On a fourth order elliptic equation with supercritical exponent

Abstract

This paper is concerned with the semi-linear elliptic problem involving nearly critical exponent $(P_{\epsilon })$: ${\mathrm{\Delta }}^{2}u={|u|}^{8/(n-4)+\epsilon }u$ in Ω, $\mathrm{\Delta }u=u=0$ on ∂ Ω, where Ω is a smooth bounded domain in ${\mathbb{R}}^n$, $n\ge 5$, and ε is a positive real parameter. We show that, for ε small, $(P_{\epsilon })$ has no sign-changing solutions with low energy which blow up at exactly three points. Moreover, we prove that $(P_{\epsilon })$ has no bubble-tower sign-changing solutions.

MSC:35J20, 35J60.

1 Introduction and results

We consider the following semi-linear elliptic problem with supercritical nonlinearity:

$$(P_{\epsilon })\{\begin{array}{c}{\mathrm{\Delta }}^{2}u={|u|}^{p-1+\epsilon }u\text{in }\mathrm{\Omega },\hfill \\ \mathrm{\Delta }u=u=0\text{on }\partial \mathrm{\Omega },\hfill \end{array}$$

where Ω is a smooth bounded domain in ${\mathbb{R}}^n$, $n\ge 5$, ε is a positive real parameter and $p+1=\frac{2n}{n-4}$ is the critical Sobolev exponent for the embedding of $H^2(\mathrm{\Omega })\cap H_0^1(\mathrm{\Omega })$ into $L^{p+1}(\mathrm{\Omega })$.

When the biharmonic operator in $(P_{\epsilon })$ is replaced by the Laplacian operator, there are many works devoted to the study of the counterpart of $(P_{\epsilon })$; see for example [1–6], and the references therein.

When $\epsilon <0$, many works have been devoted to the study of the solutions of $(P_{\epsilon })$ see for example [7–9]. In the critical case, this problem is not compact, that is, when $\epsilon =0$ it corresponds exactly to the limiting case of the Sobolev embedding $H^2(\mathrm{\Omega })\cap H_0^1(\mathrm{\Omega })$ into $L^{p+1}(\mathrm{\Omega })$, and thus we lose the compact embedding. In fact, van Der Vorst showed in [10] that $(P_0)$ has no positive solutions if Ω is a starshaped domain. Whereas Ebobisse and Ould Ahmedou proved in [11] that $(P_0)$ has a positive solution provided that some homology group of Ω is non-trivial. This topological condition is sufficient, but not necessary, as examples of contractible domains Ω on which a positive solution exists show [12].

In the supercritical case, $\epsilon >0$, the problem $(P_{\epsilon })$ becomes more delicate since we lose the Sobolev embedding which is an important point to overcome. The problem $(P_{\epsilon })$ was studied in [7] where the authors show that there is no one-bubble solution to the problem and there is a one-bubble solution to the slightly subcritical case under some suitable conditions. However, we proved in [13] that $(P_{\epsilon })$ has no sign-changing solutions which blow up exactly at two points. In this work we will show the non-existence of sign-changing solutions of $(P_{\epsilon })$ having three concentration points.

We note that problem $(P_{\epsilon })$ has a variational structure. The related functional is

$$infJ(u),\text{where }J(u):=\frac{{\int }_{\mathrm{\Omega }}{|\mathrm{\Delta }u|}^2}{{({\int }_{\mathrm{\Omega }}{|u|}^{p+1+\epsilon })}^{2/(p+1+\epsilon )}},u\in H^2(\mathrm{\Omega })\cap H_0^1(\mathrm{\Omega }),u\not\equiv 0.$$

J satisfies the Palais-Smale condition in the subcritical case, while this condition fails in the critical case. Such a failure is due to the functions

$${\delta }_{(a,\lambda )}(x)=c_0\frac{{\lambda }^{(n-4)/2}}{{(1+{\lambda }^2{|x-a|}^2)}^{(n-4)/2}},c_0={(n(n-4)(n^2-4))}^{(n-4)/8},\lambda >0,a\in {\mathbb{R}}^n.$$

(1.1)

$c_0$ is chosen so that ${\delta }_{(a,\lambda )}$ is the family of solutions of the following problem:

$${\mathrm{\Delta }}^{2}u=u^p,u>0\text{ in }{\mathbb{R}}^n.$$

(1.2)

When we study problem (1.2) in a bounded smooth domain Ω, we need to introduce the function $P{\delta }_{(a,\lambda )}$ which is the projection of ${\delta }_{(a,\lambda )}$ on $H_0^1(\mathrm{\Omega })$. It satisfies

$${\mathrm{\Delta }}^{2}P{\delta }_{(a,\lambda )}={\mathrm{\Delta }}^2{\delta }_{(a,\lambda )}\text{in }\mathrm{\Omega },\mathrm{\Delta }P{\delta }_{(a,\lambda )}=P{\delta }_{(a,\lambda )}=0\text{on }\partial \mathrm{\Omega }.$$

These functions are almost positive solutions of (1.2).

We denote by G the Green’s function defined by, $\mathrm{\forall }x\in \mathrm{\Omega }$,

$${\mathrm{\Delta }}^{2}G(x,\cdot )=c_n{\delta }_x\text{in }\mathrm{\Omega },\mathrm{\Delta }G(x,\cdot )=G(x,\cdot )=0\text{on }\partial \mathrm{\Omega },$$

where ${\delta }_x$ is the Dirac mass at x and $c_n=(n-4)(n-2)w_n$, with $w_n$ is the area of the unit sphere of ${\mathbb{R}}^n$. We denote by H the regular part of G, that is,

$$H(x_1,x_2)={|x_1-x_2|}^{4-n}-G(x_1,x_2)\text{for }(x_1,x_2)\in {\mathrm{\Omega }}^2.$$

For $x=(x_1,x_2)\in {\mathrm{\Omega }}^2\setminus \mathrm{\Gamma }$, with $\mathrm{\Gamma }=\{(y,y):y\in \mathrm{\Omega }\}$, we denote by $M(x)$ the matrix defined by

$$M(x)={(m_{ij})}_{1\le i,j\le 2},\text{where }{m}_{ii}=H(x_i,x_i),m_{12}=m_{21}=G(x_1,x_2),$$

(1.3)

and let $\rho (x)$ be its least eigenvalue.

The space $H^2(\mathrm{\Omega })\cap H_0^1(\mathrm{\Omega })$ is equipped with the norm $\parallel \cdot \parallel$ and its corresponding inner product $〈\cdot ,\cdot 〉$ defined by

$$\parallel u\parallel ={\left({\int }_{\mathrm{\Omega }}{|\mathrm{\Delta }u|}^2\right)}^{1/2}\text{and}〈u,v〉={\int }_{\mathrm{\Omega }}\mathrm{\Delta }u\mathrm{\Delta }v,u,v\in H^2(\mathrm{\Omega })\cap H_0^1(\mathrm{\Omega }).$$

(1.4)

Now, we are able to state our result.

Theorem 1.1 Let Ω be any smooth bounded domain in ${\mathbb{R}}^n$, $n\ge 6$. If 0 is a regular value of $\rho (x)$, then there exists ${\epsilon }_0>0$, such that, for each $\epsilon \in (0,{\epsilon }_0)$, problem $(P_{\epsilon })$ has no sign-changing solutions $u_{\epsilon }$ which satisfy

$$u_{\epsilon }=P{\delta }_{(a_{\epsilon ,1},{\lambda }_{\epsilon ,1})}-P{\delta }_{(a_{\epsilon ,2},{\lambda }_{\epsilon ,2})}+P{\delta }_{(a_{\epsilon ,3},{\lambda }_{\epsilon ,3})}+v_{\epsilon },$$

(1.5)

with ${|u_{\epsilon }|}_{\mathrm{\infty }}^{\epsilon }$ is bounded and

$$\{\begin{array}{c}{a}_{\epsilon ,i}\in \mathrm{\Omega },{\lambda }_{\epsilon ,i}d(a_{\epsilon ,i},\partial \mathrm{\Omega })\to \mathrm{\infty }\mathit{\text{for }}i=1,2,3,\hfill \\ 〈P{\delta }_{(a_{\epsilon ,i},{\lambda }_{\epsilon ,i})},P{\delta }_{(a_{\epsilon ,j},{\lambda }_{\epsilon ,j})}〉\to 0\mathit{\text{for }}i\ne j\mathit{\text{ and }}\parallel v_{\epsilon }\parallel \to 0\mathit{\text{ as }}\epsilon \to 0.\hfill \end{array}$$

The second result deals with the phenomenon of bubble-tower solutions for the biharmonic problem $(P_{\epsilon })$ with supercritical exponent. We will give a generalization of the result found in [13]. More precisely, we have the following.

Theorem 1.2 Let Ω be any smooth bounded domain in ${\mathbb{R}}^n$, $n\ge 5$. There exists ${\epsilon }_0>0$, such that, for each $\epsilon \in (0,{\epsilon }_0)$, problem $(P_{\epsilon })$ has no solutions $u_{\epsilon }$ of the form

$$u_{\epsilon }=\sum _{i=1}^k{\gamma }_{i}P{\delta }_{(a_{\epsilon ,i},{\lambda }_{\epsilon ,i})}+v_{\epsilon },\mathit{\text{with }}{\lambda }_{\epsilon ,1}\le {\lambda }_{\epsilon ,2}\le \cdots \le {\lambda }_{\epsilon ,k}\mathit{\text{ and }}{|u_{\epsilon }|}_{\mathrm{\infty }}^{\epsilon }\mathit{\text{ is bounded}},$$

(1.6)

where $k\ge 2$, ${\gamma }_i\in \{-1,1\}$, $a_{\epsilon ,i}\in \mathrm{\Omega }$, for each $i\le j$, ${\lambda }_{\epsilon ,i}|a_{\epsilon ,i}-a_{\epsilon ,j}|$ is bounded and as $\epsilon \to 0$, $\parallel v_{\epsilon }\parallel \to 0$, ${\lambda }_{\epsilon ,i}d(a_{\epsilon ,i},\partial \mathrm{\Omega })\to +\mathrm{\infty }$, $〈P{\delta }_{(a_{\epsilon ,i},{\lambda }_{\epsilon ,i})},P{\delta }_{(a_{\epsilon ,j},{\lambda }_{\epsilon ,j})}〉\to 0$ for $i\ne j$, and if $l\notin \{k-1,k\}$, ${\lambda }_{\epsilon ,l}|a_{\epsilon ,l}-a_{\epsilon ,l+1}|\to 0$, where $l=min\{q:{\gamma }_q=\cdots ={\gamma }_k\}$.

The proof of our results will be by contradiction. Thus, throughout this paper we will assume that there exist solutions $(u_{\epsilon })$ of $(P_{\epsilon })$ which satisfy (1.5) or (1.6). In Section 2, we will obtain some information as regards such $(u_{\epsilon })$ which allows us to develop Section 3 which deals with some useful estimates to the proof of our theorems. Finally, in Section 4, we combine these estimates to obtain a contradiction. Hence the proof of our results follows.

2 Preliminary results

In this section, we assume that there exist solutions $(u_{\epsilon })$ of $(P_{\epsilon })$ which satisfy

$$u_{\epsilon }=\sum _{i=1}^k{\gamma }_{i}P{\delta }_{(a_{\epsilon ,i},{\lambda }_{\epsilon ,i})}+v_{\epsilon },$$

(2.1)

with ${|u_{\epsilon }|}_{\mathrm{\infty }}^{\epsilon }$ is bounded, $k\ge 2$, $a_{\epsilon ,i}\in \mathrm{\Omega }$, and as $\epsilon \to 0$, $\parallel v_{\epsilon }\parallel \to 0$, ${\lambda }_{\epsilon ,i}d(a_{\epsilon ,i},\partial \mathrm{\Omega })\to +\mathrm{\infty }$, $〈P{\delta }_{(a_{\epsilon ,i},{\lambda }_{\epsilon ,i})},P{\delta }_{(a_{\epsilon ,j},{\lambda }_{\epsilon ,j})}〉\to 0$ for $i\ne j$. Arguing as in [14] and [15], we see that for $u_{\epsilon }$ satisfying (2.1), there is a unique way to choose ${\alpha }_i$, $a_i$, ${\lambda }_i$, and v such that

$$u_{\epsilon }=\sum _{i=1}^k{\gamma }_i{\alpha }_{i}P{\delta }_{(a_i,{\lambda }_i)}+v,$$

(2.2)

$$\text{with }\{\begin{array}{l}{\alpha }_i\in \mathbb{R},{\alpha }_i\to 1,\\ a_i\in \mathrm{\Omega },{\lambda }_i\in {\mathbb{R}}_{+}^{\ast },{\lambda }_{i}d(a_i,\partial \mathrm{\Omega })\to +\mathrm{\infty },\\ v\to 0\text{in }{H}^2(\mathrm{\Omega })\cap H_0^1(\mathrm{\Omega }),v\in E,\end{array}$$

(2.3)

where E denotes the subspace of $H_0^1(\mathrm{\Omega })$ defined by

$$E:=\{w:〈w,\phi 〉=0,\mathrm{\forall }\phi \in Span\{P{\delta }_i,\partial P{\delta }_i/\partial {\lambda }_i,\partial P{\delta }_i/\partial a_i^j,i\le k;j\le n\}\}.$$

(2.4)

Here, $a_i^j$ denotes the j th component of $a_i$ and in the sequel, in order to simplify the notations, we set ${\delta }_{(a_i,{\lambda }_i)}={\delta }_i$ and $P{\delta }_{(a_i,{\lambda }_i)}=P{\delta }_i$. We always assume that $u_{\epsilon }$ (which satisfies (2.1)) is written as in (2.2) and (2.3) holds. From (2.1), it is easy to see that the following remark holds.

Lemma 2.1 [13]

Let $u_{\epsilon }$ satisfying the assumption of the theorems. ${\lambda }_i$ occurring in (2.2) satisfies

$${\lambda }_i^{\epsilon }\to 1\mathit{\text{as }}\epsilon \to 0\mathit{\text{ for each }}i\le k.$$

(2.5)

Remark 2.2 [2, 16]

We recall the following estimate:

$${\delta }_i^{\epsilon }(x)-c_0^{\epsilon }{\lambda }_i^{\epsilon (n-4)/2}=O(\epsilon log(1+{\lambda }_i^2{|x-a_i|}^2))\text{in }\mathrm{\Omega }.$$

(2.6)

3 Some useful estimates

As usual in this type of problems, we first deal with the v-part of $u_{\epsilon }$, in order to show that it is negligible with respect to the concentration phenomenon.

Lemma 3.1 The function v defined in (2.2), satisfies the following estimate:

$$\parallel v\parallel \le c\epsilon +c\{\begin{array}{ll}{\sum }_i\frac{1}{{({\lambda }_i{d}_i)}^{n-4}}+{\sum }_{i\ne j}{\epsilon }_{ij}{(log{\epsilon }_{ij}^{-1})}^{(n-4)/n}& \mathit{\text{if }}n<12,\\ {\sum }_i\frac{1}{{({\lambda }_i{d}_i)}^{(n+4)/2-\epsilon (n-4)}}+{\sum }_{i\ne j}{\epsilon }_{ij}^{(n+4)/2(n-4)}{(log{\epsilon }_{ij}^{-1})}^{(n+4)/2n}& \mathit{\text{if }}n\ge 12,\end{array}$$

where $d_i:=d(a_i,\partial \mathrm{\Omega })$ for $i\le k$ and for $i\ne j$, ${\epsilon }_{ij}$ is defined by

$${\epsilon }_{ij}={(\frac{{\lambda }_i}{{\lambda }_j}+\frac{{\lambda }_j}{{\lambda }_i}+{\lambda }_i{\lambda }_j{|a_i-a_j|}^2)}^{(4-n)/2}.$$

(3.1)

Proof The proof is the same as that of Lemma 3.1 of [13], so we omit it. □

Now, we state the crucial points in the proof of our theorems.

Proposition 3.2 Assume that $n\ge 5$ and let ${\alpha }_i$, $a_i$ and ${\lambda }_i$ be the variables defined in (2.2) with $k=3$ and ${\gamma }_1=-{\gamma }_2={\gamma }_3$. We have

$$\begin{array}{c}|{\alpha }_i{c}_1\frac{n-4}{2}\frac{H(a_i,a_i)}{{\lambda }_i^{n-4}}+\sum _{j\ne i}{(-1)}^{i+j}{\alpha }_j{c}_1({\lambda }_i\frac{\partial {\epsilon }_{ij}}{\partial {\lambda }_i}+\frac{n-4}{2}\frac{H(a_i,a_j)}{{({\lambda }_i{\lambda }_j)}^{(n-4)/2}})+{\alpha }_i\frac{n-4}{2}{c}_2\epsilon |\hfill \\ \le c{\epsilon }^2+c\{\begin{array}{cc}{\sum }_k\frac{1}{{({\lambda }_k{d}_k)}^{n-2}}+{\sum }_{j\ne i}({\epsilon }_{ij}^{\frac{n}{n-4}}log{\epsilon }_{ij}^{-1}+{\epsilon }_{ij}^2{(log{\epsilon }_{12}^{-1})}^{\frac{2(n-4)}{n}})\hfill & \mathit{\text{if }}n\ge 6,\hfill \\ {\sum }_k\frac{1}{{({\lambda }_k{d}_k)}^2}+{\sum }_{j\ne i}{\epsilon }_{ij}^2{(log{\epsilon }_{12}^{-1})}^{2/5}\hfill & \mathit{\text{if }}n=5,\hfill \end{array}\hfill \end{array}$$

(3.2)

where $i,j\in \{1,2,3\}$ with $i\ne j$ and $c_1$, $c_2$ are positive constants.

Proof Let

$$c_1=c_0^{\frac{2n}{n-4}}{\int }_{{\mathbb{R}}^n}\frac{dx}{{(1+{|x|}^2)}^{(n+4)/2}}$$

and

$$c_2=\frac{n-4}{2}{c}_0^{\frac{2n}{n-4}}{\int }_{{\mathbb{R}}^n}log(1+{|x|}^2)\frac{{|x|}^2-1}{{(1+{|x|}^2)}^{n+1}}dx.$$

It suffices to prove the proposition for $i=1$. Multiplying $(P_{\epsilon })$ by ${\lambda }_1\partial P{\delta }_1/\partial {\lambda }_1$ and integrating on Ω, we obtain

$$\begin{array}{c}{\alpha }_1{\int }_{\mathrm{\Omega }}{\delta }_1^p{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}-{\alpha }_2{\int }_{\mathrm{\Omega }}{\delta }_2^p{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}+{\alpha }_3{\int }_{\mathrm{\Omega }}{\delta }_3^p{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}\hfill \\ ={\int }_{\mathrm{\Omega }}{|u_{\epsilon }|}^{p-1+\epsilon }{u}_{\epsilon }{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}.\hfill \end{array}$$

(3.3)

Using [17], we derive

$${\int }_{\mathrm{\Omega }}{\delta }_1^p{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}=\frac{n-4}{2}{c}_1\frac{H(a_1,a_1)}{{\lambda }_1^{n-4}}+O\left(\frac{log({\lambda }_1{d}_1)}{{({\lambda }_1{d}_1)}^{n-1}}\right),$$

(3.4)

$${\int }_{\mathrm{\Omega }}{\delta }_j^p{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}=c_1({\lambda }_1\frac{\partial {\epsilon }_{1j}}{\partial {\lambda }_1}+\frac{n-4}{2}\frac{H(a_1,a_j)}{{({\lambda }_1{\lambda }_j)}^{(n-4)/2}})+R_j,$$

(3.5)

where $j=2,3$ and $R_j$ satisfies

$$R_j=O(\sum _{k=1,j}\frac{log({\lambda }_k{d}_k)}{{({\lambda }_k{d}_k)}^{n-1}}+{\epsilon }_{1j}^{\frac{n}{n-4}}log{\epsilon }_{1j}^{-1}).$$

(3.6)

For the other term of (3.3), we have

$$\begin{array}{c}{\int }_{\mathrm{\Omega }}{|u_{\epsilon }|}^{p-1+\epsilon }{u}_{\epsilon }{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}\hfill \\ ={\int }_{\mathrm{\Omega }}|{\alpha }_{1}P{\delta }_1-{\alpha }_{2}P{\delta }_2+{\alpha }_{3}P{\delta }_3{|}^{p-1+\epsilon }({\alpha }_{1}P{\delta }_1-{\alpha }_{2}P{\delta }_2+{\alpha }_{3}P{\delta }_3){\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}\hfill \\ +(p+\epsilon ){\int }_{\mathrm{\Omega }}|{\alpha }_{1}P{\delta }_1-{\alpha }_{2}P{\delta }_2+{\alpha }_{3}P{\delta }_3{|}^{p-1+\epsilon }v{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}\hfill \\ +O({\parallel v\parallel }^2+\sum _{i\ne j}{\epsilon }_{ij}^{\frac{n}{n-4}}log{\epsilon }_{ij}^{-1}).\hfill \end{array}$$

(3.7)

Concerning the last integral, it can be written as

$$\begin{array}{c}{\int }_{\mathrm{\Omega }}{|{\alpha }_{1}P{\delta }_1-{\alpha }_{2}P{\delta }_2+{\alpha }_{3}P{\delta }_3|}^{p-1+\epsilon }v{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}\hfill \\ ={\int }_{\mathrm{\Omega }}{({\alpha }_{1}P{\delta }_1)}^{p-1+\epsilon }v{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}+O({\int }_{\mathrm{\Omega }\setminus A_j}P{\delta }_j^{p-1}P{\delta }_1|v|+{\int }_{A_j}P{\delta }_1^{p-1}P{\delta }_2|v|),\hfill \end{array}$$

(3.8)

where $A_j=\{x:2{\alpha }_{j}P{\delta }_j\le {\alpha }_{1}P{\delta }_1\}$ for $j=2,3$.

Observe that, for $n\ge 12$, we have $p-1=8/(n-4)\le 1$, thus

$$\begin{array}{rl}{\int }_{\mathrm{\Omega }\setminus A_j}P{\delta }_j^{p-1}P{\delta }_1|v|+{\int }_{A_j}P{\delta }_1^{p-1}P{\delta }_j|v|& \le c{\int }_{\mathrm{\Omega }}|v|{({\delta }_1{\delta }_j)}^{\frac{n+4}{2(n-4)}}\\ \le c\parallel v\parallel {\epsilon }_{1j}^{(n+4)/2(n-4)}{(log{\epsilon }_{1j}^{-1})}^{(n+4)/2n}.\end{array}$$

(3.9)

But for $n<12$, we have

$${\int }_{\mathrm{\Omega }\setminus A_j}P{\delta }_j^{p-1}P{\delta }_1|v|+{\int }_{A}P{\delta }_1^{p-1}P{\delta }_j|v|\le c{\epsilon }_{1j}{(log{\epsilon }_{1j}^{-1})}^{(n-4)/n}\parallel v\parallel .$$

(3.10)

For the other integral in (3.8), using [16, 17], and Remark 2.2, we get

$$\begin{array}{c}{\int }_{\mathrm{\Omega }}P{\delta }_1^{p-1+\epsilon }v{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}\hfill \\ =O(\parallel v\parallel [\epsilon +(\frac{1}{{({\lambda }_1{d}_1)}^{inf(n-4,(n+4)/2)}}(\text{if }n\ne 12)+\frac{log({\lambda }_1{d}_1)}{{({\lambda }_1{d}_1)}^4}(\text{if }n=12))]).\hfill \end{array}$$

(3.11)

It remains to estimate the second integral of (3.7). We have

$$\begin{array}{c}{\int }_{\mathrm{\Omega }}{|{\alpha }_{1}P{\delta }_1-{\alpha }_{2}P{\delta }_2+{\alpha }_{3}P{\delta }_3|}^{p-1+\epsilon }({\alpha }_{1}P{\delta }_1-{\alpha }_{2}P{\delta }_2+{\alpha }_{3}P{\delta }_3){\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}\hfill \\ ={\int }_{\mathrm{\Omega }}{({\alpha }_{1}P{\delta }_1)}^{p+\epsilon }{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}-{\int }_{\mathrm{\Omega }}{({\alpha }_{2}P{\delta }_2)}^{p+\epsilon }{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}+{\int }_{\mathrm{\Omega }}{({\alpha }_{3}P{\delta }_3)}^{p+\epsilon }{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}\hfill \\ -(p+\epsilon )({\int }_{\mathrm{\Omega }}{\alpha }_{2}P{\delta }_2{({\alpha }_{1}P{\delta }_1)}^{p-1+\epsilon }{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}-{\int }_{\mathrm{\Omega }}{\alpha }_{3}P{\delta }_3{({\alpha }_{1}P{\delta }_1)}^{p-1+\epsilon }{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1})\hfill \\ +O(\sum {\epsilon }_{1j}^{\frac{n}{n-4}}log{\epsilon }_{1j}^{-1}).\hfill \end{array}$$

(3.12)

Now, using Remark 2.2 and [17], we have

$$\begin{array}{c}{\int }_{\mathrm{\Omega }}P{\delta }_1^{p+\epsilon }{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}=\frac{n-4}{2}(c_2\epsilon +2c_1\frac{H(a_1,a_1)}{{\lambda }_1^{n-4}})\hfill \\ \phantom{{\int }_{\mathrm{\Omega }}P{\delta }_1^{p+\epsilon }{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}=}+O({\epsilon }^2+\frac{log({\lambda }_1{d}_1)}{{({\lambda }_1{d}_1)}^{n-1}}+\frac{1}{{({\lambda }_1{d}_1)}^2}(\text{if }n=5)),\hfill \end{array}$$

(3.13)

$${\int }_{\mathrm{\Omega }}P{\delta }_j^{p+\epsilon }{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}=c_1({\lambda }_1\frac{\partial {\epsilon }_{1j}}{\partial {\lambda }_1}+\frac{n-4}{2}\frac{H(a_1,a_j)}{{({\lambda }_1{\lambda }_j)}^{(n-4)/2}})+T_j,$$

(3.14)

$$p{\int }_{\mathrm{\Omega }}P{\delta }_{j}P{\delta }_1^{p-1+\epsilon }{\lambda }_1\frac{\partial P{\delta }_1}{\partial {\lambda }_1}=c_1({\lambda }_1\frac{\partial {\epsilon }_{1j}}{\partial {\lambda }_1}+\frac{n-4}{2}\frac{H(a_1,a_j)}{{({\lambda }_1{\lambda }_j)}^{(n-4)/2}})+T_j,$$

(3.15)

where for $i=2,3$,

$$\begin{array}{rcl}{T}_i& =& O\left(\epsilon {\epsilon }_{1j}{(log{\epsilon }_{1j}^{-1})}^{\frac{n-4}{n}}\right)+({\epsilon }_{1j}^{\frac{n}{n-4}}(log{\epsilon }_{1j}^{-1})+\frac{log({\lambda }_i{d}_i)}{{({\lambda }_i{d}_i)}^n}(\text{if }n\ge 8))\\ +(\frac{{\epsilon }_{1j}{(log{\epsilon }_{1j}^{-1})}^{\frac{n-4}{n}}}{{({\lambda }_i{d}_i)}^{n-4}}(\text{if }n<8)).\end{array}$$

Therefore, combining (3.3)-(3.15), and Lemma 3.1, the proof of Proposition 3.2 follows. □

Proposition 3.3 Let $n\ge 6$. We have the following estimate:

$$\begin{array}{c}{\alpha }_i\frac{1}{{\lambda }_i^{n-3}}\frac{\partial H(a_i,a_i)}{\partial a_i}-\frac{2}{{\lambda }_i}\sum _{j\ne i}{(-1)}^{i+j}{\alpha }_j(\frac{\partial {\epsilon }_{ij}}{\partial a_i}-\frac{1}{{({\lambda }_i{\lambda }_j)}^{(n-4)/2}}\frac{\partial H}{\partial a_i}(a_i,a_j))\hfill \\ =O(\sum _k\frac{1}{{({\lambda }_k{d}_k)}^{n-2}}+\sum _{j\ne i}{\epsilon }_{ij}^{\frac{n}{n-4}}log{\epsilon }_{ij}^{-1}+{\epsilon }_{ij}^2{(log{\epsilon }_{1j}^{-1})}^{\frac{2(n-4)}{n}}+{\epsilon }^2+\frac{\epsilon }{{({\lambda }_i{d}_i)}^{n-3}}),\hfill \end{array}$$

where $i,j\in \{1,2,3\}$ and $j\ne i$.

Proof The proof is similar to the proof of Proposition 3.2. But there exist some integrals which have different estimates. We will focus in those integrals. In fact, (3.3), (3.7)-(3.12) are also true if we change ${\lambda }_1\partial P{\delta }_1/\partial {\lambda }_1$ by $(1/{\lambda }_1)\partial P{\delta }_1/\partial a_1$. It remains to deal with the other equations. Following [17], we get

$${\int }_{\mathrm{\Omega }}{\delta }_1^p\frac{1}{{\lambda }_1}\frac{\partial P{\delta }_1}{\partial a_1}=-\frac{1}{2}\frac{c_1}{{\lambda }_1^{n-3}}\frac{\partial H(a_1,a_1)}{\partial a_1}+O\left(\frac{1}{{({\lambda }_1{d}_1)}^{n-1}}\right),$$

(3.16)

$$\begin{array}{c}{\int }_{\mathrm{\Omega }}{\delta }_j^p\frac{1}{{\lambda }_1}\frac{\partial P{\delta }_1}{\partial a_1}=\frac{c_1}{{\lambda }_1}(\frac{\partial {\epsilon }_{1j}}{\partial a_1}-\frac{1}{{({\lambda }_1{\lambda }_j)}^{(n-4)/2}}\frac{\partial H}{\partial a_1}(a_1,a_j))\hfill \\ \phantom{{\int }_{\mathrm{\Omega }}{\delta }_j^p\frac{1}{{\lambda }_1}\frac{\partial P{\delta }_1}{\partial a_1}=}+O(\sum _{k=1,j}\frac{1}{{({\lambda }_k{d}_k)}^{n-1}}+{\lambda }_j|a_1-a_j|{\epsilon }_{1j}^{(n-1)/(n-4)}),\hfill \end{array}$$

(3.17)

$${\int }_{\mathrm{\Omega }}P{\delta }_1^{p+\epsilon }\frac{1}{{\lambda }_1}\frac{\partial P{\delta }_1}{\partial a_1}=-c_0^{\epsilon }{\lambda }_1^{\epsilon (n-4)/2}\frac{c_1}{{\lambda }_1^{n-3}}\frac{\partial H(a_1,a_1)}{\partial a_1}+O(\frac{1}{{({\lambda }_1{d}_1)}^{n-2}}+\frac{\epsilon }{{({\lambda }_1{d}_1)}^{n-3}}),$$

(3.18)

$${\int }_{\mathrm{\Omega }}P{\delta }_j^{p+\epsilon }\frac{1}{{\lambda }_1}\frac{\partial P{\delta }_1}{\partial {\lambda }_1}=c_0^{\epsilon }{\lambda }_j^{\epsilon (n-4)/2}(P{\delta }_j,\frac{1}{{\lambda }_1}\frac{\partial P{\delta }_1}{\partial a_1})+O\left(\epsilon {\epsilon }_{1j}{(log{\epsilon }_{1j}^{-1})}^{(n-4)/n}\right)+T_j,$$

(3.19)

$${\int }_{\mathrm{\Omega }}P{\delta }_j\frac{1}{{\lambda }_1}\frac{\partial (P{\delta }_1^{p+\epsilon })}{\partial a_1}=c_0^{\epsilon }{\lambda }_1^{\epsilon (n-4)/2}(P{\delta }_j,\frac{1}{{\lambda }_1}\frac{\partial P{\delta }_1}{\partial a_1})+O\left(\epsilon {\epsilon }_{1j}{(log{\epsilon }_{1j}^{-1})}^{(n-4)/n}\right)+T_j.$$

(3.20)

The proof of Proposition 3.3 is thereby completed. □

4 Proof of the theorems

Proof of Theorem 1.1

Arguing by contradiction, let us assume that problem $(P_{\epsilon })$ has solutions $(u_{\epsilon })$ as stated in Theorem 1.1. Recall that $u_{\epsilon }$ is written as

$$u_{\epsilon }={\alpha }_{\epsilon ,1}P{\delta }_{(a_{\epsilon ,1},{\lambda }_{\epsilon ,1})}-{\alpha }_{\epsilon ,2}P{\delta }_{(a_{\epsilon ,2},{\lambda }_{\epsilon ,2})}+{\alpha }_{\epsilon ,3}P{\delta }_{(a_{\epsilon ,3},{\lambda }_{\epsilon ,3})}+v_{\epsilon },$$

with $v_{\epsilon }$ orthogonal to each $P{\delta }_{(a_i,{\lambda }_i)}$ and their derivatives with respect to ${\lambda }_i$ and ${(a_i)}_k$, where ${(a_i)}_k$ denotes the k th component of $a_i$ (see (2.2) and (2.3)). For simplicity, we will write ${\alpha }_i:={\alpha }_{\epsilon ,i}$, ${\lambda }_i:={\lambda }_{\epsilon ,i}$, and $a_i:=a_{\epsilon ,i}$. From Proposition 3.2, for each $i=1,2,3$, with ${\gamma }_1={\gamma }_3=1$, ${\gamma }_2=-1$. We have

$$\begin{array}{c}(E_i)c_1\frac{n-4}{2}\frac{H(a_i,a_i)}{{\lambda }_i^{n-4}}+{\gamma }_i{c}_1\sum _{j\ne i}{\gamma }_j({\lambda }_i\frac{\partial {\epsilon }_{ij}}{\partial {\lambda }_i}+\frac{n-4}{2}\frac{H(a_i,a_j)}{{({\lambda }_i{\lambda }_j)}^{(n-4)/2}})+\frac{n-4}{2}{c}_2\epsilon \hfill \\ \phantom{(E_i)}=o(\epsilon +\sum _{j=1}^3\frac{1}{{({\lambda }_j{d}_j)}^{n-4}}+\sum _{r\ne j}{\epsilon }_{rj}).\hfill \end{array}$$

Furthermore, an easy computation shows that

$${\lambda }_i\frac{\partial {\epsilon }_{ij}}{\partial {\lambda }_i}=-\frac{n-4}{2}{\epsilon }_{ij}(1-2\frac{{\lambda }_j}{{\lambda }_i}{\epsilon }_{ij}^{2/n-4})\text{for }i,j=1,2,3,j\ne i,$$

(4.1)

$$-{\lambda }_i\frac{\partial {\epsilon }_{ij}}{\partial {\lambda }_i}-2{\lambda }_j\frac{\partial {\epsilon }_{ij}}{\partial {\lambda }_j}\ge \frac{n-4}{2}{\epsilon }_{ij}\text{for }{\lambda }_i\le {\lambda }_j.$$

(4.2)

On the other hand, following the proof of Proposition 3.3, we have, for each $i=1,2,3$,

$$\begin{array}{c}(F_i)\frac{1}{{\lambda }_i^{n-3}}\frac{\partial H(a_i,a_i)}{\partial a_i}-\sum _{j\ne i}2\frac{{(-1)}^{j+i}}{{\lambda }_i}(\frac{\partial {\epsilon }_{ji}}{\partial a_i}-\frac{\partial H(a_j,a_i)}{\partial a_i}\frac{1}{{({\lambda }_j{\lambda }_i)}^{(n-4)/2}})\hfill \\ \phantom{(F_i)}=o(\sum _j\frac{1}{{({\lambda }_j{d}_j)}^{n-3}}+\sum _{r\ne j}{\epsilon }_{rj}^{\frac{n-3}{n-4}}+{\epsilon }^{\frac{n-3}{n-4}}).\hfill \end{array}$$

(4.3)

We distinguish many cases depending on the set

$$Ϝ:=\{(i,j):i\ne j\text{ and }min({\lambda }_i,{\lambda }_j)|a_i-a_j|\text{ is bounded}\}$$

and we will prove that all these cases cannot occur.

We remark that if $(i,j)\in Ϝ$ we derive ${\lambda }_i/{\lambda }_j\to 0\text{ or }\mathrm{\infty }$ and $d_i/d_j=1+o(1)$ as $\epsilon \to 0$.

Furthermore, the behavior of ${\epsilon }_{ij}$ depends on the set Ϝ. In fact we have, assuming that ${\lambda }_i\le {\lambda }_j$,

$$c{\left(\frac{{\lambda }_i}{{\lambda }_j}\right)}^{(n-4)/2}\le {\epsilon }_{ij}\le {\left(\frac{{\lambda }_i}{{\lambda }_j}\right)}^{(n-4)/2}\text{if }(i,j)\in Ϝ,$$

(4.4)

$${\epsilon }_{ij}=\frac{1}{{({\lambda }_i{\lambda }_j{|a_i-a_j|}^2)}^{(n-4)/2}}+o({\epsilon }_{ij})\text{if }(i,j)\notin Ϝ.$$

(4.5)

First we start by proving the following crucial lemmas.

Remark 4.1 Ordering the ${\lambda }_i$’s: ${\lambda }_{i_1}\le {\lambda }_{i_2}\le {\lambda }_{i_3}$, adding $(E_{i_1})+2(E_{i_2})+4(E_{i_3})$, and using (4.2), it is easy to derive a contradiction if we have ${\epsilon }_{13}=o(\sum {({\lambda }_i{d}_i)}^{4-n}+\sum {\epsilon }_{rj}+\epsilon )$.

Lemma 4.2 Let $n\ge 4$. Then there exists a positive constant ${\underline{c}}_0>0$ such that

$$\begin{array}{rl}(\mathrm{i})& {\underline{c}}_0^{-1}\le \frac{d_1}{d_3}\le {\underline{c}}_0;\\ (\mathrm{ii})& {\underline{c}}_0^{-1}\le \frac{{\lambda }_1}{{\lambda }_3}\le {\underline{c}}_0;\\ (\mathrm{iii})& {\underline{c}}_0^{-1}\le \frac{|a_1-a_3|}{d_i}\le {\underline{c}}_0^{-1}\mathit{\text{for }}i=1,3.\end{array}$$

Proof The proof will be by contradiction.

Proof of (i). Assume that $d_1/d_3\to 0$. In this case, we have

$$|a_1-a_3|\ge c{d}_3\text{and}{\epsilon }_{13}=\frac{1}{{({\lambda }_1{\lambda }_3{|a_1-a_3|}^2)}^{(n-4)/2}}+o({\epsilon }_{13}),$$

(4.6)

which implies that ${\epsilon }_{13}=o({({\lambda }_1{d}_1)}^{4-n}+{({\lambda }_3{d}_3)}^{4-n})$. Using Remark 4.1, we derive a contradiction. In the same way, we prove that $d_3/d_1\nrightarrow 0$. Hence the proof of Claim (i) is completed.

Proof of (ii). Assume that ${\lambda }_1/{\lambda }_3\to 0$. By Claim (i), we have ${({\lambda }_3{d}_3)}^{-1}=o({({\lambda }_1{d}_1)}^{-1})$. Four cases may occur.

Case 1. ${\lambda }_2/{\lambda }_3\nrightarrow 0$ or $\{(1,2),(2,3)\}\cap Ϝ=\varphi$. Using (4.5), $(E_2)$ implies that

$$\frac{H(a_2,a_2)}{{\lambda }_2^{n-4}}+{\epsilon }_{12}+{\epsilon }_{23}+\epsilon =o(\frac{1}{{({\lambda }_1{d}_1)}^{n-4}}+{\epsilon }_{13}).$$

By Claim (i) and $(E_3)$, we obtain ${\epsilon }_{13}=o({({\lambda }_1{d}_1)}^{4-n})$. By Remark 4.1, this case cannot occur.

Case 2. ${\lambda }_2/{\lambda }_3\to 0$, $\{(1,2),(2,3)\}\cap Ϝ\ne \varphi$, and ${\lambda }_2/{\lambda }_1\to +\mathrm{\infty }$. In this case, it is easy to obtain ${\epsilon }_{13}=o({\epsilon }_{12}+{\epsilon }_{23})$. Using Remark 4.1, we derive a contradiction.

Case 3. ${\lambda }_2/{\lambda }_3\to 0$, $(2,3)\in Ϝ$, $(1,2)\notin Ϝ$, and ${\lambda }_2/{\lambda }_1\nrightarrow +\mathrm{\infty }$. In this case, we see that ${\lambda }_2|a_2-a_3|$ is bounded and ${\lambda }_2|a_1-a_2|\to +\mathrm{\infty }$. Hence, we derive that ${\lambda }_2|a_1-a_3|\to +\mathrm{\infty }$, which implies that ${\lambda }_k|a_1-a_3|\to +\mathrm{\infty }$ for $k=1,3$. Thus

$${\epsilon }_{13}=\frac{1+o(1)}{{({\lambda }_1{\lambda }_3{|a_1-a_3|}^2)}^{(n-4)/2}}={\left(\frac{{\lambda }_2}{{\lambda }_3}\right)}^{(n-4)/2}\frac{1+o(1)}{{({\lambda }_1{\lambda }_2{|a_1-a_3|}^2)}^{(n-4)/2}}=o({\epsilon }_{23}).$$

Then by Remark 4.1, we get a contradiction.

Case 4. ${\lambda }_2/{\lambda }_3\to 0$, $(1,2)\in Ϝ$, and ${\lambda }_2/{\lambda }_1\nrightarrow +\mathrm{\infty }$. In this case, it is easy to get ${\epsilon }_{23}=o({\epsilon }_{12})$.

Using the formula $[(E_1)+(E_2)-(E_3)]$, we deduce that $\epsilon =o({\epsilon }_{12}+{\epsilon }_{13})$, which implies that ${\epsilon }_{13}=o({\epsilon }_{12})$. Hence by Remark 4.1, we derive a contradiction and Claim (ii) is thereby completed.

Proof of (iii). Without loss of generality, we can assume that $d_1\le d_3$. First, as in the proof of Claim (i), we get $|a_1-a_3|\le c_0{d}_1$. Now assume that $|a_1-a_3|/d_1\to 0$, which implies

$$\frac{H(a_i,a_i)}{{\lambda }_i^{n-4}}=o({\epsilon }_{13})\text{for }i=1,3.$$

Two cases may occur.

Case 1. ${\lambda }_1\le {\lambda }_2$ or $\{(1,2),(2,3)\}\cap Ϝ=\varphi$. Using $(E_2)$, we obtain

$$\frac{H(a_2,a_2)}{{\lambda }_2^{n-4}}=o({\epsilon }_{13}),{\epsilon }_{i2}=o({\epsilon }_{13})\text{for }i=1,3\text{and}\epsilon =o({\epsilon }_{13}),$$

and we derive a contradiction from $(E_1)$.

Case 2. ${\lambda }_2\le {\lambda }_1$ and $\{(1,2),(2,3)\}\cap Ϝ\ne \varphi$. Let $k\in \{1,3\}$ such that $(2,k)\in Ϝ$. Using Claim (ii) and the fact that ${\lambda }_2\le {\lambda }_1$, we derive that ${\epsilon }_{2k}\ge c{({\lambda }_2/{\lambda }_k)}^{(n-4)/2}$, which implies that $d_2\sim d_k$, ${\lambda }_2/{\lambda }_k\to 0$, and ${\lambda }_2|a_2-a_k|$ is bounded. Using (4.3) for $i=k$, we get

$$-{\lambda }_2|a_2-a_k|{\epsilon }_{2k}^{\frac{n}{n-4}}+\frac{{\lambda }_1{\lambda }_3}{{\lambda }_k}|a_1-a_3|{\epsilon }_{13}^{\frac{n}{n-4}}=o(\frac{1}{{({\lambda }_2{d}_2)}^{n-3}}+\sum _{r\ne j}{\epsilon }_{rj}^{\frac{n-3}{n-4}}+{\epsilon }^{\frac{n-3}{n-4}}).$$

(4.7)

Since ${\lambda }_2|a_2-a_k|$ is bounded and ${\epsilon }_{13}\simeq {({\lambda }_1{\lambda }_3{|a_2-a_k|}^2)}^{(4-n)/2}$, we derive that

$${\epsilon }_{13}^{\frac{n-3}{n-4}}=o(\frac{1}{{({\lambda }_2{d}_2)}^{n-3}}+{\epsilon }_{12}^{\frac{n-3}{n-4}}+{\epsilon }_{23}^{\frac{n-3}{n-4}}+{\epsilon }^{\frac{n-3}{n-4}}),$$

which implies that

$${\epsilon }_{13}=o(\frac{1}{{({\lambda }_2{d}_2)}^{n-4}}+{\epsilon }_{12}+{\epsilon }_{23}+\epsilon ).$$

(4.8)

By Remark 4.1, we get a contradiction. □

Lemma 4.3 There exists a positive constant ${\underline{c}}_0^{\prime }$ such that

$$\begin{array}{rl}(\mathrm{i})& {\underline{c}}_0^{\prime }{\lambda }_1\le {\lambda }_2;\\ (\mathrm{ii})& d_i\ge {\underline{c}}_0^{\prime }\mathit{\text{for }}i=1,3.\end{array}$$

Proof Without loss of generality, we can assume that $d_1\le d_3$.

Proof of (i). Assume that ${\lambda }_2/{\lambda }_1\to 0$. First we claim that $d_1/d_2\nrightarrow 0$. In fact, arguing by contradiction we assume that $d_1/d_2\to 0$, we get $d_1\to 0$, $|a_1-a_2|\ge c{d}_2$, and $|a_2-a_3|\ge c{d}_2$. Hence, $\{(1,2),(2,3)\}\cap Ϝ=\varphi$. From $(E_2)$, we obtain

$$\frac{H(a_2,a_2)}{{\lambda }_2^{n-4}}+{\epsilon }_{12}+{\epsilon }_{23}+\epsilon =o(\frac{1}{{({\lambda }_1{d}_1)}^{n-4}}+\frac{1}{{({\lambda }_3{d}_3)}^{n-4}}+{\epsilon }_{13}).$$

(4.9)

Let ${\nu }_i$ be the outward normal vector at $a_i$. Since $d_1$, $d_3$, and $|a_1-a_3|$ are of the same order, we have (see [18] and [19])

$$\frac{1}{{\lambda }_1^{n-3}}\frac{\partial H(a_1,a_1)}{\partial {\nu }_1}\sim \frac{c}{{({\lambda }_1{d}_1)}^{n-3}}\text{and}\frac{\partial G(a_1,a_3)}{\partial {\nu }_1}\le 0.$$

(4.10)

Using $(F_1)$, we get $1/{({\lambda }_1{d}_1)}^{n-3}=o({\epsilon }_{13}^{(n-3)/(n-4)})$, which implies that $1/{({\lambda }_1{d}_1)}^{n-4}=o({\epsilon }_{13})$. From $(E_1)$, we derive a contradiction. Hence our claim is proved.

Thus there exists a positive constant c so that $d_1\ge c{d}_2$. Now, since we have assumed that ${\lambda }_2/{\lambda }_1\to 0$, Lemma 4.2 implies that ${\epsilon }_{13}=o({({\lambda }_2{d}_2)}^{4-n})$. Finally, using Remark 4.1, we get a contradiction and the proof of Claim (i) follows.

Proof of (ii). Assume that $d_1\to 0$. Note that Claim (i) and $(E_2)$ imply that (4.9) holds.

Now, following the proof of (i), we obtain a contradiction. □

We turn now to the proof of Theorem 1.1. By the previous lemmas, we know that ${\lambda }_1$ and ${\lambda }_3$ are of the same order, $|a_1-a_3|\ge c$ and ${\lambda }_2\ge c{\lambda }_i$, for $i=1,3$ where c is a positive constant.

Hence, $(E_2)$ implies that (4.9) holds. Furthermore, for $i=1,3$ $(E_i)$ implies that

$$\frac{H(a_i,a_i)}{{\lambda }_i^{n-4}}-\frac{G(a_1,a_3)}{{({\lambda }_1{\lambda }_3)}^{n-4}}=o(\frac{1}{{({\lambda }_1{d}_1)}^{n-4}}+\frac{1}{{({\lambda }_3{d}_3)}^{n-4}}+{\epsilon }_{13}).$$

(4.11)

We denote by $r(x)$ the eigenvector associated to $\rho (x)$ whose norm is 1. We point out that we can choose $r(x)$ so that all their components are positive (see [18] and [19]).

Let ${\mathrm{\Lambda }}_i={\lambda }_i^{(4-n)/2}$, $\mathrm{\Lambda }=({\mathrm{\Lambda }}_1,{\mathrm{\Lambda }}_3)$, and $x=(a_1,a_3)$. From (4.11), we have

$$M(x)\cdot \frac{{}^t\mathrm{\Lambda }}{\parallel \mathrm{\Lambda }\parallel }=o(1).$$

(4.12)

The scalar product of (4.12) by $r(x)$ gives

$$\rho (x)r(x)\cdot \frac{{}^t\mathrm{\Lambda }}{\parallel \mathrm{\Lambda }\parallel }=o(1).$$

(4.13)

Since the components of $r(x)$ are positive and ${\lambda }_1$, ${\lambda }_3$ are of the same order, there exists a positive constant c, such that $r(x)\cdot \frac{{}^t\mathrm{\Lambda }}{\parallel \mathrm{\Lambda }\parallel }\ge c>0$. Hence, we get

$$\rho (x)=o(1).$$

(4.14)

We deduce from (4.3) and (4.11) that

$$\frac{\partial M}{\partial x_i}(x)\cdot \frac{{}^t\mathrm{\Lambda }}{\parallel \mathrm{\Lambda }\parallel }=o(1).$$

(4.15)

Observe that Λ may be written in the form

$$\mathrm{\Lambda }=\beta r(x)+\overline{r}(x),\text{with }r(x)\cdot \overline{r}(x)=0,\parallel \overline{r}\parallel =o(\beta )\text{ and }\beta \sim \parallel \mathrm{\Lambda }\parallel .$$

(4.16)

Using (4.15), we get

$$\frac{\partial M}{\partial x_i}(x)\cdot {}^{t}r(x)+\frac{\partial M}{\partial x_i}(x)\cdot \frac{\overline{r}(x)}{\parallel \mathrm{\Lambda }\parallel }=o(1).$$

(4.17)

Since $d_i\ge c_0$ for $i=1,3$ and $|a_1-a_3|\ge c_0$, the matrix $\frac{\partial M}{\partial x_i}(x)$ is bounded.

Furthermore, we have $\parallel \overline{r}\parallel =o(\parallel \mathrm{\Lambda }\parallel )$, which implies that

$$\frac{\partial M}{\partial x_i}(x)\cdot {}^{t}r(x)=o(1).$$

(4.18)

Let us consider the equality

$$M(x)\cdot {}^{t}r(x)=\rho (x)\cdot {}^{t}r(x)$$

and derivative it with respect to $x_i$; we obtain

$$\frac{\partial M}{\partial x_i}(x)\cdot {}^{t}r(x)+M(x)\frac{{\partial }^{t}r}{\partial x_i}(x)=\frac{\partial \rho }{\partial x_i}(x)\cdot {}^{t}r(x)+\rho (x)\frac{{\partial }^{t}r}{\partial x_i}(x).$$

The scalar product with $r(x)$ gives

$$r(x)\cdot \frac{\partial M}{\partial x_i}(x)\cdot {}^{t}r(x)=\frac{\partial \rho }{\partial x_i}(x).$$

(4.19)

Using (4.18), we obtain

$$\frac{\partial \rho }{\partial x_i}(x)=o(1).$$

(4.20)

Hence, we derive a contradiction from (4.14), (4.20), and the fact that 0 is a regular value of ρ. Thus the proof of our theorem follows.

Proof of Theorem 1.2

Arguing by contradiction, let us assume that problem $(P_{\epsilon })$ has solutions $(u_{\epsilon })$ as stated in Theorem 1.2. From Section 2, these solutions have to satisfy (2.2) and (2.3).

As in the proof of Proposition 3.2, we have, for each $i=1,\dots ,k$,

$$\begin{array}{c}(E_i)c_1\frac{n-4}{2}\frac{H(a_i,a_i)}{{\lambda }_i^{n-4}}+{\gamma }_i{c}_1\sum _{j\ne i}{\gamma }_j({\lambda }_i\frac{\partial {\epsilon }_{ij}}{\partial {\lambda }_i}+\frac{n-4}{2}\frac{H(a_i,a_j)}{{({\lambda }_i{\lambda }_j)}^{(n-4)/2}})+\frac{n-4}{2}{c}_2\epsilon \hfill \\ \phantom{(E_i)}=o(\epsilon +\sum _{j=1}^k\frac{1}{{({\lambda }_j{d}_j)}^{n-4}}+\sum _{r\ne j}{\epsilon }_{rj}).\hfill \end{array}$$

Observe that, if $j<i$, we have ${\lambda }_j|a_i-a_j|$ is bounded (by the assumption) which implies that

$$\begin{array}{r}|a_i-a_j|=o(d_j),d_i/d_j=1+o(1),\mathrm{\forall }i,j\text{and}\\ {\epsilon }_{ij}\ge c{({\lambda }_j/{\lambda }_i)}^{(n-4)/2},\mathrm{\forall }j<i,\end{array}$$

(4.21)

where c is a positive constant. Using (4.21), easy computations show that

$$\begin{array}{r}{\epsilon }_{(i-1)j}+{\epsilon }_{i(j+1)}=o({\epsilon }_{ij}),\mathrm{\forall }i<j,\\ \frac{H(a_i,a_j)}{{({\lambda }_i{\lambda }_j)}^{(n-4)/2}}=o\left(\frac{1}{{({\lambda }_1{d}_1)}^{n-4}}\right)\text{if }(i,j)\ne (1,1).\end{array}$$

(4.22)

Thus, using (4.22), $(E_i)$ can be written as

$$\begin{array}{rl}\left(E_1^{\prime }\right)& c_1\frac{n-4}{2}\frac{H(a_1,a_1)}{{\lambda }_1^{n-4}}+c_1{\gamma }_1{\gamma }_2{\lambda }_1\frac{\partial {\epsilon }_{12}}{\partial {\lambda }_1}+\frac{n-4}{2}{c}_2\epsilon =o(\epsilon +\frac{1}{{({\lambda }_1{d}_1)}^{n-4}}+\sum _{r\ne j}{\epsilon }_{rj}),\\ \left(E_k^{\prime }\right)& c_1{\gamma }_{k-1}{\gamma }_k{\lambda }_k\frac{\partial {\epsilon }_{(k-1)k}}{\partial {\lambda }_k}+\frac{n-4}{2}{c}_2\epsilon =o(\epsilon +\frac{1}{{({\lambda }_1{d}_1)}^{n-4}}+\sum _{r\ne j}{\epsilon }_{rj}),\end{array}$$

and for $1<i<k$,

$$\left(E_i^{\prime }\right)c_1{\gamma }_{i-1}{\gamma }_i{\lambda }_i\frac{\partial {\epsilon }_{(i-1)i}}{\partial {\lambda }_i}+c_1{\gamma }_i{\gamma }_{i+1}{\lambda }_i\frac{\partial {\epsilon }_{i(i+1)}}{\partial {\lambda }_i}+\frac{n-4}{2}{c}_2\epsilon =o(\epsilon +\frac{1}{{({\lambda }_1{d}_1)}^{n-4}}+\sum _{r\ne j}{\epsilon }_{rj}).$$

The proof will depend on the value of l which is defined in the theorem.

Case 1. $l=k$. From the definition of l we get ${\gamma }_{k-1}{\gamma }_k=-1$. Now using (4.1) and $(E_k^{\prime })$, we derive that

$$\epsilon =o(\frac{1}{{({\lambda }_1{d}_1)}^{n-4}}+\sum _{r\ne j}{\epsilon }_{ij})\text{and}{\epsilon }_{(k-1)k}=o(\frac{1}{{({\lambda }_1{d}_1)}^{n-4}}+\sum _{r\ne j}{\epsilon }_{rj}).$$

(4.23)

Now, using (4.23) and $(E_{k-1}^{\prime })$, we derive the estimate of ${\epsilon }_{(k-2)(k-1)}$ and by induction we get

$${\epsilon }_{(i-1)i}=o(\frac{1}{{({\lambda }_1{d}_1)}^{n-4}}+\sum _{r\ne j}{\epsilon }_{rj})\text{for each }i=2,\dots ,k.$$

(4.24)

Finally, using (4.22), (4.23), (4.24), and $(E_1^{\prime })$ we obtain

$$\frac{H(a_1,a_1)}{{\lambda }_1^{n-4}}=o\left(\frac{1}{{({\lambda }_1{d}_1)}^{n-4}}\right),$$

which gives a contradiction.

Case 2. $l=k-1$. Using (4.1), an easy computation implies that

$${\lambda }_{k-1}\frac{\partial {\epsilon }_{(k-1)k}}{\partial {\lambda }_{k-1}}-{\lambda }_k\frac{\partial {\epsilon }_{(k-1)k}}{\partial {\lambda }_k}\ge c{\epsilon }_{(k-1)k}.$$

(4.25)

Then from $(E_{k-1}^{\prime })$, $(E_k^{\prime })$, (4.1), (4.25), and the fact that ${\gamma }_{k-1}{\gamma }_k=1$ and ${\gamma }_{k-2}{\gamma }_{k-1}=-1$ (since $l=k-1$), we obtain

$$c{\epsilon }_{(k-1)k}+{\epsilon }_{(k-2)(k-1)}=o(\epsilon +\frac{1}{{({\lambda }_1{d}_1)}^{n-4}}+\sum _{r\ne j}{\epsilon }_{rj}).$$

(4.26)

Now using $(E_k^{\prime })$ and (4.26) we get (4.23) and as before, (4.24) is satisfied. Hence we also derive a contradiction from $(E_1^{\prime })$.

Case 3. $l\notin \{k,k-1\}$. Recall that in this case we have assumed that ${\lambda }_l|a_l-a_{l+1}|\to 0$. This implies that

$${\lambda }_l\frac{\partial {\epsilon }_{l(l+1)}}{\partial {\lambda }_l}=((n-4)/2){\epsilon }_{l(l+1)}(1+o(1)).$$

(4.27)

Hence, using $(E_l^{\prime })$, the definition of l and (4.1) we obtain the first part of (4.23). The second part follows from $(E_k^{\prime })$ and the first one. Finally, as before we derive a contradiction from $(E_1^{\prime })$.

Hence, our theorem is proved.