---
abstract: 'In this paper we develop a monotonicity formula for elliptic systems with Neumann boundary coupling, proving unique continuation and classification of blow-up profiles. As an application, we obtain strong unique continuation for some fourth order equations and higher order fractional problems.'
author:
- Veronica Felli
- Alberto Ferrero
title: 'Unique continuation and classification of blow-up profiles for elliptic systems with Neumann boundary coupling and applications to higher order fractional equations'
---

[^1]

Introduction and statement of the main results {#s:introduction}
==============================================

The present paper is devoted to the study of unique continuation from a boundary point and classification of blow-up profiles for elliptic systems with Neumann boundary coupling. Systems of such a kind arise from higher order extensions of the fractional Laplacian, as first observed in [@Y], where the well known Caffarelli-Silvestre extension procedure characterizing the fractional Laplacian as the Dirichlet-to-Neumann map in one extra spatial dimension was generalized to higher powers of the Laplacian. More precisely in [@Y] (see also [@CY]) it is proved that, if $s\in(1,2)$ and $f\in H^s({{\mathbb R}}^N)$, then $$\label{eq:44b}
(-\Delta)^s f=K_{s}\lim_{t\to 0^+}t^{b}\frac{\partial(\Delta_b U)}{\partial
  t}$$ where $b=3-2s$, $K_{s}$ is a constant depending only on $s$, $\Delta_bU=\Delta U+\frac{b}{t}\frac{\partial
  U}{\partial t}$ and $U$ is the unique solution to the problem $$\begin{cases}
  \Delta_b^2 U=0,&\text{in }{{\mathbb R}}^{N+1}_+={{\mathbb R}}^N\times(0,+\infty),\\
U(x,0)=f(x),&\text{in }{{\mathbb R}}^{N},\\
\lim_{t\to 0^+}t^{b}\frac{\partial U}{\partial
  t}=0 ,&\text{in }{{\mathbb R}}^{N}.
\end{cases}$$ Setting $V=\Delta_b U$ and taking into account , the above fourth order problem can be rewritten as the system $$\begin{cases}
  \Delta_b U=V,&\text{in }{{\mathbb R}}^{N+1}_+,\\
  \Delta_b V=0,&\text{in }{{\mathbb R}}^{N+1}_+,\\
U(x,0)=f(x),&\text{in }{{\mathbb R}}^{N},\\
\lim_{t\to 0^+}t^{b}\frac{\partial U}{\partial
  t}=0 ,&\text{in }{{\mathbb R}}^{N},\\
K_{s} \lim_{t\to 0^+}t^{b}\frac{\partial V}{\partial
  t}=(-\Delta)^s f ,&\text{in }{{\mathbb R}}^{N}.
\end{cases}$$ In [@Y] an Almgren’s frequency formula in the spirit of [@almgren] is derived for solutions to the higher order system $$\label{eq:51}
 \begin{cases}
  \Delta_b U=V,&\text{in }{{\mathbb R}}^{N+1}_+,\\
  \Delta_b V=0,&\text{in }{{\mathbb R}}^{N+1}_+,\\
\lim_{t\to 0^+}t^{b}\frac{\partial U}{\partial
  t}=0 ,&\text{in }{{\mathbb R}}^{N},\\
\lim_{t\to 0^+}t^{b}\frac{\partial V}{\partial
  t}=0 ,&\text{in }{{\mathbb R}}^{N},
\end{cases}$$ obtained by extending $s$-harmonic functions; in the spirit of Garofalo and Lin [@GL], such monotonicity formula allows proving a unique continuation property for solutions to system . In [@Y] a strong unique continuation property is also stated for $s$-harmonic functions.

The main goal of the present paper is to extend, in the case $s=\frac32$, the monotonicity formula developed in [@Y] for the homogeneous case to systems with a Neumann boundary coupling of the type $$\label{eq:52}
 \begin{cases}
  \Delta U=V,&\text{in }{{\mathbb R}}^{N+1}_+,\\
  \Delta V=0,&\text{in }{{\mathbb R}}^{N+1}_+,\\
\frac{\partial U}{\partial
  \nu}=0 ,&\text{in }{{\mathbb R}}^{N},\\
\frac{\partial V}{\partial
  \nu}=h\, U(\cdot,0),&\text{in }{{\mathbb R}}^{N},
\end{cases}$$ which arise naturally as extension of fractional equations of the form $$(-\Delta)^{3/2}u=a(x)u$$ once we put $h=-K_{3/2}^{-1}\, a=-2a$. Indeed, by [@FF-prep Proof of Lemma 3.2, Step 6] we deduce that the constant $C_b$ defined there equals $\sqrt 2$ when $b=0$ and, since it can be shown that $K_s=C_b^{-2}$ with $b=3-2s$, we deduce that $K_{3/2}=\frac 12$. The frequency function associated to problem is given by the ratio of the local energy over mass near the fixed point $0\in{{\mathbb R}}^N$ $$\label{eq:59}
\mathcal N(r)=\frac{r^{-N+1} \left[\int_{B_r^+} \left(|\nabla U|^2+|\nabla V|^2+UV\right)dz -\int_{B_r'} h(x)U(x,0)V(x,0)\, dx\right]}{r^{-N}\int_{S_r^+} (U^2+V^2)\, dS},$$ where we are denoting as $z=(x,t)\in {{\mathbb R}}^N\times{{\mathbb R}}$ the variable in ${{\mathbb R}}^{N+1}={{\mathbb R}}^N\times{{\mathbb R}}$, $dz=dx\,dt$ and, for all $r>0$, $$\begin{aligned}
&B_r=\{z\in {{\mathbb R}}^{N+1}:|z|<r\},\quad
B_r^+=\{(x,t)\in B_r:t>0\},\\
&B_r'=\{x\in {{\mathbb R}}^{N}:|x|<r\}=B_r\cap({{\mathbb R}}^N\times\{(x,0):x\in{{\mathbb R}}^N\}),\\
&S_r^+=\{(x,t)\in \partial B_r:t>0\}.\end{aligned}$$ The classical approach developed by Garofalo and Lin [@GL] to prove unique continuation through Almgren’s monotonicity formula is based on the validity of doubling type conditions, obtained as a consequence of boundedness of the quotient $\mathcal N$. We refer to [@ae; @aek; @FFFN; @FF-edinb; @KukavicaNystrom; @tz1; @tz2] for unique continuation from the boundary established via Almgren monotonicity formula.

While in the local case doubling conditions are enough to establish unique continuation, in the fractional case they provide unique continuation only for the extended local problem and not for the fractional one. Such difficulty was overcome in [@FF] for the fractional Laplacian $(-\Delta)^{s}$ with $s\in(0,1)$, by a fine blow-up analysis and a precise classification of the possible blow-up limit profiles in terms of a Neumann eigenvalue problem on the half-sphere.

The problem of unique continuation for fractional laplacians with power $s\in(0,1)$ was also studied in [@Ruland] in presence of rough potentials using Carleman estimates and in [@Yu] for fractional operators with variable coefficients using an Almgren type monotonicity formula. As far as higher fractional powers of the laplacian, the main contribution to the problem of unique continuation is due to Seo in papers [@seo0; @seo1; @seo2], through Carleman inequalities; in particular papers [@seo0; @seo1; @seo2] consider fractional Schrödinger operators with potentials in Morrey spaces and prove a *weak* unique continuation result, i.e. vanishing of solutions which are zero on an open set; we recall that the *strong* unique continuation property instead requires the weaker assumption of infinite vanishing order at some point.

We observe that the presence of a coupling Neumann term in system produces substancial additional difficulties with respect to the extension problem corresponding to the lower order fractional case $s\in(0,1)$ and consisting in a single equation associated with a Neumann boundary condition. In particular the proof of a monotonicity formula for is made quite delicate by the appearance in the derivative of the frequency $\mathcal N$ of a term of the type $$-r\int_{\partial B_r'}huv\,dS'+2\int_{B_r'}hu\,x\cdot
  \nabla_x v\,dx,$$ see Lemma \[mono\]. While in the lower order case we have only one component $u=v$ so that an integration by parts allows rewriting the above sum as an integral over $B_r'$, in the case of two components $u,v$ this is no more possible and an estimate of the integral over “the boundary of the boundary” $\int_{\partial B_r'}huv\,dS'$ is required. The method developed here to overcome this difficulty is based on estimates in terms of boundary integrals (see Lemma \[l:STIMA\_DI\_nu\_2\]) and represents one of the main technical novelty of the present paper in the context of monotonicity formulas; we think that this procedure could have future applications in the extension of some of the results of [@FF] to rough potentials, since it could avoid the integration by parts needed to write the above sum as an integral over $B_r'$, which requires differentiability of the potential $h$.

Let $N>3$, $R>0$, and $(U,V)\in H^1(B_R^+)\times H^1(B_R^+)$ be a weak solution to the system $$\label{eq:system}
  \begin{cases}
\Delta U=V,&\text{in }B_R^+,\\
\Delta V=0,&\text{in }B_R^+,\\
\frac{\partial U}{\partial \nu}= 0, &\text{in } B_R',\\
\frac{\partial V}{\partial \nu}=hu, &\text{in } B_R',
  \end{cases}$$ where $u=U(\cdot,0)$ (trace of $U$ on $B_R'$) and $h\in
C^1(B_R')$. We also denote $v=V(\cdot,0)$ (trace of $V$ on $B_R'$). By a weak solution to the system we mean a couple $(U,V)\in H^1(B_R^+)\times H^1(B_R^+)$ such that, for every $\varphi\in  H^1(B_R^+)$ having zero trace on $S_R^+$, $$\begin{cases}
\int_{B_R^+}\nabla U(z)\cdot\nabla
\varphi(z)\,dz=-\int_{B_R^+}V(z)\varphi(z)\,dz,\\
\int_{B_R^+}\nabla V(z)\cdot\nabla
\varphi(z)\,dz=\int_{B_R'}h(x)u(x)\mathop{\rm Tr}\varphi (x)\,dx,
\end{cases}$$ where $\mathop{\rm Tr}\varphi$ is the trace of $\varphi$ on $B_R'$.

Our first result is an asymptotic expansion of nontrivial solutions to ; more precisely we prove that blow-up profiles can be described as combinations of spherical harmonics symmetric with respect to the equator $t=0$.

Let $-\Delta _{\mathbb{S}^{N}}$ denote the Laplace Beltrami operator on the $N$-dimensional unit sphere $\mathbb{S}^{N}$. It is well known that the eigenvalues of $-\Delta _{\mathbb{S}^{N}}$ are given by $$\lambda _{\ell }=(N-1+\ell )\ell ,\quad \ell =0,1,2,\dots .$$ For every $\ell\in{{\mathbb N}}$, it is easy to verify that there exists a spherical harmonic on $\mathbb{S}^{N}$ of degree $\ell$ which is symmetric with respect to the equator $t=0$[^2]. Therefore the eigenvalues of the problem $$\begin{aligned}
\label{eq:sph_eig}
  \begin{cases}
    -\Delta_{{\mathbb S}^{N}}\psi=\lambda\,\psi, &\text{in }{\mathbb S}^{N}_+,\\[5pt]
\nabla_{{\mathbb
    S}^{N}}\psi\cdot {\mathbf
  e}=0,&\text{on }\partial {\mathbb S}^{N}_+,
  \end{cases}\end{aligned}$$ with $${\mathbb S}^{N}_+=\{(\theta_1,\theta_2,\dots, \theta_{N+1})\in
  {\mathbb S}^{N}:\theta_{N+1}>0\},\quad {\mathbf e}=(0,0,\dots,0,1),$$ are given by the sequence $\{\lambda_\ell:\ell=0,1,2,\dots\}$; for every $\ell$, $\lambda_\ell$ has finite multiplicity $M_\ell$ as an eigenvalue of . For every $\ell {\geqslant}0$, let $\{Y_{\ell ,m}\}_{m=1,2,\dots ,M_{\ell }}$ be a $L^{2}(\mathbb{S}^{N}_+)$-orthonormal basis of the eigenspace of associated to $\lambda _{\ell }$ with $Y_{\ell ,m}$ being spherical harmonics of degree $\ell $.

We note that, if $\Psi$ is an eigenfunction of , then $\Psi\not\equiv0$ on $\partial {\mathbb
  S}^{N}_+={\mathbb S}^{N-1}$; indeed, by unique continuation, $\Psi$ and $\nabla_{{\mathbb
    S}^{N}}\Psi\cdot {\mathbf  e}$ can not both vanish on $\partial {\mathbb
  S}^{N}_+$. In particular $Y_{\ell,m}\not\equiv0$ on $\partial {\mathbb
  S}^{N}_+={\mathbb S}^{N-1}$ for all $\ell\in{{\mathbb N}}$ and $1{\leqslant}m{\leqslant}M_\ell$.

\[t:asym\] Let $(U,V)\in H^1(B_R^+)\times H^1(B_R^+)$ be a weak solution to such that $(U,V)\not=(0,0)$. Then there exists $\ell\in{{\mathbb N}}$ such that $$\lambda^{-\ell}U(\lambda z)\to \widehat U(z),\quad
\lambda^{-\ell}V(\lambda z)\to \widehat V(z),$$ as $\lambda\to0^+$ strongly in $H^1(B_1^+)$, where $$\widehat U(z)=|z|^{\ell}
\sum_{m=1}^{M_\ell}\alpha_{\ell,m}Y_{\ell,m}\Big(\frac
z{|z|}\Big),\quad
 \widehat V(z)=|z|^{\ell}
\sum_{m=1}^{M_\ell}\alpha'_{\ell,m}Y_{\ell,m}\Big(\frac
z{|z|}\Big),$$ $$\begin{aligned}
 \label{eq:45} \alpha_{\ell,m}&=R^{-\ell}\int_{{\mathbb S}^{N}_+}U(R\,\theta)
  Y_{\ell,m}(\theta)\,dS-
\frac{R^{-N-2\ell+1}}{N+2\ell-1}\int_0^R t^{N+\ell}\left(\int_{{\mathbb S}^{N}_+}V(t\,\theta)
    Y_{\ell,m}(\theta)\,dS\right)\,dt\\
\notag&\qquad\qquad
+\int_0^R\frac{t^{-\ell+1}}{2\ell+N-1}\left(\int_{{\mathbb S}^{N}_+}V(t\,\theta)
    Y_{\ell,m}(\theta)\,dS\right)\,dt,\\
 \label{eq:46} \alpha'_{\ell,m}&=R^{-\ell}\int_{{\mathbb S}^{N}_+}V(R\,\theta)
  Y_{\ell,m}(\theta)\,dS\\
\notag&\qquad\qquad-
\frac{R^{-N-2\ell+1}}{N+2\ell-1}\int_0^R t^{N+\ell-1}\left(
\int_{{\mathbb
      S}^{N-1}}h(t\theta')U(t\theta',0) Y_{\ell,m}(\theta',0) \,dS'\right)\,dt\\
\notag&\qquad\qquad
+\int_0^R\frac{t^{-\ell}}{2\ell+N-1}\left(
\int_{{\mathbb
      S}^{N-1}}h(t\theta')U(t\theta',0)Y_{\ell,m}(\theta',0)\,dS'
\right)\,dt,\end{aligned}$$ and $$\sum_{m=1}^{M_\ell}((\alpha_{\ell,m})^2+(\alpha'_{\ell,m})^2)\neq0.$$

A first remarkable consequence of Theorem \[t:asym\] is the validity of a [*[strong unique continuation property]{}*]{} (from the boundary point $0$) for solutions to .

\[t:sun-ext\] Let $(U,V)\in H^1(B_R^+)\times H^1(B_R^+)$ be a weak solution to . If $$\label{eq:case2}
U(z)=o(|z|^n)\quad\text{as }|z|\to 0\text{ for all $n\in
  {{\mathbb N}}$}$$ then $U\equiv V\equiv 0$ in $B_R^+$.

We observe that in the case of a single equation a blow result as the one stated in Theorem \[t:asym\] directly yields the strong unique continuation: indeed, if the solution has a precise vanishing order it cannot vanish of any order. On the other hand, in the case of a system of type , the blow-up Theorem \[t:asym\] ensures that the couple of the limit profiles $(\widehat U, \widehat V)$ is not trivial, i.e. at least one of the two components $U,V$ has a precise vanishing order; hence some further analysis is needed to deduce strong unique continuation from Theorem \[t:asym\].

System is related to fourth order elliptic equations arising in Caffarelli-Silvestre type extensions for higher order fractional laplacians in the spirit of [@Y]. Let us define $\mathcal D$ as the completion of $$\label{eq:space}
\mathcal T:=\left\{U\in C^\infty_c(\overline{{{\mathbb R}}^{N+1}_+}):U_t\equiv 0 \ \text{on} \
{{\mathbb R}}^N\times\{0\}
\right\}$$ with respect to the norm $$\|U\|_{\mathcal D}=\bigg(\int_{{{\mathbb R}}^{N+1}_+} |\Delta
U(x,t)|^2\,dx\,dt\bigg)^{\!\!1/2}.$$ By [@FF-prep] there exists a well defined continuous trace map $$\mathop{\rm Tr}:\mathcal D\to \mathcal D^{3/2,2}({{\mathbb R}}^N),$$ where the space $\mathcal D^{3/2,2}({{\mathbb R}}^N)$ is defined as the completion of $C^\infty_c({{\mathbb R}}^N)$ with respect to the scalar product $$\label{eq:scalar-product}
(u,v)_{\mathcal D^{3/2,2}({{\mathbb R}}^N)}:=\int_{{{\mathbb R}}^N} |\xi|^{3}\, \widehat
u(\xi)\overline{\widehat v(\xi)} \, d\xi.$$ In $\widehat u$ denotes the Fourier transform of $u$ in ${{\mathbb R}}^N$: $$\widehat u(\xi)=\frac 1{(2\pi)^{N/2}} \int_{{{\mathbb R}}^N} e^{-i\xi x}u(x)\, dx \, .$$ Moreover in we denoted by $\overline{\widehat v(\xi)}$ the complex conjugate of $\widehat v(\xi)$.

We observe that, since $u$ and $v$ are real functions, is really a scalar product although their respective Fourier transforms are complex functions.

As a corollary of Theorem \[t:asym\] we derive sharp asymptotic estimates and a strong unique continuation principle for weak $\mathcal D$-solutions to the fourth order elliptic problem $$\label{eq:44}
  \begin{cases}
    \Delta^2 U=0,&\text{in }{{\mathbb R}}^{N+1}_+,\\[3pt]
\frac{\partial U}{\partial\nu}=0,&\text{in }{{\mathbb R}}^N \, , \\[3pt]
\frac{\partial(\Delta U)}{\partial\nu}=h\, \mathop{\rm
  Tr}(U),&\text{in }\Omega \, .
  \end{cases}$$ By a weak $\mathcal D$-solution to we mean some $U\in
\mathcal D$ such that $$\int_{{{\mathbb R}}^{N+1}_+}\Delta U(x,t) \Delta \varphi(x,t)\,dx\,dt=-\int_\Omega
h(x) \mathop{\rm Tr}U(x)
\mathop{\rm Tr}\varphi(x)\,dx$$ for all $\varphi\in\mathcal D$ such that $\mathop{\rm supp}(\mathop{\rm Tr}\varphi)\subset\Omega$.

\[t:asym-ho\]

1.  Let $U\in\mathcal D$, $U\not\equiv0$, be a nontrivial weak solution to for some $h\in C^1(\Omega)$, with $\Omega$ being an open bounded set in ${{\mathbb R}}^N$ such that $0\in\Omega$. Then there exists $\ell\in{{\mathbb N}}$ such that $$\lambda^{-\ell}U(\lambda z)\to
    |z|^{\ell}
    \sum_{m=1}^{M_\ell}\alpha_{\ell,m}Y_{\ell,m}\Big(\frac
    z{|z|}\Big), \quad
    \lambda^{-\ell}\Delta U(\lambda z)\to
    |z|^{\ell}
    \sum_{m=1}^{M_\ell}\alpha'_{\ell,m}Y_{\ell,m}\Big(\frac
    z{|z|}\Big),$$ strongly in $H^1(B_1^+)$, where $\sum_{m=1}^{M_\ell}((\alpha_{\ell,m})^2+(\alpha'_{\ell,m})^2)\neq0$ and $\alpha_{\ell,m},\alpha'_{\ell,m}$ are given in – with $V=\Delta U$.

2.  If $U\in \mathcal D$ is a weak solution to such that $$U(z)=o(|z|^n)\quad\text{as }|z|\to 0\text{ for all $n\in
      {{\mathbb N}}$},$$ then $U\equiv 0$ in $B_R^+$.

As mentioned above, a motivation for the study of higher order equations of type and consequently of systems comes from the interest in higher order fractional laplacians and their characterization as a Dirichlet-to-Neumann map in the spirit of [@CS].

Let us consider the fractional laplacian $(-\Delta)^{3/2}$ defined as $$\widehat{(-\Delta)^{\frac32} u}(\xi)=|\xi|^{3}\widehat u(\xi) \, .$$ We also consider the space $\mathcal D^{1/2,2}({{\mathbb R}}^N)$ given by the completion of $C^\infty_c({{\mathbb R}}^N)$ with respect to the scalar product $$(u,v)_{\mathcal D^{1/2,2}({{\mathbb R}}^N)}:=\int_{{{\mathbb R}}^N} |\xi|\, \widehat
u(\xi)\overline{\widehat v(\xi)} \, d\xi.$$

\[t:frac\] For $N>3$, let $\Omega\subseteq{{\mathbb R}}^N$ be open, $a\in C^1(\Omega)$, and $u\in \mathcal D^{3/2,2}({{\mathbb R}}^N)$ be a weak solution to the problem $$\label{eq:48}
(-\Delta)^{3/2}u=a\,u,\quad\text{in }\Omega,$$ i.e. $$(u,\varphi)_{\mathcal
  D^{3/2,2}({{\mathbb R}}^N)}=\int_{\Omega}a\,u\,\varphi\,dx\quad\text{for all }
\varphi\in C^\infty_c(\Omega).$$ Let us also assume that $$\label{eq:53}
  (-\Delta)^{3/2}u\in (\mathcal D^{1/2,2}({{\mathbb R}}^N))^\star,$$ where $(\mathcal D^{1/2,2}({{\mathbb R}}^N))^\star$ denotes the dual space of $\mathcal D^{1/2,2}({{\mathbb R}}^N)$, in the sense that the linear functional $\varphi\mapsto \int_{{{\mathbb R}}^N} |\xi|^{3} \widehat
  u(\xi)\overline{\widehat \varphi(\xi)} \, d\xi$, $\varphi\in C^\infty_c({{\mathbb R}}^N)$, is continuous with respect to the norm induced by $\mathcal D^{1/2,2}({{\mathbb R}}^N)$.

1.  If $u$ vanishes at some point $x_0\in\Omega$ of infinite order, i.e. if $$\label{eq:47}
    u(x)=o(|x-x_0|^n)\quad\text{as }x\to x_0\text{ for every }n\in{{\mathbb N}},$$ then $u\equiv 0$ in $\Omega$.

2.  If $u$ vanishes on a set $E\subset\Omega$ of positive Lebesgue measure, then $u\equiv 0$ in $\Omega$.

We observe that assumption is satisfied in each of the following cases:

(i) $u\in \mathcal D^{5/2,2}({{\mathbb R}}^N)$;

(ii) $u\in \mathcal D^{3/2,2}({{\mathbb R}}^N)$ solves with $\Omega={{\mathbb R}}^N$ and $a\in L^{N/2}({{\mathbb R}}^N)\cap C^1({{\mathbb R}}^N)$.

The proof of Theorem \[t:frac\] is based on Theorem \[t:asym-ho\] and the generalization of the Caffarelli-Silvestre extension to higher order fractional laplacians given in [@Y], see also [@FF-prep]. Indeed, according to [@Y], we have that if $u$ solves , then $u$ is the trace on ${{\mathbb R}}^N\times\{0\}$ of some $U\in\mathcal D$ solving with $h=-2a$.

We observe that the unique continuation result stated in Theorem \[t:frac\] does not overlap with the results in [@seo0; @seo1; @seo2]. Indeed, from one hand [@seo0; @seo1; @seo2] consider more general potentials; on the other hand we obtain here a *strong* unique continuation and a unique continuation from sets of positive measure, which are stronger results than the weak unique continuation obtained in [@seo0; @seo1; @seo2]. We also observe that we assume that equation is satisfied only on the set $\Omega$ and not in the whole ${{\mathbb R}}^N$.

The paper is organized as follows. In section \[sec:monot-argum\] we develop the monotonicity argument, proving in particular the existence of a finite limit for the frequency function as $r\to0^+$. In section \[sec:blow-up-analysis\] we carry out a careful blow-up analysis for scaled solutions, which allows proving Theorem \[t:asym\] and, as a consequence, Theorem \[t:sun-ext\]. Finally section \[sec:appl-fourth-order\] is devoted to applications of Theorem \[t:asym\] to fourth order problems and higher order fractional problems , with the proofs of Theorems \[t:asym-ho\] and \[t:frac\].

The monotonicity argument {#sec:monot-argum}
=========================

For all $r\in(0,R)$ we define the functions $$\label{eq:D(r)}
D(r)=r^{-N+1} \left[\int_{B_r^+} \left(|\nabla U|^2+|\nabla
    V|^2+UV\right)dz
-\int_{B_r'} h(x)u(x)v(x)\, dx  \right]$$ and $$\label{eq:H(r)}
H(r)=r^{-N}\int_{S_r^+} (U^2+V^2)\, dS.$$ We define the space $\mathcal
D^{1,2}({{\mathbb R}}^{N+1}_+)$ as the completion of the space $C^\infty_c(\overline{{{\mathbb R}}^{N+1}_+})$ with respect to the norm $$\|U\|_{\mathcal D^{1,2}({{\mathbb R}}^{N+1}_+)}:=\left(\int_{{{\mathbb R}}^{N+1}_+}
|\nabla U|^2 dz\right)^{1/2} \qquad \text{for any } U\in
C^\infty_c(\overline{{{\mathbb R}}^{N+1}_+}) \, .$$ From [@BCdPS], we have that there exists a constant $K>0$ such that $$\label{eq:C-b}
K \|{\rm Tr\, }U\|_{\mathcal D^{\frac12,2}({{\mathbb R}}^N)}{\leqslant}\|U\|_{\mathcal
D^{1,2}({{\mathbb R}}^{N+1}_+)} \qquad \text{for any } U\in \mathcal
D^{1,2}({{\mathbb R}}^{N+1}_+)  .$$ Here we are denoting as ${\rm Tr\, }$ the trace operator ${\rm Tr\, }:\mathcal D^{1,2}({{\mathbb R}}^{N+1}_+)\to \mathcal D^{\frac12,2}({{\mathbb R}}^N)$. We recall that, for all $\gamma<\frac N2$ the following Sobolev embedding holds: there exists a positive constant $S(N,\gamma)$ depending only on $N$ and $\gamma$ such that $$\label{eq:Sobolev-frazionario}
S(N,\gamma) \|u\|_{L^{2^*(N,\gamma)}({{\mathbb R}}^N)}^2{\leqslant}\|u\|_{\mathcal
D^{\gamma,2}({{\mathbb R}}^{N})}^2 \qquad \text{for any } u\in
C^\infty_c({{\mathbb R}}^N)$$ where $2^*(N,\gamma)=2N/(N-2\gamma)$. Moreover the following Hardy type inequality due to Herbst [@herbst] holds: there exists $\Lambda>0$ $$\label{eq:frac_hardy}
  \Lambda\int_{{{\mathbb R}}^N}\frac{\varphi^2(x)}{|x|}\,dx\leq\|\varphi\|_{\mathcal D^{1/2,2}({{\mathbb R}}^N)}^2
  ,\quad\text{for all }\varphi\in \mathcal D^{1/2,2}({{\mathbb R}}^N).$$ Combining and we obtain that $$\label{eq:Sobolev-frazionario-bis}
S\big(N,\tfrac12\big)K^2 \, \|{\rm Tr\, }U\|_{L^{\frac{2N}{N-1}}({{\mathbb R}}^N)}^2\le
\|U\|_{\mathcal D^{1,2}({{\mathbb R}}^{N+1}_+)}^2 \qquad \text{for any }
U\in \mathcal D^{1,2}({{\mathbb R}}^{N+1}_+) .$$ Similarly, combining with , we infer $$\label{eq:Hardy-frazionario}
\Lambda K^2 \int_{{{\mathbb R}}^N} \frac{|{\rm
Tr\,}U|^2}{|x|} \, dx {\leqslant}\|U\|_{\mathcal D^{1,2}({{\mathbb R}}^{N+1}_+)}^2  \qquad \text{for any } U\in \mathcal
D^{1,2}({{\mathbb R}}^{N+1}_+) .$$ We recall the following lemmas from [@FF], which provide Sobolev and Hardy type trace inequalities with boundary terms in $N+1$-dimensional half-balls.

\[l:sobolev\] For any $r>0$ and any $U\in H^1(B_r^+)$ we have $$\widetilde S \left(\int_{B_r'}
|u|^{\frac{2N}{N-1}}dx\right)^{\!\!\frac{N-1}{N}} {\leqslant}\int_{B_r^+}
 |\nabla U|^2 dz+\frac{N-1}{2r} \int_{S_r^+}  U^2 dS$$ where $u={\rm Tr\, }U$ and $\widetilde S$ is a positive constant depending only on $N$.

\[l:hardyball\] For any $r>0$ and any $U\in
  H^1(B_r^+)$ we have that $$\widetilde\Lambda \int_{B_r'} \frac{u^2}{|x|} \,
dx{\leqslant}\int_{B_r^+} |\nabla U|^2 dz+\frac{N-1}{2r}
\int_{S_r^+} U^2 dS$$ where $u={\rm Tr\,} U$ and $\widetilde \Lambda$ is a positive constant depending only on $N$.

The following Poincaré type inequality on half-balls will be useful in the sequel.

\[l:Poincare\] For every $r>0$ and $W\in H^1(B_r^+)$ we have that $$\frac{N}{r^2}\int_{B_r^+}W^2(z)\,dz{\leqslant}\frac{1}{r}\int_{S_r^+}W^2(z)\,dS+\int_{B_r^+}|\nabla
W(z)|^2\,dz.$$

From the Divergence Theorem we have that $$\begin{aligned}
(N+1)\int_{B_r^+}W^2(z)\,dz&=
\int_{B_r^+}\left(\dive(W^2z)-2W\nabla W\cdot z\right)\,dz\\
&=r\int_{S_r^+}W^2(z)\,dS-2 \int_{B_r^+}W\nabla W\cdot z\,dz\\
&{\leqslant}r\int_{S_r^+}W^2(z)\,dS+\int_{B_r^+}W^2(z)\,dz+r^2 \int_{B_r^+}|\nabla W|^2\,dz\end{aligned}$$ thus yielding the stated inequality.

The following lemma contains a Pohozaev type identity for solutions to system .

Let $(U,V)\in H^1(B_R^+)\times H^1(B_R^+)$ be a weak solution to . Then for a.e. $r\in(0,R)$ $$\label{Pohozaev-1}
\int_{B_r^+} \left(|\nabla U|^2+|\nabla
V|^2+UV\right)dz=\int_{S_r^+} \left(\frac{\partial U}{\partial
\nu} U+\frac{\partial V}{\partial \nu}V\right)dS  +\int_{B_r'}
h(x)u(x)v(x)\, dx$$ and $$\begin{aligned}
 \label{Pohozaev-2}
-\frac{N-1}2 & \int_{B_r^+}  \left(|\nabla U|^2+|\nabla
V|^2\right)dz+\int_{B_r^+}  V(z\cdot \nabla U)\, dz +\frac r2
\int_{S_r^+}  \left(|\nabla U|^2+|\nabla V|^2\right)dS \\
\notag & =\int_{B_r'} h(x)u(x)(x\cdot \nabla_x v)\,
dx+ r\int_{S_r^+}  \left(\left|\frac{\partial
U}{\partial\nu}\right|^2+\left|\frac{\partial
V}{\partial\nu}\right|^2\right)dS\end{aligned}$$ where $u(x):=U(x,0)$ and $v(x)=V(x,0)$.

Identity follows by testing the equation for $U$ with $U$ and the equation for $V$ with $V$ and by integrating by parts over $B_r^+$.

To prove we first observe that $U,V\in
H^2(B_r^+)$ for all $r\in(0,R)$. Indeed, since $\frac{\partial U}{\partial \nu}= 0$ on $B_R'$, the function $$\widetilde U(x,t)=
\begin{cases}
  U(x,t),&\text{if }t>0,\\
  U(x,-t),&\text{if }t<0,
\end{cases}$$ satisfies the equation $\Delta \widetilde U=\widetilde V$, where $\widetilde V(x,t)=
  V(x,t)$ if $t>0$ and $\widetilde V(x,t)=V(x,-t)$ if $t<0$. Since $\widetilde V\in L^2(B_R)$, by classical elliptic regularity we have that $\widetilde U\in H^2(B_r)$ and hence $U\in
H^2(B_r^+)$ for all $r\in(0,R)$. By the Gagliardo Trace Theorem we have that $u=\mathop{\rm Tr}U\in H^{1/2}(B_r')$ for all $r\in(0,R)$ . Since $h\in
C^1(B_R')$ we have that $hu \in  H^{1/2}(B_r')$ for all $r\in(0,R)$. Therefore, for all $r\in(0,R)$, $V$ satisfies $$\begin{cases}
  \Delta V=0,&\text{in }B_r^+,\\
\frac{\partial V}{\partial \nu}\in H^{1/2}(B_r').&
\end{cases}$$ From elliptic regularity under Neumann boundary conditions (see in particular [@Salsa Theorem 8.13]) we conclude that $V\in H^2(B_r^+)$ for all $r\in(0,R)$.

Since, for every $r\in(0,R)$, $U,V\in
H^2(B_r^+)$, we can test the equation for $U$ with $\nabla U\cdot z$ (which belongs to $H^1(B_r^+)$) and the equation for $V$ with $\nabla V\cdot z$ (which belongs to $H^1(B_r^+)$), thus obtaining .

\[l:Hpos\] Let $(U,V)\in H^1(B_R^+)\times H^1(B_R^+)$ be a weak solution to such that $(U,V)\not=(0,0)$ (i.e. $U$ and $V$ are not both identically null). Let $D=D(r)$ and $H=H(r)$ be the functions defined in and . Then there exists $r_0\in(0,R)$ such that $H(r)>0$ for any $r\in (0,r_0)$.

Suppose by contradiction that for any $r_0>0$ there exists $r\in (0,r_0)$ such that $H(r)=0$. Then there exists a sequence $r_n\to 0^+$ such that $H(r_n)=0$, i.e. $U=V=0$ on $S_{r_n}^+$. From it follows that $$\label{eq:2}
\int_{B_{r_n}^+} \left(|\nabla U|^2+|\nabla
V|^2+UV\right)dz=\int_{B_{r_n'}}
h(x)u(x)v(x)\, dx.$$ From , Lemma \[l:Poincare\], and Lemma \[l:hardyball\] it follows that $$\begin{aligned}
\left(1-\frac{r_n^2}{2N}\right)& \int_{B_{r_n}^+} \left(|\nabla U|^2+|\nabla
V|^2\right)dz{\leqslant}\int_{B_{r_n}^+} \left(|\nabla U|^2+|\nabla
V|^2+UV\right)dz\\
&=\int_{B_{r_n'}}h(x)u(x)v(x)\, dx{\leqslant}{\rm
  const\,}r_n\left(\int_{B_{r_n}'}\frac{u^2}{|x|}\,dx+
\int_{B_{r_n}'}\frac{v^2}{|x|}\,dx\right)
\\
&{\leqslant}{\rm
  const\,}r_n
\int_{B_{r_n}^+} \left(|\nabla U|^2+|\nabla
V|^2\right)dz.\end{aligned}$$ Since $r_n\to0^+$ as $n\to+\infty$, the above inequality implies that $\int_{B_{r_n}^+} \left(|\nabla U|^2+|\nabla
V|^2\right)dz=0$ for $n$ sufficiently large. Hence, in view of Lemma \[l:Poincare\], $U\equiv V\equiv 0$ in $B_{r_n}^+$. Classical unique continuation principles then imply that $U\equiv V\equiv 0$ in $B_R^+$ giving rise to a contradiction.

\[l:Dgeq\] Letting $(U,V)\in H^1(B_R^+)\times H^1(B_R^+)$ be as in Lemma \[l:Hpos\] and $D,H$ as in –, there holds $$\begin{aligned}
\label{eq:5}
&D(r){\geqslant}r^{1-N} \left(\int_{B_{r}^+} \left(|\nabla U|^2+|\nabla
V|^2\right)dz\right)(1+O(r))-H(r)O(r),\\
\label{eq:6}
&D(r){\geqslant}Nr^{-1-N} \left(\int_{B_{r}^+} (U^2+V^2)dz\right)(1+O(r))-H(r)O(1),\end{aligned}$$ as $r\to0^+$.

From Lemma \[l:Poincare\] we have that $$\label{eq:4}
\int_{B_r^+}(U^2+V^2)\,dz{\leqslant}\frac{r^{1+N}}{N}H(r)+\frac{r^2}{N}\int_{B_{r}^+} \left(|\nabla U|^2+|\nabla
V|^2\right)dz.$$ From it follows that $$\label{eq:1}
\left|\int_{B_r^+}UV\,dz\right|{\leqslant}\frac{r^{1+N}}{2N}H(r)+\frac{r^2}{2N}\int_{B_{r}^+} \left(|\nabla U|^2+|\nabla
V|^2\right)dz$$ whereas Lemma \[l:hardyball\] implies that, for all $r\in(0,r_0)$, $$\begin{aligned}
\label{eq:3}
\left|\int_{B_r'}huv\,dx\right|&{\leqslant}\|h\|_{L^{\infty}(B_{r_0}')}\frac{r}2\int_{B_r'}
 \frac{u^2+v^2}{|x|} \,dx\\
\notag&{\leqslant}\|h\|_{L^{\infty}(B_{r_0}')}\frac{r}{2\widetilde\Lambda}\left(\int_{B_{r}^+} \left(|\nabla U|^2+|\nabla
V|^2\right)dz\right)
+ \|h\|_{L^{\infty}(B_{r_0}')}\frac{N-1}{4\widetilde\Lambda}r^NH(r).\end{aligned}$$ From and it follows that $$\begin{aligned}
  D(r)&{\geqslant}r^{1-N} \left(\int_{B_r^+}(|\nabla U|^2+|\nabla
V|^2)dz\right)\left(1-\frac{r^2}{2N}-
        \|h\|_{L^{\infty}(B_{r_0}')}\frac{r}{2\widetilde\Lambda}\right)\\
&\qquad -rH(r)\left(\frac{r}{2N}+\|h\|_{L^{\infty}(B_{r_0}')}\frac{N-1}{4\widetilde\Lambda}\right).\end{aligned}$$ The proof of is thereby complete. Estimate follows by combination of and .

\[rem:st\] We observe that estimates and can be rewritten as $$\begin{aligned}
\label{eq:8}
&\int_{B_{r}^+} \left(|\nabla U|^2+|\nabla
V|^2\right)dz{\leqslant}D(r) r^{N-1}(1+O(r))+H(r)O(r^N),\\
\label{eq:9}
&\int_{B_{r}^+} (U^2+V^2)dz{\leqslant}\frac1N r^{N+1}D(r) (1+O(r))+H(r)O(r^{N+1}),\end{aligned}$$ as $r\to0^+$.

\[l:hprime\] We have that $H\in
  W^{1,1}_{{\rm loc}}(0,R)$ and $$\begin{aligned}
\label{H'}
 & H'(r)=2r^{-N} \int_{S_r^+}\big(U\tfrac{\partial
    U}{\partial\nu}+V\tfrac{\partial
    V}{\partial\nu}\big)
  \, dS, \quad \text{in a distributional sense and for a.e. $r\in (0,R)$},\\
 &\label{H'2} H'(r)=\frac2r D(r), \quad \text{for every } r\in (0, R).\end{aligned}$$

See the proof of [@FF Lemma 3.8].

\[l:dprime\] The function $D$ defined in belongs to $W^{1,1}_{{\rm\
      loc}}(0, R)$ and $$\begin{aligned}
\label{D'F}
  D'(r)=&\frac{2}{r^{N-1}} \int_{S_r^+}
          \left(\left|\frac{\partial U}{\partial \nu}\right|^2+
          \left|\frac{\partial V}{\partial \nu}\right|^2\right) dS
+\frac{1}{r^{N-1}} \int_{S_r^+}  UV\,dS\\
\notag&-\frac{2}{r^{N}}\int_{B_r^+}V\,\nabla U\cdot
  z\,dz-\frac{N-1}{r^{N}}\int_{B_r^+} UV\,dz\\
\notag&+\frac{N-1}{r^{N}}\int_{B_r'}huv-\frac{1}{r^{N-1}}\int_{\partial
  B_r'}huv\,dS'+2\frac1{r^{N}}\int_{B_r'}hu(x\cdot\nabla_x v)\,dx\end{aligned}$$ in a distributional sense and for a.e. $r\in (0,R)$.

For any $r\in (0,R)$ let $$\begin{aligned}
\label{I(r)}
I(r)= \int_{B_r^+} \left(|\nabla U|^2+|\nabla V|^2+UV\right)dz-
\int_{B_r'} h(x)u(x)v(x)\, dx .\end{aligned}$$ From the fact that $U,V\in  H^1(B_R^+)$ and Lemma \[l:hardyball\] it follows that $I\in
 W^{1,1}(0,R)$ and $$\label{I'(r)}
I'(r) =
\int_{S_r^+} \left(|\nabla U|^2+|\nabla V|^2+UV\right)\,dS-
\int_{\partial B_r'} h(x)u(x)v(x)\, dS'$$ for a.e. $r\in (0,R)$ and in the distributional sense. Therefore $D\in W^{1,1}_{{\rm loc}}(0,R)$ and, replacing , (\[I(r)\]), and (\[I’(r)\]) into $D'(r)=r^{-N}[-(N-1)I(r)+rI'(r)]$, we obtain (\[D’F\]).

In view of Lemma \[l:Hpos\], the function $$\label{eq:7}
\mathcal N:(0,r_0)\to{{\mathbb R}},\quad
\mathcal N(r)=\frac{D(r)}{H(r)}$$ is well defined. As a consequence of estimate we obtain the following corollary.

\[c:limpos\] Let $(U,V)\in H^1(B_R^+)\times H^1(B_R^+)$ be as in Lemma \[l:Hpos\] and let $D,H,\mathcal N$ be defined in , , and respectively. For every $\e>0$ there exists $r_\e>0$ such that $$\mathcal N(r)+\e{\geqslant}0\quad\text{for all }0<r<r_\e,$$ i.e. $$\label{eq:liminf}
\liminf_{r\to0^+}\mathcal N(r)\geq0.$$

\[mono\] The function ${\mathcal N}$ defined in belongs to $W^{1,1}_{{\rm
       loc}}(0, r_0)$ and $$\begin{aligned}
\label{formulona}
{\mathcal N}'(r)=\nu_1(r)+\nu_2(r)\end{aligned}$$ in a distributional sense and for a.e. $r\in (0,r_0)$, where $$\begin{aligned}
\nu_1(r)=\frac{2r\Big[
    \left(\int_{S_r^+}
  \left(\left|\frac{\partial U}{\partial \nu}\right|^2+
\left|\frac{\partial V}{\partial \nu}\right|^2\right) dS\right) \cdot
    \left(\int_{S_r^+}
  (U^2+V^2)\,dS\right)-\left(
\int_{S_r^+}
  \left(U\frac{\partial U}{\partial \nu}+V\frac{\partial V}{\partial \nu}\right)\, dS\right)^{\!2} \Big]}
{\left(
\int_{S_r^+}
  (U^2+V^2)\,dS\right)^2}\end{aligned}$$ and $$\begin{aligned}
\label{eq:16}
  \nu_2(r)= &\, \frac{r\int_{S_r^+}  UV\,dS-2\int_{B_r^+} V\,\nabla
              U\cdot z\,dz-(N-1)\int_{B_r^+} UV\,dz}{\int_{S_r^+}
              (U^2+V^2)\,dS}\\
\notag&\quad+ \frac{(N-1)\int_{ B_r'}huv\, dx-r\int_{\partial B_r'}huv\,dS'+2\int_{B_r'}hu\,x\cdot
  \nabla_x v\,dx} {\int_{S_r^+}
    (U^2+V^2)\,dS}.\end{aligned}$$

It follows directly from the definition of $\mathcal N$ and Lemmas \[l:hprime\] and \[l:dprime\].

We now estimate the term $\nu_2$ in . This is the most delicate point in the development of the monotonicity argument for system , due to the presence of the integral over “the boundary of the boundary” $\int_{\partial B_r'}huv\,dS'$ in the term $\nu_2$.

\[l:STIMA\_DI\_nu\_2\] Let $\nu_2$ be as in . Then $$\nu_2(r)=O\left(1+\mathcal
  N(r)+r\sqrt{\frac{B(r)}{r^NH(r)}}\right)\quad\text{as }r\to 0^+,$$ where $$\label{eq:B(r)}
B(r)=\int_{S_r^+}(|\nabla U|^2+|\nabla V|^2)\,dS.$$

We observe that $$\label{eq:10}
\frac{r\left|\int_{S_r^+}UV\,dS\right|}{\int_{S_r^+}(U^2+V^2)\,dS}=O(r)\quad\text{as
}r\to0^+.$$ From and we have that $$\begin{aligned}
\label{eq:11}
\frac{\left|\int_{B_r^+} V\,\nabla
              U\cdot z\,dz\right|}{\int_{S_r^+}(U^2+V^2)\,dS}&{\leqslant}\frac{1}{2r^NH(r)}\left(\int_{B_r^+} V^2\,dz+
r^2\int_{B_r^+} |\nabla U|^2\,dz\right)\\
&\notag\leq
\frac{N+1}{2N}\mathcal N(r)r(1+O(r))+O(r)
{\leqslant}\mathcal N(r)r(1+O(r))+O(r)\end{aligned}$$ and $$\label{eq:12}
\frac{\left|\int_{B_r^+} UV\,dz\right|}{\int_{S_r^+}
              (U^2+V^2)\,dS}{\leqslant}\frac{r}{2N}\mathcal N(r)(1+O(r)) +O(r)$$ as $r\to0^+$. From and we have that $$\label{eq:13}
  \frac{\left|\int_{ B_r'}huv\, dx\right|}
  {\int_{S_r^+}(U^2+V^2)\,dS}\leq
  \frac{\|h\|_{L^{\infty}(B_{r_0}')}}{2\widetilde \Lambda}\mathcal N(r)(1+O(r))+O(1)=\mathcal N(r)O(1)+O(1)$$ as $r\to0^+$.

Integration by parts yields $$\int_{ B_r'}hu\,x\cdot\nabla_xv\, dx=r \int_{\partial  B_r'}huv\, dS'
-\int_{ B_r'}v(Nhu+u\nabla h\cdot x+hx\cdot\nabla_xu)\, dx$$ so that $$\begin{gathered}
\label{eq:14}
-r\int_{\partial B_r'}huv\,dS'+2\int_{B_r'}hu\,x\cdot
  \nabla_x v\,dx\\
=\int_{B_r'}hu\,x\cdot
  \nabla_x v\,dx-\int_{B_r'}hv\,x\cdot
  \nabla_x u\,dx-\int_{B_r'}uv(Nh+x\cdot
  \nabla h)\,dx.\end{gathered}$$ From Lemma \[l:hardyball\] and we have that $$\begin{aligned}
\label{eq:15}
  \frac{\left|\int_{B_r'}uv(Nh+x\cdot
  \nabla h)\,dx \right|}
  {\int_{S_r^+}(U^2+V^2)\,dS}&\leq
  \frac{\|Nh+x\cdot
  \nabla_x h\|_{L^{\infty}(B_{r_0}')}}{2\widetilde \Lambda}\mathcal
                               N(r)(1+O(r))+O(1)\\
&\notag=\mathcal N(r)O(1)+O(1)\end{aligned}$$ as $r\to0^+$.

On the other hand, by the Divergence Theorem we have that $$\begin{aligned}
\label{eq:18}
\int_{B_r'}hu\,x&\cdot
  \nabla_x v\,dx= -\int_{B_r'}hu\,(x\cdot
  \nabla_x v){\mathbf e}_{N+1}\cdot \nu\,dx \\
\notag&=\int_{S_r^+}h(x)U(x,t)\,(z\cdot
  \nabla V){\mathbf e}_{N+1}\cdot \nu\,dS-
\int_{B_r^+}\frac{\partial}{\partial t}\left[ h(x)U(x,t)\,(z\cdot
  \nabla V)\right]\,dz \\
\notag&=\int_{S_r^+}h(x)U(x,t)\,(z\cdot
  \nabla V){\mathbf e}_{N+1}\cdot \nu\,dS-
\int_{B_r^+}h(x)U_t\,(z\cdot
  \nabla V)\,dz\\
\notag&\qquad-
\int_{B_r^+}h(x)U\,(V_t+z\cdot
  \nabla V_t)\,dz\\
\notag&=\int_{S_r^+}h(x)U(x,t)\,(z\cdot
  \nabla V){\mathbf e}_{N+1}\cdot \nu\,dS-
\int_{B_r^+}h(x)U_t\,(z\cdot
  \nabla V)\,dz\\
\notag&\qquad-
r \int_{S_r^+}h(x)UV_t\,dS
+
\int_{B_r^+}(Nh(x)+\nabla h\cdot x )UV_t\,dz
+
\int_{B_r^+}
hV_t(\nabla U\cdot z)\,dz.\end{aligned}$$ Hence, taking into account Lemma \[l:Poincare\], $$\begin{aligned}
 \label{eq:*}
  \left|\int_{B_r'}\!\!hu\,x\cdot
  \nabla_x v\,dx\right|\!\leq\!{\rm const\,} \!\!\left(
r\sqrt{r^N H(r)B(r)}
+r\!\!\int_{B_r^+}(|\nabla U|^2+|\nabla V|^2)\,dz\!+\!
\int_{S_r^+}(U^2+V^2)\,dS
\right)\end{aligned}$$ for some ${\rm const\,>}0$ independent of $r$. In a similar way we obtain that $$\begin{aligned}
  \left|\int_{B_r'}hv\,x\cdot
  \nabla_x u\,dx\right|\leq{\rm const\,} \left(
r\sqrt{r^N H(r)B(r)}
+r\int_{B_r^+}(|\nabla U|^2+|\nabla V|^2)\,dz+
\int_{S_r^+}(U^2+V^2)\,dS
\right).\end{aligned}$$ As a consequence, in view of we conclude that $$\label{eq:17}
\frac{\left|-r\int_{\partial B_r'}huv\,dS'+2\int_{B_r'}hu\,x\cdot
  \nabla_x v\,dx\right|} {\int_{S_r^+}
    (U^2+V^2)\,dS}{\leqslant}\mathcal N(r)O(1)+\sqrt{\frac{B(r)}{r^NH(r)}}\,O(r)+O(1)$$ as $r\to0^+$.

Inserting - into the proof of the lemma follows.

Inspired by [@FF-edinb Lemma 5.9], in the following lemma we estimate $B$ in terms of the derivative $D'$.

\[l:stima\_B\] Let $B$ be defined in . Then there exist $C_1,C_2,\bar
r>0$ such that $$B(r){\leqslant}2 r^{N-1}D'(r)+C_1 r^{N-2} (D(r)+C_2 H(r)) \quad\text{and}\quad
D(r)+C_2H(r){\geqslant}0\quad\text{for all }r\in
(0,\bar r).$$

From the definition of $D$ (see ) we have that $$\label{eq:21}
  D'(r)=r^{1-N}B(r)-(N-1)r^{-1}D(r)+r^{1-N}\int_{S_r^+}UV\,dS-r^{1-N}\int_{\partial B_r'}huv\,dS'.$$ From it follows that $$\int_{\partial B_r'}huv\,dS'=\frac1r\int_{B_r'}hu\,x\cdot
  \nabla_x v\,dx+\frac1r \int_{B_r'}hv\,x\cdot
  \nabla_x u\,dx+\frac1r \int_{B_r'}uv(Nh+x\cdot
  \nabla h)\,dx.$$ By and we deduce that, for every $\e>0$, there exists $C_\e>0$ such that $$\begin{aligned}
\bigg|\frac1r\int_{B_r'}&hu\,x\cdot
  \nabla_x v\,dx\bigg|\\
&{\leqslant}{\varepsilon}B(r)+C_{\varepsilon}r^N H(r)+O(1) \int_{B_{r}^+} \left(|\nabla U|^2+|\nabla
V|^2\right)dz+O(1)r^{N-1}H(r)\\
&{\leqslant}{\varepsilon}B(r)+O(1)r^{N-1}H(r)+O(1)r^{N-1}D(r)\quad\text{as }r\to0^+.\end{aligned}$$ An analogous estimate holds for the term $\frac1r \int_{B_r'}hv\,x\cdot
  \nabla_x u\,dx$, whereas implies that $$\frac1r \int_{B_r'}uv(Nh+x\cdot
  \nabla h)\,dx =O(1)r^{N-1}H(r)+O(1)r^{N-1}D(r) \quad\text{as }r\to0^+.$$ Therefore we conclude that $$\label{eq:20}
  \left|\int_{\partial B_r'}huv\,dS'\right|{\leqslant}2{\varepsilon}B(r)+O(1)r^{N-1}H(r)+O(1)r^{N-1}D(r)\quad\text{as }r\to0^+.$$ From Corollary \[c:limpos\], and , choosing $\e=\frac14$, we deduce that, for some constants $C_1,C_2>0$ independent of $r$, $D(r)+C_2H(r)\geq0$ and $$D'(r){\geqslant}\frac12
r^{1-N}B(r)-\frac{C_1}2r^{-1}(D(r)+C_2H(r))\quad\text{for all $r$
  sufficiently small}.$$ The proof is thereby complete.

\[l:limitN\] Let $\mathcal N:(0,r_0)\to{{\mathbb R}}$ be defined in . Then $$\label{eq:23}
  \mathcal N(r)=O(1)\quad\text{as }r\to0^+.$$ Furthermore the limit $$\gamma:=\lim_{r\to 0^+}\mathcal N(r)$$ exists, is finite and $$\gamma{\geqslant}0.$$

Let us consider the set $$\Sigma=\{r\in(0,r_0):D'(r)H(r){\leqslant}H'(r) D(r)\}$$ (which is well-defined up to a zero measure set).

If there exists $r\in(0,r_0]$ such that $|(0,r)\cap\Sigma|_1=0$ (where $|\cdot|_1$ stands for the Lebesgue measure in ${{\mathbb R}}$) we have that $\mathcal N'{\geqslant}0$ a.e. in $(0,r)$ and hence $\mathcal N$ is non-decreasing in $(0,r)$ and admits a limit as $r\to0^+$ which is necessarily finite and non-negative due to .

Let us now assume that, for all $r\in(0,r_0]$, $|(0,r)\cap\Sigma|_1>0$. In view of Lemma \[l:stima\_B\] and we have that, a.e. in $(0,r_0)\cap \Sigma$, $$\begin{aligned}
\label{eq:22}
B(r)&\leq
2 r^{N-1}\frac{H'(r)D(r)}{H(r)}+C_1 r^{N-2}  (D(r)+C_2 H(r))\\
\notag&=4 r^{N-2}\frac{D^2(r)}{H(r)}+C_1  r^{N-2}  (D(r)+C_2 H(r)).\end{aligned}$$ Schwarz inequality implies that the function $\nu_1$ appearing in Lemma \[mono\] is non-negative, hence , Lemma \[l:STIMA\_DI\_nu\_2\], and imply that $$\mathcal N'(r){\geqslant}O(1)\left(
1+\mathcal N(r)+\sqrt{4 \mathcal N^2(r)+C_1(\mathcal
  N(r)+C_2)}\right)$$ as $r\to0^+$, $r\in \Sigma$. Hence there exist $\tilde C,\tilde r>0$ such that $$\mathcal N'(r){\geqslant}-\tilde C\left(
1+\mathcal N(r)\right)\quad\text{for a.e. }r\in(0,\tilde r)\cap\Sigma.$$ Since the above inequality is obviously true in $(0,\tilde
r)\setminus\Sigma$ (provided $\tilde r$ is sufficiently small), we deduce that $$\label{eq:25}
\mathcal N'(r){\geqslant}-\tilde C\left(
1+\mathcal N(r)\right)\quad\text{for a.e. }r\in(0,\tilde r).$$ Integrating the above inequality in $(r,\tilde r)$ we obtain that $$\mathcal N(r)+1{\leqslant}e^{\tilde C \tilde r} (\mathcal N(\tilde r)+1) \quad\text{for all }r\in(0,\tilde r).$$ The above estimate together with Corollary \[c:limpos\] yield . Furthermore implies that $$\left(e^{\tilde C r}(1+\mathcal N(r))\right)'{\geqslant}0 \quad\text{a.e. in
}(0,\tilde r),$$ hence the function $r\mapsto e^{\tilde C r}(1+\mathcal N(r))$ admits a limit as $r\to0^+$. Therefore also the limit $\gamma:=\lim_{r\to 0^+}\mathcal N(r)$ exists; furthermore $\gamma$ is finite in view of and $\gamma\geq0$ in view of .

A first consequence of the previous monotonicity argument is the following estimate of the function $H$.

\[l:uppb\] Letting $\gamma$ be as in Lemma \[l:limitN\], we have that $$\label{eq:24}
H(r)=O(r^{2\gamma})\quad \text{as $r\to0^+$}.$$ Furthermore, for any $\sigma>0$ there exist $K(\sigma)>0$ depending on $\sigma$ such that $$\label{2ndest}
H(r){\geqslant}K(\sigma)\,
  r^{2\gamma+\sigma} \quad \text{for all } r\in (0, r_0).$$

See the proof of [@FF Lemma 3.16].

Blow-up analysis {#sec:blow-up-analysis}
================

\[l:blowup\] Let $(U,V)\in H^1(B_R^+)\times H^1(B_R^+)$ be a weak solution to such that $(U,V)\not=(0,0)$, let $\mathcal N$ be defined in , and let $\gamma:=\lim_{r\rightarrow 0^+} {\mathcal
    N}(r)$ be as in Lemma \[l:limitN\]. Then

-   there exists $\ell\in {{\mathbb N}}$ such that $\gamma=\ell$;

-   for every sequence $\lambda_n\to0^+$, there exist a subsequence $\{\lambda_{n_k}\}_{k\in{{\mathbb N}}}$ and $2M_\ell$ real constants $\beta_{\ell,m},\beta'_{\ell,m}$, $m=1,2,\dots,M_\ell$, such that $\sum_{m=1}^{M_\ell}((\beta_{\ell,m})^2+(\beta'_{\ell,m})^2)=1$ and $$\frac{U(\lambda_{n_k}z)}{\sqrt{H(\lambda_{n_k})}}\to
    |z|^{\ell}
    \sum_{m=1}^{M_\ell}\beta_{\ell,m}Y_{\ell,m}\Big(\frac
    z{|z|}\Big),\quad
    \frac{V(\lambda_{n_k}z)}{\sqrt{H(\lambda_{n_k})}}\to
    |z|^{\ell}
    \sum_{m=1}^{M_\ell}\beta'_{\ell,m}Y_{\ell,m}\Big(\frac
    z{|z|}\Big),$$ weakly in $H^1(B_1^+)$ and strongly in $H^1(B_r^+)$ for all $r\in(0,1)$. See Section \[s:introduction\] for the definition of $M_\ell$ and $Y_{\ell,m}$.

Let us define $$\label{eq:UVtau}
U_\lambda(z)=\frac{U(\lambda z)}{\sqrt{H(\lambda)}},
\quad
V_\lambda(z)=\frac{V(\lambda z)}{\sqrt{H(\lambda)}}.$$ We notice that $$\label{eq:26}
\Delta U_\lambda=\lambda^2V_\lambda\quad\text{and}\quad \int_{S_1^+}(U_{\lambda}^2+V_{\lambda}^2)dS=1.$$ By scaling and we have $$\begin{gathered}
\label{eq:8bis}
  \int_{B_{1}^+} \left(|\nabla U_\lambda(z)|^2 +|\nabla V_\lambda(z)|^2+\lambda^2 U_\lambda(z) V_\lambda(z)\right)\,dz -
  \lambda
\int_{B_1'}h(\lambda
x)U_\lambda(x,0) V_\lambda(x,0)\,dx
  \\={\mathcal N}(\lambda)=O(1)\end{gathered}$$ as $\lambda\to0^+$. On the other hand, Lemmas \[l:hardyball\] and \[l:Poincare\] imply $$\begin{gathered}
  \mathcal N(\lambda)\geq
\left(\int_{B_{1}^+} \left(|\nabla U_\lambda(z)|^2 +|\nabla V_\lambda(z)|^2\right)\,dz
\right)\left(1-\frac{\lambda^2}{2N}-
\frac{\lambda\|h\|_{L^{\infty}(B_{r_0}')}}{2\widetilde \Lambda}\right)\\
-\frac{\lambda^2}{2N}-\frac{\lambda\|h\|_{L^{\infty}(B_{r_0}')}(N-1)}{4\widetilde \Lambda}\end{gathered}$$ so that and Lemma \[l:Poincare\] imply that $$\{U_\lambda\}_{\lambda\in(0,\tilde \lambda)}
\text{ and }\{V_\lambda\}_{\lambda\in(0,\tilde \lambda)}
\text{ are bounded in }H^1(B_1^+)$$ for some $\tilde\lambda>0$.

Therefore, for any given sequence $\lambda_n\to 0^+$, there exists a subsequence $\lambda_{n_k}\to0^+$ such that $U_{\lambda _{n_k}}{\rightharpoonup}\widetilde U$ and $V_{\lambda _{n_k}}{\rightharpoonup}\widetilde V$ weakly in $H^1(B_1^+)$ for some $\widetilde U,\widetilde V\in H^1(B_1^+)$. From compactness of the trace embedding $H^1(B_1^+)\hookrightarrow L^2(S_1^+)$ and from we deduce that $$\label{eq:27}
\int_{S_1^+}(\widetilde U^2+\widetilde V^2)dS=1,$$ hence $(\widetilde U,\widetilde V)\neq(0,0)$, i.e. $\widetilde U$ and $\widetilde V$ can not both vanish identically. For every $\lambda\in (0,\tilde \lambda)$, the couple $(U_\lambda,V_\lambda)$ satisfies $$\label{eqlam}
\begin{cases}
   \Delta U_\lambda=\lambda^2 V_\lambda,&\text{in }B_1^+,\\
   \Delta V_\lambda=0,&\text{in }B_1^+,\\
\partial_\nu U_\lambda=0, &\text{on }B_1',\\
\partial_\nu V_\lambda=\lambda h(\lambda x)u_\lambda, &\text{on }B_1',
\end{cases}$$ in a weak sense, i.e. $$\label{eq:8tau}
\begin{cases}
\int_{B_1^+}\nabla
U_\lambda\cdot\nabla \varphi\, dz=-\lambda^2
\int_{B_1^+}V_\lambda\varphi\, dz \\
\int_{B_1^+}\nabla
V_\lambda\cdot\nabla \varphi\, dz=\lambda
\int_{B_1'}h(\lambda x)u_\lambda(x)\mathop{\rm Tr}\varphi(x) \,dx,
\end{cases}$$ for all $\varphi \in H^1(B_1^+)$ such that $\varphi=0\text{ on }S_1^+$, where $u_\lambda=\mathop{\rm Tr}U_\lambda$. From the weak convergences $U_{\lambda _{n_k}}{\rightharpoonup}\widetilde U$ and $V_{\lambda _{n_k}}{\rightharpoonup}\widetilde V$ in $H^1(B_1^+)$, we can pass to the limit in to obtain $$\begin{cases}
\int_{B_1^+}\nabla
\widetilde U\cdot\nabla \varphi\,dz=0,\\
\int_{B_1^+}\nabla \widetilde
V\cdot\nabla \varphi\,dz=0,
\end{cases}\quad \text{for all $\varphi \in H^1(B_1^+)$  such that
$\varphi=0\text{ on }S^1_+$},$$ i.e. $(\widetilde U,\widetilde V)$ weakly solves $$\label{eq:extended_limit}
\begin{cases}
   \Delta \widetilde U=0,&\text{in }B_1^+,\\
   \Delta \widetilde V=0,&\text{in }B_1^+,\\
\partial_\nu \widetilde U=0, &\text{on }B_1',\\
\partial_\nu \widetilde V=0, &\text{on }B_1'.
\end{cases}$$ From elliptic regularity under Neumann boundary conditions (see in particular [@Salsa Theorem 8.13]) we conclude that $$\label{eq:30}
  \{U_\lambda\}_{\lambda\in(0,\tilde \lambda)} \text{ and }
\{V_\lambda\}_{\lambda\in(0,\tilde \lambda)} \text{ are bounded in
$H^2(B_r^+)$ for all $r\in(0,1)$},$$ hence, by compactness, up to passing to a subsequence, $$\label{eq:strong_conv}
U_{\lambda _{n_k}}\to \widetilde U\ \text{and}\
V_{\lambda _{n_k}}\to\widetilde V\ \text{
 weakly in $H^2(B_r^+)$ and strongly in $H^1(B_r^+)$ for all $r\in(0,1)$}.$$ For any $r\in (0,1)$ and $k\in {{\mathbb N}}$, let us define the functions $$\begin{aligned}
&D_k(r)=r^{-N+1} \bigg[\int_{B_r^+} \left(|\nabla
  U_{\lambda_{n_k}}|^2+|\nabla V_{\lambda_{n_k}}|^2+\lambda_{n_k}^2 U_{\lambda_{n_k}}\!
  V_{\lambda_{n_k}}
\right)\,dz\\
&\hskip6cm-\lambda_{n_k}\int_{B_r'} h(\lambda_{n_k} x)u_{\lambda_{n_k}}\!(x)v_{\lambda_{n_k}}\!(x)\, dx  \bigg],\\
&H_k(r)=r^{-N}\int_{S_r^+} (U_{\lambda_{n_k}}^2+V_{\lambda_{n_k}}^2)\, dS,\end{aligned}$$ where we have set $v_\lambda=\mathop{\rm Tr}V_\lambda$. By direct calculations we have $$\label{NkNw}
{\mathcal
    N}_k(r):=\frac{D_k(r)}{H_k(r)}=\frac{D(\lambda_{n_k}r)}{H(\lambda_{n_k}r)}
  ={\mathcal N}(\lambda_{n_k}r) \quad \text{for all } r\in (0,1).$$ From it follows that, for any fixed $r\in (0,1)$, $$\label{convDk}
 D_k(r)\to \widetilde D(r)\quad\text{and}\quad H_k(r)\to \widetilde
 H(r)\quad\text{as }k\to+\infty$$ where $$\label{Dw(r)}
  \widetilde D (r)=
r^{-N+1} \int_{B_r^+} \left(|\nabla
  \widetilde U|^2+|\nabla \widetilde V|^2
\right)dz\quad\text{and}\quad
\widetilde H (r)=
r^{-N}\int_{S_r^+} (\widetilde U^2+\widetilde V^2)\, dS$$ for all $r\in(0,1)$. We observe that $\widetilde H (r)>0$ for all $r\in(0,1)$; indeed, if $\widetilde H (\bar r)=0$ for some $\bar r\in(0,1)$, the fact that $\widetilde
U,\widetilde V$ (and their even extension for $t<0$) are harmonic would imply that $\widetilde
U\equiv\widetilde V\equiv 0$ in $B_{\bar r}^+$, thus contradicting the classical unique continuation principle. Therefore the function $$\widetilde {\mathcal N}(r):=\frac{\widetilde  D(r)}{\widetilde  H(r)}$$ is well defined for $r\in (0,1)$. From (\[NkNw\]), (\[convDk\]), and Lemma \[l:limitN\], we deduce that $$\label{Nw(r)}
\widetilde {\mathcal    N}(r)=\lim_{k\to \infty} {\mathcal N}(\lambda_{n_k}r)=\gamma$$ for all $r\in (0,1)$. Therefore $\widetilde {\mathcal N}$ is constant in $(0,1)$ and hence $\widetilde {\mathcal
  N}'(r)=0$ for any $r\in (0,1)$. Arguing as in the proof of Lemma \[mono\] we can prove that $$\widetilde{\mathcal N}'(r)=
\frac{2r\Big[
    \Big(\int_{S_r^+}
  \Big(\Big|\frac{\partial \widetilde U}{\partial
        \nu}\Big|^2+\Big|\frac{\partial \widetilde V}
{\partial \nu}\Big|^2\Big) dS\Big) \cdot
    \left(\int_{S_r^+}
  (\widetilde U^2+\widetilde V^2)\,dS\right)-\left(
\int_{S_r^+}
  \left(\widetilde U\frac{\partial \widetilde U}{\partial
      \nu}+\widetilde V\frac{\partial \widetilde V}
{\partial \nu}\right)\, dS\right)^{\!2} \Big]}
{\left(
\int_{S_r^+}
  (\widetilde U^2+\widetilde V^2)\,dS\right)^2}$$ for all $r\in (0,1)$. Therefore for all $r\in (0,1)$ $$\Big(\int_{S_r^+}
  \Big(\Big|\tfrac{\partial \widetilde U}{\partial
        \nu}\Big|^2+\Big|\tfrac{\partial \widetilde V}
{\partial \nu}\Big|^2\Big) dS\Big) \cdot
    \left(\int_{S_r^+}
  (\widetilde U^2+\widetilde V^2)\,dS\right)-\left(
\int_{S_r^+}
  \left(\widetilde U\tfrac{\partial \widetilde U}{\partial
      \nu}+\widetilde V\tfrac{\partial \widetilde V}
{\partial \nu}\right)\, dS\right)^{\!2}=0$$ which implies that $(\widetilde U,\widetilde V)$ and $(\frac{\partial \widetilde U}{\partial \nu},\frac{\partial \widetilde
  V}{\partial \nu})$ have the same direction as vectors in $L^2(S_r^+)\times L^2(S_r^+)$. Hence there exists a function $\eta=\eta(r)$ such that $\left(\frac{\partial \widetilde U}{\partial \nu}(r\theta),\frac{\partial \widetilde
  V}{\partial \nu}(r\theta)\right)=\eta(r) (\widetilde U (r\theta),\widetilde V (r\theta))$ for all $r\in(0,1)$ and $\theta\in {\mathbb S}^N_+$. By integration we obtain $$\begin{aligned}
\label{separate}
&\widetilde U(r\theta)=e^{\int_1^r \eta(s)ds} \widetilde U(\theta)
=\varphi(r) \Psi_1(\theta), \quad  r\in(0,1), \ \theta\in {\mathbb S}^N_+,\\
\label{separate2}
&\widetilde V(r\theta)=e^{\int_1^r \eta(s)ds} \widetilde V(\theta)
=\varphi(r) \Psi_2(\theta), \quad  r\in(0,1), \ \theta\in {\mathbb S}^N_+,\end{aligned}$$ where $\varphi(r)=e^{\int_1^r \eta(s)ds}$ and $\Psi_1=\widetilde U\big|_{{\mathbb S}^{N}_+}$, $\Psi_2(\theta)=\widetilde V\big|_{{\mathbb S}^{N}_+}$. From , , and , it follows that $$\label{eq:28}
\begin{cases}
r^{-N}\big(r^{N}\varphi'\big)'\Psi_i(\theta)+
r^{-2}\varphi(r)\Delta_{{\mathbb S}^{N}_+}\Psi_i(\theta)
=0,&\text{on }{\mathbb S}^{N}_+,\\
\partial_\nu \Psi_i=0,&\text{on }\partial{\mathbb S}^{N}_+,
\end{cases}\quad i=1,2.$$ Taking $r$ fixed we deduce that $\Psi_1,\Psi_2$ are either zero or restrictions to ${\mathbb S}^{N}_+$ of eigenfunctions of $-\Delta_{{\mathbb S}^{N}}$ associated to the same eigenvalue and symmetric with respect to the equator $\partial{\mathbb S}^{N}_+$. Therefore there exist $\ell\in{{\mathbb N}}$, $\{\beta_{\ell,m},\beta'_{\ell,m}\}_{m=1}^{M_\ell}\subset {{\mathbb R}}$ such that $$\begin{cases}
-\Delta_{{\mathbb S}^{N}_+}\Psi_1=\lambda_\ell \Psi_1,&\text{on }{\mathbb S}^{N}_+,\\
\partial_\nu \Psi_1=0,&\text{on }\partial{\mathbb S}^{N}_+,
\end{cases}
\quad
\begin{cases}
-\Delta_{{\mathbb S}^{N}_+}\Psi_2=\lambda_\ell \Psi_2,&\text{on }{\mathbb S}^{N}_+,\\
\partial_\nu \Psi_2=0,&\text{on }\partial{\mathbb S}^{N}_+,
\end{cases}$$ and $$\Psi_1=\sum_{m=1}^{M_\ell}\beta_{\ell,m}Y_{\ell,m},\quad
\Psi_2=\sum_{m=1}^{M_\ell}\beta'_{\ell,m}Y_{\ell,m}.$$ In view of we have that $\int_{{\mathbb
    S}^{N}_+}(\Psi_1^2+\Psi_2^2)\,dS=1$ and hence $$\sum_{m=1}^{M_\ell}((\beta_{\ell,m})^2+(\beta'_{\ell,m})^2)=1.$$ Since $\Psi_1$ and $\Psi_2$ are not both identically zero, from it follows that $\varphi(r)$ solves the equation $$\varphi''(r)+\frac{N}r\varphi'(r)-\frac{\lambda_\ell}{r^2}\varphi(r)=0$$ and hence $\varphi(r)$ is of the form $$\varphi(r)=c_1 r^{\ell}+c_2 r^{-(N-1)-\ell}$$ for some $c_1,c_2\in{{\mathbb R}}$. Since either $|z|^{-(N-1)-\ell}\Psi_1(\frac{z}{|z|}) \notin
H^1(B_1^+)$ or $|z|^{-(N-1)-\ell}\Psi_2(\frac{z}{|z|}) \notin
H^1(B_1^+)$ (being $(\Psi_1,\Psi_2)\not\equiv(0,0)$), we have that $c_2=0$ and $\varphi(r)=c_1
r^{\ell}$. Moreover, from $\varphi(1)=1$ we deduce that $c_1=1$. Then $$\label{expw}
\widetilde U(r\theta)=r^{\ell} \Psi_1(\theta),\quad
\widetilde V(r\theta)=r^{\ell} \Psi_2(\theta),  \quad
\text{for all }r\in (0,1)\text{ and }\theta\in {\mathbb S}^N_+.$$ From and the fact that $$\int_{{\mathbb
    S}^{N}_+}(\Psi_1^2+\Psi_2^2)\,dS=1\quad\text{and}\quad
\int_{{\mathbb
    S}^{N}_+}(|\nabla_{{\mathbb
        S}^{N}}\Psi_1|^2+|\nabla_{{\mathbb
        S}^{N}}\Psi_2|^2)\,dS=\lambda_\ell$$ it follows that $$\begin{aligned}
 \widetilde D (r)&=
\frac{1}{r^{N-1}} \int_{B_r^+}
(|\nabla \widetilde U|^2+|\nabla \widetilde V|^2)\,dt\,dx  \\
&=r^{1-N}\ell^2\int_0^rt^{N+2(\ell-1)}dt+
r^{1-N}\lambda_\ell\int_0^r t^{N+2(\ell-1)}dt=\frac{\ell^2+\ell(N-1+\ell)}{N+2\ell-1}\,r^{2\ell}=
\ell \,r^{2 \ell}\end{aligned}$$ and $$\begin{aligned}
  \widetilde  H(r)=\int_{{\mathbb
      S}^{N}_+}\left(\widetilde U^2(r\theta)+\widetilde V^2(r\theta)\right)\,dS=r^{2 \ell}.\end{aligned}$$ Hence from (\[Nw(r)\]) it follows that $\gamma=\widetilde{\mathcal N}(r)=\frac{\widetilde D(r)}{\widetilde H(r)}=\ell$. The proof of the lemma is complete.

\[l:limite\] Let $(U,V)\in H^1(B_R^+)\times H^1(B_R^+)$ be a weak solution to such that $(U,V)\not=(0,0)$, let $H$ be defined in , and let $\ell$ be as in Lemma \[l:blowup\]. Then the limit $$\lim_{r\to0^+}r^{-2\ell}H(r)$$ exists and it is finite.

We recall from Lemma \[l:blowup\] that $\ell=\lim_{r\rightarrow 0^+} {\mathcal
    N}(r)$ with $\mathcal N$ as in .

In view of it is sufficient to prove that the limit exists. By and Lemma \[l:limitN\] we have $$\begin{aligned}
\label{eq:31}
\frac{d}{dr} \frac{H(r)}{r^{2\ell}}& =-2\ell r^{-2\ell-1}
H(r)+r^{-2\ell} H'(r) =2r^{-2\ell-1} (D(r)-\ell
H(r))\\
\notag&=2r^{-2\ell-1} H(r) \int_0^r {\mathcal N}'(\rho) d\rho.\end{aligned}$$ From and it follows that there exists some $c>0$ such that $\mathcal N'(r){\geqslant}-c$ for all $r\in(0,\tilde
r)$. Then we can write $\mathcal N'(r)=-c+f(r)$ for some function $f\in L^1_{\rm loc}(0,r_0)$ such that $f(r){\geqslant}0$ a.e. in $(0,\tilde r)$. Then integration of over $(r,\tilde r)$ yields $$\label{inte}
  \frac{H(\tilde r)}{\tilde r^{2\ell}}-
  \frac{H(r)}{r^{2\ell}}=2\int_r^{\tilde r} \rho^{-2\ell-1}
  H(\rho) \left( \int_0^\rho f(t) dt \right) d\rho -2c\int_r^{\tilde r} \rho^{-2\ell}
  H(\rho) d\rho.$$ Since $f\geq0$, we have that $\lim_{r\to 0^+} \int_r^{\tilde r}  \rho^{-2\ell-1} H(\rho) \left( \int_0^\rho
  f(t) dt \right) d\rho$ exists. On the other hand, implies that $\rho^{-2\ell}
H(\rho) \in L^1(0,\widetilde r)$. Therefore both terms at the right hand side of (\[inte\]) admit a limit as $r\to 0^+$ (one of which is finite) and the proof is complete.

Let $(U,V)\in H^1(B_R^+)\times H^1(B_R^+)$ be a weak solution to such that $(U,V)\not=(0,0)$. Let us expand $U$ and $V$ as $$U(z)=U(\lambda
\theta)=\sum_{k=0}^\infty\sum_{m=1}^{M_k}\varphi_{k,m}(\lambda)Y_{k,m}(\theta),\quad
V(z)=V(\lambda \theta)=\sum_{k=0}^\infty\sum_{m=1}^{M_k}\widetilde\varphi_{k,m}(\lambda)Y_{k,m}(\theta)$$ where $\lambda=|z|\in(0,R]$, $\theta=z/|z|\in{{\mathbb S}^{N}_+}$, and $$\label{eq:37}
  \varphi_{k,m}(\lambda)=\int_{{\mathbb S}^{N}_+}U(\lambda\,\theta) Y_{k,m}(\theta)\,dS,\quad
\widetilde\varphi_{k,m}(\lambda)=\int_{{\mathbb S}^{N}_+}V(\lambda\,\theta)
    Y_{k,m}(\theta)\,dS$$

\[l:coeff\_fourier\] Let $(U,V)\in H^1(B_R^+)\times H^1(B_R^+)$ be a weak solution to such that $(U,V)\not=(0,0)$, let $\ell$ be as in Lemma \[l:blowup\], and let $\widetilde\varphi_{\ell,m},
  \varphi_{\ell,m}$ be as in . Then, for all $1{\leqslant}m\leq
  M_\ell$, $$\begin{aligned}
\label{eq:24b}\varphi_{\ell,m}(\lambda)&=
\lambda^{\ell}
\bigg(c_1^{\ell,m}+\int_\lambda^R\frac{t^{-\ell+1}}{2\ell+N-1}\widetilde
                 \varphi_{\ell,m}(t)\,dt\bigg)+
\lambda^{-(N-1)-\ell}\int_0^\lambda\frac{t^{N+\ell}}{N+2\ell-1}
\widetilde \varphi_{\ell,m}(t)\,dt\\
\notag&=
\lambda^{\ell}
\bigg(c_1^{\ell,m}+\int_\lambda^R\frac{t^{-\ell+1}}{2\ell+N-1}\widetilde
                 \varphi_{\ell,m}(t)\,dt+O(\lambda^2)\bigg),\quad\text{as
                    }\lambda\to0^+,\\
\label{eq:24c}\widetilde\varphi_{\ell,m}(\lambda) &=
\lambda^{\ell}
\bigg(d_1^{\ell,m}+\int_\lambda^R\frac{t^{-\ell+1}}{2\ell+N-1}\zeta_{\ell,m}(t)\,dt\bigg)+
\lambda^{-(N-1)-\ell}\int_0^\lambda\frac{t^{N+\ell}}{N+2\ell-1}
\zeta_{\ell,m}(t)\,dt\\
\notag&=\lambda^{\ell}
\bigg(d_1^{\ell,m}+\int_\lambda^R\frac{t^{-\ell+1}}{2\ell+N-1}\zeta_{\ell,m}(t)\,dt+O(\lambda)\bigg),\quad\text{as
                    }\lambda\to0^+,\end{aligned}$$ where $$\label{eq:38ell}
  \zeta_{\ell,m}(\lambda)=\frac{1}{\lambda}\int_{{\mathbb
      S}^{N-1}}h(\lambda\theta')U(\lambda\theta',0)Y_{\ell,m}(\theta',0)\,dS',$$ and $$\begin{aligned}
\label{eq:36}&  c_1^{\ell,m}=R^{-\ell}\int_{{\mathbb S}^{N}_+}U(R\,\theta)
  Y_{\ell,m}(\theta)\,dS-
\frac{R^{-N-2\ell+1}}{N+2\ell-1}\int_0^R t^{N+\ell}\left(\int_{{\mathbb S}^{N}_+}V(t\,\theta)
    Y_{\ell,m}(\theta)\,dS\right)\,dt,\\
\label{eq:39}&  d_1^{\ell,m}=R^{-\ell}\int_{{\mathbb S}^{N}_+}V(R\,\theta)
  Y_{\ell,m}(\theta)\,dS\\
\notag&\qquad\qquad-
\frac{R^{-N-2\ell+1}}{N+2\ell-1}\int_0^R t^{N+\ell-1}\left(
\int_{{\mathbb
      S}^{N-1}}h(t\theta')U(t\theta',0)Y_{k,m}(\theta',0)\,dS'\right)\,dt.\end{aligned}$$

From the Parseval identity it follows that $$\label{eq:17bis}
H(\lambda)=\int_{{\mathbb
    S}^{N}_+}\big(U^2(\lambda\theta)+V^2(\lambda\theta)\big)\,dS=
\sum_{k=0}^\infty\sum_{m=1}^{M_k}\big(\varphi_{k,m}^2(\lambda)+\widetilde\varphi_{k,m}^2(\lambda)\big),
\quad\text{for all }0<\lambda{\leqslant}R.$$ In particular and yield, for all $k\geq0$ and $1{\leqslant}m{\leqslant}M_k$, $$\label{eq:23b}
\varphi_{k,m}(\lambda)=O(\lambda^{\ell})\quad\text{and}\quad
\widetilde\varphi_{k,m}(\lambda)=O(\lambda^{\ell})
\quad\text{as }\lambda\to0^+.$$ Equations and imply that, for every $k\geq0$ and $1{\leqslant}m{\leqslant}M_k$, $$\begin{cases}
-\varphi_{k,m}''(\lambda)-\frac{N}{\lambda}\varphi_{k,m}'(\lambda)+
\frac{k(N-1+k)}{\lambda^2} \, \varphi_{k,m}(\lambda)=\widetilde\varphi_{k,m}(\lambda),&\text{in
}(0,R),\\[10pt]
-\widetilde\varphi_{k,m}''(\lambda)-\frac{N}{\lambda}\widetilde\varphi_{k,m}'(\lambda)+
 \frac{k(N-1+k)}{\lambda^2} \, \widetilde\varphi_{k,m}(\lambda)=\zeta_{k,m}(\lambda),&\text{in
}(0,R),
\end{cases}$$ where $$\label{eq:38}
  \zeta_{k,m}(\lambda)=\frac{1}{\lambda}\int_{{\mathbb
      S}^{N-1}}h(\lambda\theta')U(\lambda\theta',0)Y_{k,m}(\theta',0)\,dS'.$$ By direct calculations we have, for some $c_1^{k,m},c_2^{k,m},d_1^{k,m},d_2^{k,m}\in{{\mathbb R}}$, $$\begin{aligned}
\label{eq:33}\varphi_{k,m}(\lambda)=\lambda^{k}
\bigg(c_1^{k,m}&+\int_\lambda^R\frac{t^{-k+1}}{2k+N-1}\widetilde
                 \varphi_{k,m}(t)\,dt\bigg)\\
\notag&+\lambda^{-(N-1)-k}
\bigg(c_2^{k,m}+\int_\lambda^R\frac{t^{N+k}}{1-N-2k}
\widetilde \varphi_{k,m}(t)\,dt\bigg),\\
\label{eq:34}\widetilde\varphi_{k,m}(\lambda)=\lambda^{k}
\bigg(d_1^{k,m}&+\int_\lambda^R\frac{t^{-k+1}}{2k+N-1}\zeta_{k,m}(t)\,dt\bigg)\\
\notag&+\lambda^{-(N-1)-k}
\bigg(d_2^{k,m}+\int_\lambda^R\frac{t^{N+k}}{1-N-2k}
\zeta_{k,m}(t)\,dt\bigg).\end{aligned}$$ We observe that $$\label{eq:32}
\zeta_{k,m}(\lambda)=\frac{2^{N-1}\sqrt{H(2\lambda)}}{\lambda}
\int_{\partial B'_{1/2}}h(2\lambda x)U_{2\lambda}(x,0)Y_{k,m}\big(\tfrac{x}{|x|},0\big)\,dS'$$ with $U_\lambda$ as in . Since $\{U_\lambda\}_\lambda$ is bounded in $H^2(B^+_{1/2})$ in view of , from continuity of the trace embedding $H^2(B^+_{1/2})\hookrightarrow
H^{3/2}(B_{1/2}')$ we deduce that $\{\mathop{\rm Tr} U_\lambda\}_\lambda$ is bounded in $H^1(B_{1/2}')$ and its trace on $\partial B_{1/2}'$ is bounded in $L^2(\partial B'_{1/2})$. Hence from and we conclude that, for all $k\geq0$ and $1{\leqslant}m{\leqslant}M_k$, $$\label{eq:zeta}
 \zeta_{k,m}(\lambda)=O(\lambda^{\ell-1})\quad\text{as }\lambda\to 0^+.$$ From and it follows that, for all $1\leq
m{\leqslant}M_\ell$, the functions $$t\mapsto t^{-\ell+1}\widetilde
                 \varphi_{\ell,m}(t),
\quad
t\mapsto t^{N+\ell}\widetilde \varphi_{\ell,m}(t),\quad
t\mapsto t^{-\ell+1}\zeta_{\ell,m}(t),
\quad
t\mapsto t^{N+\ell}\zeta_{\ell,m}(t),$$ belong to $L^1(0,R)$. Hence $$\begin{aligned}
  &\lambda^{\ell}
    \bigg(c_1^{\ell,m}+\int_\lambda^R\frac{t^{-\ell+1}}{2\ell+N-1}\widetilde
    \varphi_{\ell,m}(t)\,dt\bigg)=o(\lambda^{-(N-1)-\ell}),\quad\text{as
    }\lambda\to0^+,\\
  &\lambda^{\ell}
    \bigg(d_1^{\ell,m}+\int_\lambda^R\frac{t^{-\ell+1}}{2\ell+N-1}\zeta_{\ell,m}(t)\,dt\bigg)=o(\lambda^{-(N-1)-\ell}),
    \quad\text{as }\lambda\to0^+,\end{aligned}$$ and consequently, by , there must be $$c_2^{\ell,m}=-\int_0^R\frac{t^{N+\ell}}{1-N-2\ell}
\widetilde \varphi_{\ell,m}(t)\,dt\quad\text{and}\quad
d_2^{\ell,m}=-\int_0^R\frac{t^{N+\ell}}{1-N-2\ell}
\zeta_{\ell,m}(t)\,dt.$$ Using and , we then deduce that $$\begin{aligned}
\label{eq:12-alt}
&\lambda^{-(N-1)-\ell}
\bigg(c_2^{\ell,m}+\int_\lambda^R\tfrac{t^{N+\ell}}{1-N-2\ell}
\widetilde \varphi_{\ell,m}(t)\,dt\bigg)
=\lambda^{-(N-1)-\ell}\int_0^\lambda\tfrac{t^{N+\ell}}{N+2\ell-1}
\widetilde \varphi_{\ell,m}(t)\,dt
=O(\lambda^{\ell+2}),\\
\label{eq:12b}&\lambda^{-(N-1)-\ell}
\bigg(d_2^{\ell,m}+\int_\lambda^R\tfrac{t^{N+\ell}}{1-N-2\ell}
\zeta_{\ell,m}(t)\,dt\bigg)
=\lambda^{-(N-1)-\ell}\int_0^\lambda\tfrac{t^{N+\ell}}{N+2\ell-1}
\zeta_{\ell,m}(t)\,dt
=O(\lambda^{\ell+1}),
  \end{aligned}$$ as $\lambda\to0^+$. From , , , and we deduce and . Finally, and follow by computing and for $\lambda=R$ and recalling .

We now prove that $\lim_{r\to 0^+} r^{-2\ell} H(r)$ is strictly positive.

\[l:limitepositivo\] Under the same assumption as in Lemmas \[l:limite\], we have $$\lim_{r\to0^+}r^{-2\ell}H(r)>0.$$

Let us assume by contradiction that $\lim_{\lambda\to0^+}\lambda^{-2\ell}H(\lambda)=0$. Then, for all $1{\leqslant}m{\leqslant}M_\ell$, (\[eq:17bis\]) would imply that $$\lim_{\lambda\to0^+}\lambda^{-\ell}\varphi_{\ell,m}(\lambda)=
\lim_{\lambda\to0^+}\lambda^{-\ell}\widetilde\varphi_{\ell,m}(\lambda)=0.$$ Hence, in view of and , $$c_1^{\ell,m}+\int_0^R\frac{t^{-\ell+1}}{2\ell+N-1}\widetilde
                 \varphi_{\ell,m}(t)\,dt=0\quad\text{and}\quad
d_1^{\ell,m}+\int_0^R\frac{t^{-\ell+1}}{2\ell+N-1}\zeta_{\ell,m}(t)\,dt=0$$ which, in view of (\[eq:zeta\]) and , yields $$\begin{aligned}
\label{eq:13tris}
\lambda^{\ell}
\bigg(c_1^{\ell,m}+\int_\lambda^R\frac{t^{-\ell+1}}{2\ell+N-1}\widetilde
                 \varphi_{\ell,m}(t)\,dt\bigg)=
\lambda^{\ell}\int_0^\lambda\frac{t^{-\ell+1}}{1-2\ell-N}\widetilde
                 \varphi_{\ell,m}(t)\,dt=O(\lambda^{\ell+2})\\
\label{eq:13q}
\lambda^{\ell}
\bigg(d_1^{\ell,m}+\int_\lambda^R\frac{t^{-\ell+1}}{2\ell+N-1}\zeta_{\ell,m}(t)\,dt\bigg)=
\lambda^{\ell}\int_0^\lambda\frac{t^{-\ell+1}}{1-2\ell-N}\zeta_{\ell,m}(t)\,dt=O(\lambda^{\ell+1})\end{aligned}$$ as $\lambda\to0^+$. Estimates , , , and imply that $$\varphi_{\ell,m}(\lambda)=O(\lambda^{\ell+2})\quad\text{and}\quad
\widetilde\varphi_{\ell,m}(\lambda)=O(\lambda^{\ell+1})\quad\text{as }\lambda\to0^+
\quad\text{for every }
1{\leqslant}m{\leqslant}M_\ell,$$ namely, $$\sqrt{H(\lambda)}\,(U_\lambda,Y_{\ell,m})_{L^2({\mathbb
  S}^{N}_+)}=
O(\lambda^{\ell+2}) \quad\text{and}\quad
\sqrt{H(\lambda)}\,(V_\lambda,Y_{\ell,m})_{L^2({\mathbb
  S}^{N}_+)}=O(\lambda^{\ell+1})\quad\text{as }\lambda\to0^+,$$ for every $1{\leqslant}m{\leqslant}M_\ell$. From (\[2ndest\]), there exists $K>0$ such that $\sqrt{H(\lambda)}\geq
K\lambda^{\ell+\frac12}$ for $\lambda$ sufficiently small. Therefore $$\label{eq:26-alt}
(U_\lambda,Y_{\ell,m})_{L^2({\mathbb
  S}^{N}_+)}=
O(\lambda^{\frac32}) \quad\text{and}\quad
(V_\lambda,Y_{\ell,m})_{L^2({\mathbb
  S}^{N}_+)}=O(\lambda^{\frac12})\quad\text{as }\lambda\to0^+,$$ for every $1{\leqslant}m{\leqslant}M_\ell$. From Lemma \[l:blowup\], for every sequence $\lambda_n\to0^+$, there exist a subsequence $\{\lambda_{n_k}\}_{k\in{{\mathbb N}}}$ and $2M_\ell$ real constants $\beta_{\ell,m},\beta'_{\ell,m}$, $m=1,2,\dots,M_\ell$, such that $$\label{eq:35}
\sum_{m=1}^{M_\ell}((\beta_{\ell,m})^2+(\beta'_{\ell,m})^2)=1$$ and $$U_{\lambda_{n_k}}\to
|z|^{\ell}
\sum_{m=1}^{M_\ell}\beta_{\ell,m}Y_{\ell,m}\Big(\frac
z{|z|}\Big),\quad
V_{\lambda_{n_k}}\to
|z|^{\ell}
\sum_{m=1}^{M_\ell}\beta'_{\ell,m}Y_{\ell,m}\Big(\frac
z{|z|}\Big), \quad\text{as }k\to+\infty,$$ weakly in $H^1(B_1^+)$ and hence strongly in $L^2(S_1^+)$. It follows that, for all $m=1,2,\dots,M_\ell$, $$\beta_{\ell,m}=\lim_{k\to+\infty}(U_{\lambda_{n_k}},Y_{\ell,m})_{L^2({\mathbb
  S}^{N}_+)}\quad\text{and}\quad
\beta'_{\ell,m}=\lim_{k\to+\infty}(V_{\lambda_{n_k}},Y_{\ell,m})_{L^2({\mathbb
  S}^{N}_+)}$$ and hence, in view of , $$\beta_{\ell,m}=0\quad\text{and}\quad \beta'_{\ell,m}=0
\quad\text{for every $m=1,2,\dots,M_\ell$},$$ thus contradicting .

From Lemmas \[l:blowup\] and \[l:limitepositivo\] there exist $\ell\in {{\mathbb N}}$ such that, for every sequence $\lambda_n\to0^+$, there exist a subsequence $\{\lambda_{n_k}\}_{k\in{{\mathbb N}}}$ and $2M_\ell$ real constants $\alpha_{\ell,m},\alpha '_{\ell,m}$, $m=1,2,\dots,M_\ell$, such that $\sum_{m=1}^{M_\ell}((\alpha_{\ell,m})^2+(\alpha'_{\ell,m})^2)\neq0$ and $$\label{eq:40}
\lambda_{n_k}^{-\ell}U(\lambda_{n_k}z)\to
|z|^{\ell}
\sum_{m=1}^{M_\ell}\alpha_{\ell,m}Y_{\ell,m}\Big(\frac
z{|z|}\Big),\quad
\lambda_{n_k}^{-\ell}V(\lambda_{n_k}z)\to
|z|^{\ell}
\sum_{m=1}^{M_\ell}\alpha'_{\ell,m}Y_{\ell,m}\Big(\frac
z{|z|}\Big),$$ strongly in $H^1(B_r^+)$ for all $r\in(0,1)$, and then, by homogeneity, strongly in $H^1(B_1^+)$.

From above, , , , , , and , we deduce that $$\begin{aligned}
\alpha_{\ell,m}&=\lim_{k\to\infty}\lambda_{n_k}^{-\ell}
\int_{{\mathbb S}^{N}_+}U(\lambda_{n_k}\,\theta) Y_{\ell,m}(\theta)\,dS\\
&=\lim_{k\to\infty}\lambda_{n_k}^{-\ell}\varphi_{\ell,m}(\lambda_{n_k})=
c_1^{\ell,m}+\int_0^R\frac{t^{-\ell+1}}{2\ell+N-1}\widetilde
                 \varphi_{\ell,m}(t)\,dt\\
&=R^{-\ell}\int_{{\mathbb S}^{N}_+}U(R\,\theta)
  Y_{\ell,m}(\theta)\,dS-
\frac{R^{-N-2\ell+1}}{N+2\ell-1}\int_0^R t^{N+\ell}\left(\int_{{\mathbb S}^{N}_+}V(t\,\theta)
    Y_{\ell,m}(\theta)\,dS\right)\,dt\\
&\qquad\qquad
+\int_0^R\frac{t^{-\ell+1}}{2\ell+N-1}\left(\int_{{\mathbb S}^{N}_+}V(t\,\theta)
    Y_{\ell,m}(\theta)\,dS\right)\,dt\end{aligned}$$ and $$\begin{aligned}
\alpha'_{\ell,m}&=\lim_{k\to\infty}\lambda_{n_k}^{-\ell}
\int_{{\mathbb S}^{N}_+}V(\lambda_{n_k}\,\theta) Y_{\ell,m}(\theta)\,dS\\
&=\lim_{k\to\infty}\lambda_{n_k}^{-\ell}\widetilde\varphi_{\ell,m}(\lambda_{n_k})=
d_1^{\ell,m}+\int_0^R\frac{t^{-\ell+1}}{2\ell+N-1}\zeta_{\ell,m}(t)\,dt\\
&=R^{-\ell}\int_{{\mathbb S}^{N}_+}V(R\,\theta)
  Y_{\ell,m}(\theta)\,dS\\
\notag&\qquad\qquad-
\frac{R^{-N-2\ell+1}}{N+2\ell-1}\int_0^R t^{N+\ell-1}\left(
\int_{{\mathbb
      S}^{N-1}}h(t\theta')U(t\theta',0)Y_{k,m}(\theta',0)\,dS'\right)\,dt\\
&\qquad\qquad
+\int_0^R\frac{t^{-\ell}}{2\ell+N-1}\left(
\int_{{\mathbb
      S}^{N-1}}h(t\theta')U(t\theta',0)Y_{\ell,m}(\theta',0)\,dS'
\right)\,dt.\end{aligned}$$ We observe that the coefficients $\alpha_{\ell,m},\alpha'_{\ell,m}$ depend neither on the sequence $\{\lambda_n\}_{n\in{{\mathbb N}}}$ nor on its subsequence $\{\lambda_{n_k}\}_{k\in{{\mathbb N}}}$. Hence the convergences in hold as $\lambda\to 0^+$ and the theorem is proved.

Let us assume by contradiction that $(U,V)\not=(0,0)$. Then Theorem \[t:asym\] implies that there exist $\ell\in{{\mathbb N}}$ such that $$\label{eq:42}
\lambda^{-\ell}U(\lambda z)\to
\widehat U(\theta)
,\quad
\lambda^{-\ell}V(\lambda z)\to
\widehat  V(\theta)
,$$ strongly in $H^1(B_1^+)$, where $(\widehat  U ,\widehat  V)\neq(0,0)$.

Assumption implies that $\widehat
U\equiv0$. Hence $\widehat  V\not\equiv0$. Let us denote $\widetilde U_\lambda(z)=\lambda^{-\ell-2}U(\lambda
z)$. Then $\widetilde U_\lambda$ satisfies $$-\Delta \widetilde U_\lambda(z)=\lambda^{-\ell}V(\lambda z).$$ We have that, for all $\varphi\in C^{\infty}_{\rm c}(B_1^+)$, $$\lim_{\lambda\to0^+}\int_{B_1^+}\nabla \widetilde
U_\lambda(z)\cdot\nabla \varphi(z)\,dz=
\lim_{\lambda\to0^+}
\int_{B_1^+}\lambda^{-\ell}V(\lambda z)\varphi(z)\,dz
=\int_{B_1^+}\widehat  V(z)\varphi(z)\,dz.$$ On the other, by assumption we have that $$\begin{aligned}
\lim_{\lambda\to0^+}\int_{B_1^+}\nabla \widetilde
U_\lambda(z)\cdot\nabla \varphi(z)\,dz&=
-\lim_{\lambda\to0^+}\int_{B_1^+} \widetilde
U_\lambda(z)\Delta \varphi(z)\,dz\\
&=-\lim_{\lambda\to0^+}\lambda^{-\ell-2}\int_{B_1^+}
U(\lambda z)\Delta \varphi(z)\,dz=0.\end{aligned}$$ Therefore we obtain that $$\int_{B_1^+}\widehat  V(z)\varphi(z)\,dz=0\quad\text{for all
}\varphi\in C^{\infty}_{\rm c}(B_1^+)$$ which implies that $\widehat  V\equiv 0$ in $B_1^+$, a contradiction.

Applications to fourth order problems and higher order fractional equations {#sec:appl-fourth-order}
===========================================================================

In this section we discuss applications of Theorem \[t:asym\] to fourth order problems and higher order fractional equations, by proving Theorems \[t:asym-ho\] and \[t:frac\].

From [@FF-prep Proposition 7.2] we have that, if $U\in\mathcal D$, then $U\in H^1(B_R^+)$. Furthermore, [@FF-prep Proposition 2.4] implies that, if $U\in\mathcal D$ is a nontrivial weak solution to for some $h\in C^1(\Omega)$, then $V:=\Delta U$ belongs to $H^1(B_R^+)$ for some $R>0$ so that the couple $(U,V)\in
  H^1(B_R^+)\times H^1(B_R^+)$ is a weak solution to such that $(U,V)\not=(0,0)$. Then statement (i) follows from Theorem \[t:asym\] while (ii) comes from Theorem \[t:sun-ext\].

In view of [@Y] (see also [@FF-prep]), we have that, if $u\in
\mathcal D^{3/2,2}({{\mathbb R}}^N)$, then there exists a unique $U\in \mathcal
D$ such that $\Delta^2 U=0$ in ${{\mathbb R}}^{N+1}_+$ and $\mathop{\rm Tr}(U)=u$ on ${{\mathbb R}}^{N+1}_+$. Moreover $$\label{eq:49}
\int_{{{\mathbb R}}^{N+1}_+}\Delta U(x,t) \Delta \varphi(x,t)\,dx\,dt= 2 \, (u,
\mathop{\rm Tr}\varphi)_{\mathcal D^{3/2,2}({{\mathbb R}}^N)}$$ for all $\varphi\in\mathcal D$. In particular, if $u$ solves , we have that $U$ is a weak solution to . Let $V=\Delta U$. Since $(-\Delta)^{3/2} u\in (\mathcal D^{1/2,2}({{\mathbb R}}^N))^*$, by we have that $$\label{eq:**}
\int_{{{\mathbb R}}^{N+1}_+} V(x,t) \Delta \varphi(x,t)\,dx\,dt=2 \phantom{a}_{(\mathcal D^{1/2,2}({{\mathbb R}}^N))^\star}
\left\langle  (-\Delta)^{3/2} u,
\mathop{\rm Tr}\varphi\right\rangle_{\mathcal D^{1/2,2}({{\mathbb R}}^N)}$$ for all $\varphi\in \mathcal T$ with $\mathcal T$ as in . Applying [@FF-prep Proposition 2.4] to $V$ we deduce that $V\in H^1(B_r^+)$ for all $r>0$ and hence by and integration by parts we obtain $$\label{eq:***}
-\int_{{{\mathbb R}}^{N+1}_+} \nabla V(x,t) \cdot \nabla \varphi(x,t)\,dx\,dt=2 \phantom{a}_{(\mathcal D^{1/2,2}({{\mathbb R}}^N))^\star}
\left\langle  (-\Delta)^{3/2} u,
\mathop{\rm Tr}\varphi\right\rangle_{\mathcal D^{1/2,2}({{\mathbb R}}^N)}$$ for all $\varphi \in \mathcal T$.

Since the trace map $\mathop{\rm Tr}$ is continuous from $\mathcal D^{1,2}({{\mathbb R}}^{N+1}_+)$ into $\mathcal D^{1/2,2}({{\mathbb R}}^N)$, in view of assumption we have that $W\mapsto
\phantom{a}_{(\mathcal D^{1/2,2}({{\mathbb R}}^N))^\star}
\left\langle
  (-\Delta)^{3/2} u,
\mathop{\rm Tr}W\right\rangle_{\mathcal D^{1/2,2}({{\mathbb R}}^N)}$ belongs to $(\mathcal
  D^{1,2}({{\mathbb R}}^{N+1}_+))^\star$. Then, by classical minimization methods, we have that the minimum $$\min_{W\in \mathcal
  D^{1,2}({{\mathbb R}}^{N+1}_+)}\left[\frac12\int_{{{\mathbb R}}^{N+1}_+}|\nabla
  W(x,t)|^2\,dx\,dt+2 \phantom{a}_{(\mathcal D^{1/2,2}({{\mathbb R}}^N))^\star}
\left\langle
  (-\Delta)^{3/2} u,
\mathop{\rm Tr}W\right\rangle_{\mathcal D^{1/2,2}({{\mathbb R}}^N)}\right]$$ is attained by some $\widetilde V\in  \mathcal
  D^{1,2}({{\mathbb R}}^{N+1}_+)$ weakly solving $$\begin{aligned}
\label{eq:50}
-\int_{{{\mathbb R}}^{N+1}_+}\nabla\widetilde V(x,t) \cdot
\nabla\varphi(x,t)\,dx\,dt&=2 \,
\phantom{a}_{(\mathcal D^{1/2,2}({{\mathbb R}}^N))^\star}
\left\langle
  (-\Delta)^{3/2} u,
\mathop{\rm Tr}\varphi\right\rangle_{\mathcal D^{1/2,2}({{\mathbb R}}^N)}\\
\notag&=
2 \int_{{{\mathbb R}}^N}|\xi|^3 \widehat u\, \overline{\widehat{\mathop{\rm Tr}\varphi}} \,d\xi\end{aligned}$$ for all $\varphi\in C^\infty_c(\overline{{{\mathbb R}}^{N+1}_+})$. Combining and we infer that $$\label{eq:extension}
\int_{{{\mathbb R}}^{N+1}_+} \nabla(V(x,t)-\widetilde V(x,t))\cdot \nabla \varphi(x,t)\, dxdt=0 \qquad \text{for all } \varphi\in \mathcal T \, .$$ Actually still holds true for any $\varphi\in C^\infty_c(\overline{{{\mathbb R}}^{N+1}_+})$. Indeed, for any $\varphi\in C^\infty_c(\overline{{{\mathbb R}}^{N+1}})$, one can test with $\varphi_k(x,t)=\varphi(x,t)-\varphi_t(x,0)\, t\, \eta(kt)$, $k\in
{{\mathbb N}}$, where $\eta\in C^\infty_c({{\mathbb R}})$, $0{\leqslant}\eta{\leqslant}1$, $\eta(t)=1$ for any $t\in [-1,1]$ and $\eta(t)=0$ for any $t\in
(-\infty,-2]\cup [2,+\infty)$, and pass to the limit as $k\to
+\infty$. Therefore, if we define $$\widetilde W=
\begin{cases}
  V(x,t)-\widetilde V(x,t),&\text{if }t\geq0,\\
  V(x,-t)-\widetilde V(x,-t),&\text{if }t<0,
\end{cases}$$ we easily deduce that $\int_{{{\mathbb R}}^{N+1}}\nabla \widetilde W\cdot \nabla \varphi \, dz=0$ for all $\varphi\in C^\infty_c({{\mathbb R}}^{N+1})$. In particular $\widetilde W$ is harmonic in ${{\mathbb R}}^{N+1}$. Furthermore, since $V\in
L^2({{\mathbb R}}^{N+1}_+)$ and $\widetilde V\in  \mathcal
  D^{1,2}(\overline{{{\mathbb R}}^{N+1}_+})$, we have that $\widetilde W=W_1+W_2$ for some $W_1\in L^2({{\mathbb R}}^{N+1})$ and $W_2\in L^{\frac{2(N+1)}{N-1}}({{\mathbb R}}^{N+1})$. The mean value property for harmonic functions ensures that, for every $z\in  {{\mathbb R}}^{N+1}$ and $R>0$, $$\begin{aligned}
 |\widetilde W(z)|&= \frac1{|B(z,R)|_{N+1}}\left|\int_{B(z,R)} \widetilde
                    W(y)\,dy\right|{\leqslant}\frac{\text{\rm
                    const\,}}{R^{N+1}}
\left(\int_{B(z,R)}|W_1(y)|\,dy+\int_{B(z,R)}|W_2(y)|\,dy\right)\\
&{\leqslant}\frac{\text{\rm
                    const\,}}{R^{N+1}}\left(\|W_1\|_{L^2({{\mathbb R}}^{N+1})}R^{\frac{N+1}{2}}+
\|W_2\|_{L^{\frac{2(N+1)}{N-1}}({{\mathbb R}}^{N+1})}R^{\frac{N+3}{2}}\right)\end{aligned}$$ where $|\cdot|_{N+1}$ stands for the Lebesgue measure in ${{\mathbb R}}^{N+1}$ and $\text{\rm const\,}$ is a positive constant independent of $z$ and $R$ which could vary from line to line. Since the right hand side of the previous inequality tends to $0$ as $R\to+\infty$, we deduce that $\widetilde W\equiv0$, and then $\widetilde V=V$. In particular, in view of [@CS] and , this implies that $$(v,\varphi)_{\mathcal
  D^{1/2,2}({{\mathbb R}}^N)}=-2 \, (u,
\varphi)_{\mathcal D^{3/2,2}({{\mathbb R}}^N)}\quad\text{for all }
\varphi\in C^\infty_c({{\mathbb R}}^N),$$ where we put $v=\mathop{\rm Tr}V$. This implies that $-2 \, |\xi|^3\widehat u=|\xi|\widehat v$ and hence $v=2 \, \Delta u$ in ${{\mathbb R}}^N$.

To prove (i), it is not restrictive to assume $x_0=0$. Let us assume, by contradiction, that $u\not\equiv0$. Then the couple $(U,V)\neq(0,0)$ is a weak solution to in $H^1(B_R^+)\times H^1(B_R^+)$ for some $R>0$ with $h=-2a$.

From Theorem \[t:asym\] it follows that either $u$ or $v$ (which are the traces of $U$ and $V$ respectively) have vanishing order $\ell\in{{\mathbb N}}$ at $0$. In view of assumption we have that necessarily $V$ vanishes of order $\ell$, i.e. there exists $\Psi:{\mathbb S}^{N}_+\to{{\mathbb R}}$, a nontrivial linear combination of spherical harmonics symmetric with respect to the equator $t=0$, such that $\Psi\not\equiv0$ on $\partial{\mathbb S}^N_+$, $$\lambda^{-\ell}V(\lambda z)\to
|z|^{\ell}\Psi\Big(\frac
z{|z|}\Big) \text{ as }\lambda\to 0\text{ strongly in $H^1(B_1^+)$},$$ and consequently $$\lambda^{-\ell}v(\lambda x)\to
|x|^{\ell}\Psi\Big(\frac
x{|x|},0\Big) \text{ as }\lambda\to 0\text{ strongly in $H^{1/2}(B_1')$  }.$$ Let us denote $$v_\lambda(x)=\lambda^{-\ell}v(\lambda x)\quad\text{and}\quad
\widetilde u_\lambda(x)=\lambda^{-2-\ell}u(\lambda x),$$ so that $$\label{eq:55}
v_\lambda\to |x|^{\ell}\Psi\Big(\frac
x{|x|},0\Big) \text{ as }\lambda\to 0\text{ strongly in
  $H^{1/2}(B_1')$}$$ and $$2 \Delta \widetilde u_\lambda=v_\lambda\text{ in
}{{\mathbb R}}^N.$$ For every $\varphi\in C^{\infty}_{\rm c}(B_1')$ we have that $$\label{eq:54}
-2 \int_{{{\mathbb R}}^N} \widetilde u_\lambda(-\Delta\varphi)\,dx=-2
\int_{{{\mathbb R}}^N} \varphi(-\Delta \widetilde u_\lambda)\,dx
=\int_{{{\mathbb R}}^N} \varphi v_\lambda \,dx.$$ From one hand, assumption implies that $$\lim_{\lambda\to0^+}\int_{{{\mathbb R}}^N} \widetilde u_\lambda(-\Delta\varphi)\,dx
=0$$ whereas convergence yields $$\lim_{\lambda\to0^+}\int_{{{\mathbb R}}^N} \varphi v_\lambda \,dx=
\int_{{{\mathbb R}}^N} |x|^{\ell}\Psi\Big(\frac
x{|x|},0\Big) \varphi(x)\,dx.$$ Hence passing to the limit in we obtain that $$\int_{{{\mathbb R}}^N} |x|^{\ell}\Psi\Big(\frac
x{|x|},0\Big) \varphi(x)\,dx=0\quad\text{for every }\varphi\in
C^{\infty}_{\rm c}(B_1'),$$ thus contradicting the fact that $|x|^{\ell}\Psi\Big(\frac
x{|x|},0\Big) \not\equiv0$.

To prove (ii), let us assume by contradiction, that $u\not\equiv0$ in $\Omega$ and $u(x)=0$ a.e. in a set $E\subset \Omega$ with $|E|_N>0$, where $|\cdot|_N$ denotes the $N$-dimensional Lebesgue measure. Since $2 \Delta u=v$ and $v\in \mathcal  D^{1/2,2}({{\mathbb R}}^N)\subset
L^{2}_{\rm loc}({{\mathbb R}}^N)$, by classical regularity theory we have that $u\in H^2_{\rm loc}(\Omega)$. Since $u(x)=0$ for a.e. $x\in E$, we have that $\nabla u(x)=0$ for a.e. $x\in E$ and hence, since $\frac{\partial u}{\partial x_i}\in  H^1_{\rm loc}(\Omega)$ for every $i$, $\Delta u=0$ a.e. in $E$. In particular there exists a set $E'\subset E\subset\Omega$ with $|E'|_N>0$ such that $u(x)=\Delta u(x)=0$ a.e. in $E'$. In particular $v(x)=0$ a.e. in $E'$.

By Lebesgue’s density Theorem, a.e. point of $E'$ is a density point of $E'$. Let $x_0$ be a density point of $E'$. Hence, for all $\e>0$ there exists $r_0=r_0(\e)\in(0,1)$ such that, for all $r\in(0,r_0)$, $$\label{eq:56}
\frac{|({{\mathbb R}}^N\setminus E')\cap B'_r(x_0)|_N}{|B'_r(x_0)|_N}<\e,$$ where $B'_r(x_0)=\{x\in{{\mathbb R}}^N:|x-x_0|<r\}$. Theorem \[t:asym\] implies that there exist $\Psi_1,\Psi_2:{\mathbb S}^{N}_+\to{{\mathbb R}}$ linear combination of spherical harmonics such that either $\Psi_1\not\equiv0$ or $\Psi_2\not\equiv0$ and $$\label{eq:57}
\lambda^{-\ell}u(x_0+\lambda (x-x_0) )\to
|x-x_0|^{\ell}\Psi_1\Big(\frac{x-x_0}{|x-x_0|},0\Big)$$ and $$\label{eq:58}
\lambda^{-\ell}v(x_0+\lambda (x-x_0) )\to
|x-x_0|^{\ell}\Psi_2\Big(\frac{x-x_0}{|x-x_0|},0\Big)$$ as $\lambda\to 0$ strongly in $H^{1/2}(B_1'(x_0))$ and then, by the Sobolev embedding $H^{\frac12}(B_1'(x_0))\hookrightarrow L^{\frac{2N}{N-1}}(B_1'(x_0))$, strongly in $L^{\frac{2N}{N-1}}(B_1'(x_0))$

Since $u\equiv v\equiv 0$ in $E'$, by we have $$\begin{aligned}
\int_{B'_r(x_0)}u^2(x)\,dx&=\int_{({{\mathbb R}}^N\setminus E')\cap
  B'_r(x_0)}u^2(x)\,dx\\
&{\leqslant}\bigg(\int_{({{\mathbb R}}^N\setminus E')\cap
  B'_r(x_0)}|u(x)|^{2N/(N-1)}dx\bigg)^{\!\!\frac{N-1}{N}}|({{\mathbb R}}^N\setminus E') \cap
B'_r(x_0)|_N^{1/N}\\
&<\e^{1/N}|B'_r(x_0)|_N^{1/N}
\bigg(\int_{({{\mathbb R}}^N\setminus E')\cap
  B'_r(x_0)}|u(x)|^{2N/(N-1)}dx\bigg)^{\!\!\frac{N-1}{N}}\end{aligned}$$ and similarly $$\begin{aligned}
\int_{B'_r(x_0)}v^2(x)\,dx<\e^{1/N}|B'_r(x_0)|_N^{1/N}
\bigg(\int_{({{\mathbb R}}^N\setminus E')\cap
  B'_r(x_0)}|v(x)|^{2N/(N-1)}dx\bigg)^{\!\!\frac{N-1}{N}}\end{aligned}$$ for all $r\in(0,r_0)$. Then, letting $u^r(x):=r^{-\ell}u(x_0+r(x-x_0))$ and $v^r(x):=r^{-\ell}v(x_0+r(x-x_0))$, $$\begin{aligned}
&\int_{B'_1(x_0)}|u^r(x)|^2dx<\Big(\frac{\omega_{N-1}}N\Big)^{\!\frac1N}
\e^{\frac1N}\bigg(\int_{B'_1(x_0)}|u^r(x)|^{\frac{2N}{N-1}}dx\bigg)^{\!\!\frac{N-1}{N}},\\
&\int_{B'_1(x_0)}|v^r(x)|^2dx<\Big(\frac{\omega_{N-1}}N\Big)^{\!\frac1N}
\e^{\frac1N}\bigg(\int_{B'_1(x_0)}|v^r(x)|^{\frac{2N}{N-1}}dx\bigg)^{\!\!\frac{N-1}{N}},\end{aligned}$$ for all $r\in(0,r_0)$, where $\omega_{N-1}=\int_{{\mathbb S}^{N-1}}1\,dS'$. Letting $r\to 0^+$, from and we have that $$\begin{gathered}
\int_{B'_1(x_0)}
|x-x_0|^{2\ell}\Psi_i^2\Big(\tfrac{x-x_0}{|x-x_0|},0\Big)\,dx\\
\leq
\Big(\frac{\omega_{N-1}}N\Big)^{\!\frac1N}
\e^{\frac1N}\bigg(\int_{B'_1(x_0)}|x-x_0|^{\frac{2N \ell }{N-1}}\left|
\Psi_i\Big(\tfrac{x-x_0}{|x-x_0|},0\Big)\right|^{\frac{2N}{N-1}}dx\bigg)^{\!\!\frac{N-1}{N}}
\quad\text{for }i=1,2,\end{gathered}$$ which yields a contradiction as $\e\to 0^+$, since either $\Psi_1\not\equiv0$ or $\Psi_2\not\equiv0$.

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[^1]: The authors are partially supported by the INdAM-GNAMPA Project 2018 “Formula di monotonia e applicazioni: problemi frazionari e stabilità spettrale rispetto a perturbazioni del dominio”. V. Felli is partially supported by the PRIN 2015 grant “Variational methods, with applications to problems in mathematical physics and geometry”.\
    2000 [*Mathematics Subject Classification:*]{} 35J47, 35J30, 35B60, 35C20.\
    Neumann boundary coupling, monotonicity formula, higher order fractional problems, unique continuation.

[^2]: It is enough to take a homogeneous harmonic polynomial $P=P(x_1,x_2,\dots,x_N)$ in $N$ variables of degree $\ell$ and consider the homogeneous harmonic polynomial in $N+1$ variables $P'(x_1,x_2,\dots,x_N,x_{N+1})=P(x_1,x_2,\dots,x_N)$, whose restriction to $\mathbb{S}^{N}$ satisfies the required properties.
