Non-positivity of the semigroup generated by the Dirichlet-to-Neumann operator

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1 Non-positivity of the semigroup generated by the Dirichlet-to-Neumann operator Daniel Daners The University of Sydney Australia AustMS Annual Meeting Oct, 2013 The Dirichlet-to-Neumann operator Assume Ω R N smooth bounded open domain λ R If λ σ( ), then for every φ H 1/2 ( Ω) u + λu = 0 in Ω, u = φ on Ω has a unique solution u H 1 (Ω). Definition The Dirichlet-to-Neumann operator is defined by D λ φ := u ν H 1/2 (Ω) where ν is the outer unit normal to Ω

2 Some known facts Let 0 < λ 1 < λ 2 < λ 3 <... be the distinct eigenvalues of u = λu in Ω, u = 0 on Ω, If λ is not one of these eigenvalues, then D λ : H 1/2 ( Ω) H 1/2 ( Ω) is bounded; D λ generates an analytic semigroup on L 2 ( Ω). Moreover, if λ < λ 1, then e td λ is a positive irreducible semigroup; see Arendt and Mazzeo (2012) Aim of Talk Investigate positivity/non-positivity of e td λ for λ > λ 1. Possible conjecture In many cases, crossing a principal eigenvalue will result in loss of positivity and/or maximum principles. First Conjecture e td λ is not positive for all λ > λ 1 ; This conjecture is disproved by the simplest example, namely Ω = (0, L) R an interval.

3 The Dirichlet-to-Neumann semigroup on (0, L) Solving u + λu = 0 for λ > λ 1 = (π/l) 2 gives u(0) = a, u(l) = b u(x) = a sin λ(l x) sin λl + b sin λx sin λl. and therefore a D λ = b u (0) u = (L) λ sin λl cos λl 1 1 a cos λl b Hence we have the matrix representation where α(λ) β(λ) D λ =, β(λ) α(λ) α(λ) := λ cos λl sin λl and β(λ) := λ sin λl. The Dirichlet-to-Neumann semigroup is given by Hence e td λ e td λ cosh β(λ)t = e tα(λ) sinh β(λ)t is positive if and only if sinh β(λ)t, cosh β(λ)t sinh β(λ)t 0 for all t 0 sin λl > 0.

4 Conclusion e td λ is positive if and only if λ < π 2 or L 2kπ 2 < λ < L (2k + 1)π 2, k 1. L That is, positivity and non-positivity of e td λ alternate at each eigenvalue: λ 1 λ 2 λ 3 λ 4 λ 5 λ 6 λ 7 λ The spectrum of D λ A necessary condition for e D λ to be positive is that the minimal eigenvalue of D λ has a positive eigenvector. We have α(λ) β(λ) β(λ) 1 α(λ) ±1 = α(λ) β(λ) 1 ±1 Hence the eigenvalues/eigenvectors are λl μ 0 (λ) = α(λ) β(λ) = λ tan 2 λl μ 1 (λ) = α(λ) β(λ) = λ cot 2 e-vect e-vect 1 0; 1 1 > 0. 1

5 Hence a necessary condition for positivity of e td λ is μ 0 (λ) < μ 1 (λ) Graphs of μ 0 (λ) and μ 1 (λ) λ 1 λ 2 λ 3 λ 4 λ 5 λ Possible modifed conjectures Second conjectures Positivity and non-positivity of e td λ alternate at each eigenvalue λ k, possibly counting multiplicity. If e td λ is positive for some λ (λ k, λ k+1 ), then it is positive for all λ (λ k, λ k+1 ). If λ > λ 1, then e td λ dimensions. is not positive in higher We show that all these conjectures are disproved by the example of the disc in R 2.

6 The Dirichlet-to-Neumann operator on the disc Given φ H 1/2 ( B) solve u + λu = 0 in B, u = φ on B. (BVP) We compute u for an orthogonal basis on L 2 ( B): The solution of (BVP) is φ k = e ikθ, k Z. u k (r, θ) = J k( λr) J k ( λ) eikθ, Hence, for k Z, D λ e ikθ = u k ν = J k ( λr) r J k ( λ) eikθ = r=1 λ J k ( λ) J k ( λ) e ikθ Note that e ikθ is an eigenfunction of D λ. As J k (s) = ( 1) k J k (s) the eigenvalue μ k (λ) := λ J k ( λ) J k (, k = 0, 1, 2,... λ) has eigenfunctions e ±ikθ. Operator and semigroup on L 2 ( Ω) If φ = k= c k e ikθ H 1/2 ( B), then D λ φ = c k μ k (λ)e ikθ e td λ φ = k= k= c k e tμ k (λ) e ikθ

7 Spectrum of D λ on the disc The eigenspace to the eigenvalues μ k (λ), k = 0, 1, 2,..., is spanned by cos kθ, sin kθ. μ 0 (λ) is the only eigenvalue having a positive eigenfunction. Hence a necessary condition for e td λ to be positive is that μ 0 (λ) < μ k (λ) for all k > 0. (1) We shall show that (1) is not sufficient for e td λ be positive. to eigenvalues of D λ (the red curve is μ 0 (λ)) λ The Dirichlet eigenvalues λ n jumble the order of μ k (λ)

8 Where can e D λ be positive? From the graph we see that e td λ can only be positive in a left neighbourhood of some of the Dirichlet eigenvalues, namely where Recall that lim μ 0(λ) =. λ λ k μ 0 (λ) = λ J 0 ( λ) J 0 (. λ) These are the Dirichlet eigenvalues of determined by the zeros of J 0. Fourier series of non-negative functions Let φ = c k e ikθ 0. Then c k = c k and (c k ) is positive definite. Indeed, if ξ k C, then n j,k=1 c k j ξ k ξj = n j,k=1 = 1 2π 1 2π π π π π e ijθ e ikθ ξ k ξj φ(θ) dθ n e ikθ 2 ξ k φ(θ) dθ 0. k=1 In particular c 0 c n for all n Z. Choose ξ 0 = 1, ξ n = α with α = 1 so that αc n = c n and ξ j = 0: n c k j ξ k ξj = 2c 0 + αc 0 n + ᾱc n 0 = 2(c 0 c n ) 0. j,k=1

9 Eventual positivity & irreducibility Theorem Let μ 0 (λ) < μ k (λ) for all k 1. There exists T > 0 such that the operator e td λ positive and irreducible for all t T. It is possible that e td λ is not positive for all t > 0. If φ = k= c ke ikθ 0, then c 0 c k and e td λ φ = k= c k e tμ k (λ) e ikθ c 0 e tμ 0(λ) 2c 0 = c 0 e tμ 0(λ) 1 2 k=1 k=1 e t(μ k(λ) μ 0 (λ)) for t large enough, independently of φ 0. is e tμ k(λ) > 0 Non-positivity if μ 0 (λ) < μ k (λ) for all k The solution may dip below zero for some λ (λ 3, λ 4 ).

10 Non-positivity if μ m (λ) < μ 0 (λ) for some m > An oscillating term dominates the series e td λ φ = c k e tμ k (λ) e ikθ. k= Positivity of e td λ Theorem For each Dirichlet eigenvalue λ l such that J 0 ( λ l ) = 0 there exists β < λ l such that e td λ is a positive semigroup for all λ [β, λ l ). Write e td λ φ = c k e tμ k e ikθ π = G λ,t φ := G λ,t (θ )φ(θ) dθ. G λ,t is the heat kernel of e td λ given by G λ,t (θ) := e tμ k e ikθ t > 0. k= π

11 Positivity of e td λ... Show that G λ,t (θ) 0 for all t > 0 for λ in some interval [β, λ l ). G λ,t (θ) 0 if and only if the sequence e tμ k (λ) of Fourier coefficients is positive definite. Positive definiteness is hard to check but there is a sufficient condition, Polya s criterion: ck 0 k ck is convex, that is, c k 1 + c k+1 2c k 0 Express the Fourier series in terms of the Féjer kernels K n (θ) 0 in the form n(c k 1 + c k+1 2c k )K n 1 (θ) 0. n=1 Positivity of e td λ... Polya s criterion is only a sufficient condition. However the formula is still valid if the sequence of Fourier coefficients is eventually convex. Proposition where G λ,t (θ) = nb n (λ, t)k n 1 (θ), n=1 b n (λ, t) := e tμ n+1(λ) + e tμ n 1(λ) 2e tμ n(λ). Moreover, b n (λ, t) 0 for all n > λ and all t > 0.

12 Positivity of e td λ... If j k,l are the positive zeros of J k, then λj k μ k (λ) = ( (λ) 2λ = J k (λ) j 2 k,l λ; l=1 It is shown in Elbert and Laforgia (1984) that k j 2 k,l is concave. Hence k e μ k(λ) is eventually convex. This means almost all terms in the series are positive. G λ,t (θ) = nb n (λ, t)k n 1 (θ) n=1 If μ 0 (λ) 0, then the sum of the finitely many terms is postitive for all t > 0. Open Questions What happens on more general domains????

13 References I Arendt, W. and R. Mazzeo Friedlander s eigenvalue inequalities and the Dirichlet-to-Neumann semigroup, Commun. Pure Appl. Anal. 11, no. 6, MR Daners, D Non-positivity of the semigroup generated by the Dirichlet-to-Neumann operator. to appear. Elbert, Á. and A. Laforgia On the square of the zeros of Bessel functions, SIAM J. Math. Anal. 15, no. 1, MR (85a:33011)