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

We take the Fourier cosine transform of the differential equation.. Now we solve the differential equation with the Laplace transform... We use the Green function to find the solution of

Exercise 32.15

1 Use the sine transform to solve

y00− a2y = 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

2 Use this Green’s function to show that the solution of

y00+ 2µy0+ (β2+ µ2)y = g(x), µ > 0, β > 0, y(−∞) = y(∞) = 0,with g(±∞) = 0 in the limit as µ → 0 is

y = 1β

f (x) cos(ωx) dx

1 From the Fourier theorem show that the inverse cosine transform is given by

f (x) = 2

Z ∞ 0

3 Use the cosine transform to solve the following boundary value problem.

y00− a2y = 0 on x > 0 with y0(0) = b, y(∞) = 0Hint, Solution

f (x) sin(ωx) dx

1 Show that the inverse sine transform is given by

f (x) = 2

Z ∞ 0

3 Use this property to solve the equation

y00− a2y = 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(Fc[f (x) + f (−x)] − ıFs[f (x) − f (−x)])where F , Fc and Fs are respectively the Fourier transform, Fourier cosine transform and Fourier sine transform.Hint, Solution

x2+ b2, 0 < a < b.

Use Fourier transforms and the inverse transform Justify the choice of any contours used in the complex plane.Hint, Solution

Consider the two cases <(ω) < 0 and <(ω) > 0, closing the path of integration with a semi-circle in the lower or upperhalf plane.

Hint 32.21

Z c

−c

e−ıωx dx

= 12π

F

1

x2 + c2



= 12π

x2+ c2



= 12π2πi Res



e−ıωx(x − ıc)(x + ıc), x = ıc



= 12ce

cω

If ω > 0 then we close the path of integration in the lower half plane.

F

1

x2 + c2



= − 12π2πi Res

 e−ıωx

(x − ıc)(x + ıc), −ıc



= 12ce

−cω

Thus we have that

F

1

x2+ c2



= 12ce

y0sin(ωx) dx

= 1π

h

y sin(ωx)i

∞ 0

−ωπ

Z ∞ 0

y00sin(ωx) dx

= 1π

h

y0sin(ωx)i

∞ 0

−ωπ

Z ∞ 0

y0cos(ωx) dx

= −ωπ

y sin(ωx) dx

= −ω2ˆs(ω) + ω

πy(0).



ˆωa

.Thus

F [f (ax)] = 1

|a| ˆ

ωa



Solution 32.5

Fs[f (x)g(x)] = 1

π

Z ∞ 0

f (x)g(x) sin(ωx) dx

= 1π

Z ∞ 0

2

Z ∞ 0

ˆs(η) sin(ηx) dη

g(x) sin(ωx) dx

= 2π

Z ∞ 0

Z ∞ 0

ˆ

s(η)g(x) sin(ηx) sin(ωx) dx dη

Use the identity, sin a sin b = 12[cos(a − b) − cos(a + b)]

= 1π

Z ∞ 0

Z ∞ 0

ˆs(η)g(x)hcos((ω − η)x) − cos((ω + η)x)idx dη

=

Z ∞ 0

ˆs(η) 1π

Z ∞ 0

g(x) cos((ω − η)x) dx − 1

π

Z ∞ 0

g(x) cos((ω + η)x) dx

dη

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

Z ∞ 0

ˆs(η)Gc(|ω − η|) − Gc(ω + η) dη

Solution 32.6

Fs−1[ ˆfs(ω)Gc(ω)] = 2

Z ∞ 0

ˆ

s(ω)Gc(ω) sin(ωx) dω

= 2

Z ∞ 0

 1π

Z ∞ 0

f (ξ) sin(ωξ) dξ



Gc(ω) sin(ωx) dω

= 2π

Z ∞ 0

Z ∞ 0

f (ξ)Gc(ω) sin(ωξ) sin(ωx) dω dξ

= 1π

Z ∞ 0

Z ∞ 0

f (ξ)Gc(ω)hcos(ω(x − ξ)) − cos(ω(x + ξ))idω dξ

= 12π

Z ∞ 0

f (ξ)

2

Z ∞ 0

Gc(ω) cos(ω(x − ξ)) dω − 2

Z ∞ 0

Gc(ω) cos(ω(x + ξ)) dω)

dξ

= 12π

Z ∞ 0

f (ξ)[g(x − ξ) − g(x + ξ)] dξ

Fs−1[ ˆfs(ω)Gc(ω)] = 1

2π

Z ∞ 0

xf (x) cos(ωx) dx

= 1π

Z ∞ 0

Z ∞ 0

f (x) sin(ωx) dx

= ∂

∂ω

ˆs(ω)

Fs[xf (x)] = 1

π

Z ∞ 0

xf (x) sin(ωx) dx

= 1π

Z ∞ 0

Z ∞ 0

ω/π

ω2+ 1 +

13

f (x) sin(ωx) dx

Extend the integration because the integrand is even.

= 12π

Z ∞

−∞

f (x) sin(ωx) dxNote that R−∞∞ f (x) cos(ωx) dx = 0 as the integrand is odd

= 12π

ˆ

f (ω) sin(ωx) dωExtend the integration because the integrand is even

Fs−1h ˆf (ω) = −ıF−1h ˆf (ω) , for odd ˆf (ω).For general ˆf (ω), use the odd extension, sign(ω) ˆf (|ω|) to write the result.

xf (x) cos(ωx) dx

= 1π

Z ∞ 0

Z ∞ 0

xf (x) sin(ωx) dx

= 1π

Z ∞ 0

Z ∞ 0

f (x) cos(ωx) dx

= − ∂

∂ω

ˆc(ω)

Fc[f (cx)] = 1

π

Z ∞ 0

f (cx) cos(ωx) dx

= 1π

Z ∞ 0

f (ξ) cosω

cξ

 dξc

= 1c

ˆ

c

ωc



Fs[f (cx)] = 1

π

Z ∞ 0

f (cx) sin(ωx) dx

= 1π

Z ∞ 0

f (ξ) sinω

cξ

 dξc

= 1c

ˆ

s

ωc

Solution 32.11

p4πab/(a − b)p4πab/(a − b)e

−ω 2 (a−b)/(4ab)

Z ∞ 0

Z ∞ 0

Z ∞ c

e−zx dz

sin(ωx) dx

= 1π

Z ∞ c

Z ∞ 0

e−zxsin(ωx) dx dz

= 1π

Z ∞ c

ω

z2+ ω2 dz

= 1π

harctanz

ω

i∞ c

= 1π

y(ω) = − a

π(ω2+ a2)2

We take the inverse Fourier transform to find the solution of the differential equation.

= −ı2πa

π Res



1(ω − ıa)2(ω + ıa)2 eıxω, ω = ıa



= −ı2a lim

ω→ıa

ddω

Solution 32.14

1 We take the Fourier cosine transform of the differential equation

−ω2y(ω) −ˆ b

π − a2y(ω) = 0ˆˆ

y(ω) = − b

π(ω2+ a2)Now we take the inverse Fourier cosine transform We use the fact that ˆy(ω) is an even function

ω2+ a2Fc[δ(x − ξ)]

We express the right side as a product of Fourier cosine transforms.

ˆG(ω; ξ) = −π

f (t) g(|x − t|) + g(x + t)
dt

G(x; ξ) = −π

a

12π

Z ∞ 0

1 We take the Fourier sine transform of the differential equation

−ω2y(ω) +ˆ bω

π − a2y(ω) = 0ˆˆ

π(ω2+ a2)Now we take the inverse Fourier sine transform We use the fact that ˆy(ω) is an odd function

y(x) = Fs−1

bωπ(ω2+ a2)



= −ıF−1

bωπ(ω2+ a2)



= −ıb

πı2π Res

ω

y(x) = b e−ax

2 Now we solve the differential equation with the Laplace transform

y00− a2y = 0

s2y(s) − sy(0) − yˆ 0(0) − a2y(s) = 0ˆ

We don’t know the value of y0(0), so we treat it as an unknown constant

ˆy(s) = bs + y

ω2+ a2Fs[δ(x − ξ)]

We write the right side as a product of Fourier cosine transforms and sine transforms.

ˆG(ω; ξ) = −π

f (t) g(|x − t|) − g(x + t)
dt

G(x; ξ) = −π

a

12π

Z ∞ 0

the integrand is analytic there, the integral is zero G(x; ξ) = 0 for x < ξ For x > ξ we have

2 We use the Green function to find the solution of the inhomogeneous equation

y00+ 2µy0+ β2+ µ2
y = g(x), y(−∞) = y(∞) = 0

y(x) =

Z ∞

−∞

g(ξ)G(x; ξ) dξy(x) =

Z x

−∞

g(ξ) sin(β(x − ξ)) dξ

x2+ c2



= 12π

Z ∞

−∞

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

x2+ c2



= 12ce

1

x2+ b2 0 < a < b

We take the Fourier transform, utilizing the convolution theorem.

u(ω) = a e

−(b−a)|ω|

2πbu(x) = a

ˆc(ω) cos(ωx) dω

Fc[y00] = 1

π

Z ∞ 0

y cos(ωx) dx

Fc[y00] = −ω2ˆc(ω) − y

0(0)π

3 We take the Fourier cosine transform of the differential equation

−ω2y(ω) −ˆ b

π − a2y(ω) = 0ˆˆ

y(ω) = − b

π(ω2+ a2)Now we take the inverse Fourier cosine transform We use the fact that ˆy(ω) is an even function

y(x) = −b

ae

−ax

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) = 2

Z ∞ 0

ˆs(ω) sin(ωx) dω, ˆs(ω) = 1

π

Z ∞ 0

f (x) sin(ωx) dx

Fs[y00] = 1

π

Z ∞ 0

y00sin(ωx) dx

= 1π

h

y0sin(ωx)i

∞ 0

−ωπ

Z ∞ 0

y0cos(ωx) dx

= −ωπ

π(ω2+ a2)Now we take the inverse Fourier sine transform We use the fact that ˆy(ω) is an odd function

y(x) = Fs−1

bωπ(ω2+ a2)



= −ıF−1

bωπ(ω2+ a2)



= −ıb

πı2π Res

ω

4 Now we solve the differential equation with the Laplace transform.

y00− a2y = 0

s2y(s) − sy(0) − yˆ 0(0) − a2y(s) = 0ˆ

We don’t know the value of y0(0), so we treat it as an unknown constant

ˆy(s) = bs + y

f (x) cos(ωx) dx,

F [f (x)]s= 1

π

Z ∞ 0

f (−x) cos(ωx) dx − ı

Z ∞ 0

f (x) sin(ωx) dx + ı

Z ∞ 0

f (−x) sin(ωx) dx1

x2+ a2



= F

1

x2+ b2



We find the Fourier transform of f (x) = 1

x 2 +c 2 Note that since f (x) is an even, real-valued function, ˆf (ω) is aneven, real-valued function

F

1

x2+ c2



= 12π

Z ∞

−∞

1

x2+ c2 e−ıωx dx

For x > 0 we close the path of integration in the upper half plane and apply Jordan’s Lemma to evaluate the integral

in terms of the residues

= 12πı2π Res



e−ıωx(x − ıc)(x + ıc), x = ıc

−cω

Since ˆf (ω) is an even function, we have

F

1

x2+ c2



= 12ce

−c|ω|

.Our equation for ˆu(ω) becomes,

2π ˆu(ω) 1

2ae

−a|ω|

= 12be

−b|ω|

ˆu(ω) = a

We would like to extend the factorial function so it is defined for all complex numbers.

Consider the function Γ(z) defined by Euler’s formula

Γ(z) =

Z ∞ 0

e−ttz−1dt

(Here we take the principal value of tz−1.) The integral converges for <(z) > 0 If <(z) ≤ 0 then the integrand will

be at least as singular as 1/t at t = 0 and thus the integral will diverge

Difference Equation Using integration by parts,

Γ(z + 1) =

Z ∞ 0

e−ttzdt

=h− e−ttzi

∞ 0

−

Z ∞ 0

− e−tztz−1dt

Since <(z) > 0 the first term vanishes

= z

Z ∞ 0

e−ttz−1dt

= zΓ(z)Thus Γ(z) satisfies the difference equation

· · · = · · ·Γ(n + 1) = n!

Thus the Gamma function, Γ(z), extends the factorial function to all complex z in the right half-plane For negative, integral n we have

non-Γ(n + 1) = n!

Analyticity The derivative of Γ(z) is

Γ0(z) =

Z ∞ 0

on the negative real axis

The integral in Hankel’s formula converges for all complex z For non-positive, integral z the integral does notvanish Thus because of the sine term the Gamma function has simple poles at z = 0, −1, −2, For positive,integral z, the integrand is entire and thus the integral vanishes Using L’Hospital’s rule you can show that the points,

z = 1, 2, 3, are removable singularities and the Gamma function is analytic at these points Since the only zeroes ofsin(πz) occur for integral z, Γ(z) is analytic in the entire plane except for the points, z = 0, −1, −2,

Figure 33.1: The Hankel Contour.

Difference Equation Using integration by parts we can derive the difference equation from Hankel’s formula

Both the numerator and denominator vanish Using L’Hospital’s rule,

= lim

z→1

R

Cettz−1log t dtı2π cos(πz)

=

R

Cetlog t dtı2πLet Cr be the circle of radius r starting at −π radians and going to π radians

= 1ı2π

Z −∞

0

et dt

= 1

Thus we obtain the same value as with Euler’s formula It can be shown that Hankel’s formula is the analytic continuation

of the Gamma function into the left half-plane

n

Substituting this into Euler’s formula,

Γ(z) =

Z ∞ 0

e−ttz−1dt

= lim

n→∞

Z n 0



1 − tn

(1 − τ )nnz−1τz−1n dτ

= lim

n→∞nz

Z 1 0

(1 − τ )nτz−1dτ

Let n be an integer Using integration by parts we can evaluate the integral

Z 1 0

(1 − τ )nτz−1dτ = (1 − τ )nτz

z

1 0

−

Z 1 0

−n(1 − τ )n−1τz

z dτ

= nz

Z 1 0

(1 − τ )n−1τzdτ

= n(n − 1)z(z + 1)

Z 1 0

(1 − τ )n−2τz+1dτ

= n(n − 1) · · · (1)z(z + 1) · · · (z + n − 1)

Z 1 0

τz+n−1dτ

= n(n − 1) · · · (1)z(z + 1) · · · (z + n − 1)

 τz+n

z + n

1 0

z(z + 1) · · · (z + n)

Thus we have that

z

= 1

z n→∞lim

1(1 + z)(1 + z/2) · · · (1 + z/n)n

z

= 1

z n→∞lim

1(1 + z)(1 + z/2) · · · (1 + z/n)

2z3z· · · (n + 1)z

1z2z· · · nz

= 1z

∞

Y

n=1

1

z



1 + zn

−1

We derived this formula from Euler’s formula which is valid only in the left half-plane However, the product formula

is valid for all z except z = 0, −1, −2,

33.4 Weierstrass’ Formula

The Euler-Mascheroni Constant Before deriving Weierstrass’ product formula for the Gamma function we willneed to define the Euler-Mascheroni constant



Weierstrass’ formula for the Gamma function is then

1Γ(z) = z e



e−z/ni

Since the product is uniformly convergent, 1/Γ(z) is an entire function Since 1/Γ(z) has no singularities, we seethat Γ(z) has no zeros