First write the integral on the left as a contour integral. Thus z = c + iy and dz = idy and this is just
the contour integral
∫ c+iR
1-- eutℒg (u)du
2πi c−iR
where the contour is the straight line from c − iR to c + iR. Indeed, if you parametrize this contour as
z = c + iy and use the procedures for evaluation of contour integrals, you get the integral in 16.16. Then
taking the limit as R →∞ it is customary to write this limit as
∫ c+i∞
-1- eutℒg (u)du
2πi c− i∞
This is called the Bromwich integral and as shown earlier it recovers the mid point of the jump of g at t for
every point t where g is Holder continuous from the right and from the left. Remember t ≥ 0. Now
u → e^{ut}ℒg
(u)
is analytic for Re
(u)
> η and in particular for Re
(u)
≥ c therefore, all of the poles of
u →ℒg
(u)
are contained in the set Re
(u)
< c. Indeed, in practice, u →ℒg
(u)
ends up being represented
by a formula which is clearly a meromorphic function, one which is analytic except for isolated
poles.
So how do you compute this Bromwich integral? This is where the method of residues is very useful.
Consider the following contour.
PICT
Let γ_{R} be the above contour oriented as shown. The radius of the circular part is R. Let C_{R} be the
curved part. Then one can show that under suitable assumptions
1 ∫
Rli→m∞ --- eutF (u)du = 0 (16.17)
2πi CR
(16.17)
The needed condition is that for all
|z|
large enough,
|F (z)| ≤ -C-, some α > 0. (16.18)
|z|α
(16.18)
Note that this assumption implies there are finitely many poles for F
(z)
because if α is a pole, you have
lim_{z→α}
|F (z)|
= ∞.
Lemma 16.9.2Let the contour be as shown and assume 16.18. Then the above limit in 16.17exists.
Proof:Let a
(R)
be the angle between the positive x axis and the top dotted line while b
(R)
is defined
the same way. Thus lim_{R→∞}a
(R)
=
π
2
and lim_{R→∞}b
(R )
=
3π
2
. Consider the part of C_{R} corresponding to
the angle θ measured from the positive x axis is between
π
2
and
3π
2
. Parametrizing this part of the curve
gives
∫ 3π∕2 ( )
e(Rcosθ+iRsinθ)tF Reiθ Rieiθdθ
π∕2
∫ I1
π (Rcosθ+iR sinθ)t ( iθ) iθ
= π∕2 e F Re Rie dθ
I2
∫ 3π∕2 (R cosθ+iR sinθ)t ( iθ) iθ
+ e F Re Rie dθ (16.19)
π
Then I_{1} is no more than
∫ π ∫ π∕2
CαeRtcosθRd θ = -Cα-eRtcos(θ+π∕2)Rd θ
π∕2 R 0 R
CR- ∫ π∕2 − Rtsin(θ)
= R α 0 e dθ
Now there is δ > 0 such that if 0 < θ ≤ δ,sinθ >
(1∕2)
θ. Therefore, the above is no larger
than
( )
CR ∫ δ −(R∕2)tθ ∫ π∕2 −Rtsinδ CR ( 2 π − Rtsinδ)
R-α e dθ+ e dθ ≤ R-α Rt-+ 2e
0 δ
which converges to 0 for each t > 0. If you consider t > ε > 0, then this integral has an upper bound which
is independent of such t, namely
CR ( 2 π )
R-α R-ε + 2-e− Rεsinδ
Now consider I_{2} in 16.19. Its magnitude is no more than
∫ 3π∕2 ∫ 0
CαR e(R cosθ)tdθ = -CαR e(Rcos(θ+3π∕2))tdθ
R π R −π∕2
-C- ∫ 0 Rtsin(θ)
= R αR −π∕2e dθ
∫ π∕2
= -C-R e−Rtsin(θ)dθ
R α 0
which was shown to converge to 0 in the above. Also there exists a uniform upper bound for this part of the
line integral for t > ε.
It remains to consider those parts of C_{R} which are from a
(R)
to π∕2 and from 3π∕2 to b
(R)
.
Consider
∫ π∕2
e(R cosθ+iRsinθ)tF (Reiθ)Rieiθdθ
a(R)
Its magnitude is dominated by
∫ ∫
-C- π∕2 (Rcosθ)t -C- π∕2 (Rcos(arccos(cR )))t
R αR a(R) e dθ ≤ R αR a(R) e dθ
∫ π∕2
= -C-R ectdθ
R α a(R)
-C- ct(π- ( c) )
= R αRe 2 − arccos R
because R cosa = c and so a = arccos
( )
Rc
.
Now lim_{R→∞}R^{1−α}
( ( ))
π2 − arccos cR-
= 0 and so this contour integral converges to 0 as R →∞. Also,
there is a uniform upper bound for t <
(1∕ε)
namely
( ( ))
-C-Rec(1∕ε) π-− arccos c-
R α 2 R
if c > 0. If c < 0 a uniform bound exists for all t ≥ 0. The other integral from
3π-
2
to b
(R )
is similar.
■
Corollary 16.9.3Let the contour be as shown and assume 16.18for meromorphic F
, given by the Bromwich integral, is continuous on
(0,∞)
andits Laplace transform is F
(s)
.
Proof: It only remains to verify continuity. Let R be so large that the above contour γ_{R}^{∗} encloses all
poles of F. Then for such large R, the contour integrals are not changing because all the poles are enclosed.
Thus
= -1-2πi(sum of residues of the poles of eztF (z)) = sum of residues.
2πi
The following procedure shows how the Bromwich integral can be computed to obtain an actual formula
for a function. However, the integral itself will make sense and could be numerically computed to solve for
the inverse Laplace transform.
Procedure 16.9.4Suppose F
(s)
is a Laplace transform and is meromorphic on ℂ and satisfies16.18. (This situation is quite typical) Then to compute the function of t, f
(t)
whose Laplacetransform gives F
(s)
, do the following. Find the sum of the residues of e^{zt}F
(z)
forRez < c. Thisyields the midpoint of the jump of f
(t)
at each t where f is Holder continuous from the left andright.
Example 16.9.5Suppose F
(s)
=
--s---
(s2+1)2
. Find f
(t)
such that F
(s)
is the Laplace transform off
(t)
.
There are two residues of this function, one at i and one at −i. At both of these points the poles are of
order two and so we find the residue at i by
You should verify that this actually works giving ℒ
(f)
=
--s-2-
(s2+1)
.
Example 16.9.6Find f
(t)
if F
(s)
, the Laplace transform is e^{−s}∕s.
You need to compute the residues of
este−s
s
. The function equals
1
s
∑_{k=0}^{∞}
(−1)k(t−1)ksk-
k!
. Thus the
residue is 1. However, this fails to be the function whose Laplace transform is F
(s)
. What is wrong? The
problem with this is the failure of the estimate on F
(s)
to hold for large s. Indeed, if s = −n, you would
have e^{n}∕n but it would need to be less than C∕n^{α} which is not possible. The estimate requires F
(s)
→ 0 as
|s|
→∞ and this does not happen here. You can verify directly that the function which works is u_{1}
(t)
which is 0 for t < 1 and 1 for t ≥ 1. Thus if the estimate does not hold, the procedure does not necessarily
hold either.
If Rep < c for all p a pole of F
(s)
and if F
(s)
is meromorphic and satisfies the growth condition 16.18,
and if f
(t)
is defined by that Bromwich integral, is it true that F
(s)
is the Laplace transform of f
(t)
for
large s? Thus
∫ ∫
1-- R (c+iy)t -1- c+i∞ zt
f (t) ≡ Rli→m∞ 2π −R e F (c + iy)dy = 2πi c−i∞ e F (z)dz
The limit must exist because, as discussed above,
( )
-1- ∫ c+iR zt ∫ zt
lRim→∞ 2πi c−iR e F (z)dz + C e F (z)dz
R
is eventually constant because the contour will have enclosed all poles of F
(z)
, but as R continues to
increase, the integral over the curved part, C_{R} converges to 0. Let Res be larger than c. One needs to
consider