C||\tilde{h}||^{2}_{\mathcal{L}_{\Theta}\times\mathcal{L}_{X}\times\mathcal{L}_{\sigma}}.\\\
$

  3. 3.

Furthermore,

$\displaystyle(\partial_{x}\mathscr{E}_{u}(h(\rho,z)+\tilde{h}(\rho,z),(\rho,z)))^{t}(h^{\prime}(\rho,z)+\tilde{h}^{\prime}(\rho,z))-(\partial_{x}\mathscr{E}_{u}(h(\rho,z),(\rho,z)))^{t}(h^{\prime}(\rho,z))$
$\displaystyle-(\partial_{x}\mathscr{E}_{u}(h(\rho,z),(\rho,z)))^{t}\tilde{h}^{\prime}(\rho,z)-\tilde{h}(\rho,z)\partial_{xx}\mathscr{E}_{u}(h(\rho,z),(\rho,z))h^{\prime}(\rho,z)=$
$\displaystyle(\partial_{x}\mathscr{E}_{u}(h(\rho,z)+\tilde{h}(\rho,z),(\rho,z))-\partial_{x}\mathscr{E}_{u}(h(\rho,z),(\rho,z))-\partial_{xx}\mathscr{E}_{u}(h(\rho,z),(\rho,z))(\tilde{h}(\rho,z))^{t})^{t}(h^{\prime}(\rho,z))$
$\displaystyle+\left(\partial_{x}\mathscr{E}_{u}(h(\rho,z)+\tilde{h}(\rho,z),(\rho,z))-\partial_{x}\mathscr{E}_{u}(h(\rho,z),(\rho,z)))^{t}(h^{\prime}(\rho,z))\right)$
$\displaystyle(\partial_{x}\mathscr{E}_{u}(h(\rho,z)+\tilde{h}(\rho,z),(\rho,z))-\partial_{x}\mathscr{E}_{u}(h(\rho,z),(\rho,z))-\partial_{xx}\mathscr{E}_{u}(h(\rho,z),(\rho,z))(\tilde{h}(\rho,z))^{t})^{t}(\tilde{h}^{\prime}(\rho,z))$
$\displaystyle+(\tilde{h}(\rho,z)\partial_{x}\mathscr{E}_{u}(h(\rho,z),(\rho,z)))^{t}\tilde{h}^{\prime}(\rho,z)$
$\displaystyle=(h^{\prime}(\rho,z))O((\tilde{h}(\rho,z))^{2})+O(|\tilde{h}^{\prime}(\rho,z)||\tilde{h}(\rho,z)|)$

Thus, there exists $C=C(u,\delta_{0})>0$ such that $\forall(\rho,z)$, we have

$\displaystyle\left|\left(E_{u}(h+\tilde{h})(\rho,z)-E_{u}(h)(\rho,z)-DE_{u}(h)[\tilde{h}](\rho,z)\right)^{\prime}\right|\leq
C||\tilde{h}||^{2}_{\mathcal{L}_{\Theta}\times\mathcal{L}_{X}\times\mathcal{L}_{\sigma}}.$

  4. 4.

The $C^{2,\alpha}$ estimates and the Hölder semi-norm of the second
derivatives are obtained as above: we use the smoothness ad boundedness of
$\mathscr{E}_{u}$ as well as its derivatives.

Therefore, $E_{u}$ is Fréchet-differentiable on $B_{\delta_{0}}$ and its
Fréchet differential is given by It remains to show that $DE_{u}$ is
continuous on $B_{\delta_{0}}$. Let $h_{1},h_{2}\in B_{\delta_{0}}$ and let
$\tilde{h}\in\mathcal{L}_{\Theta}\times\mathcal{L}_{X}\times\mathcal{L}_{\sigma}$
such that
$||\tilde{h}||_{\mathcal{L}_{\Theta}\times\mathcal{L}_{X}\times\mathcal{L}_{\sigma}}\leq
1$. We have

$\displaystyle||DE_{u}(h_{1})[\tilde{h}]-DE_{u}(h_{2})[\tilde{h}]||_{\hat{C}^{2,\alpha}(\overline{\mathscr{B}})}$
$\displaystyle=||\tilde{h}\cdot(\partial_{x}\mathscr{E}_{u}(h_{1}(\cdot),\cdot)-\partial_{x}\mathscr{E}_{u}(h_{2}(\cdot),\cdot))||_{\hat{C}^{2,\alpha}(\overline{\mathscr{B}})}$
$\displaystyle\leq||\tilde{h}||_{\mathcal{L}_{\Theta}\times\mathcal{L}_{X}\times\mathcal{L}_{\sigma}}\;||(\partial_{x}\mathscr{E}_{u}(h_{1}(\cdot),\cdot)-\partial_{x}\mathscr{E}_{u}(h_{2}(\cdot),\cdot))||_{\hat{C}^{2,\alpha}(\overline{\mathscr{B}})}$

Since $\mathscr{E}_{u}$ is smooth, we have
$\forall(\rho,z)\in\overline{\mathscr{B}},$

$\partial_{x}\mathscr{E}_{u}(h_{1}(\rho,z),(\rho,z))-\partial_{x}\mathscr{E}_{u}(h_{2}(\rho,z),(\rho,z))=\partial_{xx}\mathscr{E}_{u}(h_{1}(\rho,z),(\rho,z))\cdot(h_{1}(\rho,z)-h_{2}(\rho,z))^{t}+O(|h_{1}(\rho,z)-h_{2}(\rho,z)|^{2})$

Now, since $h_{1}\in B_{\delta_{0}}$ and by Proposition 4, there exists
$C=C(u,\delta_{0})$ such that $\forall(\rho,z)\in\overline{\mathscr{B}}$

$|\partial_{xx}\mathscr{E}_{u}(h_{1}(\rho,z),(\rho,z))|\leq
C\quad\text{and}\quad
O(|h_{1}(\rho,z)-h_{2}(\rho,z)|^{2})=O(||h_{1}-h_{2}||^{2}).$

Consequently,

$||(\partial_{x}\mathscr{E}_{u}(h_{1}(\cdot),\cdot)-\partial_{x}\mathscr{E}_{u}(h_{2}(\cdot),\cdot))||_{\hat{C}^{0}(\overline{\mathscr{B}})}\leq
C(u,\delta_{0})||h_{1}-h_{2}||.$

In the same way, we estimate the $\hat{C}^{1}$, $\hat{C}^{2}$ norms and the
Hölder part. We conclude that $E_{u}$ is continuously Fréchet differentiable
on $B_{\delta_{0}}$ ∎

######  Remark 29.

As in the previous lemma, we will show the Fréchet differentiabilty of
functionals of
$(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma})$ defined
on
$B_{\delta_{0}}(0)\subset\mathcal{L}_{\Theta}\times\mathcal{L}_{X}\times\mathcal{L}_{\sigma}$
which also depend (smoothly) on $(\rho,z)\in\overline{\mathscr{B}}$. The steps
are similar to what has been done in the previous lemma and we only compute
the differential in the proofs.

######  Lemma 61.

Let $E\in\mathbb{R}$ and consider the mapping
$\tilde{L}(E,\cdot,\cdot)\,:\,B_{\delta_{0}}\subset(\mathcal{L}_{X}\times\mathcal{L}_{\sigma})\to\widehat{C}^{2,\alpha}(\overline{\mathscr{B}_{A}})$
defined by

$\tilde{L}(E,h,\rho,z):=\frac{\sqrt{X_{K}}}{\rho}\sqrt{1+\overset{\circ}{X}}\left(-1+\frac{1+\overset{\circ}{X}}{(1+\overset{\circ}{\sigma})^{2}}\frac{X_{K}}{\rho^{2}}E^{2}\right)^{\frac{1}{2}}.$

Then, $\tilde{L}(E,\cdot,\cdot)$ is well-defined and it is continuously
Fréchet differentiable on $B_{\delta_{0}}$.

###### Proof.

Let $E\in\mathbb{R}$.

  1. 1.

First, we introduce the mapping
$g_{E}:]-\delta_{0},\delta_{0}[^{2}\times\overline{\mathscr{B}_{A}}$

$g_{E}(x,y,\rho,z):=\frac{\sqrt{X_{K}}}{\rho}\sqrt{1+x}\left(-1+\frac{1+x}{(1+y)^{2}}\frac{X_{K}}{\rho^{2}}E^{2}\right)^{\frac{1}{2}}.$

By Proposition 4,
$\forall(x,y)\in]-\delta_{0},\delta_{0}[^{2}\,;\,g_{E}(x,y,\cdot)$ is well-
defined on $\overline{\mathscr{B}_{A}}$ and it is smooth. Moreover,
$\forall(\rho,z)\in\overline{\mathscr{B}_{A}}$, $g_{E}(\cdot,\rho,z)$ is
smooth on $]-\delta_{0},\delta_{0}[^{2}$ and we have,

$\forall(\rho,z)\in\overline{\mathscr{B}_{A}}\;,\;\nabla_{(x,y)}g_{E}(x,y,\rho,z)=\frac{\sqrt{X_{K}}}{\rho}\frac{(\nabla_{(x,y)}\left(-(1+x)+\frac{(1+x)^{2}}{(1+y)^{2}}E^{2}\right)}{2\left(-(1+x)^{2}+\frac{(1+x)^{2}}{(1+y)^{2}}E^{2}\right)}.$
$\forall(x,y)\in]-\delta_{0},\delta_{0}[^{2}\;,\;\nabla_{(\rho,z)}g_{E}(x,y,\rho,z)=\sqrt{1+x}\frac{\nabla_{(\rho,z)}\left(-\frac{X_{K}}{\rho^{2}}+\frac{1+x}{(1+y)^{2}}\frac{X_{K}^{2}}{\rho^{4}}E^{2}\right)}{2\left(-\frac{X_{K}}{\rho^{2}}+\frac{1+x}{(1+y)^{2}}\frac{X_{K}^{2}}{\rho^{4}}E^{2}\right)}.$

  2. 2.

Let $h=(\overset{\circ}{X},\overset{\circ}{\sigma})\in B_{\delta_{0}}$. First,
we show that
$\tilde{L}(E,h,\cdot)\in\widehat{C}^{2,\alpha}(\overline{\mathscr{B}_{A}})$.
We have

$\tilde{L}(E,h,\rho,z)=g_{E}(\overset{\circ}{X}(\rho,z),\overset{\circ}{\sigma}(\rho,z),\rho,z).$

$g_{E}(x,\cdot)$ is smooth on $\overline{\mathscr{B}_{A}}$ and
$\overset{\circ}{X},\overset{\circ}{\sigma}\in\widehat{C}^{2,\alpha}(\overline{\mathscr{B}_{A}})$.
Thus, $\tilde{L}(E,h,\cdot)$ is well-defined and it also lies in
$\widehat{C}^{2,\alpha}(\overline{\mathscr{B}_{A}})$.

  3. 3.

Now, $\tilde{L}(E,\cdot,\cdot)$ is Fréchet differentiable on $B_{\delta_{0}}$
with derivative $\forall h\in B_{\delta_{0}}$,
$\forall\tilde{h}=(\tilde{X},\tilde{\sigma})\in\mathcal{L}_{X}\times\mathcal{L}_{\sigma}$

$D_{h}\tilde{L}(E,h,\cdot)[\tilde{h}](\rho,z)=(\tilde{X}\;\;\;\tilde{\sigma})\nabla_{(x,y)}g_{E}(\overset{\circ}{X}(\rho,z),\overset{\circ}{\sigma}(\rho,z),\rho,z).$
(9.33)

Moreover , by smoothness of $g_{E}$ with respect to $(x,y)$, there exists
$C=C(\delta_{0},E)$ such that
$\forall\overline{h}\in\mathcal{L}_{\Theta}\times\mathcal{L}_{\sigma}\times\mathcal{L}_{X}$

$\left|\left|D_{h}\tilde{L}(E,h,\cdot)[\overline{h}]\right|\right|_{\widehat{C}^{2,\alpha}(\overline{\mathscr{B}_{A}})}\leq
C(\delta_{0})||h||.$

The latter follows by similar arguments in the proof of Lemma 60

∎

######  Lemma 62.

Let $s\in\mathbb{R}$ and assume that

  * •

$\Phi_{\pm}$ is compactly supported on $\mathcal{B}_{bound}^{\pm}$,

  * •

$\forall\xi\in\mathbb{R}\;,\;\ell_{z}\to\Phi_{\pm}(\xi,\ell_{z})$ lies in
$C^{2,\alpha}(\mathbb{R})$,

  * •

$\forall
h=(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma})\in\mathcal{L}_{\Theta}\times\mathcal{L}_{X}\times\mathcal{L}_{\sigma},\forall\xi\in\mathbb{R}\;,\;(\rho,\ell_{z})\to\Psi_{\eta}(\rho,(\xi,\ell_{z}),h)$
lies in $C^{2,\alpha}(\mathbb{R}^{2})$.

Then, $\forall i\in\left\\{1,2\right\\}$,

$\displaystyle f^{i,\pm}_{\Phi,\eta}(s,\cdot,\cdot):$ $\displaystyle
B_{\delta_{0}}\to\hat{C}^{2,\alpha}(\overline{\mathscr{B}})$ (9.34)
$\displaystyle
h=(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma})\mapsto
f^{i,\pm}_{\Phi,\eta}(s,h,\cdot)$

and $\forall j\in\left\\{3,4\right\\}$,

$\displaystyle f^{j,\pm}_{\Phi,\eta}(s,\cdot,\cdot):$ $\displaystyle
B_{\delta_{0}}\to\hat{C}^{2,\alpha}(\overline{\mathscr{B}_{A}})$ (9.35)
$\displaystyle
h=(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma})\mapsto
f^{j,\pm}_{\Phi,\eta}(s,h,\cdot)$

are well-defined and are continuously Fréchet differentiable mappings on
$B_{\delta_{0}}$ with Fréchet differentials: $\forall
h=(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma})\in
B_{\delta_{0}}\;,\;\forall\tilde{h}=(\tilde{\Theta},\tilde{X},\tilde{\sigma})\in\mathcal{L}_{\Theta}\times\mathcal{L}_{X}\times\mathcal{L}_{\sigma}$
we have

$\displaystyle
Df^{i,\pm}_{\phi_{\pm},\eta}(s,h)[\tilde{h}](\rho,z)=-DE_{s}(h)[\tilde{h}](\rho,z)P^{i}(h,s,\xi)(\rho,z)+\int^{\varepsilon_{2}^{\pm}}_{E_{\rho
s}(h,\rho,z)}\,D_{h}P^{i}(h,s,\xi)[\tilde{h}](\rho,z)\,d\xi$ (9.36)

where $\displaystyle
P^{i}:B_{\delta_{0}}\times\mathbb{R}^{2}\to\hat{C}^{2,\alpha}(\overline{\mathscr{B}})$
and $\displaystyle
P^{j}:B_{\delta_{0}}\times\mathbb{R}^{2}\to\hat{C}^{2,\alpha}(\overline{\mathscr{B}_{A}})$
are the continuously Fréchet différentiable mappings defined by

$P^{1}(h,s,\xi)(\rho,z):=(X_{K}(1+\overset{\circ}{X})+2(\rho
s)^{2})\phi_{\pm}(\xi,\rho s)\Psi_{\eta}(\rho,(\xi,\rho s),h),$ (9.37)
$P^{2}(h,s,\xi)(\rho,z):=\rho
s(\xi-\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)s)\phi_{\pm}(\xi,\rho
s)\Psi_{\eta}(\rho,(\xi,\rho s),h),$ (9.38)
$P^{3}(h,s,\xi)(\rho,z):=\left(\tilde{L}^{2}\left(\xi-\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)s,X_{K}(1+\overset{\circ}{X}),\sigma,\rho,z\right)-s^{2}\right)\phi_{\pm}(\xi,\rho
s)\Psi_{\eta}(\rho,(\xi,\rho s),h)$ (9.39)

where $\tilde{L}$ is defined by (9.10) and

$\displaystyle P^{4}(h,s,\xi)(\rho,z)$
$\displaystyle:=\left(\frac{X_{K}^{2}(1+\overset{\circ}{X})^{2}}{\rho^{4}\left(1+\overset{\circ}{\sigma}\right)^{2}}\left(\xi-\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)s\right)^{2}+\left(1-\frac{X_{K}(1+\overset{\circ}{X})}{\rho^{2}}\right)\left(1+\frac{\rho^{2}}{X_{K}(1+\overset{\circ}{X})}s^{2}\right)\right)$
(9.40) $\displaystyle\phi_{\pm}(\xi,\rho s)\Psi_{\eta}(\rho,(\xi,\rho s),h),$

with derivatives:

$\displaystyle D_{h}P^{1}(h,s,\xi)[\tilde{h}](\rho,z)$
$\displaystyle=-D_{h}\rho_{1}(h,\xi,\rho
s)[\tilde{h}]\mbox{\Large$\chi$}_{\eta}^{\prime}(\rho-\rho_{1}(h,\xi,\rho
s))(X_{K}(1+\overset{\circ}{X})+2(\rho s)^{2})\phi_{\pm}(\xi,\rho s)$ (9.41)
$\displaystyle+X_{K}\tilde{X}\phi_{\pm}(\xi,\rho s)\Psi_{\eta}(\rho,(\xi,\rho
s),h),$

$\displaystyle D_{h}P^{2}(h,s,\xi)[\tilde{h}](\rho,z)$ $\displaystyle=-\rho
sD_{h}\rho_{1}(h,\xi,\rho
s)[\tilde{h}]\mbox{\Large$\chi$}_{\eta}^{\prime}(\rho-\rho_{1}(h,\xi,\rho
s))(\xi-\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)s)\phi_{\pm}(\xi,\rho
s)$ (9.42) $\displaystyle+\rho^{2}s\tilde{\Theta}\phi_{\pm}(\xi,\rho
s)\Psi_{\eta}(\rho,(\xi,\rho s),h),$

$\displaystyle D_{h}P^{3}(h,s,\xi)[\tilde{h}](\rho,z)=$ (9.43) $\displaystyle-
D_{h}\rho_{1}(h,\xi,\rho
s)[\tilde{h}]\mbox{\Large$\chi$}_{\eta}^{\prime}(\rho-\rho_{1}(h,\xi,\rho
s))\left(\tilde{L}^{2}\left(\xi-\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)s,X_{K}(1+\overset{\circ}{X}),\sigma,\rho,z\right)-s^{2}\right)\phi_{\pm}(\xi,\rho
s)$
$\displaystyle+2\tilde{L}\left(\xi-\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)s,X_{K}(1+\overset{\circ}{X}),\sigma,\rho,z\right)\phi_{\pm}(\xi,\rho
s)\Psi_{\eta}(\rho,(\xi,\rho s),h)$ $\displaystyle\left(\rho
s\tilde{\Theta}\frac{\partial\tilde{L}}{\partial
E}\left(\xi-\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)s,X_{K}(1+\overset{\circ}{X}),\sigma,\rho,z\right)+X_{K}\tilde{X}\frac{\partial\tilde{L}}{\partial
X}\left(\xi-\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)s,X_{K}(1+\overset{\circ}{X}),\sigma,\rho,z\right)\right.$
$\displaystyle\left.+\rho\tilde{\sigma}\frac{\partial\tilde{L}}{\partial\sigma}\left(\xi-\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)s,X_{K}(1+\overset{\circ}{X}),\sigma,\rho,z\right)\right)$

and

$\displaystyle
D_{h}P^{4}(h,s,\xi)[\tilde{h}](\rho,z)=-D_{h}\rho_{1}(h,\xi,\rho
s)[\tilde{h}]\mbox{\Large$\chi$}_{\eta}^{\prime}(\rho-\rho_{1}(h,\xi,\rho s))$
(9.44)
$\displaystyle\left(\frac{X_{K}^{2}(1+\overset{\circ}{X})^{2}}{\rho^{4}\left(1+\overset{\circ}{\sigma}\right)^{2}}\left(\xi-\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)s\right)^{2}+\left(1-\frac{X_{K}(1+\overset{\circ}{X})}{\rho^{2}}\right)\left(1+\frac{\rho^{2}}{X_{K}(1+\overset{\circ}{X})}s^{2}\right)\right)\phi_{\pm}(\xi,\rho
s)$ $\displaystyle+\phi_{\pm}(\xi,\rho s)\Psi_{\eta}(\rho,(\xi,\rho s),h)$
$\displaystyle\left(\frac{X_{K}^{2}(1+\overset{\circ}{X})^{2}}{\rho^{3}\left(1+\overset{\circ}{\sigma}\right)^{2}}s\tilde{\Theta}\left(\xi-\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)s\right)-2\tilde{\sigma}\left(1+\overset{\circ}{\sigma}\right)\frac{X_{K}^{2}(1+\overset{\circ}{X})^{2}}{\rho^{4}\left(1+\overset{\circ}{\sigma}\right)^{4}}\left(\xi-\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)s\right)^{2}\right.$
$\displaystyle\left.+2\frac{X_{K}^{2}\tilde{X}(1+\overset{\circ}{X})}{\rho^{4}\left(1+\overset{\circ}{\sigma}\right)^{2}}\left(\xi-\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)s\right)^{2}-\frac{X_{K}\tilde{X}}{\rho^{2}}\left(1+\frac{\rho^{2}}{X_{K}(1+\overset{\circ}{X})}s^{2}\right)-\frac{\tilde{X}\rho^{2}s^{2}}{X_{K}(1+\overset{\circ}{X})^{2}}\left(1-\frac{X_{K}(1+\overset{\circ}{X})}{\rho^{2}}\right)\right).$

###### Proof.

The proof follows from Lemma 60, Lemma 61 and the dominated convergence
theorem. We give details for $f^{1,\pm}_{\Phi,\eta}(s,\cdot,\cdot)$.
Regularity for the remaining functions is proved in the same manner.
First of all, let $s\in\mathbb{R}$ and recall the definition of $\displaystyle
f^{1,\pm}_{\Phi,\eta}(s,\cdot,\cdot)$:

$f_{\Phi,\eta}^{1,\pm}(s,h,\rho,z):=\int_{E_{\rho
s}(h,\rho,z)}^{\varepsilon^{\pm}_{2}}P^{1}(h,s,\xi)(\rho,z)\,d\xi$

where

$P^{1}(h,s,\xi)(\rho,z):=(X_{K}(1+\overset{\circ}{X})+2(\rho
s)^{2})\phi_{\pm}(\xi,\rho s)\Psi_{\eta}(\rho,(\xi,\rho s),h),$

  * •

We recall

$\Psi_{\eta}(\cdot,(\varepsilon,\ell),h):=\left\\{\begin{aligned}
&\mbox{\Large$\chi$}_{\eta}(\cdot-\rho_{1}(h,(\varepsilon,\ell_{z}))),\quad(\varepsilon,\ell_{z})\in\mathcal{B}_{bound}\\\
&0\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(\varepsilon,\ell_{z})\notin\mathcal{A}_{bound},\end{aligned}\right.$

By Lemma 48,
$\forall(\varepsilon,\ell_{z})\in\mathcal{B}_{bound}\,,\,\rho_{1}(\cdot,(\varepsilon,\ell_{z}))$
is continuously Fréchet differentiable on $B_{\delta_{0}}$. By smoothness of
$\mbox{\Large$\chi$}_{\eta}$, $\Psi_{\eta}$ is continuously Fréchet
differentiable with respect to $h$ with derivative
$\forall\tilde{h}\in\mathcal{L}_{\Theta}\times\mathcal{L}_{X}\times\mathcal{L}_{\sigma}$

$D_{h}\Psi_{\eta}(\rho,(\xi,\rho s),h)=\left\\{\begin{aligned}
&-D_{h}\rho_{1}(h,\xi,\rho
s)[\tilde{h}]\mbox{\Large$\chi$}_{\eta}^{\prime}(\rho-\rho_{1}(h,\xi,\rho
s))\quad\text{if}\;(\xi,\rho s)\in\mathcal{B}_{bound},\\\
&0\quad\text{otherwise.}\end{aligned}\right.$

  * •

Hence, $P^{1}$ is Fréchet differentiable with respect to $h$ with derivative
given by (9.41).

  * •

Now, let $\overline{P}^{1}(h,s,\cdot)(\rho,z)$ be a primitive of
$P^{1}(h,s,\cdot)(\rho,z)$. We write

$f_{\Phi,\eta}^{1,\pm}(s,h,\rho,z)=\overline{P}^{1}(h,s,\varepsilon_{2}^{\pm})(\rho,z)-\overline{P}^{1}(h,s,E_{\rho
s}(h,\rho,z))(\rho,z).$

Hence,
$\forall\tilde{h}\in\mathcal{L}_{\Theta}\times\mathcal{L}_{X}\times\mathcal{L}_{\sigma}$,
we have

$\displaystyle D_{h}f_{\Phi,\eta}^{1,\pm}(s,h,\rho,z)$
$\displaystyle=D_{h}\overline{P}^{1}(h,s,\varepsilon_{2}^{\pm})[\tilde{h}](\rho,z)-D_{h}\overline{P}^{1}(h,s,E_{\rho
s}(h,\rho,z))[\tilde{h}](\rho,z)$ $\displaystyle-
D_{h}E_{s}(h)[\tilde{h}](\rho,z)\partial_{\xi}\overline{P}^{1}(h,s,E_{\rho
s}(h,\rho,z))(\rho,z)$ $\displaystyle=\int^{\varepsilon_{2}^{\pm}}_{E_{\rho
s}(h,\rho,z)}\,D_{h}P^{1}(h,s,\xi)[\tilde{h}](\rho,z)\,d\xi-
DE_{s}(h)[\tilde{h}](\rho,z)P^{1}(h,s,\xi)(\rho,z).$

The first term of the latter expression was obtained by the dominated
convergence theorem and the second term is obtained by the definition of
$\overline{P}^{1}$.

  * •

It remains to show that the derivative of $f_{\Phi,\eta}^{1,\pm}$ with respect
to $h$ is continuous. This follows using similar arguments.

∎

######  Proposition 24.

  1. 1.

Let
$(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma},\overset{\circ}{\lambda})\in\mathcal{L}_{\Theta}\times\mathcal{L}_{X}\times\mathcal{L}_{\sigma}\times\mathcal{L}_{\lambda}$.
Then,
$F_{1}(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma})$,
$F_{2}(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma})$,
$F_{3}(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma})$
lie in $\hat{C}^{2,\alpha}(\overline{\mathscr{B}})$ and
$F_{4}(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma},\overset{\circ}{\lambda})$
lies in $\hat{C}^{1,\alpha}(\overline{\mathscr{B}})$. Moreover, they are
compactly supported in $\overline{\mathscr{B}_{A}}$.

  2. 2.

Let $\delta_{0}>0$ and let $F_{i}$ be defined on
$B_{\delta_{0}}$242424$\displaystyle
B_{\delta_{0}}\subset\mathcal{L}_{\Theta}\times\mathcal{L}_{X}\times\mathcal{L}_{\sigma}\times\mathcal{L}_{\lambda}$
is the closed ball of radius $\delta_{0}$ centred in $(0,0,0,0)$. Then
$F_{i}$, $i\in\left\\{1,2,3\right\\}$, are a well-defined mappings from
$B_{\delta_{0}}$ to $\hat{C}^{2,\alpha}(\overline{\mathscr{B}})$ and $F_{4}$
is a well-defined mapping from $B_{\delta_{0}}$ to
$\hat{C}^{1,\alpha}(\overline{\mathscr{B}})$. Furthermore, $F_{i}$,
$i\in\left\\{1,2,3,4\right\\}$, are continuously Fréchet differentiable on
$B_{\delta_{0}}$ with Fréchet differential given by
$\forall(h=(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma}),\overset{\circ}{\lambda})\in
B_{\delta_{0}},(\forall\tilde{h}=(\tilde{\Theta},\tilde{X},\tilde{\sigma}),\tilde{\lambda})\in\mathcal{L}_{\Theta}\times\mathcal{L}_{X}\times\mathcal{L}_{\sigma}\times\mathcal{L}_{\lambda}$,
$\forall(\rho,z)\in\overline{\mathscr{B}}:$

$\displaystyle DF_{1}(h)[\tilde{h}](\rho,z)=$ (9.45)
$\displaystyle-\frac{2\pi}{1+\overset{\circ}{\sigma}}\left(D\tilde{L}(\varepsilon_{2}^{+},h,\rho,z)[\tilde{h}]f_{\phi_{+},\eta}^{1,+}\left(\tilde{L}(\varepsilon_{2}^{+},h,\rho,z)h,\rho,z\right)+D\tilde{L}(\varepsilon_{2}^{-},h,\rho,z)[\tilde{h}]f_{\phi_{-},\eta}^{1,-}\left(\tilde{L}(\varepsilon_{2}^{-},h,\rho,z)h,\rho,z\right)\right)$
$\displaystyle+\int_{\frac{\ell_{1}^{+}}{\rho}}^{\tilde{L}(\varepsilon_{2}^{+},h,\rho,z)}D_{h}\left(-\frac{2\pi}{1+\overset{\circ}{\sigma}}f_{\phi_{+},\eta}^{1,+}(s,h,\rho,z)\right)[\tilde{h}]\,ds+\int^{\frac{\ell_{1}^{-}}{\rho}}_{-\tilde{L}(\varepsilon_{2}^{-},h,\rho,z)}D_{h}\left(-\frac{2\pi}{1+\overset{\circ}{\sigma}}f_{\phi_{-},\eta}^{1,-}(s,h,\rho,z)\right)[\tilde{h}]\,ds$

$\displaystyle DF_{2}(h)[\tilde{h}](\rho,z)$ $\displaystyle=\frac{2\pi
X_{K}(1+\overset{\circ}{X})}{1+\overset{\circ}{\sigma}}\left(D\tilde{L}(\varepsilon_{2}^{+},h,\rho,z)[\tilde{h}]f_{\phi_{+},\eta}^{2,+}\left(\tilde{L}(\varepsilon_{2}^{+},h,\rho,z)h,\rho,z\right)\right.$
(9.46)
$\displaystyle\left.+D\tilde{L}(\varepsilon_{2}^{-},h,\rho,z)[\tilde{h}]f_{\phi_{-},\eta}^{2,-}\left(\tilde{L}(\varepsilon_{2}^{-},h,\rho,z)h,\rho,z\right)\right)$
$\displaystyle+\int_{\frac{\ell_{1}^{+}}{\rho}}^{\tilde{L}(\varepsilon_{2}^{+},h,\rho,z)}D_{h}\left(\frac{2\pi
X_{K}(1+\overset{\circ}{X})}{1+\overset{\circ}{\sigma}}f_{\phi_{+},\eta}^{2,+}(s,h,\rho,z)\right)[\tilde{h}]\,ds$
$\displaystyle+\int^{\frac{\ell_{1}^{-}}{\rho}}_{-\tilde{L}(\varepsilon_{2}^{-},h,\rho,z)}D_{h}\left(-\frac{2\pi
X_{K}(1+\overset{\circ}{X})}{1+\overset{\circ}{\sigma}}f_{\phi_{-},\eta}^{2,-}(s,h,\rho,z)\right)[\tilde{h}]\,ds$

$\displaystyle DF_{3}(h)[\tilde{h}](\rho,z)$
$\displaystyle=\frac{\rho^{4}}{X_{K}^{2}}\frac{2\pi(1+\overset{\circ}{\sigma})}{(1+\overset{\circ}{X})^{2}}\left(D\tilde{L}(\varepsilon_{2}^{+},h,\rho,z)[\tilde{h}]f_{\phi_{+},\eta}^{3,+}\left(\tilde{L}(\varepsilon_{2}^{+},h,\rho,z)h,\rho,z\right)\right.$
(9.47)
$\displaystyle\left.+D\tilde{L}(\varepsilon_{2}^{-},h,\rho,z)[\tilde{h}]f_{\phi_{-},\eta}^{3,-}\left(\tilde{L}(\varepsilon_{2}^{-},h,\rho,z)h,\rho,z\right)\right)$
$\displaystyle+\int_{\frac{\ell_{1}^{+}}{\rho}}^{\tilde{L}(\varepsilon_{2}^{+},h,\rho,z)}D_{h}\left(\frac{\rho^{4}}{X_{K}^{2}}\frac{2\pi(1+\overset{\circ}{\sigma})}{(1+\overset{\circ}{X})^{2}}f_{\phi_{+},\eta}^{3,+}(s,h,\rho,z)\right)[\tilde{h}]\,ds$
$\displaystyle+\int^{\frac{\ell_{1}^{-}}{\rho}}_{-\tilde{L}(\varepsilon_{2}^{-},h,\rho,z)}D_{h}\left(\frac{\rho^{4}}{X_{K}^{2}}\frac{2\pi(1+\overset{\circ}{\sigma})}{(1+\overset{\circ}{X})^{2}}f_{\phi_{-},\eta}^{3,-}(s,h,\rho,z)\right)[\tilde{h}]\,ds$

$\displaystyle
DF_{4}(h,\overset{\circ}{\lambda})[\tilde{h},\tilde{\lambda}](\rho,z)$
$\displaystyle=-\frac{8\pi\tilde{\lambda}e^{2\left(\overset{\circ}{\lambda}+\lambda_{K}\right)}}{1+\overset{\circ}{\sigma}}\left(\int_{\frac{\ell_{1}^{+}}{\rho}}^{\tilde{L}(\varepsilon_{2}^{+},h,\rho,z)}\;f_{\phi_{+},\eta}^{4,+}(s,h,\rho,z)\,ds,\right.$
(9.48)
$\displaystyle\left.+\int^{\frac{\ell_{1}^{-}}{\rho}}_{-\tilde{L}(\varepsilon_{2}^{-},h,\rho,z)}\;f_{\phi_{-},\eta}^{4,-}(s,h,\rho,z)\,ds\right)$
$\displaystyle-\frac{4\pi
e^{2\left(\overset{\circ}{\lambda}+\lambda_{K}\right)}}{1+\overset{\circ}{\sigma}}\left(D\tilde{L}(\varepsilon_{2}^{+},h,\rho,z)[\tilde{h}]f_{\phi_{+},\eta}^{4,+}\left(\tilde{L}(\varepsilon_{2}^{+},h,\rho,z)h,\rho,z\right)\right.$
$\displaystyle\left.+D\tilde{L}(\varepsilon_{2}^{-},h,\rho,z)[\tilde{h}]f_{\phi_{-},\eta}^{4,-}\left(\tilde{L}(\varepsilon_{2}^{-},h,\rho,z)h,\rho,z\right)\right)$
$\displaystyle+\int_{\frac{\ell_{1}^{+}}{\rho}}^{\tilde{L}(\varepsilon_{2}^{+},h,\rho,z)}D_{h}\left(-\frac{4\pi
e^{2\left(\overset{\circ}{\lambda}+\lambda_{K}\right)}}{1+\overset{\circ}{\sigma}}f_{\phi_{+},\eta}^{4,+}(s,h,\rho,z)\right)[\tilde{h}]\,ds$
$\displaystyle+\int^{\frac{\ell_{1}^{-}}{\rho}}_{-\tilde{L}(\varepsilon_{2}^{-},h,\rho,z)}D_{h}\left(-\frac{4\pi
e^{2\left(\overset{\circ}{\lambda}+\lambda_{K}\right)}}{1+\overset{\circ}{\sigma}}f_{\phi_{-},\eta}^{4,-}(s,h,\rho,z)\right)[\tilde{h}]\,ds$

###### Proof.

  1. 1.

Since the matter terms vanish in the regions
$\overline{\mathscr{B}_{N}},\overline{\mathscr{B}_{S}}$,
$\overline{\mathscr{B}_{H}}$ and near the axis we prove the result only in
$\overline{\mathscr{B}_{A}}\backslash\tilde{\mathscr{A}}$, where
$\tilde{\mathscr{A}}$ is some neighbourhood of the axis given by Lemma 59.

  2. 2.

By the first point of the proposition, the mappings $F_{i}$ are well defined.
In order the show the Fréchet differentiability, we apply lemmas 61, 62 and
Lebesgue dominated convergence theorem.

  3. 3.

Since $\mbox{\Large$\chi$}_{\eta}$ is either $0$ or $1$ on the support of
$\phi_{\pm}(\xi,\rho s)$, $D\Psi_{\eta}(\cdot,(\xi,\rho s),h)$ vanishes.
Hence, we can eliminate the terms including $D\Psi_{\eta}$ in the derivatives
of $P^{i}$’s.

  4. 4.

The estimates follow by using similar arguments from the previous lemmas.

∎

###  9.3 Introduce the bifurcation parameter $\delta$

Solutions to the Einstein-Vlasov system will be obtained by perturbing the
Kerr spacetime using a bifurcation parameter $\delta\geq 0$. The latter turns
on in the presence of Vlasov matter supported on
$\mathcal{B}_{bound}\subset\subset\mathcal{A}_{bound}$. In order to introduce
the latter in the equations, we adjust the ansatz for $f$ (5.2) in oder to
make the dependence on $\delta$ explicit:

$f^{\delta}(t,\phi,\rho,z,\phi,p^{\rho},p^{\phi},p^{z})=\Phi(\varepsilon,\ell_{z};\delta)\Psi_{\eta}(\rho,(\varepsilon,\ell_{z}),(X,W,\sigma))$
(9.49)

such that

$\forall(\varepsilon,\ell_{z})\in\mathcal{A}_{bound}\;,\;\quad\Phi(\varepsilon,\ell_{z};0)=0,$

where $\Phi:\mathcal{A}_{bound}\times\mathbb{R}_{+}\to\mathbb{R}_{+}$. We will
impose the regularity assumptions on $\Phi$ given in Lemma 62 so that the
solution operator is well defined. Assuming that $(g,f^{\delta})$ solves the
EV-system, we can apply Theorem (18) with the ansatz (9.49) to obtain the
explicit dependence of $F_{i}$ on the bifurcation parameter $\delta$:

$\displaystyle
F_{1}(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma};\delta)(\rho,z):=-\frac{2\pi
e^{2\left(\overset{\circ}{\lambda}+\lambda_{K}\right)}}{1+\overset{\circ}{\sigma}}$
$\displaystyle\int_{D(\rho,z)}(X_{K}(1+\overset{\circ}{X})+2(\rho
L)^{2})\Phi(E+\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)L,\rho L)$
$\displaystyle\Psi_{\eta}(\rho,(E+\rho(-\overset{\circ}{\Theta}+\omega_{K})L,\rho
L),(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma}))\,dEdL,$
$\displaystyle
F_{2}(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma};\delta)(\rho,z):=\frac{2\pi}{1+\overset{\circ}{\sigma}}X_{K}(1+\overset{\circ}{X})$
$\displaystyle\int_{D(\rho,z)}\rho
LE\Phi(E+\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)L,\rho L)$
$\displaystyle\Psi_{\eta}(\rho,(E+\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)L,\rho
L),(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma}))\,dEdL,$
$\displaystyle
F_{3}(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma};\delta)(\rho,z):=\frac{\rho^{4}}{X^{2}_{K}}\frac{2\pi(1+\overset{\circ}{\sigma})}{\left(1+\overset{\circ}{X}\right)^{2}}$
$\displaystyle\int_{D(\rho,z)}\left(\tilde{L}^{2}-L^{2}\right)\Phi(E+\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)L,\rho
L)$
$\displaystyle\Psi_{\eta}(\rho,(E+\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)L,\rho
L),(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma}))\,dEdL,$
$\displaystyle
F_{4}(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma};\delta)(\rho,z)$
$\displaystyle:=-\frac{4\pi
e^{2\left(\overset{\circ}{\lambda}+\lambda_{K}\right)}}{1+\overset{\circ}{\sigma}}\int_{D(\rho,z)}\left(\frac{X_{K}^{2}\left(1+\overset{\circ}{X}\right)^{2}}{\rho^{4}\left(1+\overset{\circ}{\sigma}\right)^{2}}E^{2}+\left(1-\frac{X_{K}}{\rho^{2}}\left(1+\overset{\circ}{X}\right)\right)\left(1+\frac{\rho^{2}}{X_{K}}\frac{1}{1+\overset{\circ}{X}}L^{2}\right)\right)$
$\displaystyle\Phi(E+\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right)L,\rho
L)\Psi_{\eta}(\rho,(E+\rho\left(-\overset{\circ}{\Theta}+\omega_{K}\right),\rho
L),(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma}))\,dEdL.$

Finally, we note that
$\forall(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma}),\forall(\rho,z)\,$,
$F_{i}(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma};\cdot)(\rho,z)$
is continuously Fréchet differentiable on $[0,\delta_{0}[$.

###  9.4 Solving for $\sigma$

In this section, we solve for $\overset{\circ}{\sigma}$ in terms of the
renormalised unkonws
$\left(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda}\right)$
and $\delta$. We recall that $\overset{\circ}{\sigma}$ verifies

$\Delta_{\mathbb{R}^{4}}{\overset{\circ}{\sigma}}=\rho^{-1}\sigma^{-1}{X}e^{2\lambda}F_{3}(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma},\delta)(\rho,z),$

where $\Delta_{\mathbb{R}^{4}}$ is the Laplacian corresponding to the flat
metric on $\mathbb{R}^{4}$ given by
$g_{\mathbb{R}^{4}}=d\rho^{2}+dz^{2}+\rho^{2}d\mathbb{S}^{2}$. To any function
$f:\overline{\mathscr{B}}\to\mathbb{R}$, we associate a function
$f_{\mathbb{R}^{4}}:\mathbb{R}^{4}\to\mathbb{R}$ defined in the following way:
let $x=(x_{i})_{i=1\dots 4}\in\mathbb{R}$ and define its cylindrical
coordinates
$(\rho,\theta,\phi,z)\in[0,\infty[\times\mathbb{S}^{2}\times\mathbb{R}$ such
that

$\displaystyle x_{1}$ $\displaystyle=\rho\sin\theta\cos\phi$ $\displaystyle
x_{2}$ $\displaystyle=\rho\cos\theta\sin\phi$ $\displaystyle x_{3}$
$\displaystyle=\rho\cos\theta$ $\displaystyle x_{4}$ $\displaystyle=z.$

Now define $f_{\mathbb{R}^{4}}$ by

$f_{\mathbb{R}^{4}}(x):=f(\rho=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}},z=x_{4}).$

In the following, we confound $f$ with $f_{\mathbb{R}}^{4}$. We start with
solving the linear problem.

####  9.4.1 Linear problem

We prove the following result

######  Proposition 25.

Let $H_{\sigma}\in\mathcal{N}_{\sigma}$. Then, there exists a unique
$\overset{\circ}{\sigma}\in\mathcal{L}_{\sigma}$ which solves the equation

$\Delta_{\mathbb{R}^{4}}\overset{\circ}{\sigma}=H_{\sigma}.$

It is given by

$\overset{\circ}{\sigma}(\rho,z)=\overset{\circ}{\sigma}_{\mathbb{R}^{4}}(x)=-\int_{\mathbb{R}^{4}}\,\frac{1}{|x-y|^{2}}(H_{\sigma})_{\mathbb{R}^{4}}(y)\,dy.$

Moreover, there exists $C(\alpha_{0})>0$ such that

$||\overset{\circ}{\sigma}||_{\mathcal{L}_{\sigma}}\leq
C(\alpha_{0})||H_{\sigma}||_{\mathcal{N}_{\sigma}}.$

The proof is based on Theorem 10 and the following Newtonian estimates

######  Lemma 63.

Let $n\geq 3$ and $F:\mathbb{R}^{n}\to\mathbb{R}$ satisfies $|F(x)|\leq
C\langle x\rangle^{-k}$ where $\displaystyle\langle
x\rangle:=(1+|x|^{2})^{\frac{1}{2}}$ for some $k>2$ and $k\neq n$. Let
$u:\mathbb{R}^{n}\to\mathbb{R}$ be the corresponding Newton potential:

$u(x):=\int_{\mathbb{R}^{n}}\;|x-y|^{2-n}F(y)\,dy.$

Then, we have

$|u(x)|\leq C\sup_{y\in\mathbb{R}^{n}}\left|\langle
y\rangle^{k}F(y)\right|\left(\langle x\rangle^{2-n}+\langle
x\rangle^{2-k}\right),$ (9.50)

and

$|\partial u(x)|\leq C\sup_{y\in\mathbb{R}^{n}}\left|\langle
y\rangle^{k}F(y)\right|\left(\langle x\rangle^{1-n}+\langle
x\rangle^{1-k}\right).$ (9.51)

###### Proof.

  1. 1.

First of all, we have $\forall x\in\mathbb{R}$

$\displaystyle|u(x)|$
$\displaystyle\leq\int_{\mathbb{R}^{n}}\,|x-y|^{2-n}|F(y)|\,dy$
$\displaystyle\leq\sup_{y\in\mathbb{R}^{n}}\left|\langle
y\rangle^{k}F(y)\right|\int_{\mathbb{R}^{n}}\,|x-y|^{2-n}\langle
y\rangle^{-k}\,dy.$

Set

$A(x):=\int_{\mathbb{R}^{n}}\;|x-y|^{2-n}\langle y\rangle^{-k}\,dy,$

we have

$\displaystyle A(x)$ $\displaystyle=\int_{y,|x-y|\leq\frac{\langle
x\rangle}{4}}\;|x-y|^{2-n}\langle
y\rangle^{-k}\,dy+\int_{y,|x-y|\geq\frac{\langle
x\rangle}{4}}\;|x-y|^{2-n}\langle y\rangle^{-k}\,dy=I+II.$

We estimate the first term in the right hand side.

$\displaystyle I$ $\displaystyle:=\int_{y,|x-y|\leq\frac{\langle
x\rangle}{4}}\;|x-y|^{2-n}\langle y\rangle^{-k}\,dy.$

By the reversed triangular inequality, we have

$|y|=|x-(x-y)|\geq||x|-|x-y||,$

Thus,

$\langle y\rangle^{-k}=(1+|y|^{2})^{\frac{-k}{2}}\leq(\langle
x\rangle^{2}-2\langle x\rangle|x-y|+|x-y|^{2})^{\frac{-k}{2}},$

We obtain,

$\displaystyle I$ $\displaystyle\leq\int_{y,|x-y|\leq\frac{\langle
x\rangle}{4}}\;|x-y|^{2-n}(\langle x\rangle^{2}-2\langle
x\rangle|x-y|+|x-y|^{2})^{\frac{-k}{2}}\,dy,$
$\displaystyle\leq\int^{\frac{\langle
x\rangle}{4}}_{0}\;\frac{r^{2-n}}{(\langle x\rangle^{2}-2\langle x\rangle
r+r^{2})^{\frac{k}{2}}}r^{n-1}\,dr,$ $\displaystyle\leq\int^{\frac{\langle
x\rangle}{4}}_{0}\;\frac{r^{2-n}}{(\langle x\rangle^{2}-2\langle x\rangle
r)^{\frac{k}{2}}}r^{n-1}\,dr,$ $\displaystyle=\langle
x\rangle^{-k}\int^{\frac{\langle x\rangle}{4}}_{0}\;\frac{r}{(1-2\langle
x\rangle^{-1}r)^{\frac{k}{2}}}r^{n-1}\,dr,$ $\displaystyle=\langle
x\rangle^{2-k}\int^{\frac{1}{4}}_{0}\;\frac{r^{\prime}}{(1-2r^{\prime})^{\frac{k}{2}}}\,dr^{\prime}.$

In the last line, we changed the variable by setting $r^{\prime}=\langle
x\rangle^{-1}r$ Hence

$I\leq C\langle x\rangle^{2-k}.$

Now we estimate the second term $II$ given by

$II:=\int_{y,|x-y|\geq\frac{\langle x\rangle}{4}}\;|x-y|^{2-n}\langle
y\rangle^{-k}\,dy.$

We will deal with two cases $k\leq n$ and $k\geq n$.

Suppose that $k\geq n$. Then

$\displaystyle II$ $\displaystyle=\int_{y,|x-y|\geq\frac{\langle
x\rangle}{4}}\;|x-y|^{2-n}\langle y\rangle^{-k}\,dy,$ $\displaystyle\leq
C\langle x\rangle^{2-n}\int_{y,|x-y|\geq\frac{\langle x\rangle}{4}}\;\langle
y\rangle^{-k}\,dy,$ $\displaystyle\leq C\langle
x\rangle^{2-n}\int_{\mathbb{R}^{n}}\;\langle y\rangle^{-k}\,dy,$
$\displaystyle\leq C\langle
x\rangle^{2-n}\int^{\infty}_{0}\;\frac{r^{n-1}}{(1+r^{2})^{\frac{k}{2}}}\,dr,$

When $r\to\infty$, we have

$\frac{r^{n-1}}{(1+r^{2})^{\frac{k}{2}}}\sim\frac{1}{r^{k-n+1}}$

Since $k>n$ the integral converges. Thus,

$I\leq C\langle x\rangle^{2-n}.$

Now suppose that $k\leq n$. We write

$II=\int_{\mathcal{D}_{1}(x)}\;|x-y|^{2-n}\langle
y\rangle^{-k}\,dy+\int_{\mathcal{D}_{2}(x)}\;|x-y|^{2-n}\langle
y\rangle^{-k}\,dy=II_{1}+II_{2},$

where

$\displaystyle\mathcal{D}_{1}(x)$
$\displaystyle:=\left\\{y,|x-y|\geq\frac{\langle
x\rangle}{4}\quad\text{and}\quad|y|\leq 2\langle x\rangle\right\\},$
$\displaystyle\mathcal{D}_{2}(x)$
$\displaystyle:=\left\\{y,|x-y|\geq\frac{\langle
x\rangle}{4}\quad\text{and}\quad|y|\geq 2\langle x\rangle\right\\}.$

Now

$\displaystyle II_{2}$
$\displaystyle:=\int_{\mathcal{D}_{1}(x)}\;|x-y|^{2-n}\langle
y\rangle^{-k}\,dy,$ $\displaystyle=\int_{|y|\geq 2\langle
x\rangle}\;|x-y|^{2-n}\langle y\rangle^{-k}\,dy,$
$\displaystyle\leq\int_{|y|\geq 2\langle x\rangle}\;||x|-|y||^{2-n}\langle
y\rangle^{-k}\,dy,\quad\quad\text{by reversed triangle inequality},$
$\displaystyle=\int_{|y|\geq 2\langle x\rangle}\;(|y|-|x|)^{2-n}\langle
y\rangle^{-k}\,dy,$ $\displaystyle\leq\int_{|y|\geq 2\langle
x\rangle}\;(|y|-\langle x\rangle)^{2-n}\langle y\rangle^{-k}\,dy,$
$\displaystyle=\int^{\infty}_{2\langle x\rangle}\;\frac{1}{(r-\langle
x\rangle)^{n-2}}\frac{1}{(1+r^{2})^{\frac{k}{2}}}r^{n-1}\,dr,$
$\displaystyle=\int^{\infty}_{2\langle x\rangle}\;\frac{1}{(1-\frac{\langle
x\rangle}{r})^{n-2}}\frac{r}{(1+r^{2})^{\frac{k}{2}}}\,dr.$

Since $r\geq 2\langle x\rangle$, we have the following estimate

$\frac{1}{(1-\frac{\langle x\rangle}{r})^{n-2}}\leq 2^{n-2}.$

Hence,

$\displaystyle II_{2}$ $\displaystyle\leq C\int^{\infty}_{2\langle
x\rangle}\;\frac{r}{(1+r^{2})^{\frac{k}{2}}}\,dr,$
$\displaystyle=C\left[\frac{1}{-\frac{k}{2}+1}(1+r^{2})^{-\frac{k}{2}+1}\right],$
$\displaystyle\leq C\langle x\rangle^{2-k}.$

Now we estimate $II_{1}$.

$\displaystyle II_{1}$ $\displaystyle=\int_{\left\\{|x-y|\geq\frac{\langle
x\rangle}{4}\right\\}\cap\left\\{|y|\leq 2\langle
x\rangle\right\\}}\;|x-y|^{2-n}\langle y\rangle^{-k}\,dy,$
$\displaystyle=\int_{B(0,2\langle x\rangle)\backslash B(x,\frac{\langle
x\rangle}{4})}\;|x-y|^{2-n}\langle y\rangle^{-k}\,dy,$ $\displaystyle\leq
C\langle x\rangle^{2-n}\int_{B(0,2\langle x\rangle)\backslash
B(x,\frac{\langle x\rangle}{4})}\;\langle y\rangle^{-k}\,dy,$
$\displaystyle\leq C\langle x\rangle^{2-n}\int^{2\langle
x\rangle}_{0}\;\frac{r^{n-1}}{(1+r^{2})^{\frac{k}{2}}}\,dr$

If $k<n$ then,

$\displaystyle II_{1}$ $\displaystyle\leq C\langle
x\rangle^{2-n}\int^{2\langle x\rangle}_{0}\;\frac{1}{r^{k-n+1}}\,dr,$
$\displaystyle\leq C\langle x\rangle^{2-n}\langle x\rangle^{n-k}=\langle
x\rangle^{2-k}.$

If $k=n$, then

$\displaystyle II_{1}$ $\displaystyle\leq C\langle
x\rangle^{2-n}\int^{2\langle
x\rangle}_{0}\;r^{n-1}(1+r^{2})^{-\frac{n}{2}}\,dr,$ $\displaystyle\leq
C\langle x\rangle^{2-n}\log\langle x\rangle.$

  2. 2.

For the first order estimates, by the integrability condition of $F$ and the
regularisation property of the convolution product $u$ is differentiable and
we have

$\displaystyle\partial_{x_{i}}u(x)$
$\displaystyle=(2-n)\int_{\mathbb{R}^{n}}\;\partial_{x_{i}}(|x-y|)|x-y|^{1-n}F(y)\,dy,$
$\displaystyle=(2-n)\int_{\mathbb{R}^{n}}\;(x_{i}-y_{i})|x-y|^{-n}F(y)\,dy.$

Thus,

$\displaystyle|\partial u(x)|$
$\displaystyle\leq(n-2)\int_{\mathbb{R}^{n}}\;|x_{i}-y_{i}||x-y|^{-n}F(y)\,dy,$
$\displaystyle\leq(n-2)\int_{\mathbb{R}^{n}}\;|x-y|^{1-n}F(y)\,dy,$
$\displaystyle\leq C\sup_{y\in\mathbb{R}^{n}}\left|\langle
y\rangle^{k}F(y)\right|\int_{\mathbb{R}^{n}}\;|x-y|^{1-n}\langle
y\rangle^{-k}\,dy.$

We use similar arguments to the first point in order to obtain the first order
estimates.

∎

######  Lemma 64.

Let $n\geq 3$ and $F:\mathbb{R}^{n}\to\mathbb{R}$ satisfies
$|F(x)|+\sup_{|x-y|\leq 1}\frac{|F(x)-F(y)|}{|x-y|^{\alpha}}\leq C\langle
x\rangle^{-k}$ for some $k>2$ and $k\neq n$. Let
$u:\mathbb{R}^{n}\to\mathbb{R}$ be the corresponding Newton potential:

$u(x):=\int_{\mathbb{R}^{n}}\;|x-y|^{2-n}F(y)\,dy.$

Then

$\displaystyle|\partial^{2}u(x)|$ $\displaystyle\leq
C\left(\sup_{y\in\mathbb{R}^{n}}\left|\langle
y\rangle^{k}F(y)\right|+\sup_{y\in\mathbb{R}^{n}}\left(\langle
y\rangle^{k}\left(\sup_{z,|z-y|\leq
1}\frac{|F(z)-F(y)|}{|z-y|^{\alpha}}\right)\right)\right)\log(4\langle
x\rangle)\left(\langle x\rangle^{-n}+\langle x\rangle^{-k}\right),$

###### Proof.

  1. 1.

We write 252525The details of computations are given in [34, Chapter 10]. the
second weak derivatives of u ,$\partial_{ij}u$:

$\partial^{2}_{x_{i}x_{j}}u(x)=\int_{\mathbb{R}^{n}}\;\partial^{2}_{x_{i}x_{j}}G(x,y)(F(y)-F(x))\,dy,$

where

$G(x,y):=\frac{1}{|x-y|^{n-2}}.$

  2. 2.

Note that

$\forall\lambda>0\;\int_{\mathbb{S}^{n-1}}\;\frac{\partial^{2}G}{\partial_{x_{i}}\partial{x_{j}}}(0,\lambda\sigma)\,d\sigma=0.$

Indeed, we compute

$\displaystyle\frac{\partial^{2}G}{\partial{x_{i}}\partial{x_{j}}}(x,y)$
$\displaystyle=(2-n)\partial_{x_{i}}(x_{j}-y_{j})|x-y|^{-n},$
$\displaystyle=(2-n)\delta_{ij}|x-y|^{-n}-n(2-n)(x_{i}-y_{i})(x_{j}-y_{j})|x-y|^{-n-2}.$

Besides, observe that

$\int_{\mathbb{S}^{n-1}}\;\sigma_{i}\sigma_{j}\,d\sigma=0,\quad\text{if $i\neq
j$},$

and

$\int_{\mathbb{S}^{n-1}}\;\sigma_{i}^{2}\,d\sigma=\frac{|\mathbb{S}^{n-1}|}{n},\quad\text{if
$i=j$}.$

Hence

$\displaystyle\int_{\mathbb{S}^{n-1}}\;\frac{\partial^{2}G}{\partial{x_{i}}\partial{x_{j}}}(0,\lambda\sigma)\,d\sigma$
$\displaystyle=(n-2)\lambda^{-n}(n\sigma_{i}\sigma_{j}-\delta_{ij}),$
$\displaystyle=0.$

  3. 3.

Now we proceed as in the previous lemma and we use Holder estimates for $F$ in
order to control the terms when $|x-y|\leq 1$.

We have

$\displaystyle\partial^{2}_{x_{i}x_{j}}u(x)$
$\displaystyle=\int_{\mathbb{R}^{n}}\;\partial^{2}_{x_{i}x_{j}}G(x,y)(F(y)-F(x))\,,$
$\displaystyle=\int_{y,|x-y|\leq\frac{\langle
x\rangle}{4}}\;\partial^{2}_{x_{i}x_{j}}|x-y|^{2-n}(F(y)-F(x))\,dy+\int_{y,|x-y|\geq\frac{\langle
x\rangle}{4}}\;\partial^{2}_{x_{i}x_{j}}|x-y|^{2-n}(F(y)-F(x))\,dy,$
$\displaystyle=I+II.$ $\displaystyle I$
$\displaystyle:=\int_{\left\\{y,|x-y|\leq\frac{\langle
x\rangle}{4}\right\\}}\;\partial^{2}_{x_{i}x_{j}}|x-y|^{2-n}(F(y)-F(x))\,dy,$
$\displaystyle=\int_{\left\\{y,|y|\leq\frac{\langle
x\rangle}{4}\right\\}}\;\frac{\partial^{2}G}{\partial{x_{i}}\partial{x_{j}}}(0,y)(F(y+x)-F(x))\,dy,$
$\displaystyle=\int_{\left\\{y,|y|\leq
1\right\\}}\;\frac{\partial^{2}G}{\partial{x_{i}}\partial{x_{j}}}(0,y)(F(y+x)-F(x))\,dy+\int_{\left\\{y,1\leq|y|\leq\frac{\langle
x\rangle}{4}\right\\}}\;\frac{\partial^{2}G}{\partial{x_{i}}\partial{x_{j}}}(0,y)(F(y+x)-F(x))\,dy,$
$\displaystyle=I_{1}+I_{2}.$

Note that the set $\left\\{y,1\leq|y|\leq\frac{\langle x\rangle}{4}\right\\}$
is empty if $\langle x\rangle<2.$

$\displaystyle|I_{2}|$
$\displaystyle:=\left|\int_{\left\\{y,1\leq|y|\leq\frac{\langle
x\rangle}{4}\right\\}}\;\frac{\partial^{2}G}{\partial{x_{i}}\partial{x_{j}}}(0,y)(F(y+x)-F(x))\,dy\right|,$
$\displaystyle=\left|\int_{\left\\{y,1\leq|y|\leq\frac{\langle
x\rangle}{4}\right\\}}\;\frac{\partial^{2}G}{\partial{x_{i}}\partial{x_{j}}}(0,y)F(y+x)\,dy\right|,$
$\displaystyle\leq\int_{\left\\{y,1\leq|y|\leq\frac{\langle
x\rangle}{4}\right\\}}\;\frac{|F(y)|}{|x-y|^{n}}\,dy,$ $\displaystyle\leq
C\sup_{y\in\mathbb{R}^{n}}\left|\langle
y\rangle^{k}F(y)\right|\int_{\left\\{y,1\leq|y|\leq\frac{\langle
x\rangle}{4}\right\\}}\;\frac{\langle y\rangle^{-k}}{|x-y|^{n}}\,dy,$

We estimate for $\langle x\rangle\geq 2$,

$\displaystyle\int_{\left\\{y,1\leq|y|\leq\frac{\langle
x\rangle}{4}\right\\}}\;\frac{\langle y\rangle^{-k}}{|x-y|^{n}}\,dy$
$\displaystyle\leq\int_{\left\\{y,1\leq|y|\leq\frac{\langle
x\rangle}{4}\right\\}}\;|x-y|^{-n}(\langle
x\rangle^{2}-2|x||x-y|+|x-y|^{2})^{-\frac{k}{2}}\,dy,$
$\displaystyle\leq\int^{\frac{\langle x\rangle}{4}}_{1}\frac{1}{r}(\langle
x\rangle^{2}-2|x|r+r^{2})^{-\frac{k}{2}}\,dr,$ $\displaystyle\leq\langle
x\rangle^{-k}\int^{\frac{1}{4}}_{\frac{1}{\langle
x\rangle}}\;\frac{1}{r}(1-2\frac{|x|}{\langle x\rangle}r)^{-\frac{k}{2}}\,dr,$
$\displaystyle\leq\langle x\rangle^{-k}+\langle x\rangle^{-k}\log(4\langle
x\rangle).$

Now we estimate $I_{1}$:

$\displaystyle|I_{1}|$ $\displaystyle\leq\int_{\left\\{y,|y|\leq
1\right\\}}\;|y|^{-n}|F(y+x)-F(x)|\,dy,$
$\displaystyle=\int_{\left\\{y,|y|\leq
1\right\\}}\;|y|^{\alpha-n}\frac{|F(y+x)-F(x)|}{|y|^{\alpha}}\,dy,$
$\displaystyle\leq C\left(\langle
x\rangle^{-k}\sup_{y\in\mathbb{R}^{n}}\left(\langle
y\rangle^{k}\left(\sup_{z,|z-y|\leq
1}\frac{|F(z)-F(y)|}{|z-y|^{\alpha}}\right)\right)\right)\int_{\left\\{y,|y|\leq
1\right\\}}\;|y|^{\alpha-n}\,dy,$ $\displaystyle=C\left(\langle
x\rangle^{-k}\sup_{y\in\mathbb{R}^{n}}\left(\langle
y\rangle^{k}\left(\sup_{z,|z-y|\leq
1}\frac{|F(z)-F(y)|}{|z-y|^{\alpha}}\right)\right)\right)\int^{1}_{0}\;r^{\alpha-1}\,dr.$

The integral of the last line converges. Hence,

$|I_{1}|\leq C\left(\langle
x\rangle^{-k}\sup_{y\in\mathbb{R}^{n}}\left(\langle
y\rangle^{k}\left(\sup_{z,|z-y|\leq
1}\frac{|F(z)-F(y)|}{|z-y|^{\alpha}}\right)\right)\right).$

Thus,

$|I|\leq C\left(\sup_{y\in\mathbb{R}^{n}}\left|\langle
y\rangle^{k}F(y)\right|+\sup_{y\in\mathbb{R}^{n}}\left(\langle
y\rangle^{k}\left(\sup_{z,|z-y|\leq
1}\frac{|F(z)-F(y)|}{|z-y|^{\alpha}}\right)\right)\right)\log(4\langle
x\rangle)\langle x\rangle^{-k}.$ (9.52)

Now we estimate the remaining term,

$\displaystyle|II|$ $\displaystyle=\left|\int_{y,|x-y|\geq\frac{\langle
x\rangle}{4}}\;\partial^{2}_{x_{i}x_{j}}|x-y|^{2-n}(F(y)-F(x))\,dy\right|,$
$\displaystyle=\left|\int_{y,|x-y|\geq\frac{\langle
x\rangle}{4}}\;\partial^{2}_{x_{i}x_{j}}|x-y|^{2-n}F(y)\,dy\right|,$
$\displaystyle\leq\sup_{y\in\mathbb{R}^{n}}\left|\langle
y\rangle^{k}F(y)\right|\int_{\mathbb{R}^{n}}\;|x-y|^{1-n}\langle
y\rangle^{-k}\,dy.$

Set

$\displaystyle II_{1}$ $\displaystyle:=\int_{\left\\{y,|x-y|\geq\frac{\langle
x\rangle}{4}\quad\text{and}\quad|y|\leq 4\langle
x\rangle\right\\}}\;|x-y|^{-n}\langle y\rangle^{-k}\,dy,$ $\displaystyle
II_{2}$ $\displaystyle:=\int_{\left\\{y,|x-y|\geq\frac{\langle
x\rangle}{4}\quad\text{and}\quad|y|\geq 4\langle
x\rangle\right\\}}\;|x-y|^{-n}\langle y\rangle^{-k}\,dy.$

Then,

$\displaystyle|II_{1}|$ $\displaystyle\leq C\langle
x\rangle^{-n}\int_{\left\\{y,|y|\leq 4\langle x\rangle\right\\}}\;\langle
y\rangle^{-k}\,dy,$ $\displaystyle\leq C\langle x\rangle^{-n}\int^{4\langle
x\rangle}_{0}\;r^{n-k-1}\,dr,$ $\displaystyle\leq C\langle
x\rangle^{-n}|B_{1}|+C\langle x\rangle{-n}\int^{4\langle
x\rangle}_{1}\;r^{n-k-1}\,dr,$ $\displaystyle\leq C\langle
x\rangle^{-n}+\left\\{\begin{aligned} &\langle x\rangle^{-k}\int^{4\langle
x\rangle}_{1}\;\frac{1}{r}\quad\text{if $n-k\geq 0$},\\\ &\langle
x\rangle^{-n}\int^{4\langle
x\rangle}_{1}\;\frac{1}{r}\quad\text{else},\end{aligned}\right.$
$\displaystyle\leq\langle x\rangle^{-n}+\log(4\langle x\rangle)\left(\langle
x\rangle^{-n}+\langle x\rangle^{-k}\right).$

Now we estimate

$\displaystyle|II_{2}|$ $\displaystyle\leq\int_{\left\\{y,|y|\geq 4\langle
x\rangle\right\\}}\;|x-y|^{-n}\langle y\rangle^{-k}\,dy,$
$\displaystyle\leq\int_{\left\\{y,|y|\geq 4\langle
x\rangle\right\\}}\;||x|-|y||^{-n}\langle y\rangle^{-k}\,dy,$
$\displaystyle=\int^{\infty}_{4\langle x\rangle}\;(r-\langle
x\rangle)^{-n}\frac{r^{n-1}}{(1+r^{2})^{\frac{k}{2}}}\,dr,$ $\displaystyle\leq
C\int^{\infty}_{4\langle x\rangle}\;\frac{1}{r(1+r^{2})^{\frac{k}{2}}}\,dr,$
$\displaystyle\leq C\int^{\infty}_{4\langle x\rangle}\;\frac{1}{r^{k+1}}\,dr,$
$\displaystyle=C\langle x\rangle^{-k}.$

Finally we obtain the second order estimate.
The Hölder part follows by using similar arguments.

∎

Now, we prove Proposition 25.

###### Proof.

Let $H_{\sigma}\in\mathcal{N}_{\sigma}$. Then,

$\forall x\in\mathbb{R}^{4}\;\;|H(x)|\leq\langle x\rangle^{-5}.$

We set

$\overset{\circ}{\sigma}:=-\int_{\mathbb{R}^{4}}|x-y|^{-2}H_{\sigma}(y)\,dy$

and we apply Lemma 63 with $n=4$ and $k=5$. We obtain:

$||r^{2}\overset{\circ}{\sigma}||_{L^{\infty}(\overline{\mathscr{B}})}\leq
C||H_{\sigma}||_{\mathcal{N}_{\sigma}}$

and

$||r^{3}\partial\overset{\circ}{\sigma}||_{L^{\infty}(\overline{\mathscr{B}})}\leq
C||H_{\sigma}||_{\mathcal{N}_{\sigma}}.$

Since $H_{\sigma}\in C^{1,\alpha}(\mathbb{R}^{4})$ and $H_{\sigma}\in
L^{1}\cap L^{\infty}(\mathbb{R}^{4})$, we obtain Theorem 10 in order to obtain
$\overset{\circ}{\sigma}\in C^{3,\alpha}(\overline{\mathscr{B}})$ and

$||\overset{\circ}{\sigma}||_{C^{3,\alpha}(\overline{\mathscr{B}})}\leq
C||H_{\sigma}||_{C^{1,\alpha}(\overline{\mathscr{B}})}\leq
C||H_{\sigma}||_{\mathcal{N}_{\sigma}}.$

Finally, we apply Lemma 64 with with $n=4$, $k=4$ and $F=H_{\sigma}$ in order
to obtain:

$||r^{4}\log^{-1}(4r)\partial^{2}f||_{L^{\infty}(\overline{\mathscr{B}})}\leq
C||H_{\sigma}||_{C^{1,\alpha}(\mathbb{R}^{4})}\leq
C||H_{\sigma}||_{\mathcal{N}_{\sigma}}.$

∎

####  9.4.2 Non-linear estimates

We apply Theorem 6 in order to obtain

######  Proposition 26.

Let $\alpha_{0}\in(0,1)$ and let $\overline{\delta}_{0}>0$. Then, there exists
$0<\delta_{0}\leq\overline{\delta}_{0}$ such that
$\displaystyle\forall\left(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda},\delta\right)\in
B_{\delta_{0}}\left(\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}\times[0,\infty[\right)$
there exists a unique one parameter family
$\overset{\circ}{\sigma}\left(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda},\delta\right)\in
B_{\delta_{0}}\left(\mathcal{L}_{\sigma}\right)$ which solves (5.49) and which
satisfies

$\left|\left|\overset{\circ}{\sigma}\left(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda},\delta\right)\right|\right|_{\mathcal{L}_{\sigma}}\leq
C(\alpha_{0})\left(||\overset{\circ}{X}||^{2}_{\mathcal{L}_{X}}+||\overset{\circ}{\Theta}||^{2}_{\mathcal{L}_{\Theta}}+||\overset{\circ}{\lambda}||^{2}_{\mathcal{L}_{\lambda}}+\delta\right)$

and
$\displaystyle\forall\left(\overset{\circ}{X}_{i},\overset{\circ}{\Theta}_{i},\overset{\circ}{\lambda}_{i},\delta_{i}\right)\in
B_{\delta_{0}}\left(\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}\times[0,\infty[\right)$,

$\displaystyle\left|\left|\overset{\circ}{\sigma}\left(\overset{\circ}{X}_{1},\overset{\circ}{\Theta}_{1},\overset{\circ}{\lambda}_{1},\delta_{1}\right)-\overset{\circ}{\sigma}\left(\overset{\circ}{X}_{2},\overset{\circ}{\Theta}_{2},\overset{\circ}{\lambda}_{2},\delta_{2}\right)\right|\right|_{\mathcal{L}_{X}}$
$\displaystyle\leq
C(\alpha_{0})\left(\left(\left|\left|\left(\overset{\circ}{X}_{1},\overset{\circ}{\Theta}_{1},\overset{\circ}{\lambda}_{1}\right)\right|\right|_{\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}}+\left|\left|\left(\overset{\circ}{X}_{2},\overset{\circ}{\Theta}_{2},\overset{\circ}{\lambda}_{2}\right)\right|\right|_{\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}}\right)\left|\left|\left(\overset{\circ}{X}_{1},\overset{\circ}{\Theta}_{1},\overset{\circ}{\lambda}_{1}\right)-\left(\overset{\circ}{X}_{2},\overset{\circ}{\Theta}_{2},\overset{\circ}{\lambda}_{2}\right)\right|\right|_{\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}}\right.$
$\displaystyle\left.+\left|\delta_{1}-\delta_{2}\right|\right).$

Before we prove the above proposition, we introduce the following notations

  1. 1.

Define $H_{\sigma}$ on $\overline{\mathscr{B}}$ to be the mapping

$H_{\sigma}(\rho,z)=\frac{X_{K}e^{2\lambda_{K}}}{\rho^{2}}\frac{1+\overset{\circ}{X}}{1+\overset{\circ}{\sigma}}e^{2\overset{\circ}{\lambda}}F_{3}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\sigma},\delta)(\rho,z).$

  2. 2.

Let $\overline{\delta}_{0}>0$. Define the nonlinear operator $N_{\sigma}$ on
$B_{\overline{\delta}_{0}}(\mathcal{L}_{\sigma})\times
B_{\overline{\delta}_{0}}(\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}\times[0,\infty[)$
by

$N_{\sigma}\left(\overset{\circ}{\sigma},(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda};\delta)\right)(\rho,z):=H_{\sigma}(\rho,z).$

In order to prove the above proposition, we will need the following lemma

######  Lemma 65.

  1. 1.

There exists $\overline{\delta}_{0}>0$ such that $\displaystyle
N_{\sigma}\left(B_{\overline{\delta}_{0}}(\mathcal{L}_{\sigma})\times
B_{\overline{\delta}_{0}}(\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}\times[0,\infty[)\right)\subset\mathcal{N}_{\sigma}$.

  2. 2.

There exist $0<\delta_{0}\leq\overline{\delta}_{0}$ and $C(\alpha_{0})>0$ such
that
$\forall\left(\overset{\circ}{\sigma},(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda};\delta)\right)\in
B_{\overline{\delta}_{0}}(\mathcal{L}_{\sigma})\times
B_{\overline{\delta}_{0}}(\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}\times[0,\infty[)$

$||H_{\sigma}||_{\mathcal{N}_{\sigma}}\leq
C(\alpha_{0})\left(||\overset{\circ}{\sigma}||^{2}_{\mathcal{L}_{\sigma}}+||(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda})||^{2}_{\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}}+\delta\right).$

###### Proof.

  1. 1.

Let $\overline{\delta}_{0}>0$ and let
$(\overset{\circ}{\sigma},\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda};\delta)\in
B_{\overline{\delta}_{0}}(\mathcal{L}_{\sigma})\times
B_{\overline{\delta}_{0}}(\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}\times[0,\infty[)$.
By Proposition 24, we have
$F_{3}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\sigma};\delta)\in\hat{C}^{1,\alpha}(\overline{\mathscr{B}})$.
Moreover, it is compactly supported in $\mathscr{B}$ and
$K:=\operatorname{supp}F_{3}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\sigma};\delta)\not\subset(\overline{\mathscr{B}_{N}}\cap\overline{\mathscr{B}_{S}})$.
Therefore, $1-\xi_{N}-\xi_{S}=1$ on $K$ and we have Hence,

$\displaystyle||r^{4}H_{\sigma}||_{C^{1,\alpha}(\overline{\mathscr{B}})}$
$\displaystyle=||r^{4}\frac{X_{K}e^{2\lambda_{K}}}{\rho^{2}}\frac{1+\overset{\circ}{X}}{1+\overset{\circ}{\sigma}}e^{2\overset{\circ}{\lambda}}F_{3}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\sigma},\delta)||_{C^{1,\alpha}(\overline{\mathscr{B}})}$
$\displaystyle=\left|\left|r^{4}\frac{X_{K}e^{2\lambda_{K}}}{\rho^{2}}F_{3}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\sigma};\delta)\right|\right|_{C^{1,\alpha}(K)}$
$\displaystyle\leq
C\left|\left|r^{4}\frac{X_{K}e^{2\lambda_{K}}}{\rho^{2}}\right|\right|_{C^{1,\alpha}(K)}\,\left|\left|\frac{1+\overset{\circ}{X}}{1+\overset{\circ}{\sigma}}e^{2\overset{\circ}{\lambda}}\right|\right|_{C^{1,\alpha}(K)}\,||F_{3}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\sigma};\delta)||_{C^{1,\alpha}(K)}$
$\displaystyle\leq
C||(1-\xi_{N}-\xi_{S})F_{3}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\sigma};\delta)||_{C^{1,\alpha}(\mathbb{R}^{3})}$
$\displaystyle\leq
C||F_{3}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\sigma};\delta)||_{\hat{C}^{1,\alpha}(\overline{\mathscr{B}})}.$

  2. 2.

By Proposition 24, we have $F_{3}$ is continuously Fréchet differentiable on
$B_{\overline{\delta}_{0}}$. Hence, there exists
$\delta_{0}\leq\overline{\delta}_{0}$ such that
$\forall(\overset{\circ}{\sigma},\overset{\circ}{X},\overset{\circ}{\Theta};\delta)\in
B_{\delta_{0}}$

$\displaystyle
F_{3}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\sigma};\delta)$
$\displaystyle=D_{h}F_{3}(0,0,0;0;0)[(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\sigma})]+D_{\delta}F_{3}(0,0,0;0)\delta+O(||(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\sigma})||^{2}+\delta^{2})$
$\displaystyle=D_{\delta}F_{3}(0,0,0;0)\delta+O(||(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda})||^{2}+\delta^{2}).$

Therefore,

$\displaystyle||r^{4}H_{\sigma}||_{C^{1,\alpha}(\overline{\mathscr{B}})}$
$\displaystyle\leq
C||F_{3}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\sigma};\delta)||_{\hat{C}^{1,\alpha}(\overline{\mathscr{B}})}$
$\displaystyle\leq
C(||D_{\delta}F_{3}(0,0,0;0)||\delta+||(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda})||^{2}+\delta^{2})$
$\displaystyle\leq
C\left(||\overset{\circ}{\sigma}||^{2}_{\mathcal{L}_{\sigma}}+||\overset{\circ}{X}||^{2}_{\mathcal{L}_{X}}+||\overset{\circ}{\Theta}||^{2}_{\mathcal{L}_{\Theta}}+\delta\right)$
$\displaystyle\leq
C(\alpha_{0})\left(||\overset{\circ}{\sigma}||^{2}_{\mathcal{L}_{\sigma}}+||(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda})||^{2}_{\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}}+\delta\right).$

∎

Now, we prove Proposition 26.

###### Proof.

We apply Theorem 6 with

$L=\Delta_{\mathbb{R}^{4}}\;,\;N=N_{\sigma},$
$\mathcal{L}=\mathcal{L}_{\sigma}\;,\;\mathcal{Q}=\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}\;,\;\mathcal{P}=[0,\delta_{0}[.$

By Lemma 86, the assumptions of Theorem 6 are satisfied. Therefore, we obtain
the desired result. ∎

###  9.5 Solving for
$\displaystyle\left(B_{\chi}^{(N)},B_{\chi^{\prime}}^{(S)},B_{z}^{(A)}\right)$

In this section, we solve the equations

$\displaystyle B_{\chi}^{(N)}(0,\chi)$ $\displaystyle=0,$ (9.53)
$\displaystyle B_{\chi}^{(N)}(s,0)$ $\displaystyle=0,$
$\displaystyle\partial_{s}B^{(N)}_{\chi}$
$\displaystyle=2\xi_{N}\sigma^{-1}e^{2\lambda}F_{2}(W,X,\sigma)(\rho,z)(s^{2}+\chi^{2}),$
$\displaystyle B^{(N)}_{\chi}(s,\chi)$ $\displaystyle=0,$ $\displaystyle
B_{\chi^{\prime}}^{(S)}(0,\chi^{\prime})$ $\displaystyle=0,$ $\displaystyle
B_{\chi^{\prime}}^{(S)}(s^{\prime},0)$ $\displaystyle=0,$
$\displaystyle\partial_{s^{\prime}}B^{(S)}_{\chi^{\prime}}$
$\displaystyle=2\xi_{S}\sigma^{-1}e^{2\lambda}F_{2}(W,X,\sigma)(s^{\prime},\chi^{\prime})((s^{\prime})^{2}+(\chi^{\prime})^{2}),$
$\displaystyle B^{(S)}_{s^{\prime}}(s^{\prime},\chi^{\prime})$
$\displaystyle=0,$ $\displaystyle B_{z}^{(A)}(0,z)$ $\displaystyle=0,$
$\displaystyle\partial_{\rho}B_{z}^{(A)}(\rho,z)$
$\displaystyle=2(1-\xi_{N}^{2}-\xi_{S}^{2})\sigma^{-1}e^{2\lambda}F_{2}(W,X,\sigma)(\rho,z)-(\partial_{\rho}\xi_{N})B_{z}^{(N)}+(\partial_{z}\xi_{N})B_{\rho}^{(N)}-(\partial_{\rho}\xi_{S})B_{z}^{(S)}+(\partial_{z}\xi_{S})B_{\rho}^{(S)},$
$\displaystyle B_{\rho}^{(A)}$ $\displaystyle=0,$

for
$\displaystyle\left(B_{\chi}^{(N)},B_{\chi^{\prime}}^{(S)},B_{z}^{(A)}\right)$
in terms of the renormalised unknowns
$\left((\overset{\circ}{X},\overset{\circ}{Y}),\overset{\circ}{\Theta},\overset{\circ}{\lambda}\right)$
and $\delta$.
By Proposition 22, we have

$\displaystyle\xi_{N}\sigma^{-1}e^{2\lambda}F_{2}(W,X,\sigma)(\rho,z)$
$\displaystyle=0,$
$\displaystyle\xi_{S}\sigma^{-1}e^{2\lambda}F_{2}(W,X,\sigma)(\rho,z)$
$\displaystyle=0.$

Therefore,

$B_{\chi}^{(N)}(s,\chi)=0\;,\;B_{\chi}^{(N)}(0,\chi)=0\;,\;B_{\chi^{\prime}}^{(S)}(s^{\prime},\chi^{\prime})=0\;,\;B_{\chi}^{(S)}(0,\chi^{\prime})=0,$

and

$B^{(N)}_{\rho}=0\;,\;B^{(N)}_{z}=0\;,\;B^{(S)}_{\rho}=0\;,\;B^{(S)}_{z}=0.$

####  9.5.1 Linear problem

In order to solve for $B_{z}^{(A)}$, we start with solving the linear problem:

$\partial_{\rho}B_{z}^{(A)}(\rho,z)=(1-\xi_{N}^{2}-\xi_{S}^{2})H^{(A)}_{B}\quad\quad\text{for}\quad(H^{(A)}_{B},0,0)\in\mathcal{N}_{B}$

with boundary condition

$B_{z}^{(A)}(0,z)=0.$

We state the following result

######  Proposition 27.

Let $\displaystyle\left((0,0,H^{(A)}_{B}\right)\in\mathcal{N}_{B}.$ Then,
there exists a unique solution
$\left(0,0,B_{z}^{(A)}\right)\in\mathcal{L}_{B}$ which solves the equation

$\partial_{\rho}B_{z}^{(A)}(\rho,z)=(1-\xi_{N}^{2}-\xi_{S}^{2})H^{(A)}_{B}\quad\quad\text{and}\quad
B_{z}^{(A)}(0,z)=0.$

It is given by

$B_{z}^{(A)}(\rho,z)=\int_{0}^{\rho}(1-\xi_{N}^{2}-\xi_{S}^{2})(\tilde{\rho},z)H^{(A)}_{B}(\tilde{\rho},z)\,d\tilde{\rho}.$

Moreover, there exists $C(\alpha_{0})>0$

$\left|\left|\left(0,0,B_{z}^{(A)}\right)\right|\right|_{\mathcal{L}_{B}}\leq
C(\delta_{0})\left|\left|\left(0,0,H^{(A)}_{B}\right)\right|\right|_{\mathcal{N}_{B}}.$

###### Proof.

  1. 1.

First of all, since

$\left|\left|\frac{(1+\rho^{15})(1+r^{10})}{\rho^{15}}H_{B}^{(A)}\right|\right|_{\hat{C}^{1,\alpha_{0}}(\overline{\mathscr{B}_{A}}\cup\overline{\mathscr{B}_{H}})}<\infty,$

$B_{z}^{(A)}$ is well-defined on
$\overline{\mathscr{B}_{A}}\cup\overline{\mathscr{B}_{H}}$.

  2. 2.

As for the estimates, we will show that

$\left((1-\xi_{N}^{2}-\xi_{S}^{2})\frac{(1+\rho^{10})(1+r^{10})}{\rho^{10}}B_{z}^{(A)}\right)_{\mathbb{R}^{3}}\in
C^{1,\alpha_{0}}(\mathbb{R}^{3}).$

First of all, we recall the following change of variables

$\displaystyle x$ $\displaystyle=\rho\cos\vartheta$ $\displaystyle y$
$\displaystyle=\rho\sin\vartheta.$

We have

$\partial_{x}=\frac{x}{\sqrt{x^{2}+y^{2}}}\partial_{\rho}\;,\;\partial_{y}=\frac{y}{\sqrt{x^{2}+y^{2}}}\partial_{\rho}.$

     * •

if $(\rho,z)\in\tilde{\mathscr{A}}$ where $\tilde{\mathscr{A}}$ is a
neighbourhood of the axis. We assume that $\rho<1$. Then

$\frac{\rho^{15}}{1+\rho^{15}}\leq\frac{\rho^{10}}{1+\rho^{10}}$

and we have

$\displaystyle\left|B_{z}^{(A)}\right|$ $\displaystyle\leq
C\int_{0}^{\rho}\frac{\tilde{\rho}^{10}}{1+\tilde{\rho}^{10}}\frac{1}{1+(1+\tilde{\rho}^{2}+z^{2})^{5}}\,d\tilde{\rho}$
$\displaystyle\leq
C\frac{\rho^{10}}{1+\rho^{10}}\int_{0}^{\rho}\frac{1}{1+(1+\tilde{\rho}^{2}+z^{2})^{5}}\,d\tilde{\rho}$
$\displaystyle\leq
C\frac{\rho^{10}}{(1+\rho^{10})(1+r^{10})}\int_{0}^{\rho}\frac{1+(1+\rho^{2}+z^{2})^{5}}{1+(1+\tilde{\rho}^{2}+z^{2})^{5}}\,d\tilde{\rho}.$

The integrand is bounded uniformly in $z$. Moreover, since $\rho\in[0,1]$, we
can bound the term
$\displaystyle\int_{0}^{\rho}\frac{1+(1+\rho^{2}+z^{2})^{5}}{1+(1+\tilde{\rho}^{2}+z^{2})^{5}}\,d\tilde{\rho}$
uniformly in $(\rho,z)$.

     * •

Away from the axis, say $(\rho,z)\in[\rho_{0},\infty[\times\mathbb{R}$ for
some $\rho_{0}>1$. We have

$1\leq\frac{\rho^{10}}{(1+\rho^{10})}$

and

$\displaystyle\int_{0}^{\rho}\frac{\tilde{\rho}^{15}}{1+\tilde{\rho}^{15}}\frac{1}{1+(1+\tilde{\rho}^{2}+z^{2})^{5}}\,d\tilde{\rho}$
$\displaystyle\leq
C\int_{0}^{\rho}\frac{\tilde{\rho}^{8}}{1+\tilde{\rho}^{15}}\frac{4\tilde{\rho}(1+\tilde{\rho}^{2}+z^{2})^{3}}{(1+\tilde{\rho}^{2}+z^{2})^{5}}\,d\tilde{\rho}$
$\displaystyle\leq Cr(\rho,z)^{-10}\leq C\frac{1}{1+r^{10}}$

Hence,

$\displaystyle\left|B_{z}^{(A)}\right|$ $\displaystyle\leq
C\int_{0}^{\rho}\frac{\tilde{\rho}^{15}}{1+\tilde{\rho}^{15}}\frac{1}{1+(1+\tilde{\rho}^{2}+z^{2})^{5}}\,d\tilde{\rho}$
$\displaystyle\leq C\frac{\rho^{10}}{(1+\rho^{10})(1+r^{10})}.$

     * •

Away from the axis, the $L^{\infty}$ estimate and the Hölder estimate of the
derivative are straightforward.

     * •

Near the axis, we use the above change of coordinates in order to estimate the
derivatives and the Hölder part.

∎

####  9.5.2 Non-linear estimates

We apply Theorem 6 in order to obtain

######  Proposition 28.

Let $\alpha_{0}\in(0,1)$ and let $\overline{\delta}_{0}>0$. Then, there exists
$0<\delta_{0}\leq\overline{\delta}_{0}$ such that
$\displaystyle\forall\left(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda},\delta\right)\in
B_{\delta_{0}}\left(\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}\times[0,\infty[\right)$
there exists a unique one parameter family
$\left(\overset{\circ}{\sigma},\left(0,0,B_{z}^{(A)}\right)\right)$ depending
on
$\left(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda},\delta\right)\in
B_{\delta_{0}}\left(\mathcal{L}_{\sigma}\times\mathcal{L}_{B}\right)$ which
solves (5.49) and (9.53) and which satisfies

$\left|\left|\left(\overset{\circ}{\sigma},\left(0,0,B_{z}^{(A)}\right)\right)\left(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda},\delta\right)\right|\right|_{\mathcal{L}_{\sigma}\times\mathcal{L}_{B}}\leq
C(\alpha_{0})\left(||\overset{\circ}{X}||^{2}_{\mathcal{L}_{X}}+||\overset{\circ}{\Theta}||^{2}_{\mathcal{L}_{\Theta}}+||\overset{\circ}{\lambda}||^{2}_{\mathcal{L}_{\lambda}}+\delta\right)$

and
$\displaystyle\forall\left(\overset{\circ}{X}_{i},\overset{\circ}{\Theta}_{i},\overset{\circ}{\lambda}_{i},\delta_{i}\right)\in
B_{\delta_{0}}\left(\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}\times[0,\infty[\right)$,

$\displaystyle\left|\left|\left(\overset{\circ}{\sigma},\left(0,0,B_{z}^{(A)}\right)\right)\left(\overset{\circ}{X}_{1},\overset{\circ}{\Theta}_{1},\overset{\circ}{\lambda}_{1},\delta_{1}\right)-\left(\overset{\circ}{\sigma},\left(0,0,B_{z}^{(A)}\right)\right)\left(\overset{\circ}{X}_{2},\overset{\circ}{\Theta}_{2},\overset{\circ}{\lambda}_{2},\delta_{2}\right)\right|\right|_{\mathcal{L}_{\sigma}\times\mathcal{L}_{B}}$
$\displaystyle\leq
C(\alpha_{0})\left(\left(\left|\left|\left(\overset{\circ}{X}_{1},\overset{\circ}{\Theta}_{1},\overset{\circ}{\lambda}_{1}\right)\right|\right|_{\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}}+\left|\left|\left(\overset{\circ}{X}_{2},\overset{\circ}{\Theta}_{2},\overset{\circ}{\lambda}_{2}\right)\right|\right|_{\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}}\right)\left|\left|\left(\overset{\circ}{X}_{1},\overset{\circ}{\Theta}_{1},\overset{\circ}{\lambda}_{1}\right)-\left(\overset{\circ}{X}_{2},\overset{\circ}{\Theta}_{2},\overset{\circ}{\lambda}_{2}\right)\right|\right|_{\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}}\right.$
$\displaystyle\left.+\left|\delta_{1}-\delta_{2}\right|\right).$

Before we prove the above proposition, we introduce the following notations

  1. 1.

Define $H_{B}^{(A)}$ on $\overline{\mathscr{B}}$ to be the mapping

$H_{B}^{(A)}(\rho,z)=2\sigma^{-1}e^{2\lambda}F_{2}(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma},\delta)(\rho,z).$

  2. 2.

Let $\overline{\delta}_{0}>0$ be obtained by Proposition (26). Define the
nonlinear operator $N_{B}$ on
$B_{\overline{\delta}_{0}}(\mathcal{L}_{B})\times
B_{\overline{\delta}_{0}}\left(\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}\times[0,\infty[\right)$
by

$N_{B}\left(\left(0,0,B_{z}^{(A)}\right),(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda};\delta)\right)(\rho,z):=\frac{2}{\rho}\left(\overset{\circ}{\sigma}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda},\delta)\right)^{-1}e^{2\lambda}F_{2}(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda},\delta),\delta)(\rho,z)$

where $\overset{\circ}{\sigma}$ is the mapping obtained by Proposition 26.

######  Lemma 66.

  1. 1.

$\displaystyle N_{B}\left(B_{\overline{\delta}_{0}}(\mathcal{L}_{B})\times
B_{\overline{\delta}_{0}}(\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}\times[0,\infty[)\right)\subset\mathcal{N}_{B}$.

  2. 2.

There exist $0<\delta_{0}\leq\overline{\delta}_{0}$ and $C(\alpha_{0})>0$ such
that
$\forall\left(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda};\delta\right)\in
B_{\overline{\delta}_{0}}(\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}\times[0,\infty[)$

$||(0,0,H^{(A)}_{B})||_{\mathcal{N}_{B}}\leq
C(\alpha_{0})\left(||(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda})||^{2}_{\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}}+\delta\right).$
(9.54)

###### Proof.

We use the same arguments used to prove Lemma 86: the compact support of
$F_{2}(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma};\delta)$
and the differentiability of $F_{2}$ with respect to each variable.
In order to obtain (9.72), we write

$\displaystyle
F_{2}(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda};\delta);\delta)$
$\displaystyle=D_{h}F_{3}(0,0,0;0)[(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\sigma}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda};\delta))]+D_{\delta}F_{3}(0,0,0;0)\delta$
$\displaystyle+O(||(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\sigma}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda};\delta))||^{2}+\delta^{2})$
$\displaystyle=D_{\delta}F_{3}(0,0,0;0)\delta+O(||(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\sigma}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda};\delta))||^{2}+\delta^{2}).$

By Proposition 26, we have

$\left|\left|\overset{\circ}{\sigma}\left(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda},\delta\right)\right|\right|_{\mathcal{L}_{\sigma}}\leq
C(\alpha_{0})\left(||\overset{\circ}{X}||^{2}_{\mathcal{L}_{X}}+||\overset{\circ}{\Theta}||^{2}_{\mathcal{L}_{\Theta}}+||\overset{\circ}{\lambda}||^{2}_{\mathcal{L}_{\lambda}}+\delta\right)$

Hence,

$||F_{2}(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda};\delta);\delta)||_{\hat{C}^{1,\alpha_{0}}}\leq
C(\alpha_{0})\left(||\overset{\circ}{X}||^{2}_{\mathcal{L}_{X}}+||\overset{\circ}{\Theta}||^{2}_{\mathcal{L}_{\Theta}}+||\overset{\circ}{\lambda}||^{2}_{\mathcal{L}_{\lambda}}+\delta\right).$

The estimate (9.72) follows from the latter and the compact support of
$F_{2}(\overset{\circ}{\Theta},\overset{\circ}{X},\overset{\circ}{\sigma}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda};\delta);\delta)$.
∎

Now, we prove Proposition 28.

###### Proof.

  1. 1.

First of all, by Proposition 26, there exists a solution map
$\overset{\circ}{\sigma}(\overset{\circ}{X},\overset{\circ}{\Theta},\overset{\circ}{\lambda};\delta)$
defined on $B_{\delta_{0}}$ which solves (5.49).

  2. 2.

We apply Theorem 6 with $\displaystyle
L=\partial_{\rho}\;,\;N=N_{B}\;,\;\mathcal{L}=\mathcal{L}_{B}\;,\;\mathcal{Q}=\mathcal{L}_{X}\times\mathcal{L}_{\Theta}\times\mathcal{L}_{\lambda}$
and $\displaystyle\mathcal{P}=[0,\delta_{0}[$. By the previous lemma, all the
assumptions are satisfied and we obtain the desired result.

∎

###  9.6 Solving for $(X,Y)$

####  9.6.1 Linear problem

The aim of this section is to solve the linear problem for
$\left(\overset{\circ}{X},\overset{\circ}{Y}\right)$:

$\displaystyle\Delta_{\mathbb{R}^{3}}\overset{\circ}{X}+\frac{2\partial
Y_{K}\cdot\partial\overset{\circ}{Y}}{X_{K}}-\frac{2|\partial
Y_{K}|^{2}}{X_{K}^{2}}\overset{\circ}{X}+2\frac{\partial X_{K}\cdot\partial
Y_{K}}{X_{K}^{2}}\overset{\circ}{Y}=H_{X},$ (9.55)
$\displaystyle\Delta_{\mathbb{R}^{3}}\overset{\circ}{Y}-\frac{2\partial
Y_{K}\cdot\partial\overset{\circ}{X}}{X_{K}}-\frac{(|\partial
X_{K}|^{2}+|\partial Y_{K}|^{2})}{X_{K}^{2}}\overset{\circ}{Y}=H_{Y}.$

where $\left(X_{K},Y_{K}\right)$ are the renormalised quantities for the Kerr
metric and where
$\displaystyle\left(H_{X},H_{Y}\right)\in\mathcal{N}_{X}\times\mathcal{N}_{Y}$.
More precisely, the remaining of this section is devoted to the proof of the
following result

######  Proposition 29.

Let
$\displaystyle\left(H_{X},H_{Y}\right)\in\mathcal{N}_{X}\times\mathcal{N}_{Y}$.
Then there exists a unique solution
$\left(\overset{\circ}{X},\overset{\circ}{Y}\right)\in\mathcal{L}_{X}\times\mathcal{L}_{Y}$
of (9.55). Moreover, there exists $C>0$ independent of
$\left(H_{X},H_{Y}\right)$ such that

$\left|\left|\left(\overset{\circ}{X},\overset{\circ}{Y}\right)\right|\right|_{\mathcal{L}_{X}\times\mathcal{L}_{Y}}\leq
C\left|\left|\left(H_{X},H_{Y}\right)\right|\right|_{\mathcal{N}_{X}\times\mathcal{N}_{Y}}$

##### 9.6.1.1 Existence of weak solutions to the linear non homogeneous
problem

Let $\left(H_{X},H_{Y}\right)\in C_{0}^{\infty}(\mathbb{R}^{3})$ and consider
the system (9.55). Set

$\mathcal{L}(\overset{\circ}{X},\overset{\circ}{Y}):=\int_{\mathbb{R}^{3}}\;\left(|\partial\overset{\circ}{X}+X^{-1}_{K}(\partial
Y_{K})\overset{\circ}{Y}|^{2}+|\partial\overset{\circ}{Y}-X^{-1}_{K}(\partial
Y_{K})\overset{\circ}{X}|^{2}+X^{-2}_{K}|\overset{\circ}{X}\partial
Y_{K}-\overset{\circ}{Y}\partial
X_{K}|^{2}+2\overset{\circ}{X}H_{X}+2\overset{\circ}{Y}H_{Y}\right).$ (9.56)

We state the following lemmas from [17]. We recall the proofs in order to be
self-contained.

######  Lemma 67.

  * •

$\mathcal{L}$ is well-defined on
$\displaystyle\dot{H}_{axi}^{1}(\mathbb{R}^{3})\times\left(\dot{H}_{axi}^{1}(\mathbb{R}^{3})\cap
L^{2}_{|\partial h|}(\mathbb{R}^{3})\right)$.

  * •

(9.55) are the Euler-Lagrange equations associated with the Lagrangian
$\mathcal{L}$.

Before we prove the above result, we will need the following inequalities

######  Lemma 68.

Let $f(\rho,z)\in C^{\infty}_{0}(\mathbb{R}^{3})$. Then

$\int_{\mathbb{R}^{3}}\;|\partial X_{K}|^{2}f^{2}\leq
C\int_{\mathbb{R}^{3}}\;X^{2}_{K}|\partial f|^{2}$

###### Proof.

By Lemma 55, there exist $c,C>0$ such that
$\forall(\rho,z)\in\overline{\mathscr{B}}$, we have

$ce^{h}\left|\partial h\right|\leq\left|\partial X_{K}\right|\leq
Ce^{h}\left|\partial h\right|.$

Therefore, it suffices to show that

$\int_{\mathbb{R}^{3}}\;e^{2h}|\partial h|^{2}f^{2}\leq
C\int_{\mathbb{R}^{3}}\;e^{2h}|\partial f|^{2}$

Recall that $\Delta_{\mathbb{R}^{3}}h=0$ and note that

$\text{div}_{\mathbb{R}^{3}}(e^{2h}f^{2}\partial
h)=e^{2h}f^{2}\Delta_{\mathbb{R}^{3}}h+\nabla h\cdot\nabla(e^{2h}f^{2})=\nabla
h\cdot\nabla(e^{2h}f^{2}).$

Now, let $0<\epsilon<1$, $Z>0$ large , $R>0$ large so that
$f(\rho,z)=0\;\forall\rho\geq 0$, and consider the domain
$\mathcal{U}(R,\epsilon,h)\subset\mathbb{R}^{3}$ defined by

${U}(R,\epsilon,Z):=\left\\{(r,\phi,z):\;\epsilon<r<R\;,0<\phi<2\pi\;,\;|z|<Z\right\\}.$

We apply the divergence theorem to obtain

$\int_{{U}(R,\epsilon,Z)}\text{div}_{\mathbb{R}^{3}}(e^{2h}f^{2}\partial
h)\rho d\rho d\phi
dz=-2\pi\int^{Z}_{-Z}\;(e^{2h}f^{2}\partial_{\rho}h)(\epsilon,z)\,dz.$

Therefore,

$\int_{\mathbb{R}^{3}}\text{div}_{\mathbb{R}^{3}}(e^{2h}f^{2}\partial h)\rho
d\rho d\phi dz=-2\pi\lim_{\epsilon\to
0}\int^{\infty}_{-\infty}\;(e^{2h}f^{2}\partial_{\rho}h)(\epsilon,z)\,dz.$

We have

$\displaystyle\partial_{\rho}h(e^{2h})(\rho,z)$
$\displaystyle=\left(\sqrt{\rho^{2}+(z-\gamma)^{2}}-(z-\gamma)\right)\frac{\rho}{\sqrt{\rho^{2}+(z-\gamma)^{2}}}\left(\sqrt{\rho^{2}+(z+\gamma)^{2}}+(z+\gamma)\right)^{2}$
$\displaystyle+\left(\sqrt{\rho^{2}+(z-\gamma)^{2}}-(z-\gamma)\right)^{2}\frac{\rho}{\sqrt{\rho^{2}+(z+\gamma)^{2}}}\left(\sqrt{\rho^{2}+(z+\gamma)^{2}}+(z+\gamma)\right).$

Therefore, there exists $\overline{f}(z,\gamma)>0$ which is compactly
supported in the z-variable such that $\rho\in[0,1]$

$\left|(e^{2h}f^{2}\partial h)(\rho,z)\right|\leq\overline{f}(z,\gamma).$

Moreover,

$\lim_{\epsilon\to 0}\,\partial_{\rho}h(e^{2h})(\epsilon,z)=0.$

Therefore, by the dominated convergence theorem, we obtain

$\int_{\mathbb{R}^{3}}\nabla h\cdot\nabla(e^{2h}f^{2})\rho d\rho d\phi
dz=\int_{\mathbb{R}^{3}}\text{div}_{\mathbb{R}^{3}}(e^{2h}f^{2}\partial h)\rho
d\rho d\phi dz=0.$

Now, we compute:

$\nabla h\cdot\nabla(e^{2h}f^{2})=2e^{2h}\left(f^{2}|\nabla h|^{2}+f\nabla
h\cdot\nabla f\right).$

Therefore

$\displaystyle\int_{\mathbb{R}^{3}}\,e^{2h}|\partial h|^{2}f^{2}$
$\displaystyle\leq\int_{\mathbb{R}^{3}}\,e^{2h}|f|\,|\partial f|\,|\partial
h|$ $\displaystyle\leq q\int_{\mathbb{R}^{3}}\;e^{2h}|\partial
f|^{2}+\frac{1}{4q}\int_{\mathbb{R}^{3}}\;e^{2h}f^{2}|\partial
h|^{2}\quad\forall q>0.$

Finally, we choose $q$ so that we obtain the desired ineaquality. ∎

######  Lemma 69.

Let $f\in C^{\infty}_{0}(\mathbb{R}^{3})$. Then,

$\int_{\mathbb{R}^{3}}\;\left<r\right>^{-2}f^{2}\leq
C\int_{\mathbb{R}^{3}}\;|\partial f|^{2}.$

where, $\left<r\right>$ is given by

$\left<r\right>:=(1+x^{2}+y^{2}+z^{2})^{\frac{1}{2}}.$

###### Proof.

We apply Theorem 8 with $p=2$ and $\delta$ satisfying:

$\displaystyle-2\delta-3=-2\quad\Longleftrightarrow\quad\delta=-\frac{1}{2}.$

Thus,

$\displaystyle||f_{r}||^{2}_{2,-\frac{3}{2}}$
$\displaystyle=\int_{\mathbb{R}^{3}}\;|\left<r\right>^{-1(x,y,z)^{t}\cdot\partial
f}|^{2}\left<r\right>^{3-3},$
$\displaystyle\leq\int_{\mathbb{R}^{3}}\;|\partial f|^{2}.$

Therefore,

$\int_{\mathbb{R}^{3}}\;\left<r\right>^{-2}f^{2}\leq
C\int_{\mathbb{R}^{3}}\;|\partial f|^{2}.$

∎

Now, we state the following lemma

######  Lemma 70.

There exists $C\geq 0$ such that if

$\overset{\circ}{X}\in
C^{\infty}_{0}(\mathbb{R}^{3}),\quad\overset{\circ}{Y}\in
C^{\infty}_{0}(\mathbb{R}^{3}),\quad\int_{\mathbb{R}^{3}}\;\frac{|\partial
X_{K}|^{2}}{X^{2}_{K}}\overset{\circ}{Y}^{2}<\infty,$

then

$\int_{\mathbb{R}^{3}}\;\left(|\partial\overset{\circ}{X}|^{2}+|\partial\overset{\circ}{Y}|^{2}+|\partial
h|^{2}\overset{\circ}{Y}^{2}\right)\leq
C\left(\mathcal{L}(\overset{\circ}{X},\overset{\circ}{Y})+2\int_{\mathbb{R}^{3}}\;\left(|\overset{\circ}{X}H_{X}|+|\overset{\circ}{Y}H_{Y}|\right)\right).$
(9.57)

###### Proof.

  1. 1.

Set

$\mathcal{L}_{0}(\overset{\circ}{X},\overset{\circ}{Y}):=\mathcal{L}(\overset{\circ}{X},\overset{\circ}{Y})-2\int_{\mathbb{R}^{3}}\;\left(|\overset{\circ}{X}H_{X}|+|\overset{\circ}{Y}H_{Y}|\right).$
(9.58)

We have

$\displaystyle\left|\partial\overset{\circ}{Y}-X^{-1}_{K}\partial
X_{k}\overset{\circ}{Y}\right|^{2}$
$\displaystyle=\left|\partial\overset{\circ}{Y}-X^{-1}_{K}\partial
Y_{k}\overset{\circ}{X}+X^{-1}_{K}\partial
Y_{k}\overset{\circ}{X}-X^{-1}_{K}\partial
X_{k}\overset{\circ}{Y}\right|^{2},$
$\displaystyle\leq\left|\partial\overset{\circ}{Y}-X^{-1}_{K}\partial
Y_{k}\overset{\circ}{X}\right|^{2}+X^{-2}_{K}\left|\partial
Y_{k}\overset{\circ}{X}-\partial X_{k}\overset{\circ}{Y}\right|^{2}.$

Hence,

$\displaystyle\int_{\mathbb{R}^{3}}\;\left|\partial\overset{\circ}{Y}-X^{-1}_{K}\partial
X_{k}\overset{\circ}{Y}\right|^{2}$ $\displaystyle\leq
C\mathcal{L}_{0}(\overset{\circ}{X},\overset{\circ}{Y}).$

Now, we define
$\tilde{Y}:\mathbb{R}^{3}\backslash\left\\{0\right\\}\to\mathbb{R}$ by
$\tilde{Y}:=X^{-1}_{K}\overset{\circ}{Y}$ and compute

$\displaystyle\partial\overset{\circ}{Y}-X^{-1}_{K}\partial
X_{k}\overset{\circ}{Y}$ $\displaystyle=\partial
X_{K}\tilde{Y}+X_{K}\partial\tilde{Y}-X^{-1}_{K}\partial X_{k}\tilde{Y},$
$\displaystyle=X_{K}\partial\tilde{Y}.$

Thus,

$\displaystyle\int_{\mathbb{R}^{3}}\;\left|X_{K}\partial\tilde{Y}\right|^{2}$
$\displaystyle\leq C\mathcal{L}_{0}(\overset{\circ}{X},\overset{\circ}{Y}).$

By Lemma 68, we have

$\int_{\mathbb{R}^{3}}\;|\partial X_{K}|^{2}\tilde{Y}^{2}\leq
C\int_{\mathbb{R}^{3}}\;\left|X_{K}\partial\tilde{Y}\right|^{2}\leq
C\mathcal{L}_{0}(\overset{\circ}{X},\overset{\circ}{Y}),$
$\int_{\mathbb{R}^{3}}\;|\partial
h|^{2}\overset{\circ}{Y}^{2}=\int_{\mathbb{R}^{3}}\;X^{2}_{K}|\partial
h|^{2}\tilde{Y}^{2}\leq C\int_{\mathbb{R}^{3}}\;|\partial
X_{K}|^{2}\tilde{Y}^{2}\leq
C\mathcal{L}_{0}(\overset{\circ}{X},\overset{\circ}{Y}),$

and

$\displaystyle\int_{\mathbb{R}^{3}}\;|\partial
Y_{K}|^{2}X^{-2}_{K}\overset{\circ}{X}^{2}$
$\displaystyle=\int_{\mathbb{R}^{3}}\;|\overset{\circ}{X}\partial
Y_{K}|^{2}X^{-2}_{K},$
$\displaystyle=\int_{\mathbb{R}^{3}}\;X^{-2}_{K}|\overset{\circ}{X}\partial
Y_{K}-\overset{\circ}{Y}\partial X_{K}+\overset{\circ}{Y}\partial X_{K}|^{2},$
$\displaystyle\leq\int_{\mathbb{R}^{3}}\;X^{-2}_{K}|\overset{\circ}{X}\partial
Y_{K}-\overset{\circ}{Y}\partial
X_{K}|^{2}++\int_{\mathbb{R}^{3}}\;\tilde{Y}^{2}|\partial X_{K}|^{2},$
$\displaystyle\leq C\mathcal{L}_{0}(\overset{\circ}{X},\overset{\circ}{Y}).$

Similarly, we obtain

$\displaystyle\int_{\mathbb{R}^{3}}\;|\partial\overset{\circ}{X}|^{2}\leq
C\int_{\mathbb{R}^{3}}\;|\partial\overset{\circ}{X}+X^{-1}_{K}\partial
Y_{K}\overset{\circ}{Y}|^{2}+C\int_{\mathbb{R}^{3}}\;X^{2}_{K}\overset{\circ}{Y}^{2}|\partial
Y_{K}|^{2}.$

Now, we show that

$\int_{\mathbb{R}^{3}}\;X^{2}_{K}\overset{\circ}{Y}^{2}|\partial
Y_{K}|^{2}\leq
C\int_{\mathbb{R}^{3}}\;\left<r\right>^{-2}\overset{\circ}{Y}^{2}.$

By Lemma 54,

$\rho\frac{|\partial Y_{K}|}{X^{2}_{K}}\leq
C\left<r\right>^{-4}\quad\text{on}\quad\overline{\mathscr{B}_{A}}\cup\overline{\mathscr{B}_{H}}.$

This implies

$\rho\frac{|\partial Y_{K}|^{2}}{X^{2}_{K}}\leq
C\left(\frac{X_{K}}{\rho}\right)^{2}\left<r\right>^{-8}.$

Moreover,

$\forall(\rho,z)\in\mathscr{B}\;:\;\frac{X_{K}}{\rho}=\rho\frac{\Pi}{\Sigma^{2}\Delta}=\rho^{2}\frac{(r(\rho,z)^{2}+a^{2})^{2}-a\rho^{2}}{(r(\rho,z)^{2}+\frac{z^{2}}{(r(\rho,z)-M)^{2}})(r(\rho,z)^{2}-2Mr(\rho,z)+a^{2})},$

when $\langle r\rangle$ goes to $\infty$,

$\left|\frac{X_{K}}{\rho}\right|^{2}\leq C\left<r\right>^{2}$

Thus,

$\frac{|\partial Y_{K}|^{2}}{X^{2}_{K}}\leq\left<r\right>^{-6}\leq
C\left<r\right>^{-2}.$

Therefore, by Lemma 69, we obtain

$\int_{\mathbb{R}^{3}}\;X^{2}_{K}\overset{\circ}{Y}^{2}|\partial
Y_{K}|^{2}\leq
C\int_{\mathbb{R}^{3}}\;\left<r\right>^{-2}\overset{\circ}{Y}^{2}\leq
C\int_{\mathbb{R}^{3}}\;|\partial\overset{\circ}{Y}|^{2}\leq
C\mathcal{L}_{0}(\overset{\circ}{X},\overset{\circ}{Y}).$

This finishes the proof. ∎

Now, we prove Lemma 67

###### Proof.

  1. 1.

Let
$(\overset{\circ}{X},\overset{\circ}{Y})\in\dot{H}_{axi}^{1}(\mathbb{R}^{3})\times\left(\dot{H}_{axi}^{1}(\mathbb{R}^{3})\cap
L^{2}_{|\partial h|}(\mathbb{R}^{3})\right)$. We show that

$X^{-1}_{K}(\partial Y_{K})\overset{\circ}{Y}\;,\;X^{-1}_{K}(\partial
Y_{K})\overset{\circ}{X}\in L^{2}(\mathbb{R}^{3}).$

By Lemma 54, we have

$\frac{|\partial Y_{K}|}{X_{K}}\leq
Cr^{-4}\frac{X_{K}}{\rho}\quad\text{on}\quad\overline{\mathscr{B}_{A}}\cup\overline{\mathscr{B}_{H}}$

and

${|\underline{\partial}Y_{K}|}\leq
Cs^{3}\quad\text{on}\quad\overline{\mathscr{B}_{N}}\quad\text{on}\quad{|\underline{\partial}Y_{K}|}\leq
C(s^{\prime})^{3}\quad\text{on}\quad\overline{\mathscr{B}_{S}}.$

Moreover, by Proposition 4, we have

$X_{K}(\rho,z)=s^{2}X_{N}(s^{2},\chi^{2})\quad\quad
X_{N}(0,\chi),X_{N}(s,0)>0.$

Therefore,

$\displaystyle\int_{\mathbb{R}^{3}}X^{-2}_{K}|\partial
Y_{K}|^{2}\overset{\circ}{Y}^{2}$
$\displaystyle=2\pi\int_{\overline{\mathscr{B}}}\,X^{-2}_{K}|\partial
Y_{K}|^{2}\overset{\circ}{Y}^{2}d\rho dz$
$\displaystyle=2\pi\int_{\overline{\mathscr{B}_{A}}\cup\overline{\mathscr{B}_{H}}}\,(1-\xi_{N}-\xi_{S})X^{-2}_{K}|\partial
Y_{K}|^{2}\overset{\circ}{Y}^{2}\rho d\rho dz$
$\displaystyle+2\pi\int_{\overline{\mathscr{B}_{N}}}\xi_{N}\frac{1}{s^{4}X^{2}_{N}}|\underline{\partial}Y_{K}|^{2}\overset{\circ}{Y}^{2}{s\chi}\,dsd\chi$
$\displaystyle+2\pi\int_{\overline{\mathscr{B}_{S}}}\xi_{S}\frac{1}{(s^{\prime})^{4}X^{2}_{S}}|\underline{\partial}Y_{K}|^{2}\overset{\circ}{Y}^{2}{s^{\prime}\chi^{\prime}}\,ds^{\prime}d\chi^{\prime}.$

By the decay estimates for $Y_{K}$, the above integrals are all finite.
Therefore,

$X^{-1}_{K}(\partial Y_{K})\overset{\circ}{Y}\;,\;X^{-1}_{K}(\partial
Y_{K})\overset{\circ}{X}\in L^{2}(\mathbb{R}^{3})$

and $\mathcal{L}(\overset{\circ}{X},\overset{\circ}{Y})$ is finite.

  2. 2.

Let
$(\overset{\circ}{X},\overset{\circ}{Y})\in\dot{H}^{1}_{axi}(\mathbb{R})^{3}\times\left(\dot{H}^{1}_{axi}(\mathbb{R})^{3}\cap
L^{2}_{\partial h}(\mathbb{R}^{3})\right)$ be minimiser for $\mathcal{L}$, let
$(\phi,\psi)\in C^{\infty}_{0}(\mathbb{R}^{3})$ and consider the real-valued
function

$L(\tau):=\mathcal{L}((\overset{\circ}{X},\overset{\circ}{Y})+\tau(\phi,\psi)).$

$L$ has a minimum at $\tau=0$. Therefore,

$L^{\prime}(0)=0.$

Hence,

$\left.\frac{d}{d\tau}\right|_{\tau=0}\mathcal{L}((\overset{\circ}{X},\overset{\circ}{Y})+\tau(\phi,\psi))=0.$

Now, we compute

$\displaystyle\left.\frac{d}{d\tau}\right|_{\tau=0}\mathcal{L}((\overset{\circ}{X},\overset{\circ}{Y})+\tau(\phi,\psi))$
$\displaystyle=\left.\frac{d}{d\tau}\right|_{\tau=0}\int_{\mathbb{R}^{3}}\left(|\partial(\overset{\circ}{X}+t\phi)+X^{-1}_{K}(\partial
Y_{K})(\overset{\circ}{Y}+t\psi)|^{2}\right.$
$\displaystyle+|\partial\overset{\circ}{Y}-X^{-1}_{K}(\partial
Y_{K})(\overset{\circ}{X}+t\phi)|^{2}+X^{-2}_{K}|(\overset{\circ}{X}+t\phi)\partial
Y_{K}-(\overset{\circ}{Y}+t\psi)\partial X_{K}|^{2}$
$\displaystyle\left.+2(\overset{\circ}{X}+t\phi)H_{X}+2(\overset{\circ}{Y}+t\psi)H_{Y}\right)$
$\displaystyle=2\int_{\mathbb{R}^{3}}\,\partial\overset{\circ}{X}\cdot\partial\phi-\frac{2\partial
Y_{K}\cdot\partial\overset{\circ}{Y}}{X_{K}}\phi+\frac{2|\partial
Y_{K}|^{2}}{X_{K}^{2}}\overset{\circ}{X}\phi-2\frac{\partial
X_{K}\cdot\partial Y_{K}}{X_{K}^{2}}\overset{\circ}{Y}\phi+H_{X}\phi$
$\displaystyle+\partial\overset{\circ}{Y}\cdot\partial\psi+\frac{2\partial
Y_{K}\cdot\partial\overset{\circ}{X}}{X_{K}}\psi+\frac{(|\partial
X_{K}|^{2}+|\partial Y_{K}|^{2})}{X_{K}^{2}}\overset{\circ}{Y}\psi+H_{Y}\psi$

Here, we applied the dominated convergence theorem in order to invert the
derivative and the integral. Therefore, for $(\phi,0)$ we obtain

$\int_{\mathbb{R}^{3}}\;\partial\overset{\circ}{X}\cdot\partial\phi-\frac{2\partial
Y_{K}\cdot\partial\overset{\circ}{Y}}{X_{K}}\phi+\frac{2|\partial
Y_{K}|^{2}}{X_{K}^{2}}\overset{\circ}{X}\phi-2\frac{\partial
X_{K}\cdot\partial Y_{K}}{X_{K}^{2}}\overset{\circ}{Y}\phi+H_{X}\phi=0$

and for $(0,\psi)$, we have

$\int_{\mathbb{R}^{3}}\;\partial\overset{\circ}{Y}\cdot\partial\psi+\frac{2\partial
Y_{K}\cdot\partial\overset{\circ}{X}}{X_{K}}\psi+\frac{(|\partial
X_{K}|^{2}+|\partial
Y_{K}|^{2})}{X_{K}^{2}}\overset{\circ}{Y}\psi+H_{Y}\psi=0.$

Finally, we integrate by part

$\displaystyle\int_{\mathbb{R}^{3}}\;-\Delta_{\mathbb{R}^{3}}\overset{\circ}{X}\phi-\frac{2\partial
Y_{K}\cdot\partial\overset{\circ}{Y}}{X_{K}}\phi+\frac{2|\partial
Y_{K}|^{2}}{X_{K}^{2}}\overset{\circ}{X}\phi-2\frac{\partial
X_{K}\cdot\partial Y_{K}}{X_{K}^{2}}\overset{\circ}{Y}\phi+H_{X}\phi=0,$
$\displaystyle\int_{\mathbb{R}^{3}}\;-\Delta_{\mathbb{R}^{3}}\overset{\circ}{Y}\psi+\frac{2\partial
Y_{K}\cdot\partial\overset{\circ}{X}}{X_{K}}\psi+\frac{(|\partial
X_{K}|^{2}+|\partial
Y_{K}|^{2})}{X_{K}^{2}}\overset{\circ}{Y}\psi+H_{Y}\psi=0.$

The latter holds for all $(\phi,\psi)\in C^{\infty}_{0}(\mathbb{R}^{3})$.
Therefore, (9.55) are the Euler-Lagrange equations associated to
$\mathcal{L}$.

∎

We apply classical variational methods in order to prove the existence and
uniqueness of weak solutions to (9.55). More precisely, we state

######  Lemma 71.

Let $H_{X},H_{Y}\in C^{\infty}_{0}(\mathbb{R}^{3})$. Then, the system (9.55)
has a unique weak solution
$(\overset{\circ}{X},\overset{\circ}{Y})\in\dot{H}^{1}_{axi}(\mathbb{R})^{3}\times\left(\dot{H}^{1}_{axi}(\mathbb{R})^{3}\cap
L^{2}_{\partial h}(\mathbb{R}^{3})\right)$ i.e
$(\phi_{1},\phi_{2})\in\dot{H}^{1}_{axi}(\mathbb{R})^{3}\times\left(\dot{H}^{1}_{axi}(\mathbb{R})^{3}\cap
L^{2}_{\partial h}(\mathbb{R}^{3})\right)$ we have

$\displaystyle\int_{\mathbb{R}^{3}}\;\nabla\overset{\circ}{X}\cdot\nabla\phi_{1}-\frac{2\partial
Y_{K}\cdot\partial\overset{\circ}{Y}}{X_{K}}\phi_{1}+\frac{2|\partial
Y_{K}|^{2}}{X_{K}^{2}}\overset{\circ}{X}\phi_{1}-2\frac{\partial
X_{K}\cdot\partial Y_{K}}{X_{K}^{2}}\overset{\circ}{Y}\phi_{1}+H_{X}\phi_{1}$
(9.59)
$\displaystyle+\nabla\overset{\circ}{Y}\cdot\nabla\phi_{2}+\frac{2\partial
Y_{K}\cdot\partial\overset{\circ}{X}}{X_{K}}\phi_{2}+\frac{(|\partial
X_{K}|^{2}+|\partial
Y_{K}|^{2})}{X_{K}^{2}}\overset{\circ}{Y}\phi_{2}+H_{Y}\phi_{2}=0.$

Finally, the solution is uniquely determined in the class
$\displaystyle\dot{H}^{1}_{axi}(\mathbb{R})^{3}\times\left(\dot{H}^{1}_{axi}(\mathbb{R})^{3}\cap
L^{2}_{\partial h}(\mathbb{R}^{3})\right)$.

Before we prove the above lemma, we state the following result on the
properties of $\mathcal{L}$

######  Lemma 72.

  1. 1.

There exist $C,\tilde{C}>0$ such that
$\forall\left(\overset{\circ}{X},\overset{\circ}{Y}\right)\in\mathcal{U}:=\dot{H}^{1}_{axi}(\mathbb{R})^{3}\times\left(\dot{H}^{1}_{axi}(\mathbb{R})^{3}\cap
L^{2}_{\partial h}(\mathbb{R}^{3})\right)$

$\mathcal{L}\left(\overset{\circ}{X},\overset{\circ}{Y}\right)\geq
C\left(\left|\left|\partial\overset{\circ}{X}\right|\right|^{2}_{L^{2}(\mathbb{R}^{3})}+\left|\left|\partial\overset{\circ}{Y}\right|\right|^{2}_{L^{2}(\mathbb{R}^{3})}+\left|\left|\overset{\circ}{Y}\right|\right|^{2}_{L^{2}_{\partial
h}(\mathbb{R}^{3})}\right)-\tilde{C}$

  2. 2.

$\displaystyle\mathcal{L}$ is weakly lower semi-continuous on
$\displaystyle\mathcal{U}$.

###### Proof.

  1. 1.

By Lemma 70, there exists $C>0$ such that
$\forall(\overset{\circ}{X},\overset{\circ}{Y})\in
C_{0}^{\infty}(\mathbb{R}^{3})$ and
$\displaystyle\int_{\mathbb{R}^{3}}\,\frac{|\partial
X_{K}|^{2}}{X_{K}^{2}}{\overset{\circ}{Y}}^{2}<\infty$, we have

$\int_{\mathbb{R}^{3}}\;\left(|\partial\overset{\circ}{X}|^{2}+|\partial\overset{\circ}{Y}|^{2}+|\partial
h|^{2}\overset{\circ}{Y}^{2}\right)\leq
C\left(\mathcal{L}(\overset{\circ}{X},\overset{\circ}{Y})+2\int_{\mathbb{R}^{3}}\;\left(|\overset{\circ}{X}H_{X}|+|\overset{\circ}{Y}H_{Y}|\right)\right).$

Thus, $\forall q>0$,

$\displaystyle\int_{\mathbb{R}^{3}}\;\left(|\partial\overset{\circ}{X}|^{2}+|\partial\overset{\circ}{Y}|^{2}+|\partial
h|^{2}\overset{\circ}{Y}^{2}\right)$ $\displaystyle\leq
C\mathcal{L}(\overset{\circ}{X},\overset{\circ}{Y})+2Cq||\overset{\circ}{X}||^{2}_{L^{2}(\mathbb{R}^{3})}+\frac{C}{2q}||H_{X}||^{2}_{L^{2}(\mathbb{R}^{3})}$
$\displaystyle+2Cq||\overset{\circ}{Y}||^{2}_{L^{2}(\mathbb{R}^{3})}+\frac{C}{2q}||H_{X}||^{2}_{L^{2}(\mathbb{R}^{3})}.$

By Poincaré inequality, we obtain

$\displaystyle\int_{\mathbb{R}^{3}}\;\left(|\partial\overset{\circ}{X}|^{2}+|\partial\overset{\circ}{Y}|^{2}+|\partial
h|^{2}\overset{\circ}{Y}^{2}\right)$ $\displaystyle\leq
C\mathcal{L}(\overset{\circ}{X},\overset{\circ}{Y})+2Cq||\partial\overset{\circ}{X}||^{2}_{L^{2}(\mathbb{R}^{3})}+\frac{C}{2q}||H_{X}||^{2}_{L^{2}(\mathbb{R}^{3})}$
$\displaystyle+2Cq||\partial\overset{\circ}{Y}||^{2}_{L^{2}(\mathbb{R}^{3})}+\frac{C}{2q}||H_{Y}||^{2}_{L^{2}(\mathbb{R}^{3})}.$

Finally, we choose $q>0$ so that

$\int_{\mathbb{R}^{3}}\;\left(|\partial\overset{\circ}{X}|^{2}+|\partial\overset{\circ}{Y}|^{2}+|\partial
h|^{2}\overset{\circ}{Y}^{2}\right)\leq
C\left(\mathcal{L}(\overset{\circ}{X},\overset{\circ}{Y})+||H_{X}||^{2}_{L^{2}(\mathbb{R}^{3})}+||H_{Y}||^{2}_{L^{2}(\mathbb{R}^{3})}\right)$

  2. 2.

Let $(f_{k},g_{k})_{k\in\mathbb{N}}$ be a sequence of $\mathcal{U}$ which
converges weakly in $\mathcal{U}$ to $(f,g)$. We show that

$\mathcal{L}(f,g)\leq\underline{\lim}_{k\to\infty}\mathcal{L}(f_{k},g_{k}).$

We have,

$\displaystyle\mathcal{L}_{0}(f_{k},g_{k})$
$\displaystyle=\int_{\mathbb{R}^{3}}\,|\partial f_{k}|^{2}+|\partial
g_{k}|^{2}+\frac{|\partial
Y_{K}|^{2}}{X_{K}^{2}}\left(g_{k}^{2}+2f_{k}^{2}\right)+\frac{|\partial
X_{K}|^{2}}{X_{K}^{2}}g_{k}^{2}+2g_{k}\partial f_{k}\cdot\frac{\partial
Y_{K}}{X_{K}}-2f_{k}\partial g_{k}\cdot\frac{\partial Y_{K}}{X_{K}}$
$\displaystyle-2f_{k}g_{k}\frac{\partial X_{K}\cdot\partial
Y_{K}}{X_{K}^{2}},$

where $\mathcal{L}_{0}$ is defined by (9.58). Since
$\displaystyle(f_{k},g_{k})\rightharpoonup(f,g)$ in $\mathcal{U}$, there
exists a subsequence of $(f_{k},g_{k})_{k\in\mathbb{N}}$ which converges
strongly in $L^{2}(\overline{B}_{R})$ for all $R>0$. Now, by convexity of
$x\to x^{2}$, we have

$\int_{\overline{B}_{R}}\,|\partial f_{k}|^{2}+|\partial g_{k}|^{2}\geq
2\int_{\overline{B}_{R}}\partial f\cdot(\partial f_{k}-\partial
f)+2\int_{\overline{B}_{R}}\partial g\cdot(\partial g_{k}-\partial
g)+\int_{\overline{B}_{R}}\,|\partial f^{2}+|\partial g|^{2}.$

Since $\displaystyle(\partial f_{k},\partial g_{k})\rightharpoonup(\partial
f,\partial g)$ in $L^{2}(\overline{B}_{R})$, we obtain

$\lim_{k\to\infty}\,2\int_{\overline{B}_{R}}\partial f\cdot(\partial
f_{k}-\partial f)+2\int_{\overline{B}_{R}}\partial g\cdot(\partial
g_{k}-\partial g)=0.$

Therefore,

$\underline{\lim}_{k\to\infty}\int_{\overline{B}_{R}}\,|\partial
f_{k}|^{2}+|\partial g_{k}|^{2}\geq\int_{\overline{B}_{R}}\,|\partial
f^{2}+|\partial g|^{2}.$

Moreover,

$\int_{\overline{B}_{R}}\,\frac{|\partial
Y_{K}|^{2}}{X_{K}^{2}}g_{k}^{2}\geq\int_{\overline{B}_{R}}\,\frac{|\partial
Y_{K}|^{2}}{X_{K}^{2}}g_{k}(g_{k}-g)+\int_{\overline{B}_{R}}\,\frac{|\partial
Y_{K}|^{2}}{X_{K}^{2}}g^{2}.$

Since $(g_{k})_{k}$ converges strongly in $L^{2}(\overline{B}_{R})$, we obtain

$\lim_{k\to\infty}\,\int_{\overline{B}_{R}}\,\frac{|\partial
Y_{K}|^{2}}{X_{K}^{2}}g(g_{k}-g)=0.$

We have,

$\int_{\overline{B}_{R}}\,\frac{|\partial
X_{K}|^{2}}{X_{K}^{2}}g_{k}^{2}\geq\int_{\overline{B}_{R}}\,\frac{|\partial
X_{K}|^{2}}{X_{K}^{2}}g(g_{k}-g)+\int_{\overline{B}_{R}}\,\frac{|\partial
X_{K}|^{2}}{X_{K}^{2}}g^{2}.$

Moreover, $\displaystyle\partial hg_{k}\rightharpoonup\partial hg$ in
$L^{2}(\overline{B}_{R})$ and $g\partial h\in L^{2}(\overline{B}_{R})$. Hence,

$\underline{\lim}_{k\to\infty}\int_{\overline{B}_{R}}\,\frac{|\partial
X_{K}|^{2}}{X_{K}^{2}}g_{k}^{2}\geq\int_{\overline{B}_{R}}\,\frac{|\partial
X_{K}|^{2}}{X_{K}^{2}}g^{2}.$

The remaining terms are tackled in the same manner. Therefore,

$\displaystyle l:=\underline{\lim}_{k\to\infty}\mathcal{L}_{0}(f_{k},g_{k})$
$\displaystyle\geq\int_{\overline{B}_{R}}\,|\partial f|^{2}+|\partial
g|^{2}+\frac{|\partial
Y_{K}|^{2}}{X_{K}^{2}}\left(g^{2}+2f^{2}\right)+\frac{|\partial
X_{K}|^{2}}{X_{K}^{2}}g^{2}+2g\partial f\cdot\frac{\partial
Y_{K}}{X_{K}}-2f\partial g\cdot\frac{\partial Y_{K}}{X_{K}}$ (9.60)
$\displaystyle-2fg\frac{\partial X_{K}\cdot\partial
Y_{K}}{X_{K}^{2}},\quad\quad\forall R>0.$

Now, we have,

$\displaystyle\int_{\overline{B}_{R}}\,|\partial f|^{2}+|\partial
g|^{2}+\frac{|\partial
Y_{K}|^{2}}{X_{K}^{2}}\left(g^{2}+2f^{2}\right)+\frac{|\partial
X_{K}|^{2}}{X_{K}^{2}}g^{2}+2g\partial f\cdot\frac{\partial
Y_{K}}{X_{K}}-2f\partial g\cdot\frac{\partial Y_{K}}{X_{K}}-2fg\frac{\partial
X_{K}\cdot\partial Y_{K}}{X_{K}^{2}}$
$\displaystyle=\int_{\mathbb{R}^{3}}\mathds{1}_{\left\\{x\,,\,|x|\leq
R\right\\}}\left(|\partial f|^{2}+|\partial g|^{2}+\frac{|\partial
Y_{K}|^{2}}{X_{K}^{2}}\left(g^{2}+2f^{2}\right)+\frac{|\partial
X_{K}|^{2}}{X_{K}^{2}}g^{2}+2g\partial f\cdot\frac{\partial
Y_{K}}{X_{K}}-2f\partial g\cdot\frac{\partial Y_{K}}{X_{K}}\right.$
$\displaystyle\left.-2fg\frac{\partial X_{K}\cdot\partial
Y_{K}}{X_{K}^{2}}\right).$

By positivity of the integrand, we can apply the monotone convergence theorem
to obtain

$\displaystyle\lim_{R\to\infty}\int_{\mathbb{R}^{3}}\mathds{1}_{\left\\{x\,,\,|x|\leq
R\right\\}}\left(|\partial f|^{2}+|\partial g|^{2}+\frac{|\partial
Y_{K}|^{2}}{X_{K}^{2}}\left(g^{2}+2f^{2}\right)+\frac{|\partial
X_{K}|^{2}}{X_{K}^{2}}g^{2}+2g\partial f\cdot\frac{\partial
Y_{K}}{X_{K}}-2f\partial g\cdot\frac{\partial Y_{K}}{X_{K}}\right.$
$\displaystyle\left.-2fg\frac{\partial X_{K}\cdot\partial
Y_{K}}{X_{K}^{2}}\right)$ $\displaystyle=\int_{\mathbb{R}^{3}}\left(|\partial
f|^{2}+|\partial g|^{2}+\frac{|\partial
Y_{K}|^{2}}{X_{K}^{2}}\left(g^{2}+2f^{2}\right)+\frac{|\partial
X_{K}|^{2}}{X_{K}^{2}}g^{2}+2g\partial f\cdot\frac{\partial
Y_{K}}{X_{K}}-2f\partial g\cdot\frac{\partial Y_{K}}{X_{K}}-2fg\frac{\partial
X_{K}\cdot\partial Y_{K}}{X_{K}^{2}}\right).$

Finally, we take the limit of (9.60) when $R\to\infty$, we obtain the desired
result.

∎

Now, we prove Lemma 71

###### Proof.

We begin by setting

$m:=\inf_{(f,g)\in\mathcal{U}}\;\mathcal{L}(f,g).$ (9.61)

  1. 1.

First, we claim that $m$ exists and it is finite. Indeed, the set

$\left\\{\mathcal{L}(f,g)\;,\;(f,g)\in\mathcal{U}\right\\}$

is not empty. Moreover, by Lemma 72, $\mathcal{L}$ is bounded from below.
Thus, $m$ exists. The latter is finite by Lemma 67.

  2. 2.

By the first point, there exists a minimising sequence
$\displaystyle(f_{k},g_{k})_{k\in\mathbb{N}}\in\mathcal{U}^{\mathbb{N}}$ such
that

$\lim_{k\to\infty}\mathcal{L}(f_{k},g_{k})=m.$

Therefore, $(\mathcal{L}(f_{k},g_{k}))_{k\in\mathbb{N}}$ is bounded. Moreover,
there exist $C,\tilde{C}>0$ such that $\forall k\in\mathbb{N}$,

$\left|\left|(f_{k},g_{k})\right|\right|_{\mathcal{U}}\leq
C\mathcal{L}(f_{k},g_{k})+\tilde{C}.$

Hence, $(f_{k},g_{k})_{k\in\mathbb{N}}$ is bounded in $\mathcal{U}$ and we can
extract a subsequence that converges weakly in $\mathcal{U}$.

  3. 3.

By Lemma 72, $\mathcal{L}$ is weakly lower semi-continuous. Hence, after
taking the limit, there exists a minimizer for the Lagrangian $\mathcal{L}$
and we obtain (9.59)

  4. 4.

Finally, we show that the minimizer is unique. By the linearity of the problem
(9.55), it suffices to show that if $(\overset{\circ}{X},\overset{\circ}{Y})$
solves the latter with $H_{X}=0$ and $H_{Y}=0$, then $\overset{\circ}{X}=0$
and $\overset{\circ}{Y}=0$.
If $(\overset{\circ}{X},\overset{\circ}{Y})$ solves weakly (9.55), then (9.59)
holds with $(\phi_{1},\phi_{2})=(\overset{\circ}{X},\overset{\circ}{Y})$.
Therefore, $\mathcal{L}(\overset{\circ}{X},\overset{\circ}{Y})=0$.
By Lemma 70, we obtain

$||(\overset{\circ}{X},\overset{\circ}{Y})||_{\mathcal{U}}=0.$

∎

##### 9.6.1.2 Linear estimates

We have obtained so far:

  * •

$\left(\overset{\circ}{X},\overset{\circ}{Y}\right)$ are smooth classical
solutions in the region
$\displaystyle\mathbb{R}^{3}\backslash\left\\{(0,z)\;z\in\mathbb{R}\right\\}.$

  * •

$\left(\overset{\circ}{X},\overset{\circ}{Y}\right)$ are compactly supported.

Moreover, the equations for $(\overset{\circ}{X},\overset{\circ}{Y})$ have a
singular behaviour on the the boundary because of the asymptotics for
$(X_{K},Y_{K})$ near the horizon and the axis:

  * •

the equation for $\overset{\circ}{X}$ is singular only near $p_{N}$ and
$p_{S}$, provided $\overset{\circ}{Y}$ in the right space.

  * •

the equation for $\overset{\circ}{Y}$ is singular near $p_{N}$, $p_{S}$ and
$\overline{\mathscr{B}_{A}}\cup\overline{\mathscr{B}_{H}}$.

In order to overcome this difficulty, we introduce a change of variable on
every region, work in higher dimensions in order to obtain the required
estimates for the new quantities on each region, then deduce the estimates for
the original unknowns. Let $\displaystyle(H_{X},H_{Y})\in
C_{0}^{\infty}(\mathbb{R}^{3})$ and let
$\displaystyle(\overset{\circ}{X},\overset{\circ}{Y})\in\mathcal{U}$ be the
unique weak solution of (9.55). We define $\tilde{Y}$ on
$\displaystyle\mathbb{R}^{3}\backslash\left\\{(0,z)\;z\in\mathbb{R}\right\\}$
to be the function

$\tilde{Y}:=X^{-1}_{K}{\overset{\circ}{Y}}.$ (9.62)

######  Lemma 73.

$\tilde{Y}$ is a weak solution of the equation

$\Delta_{\mathbb{R}^{3}}\tilde{Y}+2\frac{\partial
X_{K}\cdot\partial\tilde{Y}}{X_{K}}-2\frac{\partial
Y_{K}\cdot\partial\overset{\circ}{X}}{X^{2}_{K}}-2\frac{|\partial
Y_{K}|^{2}}{X^{2}_{K}}\tilde{Y}=H_{Y}X^{-1}_{K}.$ (9.63)

###### Proof.

We compute

$\displaystyle\partial\overset{\circ}{Y}$ $\displaystyle=\tilde{Y}\partial
X_{K}+X_{K}\partial\tilde{Y},$
$\displaystyle\Delta_{\mathbb{R}^{3}}\overset{\circ}{Y}$
$\displaystyle=\tilde{Y}\Delta_{\mathbb{R}^{3}}X_{K}+X_{K}\Delta_{\mathbb{R}^{3}}\tilde{Y}+2\partial
X_{K}\cdot\partial\tilde{Y}.$

We plug the latter in the equation satisfied by $\overset{\circ}{Y}$ in order
to obtain

$\tilde{Y}\Delta_{\mathbb{R}^{3}}X_{K}+X_{K}\Delta_{\mathbb{R}^{3}}\tilde{Y}+2\partial
X_{K}\cdot\partial\tilde{Y}-\frac{2\partial
Y_{K}\cdot\partial\overset{\circ}{X}}{X_{K}}-\frac{|\partial
X_{K}|^{2}}{X_{K}}\tilde{Y}-\frac{|\partial
X_{K}|^{2}}{X_{K}}\tilde{Y}=H_{Y}.$

Now, recall that the pair $(X_{K},Y_{K})$ satisfies (5.54). Therefore, on
$\mathbb{R}^{3}\backslash\left\\{(0,z),\;z\in\mathbb{R}^{3}\right\\}$, we have

$\frac{|\partial X_{K}|^{2}}{X_{K}}\tilde{Y}-\frac{|\partial
Y_{K}|^{2}}{X_{K}}\tilde{Y}+2\partial
X_{K}\cdot\partial\tilde{Y}-2\frac{\partial
Y_{K}\cdot\partial\overset{\circ}{X}}{X_{K}}-\frac{|\partial
X_{K}|^{2}}{X_{K}}\tilde{Y}+X_{K}\Delta_{\mathbb{R}^{3}}\tilde{Y}-\frac{|\partial
Y_{K}|^{2}}{X_{K}}\tilde{Y}=H_{Y}.$

Hence, $\tilde{Y}$ satisfies (9.63) .∎

######  Lemma 74.

  1. 1.

there exists a smooth vector field $\tilde{e}_{A}$ defined on
$\overline{\mathscr{B}_{A}}$ such that $\forall(\rho,z)\in\mathscr{B}:$

$\frac{\partial X_{K}}{X_{K}}=\frac{2}{\rho}\partial_{\rho}+\tilde{e}_{A},$

  2. 2.

there exist a smooth vector field $\tilde{d}_{A}$ and a smooth function
$\overline{d}_{A}$ defined on $\overline{\mathscr{B}_{A}}$ such that
$\forall(\rho,z)\in\mathscr{B}:$

$\frac{\partial
Y_{K}}{X^{2}_{K}}=\tilde{d}_{A}+\frac{\overline{d}_{A}}{\rho}\partial_{\rho},$

  3. 3.

the function $\displaystyle\frac{\left|\partial Y_{K}\right|}{X_{K}}$ extends
smoothly to $\overline{\mathscr{B}_{A}}$.

###### Proof.

This follows from Lemmas 54, 55 and 56. ∎

Therefore, $\tilde{Y}$ satisfies the equation

$\Delta_{\mathbb{R}^{3}}\tilde{Y}+\frac{4}{\rho}\partial_{\rho}\tilde{Y}+\tilde{e}_{A}\cdot\partial\tilde{Y}-2\frac{|\partial
Y_{K}|^{2}}{X^{2}_{K}}\tilde{Y}-2\tilde{d}_{A}\cdot\partial\overset{\circ}{X}-\frac{2\overline{d}_{A}}{\rho}\partial_{\rho}\overset{\circ}{X}=H_{Y}X^{-1}_{K}$
(9.64)

Moreover, by Lemma 92, we have

$\forall
z\in\mathbb{R}\quad\left.\frac{\partial_{\rho}\overset{\circ}{X}_{\mathbb{R}^{3}}(\cdot,\cdot,z)}{\rho}\right|_{(0,0)}=\left.\partial^{2}_{x}\overset{\circ}{X}_{\mathbb{R}^{3}}(\cdot,\cdot,z)\right|_{\left\\{(x,y)=(0,0)\right\\}}.$

For any $x\in\mathbb{R}^{7}$, let
$(\rho,\vartheta,\tilde{\vartheta},z)\in[0,\infty[\times(0,\pi)^{4}\times(0,2\pi)\times\mathbb{R}$
be the coordinates defined in the following way

$\displaystyle x_{1}$ $\displaystyle=\rho\cos\vartheta_{1}$ $\displaystyle
x_{2}$ $\displaystyle=\rho\sin\vartheta_{1}\cos\vartheta_{2}$ $\displaystyle
x_{3}$ $\displaystyle=\rho\sin\vartheta_{1}\sin\vartheta_{2}\cos\vartheta_{3}$
$\displaystyle x_{4}$
$\displaystyle=\rho\sin\vartheta_{1}\sin\vartheta_{2}\sin\vartheta_{3}\cos\vartheta_{4}$
$\displaystyle x_{5}$
$\displaystyle=\rho\sin\vartheta_{1}\sin\vartheta_{2}\sin\vartheta_{3}\sin\vartheta_{4}\cos\tilde{\vartheta}$
$\displaystyle x_{6}$
$\displaystyle=\rho\sin\vartheta_{1}\sin\vartheta_{2}\sin\vartheta_{3}\sin\vartheta_{4}\sin\tilde{\vartheta}$
$\displaystyle x_{7}$ $\displaystyle=z.$

Hence, $\rho$ is given by
$\displaystyle\rho=\sqrt{\sum_{i=1}^{6}\,x_{i}^{2}}$. Moreover, to any
function $f:\overline{\mathscr{B}}\mapsto\mathbb{R}$ we associate an
axisymmetric function $f_{\mathbb{R}^{7}}:{\mathbb{R}^{7}}\mapsto\mathbb{R}$
by setting

$f_{\mathbb{R}^{7}}(x):=f(\rho,z).$ (9.65)

In particular, we associate to $\tilde{Y}$ a function
$\tilde{Y}_{\mathbb{R}^{7}}$ defined on $\mathbb{R}^{7}$.
Now, note that

$\Delta_{\mathbb{R}^{7}}=\frac{4}{\rho}\partial_{\rho}+\Delta_{\mathbb{R}^{3}}.$

Therefore, $\tilde{Y}_{\mathbb{R}^{7}}$ satisfies the equation

$\Delta_{\mathbb{R}^{7}}\tilde{Y}_{\mathbb{R}^{7}}+\tilde{e}_{A}\cdot\partial\tilde{Y}-2\frac{|\partial
Y_{K}|^{2}}{X^{2}_{K}}\tilde{Y}_{\mathbb{R}^{7}}-2\tilde{d}_{A}\cdot\partial\overset{\circ}{X}-\frac{2\overline{d}_{A}}{\rho}\partial_{\rho}\overset{\circ}{X}=H_{Y}X^{-1}_{K}$
(9.66)

on $\displaystyle\mathbb{R}^{7}$. Note that this identification with
$\mathbb{R}^{7}$ only makes sense within the set
$\overline{\mathscr{B}_{A}}\cup\overline{\mathscr{B}_{H}}$. In this case, we
denote by $\mathbb{R}_{A}^{7}$ the set
$(\overline{\mathscr{B}_{A}}\cup\overline{\mathscr{B}_{H}})\times\mathbb{S}^{5}\subset\mathbb{R}^{7}$
and we introduce the following norms

$||f||_{W^{k,p}(\mathbb{R}^{7}_{A})}:=\sum_{|i|\leq
k}\left(\int\;\int_{\overline{\mathscr{B}_{A}}\cup\overline{\mathscr{B}_{H}}}\;\left|\partial^{i}f\right|^{p}\,\rho^{5}d\rho
dz\right)^{\frac{1}{p}},$
$||u||_{C^{k,\alpha}}(\mathbb{R}^{7}_{A}):=\sum_{|\alpha|\leq
k}||D^{\alpha}u||_{C^{0}(\overline{\mathscr{B}_{A}}\cup\overline{\mathscr{B}_{H}})}+\sum_{|\alpha|=k}[D^{\alpha}u]_{0,\alpha;\overline{\mathscr{B}_{A}}\cup\overline{\mathscr{B}_{H}}}$

and the semi norm

$||f||_{\dot{W}^{k,p}(\mathbb{R}^{7}_{A})}:=\left(\int\;\int_{\overline{\mathscr{B}_{A}}\cup\overline{\mathscr{B}_{H}}}\;\left|\partial^{k}f\right|^{p}\,\rho^{5}d\rho
dz\right)^{\frac{1}{p}}.$

of a function
$f:\overline{\mathscr{B}_{A}}\cup\overline{\mathscr{B}_{H}}\to\mathbb{R}$.
Now, we recall the coefficients behavior of the elliptic operator associated
to (9.55) near $p_{N}$ and $p_{S}$

######  Lemma 75.

  1. 1.

there exists a smooth vector field $\tilde{e}_{N}$ defined on
$\overline{\mathscr{B}_{N}}$ such that $\forall(s,\chi)\in\mathscr{B}_{N}:$

$\frac{\underline{\partial}X_{K}}{X_{K}}=\frac{2}{s}\underline{\partial}_{s}+\tilde{e}_{N},$

  2. 2.

there exists a smooth vector field $\tilde{e}_{S}$ defined on
$\overline{\mathscr{B}_{S}}$ such that
$\forall(s^{\prime},\chi^{\prime})\in\mathscr{B}_{S}:$

$\frac{\underline{\partial}X_{K}}{X_{K}}=\frac{2}{s^{\prime}}\underline{\partial}_{s^{\prime}}+\tilde{e}_{S},$

  3. 3.

the function
$\displaystyle\frac{\left|\underline{\partial}Y_{K}\right|}{X_{K}}$ extends
smoothly to $\overline{\mathscr{B}_{N}}$ and to $\overline{\mathscr{B}_{S}}$.

###### Proof.

This follows from Lemmas 54, 55 and 56. ∎

Now, we consider the system (9.55) near $p_{N}$ and $p_{S}$ and we write the
latter in terms of the coordinates $(s,\chi)$

######  Lemma 76.

$(\overset{\circ}{X},\tilde{Y})$ is a weak solution of the system

$\displaystyle\underline{\Delta}_{\mathbb{R}^{4}}\overset{\circ}{X}+2\frac{\underline{\partial}Y_{K}\cdot\partial\overset{\circ}{Y}}{X_{K}}-2\frac{|\underline{\partial}Y_{K}|^{2}}{X_{K}^{2}}\overset{\circ}{X}+2\frac{\underline{\partial}X_{K}\cdot\underline{\partial}Y_{K}}{X_{K}^{2}}\overset{\circ}{Y}=(s^{2}+\chi^{2})H_{X}$
(9.67)
$\displaystyle\underline{\Delta}_{\mathbb{R}^{8}}\tilde{Y}+\tilde{e}_{N}\cdot\underline{\partial}\tilde{Y}-2\frac{|\underline{\partial}Y_{K}|^{2}}{X_{K}^{2}}\tilde{Y}-2\frac{\underline{\partial}Y_{K}\cdot\underline{\partial}\overset{\circ}{X}}{X_{K}^{2}}=(s^{2}+\chi^{2})X_{K}^{-1}H_{Y},$

on $\overline{\mathscr{B}_{N}}$, where $\underline{\partial}$ is the gradient
with respect to $(s,\chi)$ coordinates and $\tilde{e}_{N}$ is given by Lemma
75. The above equations hold on $\overline{\mathscr{B}_{S}}$ provided that we
replace $\tilde{e}_{N}$ by $\tilde{e}_{S}$ and we use
$(s^{\prime},\chi^{\prime})$ coordinates.

###### Proof.

The computations are straightforward. ∎

In order to prove Proposition 29, we will prove the following estimates:

  1. 1.

First of all, we establish estimates for $\overset{\circ}{X}$ away from
$\overline{\mathscr{B}_{N}}$ and $\overline{\mathscr{B}_{S}}$ using the first
equation of (9.55).

  2. 2.

Next, we establish estimates for $\overset{\circ}{X}$ on
$\overline{\mathscr{B}_{N}}$ and $\overline{\mathscr{B}_{S}}$ using the first
equation of (9.67).

  3. 3.

We establish estimates for $\overset{\circ}{Y}$ away from the region
$\left\\{(\rho,z)\in\overline{\mathscr{B}}\;,\;\rho=0\;;\;z\in\mathbb{R}\right\\}$.

  4. 4.

In order to handle the estimates for $\overset{\circ}{Y}$ near the boundary of
$\overline{\mathscr{B}}$, we establish the estimates for $\tilde{Y}$ near
$\mathscr{A}$ using equation (9.66). Then, we use the second equation of
(9.67) so that we obtain estimates of $\tilde{Y}$ on
$\overline{\mathscr{B}_{N}}$ and $\overline{\mathscr{B}_{S}}$.

  5. 5.

Finally, we deduce the estimates for $\overset{\circ}{Y}$.

######  Lemma 77.

There exists $C>0$ such that

$\left|\left|\overset{\circ}{X}\right|\right|_{\dot{H}^{1}_{axi}(\mathbb{R}^{3})}+\left|\left|\overset{\circ}{Y}\right|\right|_{\dot{H}^{1}_{axi}(\mathbb{R}^{3})\cap
L^{2}_{\nabla h}(\mathbb{R}^{3})}\leq
C\left(\left|\left|H_{X}\overset{\circ}{X}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}+\left|\left|H_{Y}\overset{\circ}{Y}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}\right).$

###### Proof.

Since $(\overset{\circ}{X},\overset{\circ}{Y})$ is a weak solution, it
minimises the Lagrangian $\mathcal{L}$. Therefore,

$\mathcal{L}(\overset{\circ}{X},\overset{\circ}{Y})\leq\mathcal{L}(X,Y)\quad\forall(X,Y)\in\mathcal{U}.$

In particular,

$\mathcal{L}(\overset{\circ}{X},\overset{\circ}{Y})\leq\mathcal{L}(0,0)=0.$

Now, by Lemma 70, there exists $C>0$

$\left|\left|\overset{\circ}{X}\right|\right|^{2}_{\dot{H}^{1}_{axi}(\mathbb{R}^{3})}+\left|\left|\overset{\circ}{Y}\right|\right|^{2}_{\dot{H}^{1}_{axi}(\mathbb{R}^{3})\cap
L^{2}_{\nabla h}(\mathbb{R}^{3})}\leq
C\left(\mathcal{L}(\overset{\circ}{X},\overset{\circ}{Y})+2\left(\left|\left|H_{X}\overset{\circ}{X}\right|\right|_{L^{1}(\mathbb{R}^{3})}+\left|\left|H_{Y}\overset{\circ}{Y}\right|\right|_{L^{1}(\mathbb{R}^{3})}\right)\right).$

Therefore, by convexity of $x\to x^{2}$, we obtain the desired estimate. ∎

Now, we write

$\displaystyle\overset{\circ}{X}$
$\displaystyle=(1-\xi_{N}-\xi_{S})\overset{\circ}{X}+\xi_{N}\overset{\circ}{X}+\xi_{S}\overset{\circ}{X},$
$\displaystyle\overset{\circ}{Y}$
$\displaystyle=(1-\xi_{N}-\xi_{S})(1-\xi_{A})\overset{\circ}{Y}+(1-\xi_{N}-\xi_{S})\xi_{A}\overset{\circ}{Y}+\xi_{N}\overset{\circ}{Y}+\xi_{S}\overset{\circ}{Y}.$

We have

######  Lemma 78.

There exists $C>0$ such that

$\left|\left|(1-\xi_{N}-\xi_{S})\overset{\circ}{X}\right|\right|_{\dot{H}^{2}(\mathbb{R}^{3})}\leq
C\left(\left|\left|(1-\xi_{N}-\xi_{S})H_{X}\right|\right|_{L^{2}(\mathbb{R}^{3})}+\left|\left|H_{X}\overset{\circ}{X}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}+\left|\left|H_{Y}\overset{\circ}{Y}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}\right).$

###### Proof.

  1. 1.

First of all, we set
$\overset{\circ}{X}_{A}:=(1-\xi_{N}-\xi_{S})\overset{\circ}{X}$ and we claim
that $\overset{\circ}{X}_{A}$ satisfies

$\displaystyle\Delta_{\mathbb{R}^{3}}\overset{\circ}{X}_{A}$
$\displaystyle=2\partial\overset{\circ}{X}\cdot\partial(\xi_{N}+\xi_{S})+\Delta_{\mathbb{R}^{3}}(\xi_{N}+\xi_{S})\overset{\circ}{X}-2(1-\xi_{N}-\xi_{S})\frac{\partial
Y_{K}\cdot\partial\overset{\circ}{Y}}{X_{K}}+2(1-\xi_{N}-\xi_{S})\overset{\circ}{Y}\frac{\partial
X_{K}\cdot\partial Y_{K}}{X_{K}^{2}}$ (9.68) $\displaystyle+2\frac{|\partial
Y_{K}|^{2}}{X_{K}^{2}}(1-\xi_{N}-\xi_{S})\overset{\circ}{X}+(1-\xi_{N}-\xi_{S})H_{X}.$

  2. 2.

We will apply Theorem 11 with $p=2$, $n=3$ and $f$ given by

$\displaystyle f(x)$
$\displaystyle:=2\partial\overset{\circ}{X}\cdot\partial(\xi_{N}+\xi_{S})+\Delta_{\mathbb{R}^{3}}(\xi_{N}+\xi_{S})\overset{\circ}{X}-2(1-\xi_{N}-\xi_{S})\frac{\partial
Y_{K}\cdot\partial\overset{\circ}{Y}}{X_{K}}+2(1-\xi_{N}-\xi_{S})\overset{\circ}{Y}\frac{\partial
X_{K}\cdot\partial Y_{K}}{X_{K}^{2}}$ (9.69)
$\displaystyle+2\frac{|Y_{K}|^{2}}{X_{K}^{2}}(1-\xi_{N}-\xi_{S})\overset{\circ}{X}+(1-\xi_{N}-\xi_{S})H_{X}$

We recall here the identification of $f$ with $f_{\mathbb{R}^{3}}$. To this
end, we show that $f\in L^{2}(\mathbb{R}^{3})$.

     * •

First of all, we have
$\partial(\xi_{N}+\xi_{S}),\Delta_{\mathbb{R}^{3}}(\xi_{N}+\xi_{S})$ are
compactly supported in
$\overline{\mathscr{B}_{N}}\cup\overline{\mathscr{B}_{S}}$ and they vanish
near $p_{N}$ and $p_{S}$. Denote by $\overline{Q}$ the support of their
identifications on $\mathbb{R}^{3}$.

     * •

$\overset{\circ}{X}\in\dot{H}^{1}_{axi}(\mathbb{R}^{3})$. Therefore,
$\displaystyle-2\partial\overset{\circ}{X}\cdot\partial(\xi_{N}+\xi_{S})\in
L^{2}(\mathbb{R}^{3})$.

     * •

By Poincaré inequality, we have

$\displaystyle||\Delta_{\mathbb{R}^{3}}(\xi_{N}+\xi_{S})\overset{\circ}{X}||_{L^{2}(\mathbb{R}^{3})}$
$\displaystyle=||\Delta_{\mathbb{R}^{3}}(\xi_{N}+\xi_{S})\overset{\circ}{X}||_{L^{2}(\overline{Q})}$
$\displaystyle\leq C||\overset{\circ}{X}||_{L^{2}(\overline{Q})}$
$\displaystyle\leq C||\nabla\overset{\circ}{X}||_{L^{2}(\overline{Q})}$
$\displaystyle\leq
C||\overset{\circ}{X}||_{\dot{H}^{1}_{axi}(\mathbb{R}^{3})}.$

     * •

By Lemma 54 and Lemma 69, we have

$\left|\left|(1-\xi_{N}-\xi_{S})\overset{\circ}{Y}\frac{\partial
X_{K}\cdot\partial Y_{K}}{X_{K}^{2}}\right|\right|_{L^{2}(\mathbb{R}^{3})}\leq
C\left|\left|\overset{\circ}{Y}\right|\right|_{L^{2}_{\nabla
h}(\mathbb{R}^{3})}.$

     * •

Moreover, by Lemma 55 and Lemma 54, we have

$\left|\left|(1-\xi_{N}-\xi_{S})\frac{\partial
Y_{K}\cdot\partial\overset{\circ}{Y}}{X_{K}}\right|\right|_{L^{2}(\mathbb{R}^{3})}\leq
C\left|\left|\overset{\circ}{Y}\right|\right|_{\dot{H}^{1}_{axi}(\mathbb{R}^{3})}.$

     * •

Since $H_{X}\in C^{\infty}_{0}(\mathbb{R}^{3})$, the term
$\displaystyle(1-\xi_{N}-\xi_{S})H_{X}\in L^{2}(\mathbb{R}^{3})$.

Thus, $f\in L^{2}(\mathbb{R}^{3})$ and we apply Theorem 11 to obtain

$\left|\left|(1-\xi_{N}-\xi_{S})\overset{\circ}{X}\right|\right|_{\dot{H}^{2}(\mathbb{R}^{3})}\leq
C\left|\left|f\right|\right|_{L^{2}(\mathbb{R}^{3})}.$

  3. 3.

Now, we use the latter estimates to obtain

$\left|\left|f\right|\right|_{L^{2}(\mathbb{R}^{3})}\leq
C\left(\left|\left|(1-\xi_{N}-\xi_{S})H_{X}\right|\right|_{L^{2}(\mathbb{R}^{3})}+||\overset{\circ}{X}||_{\dot{H}^{1}_{axi}(\mathbb{R}^{3})}+\left|\left|\overset{\circ}{Y}\right|\right|_{\dot{H}^{1}_{axi}(\mathbb{R}^{3})}+\left|\left|\overset{\circ}{Y}\right|\right|_{L^{2}_{\nabla
h}(\mathbb{R}^{3})}\right).$

  4. 4.

Finally, we use Lemma 77 to conclude.

∎

######  Lemma 79.

There exists $C>0$ such that

$\left|\left|\xi_{N}\overset{\circ}{X}\right|\right|_{\dot{H}^{2}(\mathbb{R}^{4}_{N})}+\left|\left|\xi_{S}\overset{\circ}{X}\right|\right|_{\dot{H}^{2}(\mathbb{R}^{4}_{S})}\leq
C\left(\left|\left|\xi_{N}(s^{2}+\chi^{2})H_{X}\right|\right|_{L^{2}(\mathbb{R}^{4}_{N})}+\left|\left|H_{X}\overset{\circ}{X}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}+\left|\left|H_{Y}\overset{\circ}{Y}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}\right)$

###### Proof.

We proceed as in Lemma 78. We only mention the main steps.

  1. 1.

First of all, we multiply the first equation of (9.67) by $\xi_{N}$ and we
express
$\underline{\Delta}_{\mathbb{R}^{4}}\left(\xi_{N}\overset{\circ}{X}\right)$ in
terms of the remaining quantities:

$\displaystyle\underline{\Delta}_{\mathbb{R}^{4}}\left(\xi_{N}\overset{\circ}{X}\right)$
$\displaystyle=\underline{\Delta}_{\mathbb{R}^{4}}(\xi_{N})\overset{\circ}{X}+2\underline{\partial}(\xi_{N}+\xi_{S})\cdot\underline{\partial}\overset{\circ}{X}-2\xi_{N}\frac{\underline{\partial}Y_{K}\cdot\partial\overset{\circ}{Y}}{X_{K}}+2\xi_{N}\frac{|\underline{\partial}Y_{K}|^{2}}{X_{K}^{2}}\overset{\circ}{X}$
$\displaystyle-2\xi_{N}\frac{\underline{\partial}X_{K}\cdot\underline{\partial}Y_{K}}{X_{K}^{2}}\overset{\circ}{Y}+\xi_{N}(s^{2}+\chi^{2})H_{X}$

  2. 2.

We apply Theorem 11 with $n=4$, $p=2$ and $f$ given by the right hand side of
the above equation. To this end, we show that $\displaystyle f\in
L^{2}(\mathbb{R}^{4}_{N})$. We give details for the term
$\displaystyle\xi_{N}\frac{|\underline{\partial}Y_{K}|^{2}}{X_{K}^{2}}\overset{\circ}{X}$.
The other terms follow in the same manner.
We have

$\left|\left|\xi_{N}\frac{|\underline{\partial}Y_{K}|^{2}}{X_{K}^{2}}\overset{\circ}{X}\right|\right|^{2}_{L^{2}(\mathbb{R}^{4}_{N})}=(2\pi)^{2}\int\int_{\overline{\mathscr{B}_{N}}}\,\xi_{N}^{2}\left(\frac{|\underline{\partial}Y_{K}|^{2}}{X_{K}^{2}}\overset{\circ}{X}\right)^{2}s\chi
d\xi d\chi.$

By Lemma 4, Lemma 54, we have

$\forall(s,\chi)\in\overline{\mathscr{B}_{N}}\;:\;\frac{|\underline{\partial}Y_{K}|}{X_{K}}\leq
CsX_{N}^{-1}(s^{2},\chi^{2}).$

Therefore,

$\displaystyle\left|\left|\xi_{N}\frac{|\underline{\partial}Y_{K}|^{2}}{X_{K}^{2}}\overset{\circ}{X}\right|\right|^{2}_{L^{2}(\mathbb{R}^{4}_{N})}$
$\displaystyle\leq 2\pi
C\int\int_{\overline{\mathscr{B}_{N}}}\,\xi_{N}^{2}\overset{\circ}{X}^{2}s\chi
d\xi d\chi$ $\displaystyle\leq 2\pi
C\int_{\mathbb{R}}\int_{0}^{\infty}\,\xi_{N}^{2}\overset{\circ}{X}^{2}\rho
d\rho dz$ $\displaystyle\leq
C\left|\left|\overset{\circ}{X}\right|\right|_{\dot{H}^{1}_{axi}(\mathbb{R}^{3})}.$

Here, we used Poincaré inequality to obtain the latter estimate.

  3. 3.

We use Lemma 77 to conclude.

∎

######  Lemma 80.

There exists $C>0$ such that

$\left|\left|(1-\xi_{N}-\xi_{S})(1-\xi_{A})\overset{\circ}{Y}\right|\right|_{\dot{H}^{2}(\mathbb{R}^{3})}\leq
C\left(\left|\left|(1-\xi_{N}-\xi_{S})(1-\xi_{A})H_{Y}\right|\right|_{L^{2}(\mathbb{R}^{3})}+\left|\left|H_{X}\overset{\circ}{X}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}+\left|\left|H_{Y}\overset{\circ}{Y}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}\right).$

###### Proof.

The proof is similar to the previous lemmas. Here we apply Calderon-Zygmund
theory to the second equation of (9.55). ∎

######  Lemma 81.

There exists $C>0$ such that

$\displaystyle\left|\left|(1-\xi_{N}-\xi_{S})\xi_{A}\tilde{Y}\right|\right|_{\dot{H}^{2}(\mathbb{R}^{7}_{A})}\leq
C\left(\left|\left|(1-\xi_{N}-\xi_{S})\xi_{A}H_{Y}X_{K}^{-1}\right|\right|_{L^{2}(\mathbb{R}^{7}_{A})}+\left|\left|H_{X}\overset{\circ}{X}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}+\left|\left|H_{Y}\overset{\circ}{Y}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}\right).$

###### Proof.

Although the proof is similar to the previous lemmas, we give details leading
to $L^{2}$ estimates of the different terms.

  1. 1.

We multiply (9.66) by $\tilde{\xi}:=(1-\xi_{N}-\xi_{S})\xi_{A}$ and we write
$\Delta_{\mathbb{R}^{7}}\left(\tilde{\xi}\tilde{Y}\right)$ in terms of the
remaining quantities:

$\displaystyle\Delta_{\mathbb{R}^{7}}\left(\tilde{\xi}\tilde{Y}\right)$
$\displaystyle=\tilde{\xi}H_{Y}X_{K}^{-1}-\left(\tilde{\xi}\tilde{e}_{A}-2\partial\tilde{\xi}\right)\cdot\tilde{Y}+\left(2\frac{|\partial
Y_{K}|^{2}}{X_{K}^{2}}\tilde{\xi}+\Delta_{\mathbb{R}^{7}}\tilde{\xi}\right)\tilde{Y}.$

  2. 2.

In order to apply Calderon-Zygmund theory, we need to show that the function

$g=\tilde{\xi}H_{Y}X_{K}^{-1}-\left(\tilde{\xi}\tilde{e}_{A}-2\partial\tilde{\xi}\right)\cdot\partial\tilde{Y}+\left(2\frac{|\partial
Y_{K}|^{2}}{X_{K}^{2}}\tilde{\xi}+\Delta_{\mathbb{R}^{7}}\tilde{\xi}\right)\tilde{Y}+2\tilde{\xi}\tilde{d}_{A}\partial\overset{\circ}{X}+2\overline{d}_{A}\tilde{\xi}\frac{\partial_{\rho}\overset{\circ}{X}}{\rho}$

lies in $L^{2}(\mathbb{R}_{A}^{7})$.

     * •

Denote by $\tilde{\mathscr{A}}$ the support of $\tilde{\xi}$ and recall from
Lemma 56 the decay estimate for $\tilde{e}_{A}$:

$\left|\tilde{e}_{A}\right|=O_{r\to+\infty}\left(\langle
r\rangle^{-2}\right).$

     * •

We have

$\displaystyle\left|\left|\tilde{\xi}H_{Y}X_{K}^{-1}\right|\right|^{2}_{L^{2}(\mathbb{R}^{7}_{A})}$
$\displaystyle=(2\pi)^{5}\int\int_{\tilde{\mathscr{A}}}\,\tilde{\xi}^{2}X^{-2}_{K}H_{Y}^{2}\rho^{5}d\rho
dz$
$\displaystyle=(2\pi)^{5}\int\int_{\tilde{\mathscr{A}}}\,\tilde{\xi}^{2}X^{-2}_{\mathscr{A}}(\rho^{2},z)H_{Y}\rho
d\rho dz$

where $X_{\mathscr{A}}$ is given by Definition 4. Moreover, $X_{\mathscr{A}}$
are bounded on $\tilde{\mathscr{A}}$. Therefore,

$\displaystyle\left|\left|\tilde{\xi}H_{Y}X_{K}^{-1}\right|\right|^{2}_{L^{2}(\mathbb{R}^{7}_{A})}$
$\displaystyle\leq 2\pi
C\int\int_{\tilde{\mathscr{A}}}\,\tilde{\xi}^{2}H^{2}_{Y}\rho d\rho dz$
$\displaystyle\leq C\int_{\mathbb{R}^{3}}\tilde{\xi}^{2}H^{2}_{Y}\rho d\rho
dz$ $\displaystyle\leq
C\left|\left|\tilde{\xi}H_{Y}\right|\right|^{2}_{L^{2}(\mathbb{R}^{3})}.$

     * •

We have

$\displaystyle\left|\left|\left(\tilde{\xi}\tilde{e}_{A}-2\partial\tilde{\xi}\right)\cdot\partial\tilde{Y}\right|\right|^{2}_{L^{2}(\mathbb{R}^{7}_{A})}$
$\displaystyle=(2\pi)^{5}\int\int_{\tilde{\mathscr{A}}}\,\left(\left(\tilde{\xi}\tilde{e}_{A}-2\partial\tilde{\xi}\right)\cdot\partial\tilde{Y}\right)^{2}\rho^{5}d\rho
dz$
$\displaystyle\leq(2\pi)C\int\int_{\tilde{\mathscr{A}}}\,\left|\partial\tilde{Y}\right|^{2}\rho^{5}d\rho
dz.$

We have

$\forall(\rho,z)\in\tilde{\mathscr{A}}\;:\>\tilde{Y}=\frac{1}{\rho^{2}}X_{\mathscr{A}}^{-1}(\rho^{2},z)\overset{\circ}{Y}.$

Hence,

$\partial\tilde{Y}=-\frac{1}{\rho^{3}}X_{\mathscr{A}}^{-1}(\rho^{2},z)\overset{\circ}{Y}+\frac{1}{\rho^{2}}\left(\partial
X_{\mathscr{A}}^{-1}(\rho^{2},z)\overset{\circ}{Y}+X_{\mathscr{A}}^{-1}(\rho^{2},z)\partial\overset{\circ}{Y}\right).$

Therefore,

$\forall(\rho,z)\in\tilde{A}\;:\;\left|\partial\tilde{Y}\right|^{2}\leq
C\left(\frac{\overset{\circ}{Y}^{2}}{\rho^{6}}+\frac{\left|\partial\overset{\circ}{Y}\right|^{2}}{\rho^{4}}\right)$

and

$\displaystyle\left|\left|\left(\tilde{\xi}\tilde{e}_{A}-2\partial\tilde{\xi}\right)\cdot\partial\tilde{Y}\right|\right|^{2}_{L^{2}(\mathbb{R}^{7}_{A})}$
$\displaystyle\leq(2\pi)C\left(\int\int_{\tilde{\mathscr{A}}}\,\frac{\overset{\circ}{Y}^{2}}{\rho^{2}}\rho
d\rho
dz+\int\int_{\tilde{\mathscr{A}}}\,\left|\partial\overset{\circ}{Y}\right|^{2}\rho
d\rho dz\right).$

By Lemma 56,

$\forall(\rho,z)\in\tilde{\mathscr{A}}\;:\;\frac{1}{\rho^{2}}\leq|\partial
h|^{2}.$

This yields

$\displaystyle\int\int_{\tilde{\mathscr{A}}}\,\frac{\overset{\circ}{Y}^{2}}{\rho^{2}}\rho
d\rho dz$ $\displaystyle\leq\int\int_{\tilde{\mathscr{A}}}\,{|\partial
h|\overset{\circ}{Y}^{2}}\rho d\rho dz$
$\displaystyle\leq\left|\left|\overset{\circ}{Y}\right|\right|^{2}_{L^{2}_{\nabla
h}(\mathbb{R}^{3})}$

and the estimate

$\displaystyle\left|\left|\left(\tilde{\xi}\tilde{e}_{A}-2\partial\tilde{\xi}\right)\cdot\partial\tilde{Y}\right|\right|_{L^{2}(\mathbb{R}^{7}_{A})}$
$\displaystyle\leq
C\left|\left|\overset{\circ}{Y}\right|\right|_{\dot{H}^{1}_{axi}(\mathbb{R}^{3})\cap
L^{2}_{\nabla h}(\mathbb{R}^{3})}$ $\displaystyle\leq
C\left(\left|\left|H_{X}\overset{\circ}{X}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}+\left|\left|H_{Y}\overset{\circ}{Y}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}\right).$

     * •

In order to estimate
$\displaystyle\overline{d}_{A}\tilde{\xi}\frac{\partial_{\rho}\overset{\circ}{X}}{\rho}$,
we write

$\displaystyle\left|\left|\overline{d}_{A}\tilde{\xi}\frac{\partial_{\rho}\overset{\circ}{X}}{\rho}\right|\right|^{2}_{L^{2}(\mathbb{R}^{7}_{A})}$
$\displaystyle\leq
C\int\int_{\tilde{\mathscr{A}}}\,\tilde{\xi}^{2}(\partial_{\rho}\overset{\circ}{X})^{2}\rho
d\rho dz$ $\displaystyle\leq
C\left|\left|\overset{\circ}{X}\right|\right|^{2}_{\dot{H}^{1}_{axi}(\mathbb{R}^{3})}.$

     * •

The other terms follow in the same way.

∎

######  Lemma 82.

There exists $C>0$ such that

$\displaystyle\left|\left|\xi_{N}\tilde{Y}\right|\right|_{\dot{H}^{2}(\mathbb{R}^{8}_{N})}\leq
C\left(\left|\left|(s^{2}+\chi^{2})H_{Y}\right|\right|_{L^{2}(\overline{\mathscr{B}_{N}})}+\left|\left|H_{X}\overset{\circ}{X}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}+\left|\left|H_{Y}\overset{\circ}{Y}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}\right),$
$\displaystyle\left|\left|\xi_{S}\tilde{Y}\right|\right|_{\dot{H}^{2}(\mathbb{R}^{8}_{S})}\leq
C\left(\left|\left|((s^{\prime})^{2}+(\chi^{\prime})^{2})H_{Y}\right|\right|_{L^{2}(\overline{\mathscr{B}_{S}})}+\left|\left|H_{X}\overset{\circ}{X}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}+\left|\left|H_{Y}\overset{\circ}{Y}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}\right).$

###### Proof.

The proof is similar to the previous lemmas. ∎

Now, we derive the estimates for $\overset{\circ}{Y}$.

######  Lemma 83.

There exists $C>0$ such that

$\displaystyle\left|\left|(1-\xi_{N}-\xi_{S})\xi_{A}\overset{\circ}{Y}\right|\right|_{\dot{H}^{2}(\mathbb{R}^{3})}\leq
C\left(\left|\left|(1-\xi_{N}-\xi_{S})\xi_{A}H_{Y}\right|\right|_{L^{2}(\mathbb{R}^{3})}+\left|\left|H_{X}\overset{\circ}{X}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}+\left|\left|H_{Y}\overset{\circ}{Y}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}\right),$
$\displaystyle\left|\left|\xi_{N}\overset{\circ}{Y}\right|\right|_{\dot{H}^{2}(\mathbb{R}^{4}_{N})}\leq
C\left(\left|\left|(s^{2}+\chi^{2})H_{Y}\right|\right|_{L^{2}(\overline{\mathscr{B}_{N}})}+\left|\left|H_{X}\overset{\circ}{X}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}+\left|\left|H_{Y}\overset{\circ}{Y}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}\right),$
$\displaystyle\left|\left|\xi_{S}\overset{\circ}{Y}\right|\right|_{\dot{H}^{2}(\mathbb{R}^{4}_{S})}\leq
C\left(\left|\left|((s^{\prime})^{2}+(\chi^{\prime})^{2})H_{Y}\right|\right|_{L^{2}(\overline{\mathscr{B}_{S}})}+\left|\left|H_{X}\overset{\circ}{X}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}+\left|\left|H_{Y}\overset{\circ}{Y}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}\right).$

###### Proof.

  1. 1.

First of all, we claim that $\forall f\in C^{\infty}_{0}(\mathbb{R}^{3})$,
such that

$\forall z\in\mathbb{R}\;:\;\lim_{\rho\to\infty}f(\rho,z)=0,$

we have

$\displaystyle\int_{0}^{\infty}\,f^{2}\rho^{3}d\rho dz\leq
C\int_{0}^{\infty}\,(\partial_{\rho}f)^{2}\rho^{5}d\rho dz$ (9.70)
$\displaystyle\int_{0}^{\infty}\,f^{2}\rho d\rho dz\leq
C\int_{0}^{\infty}\,(\partial_{\rho}f)^{2}\rho^{3}d\rho dz$

Indeed, we apply the second estimate of Theorem 8 to the function $\rho\to
f(\rho,z)$ with $n=1$, $p=2$, $\delta=-1$ in order to obtain the second
inequality and $\delta=-2$ in order to obtain the first inequality.

  2. 2.

We estimate the term
$\displaystyle\left|\left|(1-\xi_{N}-\xi_{S})\xi_{A}\overset{\circ}{Y}\right|\right|_{\dot{H}^{2}(\mathbb{R}^{3})}$.
The other terms follow using similar arguments. We have

$\displaystyle\left|\left|\tilde{\xi}\overset{\circ}{Y}\right|\right|^{2}_{\dot{H}^{2}(\mathbb{R}^{3})}$
$\displaystyle=2\pi\int\int_{\tilde{\mathscr{A}}}\,\left|\partial^{2}(\tilde{\xi}\overset{\circ}{Y})\right|^{2}\rho
d\rho dz$
$\displaystyle=2\pi\int\int_{\tilde{\mathscr{A}}}\,\left|\partial^{2}(X_{K}\tilde{\xi}\tilde{Y})\right|^{2}\rho
d\rho dz$ $\displaystyle\leq
C\int\int_{\tilde{\mathscr{A}}}\,\left(X^{2}_{K}\left|\partial^{2}(\tilde{\xi}\tilde{Y})\right|^{2}+(\tilde{\xi}\tilde{Y})^{2}\left|\partial^{2}X_{K}\right|^{2}+\left|\partial
X_{K}\right|^{2}\left|\partial(\tilde{\xi}\tilde{Y})\right|^{2}\right)\rho
d\rho dz$ $\displaystyle\leq
C\int\int_{\tilde{\mathscr{A}}}\,\left(\rho^{4}\left|\partial^{2}(\tilde{\xi}\tilde{Y})\right|^{2}+(\tilde{\xi}\tilde{Y})^{2}+\rho^{2}\left|\partial(\tilde{\xi}\tilde{Y})\right|^{2}\right)\rho
d\rho dz$ $\displaystyle\leq
C\int\int_{\tilde{\mathscr{A}}}\,\left|\partial^{2}(\tilde{\xi}\tilde{Y})\right|^{2}\rho^{5}\,d\rho
dz+\int\int_{\tilde{\mathscr{A}}}\,\left|\partial(\tilde{\xi}\tilde{Y})\right|^{2}\rho^{3}\,d\rho
dz+\int\int_{\tilde{\mathscr{A}}}\,(\tilde{\xi}\tilde{Y})^{2}\rho d\rho dz.$

Now, we use (9.70) to obtain

$\displaystyle\int\int_{\tilde{\mathscr{A}}}\,\left|\partial(\tilde{\xi}\tilde{Y})\right|^{2}\rho^{3}\,d\rho
dz\leq
C\int\int_{\tilde{\mathscr{A}}}\,\left|\partial^{2}(\tilde{\xi}\tilde{Y})\right|^{2}\rho^{5}\,d\rho
dz$

and

$\displaystyle\int\int_{\tilde{\mathscr{A}}}\,(\tilde{\xi}\tilde{Y})^{2}\rho
d\rho dz\leq
C\int\int_{\tilde{\mathscr{A}}}\,\left|\partial(\tilde{\xi}\tilde{Y})\right|^{2}\rho^{3}\,d\rho
dz$ $\displaystyle\leq
C\int\int_{\tilde{\mathscr{A}}}\,\left|\partial^{2}(\tilde{\xi}\tilde{Y})\right|^{2}\rho^{5}\,d\rho
dz.$

Therefore,

$\displaystyle\left|\left|\tilde{\xi}\overset{\circ}{Y}\right|\right|^{2}_{\dot{H}^{2}(\mathbb{R}^{3})}$
$\displaystyle\leq
C\left|\left|(1-\xi_{N}-\xi_{S})\xi_{A}\tilde{Y}\right|\right|_{\dot{H}^{2}(\mathbb{R}^{7}_{A})}$
$\displaystyle\leq
C\left(\left|\left|(1-\xi_{N}-\xi_{S})\xi_{A}H_{Y}X_{K}^{-1}\right|\right|_{L^{2}(\mathbb{R}^{7}_{A})}+\left|\left|H_{X}\overset{\circ}{X}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}+\left|\left|H_{Y}\overset{\circ}{Y}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}\right).$

Here, we used Lemma 81 to obtain the latter estimate. Finally, by similar
arguments, we show that

$\left|\left|(1-\xi_{N}-\xi_{S})\xi_{A}H_{Y}\right|\right|_{L^{2}(\mathbb{R}^{7}_{A})}\leq
C\left|\left|(1-\xi_{N}-\xi_{S})\xi_{A}H_{Y}\right|\right|_{L^{2}(\mathbb{R}^{3})}$

This ends the proof.

∎

######  Lemma 84.

There exists $C>0$ such that

$\displaystyle\left|\left|\overset{\circ}{X}\right|\right|_{\hat{C}^{1}(\overline{\mathscr{B}})}$
$\displaystyle+\left|\left|\overset{\circ}{Y}\right|\right|_{\hat{C}^{1}(\overline{\mathscr{B}})}+\left|\left|\tilde{Y}\right|\right|_{\hat{C}^{1}(\overline{\mathscr{B}})}$
$\displaystyle\leq
C\left(\left|\left|H_{X}\overset{\circ}{X}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}+\left|\left|H_{Y}\overset{\circ}{Y}\right|\right|^{\frac{1}{2}}_{L^{1}(\mathbb{R}^{3})}+\left|\left|(1-\xi_{N}-\xi_{S})H_{X}\right|\right|_{L^{2}(\mathbb{R}^{3})\cap
L^{\infty}(\mathbb{R}^{3})}+\right.$
$\displaystyle\left|\left|(1-\xi_{N}-\xi_{S})(1-\xi_{A})H_{Y}\right|\right|_{L^{2}(\mathbb{R}^{3})\cap
L^{\infty}(\mathbb{R}^{3})}+\left|\left|(1-\xi_{N}-\xi_{S})\xi_{A}X^{-1}_{K}H_{Y}\right|\right|_{L^{2}(\mathbb{R}^{7}_{A})\cap
L^{\infty}(\mathbb{R}^{7}_{A})}+$
$\displaystyle\left|\left|(s^{2}+\chi^{2})\xi_{N}X_{K}^{-1}H_{Y}\right|\right|_{L^{\infty}(\mathbb{R}^{8}_{N})}+\left|\left|((s^{\prime})^{2}+(\chi^{\prime})^{2})\xi_{S}X_{K}^{-1}H_{Y}\right|\right|_{L^{\infty}(\mathbb{R}^{8}_{S})}+$
$\displaystyle\left.\left|\left|(s^{2}+\chi^{2})\xi_{N}H_{X}\right|\right|_{L^{\infty}(\mathbb{R}^{4}_{N})}+\left|\left|((s^{\prime})^{2}+(\chi^{\prime})^{2})\xi_{S}H_{X}\right|\right|_{L^{\infty}(\mathbb{R}^{4}_{S})}\right).$

###### Proof.

We will detail the $\hat{C}^{1}(\overline{\mathscr{B}})$ estimates for
$\overset{\circ}{X}$. $\hat{C}^{1}(\overline{\mathscr{B}})$ estimates for
$\overset{\circ}{Y}$ and $\tilde{Y}$ follow using similar argument.
To this end, we need to show that

$\xi_{N}\overset{\circ}{X},\xi_{S}\overset{\circ}{X}\in
C^{1}(\mathbb{R}^{4})\quad\text{and}\quad(1-\xi_{N}-\xi_{S})\overset{\circ}{X}\in
C^{1}(\mathbb{R}^{3}).$

  1. 1.

By Lemma 78,
$\overset{\circ}{X}_{A}=(1-\xi_{N}-\xi_{S})\overset{\circ}{X},\partial\overset{\circ}{X}_{A}\in\dot{H}^{1}(\mathbb{R}^{3})$.

  2. 2.

Now, we recall the continuous embedding

$\dot{H}^{1}(\mathbb{R}^{3})\subset L^{6}(\mathbb{R}^{3}).$ (9.71)

  3. 3.

Recall that $(1-\xi_{N}-\xi_{S})\overset{\circ}{X}$ satisfies (9.68).

  4. 4.

Now, we claim that $\partial^{2}\overset{\circ}{X}_{A}\in
L^{6}(\mathbb{R}^{3})$. To prove the latter, we use equation (9.68) and we
apply Theorem 11 with $f$ given by (9.69), $p=6$ and $n=3$. We need to show
that $f\in L^{6}(\mathbb{R}^{3})$.
Recall that $f$ is given by

$\displaystyle f(x)$
$\displaystyle=2\partial\overset{\circ}{X}\cdot\partial(\xi_{N}+\xi_{S})+\Delta_{\mathbb{R}^{3}}(\xi_{N}+\xi_{S})\overset{\circ}{X}-2(1-\xi_{N}-\xi_{S})\frac{\partial
Y_{K}\cdot\partial\overset{\circ}{Y}}{X_{K}}+2(1-\xi_{N}-\xi_{S})\overset{\circ}{Y}\frac{\partial
X_{K}\cdot\partial Y_{K}}{X_{K}^{2}}$
$\displaystyle+2\frac{|Y_{K}|^{2}}{X_{K}^{2}}(1-\xi_{N}-\xi_{S})\overset{\circ}{X}+(1-\xi_{N}-\xi_{S})H_{X}$

     * •

It is easy to see that the terms
$\partial\overset{\circ}{X}\cdot\partial(\xi_{N}+\xi_{S}),\Delta_{\mathbb{R}^{3}}(\xi_{N}+\xi_{S})\overset{\circ}{X}$
and $(1-\xi_{N}-\xi_{S})H_{X}$ lie in $L^{6}(\mathbb{R}^{3})$

     * •

By the decay estimates of $\displaystyle\frac{\partial Y_{K}}{X_{K}}$, we
obtain

$\left|\left|(1-\xi_{N}-\xi_{S})\frac{\partial
Y_{K}\cdot\partial\overset{\circ}{Y}}{X_{K}}\right|\right|_{L^{6}(\mathbb{R}^{3})}\leq
C\left|\left|\overset{\circ}{Y}\right|\right|_{\dot{W}^{1,6}_{axi}(\mathbb{R}^{3})}.$

     * •

We have

$\displaystyle\left|\left|(1-\xi_{N}-\xi_{S})\overset{\circ}{Y}\frac{\partial
X_{K}\cdot\partial Y_{K}}{X_{K}^{2}}\right|\right|_{L^{6}(\mathbb{R}^{3})}$
$\displaystyle\leq\left|\left|(1-\xi_{N}-\xi_{S})\overset{\circ}{Y}\frac{\partial
X_{K}}{X_{K}}\right|\right|_{L^{2}(\mathbb{R}^{3})}\left|\left|\frac{\partial
Y_{K}}{X_{K}}\right|\right|_{L^{2}(\mathbb{R}^{3})}$ $\displaystyle\leq
C\left|\left|(1-\xi_{N}-\xi_{S})\overset{\circ}{Y}\right|\right|_{L^{2}_{\nabla
h}(\mathbb{R}^{3})}$

     * •

Moreover, by Theorem 8 with $n=3$, $p=6$ and $\displaystyle\delta=\frac{1}{2}$

$\displaystyle\left|\left|(1-\xi_{N}-\xi_{S})\overset{\circ}{X}\frac{|Y_{K}|^{2}}{X_{K}^{2}}\right|\right|_{L^{6}(\mathbb{R}^{3})}$
$\displaystyle\leq C\left|\left|(1-\xi_{N}-\xi_{S})\overset{\circ}{X}\langle
r\rangle^{-8}\right|\right|_{L^{6}(\mathbb{R}^{3})}$ $\displaystyle\leq
C\left|\left|(1-\xi_{N}-\xi_{S})\overset{\circ}{X}\right|\right|_{6,\frac{1}{2}}$
$\displaystyle\leq
C\left|\left|\partial\overset{\circ}{X}_{A}\right|\right|_{6,-\frac{1}{2}}$
$\displaystyle=C\left|\left|\partial\overset{\circ}{X}_{A}\right|\right|_{L^{6}(\mathbb{R}^{3})}$

  5. 5.

Finally, we use the Sobolev embedding

$W^{2,6}(\mathbb{R}^{3})\subset C^{1,\frac{1}{2}}(\mathbb{R}^{3})$

in order to obtain

$\overset{\circ}{X}_{A}\in C^{1}(\mathbb{R}^{3}).$

Furthermore,

$\displaystyle\left|\left|\overset{\circ}{X}_{A}\right|\right|_{C^{1}(\mathbb{R}^{3})}$
$\displaystyle\leq C||f||_{L^{6}(\mathbb{R}^{3})}$ $\displaystyle\leq
C\left(\left|\left|(1-\xi_{N}-\xi_{S})\overset{\circ}{X}\right|\right|_{\dot{H}^{2}(\mathbb{R}^{3})}+\left|\left|(1-\xi_{N}-\xi_{S})\overset{\circ}{Y}\right|\right|_{\dot{H}^{2}(\mathbb{R}^{3})}+\left|\left|(1-\xi_{N}-\xi_{S})\overset{\circ}{Y}\right|\right|_{L^{2}_{\nabla
h}(\mathbb{R}^{3})}\right)$ $\displaystyle\leq