
Citation: Xichao Duan, Sanling Yuan, Kaifa Wang. Dynamics of a diffusive age-structured HBV model with saturating incidence[J]. Mathematical Biosciences and Engineering, 2016, 13(5): 935-968. doi: 10.3934/mbe.2016024
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Let R,Z, and N be the sets of all real numbers, integers, and natural numbers, respectively. Here and below, for a,b∈N with a<b, we use the notation Z(a,b):={a,a+1,⋯,b}.
In recent years, difference equations as mathematical models to describe a variety of practical problems in the fields of economy, biology, disease prevention and control, environmental protection, etc., have attracted great attention and there have been a lot of outstanding works in theory and actual application; see, for example, [1,2].
In the paper, we consider the following discrete Dirichlet boundary value problem involving the mean curvature operator
{−△ϕ(△u(k−1))+q(k)ϕ(u(k))=λf(k,u(k)),k∈Z(1,T),u(0)=u(T+1)=0, | (1.1) |
where ϕ(s)=s√1+κs2,κ>0 is a constant [3], T is a given positive integer, △ is the forward difference operator defined by △u(k)=u(k+1)−u(k),△2u(k)=△(△u(k)),q(k)≥0 for all k∈Z(1,T),λ is a real positive parameter, and f(k,⋅)∈C(R,R) for each k∈Z(1,T).
When q(k)=0 for all k∈Z(1,T), the problem (1.1) is the discrete analogy of one-dimensional prescribed mean curvature equations with Dirichlet boundary conditions
{−(ϕ(u′))′=λf(t,u(t)),t∈(0,1),u(0)=u(1)=0. |
The existence of (positive) solutions of the above problem has been studied by using the variational method, the standard techniques from bifurcation theory, fixed point index, and Kresnoselskii's fixed point theorem (see [4,5,6] and the references therein).
Note that when κ=0, the problem (1.1) degenerates into the classical second-order difference equation boundary value problem
{−△2u(k−1)+q(k)u(k)=λf(k,u(k)),k∈Z(1,T),u(0)=u(T+1)=0. | (1.2) |
The existence of (positive) solutions to the problem (1.2) has been well known with various qualitative assumptions of nonlinearity f ([7,8,9]).
When κ=1, the problem (1.1) is the ordinary discrete mean curvature problem
{−△(u(k−1)√1+u2(k−1))+q(k)u(k)√1+u2(k)=λf(k,u(k)),k∈Z(1,T),u(0)=u(T+1)=0, | (1.3) |
which has aroused the interest of many scholars recently [10,11,12]. Their approaches are variational. The existence of (positive) solutions to the problem (1.3) depends on the behavior at zero or infinity of the potential F(k,u)=∫u0f(k,s)ds. In [10,11], the oscillating behavior of F at +∞ played an important role in obtaining an infinite number of positive solutions to the problem (1.3). Chen and Zhou [12] obtained at least three positive solutions of the problem (1.3) with q(k)=0 for all k∈Z(1,T), where the primitive F on the nonlinear datum satisfies that lim sups→+∞F(k,s)|s|<α (α is a positive constant). Meanwhile, the case that the potential F satisfies lim infs→∞F(k,s)|s|>β (β is a positive constant) has been also discussed. They obtained at least two nontrivial solutions based on a two-critical-point theorem (Theorem 2.1) established in [13]. It is an important tool in obtaining at least two positive solutions of the Laplacian or the algebraic boundary value problems (see, for instance, [15,16]). However, it is relatively less used to look for positive solutions to the mean curvature boundary value problems.
Inspired by this, we will apply the two-critical-point theorem to establish the existence of at least two positive solutions for the problem (1.1) in this paper. As a special case of our main theorem, the existence of two positive solutions for the problem (1.1) with κ=1 and all q(k)=0 is obtained in Remark 4. The result improves Theorem 2 [12], where the author establishes the existence of only two nontrivial solutions without providing sign information for them under stronger hypotheses on the potential F. In Theorem 2 [12], the bilateral limit assumption on F(k,s)|s| at ∞ ensures that the energy functional of that problem is anticoercive, which consequently guarantees the establishment of the Palais-Smale condition—a pivotal requirement for applying critical point theorems. In our paper, the bilateral limit on the potential F is weakened to a unilateral limit at +∞, and then the energy functional of the problem (1.1) loses its anticoercivity. However, with the help of some inequality techniques, it is proven that the energy functional still satisfies the Palais-Smale condition. Moreover, in our main result (Theorem 3.1), the existence of at least two positive solutions is established without any asymptotic condition of the potential F at 0 and with no requiring that f(k,0)>0 for any k∈Z(1,T). In fact, the algebraic conditions in Theorem 3.1 are more general than the conditions that the potential F is subquadratic at 0 and superlinear at +∞ (see Corollary 3.1).
In 2003, Guo and Yu [17] first used the variational method to study the periodic solutions of second-order difference equations. Since then, for nonlinear difference systems, many scholars have used the variational method to study the existence of various solutions, such as periodic solutions, subharmonic solutions, and homoclinic solutions [18,19,20]. For general references on difference equations and their applications, we refer the reader to monographs [21,22] and the references therein.
This paper is organized as follows: In section 2, some definitions and results are collected. Some estimations of equivalence of the norm are provided. Moreover, Lemma 2.1 is presented to guarantee us to obtain positive solutions rather than nontrivial solutions to the problem (1.1). Section 3 is devoted to our main result. Lemma 3.1 is given to ensure the Palais-Smale condition of the functional on this basis. Some consequences of our main result are presented together with three examples.
Consider the T dimensional Banach space
S={u:Z(0,T+1)→Rsuchthatu(0)=u(T+1)=0} |
endowed with the norm ‖⋅‖ as
‖u‖:=(T∑k=0|△u(k)|2)12,∀u∈S. |
Moreover, the space S can also be equipped with the following equivalent norms:
‖u‖∞:=maxk∈Z(1,T)|u(k)|,∀u∈S, |
‖u‖1:=T∑k=1|u(k)|,∀u∈S, |
‖u‖2:=(T∑k=1|u(k)|2)12,∀u∈S, |
respectively. It is easy to know that for all u∈S,
‖u‖1≥ ‖u‖2. | (2.1) |
By Theorem 12.6.1 in [21], we have for all u∈S,
√λ1‖u‖2≤‖u‖≤√λT‖u‖2, | (2.2) |
where λj=4sin2jπ/2(T+1) for all j∈Z(1,T). By virtue of Lemma 2.2 in [23], we get for all u∈S,
‖u‖∞≤√T+12‖u‖. | (2.3) |
Note that v(0)=v(T+1)=0 for all v∈S, we have
T+1∑k=1ϕ(△u(k−1))△v(k−1)=−T∑k=1△ϕ(△u(k−1))v(k) | (2.4) |
for all u,v∈S.
Let
s+=max{s,0}ands−=max{−s,0} |
for all s∈R.
Remark 2.1. In fact, the nonnegative solution of the following problem
{−△ϕ(△u(k−1))+q(k)ϕ(u(k))=λf(k,u+(k)),k∈Z(1,T),u(0)=u(T+1)=0 | (2.5) |
is the nonnegative solution of the problem (1.1). To obtain positive solutions to the problem (1.1), we only need to search for positive solutions to the problem (2.5) now.
Consider two functionals Φ and Ψ defined respectively on S by
Φ(u)=1κT+1∑k=1(√1+κ|△u(k−1)|2−1)+1κT∑k=1q(k)(√1+κ|u(k)|2−1) | (2.6) |
and
Ψ(u)=T∑k=1F+(k,u(k)), | (2.7) |
F+(k,s)=∫s0f(k,t+)dt for all (k,s)∈Z(1,T)×R. Obviously,
F+(k,s)={∫s0f(k,t)dt, if s>0,f(k,0)s, if s≤0, |
for all k∈Z(1,T), and Φ,Ψ∈C1(S,R), which means that Φ and Ψ are two continuously Gâteaux-differentiable functionals defined on S. By (2.4), we have that for any u,v∈S,
Φ′(u)(v)=T+1∑k=1ϕ(△u(k−1))△v(k−1)+T∑k=1q(k)ϕ(u(k))v(k)=T∑k=1[−△ϕ(△u(k−1))+q(k)ϕ(u(k))]v(k) |
and
Ψ′(u)(v)=T∑k=1f(k,u+(k))v(k). |
We define Iλ:S→R by
Iλ(u)=Φ(u)−λΨ(u). |
Clearly, Iλ∈C1(S,R) and for any u,v∈S,
I′λ(u)(v)=T∑k=1[−△ϕ(△u(k−1))+q(k)ϕ(u(k))−λf(k,u+(k))]v(k). |
Remark 2.2. For any u,v∈S, I′λ(u)(v)=0 if and only if
−△ϕ(△u(k−1))+q(k)ϕ(u(k))−λf(k,u+(k))=0 |
for all u∈S and k∈Z(1,T). In other words, a critical point of Iλ on S corresponds to a solution to the problem (2.5).
For the problem (1.1), we are interested in the existence of positive solutions rather than nontrivial ones. To the end, we present an assumption on f(k,0) as follows.
(H1) f(k,0)≥0,∀k∈Z(1,T).
Lemma 2.1. If (H1) holds, then any nonzero critical point of the functional Iλ on S is a positive solution to the problem (1.1).
Proof. In fact, taking Remarks 2.1 and 2.2 into account, it is sufficient to verify that any nontrivial solution u of the problem (2.5) is positive, that is, u(k)>0 for all k∈Z(1,T).
Notice that ϕ is a strictly monotonically increasing and odd function on R. By standard computation, we can conclude that
−ϕ(△u(k−1))△u−(k−1)≥ϕ(△u−(k−1))△u−(k−1)=|△u−(k−1)|2√1+κ|△u−(k−1)|2 | (2.8) |
for all k∈Z(1,T+1). By a direct computation, we get
−q(k)ϕ(u(k))u−(k)=q(k)ϕ(u−(k))u−(k)=q(k)|u−(k)|2√1+κ|u−(k)|2 | (2.9) |
for all k∈Z(1,T). For any nontrivial solution u of the problem (2.5), combining Eqs (2.4), (2.8) and (2.9) with assumption (H1), we obtain that
0=T∑k=1[−△ϕ(△u(k−1))+q(k)ϕ(u(k))−λf(k,u+(k))](−u−(k))=−T+1∑k=1ϕ(△u(k−1))△u−(k−1)−T∑k=1[q(k)ϕ(u(k))−λf(k,u+(k))]u−(k)≥T+1∑k=1ϕ(△u−(k−1))△u−(k−1)+T∑k=1[q(k)ϕ(u−(k))+λf(k,0)]u−(k)≥T+1∑k=1|△u−(k−1)|2√1+κ|△u−(k−1)|2+T∑k=1q(k)|u−(k)|2√1+κ|u−(k)|2≥0. |
Hence △u−(0)=△u−(1)=⋯=△u−(T)=0. Recalling that u(0)=u(T+1)=0 for any u∈S, we have u−(k)=0 for all k∈Z(1,T). Thus u is nonnegative.
Next, we further prove that u is positive. Otherwise, there exists some k0∈Z(1,T) such that u(k0)=0, then
−△ϕ(△u(k0−1))=λf(k0,0)≥0. |
Clearly, ϕ(△u(k0))≤ϕ(△u(k0−1)). Because ϕ is a strictly monotonically increasing homomorphism, we get △u(k0)≤△u(k0−1). Hence, u(k0+1)+u(k0−1)≤0. As a result, u(k0±1)=0. Then iterating the process, we have that u(k)=0 for all k∈Z(1,T). In short, u is zero somewhere in Z(1,T). Then it is zero identically. This contradicts the nontriviality of u. The proof is ended.
Let (X,‖⋅‖) be a real Banach space and φ∈C1(X,R). φ is said to satisfy the Palais-Smale condition ((PS) condition), if any sequence {un}⊂X for which {φ(un)} is bounded and φ′(un)→0 as n→∞ possesses a convergent subsequence in X.
The following two-critical-point theorem is given by Bonanno and D'Aguì in 2016 [13].
Theorem 2.1. [13] Let X be a real Banach space, and let Φ,Ψ:X→R be two functionals of class C1 such that infXΦ=Φ(0)=Ψ(0)=0. Assume that there are r∈R and ˜u∈X, with 0<Φ(˜u)<r, such that
supu∈Φ−1((−∞,r])Ψ(u)r<Ψ(˜u)Φ(˜u) | (2.10) |
for each
λ∈Λ=(Φ(˜u)Ψ(˜u),rsupu∈Φ−1((−∞,r])Ψ(u)), |
the functional Iλ=Φ−λΨ satisfies the (PS) condition, and it is unbounded from below. Then for each λ∈Λ, the functional Iλ admits at least two nonzero critical points uλ,1,uλ,2 such that Iλ(uλ,1)<0<Iλ(uλ,2).
Let
L∞(k):=lim infs→+∞F(k,s)sandL∞:=mink∈Z(1,T)L∞(k), |
where F(k,s)=∫s0f(k,t)dt for all (k,s)∈Z(1,T)×R. Here and below, when L∞=0, we think 1L∞=∞. Before we come to a conclusion, let's give a lemma on the (PS) condition.
Lemma 3.1. If L∞>0 and (H1) hold, then Iλ satisfies the (PS) condition, and it is unbounded from below for all λ∈(√(T+1)λT+Q√κL∞,+∞), where Q:=(∑Tk=1|q(k)|2)12.
Proof. Because S is finite dimensional, it is sufficient to show that any (PS) sequence of Iλ is bounded on S. Let {un}⊂S be a sequence such that {Iλ(un)} is bounded and I′λ(un)→0 as n→∞.
First, we claim that {u−n} is bounded. According to (2.8) and (2.9), we have for all n∈N and k∈Z(1,T+1),
−ϕ(△un(k−1))△u−n(k−1)≥ϕ(△u−n(k−1))△u−n(k−1)=|△u−n(k−1)|2√1+κ|△u−n(k−1)|2 | (3.1) |
and
−q(k)ϕ(un(k))u−n(k)=q(k)ϕ(u−n(k))u−n(k)=q(k)|u−n(k)|2√1+κ|u−n(k)|2. | (3.2) |
By (3.1) and (3.2), we can estimate the derivative of Φ at un in the direction of −u−n that
−Φ′(un)(u−n)=T∑k=1[−△ϕ(△un(k−1))+q(k)ϕ(un(k))](−u−n(k))=−T+1∑k=1ϕ(△un(k−1))△u−n(k−1)−T∑k=1q(k)ϕ(un(k))u−n(k)≥T+1∑k=1|△u−n(k−1)|2√1+κ|△u−n(k−1)|2+T∑k=1q(k)|u−n(k)|2√1+κ|u−n(k)|2=1κT+1∑k=1κ|△u−n(k−1)|2√1+κ|△u−n(k−1)|2+1κT∑k=1q(k)κ|u−n(k)|2√1+κ|u−n(k)|2=1κT+1∑k=1κ|△u−n(k−1)|2+1−1√1+κ|△u−n(k−1)|2+1κT∑k=1q(k)κ|u−n(k)|2+1−1√1+κ|u−n(k)|2≥1κT+1∑k=1(√1+κ|△u−n(k−1)|2−1)+1κT∑k=1q(k)(√1+κ|u−n(k)|2−1). |
To put it simply,
−Φ′(un)(u−n)≥Φ(u−n),∀n∈N. | (3.3) |
And, from the condition (H1), it follows that
Ψ′(un)(u−n)=T∑k=1f(k,u+n(k))u−n(k)=T∑k=1f(k,0))u−n(k)≥0,∀n∈N. | (3.4) |
Combining (3.3) and (3.4), we have
Φ(u−n)≤−Φ′(un)(u−n)≤−Φ′(un)(u−n)+λΨ′(un)(u−n)=−I′λ(un)(u−n) | (3.5) |
for all λ>0 and n∈N. Moreover, by the definition of the functional Φ, we obtain, for all u∈S,
Φ2(u)≥[1κT+1∑k=1(√1+κ|△u(k−1)|2−1)]2≥1κ2T+1∑k=1(√1+κ|△u(k−1)|2−1)2=1κ2[T+1∑k=1(1+κ|△u(k−1)|2)+T+1∑k=11−2T+1∑k=1√1+κ|△u(k−1)|2]=1κT+1∑k=1|△u(k−1)|2−2κ2T+1∑k=1(√1+κ|△u(k−1)|2−1)≥1κ‖u‖2−2κΦ(u), |
which implies that
‖u‖2≤κΦ2(u)+2Φ(u) | (3.6) |
for all u∈S. Thus, it is clear that for all u∈S,
−1κ+√1κ‖u‖2+1κ2≤Φ(u). |
So combining (3.5) and the above inequality, we have
0≤−1κ+√1κ‖u−n‖2+1κ2≤−I′λ(un)(u−n),∀n∈N. |
Also, from limn→∞I′λ(un)=0, it follows that
limn→∞I′λ(un)(u−n)‖u−n‖=limn→∞I′λ(un)(u−n‖u−n‖)=0. |
As a result, we obtain
limn→∞−1κ+√1κ‖u−n‖2+1κ2‖u−n‖=0. |
By standard computation, the above equation is transformed into
limn→∞[√κ+1‖u−n‖2+1‖u−n‖]=+∞, |
which means that limn→∞‖u−n‖=0. Hence our claim is proved. So there is M>0 such that 0≤u−n(k)≤M for all k∈Z(1,T) and n∈N.
Next, we prove that {un} is bounded. If {un} is not bounded, we may assume, going if necessary to a subsequence, that ‖un‖→∞(n→∞). Taking L∞>0 into account, we fix λ>√(T+1)λT+Q√κL∞ and fix l=l(λ) such that for all k∈Z(1,T),
√(T+1)λT+Qλ√κ<l<L∞≤L∞(k)=lim infs→+∞F(k,s)s, |
then for all k∈Z(1,T), there is a δk>0 such that
F+(k,s)=F(k,s)>ls=l|s|,∀s>δk. |
Meanwhile, for all k∈Z(1,T) and s∈[−M,δk],
F+(k,s)≥mins∈[−M,δk]F+(k,s)≥l|s|−lmax{δk,M}+mins∈[−M,δk]F+(k,s)≥l|s|−max{lmax{δk,M}−mins∈[−M,δk]F+(k,s),0}=l|s|−η(k). |
Hence, for all k∈Z(1,T),
F+(k,s)≥l|s|−η(k),∀s≥−M. |
On account of (2.1), we obtain that for all n∈N,
Ψ(un)=T∑k=1F+(k,un(k))≥lT∑k=1|un(k)|−T∑k=1η(k)=l‖un‖1−η≥l‖un‖2−η, |
where η=∑Tk=1η(k). Using the Cauchy-Schwarz inequality, we get that for all n∈N,
Φ(un)=1κT+1∑k=1(√1+κ|△un(k−1)|2−1)+1κT∑k=1q(k)(√1+κ|un(k)|2−1)≤1κT+1∑k=1(1+√κ|△un(k−1)|2−1)+1κT∑k=1q(k)(1+√κ|un(k)|2−1)≤1√κT+1∑k=1|△un(k−1)|+1√κT∑k=1q(k)|un(k)|≤1√κ√T+1‖un‖+1√κQ‖un‖2. |
Therefore, from the previous two inequalities and (2.2), it follows that for all n∈N,
Iλ(un)=Φ(un)−λΨ(un)≤1√κ√T+1‖un‖+1√κQ‖un‖2−λl‖un‖2+λη≤1√κ√(T+1)λT‖un‖2+1√κQ‖un‖2−λl‖un‖2+λη≤(√(T+1)λT+Q√κ−λl)1√λ1‖un‖+λη. |
When ‖un‖→+∞(n→∞), by √(T+1)λT+Q√κ−λl<0, we have limn→∞Iλ(un)=−∞. This leads to a contradiction. Hence {un} is bound and Iλ satisfies the (PS) condition.
Finally, we prove that Iλ is unbounded from below. Let {un} be such that un=u+n for any n∈N and ‖u+n‖→∞(n→∞). Arguing as before, we obtain that
Iλ(un)≤(√(T+1)λT+Q√κ−λl)1√λ1‖un‖+λη,∀n∈N. |
So limn→∞Iλ(un)=−∞. Our conclusion follows.
Below, our main result is presented.
Theorem 3.1. Suppose that (H1) holds and there exist two positive constants c and d with d<c such that
(H2)∑Tk=1maxs∈[0,c]F(k,s)−1κ+√4c2κ(T+1)+1κ2<L∞√κ√(T+1)λT+Q |
and
(H3)∑Tk=1maxs∈[0,c]F(k,s)−1κ+√4c2κ(T+1)+1κ2<κ∑Tk=1F(k,d)(2+q)(√1+κd2−1). |
Then for each
λ∈Λ+:=(max{√(T+1)λT+QL∞√κ,(2+q)(√1+κd2−1)κ∑Tk=1F(k,d)},−1κ+√4c2κ(T+1)+1κ2∑Tk=1maxs∈[0,c]F(k,s)), |
the problem (1.1) possesses at least two positive solutions, where q:=∑Tk=1q(k).
Proof. By Lemma 2.1, it is enough to prove that Iλ has at least two nonzero critical points. We apply Theorem 2.1 by putting X=S,Iλ=Φ−λΨ where Φ,Ψ are the functions introduced in (2.6) and (2.7).
Clearly, infSΦ=Φ(0)=Ψ(0)=0. Also notice that F+(k,s)=F(k,s) for all (k,s)∈Z(1,T)×[0,+∞). From the conditions (H1),(H2) and (H3), it follows that L∞>0 and Λ+ is non-degenerate. Then Lemma 3.1 ensures that the function Iλ satisfies the (PS) condition and it is unbounded from below for λ∈Λ+.
For fixed λ∈Λ+, there exists c>0 such that λ≤−1κ+√4c2κ(T+1)+1κ2∑Tk=1maxs∈[0,c]F(k,s). Put r:=−1κ+√4c2κ(T+1)+1κ2. We claim that
{u∈S:Φ(u)≤r}⊂{u∈S:‖u‖∞≤c}. |
If Φ(u)≤r, by (3.6), we get that ‖u‖2≤κΦ2(u)+2Φ(u)≤κr2+2r=4c2T+1. Thus, from the inequality (2.3), it is clear that
‖u‖∞≤√T+12‖u‖≤√T+12√4c2T+1=c. |
The assertion is verified. Therefore, we obtain
supu∈Φ−1([0,r])Ψ(u)r≤∑Tk=1maxs∈[0,c]F(k,s)−1κ+√4c2κ(T+1)+1κ2. | (3.7) |
Now, we look for ˜u∈S in Theorem 2.1. Define ˜u∈RT+2 as ˜u(k)=d(0<d<c) for all k∈Z(1,T) and ˜u(0)=˜u(T+1)=0. Clearly, ˜u∈S. It is easy to see that
Φ(˜u)=1κ(2+q)(√1+κd2−1), | (3.8) |
and hence we get
Ψ(˜u)Φ(˜u)=κ∑Tk=1F(k,d)(2+q)(√1+κd2−1). | (3.9) |
Therefore, from (3.7), (3.9), and (H3), it follows that
supu∈Φ−1([0,r])Ψ(u)r≤Ψ(˜u)Φ(˜u) |
and Λ+⊂Λ. Moreover, since 0<d<c, and again by virtue of the condition (H3), we obtain that
1κ(2+q)(√1+κd2−1)<−1κ+√4c2κ(T+1)+1κ2=r. | (3.10) |
On account of (3.8) and (3.10), we get that 0<Φ(˜u)<r. Thus, (2.10) holds. Hence Theorem 2.1 ensures that Iλ admits two nonzero critical points. This completes the proof.
Remark 3.1. Indeed, condition (H1) implicitly covers both f(k,0)>0 and f(k,0)=0 cases. Theorem 3.1 holds true under either sub-case of condition (H1).
Remark 3.2. If all f(k,⋅) are nonnegative on [0,c](c>0), then maxs∈[0,c]F(k,s)=F(k,c) for all k∈Z(1,T). According to Theorem 3.1, it is enough to assume that there is a positive constant d with d<c such that
∑Tk=1F(k,c)−1κ+√4c2κ(T+1)+1κ2<min{L∞√κ√(T+1)λT+Q,κ∑Tk=1F(k,d)(2+q)(√1+κd2−1)}. |
Remark 3.3. When κ=1 and q(k)=0 for all k∈Z(1,T) (Q=0 and q=0), the conditions in Remark 3.2 are more general than those of Theorem 2 in [12], where f(k,s) is required to be positive for all (k,s)∈Z(1,T)×[−c,c] rather than nonnegative, and the potential F is assumed to possess asymptotic behavior not only at +∞ but also at −∞. Here, the two solutions we obtain are positive, while the solutions in Theorem 2 of [12] are only nontrivial. Obviously, our main result improves Theorem 2 in [12].
Now, we present two particular cases of Theorem 3.1.
Corollary 3.1. Assume that (H1) holds,
lim sups→0+F(k,s)s2=+∞ | (3.11) |
and
lims→+∞F(k,s)s=+∞ | (3.12) |
for all k∈Z(1,T). Then for each λ∈(0,λ∗), where
λ∗=supc>0−1κ+√4c2κ(T+1)+1κ2∑Tk=1maxs∈[0,c]F(k,s), |
the problem (1.1) admits at least two positive solutions.
Proof. Let λ∈(0,λ∗) and c>0 such that
λ<−1κ+√4c2κ(T+1)+1κ2∑Tk=1maxs∈[0,c]F(k,s). |
Taking (3.11) into account, we get that lim sups→0+F(k,s)√1+κs2−1=+∞. As a result, there is d>0 with d<c such that κ2+q∑Tk=1F(k,d)√1+κd2−1>1λ. Note that L∞=+∞, so Theorem 3.1 ensures the conclusion.
Corollary 3.2. Assume that f(k,⋅)=f for all k∈Z(1,T) and f is a continuous function such that
lims→0+f(s)s=+∞ | (3.13) |
and
lims→+∞f(s)=+∞. | (3.14) |
Then, for each
λ∈(0,supc>0−1κ+√4c2κ(T+1)+1κ2Tmaxs∈[0,c]∫s0f(t)dt), |
the problem (1.1) admits at least two positive solutions.
Proof. In fact, Corollary 3.2 is a consequence of Corollary 3.1. It is easy to see that (3.13) implies f(0)≥0, and (3.11) and (3.12) can be derived from (3.13) and (3.14), respectively.
Example 3.1. For each λ∈(0,−1κ+√4κ(T+1)+1κ2(3e−1)T), the problem
{−△ϕ(△u(k−1))+q(k)ϕ(u(k))=λ(e3√u(k)−1),k∈Z(1,T),u(0)=u(T+1)=0 |
admits at least two positive solutions. Here, the nonlinearity f(t)=e3√t−1 satisfies the conditions of Corollary 3.2 and
supc>0−1κ+√4c2κ(T+1)+1κ2T∫c0f(t)dt≥−1κ+√4κ(T+1)+1κ2(3e−1)T. |
The image of f(t) is shown in Figure 1.
Example 3.2. For each λ∈(0,−1κ+√4κ(T+1)+1κ2T(T+1)), the problem
{−△ϕ(△u(k−1))+q(k)ϕ(u(k))=λk√u(k)(cosu(k)+2),k∈Z(1,T),u(0)=u(T+1)=0 |
admits at least two positive solutions. It is easy to see that f(k,t)=k√t(cost+2) satisfies the assumptions of Corollary 3.1 and
λ∗≥−1κ+√4κ(T+1)+1κ2∑Tk=1∫10f(k,t)dt=−1κ+√4κ(T+1)+1κ2T(T+1)2∫10√t(cost+2)dt≥−1κ+√4κ(T+1)+1κ2T(T+1). |
Figure 2 displays the functional plot of f(k,t) for k∈(0,100).
Remark 3.4. We observe that Theorem 2 in [12] cannot be applied in the previous two examples with κ=1 and q(k)=0 for all k∈Z(1,T), since f(k,0)>0 is assumed there.
Example 3.3. Consider the boundary value problem (1.1) with κ=1,q(k)=0 and
f(k,t)=f(t)={12(4−3t)cost4,t≤83π,2π−1,t>83π |
for all k∈Z(1,T). Clearly, if t∈(0,43), then f(t)>0; and if t∈(43,2π), then f(t)<0. Such a relationship is visually apparent in Figure 3. A straightforward calculation yields
F(k,s)=F(s)={2(4sins4−3ssins4−12coss4+12),s≤83π,(2π−1)s−4(2π−1)(23π+√3),t>83π |
for all k∈Z(1,T).
Obviously, L∞(k)=lim infs→+∞F(k,s)s=2π−1 for all k∈Z(1,T), then
L∞=mink∈Z(1,T)L∞(k)=2π−1≈5.382. |
Letting T=3,c=5, and d=1, we obtain
√(T+1)λT=4sinT2(T+1)π=4sin38π≈3.695, |
T∑k=1maxs∈[0,c]F(k,s)=3∑k=1∫430f(k,t)dt=72(1−cos13)≈3.963, |
and
T∑k=1F(k,d)=3∑k=1∫10f(k,t)dt=6[sin14+12(1−cos14)]≈3.724. |
It follows that
√(T+1)λTL∞≈3.6955.382≈0.687, |
2(√1+d2−1)∑Tk=1F(k,d)≈0.8283.724≈0.222, |
and
−1+√4c2T+1+1∑Tk=1maxs∈[0,c]F(k,s)≈4.0993.963≈1.034. |
From f(k,0)=f(0)=2>0 and the preceding relations, all conditions of Theorem 3.1 hold. Thus for each λ∈(√(T+1)λTL∞,−1+√4c2T+1+1∑Tk=1maxs∈[0,c]F(k,s))≈(0.687,1.034), the boundary value problem (1.1) has at least two positive solutions.
Remark 3.5. The potential function F in the above example lacks superlinearity at +∞, failing to satisfy the conditions of Corollaries 3.1 and 3.2. Nevertheless, Theorem 3.1 still guarantees the existence of two positive solutions for the problem (1.1) in Example 3.3.
In this paper, we consider a generalized difference mean curvature problem, which includes the conventional one and the classical second-order difference equation boundary value problem. The main novelties of this research are as follows:
(1) The existence of two positive solutions rather than nontrivial solutions for our problem is established based on a two-zero critical points theorem and some inequality techniques.
(2) Under the assumption of the unilateral limit of F(k,s)|s| at +∞ on the potential F(k,s)=∫s0f(k,t)dt instead of the bilateral limit at ∞, it is proved that without anticoercivity the energy functional associated with our problem still satisfies the Palais-Smale condition that plays a key role in the critical point theorem.
(3) Our principal result can be applied to cases with nonlinear terms where f(k,0)>0 but also to cases where the nonlinear terms satisfy f(k,0)=0.
(4) It is also worth mentioning that the algebraic conditions in our main result are more general than the subquadraticity at 0 and the superlinearity at +∞ for the potential F.
In fact, due to the previous (1)–(3), our main result extends Theorem 2 in [12].
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
This work is supported by the National Nature Science Foundation of China (No.10901071, No.11501054, and No.12301186).
The authors declare there is no conflicts of interest.
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