We now aim for the proof of Theorem4.5. To do so, we start by finding a minimax formula for the eigenphases ofSk.
Theorem 4.12 ([EP95], Theorem 6.1). Let k >0 be such that k2 6∈σ(−∆D). Then for j∈N,
cotθj(k) = inf Ej+1 sup u∈EJ+1\{0} (u,Yku) (u,Jku)
where the infimum is taken over all subspaces Ej+1⊂ HΓ of dimension j+ 1.
Note that since we start indexing at j = 0 the zeroth eigenvalue will be an infimum over subspaces of dimension 1, which makes sense. To prove Theorem 4.12, we need some more con- structions. For k >0 withk26∈σ(−∆D), we introduce the (inverse) Cayley transform ofSk:
Xk=i(1 +Sk)(1−Sk)−1.
That this is well-defined follows from Theorem4.1and the factk26∈σ(−∆D), so (1−Sk) has dense range and hence Xk has dense domain (noting D(Xk) = Ran(1−Sk)). There is a subtle point here, in that we require k2 6∈ σ(−∆D) for k in some interval to apply the rest of the method of proof. This can be satisfied however, sinceDis bounded with continuous boundary and hence the discussion in Section2.1concludes that σ(−∆D) is discrete. It can be checked that unitarity ofSk implies Xk is self adjoint. Then applying the spectral mapping theorem (Theorem 1.19) provides the following lemma:
Lemma 4.13 ([EP95], Lemma 6.2). Let k > 0 with k2 6∈σ(−∆
D). Let θ∈ (0, π). Then e−2iθ is an eigenvalue of Sk of degeneracyM if and only if cotθ is an eigenvalue of Xk of degeneracy M.
This lemma combined with Theorem 4.4then shows thatXk is bounded below and has a finite number of negative eigenvalues.
With these two facts, we can explain the proof of Theorem 4.12. Since Xk is self adjoint and bounded below, its eigenvalues satisfy the common minimax formula:
cotθj(k) = inf E0 j+1 sup f∈E0 J+1\{0} (f, Xkf) (f, f)
where the infimum is over subspaces Ej0+1 ⊂D(Xk) of dimension j+ 1. This minimax formula is then combined with the representationSk = 1−2πiLkA−k1L∗k of Theorem4.10 and the factorisa- tions of Lemmas4.7and 4.11. There is a lot of simplification and density argument to interchange the set over which the infimum is taken, which then finishes the proof. For notational conve- nience, Eckmann and Pillet only prove this for the smallest eigenvaluej= 0, but their proof easily generalises.
We now have enough background material to prove Theorem 4.5.
Sketch of Proof of Theorem 4.5. We first prove one implication, that Dirichlet eigenvalues imply convergence of scattering matrix eigenvalues. Fix k0 > 0 such that k02 ∈ σ(−∆D) is an M-fold eigenvalue, and let P be the orthogonal projection onto ker(Ak0) ⊂ HΓ. Lemma 4.6 shows that
this kernel has dimension M. Now fixu∈ker(Ak0) and considerk < k0, withk
2 6∈σ(−∆
D). Due to holomorphicity ofAk, we may write
(u,Aku) = (u, PAkP u) = u, P Ak0 + (k−k0)A 0 k0+O (k−k0) 2 P u = (u, PAk0P u) + (k−k0) u, PA 0 k0P u +O (k−k0)2 ||u||2 = (k−k0) u, PA0k0P u +O (k−k0)2 ||u||2.
It is shown in Lemma 5.4 of [EP95] that PA0k0P is a positive operator. Then since ker(Ak0) is
finite dimensional there exists some C1>0 with:
u, PA0k0P u≥C1||u||2. So whenk < k0, we have (u,Yku) =<(k−k0) u, PA0k0P u +O (k−k0)2 ||u||2 ≤C1(k−k0)||u||2+O (k−k0)2 ||u||2 and (u,Jku)≥C2(k−k0)2||u||2 for someC2>0. So (u,Yku) (u,Jku) ≤ C1(k−k0)||u|| 2+O (k−k 0)2 ||u||2 C2(k−k0)2||u||2 = C1 C2(k−k0) (1 +O(k−k0)).
4.3. SPECTRAL DUALITY - HARD VERSION 43
This converges to −∞ ask↑k0, which - combined with Lemma 4.13and the minimax formula of
Theorem 4.12- shows that cotθj(k)↓ −∞for some j, and hence θj(k) ↑π. Applying this process for each eachuin a basis for ker(Ak0) shows that there are at leastM eigenphases ofSkconverging
upwards to π ask↑k0.
Now we prove the converse implication. Assume that ask↑k0, there are exactlyM eigenphases
θj(k) of Sk converging upwards to π. Then by Theorem 4.12 and Lemma 4.13, for each k there exists a subspaceEk⊂ HΓ of dimensionM, such that
sup u∈Ek\{0}
(u,Yku) (u,Jku)
≤ −λk (4.1)
for someλk which converge to ∞ ask↑k0.
Now, to avoid introducing another operator that was required for the full proof of the main theorem of [EP95], we will skip some steps and say that Equation 4.1can be used to produce an orthogonal projection of rank at leastM which maps into kerJk0|
H−
1 2 Γ
, and so dim kerJk0|
H−
1 2 Γ
≥M. Combined with Theorem4.6and Lemma4.8, this implies that dim ker(−∆D−k2)≥M, and hence the converse implication is proven.
Thus Theorem4.5 is proven.
For the interested reader, the operator we have avoided introducing in the last paragraph of the proof is Λ = (1 + (i∂s)2)12, where ∂s is the arc-length tangential derivative on Γ, which has
many significant impacts on the proofs of all preceding theorems taken from [EP95], and is used in defining the fractional Sobolev spaces HΓβ as mentioned (briefly) in Section4.3.1.
The proof of Theorem 4.5was quite technical and very challenging, and gives some interesting insights into any possible generalisations to potential scattering. The main point of the proof was the factorisation of Lemma 4.10, in which the scattering matrix is expressed in a form involving an operator with strong links to the Dirichlet eigenvalue problem on D. While this may not be possible in the potential case, this chapter illuminates many other significant spectral properties which may need to be investigated, such as one-sided accumulation and monotone evolution in k. Investigating possible generalisations of Theorem 4.5 will be the main feature of the rest of this thesis.
Chapter 5
Generalisations of Spectral Duality to
Potential Scattering
In this section, we will develop some possible extensions of the result in [EP95] to potential scatter- ing, for potentials of compact support. This of course has many difficulties, as the previous results are relating an “inside” problem and an “outside” problem, with respect to the obstacle under consideration. In the case of potential scattering, it is not clear what the correct generalisation of “inside” and “outside” should be. The result in [EP95] also had an obvious choice of boundary conditions.
Before we get to the possible generalisations, we can consider what it actually means for the Scattering matrix SV(k0) for potential V with compact support in D⊂Rn (an open bounded set with smooth boundary) to have eigenvalue 1. Recalling Remark 4.2, we find that the scattering matrix has an eigenvalue 1 at frequency k0 if and only if there exists a scattering solution of
(−∆ + V −k02)uV = 0 which agrees identically with a scattering solution of the free equation (−∆−k2
0)u0= 0 outside ofD. This leads to a few possible generalisations of the results in [EP95]
to potential scattering. Investigating these will be the main aim of this chapter.
5.1
Dirichlet to Neumann Boundary Condition Generalisation
The first attempt at a generalisation is as follows. From above, SV(k0) has an eigenvalue 1 if and
only if we have two non-zero functions uV and u0 on all of Rn satisfying (−∆ +V −k02)uV = (−∆−k20)u0 = 0, and they agree identically outside ofD.
Thenu0 restricted to D satisfies:
(
(−∆−k2
0)u0 = 0 onD
u0 = uV|∂D on∂D.
So Λ0(k0)(uV|∂D) =dnu0. But uV and u0 agree up to the boundary of D, sodnu0 =dnu0. So we
find thatuV satisfies the boundary value problem: 45
(
(−∆ +V −k20)uV = 0 on D
dnuV = Λ0(k0)uV|∂D on ∂D
(5.1) Therefore, it might be possible to use this strange boundary value problem to try to generalise the results in [EP95]. Before even considering the nature of solutions as we takek↑k0as in [EP95],
we should check that a solution of Equation 5.1 will actually provide us with an eigenvalue 1 for
SV(k0).
Suppose we are given a solution to the above boundary value problem ˜uV, with corresponding solution to the Helmholtz equation ˜u0. The issue is then finding a scattering solution on Rn\Dof (−∆−k2
0)˜u0 = 0 with boundary conditions ˜u0|∂D = ˜uV|∂D and dnu˜≡ −dnu˜V, because this could be used to extend ˜uV to a scattering solution of (−∆ +V −k20)uV = 0 on the whole space (noting the normal vectors for each region point in opposite directions). Similarly, we could extend ˜u0 to
a scattering solution of (−∆−k02)u0 = 0 on the whole space with the same extension, since the
boundary value and normal derivatives of ˜u0 and ˜uV agree.
So we can recover an eigenvalue 1 ofSV(k0) if we can find a scattering solution of
(−∆−k02)˜u0 = 0 on Rn\D ˜ u0 = u˜V on ∂D dn˜u0 = −dnu˜V.