Change of variables in polar coordinates

Question

Studying the stability at origin of the second order equation $$y''+y^2y'+y=0,$$ after a change of coordinates $u=y$ and $v=y'$ and a polar change $u=r\cos\theta$, $v=-r\sin \theta$. After some manipulations and a taylor expansion around $(0,0)$ i end up with the expansion $$\frac{dr}{d\theta}=-\frac{1}{8}\left(1-\cos(4\theta)\right)r^3+O(r^5).$$ We want to eliminate the dependece in $\theta$ for the lowest term so the autor says that after the change $r=\rho+a(\theta)\rho^3,$ where $a(\theta)$ is some $2\pi$-periodic function which we have to determinate. Now it says that $$\frac{d\rho}{d\theta}=-\frac{1}{8}\left(1-\cos(4\theta)+8a'(\theta)\right)\rho^3+O(\rho^5).$$ But i cannot see how to derivate this since using the chain rule for $\rho$ does not seems to work. Any help will be very appreciated

Answer 1

Letting $r=\rho+a(\theta)\rho^3$, your equation $$ \frac{dr}{d\theta}=-\frac{1}{8}\left(1-\cos(4\theta)\right)r^3+O(r^5) $$ becomes $$ \begin{split} \frac{d\rho}{d\theta}+a'(\theta)\rho^3+a(\theta)3\rho^2\frac{d\rho}{d\theta} &=-\frac{1}{8}\left(1-\cos(4\theta)\right)(\rho+a(\theta)\rho^3)^3+O(r^5)\\ &=-\frac{1}{8}\left(1-\cos(4\theta)\right)\rho^3+O(\rho^5) \end{split} $$ and so $$ (1+a(\theta)3\rho^2)\frac{d\rho}{d\theta}=-\frac{1}{8}\left(1-\cos(4\theta)+8a'(\theta)\right)\rho^3+O(\rho^5). $$ Therefore, $$ \begin{split} \frac{d\rho}{d\theta}&=-\frac{1}{8}\frac{\left(1-\cos(4\theta)+8a'(\theta)\right)\rho^3+O(\rho^5)}{1+a(\theta)3\rho^2}\\ &=-\frac{1}{8}(1-a(\theta)3\rho^2+\cdots)\left(1-\cos(4\theta)+8a'(\theta)\right)\rho^3+O(\rho^5)\\ &=-\frac{1}{8}\left(1-\cos(4\theta)+8a'(\theta)\right)\rho^3+O(\rho^5), \end{split} $$ using the expansion $(1-x)^{-1}=1+x+x^2+\cdots$.