Stationary states to a free boundary transmission problem for an electrostatically actuated plate

Abstract

A two-dimensional free boundary transmission problem arising in the modeling of an electrostatically actuated plate is considered and a representation formula for the derivative of the associated electrostatic energy with respect to the deflection of the plate is derived. The latter paves the way for the construction of energy minimizers and also provides the Euler–Lagrange equation satisfied by these minimizers. A by-product is the monotonicity of the electrostatic energy with respect to the deflection.

1 Introduction

We consider a model for a microelectromechanical system (MEMS) featuring an elastic, electrostatically actuated plate with positive thickness as introduced in [3]. More precisely, given a finite interval $D:=(-L,L)$ with $L>0$, let the function $u\in C({\bar{D}},[-H,\infty ))$ with $u(\pm L)=0$ measure the deflection from rest of the lower part of an elastic plate with thickness $d>0$, clamped at its boundaries and suspended above a fixed ground plate, the latter being located at $z=-H$ with $H>0$ and represented by $D\times \{-H\}$ (see Fig. 1). The deflected elastic plate is then

Fig. 1

Geometry of $\Omega (u)$ for a state $u\in {\mathcal {S}}$ with empty coincidence set

$$\begin{aligned} \Omega _2(u):= \left\{ (x,z)\in D\times {\mathbb {R}}\,:\, u(x)< z < u(x)+d\right\} \,, \end{aligned}$$

while the region between the ground plate and the deflected elastic plate is

$$\begin{aligned} \Omega _1(u) := \left\{ (x, z)\in D\times {\mathbb {R}} \,:\, -H< z< {u}(x)\right\} \,. \end{aligned}$$

The two regions are separated by the interface

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

and the subdomain of $D\times (-H,\infty )$ spanned by the MEMS device is

$$\begin{aligned} \Omega (u) := \left\{ (x,z)\in D\times {\mathbb {R}} \,:\, -H< z < u(x)+d \right\} = \Omega _1(u)\cup \Omega _2(u)\cup \Sigma (u)\,. \end{aligned}$$

The deflection of the plate being triggered by electrostatic actuation, the total energy of the device is

$$\begin{aligned} E(u):= E_\textrm{m}(u)+E_\textrm{e}(u) \end{aligned}$$

(1.1a)

with mechanical energy $E_\textrm{m}(u)$ and electrostatic energy $E_\textrm{e}(u)$. The former is given by

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

(1.1b)

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

$$\begin{aligned} E_\textrm{e}(u):=-\frac{1}{2}\int _{\Omega (u)} \sigma \vert \nabla \psi _u\vert ^2\,\textrm{d}(x,z) \end{aligned}$$

(1.1c)

involves the electrostatic potential $\psi _u$ in the domain $\Omega (u)$ with $\psi _u$ being the solution to the transmission problem

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

(1.2a)

$$\begin{aligned} \llbracket \psi _u \rrbracket =\llbracket \sigma \nabla \psi _u \rrbracket \cdot {\textbf{n}}_{ \Sigma (u)}&=0 \quad \text {on }\ \Sigma (u)\,, \end{aligned}$$

(1.2b)

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

(1.2c)

where $\llbracket \cdot \rrbracket$ denotes the (possible) jump across the interface $\Sigma (u)$; that is,

$$\begin{aligned} \llbracket f \rrbracket (x,u(x)) := f|_{\Omega _1(u)}(x,u(x)) - f|_{\Omega _2(u)}(x,u(x))\,, \qquad x\in D\,, \end{aligned}$$

whenever meaningful for a function $f:\Omega (u)\rightarrow {\mathbb {R}}$. Moreover,

$$\begin{aligned} \sigma := \sigma _1 {\textbf{1}}_{\Omega _1(u)} + \sigma _2 {\textbf{1}}_{\Omega _2(u)} \end{aligned}$$

(1.3)

involves the material dependent constant permittivities $\sigma _2,\sigma _1>0$. The unit normal vector field to $\Sigma (u)$ (pointing into $\Omega _2(u)$) is

$$\begin{aligned} {\textbf{n}}_{ \Sigma (u)}:=\frac{(-\partial _x u, 1)}{\sqrt{1+(\partial _x u)^2}}\,. \end{aligned}$$

As for the boundary values in (1.2c) we assume the particular form

$$\begin{aligned} h_u(x,z) :=\zeta (z- u(x)+1)\,, \qquad (x,z)\in {\bar{D}}\times [-H,\infty )\,, \end{aligned}$$

(1.4a)

where

$$\begin{aligned} \zeta \in C^2({\mathbb {R}})\,,\qquad \zeta |_{(-\infty ,1]}\equiv 0\,,\qquad \zeta |_{[1+d,\infty )} \equiv V\,, \end{aligned}$$

(1.4b)

with $V>0$. For instance, $\zeta (r):=V\min \{1,(r-1)^m/d^m\}$ for $r>1$ and $m>2$ and $\zeta \equiv 0$ on $(-\infty ,1]$ is a possible choice. Note that

$$\begin{aligned} h_u(x,-H)=0\,, \quad h_u(x,u(x)+d)=V\,,\qquad x\in D\,; \end{aligned}$$

that is, the ground plate and the top of the elastic plate are kept at different constant potentials. Let us emphasize that we explicitly allow that the elastic plate touches upon the ground plate when u reaches the value $-H$ somewhere, a situation corresponding to a non-empty coincidence set

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

(1.5)

as depicted in Fig. 2. In this case, the region $\Omega _1(u)$ is not connected and its boundary features cusps, so that its connected components are not Lipschitz domains.

Fig. 2

Geometry of $\Omega (u)$ for a state $u\in \bar{{\mathcal {S}}}$ with non-empty coincidence set

In this research we shall be interested in minimizers of the total energy E which correspond to stationary states of the MEMS device. More precisely, we shall show the existence of minimizers and derive the corresponding Euler–Lagrange equation they satisfy, which, due to the nature of the problem, is a variational inequality. Obviously, the main difficulty in this regard is related to the electrostatic energy $E_\textrm{e}$ and the associated transmission problem (1.2) for the electrostatic potential. The latter was investigated in [5] for deflections belonging to the set

$$\begin{aligned} \bar{{\mathcal {S}}} := \{u\in H^2(D) \cap H_0^1(D)\,:\, u\ge -H \text { in } D \;\text { and }\; \pm \llbracket \sigma \rrbracket \partial _x u(\pm L) \le 0 \} \end{aligned}$$

with $\llbracket \sigma \rrbracket =\sigma _1-\sigma _2$. More precisely, the following result is shown in [5].

Theorem 1.1

[5, Theorem 1.1] Suppose (1.4).

• (a) For each $u\in \bar{{\mathcal {S}}}$, there is a unique variational solution $\psi _u \in h_{u}+H_{0}^1(\Omega (u))$ to (1.2). Moreover,

  $$\begin{aligned} \psi _{u,1}:= \psi _{u}|_{\Omega _1(u)} \in H^2(\Omega _1(u))\,\textrm{and}\,\, \psi _{u,2} := \psi _{u}|_{\Omega _2(u)} \in H^2(\Omega _2(u)), \end{aligned}$$

  and $\psi _{u}$ is a strong solution to the transmission problem (1.2).

• (b) Given $\kappa >0$, there is $c(\kappa )>0$ such that $\psi _u$ satisfies

  $$\begin{aligned} \Vert \psi _u\Vert _{H^1(\Omega (u))} + \Vert \psi _{u,1}\Vert _{H^2(\Omega _1(u))} + \Vert \psi _{u,2}\Vert _{H^2(\Omega _2(u))} \le c(\kappa ) \end{aligned}$$

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

The $H^2$-regularity of the electrostatic potential $\psi _u$ provided by Theorem 1.1 is then the basis for deriving the existence of minimizers of the total energy E. We shall look for minimizers with clamped boundary conditions; that is, minimizers in the closed convex subset

$$\begin{aligned} \bar{{\mathcal {S}}}_0 := \{u\in H^2(D) \cap H_0^1(D)\,:\, u\ge -H \text { in } D \;\text { and }\; \partial _x u(\pm L)= 0 \} \end{aligned}$$

of $H^2(D)$. We denote by $\partial {\mathbb {I}}_{\bar{{\mathcal {S}}}_0}$ the subdifferential of the indicator function ${\mathbb {I}}_{\bar{{\mathcal {S}}}_0}$. Our main result then reads:

Theorem 1.2

Assume $a>0$ or $\llbracket \sigma \rrbracket <0$, and let (1.4) be satisfied. Then, the total energy E has at least one minimizer in $\bar{{\mathcal {S}}}_0$. Moreover, any minimizer $u_*\in \bar{{\mathcal {S}}}_0$ of E in $\bar{{\mathcal {S}}}_0$ with

$$\begin{aligned} E(u_*)=\min _{\bar{{\mathcal {S}}}_0}E \end{aligned}$$

(1.6)

is an $H^2$-weak solution to the variational inequality

$$\begin{aligned} \beta \partial _x^4u_*-(\tau +a\Vert \partial _x u_*\Vert _{L_2(D)}^2)\partial _x^2 u_*+\partial {\mathbb {I}}_{\bar{{\mathcal {S}}}_0}(u_*) \ni -g(u_*) \;\;\text { in }\;\; D\,; \end{aligned}$$

(1.7)

that is,

$$\begin{aligned} \int _D{} & {} \left\{ \beta \partial _x^2 u_*\,\partial _x^2 (w-u_*)+\big [\tau +a\Vert \partial _x u_*\Vert _{L_2(D)}^2\big ]\partial _x u_*\, \partial _x(w-u_*)\right\} \,\textrm{d}x \\ {}{} & {} \qquad \qquad \qquad \quad \ge -\int _D g(u_*) (w-u_*)\, \textrm{d}x \end{aligned}$$

for all $w\in {\bar{{\mathcal {S}}}_0}$. For $u\in \bar{{\mathcal {S}}}_0$, the function $g(u)\in L_2(D)$ is given by

$$\begin{aligned} \begin{aligned} g(u):=\,&-\frac{\llbracket \sigma \rrbracket }{2(1+(\partial _x u(x))^2)} \big (\partial _x\psi _{u,2}+\partial _x u\partial _z\psi _{u,2}\big )^2 (x,u(x))\\&-\frac{\llbracket \sigma \rrbracket \sigma _2}{2\sigma _1(1+(\partial _x u(x))^2)}\big (\partial _x u\partial _x\psi _{u,2}-\partial _z\psi _{u,2}\big )^2 (x,u(x))\\&+\frac{\sigma _2}{2}\, \big \vert \nabla \psi _{u,2}(x,u(x)+d)\big \vert ^2\,. \end{aligned} \end{aligned}$$

(1.8)

Finally, if $\llbracket \sigma \rrbracket <0$, then $u_*\le 0$ in D.

Even though the total energy E consists of two competing terms with different signs, it is not difficult to see that it is $H^2$-coercive if $a>0$ in (1.1b), see [4], and the existence of a minimizer for E in $\bar{{\mathcal {S}}}_0$ follows directly. When $a=0$, the coercivity of E is no longer obvious without additional assumptions. In fact, we are not able to prove directly that E is bounded below and thus have to proceed differently. In this case, the coercivity of the functional can be enforced by adding a penalty term which vanishes when u is bounded, an idea already used in [2]. The minimizers of the penalized energy functional on $\bar{{\mathcal {S}}}_0$ then satisfy the Euler–Lagrange equation (1.7) with an additional term. The assumption $\llbracket \sigma \rrbracket <0$ now guarantees that $g(u)\ge 0$ in D according to (1.8) which, in turn, yields an a priori bound on the minimizers by a comparison argument. This then implies that the minimizers of the penalized energy actually minimize the total energy E. It is worth emphasizing that the non-negative sign of g(u) — read off from the explicit formula (1.8) when $\llbracket \sigma \rrbracket <0$ — is essential for this approach.

The main motivation of this research is thus the derivation of an explicit formula for the electrostatic force g(u) as the (directional) derivative of the electrostatic energy $E_\textrm{e}(u)$. By definition of $E_\textrm{e}(u)$, such a computation corresponds to that of a shape derivative and thus follows the guidelines of classical results [1, 6, 7]. In fact, a computation in the same spirit is performed in [4] for a related MEMS model but with a flat transmission interface. As we shall see in Sect. 2, the non-flat transmission interface $\Sigma (u)$ in (1.2b) leads to additional terms in the electrostatic force, making the computation of the latter noticeably more involved. We first establish in  Sect. 2 differentiability properties of the electrostatic potential $\psi _u$ with respect to u which then ensure the Fréchet differentiability of the electrostatic energy $E_\textrm{e}$ on ${\mathcal {S}_0}$. The subsequent identification of g(u) as the (directional) derivative of the electrostatic energy $E_\textrm{e}(u)$ is the main contribution of  Sect. 2. It is worth already pointing out here that the derivation of the explicit formula (1.8) of g(u) does not require the explicit computation of the derivative of the electrostatic potential $\psi _u$ with respect to u. Once the formula (1.8) is established, the existence of minimizers of E in $\bar{{\mathcal {S}}}_0$ follows along the lines of [2] as described above.

As already pointed out, the electrostatic force g(u) has a sign if one assumes that $\llbracket \sigma \rrbracket <0$; that is, if $\sigma _2> \sigma _1$. For instance, this is a natural assumption when the region between the two plates is vacuumed or filled with air. We also point out that this assumption implies the monotonicity of the electrostatic energy $E_\textrm{e}$ as stated explicitly in Corollary 2.7.

Remark 1.3

The total energy E can also be minimized in $\bar{{\mathcal {S}}}$ leading then to weak solutions to (1.7) with ${\mathbb {I}}_{\bar{{\mathcal {S}}}}$ instead of ${\mathbb {I}}_{\bar{{\mathcal {S}}}_0}$ which satisfy pinned boundary conditions $u(\pm L) = \partial _x^2 u(\pm L)=0$ instead of the clamped boundary conditions involved in $\bar{{\mathcal {S}}}_0$. In this case, however, one has to be slightly more careful when computing the shape derivative of the electrostatic energy $E_\textrm{e}$ due to the different constraints on the boundary.

2 Shape derivative of the electrostatic energy

The heart of the proof of Theorem 1.2 is the differentiability of the electrostatic energy $E_\textrm{e}$ and, in particular, the identification of g(u) as its derivative at $u\in \bar{{\mathcal {S}}}_0$. On a formal level, this derivative is computed in [3] (in a three-dimensional setting). Here we provide a rigorous proof. Actually, we shall show that the electrostatic energy $E_\textrm{e}$ is Fréchet differentiable on

$$\begin{aligned} {\mathcal {S}}_0 := \{u\in H^2(D) \cap H_0^1(D)\,:\, u> -H \text { in } D \;\text { and }\; \partial _x u(\pm L) =0 \}\,, \end{aligned}$$

i.e., for points with empty coincidence set, while it admits a directional derivative at $u\in \bar{{\mathcal {S}}}_0$ in the directions $-u+{\mathcal {S}}_0$. Here and in the following, ${\mathcal {S}}_0$ and $\bar{{\mathcal {S}}}_0$ are endowed with the $H^2(D)$-topology. The precise result reads as follows:

Theorem 2.1

Assume (1.4). The electrostatic energy $E_\textrm{e}: {{\mathcal {S}}_0}\rightarrow {\mathbb {R}}$ is continuously Fréchet differentiable with

$$\begin{aligned} \partial _uE_\textrm{e}(u)[\vartheta ]= \int _D g(u)(x)\,\vartheta (x)\,\textrm{d}x \end{aligned}$$

for $u\in {\mathcal {S}}_0$ and $\vartheta \in H^2(D)\cap H_0^1(D)$, where g(u) is defined in (1.8). Moreover, if $u\in \bar{{\mathcal {S}}}_0$ and $w\in {\mathcal {S}}_0$, then

$$\begin{aligned} \begin{aligned} \lim _{t\rightarrow 0^+} \frac{1}{t}\big (E_\textrm{e}(&u+t(w-u))-E_\textrm{e}(u)\big )= \int _D g(u)(x)\,(w-u)(x)\,\textrm{d}x\,. \end{aligned} \end{aligned}$$

The function $g:\bar{{\mathcal {S}}}_0\rightarrow L_p(D)$ is continuous for each $p\in [1,\infty )$.

The proof of Theorem 2.1 follows from Proposition 2.5 and Corollary 2.7 below. We will need the following result which is contained in [5].

Proposition 2.2

[5, Theorem 1.3, Proposition 3.3] Assume (1.4). Let $u\in \bar{{\mathcal {S}}}_0$ and consider a bounded sequence $(u_n)_{n\ge 1}$ in $\bar{{\mathcal {S}}}_0$ such that

$$\begin{aligned} \lim _{n\rightarrow \infty } \Vert u_n-u\Vert _{H^1(D)} = 0\,. \end{aligned}$$

Then, for any $p\in [1,\infty )$,

$$\begin{aligned} \lim _{n\rightarrow \infty } \big \Vert \nabla \psi _{u_n,2}(\cdot ,u_n) - \nabla \psi _{u,2}(\cdot ,u) \big \Vert _{L_p(D,{\mathbb {R}}^2)}&= 0\,, \end{aligned}$$

(2.1a)

$$\begin{aligned} \lim _{n\rightarrow \infty } \big \Vert \nabla \psi _{u_n,2}(\cdot ,u_n+d) - \nabla \psi _{u,2}(\cdot ,u+d) \big \Vert _{L_p(D,{\mathbb {R}}^2)}&= 0 \,. \end{aligned}$$

(2.1b)

Moreover,

$$\begin{aligned} \lim _{n\rightarrow \infty }E_\textrm{e}(u_n)=E_\textrm{e}(u)\,. \end{aligned}$$

(2.2)

Finally, setting

$$\begin{aligned} M := d + \max \big \{ \Vert u\Vert _{L_\infty (D)} \,,\, \sup _{n\ge 1}\{\Vert u_n\Vert _{L_\infty (D)}\} \big \}\,, \end{aligned}$$

one has

$$\begin{aligned} \lim _{n\rightarrow \infty } \left\| (\psi _{u_n}-h_{u_n}) - (\psi _u - h_u) \right\| _{H_0^1(D\times (-H,M))} = 0\,. \end{aligned}$$

(2.3)

The first step of the proof of Theorem 2.1 is to show that the electrostatic energy $E_\textrm{e}$ is Fréchet differentiable on ${\mathcal {S}}_0$. The next lemma is adapted from [1, Theorem 5.3.2], see also [4, Lemma 4.1]. We include the proof for the reader’s ease.

Lemma 2.3

Assume (1.4). Let $u\in {\mathcal {S}}_0$ be fixed and define, for $v\in {\mathcal {S}}_0$, the transformation

$$\begin{aligned} \Theta _{u,v}=(\Theta _{u,v,1},\Theta _{u,v,2}):\Omega (u)\rightarrow \Omega (v) \end{aligned}$$

by

$$\begin{aligned}&\Theta _{u,v,1}(x,z):=\left( x,z+\frac{v(x)-u(x)}{H+u(x)}(z+H)\right) \,,{} & {} (x,z)\in \Omega _1(u)\,, \end{aligned}$$

(2.4a)

$$\begin{aligned}&\Theta _{u,v,2}(x,z):=(x,z+v(x)-u(x))\,,{} & {} (x,z)\in \Omega _2(u) \,. \end{aligned}$$

(2.4b)

Then there exists a neighborhood ${\mathcal {U}}$ of u in ${\mathcal {S}}_0$ such that the mapping

$$\begin{aligned} {\mathcal {U}}\rightarrow H_0^1(\Omega (u)),\quad v\mapsto \xi _v:=\big (\psi _v-h_v\big )\circ \Theta _{u,v} \end{aligned}$$

is continuously differentiable, recalling that ${\mathcal {S}}_0$ and thus also ${\mathcal {U}}$ are endowed with the $H^2(D)$-topology.

Remark 2.4

Lemma 2.3 is only an intermediate step in the computation of the Fréchet derivative of the electrostatic energy $E_\textrm{e}$. As we shall see later in the proof of Proposition 2.5, the computation does not require an explicit formula for the derivative of $v\mapsto \xi _v$. Moreover, we do not strive for optimal assumptions (e.g., the topology of ${\mathcal {S}}_0$ can be weakened, as long as it is stronger than that of $W_\infty ^1(D)$).

Proof of Lemma 2.3

The differentiability property relies on a classical approach: we shall first identify a suitable $C^1$-function

$${\mathcal {F}}: {\mathcal {S}}_0\times H_0^1(\Omega (u))\rightarrow H^{-1}(\Omega (u))$$

which vanishes at $(v,\xi _v)$ whenever $v\in {\mathcal {S}}_0$. We then show that the implicit function theorem applies to ${\mathcal {F}}$ near $(u,\xi _u)$.

To this end, set $\chi _v:=\psi _v-h_v$ for $v\in {\mathcal {S}}_0$. Owing to Theorem 1.1, the function $\chi _v$ belongs to $H_0^1(\Omega (v))$ and satisfies the integral identity

$$\begin{aligned} \int _{\Omega (v)}\sigma \nabla \chi _v\cdot \nabla \theta \,\textrm{d}(\bar{x},{{\bar{z}}}) \!\!=\!\!-\int _{\Omega (v)}\sigma \nabla h_v\cdot \nabla \theta \,\textrm{d}({{\bar{x}}},{{\bar{z}}})\,,\quad \theta \in H_0^1(\Omega (v)), \end{aligned}$$

(2.5)

which we next shall write as integrals over $\Omega (u)$. To this end, we first note that, due to $\Theta _{u,u}=\textrm{id}$,

$$\begin{aligned} \xi _u=\chi _u\,,\qquad \nabla \xi _v=D\Theta _{u,v}^T\nabla \chi _v\circ \Theta _{u,v}\,, \end{aligned}$$

(2.6)

where

$$\begin{aligned} D\Theta _{u,v,1}(x,z)=\left( \begin{matrix} 1&{} \quad 0\\ \\ \displaystyle {(z+H)\partial _x\left( \frac{v-u}{H+u}\right) (x)} &{} \quad \displaystyle {\frac{H+v(x)}{H+u(x)} }\end{matrix}\right) \,,\quad (x,z)\in \Omega _1(u)\,, \end{aligned}$$

and

$$\begin{aligned} D\Theta _{u,v,2}(x,z)=\left( \begin{matrix} 1&{} \quad 0\quad \\ \\ \partial _x(v-u)(x) &{} 1\quad \end{matrix}\right) \,,\qquad (x,z)\in \Omega _2(u)\,. \end{aligned}$$

For $\phi \in H_0^1(\Omega (u))$ we set

$$\begin{aligned} \phi _v:=\phi \circ \Theta _{u,v}^{-1}\in H_0^1(\Omega (v)) \end{aligned}$$

and note that

$$\begin{aligned} \nabla \phi _v=\big ((D\Theta _{u,v}^T)^{-1}\nabla \phi \big )\circ \Theta _{u,v}^{-1}\,. \end{aligned}$$

Performing the change of variables $({{\bar{x}}},\bar{z})=\Theta _{u,v}(x,z)$ in (2.5) with $\theta =\phi _v$ and using (1.3) give

$$\begin{aligned} \begin{aligned} \int _{\Omega (u)} \sigma \, J_v&(D\Theta _{u,v})^{-1} (D\Theta _{u,v}^T)^{-1}\nabla \xi _v\cdot \nabla \phi \,\textrm{d}(x,z)\\&= -\int _{\Omega (u)} \sigma \, J_v\, (D\Theta _{u,v})^{-1}\nabla h_v\circ \Theta _{u,v}\cdot \nabla \phi \,\textrm{d}(x,z)\,, \end{aligned} \end{aligned}$$

(2.7)

where the Jacobian $J_v:=\vert \textrm{det}(D\Theta _{u,v})\vert$ is given by

$$\begin{aligned} J_{v,1}=\frac{H+v}{H+u} \;\;\text { in }\;\; \Omega _1(u)\,,\qquad J_{v,2}= 1 \;\;\text { in }\;\; \Omega _2(u)\,. \end{aligned}$$

(2.8)

Introducing the notations

$$\begin{aligned} A(v):=\sigma \,J_v\, (D\Theta _{u,v})^{-1} (D\Theta _{u,v}^T)^{-1} \end{aligned}$$

and

$$\begin{aligned} B(v):=\textrm{div}\big (\sigma \,J_v\, (D\Theta _{u,v})^{-1}\nabla h_v\circ \Theta _{u,v}\big )\,, \end{aligned}$$

we define the function

$$\begin{aligned} {\mathcal {F}}: {\mathcal {S}}_0\times H_0^1(\Omega (u))\rightarrow H^{-1}(\Omega (u))\,,\quad (v,\xi )\mapsto -\textrm{div}\big (A(v)\nabla \xi \big )-B(v) \end{aligned}$$

and observe that (2.7) is equivalent to

$$\begin{aligned} {{\mathcal {F}}}(v,\xi _v)=0\,,\quad v\in {\mathcal {S}}_0\,. \end{aligned}$$

(2.9)

We then shall use the implicit function theorem to show that $\xi _v$ depends smoothly on v. For that purpose, let us first show that ${\mathcal {F}}$ is Fréchet differentiable in ${\mathcal {S}}_0\times H_0^1(\Omega (u))$. Indeed, by (1.4), it is readily checked that

$$\begin{aligned} \nabla h_{v} \circ \Theta _{u,v} (x,z)= \textbf{1}_{\Omega _2(u)}\zeta '\big (z- u(x)+1\big ) \left( \begin{array}{c} -\partial _x v(x) \\ 1 \end{array}\right) \end{aligned}$$

is linear in v, so that its Fréchet derivative with respect to v is

$$\begin{aligned} \partial _v\big (\nabla h_{v} \circ \Theta _{u,v}\big )[\vartheta ] (x,z)=\textbf{1}_{\Omega _2(u)}\zeta '\big (z- u(x)+1\big ) \left( \begin{array}{c} -\partial _x \vartheta (x) \\ 0 \end{array}\right) \end{aligned}$$

(2.10)

for $\vartheta \in H^2(D)\cap H_0^1(D)$. Thus,

$$\begin{aligned} \big [v\mapsto \nabla h_{v} \circ \Theta _{u,v}\big ]\in C^1\big ({\mathcal {S}}_0,L_2(\Omega (u),{\mathbb {R}}^2)\big )\,. \end{aligned}$$

Moreover, $v\mapsto J_v$ and $v\mapsto (D\Theta _{u,v})^{-1}$ are continuously differentiable from ${\mathcal {S}}_0$ to $L_\infty (\Omega (u))$ and $L_\infty (\Omega (u),{\mathbb {R}}^{2\times 2})$, respectively, and we conclude that

$$\begin{aligned} v\mapsto \sigma \,J_v\, (D\Theta _{u,v})^{-1}\nabla h_v\circ \Theta _{u,v} \end{aligned}$$

is continuously differentiable from ${\mathcal {S}}_0$ to $L_2(\Omega (u),{\mathbb {R}}^{2})$. Hence

$$B\in C^1({\mathcal {S}}_0,H^{-1}(\Omega (u))).$$

The $C^1$-smoothness of $(v,\xi )\mapsto \textrm{div}(A(v)\nabla \xi )$ is proven as in [1, Theorem 5.3.2] and we have thus established that

$$\begin{aligned} {\mathcal {F}}\in C^1\big ( {\mathcal {S}}_0\times H_0^1(\Omega (u)), H^{-1}(\Omega (u)) \big )\,. \end{aligned}$$

The Lax–Milgram theorem and the open mapping theorem imply that the mapping

$$\begin{aligned} \omega \mapsto \partial _\xi {\mathcal {F}}(u,\xi _u)[\omega ] = -\textrm{div}(\sigma \nabla \omega ) \end{aligned}$$

is an isomorphism from $H_0^1(\Omega (u))$ to $H^{-1}(\Omega (u))$. Consequently, by the implicit function theorem there is a neighborhood ${\mathcal {W}}$ of $(u,\xi _u)$ in ${\mathcal {S}}_0\times H_0^1(\Omega (u))$, a neighborhood ${\mathcal {U}}$ of u in ${\mathcal {S}}_0$, and a function $\Xi \in C^1({\mathcal {U}},H_0^1(\Omega (u)))$ with $\Xi (u)=\xi _u$ such that

$$\begin{aligned} \Big ( (v,\xi )\in {\mathcal {W}} \;\;\text { with }\;\; {\mathcal {F}}(v,\xi )=0\Big ) \ \iff \ \Big ( v\in {\mathcal {U}} \;\;\text { and }\;\; \xi =\Xi (v) \Big )\, . \end{aligned}$$

By (2.3), we may assume that $(v,\xi _v)\in {\mathcal {W}}$ for $v\in {\mathcal {U}}$. Hence, $\xi _v=\Xi (v)$ for $v\in {\mathcal {U}}$ and the proof is complete. $\square$

We next compute the Fréchet derivative of the electrostatic energy on ${\mathcal {S}}_0$ and thereby provide a proof of the first part of Theorem 2.1. This computation follows the classical approach developed in [1, 6, 7] for shape derivatives and is performed in a similar way in [4] for a geometry with a flat interface instead of $\Sigma (u)$. It is worth pointing out that the non-flat transmission interface considered herein leads to additional terms. The concise formula (1.8) that we derive for the derivative g(u) of $E_\textrm{e}(u)$ reveals the importance of these contributions to the electrostatic force, as it involves terms that counteract the contributions from the top of the elastic plate when $\llbracket \sigma \rrbracket >0$. As the identification of these additional terms and the derivation of the concise formula (1.8) do not seem to be straightforward, we give a detailed proof (see also Remark 2.6 below).

Proposition 2.5

Assume (1.4). The electrostatic energy $E_\textrm{e}: {\mathcal {S}}_0\rightarrow {\mathbb {R}}$ is continuously Fréchet differentiable with

$$\begin{aligned} \partial _uE_\textrm{e}(u)[\vartheta ]= \int _D g(u)(x)\,\vartheta (x)\,\textrm{d} x \end{aligned}$$

for $u\in {\mathcal {S}}_0$ and $\vartheta \in H^2(D)\cap H_0^1(D)$, where g(u) is defined in (1.8).

Proof

The proof is quite technical and basically includes three steps. As a starting point, we shall use Lemma 2.3 which guarantees the differentiability of $E_\textrm{e}$ and yields an abstract formula for its derivative, see (2.11) below. Computing then this derivative explicitly, we first derive in (2.20) an expression involving only the trace of the gradient of $\psi _u$ on the top of the elastic plate and the jumps of $\sigma$ and the partial derivatives of $\psi _u$ on the interface. Finally, we write the interface integrals in terms of $\psi _{u,2}$ only and thus obtain the desired formula (1.8).

To be more precise, we fix $u\in {\mathcal {S}}_0$ and use the notation introduced in Lemma 2.3. Recall that, according to Lemma 2.3, there is a neighborhood ${\mathcal {U}}$ of u in ${\mathcal {S}}_0$ such that the mapping

$$\begin{aligned} v\mapsto \xi _v=\big (\psi _v-h_v\big )\circ \Theta _{u,v} \end{aligned}$$

belongs to $C^1({\mathcal {U}},H_0^1(\Omega (u)))$, the transformation $\Theta _{u,v}:\Omega (u)\rightarrow \Omega (v)$ being defined in (2.4). Now, for $v\in {\mathcal {U}}$, we use (2.6), the relation $\chi _v=\psi _v-h_v$, and the change of variable $(\bar{x},{{\bar{z}}})=\Theta _{u,v}(x,z)$ in the integral defining $E_\textrm{e}(v)$ to obtain

$$\begin{aligned} E_\textrm{e}(v) = -\frac{1}{2}\int _{\Omega (v)} \sigma \vert \nabla \psi _v\vert ^2\,\textrm{d}({{\bar{x}}},{{\bar{z}}}) = -\frac{1}{2} \int _{\Omega (u)} \sigma |j(v)|^2 J_v\,\textrm{d}(x,z)\,, \end{aligned}$$

where

$$\begin{aligned} j(v):=(D\Theta _{u,v}^T)^{-1}\nabla \xi _v+\nabla h_v\circ \Theta _{u,v}\,. \end{aligned}$$

Owing to the differentiability of $v\mapsto \xi _v$ in ${\mathcal {U}}$, we deduce that the Fréchet derivative of $E_\textrm{e}$ at u applied to some $\vartheta \in H^2(D)\cap H_0^1(D)$ is given by

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

Taking the identity $j(u) =\nabla \chi _u + \nabla h_u=\nabla \psi _u$ into account, we infer from (2.8) that

$$\begin{aligned} \begin{aligned} \partial _u E_\textrm{e}(u)[\vartheta ]=\,&-\int _{\Omega (u)} \sigma \nabla \psi _u\cdot \big (\partial _v j(v)[\vartheta ]\big \vert _{v=u}\big )\,\textrm{d}(x,z)\\&-\frac{1}{2}\int _{\Omega _1(u)} \sigma _1 \vert \nabla \psi _{u,1}\vert ^2\, \frac{\vartheta }{H+u}\,\textrm{d}(x,z)\,. \end{aligned} \end{aligned}$$

(2.11)

We next use that $\Theta _{u,u}$ is the identity on $\Omega (u)$ and that $\xi _u=\chi _u$ to compute from the definition of j(v) that

$$\begin{aligned} \begin{aligned} \partial _v j(v)[\vartheta ]\big \vert _{v=u}=\,&-\partial _v (D\Theta _{u,v}^T)[\vartheta ]\big \vert _{v=u} \nabla \chi _u + \partial _v (\nabla \xi _v)[\vartheta ]\big \vert _{v=u}\\&+\partial _v (\nabla h_v\circ \Theta _{u,v})[\vartheta ]\big \vert _{v=u}\,. \end{aligned} \end{aligned}$$

(2.12)

Now, $\chi _{u,1}=\psi _{u,1}$ in $\Omega _1(u)$ due to (1.4), so that

$$\begin{aligned} -\partial _v (D\Theta _{u,v}^T)[\vartheta ]\big \vert _{v=u} \nabla \chi _u=-\partial _z\psi _{u}\nabla \left( \frac{\vartheta (z+H)}{H+u}\right) \quad \text { in }\ \Omega _1(u)\,, \end{aligned}$$

(2.13)

while

$$\begin{aligned} -\partial _v (D\Theta _{u,v}^T)[\vartheta ]\big \vert _{v=u} \nabla \chi _u=-\left( \begin{array}{c} \partial _z\chi _u \partial _x\vartheta \\ 0\end{array}\right) \quad \text { in }\ \Omega _2(u)\,. \end{aligned}$$

(2.14)

Also note that

$$\begin{aligned} \partial _v (\nabla \xi _v)[\vartheta ]\big \vert _{v=u}=\nabla \big (\partial _v \xi _v[\vartheta ]\big \vert _{v=u}\big )\quad \text { in }\ \Omega (u)\,. \end{aligned}$$

(2.15)

Consequently, gathering (2.11)–(2.15) and recalling (2.10) lead us to

$$\begin{aligned} \partial _u E_\textrm{e}(u)[\vartheta ] = I_0(u)[\vartheta ] + I_1(u)[\vartheta ] + I_2(u)[\vartheta ]\,, \end{aligned}$$

(2.16)

where

$$\begin{aligned} I_0(u)[\vartheta ]:= & {} -\int _{\Omega (u)} \sigma \, \nabla \psi _u\cdot \nabla \big ( \partial _v \xi _v[\vartheta ]\big \vert _{v=u}\big )\,\textrm{d}(x,z)\,,\\ I_1(u)[\vartheta ]:= & {} \int _{\Omega _1(u)} \sigma _1\, \partial _z\psi _{u,1}\, \nabla \psi _{u,1}\cdot \nabla \left( \frac{\vartheta (z+H)}{H+u}\right) \,\textrm{d}(x,z)\\{} & {} -\frac{1}{2}\int _{\Omega _1(u)} \sigma _1\, \vert \nabla \psi _{u,1}\vert ^2 \, \frac{\vartheta }{H+u} \,\textrm{d}(x,z)\,, \end{aligned}$$

and

$$\begin{aligned} I_2(u)[\vartheta ] :=\,&\int _{\Omega _2(u)} \sigma _2\, \partial _x\psi _{u,2}\, \zeta '\big (z- u+1\big )\, \partial _x\vartheta \,\textrm{d}(x,z)\\&+\int _{\Omega _2(u)} \sigma _2\, \partial _x\psi _{u,2}\,\partial _z\chi _{u,2}\, \partial _x\vartheta \, \textrm{d}(x,z)\,. \end{aligned}$$

We are left with simplifying these three integrals and begin with $I_0(u)[\vartheta ]$. We use Gauß’ theorem and (1.2a) to get

$$\begin{aligned} \begin{aligned} I_0(u)[\vartheta ]=\,&-\int _{\partial \Omega (u)} \big ( \partial _v \xi _v[\vartheta ]\big \vert _{v=u} \big )\sigma \nabla \psi _u\cdot \textbf{n}_{\partial \Omega (u)}\,\textrm{d}S\\&-\int _{\Sigma (u)}\llbracket \partial _v \xi _v[\vartheta ]\big \vert _{v=u} \sigma \nabla \psi _u\rrbracket \cdot \textbf{n}_{\Sigma (u)}\,\textrm{d}S\,. \end{aligned} \end{aligned}$$

Now, recall that $\partial _v \xi _v[\vartheta ]\big \vert _{v=u}$ belongs to $H_0^1(\Omega (u))$ according to Lemma 2.3. On the one hand, this entails that $\partial _v \xi _v[\vartheta ]\big \vert _{v=u}$ vanishes on $\partial \Omega (u)$, so that the first integral on the right-hand side of the above identity is zero. On the other hand, the $H^1$-regularity of $\partial _v \xi _v[\vartheta ]\big \vert _{v=u}$ also implies that $\llbracket \partial _v \xi _v[\vartheta ]\big \vert _{v=u} \rrbracket =0$ on $\Sigma (u)$, so that

$$\begin{aligned} \llbracket \partial _v \xi _v[\vartheta ]\big \vert _{v=u} \sigma \nabla \psi _u\rrbracket \cdot \textbf{n}_{\partial \Sigma (u)}= \partial _v \xi _v[\vartheta ]\big \vert _{v=u} \,\llbracket \sigma \nabla \psi _u\rrbracket \cdot \textbf{n}_{\Sigma (u)}=0\quad \text {on }\ \Sigma (u) \end{aligned}$$

due to (1.2b). Therefore,

$$\begin{aligned} I_0(u)[\vartheta ] = 0 \,. \end{aligned}$$

(2.17)

We next deal with $I_1(u)[\vartheta ]$. Since $\sigma _1 \Delta \psi _{u,1} = \textrm{div}(\sigma \nabla \psi _u) = 0$ in $\Omega _1(u)$ by (1.2a), it follows from Gauß’ theorem that

$$\begin{aligned} I_1(u)[\vartheta ] =\,&\int _{\Omega _1(u)} \sigma _1\, \partial _z\psi _{u,1}\, \textrm{div}\left( \left( \frac{\vartheta (z+H)}{H+u}\right) \nabla \psi _{u,1} \right) \,\textrm{d}(x,z)\\&-\frac{1}{2}\int _{\Omega _1(u)} \sigma _1\, \vert \nabla \psi _{u,1}\vert ^2 \, \frac{\vartheta }{H+u} \,\textrm{d}(x,z) \\ =\,&\int _{\partial \Omega _1(u)}\sigma _1\, \frac{\vartheta (z+H)}{H+u} \partial _z\psi _{u,1}\nabla \psi _{u,1}\cdot \textbf{n}_{\partial \Omega _1(u)}\,\textrm{d}S\\&- \int _{\Omega _1(u)} \sigma _1\, \nabla \psi _{u,1}\cdot \nabla \left( \partial _z\psi _{u,1}\,\right) \frac{\vartheta (z+H)}{H+u}\,\textrm{d}(x,z)\\&-\frac{1}{2}\int _{\Omega _1(u)} \sigma _1\, \vert \nabla \psi _{u,1}\vert ^2 \, \frac{\vartheta }{H+u} \,\textrm{d}(x,z) \,. \end{aligned}$$

Recalling that $\vartheta \in H_0^1(D)$ and noticing that

$$\nabla \psi _{u,1}\cdot \nabla \left( \partial _z\psi _{u,1}\,\right) =\partial _z\big (|\nabla \psi _{u,1}|^2)/2\,,$$

we further obtain

$$\begin{aligned} I_1(u)[\vartheta ] =\,&\int _{D} \sigma _1\, \partial _z \psi _{u,1}(x,u(x)) \big ( - \partial _x u \partial _x \psi _{u,1} + \partial _z \psi _{u,1} \big )(x,u(x)) \vartheta (x) \,\textrm{d}x \\&- \frac{1}{2} \int _D \sigma _1\, |\nabla \psi _{u,1}(x,u(x))|^2 \vartheta (x)\,\textrm{d}x \,. \end{aligned}$$

Hence,

$$\begin{aligned} \begin{aligned} I_1(u)[\vartheta ] =\,&-\frac{1}{2} \int _{D} \sigma _1\, \left( |\partial _x\psi _{u,1}|^2 - |\partial _z\psi _{u,1}|^2 \right) (x,u(x)) \vartheta (x)\,\textrm{d}x \\&- \int _{D} \sigma _1\, \partial _x u(x)\big ( \partial _x\psi _{u,1} \partial _z\psi _{u,1} \big )(x,u(x)) \vartheta (x) \,\textrm{d}x\,. \end{aligned} \end{aligned}$$

(2.18)

Finally, using (1.4a), $\chi _u=\psi _u-h_u$, and $\vartheta \in H_0^1(D)$, it follows from Green’s formula that

$$\begin{aligned} I_2(u)[\vartheta ] =\,&\int _{\Omega _2(u)} \sigma _2\, \partial _x\psi _{u,2} \partial _z\psi _{u,2} \partial _x\vartheta \,\textrm{d}(x,z) \\ =\,&-\int _{D} \sigma _2\, \big ( \partial _x\psi _{u,2} \partial _z\psi _{u,2} \big )(x,u(x)+d) \partial _x u(x)\,\textrm{d}x\\&+ \int _{D} \sigma _2\, \big ( \partial _x\psi _{u,2} \partial _z\psi _{u,2} \big )(x,u(x)) \partial _x u(x)\,\textrm{d}x\\&- \int _{\Omega _2(u)} \sigma _2\, \partial _x \big ( \partial _x\psi _{u,2} \partial _z\psi _{u,2}\big ) \vartheta \,\textrm{d}(x,z) \,. \end{aligned}$$

Owing to (1.2a), we have $\sigma _2 \partial _x^2\psi _{u,2} = -\sigma _2 \partial _z^2\psi _{u,2}$ in $\Omega _2(u)$ from which we deduce that

$$\begin{aligned}&\int _{\Omega _2(u)} \sigma _2\, \partial _x \big ( \partial _x\psi _{u,2} \partial _z\psi _{u,2}\big ) \vartheta \,\textrm{d}(x,z) \\&\quad = \int _{\Omega _2(u)} \sigma _2\, \big ( \partial _x^2\psi _{u,2} \partial _z\psi _{u,2} + \partial _x\psi _{u,2} \partial _x\partial _z\psi _{u,2} \big ) \vartheta \,\textrm{d}(x,z) \\&\quad = \int _{\Omega _2(u)} \sigma _2\, \big ( - \partial _z\psi _{u,2} \partial _z^2\psi _{u,2}+ \partial _x\psi _{u,2} \partial _x\partial _z\psi _{u,2} \big ) \vartheta \,\textrm{d}(x,z) \\&\quad = \frac{1}{2} \int _{\Omega _2(u)} \sigma _2\, \partial _z \big ( |\partial _x\psi _{u,2}|^2 - |\partial _z\psi _{u,2}|^2 \big ) \vartheta \,\textrm{d}(x,z) \\&\quad = \frac{1}{2} \int _{D} \sigma _2\, \big ( |\partial _x\psi _{u,2}|^2 - |\partial _z\psi _{u,2}|^2 \big )(x,u(x)+d) \vartheta (x) \,\textrm{d}x \\&\qquad - \frac{1}{2} \int _{D} \sigma _2\, \big ( |\partial _x\psi _{u,2}|^2 - |\partial _z\psi _{u,2}|^2 \big )(x,u(x)) \vartheta (x) \,\textrm{d}x\,. \end{aligned}$$

Consequently,

$$\begin{aligned} I_2(u)[\vartheta ] =\,&-\int _{D} \sigma _2\, \big ( \partial _x\psi _{u,2} \partial _z\psi _{u,2} \big )(x,u(x)+d) \partial _x u(x)\,\textrm{d}x\\&+ \int _{D} \sigma _2\, \big ( \partial _x\psi _{u,2} \partial _z\psi _{u,2} \big )(x,u(x)) \partial _x u(x)\,\textrm{d}x\\&- \frac{1}{2} \int _{D} \sigma _2\, \big ( |\partial _x\psi _{u,2}|^2 - |\partial _z\psi _{u,2}|^2 \big )(x,u(x)+d) \vartheta (x) \,\textrm{d}x \\&+ \frac{1}{2} \int _{D} \sigma _2\, \big ( |\partial _x\psi _{u,2}|^2 - |\partial _z\psi _{u,2}|^2 \big )(x,u(x)) \vartheta (x) \,\textrm{d}x\,. \end{aligned}$$

We finally note that

$$\begin{aligned} \partial _x\psi _{u,2}(x,u(x)+d))=-\partial _x u(x) \partial _z\psi _{u,2}(x,u(x)+d)\,, \end{aligned}$$

since $\psi _{u,2}(x,u(x)+d)=V$ owing to (1.2c) and (1.4b). This identity allows us to simplify further the formula for $I_2(u)[\vartheta ]$, so that we end up with

$$\begin{aligned} \begin{aligned} I_2(u)[\vartheta ] =\,&\frac{1}{2} \int _{D} \sigma _2\, |\nabla \psi _{u,2}(x,u(x)+d)|^2\, \textrm{d}x \\&+ \int _{D} \sigma _2\, \big ( \partial _x\psi _{u,2} \partial _z\psi _{u,2} \big )(x,u(x)) \partial _x u(x)\,\textrm{d}x\\&+ \frac{1}{2} \int _{D} \sigma _2\, \big ( |\partial _x\psi _{u,2}|^2 - |\partial _z\psi _{u,2}|^2 \big )(x,u(x)) \vartheta (x) \,\textrm{d}x\,. \end{aligned} \end{aligned}$$

(2.19)

Collecting (2.16), (2.17), (2.18), and (2.19) gives

$$\begin{aligned} \begin{aligned} \partial _u E_\textrm{e}(u)[\vartheta ]=\,&- \frac{1}{2}\int _D \llbracket \sigma (\partial _x\psi _u)^2 -\sigma (\partial _z\psi _u)^2 \rrbracket (x,u(x))\,\vartheta (x)\,\textrm{d}x\\&-\int _D \partial _x u(x)\,\llbracket \sigma \partial _x\psi _u\partial _z\psi _u \rrbracket (x,u(x))\,\vartheta (x)\,\textrm{d}x\\&+\frac{1}{2} \int _{D}\sigma _2\, \big \vert \nabla \psi _{u,2}(x,u(x)+d)\big \vert ^2\,\vartheta (x)\,\textrm{d}x\,. \end{aligned} \end{aligned}$$

(2.20)

Finally, we shall write (2.20) only in terms of $\psi _{u,2}$. To this end, we set

$$\begin{aligned} F_u := \partial _x \psi _u + \partial _x u \partial _z \psi _u\,, \qquad G_u := - \partial _x u \partial _x \psi _u + \partial _z \psi _u\,, \end{aligned}$$

(2.21)

and observe that differentiating the transmission condition $\llbracket \psi _u \rrbracket = 0$ on $\Sigma (u)$, along with the second transmission condition in (1.2b), ensures that

$$\begin{aligned} \llbracket F_u \rrbracket = \llbracket \sigma G_u \rrbracket = 0 \;\;\text { on }\;\; \Sigma (u)\,. \end{aligned}$$

These properties in turn imply that

$$\begin{aligned} \llbracket \sigma F_u^2 \rrbracket \!=\! \llbracket \sigma \rrbracket F_{u,2}^2\,, \quad \llbracket \sigma F_u G_u \rrbracket \!=\! 0\,, \qquad \llbracket \sigma G_u^2 \rrbracket \!=\! \llbracket \frac{1}{\sigma } \rrbracket \sigma _2^2 G_{u,2}^2 \;\;\text { on }\;\; \Sigma (u)\,. \end{aligned}$$

(2.22)

Guided by (2.22), we next express the jump terms in (2.20) using $F_u$ and $G_u$. Since

$$\begin{aligned} \left[ 1+ (\partial _x u)^2\right] \partial _x \psi _u = F_u - G_u \partial _x u \quad \text { and }\quad \left[ 1+ (\partial _x u)^2\right] \partial _z \psi _u = F_u \partial _x u + G_u\,, \end{aligned}$$

we compute

$$\begin{aligned}&\left[ 1+ (\partial _x u)^2\right] ^2 \left[ (\partial _x\psi _u)^2 - (\partial _z\psi _u)^2 + 2 \partial _x u \partial _x\psi _u \partial _z\psi _u \right] \\&\quad = (F_u-G_u \partial _x u)^2 - (F_u \partial _x u + G_u)^2 + 2 \partial _x u (F_u - G_u \partial _x u) (F_u \partial _x u + G_u) \\&\quad = \left[ 1+ (\partial _x u)^2\right] \left( F_u^2 - 2 F_u G_u \partial _x u - G_u^2 \right) \,. \end{aligned}$$

Therefore, by (2.22),

$$\begin{aligned}&\left[ 1+ (\partial _x u)^2\right] \llbracket \sigma (\partial _x\psi _u)^2 - \sigma (\partial _z\psi _u)^2 + 2 \sigma \partial _x u \partial _x\psi _u \partial _z\psi _u \rrbracket \\&\quad = \llbracket \sigma F_u^2 - 2 \sigma F_u G_u \partial _x u - \sigma G_u^2 \rrbracket = \llbracket \sigma \rrbracket F_{u,2}^2 - \llbracket \frac{1}{\sigma } \rrbracket \sigma _2^2 G_{u,2}^2 \\&\quad = \llbracket \sigma \rrbracket F_{u,2}^2 + \llbracket \sigma \rrbracket \frac{\sigma _2}{\sigma _1} G_{u,2}^2\,. \end{aligned}$$

Consequently, plugging this formula into (2.20) and recalling (2.21) yield

$$\begin{aligned} \begin{aligned}&\partial _u E_\textrm{e} (u)[\vartheta ]\\&\quad = - \frac{\llbracket \sigma \rrbracket }{2}\int _D\frac{1}{1+(\partial _x u(x))^2} \big (\partial _x\psi _{u,2}+\partial _xu(x)\partial _z\psi _{u,2}\big )^2 (x,u(x))\,\vartheta (x)\,\textrm{d}x\\&\qquad -\frac{\llbracket \sigma \rrbracket \sigma _2}{2\sigma _1}\int _D\frac{1}{1+(\partial _x u(x))^2}\big (\partial _x u(x)\partial _x\psi _{u,2}-\partial _z\psi _{u,2}\big )^2 (x,u(x))\,\vartheta (x)\,\textrm{d}x\\&\qquad +\frac{1}{2} \int _{D}\sigma _2\, \big \vert \nabla \psi _{u,2}(x,u(x)+d)\big \vert ^2\,\vartheta (x)\,\textrm{d}x\,; \end{aligned} \end{aligned}$$

that is,

$$\begin{aligned} \partial _uE_\textrm{e}(u)[\vartheta ]= \int _D g(u)(x)\,\vartheta (x)\,\textrm{d}x \end{aligned}$$

for $u\in {\mathcal {S}}_0$ and $\vartheta \in H^2(D)\cap H_0^1(D)$ with g(u) being defined in (1.8). It then readily follows from (2.1) that

$$\begin{aligned} \partial _uE_\textrm{e}:{\mathcal {S}}_0\rightarrow {\mathcal {L}}\big ( H^2(D)\cap H_0^1(D),{\mathbb {R}}\big ) \end{aligned}$$

is continuous. $\square$

Remark 2.6

Compared to the proof of [4, Proposition 4.2], the main difference in the proof of Proposition 2.5 is the term $I_1({u})$ stemming from the non-flatness of the interface $\Sigma ({u})$. Additionally, even though the terms $I_0({u})$ and $I_2({u})$ already appear in the flat geometry considered in [4, Proposition 4.2], they give herein different contributions to g(u) due to the specific choice (1.4) of the boundary values (1.2c).

The final step of the proof of Theorem 2.1 is to show that the electrostatic energy $E_\textrm{e}$ admits directional derivatives in the directions $-u+{\mathcal {S}}_0$.

Corollary 2.7

Assume (1.4). Let $u_0\in \bar{{\mathcal {S}}}_0$ and $u_1\in {\mathcal {S}}_0$. Then

$$\begin{aligned} \begin{aligned} \lim _{t\rightarrow 0^+} \frac{1}{t}\big [ E_\textrm{e}(&u_0+t(u_1-u_0))-E_\textrm{e}(u_0) \big ] = \int _D g(u_0)(x)\,(u_1-u_0)(x)\,\textrm{d}x\,. \end{aligned} \end{aligned}$$

Moreover, the function $g:\bar{{\mathcal {S}}}_0\rightarrow L_p(D)$ is continuous for each $p\in [1,\infty )$.

Proof

The stated continuity of g is a straightforward consequence of (2.1). Next, given $u_0\in \bar{{\mathcal {S}}}_0$ and $u_1\in {\mathcal {S}}_0$, we set

$$\begin{aligned} u_s:= u_0+s(u_1-u_0)=(1-s)u_0+su_1 \in {\mathcal {S}}_0\,,\qquad s\in (0,1]\,. \end{aligned}$$

Since $u_s\in {\mathcal {S}}_0$ for $s\in (0,1]$, we deduce from Proposition 2.5 that

$$\begin{aligned} \begin{aligned} \frac{\textrm{d}}{\textrm{d}s} E_\textrm{e}(u_s) =\,&\int _D g(u_s)(x)\, (u_1-u_0)(x)\,\textrm{d}x\,,\qquad s\in (0,1]\,. \end{aligned} \end{aligned}$$

(2.23)

Therefore, letting $s\rightarrow 0$, the continuity of g entails

$$\begin{aligned} \begin{aligned} \lim _{s\rightarrow 0^+}\frac{\textrm{d}}{\textrm{d}s} E_\textrm{e}(u_s)= \int _D g(u_0)(x)\, (u_1-u_0)(x)\,\textrm{d}x\,. \end{aligned} \end{aligned}$$

(2.24)

Now (2.2) guarantees that $E_\textrm{e}(u_s) \rightarrow E_\textrm{e}(u_0)$ as $s\rightarrow 0$, so that

$$\begin{aligned} E_\textrm{e}(u_t)-E_\textrm{e}(u_0)= \int _0^t \frac{\textrm{d}}{\textrm{d}s} E_\textrm{e}(u_s)\,\textrm{d}s\,,\quad t\in (0,1]\,, \end{aligned}$$

(2.25)

and we conclude from (2.24) that

$$\begin{aligned} \begin{aligned} \lim _{t\rightarrow 0^+} \frac{1}{t}\big (E_\textrm{e}(u_t)-E_\textrm{e}(u_0)\big )&= \lim _{t\rightarrow 0^+} \frac{1}{t}\int _0^t \frac{\textrm{d}}{\textrm{d}s} E_\textrm{e}(u_s)\,\textrm{d}s\\&=\int _D g(u_0)(x) \, (u_1-u_0)(x)\,\textrm{d}x \end{aligned} \end{aligned}$$

as claimed. $\square$

If $\llbracket \sigma \rrbracket <0$, then an obvious consequence of (1.8) is that g is non-negative on $\bar{{\mathcal {S}}}_0$. This yields the monotonicity of the electrostatic energy $E_\textrm{e}$.

Corollary 2.8

Assume $\llbracket \sigma \rrbracket <0$ and (1.4). If $u_0\in \bar{{\mathcal {S}}}_0$ and $u_1\in {\mathcal {S}}_0$ are such that $u_0\le u_1$ in D, then $E_\textrm{e}(u_0)\le E_\textrm{e}(u_1)$.

Proof

The assumption $\llbracket \sigma \rrbracket <0$ implies that $g(u_s)\ge 0$ for $s\in (0,1]$ according to (1.8), where $u_s= (1-s) u_0 + s u_1$ as in the proof of Corollary 2.7. Hence, (2.23) and (2.25) with $t=1$ imply the assertion. $\square$

3 Proof of Theorem 1.2

The proof of Theorem 1.2 now follows from  Theorem 2.1 as in [2]. Indeed, Theorem 2.1 guarantees that any minimizer of the total energy E on $\bar{{\mathcal {S}}}_0$ satisfies the Euler–Lagrange equation (1.7). In case that $a>0$, the total energy E is coercive and thus the existence of a minimizer of E in $\bar{{\mathcal {S}}}_0$ can be shown as in [2, Section 7]. In the more complex case $a=0$, the total energy E need not be coercive. But, as pointed out in the introduction, one may enforce its coercivity by adding a penalizing term and proceed along the lines of [2, Section 6], recalling that the assumption $\llbracket \sigma \rrbracket <0$ guarantees that $g(u)\ge 0$ in D, which is essential in this case (see, in particular, [2, Equation (6.4)]).