# 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 .

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 .

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.  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 , 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  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  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 )

\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  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  and take account of the new estimates since we have an equation different from the one studied in . From the idea of , 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

 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 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 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.

## References

1. Diening, L, Harjulehto, P, Hasto, P, Ruzicka, M: Lebesgue and Sobolev Spaces with Variable Exponents. Lecture Notes in Mathematics, vol. 2011. Springer, Heidelberg (2011) MR2790542

2. Ambrosetti, A, Garcia Azorero, J, Peral, A: Perturbation of $$\Delta u +u^{\frac{n+2}{n-2}}=0$$, the scalar curvature problem in $$\mathbb{R}^{n}$$ and related topics. J. Funct. Anal. 165, 117-149 (1999)

3. Bahri, A, Coron, JM: The scalar curvature problem on the standard three dimensional spheres. J. Funct. Anal. 95, 106-172 (1991)

4. Bianchi, G, Pan, XB: Yamabe equations on half spheres. Nonlinear Anal. 37, 161-186 (1999)

5. Chang, SA, Yang, P: A perturbation result in prescribing scalar curvature on $$S^{n}$$. Duke Math. J. 64, 27-69 (1991)

6. Cherrier, P: Problèmes de Neumann non linéaires sur les variétés riemaniennes. J. Funct. Anal. 57, 154-207 (1984)

7. Escobar, J: Conformal deformation of Riemannian metric to scalar flat metric with constant mean curvature on the boundary. Ann. Math. 136, 1-50 (1992)

8. Escobar, J, Schoen, R: Conformal metrics with prescribed scalar curvature. Invent. Math. 86, 243-254 (1986)

9. Han, ZC, Li, YY: The existence of conformal metrics with constant scalar curvature and constant boundary mean curvature. Commun. Anal. Geom. 8, 809-869 (2000)

10. Hebey, E: The isometry concentration method in the case of a nonlinear problem with Sobolev critical exponent on compact manifolds with boundary. Bull. Sci. Math. 116, 35-51 (1992)

11. Li, YY: Prescribing scalar curvature on $$S^{n}$$ and related topics, Part I. J. Differ. Equ. 120, 319-410 (1995); Part II. Existence and compactness. Comm. Pure Appl. Math. 49 437-477 (1996).

12. Ould Bouh, K: Blowing up of sign-changing solutions to a subcritical problem. Manuscr. Math. 146, 265-279 (2015)

13. Djadli, Z, Malchiodi, A, Ould Ahmedou, M: Prescribing the scalar and the boundary mean curvature on the three dimensional half sphere. J. Geom. Anal. 13, 233-267 (2003)

14. Ben Ayed, M, Ghoudi, R, Ould Bouh, K: Existence of conformal metrics with prescribed scalar curvature on the four dimensional half sphere. Nonlinear Differ. Equ. Appl. 19, 629-662 (2012)

15. Rey, O: The topological impact of critical points at infinity in a variational problem with lack of compactness: the dimension 3. Adv. Differ. Equ. 4, 581-616 (1999)

16. Bahri, A: An invariant for Yamabe-type flows with applications to scalar curvature problems in high dimension. A celebration of J. F. Nash jr. Duke Math. J. 81, 323-466 (1996)

17. Bahri, A, Li, YY, Rey, O: On a variational problem with lack of compactness: The topological effect of the critical points at infinity. Calc. Var. Partial Differ. Equ. 3, 67-94 (1995)

## Acknowledgements

I would like to thank Deanship of Scientific Research at Taibah University for the financial support of this research project.

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Correspondence to Iqbal H Jebril. 