Question: Obtain Eq. 9.20 directly from the wave equation by separation of variables.

Write the expression for the wave equation.

$\frac{{\partial }^{2}f}{\partial z^2}=\frac{1}{v^2}\frac{{\partial }^{2}f}{\partial t^2}$ …… (1)

Consider the equation for the function $f\left(z,t\right)$.

$f\left(z,t\right)=Z\left(z\right)T\left(t\right)$ ……. (2)

From equation (1),

$\frac{{\partial }^2}{\partial z^2}\left(Z\left(z\right)T\left(t\right)\right)=\frac{1}{v^2}\frac{{\partial }^2}{\partial t^2}\left(Z\left(z\right)T\left(t\right)\right)$ ……. (3)

Divide equation (3) by TZ.

$\frac{1}{Z}\frac{\partial Z^2}{\partial Z^2}=\frac{1}{v^{2}T}\frac{{\partial }^{2}T}{\partial t^2}$ …… (4)

From equation (4), as the L.H.S depends on Z and R.H.S depends on t, the new equation becomes,

$\frac{{\partial }^{2}Z}{\partial z^2}=-K^{2}Z$

Write the equation for $Z\left(z\right)$.

$Z\left(z\right)=A{e}^{ikz}+B{e}^{-ikz}$

Similarly,

$\frac{{\partial }^{2}T}{\partial t^2}=-{\left(kv\right)}^{2}T$

Write the equation for $T\left(t\right)$.

$T\left(t\right)=C{e}^{ikvt}+D{e}^{-ikvt}$

Substitute the known values in equation (2).

$$\begin{gathered} f\left(z,t\right)=\left(A{e}^{ikz}+B{e}^{-ikz}\right)\left(C{e}^{ikvt}+D{e}^{-ikvt}\right) \\ f\left(z,t\right)=A_1{e}^{i\left(kz+kvt\right)}+A_2{e}^{i\left(kz-kvt\right)}+A_3{e}^{i\left(kz+kvt\right)}+A_4{e}^{i\left(-kz-kvt\right)} \end{gathered}$$

Hence, the general linear combination of separable solution will be,

$f\left(z,t\right)={\int }_0^{\infty }\left[A_1{e}^{i\left(kz+kvt\right)}+A_2{e}^{i\left(kz-kvt\right)}+A_3{e}^{i\left(kz+kvt\right)}+A_4{e}^{i\left(-kz-kvt\right)}\right]dk$ ….. (5)

Since,$\omega =kv$

Substitute the above value in equation (5).

$f\left(z,t\right)={\int }_{-\infty }^{\infty }\left[A_{1}k{e}^{i\left(kz+\omega t\right)}+A_{2}k{e}^{i\left(kz+\omega t\right)}\right]dk$

Using Euler’s formula,

$e^{ix}=\cos x+i\sin x$

Rewrite the function as,

$$\begin{gathered} f\left(z,t\right)={\int }_{-\infty }^{\infty }\left[A_{1}k{e}^{i\left(kz+\omega t\right)}+A_{2}k{e}^{i\left(kz-\omega t\right)}\right]dk \\ Re\left(f\right)={\int }_{-\infty }^{\infty }\left[Re\left(A_1\right)\cos \left(kz+\omega t\right)+Im\left(A_1\right)\sin \left(kz+\omega t\right)+Re\left(A_2\right)\cos \left(kz-\omega t\right)+Im\left(A_2\right)\sin \left(kz-\omega t\right)\right]dk \end{gathered}$$

Combine the first term $\cos \left(kz+\omega t\right)=\cos \left(-kz-\omega t\right)$with the third term $\cos \left(kz-\omega t\right)$and second term $\sin \left(kz-\omega t\right)=-\sin \left(-kz-\omega t\right)$with the fourth term $\sin \left(kz-\omega t\right)$

Hence, the equation becomes,

$\tilde{f}\left(z,t\right)={\int }_{-\infty }^{\infty }\tilde{A}\left(k\right)e^{i\left(kz-\omega t\right)}dx$

Therefore, the equation is proved as $\tilde{f}\left(z,t\right)={\int }_{-\infty }^{\infty }\tilde{A}\left(k\right)e^{i\left(kz-\omega t\right)}dx$from the wave equation by separation of variables.