Multiplicity of positive solutions for a degenerate nonlocal problem with p-Laplacian

Remark 1.1

(a) All the hypotheses on a(t), namely (H₀), (H₃) and (H₄) deal only with "even" intervals [t_2k−1, t_2k] (for k ∈ {1, …, K}). The function a(t) can be arbitrary in "odd" intervals (can be negative, zero or positive without satisfying bound conditions like (H₃) or (H₄)) and these intervals can be even degenerated to a point. In other words a(t) can be any continuous function defined on a positive interval and we can split its domain into "even" intervals where the above assumptions are satisfied and "odd" intervals (possibly degenerated to a single point), where the above assumptions need not be satisfied. The number of solutions obtained in the paper will double the number of "even" intervals.

(b) Condition (H₁) implies that γ=limt→0+f(t)/tp−1 is well defined and it can be a positive number (see also assumption (H₃)) or +∞. In particular, if f (0)>0 then γ=+∞.

(c) Condition (H₃) is trivially satisfied if f(0) > 0 (see Remark 1.1(b)). If γ< +∞, (H₃) basically means that the peaks of a(t) in even intervals are controlled from above by variation of f at 0. On the other hand, hypothesis (H₄) means that the peaks of t^p a(t) in “even” intervals are controlled from below by the maximum value of tf(t) in [0, t_*] (see Fig. 1).

Fig. 1

Geometry of a(t) and t^p a(t) satisfying (H₁), (H₃) and (H₄).

(d) Finally, we wish to explicitly observe that 1 < p<+∞ and q≥1 are not required to satisfy any particular further condition. As well as, we emphasize that the nonlinear term f has a behaviour that is prescribed only on [0, t*], so that, nor asymptotic conditions at infinity, neither critical/subcritical growth are involved.

Theorem 2.1

If hypotheses (H₀), (H₁), (H₂), (H₃), and (H₄) hold, then problem (P) has at least 2K positive solutions belonging to int(C₊), with ordered L^q-norms, namely

In the proof of Theorem 2.1 we will consider a suitable auxiliary problem involving the following truncation function

Fix k ∈ {1, …, K} and α ∈ (t_2k−1, t_2k), here is the announced auxiliary problem that will play a crucial role

We start solving problem (P_k,α) by the variational approach. Observe that, after removing the nonlocal term of (P) we obtained a variational problem, so that we can look for solutions of (P_k,α) searching functions u∈W01,p(Ω) such that

Proposition 2.2

If hypotheses (H₀), (H₁) and (H₃) hold, then for each k ∈ {1, …, K} and α ∈ (t_2k−1, t_2k) fixed, problem (P_k,α) has a unique solution u_α ∈ int(C₊) such that 0 < u_α ⩽ t_* in Ω. In particular, u_α solves the problem

Proof. Fix k ∈ {1, …, K} and α ∈ (t_2k−1, t_2k). Let us define the energy functional Ik,α:W01,p(Ω)⟶R corresponding to problem (P_k,α), namely

where F∗(t)=∫0tf∗(s)ds . Since f_* is bounded and continuous, it is clear that I_{k, α} is coercive and weakly lower semicontinuous. Therefore I_{k, α} has a minimizer u_α which is a solution of (P_k,α). Since φ₁ ∈ int(C₊) (recall that φ₁ is the positive eigenfunction of the minus p-Laplacian associated to first eigenvalue λ₁ normalized in W01,p(Ω) -norm), observing that

by assumption (H₃), we can pass to the limit in the previous and obtain

namely, for t > 0 sufficiently small we obtain

and so u_α is nontrivial.

First, let us verify that 0 ⩽ u_α ⩽ t_*. Indeed, if in (2.2) we take v=−(u_α)⁻, we get

and, since a(α)>0, uα−=0 , that is 0 ⩽ u_α. Analogously, if in (2.2) we take v=(u_α−t_*)⁺, we get

so, again recalling that a(α)>0, (u_α−t_*)⁺=0 and u_α ⩽ t_*. Hence, taking in mind the definition of f_*, u_α solves (P_k,α,f).

Since u_α is bounded, by standard regularity arguments due to Lieberman [11, Theorem 1], we conclude that uα∈C01,β(Ωˉ) (for some β ∈ (0, 1)). Recalling that f ∈ C ([0, t*]) and observing that f(u_α)⩾0, one has Δ_pu_α ∈ L^∞ and Δ_pu_α⩽0. Hence, by the Maximum Principle (see Gasiński-Papageorgiou [8, Theorem 6.2.8], Vázquez [16] or Pucci-Serrin [14]) we get that u_α ∈ int(C₊).

Finally, suppose that v_α is a second solution to (P_k,α), with u_α≠ v_α. Arguing as for u_α, one can point out that 0 < v_α ⩽ t_* in Ω, as well as that v_α solves (P_{k, α, f}). Hence, in view of assumption (H₁), from [7] one can derive

and we get a contradiction. □

In order to better describe the strategy that we will follow, observe that when (H₀), (H₁) and (H₃) hold, the previous proposition allow us to define suitable maps Sk:(t2k−1,t2k)→C01(Ωˉ) for every k=1, …, K by putting

for every α ∈ (t_2k−1, t_2k), where u_α, that is a minimizer of I_{k, α}, is the unique solution of (P_k,α,) such that 0 < u_α ⩽ t_*. At this point, for every k=1, …, K, we introduce a real function Pk:(t2k−1,t2k)→R defined by

for all α ∈ (t_2k−1, t_2k), where q≥1. The role of the map Pk is revealed by the following claim

where Fix(Pk)={α∈(t2k−1,t2k): Pk(α)=α} . Indeed, let α ∈ (t_2k−1, t_2k) be such that Pk(α)=α , then, recalling that Sk(α)=uα is a solution of (P_k,α,f) one can conclude that

and the claim (ᘓ) holds.

Next lemma will be useful in the proof of Proposition 2.4 which provides the continuity of the map Pk . First observe that, since the map (0, t_*)∋ t↦ψ(t)=f(t)/t^p−1 is strictly decreasing (see hypothesis (H₁)), there exists the inverse ψ⁻¹:(0, γ) → (0, t_*), where γ is as in (H₃).

Lemma 2.3

Assume that hypotheses (H₀), (H₁) and (H₃) hold. Put

and let ε∈(0,γ−λ1aˉ) . Then, for every k=1, …, K and α ∈ (t_2k−1, t_2k) one has

where cα=Ik,α(uα)=minu∈W01,p(Ω)Ik,α(u) .

Proof. We only treat the case γ<+∞. The case γ=+∞ is analogous. First, by hypothesis (H₃), since 0<aˉ<γλ1 , it makes sense the choice of ε∈(0,γ−λ1aˉ) . Fix k ∈ {1, …, K} and α ∈ (t_2k−1, t_2k) and, observing that ε<γ−λ1aˉ≤γ−λ1a(α) , we can consider the function

From assumption (H₁), we have

Hence,

Therefore, (2.7) holds. □

Proposition 2.4

If hypotheses (H₀), (H₁), and (H₃) hold, then for each k ∈ {1, …, K} the map Pk:(t2k−1,t2k)→R defined in (2.6) is continuous.

Proof. Let {α_n}⊂(t_2k−1, t_2k) be a sequence such α_n → α_*, for some α_* ∈ (t_2k−1, t_2k). For simplicity, put un=Sk(αn) . Taking in mind (2.5),

that implies

Therefore, {u_n} is bounded in W01,p(Ω) and, passing to a subsequence if necessary, we may assume that

for some u∗∈W01,p(Ω) . Moreover, for every n∈N one has

Testing the previous with v=u_n−u_* and passing to the limsup, in view of the continuity of α and f_* as well as (2.10) and the Lebesgue’s dominated convergence theorem, we get

The (S₊) property of −Δ_pu assures that u_n → u_* in W01,p(Ω) , so that, passing to the limit in (2.11), we can conclude that u_* is a nonnegative weak solution of (P_k,α) with α=α_*. We need to show that u_*≠ 0. By Lemma 2.3, there exists ε>0, small enough, such that

So, passing to the limit and using (2.10), we obtain

Therefore u_*≠ 0. Arguing as in the proof of Proposition 2.2 we can show that u_* ∈ int(C₊). Since such a solution is unique, we conclude that u∗=uα∗=Sk(α∗) . Finally, again from Proposition (2.2) one has that 0 < u_n ⩽ t_* for every n∈N and we can use (2.10) and the Lebesgue’s dominated convergence theorem in order to conclude that Pk(αn)→Pk(α∗) . This proves the continuity of Pk . □

Remark 2.5

A deeper analysis allow us to assure a strongest convergence of the sequence {u_n} considered in the previous proof. Indeed, since every u_n solves (P_k,α), with α=α_n, one has

Recalling that f_* is bounded and a(α_n) is away from zero, we have

for some C > 0. By the regularity theory (see Lieberman [11]), we have that

for some C > 0 and β ∈ (0, 1). The compactness of the embedding C1,β(Ω‾)⊆C1(Ω‾) , passing to a subsequence if necessary, permits to obtain that

(the limit function u_* is the same as in (2.10), by the uniqueness of the limit function).

Next lemma will be needed in Proposition 2.7, where the fixed points of Pk are provided.

Lemma 2.6

If hypotheses (H₀), (H₁) and (H₃) hold, then, for every k ∈ {1, …, K} and α ∈ (t_2k−1, t_2k) one has

Proof. Fix k ∈ {1, …, K} and α ∈ (t_2k−1, t_2k). From hypothesis (H₁), the definition of ψ⁻¹ and observing that, in view of (H₃), 0 < z_α ⩽ ψ⁻¹(λ₁a(α))<t_*, we have

so

Therefore z_α is a subsolution of (P_k,α). Now, put

for every (x, t) ∈ Ω  ×  ℝ^N and

for every u∈W01,p(Ω) , where F ~ ( x , t ) = ∫ 0 t f ~ ( x , s ) d s . The direct methods assure the existence of a minimizer v_α for the functional Ĩ_{k, α}. Moreover, since φ₁, z_α ∈ int(C₊), for t > 0 small enough one has 0 < tφ₁ ⩽ z_α in Ω and

Thus v_α is a nontrivial function such that

Testing (2.15) with −(v_α)⁻ and (v_α−t_*)⁺ we obtain that

In particular, we wish to show that

Indeed, acting in (2.15) with w=(z_α−v_α)⁺ and taking in mind (2.13), we obtain

Hence,

and, taking into account [13, Lemma A.0.5], (2.16) holds. At this point, it is easy to observe that, because of the defintion of f˜ , one has

namely v_α is a solution of (P_k,α) such that 0 < v_α ⩽ t_*. Finally, Proposition 2.2 assures that v_α=u_α and the proof is complete in view of (2.16).

Proposition 2.7

If hypotheses (H₀), (H₁), (H₂), (H₃), and (H₄) hold, then, for every k ∈ {1, …, K} the map Pk has at least two fixed points t_2k−1<α_1,k<α_2,k<t_2k.

Proof. Fix k ∈ {1, …, K}. First we show that

From Lemma 2.6, we have

Hence, by hypotheses (H₁) and (H₂), we have

which proves (2.17).

Next, we show that for some α₀ ∈ (t_2k−1, t_2k) we have

For each α ∈ (t_2k−1, t_2k), let w_α be the unique (positive) solution of the problem

where uα=Sk(α) is the (unique) positive solution of (P_k,α). Multiplying by u_α, integrating by parts, using Cauchy-Schwarz inequality and Hölder inequality, we have

By the variational property of u_α, we get

thus

while, taking in mind that w_α solves (2.19), by the Hölder inequality and the variational characterization of λ₁, we have

where 1p+1p′=1 , so

Applying (2.21) and (2.22) in (2.20), we obtain

Using hypothesis (H₄), for some α₀ ∈ (t_2k−1, t_2k) one has

and we get

namely (2.18) holds.

From the continuity of Pk (see Proposition 2.4), since (2.17) implies the existence of α₁ and α₂ in (t_2k−1, t_2k) with P(α₁)>t_2k−1 and P(α₂)>t_2k, (2.18) and the intermediate value theorem for continuous real functions, we conclude that Pk has at least two fixed points in the interval (t_2k−1, t_2k). □

Proof of Theorem 2.1

Fix k ∈ {1, …, K}, recall claim (ᘓ) and observe that the two fixed points of Pk obtained in Proposition 2.7 produce two positive solutions u_k,1 and u_k,2 of problem (P), with u_k,1, u_k,2 ∈ int(C₊) and satisfying

for any k ∈ {1, …, K}. This finishes the proof.

We conclude presenting a concrete example of possible nonlocal problem that can be considered

Example 2.8

Let K be a positive integer, 1 < p<+∞, q=1 and fix t_*, M > 0 such that

where e₁ and λ₁ are as defined in Section 1, so that (H₂) holds. Let a:[0,+∞)→R be a continuous function such that

for t ∈ [0, (2K − 1)π]. Obviously a satisfies hypothesis (H₀) for the partition

Consider a continuos function f:R→R such that

It is simple to verify that

1. t↦ f(t)/t^p−1 is decreasing on (0, t_*),

2. limt→0+f(t)/tp−1=+∞ ,

3. f (0)=f(t_*)=0, f(t) > 0 for t ∈ (0, t_*).

Namely, hypotheses (H₁) and (H₃) hold. Finally, a direct computation shows that

Hence, for every k ∈ {1, …, K} one has

That is (H₄) holds too.