MATHEMATICAL METHOD IN SCIENCE AND ENGINEERING Episode 14 doc

DERIVATION OF THE FOURIER INTEGRAL 479 and it is very useful in finding solutions of systems of ordinary differential equations by converting them into a system of algebraic equations..

PROBLEMS 473

where the degeneracy, gn, and the eigenfrequencies, wn, are given as

Note: Interested students can obtain the eigenfrequencies and the degen- eracy by solving the wave equation for the massless conformal scalar field:

15.4 Using asymptotic series evaluate the logarithmic integral

so that

15.5

massless conformal scalar field with thermal spectrum can be written as

In a closed Einstein universe the renormalized energy density of a

474 INFINITE SERIES

where & is the constant radius of the universe, T is the temperature of

the radiation, and (2x21i$) is the volume of the universe The second term

(-) inside the square brackets is the well-known renormalized quantum vacuum energy, that is, the Casimir energy for the Einstein universe First find the high and low temperature limits of (p}ren and then obtain the flat spacetime limit & -+ 03

How many terms did you have to add?

15.7 Check the convergence of the series

15.8 Find the interval of convergence for the series

15.9 Evaluate the sums

cr=o an cos ne

( b ) C,"==, an sin d, a is a constant Hint: Try using complex variables

15.10 Verify the following Taylor series:

X"

00

ex = C 12? for all z n=O

and

PROBLEMS 475

15.11 Find the first three nonzero terms of the following Taylor series:

a )

f(x) = x3 + 2x + 2 about z = 2 b)

f(x) = e2= cos x about x = 0

15.12 Another important consequence of the Lorentz transformation is the formula for the addition of velocities, where the velocities measured in the K and R frames are related by the formula

dx &1

where u1 = - and El = - are the velocities measured in the K and I?

frames, respectively, and R is moving with respect to K with velocity u along

the common direction of the x- and Z-axes Using the binomial formula find

an appropriate expansion of the above formula and show that in the limit of small velocities this formula reduces to the well-known Galilean result

15.14 Given a power series

exactly for the interval 1x1 < 1, then expand your answer in powers of X

15.18 By using the Euler-Maclaurin sum formula evaluate the sum

W

C n3xn n=l Show that it agrees with the expansion found in Problem 15.17

16

INTEGRAL TRANSFORMS

Integral transforms are among the most versatile mathematical tools Their applications range from solution of differential equations to evaluation of def- inite integrals and from solution of systems of coupled differentia1 equations

to integral equations They can even he used for defining differintegrals, that

is, fractional derivatives and integrals (Chapter 14) In this chapter, after a

general introduction we mainly discuss two of the most frequently used inte- gral transforms, the Fourier and the Laplace transforms, their properties, and techniques

Commonly encountered integral transforms allow us to relate two functions through the integral

(16.1)

where g(a) is called the integral transform of f ( t ) with respect to the kernel

~ ( a , t) These transformations are also linear, that is, if the transforms

exist, then one can write

(16.3)

4 77

478 INTEGRAL TRANSFORMS

and

b

C 9 l ( a ) = 1 I d 1 (t>l4a, t ) d t , (16.4) where c is a constant Integral transforms can also be shown as an operator:

where the operator =€(a,t) is defined as

(16.6)

We can now show the inverse transform as

16.1 SOME COMMONLY ENCOUNTERED INTEGRAL

(16.10) Other frequently used kernels are

e-at, tJn(at), and ta-' (16.11) The Laplace transform is defined as

( 16.12)

DERIVATION OF THE FOURIER INTEGRAL 479

and it is very useful in finding solutions of systems of ordinary differential equations by converting them into a system of algebraic equations The Han-

kel or Fourier-Bessel transform is defined as

and it is usually encountered in potential energy calculations in cylindrical coordinates Another useful integral transform is the Mellin transform:

(16.14) The Mellin transform is useful in the reconstruction of

functions" from power series expansions Weierstrass function is defined as

"Weierstrass-type

00

(16.15)

n=O

where a and b are constants It has been proven that, provided 0 < b < 1,

a > 1, and ab > 1, the Weierstrass function has the interesting property of being continuous but nowhere differentiable These interesting functions have found widespread use in the study of earthquakes, rupture, financial crashes, etc (Gluzman and Sornette)

16.2 DERIVATION OF T H E FOURIER INTEGRAL

16.2.1 Fourier Series

Fourier series are very useful in representing a function in a finite interval, like [0,27r] or [-L, L], or a periodic function in the infinite interval (-CO,CO)

We now consider a nonperiodic function in the infinite interval (-o,co)

Physically this corresponds to expressing an arbitrary signal in terms of sine

and cosine waves We first consider the trigonometric Fourier expansion of a sufficiently smooth function in the finite interval [-L, L] as

Fourier expansion coefficients a, and b, are given as

(16.17) (16.18)

(16.23) Thus the Fourier integral is obtained as

f(.) = a 1°0 dw 1:f ( t ) cosw(t - z)dt (16.24)

In this expression we have assumed the existence of the integral

00

For the Fourier integral of a function to exist, it is sufficient for the integral

I-", If(t)l dt to be convergent

We can also write the Fourier integral in exponential form Using the fact that sin w ( t - x) is an odd function with respect to w, we can write

FOURIER AND INVERSE FOURIER TRANSFORMS 481

We now multiply Equation (16.26) by z and then add Equation (16.27) to obtain the exponential form of the Fourier integral as

f(.) = 2 1 dwe-iWx 1, f weiwt& (16.28) where w is a parameter; however, in applications to waves it is the angular frequency

27r oo

16.2.2 Dirac-Delta Function

Let us now write the Fourier integral as

where we have interchanged the order of integration The expression inside the curly brackets is nothing but the Dirac-delta function:

(16.30) which has the following properties:

6( - u ) = 0, ( 5 # a), (16.31) 6(z - a)& = 1, (16.32)

00

1,

where f(z) is continuous at 2 = a

We write the Fourier integral theorem [Eq (16.28)] as

We now define the Fourier transform of f ( t ) as

482 INTEGRAL TRANSFORMS

Fig 16.1 Wave train with N = 5

16.3.1 Fourier Sine and Cosine Transforms

When f ( t ) is an even function we can write

fc(-.) = f c ( z )

Using the identity

(16.36)

eiWt - cos wt + i sin wt, (16.37)

we can also write

gc(w> = - fc(t) (cos wt + i sin wt) dt (16.38) Considering that sin wt is an odd function with respect to t, the Fourier cosine transform is obtained as

FOURIER A N D INVERSE FOURIER TRANSFORMS 483

Example 16.1 Fourier analysis of finite wave train: We now find the

Fourier transform of a finite wave train, which is given as

(16.44)

For N = 5 this wave train is shown in Figure 16.1

Since f ( t ) is an odd function we find its Fourier sine transform as

(16.45)

gs(w) = & [ 2 (WO - W ) 2 ( w o + w ) I

2 sin (wO - w) EL sin (wo + w)

wo -

For frequencies w - wo only the first term in Equation (16.45) domi-

nates Thus gs(w) is given as in Figure 16.2

This is the diffraction pattern for a single slit It has zeroes a t

484 INTEGRAL TRANSFORMS

16.3.2 Fourier Transform of a Derivative

First we find the Fourier transform of - d f ( x ) as

d x

(16.48)

Using integration by parts we write

Assuming that f(x) -f 0 as x f &m we obtain the Fourier transform of the first derivative as

Example 16.2 Partial differential equations and Fourier transforms:

One of the many uses of integral transforms is solving partial differential equations Consider vibrations of a n infinitely long wire The equation

to be solved is the wave equation, which is given as

(16.53)

where v is the velocity of the wave and y ( x , t ) is the displacement of the wire from its equilibrium position as a function of position and time As our initial condition we take the shape of the wire at t = 0 as

FOURIER AND INVERSE FOURIER TRANSFORMS 485

where Y ( a , t ) represents the Fourier transform of y(z, t):

Y ( a , t ) = - 1, y(z,t)ei""dz (16.57)

6 -00

From Equation (16.56) we see that the effect of the integral transform on the partial differential equation is to reduce the number of independent variables Thus the differential equation to be solved for Y ( a , t ) is now

an ordinary differential equation, solution of which can be written easily

Because the last expression is nothing but the inverse Fourier transform

of F ( a ) , we can write the final solution as

This represents a wave moving to the right or left with the velocity and with its shape unchanged

16.3.3 Convolution Theorem

Let F(t) and G ( t ) be the Fourier transforms of two functions f(z) and g(z), respectively Convolution of f(z) and g(z) is defined as

16.3.4 Existence of Fourier Transforms

We can show the Fourier transform of f(x) in terms of an integral operator

16.3.5 Fourier Transforms in Three Dimensions

Fourier transforms can also be defined in three dimensions as

Substituting Equation (16.68) back in Equation (16.69) and interchanging the order of integration we obtain the three dimensional Dirac-delta function as

These formulas can easily be extended to n dimensions

SOME THEOREMS ON FOURER TRANSFORMS 487

16.4 SOME THEOREMS ON FOURIER TRANSFORMS

Proof: To prove these theorems we make the k + -k change in the Fourier

1,

tively

transform of g(x):

G(-k) = - Srn g(z)e-i"h (16.73) Multiplying the integral in Equation (16.73) with F ( k ) and integrating it over

k in the interval (-m, co) we get

converge, we can change the order of the k and x integrals as

Assuming that the inverse Fourier transform of F ( k ) exists, the second Parce-

488 INTEGRAL TRANSFORMS

which is the first Parceval theorem From this proof it is seen that pointwise existence of the inverse Fourier transform is not necessary; that is, as long as the value of the integral

does not change, the value of the integral

(16.80)

(16.81)

can be different from the value of f(z) at some isolated singular points In quantum mechanics wave functions in position and momentum spaces are each others’ Fourier transforms The significance of Parseval’s theorems is

that normalization in one space ensures normalization in the other

Example 16.3 Diffusion problem in one dimension: Let us consider

a long thin pipe filled with water and with M g of salt located at XO We would like to find the concentration of salt as a function of position and time Because we have a thin pipe, we can neglect the change in concen- tration acrms the width of the pipe The density (concentrationxmass)

satisfies the diffusion equation:

(16.83) Because a t t = 0, the density is zero everywhere except at XO, we take our initial condition as

we have

X i i W

(16.86)

SOME THEOREMS ON FOURIER TRANSFORMS 489

which is sufficient for the existence of the Fourier transform Taking the Fourier transform of the diffusion equation with respect to z we get

(16.87)

dR(k t )

dt

This is an ordinary differential equation, where R ( k , t ) is the Fourier

transform of the density The initial condition for R ( k , t ) is the Fourier

transform of the initial condition for the density, that is,

The solution of Equation (16.87) can easily be written as

490 INTEGRAL TRANSFORMS

Check that this solution satisfies the diffusion equation with the initial condition

The Laplace transform of a function is defined as

e-stF(t)dt s > 0 and real

(16.93)

(16.94)

For this transformation to exist we do not need the existence of the integral

In other words, F(t) could diverge exponentially for large values oft However,

if there exists a constant SO and a pmitive constant C, such that for sufficiently

large t , that is t > to, the inequality

fcc

.€ {t"} = lo e-sttndt (16.98) does not exist for n 5 -1 values, because it has a singular point at the origin

On the other hand, for s > 0 and n > -1, the Laplace transform is given as

n!

.€ {t"} = - Sn+ 1 (16.99)

INVERSE LAPLACE TRANSFORMS 491

Single point

t

Fig 16.3 Null function

16.6 INVERSE LAPLACE TRANSFORMS

Using operator language we can show the Laplace transform of a function as

The inverse transform of f(s) is now shown with L-' as

2-l { f ( s ) } = F(t) (16.101)

In principle, the inverse transform is not unique Two functions Fl(t) and Fz(t) could have the same Laplace transform; however, in such cases the difference of these functions is

where for all to values N ( t ) satisfies

l o N(t)dt = 0 (16.103)

N ( t ) is called a null function, and this result is also known as the Lerch theorem In practice we can take N ( t ) as zero, thus making the inverse Laplace transform unique In Figure 16.3 we show a null function

to the imaginary axis in the complex s-plane y is chosen such that all the singularities of e s t f ( s ) are to the left of the straight line For t > 0 we can close the contour with an infinite semicircle to the left-hand side of the line The above integral can now be evaluated by using the residue theorem to find the inverse Laplace transform

The Bromwich integral is a powerful tool for inverting complicated Laplace transforms when other means prove inadequate However, in practice using the fact that Laplace transforms are linear and with the help of some basic theorems we can generate many of the inverses needed from a list of elementary Laplace transforms

16.6.2 Elementary Laplace Transforms

1 Many of the discontinuous functions can be expressed in terms of the Heavyside step function (Fig 16.4)

3 The Laplace transform of F ( t ) = elCt for t > 0 is

4 Laplace transforms of hyperbolic functions

INVERSE LAPLACE TRANSFORMS 493

Fig 16.4 Heavyside step function

5 Using the relations

cos kt = cosh ikt and sin kt = -i sinh k t ,

we can find the Laplace transforms of the cosine and the sine functions

n!

L { P } = - S n + l , s > 0 , n > -1 (16.113)

494 INTEGRAL TRANSFORMS

16.6.3 Theorems About Laplace Transforms

By using the entries in a list of transforms, the following theorems are very useful in finding inverses of unknown transforms:

Theorem I: First Translation Theorem

If the function f ( t ) has the Laplace transform

L {f(t>} = F(s), then the Laplace transform of e a t f ( t ) is given as

Theorem 11: Second Translation Theorem

If F ( s ) is the Laplace transform of f ( t ) and the Heavyside step function

is shown as U ( t - a), we can write

L {V(t - u ) f ( t - u ) } = e-"'F(s) (16.118) Similarly, if L-' {F(s)} = f ( t ) is true, then we can write