# Solve differential equation by using polar coordinates

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?

• In these types of problems the usual technique is to get one equation for $\dot{r}$ and another equation for $\dot{\theta}$. – okrzysik Jun 17 '16 at 10:22
• How can I get an equation with $r$ but without $\theta$? – M6002 Jun 17 '16 at 11:04
• 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 $\dot{\theta}$. – okrzysik Jun 17 '16 at 12:11 • 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 – M6002 Jun 18 '16 at 14:49
• 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. – JJacquelin Jun 18 '16 at 17:25