On Positivity Sets for Helmholtz Solutions

We address the question of finding global solutions of the Helmholtz equation that are positive in a given set. This question arises in inverse scattering for penetrable obstacles. In particular, we show that there are solutions that are positive on the boundary of a bounded Lipschitz domain.


Introduction
The objective in this short note is to consider the following problem.
Question 1.1.Let k > 0 and let E be a subset of R n (n ≥ 2).Does there exist a solution of (∆ + k 2 )u = 0 in R n with u| E > 0?
Note that any solution of the Helmholtz equation (∆ + k 2 )u = 0 is C ∞ , and thus the condition u| E > 0 can be understood pointwise.There is a substantial literature on zero sets of solutions of elliptic equations and eigenfunctions, as discussed in the review [LM20].In our setting, any real valued solution of (∆ + k 2 )u = 0 in R n must have a zero in any closed ball of radius jn−2 2 ,1 k −1 where jn−2 2 ,1 is the first zero of the Bessel function J n−2

2
(see e.g.[SS21, Lemma 3.1]).Question 1.1 above is related to producing a global solution whose zero set avoids a given set E.
Our motivation comes from inverse scattering theory and the works [CV21, SS21, KLSS22].In these works, one considers a bounded open set D ⊂ R n (penetrable obstacle) together with a coefficient h ∈ L ∞ (R n ) with |h| ≥ c > 0 a.e.near ∂D (contrast), and asks whether it is possible to find a solution u 0 ≡ 0 of (∆ + k 2 )u 0 = 0 in R n (incident wave) such that the obstacle D with contrast h does not produce any scattering response.The last condition can be precisely formulated as the existence of a function u solving If this happens for some contrast h, then the obstacle D is called a non-scattering domain and it will be invisible with respect to probing with the incident wave u 0 .It was proved in [SS21, Theorem 2.1] that if D has real-analytic boundary and if there is an incident wave u 0 with u 0 | ∂D > 0, then D is a non-scattering domain.Similarly, the work [KLSS22] introduced the notion of quadrature domains for the Helmholtz operator ∆ + k 2 and proved that if D is such a domain, and if there is an incident wave u 0 with u 0 | ∂D > 0, then D is a non-scattering domain.On the other hand, the works [CV21,SS21] show that under a nonvanishing condition for u 0 on ∂D, the boundary of a non-scattering domain can be interpreted as a free boundary in an obstacle-type problem and hence such a domain must be either regular or have thin complement near any boundary point.
It was also proved in [SS21] that one may be able to find incident waves that are positive on the boundary of a bounded C1 domain (Lipschitz if n = 2, 3).Our first main result extends this to Lipschitz domains in any dimension.
Then there exists a Herglotz wave function u 0 (see Definition 2.1) satisfying The proof of Theorem 1.1 is done in two steps.One first constructs a solution v of (∆ + k 2 )v = 0 in D with v| ∂D > 0 by solving a Dirichlet problem.Then one approximates v in D by a suitable Herglotz wave u 0 in R n via a Runge approximation argument.This approximation needs to be done in a suitable norm to obtain the pointwise condition u 0 | ∂D > 0, but since D only has Lipschitz boundary the solution v is not very regular and this limits the choice of possible norms.We will work with fractional Sobolev spaces H s,p and invoke the theory of boundary value problems in Lipschitz domains. 1  We remark that the assumption in Theorem 1.1 that k 2 is not an eigenvalue is necessary, at least when D is a ball (see Example 2.5).For the first eigenvalue this was pointed out in [SS21, Remark 3.2].
Another instance of subsets E ⊂ R n where one can arrange u 0 | E > 0 is given in the following result.
The proof is similar to that of Theorem 1.1, except that in the first step we use the Faber-Krahn inequality to produce a solution v that is positive near E.
Remark 1.3.If E is sufficiently nice and low dimensional, it may be possible to use Theorem 1.2 to find solutions that are positive on E. For example, let E be a smooth compact manifold with This holds e.g. when m < n/2 by the Whitney embedding theorem, or when having smooth boundary ∂D and arbitrarily small measure [MT97, Theorem 9.23 and Remark 9.24] (see also [Lee13,Theorem 6.24]).Since R n \ E is connected, one can connect any two points in R n \ D by a curve γ in R n \ E. By considering the curve F (γ) where F is a continuous map on R n that fixes R n \ D and collapses D \ E to ∂D, we see that R n \ D is connected.Since D has smooth boundary, also R n \ D is connected.(See [CF77, pages 61-62] for a related discussion.)Thus we may apply Theorem 1.2 to find a Herglotz wave function u 0 satisfying (1.1).Note that the connectedness of R n \ E can fail when E has dimension n − 1.

Solutions satisfying the positivity condition
In this section we will prove Theorems 1.1 and 1.2.We begin with some preparations.
2.1.Fractional Sobolev spaces.For each s ∈ R and 1 < p < ∞, the fractional Sobolev space H s,p (R n ) is the Banach space equipped with the norm where D s is the the Bessel potential of order s, i.e. the Fourier multiplier corresponding to ξ s = (1 + |ξ| 2 ) s 2 .In particular when s = k ≥ 1 is an integer, we also have From [BL76, Corollary 6.2.8], we have the duality statement where (p ′ ) −1 + p −1 = 1.We also recall the Sobolev embedding ([BL76, Theorem 6.5.1]): (2.2) This is a Banach space equipped with the quotient norm When D is a bounded Lipschitz domain, from [JK95, Theorem 2.3] we know that there exists a bounded linear extension operator 2.2.Runge-Herglotz approximation.The next objective is to prove a result stating that solutions in H s,p (D) can be approximated in D by Herglotz waves.We first give a definition.
Definition 2.1.Let k > 0 and consider the operator The functions u = P k f with f ∈ C ∞ (S n−1 ) are called Herglotz waves, and they are particular solutions of If v is real-valued, then so are u j .
The proof of Proposition 2.2 is very similar to [SS21, Proposition 3.4] that considered approximation in W 1,p (D).Here we need to work with fractional Sobolev spaces instead.
Proof.In view of the Hahn-Banach theorem, it is enough to prove that any bounded linear functional ℓ : H s,p (D) → C that vanishes on Let ℓ be such a linear functional, and define a bounded linear functional ℓ 1 : where (•, •) is the sesquilinear distributional pairing in R n .It is easy to see that µ = 0 in R n \ D, and the condition ℓ We now define the distribution w := Φ k * µ, where is the outgoing fundamental solution of the Helmholtz operator −(∆ + k 2 ) and H (1) α is the Hankel function (see [Yaf10, §1.2.3]).Then w is a distributional solution of (2.7) Elliptic regularity yields w ∈ H 2−s,p ′ loc (R n ), and since supp(µ) ⊂ D we also have that Given any f ∈ C ∞ (S n−1 ), we write u = P k f ∈ C ∞ (R n ).Using (2.5) and the fact that µ has compact support, we have where (•, •) Br is the sesquilinear distributional pairing in the ball B r .We now consider a cut-off function χ ∈ C ∞ c (R n ) satisfying 0 ≤ χ ≤ 1 and χ = 1 near D. Using (2.7), we can write (2.8) as where ∂ |x| = x• ∇ denotes the radial derivative.Here we also used the fact that (∆+ k 2 )u = 0 in R n .
Using [Mel95, Lemma 1.2 and equation (1.18)], we know that the Herglotz function u = P k f has the following asymptotics as |x| → ∞: On the other hand, from [Yaf10, equation (2.27)], we know that w has the asymptotics is the Fourier transform of the compactly supported distribution µ.
Combining (2.9) with (2.10a)-(2.10d),we obtain By the fact that f ∈ C ∞ (S n−1 ) was arbitrary, we conclude μ(kx) = 0 for all x ∈ S n−1 .Consequently, (2.10c) becomes In other words, the far-field pattern of w is vanishing.By the Rellich uniqueness theorem [CK19,Hör73], the unique continuation principle and the connectedness of R n \ D, we conclude that From (2.4), we know that there are which is our desired result.2.3.Proof of the main result.Theorem 1.1 is an immediate consequence of the following result: Before we prove Theorem 2.3 we need the following result, which is a special case of [JK95, Theorems 1.
then there exists a unique u ∈ H s,p (D) satisfying −∆u = f in D and u = 0 on ∂D.
Proof.We first consider the case when n ≥ 3. Let p 0 be as in [JK95, Theorem 1.1] (with The case when n = 2 can be proved using identical reasoning using [JK95, Theorem 1.3] and the observation 3 p ≤ 2 p + 1 2 .Proof of Theorem 2.3.Since k 2 is not a Dirichlet eigenvalue in D, there exists a unique solution v ∈ H 1,2 (D) such that If v ∈ H s,p (D) for some 0 < s ≤ 1 and p > n/s, using Proposition 2.2, we know that there exist Herglotz waves where we used the Sobolev embedding.
It remains to show that v ∈ H s,p (D) for some s, p with s > n/p, and this follows from a standard bootstrap argument based on Proposition 2.4.We claim that where The case j = 0 follows since v ∈ H 1,2 (D).We argue by induction and assume that this holds for j.Define w := v − c 0 and note that w solves ,p j (D), w| ∂D = 0.
We have proved that v ∈ H 2 p j ,p j (D) where j is the largest integer < n−2 4 .Using the above notation, we have ∆w ∈ H 2 p j ,p j (D) and w| ∂D = 0.By Sobolev embedding we have ∆w ∈ H s−2,p (D) whenever p ≥ p j and The last condition implies that 4 − 1, using Proposition 2.4 once again we obtain that w and hence v is in H s,p for some s > n/p.On the other hand, if j = n−2 4 − 1 we iterate the argument once more to get v ∈ H s,p for some s > n/p.This concludes the proof.
The next simple example shows that the condition that k 2 is not an eigenvalue is necessary at least for balls., we have We see that (∆ + 1)u 1 = 0 in R n .The above discussion shows that u 0 must change sign on ∂B R .
The following strong maximum principle can be found in [KLSS22, Appendix A].However, for readers' convenience, here we exhibit the statement as well as its proof.
Lemma 2.6 (Strong maximum principle).Let D be a bounded Lipschitz domain in R n (n ≥ 2), and let k 2 < λ 1 (D), where λ 1 (D) > 0 denotes the smallest H 1 0 (D)-eigenvalue of −∆.If the solution u ∈ H 1 (D) satisfies Proof.It is easy to see that for each component G of D we have k 2 < λ 1 (G) and Testing the equation above by u − ∈ H 1 0 (G) and using Poincaré inequality, we have Case 2. If λ 1 (G) > k 2 , then there exists a unique solution v ∈ H 1 (G) such that Using the strong maximum principle in Lemma 2.6, we know that v > 0 in G.
Next we choose a bounded Lipschitz domain D 1 that satisfies E ⊂ D 1 , D 1 ⊂ D, and R n \ D 1 is connected.The function v| D 1 is in H 1,p (D 1 ) for any p > n and satisfies v| D 1 > 0. The approximation result in Proposition 2.2 yields a sequence of Herglotz waves u j satisfying u j | D 1 − v H 1,p (D 1 ) → 0 as j → ∞.
If j is sufficiently large, the Sobolev embedding ensures that u j | E > 0.