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 term 𝑓 satisfies sub-(𝑝1) 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.

1. Introduction

In this paper, we consider the existence of solutions for the following variable exponent integrodifferential systemΔ𝑝(𝑡)𝑢+𝛿𝑓𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢)=0,𝑡(0,+),(1.1) with the following nonlinear multipoint and integral boundary value condition𝑢(+)=0+𝑒(𝑡)𝑢(𝑡)𝑑𝑡+𝑒1,lim𝑡+||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢(𝑡)=𝑚2𝑖=1𝛼𝑖𝑤𝜉𝑖||𝑢||𝑝(𝜉𝑖)2𝑢𝜉𝑖+𝑒2,(1.2) where 𝑢[0,+)𝑁;𝑆 and 𝑇 are linear operators defined by 𝑆(𝑢)(𝑡)=𝑡0𝜓(𝑠,𝑡)𝑢(𝑠)𝑑𝑠,𝑇(𝑢)(𝑡)=0+𝜒(𝑠,𝑡)𝑢(𝑠)𝑑𝑠,(1.3) where 𝜓𝐶(𝐷,),𝜒𝐶(𝐷,),𝐷={(𝑠,𝑡)[0,+)×[0,+)};0+|𝜓(𝑠,𝑡)|𝑑𝑠 and 0+|𝜒(𝑠,𝑡)|𝑑𝑠 are uniformly bounded with 𝑡;𝑝𝐶([0,+),),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; 𝑤𝐶([0,+),) satisfies 0<𝑤(𝑡), for all 𝑡(0,+), and (𝑤(𝑡))1/(𝑝(𝑡)1)𝐿1(0,+);0<𝜉1<<𝜉𝑚2<+,𝛼𝑖0,(𝑖=1,,𝑚2) and 0𝑚2𝑖=1𝛼𝑖1;𝑒𝐿1(0,+) is nonnegative,  𝜎=0+𝑒(𝑡)𝑑𝑡 and 𝜎[0,1];𝑒1,𝑒2𝑁;𝛿 is a positive parameter.

If 𝑚2𝑖=1𝛼𝑖<1 and 𝜎<1, we say the problem is nonresonant; but if 𝑚2𝑖=1𝛼𝑖[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, [123]. 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 Ω𝑛 is a bounded domain, the Rayleigh quotient𝜆𝑝(𝑥)=inf𝑢𝑊01,𝑝(𝑥)(Ω){0}Ω||||(1/𝑝(𝑥))𝑢𝑝(𝑥)𝑑𝑥Ω(1/𝑝(𝑥))|𝑢|𝑝(𝑥)𝑑𝑥(1.4) is zero in general, and only under some special conditions 𝜆𝑝(𝑥)>0 (see [9, 1618]), 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, [2429]. On the existence of solutions for 𝑝(𝑥)-Laplacian systems boundary value problems, we refer to [2, 4, 7, 8, 1012, 2022]. In [20], the present author deals with the existence and asymptotic behavior of solutions for (1.1) with the following linear boundary value conditions𝑢(0)=𝑚2𝑖=1𝛼𝑖𝑢𝜉𝑖+𝑒0,lim𝑡+𝑢(𝑡)=0+𝑒(𝑡)𝑢(𝑡)𝑑𝑡,(1.5) when 0𝑚2𝑖=1𝛼𝑖<1 and 00+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].

Let 𝑁1 and 𝐽=[0,+), the function 𝑓=(𝑓1,,𝑓𝑁)𝐽×𝑁×𝑁×𝑁×𝑁𝑁 is assumed to be Caratheodory, by this we mean,(i)for almost every 𝑡𝐽, the function 𝑓(𝑡,,,,) is continuous;(ii)for each (𝑥,𝑦,𝑧,𝑤)𝑁×𝑁×𝑁×𝑁, the function 𝑓(,𝑥,𝑦,𝑧,𝑤) is measurable on 𝐽;(iii)for each 𝑅>0, there is a 𝛽𝑅𝐿1(𝐽,) such that, for almost every 𝑡𝐽 and every (𝑥,𝑦,𝑧,𝑤)𝑁×𝑁×𝑁×𝑁 with |𝑥|𝑅,|𝑦|𝑅,|𝑧|𝑅,|𝑤|𝑅, one has||||𝑓(𝑡,𝑥,𝑦,𝑧,𝑤)𝛽𝑅(𝑡).(1.6)

Throughout the paper, we denote||𝑢𝑤(0)||𝑝(0)2𝑢(0)=lim𝑡0+||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢||𝑢(𝑡),𝑤(+)||𝑝(+)2𝑢(+)=lim𝑡+||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢(𝑡).(1.7)

The inner product in 𝑁 will be denoted by ,,|| will denote the absolute value and the Euclidean norm on 𝑁. Let 𝐴𝐶(0,+) denote the space of absolutely continuous functions on the interval (0,+). For 𝑁1, we set 𝐶=𝐶(𝐽,𝑁),𝐶1={𝑢𝐶𝑢𝐶((0,+),𝑁),lim𝑡0+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)|,𝑢0=(𝑁𝑖=1|𝑢𝑖|20)1/2, and 𝑢1=𝑢0+(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢0. 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,𝑁) with the norm 𝑢𝐿1=[𝑁𝑖=1(0|𝑢𝑖|𝑑𝑡)2]1/2.

We say a function 𝑢𝐽𝑁 is a solution of (1.1) if 𝑢𝐶1 with 𝑤(𝑡)|𝑢|𝑝(𝑡)2𝑢 absolutely continuous on (0,+), which satisfies (1.1) a.e. on 𝐽.

In this paper, we always use 𝐶𝑖 to denote positive constants, if it cannot lead to confusion. Denote 𝑧=inf𝑡𝐽𝑧(𝑡),𝑧+=sup𝑡𝐽𝑧(𝑡),forany𝑧𝐶(𝐽,).(1.8)

We say 𝑓 satisfies sub-(𝑝1) growth condition, if 𝑓 satisfieslim||𝑦|||𝑥|++|𝑧|+|𝑤|+𝑓(𝑡,𝑥,𝑦,𝑧,𝑤)||𝑦|||𝑥|++|𝑧|+|𝑤|𝑞(𝑡)1=0,for𝑡𝐽uniformly,(1.9) where 𝑞(𝑡)𝐶(𝐽,) and 1<𝑞𝑞+<𝑝. We say 𝑓 satisfies general growth condition, if 𝑓 does not satisfy sub-(𝑝1) growth condition.

We will discuss the existence of solutions of (1.1)-(1.2) in the following three cases.Case (i): 𝑚2𝑖=1𝛼𝑖[0,1),𝜎[0,1);Case (ii): 𝑚2𝑖=1𝛼𝑖[0,1),𝜎=1;Case (iii): 𝑚2𝑖=1𝛼𝑖=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 𝑚2𝑖=1𝛼𝑖[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 𝑚2𝑖=1𝛼𝑖[0,1),𝜎=1. In the fifth section, we will discuss the existence of solutions of (1.1)-(1.2) when 𝑚2𝑖=1𝛼𝑖=1,𝜎=1.

2. Preliminary

For any (𝑡,𝑥)𝐽×𝑁, 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 any 𝑡[0,+),  𝜑(𝑡,) is strictly monotone, that is 𝜑𝑡,𝑥1𝜑𝑡,𝑥2,𝑥1𝑥2>0,forany𝑥1,𝑥2𝑁,𝑥1x2.(2.1)(ii)There exists a function 𝛽[0,+)[0,+),𝛽(𝑠)+ as 𝑠+, such that 𝜑(𝑡,𝑥),𝑥𝛽(|𝑥|)|𝑥|,𝑥𝑁.(2.2)

It is well known that 𝜑(𝑡,) is a homeomorphism from 𝑁 to 𝑁 for any fixed 𝑡[0,+). For any 𝑡𝐽, denote by 𝜑1(𝑡,) the inverse operator of 𝜑(𝑡,), then 𝜑1(𝑡,𝑥)=|𝑥|(2𝑝(𝑡))/(𝑝(𝑡)1)𝑥,for𝑥𝑁{0},𝜑1(𝑡,0)=0.(2.3)

It is clear that 𝜑1(𝑡,) is continuous and sends bounded sets into bounded sets.

Let us now consider the following problem with boundary value condition  (1.2)𝑤(𝑡)𝜑𝑡,𝑢(𝑡)=𝑔(𝑡),𝑡(0,+),(2.4) where 𝑔𝐿1.

If 𝑢 is a solution of (2.4) with (1.2), by integrating (2.4) from 0 to 𝑡, we find that 𝑤(𝑡)𝜑𝑡,𝑢(𝑡)=𝑤(0)𝜑0,𝑢(+0)𝑡0𝑔(𝑠)𝑑𝑠.(2.5)

Define operator 𝐹𝐿1𝐶 as 𝐹(𝑔)(𝑡)=𝑡0𝑔(𝑠)𝑑𝑠,𝑡𝐽,𝑔𝐿1.(2.6)

By solving for 𝑢 in (2.5) and integrating, we find that𝜑𝑢(𝑡)=𝑢(0)+𝐹1𝑡,(𝑤(𝑡))1𝑤(0)𝜑0,𝑢(0)+𝐹(𝑔)(𝑡),𝑡𝐽.(2.7)

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): 𝑚2𝑖=1𝛼𝑖[0,1),𝜎[0,1)

We denote 𝜌=𝑤(0)𝜑(0,𝑢(0)) in (2.7). It is easy to see that 𝜌 is dependent on 𝑔(), then we find that 𝜑𝑢(𝑡)=𝑢(0)+𝐹1𝑡,(𝑤(𝑡))1(𝜌+𝐹(𝑔))(𝑡),𝑡𝐽.(2.8)

The boundary value condition (1.2) implies that 𝑢(0)=0+𝑒(𝑡)𝑡0𝜑1𝑟,(𝑤(𝑟))1(𝜌+𝐹(𝑔)(𝑟))𝑑𝑟𝑑𝑡+𝑒11𝜎0+𝜑1𝑟,(𝑤(𝑟))1(𝜌+𝐹(𝑔)(𝑟))𝑑𝑟,11𝜎𝜌=1𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝜉𝑖0𝑔(𝑠)𝑑𝑠0+𝑔(𝑠)𝑑𝑠+𝑒2.(2.9)

For fixed 𝐿1, we define 𝜌𝐿1𝑁 as1𝜌()=1𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝜉𝑖0(𝑡)𝑑𝑡0+(𝑡)𝑑𝑡+𝑒2.(2.10)

Lemma 2.2. 𝜌𝐿1𝑁 is continuous and sends bounded sets of 𝐿1 to bounded sets of 𝑁. Moreover, ||||𝜌()2𝑁1𝑚2𝑖=1𝛼𝑖𝐿1+||𝑒2||.(2.11)

Proof. Since 𝜌() consists of continuous operators, it is continuous. It is easy to see that ||||𝜌()2𝑁1𝑚2𝑖=1𝛼𝑖𝐿1+||𝑒2||.(2.12)
This completes the proof.

It is clear that 𝜌() is continuous and sends bounded sets of 𝐿1 to bounded sets of 𝑁, and hence it is a compact continuous mapping.

If 𝑢 is a solution of (2.4) with (1.2), we find that 𝜑𝑢(𝑡)=𝑢(0)+𝐹1𝑡,(𝑤(𝑡))1(𝜌+𝐹(𝑔))(𝑡),𝑡𝐽,𝑢(0)=0+𝑒(𝑡)𝑡0𝜑1𝑟,(𝑤(𝑟))1(𝜌+𝐹(𝑔)(𝑟))𝑑𝑟𝑑𝑡+𝑒11𝜎0+𝜑1𝑟,(𝑤(𝑟))1(𝜌+𝐹(𝑔)(𝑟))𝑑𝑟.1𝜎(2.13)

We denote 𝜑𝐾()(𝑡)=(𝐾)(𝑡)=𝐹1𝑡,(𝑤(𝑡))1(𝜌()+𝐹())(𝑡),𝑡(0,+).(2.14)

We say a set 𝑈𝐿1 be equi-integrable, if there exists a nonnegative 𝜌𝐿1(𝐽,), such that ||||(𝑡)𝜌(𝑡)a.e.in𝐽,forany𝑈.(2.15)

Lemma 2.3. The operator 𝐾 is continuous and sends equi-integrable sets in 𝐿1 to relatively compact sets in 𝐶1.

Proof. It is easy to check that 𝐾()(𝑡)𝐶1, for all 𝐿1. Since (𝑤(𝑡))1/(𝑝(𝑡)1)𝐿1 and 𝐾()(𝑡)=𝜑1𝑡,(𝑤(𝑡))1[(𝜌()+𝐹()),𝑡0,+),(2.16) it is easy to check that 𝐾 is a continuous operator from 𝐿1 to 𝐶1.
Let now 𝑈 be an equi-integrable set in 𝐿1, then there exists a nonnegative 𝜌𝐿1(𝐽,), such that ||||(𝑡)𝜌(𝑡)a.e.in𝐽,forany𝑈.(2.17)
We want to show that 𝐾(𝑈)𝐶1 is a compact set.
Let {𝑢𝑛} be a sequence in 𝐾(𝑈), then there exists a sequence {𝑛}𝑈 such that 𝑢𝑛=𝐾(𝑛). Since 𝑛(𝑡)=(1𝑛(𝑡),,𝑁𝑛(𝑡)), where 𝑖𝑛(𝑡)𝐿1(𝐽,)(𝑖=1,,𝑁), we have ||𝑖𝑛||||(𝑡)𝑛||(𝑡),𝑖=1,,𝑁,(2.18) then for any 𝑡1,𝑡2𝐽 with 𝑡1<𝑡2, we have ||||𝑡2𝑡1𝑛||||=(𝑡)𝑑𝑡𝑁𝑖=1𝑡2𝑡1𝑖𝑛(𝑡)𝑑𝑡2𝑁𝑖=1𝑡2𝑡1||𝑖𝑛||(𝑡)𝑑𝑡2𝑁𝑖=1𝑡2𝑡1||𝑛||(𝑡)𝑑𝑡2𝑁𝑡2𝑡1||𝑛||(𝑡)𝑑𝑡,(2.19) which together with (2.17) implies ||𝐹𝑛𝑡1𝐹𝑛𝑡2||=||||𝑡10𝑛(𝑡)𝑑𝑡𝑡20𝑛||||=||||(𝑡)𝑑𝑡𝑡2𝑡1𝑛(||||𝑡)𝑑𝑡𝑁𝑡2𝑡1||𝑛(||𝑡)𝑑𝑡𝑁𝑡2𝑡1𝜌(𝑡)𝑑𝑡.(2.20)
Hence, the sequence {𝐹(𝑛)} is uniformly bounded. According to the absolute continuity of Lebesgue integral, for any 𝜀>0, there exists a 𝛿>0 such that if 0𝑡1𝑡2<𝛿, then we have 0𝑁𝑡2𝑡1𝜌(𝑡)𝑑𝑡<𝜀. Thus, (2.20) means that {𝐹(𝑛)} is equicontinuous.
Denote Ω𝑚=[0,𝑚]. Obviously, {𝐹(𝑛)} is uniformly bounded and equicontinuous on Ω𝑚 for 𝑚=1,2,. By Ascoli-Arzela Theorem, there exists a subsequence {𝐹(𝑛(1))} of {𝐹(𝑛)} being convergent in 𝐶(Ω1), we may assume 𝐹(𝑛(1))𝑣1() in 𝐶(Ω1). Since {𝐹(𝑛(1))} is uniformly bounded and equicontinuous on Ω2, there exists a subsequence {𝐹(𝑛(2))} of {𝐹(𝑛(1))} such that {𝐹(𝑛(2))} is convergent in 𝐶(Ω2), we may assume 𝐹(𝑛(2))𝑣2() in 𝐶(Ω2). Obviously, 𝑣2(𝑡)=𝑣1(𝑡), for any 𝑡Ω1. Repeating the process, we get a subsequence {𝐹(𝑛(𝑚+1))} of {𝐹(𝑛(𝑚))} such that {𝐹(𝑛(𝑚+1))} is convergent in 𝐶(Ω𝑚+1), we may assume 𝐹(𝑛(𝑚+1))𝑣𝑚+1() in 𝐶(Ω𝑚+1). Obviously, 𝑣𝑚+1(𝑡)=𝑣𝑚(𝑡) for any 𝑡Ω𝑚. Select the diagonal element, we can see that {𝐹(𝑛(𝑛))} is a subsequence of {𝐹(𝑛)} which satisfies that {𝐹(𝑛(𝑛))} is convergent in 𝐶(Ω𝑚)(𝑚=1,2,) and 𝐹(𝑛(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 𝑡Ω𝑚, and 𝐹(𝑛(𝑛))𝑣() in 𝐶(Ω𝑚)(𝑚=1,2,).
From (2.20), it is easy to see that for any 𝑛=1,2,, lim𝑡𝐹(𝑛)(𝑡) exists (we denote the limit by 𝐹(𝑛(𝑛))(+)), and, for any 𝜀>0, there exists an integer 𝑅𝜀>0 such that 𝑅+𝜀𝜌(𝑡)𝑑𝑡<𝜀/𝑁, and then |||𝐹𝑛(𝑛)(+)𝐹𝑛|||(𝑡)𝑁𝑅+𝜀𝜌(𝑡)𝑑𝑡<𝜀,𝑡𝑅𝜀,𝑛=1,2,.(2.21)
Since {𝐹(𝑛)} is uniformly bounded, then {𝐹(𝑛(𝑛))(+)} is bounded. By choosing a subsequence, we may assume that lim𝑛𝐹𝑛(𝑛)(+)=𝑏.(2.22)
We claim that lim𝑡𝑣(𝑡)=𝑏. In fact, for any 𝑡𝑅𝜀, from (2.21), we have |||||||𝑣(𝑡)𝑏𝑣(𝑡)𝐹𝑛(𝑛)|||+|||𝐹(𝑡)𝑛(𝑛)(𝑡)𝐹𝑛(𝑛)|||+|||𝐹(+)𝑛(𝑛)||||||(+)𝑏𝑣(𝑡)𝐹𝑛(𝑛)||||||𝐹(𝑡)+𝜀+𝑛(𝑛)|||.(+)𝑏(2.23)
Since lim𝑛𝐹(𝑛(𝑛))(𝑡)=𝑣(𝑡) and lim𝑛𝐹(𝑛(𝑛))(+)=𝑏, letting 𝑛, the above inequality implies ||||𝑣(𝑡)𝑏𝜀,𝑡𝑅𝜀.(2.24)
Thus, lim𝑡𝑣(𝑡)=𝑏=lim𝑛𝐹𝑛(𝑛)(+).(2.25)
Next, we will prove that 𝐹(𝑛(𝑛)) tend to 𝑣 uniformly.
Suppose 𝑡𝑅𝜀. From (2.21) and (2.24), we have |||𝐹𝑛(𝑛)||||||𝐹(𝑡)𝑣(𝑡)𝑛(𝑛)(𝑡)𝐹𝑛(𝑛)|||+|||𝐹(+)𝑛(𝑛)|||+|||||||𝐹(+)𝑏𝑏𝑣(𝑡)𝜀+𝑛(𝑛)||||||𝐹(+)𝑏+𝜀=2𝜀+𝑛(𝑛)|||.(+)𝑏(2.26)
From (2.22), there exists a 𝑁1>0 such that |𝐹(𝑛(𝑛))(+)𝑏|𝜀 for 𝑛𝑁1. Thus, for any 𝑡𝑅𝜀, |||𝐹𝑛(𝑛)|||(𝑡)𝑣(𝑡)3𝜀,𝑛𝑁1.(2.27)
Suppose 𝑡[0,𝑅𝜀]. Since 𝐹(𝑛(𝑛))𝑣 in 𝐶([0,𝑅𝜀]), there exists a 𝑁2>0 such that |||𝐹𝑛(𝑛)|||(𝑡)𝑣(𝑡)𝜀,𝑛𝑁2.(2.28)
Thus, |||𝐹𝑛(𝑛)|||[(𝑡)𝑣(𝑡)3𝜀,𝑡0,+),𝑛𝑁1+𝑁2.(2.29)
This means that 𝐹(𝑛(𝑛)) tend to 𝑣 uniformly, that is, 𝐹(𝑛(𝑛)) tend to 𝑣 in 𝐶.
According to the bounded continuous of the operator 𝜌, we can choose a subsequence of {𝜌(𝑛)+𝐹(𝑛)} (which we still denote {𝜌(𝑛)+𝐹(𝑛)}) which is convergent in 𝐶, then 𝑤(𝑡)𝜑(𝑡,𝐾(𝑛)(𝑡))=𝜌(𝑛)+𝐹(𝑛) is convergent in 𝐶.
Since 𝐾𝑛𝜑(𝑡)=𝐹1𝑡,(𝑤(𝑡))1𝜌𝑛+𝐹𝑛[(𝑡),𝑡0,+),(2.30) it follows from the continuity of 𝜑1 and the integrability of 𝑤(𝑡)1/(𝑝(𝑡)1) in 𝐿1 that 𝐾(𝑛) is convergent in 𝐶. Thus, {𝑢𝑛} is convergent in 𝐶1. This completes the proof.

Let us define 𝑃𝐶1𝐶1 as𝑃1()=1𝜎0+𝑒(𝑡)𝐾()(𝑡)𝑑𝑡𝐾()(+)+𝑒1.(2.31) It is easy to see that 𝑃 is compact continuous.

Throughout the paper, we denote 𝑁𝑓(𝑢)[0,+)×𝐶1𝐿1 the Nemytskii operator associated to 𝑓 defined by 𝑁𝑓(𝑢)(𝑡)=𝑓𝑡,𝑢(𝑡),(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢(𝑡),𝑆(𝑢)(𝑡),𝑇(𝑢)(𝑡),a.e.on𝐽.(2.32)

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: 𝑢=𝑃𝛿𝑁𝑓(𝑢)+𝐾𝛿𝑁𝑓(𝑢).(2.33)

Proof. If 𝑢 is a solution of (1.1)-(1.2), by integrating (1.1) from 0 to 𝑡, we find that 𝑤(𝑡)𝜑𝑡,𝑢(𝑡)=𝜌𝛿𝑁𝑓(𝑢)+𝐹𝛿𝑁𝑓(𝑢)(𝑡),𝑡(0,+),(2.34) which implies that 𝜑𝑢(𝑡)=𝑢(0)+𝐹1𝑡,(𝑤(𝑡))1𝜌𝛿𝑁𝑓(𝑢)+𝐹𝛿𝑁𝑓[(𝑢)(𝑡)(𝑡),𝑡0,+).(2.35)
From 𝑢(+)=0+𝑒(𝑡)𝑢(𝑡)𝑑𝑡+𝑒1, we have 𝑢(0)=0+𝑒(𝑡)𝑡0𝜑1𝑟,(𝑤(𝑟))1𝜌+𝐹𝛿𝑁𝑓(𝑢)𝑑𝑟𝑑𝑡+𝑒11𝜎0+𝜑1𝑟,(𝑤(𝑟))1𝜌+𝐹𝛿𝑁𝑓(𝑢)𝑑𝑟1𝜎=𝑃𝛿𝑁𝑓.(𝑢)(2.36)
So we have 𝑢=𝑃𝛿𝑁𝑓(𝑢)+𝐾𝛿𝑁𝑓(𝑢).(2.37)
Conversely, if 𝑢 is a solution of (2.33), we have 𝑢(0)=𝑃𝛿𝑁𝑓(𝑢)+𝐾𝛿𝑁𝑓(𝑢)(0)=𝑃𝛿𝑁𝑓=1(𝑢)1𝜎0+𝑒(𝑡)𝐾()(𝑡)𝑑𝑡𝐾()(+)+𝑒1,(2.38) which implies that (1𝜎)𝑢(0)+𝐾()(+)=0+𝑒(𝑡)(𝑢(𝑡)𝑢(0))𝑑𝑡+𝑒1,(2.39) then 𝑢(+)=0+𝑒(𝑡)𝑢(𝑡)𝑑𝑡+𝑒1.(2.40)
From (2.33), we can obtain 𝑤(𝑡)𝜑𝑡,𝑢(𝑡)=𝜌𝛿𝑁𝑓(𝑢)+𝐹𝛿𝑁𝑓(𝑢)(𝑡).(2.41)
It follows from the condition of the mapping 𝜌 that 𝑤(+)𝜑+,𝑢(+)=𝜌𝛿𝑁𝑓(𝑢)+𝐹𝛿𝑁𝑓=1(𝑢)(+)1𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝜉𝑖0𝛿𝑁𝑓(𝑢)(𝑡)𝑑𝑡0+𝛿𝑁𝑓(𝑢)(𝑡)𝑑𝑡+𝑒2+0+𝛿𝑁𝑓(𝑢)(𝑡)𝑑𝑡,(2.42) and then 1𝑚2𝑖=1𝛼𝑖𝑤(+)𝜑+,𝑢=(+)𝑚2𝑖=1𝛼𝑖𝜉𝑖0𝛿𝑁𝑓(𝑢)(𝑡)𝑑𝑡0+𝛿𝑁𝑓(𝑢)(𝑡)𝑑𝑡+𝑒2+1𝑚2𝑖=1𝛼𝑖0+𝛿𝑁𝑓=(𝑢)(𝑡)𝑑𝑡𝑚2𝑖=1𝛼𝑖𝜉𝑖0𝛿𝑁𝑓(𝑢)(𝑡)𝑑𝑡𝑚2𝑖=1𝛼𝑖0+𝛿𝑁𝑓(𝑢)(𝑡)𝑑𝑡+𝑒2=𝑚2𝑖=1𝛼𝑖𝑤𝜉𝑖𝜑𝜉𝑖,𝑢𝜉𝑖𝑚2𝑖=1𝛼𝑖𝑤(+)𝜑+,𝑢(+)+𝑒2,(2.43) thus lim𝑡+||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢(𝑡)=𝑚2𝑖=1𝛼𝑖𝑤𝜉𝑖||𝑢||𝑝(𝜉𝑖)2𝑢𝜉𝑖+𝑒2.(2.44)
From (2.40) and (2.44), we obtain (1.2).
From (2.41), we have 𝑤(𝑡)𝜑𝑡,𝑢=𝛿𝑁𝑓(𝑢)(𝑡).(2.45)
Hence, 𝑢 is a solution of (1.1)-(1.2). This completes the proof.

2.2. Case (ii): 𝑚2𝑖=1𝛼𝑖[0,1),𝜎=1

We denote 𝜌1=𝑤(0)𝜑(0,𝑢(0)) in (2.7). It is easy to see that 𝜌1 is dependent on 𝑔(), and we have 𝜑𝑢(𝑡)=𝑢(0)+𝐹1𝑡,(𝑤(𝑡))1𝜌1+𝐹(𝑔)(𝑡),𝑡𝐽.(2.46)

The boundary value condition (1.2) implies that 0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))1𝜌1+𝐹(𝑔)(𝑟)𝑑𝑟𝑑𝑡𝑒1𝜌=0,1=11𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝜉𝑖0𝑔(𝑡)𝑑𝑡0+𝑔(𝑡)𝑑𝑡+𝑒2.(2.47)

For fixed 𝐿1, we define 𝜌1𝐿1𝑁 as𝜌11()=1𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝜉𝑖0(𝑡)𝑑𝑡0+(𝑡)𝑑𝑡+𝑒2.(2.48)

Similar to Lemma 2.2, we have the following.

Lemma 2.5. 𝜌1𝐿1𝑁 is continuous and sends bounded sets of 𝐿1 to bounded sets of 𝑁. Moreover, ||𝜌1||()2𝑁1𝑚2𝑖=1𝛼𝑖𝐿1+||𝑒2||.(2.49)

It is clear that 𝜌1() is continuous and sends bounded sets of 𝐿1 to bounded sets of 𝑁, and, hence, it is a compact continuous mapping.

Let us define𝑃1𝐶1𝐶1,𝑢𝑢(0),Θ𝐿1𝑁,0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))1𝜌1()+𝐹()(𝑟)𝑑𝑟𝑑𝑡𝑒1,𝐾1𝐾()(𝑡)=1𝜑(𝑡)=𝐹1𝑡,(𝑤(𝑡))1𝜌1[()+𝐹()(𝑡),𝑡0,+).(2.50)

Lemma 2.6. The operator 𝐾1 is continuous and sends equi-integrable sets in 𝐿1 to relatively compact sets in 𝐶1.

Proof. Similar to the proof of Lemma 2.3, 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: 𝑢=𝑃1𝑢+Θ𝛿𝑁𝑓(𝑢)+𝐾1𝛿𝑁𝑓.(𝑢)(2.51)

Proof. If 𝑢 is a solution of (1.1)-(1.2), by integrating (1.1) from 0 to 𝑡, we find that 𝑤(𝑡)𝜑𝑡,𝑢(𝑡)=𝜌1𝛿𝑁𝑓(𝑢)+𝐹𝛿𝑁𝑓(𝑢)(𝑡),𝑡(0,+),(2.52) which implies that 𝜑𝑢(𝑡)=𝑢(0)+𝐹1𝑡,(𝑤(𝑡))1𝜌1𝛿𝑁𝑓(𝑢)+𝐹𝛿𝑁𝑓[(𝑢)(𝑡)(𝑡),𝑡0,+).(2.53)
From 𝜎=1 and 𝑢(+)=0+𝑒(𝑡)𝑢(𝑡)𝑑𝑡+𝑒1, we obtain 0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))1𝜌1𝛿𝑁𝑓(𝑢)+𝐹𝛿𝑁𝑓(𝑢)(𝑡)𝑑𝑟𝑑𝑡𝑒1=Θ𝛿𝑁𝑓(𝑢)=0,(2.54) then 𝑢=𝑃1𝑢+Θ𝛿𝑁𝑓(𝑢)+𝐾1𝛿𝑁𝑓.(𝑢)(2.55)
Conversely, if 𝑢 is a solution of (2.51), then 𝑢(0)=𝑃1𝑢+Θ𝛿𝑁𝑓(𝑢)+𝐾1𝛿𝑁𝑓(𝑢)(0)=𝑢(0)+Θ𝛿𝑁𝑓.(𝑢)(2.56)
Thus, Θ(𝛿𝑁𝑓(𝑢))=0, and we have 0+𝑒(𝑡)0+𝜑1𝑟,(𝑤(𝑟))1𝜌1𝛿𝑁𝑓(𝑢)+𝐹𝛿𝑁𝑓=(𝑢)(𝑟)𝑑𝑟𝑑𝑡0+𝑒(𝑡)𝑡0𝜑1𝑟,(𝑤(𝑟))1𝜌1𝛿𝑁𝑓(𝑢)+𝐹𝛿𝑁𝑓(𝑢)(𝑟)𝑑𝑟𝑑𝑡+𝑒1,(2.57) then 0+{𝑒(𝑡)(𝑢(+)𝑢(0))}𝑑𝑡=0+{𝑒(𝑡)(𝑢(𝑡)𝑢(0))}𝑑𝑡+𝑒1.(2.58)
Thus, 𝑢(+)=0+𝑒(𝑡)𝑢(𝑡)𝑑𝑡+𝑒1.(2.59)
Similar to the proof of Lemma 2.4, we can have lim𝑡+||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢(𝑡)=𝑚2𝑖=1𝛼𝑖𝑤𝜉𝑖||𝑢||𝑝(𝜉𝑖)2𝑢𝜉𝑖+𝑒2.(2.60)
From (2.59) and (2.60), we obtain (1.2).
From (2.51), we have 𝑤(𝑡)𝜑𝑡,𝑢(𝑡)=𝜌1𝛿𝑁𝑓(𝑢)+𝐹𝛿𝑁𝑓(𝑢)(𝑡),(2.61) then 𝑤(𝑡)𝜑𝑡,𝑢=𝛿𝑁𝑓(𝑢)(𝑡).(2.62)
Hence, 𝑢 is a solution of (1.1)-(1.2). This completes the proof.

2.3. Case (iii): 𝑚2𝑖=1𝛼𝑖=1,𝜎=1

We denote 𝜌2=𝑤(0)𝜑(0,𝑢(0)) in (2.7). It is easy to see that 𝜌2 is dependent on 𝑔(), then we find that 𝜑𝑢(𝑡)=𝑢(0)+𝐹1𝑡,(𝑤(𝑡))1𝜌2+𝐹(𝑔)(𝑡),𝑡𝐽.(2.63)

The boundary value condition (1.2) implies that 0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))1𝜌2+𝐹(𝑔)(𝑟)𝑑𝑟𝑑𝑡𝑒1=0,𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝑔(𝑡)𝑑𝑡𝑒2=0.(2.64)

For fixed 𝐿1, we denote Λ𝜌2=0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))1𝜌2+𝐹()(𝑟)𝑑𝑟𝑑𝑡𝑒1.(2.65)

Throughout the paper, we denote 𝐸#=0+𝑒(𝑡)𝑡+(𝑤(𝑟))1/(𝑝(𝑟)1)𝑑𝑟𝑑𝑡.(2.66)

Lemma 2.8. The function Λ() has the following properties.(i)For any fixed 𝐿1, the equation Λ𝜌2=0(2.67)has a unique solution 𝜌2()𝑁.(ii)The function 𝜌2𝐿1𝑁, defined in (i), is continuous and sends bounded sets to bounded sets. Moreover, ||𝜌2||𝐸()3𝑁#+1𝐸#𝑝+0||𝑒+2𝑁1||𝑝#1,(2.68)where the notation 𝑀𝑝#1 means 𝑀𝑝#1=𝑀𝑝+1M,𝑀>1,𝑝1,𝑀1.(2.69)

Proof. (i) From Lemma 2.1, it is immediate that Λ(𝑥)Λ(𝑦),𝑥𝑦>0,for𝑥𝑦,(2.70) and, hence, if (2.67) has a solution, then it is unique.
Let 𝑡0=3𝑁((𝐸#+1)/𝐸#)𝑝+[0+2𝑁|𝑒1|𝑝#1]. Since(𝑤(𝑡))1/(𝑝(𝑡)1)𝐿1(0,+) and 𝐿1, if |𝜌2|>𝑡0, it is easy to see that there exists an 𝑖{1,,𝑁} such that the 𝑖th component 𝜌𝑖2 of 𝜌2 satisfies |𝜌𝑖2||𝜌2|/𝑁>3((𝐸#+1)/𝐸#)𝑝+[0+2𝑁|𝑒1|𝑝#1]. Thus, (𝜌𝑖2+𝑖(𝑡)) keeps sign on 𝐽 and ||𝜌𝑖2+𝑖||||𝜌(𝑡)𝑖2||02||𝜌2||𝐸3𝑁>2#+1𝐸#𝑝+0||𝑒+2𝑁1||𝑝#1,𝑡𝐽.(2.71)
Obviously, |𝜌2+(𝑡)|4|𝜌2|/32𝑁|𝜌𝑖2+𝑖(𝑡)|, then ||𝜌2||+(𝑡)(2𝑝(𝑡))/(𝑝(𝑡)1)||𝜌𝑖2+𝑖||>1(𝑡)||𝜌2𝑁𝑖2+𝑖||(𝑡)1/(𝑝(𝑡)1)>𝐸#+12𝑁𝐸#||𝑒1||,𝑡𝐽.(2.72)
Thus, the 𝑖th component Λ𝑖(𝜌2) of Λ(𝜌2) is nonzero and keeps sign, and then we have 0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))1𝜌2+𝐹()(𝑟)𝑑𝑟𝑑𝑡𝑒10.(2.73)
Let us consider the equation 𝜆Λ𝜌2+(1𝜆)𝜌2[]=0,𝜆0,1.(2.74)
It is easy to see that all the solutions of (2.74) belong to 𝑏(𝑡0+1)={𝑥𝑁|𝑥|<𝑡0+1}. So, we have 𝑑𝐵Λ𝜌2𝑡,𝑏0+1,0=𝑑𝐵𝑡𝐼,𝑏0+1,00,(2.75) and it means the existence of solutions of Λ(𝜌2)=0.
In this way, we define a function 𝜌2()𝐿1𝑁, which satisfies Λ𝜌2()=0.(2.76)
(ii) By the proof of (i), we also obtain 𝜌2 sends bounded sets to bounded sets, and ||𝜌2||𝐸()3𝑁#+1𝐸#𝑝+0||𝑒+2𝑁1||𝑝#1.(2.77)
It only remains to prove the continuity of 𝜌2. Let {𝑢𝑛} be a convergent sequence in 𝐿1 and 𝑢𝑛𝑢 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 that 𝜌2() is continuous and sends bounded sets of 𝐿1 to bounded sets of 𝑁, and, hence, it is a compact continuous mapping.

Let us define𝑃2𝐶1𝐶1,𝑢𝑢(0),(2.78)𝑄𝐿1𝑁,𝑚2𝑖=1𝛼𝑖𝜉+𝑖(𝑡)𝑑𝑡𝑒2,𝑄(2.79)𝐿1𝐿1,𝜏(𝑡)𝑚2𝑖=1𝛼𝑖𝜉+𝑖(𝑡)𝑑𝑡𝑒2,(2.80) where 𝜏([0,+),) and satisfies 0<𝜏(𝑡)<1,𝑡𝐽,𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝜏(𝑡)𝑑𝑡=1. We denote 𝐾2𝐿1𝐶1 as 𝐾2𝜑()(𝑡)=𝐹1𝑡,(𝑤(𝑡))1𝜌2𝐼𝑄+𝐹𝐼𝑄[(𝑡),𝑡0,+).(2.81)

Similar to Lemmas 2.3 and 2.7, we have the following

Lemma 2.9. The operator 𝐾2 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: 𝑢=𝑃2𝑢+𝑄𝛿𝑁𝑓(𝑢)+𝐾2𝛿𝑁𝑓.(𝑢)(2.82)

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 𝑚2𝑖=1𝛼𝑖[0,1),𝜎[0,1). Moreover, we give the existence of nonnegative solutions.

Theorem 3.1. In Case (i), if 𝑓 satisfies sub-(𝑝1) growth condition, then problem (1.1)-(1.2) has at least a solution for any fixed parameter 𝛿.

Proof. Denote Ψ𝑓(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 𝑢=Ψ𝑓(𝑢,𝜆),(3.1) when 𝜆=1.
It is easy to see that the operator 𝑃 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 𝑢𝑛1+ as 𝑛+ and 𝑢𝑛1>1 for any 𝑛=1,2,.
From Lemma 2.2, we have ||𝜌𝜆𝑛𝛿𝑁𝑓𝑢𝑛||𝐶1𝑁𝑓𝑢𝑛𝐿1+||𝑒2||𝐶2𝑢𝑛𝑞+11,(3.2) then we have ||𝜌𝜆𝑛𝛿𝑁𝑓𝑢𝑛𝜆+𝐹𝑛𝛿𝑁𝑓𝑢𝑛||||𝜌𝜆𝑛𝛿𝑁𝑓𝑢𝑛||+||𝐹𝜆𝑛𝛿𝑁𝑓𝑢𝑛||𝐶3𝑢𝑛𝑞+11.(3.3)
From (3.1), we have ||𝑢𝑤(𝑡)𝑛||(𝑡)𝑝(𝑡)2𝑢𝑛𝜆(𝑡)=𝜌𝑛𝛿𝑁𝑓𝑢𝑛𝜆+𝐹𝑛𝛿𝑁𝑓𝑢𝑛,𝑡𝐽,(3.4) then ||𝑢𝑤(𝑡)𝑛||(𝑡)𝑝(𝑡)1||𝜌𝜆𝑛𝛿𝑁𝑓𝑢𝑛||+||𝐹𝜆𝑛𝛿𝑁𝑓𝑢𝑛||𝐶3𝑢𝑛𝑞+11.(3.5)
Denote 𝛼=(𝑞+1)/(𝑝1), from the above inequality, we have (𝑤(𝑡))1/(𝑝(𝑡)1)𝑢𝑛(𝑡)0𝐶4𝑢𝑛𝛼1.(3.6)
It follows from (2.36) and (3.3) that ||𝑢𝑛(||0)𝐶5𝑢𝑛𝛼1𝑞,where𝛼=+1𝑝.1(3.7)
For any 𝑗=1,,𝑁, we have ||𝑢𝑗𝑛||=||||𝑢(𝑡)𝑗𝑛(0)+𝑡0𝑢𝑗𝑛||||||𝑢(𝑟)𝑑𝑟𝑗𝑛(||+||||0)𝑡0(𝑤(𝑟))1/(𝑝(𝑟)1)sup𝑡(0,+)|||(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢𝑗𝑛(|||||||𝐶𝑡)𝑑𝑟6+𝐶4𝐸𝑢𝑛𝛼1𝐶7𝑢𝑛𝛼1,(3.8) which implies that ||𝑢𝑗𝑛||0𝐶8𝑢𝑛𝛼1,𝑗=1,,𝑁,𝑛=1,2,.(3.9)
Thus, 𝑢𝑛0𝐶9𝑢𝑛𝛼1,𝑛=1,2,.(3.10)
It follows from (3.6) and (3.10) that {𝑢𝑛1} is bounded.
Thus, we can choose a large enough 𝑅0>0 such that all the solutions of (3.1) belong to 𝐵(𝑅0)={𝑢𝐶1𝑢1<𝑅0}. Thus, the Leray-Schauder degree 𝑑LS[𝐼Ψ𝑓(,𝜆),𝐵(𝑅0),0] is well defined for each 𝜆[0,1], and 𝑑LS𝐼Ψ𝑓𝑅(,1),𝐵0,0=𝑑LS𝐼Ψ𝑓𝑅(,0),𝐵0.,0(3.11)
Let 𝑢0=0+𝑒(𝑡)𝑡0𝜑1𝑟,(𝑤(𝑟))1𝜌(0)𝑑𝑟𝑑𝑡0+𝜑1𝑟,(𝑤(𝑟))1𝜌(0)𝑑𝑟+𝑒1+1𝜎𝑟0𝜑1𝑡,(𝑤(𝑡))1𝜌(0)𝑑𝑡,(3.12) where 𝜌(0) is defined in (2.10), thus 𝑢0 is the unique solution of 𝑢=Ψ𝑓(𝑢,0).
It is easy to see that 𝑢 is a solution of 𝑢=Ψ𝑓(𝑢,0) if and only if 𝑢 is a solution of the following system Δ𝑝(𝑡)𝑢=0,𝑡(0,+),𝑢(+)=0+𝑒(𝑡)𝑢(𝑡)𝑑𝑡+𝑒1,lim𝑡+||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢(𝑡)=𝑚2𝑖=1𝛼𝑖𝑤𝜉𝑖||𝑢||𝑝(𝜉𝑖)2𝑢𝜉𝑖+𝑒2.(I)
Obviously, system (I ) possesses a unique solution 𝑢0. Note that 𝑢0𝐵(𝑅0), we have 𝑑LS𝐼Ψ𝑓𝑅(,1),𝐵0,0=𝑑LS𝐼Ψ𝑓𝑅(,0),𝐵0,00.(3.13)
Therefore, (1.1)-(1.2) has at least one solution when 𝑚2𝑖=1𝛼𝑖[0,1),𝜎[0,1). This completes the proof.

Denote Ω𝜀=𝑢𝐶1max1𝑖𝑁||𝑢𝑖||0+|||(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢𝑖|||0𝜀<𝜀,𝜃=2+(1/𝐸).(3.14)

Assume the following(A1)Let positive constant 𝜀 such that 𝑢0Ω𝜀,|𝑃(0)|<𝜃, and |𝜌(0)|<(1/𝑁(2𝐸+2))inf𝑡𝐽|𝜀/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 that Ω𝜀 is an open bounded domain in 𝐶1. We have the following.

Theorem 3.2. In the Case (i), assume that 𝑓 satisfies general growth condition and (A1) is satisfied, then the problem (1.1)-(1.2) has at least one solution on Ω𝜀 when the positive parameter 𝛿 is small enough.

Proof. Denote Ψ𝑓(𝑢,𝜆)=𝑃(𝜆𝛿𝑁𝑓(𝑢))+𝐾(𝜆𝛿𝑁𝑓(𝑢)). According to Lemma 2.4, 𝑢 is a solution of Δ𝑝(𝑡)𝑢+𝜆𝛿𝑓𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢)=0,𝑡(0,+),(3.15) with (1.2) if and only if 𝑢 is a solution of the following abstract equation 𝑢=Ψ𝑓(𝑢,𝜆).(3.16)
From Lemmas 2.2 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 𝜕Ω𝜀 for any 𝜆[0,1),(2°)𝑑LS[𝐼Ψ𝑓(,0),Ω𝜀,0]0, then we can conclude that the system (1.1)-(1.2) has a solution on Ω𝜀.
(1°) If there exists a 𝜆[0,1) and 𝑢𝜕Ω𝜀 is a solution of (3.15) with (1.2), then (𝜆,𝑢) satisfies 𝑤(𝑡)𝜑𝑡,𝑢(𝑡)=𝜌𝜆𝛿𝑁𝑓𝑁(𝑢)+𝜆𝛿𝐹𝑓(𝑢)(𝑡),𝑡(0,+).(3.17)
Since 𝑢𝜕Ω𝜀, there exists an 𝑖 such that |𝑢𝑖|0+|(𝑤(𝑡))1/(𝑝(𝑡)1)(𝑢𝑖)|0=𝜀.
(i) Suppose that |𝑢𝑖|0>2𝜃, then |(𝑤(𝑡))1/(𝑝(𝑡)1)(𝑢𝑖)|0<𝜀2𝜃=𝜃/𝐸. On the other hand, for any 𝑡,𝑡𝐽, we have ||𝑢𝑖(𝑡)𝑢𝑖𝑡||=||||𝑡𝑡𝑢𝑖||||(𝑟)𝑑𝑟0+(𝑤(𝑟))1/(𝑝(𝑟)1)|||(𝑤(𝑟))1/(𝑝(𝑟)1)𝑢𝑖|||(𝑟)𝑑𝑟<𝜃.(3.18)
This implies that |𝑢𝑖(𝑡)|>𝜃 for each 𝑡𝐽.
Note that 𝑢Ω𝜀, then |𝑓(𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢))|𝛽𝐶𝜀(𝑡) (where 𝐶=𝑁+sup𝑡𝐽0+|𝜓(𝑠,𝑡)|𝑑𝑠+sup𝑡𝐽0+|𝜒(𝑠,𝑡)|𝑑𝑠), holding |𝐹(𝑁𝑓(𝑢))|0+𝛽𝐶𝜀(𝑡)𝑑𝑡. Since 𝑃() is continuous, when 0<𝛿 is small enough, from (A1), we have ||𝑢||=||𝑃(0)𝜆𝛿𝑁𝑓||(𝑢)<𝜃.(3.19)
It is a contradiction to |𝑢𝑖(𝑡)|>𝜃 for any 𝑡𝐽.
(ii) Suppose that |𝑢𝑖|02𝜃, then 𝜃/𝐸|(𝑤(𝑡))1/(𝑝(𝑡)1)(𝑢𝑖)|0𝜀. This implies that |||𝑤𝑡21/(𝑝(𝑡2)1)𝑢𝑖𝑡2|||>𝜀2(𝐸+1)forsome𝑡2𝐽.(3.20)
Since 𝑢Ω𝜀, it is easy to see that |||𝑤𝑡21/(𝑝(𝑡2)1)𝑢𝑖𝑡2|||>𝜀=2(𝐸+1)𝑁𝜀|||𝑤𝑡𝑁(2𝐸+2)21/(𝑝(𝑡2)1)𝑢𝑡2|||.𝑁(2𝐸+2)(3.21)
Combining (3.17) and (3.21), we have ||||𝜀/2(𝐸+1)𝑝(𝑡2)1<1𝑁(2𝐸+2)𝑤𝑡𝑁(2𝐸+2)2|||𝑢𝑖(𝑡2)|||𝑝(𝑡2)11𝑤𝑡𝑁(2𝐸+2)2||𝑢𝑡2||𝑝(𝑡2)1𝑡𝑤2||𝑢𝑡2||𝑝(𝑡2)2|||𝑢𝑖𝑡2|||||𝜌𝜆𝛿𝑁𝑓||||𝑁+𝜆𝛿𝐹𝑓𝑡2||.(3.22)
Since 𝑢Ω𝜀 and 𝑓 is Caratheodory, it is easy to see that ||𝑓𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢||,𝑆(𝑢),𝑇(𝑢)𝛽𝐶𝜀(𝑡),(3.23) thus ||𝑁𝛿𝐹𝑓||(𝑢)𝛿0+𝛽𝐶𝜀(𝑡)𝑑𝑡.(3.24)
From Lemma 2.2, 𝜌() is continuous, then we have ||𝜌𝜆𝛿𝑁𝑓||||𝜌||(𝑢)(0)as𝛿0.(3.25)
When 0<𝛿 is small enough, from (A1) and (3.22), we can conclude that ||||𝜀/2(𝐸+1)𝑝(𝑡2)1<||𝜌𝑁(2𝐸+2)𝜆𝛿𝑁𝑓||||𝑁(𝑢)+𝜆𝛿𝐹𝑓||<1(𝑢)(𝑡)𝑁(2𝐸+2)inf𝑡𝐽||||𝜀||||2(𝐸+1)𝑝(𝑡)1.(3.26)
It is a contradiction.
Summarizing this argument, for each 𝜆[0,1), the problem (3.15) with (1.2) has no solution on 𝜕Ω𝜀.
(2°) Since 𝑢0 (where 𝑢0 is defined in (3.12)) is the unique solution of 𝑢=Ψ𝑓(𝑢,0), and (A1) holds 𝑢0Ω𝜀, we can see that the Leray-Schauder degree 𝑑LS𝐼Ψ𝑓(,0),Ω𝜀,00.(3.27)
This completes the proof.

In the following, we will deal with the existence of nonnegative solutions of (1.1)-(1.2) when 𝑚2𝑖=1𝛼𝑖[0,1),𝜎[0,1). For any 𝑥=(𝑥1,,𝑥𝑁)𝑁, the notation 𝑥0(𝑥>0) means 𝑥𝑗0(𝑥𝑗>0) for any 𝑗=1,,𝑁. For any 𝑥,𝑦𝑁, the notation 𝑥𝑦 means 𝑥𝑦0 and the notation 𝑥>𝑦 means 𝑥𝑦>0.

Theorem 3.3. In Case (i), we assume(10)𝛿𝑓(𝑡,𝑥,𝑦,𝑧,𝑤)0,(𝑡,𝑥,y,𝑧,𝑤)𝐽×𝑁×𝑁×𝑁×𝑁, (20)𝑒10, (30)𝑒20.
Then, all the solutions of (1.1)-(1.2) are nonnegative.

Proof. If 𝑢 is a solution of (1.1)-(1.2), then 𝑤(𝑡)𝜑𝑡,𝑢(𝑡)=𝜌𝛿𝑁𝑓(+𝑢)𝑡0𝛿𝑓𝑟,𝑢,(𝑤(𝑟))1/(𝑝(𝑟)1)𝑢,𝑆(𝑢),𝑇(𝑢)𝑑𝑟,𝑡𝐽.(3.28)
It follows from (2.10), (10), and (30) that 𝑤(𝑡)𝜑𝑡,𝑢(𝑡)=𝜌𝛿𝑁𝑓(𝑢)+𝐹𝛿𝑁𝑓((=1𝑢)𝑡)1𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝐹𝛿𝑁𝑓𝜉(𝑢)𝑖𝐹𝛿𝑁𝑓(𝑢)(+)+𝑒2+𝐹𝛿𝑁𝑓=1(𝑢)(𝑡)1𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝛿𝑁𝑓(𝑢)(𝑟)𝑑𝑟1𝑚2𝑖=1𝛼𝑖𝑡+𝛿𝑁𝑓(𝑢)(𝑟)𝑑𝑟+𝑒20.(3.29)
Thus, 𝑢(𝑡)0 for any 𝑡𝐽. Holding 𝑢(𝑡) is decreasing, namely, 𝑢(𝑡1)𝑢(𝑡2) for any 𝑡1,𝑡2𝐽 with 𝑡1<𝑡2.
According to the boundary value condition (1.2) and condition (2°), we have 𝑢(+)=0+𝑒(𝑡)𝑢(𝑡)𝑑𝑡+𝑒10+𝑒(𝑡)𝑢(+)𝑑𝑡+𝑒1=𝜎𝑢(+)+𝑒1,(3.30) then 𝑒𝑢(+)11𝜎0.(3.31)
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×𝑁×𝑁×𝑁×𝑁 with 𝑥,𝑧,𝑤0,(20)𝜓(𝑠,𝑡)0,𝜒(𝑠,𝑡)0,(𝑠,𝑡)𝐷, (30)𝑒10, (40)𝑒20.
Then, we have the following.
(a) On the conditions of Theorem 3.1, then (1.1)-(1.2) has at least a nonnegative solution 𝑢.
(b) On the conditions of Theorem 3.2, then (1.1)-(1.2) has at least a nonnegative solution 𝑢.

Proof. (a) Define 𝐿𝐿(𝑢)=𝑢1,,𝐿𝑢𝑁,(3.32) where 𝐿(𝑡)=𝑡,𝑡0,0,𝑡<0.(3.33)
Denote 𝑓(𝑡,𝑢,𝑣,𝑆(𝑢),𝑇(𝑢))=𝑓(𝑡,𝐿(𝑢),𝑣,𝑆(𝐿(𝑢)),𝑇(𝐿(𝑢))),(𝑡,𝑢,𝑣)𝐽×𝑁×𝑁,(3.34) then 𝑓(𝑡,𝑢,𝑣,𝑆(𝑢),𝑇(𝑢)) satisfies Caratheodory condition and 𝑓(𝑡,𝑢,𝑣,𝑆(𝑢),𝑇(𝑢))0 for any (𝑡,𝑢,𝑣)𝐽×𝑁×𝑁.
Obviously, we have(A2)lim|𝑢|+|𝑣|+(𝑓(𝑡,𝑢,𝑣,𝑆(𝑢),𝑇(𝑢))/(|𝑢|+|𝑣|)𝑞(𝑡)1)=0,for𝑡𝐽uniformly,
where 𝑞(𝑡)𝐶(𝐽,), and 1<𝑞𝑞+<𝑝.
Then, 𝑓(𝑡,,,,) satisfies sub-(𝑝1) growth condition.
Let us consider the existence of solutions of the following system: Δ𝑝(𝑡)𝑓𝑢+𝛿𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢)=0,𝑡(0,+),(3.35) 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 𝑚2𝑖=1𝛼𝑖[0,1),𝜎=1.

Theorem 4.1. Assume that Ω is an open bounded set in 𝐶1 such that the following conditions hold.(10)For each 𝜆(0,1), the problem𝑢𝑤(𝑡)𝜑𝑡,(𝑡)𝜆=𝛿𝑓𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢),𝑡(0,+),𝑢(+)=0+𝑒(𝑡)𝑢(𝑡)𝑑𝑡+𝜆2𝑒1,lim𝑡+𝑢𝑤(𝑡)𝜑𝑡,(𝑡)𝜆=𝑚2𝑖=1𝛼𝑖𝑤𝜉𝑖𝜑𝜉𝑖,𝑢𝜉𝑖𝜆+𝑒2,(4.1)has no solution on 𝜕Ω.(20)The equation𝜔(𝑎)=0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))11𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝛿𝑓(𝑠,𝑎,0,𝑆(𝑎),𝑇(𝑎))𝑑𝑠1𝑚2𝑖=1𝛼𝑖×𝑟+𝛿𝑓(𝑠,𝑎,0,𝑆(𝑎),𝑇(𝑎))𝑑𝑠+𝑒2𝑑𝑟𝑑𝑡=0(4.2)has no solution on 𝜕Ω𝑁.(30)The Brouwer degree 𝑑𝐵[𝜔,Ω𝑁,0]0.
Then, problems (1.1)-(1.2) have a solution on Ω.

Proof. For any 𝜆(0,1], it is easy to have problem (4.1) can be written in the equivalent form 𝑢=Φ𝑓(𝑢,𝜆)=𝑃1𝑢+Θ𝜆𝛿𝑁𝑓(𝑢)+𝐾𝜆1𝛿𝑁𝑓,(𝑢)(4.3) where 𝑃1𝐶1𝐶1Θ,𝑢𝑢(0),𝜆𝐿1𝑁,0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))11𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝜉+𝑖(𝑠)𝑑𝑠1𝑚2𝑖=1𝛼𝑖×𝑟+(𝑠)𝑑𝑠+𝑒2𝑑𝑟𝑑𝑡𝜆𝑒1,𝐾𝜆1()(𝑡)=𝐹𝜆𝜑1𝑡,(𝑤(𝑡))1𝜌𝜆1[()+𝐹()(𝑡),𝑡0,+),(4.4) where 𝜌𝜆1()=𝑤(0)𝜑(0,𝑢(0)/𝜆)=𝜌1().
It is easy to see that the operator 𝑃1 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].
For 𝜆[0,1], if 𝑢 is a solution of (4.3), we have Θ𝜆𝛿𝑁𝑓(𝑢)=0.(4.5)
Thus, for 𝜆=0, it follows from (4.3) and (4.4) that 𝑢=Φ𝑓(𝑢,0)=𝑃1𝑢+Θ0𝛿𝑁𝑓,(𝑢)(4.6) it holds 𝑢𝑑, a constant.
Therefore, when 𝜆=0, by (4.5), 0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))11𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝛿𝑓(𝑠,𝑑,0,𝑆(𝑑),𝑇(𝑑))𝑑𝑠1𝑚2𝑖=1𝛼𝑖×𝑟+𝛿𝑓(𝑠,𝑑,0,𝑆(𝑑),𝑇(𝑑))𝑑𝑠+𝑒2𝑑𝑟𝑑𝑡=0,(4.7) 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[𝐼Φ𝑓(,𝜆),Ω,0]is well defined, and, from the properties of that degree, we have 𝑑LS𝐼Φ𝑓(,1),Ω,0=𝑑LS𝐼Φ𝑓.(,0),Ω,0(4.8)
Now, it is clear that problem 𝑢=Φ𝑓(𝑢,1)(4.9) 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 𝑑LS𝐼Φ𝑓(,0),Ω,00.(4.10)
Since Φ𝑓(𝑢,0)=𝑃1𝑢+Θ0𝑁𝛿𝑓(𝑢)+𝐾01𝑁𝛿𝑓,(𝑢)(4.11) then 𝑢Φ𝑓(𝑢,0)=𝑢𝑃1𝑢Θ0𝑁𝛿𝑓(𝑢)𝐾01𝑁𝛿𝑓.(𝑢)(4.12)
From (4.4), we have 𝐾01(𝑁𝛿𝑓(𝑢))0. By the properties of the Leray-Schauder degree, we have 𝑑LS𝐼Φ𝑓(,0),Ω,0=(1)𝑁𝑑𝐵𝜔,Ω𝑁,,0(4.13) 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 Theorem 4.1. As an application of Theorem 4.1, let us consider the following equation with (1.2):||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢=𝑔𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢)+𝑏𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,,𝑆(𝑢),𝑇(𝑢)(4.14) where 𝑏𝐽×𝑁×𝑁×𝑁×𝑁𝑁 is Caratheodory, 𝑔=(𝑔1,,𝑔𝑁)𝐽×𝑁×𝑁×𝑁×𝑁𝑁 is continuous and Caratheodory, and, for any fixed 𝑦0𝑁{0}, if 𝑦𝑖00, then 𝑔𝑖(𝑡,𝑦0,0,𝑆(𝑦0),𝑇(𝑦0))0, for all 𝑡𝐽, 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×𝑁×𝑁×𝑁×𝑁, where 𝑞(𝑡)𝐶(𝐽,) satisfies 1<𝑞𝑞+<𝑝,(20)lim|𝑥|+|𝑦|+|𝑧|+|𝑤|+(𝑏(𝑡,𝑥,𝑦,𝑧,𝑤)/(|𝑥|+|𝑦|+|𝑧|+|𝑤|)𝑞(𝑡)1)=0,for𝑡𝐽uniformly, (30)for large enough 𝑅0>0, the equation𝜔𝑔(𝑎)=0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))11𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝑔(𝑠,𝑎,0,,𝑆(𝑎),𝑇(𝑎))𝑑𝑠1𝑚2𝑖=1𝛼𝑖×𝑟+𝑔(𝑠,𝑎,0,𝑆(𝑎),𝑇(𝑎))𝑑𝑠+𝑒2𝑑𝑟𝑑𝑡=0(4.15)has no solution on 𝜕𝐵(𝑅0)𝑁, where 𝐵(𝑅0)={𝑢𝐶1𝑢1<𝑅0},(40)the Brouwer degree 𝑑𝐵[𝜔𝑔,𝑏(𝑅0),0]0 for large enough 𝑅0>0, where 𝑏(𝑅0)={𝑥𝑁|𝑥|<𝑅0}.
Then, problem (4.14) with (1.2) has at least one solution.

Proof. Denote 𝑁𝑓(𝑢,𝜆)=𝑓𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢),𝜆=𝑔𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢)+𝜆𝑏𝑡,𝑢,(𝑤(𝑟))1/(𝑝(𝑡)1)𝑢.,𝑆(𝑢),𝑇(𝑢)(4.16)
At first, we consider the following problem 𝑢𝑤(𝑡)𝜑𝑡,(𝑡)𝜆=𝑁𝑓(𝑢,𝜆),𝑡(0,+),𝑢(+)=0+𝑒(𝑡)𝑢(𝑡)𝑑𝑡+𝜆2𝑒1,lim𝑡+𝑢𝑤(𝑡)𝜑𝑡,(𝑡)𝜆=𝑚2𝑖=1𝛼𝑖𝑤𝜉𝑖𝜑𝜉𝑖,𝑢𝜉𝑖𝜆+𝑒2.(4.17)
For any 𝜆(0,1], it is easy to have problem (4.17) can be written in the equivalent form 𝑢=Φ𝑓(𝑢,𝜆)=𝑃1𝑢+Θ𝜆𝑁𝑓(𝑢,𝜆)+𝐾𝜆1𝑁𝑓,(𝑢,𝜆)(4.18) where Θ𝜆 and 𝐾𝜆1 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 𝑢𝑛1+ as 𝑛+ and 𝑢𝑛1>1for any 𝑛=1,2,.
Since (𝑢𝑛,𝜆𝑛) are solutions of (4.17), we have 0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))11𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝑁𝑓𝑢𝑛,𝜆𝑛𝑑𝑠1𝑚2𝑖=1𝛼𝑖×𝑟+𝑁𝑓𝑢𝑛,𝜆𝑛𝑑𝑠+𝑒2𝑑𝑟𝑑𝑡𝜆𝑛𝑒1=0,𝑡(0,+).(4.19)
It follows from Lemma 2.5 that |||𝜌𝜆𝑛1𝑁𝑓𝑢𝑛,𝜆𝑛|||2𝑁1𝑚2𝑖=1𝛼𝑖𝑁𝑓𝑢𝑛,𝜆𝑛𝐿1+||𝑒2||𝐶1𝑢𝑛𝑞+11𝑔𝑢𝑡,𝑛𝑢𝑛1,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢𝑛𝑢𝑛1,𝑆𝑢𝑛𝑢𝑛1,𝑇𝑢𝑛𝑢𝑛1+𝑜(1)𝐿1,(4.20) 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 |||𝜌𝜆𝑛1𝑁𝑓𝑢𝑛,𝜆𝑛|||𝐶2𝑢𝑛𝑞+11,𝑡(0,+),(4.21) then we have |||𝜌𝜆𝑛1𝑁𝑓𝑢𝑛,𝜆𝑛|||+||𝐹𝑁𝑓𝑢𝑛,𝜆𝑛||𝐶3𝑢𝑛𝑞+11,𝑡(0,+).(4.22)
From (4.18), we have 𝑤𝑢(𝑡)𝜑𝑡,𝑛(𝑡)𝜆𝑛=𝜌𝜆𝑛1𝑁𝑓𝑢𝑛,𝜆𝑛𝑁+𝐹𝑓𝑢𝑛,𝜆𝑛,𝑡𝐽,(4.23) then ||||𝑢𝑤(𝑡)𝑛(𝑡)𝜆𝑛||||𝑝(𝑡)1|||𝜌𝜆𝑛1𝑁𝑓𝑢𝑛,𝜆𝑛|||+||𝐹𝑁𝑓𝑢𝑛,𝜆𝑛||𝐶3𝑢𝑛𝑞+11.(4.24)
Denote 𝛼=(𝑞+1)/(𝑝1), then (𝑤(𝑡))1/(𝑝(𝑡)1)𝑢𝑛(𝑡)0𝜆𝑛𝐶4𝑢𝑛𝛼1𝐶5𝑢𝑛𝛼1.(4.25)
Since 𝛼(0,1), from (4.25), we have lim𝑛+𝑢𝑛0𝑢𝑛1=1.(4.26)
Denote 𝜇𝑛=(|𝑢1𝑛|0/𝑢𝑛0,|𝑢2𝑛|0/𝑢𝑛0,,|𝑢𝑁𝑛|0/𝑢𝑛0), then 𝜇𝑛𝑁 and |𝜇𝑛|=1(𝑛=1,2,), then {𝜇𝑛} possesses a convergent subsequence (which denoted by 𝜇𝑛), and then there exists a vector 𝜇0=(𝜇10,𝜇20,,𝜇𝑁0)𝑁 such that ||𝜇0||=1,lim𝑛+𝜇𝑛=𝜇0.(4.27)
Without loss of generality, we assume that 𝜇10>0. Since 𝑢𝑛𝐶(𝐽,), there exist 𝜂𝑖𝑛(0,+) such that ||𝑢𝑖𝑛𝜂𝑖𝑛||11𝑛||𝑢𝑖𝑛||0,𝑖=1,2,,𝑁,𝑛=1,2,,(4.28) and, then, from (4.25), we have ||𝑢01𝑛(𝑡)𝑢1𝑛𝜂1𝑛||=||||𝑡𝜂1𝑛𝑢1𝑛||||(𝑟)𝑑𝑟𝐶5𝑢𝑛𝛼10+(𝑤(𝑡))1/(𝑝(𝑡)1)𝑑𝑡.(4.29)
Since 𝑢𝑛1+ (as 𝑛+), 𝛼(0,1), and 𝜇10>0, we have lim𝑛+1||𝑢1𝑛𝜂1𝑛||𝐶5𝑢𝑛𝛼10+(𝑤(𝑡))1/(𝑝(𝑡)1)𝑑𝑡=0.(4.30)
From (4.26)–(4.30), we have lim𝑛+𝑢1𝑛(𝑡)𝑢1𝑛𝜂1𝑛=1,for𝑡𝐽uniformly.(4.31)
So we get lim𝑛+𝑢𝑛(𝑡)𝑢𝑛1=𝜇,lim𝑛+(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢𝑛(𝑡)𝑢𝑛1=0,for𝑡𝐽uniformly,(4.32) where 𝜇𝑁 satisfies |𝜇|=1,|𝜇𝑖|=𝜇𝑖0.
We denote 𝑔1=𝑢𝑛1𝑞(𝑠)1𝑔1𝑠,𝜇𝜇+𝑜(1),𝑜(1),𝑆𝜇+𝑜(1),𝑇+𝑜(1)+𝑜(1),where𝑠𝐽.(4.33)
Since 𝜇100, from (4.19) and (4.32), we have 0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))11𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝑔1𝑑𝑠1𝑚2𝑖=1𝛼𝑖×𝑟+𝑔1𝑑𝑠+𝑒2𝑑𝑟𝑑𝑡𝜆𝑛𝑒1=0.(4.34)
Since 𝑔1(𝑠,𝜇,0,𝑆(𝜇),𝑇(𝜇))0, according to the continuity of 𝑔1, we have 0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))11𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝑔1𝑑𝑠1𝑚2𝑖=1𝛼𝑖×𝑟+𝑔1𝑑𝑠+𝑒2𝑑𝑟𝑑𝑡𝜆𝑛𝑒10,(4.35) 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).
For 𝜆[0,1], if 𝑢 is a solution of (4.18), we have Θ𝜆𝑁𝑓(𝑢,𝜆)=0.(4.36)
For 𝜆=0,𝑓=𝑔, from (4.18), we have 𝑢=Φ𝑔(𝑢,0)=𝑃1𝑢+Θ0(𝑔),(4.37) it holds 𝑢𝑑, a constant.
Therefore, when 𝜆=0, we have 0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))11𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝑔(𝑠,𝑑,0,𝑆(𝑑),𝑇(𝑑))𝑑𝑠1𝑚2𝑖=1𝛼𝑖×𝑟+𝑔(𝑠,𝑑,0,𝑆(𝑑),𝑇(𝑑))𝑑𝑠+𝑒2𝑑𝑟𝑑𝑡=0,(4.38) 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[𝐼Φ𝑓(,𝜆),𝐵(𝑅0),0]is well defined for each 𝜆[0,1], which implies that 𝑑LS𝐼Φ𝑓𝑅(,1),𝐵0,0=𝑑LS𝐼Φ𝑔𝑅(,0),𝐵0.,0(4.39)
Now it is clear that problem 𝑢=Φ𝑓(𝑢,1)(4.40) 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 𝑑LS𝐼Φ𝑔𝑅(,0),𝐵0,00.(4.41)
Since Φ𝑔(𝑢,0)=𝑃1𝑢+Θ0(𝑔),(4.42) then 𝑢Φ𝑔(𝑢,0)=𝑢𝑃1𝑢Θ0(𝑔).(4.43)
By the properties of the Leray-Schauder degree, we have 𝑑LS𝐼Φ𝑔(𝑅,0),𝐵0,0=(1)𝑁𝑑𝐵𝜔𝑔𝑅,𝑏0,,0(4.44) 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. If  𝑏𝐽×𝑁×𝑁×𝑁×𝑁𝑁 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,),𝛽(𝑡),𝑞(𝑡)𝐶(𝐽,) are positive functions, and satisfies 1<𝑞𝑞+<𝑝and𝜓(𝑠,𝑡) and 𝜒(𝑠,𝑡) are nonnegative, then (4.14) with (1.2) has at least one solution.

Proof. Since 𝑔(𝑡,𝑢,𝑣,𝑆(𝑢),𝑇(𝑢))=𝛽(𝑡)|𝑢|𝑞(𝑡)2𝑢+|𝑣|𝑞(𝑡)2||||𝑣+𝑆(𝑢)𝑞(𝑡)2||||𝑆(𝑢)+𝑇(𝑢)𝑞(𝑡)2𝑇(𝑢),(4.45) then 𝜔𝑔=(𝑎)0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))11𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝑔(𝑠,𝑎,0,𝑆(𝑎),𝑇(𝑎))𝑑𝑠1𝑚2𝑖=1𝛼𝑖×𝑟+𝑔(𝑠,𝑎,0,𝑆(𝑎),𝑇(𝑎))𝑑𝑠+𝑒2=𝑑𝑟𝑑𝑡0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))11𝑚2𝑖=1𝛼𝑖𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝛽(𝑠)|𝑎|𝑞(𝑠)2||||𝑎+𝑆(𝑎)𝑞(𝑠)2+||||𝑆(𝑎)𝑇(𝑎)𝑞(𝑠)2𝑇(𝑎)𝑑𝑠1𝑚2𝑖=1𝛼𝑖×𝑟+𝛽(𝑠)|𝑎|𝑞(𝑠)2||||𝑎+𝑆(𝑎)𝑞(𝑠)2+||||𝑆(𝑎)𝑇(𝑎)𝑞(𝑠)2𝑇(𝑎)𝑑𝑠+𝑒2𝑑𝑟𝑑𝑡,(4.46) then it is easy to say that 𝜔𝑔(𝑎)=0 has only one solution in 𝑁, and 𝑑𝐵𝜔𝑔𝑅,𝑏0,0=𝑑𝐵𝑅𝐼,𝑏0,00,(4.47) 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||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢+𝑓𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢),𝛿=0,𝑡(0,+),(4.48) where 𝑓𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢),𝛿=𝑔𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢)+𝛿𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,,𝑆(𝑢),𝑇(𝑢)(4.49) where 𝑔,𝐽×𝑁×𝑁×𝑁×𝑁𝑁 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 𝑓𝜆𝛿𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢)=𝑔𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢)+𝜆𝛿𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢.,𝑆(𝑢),𝑇(𝑢)(4.50)
Let us consider the existence of solutions of the following ||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢+𝑓𝜆𝛿𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢)=0,𝑡(0,+),(4.51) with (1.2).
We know that (4.51) with (1.2) has the same solution of 𝑢=Ψ𝛿(𝑢,𝜆)=𝑃1𝑁𝑢+Θ𝑓𝜆𝛿(𝑢)+𝐾1𝑁𝑓𝜆𝛿,(𝑢)(4.52) 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)={𝑢𝐶1𝑢1<𝑅0}. Since Ψ𝛿(𝑢,0) is compact continuous from 𝐶1 to 𝐶1, we have inf𝑅𝑢𝜕𝐵0𝑢Ψ𝛿(𝑢,0)1>0.(4.53)
Since 𝑔, are Caratheodory, we have||𝜌1𝑁𝑓𝜆𝛿(𝑢)𝜌1𝑁𝑓0(||𝑢)0,for(𝑢,𝜆)𝐵𝑅0×[]𝐹𝑁0,1uniformly,as𝛿0,𝑓𝜆𝛿𝑁(𝑢)𝐹𝑓0(𝑢)00,for(𝑢,𝜆)𝐵𝑅0×[]𝐾0,1uniformly,as𝛿0,1𝑁𝑓𝜆𝛿(𝑢)𝐾1𝑁𝑓0(𝑢)10,for(𝑢,𝜆)𝐵𝑅0×[]||𝑃0,1uniformly,as𝛿0,1𝑁𝑓𝜆𝛿(𝑢)𝑃1𝑁𝑓0||(𝑢)0,for(𝑢,𝜆)𝐵𝑅0×[]0,1uniformly,as𝛿0.(4.54)
Thus, Ψ𝛿(𝑢,𝜆)Ψ0(𝑢,𝜆)10for(𝑢,𝜆)𝐵𝑅0×[]0,1uniformly,as𝛿0.(4.55)
Obviously, Ψ0(𝑢,𝜆)=Ψ𝛿(𝑢,0)=Ψ0(𝑢,0). Therefore,Ψ𝛿(𝑢,𝜆)Ψ𝛿(𝑢,0)10for(𝑢,𝜆)𝐵𝑅0×[]0,1uniformly,as𝛿0.(4.56)
Thus, when 𝛿 is small enough, from (4.53), we can conclude inf𝑅(𝑢,𝜆)𝜕𝐵0×[]0,1𝑢Ψ𝛿(𝑢,𝜆)1inf𝑅𝑢𝜕𝐵0𝑢Ψ𝛿(𝑢,0)1sup(𝑢,𝜆)𝐵𝑅0×[]0,1Ψ𝛿(𝑢,0)Ψ𝛿(𝑢,𝜆)1>0.(4.57)
Thus 𝑢=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[𝐼Ψ𝛿(,𝜆),𝐵(𝑅0),0] is well defined for any 𝜆[0,1] and 𝑑LS𝐼Ψ𝛿𝑅(𝑢,𝜆),𝐵0,0=𝑑LS𝐼Ψ𝛿𝑅(𝑢,0),𝐵0,0.(4.58)
From the proof of Theorem 4.2, 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 𝑚2𝑖=1𝛼𝑖=1,𝜎=1.

Theorem 5.1. Assume that Ω is an open bounded set in 𝐶1 such that the following conditions hold.(10)For each 𝜆(0,1), the problem||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢=𝜆𝛿𝑓𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢),𝑡(0,+),𝑢(+)=0+𝑒(𝑡)𝑢(𝑡)𝑑𝑡+𝜆𝑒1,lim𝑡+||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢(𝑡)=𝑚2𝑖=1𝛼𝑖𝑤𝜉𝑖||𝑢||𝑝(𝜉𝑖)2𝑢𝜉𝑖+𝜆𝑒2,(5.1)has no solution on 𝜕Ω.(20)The equation𝜔(𝑎)=𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝛿𝑓(𝑡,𝑎,0,𝑆(𝑎),𝑇(𝑎))𝑑𝑡𝑒2=0(5.2)has no solution on 𝜕Ω𝑁.(30)The Brouwer degree 𝑑𝐵[𝜔,Ω𝑁,0]0.
Then, problems (1.1)-(1.2) have a solution on Ω.

Proof. Let us consider the following problem: ||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢=𝜆𝛿𝑁𝑓(𝑢)+(1𝜆)𝑄𝛿𝑁𝑓(𝑢),𝑡(0,+),𝑢(+)=0+𝑒(𝑡)𝑢(𝑡)𝑑𝑡+𝜆𝑒1,lim𝑡+||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢(𝑡)=𝑚2𝑖=1𝛼𝑖𝑤𝜉𝑖||𝑢||𝑝(𝜉𝑖)2𝑢𝜉𝑖+𝜆𝑒2,(5.3) where 𝑄 and 𝑄 are defined in (2.79) and (2.80), respectively.
For any 𝜆(0,1], observe that, if 𝑢 is a solution to (5.1) or 𝑢 is a solution to (5.3), we have necessarily 𝑄𝑁𝛿𝑓(𝑢)=0,𝑄𝑁𝛿𝑓(𝑢)=0.(5.4)
It means that (5.1) and (5.3) have the same solutions for 𝜆(0,1].
We denote 𝑁(,)𝐶1×𝐽𝐿1 defined by 𝑁(𝑢,𝜆)=𝜆𝑁𝛿𝑓(𝑢)+(1𝜆)𝑄𝑁𝛿𝑓,(𝑢)(5.5) where 𝑁𝛿𝑓(𝑢) is defined by (2.32). Let Φ𝑓(𝑢,𝜆)=𝑃2𝑢+𝑄𝜆(𝑁(𝑢,𝜆))+𝐾𝜆2(𝑁(𝑢,𝜆))=𝑃2𝑁𝑢+𝑄𝛿𝑓(𝑢)+𝐾𝜆2(𝑁(𝑢,𝜆)),(5.6) where 𝑃2𝐶1𝐶1𝑄,𝑢𝑢(0),𝜆𝐿1𝑁,𝑚2𝑖=1𝛼𝑖𝜉+𝑖(𝑡)𝑑𝑡𝜆𝑒2,𝑄𝜆𝐿1𝐿1,𝜏(𝑡)𝑚2𝑖=1𝛼𝑖𝜉+𝑖(𝑡)𝑑𝑡𝜆𝑒2,𝐾𝜆2𝜑()(𝑡)=𝐹1𝑡,(𝑤(𝑡))1𝜌𝜆2𝐼𝑄𝜆+𝐹𝐼𝑄𝜆[(𝑡),𝑡0,+),(5.7) where 𝜌𝜆2 satisfies 0+𝑒(𝑡)𝑡+𝜑1𝑟,(𝑤(𝑟))1𝜌𝜆2+𝐹()(𝑟)𝑑𝑟𝑑𝑡𝜆𝑒1=0,(5.8) 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 𝑢=Φ𝑓(𝑢,𝜆).(5.9)
Since 𝑓 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 ||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢=𝑄𝛿𝑁𝑓,(𝑢)𝑢(+)=0+𝑒(𝑡)𝑢(𝑡)𝑑𝑡,lim𝑡+||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢(𝑡)=𝑚2𝑖=1𝛼𝑖𝑤𝜉𝑖||𝑢||𝑝(𝜉𝑖)2𝑢𝜉𝑖.(5.10)
If 𝑢 is a solution to this problem, we must have 𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝑄𝛿𝑁𝑓(𝑢)𝑑𝑡=𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝛿𝑓𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢)𝑑𝑡𝑒2=0.(5.11)
Hence, ||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢𝑐,(5.12) where 𝑐𝑁 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, 𝑢0, it holds 𝑢𝑑, a constant. Thus, by (5.11), we have 𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝛿𝑓(𝑡,𝑑,0,𝑆(𝑑),𝑇(𝑑))𝑑𝑡𝑒2=0,(5.13) 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[𝐼Φ𝑓(,𝜆),Ω,0]is well defined for 𝜆[0,1], and from the properties of that degree, we have 𝑑LS𝐼Φ𝑓(,1),Ω,0=𝑑LS𝐼Φ𝑓(,0),Ω,0.(5.14)
Now it is clear that problem 𝑢=Φ𝑓(𝑢,1)(5.15) 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 𝑑LS𝐼Φ𝑓(,0),Ω,00.(5.16)
Since Φ𝑓(𝑢,0)=𝑃2𝑁𝑢+𝑄𝛿𝑓(𝑢)+𝐾02𝑄𝑁𝛿𝑓,(𝑢)(5.17) then 𝑢Φ𝑓(𝑢,0)=𝑢𝑃2𝑁𝑢𝑄𝛿𝑓(𝑢)𝐾02(0).(5.18)
Similar to Lemma 2.8, we have ||𝜌𝜆2𝑄𝑁𝛿𝑓||𝐸(𝑢)3𝑁#+1𝐸#𝑝+1𝑄𝑁𝛿𝑓(𝑢)0+||𝜆𝑒1||𝑝#1.(5.19)
Thus, 𝜌02(0)=0, then 𝐾02(0)0. From (5.18), we have𝑢Φ𝑓(𝑢,0)=𝑢𝑃2𝑁𝑢𝑄𝛿𝑓.(𝑢)(5.20)
By the properties of the Leray-Schauder degree, we have 𝑑LS𝐼Φ𝑓(,0),Ω,0=(1)𝑁𝑑𝐵𝜔,Ω𝑁,0,(5.21) 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 Theorem 5.1. As an application of Theorem 5.1. Let us consider the following equation with  (1.2)||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢=𝑔𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢)+𝑏𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,,𝑆(𝑢),𝑇(𝑢)(5.22) where 𝑏𝐽×𝑁×𝑁×𝑁×𝑁𝑁 is Caratheodory, 𝑔=(𝑔1,,𝑔𝑁)𝐽×𝑁×𝑁×𝑁×𝑁𝑁 is continuous and Caratheodory, and, for any fixed 𝑦0𝑁{0}, if 𝑦𝑖00, then 𝑔𝑖(𝑡,𝑦0,0,𝑆(𝑦0),𝑇(𝑦0))0, for all 𝑡𝐽, 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×𝑁×𝑁×𝑁×𝑁, where 𝑞(𝑡)𝐶(𝐽,) satisfies 1<𝑞𝑞+<𝑝,(20)lim|𝑥|+|𝑦|+|𝑧|+|𝑤|+(𝑏(𝑡,𝑥,𝑦,𝑧,𝑤)/(|𝑥|+|𝑦|+|𝑧|+|𝑤|)𝑞(𝑡)1)=0,  for 𝑡𝐽 uniformly, (30)for large enough 𝑅0>0, the equation 𝜔𝑔(𝑎)=𝑚2𝑖=1𝛼𝑖𝜉+𝑖𝑔(𝑡,𝑎,0,𝑆(𝑎),𝑇(𝑎))𝑑𝑡𝑒2=0,(5.23)has no solution on 𝜕𝐵(𝑅0)𝑁, where 𝐵(𝑅0)={𝑢𝐶1𝑢1<𝑅0},(40)the Brouwer degree 𝑑𝐵[𝜔𝑔,𝑏(𝑅0),0]0 for large enough 𝑅0>0, where 𝑏(𝑅0)={𝑥𝑁|𝑥|<𝑅0}.
Then, problem (5.22) with (1.2) has at least one solution.

Proof. Denote 𝑁𝑓(𝑢,𝜆)=𝑓𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢),𝜆=𝑔𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢)+𝜆𝑏𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢.,𝑆(𝑢),𝑇(𝑢)(5.24)
At first, we consider the following problem ||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢=𝑓𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢),𝜆,𝑡(0,+).(5.25)
According to Lemma 2.10, we know problem (5.25) with (1.2) has the same solution of Φ𝑢=𝑓(𝑢,𝜆)=𝑃2𝑁𝑢+𝑄𝑓(𝑢,𝜆)+𝐾2𝑁𝑓.(𝑢,𝜆)(5.26)
Similar to the proof of Theorem 4.2, 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 𝑑LSΦ𝐼𝑓𝑅(,1),𝐵0,0=𝑑LSΦ𝐼𝑓𝑅(,0),𝐵0,0.(5.27)
If we prove that 𝑑LSΦ[𝐼𝑓(,0),𝐵(𝑅0),0]0, then we obtain the existence of solutions for (5.22) with (1.2).
Now, we consider the following equation ||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢=𝜆𝑁𝑔(𝑢)+(1𝜆)𝑄𝑁𝑔(𝑢),𝑡(0,+),𝑢(+)=0+𝑒(𝑡)𝑢(𝑡)𝑑𝑡+𝜆𝑒1,lim𝑡+||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢(𝑡)=𝑚2𝑖=1𝛼𝑖𝑤𝜉𝑖||𝑢||𝑝(𝜉𝑖)2𝑢𝜉𝑖+𝜆𝑒2,(5.28) where 𝑁𝑔(𝑢)=𝑔(𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢),𝑇(𝑢)).
We denote 𝐺(,)𝐶1×𝐽𝐿1 defined by 𝐺(𝑢,𝜆)=𝜆𝑁𝑔(𝑢)+(1𝜆)𝑄𝑁𝑔.(𝑢)(5.29)
Similar to the proof of Theorem 5.1, we know that (5.28) has the same solution of 𝑢=Φ𝑔(𝑢,𝜆)=𝑃2𝑁𝑢+𝑄𝑔(𝑢)+𝐾𝜆2(𝐺(𝑢,𝜆)).(5.30)
Similar to the discussions of Theorem 4.2, 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[𝐼Φ𝑔(,𝜆),𝐵(𝑅0),0]is well defined for 𝜆[0,1], which implies that 𝑑LS𝐼Φ𝑔𝑅(,1),𝐵0,0=𝑑LS𝐼Φ𝑔𝑅(,0),𝐵0.,0(5.31)
Now it is clear that Φ𝑔(𝑢,1)=Φ𝑓(𝑢,0). So 𝑑LS[𝐼Φ𝑔(,1),𝐵(𝑅0),0]=𝑑LSΦ[𝐼𝑓(,0),𝐵(𝑅0),0]. If we prove that 𝑑LS[𝐼Φ𝑔(,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 𝑑LS𝐼Φ𝑔(𝑅,0),𝐵0,0=(1)𝑁𝑑𝐵𝜔𝑔𝑅,𝑏0.,0(5.32)
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 Corollary 4.3 and Theorem 4.4, we have the following.

Corollary 5.3. If  𝑏𝐽×𝑁×𝑁×𝑁×𝑁𝑁 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,),𝛽(𝑡),𝑞(𝑡)𝐶(𝐽,) are positive functions, and satisfies 1<𝑞𝑞+<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 Δ𝑝(𝑡)𝑢𝑒𝑡|𝑢|𝑞(𝑡)2𝑢𝑆(𝑢)(𝑡)(𝑡+1)2=0,𝑡(0,+),lim𝑡+𝑢(𝑡)=0+𝑒𝑡𝑢(𝑡)𝑑𝑡,lim𝑡+||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢(𝑡)=𝑚2𝑖=1𝛼𝑖𝑤𝜉𝑖||𝑢||𝑝(𝜉𝑖)2𝑢𝜉𝑖,(S1) where 𝑝(𝑡)=6+𝑒𝑡sin𝑡, 𝑞(𝑡)=3+2𝑡cos𝑡, 𝑆(𝑢)(𝑡)=0𝑒2𝑠𝑡(sin𝑠𝑡+1)𝑢(𝑠)𝑑𝑠.
Obviously, 𝑒𝑡|𝑢|𝑞(𝑡)2𝑢+𝑆(𝑢)(𝑡)+(𝑡+1)2  is Caratheodory, 𝑞(𝑡)4<5𝑝(𝑡), 𝑚2𝑖=1𝛼𝑖<1, the conditions of Corollary 4.3 are satisfied, then (S1) has a solution.

Example 6.2. Consider the following problem: Δ𝑝(𝑡)𝑢+𝑓𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢)+𝛿𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢)+𝑒𝑡=0,𝑡(0,+),lim𝑡+𝑢(𝑡)=0+𝑒2𝑡𝑢(𝑡)𝑑𝑡,lim𝑡+||𝑢𝑤(𝑡)||𝑝(𝑡)2𝑢(𝑡)=𝑚2𝑖=1𝛼𝑖𝑤𝜉𝑖||𝑢||𝑝(𝜉𝑖)2𝑢𝜉𝑖,(S2) where is Caratheodory and 𝑓𝑡,𝑢,(𝑤(𝑡))1/(𝑝(𝑡)1)𝑢,𝑆(𝑢)=𝑒𝑡|𝑢|𝑞(𝑡)2𝑢+𝑒𝑡𝑤(𝑡)(𝑞(𝑡)1)/(𝑝(𝑡)1)||𝑢||𝑞(𝑡)2𝑢𝑝+𝑆(𝑢)(𝑡),(𝑡)=7+3𝑡cos3𝑡,𝑞(𝑡)=4+𝑒2𝑡sin2𝑡,𝑆(𝑢)(𝑡)=0𝑒𝑠2𝑡(cos𝑠𝑡+1)𝑢(𝑠)𝑑𝑠.(6.1)
Obviously, 𝑒𝑡|𝑢|𝑞(𝑡)2𝑢+𝑒𝑡𝑤(𝑡)(𝑞(𝑡)1)/(𝑝(𝑡)1)|𝑢|𝑞(𝑡)2𝑢+𝑆(𝑢)(𝑡) is Caratheodory, 𝑞(𝑡)5<6𝑝(𝑡), 𝑚2𝑖=1𝛼𝑖<1, the conditions of Theorem 3.2 are satisfied, then (S2) has a solution when 𝛿 is small enough.

Acknowledgments

The authors would like to appreciate the referees for their helpful comments and suggestions. Partly supported by the National Science Foundation of China (10701066 & 10926075 & 10971087) and China Postdoctoral Science Foundation funded project (20090460969) and the Natural Science Foundation of Henan Education Committee (2008-755-65).