ISOLATION AND SIMPLICITY FOR THE FIRST EIGENVALUE OF THE p-LAPLACIAN WITH A NONLINEAR BOUNDARY CONDITION

Here Ω is a bounded domain in RN with smooth boundary, ∆pu = div(|∇u|p−2∇u) is the p-Laplacian, and ∂/∂ν is the outer normal derivative. In the linear case, that is for p = 2, this eigenvalue problem is known as the Steklov problem (see [3]). Problems of the form (1.1) appear in a natural way when one considers the Sobolev trace inequality. In fact, the immersion W1,p(Ω) ↪→ Lp(∂Ω) is compact, hence there exists a constant λ1 such that


Introduction
In this paper, we study the first eigenvalue for the following problem: in Ω, |∇u| p−2 ∂u ∂ν = λ|u| p−2 u on ∂Ω. (1.1) and is the first eigenvalue of (1.1) in the sense that λ 1 ≤ λ for any other eigenvalue λ.
In [13] it is proved that, there exists a sequence of eigenvalues λ n of (1.1) such that λ n → +∞ as n → +∞. This is done using standard variational arguments together with the Sobolev trace immersion that provide the necessary compactness. Indeed, for solutions of (1.1) we can understand critical points of the associated energy functional This functional Ᏺ is well defined and C 1 in W 1,p (Ω) and the usual min-max techniques can be applied (see [13]). Also see [14] for similar results for the p-Laplacian with Dirichlet boundary conditions. We prove the following result.
Theorem 1.1. The first eigenvalue λ 1 is isolated and simple.
We remark that this theorem says that the extremals of the Sobolev trace inequality are unique up to multiplication by a real number. In the special case of a ball, Ω = B(0,R), our result implies that the first eigenfunction is radial. In fact, if u 1 (x) is an eigenfunction associated to λ 1 and R(x) is any rotation, then u 1 (R(x)) is also an eigenfunction, by our result we have that u 1 (x) = u 1 (R(x)). We conclude that u 1 must be radial. Also from our results it follows that any other eigenvalue has nonradial eigenfunctions as they have to change sign on the boundary (see Lemma 2.4).
The study of the eigenvalue problem when the nonlinear term is placed in the equation, that is, when one considers a quasilinear problem of the form −∆ p u = λ|u| p−2 u with Dirichlet boundary conditions, has received considerable attention (cf. [1,2,15,14,17], etc.).
However, nonlinear boundary conditions have only been considered in recent years. For the Laplace operator with nonlinear boundary conditions (cf. [5,6,8,16,19]). For elliptic systems with nonlinear boundary conditions (see [11,12]). For previous work for the p-Laplacian with nonlinear boundary conditions of different type see [7,13,18]. Also, one is led to nonlinear boundary conditions in the study of conformal deformations on Riemannian manifolds with boundary (cf. [4,9,10]).

Proof of the main result
In this section, we prove that the first eigenvalue λ 1 is isolated and simple. To clarify the exposition, we will divide the proof in several lemmas.
Lemma 2.3. The first eigenvalue λ 1 is simple. Let u, v be two eigenfunctions associated with λ 1 , then there exists c such that u = cv.
Proof. By Lemma 2.1, we can assume that u, v are positive in Ω. We perform the following calculations assuming that u, v are strictly positive inΩ, to obtain our result we can consider u + ε and v + ε and let ε → 0 at the end as in [17]. Therefore, we can take η 1 = (u p − v p )/u p−1 and η 2 = (v p − u p )/v p−1 as test functions in the weak form of (1.1) satisfied by u and v, respectively. We have Adding both equations we get The first eigenvalue we obtain that the first term of (2.4) is (2.6) We also have an analogous expression for the second term of (2.4). Using both expressions we get that (2.4) becomes (2.7) Taking ξ 1 = ∇ ln u and ξ 2 = ∇ ln v and using Lemma 2.2 we get, for p ≥ 2, This implies that u = kv, as we wanted to prove. For p < 2, we use the second part of Lemma 2.2 as above.
Now we turn our attention to the proof of the isolation of the first eigenvalue, in order to prove this we need the following nodal result. where Here |A| denotes the (N − 1)-dimensional measure of a subset A of the boundary.
Proof. Assume that w does not change sign in Ω, then we can assume that w > 0 in Ω using ideas similar to those of Lemma 2.1. Let u 1 be a positive eigenfunction associated to λ 1 . Making similar computations as the ones performed in the proof of Lemma 2.3 we arrive at Therefore, if we take kw instead of w we get that, for every k > 0, we have which is a contradiction if we take Therefore, w changes sign in Ω and by the maximum principle, [20], also w changes sign in ∂Ω.
We use w − as test function in the weak form of (1.1) satisfied by w to obtain (2.14) Hence, . Now we use the trace theorem to get that there exists a constant C such that If p ≥ N, we choose α = β = 2 and we argue as before using that W 1,p (Ω) → L 2p (∂Ω). A similar argument works for w + .
With these lemmas we can prove the isolation of λ 1 .
Lemma 2.6. The first eigenvalue λ 1 is isolated, that is, there exists a > λ 1 such that λ 1 is the unique eigenvalue in [0,a].
Proof. From the characterization of λ 1 , it is easy to see that λ 1 ≤ λ for every eigenvalue λ. Assume that λ 1 is not isolated, then there exists a sequence λ k with λ k > λ 1 , λ k λ 1 . Let w k be an eigenfunction associated to λ k , we can assume that w k W 1,p (Ω) = 1. Therefore, we can extract a subsequence (that we still denote by w k ) such that w k → u 1 in L p (∂Ω). Define φ k ∈ (W 1,p (Ω)) as φ k (u) = λ k ∂Ω w k p−2 w k u (2.18) and φ ∈ (W 1,p (Ω)) by φ(u) = λ 1 ∂Ω u 1 p−2 u 1 u. (2.19) From the L p (∂Ω) convergence of w k to u 1 we get that φ k converges to φ in (W 1,p (Ω)) . Using the continuity of A p given by Lemma 2.5 we get that the sequence w k converge strongly in W 1,p (Ω). Therefore, passing to the limit in the weak form of (1.1) we get that u 1 is an eigenfunction with eigenvalue λ 1 . By Lemma 2.1 we can assume that u 1 > 0 on ∂Ω. By Egorov's theorem we can find a subset A ε of ∂Ω such that |A ε | < ε and w k → u 1 > 0 uniformly in ∂Ω \ A ε . This contradicts the fact that, by (2.10), we have, for every k ∂Ω − k = ∂Ω ∩ w k < 0 ≥ Cλ −(N−1)/(p−1) k . (2.20) This completes the proof.