Handbook of mathematics for engineers and scienteists part 120 doc

Linear Integral Equations of the First Kind with Variable Integration Limit 16.1.1.. Function and kernel classes.. The kernel Kx, t of an integral equation is called difference kernel if

Integral Equations

16.1 Linear Integral Equations of the First Kind with

Variable Integration Limit

16.1.1 Volterra Equations of the First Kind

16.1.1-1 Some definitions Function and kernel classes

A Volterra linear integral equation of the first kind has the general form

 x

a K (x, t)y(t) dt = f (x), (16.1.1.1)

where y(x) is the unknown function (a ≤ x ≤ b ), K(x, t) is the kernel of the integral equation, and f (x) is a given function, the right-hand side of equation (16.1.1.1) The functions y(x) and f (x) are usually assumed to be continuous or square integrable on [a, b] The kernel K(x, t) is usually assumed either to be continuous on the square S ={a≤x≤b,

a≤t≤b}or to satisfy the condition

 b

a

 b

a K

2(x, t) dx dt = B2 <∞, (16.1.1.2)

where B is a constant, that is, to be square integrable on this square It is assumed

in (16.1.1.2) that K(x, t)≡ 0for t > x.

The kernel K(x, t) is said to be degenerate if it can be represented in the form K(x, t) =

g1(x)h1(t) + · · · + g n (x)h n (t) The kernel K(x, t) of an integral equation is called difference kernel if it depends only on the difference of the arguments, K(x, t) = K(x – t).

Polar kernels

K (x, t) = L(x, t)(x – t)–β + M (x, t), 0< β <1, (16.1.1.3) and logarithmic kernels (kernels with logarithmic singularity)

K (x, t) = L(x, t) ln(x – t) + M (x, t), (16.1.1.4)

where the functions L(x, t) and M (x, t) are continuous on S and L(x, x)  0, are often considered as well

Polar and logarithmic kernels form a class of kernels with weak singularity Equations

containing such kernels are called equations with weak singularity.

In case the functions K(x, t) and f (x) are continuous, the right-hand side of

equa-tion (16.1.1.1) must satisfy the following condiequa-tions:

1◦ If K(a, a)≠ 0, then f (x) must be constrained by f (a) =0

801

2◦ If K(a, a) = K 

x (a, a) = · · · = K(n–1)

x (a, a) =0, 0<K(n)

x (a, a)<∞, then the right-hand

side of the equation must satisfy the conditions f (a) = f x  (a) = · · · = f(n)

x (a) =0

3◦ If K(a, a) = K 

x (a, a) = · · · = K(n–1)

x (a, a) = 0, K(n)

x (a, a) = ∞, then the right-hand side of the equation must satisfy the conditions f (a) = f x  (a) = · · · = f(n–1)

x (a) =0

For polar kernels of the form (16.1.1.4) and continuous f (x), no additional conditions

are imposed on the right-hand side of the integral equation

16.1.1-2 Existence and uniqueness of a solution

Assume that in equation (16.1.1.1) the functions f (x) and K(x, t) are continuous together with their first derivatives on [a, b] and on S, respectively If K(x, x)≠ 0(x[a, b]) and

f (a) =0, then there exists a unique continuous solution y(x) of equation (16.1.1.1).

Remark A Volterra equation of the first kind can be treated as a Fredholm equation of the first kind whose

kernel K(x, t) vanishes for t > x (see Section 16.3).

16.1.2 Equations with Degenerate Kernel:

K(x, t) = g1(x)h1(t)+ · · ·+gn(x)hn(t)

16.1.2-1 Equations with kernel of the form K(x, t) = g1(x)h1(t) + g2(x)h2(t).

Any equation of this type can be rewritten in the form

g1(x)

 x

a h1(t)y(t) dt + g2(x)

 x

a h2(t)y(t) dt = f (x). (16.1.2.1)

It is assumed that g1(x)/g2(x) ≠ const , h1(t)/h2(t)≠const , 0< g21(a) + g22(a) < ∞, and

f (a) =0

The change of variables

u (x) =

 x

a h1(t)y(t) dt, (16.1.2.2) followed by the integration by parts in the second integral in (16.1.2.1) with regard to the

relation u(a) =0, yields the following Volterra equation of the second kind:

[g1(x)h1(x) + g2(x)h2(x)]u(x) – g2(x)h1(x)

 x

a



h2(t)

h1(t)



t

u (t) dt = h1(x)f (x) (16.1.2.3) The substitution

w (x) =

 x

a



h2(t)

h1(t)



t

u (t) dt (16.1.2.4) reduces equation (16.1.2.3) to the first-order linear ordinary differential equation

[g1(x)h1(x) + g2(x)h2(x)]w  x – g2(x)h1(x)



h2(x)

h1(x)



x

w = f (x)h1(x)



h2(x)

h1(x)



x

(16.1.2.5)

1◦ In the case g1(x)h1(x) + g2(x)h2(x)

0, the solution of equation (16.1.2.5) satisfying

the condition w(a) =0[this condition is a consequence of the substitution (16.1.2.4)] has

the form

w (x) = Φ(x)

 x

a



h2(t)

h1(t)



t

f (t)h1(t) dt

Φ(t)[g1(t)h1(t) + g2(t)h2(t)], (16.1.2.6)

Φ(x) = exp

 x

a



h2(t)

h1(t)



t

g2(t)h1(t) dt

g1(t)h1(t) + g2(t)h2(t)

(16.1.2.7)

Let us differentiate relation (16.1.2.4) and substitute the function (16.1.2.6) into the

resulting expression After integrating by parts with regard to the relations f (a) = 0and

w (a) =0, for f const g2we obtain

u (x) = g2(x)h1(x) Φ(x)

g1(x)h1(x) + g2(x)h2(x)

 x

a



f (t)

g2(t)



t

dt

Φ(t).

Using formula (16.1.2.2), we find a solution of the original equation in the form

y (x) = 1

h1(x)

d dx



g2(x)h1(x) Φ(x)

g1(x)h1(x) + g2(x)h2(x)

 x

a



f (t)

g2(t)



t

dt

Φ(t)

, (16.1.2.8)

where the functionΦ(x) is given by (16.1.2.7).

If f (x) ≡ const g2(x), the solution is given by formulas (16.1.2.8) and (16.1.2.7), in

which the subscript 1 must be changed by 2 and vice versa

2◦ In the case g1(x)h1(x) + g2(x)h2(x)≡ 0, the solution has the form

y (x) = 1

h1

d dx



(f /g2) x

(g1/g2) x



= – 1

h1

d dx



(f /g2) x

(h2/h1) x



16.1.2-2 Equations with general degenerate kernel

A Volterra equation of the first kind with general degenerate kernel has the form

n



m=1

g m (x)

 x

a h m (t)y(t) dt = f (x). (16.1.2.9) Using the notation

w m (x) =

 x

a h m (t)y(t) dt, m=1, , n, (16.1.2.10)

we can rewrite equation (16.1.2.9) as follows:

n



m=1

g m (x)w m (x) = f (x). (16.1.2.11)

On differentiating formulas (16.1.2.10) and eliminating y(x) from the resulting equations,

we arrive at the following linear differential equations for the functions w m = w m (x):

h1(x)w m  = h m (x)w 1, m=2, , n (16.1.2.12)

(the prime stands for the derivative with respect to x) with the initial conditions

w m (a) =0, m=1, , n.

Any solution of system (16.1.2.11), (16.1.2.12) determines a solution of the original integral equation (16.1.2.9) by each of the expressions

y (x) = w  m (x)

h m (x), m=1, , n,

which can be obtained by differentiating formula (16.1.2.10)

System (16.1.2.11), (16.1.2.12) can be reduced to a linear (n –1)st-order

differen-tial equation for any function w m (x) (m = 1, , n) by multiple differentiation of

equa-tion (16.1.2.11) with regard to (16.1.2.12)

16.1.3 Equations with Difference Kernel: K(x, t) = K(x–t)

16.1.3-1 Solution method based on the Laplace transform

Volterra equations of the first kind with kernel depending on the difference of the arguments have the form  x

0 K (x – t)y(t) dt = f (x). (16.1.3.1)

To solve these equations, the Laplace transform can be used (see Section 11.2) In what follows we need the transforms of the kernel and the right-hand side; they are given by the formulas

2

K (p) =

 ∞

0 K (x)e

–px dx, 2f(p) = ∞

0 f (x)e

–px dx. (16.1.3.2)

Applying the Laplace transform L to equation (16.1.3.1) and taking into account the

fact that an integral with kernel depending on the difference of the arguments is transformed

to the product by the rule (see Paragraph 11.2.2-1)

L

 x

0 K (x – t)y(t) dt

= 2K (p) 2y(p),

we obtain the following equation for the transform2y(p):

2

K (p) 2y(p) = 2f(p). (16.1.3.3) The solution of equation (16.1.3.3) is given by the formula

2y(p) = 2f(p)

2

K (p). (16.1.3.4)

On applying the Laplace inversion formula (if it is applicable) to (16.1.3.4), we obtain a solution of equation (16.1.3.1) in the form

y (x) = 1

2πi

 c+i∞

c–i∞

2f(p)

2

K (p) e

px dp. (16.1.3.5)

To evaluate the corresponding integrals, tables of direct and inverse Laplace transforms can be applied (see Sections T3.1 and T3.2), and, in many cases, to find the inverse transform, methods of the theory of functions of a complex variable are applied, including formulas for the calculation of residues and the Jordan lemma (see Subsection 11.1.2)

16.1.3-2 Case in which the transform of the solution is a rational function.

Consider the important special case in which the transform (16.1.3.4) of the solution is a rational function of the form

2y(p) = 2f(p)

2

K (p) ≡ R (p)

Q (p), where Q(p) and R(p) are polynomials in the variable p and the degree of Q(p) exceeds that

of R(p).

If the zeros of the denominator Q(p) are simple, i.e.,

Q (p)≡const (p – λ1)(p – λ2) (p – λ n),

and λ i ≠λ j for i≠j, then the solution has the form

y (x) =

n



k=1

R (λ k)

Q  (λ k) exp(λ k x),

where the prime stands for the derivatives

Example 1 Consider the Volterra integral equation of the first kind

 x

0

e–a(x–t) y (t) dt = A sinh(bx).

We apply the Laplace transform to this equation and obtain (see Subsections T3.1.1 and T3.1.4)

1

p + a 2y(p) = Ab

p2– b2 This implies

2y(p) = Ab (p + a)

p2– b2 =

Ab (p + a) (p – b)(p + b).

We have Q(p) = (p – b)(p + b), R(p) = Ab(p + a), λ1= b, and λ2= –b Therefore, the solution of the integral

equation has the form

y (x) = 12A (b + a)e bx+12A (b – a)e–bx = Aa sinh(bx) + Ab cosh(bx).

16.1.3-3 Convolution representation of a solution

In solving Volterra integral equations of the first kind with difference kernel K(x – t) by

means of the Laplace transform, it is sometimes useful to apply the following approach Let us represent the transform (16.1.3.4) of a solution in the form

2y(p) = 2 N (p)9 M (p)2 f (p), N2(p)≡ 1

2

K (p)9 M (p). (16.1.3.6)

If we can find a function 9M (p) for which the inverse transforms

L–159M (p)6

= M (x), L–15 2N (p)6

= N (x) (16.1.3.7) exist and can be found in a closed form, then the solution can be written as the convolution

y (x) =

 x

0 N (x – t)F (t) dt, F (t) =

 t

0 M (t – s)f (s) ds. (16.1.3.8)

Example 2 Consider the equation

 x

0 sin k √

x – t

y (t) dt = f (x), f(0) = 0 (16.1.3.9) Applying the Laplace transform, we obtain (see Subsections T3.1.1 and T3.1.6)

2y(p) = √2

π k p

3/2exp(α/p)2 f (p), α= 14k2 (16.1.3.10) Let us rewrite the right-hand side of (16.1.3.10) in the equivalent form

2y(p) = √2

π k p

2 

p–1/2exp(α/p)2f (p), α= 14k2, (16.1.3.11) where the factor in the square brackets corresponds to 9M (p) in formula (16.1.3.6) and 2 N (p) = const p2.

By applying the Laplace inversion formula according to the above scheme to formula (16.1.3.11) with regard to the relation (see Subsections T3.2.1 and T3.2.5)

L – 1 5

p22ϕ(p)6= d

2

dx2ϕ (x), L – 1 5

p–1/2exp(α/p)6

= 1

√

πxcosh k √

x

,

we find the solution

y (x) = 2

πk

d2

dx2

 x

0

cosh k √

x – t

√

x – t f (t) dt.

16.1.3-4 Application of an auxiliary equation

Consider the equation  x

a K (x – t)y(t) dt = f (x), (16.1.3.12)

where the kernel K(x) has an integrable singularity at x =0

Let w = w(x) be the solution of the simpler auxiliary equation with f (x)≡ 1and a =0,

 x

0 K (x – t)w(t) dt =1 (16.1.3.13) Then the solution of the original equation (16.1.3.12) with arbitrary right-hand side can be expressed as follows via the solution of the auxiliary equation (16.1.3.13):

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. (16.1.3.14)

Example 3 Consider the generalized Abel equation

 x

a

y (t) dt (x – t) μ = f (x), 0< μ <1 (16.1.3.15)

We seek a solution of the corresponding auxiliary equation

 x

0

w (t) dt (x – t) μ =1, 0< μ <1, (16.1.3.16)

by the method of indeterminate coefficients in the form

Let us substitute (16.1.3.17) into (16.1.3.16) and then perform the change of variable t = xξ in the integral.

Taking into account the relationship

B (p, q) =

 1

0 ξ

p–1 (1– ξ)1–q dξ= Γ(p)Γ(q)

Γ(p + q)

between the beta and gamma functions, we obtain

A Γ(β +1) Γ( 1– μ)

Γ( 2+ β – μ) x

β+1 –μ= 1.

From this relation we find the coefficients A and β:

β = μ –1, A= 1

Γ(μ)Γ(1– μ) =

sin(πμ)

Formulas (16.1.3.17) and (16.1.3.18) define the solution of the auxiliary equation (16.1.3.16) and make it possible to find the solution of the generalized Abel equation (16.1.3.15) by means of formula (16.1.3.14) as follows:

y (x) = sin(πμ)

π

d dx

 x

a

f (t) dt (x – t)1–μ = sin(πμ)

π



f (a) (x – a)1–μ+

 x

a

f t  (t) dt (x – t)1–μ



.

16.1.4 Reduction of Volterra Equations of the First Kind to Volterra

Equations of the Second Kind

16.1.4-1 First method

Suppose that the kernel and the right-hand side of the equation

 x

a K (x, t)y(t) dt = f (x) (16.1.4.1)

have continuous derivatives with respect to x and that the condition K(x, x)  0 holds

In this case, after differentiating relation (16.1.4.1) and dividing the resulting expression

by K(x, x), we arrive at the following Volterra equation of the second kind:

y (x) +

 x

a

K 

x (x, t)

K (x, x) y (t) dt =

f 

x (x)

K (x, x). (16.1.4.2) Equations of this type are considered in Section 16.2

If K(x, x)≡ 0, then, on differentiating equation (16.1.4.1) with respect to x twice and assuming that K x  (x, t)|t=x 0, we obtain the Volterra equation of the second kind

y (x) +

 x

a

K 

xx (x, t)

K 

x (x, t)|t=x y (t) dt =

f 

xx (x)

K 

x (x, t)|t=x.

If K x  (x, x)≡ 0, we can again apply differentiation, and so on

16.1.4-2 Second method

Let us introduce the new variable

Y (x) =

 x

a y (t) dt

and integrate the right-hand side of equation (16.1.4.1) by parts taking into account the

relation f (a) = 0 After dividing the resulting expression by K(x, x), we arrive at the

Volterra equation of the second kind

Y (x) –

 x

a

K 

t (x, t)

K (x, x) Y (t) dt =

f (x)

K (x, x), for which the condition K(x, x) 0must hold