On the existence of multiple solutions for a partial discrete Dirichlet boundary value problem with mean curvature operator

Lemma 2.1

Assume that condition (A) and the following conditions hold:

(B) Φ is convex and infXΦ=Φ(0)=Ψ(0)=0;

(C) for each λ > 0 and local minima u₁, u₂ ∈ X of I_λ := Φ − λΨ with Ψ(u₁) ≥ 0 and Ψ(u₂) ≥ 0, one has inf0≤t≤1Ψ(tu1+(1−t)u2)≥0.

Assume further that there are two positive constants ρ1,ρ2andrˉ∈Xwithρ1<Φ(rˉ)<ρ22, such that

(a₁)

(a₂)

Then, for all λ∈2Φ(rˉ)Ψ(rˉ),minρ1supu∈Φ−1(−∞,ρ1)Ψ(u),ρ2/2supu∈Φ−1(−∞,ρ2)Ψ(u), , the functional I_λ has at least three distinct

critical points which lie in Φ⁻¹(−∞, ρ₂).

We recall the following lemma (Theorem 2.2 of [39]).

Lemma 2.2

Assume that condition (A) and the following conditions hold:

(D) Φ is convex and infXΦ=Φ(0)=Ψ(0)=0;

(E) for each λ > 0 and local minima u₁, u₂ ∈ X of I_λ := Φ − λΨ with Ψ(u₁) ≥ 0 and Ψ(u₂) ≥ 0, one has inf0≤t≤1Ψ(tu1+(1−t)u2)≥0.

Assume further that there are two positive constants ρ1,ρ2andrˉ∈Xwith2ρ1<Φ(rˉ)<ρ22, such that

(a₃)

(a₄)

Then, for all λ∈32⋅Φ(rˉ)Ψ(rˉ),minρ1supu∈Φ−1(−∞,ρ1)Ψ(u),ρ2/2supu∈Φ−1(−∞,ρ2)Ψ(u), the functional I_λ has at least three distinct

critical points which lie in Φ⁻¹(−∞, ρ₂).

For each v,v1,v2>infXΦ with v₂ > v₁ and v₃ > 0, set

We have the following variant of Theorem 3.3 of [43](see also Corollary 3.1 and Remark 3.9 of [43]).

Lemma 2.3

If condition (A) holds. Assume that:

(F) Φ is convex and infXΦ=Φ(0)=Ψ(0)=0;

(G) for each λ > 0 and for every u₁, u₂ ∈ X which are local minima for the functional I_λ := Φ − λΨ and such that Ψ(u₁)≥ 0 and Ψ(u₂)≥ 0, one has inf0≤t≤1Ψ(tu1+(1−t)u2)≥0.

Further, assume that there are three positive constants ρ₁, ρ₂, ρ₃ with ρ₁ < ρ₂, such that

(a₅) φ(ρ₁)< β(ρ₁, ρ₂);

(a₆) φ(ρ₂)< β(ρ₁, ρ₂);

(a₇) γ(ρ₂, ρ₃)< β (ρ₁, ρ₂).

Then, for all λ∈1β(ρ1,ρ2),1α(ρ1,ρ2,ρ3), the functional I_λ has three distinct critical points.

Let

Clearly, γ ≥ 0 and δ ≥ 0. When γ = 0 (or δ = 0), in the sequel, we agree to read 1γ(or1δ) as +∞.

We recall the following lemma (see Theorem 2.1 of [44]).

Lemma 2.4

Assume that condition (A) holds. One has:

(i) If γ < +∞then, for each λ∈(0,1γ), the following alternative holds: either

(i₁) I_λ possesses a global minimum, or

(i₂) there is a sequence {u_t} of critical points (local minima) of I_λ such that limt→+∞Φ(ut)=+∞.

(j) If δ < +∞then, for each λ∈(0,1δ), the following alternative holds: either

(j₁) there is a global minimum of Φ which is a local minimum of I_λ, or

(j₂) there is a sequence {u_t} of pairwise distinct critical points (local minima) of I_λ, with limt→+∞Φ(ut)=infXΦ,

which weakly converges to a global minimum of Φ.

We set the mn-dimensional Banach space

endowed with the norm

Put

for all u ∈ S, where F((i,j),u)=∫0uf((i,j),ξ)dξfor each(i,j)∈Z(1,m)×Z(1,n).

Let

for each u ∈ S. Evidently, I_λ ∈ C¹(S, R) with

and

for each u, z ∈ S. Taken together, we have

Accordingly, the critical points of I_λ in S are the solutions of problem (Hλf).

The following lemma comes from Proposition 1 of [42].

Lemma 2.5

For each u ∈ S, one has the following relation

Remark 2.1

Clearly, when m = 1 and n = 1, then u = 2|u(1, 1)| and the equality in (2.1) holds.

Proposition 2.1

Suppose that there is ū ∈ S such that either

for each (i, j) ∈ Z(1, m) × Z(1, n). Then, either ū(i, j) > 0 for all (i, j) ∈ Z(1, m) × Z(1, n) or ū ≡ 0.

Proof. Fix α ∈ Z(1, m), β ∈ Z(1, n) such that

When ū(α, β) > 0, we get ū(i, j) > 0 for every (i, j) ∈ Z(1, m) × Z(1, n). The proposition is proved. When ū(α, β) ≤ 0, we have ū(α, β) = min {uˉ(i,j):i∈Z(0,m+1),j∈Z(0,n+1)}. Clearly, Δ₁ ū(α − 1, β) = ū(α, β) − ū(α − 1, β) ≤ 0, Δ₂ ū(α, β − 1) = ū(α, β) − ū(α, β − 1) ≤ 0, Δ₁ ū(α, β) = ū(α + 1, β) − ū(α, β) ≥ 0, and Δ₂ ū(α, β) = ū(α, β + 1) − ū(α, β) ≥ 0. Since ϕ_c(μ) is increasing in μ and ϕ_c(0) = 0, we see that

and

This further shows that

We obtain

Consequently, ū(α + 1, β) = ū(α, β) = ū(α − 1, β). We claim that ū(α, β) = 0 when α + 1 = m + 1. If not, then (α + 1) ∈ Z(1, m). By taking the place of α with α + 1, we have ū(α + 2, β) = ū(α + 1, β). Continuing this process (m + 1 − α) times, we get ū(α, β) = ū(α + 1, β) = ū(α + 2, β) = ··· = ū(m + 1, β) = 0. Similarly, we obtain ū(α, β) = ū(α − 1, β) = ū(α − 2, β) = ··· = ū(0, β) = 0. Then, it holds that ū(i, β) = 0 for i ∈ Z(1, m). Analogously, we get ū ≡ 0.

Theorem 3.1

Suppose that f (i, j), · : ℝ → ℝ is a non-negative continuous function for each (i, j) ∈ Z(1, m) × Z(1, n) and there are positive constants b₁, b₂ and l with

such that

(g₁) max Γ(b1),2Γ(b2)<14Λ(l).

Then, for each

problem (Hλf) admits at least two positive solutions.

Proof. Obviously, Φ and Ψ satisfy conditions (A) and (B) of Lemma 2.1 by setting Φ, Ψ, I_λ as defined in Section 2 for each u ∈ S. Furthermore, let u₁ and u₂ be two local minima of I_λ, which implies that u₁ and u₂ are solutions of (Hλf). We have u₁(i, j) ≥ 0 and u₂(i, j) ≥ 0 for every (i, j) ∈ Z(1, m) × Z(1, n). Then, tu₁(i, j) + (1 − t)u₂(i, j) ≥ 0 for all (i, j) ∈ Z(1, m) ×Z(1, n) and 0 ≤ t ≤ 1. It is clear that Ψ(tu₁ + (1− t)u₂)≥ 0 for 0 ≤ t ≤ 1 and condition (C) is satisfied.

Now, we set

For u ∈ S, let

One has

Since

and

we get

Then,

For each u ∈ S, we have

and

This implies that

Similarly, for u ∈ S, one has

It follows that

and

For y ∈ S defined by

it holds that Φ(y)=(2m+2n)(1+l2−1). It is clear that ρ1<Φ(y)<ρ22. Then,

In terms of (g₁), we can verify assumptions (a₁) and (a₂) of Lemma 2.1. Thus, for each

problem (Hλf) admits at least two positive solutions. The proof of the theorem is completed.

With a similar proof of Theorem 3.1, we have the following

Theorem 3.2

Suppose that f (i, j), · : ℝ → ℝ is a non-negative continuous function for each (i, j) ∈ Z(1, m) × Z(1, n) and there are positive constants b₁, b₂ and l with

such that

(g₁) max Γ(b1),2Γ(b2)<13Λ(l).

Then, for each

problem (Hλf) admits at least two positive solutions.

Given h₁ > 0 and h₂ > 0, we further introduce another notation Ω(h₁, h₂) as follows

Theorem 3.3

Assume that f (i, j), · : ℝ → ℝ is a non-negative continuous function for every (i, j) ∈ Z(1, m) ×

Z(1, n) and there are positive constants d, e₁, e₂ and e₃ with e₃ > e₂ such that

and

(g₄) max Γ(e1),Γ(e2),Ω(e2,e3)<14Λ(d).

Then, for each

problem (Hλf) admits at least two positive solutions.

Proof. Let

When u ∈ S and Φ(u) ≤ v₁, we have

When u ∈ S and Φ(u) ≤ v₂, we get

When u ∈ S and Φ(u) ≤ (v₂ + v₃), we obtain

Define y ∈ S by

Noticing that

and

we obtain

Therefore, we can see from (g₄) that φ(v₁) < β(v₁, v₂), φ(v₂) < β(v₁, v₂) and γ(v₂, v₃) < β(v₁, v₂). This together with Lemma 2.3 finishes the proof.

Set

When E^∞ = +∞, we agree to read 1E∞=0.

Theorem 3.4

Suppose that there are two nonnegative sequences {δ_t} and {ω_t}, with limt→+∞ωt=+∞, such that

and

Then, for all λ∈2n+2mE∞,1D∞, problem (Hλf) admits an unbounded sequence of solutions.