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Author: William A. Stein
1%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
2%% MATH 54, Lenstra, Spring 96
3%% Solutions to Problem set 14
4%% by William Stein
5
6\documentclass[12pt]{article}
7\usepackage{amsmath}
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14
15\author{For Lenstra's Math 54\\by William Stein}
16\title{Solutions to Problem Set 14 and Practice Final}
17\newcommand{\problem}[1]{\vspace{.3in}\par\noindent{\Large \bfseries #1. }}
18\newcommand{\soln}{\par {\em Solution.} }
19\newcommand{\dx}{\hspace{.5em}dx}
20\newcommand{\vx}{\mathbf{x}}
21\DeclareMathOperator{\cm}{cm}
22\DeclareMathOperator{\C}{C}
23
24\begin{document}
25\maketitle
26This document contains the solutions to the 14th homework set.
27It also contains solutions to the practice final exam.
28The final examination is on Friday, May 17, 5--8 p.m. (room to
29be announced). The final examination covers differential
30equations.
31
32\section{Homework Solutions to 10.5.1, 10.5.7, 10.6.4}
33
34\problem{10.5.1} {\em Consider the conduction of
35heat in a copper rod $100\cm$ in length whose
36ends are maintained at $0^{\circ}\C$ for all $t>0$. Find an expression for
37the temperature $u(x,t)$ if the initial temperature distribution in the
38rod is given by
39\begin{itemize}
40\item[(a)] $u(x,0) = 50, \quad 0\leq x \leq 100$
41\item[(b)] $u(x,0) = \begin{cases} x, & 0\leq x < 50\\ 42 100 - x, & 50 \leq x \leq 100\end{cases}$
43\item[(c)] $u(x,0) = \begin{cases} 0, & 0\leq x < 25 \\ 44 50, & 25\leq x \leq 75 \\ 45 0, & 75 < x \leq 100\end{cases}$
46\end{itemize}
47}
48\soln
49{\em Summary.} (See page 543.) First find the
50sine series for $f(x)=u(x,0)$. Then the
51solution is $$u(x,t)=\sum_{n=1}^{\infty}b_ne^{-(n\pi\alpha/\ell)^2 t} 52 \sin\frac{n\pi x}{\ell},$$
53where the coefficients $b_n$ are the same as in the series
54$$f(x)=\sum_{n=1}^{\infty} b_n \sin\frac{n\pi x}{\ell}.$$
55The $b_n$ are given by
56$$b_n=\frac{2}{\ell}\int_{0}^{\ell}f(x)\sin\frac{n\pi x}{\ell}\dx.$$
57
58
59(a) We compute the Fourier coefficients $b_n$.
60\begin{eqnarray*}
61b_n &=& \frac{2}{100}\int_{0}^{100} u(x,0)\sin\frac{n\pi x}{100}\dx\\
62    &=& \frac{1}{50}\int_{0}^{100} 50 \sin\frac{n\pi x}{100}\dx\\
63    &=& \Bigl[-\frac{100}{n\pi}\cos\frac{n\pi x}{100}\Bigr]_0^{100}
64        = \begin{cases}\frac{200}{n\pi}&\text{$n$ odd}\\
65                       0&\text{$n$ even}\end{cases}
66\end{eqnarray*}
67Thus the solution is
68$$u(x,t) = \sum_{\text{n odd}} \frac{200}{n\pi}e^{-(.0107n\pi)^2 t} 69 \sin\frac{n\pi x}{100}.$$
70Here we have used the value $\alpha^2=1.14$ from page 513 to find
71$$\frac{\alpha}{\ell}=\frac{\sqrt{1.14}}{100}\approx .0107.$$
72
73(b) We compute
74\begin{eqnarray*}
75b_n &=& \frac{2}{100}\int_{0}^{100}u(x,0)\sin\frac{n\pi x}{100}\dx\\
76    &=& \frac{1}{50}\int_{0}^{50}x\sin\frac{n\pi x}{100}\dx
77       +\frac{1}{50}\int_{50}^{100}100\sin\frac{n\pi x}{100}\dx
78       -\frac{1}{50}\int_{50}^{100}x\sin\frac{n\pi x}{100}\dx\\
79    &=& \frac{1}{50}\Bigl\{ \Bigl[-\frac{100}{n\pi}x\cos\frac{n\pi x}{100}
80                            \Bigr]_0^{50}
81           +\frac{100}{n\pi}\int_{0}^{50}\cos\frac{n\pi x}{100}\dx\Bigr\}
82           + 2\Bigl[-\frac{100}{n\pi}\cos\frac{n\pi x}{100}\Bigr]_{50}^{100} \\
83    && -\frac{1}{50}\Bigl\{\Bigl[-\frac{100}{n\pi}x\cos\frac{n\pi x}{100}
84                            \Bigr]_{50}^{100}
85           +\frac{100}{n\pi}\int_{50}^{100}\cos\frac{n\pi x}{100}\dx\Bigr\}\\
86    &=&-\frac{2}{n\pi}\Bigl(50\cos\frac{n\pi}{2}\Bigr)+\frac{2}{n\pi}
87           \Bigl[\frac{100}{n\pi}\sin\frac{n\pi x}{100}\Bigr]_{0}^{50}
88           -\frac{200}{n\pi}\Bigl(\cos n\pi - \cos \frac{n\pi}{2}\Bigr)\\
89    & &     +\frac{2}{n\pi}\Bigl(100\cos n\pi - 50 \cos\frac{n\pi}{2}\Bigr)
90                -\frac{2}{n\pi}\Bigl[\frac{100}{n\pi}\sin\frac{n\pi x}{100}\Bigr]^{100}_{50}\\
91    &=& -\frac{100}{n\pi}\cos\frac{n\pi}{2} +
92         \frac{200}{n^2\pi^2}\sin\frac{n\pi}{2}
93         - \frac{200}{n\pi}\cos n\pi + \frac{200}{n\pi}\cos \frac{n\pi}{2} \\
94   && + \frac{200}{n\pi}\cos n\pi - \frac{100}{n\pi}\cos \frac{n\pi}{2}
95      + \frac{200}{n^2\pi^2}\sin \frac{n\pi}{2}\\
96   &=& \frac{400}{n^2\pi^2}\sin\frac{n\pi}{2}
97\end{eqnarray*}
98Thus the solution is
99$$u(x,t) = \sum_{n=1}^{\infty} \frac{400}{n^2\pi^2}\sin\frac{n\pi}{2} 100 e^{-(.0107n\pi)^2 t} \sin\frac{n\pi x}{100}.$$
101
102(c) We compute
103\begin{eqnarray*}
104b_n&=&\frac{2}{100}\int_{0}^{100}u(x,0)\sin\frac{n\pi x}{100}\dx\\
105   &=&\frac{2}{100}\int_{25}^{75}50\sin\frac{n\pi x}{100}\dx\\
106   &=&\Bigl[-\frac{100}{n\pi}\cos\frac{n\pi x}{100}\Bigr]_{25}^{75}
107     =\frac{100}{n\pi}\Bigl(\cos\frac{n\pi}{4}-\cos\frac{3n\pi}{4}\Bigr).
108\end{eqnarray*}
109The solution is thus
110$$u(x,t) = \sum_{n=1}^{\infty} \frac{100}{n\pi} 111 \Bigl(\cos\frac{n\pi}{4}-\cos\frac{3\pi{}n}{4}\Bigr) 112 e^{-(.0107n\pi)^2 t} \sin\frac{n\pi x}{100}.$$
113
114
115\problem{10.5.7} {\em Consider a uniform rod of length $\ell$ with an
116initial temperature given by $sin(\pi x/\ell)$, $0\leq x \leq \ell$. Assume
117that both ends of the bar are insulated. Find a formal series expansion
118for the temperature $u(x,t)$. What is the steady state temperature as
119$t\rightarrow\infty$?}
120\soln
121{\em Summary.} (See page 547.) Using separation of variables as in section 1
122Boyce-DiPrima derive the solution to the heat conduction problem
123with insulated ends. [Study this derivation. Understanding this type of
124derivation is essential if you want to do well on the final exam.]
125The solution is
126$$u(x,t)=\frac{c_0}{2}+\sum_{n=1}^{\infty} c_n e^{-(n\pi\alpha/\ell)^2t} 127 \cos\frac{n\pi x}{\ell}.$$
128The coefficients $c_n$ are determined by the requirement that
129$$u(x,0)=\frac{c_0}{2} + \sum_{n=1}^{\infty} c_n\cos\frac{n\pi x}{\ell}=f(x).$$
130Thus to solve this problem we just need to find the cosine series
131expansion of $f(x)=\sin(\pi x/\ell)$.
132This is computed using equation (7) page 536. As it turns out,
133to integrate we must do the cases $n=1$ and $n\neq 1$ separately.
134First suppose $n\neq 1$. Then substituting
135$\sin\frac{\pi x}{\ell}$ for $f(x)$ we obtain
136\begin{eqnarray*}
137  c_n&=&\frac{2}{\ell}\int_0^{\ell}
138        \sin\frac{\pi x}{\ell}\cos\frac{n\pi x}{\ell}\dx\\
139     &=&\frac{1}{\ell}\int_{0}^{\ell}\sin\frac{(n+1)\pi x}{\ell}
140             +\sin\frac{(1-n)\pi x}{\ell}\dx\\
141     &=&\frac{1}{\ell}\Bigl[-\frac{\ell}{(n+1)\pi}\cos\frac{(n+1)\pi x}{\ell}
142                   -\frac{\ell}{(1-n)\pi}
143                    \cos\frac{(1-n)\pi x}{\ell}\Bigr]^{\ell}_{0}\\
144     &=&-\frac{1}{(n+1)\pi}\cos(n+1)\pi - \frac{1}{(1-n)\pi}\cos(1-n)\pi
145            -\Bigl(-\frac{1}{(n+1)\pi}+\frac{1}{\pi(n-1)}\Bigr)\\
146     &=& \begin{cases} 0&\text{if $n$ odd}\\
147                      -\frac{4}{(n^2-1)\pi}&\text{if $n$ even}
148          \end{cases}
149\end{eqnarray*}
150Next suppose $n=1$. Then
151\begin{eqnarray*}
152  c_1&=&\frac{2}{\ell}\int_0^{\ell}
153        \sin\frac{\pi x}{\ell}\cos\frac{\pi x}{\ell}\dx\\
154     &=&\frac{1}{\ell}\int_{0}^{\ell}\sin\frac{2\pi x}{\ell}\dx\\
155     &=&-\frac{1}{2\pi}\Bigl[\cos\frac{2\pi x}{\ell}\Bigr]^{\ell}_{0}=0.
156\end{eqnarray*}
157
158To integrate $\sin\frac{\pi x}{\ell}\cos\frac{n\pi x}{\ell}$
159we have used the trigonometric identity
160$$\sin A \cos B = \frac{1}{2}(\sin (A+B) + \sin(A-B)).$$
161This identity follows by adding the two identities
162$$\sin(A+B)=\sin A \cos B + \sin B \cos A$$
163$$\sin(A-B)=\sin A \cos B - \sin B \cos A.$$
164The solution to this insulated heat conduction problem is thus
165$$u(x,t)=\frac{2}{\pi}-\frac{4}{\pi}\sum_{\text{n even}} 166 \frac{1}{n^2-1} 167 e^{-(n\pi\alpha/\ell)^2t} 168 \cos\frac{n\pi x}{\ell}.$$
169Finally $\lim_{t\rightarrow\infty}u(x,t)=\frac{2}{\pi}$ so
170the steady state temperature is $\frac{2}{\pi}$.
171
172\problem{10.6.4} {\em Find the displacement $u(x,t)$ in an elastic string of
173length $\ell$, fixed at both ends, that is set in motion from its straight
174equilibrium position with the initial velocity $g$ defined by
175$$g(x)=\begin{cases} Ax,&0\leq x \leq \ell/2\\ 176 A(\ell-x),& \ell/2 < x \leq \ell.\end{cases}$$}
177\soln
178{\em Summary.} (See pages 554-559.) Using separation of variables
179the authors find solutions to the elastic string problem
180$a^2u_{xx}=u_{tt}$.
181Their solution is
182$$u(x,t)=\sum_{n=1}^{\infty}\sin\frac{n\pi x}{\ell} 183 \Bigl(c_n\sin\frac{n\pi a t}{\ell} + 184 k_n\cos\frac{n\pi a t}{\ell}\Bigr).$$
185Let $f(x)$ be the initial displacement of the string and $g(x)$
186be the initial velocity of the string. Then
187$$k_n=\frac{2}{\ell}\int_{0}^{\ell} f(x)\sin\frac{n\pi x}{\ell}\dx$$
188and
189$$\frac{n\pi a}{\ell} c_n = \frac{2}{\ell}\int_{0}^{\ell}g(x)\sin\frac{n\pi x}{\ell}\dx.$$
190Note that $c_n$ is the $n$th coefficient of the sine series for $g(x)$
191{\em divided} by $\frac{n\pi a}{\ell}$.
192In our situation $f(x)=0$ so $k_n=0$ for all $n$. Thus to find the solution
193$u(x,t)$ we just need to find the sine series expansion of $g(x)$.
194Fortunately problem (10.5.1 b) involves almost the same integral.
195Looking at our work there we see that
196$b\frac{4\ell}{n^2\pi^2}\sin\frac{n\pi}{2}$
197is the $n$th Fourier coefficient
198of the sine series expansion of
199$$h(x)=\begin{cases}x,&0\leq x < \ell/2 \\ 200 \ell-x, & \ell/2\leq x \leq \ell\end{cases}.$$
201Multiplying by $A$ we see that the $n$th Fourier coefficient of $g(x)$
202is $$b_n=\frac{4A\ell}{n^2\pi^2}\sin\frac{n\pi}{2}.$$
203Thus $$c_n=\frac{\ell}{n\pi a}b_n 204 =\frac{4\ell^2A}{n^3\pi^3a}\sin\frac{n\pi}{2}.$$
205So the solution to the elastic string problem is
206$$u(x,t)=\frac{4A\ell^2}{a\pi^3}\sum_{n=1}^{\infty} 207 \frac{1}{n^3}\sin\frac{n\pi}{2} 208 \sin\frac{n\pi x}{\ell}\sin\frac{n\pi a t}{\ell}.$$
209
210\section{Solutions to the Final Exam}
211\problem{Problem 1} Solve the system of differential equations
212$$x_1'(t)=3x_1(t)+3x_2(t),$$
213$$x_2'(t)=-2x_1(t)-4x_2(t)$$
214with the initial conditions
215$$x_1(0)=1, \quad x_2(0)=3.$$
216\soln
217Let $\vx(t)=(x_1(t),x_2(t))$. Then the system of equation becomes
218$\vx'=A\vx$ where
219 $A=\Bigl(\begin{smallmatrix}3&3\\-2&-4\end{smallmatrix} \Bigr).$
220First compute the characteristic polynomial of $A$,
221$$|A-\lambda I|=\Bigl| 222 \begin{matrix}3-\lambda&3\\-2&-4-\lambda\end{matrix}\Bigr| 223 =\lambda^2+\lambda-6=(\lambda+3)(\lambda-2).$$
224Thus the eigenvalues are $-3$ and $2$. The eigenvalue $-3$ has associated
225eigenvector $v_1=(1,-2)$ and the eigenvalue $2$ has associated eigenvector
226$v_2=(3,-1)$. It follows that the general solution is
227$$\vx(t)=c_1e^{-3t}\Bigl[\begin{matrix}1\\-2\end{matrix}\Bigr] 228 +c_2e^{2t}\Bigl[\begin{matrix}3\\-1\end{matrix}\Bigr].$$
229To find $c_1$ and $c_2$ use the initial condition $\vx(0)=(1,3)$.
230This gives $c_1=-2$ and $c_2=1$. Thus
231$$\vx(t)=\Bigl[\begin{matrix}-2e^{-3t}+3e^{2t}\\ 232 4e^{-3t}-e^{2t}\end{matrix}\Bigr]$$
233so $x_1(t)=-2e^{-3t}+3e^{2t}$ and
234$x_2(t)=4e^{-3t}-e^{2t}$.
235
236\problem{Problem 2}
237
238(a) Find three functions $y_1(x), y_2(x), y_3(x)$ defined on $(-\infty,\infty)$
239whose Wronskian is given by
240$$W(y_1,y_2,y_3)(x)=e^{4x}.$$
241
242(b) Are the functions $y_1,y_2,y_3$ that you found linearly independent
243on $(-\infty,\infty)$? Justify your answer.
244\soln
245(a) There are many ways to solve this problem. One way is to
246write down a matrix whose diagonal is $1,1,e^{4t}$
247$$\left|\begin{matrix}1&*&*\\ 248 0&1&*\\ 249 0&0&e^{4t}\end{matrix}\right|.$$
250Then integrate to fill in what the $*$'s must be to obtain
251$$\left|\begin{matrix}1&x&\frac{1}{16}e^{4t}\\ 252 0&1&\frac{1}{4}e^{4t}\\ 253 0&0&e^{4t}\end{matrix}\right|.$$
254Taking $y_1=1, y_2=x, y_3=\frac{1}{16}e^{4t}$ we see
255at once that $W(y_1,y_2,y_3)=e^{4t}$.
256
257(b) Yes, since their Wronskian is nonzero at some point.
258\problem{Problem 3}
259
260Find a homogeneous third-order linear differential equation with constant
261coefficients that has
262$$y(x)=3\cdot e^{-x}-cos(2x)$$
263as a solution. Explain how you found it.
264
265What is the general solution of that differential equation?
266
267\soln
268One way to solve this problem is to compute $y, y', y''$, and $y'''$ then
269try to find a linear relation between them. In general this amounts
270to solving a system of four equations in four unknowns.
271
272Another way to solve this is to note that $y(x)$ is a solution to a
273third order homogeneous linear differential equation which has fundamental
274solution set $e^{-x}$, $\cos(2x)$ and $\sin(2x)$. The roots of the
275characteristic equation associated to this differential equation must
276be $1$, $2i$ and $-2i$. Thus the characteristic equation is
277$$0=(r-1)(r-2i)(r+2i)=(r-1)(r^2+4)=r^3-r^2+4r-4$$
278It follows that the differential equation is
279$$y'''-y''+4y'-4y=0.$$
280
281The general solution is
282$$y=c_1e^{-x}+c_2\cos(2x)+c_3\sin(2x).$$
283
284
285\problem{Problem 4}
286For the function $f(x)=e^{x/\pi}$, draw a careful sketch of the graphs of the
287functions to which the following Fourier series converge on the interval
288$[0,4\pi]$:
289
290(a) the Fourier sine series of $f$ on $[0,\pi]$;
291
292(b) the Fourier cosine series of $f$ on $[0,\pi]$;
293
294(c) the ordinary Fourier series of $f$ on $[-\pi,\pi]$.
295\vspace{3in}
296
297\problem{Problem 5}
298
299Find a function $u=u(x,t)$, defined for $0\leq x\leq \pi$ and
300$t\geq 0$, satisfying the partial differential equation
301$$u_{xx}=-u_{tt}\quad (0<x<\pi, \quad t>0),$$
302with boundary conditions
303$$u(0,t)=u(\pi,t)=0\quad (t>0)$$
304and initial conditions
305$$u(x,0)=\sin(3x), \quad u_t(x,0)=0\quad (0<x<\pi).$$
306
307\soln
308Apply separation of variables. Let $u(x,t)=X(x)T(t)$.
309
310The initial conditions
311$$u(0,t)=u(\pi,t)=0\quad (t>0)$$
312imply $0=X(0)T(t)$ and $0=X(\pi)T(t)$ for all $t$ so, to avoid
313triviality, $X(0)=X(\pi)=0$.
314
315Substituting $u=XT$ into the differential equation yields
316$X''T=-XT''$ which becomes
317$$\frac{X''}{-X}=\frac{T''}{T}.$$
318Because the left hand side does not depend on $t$ and the right
319side does not depend on $x$ we conclude that there is a constant
320$a$ such that
321$$\frac{X''}{-X}=a \text{ and } \frac{T''}{T}=a.$$
322Thus
323$$X''+aX=0 \text{ and } T'' - aT=0.$$
324Since we are looking for a solution $u(x,t)$ for which
325$u(x,0)=\sin(3x)$
326we know that $a>0$ is probably going to be the right choice
327since it gives sine and cosine type solutions.
328Making this choice and solving yields
329 $$X(x)=c_1\cos(\sqrt{a} x)+c_2\sin(\sqrt{a} x).$$
330Since $X$ must satisfy the initial conditions $X(0)=X(\pi)=0$
331it follows that
332$c_1=0$ and $c_2\sin(\sqrt{a}\pi)=0$. Thus
333$\sqrt{a}\pi=n\pi$ for some integer $n$. Thus $\sqrt{a}=n$
334so $a=n^2$. We see that $X(x)=\sin(nx)$ after setting the
335arbitrary factor $c_2$ equal to $1$.
336
337Next consider $T''-n^2T=0$. This has general solution
338$$T(t)=c_1e^{nt}+c_2e^{-nt}.$$
339
340Multiplying $X$ and $T$ together again we see that
341$$u_n=\sin (nx)e^{nt}\text{ and } v_n=\sin(nx)e^{-nt}$$
342for various integers $n$ give a bunch of solutions to the
343differential equation $u_{xx}=-u_{tt}$ satisfying the
344initial conditions $u(0,t)=u(\pi,t)=0$. [You should verify
345this assertion by substituting $u_n$ and then $v_n$ into
346$u_{xx}=-u_{tt}$.]
347
348Can we find some linear combination of the $u_n$ and $v_n$
349which satisfies the additional conditions $u(x,0)=\sin(3x)$
350and $u_t(x,0)=0$? Yes. Setting $t=0$ shows that it will have
351to be of the form
352$$c_1\sin(3x)e^{3t}+c_2\sin(3x)e^{-3t}$$
353for some constants $c_1$ and $c_2$.
354A little algebra shows that $c_1=c_2=\frac{1}{2}$ does the trick.
355Thus the solution is
356$$u(x,t)=\frac{1}{2}\sin(3x)(e^{3t}+e^{-3t}).$$
357
358\end{document}
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