Lyapunov inequalities for Partial Differential Equations at radial higher eigenvalues

This paper is devoted to the study of $L_{p}$ Lyapunov-type inequalities ($ \ 1 \leq p \leq +\infty$) for linear partial differential equations at radial higher eigenvalues. More precisely, we treat the case of Neumann boundary conditions on balls in $\real^{N}$. It is proved that the relation between the quantities $p$ and $N/2$ plays a crucial role to obtain nontrivial and optimal Lyapunov inequalities. By using appropriate minimizing sequences and a detailed analysis about the number and distribution of zeros of radial nontrivial solutions, we show significant qualitative differences according to the studied case is subcritical, supercritical or critical.


Introduction
Let us consider the linear problem The well known L 1 Lyapunov inequality states that if a ∈ Λ, then L 0 a + (x) dx > 4/L. Moreover, the constant 4/L is optimal since 4 L = inf a∈Λ a + L 1 (0,L) and this infimum is not attained (see [1], [7] and [8]). This result is as a particular case of the so called L p Lyapunov inequalities, 1 ≤ p ≤ ∞. In fact, if for each p with 1 ≤ p ≤ ∞, we define the quantity then β 1 = 4 L and for each p with 1 ≤ p ≤ ∞, it is possible to obtain an explicit expression for β p as a function of p and L ( [1], [10]).
Let us observe that the real number zero is the first eigenvalue of the eigenvalue problem Since zero is the first eigenvalue of (1.5), it is coherent to affirm that β p is the L p Lyapunov constant for the Neumann problem at the first eigenvalue. On the other hand, the set of eigenvalues of (1.5) is given by ρ k = k 2 π 2 /L 2 , k ∈ N ∪ {0} and if for each k ∈ N ∪ {0}, we consider the set (1.10) Λ k = {a ∈ L 1 (0, L) : ρ k ≺ a and (1.1) has nontrivial solutions } then for each p with 1 ≤ p ≤ ∞, we can define the constant An explicit value for β 1,k has been obtained by the authors in [3]. The case p = ∞ is trivial (β ∞,k = ρ k+1 − ρ k ) and, to the best of our knowledge, an explicit value of β p,k as a function of p, k and L is not known when 1 < p < ∞. Nevertheless, since β 1,k > 0, we trivially deduce β p,k > 0, for each p with 1 ≤ p ≤ ∞.
With regard to Partial Differential Equations, the linear problem has been studied in [2], where Ω ⊂ R N (N ≥ 2) is a bounded and regular domain, ∂ ∂n is the outer normal derivative on ∂Ω and the function a : Ω → R belongs to the set Γ defined as (1.13) Obviously, the quantity (1.14) is well defined and it is a nonnegative real number. A remarkable novelty (see [2]) with respect to the ordinary case is that γ 1 = 0 for each N ≥ 2. Moreover, if N = 2, then γ p > 0, ∀ p ∈ (1, ∞] and if N ≥ 3, then γ p > 0 if and only if p ≥ N/2. In contrast to the ordinary case, it seems difficult to obtain an explicit expressions for γ p , as a function of p, Ω and N, at least for general domains. As in the ordinary case, the real number zero is the first eigenvalue of the eigenvalue problem (1.15) ∆u(x) + ρu(x) = 0, x ∈ Ω ∂u ∂n (x) = 0 x ∈ ∂Ω so that it is natural to say that the constant γ p defined in (1.14) is the L p Lyapunov constant at the first eigenvalue for the Neumann problem (1.12).
To our knowledge, there are no significant results concerning to L p Lyapunov inequalities for PDE at higher eigenvalues and this is the main subject of this paper where we provide some new qualitative results which extend to higher eigenvalues those obtained in [2] for the case of the first eigenvalue. We carry out a complete qualitative study of the question pointing out the important role played by the dimension of the problem.
Since in the case of ODE our proof are mainly based on an exact knowledge about the number and distribution of the zeros of the corresponding solutions ( [3]), in the PDE case we are able to study L p Lyapunov inequalities if Ω is a ball and for radial higher eigenvalues. It is not restrictive to assume that Ω = B R N (0; 1) ≡ B 1 , the open ball in R N of center zero and radius one.
In Section 2 we describe the problem in a precise way and we present the main results of this paper. In Section 3 we study the subcritical case, i.e. 1 ≤ p < N 2 , if N ≥ 3, and p = 1 if N = 2. To prove the results in this section we will construct some explicit and appropriate sequences of problems like (1.12) where Dirichlet type problems play an essential role. In this subcritical case we prove that the optimal Lyapunov constants are trivial, i.e., zero.
In Section 4, we treat with the supercritical case: p > N 2 , if N ≥ 2. By using some previous results of Section 2, about the number and distribution of the zeros of nontrivial and radial solutions, together with some compact Sobolev inclusions, we use a reasoning by contradiction to prove that the optimal Lyapunov constants are strictly positive and they are attained. In Section 5 we consider the critical case, i.e. p = N 2 , if N ≥ 3. Because in this case the Sobolev inclusions are continuous but no compact, we demonstrate that the optimal Lyapunov constants are strictly positive but we do not know if they are attained or not.
Finally, we study the case of Neumann boundary conditions but similar results can be obtained in the case of Dirichlet type problems.

main results
From now on, Ω = B 1 , the open ball in R N of center zero and radius one. It is very well known ( [4]) that the operator −∆ exhibits an infinite increasing sequence of radial Neumann eigenvalues 0 = µ 0 < µ 1 < . . . < µ k < . . . with µ k → +∞, all of them simple and with associated eigenfunctions Moreover, each eigenfunction ϕ k has exactly k simple zeros r k < r k−1 < ... < r 1 in the interval (0, 1). For each integer k ≥ 0 and number p, 1 ≤ p ≤ ∞, we can define the set a is a radial function, µ k ≺ a and (1.12) has radial and nontrivial solutions } if N ≥ 3 and a is a radial function, µ k ≺ a and (1.12) has radial and nontrivial solutions} if N = 2.
We also define the quantity The main result of this paper is the following.
The following statements hold: (2) If N ≥ 2 and N 2 < p ≤ ∞ then γ p,k is attained. A key ingredient to prove this theorem is the following proposition on the number and distribution of zeros of nontrivial radial solutions of (1.12) when a ∈ Γ k .

Proposition 2.2.
Let Ω = B 1 , k ≥ 0, a ∈ Γ k and u any nontrivial radial solution of (1.12). Then u has, at least, k + 1 zeros in (0, 1). Moreover, if k ≥ 1 and we denote by x k < x k−1 < ... < x 1 the last k zeros of u, we have that where r i denotes de zeros of the eigenfunction ϕ k of (2.1).
For the proof of this proposition we will need the following lemma. Some of the results of this lemma can be proved in a different way, by using the version of the Sturm Comparison Lemma proved in [4], Lemma 4.1, for the p-laplacian operator (see also [7]). Other results are new.
Proof. To prove i), multiplying (1.12) by ϕ k and integrating by parts in B r k (the ball centered in the origin of radius r k ), we obtain On the other hand, multiplying (2.1) by u and integrating by parts in B r k , we have Subtracting these equalities yields where ω N denotes de measure of the N -dimensional unit sphere. Assume, by contradiction, that u does not vanish in (0, r k ]. We can suppose, without loss of generality, that u > 0 in this interval. We can also assume that ϕ k > 0 in (0, r k ). Since r k is a simple zero of ϕ k , we have ϕ ′ k (r k ) < 0 and since a ≥ µ k in (0, r k ) we obtain a contradiction.
Finally, if r k is the only zero of u in (0, r k ], equation 2.3 yields To deduce ii), we proceed similarly to the proof of part i), restituting B r k by A(r i+1 , r i ) (the annulus centered in the origin of radii r i+1 and r i ) and obtaining and ii) follows easily by arguments on the sign of these quantities, as in the proof of part i).
To obtain iii), a similar analysis to that in the previous cases shows that and the lemma follows easily as previously.
Proof of Proposition 2.2. Let k = 0. If we suppose that u has no zeros in (0, 1] and we integrate the equation −∆u = a u in B 1 , we obtain B 1 a u = 0, a contradiction. Hence, for the rest of the proof we will consider k ≥ 1. Let 1 ≤ i ≤ k. By the previous lemma u vanishes in the i disjoint intervals [r i , r i−1 ),...,[r 2 , r 1 ),[r 1 , 1). Therefore u has, at least, i zeros in the interval Finally, let us prove that u has, at least, k + 1 zeros. From the previous part, taking i = k, u has at least k zeros in the interval [r k , 1], one in each of the k disjoint intervals [r k , r k−1 ),...,[r 2 , r 1 ),[r 1 , 1). Suppose, by contradiction, that these are the only zeros of u. Then u does not vanish in (0, r k ) and applying part i) of Lemma 2.3 we obtain u(r k ) = 0 and a ≡ µ k in (0, r k ]. Applying now part ii) of this lemma, we deduce u(r k−1 ) = 0 and a ≡ µ k in [r k , r k−1 ]. Repeating this argument and using part iii) of the previous lemma we conclude u(r i ) = 0, for all 1 ≤ i ≤ k and a ≡ µ k in (0,1], which contradicts a ∈ Γ k .
For the proof of Theorem 2.1, we will distinguish three cases: the subcritical case (1 ≤ p <

The subcritical case
In this section, we study the subcritical case, i.e. 1 ≤ p < N 2 , if N ≥ 3, and p = 1 if N = 2. In all those cases we will prove that γ p,k = 0.
The next lemma is related to the continuous domain dependence of the eigenvalues of the Dirichlet Laplacian. In fact, the result is valid under much more general hypothesis (see [6]). Here we show a very simple proof for this special case.
where λ 1 (A(ε, R)) and λ 1 (B R ) denotes, respectively, the first eigenvalues of the Laplacian operator with Dirichlet boundary conditions of the annulus A(ε, R) and the ball B R .
Proof. For N ≥ 3 and ε ∈ (0, R/2) define the following radial function u ε ∈ H 1 0 (A(ε, R)): where φ 1 denotes the first eigenfunction with Dirichlet boundary conditions of the ball B R . It is easy to check that lim ε→0 A(ε,2ε) In the same way it is obtained lim ε→0 A(ε,2ε) u 2 ε = 0. In addition, from the variational characterization of the first eigenvalue it follows that λ 1 (A(ε, R)) ≤ On the other hand, using that the first Dirichlet eigenvalue λ 1 (Ω) is strictly decreasing with respect to the the domain Ω, it follows that λ 1 (A(ε, R)) > λ 1 (B R ). Thus and the lemma follows for N ≥ 3.
Proof. If k = 0, this lemma follows from [2, Lem. 3.1]. In this lemma a family of bounded, positive and radial solutions were used. Hence, for the rest of the proof we will consider k ≥ 1.
To prove this lemma we will construct an explicit family a ε ∈ Γ k such that lim ε→0 a ε − µ k L p (B 1 ) = 0. To this end, for every ε ∈ (0, r k ), define u ε : B 1 → R as the radial function where φ 1 (A(ε, r k )) and φ 1 (B ε ) denotes, respectively, the first eigenfunctions with Dirichlet boundary conditions of the annulus A(ε, r k ) and the ball B ε . Moreover these eigenfunctions are chosen such that u ε ∈ C 1 (B 1 ). Then, it is easy to check that u ε is a solution of (1.12), being a ε ∈ L ∞ (B 1 ) the radial function where λ 1 (A(ε, r k )) and λ 1 (B ε ) denotes, respectively, the first eigenvalues with Dirichlet boundary conditions of de annulus A(ε, r k ) and the ball B ε . Since the first Dirichlet eigenvalue λ 1 (Ω) is strictly decreasing with respect to the the domain Ω, it follows that which gives a ε ∈ Γ k . (The equality λ 1 (B r k ) = µ k follows from the fact that ϕ k is a positive solution of −∆ϕ = µ k ϕ in B r k which vanishes on ∂B r k ). Let us estimate the L p -norm of a ε − µ k : Taking into account that λ 1 (B ε ) = λ 1 (B 1 )/ε 2 , λ 1 (B r k ) = µ k , using N > 2p, and applying Lemma 3.1, we conclude and the proof is complete. Proof. If k = 0, this lemma follows from [2,Lem. 3.2]. In this lemma a family of bounded, positive and radial solutions were used. Hence, for the rest of the proof we will consider k ≥ 1.
Similarly to the proof of the previous lemma, we will construct some explicit sequences in Γ k . In this case, this construction will be slightly more complicated. First, for every α ∈ (0, 1), define v α , A α : B 1 → R as the radial functions: where r = |x|. It is easily seen that v α ∈ C 1 (B 1 ), A α ∈ L ∞ (B 1 ), and Now, for every α ∈ (0, 1) and ε ∈ (0, r k ), define u α,ε : B 1 → R as the radial function: where the eigenfunctions ϕ k and φ 1 (A(ε, r k )) are chosen such that u α,ε ∈ C 1 (B 1 ). An easy computation shows that u α,ε is a solution of (1.12), being a α,ε ∈ L ∞ (B 1 ) the radial function Again, using that the first Dirichlet eigenvalue λ 1 (Ω) is strictly decreasing with respect to the the domain Ω, it follows that We see at once that m α > 0 for every α ∈ (0, 1). Hence, if we fix α and choose ε ∈ (0, 1) such that m α /ε 2 ≥ µ k , it is deduced that a α,ε ∈ Γ k . Let us estimate the L 1 -norm of a α,ε − µ k : Doing the change of variables x = εy in the first integral and applying Lemma 3.1 in the second one, it is obtained, for fixed α ∈ (0, 1): Thus, from the definition of γ 1,k we have A α (y)dy , ∀α ∈ (0, 1). Now we will take limit when α tends to 0 in this last expression. For this purpose we first deduce easily from the definition of A α that A α (r) ≤ 16α(1− r 2 )/(− log r) ≤ 32α if r ∈ (α, 1) and A α (r) ≤ 16α + 2/α 2 /(− log α) if r ∈ (0, α). It follows that which gives lim α→0 B 1 A α (y)dy = 0 and the lemma follows from (3.12).

The supercritical case
In this section, we study the supercritical case, i.e. p > N 2 , if N ≥ 2. In all those cases we will prove that γ p,k is strictly positive and that it is attained. We begin by studying the case p = ∞.
Next we concentrate on the case N 2 < p < ∞. For every a ∈ L p (B 1 ) satisfying a L p (B 1 ) ≤ M and every u ∈ H 1 (B 1 ) radial nontrivial solution of −∆u = a u in B 1 we have i) z > ε for every zero z of u. ii) |z 2 − z 1 | > ε for every different zeros z 1 , z 2 of u.
Proof. Let z ∈ (0, 1] be a zero of u. Hence, multiplying the equation −∆u = a u by u, integrating by parts in the ball B z and applying Hölder inequality, we obtain .

From the above it follows that
.
From the change w(x) = v(z x), it is easily deduced that where we have used the compact embedding which is the critical Sobolev exponent). Thus, taking ε 1 > 0 such that M < ε 1 N p −2 α(N, p), we conclude part i) of the lemma with ε = ε 1 .
For the second part of the lemma, consider two zeros 0 < z 1 < z 2 < 1 of u. Taking into account that z 1 ≥ ε 1 and arguing in the same manner of part i), we obtain .
On the other hand, from the one dimensional change of variable w(x) = v(z 1 + (z 2 − z 1 )x), it is immediate that It follows that M ≥ ω C p , we conclude part ii) of the lemma with ε = ε 2 . Obviously, taking ε = min{ε 1 , ε 2 }, the lemma is proved. Proof. Take a sequence {a n } ⊂ Γ k such that a n − µ k L p (B 1 ) → γ p,k . Take {u n } ⊂ H 1 (B 1 ) such that u n is a radial solution of (1.12), for a = a n , with the normalization u n 2 H 1 (B 1 ) = B 1 |∇u n | 2 + u 2 n = 1. Therefore, we can suppose, up to a subsequence, that u n ⇀ u 0 in H 1 (B 1 ) and u n → u 0 in L 2p p−1 (B 1 ) (since p > N/2, then 2 < 2p p−1 < 2N N −2 , which is the critical Sobolev exponent). On the other hand, since {a n } is bounded in L p (B 1 ), and 1 ≤ N/2 < p < ∞, we can assume, up to a subsequence, that a n ⇀ a 0 in L p (B 1 ). Taking limits in the equation (1.12), for a = a n and u = u n , we obtain that u 0 is a solution of this equation for a = a 0 . Note that u n → u 0 in L 2p p−1 (B 1 ) and a n ⇀ a 0 in L p (B 1 ) yields lim B 1 |∇u n | 2 = lim B 1 a n u 2 n = B 1 a 0 u 2 0 = B 1 |∇u 0 | 2 and consequently u n → u 0 ≡ 0 in H 1 (B 1 ). Therefore, if a 0 ≡ µ k , then a 0 ∈ Γ k and a 0 − µ k p ≤ lim n→∞ a n − µ k p = γ p,k , and the lemma follows.
On the other hand, taking into account the continuous embedding H 1 rad (A(ε 0 , 1)) ⊂ C (A(ε 0 , 1)) and u n → ϕ k in H 1 0 (B 1 ), we can assert u n → ϕ k in C (A(ε 0 , 1)). Clearly min r∈(0,1]\A |ϕ k (r)| > 0. Then, for large n we see that min r∈(0,1]\A |u n (r)| > 0, which implies that u n does not vanish in (0, 1] \ A, for large n. Since u n has, at most, k zeros in A, we conclude that u n has, at most, k zeros in (0,1], for large n. This contradicts Proposition 2.2 and the lemma follows.

The critical case
In this section, we study the critical case, i.e. p = N 2 , if N ≥ 3. We will prove that γ p,k > 0. Proof. To obtain a contradiction, suppose that γ p,k = 0. Then we could find a sequence {a n } ⊂ Γ k such that a n → µ k in L N/2 (B 1 ). Similarly to the supercritical case, we can take {u n } ⊂ H 1 (B 1 ) such that u n is a radial solution of (1.12), for a = a n , with the normalization u n 2 H 1 (B 1 ) = 1. Again, we can suppose, up to a subsequence, that u n ⇀ u 0 in H 1 (B 1 ) and taking limits in the equation (1.12), for a = a n and u = u n , we obtain that u 0 is a solution of this equation for a = µ k .
We claim that u n → u 0 in H 1 (B 1 ) and consequently, u 0 = ϕ k , for some nontrivial eigenfunction ϕ k . For this purpose, we set lim B 1 |∇u n | 2 = lim B 1 a n u 2 n = lim B 1 (a n − µ k )u 2 n + lim where we have used a n → µ k in L N/2 (B 1 ) and u 2 n is bounded in L N/(N −2) (B 1 ) (since u n is bounded in H 1 (B 1 ) ⊂ L 2N/(N −2) (B 1 )). Thus, from standard arguments, we deduce that u n → u 0 = ϕ k in H 1 (B 1 ).
In the following, we will fix ε ∈ (0, r k ). Since a n → µ k in L N/2 (A(ε, 1)) and u n → u 0 = ϕ k in H 1 rad (A(ε, 1)) ⊂ C (A(ε, 1)), we can assert that a n u n → µ k ϕ k in L N/2 (A(ε, 1)) ⊂ L 1 (A(ε, 1)). Thus −∆u n → µ k ϕ k in L 1 (A(ε, 1)), which yields u n → ϕ k in C 1 (A(ε, 1)). It follows that, for large n, the number of zeros of u n is equal to the number of zeros of ϕ k in the annulus A(ε, 1), which is exactly k. Applying Proposition 2.2 we can assert that, for large n there exists a zero ε n ∈ (0, ε] of u n . Hence, multiplying the equation −∆u n = a n u n by u n , integrating by parts in the ball B εn and applying Hölder inequality, we deduce Bε n |∇u n | 2 = Bε n a n u 2 n ≤ a n L N/2 (Bε n ) u n 2 L 2N/(N−2) (Bε n ) .
From the above it follows that a n L N/2 (Bε n ) ≥ ∇u .
Taking limits when n tends to ∞ in this expression we deduce Choosing ε > 0 sufficiently small we obtain a contradiction.