Numerical approach to solve second kind Volterra integral equation of

Applied Mathematical Sciences, Vol. 5, 2011, no. 14, 685 - 696

Numerical Approach to Solve Second Kind Volterra Integral Equation of Abel Type Using Block-Pulse Functions and Taylor Expansion by Collocation Method A. Shahsavaran Faculty of Science, Islamic Azad University Borujerd Branch, Borujerd, Iran [email protected] Abstract In present paper, we give a numerical approach for solving Abel’s integral equation of the second kind which is based on the use of BlockPulse Function (BPF) and Taylor expansion by collocation method. Also, detailed error analysis is worked out that shows efficiency of the presented method. Finally, some numerical examples with exact solution are given.

Mathematics Subject Classification: 45A05; 45D05; 45E10 Keywords: Abel’s integral equation; Taylor expansion; Block-Pulse Function; collocation method

1

Introduction

Volterra integral equations arise in many problems pertaining to mathematical physics like heat conduction problems. Various methods are available concerning their numerical solutions. Block-Pulse Functions were used for solving linear Volterra integral equation of the first kind [2], Taylor series expansion method were used to solve Volterra integral equations of second kind with convolution kernel [7], Legendre wavelet were used to solve Abel integral equation [11], and so on. In recent years, many different basis functions have been used to solve and reduce integral equations to a system of algebraic equations [1-11]. The aim of this work is to present a computational method for solving special

A. Shahsavaran

686

case of singular Volterra integral equations of the second kind as named Abel’s integral equation defined as follows: x y(t) √ dt, (1) y(x) = f (x) − x−t 0 where f ∈ L2 (R) on the interval x ∈ [0, T ]. Our method is based on the use of Taylor expansion and Block-Pulse Functions by collocation method.

2

Block-Pulse Functions and Taylor expansion

In this section, we define a k-set of Block-Pulse Functions (BPF) over the interval [0, T ) as: 1, (i−1)T ≤ t < iTk , for all i = 1, 2, . . . , k k , (2) Bi (t) = 0, elsewhere with a positive integer value for k. Also, Bi is the i-th Block-Pulse Function. In this paper, it is assumed that T = 1, so BPFs are defined over [0, 1). There are some properties for BPFs, the most important properties are disjointness, orthogonality, and completeness. The disjointness property can be clearly obtained from the definition of BPFs: Bi (t)Bj (t) (3) where, i, j = 1, 2, ..., k. The other property is orthogonality: < Bi (t), Bj (t) > (4) The third property is completeness. For every y ∈ L2 [0, 1), when k approaches to the infinity, Parseval’s identity holds: 1 ∞ 2 y (t)dt = b2i Bi (t)2 , 0

i=1

where, bi = k < y(t), Bi(t) >= k

1 0

y(t)Bi(t)dt.

Volterra integral equation of Abel type

687

A function y(t) defined over the interval [0, 1) may be expanded as: y(t) =

∞

bi Bi (t),

(5)

i=1

with bi = k < y(t), Bi (t) >. In practice, only k-term of (5) are considered, where k is a power of 2, that is, y(t) yk (t) =

k

bi Bi (t),

(6)

i=1

with matrix form: y(t) yk (t) = bt B(t)

(7)

where, b = [b1 , b2 , . . . , bk ]t and B(t) = [B1 (t), B2 (t), . . . , Bk (t)]t . Now we define function g(t) = √

1 , x−t

0 ≤ t < x ≤ 1.

Taylor series expansion of g(t) based on expansion about the point t = 0 leads to: g(t) =

∞ g (n) (0)tn n=0

n!

,

(8)

it is straightforward to show that g

(n)

(2n)!(x − t)−n−1/2 , (t) = 22n n!

then, g (n) (0) =

(2n)! 22n n!xn+1/2

,

(9)

Substituting (9) into (8) leads to g(t) =

∞ n=0

(2n)!tn . 22n (n!)2 xn+1/2

(10)

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3

Singular Volterra integral equation of the second kind

Now we consider the following singular Volterra integral equation of the second kind as named Abel’s integral equation: y(x) = f (x) −

x 0

y(t) √ dt, x−t

where f ∈ L2 (R) on the interval x ∈ [0, 1] and y(t) is the unknown function to be determined. Now by substituting (7) and (10) into (1) we obtain: t

b B(x) = f (x) − = f (x) −

∞ n=0 ∞

(2n)! 2n 2 (n!)2 xn+1/2

x

Bt (t)tn dtb

0

βn (x)α(x)b,

(11)

n=0

where, βn (x) = and

(2n)! 22n (n!)2 xn+1/2

α(x) =

x

,

Bt (t)tn dt.

0

By evaluating (11) at the collocation points xj = obtain: t

b B(xj ) = f (xj ) −

∞

j−1/2 , k

j = 1, 2, . . . , k, we

βn (xj )α(xj )b,

(12)

n=0

but B(xj ) = ej where ej is the j-th column of identity matrix. So, from (12) we have bj = f (xj ) −

∞

βn (xj )α(xj )b,

j = 1, 2, . . . , k.

n=0

Now we have to compute value of the series ∞ n=0

βn (xj )α(xj ),

j = 1, 2, . . . , k.

(13)


689

If j = 1 we can write α(x1 ) =

1 2k

0

=

0

1 2k

Bt (t)tn dt [B1 (t), . . . , Bk (t)]tn dt

1 2k

[1, 0, . . . , 0]tn dt 0 1 = [1, 0, . . . , 0]. (n + 1)(2k)n+1

=

Then, ∞

βn (x1 )α(x1 ) =

n=0

=

∞

=

(2n)!

1 n+1/2 22n (n!)2 ( 2k ) n=0 ∞

1 √ 2k

1 [1, 0, . . . , 0] (n + 1)(2k)n+1

(2n)! [1, 0, . . . , 0] 22n (n!)2 (n + 1)

n=0 2 √ [1, 0, . . . , 0], 2k

because, ∞ n=0

(2n)! 22n (n!)2 (n

+ 1)

= 2,

so, ∞ n=0

βn (x1 )α(x1 ) =

2 √ 2k

[1, 0, . . . , 0].

(14)

By substituting (14) into (13) we obtain: 2 b1 = f (x1 ) − √ [1, 0, . . . , 0]b 2k 2 = f (x1 ) − √ b1 , 2k or 2 ( √ + 1)b1 = f (x1 ). 2k

(15)

A. Shahsavaran

690 Similarly, for j = 2 we can write α(x2 ) =

3 2k

0

=

1 k

Bt (t)tn dt t

n

B (t)t dt +

0

1 k

=

3 2k 1 k

n

Bt (t)tn dt

[1, 0, . . . , 0]t dt +

0

=

3 2k 1 k

[0, 1, 0, . . . , 0]tn dt

1 [2n+1 , 3n+1 − 2n+1 , 0, . . . , 0]. (n + 1)(2k)n+1

So, ∞

βn (x2 )α(x2 ) =

n=0

∞ n=0

(2n)! 1 [2n+1 , 3n+1 − 2n+1 , 3 n+1 2n 2 n+1/2 (n + 1)(2k) 2 (n!) ( 2k )

0, . . . , 0] ∞ (2n)! 1 [2n+1 , 3n+1 − 2n+1 , 0, . . . , 0] =√ 2n 2 n+1/2 (n + 1) 2k n=0 2 (n!) 3 ∞ ∞ 1 (2n)!2n+1 (2n)!(3n+1 − 2n+1 ) =√ [ , , 2k n=0 22n (n!)2 3n+1/2 (n + 1) n=0 22n (n!)2 3n+1/2 (n + 1) 0, . . . , 0] 2 √ = √ [ 3 − 1, 1, 0, . . . , 0], 2k because, ∞ n=0

√ (2n)!2n+1 = 2( 3 − 1), 2n 2 n+1/2 2 (n!) 3 (n + 1)

and ∞ (2n)!(3n+1 − 2n+1 ) = 2. 2n (n!)2 3n+1/2 (n + 1) 2 n=0

By substituting (16) into (13) we obtain: 2 √ b2 = f (x2 ) − √ [ 3 − 1, 1, 0, . . . , 0]b 2k 2 √ = f (x2 ) − √ [( 3 − 1)b1 + b2 ], 2k

(16)


691

or 2 √ 2 √ ( 3 − 1)b1 + (1 + √ )b2 = f (x2 ). 2k 2k Generally, for j(2 ≤ j ≤ k) we can write 2j−1 2k Bt (t)tn dt α(xj ) = 0 j

=

i=2 j

=

i=2

i−1 k i−2 k i−1 k i−2 k

t

n

B (t)t dt + eti−1 tn dt

+

j−1/2 k j−1 k j−1/2 k

j−1 k

Bt (t)tn dt

etj tn dt

1 [2n+1 , 4n+1 − 2n+1 , . . . , (2j − 2)n+1 − (2j − 4)n+1 , (n + 1)(2k)n+1 (2j − 1)n+1 − (2j − 2)n+1 , 0, . . . , 0], =

because from (2) we have 0 ≤ t < k1 implies that B1 (t) = 1 and Bi (t) = 0 for i = 2, . . . , k. 1 ≤ t < k2 implies that B2 (t) = 1 and Bi (t) = 0 for i = 1, . . . , k and i = 2. k .. . k−1 k

≤ t < 1 implies that Bk (t) = 1 and Bi (t) = 0 for i = 1, . . . , k − 1. Then, t ∈ [ i−1 , ki ), i = 1, 2, . . . , k, implies that B(t) = ei . So, k ∞

βn (xj )α(xj ) =

n=0

∞

(2n)!

22n (n!)2 ( 2j−1 )n+1/2 2k n=0 n+1 n+1 n+1

1 (n + 1)(2k)n+1

[2 , 4 − 2 , . . . , (2j − 2)n+1 − (2j − 4)n+1 , (2j − 1)n+1 − (2j − 2)n+1 , 0, . . . , 0] ∞ (2n)! 1 =√ 2n 2 2k n=0 2 (n!) (2j − 1)n+1/2 (n + 1)

[2n+1 , 4n+1 − 2n+1 , . . . , (2j − 2)n+1 − (2j − 4)n+1 , (2j − 1)n+1 − (2j − 2)n+1 , 0, . . . , 0] √ 2 = √ [ 2j − 1 − 2j − 3, 2j − 3 − 2j − 5, . . . , 3 − 1, 2k 1, 0, . . . , 0], (17) because, for i = 2, 3, . . . , j, we have: ∞ (2n)!((2i − 2)n+1 − (2i − 4)n+1 ) n=0

22n (n!)2 (2j − 1)n+1/2 (n + 1)

= 2( 2j − (2i − 3) − 2j − (2i − 1)),

A. Shahsavaran

692 and

∞ (2n)!((2j − 1)n+1 − (2j − 2)n+1 ) n=0

22n (n!)2 (2j − 1)n+1/2 (n + 1)

= 2.

By substituting (17) into (13) we obtain: √ 2 bj = f (xj ) − √ [ 2j − 1 − 2j − 3, 2j − 3 − 2j − 5, . . . , 3 − 1, 1 2k , 0, . . . , 0]b 2 = f (xj ) − √ [( 2j − 1 − 2j − 3)b1 + ( 2j − 3 − 2j − 5)b2 + · · · + 2k √ ( 3 − 1)bj−1 + bj ], or 2 2 √ ( 2j − 1 − 2j − 3)b1 + √ ( 2j − 3 − 2j − 5)b2 + . . . 2k 2k 2 √ 2 + √ ( 3 − 1)bj−1 + (1 + √ )bj = f (xj ), j = 2, 3, . . . , k, 2k 2k and in abstract form for j = 2, 3, . . . , k, we have: j−1 2 2 √ [ 2j − (2i − 1) − 2j − (2i + 1)]bi + (1 + √ )bj = f (xj ). 2k i=1 2k

(18)

Therefore, from equations (15) and (18) we can find column vector b and from (7) we can approximate y(x) by yk (x) at every point x ∈ [0, 1).

4

Error Analysis

In this section we assume that y(t) is a differentiable function with bounded first derivative on (0,1), that is, ∃M > 0;

∀t ∈ (0, 1) :

|y (t)| ≤ M.

The representation error when y(t) is represented in a series of BPFs over every subinterval [ i−1 , ki ) is k ei (t) = bi Bi (t) − y(t) = bi − y(t).


693

We may proceed as follows:

2

ei = =

i k i−1 k i k i−1 k

e2i (t)dt (bi − y(t))2 dt

i−1 i ))(bi − y(t1 ))2 = ( −( k k 1 i−1 i ≤ t1 < ) = (bi − y(t1 ))2 , ( k k k

(19)

where we used mean value theorem. As before, if

y(t) =

∞

bi Bi (t)

i=1

the i-th fourier coefficient is given by bi = k < y(t), Bi(t) >. Using the mean value theorem leads to: bi = k < y(t), Bi(t) > i k y(t)dt =k i−1 k

i i−1 = k( − ( ))y(t2 ) k k i−1 i ≤ t2 < ) = y(t2), ( k k

(20)

by substituting (20) into (19) and using mean value theorem we obtain: 1 (bi − y(t1 ))2 k 1 = (y(t2) − y(t1 ))2 k 1 = (t2 − t1 )2 y 2 (t0 ) (t1 < t0 < t2 ) k 1 ≤ 3 M 2. k

ei 2 =

(21)

A. Shahsavaran

694 1

Now for i < j we have [ i−1 , ki ) ∩[ j−1 , kj ) = {}, so k k

2

e =

1

e2 (t)dt

1

(

= 0

k

i=1 k

1

= 0

=

ei (t)ej (t)dt = 0. Therefore,

0

=

0

ei (t))2 dt

e2i (t)dt

+2

i=1

k 1 i=1 k

0

i