On Positivity Sets for Helmholtz Solutions

This is a self-archived version of an original article. This version may differ from the original in pagination and typographic details. Author(s): Title: Year: Version: Copyright: Rights: Rights url: Please cite the original version: CC BY 4.0 https://creativecommons.org/licenses/by/4.0/ On Positivity Sets for Helmholtz Solutions © 2023 the Authors Published version Kow, Pu-Zhao; Salo, Mikko; Shahgholian, Henrik Kow, P.-Z., Salo, M., & Shahgholian, H. (2023). On Positivity Sets for Helmholtz Solutions. Vietnam Journal of Mathematics, 51(4), 985-994. https://doi.org/10.1007/s10013-023-00646-y 2023 Vietnam Journal of Mathematics https://doi.org/10.1007/s10013-023-00646-y ORIGINAL ARTICLE On Positivity Sets for Helmholtz Solutions Pu-Zhao Kow1·Mikko Salo2·Henrik Shahgholian3 Received: 25 October 2022 / Accepted: 2 March 2023 © The Author(s) 2023 Abstract 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. Keywords Helmholtz equation ·Acoustic equation ·Lipschitz domain · Inverse scattering problem Mathematics Subject Classification (2010) 35J05 ·35J15 ·35J20 ·35R30 ·35R35 1 Introduction The objective in this short note is to consider the following problem. Question 1.1 Let k>0andletEbe a subset of Rn(n≥2). Does there exist a solution of (Δ +k2)u=0inRnwith u|E>0? Note that any solution of the Helmholtz equation (Δ +k2)u=0isC∞, 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 [11]. In our setting, any real valued solution of (Δ +k2)u=0inRnmust have a zero in any closed ball Dedicated to Carlos E. Kenig on the occasion of his 70th birthday BHenrik Shahgholian [email protected] Pu-Zhao Kow pzko[email protected] Mikko Salo [email protected] 1Department of Mathematical Sciences, National Chengchi University, No. 64, Sec. 2, ZhiNan Rd., Wenshan District, 16302 Taipei, Taiwan 2Department of Mathematics and Statistics, University of Jyväskylä, P.O. Box 35 (MaD), FI-40014 Jyväskylä, Finland 3Department of Mathematics, KTH Royal Institute of Technology, SE-10044 Stockholm, Sweden 123 P.-Z. Kow et al. of radius jn−2 2,1k−1where jn−2 2,1is the first zero of the Bessel function Jn−2 2(see e.g. [14, 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 [2,9,14]. In these works, one considers a bounded open set D⊂Rn(penetrable obstacle) together with a coefficient h∈L∞(Rn)with |h|≥c>0 a.e. near ∂D(contrast), and asks whether it is possible to find a solution u0≡ 0of(Δ +k2)u0=0inRn(incident wave) such that the obstacle Dwith contrast hdoes not produce any scattering response. The last condition can be precisely formulated as the existence of a function usolving (Δ +k2+hχD)u=0inRn, u=u0outside some ball. If this happens for some contrast h, then the obstacle Dis called a non-scattering domain and it will be invisible with respect to probing with the incident wave u0. Itwasprovedin[14, Theorem 2.1] that if Dhas real-analytic boundary and if there is an incident wave u0with u0|∂D>0, then Dis a non-scattering domain. Similarly, the work [9] introduced the notion of quadrature domains for the Helmholtz operator Δ+k2and proved that if Dis such a domain, and if there is an incident wave u0with u0|∂D>0, then Dis a non-scattering domain. On the other hand, the works [2,14] show that under a nonvanishing condition for u0on ∂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. Itwasalsoprovedin[14] that one may be able to find incident waves that are positive on the boundary of a bounded C1domain (Lipschitz if n=2,3). Our first main result extends this to Lipschitz domains in any dimension. Theorem 1.1 Let D ⊂Rn(n≥2)be a bounded Lipschitz domain such that Rn\Dis connected. Suppose that k2>0is not a Dirichlet eigenvalue of −Δin D. Then there exists a Herglotz wave function u0(see Definition 2.1)satisfying (Δ +k2)u0=0in Rnand u0|∂D>0. The proof of Theorem 1.1 is done in two steps. One first constructs a solution vof (Δ +k2)v =0inDwith v|∂D>0 by solving a Dirichlet problem. Then one approximates vin Dby a suitable Herglotz wave u0in Rnvia a Runge approximation argument. This approximation needs to be done in a suitable norm to obtain the pointwise condition u0|∂D> 0, but since Donly has Lipschitz boundary the solution vis not very regular and this limits the choice of possible norms. We will work with fractional Sobolev spaces Hs,pand invoke the theory of boundary value problems in Lipschitz domains.1 We remark that the assumption in Theorem 1.1 that k2is not an eigenvalue is necessary, at least when Dis a ball (see Example 2.5). For the first eigenvalue this was pointed out in [14, Remark 3.2]. Another instance of subsets E⊂Rnwhere one can arrange u0|E>0 is given in the following result. Theorem 1.2 Let k >0, and let D ⊂Rn(n≥2) be a bounded Lipschitz domain such that Rn\D is connected and |D|≤|Br|where r =jn−2 2,1k−1.IfE⊂D is compact, then there 1This is one of the areas where Carlos Kenig has made pioneering contributions. 123 On Positivity Sets for Helmholtz Solutions exists a Herglotz wave function u0(see Definition 2.1)satisfying (Δ +k2)u0=0in Rnand u0|E>0.(1.1) 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 vthat is positive near E. Remark 1.3 If Eis 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 Ebe a smooth compact manifold with dim(E)=m≤n−2 embedded in Rn, which is homeomorphic to a compact submanifold E1of Rn−1∼ =Rn−1×{0}⊂Rn. This holds e.g. when m<n/2 by the Whitney embedding theorem, or when Eis homeomorphic to Sm.SinceRn\E1is connected, by [12, Corollary 7.9] one sees that Rn\Eis (pathwise) connected. One can construct a tubular neighborhood D={x∈Rn:d(x,E)<ε}of Ehaving smooth boundary ∂Dand arbitrarily small measure [12, Theorem 9.23 and Remark 9.24] (see also [10, Theorem 6.24]). Since Rn\Eis connected, one can connect any two points in Rn\Dby a curve γin Rn\E. By considering the curve F(γ ) where Fis a continuous map on Rnthat fixes Rn\Dand collapses D\Eto ∂D,weseethatRn\Dis connected. Since Dhas smooth boundary, also Rn\Dis connected. (See [5, pp. 61–62] for a related discussion.) Thus we may apply Theorem 1.2 to find a Herglotz wave function u0satisfying (1.1). Note that the connectedness of Rn\Ecan fail when Ehas dimension n−1. 2 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∈Rand 1 <p<∞, the fractional Sobolev space Hs,p(Rn)is the Banach space equipped with the norm uHs,p(Rn):= DsuLp(Rn), where Dsis 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≥1isaninteger,wealsohaveHk,p(Rn)= Wk,p(Rn),where Wk,p(Rn)={u∈Lp(Rn)|Dαu∈Lp(Rn)for all multi-indices αwith |α|≤k}. From [1, Corollary 6.2.8], we have the duality statement (Hs,p(Rn))∗=H−s,p(Rn)for all s∈Rand 1 <p<∞,(2.1) where (p)−1+p−1=1. We also recall the Sobolev embedding ([1, Theorem 6.5.1]): Hs,p(Rn)⊂Hs1,p1(Rn) whenever 1 <p≤p1<∞,−∞ <s1≤s<∞,ands−n p=s1−n p1. Let Dbe an open set in Rn.Wedefine Hs,p(D):= {u|D|u∈Hs,p(Rn)}for all s∈Rand 1 <p<∞. 123 P.-Z. Kow et al. This is a Banach space equipped with the quotient norm vHs,p(D):= inf{uHs,p(Rn)|u|D=v}. When Dis a bounded Lipschitz domain, from [8, Theorem 2.3] we know that there exists a bounded linear extension operator E:Hs,p(D)→Hs,p(Rn)with Eu =uin Dfor all u∈Hs,p(D). If F⊂Rnis closed, we define Hs,p F(Rn):= {u∈Hs,p(Rn)|supp(u)⊂F}. If Dis a bounded Lipschitz domain, the following result can be found in [8, Remark 2.7]: C∞ c(D)is dense in Hs,p D(Rn)for each s∈Rand 1 <p<∞.(2.2) 2.2 Runge–Herglotz Approximation The next objective is to prove a result stating that solutions in Hs,p(D)can be approximated in Dby Herglotz waves. We first give a definition. Definition 2.1 Let k>0 and consider the operator Pk:C∞(Sn−1)→C∞(Rn)defined by (Pkf)(x):= Sn−1 eikx·ˆzf(ˆz)dˆz,x∈Rn. The functions u=Pkfwith f∈C∞(Sn−1)are called Herglotz waves,andtheyare particular solutions of (Δ +k2)u=0inRn. Proposition 2.2 Let k >0,0<s≤1,1<p<∞, and let D ⊂Rn(n≥2) be a bounded Lipschitz domain such that Rn\D is connected. Given any v∈Hs,p(D)with (Δ+k2)v =0 in D, there exist Herglotz waves u j∈C∞(Rn)such that uj−vHs,p(D)→0as j →∞. If vis real-valued, then so are u j. The proof of Proposition 2.2 is very similar to [14, Proposition 3.4] that considered approximation in W1,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 :Hs,p(D)→Cthat vanishes on {Pkf|D|f∈C∞(Sn−1)}must also vanish on {v∈Hs,p(D)|−(Δ +k2)v =0inD}.Letbe such a linear functional, and define a bounded linear functional 1:Hs,p(Rn)→Cby 1(u):= (u|D). By duality (2.1), there exists a unique μ∈H−s,p(Rn)such that 1(u)=(u,μ) for all u∈Hs,p(Rn), where (·,·)is the sesquilinear distributional pairing in Rn. It is easy to see that μ=0in Rn\D, and the condition (Pkf|D)=0forall f∈C∞(Sn−1)implies that (Pkf,μ)=0forallf∈C∞(Sn−1). (2.3) 123 On Positivity Sets for Helmholtz Solutions We now define the distribution w:= Φk∗μ,where Φk(x)=ikn−2 2 4(2π)n−2 2 |x|−n−2 2H(1) n−2 2 (k|x|) is the outgoing fundamental solution of the Helmholtz operator −(Δ +k2)and H(1) αis the Hankel function (see [15, §1.2.3]). Then wis a distributional solution of −(Δ +k2)w =μin Rn.(2.4) Elliptic regularity yields w∈H2−s,p loc (Rn), and since supp(μ) ⊂Dwe also have that wis C∞in Rn\D. Given any f∈C∞(Sn−1),wewriteu=Pkf∈C∞(Rn).Using(2.3) and the fact that μhas compact support, we have 0=(u,μ)=lim r→∞(u,μ)Br,(2.5) where (·,·)Bris the sesquilinear distributional pairing in the ball Br.Wenowconsidera cut-off function χ∈C∞ c(Rn)satisfying 0 ≤χ≤1andχ=1 near D.Using(2.4), we can write (2.5)as 0=lim r→∞ (χu,(Δ+k2)w)Br+((1−χ)u,(Δ+k2)w)Br =lim r→∞ ((Δ +k2)(χu), w)Br+((Δ +k2)((1−χ)u), w)Br +∂Br (u∂|x|w−(∂|x|u)w) dS =lim r→∞ ∂Br (u∂|x|w−(∂|x|u)w) dS,(2.6) where ∂|x|=ˆx·∇ denotes the radial derivative. Here we also used the fact that (Δ+k2)u=0 in Rn. Using [13, Lemma 1.2 and equation (1.18)], we know that the Herglotz function u=Pkf has the following asymptotics as |x|→∞: u(x)=c n,k|x|−n−1 2eik|x|f(ˆx)+in−1e−ik|x|f(−ˆx)+O(|x|−n+1 2), (2.7a) ∂|x|u(x)=c n,k|x|−n−1 2ikeik|x|f(ˆx)−in−1e−ik|x|f(−ˆx)+O(|x|−n+1 2), (2.7b) where c n,k=kn−1 2eπ(n−1)i 4(2π)−n−1 2. On the other hand, from [15, equation (2.27)], we know that whas the asymptotics w(x)=c n,k|x|−n−1 2eik|x|ˆμ(kˆx)+O(|x|−n+1 2)as |x|→∞,(2.7c) ∂|x|w(x)=c n,k|x|−n−1 2ikeik|x|ˆμ(kˆx)+O(|x|−n+1 2)as |x|→∞,(2.7d) where c n,k=2−1e−π(n−3)i 4(2π)−n−1 2kn−3 2and ˆμ∈C∞(Rn)is the Fourier transform of the compactly supported distribution μ. Combining (2.6) with (2.7a)–(2.7d), we obtain Sn−1 f(ˆx)ˆμ(kˆx)dˆx=0. 123 P.-Z. Kow et al. By the fact that f∈C∞(Sn−1)was arbitrary, we conclude ˆμ(kˆx)=0forall ˆx∈Sn−1. Consequently, (2.7c) becomes w(x)=O(|x|−n+1 2)as |x|→∞. In other words, the far-field pattern of wis vanishing. By the Rellich uniqueness theorem [4, 7], the unique continuation principle and the connectedness of Rn\D, we conclude that w=0inRn\D. Since w∈H2−s,p loc (Rn), we also conclude that w∈H2−s,p D(Rn). Now let v∈Hs,p(D)be any solution of (Δ +k2)v =0inD,andlet ˜v∈Hs,p(Rn)be such that ˜v|D=v. We see that (v) =1(˜v|D)=(˜v,μ) =(˜v,(Δ +k2)w). From (2.2), we know that there are wj∈C∞ c(D)with wj→win H2−s,p(Rn).Since (Δ +k2)˜v=0inD, we finally conclude that (v) =lim j→∞(˜v, (Δ +k2)wj)=lim j→∞((Δ +k2)˜v, wj)=0, which is our desired result.  2.3 Proof of the main result Theorem 1.1 is an immediate consequence of the following result: Theorem 2.3 Let D be a bounded Lipschitz domain in Rn(n≥2) such that Rn\Dis connected. Suppose that k2>0is not a Dirichlet eigenvalue of −Δin D. Given any constant c0∈R, there exist Herglotz wave functions u j∈C∞(Rn)solving (Δ +k2)uj=0 in Rnsuch that lim j→∞ uj−c0L∞(∂ D)=0. Before we prove Theorem 2.3 we need the following result, which is a special case of [8, Theorems 1.1 & 1.3]. Proposition 2.4 Let D be a bounded Lipschitz domain in Rn(n≥2).If2≤p<∞and f∈Hs−2,p(D)where 1 p<s<3 p, then there exists a unique u ∈Hs,p(D)satisfying −Δu=f in D and u =0on ∂D. Proof We first consider the case when n≥3. Let p0be as in [8, Theorem 1.1] (with =D). If p 0≤p<∞, the result follows from [8, Theorem 1.1(c)]. On the other hand, if 2≤p<p 0, the result follows from [8, Theorem 1.1(a)] since s<3 p≤1+1 p.Thecasewhen n=2 can be proved using identical reasoning using [8, Theorem 1.3] and the observation 3 p≤2 p+1 2. Proof of Theorem 2.3 Since k2is not a Dirichlet eigenvalue in D, there exists a unique solution v∈H1,2(D)such that (Δ +k2)v =0inDand v=c0on ∂D. 123 On Positivity Sets for Helmholtz Solutions If v∈Hs,p(D)for some 0 <s≤1andp>n/s, using Proposition 2.2, we know that there exist Herglotz waves uj∈C∞(Rn)such that uj−c0L∞(∂ D)=uj−vL∞(∂ D)≤uj−vC(D)≤Cuj−vHs,p(D)→0, where we used the Sobolev embedding. It remains to show that v∈Hs,p(D)for some s,pwith s>n/p, and this follows from a standard bootstrap argument based on Proposition 2.4. We claim that v∈H 2 pj,pj(D)for 0 ≤j<n−2 4,(2.5) where 1 pj =1 2−j2 n−2. The case j=0 follows since v∈H1,2(D). We argue by induction and assume that this holds for j.Definew:= v−c0and note that wsolves −Δw =k2v∈H 2 pj,pj(D), w|∂D=0. We next use the Sobolev embedding H 2 pj,pj(D)⊂H2 q−2,q(D)where 2 pj>2 q−2and 2 pj −n pj =2 q−2−n q. It follows that q=pj+1and then indeed 2 pj>2 q−2. In particular −Δw ∈H 2 pj+1−2,pj+1(D) with w|∂D=0, and we may use Proposition 2.4 to conclude that w∈H 2 pj+1,pj+1(D).This completes the induction step and proves (2.5). We have proved that v∈H 2 pj,pj(D)where jis the largest integer <n−2 4.Usingthe above notation, we have Δw ∈H 2 pj,pj(D)and w|∂D=0. By Sobolev embedding we have Δw ∈Hs−2,p(D)whenever p≥pjand 2 pj −n pj =s−2−n p. The last condition implies that s−n p=2+2−n pj =2+2−n 2+2j≥0 since j≥n−2 4−1. If j>n−2 4−1, using Proposition 2.4 once again we obtain that wand hence vis in Hs,pfor some s>n/p. On the other hand, if j=n−2 4−1 we iterate the argument once more to get v∈Hs,pfor some s>n/p. This concludes the proof.  The next simple example shows that the condition that k2is not an eigenvalue is necessary at least for balls. Example 2.5 Let v(x):= |x|2−n 2Jn−2 2(|x|). We see that v∈C∞(Rn)and (Δ +1)v =0in Rn. Suppose that u1is a real-valued function satisfying (Δ +1)u1=0inRn.Since v(x)=0when|x|= jn−2 2,mfor any m≥1, 123 P.-Z. Kow et al. where jn−2 2,mdenotes the mth positive zero of Jn−2 2,wehave |x|= jn−2 2,m u1 ∂v ∂rdS =|x|<jn−2 2,m (u1Δv −vΔu1)dx =0. Since (−1)m∂v ∂r(x)>0when|x|= jn−2 2,m, it follows that u1must change sign on |x|= jn−2 2,m. Similarly, if R>0andifu0solves (Δ +k2 m)u0=0inRnwhere km=R−1jn−2 2,m, define u1via the rescaling u0(x)=u1R−1jn−2 2,mxfor x∈Rn. We see that (Δ +1)u1=0inRn. The above discussion shows that u0must change sign on ∂BR. The following strong maximum principle can be found in [9, 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 Rn(n≥ 2), and let k2<λ 1(D),whereλ1(D)>0denotes the smallest H1 0(D)-eigenvalue of −Δ.If the solution u ∈H1(D)satisfies (Δ +k2)u=0in D,u≥0on ∂D, then for each open component G of D we have either u ≡0in G or u >0in G (note that u∈C∞(G)by elliptic regularity). Proof It is easy to see that for each component Gof Dwe have k2<λ 1(G)and (Δ +k2)u=0inG,u≥0on∂G. Testing the equation above by u−∈H1 0(G)and using Poincaré inequality, we have G |u−|2dx ≤1 λ1(G)G |∇u−|2dx =k2 λ1(G)G |u−|2dx. Since k2 λ1(G)<1, then u−≡0inG,thatis, u≥0inG. (2.6) Let x0∈Gsuch that u(x0)=0. The mean value theorem for Helmholtz equation (see e.g. [9, Appendix A]) gives that Bε(x0) u(x)dx =0 (2.7) for all sufficiently small ε>0sothatBε(x0)⊂G.Sinceuis continuous in G, combining (2.6)and(2.7) we know that u=0inBε(x0), and this shows that {x∈G|u(x)=0}is both open and closed in G.SinceGis connected, then we have either {x∈G|u(x)=0}=Gor {x∈G|u(x)=0}=∅, which concludes our lemma.  123