A double-phase eigenvalue problem with large exponents

Lujuan Yu

doi:10.1515/math-2023-0138

Article Open Access

A double-phase eigenvalue problem with large exponents

Lujuan Yu

Published/Copyright: December 6, 2023

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$Open Mathematics$

From the journal Open Mathematics Volume 21 Issue 1

Abstract

In the present article, we consider a double-phase eigenvalue problem with large exponents. Let λ ( p n , q n ) 1 be the first eigenvalues and u n be the first eigenfunctions, normalized by ‖ u n ‖ ℋ n = 1 . Under some assumptions on the exponents p n and q n , we show that λ ( p n , q n ) 1 converges to Λ ∞ and u n converges to u ∞ uniformly in the space C α ( Ω ) , and u ∞ is a nontrivial viscosity solution to a Dirichlet ∞ -Laplacian problem.

Keywords: Musielak-Orlicz spaces; viscosity solution; double-phase problem; ∞-Laplacian; eigenvalue

MSC 2010: 46E30; 35D40; 35P30

1 Introduction

Let Ω ⊂ R N ( N ≥ 2 ) be a bounded domain with Lipschitz boundary ∂ Ω . In this article, as p n and q n go to infinity, we deal with the asymptotic behaviour of the first eigenvalues and eigenfunctions for the following double-phase problem:

(1.1) − div p n ∇ u K n ( u ) p n − 2 + q n a ( x ) ∇ u K n ( u ) q n − 2 ∇ u K n ( u ) = λ ( p n , q n ) S n ( u ) p n u k n ( u ) p n − 2 + q n a ( x ) u k n ( u ) q n − 2 u k n ( u ) , x ∈ Ω , u = 0 , x ∈ ∂ Ω ,

where a : Ω ¯ → [ 0 , + ∞ ) is a C 1 differentiable function,

(1.2) K n ( u ) ≔ ‖ ∇ u ‖ ℋ n , k n ( u ) ≔ ‖ u ‖ ℋ n , S n ( u ) ≔ ∫ Ω p n ∇ u K n ( u ) p n − 2 + q n a ( x ) ∇ u K n ( u ) q n − 2 d x ∫ Ω p n u k n ( u ) p n − 2 + q n a ( x ) u k n ( u ) q n − 2 d x , ℋ n ≔ t p n + a ( x ) t q n ,

and the exponents q n and p n satisfy the following conditions:

( H 1 ) q n > p n > 1 and q n p n < 1 + 1 N for every n ≥ 1 ,

( H 2 ) lim n → ∞ q n = lim n → ∞ p n = + ∞ ,

( H 3 ) there exists 1 ≤ R < + ∞ such that

(1.3) lim n → ∞ q n p n = R .

Definition 1.1

We say that u ∈ W 0 1 , ℋ n ( Ω ) \ { 0 } is a weak solution for the eigenvalue problem (1.1), if there exists λ ( p n , q n ) ∈ R such that

(1.4) ∫ Ω p n ∇ u K n ( u ) p n − 2 + q n a ( x ) ∇ u K n ( u ) q n − 2 ∇ u ⋅ ∇ v K n ( u ) d x = λ ( p n , q n ) S n ( u ) ∫ Ω p n u k n ( u ) p n − 2 + q n a ( x ) u k n ( u ) q n − 2 u v k n ( u ) d x , ∀ v ∈ W 0 1 , ℋ n ( Ω ) .

If u ≠ 0 , we say that λ ( p n , q n ) is an eigenvalue of equation (1.1), and that u is an eigenfunction corresponding to λ ( p n , q n ) , where W 0 1 , ℋ n ( Ω ) is the Musielak-Orlicz space, which will be defined in Section 2.

Define

(1.5) div ( ∣ ∇ u ∣ p − 2 ∇ u + a ( x ) ∣ ∇ u ∣ q − 2 ∇ u ) .

This operator is the double-phase operator. The corresponding energy functional is defined as follows:

(1.6) u ↦ ∫ Ω ∣ ∇ u ∣ p + a ( x ) ∣ ∇ u ∣ q d x ,

whose integrand switches two different elliptic behaviours. This energy functional belongs to the class of the functionals that Zhikov used to introduce models of strongly anisotropic materials, see for instance [1–4]. Also, according to Marcellini’s terminology [5–7], the functional (1.6) is an integral functional with nonstandard growth conditions. Our work is concerned with this operator.

It is clear that, when b ( x ) ≡ 0 , the double-phase operator (1.5) becomes the p -Laplacian operator. As p → ∞ , the asymptotic analysis of the first eigenvalue and the first eigenfunction for the p -Laplace operator

− Δ p u ≔ − div ( ∣ ∇ u ∣ p − 2 ∇ u ) ,

subject to different boundary conditions, has received a lot of attention. We refer to the works of Fukagai-Juutinen-Lindqvist [8–10] (Dirichlet eigenvalue problem), Rossi and Saintier [11] (mixed Dirichlet and Robin eigenvalue problem), and Lê [12] (Steklov eigenvalue problem). As we know, for p -Laplace operator eigenvalue problems, there is an important feature that if u is an eigenfunction, so is k u , where k is an arbitrary constant. However, it is noted that, for p ( x ) -Laplace operator

− Δ p ( x ) u ≔ − div ( ∣ ∇ u ∣ p ( x ) − 2 ∇ u ) ,

the eigenfunctions do not possess this property. Using the Rayleigh quotient of two Luxemburg norms of the variable exponent Lebesgue space, Franzina and Lindqvist [13] first introduced the definition of eigenvalues for p ( x ) -Laplace operator. It is easy to find that this definition of eigenvalue implement this property. Moreover, the asymptotic behaviour of eigenvalue for p ( x ) -Laplace operator was studied in [13–16].

It is important to stress that the loss of the property under consideration (in other words, the fact that if u is an eigenfunction, so is k u ) is not only a consequence of the dependence on x , but it can also occur in presence of unbalanced growth. For example, the double-phase operator (1.5) (that does not depend on x ) loses this property.

Quite recently, Colasuonno and Squassina [17] have considered a double-phase Dirichlet eigenvalue problem, using the Rayleigh quotient of two norms of Musielak-Orlicz space. By using the Rayleigh quotient of two norms of Musielak-Orlicz space, the author of this article has defined the eigenvalue, which has the same properties as p -Laplace operator. However, to the author’s knowledge, the asymptotic behaviour of eigenvalue for double-phase operator has remained open. Our article fits into this general field of investigation.

Let δ : Ω → [ 0 , ∞ ) be defined as follows:

δ ( x ) ≔ dist ( x , ∂ Ω ) = inf y ∈ ∂ Ω ∣ x − y ∣ .

This function is a Lipschitz continuous function and ∣ ∇ δ ∣ = 1 a.e. in Ω .

Define

(1.7) λ ( p n , q n ) 1 ≔ inf u ∈ W 0 1 , ℋ n ( Ω ) \ { 0 } ‖ ∇ u ‖ ℋ n ‖ u ‖ ℋ n ,

which is the first eigenvalue of equation (1.1), and

(1.8) Λ ∞ ≔ inf α ∈ W 0 1 , ∞ ( Ω ) \ { 0 } ‖ ∇ α ‖ L ∞ ( Ω ) ‖ α ‖ L ∞ ( Ω ) .

It follows from the study by Franzina and Lindqvist [13] that

(1.9) Λ ∞ = ‖ ∇ δ ‖ L ∞ ( Ω ) ‖ δ ‖ L ∞ ( Ω ) = 1 max x ∈ Ω { dist ( x , ∂ Ω ) } .

Define

△ ∞ u ∞ ≔ ∑ i , j = 1 N ( u ∞ ) x i ( u ∞ ) x j ( u ∞ ) x i x j

and

(1.10) k ∞ ( u ) ≔ ‖ u ‖ L ∞ ( Ω ) = ess sup x ∈ Ω ∣ u ∣ .

Note that equation (1.10) holds for all u ∈ L ∞ ( Ω ) .

Under some assumptions on the exponents p n and q n , we are prepared to give the main results of this article.

Theorem 1.1

Assume that u ∈ C ( Ω ) is a weak solution of problem (1.1), then u is also a viscosity solution to problem (3.1).

Theorem 1.2

There holds that

(1.11) lim n → ∞ λ ( p n , q n ) 1 = Λ ∞ .

Theorem 1.3

For each n ∈ N , let u n be a positive first eigenfunction with ‖ u n ‖ ℋ n = 1 . Then, we can extract a subsequence (not relabelled) such that

(1.12) u n → u ∞

in the space C α ( Ω ) ( 0 < α < 1 ) and

(1.13) k ∞ ( u ∞ ) = 1 .

Moreover, u ∞ ( x ) is a nontrivial solution to

(1.14) min { − Λ ∞ u ∞ + ∣ ∇ u ∞ ∣ , − Λ ∞ ( u ∞ ) R + ∣ ∇ u ∞ ∣ , − △ ∞ u ∞ } = 0 , x ∈ Ω , u ∞ = 0 , x ∈ ∂ Ω ,

in the viscosity sense.

The plan of this article is as follows. We introduce the Musielak-Orlicz spaces L ℋ ( Ω ) and W 0 1 , ℋ ( Ω ) in Section 2. Moreover, the proofs of Theorems 1.1–1.3 are given in the following sections.

2 Preliminaries

We recall here some facts about the spaces L ℋ ( Ω ) and W 0 1 , ℋ ( Ω ) , see, for reference, [17–20].

Let

N ( Ω ) = { u ∣ u is a generalized N-function in Ω } .

For all ( x , t ) ∈ Ω × [ 0 , + ∞ ) , we define the function ℋ : Ω × [ 0 , + ∞ ) → [ 0 , + ∞ ) as follows:

ℋ ( x , t ) ≔ t p + a ( x ) t q ,

with 1 < p < q and 0 ≤ a ( ⋅ ) ∈ L 1 ( Ω ) . It is easy to know that ℋ ∈ N ( Ω ) . This function is a locally integrable function. Moreover, note that

ℋ ( x , 2 t ) ≤ 2 q ℋ ( x , t )

for a.e. x ∈ Ω and all t ∈ [ 0 , + ∞ ) . This inequality is called condition Δ 2 .

We define

ρ ℋ ( u ) ≔ ∫ Ω ℋ ( x , ∣ u ∣ ) d x ,

which is called ℋ -modular, and the space L ℋ ( Ω ) is defined as follows:

L ℋ ( Ω ) ≔ { u ∣ u is a measurable function such that ρ ℋ ( u ) < + ∞ } .

The norm of this space is

‖ u ‖ ℋ ≔ inf γ > 0 : ρ ℋ u γ ≤ 1 .

The space W 1 , ℋ ( Ω ) is given as follows:

W 1 , ℋ ( Ω ) = { u ∈ L ℋ ( Ω ) such that ∣ ∇ u ∣ ∈ L ℋ ( Ω ) } ,

with the norm

‖ u ‖ 1 , ℋ ≔ ‖ u ‖ ℋ + ‖ ∇ u ‖ ℋ .

Moreover, the space W 0 1 , ℋ ( Ω ) is defined as the closure of C 0 ∞ ( Ω ) in W 1 , ℋ ( Ω ) .

It follows that the spaces L ℋ ( Ω ) , W 1 , ℋ ( Ω ) , and W 0 1 , ℋ ( Ω ) are separable, reflexive, and uniformly convex Banach space.

Proposition 2.1

(cf. Proposition 2.18 of [17]) If q and p satisfy q p < 1 + 1 N , then for all u ∈ W 0 1 , ℋ ( Ω ) , there is a positive constant C such that

‖ u ‖ ℋ ≤ C ‖ ∇ u ‖ ℋ .

Thus, it follows that we can use ‖ ∇ u ‖ ℋ as an equivalent norm for the space W 0 1 , ℋ ( Ω ) .

Proposition 2.2

(cf. Proposition 2.4 of [17]) If u ∈ L ℋ ( Ω ) and ρ ℋ ( u ) is the ℋ -modular, then the following properties hold:

‖ u ‖ ℋ < 1 ( = 1 ; > 1 ) if and only if ϱ ℋ ( u ) < 1 ( = 1 ; > 1 ) ;
if ‖ u ‖ ℋ ≤ 1 , then ‖ u ‖ ℋ q ≤ ρ ℋ ( u ) ≤ ‖ u ‖ ℋ p ;
if ‖ u ‖ ℋ ≥ 1 , then ‖ u ‖ ℋ p ≤ ρ ℋ ( u ) ≤ ‖ u ‖ ℋ q ;
‖ u ‖ ℋ → 0 if and only if ρ ℋ ( u ) → 0 .

Proposition 2.3

(cf. Proposition 2.15 of [17]) Let p * ≔ N p N − p if p < N and p * ≔ + ∞ otherwise. Then, the following facts hold:

for all r ∈ [ 1 , p ] , we have L ℋ ( Ω ) ↪ L r ( Ω ) and W 0 1 , ℋ ( Ω ) ↪ W 0 1 , r ( Ω ) ;
if p < N for all r ∈ [ 1 , p * ] , we have
W 0 1 , ℋ ( Ω ) ↪ L r ( Ω ) ;
if p ≥ N for all r ∈ [ 1 , ∞ ) , we have
W 0 1 , ℋ ( Ω ) ↪ L r ( Ω ) ;
if p ≤ N for all r ∈ [ 1 , p * ) , we have
W 0 1 , ℋ ( Ω ) ↪ ↪ L r ( Ω ) ;
if p > N then
W 0 1 , ℋ ( Ω ) ↪ ↪ L ∞ ( Ω ) .

Proposition 2.4

[15] Assume that a , b ∈ R N are arbitrary.

If p ≥ 2 , we obtain
(2.1) ( ∣ a ∣ p − 2 a − ∣ b ∣ p − 2 b ) ⋅ ( a − b ) ≥ 2 2 − p ∣ a − b ∣ p .
If 1 < p < 2 , we obtain
(2.2) ( ∣ a ∣ p − 2 a − ∣ b ∣ p − 2 b ) ⋅ ( a − b ) ≥ ( p − 1 ) ( 1 + ∣ a ∣ 2 + ∣ b ∣ 2 ) p − 2 2 ∣ a − b ∣ 2 .

3 The proof of Theorem 1.1

Let u ∈ C ( Ω ) ⋂ W 0 1 , ℋ ( Ω ) and ϕ ∈ C 2 ( Ω ) . Define

△ p n ϕ ≔ div ( ∣ ∇ ϕ ∣ p n − 2 ∇ ϕ ) = ∣ ∇ ϕ ∣ p n − 4 { ∣ ∇ ϕ ∣ 2 △ ϕ + ( p n − 2 ) △ ∞ ϕ } , △ q n ϕ ≔ div ( ∣ ∇ ϕ ∣ q n − 2 ∇ ϕ ) = ∣ ∇ ϕ ∣ q n − 4 { ∣ ∇ ϕ ∣ 2 △ ϕ + ( q n − 2 ) △ ∞ ϕ } ,

and

△ ∞ ϕ ≔ ∑ i , j = 1 N ∂ ϕ ∂ x i ∂ ϕ ∂ x j ∂ 2 ϕ ∂ x i ∂ x j ,

where △ ∞ ϕ is the ∞ -Laplacian.

We replace u by ϕ and keep S n , K n , and k n unchanged, then

− p n ( K n ( u ) ) 1 − p n △ p n ϕ − q n a ( x ) ( K n ( u ) ) 1 − q n △ q n ϕ − q n ( K n ( u ) ) 1 − q n ∣ ∇ ϕ ∣ q n − 2 ∇ ϕ ⋅ ∇ a − λ ( p n , q n ) S n ( u ) ( p n ( k n ( u ) ) 1 − p n ∣ ϕ ∣ p n − 2 ϕ + q n a ( x ) ( k n ( u ) ) 1 − q n ∣ ϕ ∣ q n − 2 ϕ ) = 0 , x ∈ Ω , ϕ = 0 , x ∈ ∂ Ω ,

We first recall the definition of viscosity solutions. Assume we are given a continuous function

F : R N × R × R N × S ( N ) → R ,

where S ( N ) denotes the set of N × N symmetric matrices.

Consider the problem

(3.1) F ( x , u , ∇ u , D 2 u ) = 0 ,

where

(3.2) F ( x , u , ∇ u , D 2 u ) = − p n ( K n ( u ) ) 1 − p n { ∣ ∇ u ∣ p n − 4 [ ∣ ∇ u ∣ 2 △ u + ( p n − 2 ) △ ∞ u ] } − q n a ( x ) ( K n ( u ) ) 1 − q n { ∣ ∇ u ∣ q n − 4 [ ∣ ∇ u ∣ 2 △ u + ( q n − 2 ) △ ∞ u ] } − q n ( K n ( u ) ) 1 − q n ∣ ∇ u ∣ q n − 2 ∇ u ⋅ ∇ a − λ ( p n , q n ) S n ( u ) ( p n ( k n ( u ) ) 1 − p n ∣ u ∣ p n − 2 u + q n a ( x ) ( k n ( u ) ) 1 − q n ∣ u ∣ q n − 2 u ) .

Definition 3.1

Assume that x 0 ∈ Ω , u ∈ C ( Ω ) , ψ ∈ C 2 ( Ω ) , and φ ∈ C 2 ( Ω ) .

Let u ( x 0 ) = ψ ( x 0 ) and suppose that u − ψ attains its strict maximum value at x 0 . If
F ( x 0 , ψ ( x 0 ) , ∇ ψ ( x 0 ) , D 2 ψ ( x 0 ) ) ≤ 0
for all of such x 0 , then the function u is said to be a viscosity subsolution of equation (3.1).
Let u ( x 0 ) = φ ( x 0 ) and suppose that u − φ attains its strict minimum value at x 0 . If
F ( x 0 , φ ( x 0 ) , ∇ φ ( x 0 ) , D 2 φ ( x 0 ) ) ≥ 0
for all of such x 0 , then the function u is said to be a viscosity supersolution of equation (3.1).
If u is both a subsolution and a supersolution of problem (3.1), then u is a viscosity solution of the problem (3.1).

Proof of Theorem 1.1

Assume that x 0 ∈ Ω , φ ∈ C 2 ( Ω ) , and u ( x 0 ) = φ ( x 0 ) . Moreover, the function u − φ attains its strict minimum value at the point x 0 . Next, we claim that

(3.3) F ( x 0 , u ( x 0 ) , ∇ φ ( x 0 ) , D 2 φ ( x 0 ) ) ≥ 0 .

Assume that the inequality (3.3) is wrong. By continuity, we can find a constant r > 0 and a ball B ( x 0 , 2 r ) ⊂ Ω such that u > φ for all x ∈ B ( x 0 , 2 r ) \ { x 0 } and

F ( x , u ( x ) , ∇ φ ( x ) , D 2 φ ( x ) ) < 0 , x ∈ B ( x 0 , 2 r ) .

Then, also

(3.4) − div p n ∇ φ ( x ) K n ( u ) p n − 2 + q n a ( x ) ∇ φ ( x ) K n ( u ) q n − 2 ∇ φ ( x ) K n ( u ) − λ ( p n , q n ) S n ( u ) p n u ( x ) k n ( u ) p n − 2 + q n a ( x ) u ( x ) k n ( u ) q n − 2 u ( x ) k n ( u ) < 0 , x ∈ B ( x 0 , r ) .

Let Φ ( x ) ≔ φ ( x ) + m 2 , where

m = inf x ∈ ∂ B ( x 0 , r ) ( u − φ ) ( x ) > 0 .

It follows that Φ ( x ) > φ ( x ) = u ( x ) at the point x 0 . At the same time, we have Φ ( x ) < u ( x ) for all x ∈ ∂ B ( x 0 , r ) and

(3.5) − div p n ∇ Φ ( x ) K n ( u ) p n − 2 + q n a ( x ) ∇ Φ ( x ) K n ( u ) q n − 2 ∇ Φ ( x ) K n ( u ) − λ ( p n , q n ) S n ( u ) p n u ( x ) k n ( u ) p n − 2 + q n a ( x ) u ( x ) k n ( u ) q n − 2 u ( x ) k n ( u ) < 0 , x ∈ B ( x 0 , r ) .

Define Ω 1 = { x ∈ B ( x 0 , r ) : Φ ( x ) > u ( x ) } and the function η ( x ) ≔ ( Φ − u ) + . Note that for all x ∈ B ( x 0 , r ) , we have η ( x ) ≥ 0 and for all x ∈ ∂ B ( x 0 , r ) , we obtain η ( x ) ≡ 0 . Then, we multiply equation (3.5) by η ( x ) and integrate over Ω 1 , yielding

(3.6) ∫ Ω 1 p n ∇ Φ K n ( u ) p n − 2 + q n a ( x ) ∇ Φ K n ( u ) q n − 2 ∇ Φ K n ( u ) ⋅ ∇ ( Φ − u ) d x − ∫ Ω 1 p n u k n ( u ) p n − 2 + q n a ( x ) u k n ( u ) q n − 2 ⋅ λ ( p n , q n ) S n ( u ) u k n ( u ) ( Φ − u ) d x < 0 .

Let

v = ( Φ − u ) + , x ∈ B ( x 0 , r ) , 0 , x ∉ B ( x 0 , r ) .

We choose v as a test function in formula (1.4), yielding

(3.7) ∫ Ω 1 p n ∇ u K n ( u ) p n − 2 + q n a ( x ) ∇ u K n ( u ) q n − 2 ∇ u K n ( u ) ⋅ ∇ ( Φ − u ) d x − ∫ Ω 1 λ ( p n , q n ) S n ( u ) p n u k n ( u ) p n − 2 + q n a ( x ) u k n ( u ) q n − 2 ⋅ u k n ( u ) ( Φ − u ) d x = 0 .

We subtract equation (3.7) from equation (3.6) and use equation (2.1) or (2.2), yielding

(3.8) 0 ≤ ∫ Ω 1 p n ∇ Φ K n ( u ) p n − 2 ∇ Φ K n ( u ) − ∇ u K n ( u ) p n − 2 ∇ u K n ( u ) ⋅ ∇ ( Φ − u ) d x + ∫ Ω 1 q n a ( x ) ∇ Φ K n ( u ) q n − 2 ∇ Φ K n ( u ) − ∇ u K n ( u ) q n − 2 ∇ u K n ( u ) ⋅ ∇ ( Φ − u ) d x < 0 ,

which is clearly a contradiction. Hence, equation (3.3) holds, namely, the function u is a viscosity supersolution of the equation (3.1). Similarly, we can conclude that it is also a viscosity subsolution of equation (3.1).□

4 The proof of Theorem 1.2

We are now ready to prove Theorem 1.2. However, it is worth mentioning that the statement and the proof of Lemma 4.1 are inspired by the result in the study by Colasuonno and Squassina [17].

Lemma 4.1

If u ∈ L ∞ ( Ω ) , then we have

(4.1) lim n → ∞ k n ( u ) = k ∞ ( u ) ,

where k n ( u ) and k ∞ ( u ) are given in equations (1.2) and (1.10), respectively.

Proof

First, we claim that

(4.2) limsup n → ∞ k n ( u ) ≤ k ∞ ( u ) .

To show that equation (4.2) holds, we only have to discuss those indices n for which k n ( u ) > k ∞ ( u ) . Then,

1 = ∫ Ω u k n ( u ) p n + a ( x ) u k n ( u ) q n d x 1 p n ≤ ∫ Ω k ∞ ( u ) k n ( u ) p n + a ( x ) k ∞ ( u ) k n ( u ) q n d x 1 p n ≤ k ∞ ( u ) k n ( u ) ∣ Ω ∣ + ∫ Ω a ( x ) d x 1 p n ,

and the inequality (4.2) follows.

Second, we claim that we have

(4.3) liminf n → ∞ k n ( u ) ≥ k ∞ ( u ) .

If k ∞ ( u ) = 0 , it is easy to know that equation (4.3) holds. Thus, let us suppose k ∞ ( u ) > 0 . Given ε > 0 , we can find a set Ω ε ⊂ Ω such that ∣ Ω ε ∣ > 0 and the inequality ∣ u ∣ + ε > k ∞ ( u ) holds for all x ∈ Ω ε . Ignoring those indices n for which k n ( u ) + ε ≥ k ∞ ( u ) and considering q n > p n > 1 , we obtain

1 = ∫ Ω u k n ( u ) p n + a ( x ) u k n ( u ) q n d x 1 q n ≥ ∫ Ω ε u k n ( u ) p n + a ( x ) u k n ( u ) q n d x 1 q n ≥ ∫ Ω ε k ∞ ( u ) − ε k n ( u ) p n + a ( x ) k ∞ ( u ) − ε k n ( u ) q n d x 1 q n ≥ k ∞ ( u ) − ε k n ( u ) ∣ Ω ε ∣ + ∫ Ω ε a ( x ) d x 1 q n .

Then,

liminf n → ∞ k n ( u ) + ε ≥ k ∞ ( u ) .

Thus, letting ε → 0 + , we obtain equation (4.3).□

Proof of Theorem 1.2

Without loss of generality, we can assume that

∫ Ω d x = 1 .

Step 1: We want to prove the fact that limsup n → ∞ λ ( p n , q n ) 1 ≤ Λ ∞ .

By equation (1.7), we have

λ ( p n , q n ) 1 ≤ ‖ ∇ δ ‖ ℋ n ‖ δ ‖ ℋ n .

Note that by Lemma 4.1 and equation (1.9) we have

limsup n → ∞ λ ( p n , q n ) 1 ≤ ‖ ∇ δ ‖ L ∞ ( Ω ) ‖ δ ‖ L ∞ ( Ω ) = Λ ∞ .

Step 2: We now show that u ∞ ∈ W 0 1 , ∞ ( Ω ) .

It follows from Step 1 that, for all n ∈ N sufficiently large, we can obtain λ ( p n , q n ) 1 ≤ Λ ∞ + 1 . Let { u n } be the sequence of first eigenfunctions of the problem (1.1) corresponding to the first eigenvalue λ ( p n , q n ) 1 , with ‖ u n ‖ ℋ n = 1 , then we have

Λ ∞ + 1 ≥ λ ( p n , q n ) 1 = ‖ ∇ u n ‖ ℋ n .

Observe that the sequence { ‖ ∇ u n ‖ ℋ n } is bounded.

Let r ∈ [ 1 , ∞ ) be an arbitrary real number. Then, we can find an integer n r such that p n ≥ r for all n ≥ n r , we have

W 0 1 , ℋ n ( Ω ) ↪ W 0 1 , r ( Ω ) ↪ ↪ L r ( Ω ) for all n ≥ n r .

Hence, we can find a subsequence (it is still expressed as { u n } ), and a function u ∞ ∈ W 0 1 , r ( Ω ) such that ∇ u n ⇀ ∇ u ∞ weakly in the space W 0 1 , r ( Ω ) and u n → u ∞ strongly in the space L r ( Ω ) .

By equation (1.3), if n ∈ N is large enough, we have r < p n . With the help of Hölder’s inequality and ∣ Ω ∣ = 1 , we obtain

∇ u n ‖ ∇ u n ‖ ℋ n L r ( Ω ) = ∫ Ω ∇ u n ( x ) ‖ ∇ u n ‖ ℋ n r d x 1 r ≤ ∫ Ω ∇ u n ( x ) ‖ ∇ u n ‖ ℋ n p n d x 1 p n ≤ ∫ Ω ∇ u n ( x ) ‖ ∇ u n ‖ ℋ n p n + a ( x ) ∇ u n ( x ) ‖ ∇ u n ‖ ℋ n q n d x 1 p n = 1 .

Thus, we deduce that

(4.4) ‖ ∇ u n ‖ L r ( Ω ) ≤ ‖ ∇ u n ‖ ℋ n ≤ Λ ∞ + 1 .

Assume that r 1 > 0 is an arbitrary real number and B ( x 0 , r 1 ) ⊂ Ω , where x 0 ∈ Ω is a point such that ∣ ∇ u ∞ ∣ ∈ L 1 ( B ( x 0 , r 1 ) ) . Using Hölder’s inequality, equation (4.4), and a lower semicontinuity result, we have

(4.5) 1 ∣ B ( x 0 , r 1 ) ∣ ∫ B ( x 0 , r 1 ) ∣ ∇ u ∞ ( y ) ∣ d y ≤ liminf n → ∞ 1 ∣ B ( x 0 , r 1 ) ∣ ∫ B ( x 0 , r 1 ) ∣ ∇ u n ( y ) ∣ d y ≤ liminf n → ∞ ∣ B ( x 0 , r 1 ) ∣ − 1 r ‖ ∇ u n ‖ L r ( Ω ) ≤ ∣ B ( x 0 , r 1 ) ∣ − 1 r ( Λ ∞ + 1 ) .

Letting r → ∞ , we obtain

1 ∣ B ( x 0 , r 1 ) ∣ ∫ B ( x 0 , r 1 ) ∣ ∇ u ∞ ( y ) ∣ d y ≤ Λ ∞ + 1 .

Letting r 1 → 0 + , for a.e. x 0 ∈ Ω , we obtain

∣ ∇ u ∞ ( x 0 ) ∣ ≤ Λ ∞ + 1 .

Thus, u ∞ ∈ W 0 1 , ∞ ( Ω ) , as claimed.

Step 3: We claim that u n → u ∞ in the space C α ( Ω ) for a real number α ( 0 < α < 1 ) and ‖ u ∞ ‖ L ∞ ( Ω ) = 1 .

Since r ∈ [ 1 , ∞ ) is an arbitrary real number and the space W 0 1 , r ( Ω ) ↪ ↪ C α ( Ω ) ( 0 < α < 1 ) for all r > N , we conclude that 0 ≤ u ∞ ∈ C α ( Ω ) \ { 0 } , u n → u ∞ in the space C α ( Ω ) , and the function u n converges uniformly to u ∞ in Ω . Thus, for each ε ∈ ( 0 , 1 ) , we can find N ε ∈ N such that, for any x ∈ Ω ,

(4.6) ∣ u n ( x ) − u ∞ ( x ) ∣ < ε , n ≥ N ε .

It follows that for each n ≥ N ε , we have

[ ρ ℋ n ( u n − u ∞ ) ] 1 p n = ∫ Ω ∣ u n − u ∞ ∣ p n + a ( x ) ∣ u n − u ∞ ∣ q n d x 1 p n ≤ ∫ Ω ε p n + a ( x ) ε q n d x 1 p n ≤ ε ∫ Ω ( 1 + a ( x ) ) d x 1 p n ≤ ∫ Ω ( 1 + a ( x ) ) d x 1 p n

and

[ ρ ℋ n ( u n − u ∞ ) ] 1 q n ≤ ε p n q n ∫ Ω ( 1 + a ( x ) ) d x 1 q n ≤ ∫ Ω ( 1 + a ( x ) ) d x 1 q n .

Letting n → ∞ in the above inequalities yields

(4.7) lim n → ∞ [ ρ ℋ n ( u n − u ∞ ) ] 1 p n = lim n → ∞ [ ρ ℋ n ( u n − u ∞ ) ] 1 q n = 0 .

Note that

‖ ∣ u n ‖ ℋ n − ‖ u ∞ ‖ L ∞ ( Ω ) ∣ ≤ ‖ ∣ u n ‖ ℋ n − ‖ u ∞ ‖ ℋ n ∣ + ‖ ∣ u ∞ ‖ ℋ n − ‖ u ∞ ‖ L ∞ ( Ω ) ∣ ≤ ‖ u n − u ∞ ‖ ℋ n + ‖ ∣ u ∞ ‖ ℋ n − ‖ u ∞ ‖ L ∞ ( Ω ) ∣ ≤ [ ρ ℋ n ( u n − u ∞ ) ] 1 p n + [ ρ ℋ n ( u n − u ∞ ) ] 1 q n + ‖ ∣ u ∞ ‖ ℋ n − ‖ u ∞ ‖ L ∞ ( Ω ) ∣ .

Thus, in view of Lemma 4.1 and equation (4.7), we can use the fact that ‖ u n ‖ ℋ n = 1 for all n ≥ 1 , to conclude

(4.8) ‖ u ∞ ‖ L ∞ ( Ω ) = lim n → ∞ ‖ u n ‖ ℋ n = 1 .

Step 4: We claim that we have liminf n → ∞ λ ( p n , q n ) 1 ≥ Λ ∞ .

Since ∇ u n ⇀ ∇ u ∞ in the space W 0 1 , r ( Ω ) and ‖ u n ‖ ℋ n = 1 for all n ≥ 1 , by virtue of equation (4.4), we have

‖ ∇ u ∞ ‖ L r ( Ω ) ≤ liminf n → ∞ ‖ ∇ u n ‖ L r ( Ω ) ≤ liminf n → ∞ ‖ ∇ u n ‖ ℋ n = liminf n → ∞ λ ( p n , q n ) 1

Finally, letting r → ∞ , we obtain

(4.9) Λ ∞ ≤ ‖ ∇ u ∞ ‖ L ∞ ( Ω ) ‖ u ∞ ‖ L ∞ ( Ω ) ≤ liminf n → ∞ λ ( p n , q n ) 1 ,

where we have used equation (1.8) and equality (4.8). This implies that

lim n → ∞ λ ( p n , q n ) 1 = Λ ∞ .

The proof is complete.□

Remark 4.1

Let

(4.10) K ∞ ( u ∞ ) ≔ ‖ ∇ u ∞ ‖ L ∞ ( Ω ) .

Then, from Theorem 1.2, we can deduce that

K ∞ ( u ∞ ) = lim n → ∞ ‖ ∇ u n ‖ ℋ n = lim n → ∞ ‖ ∇ u n ‖ ℋ n ‖ u n ‖ ℋ n = lim n → ∞ λ ( p n , q n ) 1 = Λ ∞ .

5 The proof of Theorem 1.3

Combining equation (1.3) with the fact that W 0 1 , ℋ n ( Ω ) ↪ W 0 1 , r ( Ω ) (see Proposition 2.3(1), here we choose n ∈ N satisfies the inequality p n ≥ r > N and using the fact that W 0 1 , r ( Ω ) ↪ ↪ C α ( Ω ) ( 0 < α < 1 ), we have u n as continuous functions (here we assume that n ∈ N is large enough). Moreover, if n ∈ N is fixed, we know that u n > 0 (see [17]).

Proof of Theorem 1.3

With the help of Theorem 1.2, we deduce that equations (1.12) and (1.13) are true. Now, we claim that the function u ∞ is a viscosity solution of problem (1.14). Consider the boundary value problem

(5.1) min { ∣ ∇ u ∞ ∣ − Λ ∞ u ∞ , ∣ ∇ u ∞ ∣ − Λ ∞ ( u ∞ ) R , − △ ∞ u ∞ } = 0 , x ∈ Ω , u ∞ = 0 , x ∈ ∂ Ω ,

We now claim that u ∞ is a viscosity subsolution of equation (5.1).

Let x 0 ∈ Ω and ψ ∈ C 2 ( Ω ) . Moreover, we assume that the function u ∞ ( x 0 ) = ψ ( x 0 ) and u ∞ − ψ attains its strict maximum value at the point x = x 0 .

First, we want to obtain

(5.2) max { Λ ∞ ψ ( x 0 ) − ∣ ∇ ψ ( x 0 ) ∣ , K ∞ ( u ∞ ) ( ψ ( x 0 ) ) R − ∣ ∇ ψ ( x 0 ) ∣ , △ ∞ ψ ( x 0 ) } ≤ 0 .

Since the proof process of Theorem 1.2 implies that the convergence of u n to u ∞ is uniform in Ω , we can find a sequence { x n } ⊂ Ω such that x n → x 0 (as n → ∞ ) and u n ( x n ) = ψ ( x n ) . Moreover, the function u n − ψ attains its strict maximum at the point x n .

As we know, for all n ∈ N sufficiently large, u n is a continuous viscosity supersolution of the problem (1.1) with λ ( p n , q n ) = λ ( p n , q n ) 1 and k n ( u n ) = ‖ u n ‖ ℋ n = 1 . Thus, we have

(5.3) − p n [ K n ( u n ) ] 1 − p n ∣ ∇ ψ ( x n ) ∣ p n − 4 [ ∣ ∇ ψ ( x n ) ∣ 2 △ ψ ( x n ) + ( p n − 2 ) △ ∞ ψ ( x n ) ] − q n a ( x n ) [ K n ( u n ) ] 1 − q n ∣ ∇ ψ ( x n ) ∣ q n − 4 [ ∣ ∇ ψ ( x n ) ∣ 2 △ ψ ( x n ) + ( q n − 2 ) △ ∞ ψ ( x n ) ] − q n [ K n ( u n ) ] 1 − q n ∣ ∇ ψ ( x n ) ∣ q n − 2 ∇ ψ ( x n ) ⋅ ∇ a ( x n ) − λ ( p n , q n ) 1 S n ( u n ) p n ∣ ψ ( x n ) ∣ p n − 2 ψ ( x n ) − λ ( p n , q n ) 1 S n ( u n ) q n a ( x n ) ∣ ψ ( x n ) ∣ q n − 2 ψ ( x n ) = − p n [ K n ( u n ) ] 1 − p n ∣ ∇ ψ ( x n ) ∣ p n − 4 [ ∣ ∇ ψ ( x n ) ∣ 2 △ ψ ( x n ) + ( p n − 2 ) △ ∞ ψ ( x n ) ] − q n a ( x n ) [ K n ( u n ) ] 1 − q n ∣ ∇ ψ ( x n ) ∣ q n − 4 [ ∣ ∇ ψ ( x n ) ∣ 2 △ ψ ( x n ) + ( q n − 2 ) △ ∞ ψ ( x n ) ] − q n [ K n ( u n ) ] 1 − q n ∣ ∇ ψ ( x n ) ∣ q n − 2 ∇ ψ ( x n ) ⋅ ∇ a ( x n ) − λ ( p n , q n ) 1 S n ( u n ) p n [ k n ( u n ) ] 1 − p n ∣ ψ ( x n ) ∣ p n − 2 ψ ( x n ) − λ ( p n , q n ) 1 S n ( u n ) q n a ( x n ) [ k n ( u n ) ] 1 − q n ∣ ψ ( x n ) ∣ q n − 2 ψ ( x n ) ≥ 0 .

Case 1: ψ ( x 0 ) = u ∞ ( x 0 ) > 0 .

We now claim that, if n ∈ N is large enough, the inequality ∣ ∇ ψ ( x n ) ∣ > 0 holds. If we assume this inequality is not true, then by equation (5.3) and continuity, we obtain ψ ( x 0 ) ≤ 0 , which is a contradiction.

Dividing both sides of equation (5.3) by p n ( p n − 2 ) [ K n ( u n ) ] 1 − p n ∣ ∇ ψ ( x n ) ∣ p n − 4 , we obtain that

(5.4) − ∣ ∇ ψ ( x n ) ∣ 2 △ ψ ( x n ) p n − 2 − △ ∞ ψ ( x n ) − q n p n ∇ ψ ( x n ) K n ( u n ) q n − p n × a ( x n ) ∣ ∇ ψ ( x n ) ∣ 2 △ ψ ( x n ) p n − 2 + a ( x n ) q n − 2 p n − 2 △ ∞ ψ ( x n ) + ∣ ∇ ψ ( x n ) ∣ 2 ∇ ψ ( x n ) ⋅ ∇ a ( x n ) p n − 2 ≥ ( λ ( p n , q n ) 1 ) 3 S n ( u n ) λ ( p n , q n ) 1 ψ ( x n ) ∇ ψ ( x n ) p n − 4 ∣ ψ ( x n ) ∣ 2 ψ ( x n ) p n − 2 + ( λ ( p n , q n ) 1 ) 3 S n ( u n ) q n p n a ( x n ) ∣ ψ ( x n ) ∣ ( q n − 4 ) ∕ ( p n − 4 ) K n ( u n ) ∣ ∇ ψ ( x n ) ∣ p n − 4 ⋅ ∣ ψ ( x n ) ∣ 2 ψ ( x n ) p n − 2 ≥ 0 .

Now, we observe that, as n → ∞ ,

− ∣ ∇ ψ ( x n ) ∣ 2 △ ψ ( x n ) p n − 2 − △ ∞ ψ ( x n ) → − △ ∞ ψ ( x 0 ) , − q n p n a ( x n ) ∣ ∇ ψ ( x n ) ∣ 2 △ ψ ( x n ) p n − 2 + a ( x n ) q n − 2 p n − 2 △ ∞ ψ ( x n ) + ∣ ψ ( x n ) ∣ 2 ∇ ψ ( x n ) ⋅ ∇ a ( x n ) p n − 2 → − R 2 a ( x 0 ) △ ∞ ψ ( x 0 ) .

From those limits, we deduce that

(5.5) − liminf n → ∞ ∇ ψ ( x n ) K n ( u n ) q n − p n R 2 a ( x 0 ) + 1 △ ∞ ψ ( x 0 ) ≥ ( Λ ∞ ) 3 liminf n → ∞ S n ( u n ) λ ( p n , q n ) 1 ψ ( x n ) ∇ ψ ( x n ) p n − 4 ∣ ψ ( x n ) ∣ 2 ψ ( x n ) p n − 2 + ( Λ ∞ ) 3 R a ( x 0 ) liminf n → ∞ S n ( u n ) ∣ ψ ( x n ) ∣ ( q n − 4 ) ∕ ( p n − 4 ) K n ( u n ) ∣ ∇ ψ ( x n ) ∣ p n − 4 ⋅ ∣ ψ ( x n ) ∣ 2 ψ ( x n ) p n − 2 ≥ 0 .

Thus, it follows from equation (5.5) and

liminf n → ∞ ∇ ψ ( x n ) K n ( u n ) q n − p n ≥ 0

that

△ ∞ ψ ( x 0 ) ≤ 0 .

We now claim that

(5.6) Λ ∞ ψ ( x 0 ) ≤ ∣ ∇ ψ ( x 0 ) ∣ .

Indeed, otherwise Λ ∞ ψ ( x 0 ) − ∣ ∇ ψ ( x 0 ) ∣ > 0 , equation (1.3) implies

(5.7) lim n → ∞ λ ( p n , q n ) 1 ψ ( x n ) ∇ ψ ( x n ) ( p n − 4 ) \ ( q n − 4 ) = Λ ∞ ψ ( x 0 ) ∣ ∇ ψ ( x 0 ) ∣ 1 R > 1 ,

and thus, if n ∈ N is large enough, we can find a constant ε > 0 such that

(5.8) λ ( p n , q n ) 1 ψ ( x n ) ∇ ψ ( x n ) ( p n − 4 ) \ ( q n − 4 ) ≥ 1 + ε .

By equation (5.8), we have

(5.9) liminf n → ∞ λ ( p n , q n ) 1 ψ ( x n ) ∇ ψ ( x n ) p n − 4 ∣ ψ ( x n ) ∣ 2 ψ ( x n ) p n − 2 = liminf n → ∞ λ ( p n , q n ) 1 ψ ( x n ) ∇ ψ ( x n ) ( p n − 4 ) \ ( q n − 4 ) q n − 4 q n − 4 ∣ ψ ( x n ) ∣ 2 ψ ( x n ) p n − 2 q n − 4 ≥ R ψ ( x 0 ) 3 lim n → ∞ ( 1 + ε ) q n − 4 q n − 4 = + ∞ ,

which is a contradiction with equation (5.5). Thus, we deduce that equation (5.6) holds.

Next, we claim that we have

(5.10) ∣ ∇ ψ ( x 0 ) ∣ − ( ψ ( x 0 ) ) R K ∞ ( u ∞ ) ≥ 0 .

If we assume the contrary, we have

lim n → ∞ ∣ ψ ( x n ) ∣ ( q n − 4 ) ∕ ( p n − 4 ) K n ( u n ) ∣ ∇ ψ ( x n ) ∣ ( p n − 4 ) ∕ ( q n − 4 ) = ( ψ ( x 0 ) ) R K ∞ ( u ∞ ) ∣ ∇ ψ ( x 0 ) ∣ 1 R > 1 .

Then, if n ∈ N is large enough, it is easy to find a constant ε > 0 such that

(5.11) ∣ ψ ( x n ) ∣ ( q n − 4 ) ∕ ( p n − 4 ) K n ( u n ) ∣ ∇ ψ ( x n ) ∣ ( p n − 4 ) ∕ ( q n − 4 ) ≥ 1 + ε .

Thus, we obtain

(5.12) liminf n → ∞ ∣ ψ ( x n ) ∣ ( q n − 4 ) ∕ ( p n − 4 ) K n ( u n ) ∣ ∇ ψ ( x n ) ∣ p n − 4 ∣ ψ ( x n ) ∣ 2 ψ ( x n ) p n − 2 ≥ liminf n → ∞ ( 1 + ε ) q n − 4 q n − 4 ∣ ψ ( x n ) ∣ 2 ψ ( x n ) p n − 2 q n − 4 = R ψ ( x 0 ) 3 lim n → ∞ ( 1 + ε ) q n − 4 q n − 4 = + ∞ ,

which is a contradiction with equation (5.5). Thus, equation (5.10) holds.

Case 2: ψ ( x 0 ) = u ∞ ( x 0 ) = 0 .

We know that if ∣ ∇ ψ ( x 0 ) ∣ = 0 is true, the inequality (5.2) holds. Hence, we can assume that ∣ ∇ ψ ( x 0 ) ∣ > 0 . Then ∣ ∇ ψ ( x n ) ∣ > 0 for n ∈ N large enough. Next, the proof of equation (5.2) is the same as in case 1. Similarly, we can prove that the function u ∞ is also a viscosity supersolution. Therefore, Theorem 1.3 is proved.□

Remark 5.1

If p n = n p and q n = n q , then, Theorems 1.2 and 1.3 are also true.

Remark 5.2

In this article, we only discuss Dirichlet boundary problem. Similarly, we can deal with other boundary problems.

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Author contributions: The author confirms sole responsibility for the manuscript.
Conflict of interest: The author states that there is no conflict of interest.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analysed during the current study.

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Received: 2023-02-08

Revised: 2023-09-09

Accepted: 2023-09-28

Published Online: 2023-12-06

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2023-0138

Keywords for this article

Musielak-Orlicz spaces; viscosity solution; double-phase problem; ∞-Laplacian; eigenvalue

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