Convergence of the uniaxial PML method for time-domain electromagnetic scattering problems
aa r X i v : . [ m a t h . A P ] F e b Convergence of the uniaxial PML method fortime-domain electromagnetic scattering problems
Changkun Wei ∗ Jiaqing Yang † Bo Zhang ‡ Abstract
In this paper, we propose and study the uniaxial perfectly matched layer (PML) methodfor three-dimensional time-domain electromagnetic scattering problems, which has a greatadvantage over the spherical one in dealing with problems involving anisotropic scatterers.The truncated uniaxial PML problem is proved to be well-posed and stable, based on theLaplace transform technique and the energy method. Moreover, the L -norm and L ∞ -normerror estimates in time are given between the solutions of the original scattering problemand the truncated PML problem, leading to the exponential convergence of the time-domainuniaxial PML method in terms of the thickness and absorbing parameters of the PML layer.The proof depends on the error analysis between the EtM operators for the original scatteringproblem and the truncated PML problem, which is different from our previous work (SIAMJ. Numer. Anal. 58(3) (2020), 1918-1940). Keywords:
Well-posedness, stability, time-domain electromagnetic scattering, uniaxialPML, exponential convergence
This paper is concerned with the time-domain electromagnetic scattering by a perfectly con-ducting obstacle which is modeled by the exterior boundary value problem: ∇ × E + µ∂ t H = in ( R \ Ω) × (0 , T ), (1.1a) ∇ × H − ε∂ t E = J in ( R \ Ω) × (0 , T ), (1.1b) n × E = on Γ × (0 , T ), (1.1c) E ( x,
0) = H ( x,
0) = in R \ Ω, (1.1d)ˆ x × ( ∂ t E × ˆ x ) + ˆ x × ∂ t H = o (cid:0) | x | − (cid:1) as | x | → ∞ , t ∈ (0 , T ). (1.1e)Here, E and H denote the electric and magnetic fields, respectively, Ω ⊂ R is a boundedLipschitz domain with boundary Γ and n is the unit outer normal vector to Γ. Throughout ∗ Research Institute of Mathematics, Seoul National University, Seoul, 08826, Republic of Korea( [email protected] ) † School of Mathematics and Statistics, Xi’an Jiaotong University, Xi’an, Shaanxi, 710049, China( [email protected] ) ‡ NCMIS, LSEC and Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing,100190, China and School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing 100049,China ( [email protected] ) ε and the magnetic permeability µ are assumed to bepositive constants. Equation (1.1e) is the well-known Silver-M¨uller radiation condition in thetime domain with ˆ x := x/ | x | .Time-domain scattering problems have been widely studied recently due to their capabilityof capturing wide-band signals and modeling more general materials and nonlinearity, includingtheir mathematical analysis (see, e.g., [1,11,26–28,31,33,39,40] and the references quoted there).The well-posedness and stability of solutions to the problem (1.1a)-(1.1e) have been proved in [16]by employing an exact transparent boundary condition (TBC) on a large sphere. Recently, aspherical PML method has been proposed in [42] to solve the problem (1.1a)-(1.1e) efficiently,based on the real coordinate stretching technique associated with [Re( s )] − in the Laplace trans-form domain with the Laplace transform variable s ∈ C + := { s = s + is ∈ C : s > , s ∈ R } ,and its exponential convergence has also been established in terms of the thickness and absorbingparameters of the PML layer.In this paper, we continue our previous study in [42] and propose and study the uniaxialPML method for the problem (1.1a)-(1.1e), based on the real coordinate stretching techniqueintroduced in [42], which uses a cubic domain to define the PML problem and thus is of great ad-vantage over the spherical one in dealing with problems involving anisotropic scatterers. We firstestablish the existence, uniqueness and stability estimates of the PML problem by the Laplacetransform technique and the energy argument and then prove the exponential convergence inboth the L -norm and the L ∞ -norm in time of the time-domain uniaxial PML method. Ourproof for the L -norm convergence follows naturally from the error estimate between the EtMoperators for the original scattering problem and its truncated PML problem established also inthe paper, which is different from [42]. The L ∞ -norm convergence is obtained directly from thetime-domain variational formulation of the original scattering problem and its truncated PMLproblem with using special test functions.The PML method was first introduced in the pioneering work [3] of B´erenger in 1994 forefficiently solving the time-dependent Maxwell’s equations. Its idea is to surround the compu-tational domain with a specially designed medium layer of finite thickness in which the scat-tered waves decay rapidly regardless of the wave incident angle, thereby greatly reducing thecomputational complexity of the scattering problem. Since then, various PML methods havebeen developed and studied in the literature (see, e.g., [4, 10, 23–25, 29, 35] and the referencesquoted there). Convergence analysis of the PML method has also been widely studied fortime-harmonic acoustic, electromagnetic, and elastic wave scattering problems. For example,the exponential convergence has been established in terms of the thickness of the PML layerin [2, 4, 8, 13, 15, 21, 30, 32] for the circular or spherical PML method and in [5–7, 14, 17, 19, 20]for the uniaxial (or Cartesian) PML method. Among them, the proof in [2] is based on theerror estimate between the electric-to-magnetic (EtM) operators for the original electromag-netic scattering problem and its truncated PML problem, while the key ingredient of the proofin [13] and [14] is the decay property of the PML extensions defined by the series solution andthe integral representation solution, respectively. On the other hand, there are also severalworks on convergence analysis of the time-domain PML method for transient scattering prob-lems. For two-dimensional transient acoustic scattering problems, the exponential convergencewas proved in [12] for the circular PML method and in [18] for the uniaxial PML method,based on the complex coordinate stretching technique. For the 3D time-domain electromagneticscattering problem (1.1a)-(1.1e), the spherical PML method was proposed in [42] based on thereal coordinate stretching technique associated with [Re( s )] − in the Laplace transform domain2ith the Laplace transform variable s ∈ C + , and its exponential convergence was establishedby means of the energy argument and the exponential decay estimates of the stretched dyadicGreen’s function for the Maxwell equations in the free space. In addition, we refer to [1] for thewell-posedness and stability estimates of the time-domain PML method for the two-dimensionalacoustic-elastic interaction problem, and to [41] for the convergence analysis of the PML methodfor the fluid-solid interaction problem above an unbounded rough surface.The remaining part of this paper is as follows. In Section 2, we introduce some basic Sobolevspaces needed in this paper. In Section 3, the well-posedness of the time-domain electromagneticscattering problem is presented, and some important properties are given for the transparentboundary condition (TBC) in the Cartesian coordinate. In Section 4, we propose the uniaxialPML method in the Cartesian coordinate, study the well-posedness of the truncated PMLproblem and establish its exponential convergence. Some conclusions are given in Section 5. We briefly introduce the Sobolev space H (curl , · ) and its related trace spaces which are used inthis paper. For a bounded domain D ⊂ R with Lipschitz continuous boundary Σ, the Sobolevspace H (curl , D ) is defined by H (curl , D ) := { u ∈ L ( D ) : ∇ × u ∈ L ( D ) } which is a Hilbert space equipped with the norm k u k H (curl ,D ) = (cid:16) k u k L ( D ) + k∇ × u k L ( D ) (cid:17) / . Denote by u Σ = n × ( u × n ) | Σ the tangential component of u on Σ, where n denotes theunit outward normal vector on Σ. By [9] we have the following bounded and surjective traceoperators: γ : H ( D ) → H / (Σ) , γϕ = ϕ on Σ ,γ t : H (curl , D ) → H − / (Div , Σ) , γ t u = u × n on Σ ,γ T : H (curl , D ) → H − / (Curl , Σ) , γ T u = n × ( u × n ) on Σ , where γ t and γ T are known as the tangential trace and tangential components trace operators,and Div and Curl denote the surface divergence and surface scalar curl operators, respectively(for the detailed definition of H − / (Div , Σ) and H − / (Curl , Σ), we refer to [9]). By [9] againwe know that H − / (Div , Σ) and H − / (Curl , Σ) form a dual pairing satisfying the integrationby parts formula( u , ∇ × v ) D − ( ∇ × u , v ) D = h γ t u , γ T v i Σ ∀ u , v ∈ H (curl , D ) , (2.2)where ( · , · ) D and h· , ·i Σ denote the L -inner product on D and the dual product between H − / (Div , Σ) and H − / (Curl , Σ), respectively.For any S ⊂ Σ, the subspace with zero tangential trace on S is denoted as H S (curl , D ) := { u ∈ H (curl , D ) : γ t u = 0 on S } . In particular, if S = Σ then we write H (curl , D ) := H Σ (curl , D ).3 The well-posedness of the scattering problem
Let Ω be contained in the interior of the cuboid B := { x = ( x , x , x ) ⊤ ∈ R : | x j | 0) = H ( x, 0) = in Ω , T [ E Γ ] = H × n on Γ × (0 , T ) . (3.6)The well-posedness of the original scattering problem (1.1a)-(1.1e) has been established in [16] byusing the transparent boundary condition on a sphere. Thus the problem (3.6) is also well-posedsince it is equivalent to the problem (1.1a)-(1.1e). However, for convenience of the subsequentuse in the following sections, we study the problem (3.6) directly by studying the property ofthe EtM operator T . For any s ∈ C + := { s = s + is ∈ C : s > , s ∈ R } letˇ E ( x, s ) = L ( E )( x, s ) = Z ∞ e − st E ( x, t ) dt, ˇ H ( x, s ) = L ( H )( x, s ) = Z ∞ e − st H ( x, t ) dt be the Laplace transform of E and H with respect to time t , respectively (for extensive studies onthe Laplace transform, the reader is referred to [22]). Let B : H − / (Curl , Γ ) → H − / (Div , Γ )be the EtM operator B [ ˇ E Γ ] = ˇ H × n on Γ , (3.7)where ˇ E and ˇ H satisfy the exterior Maxwell’s equation in the Laplace domain ∇ × ˇ E + µs ˇ H = in R \ B , ∇ × ˇ H − εs ˇ E = in R \ B , ˆ x × ( ˇ E × ˆ x ) + ˆ x × ˇ H = o (cid:16) | x | (cid:17) as | x | → ∞ . (3.8)4t is obvious that T = L − ◦ B ◦ L . For each s ∈ C + it is known that, by the Lax-Milgramtheorem the problem (3.8) has a unique solution ( ˇ E , ˇ H ) ∈ H (curl , R \ B ) . Thus the operator B is a well-defined, continuous linear operator. Lemma 3.1. For each s ∈ C + , B : H − / (Curl , Γ ) → H − / (Div , Γ ) is bounded with theestimate k B k L ( H − / (Curl , Γ ) ,H − / (Div , Γ )) . | s | − + | s | , (3.9) where L ( X, Y ) denotes the standard space of bounded linear operators from the Hilbert space X to the Hilbert space Y . Further, we have Re h B ω , ω i Γ ≥ for any ω ∈ H − / (Curl , Γ ) , (3.10) where h·i Γ denotes the dual product between H − / (Div , Γ ) and H − / (Curl , Γ ) .Proof. First, eliminating ˇ H from (3.8) and multiplying both sides of the resulting equation with V ∈ H (curl , R \ B ) yield (cid:12)(cid:12) h B [ ˇ E Γ ] , γ T V i Γ (cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z R \ B (cid:2) ( µs ) − ∇ × ˇ E · ∇ × V + εs ˇ E · V dx (cid:3)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . ( | s | − + | s | ) k ˇ E k H (curl , R \ B ) k V k H (curl , R \ B ) , which implies (3.9).Now, for any ω ∈ H − / (Curl , Γ ) suppose ( ˇ E , ˇ H ) is the solution to the problem (3.8)satisfying the boundary condition γ T ˇ E = ω on Γ . Let B R := { x ∈ R : | x | < R } contain thedomain B . Eliminating ˇ H from (3.8) and integrating by parts the resulting equation multipliedwith ˇ E over B R \ B , we obtain that Z B R \ B (cid:0) ( µs ) − |∇ × ˇ E | + εs | ˇ E | (cid:1) dx − h B ω , ω i Γ + Z ∂B R ˆ x × ( µs ) − ∇ × ˇ E · ˇ E dγ = 0 . (3.11)Taking the real part of (3.11) and noting that (cid:12)(cid:12) ˆ x × ( ˇ E × ˆ x ) − ˆ x × ( µs ) − ∇ × ˇ E (cid:12)(cid:12) = | ˆ x × ( ˇ E × ˆ x ) | + | ˆ x × ( µs ) − ∇ × ˇ E | − x × ( µs ) − ∇ × ˇ E ) · ˇ E , we have s µ | s | k∇ × ˇ E k L ( B R \ B ) + εs k ˇ E k L ( B R \ B ) − Re h B ω , ω i Γ + 12 (cid:13)(cid:13) ˆ x × ( ˇ E × ˆ x ) (cid:13)(cid:13) L ( ∂B R ) + 12 (cid:13)(cid:13) ˆ x × ( µs ) − ∇ × ˇ E (cid:13)(cid:13) L ( ∂B R ) = 12 (cid:13)(cid:13) ˆ x × ( ˇ E × ˆ x ) − ˆ x × ( µs ) − ∇ × ˇ E (cid:13)(cid:13) L ( ∂B R ) . (3.12)By the Silver-M¨uller radiation condition (1.1e) in the s -domain, it is known that the right-handside of (3.12) tends to zero as R → ∞ . This implies that Re h B ω , ω i Γ ≥ 0. The proof is thuscomplete. 5y using Lemma 3.1 and [40, Lemmas 4.5-4.6], the time-domain EtM operator T has thefollowing positive properties which will be used in the error analysis of the time-domain PMLsolution. Lemma 3.2. Given ξ ≥ and ω ( · , t ) ∈ L (0 , ξ ; H − / (Curl , Γ )) it holds that Re Z Γ Z ξ (cid:18)Z t C [ ω ]( x, τ ) dτ (cid:19) ω ( x, t ) dtdγ ≥ , where C = L − ◦ s B ◦ L . Lemma 3.3. Given ξ ≥ and ω ( · , t ) ∈ L (0 , ξ ; H − / (Curl , Γ )) with ω ( · , 0) = , it holds that Re Z Γ Z ξ (cid:18)Z t C [ ∂ τ ω ]( x, τ ) dτ (cid:19) ∂ τ ω ( x, t ) dtdγ ≥ . We now introduce the equivalent variational formulation in the Laplace transform domain tothe problem (3.6). To this end, eliminate the magnetic field H and take the Laplace transformof (3.6) to get ∇ × [( µs ) − ∇ × ˇ E ] + εs ˇ E = − ˇ J in Ω , n × ˇ E = on Γ , B [ ˇ E Γ ] = − ( µs ) − ∇ × ˇ E × n on Γ . (3.13)The variational formulation of (3.13) is then as follows: find a solution ˇ E ∈ H Γ (curl , Ω ) suchthat a ( ˇ E , V ) = − Z Ω ˇ J · V dx, ∀ V ∈ H Γ (curl , Ω ) , (3.14)where the sesquilinear form a ( · , · ) is defined as a ( ˇ E , V ) = Z Ω (cid:2) ( sµ ) − ( ∇ × ˇ E ) · ( ∇ × V ) dx + εs ˇ E · V (cid:3) dx + h B [ ˇ E Γ ] , V Γ i Γ . (3.15)By Lemma 3.1 it is easy to see that a ( · , · ) is uniformly coercive, that is,Re[ a ( ˇ E , ˇ E )] & s | s | ( k∇ × ˇ E k L (Ω ) + k s ˇ E k L (Ω ) ) ≥ s min {| s | − , }k ˇ E k H (curl , Ω ) . (3.16)Then, by the Lax-Milgram theorem the problem (3.13) is well-posed for each s ∈ C + . Thus,and by the energy argument in conjunction with the inversion theorem of the Laplace trans-form (cf. [16]) the well-posedness of the problem (3.6) follows. In particular, T [ E Γ ] ∈ L (cid:0) , T ; H − / (Div , Γ ) (cid:1) . In practical applications, the scattering problems may involve anisotropic scatterers. In thiscase, the uniaxial PML method has a big advantage over the circular or spherical PML methodas it provides greater flexibility and efficiency in solving such problems. Thus, in this section,we propose and study the uniaxial PML method for solving the time-domain electromagneticscattering problem (1.1a)-(1.1e). 6 .1 The PML equation in the Cartesian coordinates In this subsection, we derive the PML equation in the Cartesian coordinates. To this end,define B := { x = ( x , x , x ) ⊤ ∈ R : | x j | < L j / d j , j = 1 , , } with boundary Γ = ∂B B B L L d G W d G G Figure 1: Geometric configuration of the uniaxial PMLwhich is a cubic domain surrounding B . Denote by n the unit outward normal to Γ . LetΩ PML = B \ B be the PML layer and let Ω = B \ Ω be the truncated PML domain. SeeFigure 1 for the uniaxial PML geometry.For x = ( x , x , x ) ⊤ ∈ R , let s > α j ( x j ) = 1 + s − σ j ( x j ) , j = 1 , , , where σ j ( x j ) = , | x j | ≤ L j / , e σ j (cid:18) | x j | − L j / d j (cid:19) m , L j / < | x j | ≤ L j / d j , e σ j , L j / d j < | x j | < ∞ (4.17)with positive constants e σ j , j = 1 , , 3, and integer m ≥ 1. In what follows, we will take the realpart of the Laplace transform variable s ∈ C + to be s , that is, Re( s ) = s .In the rest of this paper, we always make the following assumptions on the thickness of thePML layer and the parameters e σ j , which are reasonable in our model: d = d = d := d ≥ , L = max { L , L , L } ≤ C d, (4.18) e σ = e σ = e σ := σ > C . Under the assumptions (4.18) and (4.19) we have Z L j / d j σ j ( τ ) dτ = σ dm + 1 , j = 1 , , . (4.20)We remark that the constant assumption on d j and e σ j in (4.18)-(4.19) is only to simplifythe convergence analysis but not mandatory. We now introduce the real stretched Cartesiancoordinates e x = ( e x , e x , e x ) ⊤ with e x j = Z x j α j ( τ ) dτ, j = 1 , , . (4.21)7oting that the solution of the exterior problem (3.8) in R \ B can be derived as theintegral representation [34, Theorem 12.2], we can derive the PML extension under the stretchedcoordinates e x by following [42]. For any p ∈ H − / (Div , Γ ) and q ∈ H − / (Div , Γ ), define E ( p , q )( x ) := − e Ψ SL ( q )( x ) − e Ψ DL ( p )( x ) , x ∈ R \ B , (4.22)where the stretched single- and double-layer potentials are defined as e Ψ SL ( q ) = Z Γ e G ⊤ ( s, x, y ) q ( y ) dγ ( y ) , e Ψ DL ( p ) = Z Γ (curl y e G ) ⊤ ( s, x, y ) p ( y ) dγ ( y ) . Here, the stretched dyadic Green’s function is given by e G ( s, x, y ) = e Φ s ( x, y ) I + 1 k ∇ y ∇ y e Φ s ( x, y ) , x = y, k = i √ εµs (4.23)with the stretched fundamental solution and the complex distance e Φ s ( x, y ) = e −√ εµρ s ( e x,y ) πρ s ( e x, y ) s − , ρ s ( e x, y ) = s | e x − y | . (4.24)Introduce the stretched curl operator acting on vector u = ( u , u , u ) ⊤ : g curl u = e ∇ × u := (cid:18) ∂u ∂ e x − ∂u ∂ e x , ∂u ∂ e x − ∂u ∂ e x , ∂u ∂ e x − ∂u ∂ e x (cid:19) ⊤ = A ∇ × B u with the diagonal matrices A = diag (cid:26) α α , α α , α α (cid:27) and B = diag { α , α , α } . (4.25)The PML extension in the s -domain in R \ B of γ t ( ˇ E ) | Γ and γ t (curl ˇ E ) | Γ is then defined asˇ e E ( x ) = E ( γ t ( ˇ E ) , γ t (curl ˇ E )) , x ∈ R \ B . (4.26)Define ˇ f H ( x ) := − ( µs ) − g curl ˇ e E ( x ) for x ∈ R \ B . Then it is easy to see that ( ˇ e E , ˇ f H ) satisfiesthe Maxwell equation in the s -domain: e ∇ × ˇ e E + µs ˇ f H = , e ∇ × ˇ f H − εs ˇ e E = in R \ B . (4.27)Define ( E PML , H PML ) := B ( L − ( ˇ e E ) , L − ( ˇ f H )) . Then ( E PML , H PML ) can be viewed as the extension in the region R \ B of the solution of theproblem (1.1a)-(1.1e) since, by the fact that α j = 1 on Γ for j = 1 , , E PML = E , H PML = H on Γ . If we set E PML = E and H PML = H in Ω × (0 , T ), then ( E PML , H PML )satisfies the PML problem: ∇ × E PML + µ ( BA ) − ∂ t H PML = in ( R \ Ω) × (0 , T ) , ∇ × H PML − ε ( BA ) − ∂ t E PML = J in ( R \ Ω) × (0 , T ) , n × E PML = on Γ × (0 , T ) , E PML ( x, 0) = H PML ( x, 0) = in R \ Ω . (4.28)8he truncated PML problem in the time domain is to find ( E p , H p ), which is an approximationto ( E , H ) in Ω , such that ∇ × E p + µ ( BA ) − ∂ t H p = in Ω × (0 , T ) , ∇ × H p − ε ( BA ) − ∂ t E p = J in Ω × (0 , T ) , n × E p = on Γ × (0 , T ) , n × E p = on Γ × (0 , T ) , E p ( x, 0) = H p ( x, 0) = in Ω . (4.29) We now study the well-posedness of the truncated PML problem (4.29), employing the Laplacetransform technique and a variational method. Eliminate H p and take the Laplace transformof (4.29) to obtain that ∇ × (cid:2) ( µs ) − BA ∇ × ˇ E p (cid:3) + εs ( BA ) − ˇ E p = − ˇ J in Ω , n × ˇ E p = on Γ , n × ˇ E p = on Γ . (4.30)The variational formulation of (4.30) can be derived as follows: find a solution ˇ E p ∈ H (curl , Ω )such that a p (cid:0) ˇ E p , V (cid:1) = − Z Ω ˇ J · V dx, ∀ V ∈ H (curl , Ω ) , (4.31)where the sesquilinear form a p ( · , · ) is defined as a p (cid:0) ˇ E p , V (cid:1) = Z Ω (cid:2) ( µs ) − BA ( ∇ × ˇ E p ) · ( ∇ × V ) dx + εs ( BA ) − ˇ E p · V (cid:3) dx. (4.32)We have the following result on the well-posedness of the variational problem (4.31). Lemma 4.1. For each s ∈ C + with Re( s ) = s > the variational problem (4.31) has a uniquesolution ˇ E p ∈ H (curl , Ω ) . Further, it holds that k∇ × ˇ E p k L (Ω ) + k s ˇ E p k L (Ω ) . s − (1 + s − σ ) k s ˇ J k L (Ω ) . (4.33) Proof. By the definition of the diagonal matrix BA (see (4.25)) and a direct calculation it easilyfollows that (1 + s − σ ) − ≤ | BA | ≤ (1 + s − σ ) in Ω PML , (4.34)(1 + s − σ ) − ≤ | ( BA ) − | ≤ (1 + s − σ ) , in Ω PML . (4.35)Thus, it is derived thatRe[ a p ( ˇ E p , ˇ E p )] & s − σ ) s | s | (cid:0) k∇ × ˇ E p k L (Ω ) + k s ˇ E p k L (Ω ) (cid:1) . (4.36)The existence and uniqueness of solutions to the problem (4.31) then follow from the Lax-Milgram theorem. The estimate (4.33) can be obtained by combining (4.31), (4.36) and theCauchy-Schwartz inequality. The proof is thus complete.9o show the well-posedness of the truncated PML problem (4.29) in the time domain, weneed the following lemma which is the analog of the Paley-Wiener-Schwartz theorem for theFourier transform of the distributions with compact support in the case of Laplace transform [36,Theorem 43.1]. Lemma 4.2. [36, Theorem 43.1]. Let ˇ ω ( s ) denote a holomorphic function in the half complexplane s = Re( s ) > σ for some σ ∈ R , valued in the Banach space E . Then the followingstatements are equivalent:1) there is a distribution ω ∈ D ′ + ( E ) whose Laplace transform is equal to ˇ ω ( s ) , where D ′ + ( E ) is the space of distributions on the real line which vanish identically in the open negativehalf-line;2) there is a σ with σ ≤ σ < ∞ and an integer m ≥ such that for all complex numbers s with s = Re( s ) > σ it holds that k ˇ ω ( s ) k E . (1 + | s | ) m . The well-posedness and stability of the truncated PML problem (4.29) can be proved byusing Lemmas 4.1 and 4.2 and the energy method (cf. [16, Theorem 3.1]). Theorem 4.3. Let s = 1 /T . Then the truncated PML problem (4.29) in the time domain hasa unique solution ( E p ( x, t ) , H p ( x, t )) with E p ∈ L (cid:0) , T ; H (curl , Ω ) (cid:1) ∩ H (cid:0) , T ; L (Ω ) (cid:1) , H p ∈ L (cid:0) , T ; H (curl , Ω ) (cid:1) ∩ H (cid:0) , T ; L (Ω ) (cid:1) and satisfying the stability estimate max t ∈ [0 ,T ] (cid:2) k ∂ t E p k L (Ω ) + k∇ × E p k L (Ω ) + k ∂ t H p k L (Ω ) + k∇ × H p k L (Ω ) (cid:3) . (1 + σ T ) k J k H (0 ,T ; L (Ω ) ) . (4.37)To study the convergence of the uniaxial PML method, we introduce the EtM operator b B : H − / (Curl , Γ ) → H − / (Div , Γ ) associated with the truncated PML problem (4.30) inthe s -domain. Given λ ∈ H − / (Div , Γ ), define b B ( λ × n ) := n × ( µs ) − ∇ × u on Γ , (4.38)where u satisfies the following problem in the PML layer: ( ∇ × (cid:2) ( µs ) − BA ∇ × u (cid:3) + εs ( BA ) − u = 0 in Ω PML , n × u = λ on Γ , n × u = on Γ . (4.39)We need to show that (4.39) has unique solution, so b B is well-defined. To this end, we con-sider the following general problem with the tangential trace ξ on Γ , which is needed for theconvergence analysis of the PML method: ( ∇ × (cid:2) ( µs ) − BA ∇ × u (cid:3) + εs ( BA ) − u = 0 in Ω PML , n × u = λ on Γ , n × u = ξ on Γ . (4.40)10efine the sesquilinear form a PML : H (curl , Ω PML ) × H (curl , Ω PML ) → C as a PML ( u , V ) := Z Ω PML ( µs ) − BA ( ∇ × u ) · ( ∇ × V ) dx + Z Ω PML εs ( BA ) − u · V dx. (4.41)Then the variational formulation of (4.40) is as follows: Given λ ∈ H − / (Div , Γ ) and ξ ∈ H − / (Div , Γ ), find u ∈ H (curl , Ω PML ) such that n × u = λ on Γ , n × u = ξ on Γ and a PML ( u , V ) = 0 , ∀ V ∈ H (curl , Ω PML ) . (4.42)Arguing similarly as in proving (4.36), we obtain that for any V ∈ H (curl , Ω PML ),Re (cid:2) a PML ( V , V (cid:3) & s − σ ) s | s | h k∇ × V k L (Ω PML ) + k s V k L (Ω PML ) i . (4.43)By (4.43) and the Lax-Milgram theorem it follows that the variational problem (4.42) has aunique solution. We have the following stability result for the solution to the problem (4.40). Lemma 4.4. For any λ ∈ H − / (Div , Γ ) and ξ ∈ H − / (Div , Γ ) , let u be the solution to theproblem (4.40) . Then k∇ × u k L (Ω PML ) + k s u k L (Ω PML ) . s − (1 + s − σ ) | s | (1 + | s | )( k λ k H − / (Div , Γ ) + k ξ k H − / (Div , Γ ) ) . (4.44) Proof. Let u ∈ H (curl , Ω PML ) be such that n × u = λ , n × u = ξ on Γ . Then, by (4.42)we have ω := u − u ∈ H (curl , Ω PML ) and a PML ( ω , V ) = − a PML ( u , V ) , ∀ V ∈ H (curl , Ω PML ) . (4.45)This, combined with (4.41)-(4.43) and the Cauchy-Schwartz inequality, gives1(1 + s − σ ) s | s | (cid:16) k∇ × ω k L (Ω PML ) + k s ω k L (Ω PML ) (cid:17) . Re (cid:2) a PML ( ω , ω ) (cid:3) . (1 + s − σ ) | s | p | s | (cid:16) k∇ × ω k L (Ω PML ) + k s ω k L (Ω PML ) (cid:17) / k u k H (curl , Ω PML ) , yielding (cid:16) k∇ × ω k L (Ω PML ) + k s ω k L (Ω PML ) (cid:17) / . (1 + s − σ ) | s | p | s | s k u k H (curl , Ω PML ) . This, together with the definition of ω and the Cauchy-Schwartz inequality, implies that k∇ × ˇ u k L (Ω PML ) + k s ˇ u k L (Ω PML ) . (1 + s − σ ) | s | (1 + | s | ) s k u k H (curl , Ω PML ) . The desired estimate (4.44) then follows from the trace theorem.11ow, by using b B the truncated PML problem (4.30) for the electric field ˇ E p can be equiva-lently reduced to the boundary value problem in Ω : ( ∇ × (cid:2) ( µs ) − ∇ × ˇ E p (cid:3) + εs ˇ E p = − ˇ J in Ω , n × ˇ E p = on Γ , b B [ ˇ E p Γ ] = n × ( µs ) − ∇ × ˇ E p on Γ . (4.46)Similarly, for the problem (4.46) we can derive its equivalent variational formulation: find ˇ E p ∈ H Γ (curl , Ω ) such that b a (cid:0) ˇ E p , V ) (cid:1) = − Z Ω ˇ J · V dx, ∀ V ∈ H Γ (curl , Ω ) , (4.47)where the sesquilinear form b a ( · , · ) is defined as b a (cid:0) ˇ E p , V (cid:1) := Z Ω (cid:2) ( µs ) − ( ∇ × ˇ E p ) · ( ∇ × V ) dx + εs ˇ E p · V (cid:3) dx + h b B [ ˇ E Γ ] , V Γ i Γ . (4.48)By using b B and the Laplace and inverse Laplace transform, the truncated PML problem (4.29)is equivalent to the initial boundary value problem in Ω : ∇ × E p + µ∂ t H p = in Ω × (0 , T ) , ∇ × H p − ε∂ t E p = J in Ω × (0 , T ) , n × E p = on Γ × (0 , T ) , E p ( x, 0) = H p ( x, 0) = in Ω , ˆ T [ E p Γ ] = H p × n on Γ × (0 , T ) . (4.49)Here, ˆ T = L − ◦ b B ◦ L is the time-domain EtM operator for the PML problem. In this subsection, we prove the exponential convergence of the uniaxial PML method. We beginwith the following lemma which is useful in the proof of the exponential decay property of thestretched fundamental solution e Φ s ( x, y ). Lemma 4.5. Let s = s + is with s > , s ∈ R . Then, for any x ∈ Γ and y ∈ Γ thecomplex distance ρ s defined by (4.24) satisfies | ρ s ( e x, y ) /s | ≥ d, Re[ ρ s ( e x, y )] ≥ σ dm + 1 . Proof. For x ∈ Γ and y ∈ Γ , e x j − y j = ( x j − y j ) + s − x j ˆ σ j ( x j ), whereˆ σ j ( x j ) = 1 x j Z x j σ j ( τ ) dτ. Then, by the definition of the complex distance ρ s ( e x, y ) (see (4.24)) we have | ρ s ( e x, y ) /s | = | e x − y | = p ( e x − y ) + ( e x − y ) + ( e x − y ) = X j =1 (cid:2) ( x j − y j ) + 2 s − x j ˆ σ j ( x j )( x j − y j ) + s − x j ˆ σ j ( x j ) (cid:3) / ≥ | x − y | ≥ d, x j ˆ σ j ( x j )( x j − y j ) ≥ x ∈ Γ and y ∈ Γ . In addition,Re [ ρ s ( e x, y )] = Re (cid:2) s (cid:0) ( e x − y ) + ( e x − y ) + ( e x − y ) (cid:1)(cid:3) / = s p ( e x − y ) + ( e x − y ) + ( e x − y ) ≥ X j =1 x j ˆ σ j ( x j ) / . If x j = ± ( L j / d j ) ∈ Γ , then, by (4.20) we have | x j ˆ σ j ( x j ) | = σ d/ ( m + 1). Thus,Re [ ρ s ( e x, y )] ≥ σ d/ ( m + 1). The proof is thus complete.By Lemma 4.5, and arguing similarly as in the proof of [42, Lemma 5.3], we have similarestimates as in [42, Lemma 5.3] for the stretched dyadic Green’s function e G in the PML layer.Then we have the decay property of the PML extension (cf. [42, Theorem 5.4]). Lemma 4.6. For any p , q ∈ H − / (Div , Γ ) let E ( p , q ) be the PML extension in the s -domaindefined in (4.22) . Then we have that for any x ∈ Ω PML , | E ( p , q )( x ) | (4.50) . s − d / (1 + s − σ ) e − √ εµσ dm +1 h (1 + | s | ) k q k H − / (Div , Γ ) + (1 + | s | ) k p k H − / (Div , Γ ) i and | curl e x E ( p , q )( x ) | (4.51) . d / (1 + s − σ ) e − √ εµσ dm +1 h (1 + | s | ) k q k H − / (Div , Γ ) + (1 + | s | ) k p k H − / (Div , Γ ) i . We now establish the L -norm and L ∞ -norm error estimates in time between solutions tothe original scattering problem and the truncated PML problem (4.29) in the computationaldomain Ω . Theorem 4.7. Let ( E , H ) and ( E p , H p ) be the solutions of the problems (1.1a) - (1.1e) and (4 . with s = 1 /T , respectively. If the assumptions (3 . and (3 . are satisfied, then we havethe error estimates k E − E p k L (0 ,T ; H (curl , Ω )) + k H − H p k L (0 ,T ; H (curl , Ω )) . T d (1 + σ T ) e − σ d √ εµ/ k J k H (0 ,T ; L (Ω ) ) . (4.52) and k E − E p k L ∞ (0 ,T ; H (curl , Ω )) + k H − H p k L ∞ (0 ,T ; H (curl , Ω )) . T / d (1 + σ T ) e − σ d √ εµ/ k J k H (0 ,T ; L (Ω ) ) . (4.53) Proof. We first prove (4.52). Let U = E − E p and V = H − H p and let ˇ E and ˇ E p be thesolutions to the variational problems (3.14) and (4.47), respectively. Then, by (3.14) and (4.47)we get a ( ˇ U , ˇ U ) = b a ( ˇ E , ˇ U ) − a ( ˇ E p , ˇ U ) = h ( b B − B )[ ˇ E p Γ ] , ˇ U Γ i Γ . (4.54)13his, together with the uniform coercivity (3.16) of a ( · , · ), implies that k ˇ U k H (curl , Ω ) . s − (cid:0) | s | (cid:1) k ( b B − B )[ ˇ E p Γ ] k H − / (Div , Γ ) . (4.55)From the Maxwell equations in Ω obtained by taking the Laplace transform of the problems(1.1a)-(1.1e) and (4 . k ˇ V k H (curl , Ω ) . ( | s | + | s | − ) k ˇ U k H (curl , Ω ) . This, combined with (4.55), leads to the result k ˇ U k H (curl , Ω ) + k ˇ V k H (curl , Ω ) . s − ( | s | − + | s | ) k ( b B − B )[ ˇ E p Γ ] k H − / (Div , Γ ) . (4.56)We now estimate the norm k ( b B − B )[ ˇ E p Γ ] k H − / (Div , Γ ) . For ˇ E p | Γ define its PML extensionˇ e E p in the s -domain to be the solution of the exterior problem e ∇ × [( µs ) − e ∇ × v ] + εs v = in R \ B , n × v = n × ˇ E p on Γ , ˆ x × ( µs v × ˆ x ) − ˆ x × ( e ∇ × v ) = o (cid:0) | e x | − (cid:1) as | e x | → ∞ . By [34, Theorem 12.2] it is easy to see that ˇ e E p has the integral representationˇ e E p = E ( γ t ( ˇ E p ) , γ t ( g curl ˇ e E p )) . Define ˇ f H p := − ( µs ) − g curl ˇ e E p . Then ( ˇ e E p , ˇ f H p ) satisfies the stretched Maxwell equations in (4.27)in R \ B . It is worth noting that ˇ f H p is not the extension of ˇ H p | Γ .Noting that e ∇ × v = A ∇ × B v , we know that B ˇ e E p satisfies the problem ∇ × (cid:2) ( µs ) − BA ∇ × v (cid:3) + εs ( BA ) − v = in R \ B , n × v = n × ˇ E p on Γ , ˆ x × ( µs B − v × ˆ x ) − ˆ x × ( A ∇ × v ) = o (cid:0) | e x | − (cid:1) as | e x | → ∞ , where we have used the fact that ˇ e E p is the extension of ˇ E p | Γ and B = diag { , , } on Γ . Bythe definition of B , and since A = diag { , , } on Γ , it is easy to see that B [ ˇ E p Γ ] = n × ( µs ) − e ∇ × ˇ e E p = n × ( µs ) − ∇ × B ˇ e E p . By the definition of b B in (4.38), we obtain that( b B − B )[ ˇ E p Γ ] = n × ( µs ) − ∇ × ω (4.57)where ω satisfies ∇ × (cid:2) ( µs ) − BA ∇ × ω (cid:3) + εs ( BA ) − ω = 0 in Ω PML , n × ω = on Γ , n × ω = γ t ( B ˇ e E p ) on Γ . 14y Lemma 4.4 and the estimate for BA and ( BA ) − in (4.34)-(4.35), we have k n × ( µs ) − ∇ × ω k H − / (Div , Γ ) . (1 + s − σ ) k ( µs ) − BA ∇ × ω k H (curl , Ω PML ) . (1 + s − σ ) (cid:18) (1 + s − σ ) | s | k∇ × ω k L (Ω PML ) + (1 + s − σ ) k s ω k L (Ω PML ) (cid:19) / . s − (1 + s − σ ) (1 + | s | ) k γ t ( B ˇ e E p ) k H − / (Div , Γ ) . (4.58)Since e ∇ × v = A ∇ × B v and | A − | ≤ (1 + σ ) in Ω PML , we have by the boundedness of thetrace operator γ t that k γ t ( B ˇ e E p ) k H − / (Div , Γ ) . k B ˇ e E p k H (curl , Ω PML ) . (1 + s − σ ) k ˇ e E p k H ( g curl , Ω PML ) . (4.59)By Lemma 4.6 and the boundedness of γ T and γ t it is derived that k ˇ e E p k H ( g curl , Ω PML ) ≤ ( k ˇ e E p k L ∞ (Ω PML ) + k g curl ˇ e E p k L ∞ (Ω PML ) ) | Ω PML | . s − d (1 + s − σ ) e − √ εµσ dm +1 h (1 + | s | ) k γ t ( g curl ˇ e E p ) k H − / (Div , Γ ) + (1 + | s | ) k γ t ˇ E p k H − / (Div , Γ ) i . s − d (1 + s − σ ) e − √ εµσ dm +1 h (1 + | s | ) k γ T ˇ e E p k H − / (Curl , Γ ) + (1 + | s | ) k γ t ˇ E p k H − / (Div , Γ ) i . s − d (1 + s − σ ) e − √ εµσ dm +1 X l =0 k s l ˇ E p k H (curl , Ω ) . s − d (1 + s − σ ) e − √ εµσ dm +1 X l =0 k s l ˇ J k L (Ω ) , (4.60)where we have used Lemma 4.1 and the upper bound estimate (3.9) of the EtM operator B .Combining (4.56)-(4.60) yields that k ˇ U k H (curl , Ω ) + k ˇ V k H (curl , Ω ) . s − d (1 + s − σ ) e − √ εµσ dm +1 X l =0 k s l ˇ J k L (Ω ) . (4.61)This, together with the Parseval identity for the Laplace transform (see [22, (2.46)])12 π Z ∞−∞ ˇ u ( s ) · ˇ v ( s ) ds = Z ∞ e − s t u ( t ) · v ( t ) dt, s > λ , where λ is the abscissa of convergence for ˇ u and ˇ v , gives k U k L (0 ,T ; H (curl , Ω )) + k V k L (0 ,T ; H (curl , Ω )) = Z T (cid:16) k U k H (curl , Ω ) + k V k H (curl , Ω ) (cid:17) dt ≤ e s T Z ∞ e − s t (cid:16) k U k H (curl , Ω ) + k V k H (curl , Ω ) (cid:17) dt . e s T Z ∞ s − d (1 + s − σ ) e − √ εµσ dm +1 X l =0 k s l ˇ J k L (Ω ) ds . e s T s − d (1 + s − σ ) e − √ εµσ dm +1 k J k H (0 ,T ; L (Ω ) ) , (4.62)where we have used the assumptions (3.3) and (3.4) to get the last inequality. It is obvious that m should be chosen small enough to ensure rapid convergence (thus we need to take m = 1).Since s − = T in (4.62), we obtain the required estimate (4.52) by using the Cauchy-Schwartzinequality.We now prove (4.53). Since ( E , H ) and ( E p , H p ) satisfy the equations (3.6) and (4.49),respectively, it is easy to verify that ( U , V ) satisfies the problem ∇ × U + µ∂ t V = in Ω × (0 , T ) , ∇ × V − ε∂ t U = in Ω × (0 , T ) , n × U = on Γ × (0 , T ) , U ( x, 0) = V ( x, 0) = in Ω , V × n = ( T − ˆ T )[ E p Γ ] + T [ U Γ ] on Γ × (0 , T ) . (4.63)Eliminating V yields that ∇ × ( µ − ∇ × U ) + ε∂ t U = in Ω × (0 , T ) , n × U = on Γ × (0 , T ) , U ( x, 0) = ∂ t U ( x, 0) = in Ω ,µ − ( ∇ × U ) × n + C [ U Γ ] = ( ˆ T − T )[ ∂ t E p Γ ] on Γ × (0 , T ) , (4.64)where C = L − ◦ s B ◦ L . The variational problem of (4.64) is to find U ∈ H Γ (curl , Ω ) for all t > Z Ω ε∂ t U · ω dx = − Z Ω µ − ( ∇ × U )( ∇ × ω ) dx (4.65)+ Z Γ ( ˆ T − T )[ ∂ t E p Γ ] · ω Γ dγ − Z Γ C [ U Γ ] · ω Γ dγ, ∀ ω ∈ H Γ (curl , Ω ) . For 0 < ξ < T , introduce the auxiliary function Ψ ( x, t ) = Z ξt U ( x, τ ) dτ, x ∈ Ω , ≤ t ≤ ξ. Then it is easy to verify that Ψ ( x, ξ ) = 0 , ∂ t Ψ ( x, t ) = − U ( x, t ) . (4.66)16or any φ ( x, t ) ∈ L (cid:0) , ξ ; L (Ω ) (cid:1) , using integration by parts and condition (4.66), we have Z ξ φ ( x, t ) · Ψ ( x, t ) dt = Z ξ Z t φ ( x, τ ) dτ · U ( x, t ) dt. (4.67)Taking the test function ω = Ψ in (4.65) and using (4.66) giveRe Z ξ Z Ω ε∂ t U · Ψ dxdt =Re Z Ω Z ξ ε (cid:16) ∂ t ( ∂ t U · Ψ ) + ∂ t U · U (cid:17) dtdx = 12 k√ ε U ( · , ξ ) k L (Ω ) . (4.68)By (4.67) we have the estimateRe Z ξ Z Ω µ − ( ∇ × U ) · ( ∇ × Ψ ) dxdt = Re Z Ω Z ξ µ − ( ∇ × U ) · Z ξt ( ∇ × U ( x, τ )) dτ dtdx = Z Ω µ − (cid:12)(cid:12)(cid:12) Z ξ ( ∇ × U )( x, t ) dt (cid:12)(cid:12)(cid:12) dx − Re Z ξ Z Ω µ − ( ∇ × U ) · ( ∇ × Ψ ) dxdt, which implies thatRe Z ξ Z Ω µ − ( ∇ × U ) · ( ∇ × Ψ ) dxdt = 12 Z Ω µ − (cid:12)(cid:12)(cid:12) Z ξ ∇ × U ( x, t ) dt (cid:12)(cid:12)(cid:12) dx. (4.69)Integrating (4.65) from t = 0 to t = ξ and taking the real parts yield12 k√ ε U ( · , ξ ) k L (Ω ) + 12 Z Ω µ − (cid:12)(cid:12)(cid:12) Z ξ ∇ × U ( x, t ) dt (cid:12)(cid:12)(cid:12) = Re Z ξ Z Γ ( ˆ T − T )[ ∂ t E p Γ ] · Ψ dγdt − Re Z ξ Z Γ C [ U Γ ] · Ψ dγdt. (4.70)First, using (4.67) and Lemma 3.2, we haveRe Z ξ Z Γ C [ U Γ ] · Ψ dγdt = Re Z Γ Z ξ (cid:18)Z t C [ U Γ ]( x, τ ) dτ (cid:19) · U Γ ( x, t ) dt dγ ≥ . (4.71)Then, and by (4.67) we deduce the estimate12 k√ ε U ( · , ξ ) k L (Ω ) + 12 Z Ω µ − (cid:12)(cid:12)(cid:12) Z ξ ∇ × U ( x, t ) dt (cid:12)(cid:12)(cid:12) dx ≤ Re Z ξ Z Γ ( ˆ T − T )[ ∂ t E p Γ ] · Ψ dγdt = Re Z ξ Z Γ (cid:18)Z t ( ˆ T − T )[ ∂ τ E p Γ ] dτ (cid:19) U Γ ( x, t ) dγdt . (cid:18)Z ξ k ( ˆ T − T )[ ∂ t E p Γ ]( · , t ) k H − / (Div , Γ ) dt (cid:19) (cid:18)Z ξ k U ( · , t ) k H (curl , Ω ) dt (cid:19) , (4.72)17here we have used the trace theorem to get the last inequality. The right-hand of (4.72)contains the term Z ξ k U ( · , t ) k H (curl , Ω ) dt = Z ξ (cid:16) k U ( · , t ) k L (Ω ) + k∇ × U ( · , t ) k L (Ω ) (cid:17) dt which cannot be controlled by the left-hand of (4.72). To address this issue, we consider thenew problem ∇ × ( µ − ∇ × ( ∂ t U )) + ε∂ t ( ∂ t U ) = in Ω × (0 , T ) , n × ∂ t U = on Γ × (0 , T ) ,∂ t U ( x, 0) = ∂ t U ( x, 0) = in Ω ,µ − ( ∇ × ( ∂ t U )) × n + C [ ∂ t U Γ ] = ( ˆ T − T )[ ∂ t E p Γ ] on Γ × (0 , T ) , (4.73)which is obtained by differentiating each equation of (4.64) with respect to t . By a similarargument as in deriving (4.65), we obtain the variational formulation of (4.73): find u such thatfor all ω ∈ H Γ (curl , Ω ), Z Ω ε∂ t ( ∂ t U ) · ω dx = − Z Ω µ − ( ∇ × ( ∂ t U ))( ∇ × ω ) dx + Z Γ ( ˆ T − T )[ ∂ t E p Γ ] · ω Γ dγ − Z Γ C [ ∂ t U Γ ] · ω Γ dγ. (4.74)Define the auxiliary function Ψ ( x, t ) = Z ξt ∂ τ U ( x, τ ) dτ, x ∈ Ω , ≤ t ≤ ξ. Similarly as in the derivation of (4.68)-(4.69), we conclude by integration by parts thatRe Z ξ Z Ω ε∂ t ( ∂ t U ) · Ψ dxdt = 12 k√ ε∂ t U ( · , ξ ) k L (Ω ) , (4.75)Re Z ξ Z Ω R µ − e ( ∇ × ( ∂ t U )) · ( ∇ × Ψ ) dxdt = 12 (cid:13)(cid:13)(cid:13) √ µ ∇ × U ( · , ξ ) (cid:13)(cid:13)(cid:13) L (Ω ) . (4.76)Choosing the test function ω = Ψ in (4.74), integrating the resulting equation with respect to t from t = 0 to t = ξ and taking the real parts yield12 k√ ε∂ t U ( · , ξ ) k L (Ω ) + 12 (cid:13)(cid:13)(cid:13) √ µ ∇ × U ( · , ξ ) (cid:13)(cid:13)(cid:13) L (Ω ) = Re Z ξ Z Γ ( ˆ T − T )[ ∂ t E p Γ ] · Ψ dγdt − Re Z ξ Z Γ C [ ∂ t U Γ ] · Ψ dγdt. (4.77)Similarly to (4.71), it follows from (4.67) and Lemma 3.3 thatRe Z ξ Z Γ C [ ∂ t U Γ ] · Ψ dγdt ≥ . k√ ε∂ t U ( · , ξ ) k L (Ω ) + 12 (cid:13)(cid:13)(cid:13) √ µ ∇ × U ( · , ξ ) (cid:13)(cid:13)(cid:13) L (Ω ) ≤ Re Z ξ Z Γ ( ˆ T − T )[ ∂ t E p Γ ] · Ψ dγdt = Re Z ξ Z Γ (cid:18)Z t ( ˆ T − T )[ ∂ τ E p Γ ] dτ (cid:19) ∂ t U Γ ( x, t ) dγdt = Re Z ξ Z Γ ( ˆ T − T )[ ∂ t E p Γ ] · U Γ ( x, ξ ) dγdt − Re Z ξ Z Γ ( ˆ T − T )[ ∂ t E p Γ ] U Γ ( x, t ) dγdt ≤ Z ξ k ( ˆ T − T )[ ∂ t E p Γ ] k H (Div , Γ ) · (cid:0) k U ( · , ξ ) k H (curl , Ω ) + k U ( · , t ) k H (curl , Ω ) (cid:1) dt (4.78)Combining (4.72) and (4.78) gives k U ( · , ξ ) k L (Ω ) + k ∂ t U ( · , ξ ) k L (Ω ) + k∇ × U ( · , ξ ) k L (Ω ) . (cid:18)Z ξ k ( ˆ T − T )[ ∂ t E p Γ ]( · , t ) k H − / (Div , Γ ) dt (cid:19) (cid:18)Z ξ k U ( · , t ) k H (curl , Ω ) dt (cid:19) + Z ξ k ( ˆ T − T )[ ∂ t E p Γ ] k H − / (Div , Γ ) · (cid:0) k U ( · , ξ ) k H (curl , Ω ) + k U ( · , t ) k H (curl , Ω ) (cid:1) dt. (4.79)Taking the L ∞ -norm of both sides of (4.79) with respect to ξ and using the Young inequalityyield k U k L ∞ (0 ,T ; L (Ω ) ) + k ∂ t U k L ∞ (0 ,T ; L (Ω ) ) + k∇ × U k L ∞ (0 ,T ; L (Ω ) ) . T k ( ˆ T − T )[ ∂ t E p Γ ] k L (0 ,T ; H − / (Div , Γ )) + k ( ˆ T − T )[ ∂ t E p Γ ] k L (0 ,T ; H − / (Div , Γ )) , which, together with the Cauchy-Schwartz inequality, implies that k U k L ∞ (0 ,T ; L (Ω ) ) + k ∂ t U k L ∞ (0 ,T ; L (Ω ) ) + k∇ × U k L ∞ (0 ,T ; L (Ω ) ) (4.80) . T / k ( ˆ T − T )[ ∂ t E p Γ ] k L (0 ,T ; H − / (Div , Γ )) + T / k ( ˆ T − T )[ ∂ t E p Γ ] k L (0 ,T ; H − / (Div , Γ )) . We now only need to estimate the right-hand term of (4.80). By (4.57) and the definition of ˆ T (see (4.49)) we know that ( ˆ T − T )[ ∂ t E p Γ ] = n × µ − ∇ × v , where v satisfies the problem ∇ × ( µ − BA ∇ × v ) + ε ( BA ) − ∂ t v = in Ω PML × (0 , T ) , n × v = on Γ × (0 , T ) , n × v = γ t ( B e E p ) on Γ × (0 , T ) , v ( x, 0) = ∂ t v ( x, 0) = in Ω PML . (4.81)Thus we deduce that k ( ˆ T − T )[ ∂ t E p Γ ] k L (0 ,T ; H − / (Div , Γ )) = k n × µ − ∇ × v k L (0 ,T ; H − / (Div , Γ )) ≤ e s T Z ∞ e − s t k µ − ∇ × v k H (curl , Ω PML ) dt . e s T Z ∞−∞ k µ − ∇ × ˇ v k H (curl , Ω PML ) ds . k ( ˆ T − T )[ ∂ t E p Γ ] k L (0 ,T ; H − / (Div , Γ )) . e s T s − d (1 + s − σ ) e − √ εµσ dm +1 "Z ∞−∞ X l =0 k s l ˇ J k L (Ω ) ds / . e s T s − d (1 + s − σ ) e − √ εµσ dm +1 k J k H (0 ,T ; L (Ω ) ) . Similarly, we have k ( ˆ T − T )[ ∂ t E p Γ ] k L (0 ,T ; H − / (Div , Γ )) . e s T s − d (1 + s − σ ) e − √ εµσ dm +1 k J k H (0 ,T ; L (Ω ) ) . By (4.80) and the above two estimates it follows on setting s = 1 /T and m = 1 that k U k L ∞ (0 ,T ; L (Ω ) ) + k ∂ t U k L ∞ (0 ,T ; L (Ω ) ) + k∇ × U k L ∞ (0 ,T ; L (Ω ) ) . T / d (1 + σ T ) e −√ εµσ d/ k J k H (0 ,T ; L (Ω ) ) . From this, the definition of U and Maxwell’s system (4.63) the required estimate (4.53) thenfollows. The proof is thus complete. Remark 4.8. The L -norm error estimate (4.52) can also be obtained by integrating (4.79)with respect to ξ from 0 to T . The idea of using the uniform coercivity of the variational formin our proof of the L -norm error estimate (4.52) is also known for the time-harmonic PMLmethod. This builds a connection between our proposed time-domain PML method with thereal coordinate stretching technique and the time-harmonic PML method in some sense. In this paper, by using the real coordinate stretching technique we proposed a uniaxial PMLmethod in the Cartesian coordinates for 3D time-domain electromagnetic scattering problems,which is of advantage over the spherical one in dealing with scattering problems involvinganisotropic scatterers. The well-posedness and stability estimates of the truncated uniaxialPML problem in the time domain were established by employing the Laplace transform tech-nique and the energy argument. The exponential convergence of the uniaxial PML method wasalso proved in terms of the thickness and absorbing parameters of the PML layer, based on theerror estimate between the EtM operators for the original scattering problem and the truncatedPML problem established in this paper via the decay estimate of the dyadic Green’s function.Our method can be extended to other electromagnetic scattering problems such as scatteringby inhomogeneous media or bounded elastic bodies as well as scattering in a two-layered medium.It is also interesting to study the spherical and Cartesian PML methods for time-domain elasticscattering problems, which is more challenging due to the existence of shear and compressionalwaves with different wave speeds. We hope to report such results in the near future. Acknowledgments This work was partly supported by the NNSF of China grants 11771349 and 91630309. Thefirst author was also partly supported by the National Research Foundation of Korea (NRF-2020R1I1A1A01073356). 20 eferences [1] G. Bao, Y. Gao and P. Li, Time-domain analysis of an acoustic-elastic interaction problem, Arch. Rational Mech. Anal. (2018), 835-884.[2] G. Bao and H. Wu, Convergence analysis of the perfectly matched layer problems for time-harmonic Maxwell’s equations, SIAM. J. Numer. Anal. (2005), 2121-2143.[3] J.P. B´erenger, A perfectly matched layer for the absorption of electromagnetic waves, J.Comput. Phys. (1994), 185-200.[4] J.H. Bramble and J.E. Pasciak, Analysis of a finite PML approximation for the threedimensional time-harmonic Maxwell and acoustic scattering problems, Math. Comp. (2007), 597-614.[5] J.H. Bramble and J.E. Pasciak, Analysis of a finite element PML approximation for thethree dimensional time-harmonic Maxwell problem, Math. Comput. (2008), 1-10.[6] J.H. Bramble and J.E. Pasciak, Analysis of a Cartesian PML approximation to the threedimensional electromagnetic wave scattering problem, Int. J. Numer. Anal. Model. (2012),543-561.[7] J.H. Bramble and J.E. Pasciak, Analysis of a Cartesian PML approximation to acousticscattering problems in R and R , J. Comput. Appl. Math. (2013), 209-230.[8] J.H. Bramble, J.E. Pasciak and D. Trenev, Analysis of a finite PML approximation to thethree dimensional elastic wave scattering problem, Math. Comput. (2010), 2079-2101.[9] A. Buffa, M. Costabel and D. Sheen, On traces for H (curl , Ω) in Lipschitz domains, J.Math. Anal. Appl. (2002), 845-867.[10] W.C. Chew and W.H. Weedon, A 3D perfectly matched medium from modified Maxwell’sequations with stretched coordinates, Microw. Opt. Technol. Lett. (1994), 599-604.[11] Q. Chen and P. Monk, Discretization of the time domain CFIE for acoustic scatteringproblems using convolution quadrature, SIAM J. Math. Anal. (2014), 3107-3130.[12] Z. Chen, Convergence of the time-domain perfectly matched layer method for acousticscattering problems, Int. J. Numer. Anal. Model. (2009), 124-146.[13] J. Chen and Z. Chen, An adaptive perfectly matched layer technique for 3-D time-harmonicelectromagnetic scattering problems, Math. Comput. (2007), 673-698.[14] Z. Chen, T. Cui and L. Zhang, An adaptive anisotropic perfectly matched layer methodfor 3-D time harmonic electromagnetic scattering problems, Numer. Math. (2013),639-677.[15] Z. Chen and X. Liu, An adaptive perfectly matched layer technique for time-harmonicscattering problems, SIAM J. Numer. Anal. (2005), 645-671.[16] Z. Chen and J.C. N´ed´elec, On Maxwell equations with the transparent boundary condition, J. Comput. Math. (2008), 284-296. 2117] Z. Chen and X. Wu, An adaptive uniaxial perfectly matched layer method for time-harmonicscattering problems, Numer. Math. TMA (2008), 113-137.[18] Z. Chen and X. Wu, Long-time stability and convergence of the uniaxial perfectly matchedlayer method for time-domain acoustic scattering problems, SIAM J. Numer. Anal. (2012), 2632-2655.[19] Z. Chen, X. Xiang and X. Zhang, Convergence of the PML method for elastic wave scat-tering problems, Math. Comp. (2016), 2687-2714.[20] Z. Chen and W. Zheng, Convergence of the uniaxial perfectly matched layer method fortime-harmonic scattering problems in two-layered media, SIAM J. Numer. Anal. (2010),2158-2185.[21] Z. Chen and W. Zheng, PML method for electromagnetic scattering problem in a two-layermedium, SIAM. J. Numer. Anal. (2017), 2050-2084.[22] A.M. Cohen, Numerical Methods for Laplace Transform Inversion , Springer, 2007.[23] F. Collino and P. Monk, The perfectly matched layer in curvilinear coordinates, SIAM J.Sci. Comput. (1998), 2061-2090.[24] A.T. DeHoop, P.M. van den Berg, and R.F. Remis, Absorbing boundary conditions and per-fectly matched layers–An analytic time-domain performance analysis, IEEE Trans. Magn. (2002), 657-660.[25] J. Diaz and P. Joly, A time domain analysis of PML models in acoustics, Comput. MethodsAppl. Mech. Engrg. (2006), 3820-3853.[26] Y. Gao and P. Li, Analysis of time-domain scattering by periodic structures, J. DifferentialEquations (2016), 5094-5118.[27] Y. Gao and P. Li, Electromagnetic scattering for time-domain Maxwell’s equations in anunbounded structure, Math. Models Methods Appl. Sci. (2017), 1843-1870.[28] Y. Gao, P. Li, and B. Zhang, Analysis of transient acoustic-elastic interaction in an un-bounded structure, SIAM J. Math. Anal. (2017), 3951-3972.[29] T. Hagstrom, Radiation boundary conditions for the numerical simulation of waves, ActaNumer. (1999), 47-106.[30] T. Hohage, F. Schmidt and L. Zschiedrich, Solving time-harmonic scattering problems basedon the pole condition II: Convergence of the PML method, SIAM J. Math. Anal. (2003),547-560.[31] G.C. Hsiao, F.J. Sayas and R.J. Weinacht, Time-dependent fluid-structure interaction, Math. Method. Appl. Sci. (2017), 486-500.[32] M. Lassas and E. Somersalo, On the existence and convergence of the solution of PMLequations, Computing (1998), 229-241.2233] P. Li, L. Wang and A. Wood, Analysis of transient electromagnetic scattering from a three-dimensional open cavity, SIAM J. Appl. Math. (2015), 1675-1699.[34] P. Monk, Finite Element Methods for Maxwell’s Equations , Oxford Univ. Press, New York,2003.[35] F.L. Teixeira and W.C. Chew, Advances in the theory of perfectly matched layers, in: Fast and Efficient Algorithms in Computational Electromagnetics (ed. W. C. Chew et al.),Artech House, Boston, 2001, pp. 283-346.[36] F. Tr`eves, Basic Linear Partial Differential Equations , Academic Press, New York, 1975.[37] L. Wang, B. Wang and X. Zhao, Fast and accurate computation of time-domain acousticscattering problems with exact nonreflecting boundary conditions, SIAM J. Appl. Math. (2012), 1869-1898.[38] G.N. Watson, A Treatise on The Theory of Bessel Functions, Cambridge University Press,Cambridge, UK, 1922.[39] C. Wei and J. Yang, Analysis of a time-dependent fluid-solid interaction problem above alocal rough surface, Sci. China Math. (2020), 887-906.[40] C. Wei, J. Yang and B. Zhang, A time-dependent interaction problem between an electro-magnetic field and an elastic body, Acta Math. Appl. Sin. Engl. Ser. (2020), 95-118.[41] C. Wei, J. Yang and B. Zhang, Convergence of the perfectly matched layer method fortransient acoustic-elastic interaction above an unbounded rough surface, arXiv:1907.09703,2019.[42] C. Wei, J. Yang and B. Zhang, Convergence analysis of the PML method for time-domainelectromagnetic scattering problems, SIAM J. Numer. Anal.58