issue in approximation of wave speed of propagation
0
im exploring Fisher KPP equation and diffrent numerical methods to study its behaviour specifically travelling waves and speed of propagation
here is the code in Matlab where iuse finite difference scheme:
C = 0.05; % Courant number
D = 1; % Diffusion coefficient
r = 1; % Reaction rate
Lx = 10; % Width of the domain
Ly = 10; % Height of the domain
nx = 51; % Number of grid points in x-direction
ny = 51; % Number of grid points in y-direction
nt = 5000; % Number of time steps
dx = Lx / (nx-1); % Spacing of grid in x-direction
dy = Ly / (ny-1); % Spacing of grid in y-direction
dt = C * dx; % Time step size
% Initialize concentration profile
un = zeros(ny, nx);
x = linspace(-Lx/2, Lx/2, nx);
y = linspace(-Ly/2, Ly/2, ny);
[X, Y] = meshgrid(x, y);
% Set initial condition
initial_condition = (X.^2 + 2*Y.^2 < 0.25);
un = 0.5 * initial_condition;
% Initialize matrix to save solutions over time
U = zeros(nt, ny, nx);
U(1,:,:) = un;
% Loop over time steps to solve the PDE
for n = 1:nt
uc = un;
% Update concentration profile using finite difference method
for i = 2:nx-1
for j = 2:ny-1
un(j, i) = uc(j, i) + …
dt * (uc(j-1, i) – 2 * uc(j, i) + uc(j+1, i))/dy^2 + …
dt * (uc(j, i-1) – 2 * uc(j, i) + uc(j, i+1))/dx^2 + …
dt * uc(j, i) * (1 – uc(j,i));
end
end
% Apply Dirichlet boundary conditions
un(1,:) = 0;
un(end,:) = 0;
un(:,1) = 0;
un(:,end) = 0;
% Store solution
U(n+1,:,:) = un;
end
% Initialize arrays to store pairs of (x, t)
x_values = [];
t_values = [];
% Loop over time steps
for ti = 1:nt
for tj = ti+1:nt % to avoid comparing the same time step twice
% Extract concentration profiles at times ti and tj
ui = squeeze(U(ti,:,:));
uj = squeeze(U(tj,:,:));
% Find pairs (xi, xj) where ui(xi, y) = uj(xj, y) for all y
[xi, xj] = find(all(abs(ui – uj) < 1e-2, 1)); % tolerance for numerical solution
% Store pairs (xi, ti) and (xj, tj)
for k = 1:length(xi)
x_values = [x_values; x(xi(k)), x(xj(k))];
t_values = [t_values; ti, tj];
end
end
end
% Plot pairs of (ti, tj) against (xi, xj)
figure;
plot(t_values, x_values, ‘o’);
xlabel(‘Temps’);
ylabel(‘Position x’);
title(‘Paires (ti, tj) où u(xi, y, ti) = u(xj, y, tj) pour tout y’);
grid on;
i’m expecting that the slope graph equal or large than 2 but with this code i dont get that ? what is the issue? the slope graph her is the approximate speed whixh u calculte by :c=(x_j-x_i)/(t_j-t_i) such that u(x_i,y,t_i)=u(x_j,y,t_j)can you please tell me the issue and if you have another suggestion please tell me0
im exploring Fisher KPP equation and diffrent numerical methods to study its behaviour specifically travelling waves and speed of propagation
here is the code in Matlab where iuse finite difference scheme:
C = 0.05; % Courant number
D = 1; % Diffusion coefficient
r = 1; % Reaction rate
Lx = 10; % Width of the domain
Ly = 10; % Height of the domain
nx = 51; % Number of grid points in x-direction
ny = 51; % Number of grid points in y-direction
nt = 5000; % Number of time steps
dx = Lx / (nx-1); % Spacing of grid in x-direction
dy = Ly / (ny-1); % Spacing of grid in y-direction
dt = C * dx; % Time step size
% Initialize concentration profile
un = zeros(ny, nx);
x = linspace(-Lx/2, Lx/2, nx);
y = linspace(-Ly/2, Ly/2, ny);
[X, Y] = meshgrid(x, y);
% Set initial condition
initial_condition = (X.^2 + 2*Y.^2 < 0.25);
un = 0.5 * initial_condition;
% Initialize matrix to save solutions over time
U = zeros(nt, ny, nx);
U(1,:,:) = un;
% Loop over time steps to solve the PDE
for n = 1:nt
uc = un;
% Update concentration profile using finite difference method
for i = 2:nx-1
for j = 2:ny-1
un(j, i) = uc(j, i) + …
dt * (uc(j-1, i) – 2 * uc(j, i) + uc(j+1, i))/dy^2 + …
dt * (uc(j, i-1) – 2 * uc(j, i) + uc(j, i+1))/dx^2 + …
dt * uc(j, i) * (1 – uc(j,i));
end
end
% Apply Dirichlet boundary conditions
un(1,:) = 0;
un(end,:) = 0;
un(:,1) = 0;
un(:,end) = 0;
% Store solution
U(n+1,:,:) = un;
end
% Initialize arrays to store pairs of (x, t)
x_values = [];
t_values = [];
% Loop over time steps
for ti = 1:nt
for tj = ti+1:nt % to avoid comparing the same time step twice
% Extract concentration profiles at times ti and tj
ui = squeeze(U(ti,:,:));
uj = squeeze(U(tj,:,:));
% Find pairs (xi, xj) where ui(xi, y) = uj(xj, y) for all y
[xi, xj] = find(all(abs(ui – uj) < 1e-2, 1)); % tolerance for numerical solution
% Store pairs (xi, ti) and (xj, tj)
for k = 1:length(xi)
x_values = [x_values; x(xi(k)), x(xj(k))];
t_values = [t_values; ti, tj];
end
end
end
% Plot pairs of (ti, tj) against (xi, xj)
figure;
plot(t_values, x_values, ‘o’);
xlabel(‘Temps’);
ylabel(‘Position x’);
title(‘Paires (ti, tj) où u(xi, y, ti) = u(xj, y, tj) pour tout y’);
grid on;
i’m expecting that the slope graph equal or large than 2 but with this code i dont get that ? what is the issue? the slope graph her is the approximate speed whixh u calculte by :c=(x_j-x_i)/(t_j-t_i) such that u(x_i,y,t_i)=u(x_j,y,t_j)can you please tell me the issue and if you have another suggestion please tell me 0
im exploring Fisher KPP equation and diffrent numerical methods to study its behaviour specifically travelling waves and speed of propagation
here is the code in Matlab where iuse finite difference scheme:
C = 0.05; % Courant number
D = 1; % Diffusion coefficient
r = 1; % Reaction rate
Lx = 10; % Width of the domain
Ly = 10; % Height of the domain
nx = 51; % Number of grid points in x-direction
ny = 51; % Number of grid points in y-direction
nt = 5000; % Number of time steps
dx = Lx / (nx-1); % Spacing of grid in x-direction
dy = Ly / (ny-1); % Spacing of grid in y-direction
dt = C * dx; % Time step size
% Initialize concentration profile
un = zeros(ny, nx);
x = linspace(-Lx/2, Lx/2, nx);
y = linspace(-Ly/2, Ly/2, ny);
[X, Y] = meshgrid(x, y);
% Set initial condition
initial_condition = (X.^2 + 2*Y.^2 < 0.25);
un = 0.5 * initial_condition;
% Initialize matrix to save solutions over time
U = zeros(nt, ny, nx);
U(1,:,:) = un;
% Loop over time steps to solve the PDE
for n = 1:nt
uc = un;
% Update concentration profile using finite difference method
for i = 2:nx-1
for j = 2:ny-1
un(j, i) = uc(j, i) + …
dt * (uc(j-1, i) – 2 * uc(j, i) + uc(j+1, i))/dy^2 + …
dt * (uc(j, i-1) – 2 * uc(j, i) + uc(j, i+1))/dx^2 + …
dt * uc(j, i) * (1 – uc(j,i));
end
end
% Apply Dirichlet boundary conditions
un(1,:) = 0;
un(end,:) = 0;
un(:,1) = 0;
un(:,end) = 0;
% Store solution
U(n+1,:,:) = un;
end
% Initialize arrays to store pairs of (x, t)
x_values = [];
t_values = [];
% Loop over time steps
for ti = 1:nt
for tj = ti+1:nt % to avoid comparing the same time step twice
% Extract concentration profiles at times ti and tj
ui = squeeze(U(ti,:,:));
uj = squeeze(U(tj,:,:));
% Find pairs (xi, xj) where ui(xi, y) = uj(xj, y) for all y
[xi, xj] = find(all(abs(ui – uj) < 1e-2, 1)); % tolerance for numerical solution
% Store pairs (xi, ti) and (xj, tj)
for k = 1:length(xi)
x_values = [x_values; x(xi(k)), x(xj(k))];
t_values = [t_values; ti, tj];
end
end
end
% Plot pairs of (ti, tj) against (xi, xj)
figure;
plot(t_values, x_values, ‘o’);
xlabel(‘Temps’);
ylabel(‘Position x’);
title(‘Paires (ti, tj) où u(xi, y, ti) = u(xj, y, tj) pour tout y’);
grid on;
i’m expecting that the slope graph equal or large than 2 but with this code i dont get that ? what is the issue? the slope graph her is the approximate speed whixh u calculte by :c=(x_j-x_i)/(t_j-t_i) such that u(x_i,y,t_i)=u(x_j,y,t_j)can you please tell me the issue and if you have another suggestion please tell me finite difference scheme, numerical analysis MATLAB Answers — New Questions