# Three-Order Multipoint Boundary Value Problems forp-Laplacian Operator on Time Scales

Authors: Hua Su
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, the existence of positive solutions for the nonlinear four-point singular BVP for there-order withp-Laplacian operator on time scales will be studied. By using the fixed-point theory, the existence of positive solutions for nonlinear singular boundary value problem with p-Laplacian operator on time scales is obtained.


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## Body

## 1. Introduction

In recent years, the nonlinear boundary value problems have been extensively studied. Recently, for the existence of positive solutions of multipoint boundary value problems, some authors have obtained the existence results. The differential equations offer wonderful tools for describing various natural phenomena arising from natural sciences and engineering, many numerical and analytical results, for example [1–20]. However, the multipoint boundary value problems treated in the above mentioned references do not discuss the problems with singularities and the there-order p-Laplacian operator. For the singular case of multipoint boundary value problems for higher-order p-Laplacian operator, with the author’s acknowledge, no one has studied the existence of positive solutions in this case.In this paper, we study the following equation withp-Laplacian on time scale:
(1)ϕpuΔΔ∇+gtfut,uΔt=0,0<t<T,with the following boundary value conditions:
(2)u0=0,uΔ0−M0uΔΔξ=0,uΔT+M1uΔΔη=0,where ϕps=sp−2s,p>1, ϕq=ϕp−1, 1/p+1/q=1. ξ,η∈0,T is prescribed and ξ<η, g:0,T→0,∞, M0,M1 are both nondecreasing continuous odd functions defined on −∞,+∞.A time scaleT is a nonempty subset and closed subset of R. By an internal 0,T, we always mean the intersection of the real internal 0,T with the given time scale, that is 0,T⋂T. The operators σ and ρ from T to T which defined by [21],
(3)σt=infτ∈T∣τ>t∈T,ρt=supτ∈T∣τ<t∈T.are called the forward jump operator and the backward jump operator, respectively.The pointt∈T is left-dense, left-scattered, right-dense, right-scattered if ρt=t,ρt<t,σt=t,σt>t, respectively. If T has a right scattered minimum m, define Tk=T−m; otherwise set Tk=T. If T has a left scattered maximum M, define Tk=T−M; otherwise set Tk=T.In this paper, by constructing an integral equation which is equivalent to the problem (1), (2), we research the existence of positive solutions for nonlinear singular boundary value problem (1), (2) when g and f satisfy some suitable conditions.


## 2. Preliminaries and Lemmas

For convenience, in the rest of this article,T is a closed subset of R with 0∈Tk,T∈Tk.Letting(4)B=u∈Cld0,T:u0=0.Then,B is a Banach space with the norm u=maxt∈0,T∣uΔt∣. Suppose
(5)K=u∈B:uΔt≥0,uΔtisconcavefunctionont∈0,T.Obviously,K is a cone in B and 0≤uΔt≤u on 0,T. Set Kr=u∈K:u≤r.In the rest of the paper, we make the following assumptions:(H1) f∈C0,+∞2,0,+∞;(H2) g∈Cld0,T,0,+∞ and there exists t0∈0,T which satisfy
(6)gt0>0,0<∫0Tgt∇t<∞;(H3) M0,M1∈−∞,+∞,R are both increasing, continuous, odd functions, and at least one of them satisfies the condition that there exists one b>0 which satisfy
(7)0<Miv≤bv,∀v≥0,i=0or1.By direct account, From paper [22], we can easy to obtain the following results.Lemma 1.
Suppose conditionH2 holds. Then, there exists a constant L∈0,T/2 satisfies
(8)0<∫LT−Lgt∇t<∞.
Furthermore, the function(9)At=∫Ltϕq∫stgs1∇s1Δs+∫tT−Lϕq∫tsgs1∇s1Δs,t∈L,T−L,is positive ld-continuous function on L,T−L, therefore, At has minimum on L,T−L. Hence, we suppose that there exists constant B′>0 which satisfy A≥B′ on t∈L,T−L.Lemma 2.
Suppose that conditionsH1,H2,H3 hold, ut∈K∩C20,T is a solution of boundary value problems (1), (2) if and only if ut∈B is a solution of the following integral equation
(10)ut=∫0twsΔs,where
(11)wt=M0ϕq∫ξδgsfus,uΔsΔs+∫0tϕq∫sδgrfur,uΔrΔr∇s,0≤t≤δ,M1ϕq∫δηgsfus,uΔsΔs+∫tTϕq∫δsgrfur,uΔrΔr∇s,δ≤t≤T.Here,δ is a unique solution of the equation g1t=g2t, where
(12)g1t=M0ϕq∫ξδgsfus,uΔsΔs+∫0tϕq∫sδgrfur,uΔrΔr∇s,g2t=M1ϕq∫δηgsfus,uΔsΔs+∫tTϕq∫δsgrfur,uΔrΔr∇s.Equationg1t=g2t has a unique solution in 0,T. Because g1t is strictly monotone increasing on 0,T, and g10=0,g2t is strictly monotone decreasing on 0,T, and g2T=0.Proof.
Necessity. By the equation of the boundary condition and H3, we have
(13)uΔΔξ≥0,uΔΔη≤0.
Then, there exist a constantδ∈ξ,η⊂0,T which satisfy uΔΔδ=0. Firstly, by integrating the equation of the problems (1) on δ,t, we have
(14)ϕpuΔΔt=ϕpuΔΔδ−∫δtgsfus,uΔsΔs,then
(15)uΔΔt=−ϕq∫δtgsfus,uΔsΔs,(16)uΔt=uΔδ−∫δtϕq∫δsgrfur,uΔrΔr∇s.
ByuΔΔδ=0 and condition (2), let t=η on (15), we have
(17)uΔΔ1=−M1uΔΔη=M1ϕq∫δηgsfus,uΔsΔs.
Then, we have(18)uΔδ=M1ϕq∫δηgsfus,uΔsΔs+∫δTϕq∫δsgrfur,uΔrΔr∇s.
Then(19)uΔΔt=M1ϕq∫δηgsfus,uΔsΔs+∫tTϕq∫δsgrfur,uΔrΔr∇s.
Therefore, by integrating the above equation (19) on 0,t, we can east to have
(20)ut=∫0tM1ϕq∫δηgsfus,uΔsΔs∇t+∫0t∫s1Tϕq∫δsgrfur,uΔrΔr∇s∇s1.
Similarly, fort∈0,δ, by integrating the equation of problems (1) on 0,δ, we have
(21)ut=∫0tM0ϕq∫ξδgsfus,uΔsΔs∇t+∫0t∫s1Tϕq∫sδgrfur,uΔrΔr∇s∇s1.
Therefore, for anyt∈0,T, ut can be expressed as equation ut=∫0twsΔs, where wt is expressed as Lemma 3.
Sufficiency. Suppose ut=∫0twsΔs. Then we have
(22)uΔΔt=ϕq∫tδgsfus,uΔsΔs≥0,0≤t<δ,−ϕq∫δtgsfus,uΔsΔs≤0,δ<t≤T,
So,ϕpuΔΔ∇+gtfut,uΔt=0,0<t<T,t≠δ. These imply that the equation (1) holds. Furthermore, we can easily obtain the boundary value equations of (2). This completes the proof of Lemma 3.
Now, we define an operatorT:K→C20,T given by
(23)Tut=∫0twsΔs,where wt is given by (15). And we can easily obtain the following Lemma.Lemma 3.
Letu∈K and L in Lemma 1. Then
(24)uΔt≥Lu,t∈L,T−L.Lemma 4.
Suppose that conditionsH1,H2 hold, then for L∈0,T/2 in Lemma 1, we have
(25)ut≤1LuΔt,t∈L,T−L.Proof.
Ifut is the solution of problem (1), (2), then uΔt is a concave function, and ut≥0,uΔt≥0,t∈0,T.
Thus fort∈L,T−L, we have uΔt≥LuΔ. Then by ut=∫0tuΔsΔs≤uΔ, we have
(26)ut≤1LuΔt,t∈L,1−L.
The proof is complete.Remark.
: Obviously, we can obtain the following results,(27)w0−M0w′ξ=0,w1+M1w′η=0.
Furthermore, by Arzela-Ascoli Theorem, it is easy to obtain the following Lemma.Lemma 5.
T:K→K is completely continuous.
For convenience, we set(28)R∗=2B′−1,R∗=b+1ϕq∫0TgrΔr−1.where B′ and L are given as Lemma 1.


## 3. The Existence of Single and Many Positive Solution

In this section, we present the following five main results.Theorem 6.
Suppose that condition (H1), (H2), (H3) hold. Assume that f also satisfy.
(A1) For Lr≤u2≤r,0≤u1≤1/Lu2, we have fu1,u2≥mrp−1;
(A2) For 0≤u2≤R,0≤u1≤1/Lu2, we have fu1,u2≤MRp−1,where m∈R∗,∞,M∈0,R∗. Then, the boundary value problem (1), (2) has at last one solution u such that u lies between r and R.The proof of Theorem6. From Condition H3, for ∀v≥0, we can suppose that 0<M0v≤bv, and r<R. By Lemma 3, for any u∈K, we can obtain that
(29)uΔt≥Lu,t∈L,T−L.We define the following two open subsetΩ1 and Ω2 of E:
(30)Ω1=u∈K:u<r,Ω2=u∈K:u<R.For anyu∈∂Ω1, by (29), we have
(31)r=u≥uΔt≥Lu=Lr,t∈L,T−L.Fort∈L,T−L and u∈∂Ω1, we shall discuss it from three perspectives.
(i)
Ifδ∈L,T−L, thus for u∈∂Ω1, by (A1) and Lemma 3, we have(32)2TuΔδ≥∫0δϕq∫sδgrfur,uΔr∇rΔs+∫δTϕq∫δsgrfur,uΔr∇rΔs≥∫Lδϕq∫sδgrfur,uΔr∇rΔs+∫δT−Lϕq∫δsgrfur,uΔr∇rΔs≥mrAδ≥mrB′>2r=2u.Then, with the case ofδ∈L,T−L and u∈∂Ω1, we have Tu≥u.(ii)
Ifδ∈T−L,T, thus for u∈∂Ω1, by (A1) and Lemma 3, we have(33)TuΔδ≥M0ϕq∫ξδgrfur,uΔr∇r+∫0δϕq∫sδgrfur,uΔr∇rΔs≥∫LT−Lϕq∫sT−Lgrfur,uΔr∇rΔs≥mrA1−L≥mrB′>2r>r=u.Then, with the case ofδ∈T−L,T and u∈∂Ω1, we have Tu≥u.(iii)
Ifδ∈0,L, thus for u∈∂Ω1, by (A1) and Lemma 3, we have(34)TuΔδ≥M1ϕq∫δηgrfur,uΔr∇r+∫δ1ϕq∫δsgrfur,uΔr∇rΔs≥∫LT−Lϕq∫Lsgrfur,uΔr∇rΔs≥mrAL≥mrB′>2r>r=u.Then, with the case ofδ∈0,L and u∈∂Ω1, we have Tu≥u.Therefore, for any case ofδ∈0,T−L, we all easy to obtain that
(35)Tu>u,∀u∈∂Ω1.Then, by fixed point theorem of cone expansion-compression type in [23, 24], we have
(36)iT,Ω1,K=0.Secondly, foru∈∂Ω2, using uΔt≤u=R, from (A2), we can easily know that
(37)TuΔδ≤M0ϕq∫0Tgrfur,uΔrΔr+∫0Tϕq∫sδgrfur,uΔr∇rΔs≤bMRϕq∫0TgrΔr+MRϕq∫0TgrΔr=b+1MRϕq∫0TgrΔr≤R=u.Thus, we have(38)Tu<u,∀u∈∂Ω2.Then, by fixed point theorem of cone expansion-compression type in [23, 24], we have
(39)iT,Ω2,K=1.Therefore, by (36), (39), r<R we have
(40)iT,Ω2Ω¯1,K=1.Then, operatorT has at last one fixed point u∈Ω2\Ω¯1, and r≤u≤R. This completes the proof of Theorem 6.Theorem 7.
Suppose that condition (H1), (H2), (H3) hold. Assume that f also satisfy.
(A3) f0=limu2→0max0≤u1≤1/Lu2fu1,u2/u2p−1=φ∈0,R∗/4p−1;
(A4) f∞=limu2→∞min0≤u1≤1/Lu2fu1,u2/u2p−1=ψ∈2R∗/Lp−1,∞.Then, the boundary value problem (1), (2) has at last one solution u.The proof of Theorem7.First, by0≤limu2→0max0≤u1≤1/Lu2fu1,u2/u2p−1=φ<R∗/4p−1, letting ε=R∗/4p−1−φ>0, we know that there exists an appropriately small positive number ρ which satisfy as 0≤u2≤ρ,u2≠0, and we have
(41)fu1,u2≤φ+εu2p−1≤R∗/4p−1ρp−1=R∗/4ρp−1.Then, lettingR=ρ,0<M=R∗/4<R∗, thus by the above equation, we can have
(42)fu1,u2≤MRp−1,0≤u2≤R,0≤u1≤1Lu2.So condition (A2) in Theorem 6 holds.Next, by2R∗/Lp−1<limu2→∞min0≤u1≤1/Lu2fu1,u2/u2p−1=ψ<∞, letting ε=ψ−2R∗/Lp−1, we know that there exists an adequately big positive number r≠R which satisfy as u2≥Lr,0≤u1≤1/Lu2, and we have
(43)fu1,u2≥ψ−εu2p−1≥2R∗Lp−1Lrp−1=2R∗rp−1,Lettingm=2R∗>R∗, thus by the above equation, we have that (A1) in Theorem 6 holds. Therefore, by Theorem 6, we can easily obtain the results of Theorem 7 holds. The proof of Theorem 3.2 is complete.Corollary 8.
Suppose that condition (H1), (H2), (H3) hold. Assume that f also satisfy.
(A5) f∞=limu2→∞max0≤u1≤1/Lu2fu1,u2/u2p−1=λ∈0,R∗/4p−1;
(A6) f0=limu2→0min0≤u1≤1/Lu2fu1,u2/u2p−1=φ∈2R∗/Lp−1,∞.
Then, the boundary value problem (1), (2) has at last one solution u.The proof of Corollary8. Similar to the proof of Theorem 7, we can obtain Corollary 8.Theorem 9.
Suppose that conditions (H1), (H2), (H3), and A2 in Theorem 6 hold. Assume that f also satisfy.
(A7) f0=limu2→0min0≤u1≤1/Lu2fu1,u2/u2p−1=+∞;
(A8) f∞=limu2→∞min0≤u1≤1/Lu2fu1,u2/u2p−1=+∞.
Then, the boundary value problem (1), (2) have at least two solutions u1,u2 which satisfy
(44)0<u1<R<u2.
The proof of Theorem9.Firstly, bylimu2→0min0≤u1≤1/Lu2fu1,u2/u2p−1=+∞, for any M>2/B′L, there exists a constant ρ∗∈0,R which satisfy
(45)fu1,u2≥Mu2p−1,0<u2≤ρ∗,0≤u1≤1Lu2.SetΩρ∗=u∈K:u<ρ∗, similar to the previous proof of Theorem 6, for any u∈∂Ωρ∗, from the above discussion and Lemma 2, we can have from three perspectives
(46)Tu≥u,∀u∈∂Ωρ∗.Then, by fixed point theorem of cone expansion-compression type, we can have(47)iT,Ωρ∗,K=0.Secondly, for anyM¯>2/B′L, by limu2→∞min0≤u1≤1/Lu2fu1,u2/u2p−1=+∞, there exists a constant ρ0>0 which satisfy
(48)fu1,u2≥M¯u2p−1,u2>ρ0,0≤u1≤1Lu2.Therefore, we choose a constantρ∗>maxR,ρ0/L, obviously ρ∗<R<ρ∗. Set Ωρ∗=u∈K:u<ρ∗. For any u∈∂Ωρ∗, by Lemma 2, we can easily obtain
(49)uΔt≥Lu=Lρ∗>ρ0,t∈L,T−L.Then, by the above discussion and also similar to the previous proof of Theorem6, we can also have from three perspectives
(50)Tu≥u,∀u∈∂Ωρ∗.Then, by fixed point theorem of cone expansion-compression type, we have(51)iT,Ωρ∗,K=0.Finally, imitating the latter proof of Theorem6, for any u∈∂ΩR, by A2, setting ΩR=u∈K:u<R, we can also easy to have
(52)Tu≤u,∀u∈∂ΩR.Then, by fixed point theorem of cone expansion-compression type, we have(53)iT,ΩR,K=1.Therefore, by (47), (51), (53), ρ∗<R<ρ∗ we have
(54)iT,ΩR\Ω¯ρ∗,K=1,iT,Ωρ∗\Ω¯R,K=−1.Then,T have fixed point u1∈ΩR\Ω¯ρ∗, and fixed point u2∈Ωρ∗\Ω¯R.Obviously,u1,u2 are all positive solutions of problem (1), (2) and 0<u1<R<u2. The proof of Theorem 9 is complete.Theorem 10.
Suppose that conditions (H1), (H2), (H3) and A1 in Theorem 6 hold. Assume that f also satisfy.
(A9) f0=limu2→0max0≤u1≤1/Lu2fu1,u2/u2p−1=0;
(A10) f∞=limu2→∞max0≤u1≤1/Lu2fu1,u2/u2p−1=0.
Then, the boundary value problem (1), (2) have at least two solutions u1,u2 which satisfy 0<u1<r<u2.
The proof of Theorem10.Firstly, bylimu2→∞max0≤u1≤1/Lu2fu1,u2/u2p−1=0, for η1∈0,R∗, there exists a constant ρ∗∈0,r which satisfy
(55)fu1,u2≤η1u2p−1,0<u2≤ρ∗,0≤u1≤1Lu2.SetΩρ∗=u∈K:u<ρ∗, for any u∈∂Ωρ∗, by (23), we have
(56)Tu=Tun−2δ≤M0ϕq∫0Tgrfur,uΔr∇r+∫0Tϕq∫sδgrfur,uΔr∇rΔs≤M0ϕq∫0Tgrfur,uΔr∇r+ϕq∫0Tgrfur,uΔr∇r≤b+1η1ρ∗ϕq∫0Tgrdr≤ρ∗=u.Then, by fixed point theorem of cone expansion-compression type, we have(57)iT,Ωρ∗,K=1.Secondly, lettingf∗x=max0≤un−1≤x,0≤u1≤1/Lu2fu1,u2, we can easy to know that f∗x is monotone increasing with respect to x≥0.Therefore bylimu2→∞max0≤u1≤1/Lu2fu1,u2/u2p−1=0, we can easy to have limx→∞f∗x/xp−1=0.Therefore, for anyη2∈0,R∗, there exists a constant ρ∗>r which satisfy
(58)f∗x≤η2xp−1,x≥ρ∗.SetΩρ∗=u∈K:u<ρ∗, for any u∈∂Ωρ∗, by (4.8), we have
(59)Tu=TuΔδ≤M0ϕq∫0Tgrfur,uΔr∇r+∫0Tϕq∫sδgrfur,uΔr∇rΔs≤M0ϕq∫0Tgrfur,uΔr∇r+ϕq∫0Tgrfur,uΔr∇r≤b+1ϕq∫0Tgrf∗ρ∗dr≤b+1η2ρ∗ϕq∫0Tgrdr≤ρ∗=u.Then, by fixed point theorem of cone expansion-compression type, we have(60)iT,Ωρ∗,K=1.Finally, imitating the previous proof of Theorem6, for any u∈∂Ωr, by A1, setting Ωr=u∈K:u<r, For any u∈∂Ωr, we can also easy to have
(61)Tu≥u,∀u∈∂Ωr.Then, by fixed point theorem of cone expansion-compression type, we have(62)iT,Ωr,K=0.Therefore, by (57), (60), (62), ρ∗<r<ρ∗, we have
(63)iT,Ωr\Ω¯ρ∗,K=−1,iT,Ωρ∗\Ω¯r,K=1.Then,T have fixed point u1∈Ωr\Ω¯ρ∗, and fixed point u2∈Ωρ∗\Ω¯r.Obviously,u1,u2 are all positive solutions of problem (1),(2) and 0<u1<r<u2. The proof of Theorem 10 is complete.


## 4. Application

Example.
Consider the following three-order BVP withp-Laplacian
(64)ϕpu′′′+164π4t−121−tlnu5+eu′2=0,0<t<1,u0=0,u′0−u′′0.25=0,u′1+5u′′0.3=0,where p=4,ξ=0.25,η=0.3,B′=1/4,
(65)gt=164π4t−121−t,fu1,u2=lnu5+eu′2.
Then obviously,q=4/3,∫01gtdt=1/64π3,f∞=+∞,f0=+∞,
(66)M0v=v<2v=bv,M1v=5v,∀v≥0,so conditions (H1), H2, H3, (A7), A8 hold.
Next,ϕq∫01gtdt=1/4π,R∗=4π/3, we choose R=3,M=2 and for B′=1/4, because of the monotone increasing of fu1,u2 on 0,∞×0,∞, then
(67)fu1,u2≤f12,3=286,0≤u2≤3,0≤u1≤4u2.
Therefore, using0<M<R∗, we have MRp−1=328, we know
(68)fu1,u2≤MRp−1,0≤u2≤3,0≤u1≤4u2,so conditions A2 holds. Then, by Theorem 9, the Example has at least two positive solutions v1,v2 and 0<v1<3<v2.



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