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\begin{center}
{\bf Solutions to Homework 11}
\end{center}
\vspace*{.3in}

\noindent{\bf Problem 1.} 
\vspace*{.1in}
Solve the following initial value problems
\begin{enumerate}
\item $y' = \frac{\sin t}{y}$, $y(\pi/2) = 1$.

\vspace*{.1in}
\noindent{\bf Solution.}
Rearranging and integrating the equation gives
$$
\log|y| = \int \frac{dy}{y} = \int\frac{y'}{y}\,dt = \int \sin t\,dt 
	= -\cos t + C,
$$
where the constant $C$ is determined by plugging in the initial condition:
$$
\log|1| = -0 + C.
$$
So $C=0$.  Note that this also implies that the sign $y$ is positive when we
drop the absolute value inside the logarithm (why?).  We conclude that
$$
\log y(t) = -\cos t \quad\Rightarrow \quad y(t) = e^{-\cos t}.
$$


\vspace*{.2in}
\item $(1+t^2)y' + 4ty = (1+t^2)^{-2}$, $y(1) = 0$.

\vspace*{.1in}
\noindent{\bf Solution.}
First we replace the right side with zero and solve the 'homogenized'
version of the equation, obtaining
$$
y_h(t) = \frac{A}{(1+t^2)^2}.
$$
Then we assume that $y$ has the form $y(t) = A(t)(1+t^2)^2$ for some
unknown function $A(t)$.  Plugging this back into the original differential
equation leads to the following formula for $A'$:
$$
A'(t) = \frac{1}{1+t^2}.
$$
Antidifferentiating both sides yields
$
A(t) = \tan^{-1} t + C.
$
So 
$$
y(t) = \frac{C + \tan^{-1} t}{(1+t^2)^2}.
$$
Finally, we plug in the initial condition to compute that $C = -\pi/4.$
In conclusion,
$$
y(t) = \frac{-\pi/4 + \tan^{-1} t}{(1+t^2)^2}.
$$
\end{enumerate}

\newpage
\noindent{\bf Problem 2.} 
Disillusioned about mathematics, you descend the ivory tower and
set yourself the task of becoming a millionaire by age 50.  The plan is
simple.  You will stash away money continuously from now til then at a 
constant rate of $R$\$ per year.  You figure that in your new life as day
trader, you can make a reliable 8\% annual interest (compounded continuously,
of course) on your savings.  Unfortunatly, college has left you broke
as well as disillusioned, so you're starting from nothing.  At what rate
$R$ will you need to be saving your money?

\vspace*{.1in}
\noindent{\bf Solution.}
If $y(t)$ is the amount of money saved up (with interest) at time $t$, then
then $y$ satisfies the following conditions ($t=0$ is now, $t=30$ is
age 50).
$$
y(0) = 0,\quad y(30) = 10^6,\quad y' = .08y + R.
$$
The differential equation is separable, and solving it gives
$$
y(t) = -\frac{R}{.08} + A e^{.08t}
$$
for some constant $A$.  Using the two additional conditions gives us
$$
0 = -R/.08 + A,\quad 10^6 = -R/.08 + Ae^{2.4},
$$
and these can be solved together to give $R = .08\cdot 10^6(e^{2.4}-1)
\approx \mathrm{\$} 7981.5$ per year. 

\vspace*{.3in}
\noindent{\bf Problem 3.}
\emph{This problem was mis-stated:  the function $f$ should have been $C^2$
instead of $C^1$, and the second item should have said 
$N(N_\delta(r))\subset N_\delta(r)$ rather than $f(N_\delta(r))\subset
N_\delta(r)$.  The upshot is that I'll let this one go without grading.  The
correct (I hope) statement and solution of the problem are as follows.}

Remember Newton's method?  The idea is that you have a $C^2$ function 
$f:\R\to\R$.  You know that $f(r) = 0$ for some point $r\in U$, and you have 
a decent initial guess $x_0$ at the location of $r$.  Beginning with this 
guess, you then produce a sequence of (hopefully better) approximations of 
$r$ by setting
$$
x_{n+1} = N(x_n)
$$
for every $n\in\N$, where $N(x) = x - f(x)/f'(x)$.  Now assume that $r$ is 
a \emph{non-degenerate} root of $f$---i.e. that $f'(r) \neq 0$.  Prove the 
following. 
\begin{itemize}
\item $r$ is a fixed point of $N$.

\vspace*{.1in}
\noindent{\bf Solution.} $N(r) = r - f(r)/f'(r) = r - 0/f'(r) = r$.
\qed
\vspace*{.2in}

\item There exists $\delta>0$ such that $N(N_\delta(r)) \subset N_\delta(r)$

\vspace*{.1in}
\noindent{\bf Solution.} Given $x$, the mean value theorem gives us a number
$c$ between $x$ and $r$ such that
$$
N(x) - N(r) = N'(c)(x-r) = \frac{f''(c)f(c)}{f(c)^2}(x-r).
$$
Moreover, since $f$ is $C^2$, $f'(r)\neq 0$, and $f(r) = 0$, the function
$N' = ff''/(f')^2$ is continuous in some neighborhood $N_{\delta_0}(r)$
(i.e. on any open set where $f'(r)$ does not vanish)
and satisfies $N'(r) = 0$.   Hence there exists $0<\delta<\delta_0$
such that $|x-r|<\delta$ implies that 
$$
|N'(x)| = |N'(x) - N'(r)| < \frac12.
$$
Putting the two displayed equations together allows us to conclude that
$$
|N(x) - r| = |N(x) - N(r)|< \frac12 |x-r|
$$
for all $x\in N_\delta(r)$.  In particular $x\in N_\delta(r)$ implies that
$N(x) \in N_{\delta/2}(r)$.  That is, 
$$
N(N_\delta(r))\subset N_{\delta/2}(r) \subset N_\delta(r)
$$
as asserted.
\qed

\vspace*{.2in}
\item The (restricted) function $N:N_\delta(r) \to N_\delta(r)$ is a 
      contraction mapping.

\vspace*{.1in}
\noindent{\bf Solution.} Let $\delta$ be as in the solution to the previous
item.  Then for any points $x_1,x_2\in N_\delta(r)$, we
have
$$
|N(x_1) - N(x_2)| = |N'(c)||x_1-x_2| \leq \frac12|x_1 - x_2|,
$$
where $c$ is between $x_1$ and $x_2$ (and therefore belongs to $N_\delta(r)$).
\qed
\vspace*{.2in}

\end{itemize}
(The mean value theorem will be useful in the second and third items.)
What can you conclude from all this about how well Newton's method works?

\vspace*{.1in}
\noindent{\bf Answer.} 
$r$ is the unique fixed point of $N$ in the open interval $N_\delta(r)$,
and if the initial guess $x_0$ happens to be in $N_\delta(r)$, then the
sequence $x_1,x_2,\dots$ of subsequent guesses will remain in $N_\delta(r)$
and converge to $r$.  In short, Newton's method works if the initial
guess is good enough.
\vspace*{.2in}

\vspace*{.3in}
\noindent{\bf Problem 4.} 
Suppose that $f,g:\R^2\to\R$ are continuous functions and that
$f(y,t) > g(y,t)$ for all points $(y,t)\in\R^2$.  Let $y_1, y_2:\R\to\R$ 
be functions satisfying 
$$
y_1' = f(y_1,t), \quad y_2' = g(y_2,t) 
$$
for all $t\in\R$.  Show that $y_1(t_0) = y_2(t_0)$ for at most one point
$t_0\in\R$.  

\vspace*{.1in}
\noindent{\bf Solution.}
Let $h:\R\to\R$ be the function $h(t) = y_1(t) - y_2(t)$.  Then in
particular, $h'(t) = f(y_1(t),t) - g(y_2(t),t)$ exists and is continuous
and every point $t\in\R$.   Also, if $t_0\in\R$ is any point where
$y_0 := y_1(t_0) = y_2(t_0)$, then 
$$
h(t_0) = 0, \quad h'(t_0) = f(y_0,t_0) - g(y_0,t_0) > 0.
$$
By continuity, we see that there exists $\delta_0>0$ such that
$h'(t) > 0$ for all $|t-t_0| < \delta_0$.  That is, $h$ is strictly
increasing on $(t_0-\delta_0,t_0+\delta_0)$. So $h(t) > h(t_0)$ for all 
$t \in (t_0,t_0+\delta_0)$ and $h(t) < h(t_0)$ for all 
$t\in (t_0-\delta_0,t_0).$

Now suppose, for the sake of obtaining a contradiction, that there
is a second point at which $y_1$ equals $y_2$.  We can suppose without
loss of generality that this point is larger than $t_0$.  Then it is
meaningful to define 
$$
t_1 := \inf\{t>t_0: h(t) = y_1(t) - y_2(t) = 0\}
$$
(i.e. we're not taking the infimum of the empty set).
By continuity $h(t_1) = 0$.  Clearly, $t_1 \geq t_0$, so
the work above implies in fact that $t_1 \geq t_0 + \delta_0 > t_0$.
Repeating the arguments used on for $t_0$, we see that there exists 
$\delta_1>0$ such that, among other things, $h(t) < 0$ for 
$t\in (t_1-\delta_1,t_1)$.  

So to be perfectly specific, let us consider for example the points
$s_0 = t_0 + \delta_0/2$ and $s_1 = t_1 - \delta_1/2$.  Then
$$
t_0 < s_0 < s_1 < t_1, \quad\mathrm{and}\quad h(s_0) > 0 > h(s_1).
$$
The intermediate value theorem therefore gives us a point 
$s\in (s_0,s_1) \subset (t_0,t_1)$ such that $h(s) = y_1(s) - y_2(s) = 0$.
This contradicts the fact that $t_1$ was supposed to be the smallest
root of $h$ larger than $t_0$.  It follows that there is \emph{no}
point other than $t_0$ at which $y_1 = y_2$.
\qed

\vspace*{.3in}
\noindent{\bf Problem 5.} 
Let $f:\R\to\R$ be a function.  The \emph{support} of $f$ is the
set 
$$
K := \overline{\{x\in\R:f(x) \neq 0\}}
$$
Show that if $f$ is real analytic (and not the zero function), then $K$ 
cannot be compact.  Show by giving an example that $K$ can be compact if 
$f$ is merely $C^\infty$.
\vspace*{.1in}

\noindent{\bf Solution.}
Suppose that $f$ is analytic and let $g(x) \equiv 0$ be the
zero function, which is also analytic.  If the support $K$ of $f$ is compact,
then the set $E = \{x\in\R:f(x) = g(x)\}$ contains all points in the 
non-empty open set $\R - K$.  This directly contradicts Theorem 8.5, which
says that $E$ has no limit points unless $f$ and $g$ are equal.  Hence
$K$ is not compact.
\qed

To see that $C^\infty$ functions \emph{can} have compact support, let
$h:\R\to\R$ be the function considered in class
$$
h(x) = \left\{
\begin{array}{lcr}
0 & \mathrm{ if } & x \leq 0 \\
e^{-1/x} & \mathrm{ if } & x > 0 
\end{array}
\right.
.$$
Set $f(x) = h(1-x)h(x+1)$.  Then it is not hard to see that $f(x) = 0$
if $|x| \geq 1$ but $f(x) > 0$ if $|x| < 1$.  So the support of
$f$ is $[-1,1]$ which is compact.

\vspace*{.3in}
\noindent{\bf Problem 6.} 
 Redo supplementary problem 2 from the homework assigned on 11/3/03:
\vspace*{.1in}

\noindent{\bf Solution.}
Given $\epsilon>0$, we seek to construct a partition $P$ of $[0,1]$ 
satisfying
$$
U(P,f) - L(P,f) < \epsilon.
$$

This time, however, we let 
$$
S = \{x\in[0,1]: f(x) \geq \epsilon/2\}. 
$$
I claim that there are only finitely many points in $S$.  Indeed any
$x\in S$ is rational, and can be written $x = p/q$ where $\gcd(p,q) = 1$
and $q\leq 2/\epsilon$ (since $f(x) = 1/q$).  So if $N\in\N$ exceeds 
$2/\epsilon$, then the denominator $q$ is $x$ can only range from $1$ to $N$.
Moreover, for fixed denominator $q$, the numerator of $x$ must range between
$0$ and $q$.  So all told, $S$ contains at most 
$\sum_{q=1}^N (q+1) = (q+1)(q+2)/2$ elements.

Clearly then, we can cover $S$ with disjoint open intervals $I_j = (a_j,b_j)$,
$j=1,\dots,n:=\#S$ such that $\sum_{j=1}^nb_j-a_j < \epsilon/2$.  Putting the 
intervals in order and intersecting them with $[0,1]$ we obtain a partition $P$
of $[0,1]$ consisting of the points
$$
a = a_1 < b_1 \leq a_2 < b_2 \leq \dots \leq a_n < b_n=b.
$$
Since $0\leq f(x)\leq 1$ for all $x$, we have
$$
0 \leq L(P,f) \leq U(P,f) \leq \sum_{j=1}^n 1(b_j - a_j) + 
	\sum_{j=1}^{n-1} \epsilon/2 (a_{j+1} - b_j).
$$
The first sum is equal to the sum of the lengths of the intervals $I_j$ and
the second is no larger than $\epsilon/2$ times the length of the full
interval $[0,1]$.  Hence
$$
U(P,f) - L(P,f) < \epsilon
$$
and we conclude that $f$ is integrable.
\qed
\vspace*{.3in}


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