Science:Math Exam Resources/Courses/MATH257/December 2011/Question 02 (a)/Solution 1

For the one dimensional wave equation

${\begin{cases}u_{tt}=c^{2}u_{xx}\\u(x,0)=f(x)\\u_{t}(x,0)=g(x)\end{cases}}$

where $-\infty <x<\infty$ and $t>0\,$, the solution $u(x,t)$ is given by d'Alembert's formula

$u(x,t)={\frac {1}{2}}[f(x-ct)+f(x+ct)]+{\frac {1}{2c}}\int _{x-ct}^{x+ct}g(s)ds.$

For this question, we have $c=1\,$,

$f(x)={\begin{cases}2\sin(2x)&0<x<\pi \\0&{\text{otherwise}}\end{cases}}$

and $g(x)=0\,$. Therefore, the solution is given by

$u(x,t)={\frac {1}{2}}[f(x-t)+f(x+t)].\,$

This equation says that at time $t=0$, the original wave form splits into two parts, each with half the amplitude of the original. One part moves to the left at speed 1 and the other moves to the right at the same speed. At any time $t>0$, the solution of the wave equation is the sum of the left-traveling part and the right-traveling part.

At $t=0$, $u(x,0)$ is the given function $f(x)$.

The graph of $u(x,0)$

To find $u\left(x,{\frac {\pi }{4}}\right)$, one imagines that at $t=0$, the original wave form $f(x)$ splits into two equal pieces but with half the original amplitude. Hence, in this case, each part will be like $\sin(2x)$. One part will move towards the left at speed 1 and the other moves to the right at speed 1. At $t={\frac {\pi }{4}}$, one part has traveled ${\frac {\pi }{4}}$ units to the left and the other has traveled ${\frac {\pi }{4}}$ units to the right. The solution $u\left(x,{\frac {\pi }{4}}\right)$ is the sum of those two parts.

The graph of $u\left(x,{\frac {\pi }{4}}\right)$

Notice that in the above, the left traveling wave cancels with the right travelling wave in the interval from $x={\frac {\pi }{4}}$ to $x={\frac {3\pi }{4}}$.

Similarly, one finds the solutions for $u\left(x,{\frac {\pi }{2}}\right)$ and $u\left(x,{\frac {3\pi }{4}}\right)$.

The graph of $u\left(x,{\frac {\pi }{2}}\right)$

The graph of $u\left(x,{\frac {3\pi }{4}}\right)$