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
+
