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

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MATH257 December 2011

• Q1 • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) •

Other MATH257 Exams

• December 2011 •

Question 02 (a)
Consider the wave equation $u_{tt}=u_{xx},\,$ with initial conditions $u(x,0)=f(x)={\begin{cases}2\sin(2x)&0<x<\pi \\0&{\text{otherwise}}\end{cases}},\;\;u_{t}(x,0)=0.$ (a) Suppose the domain is infinite, $-\infty <x<\infty .$ Write down d'Alembert's solution, and sketch $u(x,0),\;u(x,\pi /4),\;u(x,\pi /2),\;{\text{and}}\;u(x,3\pi /4).\,$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Start by writing down the d'Alembert's formula.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution
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 $u\left(x,{\frac {3\pi }{4}}\right)$