A system of PDEs which have one PDE with a spatial variable fixed term, u(x0, t)

4 次查看（过去 30 天）
Xinyu Gu
Xinyu Gu 2020-10-1
评论： Xinyu Gu 2020-10-2
Dear all,
I want to solve one kind of systems of PDEs (as mentioned in the title) using pdepe. Here is an example:
(1)
(2)
where u_1, u2 are functions of spatial variable x and time t; x_0 is a constant; f(u_2, dudx_2) is a function of u_2, dudx_2; s(u_1, u_2) is a function of u_1, u_2.
Bascially, u_1 is independent of spatial variable x so I tried to using discontinuity setup to solve a simple example:
(1)
(2)
I choose x_0 = 0. Specific ic and bc are applied so that the analytical solution is: . Below code is trying to solve eq(1) when when x==0 (x==x_0) and when x!=0, we let . Then, u_1 is independent of spatial variable x, and also satisfies eq(1). But this code doesn't work well.
x = -0.2:0.0001:0.2;
t = [0 0.001 0.005 0.01 0.05 0.1 0.5 1];
m = 0;
sol = pdepe(m,@pdex2pde,@pdex2ic,@pdex2bc,x,t);
u1 = sol(:,:,1);
u2 = sol(:,:,2);
surf(x,t,u1)
title('Numerical solution with nonuniform mesh')
xlabel('Distance x')
ylabel('Time t')
zlabel('Solution u')
function [c,f,s] = pdex2pde(x,t,u,dudx) % Equation to solve
if x == 0 %% Outpot : Error: Spatial discretization has failed. Discretization supports only parabolic and elliptic equations, with flux term involving spatial derivative.
%if (x < 0.0005 & x > -0.0005) %% Outpot : Code could be run but the result is wrong
%if (x < 0.2 & x > -0.2) %% Outpot : Error: using surf (line 71) Z must be a matrix, not a scalar or vector.
c = [1; 1];
f = [0; u(2)];
s = [u(2)-u(1); u(2)];
else
c = [0; 1];
f = [u(1); u(2)];
s = [0; u(2)];
end
end
%----------------------------------------------
function u0 = pdex2ic(x) %Initial conditions
u0 = [0; exp(-2*x)];
end
%----------------------------------------------
function [pl,ql,pr,qr] = pdex2bc(xl,ul,xr,ur,t) % Boundary conditions
pl = [ul(1)-ur(1); ul(2)-exp(-t-2*xl)];
ql = [0; 0];
pr = [ul(1)-ur(1); ur(2)-exp(-t-2*xr)];
qr = [0; 0];
end
%------------

请先登录，再回答此问题。

采纳的回答

Bill Greene
Bill Greene 2020-10-2
pdepe is not really designed to handle systems of coupled PDE and ODE. However, I have written a PDE solver for MATLAB that has similar syntax to pdepe but allows an arbitrary number of coupled ODE. If you want to try it, it can be downloaded here.
This script shows the solution of your example problem using this solver:
function matlabAnswers_10_1_2020
x = linspace(-0.2,0.2,11);
t=linspace(0,1,10);
m = 0;
x0=0;
xOde = x0; % location where ODE and PDE are coupled
[u,uode] = pde1dM(m, @pdeFunc,@icFunc,@bcFunc,x,t,...
@odeFunc, @odeIcFunc, xOde);
u1a=u1Anal(t);
figure; plot(t, uode(:,end), t, u1a, 'o');
legend('numerical', 'analytical');
xlabel('t'); ylabel('u1');
figure; plot(x, u(end,:), x, u2Anal(t(end), x), 'o'); grid on;
xlabel('x'); ylabel('u2');
title('Solution at Final Time');
legend('numerical', 'analytical');
end
function [c,f,s] = pdeFunc(x,t,u,DuDx,v,vdot)
c = 1;
f = u;
s = u;
end
function u0 = icFunc(x, t0)
u0=exp(-2*x);
end
function [pl,ql,pr,qr] = bcFunc(xl,ul,xr,ur,t,v,vdot)
pl = ul-exp(-t-2*xl);
ql = 0;
pr = ur-exp(-t-2*xr);
qr = 0;
end
function f=odeFunc(t,v,vdot,x,u,DuDx,~,~,~)
f=vdot-u+v;
end
function v0=odeIcFunc()
v0=0;
end
function u1=u1Anal(t)
u1=t.*exp(-t);
end
function u2=u2Anal(t,x)
nt=length(t);
nx=length(x);
u2=zeros(nt,nx);
for i=1:nt
u2(i,:) = exp(-t(i)-2*x);
end
end
  1 个评论
Xinyu Gu
Xinyu Gu 2020-10-2
Dear Bill,
Thank you so much for your reply, I really appreciate it!! The solver looks very helpful for my problem! I'll download and try it :-)
Many thanks,
Xinyu

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 PDE Solvers 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by