I'm stuck with the following initial value problem \begin{equation} \begin{aligned} \frac{\partial}{\partial t}u(t,x,y) &= \Delta_{x,y} \int_y^\infty u(t,x,\eta) \, d\eta \\ u(0,x,y) &= u_0(x,y) \end{aligned}\tag{1} \end{equation} After substituting \begin{equation*} w(t,x,y) = \int_y^\infty u(t,x,\eta) \, d\eta \end{equation*} I get \begin{equation} \begin{aligned} \frac{\partial^2}{\partial t \partial y} w + \Delta_{x,y} w &= 0 \\ w(0,x,y) &= \int_y^\infty u_0(x,\eta) \, d\eta \end{aligned}\tag{2} \end{equation} With the change of variables \begin{align*} t' &= x \\ x' &=\frac{1}{\sqrt{8 (\sqrt{2}-1})} \cdot \left( (2-\sqrt{2}) \cdot y -t \right) \\ y' &= \frac{1}{\sqrt{8 (\sqrt{2}+1})} \cdot \left((2+\sqrt{2}) \cdot y -t\right) \end{align*} (2) becomes for $w=w(t',x',y')$ \begin{equation} \frac{\partial^2}{\partial x'^2} w = \frac{\partial^2}{\partial t'^2} w + \frac{\partial^2}{\partial y'^2} w \tag{3} \end{equation}
For given initials $w(t',0,y')$ and $\partial_{x'}w(t',0,y')$, Polyanin (2002, Handbook of Linear Partial Differential Equations for Engineers and Scientists) gives an explicit solution of (3).
However I can't seem to see a way to connect my original initial conditions in (1) / (2) at $t=0$ to those required at $x'=0$... Is there a way?
Many thanks in advance