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Blow up solutions for two-dimensional semilinear elliptic problem of Liouville type with nonlinear gradient terms

  • Sami Baraket EMAIL logo , Anis Ben Ghorbal , Fethi Mahmoudi and Foued Mtiri
Published/Copyright: April 15, 2025
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Abstract

This work investigates the existence of singular limit solutions for nonlinear elliptic systems. Our main approach focuses on using the nonlinear domain decomposition method to establish a new Liouville-type result.

MSC 2010: 35J65; 35B25; 46E35; 65M55

1 Introduction and statement of the results

We consider the following problem:

(1) div ( a ( u 1 ) u 1 ) = ρ 2 a ( u 1 ) e γ u 1 + ( 1 γ ) u 2 in Ω div ( a ( u 2 ) u 2 ) = ρ 2 a ( u 2 ) e ξ u 2 + ( 1 ξ ) u 1 in Ω u 1 = u 2 = 0 on Ω ,

where Ω is a regular bounded open domain in R 2 , γ , ξ , and ρ are positive constants. We assume that γ , ξ ( 0 , 1 ) such that γ + ξ > 1 , and the function a is positive and smooth. We take a ( u ) = e λ u for a small parameter λ > 0 satisfying a suitable assumption, which will be fixed later. Then the problem (1) takes the form

(2) Δ u 1 λ u 1 2 = ρ 2 e γ u 1 + ( 1 γ ) u 2 in Ω Δ u 2 λ u 2 2 = ρ 2 e ξ u 2 + ( 1 ξ ) u 1 in Ω u 1 = u 2 = 0 on Ω .

This article aims to establish the existence of solutions ( u 1 , u 2 ) to the previous problem. More precisely, our interest lies in studying the existence of solutions with singular limits since the parameters ρ and λ tend to 0.

In all the following, we can use by deduction the fact that

(3) 1 ξ γ and 1 γ ξ ( 0 , 1 ) .

Using the transformation

ω 1 = ( λ ρ 2 e γ u 1 ) λ and ω 2 = ( λ ρ 2 e ξ u 2 ) λ ,

then ( ω 1 , ω 2 ) satisfies the following problem:

(4) Δ ω 1 = 1 ω 1 λ + 1 λ ω 2 1 γ ξ λ in Ω Δ ω 2 = 2 ω 2 λ + 1 λ ω 1 1 ξ γ λ in Ω ω 1 = ω 2 = ( λ ρ 2 ) λ on Ω ,

where 1 = γ 1 ( λ ρ 2 ) γ 1 ξ and 2 = ξ 1 ( λ ρ 2 ) ξ 1 γ .

Yamabe system has attracted considerable interest in recent years due to its presence in various physical phenomena, such as nonlinear optics. Here, the coupled solution ( ω 1 , ω 2 ) represents the components of the beam within Kerr-like photorefractive media. We have self-focusing in both components of the beam, with the nonlinear coupling constant Li representing their interaction. The case of nonlinear coupling has been studied extensively, motivated by its relevance to applications in nonlinear optics and Bose-Einstein condensation. For further details, consider, for instance, [16].

In [7], Chanillo and Kiessling established a strict isoperimetric inequality and a Pohozaev-Rellich identity for the system

(5) Δ u i = exp j I γ i , j u j , in R 2 ; i I = { 1 , , N } ,

under the finite mass conditions

(6) R 2 e u i d z < , i I ,

where { γ i , j } γ G L N ( R ) is a symmetric matrix such that γ i , j 0 and γ i , i > 0 , satisfying

(7) j I γ i , j = 1 , i I .

They proved that all solutions u i are radially symmetric and decreasing about some point. This system of nonlinear elliptic partial differential equations of Liouville type, referred to as “L-systems,” represents a natural extension of Liouville’s equation [8].

(8) Δ u = e u in R 2 .

Incidentally, equation (8) is the simplest special case of the L-system, which is invariant under translation, rotation and dilation in the Euclidean plane. Moreover, it shares invariance under the Kelvin transform. In [9], Chen and Li proved the following important classification result.

Theorem 1

[9] Let u L loc 1 ( R 2 ) be a weak solution of (8), which satisfies the finite mass condition:

(9) R 2 e u d z < .

Then u is radially symmetric and decreasing about some point in R 2 .

This result plays a crucial role in obtaining the complete solution of (8) under (9) as it simplifies the problem to a straightforward ordinary differential equation (ODE) problem. Thus, Chen and Li concluded

that all solutions of (8), which verify (9), are given by

(10) u ( z ) = 2 log 1 + λ 2 z z 0 2 2 2 λ ,

where λ > 0 and z 0 R 2 . The system of equations (5), under slightly more general conditions, which include (7) as a special case, finds its applications in the physics of charged particle beams [1012].

Furthermore, similar to Liouville’s equation, the system (5) bears significant geometric implications. A solution ( u 1 , , u N ) of (5)–(7) defines a set of N metrics, each conformally equivalent to the Euclidean metric on R 2 .

The system (5)–(7) raises some interesting problems. First, there is always the solution family u i u for all i with u given by (10). In addition, the L-system has also been studied by several authors [1316].

In higher dimensions ( N 3 ) , the analogue of the critical case of the Liouville equation in dimension two is often linked with the Yamabe equation. We consider the corresponding Dirichlet problem on a bounded domain in R 2 ,

(11) Δ u = ρ 2 e u in Ω , u = 0 on Ω .

It is well known that as the parameter ρ tends to 0, nonminimal solutions of (11) exist, and exhibit a singular limit. Let G ( z , z ) the Green’s functions defined on Ω × Ω , be a solution of

(12) Δ G ( z , z ) = 8 π δ z = z in Ω G ( z , z ) = 0 on Ω ,

and let H ( z , z ) = G ( z , z ) + 4 log z z be its regular part. Baraket and Pacard [17] proved the following result.

Theorem 2

[17] Let Ω be a smooth open subset of R 2 , α ( 0 , 1 ) , and S = { z 1 , , z m } be a nonempty set of distinct points of Ω . Suppose that ( z 1 , , z m ) is a nondegenerate critical point of the function

E : ( z 1 , , z m ) C m j H ( z j , z j ) + j l G ( z j , z l ) ,

then there exists ρ 0 > 0 and ( u ρ ) ρ ( 0 , ρ 0 ) a one parameter family of solutions of (11), such that

lim ρ 0 u ρ = i = 1 m G ( , z i ) in C loc 2 , α ( Ω { z 1 , , z m } ) .

Several generalizations are available in [1821].

The blow-up analysis for the equation (11) is well understood, thanks to the results of Suzuki [22] and Nagasaki and Suzuki [23]. Their work enabled the localisation of the blow-up set of singular limit solutions (up to a subsequence) as critical points of a function given by Green’s functions. In several cases, if the solution is blow-up, then it must have a concentration [24]. In contrast, the blow-up analysis for the Liouville system (1) is almost open.

In [25], Abid et al. considered the following problem:

(13) Δ u 1 + λ u 1 2 = ρ 2 e u 1 + γ 1 u 2 in Ω R 2 Δ u 2 + λ u 2 2 = ρ 2 e u 2 + γ 2 u 1 in Ω u 1 = u 2 = 0 on Ω ,

where γ i , i = 1, 2, λ , and ρ are constants. They proved the existence of singular limit solutions ( u 1 , u 2 ) , which blow up in disjoint sets as λ and ρ tend to 0 . Similar results are proved in [26].

A natural question that arises is as follows: Can a singular limit ( u 1 , u 2 ) be constructed such that it blows up on sets that are not necessarily disjoint? Is it possible to construct a solution u 1 respectively u 2 that concentrates in z 1 and z 2 respectively in z 2 and z 3 ? This question was affirmatively answered in [27], where the authors studied the following problem:

(14) Δ u 1 = ρ 2 e γ u 1 + ( 1 γ ) u 2 in Ω Δ u 2 = ρ 2 e ξ u 2 + ( 1 ξ ) u 1 in Ω u 1 = u 2 = 0 on Ω ,

where Ω is a regular bounded open domain in R 2 , γ , ξ , and ρ are constants. Similar results are proved in dimension 4. Indeed, Baraket et al. in [28] studied the existence of a singular limit solution for the following system in cases where the singular sets are not necessarily disjoint

(15) Δ 2 u 1 = ρ 4 e γ u 1 + ( 1 γ ) u 2 in Ω R 4 Δ 2 u 2 = ρ 4 e ξ u 2 + ( 1 ξ ) u 1 in Ω Δ u 1 = Δ u 2 = 0 on Ω ,

where γ , ξ and ρ are constants.

In this article, we will prove the existence of singular limit solutions of (2). More precisely, we will construct a singular limit ( u 1 , u 2 ) that blows up on sets that are not necessarily disjoint. Specifically, u 1 blows up in z 1 and z 2 , while u 2 blows up in z 2 and z 3 , where { z 1 , z 2 , z 3 } Ω , to be chosen later.

In the following, we will assume that λ satisfies:

( A 1 ) If 0 < ε < λ , then λ 1 + μ 2 ε μ 0 as λ 0 , for any μ ( 1 , 2 ) . ( A 2 ) If 0 < ε < λ , then λ 1 + δ 2 ε δ 0 as λ 0 , for any δ 0 , min γ + ξ 1 γ , ( γ + ξ 1 ξ ) .

In the following lemma, we give a necessary condition on the position of the points ( z 1 , z 2 , z 3 ) , using techniques inspired by the work of Suzuki [22] and Baraket et al. [27].

Lemma 1

Let Ω be a regular open subset of R 2 and z 1 , z 2 , z 3 Ω be given disjoint points. Let H and G be as above. Let ( u 1 ρ , λ , u 2 ρ , λ ) be a solution of (2) such that

lim ρ , λ 0 u 1 ρ , λ = 1 γ G ( , z 1 ) + G ( , z 2 ) = u 1 * in C loc 2 , α ( Ω \ { z 1 , z 2 } )

and

lim ρ , λ 0 u 2 ρ , λ = 1 ξ G ( , z 3 ) + G ( , z 2 ) = u 2 * in C loc 2 , α ( Ω \ { z 2 , z 3 } ) ,

then ( z 1 , z 2 , z 3 ) is the critical point of the functional

(16) ( z 1 , z 2 , z 3 ) = 1 ξ γ H ( z 1 , z 1 ) + ( 2 γ ξ ) H ( z 2 , z 2 ) + 1 γ ξ H ( z 3 , z 3 ) + ( 1 γ ) ξ ( 1 ξ ) γ G ( z 1 , z 3 ) + 1 ξ γ G ( z 1 , z 2 ) + 1 γ ξ G ( z 2 , z 3 ) .

Before stating our main result, we define an auxiliary function φ , which is a cut-off function in C 0 ( Ω ) such that φ 1 in B ( z 1 , r 0 ) B ( z 3 , r 0 ) and φ 0 in Ω \ B ( z 1 , 2 r 0 ) B ( z 3 , 2 r 0 ) , where r 0 > 0 and B ( z i , 2 r 0 ) are in Ω and disjoints for i = 1, 2, 3. We will prove the following result.

Theorem 3

Let Ω be a regular open subset of R 2 , λ > 0 satisfying ( A 1 ) ( A 2 ) and z 1 , z 2 , z 3 Ω be given disjoint points. Let φ , H and G be as mentioned earlier. Let ( z 1 , z 2 , z 3 ) be a nondegenerate critical point of the functional defined by (16). Then there exists θ in ( 0 , 1 ) such that for γ and ξ ( θ , 1 ) , there exist ρ 0 > 0 , λ 0 > 0 and ( u 1 ρ , λ , u 2 ρ , λ ) ρ ρ 0 , λ λ 0 is a one parameter family of solutions of (2) such that

lim ρ , λ 0 φ u 1 ρ , λ = φ γ G ( , z 1 ) i n C loc 2 , α ( Ω \ { z 1 } ) , lim ρ , λ 0 φ u 2 ρ , λ = φ ξ G ( , z 3 ) i n C loc 2 , α ( Ω \ { z 3 } ) ,

and

lim ρ , λ 0 ( ( 1 ξ ) u 1 ρ , λ + ( 1 γ ) u 2 ρ , λ ) = ( 1 ξ ) γ G ( , z 1 ) + ( 1 γ ) ξ G ( , z 3 ) + ( 2 γ ξ ) G ( , z 2 ) in C loc 2 , α ( Ω \ { z 1 , z 2 , z 3 } ) .

By introducing a new assumption on z 1 , z 2 , and z 3 , we can provide an affirmative answer to the inverse problem. More precisely, we give the asymptotic behavior of u 1 ρ , λ and u 2 ρ , λ separately in the entire Ω .

Theorem 4

Let Ω be a regular open subset of R 2 , λ > 0 satisfying ( A 1 ) ( A 2 ) and z 1 , z 2 , z 3 Ω be given disjoint points. Let H and G be as mentioned earlier. Let ( z 1 , z 2 , z 3 ) be a nondegenerate critical point of the functional defined by (16) such that

(17) 1 γ G ( z 2 , z 1 ) = 1 ξ G ( z 2 , z 3 ) and 1 γ G ( , z 1 ) ( z 2 ) = 1 ξ G ( , z 3 ) ( z 2 ) .

Then there exists θ in ( 0 , 1 ) such that for all γ and ξ ( θ , 1 ) , there exist ρ 0 > 0 , λ 0 > 0 and ( u 1 ρ , λ , u 2 ρ , λ ) ρ ρ 0 , λ λ 0 is a one parameter family of solutions of (2) such that

lim ρ , λ 0 u 1 ρ , λ = 1 γ G ( , z 1 ) + G ( , z 2 ) in C loc 2 , α ( Ω \ { z 1 , z 2 } )

and

lim ρ , λ 0 u 2 ρ , λ = 1 ξ G ( , z 3 ) + G ( , z 2 ) in C loc 2 , α ( Ω \ { z 2 , z 3 } ) .

Remark 1

The conditions of Theorem 4 are certainly not valid on all domains of R 2 . It is thought that a certain symmetry of the domains must be imposed for the condition (17) to hold.

The outline of this article is as follows: In Section 2, we establish a necessary condition on the position of the points ( z 1 , z 2 , z 3 ) thanks to the Pohozaev identity. By using the techniques inspired by [22], we determine the appropriate functional which proves Lemma 1. Next, we want to show the inverse result of Lemma 1 without imposing any additional conditions, which is a priori impossible. To overcome this difficulty, we introduce an auxiliary function, the proof of which is detailed in Section 3. In that section, on the basis of the technique developed in [29], we establish Theorem 3. In fact, we introduce some crucial results and definitions concerning the weighted Hölder spaces, linearized operators, and the harmonic extensions. In addition, we recall some studies on the analysis of the Laplace operator in weighted spaces. We construct our approximate solution of (2) by using the appropriate transformation in a large balls. Further, we consider a nonlinear interior problem, where we establish the existence of a family of solutions of (2), that closely approximate the given solution and which are defined on Ω with small balls removed. Finally, we show how parameters of these families can be combined to produce solutions of (2). In fact, we patch these pieces together via a nonlinear version of Cauchy data matching. In Section 4, we prove Theorem 4 by considering the topological condition (17) and by using the same techniques as in the previous section.

2 Proof of Lemma 1

Let ω be a subset of Ω . By multiplying the equation Δ u 1 = λ u 1 2 + ρ 2 e γ u 1 + ( 1 γ ) u 2 by ( γ u 1 + ( 1 γ ) u 2 ) and then integrating over ω , we derive the following Pohozaev-type identity

(18) γ ω u 1 2 2 ν u 1 ν u 1 d σ ( 1 γ ) ω Δ u 1 u 2 = λ ω u 1 2 ( γ u 1 + ( 1 γ ) u 2 ) + ρ 2 ω ( e γ u 1 + ( 1 γ ) u 2 1 ) ν d σ .

Similarly, by multiplying the equation Δ u 2 = λ u 2 2 + ρ 2 e ξ u 2 + ( 1 ξ ) u 1 by ( ξ u 2 + ( 1 ξ ) u 1 ) and then integrating over ω , we obtain a Pohozaev-type identity as follows:

(19) ξ ω u 2 2 2 ν u 2 ν u 2 d σ ( 1 ξ ) ω Δ u 2 u 1 = λ ω u 2 2 ( ξ u 2 + ( 1 ξ ) u 1 ) + ρ 2 ω ( e ξ u 2 + ( 1 ξ ) u 1 1 ) ν d σ .

By making use of the identity, we obtain the following:

ω Δ u 1 u 2 + ω Δ u 2 u 1 = ω ( u 2 . u 1 ) + ω ( u 2 . ν ) u 1 d σ + ω ( u 1 . ν ) u 2 d σ = ω ( u 2 . u 1 ) ν d σ + ω ( u 2 . ν ) u 1 d σ + ω ( u 1 . ν ) u 2 d σ .

Then, by multiplying (18) by ( 1 ξ ) and (19) by ( 1 γ ) , and adding them together, we obtain

(20) I lhs = γ ( 1 ξ ) ω u 1 2 2 ν u 1 ν u 1 d σ + ξ ( 1 γ ) ω u 2 2 2 ν u 2 ν u 2 d σ ( 1 γ ) ( 1 ξ ) ω ( u 2 . u 1 ) ν d σ + ω ( u 2 . ν ) u 1 d σ + ω ( u 1 . ν ) u 2 d σ = λ ( 1 ξ ) ω u 1 2 ( γ u 1 + ( 1 γ ) u 2 ) + λ ( 1 γ ) ω u 2 2 ( ξ u 2 + ( 1 ξ ) u 1 ) + ρ 2 ( 1 ξ ) ω ( e γ u 1 + ( 1 γ ) u 2 1 ) ν d σ + ρ 2 ( 1 γ ) ω ( e ξ u 2 + ( 1 ξ ) u 1 1 ) ν d σ = I rhs .

We insert the profile of the limits of the solutions in the identity (20) as ρ and λ tend to 0 and η remains fixed small enough. We choose ω = B ( z i , η ) = B i , for i = 1, 2, 3. Then, we obtain, thanks to the regularity of solutions of (2) on Ω { z 1 , z 2 , z 3 } ,

lim ρ , λ 0 I rhs = lim ρ , λ 0 λ ( 1 ξ ) ω u 1 2 ( γ u 1 + ( 1 γ ) u 2 ) + λ ( 1 γ ) ω u 2 2 ( ξ u 2 + ( 1 ξ ) u 1 ) + ρ 2 ( 1 ξ ) B i ( e γ u 1 + ( 1 γ ) u 2 1 ) ν d σ + ρ 2 ( 1 γ ) B i ( e ξ u 2 + ( 1 ξ ) u 1 1 ) ν d σ = 0 .

So we obtain

lim ρ , λ 0 I lhs = γ ( 1 ξ ) B i u 1 * 2 2 ν u 1 * ν u 1 * d σ + ξ ( 1 γ ) B i u 2 * 2 2 ν u 2 * ν u 2 * d σ ( 1 γ ) ( 1 ξ ) B i ( u 2 * . u 1 * ) ν d σ + B i ( u 2 * . ν ) u 1 * d σ + B i ( u 1 * . ν ) u 2 * d σ = 0 .

By applying the same argument as in the proof of Theorem 3 in [27], we conclude that ( z 1 , z 2 , z 3 ) is a critical point of the functional defined by (16).

3 Proof of Theorem 3

3.1 Construction of the approximate solution

We denote by ε the smallest positive parameter satisfying

ρ 2 = 8 ε 2 ( 1 + ε 2 ) 2 .

Let

(21) u ε ( z ) 2 log 1 + ε 2 ε 2 + z 2 ,

which is a solution of

(22) Δ u = ρ 2 e u

in R 2 . Hence, for all τ > 0 , the function

(23) u ε , τ ( z ) 2 log τ ( 1 + ε 2 ) ε 2 + τ z 2 ,

is also a solution of (22). Finally, we denote by r ε ε 1 2 and R ε r ε τ , τ > 0 .

3.1.1 Linearized operators

Initially, we establish certain definitions and notation.

Definition 1

Given k N , α ( 0 , 1 ) , μ R and z = r , let C μ k , α ( R 2 ) be the space of functions in C loc k , α ( R 2 ) for which the following norm

u C μ k , α ( R 2 ) = u C k , α ( B ¯ 1 ) + sup r 1 ( r μ u ( r ) C k , α ( B ¯ 1 \ B 1 2 ) )

is finite. Similarly, for a given r ¯ 1 , let C μ k , α ( B r ¯ ) be the space of function in C k , α ( B r ¯ ) for which the following norm

u C μ k , α ( B r ¯ ) = u C k , α ( B 1 ) + sup 1 r r ¯ ( r μ u ( r ) C k , α ( B ¯ 1 \ B 1 2 ) ) ,

is finite. Finally, set B r * ( z ) = B r ( z ) { z } , let C μ k , α ( B ¯ 1 * ) be the space of functions in C loc k , α ( B ¯ 1 * ) for which the following norm

u C μ k , α ( B ¯ 1 * ) = sup r 1 2 ( r μ u ( r ) C k , α ( B ¯ 2 \ B 1 ) )

is finite.

We define the linear second-order elliptic operator L by

L Δ 8 ( 1 + r 2 ) 2 ,

which is the linearized operator of Δ u ρ 2 e u = 0 about the radial symmetric solutions u ε = 1 , τ = 1 defined by (23). When k 2 , we let [ C μ k , α ( Ω ¯ ) ] 0 to be the subspace of functions w C μ k , α ( Ω ¯ ) satisfying w = 0 on Ω .

For all ε , λ , τ i > 0 , i = 1, 2, 3 and γ , ξ ( 0 , 1 ) , we define

(24) r ε , λ max ( λ 1 2 , ε 1 2 , ε γ + ξ 1 ξ , ε γ + ξ 1 γ ) and R ε , λ i τ i r ε , λ ε .

Proposition 1

[17] All bounded solutions of L w = 0 on R 2 are linear combination of

ϕ 0 ( z ) = 1 r 2 1 + r 2 a n d ϕ i ( z ) = 2 z i 1 + r 2 f o r i = 1 , 2 .

Moreover, for μ > 1 , μ Z , L : C μ 2 , α ( R 2 ) C μ 2 0 , α ( R 2 ) is surjective.

We denote by G μ a right inverse of L .

Similarly, by using the fact that any bounded harmonic function in R 2 is constant, we claim

Proposition 2

Let δ > 0 , δ Z , then Δ is surjective from C δ 2 , α ( R 2 ) to C δ 2 0 , α ( R 2 ) .

We denote by K δ : C δ 2 0 , α ( R 2 ) C δ 2 , α ( R 2 ) a right inverse of Δ for δ > 0 , δ Z .

Finally, we consider punctured domains. Given that z ˜ 1 , z ˜ 2 , and z ˜ 3 are distinct in Ω , we define z ˜ ( z ˜ 1 , z ˜ 2 , z ˜ 3 ) and Ω ¯ * ( z ˜ ) Ω ¯ \ { z ˜ 1 , z ˜ 2 , z ˜ 3 } . Let r 0 > 0 be small such that B ¯ r 0 ( z ˜ i ) are disjoint and included in Ω . For all r ( 0 , r 0 ) , we define

Ω ¯ r ( z ˜ ) Ω ¯ \ i = 1 3 B r ( z i ˜ ) .

Definition 2

Let k R , α ( 0 , 1 ) and ν R , let C ν k , α ( Ω ¯ * ( z ˜ ) ) the set of u C loc k , α ( Ω ¯ * ( z ˜ ) ) endowed with the following norm:

w C ν k , α ( Ω ¯ * ( z ˜ ) ) w C k , α ( Ω ¯ r 0 2 ( z ˜ ) ) + i = 1 3 sup 0 < r r 0 2 ( r ν w ( z ˜ i + r . ) C k , α ( B ¯ 2 B 1 ) ) .

Furthermore, for k 2 , let [ C ν k , α ( Ω ¯ * ( z ˜ ) ) ] 0 be the set of w C k , α ( Ω ¯ * ( z ˜ ) ) satisfying w = 0 on Ω .

We recall the following result.

Proposition 3

[18] Let ν < 0 , ν Z , then Δ is surjective from [ C ν 2 , α ( Ω ¯ * ( z ˜ ) ) ] 0 to C ν 2 0 , α ( Ω ¯ * ( z ˜ ) ) .

We denote by K ˜ ν : C ν 2 0 , α ( Ω ¯ * ( z ˜ ) ) [ C ν 2 , α ( Ω ¯ * ( z ˜ ) ) ] 0 a right inverse of Δ for ν < 0 , ν Z .

Before proceeding to the various stages of constructing the solution to the problem (2), we will provide a brief overview of the technique employed and its intended purpose. First, we construct an approximate solution of the system (2) excluding the gradient term. Then, prior to constructing an approximate solution of problem (2), we perform a suitable transformation to work in a larger ball. To give a certain degree of freedom to the edge data, we establish an exact solution of our problem inside small balls centered on singularities with arbitrary data at the edge of each small ball. Next, we build an exact solution outside the balls with arbitrary data at the edge of the balls. Finally, with a suitable choice of these data at the edges, we combine interior and exterior solutions to derive a global solution across the whole domain Ω .

3.1.2 Ansatz and first estimates

For all σ 1 , we denote by ξ σ : C μ 0 , α ( B ¯ σ ) C μ 0 , α ( R 2 ) the extension operator defined by

(25) ξ σ ( f ) ( z ) = f ( z ) for z σ χ z σ f σ z z for z σ .

Here, χ is a cut-off function over R + . It is equal to 1 for t 1 and equal to 0 for t 2 . It is easy to check that there exists a constant c = c ¯ ( μ ) > 0 , independent of σ 1 , such that

(26) ξ σ ( w ) C μ 0 , α ( R 2 ) c ¯ w C μ 0 , α ( B ¯ σ ) .

Now, we define an ansatz solution for the system (2) without the gradient term. More precisely, we define an ansatz solution for the following problem:

(27) Δ u 1 + ρ 2 e γ u 1 + ( 1 γ ) u 2 = 0 and Δ u 2 + ρ 2 e ξ u 2 + ( 1 ξ ) u 1 = 0 in B ( z i , r ε , λ ) .

Let τ i > 0 , for i = 1, 2, 3, then we define

(28) u ˜ 1 ( z ) = 1 γ u ε , τ 1 ( z z 1 ) 1 γ γ G ( z , z 2 ) 1 γ γ ξ G ( z , z 3 ) log γ γ z B ( z 1 , r ε ) u ε , τ 2 ( z z 2 ) z B ( z 2 , r ε ) 1 γ G ( z , z 1 ) + G ( z , z 2 ) z Ω \ i = 1 , 2 B ( z i , r ε ) ,

and

(29) u ˜ 2 ( z ) = 1 ξ u ε , τ 3 ( z z 3 ) 1 ξ ξ G ( z , z 2 ) 1 ξ γ ξ G ( z , z 1 ) log ξ ξ z B ( z 3 , r ε ) u ε , τ 2 ( z z 2 ) z B ( z 2 , r ε ) 1 ξ G ( z , z 3 ) + G ( z , z 2 ) z Ω \ i = 2,3 B ( z i , r ε ) .

Therefore, in B ( z 1 , r ε ) , we have

(30) Δ u ˜ 1 + ρ 2 e γ u ˜ 1 + ( 1 γ ) u ˜ 2 = 0 Δ u ˜ 2 + ρ 2 e ξ u ˜ 2 + ( 1 ξ ) u ˜ 1 = 8 τ 1 2 1 ξ γ ε 2 ( 1 + ε 2 ) 2 1 ξ γ 2 γ 1 ξ γ ( ε 2 + τ 1 2 z z 1 2 ) 2 ( 1 ξ ) γ e γ + ξ 1 γ G ( z , z 2 ) + γ + ξ 1 γ ξ G ( z , z 3 ) .

Then, for r = z z 1 and 0 < δ < γ + ξ 1 γ , we have

Δ u ˜ 2 + ρ 2 e ξ u ˜ 2 + ( 1 ξ ) u ˜ 1 C δ 2 0 , α ( B r ε ( z 1 ) ) C sup r < r ε τ 1 2 1 ξ γ ε 2 γ 1 ξ γ ( 1 + ε 2 ) 2 2 ( 1 ξ γ ) r 2 δ ε 4 ( 1 ξ γ ) ( 1 + ( τ 1 ε r ) 2 ) 2 1 ξ γ C sup r < R ε ε 4 4 1 ξ γ δ ( 1 + ε 2 ) 2 2 ( 1 ξ γ ) r 2 δ ( 1 + r 2 ) 2 1 ξ γ C sup r < R ε ε 4 4 1 ξ γ δ S ( r ) ,

where S ( r ) = r 2 δ ( 1 + r 2 ) 2 1 ξ γ .

If 2 4 1 ξ γ δ < 0 , then S is bounded on R + , hence,

Δ u ˜ 2 + ρ 2 e ξ u ˜ 2 + ( 1 ξ ) u ˜ 1 C δ 2 0 , α ( B r ε ( z 1 ) ) C ε 4 4 1 ξ γ δ C r ε 2 .

If 2 4 1 ξ γ δ > 0 , sup 0 , r ε ε S ( r ) = S r ε ε , then

Δ u ˜ 2 + ρ 2 e ξ u ˜ 2 + ( 1 ξ ) u ˜ 1 C δ 2 0 , α ( B r ε ( z 1 ) ) C ε 4 4 1 ξ γ δ S r ε ε C r ε 2 r ε 4 γ + ξ 1 γ δ C r ε 2 .

On the other hand, in B ( z 2 , r ε ) , there holds

(31) Δ u ˜ 1 + ρ 2 e γ u ˜ 1 + ( 1 γ ) u ˜ 2 = 0 Δ u ˜ 2 + ρ 2 e ξ u ˜ 2 + ( 1 ξ ) u ˜ 1 = 0 .

Finally, in B ( z 3 , r ε ) , we derive

(32) Δ u ˜ 1 + ρ 2 e γ u ˜ 1 + ( 1 γ ) u ˜ 2 = 8 ε 2 τ 3 2 ( 1 γ ξ ) ( 1 + ε 2 ) 2 ( 1 γ ξ ) 2 ξ 1 γ ξ ( ε 2 + τ 3 2 z z 3 2 ) 2 ( 1 γ ) ξ e ( γ + ξ 1 ξ ) G ( z , z 2 ) + γ + ξ 1 γ ξ G ( z , z 1 ) Δ u ˜ 2 + ρ 2 e ξ u ˜ 2 + ( 1 ξ ) u ˜ 1 = 0 .

Then, for r = z z 3 , μ ( 1 , 2 ) and 0 < δ < γ + ξ 1 ξ . Similarly, as in B ( z 1 , r ε ) , we have

Δ u ˜ 1 + ρ 2 e γ u ˜ 1 + ( 1 γ ) u ˜ 2 C δ 2 0 , α ( B r ε ( z 3 ) ) C r ε 2 .

Now, we define an approximate solution for the following system near B ( z i , r ε , λ ) , i = 1, 2, 3:

(33) Δ u 1 + λ u 1 2 + ρ 2 e γ u 1 + ( 1 γ ) u 2 = 0 and Δ u 2 + λ u 2 2 + ρ 2 e ξ u 2 + ( 1 ξ ) u 1 = 0 .

By using the following transformations, we obtain

v 1 ( z ) = u 1 ε τ 1 z + 4 γ log ε + 2 γ log 2 τ 1 ( 1 + ε 2 ) in B ( z 1 , r ε , λ ) v 2 ( z ) = u 2 ε τ 1 z in B ( z 1 , r ε , λ ) ,

v 1 ( z ) = u 1 ε τ 2 z + 4 log ε + 2 log 2 τ 2 ( 1 + ε 2 ) in B ( z 2 , r ε , λ ) v 2 ( z ) = u 2 ε τ 2 z + 4 log ε + 2 log 2 τ 2 ( 1 + ε 2 ) in B ( z 2 , r ε , λ ) ,

and

v 1 ( z ) = u 1 ε τ 3 z in B ( z 3 , r ε , λ ) v 2 ( z ) = u 2 ε τ 3 z + 4 ξ log ε + 2 ξ log 2 τ 3 ( 1 + ε 2 ) in B ( z 3 , r ε , λ ) .

System (33) can be written as follows:

(34) Δ v 1 + λ v 1 2 + 2 e γ v 1 + ( 1 γ ) v 2 = 0 in B ( z 1 , R ε , λ 1 ) Δ v 2 + λ v 2 2 + 2 C 1 , ε 2 γ + ξ 1 γ ε 4 γ + ξ 1 γ e ξ v 2 + ( 1 ξ ) v 1 = 0 in B ( z 1 , R ε , λ 1 ) ,

(35) Δ v 1 + λ v 1 2 + 2 e γ v 1 + ( 1 γ ) v 2 = 0 in B ( z 2 , R ε , λ 2 ) Δ v 2 + λ v 2 2 + 2 e ξ v 2 + ( 1 ξ ) v 1 = 0 in B ( z 2 , R ε , λ 2 ) ,

and

(36) Δ v 1 + λ v 1 2 + 2 C 3 , ε 2 γ + ξ 1 ξ ε 4 γ + ξ 1 ξ e γ v 1 + ( 1 γ ) v 2 = 0 in B ( z 3 , R ε , λ 3 ) Δ v 2 + λ v 2 2 + 2 e ξ v 2 + ( 1 ξ ) v 1 = 0 in B ( z 3 , R ε , λ 3 ) ,

with C i , ε = 2 τ i ( 1 + ε 2 ) , for i = 1, 3 and τ i > 0 is a constant, which will be fixed later. Recall that for i = 1, 2, and 3, R ε , λ i = τ i r ε , λ ε . Fix μ ( 1 , 2 ) and δ 0 , min γ + ξ 1 γ , γ + ξ 1 ξ . We recall that ξ μ , ξ δ are defined in (25), G μ and K δ are defined in Propositions 1 and 2.

In B ( z 1 , R ε , λ 1 ) , we denote by u ¯ = u ε = 1 , τ = 1 , we look for a solution of (34) of the form

(37) v 1 ( z ) = 1 γ u ¯ ( z z 1 ) 1 γ γ G ε z τ 1 , z 2 1 γ γ ξ G ε z τ 1 , z 3 log γ γ + h 1 1 ( z ) v 2 ( z ) = 1 ξ G ε z τ 1 , z 3 + G ε z τ 1 , z 2 + h 2 1 ( z ) .

This amounts to solving the equation

(38) L h 1 1 = 8 γ ( 1 + r 2 ) 2 [ e γ h 1 1 + ( 1 γ ) h 2 1 γ h 1 1 1 ] + λ 1 γ u ¯ ( z z 1 ) 1 γ γ G ε z τ 1 , z 2 1 γ γ ξ G ε z τ 1 , z 3 log γ γ + h 1 1 ( z ) 2 Δ h 2 1 = 2 C 1 , ε 2 γ + ξ 1 γ 4 1 ξ γ ε 4 γ + ξ 1 γ γ 1 ξ γ ( 1 + r 2 ) 2 1 ξ γ e ξ h 2 1 + ( 1 ξ ) h 1 1 + γ + ξ 1 γ G ε z τ 1 , z 2 + γ + ξ 1 γ ξ G ε z τ 1 , z 3 + λ 1 ξ G ε z τ 1 , z 3 + G ε z τ 1 , z 2 + h 2 1 ( z ) 2 .

We denote by

L h 1 1 = T 1 ( h 1 1 , h 2 1 ) and Δ h 2 1 = 1 ( h 1 1 , h 2 1 ) .

To find a solution of (38), it is enough to find a fixed point ( h 1 1 , h 2 1 ) in a small ball of C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) , which is solution of

(39) h 1 1 = G μ ξ μ T 1 ( h 1 1 , h 2 1 ) = N 1 ( h 1 1 , h 2 1 ) h 2 1 = K δ ξ δ 1 ( h 1 1 , h 2 1 ) = 1 ( h 1 1 , h 2 1 ) .

Then, we have the following result.

Lemma 2

Given κ > 0 , there exist ε κ > 0 , λ κ > 0 , c κ > 0 , c ¯ κ > 0 and θ ( 0 , 1 ) such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , γ ( θ , 1 ) , μ ( 1 , 2 ) and δ 0 , min { γ + ξ 1 γ , γ + ξ 1 ξ } . We have

N 1 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 , 1 ( 0 , 0 ) C δ 2 , α ( R 2 ) c κ r ε , λ 2 , N 1 ( h 1 1 , h 2 1 ) N 1 ( k 1 1 , k 2 1 ) C μ 2 , α ( R 2 ) c ¯ κ r ε , λ 2 h 1 1 k 1 1 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 γ ) h 2 1 k 2 1 C δ 2 , α ( R 2 ) , 1 ( h 1 1 , h 2 1 ) 1 ( k 1 1 , k 2 1 ) C δ 2 , α ( R 2 ) c ¯ κ r ε , λ 2 ( h 1 1 , h 2 1 ) ( k 1 1 , k 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) .

Provided ( h 1 1 , h 2 1 ) , ( k 1 1 , k 2 1 ) in C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) satisfying

(40) ( h 1 1 , h 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) 2 c κ r ε , λ 2 a n d ( k 1 1 , k 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) 2 c κ r ε , λ 2 .

Proof

We have

sup r R ε , λ 1 r 2 μ T 1 ( 0 , 0 ) λ sup r R ε , λ r 2 μ 1 γ u ¯ ( z z 1 ) 1 γ γ G ε z τ 1 , z 2 1 γ γ ξ G ε z τ 1 , z 3 log γ γ 2

By making use of Proposition 1 together with (26), for μ ( 1 , 2 ) , we find that there exists c ¯ κ such that

(41) N 1 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 .

For the second estimate, we have

sup r R ε , λ 1 r 2 δ 1 ( 0 , 0 ) sup r R ε , λ 1 2 C 1 , ε 2 γ + ξ 1 γ γ 1 ξ γ ε 4 γ + ξ 1 γ r 2 δ 4 ( 1 + r 2 ) 2 1 ξ γ e γ + ξ 1 γ G ε z τ 1 , z 2 e γ + ξ 1 γ ξ G ε z τ 1 , z 3

+ λ sup r R ε , λ 1 r 2 δ 1 ξ G ε z τ 1 , z 3 + G ε z τ 1 , z 2 2 c κ sup r R ε , λ 1 ε 4 γ + ξ 1 γ r 2 δ 4 ( 1 + r 2 ) 2 1 ξ γ + c κ λ ε 2 sup r R ε , λ 1 r 2 δ c κ sup r R ε , λ 1 ε 4 γ + ξ 1 γ F ( r ) + c κ λ ε 2 sup r R ε , λ 1 r 2 δ ,

where F ( r ) = r 2 δ ( 1 + r 2 ) 2 ( 1 ξ γ ) .

If 2 δ 4 1 ξ γ < 0 , then F is bounded on R + . If 2 δ 4 1 ξ γ > 0 , then sup 0 , r ε , λ ε F ( r ) = F r ε , λ ε . We obtain

1 ( 0 , 0 ) C δ 2 , α ( R 2 ) c κ r ε , λ 2 .

To derive the third estimate, for ( h 1 1 , h 2 1 ) , ( k 1 1 , k 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) verifying (40), we have

sup r R ε , λ 1 r 2 μ T 1 ( h 1 1 , h 2 1 ) T 1 ( k 1 1 , k 2 1 ) sup r R ε , λ 1 8 r 2 μ ( 1 + r 2 ) 2 1 γ e γ h 1 1 + ( 1 γ ) h 2 1 h 1 1 1 γ 1 γ e γ k 1 1 + ( 1 γ ) k 2 1 k 1 1 1 γ + λ sup r R ε , λ 1 r 2 μ 1 γ u ¯ ( z z 1 ) 1 γ γ G ε z τ 1 , z 2 1 γ γ ξ G ε z τ 1 , z 3 log γ γ + h 1 1 ( z ) 2 1 γ u ¯ ( z z 1 ) 1 γ γ G ε z τ 1 , z 2 1 γ γ ξ G ε z τ 1 , z 3 log γ γ + k 1 1 ( z ) 2 c κ sup r R ε , λ 1 8 r 2 μ ( 1 + r 2 ) 2 1 γ [ γ 2 ( ( h 1 1 ) 2 ( k 1 1 ) 2 ) + ( 1 γ ) h 2 1 k 2 1 ] + c κ λ sup r R ε , λ 1 r 2 μ × 2 1 γ u ¯ ( z z 1 ) 1 γ γ G ε z τ 1 , z 2 1 γ γ ξ G ε z τ 1 , z 3 log γ γ + ( h 1 1 + k 1 1 ) ( h 1 1 k 1 1 ) c κ sup r R ε , λ 1 8 r 2 μ ( 1 + r 2 ) 2 1 γ [ γ 2 r 2 μ ( h 1 1 C μ 2 , α ( R 2 ) + k 1 1 C μ 2 , α ( R 2 ) ) h 1 1 k 1 1 C μ 2 , α ( R 2 ) + ( 1 γ ) r δ h 2 1 k 2 1 C δ 2 , α ( R 2 ) ] + c κ λ sup r R ε , λ 1 r 2 μ 8 r 1 + r 2 + C ε + r μ 1 ( h 1 1 C μ 2 , α ( R 2 ) + k 1 1 C μ 2 , α ( R 2 ) ) r μ 1 h 1 1 k 1 1 C μ 2 , α ( R 2 ) c κ ( h 1 1 C μ 2 , α ( R 2 ) + k 1 1 C μ 2 , α ( R 2 ) ) h 1 1 k 1 1 C μ 2 , α ( R 2 ) + c κ ( 1 γ ) h 2 1 k 2 1 C δ 2 , α ( R 2 ) + c κ λ ( 1 + r ε , λ + ε μ r ε , λ 2 + μ ) h 1 1 k 1 1 C μ 2 , α ( R 2 ) .

Using the following estimates

c κ ε μ r ε , λ 2 + μ c κ ε 1 μ 2 for ε > λ c κ λ 1 + μ 2 ε μ for λ > ε ,

together with condition ( A 1 ) , this yields c κ λ ( 1 + r ε , λ + ε μ r ε , λ 2 + μ ) c κ r ε , λ 2 .

By making use of Proposition 1 together with (26) and using the condition ( A 1 ) for μ ( 1 , 2 ) , we conclude that there exists c ¯ κ > 0 such that

(42) N 1 ( h 1 1 , h 2 1 ) N 1 ( k 1 1 , k 2 1 ) C μ 2 , α ( R 2 ) c ¯ κ r ε , λ 2 h 1 1 k 1 1 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 γ ) h 2 1 k 2 1 C δ 2 , α ( R 2 ) .

On the other hand, we have

sup r R ε , λ 1 r 2 μ T 2 ( h 1 1 , h 2 1 ) T 2 ( k 1 1 , k 2 1 ) c ε 4 γ + ξ 1 γ sup r R ε , λ 1 r 2 δ ( 1 + r 2 ) 2 1 ξ γ e γ + ξ 1 γ G ε z τ 1 , z 2 + γ + ξ 1 γ ξ G ε z τ 1 , z 3 e ξ h 2 1 + ( 1 ξ ) h 1 1 e ξ k 2 1 + ( 1 ξ ) k 1 1 + λ sup r R ε , λ 1 r 2 μ 1 ξ G ε z τ 1 , z 3 + G ε z τ 1 , z 2 + h 2 1 ( z ) 2 1 ξ G ε z τ 1 , z 3 + G ε z τ 1 , z 2 + k 2 1 ( z ) 2 c ε 4 γ + ξ 1 γ sup r R ε , λ 1 r 2 δ ( 1 + r 2 ) 2 1 ξ γ [ ξ h 2 1 k 2 1 + ( 1 ξ ) h 1 1 k 1 1 ] + λ sup r R ε , λ 1 r 2 δ 2 ξ G ε z τ 1 , z 3 + 2 G ε z τ 1 , z 2 + r δ 1 ( h 2 1 C δ 2 , α ( R 2 ) + k 2 1 C δ 2 , α ( R 2 ) ) r δ 1 h 2 1 k 2 1 C δ 2 , α ( R 2 ) c κ r ε , λ 2 h 1 1 k 1 1 C δ 2 , α ( R 2 ) + c κ r ε , λ 2 h 2 1 k 2 1 C δ 2 , α ( R 2 ) + c κ λ ( r ε , λ + ε δ r ε , λ 2 + δ ) h 2 1 k 2 1 C δ 2 , α ( R 2 ) .

Using the following estimates

c κ ε δ r ε , λ 2 + δ c κ ε 1 δ 2 for ε > λ c κ λ 1 + δ 2 ε δ for λ > ε ,

in conjunction with condition ( A 2 ) , this results c κ λ ( r ε , λ + ε δ r ε , λ 2 + δ ) c κ r ε , λ 2 .

Applying Proposition 1 together with (26) and using the condition ( A 2 ) for δ 0 , min { γ + ξ 1 γ , γ + ξ 1 ξ } , we derive that there exists c ¯ κ > 0 such that

(43) 1 ( h 1 1 , h 2 1 ) 1 ( k 1 1 , k 2 1 ) C δ 2 , α ( R 2 ) c ¯ κ r ε , λ 2 ( h 1 1 , h 2 1 ) ( k 1 1 , k 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) ,

which proves Lemma 2.□

Reducing ε κ and λ κ if necessary, we can assume that c ¯ κ r ε , λ 2 1 2 for all ε ( 0 , ε κ ) and λ ( 0 , λ κ ) . Then, there exists θ ( 0 , 1 ) such that c ¯ κ ( 1 γ ) 1 2 for all γ ( θ , 1 ) . Therefore, inequalities (42) and (43) are enough to show that

( h 1 1 , h 2 1 ) ( N 1 ( h 1 1 , h 2 1 ) , 1 ( h 1 1 , h 2 1 ) )

is a contraction from the ball

{ ( h 1 1 , h 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) : ( h 1 1 , h 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) 2 c κ r ε , λ 2 } ,

into itself. Hence, within this set, a unique fixed point ( h 1 1 , h 2 1 ) exists, providing a solution of (39).

Proposition 4

Given κ > 0 , there exist ε κ , λ κ , c κ > 0 and θ ( 0 , 1 ) such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) and γ ( θ , 1 ) , there exists a unique solution ( h 1 1 , h 2 1 ) of (39) such that

( h 1 1 , h 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) 2 c κ r ε , λ 2 .

Hence, ( v 1 , v 2 ) , which is defined by (37), solves (34) in B ( z 1 , R ε , λ 1 ) .

In B ( z 2 , R ε , λ 2 ) , we look for a solution of (35) in the form

(44) v 1 ( z ) = u ¯ ( z z 2 ) + h 1 2 ( z ) v 2 ( z ) = u ¯ ( z z 2 ) + h 2 2 ( z ) .

This amounts to solve the equation

(45) L h 1 2 = 8 ( 1 + r 2 ) 2 [ e γ h 1 2 + ( 1 γ ) h 2 2 h 1 2 1 ] + λ ( u ¯ ( z z 2 ) + h 1 2 ( z ) ) 2 L h 2 2 = 8 ( 1 + r 2 ) 2 [ e ξ h 2 2 + ( 1 ξ ) h 1 2 h 2 2 1 ] + λ ( u ¯ ( z z 2 ) + h 2 2 ( z ) ) 2 .

We denote

L h 1 2 = T 2 ( h 1 2 , h 2 2 ) and L h 2 2 = 2 ( h 1 2 , h 2 2 ) .

To find a solution of (45), it is enough to find a fixed point ( h 1 2 , h 2 2 ) in a small ball of C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) , solutions of

(46) h 1 2 = G μ ξ μ T 2 ( h 1 2 , h 2 2 ) = N 2 ( h 1 2 , h 2 2 ) h 2 2 = K μ ξ μ 2 ( h 1 2 , h 2 2 ) = 2 ( h 1 2 , h 2 2 ) .

Then, we have the following result.

Lemma 3

Given κ > 0 , there exist ε κ > 0 , λ κ > 0 , θ in ( 0 , 1 ) , c ¯ κ > 0 and c κ > 0 such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , γ , ξ ( θ , 1 ) and μ ( 1 , 2 ) . We have

N 2 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 , 2 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 , N 2 ( h 1 2 , h 2 2 ) N 2 ( k 1 2 , k 2 2 ) C μ 2 , α ( R 2 ) c ¯ κ ( 1 γ + r ε , λ 2 ) h 1 2 k 1 2 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 γ ) h 2 2 k 2 2 C μ 2 , α ( R 2 ) , 2 ( h 1 2 , h 2 2 ) 2 ( k 1 2 , k 2 2 ) C μ 2 , α ( R 2 ) c ¯ κ ( 1 ξ ) h 1 2 k 1 2 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 ξ + r ε , λ 2 ) h 2 2 k 2 2 C μ 2 , α ( R 2 ) .

Provided ( h 1 2 , h 2 2 ) , ( k 1 2 , k 2 2 ) in C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) satisfying

(47) ( h 1 2 , h 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 a n d ( k 1 2 , k 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 .

Proof

We have

sup r R ε , λ 2 r 2 μ T 2 ( 0 , 0 ) λ sup r R ε , λ 2 r 2 μ u ¯ ( z z 2 ) 2 c κ λ .

For μ ( 1 , 2 ) , we use of Proposition 1 together with (26), we derive that there exists c κ such that

(48) N 2 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 .

For the second estimate, we have 2 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 .

To establish the third estimate, for ( h 1 1 , h 2 1 ) , ( k 1 1 , k 2 1 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) verifying (47), we have

sup r R ε , λ 2 r 2 μ T 2 ( h 1 2 , h 2 2 ) T 2 ( k 1 2 , k 2 2 ) sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 ( e γ h 1 1 + ( 1 γ ) h 2 1 h 1 2 e γ k 1 2 + ( 1 γ ) k 2 2 + k 1 2 ) + λ sup r R ε , λ 2 r 2 μ ( ( u ¯ ( z z 2 ) + h 1 2 ( z ) ) 2 ( u ¯ ( z z 2 ) + k 1 2 ( z ) ) 2 ) c κ sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 [ ( γ 1 ) ( h 1 2 k 1 2 ) + ( 1 γ ) ( h 2 2 k 2 2 ) ] + c κ λ sup r R ε , λ 2 r 2 μ ( 2 ( u ¯ ( z z 2 ) ) + ( h 1 2 + k 1 2 ) ) ( h 1 2 k 1 2 ) c κ sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 ( 1 γ ) [ r μ h 1 2 k 1 2 C μ 2 , α ( R 2 ) + r μ h 2 2 k 2 2 C μ 2 , α ( R 2 ) ] + c κ λ sup r R ε , λ 2 r 2 μ 8 r 1 + r 2 + r μ 1 ( h 1 2 C μ 2 , α ( R 2 ) + k 1 2 C μ 2 , α ( R 2 ) ) r μ 1 h 1 2 k 1 2 C μ 2 , α ( R 2 ) .

By using Proposition 1 along with (26) and considering condition ( A 1 ) for μ ( 1 , 2 ) , we find that there exists c ¯ κ > 0 such that

(49) N 2 ( h 1 2 , h 2 2 ) N 2 ( k 1 2 , k 2 2 ) C μ 2 , α ( R 2 ) c ¯ κ ( 1 γ + r ε , λ 2 ) h 1 2 k 1 2 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 γ ) h 2 2 k 2 2 C μ 2 , α ( R 2 ) .

Similarly, we obtain the estimate for

(50) 2 ( h 1 2 , h 2 2 ) 2 ( k 1 2 , k 2 2 ) C μ 2 , α ( R 2 ) c ¯ κ ( 1 ξ ) h 1 2 k 1 2 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 ξ + r ε , λ 2 ) h 2 2 k 2 2 C μ 2 , α ( R 2 ) ,

which proves the Lemma 3.□

We note that there exists θ in ( 0 , 1 ) , reducing ε κ and λ κ if necessary. We can assume that c ¯ κ ( 1 γ + r ε , λ 2 ) 1 2 and c ¯ κ ( 1 ξ + r ε , λ 2 ) 1 2 for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , γ and ξ ( θ , 1 ) . Therefore, inequalities (49) and (50) are enough to show that

( h 1 2 , h 2 2 ) ( N 2 ( h 1 2 , h 2 2 ) , 2 ( h 1 2 , h 2 2 ) )

is a contraction from the ball

{ ( h 1 2 , h 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) : ( h 1 2 , h 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 } ,

(46) as the unique fixed point ( h 1 2 , h 2 2 ) .

Proposition 5

Given κ > 0 , there exist ε κ , λ κ , c κ > 0 and θ ( 0 , 1 ) such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , γ and ξ ( θ , 1 ) . Thus, there is a unique solution ( h 1 2 , h 2 2 ) of (46) such that

( h 1 2 , h 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 .

Hence, ( v 1 , v 2 ) defined by (44) solves (35) in B ( z 2 , R ε , λ 2 ) .

Our objective is to obtain a solution of (36) in B ( z 3 , R ε , λ 3 ) , having the form

(51) v 1 ( z ) = 1 γ G ε z τ 3 , z 1 + G ε z τ 3 , z 2 + h 1 3 ( z ) v 2 ( z ) = 1 ξ u ¯ ( z z 3 ) 1 ξ ξ G ε z τ 3 , z 2 1 ξ γ ξ G ε z τ 3 , z 1 log ξ ξ + h 2 3 ( z ) .

This enables to solve the equation

(52) Δ h 1 3 = 2 C 3 , ε 2 γ + ξ 1 ξ 4 1 γ ξ ε 4 γ + ξ 1 ξ ξ 1 γ ξ ( 1 + r 2 ) 2 1 γ ξ e γ h 1 3 + ( 1 γ ) h 2 3 + γ + ξ 1 ξ G ε z τ 3 , z 2 + γ + ξ 1 γ ξ G ( ε z τ 3 , z 1 ) + λ 1 γ G ε z τ 3 , z 1 + G ε z τ 3 , z 2 + h 1 3 ( z ) 2 L h 2 3 = 8 ξ ( 1 + r 2 ) 2 [ e ξ h 2 3 + ( 1 ξ ) h 1 3 ξ h 2 3 1 ] + λ 1 ξ u ¯ ( z z 3 ) 1 ξ ξ G ε z τ 3 , z 2 1 ξ γ ξ G ε z τ 3 , z 1 log ξ ξ + h 2 3 ( z ) 2 .

We denote

Δ h 1 3 = T 3 ( h 1 3 , h 2 3 ) and L h 2 3 = 3 ( h 1 3 , h 2 3 ) .

To obtain a solution of (52), it suffices to identify a fixed point ( h 1 3 , h 2 3 ) in a small ball of C δ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) , solutions of

(53) h 1 3 = G δ ξ δ T 3 ( h 1 3 , h 2 3 ) = N 3 ( h 1 3 , h 2 3 ) h 2 3 = K μ ξ μ 3 ( h 1 3 , h 2 3 ) = 3 ( h 1 3 , h 2 3 ) .

By using the same reasoning as in the proof of Proposition 4, we establish

Proposition 6

Given κ > 0 , there exist ε κ , λ κ , c κ > 0 , and θ ( 0 , 1 ) such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) and ξ ( θ , 1 ) . Then, there exists a unique solution ( h 1 3 , h 2 3 ) of (53) such that

( h 1 3 , h 2 3 ) C δ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 .

Therefore, ( v 1 , v 2 ) defined by (51) solves (36) in B ( z 3 , R ε , λ 3 ) .

3.1.3 Harmonic extensions

In this subsection, we will study the properties of interior and exterior harmonic extensions. Given φ C 2 , α ( S 1 ) , we, respectively, define H int = H int ( φ ; ) = H φ int and H ext = H ext ( φ ˜ ; ) = H φ ˜ ext to represent the solution of

(54) Δ H int ( φ ; ) = 0 in B 1 , H int ( φ ; ) = φ on B 1 .

(55) Δ H ext ( φ ˜ ; ) = 0 in R 2 \ B 1 , H ext ( φ ˜ ; ) = φ ˜ on B 1 , lim z H ext ( φ ˜ ; z ) = 0 .

We will use also the following definition.

Definition 3

Given k N , α ( 0 , 1 ) and ν R , let C ν k , α ( R 2 \ B 1 ) as the space of functions w C loc k , α ( R 2 \ B 1 ) for which the following norm

w C ν k , α ( R 2 \ B 1 ) = sup r 1 ( r ν w ( r ) C ν k , α ( B ¯ 2 \ B 1 ) ) ,

is finite.

We denote e 1 ( θ ) = cos θ , e 2 ( θ ) = sin θ .

Lemma 4

[18] There exists c > 0 such that, for ϕ C 2 , α ( S 1 ) ,

(56) S 1 φ d v S 1 = 0 a n d S 1 φ e d θ = 0 f o r = 1 , 2 ,

we have H int ( φ ; ) C 2 2 , α ( B ¯ 1 * ) c φ C 2 , α ( S 1 ) . Similarly, there exists c > 0 such that if ϕ ˜ C 2 , α ( S 1 ) with

(57) S 1 φ ˜ d θ = 0 ,

then H ext ( φ ˜ ; ) C 1 2 , α ( R 2 \ B 1 ) c φ ˜ C 2 , α ( S 1 ) .

If F L 2 ( S 1 ) is a subspace, we denote F to be the subspace of F which is L 2 ( S 1 ) -orthogonal to e 1 , e 2 . We recall the following result.

Lemma 5

[18] The mapping P : C 2 , α ( S 1 ) C 1 , α ( S 1 ) defined by

P ( φ ) = r H int ( φ ) r H ext ( φ ) ,

is an isomorphism.

3.2 The nonlinear interior problem

Here, we are interested to study the system

(58) Δ v 1 + λ v 1 2 + 2 e γ v 1 + ( 1 γ ) v 2 = 0 in B ( z 1 , R ε , λ 1 ) Δ v 2 + λ v 2 2 + 2 C 1 , ε 2 γ + ξ 1 γ ε 4 γ + ξ 1 γ e ξ v 2 + ( 1 ξ ) v 1 = 0 in B ( z 1 , R ε , λ 1 ) ,

(59) Δ v 1 + λ v 1 2 + 2 e γ v 1 + ( 1 γ ) v 2 = 0 in B ( z 2 , R ε , λ 2 ) Δ v 2 + λ v 2 2 + 2 e ξ v 2 + ( 1 ξ ) v 1 = 0 in B ( z 2 , R ε , λ 2 ) ,

and

(60) Δ v 1 + λ v 1 2 + 2 C 3 , ε 2 γ + ξ 1 ξ ε 4 γ + ξ 1 ξ e γ v 1 + ( 1 γ ) v 2 = 0 in B ( z 3 , R ε , λ 3 ) Δ v 2 + λ v 2 2 + 2 e ξ v 2 + ( 1 ξ ) v 1 = 0 in B ( z 3 , R ε , λ 3 ) .

With C i , ε = 2 τ i ( 1 + ε 2 ) , for i = 1, 3 and τ i > 0 is a constant which will be fixed later. Note that for i = 1, 2 and 3, R ε , λ i = τ i r ε , λ ε . Fix μ ( 1 , 2 ) and δ 0 , min { γ + ξ 1 γ , γ + ξ 1 ξ } . We recall that ξ μ , ξ δ are defined in equality (25), G μ and K δ are defined in Propositions 1 and 2, respectively.

In B ( z 1 , R ε , λ 1 ) : Given φ 1 ( φ 1 1 , φ 2 1 ) ( C 2 , α ( S 1 ) ) 2 satisfying (56), we seek a solution of (58) in the form

(61) v 1 ( z ) = 1 γ u ¯ ( z z 1 ) 1 γ γ G ε z τ 1 , z 2 1 γ γ ξ G ε z τ 1 , z 3 log γ γ + H int φ 1 1 , z z 1 R ε , λ 1 + h 1 1 ( z ) + v 1 1 ( z ) v 2 ( z ) = 1 ξ G ε z τ 1 , z 3 + G ε z τ 1 , z 2 + H int φ 2 1 , z z 1 R ε , λ 1 + h 2 1 ( z ) + v 2 1 ( z ) .

This involves solving the equation

(62) L v 1 1 = 8 γ ( 1 + r 2 ) 2 [ e γ ( h 1 1 + H 1 int , 1 + v 1 1 ) + ( 1 γ ) ( h 2 1 + H 2 int , 1 + v 2 1 ) γ v 1 1 1 ] + λ 1 γ u ¯ ( z z 1 ) 1 γ γ G ε z τ 1 , z 2 1 γ γ ξ G ε z τ 1 , z 3 log γ γ + H int φ 1 1 , z z 1 R ε , λ 1 + h 1 1 ( z ) + v 1 1 ( z ) 2 + Δ h 1 1 Δ v 2 1 = 2 C 1 , ε 2 γ + ξ 1 γ 4 1 ξ γ ε 4 γ + ξ 1 γ γ 1 ξ γ ( 1 + r 2 ) 2 1 ξ γ e ξ ( h 2 1 + H 2 int , 1 + v 2 1 ) + ( 1 ξ ) ( h 1 1 + H 1 int , 1 + v 1 1 ) + γ + ξ 1 γ G ε z τ 1 , z 2 + γ + ξ 1 γ ξ G ε z τ 1 , z 3 + λ 1 ξ G ε z τ 1 , z 3 + G ε z τ 1 , z 2 + H int φ 2 1 , z z 1 R ε , λ 1 + h 2 1 ( z ) + v 2 1 ( z ) 2 + Δ h 2 1 .

We denote

L v 1 1 = T 1 ( v 1 1 , v 2 1 ) and Δ v 2 1 = R 1 ( v 1 1 , v 2 1 ) .

To find a solution of (62), it is sufficient to locate a fixed point ( v 1 1 , v 2 1 ) in a small ball of C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) , solutions of

(63) v 1 1 = G μ ξ μ T 1 ( v 1 1 , v 2 1 ) = N 1 ( v 1 1 , v 2 1 ) v 2 1 = K δ ξ δ R 1 ( v 1 1 , v 2 1 ) = M 1 ( v 1 1 , v 2 1 ) .

Given κ > 0 (whose value will be fixed later on), we further assume that the functions ( φ 1 1 , φ 2 1 ) satisfy

(64) ( φ 1 1 , φ 2 1 ) ( C 2 , α ( S 1 ) ) 2 κ r ε , λ 2 .

In consequence, we obtain the following result.

Lemma 6

Given κ > 0 , there exist ε κ > 0 , λ κ > 0 , c κ > 0 , c ¯ κ > 0 and θ ( 0 , 1 ) such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , γ ( θ , 1 ) , μ ( 1 , 2 ) and δ 0 , min γ + ξ 1 γ , γ + ξ 1 ξ , we have

N 1 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 , M 1 ( 0 , 0 ) C δ 2 , α ( R 2 ) c κ r ε , λ 2 , N 1 ( v 1 1 , v 2 1 ) N 1 ( t 1 1 , t 2 1 ) C μ 2 , α ( R 2 ) c ¯ κ r ε , λ 2 v 1 1 t 1 1 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 γ ) v 2 1 t 2 1 C δ 2 , α ( R 2 ) , M 1 ( v 1 1 , v 2 1 ) M 1 ( t 1 1 , t 2 1 ) C δ 2 , α ( R 2 ) c ¯ κ r ε , λ 2 ( v 1 1 , v 2 1 ) ( t 1 1 , t 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) .

Given ( v 1 1 , v 2 1 ) , ( t 1 1 , t 2 1 ) in C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) satisfying

(65) ( v 1 1 , v 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) 2 c κ r ε , λ 2 a n d ( t 1 1 , t 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) 2 c κ r ε , λ 2 .

Proof

The initial estimate stems from Lemma 4, along with the assumption regarding the norm of the boundary data φ j 1 given by (64). SO, we have

H int ( φ j 1 ; R ε , λ ) C 2 2 , α ( B ¯ R ε , λ ) c κ R ε , λ 2 φ j 1 C 2 , α ( S 1 ) c κ ε 2 .

On the other hand, for μ ( 1 , 2 ) , we obtain

sup r R ε , λ 1 r 2 μ T 1 ( 0 , 0 ) sup r R ε , λ 1 8 r 2 μ γ ( 1 + r 2 ) 2 e γ h 1 1 + γ H 1 int , 1 + ( 1 γ ) h 2 1 + ( 1 γ ) H 2 int , 1 1 + λ sup r R ε , λ 1 r 2 μ × 1 γ u ¯ ( z z 1 ) 1 γ γ G ε z τ 1 , z 2 1 γ γ ξ G ε z τ 1 , z 3 log γ γ + H int φ 1 1 , z z 1 R ε , λ 1 + h 1 1 ( z ) 2 + Δ h 1 1 c κ sup r R ε , λ 1 8 r 2 μ γ ( 1 + r 2 ) 2 ( γ r μ h 1 1 C μ 2 , α ( R 2 ) + γ r 2 H 1 int , 1 C 2 2 , α + ( 1 γ ) r δ h 2 1 C δ 2 , α ( R 2 ) + ( 1 γ ) r 2 H 2 int , 1 C 2 2 , α ) + λ sup r R ε , λ 1 r 2 μ 1 γ u ¯ ( z z 1 ) 1 γ γ G ε z τ 1 , z 2 1 γ γ ξ G ε z τ 1 , z 3 log γ γ + H int φ 1 1 , z z 1 R ε , λ 1 + h 1 1 ( z ) 2 + Δ h 1 1 .

Using Proposition 1 and (26), according to condition ( A 1 ) for μ ( 1 , 2 ) , we find that there exists c κ > 0 such that

(66) N 1 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 .

For the second estimate, we have

sup r R ε , λ 1 r 2 δ R 1 ( 0 , 0 ) c κ sup r R ε , λ 1 C 1 , ε 2 γ + ξ 1 γ ε 4 γ + ξ 1 γ r 2 δ 4 ( 1 + r 2 ) 2 1 ξ γ e ξ h 2 1 + ξ H 2 int , 1 + ( 1 ξ ) h 1 1 + ( 1 ξ ) H 1 int , 1 e γ + ξ 1 γ G ε z τ 1 , z 2 + γ + ξ 1 γ ξ G ε z τ 1 , z 3 + λ sup r R ε , λ 1 r 2 δ 1 ξ G ε z τ 1 , z 3 + G ε z τ 1 , z 2 + H int φ 2 1 , z z 1 R ε , λ 1 + h 2 1 ( z ) 2 + Δ h 2 1 .

By using the same argument as previously mentioned, we obtain M 1 ( 0 , 0 ) C δ 2 , α ( R 2 ) c κ r ε , λ 2 .

To derive the third estimate, for ( v 1 1 , v 2 1 ) , ( t 1 1 , t 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) verifying (65), we have

sup r R ε , λ 1 r 2 μ T 1 ( v 1 1 , v 2 1 ) T 1 ( t 1 1 , t 2 1 ) sup r R ε , λ 1 8 r 2 μ ( 1 + r 2 ) 2 1 γ e γ h 1 1 + γ H 1 int , 1 + γ v 1 1 + ( 1 γ ) h 2 1 + ( 1 γ ) H 2 int , 1 + ( 1 γ ) v 2 1 v 1 1 1 γ 1 γ e γ h 1 1 + γ H 1 int , 1 + γ t 1 1 + ( 1 γ ) h 2 1 + ( 1 γ ) H 2 int , 1 + ( 1 γ ) t 2 1 t 1 1 1 γ + λ sup r R ε , λ 1 r 2 μ × 1 γ u ¯ ( z z 1 ) 1 γ γ G ε z τ 1 , z 2 1 γ γ ξ G ε z τ 1 , z 3 log γ γ + H int φ 1 1 , z z 1 R ε , λ 1 + h 1 1 ( z ) + v 1 1 ( z ) 2 1 γ u ¯ ( z z 1 ) 1 γ γ G ε z τ 1 , z 2 1 γ γ ξ G ε z τ 1 , z 3 log γ γ + H 1 int , 1 + h 1 1 ( z ) + t 1 1 ( z ) 2 c κ sup r R ε , λ 1 8 r 2 μ ( 1 + r 2 ) 2 1 γ [ γ 2 ( ( v 1 1 ) 2 ( t 1 1 ) 2 ) + ( 1 γ ) v 2 1 t 2 1 ] + c κ λ sup r R ε , λ 1 r 2 μ 2 1 γ u ¯ ( z z 1 ) 1 γ γ G ε z τ 1 , z 2 1 γ γ ξ G ε z τ 1 , z 3 log γ γ + H 1 int , 1 + h 1 1 ( z ) + ( v 1 1 + t 1 1 ) ( v 1 1 t 1 1 ) c κ sup r R ε , λ 1 8 r 2 μ ( 1 + r 2 ) 2 1 γ [ γ 2 r 2 μ ( v 1 1 C μ 2 , α ( R 2 ) + t 1 1 C μ 2 , α ( R 2 ) ) v 1 1 t 1 1 C μ 2 , α ( R 2 ) + ( 1 γ ) r δ v 2 1 t 2 1 C δ 2 , α ( R 2 ) ] + c κ λ sup r R ε , λ 1 r 2 μ 8 r 1 + r 2 + C ε + r ε H 1 int , 1 C 2 2 , α + r μ 1 h 1 1 C μ 2 , α ( R 2 ) + r μ 1 ( v 1 1 C μ 2 , α ( R 2 ) + t 1 1 C μ 2 , α ( R 2 ) ) × r μ 1 v 1 1 t 1 1 C μ 2 , α ( R 2 ) .

By making use of Proposition 1 together with (26) and using the condition ( A 1 ) for μ ( 1 , 2 ) , we conclude that there exists c ¯ κ > 0 such that

(67) N 1 ( v 1 1 , v 2 1 ) N 1 ( t 1 1 , t 2 1 ) C μ 2 , α ( R 2 ) c ¯ κ r ε , λ 2 v 1 1 t 1 1 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 γ ) v 2 1 t 2 1 C δ 2 , α ( R 2 ) .

Similarly, using the condition ( A 2 ) for δ 0 , min γ + ξ 1 γ , γ + ξ 1 ξ , we obtain

(68) M 1 ( v 1 1 , v 2 1 ) M 1 ( t 1 1 , t 2 1 ) C δ 2 , α ( R 2 ) c ¯ κ r ε , λ 2 ( v 1 1 , v 2 1 ) ( t 1 1 , t 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) .

The proof of Lemma 6 is achieved.□

By reducing ε κ and λ κ if needed, we can assume that c ¯ κ r ε , λ 2 1 2 for all ε ( 0 , ε κ ) and λ ( 0 , λ κ ) . Then, there exists θ ( 0 , 1 ) such that c ¯ κ ( 1 γ ) 1 2 for all γ ( θ , 1 ) . Therefore, (67) and (68) are enough to show that

( v 1 1 , v 2 1 ) ( N 1 ( v 1 1 , v 2 1 ) , M 1 ( v 1 1 , v 2 1 ) )

is a contraction from the ball

{ ( v 1 1 , v 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) : ( v 1 1 , v 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) 2 c κ r ε , λ 2 } ,

into itself and hence a unique fixed point ( v 1 1 , v 2 1 ) exists in this set, which is a solution of (63).

Proposition 7

Given κ > 0 , there exist ε κ , λ κ , c κ > 0 , and θ ( 0 , 1 ) such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , γ ( θ , 1 ) and for all τ 1 in some fixed compact subset of [ τ 1 , τ 1 + ] ( 0 , + ) , there exists a unique solution of (63),

that is ( v 1 1 , v 2 1 ) ( ( v 1 , ε , τ 1 , φ 1 , v 2 , ε , τ 1 , φ 1 ) ) such that

( v 1 1 , v 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) 2 c κ r ε , λ 2 .

Thus, ( v 1 , v 2 ) given by (61) solves (58) in B ( z 1 , R ε , λ 1 ) .

In B ( z 2 , R ε , λ 2 ) : Given φ 2 ( φ 1 2 , φ 2 2 ) ( C 2 , α ( S 1 ) ) 2 satisfying (56), we seek a solution of (59) in the form

(69) v 1 ( z ) = u ¯ ( z z 2 ) + H int φ 1 2 , z z 2 R ε , λ 2 + h 1 2 ( z ) + v 1 2 ( z ) v 2 ( z ) = u ¯ ( z z 2 ) + H int φ 2 2 , z z 2 R ε , λ 2 + h 2 2 ( z ) + v 2 2 ( z ) .

This is equivalent to solving the equation

(70) L v 1 2 = 8 ( 1 + r 2 ) 2 [ e γ ( H 1 int , 2 + h 1 2 + v 1 2 ) + ( 1 γ ) ( H 2 int , 2 + h 2 2 + v 2 2 ) v 1 2 1 ] + λ u ¯ ( z z 2 ) + H int φ 1 2 , z z 2 R ε , λ 2 + h 1 2 ( z ) + v 1 2 ( z ) 2 + Δ h 1 2 L v 2 2 = 8 ( 1 + r 2 ) 2 [ e ξ ( H 2 int , 2 + h 2 2 + v 2 2 ) + ( 1 ξ ) ( H 1 int , 2 + h 1 2 + v 1 2 ) v 2 2 1 ] + λ u ¯ ( z z 2 ) + H int φ 2 2 , z z 2 R ε , λ 2 + h 2 2 ( z ) + v 2 2 ( z ) 2 + Δ h 2 2 .

We designate

L v 1 2 = T 2 ( v 1 2 , v 2 2 ) and L v 2 2 = R 2 ( v 1 2 , v 2 2 ) .

Finding a fixed point ( v 1 2 , v 2 2 ) in a small ball of C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) suffices to establish the following solution of (70).

(71) v 1 2 = G μ ξ μ T 2 ( v 1 2 , v 2 2 ) = N 2 ( v 1 2 , v 2 2 ) v 2 2 = K μ ξ μ R 2 ( v 1 2 , v 2 2 ) = M 2 ( v 1 2 , v 2 2 ) .

Given κ > 0 (whose value will be fixed subsequently), we further assume that the functions ( φ 1 2 , φ 2 2 ) satisfy

(72) ( φ 1 2 , φ 2 2 ) ( C 2 , α ( S 1 ) ) 2 κ r ε , λ 2 .

Then, we have the following result.

Lemma 7

Given κ > 0 , there exist ε κ > 0 , λ κ > 0 , θ in ( 0 , 1 ) , c ¯ κ > 0 and c κ > 0 such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , γ , ξ ( θ , 1 ) and μ ( 1 , 2 ) , we have

N 2 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 , M 2 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 , N 2 ( v 1 2 , v 2 2 ) N 2 ( t 1 2 , t 2 2 ) C μ 2 , α ( R 2 ) c ¯ κ ( 1 γ + r ε , λ 2 ) v 1 2 t 1 2 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 γ ) v 2 2 t 2 2 C μ 2 , α ( R 2 ) , M 2 ( v 1 2 , v 2 2 ) M 2 ( t 1 2 , t 2 2 ) C μ 2 , α ( R 2 ) c ¯ κ ( 1 ξ ) v 1 2 t 1 2 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 ξ + r ε , λ 2 ) v 2 2 t 2 2 C μ 2 , α ( R 2 ) .

Provided ( v 1 2 , v 2 2 ) , ( t 1 2 , t 2 2 ) in C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) satisfying

(73) ( v 1 2 , v 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 a n d ( t 1 2 , t 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 .

Proof

We have

sup r R ε , λ 2 r 2 μ T 2 ( 0 , 0 ) sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 e γ h 1 2 + γ H 1 int , 2 + ( 1 γ ) h 2 2 + ( 1 γ ) H 2 int , 2 1 + λ sup r R ε , λ 2 r 2 μ u ¯ ( z z 2 ) + H int φ 1 2 , z z 2 R ε , λ 2 + h 1 2 ( z ) 2 + Δ h 1 2 c κ sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 ( γ r μ h 1 2 C μ 2 , α ( R 2 ) + γ r 2 H 1 int , 2 C 2 2 , α + ( 1 γ ) r δ h 2 2 C δ 2 , α ( R 2 ) + ( 1 γ ) r 2 H 2 int , 2 C 2 2 , α ) + λ sup r R ε , λ 2 r 2 μ u ¯ ( z z 2 ) + H int φ 1 2 , z z 2 R ε , λ 2 + h 1 2 ( z ) 2 + Δ h 1 2 .

By using Proposition 1, (26) and condition ( A 1 ) for μ ( 1 , 2 ) , we derive that there exists c κ > 0 such that

(74) N 2 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 .

For the second estimate, we have M 2 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 .

To establish the third estimate, for ( v 1 2 , v 2 2 ) , ( t 1 2 , t 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) verifying (73), we have

sup r R ε , λ 2 r 2 μ T 2 ( v 1 2 , v 2 2 ) T 2 ( t 1 2 , t 2 2 ) sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 ( e γ ( H 1 int , 2 + h 1 2 + v 1 2 ) + ( 1 γ ) ( H 2 int , 2 + h 2 2 + v 2 2 ) v 1 2 e γ ( H 1 int , 2 + h 1 2 + t 1 2 ) + ( 1 γ ) ( H 2 int , 2 + h 2 2 + t 2 2 ) + t 1 2 ) + λ sup r R ε , λ 2 r 2 μ u ¯ ( z z 2 ) + H int φ 1 2 , z z 2 R ε , λ 2 + h 1 2 ( z ) + v 1 2 ( z ) 2 u ¯ ( z z 2 ) + H int φ 1 2 , z z 2 R ε , λ 2 + h 1 2 ( z ) + t 1 2 ( z ) 2 c κ sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 [ ( γ 1 ) ( v 1 2 t 1 2 ) + ( 1 γ ) ( v 2 2 t 2 2 ) ] + c κ λ sup r R ε , λ 2 r 2 μ 2 u ¯ ( z z 2 ) + H int φ 1 2 , z z 2 R ε , λ 2 + h 1 2 ( z ) + ( v 1 2 + t 1 2 ) ( v 1 2 t 1 2 ) c κ sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 ( 1 γ ) [ r μ v 1 2 t 1 2 C μ 2 , α ( R 2 ) + ( 1 γ ) r μ v 2 2 t 2 2 C μ 2 , α ( R 2 ) ] + c κ λ sup r R ε , λ 2 r 2 μ × 8 r 1 + r 2 + r μ 1 h 1 2 C μ 2 , α ( R 2 ) + r H 1 int , 2 C 2 2 , α + r μ 1 ( v 1 2 C μ 2 , α ( R 2 ) + t 1 2 C μ 2 , α ( R 2 ) ) r μ 1 v 1 2 t 1 2 C μ 2 , α ( R 2 ) .

On the basis of Proposition 1, inequality (26), and condition ( A 1 ) for μ ( 1 , 2 ) , we show that there exists c ¯ κ > 0 such that

(75) N 2 ( v 1 2 , v 2 2 ) N 2 ( t 1 2 , t 2 2 ) C μ 2 , α ( R 2 ) c ¯ κ ( 1 γ + r ε , λ 2 ) v 1 2 t 1 2 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 γ ) v 2 2 t 2 2 C μ 2 , α ( R 2 ) .

Similarly, we obtain the estimate for

(76) M 2 ( v 1 2 , v 2 2 ) M 2 ( t 1 2 , t 2 2 ) C μ 2 , α ( R 2 ) c ¯ κ ( 1 ξ ) v 1 2 t 1 2 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 ξ + r ε , λ 2 ) v 2 2 t 2 2 C μ 2 , α ( R 2 ) .

This concludes the proof of Lemma 7.□

We note that there exists θ in ( 0 , 1 ) . By reducing ε κ and λ κ if needed, we can assume that c ¯ κ ( 1 γ + r ε , λ 2 ) 1 2 and c ¯ κ ( 1 ξ + r ε , λ 2 ) 1 2 for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , γ and ξ ( θ , 1 ) . Therefore, inequalities (75) and (76) are enough to show that

( v 1 2 , v 2 2 ) ( N 2 ( v 1 2 , v 2 2 ) , M 2 ( v 1 2 , v 2 2 ) )

is a contraction from the ball

{ ( v 1 2 , v 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) : ( v 1 2 , v 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 } ,

into itself. Consequently, a unique fixed point ( v 1 2 , v 2 2 ) exists in this set, serving as a solution to (71).

Proposition 8

Given κ > 0 , there exist ε κ > 0 , λ κ , c κ > 0 and θ ( 0 , 1 ) such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , γ , ξ ( θ , 1 ) , and for all τ 2 in some fixed compact subset of [ τ 2 , τ 2 + ] ( 0 , + ) , there exists a unique ( v 1 2 , v 2 2 ) ( ( v 1 , ε , τ 2 , φ 2 , v 2 , ε , τ 2 , φ 2 ) ) solution of (71) such that

( v 1 2 , v 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 .

Therefore, ( v 1 , v 2 ) defined by (69) solves (59) in B ( z 2 , R ε , λ 2 ) .

In B ( z 3 , R ε , λ 3 ) : Given φ 3 ( φ 1 3 , φ 2 3 ) ( C 2 , α ( S 1 ) ) 2 satisfying (56), we look for a solution of (60) in the following manner:

(77) v 1 ( z ) = 1 γ G ε z τ 3 , z 1 + G ε z τ 3 , z 2 + H int φ 1 3 , z z 3 R ε , λ 3 + h 1 3 ( z ) + v 1 3 ( z ) v 2 ( z ) = 1 ξ u ¯ ( z z 3 ) 1 ξ ξ G ε z τ 3 , z 2 1 ξ γ ξ G ε z τ 3 , z 1 log ξ ξ + H int φ 2 3 , z z 3 R ε , λ 3 + h 2 3 ( z ) + v 2 3 ( z ) .

This facilitates solving the equation

(78) Δ v 1 3 = 2 C 3 , ε 2 γ + ξ 1 ξ 4 1 γ ξ ε 4 γ + ξ 1 ξ ξ 1 γ ξ ( 1 + r 2 ) 2 1 γ ξ e γ ( h 1 3 + H 1 int , 3 + v 1 3 ) + ( 1 γ ) ( h 2 3 + H 2 int , 3 + v 2 3 ) + γ + ξ 1 ξ G ε z τ 3 , z 2 + γ + ξ 1 γ ξ G ( ε z τ 3 , z 1 ) + λ 1 γ G ε z τ 3 , z 1 + G ε z τ 3 , z 2 + H int φ 1 3 , z z 3 R ε , λ 3 + h 1 3 ( z ) + v 1 3 ( z ) 2 + Δ h 1 3 L v 2 3 = 8 ξ ( 1 + r 2 ) 2 [ e ξ ( h 2 3 + H 2 int , 3 + v 2 3 ) + ( 1 ξ ) ( h 1 3 + H 1 int , 3 + v 1 3 ) ξ v 2 3 1 ] + λ 1 ξ u ¯ ( z z 3 ) 1 ξ ξ G ε z τ 3 , z 2 1 ξ γ ξ G ε z τ 3 , z 1 log ξ ξ + H int φ 2 3 , z z 3 R ε , λ 3 + h 2 3 ( z ) + v 2 3 ( z ) 2 + Δ h 2 3 .

We denote

Δ v 1 3 = T 3 ( v 1 3 , v 2 3 ) and L v 2 3 = R 3 ( v 1 3 , v 2 3 ) .

To find a solution of (78), it is enough to find a fixed point ( v 1 3 , v 2 3 ) in a small ball of C δ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) , solutions of

(79) v 1 3 = G δ ξ δ T 3 ( v 1 3 , v 2 3 ) = N 3 ( v 1 3 , v 2 3 ) v 2 3 = K μ ξ μ R 3 ( v 1 3 , v 2 3 ) = M 3 ( v 1 3 , v 2 3 ) .

Given κ > 0 (whose value will be fixed later on), we further assume that the functions ( φ 1 3 , φ 2 3 ) satisfy

(80) ( φ 1 3 , φ 2 3 ) ( C 2 , α ( S 1 ) ) 2 κ r ε , λ 2 .

On the basis of the same arguments of the proof of Proposition 7, we prove the following proposition.

Proposition 9

Given κ > 0 , there exist ε κ > 0 , λ κ , c κ > 0 and θ ( 0 , 1 ) such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , ξ ( θ , 1 ) , and for all τ 3 in some fixed compact subset of [ τ 3 , τ 3 + ] ( 0 , + ) , there exists a unique solution of (79), that is ( v 1 3 , v 2 3 ) ( ( v 1 , ε , τ 3 , φ 3 , v 2 , ε , τ 3 , φ 3 ) ) , such that

( v 1 3 , v 2 3 ) C δ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 .

Thus, ( v 1 , v 2 ) defined by (77) solves (60) in B ( z 3 , R ε , λ 3 ) .

Also, we remark that the functions ( v 1 , ε , τ i , φ 1 i i v 2 , ε , τ i , φ 2 i i ) , for i { 1 , 2 , 3 } , derived from the aforementioned propositions, exhibit continuous dependency on the parameter τ .

3.3 The nonlinear exterior problem

Given z ˜ ( z ˜ 1 , z ˜ 2 , z ˜ 3 ) Ω 3 close to z ( z 1 , z 2 , z 3 ) and λ ( λ 1 , λ 2 , λ 3 ) R 3 close to 0. For i = 1 , 2 , 3 , ( φ ˜ 1 i , φ ˜ 2 i ) ( C 2 , α ( S 1 ) ) 2 , satisfying (57), define

(81) w ˜ 1 1 + λ 1 γ G ( , z ˜ 1 ) + ( 1 + λ 2 ) G ( , z ˜ 2 ) + i = 1 3 χ r 0 ( z ˜ i ) H ext ( φ ˜ 1 i ; ( z ˜ i ) r ε , λ ) w ˜ 2 1 + λ 3 ξ G ( , z ˜ 3 ) + ( 1 + λ 2 ) G ( , z ˜ 2 ) + i = 1 3 χ r 0 ( z ˜ i ) H ext ( φ ˜ 2 i ; ( z ˜ i ) r ε , λ ) .

Here, χ r 0 is a cut-off function taking the value of 1 inside B r 0 2 and 0 outside B r 0 . We aim to find a solution of the system

(82) Δ u 1 = ρ 2 e γ u 1 + ( 1 γ ) u 2 + λ u 1 2 , Δ u 2 + λ u 2 2 = ρ 2 e ξ u 2 + ( 1 ξ ) u 1 + λ u 2 2 ,

in the domain Ω ¯ r ε , λ ( z ˜ ) Ω ¯ \ i = 1 3 B r ( z ˜ i ) , where u k = w ˜ k + v ˜ k represents a perturbation of w ˜ k , k = 1 , 2 .

This amounts to solve in Ω ¯ r ε , λ ( z ˜ )

(83) Δ v ˜ 1 = ρ 2 e γ ( w ˜ 1 + v ˜ 1 ) + ( 1 γ ) ( w ˜ 2 + v ˜ 2 ) + λ ( w ˜ 1 + v ˜ 1 ) 2 + Δ w ˜ 1 Δ v ˜ 2 = ρ 2 e ξ ( w ˜ 2 + v ˜ 2 ) + ( 1 ξ ) ( w ˜ 1 + v ˜ 1 ) + λ ( w ˜ 2 + v ˜ 2 ) 2 + Δ w ˜ 2 .

For all σ ( 0 , r 0 2 ) and all Y = ( y 1 , y 2 , y 3 ) Ω 3 such that Z Y r 0 2 , where Z = ( z 1 , z 2 , z 3 ) , we denote by ξ ˜ σ , Y : C ν 0 , α ( Ω ¯ σ , Y ) C ν 0 , α ( Ω ¯ * ( Y ) ) , the extension operator defined by

ξ ˜ σ , Y ( f ) ( z ) = f ( z ) in Ω ¯ ( Y ) , χ ˜ z y i σ f y j + σ z y i z y i in B σ ( y j ) B σ 2 ( y j ) , 1 j 3 , 0 in B σ 2 ( y 1 ) B σ 2 ( y 2 ) B σ 2 ( y 3 ) .

Here, χ ˜ is a cut-off function over R + which equals 1 for t 1 and equals 0 for t 1 2 . Clearly, there exists a constant c ¯ = c ¯ ( ν ) > 0 only depending on ν , such that

(84) ξ ˜ σ , Y ( w ) C ν 0 , α ( Ω ¯ * ( Y ) ) c ¯ w C ν 0 , α ( Ω ¯ σ ( Y ) ) .

We fix ν ( 1 , 0 ) and we recall that K ˜ ν is defined in Proposition 3. To solve (83), it is enough to find a solution ( v ˜ 1 , v ˜ 2 ) ( C ν 2 , α ( Ω ¯ * ( z ˜ ) ) ) 2 to

(85) v ˜ 1 = K ˜ ν ξ ˜ r ε , λ T ˜ ( v ˜ 1 , v ˜ 2 ) , v ˜ 2 = K ˜ ν ξ ˜ r ε , λ ˜ ( v ˜ 1 , v ˜ 2 ) ,

where

(86) T ˜ ( v ˜ 1 , v ˜ 2 ) = ρ 2 e γ ( w ˜ 1 + v ˜ 1 ) + ( 1 γ ) ( w ˜ 2 + v ˜ 2 ) + λ ( w ˜ 1 + v ˜ 1 ) 2 + Δ w ˜ 1 ˜ ( v ˜ 1 , v ˜ 2 ) = ρ 2 e ξ ( w ˜ 2 + v ˜ 2 ) + ( 1 ξ ) ( w ˜ 1 + v ˜ 1 ) + λ ( w ˜ 2 + v ˜ 2 ) 2 + Δ w ˜ 2 .

We denote

N ˜ ( v ˜ 1 , v ˜ 2 ) = K ˜ ν ξ ˜ r ε , λ T ˜ ( v ˜ 1 , v ˜ 2 ) and ˜ ( v ˜ 1 , v ˜ 2 ) = K ˜ ν ξ ˜ r ε , λ ˜ ( v ˜ 1 , v ˜ 2 ) .

Given κ > 0 (whose value will be fixed later on), assume that the functions ( φ ˜ 1 i , φ ˜ 2 i ) , i =  1, 2, 3, the parameters λ i and the point z ˜ = ( z 1 ˜ , z 2 ˜ , z 3 ˜ ) satisfy

(87) ( φ ˜ 1 i , φ ˜ 2 i ) ( C 2 , α ) 2 κ r ε , λ 2

(88) λ i κ r ε , λ 2 , z i ˜ z i κ r ε , λ .

Then, the following result holds.

Lemma 8

Under the aforementioned assumptions, there exists a constant c κ > 0 such that

N ˜ ( 0 , 0 ) C ν 2 , α ( Ω ¯ * ( z ˜ ) ) c κ r ε , λ 2 , ˜ ( 0 , 0 ) C ν 2 , α ( Ω ¯ * ( z ˜ ) ) c κ r ε , λ 2 , N ˜ ( v ˜ 1 , v ˜ 2 ) N ˜ ( v ˜ 1 , v ˜ 2 ) C ν 2 , α ( Ω ¯ * ( z ˜ ) ) c r ε , λ 2 ( v ˜ 1 , v ˜ 2 ) ( v ˜ 1 , v ˜ 2 ) ( C ν 2 , α ( Ω ¯ * ( z ˜ ) ) ) 2 ,

and

˜ ( v ˜ 1 , v ˜ 2 ) ˜ ( v ˜ 1 , v ˜ 2 ) C ν 2 , α ( Ω ¯ * ( z ˜ ) ) c r ε , λ 2 ( v ˜ 1 , v ˜ 2 ) ( v ˜ 1 , v ˜ 2 ) ( C ν 2 , α ( Ω ¯ * ( z ˜ ) ) ) 2 .

Provided that ( v ˜ 1 , v ˜ 2 , v ˜ 1 , v ˜ 2 ) ( C ν 2 , α ( Ω ¯ * ( z ˜ ) ) ) 4 satisfy

(89) ( v ˜ 1 , v ˜ 2 ) ( C ν 2 , α ( Ω ¯ * ( z ˜ ) ) ) 2 2 c κ r ε , λ 2 , ( v ˜ 1 , v ˜ 2 ) ( C ν 2 , α ( Ω ¯ * ( z ˜ ) ) ) 2 2 c κ r ε , λ 2 .

Proof

Similarly for the interior problem, the initial two estimates stem from the asymptotic behavior of H ext in conjunction with the assumption regarding the norm of the boundary data φ ˜ j i given by (87). In fact, let c κ be a constant depending only on κ , by Lemma 2, we obtain

(90) H ext ( φ ˜ j i ; ( z z ˜ i ) r ε , λ ) c κ r ε , λ 3 r 1 .

On the other hand,

T ˜ ( 0 , 0 ) = ρ 2 e γ w ˜ 1 + ( 1 γ ) w ˜ 2 + λ w ˜ 1 2 + Δ w ˜ 1 and ˜ ( 0 , 0 ) = ρ 2 e ξ w ˜ 2 + ( 1 ξ ) w ˜ 1 + λ w ˜ 2 2 + Δ w ˜ 2 .

We will estimate T ˜ ( 0 , 0 ) in different subregions of Ω ¯ * ( z ˜ ) .

In B r 0 2 ( z ˜ 1 ) \ B r ε , λ ( z ˜ 1 ) , we have χ r 0 ( z z ˜ 1 ) = 1 , χ r 0 ( z z ˜ 2 ) = 0 , χ r 0 ( z z ˜ 3 ) = 0 and Δ w ˜ 1 = 0 . Thus, T ˜ ( 0 , 0 ) = ρ 2 e γ w ˜ 1 + ( 1 γ ) w ˜ 2 + λ w ˜ 1 2 . Then

T ˜ ( 0 , 0 ) c κ ε 2 z z 1 ˜ 4 ( 1 + λ 1 ) + c κ λ 1 + λ 1 γ ( r 1 + H ( z , z 1 ˜ ) ) + H ext ( φ ˜ 1 1 , ( z z 1 ˜ ) r ε , λ ) 2 c κ ε 2 z z 1 ˜ 4 ( 1 + λ 1 ) + c κ λ 1 + λ 1 γ ( r 1 + log r ) + r ε , λ 2 r 2 2 .

Hence, for ν ( 1 , 0 ) and λ 1 small enough, we obtain

T ˜ ( 0 , 0 ) C ν 2 0 , α ( B r 0 ( z ˜ 1 ) ) sup r ε , λ r r 0 2 r 2 ν T ˜ ( 0 , 0 ) c κ ε 2 r ε , λ 2 + c κ λ .

In B r 0 ( z ˜ 1 ) \ B r 0 2 ( z ˜ 1 ) , using the estimate (90), there holds

T ˜ ( 0 , 0 ) c κ ε 2 r 4 ( 1 + λ 1 ) + c κ λ 1 + λ 1 γ ( r 1 + log r ) + r ε , λ 2 r 2 2

+ [ Δ , χ r 0 ( z z ˜ 1 ) ] H ext ( φ ˜ 1 1 ; ( z z ˜ 1 ) r ε , λ ) c κ ε 2 + c κ λ 1 + λ 1 γ ( r 1 + log r ) + r ε , λ 2 r 2 2 + c κ r 1 r ε , λ 3 ,

where [ Δ , χ r 0 ] w = Δ w χ r 0 + w Δ χ r 0 + 2 w . χ r 0 . Hence, for ν ( 1 , 0 ) and λ 1 small enough, we obtain

T ˜ ( 0 , 0 ) C ν 2 0 , α ( B r 0 ( z ˜ 1 ) ) sup r 0 2 r r 0 r 2 ν T ˜ ( 0 , 0 ) c κ ε 2 r ε , λ 2 + c κ λ .

In B r 0 2 ( z ˜ 2 ) \ B r ε , λ ( z ˜ 2 ) , we have χ r 0 ( z z ˜ 1 ) = 0 , χ r 0 ( z z ˜ 2 ) = 1 , χ r 0 ( z z ˜ 3 ) = 0 , Δ w ˜ 1 = 0 , and T ˜ ( 0 , 0 ) = ρ 2 e γ w ˜ 1 + ( 1 γ ) w ˜ 2 + λ w ˜ 1 2 . Then

T ˜ ( 0 , 0 ) c κ ε 2 z z 2 ˜ 4 ( 1 + λ 2 ) + c κ λ ( ( 1 + λ 2 ) ( r 1 + log r ) + r ε , λ 2 r 2 ) 2 c κ ε 2 r 4 ( 1 + λ 2 ) + c κ λ ( ( 1 + λ 2 ) ( r 1 + log r ) + r ε , λ 2 r 2 ) 2 .

Thus, for ν ( 1 , 0 ) and λ 2 small enough, we obtain

T ˜ ( 0 , 0 ) C ν 2 0 , α ( B r 0 ( z ˜ 2 ) ) sup r ε , λ r r 0 2 r 2 ν T ˜ ( 0 , 0 ) c κ ε 2 r ε , λ 2 + c κ λ .

In B r 0 ( z ˜ 2 ) \ B r 0 2 ( z ˜ 2 ) , using the estimate (90), there holds

T ˜ ( 0 , 0 ) c κ ε 2 r 4 ( 1 + λ 2 ) + c κ λ ( ( 1 + λ 2 ) ( r 1 + log r ) + r ε , λ 2 r 2 ) 2 + [ Δ , χ r 0 ( z z ˜ 2 ) ] H ext ( φ ˜ 1 2 ; ( z z ˜ 2 ) r ε , λ ) c κ ε 2 r 4 ( 1 + λ 2 ) + c κ λ ( ( 1 + λ 2 ) ( r 1 + log r ) + r ε , λ 2 r 2 ) 2 + c κ r 1 r ε , λ 3 ,

where [ Δ , χ r 0 ] w = Δ w χ r 0 + w Δ χ r 0 + 2 w . χ r 0 . Hence, for ν ( 1 , 0 ) and λ 2 small enough, we obtain

T ˜ ( 0 , 0 ) C ν 2 0 , α ( B r 0 ( z ˜ 2 ) ) sup r 0 2 r r 0 r 2 ν T ˜ ( 0 , 0 ) c κ r ε , λ 2 + c κ λ .

In B r 0 2 ( z ˜ 3 ) \ B r ε , λ ( z ˜ 3 ) , we have χ r 0 ( z z ˜ 3 ) = 1 , χ r 0 ( z z ˜ 1 ) = 0 , χ r 0 ( z z ˜ 2 ) = 0 and Δ w ˜ 1 = 0 , so that T ˜ ( 0 , 0 ) = ρ 2 e γ w ˜ 1 + ( 1 γ ) w ˜ 2 + λ w ˜ 1 2 . Then

T ˜ ( 0 , 0 ) c κ ε 2 z z 3 ˜ 4 ( 1 γ ) ( 1 + λ 3 ) ξ + c κ λ H ext ( φ ˜ 1 3 , ( z z 3 ˜ ) r ε , λ ) 2 c κ ε 2 r 4 ( 1 γ ) ( 1 + λ 3 ) ξ + c κ λ r ε , λ 4 r 4 .

Hence, for ν ( 1 , 0 ) and λ 3 small enough, we obtain

T ˜ ( 0 , 0 ) C ν 2 0 , α ( B r 0 ( z ˜ 3 ) ) sup r ε , λ r r 0 2 r 2 ν T ˜ ( 0 , 0 ) c κ ε 2 r ε , λ 2 + c κ λ .

In B r 0 ( z ˜ 3 ) \ B r 0 2 ( z ˜ 3 ) , using the estimate (90), there holds

T ˜ ( 0 , 0 ) c κ ε 2 r 4 ( 1 γ ) ( 1 + λ 3 ) ξ + c κ λ r ε , λ 4 r 4 + [ Δ , χ r 0 ( z z ˜ 3 ) ] H ext ( φ ˜ 1 3 ; ( z z ˜ 3 ) r ε ) c κ ε 2 + c κ λ r ε , λ 4 r 4 + c κ r 1 r ε , λ 3 ,

where [ Δ , χ r 0 ] w = Δ w χ r 0 + w Δ χ r 0 + 2 w . χ r 0 . Therefore, for ν ( 1 , 0 ) and λ 3 small enough, we obtain

T ˜ ( 0 , 0 ) C ν 2 0 , α ( B r 0 ( z ˜ 3 ) ) sup r 0 2 r r 0 r 2 ν T ˜ ( 0 , 0 ) c κ ε 2 r ε , λ 2 + c κ λ .

In Ω \ i = 1 , 2 , 3 B ¯ r 0 ( z ˜ i ) , we have χ r 0 ( z z ˜ i ) = 0 for i = 1 , 2 , 3 and Δ w ˜ 1 = 0 . Thus,

T ˜ ( 0 , 0 ) c κ ε 2 .

For ν ( 1 , 0 ) , we have

T ˜ ( 0 , 0 ) C ν 2 0 , α ( Ω ¯ i = 1 3 B r 0 ( z i ˜ ) ) sup r 0 r r 2 ν T ˜ ( 0 , 0 ) c κ r ε , λ 2 .

Then, T ˜ ( 0 , 0 ) C ν 2 0 , α ( Ω ¯ r 0 ( z ˜ ) ) c κ r ε , λ 2 .

Now, let’s focus on the second equation of the system (86).

In B r 0 2 ( z ˜ 1 ) \ B r ε , λ ( z ˜ 1 ) , we have χ r 0 ( z z ˜ 1 ) = 1 , χ r 0 ( z z ˜ 2 ) = 0 , χ r 0 ( z z ˜ 3 ) = 0 and Δ w ˜ 2 = 0 , so that ˜ ( 0 , 0 ) = ρ 2 e ξ w ˜ 2 + ( 1 ξ ) w ˜ 1 + λ w ˜ 2 2 . Then

˜ ( 0 , 0 ) c κ ε 2 z z 1 ˜ 4 ( 1 + λ 1 ) ( 1 ξ ) γ + c κ λ H ext ( φ ˜ 2 1 , ( z z 1 ˜ ) r ε , λ ) 2 c κ ε 2 r 4 ( 1 + λ 1 ) ( 1 ξ ) γ + c κ λ r ε , λ 4 r 4 .

Hence, for ν ( 1 , 0 ) and λ 1 small enough, we obtain

˜ ( 0 , 0 ) C ν 2 0 , α ( B r 0 ( z ˜ 1 ) ) sup r ε , λ r r 0 2 r 2 ν ˜ ( 0 , 0 ) c κ r ε , λ 2 + c κ λ .

In B r 0 ( z ˜ 1 ) \ B r 0 2 ( z ˜ 1 ) , using the estimate (90), there holds

˜ ( 0 , 0 ) c κ ε 2 r 4 ( 1 + λ 1 ) ( 1 ξ ) γ + c κ λ r ε , λ 4 r 4 + [ Δ , χ r 0 ( z z ˜ 1 ) ] H ext ( φ ˜ 2 1 ; ( z z ˜ 1 ) r ε , λ ) c κ ε 2 + c κ λ r ε , λ 4 r 4 + c κ r 1 r ε , λ 3 ,

where [ Δ , χ r 0 ] w = Δ w χ r 0 + w Δ χ r 0 + 2 w . χ r 0 . Then, for ν ( 1 , 0 ) and λ 1 small enough, we obtain

˜ ( 0 , 0 ) C ν 2 0 , α ( B r 0 ( z ˜ 1 ) ) sup r 0 2 r r 0 r 2 ν ˜ ( 0 , 0 ) c κ r ε , λ 2 + c κ λ .

In B r 0 2 ( z ˜ 2 ) \ B r ε , λ ( z ˜ 2 ) , we have χ r 0 ( z z ˜ 1 ) = 0 , χ r 0 ( z z ˜ 2 ) = 1 , χ r 0 ( z z ˜ 3 ) = 0 and Δ w ˜ 2 = 0 , so that ˜ ( 0 , 0 ) = ρ 2 e ξ w ˜ 2 + ( 1 ξ ) w ˜ 1 + λ w ˜ 2 2 . Then

˜ ( 0 , 0 ) c κ ε 2 r 4 ( 1 + λ 2 ) + c κ λ ( ( 1 + λ 2 ) ( r 1 + log r ) + r ε , λ 2 r 2 ) 2 .

Therefore, for ν ( 1 , 0 ) and λ 2 small enough, we obtain

˜ ( 0 , 0 ) C ν 2 0 , α ( B r 0 ( z ˜ 2 ) ) sup r ε r r 0 2 r 2 ν ˜ ( 0 , 0 ) c κ r ε , λ 2 + c κ λ .

In B r 0 ( z ˜ 2 ) \ B r 0 2 ( z ˜ 2 ) , using the estimate (90), there holds

˜ ( 0 , 0 ) c κ ε 2 r 4 ( 1 + λ 2 ) + c κ λ ( ( 1 + λ 2 ) ( r 1 + log r ) + r ε , λ 2 r 2 ) 2 + [ Δ , χ r 0 ( z z ˜ 2 ) ] H ext ( φ ˜ 2 2 ; ( z z ˜ 2 ) r ε , λ ) c κ ε 2 r 4 ( 1 + λ 2 ) + c κ λ ( ( 1 + λ 2 ) ( r 1 + log r ) + r ε , λ 2 r 2 ) 2 + c κ r 1 r ε , λ 3 ,

where [ Δ , χ r 0 ] w = Δ w χ r 0 + w Δ χ r 0 + 2 w . χ r 0 . Consequently, for ν ( 1 , 0 ) and λ 2 small enough, we obtain

˜ ( 0 , 0 ) C ν 2 0 , α ( B r 0 ( z ˜ 2 ) ) sup r 0 2 r r 0 r 2 ν ˜ ( 0 , 0 ) c κ ε 2 r ε , λ 2 + c κ λ .

In B r 0 2 ( z ˜ 3 ) \ B r ε , λ ( z ˜ 3 ) , we have χ r 0 ( z z ˜ 3 ) = 1 , χ r 0 ( z z ˜ 1 ) = 0 , χ r 0 ( z z ˜ 2 ) = 0 and Δ w ˜ 2 = 0 , so that ˜ ( 0 , 0 ) = ρ 2 e ξ w ˜ 2 + ( 1 ξ ) w ˜ 1 + λ w ˜ 2 2 . Then

˜ ( 0 , 0 ) c κ ε 2 z z 3 ˜ 4 ( 1 + λ 3 ) + c κ λ 1 + λ 3 ξ ( r 1 + log r ) + r ε , λ 2 r 2 2 c κ ε 2 r 4 ( 1 + λ 3 ) + c κ λ 1 + λ 3 ξ ( r 1 + log r ) + r ε , λ 2 r 2 2 .

Accordingly, for ν ( 1 , 0 ) and λ 3 small enough, we obtain

˜ ( 0 , 0 ) C ν 2 0 , α ( B r 0 ( z ˜ 3 ) ) sup r ε , λ r r 0 2 r 2 ν ˜ ( 0 , 0 ) c κ r ε , λ 2 + c κ λ .

In B r 0 ( z ˜ 3 ) \ B r 0 2 ( z ˜ 3 ) , using the estimate (90), there holds

˜ ( 0 , 0 ) c κ ε 2 r 4 ( 1 + λ 3 ) + c κ λ 1 + λ 3 ξ ( r 1 + log r ) + r ε , λ 2 r 2 2 + [ Δ , χ r 0 ( z z ˜ 3 ) ] H ext ( φ ˜ 2 3 ; ( z z ˜ 3 ) r ε , λ ) c κ ε 2 + c κ λ 1 + λ 3 ξ ( r 1 + log r ) + r ε , λ 2 r 2 2 + c κ r 1 r ε , λ 3 ,

where [ Δ , χ r 0 ] w = Δ w χ r 0 + w Δ χ r 0 + 2 w . χ r 0 . Therefore, for ν ( 1 , 0 ) and λ 3 small enough, we obtain

˜ ( 0 , 0 ) C ν 2 0 , α ( B r 0 ( z ˜ 3 ) ) sup r 0 2 r r 0 r 2 ν ˜ ( 0 , 0 ) c κ r ε , λ 2 + c κ λ .

In Ω \ i = 1 , 2 , 3 B ¯ r 0 ( z ˜ i ) , we have χ r 0 ( z z ˜ i ) = 0 for i = 1 , 2 , 3 and Δ w ˜ 2 = 0 . Thus,

˜ ( 0 , 0 ) c κ ε 2 .

So for ν ( 1 , 0 ) , we have

˜ ( 0 , 0 ) C ν 2 0 , α ( Ω ¯ i = 1 3 B r 0 ( z i ˜ ) ) sup r 0 r r 2 ν ˜ ( 0 , 0 ) c κ r ε , λ 2 .

Then, ˜ ( 0 , 0 ) C ν 2 0 , α ( Ω ¯ r 0 ( z ˜ ) ) c κ r ε , λ 2 .

By using Proposition 3 and the inequality (84), we conclude that

(91) N ˜ ( 0 , 0 ) C ν 2 , α ( Ω ¯ * ( z ˜ ) ) c κ r ε , λ 2 , ˜ ( 0 , 0 ) C ν 2 , α ( Ω ¯ * ( z ˜ ) ) c κ r ε , λ 2 .

For proving of the third estimate, let v ˜ 1 , v ˜ 2 , v ˜ 1 and v ˜ 2 C ν 2 , α ( Ω ¯ * ) satisfy (89), we have

T ˜ ( v ˜ 1 , v ˜ 2 ) T ˜ ( v ˜ 1 , v ˜ 2 ) c κ ε 2 e γ w ˜ 1 + ( 1 γ ) w ˜ 2 e γ v ˜ 1 + ( 1 γ ) v ˜ 2 e γ v ˜ 1 + ( 1 γ ) v ˜ 2 + c κ λ ( v ˜ 1 v ˜ 1 ) ( v ˜ 1 + v ˜ 1 + 2 w ˜ 1 ) c κ ε 2 ( v ˜ 1 v ˜ 1 + v ˜ 1 v ˜ 1 ) + c κ λ ( v ˜ 1 v ˜ 1 ) ,

and

˜ ( v ˜ 1 , v ˜ 2 ) ˜ ( v ˜ 1 , v ˜ 2 ) c κ ε 2 e ξ w ˜ 2 + ( 1 ξ ) w ˜ 1 e ξ v ˜ 2 + ( 1 ξ ) v ˜ 1 e ξ v ˜ 2 + ( 1 ξ ) v ˜ 1 + c κ λ ( v ˜ 2 v ˜ 2 ) ( v ˜ 2 + v ˜ 2 + 2 w ˜ 2 ) c κ ε 2 ( v ˜ 2 v ˜ 2 + v ˜ 2 v ˜ 2 ) + c κ λ ( v ˜ 2 v ˜ 2 ) .

Then, for i = 1 , 2 , 3 , λ i small enough, using the estimate (84), there exist c ¯ κ (depending on κ ) such that

(92) N ˜ ( v ˜ 1 , v ˜ 2 ) N ˜ ( v ˜ 1 , v ˜ 2 ) C ν 2 , α ( Ω ¯ * ( z ˜ ) ) c ¯ κ r ε , λ 2 ( v ˜ 1 v ˜ 1 C ν 2 , α ( Ω ¯ * ( z ˜ ) ) + v ˜ 2 v ˜ 2 C ν 2 , α ( Ω ¯ * ( z ˜ ) ) ) ,

and

(93) ˜ ( v ˜ 1 , v ˜ 2 ) ˜ ( v ˜ 1 , v ˜ 2 ) C ν 2 , α ( Ω ¯ * ( z ˜ ) ) c ¯ κ r ε , λ 2 ( v ˜ 1 v ˜ 1 C ν 2 , α ( Ω ¯ * ( z ˜ ) ) + v ˜ 2 v ˜ 2 C ν 2 , α ( Ω ¯ * ( z ˜ ) ) ) .

The proof is completed.□

Reducing ε κ and λ κ if necessary, we can assume that c ¯ κ r ε , λ 2 1 2 for all ε ( 0 , ε κ ) and λ ( 0 , λ κ ) . Then, (92) and (93) suffice to show that

( v ˜ 1 , v ˜ 2 ) ( N ˜ ( v ˜ 1 , v ˜ 2 ) , ˜ ( v ˜ 1 , v ˜ 2 ) )

is a contraction from the ball

{ ( v ˜ 1 , v ˜ 2 ) ( C ν 2 , α ( R 2 ) ) 2 : ( v ˜ 1 , v ˜ 2 ) ( C ν 2 , α ( R 2 ) ) 2 2 c ¯ κ r ε , λ 2 }

into itself. Hence, there exists a unique fixed point ( v ˜ 1 , v ˜ 2 ) in this set, which is a solution of (85).

We summarize the previous results to derive the following proposition.

Proposition 10

Given κ > 0 , there exist ε κ > 0 (depending on κ ) such that for any ε ( 0 , ε κ ) , λ i , and z ˜ i satisfying (88), and any functions ( φ ˜ 1 i , φ ˜ 2 i ) satisfying (57) and (87), there exists a unique ( v ˜ 1 , v ˜ 2 ) ( ( v ˜ 1 , ε , λ , z ˜ , φ ˜ , v ˜ 2 , ε , λ , z ˜ , φ ˜ ) ) solution of (85) so that for ( v ˜ 1 , v ˜ 2 ) defined by

w ˜ 1 1 + λ 1 γ G ( , z ˜ 1 ) + ( 1 + λ 2 ) G ( , z ˜ 2 ) + i = 1 3 χ r 0 ( z ˜ i ) H ext ( φ ˜ 1 i ; ( z ˜ i ) r ε , λ ) + v ˜ 1 w ˜ 2 1 + λ 3 ξ G ( , z ˜ 3 ) + ( 1 + λ 2 ) G ( , z ˜ 2 ) + i = 1 3 χ r 0 ( z ˜ i ) H ext ( φ ˜ 2 i ; ( z ˜ i ) r ε , λ ) + v ˜ 2 ,

solve (82) in Ω ¯ r ε , λ ( z ˜ ) . In addition, we have ( v ˜ 1 , v ˜ 2 ) C ν 2 , α ( Ω ¯ * ( z ˜ ) ) 2 c ¯ κ r ε , λ 2 .

3.4 The nonlinear Cauchy-data matching

We will gather the results of previous sections. By using the previous notations, we assume that z ˜ ( z ˜ 1 , z ˜ 2 , z ˜ 3 ) Ω 3 are given close to z ( z 1 , z 2 , z 3 ) . Assume also that

τ ( τ 1 , τ 2 , τ 3 ) [ τ 1 , τ 1 + ] × [ τ 2 , τ 2 + ] × [ τ 3 , τ 3 + ] ( 0 , ) 3

are given (the values of τ i and τ i + will be fixed later). First, we consider some set of boundary data φ i ( φ 1 i , φ 2 i ) ( C 2 , α ( S 1 ) ) 2 , for i = 1, 2, 3. According to propositions 7, 8, 9 and provided ε ( 0 , ε κ ) and λ ( 0 , λ κ ) , we can find a solution u int ( u int , 1 , u int , 2 ) of

(94) Δ u 1 = ρ 2 e γ u 1 + ( 1 γ ) u 2 + λ u 1 2 , Δ u 2 = ρ 2 e ξ u 2 + ( 1 ξ ) u 1 + λ u 2 2 ,

in B r ε , λ ( z ˜ 1 ) B r ε , λ ( z ˜ 2 ) B r ε , λ ( z ˜ 3 ) , which can be decomposed as follows:

u int , 1 ( z ) 1 γ u ε , τ 1 ( z z 1 ˜ ) 1 γ γ G ( z , z 2 ˜ ) 1 γ γ ξ G ( z , z 3 ˜ ) log γ γ + H φ 1 1 int z z ˜ 1 r ε , λ + h 1 1 R ε , λ 1 ( z z ˜ 1 ) r ε , λ + v 1 1 R ε , λ 1 ( z z ˜ 1 ) r ε , λ in B r ε , λ ( z ˜ 1 ) u ε , τ 2 ( z z 2 ˜ ) + H φ 1 2 int z z ˜ 2 r ε , λ + h 1 2 R ε , λ 2 ( z z ˜ 2 ) r ε , λ + v 1 2 R ε , λ 2 ( z z ˜ 2 ) r ε , λ in B r ε , λ ( z ˜ 2 ) 1 γ G ( z , z 1 ˜ ) + G ( z , z 2 ˜ ) + H φ 1 3 int z z ˜ 3 r ε , λ + h 1 3 R ε , λ 3 ( z z ˜ 3 ) r ε , λ + v 1 3 R ε , λ 3 ( z z ˜ 3 ) r ε , λ in B r ε , λ ( z ˜ 3 ) ,

and

u int , 2 ( z ) 1 ξ G ( z , z 3 ˜ ) + G ( z , z 2 ˜ ) + H φ 2 1 int z z ˜ 1 r ε , λ + h 2 1 R ε , λ 1 ( z z ˜ 1 ) r ε , λ + v 2 1 R ε , λ 1 ( z z ˜ 1 ) r ε , λ in B r ε , λ ( z ˜ 1 ) u ε , τ 2 ( z z 2 ˜ ) + H φ 2 2 int z z ˜ 2 r ε , λ + h 2 2 R ε , λ 2 ( z z ˜ 2 ) r ε , λ + v 2 2 R ε , λ 2 ( z z ˜ 2 ) r ε , λ in B r ε , λ ( z ˜ 2 ) 1 ξ u ε , τ 3 ( z z 3 ˜ ) 1 ξ ξ G ( z , z 2 ˜ ) 1 ξ γ ξ G ( z , z 1 ˜ ) log ξ ξ + H φ 2 3 int z z ˜ 3 r ε , λ + h 2 3 R ε , λ 3 ( z z ˜ 3 ) r ε , λ + v 2 3 R ε , λ 3 ( z z ˜ 3 ) r ε , λ in B r ε , λ ( z ˜ 3 ) ,

where, for i = 1, 2, 3 and j = 1 , 2 , H φ j i int z z ˜ i r ε , λ C 2 2 , α ( B ¯ 1 * ) c κ r ε , λ 2 , R ε , λ i = τ i r ε , λ ε , the functions h j i and v j i satisfying

( h 1 1 , h 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) 2 c κ r ε , λ 2 , ( h 1 2 , h 2 2 ) ( C μ 2 , α ( R 2 ) ) 2 2 c κ r ε , λ 2 , ( h 1 3 , h 2 3 ) C δ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 , ( v 1 1 , v 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) 2 c κ r ε , λ 2 , ( v 1 2 , v 2 2 ) ( C μ 2 , α ( R 2 ) ) 2 2 c κ r ε , λ 2 , and ( v 1 3 , v 2 3 ) C δ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 .

Similarly, given boundary data ( φ ˜ 1 i , φ ˜ 2 i ) ( C 2 , α ( S 1 ) ) 2 ( i = 1 , 2 , 3 ) verifying (57), ( λ 1 , λ 2 , λ 3 ) R 3 satisfying (88), provided ε ( 0 , ε κ ) . By Proposition 10, we find a solution u ext ( u ext , 1 , u ext , 2 ) of (94) in Ω ¯ \ i = 1 , 2 , 3 B r ε , λ ( z ˜ i ) , which can be decomposed as follows:

u ext , 1 ( z ) 1 + λ 1 γ G ( z , z ˜ 1 ) + ( 1 + λ 2 ) G ( z , z ˜ 2 ) + i = 1 3 χ r 0 ( z z ˜ i ) H ext ( φ ˜ 1 i ; ( z z ˜ i ) r ε , λ ) + v 1 ˜ ( z ) u ext , 2 ( z ) 1 + λ 3 ξ G ( z , z ˜ 3 ) + ( 1 + λ 2 ) G ( z , z ˜ 2 ) + i = 1 3 χ r 0 ( z z ˜ i ) H ext ( φ ˜ 2 i ; ( z z ˜ i ) r ε , λ ) + v 2 ˜ ( z ) ,

with v ˜ 1 , v ˜ 2 C ν 2 , α ( Ω ¯ * ( z ˜ ) ) satisfying ( v ˜ 1 , v ˜ 2 ) ( C ν 2 , α ( Ω ¯ * ( z ˜ ) ) ) 2 2 c ¯ κ r ε , λ 2 . It remains to determine the parameters and the boundary data such that the function equals u int in i = 1 , 2 , 3 B r ε , λ ( z ˜ i ) and equals u ext in Ω ¯ r ε , λ ( z ˜ ) is a smooth function. This amounts to find the boundary data and the parameters so that for j = 1 , 2 ,

(95) u int , j = u ext , j and r u int , j = r u ext , j

on B r ε , λ ( z ˜ 1 ) , B r ε , λ ( z ˜ 2 ) and B r ε , λ ( z ˜ 3 ) .

Suppose equation (95) is satisfied, indicating that for sufficiently small ε , u ε , λ belongs to C 2 , α (obtained by combining the functions u int and the function u ext ), serving as a weak solution to our system. Furthermore, by elliptic regularity theory, we deduce that this solution is indeed smooth. This conclusion finalizes our proof because, as ε and λ tend to 0, the obtained sequence of solutions satisfies the required singular limit behaviors.

Before advancing, let’s address the following remarks. First, it is worth noting that the function u ε , τ i can be expanded as follows:

(96) u ε , τ i ( z ) = 2 log τ i 4 log z + O ε 2 τ i 2 z 2 , z B r ε , λ ( 0 ) .

On B r ε , λ ( z ˜ 1 ) , we have

(97) ( u int , 1 u ext , 1 ) ( z ) = 2 γ log τ 1 + 4 λ 1 γ log z z ˜ 1 1 γ γ ξ G ( z , z 3 ˜ ) log γ γ + h 1 1 ( R ε , λ 1 ( z z ˜ 1 ) r ε , λ ) + v 1 1 ( R ε , λ 1 ( z z ˜ 1 ) r ε , λ ) + H int ( φ 1 1 , ( z z ˜ 1 ) r ε , λ ) H ext ( φ ˜ 1 1 ; ( z z ˜ 1 ) r ε , λ ) v ˜ 1 ( z ) 1 + λ 1 γ H ( z , z ˜ 1 ) ( 1 + λ 2 + 1 γ γ ) G ( z , z ˜ 2 ) + O ε 2 τ 1 2 z z ˜ 1 2 + O ( r ε , λ 2 ) .

Next, although all functions are defined on B r ε , λ ( z ˜ 1 ) in (95), it will be more convenient to solve the following set of equations on S 1 :

(98) ( u int , 1 u ext , 1 ) ( z ˜ 1 + r ε , λ . ) = 0 and r ( u int , 1 u ext , 1 ) ( z ˜ 1 + r ε , λ . ) = 0 .

Since the boundary data are chosen to satisfy either (57) or (56). We make the following decomposition:

φ 1 1 = φ 1 , 0 1 + φ 1,1 1 + φ 1 , 1 and φ ˜ 1 1 = φ ˜ 1 , 0 1 + φ ˜ 1,1 1 + φ ˜ 1 , 1 ,

where φ 1 , 0 1 , φ ˜ 1 , 0 1 E 0 = R are constant on S 1 , φ 1,1 1 and φ ˜ 1,1 1 belong to E 1 = Span { e 1 , e 2 } and φ 1 , 1 , φ ˜ 1 , 1 are L 2 ( S 1 ) orthogonal to E 0 and E 1 .

Using equality (97), for z S 1 , we have

(99) ( u int , 1 u ext , 1 ) ( z ˜ 1 + r ε , λ z ) = 2 γ log τ 1 + 4 λ 1 γ log r ε , λ z 1 γ H ( z ˜ 1 , z ˜ 1 ) + G ( z ˜ 1 , z ˜ 2 ) + 1 γ ξ G ( z 1 ˜ , z ˜ 3 ) log γ γ λ 1 γ H ( z ˜ 1 , z ˜ 1 ) λ 2 G ( z ˜ 1 , z ˜ 2 ) + O ( r ε , λ 2 ) .

Then, the projection of the equations (98) over E 0 yield

(100) 4 λ 1 + O ( r ε , λ 2 ) = 0 , 2 log τ 1 + 4 λ 1 log r ε , λ log γ 1 ( z ˜ 1 , z ˜ ) + O ( r ε , λ 2 ) = 0 ,

where

1 ( , z ˜ ) H ( , z ˜ 1 ) + G ( , z ˜ 2 ) + 1 γ ξ G ( , z ˜ 3 ) .

The system (100) can be simply written as follows:

1 log r ε , λ [ 2 log τ 1 + log γ + 1 ( z ˜ 1 , z ˜ ) ] = O ( r ε , λ 2 ) and λ 1 = O ( r ε , λ 2 ) .

We are now in a position to define τ 1 and τ 1 + . In fact, according to the above analysis, as ε and λ tend to 0, we expect that z ˜ i will converge to z i for i { 1 , 2 , 3 } and τ 1 will converge to τ 1 * satisfying

2 log τ 1 * = log γ 1 ( z 1 , z ) .

Hence, it is enough to choose τ 1 and τ 1 + such that

2 log ( τ 1 ) < log γ 1 ( z 1 , z ) < 2 log ( τ 1 + ) .

Consider now the projection of (98) over E 1 . Given a smooth function f defined in Ω , we identify its gradient f = ( z 1 f , z 2 f ) with the following element of E 1 :

¯ f = i = 1 2 z i f e i .

With these notations in mind, we obtain the system

(101) ¯ 1 ( z ˜ 1 , z ˜ ) = O ( r ε , λ 2 ) , φ 1,1 1 = O ( r ε , λ 2 ) .

Finally, we consider the projection onto ( L 2 ( S 1 ) ) . This yields the system

φ 1 , 1 φ ˜ 1 , 1 + O ( r ε , λ 2 ) = 0 , r ( H 1 , int , 1 H 1 , ext ) + O ( r ε , λ 2 ) = 0 .

Thanks to Lemma 5, the above system can be rewritten as follows:

φ 1 , 1 = O ( r ε , λ 2 ) and φ ˜ 1 , 1 = O ( r ε , λ 2 ) .

If we define the parameters t 1 R by

t 1 = 1 log r ε , λ [ 2 log τ 1 + log γ + 1 ( z ˜ 1 , z ˜ ) ] ,

then the system we need to solve reads

(102) T 1 , ε 1 = ( t 1 , λ 1 , φ 1 , 0 1 , φ ˜ 1 , 0 1 , φ 1,1 1 , φ ˜ 1,1 1 , ¯ 1 ( z ˜ 1 , z ˜ ) , φ 1 , 1 , φ ˜ 1 , 1 ) = O ( r ε , λ 2 ) .

As usual, the terms O ( r ε , λ 2 ) depend nonlinearly on all the variables on the left side, but is bounded (in the appropriate norm) by a constant (independent of ε and κ ) times r ε , λ 2 , provided ε ( 0 , ε κ ) and λ ( 0 , λ κ ) . On B r ε , λ ( z ˜ 1 ) , we have

( u int , 2 u ext , 2 ) ( z ) = λ 3 ξ G ( z , z ˜ 3 ) + G ( z , z ˜ 2 ) + h 2 1 ( R ε , λ 1 ( z z ˜ 1 ) r ε , λ ) + H 2 int , 1 ( φ 2 1 , ( z z ˜ 1 ) r ε , λ ) ( 1 + λ 2 ) G ( z , z ˜ 2 ) H 2 ext ( φ ˜ 2 1 , ( z z ˜ 1 ) r ε , λ ) + O ( r ε , λ 2 ) .

By using the fact that

φ 2 1 = φ 2,0 1 + φ 2,1 1 + φ 2 , 1 and φ ˜ 2 1 = φ ˜ 2,0 1 + φ ˜ 2,1 1 + φ ˜ 2 , 1

with φ 2,0 1 , φ ˜ 2,0 1 E 0 , φ 2,1 2 , φ ˜ 2,1 1 E 1 and φ 2 , 1 , φ ˜ 2 , 1 ( L 2 ( S 1 ) ) , we can prove that

( u int , 2 u ext , 2 ) ( z ˜ 1 + r ε , λ z ) and ( r u int , 2 r u ext , 2 ) ( z ˜ 1 + r ε , λ z ) on B r ε , λ ( z ˜ 1 )

yield to

(103) T 2 , ε 1 = ( φ 2,0 1 , φ ˜ 2,0 1 , φ 2,1 1 , φ ˜ 2,1 1 , φ 2 , 1 , φ ˜ 2 , 1 ) = O ( r ε , λ 2 ) .

On B r ε , λ ( z ˜ 2 ) , we obtain

(104) ( 1 ξ ) ( u int , 1 u ext , 1 ) ( z ) + ( 1 γ ) ( u int , 2 u ext , 2 ) ( z ) = 2 ( 2 γ ξ ) log τ 2 + 4 ( 2 γ ξ ) λ 2 log z z ˜ 2 + ( 1 ξ ) h 1 2 ( R ε , λ 2 ( z z ˜ 2 ) r ε , λ ) + ( 1 γ ) h 2 2 ( R ε , λ 2 ( z z ˜ 2 ) r ε , λ ) + ( 1 ξ ) v 1 2 ( R ε , λ 2 ( z z ˜ 2 ) r ε , λ ) + ( 1 γ ) v 2 2 ( R ε , λ 2 ( z z ˜ 2 ) r ε , λ ) + ( 1 ξ ) H int ( φ 1 2 , ( z z ˜ 2 ) r ε , λ ) + ( 1 γ ) H int ( φ 2 2 , ( z z ˜ 2 ) r ε , λ ) ( 1 ξ ) H ext ( φ ˜ 1 2 , ( z z ˜ 2 ) r ε , λ ) ( 1 γ ) H ext ( φ ˜ 2 2 , ( z z ˜ 2 ) r ε , λ ) ( 2 γ ξ ) H ( z , z ˜ 2 ) + 1 ξ γ G ( z , z ˜ 1 ) + 1 γ ξ G ( z , z ˜ 3 ) + O ε 2 τ 2 2 z z ˜ 2 2 + O ( r ε , λ 2 ) .

We denote

φ 2 = ( 1 ξ ) φ 1 2 + ( 1 γ ) φ 2 2 , φ ˜ 2 = ( 1 ξ ) φ ˜ 1 2 + ( 1 γ ) φ ˜ 2 2 h 2 = ( 1 ξ ) h 1 2 + ( 1 γ ) h 2 2 , v 2 = ( 1 ξ ) v 1 2 + ( 1 γ ) v 2 2 .

Then, we have

(105) ( 1 ξ ) ( u int , 1 u ext , 1 ) ( z ) + ( 1 γ ) ( u int , 2 u ext , 2 ) ( z ) = 2 ( 2 γ ξ ) log τ 2 + 4 ( 2 γ ξ ) λ 2 log z z ˜ 2 + h 2 ( R ε , λ 2 ( z z ˜ 2 ) r ε , λ ) + v 2 ( R ε , λ 2 ( z z ˜ 2 ) r ε , λ ) + H int ( φ 2 , ( z z ˜ 2 ) r ε , λ ) H ext ( φ ˜ 2 , ( z z ˜ 2 ) r ε , λ ) ( 2 γ ξ ) H ( z , z ˜ 2 ) + 1 ξ γ G ( z , z ˜ 1 ) + 1 γ ξ G ( z , z ˜ 3 ) + O ε 2 τ 2 2 z z ˜ 2 2 + O ( r ε , λ 2 ) .

Next, despite all functions being defined on B r ε ( z ˜ 2 ) in (95), it will be more convenient to solve the following set of equations on S 1 .

(106) ( ( 1 ξ ) ( u int , 1 u ext , 1 ) + ( 1 γ ) ( u int , 2 u ext , 2 ) ) ( z ˜ 2 + r ε , λ . ) = 0 r ( ( 1 ξ ) ( u int , 1 u ext , 1 ) + ( 1 γ ) ( u int , 2 u ext , 2 ) ) ( z ˜ 2 + r ε , λ . ) = 0 .

Since the boundary data are chosen to satisfy (57) or (56). We decompose as follows:

φ 2 = φ 0 2 + φ 1 2 + φ 2 and φ ˜ 2 = φ ˜ 0 2 + φ ˜ 1 2 + φ ˜ 2 ,

where φ 0 2 , φ ˜ 0 2 E 0 = R are constant on S 1 , φ 1 2 , φ ˜ 1 2 belong to E 1 = Span { e 1 , e 2 } and φ 2 , φ ˜ 2 are L 2 ( S 1 ) orthogonal to E 0 and E 1 .

We insist that, for z S 1 , both equations (105) involve the same relationship between the parameter τ 2 and the appropriate energy 2 ,

(107) ( ( 1 ξ ) ( u int , 1 u ext , 1 ) + ( 1 γ ) ( u int , 2 u ext , 2 ) ) ( z ˜ 2 + r ε , λ z ) = 2 ( 2 γ ξ ) log τ 2 + 4 λ 2 ( 2 γ ξ ) log r ε , λ z ( 2 γ ξ ) H ( z , z ˜ 2 ) + 1 ξ γ G ( z , z ˜ 1 ) + 1 γ ξ G ( z , z ˜ 3 ) + O ( r ε , λ 2 ) .

Then, the projection of the equations (106) over E 0 results in

(108) 4 ( 2 γ ξ ) λ 2 + O ( r ε , λ 2 ) = 0 , 2 ( 2 γ ξ ) log τ 2 + 4 ( 2 γ ξ ) λ 2 log r ε , λ 2 ( z ˜ 2 , z ˜ ) + O ( r ε , λ 2 ) = 0 ,

where

2 ( , z ˜ ) ( 2 γ ξ ) H ( , z ˜ 2 ) + 1 ξ γ G ( , z ˜ 1 ) + 1 γ ξ G ( , z ˜ 3 ) .

The system (108) can be simplified to:

(109) 1 log r ε , λ 2 log τ 2 + 2 ( z ˜ 2 , z ˜ ) 2 γ ξ = O ( r ε , λ 2 ) and λ 2 = O ( r ε , λ 2 ) .

We are now in a position to define τ 2 and τ 2 + . In fact, on the basis of the preceding analysis, as ε and λ tend to 0, we expect that ( z ˜ 1 , z ˜ 2 , z ˜ 3 ) will converge to ( z 1 , z 2 , z 3 ) and τ 2 will converge to τ 2 * , satisfying

2 log τ 2 * = 2 ( z 2 , z ) 2 γ ξ .

Hence, it is enough to choose τ 2 and τ 2 + such that

2 log ( τ 2 ) < 2 ( z 2 , z ) 2 γ ξ < 2 log ( τ 2 + ) .

Consider now the projection of (106) over E 1 . Given a smooth function f defined in Ω , we identify its gradient f = ( z 1 f , z 2 f ) with the element of E 1 , as follows:

¯ f = i = 1 2 z i f e i .

Taking into account these notations, we derive the system

(110) ¯ 2 ( z ˜ 2 , z ˜ ) = O ( r ε , λ 2 ) , φ 1 2 = O ( r ε , λ 2 ) .

Finally, we consider the projection onto ( L 2 ( S 1 ) ) . This results in the system

φ 2 φ ˜ 2 + O ( r ε , λ 2 ) = 0 , r ( H int , 2 H ext ) + O ( r ε , λ 2 ) = 0 .

By Lemma 5, we can express this final system as follows:

φ 2 = O ( r ε , λ 2 ) and φ ˜ 2 = O ( r ε , λ 2 ) .

If we define the parameters t 2 R by

t 2 = 1 log r ε , λ 2 log τ 2 + 2 ( z ˜ 2 , z ˜ ) 2 γ ξ .

Thus, the system to be solved can be restated as follows:

(111) T c , ε 2 = ( t 2 , λ 2 , φ 0 2 , φ ˜ 0 2 , φ 1 2 , φ ˜ 1 2 , ¯ 2 ( z ˜ 2 , z ˜ ) , φ 2 , φ ˜ 2 ) = O ( r ε , λ 2 ) ,

where the terms O ( r ε , λ 2 ) exhibit nonlinear dependence on all variables on the left side. However, they are bounded (in the appropriate norm) by a constant (independent of ε and κ ) multiplied by r ε , λ 2 , given that ε ( 0 , ε κ ) and λ ( 0 , λ κ ) . On B r ε , λ ( z ˜ 3 ) , we obtain

( u int , 1 u ext , 1 ) ( z ) = λ 1 γ G ( z , z ˜ 1 ) λ 2 G ( z , z ˜ 2 ) + h 1 3 ( R ε , λ 3 ( z z ˜ 3 ) r ε , λ ) + v 1 3 ( R ε , λ 3 ( z z ˜ 3 ) r ε , λ ) + H int ( φ 1 3 , ( z z ˜ 3 ) r ε , λ ) H ext ( φ ˜ 1 3 , ( z z ˜ 3 ) r ε , λ ) + O ( r ε , λ 2 ) .

We use the fact that

φ 1 3 = φ 1 , 0 3 + φ 1,1 3 + φ 1 , 3 and φ ˜ 1 3 = φ ˜ 1 , 0 3 + φ ˜ 1,1 3 + φ ˜ 1 , 3 ,

with φ 1 , 0 3 , φ ˜ 1 , 0 3 E 0 , φ 1,1 3 , φ ˜ 1,1 3 E 1 and φ 1 , 3 , φ ˜ 1 , 3 ( L 2 ( S 1 ) ) . We can prove that

( u int , 1 u ext , 1 ) ( z ˜ 3 + r ε , λ z ) and ( r u int , 1 r u ext , 1 ) ( z ˜ 3 + r ε , λ z ) on B r ε , λ ( z ˜ 3 )

implies that

(112) T 1 , ε 3 = ( φ 1 , 0 3 , φ ˜ 1 , 0 3 , φ 1,1 3 , φ ˜ 1,1 3 , φ 1 , 3 , φ ˜ 1 , 3 ) = O ( r ε , λ 2 ) .

On B r ε , λ ( z ˜ 3 ) , we have

(113) ( u int , 2 u ext , 2 ) ( z ) = 2 ξ log τ 3 + 4 λ 3 ξ log z z ˜ 3 1 ξ γ ξ G ( z , z 1 ˜ ) log ξ ξ + h 2 3 ( R ε , λ 3 ( z z ˜ 3 ) r ε , λ ) + v 2 3 ( R ε , λ 3 ( z z ˜ 3 ) r ε , λ ) + H int ( φ 2 3 , ( z z ˜ 3 ) r ε , λ ) H ext ( φ ˜ 2 3 ; ( z z ˜ 3 ) r ε , λ ) 1 + λ 3 ξ H ( z , z ˜ 3 ) ( 1 + λ 2 + 1 ξ ξ ) G ( z , z ˜ 2 ) + O ε 2 τ 3 2 z z ˜ 2 2 + O ( r ε , λ 2 ) .

Alternatively, although all functions are defined on the boundary of B r ε , λ ( z ˜ 3 ) in (95), it will be more convenient to solve on S 1 the following set of equations

(114) ( u int , 2 u ext , 2 ) ( z ˜ 3 + r ε , λ . ) = 0 and r ( u int , 2 u ext , 2 ) ( z ˜ 3 + r ε , λ . ) = 0 .

Given that the boundary data are chosen to fulfill (57) or (56). We decompose as follows:

φ 2 3 = φ 2,0 3 + φ 2,1 3 + φ 2 , 3 and φ ˜ 2 3 = φ ˜ 2,0 3 + φ ˜ 2,1 3 + φ ˜ 2 , 3 ,

where φ 2,0 3 , φ ˜ 2,0 3 E 0 = R are constant on S 1 , φ 2,1 3 , φ ˜ 2,1 3 belong to E 1 = Span { e 1 , e 2 } and φ 2 , 3 , φ ˜ 2 , 3 are L 2 ( S 1 ) orthogonal to E 0 and E 1 .

By using (113), we have for z S 1

(115) ( u int , 2 u ext , 2 ) ( z ˜ 3 + r ε , λ z ) = 2 ξ log τ 3 + 4 λ 3 ξ log r ε , λ z 1 ξ H ( z ˜ 3 , z ˜ 3 ) + G ( z ˜ 3 , z ˜ 2 ) + 1 ξ γ G ( z ˜ 3 , z ˜ 1 ) log ξ ξ λ 3 ξ H ( z ˜ 3 , z ˜ 3 ) λ 2 G ( z ˜ 3 , z ˜ 2 ) + O ( r ε , λ 2 ) .

Then, the projection of the equations (114) over E 0 yields

(116) 4 λ 3 + O ( r ε , λ 2 ) = 0 , 2 log τ 3 + 4 λ 3 log r ε , λ log ξ 1 ( z ˜ 3 , z ˜ ) + O ( r ε , λ 2 ) = 0 ,

where

3 ( , z ˜ ) H ( , z ˜ 3 ) + G ( , z ˜ 2 ) + 1 ξ γ G ( , z ˜ 1 ) .

The system (116) can be simply expressed as follows:

1 log r ε , λ [ 2 log τ 3 + log ξ + 3 ( z ˜ 3 , z ˜ ) ] = O ( r ε , λ 2 ) and λ 3 = O ( r ε , λ 2 ) .

We are now able to define τ 3 and τ 3 + . Indeed, on the basis of the above analysis, as ε and λ tend to 0, we expect that z ˜ i will converge to z i for i { 1 , 2 , 3 } and τ 3 will converge to τ 3 * satisfying

2 log τ 3 * = log ξ 3 ( z 3 , z ) .

Therefore, it suffices to select τ 3 and τ 3 + in such a way that

2 log ( τ 3 ) < log ξ 3 ( z 3 , z ) < 2 log ( τ 3 + ) .

Consider now the projection of (114) over E 1 . Given a smooth function f defined in Ω , we identify its gradient f = ( z 1 f , z 2 f ) with the element of E 1

¯ f = i = 1 2 z i f e i .

Taking these notations into consideration, we derive the system.

(117) ¯ 3 ( z ˜ 3 , z ˜ ) = O ( r ε , λ 2 ) , φ 2,1 3 = O ( r ε , λ 2 ) .

Finally, we consider the projection onto ( L 2 ( S 1 ) ) . This yields the system

φ 2 , 3 φ ˜ 2 , 3 + O ( r ε , λ 2 ) = 0 , r ( H 2 , int , 3 H 2 , ext ) + O ( r ε , λ 2 ) = 0 .

Thanks to Lemma 5, we can rewrite this final system as follows:

φ 2 , 3 = O ( r ε , λ 2 ) and φ ˜ 2 , 3 = O ( r ε , λ 2 ) .

If we define the parameters t 3 R by

t 3 = 1 log r ε , λ [ 2 log τ 3 + log ξ + 3 ( z ˜ 3 , z ˜ ) ] ,

then the system we need to solve can be stated as follows:

(118) T 2 , ε 3 = ( t 3 , λ 3 , φ 2,0 3 , φ ˜ 2,0 3 , φ 2,1 3 , φ ˜ 2,1 3 , ¯ 3 ( z ˜ 3 , z ˜ ) , φ 2 , 3 , φ ˜ 2 , 3 ) = O ( r ε , λ 2 ) .

As usual, the terms O ( r ε , λ 2 ) depend nonlinearly on all the variables on the left side but are bounded (in the appropriate norm) by a constant (independent of ε and κ ) times r ε , λ 2 , provided ε ( 0 , ε κ ) and λ ( 0 , λ κ ) .

Recalling that x = r ε , λ ( z ˜ z ) , the previous systems can be written as for i = 1 , 2 , 3

( x , t i , λ i , φ i , φ ˜ i , ¯ i ) = O ( r ε , λ 2 ) .

By combining (102), (103), (111), (112), and (118), we obtain

(119) T i , ε = ( T i , ε 1 , T c , ε 2 , T i , ε 3 ) = ( O ( r ε , λ 2 ) , O ( r ε , λ 2 ) , O ( r ε , λ 2 ) ) , for i = 1 , 2 .

Furthermore, the nonlinear mapping appearing on the right-hand side of (119) is continuous and compact. Moreover, by reducing ε κ and λ κ , if needed, it sends the ball of radius κ r ε , λ 2 (considering the natural product norm) into itself, provided κ sufficiently large and fixed. By applying Schauder’s fixed-point theorem in the ball of radius κ r ε , λ 2 in the product space where the entries reside, we obtain the existence of a solution of the equation (119).

4 Proof of Theorem 4

4.1 Construction of the approximate solution

Here, we will construct another approximate solution for

(120) Δ u 1 + λ u 1 + ρ 2 e γ u 1 + ( 1 γ ) u 2 = 0 , Δ u 2 + λ u 2 + ρ 2 e ξ u 2 + ( 1 ξ ) u 1 = 0 in B ( z i , r ε , λ ) .

Using the following transformations

v 1 ( z ) = u 1 ε τ 1 z + 4 γ log ε + 2 γ log 2 τ 1 ( 1 + ε 2 ) in B ( z 1 , r ε , λ ) v 2 ( z ) = u 2 ε τ 1 z in B ( z 1 , r ε , λ ) ,

v 1 ( z ) = u 1 ε τ 2 z + 4 log ε + 2 log 2 τ 2 ( 1 + ε 2 ) in B ( z 2 , r ε , λ ) v 2 ( z ) = u 2 ε τ 2 z + 4 log ε + 2 log 2 τ 2 ( 1 + ε 2 ) in B ( z 2 , r ε , λ ) ,

and

v 1 ( z ) = u 1 ε τ 3 z in B ( z 3 , r ε , λ ) v 2 ( z ) = u 2 ε τ 3 z + 4 ξ log ε + 2 ξ log 2 τ 3 ( 1 + ε 2 ) in B ( z 3 , r ε , λ ) .

The system (120) can be expressed as follows:

(121) Δ v 1 + λ v 1 2 + 2 e γ v 1 + ( 1 γ ) v 2 = 0 in B ( z 1 , R ε , λ 1 ) Δ v 2 + λ v 2 2 + 2 C 1 , ε 2 γ + ξ 1 γ ε 4 γ + ξ 1 γ e ξ v 2 + ( 1 ξ ) v 1 = 0 in B ( z 1 , R ε , λ 1 ) ,

(122) Δ v 1 + λ v 1 2 + 2 e γ v 1 + ( 1 γ ) v 2 = 0 in B ( z 2 , R ε , λ 2 ) Δ v 2 + λ v 2 2 + 2 e ξ v 2 + ( 1 ξ ) v 1 = 0 in B ( z 2 , R ε , λ 2 ) ,

and

(123) Δ v 1 + λ v 1 2 + 2 C 3 , ε 2 γ + ξ 1 ξ ε 4 γ + ξ 1 ξ e γ v 1 + ( 1 γ ) v 2 = 0 in B ( z 3 , R ε , λ 3 ) Δ v 2 + λ v 2 2 + 2 e ξ v 2 + ( 1 ξ ) v 1 = 0 in B ( z 3 , R ε , λ 3 ) ,

with C i , ε = 2 τ i ( 1 + ε 2 ) , for i = 1, 3 and τ i > 0 is a constant, which will be fixed later. Recall that for i = 1, 2 and 3, R ε , λ i = τ i r ε , λ ε . Fix μ ( 1 , 2 ) and δ 0 , min { γ + ξ 1 γ , γ + ξ 1 ξ } . We remember that that ξ μ , ξ δ are defined in (25), G μ and K δ are defined in propositions 1 and 2.

In B ( z 1 , R ε , λ 1 ) and B ( z 3 , R ε , λ 3 ) , We replicate exactly the same process as in the proof of Theorem 3, thereby establishing the following propositions.

Proposition 11

Given κ > 0 , there exist ε κ > 0 , λ κ , c κ > 0 and θ ( 0 , 1 ) such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) and γ ( θ , 1 ) , then there exists a unique solution ( h 1 1 , h 2 1 ) of (39) such that ( h 1 1 , h 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) 2 c κ r ε , λ 2 . Hence,

v 1 ( z ) = 1 γ u ¯ ( z z 1 ) 1 γ γ G ε z τ 1 , z 2 1 γ γ ξ G ε z τ 1 , z 3 log γ γ + h 1 1 ( z ) v 2 ( z ) = 1 ξ G ε z τ 1 , z 3 + G ε z τ 1 , z 2 + h 2 1 ( z ) .

solves (121) in B ( z 1 , R ε , λ 1 ) .

Proposition 12

Given κ > 0 , there exist ε κ > 0 , λ κ , c κ > 0 and θ ( 0 , 1 ) such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) and ξ ( θ , 1 ) , then there exists a unique solution ( h 1 3 , h 2 3 ) of (53) such that ( h 1 3 , h 2 3 ) C δ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 . Then

v 1 ( z ) = 1 γ G ε z τ 3 , z 1 + G ε z τ 3 , z 2 + H int φ 1 3 , z z 3 R ε , λ 3 + h 1 3 ( z ) + v 1 3 ( z ) v 2 ( z ) = 1 ξ u ¯ ( z z 3 ) 1 ξ ξ G ε z τ 3 , z 2 1 ξ γ ξ G ε z τ 3 , z 1 log ξ ξ + H int φ 2 3 , z z 3 R ε , λ 3 + h 2 3 ( z ) + v 2 3 ( z ) .

solves (123) in B ( z 3 , R ε , λ 3 ) .

In B ( z 2 , R ε , λ 2 ) , we seek a solution to system (122) in the form

(124) v 1 ( z ) = u ¯ ( z z 2 ) + 1 γ γ ( 2 γ ξ ) G ε z τ 2 , z 1 1 γ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 + h 1 2 ( z ) v 2 ( z ) = u ¯ ( z z 2 ) 1 ξ γ ( 2 γ ξ ) G ε z τ 2 , z 1 + 1 ξ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 + h 2 2 ( z ) .

This involves solving the equation

(125) L h 1 2 = 8 ( 1 + r 2 ) 2 e ( 1 γ ) ( γ + ξ 1 ) ( 2 γ ξ ) 1 γ G ε z τ 2 , z 1 1 ξ G ε z τ 2 , z 3 + γ h 1 2 + ( 1 γ ) h 2 2 h 1 2 1 + λ u ¯ ( z z 2 ) + 1 γ γ ( 2 γ ξ ) G ε z τ 2 , z 1 1 γ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 + h 1 2 ( z ) 2 L h 2 2 = 8 ( 1 + r 2 ) 2 e ( 1 ξ ) ( γ + ξ 1 ) ( 2 γ ξ ) ( 1 γ G ε z τ 2 , z 1 + 1 ξ G ε z τ 2 , z 3 ) + ξ h 2 2 + ( 1 ξ ) h 1 2 h 2 2 1 + λ u ¯ ( z z 2 ) 1 ξ γ ( 2 γ ξ ) G ε z τ 2 , z 1 + 1 ξ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 + h 2 2 ( z ) 2 .

We denote

L h 1 2 = T 2 ( h 1 2 , h 2 2 ) and L h 2 2 = 2 ( h 1 2 , h 2 2 ) .

To find a solution of (125), it suffices to find a fixed point ( h 1 2 , h 2 2 ) in a small ball of C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) , which are solutions of

(126) h 1 2 = G μ ξ μ T 2 ( h 1 2 , h 2 2 ) = N 2 ( h 1 2 , h 2 2 ) h 2 2 = K μ ξ μ 2 ( h 1 2 , h 2 2 ) = 2 ( h 1 2 , h 2 2 ) .

Then, we obtain the following result.

Lemma 9

Given κ > 0 , there exist ε κ > 0 , λ κ > 0 , θ in ( 0 , 1 ) , c ¯ κ > 0 and c κ > 0 such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , γ , ξ ( θ , 1 ) and μ ( 1 , 2 ) . We have

N 2 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 , 2 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 , N 2 ( h 1 2 , h 2 2 ) N 2 ( k 1 2 , k 2 2 ) C μ 2 , α ( R 2 ) c ¯ κ ( 1 γ + r ε , λ 2 ) h 1 2 k 1 2 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 γ ) h 2 2 k 2 2 C μ 2 , α ( R 2 ) , 2 ( h 1 2 , h 2 2 ) 2 ( k 1 2 , k 2 2 ) C μ 2 , α ( R 2 ) c ¯ κ ( 1 ξ ) h 1 2 k 1 2 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 ξ + r ε , λ 2 ) h 2 2 k 2 2 C μ 2 , α ( R 2 ) .

Provided ( h 1 2 , h 2 2 ) , ( k 1 2 , k 2 2 ) in C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) satisfying

(127) ( h 1 2 , h 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 a n d ( k 1 2 , k 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 .

Proof

We have

sup r R ε , λ 2 r 2 μ T 2 ( 0 , 0 ) sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 e ( 1 γ ) ( γ + ξ 1 ) ( 2 γ ξ ) 1 γ G ε z τ 2 , z 1 1 ξ G ε z τ 2 , z 3 1 + λ sup r R ε , λ 2 r 2 μ u ¯ ( z z 2 ) + 1 γ γ ( 2 γ ξ ) G ε z τ 2 , z 1 1 γ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 2 .

Using Proposition 1 in conjunction with (26) and condition (17), for μ ( 1 , 2 ) , we establish the existence of c κ such that

(128) N 2 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 .

For the second estimate, we have 2 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 .

To obtain the third estimate, for ( h 1 2 , h 2 2 ) , ( k 1 2 , k 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) verifying (127), we have

sup r R ε , λ 2 r 2 μ T 2 ( h 1 2 , h 2 2 ) T 2 ( k 1 2 , k 2 2 ) sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 e ( 1 γ ) ( γ + ξ 1 ) ( 2 γ ξ ) 1 γ G ε z τ 2 , z 1 1 ξ G ε z τ 2 , z 3 e γ h 1 2 + ( 1 γ ) h 2 2 h 1 2 e ( 1 γ ) ( γ + ξ 1 ) ( 2 γ ξ ) 1 γ G ε z τ 2 , z 1 1 ξ G ε z τ 2 , z 3 e γ k 1 2 + ( 1 γ ) k 2 2 + k 1 2 + λ sup r R ε , λ 2 r 2 μ u ¯ ( z z 2 ) + 1 γ γ ( 2 γ ξ ) G ε z τ 2 , z 1 1 γ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 + h 1 2 ( z ) 2 u ¯ ( z z 2 ) + 1 γ γ ( 2 γ ξ ) G ε z τ 2 , z 1 1 γ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 + k 1 2 ( z ) 2 c κ sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 [ ( γ 1 ) ( h 1 2 k 1 2 ) + ( 1 γ ) ( h 2 2 k 2 2 ) ] + c κ λ sup r R ε , λ 2 r 2 μ × 2 u ¯ ( z z 2 ) + 1 γ γ ( 2 γ ξ ) G ε z τ 2 , z 1 1 γ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 + ( h 1 2 + k 1 2 ) ( h 1 2 k 1 2 ) c κ sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 ( 1 γ ) [ r μ h 1 2 k 1 2 C μ 2 , α ( R 2 ) + r μ h 2 2 k 2 2 C μ 2 , α ( R 2 ) ] + c κ λ sup r R ε , λ 2 r 2 μ 8 r 1 + r 2 + C ε + r μ 1 ( h 1 2 C μ 2 , α ( R 2 ) + k 1 2 C μ 2 , α ( R 2 ) ) r μ 1 h 1 2 k 1 2 C μ 2 , α ( R 2 ) .

By making use of Proposition 1 together with (26) and using the condition ( A 1 ) for μ ( 1 , 2 ) , we obtain that there exists c ¯ κ such that

(129) N 2 ( h 1 2 , h 2 2 ) N 2 ( k 1 2 , k 2 2 ) C μ 2 , α ( R 2 ) c ¯ κ ( 1 γ + r ε , λ 2 ) h 1 2 k 1 2 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 γ ) h 2 2 k 2 2 C μ 2 , α ( R 2 ) .

Similarly, we obtain the estimate for

(130) 2 ( h 1 2 , h 2 2 ) 2 ( k 1 2 , k 2 2 ) C μ 2 , α ( R 2 ) c ¯ κ ( 1 ξ ) h 1 2 k 1 2 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 ξ + r ε , λ 2 ) h 2 2 k 2 2 C μ 2 , α ( R 2 ) .

This completes the proof of Lemma 9.□

There exists θ in ( 0 , 1 ) , reducing ε κ and λ κ if needed, we can assume that c ¯ κ ( 1 γ + r ε , λ 2 ) 1 2 and c ¯ κ ( 1 ξ + r ε , λ 2 ) 1 2 for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , γ and ξ ( θ , 1 ) . Therefore, (129) and (130) are sufficient to show that

( h 1 2 , h 2 2 ) ( N 2 ( h 1 2 , h 2 2 ) , 2 ( h 1 2 , h 2 2 ) )

is a contraction from the ball

{ ( h 1 2 , h 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) : ( h 1 2 , h 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 } ,

into itself. Consequently, a unique fixed point ( h 1 2 , h 2 2 ) exists in this set, serving as a solution to (126).

Proposition 13

Given κ > 0 , there exist ε κ > 0 , λ κ , c κ > 0 and θ ( 0 , 1 ) such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , γ , ξ ( θ , 1 ) and for all τ 2 in some fixed compact subset of [ τ 2 , τ 2 + ] ( 0 , + ) , then there exists a unique solution

( h 1 2 , h 2 2 ) of (126) such that

( h 1 2 , h 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 .

Therefore, ( v 1 , v 2 ) defined by (124) solves (122) in B ( z 2 , R ε , λ 2 ) .

4.2 The nonlinear interior problem

Here, we are interested in studying the system

(131) Δ v 1 + λ v 1 2 + 2 e γ v 1 + ( 1 γ ) v 2 = 0 in B ( z 1 , R ε , λ 1 ) Δ v 2 + λ v 2 2 + 2 C 1 , ε 2 γ + ξ 1 γ ε 4 γ + ξ 1 γ e ξ v 2 + ( 1 ξ ) v 1 = 0 in B ( z 1 , R ε , λ 1 ) ,

(132) Δ v 1 + λ v 1 2 + 2 e γ v 1 + ( 1 γ ) v 2 = 0 in B ( z 2 , R ε , λ 2 ) Δ v 2 + λ v 2 2 + 2 e ξ v 2 + ( 1 ξ ) v 1 = 0 in B ( z 2 , R ε , λ 2 )

and

(133) Δ v 1 + λ v 1 2 + 2 C 3 , ε 2 γ + ξ 1 ξ ε 4 γ + ξ 1 ξ e γ v 1 + ( 1 γ ) v 2 = 0 in B ( z 3 , R ε , λ 3 ) Δ v 2 + λ v 2 2 + 2 e ξ v 2 + ( 1 ξ ) v 1 = 0 in B ( z 3 , R ε , λ 3 ) ,

with C i , ε = 2 τ i ( 1 + ε 2 ) , for i = 1, 3 and τ i > 0 is a constant, which will be specified later. For i = 1 , 2 , 3 , let φ i ( φ 1 i , φ 2 i ) ( C 2 , α ( S 1 ) ) 2 satisfying (56). In the following, we also denote by u ¯ = u ε = 1 , τ = 1 . Fix μ ( 1 , 2 ) and δ 0 , min { γ + ξ 1 γ , γ + ξ 1 ξ } . We recall that ξ μ is defined in (25), G μ and K δ are defined in Propositions 1 and 2.

In B ( z 1 , R ε , λ 1 ) and B ( z 3 , R ε , λ 3 ) , We replicate precisely the same steps as in the proof of Theorem 3, and hence, we obtain the following propositions.

Proposition 14

Given κ > 0 , there exist ε κ , λ κ , c κ > 0 and θ ( 0 , 1 ) such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , γ ( θ , 1 ) , and for all τ 1 in some fixed compact subset of [ τ 1 , τ 1 + ] ( 0 , + ) , then there exists a unique solution ( v 1 1 , v 2 1 ) ( ( v 1 , ε , τ 1 , φ 1 , v 2 , ε , τ 1 , φ 1 ) ) of (63) such that ( v 1 1 , v 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) 2 c κ r ε , λ 2 , hence,

v 1 ( z ) = 1 γ u ¯ ( z z 1 ) 1 γ γ G ε z τ 1 , z 2 1 γ γ ξ G ε z τ 1 , z 3 log γ γ + H int φ 1 1 , z z 1 R ε , λ 1 + h 1 1 ( z ) + v 1 1 ( z ) v 2 ( z ) = 1 ξ G ε z τ 1 , z 3 + G ε z τ 1 , z 2 + H int φ 2 1 , z z 1 R ε , λ 1 + h 2 1 ( z ) + v 2 1 ( z ) .

solves (131) in B ( z 1 , R ε , λ 1 ) .

Proposition 15

Given κ > 0 , there exist ε κ > 0 , λ κ , c κ > 0 , and θ ( 0 , 1 ) such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , ξ ( θ , 1 ) and for all τ 3 in some fixed compact subset of [ τ 3 , τ 3 + ] ( 0 , + ) , then there exists a unique solution ( v 1 3 , v 2 3 ) ( ( v 1 , ε , τ 3 , φ 3 , v 2 , ε , τ 3 , φ 3 ) ) of (79) such that ( v 1 3 , v 2 3 ) C δ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 , hence,

v 1 ( z ) = 1 γ G ε z τ 3 , z 1 + G ε z τ 3 , z 2 + H int φ 1 3 , z z 3 R ε , λ 3 + h 1 3 ( z ) + v 1 3 ( z ) v 2 ( z ) = 1 ξ u ¯ ( z z 3 ) 1 ξ ξ G ε z τ 3 , z 2 1 ξ γ ξ G ε z τ 3 , z 1 log ξ ξ + H int φ 2 3 , z z 3 R ε , λ 3 + h 2 3 ( z ) + v 2 3 ( z ) .

solves (133) in B ( z 3 , R ε , λ 3 ) .

In B ( z 2 , R ε , λ 2 ) : Given φ 2 ( φ 1 2 , φ 2 2 ) ( C 2 , α ( S 1 ) ) 2 satisfying (56), we look for a solution of (132) of the form

(134) v 1 ( z ) = u ¯ ( z z 2 ) + 1 γ γ ( 2 γ ξ ) G ε z τ 2 , z 1 1 γ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 + H int φ 1 2 , z z 2 R ε , λ 2 + h 1 2 ( z ) + v 1 2 ( z ) v 2 ( z ) = u ¯ ( z z 2 ) 1 ξ γ ( 2 γ ξ ) G ε z τ 2 , z 1 + 1 ξ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 + H int φ 2 2 , z z 2 R ε , λ 2 + h 2 2 ( z ) + v 2 2 ( z ) .

This entails solving the equation

(135) L v 1 2 = 8 ( 1 + r 2 ) 2 e ( 1 γ ) ( γ + ξ 1 ) ( 2 γ ξ ) 1 γ G ε z τ 2 , z 1 1 ξ G ε z τ 2 , z 3 + γ ( H 1 int , 2 + h 1 2 + v 1 2 ) + ( 1 γ ) ( H 2 int , 2 + h 2 2 + v 2 2 ) v 1 2 1 + λ u ¯ ( z z 2 ) + 1 γ γ ( 2 γ ξ ) G ε z τ 2 , z 1 1 γ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 + H int φ 1 2 , z z 2 R ε , λ 2 + h 1 2 ( z ) + v 1 2 ( z ) 2 + Δ h 1 2 L v 2 2 = 8 ( 1 + r 2 ) 2 e ( 1 ξ ) ( γ + ξ 1 ) ( 2 γ ξ ) ( 1 γ G ε z τ 2 , z 1 + 1 ξ G ε z τ 2 , z 3 ) + ξ ( H 2 int , 2 + h 2 2 + v 2 2 ) + ( 1 ξ ) ( H 1 int , 2 + h 1 2 + v 1 2 ) v 2 2 1 + λ u ¯ ( z z 2 ) 1 ξ γ ( 2 γ ξ ) G ε z τ 2 , z 1 + 1 ξ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 + H int φ 2 2 , z z 2 R ε , λ 2 + h 2 2 ( z ) + v 2 2 ( z ) 2 + Δ h 2 2 .

We denote by

L v 1 2 = T 2 ( v 1 2 , v 2 2 ) and L v 2 2 = R 2 ( v 1 2 , v 2 2 ) .

To find a solution of (135), it suffices to locate a fixed point ( v 1 2 , v 2 2 ) in a small ball of C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) , solutions of

(136) v 1 2 = G μ ξ μ T 2 ( v 1 2 , v 2 2 ) = N 2 ( v 1 2 , v 2 2 ) v 2 2 = K μ ξ μ R 2 ( v 1 2 , v 2 2 ) = M 2 ( v 1 2 , v 2 2 ) .

Given κ > 0 (whose value will be fixed later on). In addition, we assume that the functions ( φ 1 2 , φ 2 2 ) verify

(137) ( φ 1 2 , φ 2 2 ) ( C 2 , α ( S 1 ) ) 2 κ r ε , λ 2 .

Then, the following result holds.

Lemma 10

Given κ > 0 , there exist ε κ > 0 , λ κ > 0 , θ in ( 0 , 1 ) , c ¯ κ > 0 and c κ > 0 such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , γ , ξ ( θ , 1 ) and μ ( 1 , 2 ) . We have

N 2 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 , M 2 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 , N 2 ( v 1 2 , v 2 2 ) N 2 ( t 1 2 , t 2 2 ) C μ 2 , α ( R 2 ) c ¯ κ ( 1 γ + r ε , λ 2 ) v 1 2 t 1 2 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 γ ) v 2 2 t 2 2 C μ 2 , α ( R 2 ) , M 2 ( v 1 2 , v 2 2 ) M 2 ( t 1 2 , t 2 2 ) C μ 2 , α ( R 2 ) c ¯ κ ( 1 ξ ) v 1 2 t 1 2 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 ξ + r ε , λ 2 ) v 2 2 t 2 2 C μ 2 , α ( R 2 ) .

Provided ( v 1 2 , v 2 2 ) , ( t 1 2 , t 2 2 ) in C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) satisfying

(138) ( v 1 2 , v 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 a n d ( t 1 2 , t 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 .

Proof

The first estimate arises from Lemma 4, alongside the assumption regarding the norm of the boundary data φ j 2 given by (137). Then

H int ( φ j 2 ; R ε , λ ) C 2 2 , α ( B ¯ R ε , λ ) c κ R ε , λ 2 φ j 2 C 2 , α ( S 1 ) c κ ε 2 .

On the other hand for μ ( 1 , 2 ) , using the condition (17) given in Theorem 4, we obtain

sup r R ε , λ 2 r 2 μ T 2 ( 0 , 0 ) sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 e ( 1 γ ) ( γ + ξ 1 ) ( 2 γ ξ ) 1 γ G ε z τ 2 , z 1 1 ξ G ε z τ 2 , z 3 e γ h 1 2 + γ H 1 int , 2 + ( 1 γ ) h 2 2 + ( 1 γ ) H 2 int , 2 1 + λ sup r R ε , λ 2 r 2 μ u ¯ ( z z 2 ) + 1 γ γ ( 2 γ ξ ) G ε z τ 2 , z 1 1 γ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 + H int φ 1 2 , z z 2 R ε , λ 2 + h 1 2 ( z ) 2 + Δ h 1 2 c κ sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 ( ε 2 r 2 + γ r μ h 1 2 C μ 2 , α ( R 2 ) + γ r 2 H 1 int , 2 C 2 2 , α + ( 1 γ ) r δ h 2 2 C δ 2 , α ( R 2 ) + ( 1 γ ) r 2 H 2 int , 2 C 2 2 , α ) + λ sup r R ε , λ 2 r 2 μ u ¯ ( z z 2 ) + 1 γ γ ( 2 γ ξ ) G ε z τ 2 , z 1 1 γ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 + H int φ 1 2 , z z 2 R ε , λ 2 + h 1 2 ( z ) 2 + Δ h 1 2 .

By Proposition 1, inequality (26) and condition ( A 1 ) for μ ( 1 , 2 ) , we establish that there exists c κ such that

(139) N 2 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 .

For the second estimate, we have M 2 ( 0 , 0 ) C μ 2 , α ( R 2 ) c κ r ε , λ 2 .

To obtain the third estimate, for ( v 1 2 , v 2 2 ) , ( t 1 2 , t 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) verifying (138), we derive

sup r R ε , λ 2 r 2 μ T 2 ( v 1 2 , v 2 2 ) T 2 ( t 1 2 , t 2 2 ) sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 e ( 1 γ ) ( γ + ξ 1 ) ( 2 γ ξ ) 1 γ G ε z τ 2 , z 1 1 ξ G ε z τ 2 , z 3 e γ ( H 1 int , 2 + h 1 2 + v 1 2 ) + ( 1 γ ) ( H 2 int , 2 + h 2 2 + v 2 2 ) v 1 2 e ( 1 γ ) ( γ + ξ 1 ) ( 2 γ ξ ) 1 γ G ε z τ 2 , z 1 1 ξ G ε z τ 2 , z 3 e γ ( H 1 int , 2 + h 1 2 + t 1 2 ) + ( 1 γ ) ( H 2 int , 2 + h 2 2 + t 2 2 ) + t 1 2 + λ sup r R ε , λ 2 r 2 μ × u ¯ ( z z 2 ) + 1 γ γ ( 2 γ ξ ) G ε z τ 2 , z 1 1 γ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 + H int φ 1 2 , z z 2 R ε , λ 2 + h 1 2 ( z ) + v 1 2 ( z ) 2 u ¯ ( z z 2 ) + 1 γ γ ( 2 γ ξ ) G ε z τ 2 , z 1 1 γ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 + H int φ 1 2 , z z 2 R ε , λ 2 + h 1 2 ( z ) + t 1 2 ( z ) 2 c κ sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 [ ( γ 1 ) ( v 1 2 t 1 2 ) + ( 1 γ ) ( v 2 2 t 2 2 ) ] + c κ λ sup r R ε , λ 2 r 2 μ × 2 u ¯ ( z z 2 ) + 1 γ γ ( 2 γ ξ ) G ε z τ 2 , z 1 1 γ ξ ( 2 γ ξ ) G ε z τ 2 , z 3 + H int φ 1 2 , z z 2 R ε , λ 2 + h 1 2 ( z ) + ( v 1 2 + t 1 2 ) ( v 1 2 t 1 2 ) c κ sup r R ε , λ 2 8 r 2 μ ( 1 + r 2 ) 2 ( 1 γ ) [ r μ v 1 2 t 1 2 C μ 2 , α ( R 2 ) + ( 1 γ ) r μ v 2 2 t 2 2 C μ 2 , α ( R 2 ) ] + c κ λ sup r R ε , λ 2 r 2 μ × 8 r 1 + r 2 + C ε + r μ 1 h 1 2 C μ 2 , α ( R 2 ) + ε r H 1 int , 2 C 2 2 , α + r μ 1 ( v 1 2 C μ 2 , α ( R 2 ) + t 1 2 C μ 2 , α ( R 2 ) ) r μ 1 v 1 2 t 1 2 C μ 2 , α ( R 2 ) .

By making use of Proposition 1 together with (26) and using the condition ( A 1 ) for μ ( 1 , 2 ) , we deduce the existence of a c κ such that

(140) N 2 ( v 1 2 , v 2 2 ) N 2 ( t 1 2 , t 2 2 ) C μ 2 , α ( R 2 ) c ¯ κ ( 1 γ + r ε , λ 2 ) v 1 2 t 1 2 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 γ ) v 2 2 t 2 2 C μ 2 , α ( R 2 ) .

Similarly, we obtain the estimate

(141) M 2 ( v 1 2 , v 2 2 ) M 2 ( t 1 2 , t 2 2 ) C μ 2 , α ( R 2 ) c ¯ κ ( 1 ξ ) v 1 2 t 1 2 C μ 2 , α ( R 2 ) + c ¯ κ ( 1 ξ + r ε , λ 2 ) v 2 2 t 2 2 C μ 2 , α ( R 2 ) .

This concludes the proof of Lemma 10.□

There exist θ in ( 0 , 1 ) , reducing ε κ and λ κ if necessary, we can assume that c ¯ κ ( 1 γ + r ε , λ 2 ) 1 2 and c ¯ κ ( 1 ξ + r ε , λ 2 ) 1 2 for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , γ and ξ ( θ , 1 ) . Therefore, (140) and (141) are enough to show that

( v 1 2 , v 2 2 ) ( N 2 ( v 1 2 , v 2 2 ) , M 2 ( v 1 2 , v 2 2 ) )

is a contraction from the ball

{ ( v 1 2 , v 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) : ( v 1 2 , v 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 } ,

into itself, and hence, a unique fixed point ( v 1 2 , v 2 2 ) exists in this set, which is a solution of (136).

Proposition 16

Given κ > 0 , there exist ε κ > 0 , λ κ , c κ > 0 and θ ( 0 , 1 ) such that for all ε ( 0 , ε κ ) , λ ( 0 , λ κ ) , γ , ξ ( θ , 1 ) , and for all τ 2 in some fixed compact subset of [ τ 2 , τ 2 + ] ( 0 , + ) , then there exists a unique solution ( v 1 2 , v 2 2 ) ( ( v 1 , ε , τ 2 , φ 2 , v 2 , ε , τ 2 , φ 2 ) ) of (136) such that

( v 1 2 , v 2 2 ) C μ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 .

Hence, ( v 1 , v 2 ) defined by (134) solves (132) in B ( z 2 , R ε , λ 2 ) .

Also, we remark that the functions ( v 1 , ε , τ i , φ 1 i i , v 2 , ε , τ i , φ 2 i i ) , for i { 1 , 2 , 3 } , obtained in the aforementioned proposition, depend continuously on the parameter τ .

4.3 The nonlinear exterior problem

By precisely reproducing the same nonlinear exterior problem as in the proof of Theorem 3, we establish the following proposition using the same notations.

Proposition 17

Given κ > 0 , there exist ε κ > 0 (depending on κ ) such that for any ε ( 0 , ε κ ) , λ i and z ˜ i satisfying (88), any functions ( φ ˜ 1 i , φ ˜ 2 i ) satisfying (57) and (87), there exists a unique solution ( v ˜ 1 , v ˜ 2 ) ( ( v ˜ 1 , ε , λ , z ˜ , φ ˜ , v ˜ 2 , ε , λ , z ˜ , φ ˜ ) ) (85)so that for ( v ˜ 1 , v ˜ 2 ) defined by

w ˜ 1 1 + λ 1 γ G ( , z ˜ 1 ) + ( 1 + λ 2 ) G ( , z ˜ 2 ) + i = 1 3 χ r 0 ( z ˜ i ) H ext ( φ ˜ 1 i ; ( z ˜ i ) r ε , λ ) + v ˜ 1 w ˜ 2 1 + λ 3 ξ G ( , z ˜ 3 ) + ( 1 + λ 2 ) G ( , z ˜ 2 ) + i = 1 3 χ r 0 ( z ˜ i ) H ext ( φ ˜ 2 i ; ( z ˜ i ) r ε , λ ) + v ˜ 2 ,

which solves (82) in Ω ¯ r ε , λ ( z ˜ ) . In addition, we have ( v ˜ 1 , v ˜ 2 ) C ν 2 , α ( Ω ¯ * ( z ˜ ) ) 2 c ¯ κ r ε , λ 2 .

4.4 The nonlinear Cauchy-data matching

We will gather the results from previous sections. Employing the prior notations, suppose z ˜ ( z ˜ 1 , z ˜ 2 , z ˜ 3 ) Ω 3 are provided, close to z ( z 1 , z 2 , z 3 ) . In addition, assume that

τ ( τ 1 , τ 2 , τ 3 ) [ τ 1 , τ 1 + ] × [ τ 2 , τ 2 + ] × [ τ 3 , τ 3 + ] ( 0 , ) 3

are given (the values of τ i and τ i + will be fixed later). First, we consider some set of boundary data φ i ( φ 1 i , φ 2 i ) ( C 2 , α ( S 1 ) ) 2 , for i = 1, 2, 3. According to the Propositions 14, 16, and 15, provided ε ( 0 , ε κ ) and λ ( 0 , λ κ ) , we can find, u int ( u int , 1 , u int , 2 ) a solution of

(142) Δ u 1 = ρ 2 e γ u 1 + ( 1 γ ) u 2 + λ u 1 2 , Δ u 2 = ρ 2 e ξ u 2 + ( 1 ξ ) u 1 + λ u 2 2 ,

in B r ε , λ ( z ˜ 1 ) B r ε , λ ( z ˜ 2 ) B r ε , λ ( z ˜ 3 ) , which can be decomposed as follows:

u int , 1 ( z ) 1 γ u ε , τ 1 ( z z 1 ˜ ) 1 γ γ G ( z , z 2 ˜ ) 1 γ γ ξ G ( z , z 3 ˜ ) log γ γ + H φ 1 1 int z z ˜ 1 r ε , λ + h 1 1 R ε , λ 1 ( z z ˜ 1 ) r ε , λ + v 1 1 R ε , λ 1 ( z z ˜ 1 ) r ε , λ in B r ε , λ ( z ˜ 1 ) u ε , τ 2 ( z z 2 ˜ ) + 1 γ γ ( 2 γ ξ ) G ( z , z 1 ˜ ) 1 γ ξ ( 2 γ ξ ) G ( z , z 3 ˜ ) + H φ 1 2 int z z ˜ 2 r ε , λ + h 1 2 R ε , λ 2 ( z z ˜ 2 ) r ε , λ + v 1 2 R ε , λ 2 ( z z ˜ 2 ) r ε , λ in B r ε , λ ( z ˜ 2 ) 1 γ G ( z , z 1 ˜ ) + G ( z , z 2 ˜ ) + H φ 1 3 int z z ˜ 3 r ε , λ + h 1 3 R ε , λ 3 ( z z ˜ 3 ) r ε , λ + v 1 3 R ε , λ 3 ( z z ˜ 3 ) r ε , λ in B r ε , λ ( z ˜ 3 ) ,

and

u int , 2 ( z ) 1 ξ G ( z , z 3 ˜ ) + G ( z , z 2 ˜ ) + H φ 2 1 int z z ˜ 1 r ε , λ + h 2 1 R ε , λ 1 ( z z ˜ 1 ) r ε , λ + v 2 1 R ε , λ 1 ( z z ˜ 1 ) r ε , λ in B r ε , λ ( z ˜ 1 ) u ε , τ 2 ( z z 2 ˜ ) 1 ξ γ ( 2 γ ξ ) G ( z , z 1 ˜ ) + 1 ξ ξ ( 2 γ ξ ) G ( z , z 3 ˜ ) + H φ 2 2 int z z ˜ 2 r ε , λ + h 2 2 R ε , λ 2 ( z z ˜ 2 ) r ε , λ + v 2 2 R ε , λ 2 ( z z ˜ 2 ) r ε , λ in B r ε , λ ( z ˜ 2 ) 1 ξ u ε , τ 3 ( z z 3 ˜ ) 1 ξ ξ G ( z , z 2 ˜ ) 1 ξ γ ξ G ( z , z 1 ˜ ) log ξ ξ + H φ 2 3 int z z ˜ 3 r ε , λ + h 2 3 R ε , λ 3 ( z z ˜ 3 ) r ε , λ + v 2 3 R ε , λ 3 ( z z ˜ 3 ) r ε , λ in B r ε , λ ( z ˜ 3 ) ,

where, for i = 1, 2, 3 and j = 1 , 2 , H φ j i int z z ˜ i r ε , λ C 2 2 , α ( B ¯ 1 * ) c κ r ε , λ 2 , R ε , λ i = τ i r ε , λ ε , the functions h j i and v j i satisfy

( h 1 1 , h 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) 2 c κ r ε , λ 2 , ( h 1 2 , h 2 2 ) ( C μ 2 , α ( R 2 ) ) 2 2 c κ r ε , λ 2 , ( h 1 3 , h 2 3 ) C δ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 , ( v 1 1 , v 2 1 ) C μ 2 , α ( R 2 ) × C δ 2 , α ( R 2 ) 2 c κ r ε , λ 2 , ( v 1 2 , v 2 2 ) ( C μ 2 , α ( R 2 ) ) 2 2 c κ r ε , λ 2 , and ( v 1 3 , v 2 3 ) C δ 2 , α ( R 2 ) × C μ 2 , α ( R 2 ) 2 c κ r ε , λ 2 .

Similarly, given boundary data ( φ ˜ 1 i , φ ˜ 2 i ) ( C 2 , α ( S 1 ) ) 2 ( i = 1 , 2 , 3 ) satisfying (57), ( λ 1 , λ 2 , λ 3 ) R 3 satisfying (88), provided ε ( 0 , ε κ ) , Proposition 17 allows us to find a solution u ext ( u ext , 1 , u ext , 2 ) of (142) in Ω ¯ \ i = 1 , 2 , 3 B r ε , λ ( z ˜ i ) , with the following decomposition:

u ext , 1 ( z ) 1 + λ 1 γ G ( z , z ˜ 1 ) + ( 1 + λ 2 ) G ( z , z ˜ 2 ) + i = 1 3 χ r 0 ( z z ˜ i ) H ext ( φ ˜ 1 i ; ( z z ˜ i ) r ε , λ ) + v 1 ˜ ( z ) u ext , 2 ( z ) 1 + λ 3 ξ G ( z , z ˜ 3 ) + ( 1 + λ 2 ) G ( z , z ˜ 2 ) + i = 1 3 χ r 0 ( z z ˜ i ) H ext ( φ ˜ 2 i ; ( z z ˜ i ) r ε , λ ) + v 2 ˜ ( z ) .

Here, v ˜ 1 , v ˜ 2 C ν 2 , α ( Ω ¯ * ( z ˜ ) ) satisfy ( v ˜ 1 , v ˜ 2 ) ( C ν 2 , α ( Ω ¯ * ( z ˜ ) ) ) 2 2 c ¯ κ r ε , λ 2 . It remains to determine the parameters and the boundary data such that the function equal to u int in i = 1 , 2 , 3 B r ε , λ ( z ˜ i ) and equal to u ext in Ω ¯ r ε , λ ( z ˜ ) is a smooth function. This involves finding the boundary data and the parameters so that for j = 1 , 2

(143) u int , j = u ext , j and r u int , j = r u ext , j

on B r ε , λ ( z ˜ 1 ) , B r ε , λ ( z ˜ 2 ) and B r ε , λ ( z ˜ 3 ) .

Suppose that (143) is verified, this implies that for each ε sufficiently small, u ε , λ C 2 , α (which is obtained by patching together the functions u int and the function u ext ) constitutes a weak solution of our system. Elliptic regularity theory then implies that this solution is indeed smooth. This achieves our proof since ε and λ tend to 0, the sequence of solutions we obtain satisfies the required singular limit behaviors.

Before we proceed, it is worth noting the following remarks. First, it will be convenient to observe that the function u ε , τ i can be expanded as follows:

(144) u ε , τ i ( z ) = 2 log τ i 4 log z + O ε 2 τ i 2 z 2 , z B r ε , λ ( 0 ) .

On B r ε , λ ( z ˜ 1 ) , according to the proof of Theorem 3 and given that when ε and λ tend to 0, it is enough to choose τ 1 and τ 2 + in such a way that

2 log ( τ 1 ) < log γ 1 ( z 1 , z ) < 2 log ( τ 1 + ) ,

where

1 ( , z ˜ ) H ( , z ˜ 1 ) + G ( , z ˜ 2 ) + 1 γ ξ G ( , z ˜ 3 ) .

Also, using the fact that

φ 1 1 = φ 1 , 0 1 + φ 1,1 1 + φ 1 , 1 and φ ˜ 1 1 = φ ˜ 1 , 0 1 + φ ˜ 1,1 1 + φ ˜ 1 , 1 ,

where φ 1 , 0 1 , φ ˜ 1 , 0 1 E 0 = R are constant on S 1 , φ 1,1 1 , φ ˜ 1,1 1 belong to E 1 = Span { e 1 , e 2 } and φ 1 , 1 , φ ˜ 1 , 1 are L 2 ( S 1 ) orthogonal to E 0 and E 1 . We can prove that

( u int , 1 u ext , 1 ) ( z ˜ 1 + r ε , λ . ) = 0 and ( r u int , 1 r u ext , 1 ) ( z ˜ 1 + r ε , λ . ) = 0

on S 1 , which implies

(145) T 1 , ε 1 = ( t 1 , λ 1 , φ 1 , 0 1 , φ ˜ 1 , 0 1 , φ 1,1 1 , φ ˜ 1,1 1 , ¯ 1 ( z ˜ 1 , z ˜ ) , φ 1 , 1 , φ ˜ 1 , 1 ) = O ( r ε , λ 2 ) ,

where

t 1 = 1 log r ε , λ [ 2 log τ 1 + log γ + 1 ( z ˜ 1 , z ˜ ) ] ,

Finally, by using the following equalities

φ 2 1 = φ 2,0 1 + φ 2,1 1 + φ 2 , 1 and φ ˜ 2 1 = φ ˜ 2,0 1 + φ ˜ 2,1 1 + φ ˜ 2 , 1 ,

with φ 2,0 1 , φ ˜ 2,0 1 E 0 , φ 2,1 2 , φ ˜ 2,1 1 E 1 and φ 2 , 1 , φ ˜ 2 , 1 ( L 2 ( S 1 ) ) , we can establish that

( u int , 2 u ext , 2 ) ( z ˜ 1 + r ε , λ . ) = 0 and ( r u int , 2 r u ext , 2 ) ( z ˜ 1 + r ε , λ . ) = 0

on S 1 , which enables us to obtain

(146) T 2 , ε 1 = ( φ 2,0 1 , φ ˜ 2,0 1 , φ 2,1 1 , φ ˜ 2,1 1 , φ 2 , 1 , φ ˜ 2 , 1 ) = O ( r ε , λ 2 ) ,

The terms O ( r ε , λ 2 ) depend nonlinearly on all the variables on the left side, but they are bounded (in the appropriate norm) by a constant (independent of ε and κ ) multiplied by r ε , λ 2 , provided ε ( 0 , ε κ ) and λ ( 0 , λ κ ) .

On B r ε , λ ( z ˜ 2 ) , we have

(147) ( u int , 1 u ext , 1 ) ( z ) = 2 log τ 2 + 4 λ 2 log z z ˜ 2 + 1 γ γ ( 2 γ ξ ) G ( z , z ˜ 1 ) 1 γ ξ ( 2 γ ξ ) G ( z , z ˜ 3 ) + h 1 2 ( R ε , λ 2 ( z z ˜ 2 ) r ε , λ ) + v 1 2 ( R ε , λ 2 ( z z ˜ 2 ) r ε , λ ) + H int ( φ 1 2 , ( z z ˜ 2 ) r ε , λ ) H 1 ext ( φ ˜ 1 2 ; ( z z ˜ 2 ) r ε , λ ) 1 γ G ( z , z ˜ 1 ) H ( z , z ˜ 2 ) + O ε 2 τ 2 2 z z ˜ 2 2 + O ( r ε , λ 2 ) ,

and

(148) ( u int , 2 u ext , 2 ) ( z ) = 2 log τ 2 + 4 λ 2 log z z ˜ 2 + 1 ξ γ ( 2 γ ξ ) G ( z , z ˜ 1 ) 1 ξ ξ ( 2 γ ξ ) G ( z , z ˜ 3 ) + h 2 2 ( R ε , λ 2 ( z z ˜ 2 ) r ε , λ ) + v 2 2 ( R ε , λ 2 ( z z ˜ 2 ) r ε , λ ) + H int ( φ 2 2 , ( z z ˜ 2 ) r ε , λ ) H ext ( φ ˜ 2 2 ; ( z z ˜ 2 ) r ε , λ ) 1 ξ G ( z , z ˜ 3 ) H ( z , z ˜ 2 ) + O ε 2 τ 2 2 z z ˜ 2 2 + O ( r ε , λ 2 ) .

Subsequently, despite all functions being defined on B r ε , λ ( z ˜ 2 ) in (143), it will be more convenient to solve on S 1 , for i = 1 , 2 , the following set of equations

(149) ( u int , i u ext , i ) ( z ˜ 2 + r ε , λ . ) = 0 and r ( u int , i u ext , i ) ( z ˜ 2 + r ε , λ . ) = 0 .

Since the boundary data are chosen to satisfy (57) or (56), we perform the following decomposition:

φ i 2 = φ i , 0 2 + φ i , 1 2 + φ i , 2 and φ ˜ i 2 = φ ˜ i , 0 2 + φ ˜ i , 1 2 + φ ˜ i , 2 ,

where φ i , 0 2 , φ ˜ i , 0 2 E 0 = R are constant on S 1 , φ i , 1 2 , φ ˜ i , 1 2 belong to E 1 = Span { e 1 , e 2 } and φ i , 2 , φ ˜ i , 2 are L 2 ( S 1 ) orthogonal to E 0 and E 1 .

We insist that, for z S 1 , both equations (147) and (148) involve the same relationship between the parameter τ 2 and the appropriate energy 2 ,

(150) ( u int , i u ext , i ) ( z ˜ 2 + r ε , λ z ) = 2 log τ 2 + 4 λ 2 log r ε , λ z H ( z ˜ 2 , z ˜ 2 ) + 1 ξ γ ( 2 γ ξ ) G ( z ˜ 2 , z ˜ 1 ) + 1 γ ξ ( 2 γ ξ ) G ( z ˜ 2 , z ˜ 3 ) + O ( r ε , λ 2 ) .

Then, projecting equations (149) over E 0 will result in

(151) 4 λ 2 log r ε , λ + O ( r ε , λ 2 ) = 0 , 2 log τ 2 + 4 λ 2 log r ε , λ 2 ( z ˜ 2 , z ˜ ) + O ( r ε , λ 2 ) = 0 ,

where

2 ( , z ˜ ) H ( , z ˜ 2 ) + 1 ξ γ ( 2 γ ξ ) G ( , z ˜ 1 ) + 1 γ ξ ( 2 γ ξ ) G ( , z ˜ 3 ) .

The system (151) can be simply written as follows:

1 log r ε , λ [ 2 log τ 2 + 2 ( z ˜ 2 , z ˜ ) ] = O ( r ε , λ 2 ) , λ 2 = O ( r ε , λ 2 ) .

We are now in a position to define τ 2 and τ 2 + . In fact, according to the aforementioned analysis, as ε and λ tend to 0, we expect that ( z ˜ 1 , z ˜ 2 , z ˜ 3 ) will converge to ( z 1 , z 2 , z 3 ) and τ 2 will converge to τ 2 * satisfying

2 log τ 2 * = 2 ( z 2 , z ) .

Therefore, it suffices to choose τ 2 and τ 2 + such as

2 log ( τ 2 ) < 2 ( z 2 , z ) < 2 log ( τ 2 + ) .

Consider now the projection of (149) over E 1 . Given a smooth function f defined in Ω , we identify its gradient f = ( z 1 f , z 2 f ) with the element of E 1

¯ f = i = 1 2 z i f e i .

Keeping these notations, we derive the system

(152) ¯ 2 ( z ˜ 2 , z ˜ ) = O ( r ε , λ 2 ) , φ i , 1 2 = O ( r ε , λ 2 ) .

Finally, we consider the projection onto ( L 2 ( S 1 ) ) . This yields the system

φ i , 2 φ ˜ i , 2 + O ( r ε , λ 2 ) = 0 , r ( H i , int , 2 H i , ext ) + O ( r ε , λ 2 ) = 0 .

Thanks to Lemma 5, we can rewrite this final system as follows:

φ i , 2 = O ( r ε , λ 2 ) and φ ˜ i , 2 = O ( r ε , λ 2 ) .

If we define the parameters t 2 R by

t 2 = 1 log r ε [ 2 log τ 2 + 2 ( z ˜ 2 , z ˜ ) ] .

Then, the system we need to solve reads:

(153) T i , ε 2 = ( t 2 , λ 2 , φ i , 0 2 , φ ˜ i , 0 2 , φ i , 1 2 , φ ˜ i , 1 2 , ¯ 2 ( z ˜ 2 , z ˜ ) , φ i , 2 , φ ˜ i , 2 ) = O ( r ε , λ 2 ) , for i = 1 , 2 .

The terms O ( r ε , λ 2 ) depend nonlinearly on all the variables on the left side. However, they are bounded (in the appropriate norm) by a constant (independent of ε and κ ) times r ε , λ 2 , provided ε ( 0 , ε κ ) and λ ( 0 , λ κ ) .

On B r ε , λ ( z ˜ 3 ) , by referring to the proof of Theorem 3, using the fact that

φ 1 3 = φ 1 , 0 3 + φ 1,1 3 + φ 1 , 3 and φ ˜ 1 3 = φ ˜ 1 , 0 3 + φ ˜ 1,1 3 + φ ˜ 1 , 3 ,

with φ 1 , 0 3 , φ ˜ 1 , 0 3 E 0 , φ 1,1 3 , φ ˜ 1,1 3 E 1 and φ 1 , 3 , φ ˜ 1 , 3 ( L 2 ( S 1 ) ) , we can prove that

(154) T 1 , ε 3 = ( φ 1 , 0 3 , φ ˜ 1 , 0 3 , φ 1,1 3 , φ ˜ 1,1 3 , φ 1 , 3 , φ ˜ 1 , 3 ) = O ( r ε , λ 2 ) .

On other hand, according to the same proof and since when ε and λ tend to 0, it is enough to choose τ 3 and τ 3 + in such a way that

2 log ( τ 3 ) < log ξ 3 ( z 3 , z ) < 2 log ( τ 3 + ) ,

where

3 ( , z ˜ ) H ( , z ˜ 3 ) + G ( , z ˜ 2 ) + 1 ξ γ G ( , z ˜ 1 ) .

Also, by using the fact that

φ 2 3 = φ 2,0 3 + φ 2,1 3 + φ 2 , 3 and φ ˜ 2 3 = φ ˜ 2,0 3 + φ ˜ 2,1 3 + φ ˜ 2 , 3 ,

where φ 2,0 3 , φ ˜ 2,0 3 E 0 = R are constant on S 1 , φ 2,1 3 , φ ˜ 2,1 3 belong to E 1 = Span { e 1 , e 2 } and φ 2 , 3 , φ ˜ 2 , 3 are L 2 ( S 1 ) orthogonal to E 0 and E 1 . We can prove that

( u int , 2 u ext , 2 ) ( z ˜ 3 + r ε , λ . ) = 0 and ( r u int , 2 r u ext , 2 ) ( z ˜ 3 + r ε , λ . ) = 0

on S 1 , which yields to

(155) T 2 , ε 3 = ( t 3 , λ 3 , φ 2,0 3 , φ ˜ 2,0 3 , φ 2,1 3 , φ ˜ 2,1 3 , ¯ 3 ( z ˜ 3 , z ˜ ) , φ 2 , 3 , φ ˜ 2 , 3 ) = O ( r ε , λ 2 ) ,

where

t 3 = 1 log r ε , λ [ 2 log τ 3 + log ξ + 3 ( z ˜ 3 , z ˜ ) ] .

The terms O ( r ε , λ 2 ) exhibit nonlinear dependence on all variables on the left side. However, they are bounded (in the appropriate norm) by a constant (independent of ε and κ ) times r ε , λ 2 , provided ε ( 0 , ε κ ) and λ ( 0 , λ κ ) .

Recall that x = r ε , λ ( z ˜ z ) . In addition, the previous systems can be expressed as for i = 1 , 2 , 3

( x , t i , λ i , φ i , φ ˜ i , ¯ i ) = O ( r ε , λ 2 ) .

By combining (145), (146), (153), (154), and (155), we obtain

(156) T i , ε = ( T i , ε 1 , T c , ε 2 , T i , ε 3 ) = ( O ( r ε , λ 2 ) , O ( r ε , λ 2 ) . O ( r ε , λ 2 ) ) , for i = 1 , 2 .

The nonlinear mapping appearing on the right-hand side of (156) is continuous and compact. In addition, reducing ε κ and λ κ if necessary, this nonlinear mapping sends the ball of radius κ r ε , λ 2 (for the natural product norm) into itself, provided κ is fixed large enough. By applying Schauder’s fixed-point theorem in the ball of radius κ r ε , λ 2 in the product space where the entries live, we establish the existence of a solution of equation (156).

  1. Funding information: This work was supported and funded by the Deanship of Scientific Research at Imam Mohammad Ibn Saud Islamic University (IMSIU) (Grant Number IMSIU-DDRSP2503).

  2. Author contributions: All authors contributed equally to this work.

  3. Conflict of interest: The authors state that there is no conflict of interest.

  4. Ethical approval: Not applicable.

  5. Data availability statement: Not applicable.

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Received: 2023-04-08
Revised: 2024-10-15
Accepted: 2025-02-03
Published Online: 2025-04-15

© 2025 the author(s), published by De Gruyter

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

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