Let be a normed finite-dimensional real vector space and let be an open set. A vector field on is a function . (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 , and then let it move freely, guided by the arrows in the vector field. (There are plenty of good pictures online.) Intuitively, for nice 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:
Let be a continuous path. We say is a solution to the differential equation defined by if for each we have
Example 2 (Examples of DE’s)
- Consider the vector field . Then the solutions are just .
- Consider the vector field . Then is a solution exactly when . It’s well-known that .
Of course, you may be used to seeing differential equations which are time-dependent: i.e. something like , 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 . Then we instead consider
and solve the resulting differential equation over . 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:
The vector field satisfies the Lipschitz condition if
holds identically for some fixed constant .
Note that continuously differentiable implies Lipschitz.
Theorem 4 (Picard-Lindelöf)
Let be a finite-dimensional real vector space, and let be a vector field on a domain which satisfies the Lipschitz condition.
Then for every there exists and such that and . Moreover, if and are two solutions and for some , then .
In fact, Peano’s existence theorem says that if we replace Lipschitz continuity with just continuity, then exists but need not be unique. For example:
Example 5 (Counterexample if is not differentiable)
Let and consider , with . Then and are both solutions to the differential equation
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 be a complete metric space. Let be a map such that for any . Then has a unique fixed point.
For the proof of the main theorem, we are given . Let be the metric space of continuous functions from to the complete metric space which is the closed ball of radius centered at . (Here can be arbitrary, so long as it stays in .) It turns out that is itself a complete metric space when equipped with the sup norm
This is well-defined since is compact.
We wish to use the Banach theorem on , so we’ll rig a function with the property that its fixed points are solutions to the differential equation. Define it by, for every ,
This function is contrived so that and is both continuous and differentiable. By the Fundamental Theorem of Calculus, the derivative is exhibited by
In particular, fixed points correspond exactly to solutions to our differential equation.
A priori this output has signature , so we need to check that . We can check that
where ; we have since is compact. Hence by selecting , the above is bounded by , so indeed maps into . (Note that at this point we have not used the Lipschitz condition, only that is continuous.)
It remains to show that is contracting. Write
Hence once again for sufficiently small we get . Since the above holds identically for , this implies
This is a cleaned-up version of a portion of a lecture from Math 55b in Spring 2015, instructed by Dennis Gaitsgory.
2 thoughts on “Uniqueness of Solutions for DiffEq’s”
Theorem 4 should read “x_0 in V”
never mind, it shouldn’t