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Article

Interior Peak Solutions for a Semilinear Dirichlet Problem

1
Department of Mathematics, College of Science, Qassim University, Buraydah 51452, Saudi Arabia
2
Faculté des Sciences et Techniques, Université de Nouakchott, Nouakchott 2373, Mauritania
*
Author to whom correspondence should be addressed.
Axioms 2025, 14(1), 58; https://doi.org/10.3390/axioms14010058
Submission received: 7 December 2024 / Revised: 9 January 2025 / Accepted: 10 January 2025 / Published: 13 January 2025

Abstract

:
In this paper, we consider the semilinear Dirichlet problem ( P ε ) : Δ u + V ( x ) u = u n + 2 n 2 ε , u > 0 in Ω , u = 0 on ∂ Ω , where Ω is a bounded regular domain in R n , n 4 , ε is a small positive parameter, and V is a non-constant positive C 2 -function on Ω ¯ . We construct interior peak solutions with isolated bubbles. This leads to a multiplicity result for ( P ε ) . The proof of our results relies on precise expansions of the gradient of the Euler–Lagrange functional associated with ( P ε ) , along with a suitable projection of the bubbles. This projection and its associated estimates are new and play a crucial role in tackling such types of problems.

1. Introduction and Main Results

Let us consider the following semilinear Dirichlet problem
( P V , q ) : Δ u + V u = u q in Ω , u > 0 in Ω , u = 0 on Ω ,
where Ω is a bounded regular domain in R n , n 3 , 1 < q < and V is a smooth positive function on Ω ¯ .
The problem in the form of ( P V , q ) appears in various physical models, including quantum transport and non-relativistic Newtonian gravity, as discussed in [1,2,3] and the references cited therein. It is also related to the Yamabe problem in differential geometry; see, for example, [4] and the references therein.
In the subcritical case, that is, 1 < q < n + 2 n 2 , proving the existence of a solution to ( P V , q ) is relatively straightforward. This can be demonstrated by noting that the following infimum
inf { Ω | u | 2 + V u 2 : u H 0 1 ( Ω ) and Ω | u | q + 1 = 1 }
is attained due to the compactness of the embedding of H 0 1 ( Ω ) L q ( Ω ) .
In the critical case, where q = n + 2 n 2 , it is well known that the existence of solutions to ( P V , q ) depends on the geometry of the domain Ω , the properties of V, and the dimension n. For example, if V is constant and Ω is starshaped, the problem ( P V , q ) has no solutions. Given the extensive literature on this topic, we will only highlight the pioneering papers by Brezis–Nirenberg [5] and Bahri–Coron [6].
The supercritical case, q > n + 2 n 2 , is significantly more challenging to address, as there is no embedding of H 0 1 ( Ω ) L q ( Ω ) . Notable existence and nonexistence results in specific domains are due to Passaseo [7,8]. We also refer to the paper [9], which addresses domains with small circular holes. For the case of the ball, we refer to the results in [10,11,12,13]. It is important to note that the general domain case does not appear to be fully understood.
Note that, for q n + 2 n 2 , the problem ( P V , q ) is a variational problem and its Euler–Lagrange function is defined on H 0 1 ( Ω ) by
I V , q u : = 1 2 Ω u 2 + V u 2 1 q + 1 Ω u q + 1 .
When q is critical, the functional I V , q does not satisfy the Palais–Smale condition; that is, there exist sequences along which I V , q is bounded, with a gradient that goes to zero, and which do not converge. This is violated for all levels N S n n / 2 , where
S n = inf u H 0 1 ( Ω ) { 0 } Ω | u | 2 Ω | u | p + 1 2 / ( p + 1 ) where p = n + 2 n 2 .
These compactness defects correspond to a concentration in N points a i (possibly confused) of Ω ¯ sequences ( u k ) in H 0 1 ( Ω ) ; that is,
| u k | 2 S n n / 2 i = 1 N δ a i in the sense of measure ,
where δ a i denotes the Dirac mass at a i (see [14,15]).
In this paper, we are interested in slightly subcritical problems that actually admit solutions concentrating, as ε goes to zero, at certain interior points of the domain Ω . More precisely, we consider the problem
( P ε ) : Δ u + V u = u p ε i n Ω , u > 0 i n Ω , u = 0 o n Ω ,
where Ω is a bounded regular domain of R n , n 4 , p + 1 : = n + 2 n 2 + 1 = 2 n n 2 is the critical Sobolev exponent for the embedding H 0 1 ( Ω ) L q ( Ω ) , V is a C 2 positive function on Ω ¯ , and ε is a small positive parameter.
Our aim is to establish the existence and multiplicity of interior peak solutions of ( P ε ) for any positive C 2 functions V and for any smooth bounded domain Ω provided that ε is positive and sufficiently small.
To state our result, we need to fix some notation. We denote by I ε the Euler–Lagrange functional associated with ( P ε ) defined on H 0 1 ( Ω ) by
I ε u : = 1 2 Ω u 2 + V u 2 1 p + 1 ε Ω u p + 1 ε .
In this paper, we will use the following scalar product and its corresponding norm
v , w : = Ω v · w + Ω V v w ; w : = Ω | w | 2 + Ω V w 2 .
Notice that this norm is equivalent to the two classical norms of H 0 1 ( Ω ) and H 1 ( Ω ) .
Let a R n and λ > 0 ; we define
δ a , λ ( x ) : = c 0 λ ( n 2 ) / 2 1 + λ 2 | x a | 2 ( n 2 ) / 2 , with c 0 : = n ( n 2 ) ( n 2 ) / 4 .
These functions, for a R n and λ > 0 , are the only positive solutions [16] of
Δ w = | w | 4 / ( n 2 ) w in R n .
The aim of our first result is to construct interior peak solutions. More precisely, we prove the following theorem.
Theorem 1.
Let n 4 and V : Ω ¯ R be a positive C 2 -function with m non- degenerate critical points b 1 , , b m . Then, for any N m , there exists small ε 0 > 0 such that for any ε ( 0 , ε 0 ] , problem ( P ε ) admits a solution u ε , b i 1 , , b i N satisfying that u ε , b i 1 , , b i N develops exactly one peak at each point b i j and converges weakly to zero in H 0 1 ( Ω ) as ε 0 . More precisely, there exist λ 1 , ε ,..., λ N , ε with the same order as ε 1 / 2 for n 5 and as ε 1 / 2 | ln ε | 1 / 2 for n = 4 and N points a i j , ε b i j for all j such that
| | u ε , b i 1 , . . . , b i N j = 1 N δ a i j , ε , λ i j , ε | | 0 , a s ε 0 .
Theorem 1 allows us to obtain the following multiplicity result in connection with the number of critical points of V.
Theorem 2.
Let n 4 and let V : Ω ¯ R be a positive C 2 -function with m non-degenerate critical points. Then, for positive and sufficiently small ε, the number of solutions of ( P ε ) is at least equal to 2 m 1 .
The structure of the proof follows a similar approach to that in [17,18,19]. It is based on fine expansions of the gradient of the functional I ε . However, we adopt a different framework—that is, an appropriate projection—to handle our case. This projection, along with its estimates, proves instrumental in addressing such types of problems. In the next section, we introduce a two-parameter family of approximate solutions to the problem ( P ε ) and seek a true solution in a neighborhood of this set. We also introduce the neighborhood at infinity, provide precise estimates for the infinite-dimensional part, and perform the expansion of the gradient of the associated functional. Section 3 is devoted to the proof of our results. Section 4 discusses potential directions for future research. The proofs involve certain technical estimates, which are provided in the Appendix A for the reader’s convenience.

2. Analytical Frameworks

Let a Ω and λ > 0 ; we notice that the function δ a , λ does not belong to H 0 1 ( Ω ) ; thus, we need to introduce its projection onto this space. Different projections have been introduced into the literature. In our case, we will use the following projection
( Δ + V ) ( π δ a , λ ) = δ a , λ n + 2 n 2 in Ω , π δ a , λ = 0 on Ω .
In the sequel, for a Ω and λ > 0 , we denote
θ a , λ : = δ a , λ π δ a , λ .
To enhance readability, the estimates of π δ a , λ and θ a , λ are deferred to the Appendix A, allowing the main argument to be presented more clearly.
Next, we introduce the neighborhood at infinity, provide precise estimates for the infinite-dimensional part, and finally, perform the expansion of the gradient of the associated functional.
To this aim, first, let N N , a 1 , , a N be N points in Ω satisfying d ( a i , Ω ) c for some constant c > 0 , and let λ 1 , , λ N be large so that, for each i j ,
ε i j : = λ i λ j + λ j λ i + λ i λ j | a i a j | 2 ( n 2 ) / 2
is small enough, and we denote
E a , λ : = span π δ a i , λ i , ( π δ a i , λ i ) λ i , ( π δ a i , λ i ) a i , j , i { 1 , , N } , j { 1 , , n } ,
where a i , j is the jth component of a i .
Second, let μ > 0 be a small real and d 0 > 0 be a fixed small constant; we introduce the following set:
O ( N , μ , d 0 ) : = { ( α , a , λ ) ( 0 , ) N × Ω N × ( μ 1 , ) N : | α i 1 | < μ ; d ( a i , Ω ) > 2 d 0 ; M 1 | ln ε | σ n / ε λ i M | ln ε | σ n / ε ; i and | a i a j | > 2 d 0 i j }
where M is a fixed large constant, σ 4 = 1 and σ n = 0 for n 5 .
Notice that, for ( α , a , λ ) O ( N , μ , d 0 ) , since we have | a i a j | > 2 d 0 and d ( a i , Ω ) > 2 d 0 , we deduce that
B i : = B ( a i , d 0 ) Ω 0 : = { x Ω : d ( x , Ω ) d 0 } i a n d B i B j = j i .
For each ( α , a , λ ) O ( N , μ , d 0 ) and v E a , λ , we associate a function
u : = i = 1 N α i ( π δ a i , λ i ) + v : = u ̲ + v .
We notice that the orthogonality is taken with respect to the scalar product defined in (1).
Furthermore, for each i { 1 , , N } , the following estimates hold
u ̲ = α i π δ a i , λ i + O ( γ ( ε ) ) in B i , u ̲ = O ( γ ( ε ) ) in Ω ( B i ) where   γ ( ε ) : = ε 1 / 2 / | ln ε | 1 / 2 if n = 4 , ε ( n 2 ) / 4 if n 5 .
Below, our aim is to find some positive critical points u of I ε with the form (8). We start by studying the v-part of u.

2.1. Minimizing with Respect to the Infinite-Dimensional Part

Let ( α , a , λ ) O ( N , μ , d 0 ) and u be defined in (8). To study the v-part of u : = i = 1 N α i ( π δ a i , λ i ) + v : = u ̲ + v , we need to expand I ε with respect to v.
Using Lemma A1 and the fact that v , u ̲ = 0 , we obtain
I ε u = 1 2 u ̲ 2 + 1 2 v 2 1 p + 1 ε Ω u ̲ p + 1 ε Ω ( u ̲ ) p ε v 1 2 ( p ε ) Ω ( u ̲ ) p 1 ε v 2 + o | | v | | 2 = I ε ( u ̲ ) f ε ( v ) + 1 2 Q ε ( v ) + o | | v | | 2 ,
where
f ε ( v ) : = Ω ( u ̲ ) p ε v and Q ε ( v ) : = | v | 2 ( p ε ) Ω ( u ̲ ) p ε 1 v 2 .
Proposition 1.
Let N N , α 1 , , α n be N positive reals close to 1, let a 1 , , a N be N points in Ω satisfying d ( a i , Ω ) c for some constant c > 0 , and let λ 1 , , λ N be large reals. Assume that the ε i j ’s, for i j , and the ε ln λ i ’s, for i { 1 , , N } , are small enough. Then, there exists a positive constant ρ 0 > 0 such that
Q ε ( v ) ρ 0 v 2 v E a , λ .
Proof. 
Using Lemma A1 and the fact that α i = 1 + o ( 1 ) for each i, we derive that
Ω ( u ̲ ) p ε 1 v 2 = Ω ( π δ a i , λ i ) p ε 1 v 2 + o ( v 2 ) .
Observe that, since Ω is bounded, for each b Ω and λ > 0 satisfying that ε ln λ is small, it follows that
δ b , λ ε = c 0 ε λ ε ( n 2 ) / 2 1 + n 2 2 ε ln ( 1 + λ 2 | x b | 2 ) + O ε 2 ln 2 ( 1 + λ 2 | x b | 2 ) = 1 + o ( 1 ) .
Now, using (12) and Proposition A1, we deduce that
Ω ( π δ a i , λ i ) p ε 1 v 2 = Ω δ a i , λ i p 1 v 2 + o ( v 2 ) .
Therefore, the quadratic form becomes
Q ε ( v ) = v 2 p Ω δ a i , λ i p 1 v 2 + o ( v 2 ) Ω | v | 2 p Ω δ a i , λ i p 1 v 2 : = Q ¯ ( v ) + o ( v 2 ) .
Observe that Q ¯ is studied in several works, and it is proved that it is coercive on the space (see Proposition 3.1 of [20])
F a , λ : = 1 i N v H 0 1 ( Ω ) : Ω v P δ a i , λ i = Ω v P δ a i , λ i λ i = 0 , Ω v P δ a i , λ i a i = 0 ,
where, for b Ω and μ > 0 , the function P δ b , μ is defined by the following PDE:
Δ P δ b , μ = δ b , μ ( n + 2 ) / ( n 2 ) in Ω , P δ b , μ = 0 on   Ω .
Note that the coercivity of Q ¯ implies the existence of ρ 0 > 0 such that
Q ¯ ( v ) ρ 0 Ω | v | 2 v F a , λ .
Now, for b Ω and μ > 0 , observe that
v , π δ b , μ = Ω v ( π δ b , μ ) + Ω V v ( π δ b , μ ) = Ω ( Δ + V ) ( π δ b , μ ) v = Ω δ b , μ n + 2 n 2 v = Ω ( Δ ( P δ b , μ ) ) v = Ω v ( P δ b , μ ) ,
and for τ { μ , b } ,
v , ( π δ b , μ ) τ = Ω v ( π δ b , μ ) τ + Ω V v ( π δ b , μ ) τ = Ω ( Δ + V ) ( π δ b , μ ) τ v = n + 2 n 2 Ω δ b , μ 4 n 2 δ b , μ τ v = Ω ( Δ ( P δ b , μ ) τ ) v = Ω v ( P δ b , μ ) τ .
We deduce that E a , λ = F a , λ . Finally, since the norm . and the norm of H 0 1 ( Ω ) are equivalent, the proof of the proposition follows from (13) and (14). □
Now, we will estimate the linear form on v defined in (11). More precisely, we have
Lemma 1.
Let ( α , a , λ ) O ( N , μ , d 0 ) . Then, for each v E a , λ , we have
f ε ( v ) c ε v .
Proof. 
Using (9) and Lemma A1, we obtain
f ε ( v ) = i = 1 N B i α i π δ a i , λ i + γ ( ε ) p ε v + O γ ( ε ) p ε Ω ( B i ) | v | = i = 1 N B i α i π δ a i , λ i p ε v + O γ ( ε ) B i δ a i , λ i p ε 1 | v | + γ ( ε ) p ε Ω | v | = i = 1 N α i p ε Ω π δ a i , λ i p ε v + O γ ( ε ) p ε v + γ ( ε ) v B i δ a i , λ i 8 n n 2 4 ( n + 2 ) / ( 2 n )
by using the embedding Theorem of H 0 1 ( Ω ) L q ( Ω ) for each q [ 1 , 2 n / ( n 2 ) ] .
Observe that easy computations imply that
B i δ a i , λ i 8 n n 2 4 ( n + 2 ) / ( 2 n ) c 1 λ i ( n 2 ) / 2 ( if   n 5 ) ; ln ( λ i ) 2 / 3 λ i 2 ( if   n = 6 ) ; 1 λ i 2 ( if   n 7 )
c γ ( ε ) ( if   n 5 ) ; | ln ε | 2 / 3 ε ( if   n = 6 ) ; ε ( if   n 7 ) .
It remains to estimate the first integral in (15). Using Lemma A1, we obtain
Ω π δ a i , λ i p ε v = Ω δ a i , λ i θ a i , λ i p ε v = Ω δ a i , λ i p ε v + O Ω δ a i , λ i 4 n 2 ε θ a i , λ i | v | .
The last integral can be deduced from Lemma A5 by using (12).
For the other one, using (12) and the fact that v π δ a i , λ i , we obtain
Ω δ a i , λ i p ε v = c 0 ε λ i ε n 2 2 Ω δ a i , λ i p v + O ε Ω δ a i , λ i p ln ( 1 + λ i 2 | x a i | 2 ) | v | = c 0 ε λ i ε n 2 2 π δ a i , λ i , v + O ε v Ω δ a i , λ i 2 n n 2 ln 2 n n + 2 ( 1 + λ i 2 | x a i | 2 ) n + 2 2 n = O ε v .
Thus, Lemma A5 and (18) imply that
Ω π δ a i , λ i p ε v = O ε v .
Putting (17) and (19) in (15), the result follows. □
Note that Proposition 1 implies that Q ε is a positive definite quadratic form on E a , λ . Hence, combining (10), Proposition 1, and Lemma 1, we deduce that
Proposition 2.
Let ( α , a , λ ) O ( N , μ , d 0 ) and let u ̲ : = i = 1 N α i π δ a i , λ i . Then, for ε small, there exists a unique v ¯ E a , λ satisfying
I ε ( u ̲ + v ¯ ) , h = 0 h E a , λ .
Moreover, v ¯ satisfies v ¯ c ε .
Next, we perform the expansion of the gradient of the functional I ε .

2.2. Expansion of the Gradient

We begin with the expansion in terms of the gluing parameter α i . Specifically, we prove
Proposition 3.
Let ( α , a , λ ) O ( N , μ , d 0 ) , v E a , λ and u = i = 1 N α i π δ a i , λ i + v . Then, for ε small and i 1 , , N , we have
I ε ( u ) , π δ a i , λ i = α i S n 1 α i p 1 ε λ i ε ( n 2 ) / 2 + O ( ε ) w h e r e S n : = R n δ 0 , 1 2 n / ( n 2 ) .
Proof. 
Observe that, since v E a , λ , it holds that
I ε ( u ) , π δ a i , λ i = j = 1 N α j π δ a j , λ j , π δ a i , λ i Ω u p 1 ε u ( π δ a i , λ i ) .
First, for j i , we have
π δ a j , λ j , π δ a i , λ i = Ω π δ a j , λ j π δ a i , λ i + Ω V π δ a j , λ j π δ a i , λ i = Ω Δ + V π δ a j , λ j π δ a i , λ i Ω δ a j , λ j n + 2 n 2 δ a i , λ i
and we have
Ω δ a j , λ j n + 2 n 2 δ a i , λ i c λ j ( n + 2 ) / 2 B i δ a i , λ i + c λ i ( n 2 ) / 2 Ω B i δ a j , λ j n + 2 n 2 c ( λ i λ j ) ( n 2 ) / 2 c γ ( ε ) 2 .
Second, for j = i , as in the previous computation, we have
π δ a i , λ i 2 = Ω δ a i , λ i n + 2 n 2 π δ a i , λ i = Ω δ a i , λ i 2 n n 2 Ω δ a i , λ i n + 2 n 2 θ a i , λ i .
Estimate of the first integral is well known, and we have
Ω δ a i , λ i 2 n n 2 = S n + O 1 λ i n .
For the second integral, using Proposition A1, easy computations imply that
Ω δ a i , λ i n + 2 n 2 θ a i , λ i c B i R 1 ( x , a i , λ i ) δ a i , λ i 2 n n 2 + Ω B i δ a i , λ i 2 n n 2 c ε + c λ i n c ε .
Now, we focus on the last term in (21). Let u ̲ : = j = 1 N α j ( π δ a j , λ j ) , using Lemma A1, we obtain
Ω u p ε 1 u ( π δ a i , λ i ) = Ω u ̲ p ε ( π δ a i , λ i ) + ( p ε ) Ω u ̲ p ε 1 v ( π δ a i , λ i ) + O v 2 .
The first integral in the right hand side of (27) can be written as (using Lemma A1, Proposition A1, (12), (23) and (26))
Ω u ̲ p ε ( π δ a i , λ i ) = α i p ε Ω ( π δ a i , λ i ) p + 1 ε + j i O Ω δ a i , λ i p ε δ a j , λ j + Ω δ a j , λ j p ε δ a i , λ i = α i p ε Ω δ a i , λ i p + 1 ε + O Ω δ a i , λ i p θ a i , λ i + j i Ω δ a i , λ i p δ a j , λ j + Ω δ a j , λ j p δ a i , λ i = α i p ε c 0 ε λ i ε ( n 2 ) / 2 S n + O ε + 1 λ i n + γ ( ε ) 2 ,
where we have used (64) of [19]. Concerning the second integral in (27), it is computed in Lemma A6.
Combining the previous estimates, the proof follows. □
Next, we present a balancing formula that relates to the rate of concentration. Specifically, we prove
Proposition 4.
Let n 4 , ( α , a , λ ) O ( N , μ , d 0 ) , v E a , λ and u = j = 1 N α j π δ a j , λ j + v . For small ε and i 1 , . . . , N , we have
I ε ( u ) , λ i ( π δ a i , λ i ) λ i = c 0 ε λ i ε n 2 2 c 2 α i p ε ε c ( n ) α i ln σ n ( λ i ) λ i 2 V ( a i ) 2 α i p ε 1 c 0 ε λ i ε n 2 2 1 + O ( R 1 ( ε ) ) ,
where c ( n ) is defined in Lemma A3 and
R 1 ( ε ) : = ε / | ln ε | ( i f   n = 4 ) ; ε 3 / 2 ( i f   n = 5 ) ; ε 2 | ln ε | ( i f   n = 6 ) ; ε 2 ( i f   n 7 ) , c 2 : = n 2 2 2 c 0 2 n / ( n 2 ) R n ( x 2 1 ) ( 1 + x 2 ) n + 1 ln ( 1 + x 2 ) d x > 0 .
Proof. 
We will follow the proof of Proposition 3, and we need to be more precise in some integrals. Since v E a , λ , we have
u , λ i ( π δ a i , λ i ) λ i = α j π δ a j , λ j , λ i ( π δ a i , λ i ) λ i .
First, for j i , using Remark A1 and (23), we have
π δ a j , λ j , λ i ( π δ a i , λ i ) λ i = Ω δ a j , λ j n + 2 n 2 λ i ( π δ a i , λ i ) λ i = O Ω δ a j , λ j n + 2 n 2 δ a i , λ i = O ( γ ( ε ) 2 ) .
Second, for j = i , using Lemma A3, we obtain
π δ a i , λ i , λ i ( π δ a i , λ i ) λ i = Ω δ a i , λ i n + 2 n 2 λ i δ a i , λ i λ i λ i θ a i , λ i λ i = c ( n ) ln σ n λ i λ i 2 V ( a i ) + O γ ( ε ) 2 + ε 2 + ( if   n = 6 ) ε 2 | ln ε | .
We derive that
u , λ i ( π δ a i , λ i ) λ i = α i c ( n ) ln σ n λ i λ i 2 V ( a i ) + O γ ( ε ) 2 + ε 2 + ( if   n = 6 ) ε 2 | ln ε | .
Now, using Lemma A1, we obtain
Ω | u | p ε 1 u λ i ( π δ a i , λ i ) λ i = Ω ( u ̲ ) p ε λ i ( π δ a i , λ i ) λ i + ( p ε ) Ω ( u ̲ ) p ε 1 v λ i ( π δ a i , λ i ) λ i + O ( v 2 ) .
The last integral is computed in Lemma A6. For the first one, using Lemma A1 and Remark A1, we obtain
Ω ( u ̲ ) p ε λ i ( π δ a i , λ i ) λ i = α i p ε Ω ( π δ a i , λ i ) p ε λ i ( π δ a i , λ i ) λ i + j i O Ω δ a i , λ i p δ a j , λ j + Ω δ a i , λ i δ a j , λ j p .
First, we remark that the remainder term is computed in (23). Second, we focus on estimating the first integral. Using (12) and Lemma A1, we obtain
Ω ( π δ a i , λ i ) p ε λ i ( π δ a i , λ i ) λ i = Ω δ a i , λ i p ε ( p ε ) δ a i , λ i p ε 1 θ a i , λ i + O ( δ a i , λ i p 2 ε θ a i , λ i 2 ) λ i δ a i , λ i λ i λ i θ a i , λ i λ i = Ω δ a i , λ i p ε λ i δ a i , λ i λ i Ω δ a i , λ i p ε λ i θ a i , λ i λ i ( p ε ) Ω δ a i , λ i p ε 1 λ i δ a i , λ i λ i θ a i , λ i + O Ω δ a i , λ i p 1 θ a i , λ i 2 + Ω δ a i , λ i p 1 θ a i , λ i λ i θ a i , λ i λ i .
To estimate the last term in (33), using Propositions A1 and A2, we obtain
Ω δ a i , λ i p 1 θ a i , λ i 2 + Ω δ a i , λ i p 1 θ a i , λ i λ i θ a i , λ i λ i c Ω R 1 2 ( x , a i , λ i ) δ a i , λ i p + 1 c 1 λ i 4 ( if   n 5 ) , ln 3 λ i λ i 4 ( if   n = 4 ) .
Concerning the other integrals of (33), using (12) and Propositions A1 and A2, we obtain
Ω δ a i , λ i p ε λ i θ a i , λ i λ i = c 0 ε λ i ε n 2 2 Ω δ a i , λ i p λ i θ a i , λ i λ i + O ε R 1 ( x , a i , λ i ) δ a i , λ i p + 1 ln ( 1 + λ i 2 | x a i | 2 ) , Ω δ a i , λ i p ε 1 λ i δ a i , λ i λ i θ a i , λ i = c 0 ε λ i ε n 2 2 Ω δ a i , λ i p 1 λ i δ a i , λ i λ i θ a i , λ i + O ε R 1 ( x , a i , λ i ) δ a i , λ i p + 1 ln ( 1 + λ i 2 | x a i | 2 ) .
Thus, using Lemma A3, we obtain
Ω δ a i , λ i p ε λ i θ a i , λ i λ i ( p ε ) Ω δ a i , λ i p ε 1 λ i δ a i , λ i λ i θ a i , λ i = c 0 ε λ i ε n 2 2 2 c ( n ) ln σ n λ i λ i 2 V ( a i ) + O 1 λ i n 2 + 1 λ i 4 + ( if   n = 6 ) ln λ i λ i 4 + ε ln σ n λ i λ i 2 .
It remains to study the first integral of (33). Using (91) of [19], we obtain
Ω δ a i , λ i p ε λ i δ a i , λ i λ i = c 2 ε c 0 ε λ i ε n 2 2 + O ε 2 + ln λ i λ i n .
Combining the previous estimates, (33) becomes
Ω ( π δ a i , λ i ) p ε λ i ( π δ a i , λ i ) λ i = c 0 ε λ i ε n 2 2 c 2 ε + 2 c ( n ) ln σ n λ i λ i 2 V ( a i ) + O ε 2 + 1 λ i n 2 + 1 λ i 4 + ( if   n = 6 ) ln λ i λ i 4 .
This completes the estimate of the first integral of (32).
Finally, combining (30)–(32), (34) and Lemma A6, the result follows. □
We end this section by presenting a balancing condition related to the points of concentration.
Proposition 5.
Let n 4 , ( α , a , λ ) O ( N , μ , d 0 ) , v E a , λ and u = j = 1 N α j ( π δ a j , λ j ) + v . For small ε and i 1 , . . . , N , we have
I ε ( u ) , 1 λ i ( π δ a i , λ i ) a i = c 2 ( n ) α i ln σ n ( λ i ) λ i 3 V ( a i ) 2 α i p ε 1 c 0 ε λ i ε n 2 2 1 + O ( R 2 ( ε ) ) ,
where c 2 ( n ) is defined in Lemma A4 and
R 2 ( ε ) : = [ ε / | ln ε | ] 3 / 2 i f   n = 4 ; ε 2 | ln ε | i f   n = 5 ; ε 2 i f   n 6 .
Proof. 
We will follow the proof of Proposition 4. Since v E a , λ , we have
u , 1 λ i ( π δ a i , λ i ) a i = α j π δ a j , λ j , 1 λ i ( π δ a i , λ i ) a i .
First, for j i , using Remark A1 and (7), we obtain
π δ a j , λ j , 1 λ i ( π δ a i , λ i ) a i = Ω δ a j , λ j n + 2 n 2 1 λ i δ a i , λ i a i 1 λ i θ a i , λ i a i = O 1 λ i n / 2 B j δ a j , λ j n + 2 n 2 + 1 λ j ( n + 2 ) / 2 Ω B j δ a i , λ i = O 1 λ i n / 2 λ j ( n 2 ) / 2 + 1 λ j ( n + 2 ) / 2 λ i ( n 2 ) / 2 = O γ ( ε ) 2 n 2 n 2 = O [ ε / | ln ε | ] 3 / 2 ( if   n = 4 ) ; ε ( n 1 ) / 2 ( if   n 5 ) .
Second, for j = i , using Lemma A4, we obtain
π δ a i , λ i , 1 λ i ( π δ a i , λ i ) a i = Ω δ a i , λ i n + 2 n 2 1 λ i δ a i , λ i a i 1 λ i θ a i , λ i a i = c 2 ( n ) ln σ n λ i λ i 3 V ( a i ) + O 1 λ i n 1 + 1 λ i 4 + ( if   n = 5 ) ln λ i λ i 4 .
We derive that
u , 1 λ i ( π δ a i , λ i ) a i = α i c 2 ( n ) ln σ n λ i λ i 3 V ( a i ) + O [ ε / | ln ε | ] 3 / 2 ( if   n = 4 ) ; ε 2 | ln ε | ( if   n = 5 ) ; ε 2 ( if   n 6 ) .
Now, using Lemma A1, we obtain
Ω | u | p ε 1 u 1 λ i ( π δ a i , λ i ) a i = Ω ( u ̲ ) p ε 1 λ i ( π δ a i , λ i ) a i + ( p ε ) Ω ( u ̲ ) p ε 1 v 1 λ i ( π δ a i , λ i ) a i + O ( v 2 ) .
The last integral is computed in Lemma A6. For the first one, using Lemma A1, we obtain
Ω ( u ̲ ) p ε 1 λ i ( π δ a i , λ i ) a i = j = 1 N α j p ε Ω ( π δ a j , λ j ) p ε 1 λ i ( π δ a i , λ i ) a i + ( p ε ) Ω ( α i π δ a i , λ i ) p ε 1 j i α j π δ a j , λ j 1 λ i ( π δ a i , λ i ) a i + j k O Ω δ a j , λ j δ a k , λ k ( p + 1 ε ) / 2 .
To estimate the last integral in (38), using (12) and (7), we obtain
Ω δ a j , λ j δ a k , λ k ( p + 1 ε ) / 2 Ω δ a j , λ j δ a k , λ k n / ( n 2 ) c λ k n / 2 B j δ a j , λ j n / ( n 2 ) + c λ j n / 2 Ω B j δ a k , λ k n / ( n 2 ) c ln λ k + ln λ j ( λ k λ j ) n / 2 c ε 2 | ln ε | 1 ( if   n = 4 ) ; | ln ε | ε n / 2 ( if   n 5 ) .
For the first integral in (38) with j i , following the same computations as in (35), we obtain
Ω ( π δ a j , λ j ) p ε 1 λ i ( π δ a i , λ i ) a i = O γ ( ε ) ( 2 n 2 ) / ( n 2 ) .
Concerning the first integral in (38) with j = i , using Lemma A1 and (12), we obtain
Ω ( π δ a i , λ i ) p ε 1 λ i ( π δ a i , λ i ) a i = Ω δ a i , λ i p ε ( p ε ) δ a i , λ i p ε 1 θ a i , λ i + O ( δ a i , λ i p 2 ε θ a i , λ i 2 ) 1 λ i δ a i , λ i a i 1 λ i θ a i , λ i a i = Ω δ a i , λ i p ε 1 λ i δ a i , λ i a i Ω δ a i , λ i p ε 1 λ i θ a i , λ i a i ( p ε ) Ω δ a i , λ i p ε 1 1 λ i δ a i , λ i a i θ a i , λ i + O Ω δ a i , λ i p 1 θ a i , λ i 2 + Ω δ a i , λ i p 1 θ a i , λ i 1 λ i θ a i , λ i a i .
For the last two terms in (41), using Propositions A1 and A3, we obtain
Ω δ a i , λ i p 1 θ a i , λ i 2 + Ω δ a i , λ i p 1 θ a i , λ i 1 λ i θ a i , λ i a i c Ω R 1 2 ( x , a i , λ i ) + 1 λ i 4 + | x a i | 2 λ i 2 δ a i , λ i p + 1 c 1 λ i 4 ( i f n 5 ) , ln 3 λ i λ i 4 ( if   n = 4 ) .
Concerning the other integrals of (41), using (12) and Propositions A1 and A3, by oddness, we obtain
Ω δ a i , λ i p ε 1 λ i δ a i , λ i a i = B ( a i , d ) + Ω B ( a i , d ) = O 1 λ i n + 1 , Ω δ a i , λ i p ε 1 λ i θ a i , λ i a i = c 0 ε λ i ε n 2 2 Ω δ a i , λ i p 1 λ i θ a i , λ i a i + O ε R 2 ( x , a i , λ i ) δ a i , λ i p + 1 ln ( 1 + λ i 2 | x a i | 2 ) , Ω δ a i , λ i p ε 1 1 λ i δ a i , λ i a i θ a i , λ i = c 0 ε λ i ε n 2 2 Ω δ a i , λ i p 1 λ i δ a i , λ i λ i θ a i , λ i + O ε R 1 ( x , a i , λ i ) δ a i , λ i p + 1 ln ( 1 + λ i 2 | x a i | 2 ) .
Using Lemma A4, we obtain
Ω δ a i , λ i p ε 1 λ i θ a i , λ i a i ( p ε ) Ω δ a i , λ i p ε 1 1 λ i δ a i , λ i a i θ a i , λ i = 2 c 2 ( n ) c 0 ε λ i ε n 2 2 ln σ n λ i λ i 3 V ( a i ) + O 1 λ i n 1 + 1 λ i 4 + ( if   n = 5 ) ln λ i λ i 4 + ε ln σ n λ i λ i 2 .
Combining the previous estimates, (41) becomes
Ω ( π δ a i , λ i ) p ε 1 λ i ( π δ a i , λ i ) a i = 2 c 2 ( n ) c 0 ε λ i ε n 2 2 ln σ n λ i λ i 3 V ( a i ) + O 1 λ i n 1 + 1 λ i 4 + ( if   n = 5 ) ln λ i λ i 4 + ε ln σ n λ i λ i 2 .
It remains to study the second integral in the right-hand side of (38). Using Propositions A1 and A3, we obtain
Ω ( π δ a i , λ i ) p ε 1 1 λ i ( π δ a i , λ i ) a i ( π δ a j , λ j ) = B i δ a i , λ i p ε 1 + O δ a i , λ i p 2 θ a i λ i 1 λ i δ a i , λ i a i 1 λ i θ a i , λ i a i ( π δ a j , λ j ) + O Ω B i δ a i , λ i p δ a j , λ j = B i δ a i , λ i p ε 1 1 λ i δ a i , λ i a i ( π δ a j , λ j ) + O 1 λ j ( n 2 ) / 2 B i θ a i λ i δ a i , λ i p 1 + 1 λ i ( n + 2 ) / 2 Ω B i δ a j , λ j .
Observe that, using Proposition A1, we have
B i θ a i λ i δ a i , λ i p 1 c B i R 1 ( x , a i , λ i ) δ a i , λ i p c ln σ n λ i λ i 2 B i δ a i , λ i p + B i | x a i | 2 | ln | x a i | | σ n δ a i , λ i p c ln σ n + 1 λ i λ i ( n + 2 ) / 2
and, expanding π δ a j , λ j around a i , by oddness, we obtain
B i δ a i , λ i p ε 1 1 λ i δ a i , λ i a i ( π δ a j , λ j ) = O 1 λ j ( n 2 ) / 2 B i | x a i | δ a i , λ i p = O 1 λ j ( n 2 ) / 2 λ i n / 2 .
We derive that
Ω ( π δ a i , λ i ) p ε 1 1 λ i ( π δ a i , λ i ) a i ( π δ a j , λ j ) = O 1 ( λ j ( n 2 ) / 2 λ i n / 2 = O [ ε / | ln ε | ] 3 / 2 ( if   n = 4 ) ; ε ( n 1 ) / 2 ( if   n 5 ) .
Finally, combining (36)–(43) and Lemma A6, the result follows. □

3. Construction of Interior Peak Solutions

Let b 1 , b N be non-degenerate distinct critical points of V. As in [17] (see also [18,19]), the strategy of the proof of Theorem 1 is as follows. Let
O 1 ( N , ε ) = { ( α , λ , a , v ) ( R + ) N × ( R + ) N × Ω N × H 1 ( Ω ) : | α i 1 | < ε ln 2 ε , | a i b i | < ε 1 / 5 , M 1 | ln ε | σ n / ε λ i M | ln ε | σ n / ε , 1 i N , v E a , λ and v < ε }
where M is a positive constant and E a , λ is defined by (5).
Note that, for each ( α , λ , a , v ) O 1 ( N , ε ) , it follows that ( α , λ , a ) O ( N , μ , d 0 ) by taking μ > ε ln 2 ε and d 0 to satisfy d 0 min { | b i b j | / 4 : j i } and d 0 min { d ( b i , Ω ) / 4 : 1 i N } .
In addition, we consider the following function
ψ ε : O 1 ( N , ε ) R , ( α , λ , a , v ) ψ ε ( α , λ , a , v ) = I ε i = 1 N α i π δ a i , λ i + v .
We observe that ( α , λ , a , v ) is a critical point of ψ ε if and only if u = i = 1 N α i π δ i + v is a critical point of I ε . Therefore, our goal is to find the critical points of ψ ε . Since the variable v belongs to E a , λ , the Lagrange multiplier theorem enables us to derive the following proposition.
Proposition 6.
( α , λ , a , v ) O 1 ( N , ε ) is a critical point of ψ ε if and only if there exists ( A , B , C ) R N × R N × ( R n ) N such that the following holds:
ψ ε α i ( α , λ , a , v ) = 0 i 1 , . . . , N ,
ψ ε λ i ( α , λ , a , v ) = B i v , λ i 2 ( π δ a i , λ i ) λ i 2 + j = 1 n C i j v , 1 λ i 2 ( π δ a i , λ i ) λ i a i j i 1 , . . . , N ,
ψ ε a i ( α , λ , a , v ) = B i v , λ i 2 ( π δ a i , λ i ) λ i a i + j = 1 n C i j v , 1 λ i 2 ( π δ a i , λ i ) a i j a i i 1 , . . . , N ,
ψ ε v ( α , λ , a , v ) = k = 1 N A k π δ a k , λ k + B k λ k ( π δ a k , λ k ) λ k + j = 1 n C k j 1 λ k ( π δ a k , λ k ) a k j .
To prove Theorem 1, we will conduct a detailed analysis of the preceding equations. Observe that
ψ ε α i ( α , λ , a , v ) = I ε i = 1 N α i π δ a i , λ i + v , π δ a i , λ i ,
ψ ε λ i ( α , λ , a , v ) = I ε i = 1 N α i π δ a i , λ i + v , α i ( π δ a i , λ i ) λ i ,
ψ ε a i ( α , λ , a , v ) = I ε i = 1 N α i π δ a i , λ i + v , α i ( π δ a i , λ i ) a i ,
ψ ε v ( α , λ , a , v ) = I ε i = 1 N α i π δ a i , λ i + v .
Let ( α , λ , a , 0 ) O 1 ( N , ε ) , which is defined by (44). Clearly, v ¯ , as obtained in Proposition 2, satisfies Equation (48). Hence, we need to solve the system defined by Equations (45)–(47). By combining Equations (45)–(52), we observe that u = i = 1 N α i π δ a i , λ i + v ¯ is a critical point of I ε if and only if ( α , λ , a ) solves the following system: for each 1 i N
E α i I ε ( u ) , π δ a i , λ i = 0 ,
E λ i I ε ( u ) , α i ( π δ a i , λ i ) λ i = B i v ¯ , λ i 2 ( π δ a i , λ i ) λ i 2 + j = 1 n C i j v ¯ , 1 λ i 2 ( π δ a i , λ i ) λ i a i j ,
E a i I ε ( u ) , α i ( π δ a i , λ i ) a i = B i v ¯ , λ i 2 ( π δ a i , λ i ) λ i a i + j = 1 n C i j v ¯ , 1 λ i 2 ( π δ a i , λ i ) a i j a i .
To analyze the system ( E α i ) , ( E λ i ) , ( E a i ) , we need to estimate the constants A i s, B i s and C i j s that appear in the equations ( E λ i ) and ( E a i ) . This is the aim of the following lemma
Lemma 2.
Let ( α , λ , a , 0 ) O 1 ( N , ε ) . Then, for ε small, the following statements hold:
A i = O ( ε ln 2 ε ) , B i = O ( ε ) a n d C i j = O ( ε 3 / 2 | ln ε | σ n / 2 ) 1 i N , 1 j n .
Proof. 
Taking the scalar product of ψ ε v (see (48)) with the functions π δ a i , λ i , λ i ( π δ a i , λ i ) / λ i and λ i 1 ( π δ a i , λ i ) / a i j , respectively, we obtain the following quasi-diagonal system:
c A i + O ε k = 1 N ( | A k | + | B k | + j = 1 n | C k j | ) = ψ ε v , π δ a i , λ i i { 1 , , N } , c B i + O ε k = 1 N ( | A k | + | B k | + j = 1 n | C k j | ) = ψ ε v , λ i ( π δ a i , λ i ) λ i i { 1 , , N } c C i l + O ε k = 1 N ( | A k | + | B k | + j = 1 n | C k j | ) = ψ ε v , 1 λ i ( π δ a i , λ i ) a i l 1 i N , 1 n
where c, c and c are some positive universal constants.
Combining Propositions 3, 4 and 5 and the fact that ( α , λ , a , 0 ) O 1 ( N , ε ) , we see that, for all i { 1 , , N } , we have
ψ ε v , π δ a i , λ i = O ( ε ln 2 ε ) , ψ ε v , λ i π δ a i , λ i λ i = O ( ε ) , ψ ε v , 1 λ i π δ a i , λ i a i = O ( ε 3 2 | ln ε | σ n 2 ) .
Thus Lemma 2 follows. □
Next, we will analyze the equations ( E α i ) , ( E λ i ) and ( E a i ) . To simplify the system for solving, we introduce the following change of variables
β i = 1 α i p 1 , ln σ n λ i λ i 2 = c 2 c ( n ) 1 V ( b i ) ε ( 1 + i ) , z i = a i b i 1 i N ,
where c 2 and c ( n ) are defined in Proposition 4.
This change of variables enables us to express the system in the following simplified form:
Lemma 3.
For ε small, equations ( E α i ) , ( E λ i ) and ( E a i ) are equivalent to the following system:
( S ) β i = O ( ε | ln ε | ) 1 i N , i = O ( | z i | 2 + R 1 ( ε ) ε 1 ) 1 i N , D 2 V ( b i ) ( z i , . ) = O R 2 ( ε ) ε 3 / 2 | ln ε | 1 / 2 + | z i | 2 1 i N ,
where R 1 ( ε ) and R 2 ( ε ) are defined in Propositions 4 and 5.
Proof. 
Using Proposition 3, it follows that
1 c 0 ε λ i ε ( n 2 ) / 2 α i p 1 ε = O ( ε ) .
Now, observe that
α i p 1 ε = α i p 1 + O ( ε ) , λ i ε ( n 2 ) / 2 = 1 + O ( ε ln λ i ) = 1 + O ( ε | ln ε | ) ,
Thus, ( E α i ) is equivalent to
( E α i ) β i = O ( ε | ln ε | ) 1 i N .
For the second equation ( E λ i ) , Proposition 4 and Lemma 2 imply that
( 1 + O ( ε ) ) c 2 ε c ( n ) ( 1 + O ( ε ) ) V ( a i ) ln σ n λ i λ i 2 = O ( R 1 ( ε ) ) ,
where R 1 ( ε ) is defined in Proposition 4. Writing
V ( a i ) = V ( b i ) + O ( | z i | 2 ) ,
we obtain the second equation in the system ( S ) .
Lastly, writing
V ( a i ) = D 2 V ( b i ) ( z i , . ) + O ( | z i | 2 )
and using Proposition 5 and Lemma 2, we see that Equation ( E a i ) is equivalent to the third equation in the system ( S ) which completes the proof of Lemma 3. □
We are now prepared to prove our results regarding the existence of interior peak solutions. Since Theorem 2 follows directly from Theorem 1, it suffices to prove the latter.
Proof of Theorem 1.
The system ( S ) , as presented in Lemma 3, can be expressed in the following form:
( S ) β i = U 1 , i ( ε , β , , z ) = O ( ε | ln ε | ) 1 i N i = U 2 , i ( ε , β , , z ) = O ( ε | ln ε | + | z i | 2 + R 1 ( ε ) ε 1 ) 1 i N D 2 V ( b i ) ( z i , . ) = U 3 , i ( ε , β , , z ) = O R 2 ( ε ) ε 3 / 2 | ln ε | 1 / 2 + | z i | 2 1 i N
where β = ( β 1 , , β N ) , = ( 1 , , N ) and z = ( z 1 , , z N ) .
Thus, using Brouwer’s fixed point theorem, we derive that the system ( S ) has at least one solution ( β ε , ε , z ε ) for small ε . To complete the proof of the theorem, it remains to prove that the constructed function u ε = i = 1 N α i , ε π δ a i , ε , λ i , ε + v ¯ ε is positive. To this aim, we first remark, since v 0 , that u ε 0 for small ε . By construction, u ε satisfies
( Q V , ε ) : Δ u ε + V u ε = | u ε | p 1 ε u ε in Ω , u ε ν = 0 on Ω .
Multiplying ( Q V , ε ) by u ε = max ( 0 , u ε ) and integrating on Ω , we obtain
Ω | u ε | 2 + Ω V ( u ε ) 2 = Ω ( u ε ) p + 1 ε .
But we have
Ω ( u ε ) p + 1 ε c Ω ( u ε ) p + 1 p + 1 ε p + 1 C Ω | u ε | 2 + Ω V ( u ε ) 2 p + 1 ε 2
which implies that
either u ε 0 or Ω ( u ε ) p + 1 C 2 ( n 2 ) / ( 4 ε ( n 2 ) ) .
Since u ε | v ε | and v ε L p + 1 0 as ε 0 , we derive that u ε 0 and u ε 0 . Thus, using the maximum principle, u ε has to be positive. This completes the proof of the theorem. □

4. Conclusions

In this paper, we have explored the existence of solutions to a nonlinear elliptic problem with Dirichlet boundary conditions, featuring a slightly subcritical exponent for Sobolev embedding H 0 1 ( Ω ) L q ( Ω ) . By performing a detailed asymptotic analysis of the gradient of the corresponding Euler–Lagrange functional near the so-called “bubbles”, we were able to construct solutions that concentrate at isolated interior points. These techniques enabled us to establish a multiplicity result for the problem. The approach employed is specific to variational problems. While this paper addresses the existence of solutions and multiplicity results for the given problem, several natural directions for future research and open questions remain:
(i) 
Impact of the Nature of Critical Points of the Potential: The solutions we have constructed in this paper rely on the assumption that the critical points of the potential V are non-degenerate. What happens if this hypothesis is not satisfied, particularly when V satisfies some flatness condition?
(ii) 
Location of the Concentration Points: This paper focuses on constructing solutions that concentrate at isolated interior points. An interesting extension would be to study the existence of solutions that concentrate at non-isolated points, boundary points, or interior points that converge to the boundary.
(iii) 
Asymptotic Behavior of Solutions: Another important question is to fully characterize the asymptotic behavior of the solutions, providing a comprehensive understanding of their long-term behavior.
(iv) 
Impact of Subcritical Exponent: The current work focuses on a slightly subcritical exponent for Sobolev embedding. Future research could explore the problem with exponents that are slightly supercritical, i.e., when ε < 0 but close to zero.

Author Contributions

H.A. (Hissah Alharbi), H.A. (Hibah Alkhuzayyim), M.B.A. and K.E.M.: conceptualization, methodology, investigation, writing original draft preparation, writing—review and editing. All authors contributed equally to the writing of this paper. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability Statement

Data are contained within the article.

Conflicts of Interest

The authors declare no conflicts of interest.

Appendix A

In this appendix, we gather several estimates required throughout the paper. We begin with the following auxiliary analytical formulas, whose proofs follow from a Taylor expansion with the Lagrange remainder.
Lemma A1.
( 1 ) Let s , t R and γ > 2 . It holds that
s + t γ s γ γ s γ 2 s t 1 2 γ ( γ 1 ) s γ 2 t 2 c s γ 3 t 3 + c | t | γ i f γ > 3 , c t γ i f γ 3 .
( 2 ) Let β i > 0 and γ > 1 . It holds that
i = 1 N β i γ i = 1 N β i γ c i j β i β j γ / 2 i f γ 2 , c i j β i γ 1 β j i f γ > 2 .
( 3 ) Let x > 0 and y > 0 be such that y < x and γ > 1 . Then, it holds that
x y γ = x γ + O x γ 1 y a n d x y γ = x γ γ x γ 1 y + O x γ 2 y 2 .
( 4 ) Let t R , s R , z be such that | z | β | t | for some positive constant β and γ > 0 . It holds that
t + s γ 1 t + s z = | t | γ 1 t z + γ | t | γ 1 s z + O | t | γ 1 s 2 + s γ + 1 , | | t + s | γ z | t | γ z | c | t s | ( γ + 1 ) / 2 i f   γ 1 , | t | γ | s | + | s | γ | t | i f   γ > 1 .
( 5 ) Let γ > 0 and x , y R , then it holds that
x + y γ = x γ + O | x | γ 1 | y | + | y | γ .
( 6 ) Let 1 < γ 3 , t j > 0 for j { 1 , , N } , and let s be such that | s | c t i for some index i. Then, it holds that
j = 1 N t j γ s = j = 1 N t j γ s + γ t i γ 1 j i t j s + k j O ( t k t j ) ( γ + 1 ) / 2 .
Next, we provide crucial estimates of the function π δ a , λ and θ a , λ defined by (2) and (3), respectively. In particular, we show that the projection π δ a , λ is close to the initial function δ a , λ . We start by the following proposition, which is very useful throughout the paper.
Proposition A1.
Let d 0 > 0 , a Ω be such that d a : = d ( a , Ω ) 2 d 0 , and let λ be a positive large real. Then, the following statements hold
(i) 
0 θ a , λ δ a , λ ; 0 π δ a , λ δ a , λ i n Ω .
(ii) 
θ a , λ ( x ) c R 1 ( x , a , λ ) δ a , λ ( x ) x Ω 0 : = { x Ω : d ( x , Ω ) d 0 } , where
R 1 ( x , a , λ ) : = λ 2 + | x a | 2 i f n 5 , ( ln λ ) λ 2 + | x a | 2 ln | x a | i f n = 4 , λ 1 + | x a | i f n = 3 .
Proof. 
We start by proving Assertion ( i ) . We have
( Δ + V ) θ a , λ = V δ a , λ , in Ω , θ a , λ = δ a , λ on Ω . and Δ θ a , λ = V π δ a , λ , in Ω , θ a , λ = δ a , λ on Ω .
Using (A1) and Corollary 3.2 of [21], we derive that
inf Ω θ a , λ inf Ω θ a , λ = 0 w h e r e θ a , λ : = max ( 0 , θ a , λ )
which implies the first and the last inequalities of ( i ) .
In the same way, using (2) and Corollary 3.2 of [21], we derive that
inf Ω ( π δ a , λ ) inf Ω ( π δ a , λ ) = 0 .
The proof of Assertion ( i ) is thereby completed.
Now, we focus on proving Assertion ( i i ) . Using (A1), we have
c n θ a , λ ( x ) = Ω G ( x , y ) V ( y ) π δ a , λ ( y ) d y Ω G ( x , y ) ν δ a , λ ( y ) d y ,
where G denotes the Green’s function for the Laplace operator with Dirichlet boundary conditions.
For the first part in the right-hand side of (A2), using ( i ) and Lemma A2, we have
0 Ω G ( x , y ) V ( y ) π δ a , λ ( y ) d y c Ω 1 | x y | n 2 δ a , λ ( y ) d y c R 1 ( x , a , λ ) δ a , λ ( x ) .
For the second part, we have | x y | c > 0 for each x Ω 0 and for each y Ω . Hence, we obtain
G ( x , y ) ν c x Ω 0 and y Ω .
Furthermore, since d a 2 d 0 , we have
δ a , λ ( y ) c λ ( n 2 ) / 2 y Ω ,
thus, we derive that
Ω G ( x , y ) ν δ a , λ ( y ) d y c λ ( n 2 ) / 2 .
Now, observe that
1 λ ( n 2 ) / 2 c 1 λ n 2 + | x a | n 2 δ a , λ ( x ) x Ω .
Combining (A2)–(A5), the proof of Assertion ( i i ) follows. □
The next result deals with the estimates of the derivatives of δ a , λ and π a , λ with respect to λ .
Proposition A2.
Let d 0 > 0 , a Ω be such that d a : = d ( a , Ω ) 2 d 0 , and let λ be a large positive real. Then, the following statements hold:
(i) 
λ ( π δ a , λ ) λ n + 2 2 π δ a , λ i n Ω ,
(ii) 
λ θ a , λ λ n 2 2 θ a , λ i n Ω ,
(iii) 
λ θ a , λ ( x ) λ c R 1 ( x , a , λ ) δ a , λ ( x ) x Ω 0 : = { x Ω : d ( x , Ω ) d 0 } ,
where R 1 is defined in Proposition A1.
Proof. 
We start by proving Assertion ( i ) . From (2), we deduce that
( Δ + V ) λ ( π δ a , λ ) λ = n + 2 n 2 δ a , λ 4 n 2 λ δ a , λ λ in Ω , λ ( π δ a , λ ) λ = 0 on Ω .
Now, let
ψ + = n + 2 2 π δ a , λ + λ π δ a , λ λ ; ψ = n + 2 2 π δ a , λ λ π δ a , λ λ .
It follows that, ψ + and ψ are in H 0 1 ( Ω ) and
( Δ + V ) ψ + = n + 2 2 δ a , λ n + 2 n 2 + n + 2 n 2 δ a , λ 4 n 2 λ δ a , λ λ = n + 2 2 δ a , λ n + 2 n 2 2 1 + λ 2 | x a | 2 > 0 in Ω , ( Δ + V ) ψ = n + 2 2 δ a , λ n + 2 n 2 2 λ 2 | x a | 2 1 + λ 2 | x a | 2 > 0 in Ω .
Applying Corollary 3.2 of [21], we derive that
inf Ω ψ + inf Ω ψ + = 0 and inf Ω ψ inf Ω ψ = 0 ,
which completes the proof of ( i ) .
Concerning Assertion ( i i ) , from (A1), it follows that
( Δ + V ) λ θ a , λ λ = V λ δ a , λ λ in Ω , λ θ a , λ λ = λ δ a , λ λ on Ω .
Let
φ + : = n 2 2 θ a , λ + λ θ a , λ λ ; φ : = n 2 2 θ a , λ λ θ a , λ λ .
It follows that
( Δ + V ) φ + ( x ) = n 2 2 V ( x ) δ a , λ ( x ) + V ( x ) λ δ a , λ ( x ) λ = n 2 2 V ( x ) δ a , λ ( x ) 2 1 + λ 2 | x a | 2 > 0 in Ω , φ + ( x ) = n 2 2 δ a , λ ( x ) + λ δ a , λ ( x ) λ = δ a , λ ( x ) n 2 1 + λ 2 | x a | 2 > 0 x Ω .
Applying Corollary 3.2 of [21], we deduce that
inf Ω φ + inf Ω φ + = 0 .
In the same way, we derive that
( Δ + V ) φ ( x ) = n 2 2 V ( x ) δ a , λ ( x ) 2 λ 2 | x a | 2 1 + λ 2 | x a | 2 > 0 x Ω , φ ( x ) = n 2 2 δ a , λ ( x ) 2 λ 2 | x a | 2 1 + λ 2 | x a | 2 > 0 , x Ω .
Thus,
φ 0 in Ω .
(A8) and (A9) give the proof of ( i i ) .
Concerning Assertion ( i i i ) , it follows from ( i ) and ( i i ) of Proposition A1.
This completes the proof of the proposition. □
Next, we deal with the estimates of the derivatives of δ a , λ and π a , λ with respect to a.
Proposition A3.
Let d 0 > 0 , a Ω be such that d a : = d ( a , Ω ) 2 d 0 , and let λ be a large positive real. Then, for j { 1 , . . . , n } , we have
(i) 
1 λ π δ a , λ a j n + 2 2 π δ a , λ i n Ω ,
(ii) 
1 λ θ a , λ a j n 2 2 θ a , λ i n Ω ,
(iii) 
1 λ θ a , λ ( x ) a j c R 2 ( x , a , λ ) δ a , λ ( x ) x Ω 0 : = { x Ω : d ( x , Ω ) d 0 } ,
w h e r e R 2 ( x , a , λ ) : = ( ln λ ) λ 2 + λ 1 | x a | ln | x a | i f n = 3 , λ 2 + λ 1 | x a | i f n 4 .
Proof. 
The proof is identical to that of Proposition A2 utilizing the following equations:
( Δ + V ) 1 λ ( π δ a , λ ) a j = n + 2 n 2 δ a , λ 4 n 2 1 λ δ a , λ a j in Ω , 1 λ ( π δ a , λ ) a j = 0 on Ω ,
and
( Δ + V ) 1 λ θ a , λ a j = V 1 λ δ a , λ a j in Ω , 1 λ θ a , λ a j = 1 λ δ a , λ a j on Ω .
Now, we give the following important remark.
Remark A1.
As a consequence of Propositions A1, A2 and A3, we remark that
1 λ θ a , λ a c λ δ a , λ i n   Ω 0 ; 0 < π δ a , λ < δ a , λ ; λ ( π δ a , λ ) λ c δ a , λ ; 1 λ ( π δ a , λ ) a c δ a , λ i n   Ω .
Next, we give some useful integral estimates involving the functions δ a , λ and their projections.
Lemma A2
([22]). Let a Ω be such that d a : = d ( a , Ω ) c > 0 . For each x Ω , it holds that
(a) 
Ω 1 | x y | n 2 δ a , λ ( y ) d y c λ 2 + | x a | 2 δ a , λ ( x ) i f n 5 , ( ln λ ) λ 2 + | x a | 2 ln | x a | δ a , λ ( x ) i f n = 4 , λ 1 + | x a | δ a , λ ( x ) i f n = 3 .
(b) 
Ω 1 | x y | n 2 1 λ δ a , λ a i ( y ) d y c ln λ λ 2 + | x a | λ ln | x a | δ a , λ ( x ) i f n = 3 , λ 2 + λ 1 | x a | δ a , λ ( x ) i f n 4 .
Lemma A3.
Let n 4 , λ > 0 be large, and let a Ω be such that d a : = d ( a , Ω ) c > 0 . It holds that
Ω δ a , λ n + 2 n 2 λ δ a , λ λ = O 1 λ n , Ω δ a , λ n + 2 n 2 λ θ a , λ λ = c ( n ) ln σ n ( λ ) λ 2 V ( a ) + O 1 λ n 2 + 1 λ 4 + ( i f n = 6 ) ln λ λ 4 , p Ω δ a , λ p 1 λ δ a , λ λ θ a , λ = Ω δ a , λ p λ θ a , λ λ + O 1 λ n ,
where σ 4 = 1 , σ n = 0 if n 5 and
c ( 4 ) : = c 0 2 m e s ( S 3 ) ; c ( n ) : = n 2 2 c 0 2 R n | x | 2 1 ( 1 + | x | 2 ) n 1 > 0 i f   n 5 .
Proof. 
Observe that, for each b R n and μ > 0 , by using the change of variables y : = μ ( x b ) , it is easy to see that
R n δ b , μ 2 n / ( n 2 ) ( x ) d x = R n δ 0 , 1 2 n / ( n 2 ) ( y ) d y = S n .
We deduce that
R n δ b , μ n + 2 n 2 λ δ a , λ λ = 0
and therefore,
Ω δ a , λ n + 2 n 2 λ δ a , λ λ = R n Ω δ a , λ n + 2 n 2 λ δ a , λ λ c λ n
which gives the proof of the first assertion.
Concerning the second one, using (2), we obtain
Ω δ a , λ n + 2 n 2 λ θ a , λ λ = Ω Δ + V ( π δ a , λ ) λ θ a , λ λ = Ω ( π δ a , λ ) Δ + V λ θ a , λ λ Ω ν ( π δ a , λ ) λ θ a , λ λ .
Thus, (A7) and Propositions A1 and A2 imply that
Ω δ a , λ n + 2 n 2 λ θ a , λ λ = Ω V λ δ a , λ λ δ a , λ θ a , λ + O Ω ν ( π δ a , λ ) δ a , λ .
Observe that, using Proposition A1, we obtain
Ω V λ δ a , λ λ θ a , λ c B ( a , d / 2 ) R 1 ( x , a , λ ) δ a , λ 2 + Ω B ( a , d a / 2 ) δ a , λ 2
and therefore, easy computations imply that
Ω V λ δ a , λ λ θ a , λ c 1 λ n 2 ( 4 n 5 ) ; ln λ λ 4 ( n = 6 ) ; 1 λ 4 ( n 7 ) .
For the last term in (A12), using Green’s representation formula and (2), we have
c π δ a , λ ( x ) = Ω G ( x , y ) δ a , λ n + 2 n 2 ( y ) V ( y ) π δ a , λ ( y ) d y
and therefore, for x Ω such that d ( x , Ω ) is small, it follows that | y x | c for each y B ( a , d a / 2 ) and
π δ a , λ ( x ) c B ( a , d a 2 ) δ a , λ n + 2 n 2 + δ a , λ + c λ ( n 2 ) 2 Ω B ( a , d a 2 ) x G ( x , y ) c λ ( n 2 ) 2 .
Thus, we obtain
Ω ν ( π δ a , λ ) δ a , λ c λ n 2 .
Finally, expanding V around a and using (52), (53) of [23], we obtain
Ω V δ a , λ λ δ a , λ λ = V ( a ) B ( a , d a ) δ a , λ λ δ a , λ λ + O B ( a , d a ) | x a | 2 δ a , λ 2 + Ω B ( a , d a ) δ a , λ 2 = c ( n ) ln σ n ( λ ) λ 2 V ( a ) + O 1 λ n 2 + 1 λ 4 + ( i f n = 6 ) ln λ λ 4 .
The proof of the second assertion is completed.
Now let us prove the third claim. Using (A6) and the first claim of this lemma, we obtain
p Ω δ a , λ p 1 λ δ a , λ λ θ a , λ = Ω Δ + V λ ( π δ a , λ ) λ θ a , λ = Ω Δ + V λ ( π δ a , λ ) λ ( π δ a , λ ) + O 1 λ n = Ω λ ( π δ a , λ ) λ Δ + V ( π δ a , λ ) + O 1 λ n = Ω λ δ a , λ λ λ θ a , λ λ δ a , λ p + O 1 λ n = Ω δ a , λ p λ θ a , λ λ + O 1 λ n .
The proof of the lemma is thereby completed. □
Lemma A4.
Let n 4 , λ > 0 be large and let a Ω be such that d a : = d ( a , Ω ) c > 0 . It holds that
Ω δ a , λ n + 2 n 2 1 λ δ a , λ a = O 1 λ n + 1 , Ω δ a , λ n + 2 n 2 1 λ θ a , λ a = c 2 ( n ) ln σ n ( λ ) λ 3 V ( a ) + O 1 λ n 1 + 1 λ 4 + O ( n = 5 ) ln λ λ 4 , p Ω δ a , λ p 1 1 λ δ a , λ a θ a , λ = Ω δ a , λ p 1 λ θ a , λ a + O 1 λ n + 1 ,
where O ( n = 5 ) appears only when n = 5 , σ 4 = 1 , σ n = 0 if n 5 and
c 2 ( 4 ) : = 1 2 c 0 2 m e a s ( S 3 ) ; c 2 ( n ) : = n 2 n c 0 2 R n | x | 2 ( 1 + | x | 2 ) n 1 i f n 5 .
Proof. 
Let j { 1 , , n } and B : = B ( a , d a / 2 ) Ω . Notice that
δ a , λ a j = ( n 2 ) c 0 λ ( n + 2 ) / 2 ( x a ) j ( 1 + λ 2 | x a | 2 ) n / 2 .
For the first assertion, by oddness, it follows that
Ω δ a , λ n + 2 n 2 1 λ δ a , λ a j = B + Ω B = O 1 λ n + 1 Ω B 1 | x a | 2 n + 1 = O 1 λ n + 1 .
Concerning the second assertion, arguing as in the proof of (A12), using (A7) and (A11), and Propositions A1 and A3, we obtain
Ω δ a , λ n + 2 n 2 1 λ θ a , λ a j = Ω Δ + V ( π δ a , λ ) 1 λ θ a , λ a j = Ω ( π δ a , λ ) Δ + V 1 λ θ a , λ a j Ω ν ( π δ a , λ ) 1 λ θ a , λ a j = Ω ( δ a , λ θ a , λ ) V 1 λ δ a , λ a j + O Ω ν ( π δ a , λ ) 1 λ δ a , λ .
The last term in (A16) is computed in (A15). In addition, observe that | δ a , λ / a | c δ a , λ / | x a | , and therefore, using Proposition A1, we obtain
Ω V 1 λ δ a , λ a j θ a , λ c B ( a , d a / 2 ) R 1 ( x , a , λ ) 1 λ | x a | δ a , λ 2 + Ω B ( a , d a / 2 ) c λ | x a | δ a , λ 2
and therefore, easy computations imply that
Ω V 1 λ δ a , λ a j θ a , λ c 1 λ 3 ( n = 4 ) ; ln λ λ 4 ( n = 5 ) ; 1 λ 4 ( n 6 ) .
Finally, in B : = B ( a , d a / 2 ) Ω , expanding V around a, by oddness, we obtain
Ω V δ a , λ 1 λ δ a , λ a j = k = 1 n V x k ( a ) B ( x a ) k δ a , λ 1 λ δ a , λ a j + O B | x a | 3 λ | x a | δ a , λ 2 + Ω B δ a , λ 2 λ | x a | = ( n 2 ) c 0 2 V x j ( a ) B λ n 1 ( x a ) j 2 ( 1 + λ 2 | x a | 2 ) n 1 + O 1 λ n 1 + 1 λ 4 = c 2 ( n ) ln σ n ( λ ) λ 3 V x j ( a ) + O 1 λ n 1 + 1 λ 4 .
The proof of the second assertion is completed.
Now let us prove the third assertion. Using (A10) and the first assertion, we obtain
p Ω δ a , λ p 1 1 λ δ a , λ a j θ a , λ = p Ω δ a , λ p 1 1 λ δ a , λ a j ( π δ a , λ ) + p Ω δ a , λ p 1 λ δ a , λ a j = Ω Δ + V 1 λ ( π δ a , λ ) a j ( π δ a , λ ) + O 1 λ n + 1 = Ω 1 λ ( π δ a , λ ) a j Δ + V ( π δ a , λ ) + O 1 λ n + 1 = Ω 1 λ δ a , λ a j 1 λ θ a , λ a j δ a , λ p + O 1 λ n + 1 = Ω δ a , λ p 1 λ θ a , λ a j + O 1 λ n + 1 .
This completes the proof of the lemma. □
Lemma A5.
Let a Ω , λ > 0 be such that d a : = d ( a , Ω ) c > 0 . For each v H 0 1 ( Ω ) , it holds that
Ω δ a , λ p 1 θ a , λ | v | c ε v .
Proof. 
Let B : = B ( a , d a / 2 ) . Using Proposition A1, it holds that
Ω δ a , λ p 1 θ a , λ | v | B R 1 ( x , a , λ ) δ a , λ p | v | + Ω B δ a , λ p | v | .
Observe that
Ω B | v | δ a , λ p c λ ( n + 2 ) / 2 Ω | v | c v λ ( n + 2 ) / 2 .
In addition, by the definition of R 1 , for n 5 , it holds that
B R 1 ( x , a , λ ) δ a , λ p | v | c λ 2 B δ a , λ p | v | + c B | x a | 2 δ a , λ p | v | c λ 2 v B δ a , λ p + 1 n + 2 2 n + c v B | x a | 2 δ a , λ p 2 n n + 2 n + 2 2 n c v λ 2 .
The proof of the result is completed for n 5 . In the same way, the proof follows for n = 4 . □
Lemma A6.
Let ( α , a , λ ) O ( N , μ , d 0 ) . For i { 1 , , N } and j { 1 , , n } , let
φ i π δ a i , λ i , λ i ( π δ a i , λ i ) λ i , 1 λ i ( π δ a i , λ i ) ( a i ) j .
Let u ̲ : = i = 1 N α i π δ a i , λ i and v E a , λ ; it holds that
Ω ( u ̲ ) p ε 1 v φ i c ε v .
Proof. 
Observe that, from Propositions A2 and A3, we deduce that φ i c π δ a i , λ i . Hence, using Claim (4) of Lemma A1, it follows that
Ω ( u ̲ ) p ε 1 v φ i = α i p ε 1 Ω π δ a i , λ i p ε 1 v φ i + j i O Ω δ a i , λ i δ a j , λ j p 2 v if n 6 , O Ω δ a i , λ i p 1 δ a j , λ j v + Ω δ a i , λ i δ a j , λ j p 1 v if n 5 .
Note that, for n 5 (which implies that p 1 > 1 ), for each k and , it holds that
Ω δ a k , λ k p 1 δ a , λ v c v Ω δ a k , λ k p 1 δ a , λ 2 n n + 2 n + 2 2 n c v 1 λ k 4 n / ( n + 2 ) B δ a , λ 2 n n + 2 + 1 λ n ( n 2 ) / ( n + 2 ) Ω B δ a k , λ k 8 n n 2 4 n + 2 2 n c v ( λ k λ ) ( n 2 ) / 2
However, for n 6 , using (39), we derive that
Ω δ a i , λ i δ a j , λ j p 2 v c v Ω δ a i , λ i δ a j , λ j n n 2 n + 2 2 n c v ε ( n + 2 ) / 4 | ln ε | ( n + 2 ) / ( 2 n ) .
For the other integral, applying Claim (3) of Lemma A1 and (12), we obtain
Ω π δ a i , λ i p ε 1 v φ i = Ω δ a i , λ i p ε 1 v φ i + O Ω δ a i , λ i p 1 θ a i , λ i v = c 0 ε λ i ε ( n 2 2 ) Ω δ a i , λ i p 1 v φ i + O ε Ω δ a i , λ i p ln ( 1 + λ i 2 | x a i | 2 ) v + Ω δ a i , λ i p 1 θ a i , λ i v .
The last integral is computed in Lemma A5. However, the second one can be deduced by Holder’s inequality, and we obtain O ( ε v ) . It remains to estimate the first one. We distinguish three cases depending on the value of φ i .
  • If φ i = π δ a i , λ i , using Lemma A5 and the fact that v E a , λ , we obtain
    Ω δ a i , λ i p 1 v π δ a i , λ i = Ω δ a i , λ i p v + O Ω δ a i , λ i p 1 θ a i , λ i v = O ε v .
  • If φ i = λ i π δ a i , λ i λ i , using Proposition A2, Lemma A5, and the fact that v E a , λ , we obtain
    Ω δ a i , λ i p 1 v λ i π δ a i , λ i λ i = Ω δ a i , λ i p 1 λ i δ a i , λ i λ i v + O Ω δ a i , λ i p 1 v θ a i , λ i = O ε v .
  • If φ i = 1 λ i ( π δ a i , λ i ) ( a i ) j , as the previous cases, we obtain
    Ω δ a i , λ i p 1 v 1 λ i ( π δ a i , λ i ) ( a i ) j = Ω δ a i , λ i p 1 1 λ i δ a i , λ i ( a i ) j v + O Ω δ a i , λ i p 1 v θ a i , λ i = O ε v .
Hence, we obtain
Ω π δ a i , λ i p ε 1 v φ i c ε v .
The proof of the Lemma is completed. □

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Alharbi, H.; Alkhuzayyim, H.; Ben Ayed, M.; El Mehdi, K. Interior Peak Solutions for a Semilinear Dirichlet Problem. Axioms 2025, 14, 58. https://doi.org/10.3390/axioms14010058

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Alharbi H, Alkhuzayyim H, Ben Ayed M, El Mehdi K. Interior Peak Solutions for a Semilinear Dirichlet Problem. Axioms. 2025; 14(1):58. https://doi.org/10.3390/axioms14010058

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Alharbi, Hissah, Hibah Alkhuzayyim, Mohamed Ben Ayed, and Khalil El Mehdi. 2025. "Interior Peak Solutions for a Semilinear Dirichlet Problem" Axioms 14, no. 1: 58. https://doi.org/10.3390/axioms14010058

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Alharbi, H., Alkhuzayyim, H., Ben Ayed, M., & El Mehdi, K. (2025). Interior Peak Solutions for a Semilinear Dirichlet Problem. Axioms, 14(1), 58. https://doi.org/10.3390/axioms14010058

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