AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 613672 10.1155/2013/613672 613672 Research Article Triple Positive Solutions of a Nonlocal Boundary Value Problem for Singular Differential Equations with p-Laplacian Wang Jufang Yu Changlong Guo Yanping Ahmad Bashir College of Sciences Hebei University of Science and Technology Shijiazhuang, Hebei 050018 China 2013 13 3 2013 2013 22 10 2012 13 01 2013 26 01 2013 2013 Copyright © 2013 Jufang Wang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We establish the existence of triple positive solutions of an m-point boundary value problem for the nonlinear singular second-order differential equations of mixed type with a p-Laplacian operator by Leggett-William fixed point theorem. At last, we give an example to demonstrate the use of the main result of this paper. The conclusions in this paper essentially extend and improve the known results.

1. Introduction

The existence and multiplicity of positive solutions for differential equations boundary value problems (BVPs) with the p-Laplacian operator subject to Dirichlet, Sturm-Liouville, or nonlinear boundary value conditions have been extensively investigated in recent years; see  and the references therein. Particularly, the following differential equations with one-dimensional p-Laplacian (1)(ϕp(u))+q(t)f(t,u)=0,0t1 have been studied subject to different kinds of boundary conditions; see  and the references therein. The methods mainly depend on Kransnosel’skii fixed point theorem, upper and lower solution technique, Leggett-Williams fixed point theorem, and some new fixed point theorems in cones, and so forth.

Recently, in , Kong et al. have studied the existence of triple positive solutions for the following BVP: (2)(ϕp(u))(t)+q(t)f×(u(t),u(t),(Tu)(t),(Su)(t))=0,0t1,u(0)=0,u(1)=g(u(1)).

More recently, in , Hu and Ma have pointed out that the equivalent integral equation of BVP (2) is wrong in  and studied the existence of triple positive solutions for the following BVP: (3)(ϕp(u))(t)+q(t)f×(u(t),u(t),(Tu)(t),(Su)(t))=0,0t1,u(0)=βu(η),u(1)=g(u(1)).

Firstly, we confirm that the mistakes which have been pointed out in  exist. At the same time, we think that the value of M designed in Theorem  3.1 in  is not suitable, since the proof needs the condition ϕq(M)M, but in fact this condition does not always hold.

Motivated by the work above, in this paper, we will study the following more extensive second-order m-point BVP: (4)(ϕp(u))(t)+q(t)f×(u(t),u(t),(Tu)(t),(Su)(t))=0,0t1,ϕp(u(0))=i=1m-2βiϕp(u(ηi)),u(1)=g(u(1)), where ϕp(s)=|s|p-2s(p>1) is an increasing function, (ϕp)-1(s)=ϕq(s), 1/p+1/q=1;   0βi<1, i=1,2,,m-2, i=1m-2βi<1, and 0<η1<η2<<ηm-2; T and S are two linear operators defined by (5)Tu(t)=0tk(t,s)u(s)ds,Su(t)=01h(t,s)u(s)ds,uC1[0,1], in which kC[D,R+], hC[D0,R+], D={(t,s)R2:0st1}, D0={(t,s)R2:0s,t1}, R+=[0,+), R=(-,+), k0=max{k(t,s):(t,s)D}, and h0=max{h(t,s):(t,s)D0}.

Obviously, when m=3, BVP (4) reduces to BVP (3), and when m=3, β1=0, BVP (4) reduces to BVP (2), so BVP (2) and BVP (3) are special cases of BVPs (4).

Throughout this paper, we always suppose the following conditions hold:

fC(R+×R×R+×R+,(0,+));

q(t)C([0,1],R+) may be singular at t=0,1 and 0<01q(t)dt<+, so it is easy to see that there exists a constant M>0 such that 0<01q(t)dt<ϕp(M);

g:RR+ is nonincreasing and continuous, and 0g(v)|v| for vR.

2. Preliminary Results

In this section, we firstly present some definitions, theorems, and lemmas, which will be needed in the proof of the main result.

Definition 1.

Let E be a real Banach space. A nonempty closed convex set PE is called a cone if it satisfies the following two conditions:

xP, λ0 implies λxP;

xP, -xP implies x=0.

Definition 2.

Given a cone P in a real Banach space E, a continuous map α is called a concave (resp., convex) functional on P if and only if, for all x,yP and 0t1, it holds:

α(tx+(1-t)y)tα(x)+(1-t)α(y),

(resp., α(tx+(1-t)y)tα(x)+(1-t)α(y)).

We consider the Banach space E=C1[0,1] equipped with norm u=max0t1{u(t)0,u(t)0}, where u0=max0t1|u(t)|.

We denote, for any fixed constants a,b,r,

C+[0,1]={uC[0,1]:u(t)0,t[0,1]},

P = {uEu(t)  is  concave  and  nonincreasing  on  [0,1]},

Pr={uP:u<r},

P(α,a,b)={uP:aα(u),u<b}.

It’s easy to see that P is a cone in E.

Theorem 3 (Leggett-William).

Let A:P¯cP¯c be a completely continuous map and let α be a nonnegative continuous concave functional on P with α(u)u for any uP¯c. Suppose there exist constants a,b,  and  d with 0<a<b<dc such that

{uP(α,b,d):α(u)>b}ϕ and α(Au)>b for all uP(α,b,d);

Au<a for all uP¯a;

α(Au)>b for all uP(α,b,c) with Au>d.

Then A has at least three fixed points u1, u2, and u3 satisfying (6)u1<a,b<α(u2),u3>a,α(u3)<b.

Lemma 4.

Suppose yC1[0,1] with (ϕp(y))L1[0,1] satisfies (7)(ϕp(y))(t)0,0t1,ϕp(y(0))=i=1m-2βiϕp(y(ηi)),y(1)=g(y(1)). Then, y(t)0 is concave and nonincreasing on [0,1], that is, yP.

Proof.

Since (ϕp(y))(t)0, we know that ϕp(y) is nonincreasing, that is, y(t) is nonincreasing, which means y(t) is concave. At the same time, we have y(t)y(0), so y(0)=ϕq(i=1m-2βiϕp(y(ηi)))ϕq(i=1m-2βiϕp(y(0)))=ϕq(i=1m-2βi)y(0), namely y(0)0. Then, y(t)0; that is to say, y(t) is nonincreasing. So y(t)y(1)=g(y(1))0. Above all, yP. This completes the proof.

Lemma 5.

Let yC[0,1] and (ϕp(u))L1[0,1], then, BVP (8)(ϕp(u))(t)=-y(t),0t1,ϕp(u(0))=i=1m-2βiϕp(u(ηi)),u(1)=g(u(1)), has a unique solution (9)u(t)=t1ϕq(11-βi=1m-2βi0ηiy(r)dr+0sy(r)dr)ds+g(-ϕq(11-βi=1m-2βi0ηiy(r)dr+01y(r)dr)), where β=i=1m-2βi.

Define the operator A:PE by (10)(Au)(t)=t1ϕq(11-βi=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr+0sq(r)f(u,u,Tu,Su)dr11-βi=1m-2)ds+g(-ϕq(11-βi=1m-2βi0ηiq(r)×f(u,u,Tu,Su)dr+01q(r)f(u,u,Tu,Su)dr11-βi=1m-2)). Obviously, A is well defined and uE is a solution of BVP (4) if and only if u is a fixed point of A.

Lemma 6.

A : P P is completely continuous.

Proof.

It is similar to the proof of Lemma  2.2 in .

Lemma 7.

For any uP, one has Au02(Au)0, Au2(Au)0.

Proof.

From (10), we obtain (11)(Au)0=01ϕq(11-βi=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr+0sq(r)f(u,u,Tu,Su)dr11-βi=1m-2)ds+g(-ϕq(11-βi=1m-2βi0ηiq(r)×f(u,u,Tu,Su)dr+01q(r)f(u,u,Tu,Su)dr11-βi=1m-2))01ϕq(11-βi=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr+0sq(r)f(u,u,Tu,Su)dr11-βi=1m-2)ds+ϕq(11-βi=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr+01q(r)f(u,u,Tu,Su)dr11-βi=1m-2)=2ϕq(11-βi=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr+01q(r)f(u,u,Tu,Su)dr11-βi=1m-2)=2|(Au)(1)|=2(Au)0.

Since Au=max{Au0,(Au)0}, so we have Au2(Au)0, which completes the proof.

3. Main Results

For any δ(0,min{η1,1/2}), we define a nonnegative continuous concave function α:PR+ by α(u)=minδt(1-δ)u(t). Obviously, the following two conclusions hold: (12)α(u)=u(1-δ)u0,α(Au)=Au(1-δ),uP.

The main result of this paper is following.

Theorem 8.

Let m0=min0t1q(t) and β=i=1m-2βi<1. Suppose (C1), (C2), and (C3) hold. Suppose further that there exist numbers δ(0,min{η1,1/2}),a,b,c, and d such that 0<a<bm0δd/M<dc, and

f(u,v,w,l)(1-β)ϕp(a/2M),   for (u,v,w,l)[0,a] × [-a,0] × [0,k0a]×[0,h0a];

f(u,v,w,l)(1-β)ϕp(c/2M),   for (u,v,w,l)[0,c] × [-c,0] × [0,k0c]×[0,h0c];

f(u,v,w,l)>ϕp(b/δL), for (u,v,w,l)[b,d] × [-d,0]×[0,k0d] × [0,h0d], where L=ϕq(01-δq(t)dt);

min(u,v,w,l)Jf(u,v,w,l)ϕp(M/(2m0))01-δq(t)dtmax(u,v,w,l)Jf(u,v,w,l)01q(t)dt, where J=[0,c]×[-c,0] × [0,k0c]×[0,h0c].

Then, BVP (4) has at least three positive solutions u1,u2,  and u3 such that (13)u1<a,b<minδt<(1-δ)u2(t),u3>a,minδt<(1-δ)u3(t)<b.

Proof.

We divide the proof into three steps.

Step  1. We prove AP¯cP¯c, AP¯aP¯a; that is, (ii) of Theorem 3. By Lemma 6, we have AP¯cP, so uP¯c, we get 0u(t)c, -cu(t)0, 0(Tu)(t)k0c, 0(Su)(t)h0c. For t[0,1] and by (H2)(14)(Au)0=01ϕq(11-βi=1m-2βi0ηiq(r)×f(u,u,Tu,Su)dr+0sq(r)f(u,u,Tu,Su)dr11-βi=1m-2)ds+g(-ϕq(11-βi=1m-2βi0ηiq(r)×f(u,u,Tu,Su)dr+01q(r)f(u,u,Tu,Su)dr11-βi=1m-2))01ϕq(11-βi=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr+0sq(r)f(u,u,Tu,Su)dr11-βi=1m-2)ds+ϕq(11-βi=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr+01q(r)f(u,u,Tu,Su)dr11-βi=1m-2)=2ϕq(11-βi=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr+01q(r)f(u,u,Tu,Su)dr11-βi=1m-2)2ϕq(11-β01q(r)f(u,u,Tu,Su)dr)2ϕq(ϕp(c2M))ϕq(01q(r)dr)c.(Au)0=ϕq(11-βi=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr+0sq(r)f(u,u,Tu,Su)dr11-βi=1m-2)ϕq(11-βi=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr+01q(r)f(u,u,Tu,Su)dr11-βi=1m-2)ϕq(11-β01q(r)f(u,u,Tu,Su)dr)ϕq(ϕp(c2M))ϕq(01q(r)dr)c. Hence, Au<c and AP¯cP¯c. Similarly, we obtain AP¯aP¯a.

Step  2. We show (15){uP(α,b,d):α(u)>b}ϕ,(16)α(Au)>b,uP(α,b,d), that is, (i) of Theorem 3.

Let u=(b+d)/2, then uP(α,b,d),α(u)=(b+d)/2>b. Hence, (15) holds. For any uP(α,b,d), we have bu(t)d,-du(t)0,0(Tu)(t)k0d,0(Su)(t)h0d,t[0,1-δ], so by (H3), we have (17)α(Au)=mint[δ,1-δ](Au)(t)=(Au)(1-δ)=1-δ1ϕq(11-βi=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr+0sq(r)f(u,u,Tu,Su)dr11-βi=1m-2)ds+g(-ϕq(11-βi=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr+01q(r)f(u,u,Tu,Su)dr11-βi=1m-2))δϕq(11-βi=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr+01-δq(r)f(u,u,Tu,Su)dr11-βi=1m-2)+g(-ϕq(11-βi=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr+01q(r)f(u,u,Tu,Su)dr11-βi=1m-2))δϕq(01-δq(r)f(u,u,Tu,Su)dr)δϕq(ϕp(bδL))ϕq(01-δq(r)dr)=b. Hence (16) holds.

Step  3. We show that α(Au)>b for all uP(α,b,c) with Au>d, that is, (iii) of Theorem 3.

If uP(α,b,c) with Au>d, we obtain 0u(t)c,  -cu(t)0,  0(Tu)(t)k0c,  0(Su)(t)h0c, for any t[0,1], and so by (H4), we have (18)ϕp(M2m0)01-δq(r)f(u,u,Tu,Su)dr01q(r)f(u,u,Tu,Su)dr. Furthermore, we have (19)ϕp(M2m0)01-δq(r)f(u,u,Tu,Su)dr+ϕp(M2m0)11-β×i=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr01q(r)f(u,u,Tu,Su)dr+11-βi=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr. Therefore, by Lemma 7, we have (20)α(Au)=(Au)(1-δ)=1-δ1ϕq(11-βi=1m-2βi0ηiq(r)×f(u,u,Tu,Su)dr+0sq(r)f(u,u,Tu,Su)dr11-βi=1m-2)ds+g(-ϕq(11-βi=1m-2βi0ηiq(r)×f(u,u,Tu,Su)dr+01q(r)f(u,u,Tu,Su)dr11-βi=1m-2))δϕq(11-βi=1m-2βi0ηiq(r)f(u,u,Tu,Su)dr+01-δq(r)f(u,u,Tu,Su)dr11-βi=1m-2)δϕq((11-βi=1m-2βi0ηiq(r)×f(u,u,Tu,Su)dr+01q(r)f(u,u,Tu,Su)dr11-βi=1m-2βi0ηiq(r))×(ϕp(M2m0))-1(11-βi=1m-2βi0ηiq(r))=2δm0Mϕq(11-βi=1m-2βi0ηiq(r)×f(u,u,Tu,Su)dr+01q(r)f(u,u,Tu,Su)dr11-βi=1m-2)=2δm0M|(Au)(1)|=2δm0M(Au)0δm0M(Au)>δm0Mdb. Hence, by Theorem 3, the results of Theorem 8 hold. This completes the proof of Theorem 8.

4. Example

Consider the following BVP: (21)(|u|u)(t)+q(t)×f(u(t),u(t),(Tu)(t),(Su)(t))=0,f0t1,|u(0)|u(0)=14|u(25)|u(25)+14|u(12)|u(12),u(1)=g(u(1)), where (22)q(t)={t-1/2,0t1625,-18754t+93592700,1625t1.f(u,v,w,l)={10u10+2+sinv1000+w31000+l1000,0u1,v0,w,l0,10u10+2+sinv1000+w31000+l1000,1u,v0,w,l0.g(v)={v1/2,|v|1,v2,|v|<1.

Proof.

Since M=2 and m0=1/300,  β=1/2,  δ=9/25,  a=1/2,  b=1,  d=1296,  c=1600,  k(t,s)=1 and h(t,s)=1, then we can obtain 0<a<b(m0δd)/M<dc, and (23)L=01-9/25q(t)dt=016/25t-1/2dt=85,ϕp(a2M)=ϕp(1/222)=132,ϕp(c2M)=ϕp(160022)=320000,ϕp(bδL)=ϕp(1(9/25)8/5)=2529258=3125648. Next, we show that (H1)(H4) are satisfied.

If 0u1/2,-1/2v0,0w,l1/2, then (24)f(u,v,w,l)<101210+31000+21000<(1-β)ϕp(a2M)=164. So (H1) is satisfied.

If 0u1600,-1600v0,0w,l1600, then (25)f(u,v,w,l)<10160010+31000+40×21000<(1-β)ϕp(c2M)=32000. So (H2) is satisfied.

If 1u1296,-1296v0,0w,l1296, then (26)f(u,v,w,l)>10110+11000>ϕp(bδL)=3125648. So (H3) is satisfied.

For any (u,v,w,l)[0,1600]×[-1600,0]×[0,1600]×[0,1600], we have (27)minf(u,v,w,l)11000,maxf(u,v,w,l)10160010+31000+40  ×  21000,ϕp(M2m0)=2·1502,01-δq(t)dt=85,01q(t)dt=1141625. Hence, it’s easy to know that (H4) is satisfied.

So by Theorem 8, we conclude that the BVP (21) has three positive solutions u1, u2, and u3 satisfying (28)u1<12,1<minδt<(1-δ)u2(t),u3>12minδt<(1-δ)u3(t)<1.

Acknowledgments

This paper is supported by the Natural Science Foundation of China (10901045) and (11201112), the Natural Science Foundation of Hebei Province (A2009000664) and (A2011208012) and the Foundation of Hebei University of Science and Technology (XL200757).

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