To solve this integral equation, direct and inverse Laplace transforms are used.. The right-hand side f x of the integral equation must satisfy certain relations... The integral equation
Trang 11◦ To solve this integral equation, direct and inverse Laplace transforms are used The
solution can be represented in the form
y (x) = f (x) –
x
Here the resolvent R(x) is expressed via the kernel K(x) of the original equation as follows:
R (x) = 1
2πi
c+i∞
c–i∞
2
R (p)e px dp, R2(p) = K2(p)
1+ 2K (p),
2
K (p) =
∞
0 K (x)e
–px dx.
2◦ Let w = w(x) be the solution of the simpler auxiliary equation with a =0and f ≡ 1:
w (x) +
x
Then the solution of the original integral equation with arbitrary f = f (x) is expressed via
the solution of the auxiliary equation (2) as
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.
T11.3 Linear Equations of the First Kind
with Constant Limits of Integration
1.
b
a |x– t| y(t) dt = f (x), 0≤a < b < ∞.
Solution:
y (x) = 12f
xx (x).
The right-hand side f (x) of the integral equation must satisfy certain relations The general form of f (x) is as follows:
f (x) = F (x) + Ax + B,
A= –12
F
x (a) + F x (b)
, B = 12
aF
x (a) + bF x (b) – F (a) – F (b)
,
where F (x) is an arbitrary bounded twice differentiable function (with bounded first
deriva-tive)
2.
a
0
y(t)
√|x– t| dt = f (x), 0 < a≤∞.
Solution:
y (x) = – A
x1 4
d dx
a
x
dt
(t – x)1 4
t
0
f (s) ds
s1 4(t – s)1 4
√
8πΓ2(3/4).
Trang 2b
a
y(t)
|x– t| k dt = f (x), 0 < k < 1.
It is assumed that|a|+|b|<∞ Solution:
y (x) = 1
2π cot(12πk) d
dx
x
a
f (t) dt (x – t)1–k –
1
π2 cos2(12πk)
x
a
Z (t)F (t) (x – t)1–k dt,
where
Z (t) = (t – a)1+k2 (b – t)1–k2 , F (t) = d
dt
t
a
dτ
(t – τ ) k
b
τ
f (s) ds
Z (s)(s – τ )1–k
4.
b
0
y(t)
|xλ – t λ|k dt = f (x), 0 < k < 1, λ> 0.
Solution:
y (x) = –Ax λ(k–21) d
dx
" b
x
t λ(3 – 2k)–2
(t λ – x λ)1–2k
t
0
s λ(k+1 )– 2
2 f (s) ds (t λ – s λ)1–2k
# ,
A= λ
2
2π cos
πk
2
Γ(k)
Γ
1
+ k
2
–2 , whereΓ(k) is the gamma function.
5.
∞
–∞
y(t)
|x– t|1–λ dt = f (x), 0 < λ < 1.
Solution:
y (x) = λ
2π tan
πλ 2
∞
–∞
f (x) – f (t)
|x – t|1 +λ dt.
It assumed that the condition ∞
–∞|f (x)| p dx<∞ is satisfied for some p,1< p <1/λ
6.
∞
–∞
sign(x – t)
|x– t|1–λ y(t) dt = f (x), 0 < λ < 1.
Solution:
y (x) = λ
2π cotπλ
2
∞
–∞
f (x) – f (t)
|x – t|1 +λ sign(x – t) dt.
7.
∞
–∞
a + b sign(x – t)
|x– t|1–λ y(t) dt = f (x), 0 < λ < 1.
Solution:
4π
a2cos2 1
2πλ
+ b2sin2 12πλ
∞ –∞
a + b sign(x – t)
|x – t|1 +λ
f (x) – f (t)
dt
Trang 3∞
0
y(x + t) – y(x – t)
Solution: y(x) = – 1
π2
∞
0
f (x + t) – f (x – t)
In equations T11.3.9 and T11.3.10 and their solutions, all singular integrals are
under-stood in the sense of the Cauchy principal value.
9.
∞
–∞
y(t) dt
t – x = f (x).
Solution: y(x) = – 1
π2
∞ –∞
f (t) dt
t – x .
The integral equation and its solution form a Hilbert transform pair (in the asymmetric form)
10.
b
a
y(t) dt
t – x = f (x).
This equation is encountered in hydrodynamics in solving the problem on the flow of an
ideal inviscid fluid around a thin profile (a≤x≤b) It is assumed that|a|+|b|<∞.
1◦ The solution bounded at the endpoints is
y (x) = – 1
π2
(x – a)(b – x)
b
a
f (t)
√
(t – a)(b – t)
dt
t – x,
a
f (t) dt
√
(t – a)(b – t) =0
2◦ The solution bounded at the endpoint x = a and unbounded at the endpoint x = b is
y (x) = – 1
π2
x – a
b – x
b
a
b – t
t – a
f (t)
t – x dt.
3◦ The solution unbounded at the endpoints is
π2√
(x – a)(b – x)
b
a
√
(t – a)(b – t)
t – x f (t) dt + C
,
where C is an arbitrary constant The formula b
a y (t) dt =
C
π holds
11.
b
a e
λ|x–t|y(t) dt = f (x), –∞ < a < b < ∞.
Solution:
y (x) = 1
2λ
f
xx (x) – λ2f (x)
The right-hand side f (x) of the integral equation must satisfy certain relations The
general form of the right-hand side is given by
f (x) = F (x) + Ax + B,
bλ – aλ –2
F
x (a) + F x (b) + λF (a) – λF (b)
, B = –1
λ
F
x (a) + λF (a) + Aaλ + A
,
where F (x) is an arbitrary bounded, twice differentiable function.
Trang 4b
a ln|x– t| y(t) dt = f (x).
Carleman’s equation.
1◦ Solution for b – a≠ 4:
π2√
(x – a)(b – x)
b
a
√
(t – a)(b – t) f t (t) dt
1
ln1
4(b – a)
b
a
f (t) dt
√
(t – a)(b – t)
2◦ If b – a =4, then for the equation to be solvable, the condition
b
a f (t)(t – a)
– 1 2(b – t)– 1 2dt=0
must be satisfied In this case, the solution has the form
π2√
(x – a)(b – x)
b
a
√
(t – a)(b – t) f t (t) dt
,
where C is an arbitrary constant.
13.
b
a ln|x– t| + β
y(t) dt = f (x).
By setting
x = e–β z, t = e–β τ, y (t) = Y (τ ), f (x) = e–β g (z),
we arrive at an equation of the form T11.3.12:
B
A ln|z – τ|Y (τ ) dτ = g(z), A = ae β , B = be β
14.
a
–a
ln A
|x– t|
y(t) dt = f (x), –a ≤x≤ a.
Solution for 0< a <2A:
y (x) = 1
2M (a)
d da
a –a w (t, a)f (t) dt
w (x, a)
– 1 2
a
|x| w (x, ξ)
d dξ
1
M (ξ)
d dξ
ξ –ξw (t, ξ)f (t) dt
dξ
– 1 2
d dx
a
|x|
w (x, ξ)
M (ξ)
ξ –ξw (t, ξ) df (t)
dξ, where the prime stands for the derivative and
M (ξ) =
ln2A ξ
–1 , w (x, ξ) = M (ξ)
π
ξ2– x2.
Trang 5a
0
ln
x x + t – ty(t) dt = f(x).
Solution:
y (x) = – 2
π2
d dx
a
x
F (t) dt
√
t2– x2, F (t) =
d dt
t
0
sf (s) ds
√
t2– s2.
16.
∞
0
cos(xt)y(t) dt = f(x).
Solution: y(x) = 2
π
∞
0 cos(xt)f (t) dt.
Up to constant factors, the function f (x) and the solution y(t) are the Fourier cosine
transform pair
17.
∞
0
sin(xt)y(t) dt = f(x).
Solution: y(x) = 2
π
∞
0 sin(xt)f (t) dt.
Up to constant factors, the function f (x) and the solution y(t) are the Fourier sine
transform pair
18.
π/2
0
y(ξ) dt = f (x), ξ = x sin t.
Schl¨omilch equation.
Solution:
y (x) = 2
π
f(0) + x π/2
ξ (ξ) dt
, ξ = x sin t.
19.
2π
0
cot t – x
2
y(t) dt = f (x), 0 ≤x≤2π.
Here the integral is understood in the sense of the Cauchy principal value and the right-hand side is assumed to satisfy the condition 2π
0 f (t) dt =0
Solution:
y (x) = – 1
4π2
2π
0 cot
t – x 2
f (t) dt + C, where C is an arbitrary constant.
It follows from the solution that
2π
0 y (t) dt =2πC The equation and its solution form a Hilbert transform pair (in the asymmetric form)
20.
∞
0
tJ ν (xt)y(t) dt = f(x), ν > – 1 2
Here, J ν (z) is the Bessel function of the first kind.
Solution:
y (x) =
∞
0 tJ ν (xt)f (t) dt.
The function f (x) and the solution y(t) are the Hankel transform pair.
Trang 6∞
–∞ K0 |x– t| y(t) dt = f (x).
Here, K0(z) is the modified Bessel function of the second kind.
Solution:
y (x) = – 1
π2
d2
dx2 –1 ∞
–∞ K0 |x – t| f (t) dt.
22.
∞
–∞ K(x – t)y(t) dt = f (x).
The Fourier transform is used to solve this equation
1◦ Solution:
y (x) = 1
2π
∞ –∞
2f(u)
2
K (u) e iux du, 2f(u) = √1
2π
∞ –∞ f (x)e
–iux dx, K2(u) = √1
2π
∞ –∞ K (x)e
–iux dx.
The following statement is valid Let f (x) L2(–∞, ∞) and K(x) L1(–∞, ∞).
Then for a solution y(x)L2(–∞, ∞) of the integral equation to exist, it is necessary and
sufficient that 2f (u)/ 2 K (u)L2(–∞, ∞).
2◦ Let the function P (s) defined by the formula
1
P (s) =
∞ –∞ e
–stK (t) dt
be a polynomial of degree n with real roots of the form
P (s) =
1– s
a1
1– s
a2
.
1– s
a n
Then the solution of the integral equation is given by
y (x) = P (D)f (x), D= d
dx
23.
∞
0
K(x – t)y(t) dt = f (x).
Wiener–Hopf equation of the first kind This equation is discussed in the books by Gakhov
and Cherskii (1978), Mikhlin and Pr¨ossdorf (1986), Muskhelishvili (1992), and Polyanin and Manzhirov (1998) in detail
T11.4 Linear Equations of the Second Kind
with Constant Limits of Integration
1. y(x) – λ
b
a (x – t)y(t) dt = f(x).
Solution:
y (x) = f (x) + λ(A1x + A2),
Trang 7A1= 12f1+6λ (f1Δ2–2f2Δ1)
λ2Δ4
–12f2+2λ(3f2Δ2–2f1Δ3)
λ2Δ4
f1=
b
a f (x) dx, f2=
b
a xf (x) dx, Δn= b n – a n
2. y(x) + A
b
a |x– t| y(t) dt = f (x).
1◦ For A <0, the solution is given by
y (x) = C1cosh(kx) + C2sinh(kx) + f (x) + k
x
a sinh[k(x – t)]f (t) dt, k=
√
–2A, (1)
where the constants C1and C2are determined by the conditions
y
x (a) + y x (b) = f x (a) + f x (b),
y (a) + y(b) + (b – a)y x (a) = f (a) + f (b) + (b – a)f x (a). (2)
2◦ For A >0, the solution is given by
y (x) = C1cos(kx) + C2sin(kx) + f (x) – k
x
a sin[k(x – t)]f (t) dt, k=
√
2A, (3)
where the constants C1and C2are determined by conditions (2)
3◦ In the special case a = 0and A > 0, the solution of the integral equation is given by formula (3) with
C1= k Is(1+ cos λ) – Ic(λ + sin λ)
2+2cos λ + λ sin λ , C2= k
Issin λ + Ic(1+ cos λ)
2+2cos λ + λ sin λ ,
k=√
2A , λ = bk, Is=
b
0 sin[k(b – t)]f (t) dt, Ic =
b
0 cos[k(b – t)]f (t) dt.
In equations T11.4.3 and T11.4.4 and their solutions, all singular integrals are
under-stood in the sense of the Cauchy principal value.
3. Ay(x)+ B
π
1
–1
y(t) dt
t – x = f (x), –1 < x < 1.
Without loss of generality we may assume that A2+ B2=1
1◦ The solution bounded at the endpoints:
y (x) = Af (x) – B
π
1 – 1
g (x)
g (t)
f (t) dt
t – x , g (x) = (1 + x) α(1– x)1–α, (1)
where α is the solution of the trigonometric equation
on the interval0< α <1 This solution y(x) exists if and only if 1
– 1
f (t)
g (t) dt=0
... (x)
The right-hand side f (x) of the integral equation must satisfy certain relations The
general form of the right-hand side is given by
f (x) =...
Wiener–Hopf equation of the first kind This equation is discussed in the books by Gakhov
and Cherskii (1978), Mikhlin and Prăossdorf (1986), Muskhelishvili (1992), and Polyanin and Manzhirov... y (t) dt =2πC The equation and its solution form a Hilbert transform pair (in the asymmetric form)
20.
∞
0