Solution of a nonlinear PDE I'm not able to give a general solution of this pde:
$$ {\partial \over \partial t} {\Phi(x,t)}={k^2\Phi(x,t)^2}{\partial^2 \over \partial x^2}{\Phi(x,t)}$$
Can someone help me?
 A: Here's my try. It's a standard trick that is used in physics, but I can't guarantee that the solution is the most general. Actually, I'm certain that it is not.
First of all, define $\tilde t = t k^2$ so that the equation does not have $k$ anymore. Now guess an ansatz of the form $\Phi=X(x)T(t)$, where $X$ is a function of $x$ only and $T$ is a function of $T$ only. Plugging that into the equation yields
$$X(x) T'(t)-T(t)^3 X(x)^2 X''(x)=0$$
which can be transformed, assuming all $T,X$ do not vanish to 
$$\frac{T'(t)}{T(t)^3}=X(x)''X(x)$$
Since the left hand side depends on $t$ only, and the right one on $x$, they both must be a constant, which we'll denote by $g$. So you need to solve the two ODE's:
$$\begin{align}
T'&=g T^3\\
X''&=\frac{1}{X}
\end{align}
$$
The solution of the first equation is 
$$T=\pm\frac{1}{\sqrt{-2(g t+C)}}$$
where $C$ is an integration constant. If you choose it negative enough then $T$ will be real. The solution of the $X$ equation is a bit more difficult, but Mathematica finds one:
$$X=A^{-2} e^{ \operatorname{InverseErf}\left[\pm i\sqrt{A \frac{2g}{\pi } (x+B)^2}\right]^2}$$
where $A,B$ are integration constants. Is this any good?
