Lp-Lq estimates for electromagnetic Helmholtz equation

Lp-LqESTIMATES FOR ELECTROMAGNETIC HELMHOLTZ EQUATION. ANDONI GARCIA Abstract. In space dimension n≥3, we consider the electromagnetic Schr¨odinger Hamiltonian H= (∇−iA(x))2−Vand the corresponding Helmholtz equation (∇ − iA(x))2u+u−V(x)u=fin Rn. We extend the well known Lp-Lqestimates for the solution of the free Helmholtz equation to the case when the electromagnetic hamiltonian His considered. 1. Introduction This paper is devoted to the study of some estimates for the Helmholtz equation with electromagnetic potential. A very natural question is to extend the known results for the Helmholtz equation with constant coefficients to the case when we consider the perturbed Helmholtz equation by a potential. Our goal will be to extend the well known Lp-Lqestimates for the free Helmholtz equation given in [KRS], [CS], [Gut] and [Gut1] to the case when we perturb the equation with an electromagnetic potential. More precisely, conditions on the electric and the magnetic part of the potential will be given in order to ensure that the estimates remain true. The Lp-Lqestimates for the free Helmholtz equation are the following: (1.1) kukLq(Rn)≤CkfkLp(Rn), where uis a solution of (1.2) ∆u+ (τ±i)u=−f τ,  > 0. The exponents pand qin (1.1) have to verify some specific conditions that will be specified later on. Here Ccan depend on τ,p,qand nand is independent of . The investigation of the estimates (1.1) started in [KRS], where the study of uniform Sobolev estimates for constant coefficient second order differential operators was accomplished. Later on, in [Gut], [Gut1] (See also [CS]) the range of the exponents pand qwhere the estimates (1.1) hold was determined. Moreover, in the purely electric case i.e., for Schr¨odinger Hamiltonians of the type ∆ + V(x), where V:Rn→Ris the electric potential, some positive results were given in [RV]. Therefore the aim of the paper is to prove the corresponding estimates (1.1) in the case when the electromagnetic Schr¨odinger hamiltonian is considered . In the first part we will prove that the existing results for the free Helmholtz equation can be extended to the perturbed equation if we impose precise decay conditions at infinity for the electric and the magnetic potential. This can be done without assuming smallness, neither for the electric part, nor for the magnetic part. As it will be shown, for the electromagnetic case, the range for the exponents p and qwhere the estimates (1.1) are valid is not the same as the one for the free Date: November 15, 2021. 2000 Mathematics Subject Classification. 35J10, 35L05, 58J45. Key words and phrases. dispersive equations, Helmholtz equation, magnetic potential. The author is supported by the grant BFI06.42 of the Basque Government. 1 This is the accepted manuscript of the article that appeared in final form in Journal of Mathematical Analysis and Applications 384(2) : 409-420 (2011), which has been published in final form at https://doi.org/10.1016/ j.jmaa.2011.05.080. © 2011 Elsevier under CC BY-NC-ND license (http://creativecommons.org/licenses/by-ncnd/4.0/) 2 ANDONI GARCIA case, hence in order to go further, we deal with the Helmholtz equation with purely electric potential, trying to obtain results in the same region of boundedness of the free equation. Therefore, we consider the electromagnetic Schr¨odinger hamiltonian Hof the form (1.3) H= (∇ − iA(x))2−V(x), and the corresponding Helmholtz equation in dimensions n≥3, namely, (1.4) (∇ − iA(x))2u+u−V(x)u=fin Rn. Here A: (A1, . . . , An) : Rn→Rnis the magnetic potential and V(x) : Rn→Ris the electric potential. Since now on, we denote by ∇A=∇ − iA, ∆A=∇2 A. The magnetic potential Ais a mathematical construction which describes the interaction of particles with an external magnetic field. The magnetic field B, which is the physically measurable quantity, is given by (1.5) B∈ Mn×n, B =DA −(DA)t, i.e. it is the anti-symmetric gradient of the vector field A(or, in geometrical terms, the differential dA of the 1-form which is standardly associated to A). In dimension n= 3 the action of Bon vectors is identified with the vector field curlA, namely (1.6) Bv = curlA×v n = 3, where the cross denotes the vectorial product in R3. One of the most interesting facts related to the Lp-Lqestimates for the electromagnetic hamiltonian is that it seems that in order to conclude the boundedness of the solution, one should be able to bound the first order term that appears when the hamiltonian His expanded. More concretely, when the term (∇ − iA(x))2in (1.4) is expanded, a first orden term, namely A· ∇, comes out and it is well known that there are no Lp-Lqestimates for the gradient of the solution of the free Helmholtz equation, (1.7) ∆u+u=−fin Rn. We will proceed in the following way. Let us consider the modified Helmholtz equation with electromagnetic potential and fixed frequency τ= 1. It reads as follows (1.8) (∇ − iA(x))2u+ (1 ±i)u−V(x)u=fin Rn,  6= 0. Remark 1.1.For convenience we will deal only with the case τ= 1, in contrast with the case of general τ > 0. We will prove the corresponding Lp-Lqestimates, independent of , for the solution of (1.8). The independence of will imply that the result remains true for the solution of (1.4). This can be seen in [IS]. Our method is a mixture of an a priori estimate and perturbative arguments. This is what allows us to avoid smallness conditions in the potentials. Similar arguments have been used in the setting of the free Schr¨odinger equation, as can be seen in [BPST] and [DFVV]. Along the proof our basic tools will be the corresponding Lp-Lqestimates and a L2-local estimate for the solution of the free Helmholtz equation, together with an a priori estimate for the solution of the modified Helmholtz equation with electromagnetic potential (1.8). Lp-LqESTIMATES FOR ELECTROMAGNETIC HELMHOLTZ EQUATION. 3 Before we describe the results that we are going to use during our proof, let us introduce some basic notation. For f:Rn→Cwe define the Morrey-Campanato norm as (1.9) |||f|||2:= sup R>0 1 RZ|x|≤R |f|2dx. Moreover, we denote, for j∈Z, the annulus C(j) by C(j) = {x∈Rn: 2j≤ |x| ≤ 2j+1}, (1.10) N(f) := X j∈Z 2j+1 ZC(j) |f|2dx!1/2 , and we easily see the duality relation Zfgdx ≤ |||g||| · N(f). These norms were introduced by Agmon and H¨ormander in [AH]. During the exposition, the truncated version of the norms appearing above will be necessary. We will denote them respectively by (1.11) |||f|||2 0:= sup R≥1 1 RZ|x|≤R |f|2dx, (1.12) N0(f) := X j≥0 2j+1 ZC(j) |f|2dx!1/2 . Let us also denote by L2 β(Rn), for β∈R, the Hilbert space of all functions fsuch that (1 + |x|)βfis square integrable over Rn. The norm in this space is denoted by k·kβ. Trivially we have that if β > 1/2 and f∈L2 β(Rn), then N0(f)<+∞. As we have said, part of our method is perturbative, so in order to be able to start, let us remind what is known for the free Helmholtz equation. Firstly, we are going to state the result concerning the Lp-Lqestimates which appears in [Gut] and [Gut1]. Let be A=n+ 3 2n,n−1 2n, A0=n+ 1 2n,n−3 2n B=n2+ 4n−1 2n(n+ 1) ,n−1 2n, B0=n+ 1 2n,n2−2n+ 1 2n(n+ 1)  and ∆(n), the set of points of [0,1] ×[0,1] given by (1.13) ∆(n) = 1 p,1 q∈[0,1]2:2 n+ 1 ≤1 p−1 q≤2 n,1 p>n+ 1 2n,1 q<n−1 2n. The set ∆(n) is the trapezium ABB0A0with the closed line segments AB and B0A0 removed (see Figure 1). 4 ANDONI GARCIA Figure 1. ∆(n), n≥3. Now, we are in conditions to recall the existing result for the Helmholtz equation with constant coefficients. Remark 1.2.In [Gut] (See also [Gut1]), estimates for the solution of the equation perturbed with generals τ > 0 and  > 0, are given, namely, (1.14) ∆u+ (τ+i)u=−F, τ,  > 0. The special case where the point (1/p, 1/q) lies on the open segment AA0and on the duality line 1/q = 1 −1/p in Figure 1 was previously obtained in [[KRS], Theorem 2.2 and 2.3 respectively]. Recall that we will only deal with the case of fixed frequency τ= 1. The Theorem reads as follows. Theorem 1.1. Let u be a solution of (1.15) ∆u+ (1 + i)u=−F,  > 0. Then, there exists a constant C, independent of , such that (1.16) kukLq(Rn)=k(∆ + (1 + i))−1FkLq(Rn)≤CkFkLp(Rn) when (1 p,1 q)∈∆(n),n≥3. As we mentioned, another tool that will be crucial in the proof is an L2-local estimate, which bounds the truncated Morrey-Campanato norm of the solution of the free equation, defined in (1.11), in terms of the Lpnorm of the RHS data. This Theorem also appears in [RV], [Gut] and [Gut1]. The statement is the following. Theorem 1.2. Let u be a solution of (1.17) ∆u+ (1 + i)u=−F,  > 0. If (i) n= 3 or 4and 1 n+1 ≤1 p−1 2<1 2, or (ii) n≥5and 1 n+1 ≤1 p−1 2≤2 n, Lp-LqESTIMATES FOR ELECTROMAGNETIC HELMHOLTZ EQUATION. 5 then, there exists a constant C, independent of , such that (1.18) sup R≥11 RZBR |u(x)|2dx1/2 ≤CkFkLp(Rn). The last result concerns an a priori estimate for the solution of the perturbed equation. It states that, given precise conditions, without assuming smallness, on the decay at infinity for the the electric potential, the magnetic potential and the radial derivative of the electric potential, we have a precise control for k∇Auk−(1+δ) 2 and kuk−(1+δ) 2. This result appears in [IS] . The Theorem is the following one. Theorem 1.3. Let n≥3, and u∈C∞ 0be a solution of (∇ − iA(x))2u+ (1 ±i)u−V(x)u=f,  6= 0. Let us assume that: (V) V(x)can be decomposed as V(x) = V1(x) + V2(x), and there exist strictly positive constants Cand µsuch that (V1) |V1(x)| ≤ C|x|−µ,(∂rV1)(x)≤C|x|−1−µ,|x| ≥ 1, (V2) |V2(x)| ≤ C|x|−1−µ,|x| ≥ 1, (B) |B(x)| ≤ C|x|−1−µ,|x| ≥ 1. Choose δ > 0sufficiently small (so that δ≤µ/2,δ < 1). Then, there exists a constant C=C(δ), which depends uniformly in , such that the following a priori estimate holds (1.19) k∇Auk−(1+δ) 2+kuk−(1+δ) 2≤Ckfk1+δ 2. Remark 1.3.Observe that, since the electric potential Vand magnetic potential Amust satisfy the conditions of the theorem, singularities at the origin are not allowed. Remark 1.4.Notice that the unique continuation property holds for the differential operator H= (∇−iA(x))2−V(x), as can be seen in [R]. The assumptions (V), (V1), (V2) and (B), together with this observation implies that the limiting absorption principle holds. Remark 1.5.The conditions on the decay for the electric potential Vand the magnetic field Bgiven by (V1), (V2) and (B) respectively, are sufficient for us, due we have fixed the frequency τ= 1. It can be seen in [IS], that the result is true provided τand belong to the following set denoted by K (1.20) K={k=τ+i ∈C/τ ∈(τ0, τ1),  ∈(0, 1)}, where 0 < τ0< τ1<∞and 0 < 1<∞. Hence, the critical case τ= 0 is excluded. This situation requires more decay for both potentials (typically hxi−(2+), > 0, for Band Vif n= 3 and hxi−2 for n≥4, where hxi= (1 + |x|2)1/2), in order to obtain a priori estimates for the solution of the perturbed equation, as can be seen in [F], where Morrey -Campanato type estimates, uniform in , are obtained for τ≥0. Note also that this result gives an a priori estimate without assuming smallness neither for the non repulsive component of the electric field nor for the trapping component of the magnetic field, defined by Bτ:= x |x|B. 6 ANDONI GARCIA Once we have described all the tools which are going to be used, it is necessary to introduce the region where we are able to extend the known results for the free Helmholtz equation to the case when electromagnetic perturbations are considered. During the discussion, it will appear a subregion of ∆(n), for n≥3, which will be denoted by ∆0(n), given by (1.21) ∆0(n) = 1 p,1 q∈∆(n) : 1 n+ 1 ≤1 p−1 2,1 n+ 1 ≤1 2−1 q. The set ∆0(n) is the solid triangle determined by the points Q,Q0and Q00 (see Figure 2). This will be the region of boundedness for the perturbed Helmholtz equation. Figure 2. ∆0(n), n≥3. Remark 1.6.For the case of the perturbed electromagnetic equation we are not able to obtain a positive result of boundedness for the whole region ∆(n), since we have not control for the gradient term, namely A·∇, outside ∆0(n). However, when we set A≡0, and consider the electric hamiltonian, the results can be extended outside ∆(n) by imposing more decay on V. 2. Electromagnetic Helmholtz Equation. In this section we will give the precise statement and the proof of the theorem, where we extend the known result for the free Helmholtz equation to the case when electromagnetic perturbations are considered. The basic theorems which will be used along the proof were given in the section 1, as well as the basic notation. First we will announce the result for the general electromagnetic case and afterwards, by setting A≡0, the electric case will be treated by extending our previous result. Let us start by considering the solution of the Helmholtz equation with electromagnetic potential that satisfies either the ingoing or the outgoing Sommerfeld radiation condition. For n≥3, it reads, (2.1) (∇ − iA(x))2u+u−V(x)u=fin Rn, where A: (A1, . . . , An) : Rn→Rnis the magnetic potential and V(x) : Rn→Ris the electric potential. Lp-LqESTIMATES FOR ELECTROMAGNETIC HELMHOLTZ EQUATION. 7 We will assume that the magnetic potential Asatisfies the Coulomb gauge condition (2.2) ∇ · A= 0. We will prove Lp-Lqestimates for the solution of the equation (2.1). In order to do that we will consider the solution of (2.1) as the solution of the modified Helmholtz electromagnetic equation, (2.3) (∇ − iA(x))2u+ (1 ±i)u−V(x)u=fin Rn,  6= 0, via limiting absorption principle, by taking the limit of the solution of (2.3) when goes to 0. We will obtain the corresponding Lp-Lqestimates, independent of , for the solution of (2.3), so these will remain true for the solution of (2.1). This is guaranteed by the results appearing in [IS]. The goal is to determine the region of pand qwhere the solution of (2.3) satisfies Lp-Lqestimates, namely, (2.4) kukLq(Rn)≤CkfkLp(Rn). with Cindependent of . The main result of this section is the following. Theorem 2.1. Let u be a solution of (2.5) (∇ − iA(x))2u+ (1 ±i)u−V(x)u=fin Rn, n ≥3,  6= 0. Let Vand Asatisfy (V),(V1),(V2)and (B)in Theorem 1.3, and suppose that there exist constants C, µ > 0such that (2.6) |A(x)| ≤ C (1 + |x|)1+µ,|V(x)| ≤ C (1 + |x|)1+µ. Then, there exists a constant C, independent of , such that (2.7) kukLq(Rn)≤CkfkLp(Rn), when 1 p,1 q∈∆0(n). Proof. Remark 2.1.Notice that there are no smallness assumption neither for the electric potential Vnor for the magnetic potential A. Also we bound the solution only assuming short-range decay for Vand A. As we said in Remark 1.3, singularities at the origin for Vand Aare not considered. Step 1. It will be proved that, whenever 1 p,1 q∈∆0(n) then we get the following (2.8) kukLq(Rn)≤Ckfk1+δ 2. Let ube a solution of (2.5). Since ∇ · A≡0, we can expand the term (∇ − iA)2in the following form (2.9) (∇ − iA)2u= ∆u−2iA · ∇Au+|A|2u. This is the key point in order to consider the electromagnetic case as a perturbation of the free equation. As can be seen, there appear terms of order zero and order one. So by passing terms to the RHS, we have that uis solution of the following equation (2.10) ∆u+ (1 ±i)u=f+ 2iA · ∇Au− |A|2u+V u. 8 ANDONI GARCIA Now we apply the result coming from Theorem 1.2. By considering the dual estimate of (1.18), we get that, if 1 p,1 q∈∆0(n) it holds kukLq(Rn)=k(∆ + (1 ±i))−1(f+ 2iA · ∇Au− |A|2u+V u)kLq(Rn) (2.11) ≤C(N0(f) + N0(2iA · ∇Au) + N0(|A|2u) + N0(V u)). with Cindependent of and N0defined in (1.12). Now we continue by treating the terms appearing on the RHS of (2.11). First we deal with the term N0(2iA · ∇Au). From (2.6), we get that this term can be bounded as N0(2iA · ∇Au)2=CX j≥0 2j+1 ZC(j) |A· ∇Au|2dx(2.12) ≤CX j≥0 2jZC(j) |A|2|∇Au|2dx ≤CX j≥0 2j(δ−2µ)ZRn |∇Au|2 (1 + |x|)1+δdx. Therefore, we finally have (2.13) N0(2iA · ∇Au)≤Ck∇Auk−(1+δ) 2. Let us continue with the term N0(|A|2u). As before, from (2.6), we can treat this term as follows N0(|A|2u)2=X j≥0 2j+1 ZC(j) ||A|2u|2dx(2.14) ≤CX j≥0 2jZC(j) |A|4|u|2dx ≤CX j≥0 2j(−2+δ−4µ)ZRn |u|2 (1 + |x|)1+δdx. Hence, we get (2.15) N0(|A|2u)≤Ckuk−(1+δ) 2. The last term is N0(V u). Similarly, we obtain from (2.6) N0(V u)2=X j≥0 2j+1 ZC(j) |V u|2dx(2.16) ≤CX j≥0 2jZC(j) |V|2|u|2dx ≤CX j≥0 2j(δ−2µ)ZRn |u|2 (1 + |x|)1+δ.dx So, it verifies (2.17) N0(V u)≤Ckuk−(1+δ) 2. Lp-LqESTIMATES FOR ELECTROMAGNETIC HELMHOLTZ EQUATION. 9 From (2.11), (2.13), (2.15) and (2.17), we get that, whenever 1 p,1 q∈∆0(n), the Lqnorm of ucan be bounded as kukLq(Rn)=k(∆ + (1 ±i))−1(f+ 2iA · ∇Au− |A|2u+V u)kLq(Rn) (2.18) ≤C(N0(f) + N0(2iA · ∇Au) + N0(|A|2u) + N0(V u)). ≤CN0(f) + C1k∇Auk−(1+δ) 2+C2kuk−(1+δ) 2. Now we remind the a priori estimate given by Theorem 1.3, which ensures that, under the assumptions (V), (V1), (V2) and (B) for Vand Arespectively, there exists a constant C, such that the following holds (2.19) k∇Auk−(1+δ) 2+kuk−(1+δ) 2≤Ckfk1+δ 2. Remark 2.2.The constant Cwhich appears here depends uniformly in . So, from (2.18) and (2.19), we get (2.20) kukLq(Rn)≤CN0(f) + C1kfk1+δ 2. Moreover, it holds that N0(f) can be bounded as (2.21) N0(f)≤Ckfk1+δ 2. This, together with (2.20) concludes that, if uis a solution of (2.5), it verifies (2.22) kukLq(Rn)≤Ckfk1+δ 2. Therefore, we get the desired estimate. Step 2. By applying duality to the last estimate, we get that if 1 p,1 q∈∆0(n), (2.23) kuk−(1+δ) 2≤CkfkLp(Rn) Remark 2.3.The adjoint operator is the one corresponding to ∓. Since we can do the same argument for both signs, all the computations are justified. Step 3. This is the final step in the proof. As we said in the introduction the main difficulty will be to handle the first order term given by A· ∇Au, since there are no Lp-Lqestimates for the gradient of the solution of the free Helmhotz equation. Instead of considering this norm our argument will end up by treating k∇Auk−(1+δ) 2, and this norm will be under control in the region ∆0(n). Consequently, we have that if 1 p,1 q∈∆0(n), from the Lp-Lqestimates for the solution of the free equation, namely (1.16), given in Theorem 1.1, and proceeding as in the step 1 for the terms 2iA · ∇Au,|A|2uand V u, we get kukLq(Rn)=k(∆ + (1 ±i))−1(f+ 2iA · ∇Au− |A|2u+V u)kLq(Rn) (2.24) ≤CkfkLp(Rn)+C1(N0(2iA · ∇Au) + N0(|A|2u) + N0(V u)). where Cand C1do not depend on . Let us remind that, from (2.6) it holds N0(2iA · ∇Au)≤C1k∇Auk−(1+δ) 2,(2.25) N0(|A|2u)≤C2kuk−(1+δ) 2, N0(V u)≤C3kuk−(1+δ) 2. Therefore, we get (2.26) kukLq(Rn)≤CkfkLp(Rn)+C1k∇Auk−(1+δ) 2+C2kuk−(1+δ) 2.