# Existence of Three Positive Solutions to Somep-Laplacian Boundary Value Problems

Authors: Moulay Rchid Sidi Ammi; Delfim F. M. Torres
Journal: Discrete Dynamics in Nature and Society
 (2013)

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## Abstract

We obtain, by using the Leggett-Williams fixed point theorem, sufficient conditions that ensure the existence of at least three positive solutions to somep-Laplacian boundary value problems on time scales.


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## Body

## 1. Introduction

The study of dynamic equations on time scales goes back to the 1989 Ph.D. thesis of Hilger [1, 2] and is currently an area of mathematics receiving considerable attention [3–7]. Although the basic aim of the theory of time scales is to unify the study of differential and difference equations in one and the same subject, it also extends these classical domains to hybrid and in-between cases. A great deal of work has been done since the eighties of the XX century in unifying the theories of differential and difference equations by establishing more general results in the time scale setting [8–12].Boundary valuep-Laplacian problems for differential equations and finite difference equations have been studied extensively (see, e.g., [13] and references therein). Although many existence results for dynamic equations on time scales are available [14, 15], there are not many results concerning p-Laplacian problems on time scales [16–19]. In this paper we prove new existence results for three classes of p-Laplacian boundary value problems on time scales. In contrast with our previous works [17, 18], which make use of the Krasnoselskii fixed point theorem and the fixed point index theory, respectively, here we use the Leggett-Williams fixed point theorem [20, 21] obtaining multiplicity of positive solutions. The application of the Leggett-Williams fixed point theorem for proving multiplicity of solutions for boundary value problems on time scales was first introduced by Agarwal and O’Regan [22] and is now recognized as an important tool to prove existence of positive solutions for boundary value problems on time scales [23–28].The paper is organized as follows. In Section2 we present some necessary results from the theory of time scales (Section  2.1) and the theory of cones in Banach spaces (Section  2.2). We end Section  2.2 with the Leggett-Williams fixed point theorem for a cone-preserving operator, which is our main tool in proving existence of positive solutions to the boundary value problems on time scales we consider in Section 3. The contribution of the paper is Section 3, which is divided into three parts. The purpose of the first part (Section  3.1) is to prove existence of positive solutions to the nonlocal p-Laplacian dynamic equation on time scales
(1)-(ϕp(uΔ(t)))∇=λf(u(t))(∫0T‍f(u(τ))∇τ)2,t∈(0,T)𝕋,
satisfying the boundary conditions
(2)ϕp(uΔ(0))-β(ϕp(uΔ(η)))=0,u(T)-βu(η)=0,
where η∈(0,T)𝕋, ϕp(·) is the p-Laplacian operator defined by ϕp(s)=|s|p-2s, p>1, and (ϕp)-1=ϕq with q the Holder conjugate of p, that is, 1/p+1/q=1. The concrete value of p is connected with the application at hands. For p=2, problem (1)-(2) describes the operation of a device flowed by an electric current, for example, thermistors [29], which are devices made from materials whose electrical conductivity is highly dependent on the temperature. Thermistors have the advantage of being temperature measurement devices of low cost, high resolution, and flexibility in size and shape. Constant λ in (1) is a dimensionless parameter that can be identified with the square of the applied potential difference at the ends of a conductor, f(u) is the temperature-dependent resistivity of the conductor, and β in (2) is a transfer coefficient supposed to verify 0<β<1. For a more detailed discussion about the physical justification of (1)-(2) the reader is referred to [17]. Theoretical analysis (existence, uniqueness, regularity, and asymptotic results) for thermistor problems with various types of boundary and initial conditions has received significant attention in the last few years for the particular case 𝕋=ℝ [30–34]. The second part of our results (Section  3.2) is concerned with the following quasilinear elliptic problem:
(3)-(ϕp(uΔ(t)))∇=f(u(t))+h(t),t∈(0,T)𝕋,uΔ(0)=0,u(T)-u(η)=0,
where η∈(0,T)𝕋. Results on existence of infinitely many radial solutions to (3) are proved in the literature using (i) variational methods, where solutions are obtained as critical points of some energy functional on a Sobolev space, with f satisfying appropriate conditions [35, 36]; (ii) methods based on phase-plane analysis and the shooting method [37]; (iii) the technique of time maps [38]. For p=2, h≡0, and 𝕋=ℝ, problem (3) becomes a well-known boundary value problem of differential equations. Our results generalize earlier works to the case of a generic time scale 𝕋, p≠2, and h not identically zero. Finally, the third part of our contribution (Section  3.3) is devoted to the existence of positive solutions to the p-Laplacian dynamic equation
(4)(ϕp(uΔ(t)))∇+λa(t)f(u(t),u(ω(t)))=0,t∈(0,T)𝕋,u(t)=ψ(t),t∈[-r,0]𝕋,u(0)-B0(uΔ(0))=0,uΔ(T)=0
on a time scale 𝕋 such that 0,T∈𝕋κκ, -r∈𝕋 with -r≤0<T,  where λ>0. This problem is considered in [39] where the author applies the Krasnoselskii fixed point theorem to obtain one positive solution to (4). Here we use the same conditions as in [39], but applying Leggett-Williams’ theorem we are able to obtain more: we prove existence of at least three positive solutions and we are able to localize them.


## 2. Preliminaries

Here we just recall the basic concepts and results needed in the sequel. For an introduction to time scales the reader is referred to [3, 8–10, 40, 41] and references therein; for a good introduction to the theory of cones in Banach spaces we refer the reader to the book [42].


### 2.1. Time Scales

A time scale𝕋 is an arbitrary nonempty closed subset of the real numbers ℝ. The operators σ and ρ from 𝕋 to 𝕋 are defined in [1, 2] as
(5)σ(t)=inf{τ∈𝕋∣τ>t}∈𝕋,ρ(t)=sup{τ∈𝕋∣τ<t}∈𝕋
and are called the forward jump operator and the backward jump operator, respectively. A point t∈𝕋 is left-dense, left-scattered, right-dense, and right-scattered if ρ(t)=t,ρ(t)<t, and σ(t)=t,σ(t)>t, respectively. If 𝕋 has a right-scattered minimum m, define 𝕋κ=𝕋-{m}; otherwise set 𝕋κ=𝕋. If 𝕋 has a left-scattered maximum  M, define 𝕋κ=𝕋-{M}; otherwise set 𝕋κ=𝕋. Following [43], we also introduce the set 𝕋κκ=𝕋κ∩𝕋κ.Letf:𝕋→ℝ and t∈𝕋κ (assume t is not left-scattered if t=sup𝕋), then 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
(6)|f(σ(t))-f(s)-fΔ(t)(σ(t)-s)|≤|σ(t)-s|∀s∈U.
Similarly, for t∈𝕋κ (assume t is not right-scattered if t=inf𝕋), the nabla derivative of f at the point t is defined in [44] to be the number f∇(t) (provided it exists) with the property that for each ϵ>0 there is a neighborhood U of t such that
(7)|f(ρ(t))-f(s)-f∇(t)(ρ(t)-s)|≤|ρ(t)-s|∀s∈U.
If 𝕋=ℝ, then fΔ(t)=f∇(t)=f′(t). If 𝕋=ℤ, then fΔ(t)=f(t+1)-f(t) is the forward difference operator while f∇(t)=f(t)-f(t-1) is the backward difference operator.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. If f is ld-continuous, then there exists F such that F∇(t)=f(t) for any t∈𝕋κ. We then introduce the nabla integral by
(8)∫abf(t)∇t=F(b)-F(a).
We define right-dense continuous (rd-continuous) functions in a similar way. If f is rd-continuous, then there exists F such that FΔ(t)=f(t) for any t∈𝕋κ, and we define the delta integral by
(9)∫abf(t)Δt=F(b)-F(a).


### 2.2. Cones in Banach Spaces

In this paper𝕋 is a time scale with 0∈𝕋κ and T∈𝕋κ. We use ℝ+ and ℝ0+ to denote, respectively, the set of positive and nonnegative real numbers. By [0,T]𝕋 we denote the set [0,T]∩𝕋. Similarly, (0,T)𝕋=(0,T)∩𝕋. Let E=ℂld([0,T]𝕋,ℝ). It follows that E is a Banach space with the norm ∥u∥=max[0,T]𝕋|u(t)|.Definition 1.
LetE be a real Banach space. A nonempty, closed, and convex set P⊂E is called a cone if it satisfies the following two conditions:(i)
u∈P, λ≥0, implies λu∈P;(ii)
u∈P, -u∈P, implies u=0.Every coneP⊂E induces an ordering in E given by
(10)u≤viffv-u∈P.Definition 2.
LetE be a real Banach space and P⊂E  a cone. A function α:P→ℝ0+ is called a nonnegative continuous concave functional if α is continuous and
(11)α(tx+(1-t)y)≥tα(x)+(1-t)α(y)
for all x,y∈P and 0≤t≤1.Leta,b, and r be positive constants, Pr={u∈P|∥u∥<r}, P(α,a,b)={u∈P|a≤α(u),∥u∥<b}. The following fixed point theorem provides the existence of at least three positive solutions. The origin in E is denoted by ∅. The proof of the Leggett-Williams fixed point theorem can be found in Guo and Lakshmikantham [42] or Leggett and Williams [45].Theorem 3 (Leggett-Williams’ Theorem).
LetP be a cone in a real Banach space E. Let G:Pc¯→Pc¯ be a completely continuous map and α a nonnegative continuous concave functional on P such that α(u)≤∥u∥ for all u∈Pc¯. Suppose there exist positive constants a,b, and d with 0<a<b<d≤c such that(i)
{u∈P(α,b,d)|α(u)>b}≠∅ and α(Gu)>b for all u∈P(α,b,d);(ii)
∥Gu∥<a for all u∈P¯a;(iii)
α(Gu)>b for all u∈P(α,b,c) with ∥Gu∥>d.Then G has at least three fixed points u1, u2, and u3 satisfying
(12)∥u1∥<a,b<α(u2),∥u3∥>a,α(u3)<b.


## 2.1. Time Scales

A time scale𝕋 is an arbitrary nonempty closed subset of the real numbers ℝ. The operators σ and ρ from 𝕋 to 𝕋 are defined in [1, 2] as
(5)σ(t)=inf{τ∈𝕋∣τ>t}∈𝕋,ρ(t)=sup{τ∈𝕋∣τ<t}∈𝕋
and are called the forward jump operator and the backward jump operator, respectively. A point t∈𝕋 is left-dense, left-scattered, right-dense, and right-scattered if ρ(t)=t,ρ(t)<t, and σ(t)=t,σ(t)>t, respectively. If 𝕋 has a right-scattered minimum m, define 𝕋κ=𝕋-{m}; otherwise set 𝕋κ=𝕋. If 𝕋 has a left-scattered maximum  M, define 𝕋κ=𝕋-{M}; otherwise set 𝕋κ=𝕋. Following [43], we also introduce the set 𝕋κκ=𝕋κ∩𝕋κ.Letf:𝕋→ℝ and t∈𝕋κ (assume t is not left-scattered if t=sup𝕋), then 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
(6)|f(σ(t))-f(s)-fΔ(t)(σ(t)-s)|≤|σ(t)-s|∀s∈U.
Similarly, for t∈𝕋κ (assume t is not right-scattered if t=inf𝕋), the nabla derivative of f at the point t is defined in [44] to be the number f∇(t) (provided it exists) with the property that for each ϵ>0 there is a neighborhood U of t such that
(7)|f(ρ(t))-f(s)-f∇(t)(ρ(t)-s)|≤|ρ(t)-s|∀s∈U.
If 𝕋=ℝ, then fΔ(t)=f∇(t)=f′(t). If 𝕋=ℤ, then fΔ(t)=f(t+1)-f(t) is the forward difference operator while f∇(t)=f(t)-f(t-1) is the backward difference operator.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. If f is ld-continuous, then there exists F such that F∇(t)=f(t) for any t∈𝕋κ. We then introduce the nabla integral by
(8)∫abf(t)∇t=F(b)-F(a).
We define right-dense continuous (rd-continuous) functions in a similar way. If f is rd-continuous, then there exists F such that FΔ(t)=f(t) for any t∈𝕋κ, and we define the delta integral by
(9)∫abf(t)Δt=F(b)-F(a).


## 2.2. Cones in Banach Spaces

In this paper𝕋 is a time scale with 0∈𝕋κ and T∈𝕋κ. We use ℝ+ and ℝ0+ to denote, respectively, the set of positive and nonnegative real numbers. By [0,T]𝕋 we denote the set [0,T]∩𝕋. Similarly, (0,T)𝕋=(0,T)∩𝕋. Let E=ℂld([0,T]𝕋,ℝ). It follows that E is a Banach space with the norm ∥u∥=max[0,T]𝕋|u(t)|.Definition 1.
LetE be a real Banach space. A nonempty, closed, and convex set P⊂E is called a cone if it satisfies the following two conditions:(i)
u∈P, λ≥0, implies λu∈P;(ii)
u∈P, -u∈P, implies u=0.Every coneP⊂E induces an ordering in E given by
(10)u≤viffv-u∈P.Definition 2.
LetE be a real Banach space and P⊂E  a cone. A function α:P→ℝ0+ is called a nonnegative continuous concave functional if α is continuous and
(11)α(tx+(1-t)y)≥tα(x)+(1-t)α(y)
for all x,y∈P and 0≤t≤1.Leta,b, and r be positive constants, Pr={u∈P|∥u∥<r}, P(α,a,b)={u∈P|a≤α(u),∥u∥<b}. The following fixed point theorem provides the existence of at least three positive solutions. The origin in E is denoted by ∅. The proof of the Leggett-Williams fixed point theorem can be found in Guo and Lakshmikantham [42] or Leggett and Williams [45].Theorem 3 (Leggett-Williams’ Theorem).
LetP be a cone in a real Banach space E. Let G:Pc¯→Pc¯ be a completely continuous map and α a nonnegative continuous concave functional on P such that α(u)≤∥u∥ for all u∈Pc¯. Suppose there exist positive constants a,b, and d with 0<a<b<d≤c such that(i)
{u∈P(α,b,d)|α(u)>b}≠∅ and α(Gu)>b for all u∈P(α,b,d);(ii)
∥Gu∥<a for all u∈P¯a;(iii)
α(Gu)>b for all u∈P(α,b,c) with ∥Gu∥>d.Then G has at least three fixed points u1, u2, and u3 satisfying
(12)∥u1∥<a,b<α(u2),∥u3∥>a,α(u3)<b.


## 3. Main Results

We prove existence of three positive solutions to differentp-Laplacian problems on time scales: in Section  3.1 we study problem (1)-(2), in Section  3.2 problem (3), and finally (4) in Section  3.3.


### 3.1. Nonlocal Thermistor Problem

By a solutionu:𝕋→ℝ of (1)-(2) we mean a delta differentiable function such that uΔ and (|uΔ|p-2uΔ)∇ are both continuous on 𝕋κκ and u satisfies (1)-(2). We consider the following hypothesis:(H1)
f:ℝ→ℝ+ is a continuous function.Lemma 4 (see Lemma  3.1 of [17]).
Assume that hypothesis (H1) on functionf is satisfied. Then u is a solution to (1)-(2) if and only if u∈E is a solution to the integral equation
(13)u(t)=-∫0t‍ϕq(g(s))Δs+B,
where
(14)g(s)=∫0sλh(u(r))∇r-A,A=ϕp(uΔ(0))=-λβ1-β∫0ηh(u(r))∇r,h(u(t))=λf(u(t))(∫0T‍f(u(τ))∇τ)2,B=u(0)=11-β{∫0T‍ϕq(g(s))Δs-β∫0η‍ϕq(g(s))Δs}.Lemma 5.
Suppose (H1) holds. Then a solutionu to (1)-(2) satisfies u(t)≥0 for t∈(0,T)𝕋.Proof.
We haveA=-λβ/(1-β)∫0η‍h(u(r))∇r≤0. Then, g(s)=λ∫0s‍h(u(r))∇r-A≥0. It follows that ϕp(g(s))≥0. Since 0<β<1, we also have
(15)u(0)=B=11-β{∫0T‍ϕq(g(s))Δs-β∫0η‍ϕq(g(s))Δs}≥11-β{β∫0T‍ϕq(g(s))Δs-β∫0η‍ϕq(g(s))Δs}≥0,u(T)=u(0)-∫0T‍ϕq(g(s))Δs=-β1-β∫0η‍ϕq(g(s))Δs+11-β∫0T‍ϕq(g(s))Δs-∫0T‍ϕq(g(s))Δs=-β1-β∫0η‍ϕq(g(s))Δs+β1-β∫0T‍ϕq(g(s))Δs=β1-β{∫0T‍ϕq(g(s))Δs-∫0η‍ϕq(g(s))Δs}≥0.
If t∈(0,T)𝕋, then
(16)u(t)=u(0)-∫0t‍ϕq(g(s))Δs≥-∫0T‍ϕq(g(s))Δs+u(0)=u(T)≥0.
Consequently, u(t)≥0 for t∈(0,T)𝕋.On the other hand, we haveϕp(uΔ(s))=ϕp(uΔ(0))-∫0s‍λh(u(r))∇r≤0. Since A=ϕp(uΔ(0))≤0, then uΔ≤0. This means that ||u||=u(0), inft∈(0,T)𝕋u(t)=u(T). Moreover, ϕp(uΔ(s)) is nonincreasing, which implies with the monotonicity of ϕp that uΔ is a nonincreasing function on (0,T)𝕋. Hence, u is concave. In order to apply Theorem 3, let us define the cone P⊂E by
(17)P={u∈E∣uis nonnegative,decreasing on[0,T]𝕋and concave onE}.
We also define the nonnegative continuous concave functional α:P→ℝ0+ by
(18)α(u)=mint∈[ξ,T-ξ]𝕋u(t),ξ∈(0,T2),∀u∈P.
It is easy to see that problem (1)-(2) has a solution u=u(t) if and only if u is a fixed point of the operator G:P→E defined by
(19)Gu(t)=-∫0t‍ϕq(g(s))Δs+B,
where g and B are as in Lemma 4.Lemma 6.
LetG be defined by (19). Then,(i)
G(P)⊆P;(ii)
G:P→P is completely continuous.Proof.
(i)
holds clearly from above.
(ii)
Suppose thatD⊆P is a bounded set and let u∈D. Then,
(20)|Gu(t)|=|-∫0t‍ϕq(g(s))Δs+B|≤|-∫0t‍ϕq(∫0s‍λf(u(r))(∫0T‍f(u(τ))∇τ)2∇r-A)Δs|+|B|≤∫0T‍ϕq(∫0s‍λsupu∈Df(u)(Tinfu∈Df(u))2∇r-A)Δs+|B|,|A|=|λβ1-β∫0η‍h(u(r))∇r|=|λβ1-β∫0η‍f(u(r))(∫0T‍f(u(τ))∇τ)2∇r|≤λβ1-βsupu∈Df(u)(Tinfu∈Df(u))2η.
In the same way, we have(21)|B|≤11-β∫0T‍ϕq(g(s))Δs≤11-β∫0T‍ϕq(λsupu∈Df(u)(Tinfu∈Df(u))2(s+β1-βη))Δs.
It follows that
(22)|Gu(t)|≤∫0T‍ϕq(λsupu∈Df(u)(Tinfu∈Df(u))2(s+βη1-β))Δs+|B|.
As a consequence, we get
(23)∥Gu∥≤2-β1-β∫0T‍ϕq(λsupu∈Df(u)(Tinfu∈Df(u))2(s+βη1-β))Δs≤21-βϕq(λsupu∈Df(u)(Tinfu∈Df(u))2)∫0T‍ϕq(s+βη1-β)Δs.
Then G(D) is bounded on the whole bounded set D. Moreover, if t1,t2∈[0,T]𝕋 and u∈D, then we have for a positive constant c(24)|Gu(t2)-Gu(t1)|≤|∫t1t2‍ϕq(g(s))Δs|≤c|t2-t1|.
We see that the right-hand side of the above inequality goes uniformly to zero when |t2-t1|→0. Then by a standard application of the Arzela-Ascoli theorem we have that G:P→P is completely continuous.We can also easily obtain the following properties.Lemma 7.
(i)α(u)=u(T-ξ)≤∥u∥ for all u∈P;
(ii)α(Gu)=(Gu)(T-ξ);
(iii)∥Gu∥=Gu(0).We now state the main result of Section  3.1.Theorem 8.
Suppose that(H1) is verified and there exist positive constants a, b, c, and d such that 0<ζa=a1<b<d-T2cϕq(1/T)<d<ζc=c1, with ζ=(T2/(1-β)) ϕq(1/T). One further imposes f to satisfy the following hypotheses:(H2)
min0≤u≤a1f(u)≥λ2/(T(1-β)ϕp(a)) uniformly for all t∈[0,T]𝕋;(H3)
min0≤u≤c1f(u)≥λ2/(T(1-β)ϕp(c)) uniformly for all t∈[0,T]𝕋;(H4)
minb≤u≤df(u)≥ϕp(bB1) uniformly for all t∈[0,T]𝕋, where
(25)B1=(1-β)βξ|ϕp(T-ξ)|ϕp(λ(Tsupb≤u≤df(u))2).Then the boundary value problem (1)-(2) has at least three positive solutions u1, u2, and u3, verifying
(26)∥u1∥<a,b<α(u2),∥u3∥>a,α(u3)<b.Proof.
The proof passes by several lemmas. We have already seen in Lemma6 that the operator G is completely continuous. We now show the following Lemma.Lemma 9.
The following relations hold:(27)GPc1¯⊂Pc1¯,GPa1¯⊂Pa1¯.Proof.
Obviously,GPa1¯⊂P. Moreover, for all u∈Pa1¯, we have 0≤u(t)≤a1. On the other hand we have
(28)Gu(t)=-∫0t‍ϕq(g(s))Δs+11-β∫0T‍ϕq(g(s))Δs-β1-β∫0η‍ϕq(g(s))Δs,
and for all u∈GPa1¯ we have 0≤u(t)≤a1. Then,
(29)|Gu|≤11-β∫0T‍ϕq(g(s))Δs.
We have
(30)h(u(t))=λf(u(t))(∫0T‍f(u(τ))∇τ)2,g(s)=∫0s‍λh(u(r))∇r+λβ1-β∫0η‍h(u(r))∇r≤(λ+λβ1-β∫0T‍h(u(r))∇r)≤λ1-β∫0T‍h(u(r))∇r.
Using (H2) it follows that
(31)ϕq(g(s))≤aTϕq(1T).
Then we get
(32)|Gu|≤a1,GPa1¯⊂Pa1¯.
Similarly, using (H3) we get GPc1¯⊂Pc1¯.Lemma 10.
The set(33){u∈P(α,b,d)∣α(u)>b}
is nonempty, and
(34)α(Gu)>b,ifu∈P(α,b,d).Proof.
Letu=(b+d)/2. Then, u∈P,∥u∥=(b+d)/2≤d, and α(u)≥(b+d)/2>b. The first part of the lemma is proved. For u∈P(α,b,d) we have b≤u≤d. If t∈[ξ,T]𝕋, then
(35)α(Gu)=(Gu)(T-ξ)=-∫0T-ξ‍ϕq(g(s))Δs+B≥β1-β∫T-ξT‍ϕq(g(s))Δs.
Since A≤0, we have by using (H4)
(36)g(s)=λ∫0s‍h(u(r))∇r-A≥λ∫0s‍h(u(r))∇r≥λ∫0s‍f(u)(Tsupb≤u≤df(u))2∇u≥λ(bB1)p-1(Tsupb≤u≤df(u))2s.
Using the fact that ϕq is nondecreasing we get
(37)ϕq(g(s))≥ϕq(λ(bB1)p-1(Tsupb≤u≤df(u))2s)≥bB1ϕq(λ(Tsupb≤u≤df(u))2)ϕq(s).
Using the expression of B1(38)α(Gu)≥β1-βbB1ϕq(λ(Tsupb≤u≤df(u))2)∫T-ξT‍ϕq(s)Δs≥bB1β1-βϕq(λ(Tsupb≤u≤df(u))2)ϕq(T-ξ)ξ≥b.Lemma 11.
For allu∈P(α,b,c1) with ∥Gu∥>d one has
(39)α(Gu)>b.Proof.
Ifu∈P(α,b,c1) and ∥Gu∥>d, then 0≤u(t)≤c1. Using hypothesis (H3) and the fact that 0<β<1, it follows that
(40)α(Gu)=Gu(T-ξ)=-∫0T-ξ‍ϕq(g(s))Δs+B≥-∫0T‍ϕq(g(s))Δs+Gu(0)≥∥Gu∥-T2cϕq(1T)≥d-T2cϕq(1T)>b.
Gathering Lemmas 4 to 11 and applying Theorem 3, there exist at least three positive solutions u1, u2, and u3 to (1)-(2) verifying
(41)∥u1∥<a,b<α(u2),∥u3∥>a,α(u3)<b.Example 12.
Let𝕋={1-(1/2)ℕ0}∪{1}, where ℕ0 denotes the set of all nonnegative integers. Consider the p-Laplacian dynamic equation
(42)-(ϕp(uΔ(t)))∇=λf(u(t))(∫0T‍f(u(τ))∇τ)2,t∈(0,T)𝕋,
satisfying the boundary conditions
(43)ϕp(uΔ(0))-β(ϕp(uΔ(14)))=0,u(1)-βu(14)=0,
where p=3/2, q=3, η=1/4, β=1/2, λ=1, T=1, and
(44)f(u)={22,0≤u≤1,4(u-1)+22,1≤u≤32,2+22,32≤u≤10,2u+22-18,10≤u≤16.
Choose a1=1=2a, b=3/2, c1=16=2c, and d=10. It is easy to see that ζ=2, B1=1/(2(2+2)) and
(45)min{f(u):u∈[0,a1]}=22≥λ2T(1-β)ϕp(a)=2a=22,min{f(u):u∈[0,c1]}=22≥λ2T(1-β)ϕp(c)=2c=12,min{f(u):u∈[b,d]}=2+22≥ϕp(bB1)=bB1=34(2+22).
Then, hypotheses (H1)–(H4) are satisfied. Therefore, by Theorem 8, problem (42)-(43) has at least three positive solutions.


### 3.2. Quasilinear Elliptic Problem

We are interested in this section in the study of the following quasilinear elliptic problem:(46)-(ϕp(uΔ(t)))∇=f(u(t))+h(t),t∈(0,T)𝕋,uΔ(0)=0,u(T)-u(η)=0,
where η∈(0,T)𝕋. We assume the following hypotheses: (A1)
functionf:ℝ→ℝ0+ is continuous;(A2)
functionh:(0,T)𝕋→ℝ0+ is left dense continuous, that is,
(47)h∈ℂld((0,T)𝕋,ℝ0+),h∈L∞.Similarly as in Section  3.1, we prove existence of solutions by constructing an operator whose fixed points are solutions to (46). The main ingredient is, again, the Leggett-Williams fixed point theorem (Theorem 3). We can easily see that (46) is equivalent to the integral equation
(48)u(t)=ϕq(∫ηT‍(f(u(r))+h(r))∇r)+∫0t‍ϕq(∫sT‍(f(u(r))+h(r))∇r)Δs.
On the other hand, we have -(ϕp(uΔ))∇=f(u(t))+h(t). Since f, h≥0, we have (ϕp(uΔ))∇≤0 and ϕp(uΔ(t2))≤ϕp(uΔ(t1)) for any t1,t2∈[0,T]𝕋 with t1≤t2. It follows that uΔ(t2)≤uΔ(t1) for t1≤t2. Hence, uΔ(t) is a decreasing function on [0,T]𝕋. Then, u is concave. In order to apply Theorem 3 we define the cone
(49)P={u∈E∣uis  nonnegative,increasing  on[0,T]𝕋,and concave onE}.
For ξ∈(0,T/2) we also define the nonnegative continuous concave functional α:P→ℝ0+ by
(50)α(u)=mint∈[ξ,T-ξ]𝕋u(t),u∈P
and the operator F:P→E by
(51)Fu(t)=ϕq(∫ηT‍(f(u(r))+h(r))∇r)+∫0t‍ϕq(∫sT‍(f(u(r))+h(r))∇r)Δs.
It is easy to see that (46) has a solution u=u(t) if and only if u is a fixed point of the operator F. For convenience, we introduce the following notation:
(52)γ=(1+T)ϕq(T),A=a-α∥h∥∞1/(p-1)αa,whereα=ϕq(2p-2)ϕq(T)(T+1),B=ϕp(T-η).Theorem 13.
Suppose that hypotheses (A1) and (A2) are satisfied; there exist positive constantsa, b, c, and d with
(53)0<γa=a1<b<d-2p-2(T-ξ)ϕq(T-ξ)(∥h∥1/(p-1)+bB)<d<γc=c1
and, in addition to (A1) and (A2), that f satisfies (A3)
max0≤u≤af(u)≤ϕp(aA);(A4)
max0≤u≤cf(u)≤ϕp(cA);(A5)
minb≤u≤df(u)≥ϕp(bB). Then problem (46) has at least three positive solutions u1, u2, and u3, verifying
(54)∥u1∥<a,b<α(u2),∥u3∥>a,α(u3)<b.Proof.
As done for Theorem8, the proof is divided in several steps. We first show that F:P→P is completely continuous. Indeed, F is obviously continuous. Let
(55)Uδ={u∈P∣∥u∥≤δ}.
It is easy to see that for u∈Uδ there exists a constant c>0 such that |Fu(t)|≤c. On the other hand, let t1,t2∈(0,T)𝕋, u∈Uδ. Then there exists a positive constant c such that
(56)|Fu(t2)-Fu(t1)|≤c|t2-t1|,
which converges uniformly to zero when |t2-t1| tends to zero. Using the Arzela-Ascoli theorem we conclude that F:P→P is completely continuous.
We now show that(57)FPc1¯⊂Pc1¯,FPa1¯⊂Pa1¯.
For all u∈Pa1¯ we have 0≤u≤a1 and
(58)∥F(u)∥≤ϕq(∫ηT‍((aA)p-1+∥h∥∞)∇r)+∫0T‍ϕq(∫sT‍((aA)p-1+∥h∥∞)∇r)Δs≤ϕq(((aA)p-1+∥h∥∞)(T-η))+∫0T‍ϕq(((aA)p-1+∥h∥∞)(T-s))Δs≤ϕq((aA)p-1+∥h∥∞)ϕq(T)+ϕq((aA)p-1+∥h∥∞)∫0T‍ϕq(T-s)Δs≤ϕq((aA)p-1+(∥h∥∞1/(p-1))p-1)ϕq(T)+ϕq((aA)p-1+(∥h∥∞1/(p-1))p-1)ϕq(T)T≤ϕq((aA)p-1+(∥h∥∞1/(p-1))p-1)ϕq(T)(T+1).
Using the elementary inequality
(59)xp+yp≤2p-1(x+y)p
and the form of A, it follows that
(60)∥F(u)∥≤ϕq(T+1)(2p-2)(aA+∥h∥∞1/(p-1))≤ϕq(T+1)(2p-2)a=γa=a1.
Then FPa1¯⊂Pa1¯. In a similar way we prove that FPc1¯⊂Pc1¯.
Our following step is to show that(61){u∈P(α,b,d)∣α(u)>b}≠∅,α(Fu)>b,ifu∈P(α,b,d).
The first point is obvious. Let us prove the second part of (61). For u∈P(α,b,d) we have b≤u≤d, if t∈[ξ,T]𝕋. Then, using (A2) we have
(62)α(Fu)=Fu(ξ)≥ϕq(∫ηTf(u(r))∇r)+∫0ξϕq(∫sTf(u(r))∇r)Δs≥ϕq(∫ηTf(u(r))∇r)≥bBϕq(T-ξ)≥b.
Finally we prove that α(Fu)>b for all u∈P(α,b,c1) and ∥Fu∥>d:
(63)α(Fu)=Fu(ξ)=ϕq(∫ηT‍(f(u(r))+h(r))∇r)+∫0ξ‍ϕq(∫sT‍(f(u(r))+h(r))∇r)Δs=ϕq(∫ηT‍(f(u(r))+h(r))∇r)+∫0T‍ϕq(∫sT‍(f(u(r))+h(r))∇r)Δs-∫ξT‍ϕq(∫sT‍(f(u(r))+h(r))∇r)Δs≥∥Fu∥-∫ξT‍ϕq(∫sT‍(f(u(r))+h(r))∇r)Δs≥∥Fu∥-∫ξT‍ϕq(∫ξT‍(f(u(r))+h(r))∇r)Δs≥∥Fu∥-∫ξT‍ϕq((T-ξ)(∥h∥+ϕp(bB)))Δs≥∥Fu∥-(T-ξ)ϕq(T-ξ)ϕq(∥h∥+ϕp(bB)).
Using again the elementary inequality xp+yp≤2p-1(x+y)p we get that
(64)α(Fu)≥∥Fu∥-(T-ξ)ϕq(T-ξ)(∥h∥1/(p-1)+bB)≥d-2p-2(T-ξ)ϕq(T-ξ)(∥h∥1/(p-1)+bB)≥b.
By Theorem 3 there exist at least three positive solutions u1, u2, and u3 to (46) satisfying ∥u1∥<a, b<α(u2), ∥u3∥>a, and α(u3)<b.Example 14.
Let𝕋={1-(1/2)ℕ0}∪{1}, where ℕ0 denotes the set of all nonnegative integers. Consider the p-Laplacian dynamic equation
(65)[ϕp(uΔ(t))]∇+f(u(t))=0,t∈[0,1]𝕋,
satisfying the boundary conditions
(66)u(1)-u(12)=0,u∆(0)=0,
where p=3/2, q=3, a(t)≡1, h≡0, T=1, and
(67)f(u)={22,0≤u≤12,4(u-12)+22,12≤u≤32,4+22,32≤u≤25.
Choose a=1/2, b=3/2, c=25, and d=3. It is easy to see that γ=2, A=1, B=2/2, α=1, and
(68)max{f(u):u∈[0,12]}=22≤(aA)p-1=a=22,max{f(u):u∈[0,25]}=4+22≃4,707≤(cA)p-1=cp-1=c=5,min{f(u):u∈[32,3]}=4+22≃4,707≥(bB)p-1=bB≃1,02.
Therefore, by Theorem 13, problem (65)-(66) has at least three positive solutions.


### 3.3. Ap-Laplacian Functional Dynamic Equation on Time Scales with Delay

Let𝕋 be a time scale with 0,T∈𝕋κκ, -r∈𝕋 with -r≤0<T. We are concerned in this section with the existence of positive solutions to the p-Laplacian dynamic equation
(69)(ϕp(uΔ(t)))∇+λa(t)f(u(t),u(ω(t)))=0,t∈(0,T)𝕋,u(t)=ψ(t),t∈[-r,0]𝕋,u(0)-B0(uΔ(0))=0,uΔ(T)=0,
where λ>0. We define X=ℂld([0,T]𝕋,ℝ), which is a Banach space with the maximum norm ∥u∥=max[0,T]𝕋|u(t)|. We note that u is a solution to (69) if and only if
(70)u(t)={B0(ϕq(∫0Tλa(r)×f(u(r),u(ω(r)))∇r{34}){∫34})+∫0t‍ϕq(∫sT‍λa(r)×f(u(r),u(ω(r)))∇r{∫34})Δsift∈[0,T]𝕋,ψ(t)ift∈[-r,0]𝕋.
Let
(71)K={u∈X∣uisnonnegativeandconcaveonE}.
Clearly K is a cone in the Banach space X. For each u∈X, we extend u to [-r,0]𝕋 with u(t)=ψ(t) for t∈[-r,0]𝕋. We also define the nonnegative continuous concave functional α:P→ℝ0+ by
(72)α(u)=mint∈[ξ,T-ξ]𝕋u(t),ξ∈(0,T2),∀u∈K.
For t∈[0,T]𝕋, define Q:K→X as
(73)Qu(t)=B0(ϕq(∫0T‍λa(r)f(u(r),u(ω(r)))∇r))+∫0t‍ϕq(∫sT‍λa(r)f(u(r),u(ω(r)))∇r)Δs.Lemma 15.
Letu1 be a fixed point of Q in the cone K. Define
(74)u(t)={u1,t∈[0,T]𝕋,ψ(t),t∈[-r,0]𝕋.
It follows that (74) is a positive solution to (69) satisfying
(75)∥Qu∥≤(T+γ)λq-1×ϕq(∫0T‍a(r)f(u(r),u(ω(r)))∇r)fort∈[0,T]𝕋.Proof.
(76)
∥
Q
u
∥
=
(
Q
u
)
(
T
)
=
B
0
(
ϕ
q
(
∫
0
T
‍
λ
a
(
r
)
f
(
u
(
r
)
,
u
(
ω
(
r
)
)
)
∇
r
)
)
+
∫
0
T
‍
ϕ
q
(
∫
s
T
‍
λ
a
(
r
)
f
(
u
(
r
)
,
u
(
ω
(
r
)
)
)
∇
r
)
Δ
s
≤
(
T
+
γ
)
λ
q
-
1
ϕ
q
(
∫
0
T
‍
a
(
r
)
f
(
u
(
r
)
,
u
(
ω
(
r
)
)
)
∇
r
)
.From (73) and (75) it follows that (i)
Q(K)⊂K;(ii)
Q:K→K is completely continuous;(iii)
u(t)≥δ/(T+γ)∥u∥, t∈[0,T]𝕋.Depending on the signature of the delayω, we set the following two subsets of [0,T]𝕋:
(77)Y1:={t∈[0,T]𝕋∣ω(t)<0};Y2:={t∈[0,T]𝕋∣ω(t)≥0}.In the remainder of this section, we suppose thatY1 is nonempty and ∫Y1‍a(r)∇r>0. For convenience we also denote
(78)l:=ϕp(∫0T‍a(r)∇r)λq-1(T+γ),m:=ϕp(∫0T‍a(r)∇r)δλq-1.Theorem 16.
Suppose that there exist positive constantsa, b, c, and  d such that 0<a<b<δd/(T+γ)<d<c. Assume that the following hypotheses (C1)–(C8) hold: (C1)
f:ℝ0+×ℝ0+→ℝ0+ is continuous;(C2)
functiona:(0,T)𝕋→ℝ0+ is left-dense continuous;(C3)
ψ:[-r,0]𝕋→ℝ0+ is continuous;(C4)
ω:[0,T]𝕋→[-r,T]𝕋 is continuous, ω(t)≤t for all t;(C5)
B0:ℝ→ℝ is continuous and there are 0<δ≤γ such that
(79)δs≤B0(s)≤γsfors∈ℝ0+;(C6)
limx→0+f(x,ψ(s))/xp-1<lp-1, uniformly in s∈[-r,0]𝕋;(C7)
limx1→0+;x2→0+f(x1,x2)/max{x1p-1,x2p-1}<lp-1;(C8)
limx→∞f(x,ψ(s))/xp-1>mp-1, uniformly in s∈[-r,0]𝕋.Then, for each 0<λ<∞ the boundary value problem (69) has at least three positive solutions u1, u2, and u3 verifying
(80)∥u1∥<a,b<α(u2),∥u3∥>a,α(u3)<b.
Proof. The proof passes by three lemmas.Lemma 17.
The following relations hold:(81)QPa¯⊂Pa¯,QPc¯⊂Pc¯.Proof.
Using condition (C6) forε1>0 such that 0<x≤ε1, we have
(82)f(x,ψ(s))<(lx)p-1for  eachs∈[-r,0]𝕋.
Applying condition (C7) we get
(83)f(x1,x2)<max{x1p-1,x2p-1}lp-1
for ε2>0 such that 0<x1≤ε2, 0<x2≤ε2. Put ε=min{ε1,ε2}. Then for ∥u∥≤a and from (75) we have
(84)|Qu|≤∥Qu∥≤(T+γ)λq-1ϕq(∫0T‍a(r)f(u(r),u(ω(r)))∇r)=(T+γ)λq-1[ϕq(∫Y1‍a(r)f(u(r),ψ(ω(r)))∇r=(T+γ)λq-1+∫Y2‍a(r)f(u(r),u(ω(r)))∇r)]≤l(T+γ)λq-1∥u∥ϕq(∫0T‍a(r)∇r)≤l(T+γ)λq-1aϕq(∫0T‍a(r)∇r)=a.
Then QPa¯⊂Pa¯. Similarly one can show that QPc¯⊂Pc¯.Lemma 18.
The set(85){u∈P(α,b,d)∣α(u)>b}
is nonempty, and
(86)α(Qu)>b,ifu∈P(α,b,d).Proof.
Applying hypothesis(C8) we have
(87)f(u,ψ(s))>ϕp(mu)for eachs∈[-r,0]𝕋.
We also have u(t)≥δ/(T+γ)∥u∥. Let u∈P(α,b,d). Then, b≤u(t)≤d. Hence,
(88)α(Qu)=(Qu)(T-ξ)≥δT+γ∥Qu∥≥δϕq(∫0T‍λa(r)f(u(r),u(ω(r)))∇r)≥δλq-1ϕq(∫Y1‍a(r)f(u(r),ψ(ω(r)))∇rδλq-1ϕq+∫Y2‍a(r)f(u(r),ψ(ω(r)))∇r)≥δλq-1ϕq(∫Y1‍a(r)f(u(r),ψ(ω(r)))∇r)≥mδλq-1mint∈Y1{u(t)}ϕq(∫Y1‍a(r)∇r)≥bmδλq-1ϕq(∫Y1‍a(r)∇r)≥b.Lemma 19.
For allu∈P(α,b,c) and ∥Qu∥>d one has α(Qu)>b.Proof.
Using the fact thatδ≤T+γ, we have
(89)α(Qu)=(Qu)(T-ξ)≥δT+γ∥Qu∥>δdT+γ≥b.Applying the Leggett-Williams theorem (Theorem3), the proof of Theorem 16 is complete.Example 20.
Let𝕋=[-3/4,-1/4]∪{0,3/4}∪{(1/2)ℕ0}, where ℕ0 denotes the set of all nonnegative integers. Consider the following p-Laplacian functional dynamic equation on the time scale 𝕋:
(90)[Φp(uΔ(t))]∇+(u1+u2)2=0,t∈(0,1)𝕋,ψ(t)≡0,t∈[-34,0]𝕋,u(0)-B0(uΔ(14))=0,uΔ(1)=0,
where T=1, p=3/2, q=3, a(t)≡1,B0(s)=s, w(t):[0,1]𝕋→[-3/4,1]𝕋 with w(t)=t-3/4, r=3/4, η=1/4, l=1/2, m=1, and f(u,ψ(t))=u2, f(u1,u2)=(u1+u2)2. We deduce that Y1=[0,3/4)𝕋, Y2=[3/4,1]𝕋. It is easy to see that hypotheses (C1)–(C5) are verified. On the other hand, notice that limx→0+f(x,ψ(s))/xp-1=0<lp-1 and limx→∞f(x,ψ(s))/(xp-1)=+∞>mp-1. Thus, hypotheses (C6)–(C8) are obviously satisfied. Then, by Theorem 16, problem (90) has at least three positive solutions of the form
(91)u(t)={ui(t),t∈[0,1]𝕋,i=1,2,3,ψ(t),t∈[-34,0]𝕋.


## 3.1. Nonlocal Thermistor Problem

By a solutionu:𝕋→ℝ of (1)-(2) we mean a delta differentiable function such that uΔ and (|uΔ|p-2uΔ)∇ are both continuous on 𝕋κκ and u satisfies (1)-(2). We consider the following hypothesis:(H1)
f:ℝ→ℝ+ is a continuous function.Lemma 4 (see Lemma  3.1 of [17]).
Assume that hypothesis (H1) on functionf is satisfied. Then u is a solution to (1)-(2) if and only if u∈E is a solution to the integral equation
(13)u(t)=-∫0t‍ϕq(g(s))Δs+B,
where
(14)g(s)=∫0sλh(u(r))∇r-A,A=ϕp(uΔ(0))=-λβ1-β∫0ηh(u(r))∇r,h(u(t))=λf(u(t))(∫0T‍f(u(τ))∇τ)2,B=u(0)=11-β{∫0T‍ϕq(g(s))Δs-β∫0η‍ϕq(g(s))Δs}.Lemma 5.
Suppose (H1) holds. Then a solutionu to (1)-(2) satisfies u(t)≥0 for t∈(0,T)𝕋.Proof.
We haveA=-λβ/(1-β)∫0η‍h(u(r))∇r≤0. Then, g(s)=λ∫0s‍h(u(r))∇r-A≥0. It follows that ϕp(g(s))≥0. Since 0<β<1, we also have
(15)u(0)=B=11-β{∫0T‍ϕq(g(s))Δs-β∫0η‍ϕq(g(s))Δs}≥11-β{β∫0T‍ϕq(g(s))Δs-β∫0η‍ϕq(g(s))Δs}≥0,u(T)=u(0)-∫0T‍ϕq(g(s))Δs=-β1-β∫0η‍ϕq(g(s))Δs+11-β∫0T‍ϕq(g(s))Δs-∫0T‍ϕq(g(s))Δs=-β1-β∫0η‍ϕq(g(s))Δs+β1-β∫0T‍ϕq(g(s))Δs=β1-β{∫0T‍ϕq(g(s))Δs-∫0η‍ϕq(g(s))Δs}≥0.
If t∈(0,T)𝕋, then
(16)u(t)=u(0)-∫0t‍ϕq(g(s))Δs≥-∫0T‍ϕq(g(s))Δs+u(0)=u(T)≥0.
Consequently, u(t)≥0 for t∈(0,T)𝕋.On the other hand, we haveϕp(uΔ(s))=ϕp(uΔ(0))-∫0s‍λh(u(r))∇r≤0. Since A=ϕp(uΔ(0))≤0, then uΔ≤0. This means that ||u||=u(0), inft∈(0,T)𝕋u(t)=u(T). Moreover, ϕp(uΔ(s)) is nonincreasing, which implies with the monotonicity of ϕp that uΔ is a nonincreasing function on (0,T)𝕋. Hence, u is concave. In order to apply Theorem 3, let us define the cone P⊂E by
(17)P={u∈E∣uis nonnegative,decreasing on[0,T]𝕋and concave onE}.
We also define the nonnegative continuous concave functional α:P→ℝ0+ by
(18)α(u)=mint∈[ξ,T-ξ]𝕋u(t),ξ∈(0,T2),∀u∈P.
It is easy to see that problem (1)-(2) has a solution u=u(t) if and only if u is a fixed point of the operator G:P→E defined by
(19)Gu(t)=-∫0t‍ϕq(g(s))Δs+B,
where g and B are as in Lemma 4.Lemma 6.
LetG be defined by (19). Then,(i)
G(P)⊆P;(ii)
G:P→P is completely continuous.Proof.
(i)
holds clearly from above.
(ii)
Suppose thatD⊆P is a bounded set and let u∈D. Then,
(20)|Gu(t)|=|-∫0t‍ϕq(g(s))Δs+B|≤|-∫0t‍ϕq(∫0s‍λf(u(r))(∫0T‍f(u(τ))∇τ)2∇r-A)Δs|+|B|≤∫0T‍ϕq(∫0s‍λsupu∈Df(u)(Tinfu∈Df(u))2∇r-A)Δs+|B|,|A|=|λβ1-β∫0η‍h(u(r))∇r|=|λβ1-β∫0η‍f(u(r))(∫0T‍f(u(τ))∇τ)2∇r|≤λβ1-βsupu∈Df(u)(Tinfu∈Df(u))2η.
In the same way, we have(21)|B|≤11-β∫0T‍ϕq(g(s))Δs≤11-β∫0T‍ϕq(λsupu∈Df(u)(Tinfu∈Df(u))2(s+β1-βη))Δs.
It follows that
(22)|Gu(t)|≤∫0T‍ϕq(λsupu∈Df(u)(Tinfu∈Df(u))2(s+βη1-β))Δs+|B|.
As a consequence, we get
(23)∥Gu∥≤2-β1-β∫0T‍ϕq(λsupu∈Df(u)(Tinfu∈Df(u))2(s+βη1-β))Δs≤21-βϕq(λsupu∈Df(u)(Tinfu∈Df(u))2)∫0T‍ϕq(s+βη1-β)Δs.
Then G(D) is bounded on the whole bounded set D. Moreover, if t1,t2∈[0,T]𝕋 and u∈D, then we have for a positive constant c(24)|Gu(t2)-Gu(t1)|≤|∫t1t2‍ϕq(g(s))Δs|≤c|t2-t1|.
We see that the right-hand side of the above inequality goes uniformly to zero when |t2-t1|→0. Then by a standard application of the Arzela-Ascoli theorem we have that G:P→P is completely continuous.We can also easily obtain the following properties.Lemma 7.
(i)α(u)=u(T-ξ)≤∥u∥ for all u∈P;
(ii)α(Gu)=(Gu)(T-ξ);
(iii)∥Gu∥=Gu(0).We now state the main result of Section  3.1.Theorem 8.
Suppose that(H1) is verified and there exist positive constants a, b, c, and d such that 0<ζa=a1<b<d-T2cϕq(1/T)<d<ζc=c1, with ζ=(T2/(1-β)) ϕq(1/T). One further imposes f to satisfy the following hypotheses:(H2)
min0≤u≤a1f(u)≥λ2/(T(1-β)ϕp(a)) uniformly for all t∈[0,T]𝕋;(H3)
min0≤u≤c1f(u)≥λ2/(T(1-β)ϕp(c)) uniformly for all t∈[0,T]𝕋;(H4)
minb≤u≤df(u)≥ϕp(bB1) uniformly for all t∈[0,T]𝕋, where
(25)B1=(1-β)βξ|ϕp(T-ξ)|ϕp(λ(Tsupb≤u≤df(u))2).Then the boundary value problem (1)-(2) has at least three positive solutions u1, u2, and u3, verifying
(26)∥u1∥<a,b<α(u2),∥u3∥>a,α(u3)<b.Proof.
The proof passes by several lemmas. We have already seen in Lemma6 that the operator G is completely continuous. We now show the following Lemma.Lemma 9.
The following relations hold:(27)GPc1¯⊂Pc1¯,GPa1¯⊂Pa1¯.Proof.
Obviously,GPa1¯⊂P. Moreover, for all u∈Pa1¯, we have 0≤u(t)≤a1. On the other hand we have
(28)Gu(t)=-∫0t‍ϕq(g(s))Δs+11-β∫0T‍ϕq(g(s))Δs-β1-β∫0η‍ϕq(g(s))Δs,
and for all u∈GPa1¯ we have 0≤u(t)≤a1. Then,
(29)|Gu|≤11-β∫0T‍ϕq(g(s))Δs.
We have
(30)h(u(t))=λf(u(t))(∫0T‍f(u(τ))∇τ)2,g(s)=∫0s‍λh(u(r))∇r+λβ1-β∫0η‍h(u(r))∇r≤(λ+λβ1-β∫0T‍h(u(r))∇r)≤λ1-β∫0T‍h(u(r))∇r.
Using (H2) it follows that
(31)ϕq(g(s))≤aTϕq(1T).
Then we get
(32)|Gu|≤a1,GPa1¯⊂Pa1¯.
Similarly, using (H3) we get GPc1¯⊂Pc1¯.Lemma 10.
The set(33){u∈P(α,b,d)∣α(u)>b}
is nonempty, and
(34)α(Gu)>b,ifu∈P(α,b,d).Proof.
Letu=(b+d)/2. Then, u∈P,∥u∥=(b+d)/2≤d, and α(u)≥(b+d)/2>b. The first part of the lemma is proved. For u∈P(α,b,d) we have b≤u≤d. If t∈[ξ,T]𝕋, then
(35)α(Gu)=(Gu)(T-ξ)=-∫0T-ξ‍ϕq(g(s))Δs+B≥β1-β∫T-ξT‍ϕq(g(s))Δs.
Since A≤0, we have by using (H4)
(36)g(s)=λ∫0s‍h(u(r))∇r-A≥λ∫0s‍h(u(r))∇r≥λ∫0s‍f(u)(Tsupb≤u≤df(u))2∇u≥λ(bB1)p-1(Tsupb≤u≤df(u))2s.
Using the fact that ϕq is nondecreasing we get
(37)ϕq(g(s))≥ϕq(λ(bB1)p-1(Tsupb≤u≤df(u))2s)≥bB1ϕq(λ(Tsupb≤u≤df(u))2)ϕq(s).
Using the expression of B1(38)α(Gu)≥β1-βbB1ϕq(λ(Tsupb≤u≤df(u))2)∫T-ξT‍ϕq(s)Δs≥bB1β1-βϕq(λ(Tsupb≤u≤df(u))2)ϕq(T-ξ)ξ≥b.Lemma 11.
For allu∈P(α,b,c1) with ∥Gu∥>d one has
(39)α(Gu)>b.Proof.
Ifu∈P(α,b,c1) and ∥Gu∥>d, then 0≤u(t)≤c1. Using hypothesis (H3) and the fact that 0<β<1, it follows that
(40)α(Gu)=Gu(T-ξ)=-∫0T-ξ‍ϕq(g(s))Δs+B≥-∫0T‍ϕq(g(s))Δs+Gu(0)≥∥Gu∥-T2cϕq(1T)≥d-T2cϕq(1T)>b.
Gathering Lemmas 4 to 11 and applying Theorem 3, there exist at least three positive solutions u1, u2, and u3 to (1)-(2) verifying
(41)∥u1∥<a,b<α(u2),∥u3∥>a,α(u3)<b.Example 12.
Let𝕋={1-(1/2)ℕ0}∪{1}, where ℕ0 denotes the set of all nonnegative integers. Consider the p-Laplacian dynamic equation
(42)-(ϕp(uΔ(t)))∇=λf(u(t))(∫0T‍f(u(τ))∇τ)2,t∈(0,T)𝕋,
satisfying the boundary conditions
(43)ϕp(uΔ(0))-β(ϕp(uΔ(14)))=0,u(1)-βu(14)=0,
where p=3/2, q=3, η=1/4, β=1/2, λ=1, T=1, and
(44)f(u)={22,0≤u≤1,4(u-1)+22,1≤u≤32,2+22,32≤u≤10,2u+22-18,10≤u≤16.
Choose a1=1=2a, b=3/2, c1=16=2c, and d=10. It is easy to see that ζ=2, B1=1/(2(2+2)) and
(45)min{f(u):u∈[0,a1]}=22≥λ2T(1-β)ϕp(a)=2a=22,min{f(u):u∈[0,c1]}=22≥λ2T(1-β)ϕp(c)=2c=12,min{f(u):u∈[b,d]}=2+22≥ϕp(bB1)=bB1=34(2+22).
Then, hypotheses (H1)–(H4) are satisfied. Therefore, by Theorem 8, problem (42)-(43) has at least three positive solutions.


## 3.2. Quasilinear Elliptic Problem

We are interested in this section in the study of the following quasilinear elliptic problem:(46)-(ϕp(uΔ(t)))∇=f(u(t))+h(t),t∈(0,T)𝕋,uΔ(0)=0,u(T)-u(η)=0,
where η∈(0,T)𝕋. We assume the following hypotheses: (A1)
functionf:ℝ→ℝ0+ is continuous;(A2)
functionh:(0,T)𝕋→ℝ0+ is left dense continuous, that is,
(47)h∈ℂld((0,T)𝕋,ℝ0+),h∈L∞.Similarly as in Section  3.1, we prove existence of solutions by constructing an operator whose fixed points are solutions to (46). The main ingredient is, again, the Leggett-Williams fixed point theorem (Theorem 3). We can easily see that (46) is equivalent to the integral equation
(48)u(t)=ϕq(∫ηT‍(f(u(r))+h(r))∇r)+∫0t‍ϕq(∫sT‍(f(u(r))+h(r))∇r)Δs.
On the other hand, we have -(ϕp(uΔ))∇=f(u(t))+h(t). Since f, h≥0, we have (ϕp(uΔ))∇≤0 and ϕp(uΔ(t2))≤ϕp(uΔ(t1)) for any t1,t2∈[0,T]𝕋 with t1≤t2. It follows that uΔ(t2)≤uΔ(t1) for t1≤t2. Hence, uΔ(t) is a decreasing function on [0,T]𝕋. Then, u is concave. In order to apply Theorem 3 we define the cone
(49)P={u∈E∣uis  nonnegative,increasing  on[0,T]𝕋,and concave onE}.
For ξ∈(0,T/2) we also define the nonnegative continuous concave functional α:P→ℝ0+ by
(50)α(u)=mint∈[ξ,T-ξ]𝕋u(t),u∈P
and the operator F:P→E by
(51)Fu(t)=ϕq(∫ηT‍(f(u(r))+h(r))∇r)+∫0t‍ϕq(∫sT‍(f(u(r))+h(r))∇r)Δs.
It is easy to see that (46) has a solution u=u(t) if and only if u is a fixed point of the operator F. For convenience, we introduce the following notation:
(52)γ=(1+T)ϕq(T),A=a-α∥h∥∞1/(p-1)αa,whereα=ϕq(2p-2)ϕq(T)(T+1),B=ϕp(T-η).Theorem 13.
Suppose that hypotheses (A1) and (A2) are satisfied; there exist positive constantsa, b, c, and d with
(53)0<γa=a1<b<d-2p-2(T-ξ)ϕq(T-ξ)(∥h∥1/(p-1)+bB)<d<γc=c1
and, in addition to (A1) and (A2), that f satisfies (A3)
max0≤u≤af(u)≤ϕp(aA);(A4)
max0≤u≤cf(u)≤ϕp(cA);(A5)
minb≤u≤df(u)≥ϕp(bB). Then problem (46) has at least three positive solutions u1, u2, and u3, verifying
(54)∥u1∥<a,b<α(u2),∥u3∥>a,α(u3)<b.Proof.
As done for Theorem8, the proof is divided in several steps. We first show that F:P→P is completely continuous. Indeed, F is obviously continuous. Let
(55)Uδ={u∈P∣∥u∥≤δ}.
It is easy to see that for u∈Uδ there exists a constant c>0 such that |Fu(t)|≤c. On the other hand, let t1,t2∈(0,T)𝕋, u∈Uδ. Then there exists a positive constant c such that
(56)|Fu(t2)-Fu(t1)|≤c|t2-t1|,
which converges uniformly to zero when |t2-t1| tends to zero. Using the Arzela-Ascoli theorem we conclude that F:P→P is completely continuous.
We now show that(57)FPc1¯⊂Pc1¯,FPa1¯⊂Pa1¯.
For all u∈Pa1¯ we have 0≤u≤a1 and
(58)∥F(u)∥≤ϕq(∫ηT‍((aA)p-1+∥h∥∞)∇r)+∫0T‍ϕq(∫sT‍((aA)p-1+∥h∥∞)∇r)Δs≤ϕq(((aA)p-1+∥h∥∞)(T-η))+∫0T‍ϕq(((aA)p-1+∥h∥∞)(T-s))Δs≤ϕq((aA)p-1+∥h∥∞)ϕq(T)+ϕq((aA)p-1+∥h∥∞)∫0T‍ϕq(T-s)Δs≤ϕq((aA)p-1+(∥h∥∞1/(p-1))p-1)ϕq(T)+ϕq((aA)p-1+(∥h∥∞1/(p-1))p-1)ϕq(T)T≤ϕq((aA)p-1+(∥h∥∞1/(p-1))p-1)ϕq(T)(T+1).
Using the elementary inequality
(59)xp+yp≤2p-1(x+y)p
and the form of A, it follows that
(60)∥F(u)∥≤ϕq(T+1)(2p-2)(aA+∥h∥∞1/(p-1))≤ϕq(T+1)(2p-2)a=γa=a1.
Then FPa1¯⊂Pa1¯. In a similar way we prove that FPc1¯⊂Pc1¯.
Our following step is to show that(61){u∈P(α,b,d)∣α(u)>b}≠∅,α(Fu)>b,ifu∈P(α,b,d).
The first point is obvious. Let us prove the second part of (61). For u∈P(α,b,d) we have b≤u≤d, if t∈[ξ,T]𝕋. Then, using (A2) we have
(62)α(Fu)=Fu(ξ)≥ϕq(∫ηTf(u(r))∇r)+∫0ξϕq(∫sTf(u(r))∇r)Δs≥ϕq(∫ηTf(u(r))∇r)≥bBϕq(T-ξ)≥b.
Finally we prove that α(Fu)>b for all u∈P(α,b,c1) and ∥Fu∥>d:
(63)α(Fu)=Fu(ξ)=ϕq(∫ηT‍(f(u(r))+h(r))∇r)+∫0ξ‍ϕq(∫sT‍(f(u(r))+h(r))∇r)Δs=ϕq(∫ηT‍(f(u(r))+h(r))∇r)+∫0T‍ϕq(∫sT‍(f(u(r))+h(r))∇r)Δs-∫ξT‍ϕq(∫sT‍(f(u(r))+h(r))∇r)Δs≥∥Fu∥-∫ξT‍ϕq(∫sT‍(f(u(r))+h(r))∇r)Δs≥∥Fu∥-∫ξT‍ϕq(∫ξT‍(f(u(r))+h(r))∇r)Δs≥∥Fu∥-∫ξT‍ϕq((T-ξ)(∥h∥+ϕp(bB)))Δs≥∥Fu∥-(T-ξ)ϕq(T-ξ)ϕq(∥h∥+ϕp(bB)).
Using again the elementary inequality xp+yp≤2p-1(x+y)p we get that
(64)α(Fu)≥∥Fu∥-(T-ξ)ϕq(T-ξ)(∥h∥1/(p-1)+bB)≥d-2p-2(T-ξ)ϕq(T-ξ)(∥h∥1/(p-1)+bB)≥b.
By Theorem 3 there exist at least three positive solutions u1, u2, and u3 to (46) satisfying ∥u1∥<a, b<α(u2), ∥u3∥>a, and α(u3)<b.Example 14.
Let𝕋={1-(1/2)ℕ0}∪{1}, where ℕ0 denotes the set of all nonnegative integers. Consider the p-Laplacian dynamic equation
(65)[ϕp(uΔ(t))]∇+f(u(t))=0,t∈[0,1]𝕋,
satisfying the boundary conditions
(66)u(1)-u(12)=0,u∆(0)=0,
where p=3/2, q=3, a(t)≡1, h≡0, T=1, and
(67)f(u)={22,0≤u≤12,4(u-12)+22,12≤u≤32,4+22,32≤u≤25.
Choose a=1/2, b=3/2, c=25, and d=3. It is easy to see that γ=2, A=1, B=2/2, α=1, and
(68)max{f(u):u∈[0,12]}=22≤(aA)p-1=a=22,max{f(u):u∈[0,25]}=4+22≃4,707≤(cA)p-1=cp-1=c=5,min{f(u):u∈[32,3]}=4+22≃4,707≥(bB)p-1=bB≃1,02.
Therefore, by Theorem 13, problem (65)-(66) has at least three positive solutions.


## 3.3. Ap-Laplacian Functional Dynamic Equation on Time Scales with Delay

Let𝕋 be a time scale with 0,T∈𝕋κκ, -r∈𝕋 with -r≤0<T. We are concerned in this section with the existence of positive solutions to the p-Laplacian dynamic equation
(69)(ϕp(uΔ(t)))∇+λa(t)f(u(t),u(ω(t)))=0,t∈(0,T)𝕋,u(t)=ψ(t),t∈[-r,0]𝕋,u(0)-B0(uΔ(0))=0,uΔ(T)=0,
where λ>0. We define X=ℂld([0,T]𝕋,ℝ), which is a Banach space with the maximum norm ∥u∥=max[0,T]𝕋|u(t)|. We note that u is a solution to (69) if and only if
(70)u(t)={B0(ϕq(∫0Tλa(r)×f(u(r),u(ω(r)))∇r{34}){∫34})+∫0t‍ϕq(∫sT‍λa(r)×f(u(r),u(ω(r)))∇r{∫34})Δsift∈[0,T]𝕋,ψ(t)ift∈[-r,0]𝕋.
Let
(71)K={u∈X∣uisnonnegativeandconcaveonE}.
Clearly K is a cone in the Banach space X. For each u∈X, we extend u to [-r,0]𝕋 with u(t)=ψ(t) for t∈[-r,0]𝕋. We also define the nonnegative continuous concave functional α:P→ℝ0+ by
(72)α(u)=mint∈[ξ,T-ξ]𝕋u(t),ξ∈(0,T2),∀u∈K.
For t∈[0,T]𝕋, define Q:K→X as
(73)Qu(t)=B0(ϕq(∫0T‍λa(r)f(u(r),u(ω(r)))∇r))+∫0t‍ϕq(∫sT‍λa(r)f(u(r),u(ω(r)))∇r)Δs.Lemma 15.
Letu1 be a fixed point of Q in the cone K. Define
(74)u(t)={u1,t∈[0,T]𝕋,ψ(t),t∈[-r,0]𝕋.
It follows that (74) is a positive solution to (69) satisfying
(75)∥Qu∥≤(T+γ)λq-1×ϕq(∫0T‍a(r)f(u(r),u(ω(r)))∇r)fort∈[0,T]𝕋.Proof.
(76)
∥
Q
u
∥
=
(
Q
u
)
(
T
)
=
B
0
(
ϕ
q
(
∫
0
T
‍
λ
a
(
r
)
f
(
u
(
r
)
,
u
(
ω
(
r
)
)
)
∇
r
)
)
+
∫
0
T
‍
ϕ
q
(
∫
s
T
‍
λ
a
(
r
)
f
(
u
(
r
)
,
u
(
ω
(
r
)
)
)
∇
r
)
Δ
s
≤
(
T
+
γ
)
λ
q
-
1
ϕ
q
(
∫
0
T
‍
a
(
r
)
f
(
u
(
r
)
,
u
(
ω
(
r
)
)
)
∇
r
)
.From (73) and (75) it follows that (i)
Q(K)⊂K;(ii)
Q:K→K is completely continuous;(iii)
u(t)≥δ/(T+γ)∥u∥, t∈[0,T]𝕋.Depending on the signature of the delayω, we set the following two subsets of [0,T]𝕋:
(77)Y1:={t∈[0,T]𝕋∣ω(t)<0};Y2:={t∈[0,T]𝕋∣ω(t)≥0}.In the remainder of this section, we suppose thatY1 is nonempty and ∫Y1‍a(r)∇r>0. For convenience we also denote
(78)l:=ϕp(∫0T‍a(r)∇r)λq-1(T+γ),m:=ϕp(∫0T‍a(r)∇r)δλq-1.Theorem 16.
Suppose that there exist positive constantsa, b, c, and  d such that 0<a<b<δd/(T+γ)<d<c. Assume that the following hypotheses (C1)–(C8) hold: (C1)
f:ℝ0+×ℝ0+→ℝ0+ is continuous;(C2)
functiona:(0,T)𝕋→ℝ0+ is left-dense continuous;(C3)
ψ:[-r,0]𝕋→ℝ0+ is continuous;(C4)
ω:[0,T]𝕋→[-r,T]𝕋 is continuous, ω(t)≤t for all t;(C5)
B0:ℝ→ℝ is continuous and there are 0<δ≤γ such that
(79)δs≤B0(s)≤γsfors∈ℝ0+;(C6)
limx→0+f(x,ψ(s))/xp-1<lp-1, uniformly in s∈[-r,0]𝕋;(C7)
limx1→0+;x2→0+f(x1,x2)/max{x1p-1,x2p-1}<lp-1;(C8)
limx→∞f(x,ψ(s))/xp-1>mp-1, uniformly in s∈[-r,0]𝕋.Then, for each 0<λ<∞ the boundary value problem (69) has at least three positive solutions u1, u2, and u3 verifying
(80)∥u1∥<a,b<α(u2),∥u3∥>a,α(u3)<b.
Proof. The proof passes by three lemmas.Lemma 17.
The following relations hold:(81)QPa¯⊂Pa¯,QPc¯⊂Pc¯.Proof.
Using condition (C6) forε1>0 such that 0<x≤ε1, we have
(82)f(x,ψ(s))<(lx)p-1for  eachs∈[-r,0]𝕋.
Applying condition (C7) we get
(83)f(x1,x2)<max{x1p-1,x2p-1}lp-1
for ε2>0 such that 0<x1≤ε2, 0<x2≤ε2. Put ε=min{ε1,ε2}. Then for ∥u∥≤a and from (75) we have
(84)|Qu|≤∥Qu∥≤(T+γ)λq-1ϕq(∫0T‍a(r)f(u(r),u(ω(r)))∇r)=(T+γ)λq-1[ϕq(∫Y1‍a(r)f(u(r),ψ(ω(r)))∇r=(T+γ)λq-1+∫Y2‍a(r)f(u(r),u(ω(r)))∇r)]≤l(T+γ)λq-1∥u∥ϕq(∫0T‍a(r)∇r)≤l(T+γ)λq-1aϕq(∫0T‍a(r)∇r)=a.
Then QPa¯⊂Pa¯. Similarly one can show that QPc¯⊂Pc¯.Lemma 18.
The set(85){u∈P(α,b,d)∣α(u)>b}
is nonempty, and
(86)α(Qu)>b,ifu∈P(α,b,d).Proof.
Applying hypothesis(C8) we have
(87)f(u,ψ(s))>ϕp(mu)for eachs∈[-r,0]𝕋.
We also have u(t)≥δ/(T+γ)∥u∥. Let u∈P(α,b,d). Then, b≤u(t)≤d. Hence,
(88)α(Qu)=(Qu)(T-ξ)≥δT+γ∥Qu∥≥δϕq(∫0T‍λa(r)f(u(r),u(ω(r)))∇r)≥δλq-1ϕq(∫Y1‍a(r)f(u(r),ψ(ω(r)))∇rδλq-1ϕq+∫Y2‍a(r)f(u(r),ψ(ω(r)))∇r)≥δλq-1ϕq(∫Y1‍a(r)f(u(r),ψ(ω(r)))∇r)≥mδλq-1mint∈Y1{u(t)}ϕq(∫Y1‍a(r)∇r)≥bmδλq-1ϕq(∫Y1‍a(r)∇r)≥b.Lemma 19.
For allu∈P(α,b,c) and ∥Qu∥>d one has α(Qu)>b.Proof.
Using the fact thatδ≤T+γ, we have
(89)α(Qu)=(Qu)(T-ξ)≥δT+γ∥Qu∥>δdT+γ≥b.Applying the Leggett-Williams theorem (Theorem3), the proof of Theorem 16 is complete.Example 20.
Let𝕋=[-3/4,-1/4]∪{0,3/4}∪{(1/2)ℕ0}, where ℕ0 denotes the set of all nonnegative integers. Consider the following p-Laplacian functional dynamic equation on the time scale 𝕋:
(90)[Φp(uΔ(t))]∇+(u1+u2)2=0,t∈(0,1)𝕋,ψ(t)≡0,t∈[-34,0]𝕋,u(0)-B0(uΔ(14))=0,uΔ(1)=0,
where T=1, p=3/2, q=3, a(t)≡1,B0(s)=s, w(t):[0,1]𝕋→[-3/4,1]𝕋 with w(t)=t-3/4, r=3/4, η=1/4, l=1/2, m=1, and f(u,ψ(t))=u2, f(u1,u2)=(u1+u2)2. We deduce that Y1=[0,3/4)𝕋, Y2=[3/4,1]𝕋. It is easy to see that hypotheses (C1)–(C5) are verified. On the other hand, notice that limx→0+f(x,ψ(s))/xp-1=0<lp-1 and limx→∞f(x,ψ(s))/(xp-1)=+∞>mp-1. Thus, hypotheses (C6)–(C8) are obviously satisfied. Then, by Theorem 16, problem (90) has at least three positive solutions of the form
(91)u(t)={ui(t),t∈[0,1]𝕋,i=1,2,3,ψ(t),t∈[-34,0]𝕋.



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*Source: 145050-2013-02-05.xml*