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 \[ \frac{f(t, x) - f(t, \bar x)} {x-\bar x} = f_x (t, c). \]