The inside–outside duality for scattering problems by inhomogeneous media

Interior transmission eigenvalue problems occur in the study of scattering problems of plane time-harmonic fields by an inhomogeneous medium filling a bounded domain D. In scattering theory, they play about the same role as the eigenvalue problem for −Δ in D with respect to, e.g., Dirichlet boundary conditions for the scattering problem by a sound-soft obstacle. We refer to [13] for their first appearance and to [4, 5, 9, 15, 18] for an incomplete list of references. Moreover, interior transmission eigenvalue problems are interesting objects of research in themselves because they fail to be self-adjoint. There are a few results for the corresponding inverse spectral problem, see [1, 2, 17], where the task is to recover information about the index of refraction from knowledge of the interior transmission eigenvalues. From the point of view of inverse scattering problems one would like to recover this information from the far field patterns of scattered waves or from the far field operator F, the linear integral operator whose kernel is formed by the far field patterns. It has been known for a long time that the injectivity of the far field operator F corresponding to the scattering by an inhomogeneous medium D is assured if the wavenumber k is not an eigenvalue of the corresponding interior transmission eigenvalue problem in D. (This is actually easy to prove, see theorem 2.3 below.) In other words, the scattering of a superposition of plane waves of the form

$$\begin{eqnarray*} &&u^i(x)= \int _{|\hat{\theta }|=1}g(\hat{\theta })\, {\rm e}^{{\rm i}k x\cdot \hat{\theta }} \,{\rm d}s(\hat{\theta }) ,\quad x\in \mathbb {R}^3 , \end{eqnarray*}$$

cannot result in a vanishing scattered field u^s if the wave number k is not an eigenvalue of the corresponding interior transmission eigenvalue problem in D. Therefore, there exists a relationship between the injectivity of the far field operator F—a property of the exterior of D—and the eigenvalues of an eigenvalue problem—a property of the interior of D. This relationship is sometimes called the inside–outside duality in scattering theory (see, e.g., [11, 12]) and has been used to determine the interior transmission eigenvalues from the far field patterns. The computation of these eigenvalues as those wavenumbers for which the far field operator fails to be injective suffers—at least theoretically—from the fact that injectivity is only a necessary but not a sufficient condition for k being no eigenvalue. It has been shown in [19] that it is very well possible that k is an eigenvalue but at the same time F is injective. Indeed, in theorem 2.3 below we recall a result which states that non-injectivity of the far field operator F for some k is equivalent to the fact that k is an interior transmission eigenvalue plus an additional condition on the corresponding eigenfunctions. In [19] it has been shown for D being a rectangular box that the far field operator is always one-to-one. In this case it is hence—at least theoretically—not possible to determine interior transmission eigenvalues merely by checking the injectivity of far field operators in an interval of frequencies. It is the goal of this paper to weaken the condition of non-injectivity of the far field operator F for wavenumber k into a condition which is exactly equivalent to k being an interior transmission eigenvalue.

In our proof we follow the paper [12] in which the scattering by an impenetrable, sound-soft obstacle has been studied. We (AK and AL) were not aware of this paper until recently and want to point out that this impressive work already contains the factorization of the far field operator in exactly the same form as in [14]³. The notation and partly also the analytical techniques this paper uses differ from standard notation and tools used in the community interested in mathematical inverse scattering theory. This might be a possible explanation why the paper [12] and its results apparently remained largely unknown in this community.

The approach presented below strongly relies on the fact that the far field operator is normal and does not straightforwardly extend to, e.g., absorbing media. Analogous results for different scattering problems leading to normal far field operators will be presented in forthcoming papers.

The paper is organized as follows. In section 2 we introduce the scattering problem by an inhomogeneous medium and the corresponding far field operator F and scattering operator $\mathcal {S}$. We recall properties of these operators, prove the mentioned relationship between non-injectivity of F to the interior transmission eigenvalue problem and recall the factorization of F as in [16]. In section 3 we characterize the interior transmission eigenvalues by the operator T which appears in the factorization of F. Section 4 studies the phases δ_n of the eigenvalues μ_n of the scattering operator $\mathcal {S}$. The eigenvalues lie on the unit circle in $\mathbb {C}$ because of the unitarity of $\mathcal {S}$. We recall from [16] that μ_n tends to 1 from 'above' or 'below' as n → ∞, depending on the sign of the contrast. We translate this into a condition on the phases δ_n and prove a characterization of the 'first' eigenvalue⁴ using the Cayley transform and Courant's maximum–minimum principle. Finally, in sections 5 and 6 we consider this 'first' eigenvalue as a function of the wavenumber k and show that it tends to 1, too, as k approaches an interior eigenvalue k₀. However, roughly speaking, this 'first' eigenvalue approaches 1 from 'below' or 'above' if μ_n approaches 1 from 'above' or 'below' (see our main theorem 6.3 for a precise formulation). We also give the corresponding result for the far field operator F instead of the scattering operator $\mathcal {S}$ (see corollary 6.4) and finish with an illustrative numerical example for the case of scattering from a penetrable ball.

We will model time-harmonic scattering from an inhomogeneous medium by the Helmholtz equation with a refractive index n that is equal to one outside the scattering object D (compare equation (2.2) below). This motivates to introduce the contrast function q := n² − 1. Our general assumptions on the contrast and on the scatterer D in the entire paper are as follows.

Assumption 2.1. Let $D\subset \mathbb {R}^3$ be open and bounded such that the complement $\mathbb {R}^3\setminus \overline{D}$ is connected. Furthermore, we assume that the boundary ∂D of D is smooth enough such that the imbedding of H¹(D) into L²(D) is compact. Let q ∈ L^∞(D) be real-valued and satisfy

• (1)  
  There exists c₀ > 0 with 1 + q(x) ⩾ c₀ for almost all x ∈ D.
• (2)  
  Either q(x) > 0 for almost all x ∈ D or q(x) < 0 for almost all x ∈ D.
• (3)  
  |q| is locally bounded below, i.e. for every compact subset M⊂D there exists c > 0 (depending on M) such that

  $$\begin{equation} |q(x)|\ \ge \ c\qquad \mbox{ for almost all }x\in M. \end{equation} \tag{ 2.1 }$$

We extend q by zero outside of D.

We note that part (3) of this assumption is satisfied for continuous contrasts q that vanish at most on the boundary of D.

By k > 0 we denote the wavenumber. For any incident plane wave $u^i(x)=\exp ({\rm i}k\hat{\theta }\cdot x)$ of direction $\hat{\theta }\in S^2$, the (direct) scattering problem is to determine the scattered field $u^s\in H^1_{\rm loc}(\mathbb {R}^3)$ such that the total field u := u^s + uⁱ satisfies the Helmholtz equation

$$\begin{equation} \Delta u(x) + k^2 (1+q(x) )\,u(x)= 0, \quad \mbox{for } x \in \mathbb {R}^3, \end{equation} \tag{ 2.2 }$$

and such that u^s satisfies the Sommerfeld radiation condition

$$\begin{equation} \frac{\partial u^s(x)}{\partial r} - {\rm i}ku^s(x)= \mathcal {O} (r^{-2} ), \qquad r=|x|\rightarrow \infty . \end{equation} \tag{ 2.3 }$$

From now on, we call any solution v of the Helmholtz equation Δv + k²v = 0 outside some ball containing D that satisfies the radiation condition a radiating solution. If appropriate, we indicate the dependence of all fields on the incident direction by writing $u^i(x,\hat{\theta })$, $u^s(x,\hat{\theta })$ and $u(x,\hat{\theta })$.

It is well known that this direct scattering problem is uniquely solvable (see, e.g., [7] or lemma 2.4 below). Furthermore, any radiating solution v of the Helmholtz equation has the asymptotic behavior

$$\begin{equation} v(x)= \frac{\exp ({\rm i}k|x|)}{4\pi |x|}\,v^\infty (\hat{x}) + \mathcal {O}(|x|^{-2}),\quad |x|\rightarrow \infty , \end{equation} \tag{ 2.4 }$$

uniformly with respect to $\hat{x}=x/|x|\in S^2$. The complex valued function v^∞ is called the far field pattern. In the special case where v is the scattered field u^s, the far field pattern depends on the direction $\hat{x}$ of observation and the direction $\hat{\theta }$ of the incident plane wave. We indicate this dependence by writing $u^\infty (\hat{x},\hat{\theta })$. Furthermore we define the far field operator F from L²(S²) into itself as the integral operator whose kernel is this far field pattern; that is,

$$\begin{equation} (Fg)(\hat{x})= \int _{S^2}g(\hat{\theta })\,u^\infty (\hat{x},\hat{\theta })\,{\rm d}s(\hat{\theta }),\quad \hat{x}\in S^2. \end{equation} \tag{ 2.5 }$$

The far field operator F is closely related to the scattering operator (or scattering matrix) $\mathcal {S}$, namely

$$\begin{equation} \mathcal {S}= I + \frac{{\rm i}k}{8\pi ^2}\,F. \end{equation} \tag{ 2.6 }$$

We note that the factor 8π² in the denominator—instead of the more common factor 2π—stems from our definition (2.4) of the far field pattern.

It is well known (see, e.g., [16, theorem 4.4] or [6]) that $\mathcal {S}$ is unitary (that is, $\mathcal {S}^\ast \mathcal {S}=\mathcal {S}\mathcal {S}^\ast =I\,$) and that F is normal (that is, F and its adjoint F* commute).

Due to assumption 2.1, |q| is positive within D. This makes it convenient to introduce the weighted L²-space L²(D, |q| dx) which is the completion of $C^\infty _0(D)$ with respect to the inner product

$$\begin{eqnarray*} &&(u,v)_{L^2(D,|q|\,{\rm d}x)}= \int _Du(x) \overline{v(x)} |q(x)| \,{\rm d}x . \end{eqnarray*}$$

Obviously, L²(D, |q| dx) contains L²(D) as a dense subspace and coincides with L²(D) if |q| is bounded below in D by a (global) positive constant.

The starting point of this paper is the following connection between the injectivity of F and the following interior transmission eigenvalue problem.

Definition 2.2. k > 0 is an interior transmission eigenvalue if there exists a non-trivial pair (u, w) ∈ L²(D, |q| dx) × L²(D, |q| dx) such that u − w ∈ H²(D) and

$$\begin{equation} \Delta u+k^2(1+q)u=0\mbox{ in }D,\qquad \Delta w+k^2w=0\mbox{ in }D, \end{equation} \tag{ 2.7 }$$

$$\begin{equation} u=w\mbox{ on }\partial D\quad \mbox{and}\quad \frac{\partial u}{\partial \nu }= \frac{\partial w}{\partial \nu }\mbox{ on }\partial D. \end{equation} \tag{ 2.8 }$$

The differential equations are understood in the ultra-weak sense; that is,

$$\begin{eqnarray*} &&\int _Dw (\Delta \psi +k^2\psi ) \,{\rm d}x= 0\quad \mbox{for all }\psi \in C^\infty _0(D) , \end{eqnarray*}$$

and analogously for u. The boundary conditions can be reformulated as $u-w\in H^2_0(D)$.

Theorem 2.3. F fails to be one-to-one if, and only if, k > 0 is an interior transmission eigenvalue such that the corresponding solution w of Δw + k²w = 0 can be extended to all of $\mathbb {R}^3$ as a Herglotz wave function,

$$\begin{equation} w(x)= w_g(x)= \int _{S^2}g(\hat{\theta })\,{\rm e}^{{\rm i}kx\cdot \hat{\theta }}\,{\rm d}s(\hat{\theta }),\quad x\in \mathbb {R}^3, \end{equation} \tag{ 2.9 }$$

for some g ∈ L²(S²), $g\not=0$.

Proof. We sketch the proof for the convenience of the reader. Let g ∈ L²(S²) be in the null space of F; that is

$$\begin{eqnarray*} &&(Fg)(\hat{x})= \int _{S^2}g(\hat{\theta }) u^\infty (\hat{x},\hat{\theta }) \,{\rm d}s(\hat{\theta })= 0\quad \mbox{for all }\hat{x}\in S^2 . \end{eqnarray*}$$

By linearity, Fg is the far field pattern of the scattered field $u^s_g$ which corresponds to the incident field

$$\begin{eqnarray*} &&u^i_g(x)= \int _{S^2}g(\hat{\theta })\, {\rm e}^{{\rm i}kx\cdot \hat{\theta }} \,{\rm d}s(\hat{\theta }) ,\quad x\in \mathbb {R}^3 . \end{eqnarray*}$$

Rellich's lemma (see [7]) and unique continuation implies that the scattered field $u^s_g$ vanishes outside D. (Here we use the fact that the exterior of D is connected.) Therefore, the total field u_g coincides with the incident field $u^i_g$ outside D. Therefore, the Cauchy data of u_g and $u^i_g$ coincide on ∂D, and $(u_g,u^i_g)$ solves (2.7), (2.8).

If, on the other hand, (u, w) solves (2.7), (2.8) and w has the form (2.9) for some g ∈ L²(S²) then, by the same arguments, g is in the null space of F. The proof is finished by noting that $g\not=0$ is equivalent to $w_g\not=0$ (see, e.g., [7]). □

It is the aim of this paper to characterize the interior transmission eigenvalues by the far field operator F—or, equivalently, the scattering operator $\mathcal {S}$. We note that injectivity of F is equivalent to the fact that 1 is not an eigenvalue of $\mathcal {S}$. Since $\mathcal {S}$ is unitary its eigenvalues lie on the unit circle in the complex plane. This translates into the fact that the eigenvalues λ_j of F lie on the circle of radius 8π²/k centered at (8π²/k)i on the imaginary axes. They tend to zero because F is compact. We will show that they tend to zero from one side only, depending on the sign of q. To prove this we will need properties of the factorization method. We collect them in the following theorem 2.5, after introducing some notation. We denote by H: L²(S²) → L²(D, |q| dx) the linear and compact Herglotz operator, defined by

$$\begin{equation} (H\psi )(x)= \int \limits _{S^2}\psi (\theta )\, {\rm e}^{{\rm i}k\,x\cdot \theta }\,{\rm d}s(\theta ),\quad x\in D. \end{equation} \tag{ 2.10 }$$

Moreover, we introduce the constant

$$\begin{equation*} \sigma \ :=\ \left\lbrace \begin{array}{@{}l@{\quad }l@{}}1 & \hbox{if $q>0$ in $D$}, \\ -1 & \hbox{if $q<0$ in $D$}.\end{array}\right. \end{equation*}$$

Finally, T: L²(D, |q| dx) → L²(D, |q| dx) is defined by Tf = f + k²v|_D, where $v\in H^1_{\rm loc}(\mathbb {R}^3)$ is the radiating weak solution to

$$\begin{equation} \Delta v+k^2(1+q)v= -q\,f\quad \mbox{in }\mathbb {R}^3, \end{equation} \tag{ 2.11 }$$

that is,

$$\begin{equation} \int _{\mathbb {R}^3} \left( \nabla v\cdot \nabla \overline{\varphi } - k^2 (1+q)\,v\, \overline{\varphi } \right){\rm d}x= \int _{\mathbb {R}^3}q\,f\,\overline{\varphi }\,{\rm d}x \end{equation} \tag{ 2.12 }$$

for all $\varphi \in H^1(\mathbb {R}^3)$ with compact support and, additionally, v satisfies (2.3). Existence and uniqueness is formulated in the following lemma.

Lemma 2.4. For all k > 0 and all f ∈ L²(D, |q| dx) there exists a unique radiating solution $v\in H^1_{\rm loc}(\mathbb {R}^3)$ of (2.11). For every R > 0 such that $\overline{D}\subset B:=B(0,R)$ and every compact set K with $K\cap \overline{D}=\emptyset$ the solution operators f↦v|_B and f↦(v|_K, ∇v|_K) and f↦v^∞ are bounded from L²(D, |q| dx) into H¹(B), into C(K) × C(K)³, and into L²(S²), respectively, the latter two even compact. Furthermore, these operators depend continuously on k.

There are several ways to prove this lemma, e.g., by reformulating the problem into the Lippmann–Schwinger integral equation. We omit the proof but refer to, e.g., [16].

Theorem 2.5. 

• (a)  
  Let F: L²(S²) → L²(S²) be defined by (2.5). Then

  $$\begin{equation} F= k^2\sigma \,H^\ast \,T\,H. \end{equation} \tag{ 2.13 }$$
• (b)  
  The mapping f↦v|_D is compact from L²(D, |q| dx) into itself. Therefore, T is a compact perturbation of the identity.
• (c)  
  $\sigma \,{\rm Im} (Tf,f)_{L^2(D,|q|\,{\rm d}x)}\ge 0$ for all f ∈ L²(D, |q| dx).

For the proof we follow the presentation of [16], theorems 4.5 and 4.8.

• (a)  
  Note that the scattered field u^s satisfies the differential equation

  $$\begin{eqnarray*} &&\Delta u^s+k^2(1+q)u^s= -k^2q u^i\quad \mbox{in }\mathbb {R}^3 . \end{eqnarray*}$$

  Define the operator G from L²(D, |q| dx) into L²(S²) by Gf = v^∞ where v solves (2.11). Then F = k²GH by the superposition principle. The adjoint of H is given by

  $$\begin{eqnarray*} &&(H^\ast \psi )(\hat{x})= \int _D|q(y)| \psi (y) \,{\rm e}^{-{\rm i}k\hat{x}\cdot y}\,{\rm d}y ,\quad \hat{x}\in S^2 , \end{eqnarray*}$$

  which is the far field pattern w^∞ of the volume potential

  $$\begin{eqnarray*} &&w(x)= \int _D|q(y)| \psi (y) \Phi (x,y) \,{\rm d}y= \sigma \int _Dq(y) \psi (y) \Phi (x,y) \,{\rm d}y ,\quad x\in \mathbb {R}^3 . \end{eqnarray*}$$

  It satisfies Δw + k²w = −σq ψ in $\mathbb {R}^3$ and is radiating. From

  $$\begin{eqnarray*} &&\Delta v+k^2v= -q [f+k^2 v ] \end{eqnarray*}$$

  and the definition of T we conclude that Gf = v^∞ = σH*Tf. Substituting this into F = k² G H yields the factorization (2.13).
• (b)  
  This is easily seen by a regularity argument.
• (c)  
  For f ∈ L²(D, |q| dx) and the corresponding field v we have with Tf = f + k² v that

  $$\begin{eqnarray} (Tf,f)_{L^2(D,|q|\,{\rm d}x)} & = & \int \limits _D|Tf|^2\,|q|\,{\rm d}x - k^2 \int _D(f+k^2\,v)\,\overline{v}\,|q|\,{\rm d}x \nonumber \\ & = & \Vert Tf\Vert ^2_{L^2(D,|q|\,{\rm d}x)} + k^2\sigma \int _{|x|<R} (\Delta v+k^2v)\,\overline{v}\,{\rm d}x \nonumber \\ & = & \Vert Tf\Vert ^2_{L^2(D,|q|\,{\rm d}x)} + k^2\sigma \int _{|x|<R} [k^2|v|^2-|\nabla v|^2 ]\,{\rm d}x \nonumber\\ && +\, k^2\sigma \int _{|x|=R}\overline{v}\,\frac{\partial v}{\partial \nu }\,{\rm d}s. \end{eqnarray} \tag{ 2.14 }$$

  By the radiation condition we conclude that

  $$\begin{eqnarray} &&(Tf,f)_{L^2(D,|q|\,{\rm d}x)} = \Vert Tf\Vert ^2_{L^2(D,|q|\,{\rm d}x)} + k^2\sigma \int \limits _{|x|<R} [k^2|v|^2-|\nabla v|^2 ]\,{\rm d}x \nonumber \\ &&+\, {\rm i}k^3\sigma \int \limits _{|x|=R}|v|^2\,{\rm d}s + \mathcal {O}(R^{-1}) \end{eqnarray} \tag{ 2.15 }$$

  as R tends to infinity. Taking the imaginary part and the limit as R → ∞ yields

  $$\begin{equation} \sigma \,{\rm Im} (Tf,f)_{L^2(D,|q|\,{\rm d}x)}= k^3\Vert v^\infty \Vert ^2_{L^2(S^2)}\ \ge \ 0. \end{equation} \tag{ 2.16 }$$

Now we want to characterize the interior transmission eigenvalues by the operator T. To this end, we denote the closure of the range of the Herglotz operator H in L²(D, |q| dx) by X; that is,

$$\begin{eqnarray} X & = & {\rm closure}_{L^2(D,|q|\,{\rm d}x)}\mathcal {R}(H) \nonumber \\ & = & \left\lbrace w\in L^2(D,|q|\,{\rm d}x):\int _Dw(\Delta \psi +k^2\psi )\,{\rm d}x=0 \mbox{ for all }\psi \in C^\infty _0(D)\right\rbrace .\hspace{28.45274pt}\mbox{ } \end{eqnarray} \tag{ 3.1 }$$

The last set equality is due to the density of Herglotz wave functions in the set of all solutions to the Helmholtz equation (see [8]). Then we have:

Theorem 3.1. 

• (a)  
  Let k > 0 be an interior transmission eigenvalue with corresponding non-trivial pair (u, w). Then w ∈ X∖{0}, and w satisfies $(Tw,w)_{L^2(D,|q|\,{\rm d}x)}=0$.
• (b)  
  Let k > 0 and let w ∈ X∖{0} satisfy ${\rm Im} (Tw,w)_{L^2(D,|q|\,{\rm d}x)}=0$. Then there exists u ∈ L²(D) such that k is an interior transmission eigenvalue with corresponding pair (u, w). The function u − w belongs to $H^2_0(D)$ and does not vanish. In particular, $\sigma {\rm Im} (Tw,w)_{L^2(D,|q|\,{\rm d}x)}>0$ for all w ∈ X∖{0} if k > 0 is not an interior transmission eigenvalue.

Proof. 

• (a)  
  Let k > 0 be an interior transmission eigenvalue with corresponding u, w ∈ L²(D, |q| dx). We define $v=\frac{1}{k^2}(u-w)$ in D and note that $v\in H^2_0(D)$. We extend v by zero outside D and observe that v is the radiating solution of

  $$\begin{equation} \Delta v + k^2(1+q)v= -q\,w\quad \mbox{in }\mathbb {R}^3. \end{equation} \tag{ 3.2 }$$

  Therefore, $v\in H^2_{\rm loc}(\mathbb {R}^3)$ with compact support and w ∈ X satisfies (2.11). Let $w_n\in C^\infty (\overline{D})$ such that Δw_n + k²w_n = 0 in D and w_n → w in L²(D, |q| dx). Then Tw = w + k²v|_D and thus

  $$\begin{eqnarray*} (Tw,w_n)_{L^2(D,|q|\,{\rm d}x)} & = & \int _D(w+k^2v) |q| \overline{w}_n \,{\rm d}x \\ & = & -\sigma \int _D(\Delta v+k^2v) \overline{w}_n \,{\rm d}x= 0 \end{eqnarray*}$$

  by Green's second theorem. The left-hand side converges to $(Tw,w)_{L^2(D,|q|\,{\rm d}x)}$ which proves that $(Tw,w)_{L^2(D,|q|\,{\rm d}x)}=0$. Finally, $w\not=0$ because otherwise also v vanishes as a radiating solution of (3.2), too.
• (b)  
  Let now w ∈ X such that ${\rm Im} (Tw,w)_{L^2(D,|q|\,{\rm d}x)}=0$. We denote by $v\in H^2_{\rm loc}(\mathbb {R}^3)$ the radiating solution of (2.11) for f = w. From (2.16) we conclude that v^∞ = 0. Rellich's lemma yields that v vanishes outside D; that is, $v\in H^2_0(D)$. We set u = w + k²v. For $\psi \in C^\infty _0(D)$ we conclude that

  $$\begin{eqnarray*} \int _Du [\Delta \psi +k^2(1+q)\psi ] \,{\rm d}x & = & k^2\int _Dv [\Delta \psi +k^2(1+q)\psi ] \,{\rm d}x \\ & & + \int _Dw [\Delta \psi +k^2(1+q)\psi ] \,{\rm d}x \\ & = & k^2\int _D\psi [\Delta v+k^2(1+q)v] \,{\rm d}x + k^2\int _Dw q \psi \,{\rm d}x \\ & = & 0 . \end{eqnarray*}$$

  Therefore, (u, w) satisfies the conditions of the interior transmission eigenvalue problem.

 □

Remark 3.2. From part (b) of the previous proof we observe that ${\rm Im} (Tw,w)_{L^2(D,|q|\,{\rm d}x)}= 0$ for w ∈ X implies that the radiating solution $v\in H^2_{\rm loc}(\mathbb {R}^3)$ to (3.2) vanishes outside D and thus $v|_D\in H^2_0(D)$.

Corollary 3.3. Assume that $(Tw,w)_{L^2(D,|q|\,{\rm d}x)} = 0$ for some w ∈ X. Define again $v\in H^2_{\rm loc}(\mathbb {R}^3)$ to be the radiating solution of (3.2). Then

$$\begin{equation} k^2\int _{D} (|\nabla v|^2 - k^2 |v|^2)\,{\rm d}x= \sigma \,\Vert Tw \Vert _{L^2(D,|q|\,{\rm d}x)}^2. \end{equation} \tag{ 3.3 }$$

Proof. As we noted in the preceding remark 3.2, v vanishes outside D; that is, $v|_D\in H^2_0(D)$. Hence, (2.15) yields

$$\begin{equation*} 0 = (Tw,w)_{L^2(D,|q|\,{\rm d}x)} = \Vert Tw\Vert ^2_{L^2(D,|q|\,{\rm d}x)} \ + \ k^2\sigma \int _{D} [k^2|v|^2-|\nabla v|^2 ]\, {\rm d}x \ + \ \mathcal {O}(R^{-1}). \end{equation*}$$

The claim follows from passing to the limit as R → ∞. □

In this section, we collect well-known properties of the eigenvalues of the (normal) far field operator and the related scattering operator. These properties will be useful later on to prove an inside–outside duality between the interior transmission eigenvalues and the eigenvalues of the far field operator.

We recall that the eigenvalues λ_n of F lie on the circle of radius 8π²/k centered at (8π²/k)i on the imaginary axes and converge to zero.

Lemma 4.1. Let k be no interior transmission eigenvalue. Then λ_n converges from the right to zero if σ = +1 and from the left if σ = −1; that is, σReλ_n > 0 for sufficiently large n.

Proof. Let ψ_n ∈ L²(S²) be the eigenfunctions of F corresponding to the eigenvalues λ_n such that $\lbrace \psi _n:n\in \mathbb {N}\rbrace$ forms an orthonormal basis in L²(S²). Define $\phi _n\in \mathcal {R}(H)\subset L^2(D,|q|\,{\rm d}x)$ by

$$\begin{eqnarray*} &&\phi _n= \frac{k}{\sqrt{|\lambda _n|}} H\psi _n ,\quad n\in \mathbb {N}. \end{eqnarray*}$$

From the factorization we conclude that

$$\begin{eqnarray*} \sigma (T\phi _n,\phi _m)_{L^2(D,|q|\,{\rm d}x)} & = & \frac{k^2\sigma }{\sqrt{|\lambda _n||\lambda _m|}} (TH\psi _n,H\psi _m)_{L^2(D,|q|\,{\rm d}x)} \\ & = & \frac{k^2\sigma }{\sqrt{|\lambda _n||\lambda _m|}} (H^\ast TH\psi _n,\psi _m)_{L^2(S^2)} \\ & = & \frac{1}{\sqrt{|\lambda _n||\lambda _m|}} (F\psi _n,\psi _m)_{L^2(S^2)}= \frac{\lambda _n}{|\lambda _n|} \delta _{n,m}= s_n\delta _{n,m} \end{eqnarray*}$$

with s_n = λ_n/|λ_n| and δ_{n, m} = 0 for $n\not=m$ and δ_{n, m} = 1 for n = m. We note that |s_n| = 1 and Ims_n > 0. Since λ_n tends to zero, the only accumulation points of s_n can be 1 or −1. From T = I + C for some compact operator C we conclude that

$$\begin{equation} \Vert \phi _n\Vert ^2_{L^2(D,|q|\,{\rm d}x)} + (C\phi _n,\phi _n)_{L^2(D,|q|\,{\rm d}x)}= \sigma s_n. \end{equation} \tag{ 4.1 }$$

First we show that the sequence ϕ_n is bounded. Otherwise there exists a subsequence such that $\Vert \phi _n\Vert _{L^2(D,|q|\,{\rm d}x)}\rightarrow \infty$. The sequence $\hat{\phi }_n=\phi _n/\Vert \phi _n\Vert _{L^2(D,|q|\,{\rm d}x)}$ satisfies

$$\begin{eqnarray*} &&1 + (C\hat{\phi }_n,\hat{\phi }_n)_{L^2(D,|q|\,{\rm d}x)}= \frac{\sigma s_n}{\Vert \phi _n\Vert _{L^2(D,|q|\,{\rm d}x)}^2}\ \longrightarrow \ 0 . \end{eqnarray*}$$

Since it is bounded there exists a weakly convergent subsequence $\hat{\phi }_n\rightharpoonup \hat{\phi }$. The compactness of C implies $(C\hat{\phi }_n,\hat{\phi }_n)_{L^2(D,|q|\,{\rm d}x)}\rightarrow (C\hat{\phi },\hat{\phi })_{L^2(D,|q|\,{\rm d}x)}$ and thus $1+(C\hat{\phi },\hat{\phi })_{L^2(D,|q|\,{\rm d}x)}=0$. Taking the imaginary part yields ${\rm Im} (C\hat{\phi },\hat{\phi })_{L^2(D,|q|\,{\rm d}x)}=0$ and thus also

$$\begin{eqnarray*} &&{\rm Im} (T\hat{\phi },\hat{\phi })_{L^2(D,|q|\,{\rm d}x)}=0 . \end{eqnarray*}$$

theorem 3.1 shows that $\hat{\phi }=0$ which contradicts $1+(C\hat{\phi },\hat{\phi })_{L^2(D,|q|\,{\rm d}x)}=0$. Therefore, the sequence ϕ_n is bounded and contains a weakly convergent subsequence ϕ_n⇀ϕ. Again, $(C\phi _n,\phi _n)_{L^2(D,|q|\,{\rm d}x)}$ converges to $(C\phi ,\phi )_{L^2(D,|q|\,{\rm d}x)}$. Also, taking the imaginary part of (4.1), yields ${\rm Im} (C\phi _n,\phi _n)_{L^2(D,|q|\,{\rm d}x)}\rightarrow 0$ and thus again ${\rm Im} (C\phi ,\phi )_{L^2(D,|q|\,{\rm d}x)}=0$ and ${\rm Im} (T\phi ,\phi )_{L^2(D,|q|\,{\rm d}x)}=0$. We conclude again ϕ = 0 and thus $(C\phi _n,\phi _n)_{L^2(D,|q|\,{\rm d}x)}\rightarrow 0$. Equation (4.1) yields $\Vert \phi _n\Vert ^2_{L^2(D,|q|\,{\rm d}x)}- \sigma s_n\longrightarrow 0$ and enforces σ to be the accumulation point of (s_n) rather than −σ. □

We translate this into a condition on the eigenvalues of the scattering matrix $\mathcal {S}$.

Corollary 4.2. Let k be no interior transmission eigenvalue and let $\mu _n\in \mathbb {C}$ be the eigenvalues of the scattering operator $\mathcal {S}=I+\frac{{\rm i}k}{8\pi ^2}\,F$. Then |μ_n| = 1 for all n, they converge to 1 and σImμ_n > 0 for sufficiently large n.

Let k be not an interior transmission eigenvalue. Then 1 is not an eigenvalue of $\mathcal {S}$ and the far field operator F is one-to-one with dense range. We define the Cayley transform $\mathcal {T}$ by

$$\begin{eqnarray*} &&\mathcal {T}= {\rm i}(I+\mathcal {S})(I-\mathcal {S})^{-1}\ :\ L^2(S^2)\supset \mathcal {R}(F)\ \longrightarrow \ L^2(S^2) \end{eqnarray*}$$

It is easily seen that $\mathcal {T}$ is self-adjoint, its spectrum is discrete, and μ_n = exp ( − 2iδ_n) is an eigenvalue of $\mathcal {S}$ for some δ_n ∈ (0, π) if, and only if, $\cot \delta _n\in \mathbb {R}$ is an eigenvalue of $\mathcal {T}$. Since (μ_n) tends to one with σImμ_n > 0 we conclude that

$$\begin{eqnarray*} \lim \limits _{n\rightarrow \infty }\delta _n= \left\lbrace \begin{array}{@{}ll@{}}0 & \hbox{ if } \sigma =-1 , \\ \pi & \hbox{ if } \sigma =+1 ,\end{array}\right. \qquad \lim \limits _{n\rightarrow \infty }\cot \delta _n= \left\lbrace \begin{array}{@{}ll@{}}+\infty &\hbox{ if }\sigma =-1 ,\\ -\infty &\hbox{ if }\sigma =+1.\end{array}\right. \end{eqnarray*}$$

In particular, the numbers

$$\begin{equation} \delta _+ = \max _{n \in \mathbb {N}} \delta _n \quad \hbox{if } \sigma =-1 \qquad {\rm and} \qquad \delta _- = \min _{n \in \mathbb {N}} \delta _n \, \quad {\rm if }\; \sigma =+1, \end{equation} \tag{ 4.2 }$$

are well defined and belong to (0, π) if k is not an interior transmission eigenvalue. We call them the 'first' eigenvalue.

Lemma 4.3. Let k be no interior transmission eigenvalue.

• (a)  
  Let first σ = −1; that is, q < 0 in D. Then

  $$\begin{eqnarray*} &&\cot \delta _+= \inf \limits _{w\in X}\frac{{\rm Re} (Tw,w)_{L^2(D,|q|\,{\rm d}x)}}{-{\rm Im} (Tw,w)_{L^2(D,|q|\,{\rm d}x)}} . \end{eqnarray*}$$
• (b)  
  Second, let σ = +1; that is, q > 0 in D. Then

  $$\begin{eqnarray*} &&\cot \delta _-= -\inf \limits _{w\in X}\frac{{\rm Re} (Tw,w)_{L^2(D,|q|\,{\rm d}x)}}{{\rm Im} (Tw,w)_{L^2(D,|q|\,{\rm d}x)}} . \end{eqnarray*}$$

Note that in both cases the denominator is strictly positive because of the assumption on k and theorem 3.1(b).

Proof. (a) Courant's max–min principle (see [10]) yields

$$\begin{eqnarray*} \cot \delta _+ & = & \inf \limits _{f\in \mathcal {R}(F)}\frac{(\mathcal {T}f,f)_{L^2(S^2)}}{\Vert f\Vert ^2_{L^2(S^2)}}= \inf \limits _{f\in \mathcal {R}(F)}\frac{{\rm i} ((I+\mathcal {S})(I-\mathcal {S})^{-1}f,f )_{L^2(S^2)}}{\Vert f\Vert ^2_{L^2(S^2)}} \\ & = & \inf \limits _{g\in L^2(S^2)}\frac{{\rm i} ((I+\mathcal {S})g,(I-\mathcal {S})g )_{L^2(S^2)}}{\Vert (I-\mathcal {S})g\Vert ^2_{L^2(S^2)}} \\ & = & \inf \limits _{g\in L^2(S^2)}\frac{{\rm i} \big(\Vert g\Vert ^2_{L^2(S^2)}+ 2{\rm i}\, {\rm Im} (\mathcal {S}g,g)_{L^2(S^2)}-\Vert \mathcal {S}g\Vert ^2_{L^2(S^2)} \big)}{\Vert g\Vert ^2_{L^2(S^2)}-2\, {\rm Re} (\mathcal {S}g,g)_{L^2(S^2)}+ \Vert \mathcal {S}g\Vert ^2_{L^2(S^2)}} \\ & = & \inf \limits _{g\in L^2(S^2)}\frac{{\rm Im} (\mathcal {S}g,g)_{L^2(S^2)}}{{\rm Re} (\mathcal {S}g,g)_{L^2(S^2)}-\Vert g\Vert ^2_{L^2(S^2)}} . \end{eqnarray*}$$

Now we make use of the form $\mathcal {S}=I+({\rm i}k)/(8\pi ^2)F$ and the factorization (2.13). In the following we write (Th, h) instead of $(Th,h)_{L^2(D,|q|\,{\rm d}x)}$. Then

$$\begin{equation*} \cot \delta _+= \inf \limits _{g\in L^2(S^2)} \frac{{\rm Re} (Fg,g)_{L^2(S^2)}}{-{\rm Im} (Fg,g)_{L^2(S^2)}}= \inf \limits _{g\in L^2(S^2)}\frac{{\rm Re} (THg,Hg)}{-{\rm Im} (THg,Hg)}= \inf \limits _{w\in X}\frac{{\rm Re} (Tw,w)}{-{\rm Im} (Tw,w)} \end{equation*}$$

which proves the assertion. Part (b) follows the same lines by just replacing the infimum by the supremum and noting that

$$\begin{eqnarray*} &&\sup \limits _{w\in X}\frac{{\rm Re} (Tw,w)}{-{\rm Im} (Tw,w)}= -\inf \limits _{w\in X}\frac{{\rm Re} (Tw,w)}{{\rm Im} (Tw,w)} . \end{eqnarray*}$$

 □

In the following we study the dependence of the eigenvalues of the scattering operator on the wavenumber k and write X(k), T(k), δ_±(k), and so on to indicate this dependence. The eigenvalue characterization from lemma 4.3 uses k-dependent spaces over which the infimum or supremum is taken. In a first step, we will transform this k-dependence into the quadratic forms by introducing the orthogonal projection operator P(k) from L²(D, |q| dx) into X(k). This step will hence eliminate the k-dependence of the considered function spaces.

For the sake of notational simplicity, we write k↗k₀ and k→k₀ to indicate that $k\in \mathbb {R}$ tends to $k_0\in \mathbb {R}$ from below and above, respectively. More precisely, if k↗k₀ (k→k₀) the inequality k < k₀ (k > k₀) is always satisfied in the limiting process.

Recall that the numbers δ_±(k) were defined in (4.2). They allow us to state the following simple result.

Lemma 5.1. Let k₀ > 0 and w₀ ∈ X(k₀) such that $w_0\not=0$ and $(T(k_0)w_0,w_0 )_{L^2(D,|q|\,{\rm d}x)}=0$ and assume that

$$\begin{equation*} \alpha \ :=\ \left[ \frac{{\rm d}}{{\rm d}k}(T(k)P(k)w_0,P(k)w_0 )_{L^2(D,|q|\,{\rm d}x)} \right] \biggr |_{k=k_0}\in \mathbb {R}\setminus \lbrace 0\rbrace . \end{equation*}$$

• (a)  
  Let σ = −1; that is, q < 0 in D. Then

  $$\begin{eqnarray*} &&\lim \limits _{k \nearrow k_0}\delta _+(k)= \pi \mbox{ if }\alpha >0 ,\quad \lim \limits _{k \searrow k_0}\delta _+(k)= \pi \mbox{ if }\alpha <0 . \end{eqnarray*}$$
• (b)  
  Let σ = +1; that is, q > 0 in D. Then

  $$\begin{eqnarray*} &&\lim \limits _{k \nearrow k_0}\delta _-(k)= 0\mbox{ if }\alpha >0 ,\quad \lim \limits _{k \searrow k_0}\delta _-(k)= 0\mbox{ if }\alpha <0 . \end{eqnarray*}$$

Note that we will show in lemmas 5.2 and 5.3 that $k\mapsto (T(k)P(k)w_0,P(k)w_0 )_{L^2(D,|q|\,{\rm d}x)}$ is indeed a differentiable function in k₀ for every fixed w₀ ∈ X(k₀).

Proof. (a) We note from theorem 3.1 that $(T(k_0)w_0,w_0 )_{L^2(D,|q|\,{\rm d}x)}=0$ implies that k₀ is an interior transmission eigenvalue. Let I = (k₀ − ε, k₀ + ε) be an interval containing no other transmission eigenvalue. From the previous lemma we have for k ∈ I∖{k₀}

$$\begin{eqnarray*} &&\cot \delta _+(k)= \inf \limits _{g\in L^2(D,|q|\,{\rm d}x)} \frac{{\rm Re} (T(k)P(k)g,P(k)g )_{L^2(D,|q|\,{\rm d}x)}}{-{\rm Im} (T(k)P(k)g,P(k)g )_{L^2(D,|q|\,{\rm d}x)}} . \end{eqnarray*}$$

Furthermore, from Taylor's theorem we have that

$$\begin{eqnarray*} &&(T(k)P(k)w_0,P(k)w_0 )_{L^2(D,|q|\,{\rm d}x)}= \alpha (k-k_0) + r(k) \end{eqnarray*}$$

with r(k) = o(|k − k₀|) and Imr(k) < 0. Here we used that

$$\begin{eqnarray*} &&(T(k_0)P(k_0)w_0,P(k_0)w_0 )_{L^2(D,|q|\,{\rm d}x)}= (T(k_0)w_0,w_0 )_{L^2(D,|q|\,{\rm d}x)}=0 . \end{eqnarray*}$$

Therefore, we have

$$\begin{eqnarray*} &&\cot \delta _+(k)\ \le \ \frac{{\rm Re} (T(k)P(k)w_0,P(k)w_0 )_{L^2(D,|q|\,{\rm d}x)}}{-{\rm Im} (T(k)P(k)w_0,P(k)w_0 )_{L^2(D,|q|\,{\rm d}x)}}= \frac{\alpha (k-k_0)+{\rm Re} r(k)}{-{\rm Im} r(k)} \end{eqnarray*}$$

which converges to −∞ provided k → k₀ such that α(k − k₀) < 0.

Part (b) is proven in the same way. □

Now we want to investigate the derivative of $k\mapsto (T(k)P(k)w_0,P(k)w_0 )_{L^2(D,|q|\,{\rm d}x)}$ at an eigenvalue k₀. We begin with the following result.

Lemma 5.2. Let k₀ > 0 be an interior transmission eigenvalue. Due to theorem 3.1 there exists w₀ ∈ X(k₀) such that $w_0\not=0$ and $(T(k_0)w_0,w_0)_{L^2(D,|q|\,{\rm d}x)}=0$. Then $k\mapsto (T(k)w_0,w_0 )_{L^2(D,|q|\,{\rm d}x)}$ is differentiable in k₀ and

$$\begin{equation} \left[ \frac{{\rm d}}{{\rm d}k} (T(k)w_0,w_0 )_{L^2(D,|q|\,{\rm d}x)} \right] \biggr |_{k=k_0} =\ 2\sigma k_0\int _D|\nabla v_0|^2\,{\rm d}x\ \not=\ 0 \end{equation} \tag{ 5.1 }$$

where v₀ is the radiating solution of (2.11) for k = k₀ and f = w₀; that is,

$$\begin{equation} \Delta v_0+k_0^2(1+q)v_0= -q\,w_0\quad \mbox{in }\mathbb {R}^3. \end{equation} \tag{ 5.2 }$$

Proof. First we note that v₀ vanishes outside D by remark 3.2. In particular, $v_0\in H^2_0(D)$. Let v' be the unique radiating solution of

$$\begin{equation} \Delta v^\prime + k_0^2(1+q)v^\prime = -2k_0(1+q)v_0\quad \mbox{in } \mathbb {R}^3, \end{equation} \tag{ 5.3 }$$

and v_k the solution of (5.2) corresponding to k instead of k₀. Then it is easily seen that

$$\begin{eqnarray*} \frac{{\rm d}}{{\rm d}k} (T(k_0)w_0,w_0 )_{L^2(D,|q|\,{\rm d}x)} & = & \left.\frac{{\rm d}}{{\rm d}k}\int _D(w_0+k^2v_0) \overline{w_0} |q| \,{\rm d}x\right|_{k=k_0} \\ & = & 2k_0\int _Dv_0 \overline{w_0} |q| \,{\rm d}x + k_0^2\int _D v^\prime \overline{w_0} |q| \,{\rm d}x . \end{eqnarray*}$$

Now we eliminate w₀ from this equation by using (5.2)

$$\begin{eqnarray*} &&\frac{{\rm d}}{{\rm d}k} (T(k_0)w_0,w_0 )_{L^2(D,|q|\,{\rm d}x)} \\ &&= -2k_0\sigma \int _Dv_0 [\Delta \overline{v_0}+ k_0^2(1+q)\overline{v_0} ] \,{\rm d}x - \sigma k_0^2\int _Dv^\prime [\Delta \overline{v_0}+k_0^2(1+q)\overline{v_0} ] \,{\rm d}x \\ &&= 2k_0\sigma \int _D [|\nabla v_0|^2-k_0^2(1+q)|v_0|^2 ] \,{\rm d}x - \sigma k_0^2\int _D\overline{v_0} [\Delta v^\prime +k_0^2(1+q) v^\prime ] \,{\rm d}x \\ &&= 2k_0\sigma \int _D [|\nabla v_0|^2-k_0^2(1+q)|v_0|^2 ] \,{\rm d}x + 2\sigma k_0^3\int _D(1+q) |v_0|^2 \,{\rm d}x \\ &&= 2\sigma k_0\int _D|\nabla v_0|^2\,{\rm d}x \end{eqnarray*}$$

where we used the definition of v'. This term vanishes only for constant functions v₀; that is, for v₀ = 0 because of $v_0\in H^2_0(D)$. □

Now we study the dependence of the orthogonal projection P(k): L²(D, |q| dx) → X(k) on k. To give an explicit representation of P(k), let us denote by W the completion of $C_0^\infty (D)$ with respect to the semi-norm $\Vert \psi \Vert _W:= \Vert (\Delta \psi +k^2\psi )/\sqrt{|q|} \Vert _{L^2(D)}$. Then P(k) is explicitly given by

$$\begin{eqnarray*} &&P(k)g= g - \frac{1}{|q|}(\Delta \hat{w}+k^2\hat{w}) , \end{eqnarray*}$$

where $\hat{w}\in W$ solves the fourth order system

$$\begin{equation} (\Delta +k^2)\frac{1}{|q|}(\Delta +k^2)\hat{w}= (\Delta +k^2)g \end{equation} \tag{ 5.4 }$$

in the variational sense; that is, $\hat{w}$ solves the W-coercive variational problem

$$\begin{eqnarray*} &&\int _D\frac{1}{|q|} (\Delta \hat{w}+k^2\hat{w}) (\Delta \psi +k^2\psi ) \,{\rm d}x= \int _Dg (\Delta \psi +k^2\psi ) \,{\rm d}x\quad \;\;\,\hbox{for all $\psi \in W$.} \end{eqnarray*}$$

Then we have the following extension of the previous lemma:

Lemma 5.3. Let k₀ > 0 be an interior transmission eigenvalue and w₀ ∈ X(k₀) such that $w_0\not=0$ and $(T(k_0)w_0,w_0)_{L^2(D,|q|\,{\rm d}x)}=0$. Then $k\mapsto (T(k)P(k)w_0,P(k)w_0 )_{L^2(D,|q|\,{\rm d}x)}$ is differentiable in k₀ and

$$\begin{eqnarray} &&\left[ \frac{{\rm d}}{{\rm d}k}\biggl . (T(k)P(k)w_0,P(k)w_0 )_{L^2(D,|q|\,{\rm d}x)} \right] \biggr |_{k=k_0}\,{=}\, 2\sigma k_0\left[\int _D|\nabla v_0|^2\,{\rm d}x+ 2\,{\rm Re} \int _Dv_0\,\overline{w_0}\,{\rm d}x\right] \end{eqnarray} \tag{ 5.5 }$$

where v₀ is the solution of (5.2).

Proof. First we note that k↦P(k)w₀ is Frechét-differentiable and, by differentiating the characterization of P(k)w₀,

$$\begin{eqnarray*} &&P^\prime (k)w_0= -\frac{1}{|q|} (\Delta \hat{w}^\prime +k^2\hat{w}^\prime + 2k\hat{w}) \end{eqnarray*}$$

where $\hat{w}\in W$ solves (5.4) for g = w₀ and $\hat{w}^\prime \in W$ solves

$$\begin{eqnarray*} &&(\Delta +k^2)\frac{1}{|q|}(\Delta +k^2)\hat{w}^\prime = 2k w_0 - \frac{2k}{|q|}(\Delta \hat{w}+k^2\hat{w}) - 2k(\Delta +k^2)\frac{1}{|q|}\hat{w} . \end{eqnarray*}$$

By the chain rule we have

$$\begin{eqnarray*} &&\frac{{\rm d}}{{\rm d}k} (T(k)P(k)w_0,P(k)w_0 )_{L^2(D,|q|\,{\rm d}x)}\\ &&= (T^\prime (k)P(k)w_0,P(k)w_0 )_{L^2(D,|q|\,{\rm d}x)}+ (T(k)P^\prime (k)w_0,P(k)w_0 )_{L^2(D,|q|\,{\rm d}x)} \\ &&\quad +\, (T(k)P(k)w_0,P^\prime (k)w_0 )_{L^2(D,|q|\,{\rm d}x)} \\ &&= (T^\prime (k)P(k)w_0,P(k)w_0 )_{L^2(D,|q|\,{\rm d}x)} + \overline{ (T^\ast (k)P(k)w_0,P^\prime (k)w_0 )}_{L^2(D,|q|\,{\rm d}x)} \\ &&\quad +\, (T(k)P(k)w_0,P^\prime (k)w_0 )_{L^2(D,|q|\,{\rm d}x)}. \end{eqnarray*}$$

The first term on the right-hand side has been computed at k = k₀ in the previous lemma. The adjoint T* of T is given by $T^\ast g=g+k^2\overline{v_g}$ where v_g is the radiating solution of

$$\begin{eqnarray*} &&\Delta v_g + k^2(1+q)v_g= -q \overline{g}\quad \mbox{in }\mathbb {R}^3 . \end{eqnarray*}$$

Indeed, if f ∈ L²(D, |q| dx) with corresponding v_f satisfying

$$\begin{eqnarray*} &&\Delta v_f + k^2(1+q)v_f= -q f\quad \mbox{in }\mathbb {R}^3 , \end{eqnarray*}$$

then

$$\begin{eqnarray*} (Tf,g)_{L^2(D,|q|\,{\rm d}x)} & = & (f,g)_{L^2(D,|q|\,{\rm d}x)} + \sigma k^2\int _Dv_f \overline{g} q \,{\rm d}x \\ & = & (f,g)_{L^2(D,|q|\,{\rm d}x)} - \sigma k^2\int _Dv_f (\Delta v_g+ k^2(1+q)v_g ) \,{\rm d}x \\ & = & (f,g)_{L^2(D,|q|\,{\rm d}x)} - \sigma k^2\int _Dv_g (\Delta v_f+ k^2(1+q)v_f ) \,{\rm d}x \\ & = & (f,g)_{L^2(D,|q|\,{\rm d}x)} + \sigma k^2\int _Dv_g f q \,{\rm d}x \\ & = & (f,g+k^2\overline{v_g})_{L^2(D,|q|\,{\rm d}x)} . \end{eqnarray*}$$

From $0=(T(k_0)w_0,w_0)_{L^2(D,|q|\,{\rm d}x)}=(w_0,T^\ast (k_0)w_0)_{L^2(D,|q|\,{\rm d}x)}$ one concludes as in the proof of theorem 2.5 that v which corresponds to T*(k₀)w₀ vanishes outside D and, therefore, coincides with $\overline{v_0}$. This shows that T*(k₀)w₀ = T(k₀)w₀ and thus

$$\begin{eqnarray*} &&\left[ \frac{{\rm d}}{{\rm d}k}\biggl . (T(k)P(k)w_0,P(k)w_0 )_{L^2(D,|q|\,{\rm d}x)} \right] \biggr |_{k=k_0} = 2\sigma k_0\int _D|\nabla v_0|^2\,{\rm d}x \\ &&+\ 2\, {\rm Re} (T(k_0)w_0,P^\prime (k_0)w_0 )_{L^2(D,|q|\,{\rm d}x)} \end{eqnarray*}$$

because P(k₀)w₀ = w₀. Therefore also $\hat{w}=0$ and

$$\begin{eqnarray*} &&P^\prime (k_0)w_0= -\frac{1}{|q|}(\Delta \hat{w}^\prime + k_0^2\hat{w}^\prime ) \end{eqnarray*}$$

and $\hat{w}^\prime \in W$ solves

$$\begin{eqnarray*} &&(\Delta +k_0^2)\frac{1}{|q|}(\Delta +k_0^2)\hat{w}^\prime = 2k_0 w_0 . \end{eqnarray*}$$

We compute, noting that P(k₀)w₀ = w₀,

$$\begin{eqnarray*} (T(k_0)w_0,P^\prime (k_0)w_0 )_{L^2(D,|q|\,{\rm d}x)} &=& -\int _D(w_0+k_0^2v_0) \overline{(\Delta \hat{w}^\prime +k_0^2 \hat{w}^\prime )} \,{\rm d}x\\ &=& \sigma \int _D\frac{1}{|q|} (\Delta v_0+k_0^2v_0) \overline{(\Delta \hat{w}^\prime +k_0^2\hat{w}^\prime )} \,{\rm d}x \\ &=& \sigma \int _Dv_0 \overline{(\Delta +k_0^2)\frac{1}{|q|}(\Delta +k_0^2) \hat{w}^\prime } \,{\rm d}x= 2k_0\sigma \int _Dv_0 \overline{w_0} \,{\rm d}x . \end{eqnarray*}$$

This proves the assertion. □

We would like to find conditions on the contrast q ensuring that the derivative in (5.5) is either strictly positive or strictly negative, that is,

$$\begin{eqnarray} \biggl . \left[ \frac{{\rm d}}{{\rm d}k} (T(k)P(k)w_0,P(k)w_0 )_{L^2(D,|q|\,{\rm d}x)} \right] \biggr |_{k=k_0} & = 2 k_0 \sigma A\ \lessgtr \ 0 \end{eqnarray} \tag{ 6.1 }$$

where

$$\begin{eqnarray} A & = \int _D [|\nabla v_0|^2+2\, {\rm Re} (v_0 \overline{w_0}) ] {\rm d}x , \end{eqnarray} \tag{ 6.2 }$$

for all $0\not= w_0\in X(k_0)$ such that $(T(k)w_0,w_0 )_{L^2(D,|q|\,{\rm d}x)}=0$. Due to lemma 5.1, such a sign property allows, generally speaking, to characterize when k₀ > 0 is an interior transmission eigenvalue, merely knowing the far field operators F(k) in some interval around k₀. Unfortunately, we are able to prove this sign property under strong assumptions on the contrast q only, roughly speaking for constant contrast that is either positive and large enough or negative and small enough. Weakening these assumptions is an issue of ongoing research.

Theorem 6.1. Let k₀ be the smallest interior transmission eigenvalue and q(x) = q₀ > 0 for x ∈ D being constant such that

$$\begin{equation} q_0\ >\ 2\left[\left(\frac{\hat{\rho }_1}{\rho _1^2}-1\right)+ \frac{\sqrt{\hat{\rho }_1}}{\rho _1}\,\sqrt{\frac{\hat{\rho }_1}{\rho _1^2}-1} \right]. \end{equation} \tag{ 6.3 }$$

Here, $\hat{\rho }_1$ and ρ₁ denote the smallest Dirichlet eigenvalues of Δ² and −Δ, respectively, in D.⁵ Then

$$\begin{equation} \biggl . \left[ \frac{{\rm d}}{{\rm d}k} (T(k)P(k)w_0,P(k)w_0 )_{L^2(D,|q|\,{\rm d}x)} \right] \biggr |_{k=k_0}\ >\ 0 \end{equation} \tag{ 6.4 }$$

for all $0\not=w_0\in X(k_0)$ such that $(T(k)w_0,w_0 )_{L^2(D,|q|\,{\rm d}x)}=0$.

Proof. Multiplication of (5.2) with $\overline{v_0}$, integration and Green's first identity yields

$$\begin{eqnarray*} &&k_0^2(1+q_0) \Vert v_0\Vert ^2_{L^2(D)} - \Vert \nabla v_0\Vert ^2_{L^2(D)}= -q_0 (w_0,v_0)_{L^2(D)} ; \end{eqnarray*}$$

that is, taking the real part,

$$\begin{equation} 2\,{\rm Re} (w_0,v_0)_{L^2(D)}= \frac{2}{q_0}\,\Vert \nabla v_0\Vert ^2_{L^2(D)} - 2k_0^2\,\frac{1+q_0}{q_0}\,\Vert v_0\Vert ^2_{L^2(D)}. \end{equation} \tag{ 6.5 }$$

We write

$$\begin{eqnarray*} A & = & \Vert \nabla v_0\Vert ^2_{L^2(D)} + 2\, {\rm Re} (w_0,v_0)_{L^2(D)}= \frac{2+q_0}{q_0} \Vert \nabla v_0\Vert ^2_{L^2(D)} - 2k_0^2 \frac{1+q_0}{q_0} \Vert v_0\Vert ^2_{L^2(D)} \\ & = & 2 \frac{1+q_0}{q_0}\left[\frac{1+q_0/2}{1+q_0} \Vert \nabla v_0\Vert ^2_{L^2(D)} - k_0^2 \Vert v_0\Vert ^2_{L^2(D)}\right]. \end{eqnarray*}$$

Now we use that $\rho _1\,\Vert v_0\Vert ^2_{L^2(D)}\le \Vert \nabla v_0\Vert ^2_{L^2(D)}$ and arrive at

$$\begin{eqnarray*} &&A\ \ge \ 2 \frac{1+q_0}{q_0}\left[\rho _1 \frac{1+q_0/2}{1+q_0} - k_0^2 \right]\Vert v_0\Vert ^2_{L^2(D)} . \end{eqnarray*}$$

It has been shown in [15] that for q₀ satisfying (6.3) the smallest transmission eigenvalue k₀ satisfies

$$\begin{eqnarray*} &&k_0^2\ <\ \rho _1 \frac{1+q_0/2}{1+q_0} . \end{eqnarray*}$$

Therefore, for this eigenvalue we have that A > 0. □

Theorem 6.2. Let k₀ be the smallest interior transmission eigenvalue and q(x) = q₀ for q₀ ∈ ( − 1, 0). Then there exists $\hat{q}\in (-1,0)$ such that

$$\begin{equation} -1<q_0\le \hat{q}\quad \hbox{implies that}\quad \biggl . \left[ \frac{{\rm d}}{{\rm d}k} (T(k)P(k)w_0,P(k)w_0 )_{L^2(D,|q|\,{\rm d}x)} \right] \biggr |_{k=k_0}\ <\ 0 \end{equation} \tag{ 6.6 }$$

for all $0\not=w_0\in X(k_0)$ such that $(T(k)w_0,w_0 )_{L^2(D,|q|\,{\rm d}x)}=0$.

Proof. We exploit again (6.5) together with the Poincaré inequality $\rho _1 \Vert v_0 \Vert _{L^2(D)}^2 \le \Vert \nabla v_0 \Vert _{L^2(D)}^2$ involving the first Dirichlet eigenvalue ρ₁ of −Δ in D to estimate that

$$\begin{eqnarray*} A & = & \Vert \nabla v_0\Vert _{L^2(D)}^2 + 2\, {\rm Re} \int _Dv_0\overline{w}_0 \,{\rm d}x \\ & = & \left(1-\frac{2}{|q_0|}\right)\Vert \nabla v_0\Vert _{L^2(D)}^2 + 2k_0^2 \frac{1+q_0}{|q_0|} \Vert v_0\Vert ^2_{L^2(D)} \\ & \le & \left(1-\frac{2}{|q_0|} \right)\Vert \nabla v_0 \Vert _{L^2(D)}^2 + 2k_0^2 \frac{1+q_0}{|q_0|} \frac{1}{\rho _1}\Vert \nabla v_0\Vert _{L^2(D)}^2 \\ & = & \left[1-\frac{2}{|q_0|} + 2k_0^2 \frac{1+q_0}{|q_0|\rho _1}\right] \Vert \nabla v_0\Vert _{L^2(D)}^2 . \end{eqnarray*}$$

If R₊ denotes the radius of the smallest ball B = B(0, R₊) containing $\overline{D}$ then

$$\begin{eqnarray*} &&\rho _1= \inf\limits_{{{ {\begin{array}{@{}c@{}}\scriptstyle u\in H_0^1(D) \\ \scriptstyle u\not=0 \end{array}}}}}\frac{\Vert \nabla u\Vert _{L^2(D)}}{\Vert u\Vert _{L^2(D)}}\ \ge \ \inf\limits _{{{ {\begin{array}{@{}c@{}}\scriptstyle u\in H_0^1(B) \\ \scriptstyle u\not=0\end{array}}}}} \frac{\Vert \nabla u\Vert _{L^2(B)}}{\Vert u\Vert _{L^2(B)}}= \rho _{1,B}= \left(\frac{\pi }{R_+}\right)^2 \end{eqnarray*}$$

where ρ_{1, B} denotes the smallest Dirichlet eigenvalue of the ball B = B(0, R₊). Substituting this estimate into the previous estimate for A we arrive at

$$\begin{equation*} A\ \le \ \left[1-\frac{2}{|q_0|}+ \ 2k_0^2 \frac{1+q_0}{|q_0|} \frac{1}{\pi ^2} R_+^2 \right]\ \Vert \nabla v_0 \Vert _{L^2(D)}^2. \end{equation*}$$

The last term on the right is negative if, and only if,

$$\begin{equation*} 1 + 2k_0^2 \frac{1+q_0}{|q_0|} \frac{1}{\pi ^2} R_+^2\ \le \ \frac{2}{|q_0|} ;\quad \hbox{that is,}\quad k_0^2\ \le \ \pi ^2 \frac{1-|q_0|/2}{1+q_0} \frac{1}{R_+^2} . \end{equation*}$$

From the proof of theorem 3.2 and corollary 3.1 in [3] we know that the smallest transmission eigenvalue k₀ of the obstacle D for the contrast q₀ satisfies

$$\begin{equation*} k_0\ \le \ \frac{k_{B(0,1),q_0}}{R_-} \end{equation*}$$

where $k_{B(0,1),q_0}$ is the smallest transmission eigenvalue of the unit ball for the contrast q₀ and R₋ is the radius of the largest ball contained in $\overline{D}$. Hence, whenever

$$\begin{equation} k^2_{B(0,1),q_0}\ <\ \pi ^2 \, \frac{1-|q_0|/2}{1+q_0} \frac{R_-^2}{R_+^2} \end{equation} \tag{ 6.7 }$$

we can conclude that A is negative at least for the smallest transmission eigenvalue.

For constant q₀, the smallest transmission eigenvalue of the unit ball can be estimated from above by the smallest positive zero of

$$\begin{equation*} W(\lambda )= \mathrm{det}\left({\begin{array}{@{}cc@{}}j_0(\lambda ) & j_0(\lambda \sqrt{1+q_0}) \\ -j_0^{\prime }(\lambda ) & -\sqrt{1+q_0} j_0^{\prime }(\lambda \sqrt{1+q_0})\end{array}}\right) \quad \hbox{for } \lambda \ge 0 . \end{equation*}$$

It is well-known that positive roots of W are transmission eigenvalues. Setting $n_0=\sqrt{1+q_0} \ge 0$, we find that

$$\begin{eqnarray*} &&W(\lambda ,n_0)= j_0^\prime (\lambda ) j_0(\lambda n_0) - j_0(\lambda ) n_0 j_0^\prime (\lambda n_0), \qquad \lambda \ge 0, n_0 \ge 0. \end{eqnarray*}$$

We use that j₀(λ) = sin (λ)/λ and observe that j₀(0) = 1 and $j_0^\prime (0)=0$. Therefore,

$$\begin{eqnarray*} &&W(\lambda ,0)= j_0^\prime (\lambda )= \frac{\lambda \cos \lambda -\sin \lambda }{\lambda ^2}= \frac{\phi (\lambda )}{\lambda ^2} \end{eqnarray*}$$

with ϕ(λ) = λcos λ − sin λ. Elementary arguments (consider ϕ'(λ) = −λsin λ) show that the first positive zero $\hat{\lambda }_1$ of W( ·, 0) is in the interval (π, 2π). Furthermore,

$$\begin{eqnarray*} &&\left.\frac{\partial W(\lambda ,n_0)}{\partial \lambda }\right|_{n_0=0}= \frac{\phi ^\prime (\lambda )}{\lambda ^2} - 2 \frac{\phi (\lambda )}{\lambda ^3} \end{eqnarray*}$$

and thus

$$\begin{eqnarray*} &&\frac{\partial W(\hat{\lambda }_1,0)}{\partial \lambda }= -\frac{\sin \hat{\lambda }_1}{\hat{\lambda }_1}\ \not=\ 0 . \end{eqnarray*}$$

The implicit function theorem assures existence of $\hat{n}_0 > 0$ and an interval I around $\hat{\lambda }_1$ such that

$$\begin{eqnarray*} &&W(\lambda _1(n_0),n_0)= 0 \end{eqnarray*}$$

is uniquely solvable in $I\times [0, \hat{n}_0]$ and $\lambda _1(n_0)\rightarrow \hat{\lambda }_1$ as n₀ → 0. Since the limit n₀ → 0 corresponds to q₀ → −1 we obtain in particular that the smallest transmission eigenvalue $k_{B(0,1),q_0}$ remains bounded as q₀ → −1, since

$$\begin{equation*} 0\ <\ k_{B(0,1),q_0}\ \le \ \lambda _1(n_0)\ \rightarrow \ \hat{\lambda }_1 \qquad \hbox{as } q_0\rightarrow -1. \end{equation*}$$

In consequence, the left-hand side of (6.7) remains bounded as q₀ → −1 while the right-hand side obviously tends to +∞ as q₀ → −1. This shows that there exists $\hat{q}\in (-1,0)$ such that (6.7) is indeed satisfied. □

Now we are ready to prove the main result of this paper.

Theorem 6.3 (Inside–outside duality for $\mathcal {S}$). Let k₀ > 0 and I = (k₀ − ε, k₀ + ε)∖{k₀} such that no k ∈ I is an interior transmission eigenvalue. For k ∈ I let $\mu _n(k)= \exp (-2{\rm {i}}\delta _n(k) )$ be the eigenvalues of $\mathcal {S}(k)$ with phases δ_n(k) ∈ (0, π).

• (a)  
  Let k₀ be the smallest interior transmission eigenvalue and let q = q₀ satisfy (6.3). Then

  $$\begin{equation} \lim \limits _{k\in I,\ k\nearrow k_0}\delta _-(k)= 0 \end{equation} \tag{ 6.8 }$$

  where δ₋(k) = min _nδ_n(k).
• (b)  
  Let (6.8) hold and q > 0 in D. Then k₀ is an interior transmission eigenvalue.
• (c)  
  Let k₀ be the smallest interior transmission eigenvalue and let q = q₀ satisfy the assumption of (6.6) in theorem 6.2. Then

  $$\begin{equation} \lim \limits _{k\in I,\ k\nearrow k_0}\delta _+(k)= \pi \end{equation} \tag{ 6.9 }$$

  where δ₊(k) = max _nδ_n(k).
• (d)  
  Let (6.9) hold and q < 0 in D. Then k₀ is an interior transmission eigenvalue.

Proof. (a) Let k₀ be the smallest interior transmission eigenvalue and w₀ ∈ X(k₀) with $w_0\not=0$ and $(T(k_0)w_0,w_0 )_{L^2(D,|q|\,{\rm d}x)}=0$. We apply part (b) of lemma 5.1 and note that α > 0 by theorem 6.1. This yields (6.8).

(b) Let now (6.8) hold and assume, on the contrary, that k₀ is not an interior transmission eigenvalue. By lemma 4.3 the condition (6.8) is equivalent to

$$\begin{eqnarray*} &&\inf \limits _{w\in X(k)}\frac{{\rm Re} (T(k)w,w)_{L^2(D,|q|\,{\rm d}x)}}{{\rm Im} (T(k)w,w)_{L^2(D,|q|\,{\rm d}x)}}\ \longrightarrow -\infty \quad \mbox{as } k\nearrow k_0 . \end{eqnarray*}$$

Therefore, there exist sequences k_j ∈ I, k_j↗k₀, and w_j ∈ X(k_j) with $\Vert w_j\Vert _{L^2(D,|q|\,{\rm d}x)}=1$ such that

$$\begin{eqnarray*} &&{\rm Im} (T(k_j)w_j,w_j)_{L^2(D,|q|\,{\rm d}x)}\longrightarrow 0 ,\ j\rightarrow \infty ,\quad \mbox{and}\quad {\rm Re} (T(k_j)w_j,w_j)_{L^2(D,|q|\,{\rm d}x)}\ \le \ 0 \end{eqnarray*}$$

for sufficiently large j. Let $v_j\in H^1_{\rm loc}(\mathbb {R}^3)$ be the corresponding radiating solutions of (5.2). Then w_j⇀w₀ weakly in L²(D, |q| dx) for some subsequence and some w₀ ∈ L²(D, |q| dx). It is easy to see that w₀ ∈ X(k₀) and v_j⇀v₀ weakly in H¹(B(0, R)) for every ball B(0, R) where $v_0\in H^1_{\rm loc}(\mathbb {R}^3)$ is the radiating weak solution to $\Delta v_0+k_0^2(1+q)v_0= -q\,w_0$ in $\mathbb {R}^3$ (see lemma 2.4 for an existence proof). From (2.16) for f = w_j we conclude that

$$\begin{eqnarray*} &&{\rm Im} (T(k_j)w_j,w_j )_{L^2(D,|q|\,{\rm d}x)}= k_j^3\Vert v_j^\infty \Vert ^2_{L^2(S^2)} . \end{eqnarray*}$$

The left-hand side converges to zero, the right-hand side to $k_0^3\Vert v_0^\infty \Vert ^2_{L^2(S^2)}$ by lemma 2.4 again. Therefore, $v_0^\infty =0$ and thus v₀ vanishes outside D by Rellich's lemma. Because k₀ is not an interior transmission eigenvalue we conclude that w₀ and v₀ vanish everywhere; that is, w_j and v_j converge weakly to zero. Now we compute, similarly to (2.14),

$$\begin{eqnarray*} (Tw_j,w_j)_{L^2(D,|q|\,{\rm d}x)} & = & \int \limits _D|w_j|^2 q \,{\rm d}x + k_j^2 \int _Dv_j \overline{w_j} q \,{\rm d}x \nonumber \\ & = & \Vert w_j\Vert ^2_{L^2(D,|q|\,{\rm d}x)} - k_j^2\int _{|x|<R} v_j (\Delta \overline{v_j}+k_j^2(1+q)\overline{v_j}) \,{\rm d}x \nonumber \\ & = & \Vert w_j\Vert ^2_{L^2(D,|q|\,{\rm d}x)} + k_j^2\int _{|x|<R} [|\nabla v_j|^2-k_j^2(1+q) |v_j|^2 ]\, {\rm d}x \nonumber \\ & & -\ k_j^2\int _{|x|=R}v_j \frac{\partial \overline{v_j}}{\partial \nu } \,{\rm d}s \end{eqnarray*}$$

and thus, taking the real part,

$$\begin{eqnarray*} k_j^2\int \limits _{|x|<R}|\nabla v_j|^2 \,{\rm d}x & = & {\rm Re} (T(k_j)w_j, w_j )_{L^2(D,|q|\,{\rm d}x)} - \Vert w_j\Vert ^2_{L^2(D,|q|\,{\rm d}x)} \\ & & +\ k_j^4\int \limits _{|x|<R}(1+q)|v_j|^2 \,{\rm d}x + k_j^2\int _{|x|=R}v_j \frac{\partial \overline{v_j}}{\partial \nu } \,{\rm d}s \\ & \le & k_j^4\int \limits _{|x|<R}(1+q) |v_j|^2 \,{\rm d}x + k_j^2\int \limits _{|x|=R}v_j \frac{\partial \overline{v_j}}{\partial \nu } \,{\rm d}s \end{eqnarray*}$$

which converges to zero by lemma 2.4 and the compact imbedding of H¹(B(0, R)) into L²(B(0, R)). Therefore, v_j converges to zero in H¹(B(0, R)) for every B(0, R) which implies w_j → 0 in L²(D, |q| dx), a contradiction to $\Vert w_j\Vert _{L^2(D,|q|\,{\rm d}x)}=1$. This ends the proof of part (b).

(c) As for the proof of part (a), we apply lemma 5.1(a), noting that α = 2σk₀A > 0 by theorem 6.1 and the assumption q < 0 (i.e., σ = −1). This yields (6.9). The proof of part (d) follows in the same way as the proof of part (b), with obvious adaptions due to the different sign of q. □

The last theorem has been formulated for the scattering operator $\mathcal {S}$. Of course, there is an analogous result for the far field operator F. Let λ_n(k) be the eigenvalues of F(k). From lemma 4.1 we recall that the projections s_n(k) = λ_n(k)/|λ_n(k)| onto the unit circle in $\mathbb {C}$ satisfy Ims_n(k) > 0 for all n and lim_{n → ∞}s_n(k) = ±1 provided q≷0. In particular, if q > 0 there exists a unique $s_-(k)\in \mathbb {C}$ with |s₋(k)| = 1 and Ims₋(k) > 0 and

$$\begin{equation} {\rm Re}\, s_-(k)= \min \lbrace {\rm Re}\, s_n(k):n\in \mathbb {N}\rbrace , \end{equation} \tag{ 6.10 }$$

while for q < 0 there exists a unique $s_+(k)\in \mathbb {C}$ with |s₊(k)| = 1 and Ims₊(k) > 0 and

$$\begin{equation} {\rm Re}\, s_+(k)= \max \lbrace {\rm Re}\, s_n(k):n\in \mathbb {N}\rbrace . \end{equation} \tag{ 6.11 }$$

Corollary 6.4 (Inside–outside duality for F). Let k₀ > 0 and I = (k₀ − ε, k₀ + ε)∖{k₀} such that no k ∈ I is an interior transmission eigenvalue. Let λ_n(k) be the eigenvalues of F(k). Define s₋(k) and s₊(k) as above by (6.10) and (6.11), respectively. (That is, as the projection of λ_n onto the unit circle which is furthest to the left if q > 0 and furthest to the right if q < 0, respectively.)

• Let k₀ be the smallest interior transmission eigenvalue and let q = q₀ satisfy (6.3). Then
  $$\begin{equation} \lim \limits _{k\in I,\ k\nearrow k_0}s_-(k)= -1. \end{equation} \tag{ 6.12 }$$
• Let (6.12) hold and q > 0 in D. Then k₀ is an interior transmission eigenvalue.
• Let k₀ be the smallest interior transmission eigenvalue and let q = q₀ satisfy the assumption of (6.6) in theorem 6.2. Then
  $$\begin{equation} \lim \limits _{k\in I,\ k\nearrow k_0} s_+(k) = +1. \end{equation} \tag{ 6.13 }$$
• Let (6.13) hold and q < 0 in D. Then k₀ is an interior transmission eigenvalue.

Proof. We only consider the case q > 0 of (a) and (b) since the case q < 0 of (c) and (d) can be shown analogously. By $\mathcal {S}=I+({\rm i}k)/(8\pi ^2)\,F$ we observe that μ_n is an eigenvalue of $\mathcal {S}$ if, and only if, the number

$$\begin{equation*} \lambda _n= \frac{8\pi ^2 {\rm i}}{k} ( 1-\mu _n ) = \frac{8\pi ^2 {\rm i}}{k} ( 1 - \exp (- 2 {\rm i}\delta _n) ) \end{equation*}$$

is an eigenvalue of F. Furthermore, we note that

$$\begin{eqnarray*} &&{\rm Re}\, s_n(k)= -\frac{\sin (2\delta _n(k))}{\sqrt{\sin ^2(2\delta _n(k))+ (1-\cos (2\delta _n(k)))^2}} \end{eqnarray*}$$

and this is minimal for minimal δ_n (provided δ_n ∈ (0, π/2)). Furthermore, $\lim _{k\nearrow k_0}\delta _n(k)=0$ is equivalent to $\lim _{k\nearrow k_0}{\rm Re}\, s_n(k)=-1$. Now the claim follows directly from theorem 6.3. □

We want to illustrate the statements of theorem 6.3 and of lemma 4.1 with a simple and explicit numerical example. Consider the scattering from a penetrable ball $B_R \subset \mathbb {R}^3$ of radius R > 0, centered at the origin, and assume that the refractive index inside B_R equals n₀ = (1 + q₀)^1/2 for a constant q₀ > −1. For this setting it is well known that one can compute F(k): L²(S²) → L²(S²) semi-analytically. Indeed, explicit computations using spherical Bessel and Hankel functions j_l and h_l of order $l \in \mathbb {N}_0$, respectively, and spherical harmonics $Y^m_l$, one computes that the far field operator can in this special case be represented as

$$\begin{equation} F(k)g= \frac{16\pi ^2}{{\rm i}k}\sum _{l=0}^\infty \sum _{|m|\le l} \frac{n_0\,j_l^{\prime }(kn_0R)\,j_l(kR)\!-\!j_l(kn_0R)\,j_l^{\prime }(kR)}{j_l(kn_0R)\, h_l^{\prime }(kR)\!-\!n_0h_l(kR)\,j_l^{\prime }(kn_0R)}\ g^m_l\,Y^m_l, \end{equation} \tag{ 6.14 }$$

with Fourier coefficients $g^m_l$ of g ∈ L²(S²) given by

$$\begin{equation*} g^m_l = \int _{S^2} g(\hat{x}) \overline{Y^m_l}(\hat{x}) \,{\rm d}s(\hat{x}), \qquad l\in \mathbb {N}_0, |m| \le l, \end{equation*}$$

see, e.g., [7] for similar computations in the case of a sound-soft ball. Hence, the eigenvalues of F(k) are

$$\begin{equation} \lambda _l^{\mathrm{sph}}(k)= \frac{16\pi ^2}{{\rm i}k}\frac{n_0\, j_l^{\prime }(kn_0R)\, j_l(kR)-j_l(kn_0R)\,j_l^{\prime }(kR)}{j_l(kn_0R)\,h_l^{\prime }(kR)-n_0\,h_l(kR)\,j_l^{\prime }(kn_0R)}, \qquad l\in \mathbb {N}_0. \end{equation} \tag{ 6.15 }$$

Interior transmission eigenvalues are precisely those k₀ > 0 for which the numerator in the last expression vanishes. Note that the order of the eigenvalues $\lambda _l^{\mathrm{sph}}$ comes from the series expression of the far field operator in (6.14); in particular, the order is not chosen in accordance with (6.10) or (6.11).

According to lemma 4.1, as l → ∞ the eigenvalues $\lambda _l^{\mathrm{sph}}(k)$ should approach zero from the right (left) if the contrast is positive (negative). We confirm this statement for the above-described spherical setting by plotting the eigenvalues of F(k) for k = 20 and for the contrast q₀ = ±0.5. Figure 1 shows that for q₀ = 0.5 the eigenvalues converge from the right to zero, whereas for q₀ = −0.5, they converge from the left to zero. In this and in all the following examples, the radius R of the ball is equal to one.

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Figure 2 confirms the statement of theorem 6.3. For q₀ = 0.9 > 0, the function k↦δ₋(k) plotted in (a) tends to 0 as k approaches the smallest transmission eigenvalue from below (and all further ones occurring in the considered interval). For q₀ = −0.9 < 0, the function k↦δ₊(k) plotted in (b) tends to π as k approaches the smallest transmission eigenvalue from below (and all further ones occurring in the considered interval). For each k, the minimum and the maximum used to define $\delta _-(k) = \min _{n\in \mathbb {N}}\delta _n$ and $\delta _+(k) = \max _{n\in \mathbb {N}}\delta _n$, respectively, is computed using the first 300 values of $\lambda _l^{\mathrm{sph}}(k)$.

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Finally, we plot that the eigenvalue curves $k \mapsto \lambda _1^{\mathrm{sph}}(k)$ of F(k) (see (6.15)) for the special spherical setting indicated above. To this end, we use 100 equidistributed wavenumbers between k_min and k_max. Again, we consider positive and negative constant contrasts q₀ = 1.5 and q₀ = −0.9, respectively. Figure 3(a) shows that for q₀ = 1.5 the eigenvalue $\lambda _1^{\mathrm{sph}}(k)$ approaches 0 twice from the left as k increases. For q₀ = −0.9, figure 3(b) shows that $\lambda _1^{\mathrm{sph}}(k)$ approaches 0 twice from the right as k increases. Both observations fit the statement of corollary 4.1, since s₊(k) → −1 and s₋(k) → 1 as k↗k₀ means that the corresponding eigenvalue of F(k) approaches 0 from the left and right, respectively. In the special case of a spherical scatterer, zero is an eigenvalue of F(k₀) for any transmission eigenvalue k₀ and, even more, the curves $\lambda _l^{\mathrm{sph}}(k)$ depend smoothly on k > 0. This is not valid in general. We already mentioned in the introduction that it may happen that the far field operator at a transmission eigenvalue is injective.

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We would like to thank the anonymous referees for carefully reading the manuscript, in particular for finding a serious mistake in a former version of the text. The research of AL was supported through an exploratory project granted by the University of Bremen in the framework of its institutional strategy, funded by the excellence initiative of the federal and state governments of Germany.