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

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

${u_{tt}}^2u_{ttxx}=1$

$u_{xxtt}=\dfrac{1}{{u_{tt}}^2}$

Let $y=u_{tt}$ ,

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

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

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

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

$\dfrac{(y_x)^2}{2}=-\dfrac{1}{y}+c_1(t)$

$(y_x)^2=-\dfrac{2}{y}+(C_1(t))^2$

$y_x=\pm\sqrt{\dfrac{(C_1(t))^2y-2}{y}}$

$\pm\dfrac{\sqrt{y}}{\sqrt{(C_1(t))^2y-2}}dy=dx$

$\pm\int\dfrac{\sqrt{y}}{\sqrt{(C_1(t))^2y-2}}dy=\int~dx$

Let $v=\sqrt{y}$ ,

Then $y=v^2$

$dy=2v~dv$

$\therefore\int\dfrac{2v^2}{\sqrt{(C_1(t))^2v^2-2}}dv=\int\pm~dx$

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

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

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

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 http://en.wikipedia.org/wiki/Lagrange_inversion_theorem or Lagrange_reversion_theorem http://en.wikipedia.org/wiki/Lagrange_reversion_theorem.

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