Let $V$ be a normed finite-dimensional real vector space and let $U \subseteq V$ be an open set. A vector field on $U$ is a function $\xi : U \rightarrow V$. (In the words of Gaitsgory: “you should imagine a vector field as a domain, and at every point there is a little vector growing out of it.”)

The idea of a differential equation is as follows. Imagine your vector field specifies a velocity at each point. So you initially place a particle somewhere in $U$, and then let it move freely, guided by the arrows in the vector field. (There are plenty of good pictures online.) Intuitively, for nice $\xi$ it should be the case that the trajectory resulting is unique. This is the main take-away; the proof itself is just for completeness.

This is a so-called differential equation:

Definition 1. Let $\gamma : (-\varepsilon, \varepsilon) \rightarrow U$ be a continuous path. We say $\gamma$ is a solution to the differential equation defined by $\xi$ if for each $t \in (-\varepsilon, \varepsilon)$ we have $$\gamma'(t) = \xi(\gamma(t)).$$

Example 2 (Examples of DE’s)

Let $U = V = \mathbb R$.

1. Consider the vector field $\xi(x) = 1$. Then the solutions $\gamma$ are just $\gamma(t) = t+c$.
2. Consider the vector field $\xi(x) = x$. Then $\gamma$ is a solution exactly when $\gamma'(t) = \gamma(t)$. It’s well-known that $\gamma(t) = c\exp(t)$.

Of course, you may be used to seeing differential equations which are time-dependent: i.e. something like $\gamma'(t) = t$, for example. In fact, you can hack this to fit in the current model using the idea that time is itself just a dimension. Suppose we want to model $\gamma'(t) = F(\gamma(t), t)$. Then we instead consider $$\xi : V \times \mathbb R \rightarrow V \times \mathbb R \qquad\text{by}\qquad \xi(v, t) = (F(v,t), 1)$$ and solve the resulting differential equation over $V \times \mathbb R$. This does exactly what we want. Geometrically, this means making time into another dimension and imagining that our particle moves at a “constant speed through time”.

The task is then mainly about finding which conditions guarantee that our differential equation behaves nicely. The answer turns out to be:

Definition 3. The vector field $\xi : U \rightarrow V$ satisfies the Lipschitz condition if $$\left\lVert \xi(x')-\xi(x") \right\rVert \le \Lambda \left\lVert x'-x" \right\rVert$$ holds identically for some fixed constant $\Lambda$.

Note that continuously differentiable implies Lipschitz.

Theorem 4 (Picard-Lindelöf)

Let $V$ be a finite-dimensional real vector space, and let $\xi$ be a vector field on a domain $U \subseteq V$ which satisfies the Lipschitz condition.

Then for every $x_0 \in U$ there exists $(-\varepsilon,\varepsilon)$ and $\gamma : (-\varepsilon,\varepsilon) \rightarrow U$ such that $\gamma'(t) = \xi(\gamma(t))$ and $\gamma(0) = x_0$. Moreover, if $\gamma_1$ and $\gamma_2$ are two solutions and $\gamma_1(t) = \gamma_2(t)$ for some $t$, then $\gamma_1 = \gamma_2$.

In fact, Peano’s existence theorem says that if we replace Lipschitz continuity with just continuity, then $\gamma$ exists but need not be unique. For example:

Example 5 (Counterexample if $\xi$ is not differentiable)

Let $U = V = \mathbb R$ and consider $\xi(x) = x^{\frac23}$, with $x_0 = 0$. Then $\gamma(t) = 0$ and $\gamma(t) = \left( t/3 \right)^3$ are both solutions to the differential equation $$\gamma'(t) = \gamma(t)^{\frac 23}.$$

Now, for the proof of the main theorem. The main idea is the following result (sometimes called the contraction principle).

Lemma 6 (Banach Fixed-Point Theorem)

Let $(X,d)$ be a complete metric space. Let $f : X \rightarrow X$ be a map such that $d(f(x_1), f(x_2)) < \frac{1}{2} d(x_1, x_2)$ for any $x_1, x_2 \in X$. Then $f$ has a unique fixed point.

For the proof of the main theorem, we are given $x_0 \in V$. Let $X$ be the metric space of continuous functions from $(-\varepsilon, \varepsilon)$ to the complete metric space $\overline{B}(x_0, r)$ which is the closed ball of radius $r$ centered at $x_0$. (Here $r > 0$ can be arbitrary, so long as it stays in $U$.) It turns out that $X$ is itself a complete metric space when equipped with the sup norm $$d(f, g) = \sup_{t \in (-\varepsilon, \varepsilon)} \left\lVert f(t)-g(t) \right\rVert.$$ This is well-defined since $\overline{B}(x_0, r)$ is compact.

We wish to use the Banach theorem on $X$, so we’ll rig a function $\Phi : X \rightarrow X$ with the property that its fixed points are solutions to the differential equation. Define it by, for every $\gamma \in X$, $$\Phi(\gamma) : t \mapsto x_0 + \int_0^t \xi(\gamma(s)) ds.$$ This function is contrived so that $(\Phi\gamma)(0) = x_0$ and $\Phi\gamma$ is both continuous and differentiable. By the Fundamental Theorem of Calculus, the derivative is exhibited by $$(\Phi\gamma)'(t) = \left( \int_0^t \xi(\gamma(s)) ds \right)' = \xi(\gamma(t)).$$ In particular, fixed points correspond exactly to solutions to our differential equation.

A priori this output has signature $\Phi\gamma : (-\varepsilon,\varepsilon) \rightarrow V$, so we need to check that $\Phi\gamma(t) \in \overline{B}(x_0, r)$. We can check that

$$\begin{aligned} \left\lVert (\Phi\gamma)(t) - x_0 \right\rVert &=\left\lVert \int_0^t \xi(\gamma(s)) ds \right\rVert \\ &\le \int_0^t \left\lVert \xi(\gamma(s)) ds \right\rVert \\ &\le t \max_{s \in [0,t]} \left\lVert \xi\gamma(s) \right\rVert \\ &< \varepsilon \cdot A \end{aligned}$$

where $A = \max_{x \in \overline{B}(x_0,r)} \left\lVert \xi(x) \right\rVert$; we have $A < \infty$ since $\overline{B}(x_0,r)$ is compact. Hence by selecting $\varepsilon < r/A$, the above is bounded by $r$, so $\Phi\gamma$ indeed maps into $\overline{B}(x_0, r)$. (Note that at this point we have not used the Lipschitz condition, only that $\xi$ is continuous.)

It remains to show that $\Phi$ is contracting. Write

$$\begin{aligned} \left\lVert (\Phi\gamma_1)(t) - (\Phi\gamma_2)(t) \right\rVert &= \left\lVert \int_{s \in [0,t]} \left( \xi(\gamma_1(s))-\xi(\gamma_2(s)) \right) \right\rVert \\ &= \int_{s \in [0,t]} \left\lVert \xi(\gamma_1(s))-\xi(\gamma_2(s)) \right\rVert \\ &\le t\Lambda \sup_{s \in [0,t]} \left\lVert \gamma_1(s)-\gamma_2(s) \right\rVert \\ &< \varepsilon\Lambda \sup_{s \in [0,t]} \left\lVert \gamma_1(s)-\gamma_2(s) \right\rVert \\ &= \varepsilon\Lambda d(\gamma_1, \gamma_2) . \end{aligned}$$

Hence once again for $\varepsilon$ sufficiently small we get $\varepsilon\Lambda \le \frac{1}{2}$. Since the above holds identically for $t$, this implies $$d(\Phi\gamma_1, \Phi\gamma_2) \le \frac{1}{2} d(\gamma_1, \gamma_2)$$ as needed.

This is a cleaned-up version of a portion of a lecture from Math 55b in Spring 2015, instructed by Dennis Gaitsgory.