Corrigendum to ``Determining a sound-soft polyhedral scatterer by a single far-field measurement''

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by  Giovanni Alessandrini

Corrigendum to “Determining a sound-soft
polyhedral scatterer by a single far-field
measurement”
6
0
0 Giovanni Alessandrini and Luca Rondi
2
[email protected] [email protected]
n
Dipartimento di Matematica e Informatica
a
J Universit`a degli Studi di Trieste, Italy

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In the paper, [1], on the determination of a sound-soft polyhedral scatterer
]
P by a single far-fieldmeasurement,the proofofProposition3.2 is incomplete. In
A this corrigendumweprovidea newproofofthesamepropositionwhichfills the
. previous gap. In order to introduce it, we recall some definitions from [1].
h
t Let v be a nontrivial real valued solution to the Helmholtz equation
a
m (1) ∆v+k2v =0 in G,
[
in a connected open set G⊂RN, N ≥2. We denote the nodal set of v as
1
v
N ={x∈G: v(x)=0}
6 v
0
4 and we let Cv be the set of nodal critical points, that is
1
0 C ={x∈G: v(x)=0 and ∇v(x)=0}.
v
6
0 We say that Σ ⊂ N is a regular portion of N if it is an analytic open and
v v
/
h connectedhypersurfacecontainedinNv\Cv.LetusdenotebyA1,A2,...,An,...
t the nodal domains of v in G, that is the connected components of {x ∈ G :
a
m v(x)6=0}=G\Nv. Let us recall the statement of Proposition 3.2 in [1].
:
v Proposition 3.2 ([1]) We can order the nodal domains A ,A ,...,A ,... in
1 2 n
Xi such a way that for any j ≥2 there exist i, 1≤i<j, and a regular portion Σj
of N such that
r v
a
(2) Σ ⊂∂A ∩∂A .
j i j
The gap in the proof given in [1] stands in the fact that the ordering
A ,A ,...,A ,... obtained with that method might not ensure that all the
1 2 n
nodal domains are contained in the sequence. We base the new proof on the
following theorem.
Theorem 1 The set C has Hausdorff dimension not exceeding N −2.
v
1
Aproofcanbefoundin[5,Theorem2.1].Furtherdevelopmentsofthetheory
on the structure of zero sets of solutions to elliptic equations can be found, for
instance, in [2, 3] and in their references.
Let G′ = G\C . By the property of C described in the previous theorem,
v v
andbyusing[4,ChapterVII, Section4]and[4,TheoremIV4,Corollary2],we
can conclude that G′ is an open and connected set. We also remark that, for
everyx∈N \C ,there areexactly twonodaldomains,AandB, ofv suchthat
v v
x ∈ ∂A∩∂B. Finally, let us note that the nodal domains of v in G coincide
with the nodal domains of v in G′.
We shall also make use of the following elementary lemma.
Lemma 2 For any connected open set G ⊂ RN, there exists an increasing
sequence {G }∞ of bounded, connected open sets such that G = S∞ G
m m=1 m=1 m
and G ⊂⊂G for every m.
m
Proof. For every k =1,2,..., we denote
D ={x∈G: dist(x,∂G)>1/k, |x|<k}.
k
Let us assume, without loss of generality, that D 6= ∅ and let us fix y ∈ D .
1 1
For every x ∈ D , let γ be a path in G joining y to x. For every h > 0, let
k x
Uh = {z ∈ RN : dist(z,γ ) < h}. We obviously have that Uh is a connected
x x x
h(x) h(x)
open set. Let h(x)>0 be such that U ⊂⊂G. We have that {U } is
x x x∈Dk
anopencoveringofthe compactsetD .Therefore,we canfindx ,...,x ∈D
k 1 l k
such that D ⊂ Sl Uh(xj). We observe that E = Sl Uh(xj) is an open
k j=1 xj k j=1 xj
connected set such that D ⊂E ⊂⊂G. Therefore the lemma follows choosing
k k
G =Sm E . (cid:3)
m k=1 k
Proof of Proposition 3.2. We apply Lemma 2 to the connected set G′ =
G\C . We choose A such that A ∩G 6=∅ and we proceed by induction.
v 1 1 1
Let us assume that we have ordered A ,...,A in such a way that there
1 n
exist Σ ,...,Σ regular portions of N such that (2) holds for any j =2,...,n
2 n v
and for some i<j.
◦
Let Aˆ =A ∪...∪A . If G′\Aˆ = ∅, then we are done. Otherwise, let
n 1 n n
m ≥ 1 be the smallest number such that G \Aˆ 6= ∅. Since G is connected,
m n m
we can find y ∈ ∂Aˆ ∩ G and r > 0 such that B (y) ∩ ∂Aˆ is a regular
n m r n
portionofN andthereexistexactlytwonodaldomains,A˜ ⊂Aˆ andA˜ with
v 1 n 2
A˜ ∩Aˆ =∅,whoseintersectionswithB (y)arenotempty.Clearly,A˜ coincides
2 n r 1
withA ,forsomei=1,...,n,andifwepickA =A˜ andΣ =B (y)∩N ,
i n+1 2 n+1 r v
then (2) holds for j =n+1, too.
If G contains only finitely many nodal domains, then we can iterate this
constructionandafterafinitenumberofstepswerecoverallthenodaldomains,
thatisfor somel∈NwehaveG′\Aˆ =∅ andwearedone.Otherwise,we argue
l
in the following way. Since G is contained in G′, for every x ∈ G there
m m
is a neighbourhood of x intersecting at most two different nodal domains. By
compactness,weobtainthatG intersectsatmostfinitelymanydifferentnodal
m
domains. Hence, if we iterate the previous construction, after a finite number
of steps we find l ∈ N such that G \Aˆ = ∅. By repeating the argument for
m l
the smallest m′ > m such that Gm′\Aˆl 6= ∅, we conclude that for any m ∈ N
there exists l ∈ N such that G \Aˆ = ∅. Therefore the infinite sequence {A }
m l i
comprises all the nodal domains of v in G. (cid:3)
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Acknowledgements
The authors wish to express their gratitude to Hongyu Liu and Jun Zou for
pointing out to them the gap in the proof of Proposition 3.2 in [1] and for
kindly sending them their preprint [6].
References
[1] G. Alessandrini and L. Rondi, Determining a sound-soft polyhedral scat-
terer by a single far-field measurement,Proc.Amer.Math.Soc.133(2005),
pp. 1685–1691.
[2] Q. Han, R. Hardt and F. Lin, Geometric measure of singular sets of elliptic
equations, Comm. Pure Appl. Math. 51 (1998), pp. 1425–1443.
[3] R.Hardt,M.Hoffmann-Ostenhof,T.Hoffmann-OstenhofandN.Nadirashvi-
li,Critical sets of solutions to elliptic equations,J.DifferentialGeometry 51
(1999), pp. 359–373.
[4] W. Hurewicz and H. Wallman, Dimension Theory, Princeton University
Press, Princeton N.J., 1948.
[5] F.-H. Lin,Nodal sets of solutions of elliptic and parabolic equations,Comm.
Pure Appl. Math. 44 (1991), pp. 287–308.
[6] H. Liu and J. Zou, Uniqueness in an inverse acoustic obstacle scattering
problem for both sound-hard and sound-soft polyhedral scatterers, preprint
(2005).
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