Page 73 - Handbook Of Integral Equations

x

2 2
52. sin (λx) sin(µt) + sin(βx) sin (γt) y(t) dt = f(x).
a
2
This is a special case of equation 1.9.15 with g 1 (x) = sin (λx), h 1 (t) = sin(µt), g 2 (x) = sin(βx),
2
and h 2 (t) = sin (γt).
x
4
53. sin [λ(x – t)]y(t) dt = f(x).
a

It is assumed that f(a)= f (a)= ··· = f xxxx (a)=0.
x
Let us transform the kernel of the integral equation using the trigonometric formula
3
4
sin β = 1 cos 4β – 1 cos 2β + , where β = λ(x – t), and differentiate the resulting equation
8 2 8
with respect to x. Then we obtain an equation of the form 1.5.41:
x

1

λ – sin[4λ(x – t)] + sin[2λ(x – t)] y(t) dt = f (x).
2 x
a
x

n
54. sin [λ(x – t)]y(t) dt = f(x), n =2, 3, ...
a
It is assumed that f(a)= f (a)= ··· = f x (n) (a)=0.

x
Let us differentiate the equation with respect to x twice and transform the kernel of the
2
2
resulting integral equation using the formula cos β =1 – sin β, where β = λ(x – t). We have
x x

n
2 2
2

–λ n sin [λ(x – t)]y(t) dt + λ n(n – 1) sin n–2 [λ(x – t)]y(t) dt = f (x).
xx
a a
Eliminating the first term on the left-hand side with the aid of the original equation, we obtain
x 1
2 2

sin n–2 [λ(x – t)]y(t) dt = 2 f (x)+ λ n f(x) .
xx
a λ n(n – 1)
This equation has the same form as the original equation, but the degree characterizing the
kernel has been reduced by two.
By applying this technique sufficiently many times, we finally arrive at simple integral
equations of the form 1.1.1 (for even n) or 1.5.34 (for odd n).
x
√
55. sin λ x – t y(t) dt = f(x).
a
Solution: √
2 d 2 x cosh λ x – t
y(x)= √ f(t) dt.
πλ dx 2 x – t
a
x √
56. sin x – sin ty(t) dt = f(x).
a
Solution: x
2 1 d
2 cos tf(t) dt
y(x)= cos x √ .
π cos x dx a sin x – sin t
x y(t) dt
57. √ = f(x).
a sin x – sin t
Solution: x
1 d cos tf(t) dt
y(x)= √ .
π dx sin x – sin t
a
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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