# Existence and Iteration of Positive Solutions to Third-Order BVP for a Class ofp-Laplacian Dynamic Equations on Time Scales

Authors: A. Kameswara Rao
Journal: International Journal of Differential Equations
 (2015)

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## Abstract

We investigate the existence and iteration of positive solutions for the following third-orderp-Laplacian dynamic equations on time scales: (ϕp(uΔΔ(t)))∇+q(t)f(t,u(t),uΔΔ(t))=0,t∈[a,b],αu(ρ(a))-βuΔ(ρ(a))=0,γu(b)+δuΔ(b)=0,uΔΔ(ρ(a))=0, where ϕp(s) is p-Laplacian operator; that is, ϕp(s)=sp-2s,p>1,ϕp-1=ϕq, and 1/p+1/q=1. By applying the monotone iterative technique and without the assumption of the existence of lower and upper solutions, we not only obtain the existence of positive solutions for the problem, but also establish iterative schemes for approximating the solutions.


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## Body

## 1. Introduction

A time scaleT is a nonempty closed subset of R. We make the blanket assumption that a,b are points in T. By an interval [a,b], we always mean the intersection of the real interval [a,b] with the given time scale; that is, [a,b]∩T.In recent years, dynamic equations on time scales have found a considerable interest and attracted by many researchers; see, for example, [1–11]. The reason seems to be twofold. Theoretically, dynamic equations on time scales not only can unify differential and difference equations [12], but also have displayed much more complicated dynamics [13, 14]. Moreover, the study of time scales has led to several important applications in the study of insect population models, neural networks, stock market, heat transfer, economic, wound healing, and epidemic models.Recently, there is much attention paid to the existence of positive solutions for second-order nonlinear boundary value problems on time scales; see [2, 6, 15, 16] and the references therein. On the one hand, higher-order nonlinear boundary value problems have been studied extensively; for details, see [4, 10, 17, 18] and references therein. On the other hand, the boundary value problems with p-Laplacian operator have also been discussed extensively in the literature; for example, see [3, 7–9, 11, 19–21]. However, to the best of our knowledge, there are not many results concerning the third-order p-Laplacian dynamic equations on time scales.In [22], Yang and Yan studied the following third-order Sturm-Liouville boundary value with p-Laplacian: (1)ϕpu′′t′+ft,ut=0,t∈0,1,αu0-βu′0=0,γu0+δu′1=0,u′′0=0.By using the fixed point index method, they established the existence of at least one or two positive solutions for the third-order Sturm-Liouville boundary value problem with p-Laplacian.In [8], Han and Kang were concerned with the existence of multiple positive solutions of third-order p-Laplacian dynamic equation on time scales: (2)ϕpuΔΔt∇+ft,ut=0,t∈a,b,αuρa-βuΔρa=0,γub+δuΔb=0,uΔΔρa=0,where ϕp(s)=sp-2s, p>1. By using the fixed point index method, they obtained the existence of multiple positive solutions for singular nonlinear boundary value problem.In this paper, we are concerned with the existence and iteration of positive solutions to the following third-orderp-Laplacian dynamic equation on time scales:(3)ϕpuΔΔt∇+qtft,ut,uΔΔt=0,t∈a,b,(4)αuρa-βuΔρa=0,γub+δuΔb=0,uΔΔρa=0,where ϕp(s) is p-Laplacian operator; that is, ϕp(s)=sp-2s for p>1 with ϕp-1=ϕq and 1/p+1/q=1;α,β,γ,δ≥0.Throughout this paper, we assume that the following conditions are satisfied:(H1)
d=γβ+αδ+αγ(b-ρ(a))>0.(H2)
qt is a nonnegative continuous function defined on ρa,b, and qt≢0 on any subinterval of (ρ(a),b). In addition, ∫ρ(a)b‍q(t)∇t<+∞. Here, a positive solution of (3) and (4) means a solution u∗ satisfying u∗(t)>0,ρ(a)<t<b.(H3)
f:[ρ(a),b]×[0,∞)×R→[0,∞) is continuous, and there exist constants μ1,μ2>0, such that, for any t∈[ρ(a),b],u∈[0,+∞), v∈R,(5)ft,λu,v≥λμ1ft,u,v,∀0≤λ≤1,(6)ft,u,λv≥λμ2ft,u,v,∀0≤λ≤1.(H4)
f(t,0,0)≢0 and f(t,1,1)≢0, and let(7)k=maxt∈ρa,bft,1,1.Remark 1.
Inequalities (5) and (6) are equivalent to the following inequalities, respectively:(8)ft,λu,v≤λμ1ft,u,v,∀λ≥1,(9)ft,u,λv≤λμ2ft,u,v,∀λ≥1.This paper is organized as follows. In Section2, we state some basic time scale definitions and several preliminary results. Section 3 is devoted to the iteration and existence of positive solutions of for the third-order Sturm-Liouville boundary value problem with p-Laplacian on time scales (3)-(4). Finally, in Section 4, we give an example to demonstrate our main result.


## 2. Preliminaries

For convenience, we list the following definitions which can be found in [3, 5, 13–15, 23].Definition 2.
A time scaleT is a nonempty closed subset of real numbers R. For t<sup⁡T and t>inf⁡T, define the forward jump operator σ and backward jump operator ρ, respectively, by (10)σt=inf⁡τ∈T∣τ>t∈T,ρr=sup⁡τ∈T∣τ<r∈T,for all t,r∈T. If σ(t)>t, t is said to be right scattered, and if ρ(r)<r, r is said to be left scattered; if σ(t)=t, t is said to be right dense, and if ρ(r)=r, r is said to be left dense. 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 Tk=T.Throughout this paper, we make the blanket assumption thata≤b are points in T and define the interval in T:(11)a,b=t∈Tsuch  thata≤t≤b.Other types of intervals are defined similarly.Definition 3.
Forf:T→R and t∈Tk, the delta derivative of f at the point t is defined to be the number fΔ(t) (provided it exists) with the property that, for each ϵ>0, there is a neighborhood U of t such that (12)fσt-fs-fΔtσt-s≤ϵσt-s,for all s∈U.
Forf:T→R and t∈Tk, the nabla derivative of f at t is denoted by f∇(t) (provided it exists) with the property that, for each ϵ>0, there is a neighborhood U of t such that (13)fρt-fs-f∇tρt-s≤ϵρt-s,for all s∈U.Definition 4.
A functionf is left-dense-continuous (i.e., ld-continuous), if f is continuous at each left-dense point in T and its right-sided limit exists at each right-dense point in T. It is well known that if f is ld-continuous, then there is a function F(t) such that F∇(t)=f(t). In this case, it is defined that (14)∫abft∇t=Fb-Fa.Definition 5.
IfGΔ(t)=f(t), then we define the delta integral by (15)∫abftΔt=Gb-Ga.If F∇(t)=f(t), then we define the nabla integral by (16)∫abft∇t=Fb-Fa.IfuΔΔ(t)≤0 on [a,b], then we say that u is concave down on [a,b].By a functionu(t) being a positive solution of (3)-(4), we mean that u(t) is positive on (a,b) and satisfies (3)-(4). Let G(t,s) be Green’s function of uΔΔ(t)=0 with respect to boundary condition (4); then (17)Gt,s=1dγb+δ-γtβ+αρs-ρa:ρs≤t,1dγb+δ-γρsβ+αt-ρa:t≤ρs,where α,β,γ,δ≥0:d=γβ+αδ+αγ(b-ρ(a))>0.This implies thatG(t,s) satisfies(18)0≤Gt,s≤Gρs,s,fort∈ρa,b,s∈a,b.In fact, on the one hand, from Green’s function, if s is kept fixed then G(t,s) is a decreasing function of t for t>ρ(s) and increasing for t<ρ(s). This leads to the conclusion that G(t,s) has a global maximum at t=ρ(s), so G(t,s)≤G(ρ(s),s), for t∈[ρ(a),b]. On the other hand, we can easily know that G(t,s) is nonnegative from the following two equalities:(19)γb+δ-γt=δ+γb-t,γb+δ-γρs=δ+γb-ρs.Let the Banach spaceX=C2[ρ(a),b] be equipped with the norm (20)u=maxmaxt∈ρa,but,maxt∈ρa,buΔt,maxt∈ρa,buΔΔt,C2+ρa,b=u∈C2ρa,b:ut≥0,t∈ρa,b.Definition 6.
LetX be a real Banach space. Let P be a nonempty, convex closed set in X. We say that P is a cone it it satisfies the following properties:(i)
λu∈P for u∈P,λ≥0;(ii)
u,-u∈P implies u=θ (θ denotes the null element of X).Define the coneP⊂X by (21)P=u∈C2+ρa,b:utis  a  concave  functionon⁡ρa,b.We define operator A:C2+[ρ(a),b]→C[ρ(a),b] by(22)Aut=∫ρabGt,sϕq∫ρasqτfτ,uτ,uΔΔτ∇τΔs,∀u∈C2+ρa,b,and then all the fixed points of operator A are the solutions for BVP (3)-(4).Remark 7.
From the definition of the operatorA, we can easily get the following properties:(23)AuΔt=-γd∫ρatβ+αρs-ρaϕq∫ρasqτfτ,uτ,uΔΔτ∇τΔs+αd∫tbγb+δ-γρsϕq∫ρasqτfτ,uτ,uΔΔτ∇τΔs,(24)AuΔΔt=-ϕq∫ρatqτfτ,uτ,uΔΔτ∇τΔs.Lemma 8.
Assume that (H1)–(H3) hold; then the operatorA:P→P is completely continuous.Proof.
From (H1)–(H3) and the definition ofA, we deduce that, for any u∈P, there is (Au)(t)≥0. By (24), we get (25)AuΔΔt=-ϕq∫ρatqτfτ,uτ,uΔΔτ∇τΔswhich implies that (Au) is concave on [ρ(a),b]. Thus, we have A(P)⊂P.
Now we show thatA:P→P is completely continuous. The continuity of A is obvious. We prove A is compact. Let Ω0⊂P be a bounded set. Then, there exists R such that Ω0⊂{u∈P:u≤R}. We can easily prove that A(Ω0) is bounded and equicontinuous; that is, A(Ω0) is a relatively compact set according to the Ascoli-Arzela theorem. Thus A:P→P is completely continuous.For convenience we use the following notations. Let(26)B=max⁡∫ρabGρs,s=1dβγ2ρa2+b2+βδb-ρa+αδ2b-ρa2+αγ62ρa3+b3-3ρa2+3ρa2b-3ρab2,αd12γb-ρa2+δb-ρa,γd12αb-ρa2+βb-ρa,1,C=Bϕq∫ρabqτ∇τ.


## 3. The Main Results

We now give our results for the existence of positive solutions for BVP (3) and (4).Theorem 9.
Assume that (H1)–(H4) hold and there exists a positive constantl>1 such that (H5)
f(t,x1,y1)≤f(t,x2,y2) for any ρ(a)≤t≤b, 0≤x1≤x2≤l, 0≤y1≤y2≤l,(H6)
ϕp(C)≤ϕp(l)/klμ1+μ2,
whereμ1, μ2, k are defined by (5), (6), and (7), respectively. Then boundary value (3) and (4) has two positive, concave solutions u∗ and v∗ such that (27)0<u∗≤l,limn→∞un=limn→∞Anu0=u∗,limn→∞unΔΔ=limn→∞Anu0ΔΔ=u∗ΔΔ,where(28)u0t=lB∫ρabGt,sΔs,ρa≤t≤b,0<v∗≤l,limn→∞vn=limn→∞Anv0=v∗,limn→∞vnΔΔ=limn→∞Anv0ΔΔ=v∗ΔΔ,where v0(t)=0,ρ(a)≤t≤b, A is defined by (22). The iterative schemes in this theorem are u0(t), un+1=Aun=Anu0, n=0,1,2,… and v0(t), vn+1=Avn=Anv0, n=0,1,2,….Proof.
We defineP¯l={u∈P:u≤l}. Then, in what follows, we will prove that T:P¯l→P¯l. If u∈P¯l, then u≤l; we have (29)0≤ut≤maxt∈ρa,but≤u≤l,uΔΔt≤maxt∈ρa,buΔΔt≤u≤l.So by (8), (9) and (H5), we have(30)0≤ft,ut,uΔΔt≤ft,l,l≤lμ1+μ2ft,1,1≤klμ1+μ2,forρa<t<b.In fact, (31)Au=maxmaxt∈ρa,bAut,maxt∈ρa,bAuΔt,maxt∈ρa,bAuΔΔt=maxmaxt∈ρa,bAut,AuΔρa,-AuΔb,-AuΔΔb.Then, by (18), (30) and (H6), we have (32)maxt∈ρa,bAut=maxt∈ρa,b∫ρabGt,sϕq∫ρasqτfτ,uτ,uΔΔτ∇τΔs≤∫ρabGρs,sϕq∫ρasqτfτ,uτ,uΔΔτ∇τΔs≤ϕq∫ρasqτfτ,uτ,uΔΔτ∇τ∫ρabGρs,sΔs≤ϕqklμ1+μ2ϕq∫ρabqτ∇τ∫ρabGρs,sΔs≤Cϕqklμ1+μ2≤l,and from (23) and (24), we have(33)AuΔρa=αd∫ρabγb+δ-γρsϕq∫ρasqτfτ,uτ,uΔΔτ∇τΔs≤αd∫ρabγb+δ-γρsϕq∫ρabqτfτ,uτ,uΔΔτ∇τΔs≤αdϕqklμ1+μ2ϕq∫ρabqτ∇τ∫ρabγb+δ-γρsΔs=αd12γb-ρa2+δb-ρaϕqklμ1+μ2ϕq∫ρabqτ∇τ≤Cϕqklμ1+μ2≤l,-AuΔb=γd∫ρabβ+αρs-ρaϕq∫ρasqτfτ,uτ,uΔΔτ∇τΔs≤γd∫ρabβ+αρs-ρaϕq∫ρabqτfτ,uτ,uΔΔτ∇τΔs≤γρϕqklμ1+μ2ϕq∫ρabqτ∇τ∫ρabβ+αρs-ρaΔs=γd12αb-ρa2+βb-ρaϕqklμ1+μ2ϕq∫ρabqτ∇τ≤Cϕqklμ1+μ2≤l,-AuΔΔb=ϕq∫ρabqτfτ,uτ,uΔΔτ∇τ≤ϕqklμ1+μ2ϕq∫ρabqτ∇τ≤Cϕqklμ1+μ2≤l.Thus, we obtained that Au≤l. So, we have shown that TP¯l⊂P¯l.
Letu0(t)=l/B·∫ρ(a)b‍G(t,s)Δs,t∈[ρ(a),b], then (34)u0Δt=lB-γd∫ρatβ+αρs-ρaΔs+αd∫tbγb+δ-γρsΔs,u0ΔΔt=-lB.It is easy to get (35)u0t≤l∫ρabGρs,sΔs∫ρabGt,sΔs≤l,u0Δρa=lB·αd12γb-ρa2+δb-ρa≤l,-u0Δb=lB·γd12αb-ρa2+βb-ρa≤l,maxt∈ρa,bu0ΔΔt=lB≤l,so, u0∈P¯l.
Letu1=Au0, then from the above expression, we have u1∈P¯l. We define (36)un+1=Aun=An+1u0,n=0,1,2,….It follows from AP¯l⊂P¯l, that un∈P¯l,n=1,2,…. Since A is completely continuous, we can assert that {un} is a sequentially compact set.
Since(37)u1t=Au0t=∫ρabGt,sϕq∫ρasqτfτ,u0τ,u0ΔΔτ∇τΔs≤∫ρabGt,sϕq∫ρabqτfτ,u0τ,u0ΔΔτ∇τΔs≤ϕqklμ1+μ2ϕq∫ρabqτ∇τ∫ρabGt,sΔs=CBϕqklμ1+μ2∫ρabGt,sΔs≤lB∫ρabGt,sΔs=u0t,u1ΔΔt=Au0ΔΔt=ϕq∫ρatqτfτ,u0τ,u0ΔΔτ∇τ≤ϕq∫ρabqτfτ,u0τ,u0ΔΔτ∇τ≤ϕqklμ1+μ2ϕq∫ρabqτ∇τ=CBϕqklμ1+μ2≤lB=u0ΔΔtwe obtain (38)u1t≤u0t,u1ΔΔt≤u0ΔΔt,ρa≤t≤b.
By (22) and (H5), we have (39)u2t=Au1t≤Au0t=u1t,ρa≤t≤b,u2ΔΔt=Au1ΔΔt≤Au0ΔΔt=u1ΔΔt,ρa≤t≤b.Hence by induction we have (40)un+1t≤unt,un+1ΔΔt≤unΔΔt,ρa≤t≤b,n=0,1,2,….Thus, there exists u∗∈P¯l such that un→u∗. Applying the continuity of A and un+1(t)=Aun(t), we get Au∗(t)=u∗(t) which implies that u∗ is a nonnegative concave solution of the BVP (3) and (4).
On the other hand, letv0(t)=0,t∈[ρ(a),b], then v0(t)∈P¯l. Let v1=Av0=A0, then v1∈P¯l. We define vn+1=Avn=An+1v0,n=0,1,2,…. Since A:P¯l→P¯l, we have vn∈AP¯l⊆P¯l,n=1,2,…. Since A is completely continuous, we can assert that {vn} is a sequentially compact set.
Sincev1=Av0=A0∈P¯l, we have (41)v1t=Av0t=A0t≥0,ρa≤t≤b,v1ΔΔt=Av0ΔΔt=A0ΔΔt≥0,ρa≤t≤b.So, (42)v2t=Av1t≥A0t=v1t,ρa≤t≤b,v2ΔΔt=Av1ΔΔt≥A0ΔΔt=v1ΔΔt,ρa≤t≤b.By an induction argument similar to the above we obtain (43)vn+1t≥vnt,vn+1ΔΔt≥vnΔΔt,ρa≤t≤b,n=0,1,2,….Hence, there exists v∗∈P¯l such that vn→v∗. Applying the continuity of A and vn+1(t)=Avn(t), we obtain Av∗(t)=v∗(t) which implies that v∗ is a nonnegative solution of the BVP (3) and (4). From the condition (H4), we know that the zero function is not the solution of the BVP (3) and (4). Thus, maxρ(a)≤t≤bv∗t>0, and by the concavity of v∗(t) we have (44)v∗t≥min⁡t-ρa,b-tmaxt∈ρa,bv∗t>0,ρa<t<b.
Next, notingu0(t)≥v0(t),u0ΔΔt≥v0ΔΔt,t∈[ρ(a),b], it follows from (H5) and (22) that Au0≥Av0, and by induction, we have un(t)≥vn(t),n=0,1,2,… which implies that u∗≥v∗. Thus we have (45)u∗t≥v∗t≥min⁡t-ρa,b-tmaxt∈ρa,bv∗t>0.This means that u∗ is a positive concave solution of the BVP (3) and (4). Thus the BVP (3) and (4) has two positive and concave solutions u∗,v∗ such that 0≤u∗≤l,0≤v∗≤l, and, from the above proof, we know that the iterative sequences hold. The proof is completed.


## 4. Example

Now, we present an example to illustrate the main result: Theorem9.LetT={(2/5)n:n∈N0}∪{0}∪[1,2]. We consider the following third-order Sturm-Liouville boundary value problem with p-Laplacian on time scales:(46)ϕ3uΔΔt∇+1tetu3/5t+uΔΔ1/3t+1=0,t∈425,25,u8125-uΔ8125=0,u25=0,uΔΔ8125=0,where p=3, α=1, β=1, γ=1, and δ=0. Let (47)ft,u,v=etu3/5+v1/3+1,qt=1t,μ1=12,μ2=13,where, for any 0≤λ≤1 and u∈[0,∞), v∈R,t∈[8/125,2/5], we have (48)ft,λu,v=etλu3/5+v1/3+1≥λ3/5etu3/5+v1/3+1≥λ1/2ft,u,v,ft,u,λv=etu3/5+λv1/3+1≥λ1/3etu3/5+v1/3+1≥λ1/3ft,u,v,which implies that (H3) holds.On the other hand, it is clear that (H1), (H2), and (H4) are satisfied and(49)d=167125,k=maxt∈8/125,2/5ft,1,1=2e2/5+1.Since q(t)=1/t, we have (50)B=max⁡0.06,0.0425,0.2937,1=1,C=Bϕq∫8/1252/5qτ∇τ=ϕq0.759,so ϕp(C)=0.759ϕp(1)=0.759.Takel=[(2e2/5+1)]6/7; then (51)ϕplklμ1+μ2=2e2/5+16/722e2/5+12e2/5+16/75/6=1>0.759=ϕpC,which implies that (H6) holds. For l=[(2e2/5+1)]6/7, it is clear that (H5) holds. So, by Theorem 9, boundary value problem (46) has two positive and concave solutions u∗ and v∗ such that (52)0<u∗<2e2/5+16/7,0<u∗Δ<2e2/5+16/7,0<u∗ΔΔ<2e2/5+16/7,limn→∞un=limn→∞Anu0=u∗,limn→∞unΔΔ=limn→∞Anu0ΔΔ=u∗ΔΔ,where u0(t)=-1.0688t2+0.0666t+0.00308,t∈[8/125,2/5], and (53)0<v∗<2e2/5+16/7,0<v∗Δ<2e2/5+16/7,0<v∗ΔΔ<2e2/5+16/7,limn→∞vn=limn→∞Anv0=v∗,limn→∞vnΔΔ=limn→∞Anv0ΔΔ=v∗ΔΔ,where v0(t)=0,t∈[8/125,2/5].



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