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.

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.

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 $\mathbb {R}$.

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 $\Phi _{n}(x)=y_{0}+\int _{x_{0}}^{x}f(t,\Phi _{n-1}(t))dt$.

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. The proofs of the claims are below.

Proof of Uniqueness:

Suppose $\Phi$ and $\Psi$ are both solutions. Let $\mathrm {X} (x)=|\Phi (x)-\Psi (x)|$.

$\mathrm {X} (x)=|\Phi (x)-\Psi (x)|=|\int _{x_{0}}^{x}(f(x,\Phi (x))-f(x,\Psi (x)))dx|\leq \int _{x_{0}}^{x}k|\Phi (x)-\Psi (x)|dx$

We have that $\mathrm {X} \leq k\int _{x_{0}}^{x}\mathrm {X} (x)dx$ for some constant k, which means $\mathrm {X} '\leq k\mathrm {X}$, and that $\mathrm {X} (x)\geq 0$.

Let $U(x)=e^{-kx}\int _{x_{0}}^{x}\mathrm {X} (x)dx$. 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

$U'(x)=-ke^{-kx}\int _{x_{0}}^{x}\mathrm {X} (x)dx+e^{-kx}\mathrm {X} (x)=e^{-kx}(\mathrm {X} (x)-k\int _{x_{0}}^{x}\mathrm {X} (x)dx)\leq 0$

Then $U(x_{0})=0\land U'(x)=0\implies U(x)\leq 0$, and $0\leq U(x)\leq 0\implies U(x)\equiv 0\implies \mathrm {X} (x)\equiv 0\implies \Phi (x)\equiv \Psi (x)$.

$\Box$

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.

$|\Phi _{n}-y_{0}|$

$=|\int _{x_{0}}^{x}f(t,\Phi _{n-1}(t))dt|$

$\leq |\int _{x_{0}}^{x}|f(t,\Phi _{n-1}(t))|dt|$

$\leq |\int _{x_{0}}^{x}Mdt|=M|x_{0}-x|$

$\leq M\delta$

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

$=b$

$\Box$

Proof of Claim 2:

$|\Phi _{n}(x)-\Phi _{n-1}(x)|$

$=|\int _{x_{0}}^{x}f(t,\Phi _{n-1}(t))dt-\int _{x_{0}}^{x}f(t,\Phi _{n-2}(t))dt|$

$\leq |\int _{x_{0}}^{x}(f(t,\Phi _{n-1}(t)-f(t,\Phi _{n-2}(t))dt)dt|$

$\leq |\int _{x_{0}}^{x}k|\Phi _{n-1}(t)-\Phi _{n-2}(t)|dt|$

$\leq |\int _{x_{0}}^{x}k{\frac {Mk^{n-2}}{(n-1)!}}|t-x_{0}|^{n-1}dt|$

$={\frac {Mk^{n-1}}{(n-1)!}}\int _{0}^{|x-x_{0}|}t^{n-1}dt$

$={\frac {Mk^{n-1}}{n!}}|x-x_{0}|^{n}$

$\Box$

Note that the sequence $c_{n}={\frac {Mk^{n-1}}{n!}}|x-x_{0}|^{n}$ has $\sum _{n=1}^{\infty }c_{n}$ equal to some finite number.

Proof of Claim 3: Assigned in Homework 3, Task 1