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
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Licensed Under Creative Commons Attribution CC BY
Quadruple Simultaneous Fourier series Equations
Involving Heat Polynomials
Gunjan Shukla1
, K.C. Tripathi2
.1
Dr. Ambedkar Institute of Technology for Handicapped, Kanpur -208002, Uttar Pradesh, India
2
Defence 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:
n
n p, 1
n 0
A
P (x, t) f (x, t), 0 x a
1
n p
2
∞
+ σ
=
−= ≤ <
 
Γ µ + + + 
 
∑ (1.1)
n n
n
n p, 2
n 0
t A
P (x, t) f (x, t), a x b
1
n p
2
∞ −
+ ν
=
−= < <
 
Γ ν + + + 
 
∑

(1.2)
n
n p, 3
n 0
A
P (x, t) f (x, t), b x c
1
n p
2
∞
+ σ
=
−= < <
 
Γ µ + + + 
 
∑
(1.3)
n n
n
n p, 4
n 0
t A
P (x, t) f (x, t),c x
1
n p
2
∞ −
+ 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 n
n
n p, 1
n 0
t B
P (x, t) g (x, t),0 x a
1
n p
2
∞ −
+ ν
=
−= ≤ <
 
Γ ν + + + 
 
∑
 (1.5)
n
n p, 2
n 0
B
P (x, t) g (x, t), a x b
1
n p
2
∞
+ σ
=
−= < <
 
Γ µ + + + 
 
∑
(1.6)
n n
n
n p, 3
n 0
t B
P (x, t) g (x, t),b x c
1
n p
2
∞ −
+ ν
=
−= < <
 
Γ ν + + + 
 
∑
 (1.7)
n
n p, 4
n 0
B
P (x, t) g (x, t),c x
1
n p
2
∞
+ 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 b
n 1 1
0 a4(n p)
1 1
n p A
2 2
A f (x, t)+ (x, t)
1
2 (n p)! n p
2
+
   
Γ σ + Γ µ + + +   
    φ  + Γ σ + + + 
 
∫ ∫
c
3 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 n
n p,
4(n p)n 0
1 1
t e n p P (x, t)
2 2
1 1
2 (n p)! n p n p
2 2
−
∞ + ν
+=
   
Γ σ + Γ µ + + + −   
   
   
+ Γ σ + + + Γ ν + + +   
   
∑
a b c
1 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 2
1 1 3 2
0 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 2
2
1 1
2 (n p)! n p n p
2 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 2
1 1
0 a
f ( , t)S(x, , t)d ( , t)S(x, , t)dσ σ
ξ ξ ξ ξ + ξ φ ξ ξ ξ∫ ∫
c 2 2
3 2
b 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 2
1 2
a c
( , t)S(x, , t)d ( , t)S(x, , t)d
∞σ σ
ξ φ ξ ξ ξ + ξ φ ξ ξ ξ∫ ∫
where
a c2 2
1 2 1 3
0 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 3
0 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∫ ∫
(2.13)
Now using the notation given by (1.6) in
equation(2.10), we get
2
1-2 4t *b 2
1 2 1a
e a
( , t) S ( , x, y) d
x (m) ( m)
σ −ξ
σ
ων−
 ξ 
ξ φ ξ ξ ξ 
Γ Γ ν −σ +  
∫
2
2 -1 4t *
2
2 2 1c
e a
( , t) S ( , x, y) d
x (m) ( m)
ν −ξ∞ σ
ων−
 ξ 
+ ξ φ ξ ξ ξ 
Γ Γ ν−σ+  
∫
= 1F (x, t), a x b< < (2.14)
2
* x 4t
1
a
a
e ( , t)S ( , x, y)d
(m) ( m)
−ξ
ξ
 ξ φ ξ ξ ξΓ Γ ν −σ + ∫
2 2b 4t 4t
1 x 2 x
x 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 2
1
a 0
e ( , t)d (y) y x y dy
− ν−σ+ −ξ−ξ
ξ φ ξ ξ η ξ − −∫ ∫
( ) ( )
2 m 1 m 1b x4t 2 2 2 2
1
x 0
− ν−σ+ −
−ξ
+ ξ φ ξ ξ η ξ − −∫ ∫
( ) ( )
2 m 1 m 1x4t 2 2 2 2
2
c 0
− ν−σ+ −∞ −ξ
+ ξ φ ξ ξ η ξ − −∫ ∫
1* 1 2
(m) ( m)
F (x, t) a x b
a x − ν
Γ Γ ν − σ +
= < < (2.16)
Changing the order of integration in equation (2.16),
we get
( ) ( )
2
4ta b
1
1 m 1 m0 a2 2 2 2
(y)dy e ( , t)
d
x y y
−ξ
−ν+σ− −
η ξ φ ξ
ξ
− ξ −
∫ ∫

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( ) ( )
2
4tx b
1
1 m 1 ma y2 2 2 2
(y)dy e ( , t)
d
x y y
−ξ
−ν+σ− −
η ξ φ ξ
+ ξ
− ξ −
∫ ∫
( ) ( )
2
4tx
2
1 m 1 m0 c2 2 2 2
(y)dy e ( , t)
d
x y y
−ξ∞
−ν+σ− −
η ξ φ ξ
+ ξ
− ξ −
∫ ∫
1* 1 2
(m) ( m)
F (x, t) a x b
a x − ν
Γ Γ ν − σ +
< <
(2.17)
Assuming
( )
2
4tb
1
1 1 my 2 2
e ( , t)
(y) d
y
−ξ
−
ξ φ ξ
φ= ξ
ξ −
∫
(2.18)
And
( )
2
4t
2
2 1 my 2 2
e ( , t)
(y) d
y
−ξ∞
−
ξ φ ξ
φ= ξ
ξ −
∫
(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− − ν
η φ Γ Γ ν − σ +
=
−
∫
( ) ( )
2
4ta b
1
1 m 1 m0 a2 2 2 2
(y)dy e ( , t)
d
x y y
−ξ
−ν+σ− −
η ξ φ ξ
− ξ
− ξ −
∫ ∫
( ) ( )
2
4tx
2
1 m 1 m0 c2 2 2 2
(y)dy e ( , t)
d a x b
x y y
−ξ∞
−ν+σ− −
η ξ φ 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)dz
F (x, t)
a x x z
− ν −ν+σ−
Γ Γ ν − σ + η
× −
 −
∫
( ) ( )
2
4tb x
1
1 m 1 ma 02 2 2 2
e ( , t) (z)dz
d
z x z
−ξ
− −ν+σ−
ξ φ ξ η
× ξ −
ξ − −
∫ ∫
( )
2
4t
2
1 mc 2 2
e ( , t)
d , a x b
z
−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
1
1 m 1 m0 a2 2 2 2
(z)dz e ( , t)
d
x z z
−ξ
−ν+σ− −

η ξ φ ξ
× ξ
 − ξ −

∫ ∫
( ) ( )
2
4tx
2
1 m 1 m0 c2 2 2 2
e ( , t)(z)dz
d a x b
x z z
−ξ∞
−ν+σ− −

ξ φ ξη 
+ ξ < <
− ξ −

∫ ∫
(2.20)
where
1 *
Sin(1 m) (m) ( m) d
F (y, t)=
dya
− ν + σ − π Γ Γ ν − σ +
π
( )
2y
1
ma 2 2
2x F (x, t)
dx
y x
ν
ν−σ+
−
−
∫
(2.21)
we get
1 3
Sin(1 m) d
(y) (y) F (y, t) (z)dz
dy
− ν + σ − π
η φ = − η ×
π 
( ) ( ) ( )
2
4ty b
1
m 1 m 1 ma 02 2 2 2 2 2
2xdx e ( , t)
d
y 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 y
2
1 mc a2 2
e ( , t) d
d (z)dz.
dyz
−ξ∞
−
ξ φ ξ
× ξ + η
ξ −
∫ ∫
( ) ( ) ( )
2
4ty
2
m 1 m 1 mz 02 2 2 2 2 2
2xdx e ( , t)
d
y x x z z
−ξ∞
ν−σ+ −ν+σ− −

ξ φ ξ 
× ξ
− − ξ −

∫ ∫
a x b< < (2.22)
( ) ( )
y
m 1 mz 2 2 2 2
2xdx
Sin(1 m)y x x z
π
=
− ν + σ − π− −
∫
(2.23)
In equation (4.22), we get
{a
1 3
0
Sin(1 m)
(y) (y) F (y, t) (z)dz
−ν+σ− π
η φ= − η
π ∫
( )
( )( ) ( )
2
m
2 2 4tb
1
m 1 ma2 2 2 2 2 2
a z e ( , t)
d
y z y a z
ν−σ+
−ξ
ν−σ+ −
− ξ φ ξ
× ξ
− − ξ −
∫
( )
( )( ) ( )
2
m
2 2 4ta
2
m 1 m0 c2 2 2 2 2 2
a z e ( , t)
(z)dz d
y z y a z
ν−σ+
−ξ∞
ν−σ+ −
− ξ φ ξ
+ η ξ
− − ξ −
∫ ∫
( )
2
4ty
2
1 ma c 2 2
d e ( , t)
(z)dz d
dy 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 m
2 2
Sin(1 m) d 2y. (y)
e ( , t) dy
d y
−ξ
ξ
− π φ
ξ φ ξ =−
π ξ − ξ
∫
(2.25)
And
( )
2
4t 2
2 m
2 2
Sin(1 m) d 2y. (y)
e ( , t) dy
d y
∞−ξ
ξ
− π φ
ξ φ ξ =−
π ξ − ξ
∫
(2.26)
With the help of equations (2.25) and (2.26), we obtain

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( ) ( ) ( ) ( )
2
4tb b
1 1
1 m m ma a2 2 2 2 2 2 2 2
e ( , t) Sin(1 m) 2x. (x)
d dx
y a y x a x y
−ξ
− −
ξ φ ξ − π φ
ξ =
ξ − π − − −
∫ ∫
(2.27)
( ) ( ) ( ) ( )
2
4t
2 2
1 m m mc c2 2 2 2 2 2 2 2
e ( , t) Sin(1 m) 2x. (x)
d dx
y 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 m
Sin(1 m) Sin(1 m)
(y) (y) F (y, t)
(y a )ν−σ+
− ν + σ − π − π
η φ = −
π −
( )
( )( ) ( )
2m
2 2
a b
1
m2 2 2 20 a 2 2
(z) a z 2x. (x)
dz dx
y z x z x a
ν−σ+ η − φ
×
− − −

∫ ∫
( ) ( )
( )( ) ( )
2m m
2 2 2 2
a
2
m2 2 2 20 c 2 2
(z) a z c z 2x. (x)
dz dx
y z x z x a
ν−σ+
∞
η − − φ 
+ × 
− − −

∫ ∫
( )
( ) ( )
m
2 2
y
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 m
Sin(1 m) Sin(1 m)
(y) (y) F (y, t)
(y a )ν−σ+
− ν + σ − π − π
η φ = −
π −
( )
( )
( )( )
m
2 2
b a
1
m 2 2 2 2a 02 2
(z) a z2x. (x)
dx dz
y z x zx a
ν−σ+ η −φ
×
− − −

∫ ∫
( )
( ) ( )
( )( )
m m
2 2 2 2
a
2
m 2 2 2 2c 02 2
(z) a z c z2x (x)
dx dz
y z x zx a
ν−σ+
∞
η − −φ 
+ × 
− − −

∫ ∫
( )
( )
( )
m
2 2
y
2
m 2 2c a2 2
(z) c zSin(1 m) 2x (x) d
dx dz,
dy x zx c
∞ η −− π φ
− ×
π −−
∫ ∫
a x b< < (2.30)
Now, above equation can be written as
b
1 1 3 2
a 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 m
2 2
Sin(1 m)Sin(1 m) 2x
M(x, y) .
(y a ) x a
ν−σ+
− ν + σ − − π
=
π − −
( )
( )( )
m
2 2
a
2 2 2 20
(z) a z
dz
y z x z
ν−σ+
η −
×
− −
∫ (2.32)
( )
2 2 2 m m
2 2
Sin(1 m) Sin(1 m) 2x
N(x, y) .
(y a ) x c
ν−σ+
− ν + σ − π − π
=
π − −
( ) ( )
( )( )
m m
2 2 2 2
a
2 2 2 20
(z) a z c z Sin(1 m)
dz
y z x z
ν−σ+
η − − − π
× +
π− −
∫
( )
( )
( )
m
2 2
y
m 2 2a2 2
(z) c z2x d
dz
dy x zx c
η −
−−
∫
(2.33)
Again starting from equation (2.11), we have
b 2 2
1 2 2
a c
( , t)S(x, , t)d ( , t)S(x, , t)d F (x, t),
∞σ σ
ξ φ ξ ξ ξ + ξ φ ξ ξ ξ =∫ ∫
c x< < ∞ (2.34)
Using the notation given by equation (4.6), we obtain
2
1 2 4t *b 2
1 w2 1a
e a
( , t) S ( , x, y) d
x (m) ( m)
− σ −ξ
σ
ν−
 ξ 
ξ φ ξ ξ ξ 
Γ Γ ν − σ + 
∫
2
1 2 4t *
2
2 w2 1c
e a
( , t) S ( , x, y) d
x (m) ( m)
− σ −ξ∞ σ
ν−
 ξ 
+ ξ φ ξ ξ ξ 
Γ Γ ν − σ + 
∫
= 2F (x, t), c x< < ∞ (2.35)
2
* b 4t
1
a
a
e ( , t)S ( , x, y)d
(m) ( m)
−ξ
ξ
 ξ φ ξ ξ ξΓ Γ ν − σ + ∫
2 2x 4t 4t
2 2
c 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 1
2x F (x, t), c xν−
= < < ∞ (2.36)
from (2.5) in above equation, we get
( ) ( )
2 m 1 m 1b 4t 2 2 2 2
1
a 0
− ν−σ+ −ξ−ξ
ξ φ ξ ξ η ξ − −∫ ∫
( ) ( )
2 m 1 m 1x 4t 2 2 2 2
2
c 0
− ν−σ+ −ξ−ξ
+ ξ φ ξ ξ η ξ − −∫ ∫
( ) ( )
2 m 1 m 1x4t 2 2 2 2
2
x 0
− ν−σ+ −∞ −ξ
+ ξ φ ξ ξ η ξ − −∫ ∫
2* 1 2
(m) ( m)
F (x, t) c x
a x − ν
Γ Γ ν − σ +
= < < ∞ (2.37)
we get
( ) ( )
2
4ta b
1
1 m 1 m0 a2 2 2 2
(y)dy e ( , t)
d
x y y
−ξ
−ν+σ− −
η ξ φ ξ
ξ
− ξ −
∫ ∫

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( ) ( )
2
4tb b
1
1 m 1 ma y2 2 2 2
(y)dy e ( , t)
d
x y y
−ξ
−ν+σ− −
η ξ φ ξ
+ ξ
− ξ −
∫ ∫
( ) ( )
2
4tc b
2
1 m 1 m0 y2 2 2 2
(y)dy e ( , t)
d
x y y
−ξ
−ν+σ− −
η ξ φ ξ
+ ξ
− ξ −
∫ ∫
( ) ( )
2
4tx
2
1 m 1 mc y2 2 2 2
(y)dy e ( , t)
d
x y y
−ξ∞
−ν+σ− −
η ξ φ ξ
+ ξ
− ξ −
∫ ∫
2* 1 2
(m) ( m)
F (x, t) c x
a x − ν
Γ 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− − ν
η φ Γ Γ ν − σ +
=
−
∫
( ) ( )
2
4ta b
1
1 m 1 m0 a2 2 2 2
(y)dy e ( , t)
d
x y y
−ξ
−ν+σ− −
η ξ φ ξ
− ξ
− ξ −
∫ ∫
( ) ( )
2
4tb b
1
1 m 1 ma y2 2 2 2
(y)dy e ( , t)
d
x y y
−ξ
−ν+σ− −
η ξ φ ξ
− ξ
− ξ −
∫ ∫
( ) ( )
2
4tc
2
1 m 1 m0 c2 2 2 2
e ( , t)(y)dy
d c x
x 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)dz
F (x, t)
a x x y
− ν −ν+σ−
Γ Γ ν −σ + η
× −
 −
∫
( ) ( )
2
4tb b
1
1 m 1 ma a2 2 2 2
e ( , t) (z)dz
d
z x z
−ξ
− −ν+σ−
ξ φ ξ η
× ξ −
ξ − −
∫ ∫
( ) ( )
2
4tb c
1
1 m 1 mz 02 2 2 2
e ( , t) (z)dz
d
z x z
−ξ
− −ν+σ−
ξ φ ξ η
× ξ −
ξ − −
∫ ∫
( )
2
4t
2
1 mc 2 2
e ( , t)
d c x
z
−ξ∞
−

ξ φ ξ 
× ξ < < ∞
ξ −

∫ (2.40)
Let
4 *
Sin(1 m) (m) ( m) d
F (y, t)
dya
− ν + σ − π Γ Γ ν − σ +
=
π
( )
2y
2
mc 2 2
2x F (x, t)dx
dx
y x
ν
ν−σ+
−
∫ (2.41)
Using (2.41) in (2.40) and changing the order of
integration, we get
{ a
2 4
0
Sin(1 m) d
(y) (y) F (y, t) (z)dz
dy
−ν + σ − π
η φ = − η ×
π ∫
( ) ( ) ( )
2
4ty b
1
m 1 m 1 mc a2 2 2 2 2 2
2xdx e ( , t)
d
y x x z z
−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
+ η
− −
∫ ∫
( )
2
4tb c
1
1 mz 02 2
e ( , t) d
d (z)dz.
dyz
−ξ
−
ξ φ ξ
× ξ + η
ξ −
∫ ∫
( ) ( ) ( )
2
4ty
1
m 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
( ) ( )
( )
( )( )
m
2 2
y
m 1 m mc 2 2 2 2 2 2 2 2
c zd 2xdx
dy 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
{ a
2 4
0
Sin(1 m)
(y) (y) F (y, t) (z)dz
−ν + σ− π
η φ = − η
π ∫
( )
( )( ) ( )
2
m
2 2 4tb
1
m 1 ma2 2 2 2 2 2
c z e ( , t)
d
y z y c z
ν−σ+
−ξ
ν−σ+ −
− ξ φ ξ
× ξ
− − ξ −
∫
( )
( )( ) ( )
2
m
2 2 4tb b
1
m 1 ma z2 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 −
∫ ∫
( )
( )( ) ( )
2
m
2 2 4tc
2
m 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)
( )
( )
m
2 2
a
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 − −

∫

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( )
( )
( )
2
m
2 24tb b
1
1 m 2 2a a2 2
(z) c ze ( , t)
dz dz
y zz
ν−σ+
−ξ
−
η −ξ φ ξ
× +
−ξ −
∫ ∫
( )
( )
( )
2
m
2 24tb c
1
1 m 2 2z 02 2
(z) c ze ( , t)
d dz
y zz
ν−σ+
−ξ
−
η −ξ φ ξ
× ξ +
−ξ −
∫ ∫
( )
2
4t
2
1 mc 2 2
e ( , t)
d , c x
z
−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 m
Sin(1 m) Sin(1 m)
(y) (y) F (y, t)
(y c )ν−σ+
− ν + σ − π − π
η φ = −
π −
( ) ( )
( )( ) ( )
m m
2 2 2 2
a b
1
m2 2 2 20 a 2 2
(z) a z c z 2x (x)
dz dx
y z x z x a
ν−σ+ η − − φ
×
− − −

∫ ∫
( )
( )( ) ( )
2m
2 2
c
2
m2 2 2 20 c 2 2
(z) c z 2x (x)
dz dx
y z x z x c
ν−σ+
∞
η − φ 
+ 
− − −

∫ ∫
( )
( )
m
2 2
b
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
1
1 mz 2 2
e ( , t)
d , c x
z
−ξ
−

ξ φ ξ 
× ξ < < ∞
ξ −

∫ (2.46)
Changing the order of integration of equation (2.46), we get
2 4 2 2 2 m
Sin(1 m) Sin(1 m)
(y) (y) F (y, t)
(y c )ν−σ+
− ν + σ − π − π
η φ = −
π −
( )
( ) ( )
( )( )
m m
2 2 2 2
b a
1
m 2 2 2 2a 02 2
(z) a z c z2x (x)
dx dz
y z x zx a
ν−σ+ η − −φ
×
− − −

∫ ∫
( )
( )
( )( )
2m
2 2
c
2
m 2 2 2 2c 02 2
(z) c z2x (x)
dx dz
y z x zx c
ν−σ+
∞
η −φ 
+ 
− − −

∫ ∫
( )
2
4tb
1
2 2 m 1 ma 2 2
Sin(1 m) e ( , t)
d
(y c ) z
−ξ
ν−σ+ −
− ν + σ − π ξ φ ξ
− ξ
π − ξ −
∫
( )
( )
m
2 2
b
2 2z
(z) c z
dz, 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 m
Sin(1 m) Sin(1 m)
(y) (y) F (y, t)
(y c )ν−σ+
−ν + σ− π − π
η φ = −
π −
( )
( ) ( )
( )( )
m m
2 2 2 2
b a
1
m 2 2 2 2a 02 2
(z) a z c z2x (x)
dx dz
y z x zx a
ν−σ+ η − −φ
×
− − −

∫ ∫
( )
( )( )
( )( )
m
2 2 2 2
b b
1
m 2 2 2 2a z2 2
(z) a z c z2x (x)
dx dz
y z x zx a
ν−σ+ η − −φ
+
− − −

∫ ∫
( )
( )
( )( )
2m
2 2
c
2
m 2 2 2 2c 02 2
(z) c z2x (x)
dx dz c x
y z x zx c
ν−σ+
∞
η −φ 
+ < < ∞
− − −

∫ ∫
(2.48)
Equation (2.39) can now be written as
b
2 2 4 1
c 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 m
P(x, y) .
(y c ) (x c )ν−σ+
− ν + σ − π − π
=
π − −
( )
( )( )
2m
2 2
c
2 2 2 20
(z) c z
dz
y z x z
ν−σ+
η −
− −
∫
(2.50)
And
( ) ( )
( )( )
m m
2 2 2 2
a
2 2 2 20
(z) a z c z
dz
y z x z
ν−σ+ η − −

− −

∫
( ) ( )
( )( )
m m
2 2 2 2
b
2 2 2 2z
(z) a z c z
dz
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 b
n 1 2
0 a4(n p)
1 1
n p
2 2
B (x, t)+ g (x, t)
1
2 (n p)! n p
2
+
   
Γ σ + Γ µ + + +   
   = ψ  + Γ σ + + + 
 
∫ ∫
c
2 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 1
t e n p P (x, t)
2 2
1 1
2 (n p)! n p n p
2 2
−
∞ + ν
+=
   
Γ σ + Γ µ + + + −   
   
   
+ Γ σ + + + Γ ν + + +   
   
∑

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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 2
1 2 2 4
0 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 2
2 W ( , t)
1 1
2 (n p)! n p n p
2 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 2
1 2
0 a
( , t)S(x, , t)d + g ( , t)S(x, , t)dσ σ
ξ ψ ξ ξ ξ ξ ξ ξ ξ∫ ∫
c 2 2
2 4
b 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 2
0 b
( , t)S(x, , t)d + ( , t)S(x, , t)dσ σ
ξ ψ ξ ξ ξ ξ ψ ξ ξ ξ∫ ∫
where
b 2
1 1 2
a
G (x, t) g (x, t) g ( , t)S(x, , t)dσ
= − ξ ξ ξ ξ∫
2
4
c
g ( , t)S(x, , t)d
∞ σ
− ξ ξ ξ ξ∫ (2.63)
and
b 2
2 3 2
a
G (x, t) g (x, t) g ( , t)S(x, , t)dσ
= − ξ ξ ξ ξ∫
2
4
c
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
2
2 1 4t *a 2
1 1 20
e a
( , t) S ( , x, y) d
x (m) ( m)
− σ+ −ξ
σ
ω− + ν
 ξ 
ξ ψ ξ ξ ξ 
Γ Γ ν − σ +  
∫
2
1 2 4t *c 2
2 2 1b
e a
( , t) S ( , x, y) d
x (m) ( m)
− σ −ξ
σ
ων−
 ξ 
+ ξ ψ ξ ξ ξ 
Γ Γ ν − σ +  
∫
= 1G (x, t), 0 x a≤ < (2.65)
2
* x 4t
1
0
a
e ( , t)S ( , x, y)d
(m) ( m)
−ξ
ξ
 ξ ψ ξ ξ ξΓ Γ ν − σ + ∫
2 2a c4t 4t
1 x 2 x
x 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 1
1x 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 2
1
0 0
e ( , t) (y) y x y d dy
− ν−σ+ −ξ−ξ
ξ ψ ξ η ξ − − ξ∫ ∫
( ) ( )
2 m 1 m 1a x4t 2 2 2 2
1
x 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 2
2
b 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 − ν
Γ Γ ν − σ +
= ≤ < (2.67)
Inverting the order of integration of equation (2.67),
we get
( ) ( )
2
4tx a
1
1 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
2
1 m 1 m0 b2 2 2 2
(y) e ( , t)
dy d
x y x y
−ξ
−ν+σ− −
η ξ ψ ξ
+ ξ
− −
∫ ∫
1* 1 2
(m) ( m)
G (x, t), 0 x a
a x − ν
Γ Γ ν − σ +
= ≤ < (2.68)
Assuming,
( )
2
4ta
1
1 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
2
11 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 − ν
Γ Γ ν − σ +
= ≤ < (2.70)
With the help of equations (1.8), we can solve the
above equation as
( )
2
4tc
2
1 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 2xdx
dy (y x )ν−σ+
− ν + σ − π
= −
π −∫
1* 1 2
(m) ( m)
G (x, t), 0 x a
a x − ν
Γ Γ ν − σ +
= ≤ < (2.71)
Now equation (2.71) takes the form
( )
2
4tc
2
1 3 1 mb 2 2
e ( , t)
(y) (y) G (y, t) (y) d , 0 x a
y
−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)
dx
dy y x
ν
ν−σ+
×
−
∫
(2.73)
Solving the integral equation (2.69) as
2 a4t 1
1 2 2 m
Sin(1 m) d 2y (y)
e ( , t) dy
d (y )
−ξ
ξ
− π ψ
ξ ψ ξ =−
π ξ − ξ∫
(2.74)
With the help of equation (2.74), we obtain

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( ) ( ) ( )( )
2
4ta a
1 1
1 m m m0 02 2 2 2 2 2
e ( , t) Sin(1 m) 2x (x)
d
y y x y x
−ξ
− −
ξ ψ ξ − π ψ
ξ =
ξ − π − −
∫ ∫
(2.75)
Again, let
( )
2
4tc
2
2 1 my 2 2
e ( , t)
(y) d
y
−ξ
−
ξ ψ ξ
ψ= ξ
ξ −
∫
(2.76)
Solving (2.76) similarly as (2.74) and (2.75)
respectively, we get
( )
2 c4t 2
2 m
2 2
Sin(1 m) d 2y (y)
e ( , t) dy
d y
−ξ
ξ
− π ψ
ξ ψ ξ =−
π ξ − ξ
∫
(2.77)
( ) ( ) ( ) ( )
2
4tc c
2 2
1 m m mb b2 2 2 2 2 2 2 2
e ( , t) Sin(1 m) 2x (x)dx
d
y 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 m
2 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)dx
x
x b x y
ψ
− −
∫
(2.79)
Equcation (2.80) reduces to the following form
c
1 3 2
b
(y) (y) G (y, t) R(x, y) (y)dx, 0 x aη ψ= − ψ ≤ <∫ (2.80)
where
( ) ( ) ( )
m m
2 2 2 2 2 2
Sin(1 m) (y) 2x
R(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
2
2 1 4t *a 2
1 2 10
e a
( , t) S ( , x, y) d
x (m) ( m)
− σ+ −ξ
σ
ων−
 ξ 
ξ ψ ξ ξ ξ 
Γ Γ ν − σ +  
∫
2
2 1 4t *c 2
2 2 1b
e a
( , t) S ( , x, y) d
x (m) ( m)
σ+ −ξ
σ
ων−
 ξ 
+ ξ ψ ξ ξ ξ 
Γ Γ ν − σ +  
∫
= 2G (x, t), bx c< < (2.82)
2
* a 4t
1
0
a
e ( , t)S ( , x, y)d
(m) ( m)
−ξ
ξξ ψ ξ ξ ξ
Γ Γ ν − σ + ∫
2x 4t
2
b
e ( , t)S ( , x, y)d−ξ
ξ+ ξ ψ ξ ξ ξ∫
2c 4t
2 x
x
e ( , t)S ( , x, y)d−ξ
+ ξ ψ ξ ξ ξ∫
=
2 1
2x G (x, t), bx cν−
< < (2.83)
from (1.5) in above equation, we get
( ) ( )
2 m 1 m 1a 4t 2 2 2 2
1
0 0
− ν−σ+ −ξ−ξ
ξ ψ ξ ξ η ξ − −∫ ∫
( ) ( )
2 m 1 m 1x 4t 2 2 2 2
2
b 0
− ν−σ+ −ξ−ξ
+ ξ ψ ξ ξ η ξ − −∫ ∫
( ) ( )
2 m 1 m 1c x4t 2 2 2 2
2
x 0
− ν−σ+ −
−ξ
+ ξ ψ ξ ξ η ξ − −∫ ∫
2* 1 2
(m) ( m)
G (x, t),
a x − ν
Γ Γ ν − σ +
= (2.84)
Inverting the order of integration we get
( ) ( )
2
4ta a
1
1 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
4tb c
2
1 m 1 m0 b2 2 2 2
(y) e ( , t)
dy d
x y x y
−ξ
−ν+σ− −
η ξ ψ ξ
+ ξ
− −
∫ ∫
( ) ( )
2
4tx c
2
1 m 1 mb y2 2 2 2
(y) e ( , t)
dy d
x y x y
−ξ
−ν+σ− −
η ξ ψ ξ
+ ξ
− −
∫ ∫
2* 1 2
(m) ( m)
G (x, t), bx c
a x − ν
Γ 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− − ν
η ψ Γ Γ ν − σ +
=
−
∫
( ) ( )
a b
1
1 m 1 m0 02 2 2 2
(y) (y) (y)
dy dy
x y x y
−ν+σ− −ν+σ−
η ψ η
− −
− −
∫ ∫
( )
2
4tc
2
1 mb 2 2
e ( , t)
d b x c
y
−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 − π
η ψ =
π −∫
( )
a
1
2* 1 2 1 m0 2 2
(m) ( m) (z) (z)dz
G (x, t)
a x x z
− ν −ν+σ−
Γ Γ ν − σ − η ψ
× −
 −
∫
( ) ( )
2
4tb c
2
1 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 − π
η ψ −
π −∫
( ) ( ) ( )
2
4ta b c
1 2
1 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)
dx
dy y x
ν
ν−σ+
×
−
∫
(2.89)
y
2 4 2 2 mb
Sin(1 m) d 2xdx
(y) (y) G (y, t)
dy (y x )ν−σ+
−ν +σ− π
η ψ= −
π −∫
( ) ( )
2
4ta c
2
31 m 1 m0 b2 2 2 2
dz e ( , t)
G (z, t) (z) d
x z z
−ξ
−ν+σ− −
  
ξ ψ ξ × − η ξ 
 − ξ −  
∫ ∫

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( ) ( )
2
4tb c
2
1 m 1 m0 b2 2 2 2
e ( , t)(z)dz
d , b x c
x 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 − π
η ψ= −
π −∫
( ) ( ) ( )
2
4ta b c
3 2
1 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
a
2 4 3
0
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)dz
dy (y x ) (x z )ν−σ+ −ν+σ−
× + η
− −∫ ∫
( )
2
4ty c
2
2 2 m 2 2 1 m 1 mb b 2 2
d 2xdx e ( , t)
d
dy (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 2xdx
dy (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
a
2 4 3
0
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
2
2 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
2
2 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
3
2 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 m
2 2 2
S(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.

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