Nair S. Advanced Topics in Applied Mathematics: For Engineering and the Physical Sciences

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Fourier Transforms 169

3.18 Solve the wave equation for an inﬁnite string

∂

∂x

∂

∂t

with the initial conditions

u(x,0) =U

−x

∂u(x,0)

∂t

=0.

3.19 Solve the wave equation for an inﬁnite string

∂

∂x

∂

∂t

with the initial conditions

u(x,0) =

∂u(x,0)

∂t

=0.

3.20 Lateral motion of an inﬁnite beam satisﬁes

∂

∂x

∂

∂t

=0,

where EI is the bending stiffness and m is the mass per unit length.

If the initial conditions are

v(x,0) =V

−x

∂v (x,0)

∂t

=0,

and v → 0as|x|→∞, obtain v(x,t).

3.21 Consider the motion of a concentrated force on an elastically

supported taut string. The deﬂection u satisﬁes

∂

∂x

+u =

∂

∂t

−δ(x −ct),

with u →0as|x|→∞. Assume the speed of the force, c > 1. We

are interested in the steady-state distribution of the deﬂection.

Introducing the new variables,

ξ =x −ct, τ = t,

170 Advanced Topics in Applied Mathematics

rewrite the governing equation. For the steady state, let ∂u/∂τ

and ∂

u/∂τ

go to zero. Obtain the solution that is continuous at

ξ =0.

3.22 The transient heat conduction in a 1D inﬁnite bar satisﬁes

∂

∂x

∂u

∂t

with the initial condition

u(x,0) =T

−|x|

Obtain the temperature u(x,t).

3.23 The transient heat conduction in a 1D inﬁnite bar with heat

generation satisﬁes

∂

∂x

∂u

∂t

with the initial condition u(x,0) = T

. Obtain u(x, t) in the form

of a convolution integral.

3.24 The transient heat conduction in a 2D inﬁnite plate satisﬁes

∂

∂x

∂

∂y

∂u

∂t

with

u(x,y,0) = U

−(x

)

Obtain u(x,y,t).

3.25 The transient heat conduction in a 2D inﬁnite plate satisﬁes

∂

∂x

∂

∂y

∂u

∂t

with

u(x,y,0) = U

h(1 −|x|)h(1 −|y|).

Obtain u(x,y,t).

Fourier Transforms 171

3.26 A semi-inﬁnite rod has an initial temperature distribution

u(x,0) =T

−x

and the transient heat conduction is governed by

∂

∂x

∂u

∂t

If the boundary temperature is u(0, t) =0, obtain the temperature

u(x,t).

3.27 In the preceding problem, if we replace the condition at x =0by

∂u(0,t)

∂x

−hu(0, t) =0,

where h is a constant, obtain the solution using the mixed

trigonometric transform.

3.28 Solve the integral equation

u(x) =xe

−|x|

−



∞

−∞

u(t)e

−|x−t|

dt.

3.29 Find the solution of

u(x) =e

−2|x|

−



∞

−∞

u(t)e

−|x−t|

dt.

3.30 Find the solution of

u(x) =|x|

−p

−



∞

−∞

u(t)e

−|x−t|

dt,

where p is a real constant: 0 < p < 1.

3.31 Solve

u(x) =xe

−x

−

√



∞

−∞

−(x−t)

u(t)dt,

in the form of an inﬁnite series.

3.32 Solve

u(x) =

−

√



∞

−∞

−(x−t)

u(t)dt,

in the form of a series of convolution integrals.

172 Advanced Topics in Applied Mathematics

3.33 Solve the system of integral equations

(x) =



∞

−∞

(t)e

−|x−t|

dt +2e

−|x|

(x) =



∞

−∞

(t)e

−|x−t|

dt −4e

−2|x|

3.34 Show that

u =1/2andu = c −x

are solutions of

u(x) ±

√

2π



∞

−∞

−t

u(x −t)dt = 1,

corresponding to “+” and “−” signs, respectively. Here, c is a

constant.

3.35 Obtain all the solutions of the integral equation

u(x) −

√

2π



∞

−∞

−t

u(x −t)dt = 1.

3.36 Solve

u(x) −



∞

−∞

−2|x−t|

u(t)dt = e

|x|

3.37 Obtain all the solutions of the integral equation

u(x) −

√

2π



∞

−∞

−t

u(x −t)dt = x,

with λ being real and positive.

3.38 Solve

u(x) −λ



∞

−∞

−|x−t|

u(t)dt = cosx.

3.39 Consider the signal

f (t) =3 cos3t +4sin2t,

Fourier Transforms 173

which is modulated to get

g(t) =f (t)cos100t.

Show that the analytical signal

h(t) =g(t) +iH[g(τ ),τ → t]

allows the factoring

h(t) =f (t)e

i100t

3.40 Show that the ﬁnite Hilbert transform of (1 −x

)

1/2

is x and that

of (1 −x

)

−1/2

is 0. Use the change of variable,

x =

1 −t

1 +t

to simplify the integrals.

3.41 Show that the ﬁnite Hilbert transform of an even function in

(−1, 1) is odd and that of an odd function is even.

3.42 Show that



∗π

cosnφdφ

cosθ −cosφ

=−

sinnθ

sinθ

In Eq. (3.295), assuming Fourier series

f (θ) =



n=0

cosnθ , g(θ) =



n=1

sinnθ,

show that B

=−A

(n =1, 2,...). In airfoil theory, B

are known

and A

are to be found.

LAPLACE TRANSFORMS

As the name implies, this integral transform was introduced by Laplace

during his studies of probability. The English mathematician and engi-

neer Oliver Heaviside had extensively used the methods of Laplace

transforms in the name of operational calculus in solving linear dif-

ferential equations arising from electrical networks. The Laplace

transform, in its common form, applies to functions that are causal.

This limitation, compared to the Fourier transform, is more than com-

pensated by the fact that the Laplace transform can be applied to

exponentially growing functions. These two features make the Laplace

transform ideal for time-dependent functions, which are zero for t < 0

and bounded by an exponentially growing function. In this chapter, we

assume that, unless otherwise deﬁned, all functions of time are zero

for negative values of their arguments.

The Laplace transform of f (t) is deﬁned as

L[f (t),t → p]=

f (p) ≡



∞

f (t)e

−pt

dt. (4.1)

We assume f (t) is piecewise continuous and satisﬁes the generalized

integrability condition,



∞

f (t)e

−γ t

dt < M, (4.2)

where M and γ are real. The deﬁning integral, Eq. (4.1), in conjunction

with the integrability condition implies

f (p) has no singularities in the

174

Laplace Transforms 175

Re(p)

Im(p)

Figure 4.1. The complex p-plane. The hatched area shows the analytic region for

f to the

right of the line Re(p) = γ . Solid circles represent singularities of

f .

complex p-plane on the right side of the vertical line, Re(p) =γ . Thus,

f (p) is analytic in the shaded area in Fig. 4.1.

4.1 INVERSION FORMULA

We may use what we learned from the complex Fourier transform to

obtain the inversion formula for the Laplace transform. As f =0 when

t < 0, using the Heaviside step function h(t), we deﬁne

g(t) =f (t)e

−γ t

h(t), (4.3)

with γ sufﬁciently large to make g absolutely integrable. The Fourier

integral theorem when applied to g gives

g(t) =

2π



∞

−∞

−iξ t



∞

f (τ )e

−γτ+iξτ

dτ dξ. (4.4)

Replacing g by f (t)e

−γ t

f (t) =

2π



∞

−∞

(γ t−iξ)t



∞

f (τ )e

−γτ+iξτ

dτ dξ. (4.5)

176 Advanced Topics in Applied Mathematics

Let

p =γ −iξ , dp =−idξ . (4.6)

This change of variable alters our inversion integral from the line −∞<

ξ<∞ to the line  going from −i∞ to i∞ in the analytic region in

Fig. 4.1. Deﬁning the inner integral as

f (p) =



∞

f (t)e

−pt

dt, (4.7)

the inversion formula is

f (t) =

2πi





f (p)e

dp. (4.8)

The line  running vertically in the analytic region of

f is called the

Bromwich contour. From Eq. (4.1), when

f is the Laplace transform of

some function f ,asRe(p) →∞,

f →0. Now, given a function

f , we test

if there is a semi-inﬁnite region beyond Re(p) = γ where it is analytic.

Then, we also test for the limit of

f asRe(p) →∞.If

f passes both of

these tests, we may attempt to invert it using the inversion integral. For

the uniqueness of the inverse transform, we have to limit ourselves to

functions, f (t), that are piecewise continuous and discard “functions”

that are undeﬁned on certain ﬁnite intervals.

4.2 PROPERTIES OF THE LAPLACE TRANSFORM

First, let us consider a continuous function f (t) for the purpose of

illustrating the basic properties.

4.2.1 Linearity

From the deﬁnition given in Eq. (4.1), a linear combination of two

functions has the transform

L[f

(t) +f

(t)]=

(p) +

(p). (4.9)

Laplace Transforms 177

4.2.2 Scaling

Assuming a is positive, a change of scale of the time variable, t,toat

results in

L[f (at)]=



∞

f (at)e

−pt



∞

f (t)e

−pt/a

f (p/a). (4.10)

4.2.3 Shifting

If the graph of the function is shifted by an amount a to the right,

L[f (t −a)]=



∞

f (t −a)e

−pt



∞

f (t)e

−p(t+a)

−pa

f (p). (4.11)

The function f (t) is zero when the argument is negative and f (t −a) is

zero when t < a.

4.2.4 Phase Factor

Here, “phase” is used in a generalized sense. What we have is the

original f (t) multiplied by the factor e

−at

L[f (t)e

−at



∞

f (t)e

−(p+a)t

f (p +a). (4.12)

178 Advanced Topics in Applied Mathematics

4.2.5 Derivative

When considering functions of time, it is customary to denote deriva-

tives using “dots” over the variable.

f ,

(n)

. (4.13)

f (t)]=



∞

f (t)e

−pt

=f (t)e

−pt

∞



∞

f (t)e

−pt

f (p) −f (0). (4.14)

Similarly,

f (t)]=pL[

f (t)]−

f (0)

f (p) −pf (0) −

f (0). (4.15)

L[f

(n)

(t)]=p

f −p

n−1

f (0) −p

n−2

f (0) −···−f

n−1

(0). (4.16)

These are important properties that allow us to convert differential

equations into algebraic equations in the transformed form.

4.2.6 Integral

Consider the integral

F(t) =



f (τ )dτ. (4.17)

Using the relation for the derivative,

L[f (t)]=pL[F(t)]−F (0),

L[F(t)]=

f (p). (4.18)