# Chapter Part 2

```Math 334
6.3. INVERSE LAPLACE TRANSFORMS
90
theorem we get
L[cos ax] =
1
1−e
−2πs
a
Z
2π
a
e
−sx
cos ax dx =
0
=
1
1−e
−2πs
a
1
1−e
Example 6.23. Finf L[f (x)] for the function
(
1
0&lt;x61
f (x) =
,
−1 1 &lt; x 6 2
−2πs
a
2π
a
e−sx
(a sin ax − s cos ax)
2
2
s +a
0
)
( −2πs
s
e a (−s) − (−s)
= 2
.
s2 + a2
s + a2
f (x + 2n) = f (x) ∀n ∈ Z.
Solution
The function f is periodic with period 2, so we have
Z 1
Z 2
Z 2
1
1
−sx
−sx
−sx
L[f (x)] =
e
f (x) dx =
e
dx −
e
dx
1 − e−2s 0
1 − e−2s
0
1
−2s
1
(1 − e−s )2
es/2 − e−s/2
e
− 2e−s + 1
1 − e−s
1
1
= tanh( ).
=
=
=
=
−2s
−2s
−s
s/2
−s/2
1−e
s
s(1 − e )
s(1 + e
s
s
s(e + e
)
6.3
Inverse Laplace Transforms
Recall the solution procedure outlined in Figure 6.1. The final stage in that solution procedure involves
calulating inverse Laplace transforms. In this section we look at the problem of finding inverse Laplace
transforms. In other words, given F (s), how do we find f (x) so that F (s) = L[f (x)].
We begin with a simple example which illustrates a small problem on finding inverse Laplace transforms.
Example 6.24. Consider the functions
2
f (x) = x ,
and

2

x
g(x) = 48


−π
x 6= 2, 3
x=2 .
x=3
2
. Since an integral is not affected by the changing of its integrand at a few
s3
isolated points, more than one function can have the same Laplace transform.
Then L[f (x)] = L[g(x)] =
Example 6.24 illustrates that inverse Laplace transforms are not unique. However, it can be shown that, if
several functions have the same Laplace transform, then at most one of them is continuous. This prompts
us to make the following definition.
Definition 6.25. The inverse Laplace transform of F (s), denoted L−1 [F (s)], is the function f defined
on [0, ∞) which has the fewest number of discontinuities and satisfies
L[f (x)] = F (s).
◭
Math 334
6.3. INVERSE LAPLACE TRANSFORMS
91
Example 6.26.
2
1. L−1 [ 3 ] = x2 .
s
s
2. L−1 [ 2
] = cos 3x.
s +9
s−1
s−1
s
] = L−1 [
] = ex L−1 [ 2
] = ex cos 2x.
2
− 2s + 5
(s − 1) + 4
s +4
6.17 in reverse)
3. L−1 [
s2
(using property 1 of Theorem
The inverse Laplace transform is a linear operator.
Theorem 6.27.
If L−1 [F (s)] and L−1 [G(s)] exist, then L−1 [αF (s) + βG(s)] = αL−1 [F (s)] + βL−1 [G(s)].
Proof
Starting from the right hand side we have
L[αL−1 [F (s)] + βL−1 [G(s)]] = αL[L−1 [F (s)]] + βL[L−1 [G(s)]] = αF (s) + βG(s).
The result follows.
Most of the properties of the Laplace transform can be reversed for the inverse Laplace transform.
Theorem 6.28.
If L−1 [F (s)] = f (x), then the following hold:
1. L−1 [F (s + a)] = e−ax f (x);
2. L−1 [sF (s)] = fZ′ (x), if f (0) = 0;
x
1
f (t) dt;
3. L−1 [ F (s)] =
s
0
4. L−1 [e−as F (s)] = ua (x) f (x − a).
Proof
1. L[e−ax f (x)] = F (s + a) from Theorem 6.17, property 1. The result follows.
2. L[f ′ (x))] = −f (0) + sF (s) from Theorem 6.17, property 4. The result follows.
Z x
1
3. L[
f (t) dt] = F (s), from Theorem 6.17, property 5. The result follows.
s
0
4. L[ua (x)f (x − a)] = e−as L[f (x)] = e−as F (s), from Theorem 6.19. The result follows.
Example 6.29. Find L−1 [
Solution
We can write
1
].
s(s2 + 1)
1
1
1
= F (s), where F (s) = 2
. Then f (x) = L−1 [F (s)] = sin x, so we get
+ 1)
s
s +1
Z x
Z x
1
1
] = L−1 [ F (s)] =
f (t) dt =
sin t dt = 1 − cos x.
L−1 [ 2
s(s + 1)
s
0
0
s(s2
Math 334
6.3. INVERSE LAPLACE TRANSFORMS
Example 6.30. Find L−1 [
s(s2
92
1
].
+ +2s + 5)
Solution
L−1 [
s(s2
1
1
1
1
] = L−1 [
] = e−x L−1 [ 2
] = e−x sin 2x.
2
+ 2s + 5)
(s + 1) + 4
s +4
2
Example 6.31. Find L−1 [
Solution
L−1 [
1 + e−s
].
s2
−s
e−s
1 + e−s
−1 1
−1 1
−1 e
]
=
L
[
+
]
=
L
[
]
+
L
[
] = x + u1 (x)(x − 1).
s2
s2
s2
s2
s2
Example 6.32. Find L−1 [
Solution
L−1 [
4e−2s
].
s2 + 16
4e−2s
4
4
] = L−1 [e−2s &middot; 2
] = u2 (x)L−1[ 2
] = u2 (x) sin 4x.
2
s + 16
s + 16
s + 16
P (s)
, where P and Q are polynomials in s
Q(s)
P (s)
in terms of partial fractions.
with deg{Q} &gt; deg{P }. To evaluate L−1 [F (s)], one writes
Q(s)
Many transforms that one encounters are of the form F (s) =
Example 6.33 (distinct linear factors). Find L−1 [
7s − 1
].
(s + 1)(s + 2)(s − 3)
Solution
We write the expression in the form
7s − 1
A
B
C
=
+
+
.
(s + 1)(s + 2)(s − 3)
s+1 s+2 s−3
Solving for the constants yields: A = 2, B = −3, and C = 1. Thus, we get
L−1 [
2
3
1
7s − 1
] = L−1 [
] − L−1 [
] + L−1 [
] = 2e−x − 3e−2x + e3x .
(s + 1)(s + 2)(s − 3)
s+1
s+2
s−3
Example 6.34 (repeated linear factors). Find L−1 [
s2 + 9s + 2
].
(s − 1)2 (s + 3)
Solution
We write the expression in the form
C
s2 + 9s + 2
A
B
+
=
+
.
(s − 1)2 (s + 3)
s − 1 (s − 1)2 s + 3
Solving for the constants yields: A = 2, B = 3, and C = −3. Thus, we get
L−1 [
s2 + 9s + 2
1
2
1
] − L−1 [
] = L−1 [
] + 3L−1 [
] = 2e−x + 3xex − e−3x .
(s − 1)2 (s + 3)
s−1
(s − 1)2
s+3
Math 334
6.4. APPLICATIONS TO DIFFERENTIAL EQUATIONS
Example 6.35 (quadratic factors). Find L−1 [
Solution
We write the expression in the form
L−1 [
93
2s2 + 10s
(s + 1)].
s2 − 2s + 5
A(s − 1) + B
C
2s2 + 10s
(s + 1)] =
+
.
s2 − 2s + 5
(s − 1)2 + 4
s+1
Solving for the constants yields: A = 3, B = 8, and C = −1. Thus, we get
L−1 [
6.4
2s2 + 10s
s−1
2
1
(s + 1)] = 3L−1 [
] + 4L−1 [
] − L−1 [
]
s2 − 2s + 5
(s − 1)2 + 4
(s − 1)2 + 4
s+1
= 3ex cos 2x + 4ex sin 2x − e−x .
Applications to Differential Equations
The easiest way to see how to apply Laplace transforms to differential equations is to work through some
examples.
Example 6.36. Solve the following initial value problem:
y ′′ − y ′ = 2x,
y(0) = 1,
y ′ (0) = −2.
Solution
Method 1 (the old approach)
First solve the homogeneous equation: y ′′ − y = 0.
y = erx
=⇒
r2 − r = 0
=⇒
r = 0, 1
=⇒
yh (x) = c1 + c2 ex .
Now look for a particular solution: yp (x) = x(Ax + B) = Ax2 + Bx. Plug into the DE to get
(
(
2A − B = 0
A = −1
′′
′
yp − yp = 0
=⇒
=⇒
=⇒
yp (x) = −x2 − 2x.
−2A = 2
B = −2
Thus, we have
y(x) = c1 + c2 ex − x2 − 2x,
Apply the initial conditions:
y(0) = 1
=⇒
y ′ (0) = −2
(
c1 + c2 = 1
c2 − 2 = −2
=⇒
y ′ (x) = c2 ex − 2x − 2.
(
c1 = 1
c2 = 0
=⇒
y(x) = 1 − x2 − 2x .
Method 2 (using Laplace transforms)
Take Laplace transforms of the DE:
L[y ′′ ] − L[y ′ ] = 2L[x]
=⇒
=⇒
=⇒
2
2
s Y (s) − sy(0) − y ′ (0) − [sY (s) − y(0)] = 2
s
2
2
s Y (s) − s + 2 − sY (s) + 1 = 2
s
s3 − 3s2 + 2
2
(s − 1)(s2 − 2s − 2)
1
2
Y (s) =
=
= − 2− 3
s3 (s − 1)
s3 (s − 1)
s s
s
Math 334
6.4. APPLICATIONS TO DIFFERENTIAL EQUATIONS
94
Finally, taking the inverse Laplace transform, we arrive at the final solution:
y(x) = L−1 [Y (s)] = 1 − 2x − x2 .
Example 6.37. Solve the following initial value problem:
y ′′ − 2y ′ + 5y = −8e−x ,
y(0) = 2,
y ′ (0) = 12.
Solution
Take Laplace transforms of the DE:
L[y ′′ ] − 2L[y ′ ] + 5L[y] = −8L[e−x ]
=⇒
=⇒
=⇒
=⇒
2
2
s Y (s) − sy(0) − y ′ (0) − 2 [sY (s) − y(0)] + 5Y (s) = 2
s
−8
2
s Y (s) − 2s − 12 − 2sY (s) + 4 + 5Y (s) =
s+1
2s2 + 10s
Y (s) = 2
(s − 2s + 5)(s + 1)
(s − 1)
2
1
Y (s) = 3
−4
−
.
(s − 1)2 + 4
(s − 1)2 + 4 s + 1
Finally, taking the inverse Laplace transform, we arrive at the final solution:
y(x) = 3ex cos 2x + 4ex sin 2x − e−x .
Example 6.38. Solve the following initial value problem:
y ′′ − 2y ′ + 5y = −8e7−x ,
y(7) = 2,
y ′ (7) = 12.
Solution
It appears that we can not use Laplace transforms since L[y ′ ] = sY (s) − y(0), and we don’t know y(0).
But we can get around this by moving the initial point (in this case x0 = 7) to the origin by means of a
translation.
Let t = x − 7 and w(t) = y(x). Then we get
w′ (t) = y ′ (x),
w′′ (t) = y ′′ (x),
w(0) = y(7),
w′ (0) = y ′ (7),
so, the initial value problem becomes
w′′ − 2w′ + 5w = −8e−t ,
w(0) = 2,
w′ (0) = 12.
This is just the initial value problem we had in Example 6.40. The solution is
w(t) = 3et cos 2t + 4et sin 2t − e−t .
The solution to the original problem is
y(x) = w(x − 7) = 3ex−7 cos[2(x − 7)] + 4ex−7 sin[2(x − 7)] − e−(x−7) .
Next we consider an initial value problem with discontinuous forcing.
Math 334
6.4. APPLICATIONS TO DIFFERENTIAL EQUATIONS
95
Example 6.39. Solve the following initial value problem:
y ′′ + 4y = g(x),
Solution
We can re-write g as follows:
we have
y(0) = 0,


0&lt;x&lt;1
1
where g(x) = −1 1 &lt; x &lt; 2 .


0
x&gt;2
y ′ (0) = 0,
g(x) = 1[u0 (x) − u1 (x)] − 1[u1 (x) − u2 (x)] = u0 (x) − 2u1 (x) + u2 (x). Then
G(s) = L[g(x)] = L[u0 (x)] − 2L[u1 (x)] + L[u2 (x)] =
1
e−s e−2s
−2
+
.
s
s
s
Take Laplace transforms of the DE:
L[y ′′ ] + 4L[y] = L[g(x)]
where
=⇒
2
s Y (s) − sy(0) − y ′ (0) + 4Y (s) = G(s)
=⇒
Y (s) =
=⇒
Y (s) = F (s) − 2e−s F (s) + e−2s F (s),
=⇒
s2 Y (s) + 4Y (s) = G(s)
1
e−s
e−2s
G(s)
=
−
2
+
s2 + 4
s(s2 + 4)
s(s2 + 4) s(s2 + 4)
1
1
=
F (s) =
s(s2 + 4)
4
1
s
−
s s2 + 4
.
Thus,
1
(1 − cos 2x).
4
Finally, taking the inverse Laplace transform, we arrive at the final solution:
f (x) = L−1 [F (s)] =
y(x) = L−1 [Y (s)] = L−1 [F (s)] − 2L−1 [e−s F (s)] + L−1 [e−2s F (s)]
= f (x) − 2u1 (x)f (x − 1) + u2 (x)f (x − 2)
1
= {1 − cos 2x − 2u1 (x)(1 − cos[2(x − 1)]) + u2 (x)(1 − cos[2(x − 2)])} .
4
Now we consider an ODE with variable coefficients.
Example 6.40. Solve the following initial value problem:
y ′′ + 2xy ′ − 4y = 1,
y(0) = 0,
y ′ (0) = 0.
Solution
Take Laplace transforms of the DE:
L[y ′′ ] + 2L[xy ′ ] − 4L[y] = L[1]
=⇒
=⇒
=⇒
2
1
d
s Y (s) − sy(0) − y ′ (0) − 2 [sY (s) − y(0)] − 4Y (s) =
ds
s
1
s2 Y (s) − 2[sY ′ (s) + Y (s)] − 4Y (s) =
s
3
−1
s
Y ′ (s) +
Y (s) = 2 .
−
s 2
2s
Math 334
6.5. CONVOLUTION
96
This is a linear ODE in Y (s). Look for an integrating factor &micro;:
&micro;′
3 s
= −
&micro;
s 2
=⇒
ln &micro; = 3 ln s −
s2
4
&micro; = s3 e−s
=⇒
2
/4
.
The ODE for Y becomes:
2
d
s
&micro;′
&micro;
′
[&micro;Y (s)] = &micro; Y (s) + Y (s) = − 2 = − e−s /4 .
ds
&micro;
2s
2
Integrating yields:
&micro;Y (s) = −
Z
2
s −s2/4
e
ds = e−s /4 + C
2
=⇒
Y (s) =
C 2
1
+ es /4 .
s3 s3
It remains to determin the value of the constant if integration C. There is no auxiliary condition inherited
from the original ODE. To get an appropriate condition to enable us to determine C, we utilize Theorem
6.12 that states that Y (s) → 0 as s → ∞. Therfore
lim Y (s) = 0
=⇒
s→∞
C=0
=⇒
Y (s) =
1
.
s3
Finally, taking the inverse Laplace transform, we arrive at the final solution:
y(x) = L−1 [
1
x2
]=
.
3
s
2
We can summarize the application of Laplace transforms to differential equations as follows.
1. For an ODE with constant coefficients, the equation for the Laplace transform is of the form AY = B.
2. For an ODE with polynomial coefficients in x, the equation for the Laplace transform is an ODE with
polynomial coefficients in s.
3. If the coefficient functions of the ODE are liear in x, the ODE for Y is a first order ODE.
4. The auxiliary condition to use when solving a first order ODE in Y is lim Y (s) = 0.
s→∞
6.5
Convolution
Consider the following initial value problem:
y ′′ + y = g(x),
y(0) = y ′ (0) = 0.
Take the Laplace transform of the equation to get:
s2 Y (s) − sy(0) − y ′ (0) + Y (s) = G(s)
=⇒
Y (s) =
G(s)
= F (s)G(s),
s2 + 1
where F (s) =
1
.
s2 + 1
We would like to express the solution y(x) in terms of f (x) and g(x), i.e. we would like to express L−1 [F (s)G(s)]
in terms of L−1 [F (s)] and L−1 [G(s)]. To do this, we define a special type of product of functions. Let
f, g ∈ P C(0, ∞).
Math 334
6.5. CONVOLUTION
97
Definition 6.41. The convolution of f and g, denoted f ∗ g, is defined as:
Z x
(f ∗ g)(x) :=
f (x − t)g(t) dt.
0
Example 6.42. Z
x
x2
(1) 1 ∗ x =
1 &middot; t dt =
.
2
0
Z
x
x4
.
(2) x ∗ x2 =
(x − t) &middot; t2 dt =
12
0
Theorem 6.43.
The convolution product satisfies the following properties:
1. f ∗ g = g ∗ f ;
(convolution product is commutative)
2. f ∗ (g + h) = f ∗ g + f ∗ h;
(convolution product is distributive over addition)
3. f ∗ (g ∗ h) = (f ∗ g) ∗ h;
(convolution product is associative)
4. f ∗ 0 = 0.
Proof
Exercise.
Remark. While the convolution product has many of the properties of ordinary multiplication of functions,
it different in that it has no multiplicative identity element, i.e. there is no function g with the property that
g ∗ f 6= f for all funtions f . ◭
Theorem 6.44.
If
(i) f, g ∈ P C(0, ∞);
(ii) F (s) = L[f (x)] and G(s) = L[g(x)], then
L[f ∗ g] = F (s)G(s),
or equivalently
L−1 [F (s)G(s)] = (f ∗ g)(x).
Proof
Z ∞
Z x
e−sx (f ∗ g)(x) dx =
e−sx
f (x − t)g(t) dt dx
0
0
0
Z ∞Z ∞
Z ∞Z ∞
=
e−sx f (x − t)g(t) dx dt =
e−s(ξ+t) f (ξ)g(t) dξ dt
0
t
0
0
Z ∞
Z ∞
=
e−sξ f (ξ) dξ
e−stg(t) dt = F (s)G(s).
L[f ∗ g] =
∞
Z
0
Example 6.45. Find L−1 [
0
1
].
+ 4)
s(s2
Solution
Method 1 (the old approach)
L−1 [
Method 2 (using convolution)
1
1
] = L−1 [
2
s(s + 4)
4
1
1
s
] = (1 − cos 2x).
− 2
s s +4
4
Math 334
6.6. THE DELTA FUNCTION
L−1 [
98
Z
1
1 −1 1
2
1 x −1 1
2
−1
]
=
L
[
]
∗
L
[
]
=
](t) dt
L [ ](x − t)L−1[ 2
s(s2 + 4)
2
s
s2 + 4
2 0
s
s +4
Z
x
1
1 x
1
1 &middot; sin 2t dt = − cos 2t = (1 − cos 2x).
=
2 0
4
4
0
Now, returning to the origin problem:
y ′′ + y = g(x),
The solution satisfies:
y(x) = L
−1
y(0) = y ′ (0) = 0.
G(s)
[ 2
] = g ∗ sin x =
s +1
Z
x
g(t) sin(x − t) dt.
0
Example 6.46. Solve the following integro-differential equation:
Z x
′
−2x
y (x) = 1 − e
y(t)e2t dt,
y(0) = 1.
0
Solution
The equation may be written as
y′ = 1 −
Taking Laplace transforms we get
sY (s) − y(0) =
6.6
1
Y (s)
−
s s+2
=⇒
Z
x
j(t)e−2(x−t) dt = 1 − y ∗ e−2x .
0
Y (s) =
s+2
2
1
= −
s(s + 1)
s s+1
=⇒
y(x) = 2 − e−x .
The Delta Function
Mechanical systems are often acted upon by external forces that are large in magnitude but are of short
duration. Some typical example are: a hammer hitting a nail; a bat hitting a baseball. These forces are of
the form


t &lt; t0
0
F (t) = (&middot; &middot; &middot; ) t0 6 t 6 t1 .


0
t &gt; t1
y
y=F(t)
t0
t1
t
Figure 6.4: A plot of a localized force y = F (t).
Math 334
6.6. THE DELTA FUNCTION
99
The integral of the force acting over an interval of time ic called “impulse:” I =
Law, F = ma, we get
Z
Z t1
F (t) dt =
I=
t0
t1
m
t0
dv
dt = mv(t1 ) − mv(t0 )
dt
Z
t1
F (t) dt. From Newton’s
t0
(i.e. impulse = change in momentum).
Consider an impulse over shorter and shorter time intervals.
F
v
t0
t1
t0
t
t1
t
F
v
t0
t1
t
t0
t1
t
F
v
t0 t1
t
t0 t 1
t
Figure 6.5: A plot of a localized force and impulse for decreasing time intervals.
The exact nature of the force in the interval [t0 , t1 ] is frequently unknown. What usually is known is the
state of the system before and after the application of the force. These duration of the force is often so short,
that it is convenient to think of the force as acting instantaneously.
Math 334
6.6. THE DELTA FUNCTION
100
F
v
t0
t
t1
t0
t1
t
Figure 6.6: A plot of a localized force and impulse with simplified approximation.
We define the following functions:

 1
δh (t) = 2h
0
|t| 6 h
for h &gt; 0.
|t| &gt; h
h
-h
t
Figure 6.7: A plot of a δh (t).
Thes functions have the following property:
Z ∞
Z
δh (t) dt =
−∞
h
−h
1
1
dt =
(h − (−h)) = 1.
2h
2h
Definition 6.47. The Dirac Delta Function is defined implicitly by the following properties:
1. δ(t)
Z = 0 for t 6= 0;
∞
δ(t) = 1.
2.
◭
−∞
Remark. The Dirac delta function is not a function in the usual sense. However, one can “think” of it as
follows:
(
∞ t=0
δ(t) = lim δh (t) =
. ◭
h→0
0
t 6= 0
Math 334
6.6. THE DELTA FUNCTION
101
Theorem 6.48.
If f is continuous on (−∞, ∞), then for any c ∈ R,
Z ∞
f (t) δ(t − c) dt = f (c).
−∞
Proof
We have

 1
δh (t − c) = 2h
0
Therefore
Z ∞
Z
f (t)δ(t − c) dt = lim
h→0
−∞

1
|t − c| 6 h 
= 2h
|t − c| &gt; h 0
∞
1
h→0 2h
f (t)δh (t − c) dt = lim
−∞
1
f (c̄)(2h)
2h
= lim f (c̄) = f (c).
= lim
h→0
Z
c−h6 t6 c+h
.
|t − c| &gt; h
c+h
f (t) dt
c−h
for some c − h &lt; c̄ &lt; c + h (using the mean value theorem)
h→0
This is sometimes called the “sifting property” of the delta function since, from all possible values of f , it
“sifts out” the value f (c).
Does the Dirac delta function have a Laplace transform? Yes it does, since
Z ∞
L[δ(t − c)] =
e−stδ(t − c) dt = e−sc .
0
Remark. Notice that for c = 0 we get L[δ(t)] = 1. This reinforces what was said earlier, namely that δ(t)
is not a function in the usual sense, since, for a normal function we have
lim F (s) = 0,
s→∞
but for the delta function we have
lim L[δ(t)] 6= 0.
s→∞
Example 6.49. Solve the following initial value problem:
y ′′ + y = 4δ(t − 2π),
y(0) = y ′ (0) = 0.
Solution
Taking Laplace transforms we get,
s2 Y (s) − sy(0) − y ′ (0) + Y (s) = 4e−2πs
=⇒
Y (s) =
4e−2πs
.
s2 + 1
Taking inverse Laplace transforms we get
(
0
y(t) = u2π (t) sin(t − 2π) = u2π (t) sin t =
sin t
0 6 t &lt; 2π
.
t &gt; 2π
◭
```