How to make a matrix with one unknown variable, that can be in a for loop and can be used in ODE45

Hi, at the moment I have the following code. I think the problem is the matrix B is a symbolic funtion, and not a anonymous function, but I can't make an anonymous function to work within a loop.

syms t x u
l=50;         
v = 50; 
N = 5;
for n = 1:N
    for c = 1:N
        B(n,c) = sin((n*pi*v*t)/l) * sin((c*pi*v*t)/l);
    end
end

T = [t,t,t,t,t];
Td = [diff(t,t) diff(t,t) diff(t,t) diff(t,t) diff(t,t)];
tspan=[1,0.1,10];
xinit=zeros(1,2*N+2);
z=[T u Td diff(u,t)]';
[t,z]=ode45(B*z,tspan,xinit);

It's my first time using the ode45, so I'm not sure I'm using that one correct either.

So now I'm trying the mathworks-community, so I really hope you guys can help!

Answer 1

[Walter Roberson]

See the documentation for odeFunction . In particular, follow the flow of calls that is used in the first example.

[Marlene Dehn]

Thansk for the quick answer, but it does not seem like it's working for a NxN matrix. :( or at least I can't make it work.

[Walter Roberson]

What are the differential equations to be solved ?

Your T and Td are vectors of values, so your z is 12 total values, but your B matrix is 5 x 5, making it difficult to understand what you are wanting to multiply.

[Marlene Dehn]

Oh yeah I see, I made a mistake to make the question shorter. But the B matrix is actually used for a bigger matrix problem, so the final matrix that is used for the ODE45, is a 12x12.

contineuing...
K = D+B*k;

% Sinusvektor.
q = sin(i*pi*v*t/l)';

% 0-vektoren.
nulV = zeros(N,1);
% 0-matrice
nulM = zeros(N+1,N+1);
% Samling med 2. ligning
Ms = [M nulV;nulV',m];

Ks = [K -k*q; -k*q' k];
% A-matricen
A = [nulM Id; -Ms^(-1)*Ks nulM];

% T-vektoren. 
T = [T(t) T(t) T(t) T(t) T(t)];
Td = [DTt(t) DTt(t) DTt(t) DTt(t) DTt(t)];
tspan=[1,0.1,10];
xinit=zeros(1,2*N+2);
z=[T u(t) Td Dut(t)]';
Fs=[nulV',m]';
P=[nulV' ,0 , (Ms^(-1)*Fs)']';

f=A*z+P;

[t,z]=ode45(f,tspan,xinit);

And the differentiel equation I'm trying to solve is the attached picture. Which I know can be solved by the ODE45 with matrices, I just can't make a good enough script.

[Walter Roberson]

With that infinite summation, it is not clear to me that there is a finite representation.

[Marlene Dehn]

Yeah, the thing is we need to find a N, that is good enough to erased the infinite summation, that is why we need the for loop from 1:N, so the N can be a number from 1-infinite. But when we have the solution, we can figure out what the N needs to be through a convergence-analysis.

[Walter Roberson]

It is not at all obvious to me that you can justify a finite truncation. That would require that the Tn functions are decaying with increasing n, which is not an obvious conclusion. I can see that it might happen because of the increasing (n*pi/l)^4 multiplier in the first term could force the Tn smaller and smaller to match the right hand side, but the infinite summation makes things difficult.