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Message #11178
Re: Function design
On Wed, Dec 17, 2008 at 8:44 AM, Anders Logg <logg@xxxxxxxxx> wrote:
> On Wed, Dec 17, 2008 at 12:15:38AM +0000, Garth N. Wells wrote:
>> I'm running into some problems with the new Function design when trying
>> to update a solver. I'll try to sketch the issue as simply as possible.
>>
>> Function v;
>> Function u, p;
>>
>> // First form for U = (u, p)
>> FirstBilinearForm a_1(V1, V1);
>> FirstLinearForm L_1(V1);
>> LinearPDE pde_first(a_1, L_1);
>>
>> // Second form which depends on u
>> SecondBilinearForm a_2(V2, V2);
>> SecondLinearForm L_2(V2, u);
>> LinearPDE pde_second(a_2, L_2);
>>
>> Function U;
>> for(uint i = 0; i < 10; ++i)
>> {
>> pde_first.solve(U);
>>
>> // This step breaks down because FunctionSpaces don't match
>> u = U[0];
>> p = U[1];
>>
>> pde_second.solve(v);
>> }
>>
>> The problem is in assigning Functions since we now check the FunctionSpace.
>
> Which check is it that fails?
>
>> To get around this, I tried
>>
>> FunctionSpace V(mesh);
>> Function U(V);
>> Function u = U[0];
>> Function p = U[1];
>>
>> which throws an exception because U does not have a vector.
>
> What if we remove the check in operator[] and then in the constructor
> that gets a SubFunction we rely on v.v.vector() which (after Martin's
> fix from yesterday) will always return a vector with the dofs
> (possibly interpolated if there were no dofs before).
Depends...
Function u;
Vector & v = u.vector(); // error, no function space, ok
MyFunction u(V);
Vector & v = u.vector(); // calls interpolate, using MyFunction::eval
Function u(V);
Vector & v = u.vector(); // What happens here?
Also, I had another function space error yesterday.
Trying to interpolate a user function into a discrete
function space fails because the user function doesn't
have a function space, but that shouldn't be necessary.
(btw, all the "interpolate" variants makes it difficult to
discuss and remember exact signatures etc...)
Martin
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