Internat. J. Math. & Math. Sci.
Vol. 22, No. 1 (1999) 155–160
S 0161-17129922155-2
© Electronic Publishing House
NUMERICAL SOLUTION OF INTEGRAL EQUATIONS
WITH FINITE PART INTEGRALS
SAMIR A. ASHOUR
(Received 22 July 1994 and in revised form 29 March 1994)
Abstract. We obtain convergence rates for several algorithms that solve a class of
Hadamard singular integral equations using the general theory of approximations for unbounded operators.
Keywords and phrases. Singular integral equations.
1991 Mathematics Subject Classification. 65D05, 45L05.
1. Introduction. In several physical problems in aerodynamics, hydrodynamics,
and elasticity, one encounters integral equations of the form
1
Ax =
π
1 √
−1
1 − τ 2 x(τ)
1
dτ +
(τ − t)2
π
1 1 − τ 2 h(t, τ) x(τ) dτ = y,
−1
(1.1)
where the first integral in (1.1) is a finite part integral [4]. Under suitable conditions
on the kernel and the right-hand side, the convergence of Galerkin’s method and
several collocation methods, proposed by Ioakimidis [5] and Williams [9], has been
discussed by Golberg [2, 3]. This author, also, used a classical Fredholm theory to
establish the existence of a solution and likewise the basic tools necessary to discuss
convergence. In this paper, we discuss the convergence of the mechanical quadratures
method for solving (1.1) and the convergence of the least squares method for solving
the Hadamard singular integral equation
Kx =
1
π
1 √
−1
1 − τ 2 x(τ)
dτ + T (x, t) = y,
(τ − t)2
(1.2)
where T is the given continuous operator.
2. Least squares method. Let X = L2,ρ denote the space of square integrable func√
tions with respect to ρ = 1 − t 2 . The inner product on L2,ρ is given by
(φ, ψ)ρ =
2
π
1
−1
ρ(t) φ(t) ψ(t) dt
and φρ = (φ, φ)ρ .
(2.1)
Let
sin (m + 1) arccos t
√
,
Um (t) =
1 − t2
m = 0, 1, 2, . . .
(2.2)
156
SAMIR A. ASHOUR
denote the Chebyshev polynomials of the second kind. The solution x is, now, approximated by
xn (t) =
n
αk Uk−1 (t),
−1 ≤ t ≤ 1.
(2.3)
k=1
According to this method, we obtain a system of n linear algebraic equations in n
unknowns
n
i=1
αi KUi−1 , KUj−1 ρ = y, KUj−1 ρ ,
1 ≤ j ≤ n.
(2.4)
It is easy to prove that (2.4) is equivalent to
n
k=1
αk
T Uk−1 , T Uj−1
ρ −j
T Uk−1 , Uj−1
T Uj−1 , Uk−1 ρ + j 2 αj
ρ −k
= y, KUj−1 ρ ,
1 ≤ j ≤ n.
(2.5)
Theorem 2.1. If the following conditions hold
(i) y ∈ L2,ρ , T is a continuous operator in L2,ρ ;
(ii) ker K = {0};
(iii) equation (1.2) has a solution x ∗ ∈ L2,ρ for a given y ∈ L2,ρ , then for all n ∈ N
n
equation (2.5) has a unique solution α∗
k 1 ; and if
(iv) KUi−1 is closed in L2,ρ ,
then
∗
rn ρ = y − Kxn
ρ → 0
as
n → ∞,
∗
xn
=
n
k=1
α∗
k Uk−1 (t).
(2.6)
Proof. Since {Uk−1 } are linearly independent, then, from (ii), it follows that{KUk−1}
are, also, linearly independent. Therefore, the system of equations (2.5) is non-singular
and so, it has a unique solution for all n. Also, for βk ∈ R, we get
n
∗
rn ρ = y − Kxn
βi KUi−1 .
≤ y −
i=1
ρ
(2.7)
If condition (iv) is satisfied, then rn → 0 as n → ∞.
Now, we replace condition (ii) by the following condition:
(ii) K : L2,ρ → L2,ρ has a left bounded inverse operator Kl−1 .
∗
Theorem 2.2. Assume that (i), (ii) , (iii), and (iv) are satisfied, then xn
− x ∗ ρ =
O rn ρ → 0 as n → ∞.
∗
∗
∗
= Kl−1 K x ∗ − xn
= Kl−1 y − Kxn
,
Proof. From (ii) and (iii), we have x ∗ − xn
∗
then xn
− x ∗ ρ → 0 as n → ∞.
3. Mechanical quadratures methods. We introduce the following method for solving (1.1): Consider the approximation xn of x given by (2.3). Due to this method, we
get the following
157
NUMERICAL SOLUTION OF INTEGRAL EQUATIONS . . .
n
αi − iUi−1 tj +
i=1
n
1 1 − tr2 h tj , tr Ui−1 tr
= y tj ,
n + 1 r =1
1 ≤ j ≤ n,
(3.1)
where tj = cos(jπ /n + 1).
Theorem 3.1. If the following conditions
(i) y ∈ C[−1, 1], h ∈ C[−1, 1; −1, 1]
(ii) equation (1.1) has a unique solution x ∗ ∈ X for all y ∈ X
are satisfied, then, for n sufficiently large, equation (3.1) has a unique solution α∗
k,
∗
x − x ∗ = O En (y)C + E t (h)C + E τ (h)C ,
n ρ
n
n
(3.2)
where Ent (f )C = inf f − Pn C , Pn is a polynomial of degree ≤ n .
Proof. Define the projection operator Pnt : C → X by
Pnt (x) =
n
k=1
x tk Un (t)
.
t − tk Un tk
(3.3)
Thus, equation (3.1) can be written in the equivalent form
An xn = Gxn + Pnt HPnτ hxn = Pnt y,
xn , P n y ∈ X n ,
n
where Xn = xn : xn = k=1 αk Uk−1 (t), t ∈ [−1, 1] ,
√
1 1 1 − τ 2 x(τ)
dτ, x ∈ X,
Gx =
π −1
(τ − t)2
1 1 1 − τ 2 h(t, τ)x(τ) dτ.
Hhx =
π −1
Since Pnt HPnτ hxn = Pnt H (Pnτ h)xn , then, for all xn ∈ Xn , we get
Axn − An xn = Hhxn − P t HP τ hxn n
n
ρ
ρ
t t
≤ Hh − Pn Hh xn ρ + Pn H h − Pnτ h xn ρ .
According to [8], Pn ≤ C1 , x − Pn xρ ≤ C2 En (x)C , x ∈ C[−1, 1], we get
Axn − An xn ≤ C2 En Hhxn + C1 H h − P τ xn n
ρ
C
= O Ent (h)C + Enτ (h)C xn ρ ,
(3.4)
(3.5)
(3.6)
(3.7)
so that
A − An Xn →X
= O Ent (h)C + Enτ (h)C = (n .
(3.8)
According to [1], for all n such that A−1 (n < 1, An has a bounded inverse and
A−1
n = O(1), An : Xn → Xn . Since y − Pn yρ = O{En (y)C } = δn . Finally, we have
∗
x − x ∗ = A−1 y − A−1 Pn y n ρ
n
ρ
= O ( n + δn
(3.9)
= O Ent (h)C + Enτ (h)C + En (y)C .
158
SAMIR A. ASHOUR
Lemma 3.1. If Qn is an algebraic polynomial of degree n − 1, then
Qn (t) ≤ n n Qn , n ≥ 2.
C
ρ
2
(3.10)
n
Proof. One may write Qn as Qn (t) = k=1 Ck−1 (Qn )Uk−1 (t), n ∈ N, where Cj (Qn )
= (Qn , Uj )ρ . Since |Uk−1 | ≤ k it follows that
1/2 
1/2

n
n


  2
2
Qn (t) ≤
Ck−1 (Qn )
k
C
 


k=1
k=1
(3.11)
= Qn ρ n(n + 1)(2n + 1)/6
≤ n n/2 Qn ρ
= Qn ρ O n3/2 .
Define W r Hα = x : x (r −1) is absolutely continuous, x (r ) ∈ Hα .
Theorem 3.2. Assume that conditions (i) and (ii) of Theorem 3.1 are satisfied,
y(t) ∈ W r Hα , h(t, τ) ∈ W r Hα ,
then
r ≥ 0, 0 < α ≤ 1,
∗
x − x ∗ = O n−r −α ,
n ρ
∗
x − x ∗ = O n3/2−r −α .
n C
(3.12)
(3.13)
(3.14)
Proof. Since y(t) ∈ W r Hα , h(t, τ) ∈ W r Hα , then, according to [7], one has En (y)
= O(n−r −α ), Ent (h) = O(n−r −α ), Enτ (h) = O(n−r −α ), r + α > 0. This proves (3.13). It is
easy to show that
∞
∗
∗
x − x ∗ =
x
− x2∗k−1 n C
n C
2k n
k=1
≤
(3/2) ∗
x − x ∗ + x ∗ − x ∗
2k n ρ
2k−1 n ρ
∞
2k n
k=1
≤ C3
∞
(3/2) 2k n
−r −α
2k n
(3.15)
= C4 n−r −α+3/2 ,
k=1
where C3 , C4 are constants. This proves (3.14).
√
√
Define Cρ [−1, 1] = x : 1 − t 2 x ∈ C[−1, 1] and xCρ = max 1 − t 2 |x(t)| .
Lemma 3.2. If Qn is a polynomial of degree n − 1, then
Qn (t) ≤ √n Qn .
Cρ
X
Proof. Since |Um (t)| ≤ 1 − t
1 − t 2 Qn (t) ≤
2 −1/2
n
, −1 ≤ t ≤ 1, m = 0, 1, . . . , then
Ck−1 Qn k=1
≤
(3.16)
n
Ck−1 Qn 2
k=1
1/2
√
√ n = nQn ρ .
(3.17)
NUMERICAL SOLUTION OF INTEGRAL EQUATIONS . . .
159
∗
X = O(n−m ), m ∈ R, then
Theorem 3.3. If x ∗ − xn
∗
x − x ∗ n Cρ
= O n1/2−m ,
m>
1
,
2
(3.18)
Proof. Using Lemma 3.2 and the same technique as in Theorem 3.2, one obtains (3.18).
Conclusion. For the Mechanical Quadratures methods, the rate of convergence
in space Cρ [−1, 1] is better than that given in space C[−1, 1].
4. Approximation by degenerate kernels. We approximate h(t, τ) by
n
hn (t, τ) =
ak (t)bk (τ),
−1 ≤ t, τ ≤ 1,
(4.1)
k=1
where ak , bk are two sets of linearly independent functions. By substituting back
into (1.1), we obtain
An x = Gx +
1 1 − τ 2 hn (t, τ) x(τ) dτ = y.
−1
(4.2)
(4.2) can be written as
Gx +
n
αk ak = y,
(4.3)
k=1
where αk =
1 √
2
−1 1 − τ x(τ)bk (τ) dτ. Then the solution of (4.3) is given by
x = G−1 y −
n
αk G−1 ak ,
k=1
G
−1
y =−
∞
y, Uk−1 ρ
k=1
Multiplying (4.4) by
k
(4.4)
Uk−1 (t).
√
1 − t 2 bj (t) and integrating, we get the linear system of equations
αj +
n
γjk αk = yj ,
1 ≤ j ≤ n,
(4.5)
k=1
where γjk =
√
1 − τ 2.
√
1 √
1 √
−1
2
(ak ) dt, yj = −1 1 − t 2 bj G−1 (y) dt. Define q = 1 − t 2
−1 1 − t bj G
Theorem 4.1. Suppose that
(i) y ∈ L2,ρ
(ii) h ∈ L2,q [−1, 1; −1, 1]
1 1
2
2
(iii) (n
= −1 −1 q(t, τ)h(t, τ) − hn (t, τ) dt dτ → 0, n → ∞
(iv) equation (1.1) has a unique solution.
Then for all n such that qn = (n A−1 < 1, A : X → X, the linear system of equa n
∗
−1
tions (4.5) has a unique solution α∗
(y) −
k 1 and the approximate solution xn = G
n
∗ −1
∗
∗
∗
α
G
(a
)
converges
to
the
exact
solution
x
,
x
−
x
=
O((
).
k
n
k=1 k
n ρ
160
SAMIR A. ASHOUR
Proof. For x ∈ X, we have
1 2 h(t, τ) − h (t, τ) x(τ) dτ 1
−
τ
Ax − An xρ = n
−1
(4.6)
ρ
≤ xρ h − hn L2,q = (n xρ .
Then A − An X →X ≤ (n , according to [6] (4.5) has a unique solution
∗
x ∗ − xn
= O((n ).
α∗
k
n
1,
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Ashour: Department of Mathematics, Cairo University, Giza, Egypt