I have been assigned a problem with two point vortices:
Find two point vortices whose locations in the 2D plane and strengths $\gamma_1,\gamma_2$ are such that their positions remain fixed in time.
My actual Question is how to do this problem with 3 point vortices of strengths $\gamma_1,\gamma_2,\gamma_3$ whose positions remain fixed in time.
My proof for 2 vortices. (DISCLAIMER I AM NOT SURE IF THIS IS CORRECT.)
Denote the complex potential of the two point vortices as $z_1,z_2$.
The velocity induced on $z_2$ by the vortex at $z_1$ is,
$$w(z_1)=u-iv=\frac{\gamma_1}{2\pi i(z_1-z_2)}. $$The velocity induced on $z_2$ by the vortex at $z_1$ is,
$$w(z_2)=u-iv=\frac{\gamma_2}{2\pi i(z_2-z_1)}. $$Since inviscid vorticity is simply advected by the flow, the velocity is also with which the vortex moves:
$$\frac{dz_1^*}{dt}=\frac{1}{2\pi i}\frac{\gamma_1}{(z_1-z_2)}$$
$$\frac{dz_2^*}{dt}=\frac{1}{2\pi i}\frac{\gamma_2}{(z_2-z_1)}$$
Where the * denotes complex conjugations.
Adding and subtracting these two equations from each other, and got
$$\frac{d}{dt}(z_1+z_2)=\frac{1}{2\pi i}\frac{\gamma_1-\gamma_2}{z_1-z_2} \implies |z_1^*-z_2^*|=\text{ constant} =2R $$
and
$$\frac{d}{dt}(z_1^*-z_2^*)=\frac{1}{2\pi i}\frac{\gamma_1-\gamma_2}{(z_1-z_2)} \implies |z_1-z_2|=\text{ constant} =2R$$
This first result shows that the center of the two vortices is staying put while the second proves that the distance does not change. Writing then
$$z_1=R\exp(i\theta), \text{ and} z_2=-z_1$$
any of the two above equations of motion gives
$$\frac{d\theta}{dt}=\frac{\gamma}{2\pi R^2} $$
The first result shows that the centroid of the two vortices $\color{red}{\text{ is fixed within a distance } (??)}$, while the second proves
that their distance does not change. Thats the best description I've got.
**Any and all help/tips proofs would be very much appreciated. Thank you! **
Best Answer
I believe your proof of the 2 vortex is wrong, since you have not shown that the velocities are zero.
To solve the 3 vortex case lets assume that they are originally at $z_1,z_2,z_3$ and I will use a slightly different notation than you and call $w(z_j)$ the velocity of the vortex that is at $z_j$ $$0=w(z_1) = {\gamma_2 \over (z_1 -z_2)} +{\gamma_3 \over (z_1 -z_3)}$$ $$0=w(z_2) = {\gamma_1 \over (z_2 -z_1)} +{\gamma_3 \over (z_2 -z_3)}$$ $$0=w(z_3) = {\gamma_2 \over (z_3 -z_2)} +{\gamma_1 \over (z_3 -z_1)}$$ I am ignoring the $2 i \pi $ factors since the won't enter in the calculation lets call $a_{ij} = {1 \over z_i -z_j}$ and write $$ \left[\begin{array}{ccc} 0& a_{12}& a_{13} \\-a_{12}& 0& a_{23} \\=a_{13}&-a_{23}&0 \end{array}\right]\left[ \begin{array}{c}\gamma_1\\\gamma_2\\\gamma_3\end{array}\right]=\left[ \begin{array}{c}0\\0\\0\end{array}\right]$$ before I continue, notice that for the two vortice case the equations are
$$ \left[\begin{array}{cc} 0& a_{12} \\-a_{12}& 0 \end{array}\right]\left[ \begin{array}{c}\gamma_1\\\gamma_2\end{array}\right]=\left[ \begin{array}{c}0\\0\end{array}\right]$$ which can only be solved with zero vortex intensities.
in order to solve we can consider an easy case, when all vortices are on the $x$ axis so that $a$'s are real. Then we have a cross product $[-a_{23},a_{13},-a_{12}]\times[\gamma_1,\gamma_2,\gamma_3]=[0,0,0]$ so that by setting the $\vec{\gamma}$ vector to be parallel to $[-a_{23},a_{13},-a_{12}]$ we are done: $$[\gamma_1,\gamma_2,\gamma_3]=K[-a_{23},a_{13},-a_{12}]$$ for any real K. Example: $$z_1,z_2,z_3=1,2,3$$, $$a_{23}={1\over (2-3)} =-1 ; a_{13}={1\over (1-3)} =-1/2; a_{12}={1\over (1-2)} =-1 $$
$$[\gamma_1,\gamma_2,\gamma_3]=K[1,-1/2,1]$$ Any 3 co-linear vortices should work (you remove the phase by adjusting the $K$), but I am not sure you can find a non co-linear solution with 3 vortices.