The Lipschitz Uniqueness Theorem for Solutions to IVPs on F.O. ODEs

# The Lipschitz Uniqueness Theorem for Solutions to IVPs on First Order ODEs

Recall that a continuous function $f \in C(D, \mathbb{R})$ is said to satisfy a Lipschitz condition on $D$ if there exists an $L > 0$ such that for all $(t, x), (t, y) \in D$ we have that:

(1)\begin{align} \quad | f(t, x) - f(t, y) | \leq L |x - y| \end{align}

We now prove an important result for the uniqueness of solutions to the IVP $x' = f(t, x)$ with $x(\tau) = \xi$, $f \in C(D, \mathbb{R})$ which says that if $f$ satisfies a Lipschitz condition on $D$ then any solution $\phi$ to this IVP is unique on any interval $[\tau - d, \tau + d]$.

Theorem 1 (Lipschitz Uniqueness Theorem): Let $x' = f(t, x)$ with $x(\tau) = \xi$ be an IVP with $f \in C(D, \mathbb{R})$. If $f$ satisfies a Lipschitz condition then any solution $\phi$ to this IVP is unique on any interval $[\tau - d, \tau + d]$. |

**Proof:**Suppose that $f$ satisfies a Lipschitz condition on $D$ with Lipschitz constant $L$. Then for all $(t, x), (t, y) \in D$ we have that:

\begin{align} \quad | f(t, x) - f(t, y) | \leq L |x - y| \end{align}

- Let $d > 0$ and suppose that $\phi_1$ and $\phi_2$ are both solutions to the IVP $x' = f(t, x)$ with $x(\tau) = \xi$ on the interval $[\tau, \tau + d]$. Then:

\begin{align} \quad \phi_1(t) = \xi + \int_{\tau}^{t} f(s, \phi_1(s)) \: ds \quad \mathrm{and} \quad \phi_2(t) = \xi + \int_{\tau}^{t} f(s, \phi_2(s)) \: ds \end{align}

- We take the absolute value of the difference $\phi_1(t) - \phi_2(t)$:

\begin{align} \quad | \phi_1(t) - \phi_2(t) | &= \biggr \lvert \left ( \xi + \int_{\tau}^{t} f(s, \phi_1(s)) \: ds \right ) + \left ( \xi + \int_{\tau}^{t} f(s, \phi_2(s)) \: ds \right ) \biggr \rvert \\ &= \biggr \lvert \int_{\tau}^{t} [f(s, \phi_1(s)) \: ds - f(s, \phi_2(s))] \: ds \biggr \rvert \\ & \leq \int_{\tau}^{t} | f(s, \phi_1(s)) - f(s, \phi_2(s)) | \: ds \\ & \leq \int_{\tau}^{t} L|\phi_1(s) - \phi_2(s)| \: ds \end{align}

- Let $r(t) = | \phi_1(t) - \phi_2(t) |$, $\delta = 0$, and $k = L$. Then $\displaystyle{r(t) \leq \delta + \int_{\tau}^{t} kr(t) \: dt}$ where $r$ is a nonnegative continuous function (since $\phi_1, \phi_2$ are continuous), and $k, L \geq 0$. So by Gronwall's Inequality we have that:

\begin{align} \quad r(t) \leq \delta e^{k(t - \tau)} \end{align}

- Hence:

\begin{align} \quad | \phi_1(t) - \phi_2(t) | \leq 0e^{L(t - \tau)} = 0 \end{align}

- Therefore $\phi_1(t) = \phi_2(t)$ on the interval $[\tau, \tau + d]$. An analogous argument shows that $\phi_1(t) = \phi_2(t)$ on the interval $[\tau - d, \tau]$. So the solution to the IVP $x' = f(t, x)$ with $x(\tau) = \xi$ is unique on $[\tau - d, \tau + d]$. $\blacksquare$