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

MATH257 December 2011

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  •  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.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $u\left(x,{\frac {3\pi }{4}}\right)$