sub1567

International Journal of Science and Research (IJSR) ISSN (Online): 2319-7064

Index Copernicus Value (2013): 6.14 | Impact Factor (2013): 4.438

Volume 4 Issue 1, January 2015 www.ijsr.net

Licensed Under Creative Commons Attribution CC BY

Quadruple Simultaneous Fourier series Equations Involving Heat Polynomials

Gunjan Shukla1, K.C. Tripathi2

.1Dr. Ambedkar Institute of Technology for Handicapped, Kanpur -208002, Uttar Pradesh, India

2Defence Materials & Stores Research & Development Establishment, Kanpur-208013, U.P. India

Abstract: Quadruple series equations are useful in finding the solution of four part boundary value problems of electrostatics, elasticity and other fields of mathematical physics. In the present paper, we have considered the quadruple series equations involving heat polynomials and solved them. 1. Introduction Quadruple series equations are useful in finding the solution of four part boundary value problems of electrostatics, elasticity and other fields of mathematical physics. Cooke [1] devised the method for finding the solution of quadruple series equations involving Fourier–Bessel series and obtained their solution by using operator theory. Recently Dwivedi and Trivedi [2] Dwivedi and Gupta [3] Dwivedi and Singh [4] considered various types of quadruple series equations involving different polynomials. In the present paper, we have considered the quadruple series equations involving heat polynomials. 2. Quadruple Simultaneous Fourier Series Equations Involving Heat Polynomials Quadruple series equations involving heat polynomials considered here, are the generalization of dual series equations considered by Pathak [3] and corresponding triple series equations considered. Solution is obtained by reducing the problem to simultaneous Fredholm integral equations of the second kind. 3. The Equations Here we shall consider the two sets of quadruple series equations involving heat polynomials of the first kind and second kind respectively. (i) Quadruple Series Equations of the First Kind Quadruple series equations of the first kind to be studied here are given as:

nn p, 1

n 0

A P (x, t) f (x, t), 0 x a1 n p2

∞

+ σ=

− = ≤ < Γ µ + + +

∑ (1.1)

n nn

n p, 2n 0

t A P (x, t) f (x, t), a x b1 n p2

∞ −

+ ν=

− = < < Γ ν + + +

∑ (1.2)

nn p, 3

n 0

A P (x, t) f (x, t), b x c1 n p2

∞

+ σ=

− = < < Γ µ + + +

∑ (1.3)

n nn

n p, 4n 0

t A P (x, t) f (x, t),c x1 n p2

∞ −

+ ν=

− = < < ∞ Γ ν + + +

∑ (1.4)

(ii) Quadruple Series Equations of the Second Kind Quadruple series equations of the second kind to be analysed, here are given as:

n nn

n p, 1n 0

t B P (x, t) g (x, t),0 x a1 n p2

∞ −

+ ν=

− = ≤ < Γ ν + + +

∑ (1.5)

nn p, 2

n 0

B P (x, t) g (x, t), a x b1 n p2

∞

+ σ=

− = < < Γ µ + + +

∑ (1.6)

n nn

n p, 3n 0

t B P (x, t) g (x, t),b x c1 n p2

∞ −

+ ν=

− = < < Γ ν + + +

∑ (1.7)

nn p, 4

n 0

B P (x, t) g (x, t),c x1 n p2

∞

+ σ=

− = < < ∞ Γ µ + + +

∑ (1.8)

In above equations if (x, t) and

ig (x, t) (i 1, 2, 3, 4 )= are the prescribed functions

t 0≥ > and n, P (x, t)ν − is the heat polynomials.

nA and nB are the unknown coefficients to be determined. 4. The Solution

(i) Equations of the First Kind In order to solve the quadruple series equations of the first kind, we set

Paper ID: SUB1567 147

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where 1(x, t)φ and 2(x, t)φ are unknown functions. Using relation (1.1) in equations (1.2), (1.3), (2.1) and (2.2), we obtain

n a bn 1 10 a4(n p)

1 1 n p A2 2A f (x, t)+ (x, t)

12 (n p)! n p2

+

Γ σ + Γ µ + + + = φ + Γ σ + + +

∫ ∫

c3 2 n p, b c

f (x, t)+ (x, t) W (x, t)d (x)∞

+ σ+ φ Ω∫ ∫ (2.3)

Substituting this expression for nA in equations (1.2) and (1.4), we get

n nn p,

4(n p)n 0

1 1t e n p P (x, t)2 2

1 12 (n p)! n p n p2 2

−∞ + ν

+=

Γ σ + Γ µ + + + −

+ Γ σ + + + Γ ν + + +

∑

a b c1 1 3 2 n p, 0 a b c

f ( , t)+ (x, t) f ( , t)+ (x, t) W ( , t)d ( )∞

+ σ ξ φ + ξ φ ξ Ω ξ ∫ ∫ ∫ ∫

Putting the value of d ( )Ω ξ from equation (1.2) and changing the order of integration and summation in above equations, we get

a b c2 2 2 21 1 3 20 a b c

f ( , t)d + ( , t)d f ( , t)d + ( , t)d∞σ σ σ σ ξ ξ ξ ξ φ ξ ξ + ξ ξ ξ ξ φ ξ ξ ∫ ∫ ∫ ∫

n

1 n p, n p, 2

4(n p)n 0

1 n p P (x, t)W ( , t)t 22

1 12 (n p)! n p n p2 2

∞ + ν + σ−σ

+=

Γ µ + + + − ξ ×

+ Γ σ + + + Γ ν + + +

∑

(2.6)

2

4

f (x, t), a x b

f (x, t), c x < <

= < < ∞ (2.7)

Using the summation result (1.4) in equations (2.6) and (2.7), we have

a b2 21 10 a

f ( , t)S(x, , t)d ( , t)S(x, , t)dσ σξ ξ ξ ξ + ξ φ ξ ξ ξ∫ ∫

c 2 23 2b c

f ( , t)S(x, , t)d ( , t)S(x, , t)d∞σ σ+ ξ ξ ξ ξ + ξ φ ξ ξ ξ∫ ∫

⇒

b 2 21 2a c( , t)S(x, , t)d ( , t)S(x, , t)d

∞σ σξ φ ξ ξ ξ + ξ φ ξ ξ ξ∫ ∫

where

a c2 21 2 1 30 b

F (x, t) f (x, t) f ( , t)S(x, , t)d f ( , t)S(x, , t)dσ σ= − ξ ξ ξ ξ − ξ ξ ξ ξ∫ ∫ (2.12)

a c2 2

2 4 1 30 bF (x, t) f (x, t) f ( , t)S(x, , t)d f ( , t)S(x, , t)dσ σ= − ξ ξ ξ ξ − ξ ξ ξ ξ∫ ∫

(2.13) Now using the notation given by (1.6) in equation(2.10), we get

21-2 4t *b 21 2 1a

e a( , t) S ( , x, y) dx (m) ( m)

σ −ξσ

ων−

ξ ξ φ ξ ξ ξ Γ Γ ν − σ +

∫

22 -1 4t *2

2 2 1c

e a( , t) S ( , x, y) dx (m) ( m)

ν −ξ∞ σων−

ξ + ξ φ ξ ξ ξ Γ Γ ν − σ +

∫ = 1F (x, t), a x b< < (2.14)

2* x 4t1a

a e ( , t)S ( , x, y)d(m) ( m)

−ξξ

ξ φ ξ ξ ξΓ Γ ν − σ + ∫

2 2b 4t 4t1 x 2 xx c

e ( , t)S ( , x, y)d e ( , t)S ( , x, y)d∞−ξ ξ + ξ φ ξ ξ ξ + ξ φ ξ ξ ξ∫ ∫

2 1

1x F (x, t), a x bν−= < < (2.15) Putting the value of summation in terms of integral from (1.5) in equation (2.15), we obtain

( ) ( )2 m 1 m 1x 4t 2 2 2 21a 0

e ( , t)d (y) y x y dy− ν−σ+ −ξ−ξξ φ ξ ξ η ξ − −∫ ∫

( ) ( )2 m 1 m 1b x4t 2 2 2 21x 0

e ( , t)d (y) y x y dy− ν−σ+ −−ξ+ ξ φ ξ ξ η ξ − −∫ ∫

( ) ( )2 m 1 m 1x4t 2 2 2 22c 0

e ( , t)d (y) y x y dy− ν−σ+ −∞ −ξ+ ξ φ ξ ξ η ξ − −∫ ∫

1* 1 2(m) ( m) F (x, t) a x b

a x − νΓ Γ ν − σ +

= < < (2.16)

Changing the order of integration in equation (2.16), we get

( ) ( )

2 4ta b 11 m 1 m0 a2 2 2 2

(y)dy e ( , t) dx y y

−ξ

−ν+σ− −η ξ φ ξ

ξ− ξ −

∫ ∫







( ) ( )

2 4tx b 11 m 1 ma y2 2 2 2


−ξ


+ ξ− ξ −

∫ ∫

( ) ( )

2 4tx 21 m 1 m0 c2 2 2 2


−ξ∞


+ ξ− ξ −

∫ ∫

1* 1 2(m) ( m) F (x, t) a x b

a x − νΓ Γ ν − σ +

= < < (2.17)

Assuming

( )

2 4tb 11 1 my 2 2

e ( , t)(y) dy

−ξ

−ξ φ ξ

φ = ξξ −

∫ (2.18)

And

( )

2 4t2

2 1 my 2 2

e ( , t)(y) dy

−ξ∞

−ξ φ ξ

φ = ξξ −

∫ (2.19)

Now in view of the equation (4.18) equation (4.17) can be rewritten as

( )x 1

11 m * 1 2a 2 2

(y) y (m) ( m)dy F (x, t)a xx y

−ν+σ− − νη φ Γ Γ ν − σ +

=−

∫

( ) ( )

2 4ta b 11 m 1 m0 a2 2 2 2


−ξ


− ξ− ξ −

∫ ∫

( ) ( )

2 4tx 21 m 1 m0 c2 2 2 2

(y)dy e ( , t) d a x bx y y

−ξ∞


− ξ < <− ξ −

∫ ∫

The solution of the above equation is given as

( )y

1 ma 2 2

Sin(1 m) d 2xdx(y) (y)dy y x

ν−σ+− ν + σ − π

η φ =π −

∫

( )a

1* 1 2 1 m0 2 2

(m) ( m) (z)dzF (x, t)a x x z

− ν −ν+σ−Γ Γ ν − σ + η× −

−∫

( ) ( )

2 4tb x11 m 1 ma 02 2 2 2

e ( , t) (z)dzd z x z

−ξ

− −ν+σ−ξ φ ξ η

× ξ −ξ − −

∫ ∫

( )

2 4t2

1 mc 2 2

e ( , t) d , a x bz

−ξ∞

−

ξ φ ξ × ξ < <

ξ − ∫

( )y

1 3 ma 2 2

Sin(1 m) d 2xdx(y) (y) F (y, t)dy y x

ν−σ+− ν + σ − π

η φ = −π −

∫

( ) ( )

2 4ta b 11 m 1 m0 a2 2 2 2

(z)dz e ( , t) dx z z

−ξ

−ν+σ− −

η ξ φ ξ× ξ

− ξ −∫ ∫

( ) ( )

2 4tx 21 m 1 m0 c2 2 2 2

e ( , t)(z)dz d a x bx z z

−ξ∞

−ν+σ− −

ξ φ ξη + ξ < <

− ξ − ∫ ∫

(2.20)

where

1 *Sin(1 m) (m) ( m) dF (y, t)=

dya− ν + σ − π Γ Γ ν − σ +

π

( )2y 1

ma 2 2

2x F (x, t) dxy x

ν

ν−σ+ −−

∫ (2.21)


1 3Sin(1 m) d(y) (y) F (y, t) (z)dz

dy− ν + σ − π

η φ = − η ×π

( ) ( ) ( )

2 4ty b 1m 1 m 1 ma 02 2 2 2 2 2

2xdx e ( , t) dy x x z z

−ξ

ν−σ+ −ν+σ− −ξ φ ξ

ξ− − ξ −

∫ ∫

( ) ( )a y

m 1 m0 a 2 2 2 2

d 2xdx(z)dz.dy y x x z

ν−σ+ −ν+σ−+ η− −

∫ ∫

( )

2 4t y21 mc a2 2

e ( , t) dd (z)dz.dyz

−ξ∞

−ξ φ ξ

× ξ + ηξ −

∫ ∫

( ) ( ) ( )

2 4ty 2m 1 m 1 mz 02 2 2 2 2 2


−ξ∞

ν−σ+ −ν+σ− −

ξ φ ξ × ξ

− − ξ − ∫ ∫

a x b< < (2.22)

( ) ( )y

m 1 mz 2 2 2 2

2xdxSin(1 m)y x x z

ν−σ+ −ν+σ−π

=− ν + σ − π− −

∫ (2.23)

In equation (4.22), we get

a1 3 0

Sin(1 m)(y) (y) F (y, t) (z)dz− ν + σ − πη φ = − η

π ∫

( )( )( ) ( )

2m2 2 4tb 1m 1 ma2 2 2 2 2 2

a z e ( , t) dy z y a z

ν−σ+−ξ

ν−σ+ −

− ξ φ ξ× ξ

− − ξ −∫

( )( )( ) ( )

2m2 2 4ta 2m 1 m0 c2 2 2 2 2 2

a z e ( , t)(z)dz dy z y a z

ν−σ+−ξ∞

ν−σ+ −

− ξ φ ξ+ η ξ

− − ξ −∫ ∫

( )

2 4ty 21 ma c 2 2

d e ( , t)(z)dz ddy Sin(1 m) z

−ξ∞

−

π ξ φ ξ + η ξ− ν + σ − π ξ −

∫ ∫

a x b< < (2.24) Equations (4.18) and (4.19) are Able–type integral equations, hence their solutions are given by

( )2 b4t 1

1 m2 2

Sin(1 m) d 2y. (y)e ( , t) dyd y

−ξ

ξ

− π φξ φ ξ = −

π ξ − ξ∫

(2.25) And

( )2 4t 2

2 m2 2

Sin(1 m) d 2y. (y)e ( , t) dyd y

∞−ξ

ξ

− π φξ φ ξ = −

π ξ − ξ∫

(2.26)

With the help of equations (2.25) and (2.26), we obtain







( ) ( ) ( ) ( )

2 4tb b1 11 m m ma a2 2 2 2 2 2 2 2

e ( , t) Sin(1 m) 2x. (x)d dxy a y x a x y

−ξ

− −ξ φ ξ − π φ

ξ =ξ − π − − −

∫ ∫ (2.27)

( ) ( ) ( ) ( )

2 4t2 2

1 m m mc c2 2 2 2 2 2 2 2

e ( , t) Sin(1 m) 2x. (x)d dxy c y x c x y

−ξ∞ ∞

− −ξ φ ξ − π φ

ξ =ξ − π − − −

∫ ∫

(2.28) Substituting the results (2.27) and (2.28) is equation (2.24), we have

1 3 2 2 2 mSin(1 m) Sin(1 m)(y) (y) F (y, t)

(y a )ν−σ+− ν + σ − π − π

η φ = −π −

( )( )( ) ( )

2m2 2a b 1

m2 2 2 20 a 2 2

(z) a z 2x. (x)dz dxy z x z x a

ν−σ+ η − φ ×− − −

∫ ∫

( ) ( )( )( ) ( )

2m m2 2 2 2a 2

m2 2 2 20 c 2 2

(z) a z c z 2x. (x)dz dxy z x z x a

ν−σ+∞

η − − φ + × − − −

∫ ∫

( )( ) ( )

m2 2y 2

m2 2a c 2 2

(z) c zd Sin(1 m) 2x (x)dz dx,dy x z x c

∞η − − π φ− ×

π− −∫ ∫

a x b< < (2.29) Changing the order of integration of the equation (2.29), we get

1 3 2 2 2 mSin(1 m) Sin(1 m)(y) (y) F (y, t)

(y a )ν−σ+− ν + σ − π − π

η φ = −π −

( )( )

( )( )

m2 2b a1

m 2 2 2 2a 02 2

(z) a z2x. (x) dx dzy z x zx a

ν−σ+ η −φ ×− − −

∫ ∫

( )( ) ( )( )( )

m m2 2 2 2a2

m 2 2 2 2c 02 2

(z) a z c z2x (x) dx dzy z x zx a

ν−σ+∞

η − −φ + × − − −

∫ ∫

( )( )

( )

m2 2y2

m 2 2c a2 2

(z) c zSin(1 m) 2x (x) ddx dz,dy x zx c

∞ η −− π φ− ×

π −−∫ ∫

a x b< < (2.30) Now, above equation can be written as

b1 1 3 2a c

(y) (y) M(x, y) (y)dx F (y, t) N(x, y) (x)dx,∞

η φ + φ = − φ∫ ∫

a x b< < (2.31) where

( )2 2 2 m m2 2

Sin(1 m)Sin(1 m) 2xM(x, y) .(y a ) x a

ν−σ+− ν + σ − − π

=π − −

( )( )( )

m2 2a

2 2 2 20

(z) a zdz

y z x z

ν−σ+η −

×− −∫ (2.32)

( )2 2 2 m m2 2

Sin(1 m) Sin(1 m) 2xN(x, y) .(y a ) x c

ν−σ+− ν + σ − π − π

=π − −

( ) ( )( )( )

m m2 2 2 2a

2 2 2 20

(z) a z c z Sin(1 m)dzy z x z

ν−σ+η − − − π

× +π− −∫

( )( )

( )

m2 2y

m 2 2a2 2

(z) c z2x d dzdy x zx c

η −

−−∫ (2.33)

Again starting from equation (2.11), we have b 2 2

1 2 2a c( , t)S(x, , t)d ( , t)S(x, , t)d F (x, t),

∞σ σξ φ ξ ξ ξ + ξ φ ξ ξ ξ =∫ ∫ c x< < ∞ (2.34) Using the notation given by equation (4.6), we obtain

21 2 4t *b 21 w2 1a

e a( , t) S ( , x, y) dx (m) ( m)

− σ −ξσ

ν−

ξ ξ φ ξ ξ ξ Γ Γ ν − σ +

∫

21 2 4t *2

2 w2 1c

e a( , t) S ( , x, y) dx (m) ( m)

− σ −ξ∞ σν−

ξ + ξ φ ξ ξ ξ Γ Γ ν − σ +

∫

= 2F (x, t), c x< < ∞ (2.35) 2* b 4t

1a

a e ( , t)S ( , x, y)d(m) ( m)

−ξξ

ξ φ ξ ξ ξΓ Γ ν − σ + ∫

2 2x 4t 4t2 2c x

e ( , t)S ( , x, y)d e ( , t)S ( , x, y)d∞−ξ −ξ

ξ ξ+ ξ φ ξ ξ ξ+ ξ φ ξ ξ ξ∫ ∫

2 12x F (x, t), c xν−= < < ∞ (2.36)

Putting the value of summation in terms of integral from (2.5) in above equation, we get

( ) ( )2 m 1 m 1b 4t 2 2 2 21a 0

e ( , t)d (y) y x y dy− ν−σ+ −ξ−ξξ φ ξ ξ η ξ − −∫ ∫

( ) ( )2 m 1 m 1x 4t 2 2 2 22c 0

e ( , t)d (y) y x y dy− ν−σ+ −ξ−ξ+ ξ φ ξ ξ η ξ − −∫ ∫

( ) ( )2 m 1 m 1x4t 2 2 2 22x 0

e ( , t)d (y) y x y dy− ν−σ+ −∞ −ξ+ ξ φ ξ ξ η ξ − −∫ ∫

2* 1 2(m) ( m) F (x, t) c x

a x − νΓ Γ ν − σ +

= < < ∞ (2.37)


( ) ( )

2 4ta b 11 m 1 m0 a2 2 2 2


−ξ


ξ− ξ −

∫ ∫







( ) ( )

2 4tb b 11 m 1 ma y2 2 2 2


−ξ


+ ξ− ξ −

∫ ∫

( ) ( )

2 4tc b 21 m 1 m0 y2 2 2 2


−ξ


+ ξ− ξ −

∫ ∫

( ) ( )

2 4tx 21 m 1 mc y2 2 2 2


−ξ∞


+ ξ− ξ −

∫ ∫

2* 1 2(m) ( m) F (x, t) c x

a x − νΓ Γ ν − σ +

= < < ∞ (2.38)

In view of the equations (2.18) and (2.19) above equation becomes

( )x 2

21 m * 1 2c 2 2

(y) (y) (m) ( m)dy F (x, t)a xx y

−ν+σ− − νη φ Γ Γ ν − σ +

=−

∫

( ) ( )

2 4ta b 11 m 1 m0 a2 2 2 2


−ξ


− ξ− ξ −

∫ ∫

( ) ( )

2 4tb b 11 m 1 ma y2 2 2 2


−ξ


− ξ− ξ −

∫ ∫

( ) ( )

2 4tc 21 m 1 m0 c2 2 2 2

e ( , t)(y)dy d c xx y y

−ξ∞

−ν+σ− −ξ φ ξη

− ξ < < ∞− ξ −

∫ ∫ (2.39)

The solution to the equation (2.39) is given by

( )y

2 mc 2 2

Sin(1 m) d 2xdx(y) (y)dy y x

ν−σ+− ν + σ − π

η φ =π −

∫

( )a

2* 1 2 1 m0 2 2

(m) ( m) (z)dzF (x, t)a x x y

− ν −ν+σ−Γ Γ ν − σ + η× −

−∫

( ) ( )

2 4tb b11 m 1 ma a2 2 2 2

e ( , t) (z)dzdz x z

−ξ


× ξ −ξ − −

∫ ∫

( ) ( )

2 4tb c11 m 1 mz 02 2 2 2

e ( , t) (z)dzdz x z

−ξ


× ξ −ξ − −

∫ ∫

( )

2 4t2

1 mc 2 2

e ( , t) d c xz

−ξ∞

−

ξ φ ξ × ξ < < ∞

ξ − ∫ (2.40)

Let

4 *Sin(1 m) (m) ( m) dF (y, t)

dya− ν + σ − π Γ Γ ν − σ +

=π

( )2y 2

mc 2 2

2x F (x, t)dx dxy x

ν

ν−σ+−

∫ (2.41)

Using (2.41) in (2.40) and changing the order of integration, we get

a2 4 0

Sin(1 m) d(y) (y) F (y, t) (z)dzdy

− ν + σ − πη φ = − η ×

π ∫

( ) ( ) ( )

2 4ty b 1m 1 m 1 mc a2 2 2 2 2 2


−ξ

ν−σ+ −ν+σ− −ξ φ ξ

ξ− − ξ −

∫ ∫

( ) ( )b y

m 1 ma c 2 2 2 2

d 2xdx(z)dz.dy y x x z

ν−σ+ −ν+σ−+ η− −

∫ ∫

( )

2 4tb c11 mz 02 2

e ( , t) dd (z)dz.dyz

−ξ

−ξ φ ξ

× ξ + ηξ −

∫ ∫

( ) ( ) ( )

2 4ty 1m 1 m 1 mc c2 2 2 2 2 2

2xdx e ( , t) d ,y x x z z

−ξ∞

ν−σ+ −ν+σ− −

ξ φ ξ × ξ

− − ξ − ∫ ∫

c x< < ∞ (2.42) We know that

( ) ( )( )

( )( )

m2 2y

m 1 m mc 2 2 2 2 2 2 2 2

c zd 2xdxdy y x x z y z y c

ν−σ+

ν−σ+ −ν+σ− ν−σ+

−=

− − − −∫

(2.43) Using the result (2.43) in equation (2.42), we get

a2 4 0

Sin(1 m)(y) (y) F (y, t) (z)dz− ν + σ − πη φ = − η

π ∫

( )( )( ) ( )

2m2 2 4tb 1m 1 ma2 2 2 2 2 2

c z e ( , t) dy z y c z

ν−σ+−ξ

ν−σ+ −

− ξ φ ξ× ξ

− − ξ −∫

( )( )( ) ( )

2m2 2 4tb b 1m 1 ma z2 2 2 2 2 2

c z e ( , t)(z)dz dy z y c z

ν−σ+−ξ

ν−σ+ −

− ξ φ ξ+ η × ξ

− − ξ −∫ ∫

( )( )( ) ( )

2m2 2 4tc 2m 1 m0 c2 2 2 2 2 2

c z e ( , t)(z)dz d ,y z y c z

ν−σ+−ξ∞

ν−σ+ −

− ξ φ ξ+ η × ξ

− − ξ −∫ ∫

c x< < ∞ (2.44)

( )( )

m2 2a

2 4 2 2 m 2 20

(z) c zSin(1 m)(y) (y) F (y, t) dz(y a ) y z

ν−σ+

ν−σ+

η −− ν + σ − π η φ = − π − −

∫







( )( )( )

2 m2 24tb b11 m 2 2a a2 2

(z) c ze ( , t) dz dzy zz

ν−σ+−ξ

−

η −ξ φ ξ× +

−ξ −∫ ∫

( )( )( )

2 m2 24tb c11 m 2 2z 02 2

(z) c ze ( , t) d dzy zz

ν−σ+−ξ

−

η −ξ φ ξ× ξ +

−ξ −∫ ∫

( )

2 4t2

1 mc 2 2

e ( , t) d , c xz

−ξ∞

−

ξ φ ξ × ξ < < ∞

ξ − ∫

(2.45)

Substituting the values from (2.27) and (2.28) in above equation, we obtain

2 4 2 2 2 mSin(1 m) Sin(1 m)(y) (y) F (y, t)

(y c )ν−σ+− ν + σ − π − π

η φ = −π −

( ) ( )( )( ) ( )

m m2 2 2 2a b 1

m2 2 2 20 a 2 2

(z) a z c z 2x (x)dz dxy z x z x a

ν−σ+ η − − φ×− − −

∫ ∫

( )( )( ) ( )

2m2 2c 2

m2 2 2 20 c 2 2

(z) c z 2x (x)dz dxy z x z x c

ν−σ+∞

η − φ + − − −

∫ ∫

( )( )

m2 2b

2 2 m 2 2a

(z) c zSin(1 m) dz(y c ) y z

ν−σ+

ν−σ+

η −− ν + σ − π−

π − −∫

( )

2 4tb 11 mz 2 2

e ( , t) d , c xz

−ξ

−

ξ φ ξ × ξ < < ∞

ξ − ∫ (2.46)

Changing the order of integration of equation (2.46), we get

2 4 2 2 2 mSin(1 m) Sin(1 m)(y) (y) F (y, t)

(y c )ν−σ+− ν + σ − π − π

η φ = −π −

( )( ) ( )( )( )

m m2 2 2 2b a1

m 2 2 2 2a 02 2


ν−σ+ η − −φ×− − −

∫ ∫

( )( )

( )( )

2m2 2c2

m 2 2 2 2c 02 2

(z) c z2x (x) dx dzy z x zx c

ν−σ+∞

η −φ + − − −

∫ ∫

( )

2 4tb 12 2 m 1 ma 2 2

Sin(1 m) e ( , t) d (y c ) z

−ξ

ν−σ+ −− ν + σ − π ξ φ ξ

− ξπ − ξ −

∫

( )( )

m2 2b

2 2z

(z) c zdz, c x

y z

ν−σ+η −

× < < ∞−∫

(2.47)

Again using the value from equation (2.27) in equation (2.47), we get

2 4 2 2 2 mSin(1 m) Sin(1 m)(y) (y) F (y, t)

(y c )ν−σ+− ν + σ − π − π

η φ = −π −

( )( ) ( )( )( )

m m2 2 2 2b a1

m 2 2 2 2a 02 2


ν−σ+ η − −φ×− − −

∫ ∫

( )( )( )( )( )

m2 2 2 2b b1

m 2 2 2 2a z2 2


ν−σ+ η − −φ+− − −

∫ ∫

( )( )

( )( )

2m2 2c2

m 2 2 2 2c 02 2

(z) c z2x (x) dx dz c xy z x zx c

ν−σ+∞

η −φ + < < ∞− − −

∫ ∫ (2.48)

Equation (2.39) can now be written as b

2 2 4 1c a(y) (y) P(x, y) (x)dx F (y, t) Q(x, y) (x)dx,

∞η φ + φ = − φ∫ ∫ c x< < ∞ (2.49) Where

2 2 2 m 2 2 mSin(1 m) Sin(1 m) 2xP(x, y) .

(y c ) (x c )ν−σ+− ν + σ − π − π

=π − −

( )( )( )

2m2 2c

2 2 2 20

(z) c zdz

y z x z

ν−σ+η −

− −∫ (2.50)

And

( ) ( )( )( )

m m2 2 2 2a

2 2 2 20

(z) a z c zdz

y z x z

ν−σ+ η − −

− −∫

( ) ( )( )( )

m m2 2 2 2b

2 2 2 2z

(z) a z c zdz

y z x z

ν−σ+ η − − + − −

∫

(2.51)

Equations (2.31) and (2.39) are Fredholm integral equations

of the second kind which determine 1(x)φ and 2(x)φ .

Values of 1( , t)φ ξ and 2( , t)φ ξ can be determined with the help of equations (2.25) and (2.26) respectively. Finally, the coefficients nA can be computed from equation(2.3), which satisfy the quadruple series equations involving heat polynomials, of the first kind.

where 1(x, t)ψ and 2(x, t)ψ are unknown functions. Using equation (1.1) in equations (2.52), (2.53), (1.1) in equations (2.52), (2.53), (1.6) and (1.8), we obtain

a bn 1 20 a4(n p)

1 1 n p2 2B (x, t)+ g (x, t)

12 (n p)! n p2

+

Γ σ + Γ µ + + + = ψ + Γ σ + + +

∫ ∫

c2 4 n p, b c(x, t)+ g (x, t) W (x, t)d (x)

∞+ σ

+ ψ Ω∫ ∫ (2.54)

Substituting this expression for nB in equations (1.5) and (1.7), we get n n

n p,

4(n p)n 0

1 1t e n p P (x, t)2 2

1 12 (n p)! n p n p2 2

−∞ + ν

+=

Γ σ + Γ µ + + + −

+ Γ σ + + + Γ ν + + +

∑







a b c

1 2 2 4 n p, 0 a b c(x, t)+ g (x, t) (x, t)+ g (x, t) W ( , t)d ( )

∞+ σ

ψ + ψ ξ Ω ξ ∫ ∫ ∫ ∫

Using the results (1.2) in above equations and changing the order of integration and summation, we get

a b c2 2 2 21 2 2 40 a b c( , t)d + g ( , t)d ( , t)d + g ( , t)d

∞σ σ σ σ ξ ψ ξ ξ ξ ξ ξ + ξ ψ ξ ξ ξ ξ ξ ∫ ∫ ∫ ∫

n

1 n p, 2

n p, 4(n p)n 0

1 n p P (x, t)t 22 W ( , t)

1 12 (n p)! n p n p2 2

∞ + ν−σ

+ σ+=

Γ µ + + + − × ξ

+ Γ σ + + + Γ ν + + +

∑

Using summation result (1.4) in equations (2.57) and (2.58), we have

a b2 21 20 a( , t)S(x, , t)d + g ( , t)S(x, , t)dσ σξ ψ ξ ξ ξ ξ ξ ξ ξ∫ ∫

c 2 22 4b c( , t)S(x, , t)d + g ( , t)S(x, , t)d

∞σ σ+ ξ ψ ξ ξ ξ ξ ξ ξ ξ∫ ∫

a c2 2

1 20 b( , t)S(x, , t)d + ( , t)S(x, , t)dσ σξ ψ ξ ξ ξ ξ ψ ξ ξ ξ∫ ∫

where

b 21 1 2a

G (x, t) g (x, t) g ( , t)S(x, , t)dσ= − ξ ξ ξ ξ∫

24c

g ( , t)S(x, , t)d∞ σ− ξ ξ ξ ξ∫ (2.63)

and b 2

2 3 2aG (x, t) g (x, t) g ( , t)S(x, , t)dσ= − ξ ξ ξ ξ∫

24c

g ( , t)S(x, , t)d∞ σ− ξ ξ ξ ξ∫ (2.64)

Starting from equation (2.61), using notation given by (1.6), we get

22 1 4t *a 2

1 1 20

e a( , t) S ( , x, y) dx (m) ( m)

− σ+ −ξσ

ω− + ν

ξ ξ ψ ξ ξ ξ Γ Γ ν − σ +

∫

21 2 4t *c 22 2 1b

e a( , t) S ( , x, y) dx (m) ( m)

− σ −ξσ

ων−

ξ + ξ ψ ξ ξ ξ Γ Γ ν − σ +

∫

= 1G (x, t), 0 x a≤ < (2.65) 2* x 4t

10

a e ( , t)S ( , x, y)d(m) ( m)

−ξξ

ξ ψ ξ ξ ξΓ Γ ν − σ + ∫

2 2a c4t 4t1 x 2 xx b

e ( , t)S ( , x, y)d e ( , t)S ( , x, y)d−ξ −ξ + ξ ψ ξ ξ ξ+ ξ ψ ξ ξ ξ∫ ∫

= 2 11x G (x, t), 0 x aν− ≤ < (2.66)

Now putting the value of summation in terms of integral from (1.5) in equation (2.66), we obtain

( ) ( )2 m 1 m 1x 4t 2 2 2 210 0

e ( , t) (y) y x y d dy− ν−σ+ −ξ−ξξ ψ ξ η ξ − − ξ∫ ∫

( ) ( )2 m 1 m 1a x4t 2 2 2 21x 0

e ( , t) (y) y x y d dy− ν−σ+ −−ξ+ ξ ψ ξ η ξ − − ξ∫ ∫

( ) ( )2 m 1 m 1c x4t 2 2 2 22b 0

e ( , t) (y) y x y d dy− ν−σ+ −−ξ+ ξ ψ ξ η ξ − − ξ∫ ∫

1* 1 2(m) ( m) G (x, t), 0 x a

a x − νΓ Γ ν − σ +

= ≤ < (2.67)

Inverting the order of integration of equation (2.67), we get

( ) ( )

2 4tx a 11 m 1 m0 y2 2 2 2

(y) e ( , t)dy d x y x y

−ξ

−ν+σ− −η ξ ψ ξ

ξ− −

∫ ∫

( ) ( )

2 4tx c 21 m 1 m0 b2 2 2 2

(y) e ( , t)dy d x y x y

−ξ


+ ξ− −

∫ ∫

1* 1 2(m) ( m) G (x, t), 0 x a

a x − νΓ Γ ν − σ +

= ≤ < (2.68)

Assuming,

( )

2 4ta 11 1 my 2 2

e ( , t)(y) d x y

−ξ

−ξ ψ ξ

ψ = ξ−

∫ (2.69)

Now equation (2.68) can be written as

( ) ( )

2 4tx c 211 m 1 m0 b2 2 2 2

(y) e ( , t)dy (y) d x y x y

−ξ

−ν+σ− −

η ξ ψ ξ ψ + ξ

− − ∫ ∫

1* 1 2(m) ( m) G (x, t), 0 x a

a x − νΓ Γ ν − σ +

= ≤ < (2.70)

With the help of equations (1.8), we can solve the above equation as

( )

2 4tc 21 1 mb 2 2

e ( , t)(y) (y) d y

−ξ

−

ξ ψ ξ η ψ + ξ

ξ − ∫

y

2 2 m0

Sin(1 m) d 2xdxdy (y x )ν−σ+

− ν + σ − π= −

π −∫

1* 1 2(m) ( m) G (x, t), 0 x a

a x − νΓ Γ ν − σ +

= ≤ < (2.71)

Now equation (2.71) takes the form

( )

2 4tc 21 3 1 mb 2 2

e ( , t)(y) (y) G (y, t) (y) d , 0 x ay

−ξ

−ξ ψ ξ

η ψ = − η ξ ≤ <ξ −

∫ (2.72)

Where

3 *Sin(1 m) (m) ( m)G (y, t)

a− ν + σ − π Γ Γ ν − σ +

=π

( )2y 1

m0 2 2

d 2x G (x, t) dxdy y x

ν

ν−σ+×−

∫ (2.73)

Solving the integral equation (2.69) as 2 a4t 1

1 2 2 mSin(1 m) d 2y (y)e ( , t) dy

d (y )−ξ

ξ

− π ψξ ψ ξ = −

π ξ − ξ∫ (2.74)

With the help of equation (2.74), we obtain







( ) ( ) ( )( )

2 4ta a1 11 m m m0 02 2 2 2 2 2

e ( , t) Sin(1 m) 2x (x)dy y x y x

−ξ

− −ξ ψ ξ − π ψ

ξ =ξ − π − −

∫ ∫ (2.75)

Again, let

( )

2 4tc 22 1 my 2 2

e ( , t)(y) dy

−ξ

−ξ ψ ξ

ψ = ξξ −

∫ (2.76)

Solving (2.76) similarly as (2.74) and (2.75) respectively, we get

( )2 c4t 2

2 m2 2

Sin(1 m) d 2y (y)e ( , t) dyd y

−ξ

ξ

− π ψξ ψ ξ = −

π ξ − ξ∫

(2.77)

( ) ( ) ( ) ( )

2 4tc c2 21 m m mb b2 2 2 2 2 2 2 2

e ( , t) Sin(1 m) 2x (x)dxdy b y x b x y

−ξ

− −ξ ψ ξ − π ψ

ξ =ξ − π − − −

∫ ∫ (2.78)

Now using the equation (2.79) in (2.72), we get

( )1 3 m2 2

Sin(1 m) (y)(y) (y) G (y, t)b y

−− π η

η ψ = −π −

( ) ( )c 2

mb 2 2 2 2

2x (x)dxxx b x y

ψ

− −∫ (2.79)

Equcation (2.80) reduces to the following form c

1 3 2b(y) (y) G (y, t) R(x, y) (y)dx, 0 x aη ψ = − ψ ≤ <∫ (2.80)

where

( ) ( ) ( )m m2 2 2 2 2 2

Sin(1 m) (y) 2xR(x, y)b y x b x y

−− π η

=π − − −

(2.81)

Again starting from equation (2.62) and using the notation (1.6), we have

22 1 4t *a 21 2 10

e a( , t) S ( , x, y) dx (m) ( m)

− σ+ −ξσ

ων−

ξ ξ ψ ξ ξ ξ Γ Γ ν − σ +

∫22 1 4t *c 2

2 2 1b

e a( , t) S ( , x, y) dx (m) ( m)

σ+ −ξσ

ων−

ξ + ξ ψ ξ ξ ξ Γ Γ ν − σ +

∫

= 2G (x, t), bx c< < (2.82) 2* a 4t

10

a e ( , t)S ( , x, y)d(m) ( m)

−ξξξ ψ ξ ξ ξ

Γ Γ ν − σ + ∫

2x 4t2b

e ( , t)S ( , x, y)d−ξξ+ ξ ψ ξ ξ ξ∫

2c 4t2 xx

e ( , t)S ( , x, y)d−ξ+ ξ ψ ξ ξ ξ∫

= 2 12x G (x, t), bx cν− < < (2.83)

Putting the value of summation in terms of integral from (1.5) in above equation, we get

( ) ( )2 m 1 m 1a 4t 2 2 2 210 0

e ( , t)d (y) y x y dy− ν−σ+ −ξ−ξξ ψ ξ ξ η ξ − −∫ ∫

( ) ( )2 m 1 m 1x 4t 2 2 2 22b 0

e ( , t)d (y) y x y dy− ν−σ+ −ξ−ξ+ ξ ψ ξ ξ η ξ − −∫ ∫

( ) ( )2 m 1 m 1c x4t 2 2 2 22x 0

e ( , t)d (y) y x y dy− ν−σ+ −−ξ+ ξ ψ ξ ξ η ξ − −∫ ∫

2* 1 2(m) ( m) G (x, t),

a x − νΓ Γ ν − σ +

= (2.84)

Inverting the order of integration we get

( ) ( )

2 4ta a 11 m 1 m0 y2 2 2 2

(y) e ( , t)dy dx y x y

−ξ


ξ− −

∫ ∫

( ) ( )

2 4tb c 21 m 1 m0 b2 2 2 2


−ξ


+ ξ− −

∫ ∫

( ) ( )

2 4tx c 21 m 1 mb y2 2 2 2


−ξ


+ ξ− −

∫ ∫

2* 1 2(m) ( m) G (x, t), bx c

a x − νΓ Γ ν − σ +

= < < (2.85)

Using (2.69) and (2.76) the above equation becomes

( )x 2

21 m * 1 2b 2 2

(y) (y)dy (m) ( m) G (x, t)a xx y

−ν+σ− − νη ψ Γ Γ ν − σ +

=−

∫

( ) ( )a b1

1 m 1 m0 02 2 2 2

(y) (y) (y)dy dyx y x y

−ν+σ− −ν+σ−η ψ η

− −− −

∫ ∫

( )

2 4tc 21 mb 2 2

e ( , t) d b x cy

−ξ

−ξ ψ ξ

ξ < <ξ −

∫ (2.86)

The equation (2.87) can be solved as y

2 2 2 mb

Sin(1 m) d 2xdx(y) (y)dy (y x )ν−σ+

− ν + σ − πη ψ =

π −∫

( )a 1

2* 1 2 1 m0 2 2

(m) ( m) (z) (z)dzG (x, t)a x x z

− ν −ν+σ−Γ Γ ν − σ − η ψ× −

−∫

( ) ( )

2 4tb c 21 m 1 m0 b2 2 2 2

(z)dz e ( , t) d ,x z z

−ξ

−ν+σ− −

η ξ ψ ξ − ξ

− ξ − ∫ ∫

b x c< < (2.87) y

2 4 2 2 mb

Sin(1 m) d 2xdx(y) (y)G (y, t)dy (y x )ν−σ+

− ν + σ − πη ψ −

π −∫

( ) ( ) ( )

2 4ta b c1 21 m 1 m 1 m0 0 b2 2 2 2 2 2

(z) (z) (z)dz e ( , t)dz d ,x z x z z

−ξ

−ν+σ− −ν+σ− −

η ψ η ξ ψ ξ × + ξ

− − ξ − ∫ ∫ ∫

b x c< < (2.88) Where

4 *Sin(1 m) (m) ( m)G (y, t)

a− ν + σ − π Γ Γ ν − σ +

=π

( )2y 2

mb 2 2

d 2x G (x, t) dxdy y x

ν

ν−σ+×−

∫ (2.89)

y2 4 2 2 mb

Sin(1 m) d 2xdx(y) (y) G (y, t)dy (y x )ν−σ+

− ν + σ − πη ψ = −

π −∫

( ) ( )

2 4ta c 231 m 1 m0 b2 2 2 2

dz e ( , t)G (z, t) (z) dx z z

−ξ

−ν+σ− −

ξ ψ ξ × − η ξ

− ξ − ∫ ∫







( ) ( )

2 4tb c 21 m 1 m0 b2 2 2 2

e ( , t)(z)dz d , b x cx z z

−ξ

−ν+σ− −

ξ ψ ξη + ξ < <

− ξ − ∫ ∫

(2.90)

Breaking the last term of the above equation into two parts, we get y

2 4 2 2 mb

Sin(1 m) d 2xdx(y) (y) G (y, t)dy (y x )ν−σ+

− ν + σ − πη ψ = −

π −∫

( ) ( ) ( )

2 4ta b c3 21 m 1 m 1 m0 a b2 2 2 2 2 2

G (z, t)dz (z)dz e ( , t) d ,x z x z z

−ξ

−ν+σ− −ν+σ− −

η ξ ψ ξ × + × ξ

− − ξ − ∫ ∫ ∫

b x c< < (2.91) Changing the order of integration, the equation (2.92) becomes

a2 4 30

Sin(1 m)(y) (y) G (y, t) G (z, t)dz

− ν + σ − π η ψ = − π ∫

y b

2 2 m 2 2 1 mb a

d 2xdx (z)dzdy (y x ) (x z )ν−σ+ −ν+σ−× + η

− −∫ ∫

( )

2 4ty c 22 2 m 2 2 1 m 1 mb b 2 2

d 2xdx e ( , t) ddy (y x ) (x z ) z

−ξ

ν−σ+ −ν+σ− −

ξ ψ ξ × × ξ

− − ξ − ∫ ∫

b x c< < (2.92) We know that

y

2 2 m 2 2 1 mb

d 2xdxdy (y x ) (x z )ν−σ+ −ν+σ−− −∫

2 2 m

2 2 m 2 2(b z )

(y b ) (y z )

ν−σ+

ν−σ+−

=− −

(2.93)

Using the result (2.79) and (2.94) to the equation (2.93), we get

a2 4 30

Sin(1 m)(y) (y) G (y, t) G (z, t)dz

− ν + σ − π η ψ = − π ∫

2 2 m 2 2 mb

2 2 m 2 2 2 2 m 2 2a

(b z ) (b z )(z)dz(y b ) (y z ) (y b ) (y z )

ν−σ+ ν−σ+

ν−σ+ ν−σ+− −

× + η ×− − − −∫

c 22 2 m 2 2 m 2 2b

2x (x)Sin(1 m) d x b x c(b z ) (x b ) (x z )−

ψ− π× < <

π − − − ∫ (2.94)

( )2 2 ma 3

2 4 m 2 202 2

G (z, t)(b z )Sin(1 m)(y) (y) G (y, t) dz(y z )y b

ν−σ+

ν−σ+−− ν + σ − π

η ψ = −−π −

∫

( )2 2 2mb

m 2 2 2 2a2 2 2

Sin(1 m) Sin(1 m) (z)(b z ) dz(x z )(y z )y b

ν−σ+

ν−σ+− ν + σ − π − π η −

− +− −π −

∫

c 22 2 mb

2x (x) d x, b x c(x b )

ψ× < <

−∫ (2.95)

Now changing the order of integration of the last term of the equation (2.96), we get

( )2 2 ma 3

2 4 m 2 202 2

G (z, t)(b z )Sin(1 m)(y) (y) G (y, t) dz(y z )y b

ν−σ+

ν−σ+−− ν + σ − π

η ψ = −−π −

∫

( )c 2

m 2 2 mb2 2 2

Sin(1 m) Sin(1 m) 2x (x) dx(x b )y b

ν−σ+− ν + σ − π − π ψ

− ×−π −

∫

(2.96) Now equation (2.97) can be rewritten as

( )c

2 2 4 mb 2 2

Sin(1 m)(y) (y) S(x, y) (x)dx G (y, t)y b

ν−σ+− ν + σ −

η ψ + ψ = −π −

∫

2 2 ma 32 20

G (z, t)(b z ) d z, b x c(y z )

ν−σ+−< <

−∫ (2.97)

Where S(x, y) is the symmetric kernel

( ) m 2 2 m2 2 2

Sin(1 m) Sin(1 m) 2xS(x, y) .(x b )y b

ν−σ+− ν + σ − π − π

=−π −

2 2 2mb

2 2 2 2a

(z)(b z ) dz,(x z )(y z )

ν−σ+η −×

− −∫ (2.98)

Equations (2.98) and (2.81) are Fredholm integral

equations of the second kind determine 2(y)ψ and

1(y).ψ 1( , t)Ψ ξ and 2( , t)Ψ ξ can be then computed from equations (2.74) and (2.78) respectively. Finally, the coefficients nB can be calculated with the help of equation (2.54) which satisfy the equations from (1.5) to (1.8). Particular Case If we let c→∞ in equation (1.1) to (1.8), they reduce to the corresponding triple series equation involving heat polynomials and this solution can be shown to agree with that obtained earlier for triple series equations. Similarly, we can obtain the corresponding dual series equations involving heat polynomials. Acknowledgement Authors are thankful to Dr. G.K. Dubey and Dr. A.P. Dwivedi for their co-operation & support provided to me for preparing the paper. Authors are also thankful to Mr. Manish Verma, Scientist ‘C’ DMSRDE, Kanpur for his support. References [1] Cooke, J.C. (1992): The solution of triple and Quadruple

integral equations and Fourier Series Equations,Quart .J.Mech. Appl. Math., 45, pp. 247-263.

[2] Dwivedi, A.P. & Trivedi, T.N. (1972) : Quadruple series equations involving Jacobi polynomials , Proc. Nat. Acad. Sci. India , 42 (A) , pp. 203-208.

[3] Dwivedi, A.P. & Gupta, P. (1980): Quadruple series equations involving Jacobi Polynomials of different indices, Acta Cienca Indica, 6 (M) , pp. 241-247.

[4] Dwivedi, A.P. & Singh, V.B. (1997): Quadruple series equations involving series of Jacobi polynomials, Ind.J.Pure & Appl. Math. , 28, 1068-1077.



sub1567

Documents

t b p x

x b1 n p2

n nnn p

b x c1 n p2

a1 n p2

solution i equations

fourierbessel series

prescribed functions