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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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If we let
(as2 + bs + c) Y(s) — (as + b)y0 — ay0 = G(s).
(as + b) y0 + ay0 G (s )
O(s) =y- ^, n(s) = 2 ,)-, (23)
as2 + bs + c as2 + bs + c
then we can write
Consequently,
Y (s) = O(s ) + n(s). (24)
y = 0(t) + f(t), (25)
where 0(t) = C~ 1{$(s)} and y(t) = C~x{n(s)}. Observe that y = 0(t) is a solution
of the initial value problem
ay + by + cy = 0, y (0) = y,, y (0) = y0, (26)
obtained from Eqs. (21) and (22) by setting g(t) equal to zero. Similarly, y = y(t) is the solution of
ay + by + cy = g(t), y(0) = 0, / (0) = 0, (27)
in which the initial values y0 and y0 are each replaced by zero.
Once specific values of a, b, and c are given, we can find 0(t ) = C~1 {O (s)} by using Table 6.2.1, possibly in conjunction with a translation or a partial fraction expansion. To find 0 (t) = C~x{n(s)} it is convenient to write n(s) as
n(s ) = H (s)G (s ), (28)
6.6 The Convolution Integral
335
where H(s) = (as2 + bs + c)-1. The function H is known as the transfer function3 and depends only on the properties of the system under consideration; that is, H(s) is determined entirely by the coefficients a, b, and c. On the other hand, G(s) depends only on the external excitation g(t) that is applied to the system. By the convolution theorem we can write
f(t) = L-1{H(s)G(s)}=f h(t - T)g(r) dT, (29)
J0
where h(t) = L-1{H(s)}, and g(t) is the given forcing function.
To obtain a better understanding of the significance of h(t), we consider the case in which G(s) = 1; consequently, g(t) = S(t) and ^(s) = H(s). This means that y = h(t) is the solution of the initial value problem
ay" + by + cy = S(t), y(0) = 0, y (0) = 0, (30)
obtained from Eq. (27) by replacing g(t) by S(t). Thus h(t) is the response of the system to a unit impulse applied at t = 0, and it is natural to call h(t) the impulse response of the system. Equation (29) then says that y(t) is the convolution of the impulse response and the forcing function.
Referring to Example 2, we note that in that case the transfer function is H(s) = 1/(s2 + 4), and the impulse response is h(t) = (sin 2t)/2. Also, the first two terms on the right side of Eq. (20) constitute the function fi(t), the solution of the corresponding homogeneous equation that satisfies the given initial conditions.
PROBLEMS 1. Establish the commutative, distributive, and associative properties of the convolution
s integral.
(a) f * g = g* f
(b) f * (g1 + g2) = f * g1 + f * g2
(c) f * (g * h) = ( f * g) * h
2. Find an example different from the one in the text showing that (f * 1)(t) need not be equal to f (t).
3. Show, by means of the example f (t) = sin t, that f * f is not necessarily nonnegative.
In each of Problems 4 through 7 find the Laplace transform of the given function.
4. f(t) = f (t — t)2cos2t dT 5. f(t) = ( e~(t~T) sint dT
Jo Jo
6. f(t) = f (t — T)eT dt 7. f (t) = f sin(t — t) cos t dt
In each of Problems 8 through 11 find the inverse Laplace transform of the given function by using the convolution theorem.
1 s
8. F(s) = —— 9. F(s) =
s\s2 + 1) ' (s + 1)(s2 + 4)
10. F(s) = , . * 2 11. F(s) =
(s + 1)2(s2 + 4) s2 + 1
3This terminology arises from the fact that H(s) is the ratio of the transforms of the output and the input of the problem (27).
336
Chapter 6. The Laplace Transform
In each of Problems 12 through 19 express the solution of the given initial value problem in terms of a convolution integral.
12. y + o2y = g(t); y(0) = 0, /(0) = 1
13. /' + 2 y + 2 y = sin at; y(0) = 0, /(0) = 0
14. 4y" + 4y + 17y = g(t); y(0) = 0, /(0) = 0
15. y" + / + 5y = 1 — ujt); y(0) = 1, y(0) = -1
16. y + 4/ + 4y = g(t); y(0) = 2, /(0) = -3
17. y" + 3/ + 2y = cos at; y(0) = 1, y/(0) = 0
18. /v — y = g(t); y(0) = 0, y(0) = 0, /(0) = 0, /'(0) = 0
19. /v + 5y' + 4y = g(t); y(0) = 1, /(0) = 0, y"(0) = 0, /'(0) = 0
20. Consider the equation
in which f and k are known functions, and 0 is to be determined. Since the unknown function 0 appears under an integral sign, the given equation is called an integral equation; in particular, it belongs to a class of integral equations known as Volterra integral equations. Take the Laplace transform of the given integral equation and obtain an expression for L{0(t)} in terms of the transforms L{ f (t)} and C{k(t)} of the given functions f and k. The inverse transform of L{0(t)} is the solution of the original integral equation.
21. Consider the Volterra integral equation (see Problem 20)
(b) Show that the given integral equation is equivalent to the initial value problem
(c) Solve the given integral equation by using the Laplace transform.
(d) Solve the initial value problem of part (b) and verify that the solution is the same as that obtained in part (c).
22. The Tautochrone. A problem of interest in the history of mathematics is that of finding the tautochrone—the curve down which a particle will slide freely under gravity alone, reaching the bottom in the same time regardless of its starting point on the curve. This problem arose in the construction of a clock pendulum whose period is independent of the amplitude of its motion. The tautochrone was found by Christian Huygens (16291695) in 1673 by geometrical methods, and later by Leibniz and Jakob Bernoulli using analytical arguments. Bernoulli’s solution (in 1690) was one of the first occasions on which a differential equation was explicitly solved.
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