Properties of Boubaker polynomials and an application to Love's ...

Applied Mathematics and Computation 224 (2013) 74–87

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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc

Properties of Boubaker polynomials and an application to Love’s integral equation q Gradimir V. Milovanovic´ a,⇑, Dušan Joksimovic´ b,* a b

Mathematical Institute of the Serbian Academy of Sciences and Arts, Knez Mihailova 36, p.p. 367, 11001 Beograd, Serbia Graduate School of Business Studies, Megatrend University, Goce Delcˇeva 8, 11070 Belgrade, Serbia

a r t i c l e

i n f o

a b s t r a c t The paper deals with three-term recurrence relations for Boubaker and related polynomials, as well as some properties including zero distribution of such kinds of polynomials. Also, an application of these polynomials for obtaining approximate analytical solution of Love’s integral equation is presented. This Fredholm integral equation of the second kind appeared in an electrostatic problem analyzed for the first time by Love (1949) [13]. ! 2013 Elsevier Inc. All rights reserved.

Keywords: Chebyshev polynomials Boubaker polynomials Zeros Recurrence relation Fredholm integral equation Love’s integral equation

1. Introduction There are several papers on the so-called Boubaker polynomials and on their applications in different problems arising in physics and other computational and applied sciences (cf. [2,3,9,26]). These polynomials fBn ðxÞg have a close relationship with the Chebyshev polynomials offfi the first andffi second kind, T n ðxÞ and U n ðxÞ, which are orthogonal on ð#1; 1Þ with respect pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi to the weights functions 1= 1 # x2 and 1 # x2 , respectively. Solutions to several applied physics problems are based on the so-called Boubaker Polynomials Expansion Scheme (BPES), using only the subsequence fB4m ðxÞg of these polynomials (cf. [26] and references therein). Such polynomials satisfy the relation (cf. [9])

B4ðmþ1Þ ðxÞ ¼ ðx4 # 4x2 þ 2ÞB4m ðxÞ # bm B4ðm#1Þ ðxÞ;

m P 1;

ð1:1Þ

with B0 ðxÞ ¼ 1 and B4 ðxÞ ¼ x4 # 2, where b0 ¼ 0; b1 ¼ #2 and bm ¼ 1 for m P 2. Recently, Kumar [11] has presented a method for obtaining an analytical solution of Love’s integral equation (see [13,14]), with a positive parameter d,

f ðxÞ #

1

p

Z

1

#1

d 2

d þ ðx # yÞ2

f ðyÞ dy ¼ 1;

#1 6 x 6 1

ð1:2Þ

for a particular electrostatical system, based on the Boubaker polynomials expansion scheme (BPES). However, a mistake has appeared in his approach. The main goal of this paper is a correction of Kumar’s approach, as well as the proof of most important properties of the Boubaker polynomials, including the zero distribution. The paper is organized as follows.

q The authors were supported in part by the Serbian Ministry of Education, Science and Technological Development (No. #OI 174015 and No. #III 44006). A preliminary short form of some results of this paper was presented at the International Conference on Numerical Analysis and Applied Mathematics 2012 (Kos, Greece, September 19–25, 2012) and published as an extended abstract [17]. ⇑ Corresponding authors. E-mail addresses: [email protected] (G.V. Milovanovic´), [email protected] (D. Joksimovic´).

0096-3003/$ - see front matter ! 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.amc.2013.08.055

G.V. Milovanovic´, D. Joksimovic´ / Applied Mathematics and Computation 224 (2013) 74–87

75

In Section 2 some properties of the polynomials Bn ðxÞ and certain related polynomials (the three-term recurrence relation, relations with Chebyshev polynomials, properties of zeros, etc.) are presented. The proofs of statements are given in Section 3. Finally, in Section 4 an application of these polynomials Bn ðxÞ in solving Love’s integral equation is given, which appears in an electrostatic problem. 2. Properties of Boubaker polynomials The well-known Chebyshev polynomials of the first and second kind for x 2 ð#1; 1Þ are defined by (cf. [15, p. 6])

sin ððn þ 1Þ arccos xÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 1 # x2

T n ðxÞ ¼ cosðn arccos xÞ and U n ðxÞ ¼ respectively. Their explicit expressions are

T 0 ðxÞ ¼ 1;

T n ðxÞ ¼

and

U n ðxÞ ¼

n X ð#1Þk ðn # k # 1Þ! ð2xÞn#2k ; 2 k¼0 k!ðn # 2kÞ! ½n=2'

ð#1Þk ðn # kÞ! ð2xÞn#2k ; k!ðn # 2kÞ! k¼0

½n=2' X

n(1

n(0

and they satisfy the same three-term recurrence relation (cf. [15, p. 9]), i.e.,

T nþ1 ðxÞ ¼ 2xT n ðxÞ # T n#1 ðxÞ;

U nþ1 ðxÞ ¼ 2xU n ðxÞ # U n#1 ðxÞ;

n P 1;

with T 0 ðxÞ ¼ 1; T 1 ðxÞ ¼ x and U 0 ðxÞ ¼ 1; U 1 ðxÞ ¼ 2x. In a similar way, the monic Boubaker polynomials are defined as (cf. [2,3,9,26])

Bn ðxÞ ¼

" # ½n=2' X n # k n # 4k n#2k ð#1Þk x ; n#k k k¼0

n(1

ð2:1Þ

and B0 ðxÞ ¼ 1. Polynomials of even and odd orders are even and odd functions, respectively, i.e., Bn ð#xÞ ¼ ð#1Þn Bn ðxÞ; n 2 N0 . For example, for n 6 11 we have

B1 ðxÞ ¼ x;

5

B2 ðxÞ ¼ x2 þ 2; 3

B5 ðxÞ ¼ x # x # 3x;

B3 ðxÞ ¼ x3 þ x; 6

B10 ðxÞ ¼ x

8

2

B4 ðxÞ ¼ x4 # 2;

B6 ðxÞ ¼ x # 2x # 3x þ 2;

B8 ðxÞ ¼ x8 # 4x6 þ 8x2 # 2; 10

4

B7 ðxÞ ¼ x7 # 3x5 # 2x3 þ 5x;

B9 ðxÞ ¼ x9 # 5x7 þ 3x5 þ 10x3 # 7x;

6

# 6x þ 7x þ 10x4 # 15x2 þ 2;

B11 ðxÞ ¼ x11 # 7x9 þ 12x7 þ 7x5 # 25x3 þ 9x: In the sequel we give some of most important properties of these polynomials. Some of them are known. 2.1. Three-term recurrence relations and connections with Chebyshev polynomials The polynomials Bn ðxÞ can be alternatively represented by a three-term recurrence relation

Bnþ1 ðxÞ ¼ xBn ðxÞ # Bn#1 ðxÞ;

n ¼ 2; 3; . . . ; 2

ð2:2Þ

where B0 ðxÞ ¼ 1; B1 ðxÞ ¼ x; B2 ðxÞ ¼ x þ 2. As we can see, the relation (2.2) is not true for n ¼ 1. In order to provide a relation for each n 2 N, we can define a sequence fbk gk2N0 by b0 ¼ 0; b1 ¼ #2 and bk ¼ 1 for k P 2. Then, we have the three-term recurrence relation in the following form

Bkþ1 ðxÞ ¼ xBk ðxÞ # bk Bk#1 ðxÞ;

k ¼ 0; 1; . . . ;

ð2:3Þ

with B0 ðxÞ ¼ 1, B#1 ðxÞ ¼ 0. The polynomials (2.1) can be expressed in terms of Chebyshev polynomials of the first and second kind, T n ðxÞ and U n ðxÞ. Theorem 2.1. For n P 1 the following formulas

Bn ðxÞ ¼ 2T n ðx=2Þ þ 4U n#2 ðx=2Þ

ð2:4Þ

Bn ðxÞ ¼ U n ðx=2Þ þ 3U n#2 ðx=2Þ

ð2:5Þ

and


76

hold, where U #1 ðxÞ ) 0. Eliminating U n#2 ðx=2Þ from (2.4) and (2.5), we get

Bn ðxÞ ¼ 4U n ðx=2Þ # 6T n ðx=2Þ:

Now, according to Eq. (1.1.9) in [15, p. 6], we obtain the expansions of Bn ðxÞ in terms of Chebyshev polynomials of the first kind,

B2n ðxÞ ¼ 2T 2n ðx=2Þ þ 8T 2n#2 ðx=2Þ þ 8T 2n#4 ðx=2Þ þ * * * þ 8T 2 ðx=2Þ þ 4T 0 ðx=2Þ and

B2nþ1 ðxÞ ¼ 2T 2nþ1 ðx=2Þ þ 8T 2n#1 ðx=2Þ þ 8T 2n#3 ðx=2Þ þ * * * þ 8T 3 ðx=2Þ þ 8T 1 ðx=2Þ: In the following we recall a property of orthogonal polynomials on a symmetric interval ð#a; aÞ; a > 0, satisfing a threeterm recurrence relation of the form (2.3). For such kind of polynomials, which are orthogonal with respect to an even weight function on ð#a; aÞ, two new polynomial systems orthogonal on the interval ð0; a2 Þ can be defined (cf. [15, pp. 102–103]). In a quite similar way, we can also introduce here two new (nonorthogonal) systems of monic polynomials fpm ðtÞg and fqm ðtÞg via Boubaker polynomials Bm ðxÞ, so that

B2m ðxÞ ¼ pm ðx2 Þ and B2mþ1 ðxÞ ¼ xqm ðx2 Þ

ð2:6Þ

and prove the following result: Theorem 2.2. Let bm ; m ( 1, be recursive coefficients in the recurrence relation (2.3). Then,

pmþ1 ðtÞ ¼ ðt # am Þpm ðtÞ # bm pm#1 ðtÞ and qmþ1 ðtÞ ¼ ðt # cm Þqm ðtÞ # dm qm#1 ðtÞ;

ð2:7Þ

with p0 ðtÞ ¼ q0 ðtÞ ¼ 1; p#1 ðtÞ ¼ q#1 ðtÞ ¼ 0, where the recursive coefficients are given by

am ¼ b2m þ b2mþ1 ¼

$

#2; m ¼ 0; 2;

m ( 1;

bm ¼ b2m b2m#1 ¼

$

#2; m ¼ 1; 1;

m(2

and

$

cm ¼ b2mþ1 þ b2mþ2 ¼

#1; m ¼ 0; 2;

m ( 1;

dm ¼ b2m b2mþ1 ¼ 1; m ( 1:

2.2. Determinatal form of polynomials and distribution of zeros Using the recurrence relation (2.3) for k ¼ 0; 1; . . . ; n # 1 and defining the n-dimensional vector bn ðxÞ by

bn ðxÞ ¼ ½B0 ðxÞ B1 ðxÞ . . . Bn#1 ðxÞ'T ;

we obtain the equation

ðxIn # Mn Þbn ðxÞ ¼ Bn ðxÞen ;

ð2:8Þ

T

where In is the identity matrix of order n; en ¼ ½0 0 . . . 0 1' is the last coordinate vector, and M n is a tridiagonal matrix of order n, given by

2

0 6b 6 1 6 6 Mn ¼ 6 6 6 6 4 O

1 0 b2

1 ..

0 . .. .. . . bn#1

3

2 0 1 7 6 7 6 #2 0 1 7 6 7 6 .. 7¼6 1 0 . 7 6 7 6 .. .. 7 4 . . 15

O

0

O

O

1

3

7 7 7 7 7: 7 7 15 0

According to (2.8) the monic polynomials Bn ðxÞ can be expressed in the following determinant form

Bn ðxÞ ¼ detðxIn # Mn Þ;

n P 1:

It is a form which is well known in the theory of orthogonal polynomials (cf. [15, p. 100]). We also conclude that the zeros of the polynomial Bn ðxÞ are eigenvalues of the matrix M n . Now we apply a result of Veselic´ [24] to the irreducible tridiagonal matrix Mn , for which we can form, according to [24, Eq. (3)], the following sequence

1; #2; #2; . . . ; #2 : |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} n#1


77

Since the numbers of positive and negative signs in this sequence are kþ ¼ 1 and k# ¼ n # 1, respectively, and k ¼ minðkþ ; k# Þ ¼ 1, we conclude that M n for n P 3 has at least jkþ # k# j ¼ n # 2 different real spectral points. Furthermore, using Gerschgorin’s theorem, it is easy to see that all eigenvalues are in the unit circle jzj < 2 (see also [26]). Thus, Bn ðxÞ; n ( 2, can have only one pair of conjugate-complex zeros. Precisely, the following result holds: pffiffiffiffiffi Theorem 2.3. Every polynomial Bn ðxÞ; n P 2, has two complex conjugate zeros +i cn , cn > 0, and other zeros are real and symmetrically distributed in ð#2; 2Þ, where limn!þ1 cn ¼ 4=3. The corresponding representations of these polynomials for m ( 2 are

B2m ðxÞ ¼ ðx2 þ c2m Þ

m #1 Y

m¼1

ðx2 # s2m;m Þ;

B2mþ1 ðxÞ ¼ xðx2 þ c2mþ1 Þ

where 4 > sn;1 > * * * > sn;m#1 > 0 and n ¼ 2m or n ¼ 2m þ 1.

m #1 Y

m¼1

ðx2 # s2mþ1;m Þ;

ð2:9Þ

Remark 2.1. Theorem 2.3 on zero distribution of Bn ðxÞ has been mentioned in [17, Theorem 4]. Recently, a similar result has appeared on arXiv:1211.0383 (see [10]). The corresponding tridiagonal matrices (of Jacobi type) of order m for the polynomial sequences fpm ðtÞg and fqm ðtÞg are

3 3 2 #2 1 O #1 1 O 7 7 6 #2 2 1 6 #1 2 1 7 7 6 6 7 7 6 6 . . 7 7 6 6 q .. .. J pm ¼ 6 1 2 1 2 7 and J m ¼ 6 7; 7 7 6 6 . . . . 7 7 6 6 . . . . 4 4 . . . 15 . 15 2

O

1

2

O

1

2

respectively. Otherwise, these polynomial systems fpm ðtÞg and fqm ðtÞg consist of the following polynomials and

& ' 1; t þ 2; t 2 # 2; t3 # 2t 2 # 3t þ 2; t4 # 4t 3 þ 8t # 2; t5 # 6t 4 þ 7t 3 þ 10t2 # 15t þ 2; . . .

& ' 1; t þ 1; t 2 # t # 3; t 3 # 3t2 # 2t þ 5; t 4 # 5t3 þ 3t 2 þ 10t # 7; t 5 # 7t 4 þ 12t3 þ 7t 2 # 25t þ 9; . . . ;

respectively. In order to investigate zeros of the polynomials Bn ðzÞ on the imaginary axis we put z ¼ iy and consider Bn ðiyÞ=in ; n P 2, i.e., the sequence of polynomials

y2 # 2; 6

yðy2 # 1Þ; 4

2

y4 # 2;

þ 5y þ 3y # 10y # 7Þ; etc:

yðy4 þ y2 # 3Þ;

y6 þ 2y4 # 3y2 # 2; yðy6 þ 3y4 # 2y2 # 5Þ;

y8 þ 4y6 # 8y2 # 2;

yðy8

For t > 0 we define two sequences of polynomials em ðtÞ and om ðtÞ, m ¼ 1; 2; . . ., by

pffiffi pffiffi B2mþ1 ði t Þ pffiffi ; em ðtÞ ¼ ð#1Þm B2m ði tÞ and om ðtÞ ¼ ð#1Þm i t

ð2:10Þ

respectively. According to (2.1) and Theorem 2.2, it is clear that

em ðtÞ ¼ ð#1Þm pm ð#tÞ ¼

# m " X 2m # k 2m # 4k

om ðtÞ ¼ ð#1Þm qm ð#tÞ ¼

# m " X 2m # k 2m # 4k þ 1

and

k¼0

k¼0

k

k

2m # k

tm#k

2m # 2k þ 1

ð2:11Þ

t m#k :

ð2:12Þ

Theorem 2.4. For any m 2 N the polynomials em ðtÞ and om ðtÞ have only one positive zero. In Fig. 2.1 we present the graphs of polynomials em ðtÞ (left) and om ðtÞ (right) for t 2 ½0; 2' and m ¼ 2; 3; 4, and 5. Notice that em ð0Þ ¼ #2 and om ð0Þ ¼ #ð2m # 1Þ, and the unique positive zero nm of em ðtÞ (and also gm of om ðtÞ) belongs to ð1; 3=2Þ. Also, ðmÞ ðmÞ ðmÞ derivatives em ðtÞ and om ðtÞ; m ¼ 1; 2; . . . (of course for a sufficiently large m) have the unique positive zeros nðmmÞ and gm , respectively, for which it is easy to see that ðmÞ nðmmþ1Þ < nm < nm

ðmþ1Þ and gm < gðmmÞ < gm :

ð2:13Þ


78

30

15

20

10

10

5 0.5

1.0

1.5

0.5

2.0

5

1.0

1.5

2.0

10

10

20

Fig. 2.1. The graphs of polynomials em ðtÞ (left) and om ðtÞ (right) for m ¼ 2 (black line), m ¼ 3 (blue line), m ¼ 4 (red line) and m ¼ 5 (brown line). (For interpretation of the references to colour in this figure caption, the reader is referred to the web version of this article.)

Remark 2.2. Other zeros of em ðtÞ and om ðtÞ are also real, but negative. In the sequel we investigate the exact positions of these zeros and give their asymptotics when m ! þ1. First we need some auxiliary results.

Lemma 2.5. The values of polynomials em ðtÞ and om ðtÞ, as well as their derivatives of the first and second order at t ¼ 4=3 are

em ð4=3Þ ¼ #

2 ; 3m

e0m ð4=3Þ ¼

i 3 h mþ1 3 þ ð8m # 3Þ3#m ; 32

e00m ð4=3Þ ¼

i 9 h ð8m # 7Þ3mþ1 # ð32m2 þ 16m # 21Þ3#m 1024

om ð4=3Þ ¼

1 ; 3m

o00m ð4=3Þ ¼

i 9 h ð8m # 11Þ3mþ2 þ ð32m2 þ 112m þ 99Þ3#m ; 2048

and

respectively.

o0m ð4=3Þ ¼

i 3 h mþ2 3 # ð8m þ 9Þ3#m ; 64

Remark 2.3. Some interesting finite sums can be obtained from (2.11), (2.12), and Lemma 2.5. For example, we have m mþk X k¼0

2k

!

! " #k X m m þ k 4k # 2m þ 1 "4#k m # 2k 4 1 ¼ ¼ m; mþk 3 2k þ 1 3 3 2k k¼0

m mþk#1 X k¼1

2k # 1

!

ð2k # mÞ

" #k i 4 1h ¼ 3mþ1 þ ð8m # 3Þ3#m ; etc: 3 8

The following result is related with positive zeros nm and gm of the polynomials em ðtÞ and om ðtÞ, respectively. Theorem 2.6. For each m P 1, the following inequalities

0 < nm #

4 64 1 64 < * < mþ1 3 9mþ1 1 þ 8m#3 9 2mþ1

ð2:14Þ

3

and

4 2m # 1 4 64 1 1 * < gm < # * * 3 2m # 1 þ 3#m 3 3 9mþ1 1 # 8mþ9 mþ1 9

hold.

ð2:15Þ


79

3. Proofs

Proof of Theorem 2.1. First equality (2.4), which has been known earlier, can be proved, for example, by the mathematical induction, or by solving (2.2), i.e., ynþ1 # xyn þ yn#1 ¼ 0; n P 1, as a difference equation with respect to n, supposing that x is a fixed number in ð#2; 2Þ, for example, x ¼ 2 cos h. Then, the roots of the characteristic equation are e+ih , so that the general solution of this equation is yn ¼ C 1 cos nh þ C 2 sin nh. Taking the starting values

y1 ¼ B1 ðxÞ ¼ x ¼ 2 cos h and y2 ¼ B2 ðxÞ ¼ x2 þ 2 ¼ 2 cos 2h þ 4;

we get C 1 ¼ #2 and C 2 ¼ 4 cot h, so that

Bn ðxÞ ¼ #2 cos nh þ 4

cos h sin nh sinðn # 1Þh ¼ 2 cos nh þ 4 ; sin h sin h

i.e., Bn ðxÞ ¼ 2T n ðx=2Þ þ 4U n#2 ðx=2Þ. Since 2T n ðxÞ ¼ U n ðxÞ # U n#2 ðxÞ, (2.5) follows directly from (2.4).

h

Proof of Theorem 2.2. According to (2.3) we have

B2mþ3 ðxÞ ¼ xB2mþ2 ðxÞ # b2mþ2 B2mþ1 ðxÞ and B2mþ2 ðxÞ ¼ xB2mþ1 ðxÞ # b2mþ1 B2m ðxÞ; or, using (2.6),

xqmþ1 ðx2 Þ ¼ xpmþ1 ðx2 Þ # b2mþ2 xqm ðx2 Þ and pmþ1 ðx2 Þ ¼ x2 qm ðx2 Þ # b2mþ1 pm ðx2 Þ: Putting x2 ¼ t, it is easy to see that

pmþ1 ðtÞ þ b2mþ1 pm ðtÞ ¼ tqm ðtÞ

ð3:1Þ

qmþ1 ðtÞ þ b2mþ2 qm ðtÞ ¼ pmþ1 ðtÞ:

ð3:2Þ

and

If we replace k by k # 1 in (3.2), multiply it by t, and finally add it to the relation (3.1), we obtain

pmþ1 ðtÞ þ ðb2mþ1 # tÞpm ðtÞ þ b2m tqm#1 ðtÞ ¼ 0:

ð3:3Þ

In a similar way, taking k instead of k in (3.1) and combining it with (3.3) we get

pmþ1 ðtÞ þ ðb2m þ b2mþ1 # tÞpm ðtÞ þ b2m b2m#1 pm#1 ðtÞ ¼ 0; i.e., the first relation in (2.7). In a similar way we obtain the corresponding recurrence relation for polynomials fqm g.

!

Proof of Theorem 2.4. In the proof we use the number of sign variations (differences) between consecutive nonzero coefficients of a polynomial ordered by descending variable exponent. We note that the coefficients in (2.11) are positive for k < m=2 and negative for k > m=2, so that we have only one sign variation. According to Descartes’ Rule the number of positive zeros is either equal to the number of sign differences between consecutive nonzero coefficients, or lower than it by a multiple of 2. Since em ð0Þ ¼ #2 < 0 and em ðTÞ > 0 for each sufficiently large positive T, we conclude that em ðtÞ has only one positive zero. A similar proof can be done for polynomials om ðtÞ. ! Proof of Lemma 2.5. According to (2.7) and (2.11), i.e., (2.12), we have the following difference equations

emþ1 ðtÞ # ðt þ 2Þem ðtÞ þ em#1 ðtÞ ¼ 0 and omþ1 ðtÞ # ðt þ 2Þom ðtÞ þ om#1 ðtÞ ¼ 0; as well as the ones for derivatives ðmÞ

ðmÞ

ðmÞ

ðmÞ

emþ1 ðtÞ # ðt þ 2ÞeðmmÞ ðtÞ þ em#1 ðtÞ ¼ meðmm#1Þ ðtÞ and ðmÞ ðm#1Þ omþ1 ðtÞ # ðt þ 2Þom ðtÞ þ om#1 ðtÞ ¼ mom ðtÞ;

where m ¼ 1; 2; . . . . In particular, we are interested only in solutions of these difference equations for t ¼ 4=3. Since the characteristic equation k2 # ð10=4Þk þ 1 ¼ 0 (with roots k1 ¼ 3 and k2 ¼ 1=3) is the same for each of these equations, we obtain the general solutions

em ð4=3Þ ¼ C 1 3m þ C 2 3#m

and om ð4=3Þ ¼ D1 3m þ D2 3#m ;


80

where C 1 ; C 2 ; D1 ; D2 are arbitrary constants. Taking the starting values e1 ð4=3Þ ¼ #2=3; e2 ð4=3Þ ¼ #2=9; o1 ð4=3Þ ¼ 1=3; o2 ð4=3Þ ¼ 1=9, we get C 1 ¼ D1 ¼ 0; C 2 ¼ #2, and D2 ¼ 1, so that

em ð4=3Þ ¼ # For

2 3m

1 ; 3m

and om ð4=3Þ ¼

m P 1:

ð3:4Þ

m ¼ 1, the corresponding difference equations are e0mþ1 ð4=3Þ #

10 0 2 e ð4=3Þ þ e0m#1 ð4=3Þ ¼ # m 4 m 3

o0mþ1 ð4=3Þ #

10 0 1 o ð4=3Þ þ o0m#1 ð4=3Þ ¼ m 4 m 3

and

and then, with starting values e01 ð4=3Þ ¼ 1; e02 ð4=3Þ ¼ 8=3; e03 ð4=3Þ ¼ 23=3 and o01 ð4=3Þ ¼ 1; o02 ð4=3Þ ¼ 11=3; o03 ð4=3Þ ¼ 34=3, we obtain

e0m ð4=3Þ ¼

i i 3 h mþ1 3 h mþ2 3 and o0m ð4=3Þ ¼ 3 þ ð8m # 3Þ3#m # ð8m þ 9Þ3#m : 32 64

In a similar way, with starting values e001 ð4=3Þ ¼ 0; e002 ð4=3Þ ¼ 2; e003 ð4=3Þ ¼ 12; e004 ð4=3Þ ¼ 160=3 and 00 00 00 00 o1 ð4=3Þ ¼ 0; o2 ð4=3Þ ¼ 2; o3 ð4=3Þ ¼ 14; o4 ð4=3Þ ¼ 202=3, we obtain the solutions of the following difference equations

e00mþ1 ð4=3Þ #

10 00 9 m 24m # 9 #m e ð4=3Þ þ e00m#1 ð4=3Þ ¼ 3 þ 3 4 m 16 16

o00mþ1 ð4=3Þ #

10 00 27 m 24m þ 27 #m o ð4=3Þ þ o00m#1 ð4=3Þ ¼ 3 # 3 ; 4 m 32 32

and

in the form

e00m ð4=3Þ ¼

i 9 h ð8m # 7Þ3mþ1 # ð32m2 þ 16m # 21Þ3#m 1024

o00m ð4=3Þ ¼

i 9 h ð8m # 11Þ3mþ2 þ ð32m2 þ 112m þ 99Þ3#m ; 2048

and

respectively.

!

ð1Þ Proof of Theorem 2.6. Since em ð4=3Þ ¼ #2=3m < 0 we have that nm > 4=3. Also, because of nð2Þ m < nm < nm (see (2.13)) and e00m ð4=3Þ P 0; m P 1, it is clear that e00m ðtÞ P e00m ð4=3Þ > e00m ðnð2Þ Þ ¼ 0 for t P 4=3. Applying Taylor’s formula m

" # " #2 4 1 4 em ðtÞ ¼ em ð4=3Þ þ e0m ð4=3Þ t # þ e00m ðsÞ t # ; 3 2 3

4 < s < t; 3

for t ¼ nm , we conclude that

" # 4 0 ¼ em ðnm Þ > em ð4=3Þ þ e0m ð4=3Þ nm # ; 3

i.e.,

0 < nm #

4 em ð4=3Þ 0. Under similar arguments, ð1Þ ð2Þ 00 00 00 gð2Þ m < gm < gm < 4=3 and om ð4=3Þ P om ðgm Þ > om ðgm Þ ¼ 0, an application of Taylor’s formula gives

" # " #2 4 1 4 0 ¼ om ðgm Þ ¼ om ð4=3Þ þ o0m ð4=3Þ gm # ; þ o00m ðsÞ gm # 3 2 3

i.e.,

4 om ð4=3Þ # gm > 0 ¼ 3 om ð4=3Þ

3 64

h

1=3m 3

mþ2

# ð8m þ 9Þ3

#m

i¼

64 1 1 * * : 3 9mþ1 1 # 8mþ9 mþ1 9

4 3

gm < s < ;


81

On the other side, using a secant method at the points t ¼ 0 and t ¼ 4=3, we obtain a lower bound for gm . Namely, the line between the points ð0; #ð2m # 1ÞÞ and ð4=3; 1=3m Þ crosses the apscisa at

b xm ¼

4 2m # 1 * ; 3 2m # 1 þ 3#m

so that b x m < gm . Combining it with the previous upper bound we obtain (2.15). ! Finally, we can prove the result on the zero distribution of polynomials Bn ðxÞ. Proof of Theorem 2.3. Let nm and gm be unique positive zeros of em ðtÞ and om ðtÞ, respectively. Then, according to (2.10), we pffiffiffiffiffiffi pffiffiffiffiffiffi conclude that i nm (but also #i nm ) is an imaginary zero of B2m ðxÞ, so that c2m in (2.9) must be c2m ¼ nm . In a quite similar way, we prove that c2mþ1 ¼ gm . The other m # 1 negative zeros of em ðtÞ (see Remark 2.1) generate 2m # 2 real zeros of B2m ðxÞ symmetrically distributed on ð#2; 2Þ. Similarly, negative zeros of om ðtÞ generate 2m # 2 real zeros of B2mþ1 ðxÞ symmetrically distributed also on ð#2; 2Þ. Thus, the representations (2.9) hold. Using inequalities (2.14) and (2.15) from Theorem 2.6, we have

lim gm ¼ lim nm ¼

m!þ1

m!þ1

so that limn!þ1 cn ¼ 4=3.

4 ; 3

ð3:5Þ

!

4. Applications As we mentioned in Section 1, the polynomials fB4m ðxÞg play an important role in applications. These polynomials satisfy the relation (1.1). However, if we need all even polynomials B2m ðxÞ, then the corresponding recurrence relation is

B2mþ2 ðxÞ ¼ ðx2 # am ÞB2m ðxÞ # bm B2m#2 ðxÞ;

m P 0;

ð4:1Þ

where am and bm are recursive coefficients defined in Theorem 2.2. Recently, Kumar [11] has presented a method for obtaining an analytical solution of Love’s integral equation (see [13,14])

f ðxÞ #

1

p

Z

1

#1

d 2

d þ ðx # yÞ2

f ðyÞ dy ¼ 1;

#1 6 x 6 1;

ð4:2Þ

for a particular electrostatical system, based on the Boubaker polynomials expansion scheme (BPES). However, a mistake has appeared in his approach. Our goal in this section is to correct this Kumar’s approach and give a much better approximation of the solution of Love’s integral equation. In 1949 Love (1912–2001) described the electrostatic potential in space, generated by a condenser consisting of two parallel equal circular plates of the radius R, separated by a distance h (see Fig. 4.1). Taking a normalization so that h ¼ Rd, it can be considered with dimensionless variables as two unit disks, where d is a distance between them. Supposing the equal and opposite potentials at these disks, e.g., the upper at V ¼ þ1 and the lower one at V ¼ #1, and the potential at infinity being taken as zero, Love [13, Theorem 1] used a coaxial symmetry of this electrostatical system and proved that the potential in an arbitrary point Mðr; h; zÞ 2 R3 , outside the circular plates, is given by

Fig. 4.1. Electrostatical system od two parallel equal circular plates.


82

Vðr; zÞ ¼

8 9 > > < = 1 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q ffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi # ( ) ( )2 >f ðxÞ dx; #1 > : r 2 þ z # 1 d þ ix 2 r2 þ z þ 12 d þ ix ; 2

Z

1

p

1

ð4:3Þ

where both square roots in this integral have positive real part and f is the unique solution of the Fredholm integral equation of the second kind (4.2). He also proved that there exists a unique, continuous, real and even solution f of the Eq. (4.2) on the closed interval ½#1; 1'. In the case when the potentials of the plates are equal in magnitude and sign, then the corresponding integral equation (4.2) becomes

f ðxÞ þ

Z

1

p

1

#1

d 2

d þ ðx # yÞ2

#1 6 x 6 1:

f ðyÞ dy ¼ 1;

ð4:4Þ

Also, in that case the # sign between two terms on the right-hand side in (4.3) becomes a þ sign. Recently, Norgren and Jonsson [19] have calculated the capacitance of the circular parallel plate capacitor by expanding the solution to the Love integral equation (4.2) into a Fourier cosine series. For some other approaches see [8,5,25,4,22]. Love’s integral equations have the so-called difference kernel

kðx; yÞ ¼ kðx # yÞ ¼

1

d

*

p d2 þ ðx # yÞ2

;

d > 0;

ð4:5Þ

which has two complex conjugate poles x + id. We can see these poles approach the real axis when d ! 0þ , and therefore the kernel is quasi-singular. Letting

1

ðKf ÞðxÞ ¼

p

Z

1

#1

d 2

d þ ðx # yÞ2

f ðyÞ dy;

the operator form of Love’s integral Eqs. (4.2) and (4.4) are

f , Kf ¼ ðI , KÞf ¼ g;

ð4:6Þ

where I denotes the identity operator, and K is compact with (cf. [18]).

kKk1 ¼

2

p

tan#1

1 < 1: d

There are many numerical methods for solving integral equations (cf. [1,6,7,12,21,23]). Sometimes, they are developed for specific type of kernels. Numerical methods for linear integral equations of the form (4.6) lead to algebraic systems of linear equations and sometimes the conditional number of the corresponding matrices are large. The solution of an integral equation can be done in a polynomial form, as a peacewise polynomial, spline, etc. An approximation to the solution of (4.2), in the case d ¼ 1, was given by Love [14],

f ðxÞ - fL ðxÞ ¼ 1:919200 # 0:311717x2 þ 0:015676x4 þ 0:019682x6 # 0:000373x8 :

ð4:7Þ

In this section we give a correction and extensions of Kumar’s method [11], using Boubaker polynomials. In this approach we need to compute the integrals

J 2m ðx; dÞ ¼ ðKB2m ÞðxÞ ¼

1

p

Z

1

#1

d 2

d þ ðx # yÞ2

B2m ðyÞ dy

for different values of d and x, which can be done numerically for specific values of these parameters. But, we give here an analytic (symbolic) form, using the recurrence relation (4.1). In fact, these integrals J 2m are moments of the (weight) function y # kðx; yÞ, given by (4.5), with respect to the Boubaker polynomial sequence. For J 2m we can prove the following recurrence relation: Lemma 4.1. We have 2

2

J 2mþ2 þ ðd þ am # x ÞJ 2m þ bm J 2m#2 ¼

2d

p

I2m þ

xd

p

(

B2m ð1Þ log

2

d þ ð1 # xÞ2 2

d þ ð1 # xÞ2

)

þ K 2m ;

ð4:8Þ

where

* " # " #+ 1#x xþ1 tan#1 þ tan#1 ; J 2 ðx;dÞ d d p * " # " # , " # " #+ , 1 xþ1 2x 2 #1 2 #1 x # 1 2 #1 1 # x # 2xd tanh þ d þ x ¼ 2d þ 2 # d þ x2 tan#1 # 2 tan tan 2 d d d p d þ x2 þ 1

J 0 ðx;dÞ ¼

and

1


I2m ¼

Z

1

0

B2m ðyÞ dy;

K 2m ¼ K 2m ðx; dÞ ¼

Z

1

0

83

2

log

d þ ðx þ yÞ2 2

d þ ðx # yÞ2

B02m ðyÞ dy:

ð4:9Þ

The value of the first integral in (4.9) can be done in the form

I2m ¼

( ) ( ) 6 sin p6 ð4m þ 1Þ 2 cos p3 ð2m þ 1Þ þ 2m # 1 2m þ 1

and the second integral can be expressed as a linear combination of the integrals

Sk ¼ Sk ðx; dÞ ¼

Z

2

1

d þ ðx þ yÞ2

log

2

d þ ðx # yÞ2

0

y2k#1 dy;

k 2 N:

Their values are

"

# " " # " ## 2 1 d þ ð1 þ xÞ2 #1 1 þ x #1 #1 þ x þ Q 2k ðx; dÞ log 2 þ R ðx; dÞ tan # tan ; 2k#1 2k d d d þ ð#1 þ xÞ2

Sk ðx; dÞ ¼ P2k#1 ðx; dÞ þ

where P2k#1 ðx; dÞ and R2k#1 ðx; dÞ are odd polynomials of degree 2k # 1 in x, and R2k ðx; dÞ is an even polynomial of degree 2k in x. Moreover, their forms are

! k k#m X X j ðkÞ 2j P2k#1 ðx; dÞ ¼ ð#1Þ am;j d x2m#1 ; m¼1

Q 2k ðx; dÞ ¼

j¼0

k X ðkÞ 2j ð#1Þjþ1 bj d x2k#2j ; j¼0

ðkÞ

ðkÞ

R2k#1 ðx; dÞ ¼

k#1 X j¼0

ðkÞ 2jþ1 2k#2j#1

ð#1Þjþ1 cj d

x

ðkÞ

ðkÞ

; ðkÞ

with coefficients am;j ; bj ; cj , respectively, for which there is a symmetry bk#j ¼ bj

Q 2k ðx; dÞ ¼ ð#1Þk Q 2k ðd; xÞ and R2k#1 ðx; dÞ ¼ ð#1Þk#1 R2k#1 ðd; xÞ:

For example, for k ¼ 1; k ¼ 2, and k ¼ 3, we have

P1 ðx; dÞ ¼ 2x; Q 4 ðx; dÞ ¼ #

Q 2 ðx; dÞ ¼

1 2 1 2 d # x ; 2 2

1 4 3 2 2 1 4 x þ d x # d ; 4 2 4

R1 ðx; dÞ ¼ #2dx; 3

P3 ðx; dÞ ¼

ðkÞ

ðkÞ

and ck#j#1 ¼ cj , so that

" # 1 2 # 3d x þ x3 ; 3

3

R3 ðx; dÞ ¼ #2dx þ 2d x

and

P5 ðx; dÞ ¼

"

# " # 2 2 2 10 4 2 20 2 3 2 5 # d þ d xþ # d x þ x ; 15 3 3 9 3 3

Q 4 ðx; dÞ ¼

1 6 5 4 2 5 2 4 1 6 d # d x þ d x # x ; 6 2 2 6

5

R5 ðx; dÞ ¼ #2d x þ

20 3 3 5 d x # 2dx ; 3

respectively. The integrals Sk can be obtained in a symbolic form in the

MATHEMATICA

Package by the command

intS[k_]:¼Assuming[-1