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

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.


Introduction
We consider the following variational problem. 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 Ω. 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 Ω exists and if ∂Ω is smooth (at least C 2,α ), then Ω 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. where |D| denotes the n-dimensional Lebesgue measure of D, ω 0 > 0 is a given quantity and B ⊂ R n is a ball with |B| > ω 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 ⊂ R n be a ball and choose 0 < ω 0 |B|, where |B| denotes the n-dimensional Lebesgue-measure of B. For ε > 0 we define the penalization term p ε )| = ω 0 which minimizes the buckling load among all open subsets B with measure smaller or equal than ω 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 ∂Ω(u (0) ε ). 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 nondegeneracy of u (0) ε and, subsequently, proving that the free boundary has got a positive Lebesgue-density in every point. However, looking at nondegeneracy results for second order problems as in [1], [2] or [5], e.g., we see that in these settings a nondegeneracy 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 I (0) ε . Consequently, we revise the functional I (0) ε . Following an idea of N. Aguilera, H. W. Alt and L. A. Caffarelli in [1], we replace the penalization term p (0) ε by a penalization term p (1) ε which rewards volumes less than ω 0 with a negative contribution to the functional. For that purpose, we define p (1) ε : R → R by (2) p (1) ε (s) := 1 ε (s − ω 0 ), s ≥ ω 0 ε(s − ω 0 ), s ≤ ω 0 and I (1) ε : H 2,2 0 (B) → R by We will prove that for every ε > 0 there exists a minimizer u (1) ε ∈ H 2,2 0 (B) for I ε ) will be more challenging than in the case of Ω(u (0) ε ) 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 Section 2, we prove the existence of a minimizer u (k) ε for the functional I (k) ε (k = 0, 1) and show that u (k) ε ∈ C 1,α (B) for every α ∈ (0, 1). Thereby, we refine the approach presented in [11]. That paper deals with the functional I (0) ε , but considers only n = 2 and n = 3. Consequently, the C 1,α regularity of u (0) ε follows with the help of Sobolev's Embedding Theorem. In the present paper, we consider arbitrary n ≥ 2 and refine the approach to the C 1,α regularity from [11] by applying a bootstrap argument. In Section 3, we analyze the volume of the domains Ω(u (0) ε ) and Ω(u (1) ε ) separately. In Section 3.1, we consider the functional I (0) ε and prove our main theorem, Theorem 1, by scaling arguments. In Section 3.2, we consider the functional I (1) ε . Thus, we deal with a penalization term, which rewards volumes less than ω 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 Section 3.1 to show that the optimal domain's volume cannot become less than ω 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 ε is sufficiently small, 1) depends on n, ω 0 and ε. After refining the choice of ε, we will obtain the following dichotomy: there either holds ε ) is scaled to the volume ω 0 , this enlarged domain is neither a subset of B nor can it be translated into B.
In the first case, u (1) ε is a minimizer of the functional I (0) ε 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 ∂Ω(u (1) ε ) satisfies a doubling property, we will disprove the occurrence of the latter case. We assume that there exists a constant σ > 0 such that for every x 0 ∈ ∂Ω(u (1) ε ) and every 0 < R ≤ R 0 there holds This assumed doubling property enables us to establish a nondegeneracy property of the minimizing function u (1) ε and, subsequently, we can show that the latter case of the above mentioned dichotomy cannot occur. We should emphasize that for proving the nondegeneracy of u (1) ε , besides the assumption of the doubling property, the crucial incredient is the rewarding part of the penalization term p (1) ε .

The penalized problems
In this section, we analyze the penalized problems Without loss of generality, we assume u (k) ε to be normalized in the sense that ε (x) = 0} is a non-empty set. By the absolute continuity of the Lebesgue integral, the n-dimensional Lebesgue measure of O(u ε 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,α regularity of u (k) ε for every α ∈ (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
Lemma 2. Using the above notation, there exists a constant C > 0 which only depends on n such that for 0 ≤ r < R there holdŝ The next lemma will be the starting point for our bootstrap argument.
Proof. We obtain the result by comparing the I Since and the claim is proven. Now let us assume that there holds In order to avoid the penalization term while comparing I ε ) and I ε (v k ) we scalê v k . For this purpose, we set 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 * : Rearranging terms we obtain the local inequalitŷ Note that there holds and, using Taylor's expansion, we find Thus,B 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,α regularity of u (0) ε is proven in [11] (cf. Lemma 2.2 and Theorem 2.4 in [11]). Since we now consider any n ≥ 2, we need the bootstrapping. The next lemma is the essential tool for this new approach to the C 1,α regularity of u

Lemma 4. Suppose that for each
where M > 0 and µ ∈ [0, n). Then there exists a constant C(n, |O(u w dx.

Using this notation we writê
Then Young's inequality implieŝ Applying Hölder's and a local version of Poincaré's inequality, we find that where the constant C only depends on n. By assumption, we can proceed tô Now we are ready to prove the main theorem of this section.
Proof. Our aim is to show that for each x 0 ∈ B and every 0 ≤ r ≤ R ε,k there holds We prove (4) by using a bootstrap argument. Let x 0 ∈ B and 0 ≤ r ≤ R ≤ R ε,k . Then there obviously holdŝ where v k is defined in (3). Due to Lemma 2 and Lemma 3 we obtain Now we start the bootstrapping. For every 0 ≤ r ≤ R ε,k there holds Applying Lemma 4, estimate (6) yields for every 0 ≤ r ≤ R ε,k where λ 0 ∈ (0, n) if n = 2 and λ 0 = 2 if n ≥ 3. We insert this estimate in (5). This yieldsB (7) is an improvement of estimate (6). In this case, we again apply Lemma 4 and obtain Together with estimate (5) we find thatB For n ∈ {3, 4}, estimate (8) proves the claim. For n ≥ 5, we repeat the argumentation.
The continuity of u is an open set. Then classical variational arguments show that u Moreover, the C 1,α regularity of u Then Σ ε,k is part of a nodal line of u (k) ε and, consequently, L n (Σ ε,k ) = 0 for all ε > 0. We define The following lemma shows that, considering the functional I ε ) is connected. This result is a direct consequence of the strict monotonicity of the penalization term p (1) ε .

Lemma 5. For every minimizer
Proof. We prove the claim by contradiction. Without loss of generality we assume that Ω(u (1) ε ) consists of two connected components, namely Ω 1 and Ω 2 with |Ω k | = 0 for k = 1, 2. By u k we denote otherwise .
The minimality of u Since ∇u Now the strict monotonicity of p which is equivalent to ε ) and I (1) ε (u 2 ) we find Then (9) implies Obviously, this is contradictory since p (1) ε is strictly increasing.
Let us emphasize that for the proof of Lemma 5 the actual value of |Ω(u ε is strictly increasing. Considering the functional I (0) ε , and thus the only non-decreasing penalization term p (0) ε , we have to ensure that |Ω(u (0) ε )| ≥ ω 0 before we are able to copy the approach of Lemma 5 and can deduce that Ω(u (0) ε ) is connected. This will be done in Theorem 5 and Lemma 6.
The next corollary collects direct consequences of Lemma 5 is an optimal domain for minimizing the buckling eigenvalue among all domains in B with the same measure as Ω(u Thus, the minimizer u (1) ε is a buckling eigenfunction on Ω(u Hence, Ω(u (1) ε ) is an optimal domain for minimizing the buckling eigenvalue among all domains in B with the same measure as Ω(u (1) ε ).

The volume condition
In the following, we analyze the functionals I

The non-rewarding penalization term
In this section, we consider the functional I (0) ε . We will show that for every ε > 0 and every minimizer u Recall that the buckling load is decreasing and the penalization term p (0) ε is nondecreasing with respect to set inclusion. Since the penalization term grows with slope ε −1 for arguments larger than ω 0 , it seems to be natural that the optimal domain Ω(u (0) ε ) adjusts itself to the volume ω 0 provided that ε is chosen small enough. Indeed, the fact that |Ω(u (0) ε )| ≤ ω 0 for sufficiently small ε is eventually a consequence of the scaling property of the buckling load, i. e. Λ(M ) = t 2 Λ(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 Ω(u 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 ε > 0 there exists a minimizer u ε ) \ ∂B and a radius r > 0 such that and compare the I ε andv. This leads to the following local inequality:
As a consequence of Lemma 6, we get the analog to Corollary 1. The following remark will be helpful to show that for an appropriate choice of ε the n-dimensional Lebesgue measure of Ω(u (0) ε ) cannot become larger than ω 0 .

Remark 2. Note that the reference domain B compactly contains a ball
Consequently, for every ε > 0 there holds .
ε be a minimizer for I (0) ε . Then there exists a number ε 1 = ε 1 (n, ω 0 ) such that for 0 < ε ≤ ε 1 there holds Proof. We claim that the statement of the theorem holds true for .
We prove by contradiction. Thus, we choose ε ≤ ε 1 and denote by u ε . Assume that there exists a number α > 1 such that Our aim is to contradict (16).
Since Ω(u By scaling we have From (17) we then get Now let φ ∈ H 2,2 0 (B) be as in Remark 2. Then, due to the assumption (16), there holds and estimate (18) becomes With (15) this implies since α > 1 and n ≥ 2. Thus, for any ε ≤ ε 1 we get a contradiction and the assumption (16) cannot hold true for any 0 < α < 1. Hence, |Ω(u Together with Theorem 5, this proves the claim.
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 ε ≤ ε 1 , where ε 1 is given in Theorem 6 and let u 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 ∂Ω(u (0) ε ). Following [2], our next aims would by establishing a nondegeneracy result for u (0) ε . Considering second order problems (e.g. [1,2,5]), these nondegeneracy 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) ε by the rewarding penalization term p (1) ε . This will be discussed in the next section.

The rewarding penalization term
In this section, we consider the functional I (1) ε . Analog to Theorem 6 we will find that |Ω(u (1) ε )| cannot become larger than ω 0 provided that ε ≤ ε 1 . It remains to exclude that Ω(u (1) ε ) < ω 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 Ω(u (1) ε ) is still contained in the reference domain B. Choosing the parameter ε sufficiently small, we will see that one of the following two situations occurs: either |Ω(u In the first case, u ε is a minimizer of the functional I In the second case, we may think of the domain Ω(u (1) ε ) 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 ∂Ω(u (1) ε ) is needed. Exemplary, we will see that assuming that ∂Ω(u (1) ε ) satisfies a doubling condition, the domain Ω(u We begin this section with the analog result to Theorem 6.
Thereby, ε 1 is the number given in Theorem 6.
Proof. Let us assume that ε ≤ ε 1 and that |Ω(u (1) ε )| = αω 0 for an α > 1. Then arguing in exactly the same way as in the proof of Theorem 6 leads to a contradiction.
We would like to repeat the idea of the proof of Theorem 6 assuming that |Ω(u where Ω # denotes a ball in R n with the same volume as Ω. In addition, c n tends to 1 as n tends to infinity.

Moreover, we have the explicit representation
where ε 1 is given by Theorem 6 and c n is given by Remark 3. Note that lim n→∞ α 0 = 1 and lim Together with Theorem 7 this shows that for ε ≤ ε 1 the domain Ω(u (1) ε ) satisfies the volume condition asymptotically as the dimension n approaches infinity.
Consequently, the strict monotonicity of f implies that (21) can only hold true for α ∈ [α 0 , 1). This proves the theorem.

Theorem 9.
There exists a number ε 0 = ε 0 (n, ω 0 ) such that for ε ≤ ε 0 every minimizer u (1) Proof. We claim that the statement of the theorem holds true for By Ω(u (0) ε ) # we denote the ball centered in the origin with the same volume as Ω(u (0) ε ). Then From now on, we always choose ε ≤ ε 0 . Let u (1) ε ∈ H 2,2 0 (B) be a minimizer of I (1) ε and let us assume that case a) of Theorem 9 holds true. Hence, |Ω(u (1) ε )| = ω 0 and repeating the proof of Theorem 1 we gave in Section 3.1 we find that Ω(u (1) ε ) minimizes the buckling load among all open subsets of B with volume smaller or equal than ω 0 . In addition, u (1) ε minimizes the functional I (0) ε . Consequently, if we could exclude that the case b) of Theorem 9 may occur, we could treat the functionals I (0) ε and I (1) ε as equivalent. We will discuss this issue in the following section.

Discussion of case b) of Theorem 9
In this section, we always consider a minimizer u Since we do not posses any information about regularity properties of ∂Ω(u (1) ε ), we currently cannot anaylze these tentacle-like domains in more detail.
However, assuming that ∂Ω(u (1) ε ) satisfies a doubling condition, we will be able to establish a nondegeneracy result for u (1) ε . This nondegeneracy enables us to prove that, potentially after translation, the up the volume ω 0 rescaled version of Ω(u ε ) satisfies a doubling condition, the case b) of Theorem 9 cannot occur and there holds |Ω(u (1) ε )| = ω 0 From now on, we always assume that ∂Ω(u (1) ε ) satisfies the following doubling property. Assume that there exists a constant σ > 0 and a radius 0 < R 0 < 1 such that for every x 0 ∈ ∂Ω(u (1) ε ) and every 0 < R ≤ R 0 there holds Note carefully that the condition (28) does not exclude that Ω(u (1) ε ) forms thin tentacles but it determines the minimal rate at which the volume of a possible tentacle may decrease.

Nondegeneracy of u (1) ε
Our aim is to establish a nondegeneracy result for u (1) ε . For convenience, we cite a technical lemma which will be applied in the proof of Lemma 8.

and (31) becomeŝ
Although this lower bound on the density quotient is admittedly weak, it suffices to prove that Ω(u (1) ε ) can be rescaled to the volume ω 0 without leaving the reference domain B provided the radius of B is chosen sufficiently large (see Theorem 10).

The volume condition for
The next theorem is the key observation to show that Ω(u (1) ε ) has the volume ω 0 provided that ∂Ω(u (1) ε ) satisfies (28). It is a consequence of the lower bound on the density quotient according to Lemma 9.
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) ε be a minimizer of I (1) ε for ε ≤ ε 1 . In order to prove the claim, let us assume that Ω(u (0). For the sake of convenience, we abbreviate S := 2 − 1 n R B . Hence, we assume ∂Ω(u (1) ε ) ∩ ∂B S (0) = ∅. Of course, there either holds 0 ∈ Ω(u (1) ε ) or 0 ∈ Ω(u (1) ε ). At first, we will handle the case where the origin is already contained in Ω(u (1) ε ). Secondly, we will show that we may translate Ω(u (1) ε ) such that the origin becomes an inner point of Ω(u (1) ε ).
Step 1. We consider that 0 ∈ Ω(u (1) ε ). Note that for every m ∈ N with m ≥ 3 there holds Since we assume that ∂Ω(u In addition, we fix an α ∈ (0, 1). Then applying We now sum (36 Specifying the choice of m ∈ N such that and recalling that S := 2 − 1 n R B , we obtain Since this estimate is false if R B is chosen sufficiently large, the proof is finished provided that 0 ∈ Ω(u (1) ε ). Step 2.
Let us now assume that the origin is not contained in Ω(u (1) ε ). However, there exists an x 0 ∈ Ω(u (1) ε ) ∩ B. We now translate Ω(u (1) ε ) such that x 0 is translated to the origin, i.e. we consider Φ : R n → R n , x → x − x 0 .
If y 0 ∈ ∂Ω and 0 < R ≤ R 0 , then there exists a z 0 ∈ ∂Ω(u (1) ε ) such that y 0 = Φ(z 0 ). Lemma 9 together with the translational invariance of the Lebesgue measure then imply Estimate (37) enables us to repeat the approach presented in Step 1. Again we consider the segmentation (35) and assume that ∂Ω ∩ ∂B S (0) is not empty. Then there exists a smallest index i 0 (m) such that for every i 0 ≤ i ≤ m − 2 there exists an x i ∈ ∂Ω ∩ ∂B i+1 m S (0). Now applying (37)  and, by construction, contains the origin.
Considering the functional I ε : H 2,2 0 (B) → R by where p • It remains open to prove existence of an optimal domain for minimizing the buckling load among all (probably unbounded) open subsets of R n with given measure.