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\begin{document}
\centerline{Math 322 lecture for Monday, Week 7}
\bigskip

\centerline{\sc fundamental existence and uniqueness theorem}
\medskip

Our goal is to apply the contraction mapping principle to the operator
\begin{align*}
  T\colon C(I)&\to C(I)\\
  u&\mapsto x_0+\int_{s=0}^tf(u(s))\,ds
\end{align*}
in order to prove the fundamental existence and uniqueness theorem for ordinary
differential equations.
\medskip

\textbf{Derivative review.}
Let~$E\subseteq\R^n$ be an open set.  Recall from vector calculus that the
derivative of a function~$f\colon E\to\R^n$ at a point~$p\in E$ is a linear
function
\[
Df_p\colon\R^n\to\R^n
\]
approximating~$f$ near~$p$:
\[
  f(p+h)\approx f(p)+ Df_p(h)
\]
for small~$h$.  Its corresponding matrix is the Jacobian matrix for~$f$
at~$p$, whose~$j$-th column is the~$j$-th partial of~$f$ (measuring
how~$f$ is changing in the~$j$-th coordinate direction):
\[
  \frac{\partial f}{\partial x_j}(p)
  =
  \left(\begin{array}{c}
      \frac{\partial f_1}{\partial x_j}(p)\\[7pt]
      \frac{\partial f_2}{\partial x_j}(p)\\[7pt]
      \vdots\\[7pt]
      \frac{\partial f_n}{\partial x_j}(p)
  \end{array} \right).
\]

We say~$f\colon E\to\R^n$ is
{\em continuously differentiable} if it is differentiable at all points
in~$E$ and the mapping
\begin{align*}
  E&\to\mathcal{L}(\R^n)\\
  p&\mapsto Df_p
\end{align*}
is continuous. 

{\bf Explanation:} First,~$\mathcal{L}(\R^n)$ denotes the vector
space of linear functions from~$\R^n$ to itself.  Second, to talk about
continuity we define a norm on~$\mathcal{L}(\R^n)$:
for~$L\in\mathcal{L}(\R^n)$, let
\[
  \|L\|=\max_{|x|\leq 1}|L(x)|.
\]
This is the same as~$\|A\|$ if~$A$ is the matrix representing~$L$.  In that
case, since $L(x)=Ax$, the
inequality~$|Ax|\leq\|A\||x|$ can be written as
\[
  \|L(x)\|\leq\|L\|\,|x|.
\]

A theorem from calculus says that~$f$ is continuously differentiable if and only
if all of its partials~$\partial f_i/\partial x_j$ exist and are continuous.
(Also, it turns out that continuity of the partials guarantees that~$f$ is
  differentiable.)
\medskip

\textbf{Notation.} For an open subset~$E\subset\R^n$, we denote
the~$\R$-vector space of continuously differentiable functions on~$E$
by~$C^1(E)$.
\bigskip

\textbf{Lipschitz condition.}  We now introduce a condition on vector fields
that will allow the application of the contraction mapping principle to~$T$.
\medskip

\textbf{Definition.} Let~$E\subseteq\R^n$ be an open subset.  Then a
function~$f\colon E\to\R^n$ is \emph{Lipschitz} if there exists a
constant~$K$ such that
\[
  |f(x)-f(y)|\leq K|x-y|
\]
for all~$x,y\in E$.  On the other hand,~$f$ is \emph{locally Lipschitz}
on~$E$ if for each~$x_0\in E$, there exists~$\ve>0$ and a constant~$K_{x_{0}}$
such that
\[
  |f(x)-f(y)|\leq K_{x_{0}}|x-y|
\]
for all
\[
  x,y\in N_{\ve}(x_0):=\left\{ x\in\R^n : |x-x_0|<\ve\right\}.
\]
\medskip

\textbf{Proposition.} If~$f\in C^1(E)$, then~$f$ is locally Lipschitz.
\begin{proof}
 Let~$x_0\in E$. Since~$E$ is open it contains an open ball
 about~$x_0$, i.e., there exists~$\eta>0$ such that~$N_{\eta}(x_0)\subset E$.
 Define~$\ve:=\eta/2$ and consider the closed ball
 \[
   B:= B_{\ve}(x_0):=\overline{N_{\ve}(x_0)}:=\left\{ x\in\R^n: |x-x_0|\leq\ve \right\}.
 \]
 Let
 \[
   K_{x_{0}}:=\max_{x\in B}\|Df_x\|.
 \]
 The constant~$K_{x_{0}}$ exists since we're assuming~$Df$ is continuous
 ($f\in C^1(E)$).  Thus, $x\to Df_x\to\|Df_x\|$, being the composition of
 continuous functions, is also continuous.
 
 Since~$B$ is convex, given~$x,y\in B$, the line segment
 joining~$x$ to~$y$ is contained in~$B$.  Hence, it is OK to stick
 these points into~$f$. Parametrize the line segment by~$\phi(s)=x+s(y-x)$
 for~$s\in[0,1]$ and consider the composition
 \begin{align*}
   F:=f\circ\phi:\colon[0,1]&\to\R^n\\
   s&\mapsto f(x+s(y-x)),
 \end{align*}
 a curve in~$\R^n$.
 By the chain rule,
 \[
   DF_s=Df_{\phi(s)}\circ D\phi_s.
 \]
 Since~$F$ is a curve in~$\R^n$, its Jacobian matrix at~$s$ is a single column
 vector---the tangent or velocity vector~$F'(s)$---and
 \[
   DF_s(t) = tF'(s),
 \]
 a linear function of~$t$ (for fixed~$s$).  Similarly~$\phi_s$ is a curve
 in~$\R^n$, so its Jacobian matrix is its velocity at time~$s$.  It's easy to
 compute: since~$\phi(s)=x+s(y-x)$, its velocity is constant. At any
 time~$s$, we have~$\phi'(s)=y-x$.  Thus,
 \[
   D\phi_s(t)=t(y-x).
 \]
 By the chain rule,
 \[
   tF'(s)=DF_s(t)=Df_{\phi(s)}(t(y-x)).
 \]
 Setting~$t=1$, we get
 \[
   F'(s)=Df_{(x+s(y-x))}(y-x)\in\R^n.
 \]
 
 Since~$F(0)=f(x)$ and~$F(1)=f(y)$,
 \begin{align*}
   |f(y)-f(x)|
   &= |F(1)-F(0)|\\
   &= \left|\int_{s=0}^1 F'(s)\,ds\right|\\
   &\leq \int_{s=0}^1 |F'(s)|\,ds\\
   &= \int_{s=0}^1|Df_{(x+s(y-x))}(y-x)|\,ds\\
   &\leq \int_{s=0}^1\|Df(x+s(y-x))\|\,|y-x|\,ds\\
   &\leq K_{x_{0}}\int_{s=0}^1|y-x|\,ds\\
   &= K_{x_{0}}|y-x|.
 \end{align*}
 We've shown that~$f$ is locally Lipschitz.
\end{proof}
\medskip

\textbf{Theorem. (The fundamental existence and uniqueness theorem for nonlinear
systems.)} Let~$E$ be an open subset of~$\R^n$ containing~$x_0$, and
let~$f\in C^1(E)$.  Then there exists~$a>0$ such that the initial value problem
\begin{align*}
  x'&= f(x)\\
  x(0)&= x_0
\end{align*}
has a unique solution~$x(t)$ on~$[-a,a]$.
\begin{proof}
 Since~$f\in C^1(E)$, there exists an~$\ve>0$ such that
 $N_{\ve}(x_0)\subseteq E$, the open ball of radius~$\ve$ centered at~$x_0$, and
 there exists a constant~$K_{x_{0}}$ such that
 \[
   |f(x)-f(y)|\leq K_{x_{0}}|x-y|
 \]
 for all~$x,y$ in~$N_{\ve}(x_0)$.  By replacing~$\ve$ by~$\ve/2$, we may assume
 \[
   |f(x)-f(y)|\leq K_{x_{0}}|x-y|
 \]
 for all~$x,y$ in
 \[
   B:=\overline{N_{\ve}(x_0)}:=\left\{ x\in\R^n:|x-x_0|\leq \ve \right\}\subset E.
 \]
 (The point here is to get the Lipschitz condition to hold on a closed bounded
 ball rather then on the open ball, $N_{\varepsilon}(x_{0})$, in preparation for
 an application of the extreme value theorem, below.)

 Let~$I=[-a,a]$ where~$a>0$ is a constant to be determined later, and define
 \[
   X:=\left\{ u\in C(I)\colon \|u-x_0\|\leq\ve \right\},
 \]
 considering~$x_0\in C(I)$ as the constant function~$t\mapsto x_0$ for
 all~$t\in I$.
 This means that for~$u\in X$, we have
 \[
   \max_{t\in I}|u(t)-x_0|\leq\ve.
 \]
 In particular,~$u(t)\in B\subset E$ for all~$t\in I$.  Note that~$B$ is a subset
 of~$E\subseteq \R^n$ and~$X$ is a subset of the function space~$C(I)$ of
 continuous functions~$I\to\R^n$.  If~$u\in X$, then~$u(t)\in B$ for
 all~$t\in I$.
 
 Our goal is to show that~$a$ can be taken small enough so that (i)~$T(u)\in X$ for
 all~$u\in X$, i.e., so that~$T\colon X\to X$, and so that
 (ii)~$T\colon X\to X$ is a contraction mapping.
 
 For (i), since~$B$ is closed and bounded, we can define
 \[
   M=\max_{x\in B}|f(x)|.
 \]
 Suppose that~$0<a<\frac{\ve}{M}$.  Then for~$u\in X$ and~$t\in I$,
 \begin{align*}
   |T(u)(t)-x_0|
   &= \left|\left( x_0+\int_{s=0}^tf(u(s))\,ds \right)-x_0\right|\\[10pt]
   &= \left|\int_{s=0}^tf(u(s))\,ds\right|\\[10pt]
   &\leq \left|\int_{s=0}^t\left|f(u(s))\right|\,ds\right|.
 \end{align*}
 If~$s$ is in the interval between~$0$ and~$t$ and~$u\in X$, it follows that~$u(s)\in B$, and
 hence,~$|f(u(s))|\leq M$.  Therefore, continuing our calculation,
 \begin{align*}
   |T(u)(t)-x_0|
   &= \left|\int_{s=0}^t\left|f(u(s))\right|\,ds\right|\\[10pt]
   &= \left|\int_{s=0}^tM\,ds\right|\\[10pt]
   &= |t|\, M\\[10pt]
   &\leq a\, M\\[10pt]
   &< \frac{\ve}{M}\, M\\[10pt]
   &< \ve.
 \end{align*}
 Hence,
 \[
   \|T(u)-x_0\|:= \max_{t\in I}|T(u)(t)-x_0|<\ve.
 \]
 Therefore~$T(u)\in X$.  In sum: if~$0<a<\ve/M$, then~$T\colon X\to X$.
 
 We now work on~(ii): we can take~$a$ small enough so that~$T\colon X\to X$ is
 a contraction mapping.  Let~$u,v\in X$.  Then, using the Lipschitz property,
 \begin{align*}
   |T(u)-T(v)|
   &= \left|\int_{s=0}^t f(u(s))-f(v(s))\,ds\right|\\[8pt]
   &\leq \left|\int_{s=0}^t \left|f(u(s))-f(v(s))\right|\,ds\right|\\[8pt]
   &\leq K_{x_{0}}\left|\int_{s=0}^t \left|u(s)-v(s)\right|\,ds\right|\\[8pt]
   &\leq K_{x_{0}}\left|\int_{s=0}^t \max_{c\in I}\left|u(c)-v(c)\right|\,ds\right|\\[8pt]
   &= K_{x_{0}}\left|\int_{s=0}^t \|u-v\|\,ds\right|\\[8pt]
   &= K_{x_{0}}\,|t|\,\|u-v\|\\[8pt]
   &\leq aK_{x_{0}}\|u-v\|.
 \end{align*}
 To ensure~$T$ is a contraction mapping, take~$a=\frac{1}{2K_{x_{0}}}$
 (so that~$aK_{x_{0}}=\frac{1}{2}<1$). 
 
 In total, we have now shown there exists and interval~$I=[-a,a]$, a closed ball~$X\subset C(I)$ centered at the
 constant function~$x_0$, such that~$T\colon X\to X$ and~$T$ is a contraction
 mapping.  It therefore has a unique fixed point~$x\in X$.  So $x=T(x)$, i.e.,
 \[
   x(t)=T(x)(t):=x_0+\int_{s=0}^tf(x(s))\,ds.
 \]
 By the fundamental theorem of calculus and the fact that~$x(0)=x_0$, it follows
 that~$x$ is a solution to the initial value problem
 \begin{align*}
   x'&= f(x)\\
   x(0)&= x_0
 \end{align*}
 on~$I$.  For uniqueness, recall that any solution~$x$ on~$I$ will be a fixed
 point for~$T$: 
 \[
   T'(x)(t)=\left( x_0+\int_{s=0}^tf(x(s))\,ds \right)'=f(x(t))=x'(t).
 \]
 so~$T(x)$ and~$x$ differ by a constant.  However~$T(x(0))=x_0=x(0)$, so that
 constant is~$0$.  Since every solution is a fixed point of~$T$ and contraction
 mappings have unique fixed points, we are done.
\end{proof}
\end{document}
