We consider a fourth-order four-point boundary value problem for dynamic equations on time scales. By the upper and lower solution method, some results on the existence of solutions of the fourth-order four-point boundary value problem on time scales are obtained. An example is also included to illustrate our results.

1. Introduction

Let 𝕋 be a closed nonempty subset of ℝ, and let 𝕋 have the subspace topology inherited from the Euclidean topology on ℝ. In some of the current literature, 𝕋 is called a time scale (or measure chain). For notation, we shall use the convention that, for each interval J of ℝ, J will denote the time scales interval, that is, J≔J∩𝕋. Some preliminary definitions and theorems on time scales can be found in the books [1, 2], which are excellent references for calculus of time scales.

In this paper, let 𝕋 be a time scale and σ(t) the forward jump function in 𝕋. We are concerned with the following fourth-order four-point boundary value problem on time scales 𝕋:
yΔ4(t)-q(t)yΔ2(σ(t))=f(t,y(σ(t)),yΔ2(σ(t))),y(a)=0,y(σ2(b))+λyΔ(σ2(b))=0,ζyΔ2(ξ1)-ηyΔ3(ξ1)=0,γyΔ2(ξ2)+δyΔ3(ξ2)=0,
for t∈[a,b]⊂𝕋,a≤ξ1<ξ2≤σ(b). We will assume that the following conditions are satisfied.

ζ,γ,η,δ≥0,λ≥σ3(b)-σ2(b),

q(t)≥0. If q(t)≡0, then ζ+γ>0,

k=ζδ+ηγ+ζγ(ξ2-ξ1)>0,η-ζξ1≥0,δ-γ(σ(b)-ξ2)≥0.

The upper and lower solution method has been used to deal with the boundary value problems for dynamic equations in recent years. In most of these studies, two-point boundary value problem for second-order dynamic equations is considered [3–7].

Pang and Bai [8] studied the following fourth-order four-point BVP on time scales:
uΔΔΔΔ(t)=f(t,u(σ(t)),uΔΔ(t)),t∈[0,1],u(0)=u(σ4(1))=0,αyΔ2(ξ1)-βyΔ3(ξ1)=0,γyΔ2(ξ2)+ηyΔ3(ξ2)=0
for 0≤ξ1<ξ2≤σ(b), and f∈𝒞([0,1]×ℝ×ℝ,ℝ) and α,β,γ,η are nonnegative constants satisfying αη+βγ+αγ(ξ2-ξ1)>0. They establish criteria for the existence of a solution by developing the upper and lower solution method and the monotone iterative technique. Our problem is more general than the problems in [8], and our results are even new for the differential equations as well as for dynamic equations on general time scales.

2. Preliminaries

To prove the main results in this paper, we will employ several lemmas. We consider the linear boundary value problem
yΔ2(t)-q(t)y(σ(t))=h(t),t∈[ξ1,ρ(ξ2)],ζy(ξ1)-ηyΔ(ξ1)=0,γy(ξ2)+δyΔ(ξ2)=0.

Denote by φ and ψ, the solutions of the corresponding homogeneous equation
yΔ2(t)-q(t)y(σ(t))=0,t∈[ξ1,ρ(ξ2)]
under the initial conditions
φ(ξ1)=η,φΔ(ξ1)=ζ,ψ(ξ2)=δ,ψΔ(ξ2)=-γ,
so that φ and ψ satisfy the first and second boundary conditions of (2.1), respectively. Let us set
D:=ζψ(ξ1)-ηψΔ(ξ1)=δφΔ(ξ2)+γφ(ξ2).
Using the initial conditions (2.3), we can deduce from (2.2) for φ and ψ the following equations:
φ(t)=η+ζ(t-ξ1)+∫ξ1t∫ξ1τq(s)φ(σ(s))ΔsΔτ,ψ(t)=δ+γ(ξ2-t)+∫tξ2∫τξ2q(s)ψ(σ(s))ΔsΔτ.

Lemma 2.1.

Under the conditions (H1) and (H2), the following inequalities
φ(t)≥0,t∈[ξ1,σ(ξ2)];ψ(t)≥0,t∈[ξ1,ξ2];φΔ(t)≥0,t∈[ξ1,ξ2];ψΔ(t)≤0,t∈[ξ1,ξ2];
yield.

Proof.

We apply the induction principle for time scales to the statement
A(t):φ(t)≥0,φΔ(t)≥0,
where t∈[ξ1,ξ2].

(I) The statement A(ξ1) is true, since φ(ξ1)=η and φΔ(ξ1)=ζ.

(II) Let t be right-scattered and letA(t) be true, that is, φ(t)≥0 and φΔ(t)≥0. We need to show that φ(σ(t))≥0 and φΔ(σ(t))≥0. By the definition of Δ-derivative, we have
φ(σ(t))=φ(t)+[σ(t)-t]φΔ(t).
Further, by the definition of Δ-derivative and (2.2) for φ(t), we have
φΔ(σ(t))=φΔ(t)+[σ(t)-t]φΔ2(t)=φΔ(t)+[σ(t)-t]q(t)φ(σ(t)).
From (2.9), we get φ(σ(t))≥0, and then from (2.10), we get φΔ(σ(t))≥0.

(III) Let t0 be right-dense, A(t0) be true and t1∈[ξ1,ξ2] such that t1>t0 and is sufficiently close to t0. We need to prove that A(t) is true for t∈[t0,t1].

From (2.2) with y(t)=φ(t), the equations
φΔ(t)=φΔ(t0)+∫t0tq(s)φ(σ(s))Δs,φ(t)=φ(t0)+φΔ(t0)(t-t0)+∫t0t∫t0τq(s)φ(σ(s))ΔsΔτ
follow. To investigate the function φ(t) appearing in (2.12), we consider the equation
y(t)=φ(t0)+φΔ(t0)(t-t0)+∫t0t∫t0τq(s)y(σ(s))ΔsΔτ,
where y(t) is the desired solution. Our aim is to show that witht1 sufficiently close to t0, (2.13) has a unique continuous solution y(t) satisfying inequality
y(t)≥φ(t0)+φΔ(t0)(t-t0),t∈[t0,t1].
We solve (2.13) by the method of successive approximations, setting
y0(t)=φ(t0)+φΔ(t0)(t-t0),yj(t)=∫t0t∫t0τq(s)yj-1(σ(s))ΔsΔτ,j=1,2,3,….
If the series ∑j=0∞yj(t) converges uniformly with respect to t∈[t0,t1], then its sum will be, obviously, a continuous solution of (2.13). To prove the uniform convergence of this series, we let
M0=φ(t0)+φΔ(t0)(t1-t0),M1=∫t0t1∫t0τq(s)ΔsΔτ.
Then the estimate
0≤yj(t)≤M0M1j,t∈[t0,t1],j=0,1,2,…
can easily be obtained. Indeed, (2.17) evidently holds for j=0. Let it also hold for j=n. Then from (2.15), we get for t∈[t0,t1],
0≤yn+1(t)≤∫t0t1∫t0τq(s)yn(σ(s))ΔsΔτ≤∫t0t1maxt0≤s≤ρ(τ)yn(σ(s))∫t0τq(s)ΔsΔτ≤maxt0≤s≤ρ2(t1)yn(σ(s))∫t0t1∫t0τq(s)ΔsΔτ≤M0M1nM1=M0M1n+1.
Therefore, by the usual mathematical induction principle, (2.17) holds for all j=0,1,2,….

Now choosing t1 appropriately, we obtain M1<1. Then (2.13) will have a continuous solution
y(t)=∑j=0∞yj(t)fort∈[t0,t1].
Since yj(t)≥0, it follows that y(t)≥y0(t) thereby proving the validity of inequality (2.14). To prove uniqueness of solution of (2.13) for t∈[t0,t1], suppose it has two solutions y1 and y2, and passing on to the modulus, we get
|y1(t)-y2(t)|≤∫t0t∫t0τq(s)|y1(σ(s))-y2(σ(s))|ΔsΔτ≤maxt0≤s≤ρ2(t1)|y1(σ(s))-y2(σ(s))|∫t0t1∫t0τq(s)ΔsΔτ.
Thus,
|y1(t)-y2(t)|≤M1maxt0≤s≤ρ2(t1)|y1(σ(s))-y2(σ(s))|,t∈[t0,t1].
Since M1<1, hence it follows that y1(t)=y2(t) for t∈[t0,t1].

From (2.12) and (2.13) in view of the uniqueness of solution, we get that φ(t)=y(t),t∈[t0,t1]. Therefore,φ(t)≥φ(t0)+φΔ(t0)(t-t0),t∈[t0,t1].
Hence by making use of the induction hypothesis A(t0) being true, we obtain φ(t)≥0 for t∈[t0,t1]. Taking this into account, from (2.11), we also get φΔ(t)≥0 for t∈[t0,t1]. Thus, A(t) is true for all t∈[t0,t1].

(IV) Let t∈(ξ1,ξ2] and assume t is left-dense and such that A(s) is true for all s<t, that is,
φ(s)≥0,φΔ(s)≥0,∀s∈[ξ1,t).
Passing on here to the limit as s→t, we get by the continuity of φ(s) and φΔ(s) that φ(t)≥0 and φΔ(t)≥0, thereby verifying the validity of A(t).

Consequently, by the induction principle on time scales, (2.8) holds for all t∈[ξ1,ξ2].

From (2.9) and (2.8) for t=ξ2, we also get φ(σ(ξ2))≥0. So the statements (2.7) for φ are proved.

We can prove the statements of the lemma for ψ similarly applying the backward induction principle on time scales. The lemma is proved.

Lemma 2.2.

Under the conditions (H1) and (H2), the inequality D>0 holds.

Proof.

By (2.4) and (2.5), we have
D=ζδ+ηγ+ζγ(ξ2-ξ1)+δ∫ξ1ξ2q(s)φ(σ(s))Δs+γ∫ξ1ξ2∫ξ1τq(s)φ(σ(s))ΔsΔτ.
Since φ(t)≥0for t∈[ξ1,σ(ξ2)], from (2.8), we have
D≥ζδ+ηγ+ζγ(ξ2-ξ1).
If q(t)≡0, then in (2.25) the equality holds. From the condition (H2), we get D>0. This proof is completed.

Lemma 2.3.

Assume that the conditions (H1) and (H2) are satisfied. If h∈𝒞[ξ1,ρ(ξ2)], then the boundary value problem
yΔ2(t)-q(t)y(σ(t))=h(t),t∈[ξ1,ρ(ξ2)],ζy(ξ1)-ηyΔ(ξ1)=0,γy(ξ2)+δyΔ(ξ2)=0
has a unique solution
y(t)=-∫ξ1ξ2G(t,s)h(s)Δs,
where
G(t,s)=1D{ψ(σ(s))φ(t),t≤s,ψ(t)φ(σ(s)),t≥σ(s).
Here D,φ,ψ are as in (2.4), (2.5), and (2.6), respectively.

Proof.

Taking
z(t)=-1D∫ξ1t[φ(σ(s))ψ(t)-φ(t)ψ(σ(s))]h(s)Δs,
we have
zΔ2(t)=-1D[φ(σ(t))ψΔ(σ(t))-φΔ(σ(t))ψ(σ(t))]h(t)-1D∫ξ1t[φ(σ(s))ψΔ2(t)-φΔ2(t)ψ(σ(s))]h(s)Δs.
By Corollary 3.14 in [1], since the Wronskian of any two solutions of (2.2) is independent of t, we get
D=W(ψ,φ)=ψ(σ(t))φΔ(σ(t))-φ(σ(t))ψΔ(σ(t)).
Hence we get
zΔ2(t)=-1D[-Dh(t)+q(t)∫ξ1t[φ(σ(s))ψ(σ(t))-φ(σ(t))ψ(σ(s))]h(s)Δs]=h(t)+q(t)[-1D∫ξ1t[φ(σ(s))ψ(σ(t))-φ(σ(t))ψ(σ(s))]h(s)Δs]=h(t)+q(t)[-1D∫ξ1σ(t)[φ(σ(s))ψ(σ(t))-φ(σ(t))ψ(σ(s))]h(s)Δs]=h(t)+q(t)z(σ(t)).
So the general solution of equation
yΔ2(t)-q(t)y(σ(t))=h(t),t∈[ξ1,ρ(ξ2)],
has the form
y(t)=c1φ(t)+c2ψ(t)-1D∫ξ1t[φ(σ(s))ψ(t)-φ(t)ψ(σ(s))]h(s)Δs,
where c1 and c2 are arbitrary constants. Substituting this expression for y(t) in the boundary conditions of BVP (2.26), we can evaluate c1 and c2. After some easy calculations, we can get (2.27) and (2.28).

Lemma 2.4.

Under the conditions (H1) and (H2), the Green’s function of BVP (2.26) possesses the following property:
G(t,s)≥0,(t,s)∈[ξ1,ξ2]×[ξ1,ρ(ξ2)].

Proof.

The lemma follows from (2.28), Lemmas 2.1 and 2.2 immediately.

Lemma 2.5.

Assume that the conditions (H1) and (H2) are satisfied. If h∈𝒞[a,b], then the boundary value problem
yΔ4(t)-q(t)yΔ2(σ(t))=h(t),t∈[a,b],y(a)=0,y(σ2(b))+λyΔ(σ2(b))=0,ζyΔ2(ξ1)-ηyΔ3(ξ1)=0,γyΔ2(ξ2)+δyΔ3(ξ2)=0
has a unique solution
y(t)=∫aσ2(b)G1(t,ξ)∫ξ1ξ2G2(ξ,s)h(s)ΔsΔξ,
where
G1(t,s)=1σ2(b)-a+λ{(σ2(b)-σ(s)+λ)(t-a),t≤s,(σ2(b)-t+λ)(σ(s)-a),t≥σ(s),G2(t,s)=1D{ψ(σ(s))φ(t),t≤s,ψ(t)φ(σ(s)),t≥σ(s).
Here D,φ,ψ are as in (2.4), (2.5), and (2.6), respectively.

Proof.

Let us consider the following BVP:
yΔ2(t)=-∫ξ1ξ2G2(t,s)h(s)Δs,t∈[a,σ2(b)],y(a)=0,y(σ2(b))+λyΔ(σ2(b))=0.
The Green’s function associated with the BVP (2.40) is G1(t,s). This completes the proof.

Lemma 2.6.

Assume that the conditions (H1)–(H3) are satisfied. If y satisfies
yΔ4-q(t)yΔ2(σ(t))≥0,t∈[a,b],y(a)≥0,y(σ2(b))+λyΔ(σ2(b))≥0,ζyΔ2(ξ1)-ηyΔ3(ξ1)≤0,γyΔ2(ξ2)+δyΔ3(ξ2)≤0,
then y(t)≥0,t∈[a,σ2(b)] and yΔ2(t)≤0,t∈[a,σ(b)].

Proof.

Let
yΔ4(t)-q(t)yΔ2(σ(t))=h(t),t∈[a,b],y(a)=t0,y(σ2(b))+λyΔ(σ2(b))=t1,ζyΔ2(ξ1)-ηyΔ3(ξ1)=t2,γyΔ2(ξ2)+δyΔ3(ξ2)=t3,
where t0≥0,t1≥0,t2≤0,t3≤0,h≥0.

It is easy to check that y and yΔ2 can be given by the expressiony(t)=S(t)-∫aσ2(b)G1(t,ξ)R(ξ)Δξ+∫aσ2(b)G1(t,ξ)∫ξ1ξ2G2(ξ,s)h(s)ΔsΔξ,yΔ2(t)=R(t)-∫ξ1ξ2G2(t,s)h(s)Δs,
where
S(t)=1σ2(b)-a+λ[(σ2(b)+λ-t)t0+(t-a)t1],R(t)=1k[(ζ(t-ξ1)+η)t3+(γ(ξ2-t)+δ)t2],
and G1(t,s),G2(t,s)are as in (2.38) and (2.39), respectively. The hypothesis of the lemma implies that S(t)≥0 for t∈[a,σ2(b)], R(t)≤0 for t∈[a,σ(b)], G1(t,s)≥0 for (t,s)∈[a,σ2(b)]×[a,σ(b)], and G2(t,s)>0 for (t,s)∈[a,σ(b)]×[a,b]. Therefore, we get y(t)≥0 for t∈[a,σ2(b)] and yΔ2(t)≤0 for t∈[a,σ(b)]. The proof is completed.

3. Upper and Lower Solution Method

In this section, we present existence results for the BVP (1.1) by using the method of upper and lower solutions. We define the set
D:={y:yΔn∈C[a,σ4(b)]kn,k=0,1,2,3,4}.Definition 3.1.

Letting α(t)∈D on [a,σ4(b)], we say α is a lower solution for the problem (1.1) if α satisfies
αΔ4(t)-q(t)αΔ2(σ(t))≤f(t,α(σ(t)),αΔ2(σ(t))),α(a)≤0,α(σ2(b))+λαΔ(σ2(b))≤0,ζαΔ2(ξ1)-ηαΔ3(ξ1)≥0,γαΔ2(ξ2)+δαΔ3(ξ2)≥0.

Definition 3.2.

Letting β(t)∈D, on [a,σ4(b)], we say β is an upper solution for the problem (1.1) if β satisfies
βΔ4(t)-q(t)βΔ2(σ(t))≥f(t,β(σ(t)),βΔ2(σ(t))),β(a)≥0,β(σ2(b))+λβΔ(σ2(b))≥0,ζβΔ2(ξ1)-ηβΔ3(ξ1)≤0,γβΔ2(ξ2)+δβΔ3(ξ2)≤0.

We assume that the function f(t,y,yΔ2) satisfies the following condition.

f:[a,b]×ℝ×ℝ→ℝ is continuous and satisfies
f(t,y2,z)-f(t,y1,z)≥0,forα(σ(t))≤y1≤y2≤β(σ(t)),z∈R,t∈[a,b],f(t,y,z2)-f(t,y,z1)≤0,forβΔ2(σ(t))≤z1≤z2≤αΔ2(σ(t)),y∈R,t∈[a,b],

where α, β are lower and upper solutions, respectively, for the BVP (1.1), and satisfy α≤β,αΔ2≥βΔ2.Theorem 3.3.

Assume that the conditions (H1)–(H4) are satisfied. Then the problem (1.1) has a solution y(t) with
α(t)≤y(t)≤β(t),αΔ2(t)≥yΔ2(t)≥βΔ2(t)
for t∈[a,σ3(b)] and t∈[a,σ(b)], respectively.

Proof.

Consider the BVP,
yΔ4(t)-q(t)yΔ2(σ(t))=F(t,y(σ(t)),yΔ2(σ(t))),t∈[a,b],y(a)=0,y(σ2(b))+λyΔ(σ2(b))=0,ζyΔ2(ξ1)-ηyΔ3(ξ1)=0,γyΔ2(ξ2)+δyΔ3(ξ2)=0,
where
F(t,x,y)={f*(t,α(σ(t)),y),x<α(σ(t)),f*(t,x,y),α(σ(t))≤x≤β(σ(t)),f*(t,β(σ(t)),y),x>β(σ(t)),f*(t,x,y)={f(t,x,βΔ2(σ(t))),y<βΔ2(σ(t)),f(t,x,y),βΔ2(σ(t))≤y≤αΔ2(σ(t)),f(t,x,αΔ2(σ(t))),y>αΔ2(σ(t)).
By Lemma 2.5, it is clear that the solutions of the BVP (3.6) are the fixed points of the operator
Ay(t)=∫aσ2(b)G1(t,ξ)∫ξ1ξ2G2(ξ,s)F(s,y(σ(s)),yΔ2(σ(s)))ΔsΔξ,t∈[a,σ4(b)],
where G1(t,s) and G2(t,s) are as in (2.38) and (2.39), respectively. It is clear that A is continuous. Since the function f(t,y,yΔ2) satisfies the conditions (3.4),
f(t,α(σ(t)),αΔ2(σ(t)))≤F(t,y,yΔ2)≤f(t,β(σ(t)),βΔ2(σ(t)))fort∈[a,b].
Thus, there exists a positive constant M such that |F(t,y,yΔ2)|≤M, which implies that the operator A is uniformly bounded. Moreover, the operator A is equicontinuous. Therefore, from the Arzela-Ascoli theorem, the operator A is a compact operator. Thus, by Schauder’s fixed point theorem, there exists a solution y of the BVP (3.6).

Suppose y* is a solution of the BVP (3.6). Since f(t,y,yΔ2) satisfies the conditions (3.4), we know that
f(t,α(σ(t)),αΔ2(σ(t)))≤F(t,y*,y*Δ2)≤f(t,β(σ(t)),βΔ2(σ(t)))fort∈[a,b].
Thus,
(β-y*)Δ4(t)-q(t)(β-y*)Δ2(σ(t))≥f(t,β(σ(t)),β2(σ(t)))-F(t,y*(σ(t)),yΔ2(σ(t)))≥0,t∈[a,b],(β-y*)(a)≥0,(β-y*)(σ2(b))+λ(β-y*)Δ(σ2(b))≥0,ζ(β-y*)Δ2(ξ1)-η(β-y*)Δ3(ξ1)≤0,γ(β-y*)Δ2(ξ2)+δ(β-y*)Δ3(ξ2)≤0.
By virtue of Lemma 2.6, y*(t)≤β(t) for t∈[a,σ2(b)] and y*Δ2(t)≥βΔ2(t) for t∈[a,σ(b)]. If σ2(b) is right-scattered, by using the inequality (β-y*)(σ2(b))+λ(β-y*)Δ(σ2(b))≥0, we get the inequality (λ/(σ3(b)-σ2(b)))(β-y*)(σ3(b))≥((λ/(σ3(b)-σ(b)))-1)(β-y*)(σ2(b)). So we get y*(t)≤β(t) on [a,σ3(b)]. If σ2(b) is right-dense, it is trivial that the inequality y*(t)≤β(t) holds on [a,σ3(b)]. Similarly, one can show that α(t)≤y*(t) for t∈[a,σ3(b)] and αΔ2(t)≥y*Δ2(t) for t∈[a,σ(b)]. This completes the proof.

Theorem 3.4.

Assume that the conditions (H1)–(H4) are satisfied. Then there exist two monotone sequences {αn} and {βn}, nonincreasing and nondecreasing, respectively, with α0=α and β0=β, which converge to the extremal solutions in [β,α] of the problem (1.1).

Proof.

For any function ϑ which satisfies α(t)≤ϑ(t)≤β(t) for t∈[a,σ2(b)], consider the following problem:
yΔ4(t)-q(t)yΔ2(σ(t))=f(t,ϑ(σ(t)),ϑΔ2(σ(t))),t∈[a,b],y(a)=0,y(σ2(b))+λyΔ(σ2(b))=0,ζyΔ2(ξ1)-ηyΔ3(ξ1)=0,γyΔ2(ξ2)+δyΔ3(ξ2)=0.
Clearly, this problem is type (2.12). Obviously, it has a unique solution given by the expressiony(t)=∫aσ2(b)G2(t,ξ)∫ξ1ξ2G1(ξ,s)f(s,ϑ(σ(s)),ϑΔ2(σ(s)))ΔsΔξ≡Aϑ(t),t∈[a,σ4(b)].

Step 1.

α≤Aα,Aβ≤β on [a,σ3(b)] and αΔ2≥(Aα)Δ2,(Aβ)Δ2≥βΔ2 on [a,σ(b)].

Let Aα(t)=w(t). Thus(w-α)Δ4(t)-q(t)(w-α)Δ2(σ(t))≥f(t,α(σ(t)),αΔ2(σ(t)))-f(t,α(σ(t)),αΔ2(σ(t)))=0,t∈[a,b],(w-α)(a)≥0,(w-α)(σ2(b))+λ(w-α)Δ(σ2(b))≥0,ζ(w-α)Δ2(ξ1)-η(w-α)Δ3(ξ1)≤0,γ(w-α)Δ2(ξ2)+δ(w-α)Δ3(ξ2)≤0.
Using Lemma 2.6, we obtain that α≤Aα,αΔ2≥(Aα)Δ2 on [a,σ2(b)] and [a,σ(b)], respectively. In case that σ2(b) is right scattered, the assumptions λ≥σ3(b)-σ2(b) and (w-α)(σ2(b))+(w-α)Δ(σ2(b))≥0 imply that α(σ3(b))≤(Aα)(σ3(b)). Similarly, we show that (Aβ)(t)≤β(t),(Aβ)Δ2(t)≥βΔ2(t) on [a,σ3(b)] and [a,σ(b)], respectively.

Step 2.

If y1,y2∈[α,β]; y1Δ2,y2Δ2∈[βΔ2,αΔ2], y1≤y2 and y2Δ2≤y1Δ2, then we have Ay1≤Ay2 and (Ay1)Δ2≥(Ay2)Δ2.

Let Ay1=w1 and Ay2=w2. Thus(w2-w1)Δ4(t)-q(t)(w2-w1)Δ2(σ(t))=f(t,y2(σ(t)),y2Δ2(σ(t)))-f(t,y1(σ(t)),y1Δ2(σ(t)))≥0,t∈[a,b],(w2-w1)(a)=0,(w2-w1)(σ3(b))+λ(w2-w1)Δ(σ3(b))=0,ζ(w2-w1)Δ2(ξ1)-η(w2-w1)Δ3(ξ1)=0,γ(w2-w1)Δ2(ξ2)+δ(w2-w1)Δ3(ξ2)=0.
Hence we get that Ay1(t)≤Ay2(t) and (Ay1)Δ2(t)≥(Ay2)Δ2(t) on [a,σ3(b)] and [a,σ(b)], respectively.

Now, we define the sequences {αn(t)} and {βn(t)} by
α0(t)=α(t),αn+1(t)=Aαn(t),β0(t)=β(t),βn+1(t)=Aβn(t),forn≥0.
From the properties of A, we have
α=α0≤α1≤⋯≤βn≤⋯≤β1≤β0=β,αΔ2=α0Δ2≤α1Δ2≤⋯≤βnΔ2≤⋯≤β1Δ2≤β0Δ2=βΔ2.
But then α* and β* defined by
α*=limn→∞αn,β*=limn→∞βn,α*Δ2=limn→∞αnΔ2,β*Δ2=limn→∞βn
are extremal solutions of (1.1).

Example 3.5.

Consider the BVP
yΔ4(t)-t2yΔ2(σ(t))=-yΔ2(σ(t)),t∈[32,4],y(32)=0,y(σ2(4))+yΔ(σ2(4))=0,yΔ2(2)-3yΔ3(2)=0,7yΔ2(3)+8yΔ3(3)=0,
where σ(4)≤29/7 and σ3(4)-σ2(4)≤1. It is easy to check that α(t)=0, β(t)=t are lower and upper solutions of the BVP (3.19), respectively, and that all assumptions of Theorem 3.3 are fulfilled. So the BVP (3.19) has a solution y(t) satisfying 0≤y(t)≤t for t∈[3/2,σ3(4)], yΔ2(t)=0 for t∈[3/2,σ(4)].

BohnerM.PetersonA.BohnerM.PetersonA.AkinE.Boundary value problems for a differential equation on a measure chainAticiF. M.CabadaA.Existence and uniqueness results for discrete second-order periodic boundary value problemsEloeP. W.ShengQ.Approximating crossed symmetric solutions of nonlinear dynamic equations via quasilinearizationKaymakçalanB.Monotone iterative method for dynamic systems on time scalesLeelaS.SivasundaramS.Dynamic systems on time scales and superlinear convergence of iterative processPangY.BaiZ.Upper and lower solution method for a fourth-order four-point boundary value problem on time scalesAticiF. M.GuseinovG. S.On Green's functions and positive solutions for boundary value problems on time scales