Question

Show that each of the following functions is a solution of the wave equation \({u_t} = {a^2}{u_{xx}}\)

1. \(u = \sin \left( {kx} \right)\sin (axt)\)
2. \(u = \frac{t}{{\left( {{a^2}{t^2} - {x^2}} \right)}}\)
3. \(u = {\left( {x - at} \right)^6} + {(x + at)^6}\)
4. \(u = \sin \left( {x - at} \right) = \ln (x + at)\)

Short answer

The wave equation is a second order linear partial differential equation for the description of waves as they occur in classical physics as mechanical waved or electromagnetic waves.

Step-by-step solution

Double differentiation

\(u = \sin \left( {kx} \right)\sin (axt)\)

Then,

\(\begin{aligned}{l}{u_t} = ak\sin \left( {kx} \right)\cos x(axt)\\{u_t} = - {a^2}{k^2}\sin \left( {kx} \right)\sin x(axt)\\{u_x} = k\cos \left( {kx} \right)\sin (axt)\\{u_{xx}} = - {k^2}\sin \left( {kx} \right)\sin (axt)\end{aligned}\)

Now,\({u_{tt}} = - {a^2}{k^2}\sin \left( {kx} \right)\sin (axt)\)\( = {a^2}{u_{xx}}\)

Hence, given function (1) is a solution of \({u_t} = {a^2}{u_{xx}}\)

Double differentiation with repeat to “t”

Given \(u = \frac{t}{{\left( {{a^2}{t^2} - {x^2}} \right)}}\)

Then\({u_t} = \frac{t}{{\left( {{a^2}{t^2} - {x^2}} \right)}} - t{\left( {{a^2}{t^2} - {x^2}} \right)^{ - 2}}\left( {2{a^2}t} \right)\)

\({u_t} = \frac{{\left( {{a^2}{t^2} - {x^2}} \right) - 2{a^2}{t^2}}}{{{{\left( {{a^2}{t^2} - {x^2}} \right)}^2}}}\)

\({u_t} = \frac{{ - \left( {{x^2} + {a^2}{t^2}} \right)}}{{{{\left( {{a^2}{t^2} - {x^2}} \right)}^2}}}\)

\({u_{tt}} = \frac{{ - 2{a^2}{t^2}}}{{{{\left( {{a^2}{t^2} - {x^2}} \right)}^2}}} + \frac{{2\left( {{x^2} + {a^2}{t^2}} \right)(2{a^2}t)}}{{{{\left( {{a^2}{t^2} - {x^2}} \right)}^3}}}\)

\({u_{tt}} = \frac{{ - 2{a^2}{t^2}}}{{{{\left( {{a^2}{t^2} - {x^2}} \right)}^2}}} + \frac{{ - 2{a^4}{t^3} + 2{a^2} + {x^2} + 4{a^2}{x^2} + 4{a^4}{t^3}}}{{{{\left( {{a^2}{t^2} - {x^2}} \right)}^3}}}\)

\({u_{tt}} = - \frac{{2{a^2}t({a^2}{t^2} + 3{x^2})}}{{{{({a^2}{t^2} - {x^2})}^3}}}\)

Partial differentiating (2) equation

\({u_x} = \frac{{2tx}}{{{{\left( {{a^2}{t^2} - {x^2}} \right)}^2}}}\)

\({u_{xx}} = \frac{{2t}}{{{{\left( {{a^2}{t^2} - {x^2}} \right)}^2}}} - \frac{{4tx( - 2x)}}{{{{\left( {{a^2}{t^2} - {x^2}} \right)}^3}}}\)

\({u_{xx}} = \frac{{2{a^2}{t^3} - 2t{x^2} + t{x^2}}}{{{{\left( {{a^2}{t^2} - {x^2}} \right)}^3}}}\)

\({u_{xx}} = \frac{{2{a^2}{t^3} - 2t{x^2} + t{x^2}}}{{{{\left( {{a^2}{t^2} - {x^2}} \right)}^3}}}\)

\({u_{xx}} = - \frac{{2t({a^2}{t^2} + 3{x^2})}}{{{{({a^2}{t^2} - {x^2})}^3}}}\)

\({u_{tt}} = {a^2}\left( {\frac{{2t({a^2}{t^2} + 3{x^2})}}{{{{({a^2}{t^2} - {x^2})}^3}}}} \right)\)\( = {a^2}{u_{xx}}\)

Hence, the given function (2) is a solution of \({u_t} = {a^2}{u_{xx}}\)

 Step 4: Double differentiation with respect to‘t’

c) Given\(u = {\left( {x - at} \right)^6} + {(x + at)^6}\) (3)

Then,\({u_{}} = 6{\left( {x - at} \right)^5}( - a) + {(x + at)^5}(a)\)

\({u_t} = - 6a{\left( {x - at} \right)^5} + 6a{(x + at)^5}\)

\({u_t} = 30{a^2}{\left( {x - at} \right)^4} + 30{a^2}{(x + at)^4}\)

Double differentiation with respect to‘x’

\({u_x} = 6a{\left( {x - at} \right)^5} + 6{(x + at)^5}\)

\({u_{xx}} = 30{\left( {x - at} \right)^4} + 30{(x + at)^4}\)

\({u_{tt}} = {a^2}\left( {30{{\left( {x - at} \right)}^4} + 30{{(x + at)}^4}} \right)\)\( = {a^2}{u_{xx}}\)

Hence, the given solution (3) is a solution of \({u_t} = {a^2}{u_{xx}}\)

Compare Double differentiation:

D) Given\(u = \sin \left( {x - at} \right) = \ln (x + at)\) (4)

Then,\({u_t} = - a\cos (x - at) + \frac{a}{{x + at}}\)

\({u_{tt}} = - {a^2}\sin (x - at) - \frac{{{a^2}}}{{{{(x + at)}^2}}}\)

\({u_x} = \cos (x - at) + \frac{1}{{x + at}}\)

\({u_{xx}} = - \sin (x - at) - \frac{1}{{{{(x + at)}^3}}}\)

\({u_{tt}} = {a^2}\left( { - \sin (x - at) - \frac{1}{{{{(x + at)}^2}}}} \right)\)\( = {a^2}{u_{xx}}\)

Hence, the given function (4) is the solution of equation \({u_t} = {a^2}{u_{xx}}\)