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I'm trying to solve this equation $$ {u_{tt}}^2u_{ttxx} = 1. $$

First of all I did trick $u_{tt}(t,x) = y(t,x) $ and solved the ODE $y'' = \frac{1}{y^2}$. But the solution $y(t,x)$ too complicated.

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May I ask: To complicated for what purpose? – Dirk Jan 24 '12 at 19:52
@Dirk That is problem to find whole solution – Nikita Evseev Jan 25 '12 at 2:21
This question has been solved perfectly. Hope that the asker has been diving enough and accept the answer at an early date. – doraemonpaul Sep 10 '12 at 1:37
up vote 1 down vote accepted



Let $y=u_{tt}$ ,

Then $y_{xx}=\dfrac{1}{y^2}$


$\int y_xy_{xx}~dx=\int\dfrac{y_x}{y^2}dx$

$\int y_x~dy_x=\int\dfrac{1}{y^2}dy$






Let $v=\sqrt{y}$ ,

Then $y=v^2$



$\dfrac{v\sqrt{(C_1(t))^2v^2-2}}{(C_1(t))^2}+\dfrac{2}{(C_1(t))^3}\ln\left(C_1(t)v+\sqrt{(C_1(t))^2v^2-2}\right)=\pm x+C_2(t)$

$\dfrac{\sqrt{y}\sqrt{(C_1(t))^2y-2}}{(C_1(t))^2}+\dfrac{2}{(C_1(t))^3}\ln\left(C_1(t)\sqrt{y}+\sqrt{(C_1(t))^2y-2}\right)=\pm x+C_2(t)$

$\dfrac{\sqrt{u_{tt}}\sqrt{(C_1(t))^2u_{tt}-2}}{(C_1(t))^2}+\dfrac{2}{(C_1(t))^3}\ln\left(C_1(t)\sqrt{u_{tt}}+\sqrt{(C_1(t))^2u_{tt}-2}\right)=\pm x+C_2(t)$

This is just a big luck to reduce the nonlinear 4th order PDE to the nonlinear 2nd order ODE.

Make subject to $u_{tt}$ is too complicated, maybe you can try to use Lagrange_inversion_theorem or Lagrange_reversion_theorem

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