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Implementing numerical method for PDE

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schlang
schlang 2022 年 10 月 13 日
編集済み: Davide Masiello 2022 年 10 月 13 日
Hello
I am trying to solve the following PDE with intital and boundary conditions such that . I used the second order centered finite difference discrtization for and then want solve the ode system using ode15s
Here is my attempt. When I plot the solution obtained from ode15s and compare it to the exact solution they are different. I am not if I made a mistake somewhere. Help is really appreciated
clc,clear,close all
% parameters
t0 = 0;
T = 1.0;
tspan = [t0 T];
xl = 0;
xr = 1;
m = 20;
x = linspace(xl,xr,m + 1);
dx = 1/m;
Uexact = @(t,x) exp(1i*(x-t));
% initial conditions
U0 = Uexact(0,x)';
U0 = U0(2:end-1);
% solve
fn = @(t,U) ODE(t,U,m,dx);
opts = odeset('RelTol',1e-13, 'AbsTol',1e-15);
[t,U] = ode15s(fn, tspan, U0, opts);
%compare with exact solution
plot(x(2:end-1),U(end,:))
hold on
plot(x(2:end-1),Uexact(T,x(2:end-1)))
function dUdt = ODE(t,U,m,dx)
A = eye(m-1);
A = A * (-2);
A = A + diag(ones(m-2,1),1);
A = A + diag(ones(m-2,1),-1);
A = (1/dx^2) * A;
g = zeros(m-1,1);
g(1) = g(1) + (1/dx^2) * exp(1i*(-1*t));
g(end) = g(end) + (1/dx^2) * exp(1i*(1-t));
dUdt = (1i) * (A*U) + g;
end
Thanks
  2 件のコメント
Davide Masiello
Davide Masiello 2022 年 10 月 13 日
What are exactly the boundary conditions?
Davide Masiello
Davide Masiello 2022 年 10 月 13 日
Nevermind, I got it, see answer below.

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採用された回答

Davide Masiello
Davide Masiello 2022 年 10 月 13 日
編集済み: Davide Masiello 2022 年 10 月 13 日
clear,clc
tspan = [0 1];
N = 100;
x = linspace(0,1,N);
dx = 1/(N-1);
Uexact = @(t,x) exp(1i*(x-t));
U0 = Uexact(0,x);
M = eye(N);
M(1,1) = 0;
M(N,N) = 0;
opts = odeset('Mass',M,'MassSingular','yes');
[t,U] = ode15s(@(t,U)yourPDE(t,U,N,dx), tspan, U0, opts);
plot(x,real(U(end,:)),'k',x(1:4:end),real(Uexact(1,x(1:4:end))),'r.')
xlabel('x')
ylabel('U')
title('At final time')
legend('Numerical','Exact','Location','Best')
plot(x,real(U(1:3:end,:)),'k',x(1:3:end),real(Uexact(t(1:3:end),x(1:3:end))),'r.')
xlabel('x')
ylabel('U')
title('At several times')
function dUdt = yourPDE(t,U,N,dx)
dUdt(1,1) = U(1)-exp(-1i*t);
dUdt(2:N-1,1) = 1i*(U(1:end-2)-2*U(2:end-1)+U(3:end))/dx^2;
dUdt(N,1) = U(end)-exp(1i*(1-t));
end

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