# Show that each of the following functions is a solution of the wave equation u_{tt}=a^2u_{xx}....

## Question:

Show that each of the following functions is a solution of the wave equation

{eq}u_{tt}=a^2u_{xx},(\text{ with }a\ne 0). {/eq}

(a)

{eq}u= \sin (kx)\sin (akt). {/eq}

(b)

{eq}u=\frac{t}{(a^2t^2-x^2)}. {/eq}

(c)

{eq}u=(x-at)^6+(x+at)^6. {/eq}

(d)

{eq}u=\sin (x-at)+\ln(x+at). {/eq}

## Partial Differential Equations; Wave Equation and Solitons:

For any smooth function {eq}f(s) {/eq} defined over the real line and any real constant {eq}a {/eq}, the function {eq}u (x,\, t) = f(x - at) {/eq} is

a smooth function for all {eq}(x,\,t)\in\mathbb{R}^2. {/eq} Using the chain rule we can verify that {eq}u {/eq} satisfies the wave equation {eq}u_{tt} = a^2 u_{x x} {/eq} on all {eq}\mathbb{R}^2. {/eq}

We'll recall that the wave equation is a linear differential equation, which means that if {eq}\alpha\text{ and }\beta {/eq} are real constants and both {eq}u_1(x,t)\text{ and }u_2(x,t) {/eq} are solutions of the wave equation then the combination

{eq}u_3(x,t)=\alpha u_1(x,t)+\beta u_2(x,t), {/eq}

is also a solution of the wave equation.

We'll use the two important observations above to provide quick verification of the proposition in the problem statement.

We'll start considering any smooth function {eq}f(s) {/eq}, defined on some open interval {eq}I {/eq} of the real line and let {eq}a {/eq} be any real constant. We'll prove that the function {eq}u (x,\, t) = f(x - at) {/eq} satisfies the wave equation {eq}u_{tt} = a^2 u_{x x} {/eq} as long as {eq}x - at\in I. {/eq}

We can evaluate the left-hand side first:

{eq}\begin{align*} u (x,\, t) &= f(x - at)\\ &\text{differentiating with respect to }t,\\ u_t&\frac{d\, f}{ds}\cdot\frac{d\left( x-at \right)}{dt}\\ &=\frac{d\, f}{ds}\cdot(-a)\\ &\text{differentiating again with respect to }t,\\ U_{tt}&=\frac{d^2\, f}{ds^2}\cdot\frac{d\left( x-at \right)}{dt}\cdot(-a)\\ &=\frac{d^2 f }{ds^2}\cdot(-a)^2\\ &=a^2\, \frac{d^2f}{ds^2}\\ &\text{so,}\\ &\boxed{u_{tt}=a^2\,\frac{d^2 f }{ds^2}}. \end{align*} {/eq}

Now, we continue evaluating the right-hand side also:

{eq}\begin{align*} a^2\, u (x,\, t) &= a^2\, f(x - at)\\ &\text{differentiating with respect to }x,\\ a^2\, u_{x}&=a^2\,\frac{d\, f}{ds}\cdot\frac{d\left( x-at \right)}{dx}\\ &=a^2\,\frac{d\, f}{ds}\cdot(1)\\ &=a^2\,\frac{d\, f}{ds},\\ &\text{differentiating again with respect to }x,\\ a^2\,u_{xx}&=a^2\, \frac{d^2\, f}{ds^2}\cdot\frac{d\left( x-at \right)}{dx}\\ &=a^2\,\frac{d^2\, f}{ds^2}\cdot(1)\\ &\text{so,}\\ &\boxed{a^2\,u_{xx}=a^2\,\frac{d^2 f }{ds^2}}. \end{align*} {/eq}

Now, we can verify that:

{eq}u_{tt}=a^2\,\frac{d^2 f }{ds^2}=a^2\,u_{xx}, {/eq}

so the function {eq}u (x,\, t) = f(x - at) {/eq} is a solution to the wave equation {eq}u_{tt} = a^2 u_{x x} . {/eq}

Observation: we can substitute the constant {eq}-a {/eq} in place of {eq}a {/eq} in the arguments above, so we also have that {eq}u (x,\, t) = f(x + at) {/eq} ia a solution of the wave equation.

We'll use the fact that the wave equation is a linear differential equation, which means that if {eq}\alpha\text{ and }\beta {/eq} are real constants and both {eq}u_1(x,t)\text{ and }u_2(x,t) {/eq} are solutions of the wave equation then the combination

{eq}u_3(x,t)=\alpha u_1(x,t)+\beta u_2(x,t), {/eq}

is also a solution of the wave equation.

This fact, in combination with the earlier proof allows us to say that

for any pair {eq}f\text{ and }g {/eq} of smooth functions on an open interval {eq}I {/eq}, and any constants {eq}\alpha\text{ and }\beta {/eq} the function:

{eq}u(x,t)=\alpha f(x-at)+\beta g(x+at), {/eq}

is also a solution of the wave equation, as long as

{eq}\{x-at,\;x+at\}\subset I. {/eq}

Now we can verify the validity of the proposition in the problem statement.

For (a) {eq}u(x,t)= \sin (kx)\sin (akt). {/eq} We can use the product to sum trigonometric identity:

{eq}\sin(A)\sin(B)=\frac{1}{2}\cos(A-B)-\frac{1}{2}\cos(A+B), {/eq}

to rewrite the function {eq}u(x,t): {/eq}

{eq}\begin{align*} u(x,t)&= \sin (kx)\sin (akt)\\ &=\frac{1}{2}\cos(kx-akt)-\frac{1}{2}\cos(kx+akt)\\ &=\frac{1}{2}\cos(k[x-at])-\frac{1}{2}\cos(k[x+at]). \end{align*} {/eq}

So we have that

{eq}u(x,t)=\alpha f(x-at)+\beta g(x+at) {/eq}

for {eq}\alpha =\frac{1}{2}=-\beta {/eq} and {eq}f(s)=g(s)=\cos(ks) {/eq},

so {eq}u(x,t) {/eq} is a solution for the wave equation (for any {eq}(x,t)\in\mathbb{R}^2). {/eq}

For (b) {eq}u(x,t)=\frac{t}{(a^2t^2-x^2)}. {/eq} We can use the partial fraction decomposition

{eq}\frac{t}{(a^2t^2-x^2)}=\frac{1}{2a}\cdot\frac{1}{x+at}-\frac{1}{2a}\cdot\frac{1}{x-at}, {/eq}

to rewrite the function {eq}u(x,t) {/eq} as:

{eq}u(x,t)=\alpha f(x-at)+\beta g(x+at) {/eq}

for {eq}\alpha =-\frac{1}{2a}=-\beta {/eq} and {eq}f(s)=g(s)=\frac{1}{s} {/eq},

so {eq}u(x,t) {/eq} is a solution for the wave equation, for any {eq}(x,t) {/eq} with {eq}a^2t^2-x^2\ne 0. {/eq}

For (c) {eq}u(x,t)=(x-at)^6+(x+at)^6. {/eq} We can readily see that

{eq}u(x,t)=\alpha f(x-at)+\beta g(x+at) {/eq}

for {eq}\alpha =1=\beta {/eq} and {eq}f(s)=g(s)=s^6 {/eq},

so {eq}u(x,t) {/eq} is a solution for the wave equation (for any {eq}(x,t)\in\mathbb{R}^2). {/eq}

For (d) {eq}u(x,t)=\sin (x-at)+\ln(x+at). {/eq} In this case we see that

{eq}u(x,t)=\alpha f(x-at)+\beta g(x+at) {/eq}

for {eq}\alpha =1=\beta {/eq} and {eq}f(s)=\sin(s) {/eq} and {eq}g(s)=\ln(s) {/eq},

so {eq}u(x,t) {/eq} is a solution for the wave equation, for any {eq}(x,t) {/eq} such that {eq}x+at>0. {/eq}