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Why Comsol solution is mesh dependent?
Posted 5 avr. 2010, 17:18 UTC−4 4 Replies
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Hi,
I tried to solve the following simple 1D problem by COMSOL.
du/dx=-u
u(x=0)=1
u(x=1)=0
1D Comsol PDE gives a solution which is mesh dependent! If you refine the mesh, you get a different solution in 1D domain.
By the way. this is a simple 1st order ODE has general solution, u=A*exp(-u). It only has one arbitrary constant. How does Comsol deal with two point BCs in this particular case?
Thanks for your help.
Guodong
I tried to solve the following simple 1D problem by COMSOL.
du/dx=-u
u(x=0)=1
u(x=1)=0
1D Comsol PDE gives a solution which is mesh dependent! If you refine the mesh, you get a different solution in 1D domain.
By the way. this is a simple 1st order ODE has general solution, u=A*exp(-u). It only has one arbitrary constant. How does Comsol deal with two point BCs in this particular case?
Thanks for your help.
Guodong
4 Replies Last Post 9 avr. 2010, 19:05 UTC−4
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Posted:
1 decade ago
I haven't tried working in 1D yet, but aren't you seeing a basic property of the finite element method here?
In 2D graphing, for instance, if you try to fit a higher order curve solution with a finite number of lower order elements, then the solution will change with the number of elements. It's like plotting a complex curve using short straight line segments: the more segments, the better fit.
If you had an exponential order 1D element, it would solve your problem with any arbitrary number of elements.
I took a quick look at the elements available and I see linear, quadratic, .... up to quintic. No exponential there. I think you're stuck with mesh dependent solutions.
Maybe I'm missing something in your question, but I learned a little from thinking about it anyway.
-Jeff
In 2D graphing, for instance, if you try to fit a higher order curve solution with a finite number of lower order elements, then the solution will change with the number of elements. It's like plotting a complex curve using short straight line segments: the more segments, the better fit.
If you had an exponential order 1D element, it would solve your problem with any arbitrary number of elements.
I took a quick look at the elements available and I see linear, quadratic, .... up to quintic. No exponential there. I think you're stuck with mesh dependent solutions.
Maybe I'm missing something in your question, but I learned a little from thinking about it anyway.
-Jeff
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Posted:
1 decade ago
the problem you describe dont have a solution [ "not well posed" ] but comsol dont know that this is why he try and provide "garbage".. this is the old wisdom
Garbage in -> Garbage out :-)
JF
Garbage in -> Garbage out :-)
JF
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Posted:
1 decade ago
ditto -
your BC u(x=1)=0 cannot be satisfied with the DE. Another way to look at it is this- You have a first order DE but two BC's thus the problem is overspecified and not well posed.
your BC u(x=1)=0 cannot be satisfied with the DE. Another way to look at it is this- You have a first order DE but two BC's thus the problem is overspecified and not well posed.
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Posted:
1 decade ago
just wanted to point out that if the second boundary condition satisfied the solution...
i.e. analytic solution was Ae^-x and based on boundary 0, A was equal to unity...
then inserting u(x=0)=1 and u(x=1)=e^(-x)
in this manner your second boundary condition satifies the DE.
i.e. analytic solution was Ae^-x and based on boundary 0, A was equal to unity...
then inserting u(x=0)=1 and u(x=1)=e^(-x)
in this manner your second boundary condition satifies the DE.