How to solve an equation using partial differentiation?

Question

There is a equation which needs to prove using partial differentiation. But I get a different answer at the end. Can you help me to prove this?

Answer 1

The answer is correct in my opinion but there was an enormous amount of multivariable chain rule dashed with product rules all over the place. I'd be almost amazed if I haven't slipped up somewhere!

I'd be grateful if anyone wishes to check and or amend.

On a personal note I'd be interested in seeing any shortcuts for this type of calculation if they exist.

$u=u(x,y)$, $x=x(s,t)$ and $y=y(s,t)$

$\frac{\partial u}{\partial s}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}$

$\frac{\partial^2u}{\partial s^2}=\frac{\partial}{\partial s}\left[\frac{\partial u}{\partial s}\right]=\frac{\partial}{\partial s}\left[\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}\right]$

$\frac{\partial}{\partial s}\left[\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}\right]=\frac{\partial u}{\partial x}\times\frac{\partial^2x}{\partial s^2}+\frac{\partial x}{\partial s}\times\frac{\partial^2u}{\partial x^2}\frac{\partial x}{\partial s}$

$\frac{\partial}{\partial s}\left[\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}\right]=\frac{\partial u}{\partial y}\times\frac{\partial^2y}{\partial s^2}+\frac{\partial y}{\partial s}\times\frac{\partial^2u}{\partial y^2}\frac{\partial y}{\partial s}$

$\frac{\partial^2u}{\partial s^2}=\frac{\partial u}{\partial x}\times\frac{\partial^2x}{\partial s^2}+\frac{\partial x}{\partial s}\times\frac{\partial^2u}{\partial x^2}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\times\frac{\partial^2y}{\partial s^2}+\frac{\partial y}{\partial s}\times\frac{\partial^2u}{\partial y^2}\frac{\partial y}{\partial s}$

$\frac{\partial^2u}{\partial t^2}=\frac{\partial u}{\partial x}\times\frac{\partial^2x}{\partial t^2}+\frac{\partial x}{\partial t}\times\frac{\partial^2u}{\partial x^2}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\times\frac{\partial^2y}{\partial t^2}+\frac{\partial y}{\partial t}\times\frac{\partial^2u}{\partial y^2}\frac{\partial y}{\partial t}$

If $x=e^s\cos t$ we have

$\frac{\partial x}{\partial s}=e^s\cos t$

$\frac{\partial^2 x}{\partial s^2}=e^s\cos t$

$\frac{\partial x}{\partial t}=-e^s\sin t$

$\frac{\partial^2 x}{\partial t^2}=-e^s\cos t$

and $y=e^s\sin t$

$\frac{\partial y}{\partial s}=e^s\sin t$

$\frac{\partial^2 y}{\partial s^2}=e^s\sin t$

$\frac{\partial y}{\partial t}=-e^s\cos t$

$\frac{\partial^2 y}{\partial t^2}=-e^s\sin t$

$\frac{\partial^2u}{\partial s^2}=\frac{\partial u}{\partial x}\times e^s \cos t+e^s \cos t\times\frac{\partial^2u}{\partial x^2}e^s \cos t+\frac{\partial u}{\partial y}\times e^s \sin t+e^s \sin t \times\frac{\partial^2u}{\partial y^2}e^s \sin t$

$\frac{\partial^2u}{\partial s^2}=\frac{\partial u}{\partial x}\times e^s \cos t+e^{2s} \cos^2 t\times\frac{\partial^2u}{\partial x^2}+\frac{\partial u}{\partial y}\times e^s \sin t+e^{2s} \sin^2 t \times\frac{\partial^2u}{\partial y^2}$

$\frac{\partial^2u}{\partial t^2}=\frac{\partial u}{\partial x}\times-e^s\cos t+-e^s \sin t\times\frac{\partial^2u}{\partial x^2}\times-e^s \sin t+\frac{\partial u}{\partial y}\times-e^s \sin t+-e^s \cos t\times\frac{\partial^2u}{\partial y^2}-e^s \cos t$

$\frac{\partial^2u}{\partial t^2}=\frac{\partial u}{\partial x}\times-e^s\cos t+e^{2s} \sin^2 t\times\frac{\partial^2u}{\partial x^2}+\frac{\partial u}{\partial y}\times-e^s \sin t+e^{2s} \cos^2 t\times\frac{\partial^2u}{\partial y^2}$

$\frac{\partial^2u}{\partial s^2}+\frac{\partial^2u}{\partial t^2}=e^{2s}\frac{\partial^2 u}{\partial x^2}(\sin^2 t+\cos^2 t)+e^{2s}\frac{\partial^2 u}{\partial y^2}(\sin^2 t+\cos^2 t)$

$\frac{\partial^2u}{\partial s^2}+\frac{\partial^2u}{\partial t^2}=e^{2s}\left[\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right]$

$e^{-2s}\left[\frac{\partial^2u}{\partial s^2}+\frac{\partial^2u}{\partial t^2}\right]=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}$

Hope this helps.