Consider the following Helmholtz problem in the infinite triangle $y>0,\;x>y$ with parameters $Q<0$, $0\le P<|Q|$.
$$\left\{\begin{align} &\psi^{(2,0)}(x,y)+\psi^{(0,2)}(x,y)+E\psi(x,y)=0,\\ &\psi^{(0,1)}(x,0)-\frac Q2\psi(x,0)=0,\\ &\psi^{(1,0)}(x,x)-\psi^{(0,1)}(x,x)-\frac P{\sqrt2}\psi(x,x)=0\\ &|\psi(x,y)|<\infty. \end{align}\right.\tag1 $$
I'm mainly interested in the solution with lowest $E$. In the case of $P=0$ it's easy to see that
$$\psi_0(x,y)=\exp\left(\frac Q2 (x+y)\right)\tag2$$
with eigenvalue
$$E=-\frac{Q^2}2\tag3$$
solves the problem. But what about $P>0$? Can $(1)$ still be solved analytically (i.e. in terms of elementary or special functions)? If not, can the solution be given in the form of an integral or a series with explicitly specified terms?
It appears that simple product of solutions for the cases of $P=0$ and of $Q=0$ gives the full solution of $(1)$. So, if we let
$$\psi(x,y)=\exp(Ax+By),$$
by analogy with $(2)$, and then solve the boundary conditions containing $P$ and $Q$ for $A$ and $B$, we'll get:
\begin{align} A&=\frac Q2+\frac P{\sqrt2},\\ B&=\frac Q2. \end{align}
The eigenvalue will then be
$$E=-\frac{P^2+PQ\sqrt2+Q^2}2.$$
And the final eigenfunction is
$$\psi(x,y)=\exp\left(\frac P{\sqrt2}x+\frac Q2(x+y)\right).$$