Multiple positive solutions to a p-Kirchhoff equation with logarithmic terms and concave terms

Jin-Ping Liang; Ran-Ran Wang; Yue Wang

doi:10.1515/math-2025-0183

Article Open Access

Multiple positive solutions to a p-Kirchhoff equation with logarithmic terms and concave terms

Jin-Ping Liang , Ran-Ran Wang and Yue Wang

Published/Copyright: August 11, 2025

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From the journal Open Mathematics Volume 23 Issue 1

Abstract

In this article, we focus on a class of p -Kirchhoff-type equations that include logarithmic and concave terms. By applying the variational method, we establish the existence and multiplicity of positive solutions.

Keywords: p-Kirchhoff-type equation; logarithmic and concave terms; variational method; positive solution

MSC 2010: 35A15; 35B09; 35J60

1 Introduction

In this article, our aim is to investigate the multiplicity of positive solutions for the p -Kirchhoff-type equation with logarithmic and concave terms as follows:

(1.1) − a + b ∫ Ω ∣ ∇ u ∣ p d x Δ p u = ∣ u ∣ γ − 2 u ln ∣ u ∣ + λ ∣ u ∣ q − 2 u ∣ x ∣ μ , x ∈ Ω , u = 0 , x ∈ ∂ Ω ,

where Δ p u = div ( ∣ ∇ u ∣ p − 2 ∇ u ) is the p -Laplacian operator, Ω is a bounded domain in R N ( N ≥ 1 ) with smooth boundary ∂ Ω , the constants a > 0 , b ≥ 0 , 1 ≤ q < 2 ≤ p < γ < p * , 0 ≤ μ < N ( p * − q ) ⁄ p * , and the parameter λ > 0 , p * ( p * = N p ⁄ ( N − p ) if N > p and p * = ∞ if N ≤ p ) denotes the critical Sobolev exponent for the embedding W 0 1 , p ( Ω ) into L s ( Ω ) for every s ∈ [ 1 , p * ] . The Sobolev space W 0 1 , p ( Ω ) = { u ∈ L s ( Ω ) : ∇ u ∈ L s ( Ω ) , u ∣ ∂ Ω = 0 } with the norm ‖ u ‖ p = ∫ Ω ∣ ∇ u ∣ p d x and ∣ u ∣ s s = ∫ Ω ∣ u ∣ s d x denotes the norm of L s ( Ω ) with s ∈ ( 1 , + ∞ ) . Indeed, in problem (1.1), if we replace ∣ u ∣ γ − 2 u ln ∣ u ∣ + λ ∣ u ∣ q − 2 u ∣ x ∣ − μ by g ( x , u ) and p = 2 , it reduces to the following Kirchhoff-type equation with the Dirichlet boundary value condition:

(1.2) − a + b ∫ Ω ∣ ∇ u ∣ 2 d x Δ u = g ( x , u ) , x ∈ Ω , u = 0 , x ∈ ∂ Ω ,

where a , b ≥ 0 , a + b > 0 , Ω = R N or Ω is a smooth bounded domain in R N and g : Ω ¯ × R → R is a continuous function. Problem (1.2) has been extensively studied [1–6] because (1.2) is related to the stationary problem of

a model introduced by Kirchhoff [7] as follows:

ρ ∂ 2 u ∂ t 2 − ρ 0 h + E 2 L ∫ 0 L ∂ u ∂ x 2 d x ∂ 2 u ∂ x 2 = 0 ,

where ρ , ρ 0 , h , E , and L are constants, which extends the classical d’Alembert wave equation, by considering the effects of the changes in the length of the strings during the vibrations. After the pioneering work of Lions [8], Kirchhoff-type problems began to attract the attention of several researchers, where a functional analysis approach was proposed. Recently, there are many articles on the Kirchhoff-type problem involving logarithmic term, see [9–18] and the references therein. We mentioned that Bouizem [13] considered the following Kirchhoff-type problem with logarithmic terms:

(1.3) − M ∫ Ω ∣ ∇ u ∣ 2 d x Δ u = u γ − 2 u ln ∣ u ∣ + λ f ( x ) , x ∈ Ω , u = 0 , x ∈ ∂ Ω ,

where Ω is a bounded domain in R N ( N ≥ 3 ) with smooth boundary ∂ Ω , γ ∈ ( 0 , 2 N ⁄ ( N − 2 ) ) , and λ > 0 are constants, M is a continuous positive function in R + , and f ( x ) ∈ C 1 ( Ω ¯ ) changes sign. Under certain assumptions about f ( x ) and M ( ⋅ ) , the authors employ the direct variational method, Galerkin approach, and subsuper solution method to address problem (1.3) and obtain the existence results. Wen et al. [14] studied the following Kirchhoff equation with logarithmic nonlinearity:

(1.4) − a + b ∫ Ω ∣ ∇ u ∣ 2 d x Δ u = u γ − 2 u ln ∣ u ∣ 2 − k ( x ) u , x ∈ Ω , u = 0 , x ∈ ∂ Ω ,

where Ω is a bounded domain in R 3 with smooth boundary ∂ Ω , a , b are positive constants, 4 < γ < 6 , k ∈ C ( Ω , R ) and inf x ∈ Ω k ( x ) > 0 . They used the constraint variational method, topological degree theory, and some new energy estimate inequalities to prove the existence of ground state solutions and ground state sign-changing solutions with precisely two nodal domains for problem (1.4). Li et al. [15] studied the following Kirchhoff-type problem with critical and logarithmic terms:

(1.5) − a + b ∫ Ω ∣ ∇ u ∣ 2 d x Δ u = λ u γ − 2 u ln ∣ u ∣ 2 + μ ∣ u ∣ 2 u , x ∈ Ω , u = 0 , x ∈ ∂ Ω ,

where Ω is a bounded domain in R 4 with smooth boundary ∂ Ω , a , b , λ , μ > 0 and γ ∈ ( 2,4 ) . By using the Mountain Pass Lemma, Brézis-Lieb’s lemma, and other methods, they proved that either the norm of the sequence of approximation solutions goes to infinity or the problem admits a nontrivial weak solution for equation (1.5). Furthermore, when μ = u 2 , 4 < γ < 6 and Ω ⊂ R 3 , Li and Han [16] proved that problem (1.5) admits a ground state solution for any λ > 0 by using the mountain pass lemma and the concentration compactness principle. Li et al. [17] studied the following p -Kirchhoff-type problem with logarithmic nonlinearity:

(1.6) − a + b ∫ Ω ∣ ∇ u ∣ p d x Δ p u = ∣ u ∣ γ − 2 u ln ∣ u ∣ 2 , x ∈ Ω , u = 0 , x ∈ ∂ Ω ,

where Ω is a smooth bounded domain in R N , a , b > 0 , 4 ≤ 2 p < γ < p * and N > p , Δ p u denotes the p -Laplacian operator defined by Δ p u = div ( ∣ ∇ u ∣ p − 2 ∇ u ) . By using constraint variational method, topological degree theory, and the quantitative deformation lemma, they proved the existence of ground state sign-changing solutions with precisely two nodal domains. Cai et al. [18] studied the following p -Kirchhoff-type problem with

logarithmic nonlinearity:

(1.7) − M ∫ Ω ∣ ∇ u ∣ p d x Δ p u = ∣ u ∣ p * − 2 u − ∣ u ∣ γ − 2 u ln ∣ u ∣ 2 + λ ∣ u ∣ p − 2 u , x ∈ Ω , u = 0 , x ∈ ∂ Ω ,

where Ω is a bounded domain in R N with a smooth boundary ∂ Ω , 2 < p < p * < N , M ( t ) : [ 0 , + ∞ ) → R is a continuously increasing function and satisfied

M 0 : There exist k 0 > 0 and k 1 > 0 such that k 0 ≤ M ( t ) ≤ k 1 , ∀ t ≥ 0 ;

M 1 : There exist p 0 ∈ ( p , p * ) and l < p 0 p such that k 1 < l k 0 < p 0 p k 0 .

The authors proved that equation (1.7) has at least one nontrivial solution by using the Nehari manifold and the Mountain Pass Lemma without the Palais-Smale compactness condition.

Motivated by the researches mentioned earlier, especially by [13,17,18], we shall investigate the existence and multiplicity of positive solutions to problem (1.1). The proof method in this article is slightly different from those mentioned earlier, and the study of the equation is a generalization and supplement to them.

A function u ∈ W 0 1 , p ( Ω ) is called a weak solution of problem (1.1) if and only if

( a + b ‖ u ‖ p ) ∫ Ω ∣ ∇ u ∣ p − 1 ∇ φ d x − ∫ Ω ∣ u ∣ γ − 1 φ ln ∣ u ∣ d x − λ ∫ Ω ∣ u ∣ q − 1 φ ∣ x ∣ μ d x = 0 , ∀ φ ∈ W 0 1 , p ( Ω ) .

Our main result is state as follows:

Theorem 1.1

Assume a > 0 , b ≥ 0 , 1 ≤ q < 2 ≤ p < γ < p * , and 0 ≤ μ < N ( p * − q ) ⁄ p * , and, there exists T 0 > 0 such that problem (1.1) has at least two positive solutions for any 0 < λ < T 0 .

This article is organized as follows. Some preliminaries and important lemmas are given in Section 2, and in Section 3, we prove Theorem 1.1.

2 Preliminaries

To describe our results clearly, we shall introduce some notations next. We denote by B α (respectively, S α ) the closed ball (respectively, the sphere) of center zero and radius α , where B α ( u ) = { u ∈ W 0 1 , p ( Ω ) : ‖ u ‖ ≤ α } and S α ( u ) = { u ∈ W 0 1 , p ( Ω ) : ‖ u ‖ = α } . C and C i ( i = 0 , 1 , 2 , … ) denote various positive constants, which may vary from line to line. → ( respectively , ⇀ ) denotes the strong (respectively, weak) convergence. Let S be the best Sobolev constant, that is,

(2.1) S = inf ∫ Ω ∣ ∇ u ∣ p d x : u ∈ W 0 1 , p ( Ω ) \ { 0 } , ∫ Ω ∣ u ∣ p * d x = 1 .

Let α be a constant such that Ω ⊂ B ( 0 , α ) = { x ∈ R N : ∣ x ∣ < α } , and then, there exists a constant Π such that

(2.2) ∫ Ω ∣ u ∣ q ∣ x ∣ μ d x ≤ ∫ Ω ∣ u ∣ p * d x q p * ∫ Ω 1 ∣ x ∣ μ p * p * − q d x p * − q p * ≤ ∫ Ω ∣ u ∣ p * d x q p * ∫ B ( 0 , α ) 1 ∣ x ∣ μ p * p * − q d x p * − q p * ≤ S − q p ‖ u ‖ q Π ∫ 0 α r ( N − 1 ) ( p * − q ) − μ p * p * − q d r p * − q p * ≤ Π ( p * − q ) N ( p * − q ) − μ p * p * − q p * α N ( p * − q ) − μ p * p * S − q p ‖ u ‖ q

for any u ∈ W 0 1 , p ( Ω ) by the Hölder inequality, (2.1), and 0 ≤ μ < N ( p * − q ) ⁄ p * . If N = 3 , it is easy to verify that Π = 4 π . This indicates that the element x = 0 can be included in the region Ω .

Next, we provide some definitions and lemmas that are necessary for proving the result.

The energy functional I : W 0 1 , p ( Ω ) → R associated with problem (1.1) is given by

(2.3) I ( u ) = a p ‖ u ‖ p + b 2 p ‖ u ‖ 2 p + 1 γ 2 ∫ Ω ∣ u ∣ γ d x − 1 γ ∫ Ω ∣ u ∣ γ ln ∣ u ∣ d x − λ q ∫ Ω ∣ u ∣ q ∣ x ∣ μ d x , u ∈ W 0 1 , p ( Ω ) .

Clearly, I is of class C 1 ( W 0 1 , p ( Ω ) , R ) and every weak solution of problem (1.1) is the critical point of I .

Definition 2.1

Let ( K , d ) be a complete metric space and F : K → R be a continuous functional in K . For any u ∈ K , we denote by ∣ d F ∣ ( u ) the supremum of η in [ 0 , ∞ ) such that there exist α > 0 and a continuous map δ : B α ( u ) × [ 0 , α ] → K and satisfying

(2.4) F ( δ ( v , t ) ) ≤ F ( v ) − η t , ( v , t ) ∈ B α ( u ) × [ 0 , α ] , d ( δ ( v , t ) , v ) ≤ t , ( v , t ) ∈ B α ( u ) × [ 0 , α ] .

The extended real number ∣ d F ∣ ( u ) is called the weak slope of F at u .

Definition 2.2

Recalled that a function u ∈ K is a (lower) critical point of F if ∣ d F ∣ ( u ) = 0 , and a number c ∈ R is a (lower) critical value of F if there exists a (lower) critical point u ∈ K of F with F ( u ) = c .

Definition 2.3

A sequence { u n } ⊂ K is called a ( P S ) sequence of the functional F , if ∣ d F ∣ ( u n ) → 0 as n → ∞ and F ( u n ) is bounded.

Since u → ∣ d F ∣ ( u ) is lower semicontinuous, any accumulation point of a ( P S ) sequence is clearly a critical point of F . In this article, since we are looking for a positive solution to problem (1.1), the functional I is considered on the closed positive cone Λ of W 0 1 , p ( Ω ) , namely,

Λ = { u ∣ u ∈ W 0 1 , p ( Ω ) , u ( x ) ≥ 0 , a.e. x ∈ Ω } ⊂ W 0 1 , p ( Ω ) ,

where Λ is a complete metric space, and I is a continuous functional on Λ .

Lemma 2.4

If ∣ d I ∣ ( u ) < + ∞ holds, for any v ∈ Λ , we have

(2.5) λ ∫ Ω u q − 1 ( v − u ) ∣ x ∣ μ d x ≤ a ∫ Ω ∣ ∇ u ∣ p − 1 ∇ ( v − u ) d x + b ∫ Ω ∣ ∇ u ∣ p d x ∫ Ω ∣ ∇ u ∣ p − 1 ∇ ( v − u ) d x − ∫ Ω u γ − 1 ( v − u ) ln ∣ u ∣ d x + ∣ d I ∣ ( u ) ‖ v − u ‖ .

Proof

From [19, Lemma 3.1], let ∣ d I ∣ ( u ) < c , η < 1 2 ‖ v − u ‖ , v ∈ Λ , and v ≠ u . We define the mapping τ : U × [ 0 , η ] → Λ by τ ( z , t ) = z + t v − z ‖ v − z ‖ , where U is a neighborhood of u . Then, we have ‖ τ ( z , t ) − z ‖ = t , by (2.4), and there exists a pair ( z , t ) ∈ U × [ 0 , η ] such that I ( τ ( z , t ) ) > I ( z ) − c t . Therefore, we shall assume that there exists a sequences { u n } ⊂ Λ and { t n } ⊂ [ 0 , ∞ ) such that u n → u , t n → 0 + and

I u n + t n v − u n ‖ v − u n ‖ ≥ I ( u n ) − c t n ,

namely,

(2.6) I ( u n + s n ( v − u n ) ) ≥ I ( u n ) − c s n ‖ v − u n ‖ ,

where s n = t n ‖ v − u n ‖ and s n → 0 + as n → ∞ . Set H ( u ) = 1 γ 2 ∫ Ω ∣ u ∣ γ d x − 1 γ ∫ Ω ∣ u ∣ γ ln ∣ u ∣ d x , by dividing (2.6) by s n ,

we obtain the inequality

(2.7) λ q ∫ Ω [ u n + s n ( v − u n ) ] q − u n q s n ∣ x ∣ μ d x ≤ a p ⋅ ‖ ( u n + s n ( v − u n ) ) ‖ p − ‖ u n ‖ p s n + b 2 p ⋅ ‖ u n + s n ( v − u n ) ‖ 2 p − ‖ u n ‖ 2 p s n + ∫ Ω H [ u n + s n ( v − u n ) ] − H ( u n ) s n d x + c ‖ v − u n ‖ .

Now, we claim that

(2.8) lim n → ∞ ∫ Ω H [ u n + s n ( v − u n ) ] − H ( u n ) s n d x = lim n → ∞ ∫ Ω ∣ u n + s n ( v − u n ) ∣ γ − ∣ u n ∣ γ γ 2 s n d x − lim n → ∞ ∫ Ω ∣ u n + s n ( v − u n ) ∣ γ ln ∣ u n + s n ( v − u n ) ∣ − ∣ u n ∣ γ ln ∣ u n ∣ γ s n d x = 1 γ ∫ Ω ∣ u ∣ γ − 1 ( v − u ) d x − ∫ Ω ∣ u ∣ γ − 1 ( v − u ) ln ∣ u ∣ d x − 1 γ ∫ Ω ∣ u ∣ γ − 1 ( v − u ) d x = − ∫ Ω ∣ u ∣ γ − 1 ( v − u ) ln ∣ u ∣ d x .

In addition, for all 2 ≤ p < γ and m ∈ ( γ , p * ) , we obtain

lim t → 0 t γ − 1 ln ∣ t ∣ t p − 1 = 0 and lim t → ∞ t γ − 1 ln ∣ t ∣ t m − 1 = 0 .

Therefore, for any ε > 0 , there exists C ε > 0 such that

(2.9) ∣ t ∣ γ − 1 ln ∣ t ∣ ≤ ε ∣ t ∣ p − 1 + C ε ∣ t ∣ m − 1 , ∀ t ∈ R \ { 0 } .

Indeed, note that u n ( x ) → u ( x ) a.e. in Ω and u n → u n γ ln ∣ u n ∣ is continuous, there is

∣ u n ( x ) ∣ γ ln ∣ u n ( x ) ∣ → ∣ u ( x ) ∣ γ ln ∣ u ( x ) ∣ , a.e. x ∈ Ω .

By using the Lebesgue dominated convergence theorem (see [20, pp. 27]) and (2.9), we have

∫ Ω ∣ u n ∣ γ ln ∣ u n ∣ d x → ∫ Ω ∣ u ∣ γ ln ∣ u ∣ d x ,

as n → ∞ . Thus, the claim (2.8) is holds.

Set

I 1 , n = ∫ Ω [ u n + s n ( v − u n ) ] q − [ ( 1 − s n ) u n ] q s n q ∣ x ∣ μ d x and I 2 , n = [ ( 1 − s n ) ] q − 1 s n q ∫ Ω ∣ u n ∣ q ∣ x ∣ μ d x .

Then, we have

∫ Ω [ u n + s n ( v − u n ) ] q − u n q s n q ∣ x ∣ μ d x = ∫ Ω [ u n + s n ( v − u n ) ] q − [ ( 1 − s n ) u n ] q s n q ∣ x ∣ μ d x + ∫ Ω [ ( 1 − s n ) u n ] q − u n q s n q ∣ x ∣ μ d x = ∫ Ω [ u n + s n ( v − u n ) ] q − [ ( 1 − s n ) u n ] q s n q ∣ x ∣ μ d x + [ ( 1 − s n ) ] q − 1 s n q ∫ Ω ∣ u n ∣ q ∣ x ∣ μ d x = I 1 , n + I 2 , n .

Obviously,

I 1 , n = ∫ Ω ξ n q − 1 s n v s n ∣ x ∣ μ d x = ∫ Ω ξ n q − 1 v ∣ x ∣ μ d x ,

where ξ n ∈ ( u n − u n s n , u n + s n ( v − u n ) ) , which implies that ξ n → u ( u n → u ) as s n → 0 + . By applying the Fatou’s Lemma to I 1 , n , we can obtain

(2.10) liminf n → ∞ I 1 , n ≥ ∫ Ω u q − 1 v ∣ x ∣ μ d x , ∀ v ∈ Λ

since I 1 , n ≥ 0 for all n . By applying the dominated convergence theorem to I 2 , n , we obtain

(2.11) lim n → ∞ I 2 , n = − ∫ Ω u q ∣ x ∣ μ d x .

As a result, it holds that

( a + b ‖ u ‖ p ) ∫ Ω ∣ ∇ u ∣ p − 1 ∇ ( v − u ) d x − ∫ Ω u γ − 1 ( v − u ) ln ∣ u ∣ d x + c ‖ v − u ‖ ≥ liminf n → ∞ ( I 1 , n + I 2 , n ) ≥ λ ∫ Ω u q − 1 ( v − u ) ∣ x ∣ μ d x

for any v ∈ Λ since ∣ d I ∣ ( u ) < c is arbitrary. The proof is complete.□

Lemma 2.5

I satisfies the ( P S ) condition.

Proof

Let { u n } ⊂ Λ be a ( P S ) sequence of I , there is

(2.12) I ( u n ) → c and ∣ d I ∣ ( u n ) → 0 as n → ∞ .

By Lemma 2.4 and for any v ∈ Λ , it holds

(2.13) λ ∫ Ω u n q − 1 ( v − u n ) ∣ x ∣ μ d x ≤ ( a + b ‖ u ‖ p ) ∫ Ω ∣ ∇ u ∣ p − 1 ∇ ( v − u n ) d x − ∫ Ω u n γ − 1 ( v − u n ) ln ∣ u n ∣ d x + o ( 1 ) ‖ v − u n ‖ .

Taking v = 2 u n ∈ Λ in (2.13), one has

(2.14) a ‖ u n ‖ p + b ‖ u n ‖ 2 p − ∫ Ω u n γ ln ∣ u n ∣ d x + o ( 1 ) ‖ u n ‖ ≥ λ ∫ Ω u n q ∣ x ∣ μ d x .

The fact I ( u n ) → c means that

(2.15) a p ‖ u n ‖ p + b 2 p ‖ u n ‖ 2 p + 1 γ 2 ∫ Ω ∣ u n ∣ γ d x − 1 γ ∫ Ω u n γ ln ∣ u n ∣ d x − λ q ∫ Ω ∣ u n ∣ q ∣ x ∣ μ d x = c + o ( 1 ) .

By combining the inequalities (2.2), (2.14), and (2.15), there exist two positive constants C 1 and C 2 such as

(2.16) a p − a γ ‖ u n ‖ p + b 2 p − b γ ‖ u n ‖ 2 p + 1 γ 2 ∫ Ω ∣ u n ∣ γ d x ≤ 1 q − 1 γ λ ∫ Ω ∣ u n ∣ q ∣ x ∣ μ d x + c + o ( 1 ) + o ( 1 ) ‖ u n ‖ ≤ γ − q γ q λ Π ( p * − q ) N ( p * − q ) − μ p * p * − q p * α N ( p * − q ) − μ p * p * S − q p ‖ u ‖ q + c + o ( 1 ) + o ( 1 ) ‖ u n ‖ ≤ C 1 λ ‖ u n ‖ q + C 2 + o ( 1 ) ‖ u n ‖ .

From the fact (2.16), we know that { u n } is bounded in W 0 1 , p ( Ω ) . So, there exists a subsequence (still denoted by { u n } ) and u * ∈ W 0 1 , p ( Ω ) , such that

u n ⇀ u * , weakly in W 0 1 , p ( Ω ) , u n → u * , strongly in L s ( Ω ) for s ∈ [ 1 , p * ) , u n ( x ) → u * ( x ) , a.e. x ∈ Ω .

Taking v = u n in (2.13), we have

(2.17) λ ∫ Ω u n q − 1 ( u m − u n ) ∣ x ∣ μ d x ≤ a ∫ Ω ∣ ∇ u n ∣ p − 1 ∇ ( u m − u n ) d x + b ∫ Ω ∣ ∇ u n ∣ p d x ∫ Ω ∣ ∇ u n ∣ p − 1 ∇ ( u m − u n ) d x − ∫ Ω u n γ − 1 ( u m − u n ) ln ∣ u n ∣ d x + o ( 1 ) ‖ u m − u n ‖ .

Changing the role of u m and u n in (2.17), we obtain

(2.18) λ ∫ Ω u m q − 1 ( u n − u m ) ∣ x ∣ μ d x ≤ a ∫ Ω ∣ ∇ u m ∣ p − 1 ∇ ( u n − u m ) d x + b ∫ Ω ∣ ∇ u m ∣ p d x ∫ Ω ∣ ∇ u m ∣ p − 1 ∇ ( u n − u m ) d x − ∫ Ω u m γ − 1 ( u n − u m ) ln ∣ u m ∣ d x + o ( 1 ) ‖ u n − u m ‖ .

By combining inequalities (2.17) and (2.18), we obtain

a ‖ u n − u m ‖ p + b ‖ u n − u m ‖ 2 p ≤ λ ∫ Ω ( u n − u m ) u n q − 1 ∣ x ∣ μ − u m q − 1 ∣ x ∣ μ d x + ∫ Ω ( u n − u m ) ( ∣ u n ∣ γ − 1 ln ∣ u n ∣ − ∣ u m ∣ γ − 1 ln ∣ u m ∣ ) d x + o ( 1 ) ‖ u m − u n ‖ ≤ ∫ Ω ( u n − u m ) ( ∣ u n ∣ γ − 1 ln ∣ u n ∣ − ∣ u m ∣ γ − 1 ln ∣ u m ∣ ) d x + o ( 1 ) ‖ u m − u n ‖ .

In addition, we note that

lim n → ∞ ∫ Ω ∣ u n ∣ γ ln ∣ u n ∣ d x → ∫ Ω ∣ u ∣ γ ln ∣ u ∣ d x , lim m → ∞ ∫ Ω ∣ u m ∣ γ ln ∣ u m ∣ d x → ∫ Ω ∣ u ∣ γ ln ∣ u ∣ d x , lim n → ∞ ∫ Ω ( u n − u m ) ( ∣ u n ∣ γ − 1 ln ∣ u n ∣ − ∣ u m ∣ γ − 1 ln ∣ u m ∣ ) d x → 0 ,

it holds that lim n → ∞ ‖ u n − u m ‖ → 0 . Thus, u n → u in W 0 1 , p ( Ω ) as n → ∞ . The proof is complete.□

Lemma 2.6

If ∣ d I ∣ ( u ) = 0 , then, u is a weak solution of problem (1.1). That is, u q − 1 φ ∣ x ∣ μ ∈ L 1 ( Ω ) and

( a + b ‖ u ‖ p ) ∫ Ω ∣ ∇ u ∣ p − 1 ∇ φ d x = ∫ Ω ∣ u ∣ γ − 1 φ ln ∣ u ∣ d x + λ ∫ Ω ∣ u ∣ q − 1 φ ∣ x ∣ μ d x , ∀ φ ∈ W 0 1 , p ( Ω ) .

Proof

Since ∣ d I ∣ ( u ) = 0 , by Lemma 2.4, it holds

(2.19) λ ∫ Ω u q − 1 ( v − u ) ∣ x ∣ μ d x ≤ a ∫ Ω ∣ ∇ u ∣ p − 1 ∇ ( v − u ) d x + b ∫ Ω ∣ ∇ u ∣ p d x ∫ Ω ∣ ∇ u ∣ p − 1 ∇ ( v − u ) d x − ∫ Ω u γ − 1 ( v − u ) ln ∣ u ∣ d x

for every v ∈ Λ . Let t ∈ R , φ ∈ W 0 1 , p ( Ω ) and take v = ( u + t φ ) + ∈ Λ as a test function in (2.19), then

0 ≤ a ∫ Ω ∣ ∇ u ∣ p − 1 ∇ ( ( u + t φ ) + − u ) d x + b ∫ Ω ∣ ∇ u ∣ p d x ∫ Ω ∣ ∇ u ∣ p − 1 ∇ ( ( u + t φ ) + − u ) d x − ∫ Ω u γ − 1 ( ( u + t φ ) + − u ) ln ∣ u ∣ d x − λ ∫ Ω u q − 1 ( ( u + t φ ) + − u ) ∣ x ∣ μ d x = t a ∫ Ω ∣ ∇ u ∣ p − 1 ∇ φ d x + b ∫ Ω ∣ ∇ u ∣ p d x ∫ Ω ∣ ∇ u ∣ p − 1 ∇ φ d x − ∫ Ω u γ − 1 φ ln ∣ u ∣ d x − λ ∫ Ω u q − 1 φ ∣ x ∣ μ d x − a ∫ u + t φ < 0 ∣ ∇ u ∣ p − 1 ∇ ( u + t φ ) d x + λ ∫ u + t φ < 0 u q − 1 ( u + t φ ) ∣ x ∣ μ d x − b ∫ Ω ∣ ∇ u ∣ p d x ∫ u + t φ < 0 ∣ ∇ u ∣ p − 1 ∇ ( u + t φ ) d x + ∫ u + t φ < 0 u γ − 1 ( u + t φ ) ln ∣ u ∣ d x ≤ t a ∫ Ω ∣ ∇ u ∣ p − 1 ∇ φ d x + b ∫ Ω ∣ ∇ u ∣ p d x ∫ Ω ∣ ∇ u ∣ p − 1 ∇ φ d x − ∫ Ω u γ − 1 φ ln ∣ u ∣ d x − λ ∫ Ω u q − 1 φ ∣ x ∣ μ d x − a t ∫ u + t φ < 0 ∣ ∇ u ∣ p − 1 ∇ φ d x − t b ∫ u + t φ < 0 ∣ ∇ u ∣ p d x ∫ Ω ∣ ∇ u ∣ p − 1 ∇ φ d x + t ∫ u + t φ < 0 u γ − 1 φ ln ∣ u ∣ d x .

Since ∇ u ( x ) = 0 a.e. in Ω for u ( x ) = 0 and

meas { x ∈ Ω ∣ u ( x ) + t φ < 0 , u ( x ) > 0 } → 0 , as t → 0 ,

we can obtain that

∫ u + t φ < 0 ∣ ∇ u ∣ p − 1 ∇ φ d x = ∫ u + t φ < 0 , u > 0 ∣ ∇ u ∣ p − 1 ∇ φ d x → 0 , ∫ u + t φ < 0 ∣ u ∣ γ − 1 ( u + t φ ) ln ∣ u ∣ d x = ∫ u + t φ < 0 , u > 0 ∣ u ∣ γ − 1 ( u + t φ ) ln ∣ u ∣ d x → 0 .

Thus,

0 ≤ t ( a + b ‖ u ‖ p ) ∫ Ω ∣ ∇ u ∣ p − 1 ∇ φ d x − ∫ Ω u γ − 1 φ ln ∣ u ∣ d x − λ ∫ Ω u q − 1 φ ∣ x ∣ μ d x + o ( t )

as t → 0 , we obtain that

(2.20) ( a + b ‖ u ‖ p ) ∫ Ω ∣ ∇ u ∣ p − 1 ∇ φ d x − ∫ Ω u γ − 1 φ ln ∣ u ∣ d x − λ ∫ Ω u q − 1 φ ∣ x ∣ μ d x ≥ 0 .

By the arbitrariness of φ , there will be same result in (2.20) if φ is taken as − φ . Therefore, it holds that

a + b ∫ Ω ∣ ∇ u ∣ p d x ∫ Ω ∣ ∇ u ∣ p − 1 ∇ φ d x − ∫ Ω u γ − 1 φ ln ∣ u ∣ d x − λ ∫ Ω u q − 1 φ ∣ x ∣ μ d x = 0

for any φ ∈ W 0 1 , p ( Ω ) . The proof is complete.□

3 Proof of Theorem 1.1

Now, we will prove that problem (1.1) has a positive solution u * with I ( u * ) < 0 .

Lemma 3.1

There exist α , β > 0 and T 0 > 0 such that, for any λ ∈ ( 0 , T 0 ) , we have

(3.1) I ( u ) ∣ u ∈ S α ≥ β > 0 , and inf u ∈ B ¯ α I ( u ) < 0 ,

where α = ( a γ − 1 ) ( p − q ) C 3 p ( m − q ) 1 m − p , β = ( N p + p − N ) ( a γ − 1 ) N p 2 γ ( a γ − 1 ) ( p − q ) C 3 p ( m − q ) p m − p and

T 0 = q S q p p * p γ α − N ( p * − q ) − μ p * p * N ( p * − q ) − μ p * Π ( m − q ) p * − q p * ( p − q ) ( a γ − 1 ) C 3 p ( m − q ) p − q m − p .

Proof

By using the results (2.1) and (2.9), we obtain

∫ Ω ∣ u ∣ γ ln ∣ u ∣ d x ≤ 1 p ‖ u ‖ p + C 3 ‖ u ‖ m , ∀ u ∈ W 0 1 , p ( Ω ) .

Combining (2.3) and a > 1 γ , one can further obtain

I ( u ) = a p ‖ u ‖ p + b 2 p ‖ u ‖ 2 p + 1 γ 2 ∫ Ω ∣ u ∣ γ d x − 1 γ ∫ Ω ∣ u ∣ γ ln ∣ u ∣ d x − λ q ∫ Ω ∣ u ∣ q ∣ x ∣ μ d x ≥ a p ‖ u ‖ p + b 2 p ‖ u ‖ 2 p − 1 p γ ‖ u ‖ p − C 3 γ ‖ u ‖ m − λ q ∫ Ω ∣ u ∣ q ∣ x ∣ μ d x ≥ a γ − 1 p γ ‖ u ‖ p + b 2 p ‖ u ‖ 2 p − C 3 γ ‖ u ‖ m − λ α N ( p * − q ) − μ p * p * q S q p Π ( p * − q ) N ( p * − q ) − μ p * p * − q p * ‖ u ‖ q ≥ a γ − 1 p γ ‖ u ‖ p − C 3 γ ‖ u ‖ m − λ α N ( p * − q ) − μ p * p * q S q p Π ( p * − q ) N ( p * − q ) − μ p * p * − q p * ‖ u ‖ q = ‖ u ‖ q a γ − 1 p γ ‖ u ‖ p − q − C 3 γ ‖ u ‖ m − q − λ α N ( p * − q ) − μ p * p * q S q p Π ( p * − q ) N ( p * − q ) − μ p * p * − q p * .

Set

Φ ( t ) = a γ − 1 p γ t p − q − C 3 γ t m − q − λ α N ( p * − q ) − μ p * p * q S q p Π ( p * − q ) N ( p * − q ) − μ p * p * − q p *

with t ≥ 0 , and then, there exist constants

t max = ( a γ − 1 ) ( p − q ) C 3 p ( m − q ) 1 m − p > 0

and

T 0 = q S q p p * p γ α − N ( p * − q ) − μ p * p * N ( p * − q ) − μ p * Π ( m − q ) p * − q p * ( p − q ) ( a γ − 1 ) C 3 p ( m − q ) p − q m − p > 0

such that

max t > 0 Φ ( t ) = Φ ( t max ) = a γ − 1 p γ t max p − q − C 3 γ t max m − q − λ α N ( p * − q ) − μ p * p * q S q p Π ( p * − q ) N ( p * − q ) − μ p * p * − q p * = a γ − 1 p γ ( a γ − 1 ) ( p − q ) C 3 p ( m − q ) p − q m − p − λ α N ( p * − q ) − μ p * p * q S q p Π ( p * − q ) N ( p * − q ) − μ p * p * − q p * ≥ a γ − 1 p γ ( a γ − 1 ) ( p − q ) C 3 p ( m − q ) p − q m − p − T 0 α N ( p * − q ) − μ p * p * q S q p Π ( p * − q ) N ( p * − q ) − μ p * p * − q p * = ( N p + p − N ) ( a γ − 1 ) N p 2 γ ( a γ − 1 ) ( p − q ) C 3 p ( m − q ) p − q m − p > 0

for any λ ∈ ( 0 , T 0 ) . Therefore, we have

I ( u ) ≥ t max q Φ ( t max ) = ( a γ − 1 ) ( p − q ) C 3 p ( m − q ) q m − p ( N p + p − N ) ( a γ − 1 ) N p 2 γ ( a γ − 1 ) ( p − q ) C 3 p ( m − q ) p − q m − p = ( N p + p − N ) ( a γ − 1 ) N p 2 γ ( a γ − 1 ) ( p − q ) C 3 p ( m − q ) p m − p ≔ β > 0 , ∀ λ ∈ ( 0 , T 0 ) .

Choosing α = t max , then, there exists a constant β > 0 such that I ( u ) ∣ u ∈ S α ≥ β > 0 for all λ ∈ ( 0 , T 0 ) . Moreover, choosing u ∈ B ¯ α with u ≠ 0 , it holds

lim t → 0 + I ( t u ) t q = − λ q ∫ Ω ∣ u ∣ q ∣ x ∣ μ d x < 0 ,

then, we obtain that I ( t u ) < 0 for t small enough. Therefore, one has

(3.2) m 0 ≔ inf u ∈ B ¯ α I ( u ) < 0 .

The proof is complete.□

Theorem 3.2

For any 0 < λ < T 0 , then problem (1.1) has a positive solution u * with I ( u * ) < 0 .

Proof

By the definition of m 0 in (3.2), it can be inferred that there exists a minimization sequence { u n } ⊂ B α ⊂ Λ such that lim n → ∞ I ( u n ) = m 0 < 0 . Since { u n } is bounded in B α , up to a subsequence, there exists u * ∈ W 0 1 , p such that u n → u * ∈ W 0 1 , p , u n → u * in L s ( Ω ) ( 1 ≤ s < p * ) , u n ( x ) → u * , a . e . in Ω as n → ∞ . Next, we will prove that u n → W 0 1 , p as n → ∞ . Let w n = u n − u * and by the Brezis-Lieb lemma (see [21, Theorem 1]), there holds

‖ u n ‖ p = ‖ w n ‖ p + ‖ u * ‖ p + o ( 1 ) .

By Lemma 2.5, it can be inferred that

m 0 = lim n → ∞ I ( u n ) = I ( u * ) + lim n → ∞ a p ‖ w n ‖ p + b 2 p ‖ w n ‖ 2 p + b p ‖ w n ‖ p ‖ u * ‖ p ≥ I ( u * ) ≥ m 0 ,

which implies that ‖ w n ‖ → 0 as n → ∞ . Because B α is a closed convex, it holds u * ∈ B α . Hence, we can deduce that I ( u * ) = m 0 < 0 and u * ≢ 0 in Ω , which implies that u * is a local minimizer of I .

Now, we claim that u * is a critical point of I . Indeed, notice that I ( ± u ) = I ( ∣ u ∣ ) , we can take u * ≥ 0 and u * ≢ 0 . Let t > 0 such that u * + t ψ ∈ W 0 1 , p ( Ω ) for any ψ ∈ Λ ⊂ W 0 1 , p ( Ω ) , we have

(3.3) 0 ≤ I ( u * + t ψ ) − I ( u * ) = a p ‖ u * + t ψ ‖ p + b 2 p ‖ u * + t ψ ‖ 2 p + 1 γ 2 ∫ Ω ∣ u * + t ψ ∣ γ d x − 1 γ ∫ Ω ∣ u * + t ψ ∣ γ ln ∣ u * + t ψ ∣ d x − λ q ∫ Ω ∣ u * + t ψ ∣ q ∣ x ∣ μ d x − a p ‖ u * ‖ p − b 2 p ‖ u * ‖ 2 p − 1 γ 2 ∫ Ω ∣ u * ∣ γ d x + 1 γ ∫ Ω ∣ u * ∣ γ ln ∣ u * ∣ d x + λ q ∫ Ω ∣ u * ∣ q ∣ x ∣ μ d x .

From (3.3), we obtain

(3.4) λ q ∫ Ω ∣ u * + t ψ ∣ q ∣ x ∣ μ − ∣ u * ∣ q ∣ x ∣ μ d x ≤ a p ( ‖ u * + t ψ ‖ p − ‖ u * ‖ p ) + b 2 p ( ‖ u * + t ψ ‖ 2 p − ‖ u * ‖ 2 p ) + 1 γ 2 ∫ Ω [ ∣ u * + t ψ ∣ γ − ∣ u * ∣ γ ] d x − 1 γ ∫ Ω [ ∣ u * + t ψ ∣ γ ln ∣ u * + t ψ ∣ − ∣ u * ∣ γ ln ∣ u * ∣ ] d x .

According to the inequality (3.4) and passing to the limit as t → 0 + , it holds that

(3.5) λ q lim t → 0 + inf ∫ Ω ∣ u * + t ψ ∣ q − ∣ u * ∣ q t ∣ x ∣ μ d x ≤ ( a + b ‖ u * ‖ p ) ∫ Ω ∇ u * ∇ ψ d x − ∫ Ω ∣ u * ∣ γ − 1 ψ ln ∣ u * ∣ d x .

Notice that

λ q ∫ Ω ∣ u * + t ψ ∣ q − ∣ u * ∣ q t ∣ x ∣ μ d x = λ ∫ Ω ∣ u * + t ψ ∣ q − 1 ψ ∣ x ∣ μ d x + o ( t )

as t → 0 + , we can obtain that ∣ x ∣ − μ ( u * + ε t ψ ) q − 1 ψ → ∣ x ∣ − μ u * q − 1 ψ as ε t → 0 + for any ψ ∈ W 0 1 , p ( Ω ) . Owing to ∣ x ∣ − μ ( u * + ε t ψ ) q − 1 ψ ≥ 0 , we obtain

(3.6) λ q lim t → 0 + inf ∫ Ω ∣ u * + t ψ ∣ q − ∣ u * ∣ q t ∣ x ∣ μ d x ≥ λ ∫ Ω u * q − 1 ψ ∣ x ∣ μ d x .

Combining (3.5) and (3.6), we have

(3.7) ( a + b ‖ u * ‖ p ) ∫ Ω ∣ ∇ u * ∣ p − 1 ∇ ψ d x − ∫ Ω u * γ − 1 ψ ln ∣ u * ∣ d x − λ ∫ Ω u * q − 1 ψ ∣ x ∣ μ d x ≥ 0

for any ψ ∈ W 0 1 , p ( Ω ) with ψ ≥ 0 . From Lemma 3.1, the inequality I ( u * ) < 0 implies that u * ∉ S α , then, ‖ u * ‖ < α . So, there exists ϱ 1 ∈ ( 0 , 1 ) such that ( 1 + t ) u * ∈ B α with ∣ t ∣ < ϱ 1 . We define J : [ − ϱ 1 , ϱ 1 ] by J ( t ) = I ( ( 1 + t ) u * ) . Obviously, J ( t ) achieves its minimum at t = 0 , we have

(3.8) J ′ ( 0 ) = a ‖ u * ‖ p + b ‖ u * ‖ 2 p − ∫ Ω u * γ ln ∣ u * ∣ d x − λ ∫ Ω u * q ∣ x ∣ μ d x = 0 .

For any v ∈ W 0 1 , p ( Ω ) and ε > 0 , we define ϕ ∈ Λ by

ϕ = ( u * + ε v ) + ,

then, the inequalities (3.5) and (3.6) imply that

(3.9) 0 ≤ ( a + b ‖ u * ‖ p ) ∫ Ω ∣ ∇ u * ∣ p − 1 ∇ ϕ d x − ∫ Ω u * γ − 1 ϕ ln ∣ u * ∣ d x − λ ∫ Ω u * q − 1 ϕ ∣ x ∣ μ d x = ∫ u * + ε v > 0 ( a + b ‖ u * ‖ p ) ∣ ∇ u * ∣ p − 1 ∇ ( u * + ε v ) − u * γ − 1 ( u * + ε v ) ln ∣ u * ∣ − λ u * q − 1 ( u * + ε v ) ∣ x ∣ μ d x = ∫ Ω − ∫ u * + ε v ≤ 0 ( a + b ‖ u * ‖ p ) ∣ ∇ u * ∣ p − 1 ∇ ( u * + ε v ) − u * γ − 1 ( u * + ε v ) ln ∣ u * ∣ − λ u * q − 1 ( u * + ε v ) ∣ x ∣ μ d x ≤ a ‖ u * ‖ p + b ‖ u * ‖ 2 p − ∫ Ω u * γ ln ∣ u * ∣ d x − λ ∫ Ω ∣ u * ∣ q ∣ x ∣ μ d x + ε ∫ Ω [ ( a + b ‖ u * ‖ p ) ∣ ∇ u * ∣ p − 1 ∇ v − u * γ − 1 v ln ∣ u * ∣ − λ ∣ u * ∣ q − 1 v ∣ x ∣ μ d x − ∫ u * + ε v ≤ 0 ( a + b ‖ u * ‖ p ) ∣ ∇ u * ∣ p − 1 ∇ v d x + ∫ u * + ε v ≤ 0 u * γ − 1 ( u * + ε v ) ln ∣ u * ∣ + λ u * q − 1 ( u * + ε v ) ∣ x ∣ μ d x ≤ ε ∫ Ω ( a + b ‖ u * ‖ p ) ∣ ∇ u * ∣ p − 1 ∇ v − u * γ − 1 v ln ∣ u * ∣ − λ ∣ u * ∣ q − 1 v ∣ x ∣ μ d x − ∫ u * + ε v ≤ 0 ( a + b ‖ u * ‖ p ) ∣ ∇ u * ∣ p − 1 ∇ v d x .

Since ∇ u * ( x ) = 0 for a.e. in Ω with u * ( x ) = 0 and

meas { x ∈ Ω ∣ u * ( x ) + ε v < 0 , u * ( x ) > 0 } → 0 , as ε → 0 ,

we have

∫ u * + ε v < 0 ∣ ∇ u * ∣ p − 1 ∇ v d x = ∫ u + ε v < 0 , u > 0 ∣ ∇ u * ∣ p − 1 ∇ v d x → 0 , ∫ u * + ε v < 0 ∣ u * ∣ γ − 1 ( u * + ε v ) ln ∣ u * ∣ d x = ∫ u * + ε v < 0 , u > 0 ∣ u * ∣ γ − 1 ( u * + ε v ) ln ∣ u * ∣ d x → 0 .

Thus, dividing by ε and ε → 0 in (3.9), we obtain

(3.10) ( a + b ‖ u * ‖ p ) ∫ Ω ∣ ∇ u * ∣ p − 1 ∇ v d x − ∫ Ω u * γ − 1 v ln ∣ u * ∣ d x − λ ∫ Ω u * q − 1 v ∣ x ∣ μ d x ≥ 0 .

So, we have

( a + b ‖ u * ‖ p ) ∫ Ω ∣ ∇ u * ∣ p − 1 ∇ v d x − ∫ Ω u * γ − 1 v ln ∣ u * ∣ d x − λ ∫ Ω u * q − 1 v ∣ x ∣ μ d x = 0

because v is arbitrarily in (3.10). Therefore, u * is a critical point of I . Thus, u * is a nonzero negative solution of problem (1.1) with I ( u * ) = m 0 < 0 . Let

ϑ ( t ) = ln t + λ t q − γ ∣ x ∣ − μ ,

it is easy to see that

lim t → + 0 ϑ ( t ) = + ∞ , lim t → + ∞ ϑ ( t ) = + ∞

and

t * = ( γ − q ) λ ∣ x ∣ μ 1 γ + 1 − q

is the unique minimum point of function ϑ , which implies that

min t > 0 ϑ ( t ) = ϑ ( t * ) = 1 γ + 1 − q ln λ ( γ − q ) ∣ x ∣ μ + 1 γ − q ( γ − q ) λ ∣ x ∣ μ 1 γ + 1 − q ≕ C 4 > 0 .

Consequently, we have

− Δ p u * = u * γ − 1 ln u * + λ u * q − 1 a + b ‖ u * ‖ p ≥ C 4 u * γ − 1 a + b ‖ u * ‖ p ≥ 0 .

By using the strong maximum principle in [22], we obtain that u * is a positive solution of problem (1.1) and satisfies I ( u * ) = m 0 < 0 . This proof is complete.□

Theorem 3.3

For any 0 < λ < T 0 , problem (1.1) has a positive solution u * * such that I ( u * * ) > 0 .

Proof

According to Lemma 3.1, I satisfies the geometric structure of mountain pass. Set

c = inf η ∈ Γ max t ∈ [ 0 , 1 ] I ( η ( t ) ) ,

where Γ = { η ( t ) ∈ C ( [ 0 , 1 ] , W 0 1 , p ( Ω ) ) : η ( 0 ) = 0 , η ( 1 ) = e } . By applying the Mountain Pass Lemma [23] and Lemma 3.1, there exists a sequence { u n } ⊂ W 0 1 , p ( Ω ) such that

I ( u n ) → c > 0 and ∣ d I ∣ ( u n ) → 0 as n → ∞ .

By Lemma 2.5, we know that { u n } ⊂ W 0 1 , p ( Ω ) has a convergent subsequence (still denoted by itself) and there is a function u * * ∈ W 0 1 , p ( Ω ) , such that u n → u * * in W 0 1 , p ( Ω ) , I ( u * * ) = lim n → ∞ I ( u n ) = c and ∣ d I ∣ ( u n ) → 0 , which implies that u * * ≢ 0 . Similar to Theorem 3.2, we obtain u * * > 0 satisfies problem (1.1) with I ( u * * ) = c > 0 . Thus, u * * is the second positive solution of problem (1.1). This completes the proof of Theorem 1.1.□

Acknowledgments

The authors are grateful to the referees and editors for their careful reading of the manuscript and valuable comments.

Funding information: This work was supported by the Youth Science and Technology Talent Growth Project of the Department of Education of the Guizhou Province (No. Qian Jiao Ji[2024]64), the Science and Technology Plan Project of Guizhou Province (No. Qian Ke He Jichu-[2024]Qingnian 209), and the Scientific Research Staring Foundation of Guizhou Minzu University (No. GZMUZK[2024]QD06).
Author contributions: YW provided the funding support, supervised and led the planning and execution of this research, and reviewed and evaluated the manuscript. JPL and RRW proposed the research idea, formed the overall research objective, and wrote the first draft. JPL and YW revised the final draft. All authors have read and agreed to the published version of the manuscript.
Conflict of interest: The authors declare there is no conflict of interest.

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Received: 2024-10-06

Revised: 2025-05-29

Accepted: 2025-06-30

Published Online: 2025-08-11

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

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https://doi.org/10.1515/math-2025-0183

Keywords for this article

p-Kirchhoff-type equation; logarithmic and concave terms; variational method; positive solution

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