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. □