ON TIME SCALES
DOUGLAS R. ANDERSON AND RUYUN MA
Received 10 December 2004; Revised 3 November 2005; Accepted 6 November 2005
We discuss conditions for the existence of at least one positive solution to a nonlinear second-order Sturm-Liouville-type multipoint eigenvalue problem on time scales. The results extend previous work on both the continuous case and more general time scales, and are based on the Guo-Krasnosel’ski˘i fixed point theorem.
Copyright © 2006 D. R. Anderson and R. Ma. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, dis-tribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
We are interested in the second-order multipoint time-scale eigenvalue problem
py∇Δ(t)−q(t)y(t) +λh(t)f(y)=0, t1< t < tn, (1.1)
αyt1−βpt1y∇t1=n −1
i=2
aiyti, γytn+δ ptny∇tn=n −1
i=2
biyti, (1.2)
where
p,q:t1,tn−→(0,∞), p∈CΔt1,tn,q∈Ct1,tn; (1.3)
the pointsti∈Tκ
κfori∈ {1, 2,...,n}witht1< t2<···< tn;
α,β,γ,δ∈[0,∞), αγ+αδ+βγ >0, ai,bi∈[0,∞), i∈ {2,...,n−1}. (1.4)
The continuous function f : [0,∞)→[0,∞) is such that the following exist:
f0:=lim y→0+
f(y)
y , f∞:=ylim→∞
f(y)
y ; (1.5)
Hindawi Publishing Corporation Advances in Difference Equations Volume 2006, Article ID 59572, Pages1–17
and the right-dense continuous functionh: [t1,tn]→[0,∞) satisfies some suitable con-ditions to be developed. Problem (1.1), (1.2) is a generalization to time scales of the prob-lem whenTis restricted toRon the unit interval in Ma and Thompson [19], and extends the type of time-scale boundary value problem found in Anderson [2], Atici and Gu-seinov [6], Kaufmann [15], Kaufmann and Raffoul [16], and Sun and Li [21,22]. Other related three-point problems on time scales include Anderson and Avery [4], Anderson et al. [5], Peterson et al. [20], and a singular problem in DaCunha et al. [12]. Some of the work on multipoint time-scale problems includes Anderson [1,3] and Kong and Kong [17], and a recent singular multipoint problem in Bohner and Luo [8]. For more general information concerning dynamic equations on time scales, introduced by Aulbach and Hilger [7] and Hilger [14], see the excellent text by Bohner and Peterson [9] and their edited text [10].
2. Time-scale primer
Any arbitrary nonempty closed subset of the realsRcan serve as a time-scaleT; see [9,
10]. Fort∈Tdefine the forward jump operatorσ:T→Tbyσ(t)=inf{s∈T: s > t}, and the backward jump operatorρ:T→Tbyρ(t)=sup{s∈T: s < t}. The graininess operatorsμσ,μρ:T→[0,∞) are defined byμσ(t)=σ(t)−tandμρ(t)=ρ(t)−t.
A function f :T→Ris right-dense continuous (rd-continuous) provided it is con-tinuous at all right-dense points of T and its sided limit exists (is finite) at left-dense points ofT. The set of all right-dense continuous functions onTis denoted by Crd=Crd(T)=Crd(T,R).
Define the setTκbyTκ=T− {m}ifThas a right scattered minimummandTκ=T otherwise. In a similar vein,Tκ=T− {M}if Thas a left scattered maximumM and Tκ=Totherwise. We takeTκ
κ=Tκ∩Tκ.
Definition 2.1(delta derivative). Assume f :T→Ris a function and lett∈Tκ. Define fΔ(t) to be the number (provided it exists) with the property that given any>0, there is a neighborhoodU⊂Toftsuch that
f
σ(t)−f(s)−fΔ(t)σ(t)−s≤σ(t)−s ∀s∈U. (2.1)
The function fΔ(t) is the delta derivative of f att.
Definition 2.2(nabla derivative). For f :T→Randt∈Tκ, definef∇(t) to be the number (provided it exists) with the property that given any>0, there is a neighborhoodUoft such that
f
ρ(t)−f(s)−f∇(t)ρ(t)−s≤ρ(t)−s ∀s∈U. (2.2)
The function f∇(t) is the nabla derivative off att.
Definition 2.3(delta integral). Letf :T→Rbe a function, and leta,b∈T. If there exists a functionF:T→Rsuch thatFΔ(t)=f(t) for allt∈Tκ, thenFis a delta antiderivative of f. In this case the integral is given by the formula
b
a f(t)Δt=F(b)−F(a) fora,b∈T. (2.3) All right-dense continuous functions are delta integrable; see [9, Theorem 1.74].
3. Linear preliminaries
We first construct Green’s function for the second-order boundary value problem
py∇Δ(t)−q(t)y(t) +u(t)=0, t1< t < tn, (3.1) αyt1−βpt1y∇t1=0, γytn+δ ptny∇tn=0, (3.2)
whereα,β, γ,δ are real numbers such that|α|+|β| =0,|γ|+|δ| =0. The techniques here are similar to those found in [6,19].
Denote byφandψthe solutions of the corresponding homogeneous equation
py∇Δ(t)−q(t)y(t)=0, t∈t1,tn, (3.3)
under the initial conditions
ψt1=β, pt1ψ∇t1=α, (3.4)
φtn=δ, ptn)φ∇tn= −γ, (3.5)
so thatψandφsatisfy the first and second boundary conditions in (3.2), respectively. Set
d= −Wt(ψ,φ)=p(t)ψ∇(t)φ(t)−ψ(t)p(t)φ∇(t). (3.6)
Since the Wronskian of any two solutions is independent oft, evaluating att=t1,t=tn, and using the boundary conditions (3.4), (3.5) yields
d=αφt1−βpt1φ∇t1=γψtn+δ ptnψ∇tn. (3.7)
In additiond=0 if and only if the homogeneous equation (3.3) has only the trivial so-lution satisfying the boundary conditions (3.2). For the proof of the following theorem, see [6, Theorem 4.2].
Lemma3.1. Assume (1.3) and (1.4). Ifd=0, then the nonhomogeneous boundary value problem (3.1)-(3.2) has a unique solutionyfor which the formula
y(t)=
tn
t1
holds, where the functionG(t,s)is given by
Proof. The proof is very similar to the proof of [6, Lemma 5.1] and is omitted. Set dynamic equation (3.1) with boundary conditions (1.2) has a unique solutionyfor which the formula
Proof. It can be verified that for a solution y of the nonhomogeneous equation (3.1) under the nonhomogeneous boundary conditions (1.2), the formula (3.12) holds, where G(t,s) is given by (3.9). We thus show that the function ygiven in (3.12) is a solution of (3.1) with conditions (1.2) only ifAandBare given by (3.13) and (3.14), respectively. If yas in (3.12) is a solution of (3.1), (1.2), then
y(t)=1
d t
t1
φ(t)ψ(s)u(s)Δs+1 d
tn
t ψ(t)φ(s)u(s)Δs+Aψ(t) +Bφ(t) (3.15)
for some constantsAandB. Taking the nabla derivative and multiplying bypyields
py∇= pφ∇
d t
t1
ψ(s)u(s)Δs+ pψ∇ d
tn
t φ(s)u(s)Δs+Apψ
∇+Bpφ∇; (3.16)
the delta derivative of this expression is
py∇Δ=
pφ∇
d
Δσ(t)
t1
ψ(s)u(s)Δs+ pφ∇
d ψ(t)u(t) +A
pψ∇Δ+Bpφ∇Δ
+
pψ∇ d
Δtn
σ(t)φ(s)u(s)Δs− pψ∇
d φ(t)u(t).
(3.17)
Using [9, Theorem 1.75], and the fact thatψandφare solutions to (3.3), we obtain
py∇Δ(t)=q(t)
d t
t1
φ(t)ψ(s)u(s)Δs+q(t)
d φ(t)μσ(t)ψ(t)u(t) + u(t)
d p(t)φ
∇(t)ψ(t)
+q(t) d
tn
t ψ(t)φ(s)u(s)Δs− q(t)
d ψ(t)μσ(t)φ(t)u(t)
−u(t)
d p(t)ψ∇(t)φ(t) +q(t)
Aψ(t) +bφ(t).
(3.18)
Recall thatdis in terms of the Wronskian ofψandφin (3.6); it follows that
py∇Δ(t)=q(t)y(t)−u(t). (3.19)
Now
yt1=ψ
t1 d
tn
t1
φ(s)u(s)Δs+Aψt1+Bφt1,
pt1y∇t1= p
t1ψ∇t1 d
tn
t1
φ(s)u(s)Δs+Apt1ψ∇t1+Bpt1φ∇t1;
multiply the first line byαand the second by−β, and use (1.2) and (3.4) to see that
At the other end,
ytn=φ
Proof. From the previous lemmas and assumptions we know that Green’s function (3.9) satisfiesG(t,s)≥0 on [ρ(t1),tn]×[ρ(t1),tn]. Hypotheses (1.3), (1.4), and (3.25) applied
to (3.13) and (3.14) imply thatA(u),B(u)≥0.
Suppose (3.25) does not hold. For example, letn=3,p(t)≡1=α=γ,q(t)≡0=β=
δ=a2, andt1=0. Then (3.1), (1.2) becomes
Note thatψ(t)=t,d=t3, andD=t3(b2t2−t3). IfD >0, thenb2t2> t3, and there is no positive solution; see [15, Lemma 4].
Lemma3.5. Let (1.3), (1.4), and (3.25) hold, and fix
Theorem4.1. LetEbe a Banach space,P⊆Ea cone, and suppose thatΩ1,Ω2are bounded open balls ofEcentered at the origin withΩ1⊂Ω2. Suppose further thatL:P∩(Ω2\Ω1)→
Pis a completely continuous operator such that either
(i)Ly ≤ y,y∈P∩∂Ω1andLy ≥ y,y∈P∩∂Ω2, or (ii)Ly ≥ y,y∈P∩∂Ω1andLy ≤ y,y∈P∩∂Ω2 holds. ThenLhas a fixed point inP∩(Ω2\Ω1).
Assume that the right-dense continuous functionhsatisfies
h:t1,tn−→[0,∞), ∃t∗∈σt1,ρtnh(t∗)>0. (4.1)
Then there existξ1,ξ2as inLemma 3.5such that
ξ1< t∗< ξ2,
ξ2
ξ1
G(t,s)h(s)Δs >0, t∈ρt1,tn. (4.2)
In the following, letΓbe the constant defined in (3.29) with respect to such constants ξ1,ξ2. Letτ∈[ρ(t1),tn] be determined by
ξ2
ξ1
G(τ,s)h(s)Δs= max ρ(t1)≤t≤tn
ξ2
ξ1
G(t,s)h(s)Δs >0. (4.3)
ForG(t,s) in (3.9) andA,Bas in (3.13), (3.14), respectively, define the constant
K:=
tn
t1
G(s,s)h(s)Δs+A(h)ψtn+B(h)φρt1. (4.4)
LetᏮdenote the Banach spaceC[ρ(t1),tn] with the normy =supt∈[ρ(t1),tn]|y(t)|. De-fine the coneᏼ⊂Ꮾby
ᏼ=y∈Ꮾ:y(t)≥0 onρt1,tn, y(t)≥Γyonξ1,ξ2, (4.5)
whereΓis given in (3.29). Sinceyis a solution of (1.1), (1.2) if and only if
y(t)=λ tn
t1
G(t,s)h(s)fy(s)Δs+Ah f(y)ψ(t) +Bh f(y)φ(t)
define fory∈ᏼthe operatorT:ᏼ→Ꮾby
(T y)(t) :=λ tn
t1
G(t,s)h(s)fy(s)Δs+Ah f(y)ψ(t) +Bh f(y)φ(t)
. (4.7)
We seek a fixed point ofTinᏼby establishing the hypotheses ofTheorem 4.1.
Theorem4.2. Suppose (1.3), (1.4), (3.25), (4.1), and (4.3) hold. Then for eachλsatisfying
1
f∞Γξξ12G(τ,s)h(s)Δs
< λ < 1
f0K, (4.8)
there exists at least one positive solution of (1.1), (1.2) inᏼ.
Proof. Letξ1,ξ2be as inLemma 3.5, letτbe as in (4.3), letKbe as in (4.4), letλbe as in (4.8), and let>0 be such that
1
f∞−Γξξ12G(τ,s)h(s)Δs
≤λ≤ 1
f0+K. (4.9)
Consider the integral operatorTin (4.7). Ify∈ᏼ, then by (3.30) we have
(T y)(t)=λ tn
t1
G(t,s)h(s)fy(s)Δs+Ah f(y)ψ(t) +Bh f(y)φ(t)
≤λ tn
t1
G(s,s)h(s)fy(s)Δs+Ah f(y)ψtn+Bh f(y)φρt1
, (4.10)
so that fort∈[ξ1,ξ2],
(T y)(t)=λ tn
t1
G(t,s)h(s)fy(s)Δs+Ah f(y)ψ(t) +Bh f(y)φ(t)
≥λ tn
t1
G(t,s)
G(s,s)G(s,s)h(s)f
y(s)Δs+Ah f(y)ψξ1+Bh f(y)φξ2
≥λΓtn t1
G(s,s)h(s)fy(s)Δs+Ah f(y)ψtn+Bh f(y)φρt1
≥ΓT y.
(4.11)
Now consider f0. There exists anR1>0 such that f(y)≤(f0+)yfor 0< y≤R1by the definition of f0. Picky∈ᏼwithy =R1. From (3.13) and (3.14),
A
h f(y)≤A(h)f(y), Bh f(y)≤B(h)f(y). (4.12)
Using (3.30), we have
(T y)(t)=λ tn
t1
G(t,s)h(s)fy(s)Δs+Ah f(y)ψ(t) +Bh f(y)φ(t)
≤λf(y) tn
t1
G(s,s)h(s)Δs+A(h)ψtn
+B(h)φρ(t1)
≤λf0+yK≤ y
(4.13)
from the right-hand side of (4.9). As a result,T y ≤ y. Thus, take
Ω1:=y∈Ꮾ:y< R1 (4.14)
so thatT y ≤ yfory∈ᏼ∩∂Ω1.
Next consider f∞. Again by definition, there exists anR2> R1such that f(y)≥(f∞− )yfor y≥R2; takeR2=max{2R1,R2/Γ }. Ify∈ᏼwithy =R2, then fors∈[ξ1,ξ2] we have
y(s)≥Γy =ΓR2. (4.15)
DefineΩ2:= {y∈Ꮾ:y< R2}; using (4.3) and (4.15) fors∈[ξ1,ξ2], we get
(T y)(τ)=λ tn
t1
G(τ,s)h(s)fy(s)Δs+Ah f(y)ψ(τ) +Bh f(y)φ(τ)
≥λ ξ2
ξ1
G(τ,s)h(s)fy(s)Δs≥λf∞−
ξ2
ξ1
G(τ,s)h(s)y(s)Δs
≥λf∞−ΓR2
ξ2
ξ1
G(τ,s)h(s)Δs≥R2= y,
(4.16)
where we have used the left-hand side of (4.9). Hence we have shown that
T y ≥ y, y∈ᏼ∩∂Ω2. (4.17)
An application ofTheorem 4.1yields the conclusion of the theorem; in other words,T
Theorem4.3. Suppose (1.3), (1.4), (3.25), (4.1), and (4.3) hold. Then for eachλsatisfying
1 f0Γξ2
ξ1G(τ,s)h(s)Δs
< λ < 1
f∞K, (4.18)
there exists at least one positive solution of (1.1), (1.2) inᏼ. Proof. Letλbe as in (4.18) and letη >0 be such that
1
f0−ηΓξ2
ξ1G(τ,s)h(s)Δs
≤λ≤ 1
f∞+ηK. (4.19)
Again letTbe the operator defined in (4.7). We once more seek a fixed point ofT inᏼ by establishing the hypotheses ofTheorem 4.1.
First, consider f0. There exists anR1>0 such that f(y)≥(f0−η)yfor 0< y≤R1by the definition of f0. Pick y∈ᏼwithy =R1. Fors∈[ξ1, ξ2], where ξ1, ξ2 are as in
Lemma 3.5, we have
y(s)≥Γy =ΓR1. (4.20)
Using the left-hand side of (4.19) and (4.20) we get, fors∈[ξ1,ξ2],
(T y)(τ)=λ tn
t1
G(τ,s)h(s)fy(s)Δs+Ah f(y)ψ(τ) +Bh f(y)φ(τ)
≥λf0−ηξ
2
ξ1
G(τ,s)h(s)y(s)Δs≥λf0−ηR1Γ ξ2
ξ1
G(τ,s)h(s)Δs
≥R1= y.
(4.21)
ThereforeT y ≥ y. This motivates us to define
Ω1:=y∈Ꮾ:y< R1, (4.22)
whereby our work above confirms
T y ≥ y, y∈ᏼ∩∂Ω1. (4.23)
Next consider f∞. Again by definition there exists anR2> R1such that f(y)≤(f∞+η)y
fory≥R2. If f is bounded, there existsM >0 with f(y)≤Mfor ally∈(0,∞). Let
R2:=max
2R2,λM tn
t1
G(s,s)h(s)Δs+A(h)ψtn+B(h)φρt1
Ify∈ᏼwithy =R2, then we have
(T y)(t)≤λ tn
t1
G(s,s)h(s)fy(s)Δs+Ah f(y)ψtn+Bh f(y)φρt1
≤λM tn
t1
G(s,s)h(s)Δs+A(h)ψtn+B(h)φρt1
≤R2= y.
(4.25)
As a result,T y ≤ y. Thus, take
Ω2:=y∈Ꮾ:y< R2 (4.26)
so thatT y ≤ yfor y∈ᏼ∩∂Ω2. If f is unbounded, takeR2:=max{2R1,R2}such that f(y)≤ f(R2) for 0< y≤R2. Ify∈ᏼwithy =R2, then we have
(T y)(t)≤λ fR2 tn
t1
G(s,s)h(s)Δs+A(h)ψtn
+B(h)φρt1
≤λf∞+ηR2K≤R2= y,
(4.27)
where we have used the left-hand side of (4.19). Hence we have shown that
T y ≤ y, y∈ᏼ∩∂Ω2 (4.28)
if we take
Ω2:=y∈Ꮾ:y< R2. (4.29)
As before, an application ofTheorem 4.1yields the conclusion thatThas a fixed pointy
inᏼ∩(Ω2\Ω1) withR1≤ y ≤R2.
Corollary4.4. Suppose (1.3), (1.4), (3.25), and (4.1) hold. If f is sublinear (i.e., f0= ∞
andf∞=0), or if f is superlinear (i.e.,f0=0andf∞= ∞), then for anyλ >0the boundary
value problem (1.1)-(1.2) has at least one positive solution inᏼ.
Proof. For the superlinear claim, use (4.8) ofTheorem 4.2; for the sublinear claim, use
(4.18) ofTheorem 4.3.
5. Examples
Example 5.1. LetT=R, and consider the three-point boundary value problem
y−y+λ f(y)=0, −1< t <1,
y(−1)=ay(0)=y(1), (5.1)
Applying (5.4) and (5.6), we can find the interval in (4.8):
Example 5.2. LetT=hZforh=2−10, and consider the four-point boundary value prob-lem
Then direct calculation verifies that
(3.25) holds. Let [ξ1,ξ2]=[0, 1/2], so that
As in the previous example, we determineτin (4.3) from
max
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Douglas R. Anderson: Department of Mathematics and Computer Science, Concordia College, Moorhead, MN 56562, USA
E-mail address:[email protected]
