TITLE: Euler Darboux PDE solution
QUESTION [6 upvotes]: Consider the Euler-Darboux PDE
$$
u_{xy}+\frac{k}{x-y}(u_{x}-u_{y})=0
$$
What is the solution when $k>0$? All text books I have looked at give solutions
for $k<0$ and I don't seem to see how I can use that to find a solution 
for $k>0$. Hints and any help is appreciated.
Thanks, felasfa

REPLY [3 votes]: Let 
$$
\xi = x-y,
\\
\eta = x+y,
$$
then
$$
u_x - u_y = 2u_{\xi}, \quad u_{xy} = u_{\eta\eta}- u_{\xi\xi}.
$$
The equation becomes:
$$
u_{\eta\eta}- u_{\xi\xi} + \frac{2k}{\xi} u_{\xi}= 0.\tag{1}
$$
Further let $t = \xi^{\alpha}$ ($\alpha\neq -1$), then
$$
u_{\xi} = \alpha \xi^{\alpha-1}u_t ,\quad u_{\xi\xi} =\alpha^2 \xi^{2\alpha-2} u_{tt}+ \alpha(\alpha-1)\xi^{\alpha-2}u_t,
$$
plugging back to (1) gives:
$$
u_{\eta\eta} - \alpha^2 \xi^{2\alpha-2} u_{tt}- \alpha(\alpha-1)\xi^{\alpha-2}u_t
+2k\alpha \xi^{\alpha-2} u_t = 0.
$$
Let $\alpha = 2k+1>1$, first order terms get canceled:
$$
u_{\eta\eta} - \alpha^2 t^{\frac{2(\alpha-1)}{\alpha} } u_{tt} = 0.
$$
Rewrite this as:
$$
 u_{tt} - \frac{1}{(2k+1)^2}t^{\frac{-4k}{2k+1}} u_{\eta\eta} = 0.\tag{2}
$$
(2) is the Tricomi-type wave equation describing some quantity's transition from subsonic flow (elliptic region) to supersonic flow (hyperbolic region), you can try your luck with separation of variable 
$$u = F(t)G(\eta) = F\Big((x-y)^{2k+1}\Big)G(x+y).$$