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Let $A$ be differentiable on a neighbourhood of a closed radius-$r$ ball in a Banach space, with $\Vert DA\Vert \le \kappa <1$ on that ball. If $\Vert A(p_0)-p_0\Vert \le (1-\kappa )r$, the mean-value estimate gives $A(B)\subseteq B$ and $\Vert A(p)-A(q)\Vert \le \kappa \Vert p-q\Vert$. The contraction mapping theorem supplies exactly one fixed point in $B$, and its distance from $p_0$ is at most $\Vert A(p_0)-p_0\Vert /(1-\kappa )$.

// Target: dynamical-systems.bigb

A planar diffeomorphism is area-preserving if the phase-space area of every measurable set is unchanged. The change-of-variables formula proves this property when $|\det DF|=1$ everywhere, and conversely gives that determinant condition for a smooth map preserving all sufficiently small areas. Such a map cannot have an asymptotically attracting fixed point with a positive-area local basin: iterates of a small invariant neighborhood would have to enter arbitrarily small disks while retaining its phase-space area. For a map with constant positive determinant $b$, invariance of a bounded region of positive phase-space area instead forces $b=1$.

The annual pulse-breeding predator-prey model is an area-preserving map in logarithmic population coordinates: $u^{\prime }=u+\log r-\kappa e^v$, then $v^{\prime }=v+\log s+\log (1+b{e}^{u^{\prime }})$. Each step is a shear of determinant one. At a positive fixed point, the ordinary-population Jacobian matrix has determinant one and trace $2-(1-s)\log r$. Eigenvalues lie on the unit circle for $0<(1-s)\log r<4$, but this linear neutrality does not prove nonlinear stability at every resonance. Area preservation excludes an isolated attracting positive periodic cycle.

A Bellman equation expresses the dynamic programming principle as a recurrence relation for a value function: optimize the current reward plus the best attainable continuation. Let $s$ be a dynamic programming state, $A_t(s)$ the allowed actions at time $t$, $r_t(s,a)$ the immediate reward, and $F_t(s,a)$ the next dynamic programming state. With terminal time $T$ and terminal reward $g$, the deterministic finite-horizon Bellman equation isHere $V_t(s)$ measures total reward from time $t$ through the terminal reward. Start with $g$ and evaluate backwards; the terminal condition anchors the recurrence relation. For finite nonempty action sets, finite rewards, and finite terminal rewards, a maximum exists at each step. If an optimum need not be attained, use a supremum. A minimization problem instead uses a minimum or infimum. The 0-1 knapsack problem above is the same comparison between an immediate item value plus a continuation and the continuation after skipping. See Bertsekas, MIT lecture 2: the backwards recursion.

In a Markov decision process, the next dynamic programming state $S_{t+1}$ is random. If $r_t(s,a)$ denotes the expected immediate reward and $\beta$ is a discount factor, the finite-horizon Bellman equation becomesThe conditional expectation averages over the next dynamic programming state before choosing the current action; that action cannot use a future outcome that has not yet been observed. For a finite state space with transition probabilities $p_t(s^{\prime }\mid s,a)$, the continuation term is $\beta {\sum }_{s^{\prime }}{p}_t(s^{\prime }\mid s,a)V_{t+1}(s^{\prime })$. Taking $\beta =1$ gives the undiscounted finite-horizon problem. With a discount factor $0\le \beta <1$, rewards $k$ steps ahead carry a factor ${\beta }^k$, including the terminal reward. A maximizing action for each $(t,s)$ specifies a Markov policy, which contains decisions rather than just optimal scores.

For an infinite horizon with a finite state space, finite nonempty action sets, time-independent transition probabilities, bounded rewards $|r(s,a)|\le M$, and a discount factor $0\le \beta <1$, the optimal value function satisfiesIn this setting the maximizing dynamic programming operator is a contraction mapping in the supremum norm, so the Banach fixed-point theorem gives a unique bounded fixed point and convergence of repeated application to it. See Bertsekas, MIT lecture 15: bounded discounted problems. These hypotheses matter: with one state, one action, zero reward, and $\beta =1$, the Bellman equation is merely $V(s)=V(s)$, so every finite constant is a fixed point, although the actual total-reward value function is zero. The average-reward Bellman equation and Hamilton-Jacobi-Bellman equation address other settings with their own conditions.

For a strictly distance-decreasing map of a finite-dimensional Euclidean space, one bounded forward orbit suffices for existence of a fixed point. Its closure is compact by the Heine-Borel theorem and invariant by continuity, so strict distance decrease on a compact metric space applies. Strict distance decrease gives uniqueness throughout the whole space, not merely the orbit closure.

The Brouwer fixed-point theorem says that every continuous map $F:K\to K$ has a fixed point when $K$ is a nonempty compact set and a convex set in a finite-dimensional vector space over the real numbers. Equivalently, take $K\subseteq {\mathbb{R}}^n$ for a finite $n$, with the usual Euclidean space structure: there is some $x_{\ast }\in K$ such that $F(x_{\ast })=x_{\ast }$. The conclusion guarantees at least one fixed point, without asserting uniqueness or telling us how quickly to find one. ^[ref]

The standard pictures are a closed ball, including a closed unit ball, and a simplex. A filled disk qualifies; its bounding circle alone does not. A probability simplex also qualifies even though its dimension is smaller than that of its ambient Euclidean space. The hypotheses have concrete meanings: being a convex set keeps every line segment between allowed points inside $K$, and compactness prevents escape or loss of a limiting point. In Euclidean space, the Heine-Borel theorem identifies compact sets with closed sets that are bounded sets. Crucially, every output must belong to the same $K$ as the input.

For a fully computable example, let $D=\{(u,v):u^2+v^2\le 1\}$ be the filled unit disk. Apply a counterclockwise quarter-turn rotation, shrink by $0.4$, and add a translation:This affine map is a continuous map. Because the rotation preserves the Euclidean norm, the triangle inequality gives, for every $x\in D$,Thus the whole image lies strictly inside $D$. More precisely, it is a disk of radius $0.4$ centred at $(0.3,0.1)$. This containment check is what makes the Brouwer fixed-point theorem applicable; a formula being a continuous map by itself would not suffice.

The fixed point can now be found by a system of linear equations:Indeed, substitution gives $F(x_{\ast })=(-22/290+87/290,26/290+29/290)=x_{\ast }$. Its Euclidean norm is $\sqrt{290}/58\approx 0.293610$, safely inside $D$. The figure shows arrows from sample inputs to their outputs; an arrow disappears at the fixed point because the displacement $F(x)-x$ vanishes there.

Repeated iterations of a map happen to succeed for this example. Starting at $x_0=(1,0)$ and setting $x_{k+1}=F(x_k)$ gives $x_1=(0.3,0.5)$, $x_2=(0.1,0.22)$, and $x_3=(0.212,0.14)$. Subtracting the fixed point equation yields $x_{k+1}-x_{\ast }=0.4R_{90}(x_k-x_{\ast })$, so the Euclidean distance to $x_{\ast }$ shrinks by exactly $0.4$ each step. This extra fact follows from the contraction mapping property and the Banach fixed-point theorem. It is additional information beyond the Brouwer fixed-point theorem.

The distinction becomes visible with two simpler continuous maps of the same disk. The pure rotation $G(u,v)=(-v,u)$ has exactly one fixed point, $(0,0)$, since $u=-v$ and $v=u$ force both to vanish. Yet iterations of a map starting at $(1,0)$ here cycle through $(0,1)$, $(-1,0)$, and $(0,-1)$ forever. Their Euclidean distance from $(0,0)$ stays one, so there is no convergence in a metric space. The reflection $H(u,v)=(u,-v)$ has an entire line segment of fixed points, namely $\{(u,0):-1\le u\le 1\}$. Starting off that line segment produces an alternating pair. Existence, uniqueness, and convergence of iterations of a map are separate questions.

Removing hypotheses changes the answer. Restrict the same quarter-turn rotation to the circle $S^1=\{(u,v):u^2+v^2=1\}$. It remains a continuous map into that compact set, but its only possible fixed point, $(0,0)$, has been removed. The circle is not a convex set. Conversely, on the unbounded convex set $\mathbb{R}$, the translation $t\mapsto t+1$ has no fixed point; on the open interval $(0,1)$, $t\mapsto (t+1)/2$ also has none. Neither domain is a compact set. The finite-dimensional closed ball statement should likewise not be extended to arbitrary infinite-dimensional vector spaces, where a closed ball need not be a compact set.

The one-dimensional version explains the connection to the intermediate value theorem. For a continuous map $f:[a,b]\to [a,b]$, put $h(t)=f(t)-t$. Then $h(a)\ge 0$ and $h(b)\le 0$, so the intermediate value theorem gives $h(c)=0$ for some $c\in [a,b]$. In higher dimensions, the Brouwer fixed-point theorem supplies a corresponding existence principle when this scalar sign argument is unavailable. Its practical value is ensuring that a proposed equation $x=F(x)$ has a solution once a suitable domain is found; it supplies no general promise of a fast algorithm.

A useful application is positive matrix eigenvector from simplex normalization: rescale the output of a strictly positive matrix to make its entries sum to one. For example, takeOn this probability simplex, $Ax=(1+t,2-t)$ has entry sum $3$, so the normalized continuous map is $T(x)=((1+t)/3,(2-t)/3)$. Both entries are positive and sum to one, keeping the image inside the probability simplex. Its fixed point equation is $t=(1+t)/3$, hence $t=1/2$. Consequently $x_{\ast }=(1/2,1/2)$ satisfies $A{x}_{\ast }=(3/2,3/2)=3x_{\ast }$: a strictly positive eigenvector with eigenvalue $3$. In larger dimensions, the Brouwer fixed-point theorem guarantees the corresponding positive eigenvector without requiring us to solve a system of linear equations in advance. The fixed point expresses preservation of relative weights, rather than preservation of their unnormalized size.

Proofs can be approached through the no-retraction theorem or Sperner's lemma. In the latter approach, increasingly fine triangulations locate approximate fixed points; compactness supplies a convergent subsequence, and continuous maps turn its limit into an exact fixed point. ^[ref]

A primary historical reference is L. E. J. Brouwer's 1911 paper Über Abbildung von Mannigfaltigkeiten, whose digitized original is hosted by Zenodo under DOI 10.1007/BF01456812. This gives a direct route to the early literature behind the Brouwer fixed-point theorem. ^[ref]

The butterfly effect is the intuitive idea that an arbitrarily small change in an initial condition can eventually change the observed future substantially. In dynamical systems, sensitive dependence makes this precise for a map $F:X\to X$ on a metric space $(X,d)$: there is one separation scale $\delta >0$ such thatThe same $\delta$ works at every point, whereas $y$ and the separation time $n$ may depend on the point and the requested closeness. For a flow map, replace the iteration of a map $F^n$ by ${\Phi }_t$ with some $t\ge 0$. The state space matters: restricting to a single fixed point removes the alternative nearby states needed for sensitivity.

Sensitivity is compatible with deterministic evolution. Given an exact initial condition, there is still one orbit; uncertainty concerns which exact state a measurement represents. It also allows accurate short-term prediction. For a continuous function $F$, each fixed finite iterated function $F^n$ is continuous, so sufficiently close initial states remain close at any specified finite step. What fails is a single initial accuracy that controls every future step.

The logistic map illustrates both sides. At $r=2.8$, nearby orbits in the basin of attraction of its attracting fixed point approach one another. At $r=4$, arbitrarily close states can eventually separate by a macroscopic fraction of $[0,1]$. The definition asserts the existence of separating neighbours, not that every nearby pair must separate: $x$ and $1-x$ have exactly the same next state under the logistic map and hence the same subsequent orbit. Nor must separation increase at every step.

A positive Lyapunov exponent describes average exponential growth of an infinitesimal displacement along an orbit, when the relevant limit exists. It is related to sensitivity but is not a substitute for its uniform definition. For a differentiable interval map, the chain rule givesAt $r=4$, the logistic map has $\lambda =\log 2$ for Lebesgue measure-almost every initial condition, while exceptional orbits can have different exponents. Finite-separation growth is only approximately exponential while the linear approximation remains useful. In a bounded state space, a separation cannot grow exponentially forever.

Sensitivity alone is also weaker than Devaney chaos, which additionally requires topological transitivity and dense periodic points. For example, $F(x)=2x$ on the real line has sensitive dependence and a positive Lyapunov exponent, but its only periodic point is zero, so it is not Devaney-chaotic. This distinguishes growth of uncertainty from the richer organization of recurrent orbits.

Edward N. Lorenz's "Deterministic Nonperiodic Flow" (1963) studied a thermal convection model with three ordinary differential equations. Its numerical orbits exhibited persistent nonperiodic motion and instability to small changes of the initial condition. Lorenz discussed the consequences for very-long-range weather forecasting: a deterministic law need not make a detailed distant future reliably recoverable from approximate observations. The historically relevant lesson of the Lorenz system is this growth of uncertainty under deterministic evolution.^[ref]