In incompressible flow, the Kutta-Joukowski theorem can be used to relate the lift (per unit span) to the circulation around an airfoil. However, if we look at an airfoil or the flat plate (see below) in supersonic flow with shocks, how can we show that there is circulation? And, if there is circulation, then how do we calculate it?

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  • $\begingroup$ Why should it not be finite? Upper surface pressure cannot drop below vacuum, lower surface pressure cannot climb above stagnation pressure. The area is also limited. Only when you move to infinite speeds will there be a chance for infinite lift. $\endgroup$ Jun 11, 2020 at 20:10
  • $\begingroup$ I should clarify: I meant to ask: Is the circulation 0 or a finite value? $\endgroup$
    – Nick Hill
    Jun 11, 2020 at 23:02
  • $\begingroup$ Oh, the other infinite. Yes, supersonic flow can be simulated with potential flow, only what is an elliptic equation in subsonic flow becomes a hyperbolic equation so perturbations only extend in flow direction in the supersonic case. Algorithms using vortex elements exist, but for supersonic potential flow the use of doublets brings better stability, so modern potential codes don't include vortex elements. Oh, and circulation is indeed finite and not zero. $\endgroup$ Jun 12, 2020 at 5:55
  • $\begingroup$ So, you mentioned "circulation is indeed finite." How can you show this? $\endgroup$
    – Nick Hill
    Jun 12, 2020 at 15:22
  • $\begingroup$ One thing puzzles me. Supersonic flow can have no vertical component ahead of the leading-edge shock front. Clearly, air is deflected downwards in the wake. So where does the upward component of circulation come from? $\endgroup$ Jun 12, 2020 at 15:54

1 Answer 1


Maybe I'm missing something obvious to you, but why do you think there's zero circulation?

Take the simplest case of an infinitely thin 2D flat plate in linearized supersonic flow. The solution in the perturbed velocity potential ($\hat{\phi}$) is:

$$\hat{\phi}= \begin{cases} 0 & x\pm\beta z<0 \\ \frac{1}{\beta}\alpha(x-\beta z) & 0\le x-\beta z\le c, & z>0 \text{ (upper surface)}\\ -\frac{1}{\beta}\alpha(x+\beta z) & 0\le x+\beta z\le c, & z<0 \text{ (lower surface)}\\ \pm\frac{1}{\beta}(\alpha c) & x\pm\beta z>0 \end{cases} $$

where $c$ is the chord length of the flat plate, $x$ and $z$ are the horizontal and vertical coordinates, $\alpha$ is the angle of attack and $\beta=\sqrt{M_\infty^2-1}$ is the supersonic Prandtl-Glauert factor.

The velocity field ($\vec{V}$) can be obtained as:

$$\vec{V} = \begin{bmatrix}V_\infty \\ 0\end{bmatrix}+V_\infty\nabla{\hat{\phi}} = \begin{bmatrix}V_\infty\pm\frac{1}{\beta}\alpha V_\infty \\ -\alpha V_\infty\end{bmatrix}$$

where $V_\infty$ is the free-stream airspeed. The first equation is plus for upper surface and minus for lower surface.

Take an infinitely thin closed path ($C$) that wrap around the upper and lower surfaces, and calculate the circulation ($\Gamma$):

$$\Gamma = \oint_C\vec{V} \cdot \hat{n} ds=\frac{2\alpha c V_\infty}{\beta}$$

This is obviously non-zero. In fact, this can be readily seen by the potential jump in the wake.

And when we calculate the lift coefficient ($C_l$) of this flat airfoil using the Kutta-Joukowski Theorem ($L'=\rho_\infty V_\infty \Gamma$), we retrieve the classic supersonic lift relation:

$$C_l=\frac{L'}{1/2\rho_\infty V_\infty^2 c}=\frac{4\alpha}{\beta}$$


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