On the shear stress function and the critical value of the Blasius problem

Abstract

The Blasius problem has been used to describe the steady two-dimensional flow of a slightly viscous incompressible fluid past a flat plate moving at a constant speed β; and it is well known that there exists the critical value ${\beta }^{\ast }<0$ such that it has at least one solution for each $\beta \ge {\beta }^{\ast }$ and has no positive solution for $\beta <{\beta }^{\ast }$. The known numerical result shows ${\beta }^{\ast }\doteq -0.3541$. In this paper, by the study of the integral equation equivalent to the Blasius problem, we obtain the relation between the velocity function $f^{\prime }$ and the shear stress functions $f^{″}$, upper and lower bounds of $\parallel f^{″}\parallel$ and a new lower bound of ${\beta }^{\ast }$. In particular, $\sqrt[4]{27}/9\le \parallel f^{″}\parallel \le \sqrt{3}/3$, ${\beta }^{\ast }>-0.45$. Regarding ${\beta }^{\ast }$, previous results presented a lower bound −0.5 and an upper bound −0.18733.

1 Introduction

The Blasius problem [1] arising in the boundary layer problems in fluid mechanics

$$f^{‴}(\eta )+f(\eta )f^{″}(\eta )=0\text{on }[0,\mathrm{\infty })$$

(1.1)

subject to the boundary conditions

$$f(0)=0,f^{\prime }(0)=\beta \text{and}{f}^{\prime }(\mathrm{\infty })=1,$$

(1.2)

has been used to describe the steady two-dimensional flow of a slightly viscous incompressible fluid past a flat plate. It also arises in the study of the mixed convection in porous media [2], where η is the similarity boundary layer ordinate, $f(\eta )$ is the similarity stream function, $f^{\prime }(\eta )$ and $f^{″}(\eta )$ are the velocity and the shear stress functions, respectively. The case of $\beta <0$ corresponds to a flat plate moving at a steady speed opposite to that of a uniform mainstream [3].

Regarding the analytic study of the Blasius problem (1.1)-(1.2), Weyl [4] proved that (1.1)-(1.2) has one and only one solution for $\beta =0$; Coppel [5] studied the case of $\beta >0$; the cases of $0<\beta <1$ [6] and $\beta >1$ [7] were also investigated, respectively. Also, see [8]. In 1986, Hussaini and Lakin [9] indicated that there exists a critical value ${\beta }^{\ast }<0$ such that (1.1)-(1.2) has at least a solution for $\beta \ge {\beta }^{\ast }$ and no solution for $\beta <{\beta }^{\ast }$. A lower bound was presented with ${\beta }^{\ast }\ge -1/2=-0.5$ and numerical results showed ${\beta }^{\ast }\doteq -0.3541$ [9]. In 2008, Brighi, Fruchard and Sari [10] summarized historical study on the Blasius problem and analyzed the case $\beta <0$ in detail, in which the shape and the number of solutions were determined. In 2010, Yang [11] obtained an upper bound ${\beta }^{\ast }<-18733/{10}^6=-0.18733$. The Blasius problem is a special case of the Falkner-Skan equation, for $\beta =0$, we may refer to [12–15] for some recent results on the Falkner-Skan equation.

An open question is: What exactly is ${\beta }^{\ast }$? And what properties does the shear stress function $f^{″}(\eta )$ have? To our knowledge, there is little study on it. By the study of the integral equation that is equivalent to the Blasius problem, in this paper, we present the relation between $f^{\prime }$ and $f^{″}$, upper and lower bounds of $\parallel f^{″}\parallel$ and new lower bounds of ${\beta }^{\ast }$. In particular, $\sqrt[4]{27}/9\le \parallel f^{″}\parallel \le \sqrt{3}/3$, ${\beta }^{\ast }>-0.45$.

2 Upper and lower bounds of $f^{″}$

Noticing the basic fact in [10] that if f is a solution of (1.1)-(1.2), then $f^{″}>0$ for $\eta \in [0,\mathrm{\infty })$, we use the so-called Crocco transformation [9, 10], which consists of choosing $t=f^{\prime }$ as an independent variable and expressing $z=f^{″}$ as a function of t, to change (1.1)-(1.2) to the Crocco equation [10]

$$\frac{d^{2}z}{d{t}^2}=-\frac{t}{z},\beta \le t<1$$

(2.1)

with the boundary conditions

$$z^{\prime }(\beta )=0,z(1)=0.$$

(2.2)

Integrating (2.1) from β to t, we have by (2.2)

$$z^{\prime }(t)=-{\int }_{\beta }^t\frac{s}{z(s)}ds\text{on }[\beta ,1).$$

(2.3)

Integrating (2.3) from t to 1, we obtain the following integral equation that is equivalent to (1.1)-(1.2) [10, 11]:

$$z(t)={\int }_t^1\frac{s(1-s)}{z(s)}ds+(1-t){\int }_{\beta }^t\frac{s}{z(s)}ds:=Az(t)+(1-t)Bz(t),$$

(2.4)

where $z(t)\in C[\beta ,1]$ and $z(t)>0$ for $t\in [\beta ,1)$.

Since ${\beta }^{\ast }\ge -\frac{1}{2}$, our work is restricted to the case of $-\frac{1}{2}\le \beta <0$ and begins with the following lemma.

Theorem 2.1 Let z be a solution of (2.4), then

• (i) $\parallel z\parallel \le \sqrt{h(\beta )}$;

• (ii) $-\frac{{\beta }^3}{3\parallel z\parallel }\le z(0)\le \frac{3h(\beta )+{\beta }^3}{3\sqrt{h(\beta )}}$, where $\parallel z\parallel =max\{z(t):t\in [\beta ,1]\}$, $h(\beta )=\frac{1}{3}(1-3{\beta }^2-2{\beta }^3)$.

Proof Let $\tilde{t}\in [\beta ,1]$ such that $\parallel z\parallel =z(\tilde{t})=max\{z(t):t\in [\beta ,1]\}$. By (2.3), we know $z^{\prime }(t)>0$ for $t\in (\beta ,0]$, and then $z(t)$ is strictly increasing $(\beta ,0]$. This, together with $z(\tilde{t})>0$, implies $0<\tilde{t}<1$. From (2.1), we know $z^{″}(t)z(t)=-t$. Integrating this equality from β to $\tilde{t}$, we have

$${\int }_{\beta }^{\tilde{t}}{z}^{″}(s)z(s)ds=-{\int }_{\beta }^{\tilde{t}}sds=-\frac{{\tilde{t}}^2-{\beta }^2}{2}.$$

Noticing that $z^{\prime }(\tilde{t})=z^{\prime }(\beta )=0$, we obtain

$${\int }_{\beta }^{\tilde{t}}{z}^{″}(s)z(s)ds=z^{\prime }(s)z(s){|}_{\beta }^{\tilde{t}}-{\int }_{\beta }^{\tilde{t}}{z}^{\mathrm{\prime }2}(s)ds=-{\int }_{\beta }^{\tilde{t}}{z}^{\mathrm{\prime }2}(s)ds\le 0.$$

Consequently, $|\beta |\le \tilde{t}$.

• (i) From $z^{\prime }(\tilde{t})=-{\int }_{\beta }^{\tilde{t}}\frac{s}{z(s)}ds=-Bz(\tilde{t})$, we know $Bz(t)\ge Bz(\tilde{t})=0$ for $t\in [\tilde{t},1)$. This implies $Az(t)\le z(t)$, $t\in [\tilde{t},1)$.

By $A^{\prime }z(t)=-\frac{t(1-t)}{z(t)}$, we have

$$Az(t){(-Az(t))}^{\prime }\le z(t){(-Az(t))}^{\prime }\le t(1-t)\text{for }t\in [\tilde{t},1).$$

(2.5)

Noticing that $z(\tilde{t})=Az(\tilde{t})$ and $z(1)=0$, integrating (2.5) from $\tilde{t}$ to 1, we have by $|\beta |\le \tilde{t}$

$$\begin{array}{rcl}{\parallel z\parallel }^2& =& {[Az(\tilde{t})]}^2\le 2{\int }_{\tilde{t}}^{1}s(1-s)ds\le 2{\int }_{|\beta |}^{1}s(1-s)ds\\ =& \frac{1}{3}(1-3{\beta }^2-2{\beta }^3)=h(\beta ).\end{array}$$

• (ii) Integrating (2.3) from β to 0, we have

  $$z(0)=-{\int }_{\beta }^{0}Bz(t)dt+z(\beta )\ge {\int }_{\beta }^0{\int }_{\beta }^t\frac{-s}{\parallel z\parallel }dsdt+z(\beta )\ge -\frac{{\beta }^3}{3\parallel z\parallel },$$

which implies that the left inequality of (ii) holds.

Noticing that $|\beta |\le \tilde{t}$ and utilizing (2.1) and (2.2), we know

$$z(\tilde{t})-z(0)={\int }_0^{\tilde{t}}{\int }_t^{\tilde{t}}\frac{s}{z(s)}dsdt\ge {\int }_0^{|\beta |}{\int }_t^{|\beta |}\frac{s}{\parallel z\parallel }dsdt=-\frac{{\beta }^3}{3\parallel z\parallel }.$$

And then $z(0)\le \parallel z\parallel +\frac{{\beta }^3}{3\parallel z\parallel }\le \sqrt{h(\beta )}+\frac{{\beta }^3}{3\sqrt{h(\beta )}}=\frac{3h(\beta )+{\beta }^3}{3\sqrt{h(\beta )}}$. Hence, (ii) holds. □

Let

$$l(t)=\{\begin{array}{ll}\frac{1}{6\sqrt{h(\beta )}}(t+t^2-2{\beta }^3)(1-t),& t\in [0,1],\\ \frac{1}{6\sqrt{h(\beta )}}(3{\beta }^{2}t-t^3-2{\beta }^3),& t\in [\beta ,0)\end{array}$$

and

$$u(t)=\{\begin{array}{ll}3\sqrt{h(\beta )}(2ln\frac{2}{1+t}+(1-t)ln\frac{1+t}{1-t}),& t\in [0,1],\\ \frac{6h(\beta )+2{\beta }^3+3{\beta }^{2}t-t^3}{3\sqrt{h(\beta )}},& t\in [\beta ,0).\end{array}$$

Utilizing Theorem 2.1, we can obtain upper and lower bounds of z as follows.

Theorem 2.2 Let z be a solution of (2.4), then $l(t)\le z(t)\le u(t)$ for $t\in [\beta ,1]$.

Proof For $t\in [\beta ,0)$, we have by (2.3) and Theorem 2.1(i)

$$z^{\prime }(t)=-Bz(t)\ge {\int }_{\beta }^t\frac{-s}{\sqrt{h(\beta )}}ds=\frac{{\beta }^2-t^2}{2\sqrt{h(\beta )}}\text{on }[\beta ,0).$$

(2.6)

From this, we have

$$z(t)\ge {\int }_{\beta }^t\frac{{\beta }^2-s^2}{2\sqrt{h(\beta )}}ds+z(\beta )\ge \frac{1}{6\sqrt{h(\beta )}}(3{\beta }^{2}t-t^3-2{\beta }^3).$$

Hence, $z(t)\ge l(t)$ for $t\in [\beta ,0)$.

For $t\in [0,1]$, let $\epsilon >0$, we define a function $h_{\epsilon }(t)$ as follows:

$$h_{\epsilon }(t)=\frac{1}{\parallel z\parallel +\epsilon }(-\frac{{\beta }^3}{3}(1-t)+{\int }_0^{1}G(t,s)sds)\text{for }t\in [0,1],$$

where $G(t,s)$ is the Green function for $w^{″}(t)=0$ with boundary conditions $w(0)=0=w(1)$ defined by

$$G(t,s)=\{\begin{array}{ll}t(1-s),& 0\le t\le s\le 1,\\ s(1-t),& 0\le s\le t\le 1.\end{array}$$

(2.7)

Next, we prove $z(t)\ge h_{\epsilon }(t)$ for $t\in [0,1]$.

In fact, if there exists $t_0\in [0,1]$ such that $z(t_0)<h_{\epsilon }(t_0)$, since $h_{\epsilon }(1)=0=z(1)$, $h_{\epsilon }(0)=-\frac{{\beta }^3}{3(\parallel z\parallel +\epsilon )}<-\frac{{\beta }^3}{3\parallel z\parallel }\le z(0)$ by Theorem 2.1(ii), then $t_0\in (0,1)$ and there exists an interval $(a,b)⫅(0,1)$ such that

$$t_0\in (a,b),z(t)<h_{\epsilon }(t)\text{for }t\in (a,b),z(a)=h_{\epsilon }(a),z(b)=h_{\epsilon }(b).$$

Let $\phi (t)=h_{\epsilon }(t)-z(t)$, then $\phi (a)=0=\phi (b)$ and $\phi (t_0)>0$. Let $\xi \in (a,b)$ such that $\phi (\xi )=max\{\phi (t):t\in [a,b]\}$, then ${\phi }^{″}(\xi )\le 0$. On the other hand, we know easily that

$${\phi }^{″}(t)=h_{\epsilon }^{″}(t)-z^{″}(t)=-\frac{t}{\parallel z\parallel +\epsilon }+\frac{t}{z(t)}>0\text{for }t\in (a,b).$$

Then ${\phi }^{″}(\xi )>0$, a contradiction.

Taking $\epsilon \to 0$, we have

$$z(t)\ge \frac{1}{\parallel z\parallel }(-\frac{{\beta }^3}{3}(1-t)+{\int }_0^{1}G(t,s)sds)\text{for }t\in [0,1].$$

Since ${\int }_0^{1}G(t,s)sds=\frac{(t+t^2)(1-t)}{6}$, we have

$$z(t)\ge \frac{1}{6\parallel z\parallel }(t+t^2-2{\beta }^3)(1-t).$$

(2.8)

Theorem 2.1(i) leads to

$$z(t)\ge \frac{1}{6\sqrt{h(\beta )}}(t+t^2-2{\beta }^3)(1-t)=l(t)\text{for }t\in [0,1].$$

Finally, we prove $z(t)\le u(t)$. For $t\in [\beta ,0)$, integrating (2.6) from t to 0, we obtain $z(0)-z(t)\ge \frac{t^3-3{\beta }^{2}t}{6\sqrt{h(\beta )}}$. Then by Theorem 2.1(ii),

$$z(t)\le z(0)-\frac{t^3-3{\beta }^{2}t}{6\sqrt{h(\beta )}}\le \frac{3h(\beta )+{\beta }^3}{3\sqrt{h(\beta )}}-\frac{t^3-3{\beta }^{2}t}{6\sqrt{h(\beta )}}=u(t).$$

For $t\in [0,1)$, since $z(t)\ge l(t)\ge \frac{1}{6\sqrt{h(\beta )}}(t+t^2)(1-t):=l_0(t)$, we have

$$\begin{array}{rcl}z(t)& \le & {\int }_t^1\frac{s(1-s)}{z(s)}ds+(1-t){\int }_0^t\frac{s}{z(s)}ds\\ \le & {\int }_t^1\frac{s(1-s)}{l_0(s)}ds+(1-t){\int }_0^t\frac{s}{l_0(s)}ds\\ =& 6\sqrt{h(\beta )}({\int }_t^1\frac{1}{1+s}ds+(1-t){\int }_0^t\frac{1}{1-s^2}ds)\\ =& 3\sqrt{h(\beta )}(2ln\frac{2}{1+t}+(1-t)ln\frac{1+t}{1-t})=u(t).\end{array}$$

 □

Combining Theorems 2.1 and 2.2, we obtain

Corollary 2.1 Let z be a solution of (2.4), then $\sqrt[4]{h(\beta )}\sqrt{l(\frac{\sqrt{3+6{\beta }^3}}{3})}\le \parallel z\parallel \le \sqrt{h(\beta )}$. In particular, $\frac{\sqrt[4]{27}}{9}\le \parallel z\parallel \le \frac{\sqrt{3}}{3}$.

Proof Let $g_{\beta }(t)=(t+t^2-2{\beta }^3)(1-t)$, by (2.8), we have $z(t)\ge \frac{1}{6\parallel z\parallel }{g}_{\beta }(t)$ on $[0,1]$. From $g_{\beta }^{\prime }(\hat{t})=0$, we obtain $\hat{t}=\frac{\sqrt{3+6{\beta }^3}}{3}$, and then $max\{g_{\beta }(t):t\in [0,1]\}=g_{\beta }(\frac{\sqrt{3+6{\beta }^3}}{3})$. Hence, $\parallel z\parallel \ge \frac{1}{6\parallel z\parallel }{g}_{\beta }(\frac{\sqrt{3+6{\beta }^3}}{3})$. This, together with $g_{\beta }(\frac{\sqrt{3+6{\beta }^3}}{3})=6\sqrt{h(\beta )}l(\frac{\sqrt{3+6{\beta }^3}}{3})$, implies $\parallel z\parallel \ge \sqrt[4]{h(\beta )}\sqrt{l(\frac{\sqrt{3+6{\beta }^3}}{3})}$. The right hand is from Theorem 2.1(i).

Since $h(\sigma )\le h(0)=\frac{1}{3}$ for $\sigma \ge 0$, hence $\parallel z\parallel \le \sqrt{h(0})=\frac{\sqrt{3}}{3}$.

Since $g_{\beta }(t)\ge t(1+t)(1-t):=g_0(t)$ for $t\in [0,1]$, by $g_0^{\prime }(\hat{t})=0$, we have $\hat{t}=\frac{\sqrt{3}}{3}$. Hence, $max\{g_0(t):t\in [0,1]\}=g_0(\hat{t})=\frac{2\sqrt{3}}{9}$. By (2.8), we obtain

$$\parallel z\parallel \ge z(t)\ge \frac{g_0(t)}{6\parallel z\parallel }\text{on }[0,1].$$

From this, we have $\parallel z\parallel \ge \frac{g_0(\hat{t})}{6\parallel z\parallel }=\frac{\sqrt{3}}{27\parallel z\parallel }$, i.e., $\parallel z\parallel \ge \frac{\sqrt[4]{27}}{9}$. □

Based on Theorem 2.2, Corollary 2.1, $f^{\prime }(\eta )=t$ and $f^{″}(\eta )=z(t)$, we obtain the relation between the velocity function $f^{\prime }$ and the shear stress functions $f^{″}$, upper and lower bounds of $\parallel f^{″}\parallel =sup\{f^{″}(\eta ):\eta \in [0,\mathrm{\infty }\}$.

Theorem 2.3 Let f be a solution of (1.1)-(1.2), then

• (i) $l(f^{\prime })\le f^{″}\le u(f^{\prime })$ for $\eta \in [0,\mathrm{\infty })$;

• (ii) $\sqrt[4]{h(\beta )}\sqrt{l(\frac{\sqrt{3+6{\beta }^3}}{3})}\le \parallel f^{″}\parallel \le \sqrt{h(\beta )}$. Specially, $\frac{\sqrt[4]{27}}{9}\le \parallel f^{″}\parallel \le \frac{\sqrt{3}}{3}$.

Remark 2.1 There exists very little study on the upper and lower bounds of $f^{″}(\eta )$; Theorem 2.3 fills this gap. Other studies can be found in [9, 10].

3 New lower bound of ${\beta }^{\ast }$

To obtain a better lower bound of ${\beta }^{\ast }$, we first prove

Theorem 3.1 Let z be a solution of (2.4), then

$${\int }_{\beta }^1(1-s)z^{\mathrm{\prime }2}(s)ds\ge \frac{1}{1440h(\beta )}(60{\beta }^6-192{\beta }^5+7).$$

Proof Firstly, we prove

$$|z^{\prime }(t)|\ge \frac{|t^2-{\tilde{t}}^2|}{2\sqrt{h(\beta )}}\text{on }[0,1].$$

(3.1)

Integrating (2.1) from t to $\tilde{t}$, we obtain by $z^{\prime }(\tilde{t})=0$

$$z^{\prime }(t)={\int }_t^{\tilde{t}}\frac{s}{z(s)}ds\ge {\int }_t^{\tilde{t}}\frac{s}{\sqrt{h(\beta )}}ds=\frac{{\tilde{t}}^2-t^2}{2\sqrt{h(\beta )}}\text{on }[0,\tilde{t}).$$

Integrating (2.1) from $\tilde{t}$ to t, we have by $z^{\prime }(\tilde{t})=0$

$$z^{\prime }(t)={\int }_{\tilde{t}}^t\frac{-s}{z(s)}ds\le {\int }_{\tilde{t}}^t\frac{-s}{\sqrt{h(\beta )}}ds=\frac{{\tilde{t}}^2-t^2}{2\sqrt{h(\beta )}}\le 0\text{on }[\tilde{t},1].$$

Hence, (3.1) holds.

From (2.6) and (3.1), we know

$$\begin{array}{rcl}{\int }_{\beta }^1(1-s)z^{\mathrm{\prime }2}(s)ds& =& {\int }_{\beta }^0(1-s)z^{\mathrm{\prime }2}(s)ds+{\int }_0^1(1-s)z^{\mathrm{\prime }2}(s)ds\\ \ge & \frac{1}{4h(\beta )}[{\int }_{\beta }^0(1-s){({\beta }^2-s^2)}^{2}ds+{\int }_0^1(1-s){(s^2-{\tilde{t}}^2)}^{2}ds].\end{array}$$

By

$$\begin{array}{c}{\int }_{\beta }^0(1-s){({\beta }^2-s^2)}^{2}ds=\frac{1}{30}(5{\beta }^6-16{\beta }^5),\hfill \\ {\int }_0^1(1-s){(s^2-{\tilde{t}}^2)}^{2}ds=\frac{1}{30}(1-5{\tilde{t}}^2+15{\tilde{t}}^4)\hfill \end{array}$$

and

$$1-5{\tilde{t}}^2+15{\tilde{t}}^4=15{({\tilde{t}}^2-\frac{1}{6})}^2+\frac{7}{12}\ge \frac{7}{12},$$

we obtain

$$\begin{array}{rl}{\int }_{\beta }^1(1-s)z^{\mathrm{\prime }2}(s)ds& \ge \frac{1}{120h(\beta )}[(5{\beta }^6-16{\beta }^5)+\frac{7}{12}]\\ =\frac{1}{1440h(\beta )}(60{\beta }^6-192{\beta }^5+7).\end{array}$$

 □

Let

$$H(\beta )=73-480{\beta }^2+720{\beta }^4+192{\beta }^5-380{\beta }^6,\beta \in [-\frac{1}{2},0].$$

Since

$$\begin{array}{c}{H}^{\prime }(\beta )=-960\beta +2880{\beta }^3+960{\beta }^4-2280{\beta }^5>-960\beta (1-3{\beta }^2)>0\text{on }[-\frac{1}{2},0),\hfill \\ H(-\frac{9}{20})=-\frac{4396387}{3200000}<0,H(-\frac{11}{25})=\frac{55395529}{48828125}>0,\hfill \end{array}$$

then there exists a unique $\tilde{\beta }\in (-\frac{9}{20},-\frac{11}{25})=(-0.45,-0.44)$ such that $H(\tilde{\beta })=0$, $H(\beta )>0$ for $\beta \in (\tilde{\beta },0)$ and $H(\beta )<0$ for $\beta \in (-\frac{1}{2},\tilde{\beta })$.

Theorem 3.2 If $\beta \le \tilde{\beta }$, the Blasius problem (1.1)-(1.2) has no solution and then ${\beta }^{\ast }>-0.45$.

Proof The proof is by contradiction. If for some $\beta \le \tilde{\beta }$, (1.1)-(1.2) has a solution f and then (2.1) has a solution z. Rewrite (2.1) as follows:

$${(z(t)z^{\prime }(t))}^{\prime }+t=z^{\mathrm{\prime }2}(t).$$

Integrating this equality from β to t and noticing that $z^{\prime }(\beta )=0$, we obtain

$$z(t)z^{\prime }(t)+{\int }_{\beta }^{t}sds={\int }_{\beta }^t{z}^{\mathrm{\prime }2}(s)ds.$$

Integrating the last equality from β to 1 and using $z(1)=0$, we have

$$-\frac{1}{2}{z}^2(\beta )+{\int }_{\beta }^1{\int }_{\beta }^{t}sdsdt={\int }_{\beta }^1{\int }_{\beta }^t{z}^{\mathrm{\prime }2}(s)dsdt={\int }_{\beta }^1(1-s)z^{\mathrm{\prime }2}(s)ds.$$

This, together with $z(\beta )>0$ and Theorem 3.1, implies

$$\frac{1-3{\beta }^2+2{\beta }^3}{6}={\int }_{\beta }^1{\int }_{\beta }^{t}sdsdt>\frac{1}{1440h(\beta )}(60{\beta }^6-192{\beta }^5+7).$$

Since

$$\frac{1-3{\beta }^2+2{\beta }^3}{6}-\frac{1}{1440h(\beta )}(60{\beta }^6-192{\beta }^5+7)=\frac{H(\beta )}{1440h(\beta )},$$

then $H(\beta )>0$, a contradiction. Hence, ${\beta }^{\ast }>\tilde{\beta }>-0.45$. □

Remark 3.1 Theorem 3.2 improves the lower bound of ${\beta }^{\ast }$ from −0.5 in [9] to −0.45.