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After simplifying the vortex equation, I get to this equation:

$$ -\alpha y \partial_y \omega = \alpha \omega + \nu \partial_{yy} \omega $$

where the $\alpha$ and $\nu$ are constant values and $\omega(y)$. I want to solve the equation for $\omega(y)$. How can I do it? I appreciate in advance for kind helps.

Regards,

Ehsan

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2 Answers 2

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$$ -\alpha y \dfrac{d\omega}{dy} = \alpha\omega + \nu \dfrac{d^2\omega}{dy^2} $$

this is the same as

$$ \nu \dfrac{d^2\omega}{dy^2} + \alpha \dfrac{d}{dy}(y\omega) = 0 $$ thus

$$ \nu \omega ' + \alpha y \omega + C = 0 $$

now if $C=0$ then the solution is separable but if not then we have to solve $$ \frac{d\omega}{dy} + \frac{\alpha}{\nu} y \omega= -C $$ using integrating factor $$ \omega \mathrm{e}^{\frac{\alpha}{\nu}\frac{y^2}{2}} = -\int_0^y C \mathrm{e}^{\frac{\alpha}{\nu}\frac{y'^2}{2}}dy' $$

or

$$ \omega(y) = -\left(\int_0^y C \mathrm{e}^{\frac{\alpha}{\nu}\frac{y'^2}{2}}dy'\right)\mathrm{e}^{-\frac{\alpha}{\nu}\frac{y^2}{2}} $$

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  • $\begingroup$ Very well done indeed, plus one!!! $\endgroup$ Nov 27, 2014 at 5:25
  • $\begingroup$ @robertlewis Thanks! Too kind. $\endgroup$
    – Chinny84
    Nov 27, 2014 at 7:18
  • $\begingroup$ Well, I was wondering if there wasn't some clever way of solving this, and you found it! $\endgroup$ Nov 27, 2014 at 7:32
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Maple finds solutions $$ \omega(y) = \left(c_1 + c_2\; \text{erfi}\left(\sqrt{-\alpha/(2\nu)} y\right)\right) e^{-\alpha y^2/(2 \nu)}$$

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