Math 309B/C Winter 2012, Homework 8


Due March 7

Below we will frequently use the formula \[ 2\int_{0}^{L}\sin \frac{n\pi x}{L}\sin \frac{m\pi x}{L} = \int_{-L}^{L}\sin \frac{n\pi x}{L}\sin \frac{m\pi x}{L} =\left\{\begin{array}{cc} 0 & n\neq m \\ L & n=m\end{array}\right.\] whenever $n$ and $m$ are integers.
  1. Suppose $v(x,t)$ solves \[\left\{\begin{array}{cc} \alpha^{2}v_{xx}=v_{tt} & t>0, x\in[0,L] \\ v(0,t)=v(L,t)=0 \\ v(x,0)=f(x) \\ v_{t}(x,0)=0 \end{array}\right.\] and $w(x,t)$ solves \[\left\{\begin{array}{cc} \alpha^{2}w_{xx}=w_{tt} & t>0, x\in[0,L] \\ w(0,t)=w(L,t)=0 \\ w(x,0)=0 \\ w_{t}(x,0)=g(x) \end{array}\right..\] Show that $u(x,t)=v(x,t)+w(x,t)$ solves \[\left\{\begin{array}{cc} \alpha^{2}u_{xx}=u_{tt} & t>0, x\in[0,L] \\ u(0,t)=u(L,t)=0 \\ u(x,0)=f(x) \\ u_{t}(x,0)=g(x) \end{array}\right.\] Answer: We just have to check that $u$ satisfies all the conditions, so \[\alpha^{2} u_{xx} = \alpha^{2}v_{xx}+\alpha^{2} w_{xx} = v_{tt}+w_{tt}=(v+w)_{tt}=u_{tt}\] \[u(0,t)=v(0,t)+w(0,t)=0, \;\;\; u(L,t)=v(L,t)+w(L,t)=0,\] \[u(x,0)=v(x,0)+w(x,0)=f(x)+0=f(x)\] \[u_{t}(x,0)=v_{t}(x,0)+w_{t}(x,0)=0+g(x)=g(x).\]
  2. Suppose that $f(x)=\sum_{n=1}^{\infty}c_{n}\sin\frac{n\pi x}{L}$ for $x\in [0,L]$ and $u(x,t)$ is the solution to the plucked string problem $\left\{\begin{array}{cc} \alpha^{2}u_{xx}=u_{tt} & t>0, x\in [0,L] \\ u(0,t)=u(L,t)=0 & t>0 \\ u(x,0)=f(x) \\ u_{t}(x,0)=0\end{array}\right.$.
    1. Define $h(x)=\left\{\begin{array}{cc} f(x) & x\in [0,L] \\ -f(-x) & x\in [-L,0]\end{array}\right.$, and everywhere else by $h(x+2L)=h(x)$ (so $h$ is 2L-periodic). Show that the Fourier series for $h$ in $[-L,L]$ is $\sum_{n=1}^{\infty}c_{n}\sin\frac{n\pi x}{L}$. Answer: Recall that the Fourier series for $h(x)$ has the form \[h(x)=\frac{a_{0}}{2} +\sum_{n=1}^{\infty} a_{n}\cos \frac{n\pi x}{L} + b_{n}\sin \frac{n\pi x}{L}.\] Note that $h(x)$ is odd, since, for $x\in [0,L]$ (and hence $-x\in [-L,0]$), \[h(-x)=-f(-(-x))=-f(x)=-h(x)\] and similarly if $x\in [-L,0]$. Thus, $a_{0}=a_{n}=0$ always. Which means really \[h(x) =\sum_{n=1}^{\infty} b_{n}\sin \frac{n\pi x}{L}\] where \[b_{n}=\frac{1}{L}\int_{-L}^{L}h(x)\sin \frac{n\pi x}{L}=\frac{2}{L}\int_{0}^{L}h(x) \sin\frac{n\pi x}{L} = \frac{2}{L}\int_{0}^{L}f(x)\sin\frac{n\pi x}{L}\] where in the last few steps we used the fact that $h(x)\sin \frac{n\pi x}{L}$ is even and $h(x)=f(x)$ for $x\geq 0$. But this quantity above is exactly $c_{n}$.
    2. Show that $u(x,t) = \frac{h(x+\alpha t)+h(x-\alpha t)}{2}.$ Hint: Write out the sum for the right side of this equation and use the angle formulas. Answer: Note that \[ h(x+\alpha t)= \sum_{n=1}^{\infty}c_{n}\sin\left(\frac{n\pi x}{L}+\frac{\alpha t n\pi }{L}\right) = \sum_{n=1}^{\infty}c_{n}\left( \sin\frac{n\pi x}{L} \cos\frac{\alpha t n\pi }{L}+\cos\frac{n\pi x}{L} \sin\frac{\alpha t n\pi }{L}\right) \] by the law of sines, and similarly, \[ h(x-\alpha t)= \sum_{n=1}^{\infty}c_{n}\left( \sin\frac{n\pi x}{L} \cos\frac{\alpha t n\pi }{L}-\cos\frac{n\pi x}{L} \sin\frac{\alpha t n\pi }{L}\right) .\] Adding these two together and dividing by two gives \[\sum_{n=1}^{\infty}c_{n} \sin\frac{n\pi x}{L} \cos\frac{\alpha t n\pi }{L} = u(x,t).\]
    3. In the problems below, you'll be given a problem of the form \[\left\{\begin{array}{cc} \alpha^{2}u_{xx}=u_{tt} & t>0, x\in[0,L] \\ u(0,t)=u(L,t)=0 \\ u(x,0)=f(x) \\ u_{t}(x,0)=g(x) \end{array}\right.\] The trick is to split this into two problems: first solve for $v(x,t)$ where $v$ solves \[\left\{\begin{array}{cc} \alpha^{2}v_{xx}=v_{tt} & t>0, x\in[0,L] \\ v(0,t)=v(L,t)=0 \\ v(x,0)=f(x) \\ v_{t}(x,0)=0 \end{array}\right.\] This has solution \[v(x,t)= \sum_{n=1}^\infty c_{n}\sin \frac{n\pi x}{L}\cos\frac{\alpha n\pi t}{L}, \;\;\; \mbox{ where } \;\;\; c_{n}=\frac{2}{L}\int_{0}^{L} f(x)\sin \frac{n\pi x}{L}.\] Then solve for $w(x,t)$, which solves \[\left\{\begin{array}{cc} \alpha^{2}w_{xx}=w_{tt} & t>0, x\in[0,L] \\ w(0,t)=w(L,t)=0 \\ w(x,0)=0 \\ w_{t}(x,0)=g(x) \end{array}\right..\] This has solution \[w(x,t)= \sum_{n=1}^\infty c_{n}\sin \frac{n\pi x}{L}\sin\frac{\alpha n\pi t}{L}, \;\;\; \mbox{ where } \;\;\; c_{n}=\frac{2}{\alpha n\pi}\int_{0}^{L} f(x)\sin \frac{n\pi x}{L}.\] Then $u(x,t)=v(x,t)+w(x,t)$ solves the original problem.
  3. In each case below, solve for $u(x,t)$, where $u$ gives the vertical displacement at time $t$ of a string fixed at $x=0$ and $x=L$ of initial displacement $f(x)$ and initial (vertical) velocity $g(x)$:
    1. $f(x)=\left\{\begin{array}{cc} \frac{x}{K} & 0\leq x\leq \frac{K}{2} \\ \frac{K-x}{K} & \frac{K}{2}\leq x\leq K\end{array}\right.$, $g(x)=0$, $\alpha=2$, $L=K$.
    2. $f(x)=0$, $L=1$, $g(x)=\sin \pi x+\sin 2\pi x+\sin 5\pi x$, $\alpha=1$. Answer: The solution to this will be of the form \[u(x,t)=\sum_{n=1}^{\infty}c_{n}\sin n\pi x\sin n\pi t\] where \[c_{n}=\frac{2}{1\cdot \pi n}\int_{0}^{1}(\sin \pi x+\sin 2\pi x + \sin 5\pi x)\sin n\pi x = \frac{1}{n\pi}\left( \underbrace{ 2\int_{0}^{1}\sin \pi x\sin n \pi x}_{(1)} +\underbrace{2\int_{0}^{1}\sin 2\pi x\sin n \pi x}_{(2)} +\underbrace{2\int_{0}^{1}\sin 5 \pi x\sin n \pi x}_{(3)}\right).\] Note that \[(1)=\left\{\begin{array}{cc} 1 & n=1 \\ 0 & n\neq 1\end{array}\right.,\] \[(2)=\left\{\begin{array}{cc} 1 & n=2 \\ 0 & n\neq 2\end{array}\right.,\] \[(3)=\left\{\begin{array}{cc} 1 & n=5 \\ 0 & n\neq 5\end{array}\right.,\] and thus \[c_{n} = \left\{\begin{array}{cc} = \frac{1}{\pi} & n=1 \\ 0 & n\neq 1 \\ \frac{1}{2\pi} & n=2 \\ 0 & n\neq 2 \\ \frac{1}{5\pi} & n=5 \\ 0 & n\neq 5 \\ 0 & n\neq 1,2,5\end{array}\right..\] Which means that there are only 3 terms in the series giving $u$: \[u(x,t)=c_{1}\sin 1\pi x\sin 1\pi t + c_{2}\sin 2\pi x\sin 2\pi t+c_{5}\sin 5\pi x\sin 5\pi t = \frac{1}{\pi}\sin \pi x\sin \pi t + \frac{1}{2\pi}\sin 2\pi x\sin 2\pi t+\frac{1}{5\pi}\sin 5\pi x\sin 5\pi t .\]
    3. $f(x)=\sin \frac{\pi x}{3}$, $L=2$, $\alpha=3$, $g(x)=\sin \pi x$ Answer: For this one, we split into two problems and solve them independently. We first pretend that $g(x)=0$ and solve for a function $v(x,t)$ so that $v(x,0)=f(x)$ and $v_{t}(x,0)=0$. \[c_{n}=\frac{2}{2}\int_{0}^{2} \sin\frac{\pi x}{3} \sin\frac{n\pi x}{2}.\] Note that we can't use that clever formula for computing this integral of a product of sines, so we have to do integration by parts twice: \begin{multline*} c_{n}=\int_{0}^{2} \sin\frac{\pi x}{3} \sin\frac{n\pi x}{2} =-\frac{2}{n\pi}\sin\frac{\pi x}{3}\cos\frac{n\pi x}{2}|_{0}^{2} +\frac{2}{3n}\int_{0}^{2}\cos\frac{\pi x}{3} \cos\frac{n\pi x}{2}\\ =-\frac{2}{n\pi}\sin\frac{2\pi }{3} \cos n\pi +\frac{4}{3n^{2}\pi}\cos\frac{\pi x}{3} \sin\frac{n\pi x}{2}|_{0}^{2}+\frac{4}{9n^{2}}\underbrace{\int_{0}^{2}\sin\frac{\pi x}{3} \sin\frac{n\pi x}{2}}_{=c_{n}}\\ =-\frac{2}{n\pi}\sin\frac{2\pi }{3} \cos n\pi+\frac{4}{9n^{2}}c_{n} \end{multline*} and solving both sides of this long equation for $c_{n}$ gives \[c_{n}=-\frac{\frac{2}{n\pi}\sin\frac{2\pi }{3} \cos n\pi}{1-\frac{4}{9n^{2}}}.\] Thus, the solution for this partial problem is \[v(x,t)=\sum_{n=1}^{\infty}-\frac{\frac{2}{n\pi}\sin\frac{2\pi }{3}\cos n\pi}{1-\frac{4}{9n^{2}}}\sin \frac{n\pi x}{2}\cos \frac{3n\pi t}{2}.\] Now we solve the problem for $w(x,t)$ with $w(x,0)=0$ and $w_{t}(x,0)=g(x)=\sin \pi x$ (that is, we pretend this time that $f(x)$ is zero). Now the solution will have the form \[w(x,t)=\sum_{n=1}^{\infty}c_{n}\sin \frac{n\pi x}{2}\sin\frac{3n\pi t}{2}\] where \[c_{n}=\frac{2}{3 n \pi}\int_{0}^{2}\sin \pi x\sin\frac{n\pi x}{2}\] Notice that \[2\int_{0}^{2}\sin \pi x\sin\frac{n\pi x}{2}=2\int_{0}^{2}\sin \frac{2\pi x}{2}\sin\frac{n\pi x}{2}=\left\{\begin{array}{cc} 0 & n\neq 2 \\ 2 & n=2\end{array}\right.\] so that \[c_{n} = \left\{\begin{array}{cc} 0 & n\neq 2 \\ \frac{2}{3\cdot 2\pi}\cdot 2=\frac{2}{3\pi} & n=2\end{array}\right.\] so the solution to this problem is \[w(x,t)=\sum_{n=1}^{\infty}c_{n}\sin \frac{n\pi x}{2}\sin\frac{3n\pi t}{2}=\frac{2}{3\pi}\sin 2\pi x\sin 6\pi t.\] The final solution to the entire problem is \[u(x,t)=v(x,t)+w(x,t)=\sum_{n=1}^{\infty}-\frac{\frac{2}{n\pi}\sin\frac{2\pi}{3}\cos n\pi}{1-\frac{4}{9n^{2}}}\sin \frac{n\pi x}{2}\cos \frac{3n\pi t}{2} +\frac{2}{3\pi}\sin 2n\pi x\sin 6\pi t.\]
    4. $f(x)=0$, $g(x)=x(x-1)$, $L=1$, $\alpha=4$
    5. $g(x)=x+\sin x$, $f(x)=\sin 2x$, $L=\pi$, $\alpha=4$
      Answer: Again, we split this into two problems. First, we solve for $v(x,t)$ which solves the problem assuming $v_{t}(x,0)=0$ and $v(x,0)=f(x)=\sin 2x$. \[c_{n}=\frac{2}{\pi}\int_{0}^{\pi}\sin 2x\sin n x = \frac{1}{\pi}\left(2\int_{0}^{\pi}\sin\frac{2\pi x}{\pi}\sin\frac{n\pi x}{\pi}\right)=\left\{\begin{array}{cc} 0 & n\neq 2 \\ \frac{1}{\pi} \cdot \pi =1 & n=2\end{array}\right.\] And the solution to this problem is \[v(x,t)=\sum_{n=1}^{\infty} c_{n} \sin n x\cos 4nt = \sin 2x\cos 8t.\] Now we pretend that $f(x)=0$ and solve for $w(x,t)$ where $w(x,0)=0$ and $w_{t}(x,0)=g(x)=x+\sin x$. Here, it makes a little more sense to split the problem further: we'll solve for $w^{(1)}$ and $w^{(2)}$ where $w^{(1)}_{t}(x,0)=x$ and $w^{(2)}_{t}(x,t)=\sin x$. In the end, we'll add these solutions together to get $w(x,t)$ (note that $w_{t}(x,0)=w^{(1)}(x,0)+w^{(2)}(x,0)=x+\sin x$, just as before). You could solve this problem the usual way, but this way will reduce some confusion about what to do with the coefficients $c_{n}$ later. Just watch: For $w^{(1)}$, the associated coefficients are \[c_{n}=\frac{2}{4n\pi}\int_{0}^{\pi} x\sin nx = -\frac{2}{4n^2\pi}x\cos nx|_{0}^{\pi}+\int_{0}^{\pi}\frac{2}{4n^2\pi }\cos nx = -\frac{2}{4n^2}\cos n\pi +\frac{4}{4n^{3}\pi^{2}}\sin nx|_{0}^{\pi}=-\frac{1}{2n^2}\cos n\pi,\] and so \[w^{(1)}(x,t)=\sum_{n=1}^{\infty}\frac{1}{2n^2}\cos n\pi \sin nx \sin 4nt\] and $w^{(2)}$ has associated constants \[c_{n}=\frac{2}{4n\pi}\int_{0}^{\pi} \sin x \sin nx = \frac{1}{4n\pi} \left(\int_{0}^{\pi}\sin\frac{x\pi}{\pi}\sin\frac{n\pi}{\pi}\right)=\left\{\begin{array}{cc} 0 & n\neq 1 \\ \frac{1}{4\cdot 1\cdot \pi}\pi=\frac{1}{4} & n=1\end{array}\right.,\] so \[w^{(2)}(x,t)=\sum_{n=1}^{\infty}c_{n}\sin nx \sin 4nt= \frac{1}{4}\sin x\sin 4t.\] Hence \[w(x,t)=w^{(1)}(x,t)+w^{(2)}(x,t)= \sum_{n=1}^{\infty}\frac{1}{2n^2}\cos n\pi \sin nx \sin 4nt + \frac{1}{4}\sin x\sin 4t\] (notice that that last term is outside the sum), and finally \[u(x,t)=v(x,t)+w(x,t)=\sum_{n=1}^{\infty} c_{n} \sin n x\cos 4nt = \sin 2x\cos 8t + \sum_{n=1}^{\infty}\frac{1}{2n^2}\cos n\pi \sin nx \sin 4nt + \frac{1}{4}\sin x\sin 4t.\]
    6. $f(x)=0$, $g(x)=1-\cos \pi x$, $L=2$, $\alpha=3$. Answer: For this one, \[u(x,t)=\sum_{n=1}^{\infty} c_{n}\sin \frac{n\pi}{2}\sin \frac{3n\pi t}{2}\] where \[c_{n}=\frac{2}{3n\pi}\int_{0}^{2}(1-\cos\pi x)\sin \frac{n\pi x}{2}.\] Let's evaluate the integral first, and do so by splitting it into two pieces. First we evaluate \[\int_{0}^{2}1\cdot \sin\frac{n\pi x}{2} =-\frac{2}{n\pi }\cos \frac{n\pi x}{2}|_{0}^{2}=\frac{2}{n\pi}(1-\cos n\pi)\] and \begin{multline*} \int_{0}^{2}\cos \pi x\sin \frac{n\pi x}{2} = \frac{1}{\pi}\sin \pi x\sin \frac{n\pi x}{2}|_{0}^{2} -\frac{n}{2}\int_{0}^{2}\sin \pi x\cos \frac{n\pi x}{2}\\ =0+\frac{n}{2\pi}\cos \pi x\cos \frac{n\pi x}{2}|_{0}^{2}+\frac{n^{2}}{4}\int_{0}^{2}\cos \pi x\sin \frac{n\pi x}{2}\end{multline*} Now subtract $\frac{n^{2}}{4}\int_{0}^{2}\cos \pi x\sin \frac{n\pi x}{2}$ from both sides and divide by $1-\frac{n^{2}}{4}$ to get \[\int_{0}^{2}\cos \pi x\sin \frac{n\pi x}{2} = \frac{\frac{n}{2\pi}\cos \pi x\cos \frac{n\pi x}{2}|_{0}^{2}}{1-\frac{n^{2}}{4}} = \frac{\frac{n}{2\pi}(\cos n\pi-1)}{1-\frac{n^{2}}{4}}\] Hence, \begin{multline*} c_{n}=\frac{2}{3n\pi}\int_{0}^{2}(1-\cos\pi x)\sin \frac{n\pi x}{2}=\frac{2}{3n\pi}\left(\int_0^{2}1\cdot \sin\frac{n\pi x}{2} -\int_{0}^{2}\cos \pi x\sin\frac{n\pi x}{2}\right) \\ = \frac{2}{3n\pi}\left(\frac{2}{n\pi}(1-\cos n\pi)-\frac{\frac{n}{2\pi}(\cos n\pi-1)}{1-\frac{n^{2}}{4}}\right) =\frac{2}{3n\pi}(1-\cos n\pi)\left(\frac{2}{n\pi}+\frac{\frac{n}{2\pi}}{1-\frac{n^{2}}{4}}\right). \end{multline*} and so the final solution is \[u(x,t)=\sum_{n=1}^{\infty} \frac{2}{3n\pi}(1-\cos n\pi)\left(\frac{2}{n\pi}+\frac{\frac{n}{2\pi}}{1-\frac{n^{2}}{4}}\right)\sin \frac{n\pi}{2}\sin \frac{3n\pi t}{2}.\]
    7. $f(x)=1$, $g(x)=0$, $L=1$, $\alpha=1$.
  4. If $u$ models the displacement of a string from $0$ to $L$ that is free at $x=L$ where, instead of letting $u(L,t)=0$, we replace this with the condition that $u_{x}(L,t)=0$. This allows the string to move freely at one end. Use the separation of variables method to find a sequence of simple solutions like we did with the heat and wave equations, only now take head of this new boundary condition when solving for the admissible $\lambda$'s, $X$'s and $T$'s.
\