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In my study of dynamical systems I was recently met with this system of ODEs:

$ \dot{x}=\frac{\sinh{(y)}}{\cosh{(y)}+A\cos{(x)}} $

$ \dot{y}=\frac{A\sin{(x)}}{\cosh{(y)}+A\cos{(x)}} $

for a parameter A satisfying $ 0 < A < 1 $ Now clearly $ (0,0) $ and $ (2\pi,0) $ are fixed points and we may check there is a solution (heteroclinic) path (curve) from $ (0,0) $ to $ (2\pi,0) $. I need to find its representation not as a curve but via time dependency meaning (x(t),y(t)), I also know via simple integration that the solution sits on the equation curve $ \cosh{(y)} = -A\cos{(x)} + (A+1) $ but I cannot solve these equations for x,y components specifically as it seems impossible. Finally I need to determine if the following Melnikov integral has simple zeros: $ M(t_0)=\int_{-\infty}^{\infty} -\frac{A}{A+1}\sin{(x(t))}\sin{(\omega(t+t_0))} dt $, where x(t) is the x component of the solution curve mentioned earlier. I do not have $ x(t) $ explicitly let alone handle this integral. Maybe it can be handled via perturbation or numerical analysis (I only need to know if it has any simple zeros). I figured a good first step was to solve for x(t) via the equation $ \dot{x}=\frac{\sinh{(y)}}{\cosh{(y)}+A\cos{(x)}} = \frac{\sqrt{(A\cos{(x)}-A-1)^2-1}}{A+1} $ which is the equation on the solution curve.

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Croc2Alpha
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1 Answers1

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HINT :

Equation of the trajectory : $y(x)=\cosh^{-1}\left(-A\cos(x)+A+1\right)$ $$y(x)=\ln \left|\left(-A\cos(x)+A+1\right)\pm\sqrt{\left(-A\cos(x)+A+1\right)^2-1}\:\right|$$ Implicit equation of the motion : $$t=(A+1)\int \frac{dx}{\sqrt{\left(-A\cos(x)+A+1\right)^2-1}} $$ After some calculus with help of WolframAlpha : $$t=-(A+1)\sqrt{\frac{2}{A(1-\cos(x))}}\sin(\frac{x}{2})\tanh^{-1}\left(\sqrt{\frac{2}{-A\cos(x)+A+2}}\cos(\frac{x}{2}) \right)+C$$ $C=$constant. For example, if $x=\pi$ at $t=t_m$ then $C=t_m$

As far as I know there is no closed form for the inverse function $x(t)$.

IN ADDITION : $$ M(t_0)=\int_{-\infty}^{\infty} -\frac{A}{A+1}\sin{(x(t))}\sin{(\omega(t+t_0))} dt $$

$dt=(A+1)\frac{dx}{\sqrt{\left(-A\cos(x)+A+1\right)^2-1}} $

$$ M(t_0)=-A\int_0^{2\pi} \sin{(x)}\sin{(\omega(t(x)+t_0))} \frac{dx}{\sqrt{\left(-A\cos(x)+A+1\right)^2-1}} $$ where $t(x)=-(A+1)\sqrt{\frac{2}{A(1-\cos(x))}}\sin(\frac{x}{2})\tanh^{-1}\left(\sqrt{\frac{2}{-A\cos(x)+A+2}}\cos(\frac{x}{2}) \right)$

JJacquelin
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  • Thank you for this part, could you please help me in finding that improper integral as without a closed form for x I cannot really get at it – Croc2Alpha Jul 10 '16 at 08:49
  • I mean the Melnikov integral $ M(t_0)=\int_{-\infty}^{\infty} -\frac{A}{A+1}\sin{(x(t))}\sin{(\omega(t+t_0))} dt $ – Croc2Alpha Jul 10 '16 at 08:53
  • Why do you need the Melnikov integral in this case ? All is analytically known on parametric form : $t(x)$ and $y(x)$ with parameter $x$. – JJacquelin Jul 10 '16 at 09:09
  • I need the Melnikov integral in its own place it is important to me for predicting chaos this is part of my problem – Croc2Alpha Jul 10 '16 at 09:17
  • So I mean Melnikov's integral is important to me, the important part of the question – Croc2Alpha Jul 10 '16 at 09:18
  • OK. I am afraid that the integral be rather complicated. Just change of variable on the integral : $dx$ instead of $dt$ since $t(x)$ is explicit. See the integral in addition to my previous answer : horrible ! – JJacquelin Jul 10 '16 at 09:37
  • Right the integral is troy awful but I am only interested in finding if it has any simple zeros – Croc2Alpha Jul 10 '16 at 09:41
  • Well, good luck ! – JJacquelin Jul 10 '16 at 09:45