# Existence and Multiplicity of Positive Solutions of a Nonlinear Discrete Fourth-Order Boundary Value Problem

Authors: Ruyun Ma; Yanqiong Lu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

we show the existence and multiplicity of positive solutions of the nonlinear
discrete fourth-order boundary value problemΔ4ut-2=λhtfut, t∈𝕋2, u1=uT+1=Δ2u0=Δ2uT=0, where λ>0, h:𝕋2→(0,∞) is continuous, and f:ℝ→[0,∞) is continuous, T>4, 𝕋2=2,3,…,T. The main tool is the Dancer's global bifurcation theorem.


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## Body

## 1. Introduction

It's well known that the fourth order boundary value problem(1.1)u′′′′(t)=f(t,u(t)),t∈(0,1),u(0)=u(1)=u′′(0)=u′′(1)=0
can describe the stationary states of the deflection of an elastic beam with both ends hinged, (it also models a rotating shaft). The existence and multiplicity of positive solutions of the boundary value problem (1.1) have been considered extensively in the literature, see [1–10]. The existence and multiplicity of positive solutions of the parameterized boundary value problem (1.2)u′′′′(t)=λh(t)f(u(t)),t∈(0,1),u(0)=u(1)=u′′(0)=u′′(1)=0
have also been studied by several authors, see Bai and Wang [11], Cid et al. [12], and the references therein.However, relatively little is known about the corresponding discrete fourth-order problems. Let(1.3)T>4,T0={0,1,…,T+2},T1={1,2,…,T+1},T2={2,3,…,T}.Zhang et al. [13], and He and Yu [14] used the fixed point index theory in cones to study the following discrete analogue(1.4)Δ4u(t-2)=λh(t)f(u(t)),t∈T2,(1.5)u(0)=u(T+2)=Δ2u(0)=Δ2u(T)=0,
where Δ4u(t-2) denote the fourth forward difference operator and Δu(t)=u(t+1)-u(t). It has been pointed out in [13, 14] that (1.4), (1.5) are equivalent to the equation of the form: (1.6)u(t)=λ∑s=1T+1G(t,s)∑j=2TG1(s,j)h(j)f(u(j))=:A0u(t),t∈T0,
where (1.7)G(t,s)=1T+2{s(T+2-t),1≤s≤t≤T+2,t(T+2-s),0≤t≤s≤T+1,G1(s,j)=1T{(T+1-s)(j-1),2≤j≤s≤T+1,(T+1-j)(s-1),1≤s≤j≤T.Notice that two distinct Green's functions used in  (1.6) make the construction of cones and the verification of strong positivity of A0 become more complex and difficult. Therefore, Ma and Xu [15] considered (1.4) with the boundary condition (1.8)u(1)=u(T+1)=Δ2u(0)=Δ2u(T)=0,
and introduced the definition of generalized positive solutions:Definition 1.1.
A functiony:𝕋0→ℝ+ is called a generalized positive solution of (1.4), (1.8), if y satisfies (1.4), (1.8), and y(t)≥0 on 𝕋1 and y(t)>0 on 𝕋2.Remark 1.2.
Notice that the facty:𝕋0→ℝ+ is a generalized positive solution of (1.4), (1.8) does not means that y(t)≥0 on 𝕋0. In fact, y satisfies
(1)y(t)≥0 for t∈𝕋2;
(2)y(1)=y(T+1)=0;
(3)y(0)=-y(2),y(T+2)=-y(T).Ma and Xu [15] also applied the fixed point theorem in cones to obtain some results on the existence of generalized positive solutions.It is the purpose of this paper to show some new results on the existence and multiplicity of generalized positive solutions of (1.4), (1.8) by Dancer's global bifurcation theorem. To wit, we get the following.Theorem 1.3.
Leth:𝕋2→(0,∞),f∈C(ℝ,[0,∞)), and(1.9)lim⁡s→0+⁡f(s)s=f0∈(0,∞),lim⁡s→∞⁡f(s)s=f∞=+∞.
Assume that there exists B∈[0,+∞]  such that f is nondecreasing on [0,B). Then(i)
(1.4), (1.8) have at least one generalized positive solution if 0<λ<λ1/f0;(ii)
(1.4), (1.8) have at least two generalized positive solutions if
(1.10)λ1f0<λ<sup⁡s∈(0,B)sγ*f(s),
where γ*=maxt∈𝕋1∑s=2TK(t,s)h(s), K(t,s) is defined as (2.3) and λ1 is the first eigenvalue of
(1.11)Δ4u(t-2)=λh(t)u(t),t∈T2,u(1)=u(T+1)=Δ2u(0)=Δ2u(T)=0.The “dual” of Theorem1.3 is as follows.Theorem 1.4.
Leth:𝕋2→(0,∞),f∈C(ℝ,[0,∞)), and(1.12)lim⁡s→0+⁡f(s)s=f0∈(0,∞),lim⁡s→∞⁡f(s)s=f∞=0.
Assume that there exists B∈[0,+∞] such that f is nondecreasing on [0,B). Then
(i)
(1.4), (1.8) have at least a generalized positive solution provided
(1.13)λ>inf⁡s∈(0,c1B)⁡sc1γ*f(s),whereγ*=mint∈𝕋2∑s=2TK(t,s)h(s);(ii)
(1.4), (1.8) have at least two generalized positive solutions provided(1.14)inf⁡s∈(0,c1B)⁡sc1γ*f(s)<λ<λ1f0.The rest of the paper is organized as follows: in Section2, we present the form of the Green's function of (1.4), (1.8) and its properties, and we enunciate the Dancer's global bifurcation theorem ([16, Corollary  15.2]). In Section 3, we use the Dancer's bifurcation theorem to prove Theorems 1.3 and 1.4 and in Section 4, we finish the paper presenting a couple of illustrative examples.Remark 1.5.
For other results on the existence and multiplicity of positive solutions and nodal solutions for fourth-order boundary value problems based on bifurcation techniques, see [17–21].


## 2. Preliminaries and Dancer's Global Bifurcation Theorem

Lemma 2.1.
Leth:𝕋2→ℝ. Then the linear boundary value problem(2.1)Δ4u(t-2)=h(t),t∈T2,u(1)=u(T+1)=Δ2u(0)=Δ2u(T)=0
has a solution(2.2)u(t)=∑s=2TK(t,s)h(s),t∈T1,
where
(2.3)K(t,s)={(s-1)(T+1-t)(2T(t-1)-(t-1)2-(s-2)s)6T,2≤s≤t≤T+1,(t-1)(T+1-t)(2T(s-1)-(s-1)2-(t-2)t)6T,1≤t≤s≤T.Proof.
By a simple summing computation andu(1)=Δ2u(0)=0, we can obtain(2.4)u(t)=Δu(0)(t-1)+t(t-1)(t-2)6Δ3u(0)+∑s=2t-1(t-s)(t-s-1)(t-s+1)6h(s).
This together withu(T+1)=Δ2u(T)=0, it follows that
(2.5)u(t)=∑s=2T(T+1-s)(t-1)[2T(s-1)-(s-1)2-t(t-2)]6Th(s)+∑s=2t-1(t-s)(t-s-1)(t-s+1)6Th(s)=∑s=tT(T+1-s)(t-1)[2T(s-1)-(s-1)2-t(t-2)]6Th(s)+∑s=2t-1(T+1-t)(s-1)[2T(t-1)-(t-1)2-s(s-2)]6Th(s).
Therefore, (2.2) holds.Remark 2.2.
It has been pointed out in [15] that (2.1) is equivalent to the summation equation of the form(2.6)u(t)=∑s=2TG1(t,s)∑j=2TG1(s,j)h(j),t∈T1.
It is easy to verify that (2.2) and (2.6) are equivalent.
By a similar method in [9], it follows that K(t,s) satisfies(2.7)K(t,s)≤Φ(s)fors∈T1,t∈T1,K(t,s)≥c(t)Φ(s)fors∈T1,t∈T1,
where
(2.8)Φ(s)={327T(s-1)(T2-(s-2)s)3/2,1≤s≤T2+1,327T(T+1-s)(2T(s-1)-(s-2)s)3/2,T2+1<s≤T+1,(2.9)c(t)={33[T2-t(t-2)](t-1)2(T2+1)3/2,1≤t≤T2+1,33(T+1-t)[2T(t-1)-t(t-2)]2(T2+1)3/2,T2+1<t≤T+1.
Moreover, we have that(2.10)K(t,s)≥c1Φ(s),fors∈T1,t∈T2.
here c1=33T2/2(T2+1)3/2.
LetX be a real Banach space with a cone K such that X=K-K. Let us consider the equation:(2.11)x=μ(Lx+Nx),μ∈R,x∈X
under the assumptions:
(A1)
The operatorsL,N:X→X are compact. Furthermore, L is linear, ∥Nx∥X/∥x∥X→0 as ∥x∥X→0, and (L+N)(K)⊆K.(A2)
The spectral radiusr(L) of L is positive. Denote μ0=r(L)-1.(A3)
L is strongly positive.Dancer's global bifurcation theorem is the following.Theorem 2.3 (see [16, Corollary  15.2]).
Let(2.12)S+:={(μ,x)∈R×X∣(μ,x)isasolutionof(2.11)withx>0andμ>0}.
If (A1) and (A2) are satisfied, then (μ0,0) is a bifurcation point of (2.11) and S¯+ has an unbounded solution component 𝒞+ which passes through (μ0,0). Additionally, if (A3) is satisfied, then (μ,x)∈𝒞+ and μ≠μ0 always impliesx>0 and μ>0.


## 3. Proof of the Main Results

Before proving Theorem1.3, we state some preliminary results and notations. Let (3.1)ρ:=4sin⁡2π2T,e(t):=sin⁡π(t-1)T,t∈T1,(3.2)X:={u∣u:T0⟶R,u(1)=u(T+1)=Δ2u(0)=Δ2u(T)=0}.
Then X is a Banach space under the normal:(3.3)‖u‖X:=inf⁡{γρ∣-γe(t)≤-Δ2u(t-1)≤γe(t),t∈T1}.
See [22] for the detail.Let(3.4)K:={u∈X∣Δ2u(t-1)≤0,u(t)≥0,t∈T1}.
Then K is normal and has a nonempty interior and X=K-K.LetY={u∣u:𝕋2→ℝ}. Then Y is a Banach space under the norm: (3.5)‖u‖∞=max⁡t∈T2|u(t)|.
Define ℒ:X→Y by setting (3.6)Lu≔Δ4u(t-2),u∈X.It is easy to check thatℒ-1:Y→X is compact.Lemma 3.1.
Leth∈Y with h≥0 and h(t0)>0 for some t0∈𝕋2, and
(3.7)Lu-h=0.
Then u∈intK.Proof.
It is enough to show that there exist two constantsr1,r2∈(0,∞) such that
(3.8)r1e(t)≤-Δ2u(t-1)≤r2e(t),t∈T1.
In fact, we have from (3.7) that(3.9)-Δ2u(t-1)=∑s=2TG1(t,s)h(s),t∈T1.
This together with the relation ((t-1)(T+1-t)/T)G1(s,s)≤G1(t,s)≤(t-1)(T+1-t)/T implies that(3.10)[∑s=2TG1(s,s)h(s)](t-1)(T+1-t)T≤∑s=2TG1(t,s)h(s)≤‖h‖∞(t-1)(T+1-t)T.
Combining (3.9) with (3.10) and the fact that
(3.11)c1sin⁡π(t-1)T≤(t-1)(T+1-t)T≤c2sin⁡π(t-1)T,t∈T1
for some constants c1,c2∈(0,∞), we conclude that (3.8) is true.Letζ∈C(ℝ,ℝ) be such that (3.12)f(u)=f0u+ζ(u),
clearly (3.13)lim⁡|u|→0⁡ζ(u)u=0.
Let us consider (3.14)Lu=λ(h(⋅)f0u+h(⋅)ζ(u))
as a bifurcation problem from the trivial solution u≡0.By (1.4), (1.8), it follows that if u(t)∈X is one solution of (1.4), (1.8), then u(t) satisfies u(0)=-u(2),u(T+2)=-u(T). So, (u(0),0,u(2),…,u(T),0,u(T+2)) is a solution of (1.4), (1.8), if and only if, (0,u(2),…,u(T),0) solves the operator equation (3.15)u(t)=λ∑s=2TG(t,s)h(s)f(u(s)),t∈T1.Now, letJ:Y→X be the linear operator: (3.16)J(u(2),u(3),…,u(T))=(-u(2),0,u(2),u(3),…,u(T),0,-u(T)),u∈Y.LetL,N:X→X be the operators: (3.17)Lu:=(J∘L)-1(h(⋅)f0u),(3.18)Nu:=(J∘L)-1(h(⋅)ζ(u)),
respectively. Then Lemma 3.1 yields that L:X→X is strongly positive. Moreover, [16, Theorem  7.c] implies r(L)>0.Now, it follows from Theorem2.3 that there exists a continuum (3.19)C+⊆{(μ,x)∈R×X∣(μ,x)isasolutionof(1.4),(1.8)withx>0andμ>0}̅,
which joins (r(L)-1,0) with infinity in (0,∞)×K and (3.20)(μ,x)∈C+,μ≠r(L)-1⟹x>0,μ>0.
It is easy to check that (3.21)r(L)-1=λ1f0.Lemma 3.2.
Leth1,h2∈Y with h1≥h2>0. Then the eigenvalue problems
(3.22)Lu(t)=λhi(t)u(t),t∈T2,i=1,2
have the principal eigenvalue λ¯i, i=1,2 such that λ¯1≤λ¯2. Moreover, the corresponding eigenfunctions ψi are positive in 𝕋2.Proof .
LetLi:X→X be the operator
(3.23)Liu:=λ(J∘L)-1(hi(⋅)u),i=1,2.
Then Lemma3.1 yields that Li:X→X is strongly positive. By Krein-Rutman theorem [16, Theorem  7.c] the spectral radius r(Li)>0 and there exist ψi∈X with ψi>0 on 𝕋2 such that (3.24)Liψi(t)=r(Li)ψi(t),i=1,2.
That is, the eigenvalue problems (3.22) have the principal eigenvalues λ¯i=1/r(Li), and ψi(t) is the corresponding eigenfunctions of λ¯i, i=1,2.
Next, we proveλ¯1≤λ¯2. Since ∑t=2TΔ4ψ1(t-2)ψ2(t)=∑t=2Tψ1(t)Δ4ψ2(t-2), it follows that(3.25)∑t=2Th1(t)ψ1(t)ψ2(t)≥∑t=2Tλ¯2λ¯2h2(t)ψ2(t)ψ1(t)=∑t=2T1λ¯2Δ4ψ2(t-2)ψ1(t)=∑t=2T1λ¯2ψ2(t)Δ4ψ1(t-2)=∑t=2Tψ2(t)λ¯2λ¯1h1(t)ψ1(t)=λ¯1λ¯2∑t=2Th1(t)ψ1(t)ψ2(t).
Therefore, λ¯1≤λ¯2.Suppose that𝕋a={a+1,a+2,…,b-1} is a strict subset of 𝕋2 and ha denote the restriction of h on 𝕋a. Consider the linear eigenvalue problems: (3.26)Δ4u(t-2)=λha(t)f0u(t),t∈Ta,u(a)=u(b)=Δ2u(a-1)=Δ2u(b-1)=0.
Then we get the following result.Lemma 3.3.
Letλ̃1 is the principal eigenvalue of (3.17), then (3.26) has only one principal eigenvalue λa such that 0<λ̃1<λa.Proof.
It is not difficult to prove that (3.26) has only one principal eigenvalue λa>0 by Lemma 3.1, and the corresponding eigenfunction ψa>0 on 𝕋a. So we only to verify that 0<λ̃1<λa.
Letψ1 be the corresponding eigenfunction of λ̃1, we have that
(3.27)∑t=a+1b-1Δ4ψa(t-2)ψ1(t)=∑t=a+1b-1Δ4ψ1(t-2)ψa(t)-ψ1(b)Δ2ψa(b-2)-ψa(b-1)Δ2ψ1(b-1)-Δ2ψa(a)ψ1(a)-ψa(a+1)Δ2ψ1(a-1)>∑t=a+1b-1Δ4ψ1(t-2)ψa(t).
So
(3.28)∑t=a+1b-1h(t)ψa(t)ψ1(t)=∑t=a+1b-11λaΔ4ψa(t-2)ψ1(t)>∑t=a+1b-11λaΔ4ψ1(t-2)ψa(t)=∑t=a+1b-1ψa(t)λaλ̃1h(t)ψ1(t)=λ̃1λa∑t=a+1b-1h(t)ψa(t)ψ1(t).
Thus 0<λ̃1<λa.Proof of Theorem1.3.
We divide the proof into three steps.
Let{(μn,yn)}⊂𝒞+ be such that
(3.29)|μn|+‖yn‖X→∞,n→∞.
Then
(3.30)Δ4yn(t-2)=μnh(t)f(yn(t)),t∈T2,yn(1)=yn(T+1)=Δ2yn(0)=Δ2yn(T)=0.
Step 1.
We show that there exists a constantM such that μn∈(0,M] for all n.
Suppose on the contrary that(3.31)lim⁡n→∞⁡μn=∞.
Let vn=yn/∥yn∥X. Then it follows from (3.30) that (3.32)Δ4vn(t-2)=μnh(t)f(yn(t))yn(t)vn(t),t∈T2,vn(1)=vn(T+1)=Δ2vn(0)=Δ2vn(T)=0.
Since(3.33)inf⁡{f(s)s∣s>0}:=M0>0,
there exists a constant M0>0, such that
(3.34)f(yn(t))yn(t)>M0>0.
Let λ* be the principal eigenvalue of the linear eigenvalue problems:
(3.35)Δ4v(t-2)=λh(t)M0v(t),t∈T2,v(1)=v(T+1)=Δ2v(0)=Δ2v(T)=0.
Combining (3.31) and (3.34) with the relation (3.32), using Lemma 3.2, we get
(3.36)0<μn≤λ*.
This contradicts (3.31). So μn∈(0,M] for all n.
Step 2.
We show that𝒞+ joins (λ1/f0,0) with (0,∞).
Assume that there existδ>0 and {(μn,yn)}⊂𝒞+ such that
(3.37)0<δ≤μn≤M;‖yn‖X⟶∞,n⟶∞.
First, we show that(3.38)‖yn‖X⟶∞⟹‖yn‖∞⟶∞.
Suppose on the contrary that(3.39)‖yn‖∞≤M1
for some M1>0 (independent on n). Then it follows from (3.30) and 0<δ≤|μn|≤M that (3.40)‖Δ4yn‖∞≤M‖h‖∞sup⁡{f(s)∣0<s≤M1},
and subsequently, {∥yn∥X} is bounded. This is a contradiction. So, (3.38) holds.
Next, we show that(3.41)‖yn‖∞⟶∞⇒min⁡{yn(t)∣t∈T2}⟶∞.
In fact,(3.42)yn(t)=μn∑s=2TK(t,s)h(s)f(yn(s)),t∈T1.
This together with (2.7) imply that (3.41) is valid.
Finally, we have from the facts thatmin{yn(t)∣t∈𝕋2}→∞ and 0<δ≤|μn|≤M that
(3.43)μnf(yn(t))yn(t)⟶∞,n⟶∞foranyt∈Ta.
Consider the following linear eigenvalue problems:
(3.44)Δ4v(t-2)=λha(t)v(t),t∈Ta,v(a)=v(b)=Δ2v(a-1)=Δ2v(b-1)=0.
By Lemma 3.3 and (3.32), (3.44) has a positive principal eigenvalue λa, and(3.45)μnf(yn(t))yn(t)≤λa,
which contradicts (3.43). Thus limn→∞μn=0.
Step 3.
Fixedλ such that
(3.46)0<λ<sup⁡s∈(0,B)sγ*f(s).
Then there exists b∈(0,B] such that (3.47)0<λ<bγ*f(b).
We show that there is no (μ,u)∈𝒞+ such that
(3.48)‖u‖∞=b,0<μ<bγ*f(b).
In fact, if there exists (η,y)∈𝒞+ satisfying (3.48), then
(3.49)y(t)=η∑s=2TK(t,s)h(s)f(y(s))≤ηγ*f(b)=ηγ*f(b)b⋅b
for t∈𝕋1, and subsequently, η≥b/γ*f(b). Therefore, no (μ,u)∈𝒞+ satisfies (3.48).Now, combining the conclusions in Steps 2 and 3, using the fact that no (μ,u)∈𝒞+ satisfies (3.48), it concludes that for every λ∈(λ1/f0,b/γ*f(b)), (1.4), (1.8) has at least two generalized positive solutions in 𝒞+. For arbitrary λ∈(0,sups∈(0,B)(s/γ*f(s))), we may find b=b(λ) satisfying (3.47). So, for every λ∈(λ1/f0,sups∈(0,B)(s/γ*f(s))), (1.4), (1.8) has at least two generalized positive solutions in 𝒞+.Proof of Theorem1.4.
We divide the proof into three steps.
Step 1.
We show that there exists a positive constantβ>0 such that
(3.50)inf⁡{μ∣(μ,u)∈C+}=:β>0.
Suppose on the contrary that there exists {(μn,yn)}⊂𝒞+ such that
(3.51)μn→0+,asn→∞.
Then we have from (3.32), (3.51), f0∈(0,∞) and f∞=0 that
(3.52)‖vn‖X→0,asn→∞.
However, this contradicts with the fact that ∥vn∥X=1 for all n∈ℕ. Therefore, (3.50) holds.
Step 2.
We show that for any closed intervalI⊂[β,∞), there exists MI>0 such that
(3.53)sup⁡{‖u‖∣(μ,u)∈C+}≤MI.
Suppose on the contrary that there exists {(μn,yn)}⊂𝒞+ with
(3.54){μn}⊂I,‖yn‖X⟶∞asn⟶∞.
Then by (3.38),
(3.55)‖yn‖∞⟶∞,n⟶∞.
and subsequently
(3.56)min⁡t∈T2⁡yn(t)≥c1‖yn‖∞⟶∞.
This together with (3.32) and f0∈(0,∞) and f∞=0 that
(3.57)‖(vn|T2)‖∞⟶0,n⟶∞.
However, this contradicts with the fact that
(3.58)min⁡t∈T2⁡vn(t)≥c1,n∈N.
Therefore, (3.53) holds.
Step 3.
Fixedλ such that (3.59)λ>inf⁡s∈(0,c1B)⁡sc1γ*f(s).
Then there exists l∈(0,c1B) such that
(3.60)λ>lγ*c1f(l).
We show that there is no (η,y)∈𝒞+ such that
(3.61)‖y‖∞=lc1η>lγ*c1f(l).
Suppose on the contrary that there exists (η,y)∈C+ satisfying (3.61). Then for t∈𝕋2,
(3.62)y(t)=η∑s=2TK(t,s)h(s)f(y(s))≥η∑s=2TK(t,s)h(s)f(c1‖y‖∞)=η∑s=2TK(t,s)h(s)f(l)≥ηγ*f(l)=ηγ*f(l)l⋅l,
and subsequently, η≤l/c1γ*f(l). Therefore, there is no (η,y)∈𝒞+ such that (3.61) holds.
Now, combining the conclusions in Steps2 and 3, using the fact that no (μ,u)∈𝒞+ satisfying (3.61), it concludes that for every λ∈(l/c1γ*f(l),λ1/f0), (1.4), (1.8) has at least two generalized positive solutions in 𝒞+. For arbitrary λ∈(infs∈(0,c1B)(s/c1γ*f(s)),∞), we may find l=l(λ) satisfying (3.60). So, for every λ∈(infs∈(0,c1B)(s/γ*f(s)),λ1/f0), (1.4), (1.8) has at least two generalized positive solutions in 𝒞+.


## 4. Some Examples

In this section, we will apply our results to two examples.For convenience, setT=12, then 𝕋1={1,2,…,13}, 𝕋2={2,3,…,12}.Example 4.1.
Let us consider the boundary value problem(4.1)Δ4u(t-2)=λf(u(t)),t∈T2,u(1)=u(13)=Δ2u(0)=Δ2u(12)=0,
where
(4.2)f(u)={arctanu,u∈(0,1000],(u-1000)2+arctan1000,u∈(1000,∞).
Clearly, f(u) is nondecreasing, f0=1, f∞=∞. Take B=1000. By a simple computation, it follows that infs∈(0,1000)(f(s)/s)=arctan1000/1000≈0.00157, λ1=16sin4(π/24)≈0.0048 and γ*=maxt∈𝕋1∑s=212K(t,s)=1629/6, then
(4.3)0.0048≈16sin⁡4π24=λ1f0<sup⁡s∈(0,1000)⁡sf(s)γ*=61629inf⁡s∈(0,1000)(f(s)/s)≈2.34631.
So, Theorem 1.3(i) implies that (4.1) has at least one generalized positive solution for
(4.4)0<λ<λ1f0≈0.0048;
Theorem 1.3(ii) implies that (4.1) has at least two generalized positive solutions for
(4.5)λ1f0<λ<1γ*(16sin⁡4⁡(π/24)-1)≈2.34631.Example 4.2.
Let us consider the boundary value problem:(4.6)Δ4u(t-2)=λf̃(u(t)),t∈T2,u(1)=u(13)=Δ2u(0)=Δ2u(12)=0,
where
(4.7)f̃(u)={eu-1u,u∈(0,30],u-30+e30-130,u∈(30,∞).
Obviously, f̃(u) is nondecreasing in [0,∞), so f̃0=limu→0(f̃(u)/u)=1, f̃∞=limu→0(f̃(u)/u)=0. By a simple computation, it follows that λ1=16sin4(π/24) and γ*=mint∈𝕋2∑s=212K(t,s)=143/2, c1=2163/145145≈0.214. Take B=50. Since sups∈(0,10.715)(f̃(s)/s)=(e10.715-1)/10.715≈4202.0726, it follows that
(4.8)0.0000155≈inf⁡s∈(0,c1B)⁡sc1γ*f̃(s)=1sup⁡s∈(0,10.715)(f̃(s)/s)c1γ*<λ1f̃0=16sin⁡4π24≈0.0048.
Therefore, (i) of Theorem 1.4 implies that(4.6) has at least one generalized positive solution for
(4.9)λ>inf⁡s∈(0,c1B)⁡sc1γ*f̃(s)≈0.0000155;(ii)
of Theorem1.4 implies that (4.6) has at least two generalized positive solutions for(4.10)0.0000155<λ<16sin⁡4π24.



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*Source: 918082-2012-03-12.xml*