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Eigenvalue estimates and localization of the first Dirichlet - - PowerPoint PPT Presentation

Eigenvalue estimates and localization of the first Dirichlet eigenfunction Thomas Beck University of North Carolina - Chapel Hill tdbeck@email.unc.edu April 3, 2020 Thomas Beck (UNC) Localization of the first Dirichlet eigenfunction April 3,


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Eigenvalue estimates and localization of the first Dirichlet eigenfunction

Thomas Beck

University of North Carolina - Chapel Hill tdbeck@email.unc.edu

April 3, 2020

Thomas Beck (UNC) Localization of the first Dirichlet eigenfunction April 3, 2020 1 / 9

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First Dirichlet eigenfunction

u first Dirichlet eigenfunction, with eigenvalue λ:

  • (∆ + λ)u = 0

in Ω, u = 0

  • n ∂Ω.

Throughout, Ω ⊂ Rn will be a convex domain of inner radius 1. u = 0 on ∂Ω, and we normalize u > 0 inside Ω. Superlevel sets: Ωc := {x ∈ Ω : u(x) > c}.

Question

How small a subset of the domain Ω can the eigenfunction localize to? Aim: Study localization via the quantity uL2(Ω) uL∞(Ω) .

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An estimate of Chiti

Theorem (Chiti ’82)

There exists a constant c∗

n (independent of Ω) such that

uL2(Ω) uL∞(Ω) ≥ c∗

n .

Equality holds precisely if Ω is the ball of radius 1. As the diameter of Ω increases, is the eigenfunction forced to spread out along the diameter? Does there exist α > 0 (independent of the domain Ω) such that uL2(Ω) uL∞(Ω) ≥ cndiam(Ω)α.

Conjecture (van den Berg ’00)

There exists a constant cn (independent of the domain Ω) such that uL2(Ω) uL∞(Ω) ≥ cndiam(Ω)1/6.

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The two dimensional case

Two explicit examples: 1) Rectangles, Ω = [0, N] × [0, 1]: Solve via separation of variables, and u(x, y) = sin πx

N

  • sin(πy).

uL2(Ω) uL∞(Ω) is comparable to N1/2 = diam(Ω)1/2. 2) Circular sectors, Ω = {(r, θ) : 0 ≤ r ≤ α1(N), 0 ≤ θ ≤ 1

N }: Solve via

separation of variables, and u(x, y) = JN (r) sin (πNθ) . JN is the Nth Bessel function, α1(N) ∼ N is its first zero. uL2(Ω) uL∞(Ω) is comparable to N1/6 = diam(Ω)1/6. Georgiev-Mukherjee-Steinerberger ’18 showed that the sector is the most localized case and proved the conjecture in two dimensions, using work of Grieser and Jerison, ’95, ’96, ’98.

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Localization of the eigenfunction

Theorem (B. ’19)

There exists cn > 0 (independent of Ω) such that uL2(Ω) uL∞(Ω) ≥ cndiam(Ω)1/6. The 1

6 power cannot be improved in any dimension (attained by the cone).

When Ω extends in more than one direction a stronger version of the inequality holds: Let K be a John ellipsoid for Ω, with axes lengths N1 ≥ N2 ≥ · · · ≥ Nn ∼ 1. Then, uL2(Ω) uL∞(Ω) ≥ cn

n−1

  • j=1

N1/6

j

.

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Localization of the eigenfunction

The crucial estimate in the proof:

Proposition

There exists a constant Cn (independent of Ω) such that ˆ

|∂x1u|2 ≤ Cndiam(Ω)−2/3 ˆ

u2. The theorem follows from this proposition combined crucially with the convexity

  • f the superlevel sets (Brascamp and Lieb ’76).

Idea of the proof of the proposition: 1) Reduce to eigenvalue bounds by writing x = (x1, x′) and combining: ˆ

|∂x1u|2 + ˆ

|∇x′u|2 = λ ˆ

u2, ˆ

|∇x′u|2 ≥ µ∗ ˆ

u2. Here µ∗ is the minimal first eigenvalue of a (n − 1)-dimensional cross-section

  • f Ω, denoted by Ω∗.

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Localization of the eigenfunction

Proposition

There exists a constant Cn (independent of Ω) such that ˆ

|∂x1u|2 ≤ Cndiam(Ω)−2/3 ˆ

u2. 1) Combining implies ˆ

|∂x1u|2 ≤ (λ − µ∗) ˆ

u2, and it remains to estimate λ − µ∗ from above. 2) Do this by showing µ∗ ≤ λ ≤ µ∗ + Cndiam(Ω)−2/3 using the variational formulation of the first eigenvalue. Let ψ∗(x′) be the first Dirichlet eigenfunction of Ω∗, eigenvalue µ∗. Use the test function w(x1, x′) = χ(x1)ψ(x′N1/(N1 − x1)), with χ(x1) supported in an interval of length N1/3

1

∼ diam(Ω)1/3 around Ω∗.

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Future Directions

Theorem (B. ’19)

There exists cn > 0 (independent of Ω) such that uL2(Ω) uL∞(Ω) ≥ cndiam(Ω)1/6. There is also the trivial upper bound uL2(Ω) uL∞(Ω) ≤ vol(Ω)1/2.

Question

Can we use the geometry of the domain Ω to determine comparable upper and lower bounds on uL2(Ω) uL∞(Ω) ? In 2 dimensions, Jerison (’95), introduced a length scale L = L(Ω) to give a positive answer to this question. Open in higher dimensions.

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Thank you for your attention!

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