In this paper, based on the Kirchhoff transformation and the natural boundary element method, a coupled natural boundary element and curved edge finite element is applied to solve the anisotropic quasi-linear problem in an unbounded domain with a concave angle. By using the principle of the natural boundary reduction, we obtain the natural integral equation on the artificial boundary of circular arc boundary, and get the coupled variational problem and its numerical method. Then the error and convergence of coupling solution are analyzed. Finally, some numerical examples are verified to show the feasibility of our method.
In this paper, we propose the method of coupling curved edge finite element (CEFE) and natural boundary element (NBE) to solve an anisotropic quasilinear problem in an unbounded domain. The CEFE-NBE method is based on the artificial boundary method [
The standard procedure of the method of coupling curved edge finite element and natural boundary element can be described as follows. We introduce an artificial boundary to divide the original domain into two subregions, a bounded inner region and an unbounded one with special boundary, which are solved by the curved edge finite element and the boundary element method respectively.
Suppose Ω is an infinite domain with a concave angle ω , and 0 < ω ≤ 2 π ; the boundaries of Ω are disintegrated into three disjoint parts: ∂ Ω = Γ ∪ Γ 1 ∪ Γ 2 , the boundary Γ is a simple smooth curve part, Γ1 and Γ2 are two half lines. We have
Ω = { ( r , θ ) | r > R , 0 < θ < ω } , Γ = { ( r , θ ) | r = R , 0 < θ < ω } ,
Γ 1 = { ( r , θ ) | r > R , θ = 0 } , Γ 2 = { ( r , θ ) | r > R , θ = ω } .
We consider the following quasilinear problem
{ − ( ∂ ∂ x ( α a ( x , u ) ∂ u ∂ x ) + ∂ ∂ y ( β a ( x , u ) ∂ u ∂ y ) ) = f , in Ω 1 , ∂ u ∂ n = 0 , on Γ 1 ∪ Γ 2 , u = 0 , on Γ , u ( x ) isbounded , as | x | → ∞ .
Suppose that the given function a ( ⋅ , ⋅ ) satisfies:
0 < C 0 ≤ a ( x , u ) ≤ C 1 , ∀ u ∈ R , and for almost all
x ∈ Ω , (1)
with two constants C 0 , C 1 ∈ R ;
| a ( x , u ) − a ( x , v ) | ≤ C L | u − v | , ∀ u , v ∈ R , and for almost all
x ∈ Ω , (2)
with a constant C L > 0 .
This problem has many physical applications in the field of continuum mechanics, we also assume that ∂ a ∂ s , ∂ 2 a ∂ 2 s are continuous and suppose that the given function f ∈ L 2 ( Ω ) has compact support, i.e., there exists a constant R 0 > 0 , such that
s u p p f ⊂ Ω R 0 = { x ∈ R 2 | | x | ≤ R 0 } . (3)
Furthermore, we assume that
a ( x , u ) ≡ a 0 ( u ) , when | x | ≥ R 0 . (4)
The rest of the paper is organized as follows. In section 2, we obtain the exact quasilinear elliptical arc artificial boundary condition. In section 3, we give the equivalent variational problems and the finite element approximations and prove the well-posedness and the convergence results of the reduced problems. In section 4, we give some numerical examples to show the efficiency and feasibility of this method.
Now, we introduce an artificial boundary
Γ 0 = { ( R , θ ) | θ ∈ ( 0 , ω ) } , (5)
with R ≥ R 0 .
Let us introduce the artificial boundary Γ0 which divides Ω into two regions (see
{ − ( ∂ ∂ x ( α a ( x , u ) ∂ u ∂ x ) + ∂ ∂ y ( β a ( x , u ) ∂ u ∂ y ) ) = f , in Ω 1 , ∂ u ∂ n = 0 , on Γ 11 ∪ Γ 21 , u = 0 , on Γ , (6)
{ − ( ∂ ∂ x ( α a ( x , u ) ∂ u ∂ x ) + ∂ ∂ y ( β a ( x , u ) ∂ u ∂ y ) ) = 0 , in Ω 2 , ∂ u ∂ n = 0 , on Γ 12 ∪ Γ 22 , u ( x ) isbounded , as | x | → ∞ , (7)
u ( x ) and α a 0 ( u ) n x ∂ u ∂ x + β a 0 ( u ) n y ∂ u ∂ y are continuous on Γ 0 , where Γ 11 = Ω 1 ∩ Γ 1 , Γ 21 = Ω 1 ∩ Γ 2 , Γ 12 = Ω 2 ∩ Γ 1 and Γ 22 = Ω 2 ∩ Γ 2 , and n = ( n x , n y ) is the unit exterior normal vector on Γ 0 .
Particularly, a ( x , u ) ≡ a is independent of x and u when | x | ≥ R 0 > 0 , the problem (7) is simplified to the linear exterior elliptic problem [
We introduce the so-called Kirchhoff transformation: [
w ( x ) = ∫ 0 u ( x ) a 0 ( ξ ) d ξ , x ∈ Ω 2 , (8)
which gives
∇ w = a 0 ( u ) ∇ u , (9)
and
( α ∂ w ∂ x , β ∂ w ∂ y ) = ( α a 0 ( u ) n x ∂ u ∂ x , β a 0 ( u ) n y ∂ u ∂ y ) . (10)
From (7), that w satisfies the following problem:
{ − ( α ∂ 2 w ∂ x 2 + β ∂ 2 w ∂ y 2 ) = 0 , in Ω 2 , ∂ u ∂ n = 0 , on Γ 12 ∪ Γ 22 , w ( x ) isbounded , as | x | → ∞ . (11)
Assume that w ( x ) is the solution of the problem (13), and the value w | | u | = u 1 is given
w | Γ 0 = w ( R , θ ) ,
we introduce x = α ξ , y = β η , the boundary Γ 0 is changed to the elliptic arc boundary Γ 0 ¯ = { ( ξ , η ) | α ξ 2 + β η 2 = R 2 } , the unit exterior normal vector on Γ 0 ¯ is
v = − 1 α x 2 + β y 2 ( α cos φ , β sin φ ) ,
The next, we need to discuss the relationship between elliptic coordinates ( u , φ ) and cartesian coordinates ( x , y ) , the relationship can be expressed as below:
{ x = f 0 cosh u cos φ , y = f 0 sinh u sin φ , (12)
where
f 0 = β − α α β R , u 0 = ln α + β β − α ,
Ω ¯ = { ( u , φ ) | u > u 0 , 0 < φ < ω } , Γ ¯ = { ( u , φ ) | u = u 0 , 0 < φ < ω } ,
Γ ¯ 1 = { ( u , φ ) | u > u 0 , φ = 0 } , Γ ¯ 2 = { ( u , φ ) | u > u 0 , φ = ω } ,
J ( u , φ ) = | ∂ ξ ∂ u ∂ ξ ∂ φ ∂ η ∂ u ∂ η ∂ φ | = f 0 2 ( sinh 2 u cos 2 φ + cosh 2 u sin 2 φ ) .
The problem (11) is transformed into
{ − ( ∂ 2 w ∂ ξ 2 + ∂ 2 w ∂ η 2 ) = 0 , in Ω ¯ 2 , ∂ u ∂ n = 0 , on Γ ¯ 12 ∪ Γ ¯ 22 , w ( x ) isbounded , as | x | → ∞ . (13)
The based on the natural boundary reduction [
w ( r , θ ) = − 1 2 ω ( R 2 π ω − r 2 π ω ) ∫ 0 ω [ 1 r 2 π ω + R 2 π ω − 2 ( R r ) 2 π ω cos π ω ( θ − θ ′ ) + 1 r 2 π ω + R 2 π ω − 2 ( R r ) 2 π ω cos π ω ( θ + θ ′ ) ] w ( R , θ ′ ) d θ ′ , 0 < R < r , (14)
and the natural integral equation
∂ w ∂ n = − π 4 ω 2 R ∫ 0 ω ( 1 sin 2 θ − θ ′ 2 ω π + 1 sin 2 θ + θ ′ 2 ω π ) w ( R , θ ′ ) d θ ′ . (15)
From (9), we obtain
∂ w ∂ n = a 0 ( u ) ∂ u ∂ n . (16)
Combining (8), (9) and (15), we obtain the exact artificial boundary condition of u on Γ 0 ,
| α a 0 ( u ) n x ∂ u ∂ x + β a 0 ( u ) n y ∂ u ∂ y | r = R = − α β R π ∫ 0 ω ∑ n = 1 ∞ w 0 ( ∫ 0 u ( R , θ ′ ) a 0 ( y ) d y ) n cos n ( θ ′ − θ ) d θ ′ ≜ κ 1 ( u ( u 1 , φ ) ) . (17)
By the exact quasilinear artificial boundary condition (17), the original problem confines in Ω 1 can be defined as follows
{ − ( ∂ ∂ x ( α a ( x , u ) ∂ u ∂ x ) + ∂ ∂ y ( β a ( x , u ) ∂ u ∂ y ) ) = f , in Ω 1 , ∂ u ∂ n = 0 , on Γ 11 ∪ Γ 21 , u = 0 , on Γ , α a 0 ( u ) n x ∂ u ∂ x + β a 0 ( u ) n y ∂ u ∂ y = κ 1 ( u ( u 1 , φ ) ) , on Γ 0 . (18)
Now, we consider the problem (18). First, we will use W m , p ( Ω ) denoting the standard Sobolev spaces, ‖ ⋅ ‖ m . p , Ω and | ⋅ | m , p , Ω denoting the corresponding norms and semi-norms. In particular, we denote H m ( Ω ) = W m , 2 ( Ω ) , ‖ ⋅ ‖ m , Ω = ‖ ⋅ ‖ m .2 , Ω and | ⋅ | m , , Ω = | ⋅ | m , 2 , Ω .
Let us introduce the space
V = { v ∈ H 1 ( Ω 1 ) | v | Γ 0 = 0 } , (19)
and the corresponding norms
‖ v ‖ 0 , Ω 1 = ∫ Ω 1 | v | 2 d x , ‖ v ‖ 1 , Ω 1 = ∫ Ω 1 ( | v | 2 + | ∇ v | 2 ) d x .
The boundary value problem (18) is equivalent to the following variational problem
{ Find u ∈ V , suchthat A ( u ; u , v ) + B ( u ; u , v ) = F ( v ) ∀ v ∈ V , (20)
with
A ( w ; u , v ) = ∫ Ω 1 a ( x , w ) ( α ∂ u ∂ x ∂ v ∂ x + β ∂ u ∂ y ∂ v ∂ y ) d x , (21)
B ( w ; u , v ) = ∑ n = 1 ∞ α β n π ∫ 0 ω ∫ 0 ω a 0 ( w ( R , θ ′ ) ) ∂ u ( R , θ ′ ) ∂ θ ′ ∂ v ( R , θ ) ∂ θ cos n ( θ ′ − θ ) d θ ′ d θ (22)
F ( v ) = ∫ Ω 1 f ( x ) v ( x ) . (23)
Lemma 1 [
| B ( w ; u , v ) | ≤ C 0 ‖ u ‖ 1 , Ω 1 ‖ v ‖ 1 , Ω 1 , B ( u ; u , v ) ≥ C 0 ‖ u ‖ 1 , Ω 1 2 , ∀ u , v , w ∈ V .
In practice, we need to truncate the series in (17) for some nonnegative integer N, that is
| α a 0 ( u ) n x ∂ u ∂ x + β a 0 ( u ) n y ∂ u ∂ y | r = R = κ 1 N ( u ( u 1 , φ ) ) , (24)
with
κ 1 N ( u ( u 1 , φ ) ) = − α β R π ∫ 0 ω ∑ n = 1 N w 0 ( ∫ 0 u ( R , θ ′ ) a 0 ( y ) d y ) n cos n ( θ ′ − θ ) d θ ′ . (25)
Then, we consider the following approximate problem
{ − ( ∂ ∂ x ( α a ( x , u N ) ∂ u N ∂ x ) + ∂ ∂ y ( β a ( x , u N ) ∂ u N ∂ y ) ) = f , in Ω 1 , ∂ u N ∂ n = 0 , on Γ 11 ∪ Γ 21 , u N = 0 , on Γ , α a 0 ( u N ) n x ∂ u N ∂ x + β a 0 ( u N ) n y ∂ u N ∂ y = κ 1 N ( u N ( u 1 , φ ) ) , on Γ 0 . (26)
The problem (26) is equivalent to the following variational problem
{ Find u N ∈ V , suchthat A ( u N ; u N , v ) + B N ( u N ; u N , v ) = F ( v ) , ∀ v ∈ V , (27)
where
B ( w ; u , v ) = ∑ n = 1 N α β n π ∫ 0 ω ∫ 0 ω a 0 ( w ( R , θ ′ ) ) ∂ u ( R , θ ′ ) ∂ θ ′ ∂ v ( R , θ ) ∂ θ = cos ( θ ′ − θ ) d θ ′ d θ . (28)
Similar with lemma 1, we have:
Lemma 2 There exists a constant C > 0 , such that
| B N ( w ; u , v ) | ≤ C 0 ‖ u ‖ 1 , Ω 1 ‖ v ‖ 1 , Ω 1 , B N ( u ; u , v ) ≥ C 0 ‖ u ‖ 1 , Ω 1 2 , ∀ u , v , w ∈ V .
Firstly, we divide the arc Γ 0 into M parts and take an edge finite element subdivision τ in Ω 1 , so that their nodes on Γ 0 are the same. The curved edge element subdivision τ ¯ . means that we make a regular T h on Ω 1 , and then we replace the standard triangles T whose two vertices lie on the boundary Γ 0 with a curved triangles T ¯ whose one edge coincides with the boundary.
Secondly, we let T ¯ ∈ τ ¯ , and use P i = ( x i , y i ) ( i = 1 , 2 , 3 ) to denote the vertex of a curved triangles T ¯ , and P 1 P 3 to denote the curved edge. We use area coordinates ( λ 1 , λ 2 , λ 3 ) to represent a standard triangle T corresponding to the curved triangle T ¯ ,
x = ∑ i = 1 3 x i λ i , y = ∑ i = 1 3 y i λ i , (29)
where λ 1 = 1 − ξ − η , λ 2 = ξ , λ 3 = η , then, based on the linear transformation of the T, we make the following nonlinear transformation, so that the curved triangle T ¯ can be mapped to the reference element T ˜ one-by-one,
{ x = x 1 + x 2 ¯ ξ + x 3 ¯ η + ( 1 − ξ − η ) Φ ( η ) , y = y 1 + y 2 ¯ ξ + y 3 ¯ η + ( 1 − ξ − η ) Ψ ( η ) , (30)
where
Φ ( η ) = 1 1 − η ( φ ( s 1 + s 1 ¯ η ) − x 1 − x 3 ¯ η ) , (31)
Ψ ( η ) = 1 1 − η ( ψ ( s 1 + s 3 ¯ η ) − y 1 − y 3 ¯ η ) , (32)
and s 3 ¯ = s 3 − s 1 , x = φ ( s ) , y = ψ ( s ) , x 2 ¯ = x 2 − x 1 , y 2 ¯ = y 2 − y 1 , x 3 ¯ = x 3 − x 1 , y 3 ¯ = y 3 − y 1 .
In the end, we get a curved edge element subdivision by the above method in Ω 1 ¯ , and it maps a standard triangle in the plane ( ξ , η ) to a curved triangle one-by-one.
Next, we construct the coordination element, i.e. approaching space S h ⊂ H Γ 0 1 ( Ω 1 ) . Triples ( T ¯ , P 1 ( T ¯ ) , Σ T ) are used to define the finite element, and use P 1 ( T ¯ ) to denote the space of first degree polynomials and the space of zero degree polynomials on T ˜ . The following mappings exist
P 1 ( T ¯ ) = { P : T ¯ → R ; P = P ^ ∘ F T − 1 , P ^ ∈ P 1 ( T ˜ ) } , (33)
and use ∘ to denote a composite mapping, and Σ T = { N i : i = 1 , 2 , 3 } is a set of linear functions, we denote N i ( φ ) = φ ( p i ) , φ = p i ( T ¯ ) , i = 1 , 2 , 3 [
According to the reference [
Lemma 3. If τ ¯ and the edge finite element are regular, the difference error of each curved triangular element is estimated as follows
‖ v − Π T ¯ v ‖ 1 , T ¯ ≤ C h T ¯ ‖ v ‖ 2 , T ¯ ∀ v ∈ H 2 ( T ¯ ) , (34)
where C is a constant, Π is interpolation operator, Π T ¯ v ∈ P 1 ( T ¯ ) , Π T ¯ v ( P i ) = v ( P i ) ( i = 1 , 2 , 3 ) is well-determined.
Proof: we assume τ h is regular, from reference [
‖ J ( F T ( ⋅ ) ) ‖ 0 , ∞ , T ˜ = o ( h T 2 ) , ‖ F T ‖ 1 , ∞ , T ˜ = o ( h T l ) , ( l = 1 , 2 ) , ‖ F T − 1 ‖ 1 , ∞ , T ˜ = o ( h T − 1 ) .
In the affine condition, according to various properties of F T , we obtain the difference error of each curved triangular element,
‖ v − Π T ¯ v ‖ 1 , T ¯ ≤ C h T ¯ ‖ v ‖ 2 , T ¯ ∀ v ∈ H 2 ( T ¯ ) .
Let S h = { v h ∈ C ( Ω 1 ) ∩ H Γ 0 1 ( Ω 1 ) | v h | T ¯ ∈ P 1 ( T ¯ ) } , combined with (29), we obtain
inf v h ∈ S h ‖ u − v h ‖ 1 , Ω 1 ≤ C h ‖ u ‖ 2 , Ω 2 , ∀ u ∈ S h ( Ω ¯ 1 ) . (35)
The approximate problem of (27) can be written as
{ Find u h N ∈ S h ( Ω 1 ) , suchthat A ( u h N ; u h N , v h ) + B N ( u h N ; u h N , v h ) = F ( v h ) , ∀ v h ∈ S h ( Ω 1 ) . (36)
Similar with existence and uniqueness in [
Lemma 4. The variational problems (20), (27) and (36) are uniquely solvable.
From now on, let u , u N ∈ H 2 ( Ω 1 ) , and u h N ∈ S h be the solution of problems (20), (27) and (36) respectively. We also assume that
S h ⊂ V ∩ W 1 , 2 + ε ( Ω 1 ) , forsome ε ∈ ( 0 , 1 ) , (37)
and we require that { S h } h → 0 is a family of finite-dimensional subspaces of V ∩ C ( Ω 1 ) , such that
∀ v ∈ V ∩ C ( Ω 1 ) , ∃ { v h } h → 0 : v h ∈ S h , ‖ v − v h ‖ 1 , Ω 1 → 0 , as h → 0 , (38)
‖ v h ‖ 1 , 2 , Ω 1 ≤ C ( v ) , ∀ h , (39)
where C ( v ) > 0 is independent of h.
The continuous piecewise polynomial spaces, such as S h , satisfy the condition (37), and if let v h = Π h v , where Π h : V → S h is the interpolation operator, we have
‖ v h ‖ 1 , 2 + ε , Ω 1 ≤ ‖ Π h v − v ‖ 1 , 2 + ε , Ω 1 + ‖ v ‖ 1 , 2 + ε , Ω 1 ≤ C ( v ) ,
and we can obtain the following lemma.
Theorem 1 [
‖ u − u h N ‖ 1 , Ω 1 ≤ C [ h σ + e ( u 0 − u 1 ) ( N + 1 ) π ω ‖ u ‖ 1 , Ω 1 ] . (40)
Proof: For v ∈ V , we assume that
w ( u , φ ) = b 0 2 + ∑ n = 1 + ∞ b n e ( u 0 − u ) n π ω cos n π φ ω , ∀ u ≥ u 0 ,
v ( u 1 , θ ) = c 0 2 + ∑ n = 1 + ∞ c n cos n π θ ω .
Then, we have the following result
| B ( u ; u , v ) − B N ( u ; u , v ) | = | 2 n π ∫ 0 ω ∫ 0 ω ∂ w ( u 1 , φ ) ∂ φ ∂ v ( u 1 , θ ) ∂ θ ∑ n = N + 1 + ∞ sin n π φ ω sin n π θ ω d φ d θ | = | ∑ n = N + 1 + ∞ n π 2 e ( u 0 − u 1 ) n π ω b n c n | ≤ C e ( u 0 − u 1 ) ( N + 1 ) π ω ( N + 1 ) k − 1 ‖ u ‖ k − 1 2 , τ u 0 ‖ v ‖ 1 , Ω 1 .
From references [
‖ u − u N ‖ 1 , Ω 1 ≤ C | B ( u ; u , v ) − B N ( u ; u , v ) | ‖ v ‖ 1 , Ω 1 ,
then, we obtain
‖ u − u N ‖ 1 , Ω 1 ≤ C e ( u 0 − u 1 ) ( N + 1 ) π ω ‖ u ‖ k − 1 2 , τ u 0 ‖ v ‖ 1 , Ω 1 ‖ v ‖ 1 , Ω 1 = C e ( u 0 − u 1 ) ( N + 1 ) π ω ‖ u ‖ k − 1 2 , τ u 0 .
Like [
‖ v − P h v ‖ 1 , Ω 1 ≤ C 0 inf v h ∈ V h ‖ v − P h v ‖ 1 , Ω 1 ≤ C 0 h σ , 0 < σ < 1 , ∀ v ∈ V ,
for u N ∈ V , we have
‖ u N − u h N ‖ 1 , Ω 1 ≤ ‖ u N − P h u N ‖ 1 , Ω 1 + ‖ P h u N − u h N ‖ 1 , Ω 1 ≤ C h σ .
Therefore,
‖ u − u h N ‖ 1 , Ω 1 ≤ ‖ u − u N ‖ 1 , Ω 1 + ‖ u N − u h N ‖ 1 , Ω 1 = C [ h σ + e ( u 0 − u 1 ) ( N + 1 ) π ω ‖ u ‖ 1 , Ω 1 ] .
Example 1. We assume
Ω = { ( r , θ ) | r > R , 0 < θ < ω } , Γ = { ( r , θ ) | r = R , 0 < θ < ω } ,
Γ 1 = { ( r , θ ) | r > R , θ = 0 } , Ω = { ( r , θ ) | r > R , θ = ω } ,
with ω = 3 2 π , R = 0.8 , r = 1.5 , ε = α β = 0.5 . According to references [
Γ 0 = { ( R 0 , θ ) | θ ∈ ( 0 , ω ) } , R 0 = 1.5.
We take our numerical results for problem (1), with
a ( x , u ) = { 16 − r 2 + 1 1 + u 2 , 0.8 ≤ r ≤ 1.5 , 1 1 + u 2 , r > 1.5 ,
u = tan ( y x 2 + y 2 ) .
Furthermore, we let Δ r = 1 m , Δ θ = 3 π 2 M for some integer m , M ∈ N . The exact solution “u” is solved with N = 10 , Δ r = 1 16 , Δ θ = ω 64 .
The numerical results are given in
| mesh | ‖ u − u h ‖ L 2 ( Ω 1 ) | ratio ( L 2 ( Ω 1 ) ) | ‖ u − u h ‖ L ∞ ( Ω 1 ) | ratio ( L ∞ ( Ω 1 ) ) |
|---|---|---|---|---|
| h | 3.58174E-1 | 8.76106E-1 | ||
| h/2 | 1.30051E-1 | 2.75411 | 3.75232E-1 | 2.33484 |
| h/4 | 4.44990E-2 | 2.92254 | 1.50271E-1 | 2.49704 |
| h/8 | 1.46775E-2 | 3.03180 | 4.12880E-2 | 3.63954 |
| mesh | ‖ u − u h ‖ L 2 ( Ω 1 ) | ratio ( L 2 ( Ω 1 ) ) | ‖ u − u h ‖ L ∞ ( Ω 1 ) | ratio ( L ∞ ( Ω 1 ) ) |
|---|---|---|---|---|
| h | 2.44592E-1 | 7.47451E-1 | ||
| h/2 | 9.79710E-2 | 2.49660 | 3.45129E-1 | 2.165713 |
| h/4 | 3.55530E-2 | 2.75559 | 1.48746E-1 | 2.320254 |
| h/8 | 1.18910E-2 | 2.98991 | 4.12630E-2 | 3.604833 |
These are the three error analysis diagrams of Example 1.
Example 2. We assume
Ω = { ( r , θ ) | r > R , 0 < θ < ω } , Γ = { ( r , θ ) | r = R , 0 < θ < ω } ,
Γ 1 = { ( r , θ ) | r > R , θ = 0 } , Ω = { ( r , θ ) | r > R , θ = ω } ,
with ω = 15 8 π , R = 1 , r = 1.5 , ε = α β = 0.5 . Similar with example 1, ε = 0.5 . We introduce an elliptical boundary
Γ 0 = { ( R 0 , θ ) | θ ∈ ( 0 , ω ) } , R 0 = 1.5.
We take our numerical results for problem (1), with
a ( x , u ) = { 4 − r 2 + 1 1 + u 2 , 1 ≤ r ≤ 1.5 , 1 1 + u 2 , r > 1.5 ,
u = sin ( y x 2 + y 2 ) .
Furthermore, we let Δ r = 1 m , Δ θ = 15 π 8 M for some integer m , M ∈ N . The exact solution “u” is solved with N = 10 , Δ r = 1 16 , Δ θ = ω 64 .
The numerical results are given in
| mesh | ‖ u − u h ‖ L 2 ( Ω 1 ) | ratio ( L 2 ( Ω 1 ) ) | ‖ u − u h ‖ L ∞ ( Ω 1 ) | ratio ( L ∞ ( Ω 1 ) ) |
|---|---|---|---|---|
| h | 5.93620E-1 | 6.20480E-1 | ||
| h/2 | 2.45574E-1 | 2.41727 | 2.58075E-1 | 2.40426 |
| h/4 | 8.78220E-2 | 2.79628 | 9.23020E-2 | 2.79597 |
| h/8 | 2.26990E-2 | 3.86890 | 2.60230E-2 | 3.54698 |
| mesh | ‖ u − u h ‖ L 2 ( Ω 1 ) | ratio ( L 2 ( Ω 1 ) ) | ‖ u − u h ‖ L ∞ ( Ω 1 ) | ratio ( L ∞ ( Ω 1 ) ) |
|---|---|---|---|---|
| h | 5.20189E-1 | 5.65939E-1 | ||
| h/2 | 2.26124E-1 | 2.30046 | 2.46121E-1 | 2.29943 |
| h/4 | 8.49490E-2 | 2.66189 | 9.24893E-2 | 2.66108 |
| h/8 | 2.26970E-2 | 3.74275 | 2.47190E-2 | 3.74160 |
These are the three error analysis diagrams of Example 2.
The numerical examples show that the errors can be affected by the location of the artificial boundary and the order of the artificial boundary condition, and it can be reduced by refining the mesh and optimizing finite element subdivision methods. The numerical results are coincident with the theoretical analysis and show the efficiency of the coupling method.
This research was funded by the National Natural Science Foundation of China (No. 11401296). The authors are grateful for them.
The authors declare no conflicts of interest regarding the publication of this paper.
Tu, M.Y. and Liu, B.Q. (2023) On the Coupled of NBEM and CFEM for an Anisotropic Quasilinear Problem in an Unbounded Domain with a Concave Angle. American Journal of Computational Mathematics, 13, 185-198. https://doi.org/10.4236/ajcm.2023.131009