7. Fourier Integral Transforms – Differential Equations

7

Fourier Integral Transforms

7.1 INTRODUCTION

Integral transforms are useful in solving initial and boundary value problems and in evaluating certain integrals. Laplace and Fourier transforms are two important transforms which are widely used in engineering and physical applications. They are used in the solution of conduction of heat, vibration of strings, oscillations of elastic beams, transmission lines, etc.

Here we define Fourier transform together with Fourier sine and cosine transforms, their inverses and study their properties and consider evaluation of certain integrals.

7.2 INTEGRAL TRANSFORMS

A linear integral transform or simply an integral transform of a function f(x) is defined by

 

 

where K(s, x), called the Kernel of the transformation, is a known function. For different choices of the Kernel K(s, x) and the limits a and b, we obtain different transforms.

7.2.1 Laplace Transform

When K(s, x) = esx (s > 0), a = 0 and b = ∞, we obtain the Laplace transform of f(x). It is defined by

 

7.2.2 Fourier Transform

When K(s, x) = eisx, a = −∞ and b = ∞, we obtain the Fourier transform of f(x). It is defined by

 

7.3 FOURIER INTEGRAL THEOREM

Let f(x) be a function satisfying the following Dirichlet conditions in every interval (−l, l) (l > 0) however large:

  1. f(x) is periodic; f(x) and its integrals are single-valued and finite (bounded).
  2. f(x) has at most a finite number of discontinuities in any one period.
  3. f(x) has at most a finite number of maxima or minima.

Then the Fourier series expansion of f(x) in (−l, l) is given by

 

 

where

 

 

Substituting for a0, an, bn from (7.5), Eq. (7.4) becomes

 

 

If we further assume that f(x) is absolutely integrable for all x

 

i.e.,

 

is convergent, then

 

 

Now (7.6) becomes

 

 

Putting or we have, on taking limit as n → ∞ or δs → 0

 

 

which is called the Fourier Integral (FI) of f(x).

Note 7.3.1   A rigorous proof of the above theorem is beyond the scope of this book.

Note 7.3.2   Fourier integral representation for a function f(x) is useful in solving certain differential and integral equations.

7.3.1 Fourier Sine and Cosine Integrals (FSI's and FCI's)

Writing the expansion for cosine in (7.8) we have

 

 

When f(t) is an odd function in (−∞, ∞) then the integral in the first brackets in (7.9) is zero since the integrand cos st f (t) is an odd function in (−∞, ∞) and so we obtain

 

 

which is called the Fourier Sine Integral (FSI) of f(x).

When f(t) is an even function in (−∞, ∞) then the integral in the second brackets in (7.9) is zero since the integrand sin st f(t) is an odd function in (−∞, ∞) and so we obtain

 

 

which is called the Fourier Cosine Integral (FCI) of f(x).

7.4 FOURIER INTEGRAL IN COMPLEX FORM

cos s(tx) is an even function of s in (−∞, ∞) and so we can write (7.8) in the form

 

 

Also, sin s(tx) is an odd function of s in (−∞, ∞). So,we can write

 

 

Now, multiplying (7.13) by ‘i’ and adding it to (7.12) we obtain

 

 

Using Euler's formula e = cos θ + i sin α we can write the above integral as

 

 

which is called Fourier Integral of f(x) in complex form (FICF)

7.4.1 Fourier Integral Representation of a Function

By (7.9) a function f(x) may be represented by a FI as

 

 

where   

 

and   

 

If f(x) is an odd function in (−∞, ∞) then

 

 

where

 

 

and if f(x) is an even function in (−∞, ∞) then

 

 

where

 

7.5 FOURIER TRANSFORM OF f(x)

Writing the exponential function eix(tx) as a product of two exponential functions eix(tx) = eist · eisx the FI (7.14) may be put in the following form

 

 

The expression inside the brackets is a function of s. Denoting it by f(s) and replacing t by x, we have the Fourier transform (FT) of f(x). Then the inverse Fourier transform (IFT) of F(s) is given by (7.22).

Fourier Transform

The Fourier transformation of f(x), denoted by F{f(x)}, is defined by

 

Inverse Fourier Transform

The inverse FT of F(s), denoted by F−1 {F(s)} is defined by

 

 

Thus, the function F(s) defined by (7.23) is called the Fourier Transform of f(x) and is denoted by

 

 

and the function f(x) given by (7.24) is called the Inverse Fourier transform of F(s) and is denoted by

 

 

The process of obtaining the FT F{f(x)} = F(s) from a given function f(x) is called the FT method or simply Fourier Transform.

Existence of Fourier Transform

The following conditions are sufficient for the existence of the FT of a function f(x):

  1. f(x) is piecewise continuous on every finite interval.
  2. f(x) is absolutely integrable for all x.

7.5.1 Fourier Sine Transform (FST) and Fourier Cosine Transform (FCT)

The FSI in (7.10) can be written by replacing t by x as

 

 

The expression in the brackets in (7.27) is a function of s denoted by Fs(s). Writing

 

 

Equation (7.20) becomes

 

 

The function defined by (7.28) is called the Fourier sine transform (FST) of f(x) and that defined by (7.29) is called the inverse Fourier sine transform of Fs(s).

The Fourier cosine integral (7.11) can be written by replacing t by x as

 

 

The expression in the brackets (7.30) is a function of s denoted by Fc(s). Writing

 

 

 

The functions Fc(s) and Fc−1 {Fc(s)} = f(x) are, respectively, called the Fourier cosine transform and inverse Fourier cosine transform, respectively.

7.6 FINITE FOURIER SINE TRANSFORM AND FINITE FOURIER COSINE TRANSFORM (FFCT)

The finite Fourier sine transform of f(x) in 0 < x < l is defined by

 

 

where n is an integer.

The function f(x) is then called the inverse finite Fourier sine transform of Fs(n) and is given by

 

 

Similarly the finite Fourier cosine transform of f(x) in 0 < x < l is defined by

 

 

where n is an integer.

The function f(x) is then called the inverse finite Fourier cosine transform of Fc(n) and is given by

 

Note 7.6.1   The finite Fourier sine transform is useful for solving problems with boundary condition of heat distribution (say) on two parallel boundaries while the finite Fourier cosine transform is useful for solving problems in which the velocity distribution (say) normal to two parallel boundaries are prescribed.

7.6.1 FT, FST and FCT Alternative definitions

The Fourier transform, FST and FCT are alternatively defined as follows:

 

Definition 7.6.2

 

 

 

 

Definition 7.6.3

 

 

 

Note 7.6.4   Each of these notations has its advantages and disadvantages. So, one can adopt any one of these definitions. If both transformation and inverse transformation are involved completing one cycle there will not be any difference in the results. If only one of these is involved the results will be different. The student, while answering a question, should understand which notation is used and then answer, clearly stating the notation he or she is using.

7.7 CONVOLUTION THEOREM FOR FOURIER TRANSFORMS

7.7.1 Convolution

 

Definition 7.6.5   The convolution of two functions f(x) and g(x) over the interval (−∞, ∞) denoted by f * g is defined by

 

7.7.2 Convolution Theorem

Theorem 7.6.6   If F(s) = F{f(x)} and G(s) = F{g(x)} are the Fourier transforms of f(x) and g(x) then the Fourier transform of the convolution of f(x) and g(x) is the product of their transforms

 

Proof   By the definition of Fourier transforms, we have

 

7.7.3 Relation between Laplace and Fourier Transforms

 

Proof

 

 

Thus, the Fourier transform of a function f(t) defined by (7.45) is the Laplace transformation of g(t).

7.8 PROPERTIES OF FOURIER TRANSFORM

7.8.1 Linearity Property

Theorem 7.8.1   If F(s) and G(s) are the Fourier transforms of f(x) and g(x), respectively, then

 

 

where a and b are any constants.

Proof   By definition we have

 

7.8.2 Change of Scale Property or Damping Rule

Theorem 7.8.2   If F(s) is the complex Fourier transform of f(x) then

 

Proof   Since F(s) is the complex Fourier transform of f(x) we have

 

 

So, the complex Fourier transform of f(ax) is given by

 

Corollary 7.8.3   If Fs(s) and Fc(s) are the Fourier sine transform and Fourier cosine transform of f(x), respectively, then Fs(f(ax)) = (1/a)Fs(s/a) and Fc(f(ax)) = (1/a)Fc(s/a).

7.8.3 Shifting Property

Theorem 7.8.4   If F(s) is the complex Fourier transform of f(x) then

 

Proof   It is given that

By definition

 

 

where u = xa and du = dx.

7.8.4 Modulation Theorem

Theorem 7.8.5   If F(s) is the complex Fourier transform of f(x) then

Proof   It is given that

  1. By definition

     

     

  2. can be proved similarly.

Corollary 7.8.6   If Fs(s) and Fc(s) are the Fourier sine transform and the Fourier cosine transform of f(x), respectively, then

Example 7.8.7   Using Fourier integral show that

 

Solution   The Fourier cosine integral is given by

 

 

Let f(t) = eat so that f(x) = eax. Substituting in the above integral, we have

 

Example 7.8.8   Prove that

Solution   Putting a = 1 in Example 7.1, we obtain

 

Example 7.8.9   Express as a Fourier integral and hence evaluate

Solution   The Fourier integral of f(x) is given by

 

 

The funtion f(t) is defined by

 

 

∴ We obtain

 

 

which is the Fourier integral of f(x). From (7.55) we obtain

 

 

At x = ±1, f(x) is discontinuous and the value of the integral is

 

Example 7.8.10   Prove that

Solution   From Example 7.3

 

 

Putting x = 0 we have

 

 

where we have replaced the dummy variable s by x.

Example 7.8.11   Using the Fourier integral representation, show that

 

Solution   Let f(x) be defined by

 

 

We now find the Fourier integral representation of f(x) in the exponential form, which is given by

 

 

The function f(x) is discontinuous at x = 0 and its value is

 

Example 7.8.12   Express as a Fourier integral. Evaluate

  1. [JNTU 2003(2), 2004(3)]

 

Solution   The Fourier integral of f(x) is

 

 

For which are points of discontinuity the value of the integral is

 

 

Choosing x = 0 and a = 1 we obtain where we have replaced the dummy variable s by x.

Example 7.8.13   Express the function as a Fourier sine integral.

Hence, evaluate

Solution   The Fourier sine integral of f(x) is given by

 

 

 

At x = π which is a point of discontinuity for f(x) the value of the integral is

 

Example 7.8.14   Find the Fourier transform of

 

Hence, evaluate

[JNTU 2002, 2004S]

 

Solution   The Fourier transform of f(x) is given by

 

 

By the inverse Fourier transform, we have

 

 

Putting x = 1/2 we obtain from the above equation (7.62)

 

 

Equating the real parts on either side

 

 

since the integrand is even in (−∞, ∞) and we have replaced the dummy variable s by x.

Example 7.8.15   Find the Fourier transform of

 

Hence, evaluate

Solution   The Fourier transform of f(x) is given by

 

 

We have and hence F{f(x)} = F(s) = 2 when s = 0     (7.66)

By the inverse Fourier transform we have

 

 

Putting x = 0 we get

 

 

Since the integrand is an even function in (−∞, ∞) we have

 

 

where we have replaced the dummy variables s by x.

Example 7.8.16   Find the Fourier (a) cosine and (b) sine transform of f(x) = eax (x > 0, a > 0).

Deduce the values of

 

 

Solution

  1. The Fourier cosine transform of f(x) is given by

     

     

     

    The inverse Fourier cosine transform of Fc(s) is

     

     

  2. The Fourier sine transform of f(x) is given by

     

     

     

     

    The inverse Fourier sine transform of Fs(s) is

     

     

Corollary 7.8.17   Putting a = 0 in the above result we obtain

 

Example 7.8.18   Find the Fourier sine transform of Hence, show that

 

Solution   The variable x is positive in (0, ∞) so that for (0, ∞).

The Fourier sine transform of is given by

 

 

By the inverse Fourier sine transform of Fs(s) we have

 

 

If we replace ‘x’ by ‘a’ we obtain

 

 

where we have replaced the dummy variable s by x.

Hence   

Example 7.8.19   Find the Fourier sine transform of eax/x.

Solution   By definition, the Fourier sine transform of f(x) is

 

 

 

Differentiating both sides with respect to ‘s

 

 

Integrating now with respect to ‘s’ we obtain

 

Example 7.8.20   Find the Fourier cosine transform of

 

 

[JNTU2002, 2003S, 2004S]

Solution   The Fourier cosine transform of f(x) is given by

 

Example 7.8.21   Find the Fourier cosine transform of f(x) = 1/(1 + x2). Hence, derive the Fourier sine transform of

 

Solution   By definition, the Fourier cosine transform of f(x) is given by

 

 

Differentiating both sides of (7.87) with respect to ‘s

 

 

Differentiating both sides of (7.88) with respect to s again

 

 

The general solution of (7.89) is

 

 

 

When s = 0, (7.87) and (7.90) give

 

 

Also   

 

Solving (7.92) and (7.93) we get

 

 

From (7.87) and (7.90) we have

 

 

Now

 

Example 7.8.22   Find the Fourier sine and cosine transform of eax(a > 0) and deduce the inverse formula.

 

[JNTU 2002, 2004]

Solution   The Fourier sine transform of f(x) = eax is

 

 

The inverse Fourier sine transform for is

 

 

The Fourier cosine transform of f(x) = eax is

 

 

The inversion formula for

 

 

is   

Example 7.8.23   Solve the intergral equation

 

 

Hence evaluate

Solution   By definition, the Fourier cosine transform of f(x) is

 

 

By the inversion formula we have

 

 

Now

 

 

From (7.101) and (7.102) we obtain

 

 

Taking the limit as we obtain

 

 

where we have replaced the dummy variable t by x.

Example 7.8.24   Show that the Fourier transform of is reciprocal.

 

[JNTU 2002, 2004S]

Solution   Here we have to use definition 2 for the Fourier transform, viz.

 

 

Hence

 

 

Here the function f(x) and the Fourier transform F(s) are the same. Hence, is self-reciprocal.

Note 7.8.25   If we use the other definitions there will be difference of a constant factor.

Example 7.8.26   Find the Fourier sine transform of (l/x)(x2 + a2).

Hence, deduce the Fourier cosine transform of

[JNTU 2001, 2002, 2003]

Solution   The Fourier sine transform of f(x) is given by

 

 

Now, let   

 

Differentiating both sides of (7.105) with respect to s

 

 

Again differentiating both sides of (7.106) with respect to s

 

 

 

    which is a second-order linear differentialequation     (7.107)

 

The general solution of (7.107) comprises two parts: the complementary function (CF) and the particular intergal (PI).

 

 

∴ The general solution is

 

 

 

Putting s = 0 in (7.105) and (7.108) we obtain

 

 

and similarly from (7.106) and (7.109) we obtain

 

 

Solving (7.110) and (7.111) we obtain

 

 

Substituting these values in (7.108) we have

 

 

which is the required Fourier sine transform of

 

 

Now we deduce the Fourier cosine transform of

 

The Fourier cosine transform of is given by

 

 

∴ The Fourier cosine transform of g(x) = 1/(x2 + a2) is Gc(s) = (π/2a)eas.     (7.114)

EXERCISE 7.1
  1. Find the Fourier transform of ex2, (−∞ < x < ∞).

    Ans:   

  2. Find the Fourier transform of

    Ans:

    1. [(a2s2 − 2)sinax + 2ascosas]
  3. Express as a Fourier integral

    Hence, evaluate .

    Ans:   

     

  4. Using Fourier integral show that

     

     

  5. Using Fourier integral show that

     

     

  6. Using Fourier intergral representation show that
  7. Prove that

    [Hint:   Use FSI with f(t) = eatebt.]

     

  8. Find the Fourier sine transform of

    Ans:   

     

  9. Find the Fourier sine transform of

     

     

     

    Ans:   

     

  10. Find the Finite fourier sine and cosine transforms of f(x) = 2x, 0 < x < 4.

    Ans:   

     

  11. Solve

    Ans:   

     

  12. Solve

    Ans:   f(x) = (2 + 2 cos x − 4 cos 2x)/πx

     

  13. Find the Fourier cosine transform of
    1. eaxcos ax
    2. eaxsin ax.

    Ans:   

     

  14. Find the Fourier cosine transform of .

    Ans:   

     

7.9 PARSEVAL'S IDENTITY FOR FOURIER TRANSFORMS

Theorem 7.9.1   If F(s) and G(s) are the Fourier transforms of f(x) and g(x), respectively, then

where denotes the complex conjugate of g.

Proof   Let g(x) be the inverse Fourier transform of G(s) so that

 

 

  1. Taking conjugates on both sides

     

    (Changing the order of integration)

     

    by the definition of Fourier transform

  2. Setting g(x) = f(x) for all x we obtain

     

7.10 PARSEVAL'S IDENTITIES FOR FOURIER SINE AND COSINE TRANSFORMS

Theorem 7.10.1   Similarly, we can obtain the following results:

Example 7.10.2   If find the Fourier transform.

Using Parseval's identity prove that

 

   [JNTU 2003 S (1)]

Solution   The Fourier transform of f(x) is given by

 

 

 

Parsevel's identity for Fourier transform is

 

replacing the dummy variable s by t

Example 7.10.3   Using Parseval's identities prove that

 

Solution   The Fourier sine transform of is given by

 

 

By Parseval's identity for Fourier sine transform we have

 

 

 

where we have replaced the dummy variable s by x.

Example 7.10.4   Evaluate using Parseval's identity.

Solution   The Fourier cosine transform of f(x) = eax is given by

 

 

and the Fourier cosine transform of g(x) = ebx is given by

 

 

Parseval's identity for Fourier cosine transforms is

 

Example 7.10.5   Evaluate
using Parseval's identity.

Solution   The Fourier cosine transform of f(x) = eax is given by

 

 

Parseval's identity is

 

 

 

EXERCISE 7.2

Using Parseval's identities prove the following:

  1. Find the Fourier transform of

    (JNTU 2002S)

    Hence, deduce that

  2. Give that
  3. Evaluate

    Prove that

    Ans: