4.2. PROPERTIES OF THE LAPLACE TRANSFORM
4.2
101
Properties of the Laplace Transform
R1
We have de…ned the Laplace transform of a function f by L ff g (s) = 0 f (t) e st dt.
Its computation, because it involves an improper integral, is quite tedious. In
the previous section, we saw that using the linearity of the Laplace transform
made some computations easier. We see more properties which will make the
computation of the Laplace transform easier, We also see properties which will
allow us to use the Laplace transform to solve di¤erential equations.
4.2.1
An Important Property
Theorem 4.2.1 (Translation in s) If the Laplace transform L ff g (s) = F (s)
exists for s > then
L eat f (t) (s) = F (s a)
(4.1)
for s >
Proof.
+a
L eat f (t) (s)
=
Z
1
eat f (t) e
st
dt
0
=
Z
1
(s a)t
f (t) e
dt
0
= F (s
a)
Example 4.2.2 Find L feat sin btg (s)
In the previous section, we found that L fsin btg = F (s) =
hence, by the theorem,
L eat sin bt (s)
= F (s
=
(s
s2
b
if s > 0
+ b2
a)
b
2
a) + b2
for s > a.
4.2.2
Laplace Transform of Derivatives
Theorem 4.2.3 Suppose that f (t) is continuous on [0; 1) and f 0 (t) is piecewise continuous on [0; 1). Suppose further that both f (t) and f 0 (t) are of
exponential order . Then, for s > we have
L ff 0 g (s) = sL ff g (s)
f (0)
There is a similar theorem for higher order derivatives
(4.2)
102
CHAPTER 4. LAPLACE TRANSFORMS
Theorem 4.2.4 Suppose that f (t) ; f 0 (t) ; :::; f (n 1) (t) are continuous on [0; 1)
and f (n) (t) is piecewise continuous on [0; 1). Suppose further that all these
functions are of exponential order . Then, for s >
o
n
L f (n) (s) = sn L ff g (s) sn 1 f (0) sn 2 f 0 (0) sn 3 f 00 (0) ::: f (n 1) (0)
(4.3)
Example 4.2.5 Write equation 4.3 for n = 2; 3.
Case n = 2.
n 00 o
L f (s) = s2 L ff g (s)
sf (0)
f 0 (0)
Case n = 3.
n 000 o
L f
(s) = s3 L ff g (s)
sf 0 (0)
s2 f (0)
f 00 (0)
Example 4.2.6 Find L fcos btg (s).
b
0
if s > 0 and (sin bt) = b cos bt. Hence,
We know that L fsin btg = 2
s + b2
L fb cos btg (s)
0
= L (sin bt)
(s)
= sL fsin btg (s)
sb
=
2
s + b2
sin 0
Now, by linearity, L fb cos btg (s) = bL fcos btg (s). Therefore,
bL fcos btg (s) =
sb
s2 + b2
L fcos btg (s) =
s2
that is
Example 4.2.7 Show that L
nR
t
0
s
+ b2
o
1
f (u) du (s) = L ff (t)g (s), assuming f is
s
continuous and L ff (t)g exists.
Rt
We begin by de…ning g (t) = 0 f (u) du then g (0) = 0 and by the fundamental
theorem of calculus, g 0 (t) = f (t). It follows that
L ff (t)g (s)
=
=
=
L fg 0 (t)g (s)
sL fg (t)g (s) g (0)
Z t
sL
f (u) du (s)
0
hence the result.
4.2. PROPERTIES OF THE LAPLACE TRANSFORM
4.2.3
103
Derivative of the Laplace Transform
In the previous section, we learned to …nd the Laplace transform of the derivative
of a function. Another related question is if F (s) is the Laplace transform of
f (t) that is if F (s) = L ff (t)g (s) then what is F 0 (s)? Is it the Laplace
transform of some function of t? The answer is yes, more speci…cally, F 0 (s) =
L f tf (t)g (s). In fact, there is a more general result we give as a theorem.
Theorem 4.2.8 Let F (s) = L ff (t)g (s) and assume f (t) is piecewise continuous on [0; 1) and of exponential order . Then, for s >
L ftn f (t)g (s)
=
=
dn F
(s)
dsn
n
n d
( 1)
(L ff g (s))
dsn
( 1)
n
(4.4)
Remark 4.2.9 The theorem implies that if f satis…es the conditions of the
theorem then its Laplace transform has derivatives of all orders.
Example 4.2.10 Find L ft sin btg (s)
From the theorem,
L ft sin btg (s)
=
=
=
d
L fsin btg (s)
ds
d
b
2
ds s + b2
2bs
( 1)
(s2 + b2 )
2
Example 4.2.11 Find L ftg (s)
1
0
Recall that L f1g (s) = . Also, (t) = 1. Hence,
s
1
s
= L f1g (s)
= L f(t0 )g (s)
= sL ftg (s)
It follows that L ftg (s) =
4.2.4
0
1
.
s2
Summary
We summarize the properties of the Laplace transform we have studied in the
last two sections.
104
CHAPTER 4. LAPLACE TRANSFORMS
Properties of the Laplace transform
L ff + gg = L ff g + L fgg
L fcf g = cL ff g
L feat f (t)g (s) = L ff g (s a)
Ln
ff 0 go(s) = sL ff g (s) f (0)
L f
00
(s) = s2 L ff g (s)
sf (0)
f 0 (0)
L f (n) (s) = sn L ff g (s) sn 1 f (0)
n
n d
L ftn f (t)g (s) = ( 1)
(L ff g (s))
dsn
4.2.5
sn
2 0
f (0)
sn
3 00
f (0)
:::
f (n
Exercises
Do numbers 1, 2, 3, 5, 7, 9, 21, 22, 25 at the end of section 7.3 in your book.
1)
(0)
Bibliography
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fourth ed., Brooks/Cole, CENGACE Learning, 2012 (English).
[2] Charles H. Edwards, David E. Penney, and David T. Calvis, Di¤ erential
equations and boundary value problems: Computing and modeling, …fth ed.,
Pearson, 2015 (English).
[3] R. K. Nagle, Edward B. Sa¤, and Arthur D. Snider, Fundamentals of differential equations, eigth ed., Pearson/Addison-Wesley, 2012 (English).
[4] Virginia W. Noonburg, Ordinary di¤ erential equations: from calculus to
dynamical systems, The Mathematical Association of America, 2014.
201