Re: Simple PDE with side conditions

• To: mathgroup at smc.vnet.net
• Subject: [mg115431] Re: Simple PDE with side conditions
• From: schochet123 <schochet123 at gmail.com>
• Date: Tue, 11 Jan 2011 00:32:59 -0500 (EST)

```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

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?

Steve

On Sunday, January 9, 2011 9:17:44 AM UTC+2, Dave wrote:
> Hi everybody,
>
> I'm trying to have Mathematica 7 solve a simple partial differential equation
> with an initial condition and a function composition condition.
> So my input is:
>
> In[1]:= T[f_] := D[f, t]
>
> In[2]:= X[f_] := -y*D[f, x] + x*D[f, y]
>
> In[3]:= DSolve[{ T[f[t, x, y]] == X[f[t, x, y]],
>   f[0, x, y] == {x, y},
>   f[t, f[s, x, y][[1]], f[s, x, y][[2]]] == f[t + s, x, y]},
> f[t, x, y],
> {t, x, y}]
>
> However Mathematica returns with:
>
> DSolve::conarg:The arguments should be ordered consistently
> Out[3]= DSolve[{Derivative[1, 0, 0][f][t, x, y] ==
>    x*Derivative[0, 0, 1][f][t, x, y] -
>     y*Derivative[0, 1, 0][f][t, x, y], f[0, x, y] == {x, y},
>   f[t, s, x] == f[s + t, x, y]}, f[t, x, y], {t, x, y}]
>
> Now I know what the function f[t,x,y]  is and I can verify that
> it satisfies my conditions:
>
> In[4]:= f[t_, x_, y_] = {x *Cos[t] - y* Sin[t], x* Sin[t] + y* Cos[t]}
>
> In[5]:= { T[f[t, x, y]] == X[f[t, x, y]], f[0, x, y] == {x, y},
>   f[t, f[s, x, y][[1]], f[s, x, y][[2]]] ==
>    f[t + s, x, y]} // Simplify
>
> Out[5]= {True, True, True}
>
> The question is -- how can I have Mathematica solve this problem.
>
> Thanks,