# Quenching for a Non-Newtonian Filtration Equation with a Singular Boundary Condition

Authors: Xiliu Li; Chunlai Mu; Qingna Zhang; Shouming Zhou
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

This paper deals with a nonlinearp-Laplacian equation with singular boundary conditions. Under proper conditions, the solution of this equation quenches in finite time and the only quenching point thatis x=1
are obtained. Moreover, the quenching rate of this equation is established. Finally, we give an example of an application of our results.


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## Body

## 1. Introduction

In this paper, we consider the following problem:(1.1)(ψ(u))t=(|ux|p-2ux)x,0<x<1,t>0,ux(0,t)=0,ux(1,t)=-g(u(1,t)),t>0,u(x,0)=u0(x),0≤x≤1,
where ψ(u) is a monotone increasing function with ψ(0)=0, p>1, g(u)>0, g′(u)<0 for u>0, and limu→0+g(u)=+∞. The initial value u0(x) is positive and satisfying some compatibility conditions.Ifψ(u)=u1/m with m>0, (1.1) becomes the well-known non-Newtonian filtration equation, which is used to describe the non-stationary flow in a porous medium of fluids with a power dependence of the tangential stress on the velocity of the displacement under polytropic conditions (see [1, 2]).Many papers have been devoted to the study of critical exponents of non-Newtonian filtration equation, see [3–5]. There are many results on the quenching phenomenon, see, for instance [6–12]. By the quenching phenomenon we mean that the solution approaches a constant but its derivative with respect to time variable t tends to infinity as (x,t) tends to some point in the spatial-time space. The study of the quenching phenomenon began with the work of Kawarada through the famous initial boundary problem for the reaction-diffusion equation: ut=uxx+1/(1-u) (see [13]).As an example of the type of results, we wish to obtain, let us recall results for a closely related problem(1.2)ut=uxx,0<x<1,t>0,ux(0,t)=0,ux(1,t)=-u-q(1,t),t>0,u(x,0)=u0(x),0≤x≤1,
where q>0. In [14], it was shown that u quenches in finite time for all u0, and the only quenching point is x=1. Furthermore, the behavior of u near quenching was described there. It is easily seen that (1.2) is a special case of (1.1).Ifp=2, then (1.1) reduces to the following equation:(1.3)(ψ(u))t=uxx,0<x<1,t>0,ux(0,t)=0,ux(1,t)=-g(u(1,t)),t>0,u(x,0)=u0(x),0≤x≤1.
Deng and Xu proved in [15] the finite time quenching for the solution and established results on quenching set and rate for (1.3). If ψ(u)=u, then (1.1) reduces to the following equation, see [11],(1.4)ut=(|ux|p-2ux)x,0<x<1,t>0,ux(0,t)=0,ux(1,t)=-g(u(1,t)),t>0,u(x,0)=u0(x),0≤x≤1,
and they obtained that the bounds for the quenching rate, and the quenching occurs only at x=1.In this paper, we extend the equationut=(|ux|p-2ux)x, see [11], to a more general form (ψ(u))t=(|ux|p-2ux)x. We prove that quenching occurs only at x=1. We determine the bounds for the quenching rate, and present an example which shows the applicability of our results.The main results are stated as follows.Theorem 1.1.
Suppose that the initial data satisfiesu0′(x)≤0 and u0′′(x)≤0 for 0≤x≤1, and one of the following conditions holds:(i)
ψ′′(u)>0 for u>0,(ii)
ψ′′(u)<0 for u>0, limsupu→0+(g′(u)/ψ′(u))<0.Then every solution of (1.1) quenches in finite time, and the only quenching point is x=1.Next, we deal with the quenching rate. Before we establish upper bounds for the quenching rate, we introduce the following hypothesis:(H1)
ψ′′(u)g(u)≥2(p-1)ψ′(u)g′(u), ψ′(u)[(p-2)g′2(u)+g(u)g′′(u)]≥ψ′′(u)g(u)g′(u).Theorem 1.2.
Suppose that the conditions of Theorem1.1 and the hypothesis (H1) hold. Then there exists a positive constant C1 such that
(1.5)∫0u(1,t)ψ′(s)ds-gp-1(s)g′(s)≤C1(T-t).Next, we will give the lower bound on the quenching rate, the derivation of which is in the spirit of [15]. We need the following additional hypotheses: there exists a constant σ(-∞<σ≤σ0=min{1,2-1/(p-1)}) such that(H2)
(g(p-1)(σ-1)(u)g′(u)/ψ′(u))′′<0,(H3)
(5p-2σp-2σ-6)(g(p-1)(σ-1)(u)g′(u))′ψ′(u)≥(p-1)(3-σ)g(p-1)(σ-1)(u)g′(u)ψ′′(u),(H4)
(g(p-1)(σ-1)-1(u)g′(u)/ψ′(u))′<0.Theorem 1.3.
Suppose that the hypotheses of Theorem1.1 hold. Furthermore, suppose that the hypotheses (H2)–(H4) hold. Then there exists a positive constant C2 such that
(1.6)∫0u(1,t)ψ′(s)ds-gp-1(s)g′(s)≥C2(T-t).Furthermore, if (H1) holds, then the quenching rates are
(1.7)C1(T-t)≥∫0u(1,t)ψ′(s)ds-gp-1(s)g′(s)≥C2(T-t).Next, as an application of the main results of this paper, we study the following concrete example:(1.8)(um)t=(|ux|p-2ux)x,0<x<1,t>0,ux(0,t)=0,ux(1,t)=-u-q(1,t),t>0,u(x,0)=u0(x),0≤x≤1,
where q>0, p>1, and m>0. We will verify that (1.8) satisfies the hypotheses (H1)–(H4), and we give the following theorem.Theorem 1.4.
Suppose thatu0′(x)≤0 and u0′′(x)≤0 for 0≤x≤1. Then the solution of (1.8) satisfies
(1.9)C4≤u(1,t)(T-t)-1/(m+pq+1)≤C3,
where C3 and C4 are positive constants.The plan of this paper is as follows. In Section2, we prove that quenching occurs only at x=1, that is the proof of Theorem 1.1. In Section 3, we derive the estimates for the quenching rate, that is the proof of Theorems 1.2 and 1.3. In Section 4, we present results for certain ψ(u) and g(u), that is the proof of the Theorem 1.4.


## 2. Quenching on the Boundary

In this section, we prove finite time quenching. We rewrite problem (1.1) into the following form:(2.1)ut=a(u)(|ux|p-2ux)x,0<x<1,t>0,ux(0,t)=0,ux(1,t)=-g(u(1,t)),t>0,u(x,0)=u0(x),0≤x≤1,
where a(u)=1/(ψ′(u)). Clearly, ψ′(u)≠0 for u>0.Lemma 2.1.
Assume the solutionu of problem (2.1) exists in (0,T0) for some T0>0, and u0′(x)≤0, u0′′(x)≤0 for 0≤x≤1. Then ux(x,t)<0 and ut(x,t)<0 in (0,1]×(0,T0).Proof.
Letv(x,t)=ux(x,t). Then v(x,t) satisfies
(2.2)vt=a(u)(|v|p-2v)xx+a′(u)v(|v|p-2v)x,0<x<1,0<t<T0,v(0,t)=0,v(1,t)=-g(u(1,t)),0<t<T0,v(x,0)=u′0(x),0≤x≤1.The maximum principle leads to v(x,t)<0, and thus ux(x,t)<0 in (0,1]×(0,T0). Then it is easy to see that the problem (2.2) is nondegenerate in (0,1]×(0,T0). So ux(x,t) is a classical solution of (2.2). Similarly, letting w(x,t)=ut(x,t), we have
(2.3)wt=a′(u)(|ux|p-2ux)xw+(p-1)a(u)(|ux|p-2wx)x,0<x<1,0<t<T0,wx(0,t)=0,wx(1,t)=-g′(u(1,t))w(1,t),0<t<T0,w(x,0)=(p-1)a(u0(x))|u′0(x)|p-2u′′0(x),0≤x≤1.Making use of the maximum principle, we obtain ut(x,t)<0 in (0,1]×(0,T0). Hence, the solutions of problem (2.1) u∈C2,1((0,1]×(0,T0)) with ux(x,t)<0 and ut(x,t)<0 in (0,1]×(0,T0).The Proof of Theorem1.1
By the maximum principle, we know that0<u(·,t)≤M for all t in the existence interval, where M=max0≤x≤1u0(x). Define F(t)=∫01ψ(u(x,t))dx. Then F(t) satisfies
(2.4)F′(t)=∫01(ψ(u(x,t)))tdx=∫01(|ux|p-2ux)xdx=-g(u(1,t))|g(u(1,t))|p-2≤-gp-1(M).
Thus F(t)≤F(0)-gp-1(M)t, which means that F(t0)=0 for some t0>0. From the fact that ψ(u)>0 for u>0 and ux(x,t)<0 for 0<x≤1, we find that there exists a T(0<t≤T0) such that limt→T-u(1,t)=0. By virtue of the singular nonlinearity in the boundary condition, u must quench at x=1. In what follows, we only need to prove that quenching cannot occur in ((1/2),1)×(η,T) for some η(0<η<T). Consider two cases.Case 1.
ψ′′(u)>0 for u>0. Let h(x,t)=|ux|p-2ux+ε(x-(1/4))gp-1(M) in ((1/4),1)×(η,T), where ε is a positive constant. Then h(x,t) satisfies
(2.5)ht=(|ux|p-2ux)t=(p-1)|ux|p-2uxt=(p-1)|ux|p-2[a(u)(|ux|p-2ux)x]x=(p-1)|ux|p-2(a′(u)uxuta(u)+a(u)hxx)≤(p-1)a(u)|ux|p-2hxx,
for (x,t)∈((1/4),1)×(η,T), since (a′(u))/(a(u))=-(ψ′′(u))/(ψ′(u))≤0. On the parabolic boundary, h((1/4),t)=|ux|p-2ux((1/4),t)<0 for η≤t<T; if ε is sufficiently small, h(1,t)≤gp-1(M)((3ε/4)-1)<0 for η≤t<T, and h(x,η)≤-|ux((1/4),η)|p-1+(3ε/4)gp-1(M)<0 for (1/4)≤x≤1. Thus by the maximum principle, we have h(x,t)≤0 in ((1/4),1)×(η,T), which leads to
(2.6)|ux|p-2ux+ε(x-14)gp-1(M)≤0in(14,1)×(η,T).
So we have
(2.7)[ε(x-14)gp-1(M)]1/(p-1)≤-ux.
Integrating (2.7) from x to 1, we obtain
(2.8)u(x,t)≥u(1,t)+∫x1[ε(x-14)gp-1(M)]1/(p-1)dx>∫x1[ε(x-14)gp-1(M)]1/(p-1)dx>0.
It then follows that u(x,t)>0 if x<1.Case 2.
ψ′′(u)<0 for u>0. Let k(x,t)=ut-ε(x-(1/2))ux in ((1/2),1)×(η,T). Then k(x,t) satisfies
(2.9)kt=utt-ε(x-12)utx=(p-1)a(u)|ux|p-2kxx+(p-1)(p-2)a(u)|ux|p-2(-uxx)kx+a′(u)(|ux|p-2ux)xk+εp(p-1)a(u)|ux|p-2uxx≤(p-1)a(u)|ux|p-2kxx+(p-1)(p-2)a(u)|ux|p-2(-uxx)kx+a′(u)(|ux|p-2ux)xk,
for (x,t)∈((1/2),1)×(η,T). On the boundary, k((1/2),t)=ut((1/2),t)<0 for η≤t<T. Since limt→T-u(1,t)=0 and limsupu→0+((g′(u))/(ψ′(u)))<0, if η is close to T and ε is small enough, g′(u(1,t))+(εψ(u(1,t))′/2(p-1)g(p-2)u(1,t))+2≤0 for η≤t<T. Thus
(2.10)kx(1,t)=utx(1,t)-ε2uxx(1,t)-εux(1,t)=-[g′(u(1,t))+εψ′(u(1,t))2(p-1)gp-2(u(1,t))+2]ut(1,t)+2k(1,t)≤2k(1,t).
It is easily seen that k(x,η)≤0 for (1/2)≤x≤1. Hence, the maximum principle yields that k(x,t)≤0 in [(1/2),1]×[η,T). In particular,
(2.11)ut(1,t)+ε2g(u(1,t))≤0forη≤t<T.
Integrating (2.11) from t to T, we obtain
(2.12)∫0u(1,t)dsg(s)≥ε2(T-t)forη≤t<T.
Define G(u)=∫0u(1/g(s))ds. Since G′(u)=1/g(u)>0 for u>0, the inverse G-1 exists. In view of (2.12), we can see
(2.13)u(1,t)=G-1(G(u(1,t)))≥G-1(ε2(T-t))forη≤t<T.
Let H(x,t)=u(x,t)-c1(1-x2)-c2G-1((ε/2)(T-t)), where c1 and c2 are positive constants. Since g′(u)<0, ux(x,t)<0, and by (2.13), we have that in (0,1)×(η,T)(2.14)Ht=(p-1)a(u)|ux|p-2uxx+εc22g(G-1(ε2(T-t)))≥(p-1)a(u)|ux|p-2Hxx-2c1(p-1)a(u)|ux|p-2+εc22g(u(1,t))≥(p-1)a(u)|ux|p-2Hxx-2c1(p-1)|ux|p-2ψ′(M)+εc22g(M)≥(p-1)a(u)|ux|p-2Hxx,provided εc2g(M)ψ′(M)≥4c1(p-1)|ux|p-2, which is true since 0≤|ux(x,t)|≤g(u(1,t))=g(M0), where 0<M0≤M. On the other hand, Hx(0,t)=0; H(1,t)=u(1,t)-c2G-1((ε/2)(T-t))≥0 if c2≤1 and H(x,η)≥0 if c1 and c2 are small enough. Thus by the maximum principle, we find that H(x,t)≥0 in (0,1)×(η,T), which implies that u(x,t)≥c1(1-x2)>0 if x<1.


## 3. Bounds for Quenching Rate

In this section, we establish bounds on the quenching rate. We first present the upper bound.The Proof of Theorem1.2
We define a functionΦ(x,t)=|ux|p-2ux+φp-1(x)gp-1(u(x,t)) in (0,1)×(η,T), where φ(x) is given as follows:
(3.1)φ(x)={0,x∈[0,x0],(x-x0)l(1-x0)l,x∈(x0,1],
with some x0<1 and l≥max{3,(1/(p-1))} is chosen so large that φ(x)≤-(u0′(x)/g(u0(x)) for x0<x≤1. It is easy to see that Φ(0,t)=Φ(1,t)=0, and Φ(x,0)≤0. On the other hand, in (0,1)×(η,T), Φ satisfies
(3.2)Φt=(p-1)a(u)|ux|p-2Φxx+(p-1)a′(u)ux|ux|p-2Φx-R1(x,t),
where
(3.3)R1(x,t)=(p-1)2a(u)|ux|p-2φp-3(x)gp-1(u)[(p-2)ψ′2(x)+φ(x)φ′′(x)]+(p-1)2a2(u)|ux|p-2uxφp-2(x)φ′(x)gp-2(u)[-g(u)ψ′′(u)+2(p-1)g′(u)ψ′(u)]+(p-1)2a(u)|ux|pφp-1(x)gp-3(u)[-ψ′′(u)ψ′(u)g(u)g′(u)+(p-2)g′2(u)+g(u)g′′(u)].
By (H1) and the definition of φ(x), it follows that
(3.4)Φt≤(p-1)a(u)|ux|p-2Φxx+(p-1)a′(u)ux|ux|p-2Φx.
Thus, the maximum principle yields Φ(x,t)≤0, that is
(3.5)φ(x)g(u(x,t))≤-ux(x,t),for[0,1]×[η,T).
Moreover, by the definition of the limit, we see that Φx(1,t)≥0 since Φ(x,t)≤0. In fact,
(3.6)Φx(1,t)=lim⁡x→1-Φ(x,t)-Φ(1,t)x-1≥0,
which means
(3.7)ut(1,t)≥(p-1)a(u(1,t))gp-1(u(1,t))(g′(u(1,t))-φ′(1))≥c3(p-1)a(u(1,t))gp-1(u(1,t))g′(u(1,t)).
Integrating (3.7) from t to T, we get
(3.8)∫0u(1,t)ψ′(s)ds-gp-1(s)g′(s)≤C1(T-t).
We then give the lower bound.The Proof of Theorem1.3
Letd(u)=a(u)g(p-1)(σ-1)(u)g′(u). Notice that the hypotheses (H2)–(H4) are equivalent to(H̃2)
d′′(u)≤0,(H̃3)
d′(u)(5p-2σp+2σ-6)≥(d(u)/a(u))a′(u)(2p-σp+σ-3),(H̃4)
d(u)g′(u)>d′(u)g(u),respectively. Letting τ be close to T, we consider Ψ(x,t)=ut-εd(u)(-ux)(p-1)(2-σ) in (1-T+τ,1)×(τ,T), where ε is a positive constant. Through a fairly complicated calculation, we find that
(3.9)Ψt=(p-1)a(u)|ux|p-2Ψxx+(p-1)(p-2)a(u)|-ux|p-3(-uxx)Ψx+C(x,t)Ψ+R2(x,t),
where
(3.10)C(x,t)=[a′(u)a(u)+ε(2-σ)d(u)a(u)(2p-σp+σ-3)(-ux)p-σp+σ-2]ut+ε2(2-σ)d2(u)a(u)(2p-σp+σ-3)(-ux)3p-2σp+2σ-4+ε[d′(u)(5p-2σp+2σ-6)-d(u)a(u)a′(u)(2p-σp+σ-3)](-ux)2(p-1)(2-σ),R2(x,t)=ε3(2-σ)d3(u)a(u)(2p-σp+σ-3)(-ux)5p-3σp+3σ-6+ε2d(u)[d′(u)(5p-2σp+2σ-6)-d(u)a(u)a′(u)(2p-σp+σ-3)](-ux)2(p-1)(2-σ)+εa(u)(p-1)d′′(u)(-ux)(p-1)(3-σ)+1.
Since the hypotheses (H̃2)–(H̃3) hold, and d(u)<0, a(u)>0, we see that R2(x,t)<0. Thus, we have
(3.11)Ψt≤(p-1)a(u)|ux|p-2Ψxx+(p-1)(p-2)a(u)(ux)p-3(-uxx)Ψx+C(x,t)Ψ,
for (x,t)∈(1-T+τ,1)×(τ,T). On the parabolic boundary, since x=1 is the only quenching point, if ε is small enough, then both Ψ(1-T+τ,t) and Ψ(x,τ) are negative. At x=1, in view of (H̃4), we have
(3.12)Ψx(1,t)=-[1-ε(2-σ)]g′(u(1,t))Ψ(1,t)-ε{[1-ε(2-σ)]g′(u(1,t))d(u(1,t))-d′(u(1,t))g(u(1,t))}g(p-1)(2-σ)(u(1,t))≤-[1-ε(2-σ)]g′(u(1,t))Ψ(1,t),
provided ε is sufficiently small. Hence, by the maximum principle, we have Ψ(x,t)≤0 on [1-T+τ,1]×[τ,T). In particular, Ψ(1,t)≤0, that is,
(3.13)ut(1,t)≤εd(u(1,t))g(p-1)(2-σ)(u(1,t))=εa(u(1,t))gp-1(u(1,t))g′(u(1,t)).
Integration of (3.13) over (t,T), then leads to
(3.14)∫0u(1,t)ψ′(s)ds-gp-1(s)g′(s)≥C2(T-t).


## 4. Results for Certain Nonlinearities

In this section, we give the concrete quenching rate of solutions for (1.8).The Proof of Theorem1.4
We first present the upper bound. Consider two cases.Case 1.
m+q(p-1)≥2. We only need to verify the hypothesis (H1). Since
(4.1)ψ′′(u)g(u)-2(p-1)ψ′(u)g′(u)=m[m+2q(p-1)-1]um-q-2≥0,ψ′(u)[(p-2)g′2(u)+g(u)g′′(u)]-ψ′′(u)g(u)g′(u)=mq[pq-q+m-2]um-2q-3≥0,
then we have
(4.2)∫0u(1,t)ψ′(s)ds-gp-1(s)g′(s)=∫0u(1,t)msm-1-s-q(p-1)(-q)s-q-1ds=mq∫0u(1,t)sqp+mds=mq(m+qp+1)um+qp+1(1,t)≤C1(T-t).
Therefore, we get the upper bound as
(4.3)u(1,t)(T-t)-(1/(m+qp+1))≤[C1mq(m+qp+1)]1/(m+qp+1)≔C3.Case 2.
m+q(p-1)<2. We use a modification of an argument from [16]. For t∈[η,T) with some η such that u(1,η)<1, set
(4.4)y(t)=uγ(1,t)∫1-ξ(t)1um(x,t)dx,
with(4.5)ξ(t)=uq+1(1,t),
where -(m+q+1)<γ≤-(m+1)(q+1).A routine calculation shows(4.6)y′(t)=γuγ-1(1,t)ut(1,t)∫1-ξ(t)1um(x,t)dx+(q+1)uq+γ(1,t)um(1-ξ(t),t)ut(1,t)+uγ(1,t)[|ux|p-2ux(1,t)-|ux|p-2ux(1-ξ(t),t)]≥uγ-1(1,t)ut(1,t)I(t)-uγ-q(p-1)(1,t),
where I(t)=γ∫1-ξ(t)1um(x,t)dx+(q+1)uq+1(1,t)um(1-ξ(t),t). Since ux≤0 and uxx≤0 in [0,1]×[η,T), we find
(4.7)u(1,t)≤u(x,t)≤u(1-ξ(t),t)≤2u(1,t),
for any x∈[1-ξ(t),1] and t∈[η,T). By (4.4), (4.5) and (4.7), we have
(4.8)uγ+m+q+1(1,t)≤y(t)≤2muγ+m+q+1(1,t),
or equivalently,
(4.9)u(1,t)≤y1/(γ+m+q+1)(t)≤c4u(1,t)fort∈[η,T).
We now claim that I(t)≤0 on [η,T). In fact
(4.10)I(t)=(γ+q+1)∫1-ξ(t)1um(x,t)dx-(q+1)∫1-ξ(t)1(um(x,t)-um(1-ξ(t),t))dx=(γ+q+1)∫1-ξ(t)1um(x,t)dx-(q+1)∫1-ξ(t)1mum-1(ζ(t),t)ux(ζ(t),t)(x-1+ξ(t))dx,
where 1-ξ(t)<ζ(t)<1. When 0<m<1, γ≤-(((m+2)(q+1))/2), because ux,uxx≤0 and ∫1-ξ(t)1(1-ξ(t),t))dx=(1/2)u2q+2(1,t), we have
(4.11)I(t)≤(γ+q+1)∫1-ξ(t)1um(x,t)dx+m(q+1)∫1-ξ(t)1um-1(1,t)(-ux(1,t))(x-1+ξ(t))dx=[γ+q+1+m(q+1)2]um+q+1(1,t)≤0;
when 1≤m<2, γ≤-(q+1)(m+1), we have
(4.12)I(t)≤(γ+q+1)∫1-ξ(t)1um(x,t)dx+m(q+1)∫1-ξ(t)1(2u(1,t))m-1(-ux(1,t))(x-1+ξ(t))dx=[γ+q+1+2m-2m(q+1)]um+q+1(1,t)≤[γ+q+1+m(q+1)]um+q+1(1,t)≤0.
In conclusion, when 0<m<2, γ≤-(q+1)(m+1), we have
(4.13)I(t)≤0.
From (4.6), (4.9), and (4.13), it then follows that
(4.14)y′(t)≥-uγ-q(p-1)(1,t)≥-c4-γ+q(p-1)y(γ-q(p-1))/(γ+m+q+1)(1,t).
Integrating the above inequality from t to T, we obtain
(4.15)y(m+qp+1)/(γ+m+q+1)(t)≤m+qp+1γ+m+q+1c4-γ+q(p-1)(T-t),
that is,
(4.16)y1/(γ+m+q+1)(1,t)≤C3(T-t)1/(γ+m+q+1),
which in conjunction with (4.9) yields the desired upper bound.We then give the lower bound. We examine the validity of hypotheses(H2)–(H4).Firstly, for(H2), we find
(4.17)(g(p-1)(σ-1)(u)g′(u)ψ′(u))′′=-qm[q(pσ-σ-p+2)+m]⋅[q(pσ-σ-p+2)+m+1]u-q(pσ-σ-p+2)-m-2<0,
provided σ<σ1=-((m+1)/(q(p-1)))+((p-2)/(p-1)) or σ>σ2=-(m/(q(p-1)))+((p-2)/(p-1)).Secondly, for(H3), we have
(4.18)(5p-2σp+2σ-6)(g(p-1)(σ-1)(u)g′(u))′ψ′(u)-(p-1)(3-σ)g(p-1)(σ-1)(u)g′(u)ψ′′(u)=qm[-2q(p-1)2σ2+(p-1)(7pq-10q-m-1)σ+(-5p2q+16pq-12q+3pm+2p-3m-3)(p-1)2σ2]u-q(p-1)(σ-1)-q+m-3≥0,
provided σ3≤σ≤σ4 with
(4.19)σ3=(p-1)(7pq-10q-m-1)-b2-4ac4q(p-1)2,σ4=(p-1)(7pq-10q-m-1)+b2-4ac4q(p-1)2,
where a=-2q(p-1)2, b=(p-1)(7pq-10q-m-1), and c=-5p2q+16pq-12q+3pm+2p-3m-3.Thirdly, for(H4), we obtain
(4.20)(g(p-1)(σ-1)-1(u)g′(u)ψ′(u))′=-qm[-q(p-1)(σ-1)-m]u-q(p-1)(σ-1)-m-1<0,
provided σ<σ5=1-(q/(m(p-1))). Since σ2>σ5, we choose a σ such that σ3<σ<min{σ1,σ4,σ5} and hypotheses (H2)–(H4) hold. Thus the proof is completed.



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*Source: 539161-2012-04-01.xml*