**The Dirichlet problem
without the maximum principle**
W. Arendt1 and A.F.M. ter Elst2
Abstract
- Consider the Dirichlet problem with respect to an elliptic operator
[TABLE]
on a bounded Wiener regular open set Ω⊂\mathdsRd,
where akl,ck∈L∞(Ω,\mathdsR) and
bk,c0∈L∞(Ω,\mathdsC).
Suppose that the associated operator on L2(Ω) with
Dirichlet boundary conditions is invertible.
Then we show that for all φ∈C(∂Ω) there exists a
unique u∈C(Ω)∩Hloc1(Ω) such that
u∣∂Ω=φ and Au=0.
In the case when Ω has a Lipschitz boundary and
φ∈C(Ω)∩H1/2(Ω), then we
show that u coincides with the variational solution in H1(Ω).
October 2017
AMS Subject Classification: 31C25, 35J05, 31B05.
Keywords:
Dirichlet problem, Wiener regular, holomorphic semigroup.
Home institutions:
Institute of Applied Analysis
2.
Department of Mathematics
University of Ulm
University of Auckland
Helmholtzstr. 18
Private bag 92019
89081 Ulm
Auckland
Germany
New Zealand
1 Introduction
Let Ω⊂\mathdsRd be an open bounded set with boundary Γ.
Throughout this paper we assume that d≥2.
The classical Dirichlet problem is to find for each φ∈C(Γ)
a function u∈C(Ω)
such that u∣Γ=φ and Δu=0 as distribution on Ω.
The set Ω is called Wiener regular if for every
φ∈C(Γ) there exists a unique u∈C(Ω)
such that u∣Γ=φ and Δu=0 as distribution on Ω.
The Dirichlet problem has been extended naturally to more general
second-order operators.
For all k,l∈{1,…,d} let akl:Ω→\mathdsR
be a bounded measurable function and suppose that there exists a
μ>0 such that
[TABLE]
for all x∈Ω and ξ∈\mathdsCd.
Further, for all k∈{1,…,d} let bk,ck,c0:Ω→\mathdsC
be bounded and measurable.
Define the map A:Hloc1(Ω)→D′(Ω) by
[TABLE]
for all u∈Hloc1(Ω) and v∈Cc∞(Ω).
Given φ∈C(Γ), by a classical solution of the
Dirichlet problem we understand a function
u∈C(Ω)∩Hloc1(Ω) satisfying Au=0
and u∣Γ=φ.
For the pure second-order case (that is bk=ck=c0=0)
Littman–Stampacchia–Weinberger
[LSW] proved that for all φ∈C(Γ)
there exists a unique classical solution u.
Then Stampacchia [Sta] Théorème 10.2
added real valued lower order terms, under the condition
(see [Sta], (9.2’)) that there exists a
μ′>0
such that
[TABLE]
for all v∈Cc∞(Ω)+.
Gilbarg–Trudinger [GT] Theorem 8.31 merely assume that
[TABLE]
for all v∈Cc∞(Ω)+ in order to obtain the same conclusion.
A consequence of these assumptions is a weak maximum principle,
which implies that ∥u∥C(Ω)≤∥φ∥C(Γ)
for all u∈Hloc1(Ω)∩C(Ω) satisfying Au=0
and u∣Γ=φ.
We may consider (3) as a kind of submarkov condition since
it is equivalent to −A\mathds1Ω≤0 in D′(Ω).
The aim of this paper is to
show that the positivity condition (3) and the maximum principle
are not needed for the well-posedness of the Dirichlet problem.
In addition we allow the bk and c0 to be complex valued.
In order to state the main results of this paper in a more
precise way we need a few definitions.
Define the form \gothica:H1(Ω)×H1(Ω)→\mathdsC by
[TABLE]
Let AD be the operator in L2(Ω) associated with the
form \gothica∣H01(Ω)×H01(Ω).
In other words, AD is the realisation of the elliptic operator A
in L2(Ω) with Dirichlet boundary conditions.
This operator has a compact resolvent.
Moreover, if (3) is valid, then kerAD={0} by [GT]
Corollary 8.2.
Instead of (3) we assume the condition kerAD={0},
which is equivalent to the uniqueness of the Dirichlet problem
(cf. Proposition 2.3 below).
The main result of this paper is the following well-posedness result
for the Dirichlet problem.
Theorem 1.1**.**
Let Ω⊂\mathdsRd be an open bounded Wiener regular set with d≥2.
For all k,l∈{1,…,d} let akl:Ω→\mathdsR
be a bounded measurable function and suppose that there exists a
μ>0 such that
[TABLE]
for all x∈Ω and ξ∈\mathdsCd.
Further, for all k∈{1,…,d} let bk,c0:Ω→\mathdsC
and ck:Ω→\mathdsR
be bounded and measurable.
Let AD be as above.
Suppose 0∈σ(AD).
Then for all φ∈C(Γ)
there exists a unique u∈C(Ω)∩Hloc1(Ω)
such that u∣Γ=φ and Au=0.
Moreover, there exists a constant c>0 such that
[TABLE]
for all φ∈C(Γ),
where u∈C(Ω)∩Hloc1(Ω)
is such that u∣Γ=φ and Au=0.
Instead of the homogeneous equation Au=0 one can also consider
the inhomogeneous equation Au=f0+∑k=1d∂kfk.
We shall do that in Theorem 2.13.
Adopt the notation and assumptions of Theorem 1.1.
Define P:C(Γ)→C(Ω) by Pφ=u,
where u∈C(Ω)∩Hloc1(Ω)
is such that u∣Γ=φ and Au=0.
Note that Pφ is the classical solution of the Dirichlet problem.
If Ω has even a Lipschitz boundary
(which implies Wiener regularity), then there is also a variational
solution of the Dirichlet problem that we describe next.
Denote by Tr:H1(Ω)→L2(Γ) the trace operator.
Again let akl,bk,ck,c0∈L∞(Ω) and
suppose that the ellipticity condition (1) is satisfied.
Further suppose that 0∈σ(AD).
Then for each φ∈TrH1(Ω)
there exists a unique u∈H1(Ω), called the
variational solution, such that
Au=0 and Tru=φ (cf. Lemma 2.1).
Define γ:TrH1(Ω)→H1(Ω) by setting
γφ=u.
The second result of this paper says that the variational
solution and the classical solution coincide, if both are defined.
Theorem 1.2**.**
Adopt the notation and assumptions of Theorem 1.1.
Suppose that Ω has a Lipschitz boundary.
Let φ∈C(Γ)∩TrH1(Ω).
Then Pφ=γφ almost everywhere on Ω.
The last main result of this paper concerns a parabolic equation.
Let Ac denote the part of the operator AD in C0(Ω).
So
[TABLE]
and Ac=AD∣D(Ac).
Theorem 1.3**.**
Adopt the notation and assumptions of Theorem 1.1.
Then −Ac generates a holomorphic C0-semigroup on C0(Ω).
Moreover, e−tAcu=e−tADu for all u∈C0(Ω)
and t>0.
In Section 2 we prove Theorem 1.1 via
an iteration argument.
Section 3 is devoted to the comparison of the classical
and the variational solutions of the Dirichlet problem.
Theorem 1.2 is proved there with the help of a deep
result of Dahlberg [Dah].
We consider the semigroup on C0(Ω) in Section 4 and
prove Theorem 1.3.
2 The Dirichlet problem
In this section we prove Theorem 1.1 on the well-posedness
of the Dirichlet problem.
The technique is a reduction to the Stampacchia result mentioned in the
introduction.
For this reason we introduce the following two forms and operators.
Adopt the notation and assumptions of Theorem 1.1.
For all λ∈\mathdsR define the forms
\gothicaλ,\gothicbλ:H1(Ω)×H1(Ω)→\mathdsC
by
[TABLE]
where \gothica is as in (4).
Define similarly Aλ,Bλ:Hloc1(Ω)→D′(Ω)
and let BD be the operator associated with the sesquilinear form
\gothicb0∣H01(Ω)×H01(Ω).
It follows from ellipticity that there exists a λ0>0
such that
[TABLE]
for all v∈H1(Ω).
Note that Bλ satisfies the submarkovian condition
−Bλ\mathds1Ω≤0, that is (3),
and even Stampacchia’s condition (2)
for all λ>0.
So we can and will apply Stampacchia’s result (in the proof of
Lemma 2.8).
We first investigate the operator AD in L2(Ω).
Note that f0+∑k=1d∂kfk∈D′(Ω)
for all f0,f1,…,fd∈L1(Ω).
The next lemma is also valid if the akl and ck are complex valued.
Lemma 2.1**.**
Let f1,…,fd∈L2(Ω).
Let p~∈(1,∞) be such that
p~≥d+22d.
Further let f0∈Lp~(Ω).
Then there exists a unique u∈H01(Ω) such that
Au=f0+∑k=1d∂kfk.
Proof*.*
There exists a unique T∈L(H01(Ω)) such that
(Tu,v)H01(Ω)=\gothica(u,v)
for all u,v∈H01(Ω).
Then T is injective because kerAD={0}.
Moreover, the inclusion H01(Ω)↪L2(Ω) is compact.
Hence the operator T is invertible by the Fredholm–Lax–Milgram lemma,
[AEKS] Lemma 4.1.
Clearly v↦∑k=1d(fk,∂kv)L2(Ω)
is continuous from H01(Ω) into \mathdsC.
Define F:Cc∞(Ω)→\mathdsC by
F(v)=⟨f0,v⟩D′(Ω)×D(Ω).
We claim that F
extends to a continuous function from H01(Ω) into \mathdsC.
If d≥3, then H01(Ω)⊂Lr(Ω), where
r=d−22d.
So H01(Ω)⊂Lq(Ω), where q is the dual
exponent of p~.
The last inclusion is also valid if d=2.
So in any case the map F extends to a continuous function from
H01(Ω) into \mathdsC.
Then the lemma follows.
∎
The next lemma is valid for a general bounded open set Ω
and does not use the condition 0∈σ(AD).
It is an extension of [AB] Lemma 4.2.
Lemma 2.2**.**
Let u∈C0(Ω)∩Hloc1(Ω) and
f1,…,fd∈L2(Ω).
Let p~∈(1,∞) be such that
p~≥d+22d.
Further let f0∈Lp~(Ω).
Suppose that Au=f0+∑k=1d∂kfk.
Then u∈H01(Ω).
Proof*.*
As at the end of the previous proof there exists an M0>0 such that
∣∫Ωf0v∣≤M0∥v∥H1(Ω) for all
v∈H01(Ω).
Set M=M0+∑k=1d∥fk∥2.
Let ε>0.
Set vε=(Reu−ε)+.
Then suppvε⊂Ω is compact.
Hence there exists an open Ω1⊂\mathdsRd such that
suppvε⊂Ω1⊂Ω1⊂Ω.
Then vε∈H01(Ω1).
Moreover,
[TABLE]
for all v∈Cc∞(Ω1).
Since u∣Ω1∈H1(Ω1) it follows that (5)
is valid for all v∈H01(Ω1).
Choosing v=vε gives
[TABLE]
On the other hand,
∂kvε=∂k((Reu−ε)+)=\mathds1[Reu>ε]∂kReu
for all k∈{1,…,d} by [GT] Lemma 7.6.
Therefore
[TABLE]
where M′=∥c0∥∞+∑k=1d∥bk∥∞.
Since
∥vε∥2=∥(Reu−ε)+∥2≤∥u∥2≤∣Ω∣1/2∥u∥C0(Ω),
it follows that
[TABLE]
for all ε∈(0,1], where
M′′=M+M′∣Ω∣1/2(∥u∥C0(Ω)+1).
Therefore the sequence ((Reu−2−n)+)n∈\mathdsN0 is bounded in
H01(Ω).
Passing to a subsequence if necessary, we may assume without loss of generality
that there exists a w∈H01(Ω) such that
lim(Reu−2−n)+=w weakly in H01(Ω).
Then lim(Reu−2−n)+=w in L2(Ω).
But lim(Reu−2−n)+=(Reu)+ in L2(Ω).
So (Reu)+=w∈H01(Ω).
Similarly (Reu)−,(Imu)+,(Imu)−∈H01(Ω).
So u∈H01(Ω).
∎
Lemma 2.2 together with the condition 0∈σ(AD)
gives the uniqueness in Theorem 1.1.
Proposition 2.3**.**
For all φ∈C(Γ) there exists at most one
u∈C(Ω)∩Hloc1(Ω)
such that u∣Γ=φ and Au=0.
Proof*.*
Let u∈C(Ω)∩Hloc1(Ω)
and suppose that u∣Γ=0 and Au=0.
Then u∈C0(Ω).
Hence u∈H01(Ω) by Lemma 2.2.
Also Au=0.
Therefore u∈D(AD) and ADu=0.
But 0∈σ(AD).
So u=0.
∎
In the next proposition we use that Ω is Wiener regular.
Proposition 2.4**.**
Let λ>λ0 and p∈(d,∞].
Let f0∈Lp/2(Ω) and f1,…,fd∈Lp(Ω).
Then there exists a unique u∈H01(Ω)∩C0(Ω)
such that Bλu=f0+∑k=1d∂kfk.
Proof.
Since akl and ck are real valued for all k,l∈{1,…,d}
we may assume that f0,…,fd are real valued.
By [GT] Theorem 8.31 there exists a unique u∈C(Ω)∩Hloc1(Ω)
such that Bλu=f0+∑k=1d∂kfk and u∣Γ=0.
Then u∈C0(Ω) and the existence follows from Lemma 2.2.
The uniqueness follows from Proposition 2.3.
∎
Corollary 2.5**.**
Let λ>λ0 and p∈(d,∞].
Let f0∈Lp/2(Ω) and f1,…,fd∈Lp(Ω).
Let u∈H01(Ω) and suppose that Bλu=f0+∑k=1d∂kfk.
Then u∈C0(Ω).
Proof*.*
By Proposition 2.4 there exists a u~∈H01(Ω)∩C0(Ω)
such that Bλu~=f0+∑k=1d∂kfk.
Then Bλ(u−u~)=0.
So \gothicbλ(u−u~,v)=0 first for all v∈Cc∞(Ω) and
then by density for all v∈H01(Ω).
Choose v=u−u~.
Then 2μ∥u−u~∥H1(Ω)2≤Re\gothicbλ(u−u~)=0.
So u=u~∈C0(Ω).
∎
We next wish to add the other lower order terms.
Proposition 2.6**.**
There exists a c>0 such that
for all Φ∈C1(\mathdsRd) there exists a unique
u∈H1(Ω)∩C(Ω) such that
u∣Γ=Φ∣Γ and Au=0.
Moreover,
[TABLE]
For the proof we need some lemmas.
In the next lemma we introduce a parameter δ in order
to avoid duplication of the proof.
Lemma 2.7**.**
*Fix δ∈[0,λ0+1].
*
For all f∈L2(Ω) and λ>λ0
there exists a unique u∈H01(Ω) such that
[TABLE]
for all v∈H01(Ω).
For all λ>λ0 define
Rλ:L2(Ω)→L2(Ω) by Rλf=u, where
u∈H01(Ω) is as in (6).
There exists a c1>0 such that
[TABLE]
for all λ>λ0 and f∈L2(Ω),
where q1=21−4d1.
There exists a c2≥1 such that
[TABLE]
for all λ∈[λ0+1,∞),
p,q∈[2,∞] and f∈Lp(Ω) with q1=p1−4d1.
If λ>λ0,
p∈(d,∞] and f∈Lp(Ω), then Rλf∈C0(Ω).
Proof*.*
‘(a)’.
This follows from the Lax–Milgram theorem.
‘(b)’.
Define M=∥c0−δ\mathds1Ω∥L∞(Ω)+∑k=1d∥bk∥L∞(Ω).
Let λ>λ0, f∈L2(Ω) and set u=Rλf.
Then
[TABLE]
So
∥u∥H1(Ω)≤2μ−1M∥f∥L2(Ω)
and
[TABLE]
By the Sobolev embedding theorem there exists a c1>0 such that
∥v∥Lq1(Ω)≤c1∥v∥H1(Ω) for all v∈H01(Ω),
where q11=21−2d1.
(The extra factor 2 is to avoid a separate case for d=2.)
Then
∥Rλf∥Lq1(Ω)≤2μ−1c1M∥f∥L2(Ω).
Hence
[TABLE]
where c2=(2/μ)3/4c11/2M.
‘(c)’.
Apply Corollary 2.5 with p=4d and λ=λ0+1.
It follows that Rλ0+1f∈C0(Ω) for all f∈Lp(Ω).
Clearly the map Rλ0+1∣Lp(Ω):Lp(Ω)→C0(Ω)
has a closed graph.
Hence it is continuous.
In particular, there exists a c3>0 such that
∥Rλ0+1f∥L∞(Ω)=∥Rλ0+1f∥C0(Ω)≤c3∥f∥Lp(Ω)
for all f∈Lp(Ω).
Let λ≥λ0+1 and f∈L2(Ω).
Write u=Rλf and u0=Rλ0+1f.
Then \gothicbλ(u,v)=\gothicbλ0+1(u0,v)
and therefore
\gothicbλ(u−u0,v)=−(λ−λ0−1)(u,v)L2(Ω)
for all v∈H01(Ω).
Hence u−u0∈D(BD) and (BD+λI)(u−u0)=−(λ−λ0−1)u0.
Consequently
[TABLE]
for all λ≥λ0+1.
Since the semigroup generated by −BD has Gaussian bounds, there exists a
c4≥1 such that
∥(BD+λI)−1∥∞→∞≤c4λ−1
for all λ≥λ0+1.
Then ∥Rλf∥L∞(Ω)≤2c3c4∥f∥Lp(Ω)
for all λ≥λ0+1 and f∈Lp(Ω).
Finally let p′∈(2,4d) and let q′∈(2,∞) be such that
q′1=p′1−4d1.
There exists a θ∈(0,1) such that
p′1=21−θ+pθ.
Then q′1=q1−θ, where
q1=21−4d1.
Let c1>0 be as in Statement (b).
The operator Rλ is bounded from L2(Ω) into
Lq(Ω) with norm at most c1 by Statement (b),
and we just proved that the operator Rλ is bounded from Lp(Ω) into
L∞(Ω) with norm at most 2c3c4.
Hence by interpolation the operator Rλ is bounded from
Lp′(Ω) into Lq′(Ω) with norm bounded by
c11−θ(2c3c4)θ≤c1+2c3c4,
which gives Statement (c).
‘(d)’.
This is a special case of Corollary 2.5.
∎
The main step in the proof of Proposition 2.6 is
the next lemma.
Lemma 2.8**.**
There exist λ>λ0 and c>0 such that
for all Φ∈C1(Ω)∩H1(Ω) there exists a unique
u∈H1(Ω)∩C(Ω) such that
u∣Γ=Φ∣Γ and Aλu=0.
Moreover,
[TABLE]
Proof.
Choose δ=0 in Lemma 2.7.
Let c1 and c2 be as in Lemma 2.7.
Let λ∈(λ0+1,∞) be such that
c1c22d−1(λ−λ0)−1/4(1+∣Ω∣)≤21.
Let Rλ be as in Lemma 2.7.
Set φ=Φ∣Γ.
There exist unique w,w~∈H01(Ω) such that
\gothicaλ(w,v)=\gothicaλ(Φ,v) and
\gothicbλ(w~,v)=\gothicbλ(Φ,v)
for all v∈H01(Ω).
Then w~∈C0(Ω) by Corollary 2.5.
Define u=Φ−w and u~=Φ−w~.
Then u~∈H1(Ω)∩C(Ω) and
u~∣Γ=φ.
Moreover, \gothicaλ(u,v)=0 and \gothicbλ(u~,v)=0
for all v∈H01(Ω),
and ∥u~∥C(Ω)≤∥φ∥C(Γ)
by the result of Stampacchia mentioned in the introduction
([Sta] Théorème 3.8).
Let v∈H01(Ω).
Then
[TABLE]
and
u~−u=Rλu by the definition of Rλ.
For all n∈{0,…,2d} define pn=2d−n4d.
Then p0=2, p2d−1=4d, p2d=∞
and pn1=pn−11−4d1
for all n∈{1,…,2d}.
So ∥u~−u∥Lpn(Ω)≤c2∥u∥Lpn−1(Ω)
for all n∈{2,…,2d} and
∥u~−u∥Lp1(Ω)≤c1(λ−λ0)−1/4∥u∥L2(Ω)
by Lemma 2.7(c) and (b).
Then
[TABLE]
and
[TABLE]
for all n∈{2,…,2d}.
It follows by induction to n that
[TABLE]
for all n∈{2,…,2d}.
So u∈Lp2d−1(Ω)=L4d(Ω) and
u~−u=Rλu∈C0(Ω) by Lemma 2.7(d).
In particular u∈C(Ω).
Moreover,
[TABLE]
by the choice of λ.
So
[TABLE]
and the proof of the lemma is complete.
∎
We next wish to remove the λ in Lemma 2.8.
For future purposes, we consider the full inhomogeneous problem.
Proposition 2.9**.**
Let p∈(d,∞], f0∈Lp/2(Ω)
and let f1,…,fd∈Lp(Ω).
Let u∈H01(Ω) be such that
Au=f0+∑k=1d∂kfk.
Then u∈C0(Ω).
Proof*.*
Without loss of generality we may assume that p∈(d,4d).
Choose λ=δ=λ0+1 in Lemma 2.7 and in Proposition 2.4.
By Proposition 2.4 there exists a unique u~∈H01(Ω)∩C0(Ω)
such that Bλu~=f0+∑k=1d∂kfk.
If v∈Cc∞(Ω), then
[TABLE]
So
[TABLE]
and by density for all v∈H01(Ω).
Hence u−u~=Rλu, where Rλ is as in Lemma 2.7.
For all n∈{0,…,2d−1} define pn=2d−n4d.
Then u−u~∈L2(Ω)=Lp0(Ω).
It follows by induction to n that u∈Lpn−1(Ω) and
u−u~∈Lpn(Ω) for all n∈{1,…,2d−1}, where
the last part follows from Lemma 2.7(c).
Hence u−u~∈Lp2d−1(Ω)=L4d(Ω)
and u∈Lp(Ω).
Then Lemma 2.7(d) gives
u−u~=Rλu∈C0(Ω) and therefore u∈C0(Ω).
∎
Corollary 2.10**.**
Let p∈(d,∞].
Then (AD)−1(Lp(Ω))⊂C0(Ω).
Corollary 2.11**.**
There exists a c′>0 such that
∥(AD)−1f∥L∞(Ω)≤c′∥f∥L∞(Ω)
for all f∈L∞(Ω).
Proof*.*
Closed graph theorem.
∎
Proof of Proposition 2.6..
Let c,λ>0 be as in Lemma 2.8
and let c′>0 be as in Corollary 2.11.
By Lemma 2.8 there exists a unique
u~∈H1(Ω)∩C(Ω)
such that u~∣Γ=Φ∣Γ and Aλu~=0.
By Lemma 2.1 there exists a unique w∈H01(Ω) such that
\gothica(w,v)=\gothica(Φ∣Ω,v) for all v∈H01(Ω).
Set u=Φ∣Ω−w and w~=Φ∣Ω−u~.
Then
[TABLE]
for all v∈H01(Ω).
So
[TABLE]
Since u~−u∈H01(Ω) it follows that
AD(u~−u)=−λu~.
Consequently, u=u~+λ(AD)−1u~∈C0(Ω)
by Corollary 2.10.
Moreover,
[TABLE]
and the proof of Proposition 2.6 is complete.
∎
Define ∣∣∣⋅∣∣∣:Hloc1(Ω)→[0,∞] by
[TABLE]
Finally we need the following Caccioppoli inequality.
Proposition 2.12**.**
There exists a c′≥1 such that
∣∣∣u∣∣∣≤c′∥u∥L2(Ω)
for all u∈H1(Ω) such that Au=0.
Proof*.*
See [GM] Theorem 4.4.
∎
Now we are able to prove Theorem 1.1.
Proof of Theorem 1.1..
The uniqueness is already proved in Proposition 2.3.
Let c>0 and c′≥1 be as in Propositions 2.6 and 2.12.
Let Φ∈C1(\mathdsRd)∩H1(\mathdsRd).
By Proposition 2.6 there exists a unique
u∈H1(Ω)∩C(Ω) such that
u∣Γ=Φ∣Γ and Au=0.
Moreover,
[TABLE]
It follows from (7) that we can define a linear map
F:{Φ∣Γ:Φ∈C1(\mathdsRd)∩H1(\mathdsRd)}→H1(Ω)∩C(Ω)
by F(Φ∣Γ)=u, where u∈H1(Ω)∩C(Ω)
is such that
u∣Γ=Φ∣Γ and Au=0.
Now let φ∈C(Γ).
By the Stone–Weierstraß theorem there are
Φ1,Φ2,…∈C1(\mathdsRd)∩H1(\mathdsRd) such that
limΦn∣Γ=φ in C(Γ).
Set un=F(Φn∣Γ) for all n∈\mathdsN.
Then it follows from (7) that (un)n∈\mathdsN
is a Cauchy sequence in C(Ω).
Let u=limun in C(Ω).
Also (un)n∈\mathdsN is a Cauchy sequence in Hloc1(Ω) by
(7).
So u∈Hloc1(Ω).
Since Aun=0 for all n∈\mathdsN, one deduces that
Au=0.
Moreover, u∣Γ=limun∣Γ=limΦn∣Γ=φ.
This proves existence.
Finally,
[TABLE]
This completes the proof of Theorem 1.1.
∎
Theorem 1.1 has the following extension.
Theorem 2.13**.**
Adopt the notation and assumptions of Theorem 1.1.
Let φ∈C(Γ), p∈(d,∞], f0∈Lp/2(Ω)
and let f1,…,fd∈Lp(Ω).
Then there exists a unique u∈C(Ω)∩Hloc1(Ω)
such that u∣Γ=φ and Au=f0+∑k=1d∂kfk.
Proof*.*
The uniqueness follows as in the proof of Proposition 2.3.
By Lemma 2.1 there exists a u0∈H01(Ω) such that
Au0=f0+∑k=1d∂kfk.
Then u0∈C0(Ω) by Proposition 2.9.
By Theorem 1.1 there exists a u1∈C(Ω)∩Hloc1(Ω)
such that u1∣Γ=φ and Au1=0.
Define u=u0+u1.
Then u∈C(Ω)∩Hloc1(Ω).
Moreover, u∣Γ=φ and Au=f0+∑k=1d∂kfk.
∎
We conclude this section with some results for the classical solution.
They will be used in Section 3 and are of independent interest.
Recall that P:C(Γ)→C(Ω) is given by
Pφ=u, where u∈C(Ω)∩Hloc1(Ω) is
the classical solution, so u∣Γ=φ and Au=0.
Proposition 2.14**.**
Let Φ∈C(Ω)∩Hloc1(Ω).
Suppose there exists a w∈H01(Ω) such that AΦ=Aw.
Then w∈C(Ω) and P(Φ∣Γ)=Φ−w.
Proof*.*
Write w~=Φ−P(Φ∣Γ).
Then w~∈C0(Ω)∩Hloc1(Ω)
and Aw~=AΦ=Aw=f0+∑k=1d∂kfk,
where f0=c0w+∑l=1dcl∂lw∈L2(Ω)
and fk=−∑l=1dalk∂lw−bkw∈L2(Ω)
for all k∈{1,…,d}.
So w~∈H01(Ω) by Lemma 2.2.
Hence A(w~−w)= and w~−w∈kerAD={0}.
So w=w~=Φ−P(Φ∣Γ).
∎
We need the dual map of A.
Define the map At:Hloc1(Ω)→D′(Ω) by
[TABLE]
for all u∈Hloc1(Ω) and v∈Cc∞(Ω).
Corollary 2.15**.**
Suppose that akl,bk,ck∈W1,∞(Ω) for all
k,l∈{1,…,d}.
Let Φ∈C(Ω).
Suppose there exists a w∈H01(Ω) such that
[TABLE]
for all v∈Cc∞(Ω).
Then w∈C(Ω) and P(Φ∣Γ)=Φ−w.
Proof*.*
By assumption one has ⟨Φ−w,Atv⟩D′(Ω)×D(Ω)=0
for all v∈Cc∞(Ω).
Hence Φ−w∈Hloc1(Ω) by elliptic regularity.
So Φ∈Hloc1(Ω) and
[TABLE]
for all v∈Cc∞(Ω).
Therefore AΦ=Aw and the result follows from
Proposition 2.14.
∎
The last corollary takes a very simple form for the Laplacian.
Corollary 2.16**.**
Let Φ∈C(Ω).
Suppose that ΔΦ∈H−1(Ω).
Let w∈H01(Ω) be such that ΔΦ=Δw as distribution.
Then w∈C(Ω) and P(Φ∣Γ)=Φ−w.
This corollary is a special case of [AD] Theorem 1.1.
3 Variational and classical solutions: comparison
In this section we show that the variational and classical solutions
of the Dirichlet problem are the same.
For that we assume throughout this section that Ω is an open
set with Lipschitz boundary.
Moreover, we adopt the assumptions and notation of Theorem 1.1.
Recall that for all φ∈C(Γ) we denote by Pφ∈C(Ω)
the classical solution and for all φ∈H1/2(Γ),
we denote by γφ∈H1(Ω) the variational solution
of the Dirichlet problem.
We shall prove in this section that they coincide if both are defined.
The fact that they coincide for restrictions to Γ of functions in
C(Ω)∩H1(Ω) is a consequence of Proposition 2.14.
We state this as a proposition.
Proposition 3.1**.**
Let Φ∈C(Ω)∩H1(Ω).
Then P(Φ∣Γ)=γ(Φ∣Γ) almost everywhere.
So for the proof of Theorem 1.2 it suffices to show that
the map Φ↦Φ∣Γ from C(Ω)∩H1(Ω)
into C(Γ)∩H1/2(Γ) is surjective.
This is surprisingly difficult to prove.
We first prove Theorem 1.2 for the Laplacian with the help of
Proposition 3.1 and a deep result of Dahlberg.
As a consequence we obtain the desired surjectivity result.
Then as noticed earlier, Theorem 1.2 follows for our general
elliptic operator.
Theorem 3.2**.**
Assume that akl=δkl and bk=ck=c0=0
for all k,l∈{1,…,d}.
Let φ∈C(Γ)∩H1/2(Γ).
Then Pφ=γφ almost everywhere.
Proof*.*
Let x∈Ω.
By Dahlberg [Dah] Theorem 1 there exists a unique kx∈L1(Γ) such that
(Pφ)(x)=∫Γkxφdσ
for all φ∈C(Γ).
Now let φ∈C(Γ)∩H1/2(Γ).
Without loss of generality we may assume that φ is real valued.
Then there exists a u∈H1(Ω,\mathdsR) such that φ=Tru.
Since H1(Ω)∩C(Ω) is dense in H1(Ω), there
exist u1,u2,…∈H1(Ω,\mathdsR)∩C(Ω) such that
limun=u in H1(Ω).
Define vn=(−∥φ∥L∞(Γ))∨un∧∥φ∥L∞(Γ)
for all n∈\mathdsN.
Then vn∈H1(Ω)∩C(Ω).
Write φn=vn∣Γ=Trvn∈C(Γ)∩H1/2(Γ)
for all n∈\mathdsN.
Then Pφn=γφn almost everywhere for all n∈\mathdsN
by Proposition 3.1.
Note that
[TABLE]
in H1/2(Γ).
So by continuity of γ one deduces that
γφ=limγφn in H1(Ω)
and in particular in L2(Ω).
Passing to a subsequence, if necessary, we may assume that
[TABLE]
for almost all x∈Ω.
Using again that limφn=φ in H1/2(Γ)
and therefore also in L2(Γ),
we may assume that
limφn=φ almost everywhere on Γ.
Hence if x∈Ω, then
[TABLE]
by the Lebesgue dominated convergence theorem.
Since Pφn=γφn almost everywhere for all n∈\mathdsN
one concludes that (Pφ)(x)=(γφ)(x) for almost all x∈Ω.
∎
The desired surjectivity result is the following corollary
of Theorem 3.2.
Corollary 3.3**.**
Let Ω⊂\mathdsRd be a bounded open set with Lipschitz boundary.
Let φ∈C(Γ)∩H1/2(Γ).
Then there exists a u∈H1(Ω)∩C(Ω)
such that φ=u∣Γ.
Proof of Theorem 1.2..
This follows from Corollary 3.3 and Proposition 3.1.
∎
Corollary 3.4**.**
Adopt the notation and assumptions of Theorem 1.1.
Suppose that Ω has a Lipschitz boundary.
Let u∈C(Ω)∩Hloc1(Ω) and
suppose that Au=0.
Then u∈H1(Ω) if and only if u∣Γ∈H1/2(Γ).
Proof*.*
‘⇒’ is trivial.
‘⇐’.
Suppose u∣Γ∈H1/2(Γ).
Then u=P(u∣Γ)=γ(u∣Γ)∈H1(Ω) by Theorem 1.2.
∎
4 Semigroup and holomorphy on C0(Ω)
In this section we prove Theorem 1.3.
Throughout this section we adopt the notation and assumptions of Theorem 1.1.
We need several lemmas.
Lemma 4.1**.**
The operator Ac is invertible and (Ac)−1=(AD)−1∣C0(Ω).
Proof*.*
If v∈C0(Ω), then (AD)−1v∈C0(Ω) by
Corollary 2.10.
Moreover, AD((AD)−1v)=v.
So (AD)−1v∈D(Ac) and Ac((AD)−1v)=v.
Hence Ac is surjective.
Since AD is injective, also Ac is injective.
Therefore Ac is invertible and
(Ac)−1=(AD)−1∣C0(Ω).
∎
The next proof is inspired by arguments in [AB] Theorem 4.4.
Lemma 4.2**.**
The domain D(Ac) of the operator Ac is dense in C0(Ω).
Proof*.*
Let ρ∈M(Ω), the Banach space of all
complex measures on Ω and suppose that ∫Ωvdρ=0
for all v∈D(Ac).
There exist w1,w2,…∈L2(Ω) such that
sup∥wn∥L1(Ω)<∞ and
lim∫Ωvwn=∫Ωvdρ
for all v∈C0(Ω).
Choose p=d+2 and let q∈(1,2) be the dual exponent of p.
It follows from Proposition 2.9 that the operator (AD)−1
extends to a continuous operator from W−1,p(Ω) into C0(Ω).
Hence the operator (AD)−1∗
extends to a continuous operator from M(Ω) into W01,q(Ω).
In particular, there exists a c>0 such that
∥(AD)−1∗w∥W01,q(Ω)≤c∥w∥L1(Ω)
for all w∈L2(Ω).
For all n∈\mathdsN set un=(AD)−1∗wn.
We emphasise that un∈D((AD)∗).
Then sup∥un∥W01,q(Ω)<∞.
Note that W01,q(Ω) is reflexive.
Hence passing to a subsequence if necessary,
there exists a u∈W01,q(Ω)
such that limun=u weakly in W01,q(Ω).
Let v∈Cc∞(Ω).
Then (AD)−1v∈D(Ac) by Lemma 4.1.
Therefore
[TABLE]
Hence u=0.
Again let v∈Cc∞(Ω).
Then
[TABLE]
where we used (4).
So ρ=0 and D(Ac) is dense in C0(Ω).
∎
Now we prove that −Ac generates a holomorphic C0-semigroup.
Proof of Theorem 1.3..
Let S be the semigroup generated by −AD.
Then S has a kernel with Gaussian upper bounds by [Ouh] Theorem 6.10
(see also [Dan] Theorem 6.1 for operators with
real valued coefficients and [AE1] Theorems 3.1 and 4.4).
Hence the semigroup S extends consistently
to a semigroup S(p) on Lp(Ω)
for all p∈[1,∞].
Choose p∈(d,∞].
Let t>0 and u∈L2(Ω).
Since S is a holomorphic semigroup, one deduces that Stu∈D(AD)
and ADStu∈L2(Ω).
Next the Gaussian kernel bounds imply that St maps L2(Ω)
into Lp(Ω).
So ADS2tu=StADStu∈Lp(Ω)
and
[TABLE]
by Corollary 2.10.
Hence StC0(Ω)⊂C0(Ω) for all t>0.
For all t>0 let Stc=St∣C0(Ω):C0(Ω)→C0(Ω).
Then (Stc)t>0 is a semigroup on C0(Ω).
Moreover, using again the Gaussian kernel bounds
there exists an M≥1 such that
∥Stc∥≤∥St(∞)∥≤M
for all t∈(0,1].
Let t∈(0,1] and u∈D(Ac).
Then
[TABLE]
So limt↓0Stcu=u in C0(Ω).
Since D(Ac) is dense in C0(Ω) by Lemma 4.2,
one deduces that limt↓0Stcu=u in C0(Ω)
for all u∈C0(Ω).
So Sc is a C0-semigroup.
Finally, using once more the Gaussian kernel bounds,
it follows that the semigroup Sc is holomorphic (see [AE1] Theorem 5.4).
∎
We conclude this section by establishing Gaussian kernels which are continuous
up to the boundary.
For this we use the following special case of [AE2] Theorem 2.1.
Proposition 4.3**.**
Suppose that ∣∂Ω∣=0.
Let T be a semigroup in L2(Ω) such that
TtL2(Ω)⊂C(Ω) and
Tt∗L2(Ω)⊂C(Ω)
for all t>0.
Then for all t>0 there exists a unique kt∈C(Ω×Ω)
such that
[TABLE]
for all u∈L2(Ω) and x∈Ω.
We continue to denote by S the semigroup generated by −AD
and we also denote by S the holomorphic extension.
For all θ∈(0,π] let
Σ(θ)={z∈\mathdsC∖{0}:∣argz∣<θ}
be the open sector with (half)angle θ.
Theorem 4.4**.**
Adopt the notation and assumptions of Theorem 1.1.
In addition assume that bk is real valued for all k∈{1,…,d}.
Let θ be the holomorphy angle of S.
Then for all z∈Σ(θ) there exists a unique
kz∈C(Ω×Ω) such that the following is
valid.
(Szu)(x)=∫Ωkz(x,y)u(y)dy*
for all z∈Σ(θ), u∈L2(Ω) and x∈Ω.*
kz(x,y)=0* for all z∈Σ(θ) and
x,y∈Ω with x∈∂Ω or y∈∂Ω.*
The map z↦kz is holomorphic from Σ(θ) into
C(Ω×Ω).
For all θ′∈(0,θ) there exist b,c,ω>0 such that
[TABLE]
for all z∈Σ(θ′) and x,y∈Ω.
Proof*.*
It follows from (8) that SzL2(Ω)⊂C0(Ω)
for all z∈Σ(θ).
Since the coefficients bk are real, also the adjoint operator
satisfies the conditions of Theorem 1.1.
Therefore Sz∗L2(Ω)⊂C0(Ω)
for all z∈Σ(θ).
It follows from Proposition 4.3 that for all
z∈Σ(θ) there exists a unique
kz∈C(Ω×Ω) such that
(Szu)(x)=∫Ωkz(x,y)u(y)dy
for all u∈L2(Ω) and x∈Ω.
Since Szu∈C0(Ω) one deduces that kz(x,y)=0
for all z∈Σ(θ), x∈∂Ω and
y∈Ω.
Considering adjoints the same is valid with x and y interchanged.
If v,w∈C0(Ω), then the map
[TABLE]
is holomorphic on Σ(θ).
Therefore Statement (III) is a consequence of [AN] Theorem 3.1.
The Gaussian bounds of Statement (IV) follow from
[AE1] Theorem 5.4.
∎
Acknowledgements
The second-named author is most grateful for the hospitality extended
to him during a fruitful stay at the University of Ulm.
He wishes to thank the University of Ulm for financial support.
Part of this work is supported by an
NZ-EU IRSES counterpart fund and the Marsden Fund Council from Government funding,
administered by the Royal Society of New Zealand.
Part of this work is supported by the
EU Marie Curie IRSES program, project ‘AOS’, No. 318910.