I am aware that there is a very similar question posted here that asks for the same thing, however I don't really understand what they have done nor does it seem applicable to my problem. Just like the other problem I have the differential equation: where T0=300K, TL = 400K and k=2.5.
And I am given that the second derivative with central difference is:
Further information given is that N = 250. I had previously determined that matrix A would be: [2 -1 0; -1 2 -1; 0 -1 2] when N=4. However, I am unsure if this is the matrix that I should use or if I should use a general matrix such as:
diagonal = [1/h^2; 2/h^2*ones(N-1,1); 1/h^2];
A = diag( diagonal)
for i = 2:N+1
A(i-1,i) = -1/h^2; A(i,i-1)=-1/h^2;
end
A(1,2)=0; A(N+1,N)=0;
If I use the general matrix and the following code:
L = 2;
N = 250;
T0 = 300;
Tend = 400;
h = L/N;
xj = 0:h:1;
Q=@(x)(300*exp(-(xj-L/2)).^2);
diagonal = [1/h^2; 2/h^2*ones(N-1,1); 1/h^2];
A = diag( diagonal);
for i = 2:N+1
A(i-1,i)=1/h^2; A(i,i-1)=1/h^2;
end
A(1,2)=0; A(N+1,N)=0;
b = [T0/h^2; Q; Tend/h^2];
T=A\b
plot(xj,T)
I get an error message at Q saying I have too many input arguments. I don't know if that is the correct way to find Q for the different x values and then how to code that be should contain all these values. Could anyone please tell me how to define Q and if the rest of the code (using the general matrix etc.) is correct. Thank you!
