Let

• a continuous function of $n+1$ variables defined in a neighborhood $U \subseteq \reals^{n+1}$ of $(\v x_0, t_0)$
• Lipschitz continuous with respect to the first $n$ variables (i.e. with respect to $\v x$ ).

Then there exists a unique solution $\v x : I \to \reals^n$ of the Cauchy problem, defined in a neighborhood $I \subseteq \reals$ of $t_0$ .

Proof

Solving the Cauchy problem is equivalent to solving the following integral equation $$x(t) = x(t_0) + \int_{t_0}^{t} F(\v x(\tau), \tau) \d\tau$$

Let $X$ be the set of continuous functions $\v f : [t_0 - \delta, t_0 + \delta] \to B(\v x_0, \eps)$ . We'll assume $\eps$ to be chosen such that the $B(\v x_0, \eps) \subseteq U$ ¹. In this ball, therefore, $F$ is Lipschitz continuous with respect to the first $n$ variable, in other words, there exists a real number $L$ such that $${F(\v x, t) - F(\v y, t)} \leq L\norm{\v x - \v y}$$ for all points $\v x, \v y$ sufficiently near to $\v x_0$ .

Now let's define the mapping $T : X \to X$ as follows $$T\v x : t \mapsto \v x_0 + \int_{t_0}^t F(\v x(\tau), \tau) \d \tau$$ We make the following observations about $T$ .

1. Since $F$ is continuous, $\norm{F}$ attains a maximum value $M$ on the compact set $B(\v x_0, \eps) \times [t_0 \pm \delta]$ . But by hypothesis, $\norm{\v x(t) - \v x_0} \leq \eps$ , hence $$\norm{\v x(t) - \v x_0} \leq \int_{t_0}^t \norm{F(\v x(\tau), \tau)} \d \tau \leq M(t - t_0) \leq M\delta$$ for all $t \in [t_0 \pm \delta]$ .
2. The Lipschitz continuity of $F$ yields $$\norm{T\v x(t) - T\v y(t)} \leq \int_{t_0}^t \norm{F(\v x(\tau), \tau) - F(\v y(\tau), \tau)} \d \tau \leq \int_{t_0}^t L\norm{\v x(\tau) - \v y(\tau)} \d \tau \leq L \delta d_\infty(\v x, \v y)$$

• $T(X) \subseteq X$ , i.e. $T$ doesn't send us outside of $X$ .
• $T$ is a contraction mapping with respect to the uniform convergence metric $d_\infty$ on $X$ , i.e. there exists $\lambda \in \reals$ such that for all $\v x, \v y \in X$ , $$d_\infty(T\v x, T\v y) \leq \lambda d_\infty(\v x, \v x)$$

Footnotes

... $B(\v x_0, \eps) \subseteq U$¹

$B(\v x_0, \eps)$ denotes the closed ball $\{ \v x : \norm{\v x_0 - \v x} \leq \eps \}$