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I'm investigating the velocity field induced by a continuous distribution of 2D vortex points distributed along an ellipse $\{a\cos\theta,b\sin\theta\}$. I'm interested in the field inside the ellipse, and I need some help to prove whether this field is zero or not.

The intensity of each vortex point is proportional to $d\theta$ and not to the length along the ellipse. A vortex point located at a point $\boldsymbol{x'}$ induces a velocity field $\boldsymbol{u}(\boldsymbol{x})=\frac{d\theta}{2\pi |\boldsymbol{x}-\boldsymbol{x'}|} \boldsymbol{e}_\perp$ where ${e}_\perp$ is the unitary vector orthogonal to $(\boldsymbol{x}-\boldsymbol{x'})$ which is in 2D: $\boldsymbol{e}_z\times(\boldsymbol{x}-\boldsymbol{x'})/|\boldsymbol{x}-\boldsymbol{x'}|$. The total velocity field at a point $(x,y)$ inside the ellipse is obtained by integration over $\theta$. Numerical experiments seem to show that the field is zero inside the ellipse, but I cannot prove it. Dropping the factor $2\pi$, the field is in cartesian coordinates:

$$\boldsymbol{u}=\int_0^{2\pi} \left\{\frac{-y + b \sin\theta}{(x - a \cos\theta)^2 + (y - b \sin\theta)^2}, \frac{x - a \cos\theta}{(x - a \cos\theta)^2 + (y - b \sin\theta)^2}\right\} d\theta \stackrel{?}{=}\{0,0\}$$

Is the field really zero?

Maybe there is no need for the integrals to prove it. Maybe complex analysis is of help? I should mention that the following property is true in this problem: For any closed contour inside the ellipse the circulation is zero: $$\oint \boldsymbol{u}\cdot\boldsymbol{dl} =0$$

Since we are in a simply connected region, the velocity is the gradient of a potential which is single valued. But then is this potential constant?...

I'm able to prove that the velocity is zero on both the $x$ and the $y$ axis. Also $u_x$ is an even function of $x$ and an odd function of $y$. The opposite applies for $u_y$. I'm able to prove the result for a circle $a=b$. Can somebody help me for the ellipse? Any idea?

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Yes, the velocity field inside the ellipse is really zero. To convince myself of it I have run a few numeric evaluations of the integral for various $(a,b)$ and $(x,y)$.

Here is how can we obtain this result analytically.

First, we introduce the elliptical coordinate system with coordinates $\chi$, $\theta$: $$x = c \cosh \chi \cos \theta, \qquad y = c \sinh \chi \sin \theta, $$ so that the ellipse in question corresponds to a fixed value of $\chi$: $$a = c \cosh \chi_0, \qquad b = c \sinh \chi_0,$$ we also assume $a>b$.

Second, we compute the vorticity field. We start with single vortex $$ \mathop{\mathrm{curl}} \left( \frac{\Gamma}{2\pi |\mathbf{x}-\mathbf{x}_0|} \mathbf{e}_\perp \right) = \mathbf{e}_z \Gamma \delta(\mathbf{x}-\mathbf{x}_0)=\mathbf{e}_z \Gamma \delta(x-x_0)\cdot\delta(y-y_0),$$ and integrate the expression over the ellipse: \begin{multline} \mathop{\mathrm{curl}} \mathbf{u} = \mathbf{e}_z \int_0^{2\pi} \delta(\mathbf{x}-\mathbf{x}_0(\theta')) d \theta' = \mathbf{e}_z \int_0^{2\pi} \delta(x - a \cos \theta') \delta(y-b \sin \theta') d\theta' = \\= \mathbf{e}_z \int_0^{2\pi} c^{-2} (\cosh^2\chi-\cos ^2 \theta)^{-1} \delta(\chi-\chi_0)\delta(\theta - \theta') d\theta' = \\ = \mathbf{e}_z c^{-2} (\cosh^2\chi-\cos ^2 \theta)^{-1} \delta (\chi - \chi_0), \end{multline} where we used the expression for area element in elliptical coordinates: $dA = c^2 (\cosh^2\chi-\cos ^2 \theta)d\chi\, d\theta$.

Finally, we calculate the stream function $\psi$ with vorticity as a source: $$\Delta \psi = - \omega \equiv - \mathbf{e}_z \cdot \mathop{\mathrm{curl}} \mathbf{u}.$$ In elliptical coordinates this reads as: $$ c^{-2} (\cosh^2\chi-\cos ^2 \theta)^{-1} \left( \frac{\partial^2 \psi}{\partial \chi^2}+\frac{\partial^2 \psi}{\partial \theta^2}\right) = - c^{-2} (\cosh^2\chi-\cos ^2 \theta)^{-1} \delta (\chi - \chi_0). $$ Notice, that the factors on both sides are equal. So we could look for the solution depending only on $\chi$, compatible with $\mathbf{u}(\mathbf{0})=\mathbf{0}$: $$ \psi = - H(\chi - \chi_0) \cdot (\chi - \chi_0), $$ wher $H(x)$ is Heaviside function. This shows, that inside the ellipse velocity field is indeed zero. It also enables us to calculate the velocity field outside the ellipse, right away we see that the streamlines will be confocal ellipses.

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  • $\begingroup$ Thank you for your proof. I have several questions, it could be nice if you add more details on some aspects. The most difficult part to follow for me is the second one. Question1: How do you compute the curl formally. Do you use something like curl(aV)=a curl(V)+grad(a) times V ? I can see that the vorticity is purely along z and is a Dirac of the position. but I'm not sure I would be able to derive it from the curl though. I'd like to be able to do it, that could help me in the future. There are probably relations to the Green function, but I would appreciate your help on some details. $\endgroup$ – e-malito Oct 21 '13 at 10:21
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    $\begingroup$ @elmanuelito: To compute the curl of $\frac{\Gamma}{2\pi}\frac{\mathbf{e}_\perp}{|x-x_0|}$ we use the Stokes theorem: velocity circulation for any contour around the $\mathbf{x}_0$ is $\Gamma$, so contracting such contour to a point we obtain that the curl is zero everywhere except $\mathbf{x}_0$, where it is infinite, while its integral over area is finite, that means we have Dirac's $\delta$ function. $\endgroup$ – user23660 Oct 21 '13 at 13:43
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    $\begingroup$ @elmanuelito: Question 2: Sorry for the confusion, I meant the Dirac function for 2D plane, that is $\delta(x-x_0)\cdot\delta(y-y_0)$. I'll change the notation to $\delta(\mathbf{x}-\mathbf{x}_0)$ in the answer. $\endgroup$ – user23660 Oct 21 '13 at 13:47
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    $\begingroup$ @elmanueliton: Question 3. I expanded the integration over $\theta$. The factor that appears there is indeed Jacobian of coordinate transform $c^{2} (\cosh^2\chi-\cos ^2 \theta) = \mathrm{det}(\frac{\partial(x,y)}{\partial(\chi,\theta)})$. $\endgroup$ – user23660 Oct 21 '13 at 14:26
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    $\begingroup$ @elmanuelito: Last question: It is simply special case of integration by separation of variables. We have Poisson's equation in which the source depends only on one variable, so we look for the solution depending only on this variable. Boundary conditions at $\chi = 0$ (origin) could be chosen as $\psi =0$ and $\psi_\chi =0$. $\endgroup$ – user23660 Oct 21 '13 at 14:41

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