A Singularly Perturbed Convection Di usion Turning Point ...A linear singularly perturbed interior turning point problem with a continuous convection coe cient is examined in this

A Singularly Perturbed Convection Diffusion Turning Point

Problem with an Interior Layer

E. O’ Riordan, J. Quinn∗

School of Mathematical Sciences, Dublin City University, Glasnevin, Dublin 9, Ireland.

Abstract

A linear singularly perturbed interior turning point problem with a continuous convectioncoefficient is examined in this paper. Parameter uniform numerical methods composed ofmonotone finite difference operators and piecewise-uniform Shishkin meshes, are constructedand analysed for this class of problems. A refined Shishkin mesh is placed around thelocation of the interior layer and we consider disrupting the centre point of this fine meshaway from the point where the convection coefficient is zero. Numerical results are presentedto illustrate the theoretical parameter-uniform error bounds established.

Keywords: Singularly Perturbed; Shishkin mesh; Interior Turning Point

1 Introduction

Interior layers exhibiting a hyperbolic tangent profile can arise in solutions of singularly per-turbed quasilinear problems of the form

εu′′ − uu′ − b(x)u = f(x), x ∈ (0, 1), b ≥ 0, u(0) > 0, u(1) < 0; (1.1)

when the singular perturbation parameter ε > 0 can be arbitrarily small. Asymptotic expan-sions can be used to locate the interior point p, where u(p) = 0, to within an O(ε) neighbourhoodof some known point d∗. That is, p ∈ (d∗ − Cε, d∗ + Cε) (see Howes [5]).

In this paper, we examine numerical methods for a class of linear problems associated with theabove nonlinear problem. For example, problems of the form

εy′′ − 2 tanh(d− x

2ε)y′ − y = f, x ∈ (0, 1), y(0) > 0 > y(1), d ∈ (0, 1) (1.2)

will be studied. For such a problem, an interior layer forms about the interior point d. Aparameter-uniform numerical method [1], based on an upwind finite difference operator and apiecewise-uniform Shishkin mesh ([1]), will be constructed and analysed in this paper for thesetype of problems. We will also consider the effect of centring the piecewise-uniform mesh atsome point dN ∈ (d−Cε lnN, d+Cε lnN). The resulting numerical analysis may prove usefulin any future examination of the nonlinear problem (1.1), where the point p is only known tolie within some interval (d∗ − Cε, d∗ + Cε).

Singularly perturbed turning point problems of the form

−εy′′ − a(x)y′ + b(x)y = f(x), x ∈ (0, 1), a(d) = 0, d ∈ (0, 1), b > 0

∗This research was supported by the Irish Research Council for Science, Engineering and Technology

1

have been studied by several authors (Farrell [3], Berger et al.[2]). In this case, where theconvective coefficient a is independent of ε, then the nature of any interior layer is different tothe problem being considered in the current paper. Depending on the quantity b(d)/a′(d), theremay be no interior layer or a power-law layer in the vicinity of the point d. Also, the restriction|a′(x)| > 1

2 |a′(d)|, x ∈ (0, 1) is placed on this problem, whereas in (1.2), the convective coefficient

a := 2 tanh(d−x2ε ) satisfies |a′(x)| 6 |a′(d)|, x ∈ (0, 1).

Exponential interior layers can be generated by considering linear problems with discontinuouscoefficients of the form: Find y ∈ C1(0, 1) such that

εy′′ − a(x)y′ − b(x)y = f(x), x ∈ (0, d) ∪ (d, 1)

a(x) ≥ α > 0, x < d, a(x) ≤ −α < 0, a(d+) 6= a(d−).

Interior layers of exponential type form in the vicinity of the point of discontinuity in thecoefficient a (Farrell et al. [4]). The numerical analysis associated with such problems reliesheavily on the fact that the coefficient a is strictly bounded away from zero in the interval(0, d) ∪ (d, 1).

Finally, in [6], we examined a boundary turning point problem of the form: Find y ∈ C1(0, 1)such that

εy′′ + a(x)y′ = f(x), x ∈ (0, 1), y(0), y(1) given.

The presence of the coefficient

aε(x) ≥ C(1− e−αx/ε), aε(0) = 0,

generated a boundary layer of exponential type in the vicinity of the boundary point x = 0. Forthis problem, the turning point at x = 0. In the current paper, the point where the convectioncoefficient vanishes is in the interior of the domain. As a consequence, the analysis for thecurrent problem is significantly different to the analysis presented in [6]. In particular, thediscrete error analysis is more intricate. In §2 we state the class of problems examined in thispaper and derive a priori bounds on the derivatives of the solutions. In §3, we construct andanalyse a set of numerical methods for this class of problems. The numerical methods consist ofan upwind finite difference operator on piecewise uniform meshes, which are fine in the vicinityof the interior layer. In the final section, some numerical results are presented to illustrate thetheoretical error bounds. Note that throughout the paper, the notation f (k) denotes the k-thderivative of f , C denotes a generic constant that is independent of ε and N , and ‖.‖ denotesthe maximum pointwise norm.

2 Continuous Problem

Consider the following problem class on the unit interval Ω = (0, 1). Find yε such that

Lεyε(x) := (εy′′ε − aεy′ε − byε)(x) = f(x), x ∈ Ω, yε(0) = y0 > 0, yε(1) = y1 < 0,

aε, b, f ∈ C2((0, 1) \ d) ∩ C0[0, 1], b(x) > 0, x ∈ Ω,

aε(x) > 0 for x ∈ [0, d), aε(d) = 0, aε(x) < 0 for x ∈ (d, 1].

(Pε)

We will show that the solution to (Pε) exhibits an interior layer in a neighbourhood of the pointd. Additional restrictions on the function aε are listed in (2.1) below.

Assumptions on the coefficient aε in (Pε)

2

Denote Ω− := (0, d) and Ω+ := (d, 1). Define the limiting functions a−0 and a+0 as a−0 (x) :=

limε→0 aε(x), x ∈ [0, d) and a+0 (x) := limε→0 aε(x), x ∈ (d, 1] and a±0 (d) := limx→d± a

±0 (x).

Assume the following conditions on aε;

|aε(x)| > |αε(x)|, x 6= d, αε(x) := θ tanh (r(d− x)/ε), θ > 2r > 0, x ∈ Ω, (2.1a)∫ x

t=0

∣∣a′ε(t)∣∣ dt 6 C, x ∈ Ω, (2.1b)

ϕ±ε (x) := (a±0 − aε)(x) satisfies |ϕ±ε (x)| 6 |ϕ±ε (d)|e±θ2ε

(d−x), x ∈ Ω±. (2.1c)

Note that (2.1a) implies a−0 (x) > θ, x ∈ Ω− and a+0 (x) 6 −θ, x ∈ Ω+. The differential operator

Lε defined in problem (Pε) satisfies the following minimum principle.

Theorem 2.1. Let Lε be the differential operator defined in (Pε) and z ∈ C2(Ω) ∩ C0(Ω). Ifmin z(0), z(1) > 0 and Lεz(x) 6 0 for x ∈ Ω, then z(x) > 0 for all x ∈ Ω.

Proof. The proof is by contradiction. Assume that there exist a point p ∈ Ω such that z(p) < 0.

It follows from the hypotheses that p /∈ 0, 1. Define the auxiliary function u = ze12ε

∫ dt=x αε(t)dt

and note that u(p) < 0. Choose q ∈ Ω such that u(q) = minu(x) < 0. Therefore, from thedefinition of q, we have u′(q) = 0 and u′′(q) > 0. But then, using α2

ε > 2εα′ε, we have

Lεz(q) = [εu′′ +1

4ε(αε(2aε − αε)− 2εα′ε + 4εb)(−u)](q)e−

12ε

∫ dt=x αε(t)dt > 0

which is a contradiction.

Lemma 2.1. Assuming (2.1), there exists a unique solution, yε, of (Pε) such that

|y(k)ε (x)| 6 Cε−k, k = 0, 1, 2, x ∈ Ω.

Proof. Define the two barrier functions ψ±(x) := ‖f‖2θr [(x−d)αε(x)+θ]+max |y0|, |y1|±yε(x),

which are nonnegative at x = 0, 1. From (2.1) we have

εα′′ε − aεα′ε

>6

εα′′ε − αεα′ε = (2r

θ− 1)αεα

′ε

> 0, x 6 d,6 0, x > d,

, (2.2a)

and

2εα′ε − aεαε 6 2εα′ε − α2ε = (

2r

θ− 1)α2

ε − 2θr 6 −2θr, x ∈ Ω. (2.2b)

Using the definition of the problem (Pε) and the inequalities (2.2), we can deduce that

Lεψ± 6

‖f‖2θr

[(x− d)(εα′′ε − aεα′ε) + (2εα′ε − aεαε)] + ‖f‖ 6 0.

Hence, using Theorem 2.1, we have‖yε‖ 6 C on Ω.

We now bound the derivatives of yε. Using (Pε) and the assumptions in (2.1) we have for anyx ∈ Ω; ∫ x

t=0εy′′ε (t) dt = εy′ε(x)− εy′ε(0) =

∫ x

t=0(f + byε + aεy

′ε)(t) dt, (2.3a)∣∣∣∣∫ x

t=0(aεy

′ε)(t) dt

∣∣∣∣ 6 |[aεyε]x0 |+ ‖yε‖ ∫ x

t=0

∣∣a′ε(t)∣∣ dt 6 C, (2.3b)∣∣∣∣∫ x

t=0(byε + f)(t) dt

∣∣∣∣ 6 ‖b‖‖yε‖+ ‖f‖. (2.3c)

3

By the Mean Value Theorem ∃z ∈ (0, ε) s.t.

ε|y′ε(z)| 6 2‖yε‖. (2.3d)

Using (2.3) with x = z we obtain

ε|y′ε(0)| 6 ‖f‖+

(2 + ‖b‖+ 2‖aε‖+

∫ z

t=0|a′ε(t)| dt

)‖yε‖ 6 C.

Then for any x ∈ [0, 1] we have

ε|y′ε(x)| 6 2‖f‖+

(2 + 2

(‖b‖+ 2‖aε‖+

∫ x

t=0|a′ε(t)| dt

))‖yε‖.

Hence ‖y′ε‖ 6 Cε−1 and use the differential equation in (Pε) to bound y′′ε .

We split the problem (Pε) into left and right problems around x = d as follows. Define

yε(x) =

yL(x), x 6 d,

yR(x), x > d,

where the left and right problems are defined by

LεyL(x) = f(x), x ∈ Ω−,

yL(0) = y0, yL(d) = yε(d),

(PL) ;LεyR(x) = f(x), x ∈ Ω+

yR(d) = yε(d), yR(1) = y1,

(PR).

From [6], we have the existence of a unique yL and yR s.t. ‖y(k)L\R‖ 6 Cε−k, k = 0, 1, 2. We

decompose each yL\R into the sum of a regular component, vL\R, and a layer component, wL\R.If aε satisfied the bound aε > C > 0 for all x ∈ (0, d), then, as in [4], we would simply definethe left regular component as the solution of LεvL = f with suitable boundary conditions andthe left layer component as the solution of LεwL = 0, wL(0) = 0, wL(1) = (yε − vL)(d). Sinceaε does not satisfy such a lower bound, we study the problem

L−vL(x) := εv′′L + a−0 v′L − bvL = f(x), x ∈ Ω−, vL(0) = y0, vL(d) = (v0 + εv1)(d),

(2.4a)

where vL = v0 + εv1 + ε2v2 and the subcomponents v0, v1, v2 satisfy

(a−0 v′0 + bv0)(x) = −f(x), x ∈ (0, d], v0(0) = y0, (2.4b)

(a−0 v′1 + bv1)(x) =

v′′0(x), x ∈ (0, d)

limt↑d v′′0(t), x = d

, v1(0) = 0, (2.4c)

L−v2(x) = −v′′1(x), x ∈ Ω−, v2(0) = v2(d) = 0. (2.4d)

Note that in problem (2.4a), the coefficient aε, of the first derivative term, has been replacedby the strictly positive a−0 defined in (2.1). We incorporate the error (Lε−L−)vL into the layercomponent wε, which is, noting (2.1c), defined as the solution of

LεwL(x) = ϕ−ε v′L(x), x ∈ Ω−, wL(0) = 0, wL(d) = (yε − vL)(d). (2.4e)

Similarly vR and wR satisfy

L+vR(x) := εv′′R + a+0 v′R − bvR = f(x), x ∈ Ω+, vR(d) = v∗, vR(1) = y1, (2.5a)

4

LεwR(x) = ϕ+ε v′R(x), x ∈ Ω+, wR(d) = (yε − vR)(d), wR(1) = 0, (2.5b)

where v∗ is chosen in an analogous fashion to vL(d) so that we may bound the derivatives of vRappropriately.

Lemma 2.2. If vL and wL are the solutions of (2.4) and vR and wR are the solutions of (2.5)then for k = 0, 1, 2, 3, we have the following bounds on the derivatives of vL\R and wL\R:

|v(k)L (x)| 6 C(1 + ε2−k) and |w(k)

L (x)| 6 Cε−ke−θ2ε

(d−x), x ∈ Ω−,

|v(k)R (x)| 6 C(1 + ε2−k) and |w(k)

R (x)| 6 Cε−ke−θ2ε

(x−d), x ∈ Ω+,

where θ is given in (2.1a).

Proof. Analysis of problems (2.4a) can be carried out effectively in the same manner as [1, §3.3]to establish the bounds

‖v(k)0 ‖, ‖v

(k)1 ‖ 6 C and ‖v(k)

2 ‖ 6 Cε−k, k = 0, 1, 2, 3. (2.6)

Consider the barrier functions

ψ±(x) = me−12ε

∫ dt=x αε(t)dt ± wL(x), x ∈ Ω−, m := max |wL(d), 2

r‖v′L‖ε|,

which are nonnegative at x = 0, 1. Using (2.1), (2.6) we have

Lεψ±(x) 6 −mθr

2εe−

θ2ε

(d−x) + |δε(0)|‖v′L‖e−θ2ε

(d−x) 6 0.

Note that∫ dt=x αε(t) dt=

εθr ln cosh ( rε(d− x)). Thus we can use (2.1) and cosh t > 1

2et, t > 0, to

show that

e−θ2ε

(d−x) 6 e−12ε

∫ dt=x αε(t)dt 6 2

θ2r e−

θ2ε

(d−x). (2.7)

Hence using the minimum principle in Theorem 2.1 we obtain |wL(x)| 6 Ce−θ2ε

(d−x). Thebounds on the derivatives of wL are established as in [6]. Analysis of vR and wR is performedin the same manner.

3 The Discrete Problem and Error Analysis

Given the bounds in Lemma 2.2 on the solution, it is natural to refine the mesh in the vicinityof the point d. We examine such a mesh below. In addition, we consider the effect of centringthe mesh at some other point dN near the point d.

Consider the following finite difference method. Find Yε such that

LNε Yε(xi) := (εδ2 − aεD − b)Yε(xi) = f(xi), xi ∈ ΩNε , Yε(0) = y0, Yε(1) = y1,

D :=

D− if aε(xi) > 0D+ if aε(xi) < 0

, δ2Z(xi) :=D+(Z(xi)− Z(xi−1))

(hi+1 + hi)/2, hi := xi − xi−1,

(PNε )

where D± are the standard forward and backward finite difference operators. We define the

5

piecewise uniform mesh, ΩNε , with a refined mesh centred at dN by

|dN − d| < pσ, p 6 12 , µ = 1

2 min dN , 1− dN, σ := min µ, εr lnN, (3.1a)

H0 := 4N (dN − σ), h := 4

N σ, H1 := 4N (1− dN − σ), (3.1b)

ΩNε :=

xi

∣∣∣∣∣∣∣∣∣∣∣

xi = H0i, 0 6 i 6 N4 ,

xi = xN4

+ h(i− N4 ), N

4 < i 6 3N4 ,

xi = x 3N4

+H1(i− 3N4 ), 3N

4 < i 6 N.

, ΩN

ε := ΩNε \ x0, xN. (3.1c)

Note, if dN = d then set p = 0. If dN 6= d, then the mesh is not aligned to the point d.

To prove the existence of a discrete solution to (PNε ), we construct discrete analogues of thebarrier functions used in Lemma 2.1. However, in place of αε in these barrier functions, we willconstruct and use a discrete analogue Aε.

We identify the nearest mesh point to the left of d in this non-aligned mesh as xQ. That is

xQ := max xi| xi 6 d. (3.2)

Define the following mesh functions

S(xi) := (1 + ρ)Q−i − (1− ρ)Q−i, C(xi) := (1 + ρ)Q−i + (1− ρ)Q−i, (3.3a)

T (xi) := S(xi)C(xi)

, ρ = rhε−1, 0 6 i 6 N. (3.3b)

The mesh functions S, C and T can be thought of as discrete analogues of the 2 sinh, 2 cosh andtanh functions centred at xQ respectfully. Note that from (3.1), we have ρ 6 CN−1 lnN → 0as N →∞ ∀ ε. Some properties of these mesh functions are given in Lemma A.1 in AppendixA.

Define the operators

D−h Z(xi) := 1h(Z(xi)− Z(xi−1)), D+

h := D−h Z(xi+1), δ2h := 1

h [(D+h −D

−h )Z(xi)]. (3.4)

Observe that in the fine mesh; D−h = D− and in the coarse mesh; D−h 6= D−. The purposeof defining such operators is for convenience. We present identities and inequalities that resultwhen these operators are applied to the mesh function T in Lemma A.2 in Appendix A.

We now present the discrete function Aε which will replace αε in discrete analogues of thebarrier functions in Lemma 2.1. We distinguish between the cases σ < µ (σ ≡ Cε lnN) andσ = µ (σ ≡ C). When σ < µ, we define Aε as follows

Aε(xi) :=

Aε(xN4

) + (xN4− xi) 5r

2µN−(1−2p), 0 6 i < N

4 ,

θT (xi),N4 6 i 6 3N

4 ,

Aε(x 3N4

)− (xi − x 3N4

) 5r2µN

−(1−2p), 3N4 h = H1) and define Aε as follows

Aε(xi) :=

Aε(xN

4) + H0

h θ(T (xi)− T (xN4

)), 0 6 i < N4 ,

θT (xi),N4 6 i 6 N.

(3.5b)

6

When µ = dN

2 (where H0 = h 6 H1) we can define Aε in the same manner. We proveconvergence of Aε to αε in the following lemma, the proof of which is given in Appendix B.

Lemma 3.1. For sufficiently large N , the function αε defined in (2.1) and the mesh functionAε defined in (3.5) satisfies:

|(αε −Aε)(xi)| 6 CN−1(lnN)2 + CN−(1−2p).

By the assumption (2.1a), and assuming N is sufficiently large, we have

|aε(xi)| > |Aε(xi)|, xi ∈ ΩNε . (3.6)

The finite difference operator LNε satisfies the following discrete minimum principle.

Theorem 3.1. Let LNε be the difference operator defined in (PNε ) and Z be a mesh function onΩNε . If min Z(x0), Z(xN ) > 0 and LNε Z(xi) 6 0 for xi ∈ ΩN

ε , then Z(xi) > 0 for all xi ∈ ΩNε .

Proof. Suppose that Z(xq) = mini Z(xi) < 0. If follows from the hypotheses that q /∈ 0, Nand xq ∈ ΩN

ε . Since Z(xq) is the minimum value we have D−Z(xq) 6 0, D+Z(xq) > 0 andδ2Z(xq) > 0. If xq < d, to avoid a contradiction we must have LNε Z(xq) 6 0 but if b(xq) > 0then LNε Z(xq) > 0 or if b(xq) = 0 then Z(xq−1) = Z(xq) = Z(xq+1) < 0. Repeat this argumentwill eventually lead us to conclude that either LNε Z(xq) > 0 or Z(x0) < 0 for xq < d which is acontradiction. If xq > d the same argument leads to a contradiction and finally a contradictioncan also be established for xq = d.

Lemma 3.2. There exists a unique solution, Yε, of (PNε ) such that

|Yε(xi)| 6 C, xi ∈ ΩNε .

Proof. We first establish the following list of inequalities;

εδ2Aε(xi)−Aε(xi)D−Aε(xi) > 0, 0 θr, 0 θr, Q 2r. We firstconsider the case where σ < µ. Using (3.5a) we can show that

D−Aε(xi) = − 5r2µN

−(1−2p), 0 < i 6 N4 ,

3N4 < i 6 N, θD−h Ti,

N4 < i 6 3N

4 (3.8a)

and

δ2Aε(xi) =

0, max i,N − i < N

4 ,

2(D+−D−)(Ai)h+H0,1

, i = N4 ,

3N4 ,

θδ2hTi,

3N4 5rθ4µ N

−(1−2p) − N2dN

(5rθN−2(1−p)) > 0.

7

For N4 AiAi−1 > A2

N4

> (1− Cρ)2θ2 > 12θ

2 > θr.

For N4 < i 6 Q, use (A.6h). The bound (3.7d) is established in the same manner.

Next, consider the case where σ = µ = 12(1− dN ). Using (3.5b) we can show that

D−Aε(xi) = θD−h Ti, δ2Aε(xi) = hH0θδ2hTi, i <

N4 ,

2hH0+hθδ

2hTi, i = N

4 , θδ2hTi, i >

N4 .

(3.9)

Using Lemma A.2 and H0 > h = H1, we can prove (3.7a)-(3.7d), in the same manner as the

case σ < µ. The case σ = µ = dN

2 follows suit.

Define the discrete barrier functions

ψ±(xi) :=‖f‖θr

[(xi − xQ)Aε(xi) + maxxi∈ΩNε

(xi − xQ)Aε(xi)] + max |A|, |B| ± Yε(xi), (3.10)

which are nonnegative at x = 0, 1. Using (3.5a) and Lemma A.2 we can find

D±[(xi − xQ)Aε(xi)] = (xi − xQ)D±Aε(xi) +Aε(xi±1)

> 0, i 6 Q6 0, i > Q

. (3.11)

Using this, we can show that

LNε ψ±(xi) 6‖f‖θr [(xi − xQ)(εδ2Ai −AiD−Ai)− (AiAi−1 − 2ε

Σh(hiD− + hi+1D

+)Ai)] + ‖f‖, 0 < i 6 Q,‖f‖θr [(xi − xQ)(εδ2Ai −AiD+Ai)− (Ai+1Ai − 2ε

Σh(hiD− + hi+1D

+)Ai)] + ‖f‖, Q < i < N.

Using the bounds in (3.7), we have LNε ψ±(xi) 6 0 for xi ∈ ΩN

ε . Thus using Theorem 3.1, wehave |Yε| 6 C on ΩN

ε .

As with the continuous problem (Pε), we split the discrete problem (PNε ) into left and rightdiscrete problems centred around x = xQ as follows. Define

Yε(xi) =

YL(xi), xi 6 xQ,

YR(xi), xi > xQ,

where the discrete left and right problems are defined by

LNε YL(xi) = f(xi),

xi ∈ ΩNε ∩ (0, xQ),

YL(0) = y0, YL(xQ) = Yε(xQ),

(PNL )

LNε YR(xi) = f(xi),

xi ∈ ΩNε ∩ (0, xQ),

YR(xQ) = Yε(xQ), YR(1) = y1.

(PNR )

We decompose each YL\R into the sum of a regular component, VL\R, and a layer component,WL\R. We define VL as the solution of

LN,−VL(xi) := (εδ2 + a−0 D− − b)VL(xi) = f(xi), xi ∈ ΩN

ε ∩ (0, xQ)

VL(0) = y0, VL(xQ) = (V0 + εV1)(xQ),(3.12a)

8

where VL = V0 + εV1 + ε2V2 and V0, V1, V2 satisfy

(a−0 D− + b)V0(xi) = −f(xi), xi ∈ ΩN

ε ∩ (0, xQ], V0(0) = y0, (3.12b)

(a−0 D− + b)V1(xi) =

δ2V0(xi), xi ∈ ΩN

ε ∩ (0, xQ)δ2V0(xi−1), xi = xQ

, V1(0) = 0, (3.12c)

LN,−V2(xi) = −δ2V1(xi), xi ∈ ΩNε ∩ (0, xQ), V2(0) = V2(xQ) = 0. (3.12d)

We incorporate the error (LNε −LN,−)VL into the discrete layer component, WL, which is, noting(2.1c), defined as the solution of the following discrete problem

LNε WL(xi) = ϕ−ε D−VL(xi), xi ∈ ΩN

ε ∩ Ω−, (3.12e)

WL(0) = 0, WL(xQ) = (YL − VL)(xQ). (3.12f)

Similarly VR and WR satisfy

LN,+VR(xi) := (εδ2 + a+0 D

+ − b)VR(xi) = f(xi), xi ∈ ΩNε ∩ (xQ, 1), (3.13a)

VR(xQ) = v∗, VR(1) = y1, (3.13b)

LNε WR(xi) = ϕ+ε D

+VR(xi), xi ∈ ΩNε ∩ (xQ, 1), (3.13c)

WL(xQ) = (YR − VR)(xQ), WR(1) = 0. (3.13d)

where v∗ is defined analogously to VL(xQ) in (3.12). We now determine bounds on VL\R, WL\R,D±VL\R and on the error VL\R − vL\R. But first we present a mesh function that we will usein the barrier functions for the layer component, along with some of its properties. The proofof this Lemma is given in Appendix C.

Lemma 3.3. For sufficiently large N , the mesh function defined as

W (xi) :=

Q∏j=i+1

(1 +

αε(xj)

2εhj

)−1

, 0 6 i < Q, W (xQ) := 1

satisfies

W (xi) > (1− Cρ)e−θ2ε

(d−xi), 0 6 i 6 Q, (3.14a)

W (xi) 6 Ce−θ2ε

(d−xi), N4 6 i 6 Q (σ < µ), 0 6 i 6 Q (σ = µ). (3.14b)

|W (xi)| 6 CN−(1−p), i 6 N4 (σ < µ). (3.14c)

Lemma 3.4. If vL, wL are the solutions of (2.4), vR, wR are the solutions of (2.5), VL, WL

are the solutions of (3.12) and VR, WR are the solutions of (3.13) then we have the followingbounds

|D−VL(xi)| 6 C, xi ∈ ΩNε ∩ (0, xQ]; |(VL − vL)(xi)| 6 CN−1xi, xi ∈ ΩN

ε ∩ [0, xQ],

|D+VR(xi)| 6 C, xi ∈ ΩNε ∩ [xQ, 1); |(VR − vR)(xi)| 6 CN−1(1− xi), xi ∈ ΩN

ε ∩ [xQ, 1],

|WL(xi)| 6

Ce−

θ2ε

(d−xi) + CN−(1−p), N4 6 i 6 Q (σ < µ), 0 6 i 6 Q (σ = µ),

CN−(1−p), 0 6 i 6 N4 (σ < µ),

|WR(xi)| 6

Ce−

θ2ε

(xi−d) + CN−(1−p), Q 6 i 6 3N4 (σ < µ), Q 6 i 6 N (σ = µ),

CN−(1−p), 3N4 6 i 6 N (σ < µ).

9

Proof. If Z is a mesh function on ΩNε ∩ [0, xQ] then we can easily prove that:

if Z|x0 > 0 and (a−0 D− + b)Z

∣∣ΩNε ∩[0,xQ]\x0 > 0 then Z|ΩNε ∩[0,xQ] > 0. (3.15)

Consider the functions ψ±(xi) = y0 + ‖f‖θ xi ± V0(xi). Using (2.1), (3.12b) and (3.15) we can

show that ‖V0‖, ‖D−V0‖, ‖δ2V0‖ 6 C. Similarly we can show that ‖V1‖, ‖D−V1‖, ‖δ2V1‖ 6 C.For (3.12d), using the results in [1, §3.3] we have |V2(xi)| 6 Cxi, xi ∈ ΩN

ε ∩ [0, xQ]. Repeat [1,§3.5,Lemma 3.14] and obtain ε‖D−V2‖ 6 C. Thus it follows that ‖D−VL‖ 6 C.

Using (2.4) with (3.13) we can define the errors E0 := V0 − v0, E1 := V1 − v1 and E := Vε − vεas the solutions of

(a−0 D− + b)E0(xi) = a−0 (v′0 −D−v0)(xi), x ∈ ΩN

ε ∩ (0, xQ], E0(0) = 0, (3.16a)

(a−0 D− + b)E1(xi) = (a−0 (v′1 −D−v1)− (v′′0 − δ2V0))(xi),

xi ∈ ΩNε ∩ (0, xQ], E1(0) = 0,

(3.16b)

LN,−E(xi) = (L− − LN,−)vL(xi),

xi ∈ ΩNε ∩ (0, xQ), E(0) = 0, E(xQ) = (E0 + εE1)(xQ).

(3.16c)

Using Lemma 2.2, we can use standard truncation error estimates to show that

‖v′0 −D−v0‖ 6 C‖v′′0‖maxhi 6 CN−1, (3.17a)

‖v′′0 − δ2v0‖ 6 C‖v′′′0 ‖maxhi 6 CN−1, (3.17b)

‖v′1 −D−v1‖ 6 C‖v′′1‖maxhi 6 CN−1, (3.17c)

‖(L− − LN,−)vL‖ 6 C(ε‖v′′′L ‖+ ‖v′′L‖) maxhi 6 CN−1. (3.17d)

Using suitable barrier functions, we can easily show that |E0(xi)| 6 CN−1xi. Using (v′0 −D−v0)(xi) = 1

hi

∫ xit=xi−1

v′0(xi)− v′0(t) dt we can show that

|(v′′0 − δ2V0)(xi)| 6 |(v′′0 − δ2v0)(xi)|+ |δ2E0(xi)| 6 CN−1 +

CN−1, i 6= N

4 ,

C, i = N4 .

Rewrite (3.16b) as

E1(xi) = [E1(xi−1) +hi

a−0(a−0 (v′1 −D−v1)− (v′′0 − δ2V0))(xi)](1 +

bhia0

(xi))−1

=

i∑j=1

[

i∏k=j

(1 +bhi

a−0(xk))

−1]hj

a−0(a−0 (v′1 −D−v1)− (v′′0 − δ2V0))(xj).

Using (1 + hib(xi)

a−0 (xi))−1 < 1 and using (3.17) we have

|E1(xi)| 6 i‖hi‖CN−1 + C

i∑j=1

|hj(v′′0 − δ2V0)(xj)|.

For i < N4 , we clearly have |E1(xi)| 6 CN−1 + C‖hi‖iCN−1 6 CN−1 and for i > N

4 we have|E1(xi)| 6 CN−1 + ChN

4+ C‖hi‖(i − N

4 )CN−1 6 CN−1. Use suitable barrier functions with

(3.16c) to find that |E(xi)| 6 CN−1xi for xi ∈ ΩNε ∩ [0, xQ]. The proof for the bounds on VR

can be established in an analogous manner.

We now bound the layer component WL. We first establish a few inequalities. Using 12et 6

10

cosh t 6 et, t > 0 and using (2.1), (A.5) and Lemma A.2, bound ϕ−ε and aε for sufficiently largeN as follows

W (xi−1) > e−θ2ε

(d−xi+h) > 12e− θ

2ε(d−xi), N

4 θ2 , 0 6 i 6 N

4 , (σ < µ) (3.18b)

|ϕ−ε (xi)| 6 N−(1−p), 0 6 i 6 N4 , (σ < µ). (3.18c)

Consider the functions ψ±(xi) = m1W (xi) +m2N−(1−p)xi ±WL(xi), m2 := 2

θ‖ϕ−ε (d)||D−VL‖,

m1 := max |WL(d)|, 2εr m2 which are nonnegative at x = 0, 1. Consider the σ < µ case. Using

Lemma 3.3, (3.18) and (3.13) and noting for any xi, we have α′ε(xi+1) 6 D+αε(xi) 6 α′ε(xi).Then for sufficiently large N we have

LNε ψ±(xi) 6 m1(εδ2W − αεD−W )(xi)− αε(xi)m2N

−(1−p) + ‖D−VL‖|ϕ−ε (x)|

6 −m1hi+1

2ε(hi+1 + hi)(α2

ε − 2εα′ε)W (xi−1)− αε(xi)m2N−(1−p) + ‖D−VL‖|ϕ−ε (x)|

6

−m2

θ2N−1 + |ϕ−ε (d)|‖D−VL‖N−(1−p) 6 0, 0 < i 6 N

4 ,

−m1θr4εe− θ

2ε(d−xi) + |ϕ−ε (d)|‖D−VL‖e−

θ2ε

(d−xi) 6 0, N4 < i < Q.

Hence using Theorem 3.1, we have |WL(xi)| 6 C(W (xi) +N−(1−p)). The case σ = µ is provedin the same manner as for N

4 < i < Q above. The proof of the bound on WR can be performedin a similar way.

A bound on the error Yε − yε is given in the following lemma.

Lemma 3.5. If yε is the solution of (Pε) and Yε is the solution of (PNε ) then

|(Yε − yε)(xi)| 6 CN−1(lnN)2 + CN−(1−p), ∀xi ∈ ΩNε ,

where |dN − d| < pσ and p 6 12 .

Proof. First, if σ < µ then for i 6 N4 , using Lemma 2.2, Lemma 3.4 and (3.1), (A.5) we have

|(Yε − yε)(xi)| 6 |WL(xi)|+ |wL(xi)|+ ‖VL − vL‖ 6 CN−(1−p).

Similarly, if σ < µ then for i > 3N4 , we have |(Yε − yε)(xi)| 6 CN−(1−p).

We now need to examine the error over all mesh points xi ∈ [xL, xR] where xL = 0 if σ = µ orxL = xN

4if σ < µ and xR = 1 if σ = µ or xR = x 3N

4if σ < µ. Note the implication that for

any xi ∈ (xL, xR), we have hi/ε 6 CN−1 lnN . The error E(xi) := (Yε − yε)(xi) is defined as asolution of the following

LNε E(xi) = (Lε − LNε )yε, xi ∈ ΩNε ∩ (xL, xR), |E(xL\R)| 6 CN−(1−p). (3.19)

Note the following truncation error estimate

|ε(δ2yε − y′′ε )(xi)| 62ε

hi+1 + hi

i+1∑j=i

1

hj

∣∣∣∣∣∫ xj

t=xj−1

∫ t

s=xi

y′′ε (s)− y′′ε (xi) ds dt

∣∣∣∣∣. (3.20)

11

For any s ∈ [xi−1, xi+1] ⊂ [xL, xR] ∩ ΩNε , using (Pε), (2.1) and Lemma 2.1, we can bound

ε|y′′ε (s)− y′′ε (xi)| as follows

ε|y′′ε (s)− y′′ε (xi)| 6 (‖aε‖‖y′′ε‖+ (‖b‖+

∫ 1

0a′ε dt)‖y′ε‖+ ‖b′‖‖yε‖+ ‖f ′‖)|s− xi| 6 C hi

ε2. (3.21)

Thus using (3.21) with (3.20) we have

|(Lε − LNε )yε(xi)| 6 Chiε−2 6 Cε−1N−1 lnN, xi ∈ ΩN

ε ∩ (xL, xR). (3.22)

Hence, using the barrier functions in (3.10) we can show that

|E(xi)| 6 Cε−1N−1 lnN [(xi − xQ)Aε(xi) + maxxi∈ΩNε ∩[xL,xR]

(xi − xQ)Aε(xi)] + CN−(1−p) (3.23)

(3.24)

6

Cσε N

−1 lnN + CN−(1−p) 6 CN−1(lnN)2 + CN−(1−p), σ < µ,

CεN−1 lnN + CN−(1−p) 6 CN−1(lnN)2 + CN−(1−p), σ = µ.

(3.25)

Complete the proof using Theorem 3.1.

The nodal bound on the error is easily extended to a global error bound.

Lemma 3.6. If yε is the solution of (Pε) and Yε is the solution of (PNε ) then

|(Y ε − yε)(t)| 6 CN−1(lnN)2 + CN−(1−p), t ∈ [0, 1].

where Y ε is the piecewise linear interpolant of Yε on [0, 1].

Proof. Proof follows the corresponding proof in [1, Thm 3.12].

4 Numerical Examples

Example 1

In this example from the class (P1) we consider aε(x) = (2.25 + x2) tanh (1.1ε (0.6− x)), b(x) =

e−5x, f(x) = cos(3x), A = 2, B = −3, d = 0.6. We choose r = 1 and θ = 2.1. We considervarious values of p.

To generate a value of dN , we start with a scale factor of κ = 0.99 and reduceby 0.01 until dN = d± κpmin 1

2 min (d, 1− d), εr lnN satisfies |d− dN | < pσ.(4.1)

We consider p = 0, p = 0.5 and p = 1 and using (4.1) we centre the mesh at dN = d,dN ≈ d− 0.5σ and dN ≈ d− σ. We compute the approximate errors

ENε = maxΩNε ∪Ω8192

ε

|UNε − U8192ε | (4.2)

where UMε is UMε , the numerical solution of (PNε ) using N = M mesh points, interpolated onto

the mesh ΩNε ∪ Ω8192

ε . Table 4.1 displays the approximate errors ENε and the uniform errorsEN = maxεE

Nε , using (4.2) . Shifting the mesh off-centre within the limit of p 6 1

2 has littleto no effect on the differences. We further test for an effect on the value of p by producing thecomputed rates of convergence pNε and the uniform rates of convergence pN , computed usingthe double mesh principle (see [1]), as shown in Table 4.2. The N−(1−p) factor established inLemma 3.5 is not evident for p 6 1

2 . However, for p = 1 we see a collapse in the computed ratesof convergence.

12

ENε

dN = d dN ≈ d+ 0.5σ dN ≈ d+ σ

HHHHεN

32 64 128 256 512 32 64 128 256 512 32 64 128 256 512

2−0 0.004 0.002 0.001 0 0 0.005 0.002 0.001 0 0 0.006 0.003 0.001 0.001 02−1 0.027 0.013 0.007 0.003 0.002 0.04 0.02 0.01 0.005 0.002 0.047 0.023 0.011 0.006 0.0032−2 0.085 0.044 0.022 0.011 0.005 0.15 0.076 0.038 0.019 0.009 0.181 0.092 0.046 0.023 0.0112−3 0.118 0.061 0.031 0.015 0.007 0.282 0.147 0.073 0.036 0.018 0.41 0.207 0.104 0.051 0.0252−4 0.218 0.116 0.059 0.03 0.014 0.245 0.133 0.07 0.036 0.018 0.95 0.421 0.21 0.104 0.0512−5 0.24 0.147 0.088 0.05 0.028 0.271 0.15 0.088 0.047 0.023 1.748 0.957 0.428 0.212 0.1022−6 0.241 0.147 0.087 0.05 0.027 0.29 0.157 0.086 0.049 0.027 2.603 1.805 0.992 0.441 0.2112−7 0.242 0.147 0.087 0.05 0.027 0.303 0.163 0.088 0.049 0.027 3.267 2.635 1.822 1.001 0.4432−8 0.243 0.147 0.087 0.05 0.027 0.311 0.167 0.09 0.049 0.027 3.711 3.274 2.628 1.808 0.9832−9 0.243 0.147 0.087 0.05 0.027 0.315 0.17 0.091 0.049 0.027 3.981 3.697 3.243 2.586 1.7592−10 0.244 0.147 0.087 0.05 0.027 0.317 0.171 0.092 0.049 0.027 4.136 3.953 3.648 3.172 2.4922−11 0.244 0.147 0.087 0.05 0.027 0.318 0.171 0.092 0.049 0.027 4.22 4.098 3.891 3.552 3.0382−12 0.244 0.147 0.087 0.05 0.027 0.318 0.172 0.092 0.05 0.027 4.269 4.176 4.029 3.765 3.4072−13 0.244 0.147 0.087 0.05 0.027 0.319 0.172 0.092 0.05 0.027 4.295 4.222 4.101 3.918 3.592−14 0.244 0.147 0.087 0.05 0.027 0.319 0.172 0.093 0.05 0.027 4.306 4.252 4.154 3.981 3.6862−15 0.244 0.147 0.087 0.05 0.027 0.319 0.172 0.093 0.05 0.027 4.321 4.269 4.174 4.006 3.8242−16 0.244 0.147 0.087 0.05 0.027 0.319 0.172 0.093 0.05 0.027 4.326 4.275 4.181 4.073 3.9042−17 0.244 0.147 0.087 0.05 0.027 0.319 0.172 0.093 0.05 0.027 4.327 4.277 4.212 4.114 3.9462−18 0.244 0.147 0.087 0.05 0.027 0.319 0.172 0.093 0.05 0.027 4.328 4.292 4.232 4.135 3.9692−19 0.244 0.147 0.087 0.05 0.027 0.319 0.172 0.093 0.05 0.027 4.335 4.302 4.243 4.146 3.982−20 0.244 0.147 0.087 0.05 0.027 0.319 0.172 0.093 0.05 0.027 4.34 4.307 4.248 4.151 3.986

EN 0.244 0.147 0.088 0.05 0.028 0.319 0.172 0.093 0.05 0.027 4.34 4.307 4.248 4.151 3.986

Table 4.1: Approximate errors for Example 1.

pNε

dN = d dN ≈ d+ 0.5σ dN ≈ d+ σ

HHHHεN

32 64 128 256 512 32 64 128 256 512 32 64 128 256 512

2−0 0.99 0.99 0.99 1 1 1.06 1.01 1 1 1 1.11 1.01 1 1 12−1 0.9 0.95 0.97 0.99 0.99 0.91 0.95 0.98 0.99 0.99 0.9 0.95 0.97 0.98 0.992−2 0.84 0.92 0.96 0.98 0.99 0.84 0.91 0.96 0.98 0.99 0.84 0.91 0.95 0.98 0.992−3 0.84 0.92 0.96 0.98 0.99 0.7 0.84 0.92 0.96 0.98 0.84 0.89 0.93 0.96 0.982−4 0.72 0.85 0.92 0.96 0.98 0.71 0.82 0.89 0.94 0.97 0.85 0.85 0.89 0.94 0.972−5 0.33 0.56 0.61 0.63 0.87 0.8 0.62 0.68 0.93 0.96 0.37 0.93 0.9 0.85 0.92−6 0.36 0.57 0.61 0.75 0.8 0.84 0.85 0.69 0.73 0.79 -0.24 0.36 0.94 0.89 0.912−7 0.38 0.57 0.61 0.75 0.8 0.85 0.87 0.77 0.73 0.78 -0.57 -0.22 0.36 0.92 0.922−8 0.4 0.58 0.61 0.75 0.8 0.85 0.88 0.78 0.77 0.78 -0.74 -0.54 -0.22 0.35 0.92−9 0.4 0.58 0.61 0.75 0.8 0.85 0.88 0.79 0.79 0.78 -0.83 -0.69 -0.52 -0.21 0.352−10 0.41 0.58 0.61 0.75 0.8 0.85 0.88 0.79 0.81 0.78 -0.85 -0.74 -0.66 -0.52 -0.212−11 0.41 0.58 0.61 0.75 0.8 0.85 0.88 0.79 0.81 0.78 -0.84 -0.72 -0.7 -0.65 -0.512−12 0.41 0.58 0.61 0.75 0.8 0.85 0.88 0.79 0.82 0.78 -0.8 -0.64 -0.65 -0.67 -0.642−13 0.41 0.58 0.61 0.75 0.8 0.85 0.88 0.79 0.82 0.78 -0.76 -0.53 -0.54 -0.61 -0.662−14 0.41 0.58 0.61 0.75 0.8 0.84 0.88 0.79 0.82 0.78 -0.46 -0.43 -0.42 -0.49 -0.592−15 0.41 0.58 0.61 0.75 0.8 0.84 0.88 0.79 0.82 0.78 -0.27 -0.36 -0.31 -0.36 -0.462−16 0.41 0.58 0.61 0.75 0.8 0.84 0.88 0.79 0.82 0.78 -0.17 -0.32 -0.23 -0.24 -0.332−17 0.41 0.58 0.61 0.75 0.8 0.84 0.88 0.79 0.82 0.78 -0.11 -0.3 -0.19 -0.17 -0.212−18 0.41 0.58 0.61 0.75 0.8 0.84 0.88 0.79 0.82 0.78 -0.09 -0.28 -0.16 -0.12 -0.132−19 0.41 0.58 0.61 0.75 0.8 0.84 0.88 0.79 0.82 0.78 -0.07 -0.28 -0.15 -0.1 -0.092−20 0.41 0.58 0.61 0.75 0.8 0.84 0.88 0.79 0.82 0.78 -0.07 -0.27 -0.14 -0.08 -0.06

pN 0.72 0.62 0.61 0.63 0.87 0.84 0.88 0.76 0.82 0.79 -0.11 -0.05 -0.02 0 0

Table 4.2: Computed orders of convergence for Example 1.

13

In Figure 4.1, we present comparisons between numerical solutions using N = 32 and the meshcentred at dN = d and ε = 2−0, ε = 2−2, ε = 2−5 and ε = 2−10.

Fig. 4.1: Solution of (PNε ) with N = 32, dN = d for ε = 2−0, ε = 2−2, ε = 2−5 and ε = 2−10.

In Figure 4.2, we present comparisons between numerical solutions using N = 32, ε = 2−8 andthe mesh centred at dN = d, dN ≈ d − 0.5σ and dN ≈ d − σ. The location of the meshes foreach case are superimposed on the figure below the graphs.

Fig. 4.2: Solution of (PNε ) with N = 32, ε = 2−8 and dN ≈ d− pσ for p = 0, p = 12 and p = 1.

14

References

[1] P.A. Farrell, A.F. Hegarty, J.J.H. Miller, E. O’Riordan, G.I. Shishkin, Robust Computa-tional Techniques for Boundary Layers, Chapman and Hall/CRC, Boca Raton, FL. (2000)

[2] A.E. Berger, H. Han, R.B. Kellogg, A Priori Estimates and Analysis of a Numerical Methodfor a Turning Point Problem, Mathematics of Computation, 42(166), 465-492, (1984)

[3] P.A. Farrell, Sufficient Conditions for the Uniform convergence of a difference scheme fora Singularly Perturbed Turning Point Problem, SIAM J. Numer. Anal., 25(3), 618-643,(1988)

[4] P.A. Farrell, A.F. Hegarty, J.J.H. Miller, E. O’Riordan, G.I. Shishkin, Global Maxi-mum Norm Parameter-Uniform Numerical Method for a Singularly Perturbed Convection-Diffusion Problem with Discontinuous Convection Coefficient, Math. and Computer Mod-elling, 40, 1375-1392, (2000)

[5] F.A. Howes, Boundary-interior layer interactions in nonlinear singular perturbation theory,Memoirs of the American Mathematical Society, 15(203), (1978)

[6] E. O’Riordan, J. Quinn, Parameter-uniform numerical methods for some linear and non-linear singularly perturbed convection diffusion boundary turning point problems, BITNumer. Math., 51(2), 317-337, (2011)

A Properties of the discrete functions S, C and T .

Lemma A.1. For sufficiently large N , the mesh functions S and C defined in (3.3) satisfy:

S(xi−1) > S(xi), ∀i; S(xi) > 0, i < Q; S(xQ) = 0; S(xi) < 0, i > Q; (A.1a)

C(xi−1) > C(xi), i < Q; C(xi−1) < C(xi), i > Q; C(xi) > C(xQ) = 2, ∀i; (A.1b)

i 6 Q− 2, C(xi)1+ρ

Q− 1 6 i, C(xi)

6 C(xi+1) 6

C(xi), i < QC(xi)1−ρ , Q 6 i

; (A.1c)

i 6 Q, C(xi)Q < i, (1− ρ)C(xi)

6 C(xi−1) 6

(1 + ρ)C(xi), i < Q

C(xi), Q 6 i; (A.1d)

T (xi) > 0, i < Q; T (xQ) = 0; T (xi) < 0, i > Q; (A.1e)

|T (xi)| < 1, |T (xi)− tanh( rε(Q− i)h)| 6 CNρ2, ∀i; T (xN4

), |T (x 3N4

)| > 1− Cρ > 12 . (A.1f)

Proof. For convenience in this proof, we write any mesh function Z(xi) as Zi. Clearly Ci > 2,∀ i and Si > 0, i 6 Q, Si 6 0, i > Q. Use Si − Si−1 = −ρCi and Ci −Ci−1 = −ρSi to establishthe bounds in (A.1a) and (A.1b). For (A.1c), we can show Ci−1 = Ci + ρSi. We can use

Ci < Si < Ci (A.2)

to show that (1 − ρ)Ci < Ci−1 < (1 + ρ)Ci. Combine these with (A.1b) to establish (A.1c).For (A.1d), we can show that Ci+1 = 1

1−ρ2 (Ci − ρSi). Thus from (A.2), we have Ci+1 <1+ρ1−ρ2Ci = 1

1−ρCi. Similarly Ci+1 >1

1+ρCi. Combine these with (A.1b) to establish (A.1d). The

expressions in (A.1e) follow from (A.1a) and (A.1b). The first bound in (A.1f) follows from(A.2). Using the inequalities

e(1−t/2)t 6 1 + t 6 et 6 1 + 2t, t ∈ [0, 0.5], (A.3a)

e−(1+t)t 6 1− t 6 e−t 6 1− t/2, t ∈ [0, 0.5], (A.3b)

15

we can show that for any sufficiently small ρ > 0 and 0 6 j 6 N , we have

(1− CNρ2) sinh ρj − CNρ2 6 12 [(1 + ρ)j − (1− ρ)j ] 6 (1 + CNρ2) sinh ρj + CNρ2 (A.4a)

(1− CNρ2) cosh ρj 6 12 [(1 + ρ)i + (1− ρ)i] 6 cosh ρj. (A.4b)

With a change of variables, we can prove the second bound in (A.1f) using (A.3), (A.4) with(3.3) for sufficiently large N . Note that using (3.1) and (3.2), for sufficiently large N we have

xQ − xN4> (1− p)σ ⇒ Q− N

4 > (1− p)N4 , (A.5a)

x 3N4− xQ > (1− p)σ ⇒ 3N

4 −Q > (1− p)N4 . (A.5b)

For i = N4 and for sufficiently large N , using (A.5), we can write Ci 6 (1 +Cρ)(1 + ρ)Q−

N4 and

Si > (1 − Cρ)(1 + ρ)Q−N4 . Hence Si/Ci > 1 − Cρ > 1

2 . We can write a similar statement fori = 3N

4 .

Lemma A.2. Assuming (2.1), then for sufficiently large N , the operators in (3.4) and themesh function T defined in (3.3) satisfy:

D−h T (xi) < 0, 0 < i 6 N, D+h T (xi) < 0, 0 6 i < N, (A.6a)

1 + ρ

1− ρD−h T (xi) 6 D+

h T (xi) 6 D−h T (xi), 0 3N

4 , (A.6e)

ε|D+h T (xN

4)|, ε|D−h T (x 3N

4)| 6 5rθN−2(1−p), when σ = ε

r lnN, (A.6f)

εδ2hT (xi)− θT (xi)D

−h T (xi) > 0, 0 2r, 0 2r, Q < i < N. (A.6j)

Proof. To ease notation, we write U(xi) as Ui for any mesh function U . Using (3.3) with (3.4),we can show that

D−h T (xi) = rε(TiTi−1 − 1) =

−4r/ε(1− ρ2)Q−i

CiCi−1, (A.7a)

δ2hT (xi) =

r

εTi(D

+ +D−)Ti. (A.7b)

Use Lemma A and (A.7) to establish (A.6a), (A.6c) and (A.6e). For (A.6b), using Lemma Awe have Ci+1Ci > 1

(1+ρ)2CiCi−1. Thus

D+h T (xi) =

−4r/ε(1− ρ2)Q−(i+1)

Ci+1Ci>

(1 + ρ)2

1− ρ2

−4r/ε(1− ρ2)Q−i

CiCi−1=

1 + ρ

1− ρD−h T (xi).

Also, from Lemma A.1 and (A.7) we have δ2hT (xi) 6 0, i 6 Q. Thus using (3.4), we have

D+h T (xi) 6 D−h T (xi), i 6 Q. Verify (A.6d) in the same manner. For (A.6f), when i = N

4 thenusing Lemma A.1 and (3.3) we have

Ci+1Ci > C2i+1 = [(1 + ρ)Q−(i+1) + (1− ρ)Q−(i+1)]2 > (1 + ρ)2(Q−(i+1)). (A.8)

16

Using (A.5), (A.8) and the inequality (1 + t)/(1 − t) > e2t, 0 < t 6 0.5, assuming σ = εr lnN ,

then we have

D+h T (xi) =

−4rε−1(1− ρ2)Q−(i+1)

Ci+1Ci> −4rε−1 (1 + ρ)Q−(i+1)(1− ρ)Q−(i+1)

(1 + ρ)2(Q−(i+1))

= −1+ρ1−ρ4rε−1

(1− ρ1 + ρ

)Q−N4

> −5rε−1e−2ρ(Q−N4

) = −5rε−1N−2(1−p).

Use (A.6a) to bound D+h T (xi) from above. Bound D−T (x 3N

4) in the same manner. For (A.6g),

using (2.1), (A.7) and (A.6b) with Lemma A.1 and (A.6a) for i < Q, we have

εδ2hTi − θTiD−h Ti = Ti(r(D

+ +D−)Ai − θD−h Ti) > Ti(2r

(1−ρ) − θ)D−h Ti > 0.

The bound (A.6h) is established in the same manner. For (A.6i) on 0 TiTi−1 − 2εθ D−h Ti = 2r

θ + (1− 2rθ )TiTi−1 > 2r

θ .

The bound (A.6j) is established in the same manner.

B Proof of Lemma 3.1

Proof. First, using (3.1), (3.2) and the identity tanh (X + Y ) = tanhX+tanhY1+tanhX tanhY we have

tanh( rε(xQ − xi)) = tanh[ rε(d− xi) + rε(xQ − d)] =

tanh( rε(d− xi)) + tanh( rε(xQ − d))

1 + tanh( rε(d− xi)) tanh( rε(xQ − d))

6tanh( rε(d− xi)) + ρ

1− ρ6 tanh( rε(d− xi)) + Cρ.

Construct a corresponding lower bound to give

| tanh( rε(d− xi))− tanh( rε(xQ − xi))| 6 Cρ 6 CN−1 lnN, ∀i. (B.1)

Consider i ∈ S := [N4 ,3N4 ] if σ < µ, [N4 , N ] if σ = 1

2(1 − dN ), [0, 3N4 ] if σ = dN

2 ∩ Z. Using(3.1), we have xQ−xi = xN

4+h(Q− 3N

4 )−xN4−h(i− 3N

4 ) = (Q− i)h. Now using Lemma A.2

we have|Aε(xi)− θ tanh( rε(xQ − d))| 6 CNρ2.

For all other i /∈ S, using Lemma A.2 and (3.5) we have

|Aε(xi)− θ tanh( rε(xQ − xi))|6 θ|TN

4 ,3N4

+ CN−(1−2p) + C|Ti − TN4 ,

3N4| − θ tanh( rε(xQ − xi))|

6 C|TN4 ,

3N4− 1|+ CN−(1−2p) + Cρ+ C|1− θ tanh( rε(xQ − xi))|

6 Cρ+ CN−(1−2p) + Cρ+ C|1− θ tanh( rε(xQ − xN4 ,

3N4

))|

6 Cρ+ CN−(1−2p) + C|1− θ tanh( rε(Q− (N4 ,3N4 ))h)|

6 Cρ+ CN−(1−2p) + C|1− TN4 ,

3N4|+ C|TN

4 ,3N4− θ tanh( rε(Q− (N4 ,

3N4 ))h)|

6 Cρ+ CN−(1−2p) + Cρ+ CNρ2.

Thus, ∀ i, we have

|Aε(xi)− θ tanh( rε(xQ− xi))| 6 Cρ+CN−(1−2p) +CNρ2 6 CN−(1−2p) +CN−1(lnN)2. (B.2)

Combine (B.1) and (B.2) to complete the proof.

17

C Proof of Lemma 3.3

Proof. We can easily bound W from below using 1 + t 6 et, t > 0 with (2.1), (3.1) and (A.5)as follows

W (xi) >Q∏

j=i+1

e−αε(xj)

2εhj = e−

12ε

∑Qj=i+1 αε(xj)hj > e−

θ2ε

∑Qj=i+1 hj = e−

θ2ε

(xQ−xi) > 12e− θ

2ε(d−xi).

Note the Rectangle Rule for Numerical Integration: Partition the interval [a, b] into N∗ subin-tervals with width h∗ = 1/N∗, where tj = a+ jh∗. Then

∫ b

af(s) ds = h∗

N∗∑j=1

f(tj) +O((b− a)h∗2‖f ′′‖).

Using (A.5) with the rectangle rule on N4 6 i 6 Q when (σ < µ) or on 0 6 i 6 Q when (σ = µ),

we have

1

2ε

Q∑j=i+1

αε(xj)hj >1

2ε

∫ xQ

xi

αε(t) dt− Cσε (h/ε)2 >

1

2ε

∫ xQ

xi

αε(t) dt− CN−2(lnN)3. (C.1)

Using the inequality 1 + t > et−t2/2, t > 0 with (2.1), (2.7), (3.1), (A.5) and (C.1) we have

W (xi) 6Q∏

j=i+1

e−αε(xj)

2εhje

12

(αε(xj)

2εhj)

26 eCN(N−1 lnN)2e−

12ε

∑Qj=i+1 αε(xj)hj

6 Ce−12ε

∫ xQxi

αε(t) dt 6 Ce− 1

2ε

∫ dxiαε(t) dt 6 Ce−

θ2ε

(d−xi).

Finally, we have W (xi) − W (xi−1) = αε(xi)hi2ε W (xi) > 0. Thus for i 6 N

4 when (σ < µ), using(2.1) and (A.5) we have

W (xi) 6 W (xN4

) 6 Ce−θ2ε

(d−xN/4) 6 CN−(1−p).

18

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