$d^2u/dx^2 - d^2u/dy^2 = f(x,y)$ $\rightarrow$ $d^2u/dξdη = \frac{1}{4}f(\frac{1}{2}(ξ+η),\frac{1}{2}(η-ξ))$ $$\frac{\partial^2u}{\partial x^2} - \frac{\partial^2u}{\partial y^2} = f(x,y) \implies \frac{\partial^2u}{\partial ξ\partial η} = \frac{1}{4}f\left(\frac{1}{2}(ξ+η),\frac{1}{2}(η-ξ)\right)$$
Setting $ξ = x − y, η = x + y$ I get $x = \frac{1}{2}(ξ+η)$ and $y=\frac{1}{2}(η-ξ)$
I cant seem to show that $\frac{\partial^2u}{\partial x^2} - \frac{\partial^2u}{\partial y^2} = 4\times \frac{\partial^2u}{\partial ξ\partial η}$
My working is:
$$\partial u/\partial x = \frac{1}{2}\frac{\partial u} {\partial \xi} + \frac{1}{2}\frac{\partial u}{\partial \eta}$$
$$\partial u/\partial x = -\frac{1}{2}\frac{\partial u} {\partial \xi} + \frac{1}{2}\frac{\partial u}{\partial \eta}$$
$$\frac{\partial^2u}{\partial x^2} = \frac{1}{4}\frac{\partial^2u}{\partial \xi^2} + \frac{1}{4}\frac{\partial^2u}{\partial \eta^2}+ \frac{1}{2}\frac{\partial^2u}{\partial ξ\partial η}$$
$$\frac{\partial^2u}{\partial y^2} = \frac{1}{4}\frac{\partial^2u}{\partial \xi^2} + \frac{1}{4}\frac{\partial^2u}{\partial \eta^2}- \frac{1}{2}\frac{\partial^2u}{\partial ξ\partial η}$$
Where am I going wrong?
Could someone please help?
 A: You applied the chain rule incorrectly.
Applying the chain rule twice, we get
\begin{align}
\frac{\partial^2u}{\partial x^2} &= \frac{\partial}{\partial x} \left( \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial x}+\frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial x} \right)
= \frac{\partial}{\partial x} \left( \frac{\partial u}{\partial \xi}+\frac{\partial u}{\partial \eta} \right)\\
&= \frac{\partial^2 u}{\partial \xi^2}\frac{\partial \xi}{\partial x}
+ \frac{\partial^2 u}{\partial \xi \partial \eta}\frac{\partial \eta}{\partial x}
+ \frac{\partial^2 u}{\partial \eta \partial \xi}\frac{\partial \xi}{\partial x}
+ \frac{\partial^2 u}{\partial \eta^2}\frac{\partial \eta}{\partial x}\\
&= \frac{\partial^2 u}{\partial \xi^2} + 2\frac{\partial^2 u}{\partial \xi \partial \eta} + \frac{\partial^2 u}{\partial \eta^2},
\end{align}
since 
$$ \frac{\partial \xi}{\partial x} = \frac{\partial \eta}{\partial x} = 1. $$
Analogously,
\begin{align}
\frac{\partial^2u}{\partial y^2} &= \frac{\partial}{\partial y} \left( \frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial y}+\frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial y} \right)
= \frac{\partial}{\partial x} \left( -\frac{\partial u}{\partial \xi}+\frac{\partial u}{\partial \eta} \right)\\
&= -\frac{\partial^2 u}{\partial \xi^2}\frac{\partial \xi}{\partial y}
- \frac{\partial^2 u}{\partial \xi \partial \eta}\frac{\partial \eta}{\partial y}
+ \frac{\partial^2 u}{\partial \eta \partial \xi}\frac{\partial \xi}{\partial y}
+ \frac{\partial^2 u}{\partial \eta^2}\frac{\partial \eta}{\partial y}\\
&= \frac{\partial^2 u}{\partial \xi^2} - 2\frac{\partial^2 u}{\partial \xi \partial \eta} + \frac{\partial^2 u}{\partial \eta^2},
\end{align}
since 
$$ \frac{\partial \xi}{\partial y} = -1, \quad\text{and}\quad = \frac{\partial \eta}{\partial y} = 1. $$
Subtracting the two gives the result.
