The Sign-Changing Solution for Fractional ( p , q ) -Laplacian Problems Involving Supercritical Exponent

: In this article, we consider the following fractional ( p , q ) -Laplacian problem ( − ∆ ) s 1 p u + ( − ∆ ) s 2 q u + V ( x )( | u | p − 2 u + | u | q − 2 u ) = f ( u ) + λ | u | r − 2 u , where x ∈ R N , ( − ∆ ) s 1 p is the fractional p - Laplacian operator ( ( − ∆ ) s 2 q is similar), 0 < s 1 < s 2 < 1 < p < q < Ns 2 , q ∗ s 2 = NqN − s 2 q , r ≥ q ∗ s 2 , f is a C 1 real function and V is a coercive function. By using variational methods, we prove that the above problem admits a sign-changing solution if λ > 0 is small.


Introduction
For the fractional (p, q)-Laplacian problem we will prove it admits a sign-changing solution, where x ∈ R N , 0 < s 1 < s 2 < 1 < p < q < N s 2 , q * s 2 = Nq N−s 2 q , r ≥ q * s 2 , V : R N → R is a positive continuous function and f : R → R is a continuous real function.

Physical Background
The definition of the fractional p-Laplacian can be seen in [1].For the physical background, we refer the reader to [2][3][4][5][6][7].These papers tell us that the fractional p-Laplacian can describe financial markets, optimization, phase transformation, semi-permeable film, anomalous diffusion and minimal surface problems.Problem (1) models two different materials and it is called the double-phase equation (see e.g., [8]).

Related Works and Our Main Results
Recently, many authors have been concerned with the fractional (p, q)-Laplacian equations.For the critical and supercritical cases, the existence of multiple solutions is obtained in [8].In the meantime, for the problem where s ∈ (0, 1), 1 < p < q < N s , V : R N → R, Isernian [9] showed that it admits a positive ground state solution.In 2022, the existence of the least energy sign-changing solution of problem (2) was given by Cheng et al. in [10].We also quote the papers [11][12][13] for the p-Laplacian or fractional p-Laplacian in a bounded domain.For other results, please see [1,8,[14][15][16][17][18][19][20] and the references therein.
The main goal of the present paper is to investigate the problem in (1).We assume that ). ( f 3 ) there exists θ ∈ (q, q * s 2 ) such that 0 < θF(t) ≤ f (t)t for all |t| > 0, where |t| q−1 is strictly increasing for all |t| > 0.

Our Motivations and Novelties
Like [21], a natural question for us is For the (p, q)-Laplacian problem (1), does there exist a sign-changing solution?
Our motivation in this paper is to give this question an affirmative answer.It is different from [22,23] since (10), ( 11) and ( 15) are new and crucial.

Methods
We summarize our methods here.We adopt the idea from [22] or [24] to cut off the functional (see ( 4)).Then we shall prove (6) admits a minimizer w 0 .Furthermore, we need to prove that the minimizer w 0 is a critical point of I λ,K (see (5)).Finally, we borrow the idea from [22] to make a L ∞ -estimation such that ∥w 0 ∥ L ∞ (R N ) ≤ K, which implies that we do not make any truncations.

Organization
This paper is organized as follows.Section 2 provides some preliminaries.Section 3 is divided into two parts, which will prove Theorem 1.The last Section is the conclusions and our future direction.Throughout this paper, we use the standard notations.

•
C or C i (i = 1, 2, ...) denote some positive constants (possibly different from line to line) and C(•) denotes some positive constant only dependent on •.

Preliminary Results
From now on, we always assume that (V) and ( f 1 ) − ( f 4 ) hold unless a special statement is made.We continue to use the notations and work space W s,p (R N ) as in [8].Since the potential V is coercive, we introduce the subspace equipped with the following norm ∥u∥ = ∥u∥ 1 + ∥u∥ 2 , where [8].Formally, the corresponding energy functional of (1) is It is well-known that the functional I is not well-defined on E. In order to overcome this difficulty, similar to [22], where K > 0, σ ∈ (θ, q * s 2 ).Thus, it holds that |t| p−1 = 0, and for any ε > 0, there exists a positive constant C(ε) > 0 such that |t| q−1 is strictly increasing for all |t| > 0. For the auxiliary functional (h 1 ) implies that I λ,K ∈ C 1 (E, R).We want to prove that admits a minimizer, where the corresponding Nehari manifold M λ,K is given by

Some Lemmas
To begin with, we give several lemmas that will be used in the sequel.
Lemma 1.For any u ∈ E with u ± ̸ = 0, there is a unique pair (s u , t u ) of positive numbers such that s u u + + t u u − ∈ M λ,K .Moreover, Proof.The proof is standard (see e.g., [21]).For u ∈ E with u ± ̸ = 0, we can deduce from (h 1 ) and (h 2 ) that there exist C 1 (λ, K), C 2 (λ, K) > 0 such that If s > 1, in view of (h 3 ), we obtain This is a contradiction.Similarly, we can have t ≥ 1.Therefore, s = t = 1.Case 2: For u / ∈ M λ,K .Using the method in ( [25], page 90), the desired conclusion is obtained.
Proof.For u ∈ M λ,K , we only prove the result for u + .It is easy to check that Combining with (h 1 ), it is shown that Choosing ε = V 0 2 , we obtain which implies the desired conclusion.

Proof of Theorem 1
We are devoted to proving Theorem 1 in this section.From Lemmas 1-4, we find that (6) possesses a minimizer w 0 := s 0 u 0 + + t 0 u 0 − .There are two methods to ensure that the minimizer w 0 is a critical point of I λ,K .One method can be seen in ( [21], Section 3).The other method can be used as ( [28], lemma 3.6).Using a standard Moser iteration (see e.g., [22] or [23]), we can draw the conclusion that λ 0 > 0 such that ∥w 0 ∥ L ∞ (R N ) ≤ K when λ ∈ (0, λ 0 ).Thus, w 0 is a sign-changing solution of the initial problem (1), which means that in (4), we do not make any truncations.

Conclusions and Future Studies
With the above analysis made, the following conclusions can be drawn.Under the assumptions of Theorem 1, problem (1) admits a sign-changing solution.As mentioned in Remark 1, the sign-changing solution w 0 does not necessarily mean that it is a least energy sign-changing solution.Our future work will study the ground state or least energy signchanging solution to (1).Maybe it is an open problem since it appears as the supercritical term |u| r−2 u.