On the optimal domain for minimizing the buckling load of a clamped plate

Abstract

We prove the existence of an optimal domain for minimizing the buckling load among all, possibly unbounded, open subsets of ${\mathbb {R}}^n$ ($n\ge 2$) with given measure. Our approach is based on the extension of a two-dimensional existence result of Ashbaugh and Bucur and on the idea of Alt and Caffarelli to focus on the eigenfunction.

1 Introduction

We consider the following variational problem. Let $\Omega \subset {\mathbb {R}}^n$ be an open set and define

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

for $v \in W^{2,2}_0(\Omega )$ (see (2)). If the denominator vanishes, we set ${\mathcal {R}}(v,\Omega ) = \infty$. The buckling load of the clamped plate $\Omega$ is defined as

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

In 1951, Polya and Szegö conjectured that the ball minimizes the buckling load among all open sets of given measure (see [12]). It is still an open question to confirm their conjecture. Up to now, there are only partial results known.

If there exists a smooth, bounded, connected and simply connected open set $\Omega$ which minimizes the buckling load among all open sets of given measure in ${\mathbb {R}}^n$, it is known that $\Omega$ is a ball (see [14, 15]).

In [13], the existence of an optimal domain for minimizing the buckling load among all opens sets of a given measure which are contained in a sufficiently large ball $B \subset {\mathbb {R}}^{n}$, $n=2,3$, is proven. However, [13] does not provide any information about the regularity of the achieved optimal domain.

Ashbaugh and Bucur proved the existence of a plane optimal domain for minimizing $\Lambda$ in two different settings [3]. On the one hand, they prove the existence of a minimizer in the family of connected and simply connected open sets of given measure in ${\mathbb {R}}^2$. On the other hand, they find an optimal set ${\tilde{\Omega }}$ for minimizing a relaxed version of the buckling load among all open sets of given measure in ${\mathbb {R}}^2$.

In the present paper, we will adapt a part of the approach by Ashbaugh and Bucur. Therefore, let us briefly summarize their idea. For $\omega _0>0$ let us denote

$$\begin{aligned} {\mathcal {O}}_{\omega _0}:= \{\Omega \subset {\mathbb {R}}^2 : \Omega \text { open}, |\Omega |\le \omega _0\}, \end{aligned}$$

where $|\Omega |$ denotes the n-dimensional Lebesgue measure of $\Omega \subset {\mathbb {R}}^n$. Ashbaugh and Bucur start from a minimizing sequence $(\Omega _k)_k \subset {\mathcal {O}}_{\omega _0}$ and the sequence $(u_k)_k$ of corresponding normalized buckling eigenfunctions $u_k \in W^{2,2}_0(\Omega _k)$. Applying a concentration-compactness lemma, they deduce the existence of a limit function $u \in W^{2,2}({\mathbb {R}}^2)$ such that

$$\begin{aligned} {\mathcal {R}}(u,{\mathbb {R}}^2) \le \liminf _{k\rightarrow \infty } {\mathcal {R}}(u_k,\Omega _k) = \liminf _{k\rightarrow \infty } \Lambda (\Omega _k) = \inf _{\Omega \in {\mathcal {O}}_{\omega _0}}\Lambda (\Omega ). \end{aligned}$$

(1)

Since $u\in W^{2,2}({\mathbb {R}}^2)$, Sobolev’s embedding theory implies that u is continuous and the set

$$\begin{aligned} {\tilde{\Omega }} := \{x \in {\mathbb {R}}^2: u(x)\ne 0\} \end{aligned}$$

is an open set. Moreover, the strong $L^2$-convergence of $u_k$ to u implies that $|{\tilde{\Omega }}|\le \omega _0$. Hence, ${\tilde{\Omega }}\in {\mathcal {O}}_{\omega _0}$. At this point the authors face the difficulty that their ansatz does not provide any further information about ${\tilde{\Omega }}$ and u except that ${\tilde{\Omega }}\in {\mathcal {O}}_{\omega _0}$ and u is continuous. In particular, they cannot conclude that $u \in W^{2,2}_0({\tilde{\Omega }})$. They circumvent that problem by introducing the relaxed Sobolev space ${\tilde{W}}^{2,2}_0({\tilde{\Omega }})$ by

$$\begin{aligned} {\tilde{W}}^{2,2}_0({\tilde{\Omega }}) := \{ v \in W^{2,2}({\mathbb {R}}^2): v =0 \text { a.e. in } {\mathbb {R}}^2\setminus \Omega \} \end{aligned}$$

and the relaxed buckling load by

$$\begin{aligned} {\tilde{\Lambda }}(\Omega ) := \min _{v \in {\tilde{W}}^{2,2}_0(\Omega )}{\mathcal {R}}(v,\Omega ). \end{aligned}$$

By construction, $u \in {\tilde{W}}^{2,2}_0({\tilde{\Omega }})$ and, consequently, ${\tilde{\Omega }}$ minimizes ${\tilde{\Lambda }}$ in ${\mathcal {O}}_{\omega _0}$, i.e.,

$$\begin{aligned} {\tilde{\Lambda }}({\tilde{\Omega }}) {\mathop {\le }\limits ^{(1)}} {\mathcal {R}}(u,{\tilde{\Omega }}) \le \liminf _{k\rightarrow \infty } {\mathcal {R}}(u_k,\Omega _k) = \inf _{\Omega \in {\mathcal {O}}_{\omega _0}}\Lambda (\Omega ) = \inf _{\Omega \in {\mathcal {O}}_{\omega _0}}{\tilde{\Lambda }}(\Omega ), \end{aligned}$$

where [3, Theorem 3.1] provides the last equation.

In this paper, we will adapt the idea of Ashbaugh and Bucur in [3] and extend it to arbitrary dimension. Contrary to their construction via ${\tilde{W}}^{2,2}$ we prove higher regularity of the limit function u. Thereby we follow the idea of Alt and Caffarelli in [2]. We will find that the first-order derivatives of u are $\alpha$-Hölder continuous in ${\mathbb {R}}^n$ for every $\alpha \in (0,1)$.

Recall (c.f. [1, Th. 9.1.3] or [8, Sec. 3.3.5]) that for an open set $\Omega \subset {\mathbb {R}}^n$ and $v \in W^{2,2}({\mathbb {R}}^n)$ there holds

$$\begin{aligned} v\in W^{2,2}_0(\Omega ) \text { if and only if }v =|\nabla \! v| = 0 \text { quasi-everywhere in }{\mathbb {R}}^n\setminus \Omega . \end{aligned}$$

(2)

Thereby, a property p(x) is said to hold ’quasi-everywhere’ if the set of points in which p(x) does not hold true has got zero capacity. Since the concept of capacity of a set will not play any further role in the sequel of the present paper, we refer to [8] for more details on the concept of capacity. However, if a property holds pointwise, it holds quasi-everywhere, as well.

Consequently, the Hölder continuity of the first-order derivatives of u implies that $u \in W^{2,2}_0(\Omega ^*)$ for

$$\begin{aligned} \Omega ^*:= \{x \in {\mathbb {R}}^n: u(x)\ne 0\ \text { and } \nabla \! u(x)\ne 0\}. \end{aligned}$$

In addition, $\Omega ^*$ satisfies $|\Omega ^*| =\omega _0$ and we deduce that $\Omega ^*$ minimizes the buckling load among all open sets of given measure in ${\mathbb {R}}^n$.

Moreover, we will show that the minimizer $\Omega ^*$ is connected.

2 Existence of a minimizer

For $\omega _0>0$ we denote the class of admissible sets by

$$\begin{aligned} {\mathcal {O}}_{\omega _0}:= \{\Omega \subset {\mathbb {R}}^n : \Omega \text { open}, |\Omega |\le \omega _0\}, \end{aligned}$$

where $|\Omega |$ denotes the n-dimensional Lebesgue measure of $\Omega \subset {\mathbb {R}}^n$, $n\ge 2$.

Our aim is to prove the existence of a set $\Omega ^*\in {\mathcal {O}}_{\omega _0}$ which minimizes $\Lambda$ in ${\mathcal {O}}_{\omega _0}$. In the beginning, we follow the idea of [3].

Let $\left( \Omega _k\right) _k \in {\mathcal {O}}_{\omega _0}$ be a minimizing sequence for the buckling load, i.e.,

$$\begin{aligned} \lim _{k\rightarrow \infty }\Lambda (\Omega _k) = \inf _{\Omega \in {\mathcal {O}}_{\omega _0}}\Lambda (\Omega ) =: \Lambda _{\omega _0}. \end{aligned}$$

By $u_k \in W^{2,2}_0(\Omega _k)$ we denote the normalized buckling eigenfunction on $\Omega _k$. Hence, $u_k$ satisfies

$$\begin{aligned} \int \limits _{\Omega _k}|\nabla \! u_k|^2\textrm{d}x =1 \text { and } \Lambda (\Omega _k) = \int \limits _{\Omega _k}|\Delta u _k|^2\textrm{d}x \end{aligned}$$

We now apply the approach by Ashbaugh and Bucur from [3] to show that $(u_k)_k$ converges weakly to a limit function u in $W^{2,2}({\mathbb {R}}^n)$.

We will use the following concentration-compactness lemma (see [3, 10]) adapted to our setting.

Lemma 1

Let $(\Omega _k)_k \subset {\mathcal {O}}_{\omega _0}$ be a minimizing sequence for the buckling load in ${\mathcal {O}}_{\omega _0}$ and $(u_k)_k$ be the sequence of corresponding eigenfunctions. Then there exists a subsequence $(u_k)_k$ such that one of the three following situations occurs.

• 1. Compactness $\exists (y_k)_k\subset {\mathbb {R}}^n$ such that $\forall \varepsilon >0$, $\exists R <\infty$ and

  $$\begin{aligned} \forall k \in {\mathbb {N}} \quad \int \limits _{B_R(y_k)}|\nabla \! u_k|^2\textrm{d}x \ge 1-\varepsilon . \end{aligned}$$

• 2. Vanishing $\forall R \in (0,\infty )$

  $$\begin{aligned} \lim _{k\rightarrow \infty }\sup _{y \in {\mathbb {R}}^n}\int _{B_R(y)}|\nabla \! u_k|^2dx =0. \end{aligned}$$

• 3. Dichotomy There exists an $\beta \in (0,1)$ such that $\forall \varepsilon >0$ there exist two bounded sequences $(u_k^1)_k$, $(u_k^2)_k \subset H^{2,2}({\mathbb {R}}^n)$ such that:

  $$\begin{aligned} \Vert \nabla \! u_k - \nabla \! u_k^1 - \nabla \! u_k^2\Vert _{L^2({\mathbb {R}}^2,{\mathbb {R}}^2)} \le \delta (\varepsilon ) {\mathop {\longrightarrow }\limits ^{k\rightarrow \infty }} 0^+, \end{aligned}$$

  (a)
  $$\begin{aligned} |\,\int \limits _{{\mathbb {R}}^n}|\nabla \! u_k^1|^2\textrm{d}x - \beta | \rightarrow 0 \quad \text {and } \quad |\,\int \limits _{{\mathbb {R}}^n}|\nabla \! u_k^2|^2\textrm{d}x - (1-\beta )| \rightarrow 0, \end{aligned}$$

  (b)
  $$\begin{aligned} {\text {dist}}({\text {supp}}(u_k^1),{\text {supp}}(u_k^2)) {\mathop {\longrightarrow }\limits ^{k\rightarrow \infty }} \rightarrow \infty , \end{aligned}$$

  (c)
  $$\begin{aligned} \liminf _{k\rightarrow \infty }\left[ \int \limits _{{\mathbb {R}}^n} |\Delta u _k|^2 - |\Delta u_k^1 |^2-|\Delta u _k^2|^2\textrm{d}x \right] \ge 0. \end{aligned}$$

  (d)

Proof

As mentioned in the proof of [3, Lemma 3.5], the proof is done by considering the concentration function

$$\begin{aligned} R \rightarrow Q_k(R) := \sup _{y\in {\mathbb {R}}^n}\,\int \limits _{B_R(y)}|\nabla \! u_k|^2\textrm{d}x \end{aligned}$$

for $R \in [0,\infty )$ and following the same steps as in [10]. $\square$

We will see that for the sequence of eigenfunctions $(u_k)_k$ the case of vanishing and dichotomy cannot occur. Hence, $(u_k)_k$ contains a subsequence, which we again denote by $(u_k)_k$, for which the case of compactness holds true. This compactness will imply the weak convergence of $u_k$ to a limit function u in $W^{2,2}({\mathbb {R}}^2)$. Moreover, the compactness yields that $u_k$ converges to u strongly in $W^{1,2}({\mathbb {R}}^n)$.

The case of dichotomy can be disproved in exactly the same way as in [3]. For the sake of brevity, we forgo the repetition of this argument.

In order to disprove the case of vanishing, we slightly differ from [3]. Nevertheless, we adopt the following lemma [4, Lemma 3.3] (or [9, Lemma 6]) which is used in [3] and which we will apply to disprove the vanishing, as well.

Lemma 2

Let $(w_k)_k$ be a bounded sequence in $W^{1,2}({\mathbb {R}}^n)$ such that $\Vert w_k\Vert _{L^2({\mathbb {R}}^n)} = 1$ and $w_k\in W^{1,2}_0(D_k)$ for a $D_k \in {\mathcal {O}}_{\omega _0}$. There exists a sequence of vectors $(y_k)_k \subset {\mathbb {R}}^n$ such that the sequence $(w_k(\cdot +y_k))_k$ does not possess a subsequence converging weakly to zero in $W^{1,2}_0({\mathbb {R}}^n)$.

Now let us assume that for a subsequence of $(u_k)_k$, again denoted by $(u_k)_k$, the case of vanishing occurs. Hence, for every $R>0$ there holds

$$\begin{aligned} \lim _{k\rightarrow \infty } \sup _{y \in {\mathbb {R}}^n} \int \limits _{B_R(y)}|\nabla \! u_k|^2\textrm{d}x =0. \end{aligned}$$

(3)

Since there holds $\Vert \nabla \! u\Vert _{L^2({\mathbb {R}}^n)}=1$ for every $k \in {\mathbb {N}}$, we obtain for at least one $1\le l_k \le n$

$$\begin{aligned} \int \limits _{{\mathbb {R}}^n} |\partial _{l_k} u_k|^2\textrm{d}x \ge \frac{1}{n}. \end{aligned}$$

We now consider the sequence $(\partial _{l_k} u_k)_k$. Then $\partial _{l_k} u_k \in W^{1,2}_0(\Omega _k)$ and

$$\begin{aligned} \frac{1}{\sqrt{n}} \le \Vert \partial _{l_k} u_k\Vert _{L^2({\mathbb {R}}^n)} := c_k. \end{aligned}$$

The sequence $(v_k)_k$ given by $v_k := c_k^{-1}\partial _{l_k}u_k$ then satisfies the assumptions of Lemma 2. Consequently, there exists a sequence $(y_k)_k \subset {\mathbb {R}}^n$ such that the sequence $(v_k(\cdot +y_k))_k \subset W^{1,2}({\mathbb {R}}^n)$ does not possess a subsequence which converges weakly to zero in $W^{1,2}({\mathbb {R}}^n)$. However, the sequence $(v_k(\cdot +y_k))_k$ is uniformly bounded in $W^{1,2}({\mathbb {R}}^n)$ because of the normalization. Hence, there exists a $v \in W^{1,2}({\mathbb {R}}^n)$ such that a subsequence of $(v_k(\cdot +y_k))_k$ converges weakly in $W^{1,2}({\mathbb {R}}^n)$ to v. In particular, there holds

$$\begin{aligned} v_k(\cdot +y_k) {\mathop {\rightharpoonup }\limits ^{k\rightarrow \infty }} v \text { in } W^{1,2}(B_R(0)) \text { for every } R>0 \end{aligned}$$

and

$$\begin{aligned} v_k(\cdot +y_k) {\mathop {\longrightarrow }\limits ^{k\rightarrow \infty }} v \text { in }L^2(B_R(0)) \text { for every } R>0. \end{aligned}$$

Thus, we obtain

$$\begin{aligned} \Vert v\Vert ^2_{L^2(B_R(0))}&= \lim _{k\rightarrow \infty }\Vert v_k(\cdot +y_k)\Vert ^2_{L^2 (B_R(0))}=\lim _{k\rightarrow \infty }\frac{1}{c_k^2}\int \limits _{B_R(0)}|\partial _l u_k(x+y_k)|^2\textrm{d}x \\&\le n \lim _{k\rightarrow \infty }\int \limits _{B_R(y)}|\nabla \! u_k|^2dx \\&\le n \lim _{k\rightarrow \infty } \sup _{y \in {\mathbb {R}}^n} \int \limits _{B_R(y)}|\nabla \! u_k|^2\textrm{d}x {\mathop {=}\limits ^{(3)}} 0. \end{aligned}$$

Hence, $v =0$ in $L^2(B_R(0))$ and since v is the weak limit of $v_k(\cdot +y_k)$ this is a contradiction to Lemma 2. Therefore, the case of vanishing cannot occur.

Consequently, the case of compactness must occur. Following the lines of [3], we find that there exists a sequence $(y_k)_k \subset {\mathbb {R}}^n$ and an $u \in W^{2,2}({\mathbb {R}}^n)$ such that

$$\begin{aligned} u_k(\cdot + y_k) \rightharpoonup u \text { in } W^{2,2}({\mathbb {R}}^n) \end{aligned}$$

(4)

and, since we are in the compactness case of Lemma 1,

$$\begin{aligned} \int \limits _{{\mathbb {R}}^n}|\nabla \! u|^2\textrm{d}x =1. \end{aligned}$$

(5)

From now on, we set

$$\begin{aligned} u_k = u_k(\cdot +y_k)\quad \text { and } \quad \Omega _k = \Omega _k + y_k, \end{aligned}$$

where $(y_k)_k$ is given above. This is possible without loss of generality because of the translational invariance of the buckling load.

We now show that $u_k$ converges strongly to u in $W^{1,2}({\mathbb {R}}^n)$. Since this observation will be crucial for constructing an optimal domain in Sect. 2.2, we give a detailed proof although we follow the lines of [3].

Lemma 3

There holds

$$\begin{aligned} u_k {\mathop {\longrightarrow }\limits ^{k\rightarrow \infty }} u \text { in } W^{1,2}({\mathbb {R}}^n). \end{aligned}$$

Proof

We use the notation above. Recall, that $u_k = u_k(\cdot +y_k)$ and $\Omega _k = \Omega _k + y_k$. Then we get from (5)

$$\begin{aligned} \int \limits _{{\mathbb {R}}^n}|\nabla \! u -\nabla \! u_k|^2\textrm{d}x = 2- 2\int \limits _{{\mathbb {R}}^n}\nabla \! u.\nabla \! u_k\,\textrm{d}x \end{aligned}$$

and the weak convergence of $(u_k)_k$ to u in $W^{2,2}({\mathbb {R}}^n)$ yields

$$\begin{aligned} \int \limits _{{\mathbb {R}}^n}|\nabla \! u -\nabla \! u_k|^2\textrm{d}x {\mathop {\longrightarrow }\limits ^{k\rightarrow \infty }} 0. \end{aligned}$$

Thus, $(\nabla \! u_k)_k$ converges to $\nabla \! u$ in $L^2({\mathbb {R}}^n)$ and, in particular, $(\nabla \! u_k)_k$ is a Cauchy sequence in $L^2({\mathbb {R}}^n,{\mathbb {R}}^n)$. Now let $l,k \in {\mathbb {N}}$. Then $u_l-u_k \in W^{2,2}_0(\Omega _l\cup \Omega _k)$ and applying Poincaré’s inequality we obtain

$$\begin{aligned} \int \limits _{\Omega _l\cup \Omega _k}(u_l-u_k)^2\textrm{d}x&\le \left( \frac{|\Omega _l\cup \Omega _k|}{\omega _n}\right) ^\frac{2}{n}\int \limits _{\Omega _l\cup \Omega _k}|\nabla \! (u_l-u_k)|^2dx \\&\le \left( \frac{2\omega _0}{\omega _n}\right) ^\frac{2}{n}\int \limits _{\Omega _l\cup \Omega _k}|\nabla \! (u_l-u_k)|^2\textrm{d}x {\mathop {\longrightarrow }\limits ^{k\rightarrow \infty }} 0. \end{aligned}$$

Thus, $(u_k)_k$ is a Cauchy sequence in $L^2({\mathbb {R}}^n)$, which converges weakly in $L^2({\mathbb {R}}^n)$ to u. Consequently, $u_k {\mathop {\longrightarrow }\limits ^{k\rightarrow \infty }} u$ in $L^2({\mathbb {R}}^n)$. This proves the claim. $\square$

As a consequence of (4) and Lemma 3, we obtain that

$$\begin{aligned} {\mathcal {R}}(u,{\mathbb {R}}^n) \le \liminf _{k\rightarrow \infty } {\mathcal {R}}(u_k,\Omega _k) = \inf _{\Omega \in {\mathcal {O}}_{\omega _0}}\Lambda (\Omega ). \end{aligned}$$

(6)

The following proposition summarizes what we have achieved so far.

Proposition 1

Let $(\Omega _k)_k \subset {\mathcal {O}}_{\omega _0}$ be a minimizing sequence for the buckling load in ${\mathcal {O}}_{\omega _0}$ and $(u_k)_k$ be the sequence of corresponding normalized eigenfunctions. Then there exists a sequence $(y_k)_k \subset {\mathbb {R}}^n$ such that $u_k(\cdot + y_k)$ is a normalized eigenfunction on $\Omega _k+y_k$ and, denoting $u_k = u_k(\cdot +y_k)$ and $\Omega _k = \Omega _k+y_k$, there exists a subsequence, again denoted by $(u_k)_k$, and an $u \in W^{2,2}({\mathbb {R}}^n)$ with

• 1. u is normalized by

  $$\begin{aligned} \int \limits _{{\mathbb {R}}^n}|\nabla \! u|^2\textrm{d}x =1. \end{aligned}$$

• 2. $u_k \rightharpoonup u$ in $W^{2,2}({\mathbb {R}}^n)$ as k tends to $\infty$.

• 3. $u_k \longrightarrow u$ in $W^{1,2}({\mathbb {R}}^n)$ as k tends to $\infty$.

• 4. There holds ${\mathcal {R}}(u,{\mathbb {R}}^n) \le \inf _{\Omega \in {\mathcal {O}}_{\omega _0}}\Lambda (\Omega )$.

Recall that in [3] only the two-dimensional case is considered. Consequently, the limit function u is continuous due to Sobolev’s embedding theory. Hence, the set

$$\begin{aligned} {\tilde{\Omega }} := \{x \in {\mathbb {R}}^2: u(x) \ne 0\} \end{aligned}$$

is an open set and the strong $L^2$-convergence of $u_k$ to u implies that ${\tilde{\Omega }}\in {\mathcal {O}}_{\omega _0}$.

Here, we consider arbitrary dimension. Hence, we need another method to prove regularity of the function u. Inspired by [2], our approach is based on a careful analysis of the function u. This will be done in the next section.

2.1 Regularity of the limit function

Our first aim is to show that u has got Hölder continuous first-order derivatives. This will be done by using Morrey’s Dirichlet Growth Theorem (see Theorem 1) and a bootstrapping argument based on ideas of Q. Han and F. Lin in [7].

From now on, we consider a minimizing sequence $(\Omega _k)_k \subset {\mathcal {O}}_{\omega _0}$ such that there holds

$$\begin{aligned} \Lambda _{\omega _0}:= \inf _{\Omega \in {\mathcal {O}}_{\omega _0}}\Lambda (\Omega ) \le \Lambda (\Omega _k) \le \Lambda _{\omega _0}+ \frac{1}{k} \text { for every } k \in {\mathbb {N}}. \end{aligned}$$

(7)

We want to apply the following version of Morrey’s Dirichlet Growth Theorem to the first-order derivatives of u.

Theorem 1

Let $v \in W^{1,2}({\mathbb {R}}^n)$ and $0<\alpha \le 1$ such that for every $x_0 \in {\mathbb {R}}^n$ and every $0<r\le r_0$ there holds

$$\begin{aligned} \int \limits _{B_r(x_0)}|\nabla \! v|^2\textrm{d}x \le M\cdot r^{n-2+2\alpha }. \end{aligned}$$

Then v is $\alpha$-Hölder continuous almost everywhere in ${\mathbb {R}}^n$ and for almost every $x_1,x_2 \in {\mathbb {R}}^n$ there holds

$$\begin{aligned} \frac{|v(x_1)-v(x_2)|}{|x_1-x_2|^\alpha }\le C(\alpha )\cdot M. \end{aligned}$$

For a proof of this theorem we refer to [11, Theorem 3.5.2], for example. Hence, we need a $L^2$-estimate for the second-order derivatives of u in every ball $B_r(x_0) \subset {\mathbb {R}}^n$.

The following lemmata are preparatory for the proof of Theorem 2, which is the main theorem of this section. Before we start, note that by scaling there holds

$$\begin{aligned} \Lambda _{\omega _0}\le \left( \frac{\omega _n}{\omega _0}\right) ^\frac{2}{n}\Lambda (B_1) \le C(n,\omega _0), \end{aligned}$$

(8)

where $B_1$ denotes the unit ball in ${\mathbb {R}}^n$.

Lemma 4

Let $u \in W^{2,2}({\mathbb {R}}^n)$ be the limit function according to Proposition 1 and $0<R\le 1$. There exists a constant $C=C(n,\omega _0)>0$ such that for every $x_0 \in {\mathbb {R}}^n$ there holds

$$\begin{aligned} \int \limits _{B_R(x_0)}|\Delta ( u-v_0)|^2\textrm{d}x \le C(n,\omega _0)\left( R^n + \int \limits _{B_R(x_0)}|\nabla \! u|^2dx\right) , \end{aligned}$$

where $v_0 \in W^{2,2}(B_R(x_0))$ with $v_0-u\in W^{2,2}_0(B_R(x_0))$ and $\Delta ^{\!2}v _0=0$ in $B_R(x_0)$.

Proof

The proof is done in three steps.

Step 1    We choose $x_0 \in {\mathbb {R}}^n$ arbitrary, but fixed. Let $v_k \in W^{2,2}(B_R(x_0))$ with $v_k-u_k \in W^{2,2}_0(B_R(x_0))$ and $\Delta ^{\!2}v _k =0$ in $B_R(x_0)$. If $B_R(x_0)\cap \Omega _k =\emptyset$, $u_k$ and $v_k$ vanish in $B_R(x_0)$. Consequently, we obtain

$$\begin{aligned} \int \limits _{B_R(x_0)}|\Delta ( u_k-v_k)|^2\textrm{d}x =0. \end{aligned}$$

(9)

If $B_R(x_0)\cap \Omega _k \ne \emptyset$, we set

$$\begin{aligned} {\hat{u}}_k = {\left\{ \begin{array}{ll} u_k, &{}\text { in } {\mathbb {R}}^n\setminus B_R(x_0) \\ v_k, &{}\text { in } B_R(x_0) \end{array}\right. }. \end{aligned}$$

Note that $\Omega _k\cup B_R(x_0)$ is an open set and that ${\hat{u}}_k \in W^{2,2}_0(\Omega _k\cup B_R(x_0))$. Let us first consider the case $|\Omega _k \cup B_R(x_0)| \le \omega _0$. Hence, $\Omega _k\cup B_R(x_0) \in {\mathcal {O}}_{\omega _0}$ and there holds

$$\begin{aligned} \Lambda _{\omega _0}= \inf _{\Omega \in {\mathcal {O}}_{\omega _0}}\Lambda (\Omega ) \le \Lambda (\Omega _k\cup B_R(x_0)) \le {\mathcal {R}}({\hat{u}}_k,{\mathbb {R}}^n) \end{aligned}$$

since ${\hat{u}}_k \in W^{2,2}_0(\Omega _k\cup B_R(x_0))$. Rearranging terms and applying the definition of ${\hat{u}}_k$ yields

$$\begin{aligned} \Lambda _{\omega _0}\left( 1-\int \limits _{B_R(x_0)}|\nabla \! u_k|^2\textrm{d}x\right) \le \Lambda (\Omega _k) - \int \limits _{B_R(x_0)}|\Delta u _k|^2 - |\Delta v _k|^2\textrm{d}x. \end{aligned}$$

(10)

Since $v_k - u_k \in W^{2,2}_0(B_R(x_0))$ and $v_k$ is biharmonic in $B_R(x_0)$, there holds

$$\begin{aligned} \int \limits _{B_R(x_0)}|\Delta u _k|^2 - |\Delta v _k|^2\textrm{d}x = \int \limits _{B_R(x_0)}|D^2(u_k-v_k)|^2\textrm{d}x, \end{aligned}$$

where $D^2v$ denotes the Hessian matrix of a function v. Moreover, we denote by

$$\begin{aligned} |D^2v|^2 := \sum _{i,j=1}^n\partial _{ij}v \partial _{ij}v \end{aligned}$$

the Euclidian norm of the matrix $D^2v$ as a vector in ${\mathbb {R}}^{n^2}$. We rearrange terms in (10) and obtain

$$\begin{aligned} \int \limits _{B_R(x_0)}|D^2(u_k-v_k)|^2\textrm{d}x \le \Lambda (\Omega _k) - \Lambda _{\omega _0}+ \Lambda _{\omega _0}\int \limits _{B_R(x_0)}|\nabla \! u_k|^2\textrm{d}x. \end{aligned}$$

(11)

Let us now assume that $|\Omega _k\cup B_R(x_0)|> \omega _0$. Then we set

$$\begin{aligned} \mu _k := \left( \frac{|\Omega _k|+|B_R|}{|\Omega _k|}\right) ^\frac{1}{n}. \end{aligned}$$

(12)

and find that $\mu _k^{-1}\cdot (\Omega _k\cup B_R(x_0)) \in {\mathcal {O}}_{\omega _0}$. Recall that for every $M \subset {\mathbb {R}}^n$ and $t>0$ the buckling load satisfies

$$\begin{aligned} \Lambda (M) = t^2\Lambda (t M). \end{aligned}$$

(13)

Hence, we obtain

$$\begin{aligned} \Lambda _{\omega _0}\le \Lambda (\mu _k^{-1}(\Omega _k\cup B_R(x_0))) = \mu _k^2\,\Lambda (\Omega _k\cup B_R(x_0)) \le \mu _k^2\,{\mathcal {R}}({\hat{u}}_k,{\mathbb {R}}^n). \end{aligned}$$

and, subsequently,

$$\begin{aligned} \mu _k^2\int \limits _{B_R(x_0)}|D^2(u_k-v_k)|^2\textrm{d}x \le \mu _k^2\,\Lambda (\Omega _k) -\Lambda _{\omega _0}+ \Lambda _{\omega _0}\int \limits _{B_R(x_0)}|\nabla \! u_k|^2\textrm{d}x. \end{aligned}$$

Since $\mu _k>1$, we proceed to

$$\begin{aligned} \int \limits _{B_R(x_0)}|D^2(u_k-v_k)|^2\textrm{d}x \le \mu _k^2\,\Lambda (\Omega _k) -\Lambda _{\omega _0}+ \Lambda _{\omega _0}\int \limits _{B_R(x_0)}|\nabla \! u_k|^2\textrm{d}x. \end{aligned}$$

(14)

Consequently, we can collect the estimates (9), (11) and (14) in just one estimate: for every $x_0 \in {\mathbb {R}}^n$ and every $k \in {\mathbb {N}}$ there holds

$$\begin{aligned} \int \limits _{B_R(x_0)}|D^2(u_k-v_k)|^2\textrm{d}x \le \mu _k^2\,\Lambda (\Omega _k) - \Lambda _{\omega _0}+ \Lambda _{\omega _0}\int \limits _{B_R(x_0)}|\nabla \! u_k|^2\textrm{d}x. \end{aligned}$$

(15)

Step 2 We want to understand the limit as k tends to $\infty$ on both sides of (15). This needs some preparation. First, recall that we choose a minimizing sequence $(\Omega _k)_k$ such that (7) holds. Then applying (13) yields

$$\begin{aligned} \Lambda _{\omega _0}&\le \Lambda \left( \left( \frac{\omega _0}{|\Omega _k|}\right) ^\frac{1}{n}\Omega _k\right) = \left( \frac{|\Omega _k|}{\omega _0}\right) ^\frac{2}{n}\Lambda (\Omega _k) {\mathop {\le }\limits ^{(7)}} \left( \frac{|\Omega _k|}{\omega _0}\right) ^\frac{2}{n}\left( \Lambda _{\omega _0}+ \frac{1}{k}\right) . \end{aligned}$$

Rearranging terms yields

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

Thus, there holds $|\Omega _k|\rightarrow \omega _0$ as k tends to $\infty$. This immediately implies that

$$\begin{aligned} \mu _k {\mathop {\longrightarrow }\limits ^{k\rightarrow \infty }} \left( 1+\frac{|B_R|}{\omega _0}\right) ^\frac{1}{n}, \end{aligned}$$

where $\mu _k$ is given in (12). In addition, recall that $u_k \rightharpoonup u$ in $W^{2,2}({\mathbb {R}}^n)$ and, therefore, $u_k \rightharpoonup u$ in $W^{2,2}(B_R(x_0))$. In order to prove the weak convergence of the sequence $(v_k)_k$ in $W^{2,2}(B_R(x_0))$, we estimate the $W^{2,2}(B_R(x_0))$-norm of $v_k$ independently of k for every $k\in {\mathbb {N}}$. We start with the $W^{2,2}(B_R(x_0))$-norm of $u_k - v_k$ and apply Poincaré’s inequality (as stated in [6, Formula (7.44)]) on the first summand in the integral on the right-hand side. This yields

$$\begin{aligned} \Vert u_k-v_k\Vert _{W^{2,2}(B_R(x_0))}^2&= \int \limits _{B_R(x_0)} |u_k-v_k|^2 + |\nabla \! (u_k-v_k)|^2 + |D^2(u_k-v_k)|^2\textrm{d}x \\&\le (R^2+1) \int \limits _{B_R(x_0)}|\nabla \! (u_k-v_k)|^2dx + \int \limits _{B_R(x_0)}|D^2(u_k-v_k)|^2\textrm{d}x \\&= (R^2+1) \sum _{i=1}^n\int \limits _{B_R(x_0)} |\partial _i(u_k-v_k)|^2\textrm{d}x + \int \limits _{B_R(x_0)}|D^2(u_k-v_k)|^2\textrm{d}x. \end{aligned}$$

Applying Poincaré’s inequality on the integrals containing the first-order derivatives of $u_k-v_k$, we obtain

$$\begin{aligned} \Vert u_k-v_k\Vert _{W^{2,2}(B_R(x_0))}^2 \le n (R^2+1) \int \limits _{B_R(x_0)}|D^2(u_k-v_k)|^2\textrm{d}x \le 2n \int \limits _{B_R(x_0)}|D^2(u_k-v_k)|^2\textrm{d}x, \end{aligned}$$

where we additionally used that $R<1$. Now Young’s inequality applied in the last integral yields

$$\begin{aligned} \Vert u_k-v_k\Vert _{W^{2,2}(B_R(x_0))}^2 \le 4n \left( \int \limits _{B_R(x_0)}|D^2u_k|^2+ |D^2v_k|^2\textrm{d}x\right) . \end{aligned}$$

Recall that $u_k-v_k \in W^{2,2}_0(B_R(x_0))$ and $\Delta ^{\!2}v _k =0$ in $B_R(x_0)$. Hence, there holds

$$\begin{aligned} \int \limits _{B_R(x_0)}|D^2v_k|^2\textrm{d}x = \min \left\{ \int \limits _{B_R(x_0)}|D^2w|^2\textrm{d}x : w - v_k \in W^{2,2}_0(B_R(x_0))\right\} \le \int \limits _{B_R(x_0)}|D^2u_k|^2\textrm{d}x \end{aligned}$$

and, consequently,

$$\begin{aligned} \Vert u_k-v_k\Vert _{W^{2,2}(B_R(x_0))}^2 \le 8n \int \limits _{B_R(x_0)}|D^2u_k|^2\textrm{d}x \le 8n \Vert u_k\Vert _{W^{2,2}(B_R(x_0))}. \end{aligned}$$

Since $u_k \rightharpoonup u$ in $W^{2,2}(B_R(x_0))$, there exists a constant C(n), which is only depended on the dimension n, such that

$$\begin{aligned} \Vert u_k-v_k\Vert _{W^{2,2}(B_R(x_0))}^2 \le 8n \int \limits _{B_R(x_0)}|D^2u_k|^2\textrm{d}x \le 8n \Vert u_k\Vert _{W^{2,2}(B_R(x_0))} \le C(n). \end{aligned}$$

Consequently, for all $k\in {\mathbb {N}}$ there holds

$$\begin{aligned} \Vert v_k\Vert _{W^{2,2}(B_R(x_0))} \le \Vert u_k-v_k\Vert _{W^{2,2}(B_R(x_0))} + \Vert u_k\Vert _{W^{2,2}(B_R(x_0))} \le C(n) \end{aligned}$$

and there exists a $v_0 \in W^{2,2}(B_R(x_0))$ such that

$$\begin{aligned} v_k \rightharpoonup v_0 \text { in } W^{2,2}(B_R(x_0)) \text { and } v_0-u \in W^{2,2}_0(B_R(x_0)). \end{aligned}$$

Moreover, for every $\phi \in C^\infty _c(B_R(x_0))$ there holds

$$\begin{aligned} 0 = \lim _{k\rightarrow \infty }\int \limits _{B_R(x_0)}\Delta v _k\Delta \phi \,\textrm{d}x = \int \limits _{B_R(x_0)}\Delta v _0 \Delta \phi \,\textrm{d}x \end{aligned}$$

because $v_k$ is biharmonic in $B_R(x_0)$ for every $k\in {\mathbb {N}}$ and the weak convergence of $v_k$ to $v_0$ in $W^{2,2}(B_R(x_0))$. Hence, $v_0$ is biharmonic in $B_R(x_0)$.

Step 3 We take the $\liminf$ on both sides of (15). Since $u_k \rightharpoonup u$ in $W^{2,2}_0(B_r(x_0))$, this leads to

$$\begin{aligned} \int \limits _{B_R(x_0)}|D^2(u-v_0)|^2\textrm{d}x&\le \liminf _{k\rightarrow \infty }\int \limits _{B_R(x_0)}|D^2(u_k-v_k)|^2\textrm{d}x \\&\le \liminf _{k\rightarrow \infty }\left( \mu _k^2\,\Lambda (\Omega _k) - \Lambda _{\omega _0}+ \Lambda _{\omega _0}\int \limits _{B_R(x_0)}|\nabla \! u_k|^2dx\right) \\&= \left( 1+\frac{|B_R|}{\omega _0}\right) ^\frac{2}{n} \Lambda _{\omega _0}-\Lambda _{\omega _0}+\Lambda _{\omega _0}\int \limits _{B_R(x_0)}|\nabla \! u|^2\textrm{d}x \\&\le C(n,\omega _0)\left( R^n + \int \limits _{B_R(x_0)}|\nabla \! u|^2dx\right) . \end{aligned}$$

This proves the claim. $\square$

Now let $v_0 \in W^{2,2}(B_R(x_0))$ be the function from Lemma 4 and $0<r\le R$. Then there obviously holds

$$\begin{aligned} \int \limits _{B_r(x_0)}|D^2u|^2dx \le 2 \int \limits _{B_r(x_0)}|D^2v_0|^2\textrm{d}x + 2 \int \limits _{B_R(x_0)}|D^2(u-v_0)|^2\textrm{d}x \end{aligned}$$

(16)

and applying Lemma 4 yields

$$\begin{aligned} \int \limits _{B_r(x_0)}|D^2u|^2\textrm{d}x \le 2 \int \limits _{B_r(x_0)}|D^2v_0|^2\textrm{d}x +C(n,\omega _0)\left( R^n + \int \limits _{B_R(x_0)}|\nabla \! u|^2\textrm{d}x\right) . \end{aligned}$$

(17)

In order to estimate the first summand on the right-hand side of the above inequality, we cite Lemma 2.1 from [13].

Lemma 5

Using the notation above, there exists a constant $C=C(n)>0$ such that for $0< r\le R$ there holds

$$\begin{aligned} \int \limits _{B_r(x_0)}|D^2v_0|^2 \le C(n)\left( \frac{r}{R}\right) ^n\int \limits _{B_R(x_0)}|D^2u|^2\textrm{d}x. \end{aligned}$$

The constant C does not depend on r, R or $x_0$, but on the dimension n.

Thus, (17) becomes

$$\begin{aligned} \begin{aligned} \int \limits _{B_r(x_0)}|D^2u|^2\textrm{d}x \le C(n)\left( \frac{r}{R}\right) ^n&\int \limits _{B_R(x_0)}|D^2u|^2\textrm{d}x&+C(n,\omega _0)\left( R^n + \int \limits _{B_R(x_0)}|\nabla \! u|^2\textrm{d}x\right) . \end{aligned}\end{aligned}$$

(18)

This estimate will be the starting point for the bootstrapping argument which will lead to the Hölder-continuity of the first-order derivatives of u.

From [5, Chapter III, Lemma 2.1] we cite the next lemma.

Lemma 6

Let $\Phi$ be a nonnegative 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

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

The following lemma is based on ideas of [7, Chapter 3]. It will be the crucial observation for the bootstrapping.

Lemma 7

Suppose that for each $0\le r\le 1$ there holds

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

where $M>0$ and $\mu \in [0,n)$. Then there exists a constant $C(n)>0$ such that for each $0\le r\le 1$

$$\begin{aligned} \int \limits _{B_r(x_0)}|\nabla \! u|^2\textrm{d}x \le C(n,M)\,r^\lambda , \end{aligned}$$

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

Proof

Let $0\le r\le s \le 1$. For a function $w \in W^{1,2}({\mathbb {R}}^n)$ we set

Using this notation, we write

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

Then Young’s inequality implies

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

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

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

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

Now Lemma 6 implies that for each $0\le r\le s\le 1$ there holds

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

where $\lambda = \mu +2$ if $\mu < n-2$ and $\lambda$ is arbitrary in (0, n) if $n-2 \le \mu < n$. Choosing $s=1$, we deduce

$$\begin{aligned} \int \limits _{B_r(x_0)}|\nabla \! u|^2\textrm{d}x&\le C(n,M)\,r^\lambda . \end{aligned}$$

$\square$

Now we are able to prove the Hölder continuity of the first-order derivatives of u.

Theorem 2

Let u be the limit function according to Proposition 1. The first-order derivatives of u are $\alpha$-Hölder continuous almost everywhere on ${\mathbb {R}}^n$ for every $\alpha \in (0,1)$.

Proof

Our aim is to show that for every $x_0 \in {\mathbb {R}}^n$ and every $0<r\le 1$ there holds

$$\begin{aligned} \int \limits _{B_r(x_0)}|D^2u|^2\textrm{d}x \le C(n,\omega _0)\,r^{n-2+2\alpha }. \end{aligned}$$

(19)

Then Theorem 1 finishes the proof. Let us choose $x_0 \in {\mathbb {R}}^n$, $0<r\le R\le 1$ and recall estimate (18):

$$\begin{aligned} \int \limits _{B_r(x_0)}|D^2u|^2\textrm{d}x \le C(n)\left( \frac{r}{R}\right) ^n\int \limits _{B_R(x_0)}|D^2u|^2\textrm{d}x +C(n,\omega _0)\left( R^n + \int \limits _{B_R(x_0)}|\nabla \! u|^2\textrm{d}x\right) . \end{aligned}$$

We will improve this estimate using a bootstrap argument based on Lemma 7. Note that for every $0<r\le 1$ there holds

$$\begin{aligned} \int \limits _{B_r(x_0)}|D^2u|^2\textrm{d}x \le \Vert u\Vert _{W^{2,2}({\mathbb {R}}^n)}^2 = \Vert u\Vert _{W^{2,2}({\mathbb {R}}^n)}^2\,r^0. \end{aligned}$$

(20)

Then Lemma 7 implies that for every $0<r\le 1$ there holds

$$\begin{aligned} \int \limits _{B_r(x_0)}|\nabla \! u|^2\textrm{d}x \le C(n, \Vert u\Vert _{W^{2,2}({\mathbb {R}}^n)})\,r^{\lambda _0}, \end{aligned}$$

(21)

where $\lambda _0 \in (0,n)$ if $n=2$ and $\lambda _0 = 2$ if $n\ge 3$. We insert this estimate (18). Since $R\le 1$ we obtain

$$\begin{aligned} \int \limits _{B_r(x_0)} |D^2u|^2dx\le C(n)\left( \frac{r}{R}\right) ^2\int \limits _{B_R(x_0)}|D^2u|^2\textrm{d}x + C(n,\omega _0, \Vert u\Vert _{W^{2,2}({\mathbb {R}}^n)})\,R^{\lambda _0} \end{aligned}$$

for every $0< r\le R$. Applying Lemma 6, we obtain

$$\begin{aligned} \int \limits _{B_r(x_0)}|D^2u|^2\textrm{d}x \le C\,\left( \frac{r}{R}\right) ^{\lambda _0}\left( \int \limits _{B_R(x_0)}|D^2u|^2\textrm{d}x + C(n,\omega _0, \Vert u\Vert _{W^{2,2}({\mathbb {R}}^n)})\,R^{\lambda _0}\right) \end{aligned}$$

for every $0<r\le R$. Choosing $R=1$ leads to

$$\begin{aligned} \int \limits _{B_r(x_0)}|D^2u|^2\textrm{d}x \le C(n,\omega _0, \Vert u\Vert _{W^{2,2}({\mathbb {R}}^n)})r^{\lambda _0} \end{aligned}$$

(22)

for every $0<r\le 1$. If $n=2$, this is (19).

If $n \ge 3$, (22) is an improvement of estimate (20). Recall that here holds $\lambda _0=2$. We again apply Lemma 7 and obtain for every $0<r\le 1$

$$\begin{aligned} \int \limits _{B_r(x_0)}|D^2u|^2dx \le C(n,\omega _0, \Vert u\Vert _{W^{2,2}({\mathbb {R}}^n)})r^{\lambda _1}, \end{aligned}$$

where $\lambda _1 \in (0,n)$ if $n\in \{3,4\}$ and $\lambda _1 = 4$ if $n\ge 5$. Together with estimate (18) we find that

$$\begin{aligned} \int \limits _{B_r(x_0)}|D^2u|^2\textrm{d}x \le C(n)\left( \frac{r}{R}\right) ^n\int \limits _{B_R(x_0)}|D^2u|^2\textrm{d}x + C(n,\omega _0, \Vert u\Vert _{W^{2,2}({\mathbb {R}}^n)}) R^{\lambda _1} \end{aligned}$$

for every $0<r\le R\le 1$. Then Lemma 6 implies

$$\begin{aligned} \int \limits _{B_r(x_0)}|D^2u|^2\textrm{d}x \le C\,\left( \frac{r}{R}\right) ^{\lambda _1}\left( \int \limits _{B_R(x_0)}|D^2u|^2\textrm{d}x + C(n,\omega _0, \Vert u\Vert _{W^{2,2}({\mathbb {R}}^n)})\,R^{\lambda _1}\right) \end{aligned}$$

and choosing $R=1$ there holds

$$\begin{aligned} \int \limits _{B_r(x_0)}|D^2u|^2\textrm{d}x \le C(n,\omega _0, \Vert u\Vert _{W^{2,2}({\mathbb {R}}^n)})r^{\lambda _1} \end{aligned}$$

(23)

for every $0<r\le 1$. For $n \in \{3,4\}$, estimate (23) and Theorem 1 proves the claim.

If $n\ge 6$, we repeat the argumentation since (23) is an improvement of (22). Repeating this process proves the claim after finite many steps for every $n\ge 2$. $\square$

Due to Theorem 2 the limit function u has a unique representative in $W^{2,2}({\mathbb {R}}^n)$ which is continuous in ${\mathbb {R}}^n$ and which has $\alpha$-Hölder continuous first-order derivatives in ${\mathbb {R}}^n$ for every $\alpha \in (0,1)$. From now on, we rename this representative as u and focus on this function.

2.2 The minimizing domain

The regularity of u, which we achieved in the previous section, enables us to construct an optimal domain for minimizing the buckling load in ${\mathcal {O}}_{\omega _0}$. Recall that there holds (see (6))

$$\begin{aligned} {\mathcal {R}}(u,{\mathbb {R}}^n) \le \inf _{\Omega \in {\mathcal {O}}_{\omega _0}}\Lambda (\Omega ). \end{aligned}$$

If $u \in W^{2,2}_0(\Omega ^*)$ for a suitable set $\Omega ^*\in {\mathcal {O}}_{\omega _0}$, this set $\Omega ^*$ is the desired minimizer. Thus, the challenge is to construct a suitable $\Omega ^*$.

Let us define

$$\begin{aligned} {\tilde{\Omega }} := \{ x \in {\mathbb {R}}^n: u(x)\ne 0 \} \end{aligned}$$

and let $(\Omega _k)_k\subset {\mathcal {O}}_{\omega _0}$ be a minimizing sequence and $(u_k)_k \subset W^{2,2}({\mathbb {R}}^n)$ the corresponding sequence of eigenfunctions according to Proposition 1. Since $u_k$ converges strongly to u in $L^2({\mathbb {R}}^n)$, $u_k$ converges locally in measure to u, i.e., for every compact set $C \subset {\mathbb {R}}^n$ and every $\varepsilon >0$ there holds

$$\begin{aligned} |\{x \in C: |u_k(x)-u(x)| \ge \varepsilon \}| {\mathop {\longrightarrow }\limits ^{k\rightarrow \infty }} 0. \end{aligned}$$

Now let C be any compact subset of ${\tilde{\Omega }}$. Since u is a continuous function and cannot vanish in ${\tilde{\Omega }}$, there exists a constant $m_C>0$ such that

$$\begin{aligned} |u(x)|\ge m_C \text { for every }x \in C. \end{aligned}$$

Recall that $u_k$ vanishes pointwise in $\Omega _k^c$. Thus, there holds

$$\begin{aligned} |u(x)| = |u_k(x)-u(x)| \text { for every } x \in \Omega _k^c. \end{aligned}$$

Hence, we obtain

$$\begin{aligned} |\Omega _k^c \cap C| \le |\{x \in C: |u_k(x)-u(x)|\ge m_C\}| {\mathop {\longrightarrow }\limits ^{k\rightarrow \infty }} 0. \end{aligned}$$

Consequently, for every compact $C \subset {\tilde{\Omega }}$ and every $k\in {\mathbb {N}}$ there holds

$$\begin{aligned} |C|&= |C \cap \Omega _k| + |C \cap \Omega _k^c| \le |\Omega _k| +|C \cap \Omega _k^c| \le \omega _0 + |C \cap \Omega _k^c| \end{aligned}$$

and letting k tend to infinity we find that $|C| \le \omega _0$ for every compact subset C of ${\tilde{\Omega }}$. Hence,

$$\begin{aligned} |{\tilde{\Omega }}| = \sup \{|C|: C\subset {\tilde{\Omega }}, C \text { compact}\}\le \omega _0. \end{aligned}$$

We now denote

$$\begin{aligned} {\hat{\Omega }} = \{x \in {\mathbb {R}}^n: u(x)=0 \text { and } |\nabla \! u(x)|\ne 0\}. \end{aligned}$$

Note that ${\hat{\Omega }}$ is part of a nodal line of u. Since the first-order derivatives of u are continuous, the Implicit Function Theorem implies that the n-dimensional Lebesgue measure of ${\hat{\Omega }}$ is zero. Consequently, the set

$$\begin{aligned} \Omega ^*:= {\tilde{\Omega }} \cup {\hat{\Omega }} = \{x\in {\mathbb {R}}^n: u(x)\ne 0 \text { or } |\nabla \! u(x)|\ne 0\} \end{aligned}$$

(24)

is an open set and there holds

$$\begin{aligned} |\Omega ^*| = |{\tilde{\Omega }}\cup {\hat{\Omega }}| \le |{\tilde{\Omega }}| + |{\hat{\Omega }}| \le \omega _0. \end{aligned}$$

Thus, $\Omega ^*\in {\mathcal {O}}_{\omega _0}$ and by construction, u and $\nabla \! u$ vanish in every point in ${\mathbb {R}}^n\setminus \Omega ^*$.

The following corollary guarantees that $u \in W^{2,2}_0(\Omega ^*)$. For the proof of this corollary, we refer to [1, Th. 9.1.3] or [8, Sec. 3.3.5].

Corollary 1

Let $\Omega \subset {\mathbb {R}}^n$ be an arbitrary open set and $v \in W^{2,2}({\mathbb {R}}^n)$. If v and its first-order derivatives vanish pointwise in ${\mathbb {R}}^n\setminus \Omega$, then $u \in W^{2,2}_0(\Omega )$.

Now we can prove our main theorem.

Theorem 3

The set $\Omega ^*$ given by (24) minimizes the buckling load $\Lambda$ in ${\mathcal {O}}_{\omega _0}$.

Proof

Recall that there holds

$$\begin{aligned} {\mathcal {R}}(u,{\mathbb {R}}^n) {\mathop {\le }\limits ^{(6)}} \liminf _{k\rightarrow \infty }{\mathcal {R}}(u_k,\Omega _k) =\inf _{\Omega \in {\mathcal {O}}_{\omega _0}}\Lambda (\Omega ). \end{aligned}$$

Since $\Omega ^*\in {\mathcal {O}}_{\omega _0}$ and $u \in W^{2,2}_0(\Omega ^*)$, there holds

$$\begin{aligned} \inf _{\Omega \in {\mathcal {O}}_{\omega _0}}\Lambda (\Omega ) \le \Lambda (\Omega ^*) \le {\mathcal {R}}(u,\Omega ^*) = {\mathcal {R}}(u,{\mathbb {R}}^n). \end{aligned}$$

Obviously, this means that

$$\begin{aligned} \inf _{\Omega \in {\mathcal {O}}_{\omega _0}}\Lambda (\Omega )= \Lambda (\Omega ^*) = {\mathcal {R}}(u,{\mathbb {R}}^n). \end{aligned}$$

$\square$

Due to the scaling property of the buckling load, the following corollary holds true.

Corollary 2

Let $\Omega ^*\in {\mathcal {O}}_{\omega _0}$ minimize the buckling load $\Lambda$ in ${\mathcal {O}}_{\omega _0}$. Then $\Omega ^*$ satisfies $|{\tilde{\Omega }}|=\omega _0$.

As a consequence of Corollary 2, the set $\Omega ^*$ is connected.

Corollary 3

The set $\Omega ^*$ given by (24) is connected.

Proof

Let us assume that $\Omega ^*$ consists of the two connected components $\Omega _1$ and $\Omega _2$ with $|\Omega _k|>0$ for $k=1,2$. By $u_k$ we denote the eigenfunction u restricted to $\Omega _k$, i.e.,

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

Since $\Omega _k \in {\mathcal {O}}_{\omega _0}$, the minimality of $\Omega ^*$ for $\Lambda$ implies

$$\begin{aligned} \Lambda (\Omega ^*) = {\mathcal {R}}(u,\Omega ^*)\le \Lambda (\Omega _1) \le {\mathcal {R}}(u_1,\Omega _1). \end{aligned}$$

Rearranging terms and using that $\Vert \nabla \! u\Vert _{L^2(\Omega ^*)}=1$ we obtain

$$\begin{aligned} \left( \int \limits _{\Omega _1}|\Delta u _1|^2\textrm{d}x + \int \limits _{\Omega _2}|\Delta u _2|^2\textrm{d}x\right) \left( 1-\int \limits _{\Omega _2}|\nabla \! u_2|^2\textrm{d}x\right) \le \int \limits _{\Omega _1}|\Delta u _1|^2\textrm{d}x. \end{aligned}$$

Hence,

$$\begin{aligned} \int \limits _{\Omega _2}|\Delta u _2|^2 \le \Lambda (\Omega ^*)\,\int \limits _{\Omega _2}|\nabla \! u_2|^2\textrm{d}x \; \Leftrightarrow \; {\mathcal {R}}(u_2,\Omega _2) \le \Lambda (\Omega ^*). \end{aligned}$$

Then there holds

$$\begin{aligned} \Lambda (\Omega _2) \le {\mathcal {R}}(u_2,\Omega _2) \le \Lambda (\Omega ^*) \end{aligned}$$

and $\Omega _2$ is a minimizer of $\Lambda$ in ${\mathcal {O}}_{\omega _0}$. However, since $|\Omega _2| < \omega _0$, this is a contradiction to Corollary 2. $\square$

Summing up, we found an optimal domain $\Omega ^*\in {\mathcal {O}}_{\omega _0}$ for minimizing the buckling load in ${\mathcal {O}}_{\omega _0}$. The set $\Omega ^*$ is open, connected and satisfies $|\Omega ^*|=\omega _0$. Classical variational arguments show that u solves

$$\begin{aligned} \Delta ^{\!2}u + \Lambda (\Omega ^*) \Delta u =0 \text { in } \Omega ^*. \end{aligned}$$