Consider the set of differential equations given by $\dot{z}=az^m\bar{z}^n$, with $z\in\mathbb{C}$. What's the phase portrait?

Question

I don't know how to attack this problem:

Consider the set of differential equations given by $\dot{z}=az^m\bar{z}^n$, with $z\in\mathbb{C}$, $a\neq 0$. What's the phase portrait on $\mathbb{R}^2$?

I've never done a phase portrait of a differential equation involving complex numbers before. I've tried to change the set into polar coordinates:

$$ire^{it}=a\cdot re^{itm} \cdot re^{-itn}$$

but I don't know if this will lead me to a correct answer.

Thanks.

Answer 1

You can re-parametrize each solution as $w(s)=z(t(s))$ where $t(s)$ is monotonically increasing without changing the phase portrait. The modified differential equation is $$ w'(s)= z'(t(s))\,t'(s)=a\,z(t(s))^m\,\bar z(t(s))^n\,t'(s)=a\,z(t(s))^{m-n}\,|z(t(s))|^{2n}\,t'(s) $$ where $t'(s)$ is any positive function. Take $t'(s)=|z(t(s))|^{-2n}=|w(s)|^{-2n}$ so that $$ w'(s)=a\,w(s)^{m-n} $$ which can be solved via separation and should also give insight into the local solution structure. For $n+1\ne m$ it should look like $$ w(s)=w(0)\Bigl(1+(n-m+1)\,a\,w(0)^{-n+m-1}\,s\Bigr)^{1/(n-m+1)}. $$