Hello,
I want to apply periodic boundary condition on a 3D thin square plate with
one element in the thickness direction. I have already used the following
for meshing
std::vector< unsigned int > repetitions(dim, 100);
if (dim == 3)
repetitions[dim-1] = 1;
GridGenerator::subdivided_hype
Hi Oliver,
I think that this line indicates the problem:
.../dealII/8.5.0/examples/test02/tril/test02_tril.cc:129:3: error:
> ‘TrilionsWrappers’ does not name a type
> TrilionsWrappers::MPI::Vector system_rhs;
You've misspelt the namespace name "TrilinosWrappers" :-)
Best,
Jean-Paul
On Frida
Hello everyone,
presently, I am very new to deal.II and came across a problem which I spent
some time with but which I still do not fully understand.
Let me supply some background first:
Roughly, I wrote a program much in flavor of the examples in steps 4 to 6,
effectively solving an elliptic p
2017-05-12 9:32 GMT-04:00 Kartik Jujare :
> The graph shows the steady state solution given a point source at (0.6,0.6).
Why would you expect the _solution_ to be one? You set the _source_ to
one at (0.6, 0.6) What you are essentially saying is that the
divergence at this point should be one at thi
The graph shows the steady state solution given a point source at (0.6,0.6).
On Friday, May 12, 2017 at 3:21:02 PM UTC+2, Bruno Turcksin wrote:
>
> 2017-05-12 9:06 GMT-04:00 Kartik Jujare >:
>
> > Yes, as mentioned in the cited thread with the dirac delta function, my
> > \phi(x,y) is just a con
2017-05-12 9:06 GMT-04:00 Kartik Jujare :
> Yes, as mentioned in the cited thread with the dirac delta function, my
> \phi(x,y) is just a constant. So my question is, shouldn't the value on the
> graph attached show a value of 1 without multiplying this constant \phi
> value?
What I am looking at?
Hi Bruno,
Thank you for the reference.
>> If I understand correctly, your right-hand-side is similar to a Dirac
delta. So when you write your weak form, your right-hand-side is something
like \int \int \phi(x,y)
>> \delta(x_0,y_0) dx dy = \phi(x_0,y_0) So to build your right-hand-side,
you j
On 05/12/2017 02:22 AM, Jean-Paul Pelteret wrote:
By Vertex, I meant the geometric quantity only. As I have used first order
scalar basis functions i.e FE_Q<2> fe(1) to distribute the degrees of
freedom by dof_handler.distribute_dofs(fe);. I expect 'n_dofs =
n_vertices' and also
Kartik,
On Thursday, May 11, 2017 at 7:05:35 PM UTC-4, Kartik Jujare wrote:
>
> I added a create_point_source function after the assemble function in step
> 12. I have also changed the beta value to (1, 1) and g=0 everywhere on the
> boundary. Although the solution is similar to what I expected.
Dear Bhanu,
By Vertex, I meant the geometric quantity only. As I have used first order
> scalar basis functions i.e FE_Q<2> fe(1) to distribute the degrees of
> freedom by dof_handler.distribute_dofs(fe);. I expect 'n_dofs = n_vertices'
> and also the finite element nodes to be residing direct
Thanks Bruno.
By Vertex, I meant the geometric quantity only. As I have used first order
scalar basis functions i.e FE_Q<2> fe(1) to distribute the degrees of
freedom by dof_handler.distribute_dofs(fe);. I expect 'n_dofs = n_vertices'
and also the finite element nodes to be residing directly on
11 matches
Mail list logo
