Laplace transforms

13.2 LAPLACE TRANSFORMS
A similar result may be obtained for two-dimensional Fourier transforms in
which f(r) = f(ρ), i.e. f(r) is independent of azimuthal angle φ. In this case, using
the integral representation of the Bessel function J0 (x) given at the very end of
subsection 18.5.3, we find
∞
1
3
2πρf(ρ)J0 (kρ) dρ.
(13.52)
f(k) =
2π 0
13.2 Laplace transforms
Often we are interested in functions f(t) for which the Fourier transform does not
exist because f → 0 as t → ∞, and so the integral defining 3
f does not converge.
This would be the case for the function f(t) = t, which does not possess a Fourier
transform. Furthermore, we might be interested in a given function only for t > 0,
for example when we are given the value at t = 0 in an initial-value problem.
¯ or L [ f(t)], of f(t), which
This leads us to consider the Laplace transform, f(s)
is defined by
∞
¯ ≡
f(t)e−st dt,
(13.53)
f(s)
0
provided that the integral exists. We assume here that s is real, but complex values
would have to be considered in a more detailed study. In practice, for a given
function f(t) there will be some real number s0 such that the integral in (13.53)
exists for s > s0 but diverges for s ≤ s0 .
Through (13.53) we define a linear transformation L that converts functions
of the variable t to functions of a new variable s:
L [af1 (t) + bf2 (t)] = aL [ f1 (t)] + bL [ f2 (t)] = af¯1 (s) + bf¯2 (s).
(13.54)
Find the Laplace transforms of the functions (i) f(t) = 1, (ii) f(t) = eat , (iii) f(t) = tn ,
for n = 0, 1, 2, . . . .
(i) By direct application of the definition of a Laplace transform (13.53), we find
∞
∞
1
−1 −st
L [1] =
e−st dt =
= ,
e
if s > 0,
s
s
0
0
where the restriction s > 0 is required for the integral to exist.
(ii) Again using (13.53) directly, we find
∞
∞
¯ =
eat e−st dt =
e(a−s)t dt
f(s)
0
0
(a−s)t ∞
1
e
if s > a.
=
=
a−s 0
s−a
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INTEGRAL TRANSFORMS
(iii) Once again using the definition (13.53) we have
∞
tn e−st dt.
f¯n (s) =
0
Integrating by parts we find
n −st ∞
n ∞ n−1 −st
−t e
+
t e dt
f¯n (s) =
s
s 0
0
n¯
= 0 + f n−1 (s),
if s > 0.
s
We now have a recursion relation between successive transforms and by calculating
f¯0 we can infer f¯1 , f¯2 , etc. Since t0 = 1, (i) above gives
1
if s > 0,
(13.55)
f¯0 = ,
s
and
1
2!
n!
f¯2 (s) = 3 ,
...,
f¯n (s) = n+1
if s > 0.
f¯1 (s) = 2 ,
s
s
s
Thus, in each case (i)–(iii), direct application of the definition of the Laplace transform
(13.53) yields the required result. Unlike that for the Fourier transform, the inversion of the Laplace transform
¯
is not an easy operation to perform, since an explicit formula for f(t), given f(s),
is not straightforwardly obtained from (13.53). The general method for obtaining
an inverse Laplace transform makes use of complex variable theory and is not
discussed until chapter 25. However, progress can be made without having to find
an explicit inverse, since we can prepare from (13.53) a ‘dictionary’ of the Laplace
transforms of common functions and, when faced with an inversion to carry out,
hope to find the given transform (together with its parent function) in the listing.
Such a list is given in table 13.1.
When finding inverse Laplace transforms using table 13.1, it is useful to note
that for all practical purposes the inverse Laplace transform is unique§ and linear
so that
(13.56)
L −1 af¯1 (s) + bf¯2 (s) = af1 (t) + bf2 (t).
In many practical problems the method of partial fractions can be useful in
producing an expression from which the inverse Laplace transform can be found.
Using table 13.1 find f(t) if
¯ = s+3 .
f(s)
s(s + 1)
¯ may be written
Using partial fractions f(s)
¯ = 3− 2 .
f(s)
s
s+1
§
This is not strictly true, since two functions can differ from one another at a finite number of
isolated points but have the same Laplace transform.
454
13.2 LAPLACE TRANSFORMS
f(t)
¯
f(s)
s0
c
ctn
sin bt
cos bt
eat
tn eat
sinh at
cosh at
eat sin bt
eat cos bt
t1/2
t−1/2
δ(t − t0 )
c/s
cn!/sn+1
b/(s2 + b2 )
s/(s2 + b2 )
1/(s − a)
n!/(s − a)n+1
a/(s2 − a2 )
s/(s2 − a2 )
b/[(s − a)2 + b2 ]
(s − a)/[(s − a)2 + b2 ]
1
(π/s3 )1/2
2
(π/s)1/2
e−st0
0
0
0
0
a
a
|a|
|a|
a
a
0
0
0
e−st0 /s
0
H(t − t0 ) =
#
1 for t ≥ t0
0 for t < t0
Table 13.1 Standard Laplace transforms. The transforms are valid for s > s0 .
Comparing this with the standard Laplace transforms in table 13.1, we find that the inverse
transform of 3/s is 3 for s > 0 and the inverse transform of 2/(s + 1) is 2e−t for s > −1,
and so
f(t) = 3 − 2e−t ,
if s > 0. 13.2.1 Laplace transforms of derivatives and integrals
One of the main uses of Laplace transforms is in solving differential equations.
Differential equations are the subject of the next six chapters and we will return
to the application of Laplace transforms to their solution in chapter 15. In
the meantime we will derive the required results, i.e. the Laplace transforms of
derivatives.
The Laplace transform of the first derivative of f(t) is given by
∞
df −st
df
e dt
=
L
dt
dt
0
∞
∞
= f(t)e−st 0 + s
f(t)e−st dt
0
¯
= −f(0) + sf(s),
for s > 0.
(13.57)
The evaluation relies on integration by parts and higher-order derivatives may
be found in a similar manner.
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INTEGRAL TRANSFORMS
Find the Laplace transform of d2 f/dt2 .
Using the definition of the Laplace transform and integrating by parts we obtain
2 ∞ 2
d f −st
df
L
e dt
=
dt2
dt2
0
∞
∞
df −st
df −st
+s
=
e
e dt
dt
dt
0
0
df
¯ − f(0)],
= − (0) + s[sf(s)
for s > 0,
dt
where (13.57) has been substituted for the integral. This can be written more neatly as
2 df
¯ − sf(0) − df (0),
= s2 f(s)
L
for s > 0. dt2
dt
In general the Laplace transform of the nth derivative is given by
n d f
df
dn−1 f
for s > 0.
L
= sn f¯ − sn−1 f(0) − sn−2 (0) − · · · − n−1 (0),
n
dt
dt
dt
(13.58)
We now turn to integration, which is much more straightforward. From the
definition (13.53),
∞
t
t
f(u) du =
dt e−st
f(u) du
L
0
0
0
∞ ∞
t
1 −st
1
e f(t) dt.
f(u) du +
= − e−st
s
s
0
0
0
The first term on the RHS vanishes at both limits, and so
t
1
f(u) du = L [ f] .
L
s
0
(13.59)
13.2.2 Other properties of Laplace transforms
From table 13.1 it will be apparent that multiplying a function f(t) by eat has the
effect on its transform that s is replaced by s − a. This is easily proved generally:
∞
f(t)eat e−st dt
L eat f(t) =
0
∞
=
f(t)e−(s−a)t dt
0
¯ − a).
= f(s
As it were, multiplying f(t) by eat moves the origin of s by an amount a.
456
(13.60)
13.2 LAPLACE TRANSFORMS
¯ by
We may now consider the effect of multiplying the Laplace transform f(s)
(b > 0). From the definition (13.53),
e
∞
¯ =
e−bs f(s)
e−s(t+b) f(t) dt
0
∞
=
e−sz f(z − b) dz,
−bs
0
on putting t + b = z. Thus e
defined by
−bs
g(t) =
¯ is the Laplace transform of a function g(t)
f(s)
#
0
for 0 < t ≤ b,
f(t − b) for t > b.
In other words, the function f has been translated to ‘later’ t (larger values of t)
by an amount b.
Further properties of Laplace transforms can be proved in similar ways and
are listed below.
1 s
,
(13.61)
(i)
L [ f(at)] = f¯
a
a
¯
dn f(s)
,
for n = 1, 2, 3, . . . ,
(13.62)
(ii)
L [tn f(t)] = (−1)n
dsn
∞
f(t)
¯ du,
(iii)
L
f(u)
(13.63)
=
t
s
provided limt→0 [ f(t)/t] exists.
Related results may be easily proved.
Find an expression for the Laplace transform of t d2 f/dt2 .
From the definition of the Laplace transform we have
2 ∞
df
d2 f
L t 2 =
e−st t 2 dt
dt
dt
0
∞
d
d2 f
e−st 2 dt
=−
ds 0
dt
d 2¯
= − [s f(s) − sf(0) − f (0)]
ds
df¯
= −s2
− 2sf¯ + f(0). ds
Finally we mention the convolution theorem for Laplace transforms (which is
analogous to that for Fourier transforms discussed in subsection 13.1.7). If the
¯ and ḡ(s) then
functions f and g have Laplace transforms f(s)
t
¯
f(u)g(t − u) du = f(s)ḡ(s),
(13.64)
L
0
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INTEGRAL TRANSFORMS
Figure 13.7 Two representations of the Laplace transform convolution (see
text).
where the integral in the brackets on the LHS is the convolution of f and g,
denoted by f ∗ g. As in the case of Fourier transforms, the convolution defined
above is commutative, i.e. f ∗ g = g ∗ f, and is associative and distributive. From
(13.64) we also see that
¯
L −1 f(s)ḡ(s)
=
t
f(u)g(t − u) du = f ∗ g.
0
Prove the convolution theorem (13.64) for Laplace transforms.
From the definition (13.64),
∞
e−su f(u) du
e−sv g(v) dv
0
0 ∞ ∞
du
dv e−s(u+v) f(u)g(v).
=
∞
¯
f(s)ḡ(s)
=
0
0
Now letting u + v = t changes the limits on the integrals, with the result that
∞
∞
¯
du f(u)
dt g(t − u) e−st .
f(s)ḡ(s)
=
u
0
As shown in figure 13.7(a) the shaded area of integration may be considered as the sum
of vertical strips. However, we may instead integrate over this area by summing over
horizontal strips as shown in figure 13.7(b). Then the integral can be written as
t
∞
¯
f(s)ḡ(s)
=
du f(u)
dt g(t − u) e−st
0
0
t
∞
=
dt e−st
f(u)g(t − u) du
0
0
t
f(u)g(t − u) du . =L
0
458