CHAPTER 4
The Laplace Transform
4.1 Introduction
The Laplace transform provides an effective method of solving initial-value problems for
linear differential equations with constant coefficients. However, the usefulness of Laplace
transforms is by no means restricted to this class of problems. Some understanding of the
basic theory is an essential part of the mathematical background of engineers, scientists and
mathematicians.
The Laplace transform is defined in terms of an integral over the interval [0, ∞). Integrals over an infinite interval are called improper integrals, a topic studied in Calculus
II.
DEFINITION Let f be a continuous function on [0, ∞). The Laplace transform of f ,
denoted by L[(f (x)], or by F (s), is the function given by
Z ∞
L[f (x)] = F (s) =
e−sx f (x) dx.
(1)
0
The domain of F is the set of all real numbers s for which the improper integral converges.
In more advanced treatments of the Laplace transform the parameter s assumes complex values, but the restriction to real values is sufficient for our purposes here. Note that
L transforms a function f = f (x) into a function F = F (s) of the parameter s. The
continuity assumption on f will hold throughout the first three sections. It is made for
convenience in presenting the basic properties of L and for applying the Laplace transform
method to solving initial-value problems. In the last two sections of this chapter we extend
the definition of L to a larger class of functions, the piecewise continuous functions on
[0, ∞). There we will apply L to the problem of solving nonhomogeneous equations in
which the nonhomogeneous term is piecewise continuous. This will involve some extension
of our concepts of differential equation and solution.
As indicated above, the primary application of Laplace transforms of interest to us is
solving linear differential equations with constant coefficients. Referring to our work in
Chapter 3, the functions which arise naturally in the treatment of these equations are:
p(x)erx,
p(x) cos βx,
p(x) sin βx,
p(x)erx cos βx,
p(x)erx sin βx
where p is a polynomial.
We begin by calculating the Laplace transforms of some simple cases of these functions.
115
Example 1. Let f (x) = 1 · e0x ≡ 1 on [0, ∞). By the Definition,
Z ∞
Z b
−sx
L[1] =
e
· 1 dx = lim
e−sx dx
b→∞ 0
0
=
lim
"
b→∞
b #
−sb e−sx e
−1
1
1
= lim
+ .
+ = lim
sb
b→∞
−s 0
−s
s b→∞ se
s
Now, lim −1/sesb exists if and only if s > 0, and in this case
b→∞
lim
b→∞
−1
= 0.
sesb
Thus,
L[1] =
1
,
s
s > 0.
Example 2. Let f (x) = erx on [0, ∞). Then,
Z b
Z ∞
rx
−sx
rx
L[e ] =
e
· e dx = lim
e−(s−r)x dx
b→∞ 0
0

b 
"
#
−(s−r)x −(s−r)b
e
e
1

= lim 
= lim
+
.
b→∞ −(s − r) b→∞ −(s − r)
s−r
0
The limit exists (and has the value 0) if and only if s − r > 0. Therefore
L[erx ] =
1
,
s−r
s > r.
Note that if r = 0, then we have the result in Example 1.
Example 3. Let f (x) = cos βx on [0, ∞). Then,
Z ∞
Z b
−sx
L[cos βx] =
e
· cos βx dx = lim
e−sx cos βx dx
b→∞ 0
0
b
e−sx [−s cos βx − β sin βx] = lim
.
b→∞
s2 + β 2
0
(Note the integral was calculated using integration by parts; also, it is a standard entry in
a table of integrals.)
Now,
L[cos βx] = − lim
b→∞
1
e−sb
s cos βb + β sin βb
s
.
·
+ 2
s2 + β 2
s + β2
Since [s cos βb + β sin βb]/(s2 + β 2 ) is bounded, the limit exists (and has the value 0)
if and only if s > 0. Therefore,
s
L[cos βx] = 2
, s > 0.
s + β2
116
The following table gives a basic list of the Laplace transforms of functions that we will
encounter in this chapter. While the entries in the table can be verified using the Definition,
some of the integrations involved are complicated. The properties of the Laplace transform
presented in the next section provide a more efficient way to obtain many of the entries in the
table. Handbooks of mathematical functions, for example the CRC Standard Mathematical
Tables, give extensive tables of Laplace transforms.
Table of Laplace Transforms
f (x)
F (s) = L[f (x)]
1
1
,
s
1
,
s−α
s
,
s2 + β 2
eαx
cos βx
sin βx
eαx cos βx
eαx sin βx
xn ,
n = 1, 2, . . .
xn erx ,
s>0
n = 1, 2, . . .
x cos βx
x sin βx
s>α
s>0
β
,
s>0
s2 + β 2
s−α
,
s>α
(s − α)2 + β 2
β
,
(s − α)2 + β 2
n!
sn+1
,
s>α
s>0
n!
,
(s − r)n+1
s>r
s2 − β 2
,
(s2 + β 2 )2
s>0
2βs
,
+ β 2 )2
s>0
(s2
Exercises 4.1
Use the definition of the Laplace transform to find the Laplace transform of the given
function.
1. f (x) = x.
2. f (x) = x2.
117
3. f (x) = sin x.
4. f (x) = xerx .
5. f (x) = sinh x.
6. f (x) = cosh x.
7. Use the fact that
Z
eax cos bx dx =
eax
[a cos bx + b sin bx] to find L[erx cos βx]
a2 + b 2
Z
eax sin bx dx =
eax
[a sin bx − b cos bx] to find L[erx sin βx]
a2 + b 2
by the definition.
8. Use the fact that
by the definition.
Z
9. Show that L[sin x] =
2π
e−sx sin x dx
0
1 − e−2πs
.
10. Let f be a continuous function on [0, ∞) and suppose that f is periodic with
period p. That is f is continuous and f (x + p) = f (x), p > 0, for all x. Show
that
Z p
e−sx f (x) dx
L[f (x)] = 0
.
1 − e−ps
118
4.2 Basic Properties of the Laplace Transform
In the preceding section we defined the Laplace transform and calculated the Laplace transforms of some of the functions that occur in solving linear differential equations with constant coefficients. In this section we consider the basic question of the existence of the
Laplace transform of a function f , and we develop the properties of the Laplace transform
that will be used in solving initial value problems.
To motivate the material in this section, consider the differential equation
y 00 + ay 0 + by = f (x)
(2)
where a and b are constants and f is a continuous function on [0, ∞). If we assume
that y = y(x) is a solution of (1) and formally apply L, we obtain
L y 00(x) + ay 0 (x) + by(x) = L[f (x)].
(3)
The right-hand side of this equation suggests the basic question of the existence of L[f (x)].
That is, for what functions f does L[f ] exist?
DEFINITION A function f , continuous on [0, ∞), is said to be of exponential order
λ, λ a real number, if there exists a positive number M and a nonnegative number A
such that
| f (x)| ≤ M eλx
on [A, ∞).
Example 1. (a) If f is a bounded function on [0, ∞) [for example, f (x) = cos βx
or f (x) = sin βx], then f is of exponential order 0.
f bounded implies that there exists a positive number M such that | f (x)| ≤ M
for all x ∈ [0, ∞). Therefore,
| f (x)| ≤ M = M e0x
on [0, ∞).
[Note: if f (x) = cos βx or f (x) = sin βx, then we could take M = 1.]
(b) Let f (x) = x on [0, ∞). For any positive number λ,
x
lim λx = 0
x→∞ e
by L’Hôpital’s rule. Therefore, there exists a nonnegative number A such that
x
≤ 1 on [A, ∞).
eλx
This implies that
x ≤ eλx = 1 · eλx on [A, ∞)
and f (x) = x is of exponential order λ. The same argument can be used to show
that f (x) = xα , α any real number, is of exponential order λ for any positive
number λ. In general, if p = p(x) is a polynomial, then p is of exponential order
λ for any positive number λ.
119
(c) If f (x) = erx , then f is of exponential order λ for any λ ≥ r.
2
(d) Consider the function f (x) = ex . If f is of exponential order λ for some λ, then
there exists a positive number M and a nonnegative number A such that
2
ex ≤ M eλx
2
which implies e−λx ex ≤ M
on [A, ∞)
on
[A, ∞).
But,
2
lim e−λx ex = lim ex(x−λ) = ∞,
x→∞
a contradiction. Thus f (x) = e
number λ.
x→∞
x2
is not of exponential order λ for any positive
Our first property of L is a sufficient condition for L[f (x)] to exist. We shall omit
the proof.
THEOREM 1. Let f be a continuous function on [0, ∞). If f is of exponential order
λ, then the Laplace transform L[f (x)] = F (s) exists for s > λ.
We now turn to the left-hand side of equation (2) where we have L applied to the
linear combination y 00 (x) + ay 0 (x) + by(x).
THEOREM 2. The operator L is a linear operator. That is, if g and h are continuous
functions on [0, ∞), and if each of L[g(x)] and L[h(x)] exists for s > λ, then
L[g(x) + h(x)] and L[c g(x)], c constant, each exist for s > λ, and
L[g(x) + h(x)] = L[g(x)] + L[h(x)]
L[c g(x)] = cL[g(x)].
Proof:
L[g(x) + h(x)] =
Z
∞
e
−sx
[g(x) + h(x)] dx =
0
=
=
=
lim
Z
b→∞
lim
b
e−sx [g(x) + h(x)] dx
0
Z
b→∞
lim
b→∞
Z
b
e
−sx
g(x) dx +
0
Z
e−sx g(x) dx + lim
−sx
e−sx g(x) dx +
L[c g(x)] = cL[g(x)] for any constant c
Z
b→∞ 0
0
∞
The proof that
120
e
b
= L[g(x)] + L[h(x)]
b
h(x) dx
0
0
is left as an exercise.
Z
Z
b
e−sx h(x) dx
∞
e−sx h(x) dx
0
COROLLARY Let g1(x), g2(x), . . . , gn (x) be continuous functions on [0, ∞). If
L[g1(x)], L[g2(x)], . . . , L[gn (x)] all exist for s > λ, and if c1, c2, . . . , cn are real
numbers, then
L [c1g1(x) + c2 g2(x) + · · · + cn gn (x)]
exists for s > λ and
L [c1 g1(x) + c2g2 (x) + · · · + cn gn (x)] = c1 L[g1(x)] + c2 L[g2(x)] + · · · + cn L[gn (x)].
According to the Corollary, if L[y 00], L[y 0 ], and L[y] all exist for s > λ for some λ,
then
L[y 00 + ay 0 + by] = L[y 00] + aL[y 0] + bL[y].
The next property gives a relationship between the Laplace transform of the derivative
of a function and the Laplace transform of the function itself.
THEOREM 3. Let g be a continuously differentiable function on [0, ∞). If g is of
exponential order λ, then L[g 0(x)] exists for s > λ and
L[g 0(x)] = sL[g(x)] − g(0).
Proof: By the definition of L, we have
Z
Z ∞
0
−sx 0
L[g (x)] =
e g (x) dx = lim
b→∞
0
Now, using integration by parts,
Z b
e−sx g 0(x) dx =
0
b
e−sx g(x)0 + s
= e
−sb
b
e−sx g 0(x) dx.
0
Z
b
e−sx g(x) dx
0
g(b) − g(0) + s
Z
b
e−sx g(x) dx
0
Therefore,
0
L[g (x)] =
=
=
lim e
b→∞
−sb
g(b) − g(0) + s
Z
b
e
−sx
Z
b
g(x) dx
0
lim e−sb g(b) − g(0) + s lim
b→∞
b→∞
e−sx g(x) dx
0
lim e−sb g(b) − g(0) + sL[g(x)]
b→∞
provided lim e−sb g(b) exists.
b→∞
Since g is of exponential order λ, there exist constants M and A such that
| g(x)| ≤ M eλx
121
on [A, ∞). Therefore,
−sb
e g(b) ≤ e−sb M eλb = M e−(s−λ)b
for all b > A, and it follows that
lim e−sb g(b) = 0 for s > λ.
b→∞
We can now conclude that
L[g 0(x)] = sL[g(x)] − g(0).
COROLLARY Let g be a function which is n-times continuously differentiable on
[0, ∞). If each of the functions g, g 0, . . . , g (n−1) is of exponential order λ, then L[g (n)]
exists for s > λ and
L[g (n)(x)] = sn L[g(x)] − sn−1 g(0) − sn−2 g 0(0) − · · · − g (n−1)(0).
In particular, if g 00 is continuous on [0, ∞), and if g and g 0 are of exponential order λ,
then L[g 00(x)] exists for s > λ and
L[g 00(x)] = s2 L[g(x)] − sg(0) − g 0(0).
The proof of the Corollary follows by induction.
Application to Initial-Value Problems
Consider the first order initial-value problem
y 0 + ay = f (x);
y(0) = α
where α is a real number and the function f = f (x) has Laplace transform F = F (s).
Let y = y(x) be the solution of the problem. Applying the properties of L established
above, we have
L y 0 (x) + ay(x) = L[f (x)]
L[y 0 (x)] + aL[y(x)] = F (s)
sL[y(x)] − y(0) + a L[y(x)] = F (s)
(by linearity)
(by Theorem 3)
(s + a)L[y(x)] − y(0) = F (s)
Applying the initial condition y(0) = α, and solving for L[y(x)] = Y (s), we get
Y (s) =
F (s) + α
.
s+a
122
Next consider the second order initial-value problem
y 00 + ay 0 + by = f (x);
y(0) = α, y 0(0) = β.
where α and β are real numbers, and f is a function with Laplace transform L[f (x)] =
F (s).
Let y = y(x) be the solution of this problem. Then
L y 00(x) + ay 0 (x) + by(x) = L[f (x)] (equation (2))
L[y 00(x)] + aL[y 0(x)] + bL[y(x)] = F (s) (by linearity)
s2 L[y(x)] − sy(0) − y 0 (0) + a {sL[y(x)] − y(0)} + bL[y(x)] = F (s) (by the Corollary to Theorem 3)
(s2 + as + b)L[y(x)] − sy(0) − y 0 (0) − ay(0) = F (s)
Applying the initial conditions y(0) = α, y 0(0) = β and solving for L[y(x)] = Y (s), we
get
Y (s) =
F (s) + αs + β + aα
.
s2 + as + b
Implicit in these derivations is the assumption that the Laplace transform of y and
its derivatives exist. Assuming that this is the case, the importance of these results is that
it gives us the Laplace transform of the solution of an initial-value problem directly. The
question now is: Knowing Y (s), what is y(x)? This is the topic of the next section,
“inverting” the Laplace transform.
Example 2. Find the Laplace transform L[y(x)] = Y (s) of the solution of the initial-value
problem
y 0 − 2y = 2e−3x ; y(0) = −2.
SOLUTION If y = y(x) is the solution, then
L[y 0 (x) − 2y(x)] = L[2e−3x ] = 2L[e−3x ] =
L[y 0(x)] − 2L[y(x)] =
2
s+3
sL[y(x)] − y(0) − 2L[y(x)] =
2
s+3
(s − 2)L[y(x)] + 2 =
2
s+3
(s − 2)L[y(x)] =
2
−2
s+3
Therefore,
L[y(x)] = Y (s) =
2
2
−
.
(s − 2)(s + 3) s − 2
123
2
s+3
Example 3. Find the Laplace transform L[y(x)] = Y (s) of the solution of the initial-value
problem
y 00 + 4y = 3e2x; y(0) = 5, y 0 (0) = −2.
SOLUTION If y = y(x) is the solution, then
L[y 00(x) + 4y(x)] = L[3e2x ] = 3L[e2x] =
L[y 00(x)] + 4L[y(x)] =
3
s−2
s2 L[y(x)] − sy(0) − y 0 (0) + 4L[y(x)] =
3
s−2
(s2 + 4)L[y(x)] − 5s − (−2) =
3
s−2
(s2 + 4)L[y(x)] =
3
s−2
3
+ 5s − 2
s−2
Therefore,
L[y(x)] = Y (s) =
3
5s − 2
+ 2
.
2
(s − 2)(s + 4) s + 4
In the next section we will see how to go from Y (s) to y(x).
Additional Applications and Properties
Although the main use of Theorem 3 and its Corollary is in solving initial-value problems,
the results can also be used to determine entries in a table of Laplace transforms. For
example, if f is a continuously differentiable function and L[f (x)] is known, then L[f 0 (x)]
can be determined: L[f 0(x)] = sL[f (x)] − f (0).
Example 4. In Example 3, Section 4.1, we showed that
s
.
s2 + β 2
L[cos βx] =
We could use essentially the same calculations to obtain L[sin βx], but recall that the
integrations involved are a little “messy.”
Here is a simpler way to obtain L[sin βx]. Since [cos βx]0 = −β sin βx, we can write
L[−β sin βx] = L[(cos βx)0 ] = sL[cos βx] − cos(0) =
−βL[sin βx] =
−β 2
.
+ β2
s2
Therefore,
L[sin βx] =
s2
β
.
+ β2
124
s2
s2
−1
+ β2
Here are two more properties of the Laplace transform that are useful in determining
entries in a table of transforms.
THEOREM 4. Let f be a continuous function on [0, ∞), and be of exponential order
λ. Then F (s) = L[f (x)] has derivatives of all orders, and for s > λ,
dn F (s)
= (−1)n L [xn f (x)] ,
dsn
n = 1, 2, . . . .
Justification By the definition of L[f (x)] = F (s), we have
Z ∞
F (s) =
e−sx f (x) dx.
0
Differentiation of this equation with respect to s can be justified and yields
Z ∞
Z ∞
dF
d −sx
e−sx [−xf (x)] dx
=
[e f (x)] dx =
ds
ds
0
0
= L[−xf (x)] = −L[xf (x)]
Therefore, L[xf (x)] = −
dF
.
ds
Differentiating a second time, we have
Z ∞
d2F
d −sx
d dF
=
=
e [−xf (x)] dx
2
ds
ds ds
ds
0
Z ∞
=
e−sx [x2 f (x)] dx = L[x2f (x)]
0
The result stated in the theorem follows by mathematical induction.
Example 5. From the table at the end of Section 4.1, L[erx ] =
1
, s > r. Therefore,
s−r
1
1
, s > r,
=
s−r
(s − r)2
1
d2
2
2 rx
00
L[x e ] = F (s) = 2
, s > r,
=
ds
s−r
(s − r)3
1
d3
6
L[x3erx ] = −F 000(s) = − 3
, s > r,
=
ds
s−r
(s − r)4
d
L[xe ] = −F (s) = −
ds
rx
and so on.
0
The final property we’ll consider is called the translation property of L.
THEOREM 5. If f is a continuous function on [0, ∞), and if L[f (x)] = F (s) exists
for s > λ, then for any real number r,
L[erx f (x)] = F (s − r) for s > λ + r.
125
Proof: From the definition of the Laplace transform,
Z ∞
Z
−(s−r)x
F (s − r) =
e
f (x) dx =
0
∞
e−sx erx f (x) dx
0
= L[erx f (x)].
Example 6.
(a) Since L[x] = 1/s2, s > 0, we have
L[xerx ] = F (s − r) =
(b) Since L[cos βx] =
s2
1
, s>r
(s − r)2
(c.f. Example 5)
s
, s > 0, we have
+ β2
L[erx cos βx] = F (s−r) =
s−r
, s>r
(s − r)2 + β 2
(as indicated in the Table, Section 4.1).
Summary of the Properties of L
1. If f is continuous on [0, ∞), and of exponential order λ, then L[f (x)] = F (s)
exists for s > λ. (Theorem 1)
2. L is a linear operator (Theorem 2): L[f + g] = L[f ] + L[g]; L[cf ] = cL[f ].
3. If f is continuously differentiable on [0, ∞), and of exponential order λ, then
L[f 0(x)] exists for s > λ and
L[f 0 (x)] = sL[f (x)] − f (0).
If f is twice continuously differentiable on [0, ∞), and of exponential order λ, then
L[f 00(x)] exists for s > λ and
L[f 00(x)] = s2 L[f (x)] − sf (0) − f 0 (0).
And so on. (Theorem 3)
4. If f is continuous on [0, ∞), and of exponential order λ, then F (s) = L[f (x)]
has derivatives of all orders, and for s > λ,
dn F (s)
= (−1)n L [xn f (x)] ,
dsn
n = 1, 2, . . . . (Theorem 4)
5. If f is continuous on [0, ∞), and if L[f (x)] = F (s) exists for s > λ, then for any
real number r,
L[erx f (x)] = F (s − r) for s > λ + r. (Theorem 5)
126
Exercises 4.2
Use the properties of the Laplace transform and the Table to find L[f ].
1. f (x) = 3 − 2x + x2 .
2. f (x) = 2e−x − 3 sin 4x.
3. f (x) = 3 + 4e3x − 2 cos 2x.
4. f (x) = 2xe−2x + 5e2x cos 3x.
5. f (x) = 5x2 − 2e−3x sin 2x.
6. f (x) = 2 − 3x + 4x2 e2x.
7. f (x) = x sin x + 2x cos 2x.
s2
s
.
− β2
s2
β
.
− β2
8. Show that L[cosh βx] =
9. Show that L[sinh βx] =
10. Find L[3 cosh 2x − 5 sinh 3x].
11. Find L[e2x sinh x + e−x cosh 3x].
12. Find L[x cosh 2x].
13. Show that L[xerx ] =
1
by:
(s − r)2
(a) Using Property 4.
(b) Using Property 5.
2βs
.
+ β 2 )2
Find the Laplace transform Y (s) of the solution of the given initial-value problem.
14. Use Property 4 to show that L[x sin βx] =
15. y 0 − 2y = 0;
y(0 = 1.
16. y 0 − 2y = x;
y(0) = 1.
17. y 0 + 4y = 2e2x − 3 sin 3x;
y(0) = −3.
18. y 00 + 2y 0 − 8y = 0;
y(0) = 4, y 0 (0) = −2.
19. y 00 + 6y 0 + 9y = 0;
y(0) = 0, y 0 (0) = 2.
20. y 00 − 2y 0 + 5y = 2x + e−x ;
(s2
y(0) = −2, y 0(0) = 0.
127
21. y 00 − 2y 0 − 15y = 3 + 4e−3x ;
22. y 00 − 4y 0 + 4y = 5e2x ;
y(0) = 1, y 0(0) = −3.
y(0) = −3, y 0(0) = 2.
Here is another property of the Laplace transform.
23. Let f be a continuous function on [0, ∞) and assume that both f and
Z
x
f (t) dt
0
are of exponential order λ on [0, ∞). Show that if F (s) = L[f (x)], then
Z x
1
L
f (t) dt = F (s).
s
0
24. Find L
Z
x
sin 2t dt by using:
0
(a) The property given in Exercise 23.
(b) By first calculating the integral and then taking the Laplace transform of the
result.
128
4.3 Inverse Laplace Transforms and Initial-Value Problems
In Section 4.2 we saw that the Laplace transform of the solution y = y(x) of the initial-value
problem
y 00 + ay 0 + by = f (x); y(0) = α, y 0 (0) = β
is given by
F (s) + αs + β + aα
s2 + as + b
where F (s) = L[f (x)] is the Laplace transform of f .
L[y(x)] = Y (s) =
Now that we know L[y(x)], the obvious question is: What is y(x)? The general problem
of finding a function with a given Laplace transform is called the inversion problem. The
inversion problem and its application to solving initial-value problems is the topic of this
section.
If f is continuous on [0, ∞), and if the Laplace transform, L[f (x)] = F (s) exists
for s > λ, then the function F is uniquely determined by f ; that is, the operator L
is itself a function. Our first result states that L is a one-to-one function. A proof of this
result is beyond the scope of this introductory treatment.
THEOREM 1. If f and g are continuous functions on [0, ∞), and if L[f (x)] = L[g(x)],
then f ≡ g; that is f (x) = g(x) for all x ∈ [0, ∞).
The following definition gives the terminology and notation used in treating the inversion
problem.
DEFINITION If F (s) is a given transform and if the function f , continuous on [0, ∞),
has the property that L[f (x)] = F (s), then f is called the inverse Laplace transform of
F (s), and is denoted by
f (x) = L−1 [F (s)].
The operator L−1 is called the inverse operator of L.
There is a general formula for the inverse operator L−1 corresponding to (1), Section
4.1, but use of the formula requires a knowledge of complex-valued functions, a topic which
is treated in more advanced courses.
The relationship between L and L−1 is given by the following equations:
L−1 {L[f (x)]} = f (x)
L L−1 [F (s)] = F (s)
for all functions f , continuous on [0, ∞), such that L[f (x)] = F (s).
For convenience here, we reproduce the table of Laplace transforms given at the end of
Section 4.1.
129
Table of Laplace Transforms
f (x)
F (s) = L[f (x)]
1
1
,
s
1
,
s−α
s
,
2
s + β2
eαx
cos βx
s>α
s>0
β
,
s>0
+ β2
s−α
,
s>α
(s − α)2 + β 2
sin βx
s2
eαx cos βx
β
,
(s − α)2 + β 2
eαx sin βx
xn ,
s>0
n!
n = 1, 2, . . .
sn+1
,
s>α
s>0
n!
,
(s − r)n+1
s>r
x cos βx
s2 − β 2
,
(s2 + β 2 )2
s>0
x sin βx
2βs
,
(s2 + β 2 )2
s>0
xn erx ,
n = 1, 2, . . .
A simple way to interpret Theorem 1 is that the table can be read either from left to
right or from right to left. That is, the table is simultaneously a table of Laplace transforms
and of inverse Laplace transforms.
Example 1.
(a) If L[f (x)] = F (s) =
(b) If L[f (x)] = F (s) =
(c) If L[f (x)] = F (s) =
1
, then f (x) = e4x .
s−4
s
, then f (x) = cos 3x.
s2 + 9
s2
s+2
s+2
s − (−2)
=
=
, then f (x) =
2
+ 4s + 13
(s + 2) + 9
[s − (−2)]2 + 9
e−2x cos 3x.
The properties of the Laplace transform operator L can be used to derive corresponding
properties of its inverse operator L−1 . For our purposes, the most important property is
that of linearity.
THEOREM 2. The operator L−1 is linear; that is
L−1 [F (s) + G(s)] = L−1 [F (s)] + L−1 [G(s)],
130
and
L−1 [cF (s)] = cL−1 [F (s)],
c any constant.
The proof is left as an exercise.
Example 2. Find L−1 [F (s)] if
F (s) =
7
6
−
.
s + 2 s2 + 4
SOLUTION Since L−1 is a linear operator,
7
6
−1
−1
L [F (s)] = L
−
s + 3 s2 + 4
1
2
−1
−1
− 3L
= 7L
s+3
s2 + 4
Now, reading the table from right to left, we see that
1
2
−1
−3x
−1
L
and L
=e
= sin 2x.
s+3
s2 + 4
Therefore,
L
−1
7
6
−
= 7e−3x − 3 sin 2x.
s + 3 s2 + 4
The translation property of L (Theorem 5, Section 4.2) is also useful in finding inverse
transforms. The analog of Theorem 5 for inverse transforms is:
THEOREM 3. If f is continuous on [0, ∞), and if L[f (x)] = F (s) exists for s > λ,
then, for any real number r,
L−1 [F (s − r)] = erx f (x).
The following examples illustrate the kinds of manipulations that typically occur in
calculating inverse Laplace transforms. The basic strategy is to try to re-write a given
expression F (s) as sum of terms which appear in the table.
Example 3. Find L−1 [F (s)] if
F (s) =
4
1
+ 2
.
2
(s − 3)
s − 2s + 10
SOLUTION
L
−1
[F (s)] = 4L
−1
1
1
−1
+L
(s − 3)2
s2 − 2s + 10
−1
1
= xe3x .
(s − 3)2
From the table,
L
131
1
in a form in the table, we complete the square in the denominator
− 2s + 10
and “adjust” the numerator:
To put
s2
s2
1
1
1
1
3
= 2
=
=
.
2
− 2s + 10
s − 2s + 1 + 9
(s − 1) + 9
3 (s − 1)2 + 9
From the table,
−1
L
1
3
1 −1
= L
=
s2 − 2s + 10
3
(s − 1)2 + 9
Putting the two results together, we have
4
1
−1
L
+
= 4xe3x +
(s − 3)2 s2 − 2s + 10
1
3
1 x
3e
sin 3x.
ex sin 3x.
Example 4. Find L−1 [F (s)] if
F (s) =
s2
2s + 1
.
− 2s − 8
SOLUTION By factoring the denominator, we can write
F (s) =
s2
2s + 1
2s + 1
=
.
− 2s − 8
(s + 2)(s − 4)
Now, by partial fraction decomposition,
3
1
2s + 1
2
=
+ 2 .
(s + 2)(s − 4)
s−4 s+2
Therefore
L
−1
2s + 1
1
1
3 −1
1 −1
= L
+ L
=
s2 − 2s − 8
2
s−4
2
s+2
3 4x
2e
+
1
2
e−2x .
Example 5. Find L−1 [F (s)] if
F (s) =
2s + 4
.
(s − 2)(s2 − 4s + 8)
SOLUTION The quadratic factor in the denominator cannot be factored into linear factors.
By partial fraction decomposition
F (s) =
2s + 4
2
−2s + 6
=
+ 2
.
2
(s − 2)(s − 4s + 8)
s − 2 s − 4s + 8
Next, we complete the square in the denominator of the second term:
2
−2s + 6
2
−2s + 6
2
−2s + 6
+ 2
=
+ 2
=
+
.
s − 2 s − 4s + 8
s − 2 s − 4s + 4 + 4
s − 2 (s − 2)2 + 4
132
Finally, we “adjust” the numerator of the second term so that we can use the Table:
2
−2s + 6
2
−2(s − 2) + 2
2
−2(s − 2)
2
+
=
+
=
+
+
.
2
2
2
s − 2 (s − 2) + 4
s−2
(s − 2) + 4
s − 2 (s − 2) + 4 (s − 2)2 + 4
L
Now
−1
2s + 4
(s − 2)(s2 − 4s + 8)
1
s−2
2
2
= L
−2
+
s−2
(s − 2)2 + 4 (s − 2)2 + 4
1
s−2
2
−1
−1
−1
= 2L
−2L
+L
s−2
(s − 2)2 + 4
(s − 2)2 + 4
−1
= 2 e2x − 2 e2x cos 2x + e2x sin 2x.
Solution of Initial-Value Problems
Here we complete the application of Laplace transforms to the solution of initial-value
problems. For our first example, we finish Example 2, Section 4.2.
Example 6. Find the solution of the initial-value problem
y 0 − 2y = 2e−3x ;
y(0) = −2.
SOLUTION From the Example, if y = y(x) is the solution, then
L[y 0 (x) − 2y(x)] = L[2e−3x ] = 2L[e−3x ] =
L[y 0(x)] − 2L[y(x)] =
2
s+3
sL[y(x)] − y(0) − 2L[y(x)] =
2
s+3
(s − 2)L[y(x)] + 2 =
2
s+3
(s − 2)L[y(x)] =
2
s+3
2
−2
s+3
Therefore,
L[y(x)] = Y (s) =
2
2
−
.
(s − 2)(s + 3) s − 2
Now, by partial fraction decomposition
2
2/5
2/5
=
−
.
(s − 2)(s + 3)
s−2 s+3
Therefore,
2/5
2/5
2
−8/5
2/5
−
−
=
−
s−2 s+3 s−2
s−2 s+3
2/5
−1 −8/5
y(x) = L
−
= − 85 e2x − 25 e−3x .
s−2 s+3
Y (s) =
and
133
Next, we finish Example 3 of Section 4.2.
Example 7. Find the solution of the initial-value problem
y 00 + 4y = 3e2x;
y(0) = 5, y 0 (0) = −2.
SOLUTION From the Example, if y = y(x) is the solution, then
L[y 00(x) + 4y(x)] = L[3e2x ] = 3L[e2x] =
L[y 00(x)] + 4L[y(x)] =
3
s−2
s2 L[y(x)] − sy(0) − y 0 (0) + 4L[y(x)] =
3
s−2
(s2 + 4)L[y(x)] − 5s − (−2) =
3
s−2
(s2 + 4)L[y(x)] =
3
s−2
3
+ 5s − 2
s−2
Therefore,
L[y(x)] = Y (s) =
3
5s − 2
+ 2
.
2
(s − 2)(s + 4) s + 4
Now, by partial fraction decomposition,
3
3
3
3
8
8 s+ 4
=
−
.
(s − 2)(s2 + 4)
s−2
s2 + 4
Therefore,
Y (s) =
3
8
s−2
−
3
8
s + 34
5s − 2
3 1
37 s
11 2
+ 2
=
+
−
2
2
s +4
s +4
8 s−2
8 s +4
8 s2 + 4
and
y(x) = L
−1
[Y (s)] =
=
1
s
2
3 −1
37 −1
11 −1
L
+
L
−
L
8
s−2
8
s2 + 4
8
s2 + 4
3 2x
8e
+
37
8
cos 2x −
11
8
sin 2x.
Example 8. Find the solution of the initial-value problem
y 00 − 5y 0 + 6y = x − 1;
134
y(0) = 0, y 0 (0) = 1.
SOLUTION If y = y(x) is the solution, then
L[y 00(x) − 5y 0 (x) + 6y(x)] = L[x − 1] = L[x] − L[1] =
L[y 00(x)] − 5L[y 0(x)] + 6L[y(x)] =
1−s
s2
s2 L[y(x)] − sy(0) − y 0 (0) − 5 {sL[y(x)] − y(0)} + 6L[y(x)] =
1−s
s2
(s2 − 5s + 6)L[y(x)] − 1 =
1−s
s2
(s2 − 5s + 6)L[y(x)] =
1
1
−
2
s
s
1−s
+ 1.
s2
Therefore,
L[y(x)] = Y (s) =
1−s
1
1−s
1
+
= 2
+
.
s2 (s2 − 5s + 6) s2 − 5s + 6
s (s − 2)(s − 3) (s − 2)(s − 3)
Now, by partial fraction decomposition,
1−s
1
1 1
1 1
1
2
1
+
=
−
+
−
s2 (s − 2)(s − 3)
36 s
6 s2
4 s−2
9 s−3
and
1
−1
1
=
+
.
(s − 2)(s − 3)
s−2 s−3
Therefore,
1
Y (s) = −
36
3
1
1 1
1
7
1
−
+
+
s
6 s2
4 s−2
9 s−3
and
1
y(x) = − 36
+
1
6
x − 34 e2x +
7 3x
9e .
In many applications of differential equations it is not required to determine the solutions
explicitly. Instead what is needed is information about the solutions. Often such information
can be obtained by analyzing their Laplace transforms. The next example illustrates this
type of application.
Example 9. Consider the differential equation
y 00 − y 0 − 6y = 2e−x
on [0, ∞)
together with the condition y(0) = −1. From Chapter 3, we know that the general solution
of the differential equation has the form
y(x) = C1 e−2x + C2 e3x + Ae−x
where C1 , C2 are arbitrary constants and A is a constant which can be determined
135
(∗)
The question we want to examine is: Can we choose a value for α = y 0(0) so that
solution of the resulting initial value-problem
y 00 − y 0 − 6y = 2e−x ;
y(0) = −1, y 0 (0) = α
has limit 0 as x → ∞? Since e−2x → 0 and e−x → 0 as x → ∞, we want to choose
α so that the coefficient of the e3x term is 0.
If y = y(x) is the solution of the initial-value problem, then
L[y 00(x) − y 0 (x) − 6y(x)] = L[2e−x ] =
L[y 00(x)] − L[y 0(x)] − 6L[y(x)] =
2
s+1
s2 L[y(x)] − sy(0) − y 0(0) − {sL[y(x)] − y(0)} − 6L[y(x)] =
2
s+1
(s2 − s − 6)L[y(x)] + s − α − 1 =
2
s+1
(s2 − s − 6)L[y(x)] =
2
s+1
2
+ 1 + α − s.
s+1
Therefore,
L[y(x)] = Y (s) =
=
2
1+α−s
+
(s + 1)(s2 − s − 6) s2 − s − 6
2
1+α−s
+
.
(s + 1)(s + 2)(s − 3) (s + 2)(s − 3)
Now, by partial fraction decomposition,
2
1
− 12
2
A
B
C
=
+
+
=
+ 5 + 10
(s + 1)(s + 2)(s − 3)
s+1 s+2 s−3
s+1 s+2 s−3
and
α−2
− α+3
1+α−s
D
E
5
=
+
=
+ 5 .
(s + 2)(s − 3)
s−2 s−3
s+2
s−3
Combining these results, we have
α+1
1
2α − 3
1
1
1
Y (s) = −
+
−
.
5
s+2
10
s−3
2 s+1
Clearly, if 2α − 3 = 0, that is, if α = 3/2, then the e3x term in (∗) is eliminated. The
resulting solution is:
y(x) = − 12 e−2x − 12 e−x .
This solution has initial values: y(0) = −1, y 0 (0) = 32 , and lim y(x) = 0.
x→∞
136
Exercises 4.3
Find L−1 [F (s)]
1. F (s) =
6
.
s+7
2. F (s) =
1
.
2s + 2
3. F (s) =
s2
1
.
+ 25
4. F (s) =
4
3
−
.
s s−4
5. F (s) =
s+4
.
s2 + 8s + 17
6. F (s) =
4
.
s2 − 6s + 13
7. F (s) =
s2
s+4
.
+ 4s + 8
8. F (s) =
2
1
5
− 2 + 3.
s s
s
9. F (s) =
2
s
− 2
.
2
(s + 2)
s − 2s + 2
10. F (s) =
s2
s+3
.
− 2s + 9
Use partial fraction decomposition to find the inverse Laplace transform.
11. F (s) =
12. F (s) =
13. F (s) =
14. F (s) =
1
.
(s + 1)(s2 + 1)
s2
s+3
.
−s−2
1
.
+ 4)
s(s2
s2
1
.
−1
s2 − 3s − 1
.
s3 + s2 − 2s
s
16. F (s) = 4
.
s −1
15. F (s) =
17. F (s) =
s2 (s
4
.
− 1)(s − 2)
Find the solution of the initial-value problem.
137
18. y 0 − 4y = 0;
y(0) = −2.
19. y 0 + 2y = ex ;
y(0) = 1.
20. y 0 − 3y = e2x ;
y(0) = 2.
21. y 0 + y = sin x;
y(0) = 1.
22. y 00 + 4y = 0;
y(0) = 2, y 0(0) = 2.
23. y 00 − 2y 0 + 2y = 0;
24. y 00 − y = sin x;
25. y 00 − y = ex ;
y(0) = 0, y 0 (0) = 1.
y(0) = 1, y 0 (0) = 1.
y(0) = 1, y 0 (0) = 0.
26. y 00 − y 0 − 2y = sin 2x;
y(0) = 1, y 0(0) = 1.
27. y 00 + 2y 0 + y = x + ex ;
y(0) = − 74 , y 0(0) = 94 .
28. y 00 + 2y 0 + y = 4e−x ;
y(0) = 2, y 0 (0) = −1.
29. y 00 + 3y 0 + 2y = 6ex ;
y(0) = 2, y 0(0) = −1.
30. y 00 − 2y 0 + 5y = 3e−2x ;
y(0) = 1, y 0(0) = 1.
31. Given the initial-value problem
y 00 − y 0 − 6y = 2e−x ;
y(0) = α, y 0(0) = −1.
(See Example 9)
What value should be assigned to α so that the resulting solution will have limit 0
as x → ∞?
32. What initial conditions should be assigned with the differential equation
y 00 + y = e−x
so that lim y(x) = 0 where y = y(x) is the solution.
x→∞
33. Consider the differential equation: y 00 + y 0 − 2y = 3 sin 2x together with the initial
value y(0) = 2. For what value(s) of β = y 0(0) will the resulting solution(s) be
bounded?
34. Consider the differential equation: y 00 + 3y 0 + 2y = x together with the initial value
y 0(0) = −2. For what value(s) of α = y(0) will the resulting solution(s) be bounded?
The Laplace transform method applies to initial-value problems in which the initial
values are specified at x = 0. Actually, the method can be applied when the initial
138
conditions are specified at some point a 6= 0. All that is required is a simple change
of independent variable; a translation. For example, the initial-value problem
d2 y
dy
−3
+ 2y = x;
dx2
dx
y(1) = 0, y 0 (1) = 1
is transformed into
d2y
dy
− 3 + 2y = t + 1,
2
dt
dt
y(0) = 0, y 0(0) = 1
by setting t = x−1. With this transformation, we have: t = 0 when x = 1; x = t+1;
and
dy
dx
=
d2 y
dx2
=
dy dt
dy
dy
=
·1 =
dt dx
dt
dt
d dy
d dy dt
d dy
d2 y
=
=
·1= 2.
dx dx
dt dx dx
dt dt
dt
35. Find the solution of the initial-value problem: y 00 − 3y 0 + 2y = x; y(1) = 0, y 0 (1) = 1
by first solving the transformed problem.
36. Use the Laplace transform method to solve the initial-value problem
y 0 − 2y = 1 + x;
139
y(2) = −1.
4.4 Applications to Discontinuous Functions
In the preceding sections we assumed that the functions under consideration were continuous
on [0, ∞). In this section we consider Laplace transforms of certain types of discontinuous
functions that occur in various applications.
A particular example of the type of discontinuous function that we will be considering
in this section is the unit step function, or Heaviside function u. This function is defined
on (−∞, ∞) by
(
0
x<0
(1)
u(x) =
1
x ≥ 0.
The graph of u is
y
3
2
1
-10 -5
5
10
x
-1
It is clear that u is continuous on (−∞, ∞) except at x = 0. At x = 0, lim u(x)
x→0
does not exist. However, note that the left-hand and right-hand limits of u at x = 0 do
exist:
lim u(x) = 0
and
lim u(x) = 1.
x→0−
x→0+
Recall from calculus that a function f is continuous at a point c if and only if
lim f (x) = f (c), and lim f (x) exists if and only if the left-hand and right-hand limits of
x→c
x→c
f at c both exist and are equal.
DEFINITION 1. (Jump Discontinuity) Let the function f = f (x) be defined on
an interval I and continuous except at a point c ∈ I, c not an endpoint of I. If the
left-hand and right-hand limits of f at c both exist but are not equal, then f is said to
have a jump (or finite ) discontinuity at c.
Example 1. (a) The unit step function u is continuous on (−∞, ∞) except at x = 0
where it has a jump discontinuity.
(b) Let f be defined on [−5, 5] by

x+5


 1


−
f (x) =
x


2x



−(x − 5)
140
−5 ≤ x < −2
−2 ≤ x < 0
0≤x≤2
2<x≤5
The graph of f is
y
4
3
1
-5
-2
5
2
x
-1
Clearly f is discontinuous at x = −2, x = 0, and x = 2. Since
lim f (x) = 3 and
x→−2−
lim f (x) = 12 ,
x→−2+
and
lim f (x) = 4
x→2−
and
lim f (x) = 3,
x→2+
f has jump discontinuities at x = −2 and x = 2.
The discontinuity at x = 0 is not a jump discontinuity because
exist (f (x) → ∞ as x → 0− ).
lim f (x) does not
x→0−
(c) Let g be defined on [0, ∞) by

x



 3
g(x) =
 2


 −(x−5)
e
0≤x<1
x=1
1≤x<5
x≥5
The graph of g is
y
3
2
1
1
x
5
-1
This function has jump discontinuities at x = 1 and x = 5. Note that the value of g
at x = 1 is independent of the left-hand and right-hand limits of g at x = 1.
DEFINITION 2. (Piecewise Continuous) A function f defined on an interval I is
piecewise continuous on I if it is continuous on I except for at most a finite number of
points c1, c2, . . . , cn of I at which it has a jump discontinuity.
141
Remark A continuous function is also piecewise continuous.
Referring to Example 1, the unit step function u is piecewise continuous on (−∞, ∞);
the function g in (c) is piecewise continuous on [0, ∞). the function f in (b) is
not piecewise continuous on [−5, 5] because the discontinuity at x = 0 is not a jump
discontinuity.
In this section we want to apply Laplace transform methods to piecewise continuous
functions. Before we can calculate the Laplace transform of a piecewise function we have
to know how to integrate such a function.
Suppose that f is piecewise continuous on [a, b] with jump discontinuities at c1 <
c2 < · · · < cn . It follows from the definition of the definite integral that
Z
b
f (x) dx =
Z
c1
f (x) dx +
Z
a
a
c2
f (x) dx + · · · +
Z
c1
b
f (x) dx.
cn
Example 2. Let f be defined on [0, 3] by
(
x
f (x) =
4 − (x − 1)2
0≤x≤1
1<x≤3
The graph of f is
y
4
1
1
2
3
x
-1
We have
Z
f (x) dx =
Z
=
3
0
1
x dx +
0
Z
3
(4 − (x − 1)2) dx
1
1 3
x2
(x − 1)3
35
+ 4x −
= .
2 0
3
6
1
The integration of piecewise continuous functions can be extended to include (improper)
integrals on an infinite interval.
142
Example 3. Consider the function g of Example 1:
Z ∞
Z 5
Z 1
Z ∞
g(x) dx =
x dx +
2 dx +
e−(x−5) dx
0
0
1
5
=
x2
2
1
+
[2x]51
0
+ lim
Z
b
b→∞ 5
e−(x−5) dx
=
h
ib
1
+ 8 + lim −e−(x−5)
b→∞
2
5
=
17
19
+ [0 + 1] = .
2
2
The following theorem is an extension of Theorem 1, Section 4.2.
THEOREM 1. If the function f is piecewise continuous on [0, ∞), and of exponential
order λ, then the Laplace transform L[f (x)] exists for s > λ.
Let c be a real number. The translation of the unit step function u by c is the
function uc = u(x − c) defined on (−∞, ∞) by
(
0
x<c
u(x − c) =
(2)
1
x ≥ c.
The graph of uc for c > 0 is
y
2
1
x
c
-1
Example 4. The Laplace transform of uc , c > 0 is
L[u(x − c)] =
Proof: By definition
L[u(x − c)] =
Z
e−cs
, s > 0.
s
(3)
∞
e−sx u(x − c) dx
0
=
Z
c
e−sx · 0 dx +
0
=
=
lim
b→∞
∞
e−sx · 1 dx
c
Z
b
b→∞ c
lim
Z
e
−sx
dx = lim
b→∞
e−sx
−s
b
c
e−sb e−sc
e−cs
+
=
, s > 0.
−s
s
s
143
Note that if c = 0, then u0 = u(x − 0) = u(x) ≡ 1 on [0, ∞), and L[u(x)] = L[1] =
1/s, s > 0 as we saw in Section 4.1.
Example 5.
(a) Let f be the function defined on [0, ∞) by
(
1
0≤x<5
f (x) =
2
5 ≤ x < ∞.
y
3
2
1
1
5
3
x
7
-1
Then
L[f (x)] =
Z
∞
e
−sx
f (x) dx =
Z
0
=
=
5
e
−sx
· 1 dx +
0
−e−sx
s
5
Z
+ lim
b→∞
0
Z
∞
e−sx · 2 dx
5
b
2e−sx dx
0
−e−5s 1
2e−sx
+ + lim
s
s b→∞ −s
b
5
=
2e−sb 2e−5s
−e−5s 1
+ + lim
+
s
s b→∞ −s
s
=
−e−5s 1 2e−5s
1 e−5s
+ +
= +
,
s
s
s
s
s
s > 0.
(b) We calculate L[g(x)] where g is the function given in Example 1(c).
L[g(x)] =
Z
=
∞
e
−sx
g(x) dx =
0
=
Z
1
e
−sx
· x dx +
0
−sx 1
−xe−sx
e
− 2
s
s
0
+
Z
5
e
−sx
1
−sx 5
−2e
s
1
· 2 dx +
Z
∞
e−sx e−(x−5) dx
5
+ lim
b→∞
Z
b
e−[x(s+1)−5] dx
0
"
#∞
−e−s
−e−[x(s+1)−5]
1
−2e−5s 2e−s
−e−s
+ 2 + 2+
+
+ lim
b→∞
s
s
s
s
s
s+1
5
= −
=
e−s
1
2e−5s 2e−s
e−s
e−5s
− 2 + 2−
+
+
s
s
s
s
s
s+1
1
e−s e−s
2e−5s
e−5s
+
−
−
+
.
s2
s
s2
s
s+1
144
The unit step function and its translations can be used to obtain translations of arbitrary
functions. For example, if f is defined on [0, ∞) and c > 0, then the function
(
0
x<c
fc (x) = f (x − c)u(x − c) =
f (x − c)u(x − c)
x≥c
is the function f shifted c units to the right as illustrated in the figure below.
y
y
x
x
c
c
THEOREM 2. Let f be defined on [0, ∞) and suppose L[f (x)] = F (s) exists for
s > λ. If c > 0, then L[fc (x)] exists for s > λ and is given by
L[fc (x)] = L[f (x − c)u(x − c)] = e−cs F (s).
Proof: By the definition,
L[fc (x)] = L[f (x − c)u(x − c)] =
Z
∞
e−sx f (x − c)u(x − c) dx
0
=
lim
Z
b→∞ c
b
e−sx f (x − c) dx.
Now let t = x − c. Then
x = t + c,
dx = dt,
and
With this change of variable,
Z b
lim
e−sx f (x − c) dx =
b→∞
c
lim
t = 0 when
Z
= e
= e−cs
lim
b→∞
Z
145
e−s(t+c) f (t) dt
Z
b−c
e
−st
f (t) dt
0
∞
e−st f (t) dt = e−cs F (s)
0
since b − c → ∞ as b → ∞.
b−c
b→∞ 0
−cs
x = c.
This theorem can be expressed in an equivalent manner using the inverse Laplace transform.
COROLLARY
If L−1 [F (s)] = f (x) and c > 0, then
L−1 [e−cs F (s)] = f (x − c)u(x − c) = fc (x).
We now illustrate how to use translations of the unit step function and Theorem 2 to
calculate Laplace transforms of piecewise continuous functions.
Example 6. We calculate L[f (x)] where f is the function given in Example 4:
(
1
0≤x<5
f (x) =
2
5 ≤ x < ∞.
Since f has a jump discontinuity at x = 5, we’ll write f in terms of u(x − 5). Define
f1 (x) by
(
1
0≤x<5
f1 (x) =
= 1 − u(x − 5)
0
5≤x<∞
and f2 (x) by
f2 (x) =
(
0
2
0≤x<5
= 2u(x − 5).
5≤x<∞
Then
f (x) = f1 (x) + f2 (x) = 1 − u(x − 5) + 2u(x − 5) = 1 + u(x − 5)
and
L[f (x)] = L[1 + u(x − 5)] = L[1] + L[u(x − 5)] =
Example 7. Let h(x) =
(
x2
3
1 e−5s
+
,
s
s
s > 0.
0≤x<2
. Calculate L[h(x)].
2 ≤ x < ∞.
SOLUTION The graph of h is
y
4
3
2
1
1
2
x
3
-1
Let h1 (x) = x2 − x2 u(x − 2) and h2 (x) = 3u(x − 2). Then
h(x) = h1 (x) + h2 (x) = x2 − x2 u(x − 2) + 3u(x − 2)
146
and
L[h(x)] = L[x2 ] − L[x2 u(x − 2)] + 3L[u(x − 2)] =
2
e−2s
2
−
L[x
u(x
−
2)]
+
3
.
s3
s
Before we can apply Theorem 2 to calculate L[x2 u(x − 2)] we must have the coefficient
x2 of u(x − 2) expressed as a function of (x − 2). Since
x2 = [(x − 2) + 2]2 = (x − 2)2 + 4(x − 2) + 4
we have
L[x2 u(x − 2)] = L (x − 2)2u(x − 2) + 4(x − 2)u(x − 2) + 4u(x − 2)
= L[(x − 2)2u(x − 2)] + 4L[(x − 2)u(x − 2)] + 4L[u(x − 2)]
= e−2s
2
1
1
+ 4e−2s 2 + 4e−2s .
s3
s
s
It now follows that
2
2e−2s 4e−2s e−2s
e−2s
2
−2s 2
−2s 1
−2s 1
L[h(x)] = 3 − e
+
4e
+
4e
−
− 2 −
+
3
=
.
s
s3
s2
s
s
s3
s3
s
s
Example 8. Let


 2
g(x) =
x


4
0≤x<2
2≤x<4 .
x≥4
The graph of g is
y
4
3
2
1
2
x
4
Set
g1(x) = 2 − 2 u(x − 2)
g2(x) = x u(x − 2) − x u(x − 4) = [(x − 2) + 2)]u(x − 2) − [(x − 4) + 4]u(x − 4)
= (x − 2)u(x − 2) + 2u(x − 2) − (x − 4)u(x − 4) − 4u(x − 4)
g3(x) = 4 u(x − 4).
Then
g(x) = g1 (x) + g2 (x) + g3(x) = 2 + (x − 2)u(x − 2) − (x − 4)u(x − 4).
147
Therefore,
L[g(x)] =
2
1
1
+ e−2s 2 − e−4s 2
s
s
s
The Corollary to Theorem 2 is used to calculate inverse Laplace transforms.
Example 9. Suppose
1
e−4s
.
+
s2 s − 2
F (s) =
Then
−4s 1
e−4s
−1 1
−1 e
L
+L
+
=L
s2 s − 2
s2
s−2
From the Table of Laplace transforms L−1 1/s2 = x. To calculate L−1 e−4s /(s − 2) ,
let F (s) = 1/(s − 2). Then L−1 [F (s)] = e2x . Therefore, by the Corollary to Theorem 2,
−4s −1 e
L
= e2(x−4) u(x − 4).
s−2
−1
It now follows that
L
−1
[F (s)] = x + e
2(x−4)
u(x − 4) =
(
x
x+
e2(x−4)
0≤x<4
.
4≤x<∞
Example 10. Suppose
F (s) =
Then
L
−1
[F (s)] = L
−1
e−s
e−πs
+
.
(s + 1)2 s2 + 4
−πs e−s
e
−1
+L
.
2
(s + 1)
s2 + 4
To calculate L−1 [e−s /(s + 1)2], let F1 (s) = 1/(s + 1)2. Then, from the Table of Laplace
transforms, L−1 [F1 (s)] = xe−x . Therefore,
e−s
−1
L
= (x − 1)e−(x−1) u(x − 1).
(s + 1)2
To calculate L−1 [e−πs /(s2 + 4)], let
F2 (s) =
s2
1
1 2
=
.
+4
2 s2 + 4
From the Table of Laplace transforms, L−1 [F2(s)] = 12 sin 2x. Therefore,
−πs e
L−1 2
= 12 sin 2(x − π) u(x − π) = 12 sin 2x u(x − π). (sin 2(x − π) = sin 2x)
s +4
Finally,
L−1 [F (s)] = (x − 1)e−(x−1) u(x − 1) + 12 sin


0

−(x−1)
=
(x − 1)e


−(x−1)
(x − 1)e
+ 12 sin 2x
148
2x u(x − π)
0≤x<1
1≤x<π
x≥π
Exercises 4.4
Use the definition of the Laplace transform to find L[f ].
1.
2.
3.
4.
5.
6.


x
0≤x<1


f (x) =
x−1
1≤x<2 .



0
x≥2
(
−1
0≤x<1
f (x) =
.
x−1
x≥1
(
0
0≤x<5
f (x) =
.
2
x≥5
(
2x
0≤x<3
f (x) =
.
1
x≥3

1
0≤x<2



f (x) =
x−2
2≤x<4 .


 −(x−4)
e
x≥4
(
1
0≤x<1
f (x) =
−1
1≤x<2
and extended periodically on [2, ∞). See Exercises 4.1, Problem 10.
(
2x
0≤x<2
7. f (x) =
2
2≤x<4
and extended periodically on [4, ∞). (Sketch the graph of f .)
8. f (x) = x − n on [n, n + 1), n = 0, 1, 2, . . .. (Sketch the graph of f .)
Express the function f in terms of the unit step function and its translations. Then find
L[f ].
9. f (x) =
(
x2
0≤x<3
3x
x≥3


sin x


10. f (x) =
sin 2x



sin 3x
.
0≤x<π
π ≤ x < 2π . Sketch the graph of f .
x ≥ 2π
149


0
0≤x<π


11. f (x) =
1 + cos x
π ≤ x < 2π .



2 cos x
x ≥ 2π


x
0≤x<2


12. f (x) =
0
2≤x<4 .



x≥4
(x − 4)2


x
0≤x<1




 2−x
1≤x<3
. Sketch the graph of f .
13. f (x) =

x−4
3≤x<4





0
x≥4
Calculate the inverse Laplace transform of F (s).
14. F (s) =
e−s
2e−3s 6e−4s
+
−
.
s
s
s
15. F (s) =
1 + e−πs
.
s2 + 1
16. F (s) =
1
e−2s
−
.
s+1 s+1
17. F (s) =
s + (s − 1)e−πs
.
s2 + 1
18. F (s) =
2
e−2s 2e−2s 4e−3s
+
+
−
.
s2
s2
s3
s
19. F (s) =
e−2s
.
s(s + 1)
20. F (s) =
1 − e−2s
.
s2 + π 2
150
4.5 Initial-Value Problems with Piecewise Continuous Nonhomogeneous
Terms
In this section we illustrate the application of Laplace transforms to the solution of nonhomogeneous initial-value problems in which the forcing function f is piecewise continuous.
Example 1. Find the solution of the initial-value problem
y 0 + 2y = f (x);
where f (x) =
(
x
5
y(0) = 4
0≤x<3
x ≥ 3.
SOLUTION The function f has a jump discontinuity at x = 3. Therefore, the first step
is to express f in terms of u3 (x) = u(x − 3). You should verify that
f (x) = x−x u(x−3)+5u(x−3) = x−[(x−3)+3]u(x−3)+5u(x−3) = x−(x−3)u(x−3)+2u(x−3).
Now, taking the Laplace transform of the equation, we get
L[y 0 + 2y] = sL[y] − y(0) + 2L[y] =
1
e−3s
e−3s
−
+
2
s2
s2
s
Applying the initial condition and solving for L[y], we find that
L[y] = Y (s) =
=
1
e−3s
e−3s
4
+
+
2
+
2
2
s (s + 2) s (s + 2)
s(s + 2) s + 2
4s2 + 1
(2s + 1)e−3s
+
.
s2 (s + 2)
s2 (s + 2)
By partial fraction decomposition
4s2 + 1
s2 (s + 2)
(2s + 1)e−3s
s2 (s + 2)
= −
=
17 1
11 1 1
+
+
2
4 s 2s
4 s+2
3 e−3s 1 e−3s 3 e−3s
−
+
4 s
2 s2
4 s+2
Therefore,
Y (s) = −
11 1 1
17 1
3 e−3s 1 e−3s 3 e−3s
+
−
+
+
+
4 s 2 s2
4 s+2 4 s
2 s2
4 s+2
and
−2x + 3 u(x − 3) + 1 (x − 3) u(x − 3) − 3 e−2(x−3) u(x − 3)
y(x) = L−1 [Y (s)] = − 14 + 12 x + 17
4 e
4
2
4
( 1 1
−2x
− 4 + 2 x + 17
0≤x<3
4e
=
17 −2x
3 −2(x−3)
x−1+ 4 e
− 4e
x≥3
151
The graph of y is
y
x
3
Example 2. Find the solution of the initial-value problem
y 00 + 4y = f (x);
where f (x) =
(
1
−2
y(0) = 1, y 0(0) = 0
0 ≤ x < π/2
x ≥ π/2.
SOLUTION The function f has a jump discontinuity at x = π/2. Therefore, we express
f in terms of u(x − π/2):
f (x) = 1 − 3u(x − π/2).
Applying the Laplace transform to the equation and using the initial conditions, we get
L[y 00 + 4y] = s2 L[y] − sy(0) − y 0 (0) + 4L[y] =
(s2 + 4)L[y] − s =
1 3e−πs/2
−
s
s
1 3e−πs/2
−
.
s
s
Solving for L[y] = Y (s), we have
Y (s) =
1
1
s
− 3e−πs/2 2
+ 2
.
+ 4)
s(s + 4) s + 4
s(s2
By partial fraction decomposition,
1
11 1 s
=
−
.
+ 4)
4 s 4 s2 + 4
s(s2
Therefore,
Y (s) =
11 1 s
3 e−πs/2 3 −πs/2 s
s
−
−
+ e
+ 2
2
2
4 s 4 s +4 4 s
4
s +4 s +4
and
y(x) = L−1 [Y (s)] =
=
1
4
− 14 cos 2x − 34 u(x − π/2) + 34 cos 2(x − π/2)u(x − π/2) + cos 2x

 14 + 34 cos 2x
0 ≤ x < π/2

− 12
x ≥ π/2.
152
The graph of y is
y
1
x
Π
2
-0.5
-1
Exercises 4.5
Solve the initial-value problem.
1. y 0 − 2y = f (x); y(0) = 2
where
f (x) =
2. y 0 + 3y = f (x); y(0) = 1
4. y 00 + 4y = f (x);
1
0≤x<1
2
x≥1
.
where
f (x) =
3. y 00 + y = f (x);
(
(
sin x
0≤x<π
0
x≥π
.
y(0) = 0, y 0(0) = 1 where
(
1
0≤x<1
f (x) =
.
0
x≥1
y(0) = 1, y 0 (0) = 0
(
where
0
0≤x<2
1
x≥2
f (x) =
5. y 00 + 2y 0 + y = f (x);
6. y 00 + 4y = f (x);
.
y(0) = 0, y 0(0) = 0
(
1
f (x) =
x−1
where
0≤x<2
.
x≥2
y(0) = 0, y 0 (0) = 0 where
(
sin x
0 ≤ x < 2π
f (x) =
.
0
x ≥ 2π
153
7. y 00 − 4y 0 + 3y = f (x);
8. y 00 − 3y 0 + 2y = f (x);
9. y 00 + 2y 0 + y = f (x);
10. y 00 − 4y 0 + 4y = f (x);
y(0) = 0, y 0 (0) = 0
(
−1
f (x) =
1
where
0≤x<1
.
x≥1
y(0) = 0, y 0 (0) = 0
(
0
f (x) =
2x − 4
where
y(0) = 3, y 0(0) = −1
(
ex
f (x) =
ex − 1
where
0≤x<2
.
x≥2
0≤x<1
.
x≥1
y(0) = 1, y 0 (0) = −1 where
(
e2x
0≤x<2
f (x) =
.
−e2x
x≥2
154