Potential flow around ellipse

Question

I stuck at the following problem: Consider a planar incompressible potential flow (i.e., an irrotational flow) around an elliptic profile \begin{equation*} P = \lbrace (x,y) \in \mathbb{R}^2 | \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1 \rbrace \end{equation*} for $a > b > 0$ such that the stream is going into the $x$-direction ; this is described by the following PDE for the stream function $\psi$: \begin{equation*} \begin{cases} \Delta \psi = 0 &, \mbox{ on } \mathbb{R}^2 \setminus P, \\ \psi = 0 &, \mbox{ on } \partial P, \\ \psi = y &, \mbox{ on } x^2 + y^2 \to \infty. \end{cases} \end{equation*} Now I want to find the solution of which is symmetric about $x$.
I proceeded as follows:
First we will use the Laplacian equation for ellpitic coordinates $(\xi, \eta ) \in [0, \infty) \times [0, 2\pi)$ \begin{equation*} \begin{cases} x = K \cosh \xi \cos \eta \\ y = K \sinh \xi \sin \eta \end{cases} \end{equation*} with $K = \sqrt{a^2 - b^2}$. Using the chain rule we get \begin{equation*} \Delta \psi(\xi, \eta) = \frac{1}{K^2(\sinh^2 \xi + \sin^2 \eta)}(\partial^2_{\xi} + \partial^2_{\eta})\psi(\xi, \eta). \end{equation*} Using separation of variables $\psi(\xi, \eta) = u(\xi)v(\eta)$ we get \begin{align*} \frac{1}{K^2(\sinh^2 \xi + \sin^2 \eta)}(\partial^2_{\xi} + \partial^2_{\eta})\psi(\xi, \eta) &= 0 \\ & \implies (\partial^2_{\xi} + \partial^2_{\eta})u(\xi)v(\eta) = 0 \\ &\implies u''(\xi)v(\eta) + u(\xi)v''(\eta) = 0 \\ &\implies u''(\xi) v(\eta) = - u(\xi)v''(\eta) \\ &\implies \frac{u''}{u}(\xi) = \alpha = -\frac{v''}{v}(\eta). \end{align*} Now both sides depend on different variables therefore they are constant $\alpha \in \mathbb{R}$ and we get \begin{align*} u'' &= \alpha u, \\ v'' &= -\alpha v. \end{align*} which gives the solutions \begin{align*} u_{\alpha}(\xi) &= A e^{\sqrt{\alpha}\xi} + B e^{-\sqrt{\alpha} \xi} \\ v_{\alpha}(\eta) &= C \cos(\sqrt{\alpha} \eta) + D \sin(\sqrt{\alpha} \eta). \end{align*} where $\alpha \in \mathbb{Z}$ for the continuity of the solution. Now $\psi = 0$ on $\partial P$ or if $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ which gives since the $u$ coordinate lines prescribe ellipses and the $v$ hyperboles \begin{align*} \frac{K^2 \cosh^2 \xi \cos^2 \eta}{a^2} + \frac{K^2 \sinh^2 \xi \sin^2 \eta}{b^2} &= 1 \\ &\implies b^2 \cosh^2 \xi \cos^2 \eta + a^2 \sinh^2 \xi \sin^2 \eta = \frac{a^2 b^2}{K^2} \\ &\implies b^2 (1 + \sinh^2 \xi) \cos^2 \eta + a^2 \sinh^2 \xi \sin^2 \eta = \frac{a^2 b^2}{K^2} \\ &\implies b^2 \cos^2 \eta + \sinh^2 \xi \left( a^2 \sin^2 \eta + b^2 \cos^2 \eta \right) = \frac{a^2 b^2}{K^2} \\ &\implies \sinh^2 \xi = \frac{\frac{ a^2 b^2 }{K^2} - b^2 \cos^2 \eta}{a^2 \sin^2 \eta + b^2 \cos^2 \eta} \\ &\implies \sinh^2 \xi = b^2 \frac{a^2 - K^2 \cos^2 \eta }{K^2(a^2 \sin^2 \eta + b^2 \cos^2 \eta)} \\ &\implies \sinh \xi = b \sqrt{ \frac{a^2 - K^2 \cos^2 \eta }{K^2(a^2 \sin^2 \eta + b^2 \cos^2 \eta)} } \\ &\implies \xi_0 = \mbox{arsinh} \left( b \sqrt{ \frac{a^2 - K^2 \cos^2 \eta }{K^2(a^2 \sin^2 \eta + b^2 \cos^2 \eta)} } \right) \end{align*} for $\eta \in [0, 2\pi)$. Also \begin{align*} x^2 + y^2 &= K^2 \left( \cosh^2 \xi \cos^2 \eta + \sinh^2 \xi \sin^2 \eta \right) \\ &= K^2 \left( \cos^2 \eta + \sinh^2 \xi (\cos^2 \eta + \sin^2 \eta) \right) \\ &= K^2 \left( \cos^2 \eta + \sinh^2 \xi \right) \\ &= K^2 \left( 1 - \sin^2 \eta + \sinh^2 \xi \right) \\ &= K^2 \left(1 + \sinh^2 \xi - \sin^2 \eta \right) \\ &= K^2 \left( \cosh^2 \xi - \sin^2 \eta \right) \end{align*} which diverges to $+\infty$ if $\xi \to +\infty$ or $\xi \to -\infty$ and $\eta \in [0, 2\pi )$. But we only look at $\xi \to +\infty$ since $\xi \in [0, \infty)$.Then we have \begin{equation*} \psi(\xi, \eta) = u_{\alpha}(\xi)v_{\alpha}(\eta) \approx K \sinh \xi \sin \eta = K \frac{1}{2} e^{\xi} \sin \eta \end{equation*} We also have that $\xi = \xi_0$ describes the ellipse $P$ in elliptic coordinates with $\eta \in [0, 2\pi)$ \begin{equation*} \psi(\xi_0, \eta) = u_{\alpha}(\xi_0) v_{\alpha}(\eta) = 0. \end{equation*} Somehow the second boundary condition would lead to the solution being trivial. Im not sure where I did make a mistake or why my way of solving this is wrong.
Edit: From the above boundary condition and the x-axis symmetrie we can write the solution as a sum for $\alpha $ in $\mathbb{N}$ and solve $A,B$ and $D$ from the boundary conditions. The symmetrie gives $C = 0$ and only one summand $\alpha = 1$ from the condition in the farfield is needed. Then one gets a solution for $\psi$.