Cauchy initial value problemCauchyInitialValueProblem2005-01-19 06:57:402008-03-14 05:19:59Definitionsolution to the Cauchy problemsolution to the initial value problem% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\R}{\mathbb R}Let $D$ be a subset of $\R^n\times \R$, $(x_0,t_0)$ a point of $D$, and $f\colon D\to \R$ be a function.
We say that a function $x(t)$ is a solution to the Cauchy (or initial value) problem
\begin{equation}
\begin{cases}
x'(t)=f(x(t),t)\\
x(t_0)=x_0
\end{cases}
\end{equation}
if
\begin{enumerate}
\item $x$ is a differentiable function $x\colon I\to \R^n$ defined on a interval $I\subset \R$;
\item one has $(x(t),t)\in D$ for all $t\in I$ and $t_0\in I$;
\item one has $x(t_0)=x_0$ and $x'(t)=f(x(t),t)$ for all $t\in I$.
\end{enumerate}
We say that a solution $x\colon I\to\R^n$ is a \emph{maximal solution} if it cannot be extended to a bigger interval. More precisely given any other solution $y\colon J\to \R^n$ defined on an interval $J\supset I$ and such that $y(t)=x(t)$ for all $t\in I$, one has $I=J$ (and hence $x$ and $y$ are the same function).
We say that a solution $x\colon I\to\R^n$ is a \emph{global solution} if $D\subset=\R^n \times I$.
We say that a solution $x\colon I\to\R^n$ is \emph{unique} if given any other solution $y\colon I\to\R^n$ one has $x(t)=y(t)$ for all $t\in I$ (i.e.\ $x$ is the unique solution defined on the interval $I$).
\subsection{Notation}
Usually the differential equation in (1) is simply written as $x'=f(x,t)$.
Also, depending on the topics, the name chosen for the function and for the variable, can change. Other common choices are $y'=f(y,t)$ or $y'=f(y,x)$.
It is also common to write $\dot x=f(x,t)$ when the independent variable represents a time value.
\subsection{Examples}
\begin{enumerate}
\item
The function $x(t)=\log t$ defined on $I=(0,+\infty)$ is the unique maximal solution to the Cauchy problem:
\[
\begin{cases}
x'(t) = 1/t\\
x(1)=0.
\end{cases}
\]
In this case $f(x,t)=1/t$, $D=\{(x,t)\colon t\neq 0\}$, $t_0=1$, $x_0=0$.
\item
The function $x(t)=e^x$ is a global (and hence maximal), unique solution to the Cauchy problem:
\[
\begin{cases}
x'(t) = x(t)\\
x(0)=1.
\end{cases}
\]
\item
Consider the Cauchy problem
\[
\begin{cases}
x'(t) = \frac 3 2 \sqrt[3] x\\
x(0)=0.
\end{cases}
\]
The function $x(t)=0$ defined on $I=\R$ is a global solution.
However the function $y(t)=\sqrt{t^3}$ defined on $I=[0,+\infty)$ is also a solution and so are the functions
\[
z(t)=\begin{cases} \sqrt{(t-c)^3}&\text{if $t\ge c$} \\ 0 &\text{if $t < c$}.\end{cases}
\]
for every $c\ge 0$.
So there are no unique solutions. Moreover $y$ is not a maximal solution.
\end{enumerate}