# A Nonoverlapping Domain Decomposition Method for an Exterior Anisotropic Quasilinear Elliptic Problem in Elongated Domains

Authors: Baoqing Liu; Qikui Du
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

Based on the Kirchhoff transformation, a nonoverlapping domain decomposition method is discussed for solving exterior anisotropic quasilinear problems with elliptic artificial boundary. By the principle of the natural boundary reduction, we obtain the natural integral equation for the anisotropic quasilinear problem on elliptic artificial boundaries and construct the algorithm and analyze its convergence. Moreover, we give the existence and uniqueness result for the original problem. Finally, some numerical examples are presented to illustrate the feasibility of the method.


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## Body

## 1. Introduction

When solving a problem modelled by a linear or nonlinear partial differential equation in the bounded or unbounded domain, domain decomposition methods are one of the most efficient techniques. one can refer to [1–4] and references therein for more details. Based on natural boundary reduction [5, 6], the overlapping and nonoverlapping domain decomposition methods can be viewed as effective ways to solve the problems in the unbounded domains. These techniques have been used to solve many linear problems [6–8], and they have also been generalized to linear or nonlinear wave problems [5, 9, 10]. In this paper, we consider a nonoverlapping domain decomposition method for an exterior anisotropic quasilinear elliptic problem with elliptical artificial boundary. By the Kirchhoff transformation, we will discuss some exterior anisotropic quasilinear elliptic problems [11–15] by virtue of the nonoverlapping domain decomposition method.LetΩ be an elongated, bounded, and simple connected domain in ℝ2 with sufficiently smooth boundary ∂Ω=Γ0, Ωc=ℝ2/Ω-. We consider the numerical solution to the exterior anisotropic quasilinear problem
(1)-(∂∂x(αa(x,u)∂u∂x)+∂∂y(βa(x,u)∂u∂y))=f(x),-(∂∂x(αa(x,u)∂u∂x)+∂∂y(βa(x,u)∂u∂y))=inΩc,u=0,onΓ0,(2)u(x)=𝒪(1),as|x|→∞,
with β>α>0 or α=β=1, x=(x,y), a(·,·), and f(x) are the given functions which will be ranked as below.Problem (1) has many physical applications in, for example, the field of heat transfer, where a is the thermal conductivity of the medium and u is the temperature field; the field of compressible flow, where a is the density and u is the velocity potential. Problem (1) can also describe a temperature distribution in large transformers whose magnetic cores (consisting of iron tins) are nonlinear anisotropic media where a is the heat conductivity. One can also refer to [11–14, 16] for more details.Following [11, 13], suppose that the given function a(·,·) satisfies
(3)0<C0≤a(x,u)≤C1,∀u∈ℝ,and for almost allx∈Ωc,
Where the two constants C0,C1∈ℝ. In the following, we suppose that the function f(x)∈L2(Ωc) has compact support, that is, there exists a constant  γ0>0, such that
(4)suppf⊂Ωγ0={x∈ℝ2∣|x|≤γ0}.We also assume that(5)a(x,u)=˙a0(u),when|x|≥γ0.
Now, we introduce an elliptical artificial boundary Γ1 enclosing Γ0 such that
(6)Γ1={(μ,φ)∣μ=μ1>γ0,0≤φ≤2π},and dist(Γ1,Γ0)=δ0>0.Then,Ωc is divided into two nonoverlapping subdomains Ω1 and Ω2 (see Figure 1), where Ω1 denotes the bounded domain between Γ0 and Γ1, and Ω2 refers to the unbounded domain outside Γ1. The original problem (1) is decomposed into two subproblems in domains Ω1 and Ω2 with Ω1∩Ω2=∅. We have the nonoverlapping domain decomposition algorithm.Figure 1
The illustration of domainsΩ1 and Ω2.Step 1.
Choose an initial valueλ0∈H1/2(Γ1), and put k=0.Step 2.
Solve a Dirichlet boundary value problem in the exterior domainΩ2:
(7)-(∂∂x(αa(x,u2k)∂u2k∂x)+∂∂y(βa(x,u2k)∂u2k∂y))=0,inΩ2,u2k=λk,onΓ1,u2k(x)=𝒪(1),as|x|→∞.Step 3.
Solve a mixed boundary value problem in the interior domainΩ1:
(8)-(∂∂x(αa(x,u1k)∂u1k∂x)+∂∂y(βa(x,u1k)∂u1k∂y))=f(x),-(∂∂x(αa(x,u1k)∂u1k∂x)+∂∂y(βa(x,u1k)∂u1k∂y))inΩ1,∂u1k∂n1=-∂u2k∂n2,onΓ1,u1k=0,onΓ0.Step 4.
Update the boundary value0<θk<1,
(9)λk+1=θku1k+(1-θk)λk,onΓ1.Step 5.
Putk=k+1, turn to Step 2.
The relaxation factorθk is a suitably chosen real number. Notice that, in Step 3, we solve problem (8) by the standard finite element method and only need the normal derivative of the solution to problem (7) in Step 2. So, we need not to solve (7) directly. Based on the Kirchhoff transformation, the natural integral equation for the quasilinear problem can be obtained by the natural boundary reduction [6, 17, 18]. Particularly, when a(x,u)=c which is independent of x and u, [6, 8, 19] have discussed the corresponding problems by this technique. Now, we introduce the so-called Kirchhoff transformation [20]
(10)w(x)=∫0u(x)‍a(ξ)dξ,x∈Ωc.
Then, we have
(11)∇w=a(u)∇u,(12)(α∂w∂x,β∂w∂y)=(αa(u)∂u∂x,βa(u)∂u∂y).
From (7), we have that w satisfies the following problem
(13)-(α∂2wk∂x2+β∂2wk∂y2)=0,inΩ2,wk=∫0λk‍a0(ξ)dξ,onΓ1.
The rest of the paper is organized as follows. In Section 2, we obtain the natural integral equation for the quasilinear problem in the elliptical unbounded domain. In Section 3, we state the nonoverlapping domain decomposition method and discuss the convergence of the algorithm. We also show the existence and uniqueness of the original problem. At last, in Section 4, we present some numerical examples to illustrate the efficiency and feasibility of our method.


## 2. Natural Boundary Reduction

In this section, by virtue of the Poisson integral formula and natural integral equation for the linear problem, we will obtain the corresponding results for the quasilinear problem inΩ2. For this purpose, we need to discuss some properties between elliptic coordinates (μ,φ) and Cartesian coordinates (x,y) first. The relationship between the two coordinates can be expressed as below
(14)x=f0coshμcosφ,y=f0sinhμsinφ,
with f0=a2-b2, a=f0coshμ1, and b=f0sinhμ1. Following from [18], we have the following.Theorem 1.
The transformation between elliptic coordinates and Cartesian coordinates in (14) possesses the following property.(1)
The Jacobian determinant of (14) is
(15)J=f02cosh2μsin2φ+f02sinh2μcos2φ=f02(cosh2μ-cos2φ),J=0 if and only if (x,y)=(±f0,0).(2)
Foru∈C2(ℝ2), the following holds
(16)∂2u∂μ2+∂2u∂φ2=J(∂2u∂x2+∂2u∂y2).(3)
For the exterior domainΩ2, we have
(17)∂u∂ν=-1J∂u∂μ,where ν refers to the unit exterior normal vector on Γμ1(regarded as the inner boundary of Ω2).Proof.
The conclusions 1 and 2 can be obtained by direct computation. And 3 follows from the property(18)ν=-1J(f0sinhμcosφ,f0coshμsinφ).


### 2.1. Natural Integral Equation forα=β=1

Assume thatw(x) is the solution to problem (13) and the value w||μ|=μ1 is given, that is,
(19)w||μ|=μ1=w0(φ).
Then, based on the natural boundary reduction, there are the Poisson integral formulas
(20)w(μ,φ)=e2μ-e2μ12π×∫02π‍w0(φ′)e2μ+e2μ1-2eμ+μ1cos(φ-φ′)dφ′,μ>μ1,
or
(21)w(μ,φ)=1π∑j=1∞‍ej(μ1-μ)×∫02π‍cosj(φ-φ′)w0(φ′)+12π∫02π‍w0(φ′)dφ′,μ>μ1,
and the natural integral equation
(22)∂w∂n=1J0[-14πsin2(φ/2)*w0(φ)],μ=μ1,
or
(23)∂w∂n=1πJ0∑j=1∞‍j×∫02π‍cosj(φ-φ′)w0(μ1,φ′)dφ,μ=μ1,
the definition of J0 can be found in the following. The Poisson integral formulas (20)-(21) and the natural integral equation (22)-(23) can also be expressed in the Fourier series forms
(24)w(μ,φ)=∑j=-∞+∞‍aje|j|(μ1-μ)+ijφ,μ>μ1,(25)∂w∂n=1J0∑j=-∞+∞‍|j|ajeijφ,μ=μ1,
with aj=(1/2π)∫02π‍w0(φ′)e-ijφ′dφ′, J0=f02(cosh2μ1sin2φ+sinh2μ1cos2φ)=f02(cosh2μ1-cos2φ), i=-1.From (12), we obtain
(26)∂w∂n=a0(u)∂u∂n.
Combining (11), (25), and (26), we get the exact artificial boundary condition of u on Γμ1(27)a0(u)∂u(μ,φ)∂n|μ=μ1=1J0∑j=-∞+∞‍|j|ajeijφ=1πJ0∑j=1+∞‍j∫02π‍cosj(φ-φ′)(∫0u(μ1,φ)‍a0(y)dy)dφ=˙𝒦1(u(μ1,φ)),
with aj=(1/2π)∫02π‍(∫0u(μ1,φ′)‍a0(y)dy)e-ijφ′dφ′, J0=f02(cosh2μ1-cos2φ), i=-1.


### 2.2. Natural Integral Equation forβ>α>0

Now, we assume thatΓμ1 can be expressed in the form: Γμ1={(x,y)|px2+qy2=R2}, with βq>αp>0. We also assume that w(x) is the solution to problem (13) and the value w||μ|=μ1 is given, namely,
(28)w||μ|=μ1=w0(φ).
Let x=αξ and y=βη. Then, the boundary Γμ1 is changed by the elliptic boundary Γ~1={(ξ,η)|αpξ2+βqη2=R2}; the unit exterior normal vector on Γ~1 is
(29)ν=-(αpcosθ,βqsinθ)αpcos2θ+βqsin2θ.
By the above transformation, problem (13) changes into
(30)-(∂2w∂ξ2+∂2w∂η2)=0,inΩ~2,(31)w=w0,onΓ~1.This is the right problem we talked in Section2.1. Similar with (14), we let
(32)ξ=f0coshμcosφ,η=f0sinhμsinφ,
with
(33)f0=βq-αpαpβqR,μ0=ln(βq+αpβq-αp),Γ~1={(μ,φ)∣μ=μ0,φ∈[0,2π]},Ω~2={(μ,φ)∣μ>μ0,φ∈[0,2π]}.
Then, just the same as the problem discussed in Section 2.1, we have the natural integral equation on Γμ1(34)αnx∂w∂x+βny∂w∂y=-αpβqpcos2φ+qsin2φ[14πRsin2(φ/2)*w0(φ)],
where (nx,ny)=(x/R,y/R) is the unit exterior normal vector on Γμ1. From (13), we obtain
(35)αnx∂w∂x+βny∂w∂y=αnxa0(u)∂u∂x+βnya0(u)∂u∂y.
Combining (11), (34), and (35), we obtain the exact artificial boundary condition of u on Γμ1(36)(αnxa0(u)∂u∂x+βnya0(u)∂u∂y)|μ=μ1=-αpβqpcos2φ+qsin2φ×[14πRsin2(φ/2)*(∫0u(μ1,φ)‍a0(y)dy)]=αpβqpcos2φ+qsin2φ∑j=-∞+∞‍j2πR×∫02π‍cosj(φ-φ′)(∫0u(μ1,φ)‍a0(y)dy)dφ=˙𝒦1(u(μ1,φ)).


## 2.1. Natural Integral Equation forα=β=1

Assume thatw(x) is the solution to problem (13) and the value w||μ|=μ1 is given, that is,
(19)w||μ|=μ1=w0(φ).
Then, based on the natural boundary reduction, there are the Poisson integral formulas
(20)w(μ,φ)=e2μ-e2μ12π×∫02π‍w0(φ′)e2μ+e2μ1-2eμ+μ1cos(φ-φ′)dφ′,μ>μ1,
or
(21)w(μ,φ)=1π∑j=1∞‍ej(μ1-μ)×∫02π‍cosj(φ-φ′)w0(φ′)+12π∫02π‍w0(φ′)dφ′,μ>μ1,
and the natural integral equation
(22)∂w∂n=1J0[-14πsin2(φ/2)*w0(φ)],μ=μ1,
or
(23)∂w∂n=1πJ0∑j=1∞‍j×∫02π‍cosj(φ-φ′)w0(μ1,φ′)dφ,μ=μ1,
the definition of J0 can be found in the following. The Poisson integral formulas (20)-(21) and the natural integral equation (22)-(23) can also be expressed in the Fourier series forms
(24)w(μ,φ)=∑j=-∞+∞‍aje|j|(μ1-μ)+ijφ,μ>μ1,(25)∂w∂n=1J0∑j=-∞+∞‍|j|ajeijφ,μ=μ1,
with aj=(1/2π)∫02π‍w0(φ′)e-ijφ′dφ′, J0=f02(cosh2μ1sin2φ+sinh2μ1cos2φ)=f02(cosh2μ1-cos2φ), i=-1.From (12), we obtain
(26)∂w∂n=a0(u)∂u∂n.
Combining (11), (25), and (26), we get the exact artificial boundary condition of u on Γμ1(27)a0(u)∂u(μ,φ)∂n|μ=μ1=1J0∑j=-∞+∞‍|j|ajeijφ=1πJ0∑j=1+∞‍j∫02π‍cosj(φ-φ′)(∫0u(μ1,φ)‍a0(y)dy)dφ=˙𝒦1(u(μ1,φ)),
with aj=(1/2π)∫02π‍(∫0u(μ1,φ′)‍a0(y)dy)e-ijφ′dφ′, J0=f02(cosh2μ1-cos2φ), i=-1.


## 2.2. Natural Integral Equation forβ>α>0

Now, we assume thatΓμ1 can be expressed in the form: Γμ1={(x,y)|px2+qy2=R2}, with βq>αp>0. We also assume that w(x) is the solution to problem (13) and the value w||μ|=μ1 is given, namely,
(28)w||μ|=μ1=w0(φ).
Let x=αξ and y=βη. Then, the boundary Γμ1 is changed by the elliptic boundary Γ~1={(ξ,η)|αpξ2+βqη2=R2}; the unit exterior normal vector on Γ~1 is
(29)ν=-(αpcosθ,βqsinθ)αpcos2θ+βqsin2θ.
By the above transformation, problem (13) changes into
(30)-(∂2w∂ξ2+∂2w∂η2)=0,inΩ~2,(31)w=w0,onΓ~1.This is the right problem we talked in Section2.1. Similar with (14), we let
(32)ξ=f0coshμcosφ,η=f0sinhμsinφ,
with
(33)f0=βq-αpαpβqR,μ0=ln(βq+αpβq-αp),Γ~1={(μ,φ)∣μ=μ0,φ∈[0,2π]},Ω~2={(μ,φ)∣μ>μ0,φ∈[0,2π]}.
Then, just the same as the problem discussed in Section 2.1, we have the natural integral equation on Γμ1(34)αnx∂w∂x+βny∂w∂y=-αpβqpcos2φ+qsin2φ[14πRsin2(φ/2)*w0(φ)],
where (nx,ny)=(x/R,y/R) is the unit exterior normal vector on Γμ1. From (13), we obtain
(35)αnx∂w∂x+βny∂w∂y=αnxa0(u)∂u∂x+βnya0(u)∂u∂y.
Combining (11), (34), and (35), we obtain the exact artificial boundary condition of u on Γμ1(36)(αnxa0(u)∂u∂x+βnya0(u)∂u∂y)|μ=μ1=-αpβqpcos2φ+qsin2φ×[14πRsin2(φ/2)*(∫0u(μ1,φ)‍a0(y)dy)]=αpβqpcos2φ+qsin2φ∑j=-∞+∞‍j2πR×∫02π‍cosj(φ-φ′)(∫0u(μ1,φ)‍a0(y)dy)dφ=˙𝒦1(u(μ1,φ)).


## 3. Variational Problem and Convergence Analysis of the Algorithm

### 3.1. The Equivalent Variational Problem

Now, we consider problem (8). We will use Wm,p denoting the standard Sobolev spaces. ∥·∥ and |·| refer to the corresponding norms and seminorms. Especially, we define Hm(Ω)=Wm,2(Ω), ∥·∥m,Ω=∥·∥m,2,Ω, and |·|m,Ω=|·|m,2,Ω. Let us introduce the space
(37)V={v∈H1(Ω1)|v|Γ0=0},
and the corresponding norms
(38)∥v∥0,Ω1=∫Ω1‍|v|2dx,∥v∥1,Ω1=∫Ω1‍(|v|2+|∇v|2)dx.The boundary value problem (8) is equivalent to the following variational problem. 								(39)Findu∈V,such thatD(u;u,v)+D^(u;u,v)=F(v),∀v∈V,
with
(40)D(t;u,v)=∫Ω1‍a(x,t)(α∂u∂x∂v∂x+β∂u∂y∂v∂y)dx,(41)D^(t;u,v)=∑j=1+∞‍αβjπ×∫02π‍∫02π‍a0(t(μ1,φ′))×∂u(μ1,φ′)∂φ′∂v(μ1,φ)∂φ×cosj(φ′-φ)dφ′dφ,
where D^(t;u,v) follows from Green’s formula, (27) with ds=J0dφ, and (36) with ds=(R/pq)pcos2φ+qsin2φdφ.And,(42)F(v)=∫Ω1‍f(x)v(x)dx.


### 3.2. Convergence Analysis of the Method in Continuous Case

From (10)-(11), the original problem (1) can be changed to
(43)-(α∂2w∂x2+β∂2w∂y2)=f(x),inΩc,w=0,onΓ0,w(x)=𝒪(1), as|x|→∞.
Then, we let x=αξ and y=βη. Problem (43) becomes
(44)-Δw=-(∂2w∂ξ2+∂2w∂η2)=f(x),inΩ~c,w=0,onΓ~0,w(x)=𝒪(1),as|x|→∞,
where Ω~c and Γ~0 are the corresponding changes from Ωc and Γ0, respectively. Let g be extended to Ω~c, w=u-g, f=Δg. Then, problem (44) is equivalent to
(45)-Δu=0,inΩ~c,u=g,onΓ~0,u(x)=𝒪(1),as|x|→∞.Since it is difficult to estimate the convergence rate for a general unbounded domainΩc, we here let Ωc be an exterior domain of an ellipse Γ0, with μ=μ0, and Γ1 is taken as stated in Section 1. We introduce the following conclusion first.Lemma 2.
Ifu is the solution of
(46)-Δu=0,inΩ1,u=u0,onΓ0,∂u∂n=un,onΓ1,
where Ω1 is the elliptical ring domain between Γ0 and Γ1,
(47)u0=∑j=-∞+∞‍cjeijϕ∈H1/2(Γ0),un=1J0(∑-∞,j≠0+∞‍dj|j|eijϕ+d0),∈H-1/2(Γ1),
then there exist a unique u∈H1(Ω1) and
(48)u(μ,ϕ)=∑-∞,j≠0+∞‍({(e|j|(μ1-μ0)+e|j|(μ0-μ1))-1}(cj(e|j|(μ-μ1)+e|j|(μ1-μ))+dj×(e|j|(μ-μ0)-e|j|(μ0-μ)))×(e|j|(μ1-μ0)+e|j|(μ0-μ1))-1)×eijϕ+c0+d0(μ-μ0).Proof.
The result can be obtained directly from (46) by the separation of variables.Theorem 3.
If0<θk<1, then the nonoverlapping domain decomposition method (7)–(9) is convergent.Proof.
We assume that the exact solution to problem (1) is u, and we let λ=u|Γ1, uk=u|Ωk, and  k=1,2. Then, following (7)–(8), we let e1k=λ-u1k and e1k|Γ1=λ-λk=˙e2k. From (24), we assume that e2k=∑j=-∞+∞ajeijϕ∈H1/2(Γ1). By the natural integral equation (27), we have
(49)∂e1k∂n=-𝒦1(e2k)=-1J0∑j=-∞+∞‍|j|ajeijϕ.
So, e1k satisfies
(50)-Δe1k=0,inΩ1,e1k=0,onΓ0,∂e1k∂n=1J0∑j=-∞+∞‍|j|ajeijϕ,onΓ1.
By Lemma 2, one obtains
(51)e1k=-∑j=-∞+∞‍ajHj(μ)eijϕ,
with Hj(μ)=(e|j|(μ-μ0)-e|j|(μ0-μ))/(e|j|(μ1-μ0)+e|j|(μ0-μ1)). From (51), e1k confines on Γ1 can be expressed as
(52)e1k∣Γ1=-∑j=-∞+∞‍ajHj(μ1)eijϕ,𝒦1(e1k)=-1J0∑j=-∞+∞‍|j|ajHj(μ1)eijϕ.
Then, we have(53)∂e1k+1∂n=-𝒦1(λ-λk+1)=𝒦1(θku1k+(1-θk)λk-λ)=-θk𝒦1(e1k)-(1-θk)𝒦1(e2k)=1J0∑j=-∞+∞‍|j|aj(θkHj(μ1)-1+θk)eijϕ.
LetEm=˙∥∂e1k/∂n∥-1/2,Γ12, then Em=2π∑j=-∞+∞(j2/1+j2)|aj|2 and
(54)Em+1=2π∑j=-∞+∞‍j21+j2|aj|2(θkHj(μ1)-1+θk)2=(1-θk)2Em+2π∑j=-∞+∞‍j21+j2|aj|2×θkHj(μ1)[θk(Hj(μ1)+2)-2].
Assuming thatδ1=infj∈ℤ/{0}2/(2+Hj(μ1)), then 1>δ1≥2/3.
If0<θk≤δ1, k=0,1,2,…, then
(55)Em+1<(1-θk)2Em,
or equally
(56)Em+1<∏j=1m‍(1-θj)2E1≤rmE1,19≤r<1.
By the trace theorem, we have(57)∥e1k∥1,Ω12≤CEm→0,m→∞.
From (54), one also has
(58)Em+1=2π∑j=-∞+∞‍j21+j2|aj|2(θkHj(μ1)-1+θk)2=(1-2θk)2Em+2π∑j=-∞+∞‍j21+j2|aj|2×θkIj(μ1)[θk(Ij(μ1)-2)+1],
with Ij(μ1)=(1-Hj(μ1))/2. Assuming that δ2=supj∈ℤ/{0}1/(2-Ij(μ1)), then 0<δ2≤2/3.
Forδ2≤θk<1, k=0,1,2,…, the convergence result can be obtained similarly with (55)–(57). Therefore, for 0<θk<1, the nonoverlapping domain decomposition method is convergent.


### 3.3. Discrete Nonoverlapping Alternating Algorithm and Convergence Analysis

Divide the arcΓμ1 into M parts, and take a finite element subdivision in Ω1 such that their nodes on Γμ1 are coincident. That is, we make a regular and quasiuniform triangulation 𝒯h on Ω1, such that
(59)Ω1=⋃K∈𝒯h‍K,
where K is a (curved) triangle and h is the maximal diameter of triangles. Let
(60)Vh={vh∈V|v|Kis a linear polynomial,∀K∈𝒯h}.
Then, the approximate problem of (39) can be written as follows
(61)Finduh∈Vh,such  thatD(uh;uh,vh)+D^(uh;uh,vh)=F(vh),∀vh∈Vh,
with
(62)D(th;uh,vh)=∫Ω1‍a(x,th)(α∂uh∂x∂vh∂x+β∂uh∂y∂vh∂y)dx,(63)D^(th;uh,vh)=∑j=1+∞‍αβjπ∫02π‍∫02π‍a0(th(μ1,φ′))×∂uh(μ1,φ′)∂φ′∂vh(μ1,φ)∂φ·cosj(φ′-φ)dφ′dφ.
In practice, the sum of (63) is truncated to a finite number of terms N. By the hypothesis of a(·,·), it is not difficult to know the following result.Lemma 4.
There exists a constantC2>0 which has a different meaning in a different place and is related to α and β, such that
(64)|D(u;u,v)+D^(u;u,v)|≤C2∥u∥1,Ω1∥v∥1,Ω1,D(u;u,u)+D^(u;u,u)≥C0∥u∥1,Ω12,∀u,v∈V.Proof.
From (3) and (40), one can obtain that
(65)|D(u;u,v)|≤C2′∥u∥1,Ω1∥v∥1,Ω1,D(u;u,u)≥C0∥u∥1,Ω12,∀u,v∈V.
For the natural integral equation, one can obtain the following result
(66)|D^(u;u,v)|≤C2′′∥u∥1,Ω1∥v∥1,Ω1,D^(u;u,u)≥0,∀u,v∈V.
For this purpose, foru,v∈V, we assume that
(67)u(μ1,φ′)=∑j=-∞+∞‍ujeijφ′,v(μ1,φ)=∑j=-∞+∞‍vjeijφ,
with uj=(1/2π)∫02π‍u(μ1,φ′)e-ijφ′dφ′, vj=(1/2π)∫02π‍v(μ1,φ)e-ijφdφ. Then, we have
(68)∂u∂φ′(μ1,φ′)=∑j=-∞+∞‍ijujeijφ′,∂v∂φ(μ1,φ)=∑j=-∞+∞‍ijvjeijφ.
Combining property (3), Cauchy’s inequality, and the trace theorem, we have
(69)|D^(u;u,v)|≤C2′′(∑j=-∞+∞‍|j|·|uj|2)1/2×(∑j=-∞+∞‍|j|·|vj|2)1/2≤C2′′(∑j=-∞+∞‍(1+j2)1/2·|uj|2)1/2×(∑j=-∞+∞‍(1+j2)1/2·|vj|2)1/2≤C2′′∥u∥1/2,Γμ1∥v∥1/2,Γμ1≤C2′′∥u∥1,Ω1∥v∥1,Ω1,∀u,v∈V.
Next, we show thatD^(u;u,u)≥0, for any u∈V. For any given v∈V, let us consider the following auxiliary problem in Ω1:
(70)-∇·(a(x,u)∇u)=0,inΩ1,u=0,onΓ0,u=v,onΓμ1.
From the analysis in Section2.1, we know that the solution u to problem (70) satisfies
(71)a0(u)∂u∂n|Γμ1=𝒦1(u(μ1,φ))=𝒦1(v(μ1,φ)).
Multiplying (70) by u and integrating over Ω1, we have
(72)D^(u;u,u)=∫Ω1‍a0(u)|∇u|2dx≥0.
LettingC2=C2′+C2′′, we can get the desired result.We are now in the position to show the existence and uniqueness result for this type of problem (1). We begin with the following estimate.Lemma 5.
Any solutionu to problem (39) satisfies
(73)∥u∥1,Ω1≤C3∥f∥L2(Ω1),
with C3 a positive constant.Proof.
Takingv equal to u in (39), by Lemma 4, one has
(74)C0∥u∥1,Ω12≤D(u;u,u)+D^(u;u,u)=L(u).
Then, the desired result follows from Cauchy-Schwarz and Poincaré-Friedrichs inequalities.Theorem 6.
Problem (39) is uniquely solvable.Proof.
Since the spaceV is separable (indeed, it is a closed subspace of the space H1(Ω1) which is separable), there exist increasing sequences of finite-dimensional Hilbert subspaces Vm of V such that
(75)V=⋃m≥0‍Vm.
We define a mappingΦm:Vm→Vm by
(76)(Φm(u),v)=D(u;u,v)+D^(u;u,v)-L(v),∀u,v∈Vm,
where (·,·) stands for the scalar product on Vm×Vm. Since the function a(x,u) is bounded and the trace of the function in V on Γ1 belongs to H1/2(Γ1), hence at least to L4(Γ1) from the Sobolev embedding, each mapping Φm is well defined and continuous on Vm. What’s more, by Lemmas 4 and 5, we obtain
(77)(Φm(u),u)=D(u;u,u)+D^(u;u,u)-L(u)≥C0∥u∥1,Ω12-C3∥f∥L2(Ω1)∥u∥1,Ω1.
So, the right-hand side is nonnegative on the circle of radius μ which is defined by
(78)μ=C3C0∥f∥L2(Ω1).
Applying Brouwer’s fixed point theorem [21] yields the existence of um in Vm, with norm less than μ, such that
(79)Φm(um)=0.
Since the sequence {um}m is bounded by μ in V, there exists a subsequence which converges weakly to u in Vm. Using the compactness of the imbedding of H1/2(Γ1) into L3(Γ1), we obtain that u is a solution to problem (39).
Now, we show the uniqueness of the solution. Letu and u- be two solutions to this problem. Then, taking v equal to u and u- in (39), respectively, and combining with Lemma 4, one obtains
(80)C2∥u∥1,Ω1∥u-∥1,Ω1≥D(u-;u-,u)+D^(u-;u-,u)=D(u;u,u)+D^(u;u,u)⩾C0∥u∥1,Ω12,(81)C2∥u∥1,Ω1∥u-∥1,Ω1≥D(u;u,u-)+D^(u;u,u-)=D(u-;u-,u-)+D^(u-;u-,u-)⩾C0∥u-∥1,Ω12.
Since C2>C0, combining (80) with (81), we deduce the desired result.From the discrete problem (61), similar to the discussion for the linear problem in [6, 8], we can get a system of algebraic equations for our quasilinear problem with the following form:
(82)(A11+KhA12A21A22)(UV)=(0b),
where U is a vector whose components are function values at nodes on Γ1, and V is a vector whose components are function values at interior nodes of Ω1. The matrix A=˙A(u)=(A11A12A21A22) is the stiffness matrix obtained from finite element in Ω1, while Kh=˙Kh(u|Γ1) follows from the natural boundary element method on Γ1.Problem (82) can also be rewritten as follows:
(83)(A11A12A21A22)(UV)=(-KhUb).
Then, we have the iterative algorithm
(84)(A11A12A21A22)(UkVk)=(-KhΛkb),
with
(85)Λk+1=θkUk+(1-θk)Λk,k=0,1,….By condition (3), one obtains that A is a positive definite matrix, so we know that A22-1 exists. Now, we let Sh=Sh(1)+Kh be the discrete analogue of the Steklov-Poincaré operator on Γ1, with Sh(1)=A11-A12A22-1A21 and B=-A12A22-1b. Then, similar to the proof of Theorems 7.6 and  7.7 in [6], we have that the alternating algorithm (84)-(85) is equivalent to the preconditioned Richardson iteration:
(86)Sh(1)(Λk+1-Λk)=θk(B-ShΛk).And we also have the following convergence result.Theorem 7.
If0<minθk≤maxθk<1, then the discrete nonoverlapping alternating method (84)-(85) is convergent, and both the convergence rate and the condition number of [Sh(1)]-1Sh are independent of the mesh size h.


## 3.1. The Equivalent Variational Problem

Now, we consider problem (8). We will use Wm,p denoting the standard Sobolev spaces. ∥·∥ and |·| refer to the corresponding norms and seminorms. Especially, we define Hm(Ω)=Wm,2(Ω), ∥·∥m,Ω=∥·∥m,2,Ω, and |·|m,Ω=|·|m,2,Ω. Let us introduce the space
(37)V={v∈H1(Ω1)|v|Γ0=0},
and the corresponding norms
(38)∥v∥0,Ω1=∫Ω1‍|v|2dx,∥v∥1,Ω1=∫Ω1‍(|v|2+|∇v|2)dx.The boundary value problem (8) is equivalent to the following variational problem. 								(39)Findu∈V,such thatD(u;u,v)+D^(u;u,v)=F(v),∀v∈V,
with
(40)D(t;u,v)=∫Ω1‍a(x,t)(α∂u∂x∂v∂x+β∂u∂y∂v∂y)dx,(41)D^(t;u,v)=∑j=1+∞‍αβjπ×∫02π‍∫02π‍a0(t(μ1,φ′))×∂u(μ1,φ′)∂φ′∂v(μ1,φ)∂φ×cosj(φ′-φ)dφ′dφ,
where D^(t;u,v) follows from Green’s formula, (27) with ds=J0dφ, and (36) with ds=(R/pq)pcos2φ+qsin2φdφ.And,(42)F(v)=∫Ω1‍f(x)v(x)dx.


## 3.2. Convergence Analysis of the Method in Continuous Case

From (10)-(11), the original problem (1) can be changed to
(43)-(α∂2w∂x2+β∂2w∂y2)=f(x),inΩc,w=0,onΓ0,w(x)=𝒪(1), as|x|→∞.
Then, we let x=αξ and y=βη. Problem (43) becomes
(44)-Δw=-(∂2w∂ξ2+∂2w∂η2)=f(x),inΩ~c,w=0,onΓ~0,w(x)=𝒪(1),as|x|→∞,
where Ω~c and Γ~0 are the corresponding changes from Ωc and Γ0, respectively. Let g be extended to Ω~c, w=u-g, f=Δg. Then, problem (44) is equivalent to
(45)-Δu=0,inΩ~c,u=g,onΓ~0,u(x)=𝒪(1),as|x|→∞.Since it is difficult to estimate the convergence rate for a general unbounded domainΩc, we here let Ωc be an exterior domain of an ellipse Γ0, with μ=μ0, and Γ1 is taken as stated in Section 1. We introduce the following conclusion first.Lemma 2.
Ifu is the solution of
(46)-Δu=0,inΩ1,u=u0,onΓ0,∂u∂n=un,onΓ1,
where Ω1 is the elliptical ring domain between Γ0 and Γ1,
(47)u0=∑j=-∞+∞‍cjeijϕ∈H1/2(Γ0),un=1J0(∑-∞,j≠0+∞‍dj|j|eijϕ+d0),∈H-1/2(Γ1),
then there exist a unique u∈H1(Ω1) and
(48)u(μ,ϕ)=∑-∞,j≠0+∞‍({(e|j|(μ1-μ0)+e|j|(μ0-μ1))-1}(cj(e|j|(μ-μ1)+e|j|(μ1-μ))+dj×(e|j|(μ-μ0)-e|j|(μ0-μ)))×(e|j|(μ1-μ0)+e|j|(μ0-μ1))-1)×eijϕ+c0+d0(μ-μ0).Proof.
The result can be obtained directly from (46) by the separation of variables.Theorem 3.
If0<θk<1, then the nonoverlapping domain decomposition method (7)–(9) is convergent.Proof.
We assume that the exact solution to problem (1) is u, and we let λ=u|Γ1, uk=u|Ωk, and  k=1,2. Then, following (7)–(8), we let e1k=λ-u1k and e1k|Γ1=λ-λk=˙e2k. From (24), we assume that e2k=∑j=-∞+∞ajeijϕ∈H1/2(Γ1). By the natural integral equation (27), we have
(49)∂e1k∂n=-𝒦1(e2k)=-1J0∑j=-∞+∞‍|j|ajeijϕ.
So, e1k satisfies
(50)-Δe1k=0,inΩ1,e1k=0,onΓ0,∂e1k∂n=1J0∑j=-∞+∞‍|j|ajeijϕ,onΓ1.
By Lemma 2, one obtains
(51)e1k=-∑j=-∞+∞‍ajHj(μ)eijϕ,
with Hj(μ)=(e|j|(μ-μ0)-e|j|(μ0-μ))/(e|j|(μ1-μ0)+e|j|(μ0-μ1)). From (51), e1k confines on Γ1 can be expressed as
(52)e1k∣Γ1=-∑j=-∞+∞‍ajHj(μ1)eijϕ,𝒦1(e1k)=-1J0∑j=-∞+∞‍|j|ajHj(μ1)eijϕ.
Then, we have(53)∂e1k+1∂n=-𝒦1(λ-λk+1)=𝒦1(θku1k+(1-θk)λk-λ)=-θk𝒦1(e1k)-(1-θk)𝒦1(e2k)=1J0∑j=-∞+∞‍|j|aj(θkHj(μ1)-1+θk)eijϕ.
LetEm=˙∥∂e1k/∂n∥-1/2,Γ12, then Em=2π∑j=-∞+∞(j2/1+j2)|aj|2 and
(54)Em+1=2π∑j=-∞+∞‍j21+j2|aj|2(θkHj(μ1)-1+θk)2=(1-θk)2Em+2π∑j=-∞+∞‍j21+j2|aj|2×θkHj(μ1)[θk(Hj(μ1)+2)-2].
Assuming thatδ1=infj∈ℤ/{0}2/(2+Hj(μ1)), then 1>δ1≥2/3.
If0<θk≤δ1, k=0,1,2,…, then
(55)Em+1<(1-θk)2Em,
or equally
(56)Em+1<∏j=1m‍(1-θj)2E1≤rmE1,19≤r<1.
By the trace theorem, we have(57)∥e1k∥1,Ω12≤CEm→0,m→∞.
From (54), one also has
(58)Em+1=2π∑j=-∞+∞‍j21+j2|aj|2(θkHj(μ1)-1+θk)2=(1-2θk)2Em+2π∑j=-∞+∞‍j21+j2|aj|2×θkIj(μ1)[θk(Ij(μ1)-2)+1],
with Ij(μ1)=(1-Hj(μ1))/2. Assuming that δ2=supj∈ℤ/{0}1/(2-Ij(μ1)), then 0<δ2≤2/3.
Forδ2≤θk<1, k=0,1,2,…, the convergence result can be obtained similarly with (55)–(57). Therefore, for 0<θk<1, the nonoverlapping domain decomposition method is convergent.


## 3.3. Discrete Nonoverlapping Alternating Algorithm and Convergence Analysis

Divide the arcΓμ1 into M parts, and take a finite element subdivision in Ω1 such that their nodes on Γμ1 are coincident. That is, we make a regular and quasiuniform triangulation 𝒯h on Ω1, such that
(59)Ω1=⋃K∈𝒯h‍K,
where K is a (curved) triangle and h is the maximal diameter of triangles. Let
(60)Vh={vh∈V|v|Kis a linear polynomial,∀K∈𝒯h}.
Then, the approximate problem of (39) can be written as follows
(61)Finduh∈Vh,such  thatD(uh;uh,vh)+D^(uh;uh,vh)=F(vh),∀vh∈Vh,
with
(62)D(th;uh,vh)=∫Ω1‍a(x,th)(α∂uh∂x∂vh∂x+β∂uh∂y∂vh∂y)dx,(63)D^(th;uh,vh)=∑j=1+∞‍αβjπ∫02π‍∫02π‍a0(th(μ1,φ′))×∂uh(μ1,φ′)∂φ′∂vh(μ1,φ)∂φ·cosj(φ′-φ)dφ′dφ.
In practice, the sum of (63) is truncated to a finite number of terms N. By the hypothesis of a(·,·), it is not difficult to know the following result.Lemma 4.
There exists a constantC2>0 which has a different meaning in a different place and is related to α and β, such that
(64)|D(u;u,v)+D^(u;u,v)|≤C2∥u∥1,Ω1∥v∥1,Ω1,D(u;u,u)+D^(u;u,u)≥C0∥u∥1,Ω12,∀u,v∈V.Proof.
From (3) and (40), one can obtain that
(65)|D(u;u,v)|≤C2′∥u∥1,Ω1∥v∥1,Ω1,D(u;u,u)≥C0∥u∥1,Ω12,∀u,v∈V.
For the natural integral equation, one can obtain the following result
(66)|D^(u;u,v)|≤C2′′∥u∥1,Ω1∥v∥1,Ω1,D^(u;u,u)≥0,∀u,v∈V.
For this purpose, foru,v∈V, we assume that
(67)u(μ1,φ′)=∑j=-∞+∞‍ujeijφ′,v(μ1,φ)=∑j=-∞+∞‍vjeijφ,
with uj=(1/2π)∫02π‍u(μ1,φ′)e-ijφ′dφ′, vj=(1/2π)∫02π‍v(μ1,φ)e-ijφdφ. Then, we have
(68)∂u∂φ′(μ1,φ′)=∑j=-∞+∞‍ijujeijφ′,∂v∂φ(μ1,φ)=∑j=-∞+∞‍ijvjeijφ.
Combining property (3), Cauchy’s inequality, and the trace theorem, we have
(69)|D^(u;u,v)|≤C2′′(∑j=-∞+∞‍|j|·|uj|2)1/2×(∑j=-∞+∞‍|j|·|vj|2)1/2≤C2′′(∑j=-∞+∞‍(1+j2)1/2·|uj|2)1/2×(∑j=-∞+∞‍(1+j2)1/2·|vj|2)1/2≤C2′′∥u∥1/2,Γμ1∥v∥1/2,Γμ1≤C2′′∥u∥1,Ω1∥v∥1,Ω1,∀u,v∈V.
Next, we show thatD^(u;u,u)≥0, for any u∈V. For any given v∈V, let us consider the following auxiliary problem in Ω1:
(70)-∇·(a(x,u)∇u)=0,inΩ1,u=0,onΓ0,u=v,onΓμ1.
From the analysis in Section2.1, we know that the solution u to problem (70) satisfies
(71)a0(u)∂u∂n|Γμ1=𝒦1(u(μ1,φ))=𝒦1(v(μ1,φ)).
Multiplying (70) by u and integrating over Ω1, we have
(72)D^(u;u,u)=∫Ω1‍a0(u)|∇u|2dx≥0.
LettingC2=C2′+C2′′, we can get the desired result.We are now in the position to show the existence and uniqueness result for this type of problem (1). We begin with the following estimate.Lemma 5.
Any solutionu to problem (39) satisfies
(73)∥u∥1,Ω1≤C3∥f∥L2(Ω1),
with C3 a positive constant.Proof.
Takingv equal to u in (39), by Lemma 4, one has
(74)C0∥u∥1,Ω12≤D(u;u,u)+D^(u;u,u)=L(u).
Then, the desired result follows from Cauchy-Schwarz and Poincaré-Friedrichs inequalities.Theorem 6.
Problem (39) is uniquely solvable.Proof.
Since the spaceV is separable (indeed, it is a closed subspace of the space H1(Ω1) which is separable), there exist increasing sequences of finite-dimensional Hilbert subspaces Vm of V such that
(75)V=⋃m≥0‍Vm.
We define a mappingΦm:Vm→Vm by
(76)(Φm(u),v)=D(u;u,v)+D^(u;u,v)-L(v),∀u,v∈Vm,
where (·,·) stands for the scalar product on Vm×Vm. Since the function a(x,u) is bounded and the trace of the function in V on Γ1 belongs to H1/2(Γ1), hence at least to L4(Γ1) from the Sobolev embedding, each mapping Φm is well defined and continuous on Vm. What’s more, by Lemmas 4 and 5, we obtain
(77)(Φm(u),u)=D(u;u,u)+D^(u;u,u)-L(u)≥C0∥u∥1,Ω12-C3∥f∥L2(Ω1)∥u∥1,Ω1.
So, the right-hand side is nonnegative on the circle of radius μ which is defined by
(78)μ=C3C0∥f∥L2(Ω1).
Applying Brouwer’s fixed point theorem [21] yields the existence of um in Vm, with norm less than μ, such that
(79)Φm(um)=0.
Since the sequence {um}m is bounded by μ in V, there exists a subsequence which converges weakly to u in Vm. Using the compactness of the imbedding of H1/2(Γ1) into L3(Γ1), we obtain that u is a solution to problem (39).
Now, we show the uniqueness of the solution. Letu and u- be two solutions to this problem. Then, taking v equal to u and u- in (39), respectively, and combining with Lemma 4, one obtains
(80)C2∥u∥1,Ω1∥u-∥1,Ω1≥D(u-;u-,u)+D^(u-;u-,u)=D(u;u,u)+D^(u;u,u)⩾C0∥u∥1,Ω12,(81)C2∥u∥1,Ω1∥u-∥1,Ω1≥D(u;u,u-)+D^(u;u,u-)=D(u-;u-,u-)+D^(u-;u-,u-)⩾C0∥u-∥1,Ω12.
Since C2>C0, combining (80) with (81), we deduce the desired result.From the discrete problem (61), similar to the discussion for the linear problem in [6, 8], we can get a system of algebraic equations for our quasilinear problem with the following form:
(82)(A11+KhA12A21A22)(UV)=(0b),
where U is a vector whose components are function values at nodes on Γ1, and V is a vector whose components are function values at interior nodes of Ω1. The matrix A=˙A(u)=(A11A12A21A22) is the stiffness matrix obtained from finite element in Ω1, while Kh=˙Kh(u|Γ1) follows from the natural boundary element method on Γ1.Problem (82) can also be rewritten as follows:
(83)(A11A12A21A22)(UV)=(-KhUb).
Then, we have the iterative algorithm
(84)(A11A12A21A22)(UkVk)=(-KhΛkb),
with
(85)Λk+1=θkUk+(1-θk)Λk,k=0,1,….By condition (3), one obtains that A is a positive definite matrix, so we know that A22-1 exists. Now, we let Sh=Sh(1)+Kh be the discrete analogue of the Steklov-Poincaré operator on Γ1, with Sh(1)=A11-A12A22-1A21 and B=-A12A22-1b. Then, similar to the proof of Theorems 7.6 and  7.7 in [6], we have that the alternating algorithm (84)-(85) is equivalent to the preconditioned Richardson iteration:
(86)Sh(1)(Λk+1-Λk)=θk(B-ShΛk).And we also have the following convergence result.Theorem 7.
If0<minθk≤maxθk<1, then the discrete nonoverlapping alternating method (84)-(85) is convergent, and both the convergence rate and the condition number of [Sh(1)]-1Sh are independent of the mesh size h.


## 4. Numerical Examples

In this section, we will give some examples to confirm our theoretical results. In the following, we choose the finite element space as given in (60). For simplicity, we let
(87)Δμ=μ1-μ0m,Δφ=2πM,e0(k)=∥u-uh,Nk∥L2(Ωi),e(k)=∥u-uh,Nk∥L∞(Ωi).
And we let eh(k) denote the maximal error between the iteration k-1 and k, that is, eh(k)=∥uh,Nk-uh,Nk-1∥L∞(Ωi) and qh(k)=eh(k-1)/eh(k) simulate the convergence rate.We give the following four examples. Examples8 and 11 show the relationship between meshes and convergence rate for the cases α=β=1 and β>α>0, respectively. Example 9 focuses on the effect of the relaxation factor θk to the convergence rate. And Example 10 wants to show the relationship between the coefficients α,β and the convergence rate.Example 8.
We assume the exterior domainΩc with elliptical boundary Γ0={(μ0,φ)|μ0=0.8,0≤φ≤2π}, Γμ1={(μ1,φ)|μ1>μ0,0≤φ≤2π}. Now, we consider the problem
(88)-∇·(a(x,u)∇u)=f,inΩ1,u=0,onΓ0,a0(u)∂u∂n=𝒦1(u(μ1,φ)),onΓμ1,
when a(x,u)=1/(1+u2), f=0, and f0=1.35.
The exact solution of Example8 is u=tan(2sinhμsinφ/f0(cosh2μ+cos2φ)). For different meshes and μ1, we have the following results ranked in Tables 1 and 2.Table 1
The relationship between meshes and convergence rate (N=10, θk=0.58, and μ1=1.5).
(
m
,
M)
Error
Iteration number
0
1
2
3
4
5
6
9
(2,8)
e
4.9212
E
-
01
3.7241
E
-
01
3.0583
E
-
01
2.7007
E
-
01
2.5029
E
-
01
2.3880
E
-
01
2.3175
E
-
01
2.2183
E
-
01
e
h
—
1.1971
E
-
01
6.6581
E
-
02
3.5758
E
-
02
1.9781
E
-
02
1.1487
E
-
02
7.0490
E
-
03
2.2318
E
-
03
q
h
—
—
1.7980
1.8620
1.8077
1.7220
1.6296
1.3992
(4,16)
e
3.4870
E
-
01
1.9037
E
-
01
1.1902
E
-
01
8.8599
E
-
02
7.5058
E
-
02
7.0101
E
-
02
6.9010
E
-
02
6.8244
E
-
02
e
h
—
1.5833
E
-
01
7.1346
E
-
02
3.0423
E
-
02
1.3541
E
-
02
6.3774
E
-
03
3.2007
E
-
03
6.2995
E
-
04
q
h
—
—
2.2192
2.3452
2.2467
2.1233
1.9925
1.5865Table 2
The relationship between meshes and convergence rate (N=10, θk=0.58, and μ1=1.0).
(
m, M)
Error
Iteration number
0
1
2
3
4
5
6
9
(2,8)
e
3.7701
E
-
01
3.2871
E
-
01
2.9909
E
-
01
2.8081
E
-
01
2.6906
E
-
01
2.6116
E
-
01
2.5562
E
-
01
2.4614
E
-
01
e
h
—
4.8299
E
-
02
2.9614
E
-
02
1.8287
E
-
02
1.1749
E
-
02
7.8958
E
-
03
5.5417
E
-
03
2.3769
E
-
03
q
h
—
—
1.6309
1.6194
1.5564
1.4880
1.4248
1.2857
(4,16)
e
3.0912
E
-
01
2.0526
E
-
01
1.4917
E
-
01
1.1931
E
-
01
1.0260
E
-
01
9.2674
E
-
02
8.6414
E
-
02
7.7207
E
-
02
e
h
—
1.0386
E
-
01
5.6091
E
-
02
2.9857
E
-
02
1.6710
E
-
02
9.9296
E
-
03
6.2607
E
-
03
2.1229
E
-
03
q
h
—
—
1.8516
1.8787
1.7868
1.6828
1.5860
1.3728Example 9.
Similar with Example8, Γ0 and a(x,u) are replaced by Γ0={(μ0,φ)|μ0=0.5,0≤φ≤2π}, Γ1={(μ1,φ)μ1=1.0,0≤φ≤2π}, and a(x,u)=1/1-u2, respectively.
The exact solution of Example9 is u=sin(2coshμcosφ/f0(cosh2μ+cos2φ)). The numerical results are given in Table 3 and Figure 2(a).Table 3
The relationship betweenθk and convergence rate (N=10, m=4, and M=16).
θ
k
k
0
1
2
3
4
5
6
9
0.18
e
4.4218
E
-
02
3.8978
E
-
02
3.6443
E
-
02
3.4711
E
-
02
3.3392
E
-
02
3.2341
E
-
02
3.1481
E
-
02
2.9668
E
-
02
e
h
—
5.2395
E
-
03
2.5350
E
-
03
1.7316
E
-
03
1.3185
E
-
03
1.0519
E
-
03
8.5974
E
-
04
5.0325
E
-
04
q
h
—
—
2.0668
1.4640
1.3133
1.2535
1.2235
1.1860
0.38
e
4.4218
E
-
02
3.6741
E
-
02
3.3193
E
-
02
3.1073
E
-
02
2.9709
E
-
02
2.8795
E
-
02
2.8167
E
-
02
2.7174
E
-
02
e
h
—
7.4769
E
-
03
3.5473
E
-
03
2.1199
E
-
03
1.3646
E
-
03
9.1335
E
-
04
6.2830
E
-
04
2.3314
E
-
04
q
h
—
—
2.1078
1.6733
1.5535
1.4941
1.4537
1.3635
0.50
e
4.4218
E
-
02
3.5398
E
-
02
3.1599
E
-
02
2.9601
E
-
02
2.8465
E
-
02
2.7784
E
-
02
2.7357
E
-
02
2.6751
E
-
02
e
h
—
8.8194
E
-
03
3.7989
E
-
03
1.9979
E
-
03
1.1359
E
-
04
6.8102
E
-
04
4.2741
E
-
04
1.3535
E
-
05
q
h
—
—
2.3216
1.9014
1.7588
1.6680
1.5934
1.4114
0.58
e
4.4218
E
-
02
3.4503
E
-
02
3.0685
E
-
02
2.8865
E
-
02
2.7913
E
-
02
2.7377
E
-
02
2.7055
E
-
02
2.6609
E
-
02
e
h
—
9.7143
E
-
03
3.8180
E
-
03
1.8204
E
-
03
9.5190
E
-
04
5.3571
E
-
04
3.2246
E
-
04
9.9192
E
-
05
q
h
—
—
2.5443
2.0974
1.9124
1.7769
1.6613
1.4090
0.65
e
4.4218
E
-
02
3.3720
E
-
02
2.9983
E
-
02
2.8358
E
-
02
2.7564
E
-
02
2.7135
E
-
02
2.6882
E
-
02
2.6530
E
-
02
e
h
—
1.0497
E
-
02
3.7371
E
-
03
1.6248
E
-
03
7.9382
E
-
04
4.2892
E
-
04
2.5372
E
-
04
7.9261
E
-
05
q
h
—
—
2.8089
2.3000
2.04687
1.8507
1.6905
1.3951(a)   The relationship betweenθk and convergence rate (N=10, m=4 and M=16). (b) The relationship between ε and convergence rate (N=10, m=4, M=16 and θk=0.58). (c) The relationship between meshes and convergence rate (N=10, θk=0.65).
(a)
(b)
(c)Example 10.
We assume the exterior domainΩc with elliptical boundary Γ0={(μ0,φ)|μ0=0.8,0≤φ≤2π},  Γμ1={(μ1,φ)|μ1=2.0,0≤φ≤2π}. Now, we consider the problem
(89)-(∂∂x(εa(x,u)∂u∂x)+∂∂y(a(x,u)∂u∂y))=f(x),-(∂∂x(εa(x,u)∂u∂x)+∂∂y(a(x,u)∂u∂y))=inΩi,u=0,onΓ0,εnxa0(u)∂u∂x+nya0(u)∂u∂y=𝒦1(u(μ1,φ)),onΓμ1,
when a(x,u)=1/(1+u2), f0=1.25 and
(90)f=2(1-ε)sinhμsinφ(3cosh2μcos2φ-sinh2μsin2φ)(cosh2μ+cos2φ)3.
The exact solution of Example10 is u=tan(2sinhμsinφ/f0(cosh2μ+cos2φ)). The numerical results are given in Table 4 and Figure 2(b).Table 4
The relationship betweenε and convergence rate (N=10, m=4, M=16, and θk=0.58).
ε
k
0
1
2
3
4
5
6
9
0.75
e
2.3433
E
-
01
1.2450
E
-
01
7.4989
E
-
02
5.2822
E
-
02
4.7623
E
-
02
4.6278
E
-
02
4.5525
E
-
02
4.4564
E
-
02
e
h
—
1.0983
E
-
01
4.9507
E
-
02
2.2168
E
-
02
1.0239
E
-
02
4.9334
E
-
03
2.5048
E
-
03
4.9131
E
-
04
q
h
—
—
2.2185
2.2333
2.1651
2.0754
1.9696
1.5930
0.5
e
2.3120
E
-
01
1.3855
E
-
01
9.5116
E
-
02
7.4367
E
-
02
6.3994
E
-
02
5.8518
E
-
02
5.6670
E
-
02
5.5578
E
-
02
e
h
—
9.2654
E
-
02
4.3431
E
-
02
2.0749
E
-
02
1.0373
E
-
02
5.4758
E
-
03
3.0715
E
-
03
7.8777
E
-
04
q
h
—
—
2.1334
2.0931
2.0003
1.8943
1.7828
1.4837
0.05
e
3.1906
E
-
01
2.6051
E
-
01
2.3278
E
-
01
2.1796
E
-
01
2.0941
E
-
01
2.0414
E
-
01
2.0068
E
-
01
1.9529
E
-
01
e
h
—
7.1017
E
-
02
2.7730
E
-
02
1.4816
E
-
02
8.5529
E
-
03
5.2765
E
-
03
3.4545
E
-
04
1.2856
E
-
04
q
h
—
—
2.5610
1.8716
1.7323
1.6209
1.5274
1.3363
0.005
e
3.6584
E
-
01
3.2417
E
-
01
3.0784
E
-
01
2.9940
E
-
01
2.9434
E
-
01
2.9100
E
-
01
2.8867
E
-
01
2.8468
E
-
01
e
h
—
1.9591
E
-
01
1.6329
E
-
02
1.8640
E
-
02
1.2301
E
-
02
4.9778
E
-
03
2.3344
E
-
03
9.9644
E
-
04
q
h
—
—
11.9976
0.8760
1.5153
2.4712
2.1324
1.2896
0.0005
e
3.7328
E
-
01
3.4165
E
-
01
3.3061
E
-
01
3.2510
E
-
01
3.2187
E
-
01
3.1978
E
-
01
3.1833
E
-
01
3.1583
E
-
01
e
h
—
1.1714
E
-
01
3.6606
E
-
02
1.1542
E
-
02
3.2283
E
-
03
2.0883
E
-
03
1.4491
E
-
03
6.3191
E
-
04
q
h
—
—
3.2001
3.1715
3.5754
1.5459
1.4411
1.2749Example 11.
Similar with Example10, Γ0, Γ1, and a(x,u) are replaced by Γ0={(μ0,φ)|μ0=0.5,0≤φ≤2π}, Γ1={(μ1,φ)μ1=1.5,0≤φ≤2π}, and a(x,u)=1/1-u2, respectively. And we take
(91)f=2(1-ε)coshμcosφ(cosh2μcos2φ-3sinh2μsin2φ)(cosh2μ+cos2φ)3.
The exact solution of Example 11 is u=sin(2coshμcosφ/f0(cosh2μ+cos2φ)). The numerical results are given in Tables 5 and 6 and Figure 2(c).
From the numerical results, one obtains that the numerical errors can be affected by the choice of relaxation factorθk, the coefficient ε, and the location of the artificial boundary. In Tables 1, 2, 5, and 6 and Figure 2(c), the relationship between the meshes and the convergence rate is presented. We obtain that the convergence rate is independent of the finite element mesh size. In Table 3 and Figure 2(c), the convergence rates for different relaxation factor θk are compared. The results indicate that the choice of the relaxation factor is very important for the performance of the nonoverlapping alternating method. On the other hand, the convergence rate is not sensitive to the relaxation factor θk between the interval (0.5,0.67). What’s more, as it is shown in Table 4 and Figure 2(c), for the anisotropic problems, the choice of ε can also affect the convergence rate.Table 5
The relationship between meshes and convergence rate (N=10, θk=0.65, and ε=0.75).
(
m, M)
Error
Iteration number
0
1
2
3
4
5
6
9
(2,8)
e
0
6.6354
E
-
01
3.8411
E
-
01
2.8271
E
-
01
2.3876
E
-
01
2.1771
E
-
01
2.0659
E
-
01
2.0013
E
-
01
1.9132
E
-
01
e
3.2474
E
-
01
2.0835
E
-
01
1.6359
E
-
01
1.4308
E
-
01
1.3278
E
-
01
1.2713
E
-
01
1.2375
E
-
01
1.1900
E
-
01
e
h
1.5452
E
-
01
5.6859
E
-
02
2.5207
E
-
02
1.2187
E
-
02
6.4127
E
-
03
3.6744
E
-
03
1.0834
E
-
03
q
h
—
—
2.7176
2.2557
2.0684
1.9004
1.7452
1.3973
(4,16)
e
0
4.4653
E
-
01
1.7098
E
-
01
8.8097
E
-
02
5.8527
E
-
02
4.7483
E
-
02
4.2925
E
-
02
4.0759
E
-
02
3.8380
E
-
02
e
2.5970
E
-
01
9.5295
E
-
02
5.1105
E
-
02
3.9408
E
-
02
3.4740
E
-
02
3.2516
E
-
02
3.1348
E
-
02
2.9955
E
-
02
e
h
—
1.7293
E
-
01
5.2160
E
-
02
2.0094
E
-
02
8.4650
E
-
03
3.8880
E
-
03
1.9662
E
-
03
4.5470
E
-
04
q
h
—
—
3.3154
2.5958
2.3738
2.1772
1.9774
1.5007Table 6
The relationship between meshes and convergence rate (N=10, θk=0.65, and ε=0.50).
(m, M)
Error
Iteration number
0
1
2
3
4
5
6
9
(2,8)
e
0
8.2535
E
-
01
4.8503
E
-
01
3.7829
E
-
01
3.3445
E
-
01
3.1417
E
-
01
3.0370
E
-
01
2.9771
E
-
01
2.8963
E
-
01
e
4.0734
E
-
01
2.6879
E
-
01
2.2125
E
-
01
2.0008
E
-
01
1.8968
E
-
01
1.8407
E
-
01
1.8075
E
-
01
1.7609
E
-
01
e
h
2.0397
E
-
01
6.6268
E
-
02
2.7907
E
-
02
1.2785
E
-
02
6.3537
E
-
03
3.4410
E
-
03
1.0625
E
-
03
q
h
—
—
3.0780
2.3746
2.1829
2.0122
1.8465
1.3787
(4,16)
e
0
5.3524
E
-
01
1.8866
E
-
01
1.1017
E
-
01
8.6488
E
-
02
7.8123
E
-
02
7.4538
E
-
02
7.2733
E
-
02
7.0595
E
-
02
e
3.0655
E
-
01
1.1062
E
-
01
7.6720
E
-
02
6.5692
E
-
02
6.0842
E
-
02
5.8486
E
-
02
5.7219
E
-
02
5.5639
E
-
02
e
h
2.3531
E
-
01
5.4156
E
-
02
1.7431
E
-
02
6.7638
E
-
03
2.9679
E
-
03
1.4447
E
-
03
3.7294
E
-
04
q
h
—
—
4.3449
3.1069
2.5771
2.2790
2.0543
1.4481



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*Source: 828615-2013-02-14.xml*