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It is well-known that the following Hamiltonian system \begin{eqnarray} \left\{\begin{array}{rcl} \frac{dx}{dt}&=&y,\\ \frac{dy}{dt}&=&x(-1+x^2), \end{array}\right. \end{eqnarray} with $$ H(x,y)=\frac{x^2}2+\frac{y^2}{2}-\frac{x^4}{4}$$ has the solution $$ x=\tanh\left(\frac{t}{\sqrt2}\right),\qquad y=\frac1{\sqrt2}\text{sech}^2\left(\frac{t}{\sqrt2}\right) $$ such that $H(x,y)=\frac14$.

Now consider the following Hamiltonian system \begin{eqnarray} \left\{\begin{array}{rcl} \frac{dx}{dt}&=&y,\\ \frac{dy}{dt}&=&x(-1+x^4), \end{array}\right.\tag{1} \end{eqnarray} with $$ H(x,y)=\frac{x^2}2+\frac{y^2}{2}-\frac{x^6}{6}$$ and I want to find the explicit solution such that $H(x,y)=\frac{1}{3}$. I tried many ways to get the solution but failed. Any help is appreciated.

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Edited.

Surprisingly, there is an elementary solution, if I made no mistake in the following computation.

Your equation is equivalent to $$\left(\frac{dx}{dt}\right)^2=\frac{1}{3}(x^6-3x^2+2),$$ (I just plugged $y=dx/dt$ to your Hamiltonian, and used its value $H=1/3$.) This equation is separable, $$\frac{t}{\sqrt{3}}=\int\frac{dx}{\sqrt{x^6-3x^2+2}}=:I$$ and requires inversion of the integral.

To reduce it to a standard integral, change $x^2=1/(u+1), \; dx=-(1/2)(u+1)^{-3/2}du,$ and the integral becomes $$I=-\frac{1}{\sqrt{2}}\int\frac{du}{u\sqrt{2u+3}}.$$ The inverse function to this integral is elementary. Integrating and returning to the original variables, I obtained the general solution $$x(t)=\sqrt{\frac{1+\cosh(2t+c)}{2+\cosh(2t+c)}}.$$ Can you check this computation?

Of course, your value $1/3$ for the Hamiltonian is crucial here. With some other value you obtain a much more complicated integral which cannot be expressed in elementary functions. Where did this $1/3$ come from?

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    $\begingroup$ Thanks a lot. This is what I wanted. $\endgroup$ – xpaul Mar 28 at 21:29
  • $\begingroup$ Note it is much clearer than before. I really appreciate your help. $\endgroup$ – xpaul Mar 29 at 16:02
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    $\begingroup$ @xpaul: tell me where your $1/3$ comes from. $\endgroup$ – Alexandre Eremenko Mar 29 at 16:11
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    $\begingroup$ 1/3 is from fact that when $H(x,y)=\alpha$ passes through the separatrices $(\pm1,0)$, $\alpha=1/3$. $\endgroup$ – xpaul Mar 29 at 16:16
  • $\begingroup$ If I plug in your $x(t)$, I get $y=x'(t)=\frac{\sinh (c+2 t)}{\sqrt{\frac{\cosh (c+2 t)+1}{\cosh (c+2 t)+2}} (\cosh (c+2 t)+2)^2}$ but I find $\frac{x^2}2+\frac{y^2}{2}-\frac{x^6}6\neq\frac13$. $\endgroup$ – xpaul Mar 29 at 16:25

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