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For the logistic differential equation, separate variables to obtain $\log |y/(1-ay)|=rt+C$ away from the constant solutions. Applying the positive initial condition givesThe formula also includes the constant solution $y_0=1/a$. Below $1/a$ positive solutions increase; above $1/a$ they decrease. They approach $1/a$ as $t\to \infty$. Equivalently, linearizing $F(y)=ry(1-ay)$ gives $F^{\prime }(0)=r>0$ and $F^{\prime }(1/a)=-r<0$. Thus zero is unstable and $1/a$ is asymptotically stable.

The Forward Euler method gives $y_{n+1}=(1+r\delta t)y_n-ra\delta t{y}_n^2$. Set $\lambda =1+r\delta t$ and $u_n=a(\lambda -1)y_n/\lambda$; multiplying the recurrence by this scaling factor gives the logistic mapIts fixed points are $u=0$ and $u_{\ast }=1-1/\lambda$. The fixed-point multipliers are $F^{\prime }(0)=\lambda$ and $F^{\prime }(u_{\ast })=2-\lambda$. Since $\lambda >1$, zero is always unstable. The positive fixed point is stable for $1<\lambda <3$ and unstable for $\lambda >3$. For $\lambda >3$, a small displacement obeys ${ϵ}_{n+1}\simeq (2-\lambda ){ϵ}_n$: its sign alternates and its amplitude grows, giving an oscillatory instability whose period is $2\delta t$.

This is an artifact of Euler discretization of logistic growth. Exact small perturbations decay by the positive factor $e^{-r\delta t}$ per time step; a one-dimensional autonomous differential equation cannot have a nonconstant periodic orbit, since its solutions move monotonically between equilibrium points. The artificial instability occurs when $r\delta t>2$, beyond the stable step range of this discretization.

For $\lambda =3.2$, remove the fixed point factors from $F(F(u))-u$ to obtain ${\lambda }^2{u}^2-\lambda (\lambda +1)u+\lambda +1=0$. The resulting logistic map two-cycle isThese points map to each other. Their periodic-orbit multiplier is $F^{\prime }(u_{-})F^{\prime }(u_{+})=4+2\lambda -{\lambda }^2=0.16$, so the two-cycle attracts nearby iterates. The cobweb shows a small initial departure from $u_{\ast }=0.6875$ growing with alternating sign and then settling into the finite rectangle between $u_{-}$ and $u_{+}$. The time trace shows the corresponding finite-amplitude oscillation.

The contraction mapping theorem says that a self-map $g$ of a nonempty complete metric space satisfying $d(gx,gy)\le kd(x,y)$ for a constant $0\le k<1$ has exactly one fixed point. Apply it to $g=f^N$ to obtain $p$ with $f^{N}p=p$.

No continuity of $f$ itself is needed: composition givesThus $fp$ is also a fixed point of $f^N$, whose uniqueness forces $fp=p$. Conversely any fixed point of $f$ is fixed by $f^N$, so it must equal $p$. The unique fixed point of the contracting iterate is also the unique fixed point of $f$. The usual nonempty-space hypothesis is necessary; an empty space has no fixed point.

The strict distance inequality implies $d(fx,fy)\le d(x,y)$ for all points, so $f$ is Lipschitz continuous. Henceis a continuous function, since $|g(x)-g(y)|\le d(fx,fy)+d(x,y)\le 2d(x,y)$. On the nonempty compact metric space, $g$ attains its minimum at some $p$.

If $fp\ne p$, the strict distance inequality applied to $fp,p$ givesa contradiction. Thus $fp=p$. If $p,q$ were different fixed points, then $d(p,q)=d(fp,fq)<d(p,q)$, again impossible. There is exactly one fixed point. This proves strict distance decrease on a compact metric space without assuming a uniform contraction constant.

Let $K=\overline{O(x)}$. It is nonempty, closed and bounded in finite-dimensional Euclidean space, hence compact by the Heine-Borel theorem. The strict distance inequality again makes $f$ Lipschitz continuous. If $y\in K$, choose a sequence $y_j\in O(x)$ converging to $y$. Then $f{y}_j\in O(x)$, and continuity gives $f{y}_j\to fy$, so $fy\in K$. This proves $f(K)\subseteq K$.

The restriction to $K$ satisfies strict distance decrease on a compact metric space, so part (b) produces a fixed point there. If a second fixed point existed anywhere in ${\mathbb{R}}^n$, the same strict distance inequality applied to the two points would contradict equality of their distance. Therefore the fixed point is unique in all of ${\mathbb{R}}^n$. This is the bounded-orbit fixed point for a strictly distance-decreasing Euclidean map.

The function is holomorphic on a neighborhood of the closed unit disk. On $|z|=1$, the mapping assumption gives $|g(z)|<1=|z|$. Apply Rouché's theorem to $z$ and $-g(z)$. Their sum $z-g(z)$ has the same number of interior zeros as $z$, namely one counting multiplicity. ThusThe zero is simple since its total multiplicity is one. No boundary fixed point is possible because $g$ maps the boundary strictly inside.

Since every eigenvalue has modulus less than one, $1$ is not an eigenvalue and $I-H$ is invertible. Hence the iteration has a unique fixed point $\hat{x}=(I-H{)}^{-1}v$. The stated consistency implication gives $A\hat{x}=b$, and nonsingularity of $A$ identifies $\hat{x}=x^{\ast }$. This justifies using the fixed-point relation even though consistency was stated in only one direction.

Expand the initial error in the eigenvector basis, $x^0-x^{\ast }={\sum }_i{c}_i{w}_i$. Subtract the fixed-point equation from the stationary iterative method for a linear system and iterate:With $q={\max }_i|{\lambda }_i|<1$, any norm obeys $\Vert x^k-x^{\ast }\Vert \le q^k{\sum }_i|c_i|\Vert w_i\Vert \to 0$. ThusThe argument uses the given diagonalizable eigenbasis; the more general convergence criterion is spectral radius less than one.

A global bifurcation joins two mechanisms: a long passage near a hyperbolic equilibrium point and a finite excursion returning the unstable manifold to its neighborhood. Eigenvalue ratios control the local passage, while the geometry of the global return decides reinjection, folding and trapping. In the real symmetric case there are two returning branches and a power-law cusp. In the saddle-focus case rotation adds infinitely many oscillations to that cusp. The following maps display both mechanisms explicitly.

For the Lorenz configuration, enter on $z=h$ at $(x,y,h)$ and leave on $x=sh$, where $s=\mathrm{sgn}(x)$. Solving the linear flow gives a flight timePut $\gamma =-\hat{\lambda }/{\lambda }_{+}>\delta =-{\lambda }_{-}/{\lambda }_{+}$. The outgoing coordinates areThe smooth global map from each outgoing section to the incoming section has, to leading order,up to the choice of which unstable branch carries the sign of $\nu$. Reflection symmetry makes the offsets opposite and the leading coefficients equal. Since the stable transverse coordinate is multiplied by the higher power $|x{|}^{\gamma }$, it contributes only subleading terms to the leading scalar map. Absorbing powers of $h$ into a positive constant gives the Lorenz power return mapThe line $x=0$ lies on the stable manifold and has no finite return; the discontinuity is therefore a genuine missing return, not an arbitrary value to assign at zero.

If $\delta >1$, the derivative $A\delta |x{|}^{\delta -1}$ tends to zero. For $\mu <0$, there are two attracting fixed points, $x=\pm s$ with $s=-\mu +A{s}^{\delta }\simeq -\mu$. They describe two symmetry-related one-lobe attracting flow periodic orbits. For $\mu >0$, the orbit alternates between $s$ and $-s$, where $s=\mu -A{s}^{\delta }\simeq \mu$; its multiplier is $(A\delta s^{\delta -1}{)}^2<1$. This represents one attracting flow periodic orbit traversing both lobes. At $\mu =0$ the two homoclinic loops meet: this is the contracting gluing scenario, with diverging flow period at the connection.

If $\delta <1$, the local derivative is unbounded, and the small periodic orbits are repelling in the expanding return direction. For $\mu >0$, the small positive fixed point has $s\simeq (\mu /A{)}^{1/\delta }$, with a multiplier tending to infinity; there is a symmetry-related negative fixed point as well. More complex motion follows from the two branches, not merely from a large derivative at one fixed point. Choose $I=[-c\mu ,c\mu ]$ with $0<c<1$. The inverse branches areFor sufficiently small positive $\mu$, both lie inside $I$, their images are disjoint, and their derivatives are $O({\mu }^{1/\delta -1})\to 0$. Nested inverse images for every infinite sign sequence therefore define a Cantor invariant set. Iterating the forward map shifts the sequence. This constructs periodic itineraries of every symbolic period and nonperiodic itineraries; transverse contraction in the full return turns the expanding two-strip mechanism into a horseshoe.

This is the local chaotic Lorenz mechanism. A robust attracting Lorenz set additionally needs a global trapping region and the appropriate return properties. The scalar asymptote and $\delta <1$ alone do not prove that every nearby orbit is attracted to one chaotic set; periodic windows, escaping orbits and other global returns must be distinguished. Symmetry is crucial to obtaining the two homoclinic branches together in a single parameter variation.

In cylindrical coordinates the saddle-focus linearization isUse a small incoming neighborhood of the stated section point $(\rho ,0,-\mu )$. A nearby incoming cylinder $r=\rho$ is smoothly equivalent to that section for this calculation. Denote the positive unstable passage coordinate by $x=z_{\mathrm{in}}>0$ and leave at $z=h$. ThenThe global map is smooth in the outgoing Cartesian coordinates $r_{\mathrm{out}}\cos {\theta }_{\mathrm{out}}$ and $r_{\mathrm{out}}\sin {\theta }_{\mathrm{out}}$. Its returned unstable coordinate therefore has the leading formCombining the two trigonometric terms, absorbing the fixed scales into the amplitude and phase, and using the evenness of cosine givesThe returned radial section coordinate is $\rho +O(x^{\delta })$, supplying the thin transverse direction of the full two-dimensional return. In particular its determinant has order $x^{2\delta -1}$; the exponent is also obtained by integrating the flow divergence ${\lambda }_{+}+2{\lambda }_{-}$ over the local flight. Thus $\delta >1/2$ is the volume-contracting range. Corrections from the returned radial deviation enter at order $x^{2\delta }$, smaller than the fixed-point term $x$ in this range. Only positive returned $x$ on the chosen branch belong to the scalar return domain.

For $\delta >1$,For small $\mu <0$, there is one small attracting fixed point $x=-\mu +O(|\mu {|}^{\delta })$, representing a stable long-period periodic orbit. It ends in the homoclinic connection at $\mu =0$; for $\mu >0$ there is no sufficiently small positive fixed point. The oscillations alone do not create expanding chaos when their derivative tends to zero.

For $1/2<\delta <1$, there is both net volume contraction and a positive saddle equilibrium value ${\lambda }_{+}+{\lambda }_{-}>0$. At the connection, fixed points satisfyThe right side tends to zero, while the phase runs through infinitely many rotations. There are therefore infinitely many positive fixed points accumulating at zero near successive cosine zeros. Their successive size ratios tend to $e^{-\pi /q}$ and their derivative magnitudes diverge. In the full return they are saddle equilibrium periodic orbits: one multiplier expands and the thin transverse direction contracts. Their flow periods grow as ${\lambda }_{+}^{-1}\log (h/x)$.

The oscillatory graph also supplies the complicated itineraries. Take a sufficiently small positive return interval. Near successive zero crossings, narrow monotone subintervals map across that interval with arbitrarily large slope. Two or more inverse branches contract into it, so repeated inverse choices give symbolic invariant sets just as in part (a); lifting the strips to the thin two-dimensional return gives horseshoes and infinitely many periodic orbits. This explains the Shilnikov chaotic mechanism rather than only naming it.

For nonzero splitting, the graph shifts vertically. Fixed-point tangencies with slope $+1$ create saddle-node bifurcations of periodic orbits; crossings of slope $-1$ give period doubling. Near an oscillatory extremum the derivative can be small, allowing stable periodic windows. Such thresholds accumulate geometrically toward the homoclinic parameter, interspersed with expanding returns. Hence the unfolding can contain complicated switching, arbitrarily long periods and chaos, rather than a single attracting periodic branch. This is log-periodic accumulation of Shilnikov cycles. The conclusion concerns invariant sets and possible attractors allowed by the return geometry; it does not assert that a homoclinic connection alone makes all nearby trajectories chaotic.

A loop integral in a theory with dimensionless classical couplings can contain a logarithmic ultraviolet divergence. After subtraction, an expression such as $\log (-p^2/{\mu }^2)$ remains; one must specify the momentum scale at which the measured or renormalized coupling is fixed. A logarithm of a dimensionful momentum alone is not meaningful. In dimensional regularization, the same necessity appears through ${\lambda }_{\mathrm{bare}}={\mu }^{d_{\lambda }}{Z}_{\lambda }\lambda (\mu )$ in the continued dimension, with $d_{\lambda }$ tending to zero in the physical dimension.

Thus the renormalization prescription introduces an arbitrary renormalization scale, even when no mass or length occurs in the classical action. Physical quantities cannot depend on its arbitrary choice: explicit logarithms cancel the implicit change of the running coupling, governed by $\mu d\lambda /d\mu =\beta (\lambda )$. A nonzero renormalization-group beta function is a quantum scale anomaly. An integration constant may encode the same data as a dimensionful scale, which is dimensional transmutation. The scale labels a renormalization condition, not a new arbitrary measurable parameter. At a true fixed point the coupling need not run; the use of a subtraction scale is then a convention rather than an intrinsic generated physical scale.

The contraction mapping theorem states that a contraction mapping $f$ of a nonempty complete metric space into itself has a unique fixed point. More explicitly, if $d(f(x),f(y))\le qd(x,y)$ with $0\le q<1$, iteration from any $x_0$ converges to that fixed point.

To prove this, put $x_{n+1}=f(x_n)$. Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$, so for $m>n$ the triangle inequality givesThus $(x_n)$ is a Cauchy sequence, and completeness supplies a limit $x_{\ast }$. A contraction mapping is continuous, so $f(x_{\ast })=\lim f(x_n)=\lim x_{n+1}=x_{\ast }$. Two fixed points satisfy $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing equality of the points. Passing to the limit in the bound also gives the useful error estimate $d(x_n,x_{\ast })\le q^{n}d(x_1,x_0)/(1-q)$.

For the function space, take $X$ nonempty; otherwise the supremum in the printed definition is undefined without a convention. Boundedness of $X$ makes the uniform metric $\rho (f,g)$ finite. Nonnegativity and symmetry follow from $d$; $\rho (f,g)=0$ means $f(x)=g(x)$ at every $x$. Taking suprema in $d(f(x),h(x))\le d(f(x),g(x))+d(g(x),h(x))$ proves the triangle inequality. Hence $\rho$ is a metric.

Suppose $(f_n)$ is a Cauchy sequence for this uniform metric. At each $x$, $d(f_n(x),f_m(x))\le \rho (f_n,f_m)$, so completeness of $X$ supplies a value $f(x)={\lim }_n{f}_n(x)$. Given $\epsilon >0$, choose $N$ with $\rho (f_n,f_m)<\epsilon /2$ for all $n,m\ge N$. Fix $n\ge N$ and let $m\to \infty$ pointwise; then $d(f_n(x),f(x))\le \epsilon /2$ for every $x$. Thus $\rho (f_n,f)\le \epsilon /2<\epsilon$. The uniform limit theorem makes this limit continuous, so $f\in F$. This proves the completeness of the continuous-map space in the uniform metric.

Finally fix $f\in C$ with contraction constant $q_f<1$, and write $x_f=\theta (f)$ and $x_g=\theta (g)$. Using the fixed point equations,Therefore the fixed-point stability in the uniform metric estimate isChoosing $\rho (f,g)<(1-q_f)\epsilon$ proves continuity of $\theta$ at every $f$. The estimate uses the constant of the fixed map $f$ only; it does not require a common contraction constant for all nearby $g$.