In my Fluid Mechanics class, we arrived at an equation for 2D compressible flow (using perturbation theory):

$$\left(1-M^{2}\right)\frac{\partial^{2}\phi}{\partial x^{2}}+\frac{\partial^{2}\phi}{\partial y^{2}}=0$$

where $\phi$ is the perturbed fluid potential ($\vec v=\vec U+\vec\nabla(\phi)$), $\vec U=U\vec e_{x}$ is a constant flow and $M=U/c_s$ is the Mach number.

I wanted to resolve this equation for an arbitrary contour defined by $$ f_\pm=\begin{cases}f(x) & \quad\text{if }x\in[0,l] \\ 0 & \quad \text{if } x\in]l,+\infty[\end{cases} $$

where $f_{+}$ is for $y>0$ and $f_{-}$ for $y<0$. and you also asume $\frac{df}{dx}\ll(\text{the other non pertubed quantities})$

enter image description here

My teacher only solved the supersonic case using the fact that for $M>1$ you get the wave equation. He found his boundary conditions like this:

$$\frac{df}{dx}=\tan(\alpha)=\frac{\nabla\phi|_{y}}{U+\nabla\phi|_{y}}$$ enter image description here

wich gives neglecting the quadratic perturbed quantities:

$$\frac{\partial\phi}{\partial y}|_{y=f'(x)}\simeq\frac{\partial\phi}{\partial y}_{y=0}\simeq Uf'(x).$$

My problem is that I have tried solving this equation (using Fourier transform) for a subsonic flow using the same boundary condition (and a bounded solution). And while I found a solution, this solution doesn't seem very physical or accurate when I plot it on python.

So I was wondering if my boundary conditions are incorrect and if they are what are the correct one?


1 Answer 1


For the subsonic case you get Laplace's equation rather than the wave equation. The solution method is quite different and too complicated to describe here. It comes down to solving a singular integral equation, for which some slick techniques were devised, explained in any undergrad aerodynamics text

  • $\begingroup$ are the boundary conditions the same ( $\frac{\partial\phi}{\partial y}_{y=0}\simeq Uf'(x)$ )? and can you solve for those boundaries using the fourrier transform? $\endgroup$
    – Jhon Jack
    Jun 21, 2017 at 23:19
  • $\begingroup$ The boundary conditions are the same, but there is no transformation from a subsonic flow to a supersonic flow. If you look for solutions by separation of variables you will see they are quite different. $\endgroup$
    – Philip Roe
    Jun 22, 2017 at 19:18

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