INTRODUCTION

Up to now we have studied and used only one integral transform: the Laplace transform. But in Section 15.3 we saw that the Fourier integral had three alternative forms: the cosine integral, the sine integral, and the complex or exponential form. In the present section we shall take these three forms of the Fourier integral and develop them into three new integral transforms naturally called Fourier transforms. In addition, we shall expand on the concept of a transform pair, that is, an integral transform and its inverse. We shall also see that the inverse of an integral transform is itself another integral transform.

Transform Pairs

The Laplace transform F(s) of a function f(t) is defined by an integral, but up to now we have been using the symbolic representation f(t) = ⁻¹{F(s)} to denote the inverse Laplace transform of F(s). Actually, the inverse Laplace transform is also an integral transform. If

(1)

then the inverse Laplace transform is

(2)

The last integral is called a contour integral; its evaluation requires the use of complex variables and is beyond the scope of this discussion. The point here is this: Integral transforms appear in transform pairs. If f(x) is transformed into F(α) by an integral transform

(3)

then the function f can be recovered by another integral transform

(4)

called the inverse transform. The functions K and H in the integrands of (3) and (4) are called the kernels of their respective transforms. We identify K(s, t) = e^−st as the kernel of the Laplace transform and H(s, t) = e^st/2πi as the kernel of the inverse Laplace transform.

Fourier Transform Pairs

The Fourier integral is the source of three integral transforms. From (19) and (18), (9) and (8), and (11) and (10) of Section 15.3, we are prompted to define the following Fourier transform pairs.

Existence

The conditions under which (5), (7), and (9) exist are more stringent than those for the Laplace transform. For example, do not exist. Sufficient conditions for existence are that f be absolutely integrable on the appropriate interval and that be piecewise continuous on every finite interval.

Linearity of the Fourier Transforms

Like the Laplace transform, each of the six transforms given in Definition 15.4.1 are linear transforms. For example, in the case of the Fourier transform (5), under conditions such that the Fourier transform of each of the functions f and g exists, then for constants

Operational Properties

Since our immediate goal is to apply these new transforms to boundary-value problems, we need to examine the transforms of derivatives.

Fourier Transform

Suppose that f is continuous and absolutely integrable on the interval (−, ) and f′ is piecewise continuous on every finite interval. If f(x) → 0 as x → ±, then integration by parts gives

that is, (11)

Similarly, under the added assumptions that f′ is continuous on (−, ), f″ (x) is piecewise continuous on every finite interval, and f′ (x) → 0 as x → ±, we have

(12)

In general, under conditions analogous to those leading to (12), we have

where n = 0, 1, 2, … .

It is important to be aware that the cosine and sine transforms are not suitable for transforming the first derivative of a function (or for that matter, any derivative of odd order). For example, it is readily shown that

The difficulty is apparent: The transform of is not expressed in terms of the original integral transform.

Fourier Cosine Transform

Suppose that f and are continuous, f is absolutely integrable on the interval and is piecewise continuous on every finite interval. If and as then the Fourier cosine transform of is

that is, (13)

Fourier Sine Transform

Under the same assumptions that lead to (13), we find the Fourier sine transform of to be

(14)

Choosing the Transform

The transform properties (12), (13), and (14) indicate, in contrast to the Laplace transform, that Fourier transforms are suitable for problems in which the spatial variable x (or y) is defined on an infinite or semi-infinite interval. But a natural question then arises:

Clearly, to use the Fourier transform (5), the domain of the variable to be eliminated must be (−, ). To utilize a cosine or a sine transform, the domain of one of the spatial variables in the problem must be However, the determining factor in choosing between the cosine transform (7) and the sine transform (9) is the type of boundary condition specified at that is, whether a first partial derivative or u is given at this boundary.

In solving boundary-value problems using integral transforms most solutions are formal. In the language of mathematics, this means assumptions about the solution u and its partial derivatives go unstated. But one assumption should be kept in the back of your mind. In the examples that follow, it will be assumed without further mention that u and (or ) approach 0 as x → ± (or y → ±). These are not major restrictions since these conditions hold in most applications.

EXAMPLE 1 Using the Fourier Transform

Solve the heat equation , − < x < , t > 0, subject to

SOLUTION

The problem can be interpreted as finding the temperature u(x, t) in an infinite rod. Because the domain of x is the infinite interval (−,) we use the Fourier transform (5) and define the transform of u(x, t) to be

If we write

then the Fourier transform of the partial differential equation,

becomes the ordinary differential equation

Solving the last equation by the method of Section 2.3 gives . The initial temperature u(x, 0) = f(x) in the rod is shown in FIGURE 15.4.1 and its Fourier transform is

By Euler’s formula

Subtracting these two results and solving for sin α gives . Hence we can rewrite the transform of the initial condition as . Applying this condition to the solution gives and so

It then follows from the inverse Fourier transform (6) that

This integral can be simplified somewhat by using Euler’s formula again as e^−iαx = cos αx − i sin αx and noting that

because the integrand is an odd function of α. Hence we finally have the solution

(15) ≡

It is left to the reader to show that the solution (15) in Example 1 can be expressed in terms of the error function. See Problem 26 in Exercises 15.4.

EXAMPLE 2 Two Useful Fourier Transforms

EXAMPLE 3 Using the Cosine Transform

The steady-state temperature in the semi-infinite plate shown in FIGURE 15.4.2 is determined from

Solve for u(x, y).

SOLUTION

The domain of the variable y and the prescribed condition at y = 0 indicate that the Fourier cosine transform is suitable for the problem. We define

In view of (13),

becomes

Since the domain of x is a finite interval, we choose to write the solution of the ordinary differential equation as

U(x, α) = c₁ cosh αx + c₂ sinh αx. (18)

Now are in turn equivalent to

(19)

Note that the value U(π, α) in (19) is (16) of Example 2 with b = 1. When we apply the two conditions in (19) to the solution (18) we obtain c₁ = 0 and c₂ = 1/[(1 + α²) sinh απ]. Therefore,

and so from (8) we arrive at

. (20) ≡

Had u(x, 0) been given in Example 3 rather than u_y(x, 0), then the sine transform would have been appropriate.

15.4 Exercises Answers to selected odd-numbered problems begin on page ANS-39.

In Problems 1–24, use a Fourier transform in this section to solve the given boundary-value problem. Make assumptions about boundedness where necessary.

1. 
2. 
   
3. Find the temperature u(x, t) in a semi-infinite rod if u(0, t) = u₀, t > 0 and u(x, 0) = 0, x > 0.
4. Use the result to show that the solution in Problem 3 can be written as

   

5. Find the temperature u(x, t) in a semi-infinite rod if u(0, t) = 0, t > 0, and

   

6. Solve Problem 3 if the condition at the left boundary is

   

7. Solve Problem 5 if the end x = 0 is insulated.
8. Find the temperature u(x, t) in a semi-infinite rod if u(0, t) = 1, t > 0, and u(x, 0) = e^−x x > 0.
9. 1. 
   2. If g(x) = 0, show that the solution of part (a) can be written as u(x, t) = [f(x + at) + f(x − at)].
10. Find the displacement u(x, t) of a semi-infinite string if

    

11. Solve the problem in Example 3 if the boundary conditions at x = 0 and x = π are reversed:

    u(0, y) = e^−y, u(π, y) = 0, y > 0.

12. Solve the problem in Example 3 if the boundary condition at y = 0 is u(x, 0) = 1, 0 < x < π.
13. Find the steady-state temperature u(x, y) in a plate defined by x ≥ 0, y ≥ 0 if the boundary x = 0 is insulated and, at y = 0,

    

14. Solve Problem 13 if the boundary condition at x = 0 is u(0, y) = 0, y > 0.

15. 

16. 

In Problems 17 and 18, find the steady-state temperature u(x, y) in the plate given in the figure. [Hint: One way of proceeding is to express Problems 17 and 18 as two and three boundary-value problems, respectively. Use the superposition principle (see Section 13.5).]

17. 

18. 

19. Use the result to solve the boundary-value problem

    

20. If {f(x)} = F(α) and {g(x)} = G(α), then the convolution theorem for the Fourier transform is given by

    

    Use this result and the transform given in Problem 19 to show that a solution of the boundary-value problem

    

    is

21. Use the transform given in Problem 19 to find the steady-state temperature u(x, y) in the infinite strip shown in FIGURE 15.4.5.

    

22. Find the steady-state temperature in a semi-infinite cylinder described by the boundary-value problem

    

    [Hint: Use the integral in Problem 4 and the parametric form of the modified Bessel equation on page 293.]

23. Find the steady-state temperature in the semi-infinite cylinder in Problem 22 if the base of the cylinder is insulated and

    

24. Find the steady-state temperature in the infinite plate defined by if the boundary condition at is

    

    [Hint: Consider the two cases when you solve the resulting ordinary differential equation.]

25. The solution of Problem 14 can be integrated. Use entries 50 and 51 of the table in Appendix C to show that

    

26. Use Problem 20, the change of variables v = (x − τ)/2, and Problem 11 in Exercises 15.1 to show that the solution of Example 1 can be expressed as

    

Computer Lab Assignment

27. Assume u₀ = 100 and k = 1 in the solution of Problem 26. Use a CAS to graph u(x, t) over the rectangular region −4 ≤ x ≤ 4, 0 ≤ t ≤ 6. Use a 2D plot to superimpose the graphs of u(x, t) for t = 0.05, 0.125, 0.5, 1, 2, 4, 6, and 15 for −4 ≤ x ≤ 4. Use the graphs to conjecture the values of lim u(x, t) and lim u(x, t). Then prove these results analytically using the properties of erf (x).

Discussion Problems

28. 1. Suppose

       

       where

       

       Find f(x).

    2. Use part (a) to show that