Suppose that $\mathit{w}\mathbf{=}\mathit{f}\left(x,y\right)$satisfies

$\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{w}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{w}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}$.

Put $\mathit{x}\mathbf{=}\mathit{u}\mathbf{+}\mathit{v}\mathbf{,}\mathit{y}\mathbf{=}\mathit{u}\mathbf{-}\mathit{v}$, and show that w satisfies $\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{w}}{\mathbf{\partial }\mathbf{u}\mathbf{\partial }\mathbf{v}}\mathbf{=}\mathbf{1}$. Hence solve the equation.

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:

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

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

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

Solve as:

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

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

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

Also:

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

Solve for the double differential as:

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

Similarly,

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

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

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

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