Math 309B/C Winter 2012, Homework 7


Due February 29

  1. Solve $\left\{\begin{array}{cc} \alpha^{2} u_{xx}=u_{tt} & x\in [0,L], t>0 \\ u(0,t)=u(L,t)=0 & t>0 \\ u(x,0)=f(x) \\ u_{t}(x,0)=0\end{array}\right)$ for the following cases:
    1. $\alpha=1$, $L=2$, $f(x)=2-|x|$.
      2. Answer: Note that since, in our problem, $x$ is always at least zero, $|x|=x$ and so $f(x)=2-x$. Hence, \begin{align*} c_{n} & =\frac{2}{2}\int_{0}^{2}(2-x)\sin \frac{n\pi x}{2} =-\frac{2}{n\pi}(2-x)\cos\frac{n\pi x}{2}|_{0}^{2} -\int_{0}^{2} \frac{2}{n\pi}\cos \frac{n\pi x}{2} \\ & = \frac{4}{n\pi}\cos 0-\frac{4}{n^{2}\pi^{2}}\sin \frac{n\pi x}{2}|_{0}^{2} = \frac{4}{n\pi} \end{align*} and therefore, \[u(x,t)=\sum_{n=1}^{\infty} \frac{4}{n\pi}\sin \frac{n\pi x}{2}\cos \frac{n\pi t}{2}.\]
    2. $\alpha=2$, $L=\pi$, $f(x)=\sin x$
      3. Answer: Note that \[c_{n} = \frac{2}{\pi}\int_{0}^{\pi} \sin x\sin nx.\] Recall the formula \[\int_{-L}^{L}\sin\frac{n\pi x}{L}\sin \frac{m\pi x}{L} = \left\{\begin{array}{cc} 0 & n\neq m \\ L & m =n\end{array}\right.\] Now, since the integrand is an even function (since it is a product of two odd ones), this implies the formula \[\int_{0}^{L}\sin\frac{n\pi x}{L}\sin \frac{m\pi x}{L} = \left\{\begin{array}{cc} 0 & n\neq m \\ \frac{L}{2} & m =n\end{array}\right.\] Applying this to our earlier computation (with $L=\pi$ and $m=1$ in our case), this gives \[c_{n} = \frac{2}{\pi}\int_{0}^{\pi} \sin x\sin nx=\left\{\begin{array}{cc} 0 & n\neq 1 \\ \frac{2}{\pi}\cdot \frac{\pi}{2}=1 & n=1 \end{array}\right.\] Thus, \[u(x,t)=\sum_{n=1}^{\infty} c_{n} \sin n x \cos 2nt = \sin x \cos 2t.\]
    3. $\alpha=\pi$, $L=1$, $x(1-x)$
    4. $\alpha=3$, $L=3$, $f(x)=\left\{\begin{array}{cc} x & 0\leq x\leq 1 \\ -\frac{1}{2}x+\frac{3}{2} & 1\leq x \leq 3\end{array}\right.$.
    5. $\alpha=4$, $L=1$, $\sin x$ Answer: In this case, \begin{align*} c_{n} & =2\int_{0}^{1}\sin x\sin n\pi x = -2\cos x\sin n\pi x|_{0}^{1}+ 2n\pi \int_{0}^{1}\cos x\cos n\pi x \\ & = 0+2n\pi \sin x \cos n\pi|_{0}^{1} +2n^{2}\pi^{2}\int_{0}^{1}\sin x\sin n\pi x. \end{align*} Now subtract the final integral from both sides to get \[2(1-n^{2}\pi^{2})\int_{0}^{1}\sin x\sin n\pi x = 2n\pi \sin x \cos n\pi|_{0}^{1}=2n\pi \sin 1 \cos n\pi \] and hence \[c_{n}=2\int_{0}^{1}\sin x\sin n\pi x = \frac{2n\pi \sin 1 \cos n\pi}{1-n^{2}\pi^{2}}\] and thus the final solution is \[u(x,t)=\sum_{n=1}^{\infty}\frac{2n\pi \sin 1 \cos n\pi}{1-n^{2}\pi^{2}} \sin n\pi x\cos 4n\pi t.\]
  2. This problem is not correct, don't worry about it, my bad. Suppose $f$ is a function on $[0,L]$ such that $f^{(j)}(0)=f^{(j)}(L)=0$ for all $j$. Each derivative function has a sine series representation $f^{(j)}(x)=\sum_{n=1}^{\infty} b_{n}^{(j)}\sin \frac{ n\pi x}{L}$. What is $b_{n}^{(j)}$ in terms of $b_{n}^{(0)}$ for $j=1$ and $j=2$ (recall $f^{(0)}$ is just $f$). Hint: Integration by parts.
  3. Show that if $f$ and $g$ are $L$-periodic functions. Show that $f( x-\alpha t)+g( x+\alpha t)$ solves $\alpha^{2}u_{xx}=u_{tt}$. A function like $f( x +\alpha t)$ is called a traveling wave , since if you plot the graph for different values of $t$, it looks like the function is just shifting horizontally. Answer: Let $u(x,t)=f(x-\alpha t)+g(x+\alpha t)$. Note that by chain rule, \[u_{xx}(x,t)=f''(x-\alpha t)+g''(x+\alpha t)\] and \[u_{tt}(x,t)=\alpha^{2}f''(x-\alpha t)+\alpha^{2} g''(x+\alpha t),\] so that clearly, $\alpha^{2} u_{xx}=u_{tt}$.
  4. If $u(x,t)$ is the solution to the wave equation (with parameters $\alpha,L,$ and $f$, say), compute the period of $u(x,t)$ as a function of $t$, that is, find the smallest $P$ such that $u(x,t+P)=u(x,t)$ for all $t$ and all $x\in [0,L]$. Answer: Note that \[u(x,t)=\sum_{n=1}^{\infty} c_{n} \sin \frac{n\pi x}{L}\cos \frac{\alpha n\pi t}{L}.\] For $u(x,t)$ to be $P$-periodic in $t$, we need each term to also be $P$-periodic. The period of the $n$th term is $\frac{2L}{\alpha n}$ (since the period of $\cos x$ is $2\pi$ and hence the period of $\cos Mx$, for example, is $\frac{2\pi}{M}$). Hence, the minimal period of the whole series must be at least the period of the term with the largest period, namely, the first one $n=1$, which has period $\frac{2L}{\alpha}$. But this is a period for $u(x,t)$ since \[u(x,t+\frac{2L}{\alpha})=\sum_{n=1}^{\infty} c_{n} \sin \frac{n\pi x}{L}\cos \left(\frac{\alpha n\pi t}{L}+2n\pi \right)=\sum_{n=1}^{\infty} c_{n} \sin \frac{n\pi x}{L}\cos \frac{\alpha n\pi t}{L}=u(x,t).\] Thus, $\frac{2L}{\alpha}$ is the minimal period.