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Consider a planar ODE (e.g., $\dot{x}=f_1(x,y),\dot{y}=f_2(x,y)$). According to page $68$ of ODE with Applications by Chicone, a change to polar coordinates (i.e., $x=r\cos(\theta),y=r\sin(\theta)$) in this system introduces a singularity on the line $\{(r,\theta): r=0 \}$. Furthermore, if the original ODE has a rest point at the origin, this singularity is removable. Why is this true? What is the intuition? How can we prove this result? Thank you.

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From the equations $x= r\cos\theta$ and $y = r\sin \theta$ we see that the derivatives of $r$ and $\theta$ are related to the derivatives of $x$ and $y$ by the equations \begin{eqnarray*} \dot x & = & \dot r \cos \theta - \dot \theta r\sin \theta\\ \dot y & = & \dot r \sin \theta + \dot \theta r\cos \theta. \end{eqnarray*} Thus, using the notation $g_i(r, \theta) = f_i(r\cos \theta, r\sin\theta)$ (i = 1, 2), the given system of ODEs reads \begin{eqnarray*} \dot r \cos \theta - \dot \theta r\sin \theta & = & g_1(r, \theta) \\ \dot r \sin \theta + \dot \theta r\cos\theta & = & g_2(r, \theta) . \end{eqnarray*} Solving this for $\dot r $ and $\dot \theta$ gives \begin{eqnarray*} \dot r & = & g_1(r, \theta)\cos \theta + g_2(r, \theta) \sin \theta\\ \dot \theta & = & \frac{g_2(r, \theta)}{r}\cos\theta - \frac{g_1(r, \theta)}{r} \sin \theta. \end{eqnarray*} Evidently, the right-hand side of the second equation has a problem at $r = 0$.

To see why the problem at $r = 0$ is harmless when $(x,y) = (0,0)$ is a rest point, consider for $i = 1, 2$ the Taylor expansion of $g_i$ about $(r,\theta) = (0,0)$. We have \begin{eqnarray*} g_i(r, \theta) & = & g_i(0,0) + \partial_rg_i(0,0)r + \partial_\theta g_i(0,0)\theta + {\rm HOT’s}\\ & = & f_i(0,0) + \left(\partial_x f_i(0,0) \cos\theta + \partial_y f_i(0,0) \sin\theta\right)r + \left(- \partial_x f_i(0,0) r\sin \theta + \partial_y f_i(0,0) r\cos\theta\right)\theta + {\rm HOT’s}\\ & = & f_i (0,0) + r\big[\partial_x f_i(0,0) \cos\theta + \partial_y f_i(0,0) \sin\theta + \left(- \partial_x f_i(0,0)\sin \theta + \partial_y f_i(0,0)\cos\theta\right)\theta\big] + {\rm HOT’s} \end{eqnarray*} From this formula, we see that if $f_i(0,0) = 0$ then $g_i$ is of the form \begin{equation*} g_i(r, \theta) = rh_i(r, \theta) \end{equation*} for some smooth function $h_i$. In particular, $\frac{g_i(r, \theta)}{r}$ causes no trouble.

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  • $\begingroup$ Thanks for the proof. Additionally, if the original vector field $f$ is in $C^r$, then the desingularized vector field is in $C^{r-1}$. $\endgroup$ – Arthur Jul 22 '18 at 16:15

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