# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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t 0

To evaluate the limit in Eq. (11) we first observe that if t which must eventually be the case as t 0, then t0 t > 0. Since dT(t t0) is nonzero only in the interval from t0 t to t0 + t , we have

C TO

L{dT(t t0)} = e~stdT(t t0) dt

0

f,

t0+T

e~std(t to) dt.

Substituting for dT(t t0) from Eq. (4), we obtain

to+T

1

C[dT(t to)} = \ e~st dt = est

t0

2t If _T 2st

tto+T tt0 T

esto (eST e~ST)

2s t

or

sinh s t

C{dx(t to)}---------------esto. (12)

^ s T

The quotient (sinh st)/st is indeterminate as t ^ o, but its limit can be evaluated by LHospitals rule. We obtain

sinh s t s cosh s t

lim lim-------------------------- 1.

t ^o s T T ^o s

Then from Eq. (11) it follows that

L{S(t to)} esto. (13)

2Paul A. M. Dirac (1902-1984), English mathematical physicist, received his Ph.D. from Cambridge in 1926 and was professor of mathematics there until 1969. He was awarded the Nobel prize in 1933 (with Erwin Schrodinger) for fundamental work in quantum mechanics. His most celebrated result was the relativistic equation for the electron, published in 1928. From this equation he predicted the existence of an anti-electron, or positron, which was first observed in 1932. Following his retirement from Cambridge, Dirac moved to the United States and held a research professorship at Florida State University.

6.5 Impulse Functions

Equation (13) defines C{8(t t0)} for any t0 > 0. We extend this result, to allow t0 to

be zero, by letting t0 ^ 0 on the right side of Eq. (13); thus

L{S(f)} = lim esto = 1. (14)

to

In a similar way it is possible to define the integral of the product of the delta function and any continuous function f. We have

/00 r CO

S(t to) f(t) dt = lim / dx(t to) f (t) dt. (15)

-TO t^0 J o

Using the definition (4) of dT (t) and the mean value theorem for integrals, we find that

f0 1 f t0+T

dr (t t0) f (t) dt =\ f (t) dt

J-oo 2 t Jl-t

= ^ 2 t f(t*) = f(t*),

2 t

where t0 t < t* < t0 + t. Hence, t* ^ t0 as t ^ 0, and it follows from Eq. (15) that 0 0 0

* O

S(t t0) f(t) dt = f(t0). (16)

f

J a

It is often convenient to introduce the delta function when working with impulse problems, and to operate formally on it as though it were a function of the ordinary kind. This is illustrated in the example below. It is important to realize, however, that the ultimate justification of such procedures must rest on a careful analysis of the limiting operations involved. Such a rigorous mathematical theory has been developed, but we do not discuss it here.

Find the solution of the initial value problem

2/ + / + 2 y = S(t 5), (17)

y(0) = 0, y (0) = 0. (18)

This initial value problem arises from the study of the same electrical circuit or me-

chanical oscillator as in Example 1 of Section 6.4. The only difference is in the forcing term.

To solve the given problem we take the Laplace transform of the differential equation and use the initial conditions, obtaining

(2s2 + s + 2) Y (s) = e~5s.

Thus

e~5s e~5s

Y(s) = 2---------------------=----------------^---------------tt--ir (19)

2s + s + 2 2 (s + 1) + jg

By Theorem 6.3.2 or from line 9 of Table 6.2.1

H (T+vi+g =A e"4-5F » (20)

328

Chapter 6. The Laplace Transform

Hence, by Theorem 6.3.1, we have

y = C~l[Y(s)} = -j= u5(t)e-(t-5)/4sin (t - 5), (21)

which is the formal solution of the given problem. It is also possible to write y in the form

f 0, t < 5,

y =j 2== e-(t-5)/4 sin 5 (t - 5), t > 5. (22)

The graph of Eq. (22) is shown in Figure 6.5.3. Since the initial conditions at t = 0 are homogeneous, and there is no external excitation until t = 5, there is no response in the interval 0 < t < 5. The impulse at t = 5 produces a decaying oscillation that persists indefinitely. The response is continuous at t = 5 despite the singularity in the forcing function at that point. However, the first derivative of the solution has a jump discontinuity at t = 5 and the second derivative has an infinite discontinuity there. This is required by the differential equation (17), since a singularity on one side of the equation must be balanced by a corresponding singularity on the other side.

FIGURE 6.5.3 Solution of the initial value problem (17), (18).

PROBLEMS In each of Problems 1 through 12 find the solution of the given initial value problem and draw

s its graph.

?/+ 2y = S(t - n); y(0) = 1, y (0) = 0

\y = S(t - n) - S(t - 2n); y(0) = 0, /(0) = 0

;/ + 2y = S(t - 5) + u10(t); y(0) = 0, /(0) = 1/2

r = 20S(t - 3); y(0) = 1, y(0) = 0

y + 3y = sin t + S(t - 3n); y(0) = 0, /(0) = 0

y = S(t - 4n); y(0) = 1/2, /(0) = 0

r = S(t - 2n) cos t; y(0) = 0, y (0) = 1

? 8. y'+ 4y = 2S(t - n/4); y(0) = 0, /(0) = 0

? 1. y

? 2. y

? 3. y

? 4. y

? 5. y

? 6. y

? 7. y

? 8. y

6.5 Impulse Functions

329

? 9. / + y = un/2(t) + 3S(t 3n/2) - u2n(t); y(0) = 0, /(0) = 0

? 10. 2y" + / + 4y = S(t n/6) sin t; y(0) = 0, ?(0) = 0

? 11. y" + 2y7 + 2y = cos t + 5(t n/2); y(0) = 0, /(0) = 0

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