Chapter 16, Problem 16.4P

To determine

Show that $\frac{{\partial }^{2}u}{\partial t^2}-c^2\frac{{\partial }^{2}u}{\partial x^2}=-4c^2\frac{\partial }{\partial \zeta }\frac{\partial u}{\partial \eta }$.

Answer to Problem 16.4P

Showed that $\frac{{\partial }^{2}u}{\partial t^2}-c^2\frac{{\partial }^{2}u}{\partial x^2}=-4c^2\frac{\partial }{\partial \zeta }\frac{\partial u}{\partial \eta }$.

Explanation of Solution

Given that two equations $\zeta =x-ct$ and $\eta =x+ct$.

Differentiate the equation $\zeta =x-ct$ with respect to $x$.

    $\begin{array}{c}\frac{\partial \zeta }{\partial x}=\frac{\partial }{\partial x}\left(x-ct\right)\\ =1\end{array}$        (I)

Differentiate the equation $\eta =x+ct$ with respect to $x$.

    $\begin{array}{c}\frac{\partial \zeta }{\partial x}=\frac{\partial }{\partial x}\left(x+ct\right)\\ =1\end{array}$        (II)

The variable $u$ is a function of  $\zeta$ and $\eta$. Then it can be written as $u=u\left(\zeta ,\eta \right)$.

Take the differential of $u=u\left(\zeta ,\eta \right)$ with respect to $x$.

    $\begin{array}{c}\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}\left[u\left(\zeta ,\eta \right)\right]\\ =\frac{\partial u}{\partial \zeta }\frac{\partial \zeta }{\partial x}+\frac{\partial u}{\partial \eta }\frac{\partial \eta }{\partial x}\end{array}$        (III)

Use equation (I) and (II) in (III).

    $\frac{\partial u}{\partial x}=\frac{\partial u}{\partial \zeta }+\frac{\partial u}{\partial \eta }$        (IV)

Differentiate equation (IV) with respect to $x$.

    $\begin{array}{c}\frac{{\partial }^{2}u}{\partial x}=\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial \zeta }+\frac{\partial u}{\partial \eta }\right)\\ =\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial \zeta }\right)+\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial \eta }\right)\\ =\frac{\partial }{\partial \zeta }\left(\frac{\partial u}{\partial \zeta }\right)\frac{\partial \zeta }{\partial x}+\frac{\partial }{\partial \eta }\left(\frac{\partial u}{\partial \zeta }\right)\frac{\partial \eta }{\partial x}+\frac{\partial }{\partial \zeta }\left(\frac{\partial u}{\partial \eta }\right)\frac{\partial \zeta }{\partial x}+\frac{\partial }{\partial \eta }\left(\frac{\partial u}{\partial \eta }\right)\frac{\partial \eta }{\partial x}\end{array}$        (V)

Use equation (I) and (II) in (V).

    $\begin{array}{c}\frac{{\partial }^{2}u}{\partial x}=\frac{\partial }{\partial \zeta }\left(\frac{\partial u}{\partial \zeta }\right)+\frac{\partial }{\partial \eta }\left(\frac{\partial u}{\partial \zeta }\right)+\frac{\partial }{\partial \zeta }\left(\frac{\partial u}{\partial \eta }\right)+\frac{\partial }{\partial \eta }\left(\frac{\partial u}{\partial \eta }\right)\\ =\frac{{\partial }^{2}u}{\partial \zeta }+\frac{{\partial }^{2}u}{\partial \eta \partial \zeta }+\frac{{\partial }^{2}u}{\partial \eta }+\frac{{\partial }^{2}u}{\partial \zeta \partial \eta }\\ =\frac{{\partial }^{2}u}{\partial \zeta }+2\frac{{\partial }^{2}u}{\partial \eta \partial \zeta }+\frac{{\partial }^{2}u}{\partial \eta }\end{array}$        (VI)

Differentiate the equation $\zeta =x-ct$ with respect to $t$.

    $\begin{array}{c}\frac{\partial \zeta }{\partial t}=\frac{\partial }{\partial t}\left(x-ct\right)\\ =-c\end{array}$        (VII)

Differentiate the equation $\eta =x+ct$ with respect to $t$.

    $\begin{array}{c}\frac{\partial \zeta }{\partial t}=\frac{\partial }{\partial t}\left(x+ct\right)\\ =c\end{array}$        (VIII)

Take the differential of $u=u\left(\zeta ,\eta \right)$ with respect to $t$.

    $\begin{array}{c}\frac{\partial u}{\partial t}=\frac{\partial }{\partial t}\left[u\left(\zeta ,\eta \right)\right]\\ =\frac{\partial u}{\partial \zeta }\frac{\partial \zeta }{\partial t}+\frac{\partial u}{\partial \eta }\frac{\partial \eta }{\partial t}\end{array}$        (IX)

Use equation (VII) and (VIII) in (IX).

    $\begin{array}{c}\frac{\partial u}{\partial t}=\frac{\partial u}{\partial \zeta }\left(-c\right)+\frac{\partial u}{\partial \eta }\left(c\right)\\ =c\left[-\frac{\partial u}{\partial \zeta }+\frac{\partial u}{\partial \eta }\right]\end{array}$        (X)

Differentiate equation (X) with respect to $t$.

    $\begin{array}{c}\frac{{\partial }^{2}u}{\partial t}=\frac{\partial }{\partial t}c\left(-\frac{\partial u}{\partial \zeta }+\frac{\partial u}{\partial \eta }\right)\\ =c\left[\frac{\partial }{\partial t}\left(-\frac{\partial u}{\partial \zeta }\right)+\frac{\partial }{\partial t}\left(\frac{\partial u}{\partial \eta }\right)\right]\\ =c\left[\frac{\partial }{\partial \zeta }\left(-\frac{\partial u}{\partial \zeta }\right)\frac{\partial \zeta }{\partial t}+\frac{\partial }{\partial \eta }\left(-\frac{\partial u}{\partial \zeta }\right)\frac{\partial \eta }{\partial t}+\frac{\partial }{\partial \zeta }\left(\frac{\partial u}{\partial \eta }\right)\frac{\partial \zeta }{\partial t}+\frac{\partial }{\partial \eta }\left(\frac{\partial u}{\partial \eta }\right)\frac{\partial \eta }{\partial t}\right]\end{array}$        (XI)

Use equation (VII) and (VIII) in (XI).

    $\begin{array}{c}\frac{{\partial }^{2}u}{\partial t^2}=c\left[\frac{\partial }{\partial \zeta }\left(-\frac{\partial u}{\partial \zeta }\right)\left(-c\right)+\frac{\partial }{\partial \eta }\left(-\frac{\partial u}{\partial \zeta }\right)c+\frac{\partial }{\partial \zeta }\left(\frac{\partial u}{\partial \eta }\right)\left(-c\right)+\frac{\partial }{\partial \eta }\left(\frac{\partial u}{\partial \eta }\right)c\right]\\ =c^2\left[\frac{{\partial }^{2}u}{\partial \zeta }-\frac{{\partial }^{2}u}{\partial \eta \partial \zeta }+\frac{{\partial }^{2}u}{\partial \eta }-\frac{{\partial }^{2}u}{\partial \zeta \partial \eta }\right]\\ =c^2\left[\frac{{\partial }^{2}u}{\partial \zeta }-2\frac{{\partial }^{2}u}{\partial \eta \partial \zeta }+\frac{{\partial }^{2}u}{\partial \eta }\right]\\ \frac{1}{c^2}\frac{{\partial }^{2}u}{\partial t^2}=\frac{{\partial }^{2}u}{\partial \zeta }-2\frac{{\partial }^{2}u}{\partial \eta \partial \zeta }+\frac{{\partial }^{2}u}{\partial \eta }\end{array}$        (XII)

Subtract equation (XII) from (VI), and solve.

    $\begin{array}{c}\frac{{\partial }^{2}u}{\partial x^2}-\frac{1}{c^2}\frac{{\partial }^{2}u}{\partial t^2}=\frac{{\partial }^{2}u}{\partial \zeta }+2\frac{{\partial }^{2}u}{\partial \eta \partial \zeta }+\frac{{\partial }^{2}u}{\partial \eta }-\left[\frac{{\partial }^{2}u}{\partial \zeta }-2\frac{{\partial }^{2}u}{\partial \eta \partial \zeta }+\frac{{\partial }^{2}u}{\partial \eta }\right]\\ =4\frac{{\partial }^{2}u}{\partial \eta \partial \zeta }\\ c^2\frac{{\partial }^{2}u}{\partial x^2}-\frac{{\partial }^{2}u}{\partial t^2}=4c^2\frac{{\partial }^{2}u}{\partial \eta \partial \zeta }\\ \frac{{\partial }^{2}u}{\partial t^2}-c^2\frac{{\partial }^{2}u}{\partial x^2}=-4c^2\frac{\partial }{\partial \zeta }\left(\frac{\partial u}{\partial \eta }\right)\end{array}$        (XIII)

Conclusion:

Therefore, $\frac{{\partial }^{2}u}{\partial t^2}-c^2\frac{{\partial }^{2}u}{\partial x^2}=-4c^2\frac{\partial }{\partial \zeta }\frac{\partial u}{\partial \eta }$.