7.5 Homogeneous Linear Systems with Constant Coefficients Notes

Observe that we have used Theorem 6.3.1 to write the inverse transforms of e−5s H (s) and e−20s H (s), respectively. Finally, to determine h(t) we use the partial fraction expansion of H (s): H (s) = a + bs + c .

(8) s

2s2 + s + 2

Upon determining the coefficients we find that a = 1 , b = −1, and c = − 1 . Thus

2

2 s + 1 H (s) = 1/2 −

2 s

2s2 + s + 2

(s + 1) + 1

= 1/2 − 1

4

4 ,

(9) s

2 (s + 1)2 + 15

4

16

so that, by referring to lines 9 and 10 of Table 6.2.1, we obtain

√

√ h(t) = 1 − 1 [e−t/4 cos( 15 t/4) + ( 15/15)e−t/4 sin( 15 t/4)].

(10)

2

In Figure 6.4.1 the graph of y = φ(t) from Eqs. (7) and (10) shows that the solution

consists of three distinct parts. For 0 < t < 5 the differential equation is

2y + y + 2y = 0

(11)

and the initial conditions are given by Eq. (3). Since the initial conditions impart no energy to the system, and since there is no external forcing, the system remains at rest; that is, y = 0 for 0 < t < 5. This can be confirmed by solving Eq. (11) subject to the initial conditions (3). In particular, evaluating the solution and its derivative at t = 5, or more precisely, as t approaches 5 from below, we have y(5) = 0, y(5) = 0.

(12)

Once t > 5, the differential equation becomes

2y + y + 2y = 1,

(13)

whose solution is the sum of a constant (the response to the constant forcing function) and a damped oscillation (the solution of the corresponding homogeneous equation). The plot in Figure 6.4.1 shows this behavior clearly for the interval 5 ≤ t ≤ 20. An expression for this portion of the solution can be found by solving the differential equation (13) subject to the initial conditions (12). Finally, for t > 20 the differential equation becomes Eq. (11) again, and the initial conditions are obtained by evaluating the solution of Eqs. (13), (12) and its derivative at t = 20. These values are y(20) ∼

= 0.50162, y(20) ∼

= 0.01125.

(14)

The initial value problem (11), (14) contains no external forcing, so its solution is a damped oscillation about y = 0, as can be seen in Figure 6.4.1. 6.4 Differential Equations with Discontinuous Forcing Functions y

0.8

0.6

0.4

0.2

10

20

30

40 t

– 0.2 FIGURE 6.4.1

Solution of the initial value problem (1), (2), (3).

While it may be helpful to visualize the solution shown in Figure 6.4.1 as composed

of solutions of three separate initial value problems in three separate intervals, it is somewhat tedious to find the solution by solving these separate problems. Laplace transform methods provide a much more convenient and elegant approach to this problem and to others having discontinuous forcing functions.

The effect of the discontinuity in the forcing function can be seen if we examine the

solution φ(t) of Example 1 more closely. According to the existence and uniqueness Theorem 3.2.1 the solution φ and its first two derivatives are continuous except possibly at the points t = 5 and t = 20 where g is discontinuous. This can also be seen at once from Eq. (7). One can also show by direct computation from Eq. (7) that φ and φ are continuous even at t = 5 and t = 20. However, if we calculate φ, we find that

lim φ(t) = 0,

lim φ(t) = 1/2. t →5− t →5+

Consequently, φ(t) has a jump of 1/2 at t = 5. In a similar way one can show that φ(t) has a jump of −1/2 at t = 20. Thus the jump in the forcing term g(t) at these points is balanced by a corresponding jump in the highest order term 2y on the left side of the equation.

Consider now the general second order linear equation y + p(t)y + q(t)y = g(t),

(15)

where p and q are continuous on some interval α < t < β, but g is only piecewise continuous there. If y = ψ(t) is a solution of Eq. (15), then ψ and ψ are continuous on α < t < β, but ψ has jump discontinuities at the same points as g. Similar remarks apply to higher order equations; the highest derivative of the solution appearing in the differential equation has jump discontinuities at the same points as the forcing function, but the solution itself and its lower derivatives are continuous even at those points. Chapter 6. The Laplace Transform

Describe the qualitative nature of the solution of the initial value problem E X A M P L E 2 y + 4y = g(t),

(16) y(0) = 0, y(0) = 0,

(17)

where



 0,

0 ≤ t < 5, g(t) =  (t − 5)/5,

5 ≤ t < 10,

(18)

1, t ≥ 10,

and then find the solution.

In this example the forcing function has the graph shown in Figure 6.4.2 and is

known as ramp loading. It is relatively easy to identify the general form of the solution. For t < 5 the solution is simply y = 0. On the other hand, for t > 10 the solution has the form y = c cos 2t + c sin 2t + 1/4.

(19)

1

2

The constant 1/4 is a particular solution of the nonhomogeneous equation while the other two terms are the general solution of the corresponding homogeneous equation. Thus the solution (19) is a simple harmonic oscillation about y = 1/4. Similarly, in the intermediate range 5 < t < 10, the solution is an oscillation about a certain linear function. In an engineering context, for example, we might be interested in knowing the amplitude of the eventual steady oscillation. y 1 y = g(t)

0.5

5

10

15

20 t FIGURE 6.4.2

Ramp loading; y = g(t) from Eq. (18).

To solve the problem it is convenient to write g(t) = [u (t)(t − 5) − u (t)(t − 10)]/5,

(20)

5

10

as you may verify. Then we take the Laplace transform of the differential equation and use the initial conditions, thereby obtaining (s2 + 4)Y (s) = (e−5s − e−10s)/5s2,

or Y (s) = (e−5s − e−10s)H (s)/5,

(21)

where H (s) = 1/s2(s2 + 4).

(22) 6.4 Differential Equations with Discontinuous Forcing Functions

Thus the solution of the initial value problem (16), (17), (18) is y = φ(t) = [u (t)h(t − 5) − u (t)h(t − 10)]/5,

(23)

5

10

where h(t) is the inverse transform of H (s). The partial fraction expansion of H (s) is H (s) = 1/4 − 1/4 ,

(24) s2 s2 + 4

and it then follows from lines 3 and 5 of Table 6.2.1 that h(t) = 1 t − 1 sin 2t.

(25)

4

8

The graph of y = φ(t) is shown in Figure 6.4.3. Observe that it has the qualitative form that we indicated earlier. To find the amplitude of the eventual steady oscillation it is sufficient to locate one of the maximum or minimum points for t > 10. Setting the derivative of the solution (23) equal to zero, we find that the first maximum is located approximately at (10.642, 0.2979), so the amplitude of the oscillation is approximately 0.0479. y

0.30

0.20

0.10

5

10

15

20 t FIGURE 6.4.3

Solution of the initial value problem (16), (17), (18).

Note that in this example the forcing function g is continuous but g is discontinuous

at t = 5 and t = 10. It follows that the solution φ and its first two derivatives are continuous everywhere, but φ has discontinuities at t = 5 and at t = 10 that match the discontinuities in g at those points. PROBLEMS

In each of Problems 1 through 13 find the solution of the given initial value problem. Draw the graphs of the solution and of the forcing function; explain how they are related.

䉴

䉴 Chapter 6. The Laplace Transform

䉴

䉴

䉴 2

䉴

䉴 4

䉴

䉴 4

䉴

䉴 1 2

䉴 c 0 0 c 0 0 0

䉴 16. A certain spring–mass system satisfies the initial value problem

4

䉴 by

5

䉴 18. Consider the initial value problem

3 k where k and 6.4 Differential Equations with Discontinuous Forcing Functions k k k problem. 2 Resonance and Beats. In Section 3.9 we observed that an undamped harmonic oscillator

䉴 19. Consider the initial value problem

where n 0 kπ (t ).

(b) Find the solution of the initial value problem.

䉴 20. Consider the initial value problem

䉴 21. Consider the initial value problem

where n 0 kπ (t ).

䉴 22. Consider the initial value problem

Chapter 6. The Laplace Transform

(c) Compare the results of part (b) with those from Problem 20 and from Section 3.9 for a sinusoidally forced oscillator.

䉴 23. Consider the initial value problem

where n (t). 0

(a) Find the solution of this initial value problem.

(c) From the graph in part (b) estimate the “slow period” and the “fast period” for this 0 0 0 6.5 Impulse Functions

In some applications it is necessary to deal with phenomena of an impulsive nature, for example, voltages or forces of large magnitude that act over very short time intervals. Such problems often lead to differential equations of the form ay + by + cy = g(t),

(1)

where g(t) is large during a short interval t − τ < t < t + τ and is otherwise zero.

0

The integral I (τ ), defined by

t +τ

0 I (τ ) = g(t) dt,

(2) t −τ

0

or, since g(t) = 0 outside of the interval (t − τ, t + τ),

0

∞ I (τ ) = g(t) dt,

(3)

−∞

is a measure of the strength of the forcing function. In a mechanical system, where g(t) is a force, I (τ ) is the total impulse of the force g(t) over the time interval (t − τ, t + τ). Similarly, if y is the current in an electric circuit and g(t) is the time

0

derivative of the voltage, then I (τ ) represents the total voltage impressed on the circuit during the interval (t − τ, t + τ).

0

0 6.5 Impulse Functions

In particular, let us suppose that t is zero, and that g(t) is given by

0

1/2τ,

−τ < t < τ, g(t) = dτ (t) =

0, t ≤ −τ

or t ≥ τ,

(4)

where τ is a small positive constant (see Figure 6.5.1). According to Eq. (2) or (3) it follows immediately that in this case I (τ ) = 1 independent of the value of τ , as long as τ = 0. Now let us idealize the forcing function dτ by prescribing it to act over shorter

and shorter time intervals; that is, we require that τ → 0, as indicated in Figure 6.5.2. As a result of this limiting operation we obtain

lim d τ→ τ (t) = 0, t = 0.

(5)

0

Further, since I (τ ) = 1 for each τ = 0, it follows that

lim I (τ ) = 1.

(6) τ→0

Equations (5) and (6) can be used to define an idealized unit impulse function δ, which imparts an impulse of magnitude one at t = 0, but is zero for all values of t other than zero. That is, the “function” δ is defined to have the properties δ(t) = 0, t = 0;

(7)

∞ δ(t) dt = 1.

(8)

−∞ y

1

2τ

–τ

τ t FIGURE 6.5.1

Graph of y = dτ (t). y t FIGURE 6.5.2

Graphs of y = dτ (t) as τ → 0. Chapter 6. The Laplace Transform

There is no ordinary function of the kind studied in elementary calculus that satisfies both of Eqs. (7) and (8). The “function” δ, defined by those equations, is an example of what are known as generalized functions and is usually called the Dirac2 Since δ(t) corresponds to a unit impulse at t = 0, a unit impulse at an arbitrary point t = t is given by δ(t − t ). From Eqs. (7) and (8) it follows that

0

0 δ(t − t ) = 0, t = t ;

(9)

0

∞ δ(t −t ) dt = 1.

(10)

0

−∞

The delta function does not satisfy the conditions of Theorem 6.1.2, but its Laplace

transform can nevertheless be formally defined. Since δ(t) is defined as the limit of dτ (t) as τ → 0, it is natural to define the Laplace transform of δ as a similar limit of

the transform of dτ . In particular, we will assume that t > 0, and define L{δ(t − t )}

0

by the equation L{δ(t − t )} = lim L{d )}.

(11)

0 τ→ τ (t − t0

0

To evaluate the limit in Eq. (11) we first observe that if τ < t , which must eventually

0

be the case as τ → 0, then t − τ > 0. Since d ) is nonzero only in the interval

0 τ (t − t0

from t − τ to t + τ , we have

0

∞ L{dτ(t − t )} = e−st d ) dt

0 τ (t − t0

0

t +τ

0

= e−st dτ (t − t ) dt.

0 t −τ

0

Substituting for dτ (t − t ) from Eq. (4), we obtain

0

t +τ

0 t =t +τ L{

0 dτ (t − t )} = 1 e−st dt = − 1

0

2τ t −τ

2sτ e−st t=t −τ

0

= 1

2sτ e−st0(esτ − e−sτ )

or L{dτ(t − t )} = sinh sτ

0 sτ e−st0 .

(12)

The quotient (sinh sτ)/sτ is indeterminate as τ → 0, but its limit can be evaluated by L’Hospital’s rule. We obtain

sinh sτ s cosh sτ

lim

= lim

= 1. τ→0 sτ τ→0 s

Then from Eq. (11) it follows that L{δ(t − t )} = e−st0.

(13)

0

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 Schr¨odinger) 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 L{δ(t − t )} for any t > 0. We extend this result, to allow t to

0

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

0 L{δ(t)} = lim e−st0 = 1.

(14) t →0

0

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

∞

∞ δ(t − t ) f (t) dt = lim d ) f (t) dt.

(15)

0 τ (t − t0

−∞ τ→0 −∞

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

∞

t +τ

0 dτ (t − t ) f (t) dt = 1 f (t) dt

0

−∞

2τ t −τ

= 1 · 2τ · f (t∗) = f (t∗),

2τ

where t − τ < t∗ < t + τ . Hence, t∗ → t as τ → 0, and it follows from Eq. (15)

0

that

∞ δ(t −t )f(t) dt = f(t ).

(16)

0

−∞

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 E X A M P L E 1

2y + y + 2y = δ(t − 5),

(17) y(0) = 0, y(0) = 0.

(18)

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

1 Y (s) = e−5s

= e−5s .

(19)

2s2 + s + 2

2 (s + 1 )2 + 15

4

16

By Theorem 6.3.2 or from line 9 of Table 6.2.1

√ L−

1

= 4

√ e−t/4 sin

15 t.

(20) (s + 1)2 + 15

15

4

16 Chapter 6. The Laplace Transform

Hence, by Theorem 6.3.1, we have

√ y = L−1{Y (s)} = 2

√ u (t)e−(t−5)/4 sin 15 (t − 5),

(21)

15

5

4

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

0, t < 5, y =

√

2

√ e−(t−5)/4 sin 15 (t − 5), t ≥ 5.

(22)

15

4

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. y

0.3

0.2

0.1

5

10

15

20 t

– 0.1 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 its graph.

䉴

䉴 10

䉴

䉴 6.5 Impulse Functions

䉴 2

䉴

䉴

䉴 13. Consider again the system in Example 1 in the text in which an oscillation is excited by

0 0

䉴 14. Consider the initial value problem

2 1 1 4 1 1 1 1

䉴 15. Consider the initial value problem

2 1 4 1 1

䉴 16. Consider the initial value problem

k k k 0 k 0 0 0

(a) Try to predict the nature of the solution without solving the problem. (b) Test your prediction by finding the solution and drawing its graph. (c) Determine what happens after the sequence of impulses ends. Chapter 6. The Laplace Transform 20 20

䉴 20 20

䉴 15 40

䉴

䉴 23. The position of a certain lightly damped oscillator satisfies the initial value problem 20 Observe that, except for the damping term, this problem is the same as Problem 18. (a) Try to predict the nature of the solution without solving the problem. (b) Test your prediction by finding the solution and drawing its graph. (c) Determine what happens after the sequence of impulses ends.

䉴 24. Proceed as in Problem 23 for the oscillator satisfying 15 Observe that, except for the damping term, this problem is the same as Problem 21. 25. (a) By the method of variation of parameters show that the solution of the initial value problem

0

and confirm that the solution agrees with the result of part (b). 6.6 The Convolution Integral

Sometimes it is possible to identify a Laplace transform H (s) as the product of two other transforms F (s) and G(s), the latter transforms corresponding to known functions f and g, respectively. In this event, we might anticipate that H (s) would be the transform of the product of f and g. However, this is not the case; in other words, the Laplace transform cannot be commuted with ordinary multiplication. On the other hand, if an appropriately defined “generalized product” is introduced, then the situation changes, as stated in the following theorem. 6.6 The Convolution Integral Theorem 6.6.1

If F (s) = L{ f (t)} and G(s) = L{g(t)} both exist for s > a ≥ 0, then H (s) = F(s)G(s) = L{h(t)}, s > a,

(1)

where t t h(t) = f (t − τ)g(τ ) dτ = f (τ )g(t − τ) dτ.

(2)

0

The function h is known as the convolution of f and g; the integrals in Eq. (2) are known as convolution integrals.

The equality of the two integrals in Eq. (2) follows by making the change of variable t − τ = ξ in the first integral. Before giving the proof of this theorem let us make some observations about the convolution integral. According to this theorem, the transform of the convolution of two functions, rather than the transform of their ordinary product, is given by the product of the separate transforms. It is conventional to emphasize that the convolution integral can be thought of as a “generalized product” by writing h(t) = ( f ∗ g)(t).

(3)

In particular, the notation ( f ∗ g)(t) serves to indicate the first integral appearing in Eq. (2).

The convolution f ∗ g has many of the properties of ordinary multiplication. For

example, it is relatively simple to show that f ∗ g = g ∗ f

(commutative law)

(4) f ∗ (g + g ) = f ∗ g + f ∗ g

(distributive law)

(5)

1

2

1

2 ( f ∗ g) ∗ h = f ∗ (g ∗ h)

(associative law)

(6) f ∗ 0 = 0 ∗ f = 0.

(7)

The proofs of these properties are left to the reader. However, there are other properties of ordinary multiplication that the convolution integral does not have. For example, it is not true in general that f ∗ 1 is equal to f . To see this, note that

t t ( f ∗ 1)(t) = f (t − τ) · 1 dτ = f (t − τ) dτ.

0

If, for example, f (t) = cos t, then

t τ=t ( f ∗ 1)(t) =

cos(t − τ) dτ = − sin(t − τ)

0 τ=0

= − sin 0 + sin t = sin t.

Clearly, ( f ∗ 1)(t) = f (t). Similarly, it may not be true that f ∗ f is nonnegative. See Problem 3 for an example.

Convolution integrals arise in various applications in which the behavior of the

system at time t depends not only on its state at time t, but on its past history as well. Systems of this kind are sometimes called hereditary systems and occur in such diverse fields as neutron transport, viscoelasticity, and population dynamics. Chapter 6. The Laplace Transform

Turning now to the proof of Theorem 6.6.1, we note first that if

∞ F (s) = e−sξ f (ξ ) dξ

0

and

∞ G(s) = e−sηg(η) dη,

0

then

∞

∞ F (s)G(s) = e−sξ f (ξ ) dξ e−sηg(η) dη.

(8)

0

Since the integrand of the first integral does not depend on the integration variable of the second, we can write F (s)G(s) as an iterated integral,

∞

∞ F (s)G(s) = g(η) dη e−s(ξ+η) f (ξ ) dξ.

(9)

0

This expression can be put into a more convenient form by introducing new variables of integration. First let ξ = t − η, for fixed η. Then the integral with respect to ξ in Eq. (9) is transformed into one with respect to t; hence

∞

∞ F (s)G(s) = g(η) dη e−st f (t − η) dt.

(10)

0 η

Next let η = τ ; then Eq. (10) becomes

∞

∞ F (s)G(s) = g(τ ) dτ e−st f (t − τ) dt.

(11)

0 τ

The integral on the right side of Eq. (11) is carried out over the shaded wedge-shaped region extending to infinity in the tτ -plane shown in Figure 6.6.1. Assuming that the order of integration can be reversed, we finally obtain

∞

t F (s)G(s) = e−st dt f (t − τ)g(τ ) dτ,

(12)

0

τ

τ = t t = τ t → ∞

τ = 0 t FIGURE 6.6.1

Region of integration in F(s)G(s). 6.6 The Convolution Integral

or

∞ F (s)G(s) = e−st h(t) dt

0

= L{h(t)},

(13)

where h(t) is defined by Eq. (2). This completes the proof of Theorem 6.6.1.

Find the inverse transform of E X A M P L E 1 H (s) = a .

(14) s2(s2 + a2)

It is convenient to think of H (s) as the product of s−2 and a/(s2 + a2), which,

according to lines 3 and 5 of Table 6.2.1, are the transforms of t and sin at, respectively. Hence, by Theorem 6.6.1, the inverse transform of H (s) is

t h(t) = (t − τ) sin aτ dτ = at − sin at .

(15)

0 a2

You can verify that the same result is obtained if h(t) is written in the alternate form

t h(t) = τ sin a(t − τ) dτ,

0

which confirms Eq. (2) in this case. Of course, h(t) can also be found by expanding H (s) in partial fractions.

Find the solution of the initial value problem E X A M P L E 2 y + 4y = g(t),

(16) y(0) = 3, y(0) = −1.

(17)

By taking the Laplace transform of the differential equation and using the initial

conditions, we obtain s2Y (s) − 3s + 1 + 4Y (s) = G(s),

or Y (s) = 3s − 1 + G(s) .

(18) s2 + 4 s2 + 4

Observe that the first and second terms on the right side of Eq. (18) contain the dependence of Y (s) on the initial conditions and forcing function, respectively. It is convenient to write Y (s) in the form s

2

2 Y (s) = 3

− 1

+ 1 G(s).

(19) s2 + 4

2 s2 + 4

2 s2 + 4

Then, using lines 5 and 6 of Table 6.2.1 and Theorem 6.6.1, we obtain

t y = 3 cos 2t − 1 sin 2t + 1

sin 2(t − τ)g(τ ) dτ.

(20)

2

0 Chapter 6. The Laplace Transform

If a specific forcing function g is given, then the integral in Eq. (20) can be evaluated (by numerical means, if necessary). Example 2 illustrates the power of the convolution integral as a tool for writing

the solution of an initial value problem in terms of an integral. In fact, it is possible to proceed in much the same way in more general problems. Consider the problem consisting of the differential equation ay + 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) = y , y(0) = y .

(22)

0

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 y and y describe the initial state, and the solution

0

0 y is the output at time t.

By taking the Laplace transform of Eq. (21) and using the initial conditions (22), we

obtain (as2 + bs + c)Y (s) − (as + b)y − ay = G(s).

0

If we let (as + b)y + ay (s) =

0

0 , (s) = G(s) ,

(23) as2 + bs + c as2 + bs + c

then we can write Y (s) = (s) + (s).

(24)

Consequently, y = φ(t) + ψ(t),

(25)

where φ(t) = L−1{(s)} and ψ(t) = L−1{(s)}. Observe that y = φ(t) is a solution of the initial value problem ay + by + cy = 0, y(0) = y , y(0) = y ,

(26)

0

obtained from Eqs. (21) and (22) by setting g(t) equal to zero. Similarly, y = ψ(t) is the solution of ay + by + cy = g(t), y(0) = 0, y(0) = 0,

(27)

in which the initial values y and y are each replaced by zero.

0

Once specific values of a, b, and c are given, we can find φ(t) = L−1{(s)} by using Table 6.2.1, possibly in conjunction with a translation or a partial fraction expansion. To find ψ(t) = L−1{(s)} it is convenient to write (s) as (s) = H(s)G(s),

(28)

where H (s) = (as2 + bs + c)−1. The function H is known as the transfer functionand 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

t ψ(t) = L−1{H(s)G(s)} = h(t − τ)g(τ ) dτ,

(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) = δ(t) and (s) = H (s). This means that y = h(t) is the solution of the initial value problem ay + by + cy = δ(t), y(0) = 0, y(0) = 0,

(30)

obtained from Eq. (27) by replacing g(t) by δ(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 ψ(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 φ(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. 1 2 1 2

In each of Problems 4 through 7 find the Laplace transform of the given function. t t 4. 5. 0 0 t t 6. 7. 0 0 1 s 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). Chapter 6. The Laplace Transform 4 20. Consider the equation

0 21. Consider the Volterra integral equation (see Problem 20)

0 (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.

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 (1629– 1695) 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.

The geometrical configuration is shown in Figure 6.6.2. The starting point P(a, b) is

joined to the terminal point (0, 0) by the arc C. Arc length s is measured from the origin, and f (y) denotes the rate of change of s with respect to y:

1/2 d x 2 f (y) = ds = 1 + .

(i) d y d y

Then it follows from the principle of conservation of energy that the time T (b) required for a particle to slide from P to the origin is b f(y) T (b) =

1

√

√ d y.

(ii)

2g 0 b − y 6.6 The Convolution Integral y P(a, b) C s x FIGURE 6.6.2

The tautochrone.

(a) Assume that T (b) = T , a constant, for each b. By taking the Laplace transform of

0

Eq. (ii) in this case and using the convolution theorem, show that

2g T F(s) =

0

√ ; π

(iii) s

then show that

2g T f (y) =

0 . π √

(iv) y Hint: See (b) Combining Eqs. (i) and (iv), show that d x = 2α − y ,

(v) d y y

where α = gT 2/π2.

0

(c) Use the substitution y = 2α sin2(θ/2) to solve Eq. (v), and show that x = α(θ + sin θ), y = α(1 − cos θ).

(vi)

Equations (vi) can be identified as parametric equations of a cycloid. Thus the tautochrone is an arc of a cycloid. REFERENCES

The books listed below contain additional information on the Laplace transform and its applications:

Churchill, R. V., Operational Mathematics (3rd ed.) (New York: McGraw-Hill, 1971).

Doetsch, G., Nader, W. (tr.), Introduction to the Theory and Application of the Laplace Transform

(New York: Springer-Verlag, 1974).

Kaplan, W., Operational Methods for Linear Systems (Reading, MA: Addison-Wesley, 1962).

Kuhfittig, P. K. F., Introduction to the Laplace Transform (New York: Plenum, 1978).

Miles, J. W., Integral Transforms in Applied Mathematics (London: Cambridge University Press,

1971).

Rainville, E. D., The Laplace Transform: An Introduction (New York: Macmillan, 1963).

Each of the books just mentioned contains a table of transforms. Extensive tables are also available; see,

for example:

Erdelyi, A. (ed.), Tables of Integral Transforms (Vol. 1) (New York: McGraw-Hill, 1954). Chapter 6. The Laplace Transform

Roberts, G. E., and Kaufman, H., Table of Laplace Transforms (Philadelphia: Saunders, 1966).

A further discussion of generalized functions can be found in:

Lighthill, M. J., Fourier Analysis and Generalized Functions (London: Cambridge University Press,

1958). </body></html>