Quoting John Peterson <peterson@...>:
> On Sat, Sep 12, 2009 at 4:08 PM, Ted Kord <teddy.kord@...> wrote:
>> 2009/9/12 David Knezevic <dknez@...>
>>
>>> Roy Stogner wrote:
>>>
>>>>
>>>> On Sat, 12 Sep 2009, David Knezevic wrote:
>>>>
>>>> Ted Kord wrote:
>>>>>
>>>>
>>>> How do I apply a Neumann B.C at an interelement boundary?
>>>>>>
>>>>>
>>>>> The same way as a usual Neumann BC... the only trick is that you have to
>>>>> find which internal element to apply it to. One way to do this would be
>>>>> to set the subdomain_id of elements on one side of the interelement
>>>>> boundary to 1 and on the other side to 2, and then search for elements
>>>>> with subdomain_id = 1 that have a neighbor with subdomain_id = 2, and
>>>>> apply the Neumann BC to the appropriate side of those elements.
>>>>>
>>>>
>>>> The trouble with this is that you'll still have the entries in your
>>>> matrix from the shape functions which stretch between the element on
>>>> one side of the boundary and on the other. If you have a slit in your
>>>> domain on which you want to weakly impose boundary conditions, you
>>>> need to make it an actual topologically broken slit, and then it's
>>>> just another set of exterior boundaries.
>>>>
>>>
>>> I was thinking of imposing an internal flux between internal elements (e.g.
>>> as a type of forcing, but inside the domain rather than on the
>>> boundary). In
>>> that situation an "internal" Neumann condition does the job  the
>>> variational formulation takes care of everything for you...
>>>
>>>  Dave
>>>
>>
>> The problem I actually have is that there's a concentrated load at a single
>> point, say x = 16 (domain: 0 < x < 20) which is represented mathematically
>> as :
>>
>> 0.5  30 * diracdelta(x16)
>>
>> As far as I know, this, i.e., 30 will have to be applied as a Neumann B.C
>> at that point.
>
> I wouldn't think of a pointload as a boundary condition ... it's not
> a boundary condition.
>
> Assuming the delta function falls on a node in the mesh, you can just
> modify the load vector entry for the basis function associated to that
> row.
>
> 
> John
>
Yeah, this is equivalent to what I described above... multiplying the delta
function by a test function and integrating is equivalent to sampling the test
function, so it ends up looking like an "internal Neumann condition" which
imposes an interelement flux...
