# Existence of Solutions for a Class of Variable Exponent Integrodifferential System Boundary Value Problems

Authors: Guizhen Zhi; Yunrui Guo; Yan Wang; Qihu Zhang
Journal: Journal of Applied Mathematics
 (2011)

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## Abstract

This paper investigates the existence of solutions for a class of variable exponent integrodifferential system with multipoint and integral boundary value condition in half line. When the nonlinearity termf satisfies sub-(p−−1) growth condition or general growth condition, we give the existence of solutions and nonnegative solutions via Leray-Schauder degree at nonresonance, respectively. Moreover, the existence of solutions for the problem at resonance has been discussed.


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## Body

## 1. Introduction

In this paper, we consider the existence of solutions for the following variable exponent integrodifferential system(1.1)-Δp(t)u+δf(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u))=0,t∈(0,+∞),
with the following nonlinear multipoint and integral boundary value condition(1.2)u(+∞)=∫0+∞e(t)u(t)dt+e1,lim⁡t→+∞w(t)|u′|p(t)-2u′(t)=∑i=1m-2αiw(ξi)|u′|p(ξi)-2u′(ξi)+e2,
where u:[0,+∞)→ℝN;S and T are linear operators defined by (1.3)S(u)(t)=∫0tψ(s,t)u(s)ds,T(u)(t)=∫0+∞χ(s,t)u(s)ds,
where ψ∈C(D,ℝ),χ∈C(D,ℝ),D={(s,t)∈[0,+∞)×[0,+∞)};∫0+∞|ψ(s,t)|ds and ∫0+∞|χ(s,t)|ds are uniformly bounded with t;p∈C([0,+∞),ℝ),p(t)>1,limt→+∞p(t) exists and limt→+∞p(t)>1;-Δp(t)u∶=-(w(t)|u′|p(t)-2u′)′ is called the weighted p(t)-Laplacian; w∈C([0,+∞),ℝ) satisfies 0<w(t), for all t∈(0,+∞), and (w(t))-1/(p(t)-1)∈L1(0,+∞);0<ξ1<⋯<ξm-2<+∞,αi≥0,(i=1,…,m-2)  and 0≤∑i=1m-2αi≤1;e∈L1(0,+∞) is nonnegative,  σ=∫0+∞e(t)dt and σ∈[0,1];e1,e2∈ℝN;δ is a positive parameter.If∑i=1m-2αi<1 and σ<1, we say the problem is nonresonant; but if ∑i=1m-2αi∈[0,1] and σ=1, we say the problem is resonant.The study of differential equations and variational problems with variable exponent growth conditions is a new and interesting topic. Many results have been obtained on these problems, for example, [1–23]. We refer to [3, 19, 23], for the applied background on these problems. If w(t)≡1 and p(t)≡p (a constant), -Δp(t) becomes the well-known p-Laplacian. If p(t) is a general function, -Δp(t) represents a nonhomogeneity and possesses more nonlinearity, and thus -Δp(t) is more complicated than -Δp. For example, if Ω⊂ℝn is a bounded domain, the Rayleigh quotient(1.4)λp(x)=inf⁡u∈W01,p(x)(Ω)∖{0}⁡∫Ω(1/p(x))|∇u|p(x)dx∫Ω(1/p(x))|u|p(x)dx
is zero in general, and only under some special conditions λp(x)>0 (see [9, 16–18]), but the fact that λp>0 is very important in the study of p-Laplacian problems.Integral boundary conditions for evolution problems have been applied variously in chemical engineering, thermoelasticity, underground water flow, and population dynamics. There are many papers on the differential equations with integral boundary value conditions, for example, [24–29]. On the existence of solutions for p(x)-Laplacian systems boundary value problems, we refer to [2, 4, 7, 8, 10–12, 20–22]. In [20], the present author deals with the existence and asymptotic behavior of solutions for (1.1) with the following linear boundary value conditions(1.5)u(0)=∑i=1m-2αiu(ξi)+e0,lim⁡t→+∞u(t)=∫0+∞e(t)u(t)dt,
when 0≤∑i=1m-2αi<1 and 0≤∫0+∞e(t)≤1. But results on the existence of solutions for variable exponent integrodifferential systems with nonlinear boundary value conditions are rare. In this paper, when p(t) is a general function, we investigate the existence of solutions and nonnegative solutions for variable exponent integrodifferential systems with nonlinear multipoint and integral boundary value conditions, when the problem is at nonresonance. Moreover, we discuss the existence of solutions for the problem at resonance. Since the nonlinear multipoint boundary value condition is on the derivative of solution u, we meet more difficulties than [20].LetN≥1 and J=[0,+∞), the function f=(f1,…,fN):J×ℝN×ℝN×ℝN×ℝN→ℝN is assumed to be Caratheodory, by this we mean,(i)
for almost everyt∈J, the function f(t,·,·,·,·) is continuous;(ii)
for each(x,y,z,w)∈ℝN×ℝN×ℝN×ℝN, the function f(·,x,y,z,w) is measurable on J;(iii)
for eachR>0, there is a βR∈L1(J,ℝ) such that, for almost every t∈J and every (x,y,z,w)∈ℝN×ℝN×ℝN×ℝN with |x|≤R,|y|≤R,|z|≤R,|w|≤R, one has(1.6)|f(t,x,y,z,w)|≤βR(t).Throughout the paper, we denote(1.7)w(0)|u′|p(0)-2u′(0)=lim⁡t→0+w(t)|u′|p(t)-2u′(t),w(+∞)|u′|p(+∞)-2u′(+∞)=lim⁡t→+∞w(t)|u′|p(t)-2u′(t).The inner product inℝN will be denoted by 〈·,·〉,|·| will denote the absolute value and the Euclidean norm on ℝN. Let AC(0,+∞) denote the space of absolutely continuous functions on the interval (0,+∞). For N≥1, we set C=C(J,ℝN),C1={u∈C∣u′∈C((0,+∞),ℝN),limt→0+w(t)1/(p(t)-1)u′(t)exists}. For any u(t)=(u1(t),…,uN(t))∈C, we denote |ui|0=supt∈(0,+∞)|ui(t)|,∥u∥0=(∑i=1N|ui|02)1/2, and ∥u∥1=∥u∥0+∥(w(t))1/(p(t)-1)u′∥0. Spaces C and C1 will be equipped with the norm ∥·∥0 and ∥·∥1, respectively. Then, (C,∥·∥0) and (C1,∥·∥1) are Banach spaces. Denote L1=L1(J,ℝN) with the norm ∥u∥L1=[∑i=1N(∫0∞|ui|dt)2]1/2.We say a functionu:J→ℝN is a solution of (1.1) if u∈C1 with w(t)|u′|p(t)-2u′ absolutely continuous on (0,+∞), which satisfies (1.1) a.e. on J.In this paper, we always useCi to denote positive constants, if it cannot lead to confusion. Denote (1.8)z-=inf⁡t∈Jz(t),z+=sup⁡t∈Jz(t),foranyz∈C(J,R).We sayf satisfies sub-(p--1) growth condition, if f satisfies(1.9)lim⁡|x|+|y|+|z|+|w|→+∞f(t,x,y,z,w)(|x|+|y|+|z|+|w|)q(t)-1=0,fort∈Juniformly,
where q(t)∈C(J,ℝ) and 1<q-≤q+<p-. We say f satisfies general growth condition, if f does not satisfy sub-(p--1) growth condition.We will discuss the existence of solutions of (1.1)-(1.2) in the following three cases.Case (i):∑i=1m-2αi∈[0,1),σ∈[0,1);Case (ii):∑i=1m-2αi∈[0,1),σ=1;Case (iii):∑i=1m-2αi=1,σ=1.This paper is divided into five sections. In the second section, we present some preliminary and give the operator equations which have the same solutions of (1.1)-(1.2) in the three cases, respectively. In the third section, we will discuss the existence of solutions of (1.1)-(1.2) when ∑i=1m-2αi∈[0,1),σ∈[0,1), and we give the existence of nonnegative solutions. In the fourth section, we will discuss the existence of solutions of (1.1)-(1.2) when ∑i=1m-2αi∈[0,1),σ=1. In the fifth section, we will discuss the existence of solutions of (1.1)-(1.2) when ∑i=1m-2αi=1,σ=1.


## 2. Preliminary

For any(t,x)∈J×ℝN, denote φ(t,x)=|x|p(t)-2x. Obviously, φ has the following properties.Lemma 2.1 (see [7]).
φ is a continuous function and satisfies the following.(i)
For anyt∈[0,+∞),  φ(t,·) is strictly monotone, that is
(2.1)〈φ(t,x1)-φ(t,x2),x1-x2〉>0,foranyx1,x2∈RN,x1≠x2.(ii)
There exists a functionβ:[0,+∞)→[0,+∞),β(s)→+∞ as s→+∞, such that
(2.2)〈φ(t,x),x〉≥β(|x|)|x|,∀x∈RN.It is well known thatφ(t,·) is a homeomorphism from ℝN to ℝN for any fixed t∈[0,+∞). For any t∈J, denote by φ-1(t,·) the inverse operator of φ(t,·), then (2.3)φ-1(t,x)=|x|(2-p(t))/(p(t)-1)x,forx∈RN∖{0},φ-1(t,0)=0.It is clear thatφ-1(t,·) is continuous and sends bounded sets into bounded sets.Let us now consider the following problem with boundary value condition  (1.2)(2.4)(w(t)φ(t,u′(t)))′=g(t),t∈(0,+∞),
where g∈L1.Ifu is a solution of (2.4) with (1.2), by integrating (2.4) from 0 to t, we find that
(2.5)w(t)φ(t,u′(t))=w(0)φ(0,u′(0))+∫0tg(s)ds.Define operatorF:L1→C as (2.6)F(g)(t)=∫0tg(s)ds,∀t∈J,∀g∈L1.By solving foru′ in (2.5) and integrating, we find that(2.7)u(t)=u(0)+F{φ-1[t,(w(t))-1(w(0)φ(0,u′(0))+F(g))]}(t),t∈J.In the following, we will give the operator equations which have the same solutions of (1.1)-(1.2) in three cases, respectively.


### 2.1. Case (i):∑i=1m-2αi∈[0,1),σ∈[0,1)

We denoteρ=w(0)φ(0,u′(0)) in (2.7). It is easy to see that ρ is dependent on g(·), then we find that (2.8)u(t)=u(0)+F{φ-1[t,(w(t))-1(ρ+F(g))]}(t),t∈J.The boundary value condition (1.2) implies that (2.9)u(0)=∫0+∞{e(t)∫0tφ-1[r,(w(r))-1(ρ+F(g)(r))]dr}dt+e11-σ-∫0+∞φ-1[r,(w(r))-1(ρ+F(g)(r))]dr1-σ,ρ=11-∑i=1m-2αi(∑i=1m-2αi∫0ξig(s)ds-∫0+∞g(s)ds+e2).For fixedh∈L1, we define ρ:L1→ℝN as(2.10)ρ(h)=11-∑i=1m-2αi(∑i=1m-2αi∫0ξih(t)dt-∫0+∞h(t)dt+e2).Lemma 2.2.
ρ:L1→ℝN is continuous and sends bounded sets of L1 to bounded sets of ℝN. Moreover,
(2.11)|ρ(h)|≤2N1-∑i=1m-2αi⋅(‖h‖L1+|e2|).Proof.
Sinceρ(·) consists of continuous operators, it is continuous. It is easy to see that
(2.12)|ρ(h)|≤2N1-∑i=1m-2αi⋅(‖h‖L1+|e2|).
This completes the proof.It is clear thatρ(·) is continuous and sends bounded sets of L1 to bounded sets of ℝN, and hence it is a compact continuous mapping.Ifu is a solution of (2.4) with (1.2), we find that (2.13)u(t)=u(0)+F{φ-1[t,(w(t))-1(ρ+F(g))]}(t),t∈J,u(0)=∫0+∞{e(t)∫0tφ-1[r,(w(r))-1(ρ+F(g)(r))]dr}dt+e11-σ-∫0+∞φ-1[r,(w(r))-1(ρ+F(g)(r))]dr1-σ.We denote(2.14)K(h)(t)∶=(K∘h)(t)=F{φ-1[t,(w(t))-1(ρ(h)+F(h))]}(t),∀t∈(0,+∞).We say a setU⊂L1 be equi-integrable, if there exists a nonnegative ρ*∈L1(J,ℝ), such that (2.15)|h(t)|≤ρ*(t)a.e.inJ,foranyh∈U.Lemma 2.3.
The operatorK is continuous and sends equi-integrable sets in L1 to relatively compact sets in C1.Proof.
It is easy to check thatK(h)(t)∈C1, for all h∈L1. Since (w(t))-1/(p(t)-1)∈L1 and
(2.16)K(h)′(t)=φ-1[t,(w(t))-1(ρ(h)+F(h))],∀t∈[0,+∞),
it is easy to check that K is a continuous operator from L1 to C1.
Let nowU be an equi-integrable set in L1, then there exists a nonnegative ρ*∈L1(J,ℝ), such that
(2.17)|h(t)|≤ρ*(t)a.e.inJ,foranyh∈U.
We want to show thatK(U)¯⊂C1  is a compact set.
Let{un} be a sequence in K(U), then there exists a sequence {hn}⊂U such that un=K(hn). Since hn(t)=(hn1(t),…,hnN(t)), where hni(t)∈L1(J,ℝ)(i=1,…,N), we have
(2.18)|hni(t)|≤|hn(t)|,∀i=1,…,N,
then for any t1,t2∈J with t1<t2, we have
(2.19)|∫t1t2hn(t)dt|=∑i=1N(∫t1t2hni(t)dt)2≤∑i=1N(∫t1t2|hni(t)|dt)2≤∑i=1N(∫t1t2|hn(t)|dt)2≤N∫t1t2|hn(t)|dt,
which together with (2.17) implies
(2.20)|F(hn)(t1)-F(hn)(t2)|=|∫0t1hn(t)dt-∫0t2hn(t)dt|=|∫t1t2hn(t)dt|≤N∫t1t2|hn(t)|dt≤N∫t1t2ρ*(t)dt.
Hence, the sequence{F(hn)} is uniformly bounded. According to the absolute continuity of Lebesgue integral, for any ɛ>0, there exists a δ>0 such that if 0≤t1-t2<δ, then we have 0≤N∫t1t2ρ*(t)dt<ɛ. Thus, (2.20) means that {F(hn)} is equicontinuous.
DenoteΩm=[0,m]. Obviously, {F(hn)} is uniformly bounded and equicontinuous on Ωm for m=1,2,…. By Ascoli-Arzela Theorem, there exists a subsequence {F(hn(1))} of {F(hn)} being convergent in C(Ω1), we may assume F(hn(1))→v1(·) in C(Ω1). Since {F(hn(1))} is uniformly bounded and equicontinuous on Ω2, there exists a subsequence {F(hn(2))} of {F(hn(1))} such that {F(hn(2))} is convergent in C(Ω2), we may assume F(hn(2))→v2(·) in C(Ω2). Obviously, v2(t)=v1(t), for any t∈Ω1. Repeating the process, we get a subsequence {F(hn(m+1))} of {F(hn(m))} such that {F(hn(m+1))} is convergent in C(Ωm+1), we may assume F(hn(m+1))→vm+1(·) in C(Ωm+1). Obviously, vm+1(t)=vm(t) for any t∈Ωm. Select the diagonal element, we can see that {F(hn(n))} is a subsequence of {F(hn)} which satisfies that {F(hn(n))} is convergent in C(Ωm)(m=1,2,…) and F(hn(n))→vm(·) in C(Ωm)(m=1,2,…). Thus, we get a function v which is defined on [0,+∞) such that v(t)=vm(t) for any t∈Ωm, and F(hn(n))→v(·) in C(Ωm)(m=1,2,…).
From (2.20), it is easy to see that for any n=1,2,…, limt→∞F(hn)(t) exists (we denote the limit by F(hn(n))(+∞)), and, for any ɛ>0, there exists an integer Rɛ>0 such that ∫Rɛ+∞ρ*(t)dt<ɛ/N, and then
(2.21)|F(hn(n))(+∞)-F(hn)(t)|≤N∫Rɛ+∞ρ*(t)dt<ɛ,∀t≥Rɛ,∀n=1,2,….
Since{F(hn)} is uniformly bounded, then {F(hn(n))(+∞)} is bounded. By choosing a subsequence, we may assume that
(2.22)lim⁡n→∞F(hn(n))(+∞)=b.
We claim thatlimt→∞v(t)=b. In fact, for any t≥Rɛ, from (2.21), we have
(2.23)|v(t)-b|≤|v(t)-F(hn(n))(t)|+|F(hn(n))(t)-F(hn(n))(+∞)|+|F(hn(n))(+∞)-b|≤|v(t)-F(hn(n))(t)|+ɛ+|F(hn(n))(+∞)-b|.
Sincelimn→∞F(hn(n))(t)=v(t) and limn→∞F(hn(n))(+∞)=b, letting n→∞, the above inequality implies
(2.24)|v(t)-b|≤ɛ,∀t≥Rɛ.
Thus,(2.25)lim⁡t→∞v(t)=b=lim⁡n→∞F(hn(n))(+∞).
Next, we will prove thatF(hn(n)) tend to v uniformly.
Supposet≥Rɛ. From (2.21) and (2.24), we have
(2.26)|F(hn(n))(t)-v(t)|≤|F(hn(n))(t)-F(hn(n))(+∞)|+|F(hn(n))(+∞)-b|+|b-v(t)|≤ɛ+|F(hn(n))(+∞)-b|+ɛ=2ɛ+|F(hn(n))(+∞)-b|.
From (2.22), there exists a N1>0 such that |F(hn(n))(+∞)-b|≤ɛ for n≥N1. Thus, for any t≥Rɛ,
(2.27)|F(hn(n))(t)-v(t)|≤3ɛ,∀n≥N1.
Supposet∈[0,Rɛ]. Since F(hn(n))→v in C([0,Rɛ]), there exists a N2>0 such that
(2.28)|F(hn(n))(t)-v(t)|≤ɛ,∀n≥N2.
Thus,(2.29)|F(hn(n))(t)-v(t)|≤3ɛ,∀t∈[0,+∞),∀n≥N1+N2.
This means thatF(hn(n)) tend to v uniformly, that is, F(hn(n)) tend to v in C.
According to the bounded continuous of the operatorρ, we can choose a subsequence of {ρ(hn)+F(hn)} (which we still denote {ρ(hn)+F(hn)}) which is convergent in C, then w(t)φ(t,K(hn)′(t))=ρ(hn)+F(hn) is convergent in C.
Since(2.30)K(hn)(t)=F{φ-1[t,(w(t))-1(ρ(hn)+F(hn))]}(t),∀t∈[0,+∞),
it follows from the continuity of φ-1 and the integrability of w(t)-1/(p(t)-1) in L1 that K(hn) is convergent in C.  Thus, {un} is convergent in C1. This completes the proof.Let us defineP:C1→C1 as(2.31)P(h)=11-σ(∫0+∞e(t)K(h)(t)dt-K(h)(+∞)+e1).
It is easy to see that P is compact continuous.Throughout the paper, we denoteNf(u):[0,+∞)×C1→L1 the Nemytskii operator associated to f defined by (2.32)Nf(u)(t)=f(t,u(t),(w(t))1/(p(t)-1)u′(t),S(u)(t),T(u)(t)),a.e.on⁡J.Lemma 2.4.
In the Case (i),u is a solution of (1.1)-(1.2) if and only if u is a solution of the following abstract equation:
(2.33)u=P(δNf(u))+K(δNf(u)).Proof.
Ifu is a solution of (1.1)-(1.2), by integrating (1.1) from 0 to t, we find that
(2.34)w(t)φ(t,u′(t))=ρ(δNf(u))+F(δNf(u))(t),∀t∈(0,+∞),
which implies that
(2.35)u(t)=u(0)+F{φ-1[t,(w(t))-1(ρ(δNf(u))+F(δNf(u))(t))]}(t),∀t∈[0,+∞).
Fromu(+∞)=∫0+∞e(t)u(t)dt+e1, we have
(2.36)u(0)=∫0+∞{e(t)∫0tφ-1[r,(w(r))-1(ρ+F(δNf(u)))]dr}dt+e11-σ-∫0+∞φ-1[r,(w(r))-1(ρ+F(δNf(u)))]dr1-σ=P(δNf(u)).
So we have(2.37)u=P(δNf(u))+K(δNf(u)).
Conversely, ifu is a solution of (2.33), we have
(2.38)u(0)=P(δNf(u))+K(δNf(u))(0)=P(δNf(u))=11-σ(∫0+∞e(t)K(h)(t)dt-K(h)(+∞)+e1),
which implies that
(2.39)(1-σ)u(0)+K(h)(+∞)=∫0+∞e(t)(u(t)-u(0))dt+e1,
then
(2.40)u(+∞)=∫0+∞e(t)u(t)dt+e1.
From (2.33), we can obtain
(2.41)w(t)φ(t,u′(t))=ρ(δNf(u))+F(δNf(u))(t).
It follows from the condition of the mappingρ that
(2.42)w(+∞)φ(+∞,u′(+∞))=ρ(δNf(u))+F(δNf(u))(+∞)=11-∑i=1m-2αi(∑i=1m-2αi∫0ξiδNf(u)(t)dt-∫0+∞δNf(u)(t)dt+e2)+∫0+∞δNf(u)(t)dt,
and then
(2.43)(1-∑i=1m-2αi)w(+∞)φ(+∞,u′(+∞))=∑i=1m-2αi∫0ξiδNf(u)(t)dt-∫0+∞δNf(u)(t)dt+e2+(1-∑i=1m-2αi)∫0+∞δNf(u)(t)dt=∑i=1m-2αi∫0ξiδNf(u)(t)dt-∑i=1m-2αi∫0+∞δNf(u)(t)dt+e2=∑i=1m-2αiw(ξi)φ(ξi,u′(ξi))-∑i=1m-2αiw(+∞)φ(+∞,u′(+∞))+e2,
thus
(2.44)lim⁡t→+∞w(t)|u′|p(t)-2u′(t)=∑i=1m-2αiw(ξi)|u′|p(ξi)-2u′(ξi)+e2.
From (2.40) and (2.44), we obtain (1.2).
From (2.41), we have
(2.45)(w(t)φ(t,u′))′=δNf(u)(t).
Hence,u is a solution of (1.1)-(1.2). This completes the proof.


### 2.2. Case (ii):∑i=1m-2αi∈[0,1),σ=1

We denoteρ1=w(0)φ(0,u′(0)) in (2.7). It is easy to see that ρ1 is dependent on g(·), and we have (2.46)u(t)=u(0)+F{φ-1[t,(w(t))-1(ρ1+F(g))]}(t),t∈J.The boundary value condition (1.2) implies that (2.47)∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-1(ρ1+F(g)(r))]dr}dt-e1=0,ρ1=11-∑i=1m-2αi(∑i=1m-2αi∫0ξig(t)dt-∫0+∞g(t)dt+e2).For fixedh∈L1, we define ρ1:L1→ℝN as(2.48)ρ1(h)=11-∑i=1m-2αi(∑i=1m-2αi∫0ξih(t)dt-∫0+∞h(t)dt+e2).Similar to Lemma2.2, we have the following.Lemma 2.5.
ρ1:L1→ℝN is continuous and sends bounded sets of L1 to bounded sets of ℝN. Moreover,
(2.49)|ρ1(h)|≤2N1-∑i=1m-2αi⋅(‖h‖L1+|e2|).It is clear thatρ1(·) is continuous and sends bounded sets of L1 to bounded sets of ℝN, and, hence, it is a compact continuous mapping.Let us define(2.50)P1:C1⟶C1,u⟼u(0),Θ:L1⟶RN,h⟼∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-1(ρ1(h)+F(h)(r))]dr}dt-e1,K1(h)(t)∶=(K1∘h)(t)=F{φ-1[t,(w(t))-1(ρ1(h)+F(h))]}(t),∀t∈[0,+∞).Lemma 2.6.
The operatorK1 is continuous and sends equi-integrable sets in L1 to relatively compact sets in C1.Proof.
Similar to the proof of Lemma2.3, we omit it here.Lemma 2.7.
In Case (ii),u is a solution of (1.1)-(1.2) if and only if u is a solution of the following abstract equation:
(2.51)u=P1u+Θ(δNf(u))+K1(δNf(u)).Proof.
Ifu is a solution of (1.1)-(1.2), by integrating (1.1) from 0 to t, we find that
(2.52)w(t)φ(t,u′(t))=ρ1(δNf(u))+F(δNf(u))(t),∀t∈(0,+∞),
which implies that
(2.53)u(t)=u(0)+F{φ-1[t,(w(t))-1(ρ1(δNf(u))+F(δNf(u))(t))]}(t),∀t∈[0,+∞).
Fromσ=1 and u(+∞)=∫0+∞e(t)u(t)dt+e1, we obtain
(2.54)∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-1(ρ1(δNf(u))+F(δNf(u))(t))]dr}dt-e1=Θ(δNf(u))=0,
then
(2.55)u=P1u+Θ(δNf(u))+K1(δNf(u)).
Conversely, ifu is a solution of (2.51), then
(2.56)u(0)=P1u+Θ(δNf(u))+K1(δNf(u))(0)=u(0)+Θ(δNf(u)).
Thus,Θ(δNf(u))=0, and we have
(2.57)∫0+∞{e(t)∫0+∞φ-1[r,(w(r))-1(ρ1(δNf(u))+F(δNf(u))(r))]dr}dt=∫0+∞{e(t)∫0tφ-1[r,(w(r))-1(ρ1(δNf(u))+F(δNf(u))(r))]dr}dt+e1,
then
(2.58)∫0+∞{e(t)(u(+∞)-u(0))}dt=∫0+∞{e(t)(u(t)-u(0))}dt+e1.
Thus,(2.59)u(+∞)=∫0+∞e(t)u(t)dt+e1.
Similar to the proof of Lemma2.4, we can have
(2.60)lim⁡t→+∞w(t)|u′|p(t)-2u′(t)=∑i=1m-2αiw(ξi)|u′|p(ξi)-2u′(ξi)+e2.
From (2.59) and (2.60), we obtain (1.2).
From (2.51), we have
(2.61)w(t)φ(t,u′(t))=ρ1(δNf(u))+F(δNf(u))(t),
then
(2.62)(w(t)φ(t,u′))′=δNf(u)(t).
Hence,u is a solution of (1.1)-(1.2). This completes the proof.


### 2.3. Case (iii):∑i=1m-2αi=1,σ=1

We denoteρ2=w(0)φ(0,u′(0)) in (2.7). It is easy to see that ρ2 is dependent on g(·), then we find that (2.63)u(t)=u(0)+F{φ-1[t,(w(t))-1(ρ2+F(g))]}(t),t∈J.The boundary value condition (1.2) implies that (2.64)∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-1(ρ2+F(g)(r))]dr}dt-e1=0,∑i=1m-2αi∫ξi+∞g(t)dt-e2=0.For fixedh∈L1, we denote (2.65)Λh(ρ2)=∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-1(ρ2+F(h)(r))]dr}dt-e1.Throughout the paper, we denote(2.66)E#=∫0+∞e(t)∫t+∞(w(r))-1/(p(r)-1)drdt.Lemma 2.8.
The functionΛh(·) has the following properties.(i)
For any fixedh∈L1, the equation
(2.67)Λh(ρ2)=0has a unique solutionρ2(h)∈ℝN.(ii)
The functionρ2:L1→ℝN, defined in (i), is continuous and sends bounded sets to bounded sets. Moreover,
(2.68)|ρ2(h)|≤3N(E#+1E#)p+[‖h‖0+2N|e1|p#-1],where the notationMp#-1 means
(2.69)Mp#-1={Mp+-1,M>1,Mp--1,M≤1.Proof.
(i) From Lemma2.1, it is immediate that
(2.70)〈Λh(x)-Λh(y),x-y〉>0,forx≠y,
and, hence, if (2.67) has a solution, then it is unique.
Lett0=3N((E#+1)/E#)p+[∥h∥0+2N|e1|p#-1]. Since(w(t))-1/(p(t)-1)∈L1(0,+∞) and h∈L1, if |ρ2|>t0, it is easy to see that there exists an i∈{1,…,N} such that the ith component ρ2i of ρ2 satisfies |ρ2i|≥|ρ2|/N>3((E#+1)/E#)p+[∥h∥0+2N|e1|p#-1]. Thus, (ρ2i+hi(t)) keeps sign on J and
(2.71)|ρ2i+hi(t)|≥|ρ2i|-‖h‖0≥2|ρ2|3N>2(E#+1E#)p+[‖h‖0+2N|e1|p#-1],∀t∈J.
Obviously,|ρ2+h(t)|≤4|ρ2|/3≤2N|ρ2i+hi(t)|, then
(2.72)|ρ2+h(t)|(2-p(t))/(p(t)-1)|ρ2i+hi(t)|>12N|ρ2i+hi(t)|1/(p(t)-1)>E#+12NE#|e1|,∀t∈J.
Thus, theith component Λhi(ρ2) of Λh(ρ2) is nonzero and keeps sign, and then we have
(2.73)∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-1(ρ2+F(h)(r))]dr}dt-e1≠0.
Let us consider the equation(2.74)λΛh(ρ2)+(1-λ)ρ2=0,λ∈[0,1].
It is easy to see that all the solutions of (2.74) belong to b(t0+1)={x∈ℝN∣|x|<t0+1}. So, we have
(2.75)dB[Λh(ρ2),b(t0+1),0]=dB[I,b(t0+1),0]≠0,
and it means the existence of solutions of Λh(ρ2)=0.
In this way, we define a functionρ2(h):L1→ℝN, which satisfies
(2.76)Λh(ρ2(h))=0.
(ii) By the proof of (i), we also obtainρ2 sends bounded sets to bounded sets, and
(2.77)|ρ2(h)|≤3N(E#+1E#)p+[‖h‖0+2N|e1|p#-1].
It only remains to prove the continuity ofρ2. Let {un} be a convergent sequence in L1 and un→u as n→+∞. Since {ρ2(un)} is a bounded sequence, then it contains a convergent subsequence {ρ2(unj)}. Let ρ2(unj)→ρ0 as j→+∞. Since Λunj(ρ2(unj))=0, letting j→+∞, we have Λu(ρ0)=0. From (i), we get ρ0=ρ2(u), it means that ρ2 is continuous. This completes the proof.It is clear thatρ2(·) is continuous and sends bounded sets of L1 to bounded sets of ℝN, and, hence, it is a compact continuous mapping.Let us define(2.78)P2:C1⟶C1,u⟼u(0),(2.79)Q:L1⟶RN,h⟼∑i=1m-2αi∫ξi+∞h(t)dt-e2,(2.80)Q*:L1⟶L1,h⟼τ(t)(∑i=1m-2αi∫ξi+∞h(t)dt-e2),
where τ∈([0,+∞),ℝ) and satisfies 0<τ(t)<1,t∈J,∑i=1m-2αi∫ξi+∞τ(t)dt=1. We denote K2:L1→C1 as (2.81)K2(h)(t)∶=F{φ-1[t,(w(t))-1(ρ2((I-Q*)h)+F((I-Q*)h))]}(t),∀t∈[0,+∞).Similar to Lemmas2.3 and 2.7, we have the followingLemma 2.9.
The operatorK2 is continuous and sends equi-integrable sets in L1 to relatively compact sets in C1.Lemma 2.10.
In Case (iii),u is a solution of (1.1)-(1.2) if and only if u is a solution of the following abstract equation:
(2.82)u=P2u+Q(δNf(u))+K2(δNf(u)).


## 2.1. Case (i):∑i=1m-2αi∈[0,1),σ∈[0,1)

We denoteρ=w(0)φ(0,u′(0)) in (2.7). It is easy to see that ρ is dependent on g(·), then we find that (2.8)u(t)=u(0)+F{φ-1[t,(w(t))-1(ρ+F(g))]}(t),t∈J.The boundary value condition (1.2) implies that (2.9)u(0)=∫0+∞{e(t)∫0tφ-1[r,(w(r))-1(ρ+F(g)(r))]dr}dt+e11-σ-∫0+∞φ-1[r,(w(r))-1(ρ+F(g)(r))]dr1-σ,ρ=11-∑i=1m-2αi(∑i=1m-2αi∫0ξig(s)ds-∫0+∞g(s)ds+e2).For fixedh∈L1, we define ρ:L1→ℝN as(2.10)ρ(h)=11-∑i=1m-2αi(∑i=1m-2αi∫0ξih(t)dt-∫0+∞h(t)dt+e2).Lemma 2.2.
ρ:L1→ℝN is continuous and sends bounded sets of L1 to bounded sets of ℝN. Moreover,
(2.11)|ρ(h)|≤2N1-∑i=1m-2αi⋅(‖h‖L1+|e2|).Proof.
Sinceρ(·) consists of continuous operators, it is continuous. It is easy to see that
(2.12)|ρ(h)|≤2N1-∑i=1m-2αi⋅(‖h‖L1+|e2|).
This completes the proof.It is clear thatρ(·) is continuous and sends bounded sets of L1 to bounded sets of ℝN, and hence it is a compact continuous mapping.Ifu is a solution of (2.4) with (1.2), we find that (2.13)u(t)=u(0)+F{φ-1[t,(w(t))-1(ρ+F(g))]}(t),t∈J,u(0)=∫0+∞{e(t)∫0tφ-1[r,(w(r))-1(ρ+F(g)(r))]dr}dt+e11-σ-∫0+∞φ-1[r,(w(r))-1(ρ+F(g)(r))]dr1-σ.We denote(2.14)K(h)(t)∶=(K∘h)(t)=F{φ-1[t,(w(t))-1(ρ(h)+F(h))]}(t),∀t∈(0,+∞).We say a setU⊂L1 be equi-integrable, if there exists a nonnegative ρ*∈L1(J,ℝ), such that (2.15)|h(t)|≤ρ*(t)a.e.inJ,foranyh∈U.Lemma 2.3.
The operatorK is continuous and sends equi-integrable sets in L1 to relatively compact sets in C1.Proof.
It is easy to check thatK(h)(t)∈C1, for all h∈L1. Since (w(t))-1/(p(t)-1)∈L1 and
(2.16)K(h)′(t)=φ-1[t,(w(t))-1(ρ(h)+F(h))],∀t∈[0,+∞),
it is easy to check that K is a continuous operator from L1 to C1.
Let nowU be an equi-integrable set in L1, then there exists a nonnegative ρ*∈L1(J,ℝ), such that
(2.17)|h(t)|≤ρ*(t)a.e.inJ,foranyh∈U.
We want to show thatK(U)¯⊂C1  is a compact set.
Let{un} be a sequence in K(U), then there exists a sequence {hn}⊂U such that un=K(hn). Since hn(t)=(hn1(t),…,hnN(t)), where hni(t)∈L1(J,ℝ)(i=1,…,N), we have
(2.18)|hni(t)|≤|hn(t)|,∀i=1,…,N,
then for any t1,t2∈J with t1<t2, we have
(2.19)|∫t1t2hn(t)dt|=∑i=1N(∫t1t2hni(t)dt)2≤∑i=1N(∫t1t2|hni(t)|dt)2≤∑i=1N(∫t1t2|hn(t)|dt)2≤N∫t1t2|hn(t)|dt,
which together with (2.17) implies
(2.20)|F(hn)(t1)-F(hn)(t2)|=|∫0t1hn(t)dt-∫0t2hn(t)dt|=|∫t1t2hn(t)dt|≤N∫t1t2|hn(t)|dt≤N∫t1t2ρ*(t)dt.
Hence, the sequence{F(hn)} is uniformly bounded. According to the absolute continuity of Lebesgue integral, for any ɛ>0, there exists a δ>0 such that if 0≤t1-t2<δ, then we have 0≤N∫t1t2ρ*(t)dt<ɛ. Thus, (2.20) means that {F(hn)} is equicontinuous.
DenoteΩm=[0,m]. Obviously, {F(hn)} is uniformly bounded and equicontinuous on Ωm for m=1,2,…. By Ascoli-Arzela Theorem, there exists a subsequence {F(hn(1))} of {F(hn)} being convergent in C(Ω1), we may assume F(hn(1))→v1(·) in C(Ω1). Since {F(hn(1))} is uniformly bounded and equicontinuous on Ω2, there exists a subsequence {F(hn(2))} of {F(hn(1))} such that {F(hn(2))} is convergent in C(Ω2), we may assume F(hn(2))→v2(·) in C(Ω2). Obviously, v2(t)=v1(t), for any t∈Ω1. Repeating the process, we get a subsequence {F(hn(m+1))} of {F(hn(m))} such that {F(hn(m+1))} is convergent in C(Ωm+1), we may assume F(hn(m+1))→vm+1(·) in C(Ωm+1). Obviously, vm+1(t)=vm(t) for any t∈Ωm. Select the diagonal element, we can see that {F(hn(n))} is a subsequence of {F(hn)} which satisfies that {F(hn(n))} is convergent in C(Ωm)(m=1,2,…) and F(hn(n))→vm(·) in C(Ωm)(m=1,2,…). Thus, we get a function v which is defined on [0,+∞) such that v(t)=vm(t) for any t∈Ωm, and F(hn(n))→v(·) in C(Ωm)(m=1,2,…).
From (2.20), it is easy to see that for any n=1,2,…, limt→∞F(hn)(t) exists (we denote the limit by F(hn(n))(+∞)), and, for any ɛ>0, there exists an integer Rɛ>0 such that ∫Rɛ+∞ρ*(t)dt<ɛ/N, and then
(2.21)|F(hn(n))(+∞)-F(hn)(t)|≤N∫Rɛ+∞ρ*(t)dt<ɛ,∀t≥Rɛ,∀n=1,2,….
Since{F(hn)} is uniformly bounded, then {F(hn(n))(+∞)} is bounded. By choosing a subsequence, we may assume that
(2.22)lim⁡n→∞F(hn(n))(+∞)=b.
We claim thatlimt→∞v(t)=b. In fact, for any t≥Rɛ, from (2.21), we have
(2.23)|v(t)-b|≤|v(t)-F(hn(n))(t)|+|F(hn(n))(t)-F(hn(n))(+∞)|+|F(hn(n))(+∞)-b|≤|v(t)-F(hn(n))(t)|+ɛ+|F(hn(n))(+∞)-b|.
Sincelimn→∞F(hn(n))(t)=v(t) and limn→∞F(hn(n))(+∞)=b, letting n→∞, the above inequality implies
(2.24)|v(t)-b|≤ɛ,∀t≥Rɛ.
Thus,(2.25)lim⁡t→∞v(t)=b=lim⁡n→∞F(hn(n))(+∞).
Next, we will prove thatF(hn(n)) tend to v uniformly.
Supposet≥Rɛ. From (2.21) and (2.24), we have
(2.26)|F(hn(n))(t)-v(t)|≤|F(hn(n))(t)-F(hn(n))(+∞)|+|F(hn(n))(+∞)-b|+|b-v(t)|≤ɛ+|F(hn(n))(+∞)-b|+ɛ=2ɛ+|F(hn(n))(+∞)-b|.
From (2.22), there exists a N1>0 such that |F(hn(n))(+∞)-b|≤ɛ for n≥N1. Thus, for any t≥Rɛ,
(2.27)|F(hn(n))(t)-v(t)|≤3ɛ,∀n≥N1.
Supposet∈[0,Rɛ]. Since F(hn(n))→v in C([0,Rɛ]), there exists a N2>0 such that
(2.28)|F(hn(n))(t)-v(t)|≤ɛ,∀n≥N2.
Thus,(2.29)|F(hn(n))(t)-v(t)|≤3ɛ,∀t∈[0,+∞),∀n≥N1+N2.
This means thatF(hn(n)) tend to v uniformly, that is, F(hn(n)) tend to v in C.
According to the bounded continuous of the operatorρ, we can choose a subsequence of {ρ(hn)+F(hn)} (which we still denote {ρ(hn)+F(hn)}) which is convergent in C, then w(t)φ(t,K(hn)′(t))=ρ(hn)+F(hn) is convergent in C.
Since(2.30)K(hn)(t)=F{φ-1[t,(w(t))-1(ρ(hn)+F(hn))]}(t),∀t∈[0,+∞),
it follows from the continuity of φ-1 and the integrability of w(t)-1/(p(t)-1) in L1 that K(hn) is convergent in C.  Thus, {un} is convergent in C1. This completes the proof.Let us defineP:C1→C1 as(2.31)P(h)=11-σ(∫0+∞e(t)K(h)(t)dt-K(h)(+∞)+e1).
It is easy to see that P is compact continuous.Throughout the paper, we denoteNf(u):[0,+∞)×C1→L1 the Nemytskii operator associated to f defined by (2.32)Nf(u)(t)=f(t,u(t),(w(t))1/(p(t)-1)u′(t),S(u)(t),T(u)(t)),a.e.on⁡J.Lemma 2.4.
In the Case (i),u is a solution of (1.1)-(1.2) if and only if u is a solution of the following abstract equation:
(2.33)u=P(δNf(u))+K(δNf(u)).Proof.
Ifu is a solution of (1.1)-(1.2), by integrating (1.1) from 0 to t, we find that
(2.34)w(t)φ(t,u′(t))=ρ(δNf(u))+F(δNf(u))(t),∀t∈(0,+∞),
which implies that
(2.35)u(t)=u(0)+F{φ-1[t,(w(t))-1(ρ(δNf(u))+F(δNf(u))(t))]}(t),∀t∈[0,+∞).
Fromu(+∞)=∫0+∞e(t)u(t)dt+e1, we have
(2.36)u(0)=∫0+∞{e(t)∫0tφ-1[r,(w(r))-1(ρ+F(δNf(u)))]dr}dt+e11-σ-∫0+∞φ-1[r,(w(r))-1(ρ+F(δNf(u)))]dr1-σ=P(δNf(u)).
So we have(2.37)u=P(δNf(u))+K(δNf(u)).
Conversely, ifu is a solution of (2.33), we have
(2.38)u(0)=P(δNf(u))+K(δNf(u))(0)=P(δNf(u))=11-σ(∫0+∞e(t)K(h)(t)dt-K(h)(+∞)+e1),
which implies that
(2.39)(1-σ)u(0)+K(h)(+∞)=∫0+∞e(t)(u(t)-u(0))dt+e1,
then
(2.40)u(+∞)=∫0+∞e(t)u(t)dt+e1.
From (2.33), we can obtain
(2.41)w(t)φ(t,u′(t))=ρ(δNf(u))+F(δNf(u))(t).
It follows from the condition of the mappingρ that
(2.42)w(+∞)φ(+∞,u′(+∞))=ρ(δNf(u))+F(δNf(u))(+∞)=11-∑i=1m-2αi(∑i=1m-2αi∫0ξiδNf(u)(t)dt-∫0+∞δNf(u)(t)dt+e2)+∫0+∞δNf(u)(t)dt,
and then
(2.43)(1-∑i=1m-2αi)w(+∞)φ(+∞,u′(+∞))=∑i=1m-2αi∫0ξiδNf(u)(t)dt-∫0+∞δNf(u)(t)dt+e2+(1-∑i=1m-2αi)∫0+∞δNf(u)(t)dt=∑i=1m-2αi∫0ξiδNf(u)(t)dt-∑i=1m-2αi∫0+∞δNf(u)(t)dt+e2=∑i=1m-2αiw(ξi)φ(ξi,u′(ξi))-∑i=1m-2αiw(+∞)φ(+∞,u′(+∞))+e2,
thus
(2.44)lim⁡t→+∞w(t)|u′|p(t)-2u′(t)=∑i=1m-2αiw(ξi)|u′|p(ξi)-2u′(ξi)+e2.
From (2.40) and (2.44), we obtain (1.2).
From (2.41), we have
(2.45)(w(t)φ(t,u′))′=δNf(u)(t).
Hence,u is a solution of (1.1)-(1.2). This completes the proof.


## 2.2. Case (ii):∑i=1m-2αi∈[0,1),σ=1

We denoteρ1=w(0)φ(0,u′(0)) in (2.7). It is easy to see that ρ1 is dependent on g(·), and we have (2.46)u(t)=u(0)+F{φ-1[t,(w(t))-1(ρ1+F(g))]}(t),t∈J.The boundary value condition (1.2) implies that (2.47)∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-1(ρ1+F(g)(r))]dr}dt-e1=0,ρ1=11-∑i=1m-2αi(∑i=1m-2αi∫0ξig(t)dt-∫0+∞g(t)dt+e2).For fixedh∈L1, we define ρ1:L1→ℝN as(2.48)ρ1(h)=11-∑i=1m-2αi(∑i=1m-2αi∫0ξih(t)dt-∫0+∞h(t)dt+e2).Similar to Lemma2.2, we have the following.Lemma 2.5.
ρ1:L1→ℝN is continuous and sends bounded sets of L1 to bounded sets of ℝN. Moreover,
(2.49)|ρ1(h)|≤2N1-∑i=1m-2αi⋅(‖h‖L1+|e2|).It is clear thatρ1(·) is continuous and sends bounded sets of L1 to bounded sets of ℝN, and, hence, it is a compact continuous mapping.Let us define(2.50)P1:C1⟶C1,u⟼u(0),Θ:L1⟶RN,h⟼∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-1(ρ1(h)+F(h)(r))]dr}dt-e1,K1(h)(t)∶=(K1∘h)(t)=F{φ-1[t,(w(t))-1(ρ1(h)+F(h))]}(t),∀t∈[0,+∞).Lemma 2.6.
The operatorK1 is continuous and sends equi-integrable sets in L1 to relatively compact sets in C1.Proof.
Similar to the proof of Lemma2.3, we omit it here.Lemma 2.7.
In Case (ii),u is a solution of (1.1)-(1.2) if and only if u is a solution of the following abstract equation:
(2.51)u=P1u+Θ(δNf(u))+K1(δNf(u)).Proof.
Ifu is a solution of (1.1)-(1.2), by integrating (1.1) from 0 to t, we find that
(2.52)w(t)φ(t,u′(t))=ρ1(δNf(u))+F(δNf(u))(t),∀t∈(0,+∞),
which implies that
(2.53)u(t)=u(0)+F{φ-1[t,(w(t))-1(ρ1(δNf(u))+F(δNf(u))(t))]}(t),∀t∈[0,+∞).
Fromσ=1 and u(+∞)=∫0+∞e(t)u(t)dt+e1, we obtain
(2.54)∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-1(ρ1(δNf(u))+F(δNf(u))(t))]dr}dt-e1=Θ(δNf(u))=0,
then
(2.55)u=P1u+Θ(δNf(u))+K1(δNf(u)).
Conversely, ifu is a solution of (2.51), then
(2.56)u(0)=P1u+Θ(δNf(u))+K1(δNf(u))(0)=u(0)+Θ(δNf(u)).
Thus,Θ(δNf(u))=0, and we have
(2.57)∫0+∞{e(t)∫0+∞φ-1[r,(w(r))-1(ρ1(δNf(u))+F(δNf(u))(r))]dr}dt=∫0+∞{e(t)∫0tφ-1[r,(w(r))-1(ρ1(δNf(u))+F(δNf(u))(r))]dr}dt+e1,
then
(2.58)∫0+∞{e(t)(u(+∞)-u(0))}dt=∫0+∞{e(t)(u(t)-u(0))}dt+e1.
Thus,(2.59)u(+∞)=∫0+∞e(t)u(t)dt+e1.
Similar to the proof of Lemma2.4, we can have
(2.60)lim⁡t→+∞w(t)|u′|p(t)-2u′(t)=∑i=1m-2αiw(ξi)|u′|p(ξi)-2u′(ξi)+e2.
From (2.59) and (2.60), we obtain (1.2).
From (2.51), we have
(2.61)w(t)φ(t,u′(t))=ρ1(δNf(u))+F(δNf(u))(t),
then
(2.62)(w(t)φ(t,u′))′=δNf(u)(t).
Hence,u is a solution of (1.1)-(1.2). This completes the proof.


## 2.3. Case (iii):∑i=1m-2αi=1,σ=1

We denoteρ2=w(0)φ(0,u′(0)) in (2.7). It is easy to see that ρ2 is dependent on g(·), then we find that (2.63)u(t)=u(0)+F{φ-1[t,(w(t))-1(ρ2+F(g))]}(t),t∈J.The boundary value condition (1.2) implies that (2.64)∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-1(ρ2+F(g)(r))]dr}dt-e1=0,∑i=1m-2αi∫ξi+∞g(t)dt-e2=0.For fixedh∈L1, we denote (2.65)Λh(ρ2)=∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-1(ρ2+F(h)(r))]dr}dt-e1.Throughout the paper, we denote(2.66)E#=∫0+∞e(t)∫t+∞(w(r))-1/(p(r)-1)drdt.Lemma 2.8.
The functionΛh(·) has the following properties.(i)
For any fixedh∈L1, the equation
(2.67)Λh(ρ2)=0has a unique solutionρ2(h)∈ℝN.(ii)
The functionρ2:L1→ℝN, defined in (i), is continuous and sends bounded sets to bounded sets. Moreover,
(2.68)|ρ2(h)|≤3N(E#+1E#)p+[‖h‖0+2N|e1|p#-1],where the notationMp#-1 means
(2.69)Mp#-1={Mp+-1,M>1,Mp--1,M≤1.Proof.
(i) From Lemma2.1, it is immediate that
(2.70)〈Λh(x)-Λh(y),x-y〉>0,forx≠y,
and, hence, if (2.67) has a solution, then it is unique.
Lett0=3N((E#+1)/E#)p+[∥h∥0+2N|e1|p#-1]. Since(w(t))-1/(p(t)-1)∈L1(0,+∞) and h∈L1, if |ρ2|>t0, it is easy to see that there exists an i∈{1,…,N} such that the ith component ρ2i of ρ2 satisfies |ρ2i|≥|ρ2|/N>3((E#+1)/E#)p+[∥h∥0+2N|e1|p#-1]. Thus, (ρ2i+hi(t)) keeps sign on J and
(2.71)|ρ2i+hi(t)|≥|ρ2i|-‖h‖0≥2|ρ2|3N>2(E#+1E#)p+[‖h‖0+2N|e1|p#-1],∀t∈J.
Obviously,|ρ2+h(t)|≤4|ρ2|/3≤2N|ρ2i+hi(t)|, then
(2.72)|ρ2+h(t)|(2-p(t))/(p(t)-1)|ρ2i+hi(t)|>12N|ρ2i+hi(t)|1/(p(t)-1)>E#+12NE#|e1|,∀t∈J.
Thus, theith component Λhi(ρ2) of Λh(ρ2) is nonzero and keeps sign, and then we have
(2.73)∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-1(ρ2+F(h)(r))]dr}dt-e1≠0.
Let us consider the equation(2.74)λΛh(ρ2)+(1-λ)ρ2=0,λ∈[0,1].
It is easy to see that all the solutions of (2.74) belong to b(t0+1)={x∈ℝN∣|x|<t0+1}. So, we have
(2.75)dB[Λh(ρ2),b(t0+1),0]=dB[I,b(t0+1),0]≠0,
and it means the existence of solutions of Λh(ρ2)=0.
In this way, we define a functionρ2(h):L1→ℝN, which satisfies
(2.76)Λh(ρ2(h))=0.
(ii) By the proof of (i), we also obtainρ2 sends bounded sets to bounded sets, and
(2.77)|ρ2(h)|≤3N(E#+1E#)p+[‖h‖0+2N|e1|p#-1].
It only remains to prove the continuity ofρ2. Let {un} be a convergent sequence in L1 and un→u as n→+∞. Since {ρ2(un)} is a bounded sequence, then it contains a convergent subsequence {ρ2(unj)}. Let ρ2(unj)→ρ0 as j→+∞. Since Λunj(ρ2(unj))=0, letting j→+∞, we have Λu(ρ0)=0. From (i), we get ρ0=ρ2(u), it means that ρ2 is continuous. This completes the proof.It is clear thatρ2(·) is continuous and sends bounded sets of L1 to bounded sets of ℝN, and, hence, it is a compact continuous mapping.Let us define(2.78)P2:C1⟶C1,u⟼u(0),(2.79)Q:L1⟶RN,h⟼∑i=1m-2αi∫ξi+∞h(t)dt-e2,(2.80)Q*:L1⟶L1,h⟼τ(t)(∑i=1m-2αi∫ξi+∞h(t)dt-e2),
where τ∈([0,+∞),ℝ) and satisfies 0<τ(t)<1,t∈J,∑i=1m-2αi∫ξi+∞τ(t)dt=1. We denote K2:L1→C1 as (2.81)K2(h)(t)∶=F{φ-1[t,(w(t))-1(ρ2((I-Q*)h)+F((I-Q*)h))]}(t),∀t∈[0,+∞).Similar to Lemmas2.3 and 2.7, we have the followingLemma 2.9.
The operatorK2 is continuous and sends equi-integrable sets in L1 to relatively compact sets in C1.Lemma 2.10.
In Case (iii),u is a solution of (1.1)-(1.2) if and only if u is a solution of the following abstract equation:
(2.82)u=P2u+Q(δNf(u))+K2(δNf(u)).


## 3. Existence of Solutions in Case (i)

In this section, we will apply Leray-Schauder's degree to deal with the existence of solutions for (1.1)-(1.2) when ∑i=1m-2αi∈[0,1),σ∈[0,1). Moreover, we give the existence of nonnegative solutions.Theorem 3.1.
In Case (i), iff satisfies sub-(p--1) growth condition, then problem (1.1)-(1.2) has at least a solution for any fixed parameter δ.Proof.
DenoteΨf(u,λ)∶=P(λδNf(u))+K(λδNf(u)), where Nf(u) is defined in (2.32). We know that (1.1)-(1.2) has the same solution of
(3.1)u=Ψf(u,λ),
when λ=1.
It is easy to see that the operatorP is compact continuous. According to Lemmas 2.2 and 2.3, we can see that Ψf(·,·) is compact continuous from C1×[0,1] to C1.
We claim that all the solutions of (3.1) are uniformly bounded for λ∈[0,1]. In fact, if it is false, we can find a sequence of solutions {(un,λn)} for (3.1) such that ∥un∥1→+∞ as n→+∞ and ∥un∥1>1 for any n=1,2,….
From Lemma2.2, we have
(3.2)|ρ(λnδNf(un))|≤C1(‖Nf(un)‖L1+|e2|)≤C2‖un‖1q+-1,
then we have
(3.3)|ρ(λnδNf(un))+F(λnδNf(un))|≤|ρ(λnδNf(un))|+|F(λnδNf(un))|≤C3‖un‖1q+-1.
From (3.1), we have
(3.4)w(t)|un′(t)|p(t)-2un′(t)=ρ(λnδNf(un))+F(λnδNf(un)),∀t∈J,
then
(3.5)w(t)|un′(t)|p(t)-1≤|ρ(λnδNf(un))|+|F(λnδNf(un))|≤C3‖un‖1q+-1.
Denoteα=(q+-1)/(p--1), from the above inequality, we have
(3.6)‖(w(t))1/(p(t)-1)un′(t)‖0≤C4‖un‖1α.
It follows from (2.36) and (3.3) that
(3.7)|un(0)|≤C5‖un‖1α,whereα=q+-1p--1.
For anyj=1,…,N, we have
(3.8)|unj(t)|=|unj(0)+∫0t(unj)′(r)dr|≤|unj(0)|+|∫0t(w(r))-1/(p(r)-1)sup⁡t∈(0,+∞)|(w(t))1/(p(t)-1)(unj)′(t)|dr|≤[C6+C4E]‖un‖1α≤C7‖un‖1α,
which implies that
(3.9)|unj|0≤C8‖un‖1α,j=1,…,N,n=1,2,….
Thus,(3.10)‖un‖0≤C9‖un‖1α,n=1,2,….
It follows from (3.6) and (3.10) that {∥un∥1} is bounded.
Thus, we can choose a large enoughR0>0 such that all the solutions of (3.1) belong to B(R0)={u∈C1∣∥u∥1<R0}. Thus, the Leray-Schauder degree dLS[I-Ψf(·,λ),B(R0),0] is well defined for each λ∈[0,1], and
(3.11)dLS[I-Ψf(⋅,1),B(R0),0]=dLS[I-Ψf(⋅,0),B(R0),0].
Let(3.12)u0=∫0+∞{e(t)∫0tφ-1[r,(w(r))-1ρ(0)]dr}dt-∫0+∞φ-1[r,(w(r))-1ρ(0)]dr+e11-σ+∫0rφ-1[t,(w(t))-1ρ(0)]dt,
where ρ(0) is defined in (2.10), thus u0 is the unique solution of u=Ψf(u,0).
It is easy to see thatu is a solution of u=Ψf(u,0) if and only if u is a solution of the following system
((I))-Δp(t)u=0,t∈(0,+∞),u(+∞)=∫0+∞e(t)u(t)dt+e1,lim⁡t→+∞w(t)|u′|p(t)-2u′(t)=∑i=1m-2αiw(ξi)|u′|p(ξi)-2u′(ξi)+e2.
Obviously, system (I) possesses a unique solution u0. Note that u0∈B(R0), we have
(3.13)dLS[I-Ψf(⋅,1),B(R0),0]=dLS[I-Ψf(⋅,0),B(R0),0]≠0.
Therefore, (1.1)-(1.2) has at least one solution when ∑i=1m-2αi∈[0,1),σ∈[0,1). This completes the proof.Denote(3.14)Ωɛ={u∈C1∣max⁡1≤i≤N(|ui|0+|(w(t))1/(p(t)-1)(ui)′|0)<ɛ},θ=ɛ2+(1/E).Assume the following(A1)
Let positive constantɛ such that u0∈Ωɛ,|P(0)|<θ, and |ρ(0)|<(1/N(2E+2))inft∈J|ɛ/2(E+1)|p(t)-1, where u0 is defined in (3.12) and ρ(·) is defined in (2.10).It is easy to see thatΩɛ is an open bounded domain in C1. We have the following.Theorem 3.2.
In the Case (i), assume thatf satisfies general growth condition and (A1) is satisfied, then the problem (1.1)-(1.2) has at least one solution on Ωɛ¯ when the positive parameter δ is small enough.Proof.
DenoteΨf(u,λ)=P(λδNf(u))+K(λδNf(u)). According to Lemma 2.4, u is a solution of
(3.15)-Δp(t)u+λδf(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u))=0,t∈(0,+∞),
with (1.2) if and only if u is a solution of the following abstract equation
(3.16)u=Ψf(u,λ).
From Lemmas2.2 and 2.3, we can see that Ψf(·,·) is compact continuous from C1×[0,1] to C1. According to Leray-Schauder's degree theory, we only need to prove that(1°)
u=Ψf(u,λ) has no solution on ∂Ωɛ for any λ∈[0,1),(2°)
dLS[I-Ψf(·,0),Ωɛ,0]≠0, then we can conclude that the system (1.1)-(1.2) has a solution on Ωɛ¯.
(1°) If there exists aλ∈[0,1) and u∈∂Ωɛ is a solution of (3.15) with (1.2), then (λ,u) satisfies
(3.17)w(t)φ(t,u′(t))=ρ(λδNf(u))+λδF(Nf(u))(t),∀t∈(0,+∞).
Sinceu∈∂Ωɛ, there exists an i such that |ui|0+|(w(t))1/(p(t)-1)(ui)′|0=ɛ.
(i) Suppose that|ui|0>2θ, then |(w(t))1/(p(t)-1)(ui)′|0<ɛ-2θ=θ/E. On the other hand, for any t,t′∈J, we have
(3.18)|ui(t)-ui(t′)|=|∫t′t(ui)′(r)dr|≤∫0+∞(w(r))-1/(p(r)-1)|(w(r))1/(p(r)-1)(ui)′(r)|dr<θ.
This implies that|ui(t)|>θ for each t∈J.
Note thatu∈Ωɛ¯, then |f(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u))|≤βC*ɛ(t) (where C*∶=N+supt∈J∫0+∞|ψ(s,t)|ds+supt∈J∫0+∞|χ(s,t)|ds), holding |F(Nf(u))|≤∫0+∞βC*ɛ(t)dt. Since P(·) is continuous, when 0<δ is small enough, from (A1), we have
(3.19)|u(0)|=|P(λδNf(u))|<θ.
It is a contradiction to|ui(t)|>θ for any t∈J.
(ii) Suppose that|ui|0≤2θ, then θ/E≤|(w(t))1/(p(t)-1)(ui)′|0≤ɛ. This implies that
(3.20)|(w(t2))1/(p(t2)-1)(ui)′(t2)|>ɛ2(E+1)forsomet2∈J.
Sinceu∈Ωɛ¯, it is easy to see that
(3.21)|(w(t2))1/(p(t2)-1)(ui)′(t2)|>ε2(E+1)=NεN(2E+2)≥|(w(t2))1/(p(t2)-1)u′(t2)|N(2E+2).
Combining (3.17) and (3.21), we have
(3.22)|ε/2(E+1)|p(t2)-1N(2E+2)<1N(2E+2)w(t2)|(ui)′(t2)|p(t2)-1≤1N(2E+2)w(t2)|u′(t2)|p(t2)-1≤w(t2)|u′(t2)|p(t2)-2|(ui)′(t2)|≤|ρ(λδNf)|+λ|δF(Nf)(t2)|.
Sinceu∈Ωɛ¯ and f is Caratheodory, it is easy to see that
(3.23)|f(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u))|≤βC*ɛ(t),
thus
(3.24)|δF(Nf(u))|≤δ∫0+∞βC*ɛ(t)dt.
From Lemma2.2, ρ(·) is continuous, then we have
(3.25)|ρ(λδNf(u))|⟶|ρ(0)|asδ⟶0.
When0<δ is small enough, from (A1) and (3.22), we can conclude that
(3.26)|ε/2(E+1)|p(t2)-1N(2E+2)<|ρ(λδNf(u))|+λ|δF(Nf(u))(t)|<1N(2E+2)inf⁡t∈J|ɛ2(E+1)|p(t)-1.
It is a contradiction.
Summarizing this argument, for eachλ∈[0,1), the problem (3.15) with (1.2) has no solution on ∂Ωɛ.
(2°) Sinceu0 (where u0 is defined in (3.12)) is the unique solution of u=Ψf(u,0), and (A1) holds u0∈Ωɛ, we can see that the Leray-Schauder degree
(3.27)dLS[I-Ψf(⋅,0),Ωɛ,0]≠0.
This completes the proof.In the following, we will deal with the existence of nonnegative solutions of (1.1)-(1.2) when ∑i=1m-2αi∈[0,1),σ∈[0,1). For any x=(x1,…,xN)∈ℝN, the notation x≥0(x>0) means xj≥0(xj>0) for any j=1,…,N. For any x,y∈ℝN, the notation x≥y means x-y≥0 and the notation x>y means x-y>0.Theorem 3.3.
In Case (i), we assume(10)
δf(t,x,y,z,w)≥0,∀(t,x,y,z,w)∈J×ℝN×ℝN×ℝN×ℝN,(20)
e1≥0,(30)
e2≤0.
Then, all the solutions of (1.1)-(1.2) are nonnegative.Proof.
Ifu is a solution of (1.1)-(1.2), then
(3.28)w(t)φ(t,u′(t))=ρ(δNf(u))+∫0tδf(r,u,(w(r))1/(p(r)-1)u′,S(u),T(u))dr,∀t∈J.
It follows from (2.10), (10), and (30) that
(3.29)w(t)φ(t,u′(t))=ρ(δNf(u))+F(δNf(u))(t)=11-∑i=1m-2αi(∑i=1m-2αiF(δNf(u))(ξi)-F(δNf(u))(+∞)+e2)+F(δNf(u))(t)=11-∑i=1m-2αi(-∑i=1m-2αi∫ξi+∞δNf(u)(r)dr-(1-∑i=1m-2αi)∫t+∞δNf(u)(r)dr+e2)≤0.
Thus,u′(t)≤0 for any t∈J. Holding u(t) is decreasing, namely, u(t1)≥u(t2) for any t1,t2∈J with t1<t2.
According to the boundary value condition (1.2) and condition (2°), we have
(3.30)u(+∞)=∫0+∞e(t)u(t)dt+e1≥∫0+∞e(t)u(+∞)dt+e1=σu(+∞)+e1,
then
(3.31)u(+∞)≥e11-σ≥0.
Thus, all the solutions of (1.1)-(1.2) are nonnegative. The proof is completed.Corollary 3.4.
In Case (i), we assume(10)
δf(t,x,y,z,w)≥0,∀(t,x,y,z,w)∈J×ℝN×ℝN×ℝN×ℝN with x,z,w≥0,(20)
ψ(s,t)≥0,χ(s,t)≥0,∀(s,t)∈D,(30)
e1≥0,(40)
e2≤0.
Then, we have the following.
(a) On the conditions of Theorem3.1, then (1.1)-(1.2) has at least a nonnegative solution u.
(b) On the conditions of Theorem3.2, then (1.1)-(1.2) has at least a nonnegative solution u.Proof.
(a) Define(3.32)L(u)=(L*(u1),…,L*(uN)),
where
(3.33)L*(t)={t,t≥0,0,t<0.
Denote(3.34)f̃(t,u,v,S(u),T(u))=f(t,L(u),v,S(L(u)),T(L(u))),∀(t,u,v)∈J×RN×RN,
then f̃(t,u,v,S(u),T(u)) satisfies Caratheodory condition and f̃(t,u,v,S(u),T(u))≥0 for any (t,u,v)∈J×ℝN×ℝN.
Obviously, we have(A2)
lim|u|+|v|→+∞(f̃(t,u,v,S(u),T(u))/(|u|+|v|)q(t)-1)=0,fort∈Juniformly,
whereq(t)∈C(J,ℝ), and 1<q-≤q+<p-.
Then,f̃(t,·,·,·,·) satisfies sub-(p--1) growth condition.
Let us consider the existence of solutions of the following system:(3.35)-Δp(t)u+δf̃(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u))=0,t∈(0,+∞),
with boundary value condition (1.2). According to Theorem 3.1, (3.35) with (1.2) has at least a solution u. From Theorem 3.3, we can see that u is nonnegative. Thus, u is a nonnegative solution of (1.1)-(1.2).
(b) It is similar to the proof of (a).
This completes the proof.


## 4. Existence of Solutions in Case (ii)

In this section, we will apply Leray-Schauder's degree to deal with the existence of solutions for (1.1)-(1.2) when ∑i=1m-2αi∈[0,1),σ=1.Theorem 4.1.
Assume thatΩ is an open bounded set in C1 such that the following conditions hold.(10)
For eachλ∈(0,1), the problem(4.1)(w(t)φ(t,u′(t)λ))′=δf(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u)),t∈(0,+∞),u(+∞)=∫0+∞e(t)u(t)dt+λ2e1,lim⁡t→+∞w(t)φ(t,u′(t)λ)=∑i=1m-2αiw(ξi)φ(ξi,u′(ξi)λ)+e2,has no solution on∂Ω.(20)
The equation(4.2)ω(a)∶=∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-11-∑i=1m-2αi(-∑i=1m-2αi∫ξi+∞δf(s,a,0,S(a),T(a))ds-(1-∑i=1m-2αi)×∫r+∞δf(s,a,0,S(a),T(a))ds+e2-∑i=1m-2αi∫ξi+∞δf(s,a,0,S(a),T(a))ds)-∑i=1m-2αi∫ξi+∞δf(s,a,0,S(a),T(a))ds]dr-∑i=1m-2αi∫ξi+∞δf(s,a,0,S(a),T(a))ds}dt=0has no solution on∂Ω∩ℝN.(30)
The Brouwer degreedB[ω,Ω∩ℝN,0]≠0.
Then, problems (1.1)-(1.2) have a solution on Ω¯.Proof.
For anyλ∈(0,1], it is easy to have problem (4.1) can be written in the equivalent form
(4.3)u=Φf(u,λ)=P1u+Θλ(δNf(u))+K1λ(δNf(u)),
where
(4.4)P1:C1⟶C1,u⟼u(0),Θλ:L1⟶RN,h⟼∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-11-∑i=1m-2αi(-∑i=1m-2αi∫ξi+∞h(s)ds-(1-∑i=1m-2αi)×∫r+∞h(s)ds+e2-∑i=1m-2αi∫ξi+∞h(s)ds-(1-∑i=1m-2αi))-∑i=1m-2αi∫ξi+∞h(s)ds-(1-∑i=1m-2αi)]dr-∑i=1m-2αi∫ξi+∞h(s)ds-(1-∑i=1m-2αi)}dt-λe1,K1λ(h)(t)∶=F{λφ-1[t,(w(t))-1(ρ1λ(h)+F(h))]}(t),∀t∈[0,+∞),
where ρ1λ(·)=w(0)φ(0,u′(0)/λ)=ρ1(·).
It is easy to see that the operatorP1 is compact continuous. According to Lemmas 2.5 and 2.6, we can conclude that Φf is continuous and compact from C1×[0,1] to C1. We assume that for λ=1, (4.3) does not have a solution on ∂Ω, otherwise we complete the proof. Now from hypothesis (10), it follows that (4.3) has no solutions for (u,λ)∈∂Ω×(0,1].
Forλ∈[0,1], if u is a solution of (4.3), we have
(4.5)Θλ(δNf(u))=0.
Thus, forλ=0, it follows from (4.3) and (4.4) that
(4.6)u=Φf(u,0)=P1u+Θ0(δNf(u)),
it holds u≡d, a constant.
Therefore, whenλ=0, by (4.5), (4.7)∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-11-∑i=1m-2αi(-∑i=1m-2αi∫ξi+∞δf(s,d,0,S(d),T(d))ds-(1-∑i=1m-2αi)×∫r+∞δf(s,d,0,S(d),T(d))ds+e2-∑i=1m-2αi∫ξi+∞h(s)ds-(1-∑i=1m-2αi))-∑i=1m-2αi∫ξi+∞h(s)ds-(1-∑i=1m-2αi)]dr-∑i=1m-2αi∫ξi+∞h(s)ds-(1-∑i=1m-2αi)}dt=0,
which together with hypothesis (20) implies that u=d∉∂Ω. Thus, we have proved that (4.3) has no solution (u,λ) on ∂Ω×[0,1], then we get that for each λ∈[0,1], the Leray-Schauder degree dLS[I-Φf(·,λ),Ω,0]is well defined, and, from the properties of that degree, we have
(4.8)dLS[I-Φf(⋅,1),Ω,0]=dLS[I-Φf(⋅,0),Ω,0].
Now, it is clear that problem(4.9)u=Φf(u,1)
is equivalent to problem (1.1)-(1.2), and (4.8) tells us that problem (4.9) will have a solution if we can show that
(4.10)dLS[I-Φf(⋅,0),Ω,0]≠0.
Since(4.11)Φf(u,0)=P1u+Θ0(Nδf(u))+K10(Nδf(u)),
then
(4.12)u-Φf(u,0)=u-P1u-Θ0(Nδf(u))-K10(Nδf(u)).
From (4.4), we have K10(Nδf(u))≡0. By the properties of the Leray-Schauder degree, we have
(4.13)dLS[I-Φf(⋅,0),Ω,0]=(-1)NdB[ω,Ω∩RN,0],
where the function ω is defined in (4.2) and dB denotes the Brouwer degree. Since by hypothesis (30), this last degree is different from zero. This completes the proof.Our next theorem is a consequence of Theorem4.1. As an application of Theorem 4.1, let us consider the following equation with (1.2):(4.14)(w(t)|u′|p(t)-2u′)′=g(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u))+b(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u)),
where b:J×ℝN×ℝN×ℝN×ℝN→ℝN is Caratheodory, g=(g1,…,gN):J×ℝN×ℝN×ℝN×ℝN→ℝN is continuous and Caratheodory, and, for any fixed y0∈ℝN∖{0}, if y0i≠0, then gi(t,y0,0,S(y0),T(y0))≠0, for all t∈J, for all i=1,…,N.Theorem 4.2.
Assume that the following conditions hold:(10)
g(t,kx,ky,kz,kw)=kq(t)-1g(t,x,y,z,w) for all k>0 and all (t,x,y,z,w)∈J×ℝN×ℝN×ℝN×ℝN, where q(t)∈C(J,ℝ) satisfies 1<q-≤q+<p-,(20)
lim|x|+|y|+|z|+|w|→+∞(b(t,x,y,z,w)/(|x|+|y|+|z|+|w|)q(t)-1)=0,fort∈Juniformly,(30)
for large enoughR0>0, the equation(4.15)ωg(a)∶=∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-11-∑i=1m-2αi(-∑i=1m-2αi∫ξi+∞g(s,a,0,,S(a),T(a))ds-(1-∑i=1m-2αi)×∫r+∞g(s,a,0,S(a),T(a))ds+e2-∑i=1m-2αi∫ξi+∞g(s,a,0,,S(a),T(a))ds)-∑i=1m-2αi∫ξi+∞g(s,a,0,,S(a),T(a))ds]dr-∑i=1m-2αi∫ξi+∞g(s,a,0,,S(a),T(a))ds}dt=0has no solution on∂B(R0)∩ℝN, where B(R0)={u∈C1∥u∥1<R0},(40)
the Brouwer degreedB[ωg,b(R0),0]≠0 for large enough R0>0, where b(R0)={x∈ℝN∣|x|<R0}.
Then, problem (4.14) with (1.2) has at least one solution.Proof.
Denote(4.16)Nf(u,λ)=f(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u),λ)=g(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u))+λb(t,u,(w(r))1/(p(t)-1)u′,S(u),T(u)).
At first, we consider the following problem(4.17)(w(t)φ(t,u′(t)λ))′=Nf(u,λ),t∈(0,+∞),u(+∞)=∫0+∞e(t)u(t)dt+λ2e1,lim⁡t→+∞w(t)φ(t,u′(t)λ)=∑i=1m-2αiw(ξi)φ(ξi,u′(ξi)λ)+e2.
For anyλ∈(0,1], it is easy to have problem (4.17) can be written in the equivalent form
(4.18)u=Φf(u,λ)=P1u+Θλ(Nf(u,λ))+K1λ(Nf(u,λ)),
where Θλ and K1λ are defined in Theorem 4.1.
We claim that all the solutions of (4.17) are uniformly bounded for λ∈(0,1]. In fact, if it is false, we can find a sequence of solutions {(un,λn)} for (4.17) such that ∥un∥1→+∞ as n→+∞ and ∥un∥1>1for any n=1,2,….
Since(un,λn) are solutions of (4.17), we have
(4.19)∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-11-∑i=1m-2αi(-∑i=1m-2αi∫ξi+∞Nf(un,λn)ds-(1-∑i=1m-2αi)×∫r+∞Nf(un,λn)ds+e2-∑i=1m-2αi∫ξi+∞Nf(un,λn)ds-(1-∑i=1m-2αi))-∑i=1m-2αi∫ξi+∞Nf(un,λn)ds-(1-∑i=1m-2αi)]dr-∑i=1m-2αi∫ξi+∞Nf(un,λn)ds-(1-∑i=1m-2αi)}dt-λne1=0,∀t∈(0,+∞).
It follows from Lemma2.5 that
(4.20)|ρ1λn(Nf(un,λn))|≤2N1-∑i=1m-2αi⋅(‖Nf(un,λn)‖L1+|e2|)≤C1‖un‖1q+-1‖g[t,un‖un‖1,(w(t))1/(p(t)-1)un′‖un‖1,S(un)‖un‖1,T(un)‖un‖1]+o(1)‖L1,
where o(1) means the function which is uniformly convergent to 0 (as n→+∞). According to the property of g and (4.20), then there exists a positive constant C2 such that
(4.21)|ρ1λn(Nf(un,λn))|≤C2‖un‖1q+-1,∀t∈(0,+∞),
then we have
(4.22)|ρ1λn(Nf(un,λn))|+|F(Nf(un,λn))|≤C3‖un‖1q+-1,∀t∈(0,+∞).
From (4.18), we have
(4.23)w(t)φ(t,un′(t)λn)=ρ1λn(Nf(un,λn))+F(Nf(un,λn)),t∈J,
then
(4.24)w(t)|un′(t)λn|p(t)-1≤|ρ1λn(Nf(un,λn))|+|F(Nf(un,λn))|≤C3‖un‖1q+-1.
Denoteα=(q+-1)/(p--1),  then
(4.25)‖(w(t))1/(p(t)-1)un′(t)‖0≤λnC4‖un‖1α≤C5‖un‖1α.
Sinceα∈(0,1), from (4.25), we have
(4.26)lim⁡n→+∞‖un‖0‖un‖1=1.
Denoteμn=(|un1|0/∥un∥0,|un2|0/∥un∥0,…,|unN|0/∥un∥0), then μn∈ℝN and |μn|=1(n=1,2,…), then {μn} possesses a convergent subsequence (which denoted by μn), and then there exists a vector μ0=(μ01,μ02,…,μ0N)∈ℝN such that
(4.27)|μ0|=1,lim⁡n→+∞μn=μ0.
Without loss of generality, we assume thatμ01>0. Since un∈C(J,ℝ), there exist ηni∈(0,+∞) such that
(4.28)|uni(ηni)|≥(1-1n)|uni|0,i=1,2,…,N,n=1,2,…,
and, then, from (4.25), we have
(4.29)0≤|un1(t)-un1(ηn1)|=|∫ηn1t(un1)′(r)dr|≤C5‖un‖1α∫0+∞(w(t))-1/(p(t)-1)dt.
Since∥un∥1→+∞ (as n→+∞), α∈(0,1), and μ01>0, we have
(4.30)lim⁡n→+∞1|un1(ηn1)|C5‖un‖1α∫0+∞(w(t))-1/(p(t)-1)dt=0.
From (4.26)–(4.30), we have
(4.31)lim⁡n→+∞un1(t)un1(ηn1)=1,fort∈Juniformly.
So we get(4.32)lim⁡n→+∞un(t)‖un‖1=μ*,lim⁡n→+∞(w(t))1/(p(t)-1)un′(t)‖un‖1=0,fort∈Juniformly,
where μ*∈ℝN satisfies |μ*|=1,|μ*i|=μ0i.
We denote(4.33)g*1=‖un‖1q(s)-1{g1[s,μ*+o(1),o(1),S(μ*+o(1)),T(μ*+o(1))]+o(1)},wheres∈J.
Sinceμ01≠0, from (4.19) and (4.32), we have
(4.34)∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-11-∑i=1m-2αi(-∑i=1m-2αi∫ξi+∞g*1ds-(1-∑i=1m-2αi)×∫r+∞g*1ds+e2-∑i=1m-2αi∫ξi+∞g*1)-∑i=1m-2αi∫ξi+∞g*1]dr-∑i=1m-2αi∫ξi+∞g*1}dt-λne1=0.
Sinceg1(s,μ*,0,S(μ*),T(μ*))≠0, according to the continuity of g1, we have
(4.35)∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-11-∑i=1m-2αi(-∑i=1m-2αi∫ξi+∞g*1ds-(1-∑i=1m-2αi)×∫r+∞g*1ds+e2-∑i=1m-2αi∫ξi+∞g*1ds)-∑i=1m-2αi∫ξi+∞g*1ds]dr-∑i=1m-2αi∫ξi+∞g*1ds}dt-λne1≠0,
and it is a contradiction to (4.34). This implies that there exists a big enough R0>0 such that all the solutions of (4.18) when λ∈(0,1] belongs to B(R0).
Forλ∈[0,1], if u is a solution of (4.18), we have
(4.36)Θλ(Nf(u,λ))=0.
Forλ=0,f=g, from (4.18), we have
(4.37)u=Φg(u,0)=P1u+Θ0(g),
it holds u≡d, a constant.
Therefore, whenλ=0, we have
(4.38)∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-11-∑i=1m-2αi(-∑i=1m-2αi∫ξi+∞g(s,d,0,S(d),T(d))ds-(1-∑i=1m-2αi)×∫r+∞g(s,d,0,S(d),T(d))ds+e2-∑i=1m-2αi∫ξi+∞g(s,d,0,S(d),T(d))ds-(1-∑i=1m-2αi))-∑i=1m-2αi∫ξi+∞g(s,d,0,S(d),T(d))ds-(1-∑i=1m-2αi)]dr-∑i=1m-2αi∫ξi+∞g(s,d,0,S(d),T(d))ds-(1-∑i=1m-2αi)}dt=0,
which together with hypothesis (30) implies that u=d∉∂B(R0). Thus, we have proved that (4.18) has no solution (u,λ) on ∂B(R0)×[0,1], then we get that the Leray-Schauder degree dLS[I-Φf(·,λ),B(R0),0]is well defined for each λ∈[0,1], which implies that
(4.39)dLS[I-Φf(⋅,1),B(R0),0]=dLS[I-Φg(⋅,0),B(R0),0].
Now it is clear that problem(4.40)u=Φf(u,1)
is equivalent to problem (4.14) with (1.2), and (4.39) tells us that problem (4.40) will have a solution if we can show that
(4.41)dLS[I-Φg(⋅,0),B(R0),0]≠0.
Since(4.42)Φg(u,0)=P1u+Θ0(g),
then
(4.43)u-Φg(u,0)=u-P1u-Θ0(g).
By the properties of the Leray-Schauder degree, we have(4.44)dLS[I-Φg(⋅,0),B(R0),0]=(-1)NdB[ωg,b(R0),0],
where the function ωg is defined in (4.15) and dB denotes the Brouwer degree. By hypothesis (40), this last degree is different from zero. This completes the proof.Corollary 4.3.
Ifb:J×ℝN×ℝN×ℝN×ℝN→ℝN is Caratheodory, which satisfies the conditions of Theorem 4.2, g(t,u,v,S(u),T(u))=β(t)(|u|q(t)-2u+|v|q(t)-2v+|S(u)|q(t)-2S(u)+|T(u)|q(t)-2T(u)), where β(t)∈L1(J,ℝ),β(t),q(t)∈C(J,ℝ) are positive functions, and satisfies 1<q-≤q+<p-andψ(s,t) and χ(s,t) are nonnegative, then (4.14) with (1.2) has at least one solution.Proof.
Since(4.45)g(t,u,v,S(u),T(u))=β(t)(|u|q(t)-2u+|v|q(t)-2v+|S(u)|q(t)-2S(u)+|T(u)|q(t)-2T(u)),
then
(4.46)ωg(a)=∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-11-∑i=1m-2αi(-∑i=1m-2αi∫ξi+∞g(s,a,0,S(a),T(a))ds-(1-∑i=1m-2αi)×∫r+∞g(s,a,0,S(a),T(a))ds+e2-∑i=1m-2αi∫ξi+∞g(s,a,0,S(a),T(a))ds-(1-∑i=1m-2αi))-∑i=1m-2αi∫ξi+∞g(s,a,0,S(a),T(a))ds-(1-∑i=1m-2αi)]dr-∑i=1m-2αi∫ξi+∞g(s,a,0,S(a),T(a))ds-(1-∑i=1m-2αi)}dt=∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-11-∑i=1m-2αi(-∑i=1m-2αi∫ξi+∞β(s)(|a|q(s)-2a+|S(a)|q(s)-2S(a)+|T(a)|q(s)-2T(a))ds-(1-∑i=1m-2αi)×∫r+∞β(s)(|a|q(s)-2a+|S(a)|q(s)-2S(a)+|T(a)|q(s)-2T(a))ds+e2-∑i=1m-2αi∫ξi+∞β(s)(|a|q(s)-2a+|S(a)|q(s)-2S(a))-∑i=1m-2αi∫ξi+∞β(s)(|a|q(s)-2a+|S(a)|q(s)-2S(a)]dr-∑i=1m-2αi∫ξi+∞β(s)(|a|q(s)-2a+|S(a)|q(s)-2S(a)}dt,
then it is easy to say that ωg(a)=0 has only one solution in ℝN, and
(4.47)dB[ωg,b(R0),0]=dB[I,b(R0),0]≠0,
and, according to Theorem 4.2, we get that (4.14) with (1.2) has at least a solution. This completes the proof.In the following, let us consider(4.48)-(w(t)|u′|p(t)-2u′)′+f(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u),δ)=0,t∈(0,+∞),
where (4.49)f(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u),δ)=g(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u))+δh(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u)),
where g,h:J×ℝN×ℝN×ℝN×ℝN→ℝN are Caratheodory.We have the following.Theorem 4.4.
We assume that conditions of (10), (30), and (40) of Theorem 4.2 are satisfied, then problem (4.48) with (1.2) has at least one solution when the parameter δ is small enough.Proof.
Denote(4.50)fλδ(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u))=g(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u))+λδh(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u)).
Let us consider the existence of solutions of the following(4.51)-(w(t)|u′|p(t)-2u′)′+fλδ(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u))=0,t∈(0,+∞),
with (1.2).
We know that (4.51) with (1.2) has the same solution of
(4.52)u=Ψδ(u,λ)=P1u+Θ(Nfλδ(u))+K1(Nfλδ(u)),
where Nfλδ(u) is defined in (2.32).
Obviously,f0=g. So Ψδ(u,0)=Φg(u,1).  From the proof of Theorem 4.2, we can see that all the solutions of u=Ψδ(u,0) are uniformly bounded, then there exists a large enough R0>0 such that all the solutions of u=Ψδ(u,0) belong to B(R0)={u∈C1∣∥u∥1<R0}. Since Ψδ(u,0) is compact continuous from C1 to C1, we have
(4.53)inf⁡u∈∂B(R0)‖u-Ψδ(u,0)‖1>0.
Sinceg,h are Caratheodory, we have(4.54)|ρ1(Nfλδ(u))-ρ1(Nf0(u))|⟶0,for(u,λ)∈B(R0)¯×[0,1]uniformly,asδ⟶0,‖F(Nfλδ(u))-F(Nf0(u))‖0⟶0,for(u,λ)∈B(R0)¯×[0,1]uniformly,asδ⟶0,‖K1(Nfλδ(u))-K1(Nf0(u))‖1⟶0,for(u,λ)∈B(R0)¯×[0,1]uniformly,asδ⟶0,|P1(Nfλδ(u))-P1(Nf0(u))|⟶0,for(u,λ)∈B(R0)¯×[0,1]uniformly,asδ⟶0.
Thus,(4.55)‖Ψδ(u,λ)-Ψ0(u,λ)‖1⟶0for(u,λ)∈B(R0)¯×[0,1]uniformly,asδ⟶0.
Obviously,Ψ0(u,λ)=Ψδ(u,0)=Ψ0(u,0). Therefore,(4.56)‖Ψδ(u,λ)-Ψδ(u,0)‖1⟶0for(u,λ)∈B(R0)¯×[0,1]uniformly,asδ⟶0.
Thus, whenδ is small enough, from (4.53), we can conclude
(4.57)inf⁡(u,λ)∈∂B(R0)×[0,1]‖u-Ψδ(u,λ)‖1≥inf⁡u∈∂B(R0)‖u-Ψδ(u,0)‖1-sup⁡(u,λ)∈B(R0)¯×[0,1]‖Ψδ(u,0)-Ψδ(u,λ)‖1>0.
Thusu=Ψδ(u,λ) has no solution on ∂B(R0) for any λ∈[0,1], when δ is small enough. It means that the Leray-Schauder degree dLS[I-Ψδ(·,λ),B(R0),0] is well defined for any λ∈[0,1] and
(4.58)dLS[I-Ψδ(u,λ),B(R0),0]=dLS[I-Ψδ(u,0),B(R0),0].
From the proof of Theorem4.2, we can see that the right hand side is nonzero, then (4.48) with (1.2) has at least one solution, when δ is small enough. This completes the proof.


## 5. Existence of Solutions in Case (iii)

In this section, we will apply Leray-Schauder's degree to deal with the existence of solutions for (1.1)-(1.2) when ∑i=1m-2αi=1,σ=1.Theorem 5.1.
Assume thatΩ is an open bounded set in C1 such that the following conditions hold.(10)
For eachλ∈(0,1), the problem(5.1)(w(t)|u′|p(t)-2u′)′=λδf(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u)),t∈(0,+∞),u(+∞)=∫0+∞e(t)u(t)dt+λe1,lim⁡t→+∞w(t)|u′|p(t)-2u′(t)=∑i=1m-2αiw(ξi)|u′|p(ξi)-2u′(ξi)+λe2,has no solution on∂Ω.(20)
The equation(5.2)ω*(a):=∑i=1m-2αi∫ξi+∞δf(t,a,0,S(a),T(a))dt-e2=0has no solution on∂Ω∩ℝN.(30)
The Brouwer degreedB[ω*,Ω∩ℝN,0]≠0.
Then, problems (1.1)-(1.2) have a solution on Ω¯.Proof.
Let us consider the following problem:(5.3)(w(t)|u′|p(t)-2u′)′=λδNf(u)+(1-λ)Q*(δNf(u)),∀t∈(0,+∞),u(+∞)=∫0+∞e(t)u(t)dt+λe1,lim⁡t→+∞w(t)|u′|p(t)-2u′(t)=∑i=1m-2αiw(ξi)|u′|p(ξi)-2u′(ξi)+λe2,
where Q and Q* are defined in (2.79) and (2.80), respectively.
For anyλ∈(0,1], observe that, if u is a solution to (5.1) or u is a solution to (5.3), we have necessarily
(5.4)Q(Nδf(u))=0,Q*(Nδf(u))=0.
It means that (5.1) and (5.3) have the same solutions for λ∈(0,1].
We denoteN(·,·):C1×J→L1 defined by
(5.5)N(u,λ)=λNδf(u)+(1-λ)Q*(Nδf(u)),
where Nδf(u) is defined by (2.32). Let
(5.6)Φf*(u,λ)=P2u+Qλ(N(u,λ))+K2λ(N(u,λ))=P2u+Q(Nδf(u))+K2λ(N(u,λ)),
where
(5.7)P2:C1⟶C1,u⟼u(0),Qλ:L1⟶RN,h⟼∑i=1m-2αi∫ξi+∞h(t)dt-λe2,Qλ*:L1⟶L1,h⟼τ(t)(∑i=1m-2αi∫ξi+∞h(t)dt-λe2),K2λ(h)(t)∶=F{φ-1[t,(w(t))-1(ρ2λ((I-Qλ*)h)+F((I-Qλ*)h))]}(t),∀t∈[0,+∞),
where ρ2λ satisfies
(5.8)∫0+∞{e(t)∫t+∞φ-1[r,(w(r))-1(ρ2λ+F(h)(r))]dr}dt-λe1=0,
and the fixed point of Φf*(u,1) is a solution for (5.3). Also problem (5.3) can be written in the equivalent form
(5.9)u=Φf*(u,λ).
Sincef is Caratheodory, it is easy to see that N(·,·) is continuous and sends bounded sets into equi-integrable sets. It is easy to see that P2 is compact continuous. According to Lemmas 2.8 and 2.9, we can conclude that Φf*(·,·) is continuous and compact from C1×[0,1] to C1. We assume that for λ=1, (5.9) does not have a solution on ∂Ω, otherwise we complete the proof. Now, from hypothesis (10), it follows that (5.9) has no solutions for (u,λ)∈∂Ω×(0,1]. For λ=0, (5.3) is equivalent to the problem
(5.10)(w(t)|u′|p(t)-2u′)′=Q*(δNf(u)),u(+∞)=∫0+∞e(t)u(t)dt,lim⁡t→+∞w(t)|u′|p(t)-2u′(t)=∑i=1m-2αiw(ξi)|u′|p(ξi)-2u′(ξi).
Ifu is a solution to this problem, we must have
(5.11)∑i=1m-2αi∫ξi+∞Q*(δNf(u))dt=∑i=1m-2αi∫ξi+∞δf(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u))dt-e2=0.
Hence,(5.12)w(t)|u′|p(t)-2u′≡c,
where c∈ℝN is a constant.
It is easy to see that(ui)′ keeps the same sign of ci. From u(+∞)=∫0+∞e(t)u(t)dt, we have ∫0+∞e(t)(u(+∞)-u(t))dt=0. From the continuity of u, there exist ti∈(0,+∞), such that (ui)′(ti)=0,i=1,…,N. Hence, u′≡0, it holds u≡d, a constant. Thus, by (5.11), we have
(5.13)∑i=1m-2αi∫ξi+∞δf(t,d,0,S(d),T(d))dt-e2=0,
which together with hypothesis (20) implies that u=d∉∂Ω. Thus, we have proved that (5.9) has no solution (u,λ) on ∂Ω×[0,1], then we get that the Leray-Schauder degree dLS[I-Φf*(·,λ),Ω,0]is well defined for λ∈[0,1], and from the properties of that degree, we have
(5.14)dLS[I-Φf*(⋅,1),Ω,0]=dLS[I-Φf*(⋅,0),Ω,0].
Now it is clear that problem(5.15)u=Φf*(u,1)
is equivalent to the problem (1.1)-(1.2), and (5.14) tells us that problem (5.15) will have a solution if we can show that
(5.16)dLS[I-Φf*(⋅,0),Ω,0]≠0.
Since(5.17)Φf*(u,0)=P2u+Q(Nδf(u))+K20(Q*(Nδf(u))),
then
(5.18)u-Φf*(u,0)=u-P2u-Q(Nδf(u))-K20(0).
Similar to Lemma2.8, we have
(5.19)|ρ2λ(Q*(Nδf(u)))|≤3N(E#+1E#)p+-1[‖Q*(Nδf(u))‖0+|λe1|p#-1].
Thus,ρ20(0)=0, then K20(0)≡0. From (5.18), we have(5.20)u-Φf*(u,0)=u-P2u-Q(Nδf(u)).
By the properties of the Leray-Schauder degree, we have(5.21)dLS[I-Φf*(⋅,0),Ω,0]=(-1)NdB[ω*,Ω∩RN,0],
where the function ω* is defined in (5.2) and dB denotes the Brouwer degree. By hypothesis (30), this last degree is different from zero. This completes the proof.Our next theorem is a consequence of Theorem5.1. As an application of Theorem 5.1. Let us consider the following equation with  (1.2)(5.22)(w(t)|u′|p(t)-2u′)′=g(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u))+b(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u)),
where b:J×ℝN×ℝN×ℝN×ℝN→ℝN is Caratheodory, g=(g1,…,gN):J×ℝN×ℝN×ℝN×ℝN→ℝN is continuous and Caratheodory, and, for any fixed y0∈ℝN∖{0}, if y0i≠0, then gi(t,y0,0,S(y0),T(y0))≠0, for all t∈J, for all i=1,…,N.Theorem 5.2.
Assume that the following conditions hold.(10)
g(t,kx,ky,kz,kw)=kq(t)-1g(t,x,y,z,w) for all k>0 and all (t,x,y,z,w)∈J×ℝN×ℝN×ℝN×ℝN, where q(t)∈C(J,ℝ) satisfies 1<q-≤q+<p-,(20)
lim|x|+|y|+|z|+|w|→+∞(b(t,x,y,z,w)/(|x|+|y|+|z|+|w|)q(t)-1)=0,   for t∈J uniformly,(30)
for large enoughR0>0, the equation(5.23)ωg*(a)∶=∑i=1m-2αi∫ξi+∞g(t,a,0,S(a),T(a))dt-e2=0,has no solution on∂B(R0)∩ℝN, where B(R0)={u∈C1∥u∥1<R0},(40)
the Brouwer degreedB[ωg*,b(R0),0]≠0 for large enough R0>0, where b(R0)={x∈ℝN∣|x|<R0}.
Then, problem (5.22) with (1.2) has at least one solution.Proof.
Denote(5.24)Nf(u,λ)=f(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u),λ)=g(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u))+λb(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u)).
At first, we consider the following problem(5.25)(w(t)|u′|p(t)-2u′)′=f(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u),λ),t∈(0,+∞).
According to Lemma2.10, we know problem (5.25) with (1.2) has the same solution of
(5.26)u=Φ̃f(u,λ)=P2u+Q(Nf(u,λ))+K2(Nf(u,λ)).
Similar to the proof of Theorem4.2, we obtain that all the solutions of (5.26) are uniformly bounded for λ∈[0,1].  Then, there exists a big enough R0>0 such that all the solutions of (5.26) belong to B(R0), and then we have
(5.27)dLS[I-Φ̃f(⋅,1),B(R0),0]=dLS[I-Φ̃f(⋅,0),B(R0),0].
If we prove thatdLS[I-Φ̃f(·,0),B(R0),0]≠0, then we obtain the existence of solutions for (5.22) with (1.2).
Now, we consider the following equation(5.28)(w(t)|u′|p(t)-2u′)′=λNg(u)+(1-λ)Q*Ng(u),t∈(0,+∞),u(+∞)=∫0+∞e(t)u(t)dt+λe1,lim⁡t→+∞w(t)|u′|p(t)-2u′(t)=∑i=1m-2αiw(ξi)|u′|p(ξi)-2u′(ξi)+λe2,
where Ng(u)=g(t,u,(w(t))1/(p(t)-1)u′,S(u),T(u)).
We denoteG(·,·):C1×J→L1 defined by
(5.29)G(u,λ)=λNg(u)+(1-λ)Q*(Ng(u)).
Similar to the proof of Theorem5.1, we know that (5.28) has the same solution of
(5.30)u=Φg*(u,λ)=P2u+Q(Ng(u))+K2λ(G(u,λ)).
Similar to the discussions of Theorem4.2, we can obtain that all the solutions of (5.28) are uniformly bounded for each λ∈(0,1]. When λ=0, similar to the proof of Theorem 5.1, we can prove that (5.28) has no solution on ∂B(R0). Then, we get that the Leray-Schauder degree dLS[I-Φg*(·,λ),B(R0),0]is well defined for λ∈[0,1], which implies that
(5.31)dLS[I-Φg*(⋅,1),B(R0),0]=dLS[I-Φg*(⋅,0),B(R0),0].
Now it is clear thatΦg*(u,1)=Φ̃f(u,0). So dLS[I-Φg*(·,1),B(R0),0]=dLS[I-Φ̃f(·,0),B(R0),0]. If we prove that dLS[I-Φg*(·,0),B(R0),0]≠0, then we obtain the existence of solutions for (5.22) with (1.2). Similar to the proof of Theorem 5.1, we have
(5.32)dLS[I-Φg*(⋅,0),B(R0),0]=(-1)NdB[ωg*,b(R0),0].
According to hypothesis (40), this last degree is different from zero. We obtain that (5.22) with (1.2) has at least one solution. This completes the proof.Similar to Corollary4.3 and Theorem 4.4, we have the following.Corollary 5.3.
Ifb:J×ℝN×ℝN×ℝN×ℝN→ℝN is Caratheodory, which satisfies the conditions of Theorem 5.2, g(t,u,v,S(u),T(u))=β(t)(|u|q(t)-2u+|v|q(t)-2v+|S(u)|q(t)-2S(u)+|T(u)|q(t)-2T(u)), where β(t)∈L1(J,ℝ),β(t),q(t)∈C(J,ℝ) are positive functions, and satisfies 1<q-≤q+<p-,ψ(s,t) and χ(s,t) are nonnegative, then (5.22) with (1.2) has at least one solution.Theorem 5.4.
We assume that conditions of (10), (30), and (40) of Theorem 5.2 are satisfied, then problem (4.48) with (1.2) has at least one solution when the parameter δ is small enough.


## 6. Examples

Example 6.1.
Consider the following problem((S1))-Δp(t)u-e-t|u|q(t)-2u-S(u)(t)-(t+1)-2=0,t∈(0,+∞),lim⁡t→+∞u(t)=∫0+∞e-tu(t)dt,lim⁡t→+∞w(t)|u′|p(t)-2u′(t)=∑i=1m-2αiw(ξi)|u′|p(ξi)-2u′(ξi),
where p(t)=6+e-tsint, q(t)=3+2-tcost, S(u)(t)=∫0∞e-2s-t(sinst+1)u(s)ds.
Obviously,e-t|u|q(t)-2u+S(u)(t)+(t+1)-2  is Caratheodory, q(t)≤4<5≤p(t), ∑i=1m-2αi<1, the conditions of Corollary 4.3 are satisfied, then (S1) has a solution.Example 6.2.
Consider the following problem:((S2))-Δp(t)u+f(t,u,(w(t))1/(p(t)-1)u′,S(u))+δh(t,u,(w(t))1/(p(t)-1)u′,S(u))+e-t=0,t∈(0,+∞),lim⁡t→+∞u(t)=∫0+∞e-2tu(t)dt,lim⁡t→+∞w(t)|u′|p(t)-2u′(t)=∑i=1m-2αiw(ξi)|u′|p(ξi)-2u′(ξi),
where h is Caratheodory and
(6.1)f(t,u,(w(t))1/(p(t)-1)u′,S(u))=e-t|u|q(t)-2u+e-tw(t)(q(t)-1)/(p(t)-1)|u′|q(t)-2u′+S(u)(t),p(t)=7+3-tcos⁡3t,q(t)=4+e-2tsin⁡2t,S(u)(t)=∫0∞e-s-2t(cos⁡st+1)u(s)ds.
Obviously,e-t|u|q(t)-2u+e-tw(t)(q(t)-1)/(p(t)-1)|u′|q(t)-2u′+S(u)(t) is Caratheodory, q(t)≤5<6≤p(t), ∑i=1m-2αi<1, the conditions of Theorem 3.2 are satisfied, then  (S2) has a solution when δ is small enough.



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*Source: 814103-2011-11-22.xml*