HANDBOOK OFINTEGRAL EQUATIONS phần 2 ppt

We consider the interval [a, x] in which J0λx does not change its sign... We consider the interval [a, x] in which J νλx does not change its sign... We consider the interval [a, x] in wh

Ax λtanµ t + Bt βcotγ x y(t) dt = f (x).

This is a special case of equation 1.9.15 withg1(x) = Axλ,h1(t) = tanµ t, g2(x) = B cotγ x,

Ax λcotµ t + Bt βtanγ x y(t) dt = f (x).

This is a special case of equation 1.9.15 withg1(x) = Axλ,h1(t) = cotµ t, g2(x) = B tanγ x,

√

1 –λ2x2f x (x)





arccos(λx)– A

A+B

 x a

arccos(λt)– B

A+B f t (t) dt



arccos(λt) – arccos(λx) y(t) dt = f (x).

This is a special case of equation 1.9.38 withg(x) = 1 – arccos(λx).

Solution:

y(x) = 2

π ϕ(x)

1

ϕ(x)

d dx

d dx

ϕ(x)

d dx

 x a

1.6-2 Kernels Containing Arcsine

√

1 –λ2x2f x (x)



ϕ(x)

d dx

2 x a

d dx

ϕ(x)

d dx

2 x a

 x a

This is a special case of equation 1.9.6 withg(x) = A arcsin β(λx) and h(t) = B arcsinγ(µt)+C

1.6-3 Kernels Containing Arctangent

(1 +λ2x2)f x (x)

dx

n+1

f (x).

d dx

2 x a

d dx

ϕ(x)

d dx

2 x a

(1 +λ2x2)fx (x)arctanµ–1(λx)

 x a

1.6-4 Kernels Containing Arccotangent

(1 +λ2x2)f x (x)



arccot(λx)– A

A+B

 x a

arccot(λt)– B

A+B f t (t) dt



ϕ(x)

d dx

d dx

 x a

ϕ(x)

d dx

2 x a

(1 +λ2x2)fx (x)arccotµ–1(λx)

 x a

This is a special case of equation 1.9.6 withg(x) = A arccot β(λx) and h(t) = B arccotγ(µt)+C

1.7 Equations Whose Kernels Contain Combinations of

d dx

2 x a

 x a

e µ(x–t)(coshx – cosh t) λ y(t) dt = f (x), 0 <λ < 1.

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.3.23:

e µ(x–t)(coshλ x – cosh λ t)y(t) dt = f (x).

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.3.24:

 x a

A cosh λ x + B cosh λ t y(t) dt = f (x).

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.3.25:

 x

a

A cosh λ x + B cosh λ t w(t) dt = e–µx f (x).

2 x a

 x

a

e µ(x–t)(sinhx – sinh t) λ y(t) dt = f (x), 0 <λ < 1.

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.3.58:

 x a

(sinhx – sinh t) λ w(t) dt = e–µx f (x).

22.

 x

a

e µ(x–t)(sinhλ x – sinh λ t)y(t) dt = f (x).

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.3.59:

 x a

A sinh λ x + B sinh λ t y(t) dt = f (x).

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.3.60:

A tanh λ x + B tanh λ t y(t) dt = f (x).

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.3.77:

A tanh λ x + B tanh β t + C y(t) dt = f (x).

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.9.6 with g(x) = A tanh λ x,

A coth λ x + B coth λ t y(t) dt = f (x).

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.3.90:

A coth λ x + B coth β t + C y(t) dt = f (x).

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.9.6 with g(x) = A coth λ x,



x d dx

2 x a

e–λt f (t) dt

tln(x/t).

 x a

e–λt f (t) dt

tln(x/t).

 x a

d dx

n+1

F µ(x), F µ(x) = e–µx f (x).

d dx

d dx

2 x a

e µ(x–t)(cosλ x – cos λ t)y(t) dt = f (x).

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.5.24:

A cos λ x + B cos λ t y(t) dt = f (x).

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.5.25:

 x a

d dx

e µ(x–t)(sinλ x – sin λ t)y(t) dt = f (x).

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.5.59:

A sin λ x + B sin λ t y(t) dt = f (x).

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.5.60:

A tan λ x + B tan λ t y(t) dt = f (x).

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.5.77:

 x a

A tan λ x + B tan β t + C y(t) dt = f (x).

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.9.6:

 x

a

A tan λ x + B tan β t + C w(t) dt = e–µx f (x).

A cot λ x + B cot λ t y(t) dt = f (x).

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.5.90:

A cot λ x + B cot β t + C y(t) dt = f (x).

The substitutionw(x) = e–µx y(x) leads to an equation of the form 1.9.6:

This is a special case of equation 1.9.6 withg(x) = B ln γ(µx) and h(t) = A cothβ(λt) + C.

1.7-5 Kernels Containing Hyperbolic and Trigonometric Functions

This is a special case of equation 1.9.6 withg(x) = A tanh β(λx) and h(t) = B sinγ(µt) + C.

1.7-6 Kernels Containing Logarithmic and Trigonometric Functions

This is a special case of equation 1.9.6 withg(x) = B ln γ(µx) and h(t) = A sinβ(λt) + C

1.8 Equations Whose Kernels Contain Special



d2

dx2 +λ2

2 x a

 x

a

[AJ0 (λx) + BJ0 (λt)]y(t) dt = f (x).

ForB = –A, see equation 1.8.2 We consider the interval [a, x] in which J0(λx) does not

change its sign

Solution withB ≠ –A:

y(x) = ± 1

A + B

d dx

J

0(λx)– A A+B



d2

dt2 +λ2

3 t a

If the right-hand side of the equation is differentiable sufficiently many times and the

conditionsf (a) = f x (a) =· · · = f(2n+1)

x (a) = 0 are satisfied, then the solution of the integralequation can be written in the form

y(x) = A

 x a

If the right-hand side of the equation is differentiable sufficiently many times and the

conditionsf (a) = f x (a) =· · · = f(2n+2)

x (a) = 0 are satisfied, then the function g(t) defining thesolution can be written in the form

g(t) = A

 t a

(t – τ )5/2 J5/2[λ(t – τ )] f (τ ) dτ

 x

a

[AJ ν(λx) + BJ ν( λt)]y(t) dt = f (x).

ForB = –A, see equation 1.8.14 We consider the interval [a, x] in which J ν(λx) does not

change its sign

Solution withB ≠ –A:

y(x) = ± 1

A + B

d dx

J

ν(λx)– A A+B

Here the sign ofJ ν(λx) should be taken

where –1

2 <ν < n–12 andn = 1, 2,

If the right-hand side of the equation is differentiable sufficiently many times and the

conditionsf (a) = f x (a) =· · · = f(n–1)

x (a) = 0 are satisfied, then the solution of the integralequation can be written in the form

y(x) = A

 x a

where –1 <ν < n

2 – 1 andn = 1, 2,

If the right-hand side of the equation is differentiable sufficiently many times and the

conditionsf (a) = f x (a) =· · · = f(n–1)

x (a) = 0 are satisfied, then the function g(t) defining thesolution can be written in the form

g(t) = A

 t a

x does not change its sign

Solution withB ≠ –A:

y(x) = ± 1

A + B

d dx

J

ν

λ √ x

A A+B

 x

a

J ν

λ √ t

B A+B f t (t) dt



Here the signJ ν

d2

dx2

 x a

 x a

If the right-hand side of the equation is differentiable sufficiently many times and the

conditionsf (a) = f x (a) =· · · = f(n–1)

x (a) = 0 are satisfied, then the solution of the integralequation can be written in the form

π

d dx

d dx

J

ν(λx)– 2A A+B

 x

a

[AY ν( λx) + BY ν( λt)]y(t) dt = f (x).

ForB = –A, see equation 1.8.37 We consider the interval [a, x] in which Y ν(λx) does not

change its sign

Solution withB ≠ –A:

y(x) = ± 1

A + B

d dx

Y

ν(λx)– A A+B

[AJ ν(λx)Y µ( βt) + BJ ν( λt)Y µ( βx)]y(t) dt = f (x).

This is a special case of equation 1.9.15 with g1(x) = AYν(λx), h1(t) = Yµ(βt), g2(x) =



d2

dx2 –λ2

2 x a

I

0(λx)– A A+B

 x a

I0(λt)– B

A+B f t (t) dt

.Here the sign ofI ν(λx) should be taken



d2

dt2 –λ2

3 t a

(t – τ ) I1[λ(t – τ )] f (τ ) dτ

If the right-hand side of the equation is differentiable sufficiently many times and the

conditionsf (a) = f x (a) =· · · = f(2n+1)

x (a) = 0 are satisfied, then the solution of the integralequation can be written in the form

If the right-hand side of the equation is differentiable sufficiently many times and the

conditionsf (a) = f x (a) =· · · = f(2n+2)

x (a) = 0 are satisfied, then the function g(t) defining thesolution can be written in the form

g(t) = A

 t a

I ν(λx)– A

A+B

 x a



I ν(λt)– B

A+B f t (t) dt



where –12 <ν < n–12 andn = 1, 2,

If the right-hand side of the equation is differentiable sufficiently many times and the

conditionsf (a) = f x (a) =· · · = f(n–1)

x (a) = 0 are satisfied, then the solution of the integralequation can be written in the form

y(x) = A

 x a

where –1 <ν < n

2 – 1 andn = 1, 2,

If the right-hand side of the equation is differentiable sufficiently many times and the

conditionsf (a) = f x (a) =· · · = f(n–1)

x (a) = 0 are satisfied, then the function g(t) defining thesolution can be written in the form

g(t) = A

 t a

I ν

λ √ x

A A+B

 x a



I ν

λ √ t

B A+B f t (t) dt



d3

dx3

 x a

πλ

d2

dx2

 x a

If the right-hand side of the equation is differentiable sufficiently many times and the

conditionsf (a) = f x (a) =· · · = f(n–1)

x (a) = 0 are satisfied, then the solution of the integralequation can be written in the form

π

d dx

π

d dx

ν(λx)– 2A A+B



f (t) dt

,whereµ < 1, ν ≥ –1

2, andn = 1, 2,

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

f (t) dt

,whereµ < 1, ν ≥ –1

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

1.8-4 Kernels Containing Hypergeometric Functions

(x – t)n–b–1

Γ(b)Γ(n – b)Φ

–a, n – b; λ(x – t) f t(n)(t) dt

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

(x – t)n–c–1

Γ(c)Γ(n – c) F

–a, n – b, n – c; 1 – t

x f (t) dt

,where 0 <c < n and n = 1, 2,

If the right-hand side of the equation is differentiable sufficiently many times and the

x f

(n)

t (t) dt

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

1.9 Equations Whose Kernels Contain Arbitrary



f (x) g(x)

ForB = –A, see equation 1.9.2.

Solution withB ≠ –A:

y(x) = signg(x)

A + B

d dx

g(x)– A

A+B

 x a

g(t)– B A+B f t (t) dt



 x a

f t (t) dt

Φ(t)

, Φ(x) = exp

 x a

h  t(t) dt

g(t) + h(t)

 x

a



f (t) h(t)



t

dt Φ(t)

, Φ(t) = exp

–

 x

a

h(t) g(t) dt

d dx

(f /h) x(g/h) x

, where f = f (x), g = g(x), h = h(x).

HereAf + Bg + Ch /≡ 0, with A, B, and C being some constants.

 x

a

[Ag(x)h(t) + Bg(t)h(x)]y(t) dt = f (x).

ForB = –A, see equation 1.9.11.

Solution withB ≠ –A:

(A + B)h(x)

d dx



h(x) g(x)

A A+B  x

a



h(t) g(t)

dt





f (t) h(t)



t

dt Φ(t)

, Φ(x) = exp

 x a

g t (t)h(t) dt

Forg2/g1= const orh2/h1= const, see equation 1.9.1

1◦ Solution withg1(x)h1(x) + g2(x)h2(x) /≡ 0 and f(x) /≡ const g2(x):

y(x) = 1

h1(x)

d dx

Iff (x) ≡ const g2(x), the solution is given by formulas (1) and (2) in which the subscript 1

must be changed by 2 and vice versa

2◦ Solution withg1(x)h1(x) + g2(x)h2(x)≡ 0:

y(x) = 1

h1

d dx

(f /g2) x(g1/g2) x

= – 1

h1

d dx

(f /g2) x(h2/h1) x

,

wheref = f (x), g2=g2(x), h1 =h1(x), and h2=h2(x)

1.9-2 Equations With Difference Kernel: K(x, t) = K(x – t)

16.

 x

a

K(x – t)y(t) dt = f (x).

1◦ LetK(0) = 1 and f (a) = 0 Differentiating the equation with respect to x yields a Volterra

equation of the second kind:

p ˜ K(p) – 1

, K(p) = L˜ 

K(x),

where L and L–1are the operators of the direct and inverse Laplace transforms, respectively

 c+i∞

c–i∞ e

px R(p) dp.˜

2◦ LetK(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) witha = 0

and constant right-hand sidef ≡ 1,

It is convenient to calculate the coefficients B and C using tables of Laplace transforms

according to the formulasB = L {K(z), λ} and C = L{zK(z), λ}.

e–λx

= 12

where the constantsB kare found by the method of undetermined coefficients The solution

can also be obtained by the formula given in 1.9.17 (item 4◦)

where the constantsB kare found by the method of undetermined coefficients The solution

can also be obtained by the formula given in 1.9.19 (item 3◦)

The expression forB is the Laplace transform of the function K(–z) with parameter p = –λ and

can be calculated with the aid of tables of Laplace transforms given (e.g., see Supplement 4)

e–λx= 12

can also be obtained by the formula given in 1.9.27 (item 4◦).

where the constantsB kare found by the method of undetermined coefficients The solution

can also be obtained by the formula given in 1.9.29 (item 3◦)

8◦ For arbitrary right-hand side f = f (x), the solution of the integral equation can be

calculated by the formula

To calculate ˜f (p) and ˜k(–p), it is convenient to use tables of Laplace transforms, and to

determiney(x), tables of inverse Laplace transforms.

g x (x)

d dx

g  x(x)

d dx

2 x

a

f (t)g  t(t) dt

√ g(x) – g(t).

d dx

 x a

f (t)g t (t) dt

√ g(x) – g(t).

 x a

e–λt f (t)g t (t)

√ g(x) – g(t) dt.

g x (x)

d dx

2 x

a

g  t(t)f (t) dt[g(x) – g(t)]λ, k = sin(πλ)

d dx

 x a

f (t)g t (t) dt[g(x) – g(t)]1–λ

K(z)z λ–1 dz, I µ =

 1 0

K(z)z µ–1 dz.

a y(t) dt followed by integration by parts leads to an integral

equation of the form 1.9.15:

 x a

a e–λt y(t) dt followed by integration by parts leads to an integral

equation of the form 1.9.15:

1.10 Some Formulas and Transformations

1 Let the solution of the integral equation

g(x)



Below are formulas for the solutions of integral equations of the form (3) for some specific

functionsg(x) and h(t) In all cases, it is assumed that the solution of equation (1) is known and is

 x a

K(x, t)e λ(x–t) y(t) dt = f (x)

has the form

y(x) = e λx Fe–λx f (x)

2 Let the solution of the integral equation (1) have the form

x a

whereL1andL2are some linear differential operators

The solution of the more complicated integral equation

 x a

K

whereϕ(x) is an arbitrary monotone function (differentiable sufficiently many times, ϕ  x > 0), is

determined by the formula

  x a

R

ϕ(x), ϕ(t) ϕ  t(t)f (t) dt

(7)

Below are formulas for the solutions of integral equations of the form (6) for some specific

functions ϕ(x) In all cases, it is assumed that the solution of equation (1) is known and is

  x a

dx

  x a

1

t R

... )5 /2< /sup> J5 /2< /sub>[λ(t – τ )] f (τ ) dτ

 x... data-page="48">

Chapter 2< /b>

Linear Equations of the Second Kind

With Variable Limit of Integration

Notation: f = f(x),... Yµ(βt), g2< /sub>(x) =



d2< /small>

dx2< /small> –λ2< /small>

2< /small> x a