Chapter 16, Problem 44P

(a)

To determine

To shows: That the wave function $y\left(x,t\right)=x^2+v^2{t}^2$ is a solution to the linear wave equation.

Explanation of Solution

Any function is a solution of linear wave equation in general if it satisfies the equation completely.

The linear wave equation in general is,

$\frac{{\partial }^{2}y}{\partial x^2}=\frac{1}{v^2}\frac{{\partial }^{2}y}{\partial t^2}$

The given wave function is,

$y\left(x,t\right)=x^2+v^2{t}^2$ (I)

Differentiate equation (I) partially with respect to $x$.

$\begin{array}{c}\frac{\partial y\left(x,t\right)}{\partial x}=\frac{\partial }{\partial x}\left(x^2+v^2{t}^2\right)\\ =2x\end{array}$

Again differentiate partially with respect to $x$.

$\begin{array}{c}\frac{{\partial }^{2}y\left(x,t\right)}{\partial x^2}=\frac{\partial }{\partial x}\left(2x\right)\\ =2\end{array}$ (II)

Differentiate equation (I) partially with respect to $t$.

$\begin{array}{c}\frac{\partial y\left(x,t\right)}{\partial t}=\frac{\partial }{\partial t}\left(x^2+v^2{t}^2\right)\\ =v^2\left(2t\right)\end{array}$

Again differentiate partially with respect to $t$.

$\begin{array}{c}\frac{{\partial }^{2}y\left(x,t\right)}{\partial t^2}=\frac{\partial }{\partial t}\left\{v^2\left(2t\right)\right\}\\ =2v^2\\ \frac{1}{v^2}\frac{{\partial }^{2}y\left(x,t\right)}{\partial t^2}=2\\ =\frac{{\partial }^{2}y\left(x,t\right)}{\partial x^2}\end{array}$

Conclusion:

Therefore, the wave function $y\left(x,t\right)=x^2+v^2{t}^2$ is a solution to the linear wave equation.

(b)

To determine

To shows: That the wave function $y\left(x,t\right)=x^2+v^2{t}^2$ can be written as $f\left(x+vt\right)+g\left(x-vt\right)$ and determine the functional form of $f$ and $g$.

Answer to Problem 44P

The functional form of $f$ is $\frac{1}{2}{\left(x+vt\right)}^2$ and the functional form of $g$ is $\frac{1}{2}{\left(x-vt\right)}^2$.

Explanation of Solution

It can be proved as,

$\begin{array}{c}f\left(x+vt\right)+g\left(x-vt\right)=\frac{1}{2}{\left(x+vt\right)}^2+\frac{1}{2}{\left(x-vt\right)}^2\\ =\frac{1}{2}\left(x^2+v^2{t}^2+2xvt\right)+\frac{1}{2}\left(x^2+v^2{t}^2-2xvt\right)\\ =\frac{1}{2}\times 2\left(x^2+v^2{t}^2\right)\\ =y\left(x,t\right)\end{array}$

Therefore,

The functional form of $f=\frac{1}{2}{\left(x+vt\right)}^2$

The functional form of $g=\frac{1}{2}{\left(x-vt\right)}^2$

Conclusion:

Therefore, the functional form of $f$ is $\frac{1}{2}{\left(x+vt\right)}^2$ and the functional form of $g$ is $\frac{1}{2}{\left(x-vt\right)}^2$.

(c)

To determine

Repeat part (a) and part (b) for the function $y\left(x,t\right)=\sin \left(x\right)\cos \left(vt\right)$.

Explanation of Solution

Section 1:

Any function is a solution of linear wave equation in general if it satisfies the equation completely.

To shows: That the wave function $y\left(x,t\right)=\sin \left(x\right)\cos \left(vt\right)$ is a solution to the linear wave equation.

Introduction: Any function is a solution of linear wave equation in general if it satisfies the equation completely.

The given wave function is,

$y=\sin \left(x\right)\cos \left(vt\right)$ (III)

Differentiate equation (I) partially with respect to $x$.

$\begin{array}{c}\frac{\partial y\left(x,t\right)}{\partial x}=\frac{\partial }{\partial x}\left(\sin \left(x\right)\cos \left(vt\right)\right)\\ =\cos \left(x\right)\cos \left(vt\right)\end{array}$

Again differentiate partially with respect to $x$.

$\begin{array}{c}\frac{{\partial }^{2}y\left(x,t\right)}{\partial x^2}=\frac{\partial }{\partial x}\left(\cos \left(x\right)\cos \left(vt\right)\right)\\ =-\sin \left(x\right)\cos \left(vt\right)\end{array}$ (VI)

Differentiate equation (I) partially with respect to $t$.

$\begin{array}{c}\frac{\partial y\left(x,t\right)}{\partial t}=\frac{\partial }{\partial t}\left(\sin \left(x\right)\cos \left(vt\right)\right)\\ =-v\sin \left(x\right)\sin \left(vt\right)\end{array}$

Again differentiate partially with respect to $t$.

$\begin{array}{c}\frac{{\partial }^{2}y\left(x,t\right)}{\partial t^2}=\frac{\partial }{\partial t}\left\{-v\sin \left(x\right)\sin \left(vt\right)\right\}\\ =-v^2\sin \left(x\right)\cos \left(vt\right)\\ \frac{1}{v^2}\frac{{\partial }^{2}y\left(x,t\right)}{\partial t^2}=-\sin \left(x\right)\cos \left(vt\right)\\ =\frac{{\partial }^{2}y\left(x,t\right)}{\partial x^2}\end{array}$

Conclusion:

Therefore, the wave function $y\left(x,t\right)=\sin \left(x\right)\cos \left(vt\right)$ is a solution to the linear wave equation.

Section 2:

To show: That the wave function $y\left(x,t\right)=\sin \left(x\right)\cos \left(vt\right)$ can be written as $f\left(x+vt\right)+g\left(x-vt\right)$ and determine the functional form of $f$ and $g$.

Answer: The functional form of $f$ is $\frac{1}{2}\sin \left(x+vt\right)$ and the functional form of $g$ is $\frac{1}{2}\sin \left(x-vt\right)$.

From the trigonometry,

$\sin \left(x+vt\right)=\sin x\cos vt+\cos x\sin vt$ (I)

$\sin \left(x-vt\right)=\sin x\cos vt-\cos x\sin vt$ (II)

Add equation (I) and (II).

$\begin{array}{c}\sin \left(x+vt\right)+\sin \left(x-vt\right)=\sin x\cos vt+\cos x\sin vt+\sin x\cos vt-\cos x\sin vt\\ =2\sin x\cos vt\\ \frac{1}{2}\sin \left(x+vt\right)+\frac{1}{2}\sin \left(x-vt\right)=\sin \left(x\right)\cos \left(vt\right)\end{array}$

Therefore,

The functional form of $f=\frac{1}{2}\sin \left(x+vt\right)$

The functional form of $g=\frac{1}{2}\sin \left(x-vt\right)$

Conclusion:

Therefore, the functional form of $f$ is $\frac{1}{2}\sin \left(x+vt\right)$ and the functional form of $g$ is $\frac{1}{2}\sin \left(x-vt\right)$.