Transform of the Dirac Delta Function

Computer Lab Assignments

Theorem 4.5.1 Transform of the Dirac Delta Function

For t₀. 0, +5d(t2t0)6 e^st0. (3)

PROOF:

To begin, we can write da(t t0) in terms of the unit step function by virtue of (11) and (12) of Section 4.3:

d_a(t2 t₀) 1

2a f8(t2(t₀2 a))2 8(t2 (t₀a))g.

By linearity and (14) of Section 4.3, the Laplace transform of this last expression is +5d_a(t2t0)6 1

2ace^s(t02a)

s 2 e^s(t0a)

s d e^st0ae^sa2 e^sa

2sa b. (4)

Since (4) has the indeterminate form 0/0 as a S 0, we apply L’Hôpital’s rule:

+5d(t2 t₀)6 lim

aS0 +5d_a(t2t₀)6 e^st0 lim

aS0ae^sa2 e^sa

2sa b e^st0. Now when t0 0, it seems plausible to conclude from (3) that

+{d (t)} 1.

The last result emphasizes the fact that d(t) is not the usual type of function that we have been considering, since we expect from Theorem 4.2.3 that +{ f (t)} S 0 as s Sq.

EXAMPLE 1 Two Initial-Value Problems Solve y y 4 d (t 2p) subject to

(a) y(0) 1, y(0) 0 (b) y(0) 0, y(0) 0.

The two initial-value problems could serve as models for describing the motion of a mass on a spring moving in a medium in which damping is negligible. At t 2p the mass is given a sharp blow. In part (a) the mass is released from rest 1 unit below the equilibrium position. In part (b) the mass is at rest in the equilibrium position.

SOLUTION (a) From (3) the Laplace transform of the differential equation is s²Y(s) s Y(s) 4e^2ps or Y(s) s

s²1 4e^2ps s²1.

Using the inverse form of the second translation theorem, (15) of Section 4.3, we find y(t) cos t 4 sin(t 2p) 8(t 2p).

4.5 The Dirac Delta Function ^| 249

250 ^| CHAPTER 4 The Laplace Transform

REMARKS

(i) If d(t t0) were a function in the usual sense, then property (ii ) on page 249 would imply e^q0d(t t0) dt 0 rather than e^q0d(t t0) dt 1. Since the Dirac delta function did not “behave”

like an ordinary function, even though its users produced correct results, it was met initially with great scorn by mathematicians. However, in the 1940s Dirac’s controversial function was put on a rigorous footing by the French mathematician Laurent Schwartz in his book Théorie des distributions, and this, in turn, led to an entirely new branch of mathematics known as the theory of distributions or generalized functions. In this theory, (2) is not an accepted definition of d(t t0), nor does one speak of a function whose values are either q or 0. Although we shall not pursue this topic any further, suffice it to say that the Dirac delta function is best character- ized by its effect on other functions. If f is a continuous function, then

#

₀^{qf(t)d(t2t0)dtf(t0) (7)}

can be taken as the definition of d(t t0). This result is known as the sifting property since d(t t0) has the effect of sifting the value f (t0) out of the set of values of f on [0, q). Note that property (ii) (with f (t) 1) and (3) (with f (t) e^st) are consistent with (7).

(ii) In the Remarks in Section 4.2 we indicated that the transfer function of a general linear nth-order differential equation with constant coefficients is W(s) 1/P(s), where P(s) ansⁿ an1s^{n 1} p a0. The transfer function is the Laplace transform of function w(t), called the weight function of a linear system. But w(t) can be characterized in terms of the discussion at hand. For simplicity let us consider a second-order linear system in which the input is a unit impulse at t 0:

a2 y a1 y a0 y d(t), y(0) 0, y(0) 0.

Applying the Laplace transform and using +{d(t)} 1 shows that the transform of the response y in this case is the transfer function

Y(s) 1

a₂s²a₁s a₀ 1

P(s) W(s) and so y +¹e 1

P(s)f w(t).

From this we can see, in general, that the weight function y w(t) of an nth-order linear system is the zero-state response of the system to a unit impulse. For this reason w(t) is called as well the impulse response of the system.

Since sin(t 2p) sin t, the foregoing solution can be written as y(t) ecost,

cost4sint, 0#t, 2p

t$ 2p. (5)

In FIGURE 4.5.2 we see from the graph of (5) that the mass is exhibiting simple harmonic motion until it is struck at t 2p. The influence of the unit impulse is to increase the amplitude of vibration to !17 for t 2p.

(b) In this case the transform of the equation is simply Y(s) 4e^2ps

s²1, and so y(t) 4 sin(t 2p) 8(t 2p)

e0,

4sint, 0#t, 2p.

t$ 2p. (6)

The graph of (6) in FIGURE 4.5.3 shows, as we would expect from the initial conditions, that the mass exhibits no motion until it is struck at t 2p.

FIGURE 4.5.2 In Example 1(a), moving mass is struck at t 2p

y

t

–1 2

1

π 4π

FIGURE 4.5.3 In Example 1(b), mass is at rest until struck at t 2p

y

t –1

1

2π 4π

In Problems 1–12, use the Laplace transform to solve the given differential equation subject to the indicated initial conditions.

1. y 3y d(t 2), y(0) 0 2. y y d(t 1), y(0) 2

3. y y d(t 2p), y(0) 0, y(0) 1 4. y 16y d(t 2p), y(0) 0, y(0) 0

5. y y d(t p/2) d(t 3p/2), y(0) 0, y(0) 0 6. y y d(t 2p) d(t 4p), y(0) 1, y(0) 0 7. y 2y d(t 1), y(0) 0, y(0) 1

8. y 2y 1 d(t 2), y(0) 0, y(0) 1 9. y 4y 5y d(t 2p), y(0) 0, y(0) 0 10. y 2y y d(t 1), y(0) 0, y(0) 0

11. y 4y 13y d(t p) d(t 3p), y(0) 1, y(0) 0 12. y 7y 6y e^t d(t 2) d(t 4), y(0) 0, y(0) 0 In Problems 13 and 14, use the Laplace transform to solve the given initial-value problem. Graph your solution on the interval f0, 8pg.

13. y0 y a

q k1

d(t2 kp), y(0) 0, y9(0)1

14. y0 y a

q k1

d(t2 2kp), y(0) 0, y9(0) 1 In Problems 15 and 16, a uniform beam of length L carries a concentrated load w0 at x ¹2L. Solve the differential equation

EI d⁴y

dx⁴ w0d(x2 ¹₂L), 0, x, L, subject to the given boundary conditions.

15. y(0) 0, y(0) 0, y(L) 0, y(L) 0

FIGURE 4.5.4 Beam embedded at its left end and free at its right end

x

y

L w0

16. y(0) 0, y(0) 0, y(L) 0, y(L) 0

FIGURE 4.5.5 Beam embedded at both ends x

y

L w0

Discussion Problems

17. Someone tells you that the solutions of the two IVPs y 2y 10y 0, y(0) 0, y(0) 1 and y 2y 10y d(t), y(0) 0, y(0) 0 are exactly the same. Do you agree or disagree? Defend your

answer.

18. Reread (i) in the Remarks at the end of this section. Then use the Laplace transform to solve the initial-value problem:

y0 4y9 3y e^td(t2 1), y(0) 0, y9(0) 2.

Use a graphing utility to graph y(t) for 0#t#5.

Exercises

Answers to selected odd-numbered problems begin on page ANS-10.

4.5

4.6 Systems of Linear Differential Equations

INTRODUCTION

When initial conditions are specified, the Laplace transform reduces a system of linear differential equations with constant coefficients to a set of simultaneous algebraic equations in the transformed functions.

Coupled Springs

In our first example we solve the model m1x1k1x1 k2(x2 x1)

m2 x2k2(x2 x1) (1)

that describes the motion of two masses m₁ and m₂ in the coupled spring/mass system shown in Figure 3.12.1 of Section 3.12.

EXAMPLE 1 Example 4 of Section 3.12 Revisited Use the Laplace transform to solve

x1 10x₁ 4x₂ 0

4x₁ x2 4x₂ 0 (2) 4.6 Systems of Linear Differential Equations ^| 251

252 ^| CHAPTER 4 The Laplace Transform

subject to x1(0) 0, x1(0) 1, x2(0) 0, x2(0) 1. This is system (1) with k1 6, k2 4, m1 1, and m2 1.

SOLUTION The transform of each equation is

s²X1(s) sx1(0) x1(0) 10X1(s) 4X2(s) 0 4X1(s) s²X2(s) sx2(0) x2(0) 4X2(s) 0,

where X₁(s) +{x₁(t)} and X₂(s) +{x₂(t)}. The preceding system is the same as

(s² 10)X₁(s) 4X₂(s) 1

4X₁(s) (s² 4)X₂(s) 1. (3) Solving (3) for X₁(s) and using partial fractions on the result yields

X₁(s) s² (s²2)(s² 12)

1/5

s² 2 6/5 s²12, and therefore

x1(t) 1

5!2+¹e "2

s²2f 6

5!12+¹e !12 s² 12f

"2

10 sin!2t !3

5 sin2!3t.

Substituting the expression for X₁(s) into the first equation of (3) gives us X2(s) s²6

(s² 2)(s² 12)

2/5

s²22 3/5 s²12

and x2(t) 2

5!2 +¹e !2

s² 2f 2 3

5!12 +¹e !12 s² 12f

!2

5 sin!2t2 !3

10 sin2!3t.

Finally, the solution to the given system (2) is

x₁(t) !2

10 sin!2t !3

5 sin2!3t

(4)

x₂(t) !2

5 sin!2t2 !3

10 sin2!3t.

The solution (4) is the same as (14) of Section 3.12.

Networks

In (18) of Section 2.9 we saw that currents i₁(t) and i₂(t) in the network contain- ing an inductor, a resistor, and a capacitor shown in FIGURE 4.6.1 were governed by the system of first-order differential equations

L di1

dt Ri₂E(t) RC di2

dt i₂2 i₁0.

(5)

We solve this system by the Laplace transform in the next example.

FIGURE 4.6.1 Electrical network C L

R E

i1 i2

i3

EXAMPLE 2 An Electrical Network

Solve the system in (5) under the conditions E(t) 60 V, L 1 h, R 50 , C 10⁴ f, and the currents i₁ and i₂ are initially zero.

SOLUTION We must solve

di₁

dt 50i₂60 50(10⁴) di₂

dt i₂2i₁0 subject to i₁(0) 0, i₂(0) 0.

Applying the Laplace transform to each equation of the system and simplifying gives sI1(s) 50I2(s) 60

s 200I1(s) (s 200)I2(s) 0,

where I1(s) +{i1(t)} and I2(s) +{i2(t)}. Solving the system for I1 and I2 and decomposing the results into partial fractions gives

I₁(s) 60s12,000

s(s 100)² 6/5

s 2 6/5

s 100 2 60

(s 100)²

I₂(s) 12,000

s(s100)² 6/5

s 2 6/5

s 100 2 120 (s 100)². Taking the inverse Laplace transform, we find the currents to be

i₁(t) 6

52 6

5e^100t2 60te^100t

i₂(t) 6

5 2 6

5e^100t2120te^100t.

Note that both i₁(t) and i₂(t) in Example 2 tend toward the value E/R ⁶5 as t Sq. Furthermore, since the current through the capacitor is i₃(t) i1(t) i2(t) 60te^100t, we observe that i₃(t) S 0 as t Sq.

Double Pendulum

As shown in FIGURE 4.6.2, a double pendulum oscillates in a vertical plane under the influence of gravity. For small displacements u₁(t) and u₂(t), it can be shown that the system of differential equations describing the motion is

(m₁ m₂)l ²₁ u 1 m₂ l₁ l₂ u 2 (m₁ m₂)l₁g u₁ 0

m₂ l ²₂ u 2 m₂ l₁ l₂ u 1 m₂ l₂ g u₂ 0. (6) As indicated in Figure 4.6.2, u₁ is measured (in radians) from a vertical line extending downward from the pivot of the system and u₂ is measured from a vertical line extending downward from the center of mass m₁. The positive direction is to the right and the negative direction is to the left.

EXAMPLE 3 Double Pendulum

It is left as an exercise to fill in the details of using the Laplace transform to solve system (6) when m₁ 3, m₂ 1, l₁ l₂ 16, u₁(0) 1, u₂(0) 1, u1(0) 0, and u2(0) 0. You should find that

u₁(t) 1

4cos 2

!3t 3 4cos2t

(7)

u₂(t) 1

2cos 2

!3t2 3 2cos2t.

FIGURE 4.6.2 Double pendulum θ1

θ2

l2

l1

m1

m2

4.6 Systems of Linear Differential Equations ^| 253

254 ^| CHAPTER 4 The Laplace Transform

In Problems 1–12, use the Laplace transform to solve the given system of differential equations.

1. dx

dt x y 2. dx

dt 2y e^t dy

dt 2x dy

dt 8x t x(0) 0, y(0) 1 x(0) 1, y(0) 1 3. dx

dt x2 2y 4. dx

dt 3x dy dt 1 dy

dt 5x y dx

dt 2 x dy

dt 2ye^t x(0) 1, y(0) 2 x(0) 0, y(0) 0 5. 2 dx

dt dy

dt 2x 1 6. dx

dt x2 dy

dt y0 dx

dt dy

dt 3x 3y 2 dx dt dy

dt 2y 0 x(0) 0, y(0) 0 x(0) 0, y(0) 1 7. d²x

dt² x y 0 8. d²x dt² dx

dt dy dt 0 d²x

dt² y x 0 d²y dt² dy

dt 24dx dt 0 x(0) 0, x(0) 2, x(0) 1, x(0) 0, y(0) 0, y(0) 1 y(0) 1, y(0) 5 9. d²x

dt² d²y

dt² t² 10. dx

dt 4x d³y

dt³ 6 sin t d²x

dt² d²y

dt² 4t dx

dt 2x 2 d³y dt³ 0 x(0) 8, x(0) 0, x(0) 0, y(0) 0, y(0) 0, y(0) 0 y(0) 0, y(0) 0

11. d²x dt² 3 dy

dt 3y 0 12. dx

dt 4x 2y 2 8(t 1) d²x

dt² 3y te^t dy

dt 3x y 8(t 1) x(0) 0, x(0) 2, x(0) 0, y(0) ¹2

y(0) 0

13. Solve system (1) when k₁ 3, k₂ 2, m₁ 1, m₂ 1 and x₁(0) 0, x1(0) 1, x₂(0) 1, x2(0) 0.

14. Derive the system of differential equations describing the straight-line vertical motion of the coupled springs shown in equilibrium in FIGURE 4.6.4. Use the Laplace transform to solve the system when k₁ 1, k₂ 1, k₃ 1, m₁ 1, m₂ 1 and x₁(0) 0, x1(0) 1, x₂(0) 0, x2(0) 1.

FIGURE 4.6.4 Coupled springs in Problem 14 k1

m1

m2

x1

x2 k2

k3

x2 = 0 x1 = 0

Exercises

Answers to selected odd-numbered problems begin on page ANS-10.

4.6

With the aid of a CAS, the positions of the two masses at t 0 and at subsequent times are shown in FIGURE 4.6.3. See Problem 23 in Exercises 4.6.

FIGURE 4.6.3 Positions of masses at various times in Example 3

t = 1.4 t = 2.5

t = 0

(a) (b) (c)

(d) (e) (f)

t = 8.5 t = 6.9

t = 4.3

15. (a) Show that the system of differential equations for the currents i2(t) and i3(t) in the electrical network shown in FIGURE 4.6.5 is

L1

di2

dt Ri2 Ri3 E(t) L2

di₃

dt Ri2 Ri3 E(t).

(b) Solve the system in part (a) if R 5 , L1 0.01 h, L2 0.0125 h, E 100 V, i2(0) 0, and i3(0) 0.

(c) Determine the current i₁(t).

FIGURE 4.6.5 Network in Problem 15 R

E i1

L1 L2

i2 i3

16. (a) In Problem 14 in Exercises 2.9 you were asked to show that the currents i₂(t) and i₃(t) in the electrical network shown in FIGURE 4.6.6 satisfy

L di2

dt L di3

dt R₁i₂E(t)

R₁di2

dt R₂di3

dt 1 Ci₃0.

Solve the system if R₁ 10 , R₂ 5 , L 1 h, C 0.2 f,

E(t) e120, 0#t, 2

0, t$ 2,

i₂(0) 0, and i₃(0) 0.

(b) Determine the current i₁(t).

FIGURE 4.6.6 Network in Problem 16 C L

E

i1 i2

i3

R1

R2

17. Solve the system given in (17) of Section 2.9 when R₁ 6 , R2 5 , L₁ 1 h, L₂ 1 h, E(t) 50 sin t V, i₂(0) 0, and i3(0) 0.

18. Solve (5) when E 60 V, L ¹2 h, R 50 , C 10⁴ f, i1(0) 0, and i2(0) 0.

19. Solve (5) when E 60 V, L 2 h, R 50 , C 10⁴ f, i1(0) 0, and i2(0) 0.

20. (a) Show that the system of differential equations for the charge on the capacitor q(t) and the current i3(t) in the electrical network shown in FIGURE 4.6.7 is

R1

dq dt 1

C q R1i3E(t)

L di₃

dt R2i32 1 C q 0.

(b) Find the charge on the capacitor when L 1 h, R1 1 , R2 1 , C 1 f,

E(t) e0,

50e^t, 0,t, 1 t$ 1, i3(0) 0, and q(0) 0.

FIGURE 4.6.7 Network in Problem 20

E C L

i1 i2

i3

R1

R2