Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 10 potx

Be aware of the different definitions when reading other texts or consulting tables of Fourier transforms.32.3 Evaluating Fourier Integrals 32.3.1 Integrals that Converge If the Fourier

We take the Laplace transform of the differential equation.

s2y(s) − sy(0) − yˆ 0(0) − ˆy(s) = ˆf (s)

s2y(s) − s − ˆˆ y(s) = ˆf (s)ˆ

f (τ ) sinh(t − τ ) dτ

The solution for positive t is

y(t) = cosh(t) +

Z t 0

f (τ ) cosh(t − τ ) dτ

y0(t) = sinh t +

Z t 0

f (τ ) cosh(t − τ ) dτ

We see that the first derivative is also continuous

y = s(s + 2)(s + 1)(s2+ 1)

We expand the right side in partial fractions

ˆ

y = − 12(s + 1) +

1 + 3s2(s2+ 1)

We use a table of Laplace transforms to do the inversion

We consider the problem

Ldi1

dt + Ri1+ q/C = E0

Ldi2

dt + Ri2− q/C = 0dq

dt = i1− i2

i1(0) = i2(0) = E0

2R, q(0) = 0.

We take the Laplace transform of the system of differential equations.

L

sˆi1− E02R

+ Rˆi1+ qˆ

C =

E0sL

sˆi2− E02R

+ Rˆi2− qˆ

C = 0

sˆq = ˆi1− ˆi2

We solve for ˆi1, ˆi2 and ˆq

ˆi1 = E02

1

1

1

Rs +

1/L(s + α − ıω)(s + α + ıω)



ˆi2 = E02

1

2 = 2

LC − α2

We expand the functions in partial fractions.

ˆi1 = E02

1

Rs+

ı2ωL

1

s + α + ıω − 1

s + α − ıω



ˆi2 = E02

1

Rs− ı2ωL

1

 1

s +

ı2ω

 1

R +

ı2ωL e

(−α−ıω)t− e(−α+ıω)t



i2 = E02

 1

R − ı2ωL e

(−α−ıω)t− e(−α+ıω)t



q = CE02



1 + ı2ω (α + ıω) e

(−α+ıω)t−(α − ıω) e(−α−ıω)t


We simplify the expressions to obtain the solutions

i1 = E02

We take the Laplace transform of the differential equation and solve for ˆy(s).

s2y − sy(0) − yˆ 0(0) + 4sˆy − 4y(0) + 4ˆy = 4

s + 1

s2y − 2s + 3 + 4sˆˆ y − 8 + 4ˆy = 4

s + 1ˆ

Chapter 32

The Fourier Transform

32.1 Derivation from a Fourier Series

Consider the eigenvalue problem

y00+ λy = 0, y(−L) = y(L), y0(−L) = y0(L)

The eigenvalues and eigenfunctions are

λn=

nπL

The eigenfunctions form an orthogonal set A piecewise continuous function defined on [−L L] can be expanded in

a series of the eigenfunctions

The Fourier coefficients are

Z L

−L

e−ınπx/Lf (x) dx

in the limit as L → ∞ becomes

Of course this derivation is only heuristic In the next section we will explore these formulas more carefully

32.2 The Fourier Transform

Let f (x) be piecewise continuous and letR−∞∞ |f (x)| dx exist We define the function I(x, L)

= 12π

Z ∞

−∞

f (ξ) 1ı(ξ − x) e

ıL(ξ−x)− e−ıL(ξ−x)
dξ

= 1π

Z ∞

−∞

f (ξ + x)sin(Lξ)

ξ dξ.

In Example 32.3.3 we will show that

Z ∞ 0

sin(Lξ)

ξ dξ =

π

2.Continuous Functions Suppose that f (x) is continuous

f (x+ξ)−f (x) ξ

dξ < ∞ We use theRiemann-Lebesgue lemma to show that the integral vanishes as L → ∞

1π

Z ∞ 0

Z 0

−∞

f (x−)sin(Lξ)

 f (x + ξ) − f (x+)

ξ

sin(Lξ) dξ

If f (x) has a left and right derivative at x, then

Result 32.2.1 Let f (x) be piecewise continuous with R−∞∞ |f (x)| dx < ∞ The Fourier transform of f (x) is defined

Be aware of the different definitions when reading other texts or consulting tables of Fourier transforms.

32.3 Evaluating Fourier Integrals

32.3.1 Integrals that Converge

If the Fourier integral

Example 32.3.1 Consider the Fourier transform of e−a|x|, where a > 0 Since the integral of e−a|x| is absolutelyconvergent, we know that the Fourier transform integral converges for real ω We write out the integral.

e−ax−ıωx dx

= 12π

e(−a−ı<(ω)+=(ω))x dx

The integral converges for |=(ω)| < a This domain is shown in Figure 32.1

Re(z)Im(z)

Figure 32.1: The Domain of Convergence

Now We do the integration.

e−(a+ıω)x dx

= 12π

= 12π

1

aπ(ω2+ a2), for |=(ω)| < a

We can extend the domain of the Fourier transform with analytic continuation

Fe−a|x| = a

π(ω2+ a2), for ω 6= ±ıaExample 32.3.2 Consider the Fourier transform of f (x) = x−ıα1 , α > 0

F

1

x − ıα



= 12π

For ω > 0, we will close the path of integration in the lower half-plane Let CR be the contour from x = R to

x = −R following a semicircular path in the lower half-plane The integral along CR vanishes as R → ∞ by Jordan’sLemma

x − ıα



= 0

For ω < 0, we will close the path of integration in the upper half-plane Let CR denote the semicircular contour from

x = R to x = −R in the upper half-plane The integral along CR vanishes as R goes to infinity by Jordan’s Lemma

We evaluate the Fourier transform integral with the Residue Theorem

F

1

x − ıα



= 12π2πi Res

32.3.2 Cauchy Principal Value and Integrals that are Not Absolutely Convergent.

That the integral of f (x) is absolutely convergent is a sufficient but not a necessary condition that the Fourier transform

of f (x) exists The integralR−∞∞ f (x) e−ıωx dx may converge even ifR−∞∞ |f (x)| dx does not Furthermore, if the Fouriertransform integral diverges, its principal value may exist We will say that the Fourier transform of f (x) exists if theprincipal value of the integral exists

F [f (x)] = −

Z ∞

−∞

f (x) e−ıωx dxExample 32.3.3 Consider the Fourier transform of f (x) = 1/x

We can evaluate the Fourier transform with the Residue Theorem.

ˆ

f (ω) = 1

2π

 −12

(2πi) Res 1

Re(z) Im(z)

Figure 32.2: The Path of Integration

If ω < 0, we can close the contour in the upper half plane to obtain

We collect the results in one formula.

1

xsin(ωx) dx =

π2Thus we have evaluated an integral that we used in deriving the Fourier transform

1 for x < 0

e−ıωx dx

= 12π

Z ∞ 0

e(−ı<(ω)+=(ω))x dx

= 12π

 e−ıωx

−ıω

∞ 0

= − ı2πω for =(ω) < 0Using analytic continuation, we can define the Fourier transform of f+(x) for all ω except the point ω = 0

Z 0

−∞

e(−ı<(ω)+=(ω))x dx

= 12π

F [f−(x)] = ı

2πω

Now we are prepared to define the Fourier transform of f (x) = 1.

F [1] = F [f−(x)] + F [f+(x)]

= − ı2πω +

ı2πω

= 0, for ω 6= 0When ω = 0 the integral diverges When we consider the closure relation for the Fourier transform we will see that

F [1] = δ(ω)

32.4 Properties of the Fourier Transform

In this section we will explore various properties of the Fourier Transform I would like to avoid stating assumptions onvarious functions at the beginning of each subsection Unless otherwise indicated, assume that the integrals converge

−ıωξ

Next we take the inverse Fourier transform.

δ(x − ξ) ∼

Z ∞

−∞

12π e

Note that the integral is divergent, but it would be impossible to represent δ(x − ξ) with a convergent integral

32.4.2 Fourier Transform of a Derivative.

Consider the Fourier transform of y0(x)

−ıωx

∞

−∞

− 12π

Z ∞

−∞

(−ıω)y(x) e−ıωx dx

= ıω 12π

Example 32.4.1 The Dirac delta function can be expressed as the derivative of the Heaviside function.

H(x − c) =

(

0 for x < c,

1 for x > cThus we can express the Fourier transform of H(x − c) in terms of the Fourier transform of the delta function

F [δ(x − c)] = ıωF [H(x − c)]

12π

Z ∞

−∞

δ(x − c) e−ıωx dx = ıωF [H(x − c)]

12π e

32.4.3 Fourier Convolution Theorem.

Consider the Fourier transform of a product of two functions

= 12π

The convolution of two functions is defined

Z ∞

−∞

f (ξ) e−ıωξ dξ

ˆg(ω) eıωx dω

= 12π

= 12π

dξ

= 12π

Z ∞

−∞

f (ξ)g(x − ξ) dξThus

F−1[ ˆf (ω)ˆg(ω)] = 1

2πf ∗ g(x) =

12π

Example 32.4.2 Using the convolution theorem and the table of Fourier transform pairs in the appendix, we can findthe Fourier transform of

F

2c

Z 0

−∞

eηe−2|ω−η| dη +

Z ∞ 0

e−2ω+η dη +

Z ∞ ω

e2ω−3η dη



= 18

Now consider the case ω < 0.

e2ω−η dη +

Z ∞ 0

e2ω−3η dη



= 18

Analogous to this result is Parseval’s theorem for Fourier transforms.

Let f (x) be a complex valued function that is both absolutely integrable and square integrable

Z −∞

∞

f (x) eıωx dx

= 12π

Z ∞

−∞

f (x) e−ıω(x−c) dx

F [f (x + c)] = eıωcf (ω)ˆThe inverse Fourier transform of ˆf (ω + c) is

Z ∞

−∞

f (x) e−ıωx dx



F [xf (x)] = ı∂ ˆf

∂ω.Similarly, you can show that

F [xnf (x)] = (i)n∂

nˆ

∂ωn

32.5 Solving Differential Equations with the Fourier Transform

The Fourier transform is useful in solving some differential equations on the domain (−∞ ∞) with homogeneousboundary conditions at infinity We take the Fourier transform of the differential equation L[y] = f and solve for ˆy Wetake the inverse transform to determine the solution y Note that this process is only applicable if the Fourier transform

of y exists Hence the requirement for homogeneous boundary conditions at infinity

We will use the table of Fourier transforms in the appendix in solving the examples in this section

Example 32.5.1 Consider the problem

1

α2− 1

1

ω2+ α2 − α 1/π

ω2+ 1



y(x) = e

−α|x|−α e−|x|

α2− 1Example 32.5.2 Consider the Green function problem

y = −12

32.6 The Fourier Cosine and Sine Transform

32.6.1 The Fourier Cosine Transform

Suppose f (x) is an even function In this case the Fourier transform of f (x) coincides with the Fourier cosine transform

Z ∞

−∞

f (x)(cos(ωx) − ı sin(ωx)) dx

= 12π

Z ∞

−∞

f (x) cos(ωx) dx

= 1π

Z ∞ 0

f (x) cos(ωx) dxThe Fourier cosine transform is defined:

Fc[f (x)] = ˆfc(ω) = 1

π

Z ∞ 0

ˆc(ω) cos(ωx) dω.

Thus we have the Fourier cosine transform pair

f (x) = Fc−1[ ˆfc(ω)] = 2

Z ∞ 0

ˆ

c(ω) cos(ωx) dω, ˆc(ω) = Fc[f (x)] = 1

π

Z ∞ 0

f (x) cos(ωx) dx

32.6.2 The Fourier Sine Transform

Suppose f (x) is an odd function In this case the Fourier transform of f (x) coincides with the Fourier sine transform

Z ∞

−∞

f (x)(cos(ωx) − ı sin(ωx)) dx

= −ıπ

Z ∞ 0



−ıπ

Z ∞ 0

f (x) sin(ωx) dx

sin(ωx) dω

= 2

Z ∞ 0

 1π

Z ∞ 0

f (x) sin(ωx) dx

sin(ωx) dω

This gives us the Fourier sine transform pair

f (x) = Fs−1[ ˆfs(ω)] = 2

Z ∞ 0

ˆ

s(ω) sin(ωx) dω, ˆs(ω) = Fs[f (x)] = 1

π

Z ∞ 0

f (x) sin(ωx) dx

Result 32.6.1 The Fourier cosine transform pair is defined:

f (x) = Fc−1[ ˆ fc(ω)] = 2

Z ∞ 0

ˆ

fc(ω) cos(ωx) dω ˆ

fc(ω) = Fc[f (x)] = 1

π

Z ∞ 0

f (x) cos(ωx) dx The Fourier sine transform pair is defined:

f (x) = Fs−1[ ˆ fs(ω)] = 2

Z ∞ 0

ˆ

fs(ω) sin(ωx) dω ˆ

fs(ω) = Fs[f (x)] = 1

π

Z ∞ 0

and y0 vanish at infinity We calculate the transforms of the first and second derivatives.

Fc[y0] = 1

π

Z ∞ 0

f (x)g(x) cos(ωx) dx

= 1π

Z ∞ 0

2

Z ∞ 0

ˆ

c(η) cos(ηx) dη

g(x) cos(ωx) dx

= 2π

Z ∞ 0

Z ∞ 0

ˆc(η)g(x) cos(ηx) cos(ωx) dx dη

We use the identity cos a cos b = 12(cos(a − b) + cos(a + b))

= 1π

Z ∞ 0

Z ∞ 0

ˆ

c(η)g(x) cos((ω − η)x) + cos((ω + η)x)
dx dη

=

Z ∞ 0

ˆ

c(η) 1π

Z ∞ 0

g(x) cos((ω − η)x) dx + 1

π

Z ∞ 0

g(x) cos((ω + η)x) dx

dη

=

Z ∞ 0

ˆc(η) ˆgc(|ω − η|) + ˆgc(ω + η)
dη

Inverse Cosine Transform of a Product Now consider the inverse Fourier cosine transform of a product offunctions Let Fc[f (x)] = ˆfc(ω), and Fc[g(x)] = ˆgc(ω).

Fc−1[ ˆfc(ω)ˆgc(ω)] = 2

Z ∞ 0

ˆc(ω)ˆgc(ω) cos(ωx) dω

= 2

Z ∞ 0

 1π

Z ∞ 0

f (ξ) cos(ωξ) dξ

ˆ

gc(ω) cos(ωx) dω

= 2π

Z ∞ 0

Z ∞ 0

f (ξ)ˆgc(ω) cos(ωξ) cos(ωx) dω dξ

= 1π

Z ∞ 0

Z ∞ 0

f (ξ)ˆgc(ω) cos(ω(x − ξ)) + cos(ω(x + ξ))
dω dξ

= 12π

Z ∞ 0

f (ξ)

2

Z ∞ 0

ˆ

gc(ω) cos(ω(x − ξ)) dω + 2

Z ∞ 0

ˆ

gc(ω) cos(ω(x + ξ)) dω

dξ

= 12π

Z ∞ 0

f (ξ) g(|x − ξ|) − g(x + ξ)
dξ

Result 32.7.1 The Fourier cosine and sine transform convolution theorems are

Fc[f (x)g(x)] =

Z ∞ 0

f (ξ) g(|x − ξ|) + g(x + ξ)
dξ

Fs[f (x)g(x)] =

Z ∞ 0

f (ξ) g(|x − ξ|) − g(x + ξ)
dξ

32.7.3 Cosine and Sine Transform in Terms of the Fourier Transform

We can express the Fourier cosine and sine transform in terms of the Fourier transform First consider the Fouriercosine transform Let f (x) be an even function

Fc[f (x)] = 1

π

Z ∞ 0

f (x) cos(ωx) dx

We extend the domain integration because the integrand is even

= 12π

Z ∞

−∞

f (x) e−ıωx dx

= F [f (x)]

32.8 Solving Differential Equations with the Fourier Cosine and

Sine Transforms

Example 32.8.1 Consider the problem

y00− y = 0, y(0) = 1, y(∞) = 0

Since the initial condition is y(0) = 1 and the sine transform of y00 is −ω2ˆc(ω) + ω

πy(0) we take the Fourier sinetransform of both sides of the differential equation

1/3

ω2+ 1 − 1/3

ω2+ 4



= 13

ˆs(η) ˆgc(|ω − η|) − ˆgc(ω + η)
dη.

We can express the Fourier cosine and sine transform in terms of the Fourier transform First consider the Fouriercosine transform... class="text_page_counter">Trang 24< /span>

Example 32 .4. 2 Using the convolution theorem and the table of Fourier transform pairs in the appendix, we can findthe Fourier transform... = f and solve for ˆy Wetake the inverse transform to determine the solution y Note that this process is only applicable if the Fourier transform

of y exists Hence the requirement for homogeneous