MHB What is the solution to this challenging nonlinear PDE?

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    2015
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The discussion centers on finding a non-trivial solution to the nonlinear partial differential equation u_t = u^3 u_{xxx}. Participants note that no one has fully solved the problem, although kiwi provided a partial solution and identified the equation as Dym's equation. The original poster also struggled to solve it from scratch and plans to share their partial solution alongside kiwi's. The thread emphasizes the importance of checking solutions by substituting them back into the equation. Overall, the challenge remains unsolved, highlighting the complexity of the PDE.
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Here is this week's POTW:

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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.

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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!
 
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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.
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.
 

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