The generalized Euler-Maclaurin formula for the

numerical solution of Abel-type integral

equations.

Johannes Tausch

Abstract

An extension of the Euler-Maclaurin formula to singular integrals wasintroduced by Navot [12]. In this paper this result is applied to derivea quadrature rule for integral equations of the Abel type. We presenta stability and convergence analysis and numerical results that are ingood agreement with the theory. The method is particularly useful whencombined with fast methods for evaluating integral operators.

1 Introduction

The generalized Abel equation is a weakly singular Volterra integral equationof the first kind which usually appears in the form

∫ t

0

(t − τ)−αk(t, τ)g(τ) dτ = f(t). (1)

Here 0 ≤ t ≤ T , 0 < α < 1 and the kernel k(·, ·) is smooth and satisfiesk(t, t) = 1. The following solvability result was given by Atkinson [1].

Theorem 1.1 If f(t) = t1−α−β f(t) with f ∈ C∞[0, T ] and β < 1 then (1) hasa unique solution g which is of the form g(t) = t−β g(t) and g ∈ C∞[0, T ].

Numerical methods for (1) are usually based on collocation or product in-tegration methods. This subject has been studied extensively and is reviewedin the monographs by Linz [7] and Brunner [2]. A more recent approach todiscretizing (1) is the use of the convolution quadrature rules [10].

The aforementioned discretization methods lead to a recurrence formula forthe numerical approximations gn of g(tn) which are of the form

h1−α

n∑

j=1

wn,jk(tn, tj) gj = f(tn), n = 0, 1, 2, . . . (2)

where h is the stepsize, tn = nh and wn,j are quadrature weights.

1

The present paper examines another discretization method of (1) which isbased upon the generalization of the Euler-Maclaurin expansion to integralswith algebraic singularities. The formula was first discovered by Navot [12] andfurther generalized in a series of papers, see e.g., [6, 11, 15]. We will use thisresult to obtain a discretization of (1) in the form

h

n−1∑

j=0

(tn − tj)−αk(tn, tj)gj + h1−α

n∑

j=[n−p]+

r(p)n−jk(tn, tj) gj = f(tn), (3)

n = 0, 1, 2, . . .. The first sum, or history part, is the usual composite trapezoidalrule. The second sum contains only a very small number of terms and can beconsidered as a correction to account for the singularity of the integrand att = τ . The parameter p controls the order of the method; we will show that forp = 0 the order is O(h2−α) and the scheme is stable for α ∈ (0, 1). When p = 1and p = 2 the order is O(h3−α) and O(h4−α), respectively, but the interval of α’sfor which the method is stable is reduced. For larger values of p the importantcase α = 1

2 is no longer stable, and therefore the resulting schemes are probablyonly of limited interest for solving first-kind equations.

A complication we encounter is that the remainder in the generalized Euler-Maclaurin expansion depends in a subtle way on the interval length. To ensurethat we obtain uniform bounds in t we must initially restrict the discussion tothe case that the solution has sufficiently many vanishing derivatives at t = 0.To handle the general case, including singularities, we will discuss a correction ofthe right hand side to maintain the convergence rates. This procedure is similarto Lubich’s approach to overcome a similar issue for convolutional quadratureschemes [10]. We note however, that other techniques have been discussed tohandle singularities, see, e.g. [5].

The difference between the two recurrence formulas is that (2) containsquadrature weights that depend on n and j, whereas the history part in (3)does not. This simplifies matters significantly if fast summation methods areused to compute the history part in the recurrence formula efficiently. Fastmethods are based on separating the tn and tj variables. In (3) this can ac-complished fairly easily, for instance, by Taylor expanding the kernel in (1). Onthe other hand, recurrence (2) involves in addition a separation of the n andj-index of the weights, which does not seem to be practical. It is not our goal todiscuss fast methods here. Instead, we refer to the paper [14] where the p = 0method has been used as a time discretization for integral formulation of theheat equation.

It should be noted that the idea to use the generalized Euler- Maclaurinexpansion to obtain discretizations of singular Volterra integral equations is notnew, although it has received only very limited attention in the past. It appearsthat the first paper to use this result for Volterra equations of the second kindis Tao and Yong [13]. Later, the equation of the first kind was treated bytransformation to an equation of the second kind, see [8] and [9].

2

2 The Generalized Euler-Maclaurin Expansion

The result by Navot is a generalization of the Euler-Maclaurin formula for in-tegrals with an algebraic singularity at one endpoint of the interval of integra-tion [12]. Later, the formula was generalized and its derivation was simplifiedin various ways. Probably the most efficient tool is the Mellin transform, whichfirst used by Verlinden and Haegemans [15]. The methodology in their paperapplies for integrals over a semiinfinite interval, but can be easily extended tointegrals over [0, 1] using neutralizer functions [11]. We will briefly recall thistechnique and then discuss arbitrary length intervals.

Basic expansion for the semiinfinte interval. The Mellin transform of afunction ϕ is given by

Φ(z) =

∫ ∞

0

xz−1ϕ(x) dx

and the inverse transform is given by

ϕ(x) =1

2πi

c+i∞∫

c−i∞

x−zΦ(z) dz,

where c ∈ R is chosen such that Φ is analytic for Re(z) ≥ c.Suppose ϕ(x) = x−αϕ(x) where ϕ ∈ C∞[0,∞) decays superalgebraically,

that is,lim

x→∞xkϕ(x) = 0, for every k ≥ 0.

From the inverse transform it follows that

h

∞∑

j=1

ϕ(jh) =h

2πi

c+i∞∫

c−i∞

∞∑

j=1

(jh)−zΦ(z) dz =1

2πi

c+i∞∫

c−i∞

ζ(z)Φ(z)h1−z dz, (4)

where h > 0 is the stepsize, c > 1 and ζ(·) is the Riemann zeta function.Integration by parts leads to

Φ(z) = ωs(z − α)

∫ ∞

0

xz−α+sϕ(s+1)(x) dx, s = 0, 1, . . . (5)

where ωs(·) is defined by

ωs(z) =(−1)s+1

z · . . . · (z + s). (6)

The integral in (5) is an analytic function for Re(z) > −(s + 1 − α). Fur-thermore, Res(ζ(z), z = 1) = 1, so the integrand in (4) has residuals

Res(

ζ(z)Φ(z)h1−z, z = 1)

= Φ(1) =

∫ ∞

0

ϕ(x) dx

3

and

Res(

ζ(z)Φ(z)h1−z, z = α − s)

=ϕ(s)(0)

s!ζ(α − s)hs+1−α

for s = 0, 1, . . . Thus we may move the integration contour in (4) into the leftplane and obtain

h∞∑

j=1

ϕ(jh) =

∫ ∞

0

ϕ(x) dx +

q∑

s=0

ϕ(s)(0)

s!ζ(α − s)hs+1−α + R(q)(h) (7)

for some integer q. This is the generalized Euler-Maclaurin expansion. Theremainder is given by

R(q)(h) =1

2πi

c′+i∞∫

c′−i∞

ζ(z)Φ(z)h−z+1 dz, (8)

where α − q − 1 < c′ < α − q.The remainder R(q) can be estimated by

∣

∣

∣

∣

∣

∣

c′+i∞∫

c′−i∞

ζ(z)Φ(z)h−z+1 dz

∣

∣

∣

∣

∣

∣

≤h1−c′

2π

∫ ∞

−∞

|ζ(c′ + iy)||Φ(c′ + iy)| dy

From (5) it follows that |Φ(c′ + iy)| decays superalgebraically as y → ±∞. Onthe other hand, it is known that for a fixed real part, the zeta function increasesat most like a polynomial, see [15]. Therefore the integral on the right handside converges and the remainder is O(h1−c′).

Expansion for the interval [0, t]. We now derive a similar expansion forintegrals of the type that appear in (1). Of course, the form of the general-ized Euler-Maclaurin expansion for a finite interval is well known. However, inVolterra integral equations, the length is time dependent and enters the con-stants in the estimates. We will discuss this dependence in the following.

It turns out that we must assume that the solution is of the form g(τ) =τγ g(τ) where g is smooth in [0, T ] and γ is sufficiently large to ensure thatthe estimates are independent of time. Later, it will become clear that γ =p + 2− α, where p is the order of the method, defined in (3) . By Theorem 1.1this is equivalent to limiting the discussion to right hand sides of the formf(t) = t1−α+γ f(t). Section 5 will describe a modified method that maintainsits convergence properties for general, even singular solutions.

To simplify notations, we will write

ϕ(t, τ) = k(t, τ)g(τ) and ϕ(t, τ) = τγϕ(t, τ)

where ϕ ∈ C∞[0, T ] and γ > 0 is a not yet specified parameter. Furthermore,we introduce a neutralizer function χ(x) with the properties

χ ∈ C∞(R), χ(x) = 1, x ≤1

3, χ(x) = 0, x ≥

2

3(9)

4

and, in addition,χ(1 − x) = 1 − χ(x). (10)

An example of such a function is

χ(x) =1

2

{

1 + tanh

(

1

x − 13

−1

23 − x

)}

,1

3< x <

2

3.

Using these definitions the integrand in (1) can be written as the sum of twofunctions,

(t − τ)−αϕ(t, τ) = τγϕ0(t, τ) + (t − τ)−αϕ1(t, t − τ)

where

ϕ0(t, x) = χ(x

t

)

(t − x)−αϕ(t, x),

ϕ1(t, x) = χ(x

t

)

(t − x)γϕ(t, t − x)

Furthermore, we set ϕ0(t, x) = xγϕ0(t, x) and ϕ1(t, x) = x−αϕ1(t, x). All thesefunctions can be trivially extended to the semiinfinite interval x ≥ 0. Withthese definitions it follows that

∫ t

0

(t − τ)−αϕ(t, τ) dτ =

∫ ∞

0

ϕ0(t, x) dx +

∫ ∞

0

ϕ1(t, x) dx

andn−1∑

j=1

(tn − tj)−αϕ(tn, tj) =

∞∑

j=1

ϕ0(tn, tj) +

∞∑

j=1

ϕ1(tn, tj).

We can apply the generalized Euler-Maclaurin formula to both sums and com-bine the expansions. The definitions of the functions ϕ1 and ϕ imply that

∂sϕ1

∂xs(t, 0) =

∂sϕ

∂τs(t, t)(−1)s,

If γ > q the derivatives up to order q of ϕ0(t, x) vanish for x = 0. Thusthe function will not contribute to the residuals and will only appear in theremainder.

h

n−1∑

j=1

(tn − tj)−αϕ(tn, tj) =

∫ tn

0

(tn − τ)−αϕ(tn, τ) dτ

+

q∑

s=0

(−1)s

s!

∂sϕ

∂τs(tn, tn)ζ(α − s)hs+1−α + R

(q)E (tn, h)(11)

where the remainder is given by

R(q)E (t, h) =

1

2πi

c′+i∞∫

c′−i∞

ζ(z)(

Φ0(t, z) + Φ1(t, z))

h−z+1 dz (12)

and Φ0(t, z) and Φ1(t, z) are the Mellin transforms of ϕ0(t, x) and ϕ1(t, x) withrespect to the x-variable.

5

Lemma 2.1 If g(t) = tγ g(t) with g ∈ C∞[0, T ], then there is a smooth functionIE(t, h) such that the remainder is given by

R(q)E (t, h) = tγ+c′−αh1−c′IE(t, h), −q − 1 + α < c′ < −q + α .

Proof. As in (5) we obtain from integration by parts

Φ1(t, z) = ωs(z − α)

∫ ∞

0

xz−α+s

(

∂

∂x

)s+1

ϕ1(t, x) dx.

Changing variables x → x/t and sorting out the powers of t shows that

Φ1(t, z) = tz−α+γωs(z − α)I(s)1 (t, z)

where

I(s)1 (t, z) =

∫ 23

0

xz−α+s

(

∂

∂x

)s+1

χ(x)(1 − x)γ ϕ(t, t(1 − x)) dx.

Likewise, we find for Φ0 that

Φ0(t, z) = tz−α+γωs(z + γ)I(s)0 (t, z),

where

I(s)0 (t, z) =

∫ 23

0

xz+γ+s

(

∂

∂x

)s+1

χ(x)(1 − x)−αϕ(t, tx) dx.

By the properties of parameter dependent integrals the functions I(s)0 (t, z) and

I(s)1 (t, z) are smooth in t and in z. Furthermore, they are bounded in t ∈ [0, T ]

and Im(z), when Re(z) is fixed. Hence it follows from (12) that

R(q)E (t, h) =

1

2πitγ+c′−αh1−c′

×

∫ c′+i∞

c′−i∞

ζ(z)(

ωs(z + γ)I(s)0 (t, z) + ωs(z − α)I

(s)1 (t, z)

)

(th)Im(z) dz

The zeta function increases like a polynomial on the integration contour, whereasωs decreases like |z|−s−1. Thus we can choose s large enough such that theintegral converges absolutely with bounds independent of t and h. Then theintegral is a smooth function of all parameters. �

Quadrature Rule. To derive a quadrature rule we retain the first p terms inexpansion (11) and replace derivatives by finite difference approximations. The

weights r(p)j are determined such that

p∑

s=0

ζ(α − s)(−1)s

s!

∂sϕ

∂τs(t, t)h1−α+s = −h1−α

p∑

j=0

ϕ(t, t − hj)r(p)j + R

(p)D (t, h) .

6

Here we set ϕ = 0 if the second argument is negative. If γ > p + 1 then thisextension is p+1-times continously differentiable. Thus we can apply the Taylor

expansion to obtain a linear system for the r(p)j

p∑

j=0

jsr(p)j = −ζ(α − s), s = 0, . . . , p. (13)

Note that the weights are independent of t. We list the weights for the importantcases p = 0, 1, 2

r(0)0 = −ζ(α),

[

r(1)0

r(1)1

]

= −

[

1 −10 1

][

ζ(α)ζ(α − 1)

]

,

r(2)0

r(2)1

r(2)2

= −

1 − 32

12

0 2 −10 − 1

212

ζ(α)ζ(α − 1)ζ(α − 2)

.

In view of Lemma 3.5 below, it is necessary to express the remainder as theproduct of certain powers of h and t and a function that is sufficiently regularon the triangle

∆T := {(t, h) : 0 ≤ h ≤ t ≤ T} . (14)

Lemma 2.2 If g(t) = tp+2−αg(t) with g ∈ C∞[0, T ], the remainder R(p)D (t, h)

is of the form

R(p)D (t, h) = hp+2−αt1−αI

(p)D (t, h) (15)

where I(p)D ∈ C∞(∆T ).

Proof. Using the integral form of the remainder in the Taylor expansion, we get

R(p)D (t, h) = h1−α (−1)p

p!

p∑

j=0

r(p)j

∫ jh

0

τp ∂ϕp+1

∂τp+1(t, t − h + τ) dτ

Since ϕ(t, τ) = τp+2−αϕ(t, τ) there is a smooth ϕ such that ∂p+1τ ϕ(t, τ) =

τ1−α+ ϕ(t, τ), where τ+ is the positive part of τ . Hence

R(p)D (t, h) = h1−α (−1)p

p!

p∑

j=0

r(p)j

∫ jh

0

(t − jh + τ)1−α+ τpϕ(t, τ) dτ

Changing variables τ → hτ leads to

R(p)D (t, h) = hp+2−αt1−α (−1)p

p!

p∑

j=0

r(p)j Ij(t, h)

where

Ij(t, h) =

∫ j

0

(

1 −h

t(j − τ)

)1−α

+

τpϕ(t, hτ) dτ,

which is a smooth and bounded function when t ≥ h. �.

7

3 Stability Analysis

We first discuss the stability of discretizations of the p = 0 rule for the constantkernel k(t, τ) = 1. In this case the discretization rule leads to the semicirculantmatrix

−ζ(α)h1−αA = −ζ(α)h1−α

a0

a1 a0

a2 a1 a0

. . .. . .

. . .

(16)

whose coefficients are

an =

{

1, n = 0,

−n−α

ζ(α) , n ≥ 1.(17)

The characteristic function associated with a semicirculant matrix with entriesan is given by

a(z) =

∞∑

n=0

anzn .

The stability of recurrence (3) hinges on the properties of the inverse matrixwhich is another semicirculant matrix whose coefficients are the expansion co-efficients of a−1(z)

a−1(z) =

∞∑

n=0

Anzn . (18)

Here and in the following we will use the convention that capitalized coefficientsdenote the expansion coefficients of the reciprocal function. A frequently usedtool is the following result which goes back to Hardy [4].

Lemma 3.1 If c0 = 1, cn > 0, n > 0,∑

n cn = ∞ and

cn+1

cn

≥cn

cn−1, n > 0 (19)

then

Cn ≤ 0, n > 0 and

∞∑

n=1

Cn = −1. (20)

From the behavior of the zeta function in the interval (0, 1), shown in Figure 1,it follows that the sequence an in (17) is indeed positive. Furthermore, (19) isfulfilled when −ζ(α) ≥ 2α which implies that α must be in the interval (α, 1)with α = 0.4836 . . .. In this case the inverse is a bounded operator and we have

An < 0, n > 0, and ‖A‖∞ = 2, when α ∈ (α, 1). (21)

Unfortunately, there is no information for α ∈ (0, α). Furthermore, in thesubsequent analysis it is necessary to have asymptotic estimates of the An’s.More information on the An’s for all α can be obtained from Eggermont’s anal-ysis of the product integration method for the Abel equation. We summarizethe development in the appendix of the paper [3].

8

Lemma 3.2 Let cn be a sequence with c0 = 1, and set

gn =

{

1 − 12c1, n = 1,

12 (cn−1 − cn), n > 1.

(22)

If the gn satisfy

gn > 0,gn

gn−1<

gn+1

gn

< 1, n > 1 (23)

and, in addition, if there is s ∈ R and β ∈ [0, 1) such that

y∑

n=1

cn ∼ syβ , y → ∞, (24)

then

Cn ∼ −sin

(

βπ)

sπ

1

nβ+1. (25)

Our main stability result depends on Lemma 3.2. Note, however, no infor-mation on the sign of the Cn’s is given, thus the conclusions from Lemma 3.1will be still important later on.

Theorem 3.3 The coefficients An in (18) satisfy

An ∼1

πsin((1 − α)π)(1 − α)ζ(α)

1

n2−α, and

∞∑

n=0

An = 0. (26)

Proof. We first establish that the coefficients an of (17) satisfy the conditionsof Lemma 3.2. For this sequence, the gn’s are given by

gn =

{

1 + 12ζ(α) , n = 1,

− 12ζ(α) ((n − 1)−α − n−α) , n ≥ 1,

(27)

The positiveness of the gn’s follows easily from the fact that −ζ(α) increasesfrom 1

2 to ∞ in the interval (0, 1), see Figure 1. To verify the second hypothesisof (23), consider the case n = 2 first, which is equivalent to

1 − 2−α

−2ζ(α) − 1<

2−α − 3−α

1 − 2−α. (28)

This must be verified by computing function values. The validity of this in-equality is illustrated in Figure 1. When n > 2 the validity of (23) for the gn’sdefined in (27) follows from elementary calculus.

From (17) it also follows that

y∑

n=1

an ∼−1

ζ(α)(1 − α)y1−α

hence the exponent in (24) is β = 1 − α which implies the asymptotic estimatefor the An’s.

9

Since∑

n |An| converges, it follows that a−1(z) is continuous on the closedunit disk, and by the Abel limit theorem it follows that

∞∑

n=0

An = limx→1−

a−1(x) = 0.

�

0 0.5 10

0.5

1

1.5

2

2.5

3

3.5

4

0 0.5 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Figure 1: Graph of the function −ζ(·) (left) and the ratios in (23) when n = 2(right, bottom) and n = 3 (right, top).

The following two lemmas will be necessary for the stability analysis.

Lemma 3.4 Let Φ ∈ C1[0, T ], tj = hj, then

n∑

j=0

An−jajΦ(tj) = δn,0Φ(tn) + O(h) (29)

Proof. Write

n∑

j=0

An−jajΦ(tj) =

n∑

j=0

An−jajΦ(tn) −

n−1∑

j=0

An−jaj

(

Φ(tn) − Φ(tj))

The first term is the convolution product of {aj}j and its inverse and accountsfor the first term in the right hand side of (29). The first term in the secondsum is O(h). To estimate the remainder of the second term, note that it follows

10

from the generalized Euler-Maclaurin formula that

h

n−1∑

j=1

(tn − tj)−1+αt−α

j = O

(∫ tn

0

(tn − τ)−1+ατ−α dτ

)

= O (tn) .

Hence, using |Φ(tn) − Φ(tj)| ≤ C(tn − tj) and Theorem 3.3

∣

∣

∣

∣

∣

∣

n−1∑

j=1

An−j

(

Φ(tn) − Φ(tj))

aj

∣

∣

∣

∣

∣

∣

≤ Ch2n−1∑

j=1

(tn − tj)−1+αt−α

j = O(h).

This proves the assertion. �

Lemma 3.5 Let Φ ∈ C1(∆T ), then for any λ > 0

n∑

j=1

An−jtλj Φ(tj , h) = O

(

tλ−1+αn h1−α

)

(30)

Proof. For simplicity of notations, we omit the second argument from Φ. SettingΨ(t) = tλΦ(t) we have

∣

∣tλnΦ(tn) − tλj Φ(tj)∣

∣ = |Ψ(tn) − Ψ(tj)| =

∣

∣

∣

∣

∣

∫ tn

tj

Ψ′(τ) dτ

∣

∣

∣

∣

∣

≤ (tn−tj) maxtj≤τ≤tn

|Ψ′(τ)|.

It follows from the chain rule that if j ≤ n,

∣

∣tλnΦ(tn) − tλj Φ(tj)∣

∣ ≤

{

C(tn − tj)tλ−1n , λ ≥ 1,

C(tn − tj)tλ−1j , λ < 1.

Because of∑

j Aj = 0,

n∑

j=1

An−jtλj Φ(tj) =

n∑

j=1

An−jtλnΦ(tn) −

n−1∑

j=1

An−j

(

tλnΦ(tn) − tλj Φ(tj))

=∞∑

j=n−1

AjtλnΦ(tn) −

n−1∑

j=1

An−j

(

tλnΦ(tn) − tλj Φ(tj))

.(31)

For λ ≥ 1 we estimate using Theorem 3.3

∣

∣

∣

∣

∣

∣

n∑

j=1

An−j tλj Φ(tj)

∣

∣

∣

∣

∣

∣

≤ C1tλn

∞∑

j=n

j−2+α + C2tλ−1n h

n−1∑

j=1

(n − j)−1+α

≤ C(

tλn n−1+α + tλ−1n hnα

)

≤ Ctλ−1+αn h1−α

11

For the case 0 < λ < 1 note that

h

n−1∑

j=1

(tn − tj)−1+αtλ−1

j = O

(∫ tn

0

(tn − τ)−1+ατλ−1 dτ

)

= O(

tλ−1+αn

)

,

Hence, beginning with (31)

∣

∣

∣

∣

∣

∣

n∑

j=1

An−jtλj Φ(tj)

∣

∣

∣

∣

∣

∣

≤ C1tλn

∞∑

j=n+1

j−2+α + C2h2−α

n∑

j=1

(tn − tj)−1+αtλ−1

j

≤ Ctλ−1+αn h1−α.

Thus the assertion has been shown in both cases. �

Higher Order Methods. In the case p > 1 and a constant kernel k(t, τ) = 1discretization (3) leads to the system

Ag + R(p)g = −h−1+α

ζ(α)f

where A is defined in (16) and R(p) is a semicirculant matrix with coefficients

r(p)j =

−r(p)0

ζ(α) − 1, j = 0,

−r(p)j

ζ(α) , 1 ≤ j ≤ p,

0, j > p.

(32)

To determine stability we factor

A + R(p) = A(I + A−1R(p))

and investigate whether the second factor has a continuous inverse. To thatend, set S(p) = A−1R(p) and find the values of α and p for which the coefficientsof S(p) satisfy

‖S(p)‖∞ =∞∑

n=0

∣

∣

∣s(p)

n

∣

∣

∣< 1. (33)

If (33) is satisfied then I+A−1R(p) has a continuous inverse in l∞. The followinglemma gives a necessary condition for stability.

Lemma 3.6 For every p ≥ 0 there is an αp ∈ (0, 1) such that condition (33) issatisfied for α ∈ (αp, 1).

Proof. If p = 0 then S(0) = 0 and there is nothing to show, and hence we mayset α0 = 0. If p > 0 and α > α we have from (21) that ‖A−1‖∞ = 2. Hence

12

‖S(p)‖∞ = ‖A−1R(p)‖∞ ≤ 2‖R(p)‖∞. Furthermore, it follows from (13) and(32) that

r(p) = −1

ζ(α)

p∑

s=1

m(p)s (−1)sζ(α − s) (34)

where m(p)s are the columns of [M (p)]−1, and M

(p)s,j = js. Since the zeta function

is only singular when α = 1, it follows that for any p fixed ‖r(p)‖1 → 0 as α → 1.Hence there is an αp < 1 for which ‖S(p)‖∞ < 1. �

Using the technique of the proof we compute values of α1 and α2 that guar-antee stability. When p = 1 we have

[

r(1)0

r(1)1

]

= −1

ζ(α)

[

ζ(α − 1)−ζ(α − 1)

]

When α > α we can conclude that

‖S(1)‖∞ ≤ ‖A−1‖∞‖R(1)‖∞ = 4ζ(α − 1)

ζ(α)=: s1(α) (35)

As shown in Figure 2, s1(α) is always less than unity, but (35) is valid only ifα ∈ (α, 1), hence we have α1 = α = 0.4836 . . ..

0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

1.2

Figure 2: Graph of the functions s1 (bottom) and s2 (top).

When p = 2 we have

r(2)0

r(2)1

r(2)2

= −

1

ζ(α)

32

12

−2 −112

12

[

ζ(α − 1)−ζ(α − 2)

]

Figure 2 shows the graph of s2(α) := − 2ζ(α)‖r

(2)‖1. From the graph it is evident

that α ∈ (α2, 1) with α2 = 0.558 . . . is necessary for stability.

13

4 Discretization Error

The quadrature error consists of two parts, the error introduced by truncat-ing the Euler-Maclaurin expansion, and the error introduced by replacing thederivatives in the leading terms of the expansion by finite differences. Settingq = p + 1 in (11), the m-th time step the quadrature error is

dm :=(−1)p+1

(p + 1)!

∂p+1ϕ

∂τp+1(tm, tm)ζ(α− p− 1)hp+2−α + R

(p+1)E (tm, h) + R

(p)D (tm, h).

(36)Subtracting equations (1) and (3) shows that the discretization errors ej =g(tj) − gj and the quadrature error dm are related by

h

m−1∑

j=0

(tm − tj)−αk(tm, tj)ej + h1−α

p∑

j=0

r(p)j k(tm, tm−j) em−j = dm . (37)

Equation (37) is equivalent to

m∑

j=0

am−jk(tm, tj)ej +

m∑

j=m−p

r(p)m−jk(tm, tj) ej = −

hα−1

ζ(α)dm . (38)

where the coefficients r(p)j are defined in (32).

To estimate the error we convolve both sides of the equation with the se-quence {An}n. For the first term in the left hand side, we calculate

ϕ(1)n =

n∑

m=0

An−m

m∑

j=0

am−jk(tm, tj) ej

=

n∑

j=0

n∑

m=j

An−mam−jk(tm, tj) ej

=n

∑

j=0

n−j∑

m=0

An−j−mamk(tm+j , tj) ej

Lemma 3.4 applied to the inner sum with Φ(τ) = k(τ + tj , tj) shows that there

are coefficients κ(1)n,j = O(1) such that

ϕ(1)n = en + h

n∑

j=0

κ(1)n,jej (39)

Proceeding in a similar manner shows that the second term satisfies

ϕ(2)n =

n∑

j=0

p∑

m=0

An−j−mr(p)m k(tm+j , tj) ej .

14

In the above sum one has k(tm+j , tj) = k(tj , tj) +O(h) = 1 +O(h). Thus there

are coefficients κ(2)n,j = O(1) such that

ϕ(2)n =

n∑

j=0

s(p)n−j ej + h

n∑

j=0

κ(2)n,jej , (40)

where s(p)j are the coefficients of S(p) = A−1R(p). In matrix notation, (39) and

(40) can be combined as(

I + S(p))

e + hKe = hα−1A−1d (41)

where K is a lower triangular matrix with entries

Kn,j = κ(1)n,j + κ

(2)n,j

which are bounded independently of h. We continue with an estimate for theright hand side.

Lemma 4.1 If γ = 2 + p − α the coefficients of e = hα−1A−1d are bounded by

en = O(hp+2−α)

Proof. The coefficients of e = hα−1A−1d are given by

en = hα−1n

∑

j=1

An−jdj

In view of (36) we have to analyze the effect of the convolution on three parts,all of which are in a form that fit the hypotheses of Lemma 3.5.

For the first part note that the p+1-st derivative of ϕ is of the form t1−αI(t),where I(·) is smooth, hence

hα−1n

∑

j=1

An−j t1−αj I(tj)h

p+2−α = O(

hp+2−αt0)

= O(

hp+2−α)

The estimate of the second part follows from Lemma 2.1. In that lemma, wemay set, for instance, c′ = −p− 1, then

hα−1n

∑

j=1

An−jR(p+1)E (tj , h) = O

(

h1−c′ tγ+c′−1n

)

=

= O(

hp+2tγ−p−2n

)

= O(

hp+2−αtγ−p−2+αn

)

(

h

tn

)α

= O(

hp+2−α) 1

nα.

Hence this part is O(

hp+2−α)

multiplied by a decaying term. For the third partof (36) use Lemma 2.2

hα−1n

∑

j=1

An−jR(p)D (tj , h) = hp+1

n∑

j=1

An−jt1−αj I

(p)D (tj , h) = O(hp+2−α).

15

Adding the three parts together completes the proof. �

If α and p are such that ‖I + S(p)‖∞ is bounded independently of n, then(41) is equivalent to

e + h(

I + S(p))−1

Ke =(

I + S(p))−1

e. (42)

Here, the matrix has entries bounded independently of h and the right handside has the same asymptotic estimate as e. Our final result follows directlyfrom the discrete Gronwall lemma.

Theorem 4.2 If αp < α < 1 and γ = p + 2 − α then the discretization errorsatisfies the asymptotic estimate en = O(hp+2−α).

5 The general case

In Section 2 we have introduced a quadrature rule which has the desired orderonly if the integrand is O(τp+2−α) as τ → 0. We will write this rule in the form

∫ tn

0

(tn − τ)−αϕ(tn, τ) dτ =

n∑

j=1

ϕ(tn, tj)wn−j + O(hp+2−α)

where the weights are

wj = h1−α

r(p)0 , j = 0,

j−α + r(p)p−j , 1 ≤ j ≤ p,

j−α, j > p.

We will show how to modify this rule if ϕ is replaced by τ−βϕ(t, τ), whereβ < 1 and ϕ is a smooth function in both variables, that has no conditions onthe behavior near τ = 0. The modification depends on the Taylor expansion ofϕ at τ = 0

ϕq(t, τ) =

q∑

s=0

1

s!

∂sϕ

∂τs(t, 0)τs (43)

which will be subtracted from ϕ to obtain a function of order τ q+1−β . Thequadrature rule is applied to the difference, the rest is integrated analytically,as shown in the following calculation∫ tn

0

(tn − τ)−ατ−βϕ(tn, τ) dτ

=

∫ tn

0

(tn − τ)−ατ−β(

ϕ(tn, τ) − ϕq(tn, τ))

dτ +

∫ tn

0

(tn − τ)−ατ−βϕq(tn, τ) dτ

≈n

∑

j=1

t−βj

(

ϕ(tn, tj) − ϕq(tn, tj))

wn−j +

∫ tn

0

(tn − τ)−ατ−βϕq(tn, τ) dτ

=

n∑

j=1

t−βj ϕ(tn, tj)wn−j + cn.

16

This is the original quadrature rule with a correction term. Since only polyno-mials appear in this term, the integral can be evaluated in closed form

cn =

∫ tn

0

(tn − τ)−ατ−βϕp(tn, τ) dτ −n

∑

j=1

t−βj ϕp(tn, tj)wn−j

=

q∑

s=0

1

s!

∂sϕ

∂τs(tn, 0)

∫ tn

0

(tn − τ)−ατs−β dτ −n

∑

j=1

ts−βj wn−j

=

q∑

s=0

1

s!

∂sϕ

∂τs(tn, 0)

Eα,βs t1+s−α−β

n −

n∑

j=1

ts−βj wn−j

.

Here, Eα,βs denotes the Euler integral

Eα,βs =

∫ 1

0

(1 − x)−αxs−β dx =Γ(1 − α)Γ(1 + s − β)

Γ(2 + s − α − β).

We now give a recipe to solve the Abel integral equation with the generalhand side as it appears in Theorem 1.1.

1. Compute the coefficients gi, i = 0, . . . , q in the expansion of the solutiong(τ) = g0 + g1τ + g2τ

2 + . . .. These follow directly from the expansioncoefficients of the kernel and the right hand side. Set gq(t) = (g0 + g1τ +. . . + gqτ

q and gq(t) = g(t) − τ−β gq(t).

2. Subtract τ−β gq(t) in (1) and solve the equivalent Abel equation∫ t

0

(t − τ)−αk(t, τ)g(τ) dτ = f(t) −

∫ t

0

(t − τ)−αk(t, τ)τ−β gq(τ) dτ.

with g(τ) = O(τ1+q−β) as the new unknown. To apply rule (3) q mustsatisfy 1 + q − β ≥ p + 2 − α, hence the order of the Taylor expansion in(43) is the smallest integer q with

q ≥ p + 1 + β − α.

3. In every time step, compute the correction factors cn and evaluate theintegral on the right hand side with the corrected quadrature rule.

The additional error introduced in this procedure comes from replacing theintegral on the right hand side with the corrected quadrature rule. The analysisof this error parallels the discussion of Sections 2 and 4 and is therefore omitted.

6 Numerical Example

Our test problem is the integral equation

∫ t

0

(t − τ)−α exp(t − τ)g(τ) dτ = exp(t)t1−α−β

5∑

k=0

Eα,βk tk

17

with known solution g(t) = exp(t)∑5

k=0 tk−β . We solve this equation in theinterval [0, 1] with stepsize h = 1/N and β = 0.5.

α = 0.15 α = 0.5 α = 0.85N error order error order error order10 0.10025 0.188351 0.16207220 0.027132 1.8856 0.067565 1.4791 0.074338 1.124540 0.007534 1.8483 0.024040 1.4908 0.033800 1.137180 0.0020919 1.8487 0.008525 1.4958 0.015302 1.1433

160 0.0005805 1.8494 0.003018 1.4979 0.006913 1.1464320 0.0001611 1.8497 0.001068 1.4989 0.003119 1.1480

Table 1: Maximum errors and orders of convergence when p = 0.

α = 0.15 α = 0.5 α = 0.85N error order error order error order10 0.05261 0.01707 0.0149620 0.20980 -1.996 0.003117 2.4534 0.003496 2.097440 8.17868 -5.285 0.0005597 2.4776 0.0008020 2.123980 37781 -12.17 9.972e-05 2.4889 0.0001823 2.1370

160 2.384e+12 -25.91 1.770e-05 2.4944 4.127e-05 2.1435320 2.749e+28 -53.36 3.134e-06 2.4972 9.320e-06 2.1467

Table 2: Maximum errors and orders of convergence when p = 1.

Tables 1-3 display the convergence of the error for different values of α and p,all of which are in good agreement with the theoretical analysis. In particular,the method is unstable for small values of α when p ≥ 1, but otherwise the order-hp+2−α convergence is clearly visible. When p = 2 and α = 0.5 the method stillconverges at the expected rate even though this parameter combination is notin the range with guaranteed stability that is implied by Figure 2. In Figure 3we further illustrate the stability of α = 0.5 for different p by displaying the

coefficients of the semicirculant matrices(

A + R(p))−1

. It appears that after

some transient effects the entries have the same asymptotic behavior when p = 0,p = 1, and p = 2. For larger values of p the coefficients blow up and the methodis unstable.

Figure 4 shows the errors for α = 0.5 and p = 2 as a function of time. Eventhough the analytical solution is singular at t = 0, the convergence happens atthe expected rate and the error is small near the singularity.

To put the results presented into the right perspective, it should be notedthat it is possible to derive stable rules of even higher order based on convolu-tional quadrature. In fact, Lubich presents results for a forth-order scheme [10].However, the point here was to introduce quadrature rules of the simpler form

18

100 101 102 10310−6

10−5

10−4

10−3

10−2

10−1

100

Figure 3: Magnitude of the coefficients of(

A + R(p))−1

for α = 0.5 and p = 0,

p = 1 and p = 2.

0 0.2 0.4 0.6 0.8 110−12

10−10

10−8

10−6

10−4

10−2

Figure 4: Errors for α = 0.5, β = 0.5, p = 2, N = 10, 20, . . . , 640 as a functionof t.

19

α = 0.15 α = 0.5 α = 0.85N error order error order error order10 2.1507 0.004426 0.00159620 170.16 -6.305 0.0003550 3.6401 0.0001871 3.092740 6.022e+06 -15.11 6.273e-06 5.8223 2.148e-05 3.122580 4.267e+16 -32.72 1.013e-06 2.6301 2.442e-06 3.1369

160 1.212e+37 -67.94 1.044e-07 3.2788 2.763e-07 3.1439320 5.534e+78 -138.3 9.241e-09 3.4981 3.118e-08 3.1473

Table 3: Maximum errors and orders of convergence when p = 2.

(3) which is advantageous, for instance, for the fast evaluation of time-dependentboundary integral operators.

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[11] G. Monegato and J.N. Lyness. The Euler-Maclaurin expansion and finite-part integrals. Numer. Math., 81(2):273–291, 1998.

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