# Existence of Solutions for the Discrete Dirichlet Problem Involvingp-Mean Curvature Operator

Authors: Jianxia Wang; Zhan Zhou
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

This work is to discuss the Dirichlet boundary value problem of the difference equation withp -mean curvature operator. Under some determinate growth conditions on the nonlinear term, the existence of one solution or two nontrivial solutions is obtained via variational methods and some analysis techniques. Examples are also given to illustrate our theorems in the last section.


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## Body

## 1. Introduction

Letℕ,ℤ, and ℝ stand for the sets of natural numbers, integer numbers, and real numbers, respectively. For m,n∈ℤ, define ℤm,+∞=m,m+1,… and ℤm,n=m,m+1,…,n when m≤n.Consider the following Dirichlet boundary value problem:(1)−Δϕp,cΔuj−1=fj,uj,j∈ℤ1,b,u0=ub+1=0,where b is a given positive integer, fj,⋅∈Cℝ,ℝ for each j∈ℤ1,b, Δ is the forward difference operator defined by Δuj=uj+1−uj, and ϕp,ct≔1+t2p−2/2t,p∈0,+∞. Here, ϕp,cΔuj−1 is counted as a discretization of p -mean curvature operator.We may regard problem (1) as being a discrete analog of the following differential problem in one-dimensional space:(2)−divϕp,c▽u=fx,u,x∈Ω⊂ℝn,u=0,x∈∂Ω,where ϕp,c▽u, named p -mean curvature operator [1], is a generalization of mean curvature operator ([2], page 1212). When p=1, it reduces to mean curvature operator [3–6]. When p=2, it reduces to Laplacian operator.Difference equations arise in different research fields, for instance, computer science, mechanical engineering, control systems, artificial or biological neural networks, economics, and so on, see [7–12].In order to research discrete boundary problems, different methods have been used: fixed point theorems, upper and lower solutions techniques, see [3, 5] and the references given therein. Starting from Guo and Yu [13], various results have been achieved via variational methods, see [14–32].In 2011, Candito and Bisci [33] studied the boundary value problem:(3)−Δ2uj−1=fj,uj,j∈ℤ1,b,u0=ub+1=0,where Δ2uj=ΔΔuj, fj,t is continuous with respect to t for every j∈ℤ1,b. Through variational methods, they proved that problem (3) admits two nontrivial solutions.In 2015, Zhou and Su [30] discussed the following problem:(4)Δnrj−nϕcΔnuj−n+−1n+1fj,uj+1,uj,uj−1=0,j∈ℤ1,b,uj=0,j∈ℤ1−n,0∪ℤb+1,b+n,where b≥2 is positive integer, Δmuj=ΔΔm−1uj, and ϕct=t/1+t2. By critical point theory and some analytical skills, two nontrivial solutions were acquired.Motivated by the works of [15, 30, 34–43], we deal with problem (1). Compared to problem (4), problem (1) is much more difficult to handle because of p -mean curvature operator. It needs more analytical skills. For the convenience of analysis, we have no option but to split the problem into two types: 0<p<2 and 2≤p<+∞.This paper is arranged as follows. InSection 2, we establish an appropriate Banach space and a suitable functional corresponding to problem (1). In order to gain solutions of problem (1), some preliminary knowledge is introduced. In Section 3, under certain growth conditions on the antiderivative F of the nonlinearity f, one solution or two nontrivial solutions for problem (1) are obtained. In Section 4, we give two concrete examples to demonstrate the applicability of our main results.


## 2. Mathematical Background

For problem (1), we naturally select the b - dimensional Banach space:(5)S=u:ℤ0,b+1⟶ℝ:u0=ub+1=0,equipped with the norm(6)u≔∑j=1buj21/2,for all u∈S.Its infinity norm is(7)u∞≔maxj∈ℤ1,buj,for allu∈S.Apparently, one has(8)u∞≤u.Furthermore, we need to use the following inequality. When0<r<s,xj≥0,j∈ℤ1,n, one has(9)∑j=1nxjs1/s≤∑j=1nxjr1/r,see ([44], page 28).Putting(10)Fj,t≔∫0tfj,τdτ,the function I:S⟶ℝ is defined by(11)Iu≔Φu−Ψu,for every u∈S, where(12)Φu≔1p∑j=0b1+Δuj2p/2−1,Ψu≔∑j=1bFj,uj.It is not difficult to verify thatI∈C1S,ℝ, and the critical points of I on S are exactly the solutions of problem (1).One can see that(13)∑j=0bΔuj2=∑j=0buj+1−uj2=2∑j=1buj2−2∑j=1b−1uj+1uj=u˜Au˜tr,where u˜=u1,u2,…,ub is a b -dimensional row vector, u˜tr stands for the transpose of u˜, and(14)A=2−10⋯0−12−1⋯0⋮⋮⋮⋮⋮0⋯−12−10⋯0−12b×b.MatrixA is symmetric and positive definite. It is easy to see that A has b distinct positive eigenvalues which are given by(15)λj≔4sin2jπ2b+1,∀j∈ℤ1,b.We have the following lemma, see ([36], page 918).Lemma 1.
For everyu∈S, one has(16)λ1u2≤∑j=0bΔuj2≤λbu2.LetE be a real Banach space, the functional I∈C1E,ℝ is said to satisfy the Palais–Smale (P. S. for short) condition if any sequence un in E such that Iun is bounded and I′un⟶0asn⟶∞ contains a convergent subsequence.LetBr denote the open ball in E of radius r about 0, and let ∂Br denote its boundary.Our main tool is Mountain pass lemma, see ([45], Theorem 2.2).Lemma 2.
LetE be a real Banach space and I∈C1E,ℝ satisfying the P. S. condition. Suppose I0=0 and(i)
There are constantsρ,α>0 such that I∂Bρ≥α(ii)
There exists ane∈E/B¯ρ such that Ie≤0
Then,I admits a critical value c≥α. In addition, c can be characterized as(17)c=infγ∈Γmaxu∈γ0,1Iu,where(18)Γ=γ∈C0,1,Eγ0=0,γ1=e.


## 3. Main Results and Their Proofs

Theorem 1.
Letp∈0,2, for every j∈ℤ1,b, fj,⋅:ℝ⟶ℝ be continuous. Suppose that F1limsupt⟶∞Fj,t/tp<bp/2−1/p2sinπ/2b+1p,∀j∈ℤ1,b.Then,I is coercive, and problem (1) admits at least one solution.Proof.
It is obvious thatI∈C1S,ℝ.
Owing toF1, we choose (19)l1∈0,bp/2−1p2sinπ2b+1p,and M1>0 such that, for every j∈ℤ1,b and t>M1, Fj,t<l1tp. Since t⟶Fj,t−l1tp is continuous on −M1,M1, there exists some constant C1>0 such that Fj,t−l1tp≤C1, for all j,t∈ℤ1,b×−M1,M1. Consequently, we follow that(20)Fj,t≤l1tp+C1,for everyj,t∈ℤ1,b×ℝ.
Using (9) and (20), Hölder inequality, and Lemma 1, we follow that(21)Iu=1p∑j=0b1+△uj2p/2−∑j=1bFj,uj−b+1p≥1p∑j=0bΔuj2p/2−l1∑j=1bujp−b+1p−C1b≥1p∑j=0bΔuj2p/2−l1∑j=1bujp−b+1p−C1b≥1pλ1u2p/2−l1b1−p/2up−b+1p−C1b=b1−p/2bp/2−1p2sinπ2b+1p−l1up−b+1p−C1b.
Note that(22)bp/2−1p2sinπ2b+1p−l1>0,and we have limu⟶+∞Iu=+∞. So, I is coercive in S. Hence, I is bounded from below. Set c¯=infu∈SIu. According to the coercivity of the functional I again, there is L>0 such that Iu>c¯+1,for everyu>L. Thus, infu≤LIu=infu∈SIu. Taking into account that S is a finite dimensional space and I is continuous, I possesses a critical point u¯∈S satisfying Iu¯=infu∈SIu. The proof is completed.Theorem 2.
Letp∈0,2, for every j∈ℤ1,b, fj,⋅:ℝ⟶ℝ be continuous. Suppose that (23)F2limt⟶∞infFj,ttp>b+11−p/2p2sinbπ2b+1p,∀j∈ℤ1,b.Then,I is anticoercive, and problem (1) has at least one solution.Proof.
To avoid repetition, we give only the key steps of the proof.
ByF2, we choose(24)l2∈b+11−p/2p2sinbπ2b+1p,+∞and C2>0 such that(25)Fj,t≥l2tp−C2,for every j,t∈ℤ1,b×ℝ.
By virtue of (9) and (25), Hölder inequality, and Lemma 1, we follow that(26)Iu=1p∑j=0b1+Δuj2p/2−1−∑j=1bFj,uj≤1p∑j=0bΔuj2p/2−l2∑j=1bukp+C2b≤1pb+11−p/2∑j=0b△uj2p/2−l2up+C2b≤b+11−p/2pλbp/2−l2up+C2b=b+11−p/2p2sinbπ2b+1p−l2up+C2b.
Since(27)b+11−p/2p2sinbπ2b+1p−l2<0,we have limu⟶+∞Iu=−∞.Theorem 3.
Letp∈2,+∞, for every j∈ℤ1,b, fj,⋅:ℝ⟶ℝ be continuous. Suppose that(28)F3limt⟶∞supFj,ttp<b+11−p/2p2sinπ2b+1p,∀j∈ℤ1,b.Then,I is coercive, and problem (1) possesses at least one solution.Proof.
Here, we give only the main steps of the proof.
ByF3, we choose (29)l3∈0,b+11−p/2p2sinπ2b+1pand C3>0 such that(30)Fj,t≤l3tp+C3,for every j,t∈ℤ1,b×ℝ.
Thanks to (9) and (30), Hölder inequality, and Lemma 1, and we follow that(31)Iu=1p∑j=0b1+Δuj2p/2−1−∑j=1bFj,uj≥1p∑j=0bΔuj2p/2−l3∑j=1bujp−C3b≥1pb+11−p/2∑j=0bΔuj2p/2−l3up−C3b≥b+11−p/2pλ1p/2−l3up−C3b=b+11−p/2p2sinπ2b+1p−l3up−C3b.
Since(32)b+11−p/2p2sinπ2b+1p−l3>0,we have(33)limu⟶+∞Iu=+∞.Theorem 4.
Letp∈2,+∞, for every j∈ℤ1,b, fj,⋅:ℝ⟶ℝ be continuous. Suppose that(34)F4limt⟶∞infFj,ttp>bp/2−1p2sinbπ2b+1p,∀j∈ℤ1,b.Then,I is anticoercive, and problem (1) admits at least one solution.Proof.
In the same way, we give only the crucial steps of the proof.
ByF4, we choose(35)l4∈bp/2−1p2sinbπ2b+1p,+∞and C4>0 such that(36)Fj,t≥l4tp−C4,for every j,t∈ℤ1,T×ℝ.
Given anyu≠0, applying (9) and (36), Hölder inequality, and Lemma 1, we follow that(37)Iu=1p∑j=0b1+Δuj2p/2−∑j=1bFj,uj−b+1p≤1p∑j=0b1+Δuj2p/2−l4∑j=1bujp−b+1p+C4b≤1pb+1+λbu2p/2−l4b1−p/2up−b+1p+C4b≤b1−p/2bp/2−1pb+1u2+λbp/2−l4up−b+1p+C4b=b1−p/2bp/2−1pb+1u2+4sin2bπ2b+1p/2−l4up−b+1p+C4b.
Note that(38)limu⟶+∞bp/2−1pb+1u2+4sin2bπ2b+1p/2−l4=bp/2−1p2sinbπ2b+1p−l4<0,we have(39)limu⟶+∞Iu=−∞.Remark 1.
If moreover, in Theorems1–4, if fj,0≠0 for some j∈ℤ1,b, then the solution is nontrivial.Corollary 1.
Letp∈0,2, for every j∈ℤ1,b, fj,⋅:ℝ⟶ℝ be continuous, and satisfy one of the following conditions:(40)f1limt⟶∞supfj,ttp−2t<bp/2−12sinπ2b+1p,∀j∈ℤ1,b,f2limt⟶∞inffj,ttp−2t>b+11−p/22sinbπ2b+1p,∀j∈ℤ1,b.Then, problem (1) has at least one solution.Proof.
Here, we only prove thatf2 implies F2. The proof that f1 implies F1 refers to ([33], Corollary 2.1).
Fromf2, there exist ε>0 and L>0 such that(41)fj,ttp−2t>b+11−p/22sinbπ2b+1p+ε,for each t≥L and j∈ℤ1,b.
Sincefj,t is continuous with respect to t, we also let(42)m≔minfj,t:j,t∈ℤ1,b×−L,L,M≔maxfj,t:j,t∈ℤ1,b×−L,L.
Thus, whent≥L, one follows(43)Fj,t=∫0Lfj,sds+∫Ltfj,sds≥mL+1pb+11−p22sinbπ2b+1p+εtp−Lp=b+11−p/2p2sinbπ2b+1p+εptp−b+11−p/2p2sinbπ2b+1p+εpLp+mL.
Whent≤−L, we have(44)Fj,t=−∫t−Lfj,sds+∫−L0fj,sds≥1pb+11−p/22sinbπ2b+1p+ε−tp−Lp−ML=b+11−p/2p2sinbπ2b+1p+εp−tp−b+11−p/2p2sinbπ2b+1p+εpLp−ML.
Put(45)C=−b+11−p/2p2sinbπ2b+1p+εpLp+Lminm,−M.
Therefore,(46)Fj,t≥b+11−p/2p2sinbπ2b+1p+εptp+C.
Hence, our argument holds, and the proof is complete.Remark 2.
We observe that there are functions that satisfyF1, but f1 does not hold, see, such as example 1 in Section 4.
Similarly, we have the following result.Corollary 2.
Letp∈2,+∞, for every j∈ℤ1,b, fj,⋅:ℝ⟶ℝ be continuous, and satisfy one of the following conditions:(47)f3limt⟶∞supfj,ttp−2t<b+11−p/22sinπ2b+1p,∀j∈ℤ1,b,f4limt⟶∞inffj,ttp−2t>bp/2−12sinbπ2b+1p,∀j∈ℤ1,b.Then, problem (1) admits at least one solution.The following results guarantee the existence of two nontrivial solutions to problem (1).Theorem 5.
Letp∈0,2, for every j∈ℤ1,b, fj,⋅:ℝ⟶ℝ be continuous, and F1 holds. Additionally, assume that(48)F˜1limt⟶0infFj,tt2>2sin2bπ2b+1,∀j∈ℤ1,b.
Then, problem (1) has at least two nontrivial solutions.Proof.
Our aim is to apply Lemma2. To this purpose, take E=S and I as in (11). Let Ju≔−Iu,for allu∈S. Using Theorem 1, it is evident that J is anticoercive. Since the dimension of S is finite, we can easily deduce that J satisfies the P.S. condition. Next, we prove that J verifies the mountain pass geometry.
FromF˜1, there are ε>0 and δ1>0 such that(49)Fj,t≥2sin2bπ2b+1+εt2=λb2+εt2,for every j,t∈ℤ1,b×−δ1,δ1.
By (8), Lemma 1, and the inequality 1+xθ≤1+θx, for 0<θ≤1,x≥0, we have(50)Ju=∑j=1bFk,uj−1p∑j=0b1+Δuj2p/2−1≥λb2+ε∑j=1buj2−12∑j=0bΔuj2≥εu2,for u∈S with u≤δ1.
Writingα=εδ12 and ρ=δ1, then Ju≥α>0,∀u∈∂Bρ. Hence, J satisfies condition (i) of Lemma 2. It is obvious that J0=0. Because limu⟶+∞Ju=−∞, we may pick e∈S and e>ρ with Je<0, which means J satisfying condition (ii) of Lemma 2.
According to Lemma2, J has a critical value c≥α given by c=infγ∈Γmaxu∈γ0,1Ju, where(51)Γ=γ∈C0,1,Sγ0=0,γ1=e.
Setu∗∈S be a critical point of J and Ju∗=c≥α>0, then u∗ is a nonzero solution of problem (1). By Theorem 1, J possesses a critical point u¯∈S with Ju¯=supu∈SJu. Noting that Ju¯≥Ju∗>0, we have u¯≠0.Ifu¯≠u∗, then the proof is complete.Ifu¯=u∗, we divide the following proof into two cases.Case I.
b=1. In such a case, we may see that S=tu¯:t∈ℝ. Using limu⟶+∞Ju=−∞ again, there exists t˜<−δ1/u¯satisfying Jt˜u¯<0. Since J0=0 and J−δ1/u¯u¯>0, there exists t¯∈t˜,0 such that Jt¯u¯=maxJtu¯,t∈t˜,0. This shows that t¯u¯ is a critical point of J. Hence, problem (1) admits two nontrivial solutions.Case II.
b≥2. Through the definition of u∗, we show that maxu∈γ0,1Ju=Ju¯ for every γ∈Γ. Considering that b≥2, there exists d∈S which is linearly independent with e. For every k∈ℕ, define(52)γk=tkd+e,t∈0,12,1−tkd+te,t∈12,1,then γk∈Γ. Thus, we have tk∈0,1 satisfying Jγktk=Ju¯. Obviously, γmtm≠γntn when m≠n, and γktk is a sequence of pairwise distinct critical points of J in S. Consequently, problem (1) has infinitely many nontrivial solutions. The proof is finished.Theorem 6.
Letp∈1,2, for every j∈ℤ1,b, fj,⋅:ℝ⟶ℝ be continuous, and F2 holds. Moreover, suppose that(53)F˜2limt⟶0supFj,tt2<2sin2π2b+1,∀j∈ℤ1,b.Then, problem (1) admits at least two nontrivial solutions.Proof.
LetE=S and I as in (11). Owing to Theorem 2, I is anticoercive. Noting that S is a finite dimensional space, it is obvious that I satisfies the P.S. condition. Now, we show that I satisfies the mountain pass geometry.
ByF˜2, we pick ε∈0,1 such that(54)limt⟶0supFj,tt2<41−ε1+εp/2+1sin2π2b+1=1−ε1+εp/2+1λ1,for each j∈ℤ1,b.
Hence, there existsδ2>0, and we follow that(55)Fj,t≤1−ε1+εp/2+1λ1t2,when j,t∈ℤ1,b×−δ2,δ2.
Thanks to (8), for u∈S with u≤δ2, we get(56)Ψu=∑j=1bFj,uj≤1−ελ11+εp/2+1u2.
Foru∈S with u≤ε/λb1/2, exploiting Lemma 1, we get(57)∑j=0bΔuj2≤λbu2≤ε.
By Lemma1 and the inequality 1+xθ≥1+θx,wheneverθ≥1,x≥0, we follow that(58)Φu=1p∑j=0b1+Δuj2p−11+Δuj2p/2+1≥1p1+εp/2+1∑j=0b1+Δuj2p−1≥λ11+εp/2+1u2,for u∈S with u≤ε/λb1/2.
Letδ=minδ2,ελb−11/2, then combining (56) and (58), we have(59)Iu≥λ11+εp/2+1u2−1−ελ11+εp/2+1u2=ελ11+εp/2+1u2,when u∈S with u≤δ.
Pickingα=ελ1δ2/1+εp/2+1 and ρ=δ, then for every u∈∂Bρ, Iu≥α>0. The rest of the proof runs as the proof of Theorem 5.Theorem 7.
Letp∈2,+∞, for every j∈ℤ1,b, fj,⋅:ℝ⟶ℝ be continuous. Suppose that F3 and F˜1 are satisfied. Then, problem (1) has at least two nontrivial solutions.Proof.
TakeE=S and I as in (11) and denote Ju≔−Iu,for allu∈S. Applying Theorem 3, we can derive that J is anticoercive. Here, we only verify that J satisfies condition (i) of Lemma 2.
Since2≤p≤+∞, there exists n∈ℕ such that n≤p/2<n+1.
Using Taylor’s formula, we have the following inequality:(60)1+x2p/2−1≤p2x2+pp−2222!x4+⋯+pp−2⋅p−2n2n+1n+1!x2n+2.
From (9) and the above inequality, we follow that(61)Φu=1p∑j=0b1+Δuj2p/2−1≤12∑j=0bΔuj2+p−2222!∑j=0bΔuj4+⋯+p−2⋯p−2n2n+1n+1!∑j=0bΔuj2n+2≤12∑j=0bΔuj2+p−2222!∑j=0bΔuj22+⋯+p−2⋯p−2n2n+1n+1!∑j=0bΔuj2n+1≤λb2u2+p−22!λb22u4+⋯+p−2⋯p−2nn+1!λb2n+1u2n+2.
Owing toF˜1, there exist ε>0 and δ1>0 such that (49) is verified.
Because of the fact that(62)limt⟶0p−22!λb22t2+⋯+p−2⋯p−2nn+1!λb2n+1t2n=0,for the above ε>0, there is δ3>0 such that(63)p−22!λb22t2+⋯+p−2⋯p−2nn+1!λb2n+1t2n≤ε2,whenever t≤δ3.
Pickingδ=minδ1,δ3, we have u∞≤u≤δ for all u∈S with u≤δ.
Through (49), (61), and (63), we obtain(64)Ju≥λb2+εu2−λb2u2−p−22!λb22u4−⋯−p−2⋯p−2nn+1!λb2n+1u2n+2=u2ε−p−22!λT22u2−⋯−p−2⋯p−2nn+1!λb2n+1u2n≥ε2u2,for u∈S with u≤δ.
Takingα=ε/2δ2 and ρ=δ, we follow that Iu≥α>0 if u∈∂Bρ. The proof is complete.
Applying the method analogous to that used above, we can draw the following conclusion.Theorem 8.
Letp∈2,+∞, for every j∈ℤ1,T, fj,⋅:ℝ⟶ℝ be continuous. Suppose that F4 and F˜2 are fulfilled. Then, problem (1) has at least two nontrivial solutions.


## 4. Examples

### 4.1. Example 1

Letb=2,p=3/2, and f be a function which is defined as follows:(65)fj,t=2jsin2jtet1+tet+2125t1/2signtsint3/2+t3/2cost3/2.Since(66)Fj,t=∫0tfj,sds=2sin2jtet+1425t3/2sint3/2,through direct computations, we follow that(67)limt⟶∞supFj,tt3/2=1425<23/43=23/4−13/22sinπ63/2,limt⟶0Fj,tt2=2j2>32=2sin2π3,for every j∈ℤ1,2. Thus, the conditions F1 and F˜1 are satisfied. Theorem 5 ensures that problem (1) admits at least two nontrivial solutions.Moreover, we show that conditionf1 does not hold taking into account that(68)limt⟶+∞supfj,tt3/2−2t=+∞,for every j∈ℤ1,2.


### 4.2. Example 2

Whenp=4, consider the following problem:(69)−△ϕ4,c△uj−1=35juj3+2950uj,j∈ℤ1,3,u0=u4=0.Noting that(70)Fj,t=∫0tfj,sds=354jt4+29100t2,we have(71)limt⟶∞infFj,tt4=354j>322+12=34/2−142sin3π84,limt⟶0supFj,tt2=29100<2−22=2sin2π8,for every j∈ℤ1,3. According to Theorem 8, problem (69) has at least two nontrivial solutions.


## 4.1. Example 1

Letb=2,p=3/2, and f be a function which is defined as follows:(65)fj,t=2jsin2jtet1+tet+2125t1/2signtsint3/2+t3/2cost3/2.Since(66)Fj,t=∫0tfj,sds=2sin2jtet+1425t3/2sint3/2,through direct computations, we follow that(67)limt⟶∞supFj,tt3/2=1425<23/43=23/4−13/22sinπ63/2,limt⟶0Fj,tt2=2j2>32=2sin2π3,for every j∈ℤ1,2. Thus, the conditions F1 and F˜1 are satisfied. Theorem 5 ensures that problem (1) admits at least two nontrivial solutions.Moreover, we show that conditionf1 does not hold taking into account that(68)limt⟶+∞supfj,tt3/2−2t=+∞,for every j∈ℤ1,2.


## 4.2. Example 2

Whenp=4, consider the following problem:(69)−△ϕ4,c△uj−1=35juj3+2950uj,j∈ℤ1,3,u0=u4=0.Noting that(70)Fj,t=∫0tfj,sds=354jt4+29100t2,we have(71)limt⟶∞infFj,tt4=354j>322+12=34/2−142sin3π84,limt⟶0supFj,tt2=29100<2−22=2sin2π8,for every j∈ℤ1,3. According to Theorem 8, problem (69) has at least two nontrivial solutions.



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