Skip to main content
added 358 characters in body
Source Link
Loïc Teyssier
  • 5.4k
  • 3
  • 27
  • 40

You can transform (1) into the equation of the flow of a 2-dimensional real vector field with time $\omega$ (just differentiate it with respect to $\omega$). Then you plug the resulting system into any solver of differential equation. If $\omega$ is to take complex values then matters can get harder, but as I understand your problem the method I suggest should work flawlessly.

EDIT: to take care of the problem of multivaluedness you mention, look for a vector field whose components are the modulus $\rho$ and the argument $\theta$ of $k$ so that you obtain a relation of the form

$\dot{\rho}+i\rho\dot{\theta}=\frac{2B\omega \exp(-2i\theta)}{2\rho+A(a+1)\rho^a\exp(i(a-1)\theta))}$

where you don't see the problem anymore. Then compute the real- and imaginary-part of the right-hand side of the equation. Finding an initial condition at $\omega=0$ is no problem.

You can transform (1) into the equation of the flow of a 2-dimensional real vector field with time $\omega$ (just differentiate it with respect to $\omega$). Then you plug the resulting system into any solver of differential equation. If $\omega$ is to take complex values then matters can get harder, but as I understand your problem the method I suggest should work flawlessly.

You can transform (1) into the equation of the flow of a 2-dimensional real vector field with time $\omega$ (just differentiate it with respect to $\omega$). Then you plug the resulting system into any solver of differential equation. If $\omega$ is to take complex values then matters can get harder, but as I understand your problem the method I suggest should work flawlessly.

EDIT: to take care of the problem of multivaluedness you mention, look for a vector field whose components are the modulus $\rho$ and the argument $\theta$ of $k$ so that you obtain a relation of the form

$\dot{\rho}+i\rho\dot{\theta}=\frac{2B\omega \exp(-2i\theta)}{2\rho+A(a+1)\rho^a\exp(i(a-1)\theta))}$

where you don't see the problem anymore. Then compute the real- and imaginary-part of the right-hand side of the equation. Finding an initial condition at $\omega=0$ is no problem.

Source Link
Loïc Teyssier
  • 5.4k
  • 3
  • 27
  • 40

You can transform (1) into the equation of the flow of a 2-dimensional real vector field with time $\omega$ (just differentiate it with respect to $\omega$). Then you plug the resulting system into any solver of differential equation. If $\omega$ is to take complex values then matters can get harder, but as I understand your problem the method I suggest should work flawlessly.