Two limit cycles for quintic Liénard systems

Proof. Differentiating Equation (2.1) gives

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 (2.1) 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, Sections 2.2–2.3] 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,

so \xi>0 throughout the arc. Differentiating the endpoint equations now gives

In particular r_t>0. After changing from t to r,

The other strict inequality, |M|<r, follows directly from the endpoint inequalities.

At fixed t, differentiation with respect to h gives

Hence

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,

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

Multiplication by K and integration prove Equation (2.8). 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 (2.8) 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

The scaled equation is

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

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 (2.11) 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 (2.11). 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 (2.9) 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 (2.10) 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 (2.4) 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

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

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 (2.14) 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:

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

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 2.1.□

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

Thus any excursion in one half-plane is an arc of Equation (2.1).

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 2.1. Both endpoint signs are strict, so the arc belongs to the transverse domain of Lemma 2.4.

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 (2.16). Conversely, a zero at width r gives the same endpoints

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

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

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,

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 2.1 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 (2.6) gives Equation (3.5) and, by integration of the slope equation, the first bound in Equation (3.6).

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

by Lemma 2.1. This calculation is valid for either sign of \kappa. For \kappa=0, one has j_0(s,z)=z.

The smooth scaling in Lemma 2.3 gives, locally uniformly in the finite parameters,

Differentiating the angle equation at positive width gives

The initial sign supplied by Equation (3.8) 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,

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,

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 (3.5) gives

Equation (3.8), including its locally uniform parameter derivative, bounds the differentiated integrand near zero. Differentiation under this integral is therefore valid and yields Equation (3.7). Its integrand is strictly positive at every positive width.□

Proof. Along every reference characteristic, the definition of the field and Equation (3.5) give

Subtracting this identity from the evolution of the actual slope in Equation (2.6) proves Equation (3.11). Differentiating \theta(s)=\Theta_\kappa(s,z(s)) and subtracting the two angle equations gives Equation (3.12).

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 3.1 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

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 (3.13), 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 3.1. If the difference is positive somewhere in (0,r), continuity gives a subinterval on which f<0. Equation (3.13) 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 (2.12) 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 3.2 gives

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

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],

Variation of constants starting at a, using E_\delta(a)\le0, consequently gives

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 3.1 then gives constants m_0,B_0>0 with

throughout that set. Equations (3.12) and (3.18) yield

Integrating backwards from the common final label gives, for b\le s\le c,

Since the derivative of \sinh(2\theta) is at least 2, the integral formulas for the final L-values and Equation (3.17) now imply

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 (3.14) gives \partial_\kappa\lambda_*=-Q/P, and hence

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 3.2 gives pointwise increase of \theta_n(s) for 0<s<r. Put

Each f_n is integrable at zero by its smooth-peak expansion, and

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

Both sequences are bounded above and below: the four endpoint angle sequences are bounded, and for every subinterval I,

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

Thus j_n(b)-j_n(c) has a uniform upper bound.

On the other hand, its exact evolution is

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 (3.22). Its right-hand side therefore tends to infinity, contradicting Equation (3.21). This proof uses no uniform angle bound on the middle interval.

For the second limit, Equation (3.4) and uniqueness of midpoint fitting give

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 (2.5) becomes

which is Equation (3.23). For completeness, put D_0=r^2-H^2 and v=H_r. Differentiating at fixed \lambda,\kappa gives

proving Equation (3.24).

Let t=t(\lambda,\kappa,r)>0 be the peak of this base-zero model arc. Lemma 2.1 gives t_r>0 at fixed model parameters. Its peak slope is z=\lambda+\kappa t, so

The chain rule and Equation (3.5) yield

For \kappa\ne0, substitution into Equation (3.24) and cancellation of \kappa give

by Lemma 3.1. This cancellation is valid for either sign of \kappa. For \kappa=0, the peak slope is z=\lambda, so direct parameter differentiation gives

Finally, S=(G+QR)/P>0 by Equation (3.3) and Lemma 3.3.□

Proof. Fix (M,v,r) in the stated domain. For each curvature there is exactly one slope fit by Equation (3.14). Lemmas 3.3 and 3.4 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 (3.3) and Lemmas 3.3 and 3.5.

The Jacobian of Equation (3.26) 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 (3.27) 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 (2.5) to that profile, at the fitted data, therefore gives

For the actual profile, the chain rule gives

The expression for Dv in Equation (2.5) depends only on M,v,r, so it is identical for the two matched sets of data. Subtracting the preceding identities consequently yields

Their determinant is G=PS-QR>0. Solving this system proves Equation (3.28).□

Proof. Put B=1/n^2-1/m^2. Differentiation of Equation (4.15), using n_m=\ell, gives

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

For example, in this calculation

These two derivatives, together with w_m=\mathcal S+w^2/2, give the three coefficients in Equation (4.23). Now

and substitution of B_m shows that the coefficient of w in Equation (4.23) is

This proves its first factorization in Equation (4.19).

On the other hand, differentiating X=\kappa P+d and Z=dP+3\kappa Q with respect to r gives

The chain rule gives

This proves the other equality in Equation (4.19). It also gives

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

The chain rule for \gamma proves Equation (4.20).

It remains to compute \gamma_{mm}-\sigma\gamma_m. For a function f smooth near m and -n, write

A first differentiation gives

A second differentiation therefore gives

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

Here B\ne0 because m\ne n; no division by \beta is made.

For f(x)=1/x^2, Equation (4.25) becomes

Eliminate \rho using Equation (4.26). The resulting constant coefficient and coefficient of \beta are, respectively,

For completeness, the two numerator reductions in Equation (4.28) are

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

Multiplication by a_0 proves Equation (4.21).□

Proof. The arc equation is u_y=du+bu^2-y. On its finite interval [-m,n], define

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

In particular U(-m)=U(n)=0. Eliminating V where y\ne0 gives

We now justify the use of one projective solution label at both endpoints despite the singularity of Equation (4.34) at zero. Take (U_1,V_1)=(U,V) from Equation (4.32), and complete it to a fundamental pair (U_1,V_1),(U_2,V_2) of the regular system in Equation (4.33). Its coefficient matrix has trace zero, so

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

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 (4.34) gives

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

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

Expanding these two evaluations proves Equation (4.29).

For Equation (4.30), 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

Differentiating u(n(m);m)=0, and using u_y(n)=-n, gives the formula for \ell. Its logarithm can be written

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 (4.33) gives

Suppose d<0, d^2\ge4bm, and set s=y+m. For v_0(s)=V(-m+s)/V(-m) one has

Let

Variation of constants gives, for 0\le s\le m+n,

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

For \nu>0, the substitution q=\tanh(\nu s), with A=-d/(2\nu)>0, evaluates the last integral as

For \nu=0, direct integration of (1-ds/2)^{-2} gives -2/d, the same formula. Since \nu^2=d^2/4-bm,

Substitution into Equation (4.30) proves Equation (4.31).□

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 3.1.

Case 1: m\le n. The final angle is nonnegative. Along the characteristic leading to this data point, the driving slope satisfies

Its initial value z must be positive. Indeed, if z\le0, then j(s)<0 for every s>0, and \theta'+a(s)\theta=j(s), where a(s)=\sinh(2\theta)/(s\theta)>0, with value 2/s at \theta=0, forces \theta(s)<0. This last assertion follows by integrating from a positive \varepsilon and letting \varepsilon\downarrow0: \theta(\varepsilon)\to0, and the forward homogeneous factor is at most one. It would contradict the nonnegative final angle.

Since z>0, \theta(s)=zs/3+O(s^2)>0 initially. If it first reached zero at some earlier width s_1<r, then j(s_1)=\theta'(s_1)\le0. The strict decrease of j would give j(s)<0 thereafter, forcing \theta(s)<0 for s>s_1. Again this contradicts the final angle. Thus \theta(s)>0 for all 0<s<r, and

We also need the sign of w=(\log\ell)_m, where the derivative is taken along the fixed-profile family at base zero. Let t=t(r) be the peak height of that profile. Lemma 2.1 gives t_r>0, and its peak slope is z=d+\kappa t. The final model value L=\operatorname{arctanh}H_r equals \mathcal L_\kappa(r,z(r)), and Lemma 3.1 gives \partial_z\mathcal L_\kappa>0. Consequently

If m=n, Equation (4.17) now reduces to

Hence a zero of \mathcal W cannot have m=n.

If m<n, put z_1=m/n\in(0,1). The sign of w_* is the sign of

Since w>0, the quotient (w_*-w)/(m-n) is positive. Equations (4.21) and (4.20) prove the required sign.

Case 2: m>n and \ell\ge1. Here the final angle is negative, which forces d<0. In fact, if its final driving slope d were nonnegative, then the strictly decreasing driving slope would be positive at all earlier widths, and the same scalar angle equation would give a positive final angle.

Each group in Equation (4.29) is now nonnegative, and several are strictly positive. In particular

We justify w\ge0 on this region without assuming that m ranges to infinity.

For the fixed quadratic profile, Lemma 2.1 gives arcs at all widths r\in(0,\infty). Since m_r>0 and 0<m(r)<2r, the m-domain is an interval (0,M_*), with M_* finite or infinite. At its left end,

because d<0. Fix a point in the open set \{m-n>0,\ell>1\}, and let (a,c) be its connected component. The expansion just given implies a>0; also a<M_*, since the chosen point is to its right. Thus a is an interior point of the smooth parameter domain. On the component, (m-n)_m=1-\ell<0. Moving backwards to a therefore increases m-n, so its value at a is strictly positive. The component cannot enter through m=n. It must have \ell(a)=1, and the right derivative there satisfies \ell_m(a)\ge0, or w(a)\ge0. Equations (4.14) and (4.38) give w_m>0 throughout the component. Hence w>0 inside it.

At a point with m>n,\ell=1,w<0, points immediately to the left would have \ell>1,w<0, contrary to the result just proved. A point with m>n,\ell=1,w=0 is also impossible: there w_m=\mathcal S>0 and \ell_{mm}=\ell(w_m+w^2)=\mathcal S>0. Immediately to the left one would again have \ell>1,w<0. Thus w\ge0 at every point required in this case. The component argument applies to each chosen component separately; it makes no assumption about their number.

Writing z_1=m/n>1, the coefficient of w in Equation (4.17) satisfies

Together with (m-n)\mathcal S>0 and w\ge0, this gives D_*>0. Hence no zero of \mathcal W occurs in this case.

Case 3: m>n and 0<\ell<1. Again d<0. It suffices to prove that

Indeed, at a zero this will give w<w_*, and the denominator m-n in Equation (4.21) is positive.

We may set n=1,m=z_1>1 at the point. To specify this scaling, fix the original endpoint value N=n>0 as a constant and put y=N\widetilde y, u=N^2\widetilde u. The new model parameters are \widetilde d=Nd, \widetilde b=N^3b, while

Thus the equations and the signs needed for Equation (4.39) are preserved. We suppress the tildes and use n=1,m=z_1 from now on.

In these units,

For z_1\ge2 positivity is immediate. For 1<z_1<2, the coefficient of \ell in its numerator is negative; replacing \ell<1 by one decreases that numerator to (z_1-1)^3(z_1+1)>0.

For fixed \mathcal S, write

The derivative of this expression in w, evaluated at w_* and multiplied by z_1^2, is

This polynomial is affine in \ell. With \xi=z_1-1>0, its endpoint values are

Both are strictly positive. Therefore J>0 for 0<\ell<1. Since the derivative in w increases with w, D_* is strictly increasing for w\ge w_*.

Set

Substitution of Equation (4.40) gives the two polynomial identities

These factorizations can be checked by collecting powers of \ell. For Equation (4.43), multiplying its left side by 2z_1^4(z_1+1) gives

The bracket is AT, because the cross coefficient in the product is

For Equation (4.44), the numerator over z_1^2(z_1+1) is (z_1-1)[z_1-2-\ell z_1^2(2z_1-1)]=-(z_1-1)T. The sign of A requires no further restriction on the parameters. It too is affine in \ell, and

so A>0.

Assume w\ge w_*. First suppose T\ge0. If d^2\ge4bz_1, Equation (4.31) would imply

This contradicts Equation (4.44). Consequently d^2<4bz_1. In Equation (4.29), the difference from the baseline is

Each term on the right is positive in this subcase. Equation (4.43) is nonnegative because A>0,T\ge0. Since z_1-1>0, the strict improvement in Equation (4.45), followed by monotonicity in w, gives D_*>0.

It remains to consider T<0. Then

In particular z_1>2, and the same inequality implies z_1-2>3\ell. Equation (4.15), with \kappa P>0, gives

Rewrite Equation (4.29) as

If the first parentheses are nonnegative, then -d/z_1>2/z_1^2 immediately gives \mathcal S>0. If they are negative, put \nu=(d^2/4-bz_1)^{1/2}>0. The preceding use of Equation (4.31) still gives 2\nu<\Delta_0, while w_*>0,\ell>0 give \Delta_0<1/z_1. Hence

Together with -d/z_1>2/z_1^2, this strictly dominates the negative term -3/(2z_1^2) in Equation (4.46). Its remaining \ell^2-term is positive, so again \mathcal S>0. Since z_1>2, the coefficient 1-2/z_1+\ell(2z_1-1) is positive, and w\ge w_*>0. Therefore every term in D_* is positive. This proves Equation (4.39) also when T<0.

The three cases cover all endpoint configurations. At a zero of \mathcal W, Cases 1 and 3 give (w_*-w)/(m-n)>0, and the excluded diagonal and Case 2 contain no zeros. Equations (4.21) and (4.20) finish the proof.□

Proof of Theorem 4.1. First suppose \kappa>0. Fix any model data point (\lambda,\kappa,r). By the arc construction and Lemma 3.1, 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 (4.10) 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

At any zero, Equations (4.8) and (4.11), and Lemma 4.4, give

Since \alpha>0, Equation (4.47) 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

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 3.6.

Finally, apply the pointwise fixed-parameter identity Equation (4.3). 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. Fix h_0>0. Use the smooth width-one profile family g(h,\rho,s) of Equation (2.11), with signed auxiliary parameter \rho near zero and h near h_0. Lemma 2.3 provides smooth endpoint data at scaled width one for this family and for the finite-parameter perturbations below. In particular, the local implicit inversion that fixes the width is performed near the same transverse parabolic arc.

For a scaled profile \psi, let \mathcal A(\psi) denote the pair consisting of its midpoint and its midpoint derivative with respect to scaled width, evaluated at base height 0 and width 1. Only the smooth finite-parameter families of profiles just described are used in this notation. For the quadratic profile \psi(s)=a_1s+a_2s^2/2, the derivative of this pair at (a_1,a_2)=(0,0) is, by Equation (2.9),

It is nonsingular. The cubic profile variation s^3/6 gives the column

Taylor expansion of Equation (2.11), with a smooth remainder on the fixed scaled-height interval, gives

where \mathcal R is smooth. Write

Then g=g_2+\rho^5\eta. Apply the local inverse defined by J to the data of the smooth family g_2+\varepsilon\eta, with (h,\rho,\varepsilon) near (h_0,0,0). Denote its fitted scaled coefficients by a_1(h,\rho,\varepsilon) and a_2(h,\rho,\varepsilon). At \varepsilon=0 these are exactly \rho k(h) and \rho^3\phi''(h), respectively. The integral form of Taylor’s formula in \varepsilon shows that

for a smooth vector-valued function q. Differentiating at \rho=\varepsilon=0 in the profile-variation direction yields

Another smooth Taylor factorization in \rho therefore writes

For \rho=r>0, the scaled actual data are (M(h,r)/r,v(h,r)). Scaling the quadratic fit in Equation (3.27) gives scaled coefficients (r\lambda,r^3\kappa). Uniqueness in Theorem 3.6 identifies them with the local coefficients just constructed. Dividing the two corrections by r and r^3, respectively, proves Equation (5.2). Differentiating its second identity at fixed h gives

This proves all the locally uniform estimates in Equation (5.1).□

Proof. Assume first that \phi'''>0. Fix a fitted arc with base height h and width r, and let its peak height be t. Parametrize its characteristic by outward width s\in(0,r], writing the base height as h(s), with h(r)=h and h(s)\to t as s\downarrow0. Put

The characteristic equations in Equation (2.6) give

In particular c extends continuously to s=0 with c(0)=\phi''(t) and is strictly decreasing on [0,r].

In the following comparison, hold the curvature equal to the final fitted value \kappa=\kappa(h,r). Let J_\kappa and \Theta_\kappa(s,z) be the slope field and the reference angle family of Section 3. Define

and

Lemma 3.2, specifically Equation (3.11), gives

The integral formula in Equation (3.13) applies with this forcing f. Its positive kernel preserves the sign of f, even though a may be singular at the peak.

The strict decrease of c gives the following exhaustive sign cases. If \kappa\ge c(0), then f(s)>0 for every s>0, and Equation (3.13) gives E(s)>0. If \kappa\le c(r), then f(s)<0 for 0<s<r, and E(s)<0 for 0<s\le r. In the remaining case there is a unique s_0\in(0,r) with c(s_0)=\kappa. Here f<0 before s_0 and f>0 after it. The same integral formula gives E<0 on (0,s_0], whereas

Consequently E either remains negative up to the final width, possibly having E(r)=0, or crosses zero exactly once before r and is strictly positive thereafter. It cannot vanish on a nontrivial interval: Equation (5.5) would then force c=\kappa on that interval, contrary to Equation (5.4).

Let z(s) be the reference label determined by

By Equation (3.12),

The constant label z_* = z(r) describes the quadratic reference whose final angle agrees with the actual one. Its final slope is \lambda(h,r), by the construction of the fit. The fit also makes the final values of v, and hence of L, equal. Therefore Equation (2.6) gives

Both integrals converge: the two angles are O(s) near their smooth peaks.

If z(s) were monotone and nonconstant on (0,r], monotonicity of \Theta_\kappa in its label would order the two integrands in Equation (5.6) in one direction, with strict inequality on a nonempty open subinterval. This contradicts their equal integrals. The first two sign cases above therefore cannot occur. In the two-sign forcing case, E must cross zero strictly before r: otherwise z is strictly decreasing on (0,r), even if E(r)=0, giving the same contradiction. A constant label is also impossible, since it would give E\equiv0 and hence constant c. It follows that

It remains to propagate the sign of the width derivative of fitted curvature. As a function of independent model parameters, set

Theorem 4.1 asserts that its explicit partial derivative satisfies C_r\le0 when \lambda and \kappa are held fixed. Differentiating the second identity in Equation (3.28) at fixed h, we must also account for

Using \lambda_r=-Q\kappa_r/P gives

where

Here \alpha_r is the derivative of the actual function \alpha(h,r), whereas C_\lambda,C_\kappa,C_r are model partial derivatives evaluated at the fit. All terms in \mathcal T are smooth at positive height and width. The dependence of \alpha on M is thus included in the coefficient of \kappa_r; it creates no additional independent source.

On the fixed characteristic, write w(s)=\kappa_r(h(s),s). Equation (5.1), locally uniformly near its peak t, gives

for all sufficiently small positive s. Equations (5.7) and (5.8) give, along the characteristic,

Choose a sufficiently small positive s_1 with w(s_1)>0. On [s_1,r] all coefficients are continuous. Multiplication by the positive integrating factor

gives (\mu w)'\ge0. Thus w(r)>0, proving \kappa_r(h,r)>0. Since the chosen fit was arbitrary, both assertions hold for every h,r>0.

Finally, reflection of the endpoint coordinate changes \phi to -\phi and changes (M,v) to (-M,-v). Uniqueness of the quadratic fit changes (\lambda,\kappa) to (-\lambda,-\kappa). Applying the proved case to -\phi proves the two reversed inequalities when \phi'''<0.□

Proof. Assume the displayed derivative is positive and set

Fix a characteristic with peak height t. By Equation (5.1), its limit at width zero is

Thus I is strictly negative near the peak.

Consider any positive-height point on the characteristic at which I=0. At that point \lambda=d+h\kappa. We claim that

Indeed, if the reverse weak inequality held, then for every u>h the hypothesis would give

Integrating from h to u shows

The strict translated-profile comparison of Lemma 2.2 then makes the actual midpoint at the fixed base height and width strictly larger than the quadratic midpoint. This contradicts Equation (3.27), proving Equation (5.10). In particular \lambda=d+h\kappa>k(h) at any zero of I.

Equation (3.28) gives the exact identity

At a zero of I its right-hand side is strictly positive. The outward characteristic derivative is therefore -DI/\alpha<0. If I had a first zero after its initially negative interval, its outward derivative there would instead be nonnegative. This contradiction proves I<0 everywhere along the characteristic. Every fit belongs to such a characteristic, so the assertion holds for all h,r>0.

For a strictly negative derivative, apply the proved assertion to -\phi and -d, and use the reflection of fitted parameters from the proof of Proposition 5.2.□

Proof. Write m=\min_{[h,t]}\phi'' and b=\max_{[h,t]}\phi''. If the fitted \kappa were less than m, the actual curvature would be strictly larger than this constant reference curvature along the whole characteristic. Lemma 3.2 would then make the actual final L strictly larger than the reference L with the same final angle, contradicting the fit. A fitted curvature greater than b gives the reversed contradiction. This proves Equation (5.12).

If \phi'' is nonconstant and \kappa=m, the forcing in Equation (3.11) is nonpositive and strictly negative on a positive interior width interval. Its integral representation gives E\le0 everywhere and E<0 on a nonempty interval. Equation (3.12) makes the reference label nonincreasing and nonconstant. Relative to its final constant label, the actual earlier angles are therefore no smaller and are strictly larger on a nonempty open interval. As in Equation (5.6), their L integral is strictly larger, a contradiction. The case \kappa=b reverses all these inequalities. Thus both bounds are strict whenever the actual curvature is nonconstant.□

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.□