ESTIMATES FOR THE FIRST EIGENFUNCTION OF LINEAR EIGENVALUE PROBLEMS VIA STEINER SYMMETRIZATION

By means of Steiner symmetrization we get some estimates for the first eigenfunction of a class of linear problems, having as prototype the Laplacian with Dirichlet boundary conditions.


Introduction
The model problem we consider here is a very classical one: the fixed membrane problem, i.e. (1.1) where Ω is an open, bounded and connected subset of R n .
As well-known, the symmetrization methods have turned out to be a remarkable tool for the study of elliptic and parabolic equations.Many monographs, indeed, deal with this subject, see for instance [21], [6], [19], [17] and [18].For exhaustive references on this topic, we refer the reader to the detailed bibliographies contained in [18] and in [23].
In particular, Schwarz symmetrization has allowed to obtain various estimates for the first eigenfunction u and for the eigenvalues λ i of (1.1), see, for instance, [16].Let us briefly describe some results in this direction, due to Chiti, which are close to ours.Let u ⋆ denotes the Schwarz rearrangement of u and let S λ1 be the ball of R n centered at the origin, such that the Laplacian with Dirichlet boundary conditions has its first eigenvalue equal to λ 1 and finally z a corresponding eigenfunction.In [11], Chiti proved that, if u and z are normalized in a suitable way, then u ⋆ can be pointwise estimated in terms of z.This result relies on the Talenti's Theorem (see [22]) which ensures that (1.2) where ω n is the measure of the unit ball of R n and u * is the decreasing rearrangement of u u * (s) = sup {t ≥ 0 : µ(t) > s} , where, finally, µ is the distribution function of |u| .
Chiti's results have been generalized to nonlinear equations (see, for instance, [1] and [8]) and to the eigenvalue problem for the Hermite equation via the Gaussian symmetrization (see [7]).We finally recall that Chiti's type estimates were also used by Ashbaugh and Benguria in order to solve the well-known Payne-Pólya-Weinberger conjecture (see [3]) and its generalization on the n-dimensional sphere S n (see [4]).Now, let us consider problem (1.1) for domains Ω ⊂ R n x × R m y having n-dimensional cross sections of constant thickness.Or, more precisely, assume that for any fixed y in Ω ′′ , where Ω ′′ = {y ∈ R m : ∃ x ∈ R n : (x, y) ∈ Ω}, it holds1 In this case instead of Schwarz symmetrization, which should transform Ω into Ω ⋆ , the ball of R n+m centered at the origin having the same measure as Ω, it is more natural to use Steiner symmetrization with respect to the variables x i .In this way the symmetrized set turns out to be the cylinder B R × Ω ′′ , where B R is the ball of R n , centered at the origin whose measure is L. Being this set closer to the original domain with respect to Ω ⋆ , with this procedure one can obtain sharper estimates for u.This last point is the aim of this paper.Hence, let Ω be an open, bounded and connected subset of R n x × R m y verifying (1.3) and let λ 1 be the first eigenvalue of (1.1) and u = u(x, y) be a nonnegative corresponding eigenfunction.For y in Ω ′′ , we consider the function u(•, y), whose decreasing rearrangement will be denoted with u * (s, y).
Our starting point is, in place of (1.2), the following differential inequality proved in [2] (see also [5] and [13]) and let v be a corresponding positive eigenfunction.As we will see and moreover the following equality holds where V (s, y) = s 0 v * (σ, y) dσ.We prove that it is possible to normalize u and v in such a way to have The result above is then extended to the following class of elliptic problems (1.9) , for some ν > 0 and Θ > 1, ∀ (ξ, η) ∈ R n × R m and a.e.(x, y) ∈ Ω, iii) the coefficients b hk (y) are analytic in Ω ′′ .
The paper is organized as follows.In Section 2 we recall some definitions and properties about Steiner rearrangement.We also mention the one-dimensional Hardy inequality, since it will turn out to be a useful tool in proving that the operator appearing at the left hand side of the equation in (1.6) is compact in a weighted Sobolev space.In Section 3 we provide the comparison Theorem for the Laplacian.In that case, by simple factorization arguments, we show that the function V can be written explicitly in terms of Bessel functions and the first eigenfunction of the following problem (1.10) In the last section, the result is proved for problems of the type (1.9).In this case, via an approximation procedure, we overcome the difficulty arising from the presence of nonsmooth coefficients in the equation.

Preliminaries
Let Ω be an open, bounded and connected subset of R n x × R m y and let u = u(x, y) be a function defined in Ω.We will denote by Ω ′′ the projection of Ω on the linear manifold {x = 0} i.e.
In the sequel |•| n will stand for the n-dimensional Lebesgue measure, repeated indices mean summation, C will denote a positive constant whose value may change from line to line and the following notation will be in force For any fixed y in Ω ′′ , µ(t, y) will denote the distribution function of |u| (•, y).The Steiner rearrangement of u with respect to x is given by where Ω ♯ ⊂ R n × R m is the domain uniquely defined by the following relations x ∈ R n : (x, y) ∈ Ω ♯ = B ry , ∀ y ∈ Ω ′′ , with r y such that and finally The equimisurability of u and u ♯ ensures that Steiner symmetrization leaves the L p -norms of a function unaltered.On the other hand, the following Pólya-Szegö principle holds (see, for instance, [17] and [12]).
We end this section by recalling the simplest version of the Hardy inequality (see [20]).

The case of the Laplacian
Let Ω be an open, bounded, connected and Lipschitz subset of R n+m such that furthermore, we will assume that Ω ′′ is a C 2,α domain, for some positive α.
Let λ 1 be the first eigenvalue of the problem and u a corresponding eigenfunction.
Let us fix R > 0 such that λ 1 is the first eigenvalue of the symmetrized problem and let us denote by A result contained in [2] (see also [13]) ensures that where respectively and let us denote by j p,k the kth positive zero of the Bessel function J p .
The next lemma gives some information about the symmetrized problem (3.3).
Lemma 1.Under the assumptions and notation introduced below, it holds that and, up to a multiplicative factor, where χ = χ(y) is an eigenfunction of (3.6) corresponding to µ 1 .
The equation in (3.3) becomes Therefore S is a solution of the following eigenvalue problem By differentiating one obtains a straightforward calculation gives Therefore where R > 0 has to be chosen in such a way that (λ 1 − µ 1 ) is the first eigenvalue of (3.10); this fact implies Finally and the thesis follows.
At this point we want to introduce some functional spaces, naturally associated with problems of the type (3.3).
Let Γ be the portion of ∂ (I l × Ω ′′ ) where the Dirichlet boundary condition is prescribed for V in problem (3.3).Clearly, by (3.1), on the remaining part of the boundary, there is a Neumann condition imposed on V .
Let us consider the set of functions and the following two norms Finally, let us denote and by L 2 (I l × Ω ′′ ; g), the set of those functions such that Clearly, by the definition of g, H 1 0,Γ (I l × Ω ′′ ; g) is continuously embedded in H 1 0,Γ (I l × Ω ′′ ).The next two Lemmas show that a Poincarè inequality holds in H 1 0,Γ (I l × Ω ′′ ; g) and, furthermore, the embedding of In view of this results, the space H 1 0,Γ (I l × Ω ′′ ; g) will be equipped with the norm Lemma 2. It holds that where µ 1 is the first eigenvalue of the problem (3.6).
Proof: For any φ in H it happens and thus |D y φ| 2 dy ds.
By density the claim follows.
Remark 1.The above lemmas are trivial when n = 1, since, in that case, the function g(s) is a constant.Now we are in position to state the main result of this section.
As it is well-known, the first eigenvalue of problem (3.25) is the minimum of the Rayleigh quotient The corresponding eigenfunction satisfies in I l × Ω ′′ an Harnack inequality and therefore it has one sign within Ω.As a consequence the first eigenvalue of (3.25) is simple.Now we observe that the function V (s, y) If l = L, functions U and V are easily verified to be proportional.If, instead, l < L, then Ab absurdo, we suppose that the set (3.28) in nonempty.Then, by (3.17), it holds Z(s, y) ≡ U (s, y) − V (s, y) = 0 on ∂A + .
By setting Z = 0 outside A + , from (3.2) and (3.3), one deduces that This implies that Z is an eigenfunction corresponding to λ 1 , but, since, λ 1 is simple, there exists a constant C = 0 such that We can conclude that ∂U ∂s s=l = (C + 1) ∂V ∂s s=l = 0 which contradicts (3.26).

More general differential operators
Let Ω be as in the previous section, and let us consider the following class of operators Clearly Now we want to estimate u 1 in terms of the first eigenfunction of the following "symmetrized" problem (4.5) , where again we choose R in such a way that the first eigenvalue of (4.5) coincides with λ 1 .Let us denote For our purposes we need a Faber-Krahn type inequality, which will be an easy consequence of the following Pólya-Szegö type principle.Lemma 4. It holds that In this lemma, all the functions involved will be defined in the whole R n+m , by setting zero their value outside Ω or Ω ♯ .By the ellipticity condition (4.3) and Pólya-Szegö principle (2.1) we have that We are done once we show that (4.7) Inequality above is easily proved when the matrix b hk (y) is diagonal.In that case, indeed, by (4.3) one has b hh (y) ≥ ν > 0 in R m , and we have (see for instance Lemma 5.1 of [9]) a.e.y ∈ R m and ∀ h ∈ {1, . . ., m} .
Finally summing on h and integrating over R m one gets the claim.Note also that, if E is any measurable subset of R m , then (4.8) immediately implies that (4.9) Now, in order to reduce ourselves to the diagonal case, we argue by approximation.For any ǫ > 0, one can find a sequence of simple functions where E ǫ,p are measurable and mutually disjoint subsets of R m such that At this point we have b hk ǫ,p ∂u ∂y h ∂u ∂y k dy dx.
In each set E ǫ,p the integral can be carried out over a set of coordinates (y ǫ,p 1 , . . ., y ǫ,p m ) which diagonalizes the matrix B ǫ,p of entries b hk ǫ,p .Hence, for each ǫ and p, there exists an orthogonal matrix T ǫ,p such that y ǫ,p = T ǫ,p y, and T ǫ,p B ǫ,p (T ǫ,p ) −1 = D ǫ,p , where D ǫ,p is a diagonal matrix whose entries will be denoted with d ǫ,p h,k .Note that d ǫ,p h,h ≥ ν + o(1), ∀ p ∈ N and ∀ h ∈ {1, . . ., m} , so for ǫ small enough, say ǫ ∈ (0, ǫ 0 ), it holds By performing the above change of variables in each set E ǫ,p , we get ǫ,p ∂u ∂y ǫ,p h ∂u ∂y ǫ,p h dy ǫ,p dx.
Passing to the limit as ǫ goes to 0 + , we get (4.7) and therefore (4.6).
Observing the proof, is clear that the above result still holds true assuming that b hk (y) are just in C 0 (Ω ′′ ).
As an immediate consequence of Lemma 4, is the following Faber-Krahn inequality.
Lemma 5.It holds that and therefore where L is defined in (3.1).
Since the coefficients of A ♯ are all analytic, the already mentioned results in [2], allow to conclude that (4.12) For our purposes it will be useful to rewrite (4.12) in its weak form.To this aim, we consider the following problem The analogous of (3.8) and (3.9) still holds true.To this aim, let µ 1 be the first eigenvalue of the problem (4.16) and χ(y) a corresponding eigenfunction.Repeating the arguments used for Lemma 1, we get Lemma 6.We have that the function V (s, y), defined in (4.12), is proportional to and, finally, V (s, y) is the first eigenfunction of (4.14).
Since some coefficients of the operator A are just bounded, we can not apply directly the results in [2] in order to get an inequality analogous to (3.2).We overcome the lack of regularity by means of the following approximation procedure.
Firstly we define the coefficients a ij (x, y) on the whole R n+m as follows Finally, we will denote by a ǫ ij (x, y) the convolution of such functions with the Heat kernel ρ ǫ , where , with ǫ > 0.
Note that, for each ǫ > 0, the functions a ǫ ij (x, y) are analytic in R n+m and they satisfy, together with b hk (y), conditions (4.2) and (4.3).

Lemma 7. Let us introduce the following sequence of operators
For any ǫ > 0, denote by λ ǫ 1 and u ǫ 1 the first eigenvalue and the corresponding eigenfunction of the problem Observe that (4.17), (4.18) and (4.19) imply (4.20) u ≥ 0 in Ω Ω u 2 dx dy = 1 and, obviously, u cannot be identically zero.We claim that λ = λ 1 .Indeed, if this is not the case, one would have Ω uu 1 dx dy = 0, an absurd by the previous considerations.The fact that λ 1 is simple, in view of (4.4) and (4.20), guarantees that u = u 1 .
Now it remains to show that the convergence of u ǫ 1 is uniform.The reverse Hölder inequality (see [10]) ensures that u ǫ 1 q ≤ C(q) u ǫ 1 2 = C(q), for any q > 2, and therefore, by standard elliptic estimates (see [14] for instance), the functions u ǫ 1 are equicontinuous and equibounded in Ω.Finally, the claim follows by a direct application of Ascoli-Arzelà Theorem.Now we can state the comparison result in its full generality.Proof: Let u ǫ be the sequence given in Lemma 7. The results proved in [2] and [13] ensure that We have proved that the sequence l 0 u ǫ * (σ, y) dσ is uniformly (with respect to ǫ and y) Lipschitz continuous up to ∂Ω ′′ .On the other hand, using again the Hopf maximum principle, we get Letting ǫ goes to 0 + , from (4.28) and Lemma 7, we get (4.21).Finally, repeating the arguments of Theorem 3, we get s 0 u ǫ * (σ, y) dσ ≤ s 0 v * (σ, y) dσ, ∀ ǫ > 0 and ∀ (s, y) ∈ I l × Ω ′′ , and again passing to the limit as ǫ goes to 0 + , we get the claim (4.22).