rode is bolted in its left edge and free in its right.
Radius: r(x) = R(1 + h*x/L * (1 - x/L))
L = 1 m
E = 1e7 N/m^2
P = 10 N
R = 0.01 m
n = 101 node points

Value of h that gives U(L) = 1 mm?

h can be calculated using the file stangfunkD2Q10.m in the following way:

>> R=newton(@stangfunkD2Q10, 2, 1e-6)
R =
    5.8038

Answer: 5.8

Exercise 5

Start PDE Toolbox in Matlab with pdetool and open pde.m. The documentation/hand-in/description can be read in documentation.pdf or on Google Drive where it's also possible for anybody to leave comments.