# Uniqueness of Solutions for DiffEq’s

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 $\displaystyle \gamma'(t) = \xi(\gamma(t)).$

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 $\displaystyle \xi : V \times \mathbb R \rightarrow V \times \mathbb R \qquad\text{by}\qquad \xi(v, t) = (F(v,t), 1)$

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 $\displaystyle \left\lVert \xi(x')-\xi(x'') \right\rVert \le \Lambda \left\lVert x'-x'' \right\rVert$

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 $\displaystyle \gamma'(t) = \gamma(t)^{\frac 23}.$

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 $\displaystyle d(f, g) = \sup_{t \in (-\varepsilon, \varepsilon)} \left\lVert f(t)-g(t) \right\rVert.$

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 ${\gamma \in X}$, $\displaystyle \Phi(\gamma) : t \mapsto x_0 + \int_0^t \xi(\gamma(s)) \; ds.$

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 $\displaystyle (\Phi\gamma)'(t) = \left( \int_0^t \xi(\gamma(s)) \; ds \right)' = \xi(\gamma(t)).$

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 \displaystyle \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 \displaystyle \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 $\displaystyle d(\Phi\gamma_1, \Phi\gamma_2) \le \frac{1}{2} d(\gamma_1, \gamma_2)$

as needed.

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