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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!

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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 complexplot can handle.

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  • $\begingroup$ I understand the parameterisation, but why is the parametric equation $r(s)$ multiplied by $e^{i(s+\alpha)}$? I think I'm missing something simple there. Thank you for the answer, it's no doubt clear to most people! $\endgroup$ – Mike Miller Nov 27 '13 at 2:00
  • $\begingroup$ @BritMiller I understand that $\zeta=re^{it}$ in your notation. After introducing $s=t-\alpha$, this relation becomes $\zeta=re^{i(s+\alpha)}$. This gets plugged into the formula for $z$... By the way, I should have said that the range of $s$ can be taken $-\pi<s<\pi$. $\endgroup$ – Post No Bulls Nov 27 '13 at 2:14
  • $\begingroup$ Ah, of course! Yes I can see why it's a lot easier doing it this way now. So I should be able to plot the streamlines as a function of $s$; something like complexplot(z(s),s=-Pi..Pi), for certain values of C? $\endgroup$ – Mike Miller Nov 27 '13 at 2:37
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    $\begingroup$ @BritMiller Yes. Perhaps with a slightly smaller interval: going to $\pm$Pi amounts to going to infinity in both directions. Or maybe Maple will cut it automatically; I don't have. $\endgroup$ – Post No Bulls Nov 27 '13 at 3:17
  • $\begingroup$ Hiya. I've given this a go in Maple. The graphic I get after doing the transform is not what I expected - the streamlines cross (which I don't think should happen!) Though the basic shape I expected is there. Is there something simple I am missing do you think? I'd show a graphic if I could, but not sure how in the comments. $\endgroup$ – Mike Miller Nov 27 '13 at 16:35
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There are two additional issues here. First, your choice of $a=2$ and $b=3$ on the plot. The transformation $z=\zeta+b^2/\zeta$ is not conformal at $\zeta=\pm b$, where the derivative vanishes. This point should not be in the domain where you consider the flow; otherwise you don't have a conformal transformation (it is not even 1-1). The domain is the exterior of the disk of radius $a$. So, the allowable parameters are $0<b\le a$. For example, $a=3$ and $b=2$.

Second, the range of $s$ should not be $(-\pi , \pi)$. Indeed, $s$ is introduced as a polar angle (with some offset) on a streamline that stays in one halfplane; therefore its range is an interval of length $\pi$. It will be $(0,\pi)$ if $C>0$ and $(-\pi,0)$ if $C<0$. If this is unclear, consider the polar parametrization of horizontal line $y=C$. When $C>0$, it is $r=C/\sin t$, $0<t<\pi$; when $C<0$, it is $r=C/\sin t$ with $-\pi<t<0$.

With the above corrections, the plot comes out right.

streamlines

Maple code:

a:=3; b:=2; 
r:=1/2*(C/sin(s)+sqrt((C/sin(s))^2+4*a^2));
z:=r*exp(I*s)+b^2/r*exp(-I*s);
with(plots):
display(complexplot([seq(subs(C=i,z),i=0..20)],s=0.01..Pi-0.01,-15..15,-10.5..10.5,thickness=3), complexplot([seq(subs(C=i ,z),i=-20..-0)],s=-Pi+0.01..-0.01,-15..15,-10.5..10.5),thickness=3);
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  • $\begingroup$ Oops, the $b>a$ was a mix up - I did understand that part. I'm not so great with parameterisations, so thank you very much for the example $y=C$; it's perfectly understood now! :) $\endgroup$ – Mike Miller Nov 27 '13 at 18:39

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