Sharp Cubic Bounds and Critical-Point Transitions for arsinh (x+ λx³ + ηx⁵)

1. Introduction

Sharp polynomial bounds for elementary functions are a classical part of inequality theory. A general finite-interval formulation in terms of sharp Taylor polynomial enclosures was developed by Streeter and Dillon [14] [Definition 1 and Proposition 1]. Monotonicity principles of l’Hospital type, including one-turn variants, provide another route to the extrema of normalized remainders. For such results, see Pinelis [13] [Propositions 4.3–4.4] and the higher-order fraction rules of Bitsouni, Gialelis, and Marinescu [3] [Theorems 3.1–3.2].

The inverse hyperbolic sine has also been the subject of many sharp inequalities. Zhu [16] [Theorems 1.9–1.10 and Propositions 3.1–3.2] proved Shafer–Fink-type bounds for $\mathrm{arsinh}x$ using a power-series quotient rule. Masjed-Jamei [10] related ${\left(\arctan x\right)}^2$ to $x\mathrm{arsinh}x/\sqrt{1+x^2}$. Guo, Luo, and Qi [9] [Theorem 1] studied a parameter-dependent auxiliary quotient involving the fixed function $\mathrm{arsinh}x$ and obtained parameter conditions for strict increase or a unique minimum. Zhu and Malešević [17] [Theorems 1.1–1.3] later proved the whole-line extension and refinements of Masjed-Jamei’s inequality. Further inverse-trigonometric and inverse-hyperbolic inequalities appear in [2,4,5]. Direct arsinh bounds, quotient bounds, and series expansions appear respectively in [6,15], and [8].

In the present family, the parameter occurs inside the argument of the function.

$$F_{\lambda }\left(x\right)=\mathrm{arsinh}(x+\lambda x^3),\lambda \ge 0,0\le x\le 1.$$

(1)

For every $\lambda \ge 0$, the best coefficients are determined in

$$x+A{x}^3\le F_{\lambda }\left(x\right)\le x+B{x}^3.$$

(2)

As in [14] [Proposition 1], this amounts to finding the extrema of the continuous extension of

$$Q_{\lambda }\left(x\right)={\displaystyle \frac{F_{\lambda }\left(x\right)-x}{x^3}},0<x\le 1.$$

(3)

The quotient does not divide by the cubic Taylor coefficient $\lambda -1/6$ and remains defined when that coefficient vanishes.

The auxiliary quotient in [9] [Theorem 1] is $(\theta +\sqrt{1+t^2})\mathrm{arsinh}\left(t\right)/t$, with the inverse hyperbolic sine function itself unperturbed. Substituting $t=x+\lambda x^3$ in such bounds does not by itself determine the best coefficients in (2). The extremum problem (3) is therefore distinct. Its complete parameter classification tracks two changes in the minimizing point and a separate switch of the maximizing endpoint.

Section 6 treats the quintic perturbation $\mathrm{arsinh}(x+\lambda x^3+\eta x^5)$ with $\eta >0$. For every positive parameter pair, the normalized remainder has at most two positive critical points. Together with the pointwise strict decrease of its derivative as $\lambda$ increases, this bound gives a complete slice classification. There is a unique height ${\eta }_c>{\eta }_{\ast }$ at which the fold curve reaches $\lambda =0$. Hence a positive fold threshold occurs exactly for ${\eta }_{\ast }<\eta <{\eta }_c$. At $\eta ={\eta }_c$ the fold reaches the boundary $\lambda =0$ with a double critical point, while for $\eta >{\eta }_c$ the two-critical-point regime meets that boundary. The local two-critical-point regime emerges from the explicit point $({\lambda }_{\ast },{\eta }_{\ast })$. One analytic curve marks the appearance of an interior minimum, while a second marks when that minimum becomes global. Across the latter curve, the minimizing point changes discontinuously and the optimal lower coefficient loses differentiability.

The leading local polynomial has the classical sixth-order structure described by Ellis, Machta, and Otto [7] [Theorems A.1–A.2], and the parameter-monotonicity argument is elementary. For the present family, we obtain the parameter-uniform critical-point bound, the global slice classification with its unique $\lambda =0$ endpoint, and the sharp-bound analysis. The cubic classification holds for all $\lambda \ge 0$. The quintic critical-point ordering holds throughout the positive parameter region, while the two competing-minimum curves remain local near $({\lambda }_{\ast },{\eta }_{\ast })$.

2. The Normalized Cubic Remainder

The local expansion of (1) at the origin is

$$\begin{array}{cc}\hfill F_{\lambda }\left(x\right)=& x+\left(\lambda -{\displaystyle \frac{1}{6}}\right)x^3+\left({\displaystyle \frac{3}{40}}-{\displaystyle \frac{\lambda }{2}}\right)x^5\hfill \\ \hfill & +\left(-{\displaystyle \frac{{\lambda }^2}{2}}+{\displaystyle \frac{3\lambda }{8}}-{\displaystyle \frac{5}{112}}\right)x^7+O\left(x^9\right).\hfill \end{array}$$

(4)

Thus the function in (3) extends continuously to $x=0$ by

$$Q_{\lambda }\left(0\right)=\lambda -{\displaystyle \frac{1}{6}}.$$

(5)

At the other endpoint,

$$Q_{\lambda }\left(1\right)=\mathrm{arsinh}(1+\lambda )-1.$$

(6)

For $0\le x\le 1$, define

$$N_{\lambda }\left(x\right)=x\left(F_{\lambda }^{\prime }\left(x\right)-1\right)-3\left(F_{\lambda }\left(x\right)-x\right).$$

(7)

Then

$$Q_{\lambda }^{\prime }\left(x\right)={\displaystyle \frac{N_{\lambda }\left(x\right)}{x^4}},x>0.$$

(8)

Put

$$D_{\lambda }\left(x\right)=1+x^2{(1+\lambda x^2)}^2.$$

(9)

A direct differentiation gives the factorization

$$N_{\lambda }^{″}\left(x\right)={\displaystyle \frac{x^3(1+3\lambda x^2)}{D_{\lambda }{\left(x\right)}^{5/2}}}{P}_{\lambda }\left(x^2\right),$$

(10)

where

$$P_{\lambda }\left(z\right)=3{\lambda }^4{z}^4+4{\lambda }^3{z}^3+2{\lambda }^2{z}^2+4\lambda (1-6\lambda )z+3-20\lambda .$$

(11)

Its sign pattern follows directly from Descartes’ rule of signs.

Lemma 2.1.

For $\lambda >3/20$, the polynomial $P_{\lambda }$ has exactly one positive root r, and this root is simple. It is negative on $[0,r)$ and positive on $(r,\infty )$.

Proof. 

In descending order, the nonzero coefficients in (11) have exactly one sign change. The first three are positive, the linear coefficient may have either sign, and the constant is negative. Descartes’ rule gives exactly one positive root counted with multiplicity. See [1] [Proposition 1]. The root is therefore simple. The signs on its two sides follow from $P_{\lambda }\left(0\right)<0$ and the positive leading coefficient. □

The following derivative-chain lemma complements this sign analysis.

Lemma 2.2.

Let $u\in C^2\left([0,1]\right)$ satisfy $u\left(0\right)=u^{\prime }\left(0\right)=0$. Suppose $u^{″}$ is negative near 0 and has at most one sign change, necessarily from negative to positive. Then u has at most one zero in $(0,1]$, and any such zero t satisfies $u^{\prime }\left(t\right)>0$. Moreover,

1. 

If $u\left(1\right)>0$, then u has exactly one zero in $(0,1)$ and changes there from negative to positive.

2. 

If $u\left(1\right)\le 0$, then $u\left(x\right)<0$ for every $0<x<1$.

Proof. 

The hypothesis on $u^{″}$ gives a point $c\in (0,1]$ such that $u^{\prime }$ is nonincreasing on $[0,c]$ and nondecreasing on $[c,1]$. Since $u^{\prime }\left(0\right)=0$ and $u^{″}<0$ near the origin, both $u^{\prime }$ and u are negative immediately to the right of 0. If $u\left(t\right)=0$ for some $t>0$, choose $r\in (0,t)$ with $u\left(r\right)<0$. The mean value theorem on $[r,t]$ gives a point $s\in (r,t)$ with $u^{\prime }\left(s\right)>0$. Hence $s\in [c,1]$, where $u^{\prime }$ is nondecreasing, and $u^{\prime }\left(q\right)\ge u^{\prime }\left(s\right)>0$ for every $q\in [t,1]$. Thus the zero is unique, $u^{\prime }\left(t\right)>0$, and u is positive after it. The two assertions now follow by continuity and the sign of $u\left(1\right)$. □

3. The Cubic Shape Transition

Set

$$a={\displaystyle \frac{3}{20}}.$$

(12)

The right-endpoint value of $N_{\lambda }$ has the form

$$N_{\lambda }\left(1\right)=\varphi \left(\lambda \right):=2-3\mathrm{arsinh}(1+\lambda )+{\displaystyle \frac{3\lambda +1}{\sqrt{{\lambda }^2+2\lambda +2}}}.$$

(13)

Its parameter derivative is

$${\varphi }^{\prime }\left(\lambda \right)=-{\displaystyle \frac{(\lambda +1)(3\lambda +1)}{{({\lambda }^2+2\lambda +2)}^{3/2}}}<0.$$

(14)

Lemma 3.1.

There is a unique $\delta >a$ such that

$$2-3\mathrm{arsinh}(1+\delta )+{\displaystyle \frac{3\delta +1}{\sqrt{{\delta }^2+2\delta +2}}}=0.$$

(15)

Numerically,

$$\delta \approx 0.1516127199599987.$$

Proof. 

At $\lambda =a$, equation (11) gives $P_a\left(0\right)=0$. All its other coefficients are positive, so $P_a\left(z\right)>0$ for $z>0$. Hence (10) gives $N_a^{″}\left(x\right)>0$ for $x>0$. Since $N_a\left(0\right)=N_a^{\prime }\left(0\right)=0$, it follows that $N_a\left(1\right)>0$, so $\varphi \left(a\right)>0$.

On the other hand,

$$\mathrm{arsinh}(1+\lambda )=\log \left(1+\lambda +\sqrt{{(1+\lambda )}^2+1}\right)⟶\infty ,$$

whereas

$${\displaystyle \frac{3\lambda +1}{\sqrt{{\lambda }^2+2\lambda +2}}}⟶3.$$

Thus $\varphi \left(\lambda \right)\to -\infty$ as $\lambda \to \infty$. Strict decrease from (14) gives a unique zero $\delta >a$. The displayed decimal is a numerical approximation to this root. □

The preceding lemmas give the global shape classification.

Theorem 3.2.

Let $\lambda \ge 0$.

1. 

If $0\le \lambda \le a$, then $Q_{\lambda }$ is strictly increasing on $[0,1]$.

2. 

If $a<\lambda <\delta$, then there exists a unique $x_{\lambda }\in (0,1)$ such that $Q_{\lambda }$ is strictly decreasing on $[0,x_{\lambda }]$ and strictly increasing on $[x_{\lambda },1]$. Thus $x_{\lambda }$ is the unique global minimizer.

3. 

If $\lambda \ge \delta$, then $Q_{\lambda }$ is strictly decreasing on $[0,1]$.

At $\lambda =\delta$, one has $Q_{\delta }^{\prime }\left(x\right)<0$ for $0<x<1$ and $Q_{\delta }^{\prime }\left(1\right)=0$.

Proof. 

First suppose $0\le \lambda \le a$. Here

$$P_{\lambda }\left(0\right)=3-20\lambda \ge 0.$$

If $\lambda =0$, then $P_0\left(z\right)=3$. If $0<\lambda \le a<1/6$, then all nonconstant coefficients in (11) are positive. In both cases $P_{\lambda }\left(z\right)>0$ for $z>0$. Equation (10) gives $N_{\lambda }^{″}\left(x\right)>0$ for $x>0$. Together with $N_{\lambda }\left(0\right)=N_{\lambda }^{\prime }\left(0\right)=0$, this gives $N_{\lambda }\left(x\right)>0$ for $x>0$. By (8), $Q_{\lambda }$ is strictly increasing.

Now let $\lambda >a$. Lemma 2.1, restricted to $0<x\le 1$, and (10) show that $N_{\lambda }^{″}$ is negative near the origin and has at most one sign change, from negative to positive. Lemma 2.2 therefore applies to $N_{\lambda }$.

If $a<\lambda <\delta$, then (13) and Lemma 3.1 give $N_{\lambda }\left(1\right)>0$. Hence $N_{\lambda }$ has exactly one zero $x_{\lambda }$ in $(0,1)$, with $N_{\lambda }<0$ before it and $N_{\lambda }>0$ after it. Equation (8) gives the stated decrease–increase pattern.

If $\lambda \ge \delta$, then $N_{\lambda }\left(1\right)\le 0$, so Lemma 2.2 gives $N_{\lambda }\left(x\right)<0$ for $0<x<1$. This proves strict decrease. At $\lambda =\delta$, $N_{\delta }\left(1\right)=0$, which gives $Q_{\delta }^{\prime }\left(1\right)=0$. □

The value a is the zero of the $x^2$ coefficient in the expansion of $Q_{\lambda }$ obtained from (4). The derivative argument establishes the global classification and the second threshold $\delta$.

4. Sharp Cubic Bounds and the Endpoint Switch

The shape theorem determines the sharp coefficients in (2). A further parameter value determines which endpoint supplies the upper extremum in the middle regime.

Define $\mu$ by

$$\mathrm{arsinh}(1+\mu )=\mu +{\displaystyle \frac{5}{6}}.$$

(16)

Lemma 4.1.

Equation (16) has a unique solution $\mu \ge 0$. Moreover,

$$a<\mu <\delta ,\mu \approx 0.1506769390140933.$$

Proof. 

Let

$$\psi \left(\lambda \right)=Q_{\lambda }\left(1\right)-Q_{\lambda }\left(0\right)=\mathrm{arsinh}(1+\lambda )-\lambda -{\displaystyle \frac{5}{6}}.$$

Then

$${\psi }^{\prime }\left(\lambda \right)={\displaystyle \frac{1}{\sqrt{{(1+\lambda )}^2+1}}}-1<0.$$

At $\lambda =a$, Theorem 3.2 gives $Q_a\left(1\right)>Q_a\left(0\right)$, hence $\psi \left(a\right)>0$. At $\lambda =\delta$, the same theorem gives $Q_{\delta }\left(1\right)<Q_{\delta }\left(0\right)$, hence $\psi \left(\delta \right)<0$. Thus $\psi$ has a unique zero $\mu$ in $(a,\delta )$. The decimal is a numerical approximation to the root. □

Corollary 4.2.

For each $\lambda \ge 0$ and $A,B\in \mathbb{R}$, the double inequality

$$x+A{x}^3\le \mathrm{arsinh}(x+\lambda x^3)\le x+B{x}^3,0\le x\le 1,$$

holds if and only if

$$A\le A_3\left(\lambda \right),B\ge B_3\left(\lambda \right),$$

where

$$A_3\left(\lambda \right)=\left\{\begin{array}{cc}{\displaystyle \lambda -\frac{1}{6},}\hfill & 0\le \lambda \le a,\hfill \\ {\displaystyle Q_{\lambda }\left(x_{\lambda }\right),}\hfill & a<\lambda <\delta ,\hfill \\ {\displaystyle \mathrm{arsinh}(1+\lambda )-1,}\hfill & \lambda \ge \delta ,\hfill \end{array}\right.$$

(17)

and

$$B_3\left(\lambda \right)=\left\{\begin{array}{cc}{\displaystyle \mathrm{arsinh}(1+\lambda )-1,}\hfill & 0\le \lambda \le \mu ,\hfill \\ {\displaystyle \lambda -\frac{1}{6},}\hfill & \lambda \ge \mu .\hfill \end{array}\right.$$

(18)

All coefficients in (17)–(18) are best possible.

Proof. 

For $x>0$, subtract x from the desired inequalities and divide by $x^3$. Thus the largest admissible lower coefficient is ${min}_{[0,1]}{Q}_{\lambda }$, and the smallest admissible upper coefficient is ${max}_{[0,1]}{Q}_{\lambda }$.

Theorem 3.2 gives the minimum directly and proves (17). In the increasing and decreasing regimes the maximum is the opposite endpoint. In the middle regime there is only an interior minimum, so the maximum is again an endpoint value. Lemma 4.1 determines which endpoint is larger and proves (18). Since these coefficients are the actual extrema, they are best possible. □

The three relevant parameter values and the resulting geometry are summarized in Table 1.

5. Parameter Dependence of the Cubic Bounds

For the cubic family, the minimizing point moves continuously from 0 to 1. This motion also determines the regularity of the optimal coefficients.

Differentiating (7) in the parameter gives

$${\displaystyle \frac{\partial N_{\lambda }\left(x\right)}{\partial \lambda }}=-{\displaystyle \frac{x^5(1+\lambda x^2)(1+3\lambda x^2)}{D_{\lambda }{\left(x\right)}^{3/2}}}<0(x>0).$$

(19)

Proposition 5.1.

For $a<\lambda <\delta$, the minimizing point $x_{\lambda }$ is analytic and strictly increasing, with limits 0 and 1 at a and δ, respectively. Moreover,

$$x_{\lambda }^2=700(\lambda -a)+O\left({(\lambda -a)}^2\right)(\lambda \downarrow a).$$

(20)

The coefficients $A_3$ and $B_3$ are continuous and strictly increasing. The function $A_3$ is continuously differentiable, and

$$A_3^{\prime }\left(\lambda \right)=\left\{\begin{array}{cc}1,\hfill & 0\le \lambda \le a,\hfill \\ D_{\lambda }{\left(x_{\lambda }\right)}^{-1/2},\hfill & a<\lambda <\delta ,\hfill \\ {({(1+\lambda )}^2+1)}^{-1/2},\hfill & \lambda \ge \delta .\hfill \end{array}\right.$$

(21)

The function $B_3$ is differentiable except at μ, with

$$B_3^{\prime }\left(\lambda \right)=\left\{\begin{array}{cc}{({(1+\lambda )}^2+1)}^{-1/2},\hfill & 0\le \lambda <\mu ,\hfill \\ 1,\hfill & \lambda >\mu .\hfill \end{array}\right.$$

(22)

Derivatives at zero are right derivatives.

Proof. 

Lemma 2.2 gives $N_{\lambda }^{\prime }\left(x_{\lambda }\right)>0$. The analytic implicit function theorem and (19) give

$${\displaystyle \frac{d{x}_{\lambda }}{d\lambda }}=-{\displaystyle \frac{(\partial N_{\lambda }/\partial \lambda )\left(x_{\lambda }\right)}{N_{\lambda }^{\prime }\left(x_{\lambda }\right)}}>0.$$

The endpoint limits follow by continuity. A positive limit at a would be a positive zero of $N_a$, and a limit less than 1 at $\delta$ would be an interior zero of $N_{\delta }$, both impossible by Theorem 3.2.

For the local expansion, write $\tilde{Q}(\lambda ,y)=Q_{\lambda }\left(\sqrt{y}\right)$. Odd analyticity of $F_{\lambda }$ gives joint analyticity of $\tilde{Q}$ near $(a,0)$. If $G={\tilde{Q}}_y$, then (4) gives

$$G(a,0)=0,G_y(a,0)={\displaystyle \frac{1}{1400}},G_{\lambda }(a,0)=-{\displaystyle \frac{1}{2}}.$$

The implicit critical-point branch therefore has derivative 700 at a, proving (20).

For the coefficient functions,

$${\displaystyle \frac{\partial Q_{\lambda }\left(x\right)}{\partial \lambda }}=D_{\lambda }{\left(x\right)}^{-1/2}$$

(23)

is jointly continuous and positive, including at $x=0$. Because this derivative has a positive minimum on every compact rectangle in its variables, both extrema are strictly increasing. Uniform continuity gives their continuity. Extend the minimizing point by ${\xi }_{\lambda }=0$ for $\lambda \le a$ and ${\xi }_{\lambda }=1$ for $\lambda \ge \delta$. The endpoint limits prove continuity of $\xi$. The formula for $A_3^{\prime }$ is the minimum-value version of Milgrom and Segal’s compact-choice-set envelope formula [11] [Corollary 4]. Directly, minimality gives

$$\begin{array}{cc}\hfill Q_{\lambda +h}\left({\xi }_{\lambda +h}\right)-Q_{\lambda }\left({\xi }_{\lambda +h}\right)& \le A_3(\lambda +h)-A_3\left(\lambda \right)\hfill \\ & \le Q_{\lambda +h}\left({\xi }_{\lambda }\right)-Q_{\lambda }\left({\xi }_{\lambda }\right).\hfill \end{array}$$

Divide by h, reversing the inequalities for $h<0$, and use (23) and continuity of $\xi$. Both bounds tend to $D_{\lambda }{\left({\xi }_{\lambda }\right)}^{-1/2}$. This proves (21), including continuity at the two transitions. Finally (18) gives (22). Its left derivative at $\mu$ is less than its right derivative 1. □

6. Critical Points and Competing Minima for Quintic Perturbations

For $\lambda \ge 0$, $\eta >0$, and $x\ge 0$, put

$$p_{\lambda ,\eta }\left(x\right)=x+\lambda x^3+\eta x^5,F_{\lambda ,\eta }\left(x\right)=\mathrm{arsinh}{p}_{\lambda ,\eta }\left(x\right).$$

(24)

The normalized remainder is

$$Q_{\lambda ,\eta }\left(x\right)={\displaystyle \frac{F_{\lambda ,\eta }\left(x\right)-x}{x^3}},Q_{\lambda ,\eta }\left(0\right)=\lambda -{\displaystyle \frac{1}{6}}.$$

(25)

The critical-point bound is global for $\lambda \ge 0$ and $\eta >0$. The sharper extremum and coefficient analysis is local near an explicit positive pair. All critical points considered below have $x>0$.

Set $y=x^2$ and $q\left(y\right)=Q_{\lambda ,\eta }\left(\sqrt{y}\right)$. The even analytic extension at zero gives

$$q\left(y\right)=q_0+uy+v{y}^2+d(\lambda ,\eta )y^3+O\left(y^4\right),q_0=\lambda -{\displaystyle \frac{1}{6}},$$

(26)

where

$$\begin{array}{cc}\hfill u& =\eta -{\displaystyle \frac{\lambda }{2}}+{\displaystyle \frac{3}{40}},\hfill \end{array}$$

(27)

$$\begin{array}{cc}\hfill v& =-{\displaystyle \frac{\eta }{2}}-{\displaystyle \frac{{\lambda }^2}{2}}+{\displaystyle \frac{3\lambda }{8}}-{\displaystyle \frac{5}{112}},\hfill \end{array}$$

(28)

$$\begin{array}{cc}\hfill d(\lambda ,\eta )& =-\eta \lambda +{\displaystyle \frac{3\eta }{8}}-{\displaystyle \frac{{\lambda }^3}{6}}+{\displaystyle \frac{3{\lambda }^2}{4}}-{\displaystyle \frac{5\lambda }{16}}+{\displaystyle \frac{35}{1152}}.\hfill \end{array}$$

(29)

These identities follow by substituting $p_{\lambda ,\eta }$ into the classical series for arsinh [12] [Eq. 4.38.1]. The positive solution of $u=v=0$ is

$${\lambda }_{\ast }={\displaystyle \frac{35+\sqrt{105}}{280}},{\eta }_{\ast }={\displaystyle \frac{\sqrt{105}-7}{560}}.$$

(30)

In particular,

$${\lambda }_{\ast }\approx 0.161596252735570,{\eta }_{\ast }\approx 0.005798126367785.$$

At this point,

$$d_0:=d({\lambda }_{\ast },{\eta }_{\ast })={\displaystyle \frac{9\sqrt{105}-91}{564480}}>0,d_0\approx 2.16581082347716·{10}^{-6}.$$

(31)

The Jacobian of $(\lambda ,\eta )\mapsto (u,v)$ is $\lambda -1/8$, which is positive at (30). The local inverse is explicitly

$$\lambda (u,v)={\displaystyle \frac{1}{8}}+\sqrt{{\displaystyle \frac{3}{2240}}-u-2v},\eta (u,v)=u+{\displaystyle \frac{\lambda (u,v)}{2}}-{\displaystyle \frac{3}{40}}.$$

(32)

The notation $q(u,v,y)$ denotes the exact function obtained from (25) and (32), not the truncated polynomial in (26). Also write $q_0(u,v)=\lambda (u,v)-1/6$.

The following global critical-point bound holds throughout the positive parameter quadrant.

Proposition 6.1.

For every $\lambda \ge 0$ and $\eta >0$, the derivative of $Q_{\lambda ,\eta }$ has at most two positive zeros, counted with multiplicity. Moreover, $Q_{\lambda ,\eta }^{\prime }\left(x\right)>0$ for all sufficiently large x. Consequently,

1. 

If

$$u=\eta -{\displaystyle \frac{\lambda }{2}}+{\displaystyle \frac{3}{40}}<0,\mathit{equivalently}\lambda >2\eta +{\displaystyle \frac{3}{20}},$$

then $Q_{\lambda ,\eta }$ has exactly one positive critical point. It is simple and is the unique global minimum on $[0,\infty )$.

2. 

If $u>0$, then the positive critical set is either empty, a single double critical point, or two simple critical points. In the last case the first is a strict local maximum and the second a strict local minimum.

If $u=0$, the same two conclusions hold with the sign of v in place of the sign of u whenever $v\ne 0$.

Proof. 

Put $p=p_{\lambda ,\eta }$, $D=1+p^2$, and

$$N=x{F}_{\lambda ,\eta }^{\prime }-3F_{\lambda ,\eta }+2x.$$

Then $Q^{\prime }=N/x^4$, $N\left(0\right)=N^{\prime }\left(0\right)=0$, and differentiation gives

$$N^{″}={\displaystyle \frac{x^3}{D^{5/2}}}{\displaystyle \sum _{j=0}^{10}}{c}_j{x}^{2j}.$$

(33)

The coefficients are listed in Appendix A. When $\lambda >0$, Lemma A.1 shows that, after zero coefficients are omitted, their sequence has at most two sign changes. At $\lambda =0$ the nonzero coefficient signs are

$$+,+,+,-,-,-,-,-,+,$$

again with exactly two changes. Descartes’ rule of signs [1] [Proposition 1] therefore gives at most two positive zeros of the polynomial in $x^2$, counted with multiplicity, throughout $\lambda \ge 0$, $\eta >0$.

The same bound for $N^{″}$ also holds for N. Suppose the positive zeros of N have total multiplicity k. The zeros themselves contribute their multiplicities minus one to $N^{\prime }$, while Rolle’s theorem supplies one additional zero in each interval from 0 to the first positive zero and between consecutive positive zeros. Thus $N^{\prime }$ has at least k positive zeros counted with multiplicity. Since $N^{\prime }\left(0\right)=0$, the same argument applied once more shows that $N^{″}$ has at least k positive zeros. Hence $k\le 2$, and the same is true for positive zeros of $Q^{\prime }$.

As $x\to \infty$, one has $p\sim \eta x^5$, $F_{\lambda ,\eta }=O\left(\log x\right)$, and $x{F}_{\lambda ,\eta }^{\prime }=O\left(1\right)$. Hence

$$N\left(x\right)=2x+O\left(\log x\right)>0$$

for all sufficiently large x.

Finally, with $y=x^2$, expansion (26) gives

$$Q^{\prime }\left(x\right)=2x\{u+2v{x}^2+O\left(x^4\right)\}.$$

If $u<0$, the derivative is negative near zero and positive for large x. It therefore has a positive zero of odd multiplicity. The total multiplicity bound forces this zero to be unique and simple, so the sign pattern is $-,+$ and the critical point is the unique global minimum. If $u>0$, the derivative is positive at both ends. With total positive-zero multiplicity at most two, the only possibilities are no zero, one double zero, or two simple zeros. In the last case the sign pattern is $+,-,+$. When $u=0$ and $v\ne 0$, the same argument starts from $Q^{\prime }\left(x\right)=4v{x}^3+O\left(x^5\right)$. □

The global multiplicity bound provides the starting point for a finer classification of the critical-point regimes along each fixed-$\eta$ slice.

Theorem 6.2.

Let

$${\lambda }_0\left(\eta \right)=2\eta +{\displaystyle \frac{3}{20}},\eta >0.$$

Then the positive critical points of $Q_{\lambda ,\eta }$ are ordered as follows when η is fixed and λ varies.

1. 

If $0<\eta \le {\eta }_{\ast }$, then $Q_{\lambda ,\eta }$ is strictly increasing on $[0,\infty )$ for $0<\lambda \le {\lambda }_0\left(\eta \right)$. For $\lambda >{\lambda }_0\left(\eta \right)$ it has exactly one positive critical point, which is simple and is its unique global minimum.

2. 

If $\eta >{\eta }_{\ast }$, there is a unique number

$$\Lambda \left(\eta \right)\in [0,{\lambda }_0\left(\eta \right))$$

with the following properties.

(a) 

For $0<\lambda <\Lambda \left(\eta \right)$, when this interval is nonempty, $Q_{\lambda ,\eta }$ is strictly increasing.

(b) 

If $\Lambda \left(\eta \right)>0$, then at $\lambda =\Lambda \left(\eta \right)$ there is one positive critical point. It is a double zero of $Q^{\prime }$, and Q is still strictly increasing.

(c) 

For $\Lambda \left(\eta \right)<\lambda <{\lambda }_0\left(\eta \right)$ there are exactly two simple positive critical points, a strict local maximum followed by a strict local minimum.

(d) 

For $\lambda \ge {\lambda }_0\left(\eta \right)$ there is exactly one positive critical point, which is simple and is the unique global minimum.

If $\Lambda \left(\eta \right)=0$, the first two subcases are absent. Wherever $\Lambda \left(\eta \right)>0$, both the fold value $\Lambda \left(\eta \right)$ and its double critical point are locally real-analytic functions of η.

Proof. 

First note the parameter derivative

$${\displaystyle \frac{\partial Q_{\lambda ,\eta }}{\partial \lambda }}={\displaystyle \frac{1}{\sqrt{1+p_{\lambda ,\eta }^2}}},{\displaystyle \frac{\partial Q_{\lambda ,\eta }^{\prime }}{\partial \lambda }}=-{\displaystyle \frac{p_{\lambda ,\eta }{p}_{\lambda ,\eta }^{\prime }}{{(1+p_{\lambda ,\eta }^2)}^{3/2}}}<0(x>0).$$

(34)

Thus, for fixed $\eta$, $Q^{\prime }$ decreases strictly at every positive point as $\lambda$ increases.

Suppose first that $0<\eta \le {\eta }_{\ast }$. At $\lambda ={\lambda }_0\left(\eta \right)$ one has $u=0$ and

$$v=-{\displaystyle \frac{5600{\eta }^2+140\eta -1}{2800}}\ge 0.$$

(35)

It remains to verify positivity of $Q^{\prime }$ on the positive half-line at this boundary value. Put

$$a={\displaystyle \frac{\eta }{{\lambda }_0{\left(\eta \right)}^2}}.$$

This quantity is strictly increasing because

$${\displaystyle \frac{d}{d\eta }}\left({\displaystyle \frac{\eta }{{(2\eta +3/20)}^2}}\right)={\displaystyle \frac{3/20-2\eta }{{(2\eta +3/20)}^3}}>0(0<\eta \le {\eta }_{\ast }),$$

and at the endpoint

$$a_{\ast }:={\displaystyle \frac{{\eta }_{\ast }}{{\lambda }_{\ast }^2}}={\displaystyle \frac{13\sqrt{105}-119}{64}}<{\displaystyle \frac{57}{256}}<0.223,$$

where the first strict inequality follows from $\sqrt{105}<41/4$. The four switch values for the thresholds ${\ell }_1,\dots ,{\ell }_5$ in Lemma A.1 all exceed $0.224$. Hence in the present range

$${\ell }_1<{\ell }_2<{\ell }_3<{\ell }_4<{\ell }_5.$$

Since $a<0.223<{\rho }_5$, the threshold description in Lemma A.1 applies. As $c_0=40u=0$ and $c_1=100u+168v\ge 0$, it gives ${\lambda }_0\left(\eta \right)\le {\ell }_1$. Therefore $c_2,\dots ,c_5$ are strictly positive, and the tail comparison in the proof of Lemma A.1 gives $c_6,\dots ,c_{10}>0$ as well. Identity (33) now gives $N^{″}>0$ for $x>0$. Since $N\left(0\right)=N^{\prime }\left(0\right)=0$, it follows that $N>0$ and hence $Q^{\prime }>0$ on $(0,\infty )$. Equation (34) then implies the same strict positivity for every $0<\lambda <{\lambda }_0\left(\eta \right)$. For $\lambda >{\lambda }_0\left(\eta \right)$ one has $u<0$, so Proposition 6.1 gives the unique global minimum. This proves part (1).

Now let $\eta >{\eta }_{\ast }$. At $\lambda ={\lambda }_0\left(\eta \right)$, formula (35) gives $v<0$. Hence $Q^{\prime }<0$ for all sufficiently small positive x, while Proposition 6.1 gives a single simple positive zero. By continuity the inequality $Q^{\prime }<0$ at some fixed positive point persists when $\lambda$ is decreased slightly below ${\lambda }_0\left(\eta \right)$. For such parameters $u>0$, so $Q^{\prime }$ is positive near zero and positive for large x. The global multiplicity bound therefore gives exactly two simple positive zeros.

Define

$${\mathcal{S}}_{\eta }=\{\lambda \in (0,{\lambda }_0\left(\eta \right)):Q_{\lambda ,\eta }^{\prime }\left(x\right)<0\mathrm{for}\mathrm{some}x>0\}.$$

The preceding paragraph shows that ${\mathcal{S}}_{\eta }$ is nonempty, and (34) shows that it is an upper interval. Put

$$\Lambda \left(\eta \right)=inf{\mathcal{S}}_{\eta }.$$

If $0<\lambda <\Lambda \left(\eta \right)$, then $Q^{\prime }$ cannot vanish. A zero at one point would become negative after a sufficiently small increase of $\lambda$, contradicting the definition of the infimum. Thus $Q^{\prime }>0$ everywhere. If $\Lambda \left(\eta \right)<\lambda <{\lambda }_0\left(\eta \right)$, then $\lambda \in {\mathcal{S}}_{\eta }$. Positivity near zero and for sufficiently large arguments, together with Proposition 6.1, gives exactly two simple zeros with signs $+,-,+$.

Assume finally that $\Lambda \left(\eta \right)>0$. Choose ${\lambda }_n\downarrow \Lambda \left(\eta \right)$ from ${\mathcal{S}}_{\eta }$ and points $x_n>0$ with $Q_{{\lambda }_n,\eta }^{\prime }\left(x_n\right)<0$. Because $u(\Lambda (\eta ),\eta )>0$, the continuous extension of $Q^{\prime }\left(x\right)/\left(2x\right)$ is uniformly positive for small x when $\lambda$ is near $\Lambda \left(\eta \right)$. On the other hand, for $\lambda$ in a compact neighborhood of $\Lambda \left(\eta \right)$ there is a constant $C>0$ such that $p_{\lambda ,\eta }\left(x\right)\le C{x}^5$ for $x\ge 1$. Since $x{p}_{\lambda ,\eta }^{\prime }/\sqrt{1+p_{\lambda ,\eta }^2}\ge 0$ and $\mathrm{arsinh}t\le \log (1+2t)$ for $t\ge 0$, one has uniformly

$$N\left(x\right)\ge 2x-3\log (1+2C{x}^5)>0$$

for all sufficiently large x. Hence $Q^{\prime }>0$ uniformly on the far tail. Thus a subsequence of $\left(x_n\right)$ converges to some $x_f>0$. At the threshold $Q^{\prime }$ cannot be negative anywhere, again by continuity in $\lambda$. Hence

$$Q_{\Lambda \left(\eta \right),\eta }^{\prime }\left(x_f\right)=0.$$

Every zero there has even multiplicity. The global multiplicity bound forces a unique zero of multiplicity two, so Q remains strictly increasing. The case $\lambda \ge {\lambda }_0\left(\eta \right)$ follows from Proposition 6.1 and (35).

At a positive fold where $Q^{\prime }=Q^{″}=0$, the double-zero statement gives $Q^{‴}\ne 0$, and (34) gives ${\partial }_{\lambda }{Q}^{\prime }<0$. The Jacobian of $(Q^{\prime },Q^{″})$ with respect to $(\lambda ,x)$ is therefore nonzero. The analytic implicit function theorem gives the final local real-analyticity assertion. □

It remains to determine where the fold reaches the boundary $\lambda =0$.

Theorem 6.3.

There is a unique number ${\eta }_c>{\eta }_{\ast }$ such that

$$\Lambda \left(\eta \right)>0({\eta }_{\ast }<\eta <{\eta }_c),\Lambda \left(\eta \right)=0(\eta \ge {\eta }_c).$$

More explicitly, let $\alpha >0$ be the unique positive solution of

$$3y\cosh y=5\sinh y,$$

(36)

and, for $y>\alpha$, define

$$X\left(y\right)={\displaystyle \frac{3y\cosh y-5\sinh y}{2(\cosh y-2)}},E\left(y\right)={\displaystyle \frac{\sinh y-X\left(y\right)}{X{\left(y\right)}^5}}.$$

(37)

Then E has a unique critical point $y_c$, which is its strict global minimum, and

$${\eta }_c=E\left(y_c\right),x_c=X\left(y_c\right).$$

At $(\lambda ,\eta )=(0,{\eta }_c)$ the derivative $Q^{\prime }$ has the unique positive zero $x_c$, of multiplicity two. For $\eta >{\eta }_c$, $Q_{0,\eta }^{\prime }$ has exactly two simple positive zeros. The positive fold curve from Theorem 6.2 terminates transversely at $(0,{\eta }_c)$ and admits a real-analytic continuation through that point.

Proof. 

Positive critical points on the boundary are obtained by setting $\lambda =0$. Put

$$\Phi (x,\eta )=Q_{0,\eta }^{\prime }\left(x\right),p=x+\eta x^5.$$

If $\Phi (x,\eta )=0$ and $y=\mathrm{arsinh}p$, write $s=\sinh y$ and $c=\cosh y$. Since $p^{\prime }=1+5\eta x^4=5s/x-4$, the equation $N=0$ becomes

$${\displaystyle \frac{5s-4x}{c}}-3y+2x=0,$$

and hence $x=X\left(y\right)$ in (37). Conversely, if $y>\alpha$, $x=X\left(y\right)$ and $\eta =E\left(y\right)$, then $x+\eta x^5=s$ and the same identity gives $\Phi (x,\eta )=0$.

The range in (37) is precisely the branch where both parametrized quantities are positive. The function $g\left(y\right)=3y-5\tanh y$ has $g^{\prime }\left(y\right)=3-5{\mathrm{sech}}^{2}y$ and $g^{″}\left(y\right)>0$ for $y>0$, so it has one positive zero $\alpha$. In fact $3/2<\alpha <5/3$, since $\tanh (3/2)>9/10$ (equivalently $e^3>19$, which follows from its positive power series) and $\tanh (5/3)<1$. Thus $\cosh y>2$ for $y>\alpha$. Moreover,

$$2\sinh y\cosh y+\sinh y-3y\cosh y>0(y>0),$$

because after division by $\cosh y$ the left side is $2\sinh y+\tanh y-3y$, whose first derivative vanishes at 0 and whose second derivative is $2\sinh y(1-{\cosh }^{-3}y)>0$. It follows that $X\left(y\right)>0$ and $\sinh y-X\left(y\right)>0$ precisely on the branch $y>\alpha$ relevant here.

For the monotonicity calculation, set

$$\mathcal{D}\left(y\right)=3{\cosh }^{2}y+4\cosh y-5-6y\sinh y.$$

(38)

Direct differentiation of X gives

$$X^{\prime }\left(y\right)={\displaystyle \frac{\mathcal{D}\left(y\right)}{2{(\cosh y-2)}^2}}.$$

(39)

Using ${\cosh }^{2}y=(1+\cosh 2y)/2$, its power series is

$$\mathcal{D}\left(y\right)=2-y^2+{\displaystyle \frac{y^4}{6}}+{\displaystyle \sum _{n=3}^{\infty }}{\displaystyle \frac{\frac{3}{2}{4}^n+4-12n}{\left(2n\right)!}}{y}^{2n}.$$

Every coefficient in the sum is positive. Its numerator is 64 at $n=3$ and increases with n. Moreover,

$$2-y^2+{\displaystyle \frac{y^4}{6}}={\displaystyle \frac{{(y^2-3)}^2+3}{6}}>0.$$

Thus $\mathcal{D}\left(y\right)>0$ for all $y>0$, so $X^{\prime }\left(y\right)>0$.

Along the critical-point curve one has

$${\Phi }_{\eta }(X\left(y\right),E\left(y\right))=-{\displaystyle \frac{X\left(y\right)\mathcal{D}\left(y\right)}{{\cosh }^{2}y(\cosh y-2)}}<0.$$

(40)

Since ${\partial }_{\eta }Q=x^2/\sqrt{1+p^2}$, (40) follows after one x-derivative and substitution of (37).

To prove uniqueness of the minimum of E, differentiate

$$\Phi \left(X\right(y),E(y\left)\right)=0$$

to obtain

$${\Phi }_x{X}^{\prime }+{\Phi }_{\eta }{E}^{\prime }=0.$$

If $E^{\prime }\left(y_0\right)=0$, then ${\Phi }_x=0$ at $(X\left(y_0\right),E\left(y_0\right))$. Hence that positive zero of $\Phi (·,E\left(y_0\right))$ has multiplicity at least two. The multiplicity bound in Proposition 6.1 forces its multiplicity to be exactly two and excludes every other positive zero. Since $\Phi (·,E\left(y_0\right))$ is positive near both 0 and ∞, this double zero is a strict local minimum and ${\Phi }_{xx}>0$ there. A second differentiation at $y_0$ gives

$$E^{″}\left(y_0\right)=-{\displaystyle \frac{{\Phi }_{xx}{X}^{\prime }{\left(y_0\right)}^2}{{\Phi }_{\eta }}}>0.$$

Thus every critical point of E is a nondegenerate strict minimum. There cannot be two such points, since the maximum of E between two strict local minima would occur at an interior critical point that is not a minimum. On the other hand, $E\left(y\right)\to \infty$ as $y\downarrow \alpha$ because $X\left(y\right)\to 0$, while

$$X\left(y\right)={\displaystyle \frac{3y-5}{2}}+o\left(1\right)(y\to \infty ),$$

so $E\left(y\right)\to \infty$ also at the other end. Hence E has exactly one critical point $y_c$, and it is the strict global minimum. In particular, E is strictly decreasing on $(\alpha ,y_c)$ and strictly increasing on $(y_c,\infty )$.

The horizontal-line description of E now gives one double boundary critical point when $\eta ={\eta }_c$, two boundary critical points when $\eta >{\eta }_c$, and none when $0<\eta <{\eta }_c$. Away from $y_c$ one has $E^{\prime }\ne 0$. Then $X^{\prime }>0$ and (40), together with ${\Phi }_x{X}^{\prime }+{\Phi }_{\eta }{E}^{\prime }=0$, give ${\Phi }_x\ne 0$. Thus the two boundary critical points for $\eta >{\eta }_c$ are simple. Since Theorem 6.2 gives strict increase at $\eta ={\eta }_{\ast }$ and positive $\lambda$, while ${\partial }_{\lambda }{Q}^{\prime }<0$, the boundary $\lambda =0$ is also strictly increasing there. Hence ${\eta }_c>{\eta }_{\ast }$.

If $\eta \ge {\eta }_c$, a zero of $Q_{0,\eta }^{\prime }$ becomes strictly negative at the same x for every $\lambda >0$ by (34), so $\Lambda \left(\eta \right)=0$. If ${\eta }_{\ast }<\eta <{\eta }_c$, then $Q_{0,\eta }^{\prime }>0$ on $(0,\infty )$. Positivity persists for all sufficiently small positive $\lambda$. This follows from the local expansion near 0, the uniform tail estimate from the proof of Theorem 6.2, and compactness on the remaining middle interval. Thus $\Lambda \left(\eta \right)>0$.

Finally, put $\Psi (x,\lambda ,\eta )=Q_{\lambda ,\eta }^{\prime }\left(x\right)$. At $(x_c,0,{\eta }_c)$ one has $\Psi ={\Psi }_x=0$, ${\Psi }_{xx}>0$, ${\Psi }_{\lambda }<0$, and (40) gives ${\Psi }_{\eta }<0$. The Jacobian of $(\Psi ,{\Psi }_x)$ with respect to $(\lambda ,x)$ has determinant ${\Psi }_{\lambda }{\Psi }_{xx}\ne 0$. The analytic implicit function theorem continues the fold real-analytically through the boundary. Along this continuation,

$${\left.{\displaystyle \frac{d\lambda }{d\eta }}\right|}_{\eta ={\eta }_c}=-{\displaystyle \frac{{\Psi }_{\eta }}{{\Psi }_{\lambda }}}<0,$$

which proves transversality and shows that its positive-$\lambda$ branch is exactly the fold from Theorem 6.2 for $\eta <{\eta }_c$. □

Numerically, the defining equations give

$$y_c\approx 6.4201327966,x_c\approx 7.1769613328,{\eta }_c\approx 0.01574788007448.$$

These decimal values are not used in the proof.

In Theorem 6.2, pointwise parameter monotonicity combines with the parameter-uniform two-zero bound to order the critical-point regimes globally. The boundary analysis in Theorem 6.3 completes that classification.

For the polynomial model $u{x}^2+v{x}^4+d_0{x}^6$, the equal-minimum transition is the classical sextic transition in [7] [Theorem A.1]. The local fold follows directly from the derivative of the polynomial. The following estimate controls the full function away from the degenerate point.

Lemma 6.4.

At $({\lambda }_{\ast },{\eta }_{\ast })$, the function Q is strictly increasing on $[0,\infty )$. There are $r\in (0,1)$ and a neighborhood U of $(u,v)=(0,0)$ such that, for all $(u,v)\in U$, the following hold. Writing $G=q_y$, one has

$$\begin{array}{c}{G}_{yy}>0\mathit{on}[0,r],G_y\left(r\right)>0,\\ G>0\mathit{on}[r,\infty ),\underset{0\le y\le r}{sup}q(u,v,y)<q(u,v,1).\end{array}$$

(41)

In particular all positive critical points lie in $0<y<r$, and $x=1$ is the unique maximizer of Q on $[0,1]$.

Proof. 

Put again $p=p_{\lambda ,\eta }$, $D=1+p^2$, and $N=x{F}_{\lambda ,\eta }^{\prime }-3F_{\lambda ,\eta }+2x$. Then identity (33) applies. At (30), $c_0=c_1=0$. With $s=\sqrt{105}$ the remaining coefficients are as follows.

j	$c_j({\lambda }_{\ast },{\eta }_{\ast })$
2	$(27s-273)/3920$
3	$(149s-1505)/15680$
4	$(2717s-27305)/627200$
5	$(38201s-367045)/87808000$
6	$(162981-15473s)/87808000$
7	$(54635-5223s)/219520000$
8	$(183s-1861)/50176000$
9	$(1575-153s)/878080000$
10	$15{((s-7)/560)}^5$

They are all strictly positive. For the first eight rows this follows already from $256/25<s<41/4$. The displayed formula shows that the last row is positive. Consequently $N^{″}>0$ for $x>0$, whence $N^{\prime }>0$, $N>0$, and $Q^{\prime }>0$.

By (26), $G_{yy}(0,0,0)=6d_0>0$. Choose $r>0$ small enough that $G_{yy}>0$ on $[0,r]$ at the base point. Then $G_y(0,0,r)>0$. The strict increase already proved gives $N\left(\sqrt{r}\right)>0$, $N^{\prime }\left(\sqrt{r}\right)>0$, and $q(0,0,r)<q(0,0,1)$. These inequalities and the local derivative conditions persist in a parameter neighborhood.

For the unbounded tail, choose $C>0$ and shrink the same neighborhood so that $c_2\ge C$, $c_j>0$ for $j\ge 3$, and $|c_0|\le C{r}^2/4$, $|c_1|\le Cr/4$. This is possible because $c_0=c_1=0$ and $c_j>0$ for $j\ge 2$ at the base point. For $y\ge r$,

$${\displaystyle \sum _{j=0}^{10}}{c}_j{y}^j\ge C{y}^2-{\displaystyle \frac{Cr}{4}}y-{\displaystyle \frac{C{r}^2}{4}}\ge {\displaystyle \frac{C}{2}}{y}^2>0.$$

Thus $N^{″}>0$ for $x\ge \sqrt{r}$. Starting from the positive values of $N,N^{\prime }$ at $\sqrt{r}$, it follows that $N>0$ on the entire tail, hence $G>0$ on $[r,\infty )$. The strict gap between $q(0,0,r)$ and $q(0,0,1)$, together with uniform continuity on $[0,r]$, gives the last inequality in (41). This and tail increase prove that $x=1$ is the unique maximum on $[0,1]$. □

6.1. The Competing-Minimum Transition

Theorem 6.5.

In a sufficiently small neighborhood of $(u,v)=(0,0)$, the complete critical-point classification of Q on $[0,\infty )$ is the following.

1. 

If $v\ge 0$, then Q is strictly increasing for $u\ge 0$. For $u<0$ it has one interior critical point, a strict global minimum.

2. 

There is an analytic function h near zero satisfying

$$h\left(v\right)={\displaystyle \frac{v^2}{3d_0}}+O\left(v^3\right).$$

(42)

For $v<0$, the critical-point structure is

Parameter condition	Shape as x increases
$u>h\left(v\right)$	strictly increasing
$u=h\left(v\right)$	strictly increasing, one stationary inflection
$0<u<h\left(v\right)$	increasing, decreasing, increasing
$u\le 0$	decreasing, increasing

The middle open regime has exactly two interior critical points. They are a strict local maximum followed by a strict local minimum. All positive critical points away from $u=h\left(v\right)$ are simple zeros of $Q^{\prime }$.

3. 

There are analytic functions $m,w$ near zero satisfying

$$m\left(v\right)={\displaystyle \frac{v^2}{4d_0}}+O\left(v^3\right),w\left(v\right)=-{\displaystyle \frac{v}{2d_0}}+O\left(v^2\right).$$

(43)

For sufficiently small $v<0$, one has $0<m\left(v\right)<h\left(v\right)$ and $w\left(v\right)>0$. The global minimum for such v is attained at

$$\underset{x\ge 0}{\mathrm{arg\; min}}Q\left(x\right)=\left\{\begin{array}{cc}\left\{x_{+}(u,v)\right\},\hfill & u<m\left(v\right),\hfill \\ \{0,\sqrt{w\left(v\right)}\},\hfill & u=m\left(v\right),\hfill \\ \left\{0\right\},\hfill & u>m\left(v\right),\hfill \end{array}\right.$$

(44)

where $x_{+}$ denotes the interior minimum when it exists. On $[0,1]$ the unique maximum is $x=1$ throughout the neighborhood.

4. 

Let $A(u,v)={min}_{x\ge 0}Q$ be the optimal lower cubic coefficient. At every point $u=m\left(v\right)$ with small $v<0$, A is continuous but not differentiable as a function of u with v fixed. Its one-sided derivatives satisfy

$$A_u(m\left(v\right)-,v)-A_u(m\left(v\right)+,v)=w\left(v\right)+O\left(w{\left(v\right)}^3\right)>0.$$

(45)

The minimizing point has a discontinuity of magnitude $\sqrt{w\left(v\right)}$ across this curve.

The neighborhood can be chosen to have $\lambda (u,v)>0$ and $\eta (u,v)>0$ throughout.

Proof. 

By (26), with analytic remainders,

$$\begin{array}{cc}\hfill q-q_0& =uy+v{y}^2+d(u,v)y^3+y^{4}R(u,v,y),\hfill \end{array}$$

(46)

$$\begin{array}{cc}\hfill G& =u+2vy+3d(u,v)y^2+O\left(y^3\right),\hfill \end{array}$$

(47)

where $d(0,0)=d_0$. In particular $G\left(0\right)=u$ and $G_y\left(0\right)=2v$ exactly. Use r and U from Lemma 6.4.

If $v\ge 0$, then $G_y\left(y\right)>0$ for $0<y\le r$ because $G_{yy}>0$. Since $G\left(r\right)>0$, G has no zero when $u\ge 0$ and exactly one simple zero when $u<0$. Its sign gives assertion (1).

For $v<0$, $G_y\left(0\right)<0<G_y\left(r\right)$ and $G_{yy}>0$ give a unique minimum of G at $y=z(u,v)\in (0,r)$. The analytic implicit function theorem, applied to $G_y=0$ at $(u,v,y)=(0,0,0)$, also defines z analytically near the origin. In particular $z(u,0)=0$ and

$$z(u,v)=-{\displaystyle \frac{v}{3d_0}}+O\left(\left|v\right|\left(\right|u|+|v\left|\right)\right).$$

Set $M(u,v)=G(u,v,z(u,v\left)\right)$. One has $M(0,0)=0$ and $M_u(0,0)=1$, so the equation $M=0$ has a unique analytic solution $u=h\left(v\right)$ near the origin. Expanding gives $h\left(0\right)=h^{\prime }\left(0\right)=0$ and (42). Moreover $M_u>0$ nearby. Thus M has the sign of $u-h\left(v\right)$. Combining this with $G\left(0\right)=u$, $G\left(r\right)>0$, and strict convexity of G proves all four cases in assertion (2). At $u=h\left(v\right)$ the single zero of G has multiplicity two, so q and Q have a stationary inflection rather than an extremum. Every other zero is simple. There are no additional critical points by Lemma 6.4.

To compare the boundary value with the interior minimum, use the analytic divided differences

$$H(u,v,y)={\displaystyle \frac{q(u,v,y)-q_0(u,v)}{y}},K(u,v,y)={\displaystyle \frac{G(u,v,y)-H(u,v,y)}{y}}.$$

They extend analytically to $y=0$, with expansions

$$H=u+vy+d(u,v)y^2+O\left(y^3\right),K=v+2d(u,v)y+O\left(y^2\right).$$

The Jacobian of $(H,K)$ with respect to $(u,y)$ at the origin is $\mathrm{diag}(1,2d_0)$. The implicit function theorem therefore gives analytic solutions $u=m\left(v\right)$ and $y=w\left(v\right)$ of $H=K=0$. The expansions give (43), and consequently $0<m\left(v\right)<h\left(v\right)$ and $w\left(v\right)>0$ when $v<0$ is small. Since $G=H+yK=0$ at these points and

$$G_y(m\left(v\right),v,w\left(v\right))=-v+O\left(v^2\right)>0,$$

they represent the interior local minimum, with $q=q_0$.

For fixed $v<0$, let $y_{+}(u,v)$ be this minimum branch on $0<u<h\left(v\right)$ and write $E(u,v)=q(u,v,y_{+}(u,v))-q_0(u,v)$. Differentiation is legitimate because the critical point is simple. At the critical point $q_y=0$, and (46) gives

$$E_u=q_u(u,v,y_{+})-{\left(q_0\right)}_u=y_{+}+O\left(y_{+}^3\right)>0.$$

(48)

This positivity is uniform after decreasing r and shrinking the parameter neighborhood. As $E\left(m\right(v),v)=0$, the sign of E is the sign of $u-m\left(v\right)$. For $u\le 0$ the local profile decreases immediately to the right of zero and has a unique interior minimum. For $u\ge h\left(v\right)$ it is strictly increasing. In the two-critical-point regime the only candidates for the global minimum are 0 and $\sqrt{y_{+}}$, because q is increasing after its local minimum. This proves (44). The maximum on $[0,1]$ follows from Lemma 6.4.

All minimizing points lie in $0\le y<r$. Thus A is the minimum of q over the fixed compact interval $[0,r]$, and its continuity follows from uniform continuity on compact parameter rectangles. On the two sides of $u=m\left(v\right)$ the active branches are $q(u,v,y_{+})$ and $q_0(u,v)$. Equation (48) at $y_{+}=w\left(v\right)$ gives (45). The limiting minimizing points on these two sides are $\sqrt{w\left(v\right)}>0$ and 0, respectively. Finally (30) and continuity give positivity of both original parameters in a sufficiently small neighborhood. □

The same lower coefficient is optimal on $[0,1]$ and $[0,\infty )$, because all minimizing points in Theorem 6.5 lie in $[0,1)$. On the finite interval the sharp bounds are

$$\begin{array}{cc}\hfill x+A(u,v)x^3& \le F_{\lambda (u,v),\eta (u,v)}\left(x\right)\hfill \\ & \le x+\left(\mathrm{arsinh}(1+\lambda (u,v)+\eta (u,v\left)\right)-1\right)x^3,0\le x\le 1.\hfill \end{array}$$

At $u=m\left(v\right)$ the normalized quotient attains its minimum both at 0 and at $\sqrt{w\left(v\right)}$. Equality of the original bounds at $x=0$ does not imply equality of their coefficients.

The neighborhood can also be chosen with $\lambda <1/6$ and $\eta <1/120$. The positive Taylor coefficients of sinh then give $p_{\lambda ,\eta }\left(x\right)<\sinh x$ for $x>0$, so $Q_{\lambda ,\eta }<0$. Since $\eta >0$,

$$Q_{\lambda ,\eta }\left(x\right)=-x^{-2}+O(\log x/x^3)⟶0(x\to \infty ).$$

Consequently the sharp bounds on the entire half-line are

$$x+A(u,v)x^3\le F_{\lambda (u,v),\eta (u,v)}\left(x\right)\le x.$$

The upper coefficient is the unattained supremum 0 of Q.

On the slice $v=0$, assertion (1) also gives

$$x_{+}(u,0)\sim {\left({\displaystyle \frac{-u}{3d_0}}\right)}^{1/4}(u\uparrow 0,u<0).$$

(49)

The critical equation is $u+3d(u,0)y^2+O\left(y^3\right)=0$. Putting $u=-t^2$ and $y=tz$ and applying the implicit function theorem at $z={\left(3d_0\right)}^{-1/2}$ proves (49). It is the higher-order counterpart of (20).

6.2. An Exact Rational Example

The following exact rational example exhibits the two-critical-point regime without requiring an explicit size for the neighborhood in Theorem 6.5.

Example 6.6.

For

$$\lambda ={\displaystyle \frac{8081}{50000}}=0.16162,\eta ={\displaystyle \frac{290501}{50000000}}=0.00581002,$$

Q has exactly two positive critical points, with

$$0.10956<x_{-}<0.10957,0.50235<x_{+}<0.50236.$$

The first is a local maximum and the second is the unique global minimum on $[0,\infty )$.

Proof. 

Substitution into Appendix A gives $c_0>0$, $c_1<0$, and $c_j>0$ for $2\le j\le 10$. Descartes’ rule [1] [Proposition 1] bounds the number of positive zeros of $N^{″}$ by two, counting multiplicities. Since $N\left(0\right)=N^{\prime }\left(0\right)=0$, two applications of Rolle’s theorem give the same bound for N. Positive zeros of N of total multiplicity k, together with its zero at 0, force at least k positive zeros of $N^{\prime }$, and then of $N^{″}$. The exact bounds in Appendix B give

$$\begin{array}{cc}\hfill 6.3·{10}^{-14}& <N(1/10)<6.5·{10}^{-14},\hfill \\ \hfill -2.4·{10}^{-12}& <N(3/20)<-2.3·{10}^{-12},\hfill \\ \hfill 1.8·{10}^{-7}& <N(7/10)<1.9·{10}^{-7}.\hfill \end{array}$$

There are therefore exactly two simple roots, with derivative sign pattern $+,-,+$. The displayed root intervals are verified by the same exact estimates at their endpoints. Finally,

$$-1.40·{10}^{-8}<Q(1/2)-Q\left(0\right)<-1.38·{10}^{-8}$$

places the interior minimum below the boundary value. The function increases after its second critical point, so this minimum is global on the half-line. □

7. Conclusion

For cubic perturbations, the three thresholds a, $\mu$, and $\delta$ determine all changes in the formulas for the sharp coefficients on $[0,1]$. The minimizing point moves continuously, and the optimal lower coefficient is continuously differentiable. For the quintic family, the normalized remainder has at most two positive critical points for every $\lambda \ge 0$ and $\eta >0$, and the critical-point regimes occur in a fixed order along each $\eta$-slice. The positive fold exists exactly for ${\eta }_{\ast }<\eta <{\eta }_c$ and terminates transversely at the unique boundary point $(0,{\eta }_c)$, with ${\eta }_c$ characterized in Theorem 6.3. At $\eta ={\eta }_c$ the boundary profile has one double positive critical point, whereas for $\eta >{\eta }_c$ it has two simple positive critical points. Above $\lambda =2\eta +3/20$, the unique positive critical point is the global minimum on the positive half-line.

Near the explicit positive pair (30), an interior local minimum appears before it becomes global. Along the curve where its value equals the boundary value, the minimizing point changes discontinuously and the optimal lower coefficient is not differentiable in a transverse parameter direction. The local sixth-order polynomial pattern is classical [7] [Theorems A.1–A.2]. The competing-minimum curves remain local near $({\lambda }_{\ast },{\eta }_{\ast })$, whereas the critical-point ordering and the $\lambda =0$ endpoint are global.