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The planar Brouwer fixed-point theorem says that every continuous map from the closed disk $D$ to itself has a fixed point. If a retraction $r:D\to \partial D$ existed, $x\mapsto -r(x)$ would be a continuous self-map of $D$. At a fixed point $x=-r(x)$, $x$ lies on the boundary, where $r(x)=x$; this would give $x=-x$, impossible on the unit circle.

Conversely, suppose a continuous self-map $f:D\to D$ has no fixed point. For each $x$, take the ray starting at $f(x)$ through $x$, and let $r(x)$ be its exit point on $\partial D$. The denominator $|x-f(x)|$ never vanishes, and solving the quadratic intersection with the circle shows that the exit point depends continuously on $x$. More explicitly, with $v=x-f(x)$,The ray exits beyond or at $x$, so $\lambda \ge 1$, and when $x\in \partial D$ its exit point is $x$. Hence $r$ is a retraction. Nonexistence of a boundary retraction and the planar fixed-point theorem are equivalent.

For a positively oriented simple closed curve on which the vector field is nonzero, its Poincaré index is the total rotation of the vector field along that curve divided by $2\pi$, equivalently the winding number of its normalized direction. The Poincaré index of an isolated fixed point is that of a sufficiently small enclosing curve containing no other fixed point. The index of a planar periodic orbit is $+1$.

The map normal forms at multiplier $+1$ are given in the numbered subparts below. For the multiplier $-1$ problem, the second iterate instead exposes the emerging period-two cycle.

For the multiplier $-1$ problem, write $x=X+\alpha /2$. To the required weighted order, with $X=O({\mu }^{1/2})$, the shifted map is $X_{n+1}=-X_n+b{X}_n+\gamma X_n^2+\delta X_n^3$, where $b=\beta +\alpha \gamma$. Composing it with itself cancels the quadratic term. The cubic contribution is $-2\delta X^3-2{\gamma }^2{X}^3$, while the linear correction is $-2bX$. Consequently the quadratic-cubic flip criticality calculation givesThus the second iterate has normal form $X_{n+2}=X_n+\hat{\mu }{X}_n-A{X}_n^3$, up to terms of weighted order at least four. This is the asymptotic interpretation of the stated remainder under $X=O({\mu }^{1/2})$; a uniformly valid local expansion about the exact fixed point also retains mixed parameter remainders.

Assuming $\hat{\mu }$ crosses zero transversely and $A\ne 0$, the small nonzero fixed points of the second iterate satisfy $X^2\sim \hat{\mu }/A$. They form a period-two orbit. For $A>0$ they exist on the side $\hat{\mu }>0$, where the original fixed point has become unstable, and the second-iterate multiplier is $1-2\hat{\mu }+o(\hat{\mu })$, so the cycle is stable. The period-doubling bifurcation is therefore supercritical precisely when $\delta +{\gamma }^2>0$. Equality is a degenerate case requiring higher-order terms.

The contraction mapping theorem states that a contraction mapping $T:X\to X$ on a nonempty complete metric space has a unique fixed point. Here contraction means that some $0\le q<1$ satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$.

For a proof, choose $x_0$ and iterate $x_{r+1}=T{x}_r$. Induction gives $d(x_{r+1},x_r)\le q^{r}d(x_1,x_0)$. For $m>n$, the triangle inequality and a geometric series giveThus the iterates form a Cauchy sequence, and completeness supplies a limit $a\in X$. A contraction mapping is continuous, so $Ta=\lim T{x}_n=\lim x_{n+1}=a$. If $b$ is another fixed point, $d(a,b)=d(Ta,Tb)\le qd(a,b)$, forcing $d(a,b)=0$. This proves existence and uniqueness.

For the exponential problem, use $h=f\circ f$ on the complete metric space $\mathbb{R}$. Its derivative isPutting $t=e^{-x}>0$, the function $t{e}^{-t}$ has maximum $1/e$ at $t=1$, as its derivative is $(1-t)e^{-t}$. By the mean value theorem, $|h(x)-h(y)|\le e^{-1}|x-y|$, so $h$ is a contraction mapping on all of $\mathbb{R}$. Let $a$ be its unique fixed point. The compositions commute, so $h(f(a))=f(h(a))=f(a)$; uniqueness of the fixed point of $h$ forces $f(a)=a$. Conversely every fixed point of $f$ is fixed by $h$. Hence there is exactly one real solution of $x=e^{-x}$. It lies in $(0,1)$ because $a=e^{-a}>0$ and then $e^{-a}<1$. This uses the iterated contraction argument without claiming that $f$ itself contracts on all of $\mathbb{R}$.

The homogeneous fixed point must satisfy $v=u^2$. The singular point $u=v=0$ is outside the domain of the reaction term, so substitution in the other equation gives the unique regular spatially homogeneous equilibriumThis is the basally forced quadratic activator-inhibitor model. The linearization of its reaction terms gives the Jacobian matrixPut $a=(r-1)/(r+1)$, so $-1<a<1$. Its trace and determinant are $a-q$ and $q$, and its eigenvalues areSince $q>0$, both eigenvalues have negative real part exactly when $q>a$. HenceFor $r\le 1$ the equilibrium is always strictly linearly stable. For $r>1$, equality $q=a$ is an oscillatory threshold with ${\lambda }_{\pm }=\pm i\sqrt{a}$; the nonlinear qualification at this threshold is proved below.

Now insert a perturbation proportional to $e^{\lambda t+ikx}$. The diffusion terms subtract $k^2\mathrm{diag}(1,p)$ from $J$. With $s=k^2$ the mode trace and determinant areFor a strictly stable homogeneous state, $q>a$ makes $\tau (s)<0$ for all $s\ge 0$. A nonzero Fourier mode can therefore grow only if $\Delta (s)<0$, in which case its two real eigenvalues have opposite signs. The determinant parabola has its minimum at $s_{\ast }=(pa-q)/(2p)$, and that minimum is $q-(pa-q{)}^2/(4p)$. The two-species Turing criterion is consequently derived here asThese force $a>0$, hence $r>1$. Equivalently, solving $pa-q>2\sqrt{pq}$ for $\sqrt{p}$ givesThus the inhibitor in a reaction-diffusion system must diffuse sufficiently fast. Inside this region, the growing wavenumbers obeyAt onset, the minimum touches zero and the two determinant roots coalesce. ThereforeHere $p$ is evaluated on the onset surface. These are the Turing threshold of a basally forced quadratic activator-inhibitor model and its continuum critical wavelength. On a finite or domain with periodic boundary conditions an admissible nonzero spatial mode must additionally fall in the stated band. Equality in the diffusion criterion is marginal onset, while $q=a$ is the distinct homogeneous oscillatory threshold.

For completeness, the weak focus at the Hopf threshold of a basal activator-inhibitor model resolves the nonlinear homogeneous stability when $q=a>0$. Let $h=(1+a)/2$, $\beta =(1-a)/2$, $\omega =\sqrt{a}$, and set $u=(1+r)(1+x)$, $v=(1+r{)}^2(1+y)$, $X=x$, $Y=(hy-ax)/\omega$. Expanding the rational reaction term to cubic order giveswhereWrite $X=\varrho \cos \theta$, $Y=\varrho \sin \theta$ and evaluate these homogeneous polynomials on the unit circle. If $A=\cos \theta F_2+\sin \theta G_2$, $B=\cos \theta F_3+\sin \theta G_3$, and $C=\cos \theta G_2-\sin \theta F_2$, division of the radial and angular equations givesThe angular integral of $A$ is zero because it changes sign under $\theta \mapsto \theta +\pi$. Iterating its quadratic term once over a full cycle contributes the square of this zero integral, so the first nonzero return map term is cubic. Using $⟨{\cos }^4\theta ⟩=3/8$, $⟨{\cos }^2\theta {\sin }^2\theta ⟩=1/8$ and the corresponding sixth moments gives $⟨B⟩=a/4$ and $⟨AC⟩/\omega =(a+1)/8$. The Poincaré map isBecause $0<a<1$, sufficiently small positive radii decrease on every return. The equilibrium at $q=a$ is therefore nonlinearly asymptotically stable, with weak rather than exponential attraction, despite its neutral linear eigenvalues. The strict inequalities above describe the ordinary diffusion-driven instability from a strictly linearly stable homogeneous state.

In the squared-denominator population map, $r$ is the low-density per-generation multiplication factor, while $b$ controls crowding: $1/b$ is the characteristic population scale. The nonzero fixed point for $r>1$ and its multiplier areSince this multiplier lies strictly between minus one and one, $N_{\ast }$ is locally asymptotically stable. This holds for every finite $r>1$, even when the map is locally decreasing at its fixed point.

For the post-initial invariant interval for a squared-denominator population map, put $M=r/(4b)$ and $L=4r^2/[b(r+4{)}^2]$. The derivative $F^{\prime }(N)=r(1-bN)/(1+bN{)}^3$ shows that the global maximum is $M$ at $N=1/b$. The first two iterates of the supplied initial condition are consequentlyFor $r>4$, $1/b<L\le M$. Hence the map decreases throughout $[L,M]$, with minimum $F(M)=L$ there, and its global maximum bound ensures $F(L)\le M$. Thus $F([L,M])\subseteq [L,M]$, and induction provesBoth endpoints are attained at $t=2,3$. The lower bound printed without a time qualification is false at $t=1$, because the given value $N_1=1/b$ is strictly below $L$ when $r>4$. The invariant bounds apply after that initial step, not to it.

The cobweb plot below uses $n=bN$ and $r=9$. The first vertical move reaches the maximum, the next reaches its image, and subsequent vertical/horizontal steps remain in the invariant interval and approach the stable fixed point.

A Lyapunov function near an equilibrium is a continuously differentiable function with $V(0)=0$, $V(x)>0$ for nonzero nearby $x$, and $\dot{V}=\nabla V\cdot f\le 0$ along trajectories there. Strict negativity away from the equilibrium gives local asymptotic stability. Lyapunov stability itself means that every prescribed small neighborhood contains a smaller neighborhood whose forward trajectories never leave the prescribed one.

For the proposed quadratic form, direct differentiation givesWhen $\beta >1$, the quadratic part is negative definite, so it dominates the cubic remainder in a sufficiently small ball; $V$ is then a strict Lyapunov function. If $0<\beta <1$, take $x=y$ to make the leading derivative positive. At $\beta =1$, that same line gives $\dot{V}=-4x^3$, positive on its negative half. Values $\beta \le 0$ fail positive definiteness of $V$. ThereforeFor $\beta =2$ the full derivative factors, with no small-amplitude truncation:Every closed sublevel set $V\le c<1$ lies in $x>-1$, is forward invariant, and has strictly decreasing $V$ away from the origin. Compactness and the Lyapunov stability argument therefore show that the ellipse $x^2+2y^2<1$ belongs to the origin's basin of attraction. Boundary points with $V=1$ other than $(-1,0)$ also have negative derivative and enter the ellipse. The exception $(-1,0)$ is another fixed point, so the largest such sublevel ellipse cannot be included wholesale in the basin. This is a basin estimate from a Lyapunov sublevel set, not a claim that no additional points outside the ellipse converge to the origin.

The planar Brouwer fixed-point theorem states that every continuous map from a nonempty compact convex subset of ${\mathbb{R}}^2$ to itself has a fixed point. In particular it applies to a closed disk or a closed triangle. Convexity and compactness are essential hypotheses in this formulation.

Let $v_I,v_J,v_K$ be the vertices opposite the corresponding sides, and let ${\lambda }_I,{\lambda }_J,{\lambda }_K$ be their barycentric coordinates. We prove the equivalence by negating each assertion.

Suppose the closed covering sets have empty triple intersection. For $x\in T$ put $d_I(x)=\mathrm{dist}(x,A)$, and define $d_J,d_K$ similarly. At least one distance is positive, since $x$ cannot belong to all three sets, and at least one is zero, since they cover $T$. Consequentlyis continuous and lies in the boundary: one barycentric coordinate vanishes. On $I$ its first coefficient is zero because $I\subseteq A$, so $f(I)\subseteq I$, and likewise for the other sides. This contradicts assertion (1), proving (1) implies (2).

Conversely, suppose such a boundary-valued map exists. Define closed subsets of ${\mathbb{R}}^2$ byThey are closed because $T$ is compact and the coordinate functions are continuous. They cover $T$ because $f$ is boundary-valued; the side-preserving condition gives $I\subseteq A$, $J\subseteq B$, $K\subseteq C$. Their triple intersection is empty because barycentric coordinates sum to one. Thus failure of (1) implies failure of (2), completing the equivalence.

In fact both assertions hold. If the map in (1) existed, let $R$ rotate the equilateral triangle through $120^{\circ }$. Brouwer fixed-point theorem applied to $R\circ f:T\to T$ gives a fixed point $x$ on the boundary. On any side $I$ containing $x$, side preservation puts $x$ also on $R(I)$, so it is their common vertex. A side-preserving map fixes every vertex because that vertex lies on two sides. Therefore $x=Rf(x)=Rx$, impossible for a noncentral vertex. This also proves the covering-intersection assertion through the established equivalence.

Now $G(x)=x[{\mu }^2-(x^2-a{)}^2]$. The zero point exists for all parameters and has eigenvalue $G^{\prime }(0)={\mu }^2-a^2$, so it is stable for $|\mu |<|a|$ and unstable for $|\mu |>|a|$. At every nonzero fixed point,Both outer branches are stable and both inner branches unstable.

For $a>0$, the nonzero locations are as in part (i). At $(\mu ,x)=(\pm a,0)$ the local equation is $\dot{x}=({\mu }^2-a^2)x+2a{x}^3+O(x^5)$: these are subcritical pitchfork bifurcations, with the two unstable inner branches present on the side where zero is stable. At the bifurcation itself $\dot{x}\sim 2a{x}^3$, so zero is unstable. At $(0,\pm \sqrt{a})$ the inner and outer branches cross and exchange stability in transcritical bifurcations. The positive crossing is semistable, attracting from the right; the negative crossing is semistable, attracting from the left, because the additional factor $x$ reverses the local flow there.

For $a<0$, the outer stable pair exists only for $|\mu |>|a|$. The points $(\pm |a|,0)$ are supercritical pitchfork bifurcations, since the cubic coefficient $2a$ is negative. Zero loses stability as $|\mu |$ increases across $|a|$, and a stable symmetric pair appears. At the threshold zero remains nonlinearly stable, with $\dot{x}\sim 2a{x}^3$. There are no further bifurcations. The second row of the preceding figure gives the diagrams. The plots use $a=\pm 1$; the general locations are obtained from the formulas above.

Write $F(x)={\mu }^2-(x^2-a{)}^2$. Its fixed points satisfy $x^2=a\pm |\mu |$, retaining only nonnegative right sides. At a simple fixed point,Thus on the outer branches $x^2=a+|\mu |$, the positive point is stable and the negative point unstable. On the inner branches $x^2=a-|\mu |$, the positive point is unstable and the negative point stable.

For $a>0$, there are four simple points for $0<|\mu |<a$, two for $|\mu |>a$, and a saddle-node at each $(\mu ,x)=(\pm a,0)$, where the two inner points coalesce. At $\mu =0$, there are also bifurcations at $x=\pm \sqrt{a}$. The smooth branches $x=\pm \sqrt{a+\mu }$ and $x=\pm \sqrt{a-\mu }$ cross and exchange stability. Locally $F={\mu }^2-4a{\xi }^2+\text{higher terms}$, where $\xi =x-x_0$. After a parameter-dependent shift this is a transcritical bifurcation. It is not an ordinary saddle-node: two nearby branches exist on both sides of zero. At zero itself each crossing point is semistable, attracting from the right and repelling from the left.

For $a<0$, the only possible branch is $x=\pm \sqrt{a+|\mu |}$, existing for $|\mu |\ge |a|$. Each point $(\pm |a|,0)$ is a saddle-node bifurcation. There are no fixed points for $|\mu |<|a|$; outside that interval the positive point is stable and the negative point unstable. At either threshold $F\sim 2a{x}^2<0$, so the zero point is semistable. The first row of the diagram displays these branches; solid curves are stable and dashed curves unstable.

Put $s=\sqrt{b}$. The two fixed points on the invariant line $x=0$ are $P_{+}=(0,s)$ and $P_{-}=(0,-s)$. There are also $Q_{\pm }=(\pm \sqrt{b-a^2/4},-a/2)$ when $a^2<4b$. At equality they coalesce with $P_{-}$.

The Jacobian matrix isAt $P_{+}$ its eigenvalues are $-a-2s$ and $2s$, so it is always a saddle equilibrium. At $P_{-}$ they are $2s-a$ and $-2s$, so it is a saddle equilibrium for $a<2s$ and a stable node for $a>2s$, with a zero eigenvalue at $a=2s$.

At $Q_{\pm }$ the characteristic equation is ${\lambda }^2+a\lambda +4b-a^2=0$, givingFor $0<a^2<16b/5$ these points are stable spirals; for $16b/5<a^2<4b$ they are stable nodes. At $a^2=16b/5$ the double negative eigenvalue has only one independent eigenvector, so these are improper stable nodes. When $a=0$ they are center equilibria, as established in part (iii). Thus the collision at $\boxed{a^2=4b>0}$ changes both the number and stability of fixed points; it is the bifurcation classified below.