Solution for bb +1... Let Kx have an integrable power-law singularity at x =0.. Linear Equations of the Second Kind with Variable Limit of Integration 1... Generalized Abel equation of t
Trang 126. x
a
55
cos[λ(x – t)] + b66
y(t) dt = f (x), f (a) = 0.
For b =0, see equation T11.1.24 For b = –1, see equation T11.1.25
1◦ Solution for b(b +1) >0:
y (x) = f
x (x)
b+1 +
λ2
k (b +1)2
x
a sin[k(x – t)]f
t (t) dt, where k = λ b+b1.
2◦ Solution for b(b +1) <0:
y (x) = f x (x)
b+1 +
λ2
k (b +1)2
x
a sinh[k(x – t)]f
t (t) dt, where k = λ b –b+1.
27.
x
a sin[λ(x – t)]y(t) dt = f(x), f (a) = f x (a) = 0.
Solution: y(x) = 1
λ f
xx (x) + λf (x).
28.
x
a sin λ √
x – t
y(t) dt = f (x), f (a)= 0.
Solution: y(x) = 2
πλ
d2
dx2
x
a
cosh λ √
x – t
√
x – t f (t) dt.
29.
x
a J0 λ(x – t)
y(t) dt = f (x).
Here, J ν (z) is the Bessel function of the first kind and f (a) =0
Solution:
y (x) = 1
λ
d2
dx2 + λ2
2 x
a (x – t) J1 λ (x – t)
f (t) dt.
30.
x
a J0 λ
√
x – t
y(t) dt = f (x).
Here, J ν (z) is the Bessel function of the first kind and f (a) =0
Solution:
y (x) = d
2
dx2
x
a I0 λ
√
x – t
f (t) dt.
31.
x
a I0 λ(x – t)
y(t) dt = f (x).
Here, I ν (z) is the modified Bessel function of the first kind and f (a) =0
Solution:
y (x) = 1
λ
d2
dx2 – λ2
2 x
a (x – t) I1 λ (x – t)
f (t) dt.
Trang 2x
a I0 λ
√
x – t
y(t) dt = f (x).
Here, I ν (z) is the modified Bessel function of the first kind and f (a) =0
Solution: y(x) = d
2
dx2
x
a J0 λ
√
x – t
f (t) dt.
33.
x
a [g(x) – g(t)]y(t) dt = f(x).
It is assumed that f (a) = f x (a) =0and f x /g
x ≠const
Solution: y(x) = d
dx
f
x (x)
g
x (x)
34.
x
a [g(x) – g(t) + b]y(t) dt = f(x), f (a)= 0.
For b =0, see equation T11.1.33
Solution for b≠ 0:
y (x) = 1
b f
x (x) – 1
b2g x (x)
x
a exp
*g (t) – g(x)
b
+
f
t (t) dt.
35.
x
a [g(x) + h(t)]y(t) dt = f(x), f (a)= 0.
For h(t) = –g(t), see equation T11.1.33.
Solution:
y (x) = d
dx
Φ(x)
g (x) + h(x)
x
a
f
t (t) dt
Φ(t)
, Φ(x) = exp
x
a
h
t (t) dt
g (t) + h(t)
36.
x
a K(x – t)y(t) dt = f (x).
1◦ Let K(0) =1 and f (a) = 0 Differentiating the equation with respect to x yields a
Volterra equation of the second kind:
y (x) +
x
x (x – t)y(t) dt = f x (x).
The solution of this equation can be represented in the form
y (x) = f x (x) +
x
a R (x – t)f
Here the resolvent R(x) is related to the kernel K(x) of the original equation by
R (x) =L– 1 1
p 2 K (p) –1
, K2(p) =LK (x),
Trang 3whereL and L– 1are the operators of the direct and inverse Laplace transforms, respectively.
2
K (p) =LK (x)
=
∞
0 e
–px K (x) dx, R (x) =L– 12R (p)
2πi
c+i∞
c–i∞ e
px R2(p) dp.
2◦ Let K(x) have an integrable power-law singularity at x =0 Denote by w = w(x) the solution of the simpler auxiliary equation (compared with the original equation) with a =0
and constant right-hand side f ≡ 1,
x
Then the solution of the original integral equation with arbitrary right-hand side is expressed
in terms of w as follows:
y (x) = d
dx
x
a w (x – t)f (t) dt = f (a)w(x – a) +
x
a w (x – t)f
t (t) dt.
37.
x
a
g(x) – g(t) y(t) dt = f (x), f (a)= 0, g
x (x) > 0.
Solution:
y (x) = 2
π g
x (x)
1
g x (x)
d dx
2 x
a
f (t)g t (t) dt
√
g (x) – g(t).
38.
x
a
y(t) dt
g(x) – g(t) = f (x), g
x (x) > 0.
Solution: y(x) = 1
π
d dx
x
a
f (t)g t (t) dt
√
g (x) – g(t).
T11.2 Linear Equations of the Second Kind
with Variable Limit of Integration
1. y(x) – λ
x
a y(t) dt = f (x).
Solution: y(x) = f (x) + λ
x
a e λ(x–t) f (t) dt.
2. y(x) + λ
x
a (x – t)y(t) dt = f(x).
1◦ Solution for λ >0:
y (x) = f (x) – k
x
a sin[k(x – t)]f (t) dt, k=
√
λ
2◦ Solution for λ <0:
y (x) = f (x) + k
x
a sinh[k(x – t)]f (t) dt, k=
√
–λ.
Trang 43. y(x) + λ
x
a (x – t)2y(t) dt = f (x).
Solution:
y (x) = f (x) –
x
a R (x – t)f (t) dt,
R (x) = 23ke–2kx– 2
3ke kx
*
cos 3kx
–√
3sin 3kx +
, k= 14λ 1 3
4. y(x) + λ
x
a (x – t)3y(t) dt = f (x).
Solution:
y (x) = f (x) –
x
a R (x – t)f (t) dt,
where
R (x) =
k
cosh(kx) sin(kx) – sinh(kx) cos(kx)
, k= 32λ 1 4
for λ >0,
1
2s
sin(sx) – sinh(sx)
, s= (–6λ)1 4 for λ <0
5. y(x) + A
x
a (x – t) n y(t) dt = f (x), n = 1, 2,
1◦ Differentiating the equation n +1times with respect to x yields an (n +1)st-order linear
ordinary differential equation with constant coefficients for y = y(x):
y(n+1 )
x + An! y = f x(n+1)(x).
This equation under the initial conditions y(a) = f (a), y x (a) = f x (a), , y x(n) (a) = f x(n) (a)
determines the solution of the original integral equation
2◦ Solution:
y (x) = f (x) +
x
a R (x – t)f (t) dt,
R (x) = 1
n+1
n
k=0
exp(σ k x)
σ k cos(β k x ) – β k sin(β k x)
,
where the coefficients σ k and β kare given by
σ k=|An!|n+11 cos 2πk
n+1
, β k=|An!|n+11 sin 2πk
n+1
for A<0;
σ k=|An!|n+11 cos2πk+ π
n+1
, β k=|An!|n+11 sin2πk+ π
n+1
for A>0
6. y(x) + λ
x
a
y(t) dt
√
x – t = f (x).
Abel equation of the second kind This equation is encountered in problems of heat and
mass transfer
Solution:
y (x) = F (x) + πλ2
x
a exp[πλ
2(x – t)]F (t) dt,
where
F (x) = f (x) – λ
x
a
f (t) dt
√
x – t.
Trang 57. y(x) – λ x
0
y(t) dt (x – t) α = f (x), 0 < α < 1.
Generalized Abel equation of the second kind.
1◦ Assume that the number α can be represented in the form
α=1– m
n, where m=1,2, , n=2,3, (m < n).
In this case, the solution of the generalized Abel equation of the second kind can be written
in closed form (in quadratures):
y (x) = f (x) +
x
0 R (x – t)f (t) dt,
where
R (x) =
n–1
ν=1
λ νΓν (m/n)
Γ(νm/n) x(νm/n)–1+
b m
m–1
μ=0
ε μexp ε μ bx
+ b
m
n–1
ν=1
λ νΓν (m/n)
Γ(νm/n)
m–1
μ=0
ε μexp ε μ bx x
0 t
(νm/n)–1exp –ε
μ bt
dt
,
b = λ n/mΓn/m (m/n), ε μ= exp
2πμi
m
, i2 = –1, μ=0,1, , m –1
2◦ Solution for any α from0< α <1:
y (x) = f (x) +
x
0 R (x – t)f (t) dt, where R (x) =
∞
n=1
λΓ(1– α)x1–αn
xΓn(1– α)
8. y(x) + A
x
a e λ(x–t) y(t) dt = f (x).
Solution: y(x) = f (x) – A
x
a e
(λ–A)(x–t) f (t) dt.
9. y(x) + A
x
a
e λ(x–t)– 1
y(t) dt = f (x).
1◦ Solution for D≡λ (λ –4A) >0:
y (x) = f (x) – 2Aλ
√ D
x
a R (x – t)f (t) dt, R (x) = exp
1
2λx
sinh 12√
D x
2◦ Solution for D≡λ (λ –4A) <0:
y (x) = f (x) – 2Aλ
√
|D|
x
a R (x – t)f (t) dt, R (x) = exp
1
2λx
sin 12
|D|x
3◦ Solution for λ =4A:
y (x) = f (x) –4A2 x
a (x – t) exp
2A(x – t)
f (t) dt.
Trang 610. y(x) + A
x
a (x – t)e λ(x–t) y(t) dt = f (x).
1◦ Solution for A >0:
y (x) = f (x) – k
x
a e
λ(x–t) sin[k(x – t)]f (t) dt, k=√
A
2◦ Solution for A <0:
y (x) = f (x) + k
x
a e
λ(x–t) sinh[k(x – t)]f (t) dt, k=√
–A.
11. y(x) + A
x
a cosh[λ(x – t)]y(t) dt = f(x).
Solution:
y (x) = f (x) +
x
a R (x – t)f (t) dt,
R (x) = exp –12Ax A2
2k sinh(kx) – A cosh(kx)
, k=
λ2+ 1
4A2.
12. y(x) + A
x
a sinh[λ(x – t)]y(t) dt = f(x).
1◦ Solution for λ(A – λ) >0:
y (x) = f (x) – Aλ
k
x
a sin[k(x – t)]f (t) dt, where k=
λ (A – λ).
2◦ Solution for λ(A – λ) <0:
y (x) = f (x) – Aλ
k
x
a sinh[k(x – t)]f (t) dt, where k=
λ (λ – A).
3◦ Solution for A = λ:
y (x) = f (x) – λ2
x
a (x – t)f (t) dt.
13. y(x) – λ
x
0
J0(x – t)y(t) dt = f(x).
Here, J0(z) is the Bessel function of the first kind.
Solution:
y (x) = f (x) +
x
0 R (x – t)f (t) dt,
where
R (x) = λ cos 1– λ2x
+ √ λ2
1– λ2 sin 1– λ2x
+ √ λ
1– λ2
x
0 sin√
1– λ2(x – t) J1(t)
t dt.
Trang 714. y(x)– x
a g(x)h(t)y(t) dt = f (x).
Solution:
y (x) = f (x) +
x
a R (x, t)f (t) dt, where R (x, t) = g(x)h(t) exp
x
t g (s)h(s) ds
15. y(x)+
x
a (x – t)g(x)y(t) dt = f(x).
1◦ Solution:
y (x) = f (x) + 1
W
x
a
Y1(x)Y2(t) – Y2(x)Y1(t)
g (x)f (t) dt, (1)
where Y1= Y1(x) and Y2= Y2(x) are two linearly independent solutions (Y1/Y2 const) of
the second-order linear homogeneous differential equation Y xx + g(x)Y =0 In this case,
the Wronskian is a constant: W = Y1(Y2) x – Y2(Y1) x ≡const
2◦ Given only one nontrivial solution Y1 = Y1(x) of the linear homogeneous differential equation Y xx +g(x)Y =0, one can obtain the solution of the integral equation by formula (1) with
W =1, Y2(x) = Y1(x)
x
b
dξ
Y2
1(ξ)
,
where b is an arbitrary number.
16. y(x)+
x
a (x – t)g(t)y(t) dt = f(x).
1◦ Solution:
y (x) = f (x) + 1
W
x
a
Y1(x)Y2(t) – Y2(x)Y1(t)
g (t)f (t) dt, (1)
where Y1= Y1(x) and Y2= Y2(x) are two linearly independent solutions (Y1/Y2 const) of
the second-order linear homogeneous differential equation Y xx + g(x)Y =0 In this case,
the Wronskian is a constant: W = Y1(Y2) x – Y2(Y1) x ≡const
2◦ Given only one nontrivial solution Y1 = Y1(x) of the linear homogeneous differential equation Y xx +g(x)Y =0, one can obtain the solution of the integral equation by formula (1) with
W =1, Y2(x) = Y1(x)
x
b
dξ
Y2
1(ξ)
,
where b is an arbitrary number.
17. y(x)+
x
a K(x – t)y(t) dt = f (x).
Renewal equation.
...x – t = f (x).
Abel equation of the second kind This equation is encountered in problems of heat and< /i>
mass transfer
Solution:
y (x)... (m < n).
In this case, the solution of the generalized Abel equation of the second kind can be written
in closed form (in quadratures):
y (x) = f (x) +
... Y1(x) of the linear homogeneous differential equation Y xx +g(x)Y =0, one can obtain the solution of the integral equation by formula (1) with
W