Picards's Theorem

Banach Contraction Theorem

Definition.

Let $(X,d)$$(X,d)$ be a metric space. A map $T:X\to X$$T:X\to X$ is said to satisfy Lipschitz condition if there exists a real number $L\ge 0$$L\ge 0$ such that

$$\begin{array}{}\text{(1)}& d(T{x}_{,}Ty)\le Ld(x,y)\mathrm{\forall }x,y\in X.\end{array}$$

Proposition. If $T$$T$ satisfies Lipschitz condition then $T$$T$ is uniformly continuous.

Proof. If ${\displaystyle d(x,y)\le \frac{ϵ}{L}}$$\displaystyle d(x,y)\le\frac{\epsilon}{L}$, then $d(Tx,Ty)\le Ld(x,y)=ϵ$$d(Tx,Ty)\le Ld(x,y)=\epsilon$. $◻$$\square$

Definition. If $d(Tx,Ty)\le Ld(x,y)$$d(Tx,Ty)\le Ld(x,y)$ and $0\le L<1$$0\le L< 1$, then we say $T$$T$ is a contraction.

Theorem. Banach contraction theorem

Let $(X,d)$$(X,d)$ be a complete metric space and $T:X\to X$$T:X\to X$ be a contraction mapping, Then $T$$T$ has a unique fixed point. i.e. There exists a unique $x\in X$$x\in X$ such that $Tx=x$$Tx=x$.

Proof. Let $x_i=T^{i}x$$x_i=T^ix$. Assume $k\le n$$k\le n$, then

$$\begin{array}{}\text{(2)}& d(x_k,x_n)=d(T^{k}x,T^{n}x)=d(T^{k}x,T^k\circ T^{n-k}x)\le L^k(x_0,T^{n-K}x)=L^{k}d(x_0,x_{n-k})\end{array}$$

By the triangle inequality,

$$\begin{array}{}\text{(3)}& d(x_0,x_{n-k})\le d(x_0,x_1)+d(x_1,x_2)+...+d(x_{n-k-1},x_{n-k})=d(x_0,x_1)(1+L+L^2+...+L^{n-k-1})\end{array}$$

Hence

$$\begin{array}{}\text{(4)}& d(x_k,x_n)\le L^{k}d(x_0,x_1)\frac{1-L^{n-k-1}}{1-L}\end{array}$$

Since $L<1$$L<1$, $x_k$$x_k$ is a Cauchy sequence. By the completeness of $(X,d)$$(X,d)$, there exists $\begin{array}{c}x,\underset{n\to \mathrm{\infty }}{lim}{x}_n=x\end{array}$$x,\lim_{n\to\infty}x_n=x\\$.

This is a fixed point since $T(x)=T(lim{x}_n)=limT(x_n)=lim{x}_{n+1}=x$$T(x)=T(\lim x_n)=\lim T (x_n)=\lim x_{n+1}=x$.

To prove $x$$x$ is the unique fixed point, assume $y$$y$ is a fixed point as well, then

$$\begin{array}{}\text{(5)}& d(x,y)=d(Tx,Ty)\le Ld(x,y)⟹d(x,y)=0⟹x=y◻\end{array}$$

Picards's Theorem

Let $f(x,y)$$f(x,y)$ continuous on a closed rectangle $[x_0-a,x_0+a]\times [y_0-b,y_0+b]$$[x_0-a,x_0+a]\times [y_0-b,y_0+b]$​.

and $f(x,y)$$f(x,y)$ is Lipschitz continuous respect to $y$$y$.

Then the differential equation

$$\begin{array}{}\text{(6)}& \frac{dy}{dx}=f(x,y)\end{array}$$

has a unique solution $h$$h$ with $h(x_0)=y_0$$h(x_0)=y_0$.

Proof.

Firstly let us integrate respect to $x$$x$,

$$\begin{array}{}\text{(7)}& y(x)=y_0+{\int }_{x_0}^{x}f(t,y)dt\end{array}$$

Notice that the solution $y(x)$$y(x)$ is the fixed point under the map

$$\begin{array}{}\text{(8)}& T(y):=y_0+{\int }_{x_0}^{x}f(t,y)dt,C[y_0-b,y_0+b]\to C[y_0-b,y_0+b]\end{array}$$

Notice that $(C[y_0-b,y_0+b],d_{\mathrm{\infty }})$$(C[y_0-b,y_0+b],d_{\infty})$ is a complete metric space.

Hence we only need to make sure that $T$$T$ is a contraction map.

Let $a(x),b(x)\in C[y_0-b,y_0+b]$$a(x),b(x)\in C[y_0-b,y_0+b]$, consider

$$\begin{array}{}\text{(9)}& {\Vert Ta(x),Tb(x)\Vert }_{\mathrm{\infty }}={\Vert {\int }_{x_0}^{x}f(t,a)-f(t,b)dt\Vert }_{\mathrm{\infty }}\le {\Vert {\int }_{x_0}^{x}L(a-b)dt\Vert }_{\mathrm{\infty }}\end{array}$$

And

$$\begin{array}{}\text{(10)}& {\Vert {\int }_{x_0}^{x}L(a-b)dt\Vert }_{\mathrm{\infty }}=L(x-x_0){\Vert (a-b)\Vert }_{\mathrm{\infty }}\le q{\Vert Ta(x),Tb(x)\Vert }_{\mathrm{\infty }}\end{array}$$

Here $q<1$$q<1$. Hence we need $(x-x_0)\le \frac{1}{L}$$(x-x_0)\le \frac{1}{L}$​.

Since $f(x,y)$$f(x,y)$ is defined on a compact set, let $M=sup|f(x,y)|$$M=\sup |f(x,y)|$.

By the condition, $y\in [y_0-b,y_0+b]$$y\in[y_0-b,y_0+b]$, hence we need

$$\begin{array}{}\text{(11)}& \left|{\int }_{x_0}^x{y}^{\prime }dt\right|=|{\int }_{x_0}^{x}f(t,y)dt|\le \left|{\int }_{x_0}^{x}Mdt\right|=M|x-x_0|\le b⟹|x-x_0|\le \frac{b}{M}\end{array}$$

Hence by Banach contraction theorem,

we have the unique $y$$y$ satisfies $y(x_0)=y_0$$y(x_0)=y_0$ and $y^{\prime }=f(x,y)$$y'=f(x,y)$ for $x\in [x_0-\delta ,x_0+\delta ],\delta =min(a,\frac{1}{L},\frac{b}{M}).$$x\in [x_0-\delta,x_0+\delta],\delta=\min(a,\frac{1}{L},\frac{b}{M}).$

The solution is given by the limit of this sequence $y_0=y_0,y_n=T^n{y}_0$$y_0=y_0,y_n=T^n y_0$.