I am stuck on the last part of a problem regarding the following three differential equations:
$$(1)\; u(x,y), \; \begin{equation} \frac{\partial^2 u}{\partial x\partial y} + u = 0 \end{equation},\;x,y\gt 0$$ $$(2) \; F(s), \; \begin{equation} \frac{d F}{d s^2} + \frac{1}{s}\frac{d F}{d s}+F= 0 \end{equation}$$ $$(3) \; u(z,t), \; \begin{equation} \frac{\partial^2 u}{\partial z^2} - \frac{\partial^2 u}{\partial t^2} + u = 0 \end{equation},\;z\gt\lvert t\rvert$$
I have shown firstly that $u(x,y)=F(s)$ where $s=2\sqrt{xy}$, is a solution of $(1)$ provided $F(s)$ solves $(2)$.
I have also shown that $(3)$ is equivalent to $(2)$ with an appropriate transformation $x=az+bt,y=cz+dt$, and in that case, a solution is given by $$u(z,t)=F(\sqrt{z^2-t^2})$$ for all $F$ solving $(2).$
The last part of the question says:
Must all solutions have this form? Justify your answer.
I am a bit confused of what the question is asking me. I think it is saying 'can we find a solution to $(3)$ that, when transformed to $(x,y)$, is not a function of $2\sqrt{xy}$?' (or at least this is a sufficient condition for contradiction?).
If so, would I be right in saying that since $\sin(x+y)$ clearly solves $(1)$, and it is not a function of $2\sqrt{xy}$, we have found a solution without the given form? How could I show rigorously that it cannot be written as a function of $2\sqrt{xy}$?
I would be really appreciative for any feedback, thank you!