The Mixed Boundary Value Problem in Lipschitz Domains · The Mixed Boundary Value Problem in...

The Mixed Boundary Value Problemin Lipschitz Domains

Katharine Ott

University of Kentucky

Women and Mathematics

Institute for Advanced Study

June 9, 2009

Katharine Ott (UK) Mixed Problem 06/09/2009 1 / 18

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.


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) .


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.


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.


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.


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.


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.


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.


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).


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.


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.


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.


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.


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 | <∞.


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.


L1(∂Ω) Estimates for (MP)


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 .


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)

).


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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