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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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2n 3n
FIGURE 6.3.3 Graph of y = un(t) — u2n(t).
The Laplace transform of uc is easily determined:
/TO r TO
e—stuc(t) dt = J e~st dt
-cs
s > 0.
(2)
For a given function f, defined for t > 0, we will often want to consider the related function g defined by
0,
t < c, t c,
y = g(t) =
nslation of f
6.3.4. In terms of the unit step function we can write g(t) in the convenient form
f ( t - c),
which represents a translation of f a distance c in the positive t direction; see Figure
g(t) = uc(t) f (t — c).
The unit step function is particularly important in transform use because of the following relation between the transform of f (t) and that of its translation u (t) f (t — c).
FIGURE 6.3.4 A translation of the given function. (a) y — f (t); (b) y — uc (t) f (t - c).
s
Theorem 6.3.1 If F(s) = L{ f(t)} exists for s > a > 0, and if c is a positive constant, then
L{uc(t) f(t — c)} = e~csL{ f(t)} = e—csF(s), s > a. (3)
312
Chapter 6. The Laplace Transform
EXAMPLE
2
Conversely, if f(t) = L 1 {F(s)}, then
uc(t) f(t — c) = L~1{e~csF(s)}. (4)
Theorem 6.3.1 simply states that the translation of f (t) a distance c in the positive t direction corresponds to the multiplication of F(s) by e~cs. To prove Theorem 6.3.1 it is sufficient to compute the transform of uc(t) f (t — c):
c ?
L{uc(t) f(t — c)}= e~st?c(t) f(t — c) dt
Jo
?
e~st f (t — c) dt.
Introducing a new integration variable f = t — c, we have
r ? ? ?
L{uc(t) f(t — c)}= e—(f+c)s f (f) df = e~cs e~sf f(f) df
oo = e~cs F (s ).
Thus Eq. (3) is established; Eq. (4) follows by taking the inverse transform of both sides of Eq. (3).
A simple example of this theorem occurs if we take f (t ) = 1. Recalling that L{ 1 } = 1/s, we immediately have from Eq. (3)that L{uc(t)} = e~cs/s. This result agrees with that of Eq. (2). Examples 2 and 3 illustrate further how Theorem 6.3.1 can be used in the calculation of transforms and inverse transforms.
If the function f is defined by
sin t, 0 < t < n/4,
f(t) =
1 sin t + cos(t — n/4), t > n/4,
find L{ f (t)}. The graph of y = f (t) is shown in Figure 6.3.5. Note that f (t) = sin t + g(t), where
g(t) = ( t < n/4,
B( ) \ cos(t — n/4), t > n/4.
Thus
g(t) = ux/4(t) cos(t — n/4)
and
L{ f (t)} = L{sin t} + L{un/4(t) cos(t - n/4)}
= L{sin t} + e-s/4L{cos t}.
Introducing the transforms of sin t and cos t, we obtain
1 c 1 se-ns/4
L{ f(t)} =-2---------+----------e-s/4-2---------= .
s2 + 1 s2 + 1 s2 + 1
You should compare this method with the calculation of L{ f (t)} directly from the definition.
6.3 Step Functions
313
Find the inverse transform of
F (s ) =
1 - e
-2s
From the linearity of the inverse transform we have
f(t) = L-1 {F(s)} = L-1 (-U - C~A e
— t — u2(t )(t — 2)-The function f may also be written as
t,
f (t ) =
2,
0 < t < 2, t> 2.
2s
The following theorem contains another very useful property of Laplace transforms that is somewhat analogous to that given in Theorem 6.3.1.
Theorem 6.3.2 If F(s) — L{ f (t)} exists for s > a > 0, and if c is a constant, then
L{ectf (t )}— F (s — c), s > a + c. (5)
Conversely, if f (t) — F(s)}, then
II {-> i C/) 1 c)}. (6)
According to Theorem 6.3.2, multiplication of f (t) by ect results in a translation of the transform F(s) a distance c in the positive s direction, and conversely. The proof of this theorem requires merely the evaluation of L{ect f (t )}. Thus
r ? r c
L{ect f(t)} — e~stect f(t) dt — /
Jo Jo
— F (s — c),
e
-(s-c)t
f(t) dt
2
s
2
s
s
314
Chapter 6. The Laplace Transform
EXAMPLE
4
PROBLEMS
which is Eq. (5). The restriction s > a + c follows from the observation that, according to hypothesis (ii) of Theorem 6.1.2, | f(t)l < Keat; hence |ect f(t)l < Ke(a+c)t. Equation (6) follows by taking the inverse transform of Eq. (5), and the proof is complete.
The principal application of Theorem 6.3.2 is in the evaluation of certain inverse transforms, as illustrated by Example 4.
Find the inverse transform of
G(s) =
1
s — 4s T 5
By completing the square in the denominator we can write
1
G (s) =
= F(s - 2),
(s - 2)2 + 1
where F(s) = (s2 + 1)-1. Since ?-1{F(s)} = sin t, it follows from Theorem 6.3.2 that
g(t) = L-1{G (s)} = e2t sin t.
The results of this section are often useful in solving differential equations, particularly those having discontinuous forcing functions. The next section is devoted to examples illustrating this fact.
In each of Problems 1 through 6 sketch the graph of the given function on the interval t > 0.
2. (t - 3)u() - (t - 2)u()
4. f (t - 3)u3(t), where f (t) = sin t
1. ul(t) + 2u3(t) - 6u4(t)
3. f(t — n)un(t), where f(t) = t2
5. f(t — \)u2(t), where f(t) = 2t
6. f (t) = (t — 1)u i(t) — 2(t — 2)u2(t) + (t — 3)u3(t)
In each of Problems 7 through 12 find the Laplace transform of the given function.
7. f(t) = ? 0, (t - 2)2, t < 2 t > 2 8. f (t) = cT^ - 2t + 2, t < 1 t > 1
0, t < n
9. f(t) = ? t- n, n < t < 2n 10. f (t) = = u1(t) + 2u3(t) - 6u4(t)
0, t > 2n
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