What is the solution to this challenging nonlinear PDE?

[Ackbach]

Here is this week's POTW:

-----

Find a non-trivial solution to the following nonlinear partial differential equation:
$$u_t=u^3 u_{xxx}.$$
Check your solution by plugging it into the DE.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

No one answered this week's University POTW completely. An honorable mention goes to kiwi, who demonstrated a solution, but did not know how to get the solution initially. kiwi did mention this is Dym's equation, which is correct. After detailed looking at the problem, I found that I also was not able to solve it from scratch. So I will post my partial solution, along with kiwi's partial solution.

Spoiler

My partial solution:

Dym's equation is a rare bird: a nonlinear PDE for which you can separate variables. Suppose $u(x,t)=X(x) \, T(t)$, and I'll use primes for spatial derivatives and dots for temporal. Then Dym's equation yields
\begin{align*}
X\dot{T}&=X^3 T^4 X''' \\
\frac{\dot{T}}{T^4}&=X^2 \, X''' \\
\frac{\dot{T}}{T^4}&=k \\
X^2 \, X'''&=k.
\end{align*}
The $T$ equation yields:
\begin{align*}
T^{-4} \, \dot{T}&=k \\
\frac{T^{-3}}{-3}&=kt+C_1 \\
T^{-3}&=-3kt+C_1 \\
T&=(-3kt+C_1)^{-1/3}.
\end{align*}
Unfortunately, the $X$ equation doesn't seem tractable:
\begin{align*}
X^2 \, X'''&=k, \quad \text{or} \\
X^3 \, X'''&=k \, X.
\end{align*}
I'm stuck here. If anyone has a good idea on this ODE, I'd be grateful. In doing a bit more research, it appears that explicit solutions, in general, are difficult to obtain.

Here is kiwi's partial solution:

I don't know how to solve this PDE but it is the Dym equation which has the solution:

$$u(t,x)=[-3 \alpha (x + 4 \alpha ^2 t)]^{2/3}$$

So

Let $$y =[-3 \alpha (x + 4 \alpha ^2 t)]$$

$$u_t=\frac {-2}3 y^{-1/3}12 \alpha ^3=-8 y^{-1/3} \alpha ^3$$

$$u_x=\frac {-2}3 y^{-1/3}3 \alpha =-2 y^{-1/3} \alpha$$

$$u_{xx}=-2 \alpha \frac{-1}{3}y^{-4/3}(-3 \alpha)=-2 \alpha^2 y^{-4/3}$$

$$u_{xxx}=-2 \alpha^2 \frac{-4}{3}y^{-7/3}(-3 \alpha)=-8 \alpha^3 y^{-7/3}$$

So

$$u^3 u_{xxx}=-8 \alpha^3 y^{-7/3}y^2=-8 y^{-1/3}\alpha^3=u_t$$ as required.