Books
in black and white
Main menu
Share a book About us Home
Books
Biology Business Chemistry Computers Culture Economics Fiction Games Guide History Management Mathematical Medicine Mental Fitnes Physics Psychology Scince Sport Technics
Ads

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
Previous << 1 .. 128 129 130 131 132 133 < 134 > 135 136 137 138 139 140 .. 486 >> Next

Theorem 6.1.1
Theorem 6.1.2
If f is piecewise continuous for t > a, if | f (t)| < g(t) when t > Mfor some positive
J' TO /TO
g(t) dt converges, then / f(t) dt also converges. On
M J a
J- TO
g(t) dt diverges, then
M
TO
f(t) dt also diverges.
The proof of this result from the calculus will not be given here. It is made plausible,
TOTO
however, by comparing the areas represented by J g(t) dt and J | f (t)| dt. The
functions most useful for comparison purposes are ect and t-p, which were considered in Examples 1, 2, and 3.
We now return to a consideration of the Laplace transform L{ f (t)} or F(s), which is defined by Eq. (2) whenever this improper integral converges. In general, the parameter s may be complex, but for our discussion we need consider only real values of s. The foregoing discussion of integrals indicates that the Laplace transform F of a function f exists if f satisfies certain conditions, such as those stated in the following theorem.
Suppose that
1. f is piecewise continuous on the interval 0 < t < A for any positive A.
2. | f (t) | < Keat when t > M. In this inequality K, a, and M are real constants, K and M necessarily positive.
Then the Laplace transform L{ f (t)} = F(s), defined by Eq. (2), exists for s > a.
To establish this theorem it is necessary to show only that the integral in Eq. (2) converges for s > a. Splitting the improper integral into two parts, we have
r TO r M n TO
/ e~st f (t) dt = e-st f(t) dt + / e-st f (t) dt. (4)
Jo J0 J M
The first integral on the right side of Eq. (4) exists by hypothesis (1) of the theorem; hence the existence of F(s) depends on the convergence of the second integral. By hypothesis (2) we have, for t > M,
| e-stf(t)| < Ke-steat = Ke(a-s)t,
r TO ( _ )t
and thus, by Theorem 6.1.1, F(s) exists provided that j e(a s) dt converges. Referring to Example 1 with c replaced by a - s, we see that this latter integral converges when a s < 0, which establishes Theorem 6.1.2.
Unless the contrary is specifically stated, in this chapter we deal only with functions satisfying the conditions of Theorem 6.1.2. Such functions are described as piecewise continuous, and of exponential order as t ^to. The Laplace transforms of some important elementary functions are given in the following examples.
6.1 Definition of the Laplace Transform
297
EXAMPLE
EXAMPLE
Let f(t) = 1, t > 0. Then
1
-St - 1
fc
4 L{1} = e-st dt = -, s > 0.
J0 s
Let f(t) = eat, t > 0. Then
B EXAMPLE
5 L{eat} = J e-steat dt = J e-(s-a)t dt
s a
s > a.
Let f (t) = sin at, t > 0. Then
6 L{sin at}= F(s) = I e~st sin at dt, s > 0.
J0
Since
-A
F(s) = lim e-st sin at dt,
A^?
0
upon integrating by parts we obtain
' ?/ a J 0
F(s) = lim
A^?
st
e- cos at
A r A
s
e-st cos at dt
f
J0
0 a J 0
e-st cos at dt.
a
1 s f aa
A second integration by parts then yields
1 s2 '?
1 s2 ?
F (s ) =-----------------2 e~st sin at dt
a a2 0
= - - 4F(s).
a a2
Hence, solving for F(s), we have
a
F(s) = -----------2, s > 0.
s2 + a2
1
Now let us suppose that f1 and f2 are two functions whose Laplace transforms exist for s > a1 and s > a2, respectively. Then, for s greater than the maximum of a1 and a2,
C TO
L{c1 f() + C2 f()} = e-st[C1 f() + c2 f2(t)] dt
Jo
r TO r TO
= C1 / e~st f^t) dt + C2 I e-stf2(t) dt;
1 0 1 2 0 2
298
Chapter 6. The Laplace Transform
PROBLEMS
hence
L{C1 W) + c2 f2(t)} = c,C{ frit)} + c2C{ f2(t)}.
(5)
Equation (5) is a statement of the fact that the Laplace transform is a linear operator. This property is of paramount importance, and we make frequent use of it later.
In each of Problems 1 through 4 sketch the graph of the given function. In each case determine whether f is continuous, piecewise continuous, or neither on the interval 0 < t < 3.
' 7 0 < t < 1 t2, 0<t<1
1. fit) = 2 + t, 1 < t < 2 2. fit) = it 1)1, 1 < t < 2
6 t, 2 < t < 3 1, 2 < t < 3
' 7 0 < t < 1 t, 0 < t < 1
3. fit) = ? 1, 1 < t < 2 4. fit) = 3 - t, 1 < t < 2
3 t, 2 < t < 3 1, 2 < t < 3
5. Find the Laplace transform of each of the following functions:
(a) t
(b) t2
(c) tn, where n is a positive integer
6. Find the Laplace transform of f (t) = cos at, where a is a real constant.
Recall that cosh bt = (ebt + ebt)/2 and sinh bt = (ebt ebt)/2. In each of Problems 7 through 10 find the Laplace transform of the given function; a and b are real constants.
7. cosh bt 8. sinh bt
9. eat cosh bt 10. eat sinh bt
In each of Problems 11 through 14 recall that cos bt = (elbt + elbt)/2 and sin bt = (elbt e~lbt)/2i. Assuming that the necessary elementary integration formulas extend to this case, find the Laplace transform of the given function; a and b are real constants.
11. sin bt 12. cos bt
13. eat sin bt 14. eat cos bt
In each of Problems 15 through 20, using integration by parts, find the Laplace transform of the given function; n is a positive integer and a is a real constant.
15. te
at
16. t sin at 19. t2 sin at
17. t cosh at
Previous << 1 .. 128 129 130 131 132 133 < 134 > 135 136 137 138 139 140 .. 486 >> Next