Boundary conditions with stiff problems

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Albert Jimenez
Albert Jimenez 2016-12-8
评论： Albert Jimenez 2016-12-21
Hello, I'm trying to solve a system of PDEs that depends on space (1 dimension) and time. I want to solve it using a second order discretization in space and then using ode23s to solve the system of ODEs (method of lines).
The problem is that I have a laplacian operator and I can just integrate in the interior nodes (let's say from 2 to n-1). The values for nodes 1 and n depend on their neighbours and should be updated at each time step.
How could I set these restrictions to solver?
Here is the scheme:
function dydt = fun(t,u)
for j = 2:n-1
impose right hand side function
end
%Now I want to impose the value in y(1) and y(end)
end

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回答（3 个）

Torsten
Torsten 2016-12-8
Boundary conditions don't depend on neighbour values, but are given independently.
What are the boundary conditions for your PDE ?
Best wishes
Torsten.
  1 个评论
Albert Jimenez
Albert Jimenez 2016-12-8
编辑：Albert Jimenez 2016-12-8
The boundary conditions are du/dx = G(u), for a given function G at x=0 and x=1 (the same for two points).
I have tried to discretize them too but the solution is not as expected (in fact, I obtain a warning: matrix is singular, close to singular or badly scaled).
I will send you the paper I'm trying to simulate if needed.
Thanks

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Torsten
Torsten 2016-12-9
Use the equations
(u(2)-u(1))/(x(2)-x(1)) = G(u(1))
(u(n)-u(n-1))/(x(n)-x(n-1)) = G(u(n))
to solve for u(1) (u at 0) and u(n) (u at 1).
Then use these values for the discretization in the inner grid points.
Best wishes
Torsten.
Albert Jimenez
Albert Jimenez 2016-12-9
The file attached contains the system I want to solve (1D). Using a second order discretization in space, the laplacian looks like
(u(j-1)+u(j+1)-2u(j))/(deltax^2)
The boundary conditions are
du/dx = A(u-B) at x=0 and 1
where A and B are constants.
If I discretize the first order derivative with a second order discretization, I obtain
(u(0)-u(2))/(2*deltax) = A(u(1)-B)
where u(0) is the value of u in a ficticious node. Then I can isolate the value of u(0) and replace in the scheme equation for the first node.
Is this what you were suggesting? This way, and assuming I don't have any mistake in the code, I get constant solutions (and I shouldn't).
Here is the code I am using: http://pastebin.com/ANe9ZgUz
Thanks
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Torsten
Torsten 2016-12-13
编辑：Torsten 2016-12-13
No. The boundary conditions give you u(0) depending on u(1),u(2) and u(n+1) depending on u(n-1),u(n).
Inserting these expressions in the discretized Laplacian in x(1) and x(n) give you the time derivatives for u(1) and u(n).
Best wishes
Torsten.
Albert Jimenez
Albert Jimenez 2016-12-21
Thanks! You are right and this is what I tried at the begining. However, I do not obtain the result as shown in the paper I am trying to reproduce. I could sent you the paper if you are interested.
Thanks again.

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