Chapter 9: Q9.4P (page 388)

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

Short Answer

Expert verified

The equation is proved as $\tilde{f} (z, t) = \int_{- \infty}^{\infty} \tilde{A} (k) e^{i (k z - ω t)} d x$ from the wave equation by separation of variables

Step by step solution

Expression for the wave equation:

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 $f (z, t)$ .

$f (z, t) = Z (z) T (t)$ ……. (2)

From equation (1),

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

Determine the general linear combination of separable solutions:

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 $Z (z)$ .

$Z (z) = A e^{i k z} + B e^{- i k z}$

Similarly,

$\frac{\partial^{2} T}{\partial t^{2}} = - {(k v)}^{2} T$

Write the equation for $T (t)$ .

$T (t) = C e^{i k v t} + D e^{- i k v t}$

Substitute the known values in equation (2).

$f (z, t) = (A e^{i k z} + B e^{- i k z}) (C e^{i k v t} + D e^{- i k v t}) f (z, t) = A_{1} e^{i (k z + k v t)} + A_{2} e^{i (k z - k v t)} + A_{3} e^{i (k z + k v t)} + A_{4} e^{i (- k z - k v t)}$

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

$f (z, t) = \int_{0}^{\infty} [A_{1} e^{i (k z + k v t)} + A_{2} e^{i (k z - k v t)} + A_{3} e^{i (k z + k v t)} + A_{4} e^{i (- k z - k v t)}] d k$ ….. (5)

Determine Equation 9.20 directly from the wave equation by separation of variables:

Since, $ω = k v$

Substitute the above value in equation (5).

$f (z, t) = \int_{- \infty}^{\infty} [A_{1} k e^{i (k z + ω t)} + A_{2} k e^{i (k z + ω t)}] d k$

Using Euler’s formula,

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

Rewrite the function as,

$f (z, t) = \int_{- \infty}^{\infty} [A_{1} k e^{i (k z + ω t)} + A_{2} k e^{i (k z - ω t)}] d k R e (f) = \int_{- \infty}^{\infty} [R e (A_{1}) \cos (k z + ω t) + I m (A_{1}) \sin (k z + ω t) + R e (A_{2}) \cos (k z - ω t) + I m (A_{2}) \sin (k z - ω t)] d k$

Combine the first term $\cos (k z + ω t) = \cos (- k z - ω t)$ with the third term $\cos (k z - ω t)$ and second term $\sin (k z - ω t) = - \sin (- k z - ω t)$ with the fourth term $\sin (k z - ω t)$

Hence, the equation becomes,

$\tilde{f} (z, t) = \int_{- \infty}^{\infty} \tilde{A} (k) e^{i (k z - ω t)} d x$

Therefore, the equation is proved as $\tilde{f} (z, t) = \int_{- \infty}^{\infty} \tilde{A} (k) e^{i (k z - ω t)} d x$ from the wave equation by separation of variables.