Isogeometric analysis for singularly perturbed problems in 1-D: error estimates

ISOGEOMETRIC ANALYSIS FOR SINGULARLY PERTURBED PROBLEMS IN 1-D: ERROR ESTIMATES∗

CHRISTOS XENOPHONTOS†ANDIRENE SYKOPETRITOU†

Abstract.We consider one-dimensional singularly perturbed boundary value problems of reaction-convection-diffusion type, and the approximation of their solution using isogeometric analysis. In particular, we use a Galerkin formulation with B-splines as basis functions, defined on appropriately chosen knot vectors. We prove robust exponential convergence in the energy norm, independently of the singular perturbation parameters, and illustrate our findings through a numerical example.

Key words.singularly perturbed problem, reaction-convection-diffusion, boundary layers, isogeometric analysis, robust exponential convergence

AMS subject classifications.65N30

1. Introduction. We consider second order singularly perturbed problems (SPPs) in one-dimension, of reaction-convection-diffusion type, whose solution contains boundary layers; see, e.g., [13]. The approximation of the solution to SPPs has received a lot of attention in the last few decades, mainly using finite differences (FDs) and finite elements (FEs) onlayer adapted meshes; see, e.g., [9]. Various formulations and results are available in the literature, both theoretical and computational (cf. [9,13], and the references therein). One method that has not, to our knowledge, been applied to general SPPs isisogeometric analysis(IGA). Since the introduction of IGA by T. R. Hughes et al. [7], the method has been successfully applied to a large number of problem classes. Even though much attention has been given to convection-dominated problems [3], the method has not been applied, as far as we know, to a typical singularly perturbed problem, such as (2.1), (2.2) ahead.

Our goal in this article is to provide the theoretical justification for the application of IGA to SPPs and, in particular, for the approximation of the solution to the boundary value problem (BVP) given by (2.1), (2.2). We use a Galerkin formulation with B-splines as basis functions and select appropriate knot vectors, such that as the polynomial degree increases, the error in the approximation, measured in the energy norm, decays exponentially and independently of the singular perturbation parameters. This is the analog of performingprefinement in the FEM. Even though we focus on the one-dimensional case, our results will serve as the stepping stone to higher dimensions for two reasons: the boundary layer effect is one dimensional (in the direction normal to the boundary) and B-splines in two-dimensions are constructed via tensor products. Hence, this is the first step towards higher dimensions.

The rest of the paper is organized as follows: in Section2we describe the model problem, its regularity and the Galerkin formulation we will use. In Section3we give a brief introduction to IGA, as described in [3], and present the discrete problem. Section4contains our main result of uniform exponential convergence and finally, Section5shows the result of a numerical experiment to illustrate the theory.

WithI⊂_Ran interval with boundary∂I and measure|I|, we will denote byCk_(I)the space of continuous functions onIwith continuous derivatives up to orderk. We will use the usual Sobolev spacesWk,m_{(I)of functions onIwith0,1,2, . . . , kgeneralized derivatives} inLm_{(I), equipped with the norm and seminormk·k}

k,m,Iand|·|k,m,I, respectively. When

∗_{Received October 2, 2018. Accepted November 8, 2019. Published online on January 16, 2020. Recommended}

by Torsten Linß.

†_{Department of Mathematics and Statistics, University of Cyprus, P.O. Box 20537, 1678 Nicosia, Cyprus}

([email protected]).

m= 2, we will writeHk_{(I)instead ofW}k,2_{(I), and for the norm and seminorm, we will}

writek·k_k,Iand|·|_k,I, respectively. The usualL2_{(I)inner product will be denoted byh·,·i} I, with the subscript omitted when there is no confusion. We will also use the space

H₀1(I) =

u∈H1(I) : u|_∂I = 0 .

The norm of the spaceL∞(I)of essentially bounded functions is denoted byk · k∞,I. Finally, the letterCwill denote a generic positive constant, independent of any parameters and possibly having different values in each occurence.

2. The model problem and its regularity. We consider the following model BVP: find

usuch that

−ε1u00(x) +ε2b(x)u0(x) +c(x)u(x) =f(x), x∈I= (0,1),

(2.1)

u(0) =u(1) = 0,

(2.2)

where0< ε1, ε2≤1are given parameters that can approach zero and the functionsb, c, f are

given and sufficiently smooth. We assume that there exist constantsβ, γ, ρ, independent of

ε1, ε2,such that∀x∈I,

(2.3) b(x)≥β≥0, c(x)≥γ >0, c(x)−ε2

2b

0_{(x)≥ρ >0.}

The reason for the above assumptions is to ensure existence and uniqueness of a weak solution to (2.1), (2.2).

The structure of the solution to (2.1) depends on the roots of the characteristic equation associated with the differential operator. For this reason, we letλ0(x), λ1(x)be the solutions

of the characteristic equation and set

µ0=−max

x∈[0,1]λ0(x), µ1= minx∈[0,1]λ1(x),

or equivalently,

(2.4) µ0,1= min

x∈[0,1]

∓ε2b(x) +

p

ε2

2b2(x) + 4ε1c(x) 2ε1

.

The following holds true [14,17]:

(2.5)

     

    

1µ0≤µ1, ε2 ε2+ε11/2

≤Cε2µ0≤C, ε 1/2

1 µ0≤C,

max{µ−1

0 , ε1µ1} ≤Cε1+ε

1/2

2 , ε2≤Cε1µ1,

forε2

2≥ε1: ε−

1/2

1 ≤Cµ1≤Cε−11,

forε2

2≤ε1: ε−

1/2

1 ≤Cµ1≤Cε−

1/2

1 .

The values ofµ0, µ1determine the strength of the boundary layers and since|λ0(x)|<|λ1(x)|

the layer atx= 1is stronger than the layer atx= 0. Essentially, there are three regimes, as shown in Table 1, which is taken from [9]. Figure2.1shows the behavior of the solution to (2.1), (2.2), in all three regimes.

The above considerations suggest the following two cases:

1. ε1is large compared toε2: this is similar to a ‘regular perturbation’ of

reaction-diffusion type. If we consider the limiting caseε2= 0, then we see that there are two

boundary layers, one at each endpoint, of widthOε1₁/2 lnε

1/2 1

TABLE2.1

Different regimes based on the relationship betweenε1andε2.

µ0 µ1

convection-diffusion ε1ε2= 1 1 ε−11

convection-reaction-diffusion ε1ε221 ε

−1

2 ε2/ε1

reaction-diffusion ε2

2ε11 ε−

1/2

1 ε

−1/2 1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

x 0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

u(x)

Exact solution, b(x) = c(x) = f(x) = 1

1 = 10 -3_,

2 = 10 -1

1 = 10 -3_,

2 = 10 -3

1 = 10 -1_,

2 = 1

FIG. 2.1.Exact solution for different values ofε1andε2.

2. ε1is small compared toε2: before discussing the different regimes, it is instructive to

consider the limiting caseε1= 0. Then there is an exponential layer (of length scale

O(ε2)) at theleftendpoint. The homogeneous equation suggests that the different

regimes areε1ε22, ε1≈ε22, ε1ε22.

(a) In the regimeε1 ε22, we haveµ0 =O(ε2−1)andµ1 =O(ε2ε−11). Hence,

µ1is much larger thanµ0and the boundary layer in the vicinity ofx= 1is

stronger. Consequently, there is a layer of widthO(ε2)at the left endpoint (the

one that arose from the analysis of the caseε1= 0) and additionally, there is

another layer at the right endpoint, of widthO(ε1/ε2).

(b) In the regimeε1≈ε22, there are layers at both endpoints of widthO

ε1₁/2.

(c) In the regime ε2₂ ε1 1, there are layers at both endpoints of width

Oε1₁/2.

It was shown in [9, p. 46] (see also [14, Lemma 2.2], which is the result we quote below) that under the assumptionsb, c, f ∈Cq_{(I)for someq≥1andqkb}0_k

∞,Iε2≤C(1−`)for

someC, `∈(0,1), the solutionuto (2.1), (2.2) can be decomposed into a smooth partES, a boundary layer part at the left endpointEL, and a boundary layer part at the right endpoint

ER, viz.

u=ES+EL+ER,

with E

(n) S (x)

≤C ,

E

(n) L (x)

≤Cµ

n 0e−

`µ0x_,

E

(n) R (x)

≤Cµ

n 1e−

for allx ∈ I and forn = 0,1,2, . . . , q. If one wants to approximate the solution using a fixed order method (e.g., thehversion of the Finite Element Method (FEM)), then this regularity result is sufficient for proving convergence; for a higher order method a more refined regularity result is needed for the smooth part. To this end, we assume thatb, c, f satisfy,

∀n= 0,1,2, . . . ,

(2.6) f

(n)

_∞,I ≤Cn!γ n f ,

c

(n)

_∞,I≤Cn!γ n c ,

b

(n)

_∞,I ≤Cn!γ n b,

for some positive constantsC, γf, γc, γb, independent ofε1, ε2.In terms of classical

differen-tiability, we have that the solution to (2.1), (2.2) satisfies (see, e.g., [9, p. 39])

(2.7) kuk_∞,I ≤C.

The following lemma provides an estimate foru0.

LEMMA2.1.Letube the solution of (2.1),(2.2)and assume that(2.3),(2.6)hold. Then there exists a positive constantC, such that

ku0_k

∞,I ≤Cmax

ε−₁1, ε−₂1 .

Proof.The proof follows [11]. Let

A(x) =ε2

ε1

Z 1

x

b(t)dt

and note thatA(1) = 0andA0(x) =−ε2

ε1b(x). Then multiplying (2.1) bye

A(x)_and

integrat-ing fromxto1gives

−ε1u0(1) +ε1eA(x)u0(x) +

Z 1

x

eA(t)c(t)u(t)dt=

Z 1

x

eA(t)f(t)dt.

Multiplying bye−A(x)_yields

(2.8) u0(x) =e−A(x)u0(1)− 1 ε1

Z 1

x

eA(t)−A(x)c(t)u(t)dt+ 1

ε1

Z 1

x

eA(t)−A(x)f(t)dt.

Integrating from0to1,we further get

(2.9) 0 =u0(1)

Z 1

0

e−A(x)dx− 1 ε1

Z 1

0

Z 1

x

eA(t)−A(x)(c(t)u(t)−f(t))dtdx.

Since we wish to first estimateu0(1), we need upper and lower bounds forR1

0 e

−A(x)_{dx. From}

(2.3) we have

(2.10)

Z 1

0

e−A(x)dx≤

Z 1

0

e−

ε2

ε1β(x−1)dx≤C

ε1

ε2β

.

Similarly,

(2.11)

Z 1

0

e−A(x)dx≥

Z 1

0

e−

ε2

ε₁kbk∞,I(1−x)_dx≥ ε1

ε2kbk∞,I

1−e−

ε2 ε₁kbk∞,I

.

Also, to estimate the remaining term in (2.9), we consider

1

ε1

Z 1

0

Z 1

x

eA(t)−A(x)dtdx= 1

ε1

Z 1

0

Z 1

x

for someζbetweentandx. Hence,

1

ε1

Z 1

0

Z 1

x

eA(t)−A(x)dtdx≤ 1 ε1

Z 1

0

Z 1

x

e−

ε₂

ε1β(t−x)dtdx≤C 1

ε2 +ε1

ε2 2

.

Using (2.9)–(2.11), we get

|u0(1)| ≤C 1

R1

0 e−A(x)dx

kck_∞,Ikuk_∞,I+kfk_∞,I

₁

ε2 +ε1

ε2 2

≤Cε2

kbk_∞,I ε1

1−e−

ε2 ε1kbk∞,I

−11

ε2 +ε1

ε2 2

≤C(ε−₁1+ε−₂1).

Inserting this bound in (2.8) gives

|u0(x)| ≤C ε₁−1+ε−₂1+ 1

ε1

kck_∞,Ikuk_∞,I+kfk_∞,I

Z 1

x

eA(t)dt

≤C ε−₁1+ε−₂1+ 1

ε1

kck_∞,Ikuk_∞,I+kfk_∞,I

Z 1

x

e

ε2

ε1kbk∞,I(1−t)_dt

≤C ε−₁1+ε−₂1

+ 1

ε1

1−e−

ε2 ε1kbk∞,I

≤C ε−₁1+ε−₂1

,

as desired.

Using an inductive argument we are able to prove the following.

THEOREM2.2.Letube the solution of(2.1),(2.2). Then there exist positive constants

K, C, independent ofε1, ε2, andu, such that forn= 0,1,2, . . .

(2.12)

u

(n)

_∞,I≤CK

n_max

n, ε−₁1, ε−₂1 n.

Proof.The proof is by induction onnand follows from [10]. Equation (2.7) and Lemma

2.1give the result forn= 0,1, so we assume it holds for0≤ν ≤n+ 1and show that it holds forn+ 2. Differentiatingntimes (2.1) gives

−ε1u(n+2)=f(n)−ε2(bu0) (n)

−(cu)(n)

=f(n)−

n X

ν=0

_n

ν

ε2b(ν)u(n+1−ν)+c(ν)u(n−ν)

.

By the induction hypothesis we have

ε1

u

(n+2)

_∞,I ≤C f

(n)

_∞,I+

+C

n X

ν=0

_n

ν

ε2γνbν!K

n+1−ν_max

n+ 1−ν, ε−₁1, ε−₂1 n+1−ν

+γ_cνν!Kn−νmax

n−ν, ε−₁1, ε−₂1 n−ν.

Using the estimates below (which follow by standard considerations)

_n

ν

ν! max

n+ 1−ν, ε−₁1, ε₂−1 n+1−ν ≤max

n ν

ν! maxn−ν, ε−₁1, ε−₂1 n−ν ≤maxn+ 1, ε−₁1, ε−₂1 n+1,

f

(n)

_∞,I≤Cγ n

fn!≤Cmax

n+ 1, ε−₁1, ε−₂1 n+1,

we obtain

ε1

u

(n+2)

∞,I ≤C f

(n)

∞,I+

+CKn+2maxn+ 1, ε₁−1, ε−₂1 n+1

n X

ν=0

1

K

γ_b

K

ν

+ 1

K2

γ_c

K

ν

≤CKn+2max

n+ 1, ε−₁1, ε−₂1 n+1

₁

K2 + 1

K

1 (1−γb/K)

+ 1

K2 1 (1−γc/K)

.

Choose the constantK >max{1, γf, γb, γc}such that the expression in brackets above is bounded by 1, and we have

ε1

u

(n+2)

∞,I≤CK n+2

maxn+ 1, ε−₁1, ε−₂1 n+1.

Dividing byε1, gives the desired result.

For simplicity, we will focus on the case

ε2₁ε2.

For the remainder of the article we will make the following assumption:

Assumption 1: Assume there exist positive constantsC,K¯,K1,`, K2,δ >0,

indepen-dent ofε1, ε2,such that the solutionuof (2.1), (2.2) can be decomposed into a smooth part

uS, two boundary layers at each endpointu±BL, and a remainderuR, viz.

(2.13) u=uS+u−BL+u

+ BL+uR,

with the following estimates: for everyn∈N0there holds

(2.14)

u

(n) S

_∞,I≤C

¯

Knn!,

(2.15)

 

 u

−

BL (n)

(x) ≤CK

n 1ε

−n

2 e−`x/ε2,

u

+ BL

(n)

(x)

≤CK

n 2

ε1 ε2

−n

e−`(1−x)ε2/ε1_,

(2.16) kuRk_∞,∂I+kuRk0,I+ε 1/2

1 ku

0

Rk0,I ≤Cmax{e

−δε2/ε1_{, e}−δ/ε2_},

for allx∈I.

3. Discretization using isogeometric analysis. Isogeometric analysis may be combined with a number of formulations; we use Galerkin’s approach, i.e., we multiply (2.1) by a suitable test function, integrate by parts and use the boundary conditions (2.2). The resulting variational formulation reads: findu∈H₀1(I)such that

(3.1) B(u, v) =hf, vi_I ∀v∈H₀1(I),

where

(3.2) B(u, v) =ε1hu0, v0iI +ε2hbu0, viI+hcu, viI.

The bilinear formB(·,·)given by (3.2) iscoercive(due to (2.3)) with respect to theenergy norm

(3.3) kvk2_E,I :=ε1|v|₁2_,I+ρkvk2_0,I,

i.e.,

(3.4) B(v, v)≥ kvk2_E,I ∀v∈H₀1(I).

Next, we restrict our attention to a finite dimensional subspaceVN ⊂H01(I), that will be

selected shortly, and obtain the discrete version of (3.1) as: finduN ∈VN such that

(3.5) B(uN, v) =hf, viI ∀v∈VN.

There holds [9]

B(uN −u, v) = 0 ∀v∈VN.

In order to define the spaceVN,we first review the concept of IGA. In this article we use B-splines as basis functions and follow [3] closely. To this end letΞ ={ξ1, ξ2, . . . , ξn+p+1}

be aknot vector, whereξi∈Ris theith knot,i= 1,2, . . . , N+p+ 1,pis the polynomial order, andNis the number of basis functions. The numbers inΞare non-decreasing and may be repeated, in which case we are talking about a non-uniformknot vector. If the first and last knot values appearp+ 1times, the knot vector is calledopen(see [3] for more details). With a knot vectorΞin hand, the B-spline basis functions are defined recursively, starting with piecewise constants (p= 0):

Bi,0(ξ) =

1, ξi≤ξ < ξi+1, 0, otherwise.

Forp= 1,2, . . ., they are defined by theCox–de Boor recursion formula[5,6]:

Bi,p(ξ) =

ξ−ξi

ξi+p−ξi

Bi,p−1(ξ) +

ξ−ξi

ξi+p−ξi

Bi,p−1(ξ).

We also mention the recursive formula for obtaining the derivative of a B-spline [3]:

d

dξBi,p(ξ) = p ξi+p−ξi

Bi,p−1(ξ)−

p ξi+p+1−ξi+1

Bi+1,p−1(ξ).

We will be considering open knot vectors, havingξ1, . . . , ξmdistinct knots, each with multiplicityri. Then

Ξ = [ξ1, . . . , ξ1

| {z } r1times

, ξ2, . . . , ξ2

| {z } r2times

, . . . , ξm, . . . , ξm | {z } rmtimes

and there holdsPm

i=1ri=N+p+1. Since we are using open knots, we haver1=rm=p+1.

The regularity of the B-spline at each knotξiis determined byri, in that the B-spline has

p−ricontinuous derivatives atξi. For this reason, we defineki=p−ri+ 1as a measure of the regularity at the knotξiand setk= [k1, . . . , km].Note thatk1=km= 0in the case of an open knot vector.

B-splines form a partition of unity and they span the space of continuous piecewise polynomials of degreepon the subdivision{ξ1, . . . , ξm}.

Each basis function is positive and has support in[ξi, ξi+p+1]. In the sections that follow,

we will approximate the solution to the BVP under consideration, using the space

(3.6) S_kp= span{Bk,p}

N k=1

of dimension

N = dim (S_kp) =mp−

m X

i=1

ki.

We point out that we are using a uniform polynomial degreep, while we allow for the regularity at each knot to (possibly) vary. A more general approach would be to allowpto vary as well. We will refer toN as the number of degrees of freedom, DOF.

Returning to our problem, the spaceVN in (3.5) is chosen asVN =S p

k, given by (3.6).

Thus, we may write the approximate solution as

uN = N X

k=0

αkBk,p,

withα = [α1, . . . , αN]T unknown coefficients, and subsitute in (3.5) to obtain the linear system of equations

(3.7) (ε1A1+ε2A2+A0)

| {z }

M∈RN×N

α=f,

where

[A1]i,j= Z

I

B0_i,p(ξ)B_j,p0 (ξ)dξ , [A2]i,j= Z

I

B_i,p0 (ξ)Bj,p(ξ)dξ,

[A0]i,j= Z

I

Bi,p(ξ)Bj,p(ξ)dξ , [f]i= Z

I

Bi,p(ξ)f(ξ)dξ,

fori, j = 1, . . . , N.The linear system (3.7) has a unique solution since the matrixM is invertible (due to (3.4) and the linear independence of the basis functionsBj,p).

Ifε1andε2are large, then no boundary layers are present and approximatingumay be

done using a fixed mesh (of say one element) and increasingp. On the other hand, ifε1andε2

4. Error estimates. We begin by citing the main tool for the analysis (see Corollary 2 in [4]).

PROPOSITION4.1.Given the subdivision{0 =ξ1, . . . , ξm= 1}of the reference domain

I= (0,1),letIi= (ξi, ξi+1), hi=ξi+1−ξi, i= 1, . . . , m−1, k, pnon-negative integers withp≥2k−1andu(k) ∈Hs(I)for some0≤s≤κ=p−k+ 1.Then there exists a quasi-interpolation operatorπp,ksuch that, fori= 1, . . . , m−1, j= 1, . . . , k−1,

u

(j)

−(πp,ku) (j)

2

L2_(I i)

≤

_h i 2

2(s+k−j)_{(κ−s)!(κ−(k−j))!}

(κ+s)!(κ+ (k−j))!

u

(k)

2

Hs_(I i)

.

In [2], the above was specialized for maximal smoothness by selectingp= 2q−1, for someq≥1, and thenk=κ=q, so the estimate of Proposition4.1becomes

(4.1) u

(j)_−(π

2q−1,qu) (j)

2

L2_(Ii)≤

_h i 2

2(s+q−j) _(q−s)!j!

(q+s)!(2q−j)!

u

(q+s)

2

L2_(Ii),

for0≤s≤q, j= 1, . . . , q−1.

In view of the above, from now on we will be using the symbolqto denote the polynomial degree.

The knot vector described below, is ‘inspired’ by the so-calledspectral boundary layer meshused in thehp-FEM for such problems (see [12] and the references therein).

DEFINITION4.2.Letµ0, µ1be given by(2.4)and letλ≥1be a user specified parameter. Then, ifλqµ−₁1≥1/2,

(4.2) Ξ = [0, . . . ,0

| {z } q+1times

,1, . . . ,1

| {z } q+1times

],

and, ifλqµ−₀1<1/2,

(4.3) Ξ = [0, . . . ,0

| {z } q+1times

, λqµ−₀1,1−λqµ−₁1,1, . . . ,1

| {z } q+1times

],

whereqis the polynomial degree.

The following auxiliary result will be used in the sequel.

LEMMA4.3.There existsτ∈[1/2,2/3]such that for everyq≥2,3,4, . . ., there holds

s:=τ q∈Nand

(q−s)! (q+s)! ≤Ce

−σq_q−2τ q_e2τ q_,

where

(4.4) σ:= min

τ∈[1/2,2/3]

ln(1−τ) 1−τ

(1 +τ)1+τ

.

Proof.We first showτ q∈_N, for someτ ∈[1/2,2/2]. Ifqis even, then we simply take

τ= 1/2. Ifqis odd, i.e.,q= 2n+ 1, n∈_N, then we chooseτ= (n+ 1)/(2n+ 1). Now, withτ q∈_N, we have(q±τ q)! = Γ(q±τ q+ 1)and, asq→ ∞[1],

(q−τ q)! (q+τ q)! =

Γ(q−τ q+ 1) Γ(q+τ q+ 1) ≤C

(q(1−τ) + 1)q−τ q+1/2e−(q−τ q+1)

(q(1 +τ) + 1)q+τ q+1/2e−(q+τ q+1)

≤C (1−τ)

1−τ

(1 +τ)1+τ

!q

THEOREM 4.4. Letube the solution to(2.1),(2.2)and letπ2q−1,qu ∈ Sq2q−1be its quasi-interpolant, constructed using the knot vector of Definition4.2and satisfying(4.1). Then there exist positive constantsσ, C, independent ofε1, ε2, such that asq→ ∞there holds

ku−π2q−1,quk_E,I ≤Ce−σq.

Proof. The proof is separated into two cases. InCase1, we make use of the classical differentiability result (2.12) and construct a quasi-interpolant ofuon the entire intervalI

with the desired properties. InCase2, we use the decomposition (2.13) and construct quasi-interpolants for each piece separately, as follows: for the smooth part, the quasi-interpolant is constructed onI, while for the boundary layers, it has support only in the layer regions. No quasi-interpolant is constructed for the remainder, since it is already exponentially small. We next give the details.

Case1:λqµ−₁1≥1/2or equivalently2λq≥ε−₁1.In this case, the knot vector is given by (4.2). We make use of (2.12) and (4.1) to obtain a quasi-interpolantπ2q−1,quonI, such that

ku−π2q−1,quk 2 E,I ≤C

(q−s)! (q+s)!

_ε

1 (2q−1)! +

ρ

(2q)!

u

(q+s)

2

L2_(I)

≤C(q−s)!

(q+s)!

_2qε

1+ρ

(2q)!

K2(q+s)max{ε₁−1, q+s}2(q+s).

We chooses=τ q, withτ ∈[1/2,2/3]as asserted by Lemma4.3, and we note that, from (2.5) and the fact that in this case there holds2λq ≥ε−₁1, we have

max{ε−1

1 , q+s}

2(q+s)_{= max{ε}−1

1 ,(τ+ 1)q} 2q(τ+1)

≤max{2λq,(τ+ 1)q}2q(τ+1)_≤(2λq)2q(τ+1)

,

since2λ≥τ+ 1. Then, with the aid of Lemma4.3, we get

ku−π2q−1,quk 2 E,I ≤C

(q−τ q)! (q+τ q)!

_2qε

1+ρ

(2q)!

K2(τ+1)q(2λq)2q(τ+1)

≤Ce−σ1q_q−2τ q_e2τ q+1

_2qε

1+ρ

(2q)!

K2(τ+1)q(2λq)2q(τ+1)

≤Cq

2q+1

(2q)!K

2q_e−σ1q_(2eKλ)2τ q_,

withσ1given by (4.4). By Stirling’s formula q(Kq) 2q

(2q)! ≤

√

q(eK/2)2q, hence

ku−π2q−1,quk 2 E,I ≤C

√

q(eK/2)2qe−σ1q_(2eKλ)2τ q

≤C√qe−σ1q_(2eKλ)2τ_(eK/2)2 q

.

Selectingλsuch that(2eKλ)2τ(eK/2)2<1, gives

ku−π2q−1,quk_E,I ≤Cq1/4e−σ1q,

whereC >0is independent ofq∈Nandε1. Reducing the value ofσ1toσ˜1∈(0, σ1), there

exists a positive constantC, independent ofqandε1, such that

Case 2: λqµ−₀1 < 1/2 or equivalently, 2qε1 < 1/λ. In this case, the solution is

decomposed as in (2.13) with the bounds (2.14)–(2.16) being valid. We will approximate each term in (2.13) separately, using Proposition4.1to construct appropriate quasi-interpolants over the three intervals

(4.6) I1= (0, qµ−01), I2= (qµ−01,1−qµ

−1

1 ), I3= (qµ−11,1).

Throughout the proof, we choose

(4.7) λ≤ 2

emax

₁

K,

1

K1

, 1 K2

,

withK, K1, K2the constants in (2.14)–(2.16) . So we have, with the obvious notation,

u−π2q−1,qu=uS+u−BL+u +

B+uR−π2q−1,quS−π2q−1,qu−BL−π2q−1,qu+BL,

hence

|u−π2q−1,qu| ≤ |uS−π2q−1,quS|+|u−BL−π2q−1,qu−BL|+|u +

B−π2q−1,qu+BL|+|uR|.

For the smooth part, Proposition4.1ensures thatπ2q−1,quS, is such that

kuS−π2q−1,quSk 2 E,I ≤C

(q−s)! (q+s)!

_ε

1 (2q−1)!+

ρ

(2q)!

u

(q+s) S

2

L2_(I)

≤C(q−s)!

(q+s)! 1

λ(2q)!K

2(q+s)_((2(q+s))!)2

.

Choosings=_eτ qwithτ_e∈[1/2,2/3]as in Lemma4.3, and using Lemma 1 of [2], we get

kuS−π2q−1,quSk 2 E,I ≤C

1 4

(1 +_eτ)1+τe

(1−_eτ)eτ−1

!q

1

λ(2q)!K 2(q+_eτ q)

≤C 1

4

(1 +_eτ)1+τe

(1−_eτ)eτ−1

!q

1

λ(2q)2q+1/2_e−2qK 2(q+eτ q)

≤C 1 λ(2q)1/2

1 4

(1 +τ_e)1+eτ

(1−τ_e)τe−1

!q

eKτe+1 2q

2q

≤C

K λq

2q

,

by (4.7). Therefore, there exist constantsC, σ2>0such that for everyε1, ε2∈(0,1], and for

everyq∈N, there holds

(4.8) kuS−π2q−1,quSk_E,I ≤Ce−σ2q.

For the left boundary layeru−_BL we obtain from Proposition4.1, a quasi-interpolant

π2q−1,qu−BLon the intervalI1= (0, λqµ−01), such that

u−_BL−π2q−1,qu−BL

2 1,I1 ≤C

_λqµ−1 0 2

2(s+q−1)_(q−s)!

(q+s)! 2q

(2q−1)!

u

−

BL (q+s)

2

L2_(I1)

≤C

_λqµ−1 0 2

2(s+q−1)_(q−s)!

(q+s)! 2q

(2q−1)!K 2(q+s)

1 ε

−2(q+s)+1

Using (2.5), we obtain

u−_BL−π2q−1,qu−BL

2

1,I1 ≤Cµ0(µ0ε2)

−2(q+s)+1λq 2

2s_(q−s)!

(q+s)!

(λq)2q−2 (2q−1)!K

2(q+s) 1

≤Cε−₁1/2(q/e)2s(q−s)! (q+s)!

(λq)2q−2 (2q−1)!K

2q

1 ,

where (4.7) was used. Choosings=τ q, withτ ∈[1/2,2/3]as in Lemma4.3, we get (forσ3

given by (4.4) withτreplaced byτ),

u−_BL−π2q−1,qu−BL

2

1,I1 ≤Cε

−1/2

1 e

−1_q2τ q

e−σ3q_q−2τ q_e2τ q+1(λq) 2q−2

(2q−1)!K 2q 1

≤Cε−₁1/2e−σ3q(λq) 2q−2

(2q−1)!K 2q 1

≤Cε−₁1/2(λq)−2e−σ3q(λqK1) 2q

(2q−1)!.

Since, by (4.7) and Stirling’s formula, (λqK1)2q (2q−1)! ≤

(2qe−1)2q

(2q−1)! ≤

e−2q_(2q)2q_2q

(2q)2q+1/2_e−2q ≤(2q) 1/2_{, we}

have

u−_BL−π2q−1,qu−_{BL 1,I}

1 ≤Cε

−1/4

1 e

−σ3q_,

withC >0independent ofε1andq. OnI\I1,u−BLis already exponentially small, i.e., by (2.15), we have

u−_BL

2 1,I\I1 =

Z 1

qµ−1 0

[ u−_BL0(x)]2dx≤Cε−₂2

Z 1

qµ−1 0

e−2`qµ−01/ε2_dx

≤Cε−₂1e−`λq,

hence it will not be approximated. Thus, there exist constantsC, σ3>0such that for every

ε1, ε2∈(0,1], and for everyq∈N, there holds

u−_BL−π2q−1,qu−BL _1,I≤

u−_BL−π2q−1,qu−BL _1,I

1+

u−_BL

_1,I\I

1

≤Cε−₁1/4e−σ3q_+ε−1/2

2 e

−`λq

≤Cε−₁1/2e−σ3q_,

(4.9)

where again (2.5) was used. For theL2error, we have

u−_BL−π2q−1,qu−BL

2 0,I1 ≤

λqµ−₀1

2

2(s+q)

(q−s)! (q+s)!

1 (2q)!

u

−

BL (q+s)

2

L2_(I 1)

≤C

_λqµ−1 0 2

2(s+q)_(q−s)!

(q+s)! 1 (2q)!K

2(q+s)

1 ε

−2(q+s)+1 2

≤Cε2(µ0ε2)

−2(q+s)λq 2

2(s+q)_(q−s)!

(q+s)! 1 (2q)!K

2(q+s)

1 .

Choosing s as a non-integer multiple ofq, and repeating the same steps as for the H1

-seminorm, we see that there exist constantsC, σ4>0such that for everyε1, ε2∈(0,1], and

for everyq∈N, there holds

u−_BL−π2q−1,qu−_{BL 0,I}

1 ≤Cε

1/2

2 e

OnI\I1,u−BLis already exponentially small, hence there exist constantsC, σ5>0such that

for everyε1, ε2∈(0,1], and for everyq∈N, there holds

u−_BL−π2q−1,qu−BL _0,I ≤

u−_BL−π2q−1,qu−BL _0,I

1+

u−_BL

_0,I

\I1

≤Cε1₂/2e−σ4q_+e−`µ0λqµ−01

≤Ce−σ5q_.

(4.10)

We follow the same steps for the right boundary layeru+_BL: Proposition4.1 gives

π2q−1,qu+BLonI3, such that

u

+

BL−π2q−1,qu+BL

2 1,I3

≤

_λqµ−1 1 2

2(s+q−1)_(q−s)!

(q+s)! 2q

(2q−1)!

u

+ BL

(q+s)

2

L2_(I3)

≤C

λqµ−₁1

2

2(s+q−1)

(q−s)! (q+s)!

2q

(2q−1)!K 2(q+s) 2

ε1

ε2

−2(q+s)+1

≤Cµ1

ε2

µ1ε1

2(q+s)+1

λq

2

2(s+q−1)

(q−s)! (q+s)!

2q

(2q−1)!K 2(q+s)

2 .

Using (2.5), we obtain

u

+

BL−π2q−1,qu+BL

2

1,I3 ≤Cε

−1 1

q

e

2s(q−s)!

(q+s)!

(λq)2q−2 (2q−1)!K

2q 2 .

Choosings=τ q_e , with_eτ∈[1/2,2/3]as in Lemma4.3, and following the same steps as in the approximation of the left boundary layer (using (4.7)), we arrive at

u

+

BL−π2q−1,qu+BL _1,I

3 ≤Cε

−1/2

1 e

−σ6q_,

withσ6given by (4.4) withτreplaced byeτ. OnI\I3,u

+

BLis already exponentially small, i.e., by (2.15) we have

u+_BL

2 1,I\I3 =

Z qµ−₁1

0

[ u+_BL0

(x)]2dx≤C

_ε

1

ε2

−2Z qµ−₁1

0

e−`(1−x)ε2/ε1_dx

≤C

_ε

1

ε2

−1

e−`λq,

hence it will not be approximated. Thus

u

+

BL−π2q−1,qu+BL _1,I≤

u

+

BL−π2q−1,qu+BL _1,I

3+

u

+ BL

_1,I\I

3

≤C ε−₁1/2e−σ6q₊

_ε

1

ε2

−1/2

e−`λq

!

≤Cε−₁1e−eσ6q,

for some positive constantσ_e6,where (2.5) was used once more. For theL2error, we have

u+_BL−π2q−1,qu+BL

2 0,I3 ≤

_λqµ−1 1 2

2(s+q)_(q−s)!

(q+s)! 1 (2q)!

u

+ BL

(q+s)

2

L2_(I3)

≤C

_λqµ−1

1 2

2(s+q)_(q−s)!

(q+s)! 1 (2q)!K

2(q+s) 1

_ε

1

ε2

−2(q+s)+1

≤Cε1 ε2

_ε

2

µ1ε1

2(q+s)_λq

2

2(s+q)_(q−s)!

(q+s)! 1 (2q)!K

2(q+s) 1

≤Cε1 ε2

_λq

2

2(s+q)_(q−s)!

(q+s)! 1 (2q)!K

2(q+s)

1 ,

where we used (2.5) and (4.7). The remaining steps are the same as above, so they are omitted. We arrive at

(4.12) u

+

BL−π2q−1,qu+BL _0,I

3 ≤C

_ε

1

ε2

1/2

e−σ7q_,

for some positive constantσ7.Finally,

u

+

BL−π2q−1,qu+BL _0,I ≤

u

+

BL−π2q−1,qu+BL _0,I

3+

u

+ BL

_0,I\I

3

≤C

_ε

1

ε2

1/2

e−σ7q_+e−`λq

!

≤Ce−eσ7q,

(4.13)

for some positive constantσ_e7.

It remains to consider the remainder, which by (2.16) is exponentially small:

(4.14) kuRkE,I ≤Cmax{e−δε2/ε1, e−δ/ε2} ≤Ce−δ/ε2 ≤Ce−ζλq,

sinceλpµ−₀1<1/2(henceλpε2<1/2), whereζ >0is a constant independent ofε1, ε2.

The proof is completed by using the definition of the energy norm (3.3), along with (2.16), (4.5), (4.8)–(4.14), to get

ku−π2q−1,qukE,I ≤ε1

|uS−π2q−1,quS|1,I+

u±_BL−π2q−1,qu±BL _1,I

+

kuS−π2q−1,quSk_0,I+

u±_BL−π2q−1,qu±BL

_0,I+kuRkE,I

≤Ce−σq,

for some positive consantσ.

We next estimate the difference between the IGA solutionuN and the quasi-interpolant

π2q−1,qu.

LEMMA 4.5. LetuN ∈ Sq2q−1be the approximation tou, the solution of (2.1),(2.2), based on the knot vector of Defintion4.2, and letπ2q−1,qbe the approximation operator of Proposition4.1. Then there exist constantsC, σ >0, independent ofε1, ε2, such that ∀q∈N,

kπ2q−1,qu−uNk_E,I ≤Ce−σq.

Proof.Setξ:=π2q−1,qu−uN. Then, by coercivity of the bilinear formBε(eq. (3.4)), there holds

where we also used Galerkin orthogonality. Hence,

kξk2_E,I ≤ −ε1

(u−π2q−1,qu)0, ξ0

I+ε2

b(u−π2q−1,qu)0, ξ

I+hc(u−π2q−1,qu), ξiI.

The first and last term may be estimated using Cauchy-Schwarz:

−ε1

(u−π2q−1,qu)0, ξ0

I +

hc(u−π2q−1,qu), ξi_I

≤ε1

(u−π2q−1,qu)0

_0,Ikξ0k_0,I+kck_∞,Iku−π2q−1,quk_0,Ikξk_0,I

≤Cmax{1,kck_∞,I} ku−π2q−1,quk_E,IkξkE,I.

For the second term, we will consider the two ranges ofqseparately: in the asymptotic range ofq, i.e.,λqµ−₁1≥1/2or equivalentlyλqε1≥1/2, we have

ε2

b(u−π2q−1,qu)0, ξ

I

≤Cε2kbk_∞,I

(u−π2q−1,qu)0

_0,Ikξk_0,I

≤Cε2ε

−1/2

1 ku−π2q−1,qukE,IkξkE,I

≤Cε2(λq) 1/2

ku−π2q−1,quk_E,IkξkE,I

≤Ce−σqkξk_E,I.

In the pre-asymptotic range ofq, i.e.,λqµ−₀1<1/2, we first use integration by parts to obtain

ε2b(u−π2q−1,qu)0, ξ_I =

ε2hb(u−π2q−1,qu), ξ0i_I .

Next, we consider the three subintervals given by (4.6): on the first subinterval we have

ε2hb(u−π2q−1,qu), ξ

0_i

I1

≤Cε2kbk∞,I1

hu−π2q−1,qu, ξ

0_i

I1

≤Cε2ku−π2q−1,quk_0,I1kξ0k0,I1

≤C ε2

λqµ−₀1ku−π2q−1,quk0,I1kξk0,I1,

where we used an inverse inequality; see, e.g., [15, Thm. 3.91]. Thus

ε2hb(u−π2q−1,qu), ξ

0_i

I1

≤C

ε2µ0

λq ε2e

−σ3q_kξk

E,I ≤Ce

−βq_kξk E,I.

Similarly, on the second subinterval we have

ε2

b(u−π2q−1,qu)0, ξ

I2

≤ε2kbk∞,I2

hu−π2q−1,qu, ξ

0_i

I2

≤Cε2ku−π2q−1,quk_0,I2kξk0,I2

≤Ce−βqkξk_E,I.

Finally, on the third subinterval,

ε2

b(u−π2q−1,qu)

0 , ξ

I3

≤ε2kbk∞,I3

hu−π2q−1,qu, ξ

0_i

I3

≤Cε2ku−π2q−1,quk_0,I 3kξ

0_k

0,I3

≤Cε2

_ε

1

ε2

1/2

ε−₁1/2e−βqkξk_E,I

Therefore, by (4.12), we get

ε2b(u−π2q−1,qu)0, ξ_I

≤Ce−βqkξk_E,I

and

kξk2_E,I ≤Ce−βqkξk_E,I

which completes the proof.

We conclude with our main result.

THEOREM4.6. Letube the solution to(2.1),(2.2)and letuN ∈Sq2q−1be its approx-imation, based on the knot vector of Defintion4.2. Then there exist positive constantsC, σ, independent ofε1, ε2, such that, ∀q∈N,

ku−uNkE,I ≤Ce

−σq

.

Proof.We begin with the triangle inequality:

ku−uNkE,I ≤ ku−π2q−1,quk_E,I+kπ2q−1,qu−uNk_E,I,

whereπ2q−1,q is the approximation operator of Theorem4.4. The first term is handled by Lemma4.4and the second by Lemma4.5.

5. Numerical example. We will be considering the following BVP and we refer to [8] for more numerical experiments. Findusuch that

−ε1u00(x) +

ε2 1 +x2u

0_{(x) +e}−x_{u(x) = 1}

inI= (0,1),

u(0) =u(1) = 0.

An exact solution is not available, so we use as a reference solution the one computed with the highest polynomial degree, denoted byuREF. Instead of the error in the energy norm, we will be measuring

Error:= 100× max

k=1,...,n|uREF(xk)−uN(xk)|/k=1max,...,n|uREF(xk)|,

where{xk} n

k=1∈Iare points in(0,1),chosen uniformly in the layer region and outside – we

usen= 400in each region for our computations.

Figure5.1shows theErrorvs. the number of degrees of freedom, DOF, in a semi-log scale. The curves indicate exponential convergence and the fact that they coincide indicates robustness. The two curves that are not on top of the rest, correspond to even smaller errors and could be due to the fact that we are using a reference solution.

6. Conclusions. In this article we performed the numerical analysis of IGA for one-dimensional reaction-convection-diffusion problems with two small parameters. We estab-lished that if the knot vector is chosen appropriately and depending on the singular perturbation parameters, thenp-refinement yields robust, exponential rates of convergence, in the energy norm. We also presented one numerical example agreeing with, and even extending the theory.

This was a step towards the study of two-dimensional SPPs, especially fourth order SPPs, for which there are very few available methods for general two-dimensional domains.

5 10 15 20 25 30 DOF

100

101 102

Est. % Rel. Error in the Max Norm

Reaction-convection-diffusion problem

1 = 10 -9

, ₂ = 10-4

1 = 10 -10

, ₂ = 10-4

1 = 10 -11

, ₂ = 10-4

1 = 10 -12

, ₂ = 10-4

1 = 10 -11

, 2 = 10

-5

1 = 10 -12_,

2 = 10 -5

FIG. 5.1.Maximum norm convergence.

present here the much simpler case of constant coefficients in order to illustrate the procedure for obtaining such regularity estimates, for the benefit of the reader.

Recall that we are focusing on the caseε2

1ε2, hence we anticipate a layer of width

O(ε2)at the left endpoint and a layer of widthO(ε1/ε2)at the right endpoint. To deal with

this we define thestretched variablesx˜=x/ε2andxˆ= (1−x)ε2/ε1and make the formal

ansatz

(A.1) u∼

∞

X

i=0

∞

X

j=0

εi₂(ε1/ε22) j _u

i,j(x) + ˜uBLi,j (˜x) + ˆu BL i,j(ˆx)

,

withui,j,u˜BLi,j,uˆBLi,j to be determined. Substituting (A.1) into (2.1), separating the slow (i.e.,

x) and fast (i.e.,x,˜ xˆ)variables, and equating like powers ofε1andε2,we get

(A.2)

   

  

u0,0= f_c,

ui,0=−b_cui0−1,0, i≥1,

u0,j =u1,j = 0, j≥1,

ui,j= 1_c u00i−2,j−1−bu0i−1,j

, i≥2, j≥1,

(A.3)

(

b u˜BL i,0

0

+cu˜BL

i,0 = 0, i≥0,

b u˜BL_i,j0+cu˜BL_i,j = ˜uBL_i,j−100, i≥0, j≥1,

(A.4)

(

ˆ

uBL i,0

00

+b uˆBL i,0

0

= 0, i≥0,

ˆ

uBL i,j+1

00

+b uˆBL i,j+1

0

=cuˆBL

i,j, i, j≥0.

The last two equations are supplemented with the following boundary conditions (in order for (2.2) to be satisfied) for alli, j≥0:

(A.5)

 



˜

uBL

i,j(0) =−ui,j(0), ˆ

Note that, by (A.3) and the fact thatb, c >0, we automatically havelimx˜→∞u˜BLi,j(˜x) = 0. Next, we would like to describe the regularity of the functionsui,j,u˜BLi,j,uˆBLi,j ,defined by (A.2)–(A.5) above. We begin withui,j,and we have the following.

LEMMA A.1. Let ui,j be defined by(A.2)and assume(2.6)holds. Then there exist positive constantsC, Kand a complex neighborhoodGofIsuch that the complex extension ofu(denoted again byu) satisfies

|ui,j(z)| ≤Cδ−iKiii ∀z∈Gδ ={z∈G: dist(z, ∂G)> δ}.

Proof. The proof is by induction oni. The casei = 0holds trivially, so assume the result holds foriand establish it fori+ 1. Letτ ∈(0,1)and letK >0be a constant so that _K22 +

1 K

≤C. We have by (A.2), the induction hypothesis withG(1−τ)δ ⊃Gδ, and Cauchy’s Integral Theorem,

|ui+1,j(z)| ≤C

u00_i−1,j−1(z) +

u0_i,j(z)

≤C

₂

(τ δ)2((1−τ)δ)

−i+1

Ki−1(i−1)i−1+ 1

(τ δ)((1−τ)δ)

−i

Kiii

≤Cδ−i−1Ki+1(i+ 1)i+1 1

K2 1 (i+ 1)2

2

τ2_(1−τ)i−1

i−1

i+ 1

i−1

+

+ 1

K

1 (i+ 1)

1

τ(1−τ)i

_i

i+ 1

i!

.

Chooseτ = 1/(i+ 1).Then

|ui+1,j(z)| ≤Cδ−i−1Ki+1(i+ 1)i+1 ₂

K2 + 1

K

,

so by the choice ofKthe expression in brackets is bounded and this completes the proof. LEMMA A.2. Let ui,j be defined by(A.2)and assume(2.6)holds. Then there exist positive constantsC, K1, K2such that

ku_i,j(n)k∞,I≤Cn!K1ni!K2i ∀n∈N.

Proof.This follows immediately from LemmaA.1and Cauchy’s Integral Theorem for derivatives:

ku(_i,jn)k∞,I ≤C

n! (n+ 1)nδ

−i_Ki_ii_en_≤Cn!Kn 1i!K

i 2,

withK1=e, K2=K/δ.

The following auxiliary lemma, which is an analog of Lemma 7.3.6 in [10], will be used in the proof of LemmaA.4.

LEMMA A.3. Letλ, γ ∈ Cwith Re(λ) >0,Re(γ) >0. LetF be an entire function

satisfying, for somei, j∈N0andC >0,

|F(z)| ≤Cγi+je−Re(λz)(i+j+|z|)i+j ∀z∈C.

Letα∈_Cand letv: (0,∞)→_C,be the solution of the problem

Thenvcan be extended to an entire function (denoted again byv), which satisfies

|v(z)| ≤C γi+j(i+j+|z|)

i+j+1

(i+j+ 1) +|α|

!

e−Re(λz) ∀z∈C.

Proof.Using an integrating factor, we find

v(z) =e−Re(λz)

α+

Z z

0

eRe(λs)F(s)ds

,

from which we have

|v(z)| ≤e−Re(λz) |α|+

Z |z|

0

e

Re(λs)_F(s)

ds !

≤Ce−Re(λz) |α|+γi+j

Z |z|

0

(i+j+|s|) i+j

ds !

,

where we used the assumption onF. The result follows.

LEMMAA.4.The functionsu˜BL_i,j ,which satisfy(A.3), are entire and there exist positive constantsC,K,˜ γ˜, depending only on the data, such that

(A.6) u˜BL_i,j

(z)≤CK˜γ˜i+j(i+j+|z|)i+je−βRe(z) ∀z∈C.

Proof.The proof is by induction. We first note that from (A.3), we may calculate

˜

uBL_i,0(z) =−ui,0(0)e− c

bz, i≥0.

Thus, using LemmaA.2to bound the term|ui,0(0)|, we get

u˜BL_i,0(z)

≤Ci!K₂ie−|

c

bz| ≤_C˜γi_(i+|z|₎i_e−βRe(z)_,

where˜γ=K2, β=c/b, hence the result holds forj= 0and for alli≥0.We then proceed

by induction onj. We assume (A.6) holds forj ≥1(and for alli≥0) and show it forj+ 1.

We note that, by (A.3),u˜BL

i,j+1satisfies,∀i, j≥0,

˜

uBL_i,j+10

+c

bu˜

BL i,j+1=

1

b u˜

BL i,j

00

, u˜BL_i,j+1(0) =−ui,j+1(0).

By the induction hypothesis and Cauchy’s Integral Theorem for Derivatives (we take as the contour the unit circle centered atz), we get

u˜

BL i,j

00

(z) ≤Cγ˜

i+j_(i+j+|z|)i+j

e−βRe(z)_.

LemmaA.3is then applicable (withλ=c_b, γ= ˜γ, F =1_b u˜BL_i,j00andα=−ui,j+1(0)) and

with the aid of LemmaA.2(to bound|α|) we obtain

u˜BL_i,j+1

(z)≤C γ˜i+j

(i+j+ 1 +|z|)i+j+1 (i+j+ 1) +i!K

i 2

!

e−Re(λz)

≤Cγ˜i+j(i+j+ 1 +|z|)i+j+1×

1 (i+j+ 1)+

i!Ki 2

˜

γi_{(i+j+ 1 +|z|)}i+j+1 !

e−Re(λz)

≤C γ˜

i

(i+j+ 1)!(i+j+ 1 +|z|) i+j+1

where we used the fact thatγ˜=K2. Hence, the induction is complete and this concludes the

proof.

COROLLARY A.5. The functionsu˜BL_i,j ,which satisfy(A.3), are entire and there exist positive constantsC,K˜,K1,γ˜, depending only on the data, such that,∀n∈N0andx˜≥0, there holds

u˜

BL i,j

(n)

(˜x) ≤CK

n 1K˜˜γ

i+j_(i+j)i+j_e−βx˜_.

Proof.Cauchy’s Integral Theorem for Derivatives gives

˜u

BL i,j

(n)

(˜x) ≤Ce

−βx˜ n! (n+ 1)nγ˜

i+j_{(i+j+ ˜x)}i+j

en

≤Ce−βx˜ n!

(n+ 1)nγ˜

i+j_(i+j+n)i+j

en.

The proof is completed by observing that

(A.7) (i+j+n)i+j ≤(i+j)i+j(1 +n/(i+j))i+j≤(i+j)i+jen.

REMARKA.6. An analogous result may be proven for the functionsuˆBL_i,j, which satisfy (A.4), (A.5):

(A.8)

uˆ

BL i,j

(n)

(ˆx)

≤C

ˆ

Knγˆi+j(i+j)i+je−βxˆ ∀i, j≥0.

Indeed, from (A.4) we find

ˆ

uBL_i,0(ˆx) = ˆu_i,BL₀(1)e−bxˆ,

ˆ

uBL_i,1(ˆx) = ˆu_i,BL₁(1)e−bˆx+

Z xˆ

1

cuˆBL_i,0(s)1−eb(s−xˆ)ds

and an inductive argument, like in the proof of LemmaA.4, can be used to establish (A.8). Next, we define, for someM ∈_Z,

uM(x) = M X

i=0 M X

j=0

εi2(ε1/ε22)jui,j(x), (A.9)

˜

uBL_M (˜x) = M X

i=0 M X

j=0

εi₂(ε1/ε22) j_u˜BL

i,j(˜x), (A.10)

ˆ

uBLM (ˆx) = M X

i=0 M X

j=0

εi2(ε1/ε22)juˆBLi,j(ˆx), (A.11)

rM =u− uM+ ˜uBLM + ˆu BL M

,

and we have the following decomposition

(A.12) u=uM+ ˜uBLM + ˆu

BL M +rM.

THEOREM A.7. Assume that (2.6), (2.3) hold. Then there exist positive constants

C,K1,K2,K˜,Kˆ,˜γ,γˆ,δ, independent ofε1, ε2,such that the solutionuof (2.1),(2.2)can be decomposed as in(A.12), with

(A.13)

u

(n) M

∞,I ≤Cn!K n

(A.14) u˜ BL M

(n)

(x) ≤C

˜

Knε−₂ne−dist(x,∂I)/ε2 _∀n∈

N0,

(A.15)

uˆ BL M

(n)

(x) ≤C

ˆ

K₂n

_ε

1

ε2

−n

e−dist(x,∂I)ε2/ε1 _∀n∈

N0,

(A.16) krMk_∞,∂I+krMk_0,I+ε 1/2

1 kr

0

Mk0,I ≤Cmax{e

−δε2/ε1_{, e}−δ/ε2_},

providedε2eMmax{K2,γ,˜ γ}ˆ <1and ε_ε12 2

eMmax{γ,˜ ˆγ}<1.

Proof.We first show (A.13): from (A.9) and LemmaA.2we have

u

(n) M

_∞,I ≤

M X i=0 M X j=0

εi₂(ε1/ε22) j

u

(n) i,j

_∞,I ≤C M X i=0 M X j=0

εi₂(ε1/ε22) j_n!Kn

1i!K i 2

≤Cn!K₁n

M X

i=0

εi₂iiK₂i

!

 M X

j=0 (ε1/ε22)

j 



≤Cn!K1n

∞

X

i=0

(ε2M K2) i !  ∞ X j=0

(ε1/ε22)j





≤Cn!K₁n,

since both sums are convergent geometric series due to the assumptionsε2M K2 <1and

ε1/ε22<1.

Next, we show (A.14): by (A.10) and LemmaA.4, we have

u˜

BL M

(n)

(˜x) ≤ M X i=0 M X j=0

εi₂(ε1/ε22) j

u˜ BL i,j

(n)

(˜x) ≤C M X i=0 M X j=0

εi₂(ε1/ε22)

j_K˜n_γ˜i_(i+j)i+j_e−|βz|_.

Now,(i+j)i+j _≤ei_ii_ej_jj_{(cf. (A.7)) hence we get}

u˜

BL M

(n)

(˜x) ≤C

˜

Kne−β˜x

M X

i=0 ˜

γieiiiεi₂

!

 M X

j=0 (ε1/ε22)

j_ej_jj 



≤CK˜ne−β˜x ∞

X

i=0

(˜γeM ε2)i

!  ∞ X j=0 _ε 1 ε2 2 eM j  

≤CK˜ne−β˜x,

since both sums are convergent geometric series, due to the assumptions ˜γeM ε2 < 1, ε1

ε2 2

Similarly, we show (A.15): by (A.11) and (A.8), uˆ BL M

(n)

(ˆx) ≤ M X i=0 M X j=0

εi₂(ε1/ε22) j

uˆ BL i,j

(n)

(ˆx) ≤C M X i=0 M X j=0

εi₂(ε1/ε22)

j_Kˆn_γˆi_(i+j)i+j_e−βxˆ

≤CKˆne−βˆx ∞

X

i=0

(ˆγeM ε2) i !   ∞ X j=0 _ε 1 ε2 2 eM j  

≤CKˆne−βˆx.

It remains to show (A.16). To this end, note that

rM(0) =u(0)− M X i=0 M X j=0

εi₂(ε1/ε22) j _u

i,j(0) + ˜uBLi,j(0) + ˆu BL

i,j (ε2/ε1)

=− M X i=0 M X j=0

εi₂(ε1/ε22) j_uˆBL

i,j (ε2/ε1).

By (A.8),

|rM(0)| ≤ M X i=0 M X j=0

εi₂(ε1/ε22) j

uˆBL_i,j (ε2/ε1)

≤C M X i=0 M X j=0

εi₂(ε1/ε22)

j_γˆi_(i+j)i+j_e−βε2/ε1

≤Ce−βε2/ε1

∞

X

i=0

(ˆγM eε2) i !  ∞ X j=0

(ε1/ε22)eM

j 

≤Ce−δε2/ε1,

for some positiveδ,independent ofε1, ε2and bounded away from 0. Similarly,

|rM(1)| ≤ M X i=0 M X j=0

εi₂(ε1/ε22) j

u˜BL_i,j (1/ε2)

≤C M X i=0 M X j=0

εi₂(ε1/ε22)

j_˜γi_(i+j)i+j_e−β/ε2

≤Ce−β/ε2

∞

X

i=0

(˜γeε2M)i

!



∞

X

j=0

(ε1/ε22)M

j 

≤Ce− δ/ε2_.

Combining the two results, we have

krMk_∞,∂I ≤Cmax{e

−δε2/ε1_{, e}−δ/ε2_}.

Now, letL:=−ε1 d 2

dx2 +ε2b_dxd +cId, with Id the identity operator, and consider

L(u−uM) =f(x)− M X i=0 M X j=0

εi₂(ε1/ε22) j_Lu

i,j(x),

withui,jsatisfying (A.2). After some calculations, we find

L(u−uM) =−εM2 +1 M X j=1 _ε 1 ε2 2 j

hence

kL(u−uM)k∞,I ≤ε M+1 2

M X

j=1

_ε

1

ε2 2

j

kbk_∞,I u0_M,j

∞,I

≤CεM₂ +1

M X

j=1

ε1

ε2 2

j u0M,j

∞,I.

Using LemmaA.2, we further obtain

kL(u−uM)k_∞,I ≤CεM2 +1M!K M 2

M X

j=1

_ε

1

ε2 2

j

≤Cε2(ε2M K2)M,

since the finite sum can be bounded by a converging geometric series.

We also consider the operatorLin the stretched variablex˜:

˜

L=−ε1ε−22

d2

dx˜2 +b

d dx˜ +cId,

and we find, after some calculations,

˜

Lu˜BL_M = M X

i=0 M X

j=0

εi₂(ε1/ε22) j_L˜u˜BL

i,j

= M X

i=0 M X

j=0

εi₂(ε1/ε22) j

−ε1ε−22 u˜ BL i,j

00

+b u˜BL_i,j 0

+cu˜BL_i,j

=

_ε

1

ε2 2

M+1 M X

i=0

εi₂ u˜BL_i,M00,

where (A.3) was used. Hence, using (A.6), we have

˜

Lu˜BL_{M ∞,I} ≤

_ε

1

ε2 2

M+1 M X

i=0

εi₂ u˜

BL i,M

00

_∞,I ≤C _ε

1

ε2 2

M+1 M X

i=0

εi₂˜γi(i+M)i+M

≤C

_ε

1

ε2 2

M+1 M X

i=0

εi2˜γieiiieMMM ≤C

_ε

1

ε2 2

eM

M+1 M X

i=0

(ε2γeM˜ )i

≤C

_ε

1

ε2 2

eM

M+1

.

Similarly, in the stretched variablexˆwe have

ˆ

L=−ε1

_ε

2

ε1

2 _d2

dxˆ2 −bε2

ε2

ε1

and, with the help of (A.4),

ˆ

LuˆBL_M = M X

i=0 M X

j=0

εi₂(ε1/ε22) j_LˆuˆBL

i,j

= M X

i=0 M X

j=0

εi₂(ε1/ε22) j

−ε

2 2

ε1 ˆ

uBL_i,j 00

−bε

2 2

ε1 ˆ

uBL_i,j 0

+cˆuBL_i,j

=

ε1

ε2 2

M M X

i=0

εi2 u˜BLi,M 00

,

thus, by following the exact same steps as above,

ˆ

LuˆBL_{M ∞,I} ≤

_ε

1

ε2 2

M M X

i=0

εi₂ uˆ

BL i,M

00

_∞,I ≤C _ε

1

ε2 2

M M X

i=0

εi₂ˆγi(i+M)i+M

≤C

_ε

1

ε2₂eM

M

.

Therefore,

kLrMk_∞,I=

L u−uM −u˜BLM −uˆBLM

_∞,I

≤ kL(u−uM)k_∞,I+ Lu˜BL_M

_∞,I+

LuˆBL_M

_∞,I

≤C ε2(ε2M K2)M +

_ε

1

ε2 2

eM

M+1_ε

1

ε2 2

eM

M!

.

Under the assumptions of the theorem, we have shown that the remainderrMhas exponentially small values at the endpoints ofI,andLrM is uniformly bounded by an arbitrarily small quantity onI. By, e.g., stability (see [9]) we have the desired result.

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