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.
Change to polar coordinate in ODEs
control theorydifferential-geometryordinary differential equationspolar coordinatesreal-analysis
Related Solutions
For integrals you can generally ignore stuff like this. This is because integrals ignore what's happening on a set of measure zero, so as long as the part of the thing you're integrating over where $r = 0$ has measure zero, you can just ignore that part. For example, if you're trying to figure out something about a particle involving an integral, and the set of times that particle is at the origin has measure zero, you're fine.
We are given:
$$\dot x=-y(x^2+y^2), \dot y=x(x^2+y^2)$$
Recall, when we are doing polar coordinates, we have
$$x = r \cos \theta, y = r \sin \theta, x^2+y^2 = r^2$$
When we differentiate this, we have:
$$2 x x' + 2 y y' = 2 r r'$$
This gives us:
$$rr' = x(-y(x^2+y^2)) + y(x(x^2+y^2)) = 0 \rightarrow r' = 0$$
To find the angle, we take:
$$\dfrac{r \sin \theta}{r \cos \theta} = \tan \theta = \dfrac{y}{x}$$
Using the quotient rule gives us:
$$\theta' = \dfrac{x y' - y x'}{r^2} = \dfrac{x^2(x^2+y^2)+y^2(x^2+y^2)}{r^2}= \dfrac{r^4}{r^2} = r^2$$
Thus in polar coordinates, the original system becomes
- $r' = 0$
- $\theta' = r^2$
Can you now solve this system and draw the phase portrait?
The phase portrait should look like:
Best Answer
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.