The proof of the Kutta-Joukowski theorem for the lift acting on a body (see: Wiki) assumes that the complex velocity $w'(z)$ can be represented as a Laurent series.


It is not surprising that the complex velocity can be represented by a Laurent series. But it surprises me that it is assumed that there are no positive powers of $z$ and it is also surprising that it is assumed that all the singularities are at $z=0$.

The Wikipedia article says it is deduced from the physics of the problem, which sounds pretty dubious to me. Another source says that this representation is valid for distances far away from the body, which is problematic as the square of the complex velocity is later integrated on the contour of the body.

Is there any way to explain that this form of the complex velocity is assumed?


The physics of the situation is that the lifting body is immersed in a flow that is constant ${\bf V}(x,y)= {\bf V}_{\infty}$ when you are a long way from the body. This constant gives the $a_0$ in your expression The body perturbs this flow, but the perturbation falls off as we move away. The leading perturbtion is the circulation which falls of as 1/distance as is described by the $a_{-1}$ coeffecient. The remaining terms depend on the detailed shape of the body. There are no positive powers, becasue if there were, the flow would not become constant at large distances.

  • $\begingroup$ +1 For your answer. But what about the singularities? Why should they only appear for $z=0$? $\endgroup$ – MrYouMath Feb 16 '17 at 22:14
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    $\begingroup$ There are no actual singularities at $z=0$ as the derivation assumes that that point is inside the lifting body, so there is no fluid there. See my lecture notes at courses.physics.illinois.edu/phys509/sp2017/bmaster.pdf page 314. $\endgroup$ – mike stone Feb 17 '17 at 14:09
  • $\begingroup$ So, you are stating that path independence of the contour implies, that we can go far away from the body, with our integral contour. As there are no positive powers to ensure boundedness of the velocity only negative powers are possible. And we choose $z_0=0$ as our evaluations point for the Laurant series, that is why only $a_{-n}z^{-n}$ terms appear. And then after integration, we see that $a_{-1}$ is really nonvanishing, but theoretically, it could have happened that it would have vanished. Are my lines of reasoning correct? $\endgroup$ – MrYouMath Feb 17 '17 at 17:27
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    $\begingroup$ Yes. $a_{-1}$ is zero if there is no circulation around the lifting body. The circulation is determined by the Kutta condition, which is a separate idea from the K-J theorem. Also laurent expansion are usually only valid when you are far enough away from the expansion point. You don't really need $z=0$ to be within the body. The expansion only applies to integrals far away. $\endgroup$ – mike stone Feb 20 '17 at 23:42
  • $\begingroup$ Thank you for your answers! That really helped me in understanding this theorem. $\endgroup$ – MrYouMath Feb 21 '17 at 8:27

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