Page 199 - Handbook Of Integral Equations

x

λx µx
22. y(x)+ e g(t)+ e h(t) y(t) dt = f(x).
a
Let us differentiate the equation twice and then eliminate the integral terms from the resulting
relations and the original equation. As a result, we arrive at the second-order linear ordinary
differential equation

λx µx

λx
µx
y + e g(x)+ e h(x) – λ – µ y + e g (x)+ e h (x)
xx x x x
λx
µx

+(λ – µ)e g(x)+(µ – λ)e h(x)+ λµ y = f (x) – (λ + µ)f (x)+ λµf(x),
x
xx
which must be supplemented by the initial conditions
λa µa

y(a)= f(a), y (a)= f (a) – e g(a)+ e h(a) f(a).
x
x
Example. The Arutyunyan equation
x ∂ 1
y(x) – ϕ(t) + ψ(t) 1 – e –λ(x–t) y(t) dt = f(x),
a ∂t ϕ(t)
can be reduced to the above equation. The former is encountered in the theory of viscoelasticity for aging solids.
The solution of the Arutyunyan equation is given by
x 1 ∂ x
2
y(x)= f(x) – ϕ(t) – λψ(t)ϕ (t)e η(t) e –η(s) ds f(t) dt,
a ϕ(t) ∂t t
where
x
ϕ (t)

η(x)= λ 1+ ψ(t)ϕ(t) – dt.
ϕ(t)
a
•
Reference: N. Kh. Arutyunyan (1966).
x
λ(x–t) µx+λt λx+µt
23. y(x)+ λe + µe – λe h(t) y(t) dt = f(x).
a
µx
This is a special case of equation 2.9.17 with ϕ(x)= e λx and g(x)= e .
Solution:
x
1 d F(t) e 2λt h(t)
y(x)= Φ(x) dt ,
λx
e h(x) dx a e λt t Φ(t)
x x

F(x)= f(t) dt, Φ(x)=exp (λ – µ) e (λ+µ)t h(t) dt .
a a
x
–λ(x–t) λx+µt µx+λt
24. y(x) – λe + µe – λe h(x) y(t) dt = f(x).
a
µx
This is a special case of equation 2.9.18 with ϕ(x)= e λx and g(x)= e .
Assume that f(a) = 0. Solution:
x
x d e 2λx h(x) f(t)
y(x)= w(t) dt, w(x)= e –λx λt Φ(t) dt ,
a dx Φ(x) a e h(t) t
x

Φ(x)=exp (λ – µ) e (λ+µ)t h(t) dt .
a
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
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