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\noindent Solutions for Homework 4
\vskip .2in
  
\bigskip
\noindent {\bf Page 281/ \#2:}
\bigskip

\noindent $K = ||\partial f/\partial y||_\infty = ||\cos y||_\infty = 1$.
Also, $M = ||f||_\infty = 1$.  The proof shows that we obtain a solution
on $[t_0 - \delta,t_0+\delta]$ so long as both $K\delta$ and $M\delta$
are less than $1$.  In particular, we will be OK if we take $\delta = 1/2$.
\bigskip


\bigskip 
\noindent {\bf Page 282/ \#10:} 
\bigskip

\noindent We just need to find those points $(0,y_0)$ at which the
functions $f$ and $\partial f/\partial y$ are continuous.  So the answers
are
\begin{description}
\item[a] $y_0$ can be anything;
\item[b] $y_0 > 2$;
\item[c] $y_0 \neq 0$;
\item[d] $|y_0| < 1$;
\item[e] $y\neq 0$.
\end{description}
\bigskip

\noindent{\bf Other Problem A:}
\bigskip

\noindent{\it Part 1.} 
Given $x,y\in\mathbf{R}$, the Mean Value Theorem guarantees
a point $c$ between $x$ and $y$ such that 
$$
\frac{T(x) - T(y)}{x-y} = T'(c) = \frac{c}{\sqrt{1+c^2}}. 
$$
Hence 
$$  
|T(x) - T(y)| = \frac{|c|}{\sqrt{c^2+1}}|x-y| <  |x-y|.
$$   
\qed

\medskip
\noindent{\it Part 2.} 
A fixed point $x$ for $T$ satisfies the equation $\sqrt{x^2+1} = x$.
Squaring both sides and subtracting $x^2$ gives. 
$1 = 0$, which is impossible.  Therefore, $T$ has no fixed points.  
\qed

\medskip
\noindent{\it Part 3.}
Note that $T(x) = \sqrt{1+x^2} > \sqrt{x^2} = |x| \geq x$ for all
$x\in\mathbf{R}$.  Hence, given $x_1\in\mathbf{R}$, the sequence 
$x_1, x_2 = T(x_1), x_3 = T(x_2), \dots$ is increasing.  Either it 
diverges to infinity as $n$ increases or it converges to its least 
upper bound.  Suppose
(by way of contradiction) that 
$M = \lim_{n\to\infty} x_n = \sup\{x_n\}_{n=0}^\infty < \infty$
is the least upper bound.  Then since $T$ is a continuous function,
we have 
$\lim_{n\to\infty} T(x_n) = T(\lim_{n\to\infty} x_n) = T(M)$.  On the other
hand, $T(x_n) = x_{n+1} \leq M$ for every $n$ since $M$ is an upper
bound for the $x_n$'s.  Therefore, $T(M) = \lim_{n\to\infty} T(x_n) \leq M$.
But we already know that $T(M) > M$ for any number $M$, so $M$ must
not exist.  We conclude that the sequence $\{x_n\}$ diverges to infinity.
\qed
\medskip
\newpage

\noindent{\bf Other Problem B:}  
\bigskip

\noindent{\it Part 1.} If $y_1,y_2:[a,b]\to\mathbf{R}$ are continuous
functions, then for each $t\in[a,b]$, we have
\begin{eqnarray*}
|Ty_1(t) - Ty_2(t)| & = & \left|\int_a^t p(s)(y_1(s) - y_2(s)\,ds\right| \\
	& \leq & \int_a^t |p(s)||y_1(s) - y_2(s)|\,ds \\
	& \leq & ||y_1(s) - y_2(s)||_\infty \int_a^t |p(s)|\,ds \\
        & \leq & ||y_1(s) - y_2(s)||_\infty \int_a^b |p(s)|\,ds.
\end{eqnarray*}
That is,
$$
||Ty_1 - Ty_2||_\infty \leq 
	\left(\int_a^b |p(s)|\,ds\right) ||y_1 - y_2||_\infty,
$$
and $T$ is a contraction provided that the integral on the right side 
is less than one.
\qed

\medskip
\noindent{\it Part 2.} We can choose $b > a$ to be any number such
that 
$$
1 > \int_a^b |p(s)| \,ds = \int_0^b s\,ds = b^2/2.
$$
So we can take $b$ to be any number smaller than $\sqrt{2}$.  If $y_1(t) = 1$,
then
\begin{eqnarray*}
y_2(t) & = & \int_0^t (s \cdot 1 + e^s)\,ds + 1 = \frac{t^2}{2} + e^t \\
y_3(t) & = &  \int_0^t \left(\left(\frac{s^2}{2} + e^s\right)\cdot 1 
					+ e^s\right)\,ds + 1 
		= \frac{t^4}{6} + te^t + 1.
\end{eqnarray*}

\bigskip 

\noindent{\bf Other Problem C:}
\bigskip  

\noindent{\it Part 1.} Note that $y_2(t) - y_1(t)$
is a continuous function, so if $y_2 - y_1$ is positive at $t_2$, then
$y_2 - y_1$ is positive for $t$ slightly larger than $t_2$.  In particular,
there exist values of $t$ between $t_2$ and $t_1$ for which $y_2 - y_1$
is positive, and therefore $t_2$ is not the least upper bound for the
set of such values.  

If, on the other hand, $y_2 - y_1$ is negative at $t_2$, then 
$y_2 - y_1$ is negative for all values of $t$ slightly less than $t_2$.
In particular, $t_2$ is not the {\it least} upper bound for the set
of $t<t_1$ for which $y_2(t) - y_1(t)\geq 0$.  The only remaining
possibility is that $y_2(t) - y_1(t) = 0$.
\qed

\medskip
\noindent{\it Part 2.} Since $y_1(t_2) = y_2(t_2)$, we have
$$
y'_1(t_2) - y'_2(t_2) = f_1 (t_2,y_1(t_2)) - f_2 (t_2,y_2(t_2))
                       = f_1 (t_2,y_1(t_2)) - f_2 (t_2,y_1(t_2)) < 0
$$
by hypothesis on $f_1$ and $f_2$.

\medskip
\noindent{\it Part 3.} Let $h(t) = y_1(t) - y_2(t)$.  Since $h$ is $C^1$ 
and $h'(t_2) < 0$, we know that $h'(t) < 0$ for $t$ near $t_2$.  
Therefore, $h$ is decreasing for $t$ near $t_2$.  In particular,
for $t$ slightly greater than $t_2$, we have 
$0 < h(t) = y_1(t) - y_2(t) < y_1(t_2) - y_2(t_2)$.  That is, 
$y_1(t) < y_2(t)$.  \qed

\medskip
\noindent{\it Part 4.} The fact that $y_1(t) < y_2(t)$ for $t$ slightly
larger than $t_2$ contradicts the assumption that $t_2$ is
an upper bound for the set of such $t$.  This means that $t_2$
does not exist.  This in turn means that there is no $t_1$ for
which $y_1(t_1) > y_2(t_1)$.  We conclude that $y_1(t) \leq y_2(t)$
for all $t>t_0$. \qed

\medskip
\noindent{\it Part 5.} A similar argument shows that $y_1(t) \geq y_2(t)$
for all $t<t_0$.

\medskip
\noindent{\it Part 6.} Suppose that there is actually a point $t_1 > t_0$
for which $y_1(t_1) = y_2(t_1)$.  Then, as in part 2, we conclude that
$y'_1(t_1) - y'_2(t_1) < 0$.  And as in part 3, we conclude that 
$h(t) = y_1(t) - y_2(t)$ is decreasing for $t$ near $t_1$.  But this
means that $y_1(t) > y_2(t)$ for $t$ slightly {\it less} than $t_1$.
Such $t$ would be greater than $t_0$, and our earlier arguments have
already shown that $y_1(t) \leq y_2(t)$ for all $t\geq t_0$. \qed

\end{document} 




