Nonlinear PDE ${u_{tt}}^2u_{ttxx} = 1$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T09:39:06Z http://mathoverflow.net/feeds/question/87408 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87408/nonlinear-pde-u-tt2u-ttxx-1 Nonlinear PDE ${u_{tt}}^2u_{ttxx} = 1$ nikita2 2012-02-03T08:40:05Z 2012-02-07T14:33:33Z <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#87798 Answer by Jon for Nonlinear PDE ${u_{tt}}^2u_{ttxx} = 1$ Jon 2012-02-07T14:33:33Z 2012-02-07T14:33:33Z <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>