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I wonder if it is possible to solve analytically the following equation

$$ \dot{\alpha}_t = -\frac{2}{m} \alpha^2_t + \frac{1}{2m} (\alpha_t - \alpha_t^*)^2 $$

Where $\alpha_t$ is a complex function, $\alpha_t^*$ is its complex conjugate and $\dot{\alpha}_t$ is the time derivative.

All the best!

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  • $\begingroup$ What is the significance of the subscript? Would it not have been preferable to use a simpler notation? $\endgroup$
    – Ben McKay
    Commented Nov 9, 2021 at 10:53
  • $\begingroup$ just means that $\alpha_t = \alpha(t)$ $\endgroup$
    – user38747
    Commented Nov 9, 2021 at 12:55

1 Answer 1

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Yes, this can be integrated explicitly. First, notice that, since $m\not=0$, we can write $\alpha(t) = 2m\bigl(x(t)+iy(t)\bigr)$, in which case, the given equation becomes $$ \dot x + i\,\dot y = -4(x + iy)^2 + (2iy)^2 = -4 x^2 - i\,(8xy), $$ so $\dot x = -4x^2$ and $\dot y = -8 xy$. Thus, by standard ODE techniques, $$ x(t) = \frac{x(0)}{(1+4x(0)\,t)}\quad\text{and}\quad y(t) = \frac{y(0)}{(1+4x(0)\,t)^2}\,. $$

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  • $\begingroup$ Thanks for your reply. Could you please explain me how do you solve the second ODE $\dot{y} = -8xy$. I think because x depends on t is necessary to solve using separation of variables something like $dy/y = -8 x(t) dt$ $\endgroup$
    – user38747
    Commented Nov 9, 2021 at 15:52
  • $\begingroup$ @user38747: Yes. Separation of variables is exactly what you should do, after you have solved $\dot x = -4x^2$. Alternatively, you can check that the equations imply that $y/x^2$ is constant, since its derivative is zero. $\endgroup$
    – Robert Bryant
    Commented Nov 9, 2021 at 16:19
  • $\begingroup$ Thanks a lot Robert! $\endgroup$
    – user38747
    Commented Nov 9, 2021 at 17:38

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