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How to solve method of lines on one-dimensional heat equation using Euler's method?

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Nannthini
Nannthini 2023-1-3
评论： Nannthini 2023-1-10
I have tried but I couldn't solve it fully. Here i attach my code and the question.
% Parameters
k = 0.5;
% Evaluate BC
N = 3;
u(0)= 0;
u(1:N)= U;
u(N+1)=0;
h = 1/(N+1) ; %step size
%Initial condition
for i=1:N
u(i)=sin(pi*i);
end
% Define du/dt
dudt = zeros (N+1,1);
for i=1:N
dudt(i)=k/h^2*(u(i-1)-2*u(i)+u(i+1));
end
  1 个评论
Torsten
Torsten 2023-1-3
编辑：Torsten 2023-1-3
u(0)= 0;
Array indices start with 1, not with 0. Thus u(0) will throw an error.
u(1:N)= U;
You did not yet define U.
u(i)=sin(pi*i);
i must be replaced by x(i) if 0=x(1)<x(2)<...<x(N+2)=1 is your grid in x-direction. And using N+2 grid points, your loop limits 1 and N are wrong.
dudt = zeros (N+1,1);
You use N+2 grid points.
for i=1:N
dudt(i)=k/h^2*(u(i-1)-2*u(i)+u(i+1));
end
Loop limits are wrong.
You must advance u in time by dudt, so something like u(n+1,i) = u(n,i) + dt * dudt(i) is missing where the "n" refers to the solution at time dt*n.

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Torsten
Torsten 2023-1-3
编辑：Torsten 2023-1-4
Ran in:
% Parameters
k = 0.5;
% Evaluate BC
dx = 0.01;
x = 0:dx:1;
dt = 0.0001;
t = 0:dt:0.5;
h = 1/(length(x)-1) ; %step size
u = zeros(length(t),length(x));
%Initial condition
for i=1:length(x)
u(1,i)=sin(pi*x(i));
end
for j = 1:length(t)-1
for i=2:length(x)-1
u(j+1,i) = u(j,i) + dt* k/dx^2*(u(j,i-1)-2*u(j,i)+u(j,i+1));
end
end
figure(1)
plot(x,[u(1,:);u(500,:);u(1000,:);u(2000,:);u(5000,:)])
figure(2)
surf(x,t,u,'Edgecolor','none')
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