Given the partial differential equation $$ \frac{\partial^2 F(x,t)}{\partial t^2} = \frac{\partial^2 F(x,t)}{\partial x^2} + g(F) $$
for a given solution $F = F(x, t)$ of this equation, show that the function $$ F_1 = F(x \cosh \beta + t\sinh \beta, t \cosh \beta + x \sinh \beta), $$ where $\beta$ is an arbitrary constant, is also a solution. Note that $g(F)$ is just some function, not particularly important.
I can't even really decipher what this question means. I've tried using MV chain rule but don't understand how to apply it here. I've seen the solution, which goes: $$ \frac{\partial F_1}{\partial x} = \cosh\beta\frac{\partial F}{\partial x} + \sinh\beta\frac{\partial F}{\partial t}, $$ and similarly for $\frac{\partial F_1}{\partial t}$, which implies $$ \begin{align} \frac{\partial^2 F_1}{\partial x^2} &= \cosh^2\beta\frac{\partial^2 F}{\partial x^2} + \sinh^2\beta\frac{\partial^2 F}{\partial t^2} + 2\cosh\beta \sinh\beta\frac{\partial^2 F}{\partial x\partial t}\label{1} \tag{1}\\ \frac{\partial^2 F_1}{\partial t^2} &= \sinh^2\beta\frac{\partial^2 F}{\partial x^2} + \cosh^2\beta\frac{\partial^2 F}{\partial t^2} + 2\cosh\beta \sinh\beta\frac{\partial^2 F}{\partial x\partial t}\label{2}\tag{2} \end{align} $$ Then equations are subtracted and we get the answer, which I understand. I do not understand how we get \eqref{1} and \eqref{2} though?
Can anyone chuck us some assistance? It's not a homework problem, I have the solution to this past paper question but don't understand how to do it :-(
I think the confusion may be that when you write: $$ \frac{\partial F_1}{\partial x} = \cosh\beta\frac{\partial F}{\partial x} + \sinh\beta\frac{\partial F}{\partial t}, $$ you are leaving out the arguments of $\frac{\partial F}{\partial x}$ and $\frac{\partial F}{\partial t}$. Because of the chain rule, these have to be evaluated not at $(x,t)$, but rather at $(x \cosh \beta + t \sinh \beta, x \sinh \beta + t \cosh \beta )$. So although it may look uglier, the more correct way to write this is: $$ \begin{align*} \frac{\partial F_1}{\partial x}(x,t) &= \cosh\beta\frac{\partial F}{\partial x}(x \cosh \beta + t \sinh \beta, x \sinh \beta + t \cosh \beta ) \\ &\phantom{=} + \sinh\beta\frac{\partial F}{\partial t}(x \cosh \beta + t \sinh \beta, x \sinh \beta + t \cosh \beta ) \end {align*} $$ Now when you take the derivatives of this (to get the second partial derivatives), you will have to use the chain rule again, and you should get the formulas you mentioned.