0
$\begingroup$

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.

$\endgroup$
1
  • 3
    $\begingroup$ 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. $\endgroup$ Feb 3, 2012 at 9:44

1 Answer 1

1
$\begingroup$

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,

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.