Handbook of mathematics for engineers and scienteists part 122 ppt

Consider a linear equation, which we shall write in the following brief form: L [y] = fgx, λ, 16.2.4.1 where L is a linear operator integral, differential, etc.. Suppose that the solutio

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

Aw (x) + B

 x

0 K (x – t)w(t) dt =1 (16.2.3.9)

In this case, the solution of the original equation (16.2.3.8) with an arbitrary right-hand side can be expressed via the solution of the auxiliary equation (16.2.3.9) by the formula

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.

16.2.4 Construction of Solutions of Integral Equations with Special

Right-Hand Side

In this section we describe some approaches to the construction of solutions of integral equations with special right-hand sides These approaches are based on the application of auxiliary solutions that depend on a free parameter

16.2.4-1 General scheme

Consider a linear equation, which we shall write in the following brief form:

L [y] = fg(x, λ), (16.2.4.1)

where L is a linear operator (integral, differential, etc.) that acts with respect to the variable x

and is independent of the parameter λ, and fg(x, λ) is a given function that depends on the variable x and the parameter λ.

Suppose that the solution of equation (16.2.4.1) is known:

Let M be a linear operator (integral, differential, etc.) that acts with respect to the

parameter λ and is independent of the variable x Consider the (usual) case in which M

commutes with L We apply the operator M to equation (16.2.4.1) and find that the equation

L [w] = f M (x), f M (x) = M

fg(x, λ)

(16.2.4.3) has the solution

w= M

y (x, λ)

By choosing the operator M in a different way, we can obtain solutions for other

right-hand sides of equation (16.2.4.1) The original function fg(x, λ) is called the generating

function for the operator L.

16.2.4-2 Generating function of exponential form

Consider a linear equation with exponential right-hand side

L [y] = e λx (16.2.4.5) Suppose that the solution is known and is given by formula (16.2.4.2) In Table 16.1 we

present solutions of the equation L [y] = f (x) with various right-hand sides; these solutions

are expressed via the solution of equation (16.2.4.5)

Remark 1 When applying the formulas indicated in the table, we need not know the left-hand side of the linear equation (16.2.4.5) (the equation can be integral, differential, etc.) provided that a particular solution

of this equation for the exponential right-hand side is known It is only of importance that the left-hand side of

the equation is independent of the parameter λ.

Remark 2 When applying formulas indicated in the table, the convergence of the integrals occurring in the resulting solution must be verified.

Example 1 We seek a solution of the equation with exponential right-hand side

y (x) +

 ∞

x

K (x – t)y(t) dt = e λx (16 2 4 6 )

in the form y(x, λ) = ke λxby the method of indeterminate coefficients Then we obtain

y (x, λ) = 1

B (λ) e

λx, B (λ) =1 +

 ∞

0 K (–z)e λz dz (16 2 4 7 )

It follows from row 3 of Table 16.1 that the solution of the equation

y (x) +

 ∞

x

K (x – t)y(t) dt = Ax (16 2 4 8 ) has the form

y (x) = A

D x– AC

D2, where D= 1 +

 ∞

0 K (–z) dz, C=

 ∞

0 zK (–z) dz.

For such a solution to exist, it is necessary that the improper integrals of the functions K(–z) and zK(–z) exist This holds if the function K(–z) decreases more rapidly than z– 2as z → ∞ Otherwise a solution

can be nonexistent It is of interest that for functions K(–z) with power-law growth as z → ∞ in the case

λ< 0 , the solution of equation (16.2.4.6) exists and is given by formula (16.2.4.7), whereas equation (16.2.4.8) does not have a solution Therefore, we must be careful when using formulas from Table 16.1 and verify the convergence of the integrals occurring in the solution.

It follows from row 15 of Table 16.1 that the solution of the equation

y (x) +

 ∞

x

K (x – t)y(t) dt = A sin(λx) (16 2 4 9 )

is given by the formula

y (x) = A

B2c+ Bs2



Bcsin(λx) – Bscos(λx)

, where

Bc = 1 +

 ∞ 0

K (–z) cos(λz) dz, Bs =

 ∞ 0

K (–z) sin(λz) dz.

16.2.4-3 Power-law generating function

Consider the linear equation with power-law right-hand side

L [y] = x λ (16.2.4.10)

Suppose that the solution is known and is given by formula (16.2.4.2) In Table 16.2,

solutions of the equation L [y] = f (x) with various right-hand sides are presented, which

can be expressed via the solution of equation (16.2.4.10)

TABLE 16.1

Solutions of the equation L [y] = f (x) with generating function of the exponential form

2 A1e λ1x+· · · + A n e λ n x A1y (x, λ1) +· · · + A n (x, λ n) Follows from linearity

∂λ

*

y (x, λ)

+

λ=0+ By(x,0 ) Follows from linearity

and the results of row 4

4 n=0Ax,1,n,2, A



∂

∂λ n

*

y (x, λ)+

λ=0

Follows from the results

of row 6 for λ =0

x + a , a >0 A

 ∞

0 e

–aλ y (x, –λ) dλ Integration with respectto the parameter λ

6 Ax n e λx,

n= 0 , 1 , 2, A

∂

∂λ n

*

y (x, λ)

+ Differentiation with respect

to the parameter λ

8 A cosh(λx) 12A [y(x, λ) + y(x, –λ) Linearity and relations

to the exponential

9 A sinh(λx) 12A [y(x, λ) – y(x, –λ) Linearity and relations

to the exponential

10 Ax m cosh(λx),

m= 1 , 3 , 5, 12A

∂ m

∂λ m [y(x, λ) – y(x, –λ) Differentiation with respect

to λ and relation

to the exponential

11 m Ax=m2cosh(λx),,4,6, 1

2A

∂ m

∂λ m [y(x, λ) + y(x, –λ) Differentiation with respect

to λ and relation

to the exponential

12 m Ax=m1sinh(λx),,3,5, 1

2A

∂ m

∂λ m [y(x, λ) + y(x, –λ) Differentiation with respect

to λ and relation

to the exponential

13 m Ax=m2sinh(λx),,4,6, 1

2A

∂ m

∂λ m [y(x, λ) – y(x, –λ) Differentiation with respect

to λ and relation

to the exponential

y (x, iβ) Selection of the real

part for λ = iβ

y (x, iβ) Selection of the imaginary

part for λ = iβ

16 n Ax=n1,cos(βx),2,3, ARe



∂

∂λ n

*

y (x, λ)

+

λ=iβ

Differentiation with respect

to λ and selection of the real part for λ = iβ

17 n Ax=n1,sin(βx),2,3, AIm



∂

∂λ n

*

y (x, λ)

+

λ=iβ

Differentiation with respect

to λ and selection of the imaginary part for λ = iβ

y (x, μ + iβ) Selection of the real

part for λ = μ + iβ

y (x, μ + iβ) Selection of the imaginary

part for λ = μ + iβ

20 Ax n=n e1μx,2cos(βx),,3, ARe



∂

∂λ n

*

y (x, λ)

+

λ=μ+iβ

Differentiation with respect

to λ and selection of the real part for λ = μ + iβ

21 Ax n=n e1μx,2sin(βx),,3, AIm



∂

∂λ n

*

y (x, λ)

+

λ=μ+iβ

Differentiation with respect

to λ and selection of the imaginary part for λ = μ + iβ

TABLE 16.2

Solutions of the equation L [y] = f (x) with generating function of power-law form

k=0A k x

k=0A k y (x, k) Follows from linearity

∂λ

*

y (x, λ)

+

λ=0+ By(x,0 ) Follows from linearity and

from the results of row 4

4 n=A0,ln1n,x2, , A



∂

∂λ n

*

y (x, λ)+

λ=0

Follows from the results

of row 5 for λ =0

5 Ax λlnn x,

n= 0 , 1 , 2, A

∂

∂λ n

*

y (x, λ)

with respect to the parameter λ

y (x, iβ) Selection of the real

part for λ = iβ

y (x, iβ) Selection of the imaginary

part for λ = iβ

y (x, μ + iβ) Selection of the real

part for λ = μ + iβ

y (x, μ + iβ) Selection of the imaginary

part for λ = μ + iβ

Example 2 We seek a solution of the equation with power-law right-hand side

y (x) +

 x 0

1

x K

t

x



y (t) dt = x λ

in the form y(x, λ) = kx λby the method of indeterminate coefficients We finally obtain

y (x, λ) = 1

1+ B(λ) x

λ, B (λ) =

 1 0

K (t)t λ dt.

It follows from row 3 of Table 16.2 that the solution of the equation with logarithmic right-hand side

y (x) +

 x 0

1

x K

t

x



y (t) dt = A ln x

has the form

y (x) = A

1+ I0 ln x –

AI1

( 1+ I0)2, where I0=

 1 0

K (t) dt, I1=

 1 0

K (t) ln t dt.

Remark The cases where the generating function is defined sine or cosine are treated likewise.

16.2.5 Method of Model Solutions

16.2.5-1 Preliminary remarks.∗

Consider a linear equation that we briefly write out in the form

L [y(x)] = f (x), (16.2.5.1)

where L is a linear (integral) operator, y(x) is an unknown function, and f (x) is a known

function

* Before reading this section, it is useful to look over Subsection 16.2.4.

We first define arbitrarily a test solution

which depends on an auxiliary parameter λ (it is assumed that the operator L is independent

of λ and y0 const) By means of equation (16.2.5.1) we define the right-hand side that corresponds to the test solution (16.2.5.2):

f0(x, λ) = L [y0(x, λ)].

Let us multiply equation (16.2.5.1), for y = y0 and f = f0, by some function ϕ(λ) and integrate the resulting relation with respect to λ over an interval [a, b] We finally obtain

where

y ϕ (x) =

 b

a y0(x, λ)ϕ(λ) dλ, f ϕ (x) =

 b

a f0(x, λ)ϕ(λ) dλ. (16.2.5.4)

It follows from formulas (16.2.5.3) and (16.2.5.4) that, for the right-hand side f = f ϕ (x), the function y = y ϕ (x) is a solution of the original equation (16.2.5.1) Since the choice

of the function ϕ(λ) (as well as of the integration interval) is arbitrary, the function f ϕ (x) can be arbitrary in principle Here the main problem is how to choose a function ϕ(λ) to obtain a given function f ϕ (x) This problem can be solved if we can find a test solution

such that the right-hand side of equation (16.2.5.1) is the kernel of a known inverse integral

transform (we denote such a test solution by Y (x, λ) and call it a model solution).

16.2.5-2 Description of the method

Indeed, let P be an invertible integral transform that takes each function f(x) to the

corresponding transform F (λ) by the rule

Assume that the inverse transformP– 1has the kernel ψ(x, λ) and acts as follows:

P– 1{F (λ)}= f (x), P– 1{F (λ)} ≡ b

a F (λ)ψ(x, λ) dλ. (16.2.5.6)

The limits of integration a and b and the integration path in (16.2.5.6) may well lie in the

complex plane

Suppose that we succeeded in finding a model solution Y (x, λ) of the auxiliary problem

for equation (16.2.5.1) whose right-hand side is the kernel of the inverse transformP– 1:

L [Y (x, λ)] = ψ(x, λ). (16.2.5.7)

Let us multiply equation (16.2.5.7) by F (λ) and integrate with respect to λ within the same

limits that stand in the inverse transform (16.2.5.6) Taking into account the fact that the

operator L is independent of λ and applying the relationP– 1{F (λ)}= f (x), we obtain

L* b

a Y (x, λ)F (λ) dλ

+

= f (x).

Therefore, the solution of equation (16.2.5.1) for an arbitrary function f (x) on the

right-hand side is expressed via a solution of the simpler auxiliary equation (16.2.5.7) by the formula

y (x) =

 b

where F (λ) is the transform (16.2.5.5) of the function f (x).

For the right-hand side of the auxiliary equation (16.2.5.7) we can take, for instance, exponential, power-law, and trigonometric functions, which are the kernels of the Laplace, Mellin, and sine and cosine Fourier transforms (up to a constant factor) Sometimes it is rather easy to find a model solution by means of the method of indeterminate coefficients (by prescribing its structure) Afterward, to construct a solution of the equation with an arbitrary right-hand side, we can apply formulas written out below in Paragraphs 16.2.5-3–16.2.5-6

16.2.5-3 Model solution in the case of an exponential right-hand side

Assume that we have found a model solution Y = Y (x, λ) that corresponds to the exponential

right-hand side:

L [Y (x, λ)] = e λx (16.2.5.9) Consider two cases:

1◦ Equations on the semiaxis, 0 ≤ x < ∞ Let 2f(p) be the Laplace transform of the

function f (x):

2f(p) = L{f (x)}, L{f (x)} ≡ ∞

0 f (x)e

–pxdx. (16.2.5.10)

The solution of equation (16.2.5.1) for an arbitrary right-hand side f (x) can be expressed

via the solution of the simpler auxiliary equation with exponential right-hand side (16.2.5.9)

for λ = p by the formula

y (x) = 1

2πi

 c+i∞

c–i∞ Y (x, p)2 f (p) dp. (16.2.5.11)

2◦ Equations on the entire axis, – ∞ < x < ∞ Let 2f(u) the Fourier transform of the

function f (x):

2f(u) = F{f (x)}, F{f (x)} ≡ √1

2π

 ∞

–∞ f (x)e

–iux dx. (16.2.5.12)

The solution of equation (16.2.5.1) for an arbitrary right-hand side f (x) can be expressed

via the solution of the simpler auxiliary equation with exponential right-hand side (16.2.5.9)

for λ = iu by the formula

y (x) = √1

2π

 ∞

–∞ Y (x, iu)2 f (u) du. (16.2.5.13)

In the calculation of the integrals on the right-hand sides in (16.2.5.11) and (16.2.5.13), 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)

Remark 1. The structure of a model solution Y (x, λ) can differ from that of the kernel of the Laplace or

Fourier inversion formula.

Remark 2 When applying the method under consideration, the left-hand side of equation (16.2.5.1) need not be known (the equation can be integral, differential, functional, etc.) if a particular solution of this equation

is known for the exponential right-hand side Here only the most general information is important, namely, that

the equation is linear, and its left-hand side is independent of the parameter λ.

Example 1 Consider the following Volterra equation of the second kind with difference kernel:

y (x) +

 ∞

x

K (x – t)y(t) dt = f (x). (16 2 5 14 )

This equation cannot be solved by direct application of the Laplace transform because the convolution theorem cannot be used here.

In accordance with the method of model solutions, we consider the auxiliary equation with exponential right-hand side

y (x) +

 ∞

x

K (x – t)y(t) dt = e px (16 2 5 15 ) Its solution has the form (see Example 1 of Section 16.2.4)

Y (x, p) = 1

1 + 2K (–p) e

px, K2(–p) =

 ∞ 0

K (–z)e pz dz (16 2 5 16 ) This, by means of formula (16.2.5.11), yields a solution of equation (16.2.5.14) for an arbitrary right-hand side,

y (x) = 1 2πi

 c+i∞

c–i∞

2f(p)

1 + 2K (–p) e

px dp, (16 2 5 17 ) where 2f (p) is the Laplace transform (16.2.5.10) of the function f (x).

16.2.5-4 Model solution in the case of a power-law right-hand side

Suppose that we have succeeded in finding a model solution Y = Y (x, s) that corresponds

to a power-law right-hand side of the equation:

L [Y (x, s)] = x–s, λ = –s. (16.2.5.18) Let ˆf (s) be the Mellin transform of the function f (x):

ˆ

f (s) =M{f (x)}, M{f (x)} ≡ ∞

0 f (x)x

s–1dx. (16.2.5.19)

The solution of equation (16.2.5.1) for an arbitrary right-hand side f (x) can be expressed

via the solution of the simpler auxiliary equation with power-law right-hand side (16.2.5.18)

by the formula

y (x) = 1

2πi

 c+i∞

c–i∞ Y (x, s) ˆ f (s) ds. (16.2.5.20)

In the calculation of the corresponding integrals on the right-hand side of formula (16.2.5.20), one can use tables of inverse Mellin transforms (e.g., see Section T3.6, as well

as methods of the theory of functions of a complex variable, including formulas for the calculation of residues and the Jordan lemma (see Subsection 11.1.2)