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Feb 1, 2008 - R.B. Kellogg was supported by the Boole Centre for Research in Informatics at National University of. Ireland, Cork and by Science Foundation ...
J Sci Comput (2008) 36: 243–261 DOI 10.1007/s10915-008-9187-7

A Tailored Finite Point Method for a Singular Perturbation Problem on an Unbounded Domain Houde Han · Zhongyi Huang · R. Bruce Kellogg

Received: 2 September 2007 / Revised: 15 January 2008 / Accepted: 18 January 2008 / Published online: 1 February 2008 © Springer Science+Business Media, LLC 2008

Abstract In this paper, we propose a tailored-finite-point method for a kind of singular perturbation problems in unbounded domains. First, we use the artificial boundary method (Han in Frontiers and Prospects of Contemporary Applied Mathematics, 2005) to reduce the original problem to a problem on bounded computational domain. Then we propose a new approach to construct a discrete scheme for the reduced problem, where our finite point method has been tailored to some particular properties or solutions of the problem. From the numerical results, we find that our new methods can achieve very high accuracy with very coarse mesh even for very small . In the contrast, the traditional finite element method does not get satisfactory numerical results with the same mesh. Keywords Tailored finite point method · Singular perturbation problem · Unbounded domain · Artificial boundary condition

Han was supported by the NSFC Project No. 10471073. Z. Huang was supported by the NSFC Projects No. 10301017, and 10676017, the National Basic Research Program of China under the grant 2005CB321701. R.B. Kellogg was supported by the Boole Centre for Research in Informatics at National University of Ireland, Cork and by Science Foundation Ireland under the Basic Research Grant Programme 2004 (Grants 04/BR/M0055, 04/BR/M0055s1). H. Han · Z. Huang () Dept. of Mathematical Sciences, Tsinghua University, Beijing 100084, China e-mail: [email protected] H. Han e-mail: [email protected] R.B. Kellogg Department of Mathematics, University of South Carolina, Columbia, SC 29208, USA e-mail: [email protected]

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1 The Model Problem We consider the following model problem: −u + ux1 = 0, u = 1,

∀x ∈ ,

x ∈ ∂B,

(1.1) (1.2)

u|x2 =0,L = 0,

(1.3)

u → 0,

(1.4)

when |x| → +∞,

where  = S \ B, with S = (−∞, ∞) × (0, L), and B ⊂ S is a bounded convex domain. Using the Lax-Milgram lemma [4] (p. 197, Section 6.2.1), one can show that the problem (1.1)–(1.4) has a unique smooth solution in H 1 (). However the problem has several features that make it difficult to obtain an accurate approximation. First, the problem is posed on an unbounded domain, and must be approximated with a finite number of unknowns. Also, when   1 the solution has a boundary layer on a portion of ∂B, and the solution has 2 interior layers on the 2 horizontal rays emanating to the right from the top and bottom of ∂B. These layers are characterized by rapid transitions in the solution, and are thus difficult to capture in a numerical approximation without using a large number of unknowns. Also, such layers tend to cause spurious oscillations in a numerical solution to the problem. The problem (1.1)–(1.4) is similar to a problem suggested by P.W. Hemker [7], and was devised to represent these difficulties. The numerical solution of similar singular perturbation problems in bounded domains have been studied by many mathematicians [1, 2, 8–10, 13, 16–20]. In this paper, first we use the artificial boundary method [5] to reduce the original problem to a problem on bounded computational domain equivalently or approximately. Then we propose an effective numerical method, we called “a tailored finite point method”, to solve the reduced problem numerically. The numerical analysis of our method remains an open problem.

2 The Artificial Boundary Conditions In order to reduce the singular perturbation problem (1.1)–(1.4) to a bounded computational domain, we introduce two artificial boundaries: a = {x = (x1 , x2 ) | x1 = a, 0 ≤ x2 ≤ L} ,

(2.1)

b = {x = (x1 , x2 ) | x1 = b, 0 ≤ x2 ≤ L} ,

(2.2)

with (a, b) × (0, L) ⊃ B. a and b divide the unbounded domain  into three domains: a = {x = (x1 , x2 ) | −∞ < x1 < a, 0 ≤ x2 ≤ L} , b = {x = (x1 , x2 ) | b < x1 < +∞, 0 ≤ x2 ≤ L} , / B} ≡ (a, b) × (0, L) \ B, i = {x = (x1 , x2 ) | a < x1 < b, 0 ≤ x2 ≤ L and x ∈ where i is the bounded computational domain. Consider the restriction of u(x, ), the solution of problem (1.1)–(1.4) on b . u(x, ) satisfies: −u + ux1 = 0,

∀x ∈ b ,

(2.3)

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245

u|x2 =0,L = 0, u → 0,

x1 ∈ [b, +∞),

when |x| → +∞.

(2.4) (2.5)

The problem (2.3)–(2.5) is an incompletely posed problem, which has many solutions. The general solution of the problem is given by u(x, ) =

∞ 

bn e−μn (x1 −b) sin

n=1

with 1 μn = 2



 nπ 2 1 1 + 4 − 2 L 

nπ x2 , L

 n = 1, 2, . . . ,

> 0,

where {bn , n = 1, 2, . . .} can be determined by u(b, x2 , ) = bn =

2 L

∞ 

bn sin

n=1



nπ x2 , L

L

u(b, x2 , ) sin 0

nπ x2 dx2 . L

On the artificial boundary b , we have ∞  nπ ∂u x2 = − μn bn sin ∂x1 x1 =b L n=1 =−

 ∞  2μn n=1

L

L

u(b, ξ, ) sin 0

nπ nπ ξ sin x2 dξ L L

≡ Lb (u|x1 =b ).

(2.6)

The equality (2.6) is the exact boundary condition satisfied by the solution, u(x1 , x2 , ) of the problem (1.1)–(1.4) on the artificial boundary b . Similarly, consider the restriction of u(x1 , x2 , ), the solution of problem (1.1)–(1.4), on the domain a . Then u(x1 , x2 , ) satisfies: −u + ux1 = 0, u|x2 =0,L = 0, u → 0,

∀x ∈ a ,

(2.7)

x1 ∈ (−∞, a],

(2.8)

when |x| → −∞.

The general solution of the problem (2.7)–(2.9) on the domain a is given by u(x1 , x2 , ) =

∞ 

an eλn (x1 −a) sin

n=1

where 1 λn = 2



  nπ 2 1 1 , +4 + 2 L 

nπ x2 , L

n = 1, 2, . . .

(2.9)

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and {an | n = 1, 2, . . .} is given by u(a, x2 , ) = an =

2 L



∞ 

an sin

n=1

nπ x2 , L

L

u(a, ξ, ) sin 0

nπ ξ dξ. L

On the artificial boundary a , we have  ∞ 

nπ nπ 2λn L ∂u ξ sin x2 dξ ≡ La u|x1 =a . = u(a, ξ, ) sin ∂x1 x1 =a n=1 L 0 L L

(2.10)

Using the exact artificial boundary conditions (2.6) and (2.10), the original problem (1.1)–(1.4) is equivalently reduced to the following boundary value problem on the bounded domain i : −u + ux1 = 0,

∀x ∈ i ,

(2.11)

u|∂B = 1,

(2.12)

u|x2 =0 = 0, u|x2 =L = 0,

∂u = La u|x1 =a ∂x1 x1 =a

∂u = Lb u|x1 =b . ∂x1 x1 =b

(2.13) (2.14) (2.15)

In the practical computation, we only take the first N terms (for example, N = 10) of the infinite summation in (2.6) and (2.10). Then we obtain two series approximate artificial boundary conditions on the artificial boundary a and b :  N 

nπ ∂u nπ 2λn L ξ sin x2 dξ ≡ LN = u(a, ξ, ) sin a u|x1 =a , ∂x1 x1 =a n=1 L 0 L L

(2.16)

 N 

nπ nπ 2μn L ∂u ξ sin x2 dξ ≡ LN = − u(b, ξ, ) sin b u|x1 =b . (2.17) ∂x1 x1 =b L 0 L L n=1 By the approximate artificial boundary conditions (2.16) and (2.17), the original problem (1.1)–(1.4) is approximately reduced to the bounded computational domain i : −uN + uN x1 = 0,

∀x ∈ i ,

(2.18)

uN |∂B = 1,

(2.19)

uN |x2 =0 = 0, uN |x2 =L = 0,

N ∂uN = LN a u |x1 =a , ∂x1 x1 =a

N ∂uN = LN b u |x1 =b . ∂x1 x1 =b

(2.20) (2.21) (2.22)

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247

By the method given by Han and Bao [6], it is straightforward to show that the approximate problem is well-posed and the error between u(x, ) and uN (x, ) can be obtained for a fixed  > 0.

3 Finite Point Method In this section we propose a new approach to construct a discrete scheme for (2.18). We call the new scheme “a tailored finite point method” (TFPM), because the finite point method has been tailored to some particular properties or solutions of the problem. The finite point method [3, 11, 12, 15] is a development of finite difference method, in which the meshless technique is emphasized. For one dimensional singular perturbation problem, the method is very close to the method of “exponential fitting” discussed in [1, 8, 13, 14] and [16] (Chapter I, Section 2). First we construct a tailored five-point scheme for the Laplace equation u = 0

(3.1)

to explain the basic ideas of the tailored finite point method. Taking a point x 0 = (0, 0) and four points around it x 1 = (h1 , 0),

x 2 = (0, h2 ),

x 3 = (−h3 , 0),

x 4 = (0, −h4 ),

where hi > 0, (i = 1, 2, 3, 4) are given and 0 < hi  1 (cf. Fig. 1). We try to find a scheme at the point x 0 , i.e., some constants α0 , α1 , α2 , α3 , α4 , such that α1 u1 + α2 u2 + α3 u3 + α4 u4 + α0 u0 = 0

Fig. 1 The location of points x 0 , x 1 , . . . , x 8

(3.2)

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approximates the Laplace equation (3.1) at x = x 0 , where u0 , u1 , u2 , u3 , u4 denote the values of the solution, u(x), of (3.1) at points x0 , x1 , . . . , x4 . How should the constants αi , (i = 0, 1, 2, 3, 4) be found? The solution u(x) can be expanded into the summation of harmonic polynomials at x = 0: u(x) = c0 + c1 x1 + c2 x2 + c3 (x12 − x22 ) + c4 x1 x2 + c5 (x13 − 3x1 x22 ) + c6 (x23 − 3x12 x2 ) + · · · . We take the subspace of harmonic polynomials:

WI4 = v(x) | v(x) = c0 + c1 x1 + c2 x2 + c3 (x12 − x22 ), ∀ci ∈ R, (i = 0, 1, 2, 3) . Now take constants {αi , i = 0, 1, 2, 3, 4} such that α1 v(x 1 ) + α2 v(x 2 ) + α3 v(x 3 ) + α4 v(x 4 ) + α0 v(x 0 ) = 0,

∀v(x) ∈ WI4 .

(3.3)

Taking v(x) = 1, α1 + α2 + α3 + α4 + α0 = 0; taking v(x) = x1 , h1 α1 − h3 α3 = 0; taking v(x) = x2 , h2 α2 − h4 α4 = 0; and v(x) =

x12

− x22 , h21 α1 − h22 α2 + h23 α3 − h24 α4 = 0.

Thus, α1 + α2 + α3 + α4 + α0 = 0, h1 α1 − h3 α3 = 0, h2 α2 − h4 α4 = 0, h21 α1 − h22 α2 + h23 α3 − h24 α4 = 0, from which we obtain α2 =

h11 + h3 h1 α1 , h22 + h2 h4

α3 =

h1 α1 , h3

α4 =

h2 h11 + h3 h1 α1 . h4 h22 + h2 h4

Therefore, α1 = − α2 = −

h2 h3 h4 α0 , h2 (h1 h2 + h3 h4 ) + h1 h3 (h1 + h2 ) h1 h2 h3 h4 (h1 + h3 ) α0 , + h3 h4 ) + h1 h3 (h1 + h2 )

h22 (h1 h2

(3.4)

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249

α3 = −

h1 h2 h4 α0 , h2 (h1 h2 + h3 h4 ) + h1 h3 (h1 + h2 )

α4 = −

h2 h3 h4 h1 (h1 + h3 ) α0 . h4 (h2 + h4 ) h2 (h1 h2 + h3 h4 ) + h1 h3 (h1 + h2 )

In particular, if h1 = h2 = h3 = h4 , we have 1 α1 = − α0 , 4

1 α2 = − α0 , 4

1 α3 = − α0 , 4

1 α4 = − α0 . 4

Let α0 = −4/ h2 , we have the famous five points difference scheme for Laplace equation u1 + u2 + u3 + u4 − 4u0 = 0. h2

(3.5)

Similarly, we can get the nine points scheme for Laplace equation: α1 = −α0 h3

h24 + 3h2 h4 − h3 h1 + h23 + h22 − h21 A

α2 = −α0 h4

h23 + 3h3 h1 + h24 − h2 h4 − h22 + h21 B

α3 = −α0 h1

h24 + 3h2 h4 − h23 + h22 − h3 h1 + h21 A

α4 = −α0 h2

h23 + 3h3 h1 − h2 h4 − h24 + h21 + h22 B

α5 = −α0 h3 h4

h3 h1 − h23 + h21 + h2 h4 + h22 − h24 C

α6 = −α0 h4 h1

h3 h1 + h23 − h21 + h2 h4 + h22 − h24 C

α7 = −α0 h1 h2

h3 h1 + h23 − h21 + h2 h4 − h22 + h24 C

α8 = −α0 h3 h2

h3 h1 − h23 + h21 + h2 h4 − h22 + h24 , C

with A = h1 h24 + h3 h24 + 3h1 h4 h2 + 3h2 h4 h3 + 4h21 h3 + h3 h22 + 4h1 h23 + h31 + h33 + h1 h22 , B = h23 h4 + h23 h2 + 3h3 h1 h4 + 3h3 h1 h2 + h34 + h32 + 4h2 h24 + h2 h21 + 4h22 h4 + h21 h4 , C = h2 h33 + h3 h34 + 4h23 h1 h2 + h3 h32 + 4h23 h1 h4 + 4h3 h21 h2 + 4h22 h3 h4 + h33 h4 + 4h1 h2 h24 + h31 h4 + h1 h34 + 4h22 h1 h4 + h1 h32 + 4h2 h3 h24 + 4h3 h21 h4 + h31 h2 . Again, if h1 = h2 = h3 = h4 , and setting α0 = −20/ h2 , we have the traditional nine points difference scheme 4(u1 + u2 + u3 + u4 ) + (u5 + u6 + u7 + u8 ) − 20u0 =0 h2 for the Laplace equation.

(3.6)

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We now return to the singular perturbation equation (1.1) to construct a tailored five points scheme with h1 = h2 = h3 = h4 = h. Consider the expansion of u(x1 , x2 , ), the solution of (1.1), at x = x 0 . Let x1

u(x1 , x2 , ) = e 2 v(x1 , x2 , ).

(3.7)

Then v(x1 , x2 , ) satisfies: −v +

1 v = 0. 4 2

(3.8)

Let μ=

1 . 2

(3.9)

The solution of (3.8), v(x1 , x2 , ) can be expanded at x = x 0 as: v(x1 , x2 , ) = a0 I0 (μr) +

∞ 

(an In (μr) cos nθ + bn In (μr) sin nθ ) ,

(3.10)

n=1

where (r, θ ) is the polar coordinate of x = (x1 , x2 ) based at x 0 , In is the n-th order modified Bessel function of first kind. We take the first few terms in (3.10) to construct a function space WI4I = {v(x1 , x2 , ) | v = a0 I0 (μr) + a1 I1 (μr) cos θ + b1 I1 (μr) sin θ + a2 I2 (μr) cos 2θ, ∀a0 , a1 , a2 , b1 ∈ R}. We now construct the discrete schemes for (3.8) using the points x 0 , x 1 , x 2 , x 3 , x 4 with h1 = h2 = h3 = h4 = h. We seek α0 , α1 , α2 , α3 , α4 such that α1 v(x 1 ) + α2 v(x 2 ) + α3 v(x 3 ) + α4 v(x 4 ) + α0 v(x 0 ) = 0,

∀v ∈ WI4I .

(3.11)

Taking v(x) = I0 (μr), I1 (μr) cos θ , I1 (μr) sin θ , I2 (μr) cos 2θ , respectively, we have I0 (μh)(α1 + α2 + α3 + α4 ) + I0 (0)α0 = 0, I1 (μh)α1 − I1 (μh)α3 = 0, I1 (μh)α2 − I1 (μh)α4 = 0, I2 (μh)α1 − I2 (μh)α2 + I2 (μh)α3 − I2 (μh)α4 = 0. Hence α1 = α2 = α3 = α4 , 4I0 (μh)α1 + I0 (0)α0 = 0. Taking α0 =

4 + μ 2 h2 h2

(3.12)

J Sci Comput (2008) 36: 243–261

251

then α1 = α2 = α3 = α4 = −

I0 (0) I0 (0) 4 + μ2 h2 α0 = − . 4I0 (μh) I0 (μh) 4h2

(3.13)

We obtain the tailored five-point scheme −I0 (0)(v1 + v2 + v3 + v4 ) + 4I0 (μh)v0 = 0. For the case μh =

h 2

(3.14)

 1, then 1 I0 (μh) ≈ I0 (0) + μ2 h2 + O(μ4 h4 ). 4

From (3.14) we have −

v1 + v2 + v3 + v4 − 4v0 + μ2 v0 = 0. h2

(3.15)

This is the classical five points finite difference scheme for (3.8). Since the solution u of (1.1) satisfies u(x1 , x2 , ) = eμx1 v(x1 , x2 , ), u can be expanded at x = x 0 into:   ∞  u(x1 , x2 , ) = b0 + eμr cos θ a0 I0 (μr) + (an In (μr) cos nθ + bn In (μr) sin nθ ) , (3.16) n=1

where (r, θ ) is the polar coordinate of x = (x1 , x2 ) based at x 0 . We take the first few terms in (3.16) to construct a function space WI4I I = {u(x1 , x2 , ) | u = b0 + eμr cos θ [a0 I0 (μr) + a1 I1 (μr) cos θ + b1 I1 (μr) sin θ ], ∀a0 , a1 , b0 , b1 ∈ R}. We seek α0 , α1 , α2 , α3 , α4 such that α1 u(x 1 ) + α2 u(x 2 ) + α3 u(x 3 ) + α4 u(x 4 ) + α0 u(x 0 ) = 0,

∀u ∈ WI4I I .

(3.17)

This gives the equations α1 + α2 + α3 + α4 + α0 = 0, I0 (μh)(e α1 + α2 + e μh

−μh

(3.18)

α3 + α4 ) + α0 = 0,

(3.19)

α2 − α4 = 0,

(3.20)

e α1 − e μh

−μh

α3 = 0.

(3.21)

Setting β(μh) =

cosh(μh) − I0 (μh) , I0 (μh) − 1

(3.22)

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then α2 = α4 = β(μh)eμh α1 ,

(3.23)

α3 = e2μh α1 ,

(3.24)

so α1 = −

e−μh α0 , 2(cosh(μh) + β(μh))

(3.25)

α3 = −

eμh α0 , 2(cosh(μh) + β(μh))

(3.26)

α2 = α4 = −

β(μh) α0 . 2(cosh(μh) + β(μh))

(3.27)

In this manner we obtain the tailored five-point scheme for (1.1): α1 (μh)u1 + α2 (μh)u2 + α3 (μh)u3 + α4 (μh)u4 + α0 (μh)u0 = 0.

(3.28)

We can obtain a tailored 9-point scheme in a similar manner. Setting   3  8 μr cos θ (an In (μr) cos nθ WI I I = u(x1 , x2 , ) u = b0 + e a0 I0 (μr) + n=1   + bn In (μr) sin nθ ) , ∀an , bn ∈ R , we seek numbers αn , (n = 0, . . . , 8), such that 8 

αn u(x n ) = 0,

∀v ∈ WI8I I .

(3.29)

n=0

Let √ A(μh) = cosh(μh) − I0 ( 2μh), B(μh) = I0 (μh) − cosh2 (μh/2),



C(μh) = 4I0 (μh) cosh(μh) − 2I0 ( 2μh) 1 + cosh(μh) . After some algebra we arrive at the tailored 9-point scheme α1 = −

e−μh A(μh) α0 , C(μh)

(3.30)

A(μh) α0 , C(μh)

(3.31)

α2 = α4 = − α3 = −

eμh A(μh) α0 , C(μh)

α5 = α8 = −

e−μh B(μh) α0 , C(μh)

(3.32) (3.33)

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253

α6 = α7 = −

eμh B(μh) α0 . C(μh)

(3.34)

Remark 3.1 For given point x 0 and the points x 1 , x 2 , . . . , x k around x 0 (k ≥ 4), we can construct the k + 1 points scheme by our finite point method, including irregular meshes. Remark 3.2 It is easy to know that scheme (3.14) and scheme (3.28) satisfy the maximum principle.

4 Properties of the Solution Let (x1 , x2 ) be the point on ∂B where the second component is a minimum, and let (x1 , x2 ) be the point on ∂B where the second component is a maximum. Supposing that B is strictly convex, these points are unique, cf. Fig. 2. (If B is not strictly convex, the following considerations are easily modified.) In this case, when  is very small, it may be expected that the rays {(x1 , x2 ) : x1 > x1 )} and {(x1 , x2 ) : x1 > x1 )} are interior layers of the solution; that is, the solution has a sharp transition across these rays. Also these points divide ∂B into 2 curves, the curve to the left and the curve to the right. Denote the curve to the left by bl . When  is very small, it may be expected that the set bl is a boundary layer of the solution; that is, the solution has a sharp transition on this set. To understand these interior and boundary layers, consider the function ζ = e−d/ . A computation then gives − ζ + ζx1 = − −1 [|∇d|2 + dx1 ]ζ + ( d)ζ. Thus, if d is chosen to satisfy E(d) = 0, one has − ζ + ζx1 = O(1). This suggests that if d satisfies E(d) = |∇d|2 + dx1 = 0, the function ζ might be used to represent some solution properties. We now illustrate this idea in the case of the interior layers and the boundary layer. An example of a function which satisfies E(d) = 0 is d(x1 , x2 ) =

(x2 − x2 )2  . 2( (x1 − x1 )2 + (x2 − x2 )2 + x1 − x1 )

This suggests that near the ray (x1 , x2 ) the solution u of (1.1)–(1.4) behaves like exp{−c(x2 − x2 )2 /}, and so  2 /

ux2 x2 ≈  −1 e−c(x2 −x2 ) Fig. 2 The location of points (x1 , x2 ) and (x1 , x2 )

for x1 > x1 .

(4.1)

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A similar behavior is expected on the ray emanating from the point (x1 , x2 ). These formulas illustrate the interior layer in the solution when  is small. Let δ(x1 , x2 ) be the distance from a point (x1 , x2 ) to the curve bl . Setting d1 = d/δ, one can obtain from the equation E(d) = 0 a nonlinear first order differential equation for d1 , which we write E1 (d1 ) = 0. A solutions of the equation E1 (d1 ) = 0 can be obtained using the method of characteristics [4]. It can be seen that the curve bl is not a characteristic curve for the operator E1 , except at the end points (x1 , x2 ) and (x1 , x2 ). This means that a function d1 can be constructed which satisfies E1 (d1 ) = 0 in a neighborhood of bl and d1 = 1 on bl . Then d = δd1 satisfies E(d) = 0 in a neighborhood of bl , d = 0 on bl , and d/δ bounded and non-zero in a neighborhood of bl . Using the function ζ = e−d/ , this suggests that near bl the solution u of (1.1)–(1.4) behaves like exp{−c∂/}, and so ux1 ≈  −1 e−c∂/

near bl .

(4.2)

5 Numerical Example In this section, we give some numerical examples to show the efficiency of our new method. We will give a comparison between our finite point method (FPM) and the piecewise linear finite element method (FEM).  Suppose Th is a regular and quasi-uniform triangulation on i , s.t. i = K∈Th K, where K is a triangle, h denotes the maximum side of the triangles. Let   Uh = v ∈ H 1 (i ) v|K is a linear polynomial, ∀K ∈ Th , v|∂B = 1, v|x2 =0 = v|x2 =L = 0 ,   Vh = v ∈ H 1 (i ) v|K is a linear polynomial, ∀K ∈ Th , v|∂B = v|x2 =0 = v|x2 =L = 0 . We then will consider the approximate variational problem of (2.18)–(2.22),  Find uh ∈ Uh , such that A(uh , vh ) + B(uh , vh ) = 0, ∀vh ∈ Vh , where





A(u, v) =  B(u, v) =

(5.1)

∇u · ∇v + ux1 v dx1 dx2 ,

i L

N  LN b (u|x1 =b ) − La (u|x1 =a ) v dx2 .

0

Example 5.1 First, we consider the following problem: −u + ux1 = 0,

∀x ∈ ,

(5.2)

u|∂B1 = 1,

(5.3)

u|x2 =0,L = 0,

(5.4)

u → 0,

when |x| → +∞,

where

 = x = (x1 , x2 ) ∈ R2 −∞ < x1 < +∞, 0 < x2 < L \ B1 ,

(5.5)

J Sci Comput (2008) 36: 243–261 Table 1 Example 5.1: L2 errors of the numerical solutions for different methods with N = 10,  = 0.1

Table 2 Example 5.1: L2 errors of the numerical solutions for different methods with N = 10,  = 0.01

Table 3 Example 5.1: L2 errors of the numerical solutions for different methods with N = 10,  = 0.001

Table 4 Example 5.1: L2 errors of the numerical solutions for our nine-point scheme with N = 10, h = 1/8

255 Mesh size h

1

1/2

1/4

1/8

EhF P 9

4.73E-1

1.19E-1

4.04E-2

8.26E-3

EhF P 5

4.67E-1

1.36E-1

4.58E-2

1.06E-2

EhF E

4.64E-1

2.77E-1

6.47E-2

1.92E-2

Mesh size h

1

1/2

1/4

1/8

EhF P 9

5.25E-2

6.83E-1

2.43E-1

1.35E-1

EhF P 5

5.38E-1

2.94E-1

1.57E-1

9.48E-2

EhF E

6.53

4.19

9.95E-1

2.15E-1

Mesh size h

1

1/2

1/4

1/8

EhF P 9

6.81E-1

3.55E-1

2.06E-1

9.17E-2

EhF P 5

6.38E-1

4.12E-1

2.47E-1

1.16E-1

EhF E

5.99E+1

3.44E+1

9.47

2.56



1.0E-4

1.0E-5

1.0E-6

1.0E-7

EhF P 9

9.17E-2

9.28E-2

9.36E-2

9.73E-2

  L B1 = x = (x1 , x2 ) ∈ R2 |x1 | < 1, and x2 − < 1 , 2 with L = 8. In our numerical simulation, we choose a = −2 and b = 2, then we do the simulation on the domain i = (a, b) × (0, L) \ B1 . The numerical results are shown in Tables 1–4 and Figs. 3–6. In the following tables and figures, the ‘exact’ solution u is solved on a large domain (−5, 10) × (0, L) with very fine mesh and N = 100. uFh P 9 is the solution solved by our nine point scheme. uFh P 5 is the solution solved by our five point scheme. uFh E is the solution of (5.1). Furthermore, EhF P 9 =

uFh P 9 − uL2 (i ) , uL2 (i )

EhF P 5 =

uFh P 5 − uL2 (i ) , uL2 (i )

EhF E =

uFh E − uL2 (i ) . uL2 (i )

From Tables 1–4 and Figs. 3–6, we can see that our finite point methods have uniformly convergent rates in L2 -norm, while the standard finite element method can only have good approximations as h  . If h  , the errors in both L2 -norm and L∞ -norm of FEM are O(1).

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Fig. 3 The numerical solutions with different methods for Example 5.1: h = 1/8,  = 0.1

Fig. 4 The numerical solutions with different methods for Example 5.1: h = 1/8,  = 0.01

Fig. 5 The numerical solutions with different methods for Example 5.1: h = 1/8,  = 0.001

P 9 with our nine-point scheme for Example 5.1: h = 1/8 Fig. 6 The numerical solutions uF h

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257

Fig. 7 The errors at x2 = 0 by Example 5.2: h = 1/8,  = 0.1

Example 5.2 Second, we consider the following problem which we rotate the obstacle for π/4: −u + ux1 = 0,

where

∀x ∈ ,

(5.6)

u|∂B1 = 1,

(5.7)

u|x2 =0,L = 0,

(5.8)

u → 0,

(5.9)

when |x| → +∞,

   = x = (x1 , x2 ) ∈ R2 −∞ < x1 < +∞, 0 < x2 < L \ B1 ,  √  √ B1 = x = (x1 , x2 ) ∈ R2 L/2 − 2 < x2 ± x1 < L/2 + 2, ,

with L = 8. In our numerical simulation, we also choose a = −2 and b = 2, then we do the simulation on the domain i = (a, b) × (0, L) \ B1 . The comparison between the nine-point finite point scheme and finite element method are shown in Figs. 7–9. In these figures, we can see that the errors of the finite-point method are much smaller than that of the finite element method, especially when  becomes smaller and smaller.

258

Fig. 8 The errors at x2 = 0 by Example 5.2: h = 1/8,  = 0.01

Fig. 9 The errors at x2 = 0 by Example 5.2: h = 1/8,  = 0.001

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259

Example 5.3 Finally, we consider the following problem in which the obstacle is a circle: −u + ux1 = 0,

∀x ∈ ,

(5.10)

u|∂B1 = 1,

(5.11)

u|x2 =0,L = 0,

(5.12)

u → 0,

(5.13)

when |x| → +∞,

where    = x = (x1 , x2 ) ∈ R2 −∞ < x1 < +∞, 0 < x2 < L \ B1 ,   B1 = x = (x1 , x2 ) ∈ R2 (x2 − L/2)2 + x12 < 1, ,

with L = 8. In our numerical simulation, we also choose a = −2 and b = 2, then we do the simulation on the domain i = {x = (x1 , x2 ) | a < x1 < b, 0 ≤ x2 ≤ L and |x| > 1} ≡ (a, b) × (0, L) \ B1 . The numerical results are shown in Figs. 10–12. In these figures, we can also find that the approximate solutions of the finite-point method are much better than that of the finite element method, especially when  is very small.

Fig. 10 The numerical solutions with different methods for Example 5.3: h = 1/8,  = 0.1

Fig. 11 The numerical solutions with different methods for Example 5.3: h = 1/8,  = 0.01

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Fig. 12 The numerical solutions with different methods for Example 5.3: h = 1/8,  = 0.001

6 Conclusion In this paper, we present a tailored-finite-point method for a kind of singular perturbation problems in unbounded domains. First, we introduce two artificial boundaries and use the artificial boundary method [5] to reduce the original problem to a problem on bounded computational domain equivalently or approximately. Then we propose a tailored finite point method to solve the reduced problem numerically. Our finite point method has been tailored to some particular properties or solutions of the problem. Our numerical results show that our new methods can achieve very high accuracy with very coarse mesh whenever  is small or large. In the contrast, the traditional finite element method or finite difference method can not get the satisfactory numerical results with the same mesh, especially for small .

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