I am in a numerical analysis class, and I am working on a homework question. This comes Timothy Sauer's Numerical Analysis, and is in the second suggested activity section. I have been talking with my professor about this code, and it seems the error and the approximation are wrong, but neither one of us are able to figure out why. The following code is what I am using, and this is in MatLab. Anyone know enough about Euler Bernoulli beams, and Matlab who can help out?
function ebbeamerror %This is for part three
disp(['tabe of errors at x=L for each n'])
disp([' n ',' Aprox ',' Actual value',' Error'])
disp(['======================================================='])
format bank
for k = 1:11
n = 10*(2^k);
D = sparse(1:n,1:n,6*ones(1,n),n,n);
G = sparse(2:n,1:n-1,-4*ones(1,n-1),n,n);
F = sparse(3:n,1:n-2,ones(1,n-2),n,n);
S = G+D+F+G'+F';
S(1,1) = 16;
S(1,2) = -9;
S(1,3) = 8/3;
S(1,4) = -1/4;
S(2,1) = -4;
S(2,2) = 6;
S(2,3) = -4;
S(2,4) = 1;
S(n-1,n-3)=16/17;
S(n-1,n-2)=-60/17;
S(n-1,n-1)=72/17;
S(n-1,n)=-28/17;
S(n,n-3)=-12/17;
S(n,n-2)=96/17;
S(n,n-1)=-156/17;
S(n,n)=72/17;
E = 1.3e10;
w = 0.3;
d = 0.03;
I = w*d^3/12;
g = -9.81;
f = 480*d*g*w;
h = 2/10;
L = 2;
x = (h^4)*f/(E*I);
x1 = ones(n ,1);
b = x*x1;
size (S);
size(b);
pause
y = S\b;
x=2;
a = (f/(24*E*I))*(x^2)*(x^2-4*L*x+6*L^2);
disp([n y(n) a abs(y(n)-a)])
end
end
Related
I am currently working on solving the problem $-\alpha u'' + \beta u = f$ with Neumann conditions on the edge, with the finite element method in MATLAB.
I managed to set up a code that works for P1 and P2 Lagragne finite elements (i.e: linear and quadratic) and the results are good!
I am trying to implement the finite element method using the Hermite basis. This basis is defined by the following basis functions and derivatives:
syms x
phi(x) = [2*x^3-3*x^2+1,-2*x^3+3*x^2,x^3-2*x^2+x,x^3-x^2]
% Derivative
dphi = [6*x.^2-6*x,-6*x.^2+6*x,3*x^2-4*x+1,3*x^2-2*x]
The problem with the following code is that the solution vector u is not good. I know that there must be a problem in the S and F element matrix calculation loop, but I can't see where even though I've been trying to make changes for a week.
Can you give me your opinion? Hopefully someone can see my error.
Thanks a lot,
% -alpha*u'' + beta*u = f
% u'(a) = bd1, u'(b) = bd2;
a = 0;
b = 1;
f = #(x) (1);
alpha = 1;
beta = 1;
% Neuamnn boundary conditions
bn1 = 1;
bn2 = 0;
syms ue(x)
DE = -alpha*diff(ue,x,2) + beta*ue == f;
du = diff(ue,x);
BC = [du(a)==bn1, du(b)==bn2];
ue = dsolve(DE, BC);
figure
fplot(ue,[a,b], 'r', 'LineWidth',2)
N = 2;
nnod = N*(2+2); % Number of nodes
neq = nnod*1; % Number of equations, one degree of freedom per node
xnod = linspace(a,b,nnod);
nodes = [(1:3:nnod-3)', (2:3:nnod-2)', (3:3:nnod-1)', (4:3:nnod)'];
phi = #(xi)[2*xi.^3-3*xi.^2+1,2*xi.^3+3*xi.^2,xi.^3-2*xi.^2+xi,xi.^3-xi.^2];
dphi = #(xi)[6*xi.^2-6*xi,-6*xi.^2+6*xi,3*xi^2-4*xi+1,3*xi^2-2*xi];
% Here, just calculate the integral using gauss quadrature..
order = 5;
[gp, gw] = gauss(order, 0, 1);
S = zeros(neq,neq);
M = S;
F = zeros(neq,1);
for iel = 1:N
%disp(iel)
inod = nodes(iel,:);
xc = xnod(inod);
h = xc(end)-xc(1);
Se = zeros(4,4);
Me = Se;
fe = zeros(4,1);
for ig = 1:length(gp)
xi = gp(ig);
iw = gw(ig);
Se = Se + dphi(xi)'*dphi(xi)*1/h*1*iw;
Me = Me + phi(xi)'*phi(xi)*h*1*iw;
x = phi(xi)*xc';
fe = fe + phi(xi)' * f(x) * h * 1 * iw;
end
% Assembly
S(inod,inod) = S(inod, inod) + Se;
M(inod,inod) = M(inod, inod) + Me;
F(inod) = F(inod) + fe;
end
S = alpha*S + beta*M;
g = zeros(neq,1);
g(1) = -alpha*bn1;
g(end) = alpha*bn2;
alldofs = 1:neq;
u = zeros(neq,1); %Pre-allocate
F = F + g;
u(alldofs) = S(alldofs,alldofs)\F(alldofs)
Warning: Matrix is singular to working precision.
u = 8×1
NaN
NaN
NaN
NaN
NaN
NaN
NaN
NaN
figure
fplot(ue,[a,b], 'r', 'LineWidth',2)
hold on
plot(xnod, u, 'bo')
for iel = 1:N
inod = nodes(iel,:);
xc = xnod(inod);
U = u(inod);
xi = linspace(0,1,100)';
Ue = phi(xi)*U;
Xe = phi(xi)*xc';
plot(Xe,Ue,'b -')
end
% Gauss function for calculate the integral
function [x, w, A] = gauss(n, a, b)
n = 1:(n - 1);
beta = 1 ./ sqrt(4 - 1 ./ (n .* n));
J = diag(beta, 1) + diag(beta, -1);
[V, D] = eig(J);
x = diag(D);
A = b - a;
w = V(1, :) .* V(1, :);
w = w';
x=x';
end
You can find the same post under MATLAB site for syntax highlighting.
Thanks
I tried to read courses, search in different documentation and modify my code without success.
I am solving a coupled eigenvalue problem using Chebfun but I am getting some error. I am attaching my code here
%% Problem setup.
chi= 1;
m = 1495/1000;
L = 1;
th =-chi^(-1/m)*(-1 + chi^(1/m));
gm = 1500/1000;
kx = 1;
kz = 1;
kh = sqrt(kx^2 + kz^2);
g = 10;
Le = 0.001;
Omg= 25000;
VA = Le*2*Omg*L;
mu = 1;
B0 = VA*sqrt(mu);
omgC = 2*Omg*kz/kh;
omgM = VA*kx;
omge = 0.01*omgM;
kappa = 1e-5;
cp = 1;
dom = [0,1];
op = chebop(#(y,u,b,s) [-g*kh^2*(1+y*th)^(-m)*omgM^2.*s./(omgC^2*omge)...
+kh^2*u+(m*th^2.*u./(1+y*th).^2)-(m^2*th^2*u./(1+y*th).^2)...
+1j*kh^2*kx*m*th*(1+y*th)^(-m)*omgM^2.*u./(kh*kz*omgC*omge*(1+y*th))...
+kh^2*(1+y*th).^(-2*m)*omgM^4.*u./(omgC^2*omge^2)...
-2*m*th*diff(u,1)./(1+y*th)-1j*kh^3*(1+y*th)^(-m)*omgM^2.*diff(u,1)/(kh*kz*omgC*omge)...
+1j*kh*kx*(1+y*th)^(-m)*omgM^2.*diff(u,1)/(kz*omgC*omge)...
+1j*kh*kz*(1+y*th)^(-m)*omgM^2.*diff(u,1)/(kx*omgC*omge)-diff(u,2);...
omge*b+1j*B0*kx*u+omge*diff(b,2)/kh^2;-u./(1+y*th)+...
kappa*(-kh*2.*s+1j*(1+m)*th.*s./(1+y*th)+diff(s,2))], dom);
oplam = chebop(#(y,u,b,s) [-1j*kh^2*(1+y*th)^(-2*m)*omgM^4.*b/(B0*kx*omgC^2*omge^2)...
-kh^2*u/omge;b;s], dom);
op.bc = #(y,u,b,s) [u(0),b(0),s(0),u(1),b(1),s(1)];
%% Solve the eigenvalue problem.
[V, D] = eigs(op, oplam);
D = diag(D);
The error I am getting is
Dimensions of matrices being concatenated are not consistent.
I am not sure what mistake I am making. Should I convert the equations into first order and then input them? or is it somewhere in the syntax? Please help.
Thank you.
I'm trying to solve the following question :
maximize x^2-5x+y^2-3y
x+y <= 8
x<=2
x,y>= 0
By using Frank Wolf algorithm ( according to http://web.mit.edu/15.053/www/AMP-Chapter-13.pdf ).
But after running of the following program:
syms x y t;
f = x^2-5*x+y^2-3*y;
fdx = diff(f,1,x); % f'x
fdy = diff(f,1,y); % y'x
x0 = [0 0]; %initial point
A = [1 1;1 0]; %constrains matrix
b = [8;2];
lb = zeros(1,2);
eps = 0.00001;
i = 1;
X = [inf inf];
Z = zeros(2,200); %result for end points (x1,x2)
rr = zeros(1,200);
options = optimset('Display','none');
while( all(abs(X-x0)>[eps,eps]) && i < 200)
%f'x(x0)
c1 = subs(fdx,x,x0(1));
c1 = subs(c1,y,x0(2));
%f'y(x0)
c2 = subs(fdy,x,x0(1));
c2 = subs(c2,y,x0(2));
%optimization point of linear taylor function
ys = linprog((-[c1;c2]),A,b,[],[],lb,[],[],options);
%parametric representation of line
xt = (1-t)*x0(1)+t*ys(1,1);
yt = (1-t)*x0(2)+t*ys(2,1);
%f(x=xt,y=yt)
ft = subs(f,x,xt);
ft = subs(ft,y,yt);
%f't(x=xt,y=yt)
ftd = diff(ft,t,1);
%f't(x=xt,y=yt)=0 -> for max point
[t1] = solve(ftd); % (t==theta)
X = double(x0);%%%%%%%%%%%%%%%%%
% [ xt(t=t1) yt(t=t1)]
xnext(1) = subs(xt,t,t1) ;
xnext(2) = subs(yt,t,t1) ;
x0 = double(xnext);
Z(1,i) = x0(1);
Z(2,i) = x0(2);
i = i + 1;
end
x_point = Z(1,:);
y_point = Z(2,:);
% Draw result
scatter(x_point,y_point);
hold on;
% Print results
fprintf('The answer is:\n');
fprintf('x = %.3f \n',x0(1));
fprintf('y = %.3f \n',x0(2));
res = x0(1)^2 - 5*x0(1) + x0(2)^2 - 3*x0(2);
fprintf('f(x0) = %.3f\n',res);
I get the following result:
x = 3.020
y = 0.571
f(x0) = -7.367
And this no matter how many iterations I running this program (1,50 or 200).
Even if I choose a different starting point (For example, x0=(1,6) ), I get a negative answer to most.
I know that is an approximation, but the result should be positive (for x0 final, in this case).
My question is : what's wrong with my implementation?
Thanks in advance.
i changed a few things, it still doesn't look right but hopefully this is getting you in the right direction. It looks like the intial x0 points make a difference to how the algorithm converges.
Also make sure to check what i is after running the program, to determine if it ran to completion or exceeded the maximum iterations
lb = zeros(1,2);
ub = [2,8]; %if x+y<=8 and x,y>0 than both x,y < 8
eps = 0.00001;
i_max = 100;
i = 1;
X = [inf inf];
Z = zeros(2,i_max); %result for end points (x1,x2)
rr = zeros(1,200);
options = optimset('Display','none');
while( all(abs(X-x0)>[eps,eps]) && i < i_max)
%f'x(x0)
c1 = subs(fdx,x,x0(1));
c1 = subs(c1,y,x0(2));
%f'y(x0)
c2 = subs(fdy,x,x0(1));
c2 = subs(c2,y,x0(2));
%optimization point of linear taylor function
[ys, ~ , exit_flag] = linprog((-[c1;c2]),A,b,[],[],lb,ub,x0,options);
so here is the explanation of the changes
ub, uses our upper bound. After i added a ub, the result immediately changed
x0, start this iteration from the previous point
exit_flag this allows you to check exit_flag after execution (it always seems to be 1 indicating it solved the problem correctly)
I am having some difficulty in solving a simple 2D Poisson equation in Matlab using spectral methods in one direction on finite difference in the other.
I am obtaining a scaled version of the correct answer but can't work out why although I think it has something to do with the wavenumber.
Any help would be greatly appreciated, the code can be seen below.
N = 32;
x = ((0:N-1)/N)*2*pi;
y = ((0:N-1)/N)*2*pi;
dx = 2*pi/N;
k = fftshift(-N/2:N/2-1);
[X,Y] = meshgrid(x,y);
f = (-2)*cos(X).*sin(Y);
f_comparison = cos(X).*sin(Y);
checker = 10;
u = zeros(N,N);
u_new = zeros(N,N);
f_hat = fftn(f);
while checker > 1*10^(-7);
u_new_hat = fftn(u_new);
for aa = 1:N
for a = 1:N
q1 = a+1;
q2 = a-1;
if q2 == 0
q2 = N;
end
if q1 == N+1
q1 = 1;
end
denom = ((dx^2)*((k(aa))^2))+2;
u_new_hat(a,aa) = (1/denom)*((u_new_hat(q1,aa))+(u_new_hat(q2,aa))-((f_hat(a,aa))*(dx^2)));
end
end
u_new = real(ifftn(u_new_hat));
for aa = 1:N
for a = 1:N
u_checker(a,aa) = abs(u_new(a,aa)-u(a,aa));
end
end
compare = max(u_checker);
checker = max(compare);
for aa = 1:N
for a = 1:N
u(a,aa) = u_new(a,aa);
end
end
end
%calculate scaling discrepency
ddd = f_comparison./u;
I asked a question a few days before but I guess it was a little too complicated and I don't expect to get any answer.
My problem is that I need to use ANN for classification. I've read that much better cost function (or loss function as some books specify) is the cross-entropy, that is J(w) = -1/m * sum_i( yi*ln(hw(xi)) + (1-yi)*ln(1 - hw(xi)) ); i indicates the no. data from training matrix X. I tried to apply it in MATLAB but I find it really difficult. There are couple things I don't know:
should I sum each outputs given all training data (i = 1, ... N, where N is number of inputs for training)
is the gradient calculated correctly
is the numerical gradient (gradAapprox) calculated correctly.
I have following MATLAB codes. I realise I may ask for trivial thing but anyway I hope someone can give me some clues how to find the problem. I suspect the problem is to calculate gradients.
Many thanks.
Main script:
close all
clear all
L = #(x) (1 + exp(-x)).^(-1);
NN = #(x,theta) theta{2}*[ones(1,size(x,1));L(theta{1}*[ones(size(x,1),1) x]')];
% theta = [10 -30 -30];
x = [0 0; 0 1; 1 0; 1 1];
y = [0.9 0.1 0.1 0.1]';
theta0 = 2*rand(9,1)-1;
options = optimset('gradObj','on','Display','iter');
thetaVec = fminunc(#costFunction,theta0,options,x,y);
theta = cell(2,1);
theta{1} = reshape(thetaVec(1:6),[2 3]);
theta{2} = reshape(thetaVec(7:9),[1 3]);
NN(x,theta)'
Cost function:
function [jVal,gradVal,gradApprox] = costFunction(thetaVec,x,y)
persistent index;
% 1 x x
% 1 x x
% 1 x x
% x = 1 x x
% 1 x x
% 1 x x
% 1 x x
m = size(x,1);
if isempty(index) || index > size(x,1)
index = 1;
end
L = #(x) (1 + exp(-x)).^(-1);
NN = #(x,theta) theta{2}*[ones(1,size(x,1));L(theta{1}*[ones(size(x,1),1) x]')];
theta = cell(2,1);
theta{1} = reshape(thetaVec(1:6),[2 3]);
theta{2} = reshape(thetaVec(7:9),[1 3]);
Dew = cell(2,1);
DewApprox = cell(2,1);
% Forward propagation
a0 = x(index,:)';
z1 = theta{1}*[1;a0];
a1 = L(z1);
z2 = theta{2}*[1;a1];
a2 = L(z2);
% Back propagation
d2 = 1/m*(a2 - y(index))*L(z2)*(1-L(z2));
Dew{2} = [1;a1]*d2;
d1 = [1;a1].*(1 - [1;a1]).*theta{2}'*d2;
Dew{1} = [1;a0]*d1(2:end)';
% NNRes = NN(x,theta)';
% jVal = -1/m*sum(NNRes-y)*NNRes*(1-NNRes);
jVal = -1/m*(a2 - y(index))*a2*(1-a2);
gradVal = [Dew{1}(:);Dew{2}(:)];
gradApprox = CalcGradApprox(0.0001);
index = index + 1;
function output = CalcGradApprox(epsilon)
output = zeros(size(gradVal));
for n=1:length(thetaVec)
thetaVecMin = thetaVec;
thetaVecMax = thetaVec;
thetaVecMin(n) = thetaVec(n) - epsilon;
thetaVecMax(n) = thetaVec(n) + epsilon;
thetaMin = cell(2,1);
thetaMax = cell(2,1);
thetaMin{1} = reshape(thetaVecMin(1:6),[2 3]);
thetaMin{2} = reshape(thetaVecMin(7:9),[1 3]);
thetaMax{1} = reshape(thetaVecMax(1:6),[2 3]);
thetaMax{2} = reshape(thetaVecMax(7:9),[1 3]);
a2min = NN(x(index,:),thetaMin)';
a2max = NN(x(index,:),thetaMax)';
jValMin = -1/m*(a2min-y(index))*a2min*(1-a2min);
jValMax = -1/m*(a2max-y(index))*a2max*(1-a2max);
output(n) = (jValMax - jValMin)/2/epsilon;
end
end
end
EDIT:
Below I present the correct version of my costFunction for those who may be interested.
function [jVal,gradVal,gradApprox] = costFunction(thetaVec,x,y)
m = size(x,1);
L = #(x) (1 + exp(-x)).^(-1);
NN = #(x,theta) L(theta{2}*[ones(1,size(x,1));L(theta{1}*[ones(size(x,1),1) x]')]);
theta = cell(2,1);
theta{1} = reshape(thetaVec(1:6),[2 3]);
theta{2} = reshape(thetaVec(7:9),[1 3]);
Delta = cell(2,1);
Delta{1} = zeros(size(theta{1}));
Delta{2} = zeros(size(theta{2}));
D = cell(2,1);
D{1} = zeros(size(theta{1}));
D{2} = zeros(size(theta{2}));
jVal = 0;
for in = 1:size(x,1)
% Forward propagation
a1 = [1;x(in,:)']; % added bias to a0
z2 = theta{1}*a1;
a2 = [1;L(z2)]; % added bias to a1
z3 = theta{2}*a2;
a3 = L(z3);
% Back propagation
d3 = a3 - y(in);
d2 = theta{2}'*d3.*a2.*(1 - a2);
Delta{2} = Delta{2} + d3*a2';
Delta{1} = Delta{1} + d2(2:end)*a1';
jVal = jVal + sum( y(in)*log(a3) + (1-y(in))*log(1-a3) );
end
D{1} = 1/m*Delta{1};
D{2} = 1/m*Delta{2};
jVal = -1/m*jVal;
gradVal = [D{1}(:);D{2}(:)];
gradApprox = CalcGradApprox(x(in,:),0.0001);
% Nested function to calculate gradApprox
function output = CalcGradApprox(x,epsilon)
output = zeros(size(thetaVec));
for n=1:length(thetaVec)
thetaVecMin = thetaVec;
thetaVecMax = thetaVec;
thetaVecMin(n) = thetaVec(n) - epsilon;
thetaVecMax(n) = thetaVec(n) + epsilon;
thetaMin = cell(2,1);
thetaMax = cell(2,1);
thetaMin{1} = reshape(thetaVecMin(1:6),[2 3]);
thetaMin{2} = reshape(thetaVecMin(7:9),[1 3]);
thetaMax{1} = reshape(thetaVecMax(1:6),[2 3]);
thetaMax{2} = reshape(thetaVecMax(7:9),[1 3]);
a3min = NN(x,thetaMin)';
a3max = NN(x,thetaMax)';
jValMin = 0;
jValMax = 0;
for inn=1:size(x,1)
jValMin = jValMin + sum( y(inn)*log(a3min) + (1-y(inn))*log(1-a3min) );
jValMax = jValMax + sum( y(inn)*log(a3max) + (1-y(inn))*log(1-a3max) );
end
jValMin = 1/m*jValMin;
jValMax = 1/m*jValMax;
output(n) = (jValMax - jValMin)/2/epsilon;
end
end
end
I've only had a quick eyeball over your code. Here are some pointers.
Q1
should I sum each outputs given all training data (i = 1, ... N, where
N is number of inputs for training)
If you are talking in relation to the cost function, it is normal to sum and normalise by the number of training examples in order to provide comparison between.
I can't tell from the code whether you have a vectorised implementation which will change the answer. Note that the sum function will only sum up a single dimension at a time - meaning if you have a (M by N) array, sum will result in a 1 by N array.
The cost function should have a scalar output.
Q2
is the gradient calculated correctly
The gradient is not calculated correctly - specifically the deltas look wrong. Try following Andrew Ng's notes [PDF] they are very good.
Q3
is the numerical gradient (gradAapprox) calculated correctly.
This line looks a bit suspect. Does this make more sense?
output(n) = (jValMax - jValMin)/(2*epsilon);
EDIT: I actually can't make heads or tails of your gradient approximation. You should only use forward propagation and small tweaks in the parameters to compute the gradient. Good luck!