reduction of partial derivatives by chain rule

Question

How do we prove that $5\frac{\partial^2u}{\partial x^2} + 2\frac{\partial^2u}{\partial x \partial y} + 2\frac{\partial^2 u}{\partial y^2} = \frac{\partial^2u}{\partial s^2} + \frac{\partial^2 u}{\partial t^2}$

if we set $x = 2s+t, y = s-t$?

Also what does that equation with the partial derivatives actually mean?

Answer 1

First derivatives $$ \frac {\partial u}{\partial x} = \frac {\partial u}{\partial s} \frac {\partial s}{\partial x} + \frac {\partial u}{\partial t} \frac {\partial t}{\partial x} \\ \frac {\partial u}{\partial y} = \frac {\partial u}{\partial s} \frac {\partial s}{\partial y} + \frac {\partial u}{\partial t} \frac {\partial t}{\partial y} $$ Second derivatives \begin{align} \frac {\partial^2 u}{\partial x^2} &= \frac \partial{\partial x} \left( \frac {\partial u}{\partial x}\right ) = \frac \partial{\partial s} \left( \frac {\partial u}{\partial s} \frac {\partial s}{\partial x} + \frac {\partial u}{\partial t} \frac {\partial t}{\partial x}\right ) \frac {\partial s}{\partial x} + \frac \partial{\partial t} \left( \frac {\partial u}{\partial s} \frac {\partial s}{\partial x} + \frac {\partial u}{\partial t} \frac {\partial t}{\partial x}\right ) \frac {\partial t}{\partial x} = \\ &= \left( \frac {\partial^2 u}{\partial s^2} \frac {\partial s}{\partial x} + \frac {\partial^2 u}{\partial s \partial t} \frac {\partial t}{\partial x}\right) \frac {\partial s}{\partial x} + \left( \frac {\partial^2 u}{\partial s \partial t} \frac {\partial s}{\partial x} + \frac {\partial^2 u}{\partial t^2} \frac {\partial t}{\partial x}\right) \frac {\partial t}{\partial x} = \\ &= \frac {\partial^2 u}{\partial s^2} \left( \frac {\partial s}{\partial x} \right )^2 + 2 \frac {\partial^2 u}{\partial s \partial t} \frac {\partial s}{\partial x} \frac {\partial t}{\partial x} + \frac {\partial^2 u}{\partial t^2} \left( \frac {\partial t}{\partial x} \right )^2 \end{align} Analogously \begin{align} \frac {\partial^2 u}{\partial y^2} &=\frac {\partial^2 u}{\partial s^2} \left( \frac {\partial s}{\partial y}\right)^2 + 2 \frac {\partial^2 u}{\partial s \partial t} \frac {\partial s}{\partial y} \frac {\partial t}{\partial y} + \frac {\partial^2 u}{\partial t^2} \left( \frac {\partial t}{\partial y} \right )^2 \end{align} And last one \begin{align} \frac {\partial^2 u}{\partial x \partial y} &= \frac \partial {\partial x} \left( \frac {\partial u}{\partial y}\right ) = \frac \partial {\partial s} \left(\frac {\partial u}{\partial s} \frac {\partial s}{\partial y} + \frac {\partial u}{\partial t} \frac {\partial t}{\partial y} \right ) \frac {\partial s}{\partial x} + \frac \partial {\partial t} \left(\frac {\partial u}{\partial s} \frac {\partial s}{\partial y} + \frac {\partial u}{\partial t} \frac {\partial t}{\partial y} \right ) \frac {\partial t}{\partial x} = \\ &= \frac {\partial^2 u}{\partial s^2} \frac {\partial s}{\partial x} \frac {\partial s}{\partial y} + \frac {\partial^2 u}{\partial s \partial t} \left(\frac {\partial s}{\partial x} \frac {\partial t}{\partial y} + \frac {\partial s}{\partial y} \frac {\partial t}{\partial x} \right ) + \frac {\partial^2 u}{\partial t^2} \frac {\partial t}{\partial x} \frac {\partial t}{\partial y} \end{align} Now, taking into account that $$ s = \frac {x+y}3 \\ t = \frac {x - 2y}3 $$ you can find $$ \frac {\partial s}{\partial x} = \frac 13 \\ \frac {\partial s}{\partial y} = \frac 13 \\ \frac {\partial t}{\partial x} = \frac 13 \\ \frac {\partial t}{\partial y} = -\frac 23 $$ Substitute all these into your ralation $$ \frac 59 \left( \frac {\partial^2 u}{\partial s^2} + 2 \frac {\partial^2 u}{\partial s \partial t } +\frac {\partial^2 u}{\partial t^2}\right ) + \frac 29 \left( \frac {\partial^2 u}{\partial s^2} - \frac {\partial^2 u}{\partial s \partial t } - 2\frac {\partial^2 u}{\partial t^2} \right ) + \frac 29 \left( \frac {\partial^2 u}{\partial s^2} - 4 \frac {\partial^2 u}{\partial s \partial t } + 4\frac {\partial^2 u}{\partial t^2}\right) = \\ = \frac {\partial^2 u}{\partial s^2} + \frac {\partial^2 u}{\partial t^2} $$

Answer 2

$$\begin{align*} \frac{\partial^2u}{\partial s^2} &=\frac{\partial}{\partial s}\frac{\partial u}{\partial s}\\ &=\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}\right)\\ &=\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\right)\cdot\frac{\partial x}{\partial s}+\frac{\partial u}{\partial x}\cdot\frac{\partial^2x}{\partial s^2}+\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial y}\right)\cdot\frac{\partial y}{\partial s}+\frac{\partial u}{\partial y}\cdot\frac{\partial^2y}{\partial s^2}\\ &=\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\right)\cdot2+\frac{\partial u}{\partial x}\cdot0+\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial y}\right)\cdot1+\frac{\partial u}{\partial y}\cdot0\\ &=2\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\right)+\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial y}\right) \end{align*}$$ This next part is a helpful "trick": $$\begin{align*}\\ &=2\frac{\partial\left(\frac{\partial u}{\partial x}\right)}{\partial s}+\frac{\partial\left(\frac{\partial u}{\partial y}\right)}{\partial s}\\ &=2\left(\frac{\partial\left(\frac{\partial u}{\partial x}\right)}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial\left(\frac{\partial u}{\partial x}\right)}{\partial y}\frac{\partial y}{\partial s}\right)+\left(\frac{\partial\left(\frac{\partial u}{\partial y}\right)}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial\left(\frac{\partial u}{\partial y}\right)}{\partial y}\frac{\partial y}{\partial s}\right)\\ &=2\left(\frac{\partial\left(\frac{\partial u}{\partial x}\right)}{\partial x}\cdot2+\frac{\partial\left(\frac{\partial u}{\partial x}\right)}{\partial y}\cdot1\right)+\left(\frac{\partial\left(\frac{\partial u}{\partial y}\right)}{\partial x}\cdot2+\frac{\partial\left(\frac{\partial u}{\partial y}\right)}{\partial y}\cdot1\right)\\ &=4\frac{\partial^2u}{\partial x^2}+4\frac{\partial^2u}{\partial x\,\partial y}+\frac{\partial^2u}{\partial y^2}\\ \end{align*}$$

Now do something similar to find $$\begin{align*} \frac{\partial^2u}{\partial t^2} &=\frac{\partial^2u}{\partial x^2}-2\frac{\partial^2u}{\partial x\,\partial y}+\frac{\partial^2u}{\partial y^2} \end{align*}$$

and add them together.