# How to solve potential flow with FEM, stream function, and the Kutta condition?

I'm trying to solve two-dimensional potential flow over airfoils with the finite element method, using the stream function formulation ($$\Delta\psi = 0$$, $$u = -\partial\psi/\partial y$$, $$v = \partial\psi/\partial x$$) and superposition to account for and enforce the Kutta condition (zero velocity) at the trailing edge. Following the approach of de Vries we have [1]

$$$$\psi = \psi_1 + b*\psi_2 + c*\psi_3$$$$

where

$$$$\psi_1: \Delta\psi = 0,\quad \psi_{\Gamma_2} = -U_\infty y,\quad \psi_{\Gamma_1} = 0$$$$ $$$$\psi_2: \Delta\psi = 0,\quad \psi_{\Gamma_2} = 1,\quad \psi_{\Gamma_1} = 0$$$$ $$$$\psi_3: \Delta\psi = 0,\quad \psi_{\Gamma_2} = 0,\quad \psi_{\Gamma_1} = 1$$$$

and $$\Gamma_2$$ is the external boundary of the domain with horizontal velocity $$U_\infty$$, and $$\Gamma_1$$ the boundary of the airfoil. Using superposition, and the fact that $$u_T=v_T=0$$ at the point of the trailing edge $$T$$, we should be able to determine the constants $$b$$ and $$c$$ with (equations 16a and 16b in [1])

$$$$u_T = u_{1T} + bu_{2T} + cu_{3T} = 0$$$$ $$$$v_T = v_{1T} + bv_{2T} + cv_{3T} = 0$$$$

All this seems resonable until one considers the fact that $$\psi_2$$ and $$\psi_3$$ will be linearly dependent $$(\psi_3 = 1-\psi_2)$$ so the system $$\{u_T,v_T\}$$ is not determined. Is there something I'm missing here or have misunderstood? Clearly the author has used this approach successfully, but I just can't see how it can work in this case.

[1] G. de Vries and D. H. Norrie, "The Application of the Finite-Element Technique to Potential Flow Problems", J. Appl. Mech 38(4), 798-802, 1971, DOI: 10.1115/1.3408957

• It isn't obvious to me why $\psi_3 = 1 - \psi_2$. Can you explain this? Unfortunately, I don't have access to the paper. Is there a publicly-accessible copy somewhere? – Bill Greene Feb 9 at 19:17
• Consider Laplace equation on the unit line with Dirichlet BCs 0 and 1 on the end points, the one solution will be x and the other 1-x. This will extend problems in higher dimensions with only two boundaries. The paper is available here. – WillG Feb 10 at 0:48
• The problem that you are proposing only has as solution $u=x$, assuming that the left end is located at $x=0$. – nicoguaro Feb 10 at 7:17
• To clarify what I meant, applying BCs 0/1 for $\psi_2$ gives you the solution $\psi_2=x$, and similarly BCs 1/0 gives $\psi_3=1-x$, thus $\psi_3 = 1-\psi_2$. – WillG Feb 10 at 7:29
• Thanks for the paper link. I understand about the two solutions for the 1D case. But why do you say those solutions are "linearly dependent?" You can't get the second solution by multiplying the first by a constant. Have you actually run into a specific problem applying the authors' approach? – Bill Greene Feb 10 at 12:12