A "milder" version of Calder\'on's inverse problem for anisotropic conductivities and partial data

0.23 MB·English

by  El Maati Ouhabaz

A "milder" version of Calderón’s inverse problem
for anisotropic conductivities and partial data
El Maati Ouhabaz
6
1
0
2 Abstract

r Given a general symmetric elliptic operator
p
A d d
La := ∂k(akj∂j)+ ak∂k−∂k(ak.)+a0
3 k,j=1 k=1
X X
1
we definethe associated Dirichlet-to-Neumann (D-t-N)map with partial

] data,i.e., datasupportedin a partof theboundary. Weprovepositivity,
P Lp-estimatesanddominationpropertiesforthesemigroupassociatedwith
A
thisD-t-Noperator. GivenLa andLb oftheprevioustypewithbounded
h. measurable coefficients a={akj, ak,a0} and b={bkj, bk,b0}, we prove
t that if their partial D-t-N operators (with a0 and b0 replaced by a0−λ
a andb0−λ)coincideforallλ,thentheoperatorsLa andLb,endowedwith
m
Dirichlet, mixed or Robin boundary conditions are unitarily equivalent.
[ In the case of the Dirichlet boundary conditions, this result was proved

recentlybyBehrndtandRohleder[6]. Weprovideadifferentproof,based
2
on spectral theory,which works for other boundary conditions.
v
4
2010 AMS Subject Classification: 35P05, 35R30,47D05, 47G30.
6
3
7
1 Introduction
0
.
1
Let Ω be a bounded Lipschitz domain of Rd with boundary ∂Ω. Let Γ be a
0 0
5 closed subset of ∂Ω with Γ0 6= ∂Ω and Γ1 its complement in ∂Ω. We consider
1 the symmetric elliptic operator on L2(Ω) given by the formal expression:
:
v
d d
i
X La(λ):= ∂k(akj∂j)+ ak∂k−∂k(ak.)+a0−λ
r kX,j=1 kX=1
a
where a = a ,a ,a = a ∈ L∞(Ω) and λ is a constant. We define the
kj jk k 0 0
associated Dirichlet-to-Neumann (D-t-N) operator, N (λ), with partial data
Γ1,a
as follows:
1
for ϕ∈H2(∂Ω) with ϕ=0 on Γ0, one solves the Dirichlet problem
L (λ)u=0 weakly in W1,2(Ω)with u=ϕ on∂Ω, (1.1)
a
and defines (in the weak sense)
d d
N (λ)ϕ:= a ∂ u+a ϕ ν onΓ . (1.2)
Γ1,a kj k j j 1
!
j=1 k=1
X X
1
Here ν = (ν ,··· ,ν ) is the outer unit normal to the boundary of Ω. The
1 d
operatorN (λ)is interpretedas the conormalderivativeonthe boundary. It
Γ1,a
is an operator acting on L2(∂Ω). See Section 2 for more details.
Let us consider first the case where a = σ(x)δkj, a = 0,k = 0,1...d,
kj k
whereσ ∈L∞(Ω)isboundedfrombelow(byapositiveconstant). A.Calderón’s
well known inverse problem asks whether one could determine solely the con-
ductivity σ(x) from boundary measurements, i.e., from N (0). For the global
Γ1
boundary measurements, i.e., Γ = ∂Ω, the first global uniqueness result was
1
provedbySylvesterandUhlmann[27]foraC2-smoothconductivitywhend≥3.
This results was extended to C1+ǫ-smooth conductivity by Greenleaf, Lassas
and Uhlmann [12] and then by Haberman and Tataru [13] to C1 and Lipschitz
conductivity close to the identity. Haberman [14] proved the uniqueness for
Lipschitz conductivity when d = 3,4 and this was extended to all d ≥ 3 by
Caro and Rogers [7]. In the two-dimension case with C2-smooth conductivity,
the globaluniqueness wasprovedby Nachman[21]. This regularityassumption
was completely removed by Astala and Päivärinta [4] dealing with σ ∈L∞(Ω).
Theinverseproblemwithpartialdataconsistsinprovinguniqueness(eitherfor
the isotropic conductivity or for the potential) when the measurement is made
only on a part of the boundary. This means that the trace of the solution u
in (1.1) is supported on a set Γ and the D-t-N operator is known on Γ for
D N
some parts Γ and Γ of the boundary. This problem has been studied and
D N
there are some geometric conditions on Γ and Γ under which uniqueness is
D N
proved. WerefertoIsakov[15],Kenig,SjöstrandandUhlmann[18],DosSantos
et al. [10], Imanuvilov,Uhlmann and Yamamoto [16] andthe review paper [19]
by Kenig and Salo for more references and recent developments.
Now we move to the anisotropic case. This corresponds to the general case
where the conductivity is given by a general matrix a . As pointed out by
kj
Lee and Uhlmann in [20], it is not difficult to see that a change of variables
given by a diffeomorphism of Ω which is the identity on ∂Ω leads to different
coefficientsb withoutchangingtheD-t-Noperatorontheboundary. Therefore
kj
the single coefficients a are not uniquely determined in general. In [20], Lee
kj
and Uhlmann proved that for real-analytic coefficients the uniqueness up to a
diffeomorphismholdswhenthedimensiondis≥3. Thesameresultwasproved
by Astala, Lassas and Päivärinta [5] for the case d=2 and L∞-coefficients.
In [6], Behrndt and Rohleder considered general elliptic expressions L and
a
L as above and prove that if the corresponding D-t-N operators N (λ) and
b Γ1,a
N (λ)coincideforallλinasethavinganaccumulationpointinρ(LD)∩ρ(LD)
Γ1,b a b
then the operators LD and LD are unitarily equivalent. Here LD is the elliptic
a b a
operator L with Dirichlet boundary conditions. This can be seen as a milder
a
version of the uniqueness problem discussed above. The proof is based on the
theoryofextensionsofsymmetric operatorsandunique continuationresults. It
is assumed in [6] that the coefficients are Lipschitz continuous on Ω. We give
a different proof of this result which also works for other boundary conditions.
Our main result is the following.
Theorem 1.1. Suppose that Ω is a bounded Lipchitz domain of Rd with d≥2.
Let Γ be a closed subset of ∂Ω, Γ 6= ∂Ω and Γ its complement. Let a =
0 0 1
{a ,a ,a } and b = {b ,b ,b } be bounded functions on Ω such that a and
kj k 0 kj k 0 kj
b satisfy the usual ellipticity condition. If d ≥ 3 we assume in addition that
kj
the coefficients a ,b ,a and b are Lipschitz continuous on Ω.
kj kj k k
2
Suppose that N (λ) = N (λ) for all λ in a set having an accumulation
Γ1,a Γ1,b
point in ρ(LD)∩ρ(LD). Then:
a b
i) The operators L and L endowed with Robin boundary conditions are uni-
a b
tarily equivalent.
ii)The operators L and L endowed with mixedboundary conditions (Dirichlet
a b
on Γ and Neumann type on Γ ) are unitarily equivalent.
0 1
iii) The operators L and L endowed with Dirichlet boundary conditions are
a b
unitarily equivalent.
In addition, for Robin or mixed boundary conditions, the eigenfunctions as-
sociated to the same eigenvalue λ ∈/ σ(LD) = σ(LD) coincide on the boundary
a b
of Ω.
Note that unlike [6] we do not assume regularity of the coefficients when
d=2.
We shall restate this theorem in a more precise way after introducing some
necessary material and notation. The proof is given in Section 4. It is based
on spectral theory and differs from the one in [6]. Our strategy is to use a
relationshipbetweeneigenvaluesoftheD-t-NoperatorN (λ)andeigenvalues
Γ1,a
of the elliptic operator with Robin boundary conditions Lµ on Ω where µ is a
a
parameter. One of the main ingredients in the proof is that each eigenvalue of
thelatteroperatorisastrictlydecreasingmapwithrespecttotheparameterµ.
Next, the equality of N (λ) and N (λ) allows us to prove that the spectra
Γ1,a Γ1,b
of Lµ and Lµ are the same and the eigenvalues have the same multiplicity.
a b
The similarity of the two elliptic operators with Dirichlet boundary conditions
is obtained from the similarity of Lµ and Lµ by letting the parameter µ tend
a b
to −∞. During the proof we use some ideas from the papers of Arendt and
Mazzeo [2] and[3] whichdeal with a different subject, namelythe Friendlander
inequality for the eigenvalues of the Dirichlet and Neumann Laplacian on a
Lipschitzdomain. Theideaswhichweborrowfrom[2]and[3]arethenadapted
and extended to our general case of D-t-N operators with variable coefficients
and partial data.
In Section 2 we define the D-t-N operator with partial data using the method
ofsesquilinearforms. Inparticular,forsymmetriccoefficientsitisaself-adjoint
operator on L2(Γ ). It can be seen as an operator on L2(∂Ω) with a non-dense
1
domain and which we extend by 0 to L2(Γ ). Therefore one can associate with
0
this D-t-N operator a semigroup (TΓ1) acting on L2(∂Ω). In Section 3 we
t t≥0
provepositivity,sub-Markoviananddominationpropertiesforsuchsemigroups.
In particular, (TΓ1) extends to a contraction semigroup on Lp(∂Ω) for all
t t≥0
p∈[1,∞). Hence,forϕ ∈Lp(Γ ),oneobtainsexistenceanduniquenessofthe
0 1
solution in Lp(∂Ω) to the evolution problem
∂ ϕ+N (λ)ϕ=0, ϕ(0)=ϕ .
t Γ1,a 0
The results of Section 3 are of independent interest and are not used in the
proof of the theorem stated above.
3
2 The partial D-t-N operator
LetΩbeaboundedopensetofRd withLipschitzboundary∂Ω. Theboundary
is endowed with the (d−1)-dimensional Hausdorff measure dσ. Let
a ,a ,a˜,a :Ω→C
kj k k 0
be bounded measurable for 1 ≤ k,j ≤ d and such that there exists a constant
η >0 for which
d
Re a (x)ξ ξ ≥η|ξ|2 (2.1)
kj k j
k,j=1
X
for all ξ =(ξ ,··· ,ξ )∈Cd and a.e. x∈Ω.
1 d
Let Γ be an closed subset of ∂Ω and Γ its complement in ∂Ω.
0 1
Elliptic operators on Ω.
We consider the space
V ={u∈W1,2(Ω), Tr(u)=0 on Γ =0}, (2.2)
0
where Tr denotes the trace operator. We define the sesquilinear form
a:V ×V →C
by the expression
d d
a(u,v)= a ∂ u∂ v dx+ a ∂ uv+a˜u∂ v dx+a uv dx (2.3)
kj k j k k k k 0
k,j=1ZΩ k=1ZΩ
X X
for all u,v ∈V. Here we use the notation ∂ for the partial derivative ∂ .
j ∂xj
It follows easily from the ellipticity assumption (2.1) that the form a is
quasi-accretive,i.e., there exists a constant w such that
Rea(u,u)+wkuk2 ≥0 ∀u∈V.
2
Inaddition,sinceV isaclosedsubspaceofW1,2(Ω)theformaisclosed. There-
fore there exists an operator L associated with a. It is defined by
a
D(L )={u∈V,∃v ∈L2(Ω):a(u,φ)= vφ dx ∀φ∈V},
a
ZΩ
L u:=v.
a
Formally, L is given by the expression
a
d d
L u=− ∂ (a ∂ u)+ a ∂ u−∂ (a˜u)+a u. (2.4)
a k kj j k k k k 0
k,j=1 k=1
X X
Inaddition,following[2]or[3]wedefinetheconormalderivative ∂ intheweak
∂ν
sense (i.e. in H−1/2(∂Ω) the dual space of H1/2(∂Ω)= Tr(W1,2(Ω))), then L
a
is subject to the boundary conditions
Tr(u)=0 on Γ
0
∂u (2.5)
 =0 on Γ .
 ∂ν 1

4
The conormal derivative in our case is usually interpreted as
d d
a ∂ u+a˜u ν ,
kj k j j
!
j=1 k=1
X X
where ν = (ν ,··· ,ν ) is the outer unit normal to the boundary of Ω. For all
1 d
this see [23], Chapter 4.
The condition (2.5) is a mixed boundary condition which consists in taking
Dirichlet on Γ and Neumann type boundary condition on Γ . For this reason
0 1
we denote this operator by LM. The subscript a refers to the fact that the
a
coefficients of the operator are given by a = {a ,a ,a˜,a } and M refers to
kj k k 0
mixed boundary conditions.
WealsodefinetheellipticoperatorwithDirichletboundaryconditionTr(u)=
0 on ∂Ω. It is the operator associated with the form given by the expression
(2.3) with domain D(a)=W1,2(Ω). It is a quasi-accretiveand closed form and
0
its associated operator LD has the same expression as in (2.4) and subject to
a
the Dirichlet boundary condition Tr(u)=0 on ∂Ω.
Similarly, we define LN to be the elliptic operator with Neumann type
a
boundary conditions
∂u
=0 on ∂Ω.
∂ν
It is the operator associated with the form given by the expression (2.3) with
domain D(a)=W1,2(Ω).
Note that LD coincides with LM if Γ = ∂Ω and LN coincides with LM if
a a 0 a a
Γ =∅.
0
Finally we define elliptic operatorswith Robinboundaryconditions. Letµ∈R
be a constant and define
d d
aµ(u,v)= a ∂ u∂ v dx+ a ∂ uv+a˜u∂ v dx+a uv dx
kj k j k k k k 0
k,j=1ZΩ k=1ZΩ
X X
−µ Tr(u)Tr(v)dσ (2.6)
Z∂Ω
for all u,v ∈ D(aµ) := V. Again, Tr denotes the trace operator. Using the
standard inequality (see [2] or [3]),
|Tr(u)|2 ≤εkuk2 +c |u|2
W1,2(Ω) ε
Z∂Ω ZΩ
which is valid for all ε > 0 (c is a constant depending on ε) one obtains that
ε
for some positive constants w and δ
Reaµ(u,u)+w |u|2 ≥δkuk2 .
W1,2(Ω)
ZΩ
Fromthisitfollowsthataµisaquasi-accretiveandclosedsesquilinearform. One
can associate with aµ an operator Lµ. This operator has the same expression
a
(2.4) and it is subject to the Robin boundary conditions
Tr(u)=0 on Γ
0
∂u (2.7)
 =µ Tr(u) on Γ .
 ∂ν 1

5
Actually, the boundary conditions (2.7) are mixed Robin boundary conditions
in the sense that we have the Dirichlet condition on Γ and the Robin one on
0
Γ . For simplicity we ignore the word "mixed" and refer to (2.7) as the Robin
1
boundary conditions.
According to our previous notation, if µ=0, then a0 =a and L0 =LM.
a a
Note that we may choose here µ to be a bounded measurable function on the
boundary rather than just a constant.
The partial Dirichlet-to-Neumann operator on ∂Ω.
Suppose as before that a = {a ,a ,a˜,a } are bounded measurable and
kj k k 0
satisfy the ellipticity condition (2.1). Let Γ ,Γ ,V be as above and a is the
0 1
sesquilinear form defined by (2.3).
We define the space
V :={u∈V,a(u,g)=0 for all g ∈W1,2(Ω)}. (2.8)
H 0
Then V is a closed subspace of V. It is interpreted as the space of harmonic
H
functions forthe operatorL (givenby (2.4))withthe additionalpropertythat
a
Tr(u)=0 on Γ .
0
We start with the following simple lemma.
Lemma 2.1. Suppose that 0∈/ σ(LD). Then
a
V =V ⊕W1,2(Ω). (2.9)
H 0
Proof. We argue as in [11], Section 2 or [2]. Let us denote by aD the form
associatedwith LD, thatis,aD is givenby (2.3)withD(aD)=W1,2(Ω). There
a 0
exists an operator LD : W1,2(Ω) → W−1,2(Ω) := W1,2(Ω)′ (the anti-dual of
a 0 0
W1,2(Ω)) associated with aD in the sense
0
hLDh,gi=aD(h,g)
a
for all h,g ∈ W1,2(Ω). The notation h·,·i denotes the duality W1,2(Ω)′ −
0 0
W1,2(Ω). Since 0 ∈/ σ(LD), then LD is invertible. Therefore LD, seen as
0 a a a
operatoronW1,2(Ω)′ withdomainW1,2(Ω),isalsoinvertibleonW1,2(Ω)′ since
0 0 0
thetwooperatorsLD andLD havethesamespectrum(seee.g.,[1],Proposition
a a
3.10.3). Now we fix u∈V and consider the (anti-)linear functional
F :v 7→a(u,v).
Clearly, F ∈ W1,2(Ω)′ and hence there exists a unique u ∈ W1,2(Ω) such
0 0 0
that LDu = F, i.e., hLDu ,gi = F(g) for all g ∈ W1,2(Ω). This means
a 0 a 0 0
that a(u − u ,g) = 0 for all g ∈ W1,2(Ω) and hence u − u ∈ V . Thus,
0 0 0 H
u=u−u +u ∈V +W1,2(Ω). Finally, if u∈V ∩W1,2(Ω) then a(u,g)=0
0 0 H 0 H 0
for all g ∈ W1,2(Ω). This means that u ∈ D(LD) with LDu = 0. Since LD is
0 a a a
invertible we conclude that u=0.
As a consequence of Lemma 2.1, the trace operator Tr : V → L2(∂Ω) is
H
injective and
Tr(V )=Tr(V). (2.10)
H
6
Intherestofthissectionweassumethat0∈/ σ(LD). WedefineonL2(∂Ω,dσ)
a
the sesquilinear form
b(ϕ,ψ):=a(u,v) (2.11)
where u,v ∈ V are such that ϕ = Tr(u) and ψ = Tr(v). This means that
H
D(b)=Tr(V ) and by (2.10)
H
D(b)=Tr(V )=Tr(V). (2.12)
H
Lemma 2.2. There exist positive constants w, δ and M such that
Reb(ϕ,ϕ)+w |ϕ|2 ≥δkuk2 (2.13)
W1,2(Ω)
Z∂Ω
and
1/2 1/2
|b(ϕ,ψ)|≤M Reb(ϕ,ϕ)+w |ϕ|2 Reb(ψ,ψ)+w |ψ|2
(cid:20) Z∂Ω (cid:21) (cid:20) Z∂Ω (cid:21)
(2.14)
for all ϕ,ψ ∈D(b). In the first inequality, u∈V is such that Tr(u)=ϕ.
H
Proof. It is well knownthat Tr:W1,2(Ω)→L2(∂Ω) is a compactoperator and
since Tr : V → L2(∂Ω) is injective it follows that for every ǫ > 0 there exists
H
a constant c>0 such that
|u|2 ≤ǫkuk2 +c |Tr(u)|2 (2.15)
W1,2
ZΩ Z∂Ω
for all u∈V (see, e.g., [2]). In particular,
H
ǫ c
|u|2 ≤ |∇u|2+ |ϕ|2. (2.16)
1−ǫ 1−ǫ
ZΩ ZΩ Z∂Ω
Now, let ϕ∈D(b)= Tr(V ) and u ∈V such that ϕ =Tr(u). It follows from
H H
the ellipticity assumption (2.1) and the boundedness of the coefficients that for
some constant c >0
0
η
Rea(u,u)≥ |∇ u|2−c |u|2.
0
2
ZΩ ZΩ
Therefore, using (2.16) and the definition of b we obtain
Reb(ϕ,ϕ) = Rea(u,u)
η c ǫ cc
≥ ( − 0 ) |∇ u|2− 0 |ϕ|2.
2 1−ǫ 1−ǫ
ZΩ Z∂Ω
Taking ǫ>0 small enough we obtain (2.13).
Inorderto provethe secondinequality, weuse the definition ofb andagainthe
boundedness of the coefficients to see that
|b(ϕ,ψ)| = |a(u,v)|
≤ CkukW1,2kvkW1,2.
Thus, (2.14) follows from (2.13).
7
Corollary 2.3. The form b is continuous, quasi-accretive and closed.
Proof. Continuity of b is exactly (2.14). Quasi-accretivity means that
Reb(ϕ,ϕ)+w |ϕ|2 ≥0
Z∂Ω
for some w and all ϕ∈D(b). This follows from (2.13).
Now we provethat b is closedwhich means that D(b) is complete for the norm
1/2
kϕkb := Reb(ϕ,ϕ)+w |ϕ|2
(cid:18) Z∂Ω (cid:19)
in which w is as in (2.13). If (ϕn) is a Cauchysequence for k·kb then by (2.13)
the corresponding (u ) ∈ V with Tr(u ) = ϕ is a Cauchy sequence in V .
n H n n H
Since V is a closed subspace of V it follows that u is convergent to some u
H n
in V . Set ϕ:=Tr(u). We have ϕ∈D(b) and the definition of b together with
H
continuityofTrasanoperatorfromW1,2(Ω)toL2(∂Ω)showthatϕ converges
n
to ϕ for the norm k·kb. This means that b is a closed form.
NotethatthedomainTr(V )ofbmaynotbedenseinL2(∂Ω)sincefunctions
H
in this domain vanish on Γ . Indeed,
0
H :=D(b)L2(∂Ω) =L2(Γ )⊕{0}. (2.17)
1
Thedirectinclusionfollowsfromthefactthatifϕ ∈D(b)convergesinL2(∂Ω)
n
then after extracting a subsequence we have a.e. convergence. Since ϕ =0 on
n
Γ we obtain that the limit ϕ = 0 on Γ . The reverse inclusion can be proved
0 0
asfollows. LetΓ be aclosedsubsetofRd withΓ ⊂Γ andconsiderthe space
2 2 1
E ={u :u∈W1,∞(Rd),u =0}. ThenE ⊂C(Γ )andaneasyapplication
|Γ2 |Γ0 2
of the Stone-Weierstrass theorem shows that E is dense in C(Γ ). Now given
2
ϕ ∈ C (Γ ) and ǫ > 0 we find Γ such that k k < ǫ and u ∈ E such
c 1 2 1Γ1\Γ2 2 |Γ2
that ku −ϕk <ǫ. Finally we take χ∈C∞(Rd) such that χ=1 on Γ .
|Γ2 C(Γ2) c 2
Then (uχ) ∈V and
|Ω
kuχ−ϕkL2(Γ1) ≤ ku−ϕkL2(Γ2)+kχkL2(Γ1\Γ2)
≤ ǫ|Γ |+kχk ǫ.
2 ∞
Here |Γ | denotes the measure of Γ . These inequalities together with the fact
2 2
that C (Γ ) is dense in L2(Γ ) imply (2.17).
c 1 1
We return to the form b defined above. We associate with b an operator
N . It is defined by
Γ1
D(N ):={ϕ∈D(b),∃ψ ∈H :b(ϕ,ξ)= ψξ ∀ξ ∈D(b)}, N ϕ=ψ.
Γ1 Γ1
ZΓ1
The operator N can be interpreted as an operator on L2(∂Ω) defined as fol-
Γ1
lows: if ϕ ∈ D(N ) then there exists a unique u ∈ V such that ϕ = Tr (u)
Γ1 H
and
∂u
ϕ =0, N (ϕ)= on Γ . (2.18)
|Γ0 Γ1 ∂ν 1
8
Again ∂u is interpreted in the weak sense as the conormal derivative that is
∂ν
d d a ∂ u+a˜ϕ ν . In the particular case where a = δ and
j=1 k=1 kj k j j kj kj
aP1 =·(cid:16)··P=ad =0 the right(cid:17)hand side of (2.18) is seen as the normal derivative
on the boundary. All this can be made precise by applying the Green formula
if the boundary and the coefficients are smooth enough.
WecallN thepartialDirichlet-to-NeumannoperatoronL2(∂Ω)ortheDirichlet-
Γ1
to-Neumann operator with partial data. The term partial refers to the fact that
N is known only on the part Γ of the boundary ∂Ω.
Γ1 1
It follows from the general theory of forms that −N generates a holomorphic
Γ1
semigroup e−tNΓ1 on H. We define TΓ1 on L2(∂Ω) by
t
TtΓ1ϕ=e−tNΓ1(ϕ1Γ1)⊕0.
We shall refer to (TΓ1) as the "semigroup" generated by −N on L2(∂Ω).
t t≥0 Γ1
It is clear that
kTtΓ1kL(L2(∂Ω)) ≤e−w0t, t≥0, (2.19)
for some constant w . Note that if the form a is symmetric, then b is also
0
symmetric and hence N is self-adjoint. In this case, (2.19) holds with w =
Γ1 0
infσ(N ) which also coincides with the first eigenvalue of N . For all this,
Γ1 Γ1
see e.g. [23], Chapter 1.
3 Positivity and domination
Inthissectionwestudysomepropertiesofthesemigroup(TΓ1) . Weassume
t t≥0
throughout this section that
a =a , a˜ =a ,a ∈L∞(Ω,R). (3.1)
jk kj k k 0
We recall that LD is the elliptic operator with Dirichlet boundary conditions
a
defined in the previous section. Its associated symmetric form aD is given by
(2.3) andhas domain W1,2(Ω). We shallneed the accretivity assumptionof aD
0
(orequivalentlythe self-adjointoperatorLD is non-negative)whichmeansthat
a
aD(u,u)≥0 for all u∈W1,2(Ω). (3.2)
0
Theorem 3.1. Suppose that 0∈/ σ(LD), (3.1) and that LD is accretive.
a a
a) The semigroup (TΓ1) is positive (i.e., it maps non-negative functions of
t t≥0
L2(∂Ω) into non-negative functions).
b) Suppose in addition that a ≥ 0 and a = 0 for all k ∈ {1,··· ,d}. Then
0 k
(TΓ1) is a sub-Markovian semigroup.
t t≥0
Recall that the sub-Markovian property means that for ϕ ∈ L2(∂Ω) and
t≥0
0≤ϕ≤1⇒0≤TΓ1ϕ≤1.
t
Thispropertyimpliesinparticularthat(TΓ1) extendsfromL2(∂Ω)toLp(∂Ω)
t t≥0
for all p∈[2,∞[. Since a is symmetric then so is b and one obtains by duality
that (TΓ1) extends also to Lp(∂Ω) for p∈[1,2].
t t≥0
9
Proof. The proof follows exactly the same lines as for Theorem 2.3 in [11].
a) By the well known Beurling–Deny criteria (see [9], Section 1.3 or [23], The-
orem 2.6), it suffices to prove that ϕ+ ∈ D(b) and b(ϕ+,ϕ−) ≤ 0 for all real-
valued ϕ∈D(b). Note that the fact that D(b) is not densely defined does not
affect the the statements of the Beurling-Deny criteria.
Let ϕ ∈ D(b) be real-valued. There exists a real-valued u ∈ H such that
V
ϕ=Tr(u). Thenϕ+ =Tr(u+)∈Tr(V)=TrH =D(b). This followsfromthe
V
fact that v+ ∈V for all v ∈V (see [23], Section 4.2).
ByLemma2.1wecanwriteu+ =u +u andu− =v +v withu ,v ∈W1,2(Ω)
0 1 0 1 0 0 0
and u ,v ∈ H . Hence, u = u+ − u− = (u − v ) + (u − v ). Since
1 1 V 0 0 1 1
u,u −v ∈H it follows that u =v . Therefore,
1 1 V 0 0
b(ϕ+,ϕ−) = a(u ,v )=a(u ,v +v )=a(u +u ,v +v )−a(u ,v +v )
1 1 1 0 1 0 1 0 1 0 0 1
= a(u+,u−)−a(u ,v )=−a(u ,v )
0 0 0 0
= −a(u ,u )=−aD(u ,u ).
0 0 0 0
Here we use the fact that
d d
a(u+,u−)= a ∂ (u+)∂ (u−)+ a ∂ u+u−+a u+∂ u−
kj k j k k k k
k,j=1ZΩ k=1ZΩ
X X
+ a u+u− =0.
0
ZΩ
Byassumption(3.2)wehaveaD(u ,u )≥0andweobtainb(ϕ+,ϕ−)≤0. This
0 0
proves the positivity of (TΓ1) on L2(∂Ω).
t t≥0
b)By[22]or[23],Corollary2.17itsufficestoprovethat ∧ϕ:=inf( ,ϕ)∈
D(b)andb( ∧ϕ,(ϕ− )+)≥0forallϕ∈D(b)withϕ≥01. Letϕ∈D(1b)and
1 1
suppose thatϕ≥0. Letu∈H be real-valuedsuchthatϕ=Tr(u). Note that
V
∧u∈V (see[23],Section4.3). Wedecompose ∧u=u +u ∈W1,2(Ω)⊕H .
1 1 0 1 0 V
Then
(u− )+ =u− ∧u=(−u )+(u−u )∈W1,2(Ω)⊕H .
1 1 0 1 0 V
Therefore,
b( ∧ϕ,(ϕ− )+) = a(u ,u−u )=a(u +u ,u−u )
1 1 0 1 1
1 1
= a(u +u ,−u +u−u )+a(u +u ,u )
0 1 0 1 0 1 0
= a(u +u ,−u +u−u )+a(u ,u )
0 1 0 1 0 0
d
= a ∂ ( ∧u)∂ ((u− )+)+
kj k j
1 1
k,j=1ZΩ
X
a ( ∧u)(u− )++aD(u ,u )
0 0 0
1 1
ZΩ
= a (u− )++aD(u ,u )≥0.
0 0 0
1
ZΩ
This proves that b( ∧ϕ,(ϕ− )+)≥0.
1 1
Next we have the following domination property.
10