If $g(x, y)=f(u, v)$, where $u=x^{2}-y^{2}$ and $v=2 x y$ then

Prove that $\frac{d^{2} g}{d x^{2}}+\frac{d^{2} g}{d y^{2}}=4\left(x^{2}+y^{2}\right)\left[\frac{d^{2} f}{d u^{2}}+\frac{d^{2}f}{d v^{2}}\right]$

Solution :

$g(x, y)=f(u,v)$

$u=x^2-y^2, v=2xy$

$ \begin{aligned} \frac{\partial g}{\partial x} &=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x} +\frac{\partial f}{\partial v}\frac{\partial v}{\partial x} \\ &=\frac{\partial f}{\partial u}[2{x}]+\frac{\partial f}{\partial v}[2{y}] \\ \space \\ \frac{\partial}{\partial x} &=2x \frac{\partial}{\partial u}+2{y} \frac{\partial}{\partial v} \ldots \ldots \ldots \ldots .(1) \end{aligned} $

$\space $

$\begin{aligned} \frac{\partial^2 g}{\partial x^2} &=\frac{\partial}{\partial x}\left(\frac{\partial g}{\partial x}\right) \\ &= \left(2 {x} \frac{\partial}{\partial u} + 2y \frac{\partial}{\partial v}\right) \left(\frac{\partial f}{\partial u} (2 {x}) + \frac{\partial f}{\partial v} (2{y} )\right) \\ &=4{x^2} \frac{\partial ^{2} f}{\partial u^{2}}+4 {xy} \frac{\partial^{2} f}{\partial u \partial {v}}+4 {xy} \frac{\partial^{2} f}{\partial{v} \partial {u}} +4 {y}^{2} \frac{\partial^{2} f}{\partial v^{2}} \ldots \ldots \ldots \ldots(2) \\ \space \\ \frac{\partial g}{\partial y} &=\frac{\partial f}{\partial u}(-2 y)+\frac{\partial f}{\partial v}(2 x) \\ \frac{\partial^2 g}{\partial y^2} &=\frac{\partial}{\partial y}\left(\frac{\partial g}{\partial y}\right) \\ &=\left((-2 y) \frac{\partial}{\partial u}+2 x \frac{\partial}{\partial v}\right) \left(\frac{\partial f}{\partial u}(-2 y)+\frac{\partial f}{\partial v}(2 x)\right) \\ &=4 y^{2} \frac{\partial^{2} f}{\partial u^{2}}-4 x y \frac{\partial^{2} f}{\partial u \partial v}-4 x y \frac{\partial^{2} f}{\partial v \partial u}+4 x^{2} \frac{\partial^{2} f}{\partial v^{2}} \ldots \ldots \ldots \ldots(3) \end{aligned} $

Adding $(2)$ and $(3)$,

$ \begin{aligned} \frac{\partial^{2} g}{\partial x^{2}}+\frac{\partial^{2} g}{\partial y^{2}} =4 x^{2}\left(\frac{d^{2} f}{\partial u^{2}} + \frac {\partial^{2} f}{\partial v^{2}}\right)+ 4 y^{2}\left(\frac{\partial^{2} f}{\partial u^{2}} + \frac{\partial^{2} f}{\partial v^{2}}\right) \\ =4\left(x^{2}+y^{2}\right)\left(\frac{\partial^{2} f}{\partial u^{2}}+\frac{\partial^{2} f}{\partial v^{2}}\right) \end{aligned} $