Uniqueness of solutions to Differential Equations
The Uniqueness theorem
Theorem. Consider the differential equation
\[
\dot x = f(t, x)
\]
where $f$ is defined for $t_0\le t\le t_1$, $x\in I$, and where
$f$ satisfies the Lipschitz condition
\[
|f(t, x) - f(t, x')|\le L |x-x'|
\]
for all $t\in [t_0, t_1]$, $x, x'\in I$.
If two solutions $x(t)$ and $\bar x(t)$ of the differential
equation coincide at some value of $t\in [t_0, t_1]$, then they
coincide for all $t\in[t_0, t_1]$.
Proof
Given two solutions $x(t), \bar x(t)$, we consider the squared
difference
\[
z(t) = \bigl(x(t)-\bar x(t)\bigr)^2.
\]
By assumption there is some $t_*\in[t_0, t_1]$ with $z(t_*)=0$,
and we want to show that $z(t)=0$ for all $t$ in the interval
$[t_0,t_1]$.
The difference satisfies a linear differential equation
\[
z'(t) = 2 a(t)z,
\]
where
\[
a(t) = \frac{f(t, x(t)) - f(t, \bar x(t))} {x(t)-\bar x(t)}.
\]
At those points where $x(t) = \bar x(t)$ we can set $a(t) = 0$.
The Lipschitz condition implies that $|a(t)|\le 2L$ for all $t$,
and thus we have
\begin{align*}
\frac{d} {dt}\Bigl(e^{-2Lt}z(t)\Bigr)
&= e^{-2Lt}\Bigl\{ \frac{dz} {dt} -2Lz\Bigr\}\\
&= e^{-2Lt}\bigl(a(t)-2L\bigr)z(t)\\
& \leq 0.
\end{align*}
In other words: $e^{-2Lt}z(t)$ is nonincreasing. For $t\ge t_*$ we get
\[
z(t) \le e^{2L(t-t_*)} z(t_*).
\]
Hence, if $z(t_*)=0$, then it follows that $z(t)=0$ for all $t\ge t_*$.
One can show in the same way that $e^{+2Lt}z(t)$ is
nondecreasing, so that $z(t_*)=0$ also implies $z(t)=0$ for all
$t\le t_*$.
Exponential separation of solutions
The proof of the uniqueness theorem also shows the following:
Theorem. If $x(t)$ and $\bar x(t)$ are two solutions of
$\dot x = f(t, x)$, and if $f$ satisfies the Lipschitz condition
with constant $L$, then
\[
|x(t) - \bar x(t)| \leq e^{L|t-t_*|} |x(t_*)-\bar x(t_*)|
\]
holds for all $t, t_*\in[t_0, t_1]$.
This follows by using $|x(t)-\bar x(t)| = \sqrt{z(t)}$.
How to check the Lipschitz condition
A convenient way to check the Lipschitz condition involves using the
Mean Value Theorem. Suppose that $f(t, x)$ satisfies
\[
\left|\frac{\pd f} {\pd x}\right| \le L
\]
for all $(t, x)$. Then $f$ automatically satisfies the
Lipschitz condition with the same constant $L$. Indeed, given
$t, x, \bar x$, the MVT guarantees the existence of some $c$ between
$x$ and $\bar x$ for which
\[
\frac{f(t, x) - f(t, \bar x)} {x-\bar x} = f_x (t, c).
\]