The Mixed Boundary Value Problemin Lipschitz Domains

Katharine Ott

University of Kentucky

Women and Mathematics

Institute for Advanced Study

June 9, 2009

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Classical Boundary Value Problems for the Laplacian

Dirichlet Problem:

(D)

u ∈ C2(Ω),

4u = 0 in Ω,

u|∂Ω = f ∈ Lp(∂Ω).

Neumann Problem:

(N)

u ∈ C2(Ω),

4u = 0 in Ω,

∂u∂ν |∂Ω = f ∈ Lp

0(∂Ω),

where ν denotes the outward unit normal vector.

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Function Spaces

Definition. Lp(∂Ω), 1 < p <∞ is the Lebesgue space of p-integrablefunctions on ∂Ω,

Lp(∂Ω) :=

f :

(∫∂Ω|f |pdσ

)1/p

< +∞

,

where dσ denotes surface measure on ∂Ω.

Further, define

Lp0(∂Ω) :=

f ∈ Lp(∂Ω);

∫∂Ω fdσ = 0

,

and

Lp1(∂Ω) := f ∈ Lp(∂Ω); ∂τ f ∈ Lp(∂Ω) .

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Lipschitz Domains

A function φ : Rn → R is Lipschitz if there exists a constant M > 0such that for any x , y in the domain of φ,

|φ(x)− φ(y)| < M|x − y |.

Ω is a Lipschitz domain if ∂Ω locally given by the graph of aLipschitz function φ.

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History

B. Dahlberg [1977,1979], E. Fabes, M. Jodeit, N. Riviere [1978]:(D) is well-posed ∀ p ∈ (1,∞) in the class of smooth domains.

B. Dahlberg [1977,1979]: (D) is well-posed ∀ p ∈ [2,∞) in the classof Lipschitz domains. This range is sharp.

B. Dahlberg, C. Kenig [1987]: (N) is well-posed ∀ p ∈ (1, 2] in theclass of Lipschitz domains. This range is sharp.

C. Kenig [1984]: Counterexamples.

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The Mixed Problem for the Laplacian

Let Ω ⊂ Rn be a bounded open set.

Split the boundary of the domain ∂Ω into a Dirichlet and Neumannportion so that

∂Ω = D ∪ N, D ⊂ ∂Ω and N = ∂Ω \ D.

Assume D ⊂ ∂Ω is relatively open, denote by Λ the boundary of D(with respect of ∂Ω).

(MP)

∆u = 0 in Ω

u = fD on D

∂u∂ν = fN on N

where, as before, ν denotes the outward unit normal vector on ∂Ω.

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Motivation for Studying (MP)

Example 1: Iceberg

Consider an iceberg Ω partiallysubmerged in water.

Solution to (MP), u(x), is thetemperature at each point x ∈ Ω.

D is the portion of ∂Ω underneaththe waterline. Here, Ω behaves likea thermostat so Dirichlet boundaryconditions are imposed.

N is the portion of ∂Ω above thewaterline. Here, Ω acts like aninsulator so Neumann boundaryconditions are imposed.

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Motivation for Studying (MP)

Example 2: Metallurgical Melting

Ω is the cross section of aninfinitely long solid with thermalsources located within.

u(x) is the temperature of thesolid at each point x ∈ Ω.

∂Ω = Γ1 ∪ Γ2.

On Γ1, u is cooled to 0 by adistribution of heat sinks.

On Γ2, the heat u is leavingthrough Γ2 at a steady rate g .

Mathematical model takes theform

∆u = ρ in Ω

u|Γ1 = 0

∂u∂ν |Γ2 = g

Above, ρ is a source functioncapturing the input of energyinto Ω.

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The Mixed Problem for the Laplacian

(MP)

∆u = 0 in Ω

u = fD on D

∂u∂ν = fN on N

Goal. Given than fD is in a certain function space on D, and ∂u∂ν = fN

is in a certain function space on N, deduce information about ∇u onthe whole boundary ∂Ω.

Via trace theorems obtain results for ∇u on Ω.

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An Example

Expectation.

(MP)

∆u = h in Ω

u|D = fD ∈ L21(∂Ω)

∂u∂ν |N = fN ∈ L2(∂Ω)

⇒ ∇u ∈ L2(∂Ω).

In the setting of (MP), our intuition that a smooth boundary is betterdoes not hold.

Counterexample. Let Ω ⊂ R2, Ω := (x , y) : x2 + y2 < 1, y > 0.

Take u(x , y) = Im (x + iy)1/2.

In polar coordinates,u(x , y) = U(r , θ) = r1/2 sin(θ/2).

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An Example, continued

u(x , y) = U(r , θ) = r1/2 sin(θ/2)

Calculus:

∆u(x , y) = ∂2u∂x2 + ∂2u

∂y2 ,

= ∂2U∂r2 + 1

r∂U∂r + 1

r2∂2U∂θ2 .

Then ∆u = 0 in Ω.

More calculus: ∂u∂ν = ∂U

∂θ ·1r , so

∂u

∂ν|N = r−1/2 cos(

θ

2) · 1

2|N = 0.

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An Example, continued

u(x , y) = U(r , θ) = r1/2 sin(θ/2)

∆u = 0 in Ω

u|D = 0

∂u∂ν |N = 0

u satisfies u|D ∈ L21(D), ∂u

∂ν |N ∈ L2(N).

However. ∇u ∼ r−1/2 which is not in L2(∂Ω). Problem is at theorigin.

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More General Example

u(x , y) = U(r , θ) = rβsin(βθ). Then ∆u = 0 ∈ Ω and u|D = 0.

∂u∂ν |N = (∂U

∂θ ·1r )|N , so ∂u

∂ν |N = rβ−1 cos(βα)β. In order for ∂u∂ν |N = 0,

need βα = π2 ⇒ β = π

2α .

Further, ∇u ∼ rβ−1 = rπ

2α−1, so ∇u ∈ L2(∂Ω) whenever

( π2α − 1)2 > −1. In other words, when π > α.

Leads to the study of (MP) is creased domains.

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Some History of (MP)

Sevare [1997]: Ω a smooth domain, then solution u of (MP) lies in

the Besov space B3/2,2∞ (Ω).

Brown [1994]: Ω a creased domain, fD ∈ L21(∂Ω), fN ∈ L2(∂Ω), then

there exists a unique solution u with (∇u)∗ ∈ L2(∂Ω). Resultsextended with J. Sykes [1999] to Lp(∂Ω), 1 < p < 2.

Brown, Capgona and Lanzani [2008]: Ω a Lipschitz graph domain intwo dimensions with Lipschitz constant M < 1, solutions in Lp(∂Ω)for 1 < p < p0 with p0 = p0(M) > 1.

Venouziou and Verchota [2008]: L2(∂Ω) results for (MP) for certainpolyhedra in R3.

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The Mixed Problem with Atomic Data

Let Ω ⊂ Rn be a bounded Lipschitz domain.

(MPa)

∆u = 0 in Ω,

u = 0 on D,

∂u∂ν = a atom for N.

a is an atom for ∂Ω if:suppa ⊂ ∆r (x) for some x ∈ ∂Ω, where ∆r (x) = Br (x) ∩ ∂Ω,

||a||∞ ≤ 1/σ(∆r (x)),∫∂Ω

adσ = 0.

a is an atom for N if a is the restriction to N of a function a which isan atom for ∂Ω.

H1(N) is the collection of functions f which can be represented asΣjλjaj , where each aj is an atom for N and Σj |λj | <∞.

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The Fundamental Estimate

Recall ∆r (x) = Br (x) ∩ ∂Ω and let Σk = ∆2k r (x) \∆2k−1r (x).

Theorem 1, R. Brown, KO

Let u be a weak solution of (MPa) with data fN = a an atom for N whichis supported in ∆r (x) and fD = 0. There exists q > 1 such that thefollowing estimates hold(∫

∆r (x)|∇u|qdσ

)1/q

≤ Cσ(∆8r (x))−1/q′ ,

(∫Σk

|∇u|qdσ)1/q

≤ C2−αkσ(Σk)1/q′ , k ≥ 3.

Here, C , q and α depend only on the Lipschitz character of Ω.

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L1(∂Ω) Estimates for (MP)

Theorem 2, R. Brown, KO

Let u be a weak solution of the mixed problem with fD = 0 and fN = a,where a is an atom for the Hardy space H1(N). Then u satisfies

||(∇u)∗||L1(∂Ω) ≤ C .

Theorem 3, R. Brown, KO

Let u be a weak solution of (MP) with fD ∈ H11 (D) and fN ∈ H1(N).

Then u satisfies

||(∇u)∗||L1(∂Ω) ≤ C(||fD ||H1

1 (D) + ||fN ||H1(N)

).

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Results for (MP) in Other Function Spaces

Goal. Extend Theorem 3 to Lp(∂Ω), p ∈ [1, 1 + ε).

That is, wish to prove an estimate of the form

||(∇u)∗||Lp(∂Ω) ≤ C(||fD ||Lp

1(D) + ||fN ||Lp(N)

).

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