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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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