0
$\begingroup$

I'm attempting to solve the differential equation $$\left(i\omega+\sigma\right)\dot{y}+\mu\left|y\right|^2y=0 $$ with the initial condition $y(0)=0$ I've tried to use separation of variables which means I have the following integral to solve $$t-t_0=-\frac{i\omega+\sigma}{\mu}\int_{y_0}^{y}\frac{dz}{\left|z\right|^2z}. $$ I guess the general solution to this is $$\left|y\right|^2=y_0^2\frac{i\omega+\sigma}{y_0^2\mu (t-t_0)+i\omega+\sigma}, $$ but I know somewhere in there something isn't quite right (maybe because uniqueness isn't guaranteed). Does someone know of a different way to find the solution(s) to this equation for y instead of $|y|^2$?

$\endgroup$
2
$\begingroup$

For simplicity, we denote $a+ib=\mu/(\sigma+i\omega)$. Assume $y$ is not the trivial solution $y=0$. Letting $y(t)=\rho(t) e^{i\theta(t)}$ be its polar representation, your equation $$ \dot y+(a+ib)|y|^2y=0 $$ becomes $$ e^{i\theta}(i\rho\dot \theta+\dot\rho+(a+ib)\rho^3)=0. $$ Equating real and imaginary parts to zero gives a system for $\rho(t)$ and $\theta(t)$: \begin{align} \dot\rho+a\rho^3&=0,\\ \dot\theta+b\rho^2&=0. \end{align} The first equation is decoupled, so the second equation is linear in $\theta$ once $\rho$ is found.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.