Directory UMM :Data Elmu:jurnal:J-a:Journal of Computational And Applied Mathematics:Vol106.Issue2.1999:
Journal of Computational and Applied Mathematics 106 (1999) 245–254
www.elsevier.nl/locate/cam
Integral equations and potential-theoretic type integrals of
orthogonal polynomials
M.H. Fahmya , M.A. Abdoub , M.A. Darwishc; ∗
a
b
Department of Mathematics, Faculty of Science, Alexandria University, Alexandria, Egypt
Department of Mathematics, Faculty of Education, Alexandria University, Alexandria, Egypt
c
Mathematisches Seminar II, Universitat Kiel, D24098 Kiel, Germany
Received 8 May 1998
Abstract
Abel’s theorem is used to solve the Fredholm integral equations of the rst kind with Gauss’s hypergeometric kernel.
The case that the kernel takes general form is discussed in detail and the solution is given. Here, the works of some
authors are considered as special cases. The discussion focuses on the three dimensional contact problems in the theory
c 1999 Published by Elsevier Science B.V. All rights reserved.
of elasticity with general kernel.
MSC: 33C05; 33C25; 45B05; 45Exx; 73C99
Keywords: Abel’s theorem; Contact problem; Fredholm integral equation; Hypergeometric kernel; Weber–Sonin integral
1. Introduction
Many problems of mathematical physics, engineering, and contact problems in the theory of
elasticity lead to the Fredholm integral equations of the rst kind, (see [4,6,12]),
Z
a
k(x; y) (y) dy = f(x):
(1.1)
0
Here, f(x) and k(x; y) are given functions, and k(x; y) is the kernel of the integral equation (1.1)
which may have a singularity while the unknown function (x) denotes a potential function. The
end points (0; a) are the points of the geometric singularity. At these points the function (x) is
bounded if it is a potential function while it has an integrable singularity if it is a
ux type quantity.
∗
Corresponding author. Permanent address: Department of Mathematics, Faculty of Education, Alexandria University,
Damanhour branch, Egypt.
E-mail address: [email protected] (Mohamed Abdalla Darwish)
c 1999 Published by Elsevier Science B.V. All rights reserved.
0377-0427/99/$ - see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 9 ) 0 0 0 6 4 - 3
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M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
In [7], Arutyunyan has shown that the plane contact problem of the nonlinear theory of plasticity, in its rst approximation, can be reduced to Eq. (1.1) with the Carleman kernel, k(x; y) =
|x − y|− ; 06 ¡ 1.
The importance of Eq. (1.1) with Carleman kernel comes from the contact mixed problems in a
half space in the theory of elasticity wherein the modules of elasticity is changing according to the
power law. The plane contact problems for the ordinary half-space, as it is well known [14], are
equivelent to the integral equation of the rst kind with logarithmic kernel, k(x; y) = −ln(|x − y|).
Abdou [1], established Eq. (1.1) from the three dimensional contact problems in the theory of
elasticity with potential kernel where the coecients bed of the compressible materials are neglected.
In the present paper we use one integral representation of the Gauss’s hypergeometric function
to give a general solution of the Fredholm integral equation (1.1). This solution is given by means
of the composition representation of the general operator through two operators of the Volterra type
with Abel’s kernels. This paper is organized as follows: In Sections 2 and 3, Abel’s theorem is used
to solve the integral equation (1.1) when the kernel takes a certain case of hypergeometric function.
Also, the works of some authors are considered as special cases of this work. In Section 4, the three
dimensional contact problem in the theory of elasticity with general kernel. We apply in Section 5 the
Wiener–Hopf technique to study general potential kernel of two variables and corresponding integral
equation. In Section 6 a number of examples of the equations and their solutions is given. Also,
we obtained some integral relations between Gegenbauer polynomials and Chebyshev polynomials.
Finally, in Section 7 we list certain contact problems in the theory of elasticity, nonlinear theory
of plasticity and other problems of mathematical physics where these general equations and their
particular cases may appear or be applied.
2. Hypergeometric kernel
Consider the Fredholm integral equation (1.1) with the kernel
1
2xy
1
k(x; y) = 2
F n; n + ; m; 2
2
2n
(x + y )
2
x + y2
2 !
:
(2.1)
n
Here, F( ; ;
; z) = ∞
(x) is the Gamma
n=0 [( )n ( )n =n!(
)n ]z where (a)m = ( (m + a)= (a)) and
function. The function F( ; ;
; z) de ned by this series is the Gauss’s hypergeometric function (see
[8,18]).
P
Theorem 1. The general solution of Eq. (1:1) with the kernel of the form (2:1) is given by
(x) =
x2m−2
d
(1 + 2n − m) d x
Z
x
1
(t 2
tg(t)
dt;
− x2 )m−2n
(a = 1);
(2.2)
y2m−1 f(y)
dy:
(t 2 − y2 )m−2n
(2.3)
where
2 (2n)sin (2n − m)t 1−4n d
g(t) =
(m)
dt
Z
0
t
M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
247
Proof. To prove Theorem 1, rstly, consider the integral relation for the hypergeometric function [8],
2
2 −2n
(x + y )
1
2xy
F n; n + ; m; 2
2
x + y2
2 (m)(xy)2−2m
=
(2n) (m − 2n)
Z
2 !
t 4n−1
dt;
[(x2 − t 2 )(y2 − t 2 )]1+2n−m
min(x;y)
0
(2.4)
0¡2n¡m, and the Volterra integral equation of the rst kind with Abel’s kernel
Z
x
0
g(y)
dy = f(x);
− y 2 )
06 ¡ 1;
(2.5)
yf(y)
dy:
− y2 )1−
(2.6)
(x2
with its solution
2sin d
dx
g(x) =
Z
0
x
(x2
Secondly, multiply both sides of Eq. (2.4) by (y) and integrate with respect to y. Then the known
function of Eq. (1.1), a = 1, takes the form
2 (m)x2−2m
f(x) =
(2n) (m − 2n)
Z
x
0
t 4n−1 dt
(x2 − t 2 )1+2n−m
Z
0
1
y2−2m (y) dy
:
(y2 − t 2 )1+2n−m
(2.7)
From Eqs. (2.6) and (2.7), = m − 2n, we obtain
t 4n−1
Z
1
0
d
dt
Z
0
t
y2−2m (y)
dy =
(y2 − t 2 )1+2n−m
(2n) (m − 2n)sin (m − 2n)
(m)
y2m−1 f(y)
dy:
(t 2 − y2 )m−2n
(2.8)
Hence, the theorem is completely proved.
3. Interesting cases
Now, we are in a position to consider the following interesting cases derived from the
kernel (2.1).
1. If n = 12 and m = 23 in Eq. (2.1), we have the kernel [17] of logarithmic function k(x; y) =
1
ln(|x − y|), and the general solution of Eq. (1.1), a = 1, takes the form
2xy
2 d
− (x) = −
dx
+ (x) =
x d
dx
Z
x
Z
1
x
1
√
t dt
√
t 2 − x2
Z
t dt
t 2 − x2
t
Z
0
t
0
′
f−
(y) dy
p
;
t 2 − y2
f+′ (y) dy
p
:
y t 2 − y2
(3.1)
(3.2)
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M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
For establishing Eqs. (3.1) and (3.2) the following relations may be used:
1
|y − x|
1
;
ln(|y2 − x2 |) + ln
2
2
y+x
x
1
y
+ 2
;
= 2
2
y−x y −x
y − x2
f+ (x) = xf− (x):
ln(|y − x|) =
(3.3)
Eqs. (3.1) and (3.2) agree with the work of Mkhilarian and Abdou [15].
2. If m = n + 2 and n = p=2; 0 ¡ p ¡ 1, in Eq. (2.1). We have [16] the Carleman kernel k(x; y) =
|x2 − y2 |−p ; 0 ¡ p ¡ 1, and the solution of (1.1), in this case, takes the form
d
− (x) = −C(p)
dx
Z
x2
1− 2
t
1
x
(
+ (x) = C(p) (1 − t
2
!(p−1)
(p−1)
) 2 J (1)
2
−
d
dt
Z
x
t
Z
′
f−
(y)
0
(t 2
1
2
(t − x
2
−
y2 )
(1−p)
) 2
(1−p)
2
dy;
(3.4)
)
d p−1
(t J (t)) dt ;
dt
(3.5)
where
J (t) =
d
dt
Z
t
0
(t 2 − y2 )
(1−p)
2 f+ (y) dy;
p
(p)
2
;
1+p 2
( 2 ))
21−p cos
C(p) =
(
(a = 1):
(3.6)
The previous results agree with the work of Mkhitarian and Abdou [14].
3. If n = 14 and m = 1 the hypergeometric kernel of equation (2.1) takes the form of elliptic kernel,
√
2 xy
2
k(x; y) =
K
(x + y)
x+y
(see [17]). Kovalenko [11] developed the Fredholm integral equation of the rst kind for the
mechanics mixed problems of continuous media and obtained an approximate solution when the
elliptic kernel is reduced to Weber–Sonin integral formula
√
Z ∞
2 xy
1
J0 (xt)J0 (yt) dt;
(3.7)
= 2
K
(x + y)
x+y
0
where J0 (z) is the Bessel function of the rst kind. The general solution of Eq. (1.1), in this
case, becomes
(x) =
2
√
where
"
d
=
du
Z
0
u
1
− 2
2
1−x
Z
yf(y)
p
dy
u2 − y 2
0
#
1
√
du
d2
u2 − x2 du2
:
u=1
Z
0
u
yf(y) dy
;
u2 − y 2
p
(3.8)
(3.9)
M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
249
This result is very important in the contact problem of the zero harmonic symmetric kernel of
the potential function of zero harmonic.
4. Generalized potential kernel
In axi-symmetric contact problems for an impressing stamp of regular form of an elastic plane
into a half-space, the deformation of which obeys a power law, we consider
a three dimensional
p
2
contact problem of frictionless impression of an elastic space x; y ∈
; x + y2 = r6a; Z = 0. It
is known [4] that the integral equation of such a problem takes the form
Z Z
P(; ) d d
= f∗ (x; y); (06 ¡ 1);
(4.1)
[(x − )2 + (y − )2 ]
under the evident static condition
Z Z
P(x; y) d x dy = P (constant)¡∞:
(4.2)
Here,
is the contact domain between the stamps and the half-space, f∗ (x; y) is a known function
which will be de ned completely in the end of this section and P(x; y) is an unknown function which
describes the pressure between the stamp and the surface plane. By using the polar coordinates,
Eqs. (4.1) and (4.2) take the form
Z aZ
P(; ) d d
(4.3)
= f(r; );
2
2
− [r + − 2r cos ( − )]
0
and
Z
a
0
Z
P(; ) d d = P;
(4.4)
−
respectively. To separate the variables, one assumes
P(r; ) = Pm (r)
cos m;
sin m;
f(r; ) = fm (r)
cos m ;
sin m :
(4.5)
By using Eq. (4.5), we have from Eqs. (4.3) and (4.4)
Z
a
Z
a
Wm (r; )Pm () d = fm (r);
0
Pm () d =
0
where
P
2
0;
; m = 0;
(4.6)
(4.7)
m = 1; 2; : : : ;
cos m d
:
(4.8)
+ 2 − 2r cos ]
−
To write Eq. (4.8) in the Bessel function form. Firstly, we use the following famous relation
[8, p. 81]:
Wm (r; ) =
Z
0
2
Z
[r 2
cos m d
2( )m z m
=
F( ; m + ; m + 1; z 2 );
2
(1 − 2 cos + z )
m!
(4.9)
250
M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
in Eq. (4.8), we obtain
2 (m + ) r m
Wm (r; ) =
F(; m + ; m + 1; r 2 =2 );
m! ()2
(r ¡ );
2 (m + ) m
=
F(; m + ; m + 1; 2 =r 2 ); ( ¡ r);
m! ()r 2
r
(x) is the Gamma function. Using the famous relation ([10], p. 1070),
Wm (r; )
where
F( ; + 1=2 − ; + 1=2; z 2 ) = (1 + z)−2 F( ; ; 2 ; 4z=(1 + z)2 )
the formulas (4.10) and (4.11) take the symmetric form
(r)m
2 (m + )
Wm (r; ) =
m () (r + )2m+2
×F(m + ; m + 1=2; 2m + 1; 4r=(r + )2 ):
(4.10)
(4.11)
(4.12)
(4.13)
Secondly, we use the famous relation, ([10], p. 716),
∞
Z
J (ax)J (bx)x− d x =
0
2− a b ( + (1 − )=2)
(a + b)2 − +1 ( + 1) ((1 + )=2)
×F( + (1 − )=2; + 1=2; 2 + 1; 4ab=(a + b)2 ):
(4.14)
Hence, Eq. (4.13) becomes
Z
(1 − ) ∞
Jm (t)Jm (tr)t 2−1 dt:
(4.15)
Wm (r; ) = 2−2
2
() 0
The integral formula of Eq. (4.15) is known as Weber–Sonin
integral and
√
√ is called the symmetric
kernel of the integral equation (4.6). Let (r) = rPm (r) and f(r) = rfm (r) in Eq. (4.6), we
obtain
Z
a
Z
a
Km (r; ) () d = f(r);
√
where Km (r; ) = rWm (r; ), under the condition
(4.16)
0
√
() d =
0
P
2
0;
; m = 0;
(4.17)
m ¿ 0:
From Eq. (4.16) and taking into account the formulae (3.762) and (8.463) of [10], Eq. (4.15),
m = ±1=2, becomes
√
rW 1 (r; ) = ()cos [|r − |− ± (r + )− ]:
(4.18)
∓
2
2
Taking the limit of the last equation as → 0 using the relation
ln
1
|x − y|
|x − y|− − 1
;
→0
= lim
we obtain
√
rW 01 (r; ) = ln
∓
2
1
|r − |
± ln
1
:
r+
(4.19)
M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
251
The elliptic integral is obtained from Eq. (4.15) when m = 0 and = 12 . By putting = 21 in Eq.
(4.15) with the aid of Eq. (4.16) we obtain the Fredholm integral equation of the rst kind with
potential function kernel where
1
Km2 (x; y) = 2
Z
∞
Jm (tx)Jm (ty) dt:
(4.20)
0
5. Solution of the integral equation
To solve Eq. (4.16) with the kernel of the form (4.15) under the condition (4.17), rewrite it as
an integral equation of the Wiener–Ho type. For, setting r = ae− ; = ae− ; (ae− )ae− = ()
and f(ae− )a e− = g() in Eq. (4.15) and using Eq. (4.16), we obtain
Z
∞
L( − ) () d = g();
0
where
06 ¡ ∞;
(5.1)
L( − ) = e−(−) Km (ae− ; ae− ):
(5.2)
Popov [16] stated that in order to obtain the solution of Eq. (5.1), it suces to obtain the solution
of the most simple equation
Z
∞
L( − ) z () d = eiz ;
0
; Im z¿0:
(5.3)
Now, making use of the formulae
Z
1 ∞
G(−z) z () d z;
() =
2 −∞
G(z) =
Z
(5.4)
∞
iz
g()e d:
0
The solution of Eq. (5.3) (see [5,16]) is given by
1
z (r) =
r
z
a
ln
r
=
− (−z)r
m+
1
2
((1 + )=2)
3
Z
a−m−−1=2
m + + 21 + iz a t − 2 −m−−iz dt
×
1− +
1− ;
2
a−iz
r
2
2
2
06r ¡ 0, where
(a − x )
(t − x )
2
(5.5)
2
1
1
3
1
−1
m − + − iz
m + − iz
:
− (−z) = 2
2
2
2
2
After obtaining the solution of Eq. (5.5), one can derive the general solution of Eq. (5.2). It is easy
1
√
to see that the function r z (r) is a solution of Eq. (4.15) when f(r) = aiz r −− 2 −iz . Therefore, the
general solution of the integral equation
1−
Z
0
a
Km (r; )qm (; a) d = 1;
06r ¡ a;
(5.6)
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M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
is given by
qm (r; a) =
√
r[a−iz z (r)]z=i(1=2+) :
(5.7)
By using the principal of Krein’s method [12] the general solution of Eq. (4.15) with the aid of Eq.
(5.7) is given by
(r) =
21− r m+1=2
1+
2
d
da
Z
(a) = a−2m−
1−
2
(a)
(a2 − r 2 )
1−
2
Z
r
a
′ (u) du
(u2 − r 2 )
1−
2
1
a m+ 2
s
f(s) ds
;
(5.8)
1− :
(a2 − s2 ) 2
All the previous cases discussed in this paper can be established as special cases of Eq. (5.8).
Krein’s method (see [9,12]), is considered the best method for solving the integral equations in
applied mathematics problems where the singularity disappeared.
0
6. Examples
(1) Letting m = ± 12 , −1¡ ¡ 1 and f(x) = C2n2 (x), Cn (x) is a Gegenbaur polynomial, in (5.8)
and taking into account Eqs. (4.16) and (4.18). By using the famous formulae [8] with the aid of
(8:961) of [10],
√
√
1 (1−)
(− ;−
)
2
n+
(1 − 2t 2 ) = (−1)n
Pn 2
2
1+n+
1 (1−)
( ;−
)
2
tPn 2
(1 − 2t 2 ) = (−1)n
2
Pn( ; ) is the Jacobi polynomial, we obtain
Z
Ck=2 (y) dy
1
−1
|x − y| (1 −
1−
y2 ) 2
=
2
2
(k + )
Ck=2 (x);
cos =2 ()k!
1
=2
n+
C2n
(t);
2
n+
3
=2
C2n+1
(t);
2
|x|¡1:
(6.1)
(6.2)
(2) Letting m = ± 21 , = 0 and f(x) = Tn (x), Tn (x) is a Chebyshev polynomial in Eq. (5.8) and
taking into account Eqs. (4.16) and (4.19), we obtain
ln 2T0 (x) k = 0;
Tk (y) dy
p
(6.3)
=
1 − y2 Tk (x)
k ¿ 0:
k
√
1
(3) Letting m = 0, = 2 and f(x) = P2n ( 1 − x2 ), P2n (x) is the Legendre polynomial, in
Eq. (5.8) and using the result in Eq. (4.16), we obtain
Z 1
√
√
s ds
2
(6.4)
k(x; s)P2n ( 1 − s2 ) √
=
n P2n ( 1 − x );
1 − s2
0
1
1
ln
|x − y|
−1
Z
(n¿0; 06x61), where k(x; s) is the elliptic integral and n = 2 [(2n − 1)!]2 =[(2n)!(4n)].
M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
253
7. Conclusions
From the above discussion, the following may be concluded:
1. The displacement problems of anti-plane deformation of an in nite rigid strip with width 2a,
putting on an elastic layer of thickness h takes the form of Eq. (4.15), m = ± 21 and = 0. The
known function represents the displacement stress (see [2]).
2. The problems of in nite rigid strip, with width 2a immersed in a viscous liquid layer of thickness
h, where the strip has a velocity resulting from the impulsive force V = V0 e−i!t . Such problems
represents the Fredholm integral equation (4.16) when m = ± 21 , = 0, where V0 is the constant
velocity, ! is the angular velocity resulting from rotating the strip about z-axis. The known
function of (4.16) is given by V0 , where is the viscosity coecient.
3. Certain contact problems in the theory of elasticity is reduced to one of the forms of the integral
equation (4.15) (see [3,4,13]). These contact problems involve circular contact region, when = 12
in Eq. (4.15), in the form of a usual half-space (see [4]), or a half-space of a variable modules
of elasticity (see [12]).
4. Certain contact problems in the nonlinear theory of plasticity (in their rst approximation) is
reduced to Eq. (4.15), 6= 0, m = ± 21 for the symmetric and skew-symmetric cases, respectively
(see [13,14]).
5. The contact problems in the three dimensions with potential kernel can be obtained as a case of
Eq. (4.15), when = 12 , f∗ (x; y) = 2[ − f(x; y)], = G(1 − )−1 (see Eq. (4.1)). Here, G is
the displacement magnitude, is Poisson’s coecient, f(x; y) is a known function describes the
pressure between the stamp and the surface and is the rigid displacement under the action of
a force P (see [5]).
6. The three dimensional contact problems with potential kernel can be reduced to the Fredholm
integral equation of the rst kind of one variable. The kernel of such problems is reduced to the
form of Weber–Sonin integral,
√ Z ∞
k(u; v) = uv
Jm (tu) Jm (tv)t dt (Re(2m + 1)¿− Re ¿ 1):
0
7. The kernel of Eq. (4.15) can be expanded in the Legendre polynomial form as
Wmn (x; y)
=2
n−1
(xy)
m
∞
X
!=0
2
(1=2 + m + ! + n=2)P!m (x)P!m (y)
2 (1 + m + !)(1=2 + 2! + m + n=2)−1
(Pnm (t) is the Legendre polynomial).
8. The contact problem of zero harmonic kernel is included as a special case where m = 0. Also,
the contact problem of the rst and higher-order (m¿1) harmonic Fredholm integral is included
as a special case.
9. This paper is considered as a generalization of the work of the contact problems in continuous
media which is discussed in [4,6].
Acknowledgements
The authors thank the referee for his valuable comments and suggestions.
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M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
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[10] I.S. Gradshteyn, I.M. Ryzhik, Tables of Integrals, Series and Products, 5th ed., Academic Press, New York, 1994.
[11] E.V. Kovalenko, Some approximate methods of solving integral equations of mixed problems, Prikl Math. Mech.
53 (1989) 85–92.
[12] M.G. Krein, On a method for the e ective solution of the inverse boundary problems, Dokl. Acad. Nauk USSR 94
(6) (1954).
[13] S.M. Mkhitarian, Spectral relationships for the integral operators generated by a kernel in the form of a Weber–Sonin
integral and their application to contact problems, PMM 48 (1984) 67–74.
[14] S.M. Mkhitarian, M.A. Abdou, On di erent methods for solving the Fredholm integral equation of the rst kind
with Carleman kernel, Dokl. Acad. Nauk. Armenia 89 (1989) 125–130.
[15] S.M. Mkhitarian, M.A. Abdou, On di erent methods for solving the Fredholm integral equation of the rst kind
with logarithmic kernel, Dokl. Acad. Nauk. Armenia 90 (1989) 1–10.
[16] G.Ya. Popov, On a method of solving axisymmetric contact problems of elasticity theory, PMM 25 (1) (1961).
[17] G.Ya. Popov, Contact Problems for a Linearly Deformable Base, Odessa, Kiev, 1982.
[18] J.B. Seaborn, Hypergeometric Functions and their Applications, Texts in Applied Mathematics, vol. 8, Springer,
Berlin, 1991.
www.elsevier.nl/locate/cam
Integral equations and potential-theoretic type integrals of
orthogonal polynomials
M.H. Fahmya , M.A. Abdoub , M.A. Darwishc; ∗
a
b
Department of Mathematics, Faculty of Science, Alexandria University, Alexandria, Egypt
Department of Mathematics, Faculty of Education, Alexandria University, Alexandria, Egypt
c
Mathematisches Seminar II, Universitat Kiel, D24098 Kiel, Germany
Received 8 May 1998
Abstract
Abel’s theorem is used to solve the Fredholm integral equations of the rst kind with Gauss’s hypergeometric kernel.
The case that the kernel takes general form is discussed in detail and the solution is given. Here, the works of some
authors are considered as special cases. The discussion focuses on the three dimensional contact problems in the theory
c 1999 Published by Elsevier Science B.V. All rights reserved.
of elasticity with general kernel.
MSC: 33C05; 33C25; 45B05; 45Exx; 73C99
Keywords: Abel’s theorem; Contact problem; Fredholm integral equation; Hypergeometric kernel; Weber–Sonin integral
1. Introduction
Many problems of mathematical physics, engineering, and contact problems in the theory of
elasticity lead to the Fredholm integral equations of the rst kind, (see [4,6,12]),
Z
a
k(x; y) (y) dy = f(x):
(1.1)
0
Here, f(x) and k(x; y) are given functions, and k(x; y) is the kernel of the integral equation (1.1)
which may have a singularity while the unknown function (x) denotes a potential function. The
end points (0; a) are the points of the geometric singularity. At these points the function (x) is
bounded if it is a potential function while it has an integrable singularity if it is a
ux type quantity.
∗
Corresponding author. Permanent address: Department of Mathematics, Faculty of Education, Alexandria University,
Damanhour branch, Egypt.
E-mail address: [email protected] (Mohamed Abdalla Darwish)
c 1999 Published by Elsevier Science B.V. All rights reserved.
0377-0427/99/$ - see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 9 ) 0 0 0 6 4 - 3
246
M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
In [7], Arutyunyan has shown that the plane contact problem of the nonlinear theory of plasticity, in its rst approximation, can be reduced to Eq. (1.1) with the Carleman kernel, k(x; y) =
|x − y|− ; 06 ¡ 1.
The importance of Eq. (1.1) with Carleman kernel comes from the contact mixed problems in a
half space in the theory of elasticity wherein the modules of elasticity is changing according to the
power law. The plane contact problems for the ordinary half-space, as it is well known [14], are
equivelent to the integral equation of the rst kind with logarithmic kernel, k(x; y) = −ln(|x − y|).
Abdou [1], established Eq. (1.1) from the three dimensional contact problems in the theory of
elasticity with potential kernel where the coecients bed of the compressible materials are neglected.
In the present paper we use one integral representation of the Gauss’s hypergeometric function
to give a general solution of the Fredholm integral equation (1.1). This solution is given by means
of the composition representation of the general operator through two operators of the Volterra type
with Abel’s kernels. This paper is organized as follows: In Sections 2 and 3, Abel’s theorem is used
to solve the integral equation (1.1) when the kernel takes a certain case of hypergeometric function.
Also, the works of some authors are considered as special cases of this work. In Section 4, the three
dimensional contact problem in the theory of elasticity with general kernel. We apply in Section 5 the
Wiener–Hopf technique to study general potential kernel of two variables and corresponding integral
equation. In Section 6 a number of examples of the equations and their solutions is given. Also,
we obtained some integral relations between Gegenbauer polynomials and Chebyshev polynomials.
Finally, in Section 7 we list certain contact problems in the theory of elasticity, nonlinear theory
of plasticity and other problems of mathematical physics where these general equations and their
particular cases may appear or be applied.
2. Hypergeometric kernel
Consider the Fredholm integral equation (1.1) with the kernel
1
2xy
1
k(x; y) = 2
F n; n + ; m; 2
2
2n
(x + y )
2
x + y2
2 !
:
(2.1)
n
Here, F( ; ;
; z) = ∞
(x) is the Gamma
n=0 [( )n ( )n =n!(
)n ]z where (a)m = ( (m + a)= (a)) and
function. The function F( ; ;
; z) de ned by this series is the Gauss’s hypergeometric function (see
[8,18]).
P
Theorem 1. The general solution of Eq. (1:1) with the kernel of the form (2:1) is given by
(x) =
x2m−2
d
(1 + 2n − m) d x
Z
x
1
(t 2
tg(t)
dt;
− x2 )m−2n
(a = 1);
(2.2)
y2m−1 f(y)
dy:
(t 2 − y2 )m−2n
(2.3)
where
2 (2n)sin (2n − m)t 1−4n d
g(t) =
(m)
dt
Z
0
t
M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
247
Proof. To prove Theorem 1, rstly, consider the integral relation for the hypergeometric function [8],
2
2 −2n
(x + y )
1
2xy
F n; n + ; m; 2
2
x + y2
2 (m)(xy)2−2m
=
(2n) (m − 2n)
Z
2 !
t 4n−1
dt;
[(x2 − t 2 )(y2 − t 2 )]1+2n−m
min(x;y)
0
(2.4)
0¡2n¡m, and the Volterra integral equation of the rst kind with Abel’s kernel
Z
x
0
g(y)
dy = f(x);
− y 2 )
06 ¡ 1;
(2.5)
yf(y)
dy:
− y2 )1−
(2.6)
(x2
with its solution
2sin d
dx
g(x) =
Z
0
x
(x2
Secondly, multiply both sides of Eq. (2.4) by (y) and integrate with respect to y. Then the known
function of Eq. (1.1), a = 1, takes the form
2 (m)x2−2m
f(x) =
(2n) (m − 2n)
Z
x
0
t 4n−1 dt
(x2 − t 2 )1+2n−m
Z
0
1
y2−2m (y) dy
:
(y2 − t 2 )1+2n−m
(2.7)
From Eqs. (2.6) and (2.7), = m − 2n, we obtain
t 4n−1
Z
1
0
d
dt
Z
0
t
y2−2m (y)
dy =
(y2 − t 2 )1+2n−m
(2n) (m − 2n)sin (m − 2n)
(m)
y2m−1 f(y)
dy:
(t 2 − y2 )m−2n
(2.8)
Hence, the theorem is completely proved.
3. Interesting cases
Now, we are in a position to consider the following interesting cases derived from the
kernel (2.1).
1. If n = 12 and m = 23 in Eq. (2.1), we have the kernel [17] of logarithmic function k(x; y) =
1
ln(|x − y|), and the general solution of Eq. (1.1), a = 1, takes the form
2xy
2 d
− (x) = −
dx
+ (x) =
x d
dx
Z
x
Z
1
x
1
√
t dt
√
t 2 − x2
Z
t dt
t 2 − x2
t
Z
0
t
0
′
f−
(y) dy
p
;
t 2 − y2
f+′ (y) dy
p
:
y t 2 − y2
(3.1)
(3.2)
248
M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
For establishing Eqs. (3.1) and (3.2) the following relations may be used:
1
|y − x|
1
;
ln(|y2 − x2 |) + ln
2
2
y+x
x
1
y
+ 2
;
= 2
2
y−x y −x
y − x2
f+ (x) = xf− (x):
ln(|y − x|) =
(3.3)
Eqs. (3.1) and (3.2) agree with the work of Mkhilarian and Abdou [15].
2. If m = n + 2 and n = p=2; 0 ¡ p ¡ 1, in Eq. (2.1). We have [16] the Carleman kernel k(x; y) =
|x2 − y2 |−p ; 0 ¡ p ¡ 1, and the solution of (1.1), in this case, takes the form
d
− (x) = −C(p)
dx
Z
x2
1− 2
t
1
x
(
+ (x) = C(p) (1 − t
2
!(p−1)
(p−1)
) 2 J (1)
2
−
d
dt
Z
x
t
Z
′
f−
(y)
0
(t 2
1
2
(t − x
2
−
y2 )
(1−p)
) 2
(1−p)
2
dy;
(3.4)
)
d p−1
(t J (t)) dt ;
dt
(3.5)
where
J (t) =
d
dt
Z
t
0
(t 2 − y2 )
(1−p)
2 f+ (y) dy;
p
(p)
2
;
1+p 2
( 2 ))
21−p cos
C(p) =
(
(a = 1):
(3.6)
The previous results agree with the work of Mkhitarian and Abdou [14].
3. If n = 14 and m = 1 the hypergeometric kernel of equation (2.1) takes the form of elliptic kernel,
√
2 xy
2
k(x; y) =
K
(x + y)
x+y
(see [17]). Kovalenko [11] developed the Fredholm integral equation of the rst kind for the
mechanics mixed problems of continuous media and obtained an approximate solution when the
elliptic kernel is reduced to Weber–Sonin integral formula
√
Z ∞
2 xy
1
J0 (xt)J0 (yt) dt;
(3.7)
= 2
K
(x + y)
x+y
0
where J0 (z) is the Bessel function of the rst kind. The general solution of Eq. (1.1), in this
case, becomes
(x) =
2
√
where
"
d
=
du
Z
0
u
1
− 2
2
1−x
Z
yf(y)
p
dy
u2 − y 2
0
#
1
√
du
d2
u2 − x2 du2
:
u=1
Z
0
u
yf(y) dy
;
u2 − y 2
p
(3.8)
(3.9)
M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
249
This result is very important in the contact problem of the zero harmonic symmetric kernel of
the potential function of zero harmonic.
4. Generalized potential kernel
In axi-symmetric contact problems for an impressing stamp of regular form of an elastic plane
into a half-space, the deformation of which obeys a power law, we consider
a three dimensional
p
2
contact problem of frictionless impression of an elastic space x; y ∈
; x + y2 = r6a; Z = 0. It
is known [4] that the integral equation of such a problem takes the form
Z Z
P(; ) d d
= f∗ (x; y); (06 ¡ 1);
(4.1)
[(x − )2 + (y − )2 ]
under the evident static condition
Z Z
P(x; y) d x dy = P (constant)¡∞:
(4.2)
Here,
is the contact domain between the stamps and the half-space, f∗ (x; y) is a known function
which will be de ned completely in the end of this section and P(x; y) is an unknown function which
describes the pressure between the stamp and the surface plane. By using the polar coordinates,
Eqs. (4.1) and (4.2) take the form
Z aZ
P(; ) d d
(4.3)
= f(r; );
2
2
− [r + − 2r cos ( − )]
0
and
Z
a
0
Z
P(; ) d d = P;
(4.4)
−
respectively. To separate the variables, one assumes
P(r; ) = Pm (r)
cos m;
sin m;
f(r; ) = fm (r)
cos m ;
sin m :
(4.5)
By using Eq. (4.5), we have from Eqs. (4.3) and (4.4)
Z
a
Z
a
Wm (r; )Pm () d = fm (r);
0
Pm () d =
0
where
P
2
0;
; m = 0;
(4.6)
(4.7)
m = 1; 2; : : : ;
cos m d
:
(4.8)
+ 2 − 2r cos ]
−
To write Eq. (4.8) in the Bessel function form. Firstly, we use the following famous relation
[8, p. 81]:
Wm (r; ) =
Z
0
2
Z
[r 2
cos m d
2( )m z m
=
F( ; m + ; m + 1; z 2 );
2
(1 − 2 cos + z )
m!
(4.9)
250
M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
in Eq. (4.8), we obtain
2 (m + ) r m
Wm (r; ) =
F(; m + ; m + 1; r 2 =2 );
m! ()2
(r ¡ );
2 (m + ) m
=
F(; m + ; m + 1; 2 =r 2 ); ( ¡ r);
m! ()r 2
r
(x) is the Gamma function. Using the famous relation ([10], p. 1070),
Wm (r; )
where
F( ; + 1=2 − ; + 1=2; z 2 ) = (1 + z)−2 F( ; ; 2 ; 4z=(1 + z)2 )
the formulas (4.10) and (4.11) take the symmetric form
(r)m
2 (m + )
Wm (r; ) =
m () (r + )2m+2
×F(m + ; m + 1=2; 2m + 1; 4r=(r + )2 ):
(4.10)
(4.11)
(4.12)
(4.13)
Secondly, we use the famous relation, ([10], p. 716),
∞
Z
J (ax)J (bx)x− d x =
0
2− a b ( + (1 − )=2)
(a + b)2 − +1 ( + 1) ((1 + )=2)
×F( + (1 − )=2; + 1=2; 2 + 1; 4ab=(a + b)2 ):
(4.14)
Hence, Eq. (4.13) becomes
Z
(1 − ) ∞
Jm (t)Jm (tr)t 2−1 dt:
(4.15)
Wm (r; ) = 2−2
2
() 0
The integral formula of Eq. (4.15) is known as Weber–Sonin
integral and
√
√ is called the symmetric
kernel of the integral equation (4.6). Let (r) = rPm (r) and f(r) = rfm (r) in Eq. (4.6), we
obtain
Z
a
Z
a
Km (r; ) () d = f(r);
√
where Km (r; ) = rWm (r; ), under the condition
(4.16)
0
√
() d =
0
P
2
0;
; m = 0;
(4.17)
m ¿ 0:
From Eq. (4.16) and taking into account the formulae (3.762) and (8.463) of [10], Eq. (4.15),
m = ±1=2, becomes
√
rW 1 (r; ) = ()cos [|r − |− ± (r + )− ]:
(4.18)
∓
2
2
Taking the limit of the last equation as → 0 using the relation
ln
1
|x − y|
|x − y|− − 1
;
→0
= lim
we obtain
√
rW 01 (r; ) = ln
∓
2
1
|r − |
± ln
1
:
r+
(4.19)
M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
251
The elliptic integral is obtained from Eq. (4.15) when m = 0 and = 12 . By putting = 21 in Eq.
(4.15) with the aid of Eq. (4.16) we obtain the Fredholm integral equation of the rst kind with
potential function kernel where
1
Km2 (x; y) = 2
Z
∞
Jm (tx)Jm (ty) dt:
(4.20)
0
5. Solution of the integral equation
To solve Eq. (4.16) with the kernel of the form (4.15) under the condition (4.17), rewrite it as
an integral equation of the Wiener–Ho type. For, setting r = ae− ; = ae− ; (ae− )ae− = ()
and f(ae− )a e− = g() in Eq. (4.15) and using Eq. (4.16), we obtain
Z
∞
L( − ) () d = g();
0
where
06 ¡ ∞;
(5.1)
L( − ) = e−(−) Km (ae− ; ae− ):
(5.2)
Popov [16] stated that in order to obtain the solution of Eq. (5.1), it suces to obtain the solution
of the most simple equation
Z
∞
L( − ) z () d = eiz ;
0
; Im z¿0:
(5.3)
Now, making use of the formulae
Z
1 ∞
G(−z) z () d z;
() =
2 −∞
G(z) =
Z
(5.4)
∞
iz
g()e d:
0
The solution of Eq. (5.3) (see [5,16]) is given by
1
z (r) =
r
z
a
ln
r
=
− (−z)r
m+
1
2
((1 + )=2)
3
Z
a−m−−1=2
m + + 21 + iz a t − 2 −m−−iz dt
×
1− +
1− ;
2
a−iz
r
2
2
2
06r ¡ 0, where
(a − x )
(t − x )
2
(5.5)
2
1
1
3
1
−1
m − + − iz
m + − iz
:
− (−z) = 2
2
2
2
2
After obtaining the solution of Eq. (5.5), one can derive the general solution of Eq. (5.2). It is easy
1
√
to see that the function r z (r) is a solution of Eq. (4.15) when f(r) = aiz r −− 2 −iz . Therefore, the
general solution of the integral equation
1−
Z
0
a
Km (r; )qm (; a) d = 1;
06r ¡ a;
(5.6)
252
M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
is given by
qm (r; a) =
√
r[a−iz z (r)]z=i(1=2+) :
(5.7)
By using the principal of Krein’s method [12] the general solution of Eq. (4.15) with the aid of Eq.
(5.7) is given by
(r) =
21− r m+1=2
1+
2
d
da
Z
(a) = a−2m−
1−
2
(a)
(a2 − r 2 )
1−
2
Z
r
a
′ (u) du
(u2 − r 2 )
1−
2
1
a m+ 2
s
f(s) ds
;
(5.8)
1− :
(a2 − s2 ) 2
All the previous cases discussed in this paper can be established as special cases of Eq. (5.8).
Krein’s method (see [9,12]), is considered the best method for solving the integral equations in
applied mathematics problems where the singularity disappeared.
0
6. Examples
(1) Letting m = ± 12 , −1¡ ¡ 1 and f(x) = C2n2 (x), Cn (x) is a Gegenbaur polynomial, in (5.8)
and taking into account Eqs. (4.16) and (4.18). By using the famous formulae [8] with the aid of
(8:961) of [10],
√
√
1 (1−)
(− ;−
)
2
n+
(1 − 2t 2 ) = (−1)n
Pn 2
2
1+n+
1 (1−)
( ;−
)
2
tPn 2
(1 − 2t 2 ) = (−1)n
2
Pn( ; ) is the Jacobi polynomial, we obtain
Z
Ck=2 (y) dy
1
−1
|x − y| (1 −
1−
y2 ) 2
=
2
2
(k + )
Ck=2 (x);
cos =2 ()k!
1
=2
n+
C2n
(t);
2
n+
3
=2
C2n+1
(t);
2
|x|¡1:
(6.1)
(6.2)
(2) Letting m = ± 21 , = 0 and f(x) = Tn (x), Tn (x) is a Chebyshev polynomial in Eq. (5.8) and
taking into account Eqs. (4.16) and (4.19), we obtain
ln 2T0 (x) k = 0;
Tk (y) dy
p
(6.3)
=
1 − y2 Tk (x)
k ¿ 0:
k
√
1
(3) Letting m = 0, = 2 and f(x) = P2n ( 1 − x2 ), P2n (x) is the Legendre polynomial, in
Eq. (5.8) and using the result in Eq. (4.16), we obtain
Z 1
√
√
s ds
2
(6.4)
k(x; s)P2n ( 1 − s2 ) √
=
n P2n ( 1 − x );
1 − s2
0
1
1
ln
|x − y|
−1
Z
(n¿0; 06x61), where k(x; s) is the elliptic integral and n = 2 [(2n − 1)!]2 =[(2n)!(4n)].
M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
253
7. Conclusions
From the above discussion, the following may be concluded:
1. The displacement problems of anti-plane deformation of an in nite rigid strip with width 2a,
putting on an elastic layer of thickness h takes the form of Eq. (4.15), m = ± 21 and = 0. The
known function represents the displacement stress (see [2]).
2. The problems of in nite rigid strip, with width 2a immersed in a viscous liquid layer of thickness
h, where the strip has a velocity resulting from the impulsive force V = V0 e−i!t . Such problems
represents the Fredholm integral equation (4.16) when m = ± 21 , = 0, where V0 is the constant
velocity, ! is the angular velocity resulting from rotating the strip about z-axis. The known
function of (4.16) is given by V0 , where is the viscosity coecient.
3. Certain contact problems in the theory of elasticity is reduced to one of the forms of the integral
equation (4.15) (see [3,4,13]). These contact problems involve circular contact region, when = 12
in Eq. (4.15), in the form of a usual half-space (see [4]), or a half-space of a variable modules
of elasticity (see [12]).
4. Certain contact problems in the nonlinear theory of plasticity (in their rst approximation) is
reduced to Eq. (4.15), 6= 0, m = ± 21 for the symmetric and skew-symmetric cases, respectively
(see [13,14]).
5. The contact problems in the three dimensions with potential kernel can be obtained as a case of
Eq. (4.15), when = 12 , f∗ (x; y) = 2[ − f(x; y)], = G(1 − )−1 (see Eq. (4.1)). Here, G is
the displacement magnitude, is Poisson’s coecient, f(x; y) is a known function describes the
pressure between the stamp and the surface and is the rigid displacement under the action of
a force P (see [5]).
6. The three dimensional contact problems with potential kernel can be reduced to the Fredholm
integral equation of the rst kind of one variable. The kernel of such problems is reduced to the
form of Weber–Sonin integral,
√ Z ∞
k(u; v) = uv
Jm (tu) Jm (tv)t dt (Re(2m + 1)¿− Re ¿ 1):
0
7. The kernel of Eq. (4.15) can be expanded in the Legendre polynomial form as
Wmn (x; y)
=2
n−1
(xy)
m
∞
X
!=0
2
(1=2 + m + ! + n=2)P!m (x)P!m (y)
2 (1 + m + !)(1=2 + 2! + m + n=2)−1
(Pnm (t) is the Legendre polynomial).
8. The contact problem of zero harmonic kernel is included as a special case where m = 0. Also,
the contact problem of the rst and higher-order (m¿1) harmonic Fredholm integral is included
as a special case.
9. This paper is considered as a generalization of the work of the contact problems in continuous
media which is discussed in [4,6].
Acknowledgements
The authors thank the referee for his valuable comments and suggestions.
254
M.H. Fahmy et al. / Journal of Computational and Applied Mathematics 106 (1999) 245–254
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