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The quantity I am interested in is not the rotation rate $\Omega$, but I will use this quantity nonetheless to make the problem clearer. I am interested in a spherically-symmetric system and in the boundary condition of the rotation rate $\Omega=u/r\sin\theta$. Now, at $\theta=0$, we have an undetermined form $0/0$ and we must use de L'Hopital rule: $$ \Omega (r,0)=\text{lim}_{\theta\rightarrow 0}\frac{du/d\theta}{r\cos\theta}\neq 0 $$

which is not zero as expected. Using central differences (I assume this is still valid):

$$ \frac{du}{d\theta}|_{\theta=0}=\frac{u_1-u-1}{2\Delta\theta}, $$

where the index for radius was ignored (the minus 1 should be as an index but it somehow doesn't work within the fraction). Now, $u_{-1}=-u_1$ and:

$$ \frac{du}{d\theta}|_{\theta=0}=\frac{u_1}{\Delta\theta}=\frac{\Omega_1 r\sin\theta_1}{\Delta\theta}\simeq\Omega_1 r\quad\quad\text{[for $\theta_1=\Delta\theta$ small]}. $$

Finally: $$ \Omega_0=\Omega_1. $$

Now, if I decide to discretize the equations regularly in $\mu=\cos\theta$, so that:

$$ \sin\theta=\sqrt{1-\mu^2}\quad\quad\frac{1}{d\theta}=-\frac{\sin\theta}{d\cos\theta}=-\frac{\sqrt{1-\mu^2}}{d\mu}, $$

then:

$$ \Omega(r,1)=\text{lim}_{\mu\rightarrow 1}\frac{du/d\mu}{rd\sqrt{1-\mu^2}/d\mu}, $$

and

$$ \Omega(r,1)=\text{lim}_{\mu\rightarrow 1}-\frac{\sqrt{1-\mu^2}}{r\mu}\frac{du}{d\mu}=0 $$

What am I missing here? This should definitely not be zero at the pole (or anywhere).

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  • $\begingroup$ What is $u$ in your example? $\endgroup$ Feb 25, 2023 at 20:03
  • $\begingroup$ It's the velocity in the $\phi$-direction $u=r\sin\theta\Omega$ $\endgroup$ Feb 25, 2023 at 20:28

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