Re: Simple PDE with side conditions

• To: mathgroup at smc.vnet.net
• Subject: [mg115476] Re: Simple PDE with side conditions
• From: schochet123 <schochet123 at gmail.com>
• Date: Tue, 11 Jan 2011 19:24:52 -0500 (EST)
• References: <ighgn0\$ale\$1@smc.vnet.net>

```"NDSolve" was a misprint for "DSolve".  Sorry.

What about with DSolve in version 8?

Steve

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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