Nonlinear PDE ${u_{tt}}^2u_{ttxx} = 1$ - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T09:39:06Zhttp://mathoverflow.net/feeds/question/87408http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/87408/nonlinear-pde-u-tt2u-ttxx-1Nonlinear PDE ${u_{tt}}^2u_{ttxx} = 1$nikita22012-02-03T08:40:05Z2012-02-07T14:33:33Z
<p>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.</p>
http://mathoverflow.net/questions/87408/nonlinear-pde-u-tt2u-ttxx-1/87798#87798Answer by Jon for Nonlinear PDE ${u_{tt}}^2u_{ttxx} = 1$Jon2012-02-07T14:33:33Z2012-02-07T14:33:33Z<p>A generic linear function in $t$ as</p>
<p>$$u(x,t)=f_1(x)t+f_2(x)$$</p>
<p>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,</p>
<p>$$u(x,t)=f_1(x)t+f_2(x)+\int_1^tdt'\int_1^{t'}dt''e^{-{\rm erf}^{-1}\left[-\frac{2}{\pi }\left(e^{C_1t''} (x+C_2t'')^2\right)\right]-\frac{1}{2} C_1t''}.$$</p>