J Sci ComputDOI 10.1007/s10915-013-9798-5

Numerical Approximation of Second-Order EllipticProblems in Unbounded Domains

Tahar Z. Boulmezaoud · Samy Mziou · Tahar Boudjedaa

Received: 11 December 2012 / Revised: 22 July 2013 / Accepted: 17 October 2013© Springer Science+Business Media New York 2013

Abstract This paper deals with the numerical resolution of elliptic problems in unboundeddomains using inverted finite elements. In opposition to conventional approaches which arebased on the truncation of the domain, the suggested method keeps the domain unbounded andis based on a description of the asymptotic behavior in an appropriate functional framework.The method and its mathematical properties are presented first, and some computationalexamples are carried out. The obtained numerical results demonstrate the efficiency of themethod.

Keywords Inverted finite elements · Unbounded domains · Half-line · Approximation ·Weighted spaces

Mathematics Subject Classification 34B40 · 35J48 · 35J50 · 65D99 · 65N99

1 Introduction

Several partial differential equations arising in physics and in engineering are originallyformulated in unbounded regions of space. This is for example the case in geophysics, in

T. Z. Boulmezaoud (B)Laboratoire de Mathématiques de Versailles, Université de Versailles Saint-Quentin-en-Yvelines,45, Avenue des Etats-Unis, 78035 Versailles Cedex, Francee-mail: boulmezaoud@math.uvsq.fr

S. MziouDepartment of Mathematics and Statistics, College of Sciences, Al Imam Muhammad Ibn SaudUniversity (IMSIU), PO Box 90950, Riyadh 11623, Saudi Arabiae-mail: mzious@gmail.com

T. BoudjedaaLaboratoire de Physique Theorique, Faculty of Sciences, University of Jijel,BP 98, 18000 Ouled Aissa, Jijel, Algeriae-mail: boudjedaa@gmail.com

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solid and fluid mechanics, in quantum physics and quantum chemistry, in tomography and inelectromagnetic wave propagation. Dealing with those equations necessitates handling withthe far-field behavior of the solution. From a mathematical point of view, the unboundness ofthe geometrical domain complicates significantly the analysis and the numerical resolutionof arising PDEs.

In view of the importance of the topic, several numerical methods and strategies weredeveloped; we can mention integral equations and boundary elements (see, e.g. [5,13,15]),artificial boundary methods (see, e.g., [4,16,25] or [21]) and infinite elements (see, e.g., [6,14,18,24] or [17]). Although much progress has been made, serious problems remain unsolved.Good examples are approximation of non-linear PDEs or PDEs with varying coefficients.

In this paper, we propose to use a recent method developed in [9] called inverted finiteelements method (IFEM) to solve such problems. The fundamental idea of the IFEM is todecompose the domains into two regions; a bounded region in which finite element are usedand an infinite region in which inverted elements are used, modulo an inversion of a specialgraded mesh. Here, we focus our attention on the one dimensional elliptic equation of theform

− d

dx

(a(.)

du

dx

)(x) + b(x)

du

dx(x) + c(x)u(x) = F(x, u(x)) in �, (1.1)

where

• � is an unbounded interval of R, typically either a half-line � =]1,+∞[ (or ]η,+∞[ forsome η ∈ R), or the whole space � = R,

• a, b and c are given coefficients which satisfy the following conditions

0 < a0 ≤ a(x) ≤ A; |b(x)| ≤ B

(|x |2 + 1)1/2 , |c(x)| �C

|x |2 + 1, a. e. in �,

(1.2)

where a0, A, B and C are constants.

Equations with singular coefficients at the origin are beyond the scope of this paper and willbe discussed in a forthcoming paper [10]. When � is the half-line, � =]1,+∞[, Eq. (1.1)is completed with a Dirichlet condition

u(1) = α, (1.3)

or a Neumann condition

u′(1) = β. (1.4)

The remaining of this paper is organized as follows

• In Sect. 2, we establish a weak formulation of the problem, once the underlying functionalframework has been set. This formulation is based on the use of appropriate weightedspaces.

• In Sect. 3 we expose the inverted finite element method we use for approximating thesolution. After exposing the features of the method, we analyze carefully its convergenceand we give an error estimate.

• In Sect. 4, we present some computational results.

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2 Weak Formulation in Weighted Spaces

Consider the basic weight

〈x〉 = (x2 + 1

)1/2. (2.1)

Given m ∈ N, θ ∈ R and p ∈ [1,+∞], we denote by W m,pθ (�) the space of all the functions

u ∈ D ′(�) whose derivatives of orders less or equal to m satisfy

∀k ≤ m, 〈x〉θ+k−mu(k) ∈ L p(�).

This space is equipped with the norm

‖u‖W m,pθ (�) =

⎛⎝∑

k≤m

∫�

〈x〉(θ+k−m)p|u(k)|pdx

⎞⎠

1/p

.

It is well known (see, e.g., [20]) that D(�) is dense in W m,pθ (�). Let W m,p

θ (�) be the closureof D(�) in W m,p

θ (�). Then W m,pθ (R) = W m,p

θ (R) and

W m,pθ (�) =

{v ∈ W m,p

θ (�) | v(η) = v′(η) = · · · = v(m−1)(η) = 0}.

when � =]η,+∞[.Define W −m,p′

−θ (�) as the dual of W m,pθ (�) for m ≥ 0. When p = 2, the spaces

W m,pθ (�), W m,p

θ (�) are denoted W mθ (�), W m

θ (�) (p is dropped).Notice that the local properties of W m,p

α (�) coincide with those of the usual Sobolevspace W m,p . The reader can consult, e.g., [20,22] and [23] for more details about propertiesof these spaces. We underline, however, that spaces W m,p

α were used with success for studyingelliptic problems in the whole space (see, e.g., [2,19] and [3]), in exterior domains (see[19])and in the half-space (see, e.g., [8]). They were also used for studying some PDE’s arisingfrom fluid mechanics (see, e.g., [1,7,11] and [12]).

It causes no loss of generality to assume that � =]1,+∞[. Throughout this paper, weoften need the following weak integration by parts formula

∀u ∈ W 1θ (�), ∀v ∈ W 1

1−θ (�),

∫�

u′(x)v(x)dx = −u(1)v(1) −∫�

u(x)v′(x)dx .

where θ is an arbitrary real number. This formula holds true when � = R, provided that theterm u(1)v(1) is dropped.Let us consider Eq. (1.1) in which F depends only on x (the linear case)

− d

dx

(a(.)

du

dx

)(x) + b(x)

du

dx(x) + c(x)u(x) = f (x) in �. (2.2)

Equation (2.2) is completed at the moment by the Dirichlet condition

u(1) = 0. (2.3)

We search for solutions in W 10 (�) i.e. that satisfy

+∞∫1

|u(x)|2x2 + 1

dx < +∞,

+∞∫1

|u′(x)|2dx < +∞. (2.4)

The following assumptions are made

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(H1) a ∈ L∞(�) and for some constant a0 > 0

a(x) ≥ a0 a. e. in �.

(H2) b ∈ W 1,∞2 (�) and c ∈ W 0,∞

2 (�), that is, there exists three constants b1, b2 and c1

|b(x)| ≤ b1

〈x〉 , |b′(x)| ≤ b2

〈x〉2 , |c(x)| ≤ c1

〈x〉2 , a. e. in �,

(H3) f ∈ W −10 (�).

The last assumption (H3) is in particular valid when f ∈ W 01 (�), say

+∞∫1

(x2 + 1)| f (x)|2dx < +∞. (2.5)

So, data f is not assumed to be compactly-supported.

Define the bilinear form

A (u, v) =∫�

a(x)u′(x)v′(x)dx +∫�

b(x)u′(x)v(x)dx +∫�

c(x)u(x)v(x)dx . (2.6)

This bilinear form is continuous over W 10 (�)2, thanks to assumptions (H1) and (H2). The

proof of the following lemma is straightforward

Lemma 2.1 Suppose that assumptions (H1), (H2) and (H3) are fulfilled. A function u ∈W 1

0 (�) is solution of (2.2)and (2.3) iff

∀v ∈ W 10 (�), A (u, v) = 〈 f, v〉, (2.7)

We need the following usual Hardy’s inequality

Lemma 2.2 There exists a constant π0 > 0 such that

∀u ∈ W 10 (�),

+∞∫1

|u′(x)|2dx ≥ π0

+∞∫1

|u(x)|2|x |2 + 1

dx . (2.8)

The following assumption is made

(H4) there exists a constant π ′0 < π0, such that

c(x) − b′(x)

2≥ −π ′

0a0

〈x〉2 , a. e. in �, (2.9)

This condition is obviously valid when the LHS is nonnegative.

It follows from this assumption that

A (u, u) =∫�

a(x)|u′(x)|2dx +∫�

[c(x) − b′(x)

2] |u(x)|2dx

≥ a0

∫�

∣∣u′(x)∣∣2

dx − π ′0a0

∫�

|u(x)|2〈x〉2 dx

≥ a0(1 − π ′0

π0)

∫�

∣∣u′(x)∣∣2

dx .

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This proves the coerciveness of A and we deduce from Lax–Milgram Theorem this

Proposition 2.3 Suppose that assumptions (H1), (H2), (H3) and (H4) are valid. Then, Eq.(2.2) with Dirichlet condition (2.3) admits a unique solution u ∈ W 1

0 (�) and

‖u‖W 10 (�) � ‖ f ‖W−1

0 (�)(2.10)

Here, the notations A1 ∼ A2 (resp. A1 � A2) means that there exists two constants c1 > 0and c2 > 0 (resp. a constant c1 > 0) not depending on the involved functions such thatc2 A2 ≤ A1 ≤ c1 A2 (resp. A1 � c1 A2).

Remark 2.4 One can also consider Eq. (2.2) in the whole space R. The arguments are thesame, provided that the following inequality is used instead of Hardy inequality (2.8) whenc = 0

∀v ∈ W 10 (R), inf

k∈R

‖v − k‖W 10 (R) ≤ κ‖v′‖L2(R).

We can also impose instead of the Dirichlet condition (2.3) a Neumann condition of the form

(au′)(1) = β, (2.11)

where β ∈ R is given. Observe that this condition is meaningfull when u ∈ W 10 (�) and

f ∈ W 01 (�) since au′ is H1

loc(�) (recall that H1(]1, 2[)↪→C0([1, 2])). The variationalformulation of the problem (2.2) and (2.11) writes: find u ∈ W 1

0 (�) such that

∀v ∈ W 10 (�), A (u, v) =

∫�

f (x)v(x)dx − βv(1). (2.12)

with A (., .) given by (2.6). We know that

∀v ∈ W 10 (�), |v(1)| ≤ e‖v‖W 1

0 (�), (2.13)

for some constant e > 0. Thus,

A (v, v) ≥ a0

∫�

∣∣∣∣dv

dx(x)

∣∣∣∣2

dx +∫�

(c(x) − b′(x)

2)|v(x)|2dx

− max(b(1), 0)e2

2‖v‖2

W 10 (�)

.

Now, let us make the following assumption

(H ′4) there exists a constant α0 > 0 such that

c(x) − b′(x)

2≥ α0

|x |2 + 1, a. e. in �, (2.14)

b(1) <2 min(a0, α0)

e2 . (2.15)

Proposition 2.5 Suppose that assumptions (H1), (H2) and (H ′4) are true and that f ∈

W 01 (�). Then, problem (2.2)and (2.11) admits a unique solution u ∈ W 1

0 (�). Moreover,

‖u‖W 10 (�) � ‖ f ‖W 0

1 (�). (2.16)

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Remark 2.6 Proposition 2.5 cannot be applied directly to problems of the form

− d

dx

(a(.)

du

dx

)(x) = f (x) in �, (au′)(1) = β, (2.17)

which corresponds to the case b = 0 and c = 0. Actually, uniqueness in W 10 (�) is obviously

lost (constant functions are solutions to the corresponding homogenous equation). In addition,taking v = 1 in (2.12) leads to the necessary compatibility condition

β =∫�

f (x)dx, (2.18)

which must be satisfied by the data. This condition says that the Neumann problem for Eq.(2.2) is in general ill-posed in W 1

0 (�), except when β is given by (2.18). In that case, thesolution is the same as that of the Dirichlet problem (2.2) and (2.3), up to an additive constant.

3 Inverted Finite Elements Method

In this section focus is made on the discretization of the elliptic problem (2.2). The invertedfinite elements method [9] will be employed and the error of approximation will be analysed.Some numerical results will be shown.

3.1 The Discrete Space

Let us decompose the domain � into two subdomains, �0 a bounded domain and �∞ anunbounded one, such

� = �0 ∪ �∞, (3.1)

If � =]1,+∞[, then �0 =]1, R[ and �∞ =]R,+∞[ where R > 1 is a fixed parameterthat is not necessarily large and not destined to go to infinity. In the domain �0 =]1, R[,we intend to use standard finite elements. We consider a usual subdivision x0 = 1 < x1 <

· · · < xN = R of �0 and we set h = maxi∈I

|xi+1 − xi | and i =]xi , xi+1[ for i ∈ I , where

I = {0, . . . , N − 1}. We consider the inversion mapping

x �→ t (x) = R2

x, (3.2)

which maps the unbounded subdomain �∞ into the bounded one � =]0, R[. Given a functionw defined on �∞, we denote by w the function defined on � by

w(x) =(

R

x

)γ

w(t (x)), for all x ∈ �. (3.3)

where here and subsequently γ is a fixed real parameter.At this stage, we need for later use some new spaces. Given m ≥ 0, θ ∈ R and p ∈

[1,+∞[, we denote by V m,pθ (�) the space composed of functions satisfying

∀k ≤ m, |x |θ+k−mu(k) ∈ L p(�).

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This space is equipped with the norm

‖u‖V m,pθ (�) =

⎛⎜⎝∑

k≤m

∫

�

|x |(θ+k−m)p|u(k)|pdx

⎞⎟⎠

1/p

.

Now, let Pk, k ≥ 1, denotes the space of polynomials of degree less or equal to k and◦Pk the

subspace of Pk composed of elements satisfying p(0) = 0. Recall the Deny-Lions inequality:for w ∈ Hk+1(]0, 1[)

infp∈Pk

‖w − p‖Hk+1(]0,1[) � |w(k+1)|L2(]0,1[). (3.4)

The following inequality is an extension of (3.4) to V k+1θ (]0, 1[) (see [9], Proposition 3): for

all k ∈ N� and θ ∈ R such that k − 1/2 < θ < k + 1/2, one has

∀u ∈ V k+1θ (]0, 1[), inf

p∈◦Pk

‖u − p‖V k+1θ (]0,1[) � |u|V k+1

θ (]0,1[). (3.5)

We also have (see [9]):

Lemma 3.1 Let w ∈ W m,pα (�) for m ∈ N and 1 ≤ p < +∞. Then, w ∈ V m,p

δ (�) with

δ = γ + 2m − α − 2

p.

The discretization over �∞ =]R,+∞[ is obtained from the discretization on � =]0, R[by using the transformation defined in (3.2). Over � a graded subdivision is constructed asfollows.

Definition 3.2 Given a real number μ (0 < μ ≤ 1), we say that a family of subdivisions((xi )1≤i≤M

), x0 = 0 < x1 < · · · < xM = R, of the interval � is μ-graded if there exists

three constants κi > 0, 1 ≤ i ≤ 3, not depending on the subdivision, such that

κ1h1/μ ≤ x1 ≤ κ2h1/μ, (3.6)

∀ 1 ≤ i ≤ M − 1, xi+1 − xi ≤ κ3h x1−μi , (3.7)

where h = max1≤i≤M−1(xi+1 − xi ).

Notice that a 1-graded subdivision is a subdivision in which xi+1−xi � h for all i ≤ M−1.A way to construct a μ-graded family of meshes is as follows; consider the increasing

finite sequence (θ�i )1≤i≤M defined by θ�

1 = 1, θ�i+1 = θ�

i + (θ�i )1−μ, for 1 ≤ i < M.

Then, the subdivision of � =]0, R[

xi = θ�i

θ�M

R, for 1 ≤ i ≤ M,

is μ-graded. Indeed, one can observe that the difference θ�i+1 − θ�

i increases and

h = maxi

(xi+1 − xi ) = R

θ�M

maxi

(θ�i+1 − θ�

i ) = R

(θ�M )μ

,

xi+1 − xi = Rμ

(θ�M )μ

x1−μi = Rμ−1h x1−μ

i ,

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x1 = R

θ�M

= R1−1/μh1/μ,

In Fig. 1 we display three graded meshes for μ = 1 (no gradation), μ = 0.7 and μ = 0.5.Consider now a family of μ-graded subdvisions ((xi )0≤i≤M ), x0 = 0 < x1 < · · · <

xM = R, over �. Let J = {0, 1, . . . , M − 1} and i =]xi , xi+1[, for i ∈ J .For k ≥ 0 is a fixed integer, we define the finite dimensional spaces

Wh(�) ={v ∈ C 0(�); v|i ∈ Pk(i ), ∀i ∈ I,

v|i∈ Pk(i ), ∀i ∈ J, and v(0) = 0

}, (3.8)

Wh(�) = {vh ∈ Wh(�) | vh(1) = 0}. (3.9)

Observe that functions of Wh(�) are piecewise affine only in the finite element region �0,but not in the region �∞ in which inverted elements are used. Figure 2 in Sect. 4 illustratesan example of such functions.

Before stating approximation results, let us observe that the space Wh(�) depends mainlyon the discretization parameter h, which is supposed to tend to zero, and on three adjustmentparameters R, γ and μ which are appropriately fixed. Moreover, we have

Lemma 3.3 Suppose that γ > −3/2. Then

Wh(�) ⊂ W 10 (�). (3.10)

So, we assume henceforth that

γ > −3

2. (3.11)

Consequently, the discretization of the problem (2.7) is the following:Find uh ∈ Wh(�) such that

∀vh ∈ Wh(�), A (uh, vh) = 〈 f, vh〉. (3.12)

It follows immediately that this problem is well-posed and has a unique solution uh ∈ Wh(�).

Fig. 1 An illustration of agraded meshing of the interval]0, 2[ for μ = 1, 0.7, 0.5

μ=0.75

μ=0.5

μ=1

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0 0.5 1 1.5 2 2.5 3 3.5 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1ω

8ω

3

ω*3

ω5

ω*5

ω7

ω*7

R

Fig. 2 Some of the basic functions when N = M = 8 and k = 1. Observe that the functions w j , 1 ≤ j ≤N − 1, are piecewise affine over �. The function wN is piecewise affine only within �0

3.2 Error Estimate

We state this

Theorem 3.4 Suppose that assumptions of Proposition 2.3 are fulfilled. Suppose also thatu ∈ W k+1

k+η (�) for some real η > 0 and that

η − 3

2< γ < η − 1

2. (3.13)

Then, the following estimate holds

‖u − uh‖W 10 (�) � hk‖u‖Hk+1(�0) + hk min(μ�,μ)/μ‖u‖W k+1

k+η (�∞)(3.14)

where μ� = ηk > 0.

Proof From Céa’s lemma we know that

‖u − uh‖W 10 (�) � inf

wh∈Hh(�)‖u − wh‖W 1

0 (�).

Let Ih be the interpolation operator defined from C 0(�) into Wh(�) as follows: for allv ∈ C 0(�), Ihv is the only element of Wh(�) satisfying

Ihv(xi,�) = v(xi,�) for 0 ≤ i < N , 0 ≤ � ≤ k,

Ihv(xi,�) = v(xi,�) for 0 ≤ i < M, 1 ≤ � ≤ k,

where the points (xi,�) and (xi,�) are defined by formula (4.1) and (4.2) hereafter. Notice thatIhv(0) = 0 and that Ihv ∈ Wh(�) when v(1) = 0.In the finite element region �0, we know that

‖u − Ihu‖H1(�0) � hk‖u − Ihu‖Hk+1(�0).

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It remains to estimate the difference u − Ihu in the unbounded region �∞ where invertedelements are used. Since u ∈ W k+1

k+η (�), we know from Lemma 3.1, that u ∈ V k+1k+δ (�) with

δ = γ + 1 − η.For 1 ≤ i ≤ M − 1 (0 �∈ i ) and � ∈ {0, 1}, we have

∥∥∥(u − Ihu)(�)∥∥∥2

V 0γ+�(i )

=∫

i

x2(γ+�)∣∣∣(u − Ihu)(�)(x)

∣∣∣2dx

� (xi )2(γ+�)

∫

i

∣∣∣(u − Ihu)(�)(x)

∣∣∣2dx

� (xi )2(γ+�)hi

∫]0,1[

∣∣∣(u − Ihu)(�) ◦ φi (x)

∣∣∣2dx,

� (xi )2(γ+�)(hi )

1−2�

∫]0,1[

∣∣∣(u ◦ φi − Ihu ◦ φi )(�)(x)

∣∣∣2dx .

where hi = xi+1 − xi and φi (x) = (xi+1 − xi )x + xi . One can easily prove that

Ihu ◦ φi = I0(u ◦ φi ),

where I0 is the interpolation operator from C 0([0, 1]) on Pk defined as: for all w ∈C 0([0, 1]), I0w is the unique element of Pk satisfying

I0w

(�

k

)= w

(�

k

)for 0 ≤ � ≤ k.

Since I0 p = p for each p ∈ Pk , we get

‖(u − Ihu)(�)‖2V 0

γ+�(i )� (xi )

2(γ+�)(hi )1−2� inf

p∈Pk‖u ◦ φi − p‖2

Hk+1(]0,1[).

Using the classical Deny–Lions inequality (3.4) we deduce that

‖(u − Ihu)(�)‖2V 0

γ+�(i )� (xi )

2(γ+�)(hi )1−2�

∫]0,1[

∣∣∣∣dk+1(u ◦ φi )

dxk+1

∣∣∣∣2

dx

� (xi )2(γ+�)(hi )

2(k+1−�)+1∫

]0,1[

∣∣∣∣dk+1u

d xk+1 ◦ φi

∣∣∣∣2

dx,

� (xi )2(γ+�)(hi )

2(k+1−�)

∫

i

∣∣∣∣dk+1u

d xk+1 (x)

∣∣∣∣2

dx,

� (xi )2(γ+�−k−δ)(hi )

2(k+1−�)‖u‖2V k+1

k+δ (i ),

� x2(γ+1−δ−(k+1−�)μ)

i h2(k+1−�)‖u‖2V k+1

k+δ (i ),

� x2(μ�−μ)k−2(1−�)μi h2(k+1−�)‖u‖2

V k+1k+δ (i )

,

� h2k min(μ�

μ−1,0)h2k‖u‖2

V k+1k+δ (i )

,

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where we used the gradation assumptions (3.6)–(3.7). Thus,

‖u − Ihu‖2V 1

γ+1(i )� h2k min(

μ�

μ,1)‖u‖2

V k+1k+δ (i )

.

It remains to estimate ‖u − Ihu‖2V 1

γ+1(0). With the same calculus we have

‖(u − Ihu)(�)‖2V 0

γ+�(0)=

∫

0

x2(γ+�)∣∣∣(u − Ihu)(�)(x)

∣∣∣2dx

� (h0)2γ+1

∫]0,1[

x2(γ+�)∣∣∣(u ◦ φ0 − Ihu ◦ φ0)

(�)(x)

∣∣∣2dx .

� (h0)2γ+1 inf

p∈◦Pk

‖u ◦ φ0 − p‖V k+1k+δ (]0,1[).

since I0 p = p for each p ∈ ◦Pk . Using inequality (3.5) and conditions (3.11), (3.13) gives

‖(u − Ihu)(�)‖2V 0

γ+�(0)� (h0)

2γ+1∫

]0,1[x2(δ+k)

∣∣∣∣dk+1(u ◦ φ0)

dxk+1

∣∣∣∣2

dx

� (h0)2(γ+k+1)+1

∫]0,1[

x2(δ+k)

∣∣∣∣dk+1u

d xk+1 ◦ φ0

∣∣∣∣2

dx,

� (h0)2(γ+1−δ)

∫

0

x2(δ+k)

∣∣∣∣dk+1u

d xk+1

∣∣∣∣2

dx,

� (h0)2η‖u‖2

V k+1k+δ (�)

,

� h2kμ�/μ‖u‖2V k+1

k+δ (�).

By Lemma 3.1, we get

‖u − Ihu‖2W 1

0 (�∞)� h2k min(

μ�

μ,1)‖u‖2

W k+1k+η (�∞)

.

��

According to Theorem 3.4, an appropriate choice of γ when the regularity of the solutionand its behavior at large distances are unknown is γ = −1 (or between −3/2 and −1/2).We also state this

Corollary 3.5 Suppose that assumptions of Proposition 2.3 hold. Suppose also that u ∈W k+1

2k (�) and that let γ = k − 1. Then, for any μ ∈]0, 1]

‖u − uh‖W 10 (�) � hk‖u‖Hk+1(�0) + hk‖u‖W k+1

2k (�∞). (3.15)

Remark In [8], we present a slightly revisited version of the IFEM in which the conditionv(0) = 0 in (3.8) is dropped.

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4 Numerical Implementation and Computational Tests

The goal of this section is twofold. Firstly, we give some details about the structure and thecomputation of the stiffness matrix from a practical viewpoint. Secondly, we present somecomputational results which confirm the convergence of the method and its performance.

4.1 Basis Functions. The Stiffness Matrix

To construct a basis for the space Wh(�), we also introduce the following discretized pointsin �0 and �∞ defined respectively by

xi,r = xi + r

k(xi+1 − xi ), ∀i ∈ I, ∀0 ≤ r ≤ k. (4.1)

Similarly, we consider the points (xi,r )0≤i≤M−1, 0≤r≤k defined by

xi,r = xi + r

k(xi+1 − xi ), ∀0 ≤ r ≤ k. (4.2)

Let wi,r , i ∈ I and 0 ≤ r ≤ k − 1, be the usual finite element basis functions defined over� and satisfying: for all i ∈ I and 0 ≤ r ≤ k − 1

• wi,r belong to Wh(�),• wi,r (x j,s) = δi, jδr,s , for all j ∈ I and 0 ≤ s ≤ k − 1,• wi,r (xN ) = 0.

Obviously supp wi,r ⊂ ∪|i− j |≤1 j if r = 0 and supp wi,r ⊂ i if 0 < r ≤ k − 1. Similarly,define the second family of basis functions (w�

i,r )i∈J,0≤r≤k−1 as follows:

• w�i,r belong to Wh(�),

• w�i,r (x j,s) = δi, jδr,s , for all j ∈ I and 0 ≤ s ≤ k − 1,

• w�i,r (xM ) = 0.

Here supp w�i,r ⊂ ∪|i− j |≤1 j if r = 0 and supp w�

i,r ⊂ i if 0 < r ≤ k−1. Unless confusionarises, the functions wi,0, 0 ≤ i ≤ N − 1, and w�

j,0, 0 ≤ j ≤ M − 1, are denoted wi andw�

i respectively (the index 0 is dropped). Notice that the functions (w�i ) are not piecewise

affine (as displayed in Fig. 2). The last basis function, denoted here by wN , is a mixed one;its support is spanning FEM and IFEM regions. It is the unique function of Wh(�) satisfying

• wN (xN ) = wN (xM ) = 1,• wN (x j,s) = 0 for 0 ≤ j ≤ N − 1 and 0 ≤ s ≤ k − 1.• wN (x j,s) = 0, for all 0 ≤ j ≤ M − 1 and 0 ≤ s ≤ k − 1.

One can easily prove that the family composed of the functions(wi,r )0 ≤ i ≤ N − 1,0 ≤ r ≤ k − 1,(i,r) �= (0,0), (w�

i,r )0 ≤ i ≤ M − 1,0 ≤ r ≤ k − 1,(i,r) �= (0,0) and wN is a

basis of Wh(�). It follows that

dim Wh(�) = (N + M)k − 1. (4.3)

An element vh ∈ Wh(�) can be decomposed into the form

vh =N∑

i=1

vh(xi )wi +N−1∑i=0

k−1∑�=1

vh(xi,�)wi,�

+M−1∑i=1

vh(xi )w�i +

M−1∑i=0

k−1∑�=1

vh(xi,�)w�i,�

denoted by wi (i ∈ I ) in �0 and w�j ( j ∈ J ) in �∞ when k = 1, γ = 2 and R = 1.

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Now, we denote by (φm)1≤m≤d , d = (N + M)k − 1, this basis. The formulation (3.12) isreduced to a linear system

AX = B, (4.4)

where A = (a(φ j , φi )). Let us sketch the form of the matrix A in the case k = 1, for whichit is tridiagonal. The coefficients a(wi , w j ), 1 ≤ i, j ≤ N , can be computed easily as in theusual finite element method. The coefficients a(w�

i , w�j ), 1 ≤ i, j ≤ M − 1, corresponding

to inverted finite elements satisfy

A(w�j , w

�i ) =

∑�

∫

k

a(t (s))( s

R

)2γ ( γ

Rw�

j (s) + s

Rw�′

j (s))

×( γ

Rw�

i (s) + s

Rw�′

i (s))

ds

+∫

k

b(t (s))( s

R

)2γ−1 ( γ

Rw�

j (s) + s

Rw�′

j (s))

w�i (s)ds

+∫

k

c(t (s))( s

R

)2γ−2w�

j (s)w�i (s)ds,

where the sum is taken over all integers � such that |� − i | ≤ 1 and |� − j | ≤ 1 (that is,� ∈ {i −1, i, i +1}∩{ j −1, j, j +1}). In practice, these integrals might be calculated exactlywhen the coefficients a, b and c are constants. In other cases, one can use a Gauss–Lobattoquadrature rule on intervals k when 0 �∈ k . When 0 ∈ k , the integrals are singular, butthey converge since w�

i (0) = w�j (0) = 0 and γ satisfies (3.11). In the latter case, integrals

can be computed explicitly by developing the coefficients a, b and c near the origin.

4.2 Computational Tests

The aim here is to display some computational results obtained with a code in which P1-

elements of Lagrange (k = 1) were used in both the FEM and IFEM regions. The domainwe consider is the half-space � =]1,+∞[. The number of discretized equidistant points isN in �0 =]1, R[ and M in �. The gradation is done by the method explained in Sect. 3.1.In all the tests M = 2N and

h = max0≤i≤N−1

|xi+1 − xi |, h = max0≤i≤M−1

|xi+1 − xi |.

We test the impact of several parameters on the convergence of the method.

Example 1 Simple Dirichlet problem We consider Eq. (1.1) in the half-line � =]1,+∞[,when a(.) ≡ 1, b(.) ≡ 0, c(.) ≡ 0, and

f (x) = 6αx − 1(

1 + (x − 1)2)α+1 − 4α(α + 1)

(x − 1)3

(1 + (x − 1)2

)α+2 , x > 1,

where α > 0 is a constant exponent. This problem is completed with the dirichlet Boundarycondition u(1) = 0. The exact solution is given by

u(x) = x − 1(1 + (x − 1)2

)α , x > 1.

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Observe that

u(x) ∼ x1−2α when x → +∞.

It follows that u ∈ W 1η (�) for any η < 2α − 1/2. In particular u ∈ W 1

0 (�) iff α > 1/4.When α < 1/2, one has limx→+∞ u(x) = +∞.

Figure 3 shows the exact and approximated solution in the interval [0, 50] for N = 100 andM = 2N = 200 with following values of parameters: α = 0.25, 0.5, 0.6, R = 10, μ = 0.5and γ = −0.5. In all the cases the two solutions match both in the FEM and IFEM regions.Observe that when α = 0.25 or α = 1, u(x) does not decrease when x → +∞; indeed,limx→+∞ u(x) = +∞ when α = 0.25, and limx→+∞ u(x) = 1 when α = 1.

Figure 4 shows in a logarithmic scale, the relative weighted errors e1 and e2 as a functionof the stepsize h for different values of the parameter μ: μ = 1, 0.7, 0.5, 0.2, 0.1 (α =

5 10 15 20 25 30 35 40 45 500

1

2

3

4

5

6

7

x

Sol

utio

n

Exact solution with α=0.6 IFEM Approx (N=100, M=200) with α=0.6 Exact solution with α=0.5 IFEM Approx (N=100, M=200) with α=0.5 Exact solution with α=0.25 IFEM Approx (N=100, M=200) with α=0.25

α=0.25

α=0.5α=0.6

IFEMFEM

Fig. 3 Exact and approximate solutions of the first example for several values of α

10−310−210−1

10−4

10−3

10−2

10−1

Stepsize h

Rel

ativ

e er

ror

e 1

10−4

10−3

10−2

10−1

Rel

ativ

e er

ror

e 2

μ=1 (slope 0.08)μ=0.7 (slope 0.11)μ=0.5 (slope 0.15)μ=0.2 (slope 0.34)μ=0.1 (slope 0.45)

10−310−210−1

Stepsize h

μ=1 (slope 0.08)μ=0.7 (slope 0.11)μ=0.5 (slope 0.15)μ=0.2 (slope 0.34)μ=0.1 (slope 0.45)

Fig. 4 (Example 1) Relative weighted errors e1 (left) and e2 (right) versus the stepsize h for several valuesof μ (α = 0.6, R = 2 and γ = −0.5)

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0.6, R = 2 and γ = −0.5). The relative errors e1 and e2 are defined by

e1 =‖u − uh‖W 0−1(�)

‖u‖W 0−1(�)

, e2 = ‖∇u − ∇uh‖L2(�)

‖∇u‖L2(�)

.

We observe that the curves decrease quasi-linearly with h. It is quite clear that the gradationof the mesh significantly improves the decreasing of the error. This is in accordance withforecasts of Theorem 3.4.

In Fig. 5, the relative weighted errors e1 and e2 are displayed as function of the gradationparameter μ for different values of the stepsize h (α = 0.6, R = 2 and γ = −0.5). It becomesobvious that the gradation of the mesh has a direct impact on the precision of approximation:the error decreases continually with decrease of μ, until the optimal value μ = 0.2.

Let us now illustrate the influence of the parameter γ when all the other parameters arefixed (R = 2, μ = 0.5, α = 0.6). According to (3.11), it is desirable that γ > −1.5 (forγ ≤ −1.5, some of the arising integrals do not converge). In Fig. 6 we display errors e1 ande2 versus γ for several values of h. One can notice that γ has only a limited influence on theerror, despite the fact that values around γ = 0.5 seem to be optimal.

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

2

4

6

8

10

12

14 x 10−3

mu

hmax=2.0*10−2

hmax=1.0*10−2

hmax=2.0*10−3

hmax=1.0*10−3

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0

2

4

6

8

10

12

14

16

18x 10

−3

mu

hmax=2.0*10−2

hmax=1.0*10−2

hmax=2.0*10−3

hmax=1.0*10−3

Rel

ativ

e er

ror

e 1

Rel

ativ

e er

ror

e 2

Fig. 5 (Example 1) Relative weighted errors e1 (left) and e2 (right) as a function of the gradation parameterμ (α = 0.6, R = 2 and γ = −0.5)

−1.5 −1 −0.5 0 0.5 1 1.5 210−7

10−6

10−5

10−4

10−3

10−2

10−1

100

Gamma

h=1.0*10−3

h=5.0*10−3

h=2.0*10−2

−1.5 −1 −0.5 0 0.5 1 1.5 2

Gamma

h=1.0*10−3

h=5.0*10−3

h=2.0*10−2

Rel

ativ

e er

ror

e 1

10−7

10−6

10−5

10−4

10−3

10−2

10−1

100

Rel

ativ

e er

ror

e 1

Fig. 6 (Example 1) Relative weighted errors e1 (left) and e2 (right) versus γ for several values of h (R =2, μ = 0.5, α = 0.6

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Example 2 A Dirichlet problem with varying coefficients. We consider problem (1.1) in� =]1,+∞[ when the coefficient a oscillates at near and far distances

a(x) = 1 + 0.2 cos (40(x − 1)), b(x) = 0 and c(x) = 1.

The data f is chosen such that the exact solution is given by

u(x) = sin (20(x − 1))

1 + (x − 1)2 .

Figure 7 shows the analytical and approximated solutions for Example 2 in the interval[0, 6] for M = 2N with N = 20 or N = 100 and with R = 2 and γ = 1. We can observe

1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

x

Exact solution

IFEM approximed sol. (N = 20, M =40)IFEM approximed sol. (N = 100, M =200)

Fig. 7 (Example 2) Exact and approximated solutions on the interval [0, 6] when M = 2N = 40 and whenM = 2N = 200

10−410−310−210−1 10−410−310−210−110−7

10−6

10−5

10−4

10−3

10−2

10−1

Stepsize h

10−6

10−5

10−4

10−3

10−2

10−1

Stepsize h

Rel

ativ

e er

ror

e 1

Rel

ativ

e er

ror

e 2

Fig. 8 (Example 2) Relative weighted errors e1 (left) and e2 (right) as a function of the stepsize h (μ =0.5, R = 2 and γ = 1)

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that both in the FEM and IFEM regions the two solutions match when h is sufficiently small(here for N = 100 and M = 2N = 200).

Figure 8 shows in a logarithmic scale, the relative weighted errors e1 and e2 as a functionof the stepsize h for the choice of parameters μ = 0.5, R = 2 and γ = 1. We observe asuper-convergence phenomena since the error e1 decreases as h2, while the error e2 decreasesas h1.6.

5 Conclusion

The new innovation of this paper is proving the efficiency of the inverted finite elementmethod in solving second order elliptic problems. The method keeps all the advantages ofthe finite element method and avoids the addition of any artificial boundary, preserving bythe way the unboundness of the computational domain. Furthermore, it does not require alot of knowledge concerning the behavior of the solutions at large distances and it allows totackle equations with infinitely varying coefficients. Lastly, the method leads in practice tolinear systems with sparse matrices and whose size and shape are similar to those obtainedwith finite elements in bounded domains.

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