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I am trying to solve the 1D heat equation using the Crank-Nicholson method. I have managed to code up the method but my solution blows up. I'm using Neumann conditions at the ends and it was advised that I take a reduced matrix and use that to find the interior points and then afterwards. Would this work? My code is blowing up and it should work but for some reason it isn't and I can't see the reason why. Any suggestions? Code below:

%This program is meant to test out the Crank-Nicolson scheme using a simple

%nonlinear diffusion scheme.

n=5000;m=30;

t=linspace(0,20,n);% set time and distance scales

x=linspace(0,1,m);

dx=x(2)-x(1);dt=t(2)-t(1); %Increments

s=dt/(2*dx^2);%Useful for the solution.

u=zeros(n,m); %set up solution matrix

p=s*ones(1,m-1); q=-(1+2*s)*ones(1,m);

Z=diag(p,1)+diag(p,-1)+diag(q);

A=Z(2:m-1,2:m-1);

%Add in intial condition:

u(1,:)=exp(-5*(x-0.5).^2);

v=zeros(m-2,1);

%Solve the system

for i=2:n-1

%Construct the RHS for the solution

for j=2:m-1

v(j-1)=s*u(i-1,j+1)-(2*s+1)*u(i-1,j)-s*u(i-1,j-1);

end

%Solve for the new time step

w=A\v;

u(i,2:m-1)=w;

u(i,1)=u(i,2);

u(i,end)=u(i,end-1);

end

Torsten
on 3 Jul 2018

%This program is meant to test out the Crank-Nicolson scheme using a simple

%nonlinear diffusion scheme.

n=5000;m=150;

t=linspace(0,10,n);% set time and distance scales

x=linspace(0,1,m);

dx=x(2)-x(1);dt=t(2)-t(1); %Increments

s=dt/(2*dx^2);%Useful for the solution.

u=zeros(n,m); %set up solution matrix

A=zeros(m,m);

A(1,1) = 1;

A(1,2) = -1;

for j=2:m-1

A(j,j-1) = -s;

A(j,j) = (1+2*s);

A(j,j+1) = -s;

end

A(m,m-1)= -1;

A(m,m) = 1;

%Add in intial condition:

u(1,:)=exp(-5*(x-0.5).^2);

v=zeros(m,1);

%Solve the system

for i=2:n-1

%Construct the RHS for the solution

for j=2:m-1

v(j)=s*u(i-1,j+1)+(1-2*s)*u(i-1,j)+s*u(i-1,j-1);

end

%Solve for the new time step

w=A\v;

u(i,:)=w;

end

end

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