I was wondering if anyone could help me with the following problem, as I'm unsure on how to begin. Any suggestions would be appreciated, thanks for reading this.
I just don't understand how to apply this with the Milne Thomson theorem to get the required result.
I know that the complex potential for a line vortex is $w(z)=-iΓ/2\pi \ln(𝑧)$
Here is an image of the question: (https://i.sstatic.net/PUNDD.jpg)
Two equal line vortices of circulation $\Gamma$ lie at coordinates $x= – 2a, y = 0$ and $x=2a, y=0$ and are situated near a rigid circular cylinder of radius $a$, given by the equation $x^2 + y^2 = a^2$.
Using the Milne-Thomson circle theorem, show that the complex potential for this flow is
$$\omega (z) = – \frac{i\Gamma}{2\pi} \left[ \ln (z-2a) – \ln \left(\frac{a^2} z – 2a\right) + \ln (z + 2a) – \ln \left( \frac{a^2} z + 2a \right) \right]$$
Best Answer
For the vortex at $z=2a$ (without the cylinder), you have $$w(z) = \frac{-i\Gamma}{2\pi}\log(z-2a).$$ We use the circle theorem to account for the cylinder and so compute the new potential $w_2$: $$w_2(z) = \frac{-i\Gamma}{2\pi}\log(z-2a) + \frac{i\Gamma}{2\pi}\log\left(\frac{a^2}{z} - 2a\right),$$ where the second term in $w_2$ is, of course, $\overline{w\left(\frac{a^2}{\overline{z}}\right)}$.
For the vortex at $z=-2a$, you do the same thing. Combining these two and factoring out $\frac{-i\Gamma}{2\pi}$ will give what you want.