A Global Uniqueness for Formally Determined Inverse Electromagnetic Obstacle Scattering

0.24 MB·English

by  Hongyu Liu

A Global Uniqueness for Formally Determined
Inverse Electromagnetic Obstacle Scattering
8
0 Hongyu Liu∗
0
2

n
a
Abstract
J
7 ItisprovedthatageneralpolyhedralperfectconductingobstacleinR3,pos-
1 siblyconsisting offinitelymany solid polyhedra, isuniquelydeterminedbythe
far-field pattern corresponding to a single incident wave. This improves earlier

] results in the literature to theformally determined case.
P
A
Mathematics Subject Classification(2000). Primary. 78A46,35R30Sec-
. ondary. 35P25, 35Q60
h
t
a Keywords. Inverse electromagnetic scattering, identifiability and uniqueness,
m
polyhedralperfect conductingobstacle.
[

1 1 Introduction
v
3
Inthispaper,weshallbemainlyconcernedwiththeinverseelectromagneticobstacle
2
6 scattering, where one utilizes the time-harmonic electromagnetic far-field measure-
2 ments to identify the inaccessible unknown impenetrable objects.
1. For a brief description of the forward scattering problem, we let a perfect con-
0 ducting obstacle D ⊂ R3 be a compact set with connected Lipschitz complement
8 G:=R3\D, and
0
: i
v Ei(x):= curl curlpeikx·d =ik(d×p)×deikx·d, (1.1)
i k
X
Hi(x):=curlpeikx·d =ikd×peikx·d, (1.2)
r
a
be the incident electric and magnetic fields, where p∈R3, k >0 and d∈S2 :={x∈
R3;|x| = 1} represents respectively the polarization, wave number and direction of
propagation. Theincidentwavepropagatinginthehomogeneousbackgroundmedium
willbe perturbedwhenit encountersanobstacle,andproducesa scatteredfiled. We
denote by Es and Hs the scattered electric and magnetic fields respectively, and
define the the total electric and magnetic fields to be
E(x)=Ei(x)+Es(x), H(x)=Hi(x)+Hs(x) x∈R3. (1.3)
Thenthedirectscatteringproblemconsistsoffindingasolution(E,H)∈H1 (curl;G)
loc
∗Department of Mathematics, University of Washington, Box 354350, Seattle, WA 98195, USA
([email protected]).
1
×H1 (curl;G) that satisfies the following time-harmonic Maxwell equations
loc
curlE−ikH=0, curlH+ikE=0 in G:=R3\D, (1.4)
ν×E=0 on ∂G, (1.5)
s s
lim (H ×x−|x|E )=0, (1.6)
|x|→∞
where the last limit corresponds to the so-called Silver-Mu¨ller radiation condition
characterizing the fact that the scattered wave is radiating.
The well-posedness of the forward scattering problem (1.1)-(1.6) has been well
understood (see [3]). Particularly, the cartesian components of E and H are (real)
analyticinGandtheasymptoticbehaviorofthescatteredfield(Es,Hs)isgoverned
by (see [6])
eikx·d 1
s
E (x;D,p,k,d)= E (xˆ;D,p,k,d)+O( ) as |x|→∞, (1.7)
∞
|x| |x|
(cid:26) (cid:27)
eikx·d 1
s
H (x;D,p,k,d)= H (xˆ;D,p,k,d)+O( ) as |x|→∞, (1.8)
∞
|x| |x|
(cid:26) (cid:27)
uniformly for allxˆ=x/|x|∈S2. The functions E (xˆ) andH (xˆ) in (1.7) and(1.8)
∞ ∞
are called, respectively, the electric and magnetic far-field patterns, and both are
analyticontheunitsphere S2. Asisnotedabove,Es(x;D,p,k,d),E (xˆ;D,p,k,d),
∞
etc. will be frequently used to specify their dependence on the observation direction
xˆ, the polarization p, the wave number k and the incident direction d.
Now, the inverse scattering problem is the following. Assume that the obstacle
D is unknown or inaccessible and we aim to image the object and thereby identify it
by performing far-field measurements. That is, with the measurement of the electric
far-fieldpattern(or,equivalently,the magneticfar-fieldpattern)ofthewavewhichis
scatteredby D correspondingto a givenincident wave,for one or more choices of its
polarizationporofitswavenumberk,orofitspropagationdirectiond,wewouldlike
to recoverthe obstaclewhosescatteredwavesarecompatiblewith the measurements
performed. From the mathematical viewpoint, the inverse obstacle scattering can be
formulated as the following operator equation
F (∂G)=E (xˆ;D,p,k,d) for (xˆ,p,k,d)∈S2×U×K×S2, (1.9)
e ∞ 0 0
where S20,S20 ⊂ S2, U ⊂ R3 ,K ⊂ R+ := {x ∈ R;x > 0} and the nonlineear operator
F is defined by the forward scattering system (1.4)-(1.6). The inverse obstacle
e
scattering,ehaving its roots in the technology of radar and sonar, are also central to
many other areas of science such as medical imaging, geophysical exploration and
nondestructivetesting,etc.. Wereferto[6]foramoredetaileddiscussionandrelated
literature. As usual in most of the inverse problems, the first question to ask in this
context is the identifiability; i.e., whether an obstacle can really be identified from a
knowledgeofitsfar-fieldpattern. Mathematically,theidentifiability istheuniqueness
issue, which is the injectivity of the (nonlinear) operator F in (1.9). That is,
e
If two obstacles D and D produce the same far field data, i.e.,
E (xˆ;D,p,k,d)=E (xˆ;D,p,k,d) for (xˆ,p,k,d)∈S2×U×K×S2,
∞ ∞e 0 0
does D have to be the sameeas D ? e
e
2
We refer to [10] for a generaldiscussion of the critical role of uniqueness which plays
ininverseproblemstheorytheoreticallyaswellasnumerically. Itisobservedthatthe
uniqueness results also provide the practicalinformation on how many measurement
data one should use to identify the underlying object. As an important ingredientin
the uniqueness study and noting E is an analytic function, one sees that if S2 in
∞ 0
(1.9) is an open subset of the unit sphere, no matter how small the subset is, we can
always recover such data on the whole unit sphere by analytic continuation. Hence,
for our uniqueness study, without loss of generality, we can assume that the far-field
data are given on the whole unit sphere, i.e., in every possible observationdirection.
Thenitiseasilyseenthatthe inverseobstaclescatteringisformallydeterminedwith
fixedp ∈R3,k >0andd ∈S2,sincethefarfielddatadependonthesamenumber
0 0 0
of variables,as does the obstacle which is to be recovered.1 Due to suchobservation,
there is a widespread belief that one can establish the uniqueness by using the far
field pattern corresponding to a single incident wave. However,this has remained to
be a longstanding challenging open problem, though extensive study has been made
in this aspect (see [2] and [7]). The only previous result that we are aware of this
kind is in [11], where it is shown that a simple ball can be uniquely determined by
its far-field measurement corresponding to a single incident wave.
In the past few years, significant progress has been achieved on the unique de-
termination of general polyhedral type obstacles by several far-field measurements.
The breakthroughis first made in the inverseacousticobstacle scattering,whereone
utilizes the acoustic far-field measurement to identify the underlying scattering ob-
jects (see [1] [4] [9] [12] [13]). Among the arguments for the proofs of those results,
the new methodology developed in [12] which we call path argument is proved to
be particularly suitable for attacking such problems. Based on suitably devised path
arguments,togetherwithsomenovelreflectionprinciplesforthesolutionsofMaxwell
equations, various uniqueness results have been established in different settings with
general polyhedral type obstacles in [14] and [15], but all with the far-field measure-
mentscorrespondingtotwodifferentincidentwaves. Inthecurrentwork,weareable
to improvesignificantlyonthis resultto the formallydeterminedsetting. Itis shown
thatthe measurementofthe far-fieldpatterncorrespondingtoa singleincidentwave
uniquely determines a generalpolyhedralperfectconducting obstacle. For the proof,
we follow the general strategy in [14] and [15], but several technical new ingredients
must be developed and the path argument in this work is refined significantly. We
next state more precisely the main result.
ItisfirstrecalledthatacompactpolyhedroninR3 isasimplyconnectedcompact
set whose boundary is composed of (open) faces, edges and vertices. In the sequel,
we call D a polyhedral obstacle if it is composed of finitely many (but unknown a
priori) pairwise disjoint compact polyhedra. That is,
m
D= D , (1.10)
l
l=1
[
where m is an unknown integer but must be finite and each D ,1 ≤ l ≤ m is a
l
compact polyhedron such that
Dj ∩Dj′ =∅ if j ≤j′ and 16=j,j′ ≤m.
Clearly, the forward scattering problem (1.4)-(1.6) with such a polyhedral obstacle
D is well-posed. Moreover,we knowthat the singularbehaviorsofthe weak solution
1Here,thenumberofvariablesis2,sinceboth∂GandS2 are2-manifold.
3
only attach to the edges and vertices, that is, (E,H) satisfies (1.4) in the classical
sense in any subdomain ofG, whichdoes notmeet any corner or edge ofD (see [5]).
By the regularity of the strong solution for the forwardscattering problem, we know
thatbothEandHareatleastC0,α-continuous(0<α<1)uptotheregularpoints,
namely, points lying in the interior of the open faces of D.
The main result of this paper is the following:
Theorem1.1. LetDandDbetwoperfectpolyhedralobstacles. Foranyfixedk >0,
0
d ∈ S2 and p ∈ R3 such that d and p are linearly independent, we have D = D
0 0 0 0
as long as e
e
E (xˆ;D,p ,k ,d )=E (xˆ;D,p ,k ,d ) for xˆ∈S2. (1.11)
∞ 0 0 0 ∞ 0 0 0
Remark 1.2. As mentioned earlier, there are some uniqueness results established in
e
[14] and [15] in the unique determination of general polyhedral obstacles, but with the
far-field data corresponding to two different incident waves. However, the polyhedral
obstacles considered in [14] are more general than the present ones, and they admit
the simultaneous presence of crack-type components (namely, screens). Whereas the
uniqueness in [15] is established without knowing the a priori physical properties of
the underlying obstacle. In Section 4, we would make concluding remarks on that the
uniqueness result in Theorem 1.1 can not cover completely the ones obtained in [14]
and [15].
The rest of the paper is organized as follows. In Section 2, we introduce the
perfectsetandperfectplanes,andthenshowseveralcrucialpropertiesofthemwhich
shall play a key role in proving Theorem 1.1. Section 3 is devoted to the proof of
Theorem 1.1, and in Section 4, we give some concluding remarks.
2 Perfect Set and Perfect Planes
First, we fix some notations which shall be used throughout of the rest of the paper.
WedenoteanopenballinR3withcenterxandradiusrbyB (x),theclosureofB (x)
r r
by B¯ (x) and the boundary of B (x) by S (x). The notation T (x) is defined to be
r r r r
anopencubeofedgelengthr,centeredatx,whileT¯ (x)isitscorrespondingclosure.
r
Unless specified otherwise, ν shall always denote the inward normal to a concerned
domain, or the normal to an two-dimensional plane in R3. The distance between
two sets A andB in R3 is understood as usualto be d(A,B)=infx∈A,y∈B|x−y|.
Finally,acurveγ =γ(t)(t≥0)issaidtoberegularifitisC1-smoothand dγ(t)6=0.
dt
Henceforth, we let k > 0, d ∈ S2 and p ∈ R3 be fixed such that d and p
0 0 0 0 0
are linearly independent, and denote by E(x):=E(x;D,p ,k ,d ) the total electric
0 0 0
field in (1.4)-(1.6) corresponding to a polyhedral perfect conducting obstacle D as
described in (1.10). The following definition of a perfect set is modified from that in
[14] to fit the problem being under investigation.
Definition 2.1. PE is called a perfect set of E in G:=R3\D if
PE = x∈G;ν×E|Π∩Br(x)∩G=0 for some r >0 and plane Π passing through x ,
where ν(cid:8)is the unit normal to the plane Π. (cid:9)
For any x ∈ PE, we let Π be the plane involved in the definition of PE. Fur-
thermore, we let Π be the connected component of Π\D containing x, then by the
e
4
analyticityofE inG, wesee ν×E=0 onΠ byclassicalcontinuation. Inthe sequel,
such Π will be referred to as a perfect plane. The introduction of the prefect set and
perfect plane is motivated by the observateion that, when proving Theorem 1.1 by
contraediction, if two different obstacles produce the same far-field pattern, then out-
side one obstacle there exists a perfect plane which is extended from an open face of
the otherobstacle. Startingfromnowon,Π withanintegerl,shallalwaysrepresent
l
a perfect plane in G which lies on the plane Π in R3.
l
A very fine property of perfect planeseis the so-called reflection principle, which
constitutesanindispensableingredientinthepathargumentsforprovingtheunique-
ness results in [14] and [15]. We formulate the principle in the following theorem.
Subsequently, we use R to denote the reflection in R3 with respect to a plane Π.
Π
Theorem 2.2. For a connected polyhedral domain Ω in G := R3\D, let Π be one
of its faces that lies on some perfect plane. Furthermore, let Π be the plane in R3
containing Π and Ω∪R Ω⊂G. We have two consequences: e
Π
(i) νΠ×Ee =0 on Π∩(Ω∪RΠΩ);
(ii) SupposethatΣ⊂∂ΩisasubsetofonefaceofΩotherthanΠ,andthefollowing
condition holds
ν ×E=0 on Σ, e (2.1)
Σ
where ν is the unit normal to Σ directed to the interior of Ω. Then we have
Σ
νΣ′ ×E=0 on Σ′, (2.2)
where Σ′ = RΠΣ and νΣ′ is the unit normal to Σ′ directed to the interior of
R Ω.
Π
Proof. The verificationfor(i) canbe foundin the proofofTheorem3.2in[15], while
for (ii), is given in Theorem 2.3 in [14].
The reflection principle in item (i) of Theorem 2.2 is particularly useful when
(Π∩(Ω∪R Ω))\Π6=∅. Clearly, in such case, we can find a perfect plane also lying
Π
on the plane Π, but different from Π.
Next, we woulde classify all those perfect planes in G into two sets in R3, one
is bounded and the other is unbouended. In fact, it is verified directly that there
mightexistunboundedperfectplanes2. Inoursubsequentpathargumentforproving
Theorem1.1,theprocedureofcontinuationofperfectplanesalonganexitpathmight
be brokendownwith the presence of an unbounded perfect plane, since one may not
be able to find another perfect plane with an unbounded perfect plane by using the
reflection principle in Theorem 2.2. In the rest of this section, we shall show some
critical properties on the unbounded perfect planes.
Lemma 2.3. All the unbounded perfect planes associated with E in G are conplane.
Obviously, Lemma 2.3 is divided into the following two lemmata:
Lemma 2.4. There cannot exist two unbounded perfect planes Π and Π such that
1 2
Π ∦Π .
1 2
e e
2This constitutes one of the major differences from those perfect planes introduced in [14] and
[15]. Alltheperfectplanesdefinedthereareboundedduetotheuseoftwodifferentincidentwaves.
SeeLemma3.2in[14].
5
Lemma 2.5. There cannot exist two different unbounded perfect planes Π and Π
1 2
such that Π kΠ .
1 2
e e
Proof of Lemma 2.4. Assume contrarily that Π and Π are two unbounded perfect
1 2
planes in G such that Π ∦ Π . Let ν and ν , respectively, be the unit normals to
1 2 1 2
Π and Π . Noting that Es(x)=O(1/|x|) as |ex|→∞e, we have from
1 2
ν ×E(x)=0 on Π for l=1,2,
l l
that e
i
lim |ν ×E (x)|=0 for l=1,2.
l
e
x∈Πl:|x|→∞
Using (1.1), we further deduce
ν ×((d ×p )×d )=0 for l=1,2.
l 0 0 0
That is, ν kν since they are both parallel to a fixed vector (d ×p )×d , contra-
1 2 0 0 0
dicting to our assumption that Π ∦Π and completing the proof.
1 2
Proof of Lemma 2.5. By contradiction,we assume thatthere existtwo differentper-
fectplanesΠ andΠ suchthatΠ kΠ . LetT :=T (0)be asufficientlylargecube
1 2 1 2 r
suchthatD⊂T,andbysuitablerotation,wemaywithoutlossofgeneralityassume
thatbothΠe andΠeareperpendiculartoonefaceofT. Next,withalittlebitabuse
1 2
of notations, we still denote by Π and Π those parts of Π and Π lying outside
1 2 1 2
of T, namely, Π \T and Π \T, and the same rule applies to Π,l ∈ Z appearing
1 2 l
in the rest of the proof. Now, inethe (unbeounded) polyhedrael domaine R3\T, we can
makeuseoftheereflectionreeflectionasstatedin(ii)ofTheorem2.e2,andfromΠ and
1
Π to find that
2
ν×E=0 on Π :=R (Π ). e
3 Π2 1
e
Continuing with such argument, from Π and Π we have
2 e 3 e
ν×E=0 oen Π4 :e=RΠ3(Π2).
By repeating this reflection, we eventuallyefind a sequeence of perfect planes Πl, l =
1,2,3,...suchthatallΠ ’sareparalleltoeachother. Clearly,d(Π ,Π )=d(Π ,Π )>
l l l+1 1 2
0beingfixedforl=1,2,3,.... Hence,theremustexistsomel <∞suchthateT lies
0
entirely at one side ofeΠl0. That is, Πl0 = Πl0 is the whole plaene ein R3. Obveiousely,
D also lies at one side of Π . Using again the reflection principle in Theorem 2.2,
l0
(ii), we see ν×E=0 oen R (∂D). eFinally, let Σ and Σ be two adjacent faces of
R (∂D) andwe havefromΠlt0he extensionofΣ an1dΣ tw2onon-parallelunbounded
Πl0 1 2
perfect planes, which contradicts to Lemma 2.4. The proof is completed.
We proceed to make an important observation of the reflection principle (i) in
Theorem 2.2, when Ω ∪ R Ω is unbounded while Π is bounded. In this case, it
Π
is clear that the extension of some part of (Π∩(Ω∪R Ω))\Π gives at least one
Π
unbounded perfect plane. That is, some bounded eperfect plane might imply the
existenceofsomecorrespondinglyunboundedperfectplane. Nexet,westudycarefully
such special bounded perfect plane Π , which can be regarded as “unbounded”. To
0
localizeourinvestigation,wefixanarbitrarypointx ∈Π ∩Gandtakeasufficiently
0 0
small ball B0 := Br(x0) such that Be0 ⊂ G. B0 is divided by Π0 into two half balls,
which we respectively denote by B+ and B−. Let G+ bee the connected component
0 0 0
e
6
of G\Π containing B+ and G− be the connected component of G\Π containing
0 0 0 0
B−. We remark that it may happen that G+ =G−. Next, let Λ+ be the connected
0 0 0 0
compoenent of G+∩R (G−) containing B+, and Λ− be the connecteed component
0 Π0 0 0 0
of G−∩R (G+) containing B−. Finally, set Λ =Λ+∪Λ− and we see that Λ is
0 Π0 0 0 0 0 0 0
a polyhedral domain which symmetric with respect to Π , and moreover, B ⊂ Λ .
0 0 0
Onecaneasilyseethatthe constructionofΛ isonlydependentonthe prefectplane
0
Π . Since ∂Λ is composed of subsets lying either on ∂D or on R (∂D), by the
0 0 Π0
reflection principle in (ii) of Theorem 2.2, we have ν×E=0 on ∂Λ .
0
e Starting from now on, we shall denote by Λe the symmetric set constructed as
Πl
abovecorrespondingtoaboundedperfectplaneΠl;namely,intheabove,ΛΠe0 :=Λ0.
Clearly,incaseΛe is unbounded, wesee fromourearlierdiscussionthatthere must
Πl
existanunboundedperfectplanewhichisextendeedfromsomepartof(Πl∩ΛΠel)\Πl.
Such observation in combination with the result in Lemma 2.3 gives
e
Lemma 2.6. All the bounded perfect planes Πl with unbounded ΛΠel and all the
unbounded perfect planes are conplane.
e
Based on Lemma 2.6, we introduce the following set consisting of all the “un-
bounded” perfect planes
QE :={Π; Π is an unbounded perfect plane
or Π is a bounded perfect plane but with unbounded Λe}. (2.3)
e e Π
Since all the members in QE are conplane, one verifies directly that QE consists of
e
atmostfinitelymanyperfectplanesbynotingthefactthatDiscomposedoffinitely
many pairwisedisjoint compactpolyhedra. We further defineQ˘E to be the subsetof
QEconsistingofthoseboundedperfectplanesinQE. Next,weshowsometopological
properties of the sets QE and Q˘E.
Lemma 2.7. Let G:=R3\D, then
(i) G\Q˘E is connected;
(ii) G\QE has no bounded connected component.
Proof. We first observe that Q˘E is bounded since Q˘E ⊂ ch(D), where ch(D) is the
convex hull of D. By further noting that ∂G is bounded, we know that G\Q˘E has
exactly one unbounded connected component. Hence, if G\Q˘E is not connected, it
must have some bounded connected component, say C ⊂G. Clearly, there must be
0
one face ofthe polyhedraldomain C0 that comes fromexactly a perfect plane in Q˘E,
sayΠ0. Now,one can verifydirectly that ΛΠe0 ⊂C0∪RΠ0C0, whichis bounded since
C0 is bounded. But this contradicts to the assumption that Π0 ∈ QE, thus proving
asseretion (i). Next, assertion (ii) is readily seen from (i). In fact, if G\QE has a
bounded connected component, say D0, then one must haveeD0 ⊂ G\Q˘E, which is
certainly not true. The proof is completed.
Correspondingly,we set
SE ={Π; Π is a bounded perfect plane with bounded Λe}. (2.4)
Π
Finally, we give a lemma concerning the fundamentalproperty ofa connectedset
e e
(seee.g.,Theorem3.19.9in[8]), whichshallbeneededinthenextsectiononproving
Theorem 1.1.
7
Lemma 2.8. Let E be a metric space, A ⊂E be a subset and B ⊂E be a connected
set such that A ∩B 6=∅ and (E\A)∩B 6=∅, then ∂A ∩B 6=∅.
3 Proof of Theorem 1.1
The entire section is devoted to the proof of Theorem 1.1 by contradiction. Assume
that D6=D and
e E∞(xˆ;D,p0,k0,d0)=E∞(xˆ;D,p0,k0,d0) for xˆ∈S2. (3.1)
LetΩbetheunboundedconnectedcomponeentofR3\(D∪D).3 ByRellich’stheorem
(see Theorem 6.9, [6]), we infer from (3.1) that
e
E(x;D)=E(x;D) for x∈Ω, (3.2)
where E(x;D) and E(x;D) are, respectievely, abbreviations E(x;D,p0,k0,d0) and
E(x;D,p ,k ,d ). Next, noting that D 6= D, we see that either (R3\Ω¯)\D 6= ∅
0 0 0
or (R3\Ω¯)\D 6= ∅. Withoeut loss of generality, we assume the former case and let
D∗ :=e(R3\Ω¯)\D 6= ∅. It is easily seen that De∗ ⊂ D, so D∗ is bounded. Moreover,
by choosingeconnected component if necessary, we assume that D∗ is connected.
Clearly, D∗ is a bounded polyhedral domain in Ge= R3\D and E(x;D) is defined
over D∗. Noting ∂D∗ ⊂ ∂Ω∪∂D ⊂ ∂D∪∂D and using (3.2), we have from the
perfect boundary conditions of E(x;D) and E(x;D) on ∂D and ∂D that
e
ν×E(x;D)=0 one∂D∗. e (3.3)
Inthefollowing,inordertosimplynotations,weuseasthoseintroducedinSection2,
e.g.,wewriteE(x)todenoteE(x;D)etc.. Therestoftheproofwillbeproceededinto
three steps and a brief outline is as follows. In the first step, we will find a perfect
plane Π1 ∈ SE, and this is the starting point of the subsequent path argument.
In the second step, we would construct implicitly an exit path, which is a regular
curve leying entirely in the exterior of D and connecting to infinity. As we mentioned
earlier that the path argument might be broken down with the presence of some
“unbounded” perfect planes (namely, perfect planes in QE), in order to avoid our
subsequent argument being trapped at such “unbounded” perfect planes, the curve
is required to have at most one intersection with QE. Fortunately, this can be done
by using Lemma 2.7. Finally, using the reflection principle in Theorem 2.2, we make
continuation of (bounded) perfect planes along the exit path to find a sequence of
perfectplanes. Thenacontradictionisconstructedbyshowingthatthe continuation
must follow the exit path to infinity since we always step a length larger than a
fixed positive constant when making such continuation, but on the other hand, all
the bounded perfect planes are contained in the convex hull of D being bounded. In
this final step, we must be carefully treating the possible presence of “unbounded”
perfectplanesandthisisthemaindifferenceofthepresentpathargumentfromthose
implemented in [14] and [15].
Step I: Existence of a bounded perfect plane Π1 with bounded ΛΠe1
3Sinceboth DandDe arecompact sets, weknow that Ωiseunique. Moreover,itisobvious that
∂ΩformstheboundaryofapolyhedraldomaininG.
8
We first note that ∂D∗\∂D 6= ∅. Hence, there must be an open face say Σ on
0
∂D∗ that can be extended in G to form a perfect plane and it is denoted by Π .
0
Since Π is extended from a face of the bounded polyhedral domain D∗ in G, we
0
infer from the following Lemma 3.1 that Π is bounded. Now, if the symmetriceset
0
ΛΠe0 coerresponding to Π0 is bounded, then we are done since we can take Π0 as Π1.
So,withoutlossofgenerality,weassumetheatΛΠe0 isunbounded. Next,basedonΣ0,
we construct a boundedepolyhedral domain in G which is symmetric with reespecteto
Π0 but different from ΛΠe0. The construction procedure is similar to that for ΛΠe0,
and we nonetheless present it here for clearness.
Fix an arbitrary point x∗ ∈Σ and let B∗ :=B (x∗) with ε>0 sufficiently small
0 ε
suchthatB∗isdividedbyΣ intotwo(open)halfballsB+andB−satisfyingB+ ⊂D∗
0 ∗ ∗ ∗
and B− ⊂ G\D∗. Next, let Θ+ be the connected component of R (G\D∗)∩D∗
∗ ∗ Π0
containing B+ and Θ− be the connected component of R D∗∩(G\D∗) containing
∗ ∗ Π0
B−. SetΘ∗ =Θ+∪Σ ∪Θ−. Clearly,Θ∗ isanon-emptyboundedpolyhedraldomain
∗ ∗ 0 ∗
in G since B∗ ⊂ Θ∗ ⊂ D∗ ∪R D∗. We remark that Θ is in fact the connected
Π0 0
Tcohmeoproenmen2t.2o,fν(D×∗E∪(xR)Π=0D0∗o)n∩∂ΛΘΠe∗0. IctonistaoibnvinioguΣs0th.aBty∂Θth∗e\rDefl6=ec∅ti.oLneptrΣinci⊂ple∂Θ(i∗i)\Dof
1
be an open face. By analytic continuation, Σ is extended in G to give a perfect
1
plane Π . Since Θ∗ is symmetric with respect to Π , we know Σ ⊂\ Π and therefore
1 0 1 0
Π1 ∈\QE by Lemma 2.6, i.e., Π1 is bounded with bounded ΛΠe1.
e
Step II: Construction of the exit path γ
e e
Since both ∂G and Π are bounded, we see that G\Π has a unique unbounded
1 1
connected component, which is denoted by U. It readily has that Π ⊂ ∂U and
1
U contains the exterioreof a sufficiently large ball conteaining D. Next, we fix an
arbitrarilypointx ∈Π . Letγ :=γ(t)(t≥0)bearegularcurvesuchtehatγ(t )=x
1 1 1 1
with t = 0 and γ(t)(t > 0) lies entirely in U. Furthermore, γ connects to infinity,
1
i.e., limt→∞|γ(t)| = ∞e. The exit path γ constructed in this way might have non-
empty intersection with QE. In this case, we require that γ(t)(t > 0) has only one
intersectionpointwithQE. Infact,incaseγ(t)∩QE 6=∅,wewouldmodifythecurve
γ as follows to satisfy such requirement. Let x := γ(T) be the “first” intersection
T
point of γ(t)(t>0) and QE; that is,
T =min{t>0; γ(t)∈QE}<∞.
Then, set V be the connected component of G\QE such that xT ∈ ∂V. Let W :=
U ∩V. ItcanbeverifiedthatW isanunboundedconnectedopensetsuchthatx ∈
T
∂W. Indeed, the connectedness of W is obvious by noting that both U and V are
connected. WhereastheunboundednessofW isduetothefactsthatV isunbounded
by Lemma 2.7 and U contains the exterior of a sufficiently large ball containing D
as mentioned earlier. Next, let η(t)(t ≥ T) be a regular curve such that η(T) = x ,
T
η(t)(t > T) lies entirely in W and connects to infinity (i.e., lim |η(t)| = ∞).
t→∞
Furthermore,itistriviallyrequiredthatη(t)hasC1-connectionwithγ(t)(0≤t≤T)
at x . Now, set
T
γ(t) 0≤t≤T,
γ˜(t)=
( η(t) t>T,
then γ˜(t)(t≥0) satisfies all our requirements of an exit path.
Step III: Continuation of bounded perfect planes along γ
9
Let d = d(γ,D) > 0, which is attainable since D is compact, and r = d /2.
0 0 0
Clearly,B¯ (γ(t))⊂Gforanyt≥0. Letx+ =γ(t˜)∈S (x )∩γ, wheret˜ is taken
r0 2 2 r0 1 2
to be t˜ = max{t > 0;γ(t) ∈ S (x )}, and let x− be the symmetric point of x+
2 r0 1 2 2
with respect to Π1. Next, let G+1 be theeconnected component of G\Π1 containing
x+, and G− be the connected component of G\Πe containing x−. Then let Λ+ebe
2 1 1 2 1
the connectedcomponentofG+∩R (G−)containingx+ andΛ− be tehe connected
1 Π1 1 2 1
ceomponent of G−1 ∩RΠ1(G+1) containing x−2. SeteΛ1 = Λ+1 ∪Πe1∪Λ−1. In fact, Λ1
is the symmetric set ΛΠe1 corresponding to the perfecteplane Π1 and we present its
constructionagainforconvenienceofthesuebsequentargument.eSinceΠ1 ∈SE,Λ1 is
bounded. ByLemma2.8,itiseasytodeduce thatγ∩∂Λ 6=∅.eWeletx =γ(t )be
1 2 2
the‘last’intersectionpointofγ and∂Λ ;namely,t =max{t>0;γ(t)e∈∂Λ }<∞.
1 2 1
Thisthenimpliestheexistenceofaperfectplanepassingthroughx whichisextended
2
from an open face of ∂Λ whose closure contains x . We denote the perfect plane
1 2
by Π . Without lossof generality,we further assume that x is the ‘last’intersection
2 2
point of γ with Π . By the following result, we know Π is bounded. We shall prove
2 2
at tehe end of this section:
e e
Lemma 3.1. Suppose that Λ⊂G is a bounded polyhedral domain such that
ν×E=0 on ∂Λ.
Then every open facelyingon ∂Λ\Dcannotbeconnectedlyextendedtoanunbounded
planar domain in G.
Now, we still need to distinguish between two cases of Π2 ∈ Q˘E and Π2 ∈ SE.
But for the end of a more general discussion, we next give the induction procedure
for the above reflection argument of finding a different perfeect plane withea known
one. Suppose thatΠn ∈SE, n∈N andxn :=γ(tn)∈γ∩Πn is the ‘last’intersection
point between γ and Π .
n
Let x+ = γ(t˜e ) ∈ S (x )∩γ, where t˜ is takeen to be t˜ = max{t >
n+1 n+1 r0 n n+1 n+1
0;γ(t)∈S (x )}, ande let x− be the symmetric point of x+ with respect to Π .
r0 n n+1 n+1 1
Next, leet G+ be the connected component of G\Π containing x+ , and G− be
n n n+1 n
the connected component oef G\Π containing x− . Thenelet Λ+ be the connected
n n+1 n
componentofG+∩R (G−)containingx+ andΛe−betheconneectedcomponentof
n Πn n n+1 n
G−∩R (G+)containingx− .eSetΛ =Λ+e∪Π ∪Λ−. Byourearlierdiscussion,
n Πn n n+1 n n n n
Λn = ΛΠen, and it is bounded since Πne∈ SE. By Lemma 2.8, γ ∩∂Λn 6= ∅. We
let xn+1 := γ(tn+1) with tne+1 = max{t > 0;γ(t)e∈ ∂Λn} < ∞. This again implies
the existence of a perfect plane Π epassing through x , and Π is bounded
n+1 n+1 n+1
by Lemma 3.1. Furthermore, we can also assume that x is the ‘last’ intersection
n+1
point of γ with Π . In the folelowing, we list several important eresults that have
n+1
been achieved:
(i) xn,xn+1 ∈Pe E :={x∈PE; x∈Π with Π∈SE∪Q˘E}; (3.4)
(ii) Π is different from Π , since t and t with t > t are respectively
n+1 c n en en+1 n+1 n
the ‘last’ intersection points between γ and Π and Π ;
n n+1
e e
(iii) Both Π and Π are bounded;
n n+1 e e
(iv) Since B (x )⊂Λ , the length of γ(t) from t to t is not less than r , i.e.,
er0 n e n n n+1 0
|γ(t ≤t≤t )|≥|γ(t ≤t≤t˜ )|≥r . (3.5)
n n+1 n n+1 0
10