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MATH-6640 Spring 2005 COMPLEX VARIABLES AND INTEGRAL TRANSFORMS WITH APPLICATIONS
Assignment 4 Solutions
- Use the properties of the Laplace transform, and the short table of transforms presented in class, to find the following: (a) Transform of Si(t) =
∫ (^) t
0
sin u u du,
where Si(t) is the Sine Integral function which occurs in the study of optics.
Use L
f (t) t
s
F (u) du
to get L
sin t t
s
1 + u^2 du = π 2 − tan−^1 s.
Then, use L
∫ (^) t
0
f (u) du = F (s) s to get L
∫ (^) t
0
sin u u du =^1 s
{ (^) π 2 − tan−^1 s
(b) The Laplace inverse of 1 (s + ω 1 )^2 + ω 22
Recall that L−^1 s^2 + ω 22 = sin ω 2 t.
Then, L−^1
(s + ω 1 )^2 + ω^22 = e−ω^1 t^ sin ω 2 t.
(c) The Laplace inverse of e−^5 s (s − 3)^3
Recall that L−^1
s^3
t^2. Therefore, L−^1
(s − 3)^3
e^3 tt^2 = f (t), say.
Then, L−^1 e−^5 s (s − 3)^3 = f (t − 5)H(t − 5) =
e3(t−5)(t − 5)^2 H(t − 5).
(d) The Laplace inverse of s^2 s^2 + 1
if it exists. (Why should there be a question?) Don’t be hasty.
Since s^2 s^2 + 1 = 1^ −^
s^2 + 1 , L−^1
s^2 s^2 + 1
= δ(t) − sin t.
- Consider the differential equation
d^3 y dt^3
- ω^3 y = f (t), ω > 0 , y(0) = y′(0) = y′′(0) = 0.
(a) Show that the Laplace transform Y (s) of the solution satisfies
Y (s) = F (s) s^3 + ω^3
where F (s) is the transform of f (t).
Laplace transformation of the ODE leads to s^3 Y (s) + ω^3 Y (s) = F (s) whence Y (s) = F (s) s^3 + ω^3. (b) Deduce that the inverse transform of
Z(s) =
s^3 + ω^3 is given by z(t) = e−ωt 3 ω^2
3 ω^2 exp(ωt/2) cos
ωt − π 3
How will the above result allow you to find a representation for y(t)?
By using partial fractions we may write 1 s^3 + ω^3
3 ω^2
s + ω
√^1
3 ω^2
√ 3 ω 2 (s − ω/2)^2 + 3ω^2 / 4 −
3 ω^2
s − ω/ 2 (s − ω/2)^2 + 3ω^2 / 4 Then,
L−^1
s^3 + ω^3
3 ω^2 e
−ωt (^) + eωt/ 2
[
√^1
3 ω^2
sin
3 ωt 2 −^
3 ω^2 cos
3 ωt 2
]
3 ω^2 e−ωt^ − 2 3 ω^2 eωt/^2 cos
3 ωt 2 − π 3
We can now compute y by using convolution.
Similarly the residue at s = −iω is
R− =
2 πi s→−limiω(s^ +^ iω)^
est^ ln s s^2 + ω^2 =^
2 πi s→−limiω
est^ ln s s − iω =^
2 πi
e−iωt[ln ω − iπ/2] (− 2 iω).
We then get
2 πi(R+ + R−) = π 2 ω cos ωt + ln^ ω ω sin ωt.
Re s
Im s
L
L
1
2
i
-i
ω
ω
On the keyhole contour, the contribution from the small circle around the branch point is zero (indicate why). On the straight segment L 1 of the contour above the cut,
s = reiπ^ = −r, ln s = ln r + iπ, and r goes from 0 to ∞.
On the straight segment L 2 below the cut,
s = re−iπ^ = −r, ln s = ln r − iπ, and r goes from ∞ to 0.
The sum of the integrals from the two segments is ∫
L 1
L 2
2 πi
est^ ln s s^2 + ω^2 ds =
2 πi
0
e−rt[ln r + iπ] ω^2 + r^2 (−dr)
2 πi
0
e−rt[ln r − iπ] ω^2 + r^2 (−dr)
=
2 πi
0
e−rt[− 2 iπ] ω^2 + r^2 dr = −
0
e−rt ω^2 + r^2 dr
The addition of all the terms leads to
π 2 ω cos ωt + ln ω ω sin ωt −
0
e−rt ω^2 + r^2 dr.
