Explicit method for Allen-Cahn equation

The plot of the equation must start at x=-1 and end at x=1. but mu result did not show that?

clear all;
clc;

maxk = 1000; 
T = 0.10; 
n = 50; 
L = 2; % Length of the spatial domain [−1, 1]
Nx = 400; % Number of spatial grid points
dx = L / (Nx - 1); % Spatial step size
dt = T/maxk;
T = 1; % Final time
Nt = round(T / dt); % Number of time steps
a = 0.0001; 
r = a * dt / (dx * dx); % Diffusion factor for explicit scheme

% Initial condition
x = linspace(-1, 1, n+1);
u = zeros(n+1, maxk+1);
u(:,1) = x.^2 .* cos(pi * x);

% Implementation of the explicit method for Allen-Cahn equation
for t = 1:maxk
    % Internal points
    for i = 2:n
        u(i, t+1) = u(i, t) + r * (u(i-1, t) - 2 * u(i, t) + u(i+1, t)) ...
                    + dt * (5 * u(i, t)^3 - 5 * u(i, t));
    end
    
    % Periodic boundary conditions
    u(1, t+1) = u(end-1, t+1); % Periodic condition for first point
    u(end, t+1) = u(2, t+1);   % Periodic condition for last point
end

% Plot results
figure; % Create a new figure
xx = linspace(-1, 1, 100);
t_values = [0, 0.2, 0.4, 0.6, 0.8]; % Time values to plot
plot(x, u(:,1), '-', x, u(:,round(maxk*0.2)), '-', x, u(:,round(maxk*0.4)), '-', x, u(:,round(maxk*0.6)), '-', x, u(:,end), '-');
xlabel('x');
ylabel('u(x,t)');
grid on;
legend('t = 0', 't = 0.2', 't = 0.4', 't = 0.6', 't = 0.8');
hold off;

[Torsten]

Did you read somewhere that you need a boundary condition on the gradients of u ? In my opinion, u(-1,t) = u(1,t) suffices to fix a solution.

Answer 1

[Torsten]

xstart = -1.0;
xend = 1.0;
nx = 401;
x = linspace(xstart,xend,nx).';
dx = x(2)-x(1);
tstart = 0.0;
tend = 1.0;
nt = 10;
tspan = linspace(tstart,tend,nt);
u0 = x.^2.*cos(pi*x);
D = 1e-4;
[T,U] = ode15s(@(t,u)fun(t,u,D,nx,dx),tspan,u0);
plot(x,[U(1,:);U(end,:)])

function dudt = fun(t,u,D,nx,dx)
  ufull = [u(end-1);u;u(2)];
  dudt = D*(ufull(3:end)-2*ufull(2:end-1)+ufull(1:end-2))/dx^2-5*ufull(2:end-1).^3+5*ufull(2:end-1);
end

[Torsten]

It's not the ODE solver which gives a result different from yours. Your implementation of the periodic boundary condition is wrong.

[Erm]

exactly. my implementation of the periodic boundary condition is wrong. On your code why you wrote D = 1e-4

[Torsten]

On your code why you wrote D = 1e-4

I prefer the scientific notation for small numbers. Think about how many 0's you had to write if D was 1e-10 instead of 1e-4 :-)

[Erm]

got it. thank you.