Proof. Differentiating Equation (2) gives \[u_{yy}=\phi'(u)u_y-1.\] Thus every critical point in the profile domain is a nondegenerate maximum, with \(u_{yy}=-1\). On either side of the prescribed peak the height therefore decreases strictly as one moves away from it, until the level \(h\) is reached. A second critical point on either branch would require a minimum between two maxima, which is impossible.

These intersections occur at finite \(y\). Indeed, while a branch remains in \([h,t]\), the profile is bounded there. For sufficiently large positive \(y\), Equation (2) gives \(u_y<-1\); for sufficiently large negative \(y\), it gives \(u_y>1\). In the respective outward directions these inequalities force the branch to reach \(h\). There is also no finite-\(y\) failure of continuation while \(u\) stays in the compact interval \([h,t]\). The strict derivative signs at the intersections give the displayed endpoint inequalities.

All the endpoints depend smoothly on \((h,t)\), by smooth dependence of the differential equation and transversality; see, for example, [14] for the standard differential-equation facts. To find their peak derivatives, let \(\xi(y;t)=\partial_t u(y;t)\), with \(y\) held fixed. Differentiating the moving initial condition gives \(\xi(\phi(t);t)=1\), because \(u_y=0\) at the peak. Moreover, \[\xi_y=\phi'(u)\xi,\] so \(\xi>0\) throughout the arc. Differentiating the endpoint equations now gives \[(y_<)_t=-\frac{\xi(y_<;t)}{r-M}<0,\qquad (y_>)_t=\frac{\xi(y_>;t)}{r+M}>0.\] In particular \(r_t>0\). After changing from \(t\) to \(r\), \[v=\frac{(y_>)_t+(y_<)_t}{(y_>)_t-(y_<)_t}\in(-1,1).\] The other strict inequality, \(|M|<r\), follows directly from the endpoint inequalities.

At fixed \(t\), differentiation with respect to \(h\) gives \[(y_<)_h=\frac1{r-M},\qquad (y_>)_h=-\frac1{r+M}.\] Hence \[ \left.\frac{dr}{dh}\right|_t=-\frac{r}{r^2-M^2},\qquad \left.\frac{dM}{dh}\right|_t =\frac{M}{r^2-M^2}-\phi'(h). \tag{5}\] The first identity implies \(d(r^2)/dh\le-2\). The nondegenerate maximum gives \(r\to0\) as \(t\downarrow h\), or as \(h\uparrow t\) with the peak fixed. Integrating the inequality yields \(r^2\ge2(t-h)\). At fixed \(h\), it follows that \(r\to\infty\) as \(t\to\infty\). This proves the asserted bijection and its smooth inverse. Also, at fixed width, \[\left.\frac{\partial h}{\partial t}\right|_r =-\frac{r_t}{r_h}>0.\] The same proof applies to any compact interval \([h,t]\) in the domain of a smooth profile. For a profile on the whole real line, the bound \(r^2\ge2(t-h)\) also proves the last assertion as \(h\to-\infty\). ◻

Proof. Both endpoint coordinates have variation \(\delta Y\), since their difference \(2r\) is fixed. Let \(\delta u\) denote variation at fixed \(y\). Differentiating the differential equation and endpoint conditions gives \[(\delta u)_y=\phi'(u)\delta u+\psi(u),\qquad \delta u(y_<)=(M-r)\delta Y,\quad \delta u(y_>)=(M+r)\delta Y.\] Multiplication by \(K\) and integration prove Equation (9). Its coefficient is positive because \(K>0\), \(M+r>0\), and \(M-r<0\).

For the comparison, first translate each endpoint coordinate by its own base value \(\phi_i(h)\). This replaces its profile by \(\phi_i(u)-\phi_i(h)\), without changing \(M_i\). Interpolate linearly between the translated profiles. The variation vanishes at the base and is strictly positive in every arc interior. Equation (9) then makes the derivative of the midpoint strictly positive along the interpolation. Integrating in the interpolation parameter proves the comparison. ◻

Proof. For positive \(r\), introduce the scaled height and endpoint coordinate \[s=\frac{u-h}{r^2},\qquad \zeta=\frac{y-\phi(h)}r.\] The scaled equation is \[\frac{ds}{d\zeta}=g(s)-\zeta,\qquad g(s)=\frac{\phi(h+r^2s)-\phi(h)}r.\] Its relative midpoint at scaled width \(1\) is \(M(h,r)/r\). More generally, if the scaled profile is held fixed while its width \(w\) varies near \(1\), its midpoint is \(M(h,rw)/r\). Its width derivative at \(w=1\) is therefore \(v(h,r)\).

To justify smoothness through zero, introduce a signed auxiliary parameter \(\rho\), while keeping the scaled width positive and near \(1\). Define \[ g(h,\rho,s) =\rho s\int_0^1\phi'(h+\tau\rho^2s)\,d\tau. \tag{12}\] Choose a fixed compact scaled-height interval containing a neighborhood of \([0,1/2]\), and restrict \(h\) to a neighborhood of the given regular base height. For sufficiently small \(|\rho|\), all arguments of \(\phi'\) lie in its smooth domain. Equation (12) is then a smooth family through \(\rho=0\), equals the preceding difference quotient when \(\rho\ne0\), and is the zero profile at \(\rho=0\). Negative \(\rho\) is only a parameter in this family; no negative physical width is used.

At the zero profile, an arc of scaled peak height \(T\) has endpoints \(\pm\sqrt{2T}\), and its width derivative with respect to \(T\) equals \(1\) at \(T=1/2\). Smooth dependence of the scaled differential equation, endpoint transversality, and the implicit function theorem therefore give smooth peak, midpoint, and midpoint-width-derivative data at scaled width \(1\) for the family in Equation (12). All of this uses only a neighborhood of the compact parabolic arc, so the local profile domain is sufficient.

The derivative of \(g\) with respect to \(\rho\) at zero is \(k(h)s\). Equation (10) consequently gives first derivatives \(k(h)/3\) and \(2k(h)/3\) for the scaled midpoint and its width derivative. Their values at \(\rho=0\) are zero. The scaled peak is \(1/2+O(\rho)\). Taylor’s formula with a smooth remainder now gives Equation (11) on setting \(\rho=r>0\) and undoing the scaling. It also proves the stated smoothness of the remainder functions, including any smooth profile parameters. Finally \(t_h=1+O(r^2)\) at fixed \(r\), so the inverse derivative \(\partial h/\partial t\) tends to \(1\). ◻

Proof. At fixed peak, the endpoint derivatives in Equation (5) show that the lower endpoint decreases and the upper endpoint increases as \(h\downarrow0\). They remain bounded. To see this explicitly, put \(B=\sup_{0\le u\le t}|\phi(u)|\). If an upper endpoint lies beyond \(B\), integrate \(u_y\le B-y\) from \(B\) to that endpoint, using \(u(B)\le t\). This gives \(y_>\le B+\sqrt{2t}\). The inequality \(u_y\ge-B-y\) similarly gives \(y_<\ge-B-\sqrt{2t}\). The remaining bounds follow from \(y_<<\phi(h)<y_>\) and \(|\phi(h)|\le B\). The finite limits thus exist, and the signs follow by continuity of \(\phi\) at zero.

For two peaks \(t_2>t_1\), the positive-height endpoint order passes to the weak limiting order \[a(t_2)\le a(t_1),\qquad b(t_2)\ge b(t_1).\] Consequently \(\mathcal T_\phi\) is upward closed. We next show both its openness and the local smoothness asserted in the lemma.

Fix \(t_0\in\mathcal T_\phi\). At a sufficiently small fixed positive level \(\delta<t_0\), the two endpoint branches can be continued downwards by \[ \frac{dy}{du}=\frac1{\phi(u)-y}=:g(u,y). \tag{15}\] Choose compact neighborhoods of the limiting endpoints \(a(t_0)\) and \(b(t_0)\) and make \(\delta\) smaller if necessary. On each resulting branch rectangle, \(|\phi(u)-y|\) is bounded away from zero for \(0\le u\le\delta\). The function \(g\) is continuous in \(u\), smooth in \(y\), and has bounded derivatives of every fixed order with respect to \(y\).

At \(u=\delta\), the branch initial values are smooth functions of the peak \(t\) near \(t_0\). The integral equation for Equation (15) and its parameter-variation equations show that the continued branches and all their \(t\) derivatives are continuous down to \(u=0\), locally uniformly in \(t\). For the first derivative the formula is explicit: \[y_t(u,t)=y_t(\delta,t) \exp\left(\int_\delta^u g_y(v,y(v,t))\,dv\right).\] The exponential factor is positive and bounded away from zero. Thus the two nonzero variation signs persist uniformly near \(t_0\). Higher derivatives satisfy linear variation equations with continuous bounded coefficients and forcing built from lower derivatives, which proves the same assertion inductively. In particular, the strict endpoint signs persist for nearby peaks. This proves openness. An open upward-closed subset of \((0,\infty)\) is an open ray, or is empty.

On that ray, \(r_t=(b_t-a_t)/2>0\). Passing to the limit in \(r^2\ge2(t-h)\) gives \(r(0,t)^2\ge2t\); hence the width tends to infinity as the peak does. Its image is therefore one open ray, and its inverse is smooth.

It remains to justify convergence at fixed width, rather than only at fixed peak. Near \(t_0\), the preceding bounds give \(r_t(h,t)\ge c_0>0\) uniformly for all sufficiently small \(h\ge0\). Uniform strict monotonicity brackets the inverse peak \(t=T(h,r)\) in a common compact interval for nearby widths. This inverse is continuous in \((h,r)\), smooth in \(r\), and \[T_r=\frac1{r_t(h,T(h,r))}.\] Higher \(r\)-derivatives are rational expressions in the continuous peak derivatives, with powers of \(r_t\) in the denominator. They consequently converge locally uniformly as \(h\downarrow0\). Composing the endpoint data with \(T(h,r)\), and subtracting \(\phi(h)\) from their midpoint, proves the claimed convergence of all \(r\)-derivatives. No derivative of \(\phi\) at zero is needed. Finally the strict endpoint signs give \(|M|<r\), and the retained opposite peak derivatives give \(|M_r|<1\), exactly as in Lemma 2. ◻

Proof. On either open half-plane, \(y\) is strictly monotone in physical time. With \(u=x^2/2\), division of the two time derivatives gives \[\frac{du}{dy}=F(x)-y=\phi_\pm(u)-y.\] Thus any excursion in one half-plane is an arc of Equation (2).

Every nonconstant periodic solution has both signs of \(x\), since integrating \(\dot y=-x\) over a period gives \(\int x\,dt=0\). The only equilibrium is \((0,0)\), and uniqueness prevents a nonconstant orbit from containing it. At any axis intersection, \(\dot x=y\ne0\); the intersection is transverse. There are finitely many in a period. Otherwise an accumulation time would give an axis point at which either the crossing derivative is nonzero, contradicting accumulation, or the solution reaches the equilibrium.

A positive axis intersection enters \(x>0\), and a negative one enters \(x<0\). Between consecutive intersections the transformed height is positive, vanishes at the endpoints, and has a unique positive peak by Lemma 2. Both endpoint signs are strict, so the arc belongs to the transverse domain of Lemma 5.

For either profile, increasing the positive endpoint strictly decreases the negative endpoint, by peak order. The physical right-half-plane map from positive to negative endpoints is therefore strictly decreasing, and so is the physical left-half-plane map from negative to positive endpoints. Their composition on successive positive intersections is strictly increasing. An increasing map cannot have a finite periodic sequence of distinct points: if one iterate is larger, or smaller, than its predecessor, strict order persists under further iteration. Hence a closed orbit has one distinct positive intersection and one distinct negative intersection, and consists of one arc of each sign.

The two arcs of such an orbit have the same endpoints, so their widths and midpoints agree. This gives a zero of Equation (17). Conversely, a zero at width \(r\) gives the same endpoints \[a=M_\pm(0,r)-r<0,\qquad b=M_\pm(0,r)+r>0\] for the two arcs. Recover \(x\) as \(+\sqrt{2u_+(y)}\) on the right and \(-\sqrt{2u_-(y)}\) on the left, and introduce physical time by \(\dot y=-x\). On each open arc, \(\dot x=-u_y=y-F(x)\), as required. The one-sided endpoint derivatives are \[(u_\pm)_y(a)=-a>0,\qquad (u_\pm)_y(b)=-b<0.\] Thus \(u_\pm\) has a simple zero at each endpoint and \(dy/\sqrt{2u_\pm(y)}\) is integrable there. The physical travel times are finite. The derivatives of the reconstructed solution extend to the axis with values \((\dot x,\dot y)=(y,0)\). The two arcs consequently join as a solution of the original smooth vector field; uniqueness gives a closed orbit.

Different zeros yield different positive intersections, because \[\frac{d}{dr}\bigl(r+M_+(0,r)\bigr)=1+M_{+,r}(0,r)>0.\] They therefore yield different geometric orbits. This proves the bijection.

Finally fix a matched orbit, its positive intersection \((0,s_0)\), and its first positive return time \(T>0\). On a short positive-axis section \(S\) around \(s_0\), the first return ordinate \(P(s)\) and time \(\tau(s)\) are smooth and satisfy \(\tau(s)\to T\) as \(s\to s_0\). To see that these are the first returns, take disjoint small time windows about the reference axis crossings. On the intervening compact time segments, \(|x|\) is bounded away from zero. Smooth flow dependence and transversality give exactly the nearby crossings in those windows and no others. The function \(s=r+M_+(0,r)\) is a local smooth coordinate with positive derivative. In this coordinate, \[\Delta(r)=0\quad\Longleftrightarrow\quad P(s)=s\] near the reference root.

If distinct roots tend to the reference root, the corresponding section points tend to \(s_0\), their return times tend to \(T\), and smooth flow dependence on a common compact time interval makes their entire orbit images tend to the reference image. They are distinct by the preceding monotonicity. Hence a nonisolated root cannot give an isolated periodic orbit.

For the converse, suppose the root is isolated and shrink \(S\) so that \(s_0\) is its only return fixed point. If \(\Phi_\tau\) denotes the local flow, the map \((s,\tau)\mapsto\Phi_\tau(0,s)\) is a local diffeomorphism near each point \((s_0,\tau)\), \(0\le\tau\le T\), since the initial section is transverse. Finitely many such flow neighborhoods cover the compact reference orbit and give an open neighborhood \(U\) whose every point lies on a trajectory meeting \(S\). Any periodic orbit meeting \(U\) therefore meets \(S\). Its unique positive-axis intersection must be a fixed point of the local return map, and hence equals \(s_0\). Uniqueness then makes it the reference orbit. This proves isolation, without an assumption on hyperbolicity or a global bound on periods. ◻

Proof. The entire-profile construction in Lemma 2 gives an arc at every lower base height. Its width tends to zero at the peak and tends to infinity as the base tends to minus infinity, since \(s^2\ge 2(t-h)\). Its strictly negative base derivative gives the smooth inversion from base height to any \(s>0\), including smooth dependence on the quadratic coefficients. Thus no finite-width existence claim is being inferred from the singular equations at \(s=0\). Equation (7) gives Equation (22) and, by integration of the slope equation, the first bound in Equation (23).

For \(\kappa\ne0\), consider instead the single profile \(\kappa u^2/2\). Let \(\bar h(t,s)\) be its base height when the peak height is \(t\) and the width is \(s\). Its peak slope is \(z=\kappa t\), and its base slope is \(\kappa\bar h(t,s)\). Therefore \[\partial_z j_\kappa(s,z)=\partial_t\bar h(t,s)>0\] by Lemma 2. This calculation is valid for either sign of \(\kappa\). For \(\kappa=0\), one has \(j_0(s,z)=z\).

The smooth scaling in Lemma 4 gives, locally uniformly in the finite parameters, \[ \Theta_\kappa(s,z)=\frac{z}{3}s+O(s^2),\qquad \partial_z\Theta_\kappa(s,z)=\frac{s}{3}+O(s^2). \tag{25}\] Differentiating the angle equation at positive width gives \[\partial_s(\partial_z\Theta_\kappa) =\partial_z j_\kappa -\frac{2\cosh(2\Theta_\kappa)}s\partial_z\Theta_\kappa.\] The initial sign supplied by Equation (25) and the positive forcing prove \(\partial_z\Theta_\kappa>0\) by the scalar integrating-factor formula.

To verify the range, fix \(r>0\). For \(0<s\le r\), \[j_\kappa(s,z)\ge z-\frac{|\kappa|r^2}{2}.\] For sufficiently large positive \(z\), the angle is positive near zero and cannot cross zero downwards, since its derivative there would be positive. Given \(N>0\), on \([r/2,r]\) and whenever \(0\le\Theta_\kappa\le N\), \[\partial_s\Theta_\kappa \ge z-\frac{|\kappa|r^2}{2}-\frac{2\sinh(2N)}r.\] For large enough \(z\), this lower bound forces a crossing of \(N\) within that interval if the angle has not already exceeded \(N\); it also prohibits a downward crossing of \(N\). Thus \(\Theta_\kappa(r,z)\to+\infty\) as \(z\to+\infty\). Applying the same argument to \((-j,-\Theta,-\kappa)\) proves the negative limit as \(z\to-\infty\). Strict monotonicity and smoothness give the asserted bijection.

Integrating the last equation in Equation (22) gives \[\mathcal L_\kappa(s,z) =\int_0^s\frac{\sinh(2\Theta_\kappa(q,z))}{q}\,dq.\] Equation (25), including its locally uniform parameter derivative, bounds the differentiated integrand near zero. Differentiation under this integral is therefore valid and yields Equation (24). Its integrand is strictly positive at every positive width. ◻

Proof. Along every reference characteristic, the definition of the field and Equation (22) give \[(J_\kappa)_s+(J_\kappa)_\theta \left(J_\kappa-\frac{\sinh(2\theta)}s\right) =-s\kappa\operatorname{sech}^2\theta.\] Subtracting this identity from the evolution of the actual slope in Equation (7) proves Equation (28). Differentiating \(\theta(s)=\Theta_\kappa(s,z(s))\) and subtracting the two angle equations gives Equation (29).

The small-width expansions give \(\theta(s)/s\to k(t)/3\). For each \(\varepsilon>0\), the reference angles with labels \(k(t)-\varepsilon\) and \(k(t)+\varepsilon\) bracket this actual angle at all sufficiently small positive widths. The strict monotonicity in Lemma 8 therefore implies \(z(s)\to k(t)\). The bound \(\lvert j_\kappa(s,z(s))-z(s)\rvert\le |\kappa|s^2/2\) then proves \(E(s)\to0\).

For \(0<\varepsilon<s\), variation of constants gives \[E(s)=E(\varepsilon) \exp\left(-\int_\varepsilon^s a\right) +\int_\varepsilon^s \exp\left(-\int_q^s a\right)f(q)\,dq.\] The first term tends to zero because \(a>0\) and \(E(\varepsilon)\to0\). Also \(f(q)=O(q)\) near the smooth peak, and the exponential is between zero and one for \(0<q\le s\). Dominated convergence proves Equation (30), without any assumption that \(a\) is integrable at zero.

If the curvature difference is nonnegative, then \(f\le0\), so \(E\le0\) and \(z'\le0\). This proves the weak angle inequality by Lemma 8. If the difference is positive somewhere in \((0,r)\), continuity gives a subinterval on which \(f<0\). Equation (30) makes \(E\) strictly negative from the end of that subinterval through \(r\). Thus \(z(s)>z(r)\) for every \(s<r\), giving strict angle inequalities. The integral formula for \(L\) in Equation (13) then gives the stated strict inequality at the final width. Negating the inequalities proves the other case. ◻

Proof. Fix \(r,M,\kappa\). For \(\delta>0\), let \(\theta_\delta(s)\) be the characteristic of the midpoint-fitted model of curvature \(\kappa+\delta\); let \(\theta_0\) be the one of curvature \(\kappa\). Their final angles are equal. Lemma 9 gives \[ \theta_\delta(s)>\theta_0(s)\qquad(0<s<r). \tag{34}\] It follows already that the conditional \(\widehat L\) is strictly increasing. We prove a quantitative first-order bound to obtain the asserted strict derivative.

Use the field \(J_\kappa\) to compare the curvature-\(\kappa+\delta\) characteristic, and write its error as \(E_\delta\). Then \[E_\delta'+a_\delta E_\delta =-\delta s\operatorname{sech}^2\theta_\delta, \qquad a_\delta(s)=(J_\kappa)_\theta(s,\theta_\delta(s))>0, \qquad E_\delta\le0.\] Choose \(0<a<b<c<r\). The already established smooth midpoint fit and smooth finite-parameter arc dependence imply that, for all sufficiently small \(\delta\ge0\), there are constants \(A,q_0>0\), independent of \(\delta\), such that on \([a,r]\), \[a_\delta(s)\le A,\qquad s\operatorname{sech}^2\theta_\delta(s)\ge q_0.\] Variation of constants starting at \(a\), using \(E_\delta(a)\le0\), consequently gives \[ E_\delta(s)\le-c_0\delta\quad(b\le s\le r), \qquad c_0=q_0(b-a)e^{-Ar}>0. \tag{35}\] This estimate needs no bound uniform in \(\delta\) near \(s=0\). The sign at \(a\) follows from the from-zero comparison separately for every \(\delta>0\).

Let \(z_\delta(s)\) be the reference label defined by \(\theta_\delta(s)=\Theta_\kappa(s,z_\delta(s))\), and let \(z_0\) be the constant label of the curvature-\(\kappa\) arc. Final angle matching gives \(z_\delta(r)=z_0\). Smoothness of the inverse reference-label map puts all these labels, for \(b\le s\le r\) and small \(\delta\ge0\), in a common compact set. Enlarge the set to include the intervals between \(z_0\) and \(z_\delta(s)\). Lemma 8 then gives constants \(m_0,B_0>0\) with \[m_0\le\partial_z\Theta_\kappa(s,z)\le B_0\] throughout that set. Equations (29) and (35) yield \[z_\delta'(s)\le-\frac{c_0\delta}{B_0}\quad(b\le s\le r).\] Integrating backwards from the common final label gives, for \(b\le s\le c\), \[z_\delta(s)-z_0\ge\frac{c_0\delta(r-c)}{B_0},\qquad \theta_\delta(s)-\theta_0(s) \ge\frac{m_0c_0\delta(r-c)}{B_0}.\] Since the derivative of \(\sinh(2\theta)\) is at least \(2\), the integral formulas for the final \(L\)-values and Equation (34) now imply \[ \widehat L(r,M,\kappa+\delta)-\widehat L(r,M,\kappa) \ge \frac{2m_0c_0(r-c)}{B_0}\log\!\frac cb\;\delta. \tag{36}\] Thus ordinary smoothness of the conditional function at the final positive width proves \(\partial_\kappa\widehat L>0\). No differentiation of the improper \(L\)-integral is used here.

Finally, implicit differentiation of Equation (31) gives \(\partial_\kappa\lambda_*=-Q/P\), and hence \[\partial_\kappa\widehat v =S-R\frac QP=\frac GP =(1-\widehat v^2)\partial_\kappa\widehat L>0.\] Since every model arc is its own midpoint fit, this proves \(G>0\) everywhere. The argument used \(P>0\), but did not assume the sign or nonvanishing of \(G\). ◻

Proof. Suppose the first limit fails. Strict monotonicity then makes \(\widehat L(r,M,\kappa)\) bounded above as \(\kappa\to+\infty\). Choose positive curvatures \(\kappa_n\uparrow+\infty\), and denote the matched-final-angle characteristics by \(\theta_n(s)\), \(j_n(s)\), and their final \(L\)-values by \(L_n\). Lemma 9 gives pointwise increase of \(\theta_n(s)\) for \(0<s<r\). Put \[f_n(s)=\frac{\sinh(2\theta_n(s))}{s}.\] Each \(f_n\) is integrable at zero by its smooth-peak expansion, and \[f_n-f_1\ge0,\qquad \int_0^r(f_n-f_1)\,ds=L_n-L_1\le C\] for some finite \(C\). Monotone convergence shows that the limiting \(f_n\), and therefore the limiting \(\theta_n\), are finite almost everywhere in \((0,r)\).

Choose four such widths \(0<a<b<c<d<r\). The angle equation gives \[\begin{align*} A_n:=\int_a^b j_n(s)\,ds &=\theta_n(b)-\theta_n(a)+\int_a^b f_n(s)\,ds,\\ B_n:=\int_c^d j_n(s)\,ds &=\theta_n(d)-\theta_n(c)+\int_c^d f_n(s)\,ds. \end{align*}\] Both sequences are bounded above and below: the four endpoint angle sequences are bounded, and for every subinterval \(I\), \[0\le\int_I(f_n-f_1)\,ds\le C.\] Choose constants \(A_*,B_*\) with \(A_n\le A_*\) and \(B_n\ge B_*\). Because \(\kappa_n>0\), the slope \(j_n\) is decreasing, so \[ j_n(b)\le\frac{A_n}{b-a}\le\frac{A_*}{b-a},\qquad j_n(c)\ge\frac{B_n}{d-c}\ge\frac{B_*}{d-c}. \tag{38}\] Thus \(j_n(b)-j_n(c)\) has a uniform upper bound.

On the other hand, its exact evolution is \[ j_n(b)-j_n(c) =\kappa_n\int_b^c s\operatorname{sech}^2\theta_n(s)\,ds. \tag{39}\] The integrand is bounded by \(s\) and converges almost everywhere to a strictly positive function, because the limiting angle is finite almost everywhere. Dominated convergence gives a strictly positive limiting integral in Equation (39). Its right-hand side therefore tends to infinity, contradicting Equation (38). This proof uses no uniform angle bound on the middle interval.

For the second limit, Equation (21) and uniqueness of midpoint fitting give \[\lambda_*(r,-M,-\kappa)=-\lambda_*(r,M,\kappa),\qquad \widehat L(r,M,\kappa)=-\widehat L(r,-M,-\kappa).\] The positive-curvature result holds for every fixed midpoint, including \(-M\). It therefore proves the negative-curvature limit at the original fixed midpoint \(M\), as required. ◻

Proof. At base height \(h\), translating the quadratic and subtracting its base value changes its slope to \(\lambda+\kappa h\) and leaves its curvature unchanged. Its midpoint data are therefore \(H(\lambda+\kappa h,\kappa,r)\). At \(h=0\), the first transport identity in Equation (6) becomes \[\kappa P-\frac r{r^2-H^2}H_r =\frac H{r^2-H^2}-\lambda,\] which is Equation (40). For completeness, put \(D_0=r^2-H^2\) and \(v=H_r\). Differentiating at fixed \(\lambda,\kappa\) gives \[\begin{align*} \kappa R &=\frac{rv_r+2v}{D_0} -\frac{2(r-Hv)(rv+H)}{D_0^2}\\ &=\frac{r(1-v^2)}{D_0} \left(\frac{v_r}{1-v^2}-\frac{2H}{D_0}\right), \end{align*}\] proving Equation (41).

Let \(t=t(\lambda,\kappa,r)>0\) be the peak of this base-zero model arc. Lemma 2 gives \(t_r>0\) at fixed model parameters. Its peak slope is \(z=\lambda+\kappa t\), so \[L_{\mathrm{mod}}(\lambda,\kappa,r) =\mathcal L_\kappa(r,\lambda+\kappa t(\lambda,\kappa,r)).\] The chain rule and Equation (22) yield \[ (L_{\mathrm{mod}})_r =\frac{2H}{r^2-H^2} +\kappa t_r\partial_z\mathcal L_\kappa(r,z). \tag{42}\] For \(\kappa\ne0\), substitution into Equation (41) and cancellation of \(\kappa\) give \[R=\frac{r(1-H_r^2)}{r^2-H^2} t_r\partial_z\mathcal L_\kappa(r,z)>0\] by Lemma 8. This cancellation is valid for either sign of \(\kappa\). For \(\kappa=0\), the peak slope is \(z=\lambda\), so direct parameter differentiation gives \[R=\partial_\lambda H_r =(1-H_r^2)\partial_z\mathcal L_0(r,\lambda)>0.\] Finally, \(S=(G+QR)/P>0\) by Equation (20) and Lemma 10. ◻

Proof. Fix \((M,v,r)\) in the stated domain. For each curvature there is exactly one slope fit by Equation (31). Lemmas 10 and 11 show that its conditional \(\widehat L\) is a strictly increasing map of \(\mathbb R\) onto \(\mathbb R\). There is therefore exactly one curvature for which \(\widehat L=\operatorname{arctanh}v\), and then exactly one slope. The sign assertions have been proved in Equation (20) and Lemmas 10 and 12.

The Jacobian of Equation (43) has last row \((0,0,1)\), and its determinant is \(G>0\). The inverse-function theorem gives a smooth local inverse at every point. These inverses agree wherever their domains overlap, by the global uniqueness just proved, so the inverse is jointly smooth on the entire target domain. Its continuity gives the final convergence statement. ◻

Proof. Differentiating the first identity in Equation (44) at fixed \(h\) and using the second one gives \(P\lambda_r+Q\kappa_r=0\).

For a quadratic profile, increasing its base height changes the slope at rate \(\kappa\), leaves its curvature fixed, and changes the width at rate \(-\alpha\). Applying Equation (6) to that profile, at the fitted data, therefore gives \[\begin{aligned} \kappa P-\alpha H_r &=\frac{M}{r^2-M^2}-\lambda,\\ \kappa R-\alpha H_{rr} &=-\frac{2Mr(1-v^2)}{(r^2-M^2)^2}. \end{aligned}\] For the actual profile, the chain rule gives \[\begin{aligned} DM&=P D\lambda+Q D\kappa-\alpha H_r,\\ Dv&=R D\lambda+S D\kappa-\alpha H_{rr}. \end{aligned}\] The expression for \(Dv\) in Equation (6) depends only on \(M,v,r\), so it is identical for the two matched sets of data. Subtracting the preceding identities consequently yields \[\begin{aligned} P(D\lambda-\kappa)+Q D\kappa&=\lambda-k(h),\\ R(D\lambda-\kappa)+S D\kappa&=0. \end{aligned}\] Their determinant is \(G=PS-QR>0\). Solving this system proves Equation (45). ◻

Proof. Put \(B=1/n^2-1/m^2\). Differentiation of Equation (60), using \(n_m=\ell\), gives \[ Z_m=\frac{w\ell(m+n)}{(1+\ell)^2},\qquad X_m=\frac{\ell B}{1+\ell}+\frac{Z_m}{mn},\qquad B_m=\frac2{m^3}-\frac{2\ell}{n^3}. \tag{67}\] Differentiate the middle identity and insert the first one into the determinant. After division by \(\ell^2(m+n)/(1+\ell)^3\), the result is \[ B\mathcal S+ \left(\frac{B(1+\ell)}{m+n}-B_m\right)w +\frac{(m+n)^2}{2m^2n^2}w^2. \tag{68}\] For example, in this calculation \[\frac{d}{dm}\log\frac{\ell(m+n)}{(1+\ell)^2} =\frac{1-\ell}{1+\ell}w+\frac{1+\ell}{m+n}, \qquad \left(\frac1{mn}\right)_m=-\frac{n+m\ell}{m^2n^2}.\] These two derivatives, together with \(w_m=\mathcal S+w^2/2\), give the three coefficients in Equation (68). Now \[B=\frac{(m-n)(m+n)}{m^2n^2},\] and substitution of \(B_m\) shows that the coefficient of \(w\) in Equation (68) is \[\frac{m+n}{m^2n^2} \left[1-\frac{2n}{m} +\ell\left(\frac{2m}{n}-1\right)\right].\] This proves its first factorization in Equation (64).

On the other hand, differentiating \(X=\kappa P+d\) and \(Z=dP+3\kappa Q\) with respect to \(r\) gives \[X_rZ_{rr}-Z_rX_{rr}=3\kappa^2(RS_r-SR_r).\] The chain rule gives \[X_rZ_{rr}-Z_rX_{rr} =m_r^3\bigl(X_mZ_{mm}-Z_mX_{mm}\bigr).\] This proves the other equality in Equation (64). It also gives \[ X_m=\frac{\kappa R}{m_r}>0. \tag{69}\] Thus, at a zero, \(\sigma=X_{mm}/X_m\) is legitimate, even if \(Z_m=0\) or \(w=0\). The determinant identity then implies \(Z_{mm}=\sigma Z_m\). Using \(X_{rr}=\sigma_0X_r\) at that point gives \[\sigma m_r^2=\sigma_0m_r-m_{rr}.\] The chain rule for \(\gamma\) proves Equation (65).

It remains to compute \(\gamma_{mm}-\sigma\gamma_m\). For a function \(f\) smooth near \(m\) and \(-n\), write \[T_f=\frac{\ell f(m)+f(-n)}{1+\ell},\qquad a_0=\frac{\ell}{1+\ell},\qquad \beta=\frac{w}{1+\ell},\qquad \rho=(\log a_0)_m-\sigma.\] A first differentiation gives \[(T_f)_m=a_0\{f'(m)-f'(-n)+\beta[f(m)-f(-n)]\}.\] A second differentiation therefore gives \[\begin{align*} \frac{(T_f)_{mm}-\sigma(T_f)_m}{a_0} ={}&f''(m)+\ell f''(-n) +\beta\{f'(m)+\ell f'(-n)\}\\ &+\rho\{f'(m)-f'(-n)\} +(\beta_m+\rho\beta)\{f(m)-f(-n)\}. \tag{70}\end{align*}\] At the zero under consideration, this expression vanishes for \(f(x)=x\) and \(f(x)=1/x\), since \(T_x=Z\) and \(T_{1/x}=X\). Consequently \[\begin{align*} \beta_m+\rho\beta &=-\frac{\beta(1+\ell)}{m+n},\\ \rho B &=\frac{2\ell}{n^3}-\frac2{m^3} +\beta\left(\frac1{m^2}+\frac{\ell}{n^2} +\frac{1+\ell}{mn}\right). \tag{71}\end{align*}\] Here \(B\ne0\) because \(m\ne n\); no division by \(\beta\) is made.

For \(f(x)=1/x^2\), Equation (70) becomes \[\begin{align*} \frac{\gamma_{mm}-\sigma\gamma_m}{a_0} ={}&\frac6{m^4}+\frac{6\ell}{n^4} +\beta\left(-\frac2{m^3}+\frac{2\ell}{n^3}\right) -2\rho\left(\frac1{m^3}+\frac1{n^3}\right) +\frac{\beta(1+\ell)B}{m+n}. \tag{72}\end{align*}\] Eliminate \(\rho\) using Equation (71). The resulting constant coefficient and coefficient of \(\beta\) are, respectively, \[\begin{align*} A_0={}&\frac6{m^4}+\frac{6\ell}{n^4} -\frac{2(1/m^3+1/n^3)(2\ell/n^3-2/m^3)}{B} =\frac{2(m+n)N}{m^4n^4(m-n)},\\ A_\beta={}&-\frac2{m^3}+\frac{2\ell}{n^3} -\frac{2(1/m^3+1/n^3) (1/m^2+\ell/n^2+(1+\ell)/(mn))}{B} +\frac{(1+\ell)B}{m+n}\\ ={}&-\frac{(1+\ell)(m+n)^2}{m^2n^2(m-n)}, \tag{73}\\ N={}&\ell m^4-2\ell m^3n+2mn^3-n^4. \end{align*}\] For completeness, the two numerator reductions in Equation (73) are \[\begin{align*} &6(n^4+\ell m^4)(m^2-n^2) -4(m^3+n^3)(\ell m^3-n^3)=2(m+n)^2N,\\ &(m+n)\bigl[2(\ell m^3-n^3)(m-n) -2(m^3+n^3)(\ell m+n) +(1+\ell)mn(m-n)^2\bigr]\\ &\hspace{35mm}=-(1+\ell)mn(m+n)^3. \end{align*}\] They correspond to the common denominators \(m^4n^4(m^2-n^2)\) and \(m^3n^3(m^2-n^2)\). Substituting \(\beta=w/(1+\ell)\) into \(A_0+\beta A_\beta\) now gives \[\frac{(m+n)^2}{m^2n^2}\frac{w_*-w}{m-n}.\] Multiplication by \(a_0\) proves Equation (66). ◻

Proof. The arc equation is \(u_y=du+bu^2-y\). On its finite interval \([-m,n]\), define \[ V(y)=\exp\left[-\int_{-m}^{y}(bu(s)+d/2)\,ds\right]>0, \qquad U(y)=u(y)V(y). \tag{77}\] The integral is finite because the actual arc is bounded on a compact interval. In the following system, primes denote derivatives with respect to \(y\). Direct differentiation gives the regular linear system \[ U'=\frac d2U-yV,\qquad V'=-bU-\frac d2V. \tag{78}\] In particular \(U(-m)=U(n)=0\). Eliminating \(V\) where \(y\ne0\) gives \[ U''=\frac{U'}{y} +\left(by+\frac{d^2}{4}-\frac{d}{2y}\right)U. \tag{79}\]

We now justify the use of one projective solution label at both endpoints despite the singularity of Equation (79) at zero. Take \((U_1,V_1)=(U,V)\) from Equation (77), and complete it to a fundamental pair \((U_1,V_1),(U_2,V_2)\) of the regular system in Equation (78). Its coefficient matrix has trace zero, so \[\Omega=U_1V_2-U_2V_1\] is a nonzero constant. At either endpoint, \(U_1=0\) and \(V_1>0\); hence \(U_2=-\Omega/V_1\ne0\). Therefore the same ratio \(f=U_1/U_2\) is finite near both endpoints, and \[ f(-m)=f(n)=0,\qquad f'(y)=\frac{y\Omega}{U_2(y)^2}\ne0 \quad\hbox{at }y=-m,n. \tag{80}\] A nearby arc has coefficients that can be normalized as \((1,-q)\) in this fixed fundamental pair: continuous dependence keeps its first coefficient nonzero near the original arc. Its endpoint condition is \(U_1-qU_2=0\). Thus the nearby endpoint functions satisfy the same local identity \(f(n(m))=f(-m)\). No scalar solution is continued through the singular equation at zero. The scalar determinant \(U_1'U_2-U_1U_2'=y\Omega\) is nonzero in each endpoint neighborhood, so \(U_1,U_2\) form a scalar basis in each such neighborhood.

Writing \(U=\sqrt{|y|}\,\widetilde U\) in Equation (79) gives \[\widetilde U'' =\left(by+\frac{d^2}{4}-\frac{d}{2y} +\frac3{4y^2}\right)\widetilde U.\] For two solutions of \(a''=Q_0a\), \(c''=Q_0c\), their ratio has Schwarzian \(-2Q_0\): locally, \((a/c)''/(a/c)'=-2c'/c\), and differentiating this expression and subtracting half its square gives \(-2c''/c\). The ratio is unchanged by division of both solutions by \(\sqrt{|y|}\). Consequently \[\{f,y\}=-2by-\frac{d^2}{2}+\frac d y-\frac3{2y^2}, \qquad \{f,y\}=\frac{f'''}{f'}-\frac32\left(\frac{f''}{f'}\right)^2.\] Apply the Schwarzian chain rule to the common-label identity \(f(n(m))=f(-m)\). The affine map \(m\mapsto-m\) has zero Schwarzian and squared derivative one, so \[\mathcal S=\{f,y\}\big|_{y=-m} -\ell^2\{f,y\}\big|_{y=n}.\] Expanding these two evaluations proves Equation (74).

For Equation (75), vary the initial endpoint in \(u(-m;m)=0\). Since \(u_y(-m)=m\), the initial variation is \(u_m(-m)=m\). Its scalar variational equation yields \[u_m(y)=m\exp\left[\int_{-m}^{y}(d+2bu(s))\,ds\right] =m\,\frac{V(-m)^2}{V(y)^2}.\] Differentiating \(u(n(m);m)=0\), and using \(u_y(n)=-n\), gives the formula for \(\ell\). Its logarithm can be written \[\log\ell=\log m-\log n+\int_{-m}^{n}(d+2bu(y))\,dy.\] Differentiate in \(m\). The two boundary contributions from \(2bu\) vanish, while the constant term contributes \(d(1+\ell)\). Substituting the expression for \(u_m\) proves the formula for \(w\).

Finally, differentiating the second equation in Equation (78) gives \[V''=(by+d^2/4)V,\qquad V'(-m)=-dV(-m)/2.\] Suppose \(d<0\), \(d^2\ge4bm\), and set \(s=y+m\). For \(v_0(s)=V(-m+s)/V(-m)\) one has \[v_0''=(\nu^2+bs)v_0,\qquad v_0(0)=1,\qquad v_0'(0)=-d/2.\] Let \[f_\nu(s)= \begin{cases} \displaystyle\cosh(\nu s)-\frac d{2\nu}\sinh(\nu s),&\nu>0,\\[1mm] 1-ds/2,&\nu=0. \end{cases}\] Variation of constants gives, for \(0\le s\le m+n\), \[v_0(s)-f_\nu(s) =b\int_0^s K_\nu(s-t)\,t\,v_0(t)\,dt\ge0, \qquad K_\nu(t)= \begin{cases}\sinh(\nu t)/\nu,&\nu>0,\\t,&\nu=0.\end{cases}\] The inequality follows from the already constructed positivity of \(v_0\), and from \(K_\nu\ge0\). Also \(f_\nu>0\) on the positive half-line because \(d<0\). Therefore \[ V(-m)^2\int_{-m}^{n}V(y)^{-2}\,dy \le\int_0^\infty f_\nu(s)^{-2}\,ds =\frac1{\nu-d/2}. \tag{81}\] For \(\nu>0\), the substitution \(q=\tanh(\nu s)\), with \(A=-d/(2\nu)>0\), evaluates the last integral as \[\frac1\nu\int_0^1(1+Aq)^{-2}\,dq =\frac1{\nu(1+A)}.\] For \(\nu=0\), direct integration of \((1-ds/2)^{-2}\) gives \(-2/d\), the same formula. Since \(\nu^2=d^2/4-bm\), \[\frac{2bm}{\nu-d/2}=-d-2\nu.\] Substitution into Equation (75) proves Equation (76). ◻

Proof. We partition the possibilities for \(m,n,\ell\). The quantities \(\theta=\operatorname{arctanh}(H/r)\) and \(L=\operatorname{arctanh}(H_r)\), and the finite-peak reference characteristics used below, are those of Lemma 8.

Proof of Theorem 15. First suppose \(\kappa>0\). Fix any model data point \((\lambda,\kappa,r)\). By the arc construction and Lemma 8, it belongs to a characteristic coming from a finite peak. Along that characteristic, use \(s\) for outward-increasing width. The base slope tends to a finite peak slope as \(s\downarrow0\), so the uniform expansion in Equation (55) makes \(\mathcal W\) strictly positive for all sufficiently small positive \(s\).

On this curve \(\kappa\) is constant and \(d\lambda/ds=-\kappa/\alpha\). Thus the total outward derivative is \[ \frac{d}{ds}\mathcal W(\lambda(s),\kappa,s) =\mathcal W_r-\frac{\kappa}{\alpha}\mathcal W_\lambda =-\frac1\alpha\,\mathcal D\mathcal W. \tag{92}\] At any zero, Equations (53) and (56), and Lemma 18, give \[\mathcal D\mathcal W =-G(\gamma_{rr}-\sigma_0\gamma_r)<0.\] Since \(\alpha>0\), Equation (92) makes the outward derivative strictly positive. A first zero reached from strictly positive values would instead have a nonpositive outward derivative. This contradiction proves \(\mathcal W>0\) at every point with \(\kappa>0\).

Reflection of the endpoint coordinate gives \[H(-\lambda,-\kappa,r)=-H(\lambda,\kappa,r).\] Differentiating this symmetry shows that \(P,Q\), their \(r\)-derivatives, and \(\mathcal W\) are unchanged by simultaneous sign reversal of \(\lambda,\kappa\). Thus \(\mathcal W>0\) also when \(\kappa<0\). Smooth dependence of the model on finite parameters at every \(r>0\) now gives \(\mathcal W\ge0\) at \(\kappa=0\). Only this non-strict conclusion at zero curvature is required. Locally the passage is uniform on compact parameter sets with \(r>0\); \(G>0\) there is already known from Theorem 13.

Finally, apply the pointwise fixed-parameter identity Equation (48). It gives \((R/G)_r\le0\) for every \(\lambda,\kappa,r\), as asserted. The characteristic transport established the sign of a scalar function on the whole model domain; it did not identify its partial \(r\)-derivative with a total derivative along a characteristic. ◻

Proof. For the first statement, the sets where \(q\) is negative, zero, and positive occur in that order. The zero set of \(q\) is an interval, possibly empty or a singleton. Accordingly \(f\) strictly decreases, is constant, and strictly increases on the respective sets. Each strict part has at most one zero. If the constant part has nonzero value it contributes none. If it has positive length and value zero, every point of it and each endpoint lying in the domain are nonisolated zeros, and neither strict part contributes another zero. If it is a singleton at which \(f=0\), that point is the sole minimum zero. These possibilities prove the claim.

For the second statement, suppose \(r_1<r_2\) were two zeros. Positive derivatives give \(a,b\) with \(r_1<a<b<r_2\), \(f(a)>0\), and \(f(b)<0\). The first zero \(c\) in \([a,b]\) exists by continuity. Since \(f>0\) on \([a,c)\), its derivative at \(c\), which exists, is nonpositive. This contradicts the hypothesis. ◻