Simple PDE with side conditions

• To: mathgroup at smc.vnet.net
• Subject: [mg115365] Simple PDE with side conditions
• Date: Sun, 9 Jan 2011 02:17:43 -0500 (EST)
• References: <201101080838.DAA01905@smc.vnet.net>

```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,