Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 1 pdf

Exercise 32.14
1. Use the cosine transform to solve
y

− a
2
y = 0 on x ≥ 0 with y

(0) = b, y(∞) = 0.
2. Use the cosine transform to show that the Green function for the above with b = 0 is
G(x, ξ) = −
1
2a
e
−a|x−ξ|
−
1
2a
e
−a(x−ξ)
.
Hint, Solution
Exercise 32.15
1. Use the sine transform to solve
y

− a
2
y = 0 on x ≥ 0 with y(0) = b, y(∞) = 0.
2. Try using the Laplace transform on this problem. Why isn’t it as convenient as the Fourier transform?

3. Use the sine transform to show that the Green function for the above with b = 0 is
g(x; ξ) =
1
2a

e
−a(x−ξ)
−
e
−a|x+ξ|

Hint, Solution
Exercise 32.16
1. Find the Green function which solves the e quation
y

+ 2µy

+ (β
2
+ µ
2
)y = δ(x −ξ), µ > 0, β > 0,
in the range −∞ < x < ∞ with boundary conditions y(−∞) = y(∞) = 0.
1574
2. Use this Green’s function to show that the solution of
y

+ 2µy


+ (β
2
+ µ
2
)y = g(x), µ > 0, β > 0, y(−∞) = y(∞) = 0,
with g(±∞) = 0 in the limit as µ → 0 is
y =
1
β

x
−∞
g(ξ) sin[β(x − ξ)]dξ.
You may assume that the interchange of limits is permitted.
Hint, Solution
Exercise 32.17
Using Fourier transforms, find the solution u(x) to the integral equation

∞
−∞
u(ξ)
[(x − ξ)
2
+ a
2
]
dξ =
1
x
2

+ b
2
0 < a < b.
Hint, Solution
Exercise 32.18
The Fourer cosine transform is defin ed by
ˆ
f
c
(ω) =
1
π

∞
0
f(x) cos(ωx) dx.
1. From the Fourier theorem show that the inverse cosine transform is given by
f(x) = 2

∞
0
ˆ
f
c
(ω) cos(ωx) dω.
2. Show that the cosine transform of f

(x) is
−ω
2

ˆ
f
c
(ω) −
f

(0)
π
.
1575
3. Use the cosine transform to solve the following boundary value problem.
y

− a
2
y = 0 on x > 0 with y

(0) = b, y(∞) = 0
Hint, Solution
Exercise 32.19
The Fourier sine transform is d efine d by
ˆ
f
s
(ω) =
1
π

∞
0

f(x) sin(ωx) dx.
1. Show that the inverse sine transform is given by
f(x) = 2

∞
0
ˆ
f
s
(ω) sin(ωx) dω.
2. Show that the sine transform of f

(x) is
ω
π
f(0) − ω
2
ˆ
f
s
(ω).
3. Use this property to solve the equation
y

− a
2
y = 0 on x > 0 with y(0) = b, y(∞) = 0.
4. Try using the Laplace transform on this problem. Why isn’t it as convenient as the Fourier transform?
Hint, Solution
Exercise 32.20

Show that
F[f(x)] =
1
2
(F
c
[f(x) + f(−x)] − ıF
s
[f(x) − f(−x)])
where F, F
c
and F
s
are respectively the Fourier transform, Fourier cosine transform and Fourier sine transform.
Hint, Solution
1576
Exercise 32.21
Find u(x) as the solution to the integral equ ation:

∞
−∞
u(ξ)
(x − ξ)
2
+ a
2
dξ =
1
x
2

+ b
2
, 0 < a < b.
Use Fourier transforms and the inverse transform. Justify the choice of any contours used in the complex plane.
Hint, Solution
1577
32.10 Hints
Hint 32.1
H(x + c) − H(x −c) =

1 for |x| < c,
0 for |x| > c
Hint 32.2
Consider the two cases (ω) < 0 and (ω) > 0, closing the path of integration with a semi-circle in the lower or upper
half plane.
Hint 32.3
Hint 32.4
Hint 32.5
Hint 32.6
Hint 32.7
Hint 32.8
Hint 32.9
1578
Hint 32.10
Hint 32.11
The left side is the convolution of u(x) and
e
−ax
2
.

Hint 32.12
Hint 32.13
Hint 32.14
Hint 32.15
Hint 32.16
Hint 32.17
Hint 32.18
Hint 32.19
Hint 32.20
1579
Hint 32.21
1580
32.11 Solutions
Solution 32.1
F[H(x + c) − H(x − c)] =
1
2π

∞
−∞
(H(x + c) − H(x −c))
e
−ıωx
dx
=
1
2π

c
−c

e
−ıωx
dx
=
1
2π

e
−ıωx
−ıω

c
−c
=
1
2π

e
−ıωc
−ıω
−
e
ıωc
−ıω

F[H(x + c) − H(x − c)] =
sin(cω)
πω
Solution 32.2
F


1
x
2
+ c
2

=
1
2π

∞
−∞
1
x
2
+ c
2
e
−ıωx
dx
=
1
2π

∞
−∞
e
−ıωx
(x − ıc)(x + ıc)

dx
If (ω) < 0 then we close the path of integration with a semi-circle in the upper half plane.
F

1
x
2
+ c
2

=
1
2π
2πi Res

e
−ıωx
(x − ıc)(x + ıc)
, x = ıc

=
1
2c
e
cω
1581
If ω > 0 then we close the path of integration in the lower half plane.
F

1

x
2
+ c
2

= −
1
2π
2πi Res

e
−ıωx
(x − ıc)(x + ıc)
, −ıc

=
1
2c
e
−cω
Thus we have that
F

1
x
2
+ c
2

=

1
2c
e
−c|ω|
, for (c) = 0.
Solution 32.3
F
s
[y

] =
1
π

∞
0
y

sin(ωx) dx
=
1
π

y sin(ωx)

∞
0
−
ω
π


∞
0
y cos(ωx) dx
= −ωˆy
c
(ω)
F
s
[y

] =
1
π

∞
0
y

sin(ωx) dx
=
1
π

y

sin(ωx)

∞
0

−
ω
π

∞
0
y

cos(ωx) dx
= −
ω
π

y cos(ωx)

∞
0
−
ω
2
π

∞
0
y sin(ωx) dx
= −ω
2
ˆy
s
(ω) +

ω
π
y(0).
1582
Solution 32.4
1.
F[f(x −a)] =
1
2π

∞
−∞
f(x − a)
e
−ıωx
dx
=
1
2π

∞
−∞
f(x)
e
−ıω(x+a)
dx
=
e
−ıωa
1

2π

∞
−∞
f(x)
e
−ıωx
dx
F[f(x −a)] =
e
−ıωa
ˆ
f(ω)
2. If a > 0, then
F[f(ax)] =
1
2π

∞
−∞
f(ax)
e
−ıωx
dx
=
1
2π

∞
−∞

f(ξ)
e
−ıωξ/a
1
a
dξ
=
1
a
ˆ
f

ω
a

.
If a < 0, then
F[f(ax)] =
1
2π

∞
−∞
f(ax)
e
−ıωx
dx
=
1
2π


−∞
∞
e
−ıωξ/a
1
a
dξ
= −
1
a
ˆ
f

ω
a

.
Thus
F[f(ax)] =
1
|a|
ˆ
f

ω
a

.
1583

Solution 32.5
F
s
[f(x)g(x)] =
1
π

∞
0
f(x)g(x) sin(ωx) dx
=
1
π

∞
0

2

∞
0
ˆ
f
s
(η) sin(ηx) dη

g(x) sin(ωx) dx
=
2
π


∞
0

∞
0
ˆ
f
s
(η)g(x) sin(ηx) sin(ωx) dx dη
Use the identity, sin a sin b =
1
2
[cos(a − b) − cos(a + b)].
=
1
π

∞
0

∞
0
ˆ
f
s
(η)g(x)

cos((ω − η)x) − cos((ω + η)x)


dx dη
=

∞
0
ˆ
f
s
(η)

1
π

∞
0
g(x) cos((ω − η)x) dx −
1
π

∞
0
g(x) cos((ω + η)x) dx

dη
F
s
[f(x)g(x)] =

∞
0

ˆ
f
s
(η)

G
c
(|ω − η|) − G
c
(ω + η)

dη
1584
Solution 32.6
F
−1
s
[
ˆ
f
s
(ω)G
c
(ω)] = 2

∞
0
ˆ
f
s

(ω)G
c
(ω) sin(ωx) dω
= 2

∞
0

1
π

∞
0
f(ξ) sin(ωξ) dξ

G
c
(ω) sin(ωx) dω
=
2
π

∞
0

∞
0
f(ξ)G
c
(ω) sin(ωξ) sin(ωx) dω dξ

=
1
π

∞
0

∞
0
f(ξ)G
c
(ω)

cos(ω(x − ξ)) − cos(ω(x + ξ))

dω dξ
=
1
2π

∞
0
f(ξ)

2

∞
0
G
c

(ω) cos(ω(x −ξ)) dω − 2

∞
0
G
c
(ω) cos(ω(x + ξ)) dω)

dξ
=
1
2π

∞
0
f(ξ)[g(x − ξ) − g(x + ξ)] dξ
F
−1
s
[
ˆ
f
s
(ω)G
c
(ω)] =
1
2π

∞

0
f(ξ)

g(|x − ξ|) − g(x + ξ)

dξ
Solution 32.7
F
c
[xf(x)] =
1
π

∞
0
xf(x) cos(ωx) dx
=
1
π

∞
0
f(x)
∂
∂ω
(sin(ωx)) dx
=
∂
∂ω
1

π

∞
0
f(x) sin(ωx) dx
=
∂
∂ω
ˆ
f
s
(ω)
1585
F
s
[xf(x)] =
1
π

∞
0
xf(x) sin(ωx) dx
=
1
π

∞
0
f(x)
∂

∂ω
(−cos(ωx)) dx
= −
∂
∂ω
1
π

∞
0
f(x) cos(ωx) dx
= −
∂
∂ω
ˆ
f
c
(ω)
Solution 32.8
y

− y =
e
−2x
, y(0) = 1, y(∞) = 0
We take the Fourier sine transform of the di fferential equation.
−ω
2
ˆy
s

(ω) +
ω
π
y(0) − ˆy
s
(ω) =
2ω/π
ω
2
+ 4
ˆy
s
(ω) = −
ω/π
(ω
2
+ 4)(ω
2
+ 1)
+
ω/π
(ω
2
+ 1)
=
ω/(3π)
ω
2
+ 4
−

ω/(3π)
ω
2
+ 1
+
ω/π
ω
2
+ 1
=
2
3
ω/π
ω
2
+ 1
+
1
3
ω/π
ω
2
+ 4
y =
2
3
e
−x
+
1

3
e
−2x
Solution 32.9
Consider the Fourier sine transform. Let f(x) be an odd function.
F
s
[f(x)] =
1
π

∞
0
f(x) sin(ωx) dx
1586
Extend the integration because the integrand is even.
=
1
2π

∞
−∞
f(x) sin(ωx) dx
Note that

∞
−∞
f(x) cos(ωx) dx = 0 as the integrand is odd.
=
1

2π

∞
−∞
f(x)ı
e
−ıωx
dx
= ıF[f(x)]
F
s
[f(x)] = ıF[f(x)], for odd f(x).
For general f(x), use the odd extension, sign(x)f(|x|) to write the result.
F
s
[f(x)] = ıF[sign(x)f(|x|)]
Now consider the inverse Fourier s ine transform. Let
ˆ
f(ω) be an odd function.
F
−1
s

ˆ
f(ω)

= 2

∞
0

ˆ
f(ω) sin(ωx) dω
Extend the integration because the integrand is even.
=

∞
−∞
ˆ
f(ω) sin(ωx) dω
Note that

∞
−∞
ˆ
f(ω) cos(ωx) dω = 0 as the integrand is odd.
=

∞
−∞
ˆ
f(ω)(−i)
e
ıωx
dω
= −ıF
−1

ˆ
f(ω)


1587
F
−1
s

ˆ
f(ω)

= −ıF
−1

ˆ
f(ω)

, for odd
ˆ
f(ω).
For general
ˆ
f(ω), use the odd extension, sign(ω)
ˆ
f(|ω|) to write the result.
F
−1
s

ˆ
f(ω)

= −ıF

−1

sign(ω)
ˆ
f(|ω|)

Solution 32.10
F
c
[xf(x)] =
1
π

∞
0
xf(x) cos(ωx) dx
=
1
π

∞
0
f(x)
∂
∂ω
sin(ωx) dx
=
∂
∂ω
1

π

∞
0
f(x) sin(ωx) dx
=
∂
∂ω
ˆ
f
s
(ω)
F
s
[xf(x)] =
1
π

∞
0
xf(x) sin(ωx) dx
=
1
π

∞
0
f(x)
∂
∂ω

(−cos(ωx)) dx
= −
∂
∂ω
1
π

∞
0
f(x) cos(ωx) dx
= −
∂
∂ω
ˆ
f
c
(ω)
1588
F
c
[f(cx)] =
1
π

∞
0
f(cx) cos(ωx) dx
=
1
π


∞
0
f(ξ) cos

ω
c
ξ

dξ
c
=
1
c
ˆ
f
c

ω
c

F
s
[f(cx)] =
1
π

∞
0
f(cx) sin(ωx) dx

=
1
π

∞
0
f(ξ) sin

ω
c
ξ

dξ
c
=
1
c
ˆ
f
s

ω
c

Solution 32.11

∞
−∞
u(ξ)
e

−a(x−ξ)
2
dξ =
e
−bx
2
We take the Fourier transform and solve for U(ω).
2πU(ω)F

e
−ax
2

= F

e
−bx
2

2πU(ω)
1
√
4πa
e
−ω
2
/(4a)
=
1
√

4πb
e
−ω
2
/(4b)
U(ω) =
1
2π

a
b
e
−ω
2
(a−b)/(4ab)
Now we take the inverse Fourier transform.
U(ω) =
1
2π

a
b

4πab/(a − b)

4πab/(a − b)
e
−ω
2
(a−b)/(4ab)

1589
u(x) =
a

π(a − b)
e
−abx
2
/(a−b)
Solution 32.12
I =
1
π

∞
0
1
x
e
−cx
sin(ωx) dx
=
1
π

∞
0


∞

c
e
−zx
dz

sin(ωx) dx
=
1
π

∞
c

∞
0
e
−zx
sin(ωx) dx dz
=
1
π

∞
c
ω
z
2
+ ω
2
dz

=
1
π

arctan

z
ω

∞
c
=
1
π

π
2
− arctan

c
ω

=
1
π
arctan

ω
c


Solution 32.13
We consider the differential equation
y

− a
2
y =
e
−a|x|
on the domain −∞ < x < ∞ with boundary conditions y(±∞) = 0. We take the Fourier transform of the differential
equation and solve for ˆy(ω).
−ω
2
ˆy −a
2
ˆy =
a
π(ω
2
+ a
2
)
ˆy(ω) = −
a
π(ω
2
+ a
2
)
2

1590
We take the inverse Fourier transform to find the solution of the differential equation.
y(x) =

∞
−∞
−
a
π(ω
2
+ a
2
)
2
e
ıxω
dω
Note that since ˆy(ω) is a real-valued, even function, y(x) is a real-valued, even function. Thus we only need to evaluate
the integral for positive x. If we replace x by |x| in this expression we will have the solution that is valid for all x.
For x ≥ 0, we evaluate the integral by closing the path of integration in the upper half plane and using the Residue
Theorem and Jordan’s Lemma.
y(x) = −
a
π

∞
−∞
1
(ω − ıa)
2

(ω + ıa)
2
e
ıxω
dω
= −ı2π
a
π
Res

1
(ω − ıa)
2
(ω + ıa)
2
e
ıxω
, ω = ıa

= −ı2a lim
ω→ıa
d
dω

e
ıxω
(ω + ıa)
2

= −ı2a lim

ω→ıa

ıx
e
ıxω
(ω + ıa)
2
−
2
e
ıxω
(ω + ıa)
3

= −ı2a

ıx
e
−ax
−4a
2
−
2
e
−ax
−ı8a
3

= −
(1 + ax)

e
−ax
2a
2
The solution of the differential equation is
y(x) = −
1
2a
2
(1 + a|x|)
e
−a|x|
.
1591
Solution 32.14
1. We take the Fourier cosine transform of the diffe rential equation.
−ω
2
ˆy(ω) −
b
π
− a
2
ˆy(ω) = 0
ˆy(ω) = −
b
π(ω
2
+ a
2

)
Now we take the inverse Fourier cosine transform. We use the fact that ˆy(ω) is an even function.
y(x) = F
−1
c

−
b
π(ω
2
+ a
2
)

= F
−1

−
b
π(ω
2
+ a
2
)

= −
b
π
ı2π Res


1
ω
2
+ a
2
e
ıωx
, ω = ıa

= −ı2b lim
ω→ıa

e
ıωx
ω + ıa

, for x ≥ 0
y(x) = −
b
a
e
−ax
2. The Green function problem is
G

− a
2
G = δ(x − ξ) on x, ξ > 0, G

(0; ξ) = 0, G(∞; ξ) = 0.

We take the Fourier cosine transform and solve for
ˆ
G(ω; ξ).
−ω
2
ˆ
G − a
2
ˆ
G = F
c
[δ(x − ξ)]
ˆ
G(ω; ξ) = −
1
ω
2
+ a
2
F
c
[δ(x − ξ)]
1592
We express the right side as a product of Fourier cosine transforms.
ˆ
G(ω; ξ) = −
π
a
F
c

[
e
−ax
]F
c
[δ(x − ξ)]
Now we can apply the Fourier cosine convolution theorem.
F
−1
c
[F
c
[f(x)]F
c
[g(x)]] =
1
2π

∞
0
f(t)

g(|x − t|) + g(x + t)

dt
G(x; ξ) = −
π
a
1
2π


∞
0
δ(t − ξ)

e
−a|x−t|
+
e
−a(x+t)

dt
G(x; ξ) = −
1
2a

e
−a|x−ξ|
+
e
−a(x+ξ)

Solution 32.15
1. We take the Fourier sine transform of the di fferential equation.
−ω
2
ˆy(ω) +
bω
π
− a

2
ˆy(ω) = 0
ˆy(ω) =
bω
π(ω
2
+ a
2
)
Now we take the inverse Fourier sine transform. We use the fact that ˆy(ω) is an odd function.
y(x) = F
−1
s

bω
π(ω
2
+ a
2
)

= −ıF
−1

bω
π(ω
2
+ a
2
)


= −ı
b
π
ı2π Res

ω
ω
2
+ a
2
e
ıωx
, ω = ıa

= 2b lim
ω→ıa

ω
e
ıωx
ω + ıa

= b
e
−ax
for x ≥ 0
1593
y(x) = b
e

−ax
2. Now we solve the differential equation with the Laplace transform.
y

− a
2
y = 0
s
2
ˆy(s) −sy(0) − y

(0) − a
2
ˆy(s) = 0
We don’t know the value of y

(0), so we treat it as an unknown constant.
ˆy(s) =
bs + y

(0)
s
2
− a
2
y(x) = b cosh(ax) +
y

(0)
a

sinh(ax)
In order to satisfy the boundary condition at infinity we must choose y

(0) = −ab.
y(x) = b
e
−ax
We see that solving the differential equation with the Laplace transform is not as convenient, b e cause the boundary
condition at infinity is not automatically satisfied. We had to find a value of y

(0) so that y(∞) = 0.
3. The Green function problem is
G

− a
2
G = δ(x − ξ) on x, ξ > 0, G(0; ξ) = 0, G(∞; ξ) = 0.
We take the Fourier sine transform and solve for
ˆ
G(ω; ξ).
−ω
2
ˆ
G − a
2
ˆ
G = F
s
[δ(x − ξ)]
ˆ

G(ω; ξ) = −
1
ω
2
+ a
2
F
s
[δ(x − ξ)]
1594
We write the right side as a product of Fourier cosine transforms and sine transforms.
ˆ
G(ω; ξ) = −
π
a
F
c
[
e
−ax
]F
s
[δ(x − ξ)]
Now we can apply the Fourier sine c onvolution theorem.
F
−1
s
[F
s
[f(x)]F

c
[g(x)]] =
1
2π

∞
0
f(t)

g(|x − t|) −g(x + t)

dt
G(x; ξ) = −
π
a
1
2π

∞
0
δ(t − ξ)

e
−a|x−t|
−
e
−a(x+t)

dt
G(x; ξ) =

1
2a

e
−a(x−ξ)
−
e
−a|x+ξ|

Solution 32.16
1. We take the Fourier transform of the differential equ ation, solve for
ˆ
G and then invert.
G

+ 2µG

+

β
2
+ µ
2

G = δ(x − ξ)
−ω
2
ˆ
G + ı2µω
ˆ

G +

β
2
+ µ
2

ˆ
G =
e
−ıωξ
2π
ˆ
G = −
e
−ıωξ
2π (ω
2
− ı2µω − β
2
− µ
2
)
G =

∞
−∞
−
e
−ıωξ

e
ıωx
2π(ω
2
− ı2µω − β
2
− µ
2
)
dω
G = −
1
2π

∞
−∞
e
ıω(x−ξ)
(ω + β −ıµ)(ω − β −ıµ)
dω
For x > ξ we close the path of integration in the upp er half plane and use the Residue theorem. There are two
simple poles in the upper half plane. For x < ξ we close the path of integration in the lower half plane. Since
1595
the integrand is analytic there, the integral is zero. G(x; ξ) = 0 for x < ξ. For x > ξ we have
G(x; ξ) = −
1
2π
ı2π

Res


e
ıω(x−ξ)
(ω + β −ıµ)(ω − β −ıµ)
, ω = −β + ıµ

+ Res

e
ıω(x−ξ)
(ω + β −ıµ)(ω − β −ıµ)
, ω = −β − ıµ


G(x; ξ) = −ı

e
ı(−β+ıµ)(x−ξ)
−2β
+
e
ı(β+ıµ)(x−ξ)
2β

G(x; ξ) =
1
β
e
−µ(x−ξ)
sin(β(x − ξ)).

Thus the Green function is
G(x; ξ) =
1
β
e
−µ(x−ξ)
sin(β(x − ξ))H(x − ξ).
2. We use the Green function to find the solution of the inhomogeneous equation.
y

+ 2µy

+

β
2
+ µ
2

y = g(x), y(−∞) = y(∞) = 0
y(x) =

∞
−∞
g(ξ)G(x; ξ) dξ
y(x) =

∞
−∞
g(ξ)

1
β
e
−µ(x−ξ)
sin(β(x − ξ))H(x − ξ) dξ
y(x) =
1
β

x
−∞
g(ξ)
e
−µ(x−ξ)
sin(β(x − ξ)) dξ
We take the limit µ → 0.
y =
1
β

x
−∞
g(ξ) sin(β(x − ξ)) dξ
1596
Solution 32.17
First we consider the Fourier transform of f(x) = 1/(x
2
+ c
2
) where (c) > 0.

ˆ
f(ω) = F

1
x
2
+ c
2

=
1
2π

∞
−∞
1
x
2
+ c
2
e
−ıωx
dx
=
1
2π

∞
−∞
e

−ıωx
(x − ıc)(x + ıc)
dx
If ω < 0 then we close the path of integration with a semi-circle in the upper half plane.
ˆ
f(ω) =
1
2π
2πi Res

e
−ıωx
(x − ıc)(x + ıc)
, x = ıc

=
e
cω
2c
, for ω < 0
Note that f(x) = 1/(x
2
+ c
2
) is an even function of x so that
ˆ
f(ω) is an even function of ω. If
ˆ
f(ω) = g(ω) for ω < 0
then f(ω) = g(−|ω|) for all ω. Thu s

F

1
x
2
+ c
2

=
1
2c
e
−c|ω|
.
Now we consider the integral equation

∞
−∞
u(ξ)
[(x − ξ)
2
+ a
2
]
dξ =
1
x
2
+ b
2

0 < a < b.
1597