For $\alpha, \beta>0$ the differential equation, I am trying to solve, is given by
$$\begin{pmatrix}\dot x_1\\\dot x_2\end{pmatrix}=\alpha\sin(x_1^2+x_2^2)\begin{pmatrix}x_2\\-x_1\end{pmatrix}+\beta\sin(2x_1^2+2x_2^2)\begin{pmatrix}x_1\\x_2\end{pmatrix},$$ where $x_1,x_2:\mathbb{R}\rightarrow \mathbb{R}$ are functions depending on the time $t\in\mathbb{R}$. I tried to tranform this equation by using
$$x_1=r\cos\theta\\x_2=r\sin\theta$$ with $\theta\in\mathbb{R}$ and $r>0$. First I computed $$\dot x_1=\frac{d(r\cos\theta)}{dt}=\dot r\cos\theta-r\dot\theta\sin\theta\\ \dot x_2=\frac{d(r\sin\theta)}{dt}=\dot r\sin\theta+r\dot\theta\cos\theta.$$ Putting this in the differential equation leads to $$\dot r =2\beta r \sin(r^2)\cos(r^2)r=\beta r \sin(2r^2)\\ \dot \theta =-\alpha \sin(r^2)\\ \frac{dr}{d\theta}=\frac{dr}{dt}\Big(\frac{d\theta}{dt}\Big)^{-1}=\frac{-\beta r \sin(2r^2)}{\alpha \sin(r^2)}=-\frac{2\beta}{\alpha}r\cos(r^2).$$
The second equation leads to $$\theta(t)=\theta(t_0)-\alpha\int\limits_{t_0}^t \sin(r(t)^2)dr(t)=\theta(t_0)-\alpha\int\limits_{t_0}^t \sin(r(t)^2)\dot r (t) dt \\=\theta(t_0)-\alpha\beta\int\limits_{t_0}^t \sin(r(t)^2)\sin(2r(t)^2)r(t) dt.$$ Now I need to find all constant and time-periodic solutions and also say something about the other solutions (not constant or time-periodic).

Constant solutions: By using $\dot r=0,\dot \theta=0,r(t)\equiv r$ and $\theta(t)\equiv \theta$ I get $$\beta r \sin(2r^2)=0\Leftrightarrow r=\pm \Big( \frac{\pi k}{2}\Big)^\frac{1}{2}\\-\alpha \sin(r^2)=0\Leftrightarrow r=\pm (\pi k)^\frac{1}{2}$$ for $k\in\mathbb{Z}$. Since both equations need to hold, if $\theta$ and $r$ are constant, and $r\in\mathbb{R}_{>0}$, $r$ needs to be $(\pi k)^\frac{1}{2}$ for $k\in\mathbb{N}$. But what can I say about $\theta$ (is there any restriction)? And how can i understand time-periodic solutions?

  • $\begingroup$ In these types of problems the usual technique is to get one equation for $\dot{r}$ and another equation for $\dot{\theta}$. $\endgroup$ – okrzysik Jun 17 '16 at 10:22
  • $\begingroup$ How can I get an equation with $r$ but without $\theta$? $\endgroup$ – M6002 Jun 17 '16 at 11:04
  • $\begingroup$ You might not be able to get isolated equations for $r,\theta$, like I suggested a good place to start would be getting equations for $\dot{r}, \dot{\theta}$. You might do this by taking the equations under your statement "putting this in the differential equation leads to" and multiplying the top by $\cos(\theta)$ and the bottom by $\sin(\theta)$ and then add the two. This will give you an expression for $\dot{r}$. Similarly you can find an equation for $\dot{\theta}$. $\endgroup$ – okrzysik Jun 17 '16 at 12:11

The trouble is that there is no standard closed forms for the integrals involved. So, the solution cannot be expressed on explicite form, but only with integrals in the implicit equations.

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  • $\begingroup$ Thanks for your help! I don't understand what you wanted to express by $\rightarrow t(r)\rightarrow r(t)$ and $\rightarrow \theta(r) \rightarrow \theta(t)=\theta(r(t))$. Could you please explain this?@JJacquelin $\endgroup$ – M6002 Jun 18 '16 at 14:49
  • $\begingroup$ I mean that, if there was convenient closed form $t(r)$ would be expressed on an explicit form instead of an integral and if there was a closed form for the inverse function, $r(t)$ would be explicit. Also, $\theta(r)$ would be expressed on an explicit form instead of an integral and then, combined with $r(t)$, the function $\theta(t)$ would be explicit. But, since no convenient closed form are available. we cannot go further. Nevertheless, the implicit equations eventually can be used for numerical calculus. $\endgroup$ – JJacquelin Jun 18 '16 at 17:25

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