Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 9 docx

We use the convolution theorem to find the inverse Laplace transform of ˆy(s).
y(t) =

t
0
1
2
sin(2τ) cos(t − τ) dτ + cos t
=
1
4

t
0
sin(t + τ) + sin(3τ − t) dτ + cos t
=
1
4

−cos(t + τ) −
1
3
cos(3τ − t)

t
0
+ cos t
=
1
4


−cos(2t) + cos t −
1
3
cos(2t) +
1
3
cos(t)

+ cos t
= −
1
3
cos(2t) +
4
3
cos(t)
Alternatively, we can find the inverse Laplace transform of ˆy(s) by first finding its partial fraction expansion.
ˆy(s) =
s/3
s
2
+ 1
−
s/3
s
2
+ 4
+
s

s
2
+ 1
= −
s/3
s
2
+ 4
+
4s/3
s
2
+ 1
y(t) = −
1
3
cos(2t) +
4
3
cos(t)
Example 31.4.3 Consider the initial value problem
y

+ 5y

+ 2y = 0, y(0) = 1, y

(0) = 2.
Without taking a Laplace transform, we know that since
y(t) = 1 + 2t + O(t

2
)
the Laplace transform has the behavior
ˆy(s) ∼
1
s
+
2
s
2
+ O(s
−3
), as s → +∞.
1494
31.5 Systems of Constant Coefficient Differential Equations
The Laplace transform can be used to transform a system of constant coefficient differential equations into a system
of algebraic e quations. This should not be surprising, as a system of differential equations can be written as a single
differential equation, and vice versa.
Example 31.5.1 Consider the set of differential equations
y

1
= y
2
y

2
= y
3
y


3
= −y
3
− y
2
− y
1
+ t
3
with the initial conditions
y
1
(0) = y
2
(0) = y
3
(0) = 0.
We take the Laplace transform of this system.
sˆy
1
− y
1
(0) = ˆy
2
sˆy
2
− y
2
(0) = ˆy

3
sˆy
3
− y
3
(0) = −ˆy
3
− ˆy
2
− ˆy
1
+
6
s
4
The first two equations can be written as
ˆy
1
=
ˆy
3
s
2
ˆy
2
=
ˆy
3
s
.

1495
We substitute this into the third equation.
sˆy
3
= −ˆy
3
−
ˆy
3
s
−
ˆy
3
s
2
+
6
s
4
(s
3
+ s
2
+ s + 1)ˆy
3
=
6
s
2
ˆy

3
=
6
s
2
(s
3
+ s
2
+ s + 1)
.
We solve for ˆy
1
.
ˆy
1
=
6
s
4
(s
3
+ s
2
+ s + 1)
ˆy
1
=
1
s

4
−
1
s
3
+
1
2(s + 1)
+
1 − s
2(s
2
+ 1)
We then take the inverse Laplace transform of ˆy
1
.
y
1
=
t
3
6
−
t
2
2
+
1
2
e

−t
+
1
2
sin t −
1
2
cos t.
We can find y
2
and y
3
by differentiating the expression for y
1
.
y
2
=
t
2
2
− t −
1
2
e
−t
+
1
2
cos t +

1
2
sin t
y
3
= t − 1 +
1
2
e
−t
−
1
2
sin t +
1
2
cos t
1496
31.6 Exercises
Exercise 31.1
Find the Laplace transform of the following functions:
1. f(t) =
e
at
2. f(t) = sin(at)
3. f(t) = cos(at)
4. f(t) = sinh(at)
5. f(t) = cosh(at)
6. f(t) =
sin(at)

t
7. f(t) =

t
0
sin(au)
u
du
8. f(t) =

1, 0 ≤ t < π
0, π ≤ t < 2π
and f(t + 2π) = f(t) for t > 0. That is, f(t) is periodic for t > 0.
Hint, Solution
Exercise 31.2
Show that L[af(t) + bg(t)] = aL[f(t)] + bL[g(t)].
Hint, Solution
Exercise 31.3
Show that if f(t) is of exponential order α,
L[
e
ct
f(t)] =
ˆ
f(s − c) for s > c + α.
1497
Hint, Solution
Exercise 31.4
Show that
L[t

n
f(t)] = (−1)
n
d
n
ds
n
[
ˆ
f(s)] for n = 1, 2, . . .
Hint, Solution
Exercise 31.5
Show that if

β
0
f(t)
t
dt exists for positive β then
L

f(t)
t

=

∞
s
ˆ
f(σ) dσ.

Hint, Solution
Exercise 31.6
Show that
L


t
0
f(τ) dτ

=
ˆ
f(s)
s
.
Hint, Solution
Exercise 31.7
Show that if f(t) is periodic with period T then
L[f(t)] =

T
0
e
−st
f(t) dt
1 −
e
−sT
.
Hint, Solution

1498
Exercise 31.8
The function f(t) t ≥ 0, is periodic with period 2T ; i.e. f(t + 2T) ≡ f(t), and is also odd with period T ; i.e.
f(t + T ) = −f(t). Further,

T
0
f(t)
e
−st
dt = ˆg(s).
Show that the Laplace transform of f(t) is
ˆ
f(s) = ˆg(s)/(1 +
e
−sT
). Find f(t) such that
ˆ
f(s) = s
−1
tanh(sT/2).
Hint, Solution
Exercise 31.9
Find the Laplace transform of t
ν
, ν > −1 by two methods.
1. Assume that s is complex-valued. Make the change of variables z = st and use integration in the complex plane.
2. Show that the Laplace transform of t
ν
is an analytic function for (s) > 0. Assume that s is real-valued. Make

the change of variables x = st and evaluate the integral. Then use analytic continuation to extend the result to
complex-valued s.
Hint, Solution
Exercise 31.10 (mathematica/ode/laplace/laplace.nb)
Show that the Laplace transform of f(t) = ln t is
ˆ
f(s) = −
Log s
s
−
γ
s
, where γ = −

∞
0
e
−t
ln t dt.
[ γ = 0.5772 . . . is known as Euler’s constant.]
Hint, Solution
Exercise 31.11
Find the Laplace transform of t
ν
ln t. Write the answer in terms of the digamma function, ψ(ν) = Γ

(ν)/Γ(ν). What
is the answer for ν = 0?
Hint, Solution
1499

Exercise 31.12
Find the inverse Laplace transform of
ˆ
f(s) =
1
s
3
− 2s
2
+ s − 2
with the following methods.
1. Expand
ˆ
f(s) using partial fractions and then use the table of Laplace transforms.
2. Factor the denominator into (s − 2)(s
2
+ 1) and then use the convolution theorem.
3. Use Result 31.2.1.
Hint, Solution
Exercise 31.13
Solve the differential equation
y

+ y

+ y = sin t, y(0) = y

(0) = 0, 0 <   1
using the Laplace transform. This equation represents a weakly damped, driven, line ar oscillator.
Hint, Solution

Exercise 31.14
Solve the problem,
y

− ty

+ y = 0, y(0) = 0, y

(0) = 1,
with the Laplace transform.
Hint, Solution
Exercise 31.15
Prove the following relation between the inverse Laplace transform and the inverse Fourier transform,
L
−1
[
ˆ
f(s)] =
1
2π
e
ct
F
−1
[
ˆ
f(c + ıω)],
1500
where c is to the right of the singularities of
ˆ

f(s).
Hint, Solution
Exercise 31.16 (mathematica/ode/laplace/laplace.nb)
Show by evaluating the Laplace inversion integral that if
ˆ
f(s) =

π
s

1/2
e
−2(as)
1/2
, s
1/2
=
√
s for s > 0,
then f (t) =
e
−a/t
/
√
t. Hint: cut the s-plane along the negative real axis and deform the contour onto the cut.
Remember that

∞
0
e

−ax
2
cos(bx) dx =

π/4a
e
−b
2
/4a
.
Hint, Solution
Exercise 31.17 (mathematica/ode/laplace/laplace.nb)
Use Laplace transforms to solve the initial value problem
d
4
y
dt
4
− y = t, y(0) = y

(0) = y

(0) = y

(0) = 0.
Hint, Solution
Exercise 31.18 (mathematica/ode/laplace/laplace.nb)
Solve, by Laplace transforms,
dy
dt

= sin t +

t
0
y(τ) cos(t − τ) dτ, y(0) = 0.
Hint, Solution
Exercise 31.19 (mathematica/ode/laplace/laplace.nb)
Suppose u(t) satisfies the difference-differential equation
du
dt
+ u(t) − u(t − 1) = 0, t ≥ 0,
1501
and the ‘initial condi tion’ u(t) = u
0
(t), −1 ≤ t ≤ 0, where u
0
(t) is given. Show that the Laplace transform ˆu(s) of
u(t) satisfies
ˆu(s) =
u
0
(0)
1 + s −
e
−s
+
e
−s
1 + s −
e

−s

0
−1
e
−st
u
0
(t) dt.
Find u(t), t ≥ 0, when u
0
(t) = 1. Check the result.
Hint, Solution
Exercise 31.20
Let the function f(t) be defined by
f(t) =

1 0 ≤ t < π
0 π ≤ t < 2π,
and for all positive values of t so that f(t + 2π) = f(t). That is, f (t) is periodic with period 2π. Find the solution of
the intial value problem
d
2
y
dt
2
− y = f(t); y(0) = 1, y

(0) = 0.
Examine the continuity of the solution at t = nπ, where n is a positive integer, and verify that the solution is continuous

and has a continuous derivative at these points.
Hint, Solution
Exercise 31.21
Use Laplace transforms to solve
dy
dt
+

t
0
y(τ) dτ =
e
−t
, y(0) = 1.
Hint, Solution
1502
Exercise 31.22
An electric circuit gives rise to the system
L
di
1
dt
+ Ri
1
+ q/C = E
0
L
di
2
dt

+ Ri
2
− q/C = 0
dq
dt
= i
1
− i
2
with initial conditions
i
1
(0) = i
2
(0) =
E
0
2R
, q(0) = 0.
Solve the system by Laplace transform methods and show that
i
1
=
E
0
2R
+
E
0
2ωL

e
−αt
sin(ωt)
where
α =
R
2L
and ω
2
=
2
LC
− α
2
.
Hint, Solution
Exercise 31.23
Solve the initial value problem,
y

+ 4y

+ 4y = 4
e
−t
, y(0) = 2, y

(0) = −3.
Hint, Solution
1503

31.7 Hints
Hint 31.1
Use the differentiation and integration properties of the Laplace transform where appropriate.
Hint 31.2
Hint 31.3
Hint 31.4
If the integral is uniformly convergent and
∂g
∂s
is continuous then
d
ds

b
a
g(s, t) dt =

b
a
∂
∂s
g(s, t) dt
Hint 31.5

∞
s
e
−tx
dt =
1

x
e
−sx
Hint 31.6
Use integration by parts.
Hint 31.7

∞
0
e
−st
f(t) dt =

∞
n=0
(n+1)T

nT
e
−st
f(t) dt
1504
The sum can be put in the form of a geometric series.
∞

n=0
α
n
=
1

1 − α
, for |α| < 1
Hint 31.8
Hint 31.9
Write the answer in terms of the Gamma function.
Hint 31.10
Hint 31.11
Hint 31.12
Hint 31.13
Hint 31.14
Hint 31.15
Hint 31.16
1505
Hint 31.17
Hint 31.18
Hint 31.19
Hint 31.20
Hint 31.21
Hint 31.22
Hint 31.23
1506
31.8 Solutions
Solution 31.1
1.
L

e
at

=


∞
0
e
−st
e
at
dt
=

∞
0
e
−(s−a)t
dt
=

−
e
−(s−a)t
s − a

∞
0
for (s) > (a)
L

e
at


=
1
s − a
for (s) > (a)
2.
L[sin(at)] =

∞
0
e
−st
sin(at) dt
=
1
ı2

∞
0

e
(−s+ıa)t
−
e
(−s−ıa)t

dt
=
1
ı2


−
e
(−s+ıa)t
s − ıa
+
e
(−s−ıa)t
s + ıa

∞
0
, for (s) > 0
=
1
ı2

1
s − ıa
−
1
s + ıa

L[sin(at)] =
a
s
2
+ a
2
for (s) > 0
1507

3.
L[cos(at)] = L

d
dt
sin(at)
a

= sL

sin(at)
a

− sin(0)
L[cos(at)] =
s
s
2
+ a
2
for (s) > 0
4.
L[sinh(at)] =

∞
0
e
−st
sinh(at) dt
=

1
2

∞
0

e
(−s+a)t
−
e
(−s−a)t

dt
=
1
2

−
e
(−s+a)t
s − a
+
e
(−s−a)t
s + a

∞
0
for (s) > |(a)|
=

1
2

1
s − a
−
1
s + a

L[sinh(at)] =
a
s
2
− a
2
for (s) > |(a)|
5.
L[cosh(at)] = L

d
dt
sinh(at)
a

= sL

sinh(at)
a

− sinh(0)

L[cosh(at)] =
s
s
2
− a
2
for (s) > |(a)|
1508
6. First note that
L

sin(at)
t

(s) =

∞
s
L[sin(at)](σ) dσ.
Now we use the Laplace transform of sin(at) to compute the Laplace transform of sin(at)/t.
L

sin(at)
t

=

∞
s
a

σ
2
+ a
2
dσ
=

∞
s
1
(σ/a)
2
+ 1
dσ
a
=

arctan

σ
a

∞
s
=
π
2
− arctan

s

a

L

sin(at)
t

= arctan

a
s

7.
L


t
0
sin(aτ)
τ
dτ

=
1
s
L

sin(at)
t


L


t
0
sin(aτ)
τ
dτ

=
1
s
arctan

a
s

8.
L[f(t)] =

2π
0
e
−st
f(t) dt
1 −
e
−2πs
=


π
0
e
−st
dt
1 −
e
−2πs
=
1 −
e
−πs
s(1 −
e
−2πs
)
1509
L[f(t)] =
1
s(1 +
e
−πs
)
Solution 31.2
L[af(t) + bg(t)] =

∞
0
e
−st


af(t) + bg(t)

dt
= a

∞
0
e
−st
f(t) dt + b

∞
0
e
−st
g(t) dt
= aL[f(t)] + bL[g(t)]
Solution 31.3
If f(t) is of exponential order α, then
e
ct
f(t) is of exponential order c + α.
L[
e
ct
f(t)] =

∞
0

e
−st
e
ct
f(t) dt
=

∞
0
e
−(s−c)t
f(t) dt
=
ˆ
f(s − c) for s > c + α
Solution 31.4
First consider the Laplace transform of t
0
f(t).
L[t
0
f(t)] =
ˆ
f(s)
1510
Now consider the Laplace transform of t
n
f(t) for n ≥ 1.
L[t
n

f(t)] =

∞
0
e
−st
t
n
f(t) dt
= −
d
ds

∞
0
e
−st
t
n−1
f(t) dt
= −
d
ds
L[t
n−1
f(t)]
Thus we have a difference equation for the Laplace transform of t
n
f(t) with the solution
L[t

n
f(t)] = (−1)
n
d
n
ds
n
L[t
0
f(t)] for n ∈ Z
0+
,
L[t
n
f(t)] = (−1)
n
d
n
ds
n
ˆ
f(s) for n ∈ Z
0+
.
Solution 31.5
If

β
0
f(t)

t
dt exists for positive β and f(t) is of exponential order α then the Laplace transform of f(t)/t is defined for
s > α.
L

f(t)
t

=

∞
0
e
−st
1
t
f(t) dt
=

∞
0

∞
s
e
−σt
dσ f(t) dt
=

∞

s

∞
0
e
−σt
f(t) dt dσ
=

∞
s
ˆ
f(σ) dσ
1511
Solution 31.6
L


t
0
f(τ) dτ

=

∞
0
e
−st

t

0
f(τ) dτ dx
=

−
e
−st
s

t
0
f(τ) dτ

∞
0
−

∞
0
−
e
−st
s
d
dt


t
0
f(τ) dτ


dt
=
1
s

∞
0
e
−st
f(t) dt
=
1
s
ˆ
f(s)
1512
Solution 31.7
f(t) is periodic with period T .
L[f(t)] =

∞
0
e
−st
f(t) dt
=

T
0

e
−st
f(t) dt +

2T
T
e
−st
f(t) dt + ···
=
∞

n=0

(n+1)T
nT
e
−st
f(t) dt
=
∞

n=0

T
0
e
−s(t+nT )
f(t + nT ) dt
=

∞

n=0
e
−snT

T
0
e
−st
f(t) dt
=

T
0
e
−st
f(t) dt
∞

n=0
e
−snT
=

T
0
e
−st
f(t) dt

1 −
e
−sT
1513
Solution 31.8
ˆ
f(s) =

∞
0
e
−st
f(t) dt
=
n

0

(n+1)T
nT
e
−st
f(t) dt
=
n

0

T
0

e
−s(t+nT )
f(t + nT ) dt
=
n

0
e
−snT

T
0
e
−st
(−1)
n
f(t) dt
=

T
0
e
−st
f(t) dt
n

0
(−1)
n


e
−sT

n
ˆ
f(s) =
ˆg(s)
1 +
e
−sT
, for (s) > 0
Consider
ˆ
f(s) = s
−1
tanh(sT/2).
s
−1
tanh(sT/2) = s
−1
e
sT/2
−
e
−sT/2
e
sT/2
+
e
−sT/2

= s
−1
1 −
e
−sT
1 +
e
−sT
We have
ˆg(s) ≡

T
0
f(t)
e
−st
dt =
1 −
e
−st
s
.
1514
By inspection we see that this is satisfied for f(t) = 1 for 0 < t < T . We conclude:
f(t) =

1 for t ∈ [2nT . . . (2n + 1)T ),
−1 for t ∈ [(2n + 1)T . . . (2n + 2)T ),
where n ∈ Z.
Solution 31.9

The Laplace transform of t
ν
, ν > −1 is
ˆ
f(s) =

∞
0
e
−st
t
ν
dt.
Assume s is complex-valued. The integral converges for (s) > 0 and ν > −1.
Method 1. We make the change of variables z = st.
ˆ
f(s) =

C
e
−z

z
s

ν
1
s
dz
= s

−(ν+1)

C
e
−z
z
ν
dz
C is the path from 0 to ∞ along arg(z) = arg(s). (Shown in Figure 31.4).
Since the integrand is analytic in the domain  < r < R, 0 < θ < arg(s), the integral along the boundary of this
domain vanishes.


R

+

R
e
ı arg(s)
R
+


e
ı arg(s)
R
e
ı arg(s)
+




e
ı arg(s)

e
−z
z
ν
dz = 0
We show that the integral along C
R
, the circular arc of radius R, vanishes as R → ∞ with the maximum modulus
1515
Im(z)
Re(z)
arg(s)
Figure 31.4: The Path of Integration.
integral bound.





C
R
e
−z
z

ν
dz




≤ R|arg(s)|max
z∈C
R


e
−z
z
ν


= R|arg(s)|
e
−R cos(arg(s))
R
ν
→ 0 as R → ∞.
The integral along C

, the circular arc of radius , vanishes as  → 0. We demonstrate this with the maximum modulus
integral bound.






C

e
−z
z
ν
dz




≤ |arg(s)|max
z∈C



e
−z
z
ν


= |arg(s)|
e
− cos(arg(s))

ν
→ 0 as  → 0.

1516
Taking the limit as  → 0 and R → ∞, we see that the integral along C is equal to the integral along the real axis.

C
e
−z
z
ν
dz =

∞
0
e
−z
z
ν
dz
We can evaluate the Laplace transform of t
ν
in terms of this integral.
L[t
ν
] = s
−(ν+1)

∞
0
e
−t
t

ν
dt
L[t
ν
] =
Γ(ν + 1)
s
ν+1
In the case that ν is a non-negative integer ν = n > −1 we can write this in terms of the factorial.
L[t
n
] =
n!
s
n+1
Method 2. First note that the integral
ˆ
f(s) =

∞
0
e
−st
t
ν
dt
exists for (s) > 0. It converges uniformly for (s) ≥ c > 0. On this domain of uniform convergence we can
interchange differentiation and integration.
d
ˆ

f
ds
=
d
ds

∞
0
e
−st
t
ν
dt
=

∞
0
∂
∂s

e
−st
t
ν

dt
=

∞
0

−t
e
−st
t
ν
dt
= −

∞
0
e
−st
t
ν+1
dt
1517