Exit Time Moments and Eigenvalue Estimates Je ff rey Langford - PowerPoint PPT Presentation

Exit Time Moments and Eigenvalue Estimates Je ff rey Langford Bucknell University Lewisburg, PA April 3, 2020 Talk based on the papers: a. D. Colladay, J. J. Langford, and P. McDonald. Comparison results, exit time moments, and eigenvalues on Riemannian manifolds with a lower Ricci curvature bound. J. Geom. Anal., 28(4):3906–3927, 2018. b. E. B. Dryden, J. J. Langford, and P. McDonald. Exit time moments and eigenvalue estimates. Bull. Lond. Math. Soc., 49(3):480–490, 2017. Je ff rey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates

Motivation We recall torsional rigidity . For Ω ✓ R n a bounded C ∞ domain, let u 1 solve � ∆ u 1 = 1 in Ω , u = 0 on ∂ Ω . Put Z T 1 ( Ω ) = u 1 dx (torsional rigidity) . Ω Interpreting probabilistically, we let X t denote Brownian motion in R n , P x denote the probability measure charging Brownian paths starting at x 2 R n , and τ = inf { t � 0 : X t / 2 Ω } denote the first exit time of X t from Ω . Then Z u 1 ( x ) = E x [ τ ] E x [ τ ] dx . and T 1 ( Ω ) = Ω P´ olya’s inequality gives an estimate of the principal Dirichlet eigenvalue in terms of T 1 ( Ω ): | Ω | λ 1 ( Ω )  T 1 ( Ω ) . Definition and Motivating Question With Ω ✓ R n as above, put Z E x [ τ n ] dx T n ( Ω ) = (exit time moments) . Ω Our motivating question: How can we (sharply) estimate Dirichlet eigenvalues { λ n ( Ω ) } in terms of the exit time moments { T n ( Ω ) } ? Je ff rey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates

Background Let Ω ✓ R n denote a bounded C ∞ domain. Again, X t denote Brownian motion in R n , P x denote the probability measure charging Brownian paths starting at x 2 R n , and 2 Ω } denote the first exit time of X t from Ω . Then u ( x , t ) = P x ( τ > t ) solves τ = inf { t � 0 : X t / ∂ u = ∆ u in Ω ⇥ (0 , 1 ) , ∂ t u ( x , 0) = 1 in Ω , x → σ u ( x , t ) lim = 0 for all ( σ , t ) 2 ∂ Ω ⇥ (0 , 1 ) . Write Z H ( t ) = u ( x , t ) dx (heat content) . Ω Write { λ n ( Ω ) } for the eigenvalues of the Dirichlet Laplacian: � ∆ u = λ u in Ω u = 0 on ∂ Ω . Denote a λ = || Proj E λ 1 || 2 . If spec ∗ ( Ω ) denotes the set of Dirichlet eigenvalues (omitting multiplicity) with a λ > 0, then X a 2 λ e − λ t . H ( t ) = λ ∈ spec ∗ ( Ω ) Thus, X a 2 Vol( Ω ) = (volume partition) . λ λ ∈ spec ∗ ( Ω ) Je ff rey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates

Background Cont. The Mellin transform of the heat content takes the form of a Dirichlet series ✓ 1 ◆ s X a 2 ζ ( s ) = . λ λ λ ∈ spec ∗ ( Ω ) The moment spectrum is related to the Dirichlet spectrum via ζ ( k ) = T k ( Ω ) . k ! We have that spec ∗ ( Ω ), the volume partition { a λ } λ ∈ spec ∗ ( Ω ) , and heat content are all determined by the exit time moments { T n ( Ω ) } . For example, ✓ T n ( Ω ) ◆ 1 1 n λ 1 ( Ω ) = lim n ! n →∞ and n →∞ λ 1 ( Ω ) n T n ( Ω ) a λ 1 = lim . n ! Je ff rey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates

Main Results Theorem 1 (CLM, DLM) Say Ω ✓ R n is a bounded C ∞ domain. Then ✓ 1 ◆ 2 k − 1 T 2 k − 1 ( Ω ) X a 2 (2 k � 1)! � ν ν ν ∈ spec ∗ ( Ω ) ν < λ n ( Ω ) λ n ( Ω )  . ✓ 1 ◆ 2 k T 2 k ( Ω ) X a 2 � ν (2 k )! ν ν ∈ spec ∗ ( Ω ) ν < λ n ( Ω ) Moreover, if λ n ( Ω ) 2 spec ∗ ( Ω ), the inequality becomes an equality in the limit as k ! 1 . When n = 1, the result says λ 1 ( Ω )  2 kT 2 k − 1 ( Ω ) T 2 k ( Ω ) with 2 kT 2 k − 1 ( Ω ) λ 1 ( Ω ) = lim . T 2 k ( Ω ) k →∞ Je ff rey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates

Proof of Theorem 1 for n = 1. Proof. R Ω E x [ τ k ] dx ; put u k ( x ) = E x [ τ k ]. Then u k solve a hierarchy of Poisson Recall that T k ( Ω ) = problems: � ∆ u k = ku k − 1 in Ω u = 0 on ∂ Ω . Plugging u k into the Rayleigh quotient for λ 1 ( Ω ), we see Ω | r u k | 2 dx R λ 1 ( Ω )  . Ω u 2 R k dx Apply Green: Z Z | r u k | 2 dx = � u k ∆ u k dx Ω Ω Z = k u k u k − 1 dx Ω k Z = � ∆ u k +1 u k − 1 dx . k + 1 Ω Iterating this process yields ( k !) 2 ( k !) 2 Z Z | r u k | 2 dx = u 2 k − 1 dx = (2 k � 1)! T 2 k − 1 ( Ω ) . (2 k � 1)! Ω Ω Je ff rey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates

Proof of Theorem 1 for n = 1. Proof. Similarly, Z 1 Z u k u k dx = � u k ∆ u k +1 dx k + 1 Ω Ω k Z = u k +1 u k − 1 dx . k + 1 Ω Iterating as before, k dx = ( k !) 2 u 2 k dx = ( k !) 2 Z Z u 2 (2 k )! T 2 k ( Ω ) . (2 k )! Ω Ω Ω | r u k | 2 dx , R R Ω u 2 So combining our calculations for k dx , yields λ 1 ( Ω )  2 k T 2 k − 1 ( Ω ) T 2 k ( Ω ) . We next show k →∞ 2 k T 2 k − 1 ( Ω ) λ 1 ( Ω ) = lim T 2 k ( Ω ) . Je ff rey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates

Proof of Theorem 1 for n = 1. Proof. ✓ 1 ✓ 1 Start by estimating ◆ 2 k − 1 ◆ 2 k − 1 X X a 2 a 2 T 2 k − 1 ( Ω ) ν ν ν ν 2 k T 2 k − 1 ( Ω ) (2 k � 1)! ν ∈ spec ∗ ( Ω ) ν ∈ spec ∗ ( Ω ) = = .  λ 1 ✓ 1 ◆ 2 k ⌘ 2 k − 1 T 2 k ( Ω ) T 2 k ( Ω ) ⇣ a 2 1 X a 2 λ 1 λ 1 (2 k )! ν ν ν ∈ spec ∗ ( Ω ) We further estimate ✓ 1 ◆ 2 k − 1 X a 2 0 1 ν ✓ λ 1 ν ◆ 2 k − 1 1 ν ∈ spec ∗ ( Ω ) B C X a 2 λ 1 = λ 1 @ 1 + B C ν ⌘ 2 k − 1 a 2 B ν C ⇣ 1 a 2 λ 1 A ν ∈ spec ∗ ( Ω ) λ 1 λ 1 ν > λ 1 0 1 ✓ λ 1 ◆ 2 k − 1 1 B C X a 2  λ 1 @ 1 + B C ν a 2 B ν 2 C λ 1 A ν ∈ spec ∗ ( Ω ) ν > λ 1 ✓ λ 1 ◆ 2 k − 1 ! 1 + Vol( Ω )  λ 1 . a 2 ν 2 λ 1 Sending k ! 1 we see λ 1 ( Ω )  lim k →∞ 2 k T 2 k − 1 ( Ω )  λ 1 ( Ω ). T 2 k ( Ω ) Je ff rey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates

Results about Variance Definition With Ω ✓ R n a C ∞ bounded domain, denote Var k ( Ω ) = R Ω Var[ τ k ] dx. We compute Var[ τ k ] = E x [( τ k � E x [ τ k ]) 2 ] = E x [ τ 2 k � u 2 k ] = u 2 k ( x ) � u 2 k ( x ) . Thus, Z u 2 k � u 2 � � Var k ( Ω ) = dx . k Ω Using our standard trick from before, Z Z Z u 2 k dx = � u 2 k ∆ u 1 dx = 2 k u 2 k − 1 u 1 dx . Ω Ω Ω Repeated application yields Z u 2 k dx = (2 k )! Z u 2 k dx . ( k !) 2 Ω Ω R Ω u 2 We can now rewrite Var k ( Ω ) in terms of k dx : ( k !) 2 Z u 2 k dx = (2 k )! � ( k !) 2 Var k ( Ω ) . Ω In the proof of Theorem 1, we showed ( k !) 2 ( k !) 2 Z Z | r u k | 2 dx = u 2 k − 1 dx = (2 k � 1)! T 2 k − 1 ( Ω ) . (2 k � 1)! Ω Ω Je ff rey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates

Results about Variance Corollary (DLM) Let Ω be as in Theorem 1. For k a positive integer, let Var k ( Ω ) be the L 1 -norm of the variance of τ k : Z ( u 2 k � u 2 Var k ( Ω ) = k ) dx . Ω Then λ 1 ( Ω )  (2 k )! � ( k !) 2 T 2 k − 1 ( Ω ) Var k ( Ω ) . (2 k � 1)! Je ff rey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates