I'm having trouble understanding how to map the streamlines from one plane to another using the Joukowski transform.
In the $\zeta$ plane, I'm considering flow around a cylinder, with the complex potential given by $$w(\zeta)=U(\zeta e^{-i \alpha} + \frac {a^2}{\zeta}e^{i\alpha})$$
where $\zeta = re^{it}$.
From here I understand it is more useful to consider the streamlines first in the $\zeta$ plane and then use the Joukowski map $z=\zeta+\frac{b^2}{\zeta}$ to transform them to the $z$ plane. It's not a problem finding the streamlines in the $\zeta$ plane, as I can use the stream function given by $$\psi(r,t)=Usin(t-\alpha)(r-\frac{a^2}{r})$$ and equate it to constant values. From here though, the notes I'm looking at say it's easy to then use the transform to get the streamlines in the $z$ plane, but I'm not very sure how. I'm aware it's supposed to look like flow around an ellipse, but my ideas on Maple all draw a blank.
Thanks!
Best Answer
I would find the parametric equation for the streamlines in the $\zeta$ plane; then the transformation to the $z$-plane is indeed easy. As a parameter, take $s=t-\alpha$. The streamline equation is $U\sin s (r-a^2/r) = C$. Solving this for $r$ yields $$r=r(s)=\frac12\left(\frac{C}{U\sin s}+\sqrt{\left(\frac{C}{U\sin s}\right)^2+4a^2}\right)$$ So, the parametric equation of streamline is $\zeta=r(s)e^{i(s+\alpha)}$. Applying the Жуковский transformation (I hate that transliteration) yields $$z= r(s)e^{i(s+\alpha)} +\frac{b^2}{r(s)} e^{-i(s+\alpha)}$$ which is something that Maple's
complexplotcan handle.