Manuscript submitted to
AIMS’ Journals
Volume 3, Number 2, May 2009
Website: http://AIMsciences.org
pp. 1–XX
ON THE EXISTENCE OF TRANSMISSION EIGENVALUES
Andreas Kirsch1
University of Karlsruhe
Department of Mathematics
76128 Karlsruhe
Germany
Abstract. The investigation of the far field operator and the Factorization
Method in inverse scattering theory leads naturally to the study of corresponding interior transmission eigenvalue problems. In contrast to the classical
Dirichlet- or Neumann eigenvalue problem for −∆ in bounded domains these
interior transmiision eigenvalue problem fail to be selfadjoint. In general, existence of eigenvalues is an open problem. In this paper we prove existence of
eigenvalues for the scalar Helmholtz equation (isotropic and anisotropic cases)
and Maxwell’s equations under the condition that the contrast of the scattering
medium is large enough.
1. Introduction. The relationship, among physicists sometimes called the insideoutside duality, between the eigenvalue one of the scattering matrix and the Dirichlet
eigenvalues of the negative Laplacian is well known for a long time (see, e.g. [8] and
the references therein). For this case the underlying scattering model is the exterior
boundary value problem
∆us + k 2 us = 0 in the exterior of D ,
us = −uinc on ∂D ,
for the Helmholtz equation. Here, D denotes some bounded domain (the “scatterer”), uinc the incident wave (a plane wave), and us the scattered field which has
to satisfy the Sommerfeld radiation condition. Note that the scattering problem is
set up outside of D while the eigenvalue problem is set up inside of D.
The analogous relationship for penetrable obstacles leads to new kind of eigenvalue problems in D which are formulated as a pair of equations, coupled through
the Cauchy data on the boundary ∂D. While the Dirichlet eigenvalue problem in
D is one of the best studied problems in analysis, the corresponding interior transmission eigenvalue problem is a relatively young object of research. We refer to the
original papers [6, 12, 3, 10, 4, 5, 1, 9, 13, 14] and the monographs [2, 15] for the
relevance of the transmission eigenvalue problems in acoustic and electromagnetic
scattering theory.
The recent survey paper [7] reports on the state of the art for the interior transmission problems till the end of 2007. Since the eigenvalue problem doesn’t seem to
be treatable by standard methods in partial differential equations even some of the
basic questions such as the existence of eigenvalues are still open up to today. This
question of existence was raised for the first time in [6] and only last year (in 2008)
the problem has been partially answered by Lassi Päivärinta and John Sylvester
2000 Mathematics Subject Classification. Primary: 35P15, 35J05; Secondary: 35P25, 35Q60.
Key words and phrases. Helmholtz equation, Maxwell’s equations, eigenvalue problem.
1
2
ANDREAS KIRSCH
in [17]. They prove existence of eigenvalues for the simplest model of a penetrable
obstacle (in electromagnetics would this be the E-mode) provided the contrast is
large enough.
In this paper we will extend the analysis and treat the more complicated anisotropic
cases, the H-mode in electromagnetics and the case of Maxwell’s equations. The
paper is organized as follows.
In Section 2 we recall and, in our opinion, simplify the analysis of [17]. We
adopt the notation of Section 4.5 of [15] and show discreteness of the spectrum
and existence of eigenvalues, the latter under a condition on the contrast which is
similar to the one in [17]. In Section 3 we consider the anisotropic case. Although
the discreteness of the spectrum is well known (see [3, 2]) we suggest a different – and
more direct – approach which models the approach for the first case. This approach
makes it possible to derive a condition on the contrast such that eigenvalues exist. In
Section 4 we show that the analysis carries over to the case of Maxwell’s equations.
2. The Scalar Helmholtz Equation. We make the assumption that D ⊂ R3 is
some bounded connected domain with Lipschitz boundary ∂D. The two-dimensional
case can be treated analogously. Furthermore, let q ∈ L∞ (D) be real-valued such
that q(x) ≥ q0 almost everywhere in D for some q0 > 0. In the following all of the
spaces consist of real-valued functions. This is not a restriction since otherwise one
can pass over to the real- and imaginary parts.
Definition 2.1. k 2 > 0 is called an interior transmission eigenvalue if there exists
real-valued (u, w) ∈ L2 (D) × L2 (D) with (u, w) 6= (0, 0) such that u − w ∈ H02 (D)
and
∆u + k 2 (1 + q)u = 0 in D , ∆w + k 2 w = 0 in D ,
(2.1)
u = w on ∂D ,
∂u
∂w
=
= on ∂D .
∂ν
∂ν
(2.2)
Here and in the following, ν = ν(x) denotes the exterior unit normal vector
for x ∈ ∂D and H02 (D) denotes the Sobolev space of second order with vanishing traces v and ∂v/∂ν. We equip H02 (D) with the inner product hu, viH02 (D) =
(∆u), (∆v)/q L2 (D) and corresponding norm k · kH02 (D) which is equivalent to the
ordinary norm of H02 (D) (see, e.g., [15]). We note that the traces exist in H 2 (D).
Therefore, the transmission boundary conditions (2.2) are already included in the
space H02 (D). The differential equations (2.1) have to be understood in the ultra
weak sense, i.e.
ZZ
w ∆ψ + k 2 ψ dx = 0 for all ψ ∈ H02 (D)
D
and analogously for u.
Defining v = w − u we have the equivalent form that v ∈ H02 (D) satisfies
∆v + k 2 (1 + q)v = k 2 q w
in D ,
(2.3)
and w satisfies the Helmholtz equation in D. To eliminate w from (2.3) we devide
by q and apply the Helmholtz operator again. This yields
1
(∆ + k 2 )
∆ + k 2 (1 + q) v = 0 ,
q
EXISTENCE OF TRANSMISSION EIGENVALUES
i.e. in weak form
ZZ
dx
∆v + k 2 (1 + q)v ∆ψ + k 2 ψ
= 0
q
for all ψ ∈ H02 (D) .
3
(2.4)
D
We define the bilinear form ak for k ≥ 0 by (2.4), i.e.
ZZ
dx
ak (v, ψ) =
∆v + k 2 (1 + q)v ∆ψ + k 2 ψ
(2.5)
q
D
ZZ
ZZ
dx
2
2
2
=
∆v + k v ∆ψ + k ψ
v ∆ψ + k 2 ψ dx(2.6)
+ k
q
D
D
for all ψ ∈ H02 (D). Then k 2 is an interior transmission eigenvalue if, and only if,
there exists a non-trivial v ∈ H02 (D) with ak (v, ψ) = 0 for all ψ ∈ H02 (D). Indeed,
if k 2 is an interior transmission eigenvalue with corresponding eigenpair (u, w) ∈
L2 (D) × L2 (D) then v = w − u ∈ H02 (D) solves ak (v, ψ) = 0 for all ψ ∈ H02 (D)
as we have just seen. If, on the other hand, v ∈ H02 (D) solves ak (v, ψ) = 0 for all
ψ ∈ H02 (D) then
1 w = 2 ∆v + k 2 (1 + q)v
k q
belongs to L2 (D) and satisfies
ZZ
w ∆ψ + k 2 ψ dx = 0 for all ψ ∈ H02 (D) ,
D
which is the ultra weak form of ∆w + k 2 w = 0 in D.
We write ak in the form ak = a0 + k 2 b1 + k 4 b2 where the bilinear forms b1 and b2
are given by
ZZ
ZZ
dx
+
v ∆ψ dx ,
b1 (v, ψ) =
v ∆ψ + ψ ∆v
q
D
ZDZ
q+1
v ψ dx , v, ψ ∈ H02 (D) ,
b2 (v, ψ) =
q
D
and a0 (·, ·) is just the inner product in H02 (D).
By the representation theorem of Riesz there exist bounded operators B1 , B2 from
H02 (D) into itself with
bj (v, ψ) = hBj v, ψiH02 (D)
for all v, ψ ∈ H02 (D) , j = 1, 2 .
The equation ak (v, ψ) = 0 for all ψ ∈ H02 (D) takes the form
v + k 2 B1 v + k 4 B2 v = 0 .
(2.7)
Since bj are symmetric (easy to see by Green’s second formula!) we observe that
Bj are self adjoint for j = 1, 2. Also, as shown in [15] the operators Bj are compact
1/2
and B2 is positive. Therefore, the operator B2 has a positive square root B2 :
1/2
H02 (D) → H02 (D). Setting z = k 2 B2 v we observe that (2.7) is equivalent to the
system
! 1/2
v
v
0
B1
B2
2
+ k
=
.
(2.8)
1/2
z
z
0
−B2
0
4
ANDREAS KIRSCH
This is a non self adjoint linear eigenvalue problem for a compact matrix operator.
We conclude that the spectrum is discrete but we cannot conclude existence of any
eigenvalues.
We set Ak = Id + k 2 B1 + k 4 B2 and note that the spectrum of Ak is real and
discrete with one as the only accumulation point. The operator and therefore also
the eigenvalues depend continuously on k (see, e.g., [11]). Since A0 = Id the spectrum σ(A0 ) consists of 1 only. The following arguments for showing existence of
transmission eigenvalues have been recently suggested by Päivärinta and Sylvester
in [17]. The idea is to construct some v̂ ∈ H02 (D) and some k̂ > 0 such that
hAk̂ v̂, v̂iH02 (D) = ak̂ (v̂, v̂) < 0. By the min-max principle this implies that the smallest eigenvalue of Ak̂ is negative. Therefore, since the smallest eigenvalue depends
continuously on k, there exists k between 0 and k̂ such that Ak has zero as the
smallest eigenvalue i.e., in particular, possesses an eigenvalue at all.
To carry out this idea we estimate ak (v, v) from above. From (2.6) we have
ZZ
ZZ
2
1
ak (v, v) ≤
∆v + k 2 v dx + k 2
v ∆v dx + k 4 kvk2L2 (D)
q0
D
D
ZZ
1
(1 + q0 )k 4
=
(∆v)2 + k 2 (2 + q0 ) v ∆v dx +
kvk2L2 (D)
q0
q0
D
ZZ
(1 + q0 )k 4
1
(∆v)2 − k 2 (2 + q0 ) |∇v|2 dx +
kvk2L2 (D)
=
q0
q0
D
where we applied Green’s first theorem in the last step. Let now v̂ be an eigenfunction corresponding to the smallest eigenvalue µ1 of the bi-Laplacian
∆2 , i.e.
RR
2
2
v̂ ∈RRH0 (D) satisfies ∆ v̂ = µ1 v̂ in D. Green’s second theorem yields D (∆v̂)2 dx =
µ1 D v̂ 2 dx and thus
ak (v̂, v̂) ≤
k 2 (2 + q0 )
µ1 + k 4 (1 + q0 )
kv̂k2L2 (D) −
k∇v̂k2L2 (D) .
q0
q0
Now let ρ0 be the smallest Dirichlet eigenvalue of −∆ in D. Then Poincaré’s
2
1
inequality yields kuk2L2 (D) ≤ ρ−1
0 k∇ukL2 (D) for all u ∈ H0 (D) and thus, since
H02 (D) ⊂ H01 (D),
1 ak (v̂, v̂) ≤
µ1 + k 4 (1 + q0 ) − k 2 ρ0 (2 + q0 ) k∇v̂k2L2 (D) .
ρ0 q 0
We can now easily derive a condition on q0 such the term on the right hand side is
negative. First we write (completing the square)
2
p
(1 + q0 /2)ρ0
(1 + q0 /2)2 ρ20
µ1 +k 4 (1+q0 )−k 2 ρ0 (2+q0 ) = k 2 1 + q0 − √
.
+µ1 −
1 + q0
1 + q0
We choose k 2 such that the square vanishes. Then the expression is negative if
µ1 <
(1 + q0 /2)2 ρ20
1 + q0
which can be rewritten as (note that µ1 ≥ ρ20 by, e.g., [17])
√ r
µ1 µ1
µ1
q0 > 2
−1 +
−1 .
ρ20
ρ0
ρ20
(2.9)
EXISTENCE OF TRANSMISSION EIGENVALUES
5
Therefore, for this particular choice of k̂ 2 = (1 + q0 /2)ρ0 /(1 + q0 ) we have
inf
v∈H02 (D)
ak̂ (v, v) < 0 .
Therefore, the smallest eigenvalue of Ak̂ must be negative. Since the spectrum of A0
is positive there must be some k between 0 and k̂ such that the smallest eigenvalue
of Ak is zero. This k 2 is a transmission eigenvalue!
We summarize the result in the following theorem.
Theorem 2.2. Let ρ0 > 0 be the smallest Dirichlet eigenvalue of −∆ in D and
µ1 > 0 be the smallest eigenvalue of ∆2 with respect to the boundary conditions
v = 0 on ∂D and ∂v/∂ν = 0 on ∂D. Assume that q0 satisfies (2.9). Then there
exists at least one transmission eigenvalue k 2 .
By the same arguments as in [17] one can extend this to prove existence of at
least m eigenvalues. The basis is set by the following theorem. Let Vm be the set
of linear subspaces of H02 (D) of co-dimension m.
Theorem 2.3. For any m = 1, 2, . . . define fm : R≥0 → R by
fm (k) =
sup
inf
V ∈Vm−1 v∈V, v6=0
hAk v, viH02 (D)
kvk2H 2 (D)
,
k ≥ 0.
(2.10)
0
(a) fm is continuous on [0, ∞) with fm (0) = 1.
(b) If fm (k) < 1 for some k > 0 then there exist m eigenvalues λj = λj (k), j =
1, . . . , m, of Ak less than 1 (counted according to their multiplicities), ordered
as λ1 ≤ λ2 ≤ · · · ≤ λm < 1 and fm (k) = λm . In this case the supremum is
attained for the subspace V = span{v1 , . . . , vm−1 }⊥ where vj ∈ H02 (D) are the
eigenfunctions corresponding to the eigenvalues λj .
We do not prove this result but refer to, e.g., [18].
Let now µ1 ≤ µ2 ≤ · · · ≤ µm be the m smallest eigenvalues of ∆2 with respect
to homogeneous boundary conditions v = 0 on ∂D and ∂v/∂ν = 0 on ∂D with
corresponding eigenfunctions v̂1 , . . . , v̂m ∈ H02 (D). If we choose q0 according to (2.9)
where we replace µ1 by µm then hAk v, viH02 (D) < 0 for all v ∈ span{v̂1 , . . . , v̂m }.
Since V ∩ span{v̂1 , . . . , v̂m } 6= ∅ for every V ∈ Vm−1 we conclude that fm (k̂) ≤ 0
where k̂ is again given by k̂ 2 = (1 + q0 /2)ρ0 /(1 + q0 ). Therefore, the previous
theorem is applicable which yields fm (k̂) = λm (k̂). By the continuity of fm there
exists a largest km ∈ (0, k̂] with fm (km ) = 0. Therefore, km is a transmission
eigenvalue. Let dm ∈ {1, . . . , m} be the multiplicity of the eigenvalue zero of Akm .
If dm < m then fm−dm (km ) < 0. Therefore, there exists km−dm ∈ (0, km ) with
fm−dm (km−dm ) = 0. In this way we proceed and arrive at the following corollary.
Corollary 2.4. Let µj , j = 1, 2, 3, . . ., be the eigenvalues of ∆2 with respect to
homogeneous boundary conditions as in Theorem 2.2. We assume that they are
ordered as µ1 ≤ µ2 ≤ · · · and they appear according to their multiplicity. if q0
satisfies
√ r
µm µm
µm
q0 > 2
+
−
1
−
1
ρ20
ρ0
ρ20
then there exist at least m transmission eigenvalues (counted according to their
multiplicities).
6
ANDREAS KIRSCH
3. The Anisotropic Case. We make the assumption that Q ∈ L∞ (D, R3×3 ) is
matrix-valued such that Q(x) is real and symmetric for almost all x ∈ D ⊂ R3 .
Furthermore, we assume that there exists q0 > 0 such that z > Q(x)z ≥ q0 |z|2 for
all z ∈ R3 almost everywhere on D. Again, D is a bounded and connected domain
with Lipschitz boundary. Acoustic scattering with space dependent density leads
to the following interior transmission eigenvalue problem (cf. [3, 2, 14]).
Definition 3.1. k 2 > 0 is called an interior transmission eigenvalue if there exists
1
2
real-valued (u, w) ∈ H 1 (D)
×2H (D) with (u, w) 6= (0, 0) such that ∆w + k w = 0
in D and div (I + Q)∇u + k u = 0 in D and the Cauchy data of u and v coincide,
i.e. u = w on ∂D and ν > (I + Q)∇u = ∂w/∂ν on ∂D. The variational forms are
ZZ
∇w> ∇ψ − k 2 w ψ dx = 0 for all ψ ∈ H01 (D) ,
(3.11)
D
ZZ
>
∇u (I + Q)∇ψ − k 2 u ψ dx
D
ZZ
=
∇w> ∇ψ − k 2 w ψ dx
(3.12)
D
for all ψ ∈ H 1 (D).
If (u, w) solves (3.11), (3.12) with u = w on ∂D then v := w − u ∈ H01 (D) solves
ZZ
ZZ
>
∇v (I + Q)∇ψ − k 2 vψ dx =
∇w> Q∇ψ dx for all ψ ∈ H 1 (D) . (3.13)
D
D
RR
Note that ψ = 1 yields (provided k 6= 0) that D vdx = 0, i.e.
ZZ
v ∈ H̃01 (D) := v ∈ H01 (D) :
vdx = 0 .
D
Analogously, we define the space H̃ 1 (D) as the subspace of H 1 (D) of functions with
vanishing means.
The classical form of (3.13) is
div (I + Q)∇v + k 2 v = div(Q∇w) in D , ν > (I + Q)∇v = ν > Q∇w on ∂D ,
(3.14)
and v = 0 on ∂D. The idea of Section 2 to eliminate w explicitely does not work
here. Howewer, we can express w implicitely by v. We carry out the details and
define the operator Lk from H̃01 (D) into itself as follows.
For given v ∈ H̃01 (D) let w = wv ∈ H̃ 1 (D) be the unique solution of the Neumann
problem (3.14), i.e.
ZZ
ZZ
>
∇w> Q∇ψ dx =
∇v (I + Q)∇ψ − k 2 vψ dx for all ψ ∈ H 1 (D) . (3.15)
D
D
RR
Note that the solution w = wv ∈ H̃ 1 (D) exists and is unique because of D vdx =
0. Let z = zv ∈ H̃01 (D) be the unique representation of the linear and bounded
functional
ZZ
ψ 7→
∇wv> ∇ψ − k 2 wv ψ dx , ψ ∈ H̃01 (D) ,
(3.16)
D
i.e.
ZZ
hzv , ψiH 1 (D) =
D
∇wv> ∇ψ − k 2 wv ψ dx
for all ψ ∈ H̃01 (D) .
EXISTENCE OF TRANSMISSION EIGENVALUES
7
Then we set Lk v = zv .
Theorem 3.2. (a) Let (u, w) ∈ H 1 (D) × H 1 (D) be an eigenfunction. Then v =
w − u ∈ H̃01 (D) solves Lk v = 0.
(b) Let v ∈ H̃01 (D) satisfy Lk v = 0. Furthermore, let w = wv ∈ H̃ 1 (D) be as
in the construction of Lk , i.e. the solution of (3.15). Then there exists a
constant c ∈ R such that (u, w + c) is an eigenfunction where u = w + c − v.
Proof. (a) Formula (3.11) implies hzv , ψiH01 (D) = 0 for all ψ ∈ H̃01 (D), i.e. Lk v =
zv = 0.
(b) Let, on the other hand, Lk v = 0, i.e.
ZZ
∇w> ∇ψ − k 2 wψ dx = 0 for all ψ ∈ H̃01 (D) .
D
Here and in the following we write w for wv . Note that this
RR does not imply that
w solves the Helmholtz equation because of the restriction
ψdx = 0 on the test
D
RR
1
functions. However, fix a function φ ∈ H0 (D) with D φ dx = 1. Let ψ ∈ H01 (D)
RR
be any function. Then ψ̃ = ψ −
ψ dx φ ∈ H̃01 (D) and thus
D
ZZ
0 =
∇w> ∇ψ̃ − k 2 wψ̃ dx
ZDZ
=
∇w> ∇ψ − k 2 wψ dx −
D
ZZ
ZZ
ψdx
D
D
|
ZZ
=
∇w> ∇φ − k 2 wφ dx
{z
=: k2 c
}
∇(w + c)> ∇ψ − k 2 (w + c)ψ dx .
D
This shows that w + c solves the Helmholtz equation in D. We set u := w + c − v
and observe that the Cauchy data of w + c and u coincide. Furthermore, equation
(3.12) follows from (3.15).
Therefore, the transmission eigenvalues are just the parameters k 2 for which Lk fails
to be injective. The operator Lk has the same form as the corresponding operator
of equation (2.7) as we see from the next theorem.
Theorem 3.3. (a) Lk has the form Lk = L0 + k 2 C1 + k 4 C2 with self adjoint
compact operators Cj from H̃01 (D) into itself.
(b) L0 is self adjoint and coercive on H̃01 (D), in particular
hL0 v, viH 1 (D) ≥ k∇vk2L2 (D) ≥ c kvk2H 1 (D)
for all v ∈ H̃01 (D)
where c > 0 is independent of v.
Proof. (a) From the definition of w = wv ∈ H̃ 1 (D) we observe that w has the form
w = w1 − k 2 w2 where w1 , w2 ∈ H̃ 1 (D) solve
ZZ
ZZ
∇w1> Q∇ψ dx =
∇v > (I + Q)∇ψ dx for all ψ ∈ H 1 (D) ,
D
ZZ
D
D
∇w2> Q∇ψ dx
ZZ
=
D
vψ dx for all ψ ∈ H 1 (D) .
8
ANDREAS KIRSCH
Substituting w = w1 − k 2 w2 into the form of the functional (3.16) yields the form
Lk = L0 + k 2 C1 + k 4 C2 . We show that Lk is symmetric for every k ≥ 0. Then also
C1 and C2 are symmetric as the first and second derivative, respectively, of Lk with
respect to k 2 at zero.
For v1 , v2 ∈ H̃01 (D) we conclude
ZZ
hLk v1 , v2 iH 1 (D) =
∇w1> ∇v2 − k 2 w1 v2 dx
D
ZZ
=
∇w1> (I + Q)∇v2 − k 2 w1 v2 dx −
ZZ
D
∇w1> Q∇v2 dx
D
where wj = wvj . Now we use (3.15) twice: First for v = v2 , ψ = w1 , then for
w = w1 , ψ = v2 . This yields
ZZ
ZZ
>
∇v1 (I +Q)∇v2 −k 2 v1 v2 dx , (3.17)
∇w2> Q∇w1 dx −
hLk v1 , v2 iH 1 (D) =
D
D
and this is a symmetric expression in v1 and v2 .
The compactness of the operators C1 and C2 is easily seen by the compactness of the
imbedding of H̃01 (D) in L2 (D). We omit this proof but carry out the corresponding
proof for the – slightly more complicated – electromagnetic case in Theorem 4.4
below.
(b) For k = 0 and v1 = v2 = v equation (3.17) reduces to
ZZ
ZZ
∇v > (I + Q)∇v dx .
(3.18)
∇w> Q∇w dx −
hL0 v, viH 1 (D) =
D
D
Now we make use of the fact that, for almost all x ∈ D, there exists a unique
positive definite matrix Q1/2 ∈ L∞ (D, R3×3 ) with Q1/2 Q1/2 = Q. From (3.15) for
ψ = v and k = 0 we estimate
ZZ
ZZ
∇v > (I + Q)∇v dx =
∇w> Q∇v dx ≤ kQ1/2 ∇wkL2 (D) kQ1/2 ∇vkL2 (D)
D
D
(3.19)
and thus
hL0 v, viH 1 (D)
kQ1/2 ∇wk2L2 (D) − kQ1/2 ∇wkL2 (D) kQ1/2 ∇vkL2 (D)
= kQ1/2 ∇wkL2 (D) kQ1/2 ∇wkL2 (D) − kQ1/2 ∇vkL2 (D)
≥
=
kQ1/2 ∇wkL2 (D)
·
kQ1/2 ∇wkL2 (D) + kQ1/2 ∇vkL2 (D)
|
{z
}
=:c(v)
· kQ1/2 ∇wk2L2 (D) − kQ1/2 ∇vk2L2 (D)
ZZ
ZZ
>
>
= c(v)
∇w Q∇w − ∇v (I + Q)∇v dx + c(v)
|∇v|2 dx
D
D
|
{z
=hL0 v,viH 1 (D)
}
and thus
1 − c(v) hL0 v, viH 1 (D) ≥ c(v) k∇vk2L2 (D) .
EXISTENCE OF TRANSMISSION EIGENVALUES
9
We note that c(v) < 1 for v 6= 0, thus
hL0 v, viH 1 (D) ≥
c(v)
k∇vk2L2 (D) .
1 − c(v)
We show that c(v) ≥ 12 . From (3.19) we conclude that
ZZ
ZZ
kQ1/2 ∇vk2L2 (D) =
∇v > Q∇v dx ≤
∇v > (I + Q)∇v dx
D
≤
D
kQ1/2 ∇wkL2 (D) kQ1/2 ∇vkL2 (D) ,
i.e. kQ1/2 ∇vkL2 (D) ≤ kQ1/2 ∇wkL2 (D) , i.e. c(v) ≥ 21 . Finally, we note that v 7→
k∇vkL2 (D) is an equivalent norm on H̃01 (D) by Poincaré’s inequality which proves
the theorem.
We write the equation Lk v = 0 equation in the form
−1/2
ṽ + k 2 L0
−1/2
C1 L0
−1/2
ṽ + k 4 L0
−1/2
C2 L0
ṽ = 0
and
ṽ + k 2 B1 ṽ + k 4 B2 ṽ = 0
1/2
with obvious settings of B1 and B2 . Here, L0 denotes the (coercive) square root
1/2
of the coercive operator L0 and ṽ = L0 v. As in (2.8) of the previous section
we rewrite this as a linear eigenvalue system with a compact matrix operator and
conclude the following theorem:
Theorem 3.4. There exists at most a countable set of transmission eigenvalues,
and the only possible accumulation point is infinity.
Again, we want to show existence of some k > 0 and some v 6= 0 such that
Lk v = 0. Since the spectrum of Id + k 2 B1 + k 4 B2 for k = 0 is just {1} we follow
again the idea of the previous section and show that, for sufficiently large values of
q0 , there exists k̂ > 0 and v̂ 6= 0 such that hLk̂ v̂, v̂iH 1 (D) ≤ 0. Then,
1/2
1/2
1/2 (L0 v̂) + k̂ 2 B1 (L0 + k̂ 4 B2 v̂) , (L0 v̂) H 1 (D) ≤ 0 .
Therefore, by continuity of the smallest eigenvalue of Id + k 2 B1 + k 4 B2 with respect
to k there exists k between 0 and k̂ such that the smallest eigenvalue of Id + k 2 B1 +
k 4 B2 is zero.
To carry out this idea we have to estimate hLk v, viH 1 (D) from above. We have from
(3.17) for v1 = v2 = v
ZZ
ZZ
>
>
hLk v, viH 1 (D) =
∇w Q∇w dx −
∇v (I + Q)∇v − k 2 v 2 dx .
D
D
We set again u = w − v and note that u solves the Neumann problem (in the weak
sense)
∂v
div Q∇u = ∆v + k 2 v in D , ν > Q ∇u =
on ∂D ,
(3.20)
∂ν
10
ANDREAS KIRSCH
and thus
ZZ
hLk v, viH 1 (D)
=
(∇v + ∇u)> Q(∇v + ∇u) dx
D
ZZ
−
>
∇v (I + Q)∇v − k 2 v 2 dx
D
= −k∇vk2L2 (D) + 2 Q∇v, ∇u L2 (D) + kQ1/2 ∇uk2L2 (D) +
+ k 2 kvk2L2 (D) .
From the weak form of (3.20) we conclude that
hQ∇v, ∇uiL2 (D) = k∇vk2L2 (D) − k 2 kvk2L2 (D) ,
and
hLk v, viH 1 (D) = k∇vk2L2 (D) − k 2 kvk2L2 (D) + kQ1/2 ∇uk2L2 (D) .
We estimate the last term. Again, from the weak form of (3.20) we have that
ZZ
ZZ
>
2
Q |∇u| dx =
∇v ∇u − k 2 vu dx
D
D
k∇vkL2 (D) k∇ukL2 (D) + k 2 kvkL2 (D) kukL2 (D) .
≤
Let ρ0 > 0 be such that kψkL2 (D) ≤ ρ0 k∇ψkL2 (D) for all ψ ∈ H̃ 1 (D). Then
ZZ
Q |∇u|2 dx ≤ k∇ukL2 (D) k∇vkL2 (D) + k 2 ρ0 kvkL2 (D)
D
≤
1
√ kQ1/2 ∇ukL2 (D) k∇vkL2 (D) + k 2 ρ0 kvkL2 (D) ,
q0
i.e. after division of kQ1/2 ∇ukL2 (D) and squaring,
ZZ
2
1
k∇vkL2 (D) + k 2 ρ0 kvkL2 (D)
Q |∇u|2 dx ≤
q0
D
≤
2
k∇vk2L2 (D) + k 4 ρ20 kvk2L2 (D) .
q0
Altogether we have the estimate
hLk v, viH 1 (D)
ρ2
2
k∇vk2L2 (D) + 2k 4 0 kvk2L2 (D)
k∇vk2L2 (D) − k 2 kvk2L2 (D) +
q0
q0
2 2
2
2k
ρ
0
=
1+
k∇vk2L2 (D) − k 2 1 −
kvk2L2 (D) .
q0
q0
≤
Now let µ be an eigenvalue with corresponding eigenfunction v̂ ∈ H̃01 (D) of the
eigenvalue problem
ZZ
∇v∇ψ − µvψ dx = 0 for all ψ ∈ H̃01 (D) .
(3.21)
D
It is easily seen that such eigenvalues µj exist, that 0 is not an eigenvalue and that
they are real and positive and converge to infinity. Taking ψ = v̂ we conclude that
ZZ
ZZ
|∇v̂|2 dx = µ
v̂ 2 dx .
D
D
EXISTENCE OF TRANSMISSION EIGENVALUES
11
Substituting v = v̂ in the previous estimate we conclude that
2
2k 4 ρ20
2
1+
hLk v̂, v̂iH 1 (D) ≤
µ−k +
kv̂k2L2 (D) .
(3.22)
q0
q0
Now we derive a condition on q0 such that Lk v̂, v̂ H 1 (D) ≤ 0 for some k. Indeed,
0
we multiply the bracket [· · · ] by q0 and rewrite this as
(q0 + 2)µ + 2k 4 ρ20 − k 2 q0
2 √ 2
q0
q02
2 k ρ0 − √
+ (q0 + 2)µ −
.
=
8 ρ20
2 2 ρ0
Now we determine q0 large enough such that the second bracket is non-positive, i.e.
q0 [· · · ]
=
q02
.
8 ρ20
Then we choose k such that the first bracket vanishes, i.e.
q0
.
k̂ 2 =
4 ρ20
Then we have that Lk̂ v̂, v̂ H 1 (D) ≤ 0.
(q0 + 2)µ ≤
(3.23)
We summarize the result in the following theorem.
Theorem 3.5. Let ρ0 > 0 be a constant such that an estimate of Poincaré’s type
holds for the space H̃ 1 (D), i.e. kψkL2 (D) ≤ ρ0 k∇ψkL2 (D) for all ψ ∈ H̃ 1 (D).
Furthermore, let µ > 0 be some eigenvalue of the eigenvalue problem (3.21). Assume
that q0 satisfies (3.23). Then there exists at least one transmission eigenvalue k 2 .
Again, we can extend this to prove existence of at least m eigenvalues:
Corollary 3.6. Let µj , j = 1, 2, 3, . . ., be the eigenvalues of (3.21). We assume that
they are ordered as µ1 ≤ µ2 ≤ · · · and they appear according to their multiplicity.
If q0 satisfies
q02
(q0 + 2)µm ≤
8 ρ20
then there exist at least m transmission eigenvalues (counted according to their
multiplicities).
Remark 1. The classical formulation of the eigenvalue problem (3.21) is: Find
v ∈ H 1 (D) such that
ZZ
∆v + µv = const in D , v = 0 on ∂D ,
v dx = 0 .
D
4. Maxwell’s Equations. We make again the assumption that Q ∈ L∞ (D, R3×3 )
is matrix-valued such that Q(x) is real and symmetric for almost all x ∈ D. Furthermore, we assume that there exists q0 > 0 such that z > Q(x)z ≥ q0 |z|2 for all
z ∈ R3 almost everywhere on D. Again, D is a bounded and connected domain
with Lipschitz boundary. We consider scattering of time-harmonic electromagnetic
waves in non-magnetic materials.
We assume that the reader is familiar with the standard spaces in this context.
The space H(curl, D) is defined as the completion of C ∞ (D)3 with respect to the
norm
q
kukH(curl,D) =
hu, uiH(curl,D)
12
ANDREAS KIRSCH
where
ZZ
hu, viH(curl,D) =
curl u> curl v + u> v dx .
D
The subspace of vanishing tangential traces is denoted by H0 (curl, D), i.e.
H0 (curl, D) = u ∈ H(curl, D) : ν × u = 0 on ∂D} .
The trace is well defined, see e.g. [16]. The study of the Factorization Method or
already the question of uniqueness of the far field operator leads to the following
interior transmission eigenvalue problem (see [5, 9, 13, 15]).
Definition 4.1. k 2 > 0 is called an interior transmission eigenvalue if there exists
2
real-valued (u, w) ∈ H(curl, D)×H(curl,
D)2with (u, w) 6= (0, 0) such that curl w −
2
k w = 0 in D and curl (I + Q) curl u − k u = 0 in D and the Cauchy data of u
and v coincide, i.e. ν × u = ν × w on ∂D and ν × (I + Q) curl u = ν × curl w on
∂D. The variational forms are
ZZ
curl w> curl ψ − k 2 w> ψ dx = 0 for all ψ ∈ H0 (curl, D) ,
(4.24)
D
ZZ
curl u> (I + Q) curl ψ − k 2 u> ψ dx
D
ZZ
=
curl w> curl ψ − k 2 w> ψ dx(4.25)
D
for all ψ ∈ H(curl, D).
We define again the difference v = w − u and observe that v ∈ H0 (curl, D)
satisfies the equation
curl (I + Q) curl v − k 2 v = curl(Q curl w)
in D ,
ν × (I + Q) curl v = ν × (Q curl w)
on ∂D ,
i.e. in variational form
ZZ
ZZ
curl v > (I + Q) curl ψ − k 2 v > ψ dx =
curl w> Q curl ψ dx
D
(4.26)
D
for all ψ ∈ H(curl, D). By setting ψ = ∇ρ for some ρ ∈ H 1 (D) we note from this
equation that v ∈ V where


ZZ


V = v ∈ H0 (curl, D) :
v > ∇ρ dx = 0 for all ρ ∈ H 1 (D)


D
is the space of H(curl, D)−functions with vanishing normal and tangential traces
which are divergence free. Indeed, RR
for smooth functions
RR the integralRcan be written,
using the divergence theorem, as D v > ∇ρ dx = D ρ divv dx − ∂D ρ ν > v ds. If
this vanishes for all ρ ∈ H 1 (D) then the divergence of v vanishes in D and the
normal component of v vanishes on ∂D.
Analogously, we define


ZZ


W = w ∈ H(curl, D) :
w> ∇ρ dx = 0 for all ρ ∈ H 1 (D) .


D
EXISTENCE OF TRANSMISSION EIGENVALUES
13
It will later be necessary to know the orthogonal complement of V in H0 (curl, D).
For any ϕ ∈RRH 1 (D) let vϕ ∈ H0 (curl, D) be the Riesz representation of the functional ψ 7→ D ∇ϕ> ψ dx, i.e.
ZZ
∇ϕ> ψ dx for all ψ ∈ H0 (curl, D) , ϕ ∈ H 1 (D) .
hvϕ , ψiH(curl,D) =
D
Lemma 4.2. Let V ⊂ H0 (curl, D) be defined above. Then V is a closed subspace
with orthogonal complement
V ⊥ = vϕ ∈ H0 (curl, D) : ϕ ∈ H 1 (D) .
Proof. It is obvious that V is a closed subspace of H0 (curl, D). Let Ṽ be the space
on the right hand side of the characterization of V > . Then Ṽ and V are orthogonal
to each other. Indeed, for v ∈ V and vϕ ∈ Ṽ we have that (take ψ = v in the
definition of vϕ ):
ZZ
∇ϕ> v dx
hvϕ , viH(curl,D) =
D
and this vanishes by the definition of V .
Let now v ∈ H0 (curl, D) be orthogonal to Ṽ . Then
ZZ
0 = hv, vϕ iH(curl,D) =
∇ϕ> v dx for all ϕ ∈ H 1 (D) .
D
Therefore, v ∈ V . This ends the proof of the lemma.
Now we define the operator Lk from V into itself in the same way as in the previous
section. First we observe that the bilinear form
ZZ
p(w, ψ) =
curl w> Q curl ψ dx , w, ψ ∈ W ,
D
is coercive on W . This follows again from Corollary 3.51 of [16]. Therefore, for
every v ∈ V there exists a unique w = wv ∈ W such that
ZZ
ZZ
curl wv> Q curl ψ dx =
curl v > (I + Q) curl ψ − k 2 v > ψ dx
(4.27)
D
D
for all ψ ∈ W . Again, let z = zv ∈ V be the unique representation of the linear and
bounded functional
ZZ
ψ 7→
curl wv> curl ψ − k 2 wv> ψ dx , ψ ∈ V ,
(4.28)
D
i.e.
ZZ
hzv , ψiH(curl,D) =
curl wv> curl ψ − k 2 wv> ψ dx
for all ψ ∈ V .
D
Then we set Lk v = zv .
Analogously to Theorem 3.2 we can show:
Theorem 4.3. (a) Let (u, w) be a transmission eigenfunction corresponding to
k. Then v = w − u ∈ V solves Lk v = 0.
14
ANDREAS KIRSCH
(b) Let v ∈ V satisfy Lk v = 0. Furthermore, let w = wv ∈ W be as in the
1
construction of Lk , i.e.
the solution of (4.27). Then there exists ϕ ∈ H (D)
such that u, w + ∇ϕ is an eigenfunction where u = w + ∇ϕ − v.
Proof. Part (a) has been shown during the derivation of the operator Lk .
(b) Let v ∈ V such that Lk v = 0, i.e.
ZZ
curl w> curl ψ − k 2 w> ψ dx = 0 for all ψ ∈ V ,
(4.29)
D
where w = wv ∈ W is determined from (4.27). Let ẑ ∈ H0 (curl, D) be the Riesz
representation of the functional
ZZ
ψ 7→
curl w> curl ψ − k 2 w> ψ dx
D
on the space H0 (curl, D), i.e.
ZZ
hẑ, ψiH(curl,D) =
curl w> curl ψ − k 2 w> ψ dx
for all ψ ∈ H0 (curl, D) .
D
Equation (4.29) implies that ẑ ∈ V ⊥ . From Lemma 4.2 we conclude that there
exists ϕ ∈ H 1 (D) such that ẑ = vϕ , i.e.
ZZ
curl w> curl ψ − k 2 w> ψ dx = hẑ, ψiH(curl,D) = hvϕ , ψiH(curl,D)
D
ZZ
=
∇ϕ> ψ dx for all ψ ∈ H0 (curl, D) ,
D
i.e.
ZZ
>
> curl w+∇ϕ/k 2 curl ψ−k 2 w+∇ϕ/k 2 ψ dx = 0
for all ψ ∈ H0 (curl, D) .
D
This shows that w̃ := w + ∇ϕ/k 2 satisfies (4.24). Furthermore, from the definition
of w we conclude that w̃ and v satisfy (4.26) for all ψ ∈ W . It remains to show (4.26)
for all ψ ∈ H(curl, D). Therefore, let ψ ∈ H(curl, D). By the classical Helmholtz
decomposition there exists ψ̃ ∈ W and ρ ∈ H 1 (D) such that ψ = ψ̃ + ∇ρ. Since
(4.26) holds for test functions of the form ∇ρ trivially (note that v ∈ V !) we
conclude that w̃ and v satisfy (4.26) for all ψ ∈ H(curl, D) which is equivalent to
(4.25) for w̃ and u := w̃ − v.
Again, the transmission eigenvalues are just the parameters k 2 for which Lk fails
to be injective. Now we continue with the investigation of Lk . Analogously to
Theorem 3.3 one can show:
Theorem 4.4. (a) Lk has the form Lk = L0 + k 2 C1 + k 4 C2 with self adjoint
compact operators Cj from V into itself.
(b) L0 is self adjoint and coercive on V , in particular
hL0 v, viH(curl,D) ≥ k curl vk2L2 (D) ≥ c kvk2H(curl,D)
where c > 0 is independent of v.
for all v ∈ V
EXISTENCE OF TRANSMISSION EIGENVALUES
15
Proof. We write w = wv in the form w = w1 − k 2 w2 where w1 , w2 ∈ W satisfy
ZZ
ZZ
curl w1> Q curl ψ dx =
curl v > (I + Q) curl ψ dx , ψ ∈ W , (4.30)
D
D
ZZ
ZZ
curl w2> Q curl ψ dx =
D
v > ψ dx ,
ψ∈W.
(4.31)
D
Then z = zv has the form z = z0 + k 2 z1 + k 4 z2 where z0 , z1 , z2 ∈ V satisfy
ZZ
hz0 , ψiH(curl,D) =
curl w1> curl ψ dx , ψ ∈ V ,
D
ZZ
hz1 , ψiH(curl,D)
= −
curl w2> curl ψ + w1> ψ dx ,
ψ∈V ,
D
ZZ
hz2 , ψiH(curl,D)
=
w2> ψ dx ,
ψ∈V .
D
We have to show that v 7→ z1 and v 7→ z2 are compact in V . We show this only for
v 7→ z1 .
First we estimate
kz1 k2H(curl,D)
≤
k curl w2 kL2 (D) k curl z1 kL2 (D) + kw1 kL2 (D) kz1 kL2 (D)
≤ kz1 kH(curl,D) k curl w2 kL2 (D) + kw1 kL2 (D) ,
i.e.
kz1 kH(curl,D) ≤ k curl w2 kL2 (D) + kw1 kL2 (D) .
¿From (4.31) we conclude for ψ = w2
1
1
k curl w2 k2L2 (D) ≤
kQ1/2 curl w2 k2L2 (D) =
hv, w2 iL2 (D)
q0
q0
1
≤
kvkL2 (D) kw2 kL2 (D)
q0
i.e.
q
1 q
kvkL2 (D) kw2 kL2 (D) + kw1 kL2 (D) .
kz1 kH(curl,D) ≤ √
q0
Let now (vj ) converge weakly to zero in V . Denote the corresponding functions w1
and w2 by w1,j and w2,j , respectively. Since the solution operators of (4.30) and
(4.31) are bounded they converge weakly to zero in W . Now we use that W is compactly imbedded in L2 (D)3 . We refer to Corollary 3.51 of [16] for a proof. Therefore,
kw1,j kL2 (D) and kw2,j kL2 (D) converge to zero which implies kz1,j kH(curl,D) → 0, and
the compactness of C1 has been shown.
The self adjointness of L0 , C1 , and C2 as well as the coercivity of L0 are shown
in the same way as in the proof of Theorem 3.3. The constant c in part (b) exists because the norm k curl ψkL2 (D) is equivalent to the norm kψkH(curl,D) in the
subspace V .
Again, we write the equation Lk v = 0 in the form
−1/2
ṽ + k 2 L0
−1/2
C1 L0
−1/2
ṽ + k 4 L0
−1/2
C2 L0
i.e.
ṽ + k 2 B1 ṽ + k 4 B2 ṽ = 0
ṽ = 0
16
ANDREAS KIRSCH
with obvious settings of B1 and B2 . As in (2.8) of Section 2 we rewrite this as a
linear eigenvalue system with a compact matrix operator and conclude the following
theorem:
Theorem 4.5. There exists at most a countable set of transmission eigenvalues,
and the only possible accumulation point is infinity.
Now we continue with the proof that for sufficienly large q0 > 0 eigenvalues indeed
exist. We follow the same lines as in the previous section and write
ZZ
hLk v, viH(curl,D) = hzk , viH(curl,D) =
curl w> curl v − k 2 w> v dx
D
ZZ
=
curl w> (I + Q) curl v − k 2 w> v dx
D
ZZ
curl w> Q curl v dx
−
D
ZZ
=
curl w> Q curl w dx
D
ZZ
−
curl v > (I + Q) curl v − k 2 |v|2 dx
D
where we have used (4.27) twice (for ψ = w and for ψ = v). Here, w ∈ W is the
unique solution of (4.27).
We set again u = w − v and note that u solves the Neumann problem (in the weak
sense)
curl Q curl u = curl2 v − k 2 v in D , ν > Q curl u = ν > curl v on ∂D , (4.32)
and thus
ZZ
hLk v, viH(curl,D)
=
(curl v + curl u)> Q(curl v + curl u) dx
D
ZZ
−
curl v > (I + Q) curl v − k 2 |v|2 dx
D
= −k curl vk2L2 (D) + 2 Q curl v, curl u L2 (D)
+ kQ1/2 curl uk2L2 (D) + k 2 kvk2L2 (D) .
¿From the weak form of (4.32) we conclude that
hQ curl v, curl uiL2 (D) = k curl vk2L2 (D) − k 2 kvk2L2 (D) ,
i.e.
hLk v, viH(curl,D) = k curl vk2L2 (D) − k 2 kvk2L2 (D) + kQ1/2 curl uk2L2 (D) .
We estimate again the last term. From the weak form of (4.32) we have
kQ1/2 curl uk2L2 (D)
= hcurl v, curl uiL2 (D) − k 2 hv, uiL2 (D)
≤
≤
k curl vkL2 (D) k curl ukL2 (D) + k 2 kvkL2 (D) kukL2 (D)
k curl ukL2 (D) k curl vkL2 (D) + k 2 ρ0 kvkL2 (D)
EXISTENCE OF TRANSMISSION EIGENVALUES
17
where ρ0 > 0 is such that kψkL2 (D) ≤ ρ0 k curl ψkL2 (D) for all ψ ∈ W . Such a
constant exists by Corollary 3.51 of [16]. Therefore,
1
kQ1/2 curl uk2L2 (D) ≤ √ kQ1/2 curl ukL2 (D) k curl vkL2 (D) + k 2 ρ0 kvkL2 (D) ,
q0
and thus
kQ1/2 curl uk2L2 (D) ≤
2
k curl vk2L2 (D) + k 4 ρ20 kvk2L2 (D) .
q0
Altogether we have
hLk v, viH(curl,D)
≤
k curl vk2L2 (D) − k 2 kvk2L2 (D) +
2
k curl vk2L2 (D)
q0
2k 4 ρ20
+
kvk2L2 (D)
q0
2
2k 2 ρ20
2
2
=
1+
k curl vkL2 (D) − k 1 −
kvk2L2 (D) .
q0
q0
Now let µ be an eigenvalue with corresponding eigenfunction v̂ ∈ V , v̂ 6= 0, of the
eigenvalue problem
ZZ
curl v curl ψ − µvψ dx = 0 for all ψ ∈ V .
(4.33)
D
It is easily seen that such eigenvalues µj exist, that 0 is not an eigenvalue and that
they are real and positive and converge to infinity. Taking ψ = v̂ we conclude that
ZZ
ZZ
2
v̂ 2 dx .
| curl v̂| dx = µ
D
D
Substituting v = v̂ in the previous estimate we conclude that
2
2k 4 ρ20
hLk v̂, v̂iH(curl,D) ≤
1+
µ − k2 +
kv̂k2L2 (D) .
q0
q0
(4.34)
The bracket [· · · ] has exactly the form of the bracket in estimate (3.22) of the
previous section. Therefore, under the condition (3.23) and the choice
q0
k2 =
4ρ20
we have that Lk v̂, v̂ H(curl,D) ≤ 0.
We summarize the result in the following theorem.
Theorem 4.6. Let ρ0 > 0 be a constant such that an estimate of Poincare’s type
holds for the space W , i.e. kψkL2 (D) ≤ ρ0 k curl ψkL2 (D) for all ψ ∈ W . Furthermore, let µ > 0 be some eigenvalue of the eigenvalue problem (4.33). Assume that
q0 satisfies (3.23). Then there exists at least one transmission eigenvalue k 2 .
Again, the same perturbation arguments as for Corollary 3.6 show existence of
at least m eigenvalues:
Corollary 4.7. Let µj , j = 1, 2, 3, . . ., be the eigenvalues of (4.33). We assume that
they are ordered as µ1 ≤ µ2 ≤ · · · and they appear according to their multiplicity.
If q0 satisfies
q02
(q0 + 2)µm ≤
8ρ20
18
ANDREAS KIRSCH
then there exist at least m transmission eigenvalues (counted according to their
multiplicities).
Remark 2. The classical form of the eigenvalue problem (4.33) is to find µ and v
and ϕ such that
curl2 v − µv = ∇ϕ in D ,
divv = 0 in D ,
ν × v = 0 on ∂D ,
ν > v = 0 on ∂D .
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