A Parameter-Uniform Numerical Method for a Singularly Perturbed Robin Problem with a Discontinuous Source Term

A Parameter-Uniform Numerical Method for a Singularly Perturbed Robin Problem with a Discontinuous Source Term

Janet Rajaiah

and Valarmathi Sigamani

1

Department of Mathematics,

Bishop Heber College, Tiruchirappalli, Tamil Nadu-620 017, INDIA.

1*

[email protected],

[email protected].

(Received on: November 15, 2018) ABSTRACT

In this paper, a singularly perturbed second order ordinary differential equation of reaction-diffusion type with Robin boundary conditions is considered.

Due to the presence of singular perturbation parameter and due to the presence of a single discontinuity in the source term, the solution u of this problem exhibits boundary layers at x = 0 and x = 1 and twin interior layers at x = d. A piecewise- uniform Shishkin mesh is introduced and is used in conjunction with a classical finite difference discretisation, to construct a numerical method for solving this problem. It is proved that the numerical approximations obtained with this method are essentially first order convergent uniformly with respect to the parameter. Numerical illustration is provided to support the theory.

Keywords: Singular perturbations, boundary layers, interior layers, differential equations, Robin boundary conditions, finite difference scheme, Shishkin meshes.

1. INTRODUCTION

The following two-point boundary value problem is considered for a singularly perturbed linear second order differential equation with a discontinuous source term,

−𝜀𝑢

(𝑥) + 𝑎(𝑥)𝑢(𝑥) = 𝑓(𝑥), 𝑥 ∈ Ω

∪ Ω

(1) with

𝑢(0) − √𝜀𝑢

(0) = 𝜙

𝑢(1) + √𝜀𝑢

(1) = 𝜙

𝑓(𝑑 −) ≠ 𝑓(𝑑 +), [𝑢](𝑑) = 0; [𝑢′](𝑑) = 0,

}

(2)

Ω

= (0, 𝑑), Ω ̅̅̅̅ = [0, 𝑑], Ω

= (𝑑, 1), Ω ̅̅̅̅ = [𝑑, 1] and Ω = {𝑥: 0 < 𝑥 < 1}. A single

discontinuity in the source term is assumed to occur at a point 𝑑 𝜖 Ω. The jump at 𝑑 in any function 𝜔 is defined by

[𝜔](𝑑) = 𝜔(𝑑 +) − 𝜔(𝑑 −).

The above problem can be rewritten in the operator form as,

𝐿𝑢 = 𝑓 on Ω

∪ Ω

, (3)

𝛽

𝑢(0) = 𝜙

, 𝛽

𝑢(1) = 𝜙

, (4)

𝑓(𝑑 −) ≠ 𝑓(𝑑 +), (5)

[𝑢](𝑑) = 0; [𝑢′](𝑑) = 0, (6)

where the operators 𝐿, 𝛽

,𝛽

are defined by 𝐿 = −𝜀𝐷

+ 𝑎𝐼, 𝛽

= 𝐼 − √𝜀𝐷, 𝛽

= 𝐼 + √𝜀𝐷 where 𝐼 is the identity operator, 𝐷 =

and 𝐷

=

are the first and second order differential operators. For all 𝑥 ∈ [0,1], it is assumed that 𝑎(𝑥) satisfies the condition, 0 < 𝛼 < 𝑎(𝑥). (7)

It is also assumed, without loss of generality, that √𝜀 ≤

. (8)

Assume that the functions 𝑎 ∈ 𝐶

Ω ̅, 𝑓 ∈ 𝐶

(Ω

∪ Ω

) where Ω ̅ =[0,1]. From the above assumptions, the problem (1),(2) has a solution 𝑢 ∈ 𝐶(Ω ̅) ∩ 𝐶

(Ω) ∩ 𝐶

(Ω

∪ Ω

). The reduced problem obtained by putting each 𝜀 = 0 in (1) is defined by, 𝑎(𝑥)𝑢

(𝑥) = 𝑓(𝑥), 𝑥 ∈ Ω

∪ Ω

. (9)

The problem (1), (2) is singularly perturbed in the following sense; the solution 𝑢 is expected to have the following layer pattern. The solution 𝑢 of (1), (2) is expected to exhibit twin layers of width 𝑂(√𝜀)at 𝑥 = 0 and 𝑥 = 1. Further, due to discontinuity in 𝑓at 𝑥 = 𝑑, the solution 𝑢 is expected to exhibit twin interior layers of width 𝑂(√𝜀) at 𝑥 = 𝑑. The norm

‖𝑦‖

= 𝑠𝑢𝑝

|𝑦(𝑥)| is defined for any scalar-valued function 𝑦 and domain 𝐷. If 𝐷 = Ω, ̅ || ||

is simply written as || ||. Throughout the paper, 𝐶denotes a generic positive constant, which is independent of 𝑥and of the singular perturbation and the discretization parameters.

For a general introduction to parameter-uniform numerical methods for singular

perturbation problems, see

. In

, a unified framework for the error analysis of time-

dependent singularly perturbed problems with discontinuous Galerkin time discretisation is

presented. Its general analysis relies on spatial error estimates known from stationary problems

and the properties of the discontinuous Galerkin time discretisation. Applications of our

framework to second and fourth-order singularly perturbed problems in estimation and

simulation is also presented. In

, Numerical methods for a singularly perturbed reaction-

diffusion problem with a discontinuous source term is considered. A numerical method

consisting of a standard finite difference operator and a non-standard piecewise uniform mesh

is constructed. The numerical solutions converge in the maximum norm uniformly with

respect to the singular perturbation parameter is shown by extensive computations. In

, a

singularly perturbed reaction-diffusion problem with a discontinuous source term is considered. The problem is discretized by using a finite difference scheme on a Shishkin type mesh. A nonequidistant generalization of the Numerov scheme is used on the Shishkin-type mesh expect for the point of discontinuity, whereas a second-order difference scheme with an additional refined mesh is used for the point of discontinuity. The maximum error in the mesh points is uniformly bounded by (𝑁

𝐼𝑛 𝑁)

with a constant independent of the perturbation parameter is shown.

Theorem 1.1 Let 𝑎(𝑥)satisfy (7), (8). Then (1), (2) has a solution 𝑢 ∈ 𝐶(Ω ̅) ∩ 𝐶

(Ω) ∩ 𝐶

(Ω

∪ Ω

).

Proof: The proof is by construction. Let 𝑦

, 𝑦

be particular solutions of the differential equations

−𝜀𝑦

(𝑥) + 𝑎(𝑥)𝑦

(𝑥) = 𝑓(𝑥), 𝑥 ∈ Ω

and

−𝜀𝑦

(𝑥) + 𝑎(𝑥)𝑦

(𝑥) = 𝑓(𝑥), 𝑥 ∈ Ω

. Consider the function,

𝑦(𝑥) = { 𝑦

(𝑥) + 𝛽

(𝑢 − 𝑦

)(0)𝜙

(𝑥) + 𝐴

𝜙

(𝑥), 𝑥 ∈ Ω

𝑦

(𝑥) + 𝐵

𝜙

(𝑥) + 𝛽

(𝑢 − 𝑦

)(1)𝜙

(𝑥), 𝑥 ∈ Ω

(10) where 𝜙

(𝑥), 𝜙

(𝑥) are the solutions of the boundary value problems

−𝜀𝜙

′′(𝑥) + 𝑎(𝑥)𝜙

(𝑥) = 0, 𝑥 ∈ Ω, 𝛽

𝜙

(0) = 1, 𝛽

𝜙

(1) = 0

−𝜀𝜙

′′(𝑥) + 𝑎(𝑥)𝜙

(𝑥) = 0, 𝑥 ∈ Ω, 𝛽

𝜙

(0) = 0, 𝛽

𝜙

(1) = 1

and 𝐴

, 𝐵

are constants to be chosen so that 𝑦 ∈ 𝐶

(Ω). The constants 𝐴

and 𝐵

are found from the system of two equations in 𝐴

and 𝐵

derived from the conditions

𝑦(𝑑 −) = 𝑦(𝑑 +) and 𝑦

(𝑑 −) = 𝑦

(𝑑 +).

Note that on Ω, 0 <𝜙

, 𝜙

<1. Thus 𝜙

, 𝜙

cannot have an internal maximum or minimum and also

𝜙

< 0, 𝜙

> 0, 𝑥 ∈ Ω.

Hence 𝜙

′(𝑑)𝜙

(𝑑) − 𝜙

(𝑑)𝜙

′(𝑑) > 0 ensures the existence of 𝐴

and 𝐵

. 2. ANALYTICAL RESULTS

The operator 𝐿 satisfies the following maximum principle.

Lemma 2. 1 Let 𝑎(𝑥) satisfy (7),(8). Let 𝜓 be any function in the domain of 𝐿 such that 𝛽

𝜓 (0) ≥ 0, 𝛽

𝜓 (1) ≥ 0. Then 𝐿𝜓(𝑥) ≥ 0 on Ω

∪ Ω

and [𝜓](𝑑) = 0, [𝜓′](𝑑) ≤ 0, then 𝜓(𝑥) ≥ 0 for 𝑥 ∈ Ω ̅.

Proof : Let 𝑥

be such that 𝜓(𝑥

) = 𝑚𝑖𝑛

𝜓(𝑥) and assume that the lemma is false. Then 𝜓(𝑥

) < 0. For 𝑥

= 0, 𝛽

𝜓(0) = 𝜓(0) − √𝜀 𝜓′(0) < 0, for 𝑥

= 1, 𝛽

𝜓(1) = 𝜓(1) +

√𝜀 𝜓′(1) < 0, contradicting the hypothesis. Therefore, 𝑥

∉ {0,1}. Then either 𝑥

∈ Ω

∪ Ω

or 𝑥

= 𝑑.

If 𝑥

∈ Ω

∪ Ω

, then 𝜓′′(𝑥

) ≥ 0. So

𝐿𝜓(𝑥

) = −𝜀𝜓′′(𝑥

) + 𝑎(𝑥

)𝜓(𝑥

) < 0, a contradiction.

Or suppose that 𝑥

= 𝑑. Then there are two cases.

Case (i): If 𝜓 is not differentiable at 𝑑, then 𝜓

(𝑥

) ≠ 0.

Then 𝜓′(𝑑 −) ≤ 0 and 𝜓′(𝑑 +) > 0 imply that [𝜓′](𝑑) > 0, a contradiction.

Case (ii): Suppose that 𝜓 is differentiable at 𝑑. Then 𝜓′(𝑑) = 0 and 𝜓 ∈ 𝐶

(Ω ̅). As 𝜓(𝑑) <

0, there exists a neighbourhood of 𝑑, 𝑁

= [𝑑

, 𝑑] such that 𝜓(𝑥) < 0 for all 𝑥 ∈ 𝑁

. Let 𝑥

∈ 𝑁

such that 𝜓(𝑥

) > 𝜓(𝑑). By mean-value theorem, there exists 𝑥

∈ 𝑁

such that

𝜓

(𝑥

) =

𝑑−𝑥1

< 0

and also there exists 𝑥

∈ 𝑁

, such that 𝜓

(𝑥

) = 𝜓′(𝑑) − 𝜓′(𝑥

)

𝑑 − 𝑥

> 0.

Since 𝑥

∈ 𝑁

, 𝜓(𝑥

) < 0. Thus, 𝐿𝜓(𝑥

) = −𝜀𝜓′′(𝑥

) + 𝑎(𝑥

)𝜓(𝑥

) < 0, a contradiction.

Lemma 2.2 Let 𝑎(𝑥) satisfy (7), (8). Let 𝜓 be any function in the domain of 𝐿, then for each 𝑥 ∈ Ω ̅,

|𝜓(𝑥)| ≤ 𝑚𝑎𝑥 {‖𝛽

𝜓 (0)‖, ‖𝛽

𝜓 (1)‖, 1

𝛼 ‖𝐿𝜓(𝑥)‖

}.

Proof : Define the two functions,

𝜃

(𝑥) = 𝑚𝑎𝑥 {‖𝛽

𝜓 (0)‖, ‖𝛽

𝜓 (1)‖, 1

𝛼 ‖𝐿𝜓(𝑥)‖

} ± 𝜓(𝑥).

Using the properties of 𝑎(𝑥), it is not hard to verify that 𝛽

𝜃

(0) ≥ 0, 𝛽

𝜃

(1) ≥ 0 and 𝐿𝜃

(𝑥) ≥ 0 on Ω

∪ Ω

. Furthermore, since 𝜓 ∈ 𝐶

(Ω),

[𝜃

](𝑑) = ±[𝜓](𝑑) = 0 and [𝜃

](𝑑) = ±[𝜓′](𝑑) = 0.

It follows from Lemma 2.1 that 𝜃

(𝑥) ≥ 0 on Ω ̅ as required.

Standard estimates of the solution of (1),(2) and its derivatives are contained in the following lemma.

Lemma 2.3 Let 𝑎(𝑥)satisfy (7),(8) and let 𝑢 be the solution of (1),(2). Then for all 𝑥 ∈ Ω

∪ Ω

,

|𝑢(𝑥)| ≤ 𝐶‖𝑓‖

,

|𝑢

(𝑥)| ≤ 𝐶𝜀

(‖𝑢‖ + ‖𝑓‖

), for 𝑘 = 1,2,

|𝑢

(𝑥)| ≤ 𝐶𝜀

𝜀

(‖𝑢‖ + ‖𝑓‖

+ 𝜀

‖𝑓

‖

), for 𝑘 = 3,4.

Proof : The bound on 𝑢 is an immediate consequence of Lemma 2.2.

Rewriting the differential equation (1) gives

𝑢′′ = 𝜀

(𝑎𝑢 − 𝑓) (11)

and it is not hard to see that the bounds on 𝑢′′ follow from (11).

To bound 𝑢(𝑥), consider an interval 𝑁 = [𝑎, 𝑎 + √𝜀] ⊂ [0, 𝑑 −] ∪ [𝑑 + ,1], where 𝑎 ≥ 0, 0 ≤ √𝜀 ≤ (𝑑 −) − 𝑎 in [0, 𝑑 −] and 𝑎 ≥ 𝑑+, 0 ≤ √𝜀 ≤ 1 − 𝑎 in [𝑑 + ,1]. By the mean value theorem, for some 𝑦 ∈ 𝑁,

𝑢′(𝑦) =

√𝜀

. Therefore,

|𝑢′(𝑦)| ≤ 𝐶𝜀

‖𝑢‖ . (12) Now, for any 𝑥 ∈ 𝑁,

𝑢′(𝑥) = 𝑢′(𝑦) + ∫ 𝑢′′(𝑠)𝑑𝑠

𝑥

𝑦

Now by using (12),

|𝑢′(𝑥)| ≤ 𝐶𝜀

(‖𝑢‖ + ‖𝑓‖

).

Differentiating the equation (1) once and twice and using the bounds of 𝑢, 𝑢′ and 𝑢′′, the bounds of 𝑢

and 𝑢

follow.

Consider the Shishkin decomposition of the solution of 𝑢of the BVP (1),(2) into smooth and singular components,

𝑢 = 𝑣 + 𝑤 (13) where

𝑣 = 𝑢

+ 𝜀𝑣

(14) with 𝑢

given by (9) and 𝑣

given by

𝐿𝑣

= −𝑢

, on Ω

∪ Ω

, (15) 𝛽

𝑣

(0) = 0, [𝑣

](𝑑) = 0, [𝑣′

](𝑑) = 0, 𝛽

𝑣

(1) = 0. (16) Hence,

𝐿𝑣 = 𝑓, on Ω

∪ Ω

, (17) 𝛽

𝑣(0) = 𝛽

𝑢

(0), [𝑣](𝑑) = [𝑢

](𝑑), [𝑣′](𝑑) = [𝑢′

](𝑑), 𝛽

𝑣(1) = 𝛽

𝑢

(1) (18) The singular component of the solution 𝑢 satisfies,

𝐿𝑤 = 0, on Ω

∪ Ω

, (19) 𝛽

𝑤(0) = 𝛽

(𝑢 − 𝑣)(0), 𝛽

𝑤(1) = 𝛽

(𝑢 − 𝑣)(1). (20) [𝑤](𝑑) = −[𝑣](𝑑), [𝑤′](𝑑) = −[𝑣′](𝑑). (21) From the expressions (15), (16) and using Lemma 2.3 for 𝑣, it is found that for 𝑘 = 0,1,2,3,4,

|𝑣

| ≤ 𝐶(1 + 𝜀

). (22) From (14), (22), the following bounds for 𝑣 hold:

|𝑣

| ≤ 𝐶, for 𝑘 = 0,1,2

|𝑣

| ≤ 𝐶 (1 + 𝜀

) , for 𝑘 = 3,4.

For convenience, the singular component is rewritten as, 𝑤(𝑥) = { 𝑤

(𝑥) on Ω

𝑤

(𝑥) on Ω

(23)

then it is possible to have,

𝑤

(𝑥) = 𝛽

𝑤(0)𝜓

(𝑥) + 𝐴

𝜓

(𝑥), 𝑤

(𝑥) = 𝐵

𝜓

(𝑥) + 𝛽

𝑤(1)𝜓

(𝑥) (24)

with −𝜀𝜓

(𝑥) + 𝑎(𝑥)𝜓

(𝑥) = 0 on Ω

, 𝛽

𝜓

(0) = 1, 𝛽

𝜓

(𝑑) = 0 (25)

−𝜀𝜓

(𝑥) + 𝑎(𝑥)𝜓

(𝑥) = 0 on Ω

, 𝛽

𝜓

(0) = 0, 𝛽

𝜓

(𝑑) = 1 (26)

−𝜀𝜓

(𝑥) + 𝑎(𝑥)𝜓

(𝑥) = 0 on Ω

, 𝛽

𝜓

(𝑑) = 1, 𝛽

𝜓

(1) = 0 (27)

−𝜀𝜓

(𝑥) + 𝑎(𝑥)𝜓

(𝑥) = 0 on Ω

, 𝛽

𝜓

(𝑑) = 0, 𝛽

𝜓

(1) = 1. (28)

Here, 𝐴

and 𝐵

are constants to be chosen in such a way that the jump conditions at 𝑥 = 𝑑 are satisfied. Therefore, 𝐿𝑤

= 0 on Ω

, 𝛽

𝑤

(0) = 𝛽

𝑤(0), 𝛽

𝑤

(𝑑 −) = 𝛽

𝑤(𝑑 −) 𝐿𝑤

= 0 on Ω

, 𝛽

𝑤

(𝑑 +) = 𝛽

𝑤(𝑑 +), 𝛽

𝑤

(1) = 𝛽

𝑤(1) The layer functions 𝐵

, 𝐵

, 𝐵

, 𝐵

, 𝐵

, 𝐵

, associated with the solution 𝑢, are defined by 𝐵

(𝑥) = 𝑒

, 𝐵

(𝑥) = 𝑒

, 𝐵

(𝑥) = 𝐵

(𝑥) + 𝐵

(𝑥) on Ω

, 𝐵

(𝑥) = 𝑒

, 𝐵

(𝑥) = 𝑒

, 𝐵

(𝑥) = 𝐵

(𝑥) + 𝐵

(𝑥) on Ω

. The following elementary properties of the layer functions 𝐵

, 𝐵

, 𝐵

for all 0 ≤ 𝑥 < 𝑦 ≤ 𝑑, should be noted: 𝐵

(𝑥) = 𝐵

(1 − 𝑥), (29)

𝐵

(𝑥) > 𝐵

(𝑦), 0 < 𝐵

(𝑥) ≤ 1. (30)

𝐵

(𝑥) < 𝐵

(𝑦), 0 < 𝐵

(𝑥) ≤ 1. (31)

𝐵

(𝑥) is monotone decreasing on 𝑥 ∈ [0,

]. (32)

𝐵

(𝑥) is monotone increasing on 𝑥 ∈ [

, 𝑑]. (33)

𝐵

(𝑥) ≤ 2𝐵

(𝑥) for 𝑥 ∈ [0,

]and 𝐵

(𝑥) ≤ 2𝐵

(𝑥)for 𝑥 ∈ (

, 𝑑). (34)

𝐵

(2

ln 𝑁) = 𝑁

. (35) Similar properties for 𝐵

, 𝐵

, 𝐵

for all 𝑑 ≤ 𝑥 < 𝑦 ≤ 1, hold good.

Bounds on the singular component 𝑤 of 𝑢 and its derivatives are contained in

Lemma 2.4: Let 𝑎(𝑥) satisfy (7), (8). Then there exists a constant C, such that, for each 𝑥 ∈ Ω

,

|𝑤

(𝑥)| ≤ 𝐶𝐵

(𝑥), |𝑤

(𝑥)| ≤ 𝐶

, for 𝑘 =1,2,

|𝑤

(𝑥)| ≤ 𝐶 𝐵

(𝑥)

𝜀

, |𝜀𝑤

(𝑥)| ≤ 𝐶 𝐵

(𝑥) 𝜀 . Analogous result hold for 𝑤

and its derivatives.

Proof: To derive the bound of 𝑤

, define the two functions 𝜓

(𝑥) = 𝐶𝐵

(𝑥) ± 𝑤

(𝑥).

For a proper choice of 𝐶, 𝛽

𝜓

(0) ≥ 0, 𝛽

𝜓

(𝑑 −) ≥ 0 and for 𝑥 ∈ Ω

, 𝐿𝜓

(𝑥) = −𝐶𝜀 𝛼

𝜀 𝐵

(𝑥) + 𝐶𝑎𝐵

(𝑥) ≥ 0.

By Lemma 2.2.

|𝑤

(𝑥)| ≤ 𝐶𝐵

(𝑥). (36) Rearranging (19) yields

𝜀𝑤

(𝑥) = 𝑎𝑤

(𝑥).

Using (36),

|𝑤

(𝑥)| ≤ 𝐶𝜀

𝐵

(𝑥). (37) By using the mean-value theorem,

|𝑤

(𝑥)| ≤ 𝐶𝜀

𝐵

(𝑥). (38) Differentiating the equation (1) once and twice and using the bounds of 𝑤

, 𝑤

and 𝑤

, the bounds of 𝑤

and 𝑤

follow.

3. THE SHISHKIN MESH

A piecewise uniform Shishkin mesh with 𝑁 mesh- intervals is now constructed on Ω ̅ as follows. Let Ω

= {𝑥

}

𝑁 2−1

, Ω

= {𝑥

}

𝑗=^𝑁₂+1

𝑁−1

and 𝑥

= 𝑑 . Then Ω ̅̅̅̅

= {𝑥

}

𝑁 2

, Ω

̅̅̅̅

= {𝑥

}

𝑗=^𝑁₂

𝑁

, Ω ̅̅̅̅

⋃Ω ̅̅̅̅

= Ω ̅

= {𝑥

}

. The interval [0, 𝑑] is subdivided into three sub-intervals,

[0, 𝜏]⋃(𝜏, 𝑑 − 𝜏]⋃(𝑑 − 𝜏, 𝑑], where ,

𝜏 = 𝑚𝑖𝑛 {

, 2

√𝛼

ln 𝑁} (39) Then, on the subinterval (𝜏, 𝑑 − 𝜏] a uniform mesh with

mesh-points is placed and on each of the subintervals [0, 𝜏] and (𝑑 − 𝜏, 𝑑], a uniform mesh of

mesh-points is placed. In particular, when the parameter 𝜏 takes on its left choice, the mesh Ω ̅̅̅̅

becomes the classical uniform mesh throughout from 0 to 𝑑.

Similarly, the interval [𝑑, 1] is subdivided into three sub-intervals, [𝑑, 𝑑 + 𝜎]⋃(𝑑 + 𝜎, 1 − 𝜎]⋃(1 − 𝜎, 1],

where,

𝜎 = 𝑚𝑖𝑛 {

, 2

√𝛼

ln 𝑁}.

Then, on the subinterval (𝑑 + 𝜎, 1 − 𝜎] a uniform mesh with

4

mesh-points is placed and on each of the subintervals (𝑑, 𝑑 + 𝜎] and (1 − 𝜎, 1], a uniform mesh of

8

mesh-points is placed.

In particular, when the parameter 𝜎 takes on its left choice, the mesh Ω ̅̅̅̅

becomes the classical uniform mesh throughout from 𝑑 𝑡𝑜 1.

In practice, it is convenient to take

𝑁 = 8𝑘, 𝑘 ≥ 2 (40)

From the above construction of Ω ̅

, it is clear that the transition points {𝜏, 𝑑 − 𝜏, 𝑑 + 𝜎, 1 − 𝜎} are the only points at which the mesh size can change and that it does not necessarily change at each of these points. The following notations are introduced : ℎ

= 𝑥

−𝑥

, ℎ

= 𝑥

−𝑥

, 𝐽 = {𝜏, 𝑑 − 𝜏, 𝑑 + 𝜎, 1 − 𝜎: ℎ

≠ ℎ

}.

4. THE DISCRETE PROBLEM

In this section, a classical finite difference operator with an appropriate Shishkin mesh is used to construct a numerical method for the problem (1),(2) which is proved to be essentially first order parameter-uniform convergent.

The discrete two-point boundary value problem is now defined as,

−𝜀𝛿

𝑈(𝑥

) + 𝑎(𝑥)𝑈(𝑥

) = 𝑓(𝑥

), on Ω

⋃Ω

(41) with

𝑈(0) − √𝜀𝐷

𝑈(0) = 𝜙

, 𝑈(1) + √𝜀𝐷

𝑈(1) = 𝜙

, 𝐷

𝑈 (𝑥

) = 𝐷

𝑈 (𝑥

) (42) The problem (41), (42) can also be written in the operator form

𝐿

𝑈 = 𝑓 on Ω

⋃Ω

, 𝛽

𝑈(0) = 𝜙

, 𝛽

𝑈(1) = 𝜙

, 𝐷

𝑈 (𝑥

) = 𝐷

𝑈 (𝑥

) where

𝐿

= −𝜀𝛿

+ 𝑎𝐼, 𝛽

= 𝐼 − √𝜀𝐷

, 𝛽

𝑈⃗⃗ (1) = 𝐼 + √𝜀𝐷

, and 𝐷

, 𝐷

and 𝛿

are the difference operators

𝐷

𝑈(𝑥

) =

𝑗+1−𝑥_𝑗

, 𝐷

𝑈(𝑥

) =

𝑗−𝑥_𝑗−1

and 𝛿

𝑈(𝑥

) =

𝑗+1−𝑥_𝑗−1)/2

.

For any function 𝑍 defined on the Shishkin mesh Ω̅

, the following norm ‖𝑍‖ = 𝑚𝑎𝑥

0≤𝑗≤𝑁

|𝑍(𝑥

)| is introduced.

The following discrete results are analogous to those for the continuous problem.

Lemma 4.1 Let 𝑎(𝑥) satisfy (7),(8). Let 𝛹 be any mesh function, such that 𝛽

𝛹(0) ≥ 0, 𝛽

𝛹(1) ≥ 0. Then 𝐿

𝛹 ≥ 0 on Ω

⋃Ω

and 𝐷

𝑈 (𝑥

2

) = 𝐷

𝑈 (𝑥

2

) imply that 𝛹 ≥ 0 on Ω̅

.

Proof: Let 𝑗

be such that 𝛹(𝑥

) = 𝑚𝑖𝑛

1≤𝑖≤𝑛

𝛹(𝑥

) and assume that the lemma is false. Then, 𝛹(𝑥

) < 0. If 𝑥

= 0, then 𝛽

𝛹(0) = 𝛹(0) − √𝜀𝐷

𝛹(0) < 0, a contradiction.

Therefore, 𝑥

≠ 0 and for the same reason 𝑥

≠ 1. 𝛹(𝑥

) − 𝛹(𝑥

) ≤ 0, 𝛹(𝑥

) − 𝛹(𝑥

) ≥ 0. Also, 𝛿

𝛹(𝑥

) > 0. It follows that,

𝐿

𝛹(𝑥

) = −𝜀𝛿

𝛹(𝑥

) + 𝑎(𝑥

)𝛹(𝑥

) < 0,

which is a contradiction.

If 𝑥

𝜖 Ω

, this leads to a contradiction. The only other possibility is that 𝑥

= 𝑥

. Then 𝐷

𝛹 (𝑥

2

) ≤ 0 ≤ 𝐷

𝛹 (𝑥

) ≤ 𝐷

𝛹 (𝑥

) and

𝛹 (𝑥

2−1

) = 𝛹 (𝑥

) = 𝛹 (𝑥

2+1

) < 0.

Then, 𝐿

𝛹 (𝑥

2−1

) < 0, a contradiction.

Hence the result.

An immediate consequence of this is the following discrete stability result.

Lemma 4.2 Let 𝑎(𝑥) satisfy (7),(8). Let 𝛹 be any mesh function, then for Ω ̅

,

|𝛹(𝑥

)| = 𝑚𝑎𝑥 {‖𝛽

𝛹 (0)‖, ‖ 𝛽

𝛹 (1)‖, 1

𝛼 ‖𝐿

𝛹‖

} , 0 ≤ 𝑗 ≤ 𝑁.

Proof : Define the two mesh functions,

Ɵ

(𝑥

) = 𝑚𝑎𝑥 {‖𝛽

𝛹 (0)‖, ‖ 𝛽

𝛹 (1)‖, 1

𝛼 ‖𝐿

𝛹‖

} ± 𝛹(𝑥

).

Using the properties of 𝑎(𝑥), it is not hard to verify that 𝛽

Ɵ

(0) ≥ 0, 𝛽

Ɵ

(1) ≥ 0 and 𝐿

Ɵ

≥ 0 on Ω

.

At 𝑥

=

2

, 𝐷

Ɵ

(𝑥

2

) − 𝐷

Ɵ

(𝑥

) = 𝐷

𝛹

(𝑥

) − 𝐷

𝛹

(𝑥

) = 0.

It follows from Lemma 4.1 that Ɵ

≥ 0 on Ω ̅

. 5. THE LOCAL TRUNCATION ERROR

In order to bound the error 𝑈 − 𝑢, it suffices to bound 𝐿

(𝑈 − 𝑢). Notice that, for 𝑥

∈ Ω

⋃Ω

,

𝐿

(𝑈 − 𝑢) = 𝐿

𝑈 − 𝐿

𝑢 = 𝑓 − 𝐿

𝑢 = 𝐿𝑢 − 𝐿

𝑢 = (𝐿 − 𝐿

)𝑢 = −𝜀(𝛿

− 𝐷

)𝑢 which is the standard local truncation error.

Analogous to the continuous case, the discrete solution 𝑈 can be decomposed into 𝑉

, 𝑊

on Ω

̅̅̅̅

and 𝑉

, 𝑊

on Ω ̅̅̅̅

which are defined to be the solutions of the following discrete problems

𝐿

𝑉

(𝑥

) = 𝑓(𝑥

) on 𝑥

𝜖 Ω

, 𝛽

𝑉

(0) = 𝛽

𝑣(0), 𝛽

𝑉

(𝑥

) = 𝛽

𝑣(𝑑 −), 𝐿

𝑉

(𝑥

) = 𝑓(𝑥

) on 𝑥

𝜖 Ω

, 𝛽

𝑉

(𝑥

2

) = 𝛽

𝑣(𝑑 +), 𝛽

𝑉

(1) = 𝛽

𝑣(1),

𝐿

𝑊

(𝑥

) = 0 on 𝑥

𝜖 Ω

, 𝛽

𝑊

(0) = 𝛽

𝑤

(0), 𝐿

𝑊

(𝑥

) = 0

on 𝑥

𝜖 Ω

, 𝛽

𝑊

(1) = 𝛽

𝑤

(1), 𝑊

(𝑥

2

) + 𝑉

(𝑥

) = 𝑊

(𝑥

) + 𝑉

(𝑥

), 𝐷

𝑊

(𝑥

2

) + 𝐷

𝑉

(𝑥

) = 𝐷

𝑊

(𝑥

) + 𝐷

𝑉

(𝑥

).

The error at each point 𝑥

𝜖 Ω ̅

is denoted by 𝑒(𝑥

) = 𝑈(𝑥

) − 𝑢(𝑥

). Then the local truncation error 𝐿

𝑒(𝑥

),

for 𝑗 ≠

2

,

𝛽

𝑒(0) = 𝛽

(𝑉 − 𝑣)(0) + 𝛽

(𝑊

− 𝑤

)(0), 𝛽

𝑒(1) = 𝛽

(𝑉 − 𝑣)(1) + 𝛽

(𝑊

− 𝑤

)(1),

𝐿

𝑒(𝑥

) = 𝐿

(𝑉 − 𝑣)(𝑥

) + 𝐿

(𝑊

− 𝑤

)(𝑥

), 1≤ 𝑗 ≤

2

− 1 𝐿

𝑒(𝑥

) = 𝐿

(𝑉 − 𝑣)(𝑥

) + 𝐿

(𝑊

− 𝑤

)(𝑥

),

2

+ 1 ≤ 𝑗 ≤ 𝑁 − 1, where 𝑊

, 𝑊

, 𝑤

, 𝑤

are as in (1),(2),(41),(42).

For any smooth function 𝜓 and for each

Ω

⋃Ω

,the following expressions may be used to estimate the local truncation error.

|(𝐷 − 𝐷

)𝜓(𝑥

)| ≤ 𝐶(𝑥

− 𝑥

)𝑚𝑎𝑥

|𝜓

(𝑠)|, (43)

|(𝐷 − 𝐷

)𝜓(𝑥

)| ≤ 𝐶(𝑥

− 𝑥

)𝑚𝑎𝑥

|𝜓

(𝑠)|, (44)

|(𝛿

− 𝐷

)𝜓(𝑥

)| ≤ 𝐶𝑚𝑎𝑥

|𝜓

(𝑠)|. (45)

|(𝛿

− 𝐷

)𝜓(𝑥

)| ≤ 𝐶(𝑥

− 𝑥

)𝑚𝑎𝑥

|𝜓

(𝑠)|. (46)

Here, 𝐼

= [𝑥

, 𝑥

] 6. ERROR ESTIMATE The error in the smooth and singular components are estimated in the following theorems. Theorem 6.1 Let 𝑎(𝑥) satisfy (7),(8). Let 𝑣 denote the smooth component of the solution of the problem (1),(2) and 𝑉 be the smooth component of the solution of the problem (41),(42). Then for 𝑗 ≠

, ||𝐿

(𝑉 − 𝑣)(𝑥

) || ≤ 𝐶𝑁

ln 𝑁. (47)

Proof : From the expression (44), |𝛽

(𝑉 − 𝑣)(0)| ≤ 𝐶√𝜀(𝑥

−𝑥

)𝑚𝑎𝑥

|𝑣′′(𝑠)| ≤ 𝐶𝑁

(48)

from the expression (43),

|𝛽

(𝑉 − 𝑣)(1)| ≤ 𝐶√𝜀(𝑥

−𝑥

)𝑚𝑎𝑥

|𝑣′′(𝑠)| ≤ 𝐶𝑁

. (49)

For each mesh point 𝑥

, there are two possibilities: either 𝑥

∉ 𝐽 or 𝑥

∈ 𝐽.

If 𝑥

∉ 𝐽, then 𝑥

∈ (0, 𝜏)⋃(𝑑 − 𝜏, 𝑑)⋃(𝑑, 𝑑 + 𝜎)⋃(1 − 𝜎, 1), using the bounds of the derivatives of 𝑣 and the expression (46),

|𝐿

(𝑉 − 𝑣)(𝑥

)| ≤ 𝐶𝑁

𝑙𝑛 𝑁. (50)

If 𝑥

∉ 𝐽, then 𝑥

∈ (𝜏, 𝑑 − 𝜏)⋃(𝑑 + 𝜎, 1 − 𝜎), using the bounds of the derivatives of 𝑣 and the expression (46),

|𝐿

(𝑉 − 𝑣)(𝑥

)| ≤ 𝐶𝑁

. (51) On the other hand , if 𝑥

∈ 𝐽, then 𝑥

∈ {𝜏, 𝑑 − 𝜏}.

Consider the case 𝑥

= 𝜏 and for 𝑥

= 𝑑 − 𝜏, the proof is analogous.

If 𝑥

= 𝜏 ∈ 𝐽, using the bounds of the derivatives of 𝑣 and the expression (46),

|𝐿

(𝑉 − 𝑣)(𝑥

)| ≤ 𝐶𝑁

ln 𝑁. (52) For 𝑥

∈ {𝑑 + 𝜎, 1 − 𝜎}, the proof is analogous.

From (50),(51),(52) the required result follows.

Theorem 6.2 Let 𝑎(𝑥) satisfy (7),(8). Let 𝑤 denote the singular component of the solution of the problem (1),(2) and 𝑊 be the singular component of the solution of the problem (41),(42). Then

|| 𝐿

(𝑊 − 𝑤)(𝑥

) || ≤ 𝐶𝑁

ln 𝑁. (53) Proof : From the expression (44),

|𝛽

(𝑊

− 𝑤

)(0)| ≤ 𝐶√𝜀(𝑥

−𝑥

)𝑚𝑎𝑥

|𝑤

′′(𝑠)| ≤ 𝐶𝑁

ln 𝑁 (54) from the expression (43),

|𝛽

(𝑊

− 𝑤

)(1)| ≤ 𝐶√𝜀(𝑥

−𝑥

)𝑚𝑎𝑥

|𝑤

′′(𝑠)| ≤ 𝐶𝑁

ln 𝑁 (55) For 𝑥

∉ 𝐽 and 𝑥

∈ Ω

⋃Ω

, by using Lemma (2.4) and the expression (46),

|𝐿

(𝑊

− 𝑤

)(𝑥

)| ≤ 𝐶𝑁

ln 𝑁. (56) For 𝑥

∈ 𝐽 and 𝑥

∈ Ω

⋃Ω

by using Lemma (2.4) and the expression (46),

|𝐿

(𝑊

− 𝑤

)(𝑥

)| ≤ 𝐶𝑁

. (57) From (56),(57), the required result follows. Hence the result.

At the point 𝑥

= 𝑥

, (𝐷

− 𝐷

)𝑒 (𝑥

2

) = (𝐷

− 𝐷

)(𝑈 − 𝑢) (𝑥

2

)

= (𝐷

− 𝐷

)𝑈 (𝑥

2

) − (𝐷

− 𝐷

)𝑢 (𝑥

2

).

Recall that (𝐷

− 𝐷

)𝑈 (𝑥

) = 0. Let ℎ

= max {ℎ

, ℎ

}.

Then,

|(𝐷

− 𝐷

)𝑒 (𝑥

2

)| = |(𝐷

− 𝐷

)𝑢 (𝑥

2

)|

≤ |(𝐷

− 𝑑

𝑑𝑥 ) 𝑢 (𝑥

2

)| + |(𝐷

− 𝑑

𝑑𝑥 ) 𝑢 (𝑥

)|

≤ 1

2 ℎ

|𝑢

(𝜂)|

+ 1

2 ℎ

|𝑢

(𝜉)|

≤ 𝐶ℎ

𝑚𝑎𝑥

|𝑢

(𝑥)|

Therefore,

|(𝐷

− 𝐷

)𝑒 (𝑥

2

)| = 𝐶

. (58) Define a set of barrier functions on Ω ̅

by,

𝜔 (𝑥

) = {

П_𝑘=1^𝑗 (1+√𝛼_√2𝜀^ℎ𝑘)

П_𝑘=1^𝑁/2(1+√𝛼_√2𝜀^ℎ𝑘)

, 0 ≤ 𝑗 ≤ 𝑁/2

П_𝑘=𝑗^𝑁−1(1+√𝛼^ℎ𝑘+1_√2𝜀)

П_𝑘=𝑁/2^𝑁−1 (1+√𝛼^ℎ𝑘+1_√2𝜀)

, 𝑁/2 ≤ 𝑗 ≤ 𝑁

(59)

Note that, 𝜔 (0) = 0, 𝜔 (d) = 1, 𝜔 (1) = 0 (60) Also, 0≤ 𝜔 (𝑥

) ≤ 1 . (61) It is not hard to see that, for 𝑥

𝜖 Ω ̅̅̅̅

,

𝐷

𝜔 (𝑥

) = 𝜔 (𝑥

) − 𝜔 (𝑥

) ℎ

=

√2𝜀

𝜔 (𝑥

). (62) 𝐷

𝜔 (𝑥

) = 𝜔 (𝑥

) − 𝜔 (𝑥

)

ℎ

=

√2𝜀(1+√𝛼ℎ𝑗/√2𝜀)

𝜔 (𝑥

). (63) 𝛿

𝜔 (𝑥

) = 𝐷

𝜔 (𝑥

) − 𝐷

𝜔 (𝑥

)

(ℎ

+ ℎ

)/2

≤

𝜔 (𝑥

). (64) Similarly, for 𝑥

𝜖 Ω ̅̅̅̅

,

𝐷

𝜔 (𝑥

) = −

√2𝜀(1+√𝛼ℎ𝑗+1/√2𝜀)

𝜔 (𝑥

) (65) 𝐷

𝜔 (𝑥

) = − √𝛼

√2𝜀 𝜔 (𝑥

).

𝛿

𝜔 (𝑥

) ≤ 𝛼

𝜀 𝜔 (𝑥

).

In particular, at 𝑥

= 𝑥

, using (65),(63) and (60), (𝐷

− 𝐷

)𝜔 (𝑥

) = −

√2𝜀(1+√𝛼ℎ_𝑁/2⁺ /√2𝜀)

−

√2𝜀(1+√𝛼ℎ_𝑁/2⁻ /√2𝜀)

≤ −

√𝜀

. (66)

From the expressions (64) and (65),