Inverse Laplace transform

APPLICATIONS OF COMPLEX VARIABLES
Im s
Re s
L
λ
Figure 25.6 The integration path of the inverse Laplace transform is along
the infinite line L. The quantity λ must be positive and large enough for all
poles of the integrand to lie to the left of L.
25.5 Inverse Laplace transform
As a further example of the use of contour integration we now discuss a method
whereby the process of Laplace transformation, discussed in chapter 13, can be
inverted.
It will be recalled that the Laplace transform f̄(s) of a function f(x), x ≥ 0, is
given by
∞
f̄(s) =
e−sx f(x) dx,
Re s > s0 .
(25.24)
0
In chapter 13, functions f(x) were deduced from the transforms by means of a
prepared dictionary. However, an explicit formula for an unknown inverse may
be written in the form of an integral. It is known as the Bromwich integral and is
given by
f(x) =
1
2πi
λ+i∞
esx f̄(s) ds,
λ > 0,
(25.25)
λ−i∞
where s is treated as a complex variable and the integration is along the line L
indicated in figure 25.6. The position of the line is dictated by the requirements
that λ is positive and that all singularities of f̄(s) lie to the left of the line.
That (25.25) really is the unique inverse of (25.24) is difficult to show for general
functions and transforms, but the following verification should at least make it
884
25.5 INVERSE LAPLACE TRANSFORM
Γ
Γ
R
Γ
R
L
R
L
(a)
(b)
L
(c)
Figure 25.7 Some contour completions for the integration path L of the
inverse Laplace transform. For details of when each is appropriate see the
main text.
plausible:
f(x) =
1
2πi
λ+i∞
ds esx
λ−i∞
∞
∞
e−su f(u) du,
Re(s) > 0, i.e. λ > 0,
0
λ+i∞
1
du f(u)
es(x−u) ds
2πi 0
λ−i∞
∞
∞
1
=
du f(u)
eλ(x−u) eip(x−u) i dp,
2πi 0
−∞
∞
1
f(u)eλ(x−u) 2πδ(x − u) du
=
2π 0
#
f(x) x ≥ 0,
=
0
x < 0.
=
putting s = λ + ip,
(25.26)
Our main purpose here is to demonstrate the use of contour integration. To
employ it in the evaluation of the line integral (25.25), the path L must be made
part of a closed contour in such a way that the contribution from the completion
either vanishes or is simply calculable.
A typical completion is shown in figure 25.7(a) and would be appropriate if
f̄(s) had a finite number of poles. For more complicated cases, in which f̄(s) has
an infinite sequence of poles but all to the left of L as in figure 25.7(b), a sequence
of circular-arc completions that pass between the poles must be used and f(x) is
obtained as a series. If f̄(s) is a multivalued function then a cut plane is needed
and a contour such as that shown in figure 25.7(c) might be appropriate.
We consider here only the simple case in which the contour in figure 25.7(a)
is used; we refer the reader to the exercises at the end of the chapter for others.
885
APPLICATIONS OF COMPLEX VARIABLES
Ideally, we would like the contribution to the integral from the circular arc Γ to
tend to zero as its radius R → ∞. Using a modified version of Jordan’s lemma,
it may be shown that this is indeed the case if there exist constants M > 0 and
α > 0 such that on Γ
M
|f̄(s)| ≤ α .
R
Moreover, this condition always holds when f̄(s) has the form
f̄(s) =
P (s)
,
Q(s)
where P (s) and Q(s) are polynomials and the degree of Q(s) is greater than that
of P (s).
When the contribution from the part-circle Γ tends to zero as R → ∞, we
have from the residue theorem that the inverse Laplace transform (25.25) is given
simply by
residues of f̄(s)esx at all poles .
f(t) =
(25.27)
Find the function f(x) whose Laplace transform is
s
,
f̄(s) = 2
s − k2
where k is a constant.
It is clear that f̄(s) is of the form required for the integral over the circular arc Γ to tend
to zero as R → ∞, and so we may use the result (25.27) directly. Now
sesx
f̄(s)esx =
,
(s − k)(s + k)
and thus has simple poles at s = k and s = −k. Using (24.57) the residues at each pole can
be easily calculated as
kekx
ke−kx
and
R(−k) =
.
2k
2k
Thus the inverse Laplace transform is given by
f(x) = 12 ekx + e−kx = cosh kx.
R(k) =
This result may be checked by computing the forward transform of cosh kx. Sometimes a little more care is required when deciding in which half-plane to
close the contour C.
Find the function f(x) whose Laplace transform is
1 −as
(e − e−bs ),
s
where a and b are fixed and positive, with b > a.
f̄(s) =
From (25.25) we have the integral
f(x) =
1
2πi
λ+i∞
λ−i∞
e(x−a)s − e(x−b)s
ds.
s
886
(25.28)
25.5 INVERSE LAPLACE TRANSFORM
f(x)
1
a
b
x
Figure 25.8 The result of the Laplace inversion of f̄(s) = s−1 (e−as − e−bs ) with
b > a.
Now, despite appearances to the contrary, the integrand has no poles, as may be confirmed
by expanding the exponentials as Taylor series about s = 0. Depending on the value of x,
several cases arise.
(i) For x < a both exponentials in the integrand will tend to zero as Re s → ∞. Thus
we may close L with a circular arc Γ in the right half-plane (λ can be as small as desired),
and we observe that s × integrand tends to zero everywhere on Γ as R → ∞. With no
poles enclosed and no contribution from Γ, the integral along L must also be zero. Thus
f(x) = 0
for x < a.
(25.29)
(ii) For x > b the exponentials in the integrand will tend to zero as Re s → −∞, and
so we may close L in the left half-plane, as in figure 25.7(a). Again the integral around Γ
vanishes for infinite R, and so, by the residue theorem,
f(x) = 0
for x > b.
(25.30)
(iii) For a < x < b the two parts of the integrand behave in different ways and have to
be treated separately:
(x−a)s
(x−b)s
1
e
e
1
I1 − I 2 ≡
ds −
ds.
2πi L s
2πi L s
The integrand of I1 then vanishes in the far left-hand half-plane, but does now have a
(simple) pole at s = 0. Closing L in the left half-plane, and using the residue theorem, we
obtain
I1 = residue at s = 0 of s−1 e(x−a)s = 1.
(25.31)
The integrand of I2 , however, vanishes in the far right-hand half-plane (and also has
a simple pole at s = 0) and is evaluated by a circular-arc completion in that half-plane.
Such a contour encloses no poles and leads to I2 = 0.
Thus, collecting together results (25.29)–(25.31) we obtain

 0 for x < a,
1 for a < x < b,
f(x) =
 0 for x > b,
as shown in figure 25.8. 887