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For the cubic population iteration, put $g(u)=\lambda u(1-u^2)$. Its fixed points solveThus $u_{\ast }=0$ always exists. For $\lambda \ne 0$ there are alsowhenever the expression under the square root is positive, namely $\lambda <0$ or $\lambda >1$. At $\lambda =1$ these coincide with zero; at $\lambda =0$ only zero exists.

The criterion for fixed point stability for an iteration is $|g^{\prime }(u_{\ast })|<1$. To justify it, continuity of $g^{\prime }$ gives a small interval around the fixed point where $|g^{\prime }|\le q<1$. The mean value theorem then gives $|g(u)-u_{\ast }|\le q|u-u_{\ast }|$, so this interval is invariant and the error decreases geometrically. This proves local asymptotic stability rather than merely linear boundedness.

Here $g^{\prime }(u)=\lambda (1-3u^2)$. At zero its value is $\lambda$, so zero is locally asymptotically stable for $-1<\lambda <1$. At a nonzero fixed point it isThe strict inequality $|3-2\lambda |<1$ is equivalent to $1<\lambda <2$. Both nonzero real fixed points are therefore locally asymptotically stable on that interval. This proves the two requested existence ranges. The strict derivative criterion alone makes no conclusion at the endpoint multipliers of modulus one.

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 $d(Tx,Ty)\le qd(x,y)$ for all $x,y$, with one constant $0\le q<1$. Choose $x_0\in X$ and define $x_{n+1}=T{x}_n$. Successive distances satisfy $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$, so for every $m\ge 1$,Thus $(x_n)$ is a Cauchy sequence. Completeness gives $x_n\to x_{\ast }$. Since a contraction mapping is a continuous function, $T{x}_{\ast }=\lim T{x}_n=\lim x_{n+1}=x_{\ast }$. If $y_{\ast }$ were another fixed point, $d(x_{\ast },y_{\ast })=d(T{x}_{\ast },T{y}_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing distance zero. This proves both existence and uniqueness.

Now suppose $T=f^k$ is a contraction mapping, where $k$ is a positive integer. Let $y$ be its unique fixed point. Since $f$ commutes with its own iterate, $T(fy)=f(Ty)=fy$. Thus $fy$ is also a fixed point of $T$, so $fy=y$. Any fixed point of $f$ is a fixed point of $T$, and must equal $y$. A contractive iterate therefore gives a unique fixed point of $f$, without assuming continuity of $f$ itself.

The space $X=C([0,1],\mathbb{R})$ is complete for the uniform norm. Indeed, a uniformly Cauchy sequence converges pointwise by completeness of $\mathbb{R}$; passing to the limit in its uniform Cauchy bound shows uniform convergence, and the uniform limit of continuous functions is a continuous function. The operator $F$ maps this space to itself: $s\mapsto \varphi (h(s),s)$ is a continuous function, and its integral plus the continuous $g$ is a continuous function.

For $n=0$ the desired bound is the definition of $\Vert h-k{\Vert }_{\infty }$. Suppose it holds for $n$. The Lipschitz condition and the fact that the same $g$ cancels giveThis proves the estimate by induction. Taking the supremum over $0\le t\le 1$, the Lipschitz constant of $F^n$ is at most $M^n/n!$. These constants tend to zero, since the ratio of successive terms is $M/(n+1)$. Some iterate is consequently a contraction. The preceding result gives a unique fixed point of $F$, which is exactly $\boxed{\text{the unique continuous solution of the integral equation on }[0,1]}$. No assumption $M<1$ is needed.

For repeated small returns, ${\xi }_{n+1}=O(z_n^{\delta })$, so $\rho +{\xi }_n$ is close to the nonzero constant $\rho$. Retain its leading value in the second component. Combining $c\cos \vartheta +d\sin \vartheta$ into a single cosine, and absorbing the section size and fixed phases, gives the Shilnikov return mapChanging the sign of the logarithmic angle changes only the phase after using cosine's evenness. A generic reinjection has $A>0$; $\Phi$ can be represented modulo $2\pi$ by a positive value. The original PDF has exponent $\delta$, which the converted TeX mistakenly replaced by $2$.

This is an asymptotic scalar approximation, not an assertion that the full map has an exact one-dimensional invariant quotient. To see why it captures the thin geometry, differentiate the two components in part (a). Their determinant is of orderwhere $C$ involves $\Omega$ and the determinant of the global linear map. For $\delta >1/2$ this tends to zero, so the return is strongly area-contracting even when its longitudinal derivative is large. Its global quadratic corrections are $O(z^{2\delta })=o(z)$ in this range. For a small fixed point, also $\xi =O(z^{\delta })$, so ignoring $\xi$ contributes that same higher order. Multipassage iterates require all passage coordinates to stay in the small domain; the leading map describes their thin folded return geometry rather than controlling arbitrary points outside it.

Let $\theta (z)=\Omega \log z+\Phi$. A fixed point of the Shilnikov return map is a periodic orbit with one marked global excursion, and satisfiesIts local passage takes logarithmically long. If the global travel time tends to $T_g$, then its physical period isEliminating $z$ and absorbing section sizes and $T_g$ into constants $C_1,C_2>0$ and a phase ${\varphi }_0$ gives the leading period–parameter relationFor the idealized leading map with a constant global travel time, the displayed two terms follow exactly; for the original flow, local and global approximation errors qualify their accuracy.

If $\delta >1$, the oscillatory term is $o(z)$ and $f_S^{\prime }(z)\to 0$. In a sufficiently small interval the map is a contraction. For $\mu <0$ there is a unique small attracting fixed point $z=-\mu +O(|\mu {|}^{\delta })$; it is an attracting periodic orbit also in the transverse direction of the full return. For $\mu \ge 0$, $f_S(z)-z=-\mu -z+O(z^{\delta })<0$ for sufficiently small positive $z$, so no periodic return orbit can remain wholly in this neighborhood. As $\mu \uparrow 0$ from below, the attracting cycle becomes the homoclinic connection andContraction also excludes higher-period cycles wholly inside that small return neighborhood. This does not exclude remote periodic orbits elsewhere in the vector field.

If $1/2<\delta <1$, the oscillatory term dominates $z$. At $\mu =0$ the fixed-point equation isThere are infinitely many positive solutions approaching zero, with phases approaching successive zeros of cosine. Their leading ratio is $z_{j+1}/z_j\to e^{-\pi /\Omega }$, so their periods satisfy $T_{j+1}-T_j\to \pi /\omega$. At these roots the sine tends to $\pm 1$, and the fixed-point multiplier has magnitude asymptotic to $A\Omega z^{\delta -1}\to \infty$. Thus these are longitudinally unstable cycles, with the thin transverse direction contracting: infinitely many saddle periodic orbits accumulate on the homoclinic connection. Expanding oscillatory lobes produce interval-covering walks and horseshoe dynamics in suitable iterates; this is the positive-saddle-value case of the Shilnikov bifurcation.

Away from the connection, the one-passage branches wiggle in the period–parameter plane. To leading order,There is no single monotone logarithmic law: the phase matters, and the exponentially smaller $-C_2{e}^{-{\lambda }_{+}T}$ term matters near cosine zeros. Folds occur when $f_S^{\prime }(z)=1$, asymptotically at $\tan \theta =\delta /\Omega$. Their parameter values alternate in sign and accumulate geometrically toward zero. On a fixed-point branch, stability is $|f_S^{\prime }(z)|<1$; small stable windows occur near stationary points of the oscillatory map. Crossing $f_S^{\prime }=-1$ gives period-doubling bifurcations, while $f_S^{\prime }=1$ gives saddle-node bifurcations of periodic orbits. Subsequent doublings and chaotic invariant sets can occur; neither all periodic orbits being stable nor a particular universal sequence for every coefficient choice is implied.

For completeness, an $m$-cycle of the return map, corresponding to a periodic orbit with $m$ global excursions, obeysand its total period isIts longitudinal multiplier is ${\prod }_j{f}_S^{\prime }(z_j)$. This is the relevant period relation for multipassage cycles; substituting their total $T$ into the one-passage cosine law would generally be incorrect.

inversion in a circle with center $c$ and radius $r>0$ is $I(z)=c+r^2/(\overline{z}-\overline{c})$, with $c$ and infinity interchanged. It fixes the circle pointwise and reverses complex orientation. Each inversion is fractional-linear in $\overline{z}$, so composing two eliminates the conjugation and gives $(az+b)/(cz+d)$ with nonzero determinant: a Möbius transformation.

Translate and rotate two distinct centers to $0,d$, where $d>0$, and let the radii be $r,s$. The composition of inversion in the first circle followed by the second isIts fixed points solve $d{z}^2-(d^2+r^2-s^2)z+d{r}^2=0$. The discriminant isDisjoint circle boundaries require either $d>r+s$ or $d<|r-s|$, so both factors have the same nonzero sign and $\Delta >0$. There are two distinct fixed points on the line of centers. If the circles are concentric, their radii differ and the composition is $z\mapsto (s^2/r^2)z$, with exactly the two fixed points $0,\infty$. Thus the composition has two distinct fixed points, also for nested disjoint circles. Reversing the order inverts the Möbius transformation and leaves its fixed points unchanged.

Let $T$ be an isometry, and set $S(x)=T(x)-T(0)$. Then $S(0)=0$, and preservation of distances gives $|S(x)|=|x|$. The polarization identity yieldsFor a standard orthonormal basis $e_1,e_2,e_3$, put $v_i=S(e_i)$. These three vectors are an orthonormal basis. Since $⟨S(x),v_i⟩=x_i$, we have $S(x)={\sum }_i{x}_i{v}_i=Qx$ for an orthogonal matrix $Q$. Thusan affine map.

For a finite isometry group $G$, take $c=|G{|}^{-1}{\sum }_{g\in G}g(0)$. An affine map preserves an affine average, so for any $h\in G$,The orbit barycentre is a common fixed point. This proves the fixed point of a finite Euclidean isometry group result.

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.