Solving simultaneous PDEs

Question

Given the equations (1):$$\frac{\partial u}{\partial t}+g\frac{\partial \eta}{\partial x}=0$$ and (2):$$\frac{\partial\eta}{\partial t}+H\frac{\partial u}{\partial x}=0$$ can we combine the two together to form $$\frac{\partial^{2}\eta}{\partial t^{2}}-gH\frac{\partial^{2}\eta}{\partial x^{2}}= 0$$ by substituting $\frac{\partial}{\partial t}.(2)$ into (1). I should note this was given as an example but the working was not shown, I have tried a few manipulations but cannot get the result, am I missing something?

Answer 1

Differentiating the first equation with respect to $x$:$$\frac{\partial u}{\partial t} +g\frac{\partial \eta}{\partial x}=0\implies\frac{\partial^2 u}{\partial t\partial x}+g\frac{\partial^2 \eta}{\partial x^2}=0$$ and the second with respect to $t$: $$\frac{\partial\eta}{\partial t}+H\frac{\partial u}{\partial x}=0\implies \frac{\partial^2\eta}{\partial t^2}+H\frac{\partial^2 u}{\partial x \partial t}=0$$ Eliminating the mixed derivative $\frac{\partial^2 u}{\partial x \partial t}$ between these two gives : $$\frac{\partial^2\eta}{\partial t^2}+H\left(-g\frac{\partial^2 \eta}{\partial x^2}\right)=0$$ i.e. $$\frac{\partial^{2}\eta}{\partial t^{2}}-gH\frac{\partial^{2}\eta}{\partial x^{2}}= 0$$

Answer 2

$$\frac{\partial u}{\partial t}+g\frac{\partial \eta}{\partial x}=0$$ $$\frac{\partial\eta}{\partial t}+H\frac{\partial u}{\partial x}=0$$ This implies that $$ \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial t}+g\frac{\partial \eta}{\partial x}\right)$$ $$= \frac{\partial^2 u}{\partial t\partial x}+g\frac{\partial^2 \eta}{\partial x^2}=0$$ And $$ \frac{\partial}{\partial t} \left(\frac{\partial\eta}{\partial t}+H\frac{\partial u}{\partial x}\right)$$ $$=\frac{\partial^2\eta}{\partial t^2}+H\frac{\partial^2 u}{\partial t\partial x}=0$$ Using the fact that $$\frac{\partial^2 u}{\partial t\partial x}=-g\frac{\partial^2 \eta}{\partial x^2}$$ We have $$\frac{\partial^2\eta}{\partial t^2}+H\frac{\partial^2 u}{\partial t\partial x}$$ $$= \frac{\partial^2\eta}{\partial t^2}+H\left(-g\frac{\partial^2 \eta}{\partial x^2}\right) $$ $$=\frac{\partial^{2}\eta}{\partial t^{2}}-gH\frac{\partial^{2}\eta}{\partial x^{2}}= 0$$