Question

Show that the differential equation (on the left) is a solution of the function (on the right)

d^2u/dt^2 = a^2 * (d^2u/dx^2) u(x,t) = f(x-at) + g(x+at)

Answer 1

We have to show that

$\frac{\partial ^{2}u}{\partial t^{2}}=a^{2}\frac{\partial ^{2}u}{\partial x^{2}}$

for $\frac{\partial ^{2}u}{\partial t^{2}}$ we have

$\frac{\partial ^{2}u}{\partial t^{2}}=a^{2}\frac{\partial ^{2}u}{\partial x^{2}}$

$\frac{\partial ^{2}u}{\partial t^{2}}=\frac{\partial ^{2}[f(x-at)+g(x+at)]}{\partial t^{2}}$

$=\frac{\partial }{\partial t}[\frac{\partial[f(x-at)+g(x+at)] }{\partial t}]$

$\frac{\partial }{\partial t}[-a\cdot f'(x-at)+a\cdot g'(x+at)]$

$=a^{2}f''(x-at)+a^{2}g''(x+at)$

$=a^{2}[f''(x-at)+g''(x+at)].............(i)$

similarly,

$\frac{\partial ^{2}u}{\partial x^{2}}=\frac{\partial ^{2}[f(x-at)+g(x+at)]}{\partial x^{2}}$

$=\frac{\partial }{\partial x}[\frac{\partial[f(x-at)+g(x+at)] }{\partial x}]$

$=\frac{\partial }{\partial x}[f'(x-at)+g'(x+at)]$

$=f''(x-at)+g''(x+at).......(ii)$  

Comparing i and ii we get  

$a^{2}\frac{\partial ^{2}u}{\partial x^{2}}=\frac{\partial ^{2}u}{\partial t^{2}}$

Hence proved