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Re: Simple PDE with side conditions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg115486] Re: Simple PDE with side conditions
  • From: Oliver Ruebenkoenig <ruebenko at wolfram.com>
  • Date: Wed, 12 Jan 2011 04:07:59 -0500 (EST)

On Tue, 11 Jan 2011, schochet123 wrote:

> "NDSolve" was a misprint for "DSolve".  Sorry.
>
> What about with DSolve in version 8?
>
> Steve

Nope, not out of the box.

Oliver

>
>
> On Jan 11, 1:59 pm, Oliver Ruebenkoenig <ruebe... at wolfram.com> wrote:
>> On Tue, 11 Jan 2011, schochet123 wrote:
>>> Unfortunately, Mathematica cannot solve this (yet?).
>>
>>> Note that the functional equation is not needed since the solution is
>>> determined by the PDE plus the initial condition.
>>
>>> However, at least in version 7, NDSolve does not seem to solve
>>> first-order PDEs in three independent variables. Even the "trivial"
>>> problem
>>
>> You can solve this with NDSolve in version 7:
>>
>> L = 4;
>> sol = NDSolve[{
>>        D[u[t, x, y], t] + D[u[t, x, y], x] == 0
>>        , u[t, -L, y] == u[t, L, y]
>>        , u[t, x, -L] == u[t, x, L]
>>        , u[0, x, y] == Exp[-(x^2 + y^2)]
>>        }, u, {t, 0, 1}, {x, -L, L}, {y, -L, L}]
>>
>> Manipulate[
>>    Plot3D[u[t, x, y] /. First[sol], {x, -L, L}, {y, -L, L},
>>      PlotRange -> All], {t, 0, 1}]
>>
>>
>>
>>> DSolve[{D[u[t, x, y], t] + D[u[t, x, y], x] == 0,
>>>  u[0, x, y] == 0}, u, {t, x, y}]
>>
>>> does not get solved, although if you remove all the appearances of ",y" it is solved.
>>
>>> Can someone with version 8 check if there have been any improvements in that version?
>>
>
>


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