Integral Transforms Method

0.5. Integral Transforms Method

0.5.1. Main Integral Transforms

Various integral transforms are widely used to solve linear problems of mathematical physics. An integral transform is defined as ã

The function ✃ ( ➨ ) is called the transform of the function ✃ ( ➢ ) and ➳ ( ➢ , ➨ ) is called the kernel of the integral transform. The function ✃ ( ➢ ) is called the inverse transform of ✃ ( ➨ ). The limits of integration ❶ and ➟ are real numbers (usually, ❶ = 0, ➟ = ➯ or ❶ =− ➯ , ➟ = ➯ ).

Corresponding inversion formulas, which have the form ã

make it possible to recover ✃ ( ➢ ) if ✃ ( ➨ ) is given. The integration path ç can lie either on the real axis or in the complex plane.

The most commonly used integral transforms are listed in Table 3 (for the constraints imposed on the functions and parameters occurring in the integrand, see the references given at the end of Section 0.5).

TABLE 3 Main integral transforms

Integral Transform

Definition

Inversion Formula

Fourier sine ë transform

sin( ) è s ( õ ) ðïõ

Fourier cosine

( Ó )= ø transform 2 è ê ë cos( Ó õ ) Ò c ( õ ñ ðïõ )

c ( õ )= ø 2 ê ë cos( Ó õ ) Ò ( Ó ) ðïÓ

Notation ë : = −1 , ( ) and (

) are the Bessel functions of the first and the second kind, respectively; ÿ✄ ( Ó ) and þ ( Ó ) are the modified Bessel functions of the first and the second kind.

The Laplace transform and the Fourier transform are in most common use. These integral transforms are briefly described below.

0.5.2. Laplace Transform and Its Application in Mathematical Physics

0.5.2-1. The Laplace transform. The inverse Laplace transform. The Laplace transform of an arbitrary (complex-valued) function ã ✃ ( ➚ ) of a real variable ➚ ( ➚ ≥ 0) is

defined by

where ☎ = ➼ + ✠☛✡ is a complex variable, ✠ 2 = −1.

The Laplace transform exists for any continuous or piecewise-continuous function satisfying ã

the condition | ✃ ( ➚ )| < ☞

0 with some ☞

> 0 and ✡ 0 ≥ 0. In what follows, ✡ 0 often means the greatest lower bound of the possible values of ✝ ã ✌ ➘ ✡ 0 in this condition. For any ✃ ( ➚ ), the transform ✃ ( ☎ ) is defined in the half-plane Re ☎ > ✡ 0 and is analytic there.

Given the transform ✃ ( ☎ ), the function ✃ ( ➚ ) can be found by means of the inverse Laplace ã

transform

(2) where the integration path is parallel to the imaginary axis and lies to the right of all singularities ➬

of ✃ ( ☎ ), which corresponds to ✑ > ✡ 0 .

The integral in (2) is understood in the sense of the Cauchy principal value: ã ã

In the domain ➬

< 0, formula (2) gives ✃ ( ➚ ) ≡ 0.

TABLE 4 Main properties of the Laplace transform

No Function

Laplace transform ã ã

Differentiation of the transform

Shift in

the complex plane

Formula (2) holds for continuous functions. If ✃ ( ✕ ) has a (finite) jump discontinuity at a point

0 > 0, then the left-hand side of (2) is equal to 2 [ ✃ ( ✕ 0 − 0) + ✃ ( ✕ 0 + 0)] at this point (for ✕ 0 = 0, the first term in the square brackets must be omitted).

We will briefly denote the Laplace transform (1) and the inverse Laplace transform (2) as ã ã

0.5.2-2. Main properties of the Laplace transform. The main properties of the correspondence between functions and their Laplace transforms are

gathered in Table 4 . The Laplace transforms of some functions are listed in Table 5 . There are a number of books that contain detailed tables of direct and inverse Laplace transforms elsewhere; see the references at the end of Section 0.5. Such tables are convenient to use in solving linear differential equations.

Note the important case in which the transform is a rational function of the form ã

where ✩ (

) and ★ ( ☎ ) are polynomials in the variable ☎ and the degree of ✩ ( ☎ ) exceeds that of ( ☎ ). Assume that the zeros of the denominator are simple, i.e., ( ☎ ) ≡ const ( ☎

). Then the inverse transform can be determined by the formula

where the primes denote the derivatives.

TABLE 5 The Laplace transforms of some functions

No ✫ Function, (

Laplace transform, ✭ ( ✮ )

+ ✛✁✛✁✛ ✴ ) = 0.5772 is the Euler constant

2 ( ✽ + ) is the Bessel function ❶ 2

0.5.2-3. Solving linear problems of mathematical physics by the Laplace transform. Figure 2 shows schematically how one can utilize the Laplace transforms to solve boundary value

problems for linear parabolic or hyperbolic equations with two independent variables in the case where the equation coefficients are independent of ✷ .

It is significant that with the Laplace transform, the original problems for a partial differential equation is reduced to a simpler problem for an ordinary differential equation with parameter ✮ ; the derivatives with respect to ✷ are replaced by appropriate algebraic expressions taking into account the initial conditions (see property 5 or 6 in Table 4 ).

Example 1. Consider the following problem for the heat equation:

(initial condition),

= ❀ 0 at

(boundary condition),

0 ❇ at

(boundary condition).

We apply the Laplace transform with respect to ❀

. Setting

= ❈ { } and taking into account the relations

− | =0 ❀ = è (used are property 5 of Table 4 and the initial condition),

(used are property 1 of Table 4 and the relation ❈ {1} = 1 ❉ é ),

Figure 2. Scheme for solving linear boundary value problems by the Laplace transform.

we arrive at the following problem for a second-order linear ordinary differential equation with parameter :

(boundary condition),

0 at ✟ ✠

(boundary condition).

Integrating the equation yields the general solution ✎

. Using the boundary conditions, we

determine the constants, ☛ 1 ()= 0 and ☛ 2 ( ) = 0. Thus, we have

Let us apply the inverse Laplace transform to both sides of this relation. We refer to Table 5 , row 11 with ✏ = ✟ to find the inverse transform of the right-hand side. Finally, we obtain the solution of the original problem in the form

0 ✟ erfc

0.5.3. Fourier Transform and Its Application in Mathematical Physics

0.5.3-1. The Fourier transform and its properties. The Fourier transform is defined as follows:

This relation is meaningful for any function ✖

( ✜ ) absolutely integrable on the interval ( ✩ ,+ ✖ ✩ ✛ ). We will briefly write ✖ ✕ ( ✗ )= ✪ { ✜ ✖ ( )} to denote the Fourier transform (3). Given ( ✗ ), the function ( ✜ ) can be found by means of the inverse Fourier transform

TABLE 6 Main properties of the Fourier transform

No Function

Fourier transform

of the transform

where the integral is understood in the sense of the Cauchy principal value. We will briefly write ✛

−1 ✖ ( ✕ ✜ )= ✪ { ( ✗ )} to denote the inverse Fourier transform (4). The inversion formula (4) holds for continuous functions. If ✖

( ✜ ) has a (finite) jump discontinuity ✖

at a point ✜ = ✜ 0 , then the left-hand side of (4) is equal to 1 2 ✷ ( ✜ 0 − 0) + ( ✜ 0 + 0) ✸ at this point. The main properties of the correspondence between functions and their Fourier transforms are gathered in Table 6 .

0.5.3-2. Solving linear problems of mathematical physics by the Fourier transform. The Fourier transform is usually employed to solve boundary value problems for linear partial

differential equations whose coefficients are independent of the space variable ✜ ,− ✩ < ✜ < ✩ . The scheme for solving linear boundary value problems with the help of the Fourier transform is similar to that used in solving problems with help of the Laplace transform. With the Fourier transform, the derivatives with respect to ✜ in the equation are replaced by appropriate algebraic expressions; see property 4 or 5 in Table 6 . In the case of two independent variables, the problem for

a partial differential equation is reduced to a simpler problem for an ordinary differential equation with parameter ✗ . On solving the latter problem, one determines the transform. After that, by applying the inverse Fourier transform, one obtains the solution of the original boundary value problem.

Example 2. Consider the following Cauchy problem for the heat equation:

= ✻ ( ✟ ) at ✓ =0 (initial condition).

We apply the Fourier transform with respect to the space variable ✁ ✟ . Setting ✂ = ✼ { ✂ } and taking into account the relation

{ ✹ ✆✝✆ ✂ }=− ✽ 2 ✂ ✁ (see property 4 of Table 6 ), we arrive at the following problem for a first-order linear ordinary differential equation with parameter ✽ :

( ✁ ✽ ) at ✓ = 0, where ✁

) is defined by (3). On solving this problem for the transform ✂ , we find

Let us apply the inversion formula to both sides of this equation. After some calculations, we obtain the solution of the original problem in the form

( ) exp − − ❁

At the last stage we used the relation

exp ❉ − ✏ 2 ✽ 2 + ❊✵✽ ❋ ✽

exp

The Fourier transform admits ❁ ✯ -dimensional generalization:

(u) =

2 (x) ✢ − ✣ (u⋅x) ✧ x, (u ⋅ x) = ✗ 1 ✜ 1 ❏✱❏✱❏ + ✗ ✙ ✜ + , (5) (2 ) ✮ ● ✚ ❍ ■

where ✖ (x) = (

), (u) = ( ✗ 1 , ✰✱✰✱✰ , ✗ ✮ ), and x= ✜ 1 ✰✱✰✱✰ ✜ ✮ .

The corresponding inversion formula is

. The Fourier transform (5) is frequently used in the theory of linear partial differential equations

(x) = ✧ (u⋅x)

with constant coefficients (x ❑ ▲ ▼❖◆ ✮ ).

References for Section 0.5: H. Bateman and A. Erd´elyi (1954), V. A. Ditkin and A. P. Prudnikov (1965), J. W. Miles (1971), B. Davis (1978), Yu. A. Brychkov and A. P. Prudnikov (1989), L. H ¨ormander (1990), J. R. Hanna and J. H. Rowland (1990), W. H. Beyer (1991), D. Zwillinger (1998), A. D. Polyanin and A. V. Manzhirov (1998).