So I have this differential equation: $$ \frac{\text{d}}{\text{d} \theta}u(\theta, E) = P(E)(1 - u^2)\sqrt{E + \gamma^2 \ln(1-u^2)} \tag{1} $$

Where $E > 0$, $P(E)$ is a complicated function I know (if absolutely necessary, one can assume that $P(E) = \sqrt{E}$), $\gamma$ is a positive constant, and $u(0, E) = 0$. Also, this DE is clearly only defined for $u\leq \sqrt{1-\exp(-E/\gamma^2)}$, so there's some $\theta_{\text{max}}$ beyond which $u$ can not be integrated. What I want to do is generalize $u$ to include the possibility that $E$ is complex. I've never had to do this before, but it seems like if I write $u(\theta, E) = \text{Re}(u) + i \text{Im}(u)=v(\theta, E) + i w(\theta, E) = v + i w$ along with $P(E) = P_{R} + iP_{I}$, I can simply chip away at the right side of $(1)$ until I can write it like $f_{R}(v, w) + i f_{I}(v, w)$ and then I would have to simultaneously solve $\text{d}_{\theta}v = f_{R}(v, w)$ and $\text{d}_{\theta}w = f_{I}(v, w)$.

In principle that's not a problem with shrewd use of polar coordinates to deal with the square root, but there's some sketchy things about that procedure. For one, I needed $u\leq \sqrt{1-\exp(-E/\gamma^2)}$ before, does that carry over to the complex case? Probably not, because the only reason that's there is to prevent imaginary numbers from appearing in the real case! I guess what I want to know is whether or not what I've described above is what I should be doing, and how to know under what conditions the results of that procedure are valid.


Complex numbers create additional problems because $\ln u$ will not be well defined wthout a branch cut; the same is true for square roots.

Once you pick a branch cut, the local existence theorem requires the right hand side and its derivative with respect to $u$ to be continuous everywhere, which it won't be when the square root is zero, so a solution is still not guaranteed.

  • $\begingroup$ I only need to know $u$ for $0\leq \theta < 2\pi$, does that help at all? $\endgroup$ – Mr. G Jan 5 '14 at 18:12
  • $\begingroup$ It's hard to say. Have you tried running a simulation? That would give you an idea of how fast it blows up. $\endgroup$ – Brian Rushton Jan 6 '14 at 14:03

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