# The Schwartz Alternating Algorithm for Solving 3D Exterior Helmholtz Problems

Authors: Qing Chen; Baoqing Liu; Qikui Du
Journal: Mathematical Problems in Engineering
 (2018)

---

## Abstract

Based on the natural boundary reduction, an overlapping domain decomposition method is discussed for solving exterior Helmholtz problem over a three-dimensional (3D) domain. By introducing two different artificial boundaries, the original unbounded domain is divided into a bounded subdomain and a typical unbounded region, and a Schwartz alternating method is presented. The finite element method and natural boundary element method are alternately applied to solve the problems in the bounded subdomain and the typical unbounded subdomain. Moreover, the convergence of the Schwartz alternating algorithm is studied. Finally, some numerical experiments are presented to show the performance of this method.


---

## Body

## 1. Introduction

When solving a problem modelled by a linear partial differential equation in the bounded or unbounded domain, many existing methods can be adopted. For the problem in the bounded domain, the finite element method and the finite difference method and so on are very effective. Yet we often find them difficult to be applied to unbounded problem directly. To solve such problems in infinite region numerically, there are a variety of numerical methods (cf. [1–4]) and references therein for more details.Schwartz alternating algorithm is one of the most efficient techniques for solving problems in the unbounded domain, such as harmonic equation [5, 6], Stokes equation [7], and Helmholtz equation [8]. Over exterior 3D domain, Wu and Yu studied the natural integral equations of Helmholtz problems [9] and overlapping domain decomposition method for harmonic equation [10]. We gave a D-N alternating algorithm for solving 3D exterior Helmholtz problems [11].In this paper, a Schwartz alternating algorithm based on the natural boundary reduction is devised for the numerical solution of exterior three-dimensional Helmholtz problem. Firstly, an exterior Helmholtz problem is introduced and the corresponding variational form is obtained. Secondly, the Schwartz alternating algorithm is posed. Then, the convergence rate is discussed and the relationship between contraction factorα and overlapping extent is given. Besides, the error estimate of the algorithm is offered. Finally, some numerical examples are presented to illustrate the feasibility and efficiency of this method.


## 2. Problem and Its Equivalent Form

We consider the following 3D exterior Helmholtz problem [12]:(1)-Δu-k2u=0,inΩc,u=g,onΓ0,where Ω is a bounded domain in R3 with regular boundary ∂Ω=Γ0. u(x) is the unknown function and g is a known function, which satisfies appropriate conditions. k denotes the wave number, related to the wavelength λ of the incident wave through k=2π/λ. In order to assure the existence and uniqueness of the solution of (1), the solution u(x) satisfies the Sommerfeld radiation condition:(2)limr→∞r∂u∂r-iku=0,where (r,θ,φ) denotes the spherical coordinates, r=x=x2+y2+z2, i=-1. Condition (2) asserts that the scattered wave is outgoing at infinity.The corresponding variational form of problems (1)-(2) is(3)Findu∈H1Ωc,suchthatau,v=g,v,∀v∈H01Ωc,where(4)au,v=∫Ωc∇u·∇v-k2uvdx,g,v=∫Γ0gvds,H01Ωc=v∣v1+r2,∂v∂x,∂v∂y,∂v∂z∈L2Ωc.Particularly, ifΩ is a spherical domain with radius R whose centre is the origin of coordinates, the solution of (1)-(2) is given by the following Poisson integral formula:(5)ur,θ,φ=PuR,θ,φ=∑l=0∞∑m=-llhlkrhlkRYlmθ,φ∫02π∫0πuR,θ′,φ′Ylmθ′,φ′¯sin⁡θ′dθ′dφ′,where P is called the Poisson integral operator of Helmholtz equation in Ωc, hl(x)=π/2xHl+1/2(1)(x) is the first kind of Hankel function, Plm(·) denotes the associated Legendre function of the first kind, Ylm(·)¯ denotes the complex conjugate of Ylm(·), and (6)Ylmθ,φ=l-m!l+m!2l+14πPlmcos⁡θeimφ,Plmx=12ll!1-x2m/2dl+mdxl+mx2-1l,m⩾0,-1-ml+m!l-m!Pl-mx,m<0.By using (5), we will develop an overlapping domain decomposition method (Schwartz alternating algorithm based on the natural boundary reduction) for problems (1)-(2). Taking the normal derivative of (5), we have the following natural integral equation:(7)∂u∂n=KuR,θ,φ=-∑l=0∞∑m=-llkhl′kRhlkRYlmθ,φ∫02π∫0πuR,θ′,φ′Ylmθ′,φ′¯sin⁡θ′dθ′dφ′,where K is called the natural integral operator of Helmholtz equation in Ωc.


## 3. Schwartz Alternating Algorithm and Its Convergence

Taking two sphere surfacesΓ1={(r,θ,φ)∣r=R1,0⩽θ⩽π,0⩽φ⩽2π} and Γ2={(r,θ,φ)∣r=R2,0⩽θ⩽π,0⩽φ⩽2π}, such that R1>R2 and Γ0 is surrounded by Γ2, denote (8)Ω1=x,y,z∣x2+y2+z2<R12,x,y,z∈Ωc,Ω2=x,y,z∣x2+y2+z2>R22,x,y,z∈Ωc,Ω11=Ω1∩Ω2c¯,Ω12=Ω1∩Ω2,Ω22=Ω2∩Ω1c¯,V=v∣v∈H01Ωc,limr→∞r∂v∂r-ikv=0,V1=v∣v∈H01Ω1,vΓ1=0,V2=v∣v∈H01Ω2,vΓ2=0,limr→∞r∂v∂r-ikv=0,Vn=v∣v∈H01Ω1,vΓ1=λn.Construct the following Schwartz alternating algorithm.Step 1.
Select the initialu0∈V and let η(0)=u0Γ1, n=0.Step 2.
Solve the problem inΩ1: (9)-Δu12n+1-k2u12n+1=0,inΩ1,u12n+1=ηn,onΓ1,u12n+1=g,onΓ0.Step 3.
Solve the problem inΩ2: (10)-Δu22n+2-k2u22n+2=0,inΩ2,u22n+2=u12n+1,onΓ2,limr→∞r∂u22n+2∂r-iku22n+2=0.Step 4.
Letη(n+1)=u2(2n+2)Γ1.Step 5.
Letn=n+1, and go to Step 2.Whereu1(2n+1) and u2(2n+2) are the 2n+1th and 2n+2th approximate solutions in Ω1 and Ω2, respectively, η(0) is an arbitrary function in H1/2(Γ1). Note that, on interface Γ1, only η(n+1)=u2(2n+2)Γ1 is needed. So it is unnecessary to solve (10). Actually we can obtain η(n+1) by making use of the following Poisson integral equation: (11)ηn+1=u22n+2Γ1=Pγ2u12n+1Γ1,where γ2 is the trace operator on Γ2. Because of u2(2n)∈V2 and u1(2n+1)∈V1, V1 and V2 can be regarded as the subspaces of V, if elements of V are extended by zero. Let u1(2n+1)Ω22=u2(2n) and u2(2n)Ω11=u1(2n-1); obviously u(l)∈V,l=0,1,2,…, u1(2n+1)-u2(2n)∈V1, and u2(2n+2)-u1(2n+1)∈V2. (9) and (10) are, respectively, equivalent to the following variational forms:(12)Findu12n+1-u22n∈V1,suchthatau12n+1-u,v1=0,∀v1∈V1.Findu22n-u12n-1∈V2,suchthatau22n-u,v2=0,∀v2∈V2.PVi denotes V projected to Vi; change (12) into(13)au12n+1-u22n,v1=au-u22n,v1,∀v1∈V1,u12n+1-u22n∈V1,au22n-u12n-1,v2=au-u12n-1,v2,∀v2∈V2,u22n-u12n-1∈V2.By the definition of projection, we obtain(14)u12n+1-u22n=PV1u-u22n,n⩾0,u22n-u12n-1=PV2u-u12n-1,n⩾1.This is equivalent to(15)u-u12n+1=PV1⊥u-u22n,n⩾0,u-u22n=PV2⊥u-u12n-1,n⩾1,where Vi⊥ is the orthogonal complementary space of Vi; that is, (16)PVi⊥=I-PVi,i=1,2. Let ei(j)=u-ui(j)(i=1,2) denote the errors. Then (15) can be rewritten as(17)e12n+1=PV1⊥e22n,e22n=PV2⊥e12n-1.This yields(18)e12n+1=PV2⊥PV1⊥e12n-1,e22n+2=PV1⊥PV2e22n.It is easy to know that if ei(j)(i=1,2) are convergent, then their limits are in V1⊥∩V2⊥.Lemma 1.
V = V 1 + V 2, V1⊥∩V2⊥={0}.Lemma 2.
V = V 1 + V 2 and for any v∈V there exists a positive constant C0 such that(19)v1⩽C0PV112+PV2121/2,where ·1=a(·,·) denotes energy norm.Proof.
IfV=V1+V2, there exists a positive constant C0 such that (20)v112+v2121/2⩽C0v1,v=v1+v2,v1∈V1,v2∈V2.Simultaneously, (21)v12=av,v1+av,v2=aPV1v,v1+aPV2v,v2⩽PV1v12+PV2v121/2v112+v2121/2⩽C0v1PV1v12+PV2v121/2.Theorem 3.
Suppose thatV1⊥∩V2⊥={0}; then(22)limn→+∞e12n+11=0,limn→+∞e22n1=0.Proof.
Following (15), we have(23)e12n+11⩾e22n1⩾e12n+31⩾e22n+21⩾0, which means that(24)limn→+∞e12n+1-e22n1=0.Let {e1(2n+1)} and {e2(2n)} be weakly convergent subsequence, so (25)limn→+∞e12n+1=v1,limn→+∞e22n=v2. Following (24), we can obtain (26)v1=v2=0. Following (17), we have (27)ae12n+1,e12n+1=ae12n-1,e12n+3=⋯=ae14n+1,e11⟶0,ae22n,e22n=ae22n-2,e22n+2=⋯=ae24n,e20⟶0; namely (28)limn→+∞e12n+11=0,limn→+∞e22n1=0.Theorem 4.
There exists a constantα∈[0,1), such that(29)PV1⊥PV2⊥⩽α,PV2⊥PV1⊥⩽α,e12n+11⩽α2ne111,e22n+21⩽α2n+2e201hold true.Proof.
SubstitutingPV1⊥v into (19), we have (30)PV1⊥v12=PV2⊥PV1⊥v12+PV2PV1⊥v12⩾PV2⊥PV1⊥v12+1C02PV1⊥v12. Equivalently (31)PV2⊥PV1⊥v12⩽1-1C02PV1⊥v12⩽αv12,∀v∈V, where (32)α=1-1C02∈0,1; namely (33)PV2⊥PV1⊥⩽α. Similarly (34)PV1⊥PV2⊥⩽α. Following (17) and (18), it comes that (35)e12n+1⩽αPV1⊥e2n-11=αPV1⊥PV2⊥e2n-21⩽α2PV2⊥e2n-21=α2e2n-11.By recursion we obtain (36)e12n+11⩽α2ne111. Similarly, we obtain (37)e22n+21⩽α2n+2e201.


## 4. Analysis of Convergence Rate

Absolutely, the convergence rate of the above Schwartz alternating algorithm is closely related to the overlapping degree ofΩ1 and Ω2. Although it can be deduced intuitively that the larger the overlapping part is, the faster the convergence rate will be; yet we find it difficult to analyse the convergence rate for general unbounded domain Ωc. However, under certain assumptions, we can find out the relationship between contraction factor α and overlapping degree of Ω1 and Ω2.Consider the following boundary value problem in domainΩ1:(38)-Δu-k2u=0,inΩ1,u=f,onΓ1,u=g,onΓ0.Lemma 5.
Iff∈H1/2(Γ1) and g∈H1/2(Γ0), then u is the solution of (38).(39)u=∑l=0∞∑m=-ll1R12l+1-R2l+1rl+1R12l+1-r2l+1Rl+1hlkrhlkRGlm+r2l+1-R2l+1R1l+1hlkrhlkR1FlmYlmθ,φ,where(40)Flm=∫02π∫0πfR1,θ,φYlmθ,φ¯sin⁡θdθdφ,Glm=∫02π∫0πgR,θ,φYlmθ,φ¯sin⁡θdθdφ.Theorem 6.
If we apply the Schwartz alternating algorithm given in Section3 to problems (1)-(2) and u0∈H1/2(Γ0), then (41)u-u12nH1Ω1⩽C1αn,u-u22n+1W01Ω2⩽C2αn+1 hold true, where constants C1 and C2 are independent of n, while (42)α=R2-RR1-R.Proof.
On one hand, foru0∈H1/2(Γ0), we have (43)u0R,θ,φ=∑l=0∞∑m=-llhlkRhlkRUlmYlmθ,φ=∑l=0∞∑m=-llUlmYlmθ,φ,∑l=0∞∑m=-ll1+l21/2Ulm2<+∞,where (44)Ulm=∫02π∫0πu0R,θ,φYlmθ,φ¯sin⁡θdθdφ. Following [9], we obtain(45)ur,θ,φ=∑l=0∞∑m=-llhlkrhlkRUlmYlmθ,φ,r>R.On the other hand, for u~∈W01(Ωc), we have u~Γ1∈H1/2(Γ1), while (46)u~Γ1=∑l=0∞∑m=-llU~lmYlmθ,φ,∑l=0∞∑m=-ll1+l2-1/2U~lm2<+∞,where (47)U~lm=∫02π∫0πu~R1,θ,φYlmθ,φ¯sin⁡θdθdφ. According to Lemma 5, it comes that(48)u12n=∑l=0∞∑m=-llRl+1Ulmrl+1-αlnr2l+1-R2l+1R12l+1-R2l+1·Rl+1Ulm-R1l+1U~lmrl+1hlkrhlkRYlmθ,φ,where (49)αl=R22l+1-R2l+1R12l+1-R2l+1.Suppose that n∈Z-¯; we have(50)u12nΓ2=∑l=0∞∑m=-llRl+1UlmR2l+1-αln+1Rl+1Ulm-R1l+1U~lmR2l+1hlkR2hlkRYlmθ,φ,u22n+1=∑l=0∞∑m=-llRl+1UlmR2l+1-αln+1·Rl+1Ulm-R1l+1U~lmR2l+1R2l+1rl+1hlkrhlkRYlmθ,φ.Following trace theorem, we have(51)u22n+1Γ1=∑l=0∞∑m=-llRl+1UlmR1l+1-αln+1Rl+1Ulm-R1l+1U~lmR1l+1hlkR1hlkRYlmθ,φ.By Lemma 5 again, we obtain(52)u12n+2=∑l=0∞∑m=-llRl+1Ulmrl+1-αln+1r2l+1-R2l+1R12l+1-R2l+1·Rl+1Ulm-R1l+1U~lmrl+1R2l+1rl+1hlkrhlkRYlmθ,φ.By induction, it comes that(53)uΩ1-u12n=∑l=0∞∑m=-llαlnr2l+1-R2l+1R12l+1-R2l+1·Rl+1Ulm-R1l+1U~lmrl+1R2l+1rl+1hlkrhlkR1Ylmθ,φ.Denoting(54)eN2n=∑l=0N∑m=-llαlnr2l+1-R2l+1R12l+1-R2l+1Rl+1Ulm-R1l+1U~lmrl+1R2l+1rl+1hlkrhlkR1Ylmθ,φ,(53) is changed into(55)limN→+∞eN2n-u-u12nH1Ω1=0.Clearly(56)ΔeN2n=0,noting that(57)0<αl=R22l+1-R2l+1R12l+1-R2l+1⩽R2-RR1-R=α,l=0,1,….Following (56) and (57), there exists C1 independent of n, such that (58)eN2nH1Ω1⩽C1αn. Similarly, we obtain (59)eN2nW01Ω2⩽C2αn+1.It can be seen from Theorem 6 that the larger the overlapping part of Ω1 and Ω2 is, the smaller the contraction factor α is and, consequently, the faster the Schwartz alternating algorithm converges.


## 5. Error Estimates of Algorithms

SubdivideΩ1 into hexahedrons. Let Sh(Ω1) denote the linear finite element space over Ω1. Putting (60)S°hΩ1=vh∣vh∈ShΩ1,vhΓ0∪Γ1=0,S°h(Ω1) can be regarded as the subspace of V if its elements are extended by zero. We first establish the following discrete Schwartz alternating algorithm:(61)Findw1h2n+1=u1h2n+1-uh2n∈S°hΩ1,suchthataw1h2n+1,vh=-auh2n,vh,∀vh∈S°hΩ1,(62)Findw2h2n+2=u2h2n+2-uh2n+1∈V2,suchthataw2h2n+2,v=-auh2n+1,v,∀vh∈V2,with(63)uh2n+1=u1h2n+1,inΩ1,uh2n,inΩ∖Ω1,uh2n+2=uh2n+1,inΩ∖Ω2,u2h2n+2,inΩ2,uh0=u~.By the Poisson integral formula, the solution of (62) can be given as(64)u2h2n+2=Pγ2uh2n+1,where γ2:H1(Ω1)→H1/2(Γ2) denotes the Dirichlet trace operator and P:H1/2(Γ2)→H01(Ω2) denotes the Poisson integral operator. It is easy to verify that (65)uh2n+1=u~+∑i=0nw1h2i+1,onΩ¯∖Ω2,∑i=0nw1h2i+1+∑j=0n-1Pγw1h2j+1-w1h2j+1+εnPγu~-u~,inΩ1∖Ω¯∖Ω2,∑j=0n-1Pγw1h2j+1+εnPγu~-u~,onΩ∖Ω1,uh2n+2=u~+∑i=0nw1h2i+1,onΩ¯∖Ω2,∑i=0nw1h2i+1+∑j=0n-1Pγw1h2j+1-w1h2j+1+Pγu~-u~,inΩ1∖Ω¯∖Ω2,∑j=0n-1Pγw1h2j+1+Pγu~-u~,onΩ∖Ω1,εn=0,n=0,1,n>0, and the term corresponding to ∑j=0n-1‍ vanishes while n=0. Define (66)AhΩ2=Pγvh+β1u~+β2w-vh+β1u~+β2wΩ¯2∣vh∈S°hΩ1,β1,β2∈R,w=u-u~. Then, extending the elements of Ah(Ω2) by zero, we have (67)AhΩ2⊂V2⊂V. Now we introduce the following variational problem:(68)Findvh∗∈S°hΩ1+AhΩ2,suchthatavh∗,vh=-au~,vh,∀vh∈S°hΩ1+AhΩ2.Obviously, (68) exists as a unique solution. Assume that(69)uh∗=vh∗+u~. Similarly in [10], we have the following error estimates.Theorem 7.
For the discrete Schwartz alternating algorithms (61)-(62) and the constant α in Theorem 4, there hold the following error estimates(70)u-uh2n+11⩽Ch+α2nuh∗-uh11,u-uh2n+21⩽Ch+α2n+2uh∗-uh01.
Sinceu-uh(n)∈V, (70) also hold true if the norm in their left hand sides is replaced by the norm of H01(Ω).


## 6. Numerical Examples

To test the effectiveness of the method in this paper, we give some numerical examples, using the discrete Schwartz alternating algorithm in Section5. In all examples, the exact solutions are known. The purpose of these examples is to check the convergence in terms of iteration and mesh size h.Suppose thatΩc={r,θ,φ∣r>a,0⩽θ⩽π,0⩽φ⩽2π} is the exterior domain of sphere x2+y2+z2=a2. The artificial boundary is Γ1={(r,θ,φ)∣r=ma,0⩽θ⩽π,0⩽φ⩽2π}(m>1,m∈Z+), Γ2={(r,θ,φ)∣r=R2,0⩽θ⩽π,0⩽φ⩽2π}. Split the bounded domain Ω1 into finite element mesh as follows. We subdivide the bounded domain Ω1 using the following family of planes θ=π/N1j(j=0,1,…,N1), φl=2π/N2l(l=0,1,…,N2), and r=ia/s(i=s,s+1,…,ms), where s is a positive integer. Now we obtain a uniform hexahedrons partition of Ω1 with mesh size of h=a/s. Let Sh(Ω1) be the piecewise trilinear finite element space on Ω1. Substitute (71)u~I=∑Pi∈∑0g0PiLix for u~ (where Li(x) denotes the interpolation basis function corresponding to node Pi).Denotee the maximum node–error on Ω¯i(72)en=supPi∈Ω¯iuPi-u1h2nPi;eh denotes the maximum node-error of the adjacent two-steps on nodes (73)ehn=supPi∈Ω¯iu1h2n+2Pi-u1h2nPi;qh is the convergence rate qh(n)=eh(n-1)/eh(n). We substitute ∑l=0M‍ for ∑l=0∞‍ in the computing of the entries of stiffness matrix.Example 1.
We take that the exact solution of problem (1) is (74)ur,θ,φ=-34πkr+ik2r2eikrcos⁡θ,where(75)g=-34πk+ik2eikcos⁡θ.Letk=2.0, a=1, m=2, R2=1.75, and M=4; the numerical results are shown in Tables 1 and 2.Table 1
( a , R 2 , m ) = ( 1,1.75,2 ).
N 1 N 2 s n 0 1 2 3 4 e ( n ) 2.5231 × 1 0 - 1 7.2245 × 1 0 - 2 6.1215 × 1 0 - 2 4.2233 × 1 0 - 2 4.0159 × 1 0 - 2 4 8 2 e h ( n )  9.8566 × 1 0 - 2 2.6452 × 1 0 - 2 6.1205 × 1 0 - 3 1.7354 × 1 0 - 3 q h ( n )   3.7270 3.6219 3.5267  e ( n ) 2.1628 × 1 0 - 1 6.8132 × 1 0 - 2 4.2354 × 1 0 - 2 2.1532 × 1 0 - 2 2.1533 × 1 0 - 2 8 16 4 e h ( n )  1.5982 × 1 0 - 1 4.5357 × 1 0 - 2 1.2986 × 1 0 - 2 3.7281 × 1 0 - 3 q h ( n )   3.5236 3.4926 3.4432  e ( n ) 1.8437 × 1 0 - 1 3.5468 × 1 0 - 2 2.3459 × 1 0 - 2 2.0245 × 1 0 - 2 1.7342 × 1 0 - 2 16 32 8 e h ( n )  1.2183 × 1 0 - 1 3.7647 × 1 0 - 2 1.1743 × 1 0 - 2 3.6502 × 1 0 - 3 q h ( n )   3.2361 3.2058 3.2172Table 2
( a , R 2 , m ) = ( 1,1.75,2 ).
N 1 N 2 s n 5 6 7 8 9 e ( n ) 4.0029 × 1 0 - 2 3.9021 × 1 0 - 2 3.9002 × 1 0 - 2 3.8876 × 1 0 - 2 3.8651 × 1 0 - 2 4 8 2 e h ( n ) 5.2202 × 10−4 1.7305 × 1 0 - 4 5.7476 × 1 0 - 5 1.9094 × 1 0 - 5 6.2964 × 1 0 - 6 q h ( n ) 3.3245 3.0166 3.0108 3.0102 3.0326  e ( n ) 2.0522 × 1 0 - 2 2.0614 × 1 0 - 2 1.8631 × 1 0 - 2 1.7532 × 1 0 - 2 1.7092 × 1 0 - 2 8 16 4 e h ( n ) 1.1646 × 1 0 - 3 3.7540 × 1 0 - 4 1.2175 × 1 0 - 4 3.9789 × 1 0 - 5 1.3206 × 1 0 - 5 q h ( n ) 3.2012 3.1823 3.0832 3.0601 3.0129  e ( n ) 6.9625 × 1 0 - 3 3.8703 × 1 0 - 3 2.7213 × 1 0 - 3 2.8173 × 1 0 - 3 2.6785 × 1 0 - 3 16 32 8 e h ( n ) 1.1663 × 1 0 - 3 3.7108 × 1 0 - 4 1.2291 × 1 0 - 4 4.0792 × 1 0 - 5 1.3538 × 1 0 - 5 q h ( n ) 3.1295 3.1432 3.0190 3.0132 3.0130Example 2.
We take that the exact solution of problem (1) is (76)ur,θ,φ=-1532π3ik3r3+3k2r2-ikreikr+φsin⁡2θ,where (77)g=-1532π3i+3k-ik2k3eik+φsin⁡2θ.Letk=2.0, a=1, m=2, R2=1.85, and M=4; the numerical results are shown in Tables 3 and 4.Table 3
( a , R 2 , m ) = ( 1,1.85,2 ).
N 1 N 2 s n 0 1 2 3 4 e ( n ) 2.5231 × 1 0 - 1 5.2541 × 1 0 - 2 5.0562 × 1 0 - 2 4.5623 × 1 0 - 2 3.8128 × 1 0 - 2 4 8 2 e h ( n )  7.5434 × 1 0 - 1 2.1411 × 1 0 - 2 6.6464 × 1 0 - 3 2.1131 × 1 0 - 3 q h ( n )   3.5231 3.2215 3.1452  e ( n ) 2.1628 × 1 0 - 1 6.2315 × 1 0 - 2 5.1532 × 1 0 - 2 3.8321 × 1 0 - 2 2.2532 × 1 0 - 2 8 16 4 e h ( n )  1.4526 × 1 0 - 1 4.4753 × 1 0 - 2 1.4856 × 1 0 - 2 5.0306 × 1 0 - 3 q h ( n )   3.2458 3.0124 2.9532  e ( n ) 1.8437 × 1 0 - 1 2.4623 × 1 0 - 2 2.3245 × 1 0 - 2 2.1244 × 1 0 - 2 2.0132 × 1 0 - 2 16 32 8 e h ( n )  1.0157 × 1 0 - 1 3.6381 × 1 0 - 2 1.4196 × 1 0 - 2 5.7773 × 1 0 - 3 q h ( n )   2.7918 2.5628 2.4572Table 4
( a , R 2 , m ) = ( 1,1.85,2 ).
N 1 N 2 s n 5 6 7 8 9 e ( n ) 3.7170 × 1 0 - 2 3.5319 × 1 0 - 2 3.4321 × 1 0 - 2 3.2112 × 1 0 - 2 2.9518 × 1 0 - 2 4 8 2 e h ( n ) 2.3888 × 1 0 - 4 9.6127 × 1 0 - 5 4.1760 × 1 0 - 5 1.8519 × 1 0 - 5 8.2215 × 1 0 - 6 q h ( n ) 2.5621 2.4851 2.3019 2.2549 2.2526  e ( n ) 2.0213 × 1 0 - 2 1.9526 × 1 0 - 2 1.9432 × 1 0 - 2 1.9542 × 1 0 - 2 1.8926 × 1 0 - 2 8 16 4 e h ( n ) 1.3314 × 1 0 - 3 4.8891 × 1 0 - 4 2.1574 × 1 0 - 4 9.5621 × 1 0 - 5 4.2462 × 1 0 - 5 q h ( n ) 2.8001 2.7232 2.2662 2.2572 2.2519  e ( n ) 1.9524 × 1 0 - 2 9.5324 × 1 0 - 3 8.4526 × 1 0 - 3 8.0219 × 1 0 - 3 7.8243 × 1 0 - 3 16 32 8 e h ( n ) 1.4861 × 1 0 - 3 6.4006 × 1 0 - 4 2.8198 × 1 0 - 4 1.2506 × 1 0 - 4 5.5502 × 1 0 - 5 q h ( n ) 2.4562 2.3218 2.2699 2.2548 2.2532As can be seen from Tables3 and 4, the discrete Schwartz alternating algorithm is geometrically convergent and qh(n) is nearly not affected by mesh parameter h. With the increase of R2, qh(n) is smaller. So the larger the overlapping part is, the faster the convergence rate will be. All these are in accord with the theoretical analyses.



---

*Source: 6029484-2018-05-06.xml*