approximation of potential integral by radial bases for solutions of helmholtz equation

Adv Comput Math (2009) 30:201–230DOI 10.1007/s10444-008-9065-8

Approximation of potential integral by radialbases for solutions of Helmholtz equation

Xin Li

Received: 26 September 2007 / Accepted: 26 January 2008 /Published online: 12 April 2008© Springer Science + Business Media, LLC 2008

Abstract For a Helmholtz equation �u(x) + κ2u(x) = f (x) in a region of Rs, s ≥ 2,where � is the Laplace operator and κ = a + ib is a complex number with b ≥ 0, aparticular solution is given by a potential integral. In this paper the potential integralis approximated by using radial bases with the order of approximation derived.

Keywords Potential integral · Radial bases · Helmholtz equation

Mathematics Subject Classifications (2000) 35J05 · 65M15 · 41A99

1 Introduction

Helmholtz equations appear and are applied in several scientific fields, such aswave propagation and vibration phenomena, and etc. For instance, [2] mainly treatsHelmholtz equations in R3 and their applications in scattering theory. Typicallyboundary value problems of Helmholtz equations are solved by traditional finiteor boundary element methods (cf. [1], etc.). Recently mesh free methods are usedby several authors [3, 6, 7], etc. to solve the related problems, which proves to besuccessful under certain circumstances.

Let D be a bounded domain in Rs, s ≥ 2. A Helmholtz equation on D is expressedas

(� + κ2 I

)u(x) = f (x), x ∈ D, (1)

Communicated by Aihui Zhou.

X. Li (B)Department of Mathematical Sciences,University of Nevada, Las Vegas,Las Vegas, NV 89154-4020, USAe-mail: [email protected]

202 X. Li

where � is the Laplace operator, I the identity operator, and κ = a + bi is acomplex number which we assume b = Im(κ) ≥ 0 as usual. It is well known thatthe fundamental solution of a Helmholtz equation is given by

G(x) = i4

( κ

2πr

)s/2−1H(1)

s/2−1(κr), s ≥ 2, (2)

where r = ‖x‖ and H(1)s/2−1 is a Hankel function (see Section 2). And thus a particular

solution of (1) is given by a potential integral

u(x) = − i4

( κ

2π

)s/2−1∫

Df (y)

H(1)s/2−1(κ‖x − y‖)‖x − y‖s/2−1

dy. (3)

The integrand of (3) involves a singularity which causes numerical difficulties inevaluation. In this paper, using the idea in [5] for Newtonian potentials, we willconstruct solutions of Helmholtz equations by radial bases that approximate thepotential integral in (3). For this purpose certain properties of Bessel functions willbe used and an approximation scheme by radial bases in [4] will be applied.

The organization of this paper is as follows. In Section 2 the properties of Besselfunctions needed in late sections will be listed out. An approximation result in [4]will be described in Section 3 and the approximate solutions of Helmholtz equationswill then be constructed. The result on approximation of the potential integral willbe established in Section 4 with the order of approximation derived. Examples in R2

and R3 will be given for compactly supported radial bases in the last Section 5.

2 Preliminary on Bessel functions

All the properties of the Bessel functions listed below can be found in the book[8] by Watson. Denote by Jν(z), Yν(z) the Bessel function of the first, second kind,respectively, where z is a complex number and ν a real number. For ν ≥ 0, Jν(z) canbe expressed in power series as follows

Jν(z) =∞∑

m=0

(−1)m(

12 z)ν+2m

m!�(ν + m + 1), (4)

where � is the usual gamma function, and the following identity is well known

�(μ)�(ν)

�(μ + ν)=∫ 1

0tμ−1(1 − t)ν−1dt, μ > 0, ν > 0. (5)

Jν(z) can also be expressed in an integral form (cf. p. 48, [8])

Jν(z) =(

12 z)ν

�(ν + 1

2

)�(

12

)∫ 1

−1

(1 − t2)ν−1/2

cos(zt)dt. (6)

Approximation of potential integral by radial bases 203

When ν = n ≥ 0, where n denotes an integer, Yn(z) can be expanded into

Yn(z) = 2

πJn(z)

(ln(z

2

)+ γ

)− 1

π

(z2

)−n n−1∑

m=0

(n − m − 1)!m!

(z2

)2m

− 1

π

∞∑

m=0

(−1)m

m!( z

2 )n+2m

(n + m)! (ηm + ηn+m), (7)

where

ηm = 1 + 1

2+ · · · + 1

m, γ = lim

m→∞(ηm − ln m),

and γ is known as the Euler’s number. In (7), if n = 0, the second term for Y0(z) isconsidered zero. For negative integers, there holds the relation

J−n(z) = (−1)n Jn(z). (8)

If ν is not an integer,

Yν(z) = 1

sin νz

[Jν(z) cos νπ − J−ν(z)

], (9)

and in case ν = n is an integer,

Yn(z) = limν→n

Yν(z). (10)

When ν = n + 1/2, n ≥ 0, Jn+1/2(z) can be expressed by a finite summation (cf.p. 53, [8])

Jn+1/2(z) = 1√π

(z2

)n+1/2[

(−1)n+1eiz2n∑

m=n

im+122n−mm!zm+1(m − n)!(2n − m)!

+ (−1)n+1e−iz2n∑

m=n

(−i)m+122n−mm!zm+1(m − n)!(2n − m)!

]

, (11)

and (cf. p. 54, [8])

J−n−1/2(z) = 1√2πz

[

eizn∑

m=0

im+n(n + m)!m!(n − m)!(2z)m

+ e−izn∑

m=0

(−i)m+n(n + m)!m!(n − m)!(2z)m

]

. (12)

In particular,

J1/2(z) =(

2

πz

)1/2

sin z, J−1/2(z) =(

2

πz

)1/2

cos z. (13)

Therefore, for ν = n + 1/2,

Yn+1/2(z) = 1

sin(n + 1/2)π

[Jn+1/2(z) cos(n + 1/2)π − J−n−1/2(z)

]

= (−1)n+1 J−n−1/2(z). (14)

204 X. Li

The Wronskian determinant of Jν(z) and Yν(z) is given by (cf. p. 76, [8])

W(Jν(z), Yν(z)) =∣∣∣∣

Jν(z) Yν(z)d

dz Jν(z) ddz Yν(z)

∣∣∣∣ = 2

πz. (15)

Since

ddz

J0(z) = −J1(z),d

dzY0(z) = −Y1(z) (16)

(cf. p. 66, [8]), with ν = 1, (15) implies

J1(z)Y0(z) − J0(z)Y1(z) = 2

πz. (17)

Bessel functions of the third kind are also called the Hankel functions, defined by

H(1)ν (z) = Jν(z) + iYν(z), H(2)

ν (z) = Jν(z) − iYν(z). (18)

It is then easy to see from (13) and (14) that

H(1)1/2(z) = −i

(2

πz

)1/2

eiz, H(2)1/2(z) = i

(2

πz

)1/2

e−iz. (19)

The Wronskian determinant of H(1)ν (z) and H(2)

ν (z) is given by (cf. p. 76, [8])

W(H(1)ν (z), H(2)

ν (z)) = − 4iπz

, (20)

and there is a formula for the derivative of H(1)ν (z) (cf. p. 74, [8])

zdH(1)

ν (z)

dz− νH(1)

ν (z) = −zH(1)ν+1(z). (21)

Use Cν(z) for a Bessel function of any kind. There hold (cf. pp. 132 and 133, [8])∫

t+1Cν(t)dt = − (2 − ν2) ∫

t−1Cν(t)dt + [t+1Cν+1(t) + ( − ν)tCν(t)

], (22)

∫tν+1Cν(t)dt = tν+1Cν+1(t) + c. (23)

Below we state some obvious and rough estimates on the behaviors of H(1)ν (z) and

Jν(z) as z approaches zero or infinity, which for the sake of simplicity we use c forany possible constants as usual. For ν = n + 1/2 ≥ 1/2 or ν = n ≥ 1, it follows fromthe expressions for Jν(z), Yν(z), and equation (18) that for any fixed d > 0 there is aconstant c depending on d and ν such that for all |z| ≤ d,

|H(1)ν (z)| ≤ c

|z|ν . (24)

In case ν = 0 it can be seen from (7) that for |z| ≤ d,

|H(1)0 (z)| ≤ c| ln(|z|)|. (25)

Furthermore, from the following estimate (cf. p. 211, [8])

|H(1)ν (z)| ≤ C|(πz/2)−1/2ei(z− 1

2 νπ− 14 π)|, (26)


where

C =

⎧⎪⎨

⎪⎩

(1 − ν−1/2

2|z|)−ν−1/2

, ν > 1/2 and 2|z| > ν − 1/2,

(1 − ν+3/2

2|z|)−ν−5/2 (

1 + 2ν+2|z|)

, 0 ≤ ν < 1/2 and 2|z| > ν + 3/2,

and also in view of (19), it follows that for any ν ≥ 0, there is a constant dν dependingon ν such that for |z| ≥ dν ,

|H(1)ν (z)| ≤ c|z|−1/2e−Im(z). (27)

It is clear to see from (4) that for any fixed d, if |z| ≤ d, then

|Jν(z)| ≤ c|z|ν . (28)

And moreover, for any z, by (6),

|Jν(z)| ≤ c|z|νe|Im(z)|. (29)

3 Solutions of Helmholtz equations by radial bases

We first need to describe an approximation scheme in [4]. Let D be a boundeddomain in Rs, s ≥ 2, and I = [−1, 1]s the unit cube. For δ > 0, let Dδ = D + δI :={x + y; x ∈ D, y ∈ δI}. For any integer n, set

In(Dδ) ={

j ∈ Z s;[

jn

,j + 1

n

]s

∩ Dδ = ∅}

,

where 1 = (1, · · · , 1) ∈ Z s. For 1 ≤ p ≤ ∞, denote by W1,p(D) the space of allfunctions f whose gradient is in Lp(D) with the usual Sobolev norm

‖ f‖W1,p(D) = ‖ f‖Lp(D) +s∑

k=1

∥∥∥

∂ f∂xk

∥∥∥Lp(D)

.

For a function f ∈ W1,p(Dδ), one can choose a smooth function χ such that χ

is identical to 1 on the closure D, and vanishes outside of Dδ . Let fχ = f · χ ,then fχ ∈ W1,p(Rs) and it is compactly supported in Dδ . Denote by W1,p

0 (Dδ) thesubspace of functions in W1,p(Rs) which vanish outside of Dδ . We then consider theapproximation of functions in W1,p

0 (Dδ) over the domain D. Suppose that φ ∈ L1(Rs)

is given with the property∫

Rsφ(x)dx = 1. (30)

Choose γ such that 0 < γ ≤ 1. For every f ∈ W1,p0 (Dδ) and an integer n ≥ 1, let

Bn f (x) = 1

ns(1−γ )

∑

j∈In(Dδ )

f(

jn

)φ(nγ x − jnγ−1) . (31)

206 X. Li

Let q satisfy 1p + 1

q = 1. If p = ∞, we consider q = 1. For α > 0, let Sα(Rs) be the set

consisting of all functions φ satisfying

|φ(x)| ≤ c(1 + ‖x‖)−α. (32)

Then the following result is shown in [4].

Theorem 1 Suppose that φ ∈ W1,p(Rs) ∩ Sα(Rs) for some α > s and 0 < η < γ <1

1+s/q . Then there exists a positive constant c such that for all f ∈ W1,p0 (Dδ) and large

n, the inequality

‖Bn f − f‖Lp(D) ≤ cnτ

‖ f‖W1,p0 (Dδ )

holds, where τ := min{η, (γ − η)(α − s), 1 − γ − sγ /q}.

Choose φ to be a radial basis, namely φ(x) = φ(r) with r = ‖x‖. Then the condition(30) becomes

∫ ∞

0rs−1φ(r)dr = 1

ωs, s ≥ 2, (33)

where

ωs = 2π s/2

�(s/2)(34)

is the surface area of the unit sphere in Rs. Consider the corresponding Helmholtzequation

(� + κ2 I

)ψ(x) = φ(x), x ∈ Rs, (35)

which a radially particular solution is derived below.

Proposition 1 For a radial basis function φ, a radially particular solution of aHelmholtz equation (35) is given by

ψ(r) = ψ1(r) + ψ2(r) + c1r1−s/2 H(1)s/2−1(κr) + c2r1−s/2 H(2)

s/2−1(κr), (36)

where c1 and c2 are arbitrary constants, and

ψ1(r) = −π i4

H(1)s/2−1(κr)r1−s/2

∫ r

0ts/2φ(t)H(2)

s/2−1(κt)dt, (37)

ψ2(r) = π i4

H(2)s/2−1(κr)r1−s/2

∫ r

0ts/2φ(t)H(1)

s/2−1(κt)dt. (38)

Proof By polar coordinates, (35) is written as

1

rs−1

(∂

∂r

(rs−1 ∂ψ(r)

∂r

))+ κ2ψ(r) = φ(r). (39)

Let

ψ(r) = r1−s/2v(r). (40)


Then (39) is changed to

∂2v(r)∂r2

+ r−1 ∂v(r)∂r

+ (κ2 − (s/2 − 1)2r−2

)v(r) = rs/2−1φ(r). (41)

For the homogeneous part of the above equation

r2 ∂2v(r)∂r2

+ r∂v(r)∂r

+ (κ2r2 − (s/2 − 1)2) v(r) = 0, (42)

it is well known that{

H(1)s/2−1(κr), H(2)

s/2−1(κr)}

constitutes a solution basis, or the

general solution of (42) is given by

vh(r) = c1 H(1)s/2−1(κr) + c2 H(2)

s/2−1(κr),

where c1 and c2 are arbitrary constants. By the standard method of variation ofparameters, a particular solution of non-homogeneous equation (41) is given by

vp(r) = w1(r)H(1)s/2−1(κr) + w2(r)H(2)

s/2−1(κr),

where

w1(r) = −∫ rs/2−1φ(r)H(2)

s/2−1(κr)

W(

H(1)s/2−1(κr), H(2)

s/2−1(κr))dr,

w2(r) =∫ rs/2−1φ(r)H(1)

s/2−1(κr)

W(

H(1)s/2−1(κr), H(2)

s/2−1(κr))dr,

and W(H(1)s/2−1(κr), H(2)

s/2−1(κr)) denotes the Wronskian determinant of H(1)s/2−1(κr)

and H(2)s/2−1(κr). By (20), we have

W(

H(1)s/2−1(κr), H(2)

s/2−1(κr))

= κ

(− 4i

πκr

)= − 4i

πr.

Hence a general solution of (41) is given by

v(r) = vp(r) + vh(r)

= −π i4

H(1)s/2−1(κr)

∫rs/2φ(r)H(2)

s/2−1(κr)dr + π i4

H(2)s/2−1(κr)

∫rs/2φ(r)H(1)

s/2−1(κr)dr

+ c1 H(1)s/2−1(κr) + c2 H(2)

s/2−1(κr). (43)

The conclusion of the Proposition then follows from (40). �

To ensure that ψ(r) is differentiable at the origin, we choose c1 = c2 = A, namely

ψ(r) = ψ1(r) + ψ2(r) + Ar1−s/2(

H(1)s/2−1(κr) + H(2)

s/2−1(κr))

. (44)

For a general Hehmholtz equation(� + κ2 I

)u(x) = f (x), x ∈ D, (45)

208 X. Li

which we assume f ∈ W1,∞0 (Dδ) below, choose a radial basis φ and approximate f

by Bn f in (31). Let ψn be a radial solution of(

� + κ2

n2γ

)ψn(x) = φ(x). (46)

And set

un(x) = 1

ns(1−γ )

∑

j∈In(Dδ )

f(

jn

)n−2γ ψn

(nγ x − jnγ−1

). (47)

Then one can easily check(� + κ2 I

)un(x) = Bn f (x),

or un serves as an approximate particular solution of (45).By Theorem 1, the following result can be derived, which we omit its proof since

it is essentially the same as the one for the Proposition 2 of [5].

Corollary 1 Suppose that a radial basis function φ ∈ W1,p(Rs) ∩ Sα(Rs), where α > s.Let un be given by (47). Then, for any f ∈ W1,p

0 (Dδ) and large n,∥∥(� + κ2 I

)un − f

∥∥Lp(Rs)

≤ cnτ

‖ f‖W1,p0 (Dδ )

,

where τ is given in Theorem 1. Moreover, for sufficiently large x,∣∣(� + κ2 I

)un(x)

∣∣ ≤ c

‖x‖α‖ f‖L∞(Dδ ).

4 Approximation of particular solutions in potential form

In view of (44) and Proposition 1, ψn(r) in (46) can be expressed as

ψn(r) = ψn,1(r) + ψn,2(r) + Ar1−s/2(

H(1)s/2−1(κn−γ r) + H(2)

s/2−1(κn−γ r))

, (48)

where

ψn,1(r) = −π i4

H(1)s/2−1(κn−γ r)r1−s/2

∫ r

0ts/2φ(t)H(2)

s/2−1(κn−γ t)dt,

ψn,2(r) = π i4

H(2)s/2−1(κn−γ r)r1−s/2

∫ r

0ts/2φ(t)H(1)

s/2−1(κn−γ t)dt.

Choose A to be

A = −π i4

∫ ∞

0ts/2φ(t)H(1)


Then

ψn(r) = − π i4

H(1)s/2−1(κn−γ r)r1−s/2

∫ r

0ts/2φ(t)

[H(1)

s/2−1(κn−γ t) + H(2)s/2−1(κn−γ t)

]dt

− π i4

[H(1)

s/2−1(κn−γ r) + H(2)s/2−1(κn−γ r)

]r1−s/2

∫ ∞

rts/2φ(t)H(1)



And thus from (18)

ψn(r) = −π i2

H(1)s/2−1(κn−γ r)r1−s/2

∫ r

0ts/2φ(t)Js/2−1(κn−γ t)dt

−π i2

Js/2−1(κn−γ r)r1−s/2∫ ∞

rts/2φ(t)H(1)

s/2−1(κn−γ t)dt. (49)

The approximate solution un in (47) is now expressed as

un(x) = −π i2

1

ns

∑

j∈In(Dδ )

f(

jn

)n−γ+γ s/2

×[

H(1)s/2−1(κ‖x − j/n‖)‖x − j/n‖s/2−1

∫ nγ ‖x−j/n‖


+ Js/2−1(κ‖x − j/n‖)‖x − j/n‖s/2−1

∫ ∞

nγ ‖x−j/n‖ts/2φ(t)H(1)

s/2−1(κn−γ t)dt]

. (50)

In view of (3), a particular solution of the Helmholtz equation (45) is given by

up(x) = − i4

( κ

2π

)s/2−1∫

Dδ

f (y)H(1)

s/2−1(κ‖x − y‖)‖x − y‖s/2−1

dy. (51)

From (27), since κ = a + ib with b = Im(κ) ≥ 0, up(x) for large x can be estimatedas follows

|up(x)| ≤ c∫

Dδ

| f (y)| e−b‖x−y‖

‖x − y‖s/2−1/2dy ≤ c

‖x‖s/2−1/2‖ f‖L∞(Dδ ). (52)

Denote by B(ρ) = {x ∈ Rs; ‖x‖ < ρ} the ball centered at the origin and with radiusρ > 0, and Bc(ρ) the complement of B(ρ) in Rs, namely Bc(ρ) = {x ∈ Rs; ‖x‖ ≥ ρ}.From now on we assume Dδ ⊂ B(ρ) for some ρ > 0.

Theorem 2 Suppose that a radial basis function φ ∈ W1,∞(Rs) ∩ Sα(Rs) for someα > s. Let un, up be given by (50), (51), respectively. Then for s ≥ 3,

‖un − up‖L∞(Rs) ≤ cOs,α(n)‖ f‖L∞(Dδ ) + cn

‖ f‖W1,∞0 (Dδ )

, (53)

where, with τ := min{γ (α − s), 2γ },

Os,α(n) :=

⎧⎪⎪⎨

⎪⎪⎩

1

nτ, α = s + 2,

ln nnτ

, α = s + 2.

(54)

And in case s = 2,

‖un − up‖L∞(R2) ≤ cO2,α(n) ln n‖ f‖L∞(Dδ ) + cn

‖ f‖W1,∞0 (Dδ )

, (55)

210 X. Li

where O2,α(n) is given by (54) with respect to s = 2. Moreover for large x,

|un(x)| ≤ c‖x‖s/2−1/2

‖ f‖L∞(Dδ ). (56)

To show above theorem, a few steps need be taken. Let us write un in (50) as

un(x) = u(1)n (x) + u(2)

n (x),

where

u(1)n (x) = − π i

2

1

ns

∑

j∈In(Dδ )

f(

jn

) H(1)s/2−1(κ‖x − j/n‖)‖x − j/n‖s/2−1

× n−γ+γ s/2∫ nγ ‖x−j/n‖

0ts/2φ(t)Js/2−1(κn−γ t)dt, (57)

u(2)n (x) = − π i

2

1

ns

∑

j∈In(Dδ )

f(

jn

)Js/2−1(κ‖x − j/n‖)

‖x − j/n‖s/2−1

× n−γ+γ s/2∫ ∞


s/2−1(κn−γ t)dt. (58)

And then

un(x) − up(x) = (u(1)n (x) − up(x)) + u(2)

n (x). (59)

We aim to estimate u(1)n (x) − up(x) first, which will be done in the Lemmas 1–4 below.

For any x, set

I(1)n (x) = {

j ∈ In(Dδ); ‖x − j/n‖ < n−γ},

I(2)n (x) = {

j ∈ In(Dδ); ‖x − j/n‖ ≥ n−γ}.

And correspondingly let

u(1)n (x) = P1(x) + P2(x), (60)

where

Pk(x) = − π i2

1

ns

∑

j∈I(k)n (x)

f(

jn

) H(1)s/2−1(κ‖x − j/n‖)‖x − j/n‖s/2−1

× n−γ+γ s/2∫ nγ ‖x−j/n‖

0ts/2φ(t)Js/2−1(κn−γ t)dt (61)

for k = 1, 2. Then

u(1)n (x) − up(x) = P1(x) + (P2(x) − up(x)). (62)

We first estimate P2(x) − up(x) in a bounded region.


Lemma 1 Under the same notation and assumptions as in Theorem 2, let M be sochosen that (27) holds for any |z| ≥ dν := |κ|M with respect to ν = s/2 − 1. Then fors ≥ 3,

‖P2 − up‖L∞(B(M+2ρ)) ≤ cOs,α(n)‖ f‖L∞(Dδ ) + cn

‖ f‖W1,∞0 (Dδ )

.

And in case s = 2,

‖P2 − up‖L∞(B(M+2ρ)) ≤ cO2,α(n) ln n‖ f‖L∞(Dδ ) + cn

‖ f‖W1,∞0 (Dδ )

.

Proof From (61) and (51) we can write

P2(x) − up(x) = Q1(x) − Q2(x), (63)

where

Q1(x) = − π i2

1

ns

∑

j∈I(2)n (x)

f(

jn

) H(1)s/2−1(κ‖x − j/n‖)‖x − j/n‖s/2−1

×[

n−γ+γ s/2∫ nγ ‖x−j/n‖

0ts/2φ(t)Js/2−1(κn−γ t)dt − 1

2π

( κ

2π

)s/2−1]

, (64)

Q2(x) = i4

( κ

2π

)s/2−1

⎡

⎣ 1

ns

∑

j∈I(2)n (x)

f(

jn

) H(1)s/2−1(κ‖x − j/n‖)‖x − j/n‖s/2−1

−∫

Dδ

f (y)H(1)

s/2−1(κ‖x − y‖)‖x − y‖s/2−1

dy

]

. (65)

Notice from (33) and (34) that

1

2π

( κ

2π

)s/2−1 = 1

2π

( κ

2π

)s/2−1[ωs

∫ ∞

0ts−1φ(t)dt

]

= 1

2π

( κ

2π

)s/2−1[

2π s/2

�(s/2)

∫ ∞

0ts−1φ(t)dt

]

= n−γ+γ s/2∫ ∞

0ts/2φ(t)

(12κn−γ t

)s/2−1

�(s/2)dt.

Hence we can further write

Q1(x) = R1(x) + R2(x), (66)

212 X. Li

where

R1(x) =−iπ2

1

ns

∑

j∈I(2)n (x)

f(

jn

) H(1)s/2−1(κ‖x − j/n‖)‖x − j/n‖s/2−1

× n−γ+γ s/2∫ nγ ‖x−j/n‖

0ts/2φ(t)

[

Js/2−1(κn−γ t) −(

12κn−γ t

)s/2−1

�(s/2)

]

dt, (67)

R2(x) = iπ2

1

ns

∑

j∈I(2)n (x)

f(

jn

) H(1)s/2−1(κ‖x − j/n‖)‖x − j/n‖s/2−1

× n−γ+γ s/2∫ ∞

nγ ‖x−j/n‖ts/2φ(t)

(12κn−γ t

)s/2−1

�(s/2)dt. (68)

For j ∈ I(2)n (x), from (32),

∣∣∣∫ ∞

nγ ‖x−j/n‖ts/2φ(t)

(12κn−γ t

)s/2−1

�(s/2)dt∣∣∣ ≤ cn−γ (s/2−1)

∫ ∞

nγ ‖x−j/n‖ts−1|φ(t)|dt

≤ cn−γ (s/2−1)

∫ ∞

nγ ‖x−j/n‖ts−α−1dt

≤ cn−γ (s/2−1)(nγ ‖x − j/n‖)s−α. (69)

Then, with

�s,α(n) ={

nγ (α−s−2), α = s + 2,

ln n, α = s + 2,

it follows from (24) that for s ≥ 3 and x ∈ B(M + 2ρ)

|R2(x)| ≤ cns

∑

j∈I(2)n (x)

∣∣∣∣ f(

jn

)∣∣∣∣c

‖x − j/n‖s−2(nγ ‖x − j/n‖)s−α

≤ cns+γ (α−s)

‖ f‖L∞(Dδ )

∑

j∈I(2)n (x)

1

‖x − j/n‖α−2

≤ cns+γ (α−s)

‖ f‖L∞(Dδ )

∑

12 n−γ ≤‖j/n‖≤M+4ρ

1

‖j/n‖α−2

≤ cnγ (α−s)

‖ f‖L∞(Dδ )

∫

14 n−γ ≤‖y‖≤M+5ρ

1

‖y‖α−2dy

≤ cnγ (α−s)

‖ f‖L∞(Dδ )

∫ 5ρ

14 n−γ

1

rα−2rs−1dr

≤ cnγ (α−s)

‖ f‖L∞(Dδ )(c + c�s,α(n))

≤ cOs,α(n)‖ f‖L∞(Dδ ). (70)


In case s = 2, we use (25) and similarly derive as for (70) that

|R2(x)| ≤ cn2

∑

j∈I(2)n (x)

∣∣∣ f(

jn

) ∣∣∣ | ln(|κ|‖x − j/n‖)|(nγ ‖x − j/n‖)2−α

≤ c ln nn2+γ (α−2)

‖ f‖L∞(Dδ )

∑

j∈I(2)n (x)

1

‖x − j/n‖α−2

≤ cO2,α(n) ln n‖ f‖L∞(Dδ ). (71)

To estimate R1(x), we use the power series expansion of Js/2−1(z) in (4) and have

Js/2−1(κn−γ t

)−(

12κn−γ t

)s/2−1

�(s/2)= −

(1

2κn−γ t

)s/2+1 ∞∑

m=1

(−1)m−1(

12κn−γ t

)2(m−1)

m!�(s/2 + m).

Hence with x ∈ B(M + 2ρ)

∣∣∣∣∣


0ts/2φ(t)

[

Js/2−1(κn−γ t) −(

12κn−γ t

)s/2−1

�(s/2)

]

dt

∣∣∣∣∣

≤ cn−γ (s/2+1)


0ts+1|φ(t)|

∞∑

m=1

(12 |κ|n−γ t

)2(m−1)

m!�(s/2 + m)dt

≤ cn−γ (s/2+1)


0ts+1|φ(t)|dt

≤ cn−γ (s/2+1)

[∫ 1

0+∫ nγ ‖x−j/n‖

1ts+1|φ(t)|dt

]

≤ cn−γ (s/2+1)

[

1 +∫ nγ ‖x−j/n‖

1ts+1−αdt

]

≤ cn−γ (s/2+1)(1 + �s,α(n)), (72)

where

�s,α(n) ={

nγ (s+2−α), α = s + 2,

ln n, α = s + 2,

Then in case s ≥ 3,

|R1(x)| ≤ cns

∑

j∈I(2)n (x)

∣∣∣∣ f(

jn

)∣∣∣∣

c‖x − j/n‖s−2

n−γ+γ s/2n−γ (s/2+1)(1 + �s,α(n))

≤ cOs,α(n)‖ f‖L∞(Dδ )

∑

j∈I(2)n (x)

1

‖x − j/n‖s−2

1

ns


∑

0<‖j/n‖≤M+4ρ

1

‖j/n‖s−2

1

ns

214 X. Li


∫

‖y‖≤M+4ρ

1

‖y‖s−2dy

≤ cOs,α(n)‖ f‖L∞(Dδ ). (73)

And for s = 2,

|R1(x)| ≤ cn2

∑

j∈I(2)n (x)

∣∣∣∣ f(

jn

)∣∣∣∣ | ln(|κ|‖x − j/n‖)|n−2γ (1 + �2,α(n))

≤ cO2,α(n)‖ f‖L∞(Dδ )

∑

j∈I(2)n (x)

(ln n)1

n2

≤ cO2,α(n) ln n‖ f‖L∞(Dδ ). (74)

We now turn to Q2(x). In view of (65) we easily have

|Q2(x)| ≤ c

∣∣∣∣∣∣

∑

j∈I(2)n (x)

∫[

jn ,

j+1n

]s

[

f(

jn

) H(1)s/2−1(κ‖x − j/n‖)‖x − j/n‖s/2−1

− f (y)H(1)

s/2−1(κ‖x − y‖)‖x − y‖s/2−1

]

dy

∣∣∣∣∣∣

+ c∫

‖y−x‖≤2n−γ

∣∣∣∣∣

f (y)H(1)

s/2−1(κ‖x − y‖)‖x − y‖s/2−1

∣∣∣∣∣dy (75)

for large n. For s ≥ 3,

∫


∣∣∣ f (y)

H(1)s/2−1(κ‖x − y‖)‖x − y‖s/2−1

∣∣∣ dy ≤ c‖ f‖L∞(Dδ )

∫ 2n−γ

0

1

rs−2rs−1dr

≤ c‖ f‖L∞(Dδ )

∫ 2n−γ

0rdr

≤ cn2γ

‖ f‖L∞(Dδ ). (76)

In case s = 2,∫


| f (y)H(1)0 (κ‖x − y‖)|dy ≤ c‖ f‖L∞(Dδ )

∫ 2n−γ

0| ln r|rdr ≤ c ln n

n2γ‖ f‖L∞(Dδ ).

(77)

To continue the estimate in (75), we let

g(y) = f (y)H(1)

s/2−1(κ‖x − y‖)‖x − y‖s/2−1

.

Its first order derivatives, by using (21), are given by

∂g(y)

∂yk= ∂ f (y)

∂yk

H(1)s/2−1(κ‖x − y‖)‖x − y‖s/2−1

+ f (y)1

‖x − y‖s−2

×⎡

⎣∂[

H(1)s/2−1(κ‖x − y‖)

]

∂(κ‖x − y‖)∂(κ‖x − y‖)

∂yk‖x − y‖s/2−1


− H(1)s/2−1(κ‖x − y‖) ∂(‖x − y‖s/2−1)

∂yk

]

= ∂ f (y)

∂yk

H(1)s/2−1(κ‖x − y‖)‖x − y‖s/2−1

+ f (y)1

‖x − y‖s−2

×[(

−H(1)s/2(κ‖x−y‖)+ s/2 − 1

κ‖x − y‖ H(1)s/2−1(κ‖x−y‖)

)κ(yk − xk)

‖x − y‖ ‖x−y‖s/2−1

− H(1)s/2−1(κ‖x − y‖)(s/2 − 1)‖x − y‖s/2−2 (yk − xk)

‖x − y‖]

(78)

for 1 ≤ k ≤ s. Hence for s ≥ 3,

∣∣∣

∂g(y)

∂yk

∣∣∣ ≤ ‖ f‖W1,∞

0 (Dδ )

c‖x − y‖s−2

+ ‖ f‖L∞(Dδ )

1

‖x − y‖s−2

×[(

c‖x − y‖s/2

+ c‖x − y‖

c‖x − y‖s/2−1

)‖x − y‖s/2−1

+ c‖x − y‖s/2−1

‖x − y‖s/2−2

]

≤ c‖ f‖W1,∞0 (Dδ )

1

‖x − y‖s−2+ c‖ f‖L∞(Dδ )

1

‖x − y‖s−1. (79)

Notice that for any j ∈ I(2)n (x) and any yj ∈

[jn ,

j+1n

]s

‖x − yj‖ ≥ ‖x − j/n‖ − ‖j/n − yj‖ ≥ n−γ −√

sn

>1

n

for large n. By applying the mean value theorem and extreme value theorem (for thereal and imaginary parts of the function), it follows that for each j ∈ I(2)

n (x), there isyj ∈ [ j

n ,j+1

n ]s such that

∣∣∣∣∣∣

∑

j∈I(2)n (x)

∫[

jn ,

j+1n

]s

[

f(

jn

) H(1)s/2−1(κ‖x − j/n‖)‖x − j/n‖s/2−1

− f (y)H(1)

s/2−1(κ‖x − y‖)‖x − y‖s/2−1

]

dy

∣∣∣∣∣∣

≤ cn

∑

j∈I(2)n (x)

[‖ f‖W1,∞

0 (Dδ )

c‖x − yj‖s−2

+ ‖ f‖L∞(Dδ )

1

‖x − yj‖s−1

]1

ns

≤ cn

‖ f‖W1,∞0 (Dδ )

∑

0<‖j/n‖≤M+4ρ

[1

‖j/n‖s−2+ 1

‖j/n‖s−1

]1

ns

≤ cn

‖ f‖W1,∞0 (Dδ )

∫

‖y‖≤M+4ρ

[1

‖y‖s−2+ 1

‖y‖s−1

]dy

≤ cn

‖ f‖W1,∞0 (Dδ )

. (80)

216 X. Li

And in case s = 2,

∣∣∣

∂g(y)

∂yk

∣∣∣ ≤ c ‖ f‖W1,∞

0 (Dδ )| ln(|κ|‖x − y‖)| + ‖ f‖L∞(Dδ )

×[

c‖x − y‖ + c| ln(|κ|‖x − y‖)| 1

‖x − y‖]

≤ c ‖ f‖W1,∞0 (Dδ )

| ln(|κ|‖x − y‖)| + c‖ f‖L∞(Dδ )

| ln(|κ|‖x − y‖)|‖x − y‖ . (81)

As for (80),

∣∣∣∣∣∣

∑

j∈I(2)n (x)

∫[

jn ,

j+1n

]2

[f(

jn

)H(1)

0 (κ‖x − j/n‖) − f (y)H(1)0 (κ‖x − y‖)

]dy

∣∣∣∣∣∣

≤ cn

∑

j∈I(2)n (x)

[‖ f‖W1,∞

0 (Dδ )| ln(|κ|‖x − yj‖)| + ‖ f‖L∞(Dδ )

| ln(|κ|‖x − yj‖)|‖x − yj‖

]1

n2

≤ cn

‖ f‖W1,∞0 (Dδ )

∑

0<‖j/n‖≤M+4ρ

[| ln(|κ|‖j/n‖)| + | ln(|κ|‖j/n‖)|

‖j/n‖]

1

n2

≤ cn

‖ f‖W1,∞0 (Dδ )

∫

‖y‖≤M+4ρ

[| ln(|κ|‖y‖)| + | ln(|κ|‖y‖)|

‖y‖]

dy

≤ cn

‖ f‖W1,∞0 (Dδ )

. (82)

Therefore, for s ≥ 3, it follows from (63), (66), (70), (73), (75), (76), and (80) that

|P2(x) − up(x)| ≤ |Q1(x)| + |Q2(x)| ≤ |R1(x)| + |R2(x)| + |Q2(x)|≤ cOs,α(n)‖ f‖L∞(Dδ ) + c

n2γ‖ f‖L∞(Dδ ) + c

n‖ f‖W1,∞

0 (Dδ )

≤ cOs,α(n)‖ f‖L∞(Dδ ) + cn

‖ f‖W1,∞0 (Dδ )

.

Similarly, for s = 2, it follows

|P2(x) − up(x)| ≤ cO2,α(n) ln n‖ f‖L∞(Dδ ) + cn

‖ f‖W1,∞0 (Dδ )

.

The proof of this lemma is then completed. �

Next we estimate P2(x) − up(x) for ‖x‖ ≥ M + 2ρ. Notice that in this case P2(x) =u(1)

n (x) since I(1)n (x) is empty. But for the sake of presentation we still use P2(x). Many

equations, estimates, and expressions in the proof of Lemma 1 will be used in thearguments for the Lemma 2 below.


Lemma 2 Under the same notation and assumptions as in Theorem 2 and Lemma 1,for any s ≥ 2,

‖P2 − up‖L∞(Bc(M+2ρ)) ≤ cOs,α(n)‖ f‖L∞(Dδ ) + cn

‖ f‖W1,∞0 (Dδ )

. (83)

And moreover, for large x,

|P2(x)| ≤ c‖x‖s/2−1/2

‖ f‖L∞(Dδ ). (84)

Proof Let Q1(x), Q2(x), R1(x), and R2(x) be given by (64), (65), (67), and (68),respectively. For ‖x‖ ≥ M + 2ρ, by using (27) and (69), R2(x) in (68) can be esti-mated as follows

|R2(x)| ≤ cns

∑

j∈I(2)n (x)

∣∣∣ f(

jn

) ∣∣∣e−b‖x−j/n‖

‖x − j/n‖s/2−1/2(nγ ‖x − j/n‖)s−α

≤ cns

∑

j∈I(2)n (x)

∣∣∣ f(

jn

) ∣∣∣

e−b‖x−j/n‖

‖x − j/n‖α−s/2−1/2nγ (s−α)

≤ cns

‖ f‖L∞(Dδ )nsnγ (s−α)

≤ cnγ (α−s)

‖ f‖L∞(Dδ ). (85)

In order to estimate R1(x) in (67), we use (72) and have

∣∣∣∣∣


0ts/2φ(t)

[

Js/2−1(κn−γ t)−(

12κn−γ t

)s/2−1

�(s/2)

]

dt

∣∣∣∣∣

=∣∣∣∣∣

∫ nγ M

0+∫ nγ ‖x−j/n‖

nγ Mts/2φ(t)

[

Js/2−1(κn−γ t) −(

12κn−γ t

)s/2−1

�(s/2)

]

dt

∣∣∣∣∣

≤cn−γ (s/2+1)(1+�s,α(n))+∣∣∣∣∣


nγ Mts/2φ(t)

[

Js/2−1(κn−γ t)−(

12κn−γ t

)s/2−1

�(s/2)

]

dt

∣∣∣∣∣.

(86)

Notice from (5)

�( s−1

2

)�(1/2)

�(s/2)=∫ 1

0v−1/2(1 − v)s/2−3/2dv =

∫ 1

−1(1 − v2)s/2−3/2dv,

and thus

(12κn−γ t

)s/2−1

�(s/2)=(

12κn−γ t

)s/2−1

�( s−1

2

)�(1/2)

∫ 1

−1(1 − v2)s/2−3/2dv.

218 X. Li

From the integral representation (6) for Js/2−1, we get

Js/2−1(κn−γ t) −(

12κn−γ t

)s/2−1

�(s/2)

=(

12κn−γ t

)s/2−1

�( s−1

2

)�(1/2)

∫ 1

−1(1 − v2)s/2−3/2

[cos(κn−γ tv) − 1

]dv

=(

12κn−γ t

)s/2−1

�( s−1

2

)�(1/2)

∫ 1

−1(1 − v2)s/2−3/2(−2) sin2

(κn−γ tv

2

)dv

=(

12κn−γ t

)s/2−1

�( s−1

2

)�(1/2)

∫ 1

−1(1 − v2)s/2−3/2(−2)

[1

2i

(e

iκn−γ tv2 − e

−iκn−γ tv2

)]2

dv,

which implies

∣∣∣∣∣

Js/2−1(κn−γ t) −(

12κn−γ t

)s/2−1

�(s/2)

∣∣∣∣∣≤ c(n−γ t)s/2−1ebn−γ t.

Therefore

∣∣∣∣∣


nγ Mts/2φ(t)

[

Js/2−1(κn−γ t) −(

12κn−γ t

)s/2−1

�(s/2)

]

dt

∣∣∣∣∣

≤ cn−γ (s/2−1)


nγ Mts−1t−αebn−γ tdt

≤ cn−γ (s/2−1)eb‖x−j/n‖(nγ M)s−α

≤ cnγ (1+s/2−α)eb‖x−j/n‖. (87)

From (67), (86), and (87),

|R1(x)| ≤ cns

∑

j∈I(2)n (x)

∣∣∣ f(

jn

) ∣∣∣

e−b‖x−j/n‖

‖x − j/n‖s/2−1/2n−γ+γ s/2

× [n−γ (s/2+1)(1 + �s,α(n)) + nγ (1+s/2−α)eb‖x−j/n‖]

≤ cns

∑

j∈I(2)n (x)

∣∣∣∣ f(

jn

) ∣∣∣∣

[e−b‖x−j/n‖

‖x − j/n‖s/2−1/2Os,α(n) + 1

‖x − j/n‖s/2−1/2nγ (s−α)

]

≤ cOs,α(n)‖ f‖L∞(Dδ ). (88)


To estimate Q2(x), notice that the second integral in (75) vanishes, and from (78)with ‖x‖ ≥ M + 2ρ and ‖y‖ < 2ρ,

∣∣∣

∂g(y)

∂yk

∣∣∣ ≤ c‖ f‖W1,∞

0 (Dδ )

e−b‖x−y‖

‖x − y‖s/2−1/2+ ‖ f‖L∞(Dδ )

c‖x − y‖s−2

×[(

e−b‖x−y‖

‖x − y‖1/2+ ce−b‖x−y‖

‖x − y‖3/2

)‖x − y‖s/2−1 + e−b‖x−y‖

‖x − y‖1/2‖x − y‖s/2−2

]

≤ c‖ f‖W1,∞0 (Dδ )

1

‖x − y‖s/2−1/2≤ c‖ f‖W1,∞

0 (Dδ ). (89)

Hence it is clear to have

|Q2(x)| ≤ c

∣∣∣∣∣∣

∑

j∈I(2)n (x)

∫[

jn ,

j+1n

]s

[

f(

jn

) H(1)s/2−1(κ‖x − j/n‖)‖x − j/n‖s/2−1

− f (y)H(1)

s/2−1(κ‖x − y‖)‖x − y‖s/2−1

]

dy

≤∑

j∈I(2)n (x)

c‖ f‖W1,∞0 (Dδ )

n1

ns

≤ cn

‖ f‖W1,∞0 (Dδ )

. (90)

It then follows from (63), (66), (85), (88) and (90) that

|P2(x) − up(x)| ≤ |R1(x)| + |R2(x)| + |Q2(x)|≤ cOs,α(n)‖ f‖L∞(Dδ ) + c

nγ (α−s)‖ f‖L∞(Dδ ) + c

n‖ f‖W1,∞

0 (Dδ )

≤ cOs,α(n)‖ f‖L∞(Dδ ) + cn

‖ f‖W1,∞0 (Dδ )

.

Finally, notice that for any x, by using (29),

∣∣∣∣∣



∣∣∣∣∣

≤ c∫ nγ ‖x−j/n‖

0ts/2|φ(t)|(n−γ t)s/2−1ebn−γ tdt

≤ cn−γ (s/2−1)eb‖x−j/n‖[∫ 1

0+∫ nγ ‖x−j/n‖

1ts/2|φ(t)|ts/2−1dt

]

≤ cn−γ (s/2−1)eb‖x−j/n‖[

1 +∫ nγ ‖x−j/n‖

1ts/2t−αts/2−1dt

]

≤ cn−γ (s/2−1)eb‖x−j/n‖. (91)

220 X. Li

Hence for large x, from (27), (61), and (91),

|P2(x)| ≤ cns

∑

j∈I(2)n (x)

∣∣∣ f(

jn

) ∣∣∣e−b‖x−j/n‖

‖x − j/n‖s/2−1/2n−γ+γ s/2n−γ (s/2−1)eb‖x−j/n‖

≤ c‖x‖s/2−1/2

‖ f‖L∞(Dδ ).

This completes the proof of the lemma. �

We now turn to P1(x) [cf. (61)]. Note that x must be in B(M + 2ρ) for I(1)n (x) to be

non-empty.

Lemma 3 Under the assumptions of Theorem 2, there holds

‖P1‖L∞(Rs) ≤ cn2γ

‖ f‖L∞(Dδ ), (92)

for s ≥ 3. And for s = 2,

‖P1‖L∞(R2) ≤ c ln nn2γ

‖ f‖L∞(Dδ ). (93)

Proof For j ∈ I(1)n (x), from (28),

∣∣∣∫ nγ ‖x−j/n‖


∣∣∣ ≤ c∫ nγ ‖x−j/n‖

0ts/2(n−γ t)s/2−1dt

= cn−γ (s/2−1)


0ts−1dt

= cn−γ (s/2−1)(nγ ‖x − j/n‖)s.

And thus, for s ≥ 3, by (24),

|P1(x)| ≤ cns

∑

j∈I(1)n (x)

∣∣∣ f(

jn

) ∣∣∣

c‖x − j/n‖s−2

n−γ+γ s/2n−γ (s/2−1)(nγ ‖x − j/n‖)s

≤ cns

∑

j∈I(1)n (x)

∣∣∣ f(

jn

) ∣∣∣ ‖x − j/n‖2nγ s ≤ cns

‖ f‖L∞(Dδ )ns(1−γ )n−2γ nγ s

≤ cn2γ

‖ f‖L∞(Dδ ). (94)


And in case s = 2, by using (25),

|P1(x)| ≤ cn2

∑

j∈I(1)n (x)

∣∣∣ f(

jn

) ∣∣∣ | ln(|κ|‖x − j/n‖)|(nγ ‖x − j/n‖)2

≤ cn2

∑

j∈I(1)n (x)

∣∣∣ f(

jn

) ∣∣∣ n2γ (‖x − j/n‖2| ln(|κ|‖x − j/n‖)|)

≤ cn2

‖ f‖L∞(Dδ )n2(1−γ )n2γ (n−2γ | ln n−γ |) ≤ c ln n

n2γ‖ f‖L∞(Dδ ). (95)

This lemma is proved. �

For clarity we summarize the results in Lemmas 1–3 in the following lemma.

Lemma 4 Under the same notation and assumptions as in Theorem 2, let u(1)n , up be

given by (57), (51), respectively. Then for s ≥ 3,

‖u(1)n − up‖L∞(Rs) ≤ cOs,α(n)‖ f‖L∞(Dδ ) + c

n‖ f‖W1,∞

0 (Dδ ). (96)

And in case s = 2,

‖u(1)n − up‖L∞(R2) ≤ cO2,α(n) ln n‖ f‖L∞(Dδ ) + c

n‖ f‖W1,∞

0 (Dδ ). (97)

Moreover for sufficiently large x and any s ≥ 2,

|u(1)n (x)| ≤ c

‖x‖s/2−1/2‖ f‖L∞(Dδ ). (98)

Proof In view of (62), it is clear that the conclusions in (96) and (97) follow from theresults in Lemmas 1–3 applied to the estimate

‖u(1)n − up‖L∞(Rs) ≤ ‖P1‖L∞(Rs) + ‖P2 − up‖L∞(B(M+2ρ)) + ‖P2 − up‖L∞(Bc(M+2ρ)).

And for large x, from Lemma 2, we have

|u(1)n (x)| = |P2(x)| ≤ c

‖x‖s/2−1/2‖ f‖L∞(Dδ ). �

Lemma 5 Let u(2)n be given by (58). Then for s ≥ 3,

‖u(2)n ‖L∞(Rs) ≤ cOs,α(n)‖ f‖L∞(Dδ ). (99)

And in case s = 2,

‖u(2)n ‖L∞(R2) ≤ cO2,α(n) ln n‖ f‖L∞(Dδ ). (100)

Moreover for large x,

|u(2)n (x)| ≤ c

‖x‖α− s+12

‖ f‖L∞(Dδ ). (101)

222 X. Li

Proof Let M be chosen as in the Lemma 1. For each x, set

I(1)n (x) = {

j ∈ In(Dδ); ‖x − j/n‖ < n−γ},

I(2)n (x) = {

j ∈ In(Dδ); n−γ ≤ ‖x − j/n‖ < M},

I(3)n (x) = {

j ∈ In(Dδ); ‖x − j/n‖ ≥ M}.

And write

u(2)n (x) = U1(x) + U2(x) + U3(x), (102)

where

Uk(x) = − π i2

1

ns

∑

j∈ I(k)n (x)

f(

jn

)Js/2−1(κ‖x − j/n‖)

‖x − j/n‖s/2−1

× n−γ+γ s/2∫ ∞


s/2−1(κn−γ t)dt (103)

for 1 ≤ k ≤ 3.We estimate U1(x) first. For j ∈ I(1)

n (x), write

∫ ∞


s/2−1(κn−γ t)dt=∫ Mnγ

nγ ‖x−j/n‖+∫ ∞

Mnγ

ts/2φ(t)H(1)s/2−1(κn−γ t)dt. (104)

For s ≥ 3, from (24),

∣∣∣∣

∫ Mnγ


s/2−1(κn−γ t)dt

∣∣∣∣ ≤ c

∫ Mnγ

nγ ‖x−j/n‖ts/2|φ(t)| 1

(n−γ t)s/2−1dt

≤ cnγ (s/2−1)

∫ Mnγ

0t|φ(t)|dt ≤ cnγ (s/2−1). (105)

And by using (27) and (32)

∣∣∣∫ ∞

Mnγ

ts/2φ(t)H(1)s/2−1(κn−γ t)dt

∣∣∣ ≤ c∫ ∞

Mnγ

ts/2t−α 1√n−γ t

e−bn−γ tdt

≤ cnγ /2∫ ∞

Mnγ

t−α+s/2−1/2dt ≤ cnγ (−α+1+s/2). (106)

Hence (104), (105), and (106) yield

∣∣∣∣

∫ ∞



∣∣∣∣ ≤ cnγ (s/2−1). (107)


By (28) we get

|U1(x)| ≤ cns

∑

j∈ I(1)n (x)

∣∣∣ f(

jn

) ∣∣∣ n−γ+γ s/2nγ (s/2−1) ≤ c

ns‖ f‖L∞(Dδ )n

s(1−γ )n−γ+γ s/2nγ (s/2−1)

≤ cn2γ

‖ f‖L∞(Dδ ). (108)

In case s = 2, from (25), the corresponding estimate for (105) becomes

∣∣∣∫ Mnγ

nγ ‖x−j/n‖tφ(t)H(1)

0 (κn−γ t)dt∣∣∣ ≤ c

∫ Mnγ

nγ ‖x−j/n‖t|φ(t)|| ln(|κ|n−γ t)|dt

≤ c∫ Mnγ

0t|φ(t)|(ln n + | ln t|)dt ≤ c ln n. (109)

And (106) remains the same, and therefore in this case

|U1(x)| ≤ cn2

∑

j∈ I(1)n (x)

∣∣∣ f(

jn

) ∣∣∣ c ln n ≤ c

n2‖ f‖L∞(Dδ )n

2(1−γ ) ln n ≤ c ln nn2γ

‖ f‖L∞(Dδ ).

(110)

We now turn to estimate U2(x). For j ∈ I(2)n (x), we use the same expression (104).

And for s ≥ 3,

∣∣∣∣

∫ Mnγ



∣∣∣∣ ≤ c

∫ Mnγ

nγ ‖x−j/n‖ts/2t−α 1

(n−γ t)s/2−1dt

≤ cnγ (s/2−1)

∫ Mnγ

nγ ‖x−j/n‖t1−αdt

≤ cnγ (s/2−1)(nγ (2−α) + (nγ ‖x − j/n‖)2−α)

≤ cnγ (s/2+1−α)(1 + ‖x − j/n‖2−α), (111)

and (106) remains the same. Therefore

∣∣∣∣

∫ ∞



∣∣∣∣ ≤ cnγ (s/2+1−α)(1 + ‖x − j/n‖2−α). (112)

224 X. Li

Hence by the similar argument in (70)

|U2(x)| ≤ cns

∑

j∈ I(2)n (x)

∣∣∣∣ f(

jn

) ∣∣∣∣n−γ+γ s/2nγ (s/2+1−α)(1 + ‖x − j/n‖2−α)

≤ cns+γ (α−s)

‖ f‖L∞(Dδ )

∑

j∈ I(2)n (x)

(1 + ‖x − j/n‖2−α)

≤ cnγ (α−s)

‖ f‖L∞(Dδ ) + cns+γ (α−s)

‖ f‖L∞(Dδ )

∑

j∈ I(2)n (x)

1

‖x − j/n‖α−2

≤ cnγ (α−s)

‖ f‖L∞(Dδ ) + cnγ (α−s)

‖ f‖L∞(Dδ )(c + c�s,α(n))

≤ cOs,α(n)‖ f‖L∞(Dδ ). (113)

In case s = 2,

∣∣∣∣

∫ Mnγ


0 (κn−γ t)dt

∣∣∣∣ ≤ c

∫ Mnγ

nγ ‖x−j/n‖t|φ(t)|| ln(|κ|n−γ t)|dt

≤ c∫ Mnγ

nγ ‖x−j/n‖t1−α| ln(|κ|n−γ t)|dt

≤ cnγ (2−α)

∫ |κ|M

|κ|‖x−j/n‖t1−α| ln t|dt

≤ cnγ (2−α)(1 + ‖x − j/n‖2−α(1 + | ln ‖x − j/n‖|))

and again (106) applies. It yields

∣∣∣∣

∫ ∞


0 (κn−γ t)dt

∣∣∣∣ ≤ cnγ (2−α)(1 + ‖x − j/n‖2−α(1 + | ln ‖x − j/n‖|)).

(114)Hence by the similar argument as for (71)

|U2(x)| ≤ cn2

∑

j∈ I(2)n (x)

∣∣∣∣ f(

jn

) ∣∣∣∣n

γ (2−α)(1 + ‖x − j/n‖2−α(1 + | ln ‖x − j/n‖|))

≤ cnγ (α−2)

‖ f‖L∞(Dδ ) + c ln nn2+γ (α−2)

‖ f‖L∞(Dδ )

∑

j∈ I(2)n (x)

‖x − j/n‖2−α

≤ cO2,α(n) ln n‖ f‖L∞(Dδ ). (115)


Finally we estimate U3(x). For j ∈ I(3)n (x),

∣∣∣∣

∫ ∞



∣∣∣∣ ≤ c

∫ ∞

nγ ‖x−j/n‖ts/2t−α 1√

n−γ te−bn−γ tdt

≤ ce−b‖x−j/n‖nγ /2∫ ∞

nγ ‖x−j/n‖ts/2−1/2−αdt

≤ ce−b‖x−j/n‖nγ /2(nγ ‖x − j/n‖)s/2+1/2−α (116)

From (29),

|U3(x)| ≤ cns

∑

j∈ I(3)n (x)

∣∣∣ f(

jn

) ∣∣∣

‖x − j/n‖s/2−1eb‖x−j/n‖

‖x − j/n‖s/2−1e−b‖x−j/n‖ nγ (s−α)

‖x − j/n‖α−s/2−1/2

≤ cns+γ (α−s)

∑

j∈ I(3)n (x)

∣∣∣ f(

jn

) ∣∣∣1

‖x − j/n‖α−s/2−1/2(117)

≤ cnγ (α−s)

‖ f‖L∞(Dδ ). (118)

For s = 2, the estimate for U3(x) is the same.Therefore for s ≥ 3, the conclusion in (99) follows from (102), (108), (113), and

(118). And (100) is shown by (102), (110), (115), and (118). It is easy to see from(117) that for large x and any s ≥ 2,

|u(2)n (x)| = |U3(x)| ≤ c

‖x‖α−s/2−1/2‖ f‖L∞(Dδ ).

The proof of this lemma is completed. �

We are finally in a position to prove Theorem 2.

Proof of Theorem 2 Notice

‖un − up‖L∞(Rs) ≤ ‖u(1)n − up‖L∞(Rs) + ‖u(2)

n ‖L∞(Rs).

The conclusions in (53) and (55) then follow from the results in Lemma 4 and 5.Furthermore for large x,

|un(x)| ≤ |u(1)n (x)| + |u(2)

n (x)| ≤ c‖x‖s/2−1/2

‖ f‖L∞(Dδ )

+ c‖x‖α−s/2−1/2

‖ f‖L∞(Dδ ) ≤ c‖x‖s/2−1/2

‖ f‖L∞(Dδ ).

This proves the theorem. �

226 X. Li

5 Examples in R2 and R3 and some remarks

Computer softwares such as Matlab and other commonly used ones are wellequipped with the programs for the evaluation of Bessel functions of all kinds.Therefore the evaluation of particular solutions in Proposition 1 for ψn in (49) doesnot propose any numerical issue. Here we like to write out the particular solutionsfor some compactly supported radial bases.

Note that for an odd integer s, Js/2−1(z) and H(1)s/2−1(z) can be expressed by finite

summations [cf. (11), (12), and (14)]. And thus the corresponding fundamentalsolution G(r) in (2) can be explicitly written out. For instance, in case s = 3, thefundamental solution [cf. (2) and (19)] is given by

G(r) = 1

4πreiκr.

Choose radial bases, which are compactly supported, given by

φ(r) ={

cs,m(1 − r2)m, 0 ≤ r ≤ 1,

0, r > 1,(119)

where m ≥ 1 is an integer and cs,m is a constant so chosen that∫

Rs φ(x)dx = 1. Suchradial bases are also examined in [5]. We first calculate ψn for s = 3.

Example 1 For φ given in (119), a direct computation shows that the condition∫R3 φ(x)dx = 1 implies

c3,m =(

4π

m∑

k=0

(mk

)(−1)m−k

2(m − k) + 3

)−1

.

From (49), (19), and (13),

ψn(r) = − 1

κn−γ reiκn−γ r

∫ r

0tφ(t) sin(κn−γ t)dt

− 1

κn−γ rsin(κn−γ r)

∫ ∞

rtφ(t)eiκn−γ tdt.

Note that for any complex number z and any integer ,∫

teztdt = t

zezt −

z

∫t−1eztdt = t

zezt −

z

[t−1

zezt − − 1

z

∫t−2eztdt

]

= ezt

⎡

⎣∑

j=0

(−1) j!t− j

( − j)!z j+1

⎤

⎦+ c. (120)

Let

η(t; , z) = ezt

⎡

⎣∑

j=0

(−1) j!t− j

( − j)!z j+1

⎤

⎦ .


Then for r ≤ 1,

c(r, n) :=∫ r

0tφ(t) sin(κn−γ t)dt = Im

[∫ r

0tφ(t)eiκn−γ tdt

]

= Im

[

c3,m

m∑

k=0

(mk

)(−1)k

∫ r

0t2k+1eiκn−γ tdt

]

= c3,m

m∑

k=0

(mk

)(−1)k Im

[η(r; 2k + 1, iκn−γ ) − η(0; 2k + 1, iκn−γ )

],

and

d(r, n) :=∫ ∞

rtφ(t)eiκn−γ tdt = c3,m

m∑

k=0

(mk

)(−1)k

∫ 1

rt2k+1eiκn−γ tdt

= c3,m

m∑

k=0

(mk

)(−1)k [η(1; 2k + 1, iκn−γ ) − η(r; 2k + 1, iκn−γ )

].

Hence

ψn(r) =⎧⎨

⎩

− 1κn−γ r [eiκn−γ rc(r, n) + sin(κn−γ r)d(r, n)], 0 ≤ r ≤ 1,

− 1κn−γ r eiκn−γ rc(1, n), r > 1.

Next we calculate ψn for s = 2.

Example 2 We use φ in (119) again, and determine c2,m first. Note ω2 = 2π , and thusfrom

1 =∫

R2φ(x)dx = c2,mω2

∫ 1

0(1 − r2)mrdr = πc2,m

m + 1,

we have

c2,m = m + 1

π.

ψn in (49) then takes in the form

ψn(r) = −π i2

H(1)0 (κn−γ r)

∫ r

0tφ(t)J0(κn−γ t)dt

−π i2

J0(κn−γ r)∫ ∞

rtφ(t)H(1)

0 (κn−γ t)dt

Now applying (22) and (23), where Cν represents a Bessel function of any kind, weeasily have

∫t+1C0(zt)dt = − 2

z2

∫t−1C0(zt)dt + t+1

zC1(zt) + t

z2C0(zt), (121)

∫tC0(zt)dt = t

zC1(zt) + c. (122)

228 X. Li

Then for any integer ≥ 0, by recursively applying (121), and also (122), we get

∫t2+1C0(zt)dt =

[∑

k=0

(−1)k(

(2)!!(2( − k))!!

)2 t2(−k)+1

z2k+1

]

C1(zt),

+[

−1∑

k=0

(−1)k(

(2)!!(2( − k))!!

)2 2( − k)t2(−k)

z2k+2

]

C0(zt) + c, (123)

where (2)!! = (2)(2( − 1)) · · · 4 · 2, and∑−1

k=0 is considered 0 for = 0. For r ≤ 1,using (123) for J0 with z = κn−γ ,

∫ r

0tφ(t)J0(κn−γ t)dt

= m+1

π

m∑

=0

(m

)(−1)

∫ r

0t2+1 J0(κn−γ t)dt

= m+1

π

m∑

=0

(m

)(−1)

{[∑

k=0

(−1)k(

(2)!!(2( − k))!!

)2 r2(−k)+1

(κn−γ )2k+1

]

J1(κn−γ r),

+[

−1∑

k=0

(−1)k(

(2)!!(2( − k))!!

)2 2( − k)r2(−k)

(κn−γ )2k+2

]

J0(κn−γ t)

}

.

(124)

And similarly

∫ ∞

rtφ(t)H(1)

0 (κn−γ t)dt

= −m+1

π

m∑

=0

(m

)(−1)

{[∑

k=0

(−1)k(

(2)!!(2( − k))!!

)2 r2(−k)+1

(κn−γ )2k+1

]

H(1)1 (κn−γ r),

+[

−1∑

k=0

(−1)k(

(2)!!(2( − k))!!

)2

× 2( − k)r2(−k)

(κn−γ )2k+2

]

H(1)0 (κn−γ t)

}

. (125)

Hence

ψn(r) = − (m + 1)i2

m∑

=0

(m

)(−1)

[∑

k=0

(−1)k(

(2)!!(2( − k))!!

)2 r2(−k)+1

(κn−γ )2k+1

]

×[

H(1)0 (κn−γ r)J1(κn−γ r) − J0(κn−γ r)H(1)

1 (κn−γ r)]. (126)


By using (17)

H(1)0 (κn−γ r)J1(κn−γ r) − J0(κn−γ r)H(1)

1 (κn−γ r)

= i[Y0(κn−γ r)J1(κn−γ r) − J0(κn−γ r)Y1(κn−γ r)]= 2i

πκn−γ r,

we have

ψn(r) =

⎧⎪⎨

⎪⎩

m+1π

m∑

=0

(m

)(−1)

[∑

k=0(−1)k

((2)!!

(2(−k))!!)2

r2(−k)

(κn−γ )2k+2

], 0 ≤ r ≤ 1,

−π i2 H(1)

0 (κn−γ r)∫ 1

0 tφ(t)J0(κn−γ t)dt, r > 1,

where∫ 1

0 tφ(t)J0(κn−γ t)dt can be evaluated by using (124).

Remark The collocation methods by RBFs have been used to find particular solu-tions of Helmholtz equations (cf. [3, 7]) in R2 or R3. In using the collocation methodsthe basis functions φ are commonly chosen to be thin plate splines, compactlysupported RBFs, or other popular ones. Here our Proposition 1 provides a generalformula for any radial basis function φ.

For instance, consider φ(r) = r2m−1, m ≥ 1, a thin plate spline in R3. In view of(36)–(38), (19) and (120),

ψ1(r) = − π i4

H(1)1/2(κr)r−1/2

∫ r

0t3/2t2m−1 H(2)

1/2(κt)dt

= − π i4

[

−i(

2

πκr

)1/2

eiκr

]

r−1/2∫ r

0t3/2t2m−1

[

i(

2

πκt

)1/2

e−iκt

]

dt

= − i2κ

eiκr

r

⎡

⎣e−iκt

⎡

⎣2m∑

j=0

(−1) j(2m)!t2m− j

(2m− j )!(−iκ) j+1

⎤

⎦

⎤

⎦

r

0

=− i2κ

1

r

⎡

⎣2m∑

j=0

(−1) j(2m)!r2m−j

(2m − j )!(−iκ) j+1

⎤

⎦

+ i2κ

eiκr

r(2m)!

(−iκ)2m+1= 1

2κr

2m∑

j=0

(−i ) j(2m)!r2m− j

(2m − j )!κ j+1+ c1

eiκr

r

where c1 is a constant. Similarly ψ2 in (38) can be calculated and is given by

ψ2(r) = 1

2κr

2m∑

j=0

i j(2m)!r2m− j

(2m − j )!κ j+1+ c2

e−iκr

r

for some constant c2. From (36), a particular solution is given by

ψ(r) = 1

r

m∑

k=0

(−1)k(2m)!r2(m−k)

(2(m − k))!κ2k+2+ c1

eiκr

r+ c2

e−iκr

r,

which includes the results in [7] for R3 (cf. p. 416, [7]), where κ = iλ or κ = λ, λ > 0, isconsidered. In [7] annihilator methods are used and the particular solutions for thin

230 X. Li

plate splines in R2 are also derived. Indeed, for a thin plate spline φ(r) = r2m ln r inR2, similar results in [7] can be derived by Proposition 1 and the computation in Ex-ample 2 by evaluating the integrals

∫t2m+1(ln t)H(1)

0 (κt)dt and∫

t2m+1(ln t)H(2)0 (κt)dt

for ψ in (36–38) through the formula for integration by parts and using (124) and(125), which it is not necessary to present all the details here. In [3], by choosingκ = iλ and φ to be the Wendland compactly supported RBFs, the particular solutionsof Helmholtz equations in R3 are derived by different methods, which again the sameresults can be derived from Proposition 1 and similar computations in Example 1.

Once the particular solutions of Helmholtz equations are available, the boundaryelement methods, or certain mesh free methods such as the method of fundamentalsolutions (cf. [6], etc.), can be used to solve the corresponding boundary valueproblems, which will be explored in our future work.

References

1. Chen, G., Zhou, J.: Boundary Element Methods. Academic Press, New York (1992)2. Colton, D., Kress, R.: Inverse Acoustic and Electromagnetic Scattering Theory. Springer-Verlag,

New York (1992)3. Golberg, M.A., Chen, C.S., Ganesh, M.: Particular solutions of 3D Helmholtz-type equations using

compactly suported radial basis functions. Eng. Anal. Bound. Elem. 24, 539–547 (2000)4. Li, X., Micchelli, C.A.: Approximation by radial bases and neural networks. Numer. Algorithms

25, 241–262 (2000)5. Li, X.: Radial basis approximation for Newtonian potentials. (preprint)6. Li, X.: Rate of convergence of the method of fundamental solutions and hyperinterpolation for

modified Helmholtz equations on the unit ball. Adv. Comput. Math. (in press)7. Muleshkov, A.S., Golberg, M.A., Chen, C.S.: Particular solutions of Helmholtz-type operators

using higher order polyharmonic splines. Comput. Mech. 23, 411–419 (1999)8. Watson, G.N.: A Treatise on the Theory of Bessel Functions. Cambridge University Press, London

(1944)

approximation of potential integral by radial bases for solutions of helmholtz equation

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