# 12-267/Existence And Uniqueness Theorem

Disclamer: This is a student prepared note based on the lecures of Friday, September 28th and Monday October 1st.

## Lipschitz

Def. $f: \mathbb{R}_y \rightarrow \mathbb{R}$ is called Lipschitz if $\exists \epsilon > 0, k > 0$ (a Lipschitz constant of f) such that $|y_1 - y_2| < \epsilon \implies |f(y_1) - f(y_2)| \leq k |y_1 - y_2|$.

Note that any function that is Lipschitz is uniformly continuous, and that if a function f and its derivative are both continous on a compact set then f is Lipschitz.

## Statement of Existence and Uniqueness Theorem

Thm. Existence and Uniqueness Theorem for ODEs

Let $f:\mathbb{R} = [x_0 - a, x_0 + a] \times [y_0 - b, y_0 + b] \rightarrow \mathbb{R}$ be continuous and uniformly Lipschitz relative to y. Then the equation $\Phi' = f(x, \Phi)$ with $\Phi(x_0) = y_0$ has a unique solution $\Phi : [x_0 - \delta, x_0 + \delta] \rightarrow \mathbb{R}$ where $\delta = min(a, ^b/_M)$ where M is a bound of f on $\mathbb{R}$.

## Proof of Existence

This is proven by showing the equation $\Phi(x) = y_0 + \int_{x_0}^x f(t, \Phi(t))dt$ exists, given the noted assumptions.

Let $\Phi_0(x) = y_0$ and let $\Phi_n(x) = y_0 + \int_{x_0}^x f(t, \Phi_{n-1}(t))dt$. IF we can prove the following three claims, we have proven the theorem. The proofs of these claims will follow below.

Claim 1: $\Phi_n$ is well-defined. More precisely, $\Phi_n$ is continuous and $\forall x \in [x_0 - \delta, x_0 + \delta]$, $|\Phi_n(x) - y_0| \leq b$ where b is as referred to above.

Claim 2: For $n \geq 1$, $|\Phi_n(x) - \Phi_{n-1}(x)| \leq \frac{Mk^{n-1}}{n!} |x-x_0|^n$.

Claim 3: if $\Phi_n(x)$ is a series of functions such that $|\Phi_n(x) - \Phi_{n-1}(x)| < c_n$, with $\sum_{n=1}^{\infty} c_n$ equal to some finite number, then $\Phi_n$ converges uniformly to some function $\Phi$

Using these three claims, we have shown that the solution $\Phi(x)$ exists.

## Proofs of Claims

Proof of Claim 1:

The statement is trivially true for $\Phi_0$. Assume the claim is true for $\Phi_{n-1}$. $\Phi_n$ is continuous, being the integral of a continuous function.

$|\Phi_n - y_0|$

$= |\int_{x_0}^x f(t, \Phi_{n-1}(t))dt|$

$\leq |\int_{x_0}^x |f(t, \Phi_{n-1}(t))|dt|$

$\leq | \int_{x_0}^x M dt | = M |x_0 - x|$

$\leq M \delta$

$\leq M \cdot \frac{b}{M}$

$= b$

$\Box$

Proof of Claim 2:

$|\Phi_n(x) - \Phi_{n-1}(x)|$

$= |\int_{x_0}^x f(t, \Phi_{n-1}(t))dt - \int_{x_0}^x f(t, \Phi_{n-2}(t))dt|$

$\leq | \int_{x_0}^x |f(t, \Phi_{n-1}(t)) - f(t, \Phi_{n-2}(t)) | dt |$

$\leq |\int_{x_0}^x k|\Phi_{n-1}(t) - \Phi_{n-2}(t)|dt|$

$\leq |\int_{x_0}^x k \frac{M k^{n-2}}{(n-1)!} |t-x_0|^{n-1}dt|$

$= \frac{M k^{n-1}}{(n-1)!} \int_0^{|x-x_0|} t^{n-1} dt$

$= \frac{M k^{n-1}}{n!} |x-x_0|^n$

$\Box$

Note that the sequence $c_n = \frac{M k^{n-1}}{n!} |x-x_0|^n$ has $\sum_{n=1}^{\infty} c_n$ equal to some finite number.

Proof of Claim 3: Assigned in Homework 3, Task 1, see page for solutions.

## Proof of Uniqueness

Suppose $\Phi$ and $\Psi$ are both solutions. Let $\Chi(x) = |\Phi(x) - \Psi(x)|$.

$\Chi(x) = |\Phi(x) - \Psi(x)| = |\int_{x_0}^x(f(x, \Phi(x)) - f(x, \Psi(x))) dx | \leq \int_{x_0}^x k|\Phi(x) - \Psi(x)| dx$

We have that $\Chi \leq k \int_{x_0}^x \Chi(x) dx$ for some constant k, which means $\Chi' \leq k\Chi$, and that $\Chi(x) \geq 0$.

Let $U(x) = e^{-kx}\int_{x_0}^x \Chi(x) dx$. Note that $U(x_0) = 0$ as in this case we are integrating over an empty set, and that U thus defined has $U(x) \geq 0$. Then

$U'(x) = -ke^{-kx}\int_{x_0}^x\Chi(x) dx + e^{-kx} \Chi(x) = e^{-kx}(\Chi(x) - k\int_{x_0}^x\Chi(x) dx) \leq 0$

Then $U(x_0) = 0 \and U'(x) = 0 \implies U(x) \leq 0$, and $0 \leq U(x) \leq 0 \implies U(x) \equiv 0 \implies \Chi(x) \equiv 0 \implies \Phi(x) \equiv \Psi(x)$.

$\Box$