Internat. J. Math. & Math. Sci.
VOL. 15 NO. 3 (1992) 553-562
553
ON DUAL INTEGRAL EQUATIONS ARISING IN PROBLEMS
OF BENDING OF ANISOTROPIC PLATES
B.O. AGGARWALA and C. NASlM
Department of Mathematics and Statistics
University of Calgary
Calgary, Alberta, Canada
T2N 1N4
(Received March 6, 1991 and in revised form April 11, 1991)
ABSTRACT. In this paper we consider dual integral equations, which arise in boundary
value problems of bending of anisotropic plates. The function involved in these equations
is a linear combination of elementary function, which turns out to be a particular case of
a class of Fourier kernels, [2]. The method used here for solving the equations is some
what similar to the method used for solving dual integral equations of Titchmarsh type,
KEY WORDS AND PHRASES. Dual integral equations, Bessel functions, Erdelyi-Kober
operators, Parseval theorem, Mellin transforms.
1991 AMS SUBJECT CLASSIFICATION CODE: 45F10.
INTRODUCTION
In this note we consider dual integral equations
where
0
0
(t)h,(xt)dt
f(x),
0<x<l
(1.1)
(t)h,(xt)dt
g(x), x>l,
is the unknown function and
h,(x)
h,(x)
e’X cosx sinx
x"h(x),
-_<_.
The method we employ for solving the system (1.1) is similar to a method developed by
Nasim & Sneddon, [1]. This procedure has been very effective for solving dual integral
equations when the functions h, and h involve Bessel functions J,,, Y,, and K,. As in
almost all the papers concerned with dual integral equations, here also, the solutions are
not derived in a rigorous fashion. But that is not to say that the analysis in this paper
cannot be made rigorous by imposing appropriate conditions on the functions involved.
It is worth noting that in our case the functions h, are a special case of Fourier
kernels defined by us elsewhere, [2]. Also we wish to point out that the functions h,(x)
B.D. AGGARWALA and C. NASIM
554
satisfy the differential equation
d_.._ h(Ax)
h(
dx4
and equations (1.1) would therefore arise in many situations where we wanted to solve a
A particular such case e.g., is the
partial Differential Equation involving the operator
.a4
bending of an anisotropic plate where the lateral deflection w satisfies the equation
o-4
+
2k
ow + 4
O.
The case k=0 is of. some importance for such plates, [3]. If we wanted to solve the
resulting equation (k=0) in x>0, y>0 with the plate clamped along x=0 and having.
discontinuous boundary nditions .along y=0, we would quickly arrive at equations (1.1).
For, in such a situation (with k=0), we seek solution w(x,y) in the form
-
+
sin
x]d,
where f() is to be determined. This form is seen to satisfy the Partial Differential
Equation, the condition of clampness along x=0 (i.e. w
0, on x=0, y>0) and the
condition that w=-0 on y=0 in x>0. If now, the plate is bent by bending moments of of
magnitude In(x) in 0<x<l (on
0) and is clamped along x>l, (on y=0), then f(,) must
satisfy
,f(,)(e-x
cos ,x
0 f(A)(e-AX-
and
+
cos Ax
sin
+
,x)dx
sin
Ax)dx
In(x)
in 0<x<l
0
in x>l
where is an appropriate (material) constant, [3]. These equations are a particular case
of equations(1.1). We propose to solve such equations in this note.
2.
PRELIMINARIES.
We shall need the following known definitions and results.
Mellin Transform theory can be found in Titchmarsh [4].
The results from the
We define the Mellin Transform and inverse Mellin Transform, under appropriate
conditions, respectively as:
M[fCx); s]
’[f*(");x]
fCx) x-’dx
f*(s), where s==+ir, a<a<b,
7i o,-i(R) t*(s)’d
f(x).
---<r<(R), and
[+i(R)
Throughout this note, we shall denote the Mellin Transform of a function f by ft.
A function k is said to be a Fourier Kernel when for
and
0
0
k(xt) f(t) dt
k(xt) g(t)dt
f(x),
DUAL INTEGRAL EQUATIONS IN PROBLEMS OF BENDING OF PLATES
for some suitable functions f & g.
k(x), then
555
,
Furthermore, if k (s) denotes the Mellin transform of
k
(s)
k (i-s)
on some line of the complex s-plane.
The Erdelyi-Kober
[5] operators
are given by,
a ’(7+a)I(x,_t,)o/ t(n’t)-
7,aI(0,x;a) f
f(t)dt,
a>0
and
Va xolxCtO-xo)a" ’
TK, a(x’(R):a)f
’-a-1)-’
f(t)dt, a>0.
It is an easy matter to see that
r(l+,/-s/) f* (s)
+ = -s/a)
M
In,=(0,x:a) f(x):
M
r( + /) f
[K7,=(x,(R);a)f(x); s] =r(+
+ sly)
r(l+n
and
The inversion form of those equations give us the following two results.
M-’
Lemma 1.
r(, +
r(l +
’7
+
-}) .();
x]
a-)
I (O,x; a)M"[f*(s); x]
I (O,x; a)f
M"
Lemma 2.
Note that
/I,0
r( +
+)
f*(s);
x
r/K,a (x,(R);
a)M-’[f*(s); x]
K,= (x,(R);
)f
I, the identity operator.
/K, 0
Now an important result from the theory of Mellin Transform.
Lemma 3. The Parseval Theorem, [4]
jof x)g(.)1T
M-’[f*(s)g*(s); x]
dt
fog,
the convolution of f and g.
Next we give a formal description of the method we shall employ in solving.dual
integral equations (1.1) for azbitrary functions h and h 2. These dual equations (1.1), are
equivalent to
0
m,( )dx
0
((u)ht(ux)du
el(t),
0<t<l
(2.1)
(R)
1 dx
m.()
(u)h,(ux)du
t>l.
B.D. AGGARWALA and C. NASIM
556
Here the functions t and 2 are defined in terms of the functions f and g and
arbitrary functions m and m2 by the equations.
It0
tCt)
and
fCx)
m
tl dx
mt()
g(x)m,()1 dx.
(t)
Supse now we find functions
(as yet)
m
and
(2.3)
such that
m,(y)h,(y)y dy
m,(y)h,() dy
(2.4)
k(z), say
0
is the solution of the single integrM equation
then the unknown function
[0 (u)k(ut) du
t)
J
(2.5)
where
t)
(t) H(1-t)+ ,(t) H(t-1),
H(t) being
the Heaviside function. To determine the arbitrary functions
exploit the theory of Mellin Transforms.
m
and m2, we
Now, due to the result of lemma 3, we can write the equation (2.4) as
mt(s)ht(s
m2(s)h2(s
whence
,
h(s)/(s
(s),
,
m(s)
m(s)
It is then possible to decompose
k
in such a way so that m t(s and
appropriately determined and eventually the functions mt and m2 and hence
then known. Next, to find
from (2.5), we have due to lemma 3,
*(l--s)k*(s)
*(s)
i.e.,
*(1--s)
k’(l-s)
l
m:(s)
and
are
are
*(s),
#*(1-s)h*(s) say
Then the last equation is equivalent to
t)h(xt)dt,
system (1.1), where
(x)
giving us the required solution of the
h(x)=
3.
M-l[l,(1._s); x].
1
THE DUAL INTEGRAL EQUATIONS I.
We consider now
(t)h(xt)dt
f(x),
0<x<l
(3.1)
(t)h2(xt)dt
where
g(x), x>l
DUAL INTEGRAL EQUATIONS IN PROBLEMS OF BENDING OF PLATES
e-X-cos x
ht(x
+
557
sin x,
and
x’(e -x- cos x +
h2(x
sin
x), -1<__I,
the functions f and g are known and ( is to be determined. Now, for an appropriate
complex s, the Mellin transforms of h and h2 are, respectively, [6, Chapter 6],
2 r(s)sin
:-,
r(1/2 + )r(3 + )
i 3
)r(
r(1
,
h l(s+u)
Then
m,(s)
mi(s)
1/4z sin1/4(s+l)
1/41
-
r(1
r(1/2 +
+
r(1/2 + )r( + 1/4)r(
3
’r(1
r(
hi(s
h2(s)
whence,
,
m l(s)
2
1/4)r( )
)
r(-- 1/4)r(
1/4)r( +
From the equation (2.2),
Io ,()
0 f(x)mt(t-)l-’xx
1 d
f(x)m
where l(x)
for -l<v<0,
-
)
+
and
m(s)
)r(
mi(x H(x-1), H being the
dx
+
Heaviside function.
1/4)
Due to lemma 3, we have
M"[mi(s)f (s); t]
-2v I
y
(0,t:4)
0,-
_,_
I
by repeated use of lemma 1.
But if 0<v<l, then
Ct(t)
M-l[22vr(1 r( 1/4)r(1
1/4)r(
I)
I)
if(S);
t]
r(s);
(0,t;4)f,
t]
(3.2)
B.D. AGGARWALA and C. NASIM
558
I
-,1
-1
v
2 i’’ I 1 1
where
Next,
F(x)
om
(0,t;4)
0, 3
(0,t; 4) I
3
-,0,x- [x-f(x)].
equation
(2.3)
,(x)
(0’t;4)M’i
(0,t; 4)F,
(3.2a)
we have
g(x)
m
g(x)
’i
0
where
v
m(x)H(1-x).
Lr(+1/4ir(+-+1/4)
K1
dx,
Then by lemmas 2 and 3, for 0<v<l,
3 s
r 1 s
(+1 r(l-%r)
M -,r
-
x dx
v(t,(R); 4)K (t,(R);
(3.3)
4)g
But if-l<u<0, then
KI" +.:..,(,,,,,, [1"()
,,.),.,i:-,
*(,,);
.,,i
K1/2, 1/4+(t,(R);
4)M -,r
[r(_+l
(2+v+s)g*(s);
,]
,l
(3.3a)
where
G()
.+’
(-.- g()).
Hence the function
i(t)H(1-t) + ll(t)H(t-1)
(t)
is completely known, with
Further, from (2.7),
Ct
and
1
as defined above by the equations
(3.2)
and
(3.3).
DUAL INTEGRAL EQUATIONS IN PROBLEMS OF BENDING OF PLATES
559
(s)
k
Then, [7 Sec. 20.5],
M-[k*(1-s) ;x
1
h(x)
-1/22
-
3
G
) :I
Finally the solution of the dual equations
0(u)
11i,(1
.
(3.1), can now be
Nt)h(ut)at, -1 <
.
written as
v
0
Note that the solution is also valid for u
(3.4)
g,, 0
g,g
<
-1.
Next we list two special cases, which are of some interest.
1, g(x)
Let v
,
f(x)
0 and
x
a>-2. Then
o
,(t)
and from
where
That is,
F(x) =1
(3.2a),
Io, 1/2 (0,t; 4)F
Ct(t)
( + )x
(2+a)
I0,1/2(0,t; 4)x=
2-1/2(2+a) t’
r(l+) =
=4.
I
(t
0
-x4)-1/2
xS/tdx
The solution of the integral equations
x=
(t)ht(xt)dt
0<x<l
(3.5)
t0(t h,(xt)dt
0, x>l
is then given by
)(x)
where from
4
I’(I+D
i,(+[
1
IO
h(xt)t=dt
(3.4),
h(x)
r
Evaluating the above integral, we obtain,
04
[7 Sec. 20.5], for
a
> -2,
B.D. AGGARWALA and C. NASIM
560
as the solution of the system
Let v
-1,
Then
and from
f(x)
0 and
x,
g(x)
<
l(t)
-.
1
0
(3.3a),
1/4 K3 l(t,(R); 4)G,
2(t)
(-9
’
(1-)x0
G()
where
Therefore,
1-) K 3 1(t,(R):4)x3
-1/2(-B)t (
tO -1/2
-
d
Hence the solution of dual equation
0(t hl(xt)dt
0,
0<x<l
(3.6)
(xt)-)(t)hCxt)dt
l<x<(R)
is given by
(x)
2
Now from (3.4),
o],
and on evaluating the above integral, we obtain,
3
(x)
16 x "1/2
as the solution of the system
4.
0
1"( )
’s
1,,- ]1
(3.6).
DUAL INTEGRAL EQUATIONS II.
Next we consider the system (1.1) with
h(x)
ex
+
cos x-sin x
’I’
DUAL INTEGRAL EQUATIONS IN PROBLEMS OF BENDING OF PLATES
h.(x)
h(x),-1<u_1.
x
Now, [6],
2V r(s)sin
1/222s-s
1/4 2-l-s)sin 1/4l-s)
r( )r( )
and h,(s)
h1(s+v).
The equation (2.7) is satisfied, if we set
,
In
r( )r(1/2 )
2v
1(s)
m:(s)
r( + )r( +
+
1/4)
and
,
k
(s)
/F
r()r(1/4 + )
2 s*"-s
Let 0<v< 1, then as before
M"[m:z(s)g (s); t]
r(1/4 + )r()
M-’[
lv
K lv(t,(R); 4) K
[,[
-I
v(t,(R);
0,
1
Jo f(x)m,() dx
$,(t)
M"[m,(s)f (s); t]
M-’
where
F(x)
2’-’x
I)
2’-’r(1/2
I 3 1
u
[,
1
I)r(
(0,x; 4)I
,,
1 3
[
I)
(1-v-s)f*(s);
FI
v,0,x:4,
--(x"vfCx)).
Now let -l<v<0. Then, as above
,(t)
M.,[2
s
1
)s
r(1 )r(9_1 v s
I v_
[)
r(--)r(-
i’* (S);
t]
t]
561
562
- -
B.D. AGGARWALA and C. NASIM
2tuI 3
4) I
4)f
-, v(O,x; -,1 {O,x;
and
r(1)r(1/4 + l)
" [r( +
’(’)
+ r)r( + +
)
(+,).();
,]
Hence the solution of the system is given by
t)h(xt)dt,
(x)
(t) Pt(t)H(1-t)+p2(t)H(t-1),
where
h(x)
and
M"[-,(l_s);
x]
9
1
r" 1/2 2
Special cases when v
’
G
04
([)
vl
;i’[
v311
’[’J"
1 can easily be derived from the general solution.
We also note that the solution of equations (1.1) where
h,(x)
(Y0(x),
2
K0(x))
and
h,(x)
xh,(x),
-I_<_1
can also be obtained along similar lines. It is well known that the functions
Fourier Kernels, [2].
hi(x) are
This research is partially supported by grants from Natural Sciences and Engineering
Research Council of Canada.
REFERENCES
Nasim, C; Sneddon; I.N. A General Procedure for Deriig Solutions of Dual Integral
Equations. J. Engg. Math. Vol. 12, No. 2, April 1978, 115128.
Aggarwala, B.D; Nasim, C. Solutions of an Ordinary Differential Equation as a Class
of Fourier Kernels, Internat. J. Math. & Math Sci. Vol. 13, No. 2, (1990),
397-404.
3.
Timo Shenko, S. Theory
of Plates
and Shells, McGraw Hill, New York, 1989.
Titchmarsh, E.C. Introduction to the Theory
Oxford University Press. Oxford, 1966.
of Fourier Integrals,
second edition,
Erddyi, A. Some Dual Integral Equations, SIAM J. of Appl. Math, Vol. 16
1338-1340.
Erddyi, A. et al.
Tables
of Integral Transforms Vot I & II, McGraw Hill, New
Tables
of Integral Transforms,
York, 1954.
Erddyi, A. et al.
1954.
(1968),
The University of Calgary
Calgary, Alberta.
Vol II., McGraw Hill, New York,