Can a single PDE govern well the propagation of the electric wave field in a heterogeneous medium in 3D?
aa r X i v : . [ phy s i c s . c l a ss - ph ] F e b CAN A SINGLE PDE GOVERN WELL THE PROPAGATION OFTHE ELECTRIC WAVE FIELD IN A HETEROGENEOUS MEDIUMIN 3D? ∗ VLADIMIR G. ROMANOV † AND
MICHAEL V. KLIBANOV ‡ Abstract.
It is demonstrated in this paper that the propagation of the electric wave field in aheterogeneous medium in 3D can sometimes be governed well by a single PDE, which is derived fromthe Maxwell’s equations. The corresponding component of the electric field dominates two othercomponents. This justifies some past results of the second author with coauthors about numericalsolutions of coefficient inverse problems with experimental electromagnetic data. In addition, sinceit is simpler to work in applications with a single PDE rather than with the complete Maxwell’ssystem, then the result of this paper might be useful to researchers working on applied issues of thepropagation of electromagnetic waves in inhomogeneous media.
Key words.
Maxwell’s equations, geodesic lines, domination of one component, experimentaldata for inverse problems
AMS subject classifications.
1. Introduction.
In some previous works of the second author with coauthorscoefficient inverse problems were solved for frequency dependent microwave exper-imental electromagnetic data using only the single Helmholtz equation, see, e.g.[6, 7, 8]. Reconstruction results were quite accurate ones. A similar observationtook place in [1, 15], although for time dependent experimental data. Thus, a naturalquestion to pose is:
Given that the propagation of the electromagnetic wave field isgoverned by the Maxwell’s equations, why the use of only a single equation has pro-vided accurate reconstruction results?
A positive heuristic answer to this questioncan be found in the classical textbook of M. Born and E. Wolf [3, pages 695,696]for the frequency domain case. In addition, this question was positively addressednumerically in [2] for the time domain case and in [10] for the frequency domain case.It was demonstrated computationally in [2, 10] that if the incident electric wave fieldhas only a single non zero component, then this component dominates two other com-ponents while propagating through the medium, and its propagation is well governedby a wave-like PDE. That PDE is either the Helmholtz equation in the frequencydomain or the corresponding hyperbolic equation in the time domain.The goal of this paper is to investigate the above question rigorously. We believethat the results of this paper might be useful not only for an analytical explanation ofthe accuracy of imaging results of [6, 7, 8] but also for applied mathematicians, physi-cists and engineers working on various topics of electromagnetic waves propagation.Indeed, it is clear that it is easier to work in applications with a single PDE ratherthan with the whole Maxwell’s system.In section 2 we work in time domain. These results are used then in section ∗ Submitted to the editors DATE.
Funding:
The work of V.G. Romanov was supported by Mathematical Center in Akadem-gorodok at Novosibirsk State University (the agreement with Ministry of Science and High Educationof the Russian Federation number 075-15-2019-1613). The work of M.V. Klibanov was supported byUS Army Research Laboratory and US Army Research Office grant W911NF-19-1-0044. † Sobolev Institute of Mathematics, Novosibirsk 630090, Russian Federation, and Mathe-matical Center in Akademgorodok, Novosibirsk State University, Russian Federation. ([email protected]) ‡ Corresponding author. Department of Mathematics and Statistics, University of North Carolinaat Charlotte, Charlotte, NC, 28223 USA ([email protected]).1
This manuscript is for review purposes only.
V. G. ROMANOV AND M. V. KLIBANOV
3, where we derive our desired conclusion for the case of the frequency domain. Insection 4, we link our main Theorem 2 of section 3 with the above cited results of[6, 7, 8]. In section 5 (Appendix) we prove a certain energy estimate.
2. Time Domain.
Consider the Maxwell’s equations in a non magnetic medium(2.1) curl H = ε ( x ) E t , curl E = − H t , div H = 0 , x ∈ R , t > , where ε ( x ) is the spatially distributed dielectric constant. We work in this paper withdimensionless variables, since variables were made dimensionless in the above citedworks about the experimental data. Thus, in vacuum ε ( x ) = 1 , and we also assumethat the magnetic permeability µ ≡
1. Let ν ∈ S = {| ν | = 1 } be a unit vector of thedirection of propagation of the incident electric wave field. If the space R is vacuum,then equations (2.1) admit the following solution [13]:(2.2) E ( x , t ) = j δ ( t + t − x · ν ) , H ( x , t ) = ( ν × j ) δ ( t + t − x · ν ) , where δ ( t ) is the Dirac delta function, t is an arbitrary number, x · ν denotes thescalar product of these two vectors and j · ν = 0. The vector j defines the polarizationof this wave,(2.3) j = ( j , j , j ) , where j , j , j are some constants. In the sequel we assume that j ∈ S . The orthog-onality of vectors j and ν is necessary to satisfy the equation div H = 0.Below j and ν are assumed to be arbitrary but fixed vectors. Therefore we donot indicate dependence of the solution and some functions on these parameters forbrevity, unless this is really necessary. Still, we use the parameter ν to indicate somedomains and a plane wave for a more clear understanding. Below vectors j , ν , E , H ,etc. are row vectors, see, e.g. (2.3).Let R > B with the center at { } and the radius R . Let the sphere S = ∂B. Then B = { x ∈ R : | x | < R } , S = { x ∈ R : | x | = R } . We assume that(2.4) 1 ≤ ε ( x ) ≤ ε in B , ε ( x ) = 1 in R (cid:31) B, where ε ≥ t = min x ∈ S ( x · ν ) = − R. Let the incident plane wave propagates in the vacuum for t < B at a moment of time t = 0. Then the propagation of theelectromagnetic wave field is governed by the following Cauchy problem:(2.6) curl H = ε ( x ) E t , curl E = − H t , ( x , t ) ∈ R , (2.7) E | t< = E ( x , t ) , H | t< = H ( x , t ) . For the sake of convenience, we reduce now problem (2.6), (2.7) to the case when onlythe vector function E ( x , t ) is unknown. Since div (curl U ) = 0 for any appropriate This manuscript is for review purposes only.
SINGLE PDE FOR THE MAXWELL’S EQUATIONS U , then applying the operator div to both sides of the first equation(2.6), we obtain(2.8) div ( ε ( x ) E t ) = 0 . Integrating (2.8) with respect to t and using (2.7), we obtain(2.9) div ( ε ( x ) E ( x , t )) − div (cid:0) ε ( x ) E ( x , ν , (cid:1) = 0 . Note that div E ( x , t ) = 0, since j · ν = 0. Indeed,div E ( x , t ) = − ( j · ν ) δ ′ ( t + t − x · ν ) = 0 . Hence,(2.10) div ( ε ( x ) E ( x , ∇ ε ( x ) · E ( x ,
0) = ( ∇ ε ( x ) · j ) δ ( t − x · ν ) = 0 . This is because by (2.5) supp { δ ( t − x · ν ) } is the tangent plane to S , namely ( x · ν ) = − R , along which ∇ ε ( x ) = 0. Hence div ( ε ( x ) E ( x , t )) = 0 for any t . Hence, problem(2.6), (2.7) is reduced to the following problem with n ( x ) = p ε ( x ) :(2.11) n ( x ) E tt − ∆ E − ∇ ( E · ∇ ln n ( x )) = 0 , ( x , t ) ∈ R , (2.12) E | t< = E ( x , t ) . Define two domains D − ( ν ) and D + ( ν ) as D − ( ν ) = { x ∈ R : x · ν + R < } ,D + ( ν ) = { x ∈ R : x · ν + R ≥ } . Note that the domain D − ( ν ) is situated outside of B , while B ⊂ D + ( ν ).To define geodesic lines, we partially follow our paper [9]. The function n ( x )generates the Riemannian metric dτ = n ( x ) | d x | , | d x | = p ( dx ) + ( dx ) + ( dx ) . For each vector ν ∈ S define the plane Σ( ν ) as(2.13) Σ( ν ) = { ξ ∈ R : ξ · ν = − R } . Observe that the plane Σ( ν ) is tangent to S at the point ξ tan = − R ν . Hence,Σ( ν ) ∩ B = ∅ and the vector ν is a normal vector to the plane Σ( ν ) . Consider anarbitrary point y ∈ Σ( ν ) . This point can be represented as(2.14) y = y ( a , a ) = − R ν + a e + a e , ( a , a ) ∈ R , where unit vectors ν , e = j , e = ν × j form an orthogonal triple. Note that vectors e , e are parallel to the plane Σ( ν ) . Let the function ϕ ( x , ν ) be the solution of the Cauchy problem for the eikonalequation,(2.15) |∇ x ϕ ( x , ν ) | = n ( x ) , ϕ ( x , ν ) | x ∈ Σ( ν ) = 0 This manuscript is for review purposes only.
V. G. ROMANOV AND M. V. KLIBANOV satisfying the following conditions:(2.16) ϕ ( x , ν ) (cid:26) < x ∈ D − ( ν ) ,> x ∈ D + ( ν ) . The number | ϕ ( x , ν ) | is the Riemannian distance between the point x and the planeΣ( ν ). From the Physics standpoint, | ϕ ( x , ν ) | is the travel time between the point x and the plane Σ( ν ). For ξ · ν < − R, i.e. in the domain D − ( ν ), the function ϕ ( x , ν )has the form ϕ ( x , ν ) = x · ν + R . To find the function ϕ ( x , ν ) in the domain D + ( ν ) , we need to solve problem (2.15), (2.16) in this domain. It is known that to do this,we need to solve the following Cauchy problem for a system of ordinary differentialequations [14]:(2.17) d x ds = p ( x , ν ) n ( x ) , d p ( x , ν ) ds = ∇ ln n ( x ) , dϕ ( x , ν ) ds = 1 , s > , (2.18) x | s =0 = y , p | s =0 = ν , ϕ | s =0 = 0 , where y ∈ Σ( ν ) is an arbitrary point of the plane Σ( ν ) , see (2.14), s is the Riemannianarc length and(2.19) p ( x , ν ) = ∇ ϕ ( x , ν ) . Note that equations (2.17), (2.18) imply that ϕ ( ξ , ν ) = s. In particular, this meansthat the second condition (2.15) is satisfied. Equations (2.17), (2.18) define a geodesicline Γ( x , Σ( ν )) in D + ( ν ) which connects the point x ∈ D + ( ν ) with the point y ∈ Σ( ν ) and orthogonal to Σ( ν ) at y .Cauchy problem (2.17), (2.18) has the unique solution(2.20) x = f ( s, a , a ) , p = g ( s, a , a ) . These equations define the bundle of geodesic lines, which go out from different points y ∈ Σ( ν ) in direction ν . To find the geodesic line Γ( x , Σ( ν )), we need to invert thefirst equation (2.20) and calculate a and a and then to find y ∈ Σ( ν ), using formula(2.14). Set D + ( ν , R ) = { x : − R ≤ x · ν ≤ R } , T = max x ∈ D + ( ν ,R ) ϕ ( x , ν ) . Hence, D + ( ν , R ) ⊂ D + ( ν ) . Note that both sets D + ( ν ) and D + ( ν , R ) are closedones, i.e. D + ( ν ) = D + ( ν ) and D + ( ν , R ) = D + ( ν , R ) . Here T is a finite number.Indeed, let C ( ν , R ) be the circular cylinder with the circle of the radius R and theaxis x = ( − R + s ′ ) ν , s ′ ≥
0, with generating lines orthogonal to the plane Σ( ν ) . Consider the intersection D ( ν , R ) of C ( ν , R ) with D + ( ν , R ) , (2.21) D ( ν , R ) = C ( ν , R ) ∩ D + ( ν , R ) . Then the function ϕ ( x , ν ) = x · ν + R ≤ R for x ∈ D + ( ν , R ) (cid:31) D ( ν , R ), since x · ν ∈ [ − R, R ] in D + ( ν , R ). On the other hand the domain D ( ν , R ) is finite.Therefore, T = max (cid:18) R, max x ∈ D ( ν ,R ) ϕ ( x , ν ) (cid:19) . This manuscript is for review purposes only.
SINGLE PDE FOR THE MAXWELL’S EQUATIONS J ( x ) = ∂ ( x , x , x ) ∂ ( s, a , a ) . From relations (2.17) and (2.18) follows that ∂ x /∂s = p = ν and ∂ x /∂a k = e k , k = 2 ,
3, at s = 0. Then the Jacobian is the determinant which rows are formed bycomponents of three unite orthogonal vectors of the positive orientation. Hence,(2.23) J ( x ) = 1 for x ∈ Σ( ν ) . Note that J ( x ) = 1 for x ∈ D ( ν , R ) as well since x = y + s ν in D ( ν , R ).Denote by c ( x ) = 1 / p ε ( x ) the speed of propagation of electromagnetic waves.By (2.4) c ≤ c ( x ) ≤ , x ∈ R , where c = 1 / √ ε .Below we use the following assumptions: Assumptions: The function ε ( x ) ∈ C ∞ ( R ) , satisfies conditions (2.4). There exists a positive constant J such that J ( x ) ≥ J for D + ( ν , R ).3. Any point x ∈ D + ( ν , R ) can be connected with the plane Σ ( ν ) by a singlegeodesic line Γ( x , Σ( ν )) such that Γ( x , Σ( ν )) is orthogonal to Σ ( ν ) at a point y ∈ Σ ( ν ).4. Any two points x and y in R can be connected by a single geodesic line. Under these Assumptions, the equality x = f ( s, a , a ) is invertible in D + ( ν , R )and defines s = s ( x , ν ) = ϕ ( x , ν ) and parameters a k = a k ( x , ν ), k = 2 ,
3, i.e. thepoint y ∈ Σ ( ν ), see (2.14). By (2.17) and (2.18) if x = y , then d x /ds = ν . The lattervector is directed along the geodesic line Γ( x , Σ( ν )) . Hence, Γ( x , Σ( ν )) is orthogonalto Σ ( ν ) at the point y .Define domains G T ( ν ) and G ( T, ν ) in R as(2.24) G T ( ν ) = { ( x , t ) : 0 ≤ t < min( | ϕ ( x , ν ) | , T ) } , (2.25) G ( T, ν ) = { ( x , t ) , | ϕ ( x , ν ) | ≤ t ≤ T } . To differentiate between notations of the Heaviside function H ( t ) and the mag-netic wave field H ( x , t ) , it is convenient to denote θ ( t ) := H ( t ) ,θ ( t ) = (cid:26) , t ≥ , , t < . Theorem 1.
Assume that the Assumption holds. Then for every vector ν ∈ S the solution of problem (2.11), (2.12) can be represented in R T = { ( x , t ) | ≤ t ≤ T } in the form (2.26) E ( x , t ) = α − ( x ) δ ( t − ϕ ( x )) + b E ( x , t ) θ ( t − | ϕ ( x ) | ) , where α − ( x ) ∈ C ∞ ( D + ( ν , R )), α − ( x ) = 0 for x ∈ D − and b E ( x , t ) ∈ C ( G ( T, ν )) and b E ( x , t ) = 0 for t = | ϕ ( x , ν ) | . This manuscript is for review purposes only.
V. G. ROMANOV AND M. V. KLIBANOV
Proof . Introduce functions θ k ( t ) as(2.27) θ − ( t ) = δ ′′ ( t ) , θ − ( t ) = δ ′ ( t ) , θ − ( t ) = δ ( t ) , θ k ( t ) = t k k ! θ ( t ) , k = 1 , , . . . . Observe that θ ′ k ( t ) = θ k − ( t ) for all k ≥ −
2. We seek the solution of problem (2.11),(2.12) in the form(2.28) E ( x , t ) = r X k = − α k ( x ) θ k ( t − ϕ ( x )) + E r ( x , t ) , where the natural number r will be chosen later. Substituting representation (2.28)in (2.11), using the eikonal equation (2.15) and equating coefficients at θ k ( t ) for k = − , − , , , . . . , r −
1, we obtain the following reqursive formulas for findingcoefficients α k ( x ):2( ∇ ϕ ( x ) · ∇ ) α k ( x ) + α k ( x )∆ ϕ ( x ) + ( α k ( x ) · ∇ ln n ( x )) ∇ ϕ ( x )(2.29) = ∆ α k − ( x ) + ∇ ( α k − ( x ) · ∇ ln n ( x )) , k = − , , , ..., r. Here we need to formally set(2.30) α − ( x ) = 0 . Since by (2.2) and (2.12) E = j δ ( t − ( x · ν + R )) for t <
0, then we obtain(2.31) α − ( x ) = 0 , in D − ; α − | Σ( ν ) = j , α k ( x ) = 0 , in D − ; α k | Σ( ν ) = 0 , k = 0 , , . . . , r. Moreover, using (2.11), we obtain the following Cauchy problem for the residual E r ( x , t ) of expansion (2.28)(2.32) n ( x ) ∂ t E r − ∆ E r − ∇ ( E r · ∇ ln n ( x )) = F r ( x , t ) , E r | t< = 0 , where(2.33) F r ( x , t ) = (cid:0) ∆ α r ( x ) + ∇ (cid:0) α r ( x ) · ∇ ln n ( x ) (cid:1)(cid:1) θ r ( t − ϕ ( x )) . We now construct solutions of equation (2.29) with the Cauchy data (2.31). We havealong the geodesic line Γ( x , Σ( ν )) :(2.34) 2( ∇ ϕ ( ξ ) · ∇ ) α k ( ξ ) = 2( p ( ξ ) · ∇ ) α k ( ξ ) = 2 n ( ξ ) (cid:18) d ξ ds · ∇ (cid:19) α k ( ξ )= 2 n ( ξ ) d α k ( ξ ) ds , where ξ ∈ Γ( x , Σ( ν )) is an arbitrary point. Moreover, it is stated in the paper [12]that the following formula valid along Γ( x , Σ( ν )) (see Lemma 1, Equation (4.2)):(2.35) d ln J ( ξ ) ds = div (cid:0) n − ( ξ ) ∇ ϕ ( ξ ) (cid:1) , This manuscript is for review purposes only.
SINGLE PDE FOR THE MAXWELL’S EQUATIONS J ( ξ ) is the Jacobian defined in (2.22), provided that x = ( x , x , x ) is replacedwith ξ = ( ξ , ξ , ξ ). Calculating the right-hand-side of (2.35) and using (2.35), wefind div (cid:0) n − ( ξ ) ∇ ϕ ( ξ ) (cid:1) = n − ( ξ )∆ ϕ ( ξ ) + (cid:0) ∇ n − ( ξ ) · p ( ξ ) (cid:1) . Hence, (2.35) implies(2.36) n − ( ξ )∆ ϕ ( ξ ) = d ln J ( ξ ) ds − (cid:0) ∇ n − ( ξ ) · p ( ξ ) (cid:1) = d ln( J ( ξ ) n ( ξ )) ds . We have used here that, similarly (2.34), − (cid:0) ∇ n − ( ξ ) · p ( ξ ) (cid:1) = − n ( ξ ) (cid:18) ∇ n − ( ξ ) · d ξ ds (cid:19) = − n ( ξ ) dds n − ( ξ ) = dds ln n ( ξ ) . Using formulae (2.34) and (2.36) and replacing in (2.29) x with ξ , we obtain2 n ( ξ ) " d α k ( ξ ) ds + α k ( ξ ) d ln( p J ( ξ ) n ( ξ )) ds + ( α k ( ξ ) · ∇ ln n ( ξ )) ∇ ϕ ( ξ )= ∆ α k − ( ξ ) + ∇ ( α k − ( ξ ) · ∇ ln n ( ξ )) , k = − , , , ..., r. (2.37)Multiplying equation (2.37) by p J ( ξ ) / (2 n ( ξ )), we transform equation (2.29) alongΓ( x , Σ( ν )) to the recursive form dds (cid:16) α k ( ξ ) n ( ξ ) p J ( ξ ) (cid:17) − n ( ξ ) p J ( ξ )2 (cid:0) α k ( ξ ) · ∇ n − ( ξ ) (cid:1) p ( ξ )= Q k ( ξ ) , k = − , , , . . . , r, (2.38)were(2.39) Q k ( ξ ) = p J ( ξ )2 n ( ξ ) (cid:2) ∆ α k − ( ξ ) + ∇ (cid:0) α k − ( ξ ) · ∇ ln n ( ξ ) (cid:1) (cid:3) . When we solve equations (2.38) going from k to k + 1, the function Q k ( ξ ) is alwaysknown from the previous step. It follows from (2.30) and (2.39) that Q − ( ξ ) = 0.Functions α k ( ξ ) satisfy on Σ( ν ) conditions (2.31). We also recall that by (2.4) and(2.21) n ( ξ ) | Σ( ν ) = 1, J | Σ( ν ) = 1 . Hence, integrating (2.38) with respect to s , we obtain a recursive integral equationalong the geodesic line Γ( x , Σ( ν )) α k ( x ) = 1 n ( x ) p J ( x ) (cid:18) A k + Z Γ( x , Σ( ν )) (cid:2) Q k ( ξ )+ n ( ξ ) p J ( ξ )2 (cid:0) α k ( ξ ) · ∇ n − ( ξ ) (cid:1) p ( ξ ) ds (cid:19) , (2.40) k = − , , , . . . , r, where ξ = f ( s, a , a ), the vector function f ( s, a , a ) is defined in (2.20) and A − = j , A k = 0 , k = 0 , , . . . , r. This manuscript is for review purposes only.
V. G. ROMANOV AND M. V. KLIBANOV
Recall that s is the Riemannian arc length of Γ( ξ , Σ( ν )). Equation (2.40) is aVolterra-type integral equation of the second kind along the curve Γ( x , Σ( ν )). There-fore, this equation can be solved by the method of successive approximations whichis rapidly converging. We can solve equation (2.40) for different k = − , ..., r step-by-step, starting from k = − α − ( x ).We now represent this function through the resolvent R ( x , ξ ) of equation (2.40) for k = −
1. First, we introduce the vector function β ( x ) β ( x ) = n ( x ) p J ( x ) α − ( x ) . Then equation for this function has the form β ( x ) = j + 12 Z Γ( x , Σ( ν )) (cid:0) β ( ξ ) · ∇ n − ( ξ ) (cid:1) p ( ξ ) ds. The more convenient form of this equation is: β ( x ) = j + Z Γ( x , Σ( ν )) β ( ξ ) K ( ξ ) ds, where K ( ξ ) is the 3 × K ( ξ ) = 12 ( ∇ n − ( ξ )) ∗ p ( ξ )and ( ∇ n − ( ξ )) ∗ is the transposed vector ( ∇ n − ( ξ )), i.e. column vector, while p ( ξ )is row vector.Represent β ( x ) as β ( x ) = ∞ X n =0 β n ( x ) , where β ( x ) = j , β n ( x ) = Z Γ( x , Σ( ν )) β n − ( ξ ) K ( ξ ) ds, n = 1 , , . . . . Then β ( x ) = j Z Γ( x , Σ( ν )) K ( ξ ) ds, Next, β ( x ) = j Z Γ( x , Σ( ν )) Z Γ( ξ , Σ( ν )) K ( ξ ′ ) ds ′ K ( ξ ) ds. Changing here the repeated integration by place and then replacing s ′ with s and vice This manuscript is for review purposes only.
SINGLE PDE FOR THE MAXWELL’S EQUATIONS β ( x ) = j Z Γ( x , Σ( ν )) K ( x , ξ ) ds, where K ( x , ξ ) = Z Γ( x , Σ( ν )) \ Γ( ξ , Σ( ν )) K ( ξ ) K ( ξ ′ ) ds ′ Similarly, β n ( x ) = j Z Γ( x , Σ( ν )) Z Γ( ξ , Σ( ν )) K n − ( ξ , ξ ′ ) ds ′ K ( ξ ) ds = j Z Γ( x , Σ( ν )) K n − ( x , ξ ) ds, n ≥ , where K n − ( x , ξ ) = Z Γ( x , Σ( ν )) \ Γ( ξ , Σ( ν )) K n − ( ξ ′ , ξ ) K ( ξ ′ ) ds ′ and K ( x , ξ ) = K ( ξ ). Thus, we obtain β ( x ) = j I + Z Γ( x , Σ( ν )) R ( x , ξ ) ds . Here I is the identity matrix and R ( x , ξ ) is defined as(2.41) R ( x , ξ ) = ∞ X n =0 K n ( x , ξ ) . Finally we obtain for α − ( x ) the following formula α − ( x ) = j n ( x ) p J ( x ) I + Z Γ( x , Σ( ν )) R ( x , ξ ) ds . Applying the above technique, similar formulae can be easily derived for α k ( x ), k =0 , , . . . , r .Denote P k ( x ) = Z Γ( x , Σ( ν )) Q k ( ξ ) ds. Then(2.42) α k ( x ) = 1 n ( x ) p J ( x ) P k ( x ) + Z Γ( x , Σ( ν )) P k ( ξ ) R ( x , ξ ) ds , k = 0 , , . . . , r. This manuscript is for review purposes only. V. G. ROMANOV AND M. V. KLIBANOV
We estimate R ( x , ξ ) later in the proof of Theorem 2. The uniform convergence ofseries (2.41) in D + ( x , R ) follows from that estimate.Since the function n ( x ) ∈ C ∞ ( R ) , then functions f ( s, a , a ) and J ( x ) belong to C ∞ ( D + ( ν , R )). Hence, all functions α k ( x ) ∈ C ∞ ( D + ( ν , R )). Moreover, α − ( x ) = j and α k ( x ) = 0 for k = 0 , , . . . , r , if x lies outside D ( ν , R ) since supp ∇ ε ( x ) ⊂ B .Therefore, functions α k ( x ) for k = 0 , , . . . , r are compactly supported in D + ( ν , R ).Hence, F r ( x , t ) = 0 for { ( x , t ) | x ∈ D ( ν , R ) , t ≥ } (see notation (2.21)). Moreover, F r ( x , t ) = 0 for ( x , t ) ∈ G T ( ν ). These two facts imply that function F r ( x , t ) iscompactly supported in domain R T ⊃ G ( T, ν ). Hence, it follows from (2.32) thatthe vector function E r ( x , t ) vanishes in G T ( ν ) and it is compactly supported in R T because the speed of electromagnetic waves is finite.We now apply the method of energy estimates to the problem (2.32) in the do-main R T ⊃ G ( T, ν ) to estimate function E r ( x , t ). . This method is a powerful toolfor investigations of various problems of mathematical physics. Applications of thismethods for boundary value problems are given in [4], [11] and many other books re-lated to partial differential equations. So, main ideas of this method are well known.At the same time, we can not give a reference of the exact result that we need for ourgoal. Therefore, for the completeness of the proof and for the reader’s convenience weformulate below a lemma related to the estimate of solution to problem (2.32). Let Y ( t, T ) = R T ∩ { t = const } . Lemma.
Let ε ( x ) satisfy the conditions (2.43) k ε k C r ( R ) ≤ µ, k ln ε k C r +1 ( R ) ≤ µ, with a positive constant µ and function F r ( x , t ) belongs to H r ( R T ) and satisfies theinequality (2.44) k F r k H r ( R T ) ≤ M. Then the solution of problem (2.32) E r ∈ H r +1 ( Y ( t, T )) for all t ∈ (0 , T ) and thereexists a positive constant C = C ( µ, T ) such that the following estimates hold: (2.45) k E r k H r +1 ( Y ( t,T )) ≤ C M, k ∂ t E r k H r ( Y ( t,T )) ≤ C M. The proof of this Lemma is given in the Appendix.In our case F r ∈ H r ( R T ) and ε ∈ C ∞ ( R ). Hence, conditions (2.43) and (2.44)valid with some positive µ and M . Applying Lemma, we obtain that the solutionof problem (2.32) is such that E r ∈ H r +1 ( Y ( t, T )) and ∂ t E r ∈ H r ( Y ( t, T )) for all t ∈ [0 , T ]. Choosing r = 4 and applying Lemma, we obtain E ∈ H ( Y ( t, T )) forall t ∈ [0 , T ] and, hence, E ∈ H ( R T ). Therefore the embedding theorem impliesthat E ∈ C ( R T ). Hence, the vector function E ∈ C (cid:16) G ( T, ν ) (cid:17) and is continuoustogether with space derivatives up to the second order across the characteristic wedge t = | ϕ ( x ) | . In particular, E ( x , t ) = 0 for t = | ϕ ( x ) | .Setting(2.46) b E ( x , t ) = X k =0 α k ( x ) ( t − ϕ ( x )) k k ! + E ( x , t ) , This manuscript is for review purposes only.
SINGLE PDE FOR THE MAXWELL’S EQUATIONS b E ∈ C (cid:16) G ( ν , T ) (cid:17) . (cid:3) Remark 1.
The equality (2.46) implies the following formula, which we usebelow:(2.47) lim t → ϕ ( x ) + b E ( x , t ) = α ( x ) .
3. Frequency domain.
Consider the Fourier transform e E ( x , k ) of the function E ( x , t ), e E ( x , k ) = ∞ Z −∞ E ( x , t ) exp ( − ikt ) dt = e E ( x , k ) + ∞ Z E ( x , t ) exp ( − ikt ) dt, (3.1)where k = 2 π/λ is the wave number, λ is the dimensionless wavelength and e E ( x , k ) = j exp( i ( x · ν + R )) θ ( − x · ν − R )is the Fourier transform of E ( x , t ). The existence of the integral in (3.1) followsfrom results of Vainberg [17] which claim that the vector function E ( x , t ) decaysexponentially together with its appropriate derivatives as t → ∞ while x runs overany bounded domain Ω ⊂ R . Next, theorem 3.3 of [16] and theorem 6 of Chapter 9of [17] guarantee that e E ( x , k ) is the solution to the equation(3.2) (∆ + k n ( x )) e E + ∇ ( e E · ∇ ln n ( x )) = 0 , x ∈ R , where the scattering field e E sc ( x , k ) = e E ( x , k ) − e E ( x , k )satisfies the radiation condition as | x | → ∞ .We now consider the vector function e E ( x , k ) in (3.1) for x ∈ D + ( ν , R ). Usingrepresentation (2.26), we obtain(3.3) e E ( x , k ) = α − ( x ) exp( − ikϕ ( x )) + ∞ Z ϕ ( x ) b E ( x , t ) exp ( − ikt ) dt. Integrating by parts in (3.3) and using formula, we obtain e E ( x , k ) = α − ( x ) exp( − ikϕ ( x )) + exp( − ikϕ ( x )) ik α ( x )+ 1 ik ∞ Z ϕ ( x ) b E t ( x , t ) exp ( − ikt ) dt, This manuscript is for review purposes only. V. G. ROMANOV AND M. V. KLIBANOV
Thus,(3.4) e E ( x , k ) = α − ( x ) exp( − ikϕ ( x )) + O (cid:18) k (cid:19) , k → ∞ , ∀ x ∈ D + ( ν , R ) . Consider now the equation(3.5) (∆ + k n ( x ) E = 0 , x ∈ R , with the incident plane wave E ( x , k ) = j exp( ik ( x · ν + R )) θ ( − x · ν − R ) and theradiation condition for (cid:16) E − E (cid:17) . Note that E · ν = 0 and E · ( ν × j ) = 0 since E · ν = 0 and E · ( ν × j ) = 0 because E is parallel to j and also j · ν = 0.Consider the function u ( x , t ) = E · j . Then(3.6) (∆ + k n ( x ) u ( x , k ) = 0 , x ∈ R , with the incident plane wave u ( x , k ) = exp( ik ( x · ν + R )) θ ( − x · ν − R ) and theradiation condition for (cid:0) u − u (cid:1) .We impose conditions below, which guarantee that the electric wave field E j = j u ( x , k ) is close to e E at the high values of the wave number k , which is equivalentto small wavelengths λ . First, suppose that these two electric wave fields are indeedclose to each other in the norm of the space C (cid:16) D + ( ν , R ) (cid:17) . This means that(3.7) e E ( x , k ) = E ( x , k ) + V ( x , k ) , x ∈ D + ( ν , R ) , (3.8) k V ( x , k ) k C ( D + ( ν ,R )) ≤ σ, where σ > e u ( x , k ) = e E ( x , k ) · j . Then(3.9) e u ( x , k ) = u ( x , k ) + V ( x , k ) · j , (3.10) k V ( x , k ) · j k C ( D + ( ν ,R )) ≤ σ, which means that the function e u ( x , k ) approximates well the solution of the Helmholtzequation (3.6) with the above incident plane wave and corresponding radiation con-ditions.Next, since j · ν = j · ( ν × j ) = 0 , then (5.1) implies that e E ( x , k ) · ν = V ( x , k ) · ν , e E ( x , k ) · ( ν × j ) = V ( x , k ) · ( ν × j ) . Hence, components e E ( x , k ) · ν and e E ( x , k ) · ( ν × j ) are sufficiently small,(3.11) (cid:13)(cid:13)(cid:13) e E ( x , k ) · ν (cid:13)(cid:13)(cid:13) C ( D + ( ν ,R )) ≤ σ, (cid:13)(cid:13)(cid:13) e E ( x , k ) · ν (cid:13)(cid:13)(cid:13) C ( D + ( ν ,R )) ≤ σ. Thus, it follows from (3.7)-(3.11) that the component e u ( x , k ) = e E ( x , k ) · j of theelectric wave field dominates two other components and it is close to the solution This manuscript is for review purposes only.
SINGLE PDE FOR THE MAXWELL’S EQUATIONS u ( x , k ) of the Helmholtz equation supplied by the above incident plane wave andradiation conditions. Remark 2 . Since experimental data have noise, then it is sufficient to obtaina good approximation u ( x , k ) for the component e u ( x , k ) = e E ( x , k ) · j of the electricwave field in the C (cid:16) D + ( ν , R ) (cid:17) − norm.What is left to do is to prove (3.7), (3.8). And this is what is done in the rest ofthis section.It is easy to derive a complete analog of formula (3.4) for the function E . To dothis, one should consider the Cauchy problem for the time dependent analog of (3.5)and repeat arguments of Theorem 1 for the case when the term ∇ ( E · ∇ ln n ( x )) isneglected in (2.11). Hence,(3.12) E ( x , k ) = b α − ( x ) exp( − ikϕ ( x )) + O (cid:18) k (cid:19) , k → ∞ , ∀ x ∈ D + ( ν , R ) , where(3.13) b α − ( x ) = j n ( x ) p J ( x ) . Theorem 2.
Suppose that the Assumptions hold. Let η > be such a constantthat (3.14) k∇ n ( x ) k C ( B ) ≤ η. Then e E ( x , k ) − E ( x , k ) = V ( x , k ) + O (cid:18) k (cid:19) , k → ∞ , x ∈ D + ( ν , R ) and (3.15) e E ( x , k ) · j = exp( − ikϕ ( x )) n ( x ) p J ( x ) + V ( x , k ) · j + O (cid:18) k (cid:19) , e E ( x , k ) · ν = V ( x , k ) · ν + O (cid:18) k (cid:19) , e E ( x , k ) · ( ν × j ) = V ( x , k ) · ( ν × j ) + O (cid:18) k (cid:19) , k → ∞ , x ∈ D + ( ν , R ) , where (3.16) V ( x , k ) = ( α − ( x ) − b α − ( x )) exp( − ik ( x · ν )) , | V ( x , k ) | ≤ √ J [exp( ηT ) − , x ∈ D + ( ν , R ) . Thus, if the number η in (3.14) is sufficiently small, then (3.7) and (3.8) hold. Proof.
Using formulae (2.26) and (3.4) we obtain the first relation (3.16). For-mulae (3.15) follow from (3.16) as well as from (3.4). Then, using (3.4) and (3.13),we obtain | V ( x , k ) | = | α − ( x ) − b α − ( x ) | = 1 n ( x ) p J ( x ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Z Γ( x , Σ) R ( x , ξ ) ds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . This manuscript is for review purposes only. V. G. ROMANOV AND M. V. KLIBANOV
We now estimate R ( x , ξ ) for x ∈ D + ( ν , R ) and ξ ∈ Γ( x , Σ( ν )). Note that R ( x , ξ ) = 0 for x ∈ D + ( ν , R ) \ D ( ν , R ) since ∇ n ( x ) = 0 and, hence, all K n ( x , ξ ) = 0in this domain. So, we need estimate R ( x , ξ ) for x ∈ D ( ν , R ) only.Introduce the matrix norm for a matrix K ( x , ξ ) = ( k ij ( x , ξ ) i,j =1 as k K ( x , ξ k = max i,j =1 , , | k ij ( x , ξ ) | . Let x ∈ D ( ν , R ) and ϕ ( x ) = s and ϕ ( ξ ) = s . Then using (2.41) and formulae for K n ( x , ξ ), n = 0 , , , . . . , we obtain k K ( x , ξ ) k ≤ η, x ∈ D ( ν , R ) , ξ ∈ Γ( x , Σ( ν )) , k K ( x , ξ ) k ≤ η ( s − s ) , x ∈ D ( ν , R ) , ξ ∈ Γ( x , Σ( ν )) , k K n ( x , ξ ) k ≤ η n +1 ( s − s ) n n ! , x ∈ D ( ν , R ) , ξ ∈ Γ( x , Σ( ν )) , n = 2 , , . . . Hence k R ( x , ξ ) k ≤ η exp( η ( s − s )) , x ∈ D ( ν , R ) , Since n − ( x ) ≤ J ( x ) ≥ J , then we arrive at the estimate: | V ( x , k ) | ≤ √ J [exp( ηs ) − ≤ √ J [exp( ηT ) − , x ∈ D ( ν , R ) . Thus, we obtain the estimate in the second line of (3.16) . This estimate concludesthe proof. (cid:3)
4. Relevance to Experimental Results of [6, 7, 8].
We now explain whyTheorem 2 at least partially justifies the validity of modeling of the propagation ofelectromagnetic waves in the frequency domain by the single Helmholtz equation (3.6)in the works of the second author with coauthors on experimental data [6, 7, 8]. Wesay “at least partially” because a completely precise explanation is unlikely possiblesince we deal here with a sort of a “mathematics-to-physics bridge”.We recall that accurate reconstruction results were obtained in [6, 7, 8] when solv-ing coefficient inverse problems. Experimental data in these references were collectedfor the cases when rather small inclusions mimicking land mines and improvised ex-plosive devices were embedded in an otherwise uniform background (dry sand). Thedielectric constant was not changing within such an inclusion, although this was notan assumption in reconstruction algorithms. Therefore, ∇ n ( x ) = (cid:26) n ( x ) at the inclu-sion/background interface. Since any solution of an elliptic equation, such as, e.g.(3.6), is sufficiently smooth outside of discontinuities of its coefficients [5], then weconjecture that the medium “percepts” the functions n ( x ) in those inclusions as asmooth function with rather non-small values of |∇ n ( x ) | near those interfaces. Infact, this has been observed in computed images of [6, 7, 8]. Now, since values of |∇ n ( x ) | were not small only in close proximities of those interfaces and volumes ofthose proximities were small, then this means that norms k|∇ n ( x ) |k L ( B ) were actu-ally small. On the other hand, since finite differences with relatively small numbers This manuscript is for review purposes only.
SINGLE PDE FOR THE MAXWELL’S EQUATIONS k|∇ n ( x ) |k L ( B ) is equivalent to the smallness of the discrete norm k|∇ n ( x ) |k C ( B ) , which is close to the smallness assumption imposed in Theorem 2 on the number η in(3.14). Note that smallness assumptions were not used in algorithms of [6, 7, 8].Thus, Theorem 2 explains, at least partially, accurate reconstructions in [6, 7, 8].
5. Appendix. Proof of the Lemma.
Denote E rj , j = 1 , ,
3, components of vector function E r . Calculating scalarproduct of both sides of equation (2.32), 2 ∂ t E r , using n ( x ) = ε ( x ) and applying theidentity 2 ∂ t E r · ∆ E r = 2div X j =1 ( ∂ t E rj ) ∇ E rj − ∂ t X j =1 |∇ E rj | , we obtain: ∂ t ε ( x ) | ∂ t E r | + X j =1 |∇ E rj | − X j =1 ( ∂ t E rj ) ∇ E rj − X i,j =1 ∂ t E ri (cid:2) ( ∂ x i E rj ) ∂ x j ε ( x ) + E rj ∂ x j x i ln ε ( x ) (cid:3) = 2( ∂ t E r ) · F r . (5.1)Integrating identity (5.1) over the domain R t , t ∈ (0 , T ] and taking into account that E r and F r are compactly supported in R T and the initial zero data, we arrive to theequality Z Y ( t,T ) ε ( x ) | ∂ t E r ( x , t ) | + X j =1 |∇ E rj ( x , t ) | d x = 2 Z R t X i,j =1 ( ∂ τ E ri ( x , τ )) (cid:0) ∂ ξ i E rj ( x , τ ) (cid:1) ∂ ξ j ln ε ( x ) d x dτ +2 Z R t X i,j =1 ( ∂ τ E ri ( x , τ )) (cid:0) E rj ( x , τ ) (cid:1) ∂ ξ i ξ j ln ε ( x ) d x dτ +2 Z R t ∂ τ E r ( x , τ ) · F r ( x , τ ) d x dτ . (5.2)Transform this equality using assumption (2.4), (2.43), the algebraic inequalities This manuscript is for review purposes only. V. G. ROMANOV AND M. V. KLIBANOV a · b ≤ | a | + | b | , we obtain ε Z Y ( t,T ) | ∂ t E r ( x , t ) | + X j =1 |∇ E rj ( x , t ) | d x ≤ C Z R t | ∂ τ E r ( x , τ ) | + X j =1 |∇ E rj ( x , τ ) | + | E r ( x , τ ) | d x dτ + Z R t | F r ( x , τ ) d x | dτ . (5.3)Using the inequality | F r ( x , t ) | = t Z ∂ τ E r ( x , τ ) dτ ≤ T t Z | ∂ τ E r ( x , τ ) | dτ , we obtain the more general inequality Z Y ( t,T ) | ∂ t E r ( x , t ) | + X j =1 |∇ E rj ( x , t ) | + | E r ( x , t ) | d x ≤ C Z R t | ∂ τ E r ( x , τ ) | + X j =1 |∇ E rj ( x , τ ) | + | E r ( x , τ ) | d x dτ + Z R t | F r ( x , τ ) | d x dτ . (5.4)Applying Gronwall-Bellman to inequality (5.4), we find Z Y ( t,T ) | ∂ t E r ( x , t ) | + X j =1 |∇ E rj ( x , t ) | + | E r ( x , t ) | d x ≤ k F r k R T exp( C T ) . Thus, we have obtained the inequalities:(5.5) k E r k H ( Y ( t,T )) ≤ C M, k ∂ t E r k L ( Y ( t,T )) ≤ C M. where M is defined in (2.44).Differentiating equation (2.32) k ≤ r times with respect to t and then calculat-ing scalar product of both sides of the resulting equation with the vector function2 ∂ kt E r , we obtain relations (5.1)-(5.5) with ∂ kt E r instead E r . Therefore, the followingestimates hold(5.6) k ∂ kt E r k H ( Y ( t,T )) ≤ C M, k ∂ k +1 t E r k L ( Y ( t,T )) ≤ C M, k ≤ r. Apply now the mathematical induction method to prove estimate (2.45). Supposethat for some n , 1 < n − < r −
1, the estimates similar (5.5), (5.6) hold:(5.7) k E r k H n − ( Y ( t,T )) ≤ C M, k ∂ t E r k H n − ( Y ( t,T )) ≤ C M, k ∂ kt E r k H n − ( Y ( t,T )) ≤ C M, k ∂ k +1 t E r k H n − ( Y ( t,T )) ≤ C M, k ≤ r − ( n − , This manuscript is for review purposes only.
SINGLE PDE FOR THE MAXWELL’S EQUATIONS n − n . Denote D α = ∂ | α | ∂ α x ∂ α x ∂ α x , where α = ( α , α , α ) is a multi index, α , α , α are integer nonnegative numbersand | α | = α + α + α . We shall use the Leibnitz formula for a product of twofunctions D α ( uv ) = X β ≤ α C βα ( D β u )( D α − β v ) , where β = ( β , β , β ), C βα = C β α C β α C β α is product of the binomial coefficients and β ≤ α means that β ≤ α , β ≤ α , β ≤ α . Applying the differential operator D α with | α | = n to equation (2.32) and using the given above formula, we obtain ε ( x ) ∂ t D α E r − ∆ D α E r + X β ≤ α, β = α C βα (cid:0) ∂ t D β E r (cid:1) (cid:0) D α − β ε ( x ) (cid:1) + X j =1 X β ≤ α C βα (cid:2) ( ∇ D β E rj ))( ∂ x j D α − β ε ( x )) + ( D β E rj ) ∇ ( ∂ x j D α − β ln ε ( x )) (cid:3) = D α F r ( x , t ) . (5.8)Calculating scalar product of both sides of equation (5.8) and 2 ∂ t D α E r , we obtainthe relation similar in the main part to (5.1), namely: ∂ t ε | ∂ t D α E r | + X j =1 |∇ D α E rj | − X j =1 ( ∂ t D α E rj ) ∇ E rj +2 X β ≤ α, β = α C βα (cid:2) ( ∂ t D α E r ) · ∂ t D β E r (cid:3) (cid:0) D α − β ε ( x ) (cid:1) − X i,j =1 X β ≤ α C βα ( ∂ t D α E ri ) (cid:2)(cid:0) ∂ x i D β E rj (cid:1) ∂ x j ln ε ( x ) + (cid:0) D β E rj (cid:1) ∂ x i x j D α − β ln ε ( x ) (cid:3) = 2( ∂ t D α E r ) · F r . (5.9) This manuscript is for review purposes only. V. G. ROMANOV AND M. V. KLIBANOV
Integrating this identity over domain R t , t ∈ (0 , T ], we arrive to the equality Z Y ( t,T ) ε ( x ) | ∂ t D α E r ( x , t ) | + X j =1 |∇ D α E rj ( x , t ) | d x = − X β ≤ α, β = α C βα Z R t (cid:2) ( ∂ t D α E r ( x , τ )) · ∂ t D β E r ( x , τ ) (cid:3) (cid:0) D α − β ε ( x ) (cid:1) d x dτ +2 X i,j =1 X β ≤ α C βα Z R t ( ∂ t D α E ri ( x , τ )) (cid:0) ∂ x i D β E rj ( x , τ )) (cid:1) ∂ x j D α − β ln ε ( x ) d x dτ +2 X i,j =1 X β ≤ α C βα Z R t ( ∂ t D α E ri ( x , τ )) (cid:0) D β E rj ( x , τ )) (cid:1) ∂ x i x j D α − β ln ε ( x ) d x dτ +2 Z R t ∂ τ D α E r ( x , τ ) · D α F r ( x , τ ) d x dτ . (5.10)Use now assumption (2.4), (2.43) and the inequality 2 a · b ≤ | a | + | b | . Then, takinginto account that P β ≤ α C βα ≤ n and C βα ≤ n , we obtain ε Z Y ( t,T ) | ∂ t D α E r ( x , t ) | + X j =1 |∇ D α E rj ( x , t ) | d x ≤ C Z R t | ∂ τ D α E r ( x , τ ) | + X j =1 |∇ D α E rj ( x , τ ) | d x dτ +2 n µ X β ≤ α, β = α Z R t | ∂ t D β E r ( x , τ ) | d x dτ + X j =1 |∇ D β E rj ( x , τ ) | + Z R t | D α F r ( x , τ ) d x | dτ , (5.11)where C = 2 n +1 µ .Since the relations β ≤ α, β = α mean that | β | ≤ | α | − n −
1, then by theinduction assumption (5.7), there exists a positive constant C such that2 n µ X β ≤ α, β = α Z R t | ∂ t D β E r ( x , τ ) | d x dτ + X j =1 |∇ D β E rj ( x , τ ) | ≤ C M . This manuscript is for review purposes only.
SINGLE PDE FOR THE MAXWELL’S EQUATIONS Z Y ( t,T ) | ∂ t D α E r ( x , t ) | + X j =1 |∇ D α E rj ( x , t ) | d x ≤ C Z R t | ∂ τ D α E r ( x , τ ) | + X j =1 |∇ D α E rj ( x , τ ) | d x dτ + C M . (5.12)Applying the Gronwall’s inequality, we obtain(5.13) Z Y ( t,T ) | ∂ t D α E r ( x , t ) | + X j =1 |∇ D α E rj ( x , t ) | d x ≤ C M , | α | = n. Differentiating equation (5.8) k ≤ r − ( n −
1) times with respect to t and thencalculating scalar product of both sides of the obtained equation and 2 ∂ kt E r , we obtainrelations (5.8)-(5.13) with ∂ kt D α E r instead D α E r . Therefore, the following estimateshold Z Y ( t,T ) ∂ k +1 t D α E r ( x , t ) | + X j =1 |∇ ∂ kt D α E rj ( x , t ) | d x ≤ C M , k ≤ r − ( n − . Thus, inequalities (5.7) hold with n − n . This justifies the mathe-matical induction method and we can set n = r + 2 in (5.7). The latter proves therequired inequalities (2.45). ✷ REFERENCES[1]
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