Energy minimizers for an asymptotic MEMS model with heterogeneous dielectric properties

Abstract

A model for a MEMS device, consisting of a fixed bottom plate and an elastic plate, is studied. It was derived in a previous work as a reinforced limit when the thickness of the insulating layer covering the bottom plate tends to zero. This asymptotic model inherits the dielectric properties of the insulating layer. It involves the electrostatic potential in the device and the deformation of the elastic plate defining the geometry of the device. The electrostatic potential is given by an elliptic equation with mixed boundary conditions in the possibly non-Lipschitz region between the two plates. The deformation of the elastic plate is supposed to be a critical point of an energy functional which, in turn, depends on the electrostatic potential due to the force exerted by the latter on the elastic plate. The energy functional is shown to have a minimizer giving the geometry of the device. Moreover, the corresponding Euler–Lagrange equation is computed and the maximal regularity of the electrostatic potential is established.

1 Introduction

The modeling and analysis of microelectromechanical systems (MEMS) has attracted a lot of interest in recent years, see, e.g., [10, 11, 19, 20, 30, 31, 35] and the references therein. Idealized devices often consist of a rigid dielectric ground plate above which an elastic dielectric plate is suspended. Applying a voltage difference between the two plates induces a competition between attractive electrostatic Coulomb forces and restoring mechanical forces, the latter resulting from the elasticity of the upper plate. When electrostatic forces dominate mechanical forces, the two plates may come into contact, a phenomenon usually referred to as pull-in instability or touchdown. From a mathematical point of view, this phenomenon may be accounted for in different ways. In fact, in most mathematical models considered so far in the MEMS literature, the pull-in instability is revealed as a singularity in the corresponding mathematical equations which coincides with a breakdown of the model, see [10, 19, 31] and the references therein. There is a close connection between the singular character of the touchdown and the fact that the modeling does not account for the thickness of the plates. Indeed, coating the ground plate with a thin insulating layer prevents a direct contact of the plates, so that a touchdown of the elastic plate on the insulating layer does not interrupt the operation of the device [6, 21, 24, 25]. Due to the presence of this layer, the MEMS device features heterogeneous dielectric properties (with a jump of the permittivity at the interface separating the coated ground plate and the free space beneath the elastic plate) and the electrostatic potential solves a free boundary transmission problem in the non-smooth domain enclosed between the two plates [21]. The shape of the domain itself is given by a partial differential equation governing the deflection of the elastic plate from rest, which, in turn, involves the electrostatic force exerted on the latter. The mathematical treatment of such a model is rather complex, see [21, Sect. 5] and [22]. It is thus desirable to derive simpler and more tractable models. As the modeling involves two small spatial scales – the aspect ratio $\varepsilon$ of the device and the thickness d of the insulating layer – a variety of reduced models may be obtained. For instance, the assumption of a vanishing aspect ratio of the device, when either the ratio $d/\varepsilon$ has a positive finite limit [2, 6, 18, 24, 25] or converges to zero, see [10, 30, 31] and the references therein, leads to a model which no longer involves a free boundary. Indeed, in that case, the electrostatic potential can be computed explicitly in terms of the deflection of the elastic plate and the model reduces to a single equation for the deflection, with the drawback that some important information on the electrostatic potential may thus be lost.

For this reason an intermediate model is derived in [16] by letting only the thickness of the insulating layer d go to zero (keeping the aspect ratio of the device of order one). Assuming an appropriate scaling of the dielectric permittivity in dependence on the layer’s thickness (in order to keep relevant information of the dielectric heterogeneity of the device) and using a Gamma convergence approach, the resulting energy, which is the building block of the model, is computed. The next step is the mathematical analysis of the thus derived model, in which stationary solutions correspond to critical points of the energy, while the dynamics is described by the gradient flow associated with the energy. The aim of the present work is to show the existence of a particular class of stationary solutions, which are additionally energy minimizers, and to identify the corresponding Euler–Lagrange equations.

Let us provide beforehand a more precise description of the MEMS configuration under study. We consider an idealized MEMS device composed of two rectangular two-dimensional dielectric plates: a fixed ground plate above which an elastic plate, with the same shape at rest, is suspended and clamped in only one direction while free in the other. We assume that the device is homogeneous in the free direction and that it is thus sufficient to consider a cross-section of the device orthogonal to the free direction. The shape of the ground plate and that of the elastic plate at rest are then represented by $D:=(-L,L)\subset {\mathbb {R}}$, the ground plate being located at $z=-H$ with $H>0$ and covered with an infinitesimally thin dielectric layer (in consistency with the aforementioned limit). The vertical deflection of the elastic plate from its rest position at $z=0$ is described by a function $u:{\bar{D}}\rightarrow [-H, \infty )$ satisfying the clamped boundary conditions

$$\begin{aligned} u(\pm L)= \partial _x u(\pm L)=0\,, \end{aligned}$$

(1.1)

so that its graph

$$\begin{aligned} {\mathfrak {G}}(u):=\{(x,u(x))\,:\, x\in {\bar{D}}\} \end{aligned}$$

represents the elastic plate and

$$\begin{aligned} \Omega (u):=\left\{ (x,z)\in D\times {\mathbb {R}}\,:\, -H< z < u(x)\right\} \end{aligned}$$

is the free space between the elastic plate and the ground plate. Since we do not exclude the possibility of contact between the two plates, we introduce the coincidence set

$$\begin{aligned} {\mathcal {C}}(u):=\{x\in D\, :\, u(x)=-H\} \end{aligned}$$

and let

$$\begin{aligned} \Sigma (u):=\{(x,-H)\,:\, x\in D,\, u(x)>-H\} = \big (D\setminus {\mathcal {C}}(u)\big ) \times \{-H\} \end{aligned}$$

be the part of the ground plate which is not in contact with the elastic plate. A touchdown of the elastic plate on the ground plate corresponds to a non-empty coincidence set, in which case $\Sigma (u)$ is a strict subset of $D\times \{-H\}$. Note that the free space $\Omega (u)$ then has a different geometry with at least two connected components, which may not be Lipschitz domains due to cusps (independent of the smoothness of the function u). In Fig. 1 the different situations with empty and non-empty coincidence sets are depicted.

Fig. 1

Geometry of $\Omega (u)$ for a state $u=v$ with empty coincidence set (green) and a state $u=w$ with non-empty coincidence set (blue) (color figure online)

As already mentioned, the building block of the model studied in this paper is the total energy E(u) of the device at a state u given by

$$\begin{aligned} E(u):= E_m(u) + E_e(u) \end{aligned}$$

and derived in [16] in the limit of an infinitesimally small insulating layer. It consists of the mechanical energy $E_m(u)$ and the electrostatic energy $E_e(u)$. The former is given by

$$\begin{aligned} E_m(u):=\frac{\beta }{2}\Vert \partial _x^2u\Vert _{L_2(D)}^2 +\left( \frac{\tau }{2}+\frac{\alpha }{4}\Vert \partial _x u\Vert _{L_2(D)}^2\right) \Vert \partial _x u\Vert _{L_2(D)}^2 \end{aligned}$$

with $\beta >0$ and $\tau ,\alpha \ge 0$, taking into account bending and external- and self-stretching effects of the elastic plate. The electrostatic energy is

$$\begin{aligned} E_e(u):=-\dfrac{1}{2}\displaystyle \int _{\Omega (u)} \big \vert \nabla \psi _u\big \vert ^2\,\mathrm {d}(x,z) -\dfrac{1}{2}\displaystyle \int _{ D} \sigma (x) \big \vert \psi _u(x,-H) - {\mathfrak {h}}_{u}(x) \big \vert ^2\,\mathrm {d}x\,, \end{aligned}$$

(1.2)

where $\psi _u$ is the electrostatic potential in the device and solves the elliptic equation with mixed boundary conditions

$$\begin{aligned} \Delta \psi _u&=0 \quad \text {in }\ \Omega (u)\,, \end{aligned}$$

(1.3a)

$$\begin{aligned} \psi _u&=h_u\quad \text {on }\ \partial \Omega (u)\setminus \Sigma (u)\,, \end{aligned}$$

(1.3b)

$$\begin{aligned} - \partial _z\psi _u +\sigma (\psi _u-{\mathfrak {h}}_u)&=0\quad \text {on }\ \Sigma (u)\,. \end{aligned}$$

(1.3c)

In (1.3), the function $\sigma$ represents the properties of the dielectric permittivity inherited from the insulating layer while the functions $h_u$ and ${\mathfrak {h}}_u$ determining the boundary values of $\psi _u$ on $\partial \Omega (u)$ are of the form

$$\begin{aligned}&h_u(x,z):=h(x,z,u(x))\,,\quad (x,z)\in \bar{D}\times [-H,\infty )\,, \\&{\mathfrak {h}}_u(x):={\mathfrak {h}}(x,u(x))\,,\quad x\in {\bar{D}}\,,\nonumber \end{aligned}$$

(1.4)

for some prescribed functions

$$\begin{aligned} h:{\bar{D}}\times [-H,\infty )\times [-H,\infty )\rightarrow {\mathbb {R}}\,,\qquad {\mathfrak {h}}: {\bar{D}}\times [-H,\infty )\rightarrow {\mathbb {R}}\,. \end{aligned}$$

The main results of this work are the existence of at least one minimizer of the total energy E and the derivation of the corresponding Euler–Lagrange equation. This requires, of course, first to study the well-posedness of the elliptic problem (1.3) subject to its mixed boundary conditions. A first step in that direction is to guarantee that the electrostatic energy $E_e$ is well-defined, which turns out to require some care. Indeed, it should be pointed out that $\Omega (u)$ is a non-smooth domain with corners and possibly features turning points, for instance when ${\mathcal {C}}(u)$ includes an interval, see Fig. 1. Thus, $\Omega (u)$ might consist of several components no longer having a Lipschitz boundary, so that traces have first to be given a meaning. Once this matter is settled, the existence of a variational solution $\psi _u$ to (1.3) readily follows from the Lax-Milgram Theorem and the electrostatic energy is then well-defined. This paves the way to the proof of the existence of minimizers of the total energy by the direct method of calculus of variations but does not yet allow us to conclude. Indeed, since E involves two contributions with opposite signs, it might be unbounded from below. We overcome this difficulty by adding a penalization term to the total energy. This additional term can be removed afterwards, thanks to an a priori upper bound on the minimizers which follows from the corresponding Euler–Lagrange equation. However, it turns out that the derivation of the latter requires additional regularity of the electrostatic potential $\psi _u$. Such a regularity is actually not obvious, as the highest expected smoothness of the boundary of $\Omega (u)$ is Lipschitz regularity (when the coincidence set ${\mathcal {C}}(u)$ is empty). Consequently, one needs to establish sufficient regularity for $\psi _u$ both for states u with empty and with non-empty coincidence sets ${\mathcal {C}}(u)$. In particular, this will ensure a well-defined normal trace of the gradient of $\psi _u$ on $\Sigma (u)$ as required by (1.3c) and on the part of ${\mathfrak {G}}(u)$ lying above $\Sigma (u)$ as required by (2.6a) below. The above mentioned difficulties are actually not the only ones that we face in the forthcoming analysis. To name but a few, the electrostatic energy $E_e(u)$ features a nonlocal and intricate dependence upon the state u and appropriate continuity properties are needed in the minimizing procedure. This requires a thorough understanding of the dependence of $\psi _u$ on the state u, this dependence being due to the domain $\Omega (u)$ as well as the functions $h_u$ and ${\mathfrak {h}}_u$. Also, due to the prescribed constraint $u\ge -H$, the Euler–Lagrange equation solved by minimizers is in fact a variational inequality.

2 Main results

Throughout this work we shall assume that

$$\begin{aligned} \sigma \in C^2(\bar{D})\,,\qquad \sigma (x)>0\,,\quad x\in {\bar{D}}\,. \end{aligned}$$

(2.1a)

As for the functions $h_u$ and ${\mathfrak {h}}_u$ appearing in (1.3) we shall assume in the following that

$$\begin{aligned} h\in C^2({\bar{D}}\times [-H,\infty )\times [-H,\infty ))\,,\qquad {\mathfrak {h}}\in C^1({\bar{D}}\times [-H,\infty ))\,, \end{aligned}$$

(2.1b)

satisfy

$$\begin{aligned} \partial _z h(x,-H,w) = \sigma (x) \big [ h(x,-H,w) - {\mathfrak {h}}(x,w) \big ]\,, \qquad (x,w)\in D\times [-H,\infty )\, . \end{aligned}$$

(2.1c)

Assumption (2.1c) allows us later to rewrite (1.3) as an elliptic equation with homogeneous boundary conditions. In the following, we shall use the notation introduced in (1.4).

A simple example of boundary functions $(h,{\mathfrak {h}})$ satisfying (2.1b) and (2.1c) may be derived from [21, Example 5.5] with the scaling from [16]:

Example 2.1

Let $V>0$ and set

$$\begin{aligned} h(x,z,w) := V \frac{1+\sigma (x)(H+z)}{1+\sigma (x)(H+w)}\,, \qquad (x,z,w)\in \bar{D}\times [-H,\infty )\times [-H,\infty )\,, \end{aligned}$$

and ${\mathfrak {h}}\equiv 0$. Then $(h,{\mathfrak {h}})$ clearly satisfies (2.1b) and (2.1c), the former being a consequence of the regularity (2.1a) of $\sigma$. Note that $h_u(x,u(x))=V, x\in D$, for a given state u; that is, in this example the electrostatic potential is kept at a constant value V along the elastic plate, see (1.3b).

2.1 The electrostatic potential

We first turn to the existence of an electrostatic potential for a given state u. To have an appropriate functional setting for u we introduce

$$\begin{aligned} \bar{S} := \{ u\in H^2(D)\cap H_0^1(D)\ :\ -H\le u \;\text { in }\; D \}\, \end{aligned}$$

(2.2)

and point out that ${\mathcal {C}}(u)=\emptyset$ if and only if u belongs to the interior of ${\bar{S}}$; that is, $u\in S$, where

$$\begin{aligned} S := \{ u\in H^2(D)\cap H_0^1(D)\ :\ -H<u \;\text { in }\; D \}\,. \end{aligned}$$

Note that $H^2(D)$ is embedded in $C({\bar{D}})$ so that $\Omega (u)$ is well-defined for $u\in {\bar{S}}$. Regarding the well-posedness of (1.3) we shall prove the following result.

Theorem 2.2

Suppose (2.1). For each $u\in {\bar{S}}$ there exists a unique strong solution $\psi _u\in H^2(\Omega (u))$ to (1.3). Moreover, given $\kappa >0$ and $r\in [2,\infty )$, there are $c(\kappa )>0$ and $c(r,\kappa )>0$ such that

$$\begin{aligned} \Vert \psi _u\Vert _{H^2(\Omega (u))} + \Vert \partial _x\psi _u(\cdot ,-H)\Vert _{L_2(D\setminus {\mathcal {C}}(u))} \le c(\kappa )\,, \qquad \Vert \partial _z\psi _u(\cdot ,u)\Vert _{L_r(D\setminus {\mathcal {C}}(u))} \le c(r,\kappa ) \end{aligned}$$

for each $u\in {\bar{S}}$ with $\Vert u\Vert _{H^2(D)} \le \kappa$.

Theorem 2.2 is an immediate consequence of Lemma 3.1, Theorems 3.2, and (3.6) below.

2.2 Existence of energy minimizers

Owing to Theorem 2.2, the total energy is well-defined on the set

$$\begin{aligned} {\bar{S}}_0 := \{ u\in H^2(D)\, :\, u(\pm L)=\partial _x u(\pm L)=0\,,\, -H\le u \;\text { in }\; D \}\subset {\bar{S}} \,, \end{aligned}$$

taking into account the clamped boundary conditions (1.1). We shall now focus on the existence of energy minimizers on ${\bar{S}}_0$. We have already observed that the total energy E is the sum of two terms $E_m$ and $E_e$ with different signs. Hence, the coercivity of E is not obvious. However, if $\alpha >0$, the first order term in the mechanical energy $E_m$ is quartic and thus dominates the negative contribution coming from the electrostatic energy $E_e$. This property allows us to follow the lines of [21, Sect. 5] to derive the coercivity of E based on the following growth assumption for h: there is a constant $K>0$ such that

$$\begin{aligned} \vert \partial _x h(x,z,w)\vert +\vert \partial _z h(x,z,w)\vert \le K \sqrt{\frac{1+w^2}{H+w}}\,,\quad \vert \partial _w h(x,z,w)\vert \le \frac{K}{\sqrt{H+w}}\,, \end{aligned}$$

(2.3a)

for $(x,z,w)\in {\bar{D}} \times [-H,\infty ) \times [-H,\infty )$ and

$$\begin{aligned} \vert h(x,-H,w)\vert +\vert {\mathfrak {h}}(x,w)\vert \le K\,, \quad (x,w)\in {\bar{D}}\times [-H,\infty )\,. \end{aligned}$$

(2.3b)

This approach no longer works if $\alpha =0$ and the coercivity of E is not granted. To remedy this drawback, we shall use a regularized energy functional (see (6.1) below), which includes a penalization term ensuring its coercivity if, in addition to (2.3), we assume that

$$\begin{aligned} \vert h(x,w,w)\vert + \vert h(\pm L,z,w)\vert \le K\,, \quad (x,z,w)\in {\bar{D}}\times [-H,\infty )\times [-H,\infty )\,, \end{aligned}$$

(2.4a)

and

$$\begin{aligned}&|\partial _x h(x,w,w)| + |\partial _z h(x,w,w)| + |\partial _w h(x,w,w)| + \vert \partial _w {\mathfrak {h}}(x,w)\vert \le K \end{aligned}$$

(2.4b)

for $(x,w) \in D \times [-H, \infty )$. We complete the analysis when $\alpha =0$ by showing that minimizers of the regularized energy functional for a suitable choice of the penalization parameter give rise to a minimizer of E, establishing indirectly that E is bounded from below in that case as well. Consequently, in both cases we can prove the existence of at least one energy minimizer as stated in the next result.

Theorem 2.3

Assume (2.1) and (2.3) and, either $\alpha >0$, or $\alpha =0$ and (2.4). Then the total energy E has at least one minimizer $u_*$ in $\bar{S_0}$; that is, $u_*\in \bar{S_0}$ and

$$\begin{aligned} E(u_*)=\min _{\bar{S_0}}E\,. \end{aligned}$$

(2.5)

At this point, no further qualitative information on energy minimizers $u_*$ is available, and a particularly interesting question, which is yet left unanswered by our analysis, is whether the coincidence set ${\mathcal {C}}(u_*)$ is empty or not. Another interesting open issue is the uniqueness of minimizers. The proof of Theorem 2.3 is given in Sect. 6 for $\alpha =0$ and in Sect. 7 for $\alpha >0$.

2.3 Euler–Lagrange equation

We next aim at deriving the Euler–Lagrange equation satisfied by minimizers of the total energy E. Recalling the prescribed constraint $u\ge -H$ for $u\in \bar{S_0}$, we are dealing with an obstacle problem and the resulting equation is actually a variational inequality. For the precise statement we introduce, for a given $u \in \bar{S}$, the function $g(u):D\rightarrow {\mathbb {R}}$ by setting

$$\begin{aligned} \begin{aligned} g(u)(x):=&\frac{1}{2} (1+\vert \partial _x u(x)\vert ^2)\,\big [\partial _z\psi _{u}-(\partial _z h)_{u}-(\partial _w h)_{u}\big ]^2(x, u(x))\\&+ \sigma (x)\big [\psi _{u}(x,-H)-{\mathfrak {h}}_{u}(x)\big ](\partial _w {\mathfrak {h}})_{u}(x)\\&-\frac{1}{2} \left[ \big \vert (\partial _x h)_u\big \vert ^2+ \big ((\partial _z h)_u+(\partial _w h)_u\big )^2 \right] (x, u(x)) \end{aligned} \end{aligned}$$

(2.6a)

for $x\in D\setminus {\mathcal {C}}(u)$ while setting

$$\begin{aligned} \begin{aligned} g(u)(x):=&\frac{1}{2} \vert (\partial _w h)_{u}\vert ^2(x, -H)+ \sigma (x)\big [h(x,-H,-H)-{\mathfrak {h}}_u(x)\big ](\partial _w {\mathfrak {h}})_{u}(x)\\&-\frac{1}{2} \left[ \big \vert (\partial _x h)_u\big \vert ^2+ \big ((\partial _z h)_u+(\partial _w h)_u\big )^2 \right] (x, -H) \end{aligned} \end{aligned}$$

(2.6b)

for $x\in {\mathcal {C}}(u)$. In fact, g(u) represents the electrostatic force exerted on the elastic plate and is computed as the differential (in a suitable sense) of the electrostatic energy $E_e(u)$ with respect to u. We emphasize here that the regularity properties of $\psi _u$ established in Theorem 2.2 are of utmost importance to guarantee that g(u) is well-defined on $D\setminus {\mathcal {C}}(u)$, since it features the trace of $\partial _z \psi _u$ on ${\mathfrak {G}}(u)$. With this notation, we are able to identify the variational inequality solved (in a weak sense) by energy minimizers.

Theorem 2.4

Assume (2.1). Assume that $u\in \bar{S_0}$ is a minimizer of E on ${\bar{S}}_0$. Then $g(u)\in L_2(D)$ and u is an $H^2$-weak solution to the variational inequality

$$\begin{aligned} \beta \partial _x^4u-(\tau +\alpha \Vert \partial _x u\Vert _{L_2(D)}^2)\partial _x^2 u+\partial {\mathbb {I}}_{\bar{S_0}}(u) \ni - g(u) \;\;\text { in }\;\; D\,, \end{aligned}$$

(2.7)

where $\partial {\mathbb {I}}_{\bar{S_0}}$ denotes the subdifferential of the indicator function ${\mathbb {I}}_{{\bar{S}}_0}$ of the closed convex subset $\bar{S_0}$ of $H^2(D)$; that is,

$$\begin{aligned} \begin{aligned} \int _D&\Big \{\beta \partial _x^2 u\,\partial _x^2 (w-u)+\big [\tau +\alpha \Vert \partial _x u\Vert _{L_2(D)}^2\big ] \partial _x u\, \partial _x(w-u)\Big \}\,\mathrm {d}x\ge -\int _D g(u) (w-u)\, \mathrm {d}x \end{aligned} \end{aligned}$$

for all $w\in \bar{S_0}$.

At this point, we do not know whether minimizers of E in $\bar{S_0}$ are the only solutions to (2.7), a question closely connected to the uniqueness issue for (2.7). It is, however, expected that the set of solutions to (2.7) exhibits a complex structure. Indeed, in the much simpler situation studied in [18], the minimizer may coexist with other steady states, depending on the boundary values of the electrostatic potential.

The proof of Theorem 2.4 is given in Sect. 6 for $\alpha =0$ and in Sect. 7 for $\alpha >0$. It relies on the computation of the shape derivative of the electrostatic energy $E_e(u)$, which is performed in Sect. 5.

Remark 2.5

It is also possible to minimize the total energy E on the set ${\bar{S}}$ (instead on ${\bar{S}}_0$). Then the corresponding minimizer in ${\bar{S}}$ satisfies instead of the clamped boundary conditions (1.1) the Navier or pinned boundary conditions $u(\pm L)=\partial _x^2 u(\pm L)=0$. With this change, the statements of Theorem 2.3 and Theorem 2.4 remain true when $\bar{S}_0$ is replaced everywhere by ${\bar{S}}$.

Now, combining Theorem 2.3 and Theorem 2.4 we obtain the existence of a stationary configuration of the MEMS device given as a solution to the force balance (2.7):

Corollary 2.6

Assume (2.1) and (2.3) and, either $\alpha >0$, or $\alpha =0$ and (2.4). Then there is a solution $u_*\in \bar{S}_0$ to the variational inequality (2.7).

The subsequent sections are dedicated to the proofs of the results stated in this section.

Throughout the paper, we impose assumptions (2.1) and set

$$\begin{aligned} \sigma _{min} := \min _{\bar{D}}\{\sigma \}>0 \,,\qquad {\bar{\sigma }} := \Vert \sigma \Vert _{C^2(\bar{D})} <\infty \,. \end{aligned}$$

(2.8)

3 Existence and $H^2$-regularity of the electrostatic potential $\psi _u$

This section is dedicated to the proof of Theorem 2.2; that is, to the existence and regularity of a unique solution $\psi _u$ to (1.3). We first recall some basic properties of the boundary function $h_v$ which are established in [21, Lemma 3.10] and rely on the properties (2.1b) and (2.1c) of h and ${\mathfrak {h}}$.

Lemma 3.1

Let $M>0$.

(a) Given $v\in {\bar{S}}$ satisfying $-H\le v(x)\le M-H$ for $x\in D$, the function $h_v$ belongs to $H^2(\Omega (v))$ and

$$\begin{aligned} \begin{aligned}&\Vert h_v\Vert _{H^2(\Omega (v))}\le C(M) \big (1 +\Vert \partial _x^2 v\Vert _{L_2(D)}^2\big )\,, \\&\Vert \partial _x h_v(\cdot ,-H)\Vert _{L_2(D)} \le C(M) \big (1+\Vert \partial _x v\Vert _{L_2(D)}\big )\,, \\&\Vert \partial _z h_v(\cdot ,v)\Vert _{L_r(D)} \le C(M)\,, \qquad r\in [1,\infty ]\,. \end{aligned} \end{aligned}$$

(3.1)

(b) Consider a sequence $(v_n)_{n\ge 1}$ in $\bar{S}$ and $v\in {\bar{S}}$ such that

$$\begin{aligned} -H\le v_n(x)\,,\, v(x)\le M-H\,, \quad x\in D\,, \qquad v_n\rightarrow v \text { in }\ H_0^1(D)\,. \end{aligned}$$

(3.2)

Let $\Omega (M) := D\times (-H,M)$. Then

$$\begin{aligned} h_{v_n}&\rightarrow h_v\quad \text {in }\ H^1(\Omega (M))\,, \end{aligned}$$

(3.3)

$$\begin{aligned} h_{v_n}(\cdot ,-H)&\rightarrow h_v(\cdot ,-H) \quad \text {in }\ L_2(D)\,, \end{aligned}$$

(3.4)

$$\begin{aligned} {\mathfrak {h}}_{v_n}&\rightarrow {\mathfrak {h}}_v \quad \text {in }\ L_2(D)\,. \end{aligned}$$

(3.5)

Proof

Integrating

$$\begin{aligned} \partial _x v(x)=\partial _x v(y)+\int _y^x\partial _x^2 v(z)\,\mathrm {d}z\,, \qquad (x,y)\in [-L,L]^2\,, \end{aligned}$$

with respect to $y\in [-L,L]$ and taking into account the boundary condition $v(\pm L)=0$, we obtain

$$\begin{aligned} 2L\partial _x v(x)=\int _{-L}^L \int _y^x\partial _x^2 v(z)\,\mathrm {d}z\,\mathrm {d}y\,,\quad x\in [-L,L]\,. \end{aligned}$$

Hence, by Hölder’s inequality we get

$$\begin{aligned} \Vert \partial _x v\Vert _{L_\infty (D)}\le \sqrt{2L}\Vert \partial _x^2v\Vert _{L_2(D)}\,. \end{aligned}$$

Using this inequality and the fact that h and its derivatives up to second order are bounded on ${\bar{D}}\times [-H,M]\times [-H,M]$ we derive

$$\begin{aligned} \Vert h_v\Vert _{H^2(\Omega (v))}&\le C(M) \big (1+\Vert \partial _x v\Vert _{L_2(D)}+\Vert \partial _x v\Vert _{L_\infty (D)} \Vert \partial _x v\Vert _{L_2(D)} + \Vert \partial _x^2 v\Vert _{L_2(D)}\big ) \\&\le C(M) \big (1+\Vert \partial _x^2 v\Vert _{L_2(D)} + \Vert \partial _x^2 v\Vert _{L_2(D)}^2\big )\,, \end{aligned}$$

which yields (a). As for (b) we first note that (3.2) and the compact embedding of $H^1(D)$ in $C(\bar{D})$ ensure that

$$\begin{aligned} v_n \rightarrow v \;\;\text { in }\;\; C(\bar{D})\,. \end{aligned}$$

Combining this convergence with (3.2) and the continuity properties (2.1b) of h and ${\mathfrak {h}}$ readily gives (3.4) and (3.5), as well as (3.3) with the additional use of (3.2), see [21, Lemma 3.10]. $\square$

We shall now prove Theorem 2.2 and thus focus on (1.3), which is more conveniently considered with homogeneous boundary conditions. To this end, we introduce

$$\begin{aligned} \chi _v:=\psi _v-h_v \end{aligned}$$

(3.6)

for a given and fixed function $v\in {\bar{S}}$. Due to assumption (2.1c), problem (1.3) (with v instead of u) is then equivalent to

$$\begin{aligned} -\Delta \chi _v&= \Delta h_v \;\text { in }\; \Omega (v)\, , \end{aligned}$$

(3.7a)

$$\begin{aligned} \chi _v&= 0 \;\text { on }\; \partial \Omega (v)\setminus \Sigma (v)\, , \end{aligned}$$

(3.7b)

$$\begin{aligned} -\partial _z \chi _v + \sigma \chi _v&= 0 \;\text { on }\; \Sigma (v)\,. \end{aligned}$$

(3.7c)

Hence, the next result can be seen as a reformulation of Theorem 2.2 in terms of $\chi _v$.

Theorem 3.2

Consider a function $v\in {\bar{S}}$ and let $\kappa >0$ be such that

$$\begin{aligned} \Vert v\Vert _{H^2(D)} \le \kappa \ . \end{aligned}$$

(3.8)

Then there exists a unique strong solution $\chi _v\in H^2(\Omega (v))$ to (3.7) and there is $C(\kappa )>0$ depending only on $\sigma$ and $\kappa$ such that

$$\begin{aligned} \Vert \chi _v\Vert _{H^2(\Omega (v))} + \Vert \partial _x\chi _v(\cdot ,-H)\Vert _{L_2(D\setminus {\mathcal {C}}(v))} \le C(\kappa )\ . \end{aligned}$$

(3.9)

Moreover, for any $r\in [2,\infty )$, there is $C(\kappa ) >0$ depending only on $\sigma$ and $\kappa$ such that

$$\begin{aligned} \Vert \partial _z\chi _v(\cdot ,v)\Vert _{L_r(D\setminus {\mathcal {C}}(v))} \le r C(\kappa ) \ . \end{aligned}$$

(3.10)

The remainder of this section is devoted to the proof of Theorem 3.2.

3.1 Variational solution to (3.7)

We first establish the existence of a variational solution to (3.7). To this end, we introduce for $v\in {\bar{S}}$ the space $H_B^1(\Omega (v))$ as the closure in $H^1(\Omega (v))$ of the set

$$\begin{aligned} C_B^1\big ( \overline{\Omega (v)} \big ) := \Big \{\theta \in C^1\big ( \overline{\Omega (v)} \big )\,:\, \theta (x,v(x))=0\,,\ x\in D\,, \,\theta (\pm L,z)=0\,,\ z\in (-H,0 ) \Big \}\,, \end{aligned}$$

and shall then minimize the functional

$$\begin{aligned} {\mathcal {G}}(v)[\vartheta ]:= & {} \frac{1}{2} \int _{\Omega (v)} |\nabla (\vartheta +h_v)|^2 \,\mathrm {d}(x,z) \nonumber \\&+ \frac{1}{2} \int _D \sigma (x) |\vartheta (x,-H)+h_v(x,-H) - {\mathfrak {h}}_v(x)|^2 \,\mathrm {d}x \end{aligned}$$

(3.11)

with respect to $\vartheta \in H_B^1(\Omega (v))$. Let us recall from [16, Lemma 2.2] that the trace $\vartheta (\cdot ,-H)\in L_2(D)$ is well-defined for $\vartheta \in H_B^1(\Omega (v))$ (see also Lemma 3.7 below for a complete statement), while Lemma 3.1 ensures that $h_v\in H^1(\Omega (v))$ and that $h_v(\cdot ,-H)$ and ${\mathfrak {h}}_v$ belong to $L_2(D)$. Thus, ${\mathcal {G}}(v)[\vartheta ]$ is well-defined for $\vartheta \in H_B^1(\Omega (v))$.

Proposition 3.3

Let $v\in \bar{S}$. There is a unique variational solution $\chi _v\in H_B^1(\Omega (v))$ to (3.7) given as the unique minimizer of the functional ${\mathcal {G}}(v)$ on $H_B^1(\Omega (v))$. Moreover, $\chi _v$ is also the unique minimizer on $H_B^1(\Omega (v))$ of the functional $G_D(v)$ defined by

$$\begin{aligned} G_D(v)[\vartheta ] := \frac{1}{2} \int _{\Omega (v)} |\nabla \vartheta |^2\ \mathrm {d}(x,z) + \frac{1}{2} \int _D \sigma |\vartheta (\cdot ,-H)|^2\ \mathrm {d}x - \int _{\Omega (v)} \vartheta \Delta h_v\ \mathrm {d}(x,z) \,. \end{aligned}$$

Proof

As noted above, ${\mathcal {G}}(v)$ and $G_D(v)$ are both well-defined on $H_B^1(\Omega (v))$. Moreover, owing to the Poincaré inequality established in [16, Lemma 2.2], the functional ${\mathcal {G}}(v)$ is coercive on $H_B^1(\Omega (v))$. It thus readily follows from the Lax-Milgram Theorem that there is a unique minimizer $\chi _v\in H_B^1(\Omega (v))$ of the functional ${\mathcal {G}}(v)$ on $H_B^1(\Omega (v))$. Let $\vartheta \in H_B^1(\Omega (v))$. Since each connected component of $\Omega (v)$ has at most two singular points, we infer from [15, Folgerung 7.5] that we may apply Gauß’ Theorem on each connected component of $\Omega (v)$ and deduce from (2.1c) that

$$\begin{aligned} {\mathcal {G}}(v)[\vartheta ]&= \frac{1}{2} \int _{\Omega (v)} |\nabla \vartheta |^2 \,\mathrm {d}(x,z) + \int _{\Omega (v)} \nabla \vartheta \cdot \nabla h_v \,\mathrm {d}(x,z)+ \frac{1}{2} \int _{\Omega (v)} |\nabla h_v|^2 \,\mathrm {d}(x,z) \\&\quad + \frac{1}{2} \int _D \sigma |\vartheta (\cdot ,-H)|^2 \,\mathrm {d}x + \int _D \sigma \vartheta (\cdot ,-H) [h_v(\cdot ,-H)-{\mathfrak {h}}_v]\,\mathrm {d}x \\&\quad + \frac{1}{2} \int _D \sigma [h_v(\cdot ,-H)-{\mathfrak {h}}_v]^2 \,\mathrm {d}x \\&= G_D(v)[\vartheta ] + \int _{\Omega (v)} \vartheta \Delta h_v\,\mathrm {d}(x,z) \\&\quad - \int _D (\vartheta \partial _z h_v)(x,-H)\,\mathrm {d}x - \int _{\Omega (v)} \vartheta \Delta h_v\,\mathrm {d}(x,z) \\&\quad + \frac{1}{2} \int _{\Omega (v)} |\nabla h_v|^2 \,\mathrm {d}(x,z) + \int _D \sigma \vartheta (\cdot ,-H) [h_v(\cdot ,-H)-{\mathfrak {h}}_v]\,\mathrm {d}x \\&\quad + \frac{1}{2} \int _D \sigma [h_v(\cdot ,-H)-{\mathfrak {h}}_v]^2 \,\mathrm {d}x \\&= G_D(v)[\vartheta ] + \frac{1}{2} \int _{\Omega (v)} |\nabla h_v|^2 \,\mathrm {d} (x,z) + \frac{1}{2} \int _D \sigma [h_v(\cdot ,-H)-{\mathfrak {h}}_v]^2 \,\mathrm {d}x \,. \end{aligned}$$

Consequently, $\chi _v$ is also the unique minimizer of the functional $G_D(v)$ on $H_B^1(\Omega (v))$. $\square$

For further use we state the following weak maximum principle.

Lemma 3.4

Let $v\in {\bar{S}}$. Then $h_v\in C(\overline{\Omega (v)})$, ${\mathfrak {h}}_v\in C(\bar{D})$, and

$$\begin{aligned} \min \Big \{\min _{\partial \Omega (v)} h_v\,,\, \min _{\bar{D}}{\mathfrak {h}}_v\Big \}\le \chi _v+h_v\le \max \Big \{\max _{ \partial \Omega (v)} h_v\,,\, \max _{{\bar{D}}}{\mathfrak {h}}_v\Big \}\,. \end{aligned}$$

Proof

We first observe that $v\in C(\bar{D})$ which ensures, together with (2.1b), that

$$\begin{aligned} \mu _* := \min \Big \{ \min _{\partial {\Omega (v)}} h_v\,,\, \min _{\bar{D}}{\mathfrak {h}}_v\Big \} \;\;\text { and }\;\; \mu ^* := \max \Big \{\max _{\partial \Omega (v)} h_v\,,\, \max _{\bar{D}}{\mathfrak {h}}_v\Big \} \end{aligned}$$

are well-defined and finite. Next, since $\chi _v$ is the minimizer of ${\mathcal {G}}(v)$ on $H_B^1(\Omega (v))$, it satisfies

$$\begin{aligned} \int _{\Omega (v)} \nabla (\chi _v+h_v)\cdot \nabla \vartheta \,\mathrm {d}(x,z) + \int _D \sigma [(\chi _v+h_v)(\cdot ,-H) - {\mathfrak {h}}_v] \vartheta (\cdot ,-H)\,\mathrm {d}x = 0 \nonumber \\ \end{aligned}$$

(3.12)

for all $\vartheta \in H_B^1(\Omega (v))$.

Now, it follows from the definition of $\mu ^*$ that $\vartheta ^* := (\chi _v + h_v - \mu ^*)_+$ belongs to $H_B^1(\Omega (v))$ with $\nabla \vartheta ^* = \mathrm {sign}_+ (\chi _v + h_v - \mu ^*) \nabla (\chi _v + h_v - \mu ^*)$. Consequently, by (3.12),

$$\begin{aligned} 0&= \int _{\Omega (v)} \nabla (\chi _v+h_v)\cdot \nabla \vartheta ^* \,\mathrm {d}(x,z) + \int _D \sigma [(\chi _v+h_v)(\cdot ,-H) - {\mathfrak {h}}_v] \vartheta ^*(\cdot ,-H) \,\mathrm {d}x \\&= \int _{\Omega (v)} |\nabla \vartheta ^*|^2 \,\mathrm {d}(x,z) + \int _D \sigma [(\chi _v+h_v)(\cdot ,-H) - \mu ^* + \mu ^*- {\mathfrak {h}}_v] \vartheta ^*(\cdot ,-H) \,\mathrm {d}x \\&\ge \int _{\Omega (v)} |\nabla \vartheta ^*|^2 \,\mathrm {d}(x,z) + \int _D \sigma [\vartheta ^*(\cdot ,-H)]^2 \,\mathrm {d}x\,, \end{aligned}$$

where we have used the non-negativity of both $\mu ^* - {\mathfrak {h}}_v$ and $\vartheta ^*$ to derive the last inequality. We have thereby proved that $\nabla \vartheta ^*=0$ in $L_2(\Omega (v))$, which implies that $\vartheta ^*=0$ in $L_2(\Omega (v))$ thanks to the Poincaré inequality established in [16, Lemma 2.2]. In other words, $\chi _v + h_v - \mu ^*\le 0$ a.e. in $\Omega (v)$ as claimed.

Finally, a similar argument with $\vartheta _* := (\mu _*-\chi _v-h_v)_+$ leads to the inequality $\mu _*-\chi _v-h_v\le 0$ a.e. in $\Omega (v)$ and completes the proof. $\square$

We now improve the regularity of $\chi _v$ as stated in Theorem 3.2 and show that $\chi _v$ belongs to $H^2(\Omega (v))$. Once this is shown, it then readily follows that $\chi _v$ is a strong solution to (3.7) (see [16, Theorem 3.5]).

As pointed out previously, for a general $v\in {\bar{S}}$, the set $\Omega (v)$ may consist of several connected components without Lipschitz boundaries when the coincidence set ${\mathcal {C}}(v)$ is non-empty. The global $H^2(\Omega (v))$-regularity of $\chi _v$ is thus clearly not obvious. The main idea is to write the open set $D\setminus {\mathcal {C}}(v)$ as a countable union of disjoint open intervals $(I_j)_{j\in J}$, see [1, IX.Proposition 1.8], and to establish the $H^2$-regularity for $\chi _v$ first locally on each component $\left\{ (x,z)\in I_j\times {\mathbb {R}}\ :\ -H< z < v(x) \right\}$. This local regularity is performed in Sect. 3.2. The global $H^2(\Omega (v))$-regularity is subsequently established in Sect. 3.3.

3.2 Local $H^2$-regularity

Let $I:=(a,b)$ be an open interval in D and consider

$$\begin{aligned} v\in H^2(I) \;\text { with }\; v(x)> -H\ , \quad x\in I\ . \end{aligned}$$

(3.13)

We define the open set ${\mathcal {O}}_I(v)$ by

$$\begin{aligned} {\mathcal {O}}_I(v) := \left\{ (x,z)\in I\times {\mathbb {R}}\ :\ -H< z < v(x) \right\} \end{aligned}$$

(3.14)

and split its boundary $\partial {\mathcal {O}}_I(v) = \partial {\mathcal {O}}_{I,D}(v) \cup \overline{\Sigma _I}$ with

$$\begin{aligned} \partial {\mathcal {O}}_{I,D}(v)&:= \big (\{a\} \times [-H,v(a)]\big ) \cup \big (\{b\}\times [-H,v(b)]\big ) \cup \overline{{\mathfrak {G}}_I(v)}\ , \end{aligned}$$

(3.15)

$$\begin{aligned} \overline{\Sigma _I}&:= [a,b]\times \{-H\}\ , \end{aligned}$$

(3.16)

where $\Sigma _I := I \times \{-H\}$, and $\overline{{\mathfrak {G}}_I(v)}$ denotes the closure of the graph ${\mathfrak {G}}_I(v)$ of v, defined by

$$\begin{aligned} {\mathfrak {G}}_I(v) := \{ (x,v(x))\ :\ x\in I \}\ . \end{aligned}$$

(3.17)

We emphasize that ${\mathcal {O}}_I(v)$ has no Lipschitz boundary when $v(a)+H= \partial _x v(a)=0$ or $v(b)+H= \partial _x v(b)=0$, as these correspond to cuspidal boundary points, see Fig. 2.

Fig. 2

Geometry of ${\mathcal {O}}_{I}(v)$ according to the boundary values of v

Let $f\in L_2({\mathcal {O}}_I(v))$ be a fixed function. The aim is to investigate the auxiliary problem

$$\begin{aligned} -\Delta \zeta _v&= f \;\text { in }\; {\mathcal {O}}_I(v)\ , \end{aligned}$$

(3.18a)

$$\begin{aligned} \zeta _v&= 0 \;\text { on }\; \partial {\mathcal {O}}_{I,D}(v)\ , \end{aligned}$$

(3.18b)

$$\begin{aligned} -\partial _z \zeta _v + \sigma \zeta _v&= 0 \;\text { on }\; \Sigma _I\ . \end{aligned}$$

(3.18c)

We shall show the existence and uniqueness of a variational solution $\zeta _v:=\zeta _{I,v}\in H^1({\mathcal {O}}_I(v))$ to (3.18) and then prove its $H^2$-regularity. The main difficulty encountered here is the just mentioned possible lack of Lipschitz regularity of ${\mathcal {O}}_I(v)$. Indeed, the trace of functions in $H^1({\mathcal {O}}_I(v))$ on $\partial {\mathcal {O}}_I(v)$ have no meaning yet in that case, and so (3.18b) and (3.18c) are not well-defined. We shall thus first give a precise meaning to traces for functions in $H^1({\mathcal {O}}_I(v))$.

Remark 3.5

Clearly, if $v\in S$, $I=D$, and $f=h_v$, then $\chi _v=\zeta _{D,v}$, so that Theorem 3.2 follows from Theorem 3.9 below in that case. Furthermore, if $I=(a,b)$ is a strict subinterval of D, $f=h_v$, and $v\in \bar{S}$ is such that $v(a)=v(b) = -H$, or $a=-L$ and $v(-L)=v(b)+H=0$, or $b=L$ and $v(a)+H=v(L)=0$, then $\zeta _{I,v}$ coincides – at least formally – with the restriction of $\chi _v$ to I and we shall also deduce Theorem 3.2 from Theorem 3.9. We thus do not impose that $v(a)=-H$ or $v(b)=-H$ in (3.13), so as to be able to handle simultaneously the above mentioned different cases also depicted in Fig. 2.

3.2.1 Traces

As already noticed in [27], one can take advantage of the particular geometry of ${\mathcal {O}}_I(v)$, which lies between the graphs of two continuous functions, in order to define traces for functions in $H^1({\mathcal {O}}_I(v))$ along these graphs. More precisely, one can derive the following result [16, Lemma 2.1].

Lemma 3.6

[16, Lemma 2.1] Assume that v satisfies (3.13) and set $M_v := \Vert H+v\Vert _{L_\infty (I)}$.

(a):

There is a linear bounded operator

$$\begin{aligned} \Gamma _{I,v} \in {\mathcal {L}}\left( H^1({\mathcal {O}}_I(v)) , L_2( I,(H+v)\mathrm {d}x) \right) \end{aligned}$$

such that $\Gamma _{I,v} \vartheta = \vartheta (\cdot ,v)$ for $\vartheta \in C^1(\overline{{\mathcal {O}}_I(v)})$ and

$$\begin{aligned} \int _I |\Gamma _{I,v} \vartheta |^2 (H+v)\ \mathrm {d}x \le \Vert \vartheta \Vert _{L_2({\mathcal {O}}_I(v))}^2 + 2 M_v \Vert \vartheta \Vert _{L_2({\mathcal {O}}_I(v))} \Vert \partial _z \vartheta \Vert _{L_2({\mathcal {O}}_I(v))}\ . \nonumber \\ \end{aligned}$$

(3.19)

(b):

There is a linear bounded operator

$$\begin{aligned} \gamma _{I,v} \in {\mathcal {L}}\left( H^1({\mathcal {O}}_I(v)) , L_2( I,(H+v)\mathrm {d}x) \right) \end{aligned}$$

such that $\gamma _{I,v} \vartheta = \vartheta (\cdot ,-H)$ for $\vartheta \in C^1(\overline{{\mathcal {O}}_I(v)})$ and

$$\begin{aligned} \int _I |\gamma _{I,v}\vartheta |^2 (H+v)\ \mathrm {d}x \le \Vert \vartheta \Vert _{L_2({\mathcal {O}}_I(v))}^2 + 2 M_v \Vert \vartheta \Vert _{L_2({\mathcal {O}}_I(v))} \Vert \partial _z \vartheta \Vert _{L_2({\mathcal {O}}_I(v))}\ . \nonumber \\ \end{aligned}$$

(3.20)

For simplicity, for $\vartheta \in H^1({\mathcal {O}}_I(v))$, we use the notation

$$\begin{aligned} \vartheta (x,v(x)) := \Gamma _{I,v}\vartheta (x)\ , \quad \vartheta (x,-H) := \gamma _{I,v}\vartheta (x)\ , \qquad x\in I\ . \end{aligned}$$

We next introduce the variational setting associated with (3.18) and define the space $H_B^1({\mathcal {O}}_I(v))$ as the closure in $H^1({\mathcal {O}}_I(v))$ of the set

$$\begin{aligned} \begin{aligned} C_B^1\left( \overline{{\mathcal {O}}_I(v)} \right) := \Big \{\theta \in C^1\left( \overline{{\mathcal {O}}_I(v)} \right)&:\, \, \theta (x,v(x))=0\,,\ x\in I\,,\\&\text {and }\theta (x,z)=0\,,\ (x,z)\in \{a,b\}\times (-H,0] \Big \}\,. \end{aligned} \end{aligned}$$

Note that this is consistent with the previous definition of $H_B^1(\Omega (v))$ when $I=D$ and $v\in {\bar{S}}$. We have already established in [16, Lemma 2.2] a Poincaré inequality in $H_B^1({\mathcal {O}}_I(v))$, as well as refined properties of the trace on $I\times \{-H\}$, which we recall now.

Lemma 3.7

[16, Lemma 2.2] Assume that v satisfies (3.13) and consider $\vartheta \in H_B^1({\mathcal {O}}_I(v))$. Setting $M_v := \Vert H+v\Vert _{L_\infty (I)}$, there holds

$$\begin{aligned} \Vert \vartheta \Vert _{L_2({\mathcal {O}}_I(v))} \le 2 M_v \Vert \partial _z \vartheta \Vert _{L_2({\mathcal {O}}_I(v))}\,, \end{aligned}$$

(3.21)

and the trace operator $\vartheta \mapsto \vartheta (\cdot ,-H)$ maps $H_B^1({\mathcal {O}}_I(v))$ to $L_2(I)$ with

$$\begin{aligned} \Vert \vartheta (\cdot ,-H)\Vert _{L_2(I)}^2 \le 2 \Vert \vartheta \Vert _{L_2({\mathcal {O}}_I(v))} \Vert \partial _z \vartheta \Vert _{L_2({\mathcal {O}}_I(v))}\ . \end{aligned}$$

(3.22)

3.2.2 Variational solution to (3.18)

Thanks to Lemma 3.7, the trace on $I\times \{-H\}$ of a function in $H_B^1({\mathcal {O}}_I(v))$ is well-defined in $L_2(I)$ and, thus, so is the functional

$$\begin{aligned} G_I(v)[\vartheta ] := \frac{1}{2} \int _{{\mathcal {O}}_I(v)} |\nabla \vartheta |^2\ \mathrm {d}(x,z) + \frac{1}{2} \int _I \sigma \vert \vartheta (\cdot ,-H)\vert ^2\ \mathrm {d}x - \int _{{\mathcal {O}}_I(v)} f \vartheta \ \mathrm {d}(x,z) \nonumber \\ \end{aligned}$$

(3.23)

for $\vartheta \in H_B^1({\mathcal {O}}_I(v))$. We now derive the existence of a unique variational solution to (3.18), or, equivalently, of a unique minimizer of $G_I(v)$ on $H_B^1({\mathcal {O}}_I(v))$.

Lemma 3.8

There is a unique variational solution $\zeta _v:= \zeta _{I,v}\in H_B^1({\mathcal {O}}_I(v))$ to (3.18) which satisfies

$$\begin{aligned} \Vert \zeta _v\Vert _{H^1({\mathcal {O}}_I(v))}^2 + 2 \Vert \sqrt{\sigma } \zeta _v(\cdot ,-H)\Vert _{L_2(I)}^2 \le 16 M_v^2 \left( 1 + 4 M_v^2 \right) \Vert f\Vert _{L_2({\mathcal {O}}_I(v))}^2 \ , \end{aligned}$$

(3.24)

where $M_v := \Vert H+v\Vert _{L_\infty (I)}$.

Proof

It readily follows from (2.8), Lemma 3.7, and the Lax-Milgram Theorem that there is a unique variational solution $\zeta _v\in H_B^1({\mathcal {O}}_I(v))$ to (3.18) in the sense that

$$\begin{aligned} G_I(v)[\zeta _v] \le G_I(v)[\vartheta ]\,, \qquad \vartheta \in H_B^1({\mathcal {O}}_I(v))\,. \end{aligned}$$

(3.25)

Taking $\vartheta \equiv 0$ in the previous inequality, we deduce from (3.21) and Hölder’s and Young’s inequalities that

$$\begin{aligned} \Vert \nabla \zeta _v\Vert _{L_2({\mathcal {O}}_I(v))}^2 + \Vert \sqrt{\sigma } \zeta _v(\cdot ,-H)\Vert _{L_2(I)}^2&\le 2 \Vert f\Vert _{L_2({\mathcal {O}}_I(v))} \Vert \zeta _v\Vert _{L_2({\mathcal {O}}_I(v))} \\&\le 4 M_v \Vert f\Vert _{L_2({\mathcal {O}}_I(v))} \Vert \nabla \zeta _v\Vert _{L_2({\mathcal {O}}_I(v))} \\&\le \frac{1}{2} \Vert \nabla \zeta _v\Vert _{L_2({\mathcal {O}}_I(v))}^2 + 8 M_v^2 \Vert f\Vert _{L_2({\mathcal {O}}_I(v))}^2\ . \end{aligned}$$

Hence,

$$\begin{aligned} \Vert \nabla \zeta _v\Vert _{L_2({\mathcal {O}}_I(v))}^2 + 2 \Vert \sqrt{\sigma } \zeta _v(\cdot ,-H)\Vert _{L_2(I)}^2 \le 16 M_v^2 \Vert f\Vert _{L_2({\mathcal {O}}_I(v))}^2\ . \end{aligned}$$

Combining the Poincaré inequality (3.21) and the above inequality completes the proof. $\square$

3.2.3 $H^2$-regularity of $\zeta _v$

We next investigate the regularity of the variational solution $\zeta _v$ to (3.18); that is, we establish a local version of Theorem 3.2.

Theorem 3.9

Consider a function v satisfying (3.13) and let $\kappa >0$ be such that

$$\begin{aligned} \Vert v\Vert _{H^2(I)} \le \kappa \ . \end{aligned}$$

(3.26)

The variational solution $\zeta _v =\zeta _{I,v}\in H_B^1({\mathcal {O}}_I(v))$ to (3.18) given by Lemma 3.8 belongs to $H^2({\mathcal {O}}_I(v))$, and there is $C_{1}(\kappa )>0$ depending only on $\sigma$ and $\kappa$ such that

$$\begin{aligned} \Vert \zeta _v\Vert _{H^2({\mathcal {O}}_I(v))} + \Vert \partial _x \zeta _v(\cdot ,-H)\Vert _{L_2(I)} \le C_{1}(\kappa ) \Vert f\Vert _{L_2({\mathcal {O}}_I(v))}\ . \end{aligned}$$

(3.27)

Moreover, there is $C_{2}(\kappa ) >0$ depending only on $\sigma$ and $\kappa$ such that, for any $r\in [2,\infty )$,

$$\begin{aligned} \Vert \partial _z\zeta _v(\cdot ,v)\Vert _{L_r(I)} \le r C_{2}(\kappa ) \Vert f\Vert _{L_2({\mathcal {O}}_I(v))}\ . \end{aligned}$$

(3.28)

Several difficulties are encountered in the proof of Theorem 3.9, due to the low regularity of the domain ${\mathcal {O}}_I(v)$ which has a Lipschitz boundary if $v(a)>-H$ and $v(b)>-H$ but may have cusps otherwise, see Fig. 2, and due to the mixed boundary conditions (3.18b) and (3.18c). As in [12, Sect. 3.3], to remedy these problems requires to construct suitable approximations of ${\mathcal {O}}_I(v)$ and to pay special attention to the dependence of the constants on v and I in the derivation of functional inequalities and estimates. To be more precise, we shall begin with the case where v satisfies

$$\begin{aligned} v\in W_\infty ^3(I) \;\text { and }\; \min _{[a,b]} v > -H\ , \end{aligned}$$

(3.29)

an assumption which is obviously stronger than (3.13). Then ${\mathcal {O}}_I(v)$ is a Lipschitz domain with a piecewise $W_\infty ^3$-smooth boundary and the $H^2$-regularity of $\zeta _v$ is guaranteed by [5, Theorem 2.2], see Lemma 3.10 below. Next, transforming ${\mathcal {O}}_I(v)$ to the rectangle ${\mathcal {R}}_I := I\times (0,1)$, we shall adapt the proof of [12, Lemma 4.3.1.3] to establish the identity

$$\begin{aligned} \begin{aligned} \int _{{\mathcal {O}}_I(v)} \partial _x^2\zeta _v \partial _z^2\zeta _v\ \mathrm {d}(x,z)&= \int _{{\mathcal {O}}_I(v)} |\partial _x\partial _z \zeta _v|^2\ \mathrm {d}(x,z) + \int _I \left( \partial _x \zeta _v \partial _x(\sigma \zeta _v) \right) (\cdot ,-H)\ \mathrm {d}x \\ {}&\quad - \frac{1}{2} \int _I \partial _x^2 v |\partial _z\zeta _v(\cdot ,v)|^2\ \mathrm {d}x \end{aligned} \end{aligned}$$

(3.30)

in Lemma 3.11. We then shall show that the last two integrals on the right-hand side of (3.30) are controlled by the $H^2$-norm of $\zeta _v$ with a sublinear dependence, a feature which will allow us to derive (3.27) when v satisfies (3.29). To this end, we shall use the embedding of the subspace

$$\begin{aligned} H_{WS}^1({\mathcal {O}}_I(v)) := \left\{ P\in H^1({\mathcal {O}}_I(v))\ :\ \begin{array}{cl} P(x,-H) &{} = 0\ , \qquad x\in I\ , \\ P(a,z) &{} = 0\ , \qquad z\in (-H,v(a)) \ , \end{array} \right\} \end{aligned}$$

(3.31)

of $H^1({\mathcal {O}}_I(v))$ in $L_r({\mathcal {O}}_I(v))$ and the continuity of the trace operator from $H_{WS}^1({\mathcal {O}}_I(v))$ to $L_r({\mathfrak {G}}_I(v))$ for $r\in [1,\infty )$, which involves constants that do not depend on $\min _{[a,b]}\{v+H\}$, see Lemmas C.1-C.3 in Appendix C. After this preparation, we will be left with relaxing the assumption (3.29) to (3.13) and this will be achieved by an approximation argument, see Sect. 3.2.5.

3.2.4 $H^2$-regularity of $\zeta _v$ when v satisfies (3.29)

Throughout this section, we assume that v satisfies (3.29) and fix $M>0$ such that

$$\begin{aligned} M \ge \max \left\{ 1 , \Vert H+v\Vert _{L_\infty (I)}, \Vert \partial _x v\Vert _{L_\infty (I)} \right\} \, . \end{aligned}$$

(3.32)

We also denote positive constants depending only on $\sigma$ by C and $(C_i)_{i\ge 3}$. The dependence upon additional parameters will be indicated explicitly.

We begin with the $H^2$-regularity of the variational solution $\zeta _v$ to (3.18), which follows from the analysis performed in [3,4,5].

Lemma 3.10

$\zeta _v\in H^2({\mathcal {O}}_I(v))$.

Proof

We first recast the boundary value problem (3.18) in the framework of [5]. Owing to (3.29), the boundary of the domain ${\mathcal {O}}_I(v)$ includes four $W_\infty ^3$-smooth edges $(\Gamma _i)_{1\le i \le 4}$ given by

$$\begin{aligned} \Gamma _1 := I \times \{-H\}\ ,&\qquad \Gamma _3 := {\mathfrak {G}}_I(v)\ , \\ \Gamma _2 := \{b\}\times (-H,v(b))\ ,&\qquad \Gamma _4 := \{a\}\times (-H,v(a))\ , \end{aligned}$$

and four vertices $(S_i)_{1\le i\le 4}$

$$\begin{aligned} S_1 := {\overline{\Gamma }}_1 \cap {\overline{\Gamma }}_2 = (b,-H)\ ,&\qquad S_3 := {\overline{\Gamma }}_3 \cap {\overline{\Gamma }}_4 = (a,v(a))\ , \\ S_2 := {\overline{\Gamma }}_2 \cap {\overline{\Gamma }}_3 = (b,v(b))\ ,&\qquad S_4 := {\overline{\Gamma }}_4 \cap {\overline{\Gamma }}_1 = (a,-H)\ . \end{aligned}$$

We set

$$\begin{aligned}&{\mathcal {D}}_\Gamma := \{2,3,4\}\ , \quad {\mathcal {N}}_\Gamma := \{1\}\ , \\&{\mathcal {D}} := \{2,3\}\ , \quad {\mathcal {M}}_{12} := \{4\}\ , \quad {\mathcal {M}}_{21} := \{1\}\ , \quad {\mathcal {N}} := \emptyset \ , \end{aligned}$$

and note that ${\mathcal {D}}_\Gamma \ne \emptyset$ as required in [5].

Since $v\in W_\infty ^3(I)$, the measure $\omega _i$ of the angle at $S_i$ taken towards the interior of ${\mathcal {O}}_I(v)$ satisfies

$$\begin{aligned} \omega _1=\omega _4=\frac{\pi }{2}\ , \qquad (\omega _2,\omega _3)\in (0,\pi )^2\ . \end{aligned}$$

(3.33)

For $1\le i\le 4$, we denote the outward unit normal vector field and the corresponding unit tangent vector field by $\varvec{\nu }_i$ and $\varvec{\tau }_i$, respectively. According to the geometry of ${\mathcal {O}}_I(v)$,

$$\begin{aligned} \varvec{\nu }_1=(0,-1)\ , \ \varvec{\nu }_2=(1,0)\ , \ \varvec{\nu }_3 = \frac{(- \partial _x v,1)}{\sqrt{1+|\partial _x v|^2}}\ , \ \varvec{\nu }_4 = (-1,0)\ , \\ \varvec{\tau }_1=(1,0)\ , \ \varvec{\tau }_2=(0,1)\ , \ \varvec{\tau }_3 = \frac{(-1, -\partial _x v)}{\sqrt{1+|\partial _x v|^2}}\ , \ \varvec{\tau }_4 = (0,-1)\ . \end{aligned}$$

We also define

$$\begin{aligned} \varvec{\mu }_1 := \varvec{\nu }_1\ , \qquad \varvec{\mu }_i := \varvec{\tau }_i\ , \ i\in \{2,3,4\}\ , \end{aligned}$$

(3.34)

and note that the measure $\Psi _i\in [0,\pi ]$ of the angle between $\varvec{\mu }_i$ and $\varvec{\tau }_i$, $1\le i \le 4$, is given by

$$\begin{aligned} \Psi _1 = \frac{\pi }{2}\ , \qquad \Psi _i=0\ , \ i\in \{2,3,4\}\ . \end{aligned}$$

(3.35)

We also set

$$\begin{aligned} \psi _1 = \phi _2 = \phi _3 = \phi _4 = 0\ . \end{aligned}$$

(3.36)

We finally define the boundary operator

$$\begin{aligned} {\mathcal {B}}_1 := - \partial _z + \sigma \mathrm {id} \;\text { on }\; I \times \{-H\}\,. \end{aligned}$$

Now, on the one hand, the regularity of $\sigma$ implies that [5, Assumption (1.5)] is satisfied, while [5, Assumption (1.6)] obviously holds since ${\mathcal {N}}=\emptyset$. On the other hand, we note that $\varvec{\mu }_1(S_1) = - \varvec{\mu }_2(S_1)$ and $\varvec{\mu }_4(S_4)=\varvec{\mu }_1(S_4)$, so that [5, Assumption (2.1)] is satisfied for $i\in \{1,4\}$ (but not for $i\in \{2,3\}$). We then set $\varepsilon _1=-1$ and $\varepsilon _4=1$. We are left with checking [5, Assumptions (2.3)-(2.4)] but this is obvious due to (3.36). We finally observe that

$$\begin{aligned} {\mathcal {K}} := \left\{ (i,m)\in \{1,\ldots ,4\}\times {\mathbb {Z}}\ :\ \lambda _{i,m}\in (-1,0) \right\} \end{aligned}$$

is empty, since

$$\begin{aligned} \lambda _{1,m}&:= \frac{\Psi _2-\Psi _1+m\pi }{\omega _1} = 2m-1 \not \in (-1,0)\ , \\ \lambda _{2,m}&:= \frac{\Psi _3-\Psi _2+m\pi }{\omega _2} = \frac{m\pi }{\omega _2}\not \in (-1,0)\ , \\ \lambda _{3,m}&:= \frac{\Psi _4-\Psi _3+m\pi }{\omega _3} = \frac{m\pi }{\omega _3} \not \in (-1,0)\ , \\ \lambda _{4,m}&:= \frac{\Psi _1-\Psi _4+m\pi }{\omega _4} = 2m+1 \not \in (-1,0)\ , \end{aligned}$$

for any $m\in {\mathbb {Z}}$. We then infer from [5, Theorem 2.2] that $\zeta _v$ has no singular part and thus belongs to $H^2({\mathcal {O}}_I(v))$. $\square$

We now investigate the quantitative dependence of the just established $H^2$-regularity of $\zeta _v$ on v and derive an $H^2$-estimate, which is related to the regularity of v. To this end, we need the following identity.

Lemma 3.11

$$\begin{aligned} \int _{{\mathcal {O}}_I(v)} \partial _x^2\zeta _v \partial _z^2\zeta _v\ \mathrm {d}(x,z)&= \int _{{\mathcal {O}}_I(v)} |\partial _x\partial _z \zeta _v|^2\ \mathrm {d}(x,z) + \int _I \left( \partial _x \zeta _v \partial _x(\sigma \zeta _v) \right) (\cdot ,-H)\ \mathrm {d}x \\&\qquad - \frac{1}{2} \int _I \partial _x^2 v |\partial _z\zeta _v(\cdot ,v)|^2\ \mathrm {d}x\ . \end{aligned}$$

The identity of Lemma 3.11 is reminiscent of [21, Lemma 3.5]. Its proof is rather technical and thus postponed to Appendix B.

The next step of the analysis is to show that the two integrals over I on the right-hand side of the identity stated in Lemma 3.11 can be controlled by the $H^2$-norm of $\zeta _v$ with a mild dependence on v. To this end, we need some auxiliary functional and trace inequalities which are established in Appendix C. With this in hand, we begin with an estimate of the last integral.

Lemma 3.12

There is $C_{3}(M)>0$ such that, for any $r\in [2,\infty )$,

$$\begin{aligned} \Vert \partial _z\zeta _v(\cdot ,v)\Vert _{L_r(I)} \le r C_{3}(M) \Vert f\Vert _{L_2({\mathcal {O}}_I(v))}^{1/r} \left( \Vert \nabla \partial _z\zeta _v\Vert _{L_2({\mathcal {O}}_I(v))} + \Vert f\Vert _{L_2({\mathcal {O}}_I(v))} \right) ^{(r-1)/r}\ . \nonumber \\ \end{aligned}$$

(3.37)

In particular, there is $C_{4}(M)>0$ such that

$$\begin{aligned} \left| \int _I \partial _x^2 v |\partial _z\zeta _v(\cdot ,v)|^2\ \mathrm {d}x \right| \le C_{4}(M) \Vert \partial _x^2 v\Vert _{L_2(I)} \left[ \Vert f \Vert _{L_2({\mathcal {O}}_I(v))}^{1/2} \Vert \nabla \partial _z \zeta _v\Vert _{L_2({\mathcal {O}}_I(v))}^{3/2} + \Vert f\Vert _{L_2({\mathcal {O}}_I(v))}^2 \right] \ . \nonumber \\ \end{aligned}$$

(3.38)

Proof

To lighten notation, we set ${\mathcal {O}} := {\mathcal {O}}_I(v)$ and introduce $P:= \partial _z \zeta _v - \sigma \zeta _v$. Since $\zeta _v\in H^2({\mathcal {O}})$ by Lemma 3.10 and $\sigma \in C^2(\bar{I})$, the function P belongs to $H^1({\mathcal {O}})$ and satisfies (C.2) by (3.18b) and (3.18c). In addition, we observe that $P(\cdot ,v)=\partial _z \zeta _v(\cdot ,v)$ by (3.18b). It then follows from Lemma C.3 that

$$\begin{aligned} \Vert \partial _z\zeta _v(\cdot ,v)\Vert _{L_r(I)}^r = \Vert P(\cdot ,v)\Vert _{L_r(I)}^r \le \left( 4r\sqrt{M} \right) ^r \Vert P\Vert _{L_2({\mathcal {O}})} \Vert \nabla P\Vert _{L_2({\mathcal {O}})}^{r-1}\ . \end{aligned}$$

Moreover, by (2.8) and Lemma 3.8,

$$\begin{aligned} \Vert P\Vert _{L_2({\mathcal {O}})}&\le \Vert \partial _z\zeta _v\Vert _{L_2({\mathcal {O}})} + {\bar{\sigma }} \Vert \zeta _v\Vert _{L_2({\mathcal {O}})} \le \left( 1 + {\bar{\sigma }} \right) \Vert \zeta _v\Vert _{H^1({\mathcal {O}})} \\&\le 4 \Vert H+v\Vert _{L_\infty (I)} \sqrt{1+4\Vert H+v\Vert _{L_\infty (I)}^2} \left( 1 + {\bar{\sigma }} \right) \Vert f\Vert _{L_2({\mathcal {O}})} \le C(M) \Vert f\Vert _{L_2({\mathcal {O}})} \end{aligned}$$

and

$$\begin{aligned} \Vert \nabla P\Vert _{L_2({\mathcal {O}})}&\le \Vert \partial _x P\Vert _{L_2({\mathcal {O}})} + \Vert \partial _z P\Vert _{L_2({\mathcal {O}})} \\&\le \Vert \partial _x\partial _z \zeta _v\Vert _{L_2({\mathcal {O}})} {+} {\bar{\sigma }} \Vert \partial _x \zeta _v\Vert _{L_2({\mathcal {O}})} {+} {\bar{\sigma }} \Vert \zeta _v\Vert _{L_2({\mathcal {O}})} {+} \Vert \partial _z^2 \zeta _v\Vert _{L_2({\mathcal {O}})} {+} {\bar{\sigma }} \Vert \partial _z \zeta _v\Vert _{L_2({\mathcal {O}})} \\&\le \sqrt{2} \Vert \nabla \partial _z \zeta _v\Vert _{L_2({\mathcal {O}})} + {\bar{\sigma }} \left( \sqrt{2} \Vert \nabla \zeta _v\Vert _{L_2({\mathcal {O}})} + \Vert \zeta _v\Vert _{L_2({\mathcal {O}})} \right) \\&\le \sqrt{2} \Vert \nabla \partial _z \zeta _v\Vert _{L_2({\mathcal {O}})} + C(M) \Vert f\Vert _{L_2({\mathcal {O}})}\ . \end{aligned}$$

Collecting the previous estimates, we end up with

$$\begin{aligned} \Vert \partial _z\zeta _v(\cdot ,v)\Vert _{L_r(I)}^r&\le \left( 4r\sqrt{M} \right) ^r C(M) \Vert f\Vert _{L_2({\mathcal {O}})} \left( \sqrt{2} \Vert \nabla \partial _z \zeta _v\Vert _{L_2({\mathcal {O}})} + C(M) \Vert f\Vert _{L_2({\mathcal {O}})} \right) ^{r-1} \\&\le (rC(M))^r \Vert f\Vert _{L_2({\mathcal {O}})} \left( \Vert \nabla \partial _z \zeta _v\Vert _{L_2({\mathcal {O}})} + \Vert f\Vert _{L_2({\mathcal {O}})} \right) ^{r-1}\ , \end{aligned}$$

from which (3.37) follows. We next deduce from (3.37) (with $r=4$) and Hölder’s inequality that

$$\begin{aligned} \left| \int _I \partial _x^2 v |\partial _z\zeta _v(\cdot ,v)|^2\ \mathrm {d}x \right|&\le \Vert \partial _x^2 v\Vert _{L_2(I)} \Vert \partial _z\zeta _v(\cdot ,v)\Vert _{L_4(I)}^2 \\&\le 16 C_{3}(M)^2 \Vert \partial _x^2 v\Vert _{L_2(I)} \Vert f\Vert _{L_2({\mathcal {O}})}^{1/2} \left( \Vert \nabla \partial _z \zeta _v\Vert _{L_2({\mathcal {O}})} + \Vert f\Vert _{L_2({\mathcal {O}})} \right) ^{3/2} \\&\le C(M) \Vert \partial _x^2 v\Vert _{L_2(I)} \Vert f\Vert _{L_2({\mathcal {O}})}^{1/2} \left( \Vert \nabla \partial _z \zeta _v\Vert _{L_2({\mathcal {O}})}^{3/2} + \Vert f\Vert _{L_2({\mathcal {O}})}^{3/2} \right) \ , \end{aligned}$$

and the proof is complete. $\square$

We are now in a position to derive quantitative estimates in $H^2$ for $\zeta _v$, which only depends on the $H^2$-norm of v, even though v is assumed to be more regular.

Lemma 3.13

There is $C_{5}(M)>0$ such that

$$\begin{aligned} \Vert \nabla \partial _z \zeta _v\Vert _{L_2({\mathcal {O}}_I(v))}^2 + \Vert \sqrt{\sigma }\partial _x\zeta _v(\cdot ,-H)\Vert _{L_2(I)}^2&\le C_{5}(M) \left( 1 + \Vert \partial _x^2 v\Vert _{L_2(I)}^4 \right) \Vert f \Vert _{L_2({\mathcal {O}}_I(v))}^2\ , \end{aligned}$$

(3.39a)

$$\begin{aligned} \Vert \partial _x^2 \zeta _v \Vert _{L_2({\mathcal {O}}_I(v))}^2&\le C_{5}(M) \left( 1 + \Vert \partial _x^2 v\Vert _{L_2(I)}^4 \right) \Vert f\Vert _{L_2({\mathcal {O}}_I(v))}^2\ . \end{aligned}$$

(3.39b)

Proof

To lighten notation, we set ${\mathcal {O}} := {\mathcal {O}}_I(v)$. We infer from (3.18a) and Lemma 3.11 that

$$\begin{aligned} - \int _{{\mathcal {O}}} f\partial _z^2\zeta _v\ \mathrm {d}(x,z)&= \int _{{\mathcal {O}}} \left( \partial _x^2\zeta _v \partial _z^2\zeta _v + |\partial _z^2\zeta _v|^2 \right) \ \mathrm {d}(x,z) \\&= \Vert \nabla \partial _z\zeta _v\Vert _{L_2({\mathcal {O}})}^2 + \int _I \partial _x\zeta _v(\cdot ,-H) \partial _x(\sigma \zeta _v)(\cdot ,-H)\ \mathrm {d}x \\&\quad - \frac{1}{2} \int _I \partial _x^2 v |\partial _z\zeta _v(\cdot ,v)|^2\ \mathrm {d}x\ . \end{aligned}$$

Hence, thanks to (2.8), Lemma 3.12, and Hölder’s and Young’s inequalities,

$$\begin{aligned} X:= & {} \Vert \nabla \partial _z\zeta _v\Vert _{L_2({\mathcal {O}})}^2 + \Vert \sqrt{\sigma }\partial _x\zeta _v(\cdot ,-H)\Vert _{L_2(I)}^2 \\= & {} - \int _{{\mathcal {O}}} f\partial _z^2\zeta _v\ \mathrm {d}(x,z) - \int _I \partial _x \sigma (\zeta _v\partial _x\zeta _v)(\cdot ,-H)\ \mathrm {d}x + \frac{1}{2} \int _I \partial _x^2 v |\partial _z\zeta _v(\cdot ,v)|^2\ \mathrm {d}x \\\le & {} \Vert f\Vert _{L_2({\mathcal {O}})} \Vert \partial _z^2\zeta _v\Vert _{L_2({\mathcal {O}})} + {\bar{\sigma }} \Vert \zeta _v(\cdot ,-H)\Vert _{L_2(I)} \Vert \partial _x\zeta _v(\cdot ,-H)\Vert _{L_2(I)} \\&+ \frac{C_{4}(M)}{2} \Vert \partial _x^2 v\Vert _{L_2(I)} \left[ \Vert f \Vert _{L_2( {\mathcal {O}})}^{1/2} \Vert \nabla \partial _z \zeta _v\Vert _{L_2( {\mathcal {O}})}^{3/2} + \Vert f\Vert _{L_2({\mathcal {O}})}^2 \right] \\\le & {} \frac{1}{4} \Vert \partial _z^2\zeta _v\Vert _{L_2({\mathcal {O}})}^2 + \Vert f\Vert _{L_2({\mathcal {O}})}^2 + \frac{{\bar{\sigma }}}{\sqrt{\sigma _{min}}} \Vert \zeta _v(\cdot ,-H)\Vert _{L_2(I)} \Vert \sqrt{\sigma }\partial _x\zeta _v(\cdot ,-H)\Vert _{L_2(I)} \\&+ \frac{1}{4} \Vert \nabla \partial _z\zeta _v\Vert _{L_2({\mathcal {O}})}^2 + C(M) \left( \Vert \partial _x^2 v\Vert _{L_2(I)}^4 + \Vert \partial _x^2 v\Vert _{L_2(I)} \right) \Vert f\Vert _{L_2({\mathcal {O}})}^2 \\\le & {} \frac{1}{2} \Vert \nabla \partial _z\zeta _v\Vert _{L_2({\mathcal {O}})}^2 + \frac{1}{2} \Vert \sqrt{\sigma }\partial _x\zeta _v(\cdot ,-H)\Vert _{L_2(I)}^2 + \frac{{\bar{\sigma }}^2}{2\sigma _{min}} \Vert \zeta _v(\cdot ,-H)\Vert _{L_2(I)}^2 \\&+ C(M) \left( 1 + \Vert \partial _x^2 v\Vert _{L_2(I)}^4 \right) \Vert f\Vert _{L_2({\mathcal {O}})}^2\ . \end{aligned}$$

Consequently, using once more Young’s inequality,

$$\begin{aligned} X \le \frac{{\bar{\sigma }}^2}{\sigma _{min}} \Vert \zeta _v(\cdot ,-H)\Vert _{L_2(I)}^2 + C(M) \left( 1 + \Vert \partial _x^2 v\Vert _{L_2(I)}^4 \right) \Vert f\Vert _{L_2({\mathcal {O}})}^2\ . \end{aligned}$$

Now, since $\zeta _v\in H^1_B({\mathcal {O}})$, it follows from (2.8), (3.32), and Lemma 3.8 that

$$\begin{aligned} 2 \sigma _{min} \Vert \zeta _v(\cdot ,-H)\Vert _{L_2(I)}^2 \le 16 M^2 (1+4M^2) \Vert f\Vert _{L_2({\mathcal {O}})}^2\ . \end{aligned}$$

Combining the above two estimates gives (3.39a).

To complete the proof of Lemma 3.13, we simply notice that (3.18a) ensures that

$$\begin{aligned} \Vert \partial _x^2\zeta _v\Vert _{L_2({\mathcal {O}})}^2 = \Vert f + \partial _z^2\zeta _v\Vert _{L_2({\mathcal {O}})}^2 \le 2 \Vert \partial _z^2\zeta _v\Vert _{L_2({\mathcal {O}})}^2 + 2 \Vert f\Vert _{L_2({\mathcal {O}})}^2 \end{aligned}$$

and deduce (3.39b) from (3.39a). $\square$

Summarizing, we have established the following result:

Proposition 3.14

Consider $v\in H^2(I)$ satisfying (3.29); that is,

$$\begin{aligned} v\in W_\infty ^3(I) \;\text { and }\; \min _{[a,b]} v > -H\, , \end{aligned}$$

and fix $\kappa >0$ such that

$$\begin{aligned} \Vert v\Vert _{H^2(I)}\le \kappa \ . \end{aligned}$$

(3.40)

Then the elliptic boundary value problem (3.18) has a unique strong solution $\zeta _v\in H^2({\mathcal {O}}_I(v))$ which satisfies

$$\begin{aligned} \Vert \zeta _v\Vert _{H^2({\mathcal {O}}_I(v))} + \Vert \partial _x\zeta _v(\cdot ,-H)\Vert _{L_2(I)}&\le C_{6}(\kappa ) \Vert f\Vert _{L_2({\mathcal {O}}_I(v))}\ , \end{aligned}$$

(3.41)

$$\begin{aligned} \Vert \partial _z \zeta _v(\cdot ,v)\Vert _{L_r(I)}&\le r C_{6}(\kappa ) \Vert f\Vert _{L_2({\mathcal {O}}_I(v))} \ , \qquad r\in [2,\infty )\ . \end{aligned}$$

(3.42)

Proof

The existence and uniqueness of a strong solution $\zeta _v\in H^2({\mathcal {O}}_I(v))$ to (3.18) are consequences of Lemma 3.8 and Lemma 3.10. Next, it readily follows from (3.40) and the continuous embedding of $H^2(I)$ in $W_\infty ^1(I)$ that there is $M\ge 1$ depending on $\kappa$ such that

$$\begin{aligned} \Vert H+v\Vert _{L_\infty (I)} + \Vert \partial _x v\Vert _{L_\infty (I)} \le M\ . \end{aligned}$$

(3.43)

Due to (3.43), we deduce (3.41) from (2.8), (3.40), Lemma 3.8, and Lemma 3.13, while (3.42) follows from (3.41) and Lemma 3.12. $\square$

We emphasize that, though derived for functions $v\in H^2(I)$ satisfying the additional assumption (3.29), the estimates stated in Proposition 3.14 only depend on the $H^2$-norm of v and, neither on its $W_\infty ^2$-norm, nor on the value of its minimum (provided that it stays above $-H$). The outcome of Proposition 3.14 is thus likely to extend to any configuration depicted in Fig. 2 under the sole assumption (3.13) and this will be shown in the next section by an approximation argument.

3.2.5 $H^2$-regularity: Proof of Theorem 3.9

We now prove the $H^2$-regularity of $\zeta _v$ as stated in Theorem 3.9. We thus assume that v satisfies (3.13); that is,

$$\begin{aligned} v\in H^2(I) \;\text { such that }\; v(x)>-H\, , \qquad x\in I\, , \end{aligned}$$

and fix $\kappa >0$ such that $\Vert v\Vert _{H^2(I)}\le \kappa$. Owing to the density of $C^\infty ([a,b])$ in $H^2(I)$ and since v satisfies (3.13), we employ classical approximation arguments to construct a sequence $(v_n)_{n\ge 1}$ of functions in $C^\infty ([a,b])$ with the following properties:

$$\begin{aligned}&\lim _{n\rightarrow \infty } \Vert v_n - v\Vert _{H^2(I)} = 0\ , \qquad \sup _{n\ge 1}\{\Vert v_n\Vert _{H^2(I)}\} \le 1+\kappa \ , \end{aligned}$$

(3.44a)

$$\begin{aligned}&v_n\ge v + \frac{1}{n}\ , \qquad n\ge 1\ . \end{aligned}$$

(3.44b)

A first consequence of (3.44a) and the continuous embedding of $H^2(I)$ in $W_\infty ^1(I)$ is that

$$\begin{aligned} \begin{aligned}&\Vert H+v_n\Vert _{L_\infty (I)} + \Vert \partial _x v_n\Vert _{L_\infty (I)} \le C(\kappa )\ , \qquad n\ge 1\ , \\&\lim _{n\rightarrow \infty } \Vert v_n - v\Vert _{W_\infty ^1(I)} = 0 \ . \end{aligned} \end{aligned}$$

(3.45)

According to (3.13) and (3.44b), the function $v_n$ satisfies (3.29) for each $n\ge 1$ and, since ${\mathcal {O}}_I(v)\subset {\mathcal {O}}_I(v_n)$, we infer from Proposition 3.14 that the strong solution $\zeta _{v_n}$ to (3.18) with $v_n$ instead of v (and f replaced by its trivial extension to ${\mathcal {O}}_I(v_n)$) satisfies

$$\begin{aligned} \Vert \zeta _{v_n}\Vert _{H^2({\mathcal {O}}_I(v_n))} + \Vert \partial _x\zeta _{v_n}(\cdot ,-H)\Vert _{L_2(I)}&\le C_{7}(\kappa ) \Vert f\Vert _{L_2({\mathcal {O}}_I(v))}\ , \end{aligned}$$

(3.46)

$$\begin{aligned} \Vert \partial _z \zeta _{v_n}(\cdot ,v_n)\Vert _{L_r(I)}&\le r C_{7}(\kappa ) \Vert f\Vert _{L_2({\mathcal {O}}_I(v))} \ , \qquad r\in [2,\infty )\ . \end{aligned}$$

(3.47)

Using again the inclusion ${\mathcal {O}}_I(v)\subset {\mathcal {O}}_I(v_n)$, we deduce from (3.46) that $(\zeta _{v_n})_{n\ge 1}$ is bounded in $H^2({\mathcal {O}}_I(v))$. Consequently, recalling that $H^1({\mathcal {O}}_I(v))$ is compactly embedded in $L_2({\mathcal {O}}_I(v))$ (despite the non-Lipschitz character of ${\mathcal {O}}_I(v)$, see [23, Theorem 11.21] or [28, I.Theorem 1.4]), there are a subsequence of $(\zeta _{v_n})_{n\ge 1}$ (not relabeled) and $\phi \in H^2({\mathcal {O}}_I(v))$ such that

$$\begin{aligned} \begin{aligned} \zeta _{v_n} \rightharpoonup \phi \;\;\text { in }\;\; H^2({\mathcal {O}}_I(v)) \ , \\ \zeta _{v_n} \longrightarrow \phi \;\;\text { in }\;\; H^1({\mathcal {O}}_I(v)) \ . \end{aligned} \end{aligned}$$

(3.48)

Let us first check that $\phi \in H_B^1({\mathcal {O}}_I(v))$. On the one hand, since both $\phi$ and $\zeta _{v_n}$ belong to $H^1({\mathcal {O}}_I(v))$, we infer from (3.19) that

$$\begin{aligned} \int _I \left| (\phi -\zeta _{v_n})(\cdot ,v) \right| ^2 (H+v)\ \mathrm {d}x \le C(\kappa ) \Vert \phi -\zeta _{v_n}\Vert _{H^1({\mathcal {O}}_I(v))}^2\ . \end{aligned}$$

Hence, by (3.48),

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _I \left| (\phi -\zeta _{v_n})(\cdot ,v) \right| ^2 (H+v)\ \mathrm {d}x = 0\ . \end{aligned}$$

On the other hand, since $\zeta _{v_n}\in H_B^1({\mathcal {O}}_I(v_n))$ and $v_n\ge v$, it follows from Lemma A.1 and (3.46) that

$$\begin{aligned} \int _I \left| \zeta _{v_n}(\cdot ,v)\right| ^2 (H+v)\ \mathrm {d}x&= \int _I \left| \zeta _{v_n}(\cdot ,v) - \zeta _{v_n}(\cdot ,v_n) \right| ^2 (H+v)\ \mathrm {d}x \\&\le \Vert (v-v_n) (H+v)\Vert _{L_\infty (I)} \Vert \partial _z \zeta _{v_n} \Vert _{L_2({\mathcal {O}}_I(v_n))} \\&\le C(\kappa ) \Vert v-v_n\Vert _{L_\infty (I)} \Vert f\Vert _{L_2({\mathcal {O}}_I(v))}\ . \end{aligned}$$

Hence, by (3.45),

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _I \left| \zeta _{v_n}(\cdot ,v)\right| ^2 (H+v)\ \mathrm {d}x = 0\ . \end{aligned}$$

Combining the previous two limits, we deduce

$$\begin{aligned} \int _I \left| \phi (\cdot ,v) \right| ^2 (H+v)\ \mathrm {d}x = 0\ , \end{aligned}$$

so that $\phi \in H_B^1({\mathcal {O}}_I(v))$. In particular, for $n\ge 1$, due to the inclusion ${\mathcal {O}}_I(v)\subset {\mathcal {O}}_I(v_n)$, the function $\phi$ also belongs to $H_B^1({\mathcal {O}}_I(v_n))$ and we infer from (3.22) and (3.48) that

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _I \left| (\zeta _{v_n}-\phi )(\cdot ,-H) \right| ^2 \mathrm {d}x = 0\ . \end{aligned}$$

(3.49)

We next recall that $\zeta _{v_n}$ is the unique solution in $H_B^1({\mathcal {O}}_I(v_n))$ to

$$\begin{aligned} \int _{{\mathcal {O}}_I(v_n)} \nabla \zeta _{v_n}\cdot \nabla \vartheta \ \mathrm {d}(x,z) + \int _I \sigma \zeta _{v_n}(\cdot ,-H) \vartheta (\cdot ,-H)\ \mathrm {d}x= \int _{{\mathcal {O}}_I(v_n)} f \vartheta \ \mathrm {d}x \end{aligned}$$

(3.50)

for all $\vartheta \in H_B^1({\mathcal {O}}_I(v_n))$. Now, since $H_B^1({\mathcal {O}}_I(v)) \subset H_B^1({\mathcal {O}}_I(v_n))$, we can take $\vartheta \in H_B^1({\mathcal {O}}_I(v))$ in (3.50) and use the convergences (3.48) and (3.49) to pass to the limit $n\rightarrow \infty$ and conclude that $\phi \in H_B^1({\mathcal {O}}_I(v))$ satisfies the variational formulation of (3.18). Therefore, Lemma 3.8 guarantees that $\phi =\zeta _v$. We have thus shown that $\zeta _v\in H^2({\mathcal {O}}_I(v))$ and it follows from (3.46) and (3.48) that

$$\begin{aligned} \Vert \zeta _{v}\Vert _{H^2({\mathcal {O}}_I(v))}\le & {} \liminf _{n\rightarrow \infty } \Vert \zeta _{v_n}\Vert _{H^2({\mathcal {O}}_I(v))} \nonumber \\\le & {} \liminf _{n\rightarrow \infty }\Vert \zeta _{v_n}\Vert _{H^2({\mathcal {O}}_I(v_n))} \le C_{7}(\kappa ) \Vert f\Vert _{L_2({\mathcal {O}}_I(v))}\ . \end{aligned}$$

(3.51)

A further consequence of (3.20) and (3.48) is that $(\partial _x\zeta _{v_n}(\cdot ,-H))_{n\ge 1}$ converges to $\partial _x\zeta _{v}(\cdot ,-H)$ in $L_2(I,(H+v)\mathrm {d}x)$, which, together with the positivity of $H+v$ in I, implies that $(\partial _x\zeta _{v_n}(\cdot ,-H))_{n\ge 1}$ converges to $\partial _x\zeta _{v}(\cdot ,-H)$ in $L_2(a+\varepsilon ,b-\varepsilon )$ for any $\varepsilon \in (0,(b-a)/2)$. Combining this convergence with (3.46) and using Fatou’s lemma to take the limit $\varepsilon \rightarrow 0$ give

$$\begin{aligned} \Vert \partial _x\zeta _{v}(\cdot ,-H)\Vert _{L_2(I)} \le C_{7}(\kappa ) \Vert f\Vert _{L_2({\mathcal {O}}_I(v))}\ . \end{aligned}$$

(3.52)

Finally, by (3.19) and (3.46),

$$\begin{aligned} \int _I \left| (\partial _z\zeta _{v_n} - \partial _z \zeta _v)(\cdot ,v) \right| ^2 (H+v)\ \mathrm {d}x \le C(\kappa ) \Vert \partial _z\zeta _{v_n} - \partial _z \zeta _v\Vert _{L_2({\mathcal {O}}_I(v))}\ . \end{aligned}$$

Hence, by (3.48),

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _I \left| (\partial _z\zeta _{v_n} - \partial _z \zeta _v)(\cdot ,v) \right| ^2 (H+v)\ \mathrm {d}x = 0\ . \end{aligned}$$

(3.53)

Moreover, owing to Lemma A.1, (3.46), and the properties $\zeta _{v_n}\in H_B^1({\mathcal {O}}_I(v_n))$ and $v_n\ge v$,

$$\begin{aligned} \int _I \left| \partial _z\zeta _{v_n}(\cdot ,v) - \partial _z\zeta _{v_n}(\cdot ,v_n) \right| ^2 (H+v)\ \mathrm {d}x&\le \Vert (v-v_n)(H+v)\Vert _{L_\infty (I)} \Vert \partial _z^2 \zeta _{v_n} \Vert _{L_2({\mathcal {O}}_I(v_n))}^{2} \\&\le C(\kappa ) \Vert v-v_n\Vert _{L_\infty (I)}\ , \end{aligned}$$

and it follows from (3.45) that

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _I \left| \partial _z\zeta _{v_n}(\cdot ,v) - \partial _z\zeta _{v_n}(\cdot ,v_n) \right| ^2 (H+v)\ \mathrm {d}x = 0\ . \end{aligned}$$

(3.54)

Gathering (3.53) and (3.54) leads us to

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _I \left| \partial _z\zeta _{v}(\cdot ,v) - \partial _z\zeta _{v_n}(\cdot ,v_n) \right| ^2 (H+v)\ \mathrm {d}x = 0\ . \end{aligned}$$

(3.55)

Since $H+v>0$ in I, we may extract a further subsequence (not relabeled) such that $(\partial _z\zeta _{v_n}(\cdot ,v_n))_{n\ge 1}$ converges a.e. in I to $\partial _z\zeta _{v}(\cdot ,v)$. We then use Fatou’s lemma to pass to the limit $n\rightarrow \infty$ in (3.47) and conclude that

$$\begin{aligned} \Vert \partial _z \zeta _{v}(\cdot ,v)\Vert _{L_r(I)} \le r C_{7}(\kappa ) \Vert f\Vert _{L_2({\mathcal {O}}_I(v))} \ , \qquad r\in [2,\infty )\ , \end{aligned}$$

thereby completing the proof of Theorem 3.9.

3.3 Global $H^2$-regularity of $\chi _v$: Proof of Theorem 3.2 and Theorem 2.2

Finally, we prove Theorem 3.2 and Theorem 2.2 for which we consider an arbitrary function v in ${\bar{S}}$ and $\kappa >0$ satisfying (3.8). According to [1, IX.Proposition 1.8] we can write the open set $D\setminus {\mathcal {C}}(v)$ as a countable union of disjoint open intervals $(I_j)_{j\in J}$; that is,

$$\begin{aligned} D\setminus {\mathcal {C}}(v)=\bigcup _{j\in J} I_j\,. \end{aligned}$$

Hence, $\Omega (v)$ is the disjoint union of the open domains ${\mathcal {O}}_{I_j}(v)$. Now recall from Proposition 3.3 that $\chi _v\in H_B^1(\Omega (v))$ is the unique minimizer on $H_B^1(\Omega (v))$ of the functional

$$\begin{aligned} G_D(v)[\vartheta ]= & {} \frac{1}{2} \int _{\Omega (v)} |\nabla \vartheta |^2\ \mathrm {d}(x,z) + \frac{1}{2} \int _D \sigma |\vartheta (\cdot ,-H)|^2\ \mathrm {d}x \\&- \int _{\Omega (v)} \vartheta \Delta h_v \ \mathrm {d}(x,z)\,, \qquad \qquad \qquad \qquad \qquad \qquad \qquad \vartheta \in H_B^1(\Omega (v))\,. \end{aligned}$$

Furthermore, since $\Delta h_v$ belongs to $L_2(\Omega (v))$ by Lemma 3.1, it follows from the definition of $H_B^1(\Omega (v))$ that

$$\begin{aligned} G_D(v)[\vartheta ]=\sum _{j\in J} G_{I_j}(v)[\vartheta ]\,, \qquad \vartheta \in H_B^1(\Omega (v))\,, \end{aligned}$$

where $G_{I_j}(v)[\vartheta ]$ is defined by (3.23) with $f :=\Delta h_v {\mathbf {1}}_{{\mathcal {O}}_{I_j}(v)}$. Restricting to $\vartheta \in H_B^1({\mathcal {O}}_{I_j}(v))$, it thus readily follows that $\chi _v {\mathbf {1}}_{{\mathcal {O}}_{I_j}(v)}$ is a minimizer of $G_{I_j}(v)$ on $H_B^1({\mathcal {O}}_{I_j}(v))$. Consequently, $\chi _v {\mathbf {1}}_{{\mathcal {O}}_{I_j}(v)} = \zeta _{I_j,v}$ by Lemma 3.8. Hence Theorem 3.9 yields

$$\begin{aligned} \Vert \chi _v\Vert _{H^2({\mathcal {O}}_{I_j}(v))} + \Vert \partial _x \chi _v(\cdot ,-H)\Vert _{L_2(I_j)} \le C_{1}(\kappa ) \Vert \Delta h_v\Vert _{L_2({\mathcal {O}}_{I_j}(v))} \end{aligned}$$

and

$$\begin{aligned} \Vert \partial _z\chi _v(\cdot ,v)\Vert _{L_r(I_j)} \le r C_{2}(\kappa ) \Vert \Delta h_v\Vert _{L_2({\mathcal {O}}_{I_j}(v))} \,, \qquad r\in [2,\infty )\,, \end{aligned}$$

with constants $C_{1}(\kappa )$ and $C_{2}(\kappa )$ not depending on $I_j$. Therefore, summing with respect to $j\in J$, we conclude that $\chi _v\in H^2(\Omega (v))$ and satisfies (3.9) and (3.10), since $\Vert \Delta h_v\Vert _{L_2(\Omega (v))}\le c(\kappa )$ by Lemma 3.1. Therefore, as in [16, Theorem 3.5], we may use the version of Gauß’ Theorem stated in [15, Folgerung 7.5] in the variational characterization of $\chi _v$ featuring ${\mathcal {G}}(v)$ to deduce that $\chi _v\in H^2(\Omega (v))$ is indeed a strong solution to (3.7). This proves Theorem 3.2. Owing to (3.6) and Lemma 3.1, this also entails Theorem 2.2.

4 Continuity of $\chi _v$ with respect to v

In this section we derive continuity properties of $\chi _v$ and its gradient trace $\partial _z\chi _v(\cdot , v)$ with respect to $v\in {\bar{S}}$. The latter will also yield the continuity of the function g defined in (2.6). Throughout this section we denote positive constants depending only on $\sigma$ by C. The dependence upon additional parameters will be indicated explicitly.

4.1 $H^1$-Continuity: $\Gamma$-convergence of ${\mathcal {G}}$

Let us recall that, according to Proposition 3.3, $\chi _v$ is the unique minimizer on $H_B^1(\Omega (v))$ of the functional ${\mathcal {G}}(v)$ introduced in (3.11) as

$$\begin{aligned} {\mathcal {G}}(v)[\vartheta ]= \frac{1}{2} \int _{\Omega (v)} |\nabla (\vartheta +h_v)|^2 \,\mathrm {d}(x,z)\! +\! \frac{1}{2} \int _D \sigma (x) |\vartheta (x,-H)+h_v(x,-H) \!-\! {\mathfrak {h}}_v(x)|^2 \,\mathrm {d}x \end{aligned}$$

for $\vartheta \in H_B^1(\Omega (v))$. Now, in order to derive continuity properties of $\chi _v$ (and $\psi _v$) with respect to $v\in {\bar{S}}$, we first prove a $\Gamma$–convergence result for the set of functionals $\{{\mathcal {G}}(v)\,,\, v\in {\bar{S}}\}$. More precisely, given $M>0$ we set as before $\Omega (M) := D\times (-H,M)$ and, for $v\in {\bar{S}}$ such that $v\le M-H$, we extend the functional ${\mathcal {G}}(v)$ to $L_2(\Omega (M))$ by defining

$$\begin{aligned} {\mathcal {G}}(v)[\vartheta ]:= \infty \,, \quad \vartheta \in L_2(\Omega (M))\setminus H_B^1(\Omega (v))\,. \end{aligned}$$

With these notations we have:

Proposition 4.1

Let $M>0$ and consider a sequence $(v_n)_{n\ge 1}$ in $\bar{S}$ and $v\in {\bar{S}}$ such that

$$\begin{aligned} -H\le v_n(x)\,,\, v(x)\le M-H\,,\quad x\in D\,, \qquad v_n\rightarrow v \text { in }\ H^1(D)\,. \end{aligned}$$

(4.1)

Then

$$\begin{aligned} \Gamma -\lim _{n\rightarrow \infty } {\mathcal {G}}(v_n) ={\mathcal {G}}(v)\quad \text {in }\ L_2(\Omega (M))\,. \end{aligned}$$

Proof

The proof is very similar to that of [21, Proposition 3.11].

(i) Asymptotic weak lower semi-continuity. Given a sequence $(\vartheta _n)_{n\ge 1}$ in $L_2(\Omega (M))$ and $\vartheta \in L_2(\Omega (M))$ satisfying

$$\begin{aligned} \vartheta _n\rightarrow \vartheta \ \text { in }\ L_2(\Omega (M))\,, \end{aligned}$$

(4.2)

we shall show that

$$\begin{aligned} {\mathcal {G}}(v)[\vartheta ]\le \liminf _{n\rightarrow \infty } {\mathcal {G}}(v_n)[\vartheta _n]\,. \end{aligned}$$

(4.3)

We may assume without loss of generality that

$$\begin{aligned} \vartheta _n\in H_B^1(\Omega (v_n))\,, \quad n\ge 1\,, \qquad {\mathcal {G}}_\infty := \sup _{n\ge 1} {\mathcal {G}}(v_n)[\vartheta _n]<\infty \,. \end{aligned}$$

(4.4)

Let $n\ge 1$ and denote the extension by zero of $\vartheta _n$ to $\Omega (M)\setminus \Omega (v_n)$ by $\vartheta _n$. Then $\vartheta _n\in H_B^1(\Omega (M))$ and it follows from (4.1), (4.2), (4.4), and Lemma 3.1 (b) that the sequence $(\vartheta _n)_{n\ge 1}$ is bounded in $H_B^1(\Omega (M))$. Since $\Omega (M)$ is a Lipschitz domain, the compactness of the embedding of $H^1(\Omega (M))$ in $H^{3/4}(\Omega (M))$ [12, Theorem 1.4.3.2], the continuity of the trace operator from $H^{3/4}(\Omega (M))$ to $L_2(\partial \Omega (M))$ (see, e.g., [12, Theorem 1.5.1.2], [26], or [34, Satz 8.7]) and (4.2) ensure that there is a subsequence of $(\vartheta _n)_{n\ge 1}$ (not relabeled) such that

$$\begin{aligned} \vartheta _n&\rightharpoonup \vartheta \quad \text {in }\ H_B^1(\Omega (M))\,, \end{aligned}$$

(4.5)

$$\begin{aligned} \vartheta _n&\rightarrow \vartheta \quad \text {in }\ L_2(\partial \Omega (M))\,. \end{aligned}$$

(4.6)

In particular, $\vartheta \in H^1(\Omega (v))$ and its trace $\vartheta (\cdot ,v)$ is well-defined in $L_2(D,(H+v)\,\mathrm {d}x)$ according to Lemma 3.6. Similarly, for each $n\ge 1$, $\vartheta \in H^1(\Omega (v_n))$ and its trace $\vartheta (\cdot ,v_n)$ is well-defined in $L_2(D,(H+v_n)\,\mathrm {d}x)$. Consequently, for $n\ge 1$,

$$\begin{aligned} \begin{aligned} \int _D (H+v) (H+v_n) |\vartheta (\cdot ,v)|^2\,\mathrm {d}x&\le 2 \int _D (H+v) (H+v_n) |\vartheta (\cdot ,v)-\vartheta (\cdot ,v_n)|^2\,\mathrm {d}x \\&\quad + 2 \int _D (H+v) (H+v_n) |\vartheta (\cdot ,v_n)|^2\,\mathrm {d}x\,. \end{aligned} \end{aligned}$$

(4.7)

On the one hand, by Lemma A.1 and (4.1),

$$\begin{aligned}&\int _D (H+v) (H+v_n) |\vartheta (\cdot ,v)-\vartheta (\cdot ,v_n)|^2\,\mathrm {d}x \nonumber \\&\quad \le \Vert (H+v) (H+v_n) (v-v_n)\Vert _{L_\infty (D)} \Vert \partial _z\vartheta \Vert _{L_2(\Omega (M))}^2 \nonumber \\&\quad \le M^2 \Vert v-v_n\Vert _{L_\infty (D)} \Vert \partial _z\vartheta \Vert _{L_2(\Omega (M))}^2\,. \end{aligned}$$

(4.8)

On the other hand, since $\vartheta _n\in H_B^1(\Omega (v_n))$, we infer from (4.1) and Lemma 3.6 that

$$\begin{aligned}&\int _D (H+v) (H+v_n) |\vartheta (\cdot ,v_n)|^2\,\mathrm {d}x \nonumber \\&\quad = \int _D (H+v) (H+v_n) |\vartheta (\cdot ,v_n)-\vartheta _n(\cdot ,v_n)|^2\,\mathrm {d}x \nonumber \\&\quad \le M \int _D (H+v_n) |\vartheta (\cdot ,v_n)-\vartheta _n(\cdot ,v_n)|^2\,\mathrm {d}x \nonumber \\&\quad \le M \left[ \Vert \vartheta -\vartheta _n\Vert _{L_2(\Omega (v_n))}^2 + 2 \Vert H+v_n\Vert _{L_\infty (D)} \Vert \vartheta -\vartheta _n\Vert _{L_2(\Omega (v_n))} \Vert \partial _z(\vartheta -\vartheta _n)\Vert _{L_2(\Omega (v_n))} \right] \nonumber \\&\quad \le M \Vert \vartheta -\vartheta _n\Vert _{L_2(\Omega (M))} \left[ \sup _{m\ge 1} \Vert \vartheta -\vartheta _m\Vert _{L_2(\Omega (M))} + 2 M \sup _{m\ge 1} \Vert \partial _z(\vartheta -\vartheta _m)\Vert _{L_2(\Omega (M))} \right] \nonumber \\&\quad \le 2M(1+M) \Vert \vartheta -\vartheta _n\Vert _{L_2(\Omega (M))} \left[ \Vert \vartheta \Vert _{H^1(\Omega (M))} + \sup _{m\ge 1} \Vert \vartheta _m\Vert _{H^1(\Omega (M))} \right] \,. \end{aligned}$$

(4.9)

Now, it readily follows from (4.1), (4.2), (4.5), (4.8), (4.9), and the continuous embedding of $H_0^1(D)$ in $C(\bar{D})$ that the right-hand side of (4.7) converges to zero as $n\rightarrow \infty$. Therefore,

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _D (H+v) (H+v_n) |\vartheta (\cdot ,v)|^2\,\mathrm {d}x = 0\,, \end{aligned}$$

and we use Fatou’s lemma to conclude that

$$\begin{aligned} \vartheta (\cdot ,v)=0 \quad \text {in }\quad L_2(D, (H+v)^2\,\mathrm {d}x)\,. \end{aligned}$$

Combining this result with (4.5) and (4.6) implies that

$$\begin{aligned} \vartheta \in H_B^1(\Omega (v))\,. \end{aligned}$$

(4.10)

Now, we infer from (3.3), (4.1), (4.5), (4.10), and the continuous embedding of $H_0^1(D)$ in $C(\bar{D})$ that

$$\begin{aligned} \int _{\Omega (v)} |\nabla (\vartheta +h_v)|^2\,\mathrm {d}(x,z)&= \int _{\Omega (M)} |\nabla (\vartheta +h_v)|^2\,\mathrm {d}(x,z) - \int _{ \Omega (M)\setminus \Omega (v)} |\nabla h_v|^2\,\mathrm {d}(x,z) \\&\le \liminf _{n\rightarrow \infty } \int _{\Omega (M)} |\nabla (\vartheta _n+h_{v_n})|^2\,\mathrm {d}(x,z) \\&\quad - \lim _{n\rightarrow \infty } \int _{\Omega (M)\setminus \Omega (v_n)} |\nabla h_{v_n}|^2\,\mathrm {d}(x,z) \\&= \liminf _{n\rightarrow \infty } \int _{\Omega (v_n)} |\nabla (\vartheta _n+h_{v_n})|^2\,\mathrm {d}(x,z)\,. \end{aligned}$$

Also, from (4.6) and Lemma 3.1 we deduce that

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _D \sigma \left| (\vartheta _n + h_{v_n})(\cdot ,-H) - {\mathfrak {h}}_{v_n} \right| ^2 \,\mathrm {d}x = \int _D \sigma \left| (\vartheta + h_{v})(\cdot ,-H) - {\mathfrak {h}}_{v} \right| ^2 \,\mathrm {d}x\,. \end{aligned}$$

Gathering the outcome of the above analysis gives (4.3).

(ii) Recovery sequence. Consider $\vartheta \in H_B^1(\Omega (v))$ and introduce the function ${\bar{\vartheta }}$ defined on

$$\begin{aligned} {\hat{\Omega }}(M) := D\times (-2H-M,M) \end{aligned}$$

by

$$\begin{aligned} {{\bar{\vartheta }}}(x,z):= \left\{ \begin{array}{ll} 0\,, &{} x\in D\,,\ v(x)<z<M\,, \\ \vartheta (x,z)\,, &{} x\in D\,,\ -H<z\le v(x)\,,\\ \vartheta (x,-2H-z)\,, &{} x\in D\,,\ -2H-v(x)<z\le -H\,,\\ 0\,, &{} x\in D\,,\ -2H-M<z\le -2H-v(x)\,,\\ \end{array} \right. \end{aligned}$$

which is the extension of $\vartheta$ by zero in $\Omega (M)\setminus \Omega (v)$ and the reflection of the thus obtained function to $D\times (-2H-M,-H)$. Then ${{\bar{\vartheta }}}\in H_0^1({{\hat{\Omega }}}(M))$, so that $F:=-\Delta {{\bar{\vartheta }}}\in H^{-1}({{\hat{\Omega }}}(M))$.

Let $n\ge 1$. Since

$$\begin{aligned} {{\hat{\Omega }}}(v_n):=\Omega (v_n) \cup \big (D\times (-2H-M,-H]\big )\subset {{\hat{\Omega }}}(M)\,, \end{aligned}$$

the distribution F can also be considered as an element of $H^{-1}({{\hat{\Omega }}}(v_n))$ by restriction. Then there is a unique variational solution ${{\hat{\vartheta }}_n\in } H_0^1({{\hat{\Omega }}}(v_n))\subset H_0^1({{\hat{\Omega }}}(M))$ to

$$\begin{aligned} -\Delta {{\hat{\vartheta }}_n}=F \quad \text {in }\ {{\hat{\Omega }}}(v_n)\,,\qquad {\hat{\vartheta }}_n=0 \quad \text {on }\ \partial {{\hat{\Omega }}}(v_n)\,. \end{aligned}$$

Owing to (4.1) and the continuous embedding of $H_0^1(D)$ in $C(\bar{D})$,

$$\begin{aligned} d_H\left( {{\hat{\Omega }}}(v_n),{{\hat{\Omega }}}(v) \right) \le \Vert v_n - v\Vert _{L_\infty (D)}\rightarrow 0\,, \end{aligned}$$

where $d_H$ stands for the Hausdorff distance in ${{\hat{\Omega }}}(M)$, see [14, Sect. 2.2.3]. Since $\overline{{{\hat{\Omega }}}(M)}\setminus {{\hat{\Omega }}}(v_n)$ has a single connected component for all $n\ge 1$, it follows from [33, Theorem 4.1] and [14, Theorem 3.2.5] that ${{{\hat{\vartheta }}_n\rightarrow }} {{\hat{\vartheta }}}$ in $H_0^1({{\hat{\Omega }}}(M))$, where ${{{\hat{\vartheta }}_n\in }} H_0^1({{\hat{\Omega }}}(M))$ is the unique variational solution to

$$\begin{aligned} -\Delta {{\hat{\vartheta }}}=F \quad \text {in }\ {{\hat{\Omega }}}(M)\,,\qquad {{\hat{\vartheta }}}=0 \quad \text {on }\ \partial {{\hat{\Omega }}}(M)\,. \end{aligned}$$

Clearly, ${{\hat{\vartheta }}}={{\bar{\vartheta }}}$ by uniqueness, so that ${\hat{\vartheta }}_n \rightarrow {{\bar{\vartheta }}}$ in $H_0^1({\hat{\Omega }}(M))$. Setting $\vartheta _n := {{\hat{\vartheta }}_n} {\mathbf {1}}_{\Omega (v_n)}\in H^1(\Omega (M))$, $n\ge 1$, this convergence implies that

$$\begin{aligned} \vartheta _n\rightarrow {\bar{\vartheta }}\quad \text { in }\ H^1(\Omega (M))\,. \end{aligned}$$

(4.11)

Since $\vartheta _n=0$ in $\Omega (M)\setminus \Omega (v_n)$ we obtain from (3.3), (4.1), and (4.11) that

$$\begin{aligned} \begin{aligned} \int _{\Omega (v)} \vert \nabla (\vartheta +h_v)\vert ^2\,\mathrm {d} (x,z)&= \int _{\Omega (M)} \left( \vert \nabla {\bar{\vartheta }}\vert ^2 + 2 \nabla {\bar{\vartheta }}\cdot \nabla h_v\right) \,\mathrm {d} (x,z) + \int _{\Omega (v)} \vert \nabla h_v\vert ^2\,\mathrm {d} (x,z) \\&= \lim _{n\rightarrow \infty } \int _{\Omega (M)} \left( \vert \nabla \vartheta _n\vert ^2 + 2 \nabla \vartheta _n\cdot \nabla h_{v_n}\right) \,\mathrm {d} (x,z) \\&\quad + \lim _{n\rightarrow \infty } \int _{\Omega (v_n)} \vert \nabla h_{v_n}\vert ^2\,\mathrm {d} (x,z)\\&= \lim _{n\rightarrow \infty } \int _{\Omega (v_n)} \vert \nabla (\vartheta _n +h_{v_n})\vert ^2\,\mathrm {d} (x,z)\,. \end{aligned} \end{aligned}$$

Moreover, the continuity of the trace from $H^1(\Omega (M))$ to $L_2(D\times \{-H\})$ and (4.11) entail that

$$\begin{aligned} \vartheta _n(\cdot ,-H)\rightarrow {{\bar{\vartheta }}}(\cdot ,-H)=\vartheta (\cdot ,-H)\quad \text {in}\quad L_2(D)\,. \end{aligned}$$

These two properties, along with (3.4) and (3.5), imply that

$$\begin{aligned} {\mathcal {G}}(v)[\vartheta ]=\lim _{n\rightarrow \infty } {\mathcal {G}}(v_n)[\vartheta _n]\,; \end{aligned}$$

that is, $(\vartheta _n)_{n\ge 1}$ is a recovery sequence for $\vartheta$ and the claim is proved. $\square$

The Fundamental Theorem of $\Gamma$-convergence, see [9, Corollary 7.20], then yields the following continuous dependence of $\chi _v$ on $v\in {\bar{S}}$:

Corollary 4.2

Suppose (4.1) and assume further that there is $\kappa >0$ such that

$$\begin{aligned} \Vert v\Vert _{H^2(D)}\le \kappa \;\text { and }\; \Vert v_n\Vert _{H^2(D)}\le \kappa \,, \qquad n\ge 1\, . \end{aligned}$$

(4.12)

Then

$$\begin{aligned} \lim _{n\rightarrow \infty } {\mathcal {G}}(v_n)[\chi _{v_n}] = {\mathcal {G}}(v)[\chi _{v}] \end{aligned}$$

(4.13)

and, for $r \in [1, \infty$),

$$\begin{aligned} \lim _{n\rightarrow \infty } \Vert \chi _{v_n}-\chi _{v}\Vert _{{H^1}(\Omega (M))}= & {} \lim _{n\rightarrow \infty } \Vert \chi _{v_n}(\cdot ,-H)-\chi _{v}(\cdot ,-H)\Vert _{L_r(D)}= 0\,. \end{aligned}$$

(4.14)

Proof

It readily follows from (4.1), (4.12), and Theorem 3.2 that

$$\begin{aligned} (\chi _{v_n})_{n\ge 1} \;\text { is bounded in }\; H^1(\Omega (M)) \end{aligned}$$

(4.15)

and thus relatively compact in $L_2(\Omega (M))$ by [12, Theorem 1.4.5.2]. According to Proposition 4.1, we deduce from the Fundamental Theorem of $\Gamma$-convergence, see [9, Corollary 7.20], that any cluster point of $(\chi _{v_n})_{n\ge 1}$ in $L_2(\Omega (M))$ is a minimizer of ${\mathcal {G}}(v)$ and thus coincides with $\chi _v$ by Proposition 3.3. Therefore,

$$\begin{aligned} \lim _{n\rightarrow \infty } \Vert \chi _{v_n} - \chi _v \Vert _{L_2(\Omega (M))} = 0\,, \end{aligned}$$

(4.16)

and, using once more [9, Corollary 7.20], we obtain (4.13).

We are left with proving (4.14). To this end, we first observe that, since $\Omega (M)$ is a Lipschitz domain, [12, Theorem 1.4.3.2, Theorem 1.4.5.2] imply that $H^1(\Omega (M))$ compactly embeds in $W_q^{3/2q}(\Omega (M))$ for $q\ge 2$. Thus, the continuity of the trace operator from $W_q^{3/2q}(\Omega (M))$ to $L_q(\partial \Omega (M))$ (see [12, Theorem 1.5.1.2] and [26]), along with (4.15) and (4.16), ensure that there is a subsequence of $(\chi _{v_n})_{n\ge 1}$ (not relabeled) such that

$$\begin{aligned} \chi _{v_n}&\rightharpoonup \chi _v \quad \text {in }\ H_B^1(\Omega (M))\,, \end{aligned}$$

(4.17)

$$\begin{aligned} \chi _{v_n}(\cdot ,-H)&\rightarrow \chi _v(\cdot ,-H) \quad \text {in }\ L_q(D)\,, \quad q\ge 2\,. \end{aligned}$$

(4.18)

Notice that (4.18) yields the second assertion of (4.14). It now follows from (3.3), (3.4), (3.5), (4.13), and (4.18) that

$$\begin{aligned} \lim _{n\rightarrow \infty } \Vert \nabla (\chi _{v_n} + h_{v_n})\Vert _{L_2(\Omega (M))}^2&= \lim _{n\rightarrow \infty } \Vert \nabla (\chi _{v_n} + h_{v_n})\Vert _{L_2(\Omega (v_n))}^2 \\&\quad + \lim _{n\rightarrow \infty } \Vert \nabla h_{v_n}\Vert _{L_2(\Omega (M)\setminus \Omega (v_n))}^2 \\&= \Vert \nabla (\chi _{v} + h_{v})\Vert _{L_2(\Omega (v))}^2 + \Vert \nabla h_{v}\Vert _{L_2(\Omega (M)\setminus \Omega (v))}^2 \\&= \Vert \nabla (\chi _{v} + h_{v})\Vert _{L_2(\Omega (M))}^2\, . \end{aligned}$$

This property, along with (3.3) and (4.17), guarantees that $(\nabla \chi _{v_n})_{n\ge 1}$ converges to $\nabla \chi _v$ in $L_2(\Omega (M))$ and the proof of (4.14) is complete. $\square$

4.2 Continuity of $\partial _z\chi _v(\cdot , v)$ with respect to v

Finally, in order to establish the continuity of the function g defined in (2.6) we need also to investigate the continuous dependence of the gradient trace $\partial _z\chi _v(\cdot , v)$ on $v\in {\bar{S}}$, the main difficulty arising when ${\mathcal {C}}(v)\not =\emptyset$. In this regard we note:

Proposition 4.3

Consider $v \in \bar{S}$ and a sequence $(v_n)_{n\ge 1}$ in $\bar{S}$ such that

$$\begin{aligned} \Vert v\Vert _{H^2(D)} + \sup _{n\ge 1} \Vert v_n\Vert _{H^2(D)} \le \kappa \;\text { and }\; \lim _{n\rightarrow \infty } \Vert v_n - v\Vert _{H^1(D)} = 0\,. \end{aligned}$$

(4.19)

Then

$$\begin{aligned} \ell (v_n)\rightarrow \ell (v) \quad \text { in }\ L_r(D) \;\ \text { for }\ \; r\in [1,\infty )\,, \end{aligned}$$

(4.20)

where $\ell (v)$ is given by

$$\begin{aligned} \ell (v)(x):=\left\{ \begin{array}{ll} \partial _z\chi _v(x,v(x))\,, &{} x\in D\setminus {\mathcal {C}}(v)\,,\\ [0.1cm] 0\,, &{}x\in {\mathcal {C}}(v)\,. \end{array} \right. \end{aligned}$$

Proof

Thanks to (4.19) and the continuous embedding of $H^2(D)$ in $L_\infty (D)$, we may fix $M>H$ (only depending on $\kappa$) such that

$$\begin{aligned} - H \le v_n(x) , v(x) \le M-H\ , \qquad x\in \bar{D}\ , \quad n\ge 1\, . \end{aligned}$$

(4.21)

Step 1. We first establish an estimate ensuring that there is no concentration of $\partial _z\chi _v(\cdot ,v)$ on small subsets of $D\setminus {\mathcal {C}}(v)$. Indeed, since $\chi _v\in H^2(\Omega (v))$ we have $\chi _v(x,\cdot )\in H^2((-H,v(x)))$ for a.a. $x\in D\setminus {\mathcal {C}}(v)$, so that it follows from the boundary conditions (3.18b) and (3.18c) that

$$\begin{aligned} \partial _z \chi _v(x,v(x))&= \partial _z \chi _v(x,-H) + \int _{-H}^{v(x)} \partial _z^2\chi _v(x,z)\, \mathrm {d}z \\&= \sigma (x) \, \chi _v(x,-H) + \int _{-H}^{v(x)} \partial _z^2\chi _v(x,z)\, \mathrm {d}z \\&=\sigma (x) \left( \chi _v(x,v(x)) - \int _{-H}^{v(x)} \partial _z\chi _v(x,z)\mathrm {d}z\right) + \int _{-H}^{v(x)} \partial _z^2\chi _v(x,z)\, \mathrm {d}z \\&= \int _{-H}^{v(x)} \left( \partial _z^2\chi _v(x,z) - \sigma (x) \partial _z\chi _v(x,z)\right) \mathrm {d}z \end{aligned}$$

for a.a. $x\in D\setminus {\mathcal {C}}(v)$. Thus, for an arbitrary measurable subset $E \subset D\setminus {\mathcal {C}}(v)$, we infer from Hölder’s inequality that

$$\begin{aligned}&\int _E \vert \partial _z \chi _v(x,v(x))\vert \, \mathrm {d}x \nonumber \\&\quad \le \int _E \int _{-H}^{v(x)} \left( \vert \partial _z^2\chi _v(x,z) \vert + \sigma (x)\vert \partial _z\chi _v(x,z)\vert \right) \mathrm {d}z \mathrm {d}x \nonumber \\&\quad \le \left( \int _E (H+v)(x) \, \mathrm {d}x \right) ^{1/2} \bigg (\int _{\Omega (v)} \left( 2 \vert \partial _z^2\chi _v(x,z) \vert ^2 + 2 \Vert \sigma \Vert _{\infty }^2 \vert \partial _z\chi _v(x,z) \vert ^2 \right) \mathrm {d}(x,z) \bigg )^{1/2} \nonumber \\&\quad \le C \left( \int _E (H+v)(x) \, \mathrm {d}x \right) ^{1/2} \Vert \chi _v \Vert _{H^2(\Omega (v))}\,. \end{aligned}$$

(4.22a)

Clearly, the same proof implies that, for any $n\ge 1$ and arbitrary measurable subset $E \subset D\setminus {\mathcal {C}}(v_n)$,

$$\begin{aligned} \int _E \vert \partial _z \chi _{v_n}(x,v_n(x))\vert \, \mathrm {d}x \le C \left( \int _E (H+v_n)(x) \, \mathrm {d}x \right) ^{1/2} \Vert \chi _{v_n} \Vert _{H^2(\Omega (v_n))}\,. \end{aligned}$$

(4.22b)

Step 2. We next handle the behavior of $\partial _z \chi _v(\cdot ,v)$ where v stays away from $-H$. To this end, let $\varepsilon \in (0,H/2)$ and define

$$\begin{aligned} \Lambda (\varepsilon ) := \{ x \in D\ :\ v(x)>-H+2\varepsilon \}\,, \end{aligned}$$

(4.23)

which is a non-empty open subset of D, since $v\in C(\bar{D})$ with $v(\pm L)=0$. We can thus write it as a countable union of disjoint open intervals $(\Lambda _j(\varepsilon ))_{j\in J}$, see [1, IX.Proposition 1.8]. Also, owing to (4.19) and the continuous embedding of $H^1(D)$ in $C(\bar{D})$, there is $n_\varepsilon \ge 1$ such that

$$\begin{aligned} v(x)-\varepsilon \le v_n(x) \le v(x)+\varepsilon \ , \qquad x\in \bar{D}\ , \quad n\ge n_\varepsilon \, . \end{aligned}$$

(4.24)

A straightforward consequence of (4.23) and (4.24) is that

$$\begin{aligned} \{ (x,z) \in \Lambda (\varepsilon )\times [-H,\infty )\ :\ -H< z < v(x)-\varepsilon \} \subset \Omega (v_n)\ , \qquad n\ge n_\varepsilon \,. \end{aligned}$$

(4.25)

Therefore, the function $X_{n,\varepsilon }$, given by

$$\begin{aligned} X_{n,\varepsilon }(x) := \partial _z \chi _v(x,v(x)-\varepsilon ) - \partial _z \chi _{v_n}(x,v(x)-\varepsilon ), \qquad x \in \Lambda (\varepsilon )\, , \quad n\ge n_\varepsilon \, , \end{aligned}$$

is well-defined. Let $j\in J$ and $n\ge n_\varepsilon$. Since $\partial _z \chi _v$ and $\partial _z \chi _{v_n}$ belong to $H^1({\mathcal {O}}_{\Lambda _j(\varepsilon )}(v-\varepsilon ))$, the set ${\mathcal {O}}_{\Lambda _j(\varepsilon )}(v-\varepsilon )$ being defined in (3.14), it follows from (3.19), (4.21), and the definition of $\Lambda (\varepsilon )$ that

$$\begin{aligned}&\varepsilon \int _{\Lambda _j(\varepsilon )} |X_{n,\varepsilon }(x)|^2\ \mathrm {d}x \\&\quad \le \int _{\Lambda _j(\varepsilon )} |X_{n,\varepsilon }(x)|^2 (H+v(x)-\varepsilon )\ \mathrm {d}x \\&\quad \le \Vert \partial _z (\chi _v - \chi _{v_n})\Vert _{L_2({\mathcal {O}}_{\Lambda _j(\varepsilon )}(v-\varepsilon ))}^2 \\&\qquad + 2 \Vert H+v-\varepsilon \Vert _{L_\infty (\Lambda _j(\varepsilon ))} \Vert \partial _z (\chi _v - \chi _{v_n})\Vert _{L_2({\mathcal {O}}_{\Lambda _j(\varepsilon )}(v-\varepsilon ))} \Vert \partial _z^2 (\chi _v - \chi _{v_n})\Vert _{L_2({\mathcal {O}}_{\Lambda _j(\varepsilon )}(v-\varepsilon ))} \\&\quad \le \Vert \partial _z (\chi _v - \chi _{v_n})\Vert _{L_2({\mathcal {O}}_{\Lambda _j(\varepsilon )}(M))}^2 \\&\qquad + C(\kappa ) \Vert \partial _z (\chi _v - \chi _{v_n})\Vert _{L_2({\mathcal {O}}_{\Lambda _j(\varepsilon )}(M))} \left( \Vert \partial _z^2 \chi _v\Vert _{L_2({\mathcal {O}}_{\Lambda _j(\varepsilon )}(v))} + \Vert \partial _z^2 \chi _{v_n}\Vert _{L_2({\mathcal {O}}_{\Lambda _j(\varepsilon )}(v_n))} \right) \,. \end{aligned}$$

Summing the above inequality over $j\in J$ and noticing that

$$\begin{aligned}&\sum _{j\in J} \Vert \partial _z (\chi _v - \chi _{v_n})\Vert _{L_2({\mathcal {O}}_{\Lambda _j(\varepsilon )}(M))} \left( \Vert \partial _z^2 \chi _v\Vert _{L_2({\mathcal {O}}_{\Lambda _j(\varepsilon )}(v))} + \Vert \partial _z^2 \chi _{v_n}\Vert _{L_2({\mathcal {O}}_{\Lambda _j(\varepsilon )}(v_n))} \right) \\&\quad \le \left( \sum _{j\in J} \Vert \partial _z (\chi _v - \chi _{v_n})\Vert _{L_2({\mathcal {O}}_{\Lambda _j(\varepsilon )}(M))}^2 \right) ^{1/2} \\&\qquad \qquad \qquad \qquad \quad \times \left( \sum _{j\in J} \left( \Vert \partial _z^2 \chi _v\Vert _{L_2({\mathcal {O}}_{\Lambda _j(\varepsilon )}(v))} + \Vert \partial _z^2 \chi _{v_n}\Vert _{L_2({\mathcal {O}}_{\Lambda _j(\varepsilon )}(v_n))} \right) ^2 \right) ^{1/2} \\&\quad \le \sqrt{2} \Vert \partial _z (\chi _v - \chi _{v_n})\Vert _{L_2(\Omega (M))} \left( \sum _{j\in J} \left( \Vert \partial _z^2 \chi _v\Vert _{L_2({\mathcal {O}}_{\Lambda _j(\varepsilon )}(v))}^2 + \Vert \partial _z^2 \chi _{v_n}\Vert _{L_2({\mathcal {O}}_{\Lambda _j(\varepsilon )}(v_n))}^2 \right) \right) ^{1/2} \\&\quad \le \sqrt{2} \Vert \partial _z (\chi _v - \chi _{v_n})\Vert _{L_2(\Omega (M))} \left( \Vert \partial _z^2 \chi _v\Vert _{L_2(\Omega (v))} + \Vert \partial _z^2 \chi _{v_n}\Vert _{L_2(\Omega (v_n))} \right) \\&\quad \le C(\kappa ) \Vert \partial _z (\chi _v - \chi _{v_n})\Vert _{L_2(\Omega (M))} \end{aligned}$$

by Cauchy-Schwarz’ inequality, (4.19), and Theorem 3.2, we obtain

$$\begin{aligned} \varepsilon \int _{\Lambda (\varepsilon )} |X_{n,\varepsilon }(x)|^2\ \mathrm {d}x \le \Vert \partial _z (\chi _v - \chi _{v_n})\Vert _{L_2(\Omega (M))}^2 + C(\kappa ) \Vert \partial _z (\chi _v - \chi _{v_n})\Vert _{L_2(\Omega (M))} \,. \end{aligned}$$

We now infer from (4.14) and the above inequality that

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _{\Lambda (\varepsilon )} |X_{n,\varepsilon }(x)|^2\ \mathrm {d}x = 0\, . \end{aligned}$$

(4.26)

We next set

$$\begin{aligned} Y_n (x) := \partial _z \chi _v(x,v(x)) - \partial _z \chi _{v_n}(x,v_n(x)), \qquad x \in \Lambda (\varepsilon )\ , \quad n\ge n_\varepsilon \, . \end{aligned}$$

Using (4.24) and Hölder’s and Young’s inequalities, we obtain, for $j\in J$,

$$\begin{aligned} \Vert Y_n \Vert _{L_1(\Lambda _j(\varepsilon ))}&\le \Vert X_{n,\varepsilon }\Vert _{L_1(\Lambda _j(\varepsilon ))} + \int _{\Lambda _j(\varepsilon )} \left| \int _{v-\varepsilon }^v \partial _z^2 \chi _v(\cdot ,z)\,\mathrm {d}z - \int _{v-\varepsilon }^{v_n}\partial _z^2 \chi _{v_n}(\cdot ,z)\,\mathrm {d}z\right| \, \mathrm {d}x \\&\le \Vert X_{n,\varepsilon }\Vert _{L_1(\Lambda _j(\varepsilon ))} +\int _{\Lambda _j(\varepsilon )} \int _{v-\varepsilon }^v \vert \partial _z^2 \chi _v(\cdot ,z)\vert \,\mathrm {d}z \mathrm {d}x\\&\quad + \int _{\Lambda _j(\varepsilon )} \int _{v-\varepsilon }^{v_n} \vert \partial _z^2 \chi _{v_n}(\cdot ,z)\vert \,\mathrm {d}z \mathrm {d}x \\&\le \Vert X_{n,\varepsilon } \Vert _{L_1(\Lambda _j(\varepsilon ))} +\sqrt{\varepsilon \vert \Lambda _j(\varepsilon )\vert }\left( \int _{\Lambda _j(\varepsilon )} \int _{v-\varepsilon }^v \vert \partial _z^2 \chi _v(\cdot ,z)\vert ^2 \,\mathrm {d}z \mathrm {d}x \right) ^{1/2} \\&\quad +\sqrt{2\varepsilon \vert \Lambda _j(\varepsilon )\vert }\left( \int _{\Lambda _j(\varepsilon )} \int _{v-\varepsilon }^{v_n} \vert \partial _z^2 \chi _{v_n}(\cdot ,z)\vert ^2 \,\mathrm {d}z \mathrm {d}x \right) ^{1/2} \\&\le \Vert X_{n,\varepsilon } \Vert _{L_1(\Lambda _j(\varepsilon ))} + \frac{\sqrt{\varepsilon }}{2} \, \vert \Lambda _j(\varepsilon ) \vert + \frac{\sqrt{\varepsilon }}{2} \int _{\Lambda _j(\varepsilon )} \int _{-H}^v \vert \partial _z^2 \chi _v(\cdot ,z)\vert ^2 \,\mathrm {d}z \mathrm {d}x \\&\quad + \frac{\sqrt{\varepsilon }}{2} \, \vert \Lambda _j(\varepsilon ) \vert + \sqrt{\varepsilon } \int _{\Lambda _j(\varepsilon )} \int _{-H}^{v_n} \vert \partial _z^2 \chi _{v_n}(\cdot ,z)\vert ^2 \,\mathrm {d}z \mathrm {d}x\,. \end{aligned}$$

Summing over $j\in J$ and using (4.19) and Theorem 3.2 give

$$\begin{aligned} \Vert Y_n \Vert _{L_1(\Lambda (\varepsilon ))}&\le \Vert X_{n,\varepsilon } \Vert _{L_1(\Lambda (\varepsilon ))} + \sqrt{\varepsilon } \vert \Lambda (\varepsilon )\vert + \sqrt{\varepsilon } \Vert \chi _v \Vert _{H^2(\Omega (v))} + \sqrt{\varepsilon } \Vert \chi _{v_n} \Vert _{H^2(\Omega (v_n))} \\&\le \Vert X_{n,\varepsilon } \Vert _{L_1(\Lambda (\varepsilon ))} + C(\kappa ) \sqrt{\varepsilon } \,. \end{aligned}$$

Owing to (4.26), we may take the limit $n\rightarrow \infty$ in the previous inequality and obtain

$$\begin{aligned} \limsup _{n\rightarrow \infty } \Vert Y_n \Vert _{L_1(\Lambda (\varepsilon ))} \le C(\kappa ) \sqrt{\varepsilon } . \end{aligned}$$

Since $\Lambda (\varepsilon )\subset \Lambda (\delta )$ for all $\delta \in (0,\varepsilon )$, we infer from the above inequality that

$$\begin{aligned} \limsup _{n\rightarrow \infty } \Vert Y_n \Vert _{L_1(\Lambda (\varepsilon ))} \le \limsup _{n\rightarrow \infty } \Vert Y_n \Vert _{L_1(\Lambda (\delta ))} \le C(\kappa ) \sqrt{\delta } \end{aligned}$$

and we may pass to the limit $\delta \rightarrow 0$ to conclude that

$$\begin{aligned} \lim _{n\rightarrow \infty } \Vert Y_n \Vert _{L_1(\Lambda (\varepsilon ))} =0, \qquad \varepsilon \in (0,H/2). \end{aligned}$$

(4.27)

Step 3. Finally, we infer from (4.19), (4.21), (4.22), and Theorem 3.2 that

$$\begin{aligned}&\Vert \ell (v_n)- \ell (v)\Vert _{L_1(D)} \\&\quad \;\le \int _{\Lambda (\varepsilon )} \vert \ell (v_n)- \ell (v)\vert \, \mathrm {d}x + \int _{D\setminus \Lambda (\varepsilon )} \vert \ell (v_n)\vert \, \mathrm {d}x +\int _{D\setminus \Lambda (\varepsilon )} \vert \ell (v)\vert \, \mathrm {d}x \\&\quad \; = \Vert Y_n \Vert _{L_1(\Lambda (\varepsilon ))} + \int _{(D\setminus \Lambda (\varepsilon ))\setminus {\mathcal {C}}(v_n)} \vert \partial _z \chi _{v_n}(\cdot , v_n)\vert \,\mathrm {d}x + \int _{(D\setminus \Lambda (\varepsilon ))\setminus {\mathcal {C}}(v)} \vert \partial _z \chi _{v}(\cdot , v)\vert \,\mathrm {d}x \\&\quad \;\le \Vert Y_n \Vert _{L_1(\Lambda (\varepsilon ))} + C \left( \int _{(D\setminus \Lambda (\varepsilon ))\setminus {\mathcal {C}}(v_n)} (H + v_n)(x) \, \mathrm {d}x \right) ^{1/2} \Vert \chi _{v_n} \Vert _{H^2(\Omega (v_n))} \\&\qquad + C \left( \int _{(D\setminus \Lambda (\varepsilon ))\setminus {\mathcal {C}}(v)} (H+v)(x) \, \mathrm {d}x \right) ^{1/2} \Vert \chi _{v} \Vert _{H^2(\Omega (v))} \\&\quad \;\le \Vert Y_n \Vert _{L_1(\Lambda (\varepsilon ))} + C(\kappa ) \left( \int _{D\setminus \Lambda (\varepsilon )} (H + v)(x) \, \mathrm {d}x \right) ^{1/2} \\&\qquad + C(\kappa ) \left( \int _{D\setminus \Lambda (\varepsilon )} (H + v_n)(x) \, \mathrm {d}x \right) ^{1/2} \,. \end{aligned}$$

Since $0\le H+v\le 2\varepsilon$ and $0\le H+v_n\le 3\varepsilon$ in $D\setminus \Lambda (\varepsilon )$ for $n\ge n_\varepsilon$ by (4.23) and (4.24), we further obtain

$$\begin{aligned} \Vert \ell (v_n)-\ell (v)\Vert _{L_1(D)} \le \Vert Y_n \Vert _{L_1(\Lambda (\varepsilon ))} + C(\kappa )\sqrt{\varepsilon }\,. \end{aligned}$$

We now first let $n\rightarrow \infty$ with the help of (4.27) and then take the limit $\varepsilon \rightarrow 0$ to conclude that

$$\begin{aligned} \lim _{n\rightarrow \infty } \Vert \ell (v_n)-\ell (v)\Vert _{L_1(D)} = 0\,. \end{aligned}$$

(4.28)

Finally, given $r\in [1,\infty )$, we infer from Hölder’s inequality, Lemma 3.1, (3.10), and (4.19) that

$$\begin{aligned} \Vert \ell (v_n)-\ell (v)\Vert _{L_r(D)}&\le \Vert \ell (v_n)-\ell (v)\Vert _{L_1(D)}^{1/(2r-1)} \Vert \ell (v_n)-\ell (v)\Vert _{L_{2r}(D)}^{2(r-1)/(2r-1)} \\&\le \Vert \ell (v_n)-\ell (v)\Vert _{L_1(D)}^{1/(2r-1)} \left( \Vert \ell (v_n)\Vert _{L_{2r}(D)}^{2(r-1)/(2r-1)} + \Vert \ell (v)\Vert _{L_{2r}(D)}^{2(r-1)/(2r-1)} \right) \\&\le C(\kappa , r) \Vert \ell (v_n)-\ell (v)\Vert _{L_1(D)}^{1/(2r-1)} \end{aligned}$$

and the assertion follows from (4.28). $\square$

Summarizing the outcome of this section, we have obtained continuity properties of the electrostatic energy $E_e$ and the function g introduced in (2.6).

Theorem 4.4

The electrostatic energy $E_e: \bar{S} \rightarrow {\mathbb {R}}$ is continuous for the weak topology of $H^2(D)$. The function $g: \bar{S} \rightarrow L_r(D)$ is continuous for each $r \in [1,\infty )$, the set $\bar{S}$ being still endowed with the weak topology of $H^2(D)$.

Proof

Let us first recall that, if $(v_n)_{n\ge 1}$ is a sequence in $\bar{S}$ converging weakly in $H^2(D)$ to $v\in \bar{S}$, then there is $\kappa >0$ such that (4.12) and (4.19) hold true. Consequently, we infer from Corollary 4.2 that

$$\begin{aligned} \lim _{n\rightarrow \infty } E_e(v_n) = - \lim _{n\rightarrow \infty } {\mathcal {G}}(v_n)[\chi _{v_n}] = - {\mathcal {G}}(v)[\chi _v] = E_e(v)\,, \end{aligned}$$

thereby establishing the stated continuity of $E_e$. Next, let $v\in \bar{S}$. Since $\partial _x v=0$ a.e. in ${\mathcal {C}}(v)$, it follows from (2.6) and Proposition 4.3 that

$$\begin{aligned} g(v)(x)&= \frac{1}{2} (1+|\partial _x v(x)|^2)\,\big [\ell (v)(x) - (\partial _w h)_{v}(x,v(x))\big ]^2\\&\quad + \sigma (x)\big [\chi _{v}(x,-H)+ h_v(x,-H)-{\mathfrak {h}}_{v}(x)\big ](\partial _w {\mathfrak {h}})_{v}(x)\\&\quad -\frac{1}{2} \left[ \big \vert (\partial _x h)_v\big \vert ^2+ \big ((\partial _z h)_v+(\partial _w h)_v\big )^2 \right] (x, v(x)) \end{aligned}$$

for $x\in D$. The stated continuity of g then readily follows from Proposition 4.3 and the $C^1$-regularity of h and ${\mathfrak {h}}$ (see also Lemma 3.1(b)). $\square$

5 Shape derivative of the electrostatic energy

In this section we investigate differentiability properties of the electrostatic energy

$$\begin{aligned} E_e(u)= & {} -\dfrac{1}{2}\displaystyle \int _{\Omega (u)} \big \vert \nabla \psi _u\big \vert ^2\,\mathrm {d}(x,z)\\&-\dfrac{1}{2}\displaystyle \int _{ D} \sigma (x) \big \vert \psi _u(x,-H)- {\mathfrak {h}}_{u}(x)\big \vert ^2\,\mathrm {d}x \end{aligned}$$

with respect to $u\in {\bar{S}}$, where $\psi _u$ is the strong solution to (1.3), see Theorem 2.2. Owing to the dependence of $\psi _u$ on the domain $\Omega (u)$ this resembles the computation of a shape derivative, a topic which has received considerable attention in recent years, see [8, 14, 32] and the references therein. Note that we may write alternatively $E_e(u)=-{\mathcal {G}}(u)[\psi _u-h_u]$, since $\chi _u= \psi _u-h_u$ is the strong solution to (3.7) (with $v=u$) given by Theorem 3.2.

As might be expected, the switch between boundary conditions for $\psi _u$ when ${\mathcal {C}}(u)\ne \emptyset$ generates additional difficulties and we begin with the differentiability of $\psi _u$ with respect to $u\in S$.

Lemma 5.1

Let $u\in S$ be fixed and define, for $v\in S$, the transformation $\Theta _v:\Omega (u)\rightarrow \Omega (v)$ by

$$\begin{aligned} \Theta _{v}(x,z):=\left( x,z+\frac{v(x)-u(x)}{H+u(x)}(z+H)\right) \,,\quad (x,z)\in \Omega (u) \,. \end{aligned}$$

Then there exists a neighborhood U of u in S such that the mapping

$$\begin{aligned} U\rightarrow H_B^1(\Omega (u)),\quad v\mapsto \chi _v\circ \Theta _v \end{aligned}$$

is continuously differentiable, where $\chi _v=\psi _v-h_v\in H_B^1(\Omega (v))$ solves (3.7), see Theorem 3.2, and S is endowed with the $H^2(D)$-topology.

Proof

The proof follows the lines of [14, Theorem 5.3.2], a similar proof is given in [21, Lemma 4.1]. We thus only provide a very brief sketch here. Let $u\in S$ and $v\in S$. Setting $\xi _v:=\chi _v\circ \Theta _v$ and performing a change of variables $({\bar{x}},{\bar{z}})=\Theta _v(x,z)$, the weak formulation (3.12) satisfied by $\chi _v$ (as a critical point of ${\mathcal {G}}(v)$) can be written in the form

$$\begin{aligned} \begin{aligned}&\int _{\Omega (u)} J_v\, \big ((D\Theta _v)^{-1} (D\Theta _v^T)^{-1}\nabla \xi _v \big )\cdot \nabla \phi \,\mathrm {d}(x,z) +\int _D \sigma \big (\xi _v \phi \big )(\cdot ,-H)\,\mathrm {d}x\\&\quad = -\int _{\Omega (u)} J_v\, \big ((D\Theta _v)^{-1} (\nabla h_v\circ \Theta _v) \big )\cdot \nabla \phi \,\mathrm {d}(x,z) \\&\qquad +\int _D \sigma \big [{\mathfrak {h}}_v-h_v(\cdot ,-H)\big ]\phi (\cdot ,-H)\, \mathrm {d}x \end{aligned} \end{aligned}$$

(5.1)

for $\phi \in H_B^1(\Omega (u))$, where $J_v:= |\mathrm {det}(D\Theta _v)|$. Therefore, (5.1) is equivalent to

$$\begin{aligned} F(v,\xi _v)=0\,,\quad v\in S\,, \end{aligned}$$

(5.2)

for some Fréchet differentiable function

$$\begin{aligned} F: S\times H_B^1(\Omega (u))\rightarrow (H_B^1(\Omega (u)))'\,. \end{aligned}$$

One then uses the Implicit Function Theorem to derive that $\xi _v$ depends smoothly on v. $\square$

As a next step we establish the Fréchet differentiability of $E_e$ on the open set S. For $u\in S$ recall that g(u) is given by (2.6a) since ${\mathcal {C}}(u)=\emptyset$ in this case.

Proposition 5.2

Let S be endowed with the $H^2(D)$-topology. Then the electrostatic energy $E_e: S \rightarrow {\mathbb {R}}$ is continuously Fréchet differentiable with

$$\begin{aligned} \partial _u E_e(u) [\vartheta ] =\int _D g(u)\vartheta \,\mathrm {d}x \end{aligned}$$

for $u\in S$ and $\vartheta \in H^2(D)\cap H^1_0(D)$.

Proof

In this proof we shall use the notation from Lemma 5.1. We fix $u\in S$ and recall from Lemma 5.1 that the mapping $v \mapsto \xi _v = \chi _v \circ \Theta _v$ is continuously differentiable with respect to v in a neighborhood U of u in S and takes values in $H^1_B(\Omega (u))$. With $\psi _v = \chi _v + h_v$, $J_v=\vert \mathrm {det}(D\Theta _v)\vert$, and the change of variables $({\bar{x}},{\bar{z}})=\Theta _v(x,z)$, we obtain that, for $v \in U$,

$$\begin{aligned} E_e(v)&= -\frac{1}{2} \int _{\Omega (v)} \vert \nabla \psi _v\vert ^2 \, \mathrm {d}({\bar{x}},{\bar{z}}) - \frac{1}{2} \int _D \sigma \big \vert \psi _v({\bar{x}}, -H) - {\mathfrak {h}}_v(\bar{x})\big \vert ^2 \, \mathrm {d}\bar{x} \\&= -\frac{1}{2} \int _{\Omega (u)} \left| (D\Theta _v^T)^{-1}\nabla \xi _v + \nabla h_v \circ \Theta _v \right| ^2 \, J_v \, \mathrm {d}(x,z) \\&\quad - \frac{1}{2} \int _D \sigma \vert (\xi _v + h_v)(x,-H) - {\mathfrak {h}}_v(x)\vert ^2 \, \mathrm {d}x\,. \end{aligned}$$

We introduce the functions

$$\begin{aligned} j(v)&:= (D\Theta _v^T)^{-1} \nabla \xi _v + \nabla h_v \circ \Theta _v \quad \text { in } \Omega (u), \\ m(v)&:= \big ( \xi _v + h_v\big )(\cdot ,-H) - {\mathfrak {h}}_v \quad \text { in } D\, . \end{aligned}$$

Then, recalling that h and ${\mathfrak {h}}$ are $C^1$-functions in all their arguments by (2.1b), we conclude that the Fréchet derivative of $E_e$ at u applied to $\vartheta \in H^2(D)\cap H^1_0(D)$ is given by

$$\begin{aligned} \partial _u E_e(u) [\vartheta ]&= \partial _v E_e(v) [\vartheta ]\vert _{v=u} = - \int _{\Omega (u)} j(u) \cdot (\partial _v j(v)[\vartheta ]\vert _{v=u}) \, J_u \, \mathrm {d}(x,z) \\&\quad - \frac{1}{2}\int _{\Omega (u)} \vert j(u)\vert ^2 \, (\partial _v J_v[\vartheta ]\vert _{v=u}) \, \mathrm {d}(x,z) - \int _D \sigma \, m(u)\, (\partial _v m(v)[\vartheta ]\vert _{v=u}) \, \mathrm {d}x\,. \end{aligned}$$

On the one hand, we argue as in the proof of [21, Equation (4.12)] to show that

$$\begin{aligned} \begin{aligned}&- \int _{\Omega (u)} j(u) \cdot (\partial _v j(v)[\vartheta ]\vert _{v=u}) \, J_u \, \mathrm {d}(x,z) - \frac{1}{2}\int _{\Omega (u)} \vert j(u)\vert ^2 \, (\partial _v J_v[\vartheta ]\vert _{v=u}) \, \mathrm {d}(x,z) \\&\quad = - \int _{\Omega (u)} \nabla \psi _u \cdot \nabla \big ( \partial _v \xi _v [\vartheta ]\vert _{v=u} + (\partial _w h)_u \vartheta \big ) \,\mathrm {d}(x,z) \\&\qquad + \int _{\Omega (u)} \nabla \psi _u \cdot \left[ \partial _z \chi _u \, \nabla \left( \frac{(z+H)\vartheta }{H+u}\right) - \frac{(z+H)\vartheta }{H+u} \nabla \big ((\partial _z h)_u \big ) \right] \mathrm {d}(x,z) \\&\qquad - \frac{1}{2} \int _{\Omega (u)} \vert \nabla \psi _u \vert ^2 \frac{\vartheta }{H+u} \, \mathrm {d}(x,z)\,. \end{aligned} \end{aligned}$$

On the other hand, since $m(u) = \psi _u(\cdot ,-H) - {\mathfrak {h}}_u$ in D and

$$\begin{aligned} \partial _v m(v)[\vartheta ]\vert _{v=u} = (\partial _v \xi _v[\vartheta ]\vert _{v=u})(\cdot ,-H) + (\partial _w h)_u(\cdot ,-H) \, \vartheta - (\partial _w {\mathfrak {h}})_u \,\vartheta \quad \text { in } D\,, \end{aligned}$$

we see that

$$\begin{aligned} \begin{aligned}&- \int _D \sigma \, m(u)\, (\partial _v m(v)[\vartheta ]\vert _{v=u}) \, \mathrm {d}x \\&\quad = - \int _D \sigma \big [ \psi _u(\cdot ,-H) - {\mathfrak {h}}_u\big ] \big [ (\partial _v \xi _v[\vartheta ]\vert _{v=u})(\cdot ,-H) + (\partial _w h)_u(\cdot ,-H) \, \vartheta - (\partial _w {\mathfrak {h}})_u \,\vartheta \big ] \, \mathrm {d}x\,. \end{aligned} \end{aligned}$$

The above two identities yield

$$\begin{aligned} \begin{aligned} \partial _u E_e(u) [\vartheta ]&=- \int _{\Omega (u)} \nabla \psi _u \cdot \nabla \big ( \partial _v \xi _v [\vartheta ]\vert _{v=u} + (\partial _w h)_u \vartheta \big ) \,\mathrm {d}(x,z) \\&\quad + \int _{\Omega (u)} \nabla \psi _u \cdot \left[ \partial _z \chi _u \, \nabla \left( \frac{(z+H)\vartheta }{H+u}\right) - \frac{(z+H)\vartheta }{H+u} \nabla \big ((\partial _z h)_u \big ) \right] \mathrm {d}(x,z) \\&\quad - \frac{1}{2} \int _{\Omega (u)} \vert \nabla \psi _u \vert ^2 \frac{\vartheta }{H+u} \, \mathrm {d}(x,z) \\&\quad - \int _D \sigma \big [ \psi _u(\cdot ,-H) - {\mathfrak {h}}_u\big ] \big [ (\partial _v \xi _v[\vartheta ]\vert _{v=u})(\cdot ,-H) \\&\quad + (\partial _w h)_u(\cdot ,-H) \, \vartheta - (\partial _w {\mathfrak {h}})_u \,\vartheta \big ] \, \mathrm {d}x\,. \end{aligned} \end{aligned}$$

(5.3)

Next we shall simplify the right-hand side of (5.3). Using Gauß’ Theorem, the fact that $\psi _u$ is a strong solution to (1.3a), $\vartheta =0$ on $\partial D$, and the fact that $\partial _v \xi _v [\vartheta ]\vert _{v=u}$ belongs to $H^1_B(\Omega (u))$, the first integral on the right-hand side of (5.3) can be rewritten in the form

$$\begin{aligned}&- \int _{\Omega (u)} \nabla \psi _u \cdot \nabla \big ( \partial _v \xi _v [\vartheta ]\vert _{v=u} + (\partial _w h)_u \vartheta \big ) \,\mathrm {d}(x,z) \\&\quad = - \int _D (\partial _w h)_u (x,u(x)) \, \vartheta (x) \big [ \partial _z \psi _u - \partial _x u \,\partial _x \psi _u \big ](x,u(x))\, \mathrm {d}x \\&\qquad + \int _D \big [ (\partial _v \xi _v[\vartheta ]\vert _{v=u})(x,-H) + (\partial _w h)_u(x,-H) \, \vartheta (x) \big ] \, \partial _z \psi _u(x,-H) \, \mathrm {d}x\,. \end{aligned}$$

Since, due to (1.3c),

$$\begin{aligned} \partial _z \psi _u(x,-H) = \sigma (x) \big [\psi _u(x,-H) - {\mathfrak {h}}_u(x)\big ]\,, \quad x \in D, \end{aligned}$$

it follows that

$$\begin{aligned}&- \int _{\Omega (u)} \nabla \psi _u \cdot \nabla \big ( \partial _v \xi _v [\vartheta ]\vert _{v=u} + (\partial _w h)_u \vartheta \big ) \,\mathrm {d}(x,z) \nonumber \\&\quad = - \int _D \vartheta (x) \Big [ (\partial _w h)_u \big ( \partial _z \psi _u - \partial _x u \,\partial _x \psi _u \big ) \Big ](x,u(x))\, \mathrm {d}x\\&\qquad + \int _D \sigma (x) \big [\psi _u(x,-H) - {\mathfrak {h}}_u(x)\big ] \big [ (\partial _v \xi _v[\vartheta ]\vert _{v=u}) (x,-H) + (\partial _w h)_u(x,-H) \, \vartheta (x) \big ] \, \mathrm {d}x\,.\nonumber \end{aligned}$$

(5.4)

We next proceed as in [21, p. 486] to simplify the second integral on the right-hand side of (5.3) and show that it can be written

$$\begin{aligned}&\int _{\Omega (u)} \nabla \psi _u \cdot \left[ \partial _z \chi _u \, \nabla \left( \frac{(z+H)\vartheta }{H+u}\right) - \frac{(z+H)\vartheta }{H+u} \nabla \big ((\partial _z h)_u \big ) \right] \mathrm {d}(x,z) \\&\quad = \frac{1}{2} \int _{\Omega (u)} \frac{\vartheta }{H+u}\, \vert \nabla \psi _u \vert ^2 \, \mathrm {d}(x,z) - \frac{1}{2} \int _D \vartheta (x) \, \vert \nabla \psi _u(x,u(x)) \vert ^2 \, \mathrm {d}x \\&\qquad + \int _D \vartheta (x) \, \Big [ \big ( \partial _z \psi _u -(\partial _z h)_u\big ) \big ( \partial _z \psi _u - \partial _x u \,\partial _x \psi _u \big ) \Big ](x,u(x))\, \mathrm {d}x\, . \end{aligned}$$

Combining this identity with (5.3) and (5.4) yields

$$\begin{aligned} \begin{aligned} \partial _u E_e(u) [\vartheta ]&= \int _D \vartheta (x) \,\Big [ \big ( \partial _z \psi _u -(\partial _z h)_u - (\partial _wh)_u\big ) \big ( \partial _z \psi _u - \partial _x u \,\partial _x \psi _u \big ) \Big ](x,u(x))\, \mathrm {d}x \\&\quad - \frac{1}{2}\int _D \vartheta (x) \, \vert \nabla \psi _u(x,u(x)) \vert ^2 \, \mathrm {d}x \\&\quad + \int _D \sigma (x) \big [\psi _u(x,-H) - {\mathfrak {h}}_u(x)\big ] \, (\partial _w {\mathfrak {h}})_u(x) \,\vartheta (x) \, \mathrm {d}x\, . \end{aligned} \end{aligned}$$

(5.5)

Since (1.3b) entails $\psi _u(x,u(x))= h(x,u(x),u(x))$, $x\in D$, we have

$$\begin{aligned} \partial _x \psi _u(x,u(x)) = (\partial _x h)_u(x,u(x)) - \partial _xu(x) \big [ \partial _z \psi _u - (\partial _z h)_u-(\partial _w h)_u \big ](x,u(x)), \quad x \in D, \end{aligned}$$

and hence, for $x\in D$,

$$\begin{aligned} \begin{aligned}&\frac{1}{2}\big \vert \nabla \psi _{u}(x,u(x))\big \vert ^2-\Big [ \big ( \partial _z\psi _{u}-(\partial _z h)_u-(\partial _w h)_u\big ) \big ( \partial _z\psi _{u} - \partial _x u\, \partial _x\psi _{u}\big ) \Big ](x, u(x))\\&\quad = -\frac{1}{2}(1+\vert \partial _x u(x)\vert ^2)\,\big [\partial _z\psi _{u}-(\partial _z h)_u-(\partial _w h)_u\big ]^2(x, u(x))\\&\qquad +\frac{1}{2}\big [\big \vert (\partial _x h)_u\big \vert ^2+ \big ((\partial _z h)_u+(\partial _w h)_u\big )^2\big ](x, u(x))\,. \end{aligned} \end{aligned}$$

Inserting this identity into (5.5) gives

$$\begin{aligned} \partial _u E_e(u) [\vartheta ]&= \frac{1}{2} \int _D (1+ \vert \partial _x u(x)\vert ^2) \big [ \partial _z \psi _u -(\partial _z h)_u - (\partial _w h)_u \big ]^2 (x,u(x)) \, \vartheta (x) \, \mathrm {d}x \\&\qquad - \frac{1}{2} \int _D \big [\vert (\partial _x h)_u\vert ^2 + \left( (\partial _z h)_u + (\partial _w h)_u\right) ^2\big ](x,u(x)) \, \vartheta (x) \, \mathrm {d}x \\&\qquad + \int _D \sigma (x) \big [\psi _u(x,-H) - {\mathfrak {h}}_u(x)\big ] \ (\partial _w {\mathfrak {h}})_u(x) \, \vartheta (x) \, \mathrm {d}x\\&=\int _D g(u)(x)\,\vartheta (x) \, \mathrm {d}x\,, \end{aligned}$$

according to (2.6a). Finally, the continuity of

$$\begin{aligned} \partial _u E_e : S \rightarrow {\mathcal {L}}\big (H^2(D)\cap H_0^1(D),{\mathbb {R}}\big ) \end{aligned}$$

readily follows from Theorem 4.4. $\square$

We finally provide the differentiability property of $E_e$ on the closed set ${\bar{S}}$. More precisely, we show that $E_e$ admits a directional derivative at a point $u\in {\bar{S}}$ in any direction of $-u+S$, which is given by g(u) defined in (2.6). Recall that ${\mathcal {C}}(u)$ may be non-empty in this case.

Proposition 5.3

Let $u\in {\bar{S}}$ and $w\in S$. Then

$$\begin{aligned} \lim _{s\rightarrow 0^+} \frac{1}{s}\Big [ E_e(u+s(w-u))-E_e(u) \Big ] =\int _D g(u) (w-u)\, \mathrm {d}x\,. \end{aligned}$$

Proposition 5.3 is a rather immediate consequence of Theorem 4.4, Proposition 5.2, and the observation that $u+s(w-u)=(1-s)u+sw \in S$ for all $u\in \bar{S}$, $w\in S$, and $s\in (0,1]$. We refer to [21, Corollary 4.3] for a detailed proof.

6 Proofs of Theorem 2.3 and Theorem 2.4 for $\alpha =0$

In this section we deal with the case $\alpha =0$ and recall that the total energy is then given by

$$\begin{aligned} E(u)= E_m(u)+E_e(u) \end{aligned}$$

with mechanical energy

$$\begin{aligned} E_m(u)=\frac{\beta }{2}\Vert \partial _x^2u\Vert _{L_2(D)}^2 + \frac{\tau }{2} \Vert \partial _x u\Vert _{L_2(D)}^2 \end{aligned}$$

and electrostatic energy

$$\begin{aligned} E_e(u)= & {} -\dfrac{1}{2}\displaystyle \int _{\Omega (u)} \big \vert \nabla \psi _u\big \vert ^2\,\mathrm {d}(x,z)\\&-\dfrac{1}{2}\displaystyle \int _{ D} \sigma (x) \big \vert \psi _u(x,-H)-{\mathfrak {h}}_{u}(x)\big \vert ^2\,\mathrm {d}x\,. \end{aligned}$$

6.1 Existence of a minimizer of a regularized energy

As already noted in [21], the boundedness from below of the functional E is a priori unclear since $\alpha =0$. To cope with this issue, we work with the regularized functional given by

$$\begin{aligned} {\mathcal {E}}_k(u):= E(u) + \frac{A}{2} \Vert (u-k)_+\Vert _{L_2(D)}^2\,,\quad u\in {\bar{S}}_0\,, \end{aligned}$$

(6.1)

for $k\ge H$, where

$$\begin{aligned} A:= 8\left( \frac{K^4}{\beta } + 2 K^2 \right) \,, \end{aligned}$$

and the constant K is introduced in (2.4).

Lemma 6.1

For each $k\ge H$, the functional ${\mathcal {E}}_k$ is bounded from below with

$$\begin{aligned} {\mathcal {E}}_k(u)\ge \frac{\beta }{4}\Vert \partial _x^2 u\Vert _{L_2(D)}^2+\frac{A}{4}\Vert (u-k)_+\Vert _{L_2(D)}^2- c(k) \end{aligned}$$

for some constant $c(k)>0$.

Proof

By (2.3), (2.8), and Proposition 3.3,

$$\begin{aligned} -E_e(u)&={\mathcal {G}}(u)[\psi _u-h_u] \le {\mathcal {G}}(u)[0] \\&=\frac{1}{2}\int _{\Omega (u)} \vert \nabla h_u\vert ^2\, \mathrm {d}(x,z) +\frac{1}{2}\int _D \sigma (x)\big [h_u(x,-H)-{\mathfrak {h}}_u(x)\big ]^2\,\mathrm {d}x\\&\le \,\,\,\int _{\Omega (u)} \left[ (\partial _x h)_u^2 + |\partial _x u|^2 (\partial _w h)_u^2 + (\partial _z h)_u^2 \right] \,\mathrm {d}x \\&\qquad \qquad \quad \qquad \qquad \,\,\,+ {\bar{\sigma }} \int _D \Big \{ \big [h_u(x,-H)\big ]^2 + \big [{\mathfrak {h}}_u(x)\big ]^2 \Big \}\,\mathrm {d}x \\&\le K^2\int _{\Omega (u)} \left( 2 \frac{1+u(x)^2}{H+u(x)}+\frac{\vert \partial _x u(x)\vert ^2}{H+u(x)}\right) \,\mathrm {d}(x,z) + 2 {{\bar{\sigma }}} K^2|D|\\&\le K^2 \left( 2|D| + 2 \Vert u\Vert _{L_2(D)}^2 + \Vert \partial _x u\Vert _{L_2(D)}^2 \right) + 2 {{\bar{\sigma }}} K^2|D|\\&=2(1+{\bar{\sigma }}) |D| K^2 + 2K^2 \Vert u\Vert _{L_2(D)}^2 + K^2 \Vert \partial _x u\Vert _{L_2(D)}^2\,. \end{aligned}$$

Now, since $u\in {\bar{S}}$,

$$\begin{aligned} \int _D \vert \partial _x u\vert ^2\,\mathrm {d}x=-\int _D u\partial _x^2u\,\mathrm {d}x\le \Vert u\Vert _{L_2(D)} \Vert \partial _x^2u\Vert _{L_2(D)}\,, \end{aligned}$$

and we further obtain with the help of Young’s inequality that

$$\begin{aligned} -E_e(u)&\le 2(1+{\bar{\sigma }}) |D| K^2 + 2 K^2 \Vert u\Vert _{L_2(D)}^2 + K^2 \Vert u\Vert _{L_2(D)} \Vert \partial _x^2u\Vert _{L_2(D)}\\&\le 2(1+{\bar{\sigma }}) |D| K^2 +\left( \frac{K^4}{\beta } +2K^2 \right) \Vert u\Vert _{L_2(D)}^2+\frac{\beta }{4} \Vert \partial _x^2u\Vert _{L_2(D)}^2 \,. \end{aligned}$$

Using this estimate in the definition of ${\mathcal {E}}_k(u)$ along with

$$\begin{aligned} \Vert u\Vert _{L_2(D)}^2&= \int _D u^2 {\mathbf {1}}_{(k,\infty )}(u)\,\mathrm {d}x + \int _D u^2 {\mathbf {1}}_{[-H,k]}(u)\,\mathrm {d}x \\&\le 2 \Vert (u-k)_+\Vert _{L_2(D)}^2 + 2 k^2 \int _D {\mathbf {1}}_{(k,\infty )}(u)\,\mathrm {d}x + k^2 \int _D {\mathbf {1}}_{[-H,k]}(u)\,\mathrm {d}x \\&\le 2 \Vert (u-k)_+\Vert _{L_2(D)}^2 + 2 k^2 |D|\,, \end{aligned}$$

we derive

$$\begin{aligned} {\mathcal {E}}_k(u)&\ge \frac{\beta }{4} \Vert \partial _x^2u\Vert _{L_2(D)}^2 - 2(1+{\bar{\sigma }}) |D| K^2 - \left( \frac{K^4}{\beta } +2K^2 \right) \Vert u\Vert _{L_2(D)}^2 \\&\quad + \frac{A}{2}\Vert (u-k)_+\Vert _{L_2(D)}^2 \\&\ge \frac{\beta }{4} \Vert \partial _x^2u\Vert _{L_2(D)}^2 - c(k) + \left[ \frac{A}{2}-2\left( \frac{K^4}{\beta } + 2 K^2 \right) \right] \Vert (u-k)_+\Vert _{L_2(D)}^2\\&\ge \frac{\beta }{4} \Vert \partial _x^2u\Vert _{L_2(D)}^2 + \frac{A}{4} \Vert (u-k)_+\Vert _{L_2(D)}^2 - c(k)\,, \end{aligned}$$

thereby completing the proof. $\square$

Due to the weak lower semicontinuity of $E_m$ in $H^2(D)$ and the continuity of $E_e$ with respect to the weak topology of $H^2(D)$ (see Theorem 4.4), Lemma 6.1 allows us to apply the direct method of the calculus of variations to derive the existence of a minimizer of ${\mathcal {E}}_k$ in $\bar{S}_0$.

Corollary 6.2

For each $k\ge H$, the functional ${\mathcal {E}}_k$ has at least one minimizer $u_k\in {\bar{S}}_0$; that is,

$$\begin{aligned} {\mathcal {E}}_k(u_k)=\min _{{\bar{S}}_0} {\mathcal {E}}_k\,. \end{aligned}$$

(6.2)

6.2 Derivation of the Euler–Lagrange equation for the regularized energy

We shall next identify the Euler–Lagrange equation satisfied by a minimizer of the regularized energy ${\mathcal {E}}_k$ on ${\bar{S}}_0$.

Proposition 6.3

Let $k\ge H$ and let $u\in \bar{S_0}$ be a minimizer of ${\mathcal {E}}_k$ on ${\bar{S}}_0$. Then u is an $H^2$-weak solution to the variational inequality

$$\begin{aligned} \beta \partial _x^4u- \tau \partial _x^2 u+ A (u-k)_++\partial {\mathbb {I}}_{\bar{S_0}}(u) \ni -g(u) \;\;\text { in }\;\; D\,, \end{aligned}$$

(6.3a)

where $\partial {\mathbb {I}}_{\bar{S_0}}$ is the subdifferential of the indicator function ${\mathbb {I}}_{{\bar{S}}_0}$ of the closed convex subset $\bar{S_0}$ of $H^2(D)$; that is,

$$\begin{aligned} \begin{aligned} \int _D&\left\{ \beta \partial _x^2 u\,\partial _x^2 (w-u) + \tau \partial _x u\, \partial _x(w-u) +A(u-k)_+ (w-u)\right\} \,\mathrm {d}x \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \, \ge -\int _D g(u) (w-u)\, \mathrm {d}x \end{aligned} \end{aligned}$$

(6.3b)

for all $w\in \bar{S_0}$.

Proof

Let $k\ge H$ be fixed. Consider a minimizer $u\in \bar{S_0}$ of ${\mathcal {E}}_k$ on $\bar{S_0}$ and fix $w\in S_0 := \bar{S}_0\cap S$. Owing to the convexity of $\bar{S_0}$, the function $u+s(w-u)=(1-s)u+sw$ belongs to $S_0$ for all $s\in (0,1]$ and the minimizing property of u guarantees that

$$\begin{aligned} 0\le \liminf _{s\rightarrow 0^+} \frac{1}{ s}\big ({\mathcal {E}}_k(u+s(w-u))-{\mathcal {E}}_k(u)\big )\,. \end{aligned}$$

Since $u\in \bar{S}_0\subset \bar{S}$ and $w\in S_0\subset S$, Proposition 5.3 implies that

$$\begin{aligned} 0&\le \int _D \left\{ \beta \partial _x^2 u\,\partial _x^2 (w-u) + \tau \partial _x u\, \partial _x (w-u)+A(u-k)_+ (w-u)\right\} \,\mathrm {d}x\\&\quad + \int _D g(u) (w-u)\, \mathrm {d}x \end{aligned}$$

for all $w\in S_0$. Since $S_0$ is dense in $\bar{S_0}$ and (u, g(u)) belongs to $H^2(D)\times L_2(D)$, this inequality also holds for any $w\in \bar{S_0}$. $\square$

Proposition 6.4

There is $\kappa _0 \ge H$ depending only on K such that, if $u\in \bar{S_0}$ is any solution to the variational inequality (6.3) with $k\ge H$, then $\Vert u\Vert _{L_\infty (D)}\le \kappa _0$.

Proof

Owing to the continuous embedding of $H_0^1(D)$ in $C(\bar{D})$, the function u belongs to $C(\bar{D})$ with $u(\pm L)=0$. Consequently, the set $\{ x \in D\,:\, u(x)>-H\}$ is a non-empty open subset of D and we can write it as a countable union of disjoint open intervals $(I_j)_{j\in J}$, see [1, IX.Proposition 1.8]. Using once more the property $u(\pm L)=0>-H$, we may assume without loss of generality that $I_0=(-L,a_0)$ and $I_1=(b_0,L)$ for some $-L<a_0<b_0<L$, and $\bar{I}_j\subset (-L,L)$ for $j\in J$ with $j\ge 2$.

Step 1: Thanks to (2.3b) and (2.4a), we infer from Lemma 3.4 that $|\psi _u|\le K$ in $\Omega (u)$. Combining this bound with (2.3), (2.4), (2.6), and (2.8) readily gives

$$\begin{aligned} g(u)(x)\ge -2{{\bar{\sigma }}} K^2 - K^2=:-G_0\,, \quad x\in D\,. \end{aligned}$$

(6.4)

Step 2: Consider first $j\in J$ with $j\ge 2$ and let $\theta \in {\mathcal {D}}(I_j)$. Since $u>-H$ in the support of $\theta$, the function $u\pm \delta \theta$ belongs to $S_0$ for $\delta >0$ small enough. We thus infer from (6.3b) that

$$\begin{aligned} \pm \delta \int _{I_j} \left\{ \beta \partial _x^2 u\,\partial _x^2 \theta +\tau \partial _x u\, \partial _x\theta +A(u-k)_+ \theta \right\} \,\mathrm {d}x \ge \mp \delta \int _{I_j} g(u) \theta \, \mathrm {d}x \,, \end{aligned}$$

hence

$$\begin{aligned} \int _{I_j} \left\{ \beta \partial _x^2 u\,\partial _x^2 \theta +\tau \partial _x u\, \partial _x\theta +A(u-k)_+ \theta \right\} \,\mathrm {d}x =-\int _{I_j} g(u) \theta \, \mathrm {d}x \,. \end{aligned}$$

Consequently, using the function $S_{I_j}$ defined in Proposition D.1, we realize that $u-S_{I_j} \in H^2(I_j)$ is a weak solution to the boundary value problem

$$\begin{aligned} \beta \partial _x^4 w-\tau \partial _x^2 w&= -G_0-g(u)-A (u-k)_+\;\;\text { in }\;\; I_j\,, \end{aligned}$$

(6.5a)

$$\begin{aligned}&w=\partial _x w=0 \;\;\text { in }\;\; \partial I_j\,, \end{aligned}$$

(6.5b)

the boundary conditions (6.5b) being a consequence of the definition of $I_j$, $j\ge 2$, the $H^2(D)$-regularity of u, and the constraint $u\ge -H$. Taking into account that $g(u)+A (u-k)_+\in L_2(I_j)$ by Theorem 4.4, classical elliptic regularity theory implies that $u-S_{I_j}\in H^4(I_j)$ is a strong solution to (6.5). Since the right hand side of (6.5a) is non-positive due to (6.4), it now follows from a version of Boggio’s comparison principle [7, 13, 17, 29] that $u-S_{I_j} < 0$ in $I_j$, so that $u(x)\le \kappa _0$ for $x\in \bar{I}_j$ and $j\ge 2$ by Proposition D.1.

Step 3: We next handle the case $j=0$ in which $I_0=(-L,a_0)$. We first argue as in the previous step to conclude that

$$\begin{aligned} \int _{I_0} \left\{ \beta \partial _x^2 u\,\partial _x^2 \theta +\tau \partial _x u\, \partial _x\theta + A(u-k)_+ \theta \right\} \,\mathrm {d}x =-\int _{I_0} g(u) \theta \, \mathrm {d}x \end{aligned}$$

(6.6)

for all $\theta \in {\mathcal {D}}(I_0)$ and that $u(-L)=\partial _x u(-L)=u(a_0)+H = \partial _x u(a_0)=0$. Consequently, we infer from (6.6) and Proposition D.1 that $u-S_{I_0}\in H^2(I_0)$ is a weak solution to the boundary value problem

$$\begin{aligned} \beta \partial _x^4 w-\tau \partial _x^2 w&= -G_0 -g(u)-A (u-k)_+\;\;\text { in }\;\; I_0\,,\\&w=\partial _x w=0 \;\;\text { on }\;\; \partial I_0\,. \end{aligned}$$

We then argue as in Step 2 to establish that $u-S_{I_0} < 0$ in $I_0=(-L,a_0)$. Hence, $u\le \kappa _0$ in $[-L,a_0]$ by Proposition D.1.

Step 4: For $j=1$ ($I_1=(b_0,L)$), we proceed as in Step 3 using Proposition D.1 to deduce that $u\le \kappa _0$ in $[b_0,L]$. This completes the proof. $\square$

6.3 Proof of Theorem 2.3 for $\alpha =0$

Let $k\ge H$ and consider a minimizer $u_k\in {\bar{S}}_0$ of the functional ${\mathcal {E}}_k$ on ${\bar{S}}_0$ as provided by Corollary 6.2. Then, $-H\le u_k\le \kappa _0$ in D according to Proposition 6.4. Therefore, if $k\ge \kappa _0$, then

$$\begin{aligned} E(u_k)={\mathcal {E}}_{\kappa _0}(u_k)={\mathcal {E}}_k(u_k)\le {\mathcal {E}}_k(v) = E(v) + \frac{A}{2} \Vert (v-k)_+\Vert _{L_2(D)}^2\,,\quad v\in {\bar{S}}_0\,. \end{aligned}$$

(6.7)

Now, it follows from Lemma 6.1 and the fact that $0\in \bar{S}_0$ that, for $k\ge \kappa _0$,

$$\begin{aligned} \frac{\beta }{4}\Vert \partial _x^2 u_k\Vert _{L_2(D)}^2 \le {\mathcal {E}}_{\kappa _0}(u_k)+ c(\kappa _0)\le {\mathcal {E}}_{k}(0)+ c(\kappa _0)=E(0)+ c(\kappa _0)\,. \end{aligned}$$

Therefore, $(u_k)_{k\ge \kappa _0}$ is bounded in $H^2(D)$ and there is a subsequence of $(u_k)_{k\ge \kappa _0}$ (not relabeled) which converges weakly in $H^2(D)$ and strongly in $H^1(D)$ towards some $u_*\in {\bar{S}}_0$. Due to the weak lower semicontinuity of $E_m$ in $H^2(D)$ and the continuity of $E_e$ with respect to the weak topology of $H^2(D)$ (see Theorem 4.4), we readily infer from (6.7) that

$$\begin{aligned} E(u_*)\le E(v)\,,\quad v\in {\bar{S}}_0\,, \end{aligned}$$

after taking into account that

$$\begin{aligned} \lim _{k\rightarrow \infty } \Vert (v-k)_+\Vert _{L_2(D)} = 0\,, \qquad v\in L_2(D)\,. \end{aligned}$$

Consequently, $u_*\in {\bar{S}}_0$ is a minimizer of E on ${\bar{S}}_0$. This proves Theorem 2.3.

6.4 Proof of Theorem 2.4 for $\alpha =0$

Let $u\in {\bar{S}}_0$ be any minimizer of E on ${\bar{S}}_0$. Proceeding as in the proof of Proposition 6.3, this implies that $u\in {\bar{S}}_0$ is an $H^2$-weak solution to the variational inequality

$$\begin{aligned} \beta \partial _x^4u- \tau \partial _x^2 u+ +\partial {\mathbb {I}}_{\bar{S_0}}(u) \ni -g(u) \;\;\text { in }\;\; D\,, \end{aligned}$$

which completes the proof of Theorem 2.4.

7 Proofs of Theorem 2.3 and Theorem 2.4 for $\alpha >0$

Consider now $\alpha >0$. In that case, the total energy is given by

$$\begin{aligned} E(u)= E_m(u)+E_e(u) \end{aligned}$$

with mechanical energy

$$\begin{aligned} E_m(u)=\frac{\beta }{2}\Vert \partial _x^2u\Vert _{L_2(D)}^2 +\left( \frac{\tau }{2}+\frac{\alpha }{4}\Vert \partial _x u\Vert _{L_2(D)}^2\right) \Vert \partial _x u\Vert _{L_2(D)}^2 \end{aligned}$$

and electrostatic energy

$$\begin{aligned} E_e(u)= & {} -\dfrac{1}{2}\displaystyle \int _{\Omega (u)} \big \vert \nabla \psi _u\big \vert ^2\,\mathrm {d}(x,z)\\&-\dfrac{1}{2}\displaystyle \int _{ D} \sigma (x) \big \vert \psi _u(x,-H)-{\mathfrak {h}}_{u}(x)\big \vert ^2\,\mathrm {d}x\,. \end{aligned}$$

Observe that, since $\alpha >0$, the mechanical energy $E_m$ features a super-quadratic term in $\Vert \partial _x u\Vert _{L_2(D)}$ which has the following far-reaching consequence, which is shown as in the proof of [21, Theorem 5.1], with the help of (2.3), (2.8), and Proposition 3.3 for the derivation of an appropriate upper bound on $-E_e(u)$, see the proof of Lemma 6.1.

Lemma 7.1

The functional E is bounded from below with

$$\begin{aligned} E(u) \ge \frac{\beta }{4} \Vert \partial _x^2 u\Vert _{L_2(D)}^2 -c \end{aligned}$$

for some constant $c>0$.

Once Lemma 7.1 is established, the existence of a minimizer of E on ${\bar{S}}_0$ follows from the weak lower semicontinuity of $E_m$ in $H^2(D)$ and the continuity of $E_e$ with respect to the weak topology of $H^2(D)$ (see Corollary 4.2) by the direct method of the calculus of variations, hence Theorem 2.3 for $\alpha >0$ (see also [21, Theorem 5.1]). As for the proof of Theorem 2.4 for $\alpha >0$, it is the same as that for $\alpha =0$, see Sect. 6.4.

Appendices

Appendix A: A technical lemma

Lemma A.1

Let I and J be two bounded intervals in ${\mathbb {R}}$, and let U be a bounded open subset of $I \times J$. Consider $\vartheta \in H^1(U)$ and functions $v\in C(\bar{I})$, $w\in C(\bar{I})$, and $\rho \in C(\bar{I})$, $\rho \ge 0$, such that

• (a) $x \mapsto \vartheta (x,v(x))$ and $x \mapsto \vartheta (x,w(x))$ are well-defined and belong to $L_2(I,\rho \, \mathrm {d}x)$;

• (b) $\{ (x,z) \in I \times J : \min \{v(x),w(x)\}< z< \max \{v(x),w(x)\}\} \subset \bar{U}$.

Then

$$\begin{aligned} \int _I \vert \vartheta (\cdot , v) - \vartheta (\cdot , w)\vert ^2 \rho \, \mathrm {d}x \le \Vert (v-w) \rho \Vert _{L_{\infty }(I)} \Vert \partial _z \vartheta \Vert _{L_2(U)}^2. \end{aligned}$$

Proof

Owing to (b) we have, for a.a. $x \in I$,

$$\begin{aligned} \vert \vartheta (x, v(x)) - \vartheta (x, w(x))\vert ^2 = \left( \int _{w(x)}^{v(x)} \partial _z \vartheta (x,z) \, \mathrm {d}z \right) ^2\,. \end{aligned}$$

Integrating with respect to $x\in I$ after multiplication by $\rho (x)$ and using Hölder’s inequality give

$$\begin{aligned} \int _I \vert \vartheta (x, v(x)) - \vartheta (x, w(x))\vert ^2 \rho (x)\, \mathrm {d}x&\le \int _I \vert v(x)-w(x)\vert \left| \int _{w(x)}^{v(x)} \vert \partial _z \vartheta (x,z)\vert ^2 \mathrm {d}z \right| \,\rho (x)\, \mathrm {d}x \\&\le \Vert \rho (v-w) \Vert _{L_{\infty }(I)} \int _U \vert \partial _z \vartheta (x,z)\vert ^2 \, \mathrm {d}(x,z) \end{aligned}$$

and the proof is complete. $\square$

Appendix B: Proof of Lemma 3.11

The proof of Lemma 3.11 relies on the following result, which can be seen as an extension of [12, Lemma 4.3.1.3] to include Robin boundary conditions.

Lemma B.1

Let $I:=(a,b)$ and set ${\mathcal {R}}_I = I\times (0,1)$. Consider $\varphi \in H^2({\mathcal {R}}_I)$ and $\mu \in C^2(\bar{I})$ such that

$$\begin{aligned} \varphi (a,\eta ) = \varphi (b,\eta )&= 0\ , \qquad \eta \in (0,1)\ , \end{aligned}$$

(B.1a)

$$\begin{aligned} \varphi (x,1) = - \partial _\eta \varphi (x,0) + \mu (x) \varphi (x,0)&= 0\ , \qquad x\in I\ . \end{aligned}$$

(B.1b)

Then

$$\begin{aligned} \int _{{\mathcal {R}}_I} \partial _x^2\varphi \partial _\eta ^2\varphi \ \mathrm {d}(x,\eta ) = \int _{{\mathcal {R}}_I} |\partial _x \partial _\eta \varphi |^2\ \mathrm {d}(x,\eta ) + \int _I \left( \partial _x\varphi \partial _x(\mu \varphi ) \right) (\cdot ,0)\ \mathrm {d}x\ . \end{aligned}$$

Proof

We put $\xi (x,\eta ) := e^{-\eta \mu (x)} \varphi (x,\eta )$ and $\rho (x,\eta ) := e^{\eta \mu (x)}$ for $(x,\eta )\in {\mathcal {R}}_I$. Owing to the regularity of $\varphi$ and $\mu$, the function $\xi$ belongs to $H^2({\mathcal {R}}_I)$ and, for $(x,\eta )\in {\mathcal {R}}_I$,

$$\begin{aligned} \partial _x \xi (x,\eta )&= e^{-\eta \mu (x)} \left[ \partial _x\varphi (x,\eta ) - \eta \partial _x \mu (x) \varphi (x,\eta ) \right] \ , \\ \partial _\eta \xi (x,\eta )&= e^{-\eta \mu (x)} \left[ \partial _\eta \varphi (x,\eta ) - \mu (x) \varphi (x,\eta ) \right] \ . \end{aligned}$$

Consequently, the functions F and G, defined for $(x,\eta )\in {\mathcal {R}}_I$ by

$$\begin{aligned} F(x,\eta )&:= \rho (x,\eta ) \partial _x \xi (x,\eta ) = \partial _x\varphi (x,\eta ) - \eta \partial _x \mu (x) \varphi (x,\eta ) \ , \\ G(x,\eta )&:= \rho (x,\eta ) \partial _\eta \xi (x,\eta ) = \partial _\eta \varphi (x,\eta ) - \mu (x) \varphi (x,\eta ) \ , \end{aligned}$$

satisfy

$$\begin{aligned} G(a,\eta ) = G(b,\eta )&= 0\ , \qquad \eta \in (0,1)\ , \\ F(x,1) = G(x,0)&= 0\ , \qquad x\in I\ , \end{aligned}$$

since, by (B.1),

$$\begin{aligned} \begin{aligned} \partial _\eta \varphi (a,\eta ) = \partial _\eta \varphi (b,\eta )&= 0\ , \qquad \eta \in (0,1)\ , \\ \partial _x\varphi (x,1)&= 0\ , \qquad x\in I\,. \end{aligned} \end{aligned}$$

(B.2)

We then infer from [12, Lemma 4.3.1.3] that

$$\begin{aligned} \int _{{\mathcal {R}}_I} \partial _x(\rho \partial _x\xi ) \partial _\eta (\rho \partial _\eta \xi )\ \mathrm {d}(x,\eta )&= \int _{{\mathcal {R}}_I} \partial _x F \partial _\eta G\ \mathrm {d}(x,\eta ) = \int _{{\mathcal {R}}_I} \partial _\eta F \partial _x G\ \mathrm {d}(x,\eta ) \\&= \int _{{\mathcal {R}}_I} \partial _\eta (\rho \partial _x\xi ) \partial _x(\rho \partial _\eta \xi )\ \mathrm {d}(x,\eta )\ ; \end{aligned}$$

that is,

$$\begin{aligned} 0 = \int _{{\mathcal {R}}_I} \left[ \partial _x^2\varphi \partial _\eta ^2\varphi - |\partial _x\partial _\eta \varphi |^2 \right] \ \mathrm {d}(x,\eta ) + \sum _{j=1}^3 I_j\ , \end{aligned}$$

(B.3)

where

$$\begin{aligned} I_1&:= \int _{{\mathcal {R}}_I} \left[ - \partial _\eta (\mu \varphi ) \partial _x^2\varphi + \partial _x(\mu \varphi ) \partial _x\partial _\eta \varphi \right] \ \mathrm {d}(x,\eta )\, , \\ I_2&:= \int _{{\mathcal {R}}_I} \left[ - \partial _x(\eta \varphi \partial _x \mu ) \partial _\eta ^2\varphi + \partial _\eta (\eta \varphi \partial _x \mu ) \partial _x\partial _\eta \varphi \right] \ \mathrm {d}(x,\eta )\ , \\ I_3&:= \int _{{\mathcal {R}}_I} \left[ \partial _x(\eta \varphi \partial _x \mu ) \partial _\eta (\tau \varphi ) - \partial _\eta (\eta \varphi \partial _x \mu ) \partial _x(\tau \varphi ) \right] \ \mathrm {d}(x,\eta )\,. \end{aligned}$$

First, integrating by parts and using the boundary values (B.1) of $\varphi$ give

$$\begin{aligned} I_3 = \int _0^1 \Big [ \eta \varphi \partial _x \mu \partial _\eta (\mu \varphi ) \Big ]_{x=a}^{x=b}\ \mathrm {d}\eta - \int _I \Big [ \eta \varphi \partial _x \mu \partial _x(\mu \varphi ) \Big ]_{\eta =0}^{\eta =1}\ \mathrm {d}x = 0 \end{aligned}$$

and

$$\begin{aligned} I_2&= - \int _I \Big [ \partial _x(\eta \varphi \partial _x \mu ) \partial _\eta \varphi \Big ]_{\eta =0}^{\eta =1}\ \mathrm {d}x + \int _0^1 \Big [ \partial _\eta (\eta \varphi \partial _x \mu ) \partial _\eta \varphi \Big ]_{x=a}^{x=b}\ \mathrm {d}\eta \\&= - \int _I \partial _x \mu \partial _x\varphi (\cdot ,1) \partial _\eta \varphi (\cdot ,1)\ \mathrm {d}x + \int _0^1 \partial _\eta (\eta \varphi \partial _x \mu )(b,\cdot ) \partial _\eta \varphi (b,\cdot )\ \mathrm {d}\eta \\&\quad \ - \int _0^1 \partial _\eta (\eta \varphi \partial _x \mu )(a,\cdot ) \partial _\eta \varphi (a,\cdot )\ \mathrm {d}\eta \ . \end{aligned}$$

Owing to (B.2) we conclude that $I_2=0$. Finally, we deduce from (B.1) and (B.2) after integrating by parts that

$$\begin{aligned} I_1&= - \int _0^1 \Big [ \partial _\eta (\mu \varphi ) \partial _x\varphi \Big ]_{x=a}^{x=b}\ \mathrm {d}\eta + \int _I \Big [ \partial _x(\mu \varphi ) \partial _x\varphi \Big ]_{\eta =0}^{\eta =1}\ \mathrm {d}x \\&= - \int _0^1 \mu (b) \partial _x\varphi (b,\cdot ) \partial _\eta \varphi (b,\cdot )\ \mathrm {d}\eta + \int _0^1 \mu (a) \partial _x\varphi (a,\cdot ) \partial _\eta \varphi (a,\cdot )\ \mathrm {d}\eta \\&\qquad + \int _I \partial _x(\mu \varphi )(\cdot ,1) \partial _x\varphi (\cdot ,1)\ \mathrm {d}x - \int _I \partial _x(\mu \varphi )(\cdot ,0) \partial _x\varphi (\cdot ,0)\ \mathrm {d}x \\&= - \int _I \partial _x(\mu \varphi )(\cdot ,0) \partial _x\varphi (\cdot ,0)\ \mathrm {d}x\ . \end{aligned}$$

Collecting (B.3) and the formulas for $I_j$, $1\le j\le 3$, completes the proof. $\square$

Proof of Lemma 3.11

For $(x,\eta )\in {\mathcal {R}}_I$, we define

$$\begin{aligned} \Phi (x,\eta ) := \zeta _v(x,-H+\eta (H+v(x)))\,, \end{aligned}$$

(B.4)

or, equivalently,

$$\begin{aligned} \zeta _v(x,z) = \Phi \left( x , \frac{H+z}{H+v(x)} \right) \ , \qquad (x,z)\in {\mathcal {O}}_I(v)\,. \end{aligned}$$

Since $\zeta _v\in H^2({\mathcal {O}}_I(v))$ by Lemma 3.10 and $v\in H^2(I)$, the function $\Phi$ belongs to $H^2({\mathcal {R}}_I)$ and we infer from (3.18b) and (3.18c) that

$$\begin{aligned} \begin{aligned} \Phi (a,\eta ) = \Phi (b,\eta )&= 0\ , \qquad \eta \in (0,1)\ , \\ \Phi (x,1) = -\partial _\eta \Phi (x,0) + \sigma (x) (H+v)(x) \Phi (x,0)&= 0\ , \qquad x\in I\ . \end{aligned} \end{aligned}$$

(B.5)

Next,

$$\begin{aligned} J := \int _{{\mathcal {O}}_I(v)} \partial _x^2 \zeta _v \partial _z^2 \zeta _v\ \mathrm {d}(x,z) = \sum _{i=1}^3 J_i\ , \end{aligned}$$

(B.6)

where

$$\begin{aligned} J_1&:= \int _{{\mathcal {R}}_I} \partial _x^2\Phi \partial _\eta ^2\Phi \frac{\mathrm {d}(x,\eta )}{H+v}\ , \\ J_2&:= \int _{{\mathcal {R}}_I} \left[ - 2 \eta \frac{\partial _x v}{H+v} \partial _x\partial _\eta \Phi + \eta ^2 \left( \frac{\partial _x v}{H+v} \right) ^2 \partial _\eta ^2\Phi \right] \partial _\eta ^2\Phi \frac{\mathrm {d}(x,\eta )}{H+v}\ , \\ J_3&:= \int _{{\mathcal {R}}_I} \eta \left[ 2 \left( \frac{\partial _x v}{H+v} \right) ^2 - \frac{\partial _x^2 v}{H+v} \right] \partial _\eta \Phi \partial _\eta ^2\Phi \frac{\mathrm {d}(x,\eta )}{H+v}\ . \end{aligned}$$

Since

$$\begin{aligned} \partial _x^2 \left( \frac{\Phi }{\sqrt{H+v}} \right) = \frac{\partial _x^2\Phi }{\sqrt{H+v}} - \frac{\partial _x v}{(H+v)^{3/2}} \partial _x \Phi - \frac{1}{2} \partial _x\left( \frac{\partial _x v}{(H+v)^{3/2}} \right) \Phi \ , \end{aligned}$$

we further obtain

$$\begin{aligned} J_1 := \sum _{i=1}^{3} J_{1,i}\ , \end{aligned}$$

where

$$\begin{aligned} J_{1,1}&:= \int _{{\mathcal {R}}_I} \partial _x^2 \left( \frac{\Phi }{\sqrt{H+v}} \right) \partial _\eta ^2\left( \frac{\Phi }{\sqrt{H+v}} \right) \ \mathrm {d}(x,\eta )\,, \\ J_{1,2}&:= \int _{{\mathcal {R}}_I} \frac{\partial _x v}{(H+v)^{3/2}} \,\partial _x \Phi \partial _\eta ^2\left( \frac{\Phi }{\sqrt{H+v}} \right) \ \mathrm {d}(x,\eta )\,, \\ J_{1,3}&:= \frac{1}{2} \int _{{\mathcal {R}}_I} \partial _x\left( \frac{\partial _x v}{(H+v)^{3/2}} \right) \Phi \partial _\eta ^2\left( \frac{\Phi }{\sqrt{H+v}} \right) \ \mathrm {d}(x,\eta )\ . \end{aligned}$$

We first infer from (B.5) and Lemma B.1 (with $\varphi =\Phi /\sqrt{H+v}$ and $\mu =\sigma (H+v)$) that

$$\begin{aligned} J_{1,1}&= \int _{{\mathcal {R}}_I} \left| \partial _x\partial _\eta \left( \frac{\Phi }{\sqrt{H+v}} \right) \right| ^2\ \mathrm {d}(x,\eta ) \\&\quad + \int _I \partial _x \left( \frac{\Phi }{\sqrt{H+v}} \right) (\cdot ,0)\, \partial _x \left( \sigma \sqrt{H+v} \Phi \right) (\cdot ,0)\ \mathrm {d}x \\&= \int _{{\mathcal {R}}_I} \frac{\left| \partial _x\partial _\eta \Phi \right| ^2}{H+v}\ \mathrm {d}(x,\eta ) - \int _{{\mathcal {R}}_I} \frac{\partial _x v}{(H+v)^2} \partial _\eta \Phi \partial _x\partial _\eta \Phi \ \mathrm {d}(x,\eta ) \\&\quad + \frac{1}{4} \int _{{\mathcal {R}}_I} \frac{(\partial _x v)^2}{(H+v)^3} |\partial _\eta \Phi |^2\ \mathrm {d}(x,\eta ) \\&\quad + \int _I \partial _x \left( \frac{\Phi }{\sqrt{H+v}} \right) (\cdot ,0) \partial _x \left( \sigma \sqrt{H+v} \Phi \right) (\cdot ,0)\ \mathrm {d}x \ . \end{aligned}$$

Next, we integrate by parts and use the boundary values (B.5) of $\Phi$ to obtain

$$\begin{aligned} J_{1,2}&= \int _{{\mathcal {R}}_I} \frac{\partial _x v}{(H+v)^2} \partial _x \Phi \partial _\eta ^2\Phi \ \mathrm {d}(x,\eta ) \\&= \int _I \frac{\partial _x v}{(H+v)^2} \Big [ \partial _x \Phi \partial _\eta \Phi \Big ]_{\eta =0}^{\eta =1}\ \mathrm {d}x - \int _{{\mathcal {R}}_I} \frac{\partial _x v}{(H+v)^2} \partial _\eta \Phi \partial _x\partial _\eta \Phi \ \mathrm {d}(x,\eta ) \\&= - \int _I \frac{\sigma \partial _x v}{(H+v)} \Phi (\cdot ,0) \partial _x \Phi (\cdot ,0)\ \mathrm {d}x - \int _{{\mathcal {R}}_I} \frac{\partial _x v}{(H+v)^2} \partial _\eta \Phi \partial _x\partial _\eta \Phi \ \mathrm {d}(x,\eta ) \end{aligned}$$

and

$$\begin{aligned} J_{1,3}&= \int _{{\mathcal {R}}_I} \left( \frac{\partial _x^2 v}{2(H+v)^2} - \frac{3}{4} \frac{(\partial _x v)^2}{(H+v)^3} \right) \Phi \partial _\eta ^2\Phi \ \mathrm {d}(x,\eta ) \\&= \int _I \left( \frac{\partial _x^2 v}{2(H+v)^2} - \frac{3}{4} \frac{(\partial _x v)^2}{(H+v)^3} \right) \Big [ \Phi \partial _\eta \Phi \Big ]_{\eta =0}^{\eta =1}\ \mathrm {d}x \\&\quad - \int _{{\mathcal {R}}_I} \left( \frac{\partial _x^2 v}{2(H+v)^2} - \frac{3}{4} \frac{(\partial _x v)^2}{(H+v)^3} \right) |\partial _\eta \Phi |^2\ \mathrm {d}(x,\eta ) \\&= - \int _I \sigma \left( \frac{\partial _x^2 v}{2(H+v)} - \frac{3}{4} \frac{(\partial _x v)^2}{(H+v)^2} \right) |\Phi (\cdot ,0)|^2\ \mathrm {d}x \\&\quad - \int _{{\mathcal {R}}_I} \left( \frac{\partial _x^2 v}{2(H+v)^2} - \frac{3}{4} \frac{(\partial _x v)^2}{(H+v)^3} \right) |\partial _\eta \Phi |^2\ \mathrm {d}(x,\eta )\ . \end{aligned}$$

Next,

$$\begin{aligned} J_3 = \int _{{\mathcal {R}}_I} 2\eta \frac{(\partial _x v)^2}{(H+v)^3} \partial _\eta \Phi \partial _\eta ^2\Phi \mathrm {d}(x,\eta ) + J_{3,2} \end{aligned}$$

with

$$\begin{aligned} J_{3,2} := - \int _{{\mathcal {R}}_I} \frac{\eta \partial _x^2 v}{2(H+v)^2} \partial _\eta \left( |\partial _\eta \Phi |^2 \right) \mathrm {d}(x,\eta ) \ . \end{aligned}$$

Integrating by parts and using (B.5) give

$$\begin{aligned} J_{3,2}&= - \int _I \frac{\partial _x^2 v}{2(H+v)^2} \Big [ \eta |\partial _\eta \Phi |^2 \Big ]_{\eta =0}^{\eta =1}\ \mathrm {d}x + \int _{{\mathcal {R}}_I} \frac{\partial _x^2 v}{2(H+v)^2} |\partial _\eta \Phi |^2 \mathrm {d}(x,\eta ) \\&= - \int _I \frac{\partial _x^2 v}{2(H+v)^2} |\partial _\eta \Phi (\cdot ,1)|^2\ \mathrm {d}x + \int _{{\mathcal {R}}_I} \frac{\partial _x^2 v}{2(H+v)^2} |\partial _\eta \Phi |^2 \mathrm {d}(x,\eta )\ . \end{aligned}$$

Consequently,

$$\begin{aligned} \begin{aligned} J&= \int _{{\mathcal {R}}_I} (H+v) \left( \frac{\partial _x\partial _\eta \Phi }{H+v} - \frac{\eta \partial _x v}{(H+v)^2} \partial _\eta ^2\Phi - \frac{\partial _x v}{(H+v)^2} \partial _\eta \Phi \right) ^2\ \mathrm {d}(x,\eta ) \\&\quad + J_4 - \frac{1}{2} \int _I \frac{\partial _x^2 v}{(H+v)^2} |\partial _\eta \Phi (\cdot ,1)|^2\ \mathrm {d}x\ , \end{aligned} \end{aligned}$$

(B.7)

where

$$\begin{aligned} J_4&:= \int _I \left( \partial _x\Phi \partial _x(\sigma \Phi ) - \frac{\sigma \partial _x v}{2(H+v)} \Phi \partial _x\Phi - \frac{\partial _x v}{2(H+v)} \Phi \partial _x(\sigma \Phi ) \right) (\cdot ,0)\ \mathrm {d}x \\&\quad + \frac{1}{2} \int _I \sigma \left( \frac{(\partial _x v)^2}{(H+v)^2} - \frac{\partial _x^2 v}{H+v} \right) |\Phi (\cdot ,0)|^2\ \mathrm {d}x\ . \end{aligned}$$

Now, since $H^2({\mathcal {R}}_I)$ is continuously embedded in $C(\overline{{\mathcal {R}}_I})$ by [12, Theorem 1.4.5.2], we infer from (B.5) that

$$\begin{aligned} \Phi (a,0)=\Phi (b,0)=0\ . \end{aligned}$$

Using this property along with an integration by parts, we obtain

$$\begin{aligned}&- \int _I \left( \frac{\sigma \partial _x v}{2(H+v)} \Phi \partial _x\Phi + \frac{\partial _x v}{2(H+v)} \Phi \partial _x(\sigma \Phi ) \right) (\cdot ,0)\ \mathrm {d}x \\&\quad = - \frac{1}{2} \int _I \frac{\partial _x v}{H+v} \partial _x(\sigma \Phi ^2)\ \mathrm {d}x \\&\quad = - \frac{1}{2} \Big [ \frac{\partial _x v}{H+v} \sigma |\Phi (\cdot ,0)|^2 \Big ]_{x=a}^{x=b} + \frac{1}{2} \int _I \left( \frac{\partial _x^2 v}{H+v} - \frac{(\partial _x v)^2}{(H+v)^2} \right) \sigma |\Phi (\cdot ,0)|^2\ \mathrm {d}x \\&\quad = \frac{1}{2} \int _I \left( \frac{\partial _x^2 v}{H+v} - \frac{(\partial _x v)^2}{(H+v)^2} \right) \sigma |\Phi (\cdot ,0)|^2\ \mathrm {d}x \ , \end{aligned}$$

so that $J_4$ reduces to

$$\begin{aligned} J_4 = \int _I \left( \partial _x\Phi \partial _x(\sigma \Phi ) \right) (\cdot ,0)\ \mathrm {d}x\ . \end{aligned}$$

(B.8)

We then infer from (B.4), (B.7), and (B.8) that

$$\begin{aligned} J= & {} \int _{{\mathcal {O}}_I(v)} |\partial _x\partial _z\zeta _v|^2\ \mathrm {d}(x,z) + \int _I \partial _x \zeta _v(\cdot ,-H) \partial _x(\sigma \zeta _v)(\cdot ,-H)\ \mathrm {d}x \\&- \frac{1}{2} \int _I \partial _x^2 v |\partial _z \zeta _v(\cdot ,v)|^2\ \mathrm {d}x\ . \end{aligned}$$

Combining (B.6) and the above identity completes the proof. $\square$

Appendix C: Some functional inequalities

Let $I=(a,b)\subset D$ be an open interval and consider $v\in W_\infty ^3(I)$ such that $\min _{[a,b]} v >-H$. Let $M>0$ be such that

$$\begin{aligned} M \ge \max \left\{ 1, \Vert H+v\Vert _{L_\infty (I)} , \Vert \partial _x v\Vert _{L_\infty (I)} \right\} \,. \end{aligned}$$

(C.1)

We derive in this section functional inequalities for functions in the subspace $H^1_{WS}({\mathcal {O}}_I(v))$ of $H^1({\mathcal {O}}_I(v))$ introduced in (3.31). Recall that $P\in H^1_{WS}({\mathcal {O}}_I(v))$ if and only if $P\in H^1({\mathcal {O}}_I(v))$ satisfies

$$\begin{aligned} P(x,-H)&= 0\ , \qquad x\in I\ , \end{aligned}$$

(C.2a)

$$\begin{aligned} P(a,z)&= 0\ , \qquad z\in (-H,v(a)) \ , \end{aligned}$$

(C.2b)

We begin with Poincaré and Sobolev inequalities and pay special attention to the dependence of the constants on v.

Lemma C.1

Let $P\in H^1_{WS}({\mathcal {O}}_I(v))$. Then

$$\begin{aligned} \Vert P\Vert _{L_2({\mathcal {O}}_I(v))}^2 \le 2 M \Vert \nabla P\Vert _{L_1({\mathcal {O}}_I(v))} \Vert \partial _z P\Vert _{L_1({\mathcal {O}}_I(v))}\ , \end{aligned}$$

where M is given by (C.1).

Proof

For $(x,\eta )\in {\mathcal {R}}_I=I\times (0,1)$, we define

$$\begin{aligned} Q(x,\eta ) := P(x,-H+\eta (H+v(x)))\ , \end{aligned}$$

(C.3)

and observe that the regularity of v and P implies that $Q\in H^1({\mathcal {R}}_I)$. In addition, we deduce from (C.2) that

$$\begin{aligned} Q(x,0)&= 0\ , \qquad x\in I\ , \end{aligned}$$

(C.4a)

$$\begin{aligned} Q(a,\eta )&= 0\ , \qquad \eta \in (0,1)\ . \end{aligned}$$

(C.4b)

On the one hand, it follows from (C.4b) that, for a.a. $(x,\eta )\in {\mathcal {R}}_I$,

$$\begin{aligned} |(H+v)(x) Q(x,\eta )|&= \left| \int _a^x \left[ (H+v(x_*)) \partial _x Q(x_*,\eta ) + \partial _x v(x_*) Q(x_*,\eta ) \right] \ \mathrm {d}x_* \right| \nonumber \\&\le {\mathcal {J}}_1(\eta ) := \int _I \left| (H+v(x_*)) \partial _x Q(x_*,\eta ) + \partial _x v(x_*) Q(x_*,\eta ) \right| \ \mathrm {d}x_*\ . \end{aligned}$$

(C.5)

On the other hand, by (C.4a), we obtain, for a.a. $(x,\eta )\in {\mathcal {R}}_I$,

$$\begin{aligned} |Q(x,\eta )| = \left| \int _0^\eta \partial _\eta Q(x,\eta _*)\ \mathrm {d}\eta _* \right| \le {\mathcal {J}}_2(x) := \int _0^1 \left| \partial _\eta Q(x,\eta _*) \right| \ \mathrm {d}\eta _*\ . \end{aligned}$$

(C.6)

We then infer from (C.3), (C.5), and (C.6) that

$$\begin{aligned} \Vert P\Vert _{L_2({\mathcal {O}}_I(v))}^2&= \int _{{\mathcal {R}}_I} (H+v(x)) |Q(x,\eta )|^2\ \mathrm {d}(x,\eta ) \nonumber \\&\le \int _{{\mathcal {R}}_I} {\mathcal {J}}_1(\eta ) {\mathcal {J}}_2(x)\ \mathrm {d}(x,\eta ) = \left( \int _I {\mathcal {J}}_2(x)\ \mathrm {d}x \right) \left( \int _0^1 {\mathcal {J}}_1(\eta )\ \mathrm {d}\eta \right) \ . \end{aligned}$$

(C.7)

Now,

$$\begin{aligned} \int _I {\mathcal {J}}_2(x)\ \mathrm {d}x = \int _{{\mathcal {R}}_I} |\partial _\eta Q(x,\eta )|\ \mathrm {d}(x,\eta ) = \int _{{\mathcal {O}}_I(v)} |\partial _z P(x,z)|\ \mathrm {d}(x,z) \end{aligned}$$

(C.8)

and

$$\begin{aligned}&\int _0^1 {\mathcal {J}}_1(\eta )\ \mathrm {d}\eta = \int _{{\mathcal {R}}_I} \left| (H+v(x)) \partial _x Q(x,\eta ) + \partial _x v(x) Q(x,\eta ) \right| \ \mathrm {d}(x,\eta ) \nonumber \\&= \int _{{\mathcal {O}}_I(v)} \left| \partial _x P(x,z) + \frac{H+z}{H+v(x)} \partial _x v(x) \partial _z P(x,z) + \frac{\partial _x v(x)}{(H+v)(x)} P(x,z) \right| \ \mathrm {d}(x,z)\ . \end{aligned}$$

(C.9)

It further follows from (C.2a) that, for a.a. $(x,z)\in {\mathcal {O}}_I(v)$,

$$\begin{aligned} |P(x,z)| = \left| \int _{-H}^z \partial _z P(x,z_*)\ \mathrm {d}z_* \right| \le \int _{-H}^{v(x)} |\partial _z P(x,z_*)|\ \mathrm {d}z_*\ . \end{aligned}$$

Hence,

$$\begin{aligned} \int _{{\mathcal {O}}_I(v)} \left| \frac{\partial _x v(x)}{(H+v)(x)} P(x,z) \right| \ \mathrm {d}(x,z) \le \int _{{\mathcal {O}}_I(v)} |\partial _x v(x)| |\partial _z P(x,z_*)|\ \mathrm {d}(x,z_*)\ . \end{aligned}$$

(C.10)

Since $0\le H+z \le H+v(x)$ for $(x,z)\in {\mathcal {O}}_I(v)$, we deduce from (C.1), (C.9), and (C.10) that

$$\begin{aligned} \int _0^1 {\mathcal {J}}_1(\eta )\ \mathrm {d}\eta\le & {} \int _{{\mathcal {O}}_I(v)} \left( |\partial _x P(x,z)| + 2 |\partial _x v(x)| |\partial _z P(x,z)| \right) \ \mathrm {d}(x,z) \nonumber \\\le & {} 2 M \Vert \nabla P\Vert _{L_1({\mathcal {O}}_I(v))}\ . \end{aligned}$$

(C.11)

Collecting (C.7), (C.8), and (C.11) completes the proof. $\square$

Since ${\mathcal {O}}_I(v)$ is a two-dimensional domain, a classical consequence of Lemma C.1 is the continuous embedding of $H_{WS}^1({\mathcal {O}}_I(v))$ in $L_r({\mathcal {O}}_I(v))$ for $r\in [1,\infty )$. We stress here once more that our main concern is the precise dependence of the embedding constant on v.

Lemma C.2

Let $P\in H^1_{WS}({\mathcal {O}}_I(v))$ and $r\in [2,\infty )$. Then

$$\begin{aligned} \Vert P\Vert _{L_r({\mathcal {O}}_I(v))}^r \le \left( 2r \sqrt{M} \right) ^{r-2} \Vert P\Vert _{L_2({\mathcal {O}}_I(v))}^2 \Vert \nabla P\Vert _{L_2({\mathcal {O}}_I(v))}^{(r-2)/2} \Vert \partial _z P\Vert _{L_2({\mathcal {O}}_I(v))}^{(r-2)/2}\ , \end{aligned}$$

where M is given by (C.1).

Proof

Step 1. Assume first that $r\ge 4$. For $n\ge 1$, we define the truncation ${\mathcal {T}}_n$ by ${\mathcal {T}}_n(s):= s$ for $s\in [-n,n]$ and ${\mathcal {T}}_n(s) := n\, \mathrm {sign}(s)$ for $s\in (-\infty ,-n)\cup (n,\infty )$. Since ${\mathcal {T}}_n$ is a Lipschitz continuous function on ${\mathbb {R}}$ with $|{\mathcal {T}}_n'|\le 1$ and vanishes at zero, the function ${\mathcal {T}}_n(P)^{r/2}$ also belongs to $H^1_{WS}({\mathcal {O}}_I(v))$. We then infer from Lemma C.1, the bound $|{\mathcal {T}}_n'|\le 1$, and Hölder’s inequality that

$$\begin{aligned} \Vert {\mathcal {T}}_n(P)\Vert _{L_r({\mathcal {O}}_I(v))}^r&\le \frac{M r^2}{2} \left\| {\mathcal {T}}_n(P)^{(r-2)/2} \nabla P \right\| _{L_1({\mathcal {O}}_I(v))} \left\| {\mathcal {T}}_n(P)^{(r-2)/2} \partial _z P \right\| _{L_1({\mathcal {O}}_I(v))} \\&\le M r^2 \Vert {\mathcal {T}}_n(P)\Vert _{L_{r-2}({\mathcal {O}}_I(v))}^{r-2} \Vert \nabla P\Vert _{L_2({\mathcal {O}}_I(v))} \Vert \partial _z P\Vert _{L_2({\mathcal {O}}_I(v))}\ . \end{aligned}$$

Using again Hölder’s inequality, as well as the property $|{\mathcal {T}}_n(s)|\le |s|$ for $s\in {\mathbb {R}}$, gives

$$\begin{aligned} \Vert {\mathcal {T}}_n(P)\Vert _{L_{r-2}({\mathcal {O}}_I(v))}^{r-2}&\le \Vert {\mathcal {T}}_n(P)\Vert _{L_r({\mathcal {O}}_I(v))}^{r(r-4)/(r-2)} \Vert {\mathcal {T}}_n(P)\Vert _{L_2({\mathcal {O}}_I(v))}^{4/(r-2)} \\&\le \Vert {\mathcal {T}}_n(P)\Vert _{L_r({\mathcal {O}}_I(v))}^{r(r-4)/(r-2)} \Vert P\Vert _{L_2({\mathcal {O}}_I(v))}^{4/(r-2)}\ , \end{aligned}$$

since $r\ge 4$. Combining the above two inequalities leads us to

$$\begin{aligned} \Vert {\mathcal {T}}_n(P)\Vert _{L_r({\mathcal {O}}_I(v))}^r \le (M r^2)^{(r-2)/2} \Vert P\Vert _{L_2({\mathcal {O}}_I(v))}^{2} \Vert \nabla P\Vert _{L_2({\mathcal {O}}_I(v))}^{(r-2)/2} \Vert \partial _z P\Vert _{L_2({\mathcal {O}}_I(v))}^{(r-2)/2}\ . \end{aligned}$$

Since the right-hand side of the above inequality does not depend on $n\ge 1$, we may take the limit $n\rightarrow \infty$ and deduce from Fatou’s lemma that $P\in L_r({\mathcal {O}}_I(v))$ and satisfies the stated bound for $r\ge 4$.

Step 2. Consider now $r\in [2,4]$. By Hölder’s inequality and Lemma C.2 for $r=4$,

$$\begin{aligned} \Vert P\Vert _{L_r({\mathcal {O}}_I(v))}^r&\le \Vert P\Vert _{L_4({\mathcal {O}}_I(v))}^{2(r-2)} \Vert P\Vert _{L_2({\mathcal {O}}_I(v))}^{4-r} \\&\le (16M)^{(r-2)/2} \Vert P\Vert _{L_2({\mathcal {O}}_I(v))}^{2} \Vert \nabla P\Vert _{L_2({\mathcal {O}}_I(v))}^{(r-2)/2} \Vert \partial _z P\Vert _{L_2({\mathcal {O}}_I(v))}^{(r-2)/2}\ , \end{aligned}$$

and we complete the proof by noticing that $4\le 2r$. $\square$

In the same vein, we derive an estimate for the trace of $P\in H_{WS}^1({\mathcal {O}}_I(v))$ on the graph ${\mathfrak {G}}_I(v)$ of v, the trace being here well-defined since the assumption $\min _{[a,b]} v >-H$ guarantees that ${\mathcal {O}}_I(v)$ is a Lipschitz domain.

Lemma C.3

Let $P\in H^1_{WS}({\mathcal {O}}_I(v))$ and $r\in [2,\infty )$. Then

$$\begin{aligned} \Vert P(\cdot ,v)\Vert _{L_r(I)}^r \le \left( 4r\sqrt{M} \right) ^r \Vert P\Vert _{L_2({\mathcal {O}}_I(v))} \Vert \nabla P\Vert _{L_2({\mathcal {O}}_I(v))}^{(r-2)/2} \Vert \partial _z P\Vert _{L_2({\mathcal {O}}_I(v))}^{r/2}\ , \end{aligned}$$

where M is given by (C.1).

Proof

By (C.2a) we have, for a.a. $x\in I$,

$$\begin{aligned} |P(x,v(x))|^r&\le r \int _{-H}^{v(x)} |P(x,z)|^{r-1} |\partial _z P(x,z)|\ \mathrm {d}z \ . \end{aligned}$$

Integrating over I and using Hölder’s inequality lead us to

$$\begin{aligned} \Vert P(\cdot ,v)\Vert _{L_r(I)}^r \le r \Vert P\Vert _{L_{2(r-1)}({\mathcal {O}}_I(v))}^{r-1} \Vert \partial _z P\Vert _{L_2({\mathcal {O}}_I(v))}\,. \end{aligned}$$

Since $2(r-1)\ge 2$ as $r\ge 2$, we deduce from Lemma C.2 and the above inequality that

$$\begin{aligned} \Vert P(\cdot ,v)\Vert _{L_r(I)}^r \le r \left( 4(r-1) \sqrt{M} \right) ^{r-2} \Vert P\Vert _{L_2({\mathcal {O}}_I(v))} \Vert \nabla P\Vert _{L_2({\mathcal {O}}_I(v))}^{(r-2)/2} \Vert \partial _z P\Vert _{L_2({\mathcal {O}}_I(v))}^{r/2}\ , \end{aligned}$$

from which Lemma C.3 follows, after using that $r \left( 4(r-1) \sqrt{M} \right) ^{r-2} \le \left( 4r\sqrt{M} \right) ^r$. $\square$

Appendix D: A uniform bound for an auxiliary stationary problem

Proposition D.1

Consider $G_0> 0$, $\beta >0$, and $\tau \ge 0$. Let $I=(a,b)\subset (-L,L)$ be an open interval and let $S_{I}$ be the unique solution to the boundary value problem

$$\begin{aligned} \beta S_{I}'''' - \tau S_{I}'' = G_0\ , \qquad x\in I\ , \end{aligned}$$

(D.1)

supplemented with inhomogeneous Dirichlet boundary conditions

$$\begin{aligned}&S_{I}(a) + H = S'_{I}(a)= S_{I}(b) + H = S'_{I}(b) = 0 \quad \text{ if } \quad -L<a<b<L\ , \end{aligned}$$

(D.2)

$$\begin{aligned}&S_{I}(-L) = S'_{I}(-L)= S_{I}(b) + H= S'_{I}(b) = 0 \quad \text{ if } \quad -L=a<b<L\ , \end{aligned}$$

(D.3)

$$\begin{aligned}&S_{I}(a)+H = S'_{I}(a)= S_{I}(L) = S'_{I}(L) = 0 \quad \text{ if } \quad -L<a<b=L\ , \end{aligned}$$

(D.4)

or clamped boundary conditions

$$\begin{aligned} S_{I}(-L) = S'_{I}(-L)= S_{I}(L) = S'_{I}(L) = 0 \quad \text{ if } \quad -L=a<b=L\ . \end{aligned}$$

(D.5)

There is $\kappa _0>0$ depending only on $G_0$, $\beta$, L, H, and $\tau$ such that

$$\begin{aligned} \left| S_{I}(x) \right| \le \kappa _0\ , \qquad x\in [a,b]\ , \qquad -L \le a < b \le L\ . \end{aligned}$$

Proof

Case 1: $-L<a<b<L$. We set $P(y) := S_{I}(a+(b-a)y)+H$ for $y\in [0,1]$ and deduce from (D.1) and (D.2) that P solves the boundary-value problem

$$\begin{aligned} \begin{aligned}&\beta P'''' - \tau (b-a)^2 P'' = (b-a)^4 G_0\ , \qquad y\in (0,1)\ , \\&P(0) = P'(0) = P(1) = P'(1) = 0\ . \end{aligned} \end{aligned}$$

(D.6)

We first infer from (D.6), the positivity of $G_0$, and a version of Boggio’s comparison principle [7, 13, 17, 29] that $P>0$ in (0, 1). We next multiply (D.6) by P and integrate over (0, 1). After integrating by parts and using the boundary conditions, we obtain

$$\begin{aligned} \beta \Vert P''\Vert _{L_2(0,1)}^2 + \tau (b-a)^2 \Vert P'\Vert _{L_2(0,1)}^2 = (b-a)^4 G_0 \int _0^1 P(y)\ \mathrm {d}y \ . \end{aligned}$$

Since

$$\begin{aligned} |P(y)| = \left| \int _0^y (y-y_*) P''(y_*) \ \mathrm {d}y_* \right| \le \Vert P''\Vert _{L_2(0,1)}\ , \qquad y\in (0,1)\ , \end{aligned}$$

by (D.6), we infer from these observations that

$$\begin{aligned} \beta \Vert P\Vert _{L_\infty (0,1)}^2 \le \beta \Vert P''\Vert _{L_2(0,1)}^2 \le (b-a)^4 G_0 \Vert P\Vert _{L_\infty (0,1)} \le 16L^4 G_0 \Vert P\Vert _{L_\infty (0,1)}\ . \end{aligned}$$

Consequently, $0\le P \le 16L^4 G_0/ \beta$ in [0, 1], hence $-H \le S_{I} \le 16 L^4 G_0/ \beta - H$ in [a, b].

Case 2: $-L=a<b<L$. Let $Q\in {\mathbb {R}}_4[X]$ be such that $Q(0)=Q'(0) = Q(1)+H = Q'(1)= 0$; that is, $Q(y) = y^2(y^2 +2(H-1)y +1- 3H)$. We set $P(y) := S_{I}(-L+(b+L)y)-Q(y)$ for $y\in [0,1]$ and deduce from (D.1) and (D.3) that P solves the boundary value problem

$$\begin{aligned} \begin{aligned}&\beta P'''' - \tau (b+L)^2 P'' = (b+L)^4 G_0 - \beta Q'''' + \tau (b+L)^2 Q'' \ , \qquad y\in (0,1)\ , \\&P(0) = P'(0) = P(1) = P'(1) = 0\ . \end{aligned} \end{aligned}$$

(D.7)

Arguing as in Case 1, we are led to

$$\begin{aligned} \beta \Vert P\Vert _{L_\infty (0,1)}^2&\le \beta \Vert P''\Vert _{L_2(0,1)}^2 + \tau (b+L)^2 \Vert P'\Vert _{L_2(0,1)}^2 \\&\le \left[ (b+L)^4 G_0 + 24\beta +14 \tau (H+1) (b+L)^2 \right] \Vert P\Vert _{L_\infty (0,1)} \\&\le \left[ 16L^4 G_0 + 24\beta + 56 \tau (H+1) L^2 \right] \Vert P\Vert _{L_\infty (0,1)}\ , \end{aligned}$$

since $Q''''=24$ and

$$\begin{aligned} -14(H+1)\le & {} -12y - 6H \le Q''(y) = 12 y^2 + 12(H-1)y + 2(1-3H) \\\le & {} 14 + 12H \le 14(H+1)\ . \end{aligned}$$

Consequently,

$$\begin{aligned} \Vert S_{I}\Vert _{L_\infty (I)} \le \Vert P\Vert _{L_\infty (0,1)} + \Vert Q\Vert _{L_\infty (0,1)} \le \frac{16L^4 G_0 + 24\beta + 56 \tau (H+1) L^2}{\beta } + \Vert Q\Vert _{L_\infty (0,1)}\ . \end{aligned}$$

Case 3: $-L<a<b=L$. We set $P(y) := S_{I}(a+y(L-a))-Q(1-y)$ for $y\in [0,1]$ and proceed as in the previous case to derive the same bound for $\Vert S_{I}\Vert _{L_\infty (I)}$.

Case 4: $-L=a<b=L$. We set $P(y) := S_{I}(-L+2Ly)$ for $y\in [0,1]$ and deduce from (D.1) and (D.5) that P solves the boundary value problem

$$\begin{aligned} \begin{aligned}&\beta P'''' - 4\tau L^2 P'' = 16L^4 G_0\ , \qquad y\in (0,1)\ , \\&P(0) = P'(0) = P(1) = P'(1) = 0\ . \end{aligned} \end{aligned}$$

We then argue as in Case 1 to conclude that $0\le S_{I} \le 16L^4 G_0/\beta$ in $[-L,L]$. $\square$