91Ó°ÊÓ

Consider the function $v=v\left(x,y\right)$ has independent variables as $x=x\left(s,t\right)$and $y=y\left(s,t\right)$. Then, the corresponding chain rule for the function is:

$$\begin{gathered} \frac{∂v}{∂s}=\frac{∂v}{∂x}\frac{∂x}{∂s}+\frac{∂v}{∂y}\frac{∂y}{∂s} \\ \frac{∂v}{∂t}=\frac{∂v}{∂x}\frac{∂x}{∂t}+\frac{∂v}{∂y}\frac{∂y}{∂t} \end{gathered}$$

The given function is $w=f\left(x,y\right)$ which satisfies:

$\frac{{∂}^{2}w}{∂x^2}-\frac{{∂}^{2}w}{∂y^2}=1$

Solve as:

$$\begin{gathered} x=u+v \\ y=u-v \\ ⇒u=\frac{x+y}{2} \\ \mathrm{and}v=\frac{x-y}{2} \end{gathered}$$

From $w=f\left(x,y\right)$solve as:

$$\begin{gathered} \frac{∂w}{∂x}=\frac{∂w}{∂u}\frac{∂u}{∂x}+\frac{∂w}{∂v}\frac{∂v}{∂x} \\ \frac{∂w}{∂x}=\frac{1}{2}\frac{∂w}{∂u}+\frac{1}{2}\frac{∂w}{∂v} \\ \frac{∂}{∂x}=\frac{1}{2}\frac{∂}{∂u}+\frac{1}{2}\frac{∂}{∂v} \end{gathered}$$ …… (1)

Also:

$$\begin{gathered} \frac{∂w}{∂y}=\frac{∂w}{∂u}\frac{∂u}{∂y}+\frac{∂w}{∂v}\frac{∂v}{∂y} \\ \frac{∂w}{∂y}=\frac{1}{2}\frac{∂w}{∂u}-\frac{1}{2}\frac{∂w}{∂v} \\ \frac{∂}{∂y}=\frac{1}{2}\frac{∂}{∂u}-\frac{1}{2}\frac{∂}{∂v} \end{gathered}$$ …… (2)

Solve for the double differential as:

$$\begin{gathered} \frac{{∂}^{2}w}{∂x^2}=\frac{∂}{∂x}\left(\frac{∂w}{∂x}\right) \\ =\left(\frac{1}{2}\frac{∂}{∂u}+\frac{1}{2}\frac{∂}{∂v}\right)\left(\frac{1}{2}\frac{∂w}{∂u}+\frac{1}{2}\frac{∂w}{∂v}\right) \\ =\frac{1}{4}\frac{{∂}^{2}w}{∂u^2}+\frac{1}{2}\frac{{∂}^{2}w}{∂u∂v}+\frac{1}{4}\frac{{∂}^{2}w}{∂v^2} \end{gathered}$$

Similarly,

$$\begin{gathered} \frac{{∂}^{2}w}{∂y^2}=\frac{∂}{∂y}\left(\frac{∂w}{∂y}\right) \\ =\left(\frac{1}{2}\frac{∂}{∂u}-\frac{1}{2}\frac{∂}{∂v}\right)\left(\frac{1}{2}\frac{∂w}{∂u}-\frac{1}{2}\frac{∂w}{∂v}\right) \\ =\frac{1}{4}\frac{{∂}^{2}w}{∂u^2}-\frac{1}{2}\frac{{∂}^{2}w}{∂u∂v}+\frac{1}{4}\frac{{∂}^{2}w}{∂v^2} \end{gathered}$$

From $\frac{{∂}^{2}w}{∂x^2}-\frac{{∂}^{2}w}{∂y^2}=1$, solve as:

$$\begin{gathered} \frac{{∂}^{2}w}{∂x^2}-\frac{{∂}^{2}w}{∂y^2}=1 \\ \frac{1}{4}\frac{{∂}^{2}w}{∂u^2}+\frac{1}{2}\frac{{∂}^{2}w}{∂u∂v}+\frac{1}{4}\frac{{∂}^{2}w}{∂v^2}-\frac{1}{4}\frac{{∂}^{2}w}{∂u^2}+\frac{1}{2}\frac{{∂}^{2}w}{∂u∂v}-\frac{1}{4}\frac{{∂}^{2}w}{∂v^2}=1 \\ \frac{1}{2}\frac{{∂}^{2}w}{∂u∂v}+\frac{1}{2}\frac{{∂}^{2}w}{∂u∂v}=1 \\ \frac{{∂}^{2}w}{∂u∂v}=1 \end{gathered}$$

Hence proved, w satisfies $\frac{{∂}^{2}w}{∂u∂v}=1$.