reduction of partial derivatives by chain rule 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?
 A: 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
$$
\frac 59 \left( \frac {\partial^2 u}{\partial s^2} + 2 \frac {\partial^2 u}{\partial s \partial t }  +\frac {\partial^2 u}{\partial t^2}\right ) + \frac 29 \left( \frac {\partial^2 u}{\partial s^2} - \frac {\partial^2 u}{\partial s \partial t } - 2\frac {\partial^2 u}{\partial t^2} \right ) + \frac 29 \left( \frac {\partial^2 u}{\partial s^2} - 4 \frac {\partial^2 u}{\partial s \partial t }  + 4\frac {\partial^2 u}{\partial t^2}\right) = \\
= \frac {\partial^2 u}{\partial s^2} + \frac {\partial^2 u}{\partial t^2}
$$
A: $$\begin{align*}
\frac{\partial^2u}{\partial s^2}
&=\frac{\partial}{\partial s}\frac{\partial u}{\partial s}\\
&=\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}\right)\\
&=\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\right)\cdot\frac{\partial x}{\partial s}+\frac{\partial u}{\partial x}\cdot\frac{\partial^2x}{\partial s^2}+\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial y}\right)\cdot\frac{\partial y}{\partial s}+\frac{\partial u}{\partial y}\cdot\frac{\partial^2y}{\partial s^2}\\
&=\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\right)\cdot2+\frac{\partial u}{\partial x}\cdot0+\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial y}\right)\cdot1+\frac{\partial u}{\partial y}\cdot0\\
&=2\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial x}\right)+\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial y}\right)
\end{align*}$$
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.
