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

Exit Time Moments and Eigenvalue Estimates

Jeffrey 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.

Jeffrey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates

Motivation

We recall torsional rigidity. For Ω ✓ Rn a bounded C ∞ domain, let u1 solve ∆u1 = 1 in Ω, u = 0

• n ∂Ω.

Put T1(Ω) = Z

Ω

u1 dx (torsional rigidity). Interpreting probabilistically, we let Xt denote Brownian motion in Rn, Px denote the probability measure charging Brownian paths starting at x 2 Rn, and τ = inf{t 0 : Xt / 2 Ω} denote the first exit time of Xt from Ω. Then u1(x) = Ex[τ] and T1(Ω) = Z

Ω

Ex[τ] dx. P´

• lya’s inequality gives an estimate of the principal Dirichlet eigenvalue in terms of T1(Ω):

λ1(Ω)  |Ω| T1(Ω) . Definition and Motivating Question With Ω ✓ Rn as above, put Tn(Ω) = Z

Ω

Ex[τ n] dx (exit time moments). Our motivating question: How can we (sharply) estimate Dirichlet eigenvalues {λn(Ω)} in terms

• f the exit time moments {Tn(Ω)}?

Jeffrey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates

Background

Let Ω ✓ Rn denote a bounded C ∞ domain. Again, Xt denote Brownian motion in Rn, Px denote the probability measure charging Brownian paths starting at x 2 Rn, and τ = inf{t 0 : Xt / 2 Ω} denote the first exit time of Xt from Ω. Then u(x, t) = Px(τ > t) solves ∂u ∂t = ∆u in Ω ⇥ (0, 1), u(x, 0) = 1 in Ω, lim

x→σ u(x, t)

= 0 for all (σ, t) 2 ∂Ω ⇥ (0, 1). Write H(t) = Z

Ω

u(x, t) dx (heat content). Write {λn(Ω)} for the eigenvalues of the Dirichlet Laplacian: ∆u = λu in Ω u = 0

• n ∂Ω.

Denote aλ = ||ProjEλ1||2. If spec∗(Ω) denotes the set of Dirichlet eigenvalues (omitting multiplicity) with aλ > 0, then H(t) = X

λ∈spec∗(Ω)

a2

λe−λt.

Thus, Vol(Ω) = X

λ∈spec∗(Ω)

a2

λ

(volume partition).

Jeffrey 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 ζ(s) = X

λ∈spec∗(Ω)

a2

λ

✓ 1 λ ◆s . The moment spectrum is related to the Dirichlet spectrum via ζ(k) = Tk(Ω) k! . We have that spec∗(Ω), the volume partition {aλ}λ∈spec∗(Ω), and heat content are all determined by the exit time moments {Tn(Ω)}. For example, 1 λ1(Ω) = lim

n→∞

✓ Tn(Ω) n! ◆ 1

n

and aλ1 = lim

n→∞ λ1(Ω)n Tn(Ω)

n! .

Jeffrey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates

Main Results

Theorem 1 (CLM, DLM) Say Ω ✓ Rn is a bounded C ∞ domain. Then λn(Ω)  T2k−1(Ω) (2k 1)! X

ν∈spec∗(Ω)

ν<λn(Ω)

a2

ν

✓ 1 ν ◆2k−1 T2k(Ω) (2k)!

• X

ν∈spec∗(Ω)

ν<λn(Ω)

a2

ν

✓ 1 ν ◆2k . Moreover, if λn(Ω) 2 spec∗(Ω), the inequality becomes an equality in the limit as k ! 1. When n = 1, the result says λ1(Ω)  2kT2k−1(Ω) T2k(Ω) with λ1(Ω) = lim

k→∞

2kT2k−1(Ω) T2k(Ω) .

Jeffrey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates

Proof of Theorem 1 for n = 1.

Proof. Recall that Tk(Ω) = R

Ω Ex[τ k] dx; put uk(x) = Ex[τ k]. Then uk solve a hierarchy of Poisson

problems: ∆uk = kuk−1 in Ω u = 0

• n ∂Ω.

Plugging uk into the Rayleigh quotient for λ1(Ω), we see λ1(Ω)  R

Ω |ruk|2 dx

R

Ω u2 k dx

. Apply Green: Z

Ω

|ruk|2 dx = Z

Ω

uk∆uk dx = k Z

Ω

ukuk−1 dx = k k + 1 Z

Ω

∆uk+1uk−1 dx. Iterating this process yields Z

Ω

|ruk|2 dx = (k!)2 (2k 1)! Z

Ω

u2k−1 dx = (k!)2 (2k 1)! T2k−1(Ω).

Jeffrey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates

Proof of Theorem 1 for n = 1.

Proof. Similarly, Z

Ω

ukuk dx = 1 k + 1 Z

Ω

uk∆uk+1 dx = k k + 1 Z

Ω

uk+1uk−1 dx. Iterating as before, Z

Ω

u2

k dx = (k!)2

(2k)! Z

Ω

u2k dx = (k!)2 (2k)! T2k(Ω). So combining our calculations for R

Ω |ruk|2 dx,

R

Ω u2 k dx, yields

λ1(Ω)  2k T2k−1(Ω) T2k(Ω) . We next show λ1(Ω) = lim

k→∞ 2k T2k−1(Ω)

T2k(Ω) .

Jeffrey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates

Proof of Theorem 1 for n = 1.

Proof. Start by estimating 2k T2k−1(Ω) T2k(Ω) = T2k−1(Ω) (2k 1)! T2k(Ω) (2k)! = X

ν∈spec∗(Ω)

a2

ν

✓ 1 ν ◆2k−1 X

ν∈spec∗(Ω)

a2

ν

✓ 1 ν ◆2k  λ1 X

ν∈spec∗(Ω)

a2

ν

✓ 1 ν ◆2k−1 a2

λ1

⇣

1 λ1

⌘2k−1 . We further estimate λ1 X

ν∈spec∗(Ω)

a2

ν

✓ 1 ν ◆2k−1 a2

λ1

⇣

1 λ1

⌘2k−1 = λ1 B B B @1 + 1 a2

λ1

X

ν∈spec∗(Ω)

ν>λ1

a2

ν

✓ λ1 ν ◆2k−1 1 C C C A  λ1 B B B @1 + 1 a2

λ1

✓ λ1 ν2 ◆2k−1 X

ν∈spec∗(Ω)

ν>λ1

a2

ν

1 C C C A  λ1 1 + Vol(Ω) a2

λ1

✓ λ1 ν2 ◆2k−1! . Sending k ! 1 we see λ1(Ω)  limk→∞ 2k T2k−1(Ω)

T2k (Ω)

 λ1(Ω).

Jeffrey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates

Results about Variance

Definition With Ω ✓ Rn a C ∞ bounded domain, denote Vark(Ω) = R

Ω Var[τ k] dx.

We compute Var[τ k] = Ex[(τ k Ex[τ k])2] = Ex[τ 2k u2

k] = u2k(x) u2 k(x).

Thus, Vark(Ω) = Z

Ω

• u2k u2

k

• dx.

Using our standard trick from before, Z

Ω

u2kdx = Z

Ω

u2k∆u1 dx = 2k Z

Ω

u2k−1u1 dx. Repeated application yields Z

Ω

u2kdx = (2k)! (k!)2 Z

Ω

u2

kdx.

We can now rewrite Vark(Ω) in terms of R

Ω u2 kdx:

Z

Ω

u2

kdx =

(k!)2 (2k)! (k!)2 Vark(Ω). In the proof of Theorem 1, we showed Z

Ω

|ruk|2 dx = (k!)2 (2k 1)! Z

Ω

u2k−1 dx = (k!)2 (2k 1)! T2k−1(Ω).

Jeffrey 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 Vark(Ω) be the L1-norm of the variance

• f τ k :

Vark(Ω) = Z

Ω

(u2k u2

k)dx.

Then λ1(Ω)  (2k)! (k!)2 (2k 1)! T2k−1(Ω) Vark(Ω) .

Jeffrey Langford Bucknell University Lewisburg, PA Exit Time Moments and Eigenvalue Estimates