In this note, we prove the local existence and uniqueness of initial value problem of ODEs

where satisfies Lipschitz condition.

### 1. Function spaces

To discuss our main topic, we need to extend some concepts that we already studied. In linear algebra, we studied vector spaces, which satisfies nine axioms. Such concepts give our concept of `vectors’ in high school as an example. One can also check that the space of real-valued continuous function on is a vector space with usual addition and scalar multiplication. We denote this space by .

In the Euclidean space , we can measure a distance of two vectors and by

Likewise, as is a vector space, one tries to measure a distance of two vectors. In this case, we want to measure a distance of two functions in . There is some issues to be considered. First, how can we define a distance on ? Next, the distance is useful?

There are many ways to define a distance on . Here we pick some special distance. Consider with , where

We check induces a distance on . Define by . To say is a distance, it should satisfies basic properties which the Euclidean distance

has.

- is nonnegative. Indeed, .
- if and only if . Indeed, implies for all . So . The converse is obvious.
- . It is also easy since
- for all : Indeed, this follows from the triangle inequality of Euclidean space:
So

This shows is a distance. So induces a distance .

To start an mathematical analysis, we need to ensure that with this kind of distance behaves likes One of the important property of is completeness. Here our completeness is Cauchy complete, i.e., for any Cauchy sequence in , there exists such that . We want to obtain such kind of completeness on with .

Now recall the definition of uniform convergence. Let be a sequence of real-valued functions on a set and a real-valued function on . We say that *converges uniformly* on if for every , there exists an such that

The function is called the *uniform limit* of the sequence . Note that

So the convergence in is a uniform convergence. Now considering Theorem 5.46 (Cauchy criterion for uniform convergence) in the textbook, for any Cauchy sequences in , converges uniformly on . Since is a sequence of continuous functions on and if we denote is a uniform limit of , then . This shows with is complete.

Complete space has various properties we can discuss in analysis. In the next section, we study one of the important property related to completeness.

### 2. Contraction Mapping Principle

Recall the Problem 1.14 in our textbook.

A sequence of real numbers is said to be \emph{contractive }if there exists a constant such that

Prove that for all . Also prove is convergent. Fianlly, if , then

In the proof of this problem, we used the Cauchy criterion to ensure is convergent. In the previous section, we already observed with is complete. So one can think that similar fact must be hold. Indeed, it is true. As an application, we study some convergence of sequences which is recursively defined. It is related to fixed points.

Theorem(Contraction mapping principle).Let be a contraction, i.e., there exists such that

Then there exists a unique function such that .

*Proof. *Uniqueness is easy. If and , then

which can only happen when . So .

Let . Define

Then . Iterate this so that

So if , then

For , choose so that . So implies . Hence is a Cauchy sequence in . So by the completeness of , there exists such that

Now from

we conclude that

in the uniform limit. This completes the proof.

**Remark. **The above theorem holds for any complete metric space, which will be studied in Topology I.

### 3. Existence and uniqueness of ODE

Now we give one application of the contraction mapping principle. We consider the following initival value problem of the first-order ODE:

(1)

where is an interval containing and is a Lipschitz function. Observe that is a solution of (1) if and only if

(2)

This follows from the fundamental theorem of calculus.

Theorem.Suppose that is a Lipschitz function. Then the equation has a unique -solution on for some .Let be a Lipschitz function with constant , i.e., there exists a constant such that

for all .

*Proof.* It suffices to prove there exists saitsfying (2). Note that as we saw before is a complete with .

Define by

If we show has a fixed point , then this is a solution of (2), which is a solution of (1).

Observe that

Now choose so that . Hence by the contraction mapping principle, there exists a unique fixed point . This completes the proof.

**Remark.** If we try to use the contraction mapping principle on , we have a trouble since is not complete under the distance . Note that converges uniformly to on .

Also, , but is not differentiable at .

**Remark.** The th order ODE given by

can be reduced to the first order ODE. Define by

Then

where is the function