158 TIME-DEPENDENT AND NON-LINEAR PROBLEMS
Example 6.2
Consider the equation
d
2
u du
dr di
m
^ +
c
^7 +
ku
=f(
t
) (
a
)
now with the initial conditions u(0) = u
0
, ύ(0) = ύ
0
.
If we now take the Laplace transform of this equation we obtain,
mK
2
U{κ)
—
mKU
0
—
mu
0
+
CKU(κ)
—
cu
0
+ k
U(K)
= F(K) (b)
Hence,
U{K) = γ- — (c)
πικ
+CK
+ k
u(t) may then be obtained by using tables of transforms, by analytical
evaluation of the Bromwich inversion integral or by a numerical
inversion process.
Note that the initial values are taken into account in equation (c),
which gives the 'transient' part of the solution.
THE BROMWICH INTEGRAL
We have,
F(
K
)= I e-*'/(t)dr (6.29)
Jo
where Re(K) > y for convergence.
(K) =
Jo
Then the use of Planchard's theorem gives'
9
f(t) =
Q
Kt
F(K)dK (6.30)
This is the Bromwich integral (see Figure 6.2). y
1
is chosen to be
such that equation (6.30) gives/(i) = 0 for t < 0. In fact y
x
is chosen
such that all the singularities of
F (κ)
lie to the left of the contour from
y
l
—
iootoyj+ioo.Bya minor distortion of the contour we may write
(for t > 0),
f(t) =
Y
j
e«j<F
j
(K)
(6.31)