Existence of an optimal domain for the buckling load of a clamped plate with prescribed volume

Abstract

We formulate the minimization of the buckling load of a clamped plate as a free boundary value problem with a penalization term for the volume constraint. As the penalization parameter becomes small, we show that the optimal shape problem with prescribed volume is solved. In addition, we discuss two different choices for the penalization term.

1 Introduction

We consider the following variational problem. For $n \ge 2$ let $\Omega \subset \mathbb {R}^n$ be an open bounded domain. Then we define

$$\begin{aligned} {\mathcal {R}}(v,\Omega ) := \frac{\int _\Omega |\Delta v |^2 {\text {d}}x }{\int _\Omega |\nabla \! v|^2{\text {d}}x} \end{aligned}$$

for $v \in H^{2,2}_0(\Omega )$ and denote ${\mathcal {R}}(v,\Omega )=\infty$ if the denominator vanishes. The quantity

$$\begin{aligned} \Lambda (\Omega ) := \inf \{{\mathcal {R}}(v,\Omega ) : v \in H^{2,2}_0(\Omega )\} \end{aligned}$$

is called the buckling load of $\Omega$. The infimum is attained by the first eigenfunction u, which solves the following Euler-Lagrange equation

$$\begin{aligned} {\left\{ \begin{array}{ll} \Delta ^{\!2}u + \Lambda (\Omega )\Delta u &{}=0\; \text{ in } \Omega \\ u = |\nabla \! u| &{}=0 \; \text{ on } \partial \Omega \end{array}\right. } \end{aligned}$$

if the boundary of $\Omega$ is smooth enough.

In 1951, G. Polya and G. Szegö conjectured that among all domains of given measure the ball minimizes the buckling load (see [10]).

Up to now, this conjecture is still open. However, some partial results are known. In [13] Szegö proved the conjecture for all smooth plane domains under the additional assumption that $u > 0$ in $\Omega$. M. S. Ashbaugh and D. Bucur proved that among all simply connected plane domains of prescribed volume there exists an optimal domain [3]. In 1995, H. Weinberger and B. Willms proved the following uniqueness result for $n=2$, see [14]. If an optimal simply connected bounded plane domain $\Omega$ exists and if $\partial \Omega$ is smooth (at least $C^{2,\alpha }$), then $\Omega$ is a disc. This result has been extended to arbitrary dimension in 2015, see [12].

In the present paper, we will prove the following main theorem.

Theorem 1

There exists a bounded domain $\Omega ^*$ with $|\Omega ^*|=\omega _0$ such that

$$\begin{aligned} \Lambda (\Omega ^*) = \min \{\Lambda (D): D\subset B , D \text{ open }, |D|\le \omega _0\}, \end{aligned}$$

where $|D|$ denotes the n-dimensional Lebesgue measure of D, $\omega _0>0$ is a given quantity and $B \subset \mathbb {R}^n$ is a ball with $|B|>\omega _0>0$.

In order to prove Theorem 1, we introduce a penalized variational problem following an idea of H. W. Alt and L. A. Caffarelli in [2]. Let $B \subset \mathbb {R}^n$ be a ball and choose $0<\omega _0 \ll |B|$, where $|B|$ denotes the n-dimensional Lebesgue-measure of B. For $\varepsilon >0$ we define the penalization term $p^{(0)}_{\varepsilon } : \mathbb {R}\rightarrow \mathbb {R}$ by

$$\begin{aligned} p^{(0)}_{\varepsilon }(s) := {\left\{ \begin{array}{ll} \frac{1}{\varepsilon }(s-\omega _0), &{} s\ge \omega _0 \\ 0, &{} s \le \omega _0 \end{array}\right. } \end{aligned}$$

(1)

and ${\mathcal {I}}^{(0)}_{\varepsilon } : H^{2,2}_0(B) \rightarrow \mathbb {R}$ by

$$\begin{aligned} {\mathcal {I}}^{(0)}_{\varepsilon }(v) := {\mathcal {R}}(v,B) + p^{(0)}_{\varepsilon }(|\mathcal {O}(v)|), \end{aligned}$$

where $\mathcal {O}(v) := \{x\in B: v(x)\ne 0 \}$. We will see that for each $\varepsilon >0$ there exists a minimizer $u^{(0)}_{\varepsilon } \in H^{2,2}_0(B)$ for ${\mathcal {I}}^{(0)}_{\varepsilon }$ and that $u^{(0)}_{\varepsilon }$ yields a domain $\Omega (u^{(0)}_{\varepsilon })$ with $|\Omega (u^{(0)}_{\varepsilon })|=\omega _0$ which minimizes the buckling load among all open subsets B with measure smaller or equal than $\omega _0$. This will prove Theorem 1.

In view of the conjecture of Polya and Szegö, the next reasonable step would be to analyze regularity properties of the free boundary $\partial \Omega (u^{(0)}_{\varepsilon })$. If we orientate ourselves on the pioneering work of Alt and Caffarelli in [2], the next way points towards qualitative properties of the free boundary would be establishing the non-degeneracy of $u^{(0)}_{\varepsilon }$ and, subsequently, proving that the free boundary has got a positive Lebesgue-density in every point. However, looking at non-degeneracy results for the second-order problems as in [1, 2] or [5], e.g., we see that in these settings a non-degeneracy result for the minimizing function is achieved by constructing suitable testfunctions which heavily rely on properties of the Sobolev space $H^{1,2}$. In our $H^{2,2}$ setting we do not possess any comparison principle. It seems that this might be the end point for our approach via the functional ${\mathcal {I}}^{(0)}_{\varepsilon }$.

Consequently, we revise the functional ${\mathcal {I}}^{(0)}_{\varepsilon }$. Following an idea of N. Aguilera, H. W. Alt and L. A. Caffarelli in [1], we replace the penalization term $p^{(0)}_{\varepsilon }$ by a penalization term $p^{(1)}_{\varepsilon }$ which rewards volumes less than $\omega _0$ with a negative contribution to the functional. For that purpose, we define $p^{(1)}_{\varepsilon }:\mathbb {R}\rightarrow \mathbb {R}$ by

$$\begin{aligned} p^{(1)}_{\varepsilon }(s) := {\left\{ \begin{array}{ll} \frac{1}{\varepsilon }(s-\omega _0), &{} s\ge \omega _0 \\ \varepsilon (s-\omega _0), &{}s\le \omega _0 \end{array}\right. } \end{aligned}$$

(2)

and ${\mathcal {I}}^{(1)}_{\varepsilon }: H^{2,2}_0(B) \rightarrow \mathbb {R}$ by

$$\begin{aligned} {\mathcal {I}}^{(1)}_{\varepsilon }(v) := {\mathcal {R}}(v,B) + p^{(1)}_{\varepsilon }(|\mathcal {O}(v)|). \end{aligned}$$

We will prove that for every $\varepsilon >0$ there exists a minimizer $u^{(1)}_{\varepsilon } \in H^{2,2}_0(B)$ for ${\mathcal {I}}^{(1)}_{\varepsilon }$ and that $u^{(1)}_{\varepsilon }$ yields a domain $\Omega (u^{(1)}_{\varepsilon })$ which minimizes the buckling load among all open subsets of B with the same measure as $\Omega (u^{(1)}_{\varepsilon })$. The analysis of the volume of $\Omega (u^{(1)}_{\varepsilon })$ will be more challenging than in the case of $\Omega (u^{(0)}_{\varepsilon })$ since the rewarding part of the penalization term counteracts the monotonicity of the buckling load with respect to set inclusion . We will give more details on this issue in the sequel.

The present paper is organized as follows. In Sect. 2, we prove the existence of a minimizer $u^{(k)}_{\varepsilon }$ for the functional ${\mathcal {I}}^{(k)}_{\varepsilon }$ $(k=0,1)$ and show that $u^{(k)}_{\varepsilon } \in C^{1,\alpha }({\overline{B}})$ for every $\alpha \in (0,1)$. Thereby, we refine the approach presented in [11]. That paper deals with the functional ${\mathcal {I}}^{(0)}_{\varepsilon }$, but considers only $n=2$ and $n=3$. Consequently, the $C^{1,\alpha }$ regularity of $u^{(0)}_{\varepsilon }$ follows with the help of Sobolev’s Embedding Theorem. In the present paper, we consider arbitrary $n\ge 2$ and refine the approach to $C^{1,\alpha }$ regularity from [11] by applying a bootstrap argument. In Sect. 3, we analyze the volume of the domains $\Omega (u^{(0)}_{\varepsilon })$ and $\Omega (u^{(1)}_{\varepsilon })$ separately. In Sect. 3.1, we consider the functional ${\mathcal {I}}^{(0)}_{\varepsilon }$ and prove our main theorem, Theorem 1, by scaling arguments. In Sect. 3.2, we consider the functional ${\mathcal {I}}^{(1)}_{\varepsilon }$. Thus, we deal with a penalization term, which rewards volumes less than $\omega _0$ with a negative contribution to the functional. Hence, the behavior of the penalization term antagonizes the monotonicity of the buckling load with respect to set inclusion and we cannot adopt the approach from Sect. 3.1 to show that the optimal domain’s volume cannot become less than $\omega _0$. In fact, we will use an inequality by M. S. Ashbaugh and R. S. Laugesen (see [4]) to balance the rewarding penalization term and the monotonicity of the buckling load. Hence, provided $\varepsilon$ is sufficiently small, $|\Omega (u^{(1)}_{\varepsilon })| \in [\alpha _0\omega _0,\omega _0]$, where $\alpha _0 \in (\frac{1}{2},1)$ depends on $n, \omega _0$ and $\varepsilon$. After refining the choice of $\varepsilon$, we will obtain the following dichotomy: there either holds

• $|\Omega (u^{(1)}_{\varepsilon })|=\omega _0$ or

• $|\Omega (u^{(1)}_{\varepsilon })|<\omega _0$ and if $\Omega (u^{(1)}_{\varepsilon })$ is scaled to the volume $\omega _0$, this enlarged domain is neither a subset of B nor can it be translated into B.

In the first case, $u^{(1)}_{\varepsilon }$ is a minimizer of the functional ${\mathcal {I}}^{(0)}_{\varepsilon }$ and we can treat both functionals as equivalent in the sense that they are minimized by the same functions. Under the additional assumption that the free boundary $\partial \Omega (u^{(1)}_{\varepsilon })$ satisfies a doubling property, we will disprove the occurrence of the latter case. We assume that there exists a constant $\sigma >0$ such that for every $x_0 \in \partial \Omega (u^{(1)}_{\varepsilon })$ and every $0<R\le R_0$ there holds

$$\begin{aligned} |B_{2R}(x_0)\cap \Omega (u^{(1)}_{\varepsilon })| \le \sigma |B_R(x_0)\cap \Omega (u^{(1)}_{\varepsilon })|. \end{aligned}$$

This assumed doubling property enables us to establish a non-degeneracy property of the minimizing function $u^{(1)}_{\varepsilon }$ and, subsequently, we can show that the latter case of the above mentioned dichotomy cannot occur. We should emphasize that for proving the non-degeneracy of $u^{(1)}_{\varepsilon }$, besides the assumption of the doubling property, the crucial incredient is the rewarding part of the penalization term $p^{(1)}_{\varepsilon }$.

2 The penalized problems

In this section, we analyze the penalized problems

$$\begin{aligned} \min \{{\mathcal {I}}^{(k)}_{\varepsilon }(v); v \in H^{2,2}_0(B)\} \end{aligned}$$

for $k\in \{0,1\}$. Proving the existence of a minimizer $u^{(k)}_{\varepsilon }$ for every $\varepsilon >0$ is our first step. For the proof we refer to [11, Theorem 2.1].

Theorem 2

For every $\varepsilon >0$ and for $k \in \{0,1\}$ there exists a function $u^{(k)}_{\varepsilon }\in H^{2,2}_0(B)$ such that

$$\begin{aligned} {\mathcal {I}}^{(k)}_{\varepsilon }(u^{(k)}_{\varepsilon }) = \min \{{\mathcal {I}}^{(k)}_{\varepsilon }(v): v\in H^{2,2}_0(B)\}. \end{aligned}$$

Without loss of generality, we assume $u^{(k)}_{\varepsilon }$ to be normalized in the sense that

$$\begin{aligned} \int \limits _B |\nabla \! u^{(k)}_{\varepsilon }|^2 {\text {d}}x = 1 \end{aligned}$$

and denote

$$\begin{aligned} \Lambda _\varepsilon ^{(k)}:= {\mathcal {R}}(u^{(k)}_{\varepsilon }) = \int \limits _B |\Delta u^{(k)}_{\varepsilon } |^2\,{\text {d}}x. \end{aligned}$$

Note that $\mathcal {O}(u^{(k)}_{\varepsilon }) = \{u^{(k)}_{\varepsilon }(x)\ne 0\}$ is a non-empty set. By the absolute continuity of the Lebesgue integral, the n-dimensional Lebesgue measure of $\mathcal {O}(u^{(k)}_{\varepsilon })$ cannot vanish. Since the minimizer $u^{(k)}_{\varepsilon }\in H^{2,2}_0(B)$, we do not know any continuity properties of $u^{(k)}_{\varepsilon }$ yet. In particular, we do not know if $\mathcal {O}(u^{(k)}_{\varepsilon })$ is an open set. To prove continuity of $u^{(k)}_{\varepsilon }$ we will apply an idea of Q. Han and F. Lin in [8], which is based on a bootstrap argument and Morrey’s Dirichlet Growth Theorem. This ansatz will lead to the $C^{1,\alpha }$ regularity of $u^{(k)}_{\varepsilon }$ for every $\alpha \in (0,1)$ in arbitrary dimension n.

In the sequel, we will apply the following version of Morrey’s Dirichlet Growth Theorem (see [9, Theorem 3.5.2]).

Theorem 3

(Morrey’s Dirichlet Growth Theorem) Suppose $\phi \in H^{1,p}_0(B)$, $1 \le p\le n$, $0< \alpha \le 1$ and suppose there exists a constant $M>0$ such that

$$\begin{aligned} \int \limits _{B_r(x_0)\cap B}|\nabla \! \phi |^p{\text {d}}x \le M \, r^{n-p+\alpha p} \end{aligned}$$

for every $B_r(x_0)$ with $x_0 \in {\overline{B}}$. Then $\phi \in C^{0,\alpha }({\overline{B}})$.

In addition, we need a technical lemma cited from [6, Chapter III, Lemma 2.1].

Lemma 1

Let $\Phi$ be a non-negative and non-decreasing function on [0, R]. Suppose that there exist positive constants $\gamma , \alpha , \kappa , \beta$, $\beta <\alpha$, such that for all $0\le r\le R \le R_0$

$$\begin{aligned} \Phi (r) \le \gamma \left[ \left( \frac{r}{R}\right) ^\alpha + \delta \right] \Phi (R)+\kappa R^\beta . \end{aligned}$$

Then there exist positive constants $\delta _0 =\delta _0(\gamma ,\alpha ,\beta )$ and $C=C(\gamma , \alpha ,\beta )$ such that if $\delta < \delta _0$, for all $0\le r\le R\le R_0$ we have

$$\begin{aligned} \Phi (r) \le C \left( \frac{r}{R}\right) ^\beta \left[ \Phi (R) + \kappa R^\beta \right] . \end{aligned}$$

Let us fix $R_{\varepsilon ,k}$ as

$$\begin{aligned} R_{\varepsilon ,k} := \min \left\{ 1, \left( \frac{|\mathcal {O}(u^{(k)}_{\varepsilon })|}{2\omega _n}\right) ^\frac{1}{n}\right\} . \end{aligned}$$

Now suppose $x_0 \in {\overline{B}}$ and choose $0<r\le R\le R_{\varepsilon ,k}$. We define ${\hat{v}}_k \in H^{2,2}_0(B)$ by

$$\begin{aligned} {\hat{v}}_k = {\left\{ \begin{array}{ll} u^{(k)}_{\varepsilon }, &{}\text{ in } B\setminus B_R(x_0) \\ v_k, &{} \text{ in } B_R(x_0) \end{array}\right. }, \end{aligned}$$

(3)

for a $v_k \in H^{2,2}(B_R(x_0)\cap B)$ such that $v_k-u^{(k)}_{\varepsilon }\in H^{2,2}_0(B_R(x_0)\cap B)$ and $\Delta ^{\!2}v _k =0$ in $B_R(x_0)\cap B$. We restrict ourselves to case $B_R(x_0)\cap \mathcal {O}(u^{(k)}_{\varepsilon })\ne \emptyset$; otherwise, $u^{(k)}_{\varepsilon }-v_k$ vanishes in $B_R(x_0)$.

As in [11, Lemma 2.1] we prove the next lemma.

Lemma 2

Using the above notation, there exists a constant $C>0$ which only depends on n such that for $0\le r < R$ there holds

$$\begin{aligned} \int \limits _{B_r(x_0)\cap B}|D^2 v_k|^2{\text {d}}x \le C\,\left( \frac{r}{R}\right) ^n \,\int \limits _{B_R(x_0)\cap B}|D^2u^{(k)}_{\varepsilon }|^2{\text {d}}x. \end{aligned}$$

The next lemma will be the starting point for our bootstrap argument.

Lemma 3

Let $u^{(k)}_{\varepsilon }$ be a minimizer of ${\mathcal {I}}^{(k)}_{\varepsilon }$ and $v_k-u^{(k)}_{\varepsilon }\in H^{2,2}_0(B_R(x_0)\cap B)$. Then there exists a constant $C = C(n,\Lambda _\varepsilon ^{(k)},|\mathcal {O}(u^{(k)}_{\varepsilon })|) > 0$ such that for each $x_0 \in {\overline{B}}$ and each $0<R\le R_{\varepsilon ,k}$ there holds

$$\begin{aligned} \int \limits _{B_R(x_0)\cap B}|D^2 (u^{(k)}_{\varepsilon }-v_k)|^2dx \,\le \, C(n,\Lambda _\varepsilon ^{(k)},|\mathcal {O}(u^{(k)}_{\varepsilon })|)\left( R^n+\int \limits _{B_R\cap B}|\nabla \! u^{(k)}_{\varepsilon }|^2{\text {d}}x\right) . \end{aligned}$$

Proof

We obtain the result by comparing the ${\mathcal {I}}^{(k)}_{\varepsilon }$-energies of $u^{(k)}_{\varepsilon }$ and ${\hat{v}}_k$, where ${\hat{v}}_k$ is defined as in (3). Note that $\mathcal {O}(u^{(k)}_{\varepsilon }) \not \subset B_R(x_0)$ due to the definition of $R_{\varepsilon ,k}$. However, we have to distinguish two different cases depending on the volume of $\mathcal {O}({\hat{v}}_k)$. First, let us consider that $|\mathcal {O}({\hat{v}}_k)| \le |\mathcal {O}(u^{(k)}_{\varepsilon })|$. Then the minimality of $u^{(k)}_{\varepsilon }$ for ${\mathcal {I}}^{(k)}_{\varepsilon }$ implies

$$\begin{aligned} \int \limits _{B_R(x_0)\cap B}|\Delta u^{(k)}_{\varepsilon } |^2-|\Delta v _k|^ 2{\text {d}}x \le \Lambda _\varepsilon ^{(k)}\int \limits _{B_R(x_0)\cap B}|\nabla \! u^{(k)}_{\varepsilon }|^2{\text {d}}x. \end{aligned}$$

Since $v_k$ is biharmonic in $B_R(x_0)\cap B$ and $v_k-u^{(k)}_{\varepsilon }\in H^{2,2}_0(B_R(x_0)\cap B)$, we obtain

$$\begin{aligned} \int \limits _{B_R(x_0)\cap B}|D^2(u^{(k)}_{\varepsilon }-v_k)|^2{\text {d}}x \le \Lambda _\varepsilon ^{(k)}\int \limits _{B_R(x_0)\cap B}|\nabla \! u^{(k)}_{\varepsilon }|^2{\text {d}}x \end{aligned}$$

and the claim is proven.

Now let us assume that there holds

$$\begin{aligned} |\mathcal {O}(u^{(k)}_{\varepsilon })| < |\mathcal {O}({\hat{v}}_k)| \le |\mathcal {O}(u^{(k)}_{\varepsilon })|+|B_R|. \end{aligned}$$

In order to avoid the penalization term while comparing ${\mathcal {I}}^{(k)}_{\varepsilon }(u^{(k)}_{\varepsilon })$ and $\mathcal {I}_\varepsilon ({\hat{v}}_k)$ we scale ${\hat{v}}_k$. For this purpose, we set

$$\begin{aligned} \mu := \left( \frac{|\mathcal {O}({\hat{v}}_k)|}{|\mathcal {O}(u^{(k)}_{\varepsilon })|}\right) ^\frac{1}{n} >1. \end{aligned}$$

Without loss of generality, we think of B as of a ball with center in the origin and radius $R_B$. Then we denote $B^*:= B_{\mu ^{-1}{R_B}}$ and $w_k(x) := {\hat{v}}_k(\mu x)$ for $x \in B^*$. Consequently, $w_k\in H^{2,2}_0(B^*) \subset H^{2,2}_0(B)$ and $|\mathcal {O}(w_k)|=|\mathcal {O}(u^{(k)}_{\varepsilon })|$. The minimality of $u^{(k)}_{\varepsilon }$ for ${\mathcal {I}}^{(k)}_{\varepsilon }$ in $H^{2,2}_0(B)$ then implies

$$\begin{aligned} \Lambda _\varepsilon ^{(k)}\int _B|\nabla \! w_k|^2 dy \le \int \limits _B|\Delta w _k|^2{\text {d}}y \; \Leftrightarrow \; \Lambda _\varepsilon ^{(k)}\mu ^{-2}\int \limits _{B}|\nabla \! \hat{v}_k|^2dx \le \int \limits _{B}|\Delta \hat{v} _k|^2{\text {d}}x. \end{aligned}$$

Rearranging terms we obtain the local inequality

$$\begin{aligned} \int \limits _{B_R\cap B}|\Delta u^{(k)}_{\varepsilon } |^2- |\Delta v _k|^2{\text {d}}x \le \Lambda _\varepsilon ^{(k)}\left( 1-\frac{1}{\mu ^2}\right) + \frac{\Lambda _\varepsilon ^{(k)}}{\mu ^2}\left[ \,\,\int \limits _{B_R\cap B}|\nabla \! u^{(k)}_{\varepsilon }|^2-|\nabla \! v_k|^2dx\right] , \end{aligned}$$

where we denote $B_R= B_R(x_0)$ for simplicity. Since $v_k$ is biharmonic in $B_R\cap B$ and $v_k-u^{(k)}_{\varepsilon }\in H^{2,2}_0(B_R\cap B)$, we obtain

$$\begin{aligned} \int \limits _{B_R\cap B}|\Delta (u^{(k)}_{\varepsilon }-v_k) |^2{\text {d}}x \,\le \, \Lambda _\varepsilon ^{(k)}\left( 1-\frac{1}{\mu ^2}\right) + \frac{\Lambda _\varepsilon ^{(k)}}{\mu ^2}\,\int \limits _{B_R\cap B}|\nabla \! u^{(k)}_{\varepsilon }|^2{\text {d}}x. \end{aligned}$$

Note that there holds

$$\begin{aligned} 1 < \mu \le \left( 1+\frac{|B_R|}{|\mathcal {O}(u^{(k)}_{\varepsilon })|}\right) ^\frac{1}{n}. \end{aligned}$$

and, using Taylor’s expansion, we find

$$\begin{aligned} 1-\mu ^{-2} \le C(n,|\mathcal {O}(u^{(k)}_{\varepsilon })|)\,R^n. \end{aligned}$$

Thus,

$$\begin{aligned} \int \limits _{B_R(x_0)\cap B}|\Delta (u^{(k)}_{\varepsilon }-v_k) |^2{\text {d}}x \,&\le \, C(n,\Lambda _\varepsilon ^{(k)},|\mathcal {O}(u^{(k)}_{\varepsilon })|)\left( R^n+\int \limits _{B_R\cap B}|\nabla \! u^{(k)}_{\varepsilon }|^2{\text {d}}x\right) . \end{aligned}$$

$\square$

Note that if we restrict ourselves to the dimensions $n=2$ and $n=3$, we could improve the statement of Lemma 3 using Sobolev’s embedding theorem. This is how the $C^{1,\alpha }$ regularity of $u^{(0)}_{\varepsilon }$ is proven in [11] (cf. Lemma 2.2 and Theorem 2.4 in [11]). Since we now consider any $n\ge 2$, we need the bootstrapping. The next lemma is the essential tool for this new approach to the $C^{1,\alpha }$ regularity of $u^{(k)}_{\varepsilon }$. It is based on ideas of [8, Chapter 3].

Lemma 4

Suppose that for each $0\le r\le R_{\varepsilon ,k}$ there holds

$$\begin{aligned} \int \limits _{B_r(x_0)} |D^2u^{(k)}_{\varepsilon }|^2{\text {d}}x \le M\,r^\mu , \end{aligned}$$

where $M>0$ and $\mu \in [0,n)$. Then there exists a constant $C(n,|\mathcal {O}(u^{(k)}_{\varepsilon })|)>0$ such that for each $0\le r\le R_{\varepsilon ,k}$

$$\begin{aligned} \int \limits _{B_r(x_0)}|\nabla \! u^{(k)}_{\varepsilon }|^2{\text {d}}x \le C(n,|\mathcal {O}(u^{(k)}_{\varepsilon })|)(1+M)\,r^\lambda , \end{aligned}$$

where $\lambda = \mu +2$ if $\mu < n-2$ and $\lambda$ is arbitrary in (0, n) if $n-2\le \mu < n$.

Proof

Let $0\le r\le s \le R_{\varepsilon ,k}$. For a function $w \in H^{2,2}(B)$ we set

$$\begin{aligned} (w)_{r,x_0} := \mathop {\fint }\limits _{B_r(x_0)}w\,{\text {d}}x = \frac{1}{|B_r(x_0)|}\int \limits _{B_r(x_0)}w\,{\text {d}}x. \end{aligned}$$

Using this notation we write

$$\begin{aligned} \int \limits _{B_r(x_0)}|\nabla \! u^{(k)}_{\varepsilon }|^2{\text {d}}x = \sum _{i=1}^n \int \limits _{B_r(x_0)}|\partial _iu^{(k)}_{\varepsilon }- (\partial _iu^{(k)}_{\varepsilon })_{s,x_0} + (\partial _iu^{(k)}_{\varepsilon })_{s,x_0}|^2{\text {d}}x. \end{aligned}$$

Then Young’s inequality implies

$$\begin{aligned} \int \limits _{B_r(x_0)}|\nabla \! u^{(k)}_{\varepsilon }|^2{\text {d}}x&\le 2 \sum _{i=1}^n \left( \int \limits _{B_r(x_0)}(\partial _iu^{(k)}_{\varepsilon })^2_{s,x_0}{\text {d}}x + \int \limits _{B_r(x_0)}|\partial _iu^{(k)}_{\varepsilon }- (\partial _iu^{(k)}_{\varepsilon })_{s,x_0}|^2{\text {d}}x \right) \\&\le 2 \sum _{i=1}^n \left( |B_r| \left( \mathop {\fint }\limits _{B_s(x_0)}\partial _iu^{(k)}_{\varepsilon }\,{\text {d}}x \right) ^2 + \int \limits _{B_s(x_0)}|\partial _iu^{(k)}_{\varepsilon }- (\partial _iu^{(k)}_{\varepsilon })_{s,x_0}|^2{\text {d}}x \right) . \end{aligned}$$

Applying Hölder’s and a local version of Poincaré’s inequality, we find that

$$\begin{aligned} \int \limits _{B_r(x_0)}|\nabla \! u^{(k)}_{\varepsilon }|^2{\text {d}}x&\le C(n) \left[ \left( \frac{r}{s}\right) ^n \int \limits _{B_s(x_0)}|\nabla \! u^{(k)}_{\varepsilon }|^2{\text {d}}x + s^2\int \limits _{B_s(x_0)}|D^2u^{(k)}_{\varepsilon }|^2{\text {d}}x \right] , \end{aligned}$$

where the constant C only depends on n. By assumption, we can proceed to

$$\begin{aligned} \int \limits _{B_r(x_0)}|\nabla \! u^{(k)}_{\varepsilon }|^2{\text {d}}x&\le C(n) \left[ \left( \frac{r}{s}\right) ^n \int \limits _{B_s(x_0)}|\nabla \! u^{(k)}_{\varepsilon }|^2{\text {d}}x +M\, s^{\mu +2} \right] . \end{aligned}$$

Now Lemma 1 implies that for each $0\le r\le s\le R_\varepsilon$ there holds

$$\begin{aligned} \int \limits _{B_r(x_0)}|\nabla \! u^{(k)}_{\varepsilon }|^2{\text {d}}x&\le C(n) \left( \frac{r}{s}\right) ^\lambda \left[ \int \limits _{B_{s}(x_0)}|\nabla \! u^{(k)}_{\varepsilon }|^2{\text {d}}x + M\,s^\lambda \right] . \end{aligned}$$

where $\lambda = \mu +2$ if $\mu < n-2$ and $\lambda$ is arbitrary in (0, n) if $n-2 \le \mu < n$. Now we choose $s=R_\varepsilon$. Recalling that $R_\varepsilon$ depends on $|\mathcal {O}(u^{(k)}_{\varepsilon })|$ we deduce

$$\begin{aligned} \int \limits _{B_r(x_0)}|\nabla \! u^{(k)}_{\varepsilon }|^2{\text {d}}x&\le C(n,|\mathcal {O}(u^{(k)}_{\varepsilon })|)(1+M)r^\lambda . \end{aligned}$$

$\square$

Now we are ready to prove the main theorem of this section.

Theorem 4

For every $\varepsilon >0$ and $k\in \{0,1\}$ every minimizer $u^{(k)}_{\varepsilon }$ of ${\mathcal {I}}^{(k)}_{\varepsilon }$ is in $C^{1,\alpha }({\overline{B}})$ for every $\alpha \in (0,1)$.

Proof

Our aim is to show that for each $x_0\in {\overline{B}}$ and every $0\le r \le R_{\varepsilon ,k}$ there holds

$$\begin{aligned} \int \limits _{B_r(x_0)\cap B}|D^2u^{(k)}_{\varepsilon }|^2 {\text {d}}x \le C(n,\Lambda _\varepsilon ^{(k)},|\mathcal {O}(u^{(k)}_{\varepsilon })|)\,r^{n-2+2\alpha } \end{aligned}$$

(4)

for each $\alpha \in (0,1)$. Then Theorem 3 finishes the proof if we choose $\phi = \partial _i u^{(k)}_{\varepsilon }$ for $i \in \{1,\ldots ,n\}$. We prove (4) by using a bootstrap argument. Let $x_0 \in {\overline{B}}$ and $0\le r\le R\le R_{\varepsilon ,k}$. Then there obviously holds

$$\begin{aligned} \int \limits _{B_r(x_0)\cap B}|D^2u^{(k)}_{\varepsilon }|^2dx \le 2 \int \limits _{B_r(x_0)}|D^2v_k|^2dx + 2 \int \limits _{B_R(x_0)}|D^2(u^{(k)}_{\varepsilon }-v_k)|^2dx, \end{aligned}$$

where $v_k$ is defined in (3). Due to Lemma 2 and Lemma 3 we obtain

$$\begin{aligned} \begin{aligned} \int \limits _{B_r(x_0)\cap B}|D^2u^{(k)}_{\varepsilon }|^2dx \le&\, C \left( \frac{r}{R}\right) ^n\int \limits _{B_R(x_0)}|D^2u^{(k)}_{\varepsilon }|^2dx \\&+ C(n,\Lambda _\varepsilon ^{(k)},|\mathcal {O}(u^{(k)}_{\varepsilon })|)\left( R^n + \int \limits _{B_R(x_0)}|\nabla \! u^{(k)}_{\varepsilon }|^2dx\right) . \end{aligned} \end{aligned}$$

(5)

Now we start the bootstrapping. For every $0\le r\le R_{\varepsilon ,k}$ there holds

$$\begin{aligned} \int \limits _{B_r(x_0)\cap B}|D^2u^{(k)}_{\varepsilon }|^2dx \le \int \limits _B|D^2u^{(k)}_{\varepsilon }|^2dx = \Lambda _\varepsilon ^{(k)}= \Lambda _\varepsilon ^{(k)}\,r^0. \end{aligned}$$

(6)

Applying Lemma 4, estimate (6) yields for every $0\le r \le R_{\varepsilon ,k}$

$$\begin{aligned} \int \limits _{B_r(x_0)}|\nabla \! u^{(k)}_{\varepsilon }|^2dx \le C(n,\Lambda _\varepsilon ^{(k)},|\mathcal {O}(u^{(k)}_{\varepsilon })|)\,r^{\lambda _0}, \end{aligned}$$

where $\lambda _0 \in (0,n)$ if $n=2$ and $\lambda _0 = 2$ if $n\ge 3$. We insert this estimate in (5). This yields

$$\begin{aligned} \int \limits _{B_r(x_0)\cap B}|D^2u^{(k)}_{\varepsilon }|^2dx&\le C \left( \frac{r}{R}\right) ^n\int \limits _{B_R(x_0)\cap B}|D^2u^{(k)}_{\varepsilon }|^2dx + C(n,\Lambda _\varepsilon ^{(k)},|\mathcal {O}(u^{(k)}_{\varepsilon })|))R^{\lambda _0} \end{aligned}$$

for every $0\le r \le R\le R_{\varepsilon ,k}$. Applying Lemma 1, we obtain

$$\begin{aligned} \int \limits _{B_r(x_0)\cap B}|D^2u^{(k)}_{\varepsilon }|^2dx&\le C \left( \frac{r}{R}\right) ^{\lambda _0}\left( \int \limits _{B_R(x_0)\cap B}|D^2u^{(k)}_{\varepsilon }|^2dx + C(n,\Lambda _\varepsilon ^{(k)},|\mathcal {O}(u^{(k)}_{\varepsilon })|)R^{\lambda _0}\right) \end{aligned}$$

for $0\le r\le R\le R_{\varepsilon ,k}$. Then choosing $R =R_{\varepsilon ,k}$ gives us

$$\begin{aligned} \int \limits _{B_r(x_0)\cap B}|D^2u^{(k)}_{\varepsilon }|^2dx \le C(n,\Lambda _\varepsilon ^{(k)},|\mathcal {O}(u^{(k)}_{\varepsilon })|)\,r^{\lambda _0} \end{aligned}$$

(7)

for every $0\le r\le R_{\varepsilon ,k}$. If $n=2$, this is (4). If $n\ge 3$, (7) is an improvement of estimate (6). In this case, we again apply Lemma 4 and obtain

$$\begin{aligned} \int \limits _{B_r(x_0)}|\nabla \! u^{(k)}_{\varepsilon }|^2dx \le C(n,\Lambda _\varepsilon ^{(k)},|\mathcal {O}(u^{(k)}_{\varepsilon })|)\,r^{\lambda _1}, \end{aligned}$$

where $\lambda _1 \in (0,n)$ if $n\in \{3,4\}$ and $\lambda _1 = 4$ if $n>4$. Together with estimate (5) we find that

$$\begin{aligned} \int \limits _{B_r(x_0)\cap B} |D^2u^{(k)}_{\varepsilon }|^2dx \le C \left( \frac{r}{R}\right) ^n \int \limits _{B_R(x_0)\cap B} |D^2u^{(k)}_{\varepsilon }|^2dx + C(n,\Lambda _\varepsilon ^{(k)},|\mathcal {O}(u^{(k)}_{\varepsilon })|)R^{\lambda _1} \end{aligned}$$

for every $0\le r\le R\le R_{\varepsilon ,k}$. Then Lemma 1 implies

$$\begin{aligned} \int \limits _{B_r(x_0)\cap B} |D^2u^{(k)}_{\varepsilon }|^2dx \le C \left( \frac{r}{R}\right) ^{\lambda _1}\left( \int \limits _{B_R(x_0)\cap B}|D^2u^{(k)}_{\varepsilon }|^2dx + C(n,\Lambda _\varepsilon ^{(k)},|\mathcal {O}(u^{(k)}_{\varepsilon })|)R^{\lambda _1}\right) \end{aligned}$$

for every $0\le r\le R\le R_{\varepsilon ,k}$. Choosing $R=R_{\varepsilon ,k}$ we gain

$$\begin{aligned} \int \limits _{B_r(x_0)\cap B} |D^2u^{(k)}_{\varepsilon }|^2dx \le C(n,\Lambda _\varepsilon ^{(k)},|\mathcal {O}(u^{(k)}_{\varepsilon })|)\,r^{\lambda _1}. \end{aligned}$$

(8)

For $n\in \{3,4\}$, estimate (8) proves the claim. For $n\ge 5$, we repeat the argumentation. In view of (8), Lemma 4 implies

$$\begin{aligned} \int \limits _{B_r(x_0)}|\nabla \! u^{(k)}_{\varepsilon }|^2dx \le C(n,\Lambda _\varepsilon ^{(k)},|\mathcal {O}(u^{(k)}_{\varepsilon })|)\,r^{\lambda _2}, \end{aligned}$$

where $\lambda _2= 6$ if $n> 6$ and $\lambda _2$ is arbitrary in (0, n) if $n\in \{5,6\}$. Again, we insert this estimate in (5) and deduce an improvement of (7). Repeating this process proves the claim after finite many steps for every $n\ge 2$. $\square$

The continuity of $u^{(k)}_{\varepsilon }$ implies that $\mathcal {O}(u^{(k)}_{\varepsilon })$ is an open set. Then classical variational arguments show that $u^{(k)}_{\varepsilon }$ solves

$$\begin{aligned} \Delta ^{\!2}u^{(k)}_{\varepsilon } + \Lambda _\varepsilon ^{(k)}\,\Delta u^{(k)}_{\varepsilon } =0 \; \text{ in } \mathcal {O}(u^{(k)}_{\varepsilon }). \end{aligned}$$

Moreover, $C^{1,\alpha }$ regularity of $u^{(k)}_{\varepsilon }$ allows us to split $\partial \mathcal {O}(u^{(k)}_{\varepsilon })$ in the following two parts

$$\begin{aligned} \Gamma _{\varepsilon ,k}:= \{ x\in \partial \mathcal {O}(u^{(k)}_{\varepsilon }): |\nabla \! u^{(k)}_{\varepsilon }(x)|=0 \} \; \text{ and } \, \Sigma _{\varepsilon ,k} := \{ x\in \partial \mathcal {O}(u^{(k)}_{\varepsilon }): |\nabla \! u^{(k)}_{\varepsilon }(x)|>0 \}. \end{aligned}$$

Then $\Sigma _{\varepsilon ,k}$ is part of a nodal line of $u^{(k)}_{\varepsilon }$ and, consequently, ${\mathcal {L}}^n(\Sigma _{\varepsilon ,k})=0$ for all $\varepsilon >0$. We define

$$\begin{aligned} \Omega (u^{(k)}_{\varepsilon }) := \mathcal {O}(u^{(k)}_{\varepsilon }) \cup \Sigma _{\varepsilon ,k} \end{aligned}$$

and call $\partial \Omega (u^{(k)}_{\varepsilon })=\Gamma _{\varepsilon ,k}$ the free boundary.

Remark 1

Note that $\Omega (u^{(k)}_{\varepsilon })$ is an open set in $\mathbb {R}^n$ and $|\Omega (u^{(k)}_{\varepsilon })| = |\mathcal {O}(u^{(k)}_{\varepsilon })|$. Moreover, the minimizer $u^{(k)}_{\varepsilon }$ solves

$$\begin{aligned} {\left\{ \begin{array}{ll} \Delta ^{\!2}u^{(k)}_{\varepsilon } + \Lambda _\varepsilon ^{(k)}\Delta u^{(k)}_{\varepsilon } =0, &{} \text{ in } \Omega (u^{(k)}_{\varepsilon }) \\ u^{(k)}_{\varepsilon }= |\nabla u^{(k)}_{\varepsilon }| =0, &{} \text{ on } \partial \Omega (u^{(k)}_{\varepsilon }). \end{array}\right. } \end{aligned}$$

The following lemma shows that, considering the functional ${\mathcal {I}}^{(1)}_{\varepsilon }$, the set $\Omega (u^{(1)}_{\varepsilon })$ is connected. This result is a direct consequence of the strict monotonicity of the penalization term $p^{(1)}_{\varepsilon }$.

Lemma 5

For every minimizer $u^{(1)}_{\varepsilon } \in H^{2,2}_0(B)$ of ${\mathcal {I}}^{(1)}_{\varepsilon }$ the set $\Omega (u^{(1)}_{\varepsilon })$ is connected.

Proof

We prove the claim by contradiction. Without loss of generality we assume that $\Omega (u^{(1)}_{\varepsilon })$ consists of two connected components, namely $\Omega _1$ and $\Omega _2$ with $|\Omega _k|\ne 0$ for $k=1,2$. By $u_k$ we denote

$$\begin{aligned} u_k := {\left\{ \begin{array}{ll} u^{(1)}_{\varepsilon }, &{} \text{ in } \Omega _k \\ 0,&{} \text{ otherwise } \end{array}\right. }. \end{aligned}$$

The minimality of $u^{(1)}_{\varepsilon }$ for ${\mathcal {I}}^{(1)}_{\varepsilon }$ implies

$$\begin{aligned} {\mathcal {I}}^{(1)}_{\varepsilon }(u^{(1)}_{\varepsilon }) = \Lambda ^{(1)}_\varepsilon + p^{(1)}_{\varepsilon }(|\Omega _1|+|\Omega _2|) \le {\mathcal {R}}(u_1) + p^{(1)}_{\varepsilon }(|\Omega _1|) = {\mathcal {I}}^{(1)}_{\varepsilon }(u_1). \end{aligned}$$

Since $\Vert \nabla \! u^{(1)}_{\varepsilon }\Vert _{L^2(B)} = 1$, there holds

$$\begin{aligned}&\left( \int \limits _{\Omega _1}|\Delta u _1| ^2dx + \int \limits _{\Omega _2}|\Delta u _2|^2dx\right) \left( 1 - \int \limits _{\Omega _2}|\nabla \! u_2|^2dx\right) \\& \quad \le \int \limits _{\Omega _1}|\Delta u _1|^2dx + \left( p^{(1)}_{\varepsilon }(|\Omega _1|)-p^{(1)}_{\varepsilon }(|\Omega _1|+|\Omega _2|)\right) \int \limits _{\Omega _1}|\nabla \! u_1|^2dx. \end{aligned}$$

Now the strict monotonicity of $p^{(1)}_{\varepsilon }$ implies

$$\begin{aligned} \int \limits _{\Omega _2}|\Delta u _2|^2dx < \Lambda ^{(1)}_\varepsilon \,\int \limits |\nabla \! u_2|^2dx, \end{aligned}$$

which is equivalent to

$$\begin{aligned} {\mathcal {R}}(u_2) < \Lambda ^{(1)}_\varepsilon . \end{aligned}$$

(9)

Comparing ${\mathcal {I}}^{(1)}_{\varepsilon }(u^{(1)}_{\varepsilon })$ and ${\mathcal {I}}^{(1)}_{\varepsilon }(u_2)$ we find

$$\begin{aligned} {\mathcal {I}}^{(1)}_{\varepsilon }(u^{(1)}_{\varepsilon }) = \Lambda ^{(1)}_\varepsilon + p^{(1)}_{\varepsilon }(|\Omega _1|+|\Omega _2|) \le {\mathcal {R}}(u_2) + p^{(1)}_{\varepsilon }(|\Omega _2|) = {\mathcal {I}}^{(1)}_{\varepsilon }(u_2). \end{aligned}$$

Then (9) implies

$$\begin{aligned} \Lambda ^{(1)}_\varepsilon + p^{(1)}_{\varepsilon }(|\Omega _1|+|\Omega _2|) < \Lambda ^{(1)}_\varepsilon +p^{(1)}_{\varepsilon }(|\Omega _2|). \end{aligned}$$

Obviously, this is contradictory since $p^{(1)}_{\varepsilon }$ is strictly increasing. $\square$

Let us emphasize that for the proof of Lemma 5 the actual value of $|\Omega (u^{(1)}_{\varepsilon })|$ is irrelevant since $p^{(1)}_{\varepsilon }$ is strictly increasing. Considering the functional ${\mathcal {I}}^{(0)}_{\varepsilon }$, and thus the only non-decreasing penalization term $p^{(0)}_{\varepsilon }$, we have to ensure that $|\Omega (u^{(0)}_{\varepsilon })|\ge \omega _0$ before we are able to copy the approach of Lemma 5 and can deduce that $\Omega (u^{(0)}_{\varepsilon })$ is connected. This will be done in Theorem 5 and Lemma 6.

The next corollary collects direct consequences of Lemma 5

Corollary 1

For every minimizer $u^{(1)}_{\varepsilon }$ of ${\mathcal {I}}^{(1)}_{\varepsilon }$ the domain $\Omega (u^{(1)}_{\varepsilon })$ satisfies $\Lambda ^{(1)}_\varepsilon =\Lambda (\Omega (u^{(1)}_{\varepsilon }))$ and $\Omega (u^{(1)}_{\varepsilon })$ is an optimal domain for minimizing the buckling eigenvalue among all domains in B with the same measure as $\Omega (u^{(1)}_{\varepsilon })$.

Proof

Let $u^{(1)}_{\varepsilon } \in H^{2,2}_0(B)$ minimizes the functional ${\mathcal {I}}^{(u^{(1)}_{\varepsilon })}_{\varepsilon }$. The minimality of $u^{(1)}_{\varepsilon }$ for ${\mathcal {I}}^{(1)}_{\varepsilon }$ then implies

$$\begin{aligned} \Lambda ^{(1)}_\varepsilon + p^{(1)}_{\varepsilon }(|\Omega (u^{(1)}_{\varepsilon })|)&= \min \{{\mathcal {I}}^{(1)}_{\varepsilon }(v): v\in H^{2,2}_0(B)\} \\&\le \min \{{\mathcal {R}}(v) + p^{(1)}_{\varepsilon }(|\Omega (v)|): v\in H^{2,2}_0(B), |\Omega (v)|=|\Omega (u^{(2)}_{\varepsilon })|\} \\&\le \min \{{\mathcal {R}}(v) :v \in H^{2,2}_0(\Omega (u^{(k)}_{\varepsilon })) \} + p^{(1)}_{\varepsilon }(|\Omega (u^{(1)}_{\varepsilon })|) \\&= \Lambda (\Omega (u^{(1)}_{\varepsilon })) + p^{(2)}_{\varepsilon }(|\Omega (u^{(1)}_{\varepsilon })|) \\&\le \Lambda ^{(1)}_\varepsilon + p^{(1)}_{\varepsilon }(|\Omega (u^{(1)}_{\varepsilon })|). \end{aligned}$$

Thus, the minimizer $u^{(1)}_{\varepsilon }$ is a buckling eigenfunction on $\Omega (u^{(1)}_{\varepsilon })$. Moreover, there holds

$$\begin{aligned} \Lambda (\Omega (u^{(1)}_{\varepsilon })) = \min \{{\mathcal {R}}(v): v\in H^{2,2}_0(B), |\Omega (v)|=|\Omega (u^{(1)}_{\varepsilon })|\}. \end{aligned}$$

Hence, $\Omega (u^{(1)}_{\varepsilon })$ is an optimal domain for minimizing the buckling eigenvalue among all domains in B with the same measure as $\Omega (u^{(1)}_{\varepsilon })$. $\square$

3 The volume condition

In the following, we analyze the functionals ${\mathcal {I}}^{(0)}_{\varepsilon }$ and ${\mathcal {I}}^{(1)}_{\varepsilon }$ seperately.

3.1 The non-rewarding penalization term

In this section, we consider the functional ${\mathcal {I}}^{(0)}_{\varepsilon }$. We will show that for every $\varepsilon >0$ and every minimizer $u^{(0)}_{\varepsilon }$ of ${\mathcal {I}}^{(0)}_{\varepsilon }$ the volume of the set $\Omega (u^{(0)}_{\varepsilon })$ cannot fall below the value $\omega _0$. This observation allows us to adopt the proof of Lemma 5 to show that $\Omega (u^{(0)}_{\varepsilon })$ is connected. Thus, analog to Corollary 1, $\Omega (u^{(0)}_{\varepsilon })$ minimizes the buckling load among all open subsets of B with the same measure as $\Omega (u^{(0)}_{\varepsilon })$.

Recall that the buckling load is decreasing and the penalization term $p^{(0)}_{\varepsilon }$ is non-decreasing with respect to set inclusion. Since the penalization term grows with slope $\varepsilon ^{-1}$ for arguments larger than $\omega _0$, it seems to be natural that the optimal domain $\Omega (u^{(0)}_{\varepsilon })$ adjusts itself to the volume $\omega _0$ provided that $\varepsilon$ is chosen small enough. Indeed, the fact that $|\Omega (u^{(0)}_{\varepsilon })| \le \omega _0$ for sufficiently small $\varepsilon$ is eventually a consequence of the scaling property of the buckling load, i.e., $\Lambda (M) = t^2\Lambda (tM)$ for $t>0$. Finally, we finish the present section with the proof of Theorem 1.

The following theorem shows that the volume of the set $\Omega (u^{(0)}_{\varepsilon })$ cannot fall below $\omega _0$.

Theorem 5

For every $\varepsilon >0$ there holds $|\Omega (u^{(0)}_{\varepsilon })| \ge \omega _0$ for every minimizer $u^{(0)}_{\varepsilon }$ of ${\mathcal {I}}^{(0)}_{\varepsilon }$.

The proof of Theorem 5 follows exactly the idea of the proof of [11, Theorem 2.2]. However, since Theorem 5 is crucial for proving our main theorem, Theorem 1, we present the proof in detail.

Proof of Theorem 5

We prove the claim by contradiction. Let us assume that for an $\varepsilon >0$ there exists a minimizer $u^{(0)}_{\varepsilon }$ of ${\mathcal {I}}^{(0)}_{\varepsilon }$ such that $|\Omega (u^{(0)}_{\varepsilon })|<\omega _0$. Now choose an $x_0\in \partial \Omega (u^{(0)}_{\varepsilon })\setminus \partial B$ and a radius $r>0$ such that

$$\begin{aligned} |B_r(x_0)\cap \{x\in B:u^{(0)}_{\varepsilon }(x)\ne 0\}|>0 \end{aligned}$$

(10)

and $|\Omega (u^{(0)}_{\varepsilon })\cup B_r(x_0)|\le \omega _0$. Note that such an $x_0 \in \partial \Omega (u^{(0)}_{\varepsilon })\setminus \partial B$ exists since we assume $|\Omega (u^{(0)}_{\varepsilon })|<\omega _0$. Let $v \in H^{2,2}_0(B)$ with $v-u^{(0)}_{\varepsilon } \in H^{2,2}_0(B_r(x_0))$ and

$$\begin{aligned} \Delta ^{\!2}v + \Lambda ^{(0)}_\varepsilon \Delta v = 0 \text{ in } B_r(x_0). \end{aligned}$$

We set

$$\begin{aligned} {\hat{v}} := {\left\{ \begin{array}{ll} u^{(0)}_{\varepsilon }, &{} \text{ in } B\setminus B_r(x_0)\\ v, &{} \text{ in } B_r(x_0) \end{array}\right. } \end{aligned}$$

and compare the ${\mathcal {I}}^{(0)}_{\varepsilon }$-energies of $u^{(0)}_{\varepsilon }$ and ${\hat{v}}$. This leads to the following local inequality:

$$\begin{aligned} \int \limits _{B_r(x_0)}|\Delta u^{(0)}_{\varepsilon }|^2-|\Delta v |^2dx \le \Lambda ^{(0)}_\varepsilon \int \limits _{B_r(x_0)}|\nabla \! u^{(0)}_{\varepsilon }|^2-|\nabla \! v|^2dx. \end{aligned}$$

(11)

Using integration by parts and the definition of v, we obtain

$$\begin{aligned} \int \limits _{B_r(x_0)}|\Delta u^{(0)}_{\varepsilon }|^2-|\Delta v |^2dx = \int \limits _{B_r(x_0)}|\Delta ( u^{(0)}_{\varepsilon }-v)|^2 + 2\Lambda ^{(0)}_\varepsilon \int \limits _{B_r(x_0)}|\nabla \! ( u^{(0)}_{\varepsilon }- v)|^2dx. \end{aligned}$$

Thus, (11) becomes

$$\begin{aligned} \int \limits _{B_r(x_0)}|\Delta u^{(0)}_{\varepsilon }|^2-|\Delta v |^2dx \le \Lambda ^{(0)}_\varepsilon \int \limits _{B_r(x_0)}|\nabla \! ( u^{(0)}_{\varepsilon }- v)|^2dx \end{aligned}$$

and Poincaré’s inequality yields

$$\begin{aligned} \int \limits _{B_r(x_0)}|\Delta u^{(0)}_{\varepsilon }|^2-|\Delta v |^2dx \le \Lambda ^{(0)}_\varepsilon r^2 \int \limits _{B_r(x_0)}|\Delta ( u^{(0)}_{\varepsilon }- v)|^2dx. \end{aligned}$$

(12)

Provided that the integral in (12) does not vanish, (12) is contradictory for sufficently small r.

If the integral in (12) vanishes, there holds $u^{(0)}_{\varepsilon }\equiv v$ in $B_r(x_0)$. Consequently, $u^{(0)}_{\varepsilon }$ is analytic in $B_\frac{r}{2}(x_0)$ since v is there analytic as a solution of an ellipitc equation. However, then $u^{(0)}_{\varepsilon }$ vanishes in $B_\frac{r}{2}(x_0)$ because of (10). This is contradictory since $x_0\in \partial \Omega (u^{(0)}_{\varepsilon })$. This proves the claim. $\square$

Now we are able to adopt the proof of Lemma 5 to show that $\Omega (u^{(0)}_{\varepsilon })$ is connected.

Lemma 6

For every $\varepsilon >0$ and every minimizer $u^{(0)}_{\varepsilon }$ of ${\mathcal {I}}^{(0)}_{\varepsilon }$ the set $\Omega (u^{(0)}_{\varepsilon })$ is connected.

Proof

We follow the idea of the proof of Lemma 5 and assume that $\Omega (u^{(0)}_{\varepsilon })=\Omega _1 {\dot{\cup }} \Omega _2$ with $|\Omega _k|>0$ for $k = 1,2$. Let us first consider that $|\Omega (u^{(0)}_{\varepsilon })|>\omega _0$. Then there holds

$$\begin{aligned} p^{(0)}_{\varepsilon }(|\Omega (u^{(0)}_{\varepsilon })|) > p^{(0)}_{\varepsilon }(\Omega _k) \end{aligned}$$

(13)

for $k= 1,2$. Hence, arguing as in the proof of Lemma 5 and replacing the strict monotonicity of $p^{(1)}_{\varepsilon }$ by (13), we arrive at a contradiction.

Now let us assume that $|\Omega (u^{(0)}_{\varepsilon })|=\omega _0$. Then

$$\begin{aligned} p^{(0)}_{\varepsilon }(|\Omega (u^{(0)}_{\varepsilon })|) = p^{(0)}_{\varepsilon }(\Omega _k) =0 \end{aligned}$$

(14)

and arguing as in the proof of Lemma 5 we obtain

$$\begin{aligned} {\mathcal {I}}^{(0)}_{\varepsilon }(u_2)={\mathcal {R}}(u_2) \le \Lambda ^{(0)}_\varepsilon = {\mathcal {I}}^{(0)}_{\varepsilon }(u^{(0)}_{\varepsilon }). \end{aligned}$$

This implies that $u_2$ minimizes the fucntional ${\mathcal {I}}^{(0)}_{\varepsilon }$. Since $\Omega (u_2)=\Omega _2$ and $|\Omega _2|<\omega _0$ this is contradictory to Theorem 5. Thus, the claim is proven. $\square$

As a consequence of Lemma 6, we get the analog to Corollary 1.

Corollary 2

For every minimizer $u^{(0)}_{\varepsilon }$ of ${\mathcal {I}}^{(0)}_{\varepsilon }$ the set $\Omega (u^{(0)}_{\varepsilon })$ is a domain and $\Lambda ^{(0)}_\varepsilon = \Lambda (\Omega (u^{(0)}_{\varepsilon }))$. In addition, $\Omega (u^{(0)}_{\varepsilon })$ minimzes the buckling load among all open subsets of B with the same measure as $\Omega (u^{(0)}_{\varepsilon })$.

The following remark will be helpful to show that for an appropriate choice of $\varepsilon$ the n-dimensional Lebesgue measure of $\Omega (u^{(0)}_{\varepsilon })$ cannot become larger than $\omega _0$.

Remark 2

Note that the reference domain B compactly contains a ball $B_R(x_0)$ with $|B_R|=\omega _0$. Let $\phi \in H^{2,2}_0(B_R(x_0))$ denote the buckling eigenfunction on $B_R(x_0)$, i.e.,

$$\begin{aligned} \Lambda (B_R(x_0)) = \min \{{\mathcal {R}}(v,B_R(x_0)): v\in H^{2,2}_0(B_R(x_0))\} = {\mathcal {R}}(\phi ,B_R(x_0)). \end{aligned}$$

Consequently, for every $\varepsilon >0$ there holds

$$\begin{aligned} {\mathcal {I}}^{(0)}_{\varepsilon }(\phi ) = {\mathcal {I}}^{(1)}_{\varepsilon }(\phi ) ={\mathcal {R}}(\phi ,B_R(x_0)) = \Lambda (B_R(x_0)) = \left( \frac{\omega _n}{\omega _0}\right) ^\frac{2}{n}\Lambda (B_1). \end{aligned}$$

Theorem 6

Let $u^{(0)}_{\varepsilon }$ be a minimizer for ${\mathcal {I}}^{(0)}_{\varepsilon }$. Then there exists a number $\varepsilon _1 = \varepsilon _1(n,\omega _0)$ such that for $0<\varepsilon \le \varepsilon _1$ there holds

$$\begin{aligned} |\Omega (u^{(0)}_{\varepsilon })| = \omega _0. \end{aligned}$$

Proof

We claim that the statement of the theorem holds true for

$$\begin{aligned} \varepsilon _1 = \varepsilon _1(n,\omega _0) = \left( \frac{\omega _0}{\omega _n}\right) ^\frac{2}{n}\,\frac{\omega _0}{\Lambda (B_1)}. \end{aligned}$$

(15)

We prove by contradiction. Thus, we choose $\varepsilon \le \varepsilon _1$ and denote by $u^{(0)}_{\varepsilon }$ a minimizer of ${\mathcal {I}}^{(0)}_{\varepsilon }$. Assume that there exists a number $\alpha >1$ such that

$$\begin{aligned} |\Omega (u^{(0)}_{\varepsilon })| = \alpha \omega _0. \end{aligned}$$

(16)

Our aim is to contradict (16).

Since $\Omega (u^{(0)}_{\varepsilon }) \subset B$ and $\alpha >1$, the scaled domain $\Omega ' := \alpha ^{-\frac{1}{n}}\Omega (u^{(0)}_{\varepsilon })$ is also contained in B and satisfies $|\Omega '|=\omega _0$. Let $\psi \in H^{2,2}_0(B)$ denote the first buckling eigenfunction on $\Omega '$. Then the minimality of $u^{(0)}_{\varepsilon }$ for ${\mathcal {I}}^{(0)}_{\varepsilon }$ implies

$$\begin{aligned} \Lambda (\Omega (u^{(0)}_{\varepsilon })) + \frac{\omega _0}{\varepsilon }(\alpha -1) = {\mathcal {I}}^{(0)}_{\varepsilon }(u^{(0)}_{\varepsilon }) \le {\mathcal {I}}^{(0)}_{\varepsilon }(\psi ) = \Lambda (\Omega '). \end{aligned}$$

(17)

By scaling we have

$$\begin{aligned} \Lambda (\Omega ') = \Lambda (\alpha ^{-\frac{1}{n}}\Omega (u^{(0)}_{\varepsilon })) = \alpha ^\frac{2}{n}\Lambda (\Omega (u^{(0)}_{\varepsilon })). \end{aligned}$$

From (17) we then get

$$\begin{aligned} \frac{\omega _0}{\varepsilon }(\alpha -1) \le \left( \alpha ^\frac{2}{n}-1\right) \Lambda (\Omega (u^{(0)}_{\varepsilon })). \end{aligned}$$

(18)

Now let $\phi \in H^{2,2}_0(B)$ be as in Remark 2. Then, due to the assumption (16), there holds

$$\begin{aligned} \Lambda (\Omega (u^{(0)}_{\varepsilon })) < {\mathcal {I}}^{(0)}_{\varepsilon }(u^{(0)}_{\varepsilon }) \le {\mathcal {I}}^{(0)}_{\varepsilon }(\phi ) = \left( \frac{\omega _n}{\omega _0}\right) ^\frac{2}{n}\Lambda (B_1) \end{aligned}$$

and estimate (18) becomes

$$\begin{aligned} \left( \frac{\omega _0}{\omega _n}\right) ^\frac{2}{n}\,\frac{\omega _0}{\Lambda (B_1)}\,\frac{\alpha -1}{\alpha ^\frac{2}{n}-1}< \varepsilon . \end{aligned}$$

Hence,

$$\begin{aligned} \frac{\alpha -1}{\alpha ^\frac{2}{n}-1} < \frac{\varepsilon }{\omega _0}\,\left( \frac{\omega _n}{\omega _0}\right) ^\frac{2}{n}\Lambda (B_1). \end{aligned}$$

With (15) this implies

$$\begin{aligned} 1 \le \frac{\alpha -1}{\alpha ^\frac{2}{n}-1} < \varepsilon \,\varepsilon _1^{-1} \end{aligned}$$

since $\alpha >1$ and $n\ge 2$. Thus, for any $\varepsilon \le \varepsilon _1$ we get a contradiction and the assumption (16) cannot hold true for any $0<\alpha <1$. Hence, $|\Omega (u^{(0)}_{\varepsilon })|\ge \omega _0$ if $\varepsilon \le \varepsilon _1$ .Together with Theorem 5, this proves the claim. $\square$

Finally, the proof of the main theorem, Theorem 1, is a direct consequence of the previous results in this section.

Proof of Theorem 1

Let us choose $\varepsilon \le \varepsilon _1$, where $\varepsilon _1$ is given in Theorem 6 and let $u^{(0)}_{\varepsilon }$ be a minimizer of ${\mathcal {I}}^{(0)}_{\varepsilon }$. Then Theorem 6 implies that $|\Omega ( u^{(0)}_{\varepsilon })|=\omega _0$ and, due to Corollary 2, there holds

$$\begin{aligned} {\mathcal {I}}^{(0)}_{\varepsilon }(u^{(0)}_{\varepsilon })=\Lambda (\Omega (u^{(0)}_{\varepsilon })). \end{aligned}$$

Now choose an open set $D \subset B$ with $|D| \le \omega _0$ and denote by $u_D \in H^{2,2}_0(D)$ the buckling eigenfunction on D. Then the minimality of $u^{(0)}_{\varepsilon }$ for ${\mathcal {I}}^{(0)}_{\varepsilon }$ implies

$$\begin{aligned} \Lambda (\Omega (u^{(0)}_{\varepsilon })) = {\mathcal {I}}^{(0)}_{\varepsilon }(u^{(0)}_{\varepsilon }) \le {\mathcal {I}}^{(0)}_{\varepsilon }(u_D) = \Lambda (D). \end{aligned}$$

Hence,

$$\begin{aligned} \Lambda (\Omega (u^{(0)}_{\varepsilon })) = \min \{\Lambda (D) : D\subset B, D \text{ open, } |D| \le \omega _0\}. \end{aligned}$$

(19)

In addition, $\Omega (u^{(0)}_{\varepsilon })$ is connected (see Lemma 6). This proves Theorem 1. $\square$

Since the existence of an optimal domain among all open subsets of B of given volume is now proven, the next reasonable step would be a qualitative analysis of the free boundary $\partial \Omega (u^{(0)}_{\varepsilon })$. Following [2], our next aims would by establishing a non-degeneracy result for $u^{(0)}_{\varepsilon }$. Considering the second-order problems (e.g., [1, 2, 5]), these non-degeneracy results are achieved by applying comparision principles, which which are not available for fourth order operators in general. One possible way out of this difficulty is to replace the penalization term $p^{(0)}_{\varepsilon }$ by the rewarding penalization term $p^{(1)}_{\varepsilon }$. This will be discussed in the next section.

3.2 The rewarding penalization term

In this section, we consider the functional ${\mathcal {I}}^{(1)}_{\varepsilon }$. Analog to Theorem 6 we will find that $|\Omega (u^{(1)}_{\varepsilon })|$ cannot become larger than $\omega _0$ provided that $\varepsilon \le \varepsilon _1$.

It remains to exclude that $\Omega (u^{(1)}_{\varepsilon })<\omega _0$. This will be more involved since adopting a scaling argument like the one we used in the proof of Theorem 6 collapses if we cannot guarantee that the scaled version of $\Omega (u^{(1)}_{\varepsilon })$ is still contained in the reference domain B.

Choosing the parameter $\varepsilon$ sufficiently small, we will see that one of the following two situations occurs: either $|\Omega (u^{(1)}_{\varepsilon })|=\omega _0$ or $|\Omega (u^{(1)}_{\varepsilon })|<\omega _0$ and the rescaled domain $\Omega '$ with $|\Omega '|=\omega _0$ cannot be translated into the reference domain B.

In the first case, $u^{(1)}_{\varepsilon }$ is a minimizer of the functional ${\mathcal {I}}^{(0)}_{\varepsilon }$ and $\Omega (u^{(1)}_{\varepsilon })$ is an optimal domain for minimizing the buckling load in the sense of Theorem 1. Thus, in this case, we can treat the functionals ${\mathcal {I}}^{(0)}_{\varepsilon }$ and ${\mathcal {I}}^{(1)}_{\varepsilon }$ as equivalent.

In the second case, we may think of the domain $\Omega (u^{(1)}_{\varepsilon })$ as of a domain with thin tentacles, which may all touch the boundary of the reference domain B. These tentacles eludes the scaling. Consequently, in this case a more local analysis of $\partial \Omega (u^{(1)}_{\varepsilon })$ is needed. Exemplary, we will see that assuming that $\partial \Omega (u^{(1)}_{\varepsilon })$ satisfies a doubling condition, the domain $\Omega (u^{(1)}_{\varepsilon })$ fulfills $|\Omega (u^{(1)}_{\varepsilon })|=\omega _0$.

We begin this section with the analog result to Theorem 6.

Theorem 7

For $\varepsilon \le \varepsilon _1$ every minimizer $u^{(1)}_{\varepsilon }$ of ${\mathcal {I}}^{(1)}_{\varepsilon }$ satisfies $|\Omega (u^{(1)}_{\varepsilon })|\le \omega _0$. Thereby, $\varepsilon _1$ is the number given in Theorem 6.

Proof

Let us assume that $\varepsilon \le \varepsilon _1$ and that $|\Omega (u^{(1)}_{\varepsilon })|=\alpha \omega _0$ for an $\alpha >1$. Then arguing in exactly the same way as in the proof of Theorem 6 leads to a contradiction. $\square$

We would like to repeat the idea of the proof of Theorem 6 assuming that $|\Omega (u^{(1)}_{\varepsilon })|=\alpha \omega _0$ for some $\alpha \in (0,1)$ and then define the scaled domain $\Omega ' = \alpha ^{-\frac{1}{n}}\Omega (u^{(1)}_{\varepsilon })$. However, it is not clear if the enlarged domain $\Omega '$ is still contained in B. A partial result can be obtained with the help of an argument of M. S. Ashbaugh and R. S. Laugesen.

Remark 3

In [4], M. S. Ashbaugh and R. S. Laugesen showed that there exists a constant $c_n \in (0,1)$, only depending on the dimension n, such that for every domain $\Omega \subset \mathbb {R}^n$ there holds

$$\begin{aligned} \Lambda (\Omega ) > c_n \, \Lambda (\Omega ^\#), \end{aligned}$$

where $\Omega ^\#$ denotes a ball in $\mathbb {R}^n$ with the same volume as $\Omega$. In addition, $c_n$ tends to 1 as n tends to infinity.

Theorem 8

Let $u^{(1)}_{\varepsilon }$ be a minimizer of ${\mathcal {I}}^{(1)}_{\varepsilon }$ for $\varepsilon \le \varepsilon _1$. Then there exists a number $\alpha _0 = \alpha _0(n,\varepsilon _1,\varepsilon )$ such that

$$\begin{aligned} |\Omega (u^{(1)}_{\varepsilon })| \ge \alpha _0 \omega _0. \end{aligned}$$

Moreover, we have the explicit representation

$$\begin{aligned} \alpha _0 = \frac{1+\varepsilon \varepsilon _1-\sqrt{1+2\varepsilon \varepsilon _1+(\varepsilon \varepsilon _1)^2-4c_n\varepsilon \varepsilon _1}}{2\varepsilon \varepsilon _1}, \end{aligned}$$

where $\varepsilon _1$ is given by Theorem 6and $c_n$ is given by Remark 3.

Note that

$$\begin{aligned} \lim _{n\rightarrow \infty }\alpha _0 = 1 \quad \text{ and } \lim _{\varepsilon \rightarrow 0}\alpha _0 = c_n. \end{aligned}$$

Together with Theorem 7 this shows that for $\varepsilon \le \varepsilon _1$ the domain $\Omega (u^{(1)}_{\varepsilon })$ satisfies the volume condition asymptotically as the dimension n approaches infinity.

Proof of Theorem 8

Let $u^{(1)}_{\varepsilon }$ be a minimizer of ${\mathcal {I}}^{(1)}_{\varepsilon }$ and let $\alpha \in (0,1)$ be such that

$$\begin{aligned} |\Omega (u^{(1)}_{\varepsilon })| = \alpha \omega _0. \end{aligned}$$

Choose $\phi$ as in Remark 2. Then the minimality of $u^{(1)}_{\varepsilon }$ for ${\mathcal {I}}^{(1)}_{\varepsilon }$ implies

$$\begin{aligned} \Lambda (\Omega (u^{(1)}_{\varepsilon })) - \varepsilon \omega _0 (1-\alpha ) = {\mathcal {I}}^{(1)}_{\varepsilon }(u^{(1)}_{\varepsilon }) \le {\mathcal {I}}^{(1)}_{\varepsilon }(\phi ) = \left( \frac{\omega _n}{\omega _0}\right) ^\frac{2}{n}\Lambda (B_1). \end{aligned}$$

(20)

By $\Omega (u^{(1)}_{\varepsilon })^\#$ we denote the ball centered in the origin, having the same volume as $\Omega (u^{(1)}_{\varepsilon })$. Then

$$\begin{aligned} \Lambda (\Omega (u^{(1)}_{\varepsilon })^\#) = \left( \frac{\omega _n}{\alpha \omega _0}\right) ^\frac{2}{n}\Lambda (B_1) \end{aligned}$$

and applying Remark 3 we obtain from (20)

$$\begin{aligned}&c_n\,\Lambda (\Omega (u^{(1)}_{\varepsilon })^\#)- \left( \frac{\omega _n}{\omega _0}\right) ^\frac{2}{n}\Lambda (B_1)< \varepsilon \,\omega _0\,(1-\alpha ) \\ \Rightarrow \quad&\left( \frac{\omega _n}{\omega _0}\right) ^\frac{2}{n}\Lambda (B_1) \left( c_n\alpha ^{-\frac{2}{n}}-1\right)< \varepsilon \,\omega _0\,(1-\alpha ) \\ \Rightarrow \quad&\frac{c_n\alpha ^{-\frac{2}{n}}-1}{1-\alpha } < \varepsilon \,\left( \frac{\omega _0}{\omega _n}\right) ^\frac{2}{n}\frac{\omega _0}{\Lambda (B_1)} = \varepsilon \,\varepsilon _1. \end{aligned}$$

Since $\alpha <1$, there holds $\alpha ^{-\frac{2}{n}}\ge \alpha ^{-1}$ and we have

$$\begin{aligned} \frac{c_n\alpha ^{-1}-1}{1-\alpha } \le \varepsilon \,\varepsilon _1. \end{aligned}$$

(21)

We set $f(\alpha ) := \frac{c_n\alpha ^{-1}-1}{1-\alpha }$. Then $f: (0,1)\rightarrow \mathbb {R}$ is smooth, strictly decreasing and

$$\begin{aligned} \lim _{\alpha \rightarrow 1}\;f(\alpha ) = -\infty \; \text{ and } \lim _{\alpha \rightarrow 0}\;f(\alpha )=\infty . \end{aligned}$$

By $\alpha _0$ we denote the (unique) solution in (0, 1) for equality in (21), i.e.

$$\begin{aligned} \alpha _0 = \frac{1+\varepsilon \varepsilon _1-\sqrt{1+2\varepsilon \varepsilon _1+(\varepsilon \varepsilon _1)^2-4c_n\varepsilon \varepsilon _1}}{2\varepsilon \varepsilon _1}. \end{aligned}$$

Consequently, the strict monotonicity of f implies that (21) can only hold true for $\alpha \in [\alpha _0,1)$. This proves the theorem. $\square$

From now on, we always consider $0<\varepsilon \le \varepsilon _1$. Consequently, there holds

$$\begin{aligned} \alpha _0\omega _0 \le |\Omega (u^{(1)}_{\varepsilon })| \le \omega _0. \end{aligned}$$

Theorem 9

There exists a number $\varepsilon _0 = \varepsilon _0(n,\omega _0)$ such that for $\varepsilon \le \varepsilon _0$ every minimizer $u^{(1)}_{\varepsilon }$ of ${\mathcal {I}}^{(1)}_{\varepsilon }$ satisfies either

• (a) $|\Omega (u^{(1)}_{\varepsilon })| = \omega _0$       or

• (b) $|\Omega (u^{(1)}_{\varepsilon })|< \omega _0$ and the rescaled domain $t\,\Omega (u^{(1)}_{\varepsilon })$ with $|t\,\Omega (u^{(1)}_{\varepsilon })|=\omega _0$ is not a subset of B. In addition, there exists no translation $\Phi : \mathbb {R}^n\rightarrow \mathbb {R}^n$ such that $\Phi (t\,\Omega (u^{(1)}_{\varepsilon }))$ is contained in B.

Proof

We claim that the statement of the theorem holds true for

$$\begin{aligned} \varepsilon _0 := \min \left\{ \varepsilon _1, c_n\frac{\Lambda (B_1)}{\omega _0}\frac{2}{n}\left( \frac{\omega _n}{\omega _0}\right) ^\frac{2}{n}\right\} = \min \left\{ \varepsilon _1, c_n\frac{2}{n}\varepsilon _1^{-1}\right\} , \end{aligned}$$

(22)

where $\varepsilon _1$ is given in Theorem 6 and $c_n$ is given in Remark 3. We prove by contradiction. Let $0<\varepsilon \le \varepsilon _0$ and $u^{(1)}_{\varepsilon }$ be a minimizer of ${\mathcal {I}}^{(1)}_{\varepsilon }$. Recall that Theorem 8 implies $|\Omega (u^{(1)}_{\varepsilon })| \in [\alpha _0\omega _0,\omega _0]$ since we assume $\varepsilon \le \varepsilon _1$. Now let us assume that there exists an $\alpha \in [\alpha _0,1)$ such that

$$\begin{aligned} |\Omega (u^{(1)}_{\varepsilon })| = \alpha \omega _0 . \end{aligned}$$

(23)

We split the proof in two steps.

Step 1.    Assume $\varepsilon \le \varepsilon _0$ and let $u^{(0)}_{\varepsilon }$ be a minimizer of ${\mathcal {I}}^{(0)}_{\varepsilon }$. Recall that Theorem 6 implies that $|\Omega (u^{(0)}_{\varepsilon })|=\omega _0$. We now show that, for every $t \in (0,1)$, the buckling eigenfunction $u_{\varepsilon }^t$ of the scaled domain $t\,\Omega (u^{(0)}_{\varepsilon })$ does not minimize the functional ${\mathcal {I}}^{(1)}_{\varepsilon }$, i.e., for every $t \in (0,1)$ there holds

$$\begin{aligned} {\mathcal {I}}^{(1)}_{\varepsilon }(u^{(1)}_{\varepsilon }) < {\mathcal {I}}^{(1)}_{\varepsilon }(u_\varepsilon ^t). \end{aligned}$$

(24)

For that purpose, let us assume that (24) does not hold true. Then there exists a $t\in (0,1)$ such that

$$\begin{aligned} {\mathcal {I}}^{(1)}_{\varepsilon }(u^{(1)}_{\varepsilon }) = {\mathcal {I}}^{(1)}_{\varepsilon }(u_\varepsilon ^t) \end{aligned}$$

and, since $|t\,\Omega (u^{(0)}_{\varepsilon })|=t^n\omega _0 < \omega _0$, we obtain

$$\begin{aligned} {\mathcal {I}}^{(1)}_{\varepsilon }(u_\varepsilon ^t) = \Lambda (t\,\Omega (u^{(0)}_{\varepsilon })) - \varepsilon \omega _0(1-t^n) \le \Lambda (\Omega (u^{(0)}_{\varepsilon })) = {\mathcal {I}}^{(1)}_{\varepsilon }(u^{(0)}_{\varepsilon }). \end{aligned}$$

(25)

By scaling there holds

$$\begin{aligned} \Lambda (t\,\Omega (u^{(0)}_{\varepsilon })) = t^{-2} \Lambda (\Omega (u^{(0)}_{\varepsilon })). \end{aligned}$$

Thus, from (25) we get

$$\begin{aligned} \Lambda (\Omega (u^{(0)}_{\varepsilon })) \left( t^{-2} - 1\right) \le \varepsilon \omega _0 (1-t^n). \end{aligned}$$

(26)

By $\Omega (u^{(0)}_{\varepsilon })^\#$ we denote the ball centered in the origin with the same volume as $\Omega (u^{(0)}_{\varepsilon })$. Then

$$\begin{aligned} \Lambda (\Omega (u^{(0)}_{\varepsilon })^\#) = \left( \frac{\omega _n}{\omega _0}\right) ^\frac{2}{n}\Lambda (B_1) \end{aligned}$$

and applying Remark 3 we obtain from (26)

$$\begin{aligned}&c_n \Lambda (\Omega (u^{(0)}_{\varepsilon })^\#) \left( t^{-2}-1\right)< \varepsilon \omega _0 \left( 1-t^n\right) \\ \Leftrightarrow \quad&c_n\,\left( \frac{\omega _n}{\omega _0}\right) ^\frac{2}{n}\Lambda (B_1)\left( t^{-2}-1\right)< \varepsilon \,\omega _0\,(1-t^n) \\ \Leftrightarrow \quad&\frac{t^{-2}-1}{1-t^n} < \varepsilon \, \frac{\omega _0}{\Lambda (B_1)}\,c_n^{-1}\,\left( \frac{\omega _0}{\omega _n}\right) ^\frac{2}{n} = \varepsilon \cdot \varepsilon _1\cdot c_n^{-1}. \end{aligned}$$

Since $t <1$ and $\varepsilon \le \varepsilon _0$ (see (22)), we obtain

$$\begin{aligned} \frac{2}{n} \le \frac{t^{-2}-1}{1-t^n} < \varepsilon \cdot \varepsilon _1\cdot c_n^{-1} \le \frac{2}{n}. \end{aligned}$$

Obviously, this statement is false and we conculde that (24) holds true for every $t \in (0,1)$.

Step 2.    Let us fix $t_*\in (0,1)$ such that $|t_*\Omega (u^{(0)}_{\varepsilon })| = |\Omega (u^{(1)}_{\varepsilon })|$. Hence, according to (23), there holds $t_*= \alpha ^\frac{1}{n}$ and applying (24) we obtain

$$\begin{aligned} {\mathcal {I}}^{(1)}_{\varepsilon }(u^{(1)}_{\varepsilon }) < {\mathcal {I}}^{(1)}_{\varepsilon }(u_\varepsilon ^{t_*}). \end{aligned}$$

By choice of $t_*$, this is equivalent to

$$\begin{aligned} \Lambda (\Omega (u^{(1)}_{\varepsilon })) < \Lambda (t_*\Omega (u^{(0)}_{\varepsilon })) = t_*^{-2} \Lambda (\Omega (u^{(0)}_{\varepsilon })), \end{aligned}$$

where we additionally used the scaling property of the buckling load. Multiplying the above inequality with $t_*^2$ and again applying the scaling property of the buckling load, we deduce

$$\begin{aligned} \Lambda (t_*^{-1}\Omega (u^{(1)}_{\varepsilon })) = t_*^{-2}\Lambda (\Omega (u^{(1)}_{\varepsilon })) < \Lambda (\Omega (u^{(0)}_{\varepsilon })). \end{aligned}$$

(27)

Note that $|t_*^{-1}\Omega (u^{(1)}_{\varepsilon })|=\omega _0$. Thus, if $t_*^{-1}\Omega (u^{(1)}_{\varepsilon }) \subset B$, (27) is contradictory to the minimality of $\Omega (u^{(0)}_{\varepsilon })$ for the buckling load among all open subsets of B with volume smaller or equal than $\omega _0$ (see (19)). Consequently, if $t_*^{-1}\Omega (u^{(1)}_{\varepsilon }) \subset B$, the assumption (23) is false and there holds $|\Omega (u^{(1)}_{\varepsilon })|=\omega _0$. This proves part a) of the claim.

If $t_*^{-1}\Omega (u^{(1)}_{\varepsilon }) \not \subset B$, but there exists a translation $\Phi : \mathbb {R}^n\rightarrow \mathbb {R}^n$ such that $\Phi (t_*^{-1}\Omega (u^{(1)}_{\varepsilon })) \subset B$, we arrive at the same contradiction to (19) as above because of the translational invariance of the buckling load (i.e., $\Lambda (D) = \Lambda (\Phi (D))$ for every translation $\Phi$).

Hence, if $|\Omega (u^{(1)}_{\varepsilon })|<\omega _0$, the scaled domain $t_*^{-1}\Omega (u^{(1)}_{\varepsilon })$ cannot be translated into the ball B. This proves part b) of the claim. $\square$

From now on, we always choose $\varepsilon \le \varepsilon _0$. Let $u^{(1)}_{\varepsilon }\in H^{2,2}_0(B)$ be a minimizer of ${\mathcal {I}}^{(1)}_{\varepsilon }$ and let us assume that case a) of Theorem 9 holds true. Hence, $|\Omega (u^{(1)}_{\varepsilon })|=\omega _0$ and repeating the proof of Theorem 1 we gave in Sect. 3.1 we find that $\Omega (u^{(1)}_{\varepsilon })$ minimizes the buckling load among all open subsets of B with volume smaller or equal than $\omega _0$. In addition, $u^{(1)}_{\varepsilon }$ minimizes the functional ${\mathcal {I}}^{(0)}_{\varepsilon }$.

Consequently, if we could exclude that the case (b) of Theorem 9 may occur, we could treat the functionals ${\mathcal {I}}^{(0)}_{\varepsilon }$ and ${\mathcal {I}}^{(1)}_{\varepsilon }$ as equivalent. We will discuss this issue in the following section.

3.3 Discussion of case (b) of Theorem 9

In this section, we always consider a minimizer $u^{(1)}_{\varepsilon } \in H^{2,2}_0(B)$ such that case b) of Theorem 9 occurs. Thus, there holds $|\Omega (u^{(1)}_{\varepsilon })|<\omega _0$ and the rescaled domain $\Omega '$ with $|\Omega '|=\omega _0$ is neither contained in B, nor can it be translated into B. Consequently, we have to think of the domain $\Omega (u^{(1)}_{\varepsilon })$ as of a domain with tentacle-like, very long and very thin parts.

Since we do not posses any information about regularity properties of $\partial \Omega (u^{(1)}_{\varepsilon })$, we currently cannot anaylze these tentacle-like domains in more detail.

However, assuming that $\partial \Omega (u^{(1)}_{\varepsilon })$ satisfies a doubling condition, we will be able to establish a non-degeneracy result for $u^{(1)}_{\varepsilon }$. This non-degeneracy enables us to prove that, potentially after translation, the up the volume $\omega _0$ rescaled version of $\Omega (u^{(1)}_{\varepsilon })$ is always contained in B. Hence, if $\partial \Omega (u^{(1)}_{\varepsilon })$ satisfies a doubling condition, the case (b) of Theorem 9 cannot occur and there holds $|\Omega (u^{(1)}_{\varepsilon })|=\omega _0$

From now on, we always assume that $\partial \Omega (u^{(1)}_{\varepsilon })$ satisfies the following doubling property. Assume that there exists a constant $\sigma >0$ and a radius $0<R_0<1$ such that for every $x_0 \in \partial \Omega (u^{(1)}_{\varepsilon })$ and every $0<R\le R_0$ there holds

$$\begin{aligned} |B_{2R}(x_0)\cap \Omega (u^{(1)}_{\varepsilon })| \le \sigma \, |B_R(x_0) \cap \Omega (u^{(1)}_{\varepsilon })|. \end{aligned}$$

(28)

Note carefully that the condition (28) does not exclude that $\Omega (u^{(1)}_{\varepsilon })$ forms thin tentacles but it determines the minimal rate at which the volume of a possible tentacle may decrease.

3.3.1 Non-degeneracy of $u^{(1)}_{\varepsilon }$

Our aim is to establish a non-degeneracy result for $u^{(1)}_{\varepsilon }$. For convenience, we cite a technical lemma which will be applied in the proof of Lemma 8.

Lemma 7

([7, Lemma 6.1]) Let Z(t) be a bounded non-negative function in the interval $[\rho ,R]$. Assume that for $\rho \le t < s \le R$ we have

$$\begin{aligned} Z(t) \le \left[ A(s-t)^{-\alpha } + B(s-t)^{-\beta }+C\right] + \vartheta Z(s) \end{aligned}$$

with $A,B,C \ge 0$, $\alpha>\beta >0$ and $0\le \vartheta <1$. Then,

$$\begin{aligned} Z(\rho ) \le c(\alpha ,\theta )\left[ A(R-\rho )^{-\alpha } + B(R-\rho )^{-\beta }+C \right] . \end{aligned}$$

Lemma 8

Let $u^{(1)}_{\varepsilon }$ be a minimizer of ${\mathcal {I}}^{(1)}_{\varepsilon }$ and let $\partial \Omega (u^{(1)}_{\varepsilon })$ satisfy (28). There exists a $c_1 = c_1(\varepsilon ,n,\omega _0,\sigma ) >0$ such that there holds

$$\begin{aligned} c_1 \,R \le \sup _{B_R(x_0)}|\nabla \! u^{(1)}_{\varepsilon }|, \end{aligned}$$

(29)

where $x_0 \in \partial \Omega (u^{(1)}_{\varepsilon })$ and $0<R\le R_0$. In particular, $c_1$ is independent of the choice of $x_0$ and R.

Proof

Let $x_0 \in \partial \Omega (u^{(1)}_{\varepsilon })$ and $0<R\le R_0$. For $\frac{R}{2} \le t < s \le R$, let $\eta \in C^\infty (\mathbb {R}^n)$ satisfy $0 \le \eta \le 1$, $\eta \equiv 0$ in $B_t(x_0)$ and $\eta \equiv 1$ in $R^n\setminus B_s(x_0)$. Note that for every $x \in B_s(x_0)\setminus B_t(x_0)$ there holds

$$\begin{aligned} |\eta (x)| \le \frac{C(n)}{s-t} \; \text{ and } |\Delta \eta (x)|\le \frac{C(n)}{(s-t)^2}, \end{aligned}$$

(30)

where C(n) only depends on n. We use $u^{(1)}_{\varepsilon }\eta$ as a comparison function for ${\mathcal {I}}^{(1)}_{\varepsilon }$. By the minimality of $u^{(1)}_{\varepsilon }$ we obtain

$$\begin{aligned} {\mathcal {I}}^{(1)}_{\varepsilon }(u^{(1)}_{\varepsilon }) \le {\mathcal {I}}^{(1)}_{\varepsilon }(u^{(1)}_{\varepsilon }\eta ). \end{aligned}$$

Since, by construction, there holds $|\mathcal {O}(\eta u^{(1)}_{\varepsilon })|= |\mathcal {O}(u^{(1)}_{\varepsilon })| - |B_t(x_0)\cap \Omega (u^{(1)}_{\varepsilon })|$, straight forward computation yields

$$\begin{aligned} \begin{aligned} \int \limits _{B_s(x_0)}|\Delta u^{(1)}_{\varepsilon }|^2dx + \varepsilon \,|B_t&(x_0)\cap \Omega (u^{(1)}_{\varepsilon })| \le \int \limits _{B_s\setminus B_t(x_0)}|\Delta ( \eta u^{(1)}_{\varepsilon })|^2dx \\&+ (\Lambda (\Omega (u^{(1)}_{\varepsilon })+\varepsilon |B_t(x_0)\cap \Omega (u^{(1)}_{\varepsilon })|)\int \limits _{B_s(x_0)}|\nabla \! u^{(1)}_{\varepsilon }|^2dx. \end{aligned} \end{aligned}$$

(31)

A detailed analysis of this inequality gives the claim. This is done in three steps.

Step 1.    We estimate the last summand on the right hand side of (31). Recall that

$$\begin{aligned} \Lambda (\Omega (u^{(1)}_{\varepsilon }))&\le \left( \frac{\omega _n}{\omega _0}\right) ^\frac{2}{n}\Lambda (B_1) + \varepsilon \,(\omega _0 - |\mathcal {O}(u^{(1)}_{\varepsilon })|)\\&\le \left( \frac{\omega _n}{\omega _0}\right) ^\frac{2}{n}\Lambda (B_1) + \varepsilon _1\,\omega _0 = C(n,\omega _0). \end{aligned}$$

Consequently,

$$\begin{aligned} \Lambda (\Omega (u^{(1)}_{\varepsilon })) + \varepsilon |B_t(x_0)\cap \Omega (u^{(1)}_{\varepsilon })| \le C(n,\omega _0) + \varepsilon _1|B_{R_0}(x_0)| \le C(n,\omega _0) \end{aligned}$$

and (31) becomes

$$\begin{aligned} \begin{aligned} \int \limits _{B_s(x_0)}|\Delta u^{(1)}_{\varepsilon }|^2dx + \varepsilon \,|B_t(x_0)&\cap \Omega (u^{(1)}_{\varepsilon })| \\&\le \int \limits _{B_s\setminus B_t(x_0)}|\Delta ( \eta u^{(1)}_{\varepsilon })|^2dx + C(n,\omega _0)\int \limits _{B_s(x_0)}|\nabla \! u^{(1)}_{\varepsilon }|^2dx. \end{aligned} \end{aligned}$$

(32)

Step 2.    Applying Young’s inequality we estimate

$$\begin{aligned} |\Delta ( u^{(1)}_{\varepsilon }\eta )|^2 \le 4\left( |\Delta u^{(1)}_{\varepsilon }|^2\eta ^2 + 2|\nabla \! u^{(1)}_{\varepsilon }.\nabla \! \eta |^2 + (u^{(1)}_{\varepsilon })^2|\Delta \eta |^2\right) \end{aligned}$$

and together with (30) we deduce in $B_s(x_0)\setminus B_t(x_0)$

$$\begin{aligned} |\Delta ( u^{(1)}_{\varepsilon }\eta )|^2 \le 4\left( |\Delta u^{(1)}_{\varepsilon }|^2\eta ^2 + 2\frac{C(n)}{(s-t)^2}|\nabla \! u^{(1)}_{\varepsilon }|^2 + \frac{C(n)}{(s-t)^4}(u^{(1)}_{\varepsilon })^2\right) . \end{aligned}$$

With the splitting

$$\begin{aligned} \int \limits _{B_s(x_0)}|\Delta u^{(1)}_{\varepsilon }|^2dx = \int \limits _{B_s\setminus B_t(x_0) }|\Delta u^{(1)}_{\varepsilon }|^2 dx + \int \limits _{B_t(x_0)}|\Delta u^{(1)}_{\varepsilon }|^2 dx \end{aligned}$$

(32) becomes

$$\begin{aligned}&\int \limits _{B_t(x_0)}|\Delta u^{(1)}_{\varepsilon }|^2dx + \varepsilon \,|B_t(x_0)\cap \Omega (u^{(1)}_{\varepsilon })| \le \int \limits _{B_s\setminus B_t(x_0)}|\Delta u^{(1)}_{\varepsilon }|^2(4\eta ^2-1)dx \\&+ \int \limits _{B_s\setminus B_t(x_0)}\frac{C(n)}{(s-t)^2}|\nabla \! u^{(1)}_{\varepsilon }|^2 + \frac{C(n)}{(s-t)^4}(u^{(1)}_{\varepsilon })^2dx + C(n,\omega _0)\int \limits _{B_s(x_0)}|\nabla \! u^{(1)}_{\varepsilon }|^2dx, \end{aligned}$$

where C(n) collects all constants only depending on n. Since $0\le \eta \le 1$, we obtain

$$\begin{aligned}&\int \limits _{B_t(x_0)}|\Delta u^{(1)}_{\varepsilon }|^2dx + \varepsilon \,|B_t(x_0)\cap \Omega (u^{(1)}_{\varepsilon })| \le 3 \int \limits _{B_s\setminus B_t(x_0)}|\Delta u^{(1)}_{\varepsilon }|^2dx \\&\quad +C(n,\omega _0)\, \int \limits _{B_R(x_0)}\left( 1+\frac{1}{(s-t)^2}\right) |\nabla \! u^{(1)}_{\varepsilon }|^2 + \frac{1}{(s-t)^4}(u^{(1)}_{\varepsilon })^2dx. \end{aligned}$$

Now we add $3\int _{B_t(x_0)}|\Delta u^{(1)}_{\varepsilon }|^2dx$ to both sides of the above inequality and divide the resulting inequality by 4. Subsequently, we add $\frac{3\,\varepsilon }{16}|B_s(x_0)\cap \Omega (u^{(1)}_{\varepsilon })|$ to the right hand side. This leads to

$$\begin{aligned} \begin{aligned}&\int \limits _{B_t(x_0)}|\Delta u^{(1)}_{\varepsilon }|^2dx + \frac{\varepsilon }{4}\,|B_t(x_0)\cap \Omega (u^{(1)}_{\varepsilon })| \le \frac{3}{4}\left( \int \limits _{B_s(x_0)}|\Delta u^{(1)}_{\varepsilon }|^2dx + \frac{\varepsilon }{4}|B_s(x_0)\cap \Omega (u^{(1)}_{\varepsilon })|\right) \\ {}&\quad +C(n,\omega _0)\, \int \limits _{B_R(x_0)}\left( 1+\frac{1}{(s-t)^2}\right) |\nabla \! u^{(1)}_{\varepsilon }|^2 + \frac{1}{(s-t)^4}(u^{(1)}_{\varepsilon })^2dx . \end{aligned} \end{aligned}$$

(33)

Setting

$$\begin{aligned} Z(t) := \int \limits _{B_t(x_0)}|\Delta u^{(1)}_{\varepsilon }|^2dx + \frac{\varepsilon }{4}\,|B_t(x_0)\cap \Omega (u^{(1)}_{\varepsilon })|, \end{aligned}$$

estimate (33) enables us to apply Lemma 7 and we obtain

$$\begin{aligned} \int \limits _{B_\frac{R}{2}(x_0)}|\Delta u^{(1)}_{\varepsilon }|^2dx + \frac{\varepsilon }{4}\,|B_\frac{R}{2}(x_0)\cap \Omega (u^{(1)}_{\varepsilon })| \le C(n,\omega _0)\, \int \limits _{B_R(x_0)}\left( 1+\frac{1}{R^2}\right) |\nabla \! u^{(1)}_{\varepsilon }|^2 + \frac{1}{R^4}(u^{(1)}_{\varepsilon })^2dx. \end{aligned}$$

Step 3. $C^{1,\alpha }$ regularity of $u^{(1)}_{\varepsilon }$ allows us to estimate

$$\begin{aligned} |u^{(1)}_{\varepsilon }(x)| \le 2\,R\,\sup _{B_R(x_0)}|\nabla \! u^{(1)}_{\varepsilon }| \end{aligned}$$

for every $x\in B_R(x_0)$. Moreover, we assume $R\le R_0<1$. Hence, we find

$$\begin{aligned} \begin{aligned} \int \limits _{B_\frac{R}{2}(x_0)}|\Delta u^{(1)}_{\varepsilon }|^2dx + \frac{\varepsilon }{4}\,|B_\frac{R}{2}(x_0)&\cap \Omega (u^{(1)}_{\varepsilon })| \\&\le C(n,\omega _0)\, R^{-2}\,\sup _{B_R(x_0)}|\nabla \! u^{(1)}_{\varepsilon }|^2\,|B_R(x_0)\cap \Omega (u^{(1)}_{\varepsilon })|. \end{aligned} \end{aligned}$$

(34)

Since we assume the doubling property (28) to hold true, omitting the non-negative integral on the left hand side we obtain

$$\begin{aligned} \frac{\varepsilon }{4} \le C(n,\omega _0)\,\sigma \, R^{-2}\,\sup _{B_R(x_0)}|\nabla \! u^{(1)}_{\varepsilon }|^2. \end{aligned}$$

This proves the claim. $\square$

Note that the assumption (28) enables us to compare $|B_\frac{R}{2}(x_0)\cap \Omega (u^{(1)}_{\varepsilon })|$ with $|B_R(x_0)\cap \Omega (u^{(1)}_{\varepsilon })|$ in estimate (34). This assumption is only needed because we currently do not have any further information about the free boundary $\partial \Omega (u^{(1)}_{\varepsilon })$ and could be replaced by regularity properties of the free boundary. However, let us emphasize that the rewarding property of the penalization term $p^{(1)}_{\varepsilon }$ is crucial for proving Lemma 8 and cannot be replaced since the rewarding term yields the strictly positive lower bound in (29).

The non-degeneracy of $u^{(1)}_{\varepsilon }$ along the free boundary according to Lemma 8 allows us to establish a lower bound on the density quotient of $\Omega (u^{(1)}_{\varepsilon })$.

Lemma 9

Let $u^{(1)}_{\varepsilon }$ be a minimizer of ${\mathcal {I}}^{(1)}_{\varepsilon }$, $\alpha \in (0,1)$ and let $\partial \Omega (u^{(1)}_{\varepsilon })$ satisfy (28). There exists a constant $c_2 = c_2(\varepsilon ,n,\omega _0,\sigma ,\alpha )>0$ such that for every $x_0 \in \partial \Omega (u^{(1)}_{\varepsilon })$ and every $0<R\le R_0$ there holds

$$\begin{aligned} c_2 \,|B_R|^\frac{1-\alpha }{\alpha } \le \frac{|\Omega (u^{(1)}_{\varepsilon })\cap B_R(x_0)|}{|B_R|}. \end{aligned}$$

Although this lower bound on the density quotient is admittedly weak, it suffices to prove that $\Omega (u^{(1)}_{\varepsilon })$ can be rescaled to the volume $\omega _0$ without leaving the reference domain B provided the radius of B is chosen sufficiently large (see Theorem 10).

Proof of Lemma 9

Let $x_0 \in \partial \Omega (u^{(1)}_{\varepsilon })$ and $0<R\le R_0$. According to Lemma 8 and the $C^{1,\alpha }$ regularity of $u^{(1)}_{\varepsilon }$ there exists an $x_1 \in \Omega (u^{(1)}_{\varepsilon })\cap \overline{B_\frac{R}{2}(x_0)}$ such that

$$\begin{aligned} c_1 \frac{R}{2} \le \sup _{B_\frac{R}{2}(x_0)}|\nabla \! u^{(1)}_{\varepsilon }| = |\nabla \! u^{(1)}_{\varepsilon }(x_1)|. \end{aligned}$$

Now choose $x_2 \in \partial \Omega (u^{(1)}_{\varepsilon })$ such that

$$\begin{aligned} d:= {\text {dist}}(x_1,\partial \Omega (u^{(1)}_{\varepsilon }) = |x_1-x_2|. \end{aligned}$$

Since $|\nabla \! u^{(1)}_{\varepsilon }(x_2)|=0$, we obtain

$$\begin{aligned} c_1 \frac{R}{2} \le |\nabla \! u^{(1)}_{\varepsilon }(x_1)-\nabla \! u^{(1)}_{\varepsilon }(x_2)| \le L_\alpha \,d^\alpha , \end{aligned}$$

where $L_\alpha =L(n,\omega _0,\alpha )$ denotes the $\alpha$-Hölder coefficient of $\nabla \! u^{(1)}_{\varepsilon }$. By construction, there holds $B_d(x_1)\subset \Omega (u^{(1)}_{\varepsilon })\cap B_R(x_0)$. Consequently, we may proceed to

$$\begin{aligned} \left( \frac{c_1}{2L_\alpha }\right) ^n\,R^n \le d^{\alpha \,n}&\Leftrightarrow \left( \frac{c_1}{2L_\alpha \omega _n^{1-\alpha }}\right) ^n\,|B_R| \le |B_d(x_1)|^\alpha \\&\Rightarrow \left( \frac{c_1}{2L_\alpha \omega _n^{1-\alpha }}\right) ^n\,|B_R| \le |\Omega (u^{(1)}_{\varepsilon })\cap B_R(x_0)|^\alpha . \end{aligned}$$

This proves the claim. $\square$

3.3.2 The volume condition for $\Omega (u^{(1)}_{\varepsilon })$

The next theorem is the key observation to show that $\Omega (u^{(1)}_{\varepsilon })$ has the volume $\omega _0$ provided that $\partial \Omega (u^{(1)}_{\varepsilon })$ satisfies (28). It is a consequence of the lower bound on the density quotient according to Lemma 9.

Theorem 10

Let $\varepsilon \le \varepsilon _1$ and let $B = B_{R_B}(0)$. Provided that $R_B$ is chosen sufficiently large, for every minimizer $u^{(1)}_{\varepsilon }$ of ${\mathcal {I}}^{(1)}_{\varepsilon }$ such that $\partial \Omega (u^{(1)}_{\varepsilon })$ satisfies (28) there holds

• (a) $\Omega (u^{(1)}_{\varepsilon })$ is compactly contained in $B_{2^{-\frac{1}{n}}R_B}(0)$ or

• (b) there exists a translation $\Phi : \mathbb {R}^n\rightarrow \mathbb {R}^n$ such that $\Phi (\Omega (u^{(1)}_{\varepsilon }))$ is compactly contained in $B_{2^{-\frac{1}{n}}R_B}(0)$.

Proof

Let us think of the reference domain B as of a ball centered at the origin and with radius $R_B$. In addition, let $u^{(1)}_{\varepsilon }$ be a minimizer of ${\mathcal {I}}^{(1)}_{\varepsilon }$ for $\varepsilon \le \varepsilon _1$. In order to prove the claim, let us assume that $\Omega (u^{(1)}_{\varepsilon })$ is not compactly contained in $B_{2^{-\frac{1}{n}}R_B}(0)$. For the sake of convenience, we abbreviate $S := 2^{-\frac{1}{n}}R_B$. Hence, we assume $\partial \Omega (u^{(1)}_{\varepsilon })\cap \partial B_{S}(0) \ne \emptyset$.

Of course, there either holds $0 \in \Omega (u^{(1)}_{\varepsilon })$ or $0\not \in \Omega (u^{(1)}_{\varepsilon })$. At first, we will handle the case where the origin is already contained in $\Omega (u^{(1)}_{\varepsilon })$. Secondly, we will show that we may translate $\Omega (u^{(1)}_{\varepsilon })$ such that the origin becomes an inner point of $\Omega (u^{(1)}_{\varepsilon })$.

Step 1.    We consider that $0\in \Omega (u^{(1)}_{\varepsilon })$.

Note that for every $m\in \mathbb {N}$ with $m\ge 3$ there holds

$$\begin{aligned} B_S(0) = \bigcup _{i=0}^{m-2} B_{\frac{i+2}{m}S}(0) \setminus B_{\frac{i}{m}S}(0). \end{aligned}$$

(35)

Since we assume that $\partial \Omega (u^{(1)}_{\varepsilon })\cap \partial B_S(0) \ne \emptyset$ and $0\in \Omega (u^{(1)}_{\varepsilon })$, there exists a smallest index $i_0=i_0(m)$ such that for each $i\ge i_0$ there exists an $x_i \in \partial \Omega (u^{(1)}_{\varepsilon })\cap \partial B_{\frac{i+1}{m}S}(0)$.

We fix $m\in \mathbb {N}$ such that $\frac{S}{m} \le R_0.$ In addition, we fix an $\alpha \in (0,1)$. Then applying Lemma 9 for $i_0\le i\le m-2$ we obtain

$$\begin{aligned} c_2 |B_\frac{S}{m}|^\frac{1}{\alpha } \le |\Omega (u^{(1)}_{\varepsilon })\cap B_\frac{S}{m}(x_i)|. \end{aligned}$$

(36)

We now sum (36) from $i=i_0(m)$ to $i=m-2$. Since $B_\frac{S}{m}(x_i)\cap B_\frac{S}{m}(x_k)=\emptyset$ for $i\ne k$ and $|\Omega (u^{(1)}_{\varepsilon })|\le \omega _0$, this implies

$$\begin{aligned} c_2\,(m-1-i_0(m)) \, |B_{\frac{S}{m}}|^\frac{1}{\alpha } \le \sum _{i=i_0}^{m-2}|\Omega (u^{(1)}_{\varepsilon })\cap B_\frac{S}{m}(x_i)| \le |\Omega (u^{(1)}_{\varepsilon })| \le \omega _0. \end{aligned}$$

Note that since $B_{\frac{i_0}{m}S}(0)\subset \Omega (u^{(1)}_{\varepsilon })$ and $|\Omega (u^{(1)}_{\varepsilon })|\le \omega _0$, $i_0(m)$ is bounded. Indeed,

$$\begin{aligned} |B_{\frac{i_0}{m}S}| \le |\Omega (u^{(1)}_{\varepsilon })| \le \omega _0 \end{aligned}$$

implies

$$\begin{aligned} i_0(m) \le \left( \frac{\omega _0}{\omega _n}\right) ^\frac{1}{n}\frac{m}{S}. \end{aligned}$$

Specifying the choice of $m\in \mathbb {N}$ such that

$$\begin{aligned} \frac{R_0}{4} \le \frac{S}{m}\le \frac{R_0}{2} \end{aligned}$$

and recalling that $S:=2^{-\frac{1}{n}}R_B$, we obtain

$$\begin{aligned} c_2\,\left( \frac{2\,R_B}{2^\frac{1}{n}R_0}-1-\frac{4}{R_0}\left( \frac{\omega _0}{\omega _n}\right) ^\frac{1}{n}\right) |B_\frac{R_0}{4}|^\frac{1}{\alpha } \le c_2 \,(m-1-i_0(m))|B_\frac{S}{m}|^\frac{1}{\alpha } \le \omega _0. \end{aligned}$$

Since this estimate is false if $R_B$ is chosen sufficiently large, the proof is finished provided that $0 \in \Omega (u^{(1)}_{\varepsilon })$.

Step 2.    Let us now assume that the origin is not contained in $\Omega (u^{(1)}_{\varepsilon })$. However, there exists an $x_0 \in \Omega (u^{(1)}_{\varepsilon })\cap B$. We now translate $\Omega (u^{(1)}_{\varepsilon })$ such that $x_0$ is translated to the origin, i.e., we consider

$$\begin{aligned} \Phi : \mathbb {R}^n \rightarrow \mathbb {R}^n,\, x \mapsto x-x_0. \end{aligned}$$

We call $\Omega ' :=\Phi (\Omega (u^{(1)}_{\varepsilon }))$ and $v_\varepsilon (x) := u^{(1)}_{\varepsilon }(\Phi ^{-1}(x))$. Thus,

$$\begin{aligned} \Omega ' = \{x\in \mathbb {R}^n : v_\varepsilon (x)\ne 0 \text{ or } (v_\varepsilon (x)=0 \wedge |\nabla \! v_\varepsilon (x)|>0)\} \end{aligned}$$

and $v_\varepsilon \in H^{2,2}_0(\Omega ')$. Moreover, $\partial \Omega ' = \Phi (\partial \Omega (u^{(1)}_{\varepsilon }))$.

Let us emphasize that, in general, $\Omega '$ may not be contained in B and, thus, $v_\varepsilon \not \in H^{2,2}_0(B)$. Note carefully that in Step 1 the minimality of $u^{(1)}_{\varepsilon }$ for ${\mathcal {I}}^{(1)}_{\varepsilon }$ is not used explicitly. However, the minimality is necessary to establish Lemma 8 and, subsequently, Lemma 9 and the application of Lemma 9 leads to estimate (36), which is the crucial observation in Step 1.

If $y_0 \in \partial \Omega '$ and $0<R\le R_0$, then there exists a $z_0 \in \partial \Omega (u^{(1)}_{\varepsilon })$ such that $y_0 = \Phi (z_0)$. Lemma 9 together with the translational invariance of the Lebesgue measure then imply

$$\begin{aligned} |\Omega ' \cap B_R(y_0)| = |\Omega (u^{(1)}_{\varepsilon })\cap B_R(z_0)| \ge c_2 |B_R|^\frac{1}{\alpha }. \end{aligned}$$

(37)

Estimate (37) enables us to repeat the approach presented in Step 1. Again we consider the segmentation (35) and assume that $\partial \Omega '\cap \partial B_S(0)$ is not empty. Then there exists a smallest index $i_0(m)$ such that for every $i_0\le i \le m-2$ there exists an $x_i \in \partial \Omega '\cap \partial B_{\frac{i+1}{m}S}(0)$. Now applying (37) we obtain for $i_0\le i \le m-2$

$$\begin{aligned} c_2\,|B_\frac{S}{m}|^\frac{1}{\alpha } \le |\Omega ' \cap B_\frac{S}{m}(x_i)| \end{aligned}$$

Since $|\Omega '| = |\Omega (u^{(1)}_{\varepsilon })| \le \omega _0$ we may repeat the argumentation from Step 1 and obtain that $\Omega '=\Phi (\Omega (u^{(1)}_{\varepsilon }))$ is compactly contained in $B_{2^{-\frac{1}{n}}R_B}$ and, by construction, contains the origin. $\square$

As a direct consequence of Theorem 10 we deduce that, if $\partial \Omega (u^{(1)}_{\varepsilon })$ satisfies the doubling condition (28), the domain $\Omega (u^{(1)}_{\varepsilon })$ satisfies $|\Omega (u^{(1)}_{\varepsilon })|=\omega _0$.

Corollary 3

Let $\varepsilon \le \varepsilon _0$ and let $u^{(1)}_{\varepsilon }\in H^{2,2}_0(B)$ minimize ${\mathcal {I}}^{(1)}_{\varepsilon }$. Provided that the radius $R_B$ of B is chosen sufficiently large and that $\partial \Omega (u^{(1)}_{\varepsilon })$ satisfies the doubling property (28), there holds $|\Omega (u^{(1)}_{\varepsilon })|=\omega _0$.

Proof

Let $\varepsilon \le \varepsilon _0$ and let $u^{(1)}_{\varepsilon }\in H^{2,2}_0(B)$ be a minimizer of ${\mathcal {I}}^{(1)}_{\varepsilon }$ such that $\partial \Omega (u^{(1)}_{\varepsilon })$ satisfies (28). In addition, we assume $|\Omega (u^{(1)}_{\varepsilon })|=\alpha \omega _0$ for an $\alpha \in [\alpha _0,1)$. Note that since $c_n\ge \frac{1}{2}$ for every $n \in \mathbb {N}$, the quantity $\alpha _0$ given in Theorem 8 satisfies

$$\begin{aligned} \alpha _0\ge \frac{1+\varepsilon _1^2-\sqrt{(1+\varepsilon _1^2)^2-4c_n\varepsilon _1^2}}{2\varepsilon _1^2} \ge \frac{1+\varepsilon _1^2-\sqrt{1+\varepsilon _1^4}}{2\varepsilon _1^2} \ge \frac{1}{2} \end{aligned}$$

(38)

for every choice of $\omega _0$, which determines $\varepsilon _1$. Applying Theorem 10 the domain $\Omega (u^{(1)}_{\varepsilon })$ is compactly contained in $B_{2^{-\frac{1}{n}}R_B}$ (possibly after translation) and estimate (38) implies that the scaled domain $\alpha ^{-\frac{1}{n}}\Omega (u^{(1)}_{\varepsilon })$ is a subset of $B_{R_B}$. This is a contradiction to Theorem 9 since we assume that $|\Omega (u^{(1)}_{\varepsilon })|<\omega _0$. $\square$

We finish with some concluding remarks.

Remark 4

The fundamental tone of a domain $\Omega \subset \mathbb {R}^n$ is defined as

$$\begin{aligned} \Gamma (\Omega ) := \min \left\{ \frac{\int _\Omega |\Delta v |^2dx}{\int _\Omega v^2dx}: v \in H^{2,2}_0(\Omega )\right\} . \end{aligned}$$

Considering the functional $\mathcal {I}_\varepsilon : H^{2,2}_0(B) \rightarrow \mathbb {R}$ by

$$\begin{aligned} {\mathcal {I}}^{(k)}_{\varepsilon }(v) := \frac{\int _\Omega |\Delta v |^2dx}{\int _\Omega v^2dx} +p^{(k)}_{\varepsilon }(|\mathcal {O}(v)|), \end{aligned}$$

where $p^{(k)}_{\varepsilon }$ and $\mathcal {O}(v)$ are defined as above, the approach presented in the Sects. 2 and 3 can be adopted to prove the existence of a connected domain $\Omega ^*$ with $|\Omega ^*|=\omega _0$ such that

$$\begin{aligned} \Gamma (\Omega ^*) = \min \{\Gamma (D): D\subset B, D \text{ open }, |D|\le \omega _0 \}. \end{aligned}$$

Remark 5

Future work should address the following issues.

• It remains an open problem how to prove regularity of the free boundaries $\partial \Omega (u^{(0)}_{\varepsilon })$ and $\partial \Omega (u^{(1)}_{\varepsilon })$, respectively.

• It remains open to prove the doubling property we assumed in Sect. 3.3 or to find another argument why the the enlarged domain $\alpha ^{-\frac{1}{n}}\Omega (u^{(1)}_{\varepsilon })$ is still contained in B.

• It remains open to prove existence of an optimal domain for minimizing the buckling load among all (probably unbounded) open subsets of $\mathbb {R}^n$ with given measure.