# A New Solution Operator of One-Dimensionalp-Laplacian with a Sign-Changing Weight and Its Application

Authors: Inbo Sim; Yong Hoon Lee
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We establish a new solution operator for the following problem-φp(u′)′=g, t∈(0,1), u(0)=0=u(1), where φp(x)=|x|p-2x, p>1. g may be singular at the boundary or change signs or may not be in L1(0,1) so that this solution operator can cover larger class of g than previously known ones. As an application, by checking complete continuity of the solution operator, we show the existence of nontrivial solutions for p-Laplacian φp(u′)′+λh(t)f(u(t))=0, t∈(0,1), u(0)=0=u(1), where λ>0 a parameter and f∈C(ℝ,ℝ) and f(0)>0 and h may change signs or may be beyond of L1(0,1).


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## Body

## 1. Introduction

The solution operator approach has been used to show the existence of solutions for wide class of nonlinear differential equations. As it is wellknown, a solution operator plays a key role when we employ nonlinear analytic methods such as degree theory, fixed point index theory, and bifurcation theory. In this paper, we are concerned with a solution operator for one-dimensionalp-Laplacian with a sign-changing singular weight.Let us summarize a brief history about solution operators for one-dimensionalp-Laplacian with Dirichlet boundary condition
(A)-φp(u′)′=g,t∈(0,1),u(0)=0=u(1),
where φp(x)=|x|p-2x, p>1, and g∈Lloc1(0,1).First, wheng∈L1(0,1), a solution operator was introduced by Manásevich and Mawhin [1, 2]. In this case, we know that all solutions of (A) are in C1[0,1] so that by a solution of problem (A), we understand a function u∈C1[0,1] with φp(u′) absolutely continuous which satisfies (A). Key step of establishing a solution operator for this case is to show that, for each g∈L1(0,1), the equation
(1.1)∫01‍φp-1(a(g)+∫0s‍g(τ)dτ)dt=0
has a unique solution a=a(g) where a:L1(0,1)→ℝ is a continuous function. Defining
(1.2)G(g)(t)≜∫0t‍φp-1(a(g)+∫0s‍g(τ)dτ)ds,
problem (A) can be equivalently written as u=G(g).Secondly, wheng≥0 and satisfies
(1.3)∫01/2‍φp-1(∫s1/2‍g(τ)dτ)ds+∫1/21‍φp-1(∫1/2s‍g(τ)dτ)ds<∞,
a solution operator was studied by Agarwal et al. [3]. We note that a function satisfying (1.3) needs not be in L1(0,1). In this case, we know that solutions of (A) may not be in C1[0,1] so that, by a solution of problem (A), we understand a function u∈C[0,1]∩C1(0,1) with φp(u′) absolutely continuous which satisfies (A). Define a function x by
(1.4)x(t)=∫0t‍φp-1(∫st‍g(τ)dτ)ds-∫t1‍φp-1(∫ts‍g(τ)dτ)ds
for 0<t<1. It is easy to see that x is continuous, strictly increasing in (0,1) and x(0+)<0<x(1-). Thus, x has a unique zero in (0,1), say Ag. Then, we see that
(1.5)∫0Ag‍φp-1(∫sAg‍g(τ)dτ)ds=∫Ag1‍φp-1(∫Ags‍g(τ)dτ)ds,
and defining
(1.6)G(g)(t)≜{∫0t‍φp-1(∫sAg‍g(τ)dτ)ds,0≤t≤Ag,∫t1‍φp-1(∫Ags‍g(τ)dτ)ds,Ag≤t≤1,
problem (A) can be equivalently written as u=G(g).Now let us consider the case thatg may change signs and belongs to ℋ⊂Lloc1(0,1) which is defined as
(1.7)H={g:∫01/2‍φp-1(∫s1/2‍|g(τ)|dτ)ds+∫1/21‍φp-1(∫1/2s‍|g(τ)|dτ)ds<∞}.
Defining g:(0,1)→ℝ by
(1.8)g(t)={t-α,0<t≤12,-1,12<t<1,
where 1<α<p, it is not hard to check that g∈ℋ, but g∉L1(0,1).Forg∈ℋ but g∉L1(0,1), existence of solution for (1.1) is not known so that Manásevich-Mawhin's approach cannot be applied directly. For g sign changing, the uniqueness of zeros of function x in (1.4) is not guaranteed so that Agarwal et al.’s approach cannot be applied directly either. The aim of this paper is to introduce a new solution operator for (A) under this situation.As an application, we will make use of newly established solution operator to show the existence of nontrivial solutions with respect to a parameter for the following type of nonlinear problem(Pλ)φp(u′)′+λh(t)f(u(t))=0,t∈(0,1),u(0)=0=u(1),
where λ>0 a parameter, h∈ℋ, f∈C(ℝ,ℝ), and f(0)>0.This kind of problems can be classified as casesf(0)>0, f(0)=0, or f(0)<0, and, under additional nonlinear growth conditions, it is easy to see that bifurcation phenomena usually happen for the case of f(0)>0. In this paper, we will employ the global continuation theorem to get an existence result. One of the major steps for proof is to show the complete continuity of corresponding solution operator for (Pλ). There have been many studies for f nonnegative with f(0)>0 and h≥0. Many authors are interested in finding positive solutions mainly employing the fixed-point index theory on a cone. Since h may change signs in our case, this method is not suitable. Hai [4] investigated semilinear boundary value problems with a sign-changing weight; however, his method is also restricted to fit our quasilinear case.The paper is organized as follow. In Section2, we establish a solution operator for (A), when g∈ℋ. In Section 3, we prove the continuity and compactness of the solution operator for problem (Pλ). In Section 4, we obtain the existence of non-trivial solutions for (Pλ).


## 2. A Solution Operator

In this section, we construct a new solution operator for(A).Theorem 2.1.
For eachg∈ℋ, there is a unique solution u∈C[0,1]∩C1(0,1) for (A).The proof follows the next two lemmas. We first show the existence of a solution for(A).Lemma 2.2.
For eachg∈ℋ, there is a unique constant α=α(g)∈ℝ such that a function u defined by
(2.1)u(t)={∫0t‍φp-1(α+∫s1/2‍g(r)dr)ds,0≤t≤12,∫t1‍φp-1(-α+∫1/2s‍g(r)dr)ds,12≤t≤1,
satisfies u∈C[0,1]∩C1(0,1) and u solves (A). Furthermore, we have
(2.2)∫01/2‍φp-1(α+∫s1/2‍g(r)dr)ds=∫1/21‍φp-1(-α+∫1/2s‍g(r)dr)ds.Remark 2.3.
Fora,b>0,
(2.3)φp-1(a+b)=(a+b)1/(p-1)≤{a1/(p-1)+b1/(p-1),p>2,2(2-p)/(p-1)(a1/(p-1)+b1/(p-1)),1<p≤2.
Thus for 0<t≤1/2,
(2.4)|u(t)|≤Cp(∫0t‍|α|1/(p-1)dt+∫0t‍φp-1(∫s1/2‍|g(r)|dr)ds)<∞,
where
(2.5)Cp={1,p>2,2(2-p)/(p-1),1<p≤2.
Similarly, we get an upper bound of |u| on (1/2,1). It is clear that u(0)=0=u(1).Proof of Lemma2.2.
Define(2.6)vα(t)≜∫0t‍φp-1(α+∫s1/2‍g(r)dr)ds,0≤t≤12,wα(t)≜∫t1‍φp-1(-α+∫1/2s‍g(r)dr)ds,12≤t≤1.
Then, by Lebesgue dominated convergence theorem, we see that vα(1/2) and wα(1/2) are continuous on α. Because of the strict monotonicity of φp, it is easy to show that vα(1/2) is strictly increasing on α and wα(1/2) is strictly decreasing on α.
We claim thatvα(1/2)→∞, wα(1/2)→-∞ as α→∞, and vα(1/2)→-∞, wα(1/2)→∞ as α→-∞. Since (α/2)+∫s1/2‍g(r)dr>0 for 1/3≤s≤1/2, for sufficiently large α>0, we have
(2.7)vα(12)=∫01/2‍φp-1(α+∫s1/2‍g(r)dr)ds=∫01/3‍φp-1(α+∫s1/2‍g(r)dr)ds+∫1/31/2‍φp-1(α+∫s1/2‍g(r)dr)ds≥16(α2)1/(p-1)+∫01/3‍φp-1(∫s1/2‍g(r)dr)ds≥16(α2)1/(p-1)-∫01/2‍φp-1(∫s1/2‍|g(r)|dr)ds⟶∞,
as α→∞. Since -α/2+∫1/2s‍g(r)dr<0 for 1/2≤s≤2/3, for sufficiently large α>0, we have
(2.8)wα(12)=∫1/21‍φp-1(-α+∫1/2s‍g(r)dr)ds=∫1/22/3‍φp-1(-α+∫1/2s‍g(r)dr)ds+∫2/31‍φp-1(-α+∫1/2s‍g(r)dr)ds≤-16(α2)1/(p-1)+∫2/31‍φp-1(∫1/2s‍g(r)dr)ds≤-16(α2)1/(p-1)+∫1/21‍φp-1(∫1/2s‍|g(r)|dr)ds⟶-∞,
as α→∞. Similarly, since α/2+∫s1/2‍g(r)dr<0 for 1/3≤s≤1/2, for negatively large α<0, we have
(2.9)vα(12)=∫01/2‍φp-1(α+∫s1/2‍g(r)dr)ds=∫01/3‍φp-1(α+∫s1/2‍g(r)dr)ds+∫1/31/2‍φp-1(α+∫s1/2‍g(r)dr)ds≤-16(-α2)1/(p-1)+∫01/3‍φp-1(∫s1/2‍g(r)dr)ds≤-16(-α2)1/(p-1)+∫01/2‍φp-1(∫s1/2‍|g(r)|dr)ds⟶-∞,
as α→-∞. Since -α/2+∫1/2s‍g(r)dr>0 for 1/2≤s≤2/3, for negatively large α<0, we have
(2.10)wα(12)=∫1/21‍φp-1(-α+∫1/2s‍g(r)dr)ds=∫1/22/3‍φp-1(-α+∫1/2s‍g(r)dr)ds+∫2/31‍φp-1(-α+∫1/2s‍g(r)dr)ds≥16(-α2)1/(p-1)+∫2/31‍φp-1(∫1/2s‍g(r)dr)ds≥16(-α2)1/(p-1)-∫1/21‍φp-1(∫1/2s‍|g(r)|dr)ds⟶∞,
as α→-∞. Thus, by intermediate value theorem, there is a unique α0∈ℝ such that vα0(1/2)=wα0(1/2). Put
(2.11)u(t)={vα0(t),0≤t≤12,wα0(t),12≤t≤1.
Then, u∈C[0,1]∩C1(0,1) and u solves (A).From the uniqueness ofα, the following is obvious.Corollary 2.4.
Suppose thatα is a function defined in Lemma 2.2. Then, α has the following property:
(2.12)α(λg)=λα(g),
for all λ∈ℝ.The following lemma guarantees the uniqueness of solutions for(A).Lemma 2.5.
Assume that, fori=1,2, ui∈C[0,1]∩C1(0,1) satisfies
(Ai)-φp(u′)′=gi,t∈(0,1),u(0)=0=u(1).
If g1≤g2 or g2≤g1 on [0,1], then u1≤u2 or u2≤u1 on [0,1], respectively.Proof.
We only prove the first case. We may use similar argument to show the second case. Letui be a solution of (Ai), i=1,2, and assume g1≤g2 for 0≤t≤1. We want to show u1(t)≤u2(t), 0≤t≤1. If this is false, then there exists t0∈(0,1) with u1(t0)>u2(t0). Hence, there is an interval (a,b)⊂(0,1) with t0∈(a,b) such that u1(t)>u2(t), a<t<b and u1(a)-u2(a)=u1(b)-u2(b)=0. Let m=u1(B)-u2(B) be the positive maximum of u1(t)-u2(t) on [a,b]. Then, B∈[a,b] and u1′(B)=u2′(B). Integrating both sides (A1) and (A2) over [s,B], a<s<B, we have
(2.13)u1′(s)=φp-1(φp(u1′(B))+∫sB‍g1(r)dr),u2′(s)=φp-1(φp(u2′(B))+∫sB‍g2(r)dr).
Integrating both sides of the above equalities over [a,B], we get
(2.14)u1(B)-u1(a)=∫aB‍φp-1(φp(u1′(B))+∫sB‍g1(r)dr)ds,u2(B)-u2(a)=∫aB‍φp-1(φp(u2′(B))+∫sB‍g2(r)dr)ds.
Thus, we obtain
(2.15)0<m=u1(B)-u2(B)≤0,
which is a contradiction. This completes the proof.Let us define(2.16)G(g)(t)={∫0t‍φp-1(α+∫s1/2‍g(r)dr)ds,0≤t≤12,∫t1‍φp-1(-α+∫1/2s‍g(r)dr)ds,12≤t≤1.
Then, problem (A) can be equivalently written as u=G(g).


## 3. Application: Compactness

In this and the following sections, as an application of new solution operator, we consider a problem of one-dimensionalp-Laplacian with a singular weight as follows:
(Pλ)φp(u′)′+λh(t)f(u(t))=0,t∈(0,1),u(0)=0=u(1),
where λ>0 a parameter, h∈ℋ, and f∈C(ℝ,ℝ). Note that we do not need any other assumptions on f except the continuity in this section. Employing the solution operator G established in Section 2, we may rewrite (Pλ) equivalently as
(3.1)u=G(λ,u),
where G:(0,∞)×C[0,1]→C[0,1] is defined by
(3.2)G(λ,u)(t)≜{∫0t‍φp-1(α(λhf(u))+∫s1/2‍λh(r)f(u(r))dr)ds,0≤t≤12,∫t1‍φp-1(-α(λhf(u))+∫1/2s‍λh(r)f(u(r))dr)ds,12≤t≤1.
The purpose of this section is to show that G is completely continuous on (0,∞)×C[0,1]. For this, we need to know the properties of function α:(0,∞)×C[0,1]→ℝ.Lemma 3.1.
α sends bounded sets in (0,∞)×C[0,1] into bounded sets in ℝ.Proof.
Assume that{λn} and {un} are bounded on (0,∞) and C[0,1], respectively. For convenience, we denote αn=α(λnhf(un)). Suppose that {αn} is unbounded in ℝ. Then, there is a subsequence {αnk} such that αnk→∞ or αnk→-∞ as k→∞. Here, we consider the first case αnk→∞. We can prove the second case by the same argument. As in (2.2) of Lemma 2.2, we have
(3.3)∫01/2φp-1(αnk+∫s1/2‍λnkh(r)f(unk(r))dr)dsds=∫1/21‍φp-1(-αnk+∫1/2s‍λnkh(r)f(unk(r))dr)ds.
As in the proof of Lemma 2.2, we get the following inequality for the left-hand side
(3.4)0≤∫01/2‍φp-1(αnk-M∫s1/2‍|h(r)|dr)ds≤∫01/2‍φp-1(αnk+∫s1/2‍λnkh(r)f(unk(r))dr)ds,
for sufficiently large k, where M=sup{|λnk|∥f(unk)∥∞}. On other hand, as in the proof of Lemma 2.2, we get the following inequality for the right-hand side
(3.5)∫1/21φp-1(-αnk+∫1/2s‍λnkh(r)f(unk(r))dr)ds≤∫1/21φp-1(-αnk+M∫1/2s‍|h(r)|dr)ds<0,
for sufficiently large k, and this is a contradiction.Lemma 3.2.
α is continuous on (0,∞)×C[0,1].Proof.
Assume thatλn→λ in ℝ and un→u in C[0,1]. Then, there exists a constant M>0 such that
(3.6)|λn|+‖un‖∞≤M,∀n.
Lemma 3.1 implies that {α(λnhf(un))} is bounded in ℝ. Hence, it has a convergent subsequence denoted by {α(λnkhf(unk))}. Let
(3.7)α(λnkhf(unk))⟶α^
as k→∞. As in (2.2) of Lemma 2.2, we have
(3.8)∫01/2φp-1(α(λnkhf(unk))+∫s1/2‍λnkh(r)f(unk(r))dr)dsds=∫1/21φp-1(-α(λnkhf(unk))+∫1/2s‍λnkh(r)f(unk(r))dr)ds.
By Lebesgue dominated convergence theorem, as k→∞, we get
(3.9)∫01/2‍φp-1(α^+∫s1/2‍λh(r)f(u(r))dr)ds=∫1/21‍φp-1(-α^+∫1/2s‍λh(r)f(u(r))dr)ds.
By the uniqueness of α, we obtain α^=α(λhf(u)), and this proves the continuity of α.We have the following corollary using similar argument in the proof of Lemma3.2.Corollary 3.3.
Letλn∈(0,∞) satisfy λn→0 as n→∞. Then, α(λn)→0 as n→∞.Theorem 3.4.
G is continuous and compact on (0,∞)×C[0,1].Proof.
First, we show the continuity ofG. Assume that λn→λ on (0,∞) and un→u in C[0,1]. Then, from the continuity of α and Lebesgue dominated convergence theorem, we have(3.10)‖G(λn,un)-G(λ,u)‖∞=sup⁡0≤t≤1|G(λn,un)(t)-G(λ,u)(t)|≤sup⁡0≤t≤1/2|∫0tφp-1(α(λnhf(un))+∫s1/2‍λnh(r)f(un(r))dr)2-φp-1-φp-1(α(λhf(u))+∫s1/2λh(r)f(u(r))dr)ds|+sup⁡1/2≤t≤1|∫t1φp-1(α(λnhf(un))+∫1/2s‍λnh(r)f(un(r))dr)sup⁡1/2≤t≤1φp-1-φp-1(α(λhf(u))+∫1/2sλh(r)f(u(r))dr)ds|⟶0.
Next, we show thatG is uniformly bounded. Let ℳ be a bounded subset of (0,∞)×C[0,1], and let (λ,u)∈ℳ, then using the fact that α sends bounded sets on (0,∞)×C[0,1] into bounded sets on ℝ and taking M=sup{|λ|∥f(u)∥∞:(λ,u)∈ℳ}, we obtain
(3.11)|G(λ,u)(t)|≤{∫0tφp-1(|α(λhf(u))|+M∫s1/2‍|h(r)|dr)ds,0≤t≤12,∫t1φp-1(|α(λhf(u))|+M∫1/2s‍|h(r)|dr)ds,12≤t≤1,≤{∫01/2φp-1(|K(M,h)|+M∫s1/2‍|h(r)|dr)ds<∞,0≤t≤12,∫1/21φp-1(|K(M,h)|+M∫1/2s‍|h(r)|dr)ds<∞,12≤t≤1,
where K(M,h) is a constant and the above upper bound is independent of (λ,u)∈ℳ.
Finally, we show thatG(ℳ) is equicontinuous. By Lemma 3.1, {α(λhf(u)):(λ,u)∈ℳ} is bounded so by taking N=sup{φp-1(|α(λhf(u))|):(λ,u)∈ℳ}, we obtain for 0≤t1,t2≤1/2,
(3.12)|G(λ,u)(t1)-G(λ,u)(t2)|≤|∫t1t2‍φp-1(α(λhf(u))+∫s1/2‍λh(r)f(u(r))dr)ds|≤CpN|t2-t1|+CpM∫t1t2‍φp-1(∫s1/2‍|h(r)|dr)ds.
Since h∈ℋ, the above upper bound tends to 0 as |t1-t2| does. For case 1/2≤t1, t2≤1, we may handle similar way. For case 0≤t1≤1/2, 1/2≤t2≤1 (or case 0≤t2≤1/2,1/2≤t1≤1), we can compute the following:
(3.13)|G(λ,u)(t1)-G(λ,u)(t2)|=|∫0t1φp-1(α(λhf(u))+∫s1/2‍λh(r)f(u(r))dr)ds-∫t21‍φp-1(-α(λhf(u))+∫1/2s‍λh(r)f(u(r))dr)ds|=|∫01/2φp-1(α(λhf(u))+∫s1/2‍λh(r)f(u(r))dr)ds-∫t11/2‍φp-1(α(λhf(u))+∫s1/2‍λh(r)f(u(r))dr)ds-[∫1/21‍φp-1(-α(λhf(u))+∫1/2s‍λh(r)f(u(r))dr)ds-∫1/2t2‍φp-1(-α(λhf(u))+∫1/2s‍λh(r)f(u(r))dr)ds]|=|∫1/2t2φp-1(-α(λhf(u))+∫1/2s‍λh(r)f(u(r))dr)ds-∫t11/2φp-1(α(λhf(u))+∫s1/2‍λh(r)f(u(r))dr)ds|≤2Cp∫t1t2‍{φp-1(∫1/2s‍|h(r)|dr)φp-1(|α(λhf(u))|)2Cp+M1[φp-1(∫1/2s‍|h(r)|dr)+φp-1(∫s1/2‍|h(r)|dr)]}ds≤2CpN|t2-t1|+2CpM1∫t1t2‍φp-1(∫1/2s‍|h(r)|dr)2CpN|t2-t1|+2CpM1+φp-1(∫s1/2‍|h(r)|dr)ds,
where M1=φp-1(M). Thus, by Arzela-Ascoli Theorem, G is compact, and the proof is complete.


## 4. Application: Existence

In this section, we show the existence of nontrivial solutions for problem(Pλ):
(Pλ)φp(u′)′+λh(t)f(u(t))=0,t∈(0,1),u(0)=0=u(1),
where λ>0 a parameter, h∈ℋ, and f∈C(ℝ,ℝ). In addition, we assume that (F)
f(0)>0 and f∞≡lims→∞f(s)/sp-1=0.Based on the solution operator and its complete continuity established in Section3, we show the existence of nontrivial solutions of (Pλ) for all λ>0. We will employ the following theorem the so-called global continuation theorem for the proof of our existence theorem.LetX be a Banach space and G:ℝ×X→ℝ completely continuous with G(0,u)=0, and consider
(4.1)u=G(λ,u).
We denote 𝒮 as the set of solutions for (4.1), ℝ+=[0,∞) and ℝ-=(-∞,0].Theorem 4.1 (see [5, Theorem 3.2]).
LetX be a Banach space and G:ℝ×X→X continuous and compact and G(0,u)=0. Then, 𝒮 contains a pair of unbounded components 𝒞+,𝒞- in ℝ+×X and ℝ-×X, respectively, and 𝒞+∩𝒞-={(0,0)}.We now state our main existence theorem.Theorem 4.2.
Assume thath∈ℋ and f holds (F). Then (Pλ) has non-trivial solutions for all λ>0.Let us takeX=C[0,1] with the norm ∥·∥∞. We know that, to solve (Pλ), it is enough to solve
(4.2)u=G(λ,u),
where G is defined in Section 3. And we have shown that G is continuous and compact in Section 3. Since G(0,u)=0 is obvious, employing Theorem 4.1, we obtain an unbounded continuum 𝒞+. If we provide a priori boundedness of solutions for (Pλ), then the unbounded continuum guarantees the existence of solutions for all λ>0. Therefore, to complete the proof of Theorem 4.2, we need to give a priori estimate of solutions for (Pλ).Lemma 4.3.
Assume thath∈ℋ and f holds (F). Let (λ,u) be a solution for (Pλ) with λ∈(0,Λ]. Then, there exists a constant C(Λ) which depends only on Λ such that ∥u∥∞≤C(Λ).Proof.
Let us define a functionK:(0,∞)×ℝ→ℝ which satisfies the following property: K(λn,h)→0 as λn→0. Since h∈ℋ, for given Λ>0, there exists ɛ>0 such that
(4.3)Λp-1max⁡{∫01/2φp-1(|K(ɛ,h)|+ɛ∫s1/2‍|h(r)|dr)ds,d∫1/21φp-1(|K(ɛ,h)|+ɛ∫1/2s‍|h(r)|dr)ds}<12.
Then, since f∞=0, for ɛ given above, there exists N>0 such that for all s with |s|≥N,
(4.4)|f(s)|≤ɛ|s|p-1.
Since f is continuous on [-N,N], for N given above, there exists a maximum value M>0 for |f| on [-N,N], that is, for all s with |s|≤N, we have |f(s)|≤M. Let us divide the interval [0,1] into I and J as
(4.5)I:={t∈[0,1]:|u(t)|≥N},J:={t∈[0,1]:|u(t)|≤N},
respectively. Then, we can estimate solution u as follows:
(4.6)|u(t)|≤{∫0tφp-1(|α(λhf(u))|+∫s1/2‍λ|h(r)f(u(r))|dr)ds,0≤t≤12,∫t1φp-1(|α(λhf(u))|+∫1/2s‍λ|h(r)f(u(r))|dr)ds,12≤t≤1,=∫I‍+∫J‍,
where ∫Γ‍ with Γ=I or J can be given as
(4.7)∫Γ‍={∫Γ‍φp-1(|α(λhf(u))|+∫s1/2λ|h(r)f(u(r))|dr)ds,0≤t≤12,∫Γ‍φp-1(|α(λhf(u))|+∫1/2s‍λ|h(r)f(u(r))|dr)ds,12≤t≤1.
By Lemma 3.1, if |λf(u)|≤M for some M>0, then there exists a constant K(M,h)>0 such that |α(λhf(u))|≤K(M,h). Thus, on J, we have
(4.8)∫J‍={∫J‍φp-1(|α(λhf(u))|+∫s1/2‍λ|h(r)f(u(r))|dr)ds,0≤t≤12,∫J‍φp-1(|α(λhf(u))|+∫1/2s‍λ|h(r)f(u(r))|dr)ds,12≤t≤1,≤{∫J‍φp-1(|K(ΛM,h)|+ΛM∫s1/2‍|h(r)|dr)ds,0≤t≤12,∫J‍φp-1(|K(ΛM,h)|+ΛM∫1/2s‍|h(r)|dr)ds,12≤t≤1,≤{∫01/2‍φp-1(|K(ΛM,h)|+ΛM∫s1/2‍|h(r)|dr)ds,0≤t≤12,∫01/2‍φp-1(|K(ΛM,h)|+ΛM∫1/2s‍|h(r)|dr)ds,12≤t≤1,≤KCΛ,
where CΛ=max{|K(ΛM,h)|,ΛM} and
(4.9)K=max⁡{∫01/2‍φp-1(1+∫s1/2‍|h(r)|dr)ds,∫1/21‍φp-1(1+∫1/2s‍|h(r)|dr)ds}.
On I, using Corollaries 2.4 and 3.3, and λ/Λ<1, we obtain
(4.10)∫I≤{∫I‍φp-1(|α(λΛhf(u)‖u‖∞p-1)|+∫s1/2‍λΛ|h(r)||f(u)‖u‖∞p-1|dr)dsΛp-1‖u‖∞,0≤t≤12,∫I‍φp-1(|α(λΛhf(u)‖u‖∞p-1)|+∫1/2s‍λΛ|h(r)||f(u)‖u‖∞p-1|dr)dsΛp-1‖u‖∞,12≤t≤1,‍≤{∫I‍φp-1(K(ɛ,h)+ɛ∫s1/2‍|h(r)|dr)dsΛp-1‖u‖∞,0≤t≤12,∫I‍φp-1(K(ɛ,h)+ɛ∫1/2s‍|h(r)|dr)dsΛp-1‖u‖∞,12≤t≤1,≤{∫01/2‍φp-1(K(ɛ,h)+ɛ∫s1/2‍|h(r)|dr)dsΛp-1‖u‖∞,0≤t≤12,∫1/21‍φp-1(K(ɛ,h)+ɛ∫1/2s‍|h(r)|dr)dsΛp-1‖u‖∞,12≤t≤1,≤12‖u‖∞.
Therefore, we have
(4.11)|u(t)|≤KCΛ+12‖u‖∞,
so that
(4.12)‖u‖∞≤2KCΛ,
and this completes the proof.



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*Source: 243740-2012-08-08.xml*