A real root of a cubic equation for a stationary point

Question

Let us consider the quartic polynomial in $x$ \begin{equation} F(x) = (2 a p +2)x^4+ (6a(1-a)p^2+(6-12a)p-6)x^3 + p(2(a-2)(a-1)a p^2 + 3(5a^2-9a+2)p +12a-18)x^2 - p^2 ((a-2)(4a^2 - 7a +1)p +9a^2-25a+18)x + (a-2)(a-1)(2a-3)p^3, (0) \end{equation} where $0< a < 2$ and $p > 0$. We are interested in stationary points of the function $F(x)$ which obey the cubic equation \begin{equation} F'(x) = 4(2 a p +2)x^3+ 3(6a(1-a)p^2+(6-12a)p-6)x^2 + 2p(2(a-2)(a-1)a p^2 + 3(5a^2-9a+2)p +12a-18)x - p^2 ((a-2)(4a^2 - 7a +1)p +9a^2-25a+18) = 0. (1) \end{equation}

It is necessary to prove that in the interval $(1, + \infty)$ there no two different stationary points of $F(x)$ obeying equation (1).

The numerical analysis supports this statetment. Moreover, the cubic equation (1) has only one root in the interval $(1, + \infty)$ if and only if \begin{equation} F'(1) = (p+1)[(a-2)(3a -1)p^2 - 4(a+2)p - 10] < 0. \end{equation} This takes place, for instance, for $1/3 < a < 2$ and all $p > 0$.

Answer 1

Here is a way, as noted, if $F'(1)<0$ we find one root in $(1,+\infty)$ (proved later).

Assume $F'(1)\ge 0$, so necessarily $0<a<\dfrac{1}{3}$, for $[(2-a)(3a -1)p^2+4(a+2)p+10]\le 0$, $p$ should be outside of its roots, that is $$p\ge \dfrac{4(a+2)+\sqrt{136a^2-216a+144}}{2(1-3a)(2-a)}$$. It is not hard to prove that the last root is increasing in $a$ for $a\in (0,\frac{1}{3})$ (for example, proving $\dfrac{\sqrt{136a^2-216a+144}}{2(1-3a)}$ is increasing...). In case $a=0$, $p\ge 5$. Now we want to prove that for $x\ge 1$, $F''(x)\ge 0$, this implies the initial claim in question.

$$F''(x)=12(2 a p +2)x^2+ 6(6a(1-a)p^2+(6-12a)p-6)x + 2p(2(a-2)(a-1)a p^2 + 3(5a^2-9a+2)p +12a-18)\ge 0.$$

Since $x\ge 1$, for some fixed $x$ it is almost straight that the expression is increasing in $p$, (for example, rearranging the terms and taking $24apx^2+36a(1-a)p^2+36p-72ap+6p^2(5a^2-9a+2)+24ap-36p$ which is increasing in $p$, $p\ge 5$ ...).

Finally, for $p=5$ and $0<a<\frac{1}{3}$ everything can be proved to be positive, which completes the proof.

The other case can be proved as follows, if $F'(0)\ge 0$ and $F'(1)<0$ then from the degree of the polynomial there should be one negative root, one in $[0;1)$ and one in $(1;+\infty)$. Otherwise consider $F'(0)<0$ (product of the roots is positive), the only case to consider is when $F'$ has three roots on $(1;+\infty)$. So assume $F'(1)<0$ and there is two extremums to $F'$ on $(1;+\infty)$:

Case $1\le a<2$: We prove that $F''(1)<0$, this implies that one extremum of $F'$ is on $(-\infty;1)$ a contradiction: $$F''(1)=-24ap-12+36ap^2(1-a)+4ap^3(a-2)(a-1)+6p^2(5a^2-9a+2).$$ This can be proved decreasing in $p$ and for $p=0$ it is negative.

Case $0<a<1$: We take the sum of the roots of $F''(x)$ and prove it is less than $2$, that is $$\dfrac{-6(6a(1-a)p^2+(6-12a)p-6)}{12(2ap+2)}<2$$

a contradiction.

Answer 2

Here I present another proof of the Proposition under consideration (called here as Lemma) which has common points with previous Toni's proof.

Lemma. Let us consider the function $$F' = 4(2ap+2)x^3+3(6(1-a)ap^2+(6-12a)p-6)x^2+ $$ $$ 2p(2(a-2)(a-1)ap^2+3(5a^2-9a+2)p+12a-18)x $$ $$ - p^2((a-2)(4a^2-7a+1)p+9a^2-25a+18). \quad (1) $$
Here $0 < a < 2$ and $p > 0$. Then, the function $F'(x)$ has no more than one root in the interval $(1, + \infty)$.
Moreover, it has one root if $$(a) \ F'(1) = (p+1)[(a-2)(3a-1)p^2 - 4(a+2)p -10] < 0 $$ and no roots if
$$(b) \ F'(1) \geq 0. $$

Proof. First, we consider the case (b), i. e. when $$(2 - a)(3a-1)p^2 + 4(a+2)p + 10 \leq 0. \quad (2)$$ This relation implies $$0 < a < 1/3 $$
since otherwise (for $1/3 \leq a < 2$) the left hand side in (2) is positive.
Due to (2) or $$(2 - a)(1 - 3a)p^2 - 4(a+2)p - 10 \geq 0. \qquad (3a) $$ we get $$p \geq p(a) \equiv \frac{4(a +2) + \sqrt{16(a+2)^2 + 40(2 - a)(1 - 3a)}}{2(1-3a)(2-a)}. \quad (4)$$

The function $p(a)$ is monotonically increasing in interval $(0,1/3)$ since $$ \frac{ \sqrt{16(a+2)^2 + 40(2 - a)(1 - 3a)}}{(1-3a)} = $$ $$ \sqrt{16(a+2)^2 (1-3a)^{-2} + 40(2 - a)(1 - 3a)^{-1}} $$ is monotonically increasing in $a \in (0,1/3)$. We note that due to the limit $p(+0) = 5$ we obtain $$ p(a) > 5, \qquad (5a)$$
for all $a$ belonging to $(0,1/3)$.
Now we prove that in case (b) $$F''(x) = 12(2ap+2)x^2+6(6(1-a)ap^2+(6-12a)p-6)x+ $$ $$ 2p(2(a-2)(a-1)ap^2+3(5a^2-9a+2)p+12a-18) > 0 \quad (5)$$ for all $x > 1$, $0 < a < 1/3$ and $p \geq p(a)$, implying $$F'(x) > 0 \quad (6)$$ for all $x > 1$, $0 < a < 1/3$, $p \geq p(a)$, which proves ``one half'' of the Lemma for the case (b).

In order to prove inequality (5) it is sufficient to prove two inequalities $$V = 6(1-a)ap^2+(6-12a)p-6 > 0, \qquad (6V)$$ and $$Z = 2(a-2)(a-1)ap^2+3(5a^2-9a+2)p+12a-18 > 0 \quad (6Z)$$ for all $x > 1$, $0 < a < 1/3$, $p \geq p(a)$. Plugging the bound $$p^2 \geq \frac{(4(a+2)p + 10)}{(2 - a)(1 - 3a)},$$ following from (3a), into relation for $V$ we obtain $$V \geq V_b \equiv \frac{6(1-a)a (4(a+2)p + 10)}{(2 - a)(1 - 3a)} (6-12a)p-6 $$ $$= \frac{6(-10a^3p+13a^2p-3ap+2p-13a^2+17a-2)}{(2-a)(1 -3a)}, \qquad (7V) $$ and
$$Z \geq Z_b \equiv \frac{2(a-2)(a-1)a(4(a+2)p + 10)}{(2 - a)(1 - 3a)} $$ $$ +3(5a^2-9a+2)p +12a-18 $$ $$ = \frac{-53a^3p+88a^2p-29ap+6p-56a^2+86a-18}{1-3a}. \quad (7Z)$$ In order to prove the inequalities (6V) and (6Z) one needs (due to (7V) and (7Z)) to prove $$v = (-10a^3+13a^2-3a+2)p-13a^2+17a-2 > 0 \qquad (8v) $$ and $$z = (-53a^3 + 88a^2 - 29a + 6)p - 56a^2 + 86a - 18 > 0, \quad (8z)$$ for all $0 < a < 1/3$ and $p \geq p(a)$.

The first bound can be readily proved. Indeed, for the bracket in (8v) we have $$(-10a^3+13a^2-3a+2) > -10 \left(\frac{1}{3}\right)^3 - 3 \left(\frac{1}{3}\right) +2 =\frac{17}{27}$$ and hence (using (5a): $ p(a) > 5 $) we get $$ v > \frac{17}{27}5 - \frac{13}{9} - 2 = \frac{5}{27} > 0 \quad (9v)$$ Now we prove the second bound (8z). For the bracket in
(8z) we use the following inequality $$z_1(a) = (-53a^3 + 88a^2 - 29a + 6) > 3 > 0, \qquad (9z1)$$ for all $0 < a < 1/3$. This can be readily obtained from graphical analysis of the function $z_1(a)$ on interval $(0,1/3)$. This function reaches a minimum at the point $$a_{min} = -(\sqrt{3133}-88)/159 \approx 0.2014 \quad (10a)$$ (which could be obtained from quadratic equation $\frac{dz_1(a)}{da} = 0$) with the value $$z_1(a_{min}) = \frac{600698 - 2(3133)^{3/2}}{75843} \approx 3.296 > 3. \quad (10z1)$$

Since $z_1(a) > 0$, we obtain for $p \geq p(a)$ $$z = (-53a^3+88a^2-29a+6)p-56a^2+86a-18 $$, $$ \geq z_b (a) = (-53a^3 + 88a^2-29a+6) p(a) $$ $$ -56a^2+86a-18 > 12. \quad (11z) $$

The last bound (11z) just follows from the graphical analysis of the function $z_b (a)$ on the interval (0, 1/3). This function is monotonically increasing
from $(12_{+0})$ to $+ \infty$. Thus, relation (8z) is proved and hence the part (b) of the Lemma is also proved.

Now we consider the case (a) $F'(1) < 0$. Here we get two subcases $$(a1) \ \ 1/3 \leq a < 2, \quad p > 0, \qquad (12a1)$$ and $$(a2) \ \ 0 < a < 1/3, \qquad 0 < p < p(a), \qquad (12a2)$$ where $ p(a)$ is defined in (4).

Since $F'(1) < 0$ and $F'(+\infty) = + \infty$, then due to Intermediate Value Theorem for all $a$ and $p$ from (12a1) and (12a2) there exists at least one root $x_{*}$ of the cubic polynomial $F'(x)$ in the interval $(1, + \infty)$, i.e. $$F'(x_{*}) = 0, \qquad x_{*} > 1. \qquad (13)$$

Let us suppose that for some $a$ and $p$ there exists another root $x_{**} \neq x_{*}$,
i.e. $F'(x_{**}) = 0$. Without loss of generality we put $x_{*} < x_{**}$.
An elementary graphical analysis leads us to three variants for our cubic polynomial $F'(x)$: $$ i) F''(x_{*}) = 0, \quad F''(x_{**}) > 0, $$ $$ii) F''(x_{*}) > 0, \quad F''(x_{**}) = 0,$$ $$iii) F''(x_{*}) > 0, \quad F''(x_{**}) < 0, \quad F''(x_{***}) > 0, \quad (14)$$

where $x_{***}$ is the third root: $F'(x_{***}) = 0$, obeying (without loss of generality) $x_{***} > x_{**} $.

By Rolle's theorem we find that in all three cases there exist two different roots $x_1$, $x_2$ of quadratic polynomial
$$ H = F''(x) = 0, \qquad (15) $$ (see (5)), which satisfy $x_1 \neq x_2$ and $$F''(x_1) = F''(x_2) = 0. \qquad (16) $$ By making the redefinition of our variable: $y = x - 1$, we rewrite the quadratic equation (15) as

$$H = H(y) = 24(ap+1)y^2+ 12(3a(1-a)p^2+(3-2a)p+1)y $$ $$ +(4(a - 1)(a - 2)ap^3+ 6(-a^2 - 3a +2)p^2 -24ap -12 = 0. \quad (17)$$

This quadratic polynomial H(y) should have two different positive roots $y_1 = x_1 - 1 >0$ and $y_2 = x_2 - 1 > 0$. Due to Vieta's formulas and $ ap+1 >0$ we obtain the following inequalities: $$3a(1-a)p^2+(3-2a)p+1 < 0, \qquad (18B)$$ $$4(a - 1)(a - 2)ap^3+ 6(-a^2- 3a+2)p^2 -24ap-12 > 0. \quad (18C)$$ Due to (18B) we get $$1 < a < 2. \qquad (19)$$ It can be readily seen that for this values of $a$ all coefficients
of the cubic polynomial in the left hand side of (18C) are negative and hence the inequality (18C) is not satisfied for all $p > 0$ and $1 < a < 2$. Thus, in case (a) we have only one root of cubic polynomial $F'(x)$ belonging to $(1, + \infty)$. This finishes the proof of the Lemma.