Page 71 - Handbook Of Integral Equations

2 . Solution for ∆ <0:
◦
x

(A 1 λ 1 + A 2 λ 2 )y(x)= f (x)+ Bf(x)+ C sin[k(x – t)]f(t) dt,

xx
a
√ 1
2
2
2
2
2 2
2
k = –∆, B = ∆ + λ + λ , C = √ ∆ +(λ + λ )∆ + λ λ .
2
1
1
2
1 2
–∆
◦
3 . Solution for ∆ =0:
x

2
2 2
2

(A 1 λ 1 + A 2 λ 2 )y(x)= f (x)+(λ + λ )f(x)+ λ λ (x – t)f(t) dt.
xx 1 2 1 2
a
4 . Solution for ∆ = ∞:
◦
2
2
2 2
f xxxx +(λ + λ )f xx + λ λ f

2
1 2
1
y(x)= – 3 3 , f = f(x).
A 1 λ + A 2 λ
1 2
In the last case, the relation A 1 λ 1 + A 2 λ 2 = 0 holds and the right-hand side of the integral
equation is assumed to satisfy the conditions f(a)= f (a)= f (a)= f (a)=0.

x xx xxx
Remark. The solution can be obtained from the solution of equation 1.3.41 in which the
2
change of variables λ k → iλ k , A k → –iA k , i = –1(k = 1, 2), should be made.
x

42. A sin[λ(x – t)] + B sin[µ(x – t)] + C sin[β(x – t)] y(t) dt = f(x).
a
It is assumed that f(a)= f (a) = 0. Differentiating the integral equation twice yields

x
x

2 2
(Aλ + Bµ + Cβ)y(x) – Aλ sin[λ(x – t)] + Bµ sin[µ(x – t)] y(t) dt
a
x

– Cβ 2 sin[β(x – t)]y(t) dt = f (x).
xx
a
Eliminating the last integral with the aid of the original equation, we arrive at an equation of
the form 2.5.18:
x
2 2
(Aλ + Bµ + Cβ)y(x)+ A(β – λ ) sin[λ(x – t)]
a
2
2

2

+ B(β – µ ) sin[µ(x – t)] y(t) dt = f (x)+ β f(x).
xx
In the special case Aλ + Bµ + Cβ = 0, this is an equation of the form 1.5.41.
x

2

43. sin [λ(x – t)]y(t) dt = f(x), f(a)= f (a)= f (a)=0.
x xx
a
Differentiation yields an equation of the form 1.5.34:
x
1

sin[2λ(x – t)]y(t) dt = f (x).
x
a λ
1
–2
Solution: y(x)= λ f (x)+2f (x).

2 xxx x
x

2 2

44. sin (λx) – sin (λt) y(t) dt = f(x), f(a)= f (a)=0.
x
a
1 d f (x)

x
Solution: y(x)= .
λ dx sin(2λx)
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
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