# Nonlinear PDE ${u_{tt}}^2u_{ttxx} = 1$

I have been trying to solve this equation during fortnight $${u_{tt}}^2u_{ttxx} = 1.$$ But I still here. The only thing is change of variables $u_{tt}(t,x) = y(t,x)$ and solved the ODE $y'' = \frac{1}{y^2}$. But the solution $y(t,x)$ too complicated. I know that there are no common methods for solving such equation. But I wonder if somebody have any experience with this kind.

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However, I think you have solved it. Note that the second order ODE reduces to a first order $(y')^2=-2/y + C$ multiplying by $y'$ and integrating, that you can solve. – Pietro Majer Feb 3 '12 at 9:44

## 1 Answer

A generic linear function in $t$ as

$$u(x,t)=f_1(x)t+f_2(x)$$

does the job but, for the sake of completeness, I give here the result of Mathematica that involves ${\rm erf}^{-1}$, the inverse of the error function,

$$u(x,t)=f_1(x)t+f_2(x)+\int_1^tdt'\int_1^{t'}dt''e^{-{\rm erf}^{-1}\left[-\frac{2}{\pi }\left(e^{C_1t''} (x+C_2t'')^2\right)\right]-\frac{1}{2} C_1t''}.$$

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