Dirichlet or Neumann Problem for Weighted 1-Laplace Equation with Application to Image Denoising

Abstract

In this paper, we mainly consider the limit behavior of a weak solution $u_{p}$ as $p\rightarrow 1$, of the Dirichlet or Neumann problem for a weighted $(1,\infty )\ni p$-Laplace equation on a Lipschitz domain, thereby giving an application of the standard weighted 1-Laplace operator to the image denoising.

Appendix A

This appendix is a weighted version of the Appendix in [34].

1.1 A.1. The Pairing $(z,\nabla u)$ and the Gauss-Green Formula

We first recall some notions and notation from [34, Appendix A.1]. Let $u\in BV(\Omega )\cap L^{\infty }(\Omega )$ & $z\in {{\mathcal {D}}}{{\mathcal {M}}}^\infty (\Omega )$. Setting $\mu :=\textrm{div}(z)$, the distribution $(z,\nabla u)$ is defined by

$$\begin{aligned} \left\langle (z,\nabla u),\phi \right\rangle :=-\int _\Omega u(x)\phi (x)\,d\mu (x) -\int _{\Omega }u(x)z(x)\cdot \nabla \phi (x)\,dx, \quad \forall \,\phi \in C_c^\infty (\Omega ). \end{aligned}$$

The following Gauss-Green’s formula is taken from [5, Proposition 1.3].

Theorem A.1

Let $z\in {{\mathcal {D}}}{{\mathcal {M}}}^\infty (\Omega )$, $u\in BV(\Omega )\cap L^{\infty }(\Omega )$ & $\mu =\textrm{div}(z)$. Then

$$\begin{aligned} \int _\Omega u(x)\,d\mu (x)+\int _\Omega z(x)\cdot \nabla u(x)\,dx =\int _{\partial \Omega }[z,\nu ](x)u(x)\,d{{\mathcal {H}}}^{n-1}(x). \end{aligned}$$

1.2 A.2 The Measures $(wz_k,\nabla T_k(u))$ & $(wz_k,\nabla 1_{\{|u|>k\}})$

In this subsection, following some ideas from [34, Appendix A.2], we investigate the properties of the distributions in (1.13), (1.14) and (3.12).

Proposition A.2

The distribution $(wz,\nabla T_k(u))$ in (3.12) is a Radon measure.

Proof

Let $u_p$ be the renormalized solution to problem (1.1). By some arguments used in Step 5 of the proof of Theorem 1.15, we obtain

$$\begin{aligned}&\int _{\{|u_p|<k\}}\left| \nabla u_p(x)\right| ^p\phi (x)w(x)\,dx\\&\qquad +\int _{\Omega }T_k(u_p(x))\left| \nabla u_p(x)\right| ^{p-2}\nabla u_p(x)\cdot \nabla \phi (x)w(x)\,dx\\&\hspace{0.25cm}=\int _\Omega T_k(u_p(x))f(x)\phi (x)w(x)\,dx. \end{aligned}$$

Letting $p\rightarrow 1$ in the above equation, we conclude

$$\begin{aligned}&\lim _{p\rightarrow 1}\int _{\{|u_p|<k\}}\left| \nabla u_p(x)\right| ^p\phi (x)w(x)\,dx\nonumber \\&\hspace{0.25cm}=\int _\Omega T_k(u(x))f(x)\phi (x)w(x)\,dx -\int _{\Omega }T_k(u(x))z(x)\cdot \nabla \phi (x)w(x)\,dx\nonumber \\&\hspace{0.25cm}=\left\langle (wz,\nabla T_k(u)),\phi \right\rangle . \end{aligned}$$

(A.1)

Moreover, we notice that

$$\begin{aligned} \left| \int _{\{|u_p|<k\}}\left| \nabla u_p(x)\right| ^p\phi (x)w(x)\,dx\right|&\le \Vert \phi \Vert _{L^{\infty }(\Omega )}\int _{\Omega }\left| \nabla T_k(u_p(x))\right| ^pw(x)\,dx\\&=\Vert \phi \Vert _{L^{\infty }(\Omega )}\int _{\Omega }f(x)T_k(u_p(x))w(x)\,dx\\&\le \Vert \phi \Vert _{L^{\infty }(\Omega )}k\int _{\Omega }|f(x)|w(x)\,dx, \end{aligned}$$

which implies

$$\begin{aligned} \left| \left\langle \left( wz,\nabla T_k(u)\right) ,\phi \right\rangle \right| \le \Vert \phi \Vert _{L^{\infty }(\Omega )}k\int _{\Omega }|f(x)|w(x)\,dx, \end{aligned}$$

as desired. $\square$

Remark A.3

In the same way, we may define the distribution $(wz,\nabla h(u))$ for any Lipschitz function h on ${{\mathbb {R}}}$ such that the support of its derivative is compact, namely, for any $\phi \in C_c^\infty (\Omega )$,

$$\begin{aligned} \left\langle \left( zw,\nabla h(u)\right) ,\phi \right\rangle :=\int _\Omega f(x)h(u(x))\phi (x)w(x)\,dx -\int _\Omega h(u(x))z(x)\cdot \nabla \phi (x)w(x)\,dx. \end{aligned}$$

Following the proof of Proposition A.2, we know that $(z,\nabla h(u))$ is a Radon measure, which implies that, for any $k\in (0,\infty )$ & $\eta \in [0,\infty )$,

$$ \begin{aligned} \left( wz,\nabla T_k\left( u-T_\eta (u)\right) ^+\right) \quad \& \quad \left( wz,\nabla T_k\left( u-T_\eta (u)\right) ^-\right) \end{aligned}$$

are Radon measures satisfying

$$\begin{aligned} {\left\{ \begin{array}{ll} |\langle (wz,\nabla T_k(u-T_\eta (u))^+),\phi \rangle | \le \Vert \phi \Vert _{L^{\infty }(\Omega )}k\int _{\{u\ge \eta \}}|f(x)|w(x)\,dx;\\ |\langle (wz,\nabla T_k(u-T_\eta (u))^-),\phi \rangle | \le \Vert \phi \Vert _{L^{\infty }(\Omega )}k\int _{\{-u\ge \eta \}}|f(x)|w(x)\,dx, \end{array}\right. } \end{aligned}$$

respectively.

Proposition A.4

The Radon measures

$$ \begin{aligned} \left( wz,\nabla T_k\left( u-T_\eta (u)\right) ^+\right) \quad \& \quad \left( wz,\nabla T_k\left( u-T_\eta (u)\right) ^-\right) \end{aligned}$$

are, respectively, concentrated on the sets

$$ \begin{aligned} \left\{ x\in \Omega :\ \eta \le u(x)\le k+\eta \right\} \quad \& \quad \left\{ x\in \Omega :\ \eta \le -u(x)\le k+\eta \right\} . \end{aligned}$$

In particular, when $\eta =0$, the Radon measure $(wz,\nabla T_k(u))$ is concentrated on the set $\{|u|\le k\}$.

Proof

Due to the similarity, we only prove that $(wz,\nabla T_k(u-T_\eta (u))^+)$ is concentrated on the set $\{\eta \le u\le k+\eta \}$. To this end, it suffices to show that, for any open set $U\Subset \Omega$ (namely, $\overline{U}$ is compact in $\Omega$),

$$\begin{aligned}&\left( wz,\nabla T_k\left( u-T_\eta (u)\right) ^+\right) \left( U\cap \{u>k+\eta \}\right) =0\\&\quad =\left( wz,\nabla T_k\left( u-T_\eta (u)\right) ^+\right) \left( U\cap \{u<\eta \}\right) . \end{aligned}$$

Fix $U\Subset \Omega$ and denote by $\{\rho _m\}_{m\in {{\mathbb {N}}}}$ a sequence of mollifiers. Define

$$ \begin{aligned} z_m:=\rho _m*(zw)\quad \& \quad f_m:=\rho _m*(fw). \end{aligned}$$

Then

$$\begin{aligned} {\left\{ \begin{array}{ll} f_m=-\textrm{div}(wz_m)\quad \textrm{in}\ \ U,\quad &{}\mathrm{for\ large\ \ }m;\\ z_m\rightarrow zw\quad \textrm{in}\ \ L^1(U;{{{{\mathbb {R}}}}^n}), \quad &{}\mathrm{as\ \ }m\rightarrow \infty ;\\ f_m\rightarrow fw\quad \textrm{in}\ \ L^1(U), \quad &{}\mathrm{as\ \ }m\rightarrow \infty . \end{array}\right. } \end{aligned}$$

By the result [5, Corollary 1.6], we obtain

$$\begin{aligned}&\int _{U\cap \{u>k+\eta \}}\,d\left| \left( z_m,\nabla T_k\left( u-T_\eta (u)\right) ^+\right) \right| \\&\hspace{0.25cm}\le \big \Vert |z_m|\big \Vert _{L^{\infty }(\Omega )}\left| \nabla T_k\left( u-T_\eta (u)\right) ^+\right| \left( U\cap \left\{ u>k+\eta \right\} \right) \\&\hspace{0.25cm}\le \Vert \rho _m\Vert _{L^{\infty }(\Omega )}\Vert wz\Vert _{L^1(\Omega )}\left| \nabla T_k\left( u-T_\eta (u)\right) ^+\right| \left( U\cap \left\{ u>k+\eta \right\} \right) , \end{aligned}$$

which, combined with the fact that

$$\begin{aligned} \left| \nabla T_k\left( u-T_\eta (u)\right) ^+\right| \left( \left\{ u>k+\eta \right\} \right) =0, \end{aligned}$$

further shows

$$\begin{aligned} \int _{U\cap \{u>k+\eta \}}\,d\left( z_m,\nabla T_k\left( u-T_\eta (u)\right) ^+\right) =0,\quad \forall \,m\in {{\mathbb {N}}}. \end{aligned}$$

Similarly, we have

$$\begin{aligned} \int _{U\cap \{u<\eta \}}\,d\left( z_m,\nabla T_k\left( u-T_\eta (u)\right) ^+\right) =0,\quad \forall \,m\in {{\mathbb {N}}}. \end{aligned}$$

Moreover, we observe that, for any $\phi \in C_c^\infty (U)$,

$$\begin{aligned}&\left| \int _U T_k\left( u-T_\eta (u)\right) ^+ w(x)z\cdot \nabla \phi \,dx\right| \\&\hspace{0.25cm}\le \left| \left\langle \left( wz,\nabla T_k(u-T_\eta (u))^+\right) ,\phi \right\rangle \right| +\left| \int _U T_k\left( u(x)-T_\eta (u(x))\right) ^+f(x)\phi (x)w(x)\,dx\right| \\&\hspace{0.25cm}\le 2\Vert \phi \Vert _{L^{\infty }(\Omega )}k\int _{\{u\ge \eta \}}|f(x)|\,dx, \end{aligned}$$

which, together with the fact that

$$\begin{aligned} \int _U T_k\left( u-T_\eta (u)\right) ^+ w(x)z\cdot \nabla \phi (x)\,dx =\lim _{m\rightarrow \infty }\int _U T_k\left( u-T_\eta (u)\right) ^+ w(x)z_m\cdot \nabla \phi (x)\,dx \end{aligned}$$

(A.2)

implies

$$\begin{aligned} \left| \int _U T_k\left( u-T_\eta (u)\right) ^+ w(x)z_m\cdot \nabla \phi (x)\,dx\right| \le 3\Vert \phi \Vert _{L^{\infty }(\Omega )}k\int _{\{u\ge \eta \}}|f(x)|w(x)\,dx. \end{aligned}$$

From this and the fact that

$$\begin{aligned} \int _U T_k\left( u-T_\eta (u)\right) ^+ f_m(x)\phi (x)\,dx =\int _U T_k\left( u-T_\eta (u)\right) ^+ w(x)z_m\cdot \nabla \phi \,dx \end{aligned}$$

(A.3)

we deduce that, for large m,

$$\begin{aligned}&\left| \left\langle \left( z_m,\nabla T_k\left( u-T_\eta (u)\right) ^+\right) ,\phi \right\rangle \right| \\&\hspace{0.25cm}\le \left| \int _U T_k\left( u-T_\eta (u)\right) ^+ w(x)z_m\cdot \nabla \phi (x)\,dx\right| \\&\qquad +\left| \int _U T_k\left( u(x)-T_\eta (u(x))\right) ^+f_m(x)\phi (x)w(x)\,dx\right| \\&\hspace{0.25cm}\le 5\Vert \phi \Vert _{L^{\infty }(\Omega )}k\int _{\{u\ge \eta \}}|f(x)|w(x)\,dx. \end{aligned}$$

Furthermore, by (A.2) and (A.3), we know that, when $m\rightarrow \infty$,

$$\begin{aligned}&\left( z_m,\nabla T_k\left( u-T_\eta (u)\right) ^+\right) \mid _U\rightarrow \left( wz,\nabla T_k\left( u-T_\eta (u)\right) ^+\right) \mid _U,\\&\qquad \qquad \qquad \qquad \qquad *-\mathrm{weakly\ as\ measures}. \end{aligned}$$

Thus,

$$\begin{aligned} {\left\{ \begin{array}{ll} \int _{U\cap \{u>k+\eta \}}\,d\left( wz,\nabla T_k\left( u-T_\eta (u)\right) ^+\right) (x)\\ \qquad =\displaystyle \lim _{m\rightarrow \infty }\int _{U\cap \{u>k+\eta \}}\,d\left( z_m,\nabla T_k\left( u-T_\eta (u)\right) ^+\right) (x)=0;\\ \int _{U\cap \{u<\eta \}}\,d\left( wz,\nabla T_k\left( u-T_\eta (u)\right) ^+\right) (x)\\ \qquad =\displaystyle \lim _{m\rightarrow \infty }\int _{U\cap \{u<\eta \}}\,d\left( z_m,\nabla T_k\left( u-T_\eta (u)\right) ^+\right) (x)=0, \end{array}\right. } \end{aligned}$$

as desired. $\square$

Corollary A.5

For any $\epsilon \in (0,\infty )$, the Radon measures

$$ \begin{aligned} \left( wz,\nabla T_\epsilon \left( u-T_k(u)\right) ^+\right) \quad \& \quad \left( wz,\nabla T_\epsilon \left( u-T_k(u)\right) ^-\right) \end{aligned}$$

are, respectively, concentrated on the sets

$$ \begin{aligned} \left\{ x\in \Omega :\ k\le u(x)\le k+\epsilon \right\} \quad \& \quad \left\{ x\in \Omega :\ k\le -u(x)\le k+\epsilon \right\} . \end{aligned}$$

Proposition A.6

Let $(wz,\nabla 1_{\{u>k\}})$ & $(wz,\nabla 1_{\{-u>k\}})$ be the Radon measures defined in (3.10) and (3.11), respectively. Then, for any $\phi \in C_c^\infty (\Omega )$,

$$\begin{aligned} {\left\{ \begin{array}{ll} \left\langle \left( wz,\nabla 1_{\{u>k\}}\right) ,\phi \right\rangle =\lim _{\epsilon \rightarrow 0}\epsilon ^{-1}\left\langle \left( wz,\nabla T_\epsilon \left( u-T_k(u)\right) ^+\right) , \phi \right\rangle ;\\ \left\langle \left( wz,\nabla 1_{\{-u>k\}}\right) ,\phi \right\rangle =\lim _{\epsilon \rightarrow 0}\frac{-1}{\epsilon }\left\langle \left( wz,\nabla T_\epsilon \left( u-T_k(u)\right) ^-\right) . \phi \right\rangle \end{array}\right. } \end{aligned}$$

Moreover,

$$ \begin{aligned} \left( wz,\nabla 1_{\{u>k\}}\right) \quad \& \quad \left( wz,\nabla 1_{\{-u>k\}}\right) \end{aligned}$$

are, respectively, concentrated on

$$ \begin{aligned} \{u=k\}\quad \& \quad \{-u=k\}. \end{aligned}$$

As a consequence, the Radon measure

$$\begin{aligned} \left( wz,\nabla 1_{\{|u|>k\}}\right) \quad \mathrm{is\ concentrated\ on\ \ } \{|u|=k\}. \end{aligned}$$

(A.4)

Proof

Let $\phi \in C_c^\infty (\Omega )$ and let $u_p$ be the renormalized solution to the problem (1.12). From some arguments used in Step 4 of the proof of Theorem 1.15, we deduce

$$\begin{aligned}&-\epsilon ^{-1}\int _{\{k\le u_p<k+\epsilon \}} \left| \nabla u_p(x)\right| ^p\phi (x)w(x)\,dx\nonumber \\&\hspace{0.25cm}=\epsilon ^{-1}\int _\Omega f(x)T_\epsilon \left( u_p(x)-T_k(u_p(x))\right) ^+\phi (x)w(x)\,dx\nonumber \\&\hspace{0.25cm}\hspace{0.25cm}-\epsilon ^{-1}\int _\Omega T_\epsilon \left( u_p(x)-T_k(u_p(x))\right) ^+ \left| \nabla u_p(x)\right| ^{p-2}\nabla u_p(x)\cdot \nabla \phi (x)w(x)\,dx. \end{aligned}$$

(A.5)

Letting $p\rightarrow 1$ and $\epsilon \rightarrow 0$ in (A.5), we have

$$\begin{aligned} \lim _{p\rightarrow 1}-\epsilon ^{-1}\int _{\{k\le u_p<k+\epsilon \}} \left| \nabla u_p(x)\right| ^p\phi (x)w(x)\,dx =\epsilon ^{-1}\left\langle \left( wz,\nabla T_\epsilon \left( u-T_k(u)\right) ^+\right) ,\phi \right\rangle \end{aligned}$$

(A.6)

and

$$\begin{aligned} \lim _{\epsilon \rightarrow 0}\lim _{p\rightarrow 1}-\epsilon ^{-1}\int _{\{k\le u_p<k+\epsilon \}} \left| \nabla u_p(x)\right| ^p\phi (x)w(x)\,dx =\left\langle \left( wz,\nabla {\textbf {1}}_{\{u>k\}}\right) ,\phi \right\rangle . \end{aligned}$$

By this and (A.6), we obtain

$$\begin{aligned} \left\langle \left( wz,\nabla {\textbf {1}}_{\{u>k\}}\right) ,\phi \right\rangle =\lim _{\epsilon \rightarrow 0}\epsilon ^{-1}\left\langle \left( wz,\nabla T_\epsilon \left( u-T_k(u)\right) ^+\right) ,\phi \right\rangle , \end{aligned}$$

which, combined with Corollary A.5, implies that the measure $(wz,\nabla 1_{\{u>k\}})$ is concentrated on the set

$$\begin{aligned} \bigcap _{m=1}\left\{ k\le u\le k+m^{-1}\right\} =\{u=k\}. \end{aligned}$$

Similarly, we can also obtain the desired result for the measure $(wz,\nabla {\textbf {1}}_{\{-u>k\}})$.

As a consequence of

$$\begin{aligned} (wz,\nabla {\textbf {1}}_{\{|u|>k\}}) =(wz,\nabla {\textbf {1}}_{\{u>k\}})+(wz,\nabla {\textbf {1}}_{\{-u>k\}}), \end{aligned}$$

(A.4) holds. $\square$

1.3 A.3 Weighted Weak Trace on $\partial \Omega$ of the Normal Component of z

In this subsection, we define $[wz,\nu ]$, which is called the weighted weak trace on $\partial \Omega$ of the normal component of z. Recall that

$$ \begin{aligned} z\in L_w^{\frac{n}{n-1},\infty }(\Omega ;{{{{\mathbb {R}}}}^n})\quad \& \quad -w^{-1}\textrm{div}(wz)=f\quad \textrm{in}\quad \left( C_c^\infty (\Omega )\right) '. \end{aligned}$$

Let $v\in W_w^{1-\frac{1}{q},q}(\partial \Omega )\cap L^\infty (\partial \Omega )$ for some $q>N$. Then there exists

$$\begin{aligned} V\in W_w^{1,q}(\Omega )\cap L^{\infty }(\Omega )\end{aligned}$$

such that $V\mid _{\partial \Omega }=v$.

Proposition A.7

Define

$$\begin{aligned} \left\langle wz,v\right\rangle _{\partial \Omega } :=\int _\Omega z(x)\nabla V(x)w(x)\,dx -\int _{\Omega }f(x)V(x)w(x)\,dx. \end{aligned}$$

(A.7)

• (i) The value $\langle wz,v\rangle _{\partial \Omega }$ defined in (A.7) is independent of the choice of the function V;

• (ii) for any $q\in (N,\infty )$, the linear map $\langle wz,\cdot \rangle _{\partial \Omega }$ on

  $$\begin{aligned} W_w^{1-\frac{1}{q},q}(\partial \Omega )\cap L^\infty (\partial \Omega )\quad \mathrm{is\ continuous\ in\ \ } W_w^{1-\frac{1}{q},q}(\partial \Omega ). \end{aligned}$$

We write

$$\begin{aligned} \int _{\partial \Omega }\left[ wz,\nu \right] (x)v(x)\,d{{\mathcal {H}}}^{n-1}(x) \end{aligned}$$

instead of $\langle wz,v\rangle _{\partial \Omega }$.

Now we define

$$ \begin{aligned} \int _{\partial \Omega }\left[ wz{\textbf {1}}_{\{u=\infty \}},\nu \right] (x)v(x)\,d{{\mathcal {H}}}^{n-1}(x) \quad \& \quad \int _{\partial \Omega }\left[ wz{\textbf {1}}_{\{u=-\infty \}},\nu \right] (x)v(x)\,d{{\mathcal {H}}}^{n-1}(x). \end{aligned}$$

It is easy to verify that

$$\begin{aligned} {\left\{ \begin{array}{ll} -\textrm{div}(wz{\textbf {1}}_{\{u=\infty \}})=wf{\textbf {1}}_{\{u=\infty \}}- (wz,\nabla {\textbf {1}}_{\{u=\infty \}});\\ -\textrm{div}(wz{\textbf {1}}_{\{u=-\infty \}})=wf{\textbf {1}}_{\{u=-\infty \}}- (wz,\nabla {\textbf {1}}_{\{u=-\infty \}});\\ -\textrm{div}(wz{\textbf {1}}_{\{|u|<\infty \}})=wf{\textbf {1}}_{\{|u|<\infty \}}- (wz,\nabla {\textbf {1}}_{\{|u|=\infty \}})\\ \end{array}\right. } \end{aligned}$$

hold in the sense of distributions, where, for any $\phi \in C_c^\infty (\Omega )$,

$$\begin{aligned} \left\langle \left( wz,\nabla {\textbf {1}}_{\{|u|=\infty \}}\right) ,\phi \right\rangle :=\int _{\{|u|=\infty \}}f(x)\phi (x)w(x)\,dx -\int _{\{|u|=\infty \}}z(x)\cdot \nabla \phi (x)w(x)\,dx, \end{aligned}$$

$$ \begin{aligned} \left\langle \left( wz,\nabla {\textbf {1}}_{\{u=\infty \}}\right) ,\phi \right\rangle \quad \& \quad \left\langle \left( wz,\nabla {\textbf {1}}_{\{u=-\infty \}}\right) ,\phi \right\rangle \end{aligned}$$

are similar to define.

Then we may write

$$\begin{aligned}&\int _\Omega \left[ wz{\textbf {1}}_{\{u=\infty \}},\nu \right] (x)v(x)w(x)\,d{{\mathcal {H}}}^{n-1}(x)\\&\quad =\int _{\Omega }w(x)z(x){\textbf {1}}_{\{u=\infty \}}\cdot \nabla V(x)\,dx-\int _{\{u=\infty \}}f(x)V(x)w(x)\,dx\\&\qquad +\int _{\Omega }V(x)\,d\left( wz,\nabla {\textbf {1}}_{\{u=\infty \}}\right) \end{aligned}$$

and

$$\begin{aligned}&\int _\Omega \left[ wz{\textbf {1}}_{\{u=-\infty \}},\nu \right] (x)v(x)w(x)\,d{{\mathcal {H}}}^{n-1}(x)\\&\quad =\int _{\Omega }w(x)z(x){\textbf {1}}_{\{u=-\infty \}}\cdot \nabla V(x)\,dx-\int _{\{u=-\infty \}}f(x)V(x)w(x)\,dx\\&\qquad +\int _{\Omega }V(x)\,d\left( wz,\nabla {\textbf {1}}_{\{u=-\infty \}}\right) . \end{aligned}$$

Moreover, we have, for any

$$\begin{aligned}&v\in W^{1-\frac{1}{q},q}_w(\partial \Omega )\cap L^\infty (\partial \Omega )\quad \textrm{with}\ \ q\in (n,\infty ), \\&\int _{\partial \Omega }\left[ wz,\nu \right] (x)v(x)\,d{{\mathcal {H}}}^{n-1}(x) =\int _{\partial \Omega }\left[ wz{\textbf {1}}_{\{|u|<\infty \}},\nu \right] (x)v(x)w(x)\,d{{\mathcal {H}}}^{n-1}(x)\\&\hspace{0.25cm}+\int _{\partial \Omega }\left[ wz{\textbf {1}}_{\{u=\infty \}},\nu \right] (x)v(x)w(x)\,d{{\mathcal {H}}}^{n-1}(x)\\&\hspace{0.25cm}+\int _{\partial \Omega }\left[ wz{\textbf {1}}_{\{u=-\infty \}},\nu \right] (x)v(x)w(x)\,d{{\mathcal {H}}}^{n-1}(x). \end{aligned}$$