Page 592 - Handbook Of Integral Equations
P. 592
◦
1 . Let the equation
b
y(x) – K(x, t)y(t) dt = f(x), a ≤ x ≤ b, (1)
a
be given. Along with (1), we consider two auxiliary equations depending on a parameter ξ (a≤ξ ≤b):
ξ
w(x, ξ) – K(x, t)w(t, ξ) dt =1,
a
ξ (2)
w (x, ξ) – K(t, x)w (t, ξ) dt =1,
∗
∗
a
where a ≤ x ≤ ξ. Assume that for any ξ the auxiliary equations (2) have unique continuous solutions
∗
w(x, ξ) and w (x, ξ), respectively, which satisfy the condition w(ξ, ξ)w (ξ, ξ) ≠ 0(a ≤ ξ ≤ b). In this
∗
case, for any continuous function f(x), the unique continuous solution of Eq. (1) can be obtained
by the formula
b 1 d ξ
y(x)= F(b)w(x, b) – w(x, ξ)F (ξ) dξ, F(ξ)= w (t, ξ)f(t) dt, (3)
∗
ξ
x m(ξ) dξ a
where
∗
m(ξ)= w(ξ, ξ)w (ξ, ξ).
Formula (3) permits one to construct a solution of Eq. (1) with an arbitrary right-hand side f(x)
by means of solutions to the two simpler auxiliary equations (2) (depending on the parameter ξ)
with a constant right-hand side equal to 1.
2 . Consider now an equation with the kernel depending on the difference of the arguments:
◦
b
y(x)+ K(x – t)y(t) dt = f(x), a ≤ x ≤ b. (4)
a
It is assumed that K(x) is an even function integrable on [a – b, b – a]. Along with (4) we consider
the following auxiliary equation depending on a parameter ξ (a ≤ ξ ≤ b):
ξ
w(x, ξ)+ K(x – t)w(t, ξ) dt =1, a ≤ x ≤ ξ. (5)
a
Assume that for an arbitrary ξ the auxiliary equation (5) has a unique continuous solution w(x, ξ).
In this case, for any continuous function f(x), a solution of Eq. (4) can be obtained from formula (3)
by setting w (x, t)= w(x, t) in this formula.
∗
Now let us indicate another useful formula for equations whose kernel depends on the difference
of the arguments:
a
y(x)+ K(x – t)y(t) dt = f(x), –a ≤ x ≤ a. (6)
–a
It is assumed that K(x) is an even function that is integrable on the segment [–2a,2a]. Along with
(6) we consider an auxiliary equation depending on a parameter ξ (0 < ξ ≤ a):
ξ
w(x, ξ)+ K(x – t)w(t, ξ) dt =1, –ξ ≤ x ≤ ξ. (7)
–ξ
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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