- I have set the value dt to be computed automatically from the value of dx (using stability criterion). In this way, you can change the value of dx at your discretion without having to manually compute the new dt.
- I have eliminated the inner loop in the PDE solving, because you can code spatial discretization making intelligent use of indexing (see code).
- I have approximated the first order spatial derivative with a first order accurate finite difference using an upwind scheme, which is notoriously better for convective terms.
Soving PDE using explicit method
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I need to solve a PDE using the explict method. I created a code by the plot didn't look god.So I could'nt move forward for the following question.
Could anyone check it and tell me what's wrong?
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采纳的回答
Davide Masiello
2022-2-7
编辑:Davide Masiello
2022-2-8
Hi, I have edited your code as shown below
%% Case 1. Numerical solution of the PDE Ut=-Ux+Uxx-U using the explicit FDM
% Written by Viridiana Salazar
% Edited by Davide Masiello
clear,clc
%% Parameters
B2 = 1/878;
%% Space steps
Lx = 1; % Max value of x
dx = 1/40; % Space step
N = Lx/dx+1; % Number of space steps
x = 0:dx:Lx;
%% Time steps
tf = 0.5; % Final time
dt = 0.9*0.5*dx^2; % Time step such that dt/dx^2<=0.5
t = 0:dt:tf;
M = length(t); % Number of time of steps
%% Initial and Boundary Conditions
U = zeros(N,M); % Create a Matrix of NxM
U(:,1) = 0; % Initial conditions
U(1,:) = 1; % Boundary conditions at the left side
%% Explicit Method
for j = 1:M-1 %Time Loop
U(2:end-1,j+1) = -(U(2:end-1,j)-U(1:end-2,j))*dt/(dx)+B2*(U(3:end,j)-2*U(2:end-1,j)+U(1:end-2,j))*dt/(dx^2)+U(2:end-1,j);
U(end,j) = U(end-1,j);
end
%% Analytical Solution
Ur = zeros(N,M);
Ur(:,1) = 0; % Initial conditions
Ur(1,:) = 1; % Boundary conditions at the left side
for j = 1:M-1 %Time Loop
Ur(:,j+1) = 0.5*erfc((x-t(j+1))/(2*B2*t(j+1)^0.5))+0.5*exp(x/B2).*erfc((x+t(j+1))/(2*B2*t(j+1)^0.5));
end
%% Error
Error = abs(Ur-U);
%% Plots
figure(1)
for idx = 1:50:length(t)
subplot(3,1,1)
plot(x,U(:,idx));
hold on
xlabel('x-values')
ylabel('U-values')
subplot(3,1,2)
plot(x,Ur(:,idx));
hold on
title('Analytical Solution')
xlabel('x-values')
ylabel('U-values')
hold on
subplot(3,1,3)
plot(x,Error(:,idx));
title('Error')
xlabel('x-values')
ylabel('Error')
hold on
end
EXPLANATION
First, you were mis-calculating B, because you wrote B = (1/878)^1/2, which first raises the content of the brackets to the power of 1 and then divides the result by 2. I avoided any mistake by just defining B^2 directly.
This was your only actual mistake, the rest of the code was fine.
However, I have changed a couple of things.
With these changes, you get the following plot
So numerical and analytical solutions do not match and there's a large error.
However, I think there might be some problem with how the analytical solution is formulated.
In fact, given the presence of the second-order spatial derivative (i.e., diffusion), I would expect a behaviour like the one yielded by the numerical integration, while the analytical solution seems to only advect the initial profile.
Could you double check the correctness of the analytical solution?
更多回答(1 个)
Hana Bachi
2022-2-8
2 个评论
Davide Masiello
2022-2-8
Regarding your question:
The stability criterion is
but of course you can't specify an inequaity on matlab, so I have just set to be slightly less than , e.g.
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