Nonlinear problem with subcritical exponent in Sobolev space

Abstract

Using Brouwer’s fixed point theorem, we prove the existence of solutions for some nonlinear problem with subcritical Sobolev exponent in $S_{+}^{4}$.

1 Introduction and the main result

The exponent Lebesgue space $L^{p}(\Omega)$ is defined by

$$L^{p}(\Omega)= \biggl\{ u\in L_{\mathrm{loc}}^{1}(\Omega): \int_{\Omega }\bigl\vert u(x)\bigr\vert ^{p}\,dx< \infty \biggr\} . $$

This space is endowed with the norm

$$\Vert u\Vert _{ L^{p}(\Omega)} =\inf \biggl\{ \lambda>0: \int_{\Omega}\biggl\vert \frac {u(x)}{\lambda} \biggr\vert ^{p}\,dx\leq1 \biggr\} . $$

The Sobolev space $W^{1,p}(\Omega)$ is defined by

$$W^{1,p}(\Omega)= \bigl\{ u\in W_{\mathrm{loc}}^{1,1}(\Omega):u \in L^{p}(\Omega) \text{ and }\vert \nabla u \vert \in L^{p}(\Omega) \bigr\} . $$

The corresponding norm for this space is

$$ \Vert u\Vert _{ W^{1,p}(\Omega )} =\Vert u\Vert _{L^{p}(\Omega)}+ \Vert \nabla u\Vert _{ L^{p}(\Omega)}. $$

Define $W_{0}^{1}(\Omega)=H_{0}^{1}(\Omega)$ as the closure of $C_{c}^{\infty}(\Omega)$ with respect to the $W^{1,p}(\Omega)$ norm which is a Hilbert space [1].

We consider the problem of the scalar curvature on the standard four dimensional half sphere under minimal boundary conditions $(S)$:

$$(S)\quad \textstyle\begin{cases} L_{g}u:=-\Delta_{g} u+2u = Ku^{3},\qquad u>0 & \mbox{in } S^{4}_{+} , \\ \frac{\partial u}{\partial\nu} = 0 & \mbox{on } \partial S^{4}_{+}, \end{cases} $$

where $S^{4}_{+}= \{ x\in \mathbb{R}^{5} / \vert x\vert = 1, x_{5} >0 \}$, g is the standard metric, and K is a $C^{3}$ positive Morse function on $\overline{S_{+}^{4}}$.

The scalar curvature problem on $S^{n}$ and also on $S^{n}_{+}$ was the subject of several works in recent years, we can cite for example [2–12].

Recall that the embedding of $H^{1}(S^{4}_{+})$ into $L^{4}(S^{4}_{+})$ is noncompact. For this reason, we have focused our study on the family of subcritical problems $(S_{\varepsilon})$

$$(S_{\varepsilon})\quad \textstyle\begin{cases} -\Delta_{g} u+2u = Ku^{3-\varepsilon},\qquad u>0 &\mbox{in }S^{4}_{+}, \\ \frac{\partial u}{\partial\nu} = 0 & \mbox{on } \partial S^{4}_{+}, \end{cases} $$

where ε is a small positive parameter.

Note that the solutions of problem $(S)$ can be the limit as $\varepsilon\to0$ of some solutions $(u_{\varepsilon})$ for $(S_{\varepsilon})$.

Djadli et al. [13] studied this problem in the case of the three dimensional half sphere. Assuming that the critical points of $K_{1}$ verify $(\partial K/\partial\nu)(a_{i})>0$ they demonstrated that there exist solutions $(u_{\varepsilon})$ concentrated at the points $(a_{1}, \ldots, a_{p})$. Moreover, in [14], we established the existence of another type of solutions $(u_{\varepsilon})$ of $(S_{\varepsilon})$ such that is concentrated at two points $a_{1}\in \partial S^{4}_{+}$ and $a_{2}\in S^{4}_{+}$.

In this work, we aim to construct some positive solutions of $(S_{\varepsilon})$ which are concentrated at two different points of the boundary. To state our result, we will give the following notations. For $a\in \overline{S_{+}^{4}}$ and $\lambda>0$, let

$$ \delta_{(a,\lambda) }(x)= c_{0}\frac{\lambda}{(\lambda ^{2}+1+(1-\lambda^{2}) \cos d(a,x))}, $$

(1)

where d is the geodesic distance on $(\overline{S_{+}^{4}},g)$ and $c_{0}$ is chosen so that $\delta_{(a,\lambda) }$ is the family of solutions of the following problem:

$$-\Delta u + 2u= u^{3}, \qquad u>0, \quad\mbox{in } S^{4}. $$

The space $H^{1}(S_{+}^{4})$ is equipped with the norm $\Vert \cdot \Vert$ and its corresponding inner product $\langle\cdot ,\cdot \rangle$:

$$\begin{aligned} \Vert f\Vert ^{2}= \int_{S^{4}_{+}} \vert \nabla f\vert ^{2} +2 \int_{S^{4}_{+}} f^{2}, \quad \mbox{and}\quad\langle f,g \rangle= \int_{S_{+}^{4}}\nabla f \nabla g + 2 \int_{S_{+}^{4}}fg,\quad f,g \in H^{1}\bigl(S_{+}^{4} \bigr). \end{aligned}$$

Theorem 1

Let $z_{1}$ and $z_{2}$ be a nondegenerate critical points of $K_{1}=K_{\mid\partial S^{4}_{+}}$ with $(\partial K/\partial\nu)(z_{i}) >0$, $i=1,2$. Then there exists $\varepsilon_{0}>0$ such that, for each $\varepsilon\in(0,\varepsilon_{0})$, problem $(S_{\varepsilon})$ has a solution $(u_{\varepsilon})$ of the form

$$\begin{aligned} u_{\varepsilon}=\alpha_{1}\delta_{(x_{1},\lambda _{1})}+\alpha_{2} \delta_{(x_{2},\lambda_{2})}+v, \end{aligned}$$

where, as $\varepsilon\rightarrow0$, $\alpha_{i}\rightarrow K(z_{i})^{-1/2}$; $\Vert v\Vert \rightarrow0$; $x_{i}\rightarrow z_{i}$; $x_{i}\in\partial S^{4}_{+}$; $\lambda_{i}\rightarrow+\infty$; $\lambda_{1}=c\lambda_{2}(1+o(1))$.

The rest of this work is summarized as follows. In Section 2, we present a classical preliminaries and we perform a useful estimations of functional $(I_{\varepsilon})$ associated to the problem $(S_{\varepsilon})$ for $(\varepsilon> 0)$ and its gradient. Section 3 is devoted to the construction of solutions and the proof of our result.

2 Useful estimations

We introduce the structure variational associated to the problem $(S_{\varepsilon})$ for $\varepsilon> 0$

$$\begin{aligned} I_{\varepsilon}(u)=\frac{1}{2} \int_{S^{4}_{+}}\vert \nabla u\vert ^{2}+ \int_{S^{4}_{+}} u^{2}-\frac{1}{4-\varepsilon} \int_{S^{4}_{+}}K\vert u\vert ^{4-\varepsilon},\quad u \in H^{1}\bigl(S^{4}_{+}\bigr). \end{aligned}$$

(2)

It is well known that there is an equivalence between the existence of solutions for $(S_{\varepsilon})$ and the positive critical point of $I_{\varepsilon}$. Moreover, in order to reduce our problem to $\mathbb{R}^{4}_{+}$ we will perform some stereographic projection. We denote $D^{1,2}(\mathbb{R}^{4}_{+})$ for the completion of $C^{\infty}_{c} (\overline{\mathbb{R}^{4}_{+}})$ with respect to the Dirichlet norm. Recall that an isometry $\i: H^{1}(S^{4}_{+}) \to D^{1,2}(\mathbb{R}^{4}_{+})$ is induced by the stereographic projection $\pi_{a}$ about a point $a \in\partial S^{4}_{+}$ following the formula

$$\begin{aligned} (\i\phi) (y)= \biggl(\frac{2}{1+\vert x\vert ^{2}} \biggr)\phi\bigl( \pi_{a}^{-1}(y)\bigr), \quad \phi\in H^{1} \bigl(S^{4}_{+}\bigr), y\in\mathbb{R}^{4}_{+}. \end{aligned}$$

(3)

For every $\phi\in H^{1}(S^{4}_{+})$, one can check that the following holds true:

$$\int_{S^{4}_{+}}\bigl(\vert \nabla\phi \vert ^{2} + 2 \phi^{2}\bigr) = \int_{\mathbb {R}^{4}_{+}}\bigl\vert \nabla(\i\phi)\bigr\vert ^{2} \quad\mbox{and}\quad \int_{S^{4}_{+}}\vert \phi \vert ^{4}= \int_{\mathbb{R}^{4}_{+}}\vert \i \phi \vert ^{4}. $$

Furthermore, using (3) with $\pi_{-a}$, it is easy to see that $\i\delta_{(a,\lambda)}$ is given by

$$\i\delta_{(a,\lambda)}=\frac{c_{0}\lambda}{1+\lambda^{2}\vert x-a\vert ^{2}}. $$

$\delta_{(a,\lambda)}$ will be written instead of $\i\delta_{(a,\lambda)}$ in the sequel.

Let

$$\begin{aligned} M_{\varepsilon} =& \biggl\{ m =(\alpha,\lambda,x,v)\in\mathbb {R}^{2} \times\bigl(\mathbb{R}_{+}^{*}\bigr)^{2}\times \bigl(\partial S^{4}_{+}\bigr)^{2}\times H^{1}\bigl(S^{4}_{+} \bigr): v\in E_{(x,\lambda)}, \Vert v\Vert < \nu_{0} ; \\ &{} \biggl\vert \frac{\alpha_{i}^{2}K(x_{i})}{\alpha _{j}^{2}K(x_{j})}-1\biggr\vert < \nu_{0}, \lambda_{i}>\frac{1}{\nu_{0}},\varepsilon\log\lambda_{i}< \nu_{0}, \forall i; c_{0}< \frac{\lambda_{1}}{\lambda_{2}}< c_{0}^{-1}; \vert x_{1}-x_{2}\vert >d_{0}; \\ &{} \biggl\vert -2c_{3}\frac{\partial K}{\partial\nu}(x_{i})\frac{1}{\lambda_{i}}+ \frac{\varepsilon K(x_{i})S_{4}}{8}\biggr\vert < \varepsilon^{1+\frac{\sigma}{2}}\biggr\} , \end{aligned}$$

where $\nu_{0}$ is a small positive constant, σ, $c_{0}$, $d_{0}$ are some suitable positive constants, and

$$E_{(x,\lambda)}=\biggl\{ w\in H^{1}\bigl(S^{4}_{+}\bigr)/ \langle w,\varphi\rangle=0\ \forall\varphi\in \operatorname{Span}\biggl\{ \delta_{i}, \frac{\partial\delta_{i}}{\partial\lambda_{i}}, \frac{\partial\delta_{i}}{\partial x^{j}_{i}}, i=1,2; j\leq4 \biggr\} \biggr\} . $$

Here, $x^{j}_{i}$ denotes the jth component of $x_{i}$. Also

$$\begin{aligned} \Psi_{\varepsilon}:M_{\varepsilon} \rightarrow \mathbb{R} ;\quad m=(\alpha,\lambda,x,v)\mapsto I_{\varepsilon}(\alpha_{1} \delta_{(x_{1},\lambda_{1})}+\alpha_{2}\delta _{(x_{2},\lambda_{2})}+v ). \end{aligned}$$

(4)

In the sequel, we will write $\delta_{i}$ instead of $\delta_{(x_{i},\lambda_{i})}$ and $u=\alpha_{1}\delta_{1}+\alpha_{2}\delta _{2}+v$ for the sake of simplicity.

In the remainder of this section, we will give expansions of the gradient of the functional $I_{\varepsilon}$ associated to $(S_{\varepsilon})$ for $\varepsilon>0$. Thus estimations are needed in Section 3. We need to recall that [15] proved the following remark when $n= 3$, but the same argument is available for the dimension 4.

Remark 2

For $\varepsilon>0$ and $\delta _{(a,\lambda)}$ defined in (1), we have

$$\delta_{(a,\lambda)}^{-\varepsilon} (x) =1-\varepsilon\log\delta _{(a,\lambda)}+O\bigl(\varepsilon^{2}\log^{2}\lambda \bigr) \quad \mbox{in } S^{4}_{+}. $$

Now, explicit computations, using Remark 2, yield the following propositions.

Proposition 3

Let $(\alpha,\lambda,x,v) \in M_{\varepsilon}$. Then, for $u=\alpha_{1}\delta_{(x_{1},\lambda _{1})}+\alpha_{2}\delta_{(x_{2},\lambda_{2})}+v$, we have the following expansion:

$$ \bigl\langle \nabla I_{\varepsilon}(u),\delta_{i}\bigr\rangle = \frac{\alpha_{i}S_{4}}{2}\bigl(1-\alpha _{i}^{2-\varepsilon}K(x_{i}) \bigr)+O \biggl(\varepsilon\log \lambda_{i}+\frac{1}{\lambda_{i}}+ \varepsilon_{12}+\Vert v\Vert ^{2} \biggr), $$

where

$$\begin{aligned} &{\varepsilon_{ij}=\frac{1}{\frac{\lambda_{i}}{\lambda _{j}}+\frac{\lambda_{j}}{\lambda_{i}}+\lambda_{i}\lambda_{j}\vert a_{i}-a_{j}\vert ^{2}},} \\ &{S_{4}=64 \int_{\mathbb{R}^{4}}\frac{dx}{(1+\vert x\vert ^{2})^{4}}.} \end{aligned}$$

Proof

We have

$$ \bigl\langle \nabla I_{\varepsilon}(u),h\bigr\rangle = \int_{S_{+}^{4}}\nabla u\nabla h+2 \int_{S_{+}^{4}}uh- \int_{S_{+}^{4}}Ku^{3-\varepsilon}h . $$

(5)

A computation similar to the one performed in [16] shows that, for $i=1,2$,

$$ \Vert \delta_{i}\Vert ^{2}= \int_{\mathbb{R}_{+}^{4}}\vert \nabla\delta_{i} \vert ^{2} = \frac{S_{4}}{2} $$

(6)

and

$$ \int_{S_{+}^{4}}\nabla\delta_{i}\nabla\delta_{j} +2 \int_{S_{+}^{4}}\delta _{i}\delta_{j}= \int_{\mathbb{R}_{+}^{4}}\nabla\delta_{i}\nabla\delta_{j} = \int_{\mathbb {R}_{+}^{4}}\delta_{i}^{3} \delta _{j}=O(\varepsilon_{12}) . $$

(7)

For the other integral, we write

$$ \int_{S_{+}^{4}}Ku^{3-\varepsilon}\delta_{i}= \int_{S_{+}^{4}}K (\alpha _{1}\delta_{1}+ \alpha_{2} \delta _{2})^{3-\varepsilon}\delta_{i}+O \bigl(\varepsilon_{12}^{2}\log \varepsilon_{12}^{-1}+ \vert v\vert ^{2} \bigr). $$

(8)

We also write

$$\begin{aligned} \int_{S_{+}^{4}}K (\alpha_{1}\delta_{1}+ \alpha_{2} \delta_{2})^{3-\varepsilon }\delta_{i} ={}& \alpha _{i}^{3-\varepsilon} \int_{S_{+}^{4}}K \delta_{i}^{4-\varepsilon}+ \alpha _{j}^{3-\varepsilon} \int_{S_{+}^{4}}K \delta _{j}^{3-\varepsilon} \delta_{i} \\ &{} +(3-\varepsilon) \alpha_{i}^{2-\varepsilon}\alpha_{j} \int_{S_{+}^{4}}K \delta _{i}^{3-\varepsilon} \delta_{j} +O \bigl(\varepsilon_{12}^{2}\log \varepsilon_{12}^{-1} \bigr). \end{aligned}$$

(9)

Expansions of K around $x_{i}$ and $x_{j}$ give

$$\begin{aligned} &{\int_{S_{+}^{4}}K \delta_{i}^{4-\varepsilon}= \int_{\mathbb{R}_{+}^{4}}K \delta _{i}^{4-\varepsilon}=K(x_{i}) \frac{S_{4}}{2}+O \biggl(\varepsilon\log \lambda_{i}+ \frac{1}{\lambda _{i} } \biggr) ,} \end{aligned}$$

(10)

$$\begin{aligned} &{\int_{S_{+}^{4}}K \delta_{j}^{3-\varepsilon} \delta_{i}= \int_{\mathbb {R}_{+}^{4}}K \delta _{j}^{3-\varepsilon} \delta_{i}=O (\varepsilon\log\lambda _{i}+ \varepsilon_{12} ),} \end{aligned}$$

(11)

$$\begin{aligned} &{\int_{S_{+}^{4}}K \delta_{i}^{3-\varepsilon} \delta_{j}= \int_{\mathbb {R}_{+}^{4}}K \delta _{i}^{3-\varepsilon} \delta_{j}=O (\varepsilon\log\lambda _{i}+ \varepsilon_{12} ).} \end{aligned}$$

(12)

Combining (5)-(12), we derive our proposition. □

Proposition 4

Let $(\alpha,\lambda,x,v) \in M_{\varepsilon}$. Then, for $u=\alpha_{1}\delta_{(x_{1},\lambda _{1})}+\alpha_{2}\delta_{(x_{2},\lambda_{2})}+v$, we have

$$\begin{aligned} \biggl\langle \nabla I_{\varepsilon}(u),\lambda_{i}\frac{\partial\delta_{i}}{\partial \lambda_{i}} \biggr\rangle ={}&\alpha_{j} \bigl(1-\alpha_{j}^{2-\varepsilon }K(x_{j})- \alpha_{i}^{2-\varepsilon}K(x_{i}) \bigr)c_{2} \lambda_{i}\frac {\partial \varepsilon_{12}}{\partial\lambda_{i}}+\alpha_{i}^{3-\varepsilon} \frac {\varepsilon S_{4}K(x_{i})}{8} \\ &{} +\alpha_{i}^{3-\varepsilon}\frac{2c_{3}}{\lambda _{i}}\frac{\partial K}{\partial\nu}(x_{i})+O \biggl(\Vert v\Vert ^{2} +\frac{1}{\lambda_{i}^{2}}+\varepsilon^{2} \log\lambda_{i}+\frac {\varepsilon\log\lambda_{i}}{\lambda_{i}} \biggr) \\ &{} +O \biggl(\varepsilon \varepsilon_{12}\bigl(\log \varepsilon_{12}^{-1} \bigr)^{1/2}+\varepsilon_{12}^{2}\log \varepsilon_{12}^{-1}+\frac{\varepsilon_{12}}{\lambda_{j}}\bigl(\log \varepsilon_{12}^{-1}\bigr)^{1/2} \biggr), \end{aligned}$$

where

$$ S_{4}=64 \int_{\mathbb{R}^{4}}\frac {dx}{(1+\vert x\vert ^{2})^{4}},\qquad c_{2}=64 \int_{\mathbb{R}^{4}}\frac{dx}{(1+\vert x\vert ^{2})^{3}},\qquad c_{3}=64 \int_{\mathbb{R}^{4}_{+}}\frac{x_{4}(\vert x\vert ^{2}-1)}{(1+\vert x\vert ^{2})^{5}}\,dx. $$

Proof

Observe that (see [16])

$$\begin{aligned} &{\int_{\mathbb{R}^{4}_{+}}\nabla\delta_{i} \nabla\biggl( \lambda_{i} \frac {\partial\delta_{i}}{\partial\lambda _{i}}\biggr) = \int_{\mathbb{R}^{4}_{+}}\delta_{i}^{3} \lambda_{i}\frac{\partial\delta _{i}}{\partial\lambda _{i}} = 0,} \end{aligned}$$

(13)

$$\begin{aligned} &{\int_{\mathbb{R}^{4}_{+}}\nabla\delta_{j} \nabla\biggl( \lambda_{i} \frac {\partial\delta_{i}}{\partial\lambda _{i}}\biggr) = \int_{\mathbb{R}^{4}_{+}}\delta_{j}^{3} \lambda_{i}\frac{\partial\delta _{i}}{\partial\lambda _{i}}=\frac{1}{2} c_{2} \lambda_{i} \frac{\partial\varepsilon _{12}}{\partial\lambda_{i}} + O \bigl( \varepsilon_{12}^{2} \log\bigl(\varepsilon_{12}^{-1}\bigr) \bigr).} \end{aligned}$$

(14)

For the other part, we have the expansions of K around $x_{i}$ and using Remark 2,

$$\begin{aligned} &{\int_{\mathbb{R}^{4}_{+}}K \delta_{i} ^{3-\varepsilon} \lambda_{i} \frac {\partial\delta _{i}}{\partial\lambda_{i}} =-\frac{\varepsilon S_{4}K(x_{i})}{8}- \frac{2c_{3}}{\lambda_{i}} \nabla K(x_{i}) e_{4} + O \biggl(\varepsilon^{2} \log\lambda_{i}+\frac{1}{\lambda_{i}^{2}}+\frac {\varepsilon}{\lambda_{i} } \biggr),} \end{aligned}$$

(15)

$$\begin{aligned} &{\int_{\mathbb{R}^{4}_{+}}KP\delta_{j} ^{3-\varepsilon }\lambda_{i} \frac{\partial \delta_{i}}{\partial\lambda_{i}} = K(x_{j}) \frac{1}{2}c_{2}\lambda_{i} \frac{\partial\varepsilon_{12}}{\partial\lambda_{i}} + O \biggl(\varepsilon\varepsilon_{12}\bigl(\log\bigl(\varepsilon _{12}^{-1}\bigr)\bigr)^{\frac{1}{2}}+\frac{1}{\lambda_{j}^{2}} \biggr)} \\ &{ \hphantom{\int_{\mathbb{R}^{4}_{+}}KP\delta_{j} ^{3-\varepsilon }\lambda_{i} \frac{\partial \delta_{i}}{\partial\lambda_{i}} = } {} + O \bigl(\varepsilon_{12}^{2}\log\bigl( \varepsilon_{12}^{-1}\bigr) \bigr),} \end{aligned}$$

(16)

$$\begin{aligned} &{(3-\varepsilon) \int_{\mathbb{R}^{4}_{+}}K\delta_{i} ^{2-\varepsilon} \delta_{j}\lambda_{i} \frac{\partial\delta_{i}}{\partial\lambda_{i}} = K(x_{i})\frac{1}{2}c_{2} \lambda_{i} \frac{\partial\varepsilon_{12}}{\partial\lambda_{i}} + O \bigl(\varepsilon\varepsilon_{12}\bigl(\log\bigl( \varepsilon _{12}^{-1}\bigr)\bigr)^{\frac{1}{2}} \bigr)} \\ &{ \hphantom{(3-\varepsilon) \int_{\mathbb{R}^{4}_{+}}K\delta_{i} ^{2-\varepsilon} \delta_{j}\lambda_{i} \frac{\partial\delta_{i}}{\partial\lambda_{i}} =} {}+O \biggl(\varepsilon_{12}^{2}\log\bigl( \varepsilon_{12}^{-1}\bigr)+\frac {\varepsilon_{12}}{\lambda_{j}}\bigl(\log\bigl( \varepsilon_{12}^{-1}\bigr)\bigr)^{\frac {1}{2}} \biggr).} \end{aligned}$$

(17)

Combining (5), (13), (14), (15), (16), and (17), the proof follows. □

Proposition 5

Let $(\alpha,\lambda,x,v) \in M_{\varepsilon}$. Then, for $u=\alpha_{1}\delta_{(x_{1},\lambda _{1})}+\alpha_{2}\delta_{(x_{2},\lambda_{2})}+v$, we have the following expansion:

$$\begin{aligned} \biggl\langle \nabla I_{\varepsilon}(u),\frac{1}{\lambda_{i}}\frac{\partial\delta _{i}}{\partial x_{i}} \biggr\rangle ={}& \biggl(\alpha_{i}c_{4} \bigl(1- \alpha_{i}^{2-\varepsilon}K(x_{i}) \bigr)+ \alpha_{i}^{3-\varepsilon}K(x_{i})\varepsilon(c_{4} \log\lambda_{i} +c_{7}) \\ &{}+2\alpha_{i}^{3-\varepsilon}\frac{c_{5}}{\lambda_{i}}\frac {\partial K}{\partial \nu}(x_{i}) \biggr)e_{4}+\alpha_{j} \Bigl(1-\sum \alpha_{i}^{2-\varepsilon }K(x_{i}) \Bigr)\frac{c_{2}}{\lambda_{i}} \frac{\partial \varepsilon_{12}}{\partial x_{i}} \\ &{}-2\alpha_{i}^{3-\varepsilon}c_{5}\frac {\nabla_{T} K(x_{i})}{\lambda_{i}}+O \biggl(\Vert v\Vert ^{2} +\lambda_{j}\vert x_{1}-x_{2}\vert \varepsilon_{12}^{\frac{5}{2}}+ \frac{\varepsilon \log\lambda_{i}}{\lambda_{i}}\bigl\vert \nabla_{T}K(x_{i})\bigr\vert \biggr) \\ &{}+O \biggl(\varepsilon\varepsilon_{12}\bigl(\log \varepsilon_{12}^{-1} \bigr)^{\frac{1}{2}}+\varepsilon_{12}^{2}\log \varepsilon_{12}^{-1}+\frac{\varepsilon_{12}}{\lambda_{j}}\bigl(\log \varepsilon_{12}^{-1}\bigr)^{\frac{1}{2}}+\frac{1}{\lambda _{i}^{2}}+ \varepsilon^{2}\log^{2}\lambda_{i} \biggr), \end{aligned}$$

where

$$ c_{4}=132 \int_{\mathbb{R}^{4}_{+}}\frac {x_{4}}{(1+\vert x\vert ^{2})^{5}}\,dx,\qquad c_{5}=16 \int_{\mathbb{R}^{4}}\frac{x_{4}^{2}}{(1+\vert x\vert ^{2})^{5}}\,dx. $$

Proof

We have

$$\begin{aligned} &{\int_{\mathbb{R}^{4}_{+}}\nabla\delta_{i}\nabla \biggl( \frac{1}{\lambda _{i}}\frac{\partial\delta _{i}}{\partial x_{i}} \biggr) = \int_{\mathbb{R}^{4}_{+}}\delta _{i}^{3} \frac{1}{\lambda_{i}}\frac{\partial\delta_{i}}{\partial x_{i}} = c_{4}e_{4},} \end{aligned}$$

(18)

$$\begin{aligned} &{\int_{\mathbb{R}^{4}_{+}}\nabla\delta_{j}\nabla \biggl( \frac{1}{\lambda _{i}}\frac{\partial\delta _{i}}{\partial x_{i}} \biggr) = \int_{\mathbb{R}^{4}_{+}}\delta _{j}^{3}\frac{1}{\lambda_{i}} \frac{\partial\delta_{i}}{\partial x_{i}}= \frac{1}{2}\frac{c_{2}}{\lambda_{i}}\frac{\partial\varepsilon _{12}}{\partial x_{i}} + O \bigl(\varepsilon _{12}^{2}\log\bigl(\varepsilon_{12}^{-1} \bigr) + \varepsilon_{12}^{\frac{5}{2}}\lambda_{j} \vert x_{1}-x_{2}\vert \bigr).} \end{aligned}$$

(19)

For the other part

$$\begin{aligned} &{ \int_{\mathbb{R}^{4}_{+}}K\delta _{i}^{3-\varepsilon} \frac{1}{\lambda_{i}}\frac{\partial\delta _{i}}{\partial x_{i}} =K(x_{i}) c_{4}e_{4} + 2\frac{c_{5}}{\lambda_{i}}\nabla K(x_{i})-\varepsilon \log \lambda_{i}K(x_{i})c_{4}e_{4}} \\ &{\hphantom{\int_{\mathbb{R}^{4}_{+}}K\delta _{i}^{3-\varepsilon} \frac{1}{\lambda_{i}}\frac{\partial\delta _{i}}{\partial x_{i}} =} {} -\varepsilon K(x_{i})c_{7}e_{4}+O \biggl( \frac{1}{\lambda_{i}^{2}}+\varepsilon ^{2}\log^{2} \lambda_{i} \biggr),} \end{aligned}$$

(20)

$$\begin{aligned} &{\int_{\mathbb{R}^{4}_{+}}K\delta _{j}^{3-\varepsilon} \frac{1}{\lambda_{i}}\frac{\partial\delta _{i}}{\partial x_{i}} =K(x_{j})\frac{1}{2}c_{2} \frac{1}{\lambda_{i}}\frac{\partial \varepsilon _{12}}{\partial a_{i}}+ O \bigl(\varepsilon_{12}^{\frac{5}{2}} \lambda _{j}\vert x_{1}-x_{2}\vert \bigr)} \\ &{\hphantom{\int_{\mathbb{R}^{4}_{+}}K\delta _{j}^{3-\varepsilon} \frac{1}{\lambda_{i}}\frac{\partial\delta _{i}}{\partial x_{i}} =} {}+O \biggl(\varepsilon_{12}^{2}\log\bigl(\varepsilon _{12}^{-1}\bigr)+\frac{1}{\lambda_{j}}\varepsilon_{12} \bigl(\log \bigl(\varepsilon _{12}^{-1}\bigr) \bigr)^{\frac{1}{2}} \biggr),} \end{aligned}$$

(21)

$$\begin{aligned} &{(3-\varepsilon)\int_{\mathbb{R}^{4}_{+}}K\delta _{i}^{2-\varepsilon} \delta_{j} \frac{1}{\lambda_{i}}\frac{\partial \delta_{i}}{\partial x_{i}} = K(x_{i}) \frac{1}{2}c_{2}\frac{1}{\lambda_{i}}\frac{\partial \varepsilon _{12}}{\partial x_{i}}+ O \bigl( \varepsilon_{12}^{\frac{5}{2}}\lambda _{j}\vert x_{1}-x_{2}\vert \bigr)} \\ &{\hphantom{(3-\varepsilon)\int_{\mathbb{R}^{4}_{+}}K\delta _{i}^{2-\varepsilon} \delta_{j} \frac{1}{\lambda_{i}}\frac{\partial \delta_{i}}{\partial x_{i}} =} {}+O \biggl(\varepsilon_{12}^{2}\log\bigl(\varepsilon _{12}^{-1}\bigr)+\frac{1}{\lambda_{i}}\varepsilon_{12} \bigl(\log \bigl(\varepsilon _{12}^{-1}\bigr) \bigr)^{\frac{1}{2}} \biggr).} \end{aligned}$$

(22)

Using (5), (18)-(22), our proposition follows. □

3 Construction of the solution

The method of this type of theorem was followed first by Bahri, Li and Rey [17] when they studied an approximation problem of the Yamabe-type problem on domains. Many authors used this idea to construct some solutions to other problems. The method becomes standard. Here we will follow the idea of [17] and take account of the new estimates since we have an equation different from the one studied in [17]. From the idea of [17], using the coefficients of Euler-Lagrange, we obtain

Proposition 6

Let A point $m=(\alpha,\lambda,x,v)\in M_{\varepsilon}$ is a critical point of the function $\Psi_{\varepsilon}$ if and only if $u=\alpha_{1}\delta_{1}+\alpha_{2}\delta_{2}+v$ is a critical point of functional $I_{\varepsilon}$, which means the existence of some $(A,B,C )\in \mathbb{R}^{2}\times\mathbb{R}^{2}\times (\mathbb{R}^{4} )^{2}$ with the following:

$$\begin{aligned} &{(E_{\alpha_{i}}) \frac{\partial\Psi_{\varepsilon }}{\partial \alpha_{i}}=0,\quad \forall i=1,2,} \end{aligned}$$

(23)

$$\begin{aligned} &{(E_{\lambda_{i}}) \frac{\partial\Psi_{\varepsilon }}{\partial \lambda_{i}}=B_{i}\biggl\langle \frac{\partial^{2}\delta_{i}}{\partial\lambda _{i}^{2}},v\biggr\rangle +\sum_{j=1}^{4} C_{ij}\biggl\langle \frac{\partial^{2}\delta_{i}}{\partial x_{i}^{j}\partial\lambda_{i}},v\biggr\rangle ,\quad \forall i=1,2,} \end{aligned}$$

(24)

$$\begin{aligned} &{(E_{x_{i}}) \frac{\partial\Psi_{\varepsilon }}{\partial x_{i}}=B_{i}\biggl\langle \frac{\partial^{2}\delta_{i}}{\partial\lambda _{i}\partial x_{i}},v\biggr\rangle + \sum_{j=1}^{4} C_{ij}\biggl\langle \frac{\partial^{2}\delta_{i}}{\partial x_{i}^{j}\partial x_{i}},v\biggr\rangle ,\quad \forall i=1,2,} \end{aligned}$$

(25)

$$\begin{aligned} &{(E_{v}) \frac{\partial\Psi_{\varepsilon}}{\partial v}=\sum _{i=1,2} \Biggl(A_{i} \delta_{i}+B_{i} \frac{\partial\delta _{i}}{\partial \lambda_{i}}+\sum_{j=1}^{4} C_{ij}\frac{\partial\delta_{i}}{\partial x_{i}^{j}} \Biggr).} \end{aligned}$$

(26)

Now, by a careful study of equation $(E_{v})$, we get the following.

Proposition 7

[12] For any $(\varepsilon,\alpha,\lambda,x)$ with $(\alpha,\lambda,x,0)\in M_{\varepsilon}$, there exists a smooth map which associates $\overline{v} \in E_{(x,\lambda)}$ with $\Vert \overline{v}\Vert <\nu_{0}$ and equation (26) in the previous proposition is verified for some $(A,B,C )\in\mathbb{R}^{2}\times\mathbb{R}^{2}\times (\mathbb{R}^{4} )^{2}$. Such a v̅ is unique, minimizes $\Psi_{\varepsilon}(\alpha ,\lambda,x,v)$ with respect to v in $\{v\in E_{(x,\lambda)}/\Vert v\Vert <\nu_{0}\}$, and

$$ \Vert \overline{v}\Vert =O \biggl(\varepsilon+ \frac {1}{\lambda_{1}}+\frac{1}{\lambda_{2}}+\varepsilon_{12}\bigl(\log \varepsilon_{12}^{-1}\bigr)^{1/2} \biggr). $$

(27)

Proof of Theorem 1

Once v̅ is defined by Proposition 7, we estimate the corresponding numbers A, B, C by taking the scalar product in $H^{1}(S^{4}_{+})$ of $(E_{v})$ with $\delta_{i}$, ${\partial\delta _{i}}/{\partial \lambda_{i}}$, ${\partial\delta_{i}}/{\partial x_{i}}$ for $i=1,2$, respectively. So we get the following coefficients of a quasi-diagonal system:

$$\begin{aligned} &{ \int_{\mathbb{R}^{4}_{+}} \vert \nabla\delta_{i}\vert ^{2} =\frac {S_{4}}{2};\qquad \int_{\mathbb{R}^{4}_{+}} \nabla\delta_{1} \nabla \delta_{2}=O \biggl(\frac{1}{\lambda_{2}\lambda_{1}} \biggr);\qquad \int _{\mathbb{R}^{4}_{+}} \nabla\delta_{i} \nabla \frac{\partial\delta _{i}}{\partial \lambda_{i}}=0 ;} \\ &{\int_{\mathbb{R}^{4}_{+}} \nabla\delta_{1} \nabla \frac{\partial\delta_{2}}{\partial \lambda_{2}}=O\biggl(\frac{1}{\lambda_{1}\lambda_{2}^{2}}\biggr), \qquad \int_{\mathbb {R}^{4}_{+}} \nabla\delta_{2} \nabla \frac{\partial\delta_{1}}{\partial \lambda_{1}}=O\biggl(\frac{1}{\lambda_{1}^{2}\lambda_{2}}\biggr);\qquad \int_{\mathbb {R}^{4}_{+}} \biggl\vert \nabla \frac{\partial\delta_{i}}{\partial\lambda_{i}} \biggr\vert ^{2}=\frac{\Gamma_{1}}{2\lambda_{i}^{2}} ;} \\ &{\int_{\mathbb {R}^{4}_{+}} \nabla\frac{\partial \delta_{1}}{\partial\lambda_{1}}\nabla\frac{\partial\delta _{2}}{\partial\lambda_{2}} =O \biggl(\frac{1}{\lambda_{1}^{2}\lambda_{2}^{2}}\biggr),\qquad \int_{\mathbb{R}^{4}_{+}} \biggl\vert \nabla \frac{\partial\delta_{i}}{\partial x_{i}}\biggr\vert ^{2}=\frac{\Gamma _{2}}{2}\lambda_{i}^{2} ;\qquad \int_{\mathbb{R}^{4}_{+}}\nabla\delta_{i} \nabla \frac{\partial\delta _{i}}{\partial x_{i}}=O(\lambda_{1}) ;} \\ &{\int_{\mathbb{R}^{4}_{+}}\nabla\delta_{1} \nabla\frac{\partial\delta _{2}}{\partial x_{2}}=O \biggl(\frac{1}{\lambda_{1}}\biggr),\qquad \int_{\mathbb{R}^{4}_{+}}\nabla\delta_{2} \nabla\frac{\partial\delta_{1}}{\partial x_{1}}=O \biggl(\frac{1}{\lambda _{2}}\biggr);} \\ &{\int_{\mathbb{R}^{4}_{+}}\nabla\frac{\partial\delta_{1}}{\partial x_{1}} \nabla\frac{\partial\delta_{2}}{\partial x_{2}}= \frac{n+2}{n-2} \int _{\mathbb{R}^{4}_{+}} \delta_{2}^{\frac{4}{n-2}} \nabla \frac{\partial\delta_{2}}{\partial x_{2}}\frac{\partial\delta_{1}}{\partial x_{1}}=O\biggl(\frac{1}{\lambda_{1}}\biggr),} \end{aligned}$$

with $\vert x_{1}-x_{2}\vert \geq c >0$ and $\Gamma_{1}$, $\Gamma_{2}$ are positive constants.

We have also

$$\begin{aligned} \biggl\langle \frac{\partial\Psi_{\varepsilon}}{\partial v},\delta_{i}\biggr\rangle = \frac{\partial\Psi_{\varepsilon}}{\partial \alpha_{i}};\qquad \biggl\langle \frac{\partial \Psi_{\varepsilon}}{\partial v},\frac{\partial\delta_{i}}{\partial \lambda_{i}}\biggr\rangle =\frac{1}{\alpha_{i}}\frac{\partial \Psi_{\varepsilon}}{\partial\lambda_{i}};\qquad \biggl\langle \frac {\partial \Psi_{\varepsilon}}{\partial v},\frac{\partial\delta_{i}}{\partial x_{i}}\biggr\rangle =\frac{1}{\alpha_{i}} \frac{\partial \Psi_{\varepsilon}}{\partial x_{i}}. \end{aligned}$$

Using Propositions 3, some computations yield

$$ \frac{\partial \Psi_{\varepsilon}}{\partial\alpha_{i}}=-S_{4}\beta_{i}+V_{\alpha _{i}}( \varepsilon,\alpha,\lambda,x), $$

(28)

with $\beta_{i}=\alpha_{i}-1/K(z_{i})^{\frac{1}{2}}$ and

$$ V_{\alpha_{i}}=O \biggl(\beta_{i}^{2}+ \varepsilon \log\lambda_{i}+ \frac{1}{\lambda_{i}}+\vert x_{i}-z_{i}\vert ^{2} \biggr). $$

(29)

In the same way, using Propositions 4, we get

$$ \frac{\partial\Psi_{\varepsilon}}{\partial \lambda_{i}}=\frac{1}{K(z_{i})} \biggl( \frac{2c_{3}}{\lambda_{i}^{2}}\frac {\partial K}{\partial\nu}(x_{i})+\frac{\varepsilon K(x_{i})S_{4}}{8\lambda_{i}} \biggr) +V_{\lambda_{i}}(\varepsilon,\alpha,\lambda,x), $$

(30)

where $c_{2}$ and $c_{3}$ are defined in Proposition 4 and

$$ V_{\lambda_{i}}=O \biggl[\frac{1}{\lambda _{i}} \biggl( \frac{1}{\lambda_{i}^{2}}+\varepsilon^{2}\log\lambda_{i}+ \frac {\varepsilon\log\lambda_{i}}{\lambda_{i}} \biggr) + \bigl(\vert \beta \vert +\varepsilon+\vert x_{i}-z_{i}\vert ^{2} \bigr) \biggl( \frac{\varepsilon }{\lambda_{i}} +\frac{1}{\lambda_{i}^{2}}\biggr) \biggr]. $$

(31)

Lastly, using Propositions 5, we have

$$ \frac{\partial\Psi_{\varepsilon }}{\partial x_{i}}=-2c_{5}\nabla_{T}K(x_{i})+V_{x_{i}}( \varepsilon,\alpha,\lambda,x), $$

(32)

where

$$ V_{x_{i}}=O \biggl(\frac{1}{\lambda_{i}}+ \bigl(\vert \beta \vert +\varepsilon\log\lambda_{i}+\vert x_{i}-z_{i} \vert ^{2}\bigr)\vert x_{i}-z_{i}\vert \biggr). $$

(33)

From these estimates, we deduce

$$\begin{aligned} &{\frac{\partial \Psi_{\varepsilon}}{\partial\alpha_{i}}=O \biggl(\vert \beta \vert +\varepsilon \log \lambda_{i} +\frac{1}{\lambda_{i}}+\vert x_{i}-z_{i} \vert ^{2} \biggr),} \\ &{\frac{\partial \Psi_{\varepsilon}}{\partial\lambda_{i}}=O \biggl(\frac{\varepsilon ^{1+\sigma/2}}{\lambda_{i}} \biggr);\qquad \frac{\partial\Psi_{\varepsilon}}{\partial x_{i}}=O \biggl(\vert x_{i}-z_{i}\vert + \frac{1}{\lambda_{i}} \biggr).} \end{aligned}$$

By solving the system in A, B, and C, we find

$$\begin{aligned} \textstyle\begin{cases} A_{i}=O (\vert \beta \vert +\varepsilon\log\lambda_{i} +\frac{1}{\lambda_{i}}+\vert x_{i}-z_{i}\vert ^{2} ), \\ B_{i}=O (\varepsilon^{1+\sigma/2}\lambda_{i} );\qquad C_{i}=O (\frac{\vert x_{i}-z_{i}\vert }{\lambda_{i}^{2}}+\frac{1}{\lambda_{i}^{3}} ). \end{cases}\displaystyle \end{aligned}$$

(34)

Now, we can evaluate the right hand side in $(E_{\lambda_{i}})$ and $(E_{x_{i}})$,

$$\begin{aligned} &{B_{i}\biggl\langle \frac{\partial^{2}\delta_{i}}{\partial\lambda _{i}^{2}},\overline{v}\biggr\rangle +\sum_{j=1}^{4} C_{ij}\biggl\langle \frac{\partial^{2}\delta_{i}}{\partial x_{i}^{j}\partial\lambda_{i}},\overline{v}\biggr\rangle =O \biggl( \biggl( \frac{\varepsilon^{1+\sigma/2}}{\lambda_{i}}+\frac {\vert x_{i}-z_{i}\vert }{\lambda_{i}^{2}} +\frac{1}{\lambda_{i}^{3}} \biggr)\Vert \overline{v}\Vert \biggr),} \end{aligned}$$

(35)

$$\begin{aligned} &{B_{i}\biggl\langle \frac{\partial^{2}\delta_{i}}{\partial\lambda_{i}\partial x_{i}},\overline{v} \biggr\rangle +\sum_{j=1}^{4} C_{ij} \biggl\langle \frac{\partial^{2}\delta_{i}}{\partial x^{j}_{i}\partial x_{i}},\overline{v}\biggr\rangle =O \biggl( \biggl( \varepsilon^{1+\sigma/2}\lambda_{i}+\vert x_{i}-z_{i} \vert +\frac {1}{\lambda_{i}} \biggr) \Vert \overline{v}\Vert \biggr),} \end{aligned}$$

(36)

where

$$\begin{aligned} \biggl\Vert \frac{\partial^{2}P\delta_{i}}{\partial \lambda_{i}^{2}}\biggr\Vert =O \biggl(\frac{1}{\lambda_{i}^{2}} \biggr);\qquad \biggl\Vert \frac{\partial^{2}P\delta_{i}}{\partial x_{i}\partial\lambda_{i}}\biggr\Vert =O(1);\qquad \biggl\Vert \frac{\partial^{2}P\delta_{i}}{\partial x_{i}^{2}}\biggr\Vert =O\bigl(\lambda_{i}^{2} \bigr). \end{aligned}$$

Now, we consider a point $(z_{1},z_{2})\in \partial S^{4}_{+}\times\partial S^{4}_{+}$ such that $z_{1}$ and $z_{2}$ are nondegenerate critical points of $K_{1}$. We set

$$\begin{aligned} &\frac{1}{\lambda_{i}}=\varepsilon\frac{S_{4}}{16c_{3}}K(z_{i}) \biggl( \frac{\partial K}{\partial\nu}(z_{i}) \biggr)^{-1}(1+ \zeta_{i});\qquad x_{i}=z_{i}+\xi_{i}, \end{aligned}$$

where $\zeta_{i} \in\mathbb{R}$ and $(\xi_{1},\xi_{2})\in\mathbb {R}^{3}\times\mathbb{R}^{3}$ are assumed to be small.

Using (28) and these changes of variables, $(E_{\alpha_{i}})$ becomes

$$\begin{aligned} \beta_{i}=V_{\alpha_{i}}(\varepsilon,\beta ,\zeta, \xi)=O\bigl(\beta^{2}+\varepsilon \vert \log\varepsilon \vert +\vert \xi \vert ^{2}\bigr). \end{aligned}$$

(37)

Also, we use (30), we have

$$\begin{aligned}& \frac{2c_{3}}{\lambda_{i}^{2}}\frac{\partial K}{\partial\nu}(z_{i}+\xi_{i})+ \frac{\varepsilon K(z_{i}+\xi _{i})S_{4}}{8\lambda_{i}} \\& \quad=\frac{\varepsilon^{2}S_{4}^{2}K(z_{i})^{2}}{128c_{3}} \biggl(\frac{\partial K}{\partial \nu}(z_{i}) \biggr)^{-2}(1+2\zeta_{i}) \biggl(-\frac{\partial K}{\partial \nu}(z_{i})+D^{2}K(z_{i}) (e_{4},\xi_{i}) \biggr) \\& \qquad{}+\frac{\varepsilon^{2}S_{4}^{2}K(z_{i})^{2}}{128c_{3}} \biggl(\frac{\partial K}{\partial \nu}(z_{i}) \biggr)^{-1}(1+\zeta_{i})+O \bigl(\varepsilon^{2} \bigl( \zeta_{i}^{2}+\vert \xi_{i}\vert ^{2}\bigr) \bigr) \\& \quad =-\frac{\varepsilon^{2}S_{4}^{2}K(z_{i})^{2}}{128c_{3}} \biggl(\frac{\partial K}{\partial \nu}(z_{i}) \biggr)^{-1}\zeta_{i} \\& \qquad{} +\frac{\varepsilon ^{2}S_{4}^{2}K(z_{i})^{2}}{128c_{3}} \biggl( \frac{\partial K}{\partial\nu}(z_{i}) \biggr)^{-2}D^{2}K(z_{i}) (e_{4},\xi_{i}) \\& \qquad{} + O \bigl(\varepsilon^{2}\bigl(\zeta_{i}^{2}+ \vert \xi_{i}\vert ^{2}\bigr) \bigr). \end{aligned}$$

Combining this with (31), then $(E_{\lambda_{i}})$ becomes

$$\begin{aligned} -\zeta_{i}+ \biggl(\frac{\partial K}{\partial \nu}(z_{i}) \biggr)^{-1}D^{2}K_{1}(z_{i}) (e_{4},\xi_{i})&= V_{\lambda _{i}}(\varepsilon,\beta,\zeta, \xi) \\ &=O\bigl( \varepsilon \vert \log\varepsilon \vert +\vert \beta \vert ^{2}+\zeta_{i}^{2}+\vert \xi \vert ^{2}\bigr). \end{aligned}$$

(38)

Using (32), (33), and (36), $(E_{x_{i}})$ is equivalent to

$$\begin{aligned} D^{2}K_{1}(z_{i}) \xi_{i}=V_{x_{i}}(\varepsilon,\beta ,\zeta,\xi)= O\bigl( \varepsilon^{1/2}+\vert \beta \vert ^{2}+\vert \zeta \vert ^{2}+\vert \xi \vert ^{2}\bigr). \end{aligned}$$

(39)

Observe that the functions $V_{\alpha_{i}}$, $V_{\lambda_{i}}$, and $V_{x_{i}}$ are smooth.

We can also write the system as

$$\begin{aligned} \textstyle\begin{cases}\beta=V(\varepsilon,\beta,\zeta,\xi),\\ L(\zeta,\xi)=W(\varepsilon,\beta,\zeta,\xi), \end{cases}\displaystyle \end{aligned}$$

(40)

where L is a fixed linear operator on $\mathbb{R}^{8}$ defined by

$$\begin{aligned} L(\zeta,\xi) ={}& \biggl(-\zeta_{1}+ \biggl(\frac{\partial K}{\partial \nu}(z_{1}) \biggr)^{-1}D^{2}K_{1}(z_{1}) (e_{4},\xi_{1}); -\zeta_{2}+ \biggl( \frac{\partial K}{\partial \nu}(z_{2}) \biggr)^{-1}D^{2}K_{1}(z_{2}) (e_{4},\xi_{2}); \\ &{} D^{2}K_{1}(z_{1})\xi _{1};D^{2}K_{1}(z_{2})\xi_{2} \biggr), \end{aligned}$$

and V, W are smooth functions satisfying

$$\begin{aligned} \textstyle\begin{cases}V(\varepsilon,\beta,\zeta,\xi)=O(\varepsilon ^{1/2}+\vert \beta \vert ^{2}+\vert \xi \vert ^{2}),\\ W(\varepsilon,\beta,\zeta,\xi)=O (\varepsilon^{\frac{1}{2}} +\vert \beta \vert ^{2}+\vert \zeta \vert ^{2}+\vert \xi \vert ^{2} ). \end{cases}\displaystyle \end{aligned}$$

Now, by an easy computation, we see that the determinant of the linear operator L is not 0. Hence L is invertible, and according to Brouwer’s fixed point theorem, there exists a solution $(\beta^{\varepsilon},\zeta^{\varepsilon},\xi^{\varepsilon})$ of (40) for ε small enough, such that

$$\begin{aligned} \bigl\vert \beta^{\varepsilon}\bigr\vert =O\bigl(\varepsilon^{1/2} \bigr);\qquad \bigl\vert \zeta^{\varepsilon}\bigr\vert =O \bigl( \varepsilon^{{1}/{2}} \bigr);\qquad\bigl\vert \xi^{\varepsilon}\bigr\vert =O \bigl(\varepsilon^{{1}/{2}} \bigr). \end{aligned}$$

Hence, we have constructed $m^{\varepsilon}= (\alpha_{1}^{\varepsilon},\alpha_{2}^{\varepsilon},\lambda_{1}^{\varepsilon}, \lambda_{2}^{\varepsilon},x_{1}^{\varepsilon},x_{2}^{\varepsilon})$ such that $u_{\varepsilon}:= \sum\alpha_{i}^{\varepsilon}\delta_{(x_{i}^{\varepsilon},\lambda _{i}^{\varepsilon})}+\overline{v_{\varepsilon}}$, verifies (23)-(27). From Proposition 6, $u_{\varepsilon}$ is a critical point of $I_{\varepsilon}$, which implies that $u_{\varepsilon}$ verify

$$ -\Delta u_{\varepsilon}+2u_{\varepsilon}=K\vert u_{\varepsilon} \vert ^{2 -\varepsilon}u_{\varepsilon}\quad\mbox{in } S^{4}_{+}, \qquad \partial u_{\varepsilon}/\partial\nu=0\quad\mbox{on }\partial S^{4}_{+}. $$

(41)

We multiply equation (41) by $u_{\varepsilon}^{-}=\max(0,-u_{\varepsilon})$ and we integrate on $S^{4}_{+}$, we get

$$\begin{aligned} \int_{S^{4}_{+}} \bigl\vert \nabla u_{\varepsilon}^{-}\bigr\vert ^{2} +2 \int_{S^{4}_{+}} \bigl(u_{\varepsilon}^{-}\bigr)^{2}= \int_{S^{4}_{+}}K\bigl(u_{\varepsilon}^{-}\bigr)^{4-\varepsilon}. \end{aligned}$$

(42)

We know also from the Sobolev embedding theorem that

$$\begin{aligned} \bigl\vert u_{\varepsilon}^{-}\bigr\vert _{4-\varepsilon}^{2}:= \biggl( \int _{S^{4}_{+}}K\bigl(u_{\varepsilon}^{-}\bigr)^{4-\varepsilon} \biggr)^{\frac {2}{4-\varepsilon}} \leq C\bigl\Vert u_{\varepsilon}^{-}\bigr\Vert ^{2}. \end{aligned}$$

(43)

Equations (42) and (43) imply that either $u_{\varepsilon}^{-}\equiv 0$, or $\vert u_{\varepsilon}^{-}\vert _{4-\varepsilon}$ is far away from zero. Since $m^{\varepsilon}\in M^{\varepsilon}$, we have $\Vert \overline{v_{\varepsilon}} \Vert <\nu_{0}$, where $\nu_{0}$ is a small positive constant (see the definition of $M_{\varepsilon}$). This implies that $\vert u_{\varepsilon}^{-}\vert _{4-\varepsilon}$ is very small. Thus, $u_{\varepsilon}^{-}\equiv0$ for ε small enough. Then $u_{\varepsilon}$ is a non-negative function which satisfies (41). Finally, the maximum principle completes the proof of our theorem. □

4 Conclusion

Thus it has been concluded that under some assumptions on the function K, there exist solutions of the nonlinear problem $(S_{\varepsilon})$ which are concentrated at two different points of the boundary.