Page 120 - Advanced engineering mathematics

100 CHAPTER 3 The Laplace Transform

Then
F(s) − f (0)M(s)
R(s) = .
sM(s)
If we can invert R(s),wehave r(t).
We will see how this model works in a specific example. Suppose we want to have f (t) =
A + Bt doses of a drug on hand at time t with A and B as positive constants. Thus, f (0) = A,

and the need increases in time at the rate f (t) = B. Suppose the mortality function is
m(t) = 1 − H(t − k)

in which H is the Heaviside function and k is a positive constant determined by how long doses
remain effective.
Now
A B 1 1
F(s) = + and M(s) = − e −ks .
s s 2 s s
The transform of the replacement function is
F(s) − F(0)M(s)
R(s) =
sM(s)
A + B 1 − e
1 −ks
s s 2 − A s s
=
1 −ks
s 1 − e
s s
A 1 B 1 A
= + −
2
s 1 − e −ks s 1 − e −ks s
in which we have omitted some routine algebra in going from the second line to the third. Now
0 < e −ks < 1for ks > 0, so we can use the geometric series to write
∞ ∞
1
= (e −ks n e −kns .
) = 1 +
1 − e −ks
n=0 n=1
Therefore,

∞
1 1
R(s) = A + e −kns
s s
n=1

∞
1 1 A
+ B + e −kns − .
s 2 s 2 s
n=1
Invert this term by term to obtain
∞ ∞

r(t) = A + A H(t − nk) + Bt + B (t − nk)H(t − nk) − A
n=1 n=1
∞

= Bt + (A + B(t − nk))H(t − nk).
n=1
Notice that t − nk < 0 (hence H(t − nk) = 0) if t/n < k. Since k is given and n increases from 1
through the positive integers, this always occurs after some time, so “most” of the terms of this
series vanish for a given time.

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October 14, 2010 14:14 THM/NEIL Page-100 27410_03_ch03_p77-120