A pure source transfer domain decomposition method for Helmholtz equations in unbounded domain
AAN IMPROVED PURE SOURCE TRANSFER DOMAINDECOMPOSITION METHOD FOR HELMHOLTZ EQUATIONS INUNBOUNDED DOMAIN
YU DU ∗ AND
HAIJUN WU ∗ Abstract.
We propose an improved pure source transfer domain decomposition method (PST-DDM) for solving the truncated perfectly matched layer (PML) approximation in bounded domainof Helmholtz scattering problem. The method is based on the the source transfer domain decompo-sition method (STDDM) proposed by Chen and Xiang. The idea is decomposing the domain intonon-overlapping layers and transferring the sources equivalently layer by layer so that the solutionin each layer can be solved using a PML method defined locally outside its two adjacent layers.Furthermore, we divide the domain into non-overlapping blocks and solve the solution in each blockby using a PML method defined locally outside its adjacent blocks. The convergence analysis of themethod is provided for the case of constant wave number. Finally, numerical examples are providedwhere the method is used as both a direct solver and an efficient preconditioner in the GMRESmethod for solving the Helmholtz equation.
Key words.
Helmholtz equation, large wave number, perfectly matched layer, source transfer,domain decomposition method
AMS subject classifications.
1. Introduction.
This paper is devoted to a domain decomposition methodbased on the STDDM method (cf. [15]) for the Helmholtz problem in the full space R with Sommerfeld radiation condition:∆ u + k u = f in R , (1.1) (cid:12)(cid:12)(cid:12)(cid:12) ∂u∂r − i ku (cid:12)(cid:12)(cid:12)(cid:12) = o ( r − / ) as r = | x | → ∞ . (1.2)where the wave number k is positive and f ∈ H ( R ) (cid:48) having compact support, where H ( R ) (cid:48) is the dual space of H ( R ). The problem is satisfied in a weak sense (cf.[33]).Helmholtz boundary value problems appear in various applications, for example,in the context of inverse and scattering problems. Since the huge number of degrees offreedom is required resulting from the pollution error and the highly indefinite natureof Helmholtz problem with large number wave [1, 2, 3, 10, 18, 21, 22, 24, 25, 26, 27,28, 34], it is challenging to solve the algebraic linear equations resulting from the finitedifference or finite element method with large wave number. Considerable efforts inthe literature have been made. One way is to find efficient and cheap methods [2, 10,17, 21, 22, 24], such as the coutinous interior penalty finite element method [9, 35, 36], which use less degrees of freedom as the same relative error reached. Another wayis to find efficient algorithms for solving discrete Helmholtz equations, e.g. Benamouand Despr´es [4], Gander et al [23] for domain decomposition techniques and Brandtand Livshit [8], Elman et at [19] for multigrid methods. Recently Engquist and Yingconstructed a new sweeping preconditioner for the interior solution [20]. Then Chenand Xiang proposed the source transfer domain decomposition method (STDDM) ∗ Department of Mathematics, Nanjing University, Jiangsu, 210093, P.R. China.( [email protected] , [email protected] ). This research was partially supported by the National MagneticConfinement Fusion Science Program under grant 2011GB105003 and by the NSF of China undergrants 11071116 and 91130004. 1 a r X i v : . [ m a t h . NA ] D ec Y. Du & H. Wu [15], in which only some local PML problems defined locally outside the union of twolayers are solved. Thus the complexity of STDDM is the sum of the complexity ofthe algorithms for solving those local problems which reduce the complexity to solvethe whole linear system. We are inspired by the key idea of STDDM, and the mainlemmas and idea of proofs also come from their work.In this paper we show the improved pure source transfer domain decompositionmethod (PSTDDM).Let B l = { x = ( x , x ) T ∈ R : | x | < l , | x | < l } . Assume that f is supportedin B l . We divide the interval ( − l , l ) into N segments with the points ζ i = − l +( i − ζ where ∆ ζ = 2 l /N . Then we denote the layers byΩ = { x = ( x , x ) T ∈ R : x < ζ } , Ω i = { x = ( x , x ) T ∈ R : ζ i < x < ζ i +1 } , i = 1 , · · · , N, Ω N +1 = { x = ( x , x ) T ∈ R : ζ N +1 < x } . Clearly, supp f ⊂ ∪ Ni =1 Ω i . Let f i = f in Ω i and f i = 0 elsewhere. Let ¯ f +1 = f and¯ f − N = f N . The key idea is that by defining the source transfer function Ψ ± i in thesense that (cid:90) Ω i ¯ f + i ( y ) G ( x, y ) dy = (cid:90) Ω i +1 Ψ + i +1 ( ¯ f + i )( y ) G ( x, y ) dy ∀ x ∈ ∪ N +1 j = i +2 Ω j , (cid:90) Ω i ¯ f − i ( y ) G ( x, y ) dy = (cid:90) Ω i − Ψ − i − ( ¯ f − i )( y ) G ( x, y ) dy ∀ x ∈ ∪ i − j =0 Ω j , and letting ¯ f ± i ± = f i ± + Ψ ± i ± ( ¯ f i ), then we have for any x ∈ Ω i u ( x ) = (cid:18) − (cid:90) Ω i f i ( y ) G ( x, y ) dy − (cid:90) Ω i − ¯ f + i − ( y ) G ( x, y ) dy (cid:19) +(1.3) (cid:18) − (cid:90) Ω i +1 ¯ f − i +1 ( y ) G ( x, y ) dy (cid:19) . Here, G ( x, y ) is the Green’s function of the problem (1.1)–(1.2). Observing (2.1), weknow that u ( x ) in Ω i consists of two independent parts. The first part only involvesthe sources in Ω i and Ω i − and the second one only involves the source in Ω i +1 . Thusthey could be solved independently by using the PML method outside only Ω i − ∪ Ω i and Ω i ∪ Ω i +1 respectively. Similar to STDDM, our PSTDDM also consists of twosteps which could run in parallel and the complexity of every step is the same asthat of STDDM. The error of PSTDDM are not larger than that of STDDM if thesame numerical algorithm, such as the finite element or difference method, was used.Besides, since every step of PSTDDM just consists of local PML problems definedin the whole space, not in the half-space with Dirichlet boundary, these local PMLproblems also can be solved by using our PSTDDM recursively. As a result, thecomputational domain is divided into blocks and the local PML problems needed tobe solved are defined outside the union of four adjacent blocks.The perfectly matched layer (PML) is a mesh termination technique of effective-ness, simplicity and flexibility in computational wave propagation. After the pioneer-ing work of B´erenger [5, 6], various constructions of PML absorbing layers have beenproposed and many theoretical results about Helmholtz problem, such as those aboutthe convergence and stability, have been studied [7, 13, 14, 16, 29, 30, 31]. In thispaper, the uniaxial PML methods will be used. ure source transfer domain decomposition method R and some important lemmas and theo-rems, which are fundamental and illuminating for the PSTDDM in truncated domain,are introduced. Section 2.3 shows the algorithm in the truncated bounded domain andthe main result, that is, the exponentially convergence of the solution of PSTDDMin the truncated domain to the solution in R . In section 3, we show that the localPML problems can be solved by out PSTDDM recursively and proof the convergenceof the method. At last, we both use the method as a direct solver and an efficientpreconditioner in the GMRES method to solve the Helmholtz equation.
2. Source transfer layer by layer.
In this section, we introduce the PSTDDMfor the PML method in the whole space. First, we recall the progress of deriving thePML method and set the medium properties of perfect matched lays which are abit different from traditional medium and would be used in the following lemmas andtheorems [15]. In subsection 2.1, we also recall some basic lemmas. Then we introducethe two steps of the pore source transfer domain decomposition method in R . In this subsection, we recall some knowledge aboutthe PML method.The exact solution of equation(1.1) with the radiation condition 1.2 can be writtenas the acoustic volume potential. Let G ( x, y ) be the fundamental solution of theHelmholtz problem ∆ G ( x, y ) + k G ( x, y ) = − δ y ( x ) in R . We know G ( x, y ) = i H (1)0 ( k | x − y | ) where H (1)0 ( z ), for z ∈ C , is the first kind Hankelfunction of order zero.Then, the solution of (1.1) is given by u ( x ) = − (cid:90) R f ( y ) G ( x, y ) dy ∀ x ∈ R . (2.1)In this paper we used the uniaxial PML method [7, 15, 12, 29] . the model mediumproperties are defined by α ( x ) = 1 + i σ ( x ) , α ( x ) = 1 + i σ ( x ) σ j ( t ) = σ j ( − t ) for t ∈ R , σ j = 0 for | t | ≤ l j , σ j = γ > | t | ≥ ¯ l j . where σ j ( x j ) ∈ C ( R ) are piecewise smooth functions, ¯ l j > l j is fixed and γ is aconstant.For x = ( x , x ) T , we define the complex coordinate as ˜ x ( x ) = (˜ x ( x ) , ˜ x ( x )),where ˜ x j ( x j ) = (cid:90) x j α j ( t ) dt = x j + i (cid:90) x j σ j ( t ) dt, j = 1 , . (2.2)We remark that this kind of definition has been proposed in [15] and recall that therequirement, σ j = γ for | t | ≥ ¯ l j , is very important because of the use of proving thelocal inf-sup condition (2.33) (cf. [15]) for the truncated PML problem by using thereflection argument of [7, 15] and estimating the dependence of the inf-sup constantson the wave number k . Y. Du & H. Wu
The complex distance is defined as ρ (˜ x, ˜ y ) = (cid:2) (˜ x ( x ) − ˜ y ( y )) + (˜ x ( x ) − ˜ y ( y )) (cid:3) / . (2.3)Here, z / denote the analytic branch of √ z such that Re( z / ) > z ∈ C \ [0 , + ∞ ).The solution to the PML problem is˜ u ( x ) = u (˜ x ) = − (cid:90) R f ( y ) G (˜ x, ˜ y ) dy ∀ x ∈ R , (2.4)Since f is supported inside B l , we know that ˜ y = y and ˜ u = u in B l .The solution (2.4) satisfies the PML equation J − ∇ · ( A ∇ ˜ u ) + k ˜ u = f in R , (2.5)which could be obtained by the fact that ˜∆˜ u + k ˜ u = f in R and using the chainrule, where A ( x ) = diag (cid:16) α ( x ) α ( x ) , α ( x ) α ( x ) (cid:17) and J ( x ) = α ( x ) α ( x ).The weak formulation of (2.5) is: Find ˜ u ∈ H ( R ) such that( A ∇ ˜ u, ∇ v ) − k ( J ˜ u, v ) = − (cid:104) Jf, v (cid:105) ∀ v ∈ H ( R ) , (2.6)where ( · , · ) is the inner product in L ( R ) and (cid:104)· , ·(cid:105) is the duality pairing between H ( R ) (cid:48) and H ( R ).We have the following inf-sup condition for the sesquilinear form associated withthe PML problem in R which has been proved (cf. [15], Lemma 3.3):sup ψ ∈ H ( R ) (cid:12)(cid:12) ( A ∇ φ, ∇ ψ ) − k ( Jφ, ψ ) (cid:12)(cid:12) (cid:107) ψ (cid:107) H ( R ) ≥ µ (cid:107) φ (cid:107) H ( R ) ∀ φ ∈ H ( R ) , (2.7)where the inf-sup condition µ − ≤ Ck / which is fundamental to our estimates.The fundamental solution of the PML equation (2.5) is (cf. [7, 31])˜ G ( x, y ) = J ( y ) G (˜ x, ˜ y ) = i J ( y ) H (1)0 ( kρ (˜ x, ˜ y )) . (2.8) R . In this subsection, weintroduce our PSTDDM for the PML equation in the whole space and give the fun-damental theorems.We recall that the domain { x = ( x , x ) : | x | ≤ l } is divided into N layers ( ?? )and that f i ( x ) = f ( x ) | Ω i for any x ∈ Ω i and f i ( x ) = 0 for any x ∈ R \ ¯Ω i . We definesmooth functions β + i ( x ) and β − i ( x ) by β + i = 1 , β − i = 0 , β + i (cid:48) = β − i (cid:48) = 0 as x ≤ ζ i , (2.9) β + i = 0 , β − i = 1 , β + i (cid:48) = β − i (cid:48) = 0 as x ≥ ζ i +1 , (cid:12)(cid:12)(cid:12) β + i (cid:48) (cid:12)(cid:12)(cid:12) ≤ C ( ∇ ζ ) − , (cid:12)(cid:12)(cid:12) β − i (cid:48) (cid:12)(cid:12)(cid:12) ≤ C ( ∇ ζ ) − , where C is a constant independent of ζ i , ζ i +1 and the subscript i . Our PSTDDMconsists of two steps 1 and 2. Clearly, the two steps 1 and 2 are independent of eachother and can be computed in parallel. ure source transfer domain decomposition method Algorithm 1
Source Transfer I for PML problem in R
1. Let ¯ f +1 = f ;2. While i = 1 , · · · , N − • Find u + i ∈ H ( R ) such that J − ∇ · ( A ∇ u + i ) + k u + i = − ¯ f + i − f i +1 in R (2.10) • Compute Ψ + i +1 ( ¯ f + i ) = J − ∇ · ( A ∇ ( β + i +1 u + i )) + k ( β + i +1 u + i ) . (2.11) • Set ¯ f + i +1 = f i +1 + Ψ + i +1 ( ¯ f + i )(2.12)in Ω i +1 and ¯ f + i +1 = 0 elsewhere.End while3. For i = N −
1, find u + N − ∈ H ( R ) such that J − ∇ · ( A ∇ u + N − ) + k u + N − = − ¯ f + N − − f N in R (2.13) Algorithm 2
Source Transfer II for PML problem in R
1. Let ¯ f − N = f N ;2. While i = N, · · · , • Find u − i ∈ H ( R ) such that J − ∇ · ( A ∇ u − i ) + k u − i = − ¯ f − i in R (2.14) • Compute Ψ − i − ( ¯ f − i ) = J − ∇ · ( A ∇ ( β − i − u − i )) + k ( β − i − u − i ) . • Set ¯ f − i − = f i − + Ψ − i − ( ¯ f − i ) in Ω i − and ¯ f + i − = 0 elsewhere.End while3. For i=2, find u − ∈ H ( R ) such that J − ∇ · ( A ∇ u − ) + k u − = − f − ¯ f − in R . (2.15)By (2.10), (2.14) and (2.15), we know that u + i is given by u + i ( x ) = (cid:90) Ω i ∪ Ω i +1 ( ¯ f + i + f i +1 ) J ( y ) G (˜ x, ˜ y ) dy ∀ x ∈ R , i = 1 , · · · , N − , (2.16)and u − i is given by u − i ( x ) = (cid:90) Ω i ¯ f − i ( y ) J ( y ) G (˜ x, ˜ y ) dy ∀ x ∈ R , i = N, · · · , , (2.17) u − ( x ) = (cid:90) Ω ∪ Ω ( ¯ f − ( y ) + f ( y )) J ( y ) G (˜ x, ˜ y ) dy. (2.18) Y. Du & H. Wu
By simple calculation, we have the equivalent form of the source transfer operatorΨ + i +1 : Ψ + i +1 ( ¯ f + i ) = J − ∇ (cid:0) A ∇ β + i +1 u + i (cid:1) + J − ∇ β + i +1 · ( A ∇ u + i ) − β + i +1 f i +1 . (2.19)and it’s easily obtained that Ψ + i +1 ( ¯ f + i ) + β + i +1 f i +1 is in L (Ω i +1 ) and supported inΩ i +1 . Similarly, we can get the equivalent form for Ψ − i − :Ψ − i − ( ¯ f − i ) = J − ∇ (cid:0) A ∇ β − i − u − i (cid:1) + J − ∇ β − i − · ( A ∇ u − i ) . (2.20)The proof of the following two lemmas is quit similar to Lemma 2.6 in [15]. Weomit the details. Lemma 2.1.
For i = 1 , · · · , N − , we have u + i ∈ H ( R ) and (cid:13)(cid:13) u + i (cid:13)(cid:13) H ( R ) ≤ C (cid:107) f (cid:107) H ( B l ) (cid:48) . Let M = l , M i = √ M + (1 + √ M i − , where l is the diameter of B l and M = max(¯ l , ¯ l ) . Then there exists a constant C > such that (cid:12)(cid:12) u + i ( x ) (cid:12)(cid:12) + (cid:12)(cid:12) ∆ u + i ( x ) (cid:12)(cid:12) ≤ Ce − kγ | x | (cid:107) f (cid:107) H ( B l ) (cid:48) ∀ | x | ≥ M i , where H ( B l ) (cid:48) is the dual space of H ( B l ) . Lemma 2.2.
For i = N, · · · , , we have u − i ∈ H ( R ) and (cid:13)(cid:13) u − i (cid:13)(cid:13) H ( R ) ≤ C (cid:107) f (cid:107) H ( B l ) (cid:48) . Let M = l , M i = √ M + (1 + √ M i − , where l is the diameterof B l and M = max(¯ l , ¯ l ) . Then there exists a constant C > such that (cid:12)(cid:12) u − i ( x ) (cid:12)(cid:12) + (cid:12)(cid:12) ∆ u − i ( x ) (cid:12)(cid:12) ≤ Ce − kγ | x | (cid:107) f (cid:107) H ( B l ) (cid:48) ∀ | x | ≥ M i , where H ( B l ) (cid:48) is the dual space of H ( B l ) . Lemma 2.1 and Lemma 2.2 show that u + i and u − i decay exponentially at infinity,which will be used in the following theorems. Theorem 2.3.
The following assertions hold: (i)
For i = 1 , · · · , N − , we have, for any x ∈ Ω( ζ i +2 , + ∞ ) , (cid:90) Ω i ¯ f + i ( y ) ˜ G ( x, y ) dy = (cid:90) Ω i +1 Ψ + i +1 ( ¯ f + i )( y ) ˜ G ( x, y ) dy. (2.21)(ii) For the solution u + i in (2.10) , we have, for any x ∈ Ω i +1 , i = 1 , · · · , N − , u + i ( x ) = (cid:90) Ω( −∞ ,ζ i +2 ) f ( y ) G (˜ x, ˜ y ) dy. (2.22) Proof . We first prove (2.21). By the property of (2.8) (cf. e.g. [[15], 2.11-2.13],[[7], Theorem 2.8] and [[31], Theorem 4.1]), we know that for any x ∈ Ω( ζ i +2 , + ∞ )and y ∈ Ω( ζ , ζ i +2 ) ∇ y · ( A ∇ y ( J − ˜ G ( x, y ))) + k J ( J − ˜ G ( x, y )) = 0 . For x ∈ Ω( ζ i +2 , + ∞ ), y ∈ Ω j , j = 1 , · · · , i + 1, ˜ G ( x, y ) decays exponentially as | y | → u + i ( y ) decays also exponentiallyat infinity. By integrating by parts, we have (cid:90) Ω i ¯ f + i ˜ G ( x, y ) dy = − (cid:90) Ω( −∞ ,ζ i +1 ) J − [ ∇ y · ( A ∇ y u + i ( y )) + k Ju + i ( y )] ˜ G ( x, y ) dy = − (cid:90) Γ i +1 (cid:104) ( A ∇ y u + i ( y ) · e ) J − ˜ G ( x, y ) − ( A ∇ y ( J − ˜ G ( x, y )) · e ) u + i ( y ) (cid:105) ds ( y ) , ure source transfer domain decomposition method e is the unit vector in the x axis. By using (2.9), we can do integration byparts to have (cid:90) Ω i ¯ f + i ˜ G ( x, y ) dy = (cid:90) ∂ Ω i +1 (cid:2) ( A ∇ y ( β + i +1 u + i ( y )) · n ) J − ˜ G ( x, y ) − ( A ∇ y ( J − ˜ G ( x, y )) · n ) β + i +1 u + i ( y ) (cid:3) ds ( y )= (cid:90) Ω i +1 J − [ ∇ y · ( A ∇ y ( β + i +1 u + i ( y ))) + k Jβ + i +1 u + i ( y )] ˜ G ( x, y ) dy = (cid:90) Ω i +1 Ψ + i +1 ( ¯ f + i )( y ) ˜ G ( x, y ) dy, where n is the unit outer normal to ∂ Ω i +1 .Since ˜ y ( y ) = y and J ( y ) = 1 for any y ∈ B l , By using (2.16) and (2.21) we couldprove (2.22). For any x ∈ Ω i +1 u + i ( x ) = (cid:90) Ω i ∪ Ω i +1 ( ¯ f + i + f i +1 ) J ( y ) G (˜ x, ˜ y ) dy = (cid:90) Ω i +1 f i +1 ( y ) G (˜ x, ˜ y ) dy + (cid:90) Ω i f i ( y ) G (˜ x, ˜ y ) dy + (cid:90) Ω i Ψ + i ( ¯ f + i − )( y ) J ( y ) G (˜ x, ˜ y ) dy = (cid:90) Ω i ∪ Ω i +1 f ( y ) G (˜ x, ˜ y ) dy + (cid:90) Ω i − ¯ f + i − ( y ) J ( y ) G (˜ x, ˜ y ) dy = · · · = (cid:90) ∪ i +1 j =1 Ω j f ( y ) G (˜ x, ˜ y ) dy. This completes the proof.The second step 2 is similar to the first one 1 of the PSTDDM for the PMLequation in the whole space. So by argument similar to the proof above, we can easilyobtain the following results.
Theorem 2.4.
The following assertions hold: (i)
For i = N, · · · , , we have, for any x ∈ Ω( −∞ , ζ i − ) , (cid:90) Ω i ¯ f − i ( y ) ˜ G ( x, y ) dy = (cid:90) Ω i − Ψ − i − ( ¯ f − i )( y ) ˜ G ( x, y ) dy. (2.23)(ii) For the solution u − i , i = N, · · · , , in (2.14) , we have, for any x ∈ Ω i − , u − i ( x ) = (cid:90) Ω( ζ i , + ∞ ) f ( y ) G (˜ x, ˜ y ) dy. (2.24)(iii) For the solution u − ( x ) in (2.15) , we have, for any x ∈ Ω , u − ( x ) = (cid:90) R f ( y ) G (˜ x, ˜ y ) dy. (2.25)Combining Theorem 2.3 and Theorem 2.4, we could obtain the main result in thissection. Theorem 2.5.
We define u +0 ( x ) = 0 and u − N +1 ( x ) = 0 for any x ∈ R . For any x ∈ Ω i , i = 1 , · · · , N , we have ˜ u ( x ) = − ( u + i − ( x ) + u − i +1 ( x )) . Y. Du & H. Wu
Proof . From (2.25), it’s easy to see that the lemma holds for i = 1. Using thedefinition of ˜ u ( x ) (2.4) and (2.22), (2.24), we have, for any x ∈ Ω i , i = 2 , · · · , N ,˜ u ( x ) = − (cid:90) R f ( y ) G (˜ x, ˜ y ) dy = − (cid:32)(cid:90) Ω( −∞ ,ζ i +1 ) f ( y ) G (˜ x, ˜ y ) dy + (cid:90) Ω( ζ i +1 , + ∞ ) f ( y ) G (˜ x, ˜ y ) dy (cid:33) = − ( u + i − ( x ) + u − i +1 ( x )) , where we have used ˜ y ( y ) = y in B l . The PSTDDM for PML equation in the truncated bounded domain and themost important results in this paper are introduced in this section. First we introducesome notation. Let U be a bounded domain in R and ∂U = Γ. Then the weightednorms are written as (cid:107) u (cid:107) H ( U ) = (cid:16) (cid:107)∇ u (cid:107) L ( U ) + (cid:107) ku (cid:107) L ( U ) (cid:17) / , (cid:107) v (cid:107) H / (Γ) = (cid:16) d − U (cid:107) v (cid:107) L (Γ) + | v | , Γ (cid:17) / , where d U = diam(U) and | v | , Γ = (cid:90) Γ (cid:90) Γ | v ( x ) − v ( x (cid:48) ) | | x − x (cid:48) | ds ( x ) ds ( x (cid:48) ) . The following inequality are given (cf. [[15], 3.1]), (cid:107) v (cid:107) H / (Γ) ≤ ( | Γ | d − U ) / (cid:107) v (cid:107) L ∞ (Γ) + | Γ | (cid:107)∇ v (cid:107) L ∞ (Γ) ∀ v ∈ W , ∞ (Γ) , , (2.26)The inequality (2.26) is easily derived from the definition of weighted norms.For simplicity, the following assumption about the medium property is adopted: H1 (cid:90) l + d l σ ( t ) dt = (cid:90) l + d l σ ( t ) dt =: ¯ σ, (cid:90) l + d l + d σ ( t ) dt ≥ ¯ σ (2.27) and l ≤ l , d = 2 d . This assumption is not essential. Those lemmas and theorems are also valid witha bit modification of the proof if the assumption is changed.Denote by B L = ( − l − d , l + d ) × ( − l − d , l + d ) where l + d > ¯ l and l + d > ¯ l . Clearly, B L contains B l . We will show how to get a approximation ofthe solution to the PML equation in the truncated domain B L .We introduce local PML problems by using the PML complex coordinate stretch-ing outside the domain ( − l , l ) × ( ζ i , ζ i +2 ). The PML stretching is ˜ x i ( x ) = (˜ x i, ( x ) , ˜ x i, ( x )) T , which has been proposed in [15], where ˜ x i, ( x ) = ˜ x ( x ) and(2.28) ˜ x i, ( x ) = x + i (cid:82) x ζ i +2 σ ( t + ζ N +1 − ζ i +2 ) dt if x > ζ i +2 ,x if ζ i ≤ x ≤ ζ i +2 ,x + i (cid:82) x ζ i σ ( t − ζ i + ζ ) dt if x < ζ i . We define A i ( x ) = diag (cid:18) ˜ x i, ( x ) (cid:48) ˜ x i, ( x ) (cid:48) , ˜ x i, ( x ) (cid:48) ˜ x i, ( x ) (cid:48) (cid:19) , J i ( x ) = ˜ x i, ( x ) (cid:48) ˜ x i, ( x ) (cid:48) . ure source transfer domain decomposition method PML i = { x = ( x , x ) ∈ B L : ζ i − d ≤ x ≤ ζ i +2 + d } . The localPML problems in truncated domains can be defined for some wave source F ∈ H (Ω PML i ) (cid:48) as: find φ ∈ H (Ω PML i ) such that( A i ∇ φ, ∇ ψ ) − k ( J i φ, ψ ) = − (cid:104) JF, ψ (cid:105) ∀ ψ ∈ H (Ω PML i ) . (2.29)Then our PSTDDM for the PML equation in a truncated bounded domain consist oftwo steps 3 and 4.Before proving the convergence of the method, we introduce some functions and animportant result which would be used often. The functions are ¯ u + i , i = 1 , · · · , N − u − i , i = N, · · · , u + i ( x ) = (cid:90) Ω i ∪ Ω i +1 ( ¯ f + i ( y ) + f i +1 ( y )) J i ( y ) G (˜ x i , ˜ y i ) dy, i = 1 , · · · , N − , (2.30) ¯ u − i ( x ) = (cid:90) Ω i ¯ f − i ( y ) J i − ( y ) G (˜ x i − , ˜ y i − ) dy, i = N, N − , · · · , , (2.31) ¯ u − ( x ) = (cid:90) Ω ∪ Ω ( ¯ f − + f ) J ( y ) G (˜ x , ˜ y ) dy. (2.32)The result is about the inf − sup condition of the PML equations in truncated do-mains. Theorem 2.6.
Let σ d be sufficiently large. There’s some constant α < suchthat sup ψ ∈ H (Ω PML i ) ( A i ∇ φ, ∇ ψ ) − k ( J i φ, ψ ) (cid:107) ψ (cid:107) H (Ω PML i ) ≥ µ (cid:107) φ (cid:107) H (Ω PML i ) ∀ φ ∈ H (Ω PML i ) , (2.33) where µ − ≤ Ck α . C is independent of k . We remark that the recent work (cf. [[15], 3.16]) of Chen and Xiang shows thatthe inequality in the theorem above holds for α = 1 / k − (cf. [32, 11]) for the Helmholtz problem (1.1)with Sommerfeld radiation condition (1.2) or Robin boundary condition.The proof of the following lemma is omitted. The reader can complete it easilyby arguments similar to Lemma 3.5–Lemma 3.7 in [15]. Lemma 2.7.
Let σ d ≥ be sufficiently large. we have (i) For i = 1 , · · · , N − , (cid:13)(cid:13) ¯ u + i (cid:13)(cid:13) H / ( ∂ Ω PML i ) ≤ Ck (1 + kL ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) . (ii) For i = N, N − , · · · , , (cid:13)(cid:13) ¯ u − i (cid:13)(cid:13) H / ( ∂ Ω PML i − ) ≤ Ck (1 + kL ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) . Theorem 2.8.
Let σ d ≥ be sufficiently large. we have (i) For i = 2 , · · · , N − , (cid:13)(cid:13)(cid:13) ¯ f + i − ˆ f + i (cid:13)(cid:13)(cid:13) H − (Ω PML i ) ≤ Ck α ( i − k (1 + kL ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) . (ii) For i = N − , · · · , , (cid:13)(cid:13)(cid:13) ¯ f − i − ˆ f − i (cid:13)(cid:13)(cid:13) H − (Ω PML i − ) ≤ Ck α ( N − i ) k (1 + kL ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) . Y. Du & H. Wu
Algorithm 3
Source Transfer I for Truncated PML problem1. Let ˆ f +1 = f ;2. While i = 1 , · · · , N −
2, do • Find ˆ u + i ∈ H (Ω PML i ), where Ω PML i = ( − l − d , l + d ) × ( ζ i − d , ζ i +2 + d ), such that( A i ∇ ˆ u + i , ∇ ψ ) − k ( J i ˆ u + i , ψ ) = (cid:68) J i ( ˆ f + i + f i +1 ) , ψ (cid:69) ∀ ψ ∈ H (Ω PML i ) , (2.34) • Compute ˆΨ + i +1 ( ˆ f + i ) ∈ H − (Ω PML i ) such thatˆΨ + i +1 ( ˆ f + i ) = J − i ∇ ( A i ∇ ( β + i +1 ˆ u + i )) + k ( β + i +1 ˆ u + i ) . • Set ˆ f + i +1 = f i +1 + ˆΨ + i +1 ( ˆ f + i ) in Ω i +1 ∩ B L and ˆ f + i +1 = 0 elsewhere.End while3. For i = N −
1, find ˆ u + N − ∈ H (Ω PML N − ) where Ω PML N − = ( − l − d , l + d ) × ( ζ N − − d , ζ N +1 + d ), such that ∀ ψ ∈ H (Ω PML N − )( A N − ∇ ˆ u + N − , ∇ ψ ) − k ( J N − ˆ u + N − , ψ ) = (cid:68) J i ( ˆ f + N − + f N ) , ψ (cid:69) . (2.35) Algorithm 4
Source Transfer II for Truncated PML problem1. Let ˆ f − N = f N ;2. While i = N, · · · , • Find ˆ u − i ∈ H (Ω PML i − ) such that( A i − ∇ ˆ u − i , ∇ ψ ) − k ( J i − ˆ u − i , ψ ) = (cid:68) J i − ˆ f − i , ψ (cid:69) ∀ ψ ∈ H (Ω PML i − ) , (2.36) • Compute ˆΨ − i − ( ˆ f − i ) ∈ H − (Ω PML i − ) such thatˆΨ − i − ( ˆ f − i ) = J − i − ∇ ( A i − ∇ ( β − i − ˆ u − i )) + k ( β − i − ˆ u − i ) . • Set ˆ f − i − = f i − + Ψ − i − ( ˆ f − i ) in Ω i − and ˆ f − i − = 0 elsewhere.End while3. For i=2, find ˆ u − ∈ H (Ω PML1 ) such that ∀ ψ ∈ H (Ω PML1 )( A ∇ ˆ u − , ∇ ψ ) − k ( J ˆ u − , ψ ) = (cid:68) J ( ˆ f − + f ) , ψ (cid:69) . (2.37) Proof . From the expressions (2.16) and (2.30), we have that u + i ( x ) = ¯ u + i ( x ) for x ∈ Ω i ∩ Ω i +1 , which impliesΨ + i +1 ( ¯ f + i ) = J − i ∇ · ( A i ∇ ( β + i +1 ¯ u + i )) + k ( β + i +1 ¯ u + i ) . ure source transfer domain decomposition method ψ ∈ H (Ω PML i +1 ), (cid:10) J i Ψ + i +1 ( ¯ f + i ) , ψ (cid:11) = − ( A i ∇ β + i +1 ¯ u + i , ∇ ψ ) Ω i +1 + ( A i ∇ ¯ u + i , ∇ β + i +1 ψ ) Ω i +1 − (cid:10) f i +1 , β + i +1 ψ (cid:11) , and (cid:68) J i ˆΨ + i +1 ( ˆ f + i ) , ψ (cid:69) = − ( A i ∇ β + i +1 ˆ u + i , ∇ ψ ) Ω i +1 + ( A i ∇ ˆ u + i , ∇ β + i +1 ψ ) Ω i +1 − (cid:10) f i +1 , β + i +1 ψ (cid:11) . Therefore,( J i − ( ¯ f + i − ˆ f + i ) , v ) = (cid:0) J i − (Ψ + i ( ¯ f + i − ) − ˆΨ + i ( ˆ f + i − )) , v (cid:1) Ω i ∩ B L = − (cid:0) A i − ∇ β + i (¯ u + i − − ˆ u + i − ) , ∇ v (cid:1) Ω i ∩ B L + (cid:0) A i − ∇ (¯ u + i − − ˆ u + i − ) , ∇ β + i v (cid:1) Ω i ∩ B L ≤ Ck − (cid:13)(cid:13) ¯ u + i − − ˆ u + i − (cid:13)(cid:13) H (Ω i ∩ B L ) (cid:107) v (cid:107) H (Ω i ∩ B L ) ≤ Ck − (cid:13)(cid:13) ¯ u + i − − ˆ u + i − (cid:13)(cid:13) H (Ω PML i ) (cid:107) v (cid:107) H (Ω PML i ) . On the other hand, ¯ u + i − − ˆ u + i − = ¯ u + i − on ∂ Ω PML i − and for any ψ ∈ H (Ω PML i − )( A i − ∇ (¯ u + i − − ˆ u + i − ) , ∇ ψ ) − k ( J i − (¯ u + i − − ˆ u + i − ) , ψ ) = (cid:68) J i − ( ¯ f + i − − ˆ f + i − ) , ψ (cid:69) . By the inf-sup condition 2.33, and Lemma 2.7, we have (cid:13)(cid:13) ¯ u + i − − ˆ u + i − (cid:13)(cid:13) H (Ω PML i − ) ≤ Ck α (cid:13)(cid:13)(cid:13) ¯ f + i − − ˆ f + i − (cid:13)(cid:13)(cid:13) H − (Ω PML i − ) + Ck α (1 + kL ) (cid:107) ¯ u i − (cid:107) H / ( ∂ Ω PML i − ) ≤ Ck α (cid:13)(cid:13)(cid:13) ¯ f + i − − ˆ f + i − (cid:13)(cid:13)(cid:13) H − (Ω PML i − ) + Ck α (1 + kL ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) . Therefore, (cid:13)(cid:13)(cid:13) ¯ f + i − ˆ f + i (cid:13)(cid:13)(cid:13) H − (Ω PML i ) ≤ Ck α (cid:13)(cid:13)(cid:13) ¯ f + i − − ˆ f + i − (cid:13)(cid:13)(cid:13) H − (Ω PML i − ) + Ck α (1 + kL ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) . (i) follows from the induction argument and the fact that ¯ f +1 − ˆ f +1 = 0. Finally, wecould prove (ii) by an argument similar to that of (i). This completes the proof. Lemma 2.9.
Let σ d ≥ be sufficiently large. (i) For i = 1 , , · · · , N − , (cid:13)(cid:13) ¯ u + i − ˆ u + i (cid:13)(cid:13) H (Ω PML i ) ≤ Ck αi +2 (1 + kL ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) . (ii) For i = N, · · · , , (cid:13)(cid:13) ¯ u − i − ˆ u − i (cid:13)(cid:13) H (Ω PML i − ) ≤ Ck α ( N − i +1)+2 (1 + kL ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) . Proof . By using the fact that ¯ u + i − ˆ u + i = ¯ u + i on ∂ Ω PML i and for any ψ ∈ H (Ω PML i )( A i ∇ (¯ u + i − ˆ u + i ) , ∇ ψ ) − k ( J i (¯ u + i − ˆ u + i ) , ψ ) = (cid:68) J i ( ¯ f + i − ˆ f + i ) , ψ (cid:69) , Y. Du & H. Wu
Lemma 2.8 and the inf–sup condition (2.33), the first result can be obtained. Thesecond result can be proved similarly.
Theorem 2.10.
We define ˆ u +0 = ˆ u − N +1 = 0 in R . Let ˆ v = − (ˆ u + i − + ˆ u − i +1 ) in Ω i ∩ B L for all i = 1 , , · · · , N . Then for sufficiently large σ d ≥ , we have (cid:107) ˜ u − ˆ v (cid:107) H ( B L ) ≤ Ck α ( N − k (1 + kL ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) , (2.38) where ˜ u is the solution to the PML problem (2.6) in the whole space.Proof . From their definitions, we know that u + i = ¯ u + i and u − i = ¯ u − i for i =1 , · · · , N . We also define ¯ u +0 = ¯ u − N +1 = 0 in R . Combining with Theorem 2.5, wehave ˜ u ( x ) = − (¯ u + i − ( x ) + ¯ u − i +1 ( x )) for any x ∈ Ω i , i = 1 , · · · , N . Then, by usingLemma 2.9, we complete the proof.Theorem 2.10 shows that the solution ˆ v obtained by our PSTDDM is a fineapproximation of the exact solution to the PML problem in the whole space. Weremark that the constant ‘C’ in (2.38) generally depends on Ω PML i and the number N of layers due to the inf-sup condition and the induction argument used in the proofswhich are omitted (cf. [[15], Theorem 3.7]).
3. Source transfer block by block.
Since every local PML problem in ourPSTDDM (cf. 1 and 2) is defined on the whole space R , we can use the PSTDDM tosolve the local PML problems (cf. (2.10) and (2.14)) recursively. As a consequence,the domain B l is divided into some small squares and we only need to solve local PMLproblems defined outside the union of four squares. R . We only show the al-gorithms (cf. 5 and 6) to solve the local PML equation (2.10). In order to showthe details of our method, Some notations are introduced. For simplicity, we set l = l , , d = d . In order to use the results obtained in the previous sections, thefollowing assumption about the medium property is adopted: H2 (cid:90) l + d/ l σ ( t ) dt ≥ ¯ σ, (cid:90) l + d/ l σ ( t ) dt ≥ ¯ σ, (3.1) and (cid:90) l + dl + d/ σ ( t ) dt ≥ ¯ σ, (cid:90) l + dl + d/ σ ( t ) dt ≥ ¯ σ. which is a direct consequence of the assumption H1 (2.27).We divide the whole space into layers:Ω = { x = ( x , x ) T ∈ R : x < ζ } , Ω i = { x = ( x , x ) T ∈ R : ζ i < x < ζ i +1 } , i = 1 , · · · , N, Ω N +1 = { x = ( x , x ) T ∈ R : ζ N +1 < x } , and denote by Ω i,j = Ω i ∩ Ω j .We define the PML complex coordinate stretching ˜ x i,j = (˜ x i,j ( x ) , ˜ x i,j ( x )) out-side the domain ( ζ j , ζ j +2 ) × ( ζ i , ζ i +2 ) by ˜ x i,j ( x ) = ˜ x i, ( x ) and˜ x i,j ( x ) = x + i (cid:82) x ζ j +2 σ ( t + ζ M +1 − ζ j +2 ) dt if x > ζ j +2 ,x if ζ j ≤ x ≤ ζ j +2 ,x + i (cid:82) x ζ j σ ( t − ζ j + ζ ) dt if x < ζ j . (3.2) ure source transfer domain decomposition method A i,j ( x ) = diag (cid:32) ˜ x i,j ( x ) (cid:48) ˜ x i,j ( x ) (cid:48) , ˜ x i,j ( x ) (cid:48) ˜ x i,j ( x ) (cid:48) (cid:33) , J i,j ( x ) = ˜ x i,j ( x ) (cid:48) ˜ x i,j ( x ) (cid:48) . Let ¯ f + i,j = ¯ f + i + f i +1 in Ω j and ¯ f + i,j = 0 elsewhere. We denote by γ + i ( x ) and γ − i ( x ) smooth functions such that γ + i ( t ) = β + i ( t ) and γ − i ( t ) = β − i ( t ) for any t ∈ R . Algorithm 5
Source Transfer I + for the i th local PML problem1. Let ¯¯ f + i, = ¯ f + i, in Ω and ¯¯ f + i, = 0 elsewhere.while j = 1 , · · · , N −
2, do • Find ¯ u + i,j ∈ H ( R ), such that −∇ ( A i,j ∇ ¯ u + i,j ) − k J i,j ¯ u + i,j = J i,j ( ¯¯ f + i,j + ¯ f + i,j +1 ) , (3.3) • Compute ¯Ψ + i,j +1 ( ¯¯ f + i,j ) ∈ H − ( R ) such that¯Ψ + i,j +1 ( ¯¯ f + i,j ) = J − i,j ∇ ( A i,j ∇ ( γ + j +1 ¯ u + i,j )) + k ( γ + j +1 ¯ u + i,j ) . • Set ¯¯ f + i,j +1 = ¯ f + i,j +1 + ¯Ψ + i,j +1 in Ω j +1 and ¯¯ f + i,j +1 = 0 elsewhere.End while2. Find ¯ u + i,N − ∈ H ( R ), such that −∇ ( A i,N − ∇ ¯ u + i,N − ) − k ( J i,N − ¯ u + i,N − ) = J i,N − ( ¯¯ f + i,N − + ¯ f + i,N ) . (3.4) Algorithm 6
Source Transfer I − for the i th local PML problem1. Let ¯¯ f − i,N = ¯ f + i,N .While j = N, · · · , • Find ¯ u − i,j ∈ H ( R ), such that −∇ ( A i,j − ∇ ¯ u − i,j ) − k J i,j − ¯ u − i,j = J i,j − ¯¯ f − i,j , (3.5) • Compute ¯Ψ − i,j − ( ¯¯ f − i,j ) ∈ H − ( R ) such that¯Ψ − i,j − ( ¯¯ f − i,j ) = J − i,j − ∇ ( A i,j − ∇ ( γ − j − ¯ u − i,j )) + k ( γ − j − ¯ u − i,j ) . • Set ¯¯ f − i,j − = ¯ f + i,j − + ¯Ψ − i,j − ( ¯¯ f − i,j ) in Ω j − and ¯¯ f − i,j − = 0 elsewhere.End while2. Find ¯ u − i, ∈ H ( R ) such that −∇ ( A i, ∇ ¯ u − i, ) − k ( J i, ¯ u − i, ) = J i, ( ¯¯ f − i, + ¯ f + i, ) . (3.6)Algorithm 5 and Algorithm 6 show the details of our PSTDDM solving the i-thPML problem in Algorithm 1. We omit the details about Algorithm 2 to save the4 Y. Du & H. Wu space. Then we can obtain some results similar to those in section 2.2, but only statebriefly them when needed.The following lemma can be proved by their definitions. We omit the details.
Lemma 3.1.
Let ¯ u + i, ≡ and ¯ u − i,N +1 ≡ . For any x ∈ Ω i,j , ¯ u + i ( x ) = ¯ u + i,j − ( x ) + ¯ u − i,j +1 ( x ) . Lemma 3.2.
Let σ d > be sufficiently large. For i = 1 , · · · , N − For j = 1 , · · · , N − , we have for any x ∈ Ω + i,j +1 := { x = ( x , x ) T : ζ j +1
2, ¯ u + i,j ( x ) satisfies¯ u + i,j ( x ) = (cid:90) x <ζ j +2 f ( y ) J ( y ) G (˜ x, ˜ y ) dy + (cid:90) x <ζ j +1 ¯ f i ( y ) J ( y ) G (˜ x, ˜ y ) dy := ¯ u +I i,j ( x ) + ¯ u +II i,j ( x ) ∀ x ∈ Ω + i,j +1 . It is clear that (cid:12)(cid:12) ¯ u +II i,j ( x ) (cid:12)(cid:12) ≤ C (cid:90) Ω ini,j (cid:12)(cid:12) ¯ f i ( y ) (cid:12)(cid:12) | G (˜ x, ˜ y ) | dy + C (cid:90) Ω outi,j (cid:12)(cid:12) ¯ f i ( y ) (cid:12)(cid:12) | G (˜ x, ˜ y ) | dy, where Ω ini,j = { x = ( x , x ) T ∈ Ω i : − l − d/ < x < ζ j +1 } and Ω outi,j = { x =( x , x ) T ∈ Ω i : x < − l − d/ } . By the standard argument (cf. [[15], 3.11]), we canget (cid:90) Ω ini,j (cid:12)(cid:12) ¯ f i ( y ) (cid:12)(cid:12) | G (˜ x, ˜ y ) | dy ≤ Ck / e − kγ ¯ σ (cid:13)(cid:13) ¯ f i (cid:13)(cid:13) H (Ω i ) (cid:48) . We recall that ¯ u i ( y ) and ∇ ¯ u i ( y ) decay exponentially as | y | → ∞ , that is (cid:12)(cid:12) ¯ u + i ( y ) (cid:12)(cid:12) ≤ Ck / e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) and (cid:12)(cid:12) ∇ ¯ u + i ( y ) (cid:12)(cid:12) ≤ Ck / e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) for | y | > l d / σ d is large enough, which implies (cid:90) Ω outi,j (cid:12)(cid:12) ¯ f i ( y ) (cid:12)(cid:12) | G (˜ x, ˜ y ) | dy ≤ C sup y ∈ Ω outi,j ( (cid:12)(cid:12) ¯ u + i − (cid:12)(cid:12) + (cid:12)(cid:12) ∇ ¯ u + i − (cid:12)(cid:12) ) (cid:90) Ω outi,j | G (˜ x, ˜ y ) | dy ≤ Cke − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) . ure source transfer domain decomposition method x ∈ Ω + i,j +1 , (cid:12)(cid:12) ¯ u +II i,j ( x ) (cid:12)(cid:12) ≤ Cke − kγ ¯ σ ( (cid:107) f (cid:107) H ( B l ) (cid:48) + (cid:13)(cid:13) ¯ f i (cid:13)(cid:13) H (Ω i ) (cid:48) ) . It’s so easy to get the estimates, (cid:12)(cid:12) ¯ u +I i,j ( x ) (cid:12)(cid:12) ≤ Ck / e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) . Therefor, we have (cid:12)(cid:12) ¯ u + i,j ( x ) (cid:12)(cid:12) ≤ Cke − kγ ¯ σ ( (cid:107) f (cid:107) H ( B l ) (cid:48) + (cid:13)(cid:13) ¯ f i (cid:13)(cid:13) H (Ω i ) (cid:48) ) for x ∈ Ω + i,j +1 . Asimilar argument implies that (cid:12)(cid:12) ∇ ¯ u + i,j ( x ) (cid:12)(cid:12) ≤ Ck e − kγ ¯ σ ( (cid:107) f (cid:107) H ( B l ) (cid:48) + (cid:13)(cid:13) ¯ f i (cid:13)(cid:13) H (Ω i ) (cid:48) ) ∀ x ∈ Ω + i,j +1 . This completes the proof.Define l N := 2 l/N +2 d and Ω PML i,j := { x = ( x , x ) : ζ i − d ≤ x ≤ ζ i +2 + d, ζ j − d ≤ x ≤ ζ j +2 + d } . We have the following lemma. Lemma 3.3.
Assume that σ d > be sufficiently large. There exists a constant C b independent of l N , k and N such that (i) For j = 1 , , · · · , N − , (cid:13)(cid:13) ¯ u + i,j (cid:13)(cid:13) H / ( ∂ Ω PML i,j ) ≤ C b k (1 + kl N ) (cid:107) f (cid:107) H ( B l ) (cid:48) ;(ii) For j = N, N − , · · · , , (cid:13)(cid:13) ¯ u − i,j (cid:13)(cid:13) H / ( ∂ Ω PML i,j − ) ≤ C b k (1 + kl N ) (cid:107) f (cid:107) H ( B l ) (cid:48) . (3.7) Proof . The two assertions can be obtained by using arguments similar to thosein Lemma 3.6 in [15]. We omit the details and show the result (cid:13)(cid:13) ¯ u + i,j (cid:13)(cid:13) H / ( ∂ Ω PML i,j ) ≤ Ck / (1 + kl N ) e − kγ ¯ σ (cid:0) (cid:107) f (cid:107) H ( B l ) (cid:48) + (cid:13)(cid:13) ¯ f i (cid:13)(cid:13) H (Ω i ) (cid:48) (cid:1) . By the definition of source transfer operator (2.19), we have (cid:13)(cid:13) ¯ f i (cid:13)(cid:13) H (Ω i ) (cid:48) ≤ C (cid:107) ¯ u i − (cid:107) H (Ω i ) (cid:48) ≤ Ck / (cid:107) f (cid:107) H ( B l ) (cid:48) . Combining the two inequalities above, we have (cid:13)(cid:13) ¯ u + i,j (cid:13)(cid:13) H / ( ∂ Ω PML i,j ) ≤ C b k (1 + kl N ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) . Clearly, the C ’s used here are independent of l N , k and N . Thus we complete theproof of the first assertion and the second one can be proved by the same way. For the ease ofpresentation, we denoteΩ tru i, = ( ζ − d, ζ ) × ( ζ i − d, ζ i +2 + d ) , Ω tru i,j = ( ζ j , ζ j +1 ) × ( ζ i − d, ζ i +2 + d ) , j = 2 , · · · , N − , Ω tru i,N = ( ζ N , ζ N +1 + d ) × ( ζ i − d, ζ i +2 + d ) . Y. Du & H. Wu
We can get the approximation ˇ u + i ( x ) of ¯ u + i ( x ), i = 1 , · · · , N −
1, in Ω
PML i byusing Algorithm 7 and Algorithm 8 , where ˇ u + i ( x ) are defined byˇ u + i = ˆ u + i, in Ω tru i, , ˇ u + i = ˆ u − i,N in Ω tru i,N (3.8) ˇ u + i = ˆ u + i,j − + ˆ u − i,j +1 in Ω tru i,j , f or j = 2 , · · · , N − . In Algorithm 7 and Algorithm 8 we have defined ˇ f + i,j = ˇ f + i + f i +1 in Ω tru i,j and ˇ f + i are defined by1. Let ˇ f +1 = f in Ω PML1 .2. Compute ˜Ψ + i +1 ∈ H − (Ω PML i ) such that˜Ψ + i +1 = J − i ∇ ( A i ∇ ( β + i +1 ˇ u + i )) + k ( β + i +1 ˇ u + i ) .
3. Set ˇ f + i +1 = f i +1 + ˜Ψ + i +1 in Ω i +1 ∩ B L and ˇ f + i +1 = 0 elsewhere.We also could obtain the approximation ˇ u − i ( x ) of ¯ u − i ( x ), i = N, · · · ,
2, in Ω
PML i − . Thedetails are omitted in order to save space. Algorithm 7
Source Transfer I + for local Truncated PML problem1. Let ˆ f + i, = ˇ f + i, .While j = 1 , · · · , N −
2, do • Find ˆ u + i,j ∈ H (Ω PML i,j ), where Ω
PML i,j = ( ζ i − d, ζ i +2 + d ) × ( ζ j − d, ζ j +2 + d ),such that ∀ ψ ∈ H (Ω PML i,j )( A i,j ∇ ˆ u + i,j , ∇ ψ ) − k ( J i,j ˆ u + i,j , ψ ) = (cid:68) J i ( ˆ f + i,j + ˇ f + i,j +1 ) , ψ (cid:69) , (3.9) • Compute ˆΨ + i,j +1 ( ˆ f + i,j ) ∈ H − (Ω PML i,j ) such thatˆΨ + i,j +1 ( ˆ f + i,j ) = J i,j ∇ ( A i,j ∇ ( γ + j +1 ˆ u + i,j )) + k ( γ + j +1 ˆ u + i,j ) . • Set ˆ f + i,j +1 = ˇ f + i,j +1 + ˆΨ + i,j +1 ( ˆ f + i,j ) in D j +1 ∩ Ω PML i,j and ˆ f + i,j +1 = 0 else-where.End while2. Find ˆ u + i,N − ∈ H (Ω PML i,N − ) where Ω PML i,N − = ( ζ i − d, ζ i +2 + d ) × ( ζ N − − d, ζ N +1 + d ), such that ∀ ψ ∈ H (Ω PML i,N − )( A i,N − ∇ ˆ u + i,N − , ∇ ψ ) − k ( J i,N − ˆ u + i,N − , ψ ) = (cid:68) J i,N − ( ˆ f + i,N − + ˇ f + i,N ) , ψ (cid:69) . (3.10)We can improve the local inf-sup condition (2.33). Lemma 3.4.
For sufficiently large σ d > , we have the inf-sup condition for any φ ∈ H (Ω PML i,j ) sup ψ ∈ H (Ω PML i,j ) ( A i,j ∇ φ, ∇ ψ ) − k ( J i,j φ, ψ ) (cid:107) ψ (cid:107) H (Ω PML i,j ) ≥ µ (cid:107) φ (cid:107) H (Ω PML i,j ) , (3.13) where µ − ≤ C is ( l N k ) / if l N k large enough, and µ − ≤ C is if l N k ≈ . C is independent of l N , k and N . ure source transfer domain decomposition method Algorithm 8
Source Transfer I − for local Truncated PML problem1. Let ˆ f − i,M = ˇ f + i,M .While j = N, · · · , • Find ˆ u − i,j ∈ H (Ω PML i,j − ) such that ∀ ψ ∈ H (Ω PML i,j − )( A i,j − ∇ ˆ u − i,j , ∇ ψ ) − k ( J i,j − ˆ u − i,j , ψ ) = (cid:68) J i,j − ˆ f − i,j , ψ (cid:69) , (3.11) • Compute ˆΨ − i,j − ( ˆ f − i,j ) ∈ H − (Ω PML i,j − ) such thatˆΨ − i,j − ( ˆ f − i,j ) = J − i,j − ∇ ( A i,j − ∇ ( γ − j − ˆ u − i,j )) + k ( γ − j − ˆ u − i,j ) . • Set ˆ f − i,j − = ˇ f + i,j − + Ψ − i,j − ( ˆ f − i,j ) in Ω i,j − and ˆ f − i,j − = 0 elsewhere.End while2. Find ˆ u − i, ∈ H (Ω PML i, ) such that ∀ ψ ∈ H (Ω PML i, )( A i, ∇ ˆ u − i, , ∇ ψ ) − k ( J i, ˆ u − i, , ψ ) = (cid:68) J i, ( ˆ f − i, + ˇ f + i, ) , ψ (cid:69) . (3.12) Proof . The inequality can be proved easily by using scaling argument. We knowthat there is a unique solution φ ∈ H (Ω PML i,j ) to the problem −∇ ( A i,j ∇ φ ) − k J i,j φ = F, for some F ∈ H (Ω PML i,j ) (cid:48) . We define a mapping m : I := [0 , × [0 , → Ω PML i,j as m ( z ) = l N z + ( ζ i − d, ζ j − d ) and denote by ˆ φ ( z ) := φ ( m ( z )) and ˆ F ( z ) := F ( m ( z )).The equation above implies ˆ φ ( z ) ∈ H ( I ) satisfying −∇ z ( A i,j ( m ( z )) ∇ z ˆ φ ( z )) − ( l N k ) J i,j ( m ( z )) ˆ φ ( z ) = l N ˆ F ( z ) . (3.14)If l N k large enough, by the local inf-sup condition (2.33), we get (cid:0) ( l N k ) (cid:13)(cid:13)(cid:13) ˆ φ ( z ) (cid:13)(cid:13)(cid:13) L ( I ) + (cid:12)(cid:12)(cid:12) ˆ φ ( z ) (cid:12)(cid:12)(cid:12) H ( I ) (cid:1) / ≤ C is ( l N k ) / (cid:13)(cid:13)(cid:13) l N ˆ F ( z ) (cid:13)(cid:13)(cid:13) H ( I ) (cid:48) , (3.15)where (cid:107)·(cid:107) H ( I ) (cid:48) is defined assup ψ ∈ H ( I ) ( · , ψ ) (cid:0) ( l N k ) (cid:107) ψ (cid:107) L ( I ) + | ψ | H ( I ) (cid:1) / , from the definition of weighted norm (cid:107) u (cid:107) H ( U ) at the beginning of section 2.3. How-ever, if l N k ≈ (cid:0) (cid:13)(cid:13)(cid:13) ˆ φ ( z ) (cid:13)(cid:13)(cid:13) L ( I ) + (cid:12)(cid:12)(cid:12) ˆ φ ( z ) (cid:12)(cid:12)(cid:12) H ( I ) (cid:1) / ≤ C is (cid:13)(cid:13)(cid:13) l N ˆ F ( z ) (cid:13)(cid:13)(cid:13) H ( I ) (cid:48) . (3.16)Clearly, C is is independent of l N , k and N . Finally, The consequence is obtained bycombining the inequalities (3.15), (3.16) and the fact that( l N k ) (cid:13)(cid:13)(cid:13) ˆ φ ( z ) (cid:13)(cid:13)(cid:13) L ( I ) = k (cid:107) φ ( x ) (cid:107) L (Ω PML i,j ) , (cid:12)(cid:12)(cid:12) ˆ φ ( z ) (cid:12)(cid:12)(cid:12) H ( I ) = | φ ( x ) | H (Ω PML i,j ) , (cid:13)(cid:13)(cid:13) l N ˆ F ( z ) (cid:13)(cid:13)(cid:13) H ( I ) (cid:48) = (cid:107) F ( x ) (cid:107) H (Ω PML i,j ) (cid:48) . Y. Du & H. Wu
In general, we can expect that l N k is less than k . If N is large enough such that l N k ≈
1, the local truncated PML problems (cf. (3.9)–(3.12)) needed to be solved areabout elliptic.
Lemma 3.5.
Let σ d > be sufficiently large. There are constants C t and C bt independent of l N , k and N such that (i) For i = 1 , , · · · , N − , (cid:13)(cid:13) ˇ u + i − ¯ u + i (cid:13)(cid:13) H (Ω PML i ) ≤ C bt C k,N,i k (1 + kl N ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) . (3.17)(ii) For i = N, N − , · · · , , (cid:13)(cid:13) ˇ u − i − ¯ u − i (cid:13)(cid:13) H (Ω PML i − ) ≤ C bt C k,N,N +1 − i k (1 + kl N ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) . (3.18) Here C k,N,j , j ∈ N are defined as C k,N,j = j (cid:88) q =1 (cid:18) N − (cid:88) p =1 (cid:0) C t µ − (cid:1) p (cid:19) q . (3.19) Proof . We only show the details of the proof for the first assertion and the secondone could be proved similarly. At the beginning, we recall the property (cf. [15],Theorem 3.7) of source transfer operators that there’s a constant C t independent of l N , k and N , such that (cid:13)(cid:13)(cid:13) ¯¯ f + i,j − ˆ f + i,j (cid:13)(cid:13)(cid:13) H (Ω PML i,j ) (cid:48) ≤ C t (cid:13)(cid:13) ¯ u + i,j − − ˆ u + i,j − (cid:13)(cid:13) H (Ω PML i,j − ) , (cid:13)(cid:13) ¯ f + i − ˇ f + i (cid:13)(cid:13) H (Ω PML i ) (cid:48) ≤ C t (cid:13)(cid:13) ¯ u + i − − ˆ u + i − (cid:13)(cid:13) H (Ω i ∩ B L ) , for i, j = 2 , · · · , N − (cid:13)(cid:13) ¯ u + i,j − ˆ u + i,j (cid:13)(cid:13) H (Ω PML i,j ) ≤ µ − (cid:13)(cid:13)(cid:13) ( ¯¯ f + i,j + ¯ f + i,j +1 ) − ( ˆ f + i,j + ˇ f + i,j +1 ) (cid:13)(cid:13)(cid:13) H (Ω PML i,j ) (cid:48) (3.20) + µ − (1 + kl N ) (cid:13)(cid:13) ¯ u + i,j (cid:13)(cid:13) H / ( ∂ Ω PML i,j ) ≤ C t µ − (cid:13)(cid:13) ¯ u + i,j − − ˆ u + i,j − (cid:13)(cid:13) H (Ω PML i,j − ) + µ − (cid:13)(cid:13) ¯ f + i − ˇ f + i (cid:13)(cid:13) H (Ω PML i ) (cid:48) + C b µ − k (1 + kl N ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) . By the induction argument and the fact that (cid:13)(cid:13) ¯ u + i, − ˆ u + i, (cid:13)(cid:13) H (Ω PML i, ) ≤ µ − (cid:13)(cid:13) ¯ f + i − ˇ f + i (cid:13)(cid:13) H (Ω PML i ) (cid:48) + C b µ − k (1 + kl N ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) , (3.20) implies for j = 1 , · · · , N − (cid:13)(cid:13) ¯ u + i,j − ˆ u + i,j (cid:13)(cid:13) H (Ω PML i,j ) ≤ j − (cid:88) p =0 (cid:0) C t µ − (cid:1) p · µ − · (3.21) (cid:2) (cid:13)(cid:13) ¯ f + i − ˇ f + i (cid:13)(cid:13) H (Ω PML i ) (cid:48) + C b k (1 + kl N ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) (cid:3) . ure source transfer domain decomposition method j = N, · · · , (cid:13)(cid:13) ¯ u − i,j − ˆ u − i,j (cid:13)(cid:13) H (Ω PML i,j − ) ≤ N − j (cid:88) p =0 (cid:0) C t µ − (cid:1) p · µ − · (3.22) (cid:2) (cid:13)(cid:13) ¯ f + i − ˇ f + i (cid:13)(cid:13) H (Ω PML i ) (cid:48) + C b k (1 + kl N ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) (cid:3) . From (3.21), (3.22), Lemma 3.1 and the definition 3.8, we obtain (cid:13)(cid:13) ¯ u + i − ˇ u + i (cid:13)(cid:13) H (Ω PML i ) ≤ N − (cid:88) p =0 (cid:0) C t µ − (cid:1) p · µ − · (3.23) (cid:2) (cid:13)(cid:13) ¯ f + i − ˇ f + i (cid:13)(cid:13) H (Ω PML i ) (cid:48) + C b k (1 + kl N ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) (cid:3) . ≤ N − (cid:88) p =1 (cid:0) C t µ − (cid:1) p (cid:2) (cid:13)(cid:13) ¯ u + i − − ˆ u + i − (cid:13)(cid:13) H (Ω i ∩ B L ) + C b C t k (1 + kl N ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) (cid:3) . Since C t and C b don’t depend on l N , k and N , we can denote C bt = C b C t . Then wecomplete the proof for the first assertion (3.17) by the induction argument and thefact (cid:13)(cid:13) ¯ u +1 − ˇ u +1 (cid:13)(cid:13) H (Ω PML i ) ≤ N − (cid:88) p =1 (cid:0) C t µ − (cid:1) p C bt k (1 + kl N ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) . The following theorem is a direct consequence of Theorem 2.5, Lemma 3.5 andthe fact that ˜ u ± i = ¯ u ± i in Ω i ∪ Ω i +1 . Theorem 3.6.
Let ˇ u +0 = ˇ u − N +1 = 0 in B L and ˇ u ( x ) = − (ˇ u + i − + ˇ u − i +1 ) in Ω i ∩ B L for all i = 1 , · · · , N . Denote C k,N = C k,N,N − . Then for sufficiently large σ d ≥ ,we have (cid:107) ˇ u − ˜ u (cid:107) H ( B L ) ≤ C bt C k,N k (1 + kl N ) e − kγ ¯ σ (cid:107) f (cid:107) H ( B l ) (cid:48) . (3.24)We remark that from the theorem above, we can know that the larger number N doesn’t mean the solution ˇ u performing better although the local problem solved maybe elliptic. However, our numerical examples in the following section show that therelative errors between ˇ u and the discrete solutions don’t increase significantly when N becomes larger.
4. Numerical examples.
In this section, we simulate the problem 1.1 and 1.2for constant and heterogeneous wave number by FEM and STDDM, where f is givenso that the exact solution is u = (cid:40) − r ( r + 3 r − r + 9) H (1)0 ( kr ) , r < , − H (1)0 ( kr ) , r > = 1 . Let d = 0 . d = 0 . l = l = 1 .
1. So the computational domain B L is( − . , . × ( − . , .
2) and the perfect matched lay is B L \ B l where B l = ( − . , . .It is easy to see that u ∈ C ( R ) and supp f ⊂ B l .0 Y. Du & H. Wu
We define the medium property [15] by setting ¯ l = ¯ l = 1 .
18 and σ j ( t ) =ˆ σ j ( t ) + ( t − l j )ˆ σ (cid:48) j ( t ) for l j < t < ¯ l j , whereˆ σ j ( t ) = γ (cid:32)(cid:90) tl j ( s − l j ) (¯ l j − s ) ds (cid:33) (cid:32)(cid:90) ¯ l j l j ( s − l j ) (¯ l j − s ) ds (cid:33) − . (4.1)The functions β ± i ( x ) , x ∈ Ω i , i = 2 , · · · , N −
1, used in the source transferalgorithm are defined as β + i ( x ) = , ζ i ≤ x < ζ i + ζ i + ∆ ζ/ ,η i ( x ) , ζ i + ∆ ζ/ ≤ x < ζ i + 3∆ ζ/ , , ζ i + 3∆ ζ/ ≤ x ≤ ζ i +1 , and β − i = 1 − β + i , where η i ( x ) = 1 + (cid:18) x − ( ζ i + ∆ ζ/ ζ/ (cid:19) − (cid:18) x − ( ζ i + ∆ ζ/ ζ/ (cid:19) . Clearly, β ± i ( x ) , i = 2 , · · · , N −
1, are in C (Ω i ) and this fact avoids the discontinuityof β ± i ( x ) (cid:48) which may make ˆ f ± i oscillate heavily.We use the finite element method to solve truncated PML problems. The numberof nodes in the x j -direction is n j = q · L j /λ, j = 1 ,
2, where q is the mesh densitywhich is the number of nodes in each wavelength λ = 2 π/k . Then the number ofdegree freedom DOF is n n . Let N be the division number in the x -direction. e i , e f and e s denote the relative error in H -seminorm of the interpolation, the FEMsolution and the PSTDDM solution bounded in B l respectively.We first test the algorithm 3 and 4 for the wave number k/ (2 π ) = 25 and k/ (2 π ) =50. The left graph of Figure 4.1 plots the relative error decay of the interpolation,FE solution and PSTDDM solution with a fixed number of lays N = 10 in termsof DOF for k/ (2 π ) = 25 ,
50 respectively. We could find that the relative errorsof PSTDDM solution is the same to that of FE solution when DOF is equal. Thisis best result about comparison between the PSTDDM and FEM which we couldexpect, since the details of the algorithms 3 and 4 show that the errors of PSTDDMsolutions can not be less than those of FE solutions under the condtion that the meshis same. In the right graph of Figure 4.1, we set DOF = 624 × and give the relativeerrors in H -seminorm of the PSTDDM 3. 4 solutions in terms of the number of laysin x -direction N = 1 , , , , , , k/ (2 π ) = 25 ,
50 respectively, where N = 1 means that this solution is the FE solution. It is shown that the error ofPSTDDM solution remains unchanged even if the number of lays in the x -directionbecomes larger. So we could choose a relatively large number of lays to reduce thecomputational complexity.Next we test our further consideration (cf. 3.8, 7, 8) about the PSTDDM for k/ (2 π ) = 25 and k/ (2 π ) = 50. The parameters about PML layers are still thoseprovided at the beginning of this section, since they’re not essential from the previousproofs.In the left graph of Figure 4.2, we set N = 10, and show the error decay ofthe FE solution and further PSTDDM solution when mesh density q increases. Thegraph is very quite similar to that of Figure 4.1, what we could like to obtain. In theright graph, we show the relative errors of the further PSTDDM (cf. 3.8, 7, 8) when ure source transfer domain decomposition method N = , , ,
25. Thus the number of the squares, which the domain B l is divided into,is N = 25 , , , N too small becauseof the fixed width of PML layer resulting in low computational efficiency in practicalapplication. −2 −1 DOF R e l a t i v e E rr o r e i of k/(2 π )=25e f of k/(2 π )=25e s of k/(2 π )=25e i of k/(2 π )=50e f of k/(2 π )=50e s of k/(2 π )=50 R e l a t i v e E rr o r s k/(2 π )=25k/(2 π )=50 Fig. 4.1 . Left graph: The relative errors e i , e f , e s for the interpolations, FE solutions and PSTDDM
3, 4 solutions with a fixed number of lays in x -direction N = 10 in terms of the numberof degree freedom DOF = n n for k/ (2 π ) = 25 and k/ (2 π ) = 50 respectively. Right graph: Therelative errors for the PSTDDM
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