# How can I calculate this contour integral (to obtain lift) after applying a Joukowski transform?

I obtained the complex potential for flow past a flat plate at angle of attack by applying the Joukowski transform $z = \zeta + 1/\zeta$ (centered at the origin, so $\zeta = e^{i\theta}$) to flow past a cylinder.

For reference, the complex potential for flow past a cylinder (at angle of attack $\alpha$, of unit radius, with freestream velocity $U_\infty$, and with circulation $\Gamma$) is $$F(\zeta) = U_\infty \left( \zeta e^{-i \alpha} + \frac{e^{i \alpha}} {\zeta} \right) -\frac{i \Gamma}{2 \pi} \log \zeta$$ and after taking the Joukowski transform, I get $$F(z) = U_\infty \left( z \cos \alpha - i (z^2 - 4)^{1/2} \sin \alpha \right) - \frac{i \Gamma}{2 \pi} \log \left[\frac{1}{2}\left( z + (z^2-4)^{1/2} \right)\right]$$

Then we should be able to calculate lift around the flat plate with the Blasius formula, $$F_x - i F_y = \frac{i \rho}{2} \oint \left( \frac{dF(z)}{dz} \right)^2 dz$$ where the integral is around the surface (flat plate from $z = -2$ to $z = 2$, or the cylinder $\zeta = e^{i \theta}$ if we integrate in the $\zeta$-plane).

What's the best way to evaluate this contour integral? I tried changing the integral to the $\zeta$-plane with something like $$F_x - i F_y = \frac{i \rho}{2} \oint \left( \frac{dF(\zeta)}{d\zeta} \frac{d\zeta}{dz} \right)^2 \frac{dz}{d\zeta} d\zeta = \frac{i \rho}{2} \oint \left( \frac{dF(\zeta)}{d\zeta} \right)^2 \frac{1}{1-1/\zeta^2} d\zeta$$ but the $\frac{1}{1-1/\zeta^2}$ factor seems to mess things up by adding poles at $z=\pm1$ and making the residue of the pole at $z=0$ go to $0$. (I know what the answer should come out to by the Kutta-Joukowski theorem, but I want to be able to verify it.) Do I need to evaluate this integral in the $z$-plane?

Thanks in advance, any help is appreciated! My complex analysis and conformal mapping is definitely a bit rusty.

EDIT 1: I decided to try and generate interest by taking the calculations further to show where the problem is, and found out that continuing as I was doing does indeed give the right answer! Valuable lesson for me there that I shouldn't claim to be stuck just because it doesn't "seem" like I will get the right answer.

In the end, you do indeed have poles at $\zeta=\pm 1$ which have nonzero residues, and the pole at $\zeta = 0$ still has a nonzero residue as well. They add up perfectly to give the same sum of residues as in the non-transformed case, so that the Kutta-Joukowski theorem holds as we expect. (I guess so it's no longer unanswered, I'll answer my own question.)

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Welcome to MSE =) –  Patrick Da Silva Dec 7 '11 at 6:40
Is $U_\infty$ a function, or a constant ? –  Sasha Dec 7 '11 at 13:43
Thanks! And $U_\infty$ is a constant, the freestream velocity. (I'll edit my original post.) –  Frecka Dec 7 '11 at 16:00
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## 1 Answer

I feel a little silly answering my own question, but that's alright.

As I said in edit 1, the approach in the $\zeta$-plane is perfectly valid. I'll show the math here real quick too. (I have to write it up since it's part of the solution set for a class I'm TAing anyway!)

Starting from where I left off in the question, $$F_x - i F_y = \frac{i \rho}{2} \oint \left( \frac{dF(\zeta)}{d\zeta} \right)^2 \frac{1}{1-1/\zeta^2} d\zeta = \frac{i \rho}{2} \oint \left( \frac{dF(\zeta)}{d\zeta} \right)^2 \frac{\zeta^2}{\zeta^2-1} d\zeta$$ We can evaluate the derivative of the complex potential in the $\zeta$ plane as $$\frac{dF(\zeta)}{d\zeta} = U_\infty \left( e^{-i \alpha} - \frac{e^{i \alpha}} {\zeta^2} \right) -\frac{i \Gamma}{2 \pi \zeta}$$ Squaring it gives $$\left( \frac{dF(\zeta)}{d\zeta} \right)^2 = U_\infty^2 \left( e^{-2 i \alpha} - \frac{2}{\zeta^2} +\frac{e^{2 i \alpha}} {\zeta^4} \right) -\frac{i U_\infty \Gamma}{\pi} \left( \frac{e^{-i \alpha}}{\zeta} - \frac{e^{i \alpha}}{\zeta^3} \right) -\frac{\Gamma^2}{4 \pi \zeta^2} = A(\zeta) + B(\zeta)$$ where we group all the terms with even powers of $\zeta$ in $A(\zeta)$, and all the terms with odd powers of $\zeta$ in $B(\zeta)$. That is, $$A(\zeta) = U_\infty^2 \left( e^{-2 i \alpha} - \frac{2}{\zeta^2} +\frac{e^{2 i \alpha}} {\zeta^4} \right) -\frac{\Gamma^2}{4 \pi \zeta^2}, \qquad B(\zeta) = -\frac{i U_\infty \Gamma}{\pi} \left( \frac{e^{-i \alpha}}{\zeta} - \frac{e^{i \alpha}}{\zeta^3} \right)$$ so $A(\zeta) = A(-\zeta)$, and $B(\zeta) = -B(-\zeta)$.

Now we're ready to evaluate our contour integral. $$F_x - i F_y = \frac{i \rho}{2} \oint \left( \frac{dF(\zeta)}{d\zeta} \right)^2 \frac{\zeta^2}{\zeta^2-1} d\zeta = \frac{i \rho}{2} \oint \frac{(A(\zeta) + B(\zeta))\zeta^2}{(\zeta+1)(\zeta-1)} d\zeta$$ Applying the Residue theorem, $$F_x - i F_y = \frac{i \rho}{2} (2 \pi i) \sum_k \text{Res} \left( \frac{(A(\zeta) + B(\zeta))\zeta^2}{(\zeta+1)(\zeta-1)},a_k \right)$$ Now we just need to calculate the residue of all the poles of that function inside the unit circle. There are 3 poles, $\zeta = 0, \pm 1$.

For $\zeta = 0$, the only term with $1/\zeta$ is $\frac{i U_\infty \Gamma}{\pi} \left(\frac{e^{i \alpha}}{{\zeta (\zeta+1) (\zeta-1)}} \right)$, which gives us $$\text{Res}(0) = -\frac{i U_\infty \Gamma e^{i \alpha}}{\pi}$$ All terms have $\zeta = \pm 1$ as a pole, so $$\text{Res}(1) = \frac{A(1) + B(1)}{2}$$ $$\text{Res}(-1) = \frac{A(-1) + B(-1)}{-2} = \frac{-A(1) + B(1)}{2}$$ Summing the residues, we find $$\sum_k \text{Res} \left( a_k \right) = B(1) - \frac{i U_\infty \Gamma e^{i \alpha}}{\pi} = -\frac{i U_\infty \Gamma}{\pi} \left( e^{-i \alpha} - e^{i \alpha} \right) - \frac{i U_\infty \Gamma e^{i \alpha}}{\pi} = -\frac{i U_\infty \Gamma e^{-i \alpha}}{\pi}$$ This is the exactly same as Res$(0)$ for the case of flow past a cylinder (without the $\zeta^2/(\zeta^2-1)$ factor), which is the only residue in that case, and so we know our lift will be the same. $$F_x - i F_y = \frac{i \rho}{2} (2 \pi i) \left(-\frac{i U_\infty \Gamma e^{-i \alpha}}{\pi}\right) = i \rho \Gamma U_\infty e^{-i \alpha}$$ so $$F_x = \rho \Gamma U_\infty \sin \alpha = \rho \Gamma U_{\infty,y}, \qquad F_y = -\rho \Gamma U_\infty \cos \alpha = -\rho \Gamma U_{\infty,x}$$ give the (lift-induced) drag and lift, respectively.

The agreement with the Kutta-Joukowski theorem is easier to see if we notice that the freestream isn't in the $x$-direction and compute $F_\perp$ and $F_\parallel$ rather than $F_x$ and $F_y$. Multiplying both sides by $e^{i \alpha}$ gives $$e^{i \alpha} (F_x - i F_y) = (F_x \cos \alpha + F_y \sin \alpha) - i(-F_x \sin \alpha + F_y \cos \alpha ) = F_\parallel - i F_\perp = i \rho \Gamma U_\infty$$ Thus we have $F_\parallel = 0$ and $F_\perp = -\rho \Gamma U_\infty$ as desired.

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You shouldn't feel silly for answering your own question; it's perfectly alright here... –  Ｊ. Ｍ. Dec 9 '11 at 0:30
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