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\section{Existence and Uniqueness of Solutions}

\subsection{Systems of Ordinary Differential Equations} Let $x:[a.b]\to
\R^n$ be a vector function defined on an interval of real
numbers. Denote by $x^{(m)}(t)$ the $m$-th derivative of $x(t)$ with
respect to~$t$. Suppose $G:[a,b]\times D\to \R^n$ is a continuous function
for some open subset $D\subset \R^{nm}$. An equation of the form
\[ x^{(m)}(t) = G(t,x(t),x'(t),\dots,x^{(m-1)}(t)) \]
is called a \emph{system of ordinary differential equations of order~$m$}.

By increasing the number of unknowns and equations, any $m$-th order
system can be written as a first order system, as follows. Define
\[ v_1(t)=x(t),\ v_2(t)=x'(t),\dots,\ v_m(t)=x^{(m-1)}(t) \]
Then the equivalent first order system is given by the equations
$$\displaylines{
v_1'(t)=v_2(t),\ v_2'(t)=v_3(t),\dots,\ v_{m-1}'(t)=v_m(t),\cr
v_m'(t) = G(t,v_1(t),\dots,v_m(t))
}$$
or more compactly as
\[ v'(t) = H(t,v(t)) \]
where $v:[a,b]\to \R^{nm}$ is $v(t) = (v_1(t),\dots,v_m(t))$
and $H:[a,b]\times D\to \R^{nm}$ is $H(t,v(t)) =
(v_2(t),\dots,v_m(t),G(t,v_1(t),\dots,v_m(t)))$.

\emph{Example:}
Consider the spring-mass system shown below. The two masses move on a
frictionless surface under the influence of external forces $F_1(t)$ and
$F_2(t)$ and they are also constrained by the three springs whose constants
are $k_1$, $k_2$, and $k_3$.
\[ \resizebox{4in}{!}{\includegraphics{Pictures/09.fig1.eps}} \]
From Newton's law of motion and Hooke's law for springs, we obtain the
following equations for the coordinates $x_1$ and $x_2$ of the two masses
\begin{eqnarray*}
m_1 x_1''(t) &=& -k_1x_1(t) + k_2(x_2(t)-x_1(t)) + F_1(t)\\
m_2 x_2''(t) &=& -k_3x_2(t) - k_2(x_2(t)-x_1(t)) + F_2(t)
\end{eqnarray*}
To put this in vector form, $x''(t) = G(t,x(t),x'(t))$, we let
\[ x(t) =
\left[\begin{array}{l}x_1(t)\\x_2(t)\end{array}\right] \]
and
\[ G(t,x(t),x'(t)) = \left[\begin{array}{l}
\big(k_2x_2(t) -(k_1+k_2)x_1(t) + F_1(t)\big)/m_1\\
\big(k_2x_1(t) -(k_2+k_3)x_2(t) + F_2(t)\big)/m_2
\end{array}\right] \]
The equivalent first order system is
$v'(t) = H(t,v(t))$, where
\[ v(t) = \left[\begin{array}{l}
v_{11}(t)\\v_{12}(t)\\v_{21}(t)\\v_{22}(t)
\end{array}\right]
= \left[\begin{array}{l}
x_1(t)\\x_2(t)\\x_1'(t)\\x_2'(t)
\end{array}\right] \]
and
\[ H(t,v(t)) = \left[\begin{array}{l}
v_{21}\\v_{22}\\
\big(k_2v_{12}(t) -(k_1+k_2)v_{11}(t) + F_1(t)\big)/m_1\\
\big(k_2v_{11}(t) -(k_2+k_3)v_{12}(t) + F_2(t)\big)/m_2
\end{array}\right] \]

The process of solving a differential equation is essentially a form of
integration and always involves undetermined constants as part of the
answer. In order to get a unique solution we must impose extra conditions
that determine the value of these constants. Typically \emph{initial
conditions} are specified, that is, a solution to $x'(t) = G(t,x(t))$
must satisfy $x(t_0) = x_0$ for some given $t_0\in[a,b]$ and $x_0\in D$.
A system of differential equations with an initial condition is called
an \emph{initial value problem}.

We shall prove, under mild assumptions about the function~$G$, that
the solution to an initial value problem always exists and is unique.
Our proof is based on the Contractive Mapping Principle which requires
converting the initial value problem into an equivalent integral equation,
then using the integral equation to define a contractive mapping of a
certain complete metric space into itself. The solution will be the
fixed point of the contractive mapping.

\subsection{Integral Equations}

We say that a function $f$ is $C^k$ if $f$ has continuous derivatives
up to order~$k$. We write $C^k(D)$ to denote the vector space of $C^k$
functions with common domain~$D$. We sometimes need to specify a common
range as well, so we let $C^k(D,E)$ denote the set of $C^k$ functions
$f:D\to E$.

For example, $C^k([a,b],\R^n)$ denotes the vector space of
continuous vector functions $f:[a,b]\to\R^n$ that have $k$
continuous derivatives. Since derivatives of vector functions
are taken component-wise, it follows that
\[ C^k([a,b],R^n) \cong C^k[a,b]\times\cdots\times C^k[a,b]\quad
\mbox{($n$ times)} \]

The translation of an initial value problem to an integral equation
follows immediately from the Fundamental Theorem of Calculus for
vector functions.

\begin{lemma}\label{09:INTEGRALEQN}
Let $G:[a,b]\times D\to\R^n$ be a continuous function defined on
some neighborhood of $(t_0,x_0)\in \R\times \R^n$. If $x(t):[a,b]\to\R^n$
is $C^1$ and $x(t)$ satisfies the initial value problem
\begin{equation}\label{09:D}
x'(t) = G(t,x(t)),\quad x(t_0) = x_0
\end{equation}
then $x(t)$ satisfies the integral equation
\begin{equation}\label{09:I}
x(t) = x_0 + \int_{t_0}^{t} G(s,x(s))\,ds
\end{equation}
Conversely, if $x(t):[a,b]\to\R^n$ is continuous and $x(t)$ satisfies
(\ref{09:I}), then $x(t)$ is $C^1$ and $x(t)$ satisfies~(\ref{09:D}).
\end{lemma}

\proof The first statement follows by integrating the
differential equation to get
\[ x(t) - x(t_0) = \int_{t_0}^t x'(s)\,ds = \int_{t_0}^t G(s,x(s))\,ds \]
The second statement follows by differentiating the integral equation to get
\[\frac{d}{dt} x(t) = \frac{d}{dt} \Big( x_0 + \int_{t_0}^t G(s,x(s))\,ds
\Big) = G(t,x(t)) \]
\qed

Notice that the integral equation has the initial condition ``built-in.''
Also, we need only assume $x(t)$ is continuous for the integral to exist,
yet any solution of the integral equation gets ``promoted'' to having
a continuous derivative.

Now we describe the type of complete metric space on which we shall
construct a contractive mapping. The sup-norm on $C([a,b],\R^n)$ is
\[ \|f\|_\infty = \sup_{x\in[a,b]}|f(x)| \]
where $|f(x)|$ is the Euclidean norm of the vector $f(x)$. Convergence
with respect to this norm is equivalent to component-wise convergence
with respect to the sup-norm on~$C[a,b]$.

\begin{lemma}\label{09:C([a,b],B)}
Let $B$ be a closed subset of $\R^n$. Then the metric space $C([a,b],B)$
is complete with respect to the sup-norm.
\end{lemma}

\proof Let $\{f_j\}\subset C([a,b],B)$ be a Cauchy sequence. Then
$\|f_j - f_k\|_\infty\to 0$ implies $\|f_{ji}-f_{ki}\|_\infty\to0$ where
$f_{ji}$ are the components of $f_j =(f_{j1},\dots,f_{jn})$, and hence
$\{f_{ji}\}\subset C[a,b]$ is a Cauchy sequence for each $i=1,\dots,n$.
Since $C[a,b]$ is complete by Theorem~\ref{07:C[a,b]}, $f_{ji}\to f_i\in
C[a,b]$, so $f_j \to f = (f_1,\dots,f_n)\in C([a,b],\R^n)$. Finally, Since
$B$ is closed, $f(x) = \lim f_j(x) \in B$, so $f\in C([a,b],B)$. \qed

The next two lemmas take care of some technical details. The first
is an extension of a standard inequality for continuous functions,
to continuous vector functions.

\begin{lemma}[Minkowski's Inequality]\label{09:MI}
If $v:[a,b]\to\R^n$ is a continuous function, then
\[ \Big|\int_a^b v(s)\,ds\Big| \le \int_a^b|v(s)|\,ds \]
\end{lemma}

\proof The statement follows from the triangle inequality for the
Euclidean metric. Integration of a vector function is done
component-wise,
\[ \int_a^b v(s)\,ds = \Big(\int_a^b
v_1(s)\,ds,\dots,\int_a^bv_n(s)\,ds\Big) \]
and for each component
\[ \int_a^b v_i(s)\,ds = \sup\Big\{\int_a^b t_i(s)\,ds\ \Big|\ t_i \
(\mbox{step fn}) \le v_i \Big\} = \lim_{j\to\infty} \int_a^b t_{ij}(s)\,ds \]
for some sequence $\{t_{ij}\}$ of step functions $\le v_i$.
Recall that the integral of a step function is a finite sum,
\[ \int_a^b t_{ij}(s)\,ds = \sum_k t_{ij}(s_k)\Delta s_k \]
Therefore, by using a common refinement for the partitions of the
step functions $t_{1j},\dots,t_{nj}$, for each $j$, we get
\begin{eqnarray*}
\Big|\int_a^b v(s)\,ds\Big| &=& \lim_{j\to\infty} \Big|\sum_k
(t_{1j}(s_k),\dots,t_{nj}(s_k))\Delta s_k \Big|\\
&\le& \lim_{j\to\infty} \sum_k |(t_{1j}(s_k),\dots,t_{nj}(s_k))|\Delta s_k
\quad\mbox{[T.I.]}\\
&\le& \sup\Big\{ \int_a^b u(s) \,ds\ \Big|\ u\ (\mbox{step fn}) \le |v| \Big\}\\
&=& \int_a^b |v(s)|\,ds
\end{eqnarray*}
\qed

Finally, we establish a crucial regularity property of $G$
that follows from the continuity of the derivatives of~$G$.

\begin{lemma}[Lipschitz Condition]\label{09:LIPSCHITZ}
Let $D$ be open an open subset of $\R^n$ and let $G:[a,b]\times
D\to\R^n$ be a $C^1$ function. Then for any $x_0\in D$ and any open ball
$B = B(x_0,\epsilon)$ such that $\overline{B}\subset D$, there exists
an $M>0$ such that
\[ |G(t,x) - G(t,y)| \le M|x - y|,\quad\forall\,(t,x),(t,y)\in[a,b]\times
\overline{B} \]
\end{lemma}

\proof
Let $(t,x),(t,y)\in[a,b]\times \overline{B}$ be arbitrary. Since the ball
$\overline{B}$ is convex, the line segment from $x$ to $y$ is contained
in $\overline{B}$,
\[ x + s(x-y) \in \overline{B},\quad \forall\,s\in[0,1] \]
Consider the continuous function $g:[0,1]\to\R^n$ defined by
\[ g(s) = G(t, x + s(y-x)) \]
Then
\[ G(t,y) - G(t,x) = g(1) - g(0) = \int_0^1 g'(s)\,ds \]
By the Chain Rule,
\[ g'(s) = \sum_{k=1}^n (y_k - x_k) \frac{\partial G}{\partial x_k}(t,x +
s(y-x)) \]
Let $m$ be the maximum of all the continuous functions
$\big|\partial G/\partial x_k\big|$, $k=1,\dots,n$, on the compact
set $[a,b]\times\overline{B}$ and let $M = m\cdot n$. Then
\begin{eqnarray*}
|G(t,y)-G(t,x)| &=& \Big|\int_0^1 g'(s)\,ds\Big|\\
&\le& \int_0^1|g'(s)|\,ds\quad\mbox{[Lemma~\ref{09:MI}]}\\
&\le& \sum_{k=1}^n|y_k-x_k|\int_0^1\Big|\frac{\partial G}{\partial
x_k}(t,x+s(y-x))\Big|\,ds\\
&\le& m \sum_{k=1}^n |y_k - x_k|\\
&\le& m\cdot n\cdot\|y - x\|_\infty\\
&\le& M\cdot|x - y|
\end{eqnarray*}
\qed

\subsection{Existence and Uniqueness of Solutions}
Now we come to the proof of the existence and uniqueness of a solution
to an initial value problem. The basic idea is to show that the mapping of
continuous vector functions defined by $Tx(t) = x_0 + \int_{t_0}^t
G(s,x(s))\,ds$ is, under certain restrictions, a \emph{contractive
mapping}. The Contractive Mapping Principle implies that $T$ has a
unique fixed point, say $T\phi(t) = \phi(t)$. The function $\phi(t)$ then
satisfies the integral equation $\phi(t) = T\phi(t) = x_0 + \int_{t_0}^t
G(s,\phi(s))\,ds$, and therefore the initial value problem $\phi'(t) =
G(t,\phi(t))$, $\phi(t_0) = x_0$.

\begin{theorem}[Existence and Uniqueness]\label{09:E+U}
Let $G:[a,b]\times D\to\R^n$ be a $C^1$ function defined on some
neighborhood of $(t_0,x_0)\in\R\times \R^n$. Then there is a subinterval
$[c,d]\subset[a,b]$ containing $t_0$ such that the initial value problem
\[ x'(t) = G(t,x(t)),\quad x(t_0) = x_0 \]
has a unique solution defined on the interval $[c,d]$.
\end{theorem}

\proof
Let $B = B(x_0,\epsilon)$ be such that $\overline{B}\subset D$.
By Lemma~\ref{09:LIPSCHITZ}, $\exists\,M>0$ such that
\[ |G(t,x) - G(t,y)| \le M|x - y|,\quad\forall\,(t,x),(t,y)\in[a,b]\times
\overline{B} \]
Let $\|G\|_\infty$ be the maximum of $|G(t,x)|$ on $[a,b]\times
\overline{B}$. Fix $0 < r < 1$ and let $[c,d]$
be the interval $[t_0-\epsilon_0,t_0+\epsilon_0]$ where
\[ \epsilon_0 = \min\{\epsilon/\|G\|_\infty, r/M\} \]
For a continuous function $x:[c,d]\to \overline{B}$ define
\[ Tx(t) = x_0 + \int_{t_0}^t G(s,x(s))\,ds \]
Then
\[ |Tx(t) - x_0| = \Big|\int_{t_0}^t G(s,x(s))\,ds\Big|
\le \|G\|_\infty|t - t_0| \le \epsilon \]
since $|t - t_0| \le \epsilon_0 \le \epsilon/\|G\|_\infty$. Thus,
$Tx: [c,d]\to \overline{B}$ showing that $T$ is a mapping of the
metric space $C([c,d],\overline{B})$ into itself. Moreover,
by Lemma~\ref{09:C([a,b],B)}, $C([c,d],\overline{B})$ with the
sup-norm is a complete metric space.

We now show that $T$ is a contractive mapping. If $x,y\in
C([c,d],\overline{B})$ then
\begin{eqnarray*}
\|Tx - Ty\|_\infty &=& \Big|\Big|x_0 + \int_{t_0}^t G(s,x(s))\,ds
- x_0 - \int_{t_0}^t G(s,y(s))\,ds\Big|\Big|_\infty\\
&=& \sup_{t\in[c,d]}\Big|\int_{t_0}^t G(s,x(s)) - G(s,y(s))\,ds\Big|\\
&\le&\sup_{t\in[c,d]}\Big|\int_{t_0}^t |G(s,x(s)) - G(s,y(s))|\,ds\Big|\quad
\mbox{[Lemma~\ref{09:MI}]}\\
&\le& \sup_{t\in[c,d]}\Big|\int_{t_0}^t M|x(s) - y(s)|\,ds\Big|
\quad\mbox{[Lemma~\ref{09:LIPSCHITZ}]}\\
&\le& M \|x - y\|_\infty \sup_{t\in[c,d]} |t-t_0| =
M \|x - y\|_\infty \epsilon_0\\
&\le& r \|x - y\|_\infty
\end{eqnarray*}
since $\epsilon_0\le r/M$.

By the Contractive Mapping Principle, Theorem~\ref{08:CMP}, $T$ has a unique
fixed point $\phi\in C([c,d],\overline{B})$. Therefore, $\phi(t)$
is the unique solution of the integral equation
\[ \phi(t) = T\phi(t) = x_0 + \int_{t_0}^tG(s,\phi(s))\,ds \]
and hence, by Lemma~\ref{09:INTEGRALEQN}, $\phi(t)$ is the unique solution
of the initial value problem
\[ \phi'(t) = G(t,\phi(t)),\quad \phi(t_0) = x_0 \]
\qed

\subsection{Picard Iteration}
Let us apply the method of the previous proof to find a solution to the
initial value problem
\[ x'(t) = 2t(1 + x(t)),\quad x(0) = 0 \]
Here $G(t,x) = 2t(1+x)$ has continuous first derivatives
and the Lipschitz Condition is easy to determine. If $|t|\le R$, then
\[ |G(t,x) - G(t,y)| = |2t + 2tx - 2t - 2ty| = 2|t|\cdot|x-y| \le 2R|x-y| \]
So $M = 2R$ independent of $x$ and~$y$. The maximum of $G$ on
$[-R,R]\times[-\epsilon,\epsilon]$ is
\[ \|G\|_\infty = \sup|2t(1+x)| = 2R(1 + \epsilon) \]
The proof of the theorem says that for some fixed $0 < r < 1$ the mapping
\[ Tx(t) = 0 + \int_0^t 2s(1+x(s))\,ds \]
is contractive if we restrict $t$ to the interval
$[-\epsilon_0,\epsilon_0]$ where
\[ \epsilon_0 =\min\Big\{\frac{\epsilon}{\|G\|_\infty},\frac{r}{M}\Big\} =
\min\Big\{\frac{\epsilon}{2R(1+\epsilon)}, \frac{r}{2R}\Big\} < \frac{1}{2R} \]
For the largest interval of $t$ we should choose $R = 1/\sqrt{2}$.

The solution $\phi(t)$ of the initial value problem is the fixed point of
$T$ which can be found, according to the Contractive Mapping Principle,
by iterating the mapping $T$ on any starting function $\phi_0(t)$. Let
$\phi_0(t) = 0$ and for $n\ge 1$ define recursively
\[ \phi_n(t) = T\phi_{n-1}(t) = \int_0^t
2s(1+\phi_{n-1}(s))\,ds \]
Then $\phi_n(t) = T^n\phi_0(t)$ and
\[ \phi(t) = \lim_{n\to\infty}T^n\phi_0(t) = \lim_{n\to\infty}\phi_n(t) \]
We obtain
\begin{eqnarray*}
\phi_1(t) &=& T\phi_0(t) = \int_0^t 2s(1 + 0)\,ds = t^2\\
\phi_2(t) &=& T\phi_1(t) = \int_0^t 2s(1 + s^2)\,ds = t^2 + \frac{t^4}{2}\\
\phi_3(t) &=& T\phi_2(t) = \int_0^t 2s(1 + s^2+ \frac{s^4}{2})\,ds =
t^2 + \frac{t^4}{2} + \frac{t^6}{3\cdot 2}
\end{eqnarray*}
These equations suggest that 
\[ \phi_n(t) = t^2 + \frac{t^4}{2!} +
\frac{t^6}{3!}+\cdots+\frac{t^{2n}}{n!} \]
which can be proved by induction:
\begin{eqnarray*}
\phi_{n}(t) &=& \int_0^t 2s(1 + \phi_{n-1}(s))\,ds\\
&=& \int_0^t 2s\big(1 + s^2 + \cdots + \frac{s^{2(n-1)}}{(n-1)!}\Big)ds\\
&=& t^2 + \frac{t^4}{2!} + \frac{t^6}{3!}+\cdots+\frac{t^{2n}}{n!}
\end{eqnarray*}
The solution to the initial value problem is thus
\[ \phi(t) = \lim_{n\to\infty}\phi_n(t) =
\sum_{n=1}^\infty\frac{t^{2n}}{n!} = e^{t^2} - 1 \]
After finding the limit we see that it converges for all $t$, and hence
the solution is also valid for all~$t$, although the \emph{a priori}
restriction $|t| \le 1/\sqrt{2}$ is needed to invoke the Contractive
Mapping Principle.

\subsection{Exercises}

\begin{enumerate}
\item Consider the initial value problem
\[ x'(t) + x(t)^2 = 0,\quad x(0) = x_0 > 0\]

\begin{itemize}
\item[\textit{a)}] Show that $x(t) = x_0/(1 + x_0 t)$ is a solution for $t
> -1/x_0$.
\item[\textit{b)}] The proof of Theorem~\ref{09:E+U} asserts that a solution
exists on an interval $[x_0-\epsilon_0,x_0+\epsilon_0]$ where
\[ \epsilon_0 =\min\Big\{\frac{\epsilon}{\|G\|_\infty},\frac{r}{M}\Big\} \]
Calculate $\epsilon_0$ for this equation and verify that $x_0-\epsilon_0 >
-1/x_0$.
\end{itemize}

\item Find two distinct solutions of the initial value problem
\[ x'(t) = x(t)^{1/3},\quad x(0) = 0 \]
Explain how this is compatible with the uniqueness of solutions asserted by
Theorem~\ref{09:E+U}.

\item Use Picard Iteration (the Contractive Mapping Principle) to compute
the solution of the initial value problem
\[ x'(t) = t x(t) + 1, \quad x(0) = 0 \]
\begin{itemize}
\item[\textit{a)}] Let $\phi_0(t) = 0$ and for $n\ge 1$ compute
recursively
\[ \phi_n(t) = T\phi_{n-1}(t) = 0 + \int_0^t (s \phi_{n-1}(t) + 1)\,ds \]
\item[\textit{b)}] Find $\phi(t) = \lim_{n\to\infty}\phi_n(t)$.
\end{itemize}

\end{enumerate}

 \end{document}
