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International Journal of Differential Equations
Volume 2010 (2010), Article ID 984671, 23 pages
http://dx.doi.org/10.1155/2010/984671
Research Article

The Second Eigenvalue of the -Laplacian as Goes to

Mathematisches Institut, Universität zu Köln, Weyertal 86-90, D 50931 Köln, Germany

Received 15 July 2009; Accepted 29 September 2009

Academic Editor: Norimichi Hirano

Copyright © 2010 Enea Parini. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

The asymptotic behaviour of the second eigenvalue of the -Laplacian operator as goes to 1 is investigated. The limit setting depends only on the geometry of the domain. In the particular case of a planar disc, it is possible to show that the second eigenfunctions are nonradial if is close enough to 1.

1. Introduction

Let be a bounded open domain with Lipschitz boundary. We consider the following nonlinear eigenvalue problem: where , and is the -Laplacian operator. Notice that for we recover the usual Laplacian. A real number is said to be an eigenvalue if there exists a function (called eigenfunction) satisfying (1.1) in the weak sense, which means

A sequence of eigenvalues can be obtained by means of a minimax principle, as shown, for instance, in [1] and explained later in Section 5. The sequence is such that and as .

In this paper we will mainly focus on the asymptotic behaviour of the second eigenvalue of the -Laplacian as goes to . Our aim is to extend the results found in [2], where it was shown that the first eigenvalue converges to the so-called Cheeger constant defined as Here is the perimeter of measured with respect to and defined in the distributional sense (see [3]), while stands for the -dimensional Lebesgue measure of . The task of finding a set for which the infimum is attained is called Cheeger problem. We will show that a similar result holds also for the second eigenvalue; moreover, we are able to state that the second eigenfunctions of the -Laplacian in a planar disc cannot be radial, if is sufficiently close to .

The paper is structured as follows. After recalling some known results about the Cheeger problem (Section 2), in Sections 3 and 4 we deal with a geometrical problem, which will turn out to be crucial in order to describe the asymptotic behaviour of as (Section 5). Finally, in Section 6 we apply the results to the particular case where is a planar disc.

2. Known Facts on the Cheeger Problem

In this section we will recall some known results about the Cheeger problem. We say that a function has bounded variation if the quantity called total variation, is finite; in this case we will write . Since we assume to have a Lipschitz boundary, it can be proved that implies (we consider equal to zero outside ). The space is compactly embedded in . Moreover, the total variation is a lower semicontinuous functional with respect to the -convergence, that is, A set has finite perimeter (measured with respect to ) if where is the characteristic function of . For the sake of simplicity, in the following we will set .

A set such that is called a Cheeger set for . The existence of a Cheeger set for every domain follows easily from the coarea formula and from the fact that actually

The following proposition is a useful approximation result, whose proof can be found in [4].

Proposition 2.1. Let be a set of finite perimeter. Then there exists a sequence of sets of finite perimeter such that: (1) is for every ; (2) for every ; (3) in as ; (4) as .

Proposition 2.2. The following equalities hold:

Proof. It is clear that Let be a Cheeger set for ; by Proposition 2.1 we can approximate with a sequence of smooth sets such that and . This yields so that the claim is proved.

In the following we will mention some geometric properties of Cheeger sets.

Proposition 2.3. Let be a Cheeger set for ; then .

Proof. Let us suppose that this is not the case. Then is compactly contained in , which means that there exists a number such that the set is contained in . But then which contradicts the fact that is a Cheeger set.

Proposition 2.4. Let be a Cheeger set for ; then (1) is analytical, up to a singular set of Hausdorff dimension ; (2)the mean curvature in every regular point of is equal to ; (3)let be a regular point for ; then is a regular point for .

Proof. The proof can be found in [5]. As a consequence, the boundary of must touch the boundary of tangentially.

Proposition 2.5. Let be a convex domain. Then there exists an unique Cheeger set for . Moreover, is convex.

Proof. A proof of the existence of a convex Cheeger set can be found in [2, Remark ]. Uniqueness has been proved in [6] for the case , and in [7] for general .

Remark 2.6. If and is convex, then according to [6] the Cheeger set is the union of balls of suitable radius contained in (where “suitable” means in this case equal to ). It seems that the hypothesis of convexity cannot be dropped; there are examples of star-shaped domains which admit infinitely many Cheeger sets (see [8]). However, it was proved that “almost all” bounded domains admit a unique Cheeger set (see [9]).

We will often make use of the following property.

Proposition 2.7. Let , and let be a ball such that . Then

Proof. The proof is a consequence of the well-known isoperimetric property of the ball (see e.g., [10]).

Remark 2.8. There are some domains whose Cheeger set coincides with the whole ; this is of course the case of balls, but also of annuli and other domains satisfying a condition on the curvature of the boundary (see [6]).

3. Higher Cheeger Constants

Let be an open bounded domain with Lipschitz boundary. We define, for , with the convention that whenever . We will call the th Cheeger constant for . Notice that, for , we recover the definition of the Cheeger constant . By Proposition 2.1 it is possible to take the infimum on sets compactly contained in , or even on sets compactly contained in with smooth boundary.

Theorem 3.1. For every , there exist pairwise disjoint subsets contained in such that

Proof. Let us consider minimizing sequences of pairwise disjoint sets for corresponding to the value , where Set for . Fix as the radius of equal disjoint balls of fixed arbitrary volume contained in . We are going to show that we can consider for every . Indeed, if we had for some values of and , then by Proposition 2.7 we would surely have where is a ball with the same volume as and so of radius . As a consequence, , which means that we can actually discard the -tuple of sets . Because of the compact embedding of in , there exist such that, up to a subsequence, almost everywhere on . Moreover, . Denote with the negligible set of nonconvergence. From the lower semicontinuity of the total variation, it follows that for every . We are going to show that the are pairwise disjoint: suppose , then , which implies definitely. This means definitely, hence , that is, . If , we can assign arbitrary values to the characteristic functions (this does not affect the total variation). Hence we obtain the claim.

Definition 3.2. Any -tuple of sets as in Theorem 3.1 will be called a -tuple of multiple Cheeger sets. If , we will also speak of coupled Cheeger sets.

Remark 3.3. The proof of the theorem shows that we can always consider a minimizing sequence of -tuples of sets for , where the volumes of the sets are uniformly bounded from below.

Remark 3.4. Proceeding as in Proposition 2.3, one can show that at least one of the minimizing sets must touch the boundary.

In the following we will give a different characterization of the higher Cheeger constants.

Proposition 3.5. Let be the set of all partitions of with subsets . Then

Proof. Set . Let us suppose ; then there exists a partition of such that which is a contradiction. Thus . On the other hand, if are a -tuple of multiple Cheeger sets (which exist by Theorem 3.1), we can find a partition of with the property that for every . Hence, for every , and consequently that is, which finally yields

Remark 3.6. The sets realizing can be supposed to be connected. Indeed, if , with , we have This follows from the fact that, for any , , , , So one connected component of has a lower or equal ratio perimeter/area. If , but , we modify on a set of measure zero (this does not affect the total variation) to get a connected set defined as

Theorem 3.7. It is possible to find multiple Cheeger sets such that the part of their boundary contained in is a piecewise smooth hypersurface of piecewise constant mean curvature.

Proof. We will give the proof for the case . Let and be two coupled Cheeger sets, which exist according to Theorem 3.1. Since minimizes perimeter (measured in ) in with a volume constraint, it will have interior regularity according to [5]. More precisely, is an analytic hypersurface up to a singular set with Hausdorff dimension , whose regular points have constant mean curvature. The same can be stated for . Then we have to consider the possibly nonempty contact surface; also in this case, [5, Theorem ] can be applied to state that the contact surface (if it exists) enjoys the same regularity as the interior boundary of the two sets and has constant mean curvature.

Definition 3.8. Let and be a pair of coupled Cheeger sets. The free boundary of is defined as (analogously for ). The contact surface between and is .

Theorem 3.9. It is possible to find two coupled Cheeger sets and such that the following holds. Suppose that . Let us denote by the mean curvature of the free boundary of , by the mean curvature of the free boundary of , and by the mean curvature of the contact surface, measured from . Then the relation holds.

Proof. We follow [11, pages 10-11]. Take , and . Suppose that the boundaries of and can be locally described by the graph of a function defined in an open subset of , where , , and are disjoint open neighborhoods of , and respectively. For , let be a function defined in with compact support and such that the following conditions are satisfied: Since and are coupled Cheeger sets, we can suppose that is such that for small . It follows that Since also the functions , and are admissible, it follows that for arbitrary , , satisfying conditions (3.17); hence we obtain

Remark 3.10. The condition on the mean curvatures is similar to the one given in [12] for the double bubble problem: find two regions in which enclose two given amounts of volume, such that they minimize the sum of the surface measures. However, in that problem the quantity to minimize is slightly different, so also the condition on the mean curvatures differs and reads .

Proposition 3.11. Let be a convex planar domain; then it is possible to find two coupled Cheeger sets , such that they satisfy condition (3.16) in Theorem 3.9 and such that, if , then their boundaries meet tangentially.

Proof. We can suppose that , ; otherwise, since is convex, it would be possible to modify the sets suitably in order to decrease their perimeter and increase their volume. As a consequence, at least one of the two sets (say ) is convex. Let us suppose that and meet each other in a nonsmooth way. Then one could consider the Cheeger set of , which is convex and has a boundary, and then find a perimeter-minimizing set in under the volume constraint . The boundaries and will meet tangentially as proved in [5]. Then one can apply again Theorem 3.9 to get the condition on the curvatures.

Proposition 3.12. Let admit a unique Cheeger set. Then

Proof. Let us suppose that ; then there exist two disjoint subsets , such that which means, by definition of , This is a contradiction to the uniqueness of the Cheeger set for .

Remark 3.13. It is worth noting that there exist nonconvex domains for which ; think, for example, of a “barbell domain” made of two identical rectangles connected by a thin rope. For instance, we could consider the planar set where is small enough.

Proposition 3.14. Let us denote by the volume of the unit ball in . Then

Proof. Let be a family of multiple Cheeger sets, so that The volume of each cannot be smaller than the volume of a ball with Cheeger constant , which is exactly In fact, let be a ball such that , and a ball such that ; if we would have, applying Proposition 2.7, which is a contradiction. So we obtain

Corollary 3.15. It holds as .

Remark 3.16. The lower bound in Proposition 3.14 for follows directly from Proposition 2.7, and is obviously optimal if is a ball. For the higher Cheeger constants, it can be easily seen that the estimate is optimal for the union of balls with equal radii. If we try to minimize among connected sets, it turns out that the infimum is the same (consider a family of balls of equal radii connected by thin strips whose width goes to ). An interesting question would be to minimize among convex sets. If we focus on , it seems that a stadium (the convex hull of two tangent balls with both radii equal to ) is very near to be a minimizer; namely, it is possible to show that The lower bound follows directly from Proposition 3.14. To obtain the upper bound, one can note that the common tangent divides into two equal convex halves, whose Cheeger set is given by the union of balls of constant radius . satisfies then the conditions and since it must be we get . This yields the estimate from above. However, it should be mentioned that the stadium does not minimize the second eigenvalue of the Laplacian among convex planar domains, as proved in [13].

4. Coupled Cheeger Sets for a Disc

In this section we will determine the coupled Cheeger sets of a disc with radius . As a first step we will compute the Cheeger set for a half-disc of same radius. According to the results in Section 2, the Cheeger set must have the geometry shown in Figure 1.

984671.fig.001
Figure 1: The candidate Cheeger set for a half-disc.

We will denote by the inner angle and by the radius of the inner arc. Thus we have the relation which gives the existence condition . Then Remember that and , since we consider . Numerical resolution of the equation gives, for , which means

This is the best configuration with convex subsets to compute ; indeed, a convex partition of a convex set can be obtained only cutting the set with hyperplanes (otherwise we would have a point of nonzero curvature which gives convexity from one side but concavity from the other one). The Cheeger sets of each of the two partitioning subsets are then convex. Conversely, two convex subsets can be separated by a hyperplane thanks to the Hahn-Banach theorem. The Cheeger constant of a circular segment strictly contained in a half-disc is then strictly higher, due to uniqueness reasons. So the above configuration is the best among convex subsets of the disc.

Observe that the two coupled Cheeger sets and must have a contact surface. If it was not the case, we can suppose without loss of generality that and that is a Cheeger set for . Notice that is automatically a Cheeger set for . Due to the properties of Cheeger sets, the free boundaries of and must be circular arcs which meet tangentially. The only possibility is that and are discs, and the best configuration is given by to equal discs with radius , which is clearly not optimal for .

We are now going to prove that the contact surface cannot be closed; if it was the case, then one of the two coupled Cheeger sets, say , would be a disc of radius , as in Figure 2. The other set will be then contained in . Suppose that has a free boundary consisting of arcs with constant curvature . An easy computation shows that the case is never optimal; so we can suppose that the arcs have constant curvature . Due to the fact that is the contact surface, these arcs cannot start on and end on ; the only possibility is that the free boundary “encloses” as the dashed line in Figure 2. But in this case, the choice would give a lower ratio perimeter/area. So the optimal choice is the pair consisting of and its complement. By modifying suitably, one can easily convince himself that the optimal configuration is achieved when the ratios perimeter/area of and are equal. This implies which yields, for , This gives a worse configuration than the one found before. As a consequence, the contact surface cannot be a closed line.

984671.fig.002
Figure 2: The contact surface cannot be closed.

We will now use the regularity results about the coupled Cheeger sets; in particular, by Proposition 3.11 we can suppose that the boundary of each of the two sets meets the boundary of the other set tangentially. Suppose that the separating surface is an arc with constant curvature . From the point two arcs of curvature and , respectively, will depart, in such a way that the centres of curvature lie on the chord orthogonal to and such that . Notice that we can suppose, without loss of generality, that , .

Let be the “candidate” Cheeger set containing the segment and such that the curvature of its free boundary is ; let be the set containing the segment and with curvature of the free boundary equal to . Without loss of generality, we can suppose that . Let be the middle point of the segment . If , it is impossible that (as in Figure 3); indeed, since would be a subset of a circular segment strictly contained in a half-disc, this would contradict the fact that the configuration of the Cheeger sets of the two half-discs is better. So it must be .

984671.fig.003
Figure 3: The case .

Let and the centers of curvature of the free boundaries of and , respectively, and , as in Figure 4 such that and . Since , from Theorem 3.9 it must be , that is . This is impossible for geometrical reasons; indeed, take a point on such that ; it follows , which means that the point must lie between and . If is the intersection of the circle with the line , it is clear that . This is a contradiction because we would have .

984671.fig.004
Figure 4: The case .
984671.fig.005
Figure 5: Second eigenfunction in a planar disc for . The value of is about (picture courtesy of Jiří Horák).

It follows that necessarily . For symmetry reasons, this implies and hence, again from Theorem 3.9, . So we recover the optimal configuration consisting of the Cheeger sets of the two half-discs.

5. The Second Eigenvalue as

Let us consider now the eigenvalue problem (1.1). The natural question which arises is how one can find eigenvalues of the -Laplacian. A possibility is to use the direct method of Calculus of Variations by minimizing the so-called Rayleigh quotient; working this way we obtain the first eigenvalue One can prove (see e.g., [14]) that there exists, up to a nonzero multiplicative constant, one and only one eigenfunction associated to . Moreover, is of only one sign.

We will now describe how higher eigenvalues of the -Laplacian can be obtained. Let be a closed, symmetric subset. The Krasnoselskii genus of is defined as Let us denote by the set Then, for every , the following numbers are eigenvalues: In the literature they are sometimes called variational eigenvalues. It can be easily seen that the two definitions of given so far coincide. It is still an open question whether other eigenvalues can exist. As shown, for instance, in [15, Lemma ], eigenfunctions associated to higher eigenvalues of the -Laplacian must be sign-changing. Moreover, there does not exist any eigenvalue between and , which means that is isolated (see [16]).

A nodal domain of a function is a connected component of the set . It is not known whether the zero set of an eigenfunction of the -Laplacian has Hausdorff dimension , or if it can be even an open subset. A generalization of Courant's nodal domain theorem states that every eigenfunction associated to has at most nodal domains ([17,Theorem ]). As an easy consequence it follows that any second eigenfunction has exactly two nodal domains.

We are now ready to prove the main result of this paper.

Lemma 5.1. Let be a set with Lipschitz boundary, and let be the -strip around defined as Then where as .

Proof. The proof can be found in [18].

Theorem 5.2. It holds

Proof. Let , be two subsets such that , and By Proposition 2.1 it is possible to find , with the property that, for , , is smooth, and Let , and let () be two functions such that: on , outside a -neighbourhood of , and on the -layer outside . should be chosen in a way that . Set Then (see also [19, Lemma ]). Thus we have as we have from Lemma 5.1 where as . The last inequality follows from (3.14). If we send , we obtain and if The claim follows if we send . The fact that and depend on does not constitute a problem, since we can estimate uniformly from below (see Remark 3.3).

Remark 5.3. The theorem can be easily generalized to the th variational eigenvalue obtaining

The so-called Cheeger's inequality, whose proof can be found in [20], yields the following lower bound for the first eigenvalue:

In the following theorem we show that a similar estimate holds also for the second eigenvalue.

Theorem 5.4. The following Cheeger-type inequality holds:

Proof. Let be a second eigenfunction of the -Laplacian. From [17, Theorem ], we know that has exactly two nodal domains , . is also a first eigenfunction on each of the two nodal domains; from Cheeger's inequality it follows, for , But as , we have due to the definition of . So we obtain the claim.

Theorem 5.5. It holds

Proof. The claim follows easily from Theorems 5.2 and 5.4.

The Second Eigenfunction as
In the following we will focus on the asymptotic behaviour of the second eigenfunctions as .

Theorem 5.6. Let , be the nodal domains of a second eigenfunction of the -Laplacian. Then

Proof. By definition of we have It remains to prove that for every , there exists such that for every , Suppose that this is not the case; then there exists such that, without loss of generality, for a subsequence . From Cheeger's inequality for large enough. But this contradicts the fact that . Hence the claim follows.

Corollary 5.7. For , the volume of each of the nodal sets is uniformly bounded from below by .

Proof. From the preceding theorem there exists such that, for every , Arguing as in Proposition 3.14, the volume of the nodal sets cannot be smaller than the volume of a ball with Cheeger constant , which is exactly . Thus, for , as claimed.

We are now going to investigate the asymptotic behaviour of the second eigenfunctions as . First, we state some technical lemmas.

Lemma 5.8. Let be a bounded set with Lipschitz boundary, as (), for every , in as . Then

Proof. Since is Lipschitz, the functions are in particular in . Let us denote by the exponent conjugate to ; by (2.2), Hölder's inequality and Young's inequality we have

Lemma 5.9. Let be a bounded set, as (), for every (), , and in as . Then

Proof. Let us denote by the exponent conjugate to . By Hölder's inequality and Young's inequality we have
On the other hand from and we have so that The last equation and (5.30) end the proof.

Lemma 5.10. Let be a second eigenfunction of the -Laplacian. Then

Proof. The proof can be found in [21].

Theorem 5.11. Let be second eigenfunctions of the -Laplacian such that . Then (after possibly passing to a subsequence) converge, as , in and hence pointwise a.e. to a function such that and . Moreover, cannot be strictly positive or strictly negative.

Proof. From Theorem 5.5, Lemma 5.10, and Hölder's inequality, are uniformly bounded in . Moreover, we have where is the exponent conjugate to . Since , the functions are uniformly bounded in ; hence there exists a subsequence converging in to a function . From Lemma 5.8 we have Finally, Lemma 5.9 yields .
The fact that cannot be strictly positive or strictly negative is a consequence of Corollary 5.7.

6. Nonradiality of the Second Eigenfunction in a Planar Disc

In this section we will apply the previously found results to the particular case where the domain is a disc, in order to establish whether a second eigenfunction can be radial or not. Let us recall that the existence of radial eigenfunctions was shown in [22]; in this case, one has to solve the ordinary differential equation

The question seems to be still an open problem in its full generality, except for the case (see [23]) where the answer is negative. In the following theorem it is shown that the answer is also negative if is sufficiently close to .

Theorem 6.1. Let be the unit disc. Then, for close to , the second eigenfunction of the -Laplacian in cannot be radial.

Proof. From the results in Section 4 and Theorem 5.5 we have Then there exists such that for . Let us suppose that there exists such that a second eigenfunction of the -Laplacian is radial; this implies that its nodal domains are a disc of radius (), compactly contained in , and an annulus . If we suppose w.l.o.g. , Cheeger's inequality allows us to state that Indeed, since the Cheeger set of is itself (see Remark 2.8), one has Then we have the following compatibility conditions: which are incompatible. Hence we obtain the claim.

The following image was obtained by an implementation of the numerical method described in [24].

Acknowledgments

The author is grateful to Professor Bernd Kawohl for suggesting the research topic of this paper and for valuable advice and helpful suggestions at various stages of its preparation. The author would also like to thank Jiří Horák for providing the image in Figure 5 and for useful discussions about this subject. This project was financially supported by the DFG-Deutsche Forschungsgemeinschaft.

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