An incompressible inviscid fluid fills a long circular cylinder of radius a. Show that if there is a line vortex of circulation $2\pi \Gamma$ parallel to the axis of the cylinder and at a distance $b (< a)$ from it, it will revolve around the axis with a constant speed $(\Gamma b)/(a^2−b^2)$.
So i know that we have a vortex and the complex potential is $-(i\Gamma/2\pi)*\ln(z)$
since we are given speed that is $w'(z)$, right? so how would we go about showing all this. where would we start? would we start by integrating $w'(z)$? or could we start with a vortex pair within the cylinder?
Orient the coordinate system so that the vortex inside the cylinder is at the point $z_b =b +i\cdot 0$ on the real axis. To satisfy the no flux condition at the surface of the cylinder the velocity field inside the cylinder can be obtained using the method of images with another vortex of equal and opposite strength placed at the point $z_c =c + i \cdot 0$ where $bc = a^2$.
There is no self-induced motion of the vortex at $z_b$. The motion of this vortex is due to the velocity field produced by the image vortex alone with potential
$$f_c(z) = \frac{i2 \pi\Gamma}{2\pi}\log (z - z_c),$$
and complex velocity
$$u-iv =-\frac{d f_c}{dz} = -i \Gamma \frac{1}{z - z_c}$$
The velocity at the location $z_b$ is
$$u(z_b) - i v(z_b) = -i \Gamma \frac{1}{z_b - z_c} = i\frac{\Gamma}{c - b} = i\frac{\Gamma b}{bc - b^2} = i \frac{\Gamma b}{a^2- b^2}$$
Hence the vortex at $z_b$ moves with speed given by the modulus
$$|u(z_b) - i v(z_b)| = \frac{\Gamma b}{a^2 - b^2}$$