HANDBOOK OFINTEGRAL EQUATIONS phần 4 docx

This equation can be rewritten in the form 3.1.2: b a |x – t| yt dt = 1 Therefore the solution of the integral equation is given by The right-hand sidef x of the equation must satisfy c

3.1-3 Kernels Containing Integer Powers ofx and t or Rational Functions

(x – t)y(t) dt + 6

 b x

(t – x)y(t) dt = f xx  (x).

This equation can be rewritten in the form 3.1.2:

 b a

|x – t| y(t) dt = 1

Therefore the solution of the integral equation is given by

The right-hand sidef (x) of the equation must satisfy certain conditions To obtain these

conditions, one must substitute solution (3) into (1) withx = a and x = b and into (2) with

x = a and x = b, and then integrate the four resulting relations by parts.

The right-hand side f (x) of the equation must satisfy certain conditions To obtain these

conditions, one must substitute solution (1) into the relations

 b

a

(t – a)2n+1 y(t) dt = f (a),

 b a

iux du, g(u) =˜ √1

2π

 ∞–∞ g(z)e

–iuz dz.

• Reference: P P Zabreyko, A I Koshelev, et al (1975).

2◦ Under some assumptions, the solution of the original equation can be represented in the

form

y(x) = lim n→∞

which is the real inversion of the Stieltjes transform

An alternative form of the solution is

y(x) = lim n→∞

To obtain an approximate solution of the integral equation, one restricts oneself to a

specific value ofn in (1) or (2) instead of taking the limit.

• Reference: I I Hirschman and D V Widder (1955).

3.1-4 Kernels Containing Square Roots

√

x f x (x)

The right-hand sidef (x) of the equation must satisfy certain conditions The general

form of the right-hand side is

y(x) = – A

x1/4

d dx

y(x) = 1

4π

 ∞–∞

f (x) – f (t)

|x – t|3/2 dt.

3.1-5 Kernels Containing Arbitrary Powers

 a x

Let us divide both sides of (2) bykx k–1and differentiate the resulting equation As a result,

we obtain the solution

y(x) = 1

2k

d dx



x1–k f x (x)

2◦ Let us demonstrate that the right-hand side f (x) of the integral equation must satisfy

certain relations By settingx = 0 and x = a, in (1), we obtain two corollaries

Substitutey(x) of (3) into (4) Integrating by parts yields the relations af x (a) = kf (a) + kf (0)

andaf x (a) = 2kf (a) + 2kf (0) Hence, the desired constraints for f (x) have the form

Conditions (5) make it possible to find the admissible general form of the right-hand side

of the integral equation:

f (x) = F (x) + Ax + B, A = –F x (a), B = 12

aF x (a) – F (a) – F (0)

,whereF (x) is an arbitrary bounded twice differentiable function with bounded first derivative.

The first derivative may be unbounded atx = 0, in which case the conditions

x1–k F x 

x=0= 0must hold

|x k–t k |w(t) dt = f(x).

Forλ = 0, see equation 3.1.2 Assume that 0 < λ < 1.

1◦ Let us remove the modulus in the integrand:

(x – t) λ–1 y(t) dt +

 b x

(t – x) λ–1 y(t) dt = 1

Rewrite equation (2) in the form

 b a

y(t) dt

|x – t| k = 1

See 3.1.29 and 3.1.30 for the solutions of equation (3) for variousa and b.

2◦ The right-hand sidef (x) of the integral equation must satisfy certain relations By setting

x = a and x = b in (1), we obtain two corollaries

 b a

(t – a)1+λ y(t) dt = f (a),

 b a

(b – t)1+λ y(t) dt = f (b). (4)

On substituting the solutiony(x) of (3) into (4) and then integrating by parts, we obtain the

desired constraints forf (x).

t1–22k dt

(t – x)1–2k

 t0

f (s) ds

s1–2k(t – s)1–2k

,

2π cos

πk2



Γ(k)

Γ

1 +k

2

–2,whereΓ(k) is the gamma function.

2◦ The transformationx = z2,t = ξ2,w(ξ) = 2ξy(t) leads to an equation of the form 3.1.31:

 √ a

0

w(ξ)

|z2–ξ2|k dξ = f

z2

 x a

f (t) dt

(x – t)1–k – 1

π2 cos2(12πk)

 x a

2 x k–1 d

dx

 a x

t λ(3–2k)–22 dt

(λ–x λ)1–2k

 t0

s λ(k+1)–22 f (s) ds

(λ–s λ)1–2k

,

1 +k

2

–2,whereΓ(k) is the gamma function.

w(τ )

|z – τ| k dτ = F (z), F (z) = mf (z1/λ)

• Reference: S G Samko, A A Kilbas, and A A Marichev (1993).

y(t)

|z – t|1–λ dt = f

z1/3

w(τ )

|z – τ|1–λ dτ = F (z), F (z) = 3f

z1/3

2πλ

 ∞–∞

coshk(z – τ ) =g(z).

–z f (z) dz, i2= –1.

For specificf (z), one can use tables of Mellin and Laplace integral transforms to calculate

the integral

• References: H Bateman and A Erd´elyi (vol 2, 1954), V A Ditkin and A P Prudnikov (1965).

3.1-6 Equation Containing the Unknown Function of a Complicated Argument

0 ξ λ y(ξ) dξ = x λ+1 f (x) Differentiating with

respect tox yields the solution

y(ξ) dξ

(x – ξ) λ =x1–λ 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

√

(t – a)(b – t)

t – x f (t) dt + C

,whereC is an arbitrary constant The formula

3.2 Equations Whose Kernels Contain Exponential

2◦ The right-hand sidef (x) of the integral equation must satisfy certain relations By setting

x = a and x = b in (1), we obtain two corollaries

On substituting the solutiony(x) of (3) into (4) and then integrating by parts, we see that

e λb f x (b) – e λa f x (a) = λe λa f (a) + λe λb f (b),

e–λb f x (b) – e–λa f x (a) = λe–λa f (a) + λe–λb f (b).

Hence, we obtain the desired constraints forf (x):

Let us remove the modulus in the integrand and differentiate the resulting equation with

respect tox twice to obtain

2(Aλ + Bµ)y(x) +

 b a



Aλ2e λ |x–t|+Bµ2e µ |x–t|

y(t) dt = f xx (x). (1)Eliminating the integral term withe µ |x–t|from (1) with the aid of the original integral equation,

we find that

2(Aλ + Bµ)y(x) + A(λ2–µ2)

 b a

e λ |x–t| y(t) dt = f xx (x) – µ2f (x). (2)ForAλ + Bµ = 0, this is an equation of the form 3.2.1, and for Aλ + Bµ≠ 0, this is an equation

of the form 4.2.15

The right-hand sidef (x) must satisfy certain relations, which can be obtained by setting

x = a and x = b in the original equation (a similar procedure is used in 3.2.1).

e–λx f x (x)

.The right-hand sidef (x) of the integral equation must satisfy certain relations (see item 2 ◦

exp[λk(x – t)]y(t) dt – λk

 b x

where the primes denote the derivatives with respect tox By comparing formulas (1) and (2),

we find the relation betweenI k andIk:

with the aid of (6), we obtain a similar equation whose right-hand side is a second-order

linear differential operator (acting ony) with constant coefficients plus the sum

n–2

k=1

B k I k If

we successively eliminateIn–2,In–3, , I1 with the aid of double differentiation, then we

finally arrive at a linear nonhomogeneous ordinary differential equation of order 2(n – 1) with

constant coefficients

3◦ The right-hand sidef (x) must satisfy certain conditions To find these conditions, one

must setx = a in the integral equation and its derivatives (Alternatively, these conditions can

be found by settingx = a and x = b in the integral equation and all its derivatives obtained by

means of double differentiation.)

For specific functionsf (z), one may use tables of inverse Laplace transforms to calculate

the integral (e.g., see Supplement 5)

2◦ For real z = x, under some assumptions the solution of the original equation can be

represented in the form

y(x) = lim

n →∞

(–1)n n!

n

x

n+1

f(n) x

n

x

,which is the real inversion of the Laplace transform To calculate the solution approximately,

one should restrict oneself to a specific value ofn in this formula instead of taking the limit.

• References: H Bateman and A Erd´elyi (vol 1, 1954), I I Hirschman and D V Widder (1955), V A Ditkin

xexp(–λx

2)f x (x)

The right-hand sidef (x) of the integral equation must satisfy certain relations (see item 2 ◦

Applying the Laplace transformation to the equation, we obtain

e–pt f (t) dt.

Substitutingp by p2and solving for the transform ˜y, we find that ˜ y(p) = p ˜ f (p2) The inverse

Laplace transform provides the solution of the original integral equation:

 ∞0exp

–t2

4 y(t) dt = F (z),where the functionF (z) is determined by the relations F = √2

cosh(λx) – cosh(λt)y(t) dt = f (x).

This is a special case of equation 3.8.3 withg(x) = cosh(λx).

Solution:

y(x) = 1

2λ

d dx



f x (x)

sinh(λx)

The right-hand sidef (x) of the integral equation must satisfy certain relations (see item 2 ◦

of equation 3.8.3)

 a

0

cosh(βx) – cosh(µt)y(t) dt = f (x), β > 0, µ > 0.

This is a special case of equation 3.8.4 withg(x) = cosh(βx) and λ = µ/β.



f x (x)

sinhx cosh k–1 x

The right-hand sidef (x) of the integral equation must satisfy certain relations (see item 2 ◦

2◦ The right-hand sidef (x) of the integral equation must satisfy certain relations By setting

x = a and x = b in (1), we obtain two corollaries

 b a

sinh[λ(t – a)]y(t) dt = f (a),

 b a

sinh[λ(b – t)]y(t) dt = f (b). (4)Substituting solution (3) into (4) and integrating by parts yields the desired conditions forf (x):

sinh[λ(b – a)]f x (b) – λ cosh[λ(b – a)]f (b) = λf (a),

sinh[λ(b – a)]f x (a) + λ cosh[λ(b – a)]f (a) = –λf (b). (5)

The general form of the right-hand side is given by

whereF (x) is an arbitrary bounded twice differentiable function, and the coefficients A and B

are expressed in terms ofF (a), F (b), F x (a), and F x (b) and can be determined by substituting

formula (6) into conditions (5)

µ |x – t|from (1) yields2(Aλ + Bµ)y(x) + A(λ2–µ2)

 b a

sinh

λ |x – t|y(t) dt = f xx (x) – µ2f (x). (2)ForAλ + Bµ = 0, this is an equation of the form 3.3.5, and for Aλ + Bµ≠ 0, this is an equation

of the form 4.3.26

The right-hand sidef (x) must satisfy certain relations, which can be obtained by setting

x = a and x = b in the original equation (a similar procedure is used in 3.3.5).

7.

 b

a

sinh(λx) – sinh(λt)y(t) dt = f (x).

This is a special case of equation 3.8.3 withg(x) = sinh(λx).

Solution:

y(x) = 1

2λ

d dx



f x (x)

cosh(λx)

The right-hand sidef (x) of the integral equation must satisfy certain relations (see item 2 ◦

of equation 3.8.3)

8.

 a

0

sinh(βx) – sinh(µt)y(t) dt = f (x), β > 0, µ > 0.

This is a special case of equation 3.8.4 withg(x) = sinh(βx) and λ = µ/β.

sinh[λk(x – t)]y(t) dt +

 b x

sinh[λk(t – x)]y(t) dt (1)

Differentiating (1) with respect tox twice yields

I k  =λk

 x a

cosh[λk(x – t)]y(t) dt – λk

 b x

where the primes denote the derivatives with respect tox By comparing formulas (1) and (2),

we find the relation betweenI k andI k:

Differentiating (6) with respect tox twice and eliminating In–1from the resulting equation

with the aid of (6), we obtain a similar equation whose right-hand side is a second-order

linear differential operator (acting ony) with constant coefficients plus the sum

n–2

k=1

B k I k

If we successively eliminateIn–2,In–3, , with the aid of double differentiation, then we

finally arrive at a linear nonhomogeneous ordinary differential equation of order 2(n – 1) with

constant coefficients

3◦ The right-hand sidef (x) must satisfy certain conditions To find these conditions, one

should setx = a in the integral equation and its derivatives (Alternatively, these conditions

can be found by settingx = a and x = b in the integral equation and all its derivatives obtained

by means of double differentiation.)

11.

 b

0

sinhk x – sinh k ty(t) dt = f (x), 0 <k < 1.

This is a special case of equation 3.8.3 withg(x) = sinh k x.

Solution:

y(x) = 1

2k

d dx



f x (x)

coshx sinh k–1 x

The right-hand sidef (x) must satisfy certain conditions As follows from item 3 ◦of equation

3.8.3, the admissible general form of the right-hand side is given by

f (x) = F (x) + Ax + B, A = –F x (b), B = 12

bF x (b) – F (0) – F (b)

,whereF (x) is an arbitrary bounded twice differentiable function (with bounded first deriva-

This is a special case of equation 3.8.6 withg(x) = k sinh(λt).

3.3-3 Kernels Containing Hyperbolic Tangent

15.

 b

a

tanh(λx) – tanh(λt)y(t) dt = f (x).

This is a special case of equation 3.8.3 withg(x) = tanh(λx).

Solution:

y(x) = 1

2λ

d dx

cosh2(λx)f x (x)

.The right-hand sidef (x) of the integral equation must satisfy certain relations (see item 2 ◦of

equation 3.8.3)

16.

 a

0

tanh(βx) – tanh(µt)y(t) dt = f (x), β > 0, µ > 0.

This is a special case of equation 3.8.4 withg(x) = tanh(βx) and λ = µ/β.

17.

 b

0

| tanhk x – tanh k t | y(t) dt = f(x), 0 <k < 1.

This is a special case of equation 3.8.3 withg(x) = tanh k x.

Solution:

y(x) = 1

2k

d dx

cosh2x coth k–1 x f x (x)

.The right-hand sidef (x) must satisfy certain conditions As follows from item 3 ◦of equation

3.8.3, the admissible general form of the right-hand side is given by

f (x) = F (x) + Ax + B, A = –F x (b), B = 12

bF x (b) – F (0) – F (b)

,whereF (x) is an arbitrary bounded twice differentiable function (with bounded first deriva-

3.3-4 Kernels Containing Hyperbolic Cotangent

21.

 b

a

coth(λx) – coth(λt)y(t) dt = f (x).

This is a special case of equation 3.8.3 withg(x) = coth(λx).

22.

 b

0

cothk x – coth k ty(t) dt = f (x), 0 <k < 1.

This is a special case of equation 3.8.3 withg(x) = coth k x.

3.4 Equations Whose Kernels Contain Logarithmic



xf x (x).The right-hand sidef (x) of the integral equation must satisfy certain relations (see item 2 ◦of

2◦ Ifb – a = 4, then for the equation to be solvable, the condition

 b a

√

(t – a)(b – t) f t (t) dt

,

whereC is an arbitrary constant.

• Reference: F D Gakhov (1977).

 a–a w(t, a)f (t) dt

w(x, a)

– 12

 a

|x| w(x, ξ)

d dξ

1

M (ξ)

d dξ

 ξ–

w(t, ξ)f (t) dt

dξ

– 12

d dx

• Reference: I C Gohberg and M G Krein (1967).

(1 +λx)f x (x)

.The right-hand sidef (x) of the integral equation must satisfy certain relations (see item 2 ◦

3.4-2 Kernels Containing Power-Law and Logarithmic Functions

 ∞0

The left-hand side of this equation is the iterated Stieltjes transform

Under some assumptions, the solution of the integral equation can be represented in the

To calculate the solution approximately, one should restrict oneself to a specific value ofn in

this formula instead of taking the limit

• Reference: I I Hirschman and D V Widder (1955).

ln|z – τ|w(τ) dτ = F (z), F (z) = µf

z1/β

3.4-3 An Equation Containing the Unknown Function of a Complicated Argument

Up to constant factors, the functionf (x) and the solution y(t) are the Fourier cosine

cos(λx) – cos(λt)y(t) dt = f (x).

This is a special case of equation 3.8.3 withg(x) = cos(λx).

Solution:

y(x) = – 1

2λ

d dx



f x (x)

sin(λx)

The right-hand sidef (x) of the integral equation must satisfy certain relations (see item 2 ◦of

equation 3.8.3)

3.

 a

0

cos(βx) – cos(µt)y(t) dt = f (x), β > 0, µ > 0.

This is a special case of equation 3.8.4 withg(x) = cos(βx) and λ = µ/β.

4.

 b

a

cosk x – cos k ty(t) dt = f (x), 0 <k < 1.

This is a special case of equation 3.8.3 withg(x) = cos k x.

Solution:

y(x) = – 1

2k

d dx



f x (x)

sinx cos k–1 x

The right-hand sidef (x) of the integral equation must satisfy certain relations (see item 2 ◦of

3.5-2 Kernels Containing Sine

Up to constant factors, the functionf (x) and the solution y(t) are the Fourier sine transform

sin[λ(x – t)]y(t) dt +

 b x

2◦ The right-hand sidef (x) of the integral equation must satisfy certain relations By setting

x = a and x = b in (1), we obtain two corollaries

 b a

sin[λ(t – a)]y(t) dt = f (a),

 b a

sin[λ(b – t)]y(t) dt = f (b). (4)

Substituting solution (3) into (4) followed by integrating by parts yields the desired conditions

forf (x):

sin[λ(b – a)]f x (b) – λ cos[λ(b – a)]f (b) = λf (a),

sin[λ(b – a)]f x (a) + λ cos[λ(b – a)]f (a) = –λf (b). (5)

The general form of the right-hand side of the integral equation is given by

whereF (x) is an arbitrary bounded twice differentiable function, and the coefficients A and B

are expressed in terms ofF (a), F (b), F x (a), and F x (b) and can be determined by substituting

formula (6) into conditions (5)

Aλ2sin

λ |x – t|+Bµ2sin

µ |x – t|y(t) dt = f xx  (x). (1)Eliminating the integral term with sin

µ |x – t|from (1) with the aid of the original equation,

of the form 4.5.29

The right-hand sidef (x) must satisfy certain relations, which can be obtained by setting

x = a and x = b in the original equation (a similar procedure is used in 3.5.7).

9.

 b

a

sin(λx) – sin(λt)y(t) dt = f (x).

This is a special case of equation 3.8.3 withg(x) = sin(λx).

Solution:

y(x) = 1

2λ

d dx



f x (x)

cos(λx)

The right-hand sidef (x) of the integral equation must satisfy certain relations (see item 2 ◦of

equation 3.8.3)

10.

 a

0

sin(βx) – sin(µt)y(t) dt = f (x), β > 0, µ > 0.

This is a special case of equation 3.8.4 withg(x) = sin(βx) and λ = µ/β.

–14A sin

sin[λk(x – t)]y(t) dt +

 b x

sin[λk(t – x)]y(t) dt (1)

Differentiating (1) with respect tox yields

I k  =λk

 x a

cos[λk(x – t)]y(t) dt – λk

 b x

where the primes denote the derivatives with respect tox By comparing formulas (1) and (2),

we find the relation betweenI k andI k:

with the aid of (6), we obtain a similar equation whose left-hand side is a second-order

linear differential operator (acting ony) with constant coefficients plus the sum

n–2

k=1 BkIk

If we successively eliminateI n–2,I n–3, , with the aid of double differentiation, then we

finally arrive at a linear nonhomogeneous ordinary differential equation of order 2(n – 1) with

constant coefficients

3◦ The right-hand sidef (x) must satisfy certain conditions To find these conditions, one

should setx = a in the integral equation and its derivatives (Alternatively, these conditions

can be found by settingx = a and x = b in the integral equation and all its derivatives obtained

by means of double differentiation.)

13.

 b

0

sink x – sin k ty(t) dt = f (x), 0 <k < 1.

This is a special case of equation 3.8.3 withg(x) = sin k x.

Solution:

y(x) = 1

2k

d dx



f x (x)

cosx sin k–1 x

The right-hand sidef (x) must satisfy certain conditions As follows from item 3 ◦of equation

3.8.3, the admissible general form of the right-hand side is given by

f (x) = F (x) + Ax + B, A = –F x (b), B = 12

bF x (b) – F (0) – F (b)

,whereF (x) is an arbitrary bounded twice differentiable function (with bounded first deriva-

3.5-3 Kernels Containing Tangent

17.

 b

a

tan(λx) – tan(λt)y(t) dt = f (x).

This is a special case of equation 3.8.3 withg(x) = tan(λx).

Solution:

y(x) = 1

2λ

d dx

cos2(λx)f x (x)

The right-hand sidef (x) of the integral equation must satisfy certain relations (see item 2 ◦of

equation 3.8.3)

18.

 a

0

tan(βx) – tan(µt)y(t) dt = f (x), β > 0, µ > 0.

This is a special case of equation 3.8.4 withg(x) = tan(βx) and λ = µ/β.

19.

 b

0

tank x – tan k ty(t) dt = f (x), 0 <k < 1.

This is a special case of equation 3.8.3 withg(x) = tan k x.

Solution:

y(x) = 1

2k

d dx

cos2x cot k–1 xf x (x)

The right-hand sidef (x) must satisfy certain conditions As follows from item 3 ◦of equation

3.8.3, the admissible general form of the right-hand side is given by

This is a special case of equation 3.8.6 withg(x) = k tan(λt).

3.5-4 Kernels Containing Cotangent

23.

 b

a

cot(λx) – cot(λt)y(t) dt = f (x).

This is a special case of equation 3.8.3 withg(x) = cot(λx).

24.

 b

a

cotk x – cot k ty(t) dt = f (x), 0 <k < 1.

This is a special case of equation 3.8.3 withg(x) = cot k x.

3.5-5 Kernels Containing a Combination of Trigonometric Functions

cos(xt) + sin(xt)

 ∞0



f (0) + x

 π/20

f ξ (ξ) dt

, ξ = x sin t.

• References: E T Whittaker and G N Watson (1958), F D Gakhov (1977).

28.

 π/2

0

y(ξ) dt = f (x), ξ = x sin k t.

Generalized Schl¨omilch equation.

This is a special case of equation 3.5.29 forn = 0 and m = 0.



x k1

 x

0sint f (ξ) dt

, ξ = x sin k t.

29.

 π/2

0

sinλ t y(ξ) dt = f (x), ξ = x sin k t.

This is a special case of equation 3.5.29 form = 0.

, ξ = x sin k t.

 π/2

0

sinλ t cos m t y(ξ) dt = f (x), ξ = x sin k t.

1◦ Letλ > –1, m > –1, and k > 0 The transformation

z = x k2, ζ = z sin2t, w(ζ) = ζ λ–12 y

ζ k2leads to an equation of the form 1.1.43:

 z

0(z – ζ) m–12 w(ζ) dζ = F (z), F (z) = 2z λ+m2 f

z k2

2◦ Solution with –1 <m < 1:

y(x) = 2k

π sin

π(1 – m)2



x k–λ–1 k d dx



x λ+1 k

 π/2

0sinλ+1 t tan m t f (ξ) dt

,

Here the integral is understood in the sense of the Cauchy principal value and the right-hand

side is assumed to satisfy the condition

t – x2



f (t) dt + C,

whereC is an arbitrary constant.

It follows from the solution that

lncosh(λx) – cosh(λt)y(t) dt = f (x).

This is a special case of equation 1.8.9 withg(x) = cosh(λx).

2.

 b

a

lnsinh(λx) – sinh(λt)y(t) dt = f (x).

This is a special case of equation 1.8.9 withg(x) = sinh(λx).

 a–a w(t, a)f (t) dt

w(x, a)

– 12

 a

|x| w(x, ξ)

d dξ

1

M (ξ)

d dξ

 ξ–

w(t, ξ)f (t) dt

dξ

– 12

d dx

sinh1

2Asinh1

lntanh(λx) – tanh(λt)y(t) dt = f (x).

This is a special case of equation 1.8.9 withg(x) = tanh(λx).

 a

–a w(t, a)f (t) dt

w(x, a)

– 12

 a

|x| w(x, ξ)

d dξ

1

M (ξ)

d dξ

d dx

• Reference: I C Gohberg and M G Krein (1967).

3.6-2 Kernels Containing Logarithmic and Trigonometric Functions

6.

 b

a

lncos(λx) – cos(λt)y(t) dt = f (x).

This is a special case of equation 1.8.9 withg(x) = cos(λx).

 b

a

lnsin(λx) – sin(λt)y(t) dt = f (x).

This is a special case of equation 1.8.9 withg(x) = sin(λx).

 a

–a w(t, a)f (t) dt

w(x, a)

– 12

 a

|x| w(x, ξ)

d dξ

1

M (ξ)

d dξ

d dx

sin1

2Asin1

• Reference: I C Gohberg and M G Krein (1967).

3.7 Equations Whose Kernels Contain Special

tJν(xt)f (t) dt.

The functionf (x) and the solution y(t) are the Hankel transform pair.

• Reference: V A Ditkin and A P Prudnikov (1965).

2.

 b

a

J ν(λx) – Jν(λt)y(t) dt = f (x).

This is a special case of equation 3.8.3 withg(x) = Jν(λx), where Jνis the Bessel function

of the first kind

3.

 b

a

Y ν(λx) – Yν(λt)y(t) dt = f (x).

This is a special case of equation 3.8.3 withg(x) = Yν(λx), where Yνis the Bessel function

of the second kind

3.7-2 Kernels Containing Modified Bessel Functions

4.

 b

a

I ν(λx) – Iν(λt)y(t) dt = f (x).

This is a special case of equation 3.8.3 withg(x) = I ν(λx), where I ν is the modified Bessel

function of the first kind

5.

 b

a

K ν(λx) – Kν(λt)y(t) dt = f (x).

This is a special case of equation 3.8.3 withg(x) = Kν(λx), where Kνis the modified Bessel

function of the second kind (the Macdonald function)

HereK νis the modified Bessel function of the second kind

Up to a constant factor, the left-hand side of this equation is the Meijer transform ofy(t)

(z is treated as a complex variable).

Solution:

y(t) = 1πi

 c+i ∞ c–i∞

√

zt Iν(zt)f (z) dz.

For specificf (z), one may use tables of Meijer integral transforms to calculate the integral.

• Reference: V A Ditkin and A P Prudnikov (1965).

dt

(1 –t2)(1 –z2t2) is the complete elliptic integral of the first kind.

Solution:

y(x) = – 4

π2

d dx

where A = Γ(β) Γ(2µ – β) sin[(β – µ)π]

• Reference: P P Zabreyko, A I Koshelev, et al (1975).

3.8 Equations Whose Kernels Contain Arbitrary

h1(t)y(t) dt = A1,

 b a

where theAk are some constants In this case, any functiony = y(x) satisfying the

normal-ization type conditions

3.8-2 Equations Containing Modulus



g(x) – g(t)

y(t) dt +

 b x

 b x

2◦ Let us demonstrate that the right-hand side f (x) of the integral equation must satisfy

certain relations By settingx = a and x = b, in (1), we obtain two corollaries

 b a



g(t) – g(a)

y(t) dt = f (a),

 b a

Let us point out a useful property of these constraints:f x (b)g x (a) + f x (a)g  x(b) = 0.

Conditions (5) make it possible to find the admissible general form of the right-hand side

of the integral equation:

whereF (x) is an arbitrary bounded twice differentiable function (with bounded first

deriva-tive), and the coefficientsA and B are given by

3◦ Ifg(x) is representable in the form g(x) = O(x – a) k with 0 <k < 1 in the vicinity of

the pointx = a (in particular, the derivative g x  is unbounded asx → a), then the solution of

the integral equation is given by formula (3) as well In this case, the right-hand side of the

integral equation must satisfy the conditions

As before, the right-hand side of the integral equation is given by (6), with

2◦ Let us demonstrate that the right-hand side f (x) of the integral equation must satisfy

certain relations By settingx = 0 in (1) and (2), we obtain two corollaries

 a0



g(λt) – g(0)

y(t) dt = f (0), g x (0)

 a0

y(t) dt = –f x (0) (4)Substitutey(x) of (3) into (4) Integrating by parts yields the desired constraints for f (x):

Conditions (5) make it possible to find the admissible general form of the right-hand side

of the integral equation:

whereF (x) is an arbitrary bounded twice differentiable function (with bounded first

deriva-tive), and the coefficientsA and B are given by

3◦ Ifg(x) is representable in the form g(x) = O(x) k with 0 <k < 1 in the vicinity of the

pointx = 0 (in particular, the derivative g  x is unbounded asx → 0), then the solution of

the integral equation is given by formula (3) as well In this case, the right-hand side of the

integral equation must satisfy the conditions

whereg–1is the inverse ofg.

2◦ Let us demonstrate that the right-hand side f (x) of the integral equation must satisfy

certain relations By settingx = 0 in (1) and (2), we obtain two corollaries

 a0

ty(t) dt = f (0), g  x(0)

 a0

Conditions (5) make it possible to find the admissible general form of the right-hand side

of the integral equation in question:

whereF (x) is an arbitrary bounded twice differentiable function (with bounded first

deriva-tive), and the coefficientsA and B are given by

3◦ Ifg(x) is representable in the vicinity of the point x = 0 in the form g(x) = O(x) kwith

0 <k < 1 (i.e., the derivative g x  is unbounded asx → 0), then the solution of the integral

equation is given by formula (3) as well In this case, the right-hand side of the integral

equation must satisfy the conditions

2◦ Let us demonstrate that the right-hand side f (x) of the integral equation must satisfy

certain relations By settingx = 0 in (1) and (2), we obtain two corollaries

of the integral equation:

ln|z – τ|w(τ) dτ = F (z),

whereF = F (z) is the function which is obtained from z = g(x) and F = f (x) by eliminating x.

3.8-3 Equations With Difference Kernel: K(x, t) = K(x – t)

 ∞–∞ xK(x).



λ=0

 ∞–∞ K(x)e–λx dx.

 ∞–∞ xK(x)e–λx dx.

 ∞–∞ K(x)e

cos(λx) + BIc–AIs

I2+I2 ssin(λx),

Ic=

 ∞–∞ K(z) cos(λz) dz, Is=

 ∞–∞ K(z) sin(λz) dz.

f (u)

˜

K(u) e iux du,

˜

f (u) = √1

2π

 ∞–∞

f (x)e–iux dx, K(u) =˜ √1

2π

 ∞–∞ K(x)e–iux dx.

The following statement is valid Letf (x) ∈ L2(–∞, ∞) and K(x) ∈ L1(–∞, ∞) Then

for a solutiony(x) ∈ L2(–∞, ∞) of the integral equation to exist, it is necessary and sufficient

that ˜f (u)/ ˜ K(u) ∈ L2(–∞, ∞).

2◦ Let the functionP (s) defined by the formula

1

P (s) =

 ∞–∞ e

–λ

a x



 ∞–∞ K(z) exp

–λ

(β+q)z dz, I m(q) =

 ∞–∞ M (z)e

3.8-5 Equations of the Formb

 b a

f (t) dt, I1=

 b a

 b a

f (t) dt, I l=

 b a

cos(lnx) + AIs

I2+I2 ssin(lnx),

3.8-5 Equations of the Formb

...

 b a

sinh[λ(b – t)]y(t) dt = f (b). (4) Substituting solution (3) into (4) and integrating by parts yields the desired conditions forf (x):

sinh[λ(b...

This is a special case of equation 3.8.3 withg(x) = coth k x.

3 .4 Equations Whose Kernels Contain Logarithmic



xf x (x).The