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Solution crushed by using second application mode

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Hi everybody,

I use COMSOL using two application modes.

1) First, I use the PDE mode / Weak Form Boundary using the eigensolver to solve an eigenvalues equation.
The solution is found and COMSOL gave me a set of eigenvalues (lambda_i) and eigenmodes (u_i) for i=1 to 10...
---> ok

2) Then, I use the second application mode PDE general Form using the time dependant solver to solve an other equation (time dependant) using the previous solutions (ie lambda_i and u_i). I would like solve this second problem for each solutions (lambda_i , u_i) found previously...

The problem is that COMSOL solve this second problem only for the current solution (I mean for only one solution lambda_2 , u_2), not for all previous solutions.

More, the different lambda are not available after this second step...

I tried to use solve / solver manager and Initial value to keep my first solutions but this does not work !

Please have you an idea to my problem ?

Thank you !

Mathias Clervoy


1 Reply Last Post 21.07.2010, 12:01 GMT-4

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Posted: 1 decade ago 21.07.2010, 12:01 GMT-4
Hi,

I have just found a solution for my previous problem.
We have use solver/solver manager and stored solution. This allowed to choose the current solution (u and lambda) to use for the second step...This second step works correctly and gives me the solution u2 for the previous (u, lambda) stored .

Now, the problem is that I need to use a third step for my calculations. This third step use the previous solution (u2) AND the old solution (u and lambda)...

After the second step lambda is crushed and become unknown for the third step (but u is always known).

How keep lambda available for the third step ?

Thank you for your help...


Best

Mathias

Hi, I have just found a solution for my previous problem. We have use solver/solver manager and stored solution. This allowed to choose the current solution (u and lambda) to use for the second step...This second step works correctly and gives me the solution u2 for the previous (u, lambda) stored . Now, the problem is that I need to use a third step for my calculations. This third step use the previous solution (u2) AND the old solution (u and lambda)... After the second step lambda is crushed and become unknown for the third step (but u is always known). How keep lambda available for the third step ? Thank you for your help... Best Mathias

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