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Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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af + by + cy = g(t), (21)
where a, b, and c are real constants and g is a given function, together with the initial conditions
y(0) = y0, /(0) = 0 (22)
The transform approach yields some important insights concerning the structure of the solution of any problem of this type.
The initial value problem (21), (22) is often referred to as an input-output problem. The coefficients a, b, and c describe the properties of some physical system, and g(t) is the input to the system. The values y0 and y0 describe the initial state, and the solution y is the output at time t.
By taking the Laplace transform of Eq. (21) and using the initial conditions (22), we obtain
If we let
(as2 + bs + c) Y(s) - (as + b)y0 - ay0 = G(s).
(as + b) y0 + ay0 G (s)
^) =K----------------2df-------^, n(s) = 2 \)-------, (23)
as + bs + c as + bs + c
then we can write
Consequently,
Y (s) = ^) + n(s). (24)
y = () + f(t), (25)
where () = ?-1{^)} and (t) = ?-1{n(s)}. Observe that y = () is a solution
of the initial value problem
ay' + by + cy = 0, y (0) = y0, /(0) = y?, (26)
obtained from Eqs. (21) and (22) by setting g(t) equal to zero. Similarly, y = ty(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 () = C~l{ (s)} by using
Table 6.2.1, possibly in conjunction with a translation or a partial fraction expansion. To find () = L-1{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 x)g(x) dx, (29)
0
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
7 = h(t) is the solution of the initial value problem
ay" + b7 + c7 = S(t), 7(0) = 0, /(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 ty(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 ofEq. (20) constitute the function (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
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 x)2cos2x dx 5. f(t) = f e~(tT'> sinx dx
J0 J0
6. f(t) = t (t x)eT dx 7. f (t) = f sin(t x) cos x dx
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