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Published Date:28-07-2017
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The Laplace Transform:
Theory and Applications
Joel L. Schiff
Springer.
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. Principles
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CHAPTER .
Ordinary and partial differential equations describe the way certain
quantities vary with time, such as the current in an electrical circuit,
the oscillations of a vibrating membrane, or the flow of heat through
an insulated conductor. These equations are generally coupled with
initial conditions that describe the state of the system at time t 0.
A very powerful technique for solving these problems is that of
the Laplace transform, which literally transforms the original differ-
ential equation into an elementary algebraic expression. This latter
can then simply be transformed once again, into the solution of the
original problem. This technique is known as the “Laplace transform
method.” It will be treated extensively in Chapter 2. In the present
chapter we lay down the foundations of the theory and the basic
properties of the Laplace transform.
1.1 The Laplace Transform
Suppose that f is a real- or complex-valued function of the (time)
variablet 0 and s is a real or complex parameter. We define the
11. Basic Principles
2
Laplace transform of f as
∞
−st
F(s) L f (t) e f (t) dt
0
τ
−st
lim e f (t) dt (1.1)
τ→∞
0
whenever the limit exists (as a finite number). When it does, the
integral (1.1) is said toconverge. If the limit does not exist, the integral
is said to diverge and there is no Laplace transform defined for f . The
notation L(f ) will also be used to denote the Laplace transform of
f , and the integral is the ordinary Riemann (improper) integral (see
Appendix).
The parameter s belongs to some domain on the real line or in
the complex plane. We will choose s appropriately so as to ensure
the convergence of the Laplace integral (1.1). In a mathematical and
technical sense, the domain of s is quite important. However, in a
practical sense, when differential equations are solved, the domain
of s is routinely ignored. When s is complex, we will always use the
notation s x + iy.
The symbol L is the Laplace transformation, which acts on
functions f f (t) and generates a new function, F(s) L f (t) .
Example 1.1. If f (t) ≡ 1for t ≥ 0, then
∞
−st
L f (t) e 1 dt
0
τ
−st
e
lim
τ→∞
−s
0
−sτ
e 1
lim + (1.2)
τ→∞ −s s
1
s
provided of course thats 0 (if s is real). Thus we have
1
L(1) (s 0). (1.3)
s1.1. The Laplace Transform
3
If s ≤ 0, then the integral would diverge and there would be no re-
sulting Laplace transform. If we had taken s to be a complex variable,
the same calculation, with Re(s) 0, would have given L(1) 1/s.
In fact, let us just verify that in the above calculation the integral
can be treated in the same way even if s is a complex variable. We
require the well-known Euler formula (see Chapter 3)
iθ
e cos θ + i sin θ, θ real, (1.4)
iθ
and the fact that e 1. The claim is that (ignoring the minus sign
as well as the limits of integration to simplify the calculation)
st
e
st
e dt , (1.5)
s
for s x + iy any complex number 0. To see this observe that
st (x+iy)t
e dt e dt
xt xt
e cos yt dt + i e sin yt dt
by Euler’s formula. Performing a double integration by parts on both
these integrals gives
xt
e
st
e dt (x cos yt + y sin yt) + i(x sin yt − y cos yt) .
2 2
x + y
Now the right-hand side of (1.5) can be expressed as
st (x+iy)t
e e
s x + iy
xt
e (cos yt + i sin yt)(x − iy)
2 2
x + y
xt
e
(x cos yt + y sin yt) + i(x sin yt − y cos yt) ,
2 2
x + y
which equals the left-hand side, and (1.5) follows.
Furthermore, we obtain the result of (1.3) for s complex if we
take Re(s) x 0, since then
−sτ −xτ
lim e lim e 0,
τ→∞ τ→∞1. Basic Principles
4
killing off the limit term in (1.3).
Let us use the preceding to calculate L(cos ωt) and L(sin ωt)
(ω real).
Example 1.2. We begin with
∞
iωt −st iωt
L(e ) e e dt
0
τ
(iω−s)t
e
lim
τ→∞
iω − s
0
1
,
s − iω
iωτ −sτ −xτ
since lim e e lim e 0, provided x Re(s)
τ→∞ τ→∞
−iωt
0. Similarly, L(e ) 1/(s + iω). Therefore, using the linearity
property of L, which follows from the fact that integrals are linear
operators (discussed in Section 1.6),
iωt −iωt iωt −iωt
L(e ) +L(e ) e + e
L L(cos ωt),
2 2
and consequently,
1 1 1 s
L(cos ωt) + . (1.6)
2 2
2 s − iω s + iω s + ω
Similarly,
1 1 1 ω
L(sin ωt) − Re(s) 0 .
2 2
2i s − iω s + iω s + ω
(1.7)
The Laplace transform of functions defined in a piecewise
fashion is readily handled as follows.
Example 1.3. Let (Figure 1.1)
t 0 ≤ t ≤ 1
f (t)
1 t 1.Exercises 1.1
5
f t
O t
FIGURE 1.1
From the definition,
∞
−st
L f (t) e f (t) dt
0
1 τ
−st −st
te dt + lim e dt
τ→∞
0 1
1 τ
−st 1 −st
te 1 e
−st
+ e dt + lim
τ→∞
−s s −s
0 0 1
−s
1 − e
Re(s) 0 .
2
s
Exercises 1.1
1. From the definition of the Laplace transform, compute L f (t)
for
2t
(a) f (t) 4t (b) f (t) e
(c) f (t) 2 cos 3t (d) f (t) 1 − cos ωt
2t t
(e) f (t) te (f) f (t) e sin t
π
sin ωt 0t
1 t ≥ a
ω
(g) f (t) (h) f (t)
π
0 ta
0 ≤ t
ω1. Basic Principles
6
2 t ≤ 1
(i) f (t)
t
e t 1.
2. Compute the Laplace transform of the function f (t) whose graph
is given in the figures below.
f t f t
a
b
O tO t
FIGURE E.1 FIGURE E.2
1.2 Convergence
Although the Laplace operator can be applied to a great many
functions, there are some for which the integral (1.1) does not
converge.
2
(t )
Example 1.4. For the function f (t) e ,
τ τ
2 2
−st t t −st
lim e e dt lim e dt ∞
τ→∞ τ→∞
0 0
for any choice of the variable s, since the integrand grows without
bound as τ →∞.
In order to go beyond the superficial aspects of the Laplace trans-
form, we need to distinguish two special modes of convergence of
the Laplace integral.
The integral (1.1) is said to be absolutely convergent if
τ
−st
lim e f (t) dt
τ→∞
0
exists. If L f (t) does converge absolutely, then
τ τ
−st −st
e f (t) dt ≤ e f (t)dt → 0
τ τExercises 1.2
7
as τ →∞, for all τ τ. This then implies thatL f (t) also converges
∗
in the ordinary sense of (1.1).
There is another form of convergence that is of the utmost im-
portance from a mathematical perspective. The integral (1.1) is said
to converge uniformly for s in some domain in the complex plane if
for anyε 0, there exists some number τ such that if τ ≥ τ , then
0 0
∞
−st
e f (t) dt ε
τ
for all s in . The point here is that τ can be chosen sufficiently
0
large in order to make the “tail” of the integral arbitrarily small,
independent of s.
Exercises 1.2
1. Suppose that f is a continuous function on 0, ∞) and f (t)≤
M ∞ for 0 ≤t ∞.
(a) Show that the Laplace transform F(s) L f (t) con-
verges absolutely (and hence converges) for any s satisfying
Re(s) 0.
(b) Show that L f (t) converges uniformly if Re(s) ≥ x 0.
0
(c) Show that F(s) L f (t) → 0as Re(s)→∞.
t
2. Let f (t) e on 0, ∞).
t
(a) Show that F(s) L(e ) converges for Re(s) 1.
t
(b) Show that L(e ) converges uniformly if Re(s) ≥ x 1.
0
∗
Convergence of an integral
∞
ϕ(t)dt
0
is equivalent to the Cauchy criterion:
τ
ϕ(t)dt →0as τ →∞,τ τ.
τ1. Basic Principles
8
t
(c) Show that F(s) L(e ) → 0as Re(s)→∞.
3. Show that the Laplace transform of the function f (t) 1/t,t 0
does not exist for any value of s.
1.3 Continuity Requirements
Since we can compute the Laplace transform for some functions and
2
(t )
not others, such as e , we would like to know that there is a large
class of functions that do have a Laplace tranform. There is such a
class once we make a few restrictions on the functions we wish to
consider.
Definition 1.5. A function f has ajump discontinuity at a point
t if both the limits
0
− +
lim f (t) f (t ) and lim f (t) f (t )
0 0
− +
t→t t→t
0 0
− + − +
exist (as finite numbers) and f (t ) f (t ). Here, t → t and t → t
0 0 0 0
mean that t → t from the left and right, respectively (Figure 1.2).
0
Example 1.6. The function (Figure 1.3)
1
f (t)
t − 3
f t
f t
f t
Ott
FIGURE 1.21.3. Continuity Requirements
9
f t
O t
FIGURE 1.3
f t
Ot FIGURE 1.4
has a discontinuity at t 3, but it is not a jump discontinuity since
− +
neither lim f (t) nor lim f (t) exists.
t→3 t→3
Example 1.7. The function (Figure 1.4)
2
t
−
2
e t 0
f (t)
0 t 0
has a jump discontinuity at t 0 and is continuous elsewhere.
Example 1.8. The function (Figure 1.5)
0 t 0
f (t)
1
cos t 0
t
is discontinuous at t 0, but lim + f (t) fails to exist, so f does not
t→0
have a jump discontinuity at t 0.1. Basic Principles
10
f t
Ot
FIGURE 1.5
f t
O bt
FIGURE 1.6
The class of functions for which we consider the Laplace
transform defined will have the following property.
Definition 1.9. A function f is piecewise continuous on the in-
+
+
terval 0, ∞) if (i) lim f (t) f (0 ) exists and (ii) f is continuous
t→0
on every finite interval (0,b) except possibly at a finite number
of points τ ,τ ,...,τ in (0,b) at which f has a jump discontinuity
1 2 n
(Figure 1.6).
The function in Example 1.6 is not piecewise continuous on
0, ∞). Nor is the function in Example 1.8. However, the function
in Example 1.7 is piecewise continuous on 0, ∞).
An important consequence of piecewise continuity is that on
each subinterval the function f is also bounded. That is to say,
f (t)≤ M,τ tτ,i 1, 2,...,n − 1,
i i i+1
for finite constants M .
iExercises 1.3
11
In order to integrate piecewise continuous functions from 0 to b,
one simply integrates f over each of the subintervals and takes the
sum of these integrals, that is,
b τ τ b
1 2
f (t) dt f (t) dt + f (t) dt +···+ f (t) dt.
0 0 τ τ
1 n
This can be done since the function f is both continuous and
bounded on each subinterval and thus on each has a well-defined
(Riemann) integral.
Exercises 1.3
Discuss the continuity of each of the following functions and locate
any jump discontinuities.
1
1. f (t)
1 + t
1
2. g(t) t sin (t 0)
t
tt ≤ 1
3. h(t)
1
t 1
2
1 + t
sinh t
t 0
4. i(t)
t
1 t 0
1 1
5. j(t) sinh (t 0)
t t
−t
1 − e
t 0
6. k(t)
t
0 t 0
12na ≤t (2n + 1)a
7. l(t) a 0, n 0, 1, 2, ...
−1(2n + 1)a ≤t (2n + 2)a
t
8. m(t) +1, for t ≥ 0,a 0, where x greatest integer ≤ x.
a1. Basic Principles
12
1.4 Exponential Order
The second consideration of our class of functions possessing a well-
defined Laplace transform has to do with the growth rate of the
functions. In the definition
∞
−st
L f (t) e f (t) dt,
0
when we takes 0 orRe(s) 0 , the integral will converge as long
as f does not grow too rapidly. We have already seen by Example 1.4
2
t
that f (t) e does grow too rapidly for our purposes. A suitable rate
of growth can be made explicit.
Definition 1.10. A function f has exponential order α if there
exist constantsM 0 and α such that for some t ≥ 0,
0
αt
f (t)≤ Me,t ≥ t .
0
at
Clearly the exponential function e has exponential order α a,
n
whereas t has exponential order α for anyα 0 and any n ∈ N
(Exercises 1.4, Question 2), and bounded functions like sin t, cos t,
−1 −t
tan t have exponential order 0, whereas e has order −1. How-
2
t
ever, e does not have exponential order. Note that ifβα, then
αt βt
exponential order α implies exponential order β, since e ≤ e ,
t ≥ 0. We customarily state the order as the smallest value of α that
works, and if the value itself is not significant it may be suppressed
altogether.
Exercises 1.4
1. If f and f are piecewise continuous functions of orders α and
1 2
β, respectively, on 0, ∞), what can be said about the continuity
and order of the functions
(i) c f + c f , c ,c constants,
1 1 2 2 1 2
(ii) f · g?
n
2. Show that f (t) t has exponential order α for anyα 0, n ∈ N.
2
t
3. Prove that the function g(t) e does not have exponential order.1.5. The ClassL
13
1.5 The Class L
We now show that a large class of functions possesses a Laplace
transform.
Theorem 1.11. If f is piecewise continuous on 0, ∞) and of exponen-
tial order α, then the Laplace transform L(f ) exists for Re(s)α and
converges absolutely.
Proof. First,
αt
f (t)≤ M e,t ≥ t ,
1 0
for some real α. Also, f is piecewise continuous on 0,t and hence
0
bounded there (the bound being just the largest bound over all the
subintervals), say
f (t)≤ M , 0tt .
2 0
αt
Since e has a positive minimum on 0,t , a constant M can be
0
chosen sufficiently large so that
αt
f (t)≤ Me,t 0.
Therefore,
τ τ
−st −(x−α)t
e f (t)dt ≤ M e dt
0 0
τ
−(x−α)t
Me
−(x − α)
0
−(x−α)τ
M Me
− .
x − α x − α
Letting τ →∞ and noting that Re(s) xα yield
∞
M
−st
e f (t)dt ≤ . (1.8)
x − α
0
Thus the Laplace integral converges absolutely in this instance (and
hence converges) for Re(s)α. 1. Basic Principles
14
at
Example 1.12. Let f (t) e , a real. This function is continuous
on 0, ∞) and of exponential order a. Then
∞
at −st at
L(e ) e e dt
0
∞
−(s−a)t
e dt
0
∞
−(s−a)t
e 1
Re(s)a .
−(s − a) s − a
0
The same calculation holds for a complex and Re(s) Re(a).
Example 1.13. Applying integration by parts to the function f (t)
t (t ≥ 0), which is continuous and of exponential order, gives
∞
−st
L(t) te dt
0
∞
−st ∞
−te 1
−st
+ e dt
s s
0
0
1
L(1) provided Re(s) 0
s
1
.
2
s
Performing integration by parts twice as above, we find that
∞
2 −st 2
L(t ) e t dt
0
2
Re(s) 0 .
3
s
By induction, one can show that in general,
n
n
L(t ) Re(s) 0 (1.9)
n+1
s
for n 1, 2, 3, ... . Indeed, this formula holds even for n 0, since
0 1, and will be shown to hold even for non-integer values of n
in Section 2.1.
Let us define the class L as the set of those real- or complex-
valued functions defined on the open interval (0, ∞) for which theExercises 1.5
15
Laplace transform (defined in terms of the Riemann integral) exists
for some value of s. It is known that whenever F(s) L f (t) exists
for some value s , then F(s) exists for all s with Re(s) Re(s ), that
0 0
is, the Laplace transform exists for all s in some right half-plane (cf.
Doetsch 2, Theorem 3.4). By Theorem 1.11, piecewise continuous
functions on 0, ∞) having exponential order belong to L. However,
there certainly are functions in L that do not satisfy one or both of
these conditions.
Example 1.14. Consider
2 2
t t
f (t) 2te cos(e ).
Then f (t) is continuous on 0, ∞) but not of exponential order.
However, the Laplace transform of f (t),
∞
2 2
−st t t
L f (t) e 2te cos(e )dt,
0
exists, since integration by parts yields
∞
∞
2 2
−st t −st t
L f (t) e sin(e ) + s e sin(e ) dt
0
0
2
t
− sin(1) + sL sin(e ) Re(s) 0 .
and the latter Laplace transform exists by Theorem 1.11. Thus we
have a continuous function that is not of exponential order yet
nevertheless possesses a Laplace transform. See also Remark 2.8.
Another example is the function
1
f (t) √ . (1.10)
t
We will compute its actual Laplace transform in Section 2.1 in the
context of the gamma function. While (1.10) has exponential order
α 0 f (t)≤ 1, t ≥ 1 , it is not piecewise continuous on 0, ∞)
+
since f (t)→∞ as t → 0 , that is, t 0 is not a jump discontinuity.
Exercises 1.5
2 2
t t
1. Consider the function g(t) te sin(e ).1. Basic Principles
16
(a) Is g continuous on 0, ∞)? Does g have exponential order?
(b) Show that the Laplace transform F(s) exists for Re(s) 0.
(c) Show that g is the derivative of some function having
exponential order.
2. Without actually determining it, show that the following func-
tions possess a Laplace transform.
sin t 1 − cos t
(a) (b)
t t
2
(c) t sinh t
3. Without determining it, show that the function f , whose graph is
given in Figure E.3, possesses a Laplace transform. (See Question
3(a), Exercises 1.7.)
f t
Oa a a at
FIGURE E.3
1.6 Basic Properties of the Laplace
Transform
Linearity. One of the most basic and useful properties of the
Laplace operatorL is that of linearity, namely, if f ∈ L forRe(s)α,
1
f ∈ L for Re(s)β, then f + f ∈ L for Re(s) maxα, β, and
2 1 2
L(c f + c f ) c L(f ) + c L(f ) (1.11)
1 1 2 2 1 1 2 21.6. Basic Properties of the Laplace Transform
17
for arbitrary constants c , c .
1 2
This follows from the fact that integration is a linear process, to
wit,
∞
−st
e c f (t) + c f (t) dt
1 1 2 2
0
∞ ∞
−st −st
c e f (t) dt + c e f (t) dt (f ,f ∈ L).
1 1 2 2 1 2
0 0
Example 1.15. The hyperbolic cosine function
ωt −ωt
e + e
cosh ωt
2
describes the curve of a hanging cable between two supports. By
linearity
1
ωt −ωt
L(cosh ωt) L(e ) +L(e )
2
1 1 1
+
2 s − ω s + ω
s
.
2 2
s − ω
Similarly,
ω
L(sinh ωt) .
2 2
s − ω
n
Example 1.16. If f (t) a + a t + ··· + a t is a polynomial of
0 1 n
degree n, then
n
k
L f (t) a L(t )
k
k0
n
a k
k
k+1
s
k0
by (1.9) and (1.11).
∞
n
Infinite Series. For an infinite series, a t , in general it is not
n
n0
possible to obtain the Laplace transform of the series by taking the
transform term by term.1. Basic Principles
18
Example 1.17.
∞
n 2n
(−1) t
2
−t
f (t) e , −∞t ∞.
n
n0
Taking the Laplace transform term by term gives
∞ ∞
n n
(−1) (−1) (2n)
2n
L(t )
2n+1
n n s
n0 n0
∞
n
1 (−1) (2n)··· (n + 2)(n + 1)
.
2n
s s
n0
Applying the ratio test,
u 2(2n + 1)
n+1
lim lim ∞,
2
n→∞ n→∞
u s
n
and so the series diverges for all values of s.
2 2
−t −t
However,L(e ) does exist since e is continuous and bounded
on 0, ∞).
So when can we guarantee obtaining the Laplace transform of an
infinite series by term-by-term computation?
Theorem 1.18. If
∞
n
f (t) a t
n
n0
converges for t ≥ 0, with
n
Kα
a ≤ ,
n
n
for all n sufficiently large andα 0,K 0, then
∞ ∞
a n
n
n
L f (t) a L(t ) Re(s)α .
n
n+1
s
n0 n0
Proof. Since f (t) is represented by a convergent power series, it is
continuous on 0, ∞). We desire to show that the difference
N N
n n
L f (t) − a L(t ) L f (t) − a t
n n
n0 n01.6. Basic Properties of the Laplace Transform
19
N
n
≤ L f (t) − a t
x n
n0
∞
−xt
converges to zero as N →∞, where L h(t) e h(t) dt, x
x
0
Re(s).
To this end,
N ∞
n n
f (t) − a t a t
n n
n0 nN+1
∞ n
(αt)
≤ K
n
nN+1
N
n
(αt)
αt
K e −
n
n0
∞
x n
since e x /n. As h ≤ g implies L (h) ≤ L (g) when the
x x
n0
transforms exist,
N N
n
(αt)
n αt
L f (t) − a t ≤ KL e −
x n x
n
n0 n0
N
n
1 α
K −
n+1
x − α x
n0
N
n
1 1 α
K −
x − α x x
n0
→ 0 Re(s) xα
as N →∞. We have used the fact that the geometric series has the
sum
∞
1
n
z , z 1.
1 − z
n0
Therefore,
N
n
L f (t) lim a L(t )
n
N→∞
n0
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