The Kato Square Root Problem
for Mixed Boundary Conditions
Moritz Egert
joint work with R. Haller-Dintelmann and P. Tolksdorf
June 2013
Setup
D
Let
I
Ω ⊆ Rd bdd. ‘rough’ domain,
I
D ⊆ ∂Ω closed,
Pd
AD u ∼ α,β=1 −∂α (aα,β ∂β u),
I
aα,β ∈ L∞ (Ω).
I
Realize AD on L2 (Ω) via form method

Z
d
X


aD (u, v ) :=
aα,β ∂β u · ∂α v ,
α,β=1 Ω



dom(aD )
:= HD1 (Ω) := u|Ω : u ∈ CD∞ (Rd )
H 1 (Ω)
.
Assume Gårding Inequality
∃λ > 0 :
<(aD (u, u)) ≥ λk∇uk2L2 (Ω)
(u ∈ HD1 (Ω)).
Setup
D
Let
I
Ω ⊆ Rd bdd. ‘rough’ domain,
I
D ⊆ ∂Ω closed,
Pd
AD u ∼ α,β=1 −∂α (aα,β ∂β u),
I
aα,β ∈ L∞ (Ω).
I
Realize AD on L2 (Ω) via form method

Z
d
X


aD (u, v ) :=
aα,β ∂β u · ∂α v ,
α,β=1 Ω



dom(aD )
:= HD1 (Ω) := u|Ω : u ∈ CD∞ (Rd )
H 1 (Ω)
.
Assume Gårding Inequality
∃λ > 0 :
<(aD (u, u)) ≥ λk∇uk2L2 (Ω)
(u ∈ HD1 (Ω)).
Properties of AD
I
I
I
−AD ∼ analytic contraction semigroup on L2 (Ω).
HD1 (Ω) ⊆ dom(AD ) * HD1 (Ω) ∩ H 2 (Ω) in general.
√
dom( AD ) = dom(aD ) = HD1 (Ω) if AD self-adjoint.
The Kato Square Root Problem
√
Does dom( AD ) = dom(aD ) = HD1 (Ω) always hold?
Properties of AD
I
I
I
−AD ∼ analytic contraction semigroup on L2 (Ω).
HD1 (Ω) ⊆ dom(AD ) * HD1 (Ω) ∩ H 2 (Ω) in general.
√
dom( AD ) = dom(aD ) = HD1 (Ω) if AD self-adjoint.
The Kato Square Root Problem
√
Does dom( AD ) = dom(aD ) = HD1 (Ω) always hold?
Positive Answers
I
Auscher-Hofmann-Lacey-Mc Intosh-Tchamitchian ’01: Ω = Rd ,
Properties of AD
I
I
I
−AD ∼ analytic contraction semigroup on L2 (Ω).
HD1 (Ω) ⊆ dom(AD ) * HD1 (Ω) ∩ H 2 (Ω) in general.
√
dom( AD ) = dom(aD ) = HD1 (Ω) if AD self-adjoint.
The Kato Square Root Problem
√
Does dom( AD ) = dom(aD ) = HD1 (Ω) always hold?
Positive Answers
I
Auscher-Hofmann-Lacey-Mc Intosh-Tchamitchian ’01: Ω = Rd ,
I
Auscher-Tchamitchian ’03: Ω strongly Lipschitz, D ∈ {∅, ∂Ω},
Properties of AD
I
I
I
−AD ∼ analytic contraction semigroup on L2 (Ω).
HD1 (Ω) ⊆ dom(AD ) * HD1 (Ω) ∩ H 2 (Ω) in general.
√
dom( AD ) = dom(aD ) = HD1 (Ω) if AD self-adjoint.
The Kato Square Root Problem
√
Does dom( AD ) = dom(aD ) = HD1 (Ω) always hold?
Positive Answers
I
Auscher-Hofmann-Lacey-Mc Intosh-Tchamitchian ’01: Ω = Rd ,
I
Auscher-Tchamitchian ’03: Ω strongly Lipschitz, D ∈ {∅, ∂Ω},
I
Axelsson-Keith-Mc Intosh ’06: Ω smooth, D̊ extension domain of
∂Ω and bi-Lipschitz images of these configurations.
Properties of AD
I
I
I
−AD ∼ analytic contraction semigroup on L2 (Ω).
HD1 (Ω) ⊆ dom(AD ) * HD1 (Ω) ∩ H 2 (Ω) in general.
√
dom( AD ) = dom(aD ) = HD1 (Ω) if AD self-adjoint.
The Kato Square Root Problem
√
Does dom( AD ) = dom(aD ) = HD1 (Ω) always hold?
Positive Answers
I
Auscher-Hofmann-Lacey-Mc Intosh-Tchamitchian ’01: Ω = Rd ,
I
Auscher-Tchamitchian ’03: Ω strongly Lipschitz, D ∈ {∅, ∂Ω},
I
Axelsson-Keith-Mc Intosh ’06: Ω smooth, D̊ extension domain of
∂Ω and bi-Lipschitz images of these configurations.
Reduction to −∆D
Minimal smoothness assumptions
I
Existence of a bounded extension operator H 1 (Ω) → H 1 (Rd ),
I
d-set property:
|B(x, r ) ∩ Ω| ∼ r d
(x ∈ Ω, r ∈ (0, r0 )).
I
No assumptions on D.
I
Includes all bounded Lipschitz domains.
Reduction to −∆D
Minimal smoothness assumptions
I
Existence of a bounded extension operator H 1 (Ω) → H 1 (Rd ),
I
d-set property:
|B(x, r ) ∩ Ω| ∼ r d
(x ∈ Ω, r ∈ (0, r0 )).
I
No assumptions on D.
I
Includes all bounded Lipschitz domains.
Reduction to −∆D
Minimal smoothness assumptions
I
Existence of a bounded extension operator H 1 (Ω) → H 1 (Rd ),
I
d-set property:
|B(x, r ) ∩ Ω| ∼ r d
(x ∈ Ω, r ∈ (0, r0 )).
I
No assumptions on D.
I
Includes all bounded Lipschitz domains.
Proposition
If dom (−∆D )
1+α
2
,→
H 1+α (Ω)
for an α > 0, then
HD1 (Ω)
√
= dom( AD ).
Reduction to −∆D
Minimal smoothness assumptions
I
Existence of a bounded extension operator H 1 (Ω) → H 1 (Rd ),
I
d-set property:
|B(x, r ) ∩ Ω| ∼ r d
(x ∈ Ω, r ∈ (0, r0 )).
I
No assumptions on D.
I
Includes all bounded Lipschitz domains.
Proposition
If dom (−∆D )
1+α
2
,→
H 1+α (Ω)
for an α > 0, then
HD1 (Ω)
√
= dom( AD ).
Extrapolate Kato property for −∆D =⇒ Kato property for general AD
Reduction to −∆D
Minimal smoothness assumptions
I
Existence of a bounded extension operator H 1 (Ω) → H 1 (Rd ),
I
d-set property:
|B(x, r ) ∩ Ω| ∼ r d
(x ∈ Ω, r ∈ (0, r0 )).
I
No assumptions on D.
I
Includes all bounded Lipschitz domains.
Proposition
If dom (−∆D )
1+α
2
,→
H 1+α (Ω)
for an α > 0, then
HD1 (Ω)
√
= dom( AD ).
Extrapolate Kato property for −∆D =⇒ Kato property for general AD
geometry ! operator theory
Our Geometric Setting
Let
I
I
Ω bounded Lipschitz domain,
D a (d − 1)-set:
|B(x, r ) ∩ D| ∼ r d−1
(x ∈ D, r ∈ (0, r0 )).
D
Introduce
n
o
1+α
I HD
(Ω) := u ∈ H 1+α (Ω) : u|D = 0 Hd−1 - a.e.
I
{HDθ (Ω)} 1 <θ< 3 complex interpolation scale.
2
2
(|α| < 12 ).
From Extrapolation to Interpolation
Goal
(♠)
dom (−∆D )
1+α
2
,→ HD1+α (Ω) for small α ∈ (0, 12 ).
From Extrapolation to Interpolation
Goal
(♠)
dom (−∆D )
1+α
2
,→ HD1+α (Ω) for small α ∈ (0, 12 ).
Lemma
One gets (♠) if
HD1−α (Ω)
,→ dom (−∆D )
1−α
2
holds for α ∈ (0, 12 ).
Observations
I Reduces extrapolation result to interpolation result.
From Extrapolation to Interpolation
Goal
(♠)
dom (−∆D )
1+α
2
,→ HD1+α (Ω) for small α ∈ (0, 12 ).
Lemma
One gets (♠) if
HD1−α (Ω)
,→ dom (−∆D )
1−α
2
holds for α ∈ (0, 12 ).
Observations
I Reduces extrapolation result to interpolation result.
p
2
1−α I dom (−∆D ) 2
= L , dom( −∆D ) 1−α
From Extrapolation to Interpolation
Goal
(♠)
dom (−∆D )
1+α
2
,→ HD1+α (Ω) for small α ∈ (0, 12 ).
Lemma
One gets (♠) if
HD1−α (Ω)
,→ dom (−∆D )
1−α
2
holds for α ∈ (0, 12 ).
Observations
I Reduces extrapolation result to interpolation result.
p
2
1−α I dom (−∆D ) 2
= L , dom( −∆D ) 1−α
From Extrapolation to Interpolation
Goal
(♠)
dom (−∆D )
1+α
2
,→ HD1+α (Ω) for small α ∈ (0, 12 ).
Lemma
One gets (♠) if
HD1−α (Ω)
,→ dom (−∆D )
1−α
2
holds for α ∈ (0, 12 ).
Observations
I Reduces extrapolation result to interpolation result.
p
2
2 1
1−α I dom (−∆D ) 2
= L , dom( −∆D ) 1−α = L , HD 1−α
From Extrapolation to Interpolation
Goal
(♠)
dom (−∆D )
1+α
2
,→ HD1+α (Ω) for small α ∈ (0, 12 ).
Lemma
One gets (♠) if
HD1−α (Ω)
,→ dom (−∆D )
1−α
2
holds for α ∈ (0, 12 ).
Observations
I Reduces extrapolation result to interpolation result.
p
2
2 1
1−α ?
I dom (−∆D ) 2
= L , dom( −∆D ) 1−α = L , HD 1−α = HD1−α .
From Extrapolation to Interpolation
Goal
(♠)
dom (−∆D )
1+α
2
,→ HD1+α (Ω) for small α ∈ (0, 12 ).
Lemma
One gets (♠) if
HD1−α (Ω)
,→ dom (−∆D )
1−α
2
holds for α ∈ (0, 12 ).
Observations
I Reduces extrapolation result to interpolation result.
p
2
2 1
1−α ?
I dom (−∆D ) 2
= L , dom( −∆D ) 1−α = L , HD 1−α = HD1−α .
I
Main ingredient for interpolation result: Fractional Hardy Inequality
Z
|u(x)|2
1−α
2
dx
.
kuk
(u
∈
H
(Ω)).
1−α
D
2(1−α)
H
(Ω)
D
Ω distD (x)
The Main Result
Summary
1+α
2
,→ HD1+α (Ω).
I
For α > 0 small: dom (−∆D )
I
Extrapolate Kato property for −∆D =⇒ Kato property for AD .
The Main Result
Summary
1+α
2
,→ HD1+α (Ω).
I
For α > 0 small: dom (−∆D )
I
Extrapolate Kato property for −∆D =⇒ Kato property for AD .
Theorem
Let Ω ⊆ Rd bounded Lipschitz domain and D a (d − 1)-set. Then
p
dom( AD ) = HD1 (Ω)
and moreover,
(
α
HDα (Ω) α ∈ ( 12 , 1]
dom((AD ) 2 ) =
.
1
α
H (Ω) α ∈ (0, 2 )
The Main Result
Summary
1+α
2
,→ HD1+α (Ω).
I
For α > 0 small: dom (−∆D )
I
Extrapolate Kato property for −∆D =⇒ Kato property for AD .
Theorem
Let Ω ⊆ Rd bounded Lipschitz domain and D a (d − 1)-set. Then
p
dom( AD ) = HD1 (Ω)
and moreover,
(
α
HDα (Ω) α ∈ ( 12 , 1]
dom((AD ) 2 ) =
.
1
α
H (Ω) α ∈ (0, 2 )
Thank you for your attention!