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Find the general solution of the equation $$u_{ttxx}(x,t)=(u_{tt}(t,x))^2$$ Let set $v(x,t)=u_{tt}(x,t)$. Then $$v_{xx}(x,t)=(v(x,t))^2$$ What should I do next? Any help would be greatly appreciated.

Empiricist
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Roman
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    if i'm not mistaken the last differential equation can be solved in terms of weierstrass elliptic functions – tired Jun 25 '15 at 07:11
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    Your last equation is now an ODE.. Multiply both sides by $v_{x}$ and integrate (remember your implicit differentiation). – Matthew Cassell Jun 25 '15 at 07:24
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    After multiplying both sides by v{x} we obtain $\frac{d}{dx}\left[\frac{1}{2} v_x^2 \right] = \frac{d}{dx}\left[\frac{1}{3}v(x,t)^3\right]$ Is that right? Sadly, but i don't understand how to integrate this. – Roman Jun 25 '15 at 08:04
  • @Roman get's me something like, $\frac12 v_x^2(x,t)=\frac13 v^3(x,t)+\phi_0(t)$ . But yet again as you say, even this is hard to integrate.

    $$\int{\frac{\partial{\left(v(x,t)\right)}}{\sqrt{\frac23 v^3(x,t)+\phi_0(t)}}}=\pm x+\lambda(t)$$

     – Someone Jun 25 '15 at 08:52
  • @tired You not mistaken. My examiner said thats right. But he didn't explain how to do it. What should I read to understand this? Will someone show me an identic example? – Roman Jun 25 '15 at 16:32
  • Maybe a duplicate of: http://math.stackexchange.com/questions/1314388/cauchy-problem-for-nolinear-pde ? – Dmoreno Jun 26 '15 at 15:14

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