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\begin{document}


\section{Basic Theory of First-Order Equations}

\vspace{1pt}The differential equations to be discussed in this section can
be expressed in the form 
\[
\frac{dy}{dx}=f(x,y)\text{.} 
\]

\vspace{1pt}Assume that the function $f$ is defined in a region $D$ of the $%
xy$ plane. An \textbf{explicit} \textbf{solution\label{Explicit Solution}}
is a function $y=g(x)$ defined on an interval $I$ such that $(x,g(x))\in D$
and $g^{\prime }(x)=f(x,g(x))$ for all $x\in I.$ An \textbf{implicit} 
\textbf{solution\label{Implicit Solution}} is a relation $\varphi (x,y)=0$
that can solved for $y$ to yield an explicit solution.

\begin{example}
For the equation $\dfrac{dy}{dx}=\dfrac{x}{y}$, $D$ $=$ the upper
half-plane, the function $x^{2}-y^{2}-1=0$ is an implicit solution. The
function $y=\sqrt{x^{2}-1}$ defined on the interval $I=\left( 0,1\right) $
satisfies $\left( x,\sqrt{x^{2}-1}\right) \in D$ and 
\[
\dfrac{d}{dx}\left( \sqrt{x^{2}-1}\right) =\allowbreak \frac{x}{\sqrt{x^{2}-1%
}}=\,f(x,g(x))
\]
so it is an explicit solution.
\end{example}

The integration techniques frequently used to solve the equation $\frac{dy}{%
dx}=f(x,y)$ introduce an arbitrary constant into the solution so that there
are infinitely many solutions on a given interval $I$. Therefore, the 
\textbf{general} \textbf{solution\label{General Solution}} will appear in
either the form $y=g(x,c)$ or the form $\varphi (x,y,c)=0$ where, for each
value of $c$, a \textbf{particular} \textbf{solution\label%
{Particular Solution}} is obtained.

The graphs of the solutions are called the \textbf{integral curves\label%
{Integral Curves}}. The slopes of these curves are determined by the
derivative function $f$. When $f$ satisfies certain conditions, each point
in $D$ is on exactly one integral curve. Therefore, in order to talk about
the uniqueness of solutions, one is led to the problem of finding a solution
passing through a given point in $D.\medskip $

\begin{description}
\item[Initial-value Problem]  \label{Initial-value Problem}Suppose $%
(x_{0},y_{0})\in D$. Find an interval $I$ containing $x_{0}$ and a solution $%
g$ defined on $I$ satisfying $g(x_{0})=y_{0}$.\medskip 
\end{description}

\subsection{Existence and Uniqueness Theorems\label%
{Existence and Uniqueness Theorems}}

The theorems below establish conditions for the existence and uniqueness of
solutions to the initial-value problem.

\begin{theorem}
Let $f$ be a real-valued function, defined and continuous on a region $D$ of
the $xy$ plane, and let $(x_{0},y_{0})\in D$. A function $g$, defined and
continuous on an interval $I$, is a solution satisfying $g(x_{0})=y_{0}$ if
and only if $(x,g(x))\in D$ and $g(x)=y_{0}+\int_{x_{0}}^{x}f(t,g(t))dt$ for
all $x\in I$.
\end{theorem}

\begin{theorem}
\textbf{(Cauchy-Peano)} \ Let $f$ be a real-valued function, defined and
continuous on a region $D$ of the $xy$ plane. Let $(x_{0},y_{0})\in D$ and
suppose the rectangle $R:|x-x_{0}|\leq a,|y-y_{0}|\leq b$ is contained in $D$
for some $a>0,b>0$. Let $M=\max |f(x,y)|$ for $(x,y)\in R$ and let $\alpha
=\min (a,\frac{b}{M})$. Then there exists a solution $g$ defined on the
interval $|x-x_{0}|\leq \alpha $ and satisfying $g(x_{0})=y_{0}$.
\end{theorem}

\begin{definition}
Let $f$ be a real-valued function defined on a region $D$ of the $xy$ plane.
Then $f$ satisfies a \textbf{Lipschitz condition\label{Lipschitz Condition}}
with respect to $y$ in $D$ if there exists a constant $k$ such that 
\[
|f(x,y_{1})-f(x,y_{2})|\leq k|x_{1}-x_{2}|\text{ }
\]
for every $(x,y_{1})$ and $(x,y_{2})$ in $D$. The constant $k$ is called a 
\textbf{Lipschitz constant} for $f$.
\end{definition}

\begin{definition}
A region $D$ in the $xy$ plane is \textbf{convex} if the line segment
connecting any two points in $D$ is contained in $D$.
\end{definition}

\begin{theorem}
A sufficient condition for a function $f$ to satisfy a Lipschitz condition
on a convex region $D$ is that the partial derivative $\dfrac{\partial f}{%
\partial y}$ be defined and bounded on $D$. In this case, a Lipschitz
constant for $f$ is given by 
\[
k=\max \left\{ \left| \frac{\partial f}{\partial y}(x,y)\right| :(x,y)\in
D\right\} \text{.}
\]
\end{theorem}

\begin{theorem}
\vspace{1pt}Let $f$ be a real-valued function, defined and continuous on a
region $D$ of the $xy$ plane. Suppose $f$ satisfies a Lipschitz condition
with respect to $y$ in $D$. If $g_{1}$ and $g_{2}$ are two solutions defined
on an open interval $I$ such that $g_{1}(x_{0})=g_{2}(x_{0})$ for some $%
x_{0}\in I$, then $g_{1}(x)=g_{2}(x)$ for all $x\in I$.
\end{theorem}

The Cauchy-Peano theorem does not provide a constructive method for
obtaining a solution. Another approach for establishing the existence of
solutions is the \textbf{method of successive approximations\label%
{Successive Approximations}. }This method involves the construction of a
sequence of functions which converges \textit{uniformly} to a solution of
the initial value problem.

\begin{definition}
Let $\{g_{n}\}$ and $g$ be functions defined on an interval $I$. Then the
sequence $\{g_{n}\}$ \textbf{converges} \textbf{uniformly\label%
{Uniform Convergence}} to $g$ on $I$ if for every $\epsilon >0$, there
exists an integer $N$ such that $n\geq N$ implies 
\[
\left| g_{n}(x)-g(x)\right| \leq \epsilon \text{.}
\]
\end{definition}

\begin{theorem}
In addition to the hypothesis of the Cauchy-Peano theorem, suppose that the
function $f$ satisfies a Lipschitz condition with respect to $y$ in $D$. \
For $\left| x-x_{0}\right| \leq \alpha $ and\quad $k=0,1,....$define the
successive approximations (Picard iterates) by 
\begin{eqnarray*}
g_{0}(x) &=&y\Sb 0 \\  \endSb  \\
g_{n+1}(x) &=&y_{0}+\int_{x_{0}}^{x}f(t,g_{n}(t))dt
\end{eqnarray*}
\quad Then each $g_{n}$ exists on $\left| x-x_{0}\right| \leq \alpha $ as a
continuous function and the sequence $\left\{ g_{n}\right\} $ converges
uniformly on this interval to the unique solution $g$ satisfying $%
g(x_{0})=y_{0}$.
\end{theorem}

The restriction that $f$ satisfy a Lipschitz condition is not only necessary
for uniqueness, but is also required in the above theorem to show that the
limit function $g$ is actually a solution.

\subsection{\ Related Scientific Notebook Topics}

{\small \hyperref{Exact solutions}{}{}{../../Help/DM10-1.tex#Differential
equation}}

{\small \hyperref{Initial value problems}{}{}{../../Help/DM10-3.tex#Initial
value problems}}

{\small \hyperref{Solve ODE, exact}{}{}{../../Help/DM10-1.tex#Solve ODE}}

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