Lack of Boundary Conditions to solve PDE (using pdepe function)

Question

Xuetao Xing 2017-6-17

0
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https://ww2.mathworks.cn/matlabcentral/answers/345123-lack-of-boundary-conditions-to-solve-pde-using-pdepe-function

回答： Bill Greene 2017-6-19

Hello,

I'm simulating the internal thermal and chemical dynamics of a fuel cell, using pdepe function (which is new to me). There are plenty of variables ( u1, u2, ..., un ) to be solved. However, I find that for each variable ui added, pdepe asks for boundary conditions (left and right) of ui, which I don't have sometimes.

For a simplified example, I don't know how to give boundary conditions of u2 for the following problem:

Its code is here (modified from MATLAB's example of pdepe):

function pde_example
m = 0;
x = linspace(0,1,20);
t = linspace(0,2,20);
sol = pdepe(m,@pdex1pde,@pdex1ic,@pdex1bc,x,t);
% Extract the first solution component as u1, u2.
u1 = sol(:,:,1);
u2 = sol(:,:,2);
% A surface plot of u1.
figure(1)
surf(x,t,u1) 
title('Solution of u1')
xlabel('Distance x')
ylabel('Time t')
% A solution profile of u1 at t=2.
figure(2)
plot(x,u1(end,:))
title('Solution of u1 at t = 2')
xlabel('Distance x')
ylabel('u1(x,2)')
% A surface plot of u2.
figure(3)
surf(x,t,u2) 
title('Solution of u2.')
xlabel('Distance x')
ylabel('Time t')
% A solution profile of u2 at t=2.
figure(4)
plot(x,u2(end,:))
title('Solution of u2 at t = 2')
xlabel('Distance x')
ylabel('u2(x,2)')
% ----------------------------*PDE part*----------------------------------
function [c,f,s] = pdex1pde(x,t,u,DuDx)
c = [pi^2;0];
f = [DuDx(1);0];
s = [0;u(2)-DuDx(1)];
% -----------------------------*IC part*---------------------------------
function u0 = pdex1ic(x)
u0 = [sin(pi*x);pi*cos(pi*x)];
% ----------------------------*BC part*----------------------------------
function [pl,ql,pr,qr] = pdex1bc(xl,ul,xr,ur,t)
pl = [ul(1); ul(2)-pi*exp(-t)];
ql = [0;0];
pr = [pi*exp(-t); ur(2)+pi*exp(-t)];
qr = [1;0];

It seems that this code can get the appropriate solution of u2 only if boundary conditions of u2 is given (i.e. ul(2)-pi*exp(-t) and ur(2)+pi*exp(-t)), which is obtained manually. If it's true, it will be a disaster for my simulation. Or am I wrong? How to get the appropriate solutions without manual pre-calculation? Thanks a lot!

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Xuetao Xing 2017-6-18

编辑：Xuetao Xing 2017-6-18

In my fuel cell case, it is like: (T-temperature, U-voltage, I-current) (DIDx is current density)

Torsten 2017-6-19

No way to solve this system using "pdepe".

Discretize the first and second equation in space and use ODE15S to solve for T.

The profile for I can be obtained directly in each time step:

dU/dx = 0 gives U=constant. To solve U = f2(dI/dx,T) for I, you will need one boundary condition for I (either at x=0 or x=1). Now if T is given, you can use forward or backward differencing (depending on where the boundary condition for I is given) to solve for I.

Best wishes

Torsten.

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Answer 1

Bill Greene 2017-6-19

0
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/345123-lack-of-boundary-conditions-to-solve-pde-using-pdepe-function#answer_271148

在 MATLAB Online 中打开

In your original code, set your boundary condition function to this:

pl = [ul(1); 0];
ql = [0;1];
pr = [pi*exp(-t); 0];
qr = [1;1];

Since f in your second PDE is zero, this boundary condition doesn't affect the solution but it does satisfy pdepe's requirement for a BC at both ends. This "trick" can often be used to get good solutions from pdepe when one or more of the PDE is only first order.