Continuity of the Free Boundary in a Problem involving the A-Laplacian

TL;DR

This paper studies a two-dimensional free boundary problem with the A-Laplacian, demonstrating that the free boundary can be locally represented by continuous graphs, advancing understanding of boundary regularity in nonlinear PDEs.

Contribution

It proves the local continuity of the free boundary in a problem involving the A-Laplacian, a novel result in nonlinear free boundary problems.

Findings

The free boundary is locally represented by continuous graphs.

The analysis advances understanding of boundary regularity for A-Laplacian problems.

Provides foundational results for further regularity studies.

Abstract

In this paper we investigate a two dimensional free boundary problem involving the A-Laplacian. We show that the free boundary is represented locally by graphs of a family of continuous functions.

Equations

$$(P)\{\begin{array}{ll}& \text{ Find }(u,\chi )\in W^{1,A}(\Omega )\times L^{\infty }(\Omega )\text{ such that}:\\ & (i)0⩽u⩽M,0⩽\chi ⩽1,u(1-\chi )=0\text{ a.e. in }\Omega \\ & (ii){\Delta }_{A}u=-div(\chi H(x))\text{in }(W_0^{1,A}(\Omega ){)}^{\prime },\end{array}$$

$$a_0⩽\frac{t{a}^{\prime }(t)}{a(t)}⩽a_1\forall t>0.$$

$$\Big{(}{a(|\xi|)\over|\xi|}\xi-{a(|\zeta|)\over|\zeta|}\zeta\Big{)}.(\xi-\zeta)>0\qquad\forall\xi,\zeta\in\mathbb{R}^{2}\setminus\{0\},\quad\xi\neq\zeta.$$

$$|H_1|⩽\overline{h},0<\underline{h}⩽H_2⩽\overline{h}\text{ in }\Omega$$

$$H\in C^{0,1}(\overline{\Omega })$$

$$div(H)⩾0\text{ a.e. in }\Omega$$

$$div(H)⩽\overline{h}\text{ a.e. in }.$$

$$W^{1,A}(\Omega )=\left\{u\in L^A(\Omega ):|\nabla u|\in L^A(\Omega )\right\},\text{where}$$

$$L^A(\Omega )=\left\{u\in L^1(\Omega ):{\int }_{\Omega }A(|u(x)|)dx<\infty \right\}$$

$$div(\chi H)-\chi (\{u>0\})div(H)⩽0\text{ in }{\mathcal{D}}^{\prime }(\Omega ).$$

$$\displaystyle{(E(w,h))\,\,\left\{\begin{array}[]{r@{\quad=\quad}l}X^{\prime}(t,w,h)&H(X(t,w,h))\\ X(0,w,h)&(w,h)\end{array}\right.}$$

$$T_h:D_h⟶T_h(D_h)$$

$$(t,w)⟼T_h(t,w)=(T_h^1,T_h^2)(t,w)=X(t,w).$$

$${\partial\over\partial t}\big{(}\chi oT_{h}\big{)}\leqslant 0\qquad\text{ in }{\cal D}^{\prime}(D_{h}).$$

$$uo{T}_h(t,w)>0\forall (t,w)\in C_{ϵ}=\{(t,w)\in D_h/|w-w_0|<ϵ,t<t_0+ϵ\}.$$

$${\varphi }_h(w)=\{\begin{array}{ll}& \sup \left\{t:(t,w)\in D_h,uo{T}_h(t,w)>0\right\}\\ & \text{ if this set is not empty }\\ & {\alpha }_{-}(w)\text{ otherwise.}\end{array}$$

$$\{uo{T}_h(t,w)>0\}\cap D_h=\{t<{\varphi }_h(w)\}.$$

$$|S(k,w)-S(k_0,w_0)|⩽C(|k-k_0|+|w-w_0|)\forall (k,w),(k_0,w_0)\in domain(S).$$

$$k=h+{\int }_0^{t_k(w)}{H}_2(X(t,w))ds\text{ and }{k}_0=h+{\int }_0^{t_{k_0}(w_0)}{H}_2(X(t,w_0))ds.$$

$$k-k_0={\int }_0^{t_k(w)}{H}_2(X(t,w))ds-{\int }_0^{t_{k_0}(w_0)}{H}_2(X(t,w_0))ds.$$

$$\underline{h}(t_k(w)-t_{k_0}(w_0))$$

$${\int }_{t_{k_0}(w_0)}^{t_k(w)}{H}_2(X(s,w))ds)$$

$$\underline{h}(t_k(w)-t_{k_0}(w_0))$$

$$t_k(w)-t_{k_0}(w_0)$$

$$t_{k_0}(w_0)-t_k(w)$$

$$(P)\left\{\begin{array}[]{ll}\text{Find }(u,\chi)\in W^{1,A}(U)\times L^{\infty}(U)\text{ such that }:\quad u=0\hbox{ on }\Gamma&\\ \quad 0\leqslant u\leqslant M,\quad 0\leqslant\chi\leqslant 1,\quad u(1-\chi)=0\,\,\text{ a.e. in }U&\hbox{}\\ \displaystyle{\int_{U}}\Big{(}{a(|\nabla u|)\over|\nabla u|}\nabla u+\chi H(x)\Big{)}.\nabla\zeta dx\leqslant 0&\hbox{}\\ \quad\forall\zeta\in W^{1,A}(U),\quad\zeta\geqslant 0\quad\hbox{ on }\Gamma,\quad\zeta=0\quad\hbox{ on }\partial U\setminus\Gamma.&\hbox{}\end{array}\right.$$

$$Z_k=T_h\left(\left\{(t,w)\in D_h,w\in (w_1,w_2),t>t_k(w)\right\}\right)=T_h(\{w_1<w<w_2\})\cap \left\{x_2>k\right\}.$$

$$\displaystyle\int_{Z_{k}}\Big{(}{a(|\nabla u|)\over|\nabla u|}\nabla u+\chi H(x)\Big{)}.\nabla\zeta dx\leqslant 0$$

$$\forall \zeta \in W^{1,A}(Z_k)\cap C^0({\overline{Z}}_k),\zeta ⩾0on{\overline{Z}}_k\setminus \{x_2=k\},\zeta =0on{\overline{Z}}_k\cap \{x_2=k\}.$$

$$\displaystyle\int_{Z_{k}}\Big{(}{a(|\nabla u|)\over|\nabla u|}\nabla u.\nabla\zeta dx-\int_{Z_{k}}\chi_{\{u>0\}}div(H).\zeta dx\leqslant 0$$

$$\forall \zeta \in W^{1,A}(Z_k)\cap C^0({\overline{Z}}_k),\zeta ⩾0on{\overline{Z}}_k\setminus \{x_2=k\},\zeta =0on{\overline{Z}}_k\cap \{x_2=k\}.$$

$${\int }_{Z_k}{F}_{ϵ}(u)\frac{a(|\nabla u|)}{|\nabla u|}\nabla u.\nabla \zeta dx+{\int }_{Z_k}H(x).\nabla (F_{ϵ}(u)\zeta )dx$$

$$⩽-{\int }_{Z_k}{F}_{ϵ}^{\prime }(u)\zeta |\nabla u|a(|\nabla u|)dx⩽0$$

$${\int }_{Z_k}{F}_{ϵ}(u)\frac{a(|\nabla u|)}{|\nabla u|}\nabla u\nabla \zeta dx-{\int }_{Z_k}div(H).F_{ϵ}(u)\zeta dx⩽0.$$

Full text

Key words : $A$-Laplacian, Free boundary, Continuity

AMS 2000 Mathematics Subject Classification: 35R35; 35J60

Introduction

The authors have considered the following problem in [15]

[TABLE]

where $\Omega$ is an open bounded domain of $\mathbb{R}^{2}$, $x=(x_{1},x_{2})$, $M$ is a positive constant, $\displaystyle{A(t)=\int_{0}^{t}a(s)ds}$, \displaystyle{\Delta_{A}=div\Big{(}{{a(|\nabla u|)}\over{|\nabla u|}}\nabla u\Big{)}} is the $A$-Laplacian, $a$ is a $C^{1}$ function from $[0,\infty)$ to $[0,\infty)$ such that $a(0)=0$ and for some positive constants $a_{0},a_{1}$

[TABLE]

As a consequence of (0.1), we have the following monotonicity inequality (see [9])

[TABLE]

For examples of functions $a(t)$, we refer to [16].

$H=(H_{1},H_{2})$ is a vector function that satisfies for some positive constants $\underline{h},\overline{h}$

[TABLE]

The Orlicz-Sobolev space $W^{1,A}(\Omega)$ is defined by:

[TABLE]

Throughout this paper, we shall denote by $B_{r}(x)$ a ball with center $x$ and radius $r$.

In [15], we showed that for any solution $(u,\chi)$, $u$ is Lipschitz continuous and that the free boundary is a union of graphs of a family of lower semi-continuous functions depending only on the vector function $H$. In this paper, we will show that these functions are actually continuous and that $\chi$ is the characteristic function of the set $\{u>0\}$.

Problem $(P)$ describes a variety of free boundary problems including the lubrication problem [1], and the dam problem [4], [5], [6], [7], [8], [14], [20], [21], [22], [25], and [26]. For a more general framework, we refer to [2], [3], [18], [9], [11], [12], [13] and [15]. Regarding the problem with a Newman boundary condition, we refer the reader to [23] and [24].

1 The free boundary

The free boundary is defined as the intersection between the two sets $\{u=0\}$ and $\overline{\{u>0\}}$. When $H_{1}=0$ and $H_{2}$ is a constant function, it is easy to show as in [13] that $\chi_{x_{2}}\leqslant 0$ in ${\cal D}^{\prime}(\Omega)$ and that the free boundary is the graph of a continuous function $x_{2}=\phi(x_{1})$. When $H$ is not a constant vector, we can show as in [11] that

[TABLE]

Actually (1.1) can be obtained from $(P)ii)$ and by adapting the proof of Lemma 1.4. As a consequence of (1.1), a weak monotonicity of the function $\chi$ (see [15]) holds, which means that $\chi$ decreases along the orbits of the following differential equation:

[TABLE]

where $h\in\pi_{x_{2}}(\Omega)$ and $w\in\pi_{x_{1}}(\Omega\cap\{x_{2}=h\})$, $\pi_{x_{1}}$ and $\pi_{x_{2}}$ are respectively the orthogonal projections on the $x_{1}$ and $x_{2}$ axes. We will denote by $X(.,w)$ the maximal solution of $E(w,h)$ defined on the interval $(\alpha_{-}(w),\alpha_{+}(w))$. We know [14] that the limits $\displaystyle{\lim_{t\rightarrow\alpha_{-}(w)^{+}}X(t,w)}\in\partial\Omega\cap\{x_{2}<h\}$ and $\displaystyle{\lim_{t\rightarrow\alpha_{+}(w)^{-}}X(t,w)}\in\partial\Omega\cap\{x_{2}>h\}$ both exit, which we shall denote respectively by $X(\alpha_{-}(w),w)$ and $X(\alpha_{+}(w),w)$. We shall also denote the orbit of $X(.,w)$ by $\gamma(w)$.

Now, we recall for the reader’s convenience a few technical properties and definitions established in [11] and [15]:

$\bullet$ $\alpha_{+}$ and $\alpha_{-}$ are uniformly bounded.

$\bullet$ For each $h\in\pi_{x_{2}}(\Omega)$, the mapping is one to one

[TABLE]

where $\displaystyle{D_{h}=\{(t,w)\,/\,w\in\pi_{x_{1}}(\Omega\cap\{x_{2}=h\}),\,t\in(\alpha_{-}(w),\alpha_{+}(w))\}}$.

$\bullet$ $\displaystyle{\Omega=\bigsqcup_{h\in\pi_{x_{2}}(\Omega)}T_{h}(D_{h})}$.

$\bullet$ $T_{h}$ and $T_{h}^{-1}$ are $C^{0,1}$. $\bullet$ The determinant of the Jacobian matrix of the mapping $T_{h}$, denoted by $Y_{h}(t,w)$, satisfies:

i)  \displaystyle{Y_{h}(t,w)=-H_{2}(w,h)\exp\big{(}\int_{0}^{t}(divH)(X(s,w))ds\big{)}}\qquad\mbox{a.e. in }D_{h}.

ii)  $\underline{h}\leqslant-Y_{h}(t,w)\leqslant C{\overline{h}},\quad C>0,\qquad\mbox{a.e. in }D_{h}.$

The following monotonicity of $\chi$ based on (1.1) (see [11], [15]) is the key point in parameterizing the free boundary:

[TABLE]

Property (1.2) means that $\chi$ decreases along the orbits of the differential equation $(E(w,h))$. The consequence of this monotonicity on $u$ is materialized in the next theorem established in [15].

Theorem 1.1**.**

Let $(u,\chi)$ be a solution of $(P)$ and $x_{0}=T_{h}(t_{0},w_{0})\in T_{h}(D_{h})$.

i) If $\quad u(x_{0})=uoT_{h}(t_{0},w_{0})>0,\quad$ then there exists $\epsilon>0$ such that

[TABLE]

*ii) If $\quad u(x_{0})=uoT_{h}(t_{0},w_{0})=0,\quad$ then *$\quad uoT_{h}(t,w_{0})=0\qquad\forall t\geqslant t_{0}.$

The proof of Theorem 1.1 is based on the following strong maximum principle:

Lemma 1.1**.**

Let $u\in W^{1,A}(U)\cap C^{1}(U)\cap C^{0}(\overline{U})$ such that $u\geqslant 0$ in $U$ and $\Delta_{A}u\leqslant 0$ in $U$. Then $u\equiv 0$ in $U$ or $u>0$ in $U$.

Thanks to Theorem 1.1, we can define for each $h\in\pi_{x_{2}}(\Omega)$, the following function $\phi_{h}$ on $\pi_{x_{1}}(\Omega\cap\{x_{2}=h\})$ (see [15]):

[TABLE]

Then one can easily establish the following proposition as in [7]:

Proposition 1.1**.**

For each $h\in\pi_{x_{2}}(\Omega)$, the function $\phi_{h}$ is lower semi-continuous at each $w\in\pi_{x_{1}}(\Omega\cap\{x_{2}=h\})$ such that $T_{h}(\phi_{h}(w),w)\in\Omega$. Moreover

[TABLE]

Lemma 1.2**.**

Let $h\in\pi_{x_{2}}(\Omega)$. For each $k\in\pi_{x_{2}}(\Omega)$ and $w\in\pi_{x_{1}}(\Omega\cap\{x_{2}=h\})$, let $t_{k}(w)$ be the unique value of $t$ at which the orbit $\gamma(w)$ intersects with the line $\{x_{2}=k\}$ if it exists. Then the function $S(k,w)=t_{k}(w)$ is Lipschitz continuous in its domain. More precisely, we have for some positive constant $C$

[TABLE]

Proof. Let $(k,w),(k_{0},w_{0})\in domain(S)$. First we have from the differential equation $(E(w,h))$

[TABLE]

If we subtract the two equalities from each other, we get

[TABLE]

Next, if we assume that $t_{k}(w)>t_{k_{0}}(w_{0})$, then we get by (0.3)

[TABLE]

Now, observe that

[TABLE]

Using (1.4), (1.6) and the fact that $H_{2}\circ X$ is Lipschitz continuous in $\overline{D_{h}}$, and since $t_{k_{0}}(w_{0})$ is bounded independently of $k$ and $w$, we obtain from (1.5), for some positive constant $C_{0}$

[TABLE]

which leads for $C=\max(1,C_{0})/\underline{h}$, to

[TABLE]

If $t_{k}(w)<t_{k_{0}}(w_{0})$, we get in a similar fashion

[TABLE]

Combining (1.7) and (1.8), the lemma follows. ∎Our main goal is to prove that for each $h\in\pi_{x_{2}}(\Omega)$, the function $\phi_{h}$ is actually continuous. Due to the local character of this result, we will confine ourselves to the following situation:

We assume that $u=0$ on an open and connected subset $\Gamma$ of $\partial\Omega$ and consider a free boundary point in a neighborhood of the form U=T_{h}\big{(}(w_{*},w^{*})\times(h,\infty)\big{)}\cap\Omega\subset T_{h}(D_{h}) such that $T_{h}\left(\left\{(w,\alpha_{+}(w)),\,\,w\in(w_{*},w^{*})\right\}\right)\subset\subset\Gamma$. So we are led to study the following problem:

[TABLE]

We observe that the free boundary $(\partial\{u>0\})\cap U$ is the graph of the lower semi-continuous function $\phi_{h}$ in $(w_{*},w^{*})$. Our objective is to prove the continuity of the function $\phi_{h}$. To do that, it is enough to show that it is upper semi-continuous. To this end, we need to generalize few lemmas previously established for a linear operator in [11]. In the sequel and without notice, we shall denote by $(u,\chi)$ a solution of the problem $(P)$.

Lemma 1.3**.**

Let $w_{1},w_{2}\in(w_{*},w^{*})$ with $w_{1}<w_{2}$, and let $k\in\pi_{x_{2}}(U)$ such that $\{x_{2}=k\}\cap\gamma(w_{i})\neq\emptyset$, $i=1,2$. Let

[TABLE]

If $uoT_{h}(t_{k}(w_{i}),w_{i})=0$ for $i=1,2$, then we have

[TABLE]

The proof of Lemma 1.3 is inspired from the one of a similar lemma in [2] obtained for the case $H(x)=(h(x),0)$. Our proof is based on the next lemma.

Lemma 1.4**.**

Under the assumptions of Lemma 1.3, we have

[TABLE]

Proof. Let $\zeta$ be as in the lemma, $\epsilon>0$, and F_{\epsilon}(u)=\min\big{(}{u^{+}\over\epsilon},1\big{)}. Using $\chi(Z_{k})F_{\epsilon}(u)\zeta$ as a test function for $(P)$, we get

[TABLE]

Integrating by parts, we obtain

[TABLE]

The lemma follows by letting $\epsilon$ go to [math] in (1.9). ∎

Proof. For $\epsilon>0$ small enough, let $\alpha_{\epsilon}(w)=\displaystyle\min\left(1,\frac{(w-w_{1})^{+}}{\epsilon},\frac{(w_{2}-w)^{+}}{\epsilon}\right)$, and observe that

[TABLE]

Since $\chi(Z_{k})(\alpha_{\epsilon}\circ T_{h}^{-1})\zeta$ is a test function for (P), we have:

[TABLE]

Now, if we apply Lemma 1.4 to the function $(1-\alpha_{\epsilon}\circ T_{h}^{-1})\zeta$, we get

[TABLE]

Taking into account (1.11)-(1.12), we obtain from (1.10)

[TABLE]

Using the change of variables $x=T_{h}(t,w)$ and arguing as in the proof of Theorem 2.1 in [11], we get

[TABLE]

Hence we derive from (1.13) and (1.14)

[TABLE]

The lemma follows by letting $\epsilon$ go to [math] in (1.15). ∎

Lemma 1.5**.**

Let $x_{0}=T_{h}(t_{0},w_{0})$ be a point in $U$.

If $uoT_{h}=0$ in $B_{r}(t_{0},w_{0})$, then

[TABLE]

where $C_{r}=\{(t,w)\in D_{h},\quad|w-w_{0}|<r,\quad t>t_{0}\}\cup B_{r}(t_{0},w_{0}).$

Proof. By Theorem 1.1 $ii)$, we have $uoT_{h}=0$ in $C_{r}$. Applying Lemma 1.2 with domains $Z_{k}=T_{h}(\{w_{1}<w<w_{2}\})\cap\{x_{2}>k\}\subset T_{h}(C_{r})$, ($k\in\pi_{x_{2}}(U)$) and taking $\zeta=x_{2}-k$, we obtain $\displaystyle{\int_{Z_{k}}\chi H_{2}dx\leqslant 0}$. Then we deduce from (0.3) that $\chi=0$ a.e. in $Z_{k}$. This holds for all domains $Z_{k}$ in $T_{h}(C_{r})$. Hence $\chi=0$ a.e. in $T_{h}(C_{r})$.∎

Lemma 1.6**.**

Let $x_{0}=T_{h}(t_{0},w_{0})\in U$ such that $B_{r}=B_{r}(t_{0},w_{0})\subset D_{h}$. Then we cannot have the following three situations

[TABLE]

Proof. Assume that $ii)$ holds. The proofs of $i)$ and $iii)$ are based on similar arguments. Let $\zeta\in\mathcal{D}(T_{h}(B_{r}))$, $\zeta\geqslant 0$. Using the fact that, by Lemma 1.3, $\chi oT_{h}=0$ a.e. in $B_{r}\cap\{w\leqslant w_{0}\}$, we obtain after using the change of variable $T_{h}$

[TABLE]

This means that $\triangle_{A}u\leqslant 0$ in $\mathcal{D}^{\prime}(T_{h}(B_{r}))$. By Lemma 1.1, either $u>0$ or $u=0$ in $T_{h}(B_{r})$, which contradicts the assumption.∎

2 Continuity of the free boundary

As mentioned in section 1, to prove the continuity of the function $\phi_{h}$, it is enough to show that it is upper semi-continuous. The main idea to do that is to compare $u$ with a suitable barrier function near a free boundary point. In the following step, we construct such a function. For this purpose, let $\epsilon>0$, $w_{1},w_{2}\in(w_{*},w^{*})$ such that $w_{1}<w_{2}$, $k\in\pi_{x_{2}}(U)$, and assume that $\epsilon$ is small enough to guarantee that $Z_{k}^{k+\epsilon}(w_{1},w_{2})=T_{h}(\{w_{1}<w<w_{2}\})\cap\{k<x_{2}<k+\epsilon\}\subset\subset U$ and $\epsilon<\underline{h}/{2\overline{h}}$.

Then the function $\overline{v}_{\epsilon}$ defined by

[TABLE]

satisfies

[TABLE]

Now let $v_{\epsilon}$ be the unique solution in $W^{1,A}(Z_{k}^{k+\epsilon}(w_{1},w_{2}))$ of

[TABLE]

Then we have:

Lemma 2.1**.**

[TABLE]

Proof. $i)$ Note that $v_{\epsilon}^{-}\in W^{1,A}(Z_{k}^{k+\epsilon}(w_{1},w_{2}))$ and $v_{\epsilon}^{-}=0$ on $\partial Z_{k}^{k+\epsilon}(w_{1},w_{2})$. Therefore we obtain from (2.2) and (0.5)

[TABLE]

Taking into account (2.4) and the fact that $ta(t)$ is an increasing function, we deduce that $\nabla v_{\epsilon}^{-}=0$ a.e. in $Z_{k}^{k+\epsilon}(w_{1},w_{2})$. Since $v_{\epsilon}^{-}=0$ on $\partial Z_{k}^{k+\epsilon}(w_{1},w_{2})$, we must have $v_{\epsilon}^{-}=0$ in $Z_{k}^{k+\epsilon}(w_{1},w_{2})$. Hence $v_{\epsilon}\geqslant 0$ in $Z_{k}^{k+\epsilon}(w_{1},w_{2})$.

$ii)$ Similarly, we observe that $(v_{\epsilon}-\overline{v}_{\epsilon})^{+}\in W^{1,A}(Z_{k}^{k+\epsilon}(w_{1},w_{2}))$ and $(v_{\epsilon}-\overline{v}_{\epsilon})^{+}=0$ on $\partial Z_{k}^{k+\epsilon}(w_{1},w_{2})$. Therefore we obtain from (2.1) and (2.2)

[TABLE]

Subtracting (2.6) from (2.5), and using (0.6), we get

[TABLE]

Taking into account (2.7) and (0.2), we obtain $\nabla(v_{\epsilon}-v_{\epsilon})^{+}=0$ a.e. in $Z_{k}^{k+\epsilon}(w_{1},w_{2})$. Since $(v_{\epsilon}-\overline{v}_{\epsilon})^{+}=0$ on $\partial Z_{k}^{k+\epsilon}(w_{1},w_{2})$, we get $(v_{\epsilon}-\overline{v}_{\epsilon})^{+}=0$ in $Z_{k}^{k+\epsilon}(w_{1},w_{2})$. Hence $v_{\epsilon}\leqslant\overline{v}_{\epsilon}$ in $Z_{k}^{k+\epsilon}(w_{1},w_{2})$.

∎

Lemma 2.2**.**

After extending $v_{\epsilon}$ by [math] to $Z_{k+\epsilon}$, we obtain

[TABLE]

Proof. First we have $\Delta_{A}v_{\epsilon}=-divH\leqslant 0$ in $Z_{k}^{k+\epsilon}(w_{1},w_{2})$, and by (2.3), $v_{\epsilon}\geqslant 0$ in $Z_{k}^{k+\epsilon}(w_{1},w_{2})$. By Lemma 1.1, we obtain $v_{\epsilon}>0$ in $Z_{k}^{k+\epsilon}(w_{1},w_{2})$.

Note that $v_{\epsilon}=0$ on $L=\partial Z_{k}^{k+\epsilon}(w_{1},w_{2})\cap\{x_{2}=k+\epsilon\}$ and $v_{\epsilon}\in C^{1,\alpha}_{loc}(Z_{k}^{k+\epsilon}(w_{1},w_{2})\cup L)$ for some $\alpha\in(0,1)$ (see [18]).

Next we claim that

[TABLE]

From Lemma 2.1, we have $v_{\epsilon}\leqslant\overline{v}_{\epsilon}$ in $Z_{k}^{k+\epsilon}(w_{1},w_{2})$. In particular we have

[TABLE]

We obtain $(2.8)$ since

[TABLE]

Now if $\nu$ is the outward unit normal vector to $L$, then we have by (2.8), since $\epsilon\in(0,\underline{h}/{2\overline{h}})$

[TABLE]

Finally, for $\zeta\in W^{1,A}(Z_{k})$, $\zeta\geqslant 0$, $\zeta=0$ on $\partial Z_{k}\cap U$, we obtain from (2.2) and (2.9)

[TABLE]

∎

Lemma 2.3**.**

Assume that

[TABLE]

Then we have

[TABLE]

Proof. For $\delta,\eta>0$, let $F_{\delta}(s)$ as in the proof of Lemma 1.4, $\,d_{\eta}(x_{2})=F_{\eta}(x_{2}-\overline{k})$ and $\overline{k}=k+\epsilon$. By applying Lemma 1.2 and Lemma 2.1 respectively for $\zeta=F_{\delta}(u-v_{\epsilon})+d_{\eta}(1-F_{\delta}(u))$ and $\zeta=F_{\delta}(u-v_{\epsilon})$, we get

[TABLE]

[TABLE]

Adding these inequalities, we get since $d_{\eta}=0$ in $\{v_{\epsilon}>0\}$

[TABLE]

Since

[TABLE]

we obtain $\displaystyle{\lim_{\eta\rightarrow 0}I_{1}^{\delta\eta}=0}$.

As for $I_{2}^{\delta\eta}$, we have

[TABLE]

since we have in $Z_{k}\cap\{u=v_{\epsilon}=0\}$

[TABLE]

Let $J=\{w\in(w_{1},w_{2})\,\,/\,\,\phi_{h}(w)>t_{\overline{k}}(w)\,\}$. Then given that $u\in C^{0,1}_{loc}(U)$, one has for some positive constant $C$

[TABLE]

Since the function $t\mapsto 1-F_{\delta}(uoT_{h}(t,w))$ is continuous, we obtain

[TABLE]

Hence

[TABLE]

The Lemma follows by letting $\delta\rightarrow 0$. ∎

Lemma 2.4**.**

Assume that the assumptions of Lemma 2.3 hold. Then we have

[TABLE]

where $\displaystyle{\mathbb{A}(\xi)=\left(\mathbb{A}_{ij}\right)}$, $\displaystyle{\mathbb{A}_{ij}={\partial\mathcal{A}^{i}\over\partial x_{j}}}$, and $\displaystyle{\mathcal{A}^{i}(\xi)={a(|\xi|)\over|\xi|}\xi_{i}}$.

Proof. First, we observe that we have for any $\zeta\in\mathcal{D}(Z_{k}^{k+\epsilon}(w_{1},w_{2}))$

[TABLE]

Next we have

[TABLE]

By Lemma 2.3, we have

[TABLE]

Regarding $I_{\delta}^{1}$, we have from $(P)ii)$ and (2.2), since $(F_{\delta}(u-v_{\epsilon})\zeta)\in W^{1,A}_{0}(Z_{k}^{k+\epsilon}(w_{1},w_{2}))$ and $\chi=1$ a.e. in $\{u>v_{\epsilon}\}$

[TABLE]

It follows from (2.11)-(2.14) that

[TABLE]

which can be written as

[TABLE]

where $\displaystyle{\mathcal{A}(\xi)=(\mathcal{A}^{1},\mathcal{A}^{2})(\xi)={a(|\xi|)\over|\xi|}\xi}$ and $w_{t}=tu+(1-t)v_{\epsilon}$.

Now observe that

[TABLE]

Hence we obtain (2.10) from (2.15) and (2.16). ∎

Lemma 2.5**.**

[TABLE]

Proof. Let $z\neq 0$ and $\xi=(\xi_{1},\xi_{2})\in\mathbb{R}^{2}$. Since $A\in C^{1}(\mathbb{R}^{2}\setminus\{(0,0)\})$, we have by direct calculation

[TABLE]

Using (0.1), we obtain

[TABLE]

Then, if $a_{0}\geqslant 1$, the left hand side of inequality (2.17) holds. When $a_{0}<1$, we use Cauchy-Schwartz inequality : $|z.\xi|\leqslant|z||\xi|$, to conclude. We proceed in the same way for the right hand side.∎

Lemma 2.6**.**

Assume that the assumptions of Lemma 2.3 hold. Then only one of the following situations holds

[TABLE]

Proof. Assume that $(i)$ is false. Then

[TABLE]

This leads by Theorem 1.1 $ii)$ to

[TABLE]

We will show in this case that $(ii)$ holds. From Lemmas 2.4 and 2.5 we know that

[TABLE]

Moreover, by Lemma 2.5, the matrix $\mathbb{A}(x)$ satisfies

[TABLE]

Next, we have $v_{\epsilon}\in C^{1,\alpha}(Z_{k}^{k+\epsilon}(w_{1},w_{2})\cup L)$, where

[TABLE]

We also have $v_{\epsilon}=0$ on $L$ and $v_{\epsilon}>0$ in $Z_{k}^{k+\epsilon}(w_{1},w_{2})$. So $v_{\epsilon}$ achieves its minimum value on the line segment $L$. By Lemma 2.2 of [15], we must have $|\nabla v_{\epsilon}|>0$ along $L$. Therefore for $\delta$ small enough such that $w_{1}+\delta<w_{2}-\delta$, there exists two positive constants $c_{0}$, $c_{1}$ such that

[TABLE]

On the other hand, $|\nabla u|$ is also bounded in $Z_{k+\epsilon-\delta}^{k+\epsilon}$ since $u\in C^{0,1}(\overline{Z}_{k}^{k+\epsilon}(w_{1},w_{2}))$ ([15] Theorem 1.1). It follows from (2.20)-(2.21) that we have for two positive constants $\lambda_{0}$ and $\lambda_{1}$

[TABLE]

and therefore we get from (2.20)

[TABLE]

Taking into account (2.18), we see that

[TABLE]

It follows from (2.19), (2.22), (2.23), and the strong maximum principle that $(u-v_{\epsilon})^{+}\equiv 0$ in $Z_{k+\epsilon-\delta}^{k+\epsilon}$. Consequently, we obtain $u\leqslant v_{\epsilon}$ in $Z_{k+\epsilon-\delta}^{k+\epsilon}$, and then $uoT_{h}(t_{k+\epsilon}(w),w)=0$ for all $w\in(w_{1}+\delta,w_{2}-\delta)$. Since $\delta$ is arbitrary small, we get $uoT_{h}(t_{k+\epsilon}(w),w)=0$ for all $w\in(w_{1},w_{2})$. Hence $(ii)$ holds by Theorem 1.1 $ii)$.

∎

Lemma 2.7**.**

Let $w_{0}\in(w_{*},w^{*})$, $x_{0}=T_{h}(t_{0},w_{0})$ such that for some $\eta>0$, $B_{\eta}(T_{h}(t_{0},w_{0}))\subset\subset U$ and $u(x_{0})=0$. Then there exists two sequences $(t_{n}^{-},w_{n}^{-})_{n}$ and $(t_{n}^{+},w_{n}^{+})_{n}$ such that for all $n$:

[TABLE]

Proof. First we observe that by Lemma 1.6 the following situations cannot hold simultaneously

[TABLE]

In fact, to prove the lemma, it is enough to show that neither $a)$ nor $b)$ holds. So assume for example that $a)$ holds. Then by Lemma 1.6 there exists a sequence $(t_{n}^{+},w_{n}^{+})_{n}$ such that $T_{h}(t_{n}^{+},w_{n}^{+})\in B_{\eta}(T_{h}(t_{0},w_{0}))\cap\{w>w_{0}\}$,

[TABLE]

Let $k=\max\{T_{h}^{2}(t_{0},w_{0}),T_{h}^{2}(t_{n}^{+},w_{n}^{+})\}$. Then since $u(x_{0})=0$ and $u$ is continuous at $x_{0}$, we may assume that for $n$ large enough, we have

[TABLE]

For $\epsilon>0$ small enough and $n$ large enough, we may assume that

[TABLE]

Using (2.24), (2.25) and Lemma 2.6, we conclude that for $\epsilon>0$ small enough and $n$ large enough, we have $u=0$ in $Z_{k+\epsilon}\cap T_{h}(\{w_{0}<w<w_{n}^{+}\})$. Now since we have assumed that $a)$ holds, we are in contradiction with Lemma 1.6.

Similarly, if we assume that $b)$ holds, we will get a contradiction as well.

∎

Finally, by using Lemma 2.6 and Lemma 2.7, we can establish the main result of the paper.

Theorem 2.1**.**

The function $\phi_{h}$ is continuous in the interval $(w_{*},w^{*})$.

Proof. Let $w_{0}\in(w_{*},w^{*})$. We will prove that $\phi_{h}$ is continuous at $w_{0}$. To this end, it is enough to show that $\phi_{h}$ is upper semi-continuous at $w_{0}$.

Let $x_{0}=T_{h}(\phi_{h}(w_{0}),w_{0})=T_{h}(t_{0},w_{0})$ and let $\epsilon>0$. Since $u(x_{0})=0$ and $u$ is continuous at $x_{0}$, there exists $\eta\in(0,\epsilon)$ such that

[TABLE]

By Lemma 2.7 there exists two sequences $(t_{n}^{-},w_{n}^{-})_{n}$ and $(t_{n}^{+},w_{n}^{+})_{n}$ such that for all $n$:

[TABLE]

Let $k=\max\{T_{h}^{2}(t_{n}^{-},w_{n}^{-}),T_{h}^{2}(t_{0},w_{0}),T_{h}^{2}(t_{n}^{+},w_{n}^{+})\}$ and let $C$ be the constant in Lemma 1.2. We observe that we can choose $\epsilon$ small enough and $n$ large enough so that

[TABLE]

Using (2.26), (2.27) and Lemma 2.6, we see that for $n$ large enough, we have

[TABLE]

Therefore we obtain

[TABLE]

From Lemma 1.2, we infer that if $\eta<\epsilon/4C$:

[TABLE]

Combining (2.28) and (2.29), we obtain

[TABLE]

which is the upper semi-continuity of $\phi_{h}$ at $w_{0}$.∎

Remark 2.1**.**

Thanks to Theorem 2.1 and (1.3), and taking into account that $\chi=1$ a.e. in $\{u>0\}$, we obtain

[TABLE]

*Acknowledgments * The authors are grateful for the facilities and excellent research conditions provided by Fields Institute where part of this research was carried out.