Fixed Point and ODE

Discrete Dynamical Systems

Discrete Dynamical Systems Let \(X\subseteq (V,\|\cdot\|), T:X\rightarrow X\) continuous. \((X,T)\) is a discrete dynamical system
Forward orbit \(\forall x\in X\), forward orbit of x is the sequence \(O(x):=\{T^n_x\}_{n\geq 0}\), \(T^n_x=T(T...T(x))\)

Fixed Point \(x^*\) is
attractive fixed point (sink) if \(\exists (a,b)\ni x^*\) s.t. \(\forall x \in (a,b). T^N_x\rightarrow x^*\)
repelling fixed point (source) if \(\forall x\neq x^*\), \(O(x)\) leaves \((a,b)\)

Contraction \(X\subseteq (V,\|\cdot\|),T:X\rightarrow X\) is a contraction on \(X\) if

\[\exists c < 1. \forall x,y\in X. \|Tx-Ty\|\leq c\|x-y\|\]

i.e. \(T\) is \(c\)-Lipschitz \(c<1\)

Banach Contraction Principle

Let \(X\subseteq (V,\|\cdot\|)\) be a closed subset of a complete normed vector space,
IF \(T:X\rightarrow X\) is a contraction on \(X\)
THEN
(i) \(T\) has unique fixed point \(x^*\)
(ii) \(\forall x\in X. x^* = \lim_{n\rightarrow\infty} T^nx\)
(iii) \(\|T^nx-x^*\|\leq c^n\|x-x^*\|\leq \frac{c^n}{1-c}\|x-Tx\|, c\) is the Lipschitz constant

proof. Let \(x_0\in X\), recursively define \(x_{n+1} = Tx_n\), hence form a sequence \(\{x_n\}\).
Then, we can show that \(\{x_n\}\) is Cauchy by observing

\[\begin{align*} \|x_{n+m} - x_n\| &\leq \sum_0^{m-1}\|x_{n+i-1}-x_{n+i}\| &\text{triangle inequality} \\ &= \sum_{i=0}^{m-1} \|Tx_{n+i}-Tx_{n+i-1}\| \\ &\leq \sum_{i=0}^{m-1} c\|x_{n+i}-x_{n+i-1}\|&\text{by contraction}\\ &\leq \sum^{m-1}c^{n+i}\|x_1-x_0\| &\text{recursively repeat such process} \\ &<\sum^\infty c^{n+i}\|x_1-x_0\| \\ &= \frac{c^n\|x_1-x_0\|}{1-c} \end{align*}\]

Since \(c^n\rightarrow 0\), we can choose \(N\) sufficiently large to have \(\|x_{n+m}-x_n\|\rightarrow 0\)

Existence By completeness of the normed space, Cauchy implies convergent to some \(x^*\in V\), by \(X\) closed, \(x^*\in X\). By Lipshitz, hence continuous of \(T\), \(Tx^* = T(\lim x_n) = \lim Tx_n = x^*\Rightarrow x^*\) is the fixed point.

Uniqueness Suppose \(y\in X\) is also fixed point, \(\|x^*-y\| = \|Tx^* - Ty\| \leq c\|x^*-y\|\Rightarrow \|x^*-y\| = 0\Rightarrow x^*=y\)

\[\|T^Nx-x^*\| = \|T^n x-T^nx^*\|\leq c^n\|x-x^*\|= c^n lim \|x-x_m\|\leq \frac{c^n}{1-c}\|Tx-x\|\]

Solving ODE

Example given the initial value problem \(f'(x)=1+x-f(x), |x|\leq 1/2\) and \(f(0)=1\)

First convert to the integral problem

\[f(x) = 1+ \int_0^x 1 + t - f(t)dt = 1+ x + \frac{x^2}{2} - \int_0^x f(t)dt, f\in C[-1/2,1/2]\]

Define \(T\) on \(C[-1/2,1/2]\) and \(Tf = f\) and the solution of the integral equation is a fixed point of \(T\) Then, to show \(T\) is a contraction

\[\begin{align*} |Tf(x)-Tg(x) &= |\int_0^x f(t)-g(t)dt| \\ &\leq \big|\int_0^x |f(t)-g(t)|dt\big| &\text{tri. ineq.}\\ &\leq \int_0^x \|f-g\|dt \\ &=\|f-g\|_\infty \int_0^{|x|}dt\\ &\leq \frac{\|f-g\|_\infty}{2} \end{align*}\]

Thus, by Banach Contraction Principle, choose \(f_0 := 1\in C[-1/2,1/2]\), then

\[\begin{align*} f_1(x) &= Tf_0(x) = 1+x+\frac{x^2}{2} - \int_0^x dt = 1+\frac{x^2}{2} \\ f_2(x) &= Tf_1(x) = 1+x+\frac{x^2}{2} - \int_0^x 1+\frac{x^2}{2} dt = 1 + \frac{x^2}{2} - \frac{x^3}{6}\\ &...\\ f_n(x) &= \sum \frac{(-x)^{n+1}}{(n+1)!} \end{align*}\]

Since \(|x|\leq 1/2\), using M-test, this power series is uniform convergence, and note that \(f_n(x):=e^{-x} + x\)

Thrm. Existence of ODE

If \(\Phi:[a,b]\times \mathbb R^n\rightarrow \mathbb R^n\) is Lipschitz on the second coordinate \(y\in\mathbb R^n\).
Then \(\exists F_*\in C([a,b],\mathbb R^n)\) s.t. \(F_*=TF_*=\Gamma + \int_a^x \Phi(t,F_*(t))dt\)

proof. (Picard iteration)

Recursively define \(F_0(x)=\Gamma, F_{k+1}(x)=TF_k = \Gamma + \int_a^x \Phi(t, F_k(t))dt\)

Consider \(F_{k+1}-F_k\), let \(M = \|\Phi(t,\Gamma)\|_\infty\), by EVT, since \([a,b]\) compact, \(M\) is the maximum.
Then for any \(x\in [a,b]\). \(\|F_1(x) - F_0(x)\| = \int_a^x \Phi(t,\Gamma)dt\leq M(x-a)\)

\[\begin{align*} \|F_{k+1}(x) - F_k(x)\| &= \int_a^x \Phi(t, F_{k}(t))-\Phi(t, F_{k-1}(t))dt \\ &\leq L\int_a^x\|F_k - F_{k-1}\|_\infty dt &L \text{ is the Lipschitz constant} \\ &\leq L \int_a^x \frac{L^{k-1}M(x-a)^k}{k!} &\text{induction hypothesis}\\ &= L^k \frac{M(x-a)^{k+1}}{(k+1)!} \end{align*}\]

Then, for \(n\geq N\), for some large \(N\)

\[\begin{align*} \|F_n - F_0\| &\leq \sum^{n-1} \|F_{k+1}-F_k\| \\ &\leq \sum^{n-1} \frac{L^k M(x-a)^{k+1}}{(k+1)!} \\ &\leq \frac{M}{L} \sum^\infty \frac{(L(x-a))^{k+1}}{(k+1)!}\\ &\leq \frac{M}{L} \sum^\infty \frac{(L(b-a))^{k+1}}{(k+1)!} < \infty \end{align*}\]

Since the infinite sum is finite, the series is Cauchy. so that \(F_n\rightarrow^{u.c.}F_*\in C([a,b], \mathbb R^n)\) by compactness and closeness.

Therefore, \(TF_* = T\lim F_n = \lim TF_n = F_*\) using continuity

Uniqueness Suppose exists \(G_*\), then

\[\begin{align*} TF_* - G_* &= TTF_* - TG_* &\text{by fixed point assumption} \\ &= T^kF_* - T^kG_* \\ &\leq \frac{(L(x-a))^{k+1}M}{(k+1)!L} \\ &\sim 0 &\text{Sterling's formula} \end{align*}\]

Example

\(f'(x)=f(x), f(0)=1, x\in [0,1]\)

proof. Let \(F_0(x)=1, F_1(x) = 1 + \int_0^x F_0(t)dt = 1 + \int_0^x dt = 1 + x\)
Assume \(F_n(x) = \sum_0^k \frac{x^m}{m!}\), then

\[F_{k+1}(x) = 1 + \int_0^x F_n(t)dt = 1 + \int_0^x \sum_0^k \frac{x^m}{m!} = \sum_0^{k+1}\frac{x^m}{m!}\]

Therefore by induction, \(F_*(x) = \lim_nF_n(x) = \sum^\infty x^m/m! = e^x\)