# Necessary and Sufficient Condition for the Existence of Solutions to a Discrete Second-Order Boundary Value Problem

Authors: Chenghua Gao
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

This paper is concerned with the existence of solutions for the discrete second-order boundary value problemΔ2u(t-1)+λ1u(t)+g(Δu(t))=f(t), t∈{1,2,…,T}, u(0)=u(T+1)=0, where T>1 is an integer, f:{1,…,T}→ℝ, g:ℝ→ℝ is bounded and continuous, and λ1 is the first eigenvalue of the eigenvalue problem Δ2u(t-1)+λu(t)=0, t∈𝕋, u(0)=u(T+1)=0.


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## Body

## 1. Introduction

Letg:ℝ→ℝ and p:[0,π]→ℝ be continuous. The nonlinear two-point boundary value problem of ordinary differential equation(1.1)u′′(t)+u(t)+g(u′)=p(t),t∈(0,π),u(0)=u(π)=0,
is very important in applications. Let us mention the problems arising in viscosity, nonlinear oscillations, electric circuits, and so forth. The term g(u′) may be regarded as a nonlinear damping term in resonance problems and its appears, for example, in Rayleigh’s equation (which is closely connected with a theory of oscillation of violin string), in oscillations of a simple pendulum under the action of viscous damping, in dry (Coulomb) friction (which occurs when the surfaces of two solids are contact and relative motion without lubrication), and in some cases of van der Pol oscillator, see [1–4] and the references therein.Since the pioneer work of Landesman and Lazer [5], the problems of the type(1.2)u′′(t)+u(t)+g(u)=f(t),t∈(0,π),u(0)=u(π)=0,
(where g is independent of u′) have been extensively studied in the past forty years, see Iannacci and Nkashama [6] and the references therein.It has been remarked (see [7, 8]) that conditions of the Landesmen-Lazer type are not appropriated to yield the existence of solutions to (1.1). Thus, it is usually much more difficult to deal with (1.1) than to deal with (1.2), see Kannan et al. [7], Cañada and Drábek [8], Habets and Sanchez [9], Drábek et al. [10], and Del Toro and Roca [11].In [8], Cañada and Drábek used the well-known Lyapunov-Schmidt method and the Schauder fixed point theorem to find a necessary and sufficient condition for the existence of solutions of (1.1). To wit, they provedTheorem A (See [8, Theorem  3.1]).
Letp:[0,π]→ℝ be continuous and let
(1.3)p(t)=s2πsin⁡t+p̃(t),s∈R,∫0πp̃(t)sin⁡tdt=0.Letg:ℝ→ℝ be continuous and bounded with g(-∞)=g(+∞) and g(ξ)<g(+∞) for ξ∈ℝ, where(1.4)g(-∞):=lim⁡s→-∞g(s),g(+∞):=lim⁡s→+∞g(s).Then for anyp̃∈C[0,π] with ∫0πp̃(t)sintdt=0, there exists a real number gp̃<22/πg(+∞) such that (1.1) has at least one solution u∈C2[0,π] if and only if(1.5)s∈[gp̃,22πg(+∞)).It is the purpose of this paper to establish the similar results for the discrete analogue of (1.1) of the form(1.6)Δ2u(t-1)+λ1u(t)+g(Δu(t))=f(t),t∈T,u(0)=u(T+1)=0,
where T>1 is an integer, 𝕋:={1,…,T}, g:ℝ→ℝ is bounded and continuous, f:𝕋→ℝ, λ1 is the first eigenvalue of the linear eigenvalue problem(1.7)Δ2u(t-1)+λu(t)=0,t∈T,u(0)=u(T+1)=0.
Finally, it is worth remarking that the existence of solutions for nonlinear problem(1.8)Δ2u(t-1)+λ1u(t)+g(u(t))=f(t),t∈T,u(0)=u(T+1)=0,
which is a discrete analogue of (1.2), has been studied by Rodriguez [12] and Ma [13]. For other recent results on the existence of solutions of discrete problems, see [14–21] and the reference therein.The rest of this paper is arranged as follows. In Section2, we give some preliminaries and develop the methods of lower and upper solutions for the more generalized problems, that is, the case of the nonlinearity g=g(t,u,Δu); in Section 3, we state our main result and provide the proof.


## 2. Preliminaries

Recall that𝕋={1,2,…,T}. Let 𝕋̂={0,1,…,T+1}. Let X:={u∣u:𝕋̂→ℝ}, Y:={u∣u:𝕋→ℝ} be equipped with the norm(2.1)‖u‖X=max⁡k∈T̂|u(k)|,‖u‖Y=max⁡k∈T|u(k)|,respectively. It is easy to see that(X,∥·∥X) and (Y,∥·∥Y) are Banach spaces.Assume thatg0:𝕋×ℝ2→ℝ is a continuous function, bounded by a constant M>0:(2.2)|g0(t,η,ξ)|≤Mfort∈𝕋 and (η,ξ)∈ℝ2. Consider the following problem:(2.3)Δ2u(t-1)+λ1u(t)+g0(t,u(t),Δu(t))=f(t),t∈T,(2.4)u(0)=u(T+1)=0.Definition 2.1.
Ifx∈X satisfies
(2.5)Δ2x(t-1)+λ1x(t)≥f(t)-g0(t,x(t),Δx(t)),t∈T,x(0)≤0,x(T+1)≤0,
then one says x(t) is a lower solution of (2.3), (2.4). If y∈X satisfies
(2.6)Δ2y(t-1)+λ1y(t)≤f(t)-g0(t,y(t),Δy(t)),t∈T,y(0)≥0,y(T+1)≥0,
then one says y(t) is an upper solution of (2.3), (2.4).Theorem 2.2.
Suppose thatx(t), y(t) are the lower and upper solutions of (2.3), (2.4), respectively, and x(t)≤y(t), t∈𝕋. Then BVP (2.3) and (2.4) have at least one solution u(t) satisfies
(2.7)x(t)≤u(t)≤y(t).Proof.
Define the functionp:𝕋×ℝ→ℝ by
(2.8)p(t,u(t))={x(t),u(t)<x(t),u(t),x(t)≤u(t)≤y(t),y(t),u(t)>y(t).
Set f*(t,u,v)=f(t)-g0(t,u,v)-λ1u. Consider the auxiliary problems:
(2.9)Δ2u(t-1)=f*(t,p(t,u),Δp(t,u)),t∈T,u(0)=u(T+1)=0.
From (2.8) and the boundness of g0, we know f*(t,p(t,u),Δp(t,u)) is bounded. So, by the Schauder fixed point theorem, (2.9) has a solution u∈X.
Now, we only proveu(t)≤y(t), the other case u(t)≥x(t) is similar.
Setz(t)=u(t)-y(t). Suppose that z(t)>0, for t∈{t0+1,t0+2,…,t0+p}, and z(t0)≤0, z(t0+p+1)≤0, where t0∈{0,1,…,T}, p∈{1,2,…,T}.
On the other hand, by the definition of upper solution, fort∈{t0+1,t0+2,…,t0+p},(2.10)Δ2y(t-1)≤f*(t,y(t),Δy(t))=f(t)-λ1y(t)-g0(t,y(t),Δy(t))=f(t)-λ1p(t,u)-g0(t,p(t,u(t)),Δp(t,u(t)))=Δ2u(t-1).
Then
(2.11)Δ2z(t-1)≥0,t∈{t0+1,t0+2,…,t0+p},z(t0)≤0,z(t0+p+1)≤0.
Now, by the convexity of z on {t0+1,t0+2,…,t0+p}, we get z(t)≤0, t∈{t0+1,t0+2,…,t0+p}, that is, u(t)≤y(t), t∈{t0+1,t0+2,…,t0+p}. This contradicts u(t)>y(t), t∈{t0+1,t0+2,…,t0+p}. Thus, u(t)≤y(t), t∈𝕋̂.Lemma 2.3.
See∑t=1Tsin2(πt/(T+1))=(T+1)/2.Proof.
Letω=cos(2π/(T+1))+isin(2π/(T+1)). Then ωT+1=1 and (1-ω)(1+ω+ω2+⋯+ωT)=0. Since 1-ω≠0, we have ∑t=1Tcos(2πt/(T+1))=-1. This together with the fact that ∑t=1Tsin2(πt/(T+1))=∑t=1T(1-cos(2πt/(T+1)))/2 implies the assertion holds.Now, letψ1(t):=2/(T+1)sin(πt/(T+1)), t∈𝕋, denote the positive eigenfunction corresponding to the first eigenvalue λ1=4sin2(π/2(T+1)) of (1.7). Then by Lemma 2.3, ∑t=1Tψ12(t)=1.Sinceψ1(t) is located on 2/(T+1)sint,t∈[0,π], by the direct computation, we can obtain the following result.Lemma 2.4.
IfT is an odd number, then
(2.12)Δψ1(t)>0,fort∈{0,…[T2]},Δψ1(t)<0,fort∈{[T2]+1,…,T},
if T is an even number, then Δψ1(T/2)=0,
(2.13)Δψ1(t)>0,fort∈{0,…T2-1},Δψ1(t)<0,fort∈{T2+1,…,T}.Define the operatorL:D(L)⊂X→Y by(2.14)Lu(t)=Δ2u(t-1)+λ1u(t),whereD(L)={u∈X∣u(0)=u(T+1)=0}.DefineN:X→Y by(2.15)(Nu)(t)=f(t)-g0(t,u(t),Δu(t)).Then (2.3), (2.4) is equivalent to the operator equation Lu=Nu.In Theorem2.2, we established the methods of lower and upper solutions under well order. Now, we can also develop the methods of lower and upper solutions for (2.3), (2.4) when x(t)≤y(t) is not necessary, its proofs are based on the following lemma, that is, the connectivity properties of the solution sets of parameterized families of compact vector fields, they are a direct consequence of Mawhin [22, Lemma  2.3].Lemma 2.5 (see [22, Lemma  2.3]).
LetE be a Banach space and C⊂E a nonempty, bounded, closed convex subset. Suppose that T:[a,b]×C→C is completely continuous. Then the set
(2.16)S={(λ,x)∣T(λ,x)=x,λ∈[a,b]}
contains to be a closed connected subset Σ which connects {a}×C to {b}×C.Theorem 2.6.
Assume thatx(t), y(t) are the lower solution and the upper solution of (2.3), (2.4), respectively. Then (2.3) and (2.4) have at least one solution.Proof.
Define the projectionsP:X→X, Q:Y→ℝ by
(2.17)(Pu)(t)=[∑t=1Tu(t)ψ1(t)]ψ1(t),(Qy)(t)=∑t=1Ty(t)ψ1(t).
Then ImP=KerL, KerQ=ImL, and X=(KerP⊕KerL),Y=(ImL⊕ImQ). Now, the operator equation Lu=Nu is equivalent to the alternative system
(2.18)u-Pu=K(I-Q)Nu,QNu=0,
where K is the inverse of mapping L:(D(L)∩KerP)→ImL.
Writingu∈D(L) in the form u(t)=cψ1(t)+w(t), c∈ℝ, ∑t=1Tψ1(t)w(t)=0, (2.3) and (2.4) are equivalent to the system
(2.19)w=K(I-Q)N(cψ1(⋅)+w),(2.20)QN(cψ1(⋅)+w)=0.
Since X is finite dimensional, it is easy to see that K(I-Q)N is completely continuous, by the Schauder fixed point theorem and the fact N(cψ1(t)+w(t)) is bounded, we get that for any fixed c∈ℝ, W(c):={w∈X∩KerP∣(c,w)satisfies(2.19)}≠∅ and W(c) is bounded. Then there exist positive constants α>τ such that -τψ1≤w≤τψ1 for all w∈W(c), (α-τ)ψ1≥x(t) and -(α-τ)ψ1≤y(t). Let
(2.21)γ(c,w)=∑t=1TN(cψ1(t)+w(t))ψ1(t),
for all (c,w)∈ℝ×W(c). Observe that Lemma 2.5 is applicable. Hence there exists a connected subset of {(c,w)∈R×(X∩KerP)∣(c,w)satisfies(2.19)}, Σα(ψ1), which connected {-α}×W(-α) and {α}×W(α). Since γ:Σα(ψ1)→ℝ is continuous, I:=γ(Σα(ψ1)) is an interval. If 0∈I, then (2.3) and (2.4) have a solution. If I⊂(0,∞), then every cψ1+w with (c,w)∈Σα(ψ1) is an upper solution. Indeed, it is obvious that
(2.22)L(cψ1(t)+w(t))-N(cψ1(t)+w(t))=-γ(c,w)≤0,t∈T,(cψ1+w)(0)=(cψ1+w)(T+1)=0.
By construction, αψ1+w with (α,w)∈Σα(ψ1) satisfies x(t)≤αψ1+w. Hence, from Theorem 2.2, (2.3) and (2.4) have a solution. A similar argument applies if I⊂(-∞,0).Theorem 2.7.
Suppose thatf satisfies
(2.23)f(t)=sψ1(t)+f̃(t),s∈R,
where f̃ satisfies
(2.24)∑t=1Tf̃(t)ψ1(t)=0.
Then, there exists a nonempty, connected, and bounded set Jf̃⊂ℝ such that (2.3) and (2.4) have at least one solution u∈X if and only ifs∈Jf̃.Proof.
As the proof of Theorem2.6, (2.3) and (2.4) are equivalent to the system (2.19), (2.20). Since N is bounded, applying the Schauder fixed point theorem we obtain that for any fixed c∈ℝ, there exits at least one wc∈X such that (2.19) holds.
Now, (2.20) becomes
(2.25)∑t=1Tg0(t,cψ1(t)+wc(t),cΔψ1(t)+Δwc(t))ψ1(t)=s.
Hence, for a given f̃, ∑t=1Tf̃(t)ψ1(t)=0, (2.3), (2.4)  with f(t)=cψ1(t)+f̃(t) has at least one solution if and only if s belongs to the range of the (multivalued, in general) function Γf̃:ℝ→Γf̃(ℝ),
(2.26)Γf̃(c)=∑t=1Tg0(t,cψ1(t)+wc(t),cΔψ1(t)+Δwc(t))ψ1(t),
where wc∈{w∈D(L):wisasolutionof(2.19)forfixedc}. But Jf̃≡Γf̃(ℝ) is a connected set. In fact, let s1 and s2 belong to Jf̃ and s1≤s2. Then (2.3), (2.4) with f1=s1ψ1+f̃ and f2=s2ψ1+f̃ has solutions u1 and u2, respectively. If we consider (2.3), (2.4) with f=sψ1+f̃, where s∈[s1,s2], then u1 is an upper solution and u2 is a lower solution to this problem. By Theorem 2.6, there exists at least one solution, that is, s belongs to Jf̃. Moreover, since g is bounded, the range of Γf̃ is bounded.


## 3. Main Results

In this section, we deal with (1.6). First, let us make the following assumptions:(H1)g:ℝ→ℝ is a bounded and continuous function and satisfies g(+∞)=g(-∞) and g(ξ)<g(+∞) for any ξ∈ℝ,(H2)f:𝕋→ℝ satisfies(3.1)f(t)=sψ1(t)+f̃(t),s∈R,∑t=1Tf̃(t)ψ1(t)=0.Theorem 3.1.
Suppose that (H1),  (H2) hold. Then there exists a real numbergf̃, gf̃<2/(T+1)g(+∞)∑t=1Tsin(πt/(T+1)), such that (1.6) has at least one solution u∈X if and only if
(3.2)s∈[gf̃,2T+1g(+∞)∑t=1Tsin⁡πtT+1).Proof.
Note thatψ1(t)=2/(T+1)sin(πt/(T+1)). Due to the consideration in the the proof of Theorem 2.7. It is sufficient to show that for a given f̃ with ∑t=1Tf̃(t)ψ1(t)=0, we have
(3.3)Γf̃(R)=[gf̃,2T+1g(+∞)∑t=1Tsin⁡πtT+1).
The (possibly multivalued) function Γf̃ has the following form:
(3.4)Γf̃(c)=2T+1∑t=1Tg(cΔψ1(t)+Δwc(t))sin⁡πtT+1,
where c∈ℝ and wc verify (2.19). From the boundedness of g and (2.19), there exists a constant D>0 (independent of c) such that ∥wc∥X≤D for any c∈ℝ, furthermore,
(3.5)max⁡t∈{0,1,…,T}|Δwc(t)|≤2D.
Now, we divide the proof into two cases.Case 1.
T is an odd number. By Lemma 2.4, we obtain that
(3.6)∑t=1Tg(2c2T+1sin⁡π2(T+1)cos⁡π(2t+1)2(T+1)+Δwc(t))sin⁡πtT+1=∑t=1[T/2]g(2c2T+1sin⁡π2(T+1)cos⁡π(2t+1)2(T+1)+Δwc(t))sin⁡πtT+1+∑t=[T/2]+1Tg(2c2T+1sin⁡π2(T+1)cos⁡π(2t+1)2(T+1)+Δwc(t))sin⁡πtT+1⟶∑t=1[T/2]g(±∞)sin⁡πtT+1+∑s=[T/2]+1Tg(∓∞)sin⁡πtT+1,
as c→±∞. Due to g(+∞)=g(-∞), we get
(3.7)Γf̃(c)→2T+1g(+∞)∑t=1Tsin⁡πtT+1.
The assumption g(ξ)<g(+∞), ξ∈ℝ, and (3.5) yields
(3.8)Γf̃(c)<2T+1g(+∞)∑t=1Tsin⁡πtT+1
for any c∈ℝ.Case 2.
T is an even number. By Lemma 2.4, we know that
(3.9)Δψ1(T2)=2T+1Δsin⁡πT2(T+1)=22T+1sin⁡π2(T+1)cos⁡π(2T/2+1)2(T+1)=0.
Hence,
(3.10)∑t=1Tg(2c2T+1sin⁡π2(T+1)cos⁡π(2t+1)2(T+1)+Δwc(t))sin⁡πtT+1=∑t=1(T/2)-1g(2c2T+1sin⁡π2(T+1)cos⁡π(2t+1)2(T+1)+Δwc(t))sin⁡πtT+1+∑t=(T/2)+1Tg(2c2T+1sin⁡π2(T+1)cos⁡π(2t+1)2(T+1)+Δwc(t))sin⁡πtT+1+g(Δwc(T2))sin⁡πT2(T+1).By (3.5) and the assumption g(ξ)<g(+∞), ξ∈ℝ, we know that for any c∈ℝ, g(Δwc(T/2))<g(+∞). Thus, for any c∈ℝ,
(3.11)Γf̃(c)<2T+1g(+∞)∑t=1Tsin⁡πtT+1.
It is sufficient to prove that this infimum is achieved. Let us denote(3.12)gf̃=inf⁡c∈RΓf̃(c).
Suppose that {sn}⊂Γf̃(ℝ) satisfies sn→gf̃ and {cn} is the corresponding minimizing sequence, that is, un(t)=cn2/(T+1)sin(πt/(T+1))+wcn(t), are the solution of (1.1), with the right-hand sides fn(t)=sn2/(T+1)sin(πt/(T+1))+f̃(t).
We claim that{cn} is bounded. In fact, if cn→∞ as n→∞, then we can get two contradictions in the following two cases.
Case 1.
IfT is an odd number, then by (2.20),
(3.13)2T+1∑t=1Tg(2cn2T+1sin⁡π2(T+1)cos⁡π(2t+1)2(T+1)+Δwcn(t))sin⁡πtT+1=sn,
letting n→∞ in (3.13), we get
(3.14)2T+1∑t=1Tg(+∞)sin⁡πtT+1=gf̃.From (H1), we arrive for any ξ∈ℝ, g(ξ)<g(+∞), which together with (3.14) implies that
(3.15)2T+1∑t=1Tg(2c2T+1sin⁡π2(T+1)cos⁡π(2t+1)2(T+1)+Δwc(t))sin⁡πtT+1<gf̃.
This contradicts (3.12).Case 2.
IfT is an even number, then by (2.20) and Δψ1(T/2)=0, we get
(3.16)sn=2T+1∑t=1(T/2)-1g(2cn2T+1sin⁡π2(T+1)cos⁡π(2t+1)2(T+1)+Δwcn(t))sin⁡πtT+1+2T+1∑t=(T/2)+1Tg(2cn2T+1sin⁡π2(T+1)cos⁡π(2t+1)2(T+1)+Δwcn(t))sin⁡πtT+1+2T+1g(Δwcn(T2))sin⁡πT2(T+1).This implies that
(3.17)gf̃≥2T+1∑t=1(T/2)-1g(+∞)sin⁡πtT+1+2T+1∑t=(T/2)+1Tg(+∞)sin⁡πtT+1+2T+1inf⁡wcn∈Xg(Δwcn(T2))sin⁡πT2(T+1).
On the other hand, by (3.12) and (H1), we get that for any fixed n∈ℕ,
(3.18)gf̃≤2T+1∑t=1(T/2)-1g(2cn2T+1sin⁡π2(T+1)cos⁡π(2t+1)2(T+1)+Δwcn(t))sin⁡πtT+1+2T+1∑t=(T/2)+1Tg(2cn2T+1sin⁡π2(T+1)cos⁡π(2t+1)2(T+1)+Δwcn(t))sin⁡πtT+1+2T+1inf⁡wcn∈Xg(Δwcn(T2))sin⁡πT2(T+1)<2T+1∑t=1(T/2)-1g(+∞)sin⁡πtT+1+2T+1∑t=(T/2)+1Tg(+∞)sin⁡πtT+1+2T+1inf⁡wcn∈Xg(Δwcn(T2))sin⁡πT2(T+1).
Now, we obtain a contradiction. Thus, cn is bounded.
SinceX is finite dimensional and wcn is bounded, we obtain that cn→c, wcn→wc (at least for a subsequence), and u(t)=c2/(T+1)sin(πt/(T+1))+wc(t) is a solution of (1.1) with f(t)=gf̃2/(T+1)sin(πt/(T+1))+f̃(t). Hence, the infimum is achieved in c.



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*Source: 951251-2012-05-09.xml*