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3.1-5. Kernels Containing Arbitrary Powers
a
k
k
24. |x – t | y(t) dt = f(x), 0 < k <1, 0 < a < ∞.
0
1 . Let us remove the modulus in the integrand:
◦
x a
k k k k
(x – t )y(t) dt + (t – x )y(t) dt = f(x). (1)
0 x
Differentiating (1) with respect to x yields
x a
kx k–1 y(t) dt – kx k–1 y(t) dt = f (x). (2)
x
0 x
Let us divide both sides of (2) by kx k–1 and differentiate the resulting equation. As a result,
we obtain the solution
1 d 1–k
y(x)= x f (x) . (3)
x
2k dx
2 . Let us demonstrate that the right-hand side f(x) of the integral equation must satisfy
◦
a
k
certain relations. By setting x=0 and x=a, in (1), we obtain two corollaries t y(t) dt=f(0)
0
a
k
k
and (a – t )y(t) dt = f(a), which can be rewritten in the form
0
a a
k k
t y(t) dt = f(0), a y(t) dt = f(0) + f(a). (4)
0 0
Substitute y(x) of (3) into (4). Integrating by parts yields the relations af (a)= kf(a)+kf(0)
x
and af (a)=2kf(a)+2kf(0). Hence, the desired constraints for f(x) have the form
x
f(0) + f(a)=0, f (a)=0. (5)
x
Conditions (5) make it possible to find the admissible general form of the right-hand side
of the integral equation:
f(x)= F(x)+ Ax + B, A = –F (a), B = 1 2 aF (a) – F(a) – F(0) ,
x
x
where F(x) is an arbitrary bounded twice differentiable function with bounded first derivative.
1–k
The first derivative may be unbounded at x = 0, in which case the conditions x F =0
x x=0
must hold.
a
k
k
25. |x – βt | y(t) dt = f(x), 0 < k <1, β >0.
0
k
k
This is a special case of equation 3.8.4 with g(x)= x and β = λ .
a
k m
26. |x t – t k+m | y(t) dt = f(x), 0 < k <1, 0 < a < ∞.
0
m
The substitution w(t)= t y(t) leads to an equation of the form 3.1.24:
a
k k
|x – t |w(t) dt = f(x).
0
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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