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.
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 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
I am not sure if what I have done so far is correct, and I need help with the iterative step as I don't understand what is going on in the algorithm. Here is my code. Help to finish this would be much appreciated. Thank you
function x_opt = out(A,b)
%UNTITLED2 Summary of this function goes here
% Detailed explanation goes here
b_vect = b';
m = size(A,1);
n = size(1,A);
set_B = 1:n;
set_B_Comp = n+1:m;
M = inv(A(set_B, :));
is_opt = 0;
x_temp = M*b_vect(set_B);
h = A*x_temp - b_vect;
y_vect = zeros(m, 1);
y_vect(set_B_Comp) = sign(h(set_B_Comp));
y_vect(set_B) = -inv(A(set_B, :))*(A(set_B_Comp, :))'*y_vect(set_B_Comp);
abs_y_B = abs(y_vect(set_B));
if all(abs_y_B <= 1)
x_opt = x_temp;
...
else
all_index_y_vect_more_than_1 = find(abs(y_vect) >= 1);
set_B_index_y_vect_more_than_1 = intersect(set_B, all_index_y_vect_more_than_1);
s = set_B_index_y_vect_more_than_1(1);
y_s = y(s)
t_vect = zeros(m, 1);
temp = inv(A(set_B,:));
t_vect(set_B_Comp) = -(sign(y_s))*(y(set_B_Comp)).*(A(set_B_Comp, :)*temp(:, s));
cur_min = h(set_B_Comp(1))/t_vect(set_B_Comp(1)) + 1;
cur_r = set_B_Comp(1);
for j = set_B_Comp
h_j = h(j);
t_j = t_vect(j);
temp1 = abs(h_j)/t_j;
if (temp1 < cur_min) && (temp1 > 0)
cur_min = temp1;
cur_r = j;
end
end
r = cur_r;
set_B_new = union(setdiff(set_B, s), r);
set_B_Comp_new = setdiff(1:m,set_B_new);
x_new = inv(A(set_B_new, :))*b_vect(set_B_new);
end
x_opt = x_temp;
end
I don't understand what's going on in your algorithm either. It's written without comments or explanations.
However, you can model your problem as a convex optimization problem. A formulation in Python using cvxpy is then quite simple and readable:
#!/usr/bin/env python3
import cvxpy as cp
import numpy as np
# Coefficient of regularization
alpha = 1e-4
# Dimensionality
N = 400
d = 20
# Synthetic data
A = np.random.randn(N, d)
b = np.random.randn(N)
# Define and solve the CVXPY problem.
x = cp.Variable(d)
objective = cp.sum_squares(A # x - b) + alpha * cp.norm1(x)
prob = cp.Problem(cp.Minimize(objective))
optval = prob.solve()
# Print result.
print("Optimal value ", optval)
print("The optimal x is")
print(x.value)
print("The norm of the residual is ", cp.norm(A # x - b, p=2).value)
and gives
Optimal value 328.41957961297607
The optimal x is
[-0.02041302 -0.16156503 0.07215877 0.00505087 0.01188415 -0.01029848
-0.0237066 0.0370556 0.02205413 0.00137185 0.04055319 -0.01157271
0.00369032 0.06349145 0.07494259 -0.04172275 0.04376864 0.02698337
-0.04696984 0.05245699]
The norm of the residual is 18.122348149231115
I'm currently learning Matlab's ODE-functions to solve simple vibration-problems.
For instance mx''+cx'+kx=F*sin(wt) can be solved using
function dx = fun(t,x)
m=0.02; % Mass - kg
k=25.0; % Stiffness - N/m
c=0.0125; % System damping - Ns/m
f=10; % Frequency
F=5;
dx= [x(2); (F*sin(2*pi*f*t)-c*x(2)-k*x(1))/m]
And then calling the ode45 function to get displacement and velocity
[t,x]=ode45(#fun,[0 10],[0.0;0.0])
My question, which I have not fully understood searching the web, is if it is possible to use ODE-function for a multiple degree of freedom system? For instance, if we have two masses, springs and dampers, which we excite att mass 1, we get the following equations:
m1*x1''+c1*x1'-c2*x2'+(k1+k2)*x1-k2*x2 = f1(t)
m2*x2''-c2*x1'+(c1+c2)*x2'-k2*x1+k2*x2 = 0
Here, the displacements x1 & x2 depend on each other, my question is how one should go about to solve these ODE's in Matlab?
There is no restriction that the inputs to the function solved by ODE45 be scalar. Just pass in an input matrix and expect out an output matrix. For example here is a function that solves the position of a 6 bar mechanism.
function zdot = cp_solve(t,z)
%% Constants
g = -9.81;
L1 = .2;
m0 = 0;
I0 = 0;
m1 = 1;
I1 = (1/3) * m1 * L1^2;
%% Inputs
q = z(1:6);
qdot = z(7:12);
%% Mass Matrix
M = zeros(6,6);
M(1,1) = m0;
M(2,2) = m0;
M(3,3) = I0;
M(4,4) = m1;
M(5,5) = m1;
M(6,6) = I1;
%% Constraint Matrix
Phiq = zeros(5,6);
Phiq(1,1) = 1;
Phiq(2,2) = 1;
Phiq(3,3) = 1;
Phiq(4,1) = 1;
Phiq(4,4) = -1;
Phiq(4,6) = (-L1/2)*sin(q(6));
Phiq(5,2) = 1;
Phiq(5,5) = -1;
Phiq(5,6) = (L1/2)*cos(q(6));
%% Generalized Forces
Q = zeros(6,1);
Q(5) = m1 * g;
%% Right Side Vector
rs = zeros(5,1);
rs(4) = (L1/2) * cos(q(6)) * qdot(6)^2;
rs(5) = (L1/2) * sin(q(6)) * qdot(6)^2;
%% Coefficient Matrix
C = [M Phiq'; Phiq zeros(5,5)];
R = [Q; rs];
%% Solution
Sol = C \ R;
zdot = [qdot; Sol(1:6)];
end
The inputs are the positions and velocities of the members. The outputs are the new positions and velocities.
You use it the same way you would any ODE45 problem. Setup the initial conditions, define a time and solve the problem.
%% Constants
L1 = .2;
C1 = L1/2;
theta1 = 30*pi/180;
theta_dot1 = 0;
tspan = 0:.001:2;
%% Initial Conditions
q = zeros(6,1);
q(6) = theta1;
q(4) = C1 * cos(q(6));
q(5) = C1 * sin(q(6));
qdot = zeros(6,1);
qdot(6) = theta_dot1;
z0 = [q; qdot];
%% Solve the problem
options = odeset('RelTol', 1.0e-9, 'AbsTol', 1.0e-6);
[Tout, Zout] = ode45(#cp_solve, tspan, z0, options);
In your case you have 2 equations and 2 unknowns. Set the problem up as a matrix problem and solve it simultaneously in your function. I would recommend the modal approach for your case.
I'm trying to solve a differential equation like below:
%% Parameters Initialization
C = 0.01;
gfi = 0.2308;
gso = 0.769;
Efi = 17.333;
Eso = 0;
km = 0.842;
v12m = 5.85;
kn = 4.4;
v12n = 9.667;
tauni = 3;
%% defining Functions
syms v n;
n_v_e = 1 / (1+exp((v12n-v)/kn));
m_v = 1 / (1+exp((v12m-v)/km));
gfi_v = gfi * m_v;
Dn = tauno*(n_v_e-n);
Ifi = gfi_v*(v-Efi);
Iso = gso*n*(v-Eso);
Dv = -1/C * (Ifi + Iso);
%% Solving ODE
v0 = -10;
n0 = 0.1;
te = 2000;
Dv = subs(Dv,'v','y(1)');
Dv = subs(Dv,'n','y(2)');
Dn = subs(Dn,'v','y(1)');
Dn = subs(Dn,'n','y(2)');
[t,V] = ode45(#(t,y)[Dv;Dn],[0 te],[v0;n0]);
the problem occurs right here at ode45 saying "Inputs must be floats, namely single or double."
I know the problem is due to symbols (n and v) but I don't know how to get rid of them!
I don't also want to write a seperate function and use it instead of [Dv;Dn].
Thank you for your help