Quadrature identities with a background PDE

Abstract

Our prime goal with this text is to introduce a nonlinear version of quadrature identities, related to semilinear PDEs, and discuss a few basic properties.

1 Introduction

1.1 Background

For any given (bounded) open set $D \subset {\mathbb R}^n$ $(n \ge 2)$, let $HL^1 (D)$, and $SL^1 (D)$ denote the set of integrable harmonic, respectively subharmonic functions over D.

It is known (see [13, 15]) that for a given positive function $f \ge c_0> 0$, and a non-negative bounded function (generally a measure) $\mu$, concentrated enough, one can find a bounded open set $D \Supset \hbox {supp} (\mu )$ such that the quadrature identity (inequality) holds

$$\begin{aligned} \int _{D } h(y) (f (y) -\mu ) dy \ge 0, \qquad \forall \ h \in SL^1(D). \end{aligned}$$

(1.1)

Here, the open set D, with this property, is called a Quadrature Domain.¹ Now, if a QI (for the harmonic class)

$$\begin{aligned} \int _{D } h(y) (f (y) -\mu ) dy = 0 , \qquad \forall \ h \in HL^1(D), \end{aligned}$$

(1.2)

is given then, a priori it is not obvious whether the triple $(\mu , f, D)$ also constitutes a quadrature identity for the class $SL^1(D)$. This depends on the fact that generally the theory of quadrature domains does not offer us uniqueness results. We refer to [7, 8] for further background.

Consider now a hybrid quadrature identity which is modeled by adding a boundary integral to the left hand side of (1.2)

$$\begin{aligned} \int _{D } h(y) (f (y) -\mu ) dy + \int _{ \partial D } \lambda _0 h(y) d\sigma _y = 0 , \qquad \forall h \in H( {\overline{D}}), \end{aligned}$$

(1.3)

where $\lambda _0 \ge 0$ is a given constant (or a smooth function) and, to avoid technicalities, we have reduced the class to $H( {\overline{D}})$, which denotes harmonic functions over ${\overline{D}}$.

Equation (1.3) in terms of Newtonian kernels K(x) is expressed as

$$\begin{aligned} U^{f\chi _{D} -\mu , \lambda _0} (x) = 0 \qquad \forall \ x \in D^c. \end{aligned}$$

(1.4)

Here for any functions (or bounded measures) $a ,b \ge 0$ we have used the notation

$$\begin{aligned} U^{a,b} (x) = U^{a,b,D} (x)= \int _D a (y) K(x-y) dy + \int _{\partial D } b (y) K(x -y) d\sigma _y, \end{aligned}$$

where $K(x) = c_n |x|^{2-n}$, and in ${\mathbb R}^2$ we have $K(x) = c_2\log |x|$; i.e., the fundamental solution to the Laplace operator normalised such that $\Delta K(x) = - \delta _0(x)$.

By standard approximation theory (see [14]) one can show the equivalence between (1.3) and (1.4).

1.2 Nonlinear quadrature identities

We shall now introduce a nonlinear version of quadrature identities, related to semilinear problems. To do so, let us consider a bounded domain D, and a solution u to the following (background) PDE²

$$\begin{aligned} \Delta u = g(x,u) \quad \hbox {in } D, \qquad u= 0 \quad \hbox {on } \partial D, \end{aligned}$$

(1.5)

where, unless otherwise stated, $g (x,u) = g_2(x,u) - \mu$, with $g_2 (x,u) \approx c_0 (u_+)^a$ for $u \approx 0$ with $-1< a < 1$, and $\mu \ge 0$ is a bounded function (or a Radon measure) with compact support. The equation above is in the sense of distributions.

Suppose now that for $\lambda _0 \ge 0$, the following quadrature identity holds

$$\begin{aligned} \int _{D } g(y,u (y) ) h(y) dy + \int _{ \partial D} \lambda _0 h(y) d\sigma _y = 0 , \qquad \forall h \in H( {\overline{D}}). \end{aligned}$$

(1.6)

Using the Newtonian kernel K(x) as in (1.4), we have

$$\begin{aligned} \Delta U^{g (\cdot , u)\chi _D ,\lambda _0} = g (x,u) \quad \hbox {in } D, \end{aligned}$$

and

$$\begin{aligned} U^{g(\cdot , u) \chi _D,\lambda _0} (x) = 0 , \qquad \partial _\nu U^{g(\cdot , u) \chi _D,\lambda _0} = \lambda _0 \quad x \in \partial D, \end{aligned}$$

(1.7)

where $\nu$ is the inward unit normal vector on $\partial D$.

From (1.5)–(1.7) we conclude $u= U^{g (\cdot , u)\chi _D ,\lambda _0}$. Therefore u solves the semilinear free boundary problem

$$\begin{aligned} \Delta u = g(x,u) \quad \hbox {in } D, \qquad u= 0 ,\quad \partial _\nu u = \lambda _0 \quad \hbox {on } \partial D. \end{aligned}$$

(1.8)

By Green’s theorem the converse also holds true. I.e., if D admits a solution to (1.8), then D is a quadrature domain in the sense of (1.6).

The classical QI (1.2) in light of this new definition has the background PDE

$$\begin{aligned} \Delta u = f -\mu \quad \hbox {in } D, \qquad u= 0 \quad \hbox {on } \partial D, \end{aligned}$$

where $f-\mu$ plays the role of g(x, u(x) ), and $\lambda _0 =0$.

Definition 1

(Nonlinear QD) We say a bounded open set D is a QD for the class $H({\bar{D}})$, and with respect to the pair $(\lambda _0, g(x,u))$ if u is a solution to equation (1.5) and the QI (1.6) is satisfied. We call u the background potential, and equation (1.5) the background PDE.

As just discussed the equivalent definition for a nonlinear QD is given by the semilinear free boundary problem (1.8).

Remark 1

It is noteworthy, that even though our definition of nonlinear QD may seem slightly unorthodox, for QD-community, such types of considerations/definitions are common practices in shape optimization problems; see [3] and the references therein. In domain variational approach for shape optimization one gives a domain and a solution to a PDE inside the domain, with boundary values (usually zero), and asks to minimize certain functionals among all such domains. The optimal shape is then a solution to a free boundary problem, where the free boundary condition depends on the functional to be minimized.

Example 1

We mention some examples of g(x, u) that are connected to standard free boundary problems.

• a) For $-1<q < 1$, and $\mu$ concentrated enough (say an smooth approximation of Dirac masses) one can take

  $$\begin{aligned} g(x,u) = u^q - \mu , \qquad \lambda _0 = 0, \qquad u \ge 0, \end{aligned}$$

  which resembles semilinear free boundary problem, that has been well studied; see [2], and [4].

• b) The well-known Pompeiu problem³ with $\mu =0$, and

  $$\begin{aligned} g(x,u) = -a u + 1, \qquad \lambda _0 = 0 . \end{aligned}$$

• c) Schiffer’s conjecture⁴ (relates also to Pompeiu problem) with $\mu =0$

  $$\begin{aligned} g(x,u) = -au , \qquad \lambda _0 >0 . \end{aligned}$$

In b)-c) u may change sign, and thus the famous conjectures of solutions to b)-c) being spherical, becomes very difficult.

2 Main results

In this section we shall state the main results, that concern the nonlinear quadrature domains. Our intention is primarily the introduction of the concept and statement of some basic facts, that link to already existing results for the standard quadrature domains.

Therefore, rather than getting involved with various technical statements and tools, we shall state and prove existence, in a general framework, and prove standard uniqueness results, along with a symmetry problem for quadrature domains. The latter is also new in the context of standard QD. These results, in the eyes of experts, are obvious but we hope it can tease the appetite of others new to the topic.

It is noteworthy, that the topic of quadrature domains with a Helmholtz PDE background arises in non-scattering phenomena in recent literature. This partly motivates our work here. We also refer to recent works [10, 11, 16] for further background in terms of minimization and partial balayage in potential theory.

To state an existence theory we need to define the corresponding functional, whose minimizers give us possible solutions to the quadrature domain problem, in terms of the PDE (1.8).

Define G(x, u) to be the function satisfying $\frac{\partial G (x,u)}{\partial u}=g(x,u)$ and assume $\lambda _0 \ge 0$,⁵ and

$$\begin{aligned} J(u) = \int _{{\mathbb R}^n} |\nabla u|^2 + 2G(x,u) + \lambda _0^2 \chi _{\{u>0\}}, \end{aligned}$$

(2.1)

with its admissible class

$$\begin{aligned} {\mathbb {K}} =\{u \in W_{0}^{1,2}({\mathbb {R}}^n): u \ge 0\} . \end{aligned}$$

Standing Assumptions (on G , and g): To assure the existence of minimizers, we would need the functional to be bounded from below. Indeed, even simple examples such as $G(x,u) = au + bu^2$ with |b| large may cause the functional not to be bounded from below; see e.g. [11]. Hence the simplest way to avoid this technical point of view would be to assume

$$\begin{aligned} \int _{{\mathbb R}^n} |\nabla u|^2 + 2G(x,u) \ge -C > -\infty , \qquad \hbox { for all } u \in {\mathbb {K}}. \end{aligned}$$

This encompasses a large class of functions G(x, u), including classical QD, and also the Helmholtz QD, for small frequencies (see [11]).

We further assume that for some positive constants $0<c_1< c_2 < \infty$, and a bounded continuous function $0< c_0< c_0 (x) < c_0^{-1}$, the function $g = g_2 -\mu$ satisfies

$$\begin{aligned} g_2 (x, t) = c_0 (x) t_+^{a}, \quad -1<a<1, \quad t \approx 0, \\ c_1 t^a \le g_2(x,t) \le c_2 t^a \quad \hbox {for } \, t> t_1 >0 , \hbox {where}\,\, t_1 \approx 0, \end{aligned}$$

and

$$\begin{aligned} \mu \ge 0 , \qquad \hbox {is a bounded function with compact support. } \end{aligned}$$

These assumptions, in particular, imply that

$$\begin{aligned} G(x,u) = G_2 (x,u) - \mu u, \qquad \hbox {where }\ \partial _u G_2 = g_2. \end{aligned}$$

(2.2)

The assumption on $g_2$ (which plays the role of a sink) falls under conditions considered by [2], which forces the support of the minimizer to be bounded, given that $\mu$ (the source) has compact support.

We further assume the following concentration condition for $\mu$: For each $z\in \hbox {supp}(\mu )$, there is a radius $r>0$, such that

$$\begin{aligned} u_r (z) > 0, \qquad \hbox { for any minimizer } u_r \text{ of } J_r \end{aligned}$$

(2.3)

with

$$\begin{aligned} J_r(v) = \int _{{\mathbb R}^n } |\nabla v|^2 + 2G_r (x,v) + \lambda _0^2 \chi _{\{v>0\}}, \quad G_r (x,v) = G_2(x,v) - v \mu \chi _{B_r(z)}.\nonumber \\ \end{aligned}$$

(2.4)

Theorem 1

(Existence) For the functional J, under Standing Assumption above and with $\lambda _0 \ge 0$, there is a minimizer u to J solving (1.8), with $D = \{ u > 0\}$. Moreover, if $\lambda _0=0$, and condition (2.3)-(2.4) is satisfied for some $B_r(z)$ then $z \in \{u > 0\}$. In particular if this holds for all $z \in supp(\mu )$, then supp$(\mu ) \subset D$.

This theorem may be sharpened substantially using Sakai’s concentration principle but (probably) would need more restrictions on g(x, u), see [7, 15]. Our assumptions somehow suggests a localization property of concentration of the source versus that of the sink.

When $\partial _u G$ is non-Lipschitz, e.g. $\partial _u G (x,u) \approx u^{ a }$ ($-1< a <0$), or $\lambda _0 > 0$, one would require further restrictions on the measure $\mu$, along with tools from geometric measure theory to obtain results that hint towards supp$(\mu ) \subset D$. This is outside the scope of this paper.

We now state a uniqueness result for the PDE formulation (1.8) of the quadrature domain.

Theorem 2

(Uniqueness) Let u be a non-negative solution to (1.8), or equivalently, let D be a quadrature domain according to (1.6), with background PDE (1.5), and with non-negative potential $U^{g(\cdot , u) \chi _D,\lambda _0} (x) \ge 0$. Suppose further

$$\begin{aligned} g_2 \ge 0, \qquad \partial _u g_2 (x,u) \ge 0 . \end{aligned}$$

(2.5)

For $\lambda _0 =0$ assume D is solid,⁶ and for $\lambda _0 > 0$ assume D is convex and $\partial D$ is $C^{1,dini}$.

Then the following hold:

• (1) D is the largest QD, satisfying the QI (1.6). More precisely, if there is another QD, $D_1$ with the background potential $u_1$, then $D_1 \subset D$ and $u_1 \le u$.

• (2) Moreover, in case $\lambda _0 = 0$ we have D is unique (i.e $D_1 = D$), and in case $\lambda _0 > 0$ the domain D is unique among all convex QD, with $C^{1,dini}$ boundary.

Our next result is a symmetry problem, that is enforced through imposing an extra geometric condition on the QI (1.6), for a particular g(x, u). Similar problems, for potentials, have been considered earlier in the literature. See e.g., [6], and for Bessel potentials see [9], and [12], where the authors, under certain mild assumptions, show the only bounded domains whose Bessel potential is constant on the boundary are balls.

To state our rigidity theorem, we need certain assumptions on the function $g_2 (t)$, that falls within the range of our earlier work [4].

Further assumptions on g :

For the next theorem we assume $g(x,u) = g_2 (u) - g_1 (u) \chi _{D_1}$, for some bounded domain $D_1$. We also assume that for some $-1< a < 1$, $\kappa _0 > 0$ and $t_1 >0$ (small) g has the following properties:

$$\begin{aligned} {\left\{ \begin{array}{ll} a)\ &{} C_t : = \limsup _{\epsilon> 0} \frac{g(t+ \epsilon ) - g(t)}{\epsilon } \le \kappa _0 t^{a-1},\quad t>0 , \\ b)\ &{} g_2(t) = 0 \qquad \hbox { for } t \le 0, \\ c)\ &{} |g(t)| \le C_0 \qquad \hbox { for all } t > t_1 ,\\ d)\ &{} g_i(t) \ge 0 \qquad i =1, 2 .\\ \end{array}\right. } \end{aligned}$$

(2.6)

Finally when $-1<a<0$ and $\lambda _0 =0$, we also assume the following asymptotic expansion for u: At any free boundary point $x^0$ the weak second-order asymptotic expansion holds

$$\begin{aligned} u(x) = A_0 ((x-x^0)\cdot \nu _0)_+^\beta + O(|x-x^0|^{2+ \delta _\beta } ), \end{aligned}$$

(2.7)

for some fixed $A_0 > 0$, a unit normal $\nu _0= \nu _0 (x^0)$, and some $0< \delta _\beta < \beta - 1$, with $\beta = 2/(1-a)$; see discussion in [4].

Theorem 3

(Symmetry and Rigidity) Let $\lambda _0=0$, and D be a bounded $C^1$-domain, which is a QD, in the sense of Definition 1, with background PDE (1.5) such that

$$\begin{aligned} g(x,u) = g_2 (u) - g_1 (u) \chi _{D_1}, \qquad D_1 \Subset D, \end{aligned}$$

(2.8)

where $u \ge 0$, the inclusion $D_1$ is $C^{1,dini}$ domain, and condition (2.6) is satisfied. Assume also (2.7) is satisfied, on D when $a<0$. Suppose further that there is a constant $m>0$ such that

$$\begin{aligned} U^{ g\chi _D, 0 } (x) = m , \qquad \hbox {for all }\ x \in \partial D_1. \end{aligned}$$

(2.9)

Then D and $D_1$ are concentric balls, and $U^{g\chi _D, 0 } (x)$ is spherically symmetric around the center of D.

We remark that similar results can be obtained when $\lambda _0 > 0$, provided one assume that $\partial D$ is $C^2$, and u is $C^2$ in a neighborhood, and up to the the free boundary. For clarity of exposition we have not included this in the statement of the theorem. The reader may easily deduce such a result.

3 Proofs of main theorems

3.1 Proof of Theorem 1

The proof of this theorem follows in the same spirit as that of Theorem 1.4 in [7] (for classical setting), and [11] for the case of Helmholtz operator. We shall sketch in a few lines the steps needed, leaving out the details to readers to check out.

The approach for existence of a minimizer uses a few steps, the first being a comparison argument, showing that smaller $\lambda _0$, and g (i.e., larger $\mu$, and smaller $g_2$) gives rise to larger solutions with larger support; see Lemma 1.1 in [7]. Using this comparison argument, one can compare the minimizer of the functional to another functional, with symmetric ingredients and such that $G^* (x,\cdot )\le G (x,\cdot )$. Here $G^*$ is a symmetrization in the x variable of G. Next one can use symmetric rearrangement to show that solutions have compact support for the corresponding symmetric functional. This part usually is based on explicit computation. From here one may (using again the comparison argument) consider a restriction of the admissible class to those with compact supports.

Next by lower-semicontinuity, boundedness of J(u) from below, and Fatou type lemma we shall obtain a global minimizer u to the functional, with support $D=\{u >0\}$ being compact.

When $\lambda _0 =0$, one easily checks that a minimizer of this functional over ${\mathbb {K}}$ satisfies equations (1.5) and (1.6).

Now, let z be the point as stated in the theorem, and $u_r$ the minimizer obtained in (2.3)–(2.4). By Lemma 1.1 in [7], we know that $\max (u, u_r)$ is also a minimizer of the function J, and hence $u \ge u_r$. Hence $u(z) >0$. This completes the proof of the theorem. $\square$

It is noteworthy that in the second statement of Theorem 1, the assumption (2.3)–(2.4) is equivalent to the PDE

$$\begin{aligned} \exists \ h : \qquad \Delta h = g(x,h) \hbox { and } \ h >0 \quad \hbox {in } B_r(z), \qquad h= 0 \quad \hbox {on } \partial B_r(z), \end{aligned}$$

(3.1)

admitting a solution h, with $h (z) > 0$. Then $u(z) >0$. We leave the obvious proof to the readers.

E.g., for classical QD theory, that is when $g(x,h) = 1 - \mu$, it suffices that $\mu = A \chi _S$, with $A > 1$. Then for $B_r (z) \subset \hbox {supp}(\mu )$ we have that the PDE above has a positive solution $h= (A-1) (r^2 - |x-z|^2)/2n$. Hence $B_r(z) \subset \{ u > 0\}$, and hence $\hbox {supp}(\mu )\subset \{u >0\}$.

3.2 Proof of Theorem 2

Aiming at a contradiction, suppose there exists another QD, $D_1$ with potential $u_1$ satisfying

$$\begin{aligned} \Delta u_1 = g(x,u_1)\,\,\text { in } D_1, \,\,\,\, u_1=0 \,\, \text { on } \partial D_1 \end{aligned}$$

and

$$\begin{aligned} U^{g(\cdot , u_1) \chi _{D_1},\lambda _0} (x) = 0 , \qquad x \in D_1^c. \end{aligned}$$

The aim is to prove that $D_1 \subset D$. Define $w:=u_1 -u + \lambda _0 \left( d(x) -d_1\right)$, where

$$\begin{aligned} d(x) = \hbox {distance}(x, D), \qquad d_1 = \sup _{D_1 } d(x). \end{aligned}$$

Since $u \ge 0$, we obviously have $w\le 0$ on $\partial D_1$. When $\lambda _0 > 0$, by the assumption that D is convex, we have d(x) is subharmonic in ${\mathbb R}^n$, and $\Delta d (x) \ge {{\mathcal {H}}}^{n-1}_{\partial _D}$ (Hausdorff measure on $\partial D$). The function u (after extending it by zero outside D) also satisfies $\Delta u (x) = g(x,u) \chi _D + \lambda _0 {{\mathcal {H}}}^{n-1}_{\partial D}$. When $\lambda _0 = 0$, the distance function is not involved anymore. We thus have in $D_1$

$$\begin{aligned} \Delta w= & {} g_2(x,u_1)\chi _{D_1 } - g_2(x,u)\chi _{D} - \lambda _0 {{\mathcal {H}}}^{n-1}_{\partial D} + \lambda _0 \Delta d(x) \ge c w\chi _{D_1 \cap D} \nonumber \\&+ g_2 (x,u_1) \chi _{D_1 \setminus D} , \end{aligned}$$

(3.2)

where $c =c(x)= (g_2(x,u_1) -g_2(x,u) )/(u_1 - u) \ge 0$ (by the monotonicity assumption). In case $D_1 \setminus D \ne \emptyset$, we have $w= u_1$ in $D_1 \setminus D$ and hence $c(x) = g_2(x,u_1)/u_1 \ge 0$ for $x\in D_1 \setminus D$ that is, $g_2 (x,u_1) \chi _{D_1 \setminus D} =c(x) w\chi _{D_1 \setminus D}$. We thus arrive at

$$\begin{aligned} \Delta w - c w \ge 0 \qquad \hbox {in } D_1, \end{aligned}$$

and conclude by the maximum principle that $w \le 0$ in $D_1$.

Consider now two cases.

Case $\lambda _0 = 0$: By the conclusion above $w \le 0$ in $D_1$, and since $\lambda _0 = 0$ we must have $u_1 \le u$ in $D_1$. Using that D is solid, we should have $D_1 \setminus D=\emptyset$, since otherwise $\Delta u_1 \ge 0$ on $D_1 \setminus D$, and this set contains points $z \in \partial D_1$ (where $u_1 (z) =0$), at the same time that $u_1 \le u \le 0$ on this set. Let r be small enough so that $B_r (z) \subset D^c$. Then $u_1$ would obtain a local maximum in $B_r (z)$, at z, contradicting the strong maximum principle. This proves case (1) in the statement of the theorem, when $\lambda _0=0$.

We consider now case (2) in the statement of the theorem, when $\lambda _0=0$. If $D\setminus D_1 \ne \emptyset$ then by positivity and monotonicity of g

$$\begin{aligned} 0 = \int _{D_1 }\Delta u_1 = \int _{D_1} g(x,u_1) \le \int _{D_1} g(x,u ) < \int _{D} g(x,u) = \int _{D }\Delta u = 0, \end{aligned}$$

(3.3)

which is a contradiction.

Case $\lambda _0 > 0$:

We first observe that $w= -u + \lambda _0 (d(x) - d_1)$ on $\partial D_1$. Obviously $w< 0$ on $\partial D_1 \cap D$, and $w= \lambda _0 (d(x) - d_1)$ on $\partial D_1 \setminus D$. Let $z \in \partial D_1$ be any point that realizes $d_1 = d(z)$. Then $w(z) = 0$, and the normal vector $\nu$ at $\partial D_1$, is also normal to the level sets of d(x) at z. Since $\partial D_1$ is $C^{1,dini}$, by Hopf boundary point lemma we must have

$$\begin{aligned} \frac{\partial w}{\partial \nu }(z)> 0 , \qquad \ \hbox { where}\, \nu \,\hbox {is the unit outward normal vector}, \end{aligned}$$

and consequently we arrive at the following contradiction

$$\begin{aligned} -\lambda _0 = \frac{\partial u_1}{\partial \nu }(z) > - \lambda _0\frac{\partial d}{\partial \nu }(z) = - \lambda _0. \end{aligned}$$

(3.4)

Here we have used that $u\equiv 0$ in $D_1 \setminus D$, and that at z we have $\frac{\partial d}{\partial \nu }(z) = |\nabla d(z) | = 1$. The latter depends on the fact that the level surface $d(x) = d_1$ and $\partial D_1$ have the same normal vector at z. Since D is convex, and hence solid, we conclude that $D_1 \setminus D = \emptyset$, and that $D_1 \subset D$. This proves the first statement in the theorem in the case $\lambda _0>0$. The second statement follows straightforwardly, by replacing the role of $D_1$ and D, whenever $D_1$ is another convex QD, with a solution $u_1 \ge 0$. $\square$

The proof of our symmetry theorem, is very standard, and as mentioned earlier uses moving plane-techniques. We thus give a sketch of the proof without much details. Readers may consult [4] for further detail on the technical issues, and for an interior symmetry problem.

3.3 Proof of Theorem 3

The function $u(x) := U^{g\chi _D, \lambda _0 } (x)$, satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \Delta u = g_2(u) - g_1 (u) \chi _{D_1} &{}\quad \hbox {in } D, \\ u(x) = m &{}\quad x \in \partial D_1,\\ u= \partial _\nu u =0 &{}\quad \hbox {on } \partial D ,\\ \end{array}\right. } \end{aligned}$$

(3.5)

We show that u is spherically symmetric, around a center, and the level subsets $\{ u > l\}$ are concentric balls.

The argument uses standard symmetry argument, by the moving plane technique; see [1, 17]. We shall however follow the lines of arguments in [4]. Since our problem is invariant under rigid motion, we may fix a direction, and prove symmetry in that direction. In doing so we choose the $x_1$-direction, and consider the following set up

$$\begin{aligned} {\left\{ \begin{array}{ll} T^{\tau } = \{x \in {\mathbb R}^n | \ x_1=\tau \}\, &{}(\text {the hyperplane}),\\ x^{\tau }=(2\tau -x_1,x_2,x_3,...)\,&{}(\text {the reflection of } x \,\, \text {at} \,\, T^{\tau }),\\ \Sigma ^\tau =\{x \, | x_1 < \tau , x^{\tau } \in D\}\,\,&{} \\ u^{\tau }(x)=u(x^{\tau }) \,&{}(\text {the reflection of}\,\,\, u \,\,\text {at}\,\, T^{\tau }).\ \end{array}\right. } \end{aligned}$$

(3.6)

Start the moving plane from far away ($x_1=$ large) until the plane touches the boundary $\partial D$, for some value $\tau = \tau _1$. Since $\partial D$ is $C^1$, for $\tau _1 -\epsilon< \tau < \tau _1$, and $\epsilon$ small enough, we have that $\Sigma ^\tau$ is inside D. For such values of $\tau$, consider $w^\tau := u - u^\tau$ in $\Sigma ^\tau$. When $a \ge 0$, we have $g' (u) \ge 0$, for $u \approx 0$ (see Standing Assumption), and hence on the set $\{w^\tau <0 \} \cap \Sigma ^\tau$, we have $\Delta w^{\tau } \le 0$, so the minimum principle applied to $w^\tau$ on that set gives us a contradiction. Therefore, $\{w^\tau <0 \} \cap \Sigma ^\tau = \emptyset$ for $\tau _1 -\epsilon< \tau < \tau _1$. For $a<0$, by the asymptotic property (2.7) we obviously have $w^\tau > 0$ in $\Sigma ^\tau$.⁷

Next we move the plane further in the negative $x_1$-direction, as long as $w^{\tau } > 0$. Let

$$\begin{aligned} \tau _0 = \inf \{\tau : \ w^t >0 \ \hbox {in } \Sigma _t , \quad \forall \tau< t < \tau _1 \}. \end{aligned}$$

(3.7)

We first note that

$$\begin{aligned} (D_1)^{\tau _0}\setminus D_1 = \emptyset . \end{aligned}$$

(3.8)

To prove this, We first notice that since $g_2 \ge 0$, the function u is a subsolution in $D\setminus D_1$, and hence maximum value for u is attained on $\partial D_1$. The latter being a $C^{1,dini}$ allows invoking the Hopf’s boundary principle to conclude $\partial _\nu u >0$ on the boundary $\partial D_1$, implying that $u > m$ in a vicinity of $\partial D_1$ inside $D_1$. Here $\nu$ is the interior unit normal direction on $\partial D_1$.

Now, if $(D_1)^{\tau _0}\setminus D_1 \ne \emptyset$, it means that we can find some value $\tau _2 > \tau _0$ such that $(D_1){\tau _2} \setminus D_1 \ne \emptyset$, and this set is very close to $\partial D_1$. By the discussion just made we have $u < m$, and $u^{\tau _2 } > m$ in this set, implying $w^{\tau _2} < 0$ in this set, contradicting the definition of $\tau _0$, and the construction of $w^{\tau _0}$, see (3.7).

To proceed, we may assume $\Sigma ^{\tau _0} \not \equiv D \cap \{x_1 < \tau _0 \}$, or equivalently $w^{\tau _0 } \not \equiv 0$, since otherwise we are done. From here on, the argument follows that of [4], by moving the plane slightly further and call it $T^{\tau _i }$ with $\tau _i = \tau _0 - \epsilon _i >0$ and $\epsilon _i$ being very small, so that a small set $D_{i} = \{ w^{\tau _i} < 0\}$ appears. As in [4], we slide back the plane by letting $\epsilon _i \rightarrow 0$, so that the limit domain $D_0 = \lim D_i$ consists of points z where $w^{\tau _0 } (z) = |\nabla w^{\tau _0} | (z) = 0$.

The key point is to show $D_0 = \emptyset$. This set, may appear at several places, and one needs to prove each of these possibilities would result in a contradiction. Indeed, one needs to consider three different possibilities:

$$\begin{aligned} D_0 \cap \Sigma ^{\tau _0}, \qquad D_0 \cap T^{\tau _0} , \qquad D_0 \cap \partial D , \end{aligned}$$

and prove each of these sets are void. For this, we need to calculate the Laplacian of $w^{\tau _0}$, in order to use the minimum principle for dealing with the above cases, in an exact way as done in [4]. We have

$$\begin{aligned} \Delta w^{\tau _0} = \left( g_2 (u) - g_1 (u) \chi _{D_1} \right) - \left( g_2 (u^{\tau _0}) - g_1 (u^{\tau _0}) \chi _{(D_1)^{\tau _0}} \right) , \end{aligned}$$

which in light of (3.8) can be rephrased as

$$\begin{aligned} \Delta w^{\tau _0} \le \left( g_2 (u) - g_2 (u^{\tau _0}) \right) - \left( g_1 (u) - g_1 (u^{\tau _0}) \right) \chi _{D_1} . \end{aligned}$$

By assumptions (2.6) we conclude that $\Delta w^{\tau _0} \le C w^{\tau _0}$, for some c, as in a) in (2.6). Since also $w^{\tau _0} \ge 0$ in $\Sigma ^{\tau _0}$, we have the scene ready for applying the comparison principle in a more elaborated way as done in [4]. Indeed, from here, we may follow exactly the same lines of proofs mutatis-mutandis, as in [4], to arrive at a contradiction. See also [5] for the case of fully nonlinear operators. $\square$