CHAPTER 6Laplace Transforms

CHAPTER 6 Laplace Transforms

In the ninth edition, this chapter underwent major changes, which have been retained and further extended. This concerns a more natural order of the material and changing emphasis placed on the various topics. This has made the chapter more teachable and simpler, a goal reached by placing Dirac’s delta function in a separate section, discussing partial fractions earlier, and in terms of practical problems rather than in terms of impractical and lengthy general formulas.

Also, to have a better flow of ideas, convolution and nonhomogeneous linear ODEs now appear earlier, whereas differentiation and integration of transforms (not of functions!) are now presented near the end of the chapter and with less emphasis.

SECTION 6.1. Laplace Transform. Linearity. First Shifting Theorem (s-Shifting), page 204

Purpose. To explain the basic concepts, to present a short list of basic transforms, and to show how these are derived from the definition.

Main Content, Important Concepts Transform, inverse transform, linearity First shifting theorem

Table 6.1

Existence and its practical significance Comment on Table 6.1

After working for a while in this chapter, the student should be able to memorize these transforms. Further transforms in Sec. 6.9 are derived as we go along, many of them from Table 6.1.

Problem Set 6.1

The problem set addresses the two main tasks in connection with Laplace transform, namely, to find transforms of given functions (Probs. 1–16 and 33–36), and to find functions for given transforms (that is, to find inverse transforms, Probs. 25–32 and 37–45).

This includes the task of finding formulas for graphically given functions, and, as techniques, integration, reduction by partial fractions, the use of Table 6.1, and s-shifting (in Probs. 33–45).

Problems 17–24 include a modest amount of theory compatible with the level of this book.

SOLUTIONS TO PROBLEM SET 6.1, page 210 1.

2.

3.

4. l(cos² vt)⫽ l a1 2⫹ 1

2 cos 2vtb ⫽ 1 2s⫹ 1

2

# ^s

s²⫹ 4v² s>(s²⫹ 4p²)

a²>s ⫺ 2ab>s²⫹ 2b²>s³ 2>s²⫹ 8>s

99

5.

6. ; transform

This can be checked by the first shifting theorem.

7.

8. transform

9.

10.

12.

14.

16. . Integration by parts gives

18. Use Prob. 10 with and to obtain

20. No matter how large we choose M and k, we have for all t greater than some because for all sufficiently large t (and fixed positive M and k).

t²⬎ ln M ⫹ kt t0

e^t2⬎ Me^kt l( f₁)⫽ l(1) ⫺ l( f ) ⫽ 1

s ⫺ 1

s (1⫺ e^ⴚ2s)⫽ e^ⴚ2s s . c⫽ 2

k⫽ 1

⫽ 1

s² (⫺e^ⴚs⫹ 1 ⫹ e^ⴚ2s⫺ e^ⴚs)⫽(1⫺ e^ⴚs)² s² .

⫽ ⫺e^ⴚs s ⫺ 1

s² (e^ⴚs⫺ 1) ⫹ e^ⴚs s ⫹ 1

s² (e^ⴚ2s⫺ e^ⴚs)

⫺e^ⴚstt s `¹

0 ⫹ 1

s冮₀^{1eⴚst dt⫺eⴚst(2s⫺ t)`2}1 ⫺ 1

s冮₁^{2eⴚst dt} 冮₀^{1eⴚstt dt⫹} 冮₁^{2eⴚst(2⫺ t) dt}

k冮_a^{beⴚst dt⫽ ks (eⴚas⫺ eⴚbs)}

1⫺ e^ⴚs

s² ⫺ 1

se^2s k

s (1⫺ e^ⴚcs)

1 0.8 0.6 0.4 0.2 0

0 0.5

t

1 15

Problem 9

1⫺ (s ⫹ 1)e^ⴚs s²

⫺1.5s>(s²⫹ 9)

1.5 sin (3t⫺p>2) ⫽ 1.5(sin 3t cos p>2 ⫺ cos 3t sin p>2) ⫽ ⫺1.5 cos 3t;

cos(u)s⫺ sin(u)v s²⫹ v²

1 2a 1

s⫺ 3 ⫺ 1

s⫹ 5b⫽ 4

(s⫹ 1)²⫺ 16 e^ⴚt sinh 4t⫽1

2 (e^3t⫺ e^ⴚ5t)

1 (s⫺ 4)(s ⫺ 2)

22. Let . Then

24. Let . Since the transform is linear, we obtain

Now apply on both sides to get the desired result,

. Note that we have proved much more than just the claim, namely, the following.

Theorem. If a linear transformation has an inverse, the inverse is linear.

25.

26. The inverse transform is

27.

28. In terms of partial fractions, the given function is

Hence the inverse transform is

. 29.

30. The inverse transform is

. 31.

32. If , then . If , then by the shifting theorem (or from the first result by l’Hôpital’s rule, taking derivatives with respect to a).

33.

34.

35. 32_{(2s⫹ 1)}^p2⫹ 64p²

k(s⫹ a) (s⫹ a)²⫹ v²

6 (s⫹ 2)⁴

te^ⴚbt a⫽ b

1

b⫺ a (e^ⴚat⫺ e^ⴚbt) a⫽ b

2e^3t⫺ 3e^ⴚt

4 cosh 4t⫹ 8 sinh 4t ⫽ 6e^4t⫺ 2e^ⴚ4t

⫺1>15t³(⫺5 ⫹ 6t²)

1

13 ⫹ 12 (e^t13⫺ e^ⴚt12) 1

13 ⫹ 12a 1

s⫺ 13 ⫺ 1

s⫹ 12b. cos(¹₂^p_L^t)

L²

5 cosh 5t⫹¹5 sinh 5t⫽¹³5e^5t⫹¹²5e^ⴚ5t. 0.2 cos (1.4t)⫹ sin (1.4t)

l^ⴚ1(aF⫹ bG) ⫽ l^ⴚ1l(af⫹ bg) ⫽ af ⫹ bg ⫽ al^ⴚ1(F)⫹ bl^ⴚ1(G) l^ⴚ1

aF⫹ bG ⫽ al( f ) ⫹ bl(g) ⫽ l(af ⫹ bg).

f⫽ l^ⴚ1(F), g⫽ l^ⴚ1(G)

⫽ s^ⴚ1>2⌫(¹2)⫽ 1p_>s.

⫽ s^ⴚ1>2冮₀^{⬁eⴚttⴚ1>2 dt}

⫽ 冮₀^{⬁eⴚtats bⴚ1>21s dt}

l(1> 1t) ⫽ 冮₀^{⬁eⴚsttⴚ1>2 dt}

st⫽ t, t ⫽ t>s, dt ⫽ dt>s

36.

37.

38.

39.

40.

41.

42.

43.

44.

45.

SECTION 6.2. Transforms of Derivatives and Integrals. ODEs, page 211 Purpose. To get a first impression of how the Laplace transform solves ODEs and initial value problems, the task for which it is designed.

Main Content, Important Concepts (1)

Extension of (1) to higher derivatives [(2), (3)]

Solution of an ODE, subsidiary equation

Transform of the integral of a function (Theorem 3) Transfer function (6)

Shifted data problems (Example 6) Comment on ODEs

The last of the three steps of solution is the hardest, but we shall derive many general properties of the Laplace transform (collected in Sec. 6.8) that will help, along with formulas in Table 6.1 and those in Sec. 6.9, so that we can proceed to ODEs for which the present method is superior to the classical one.

Problem Set 6.2

Problems 1–15 concern IVPs, including those with shifted data (12–15).

Problems 16–22 show how to obtain new transforms from known ones by differentiation, the basic formulas being (1) and (2).

Conversely, Probs. 23–29 show how to obtain new inverse transforms by integration (Theorem 3).

Project 30 extends Theorem 1 to the case when a discontinuity (finite jump) occurs.

l( f

r

e^ⴚkt(a cos pt⫹ b sin pt)

1

4e^ⴚt(12 cos (2t)⫹ sin (2t)) (a₀⫹ a1t⫹¹2 a₂t²)e^ⴚt e^ⴚ2ptsinh (pt)

2e^t sinh 2t⫽ e^3t⫺ e^ⴚt

3 4t⁵e^ⴚ23t 3t²e^ⴚt pt²e^ⴚpt

⫽ s²⫺ 2 s⁴⫹ 4

⫽ 1

2a2s²⫺ 4 s⁴⫹ 4 b l a1

2 (e^t⫺ e^ⴚt) cos tb ⫽ 1

2a s⫺ 1

(s⫺ 1)²⫹ 1⫺ s⫹ 1 (s⫹ 1)²⫹ 1b

SOLUTIONS TO PROBLEM SET 6.2, page 216 1.

2. The subsidiary equation is . Hence

. 3.

4. We obtain

. The solution of this subsidiary equation is

.

This gives the solution .

6. The subsidiary equation is

The solution Y is

. In terms of partial fractions, this becomes

. The inverse transform of this gives the solution

. 8. The subsidiary equation is

. Its solution is

. This can be written

. The inverse transform (the solution of the IVP) is

.

It has the form expected in the case of a double root of the characteristic equation.

9. y⫽ 2e^2t⫹ e^t⫺ 1 ⫹ 2t

y⫽ 8.1e^2t⫺ 12.3te^2t Y⫽ 8.1

s⫺ 2 ⫺ 12.3 (s⫺ 2)² Y⫽ 8.1s⫺ 28.5

(s⫺ 2)² ⫽8.1(s⫺ 2) ⫹ 8.1#

2⫺ 28.5 (s⫺ 2)²

(s⫺ 2)²Y⫽ 8.1s ⫹ 3.9 ⫺ 4# _8.1 y⫽ e^t⫹ 2e^5t⫹ 0.2 cos 2t ⫺ 2.4 sin 2t

Y⫽ 1

s⫺ 1⫹ 2

s⫺ 5⫹ 0.2s⫺ 2.4 # 2 s²⫹ 4 Y⫽ (3.2s⫺ 13)(s²⫹ 4) ⫹ 29s

(s⫺ 1)(s ⫺ 5)(s²⫹ 4)

(s²⫺ 6s ⫹ 5)Y ⫽ 3.2s ⫹ 6.2 ⫺ 6 # _3.2⫹ 29s>(s²⫹ 4).

y⫽ e^ⴚt⫺ cos 3t ⫹¹3 sin 3t

Y⫽ 10

(s⫹ 1)(s²⫹ 9) ⫽ 1

s⫹ 1⫺ s⫺ 1 s²⫹ 9 (s²⫹ 9)Y ⫽ 10

s⫹ 1

s²Y⫹ (Y ⫺ 1)s ⫺ 6Y ⫺ 2 ⫽ 0, Y ⫽s^2s⫹ s ⫺ 6^{⫹ 2} , y⫽ 4>5 e^2t⫹ 1>5 e^ⴚ3t. (s⫹ 2)Y ⫽ 1.5, Y ⫽ 1.5>(s ⫹ 2), y ⫽ 1.5e^ⴚ2t

sY⫺ 1.5 ⫹ 2Y ⫽ 0

y⫽ ⫺¹⁸5sin(x) cos (x)⫹ 3>5 ⫺ 6>5 (cos (x))²⫹ 3>5e^ⴚ2>3x

10. . The solution is . The inverse transform of Y (the solution of the IVP) is

.

Note that the initial values are such that the general solution of the homogeneous ODE does not contribute to the solution of the IVP.

12. so that the “shifted problem” is

The subsidiary equation is

Its solution is

Inversion gives,

Hence the solution to the given problem is

14. , so that the “shifted equation” is

. The corresponding subsidiary equation is

. Its solution is

. The inverse transform is

Hence the given problem has the solution

.

16. . Hence

and thus

. (s²⫹ 16)l( f ) ⫽ ⫺32

s²⫹ 16⫹ 1 ⫽ s²⫺ 16 s²⫹ 16 l( f

s

s²⫹ 16⫺ 16l( f ) ⫽ s²l( f )⫺ s # _{0⫺ 1} f⫽ t cos 4t, f

r

s

y⫽ 10(t ⫺ 2) ⫺ 4 ⫹ 2e^ⴚ(tⴚ2) sin 2(t⫺ 2) y~ ⫽ 10t~ ⫺ 4 ⫹ 2e^ⴚt~ sin 2t~.

Y~ ⫽ 10 s² ⫺4

s ⫹ 4

(s⫹ 1)²⫹ 4

3(s ⫹ 1)²⫹ 44Y~ ⫽ ⫺4s ⫹ 14 ⫺ 2 # _{4⫹ 50>s}2

~y

s

r

t⫽ t~ ⫹ 2

y⫽ ⫺7>2e^(tⴚ2)⫹ 1>2e^ⴚ3(tⴚ2)

~ ⫽ ⫺7>2ey ^t~⫹ 1>2e^ⴚ3t~ Y~ ⫽s^{⫺11 ⫹ 3s2}⫹ 2s ⫺ 3⫽2(s¹⫹ 3)⫺2(s⁷⫺ 1)

(s²⫹ 2s ⫺ 3) Y~ ⫹ 11 ⫹ 3s ⫽ 0

~y

s

r

r

t⫽ t~ ⫹ 2,

yh⫽ c1 cos 0.2t⫹ c2 sin 0.2t

y⫽ ⫺25 ⫹ 0.5t² Y⫽ ⫺25

s ⫹ 1 s³ (s²⫹ 0.04)Y ⫽ ⫺25s ⫹ 0.04>s³

Hence the answer is

17.

18. . Hence

. Solving for gives

. Division by s gives the answer

. 19.

20. . From this and (2), it

follows that

.

Collecting terms and using Prob. 19 with and , we obtain .

Answer

. 21.

22. Project. We derive (a). We have and

By (2),

. Collecting -terms, we obtain

. Division by on both sides gives (a).

In (b) on the right we get from (a)

l(sin vt⫺ vt cos vt) ⫽ v .

s²⫹ v²⫺ v s²⫺ v² (s²⫹ v²)² s²⫹ v²

l( f )(s²⫹ v²)⫽ ⫺2v²

s²⫹ v²⫹ 1 ⫽ s²⫺ v² s²⫹ v² l( f )

l( f

s

s²⫹ v² ⫺ v²l( f )⫽ s²l( f )⫺ 1 f

s

f

r

r

f(0)⫽ 0

l( f

r

l( f )⫽ 24

s(s²⫹ 4)(s²⫹ 16) (s²⫹ 4)l( f ) ⫽ 3l(sin² 2t)⫽ 3 # ₈

s(s²⫹ 16)

sin 2a⫽ 2 sin a cos a v⫽ 2

l( f

s

r

s

2v²⫹ s² s(s²⫹ 4v²)

l( f )⫽ s²⫹ 8 s(s²⫹ 16) sl( f )⫽ s²⫹ 8

s²⫹ 16 sl( f )

l( f

r

s²⫹ 16⫽ sl( f ) ⫺ 1 f⫽ cos² 2t, f

r

1 (s⫺ a)²

l( f )⫽ s²⫺ 16 (s²⫹ 16) ².

Taking the common denominator and simplifying the numerator,

we get (b).

(c) is shown in Example 1.

(d) is derived the same way as (b), with instead of , so that the numerator is

which gives (d).

(e) is similar to (a). We have and obtain

By (2) we obtain

. Hence

. Division by gives (e).

(f ) follows similarly. We have and, furthermore,

. Division by gives formula (f ).

23.

24. We start from

and integrate twice. The first integration gives .

Multiplication by 20 and another integration from 0 to t gives the answer

25. 4¹⫺ cos(1>2vt)

v² .

20 4p^{2 (e}

2pt⫺ 1) ⫺ 20t 2p^⫽

1 5p^{2 (e}

2pt⫺ 1) ⫺10t p ^. 1

2p ^(e

2pt⫺ 1) l^ⴚ1 a 1

s⫺ 2p^{b ⫽ e}

2pt

6⫺ 6e^ⴚ1>3t. s²⫺ a²

l( f )(s²⫺ a²)⫽ 2as s²⫺ a² l( f

s

s²⫺ a² ⫹ a²l( f )⫽ s²l( f ) f

s

f

r

r

f(0)⫽ 0 s²⫺ a²

l( f )(s²⫺ a²)⫽ 2a²

s²⫺ a² ⫹ 1 ⫽s²⫹ a² s²⫺ a² l( f

s

2

s²⫺ a² ⫹ a²l( f )⫽ s²l( f )⫺ 1 f

s

f

r

r

f(0)⫽ 0

v(s²⫹ v²)⫹ v(s²⫺ v²)⫽ 2vs²,

⫺

⫹

v(s²⫹ v²)⫺ v(s²⫺ v²)⫽ 2v³

26. The inverse of is . A first integration from 0 to t gives . Another integration from 0 to t yields . This can be confirmed as follows.

27.

28. The transform of

is . A first integration from 0 to t gives

. Another integration from 0 to t gives the answer

30. Project. (a) Theorems 1 and 2 are crucial in solving ODEs, whereas Theorem 3 serves as a tool for obtaining new transforms, that is, as one of various tools for this purpose.

(b) In the integration by parts shown in the proof of Theorem 1 we now have to integrate from 0 to a and then from a to , thus obtaining from the upper limit of integration of the first integral and from the lower limit of integration of the second integral.

(c) Direct integration of the defining integral formulas gives

and

Formula (1*) confirms this by straightforward calculation and simplification since , and

.

SECTION 6.3. Unit Step Function (Heaviside Function). Second Shifting Theorem (t-Shifting), page 217

Purpose

1. To introduce the unit step function which, together with Dirac’s delta (Sec. 6.4), greatly increases the usefulness of the Laplace transform.

2. To find the transform of

if and if

where the transform of is known (Theorem 1, called Second Shifting Theorem).

f(t) (0⬍ t ⬍ ⬁)

t⬎ a f~

(t)⫽ f(t⫺ a) f⬍ a

f~ (t)⫽ 0

u(t⫺ a),

f(a⫹ 0) ⫺ f(a⫺ 0) ⫽ 0 ⫺ e^ⴚ1 f(0)⫽ 1, a ⫽ 1

l( f

r

l( f )⫽ (1 ⫺ e^ⴚ(s⫹1))>(s ⫹ 1)

⫺f(a⫹ 0)e^ⴚasf(a⫺ 0)e^ⴚas

⬁

⫺ 3

k² (cos kt⫺ 1) ⫺ 4 k²a1

k sin kt⫺ tb. 3

k sin kt⫺ 4

k² (cos kt⫺ 1) 3 cos kt⫹ 4

k sin kt

3s⫹ 4 s²⫹ k²

⫺¹4 cos 2t⫺ sin 2t ⫹¹4⫹ 2t.

l a 1

s⁴⫺ s²b ⫽ l a 1 s²⫺ 1⫺ 1

s²b ⫽ sinh t ⫺ t.

sinh t⫺ t cosh t⫺ 1

sinh t 1>(s²⫺ 1)

Main Content, Important Concepts Unit step function (1), its transform (2) Second shifting theorem (Theorem 1) Comment on the Unit Step Function

Problem Set 6.3 shows that is the basic function for representing discontinuous functions.

Problem Set 6.3

Problems 2–11 concern the role of the unit step function in t-shifting in the second shifting theorem.

Problems 12–17 show the application of that theorem in finding inverse transforms.

IVPs, some with discontinuous inputs (right sides of the ODEs) are solved in Probs.

18–27; this is a typical task using the Laplace transform.

Models of electric circuits follow in Probs. 28–40; these have a discontinuous EMF (electromotive force); the most general of these circuits are RLC-circuits (Probs. 38–40).

SOLUTIONS TO PROBLEM SET 6.3, page 223

2. Hence the transform is

3.

4.

5.

6. because of periodicity. The representation in

terms of unit step functions is

and thus gives the transform

7.

8. The given function is represented by

and thus has the transform

9.

10.

Hence the transform is 1 2

# ¹

s⫺ 1 ⫺ 1 2

# ¹

s⫹ 1 ⫺ 1 2a e²

s⫺ 1 ⫺ e^ⴚ2 s⫹ 1b e^ⴚ2s.

1

2(e^t⫺ e^ⴚt)(1⫺ u(t ⫺ 2)) ⫽¹2(e^t⫺ e^ⴚt)⫺¹2(e^(tⴚ2)⫹2⫺ e^{(ⴚt⫹2)ⴚ2})u(t⫺ 2).

1>2^{eⴚ5>2s(25s2}s^{⫹ 20s ⫹ 8)3}

a2 s³⫹ 2

s²⫹ 1

sb e^ⴚs⫺a2 s³ ⫹ 4

s² ⫹4 sb e^ⴚ2s

⫺3(t ⫺ 2)²⫹ 4(t ⫺ 2) ⫹ 44u(t ⫺ 2) t²(u(t⫺ 1) ⫺ u(t ⫺ 2)) ⫽3(t ⫺ 1)²⫹ 2(t ⫺ 1) ⫹ 14u(t ⫺ 1) 2^eⴚsⴚ1>2_2s^{p⫺ e}_⫹p^ⴚ3>2pⴚ3s

p s²⫹p^{2 (e}

ⴚ2s⫺ e^ⴚ4s).

(sin pt)(u(t⫺ 2) ⫺ u(t ⫺ 4)) sin (t⫺ 2)p⫽ sin (t ⫺ 4)p_{⫽ sin} pt

e^ⴚt(1⫺ u(p>2)) ⫽ (1 ⫺ e^{ⴚ12p(1⫹s)})>(1 ⫹ s) cos 2t(1⫺ u(t ⫺p))⫽^s(1s^{2⫺ e}⫹ 4^ⴚsp)

e^ⴚ3s s²

1

s² ⫺ e^ⴚ2s

s² ⫺ 2e^ⴚ2s s . t(1⫺ u(t ⫺ 2)) ⫽ t ⫺3(t ⫺ 2) ⫹ 24u(t ⫺ 2).

u(t⫺ a)

Alternatively, by using the addition formula (22) in App. 3.1 we obtain the transform in the form

12. has the inverse , hence has the inverse (first shifting), and has the inverse (second shifting).

13.

14.

15.

16.

18. After applying the inverse Laplace

transform, we get 19.

20. Division by

and expansion in terms of partial fractions gives

Hence the answer is Note that it does not contain a contribution from the general solution of the homogeneous ODE.

21.

22. In terms of unit step functions, the function on the right is

Answer:

23.

24. The subsidiary equation is

It has the solution

This gives the answer

y⫽¹2⫹¹2e^ⴚ2t⫺ e^ⴚt⫺ (¹2⫹¹2e^ⴚ2(tⴚ1)⫺ e^ⴚ(tⴚ1))u(t⫺ 1);

Y⫽ 1⫺ e^ⴚs

s(s²⫹ 3s ⫹ 2) ⫽a1

2s⫹ 1

2(s⫹ 2) ⫺ 1

s⫹ 1b (1⫺ e^ⴚs).

(s²⫹ 3s ⫹ 2)Y ⫽1 s ⫺

e^ⴚs s .

⫺sin t, (0 ⬍ t ⬍ 2p), ⫺¹2 sin 2t, (t⬎ 2p).

y⫽ •4e^ⴚt⫺ e^ⴚ2t⫹ 2t ⫺ 3

(4⫺ 8e)e^ⴚt⫹ (3e²⫺ 1)e^ⴚ2t⫹ 4

if 0 ⬍ t ⬍ 1 if t⬎ 1.

r(t)⫽ 4t31 ⫺ u(t ⫺ 1)4 ⫹ 8u(t ⫺ 1) ⫽ 4t ⫺ 34(t ⫺ 1) ⫺ 44u(t ⫺ 1).

1

2 sin 2t⫹⁴3 cos t⫹⁴3 cos 2t, (0⬍ t ⬍p); ¹₂ sin 2t, (t⬎p).

y⫽ 6t²⫺ 5t ⫹¹⁹12. Y⫽ 12

s³ ⫺ 5 s² ⫹

19 12

s . 10s⫹ 24

s²⫹

(s²⫹ 10s ⫹ 24)Y ⫽ (s ⫹ 4)(s ⫹ 6)Y ⫽ 288 s³ ⫹ 19

12 s⫺ 5 ⫹190 12. 1>15e^ⴚx⫹⁸³80e^4x⫺ 1>48e^ⴚ4x⫺ 1>12e^2x

2 3e^3t2.

(4s²⫺ 12s ⫹ 9) Y ⫽⁸3s⫺ 4, Y ⫽3(⫺3 ⫹ 2s)⁴ . (sinh (2t⫺ 2))u(t ⫺ 1) ⫺ (sinh (2t ⫺ 6))u(t ⫺ 3)

1

120(t⫺ 2)⁵u(t⫺ 2) 4u(t⫺ 2) ⫺ 8u(t ⫺ 5) sin 2tu(p⫺ t) sin 2t

1

2u(t⫺ 2)(t ⫺ 2)²e^tⴚ2

e^ⴚ2s>(s ⫺ 1)³ t²>2 (s⫺ 1)^ⴚ3 e^tt²>2 s^ⴚ3

1

s²⫺ 1 ⫺acosh 2

s²⫺ 1 ⫹ s sinh 2 s²⫺ 1b e^ⴚ2s.

that is,

25.

26.

The solution in terms of is

In terms of t,

28. The subsidiary equation is

Its solution is

The inverse transform is

;

hence if and if .

30. The subsidiary equation is

Its solution is

The solution of the problem (the current in the circuit) is

32. We obtain

hence

i⫽ u(t ⫺ 4) 32 # ₁₀6eⴚ10(tⴚ4)ⴚ12⫺ 6 # ₁₀5e^{ⴚ3(tⴚ4)ⴚ12}4.

I⫽ 14 # ₁₀5 se^ⴚ4sⴚ12

(s⫹ 10)(s ⫹ 3)⫽a2# 10⁶ s⫹ 10 ⫺ 6#

10⁵

s⫹ 3 b e^ⴚ4sⴚ12 i⫽ e^ⴚ20t⫹ 20t ⫺ 1 ⫹ u(t ⫺ 2)3⫺20t ⫹ 1 ⫹ 39e^ⴚ20(tⴚ2)4.

I⫽ 400(⫺1 ⫹ e^ⴚ2s⫹ 2se^ⴚ2s)>(s²(s⫹ 20)).

(0.5s⫹ 10)I(s) ⫽ 200(1 ⫺ e^ⴚ2s(1⫹ 2s))>s². (0.5i

r

t⬎p i⬇ 0.04 sin t

t⬍p, i⫽ 0

i⫽ u(t ⫺p)c⫺ 40

1,000,001 (cos t⫹ e^{ⴚ1000(tⴚp)})⫹ 40,000 1,000,001 sin td I⫽ ⫺40e^ⴚps

(s²⫹ 1)(s ⫹ 1000)⫽ ⫺40e^ⴚps 1,000,001a 1

s⫹ 1000⫺ s⫺ 1000 s⫹ 1 b. sI⫹ 1000I ⫽ ⫺40 e^ⴚps

s²⫹ 1. i

r

⫹ u(t ⫺ 2p)3e^ⴚt⫹2p(⫺cos 2t ⫹¹2 sin 2t)⫹ cos t ⫺ 2 sin t4.

y(t)⫽ e^ⴚt sin 2t⫺ cos t ⫹ 2 sin t

⫹ u(t~ ⫺p)3e^ⴚt~⫹p(⫺cos 2t~⫹¹2sin 2t~)⫺ cos t~ ⫹ 2 sin t~4.

~(t~) ⫽ e y ^ⴚtⴚpsin 2t~⫹ cos t~ ⫺ 2 sin t~

⫺2 ⫹ 2e^ⴚp. t~

y~

r

t⫽ p_{⫹ t~, y~}

s

r

y⫽ •

1

2⫹¹2e^ⴚ2t⫺ e^ⴚt

1

2e^ⴚ2t(1⫺ e²)⫺ e^ⴚt(1⫺ e)

if 0 ⬍ t ⬍ 1 if t⬎ 1.

34. Divide by 10 and take the transform, using Theorem 3 in Sec. 6.2,

Solving for gives

The inverse transform is

Hence

Jumps occur at (upward) and at (downward) because the right side has those jumps and the term involving the integral (representing the charge on the capacitor) cannot change abruptly; hence the first term, Ri(t), must jump by the amounts of the jumps on the right, which have size 100, and since the current has jumps of size 10.

36. Observing that

we obtain the subsidiary equation

Its solution is

Its inverse transform is

⫹ u(t ⫺ 1)3⫺75 ⫹ 50t²⫹ 25 cos (2t ⫺ 2) ⫺ 50 sin (2t ⫺ 2)4.

i⫽ 75 ⫺ 50t²⫺ 75 cos 2t

⫹ 25e^ⴚs a⫺1 s ⫹ 4

s² ⫹ 4

s³⫹ s⫺ 4 s²⫹ 4b. I⫽ 200 s²⫺ 2 ⫹ e^ⴚs(2s⫹ 2)

s³(s²⫹ 4) ⫽ 25 a3 s ⫺ 4

s³ ⫺ 3s s²⫹ 4b (s²⫹ 4)I ⫽ 200a1

s⫺ 2

s³b ⫹ 200e^ⴚs a2 s²⫹ 2

s³b. t²⫽ (t ⫺ 1)²⫹ 2(t ⫺ 1) ⫹ 1,

i

s

R⫽ 10, t⫽ 0.6

t⫽ 0.5

i(t)⫽ 0 if t⬍ 0.5

i(t)⫽ 10e^{ⴚ10(tⴚ0.5)} if 0.5 ⬍ t ⬍ 0.6 i(t)⫽ 10(e^{ⴚ10(tⴚ0.5)}⫺ e^{ⴚ10(tⴚ0.6)})

⫽ 10e^ⴚ10t(e⁵⫺ e⁶)

⫽ ⫺2550e^ⴚ10t if t⬎ 0.6.

i(t)⫽ 10(e^{ⴚ10(tⴚ0.5)} u(t⫺ 0.5) ⫺ e^{ⴚ10(tⴚ0.6)} u(t⫺ 0.6)).

I⫽ 10

s⫹ 10 (e^ⴚ0.5s⫺ e^ⴚ0.6s).

I⫽ l(i)

I⫹ 10 s I⫽ 10

s (e^ⴚ0.5s⫺ e^ⴚ0.6s).

10i⫹ 100冮₀^{ti(t) dt}⫽ 100(u(t ⫺ 0.5) ⫺ u(t ⫺ 0.6)).

38. The subsidiary equation is

Its solution is

The inverse transform is

40. The model

This gives the subsidiary equation

Its solution is

The inverse transform (the solution of the problem) is

SECTION 6.4. Short Impulses. Dirac s Delta Function. Partial Fractions, page 225

Purpose. Modeling of short impulses by Dirac’s delta function (also known as unit impulse function). The text includes a remark that this is not a function in the usual sense of calculus but a “generalized function” or “distribution.” Details cannot be discussed on the level of this book; they can be found in books on functional analysis or on PDEs. See, e.g., L. Schwartz, Mathematics for the Physical Sciences, Paris: Hermann, 1966. The French mathematician LAURENT SCHWARTZ (1915–2002) created and popularized the theory of distributions. See also footnote 2.

Main Content

Definition of Dirac’s delta (3) Sifting property (4)

Transform of delta (5)

⫺ (27 cos t ⫹ 6 sin t)4.

⫹ u(t ⫺ 2p)3e^ⴚ(tⴚ2p)(27 cos 3t⫹ 11 sin 3t) i⫽ 27 cos t ⫹ 6 sin t ⫺ e^ⴚt(27 cos 3t⫹ 11 sin 3t)

⫽a27s⫹ 6

s²⫹ 1 ⫺27(s⫹ 1) ⫹ 33

(s⫹ 1)²⫹ 9 b (1 ⫺ e^ⴚ2ps).

I⫽ 255s(1⫺ e^ⴚ2ps) (s²⫹ 2s ⫹ 10)(s²⫹ 1) c (s ⫹ 2) ⫹10

s dI⫽ 255

s²⫹ 1(1⫺ e^ⴚ2ps).

i

r

⫹ u(t ⫺ 4)32e^ⴚt⫺ e^2(2ⴚt)(9 sin 4(t⫺ 4) ⫹ 2 cos 4(t ⫺ 4))4.

i⫽ ⫺2e^ⴚt⫹ e^ⴚ2t(2 cos 4t⫹ 9 sin 4t)

I⫽ 34s

(s⫹ 1)(s²⫹ 4s ⫹ 20) (1⫺ e^ⴚ4sⴚ4).

as ⫹ 4 ⫹20

s b I⫽ 34

s⫹ 1 (1 ⫺ e^ⴚ4sⴚ4).

i

r

Application to mass–spring systems and electric networks More on partial fractions (Example 4)

For the beginning of the discussion of partial fractions in the present context, see Sec. 6.2.

Problem Set 6.4

CAS Project 1 involves exploring the nature of solutions with the damping or spring constant being changed in the presence of one or two Dirac delta functions on the right.

CAS Experiment 2 models a wave of constant area acting for shorter and shorter times.

Problems 3–12 model vibrating systems with driving forces consisting of unit step functions, Dirac’s delta functions, and other functions.

Problems 14–15 concern rectifiers, sawtooth waves, and staircase functions.

SOLUTIONS TO PROBLEM SET 6.4, page 230

2. CAS Experiment. Students should become aware that careful observation of graphs may lead to discoveries or to more information about conjectures that they may want to prove or disprove. The curves branch from the solution of the homogeneous ODE at the instant at which the impulse is applied, which by choosing, say

gives an interesting joint graph.

3.

4. The subsidiary equation is

where 2s comes from The solution is

The inverse transform (the solution of the IVP) is

y⫽ 2 cos 4t ⫹ u(t ⫺ 3p) sin 4t.

Y⫽ 2s⫹ 4e^ⴚ3ps s²⫹ 16 . y(0).

(s²⫹ 16)Y ⫽ 2s ⫹ 4e^ⴚ3ps y⫽ 2 cos (3t) ⫹ 1>3u(t ⫺ 1>2p) cos (3t)

a⫽ 1, 2, 3, Á ,

2 1

0 2 4 6 8 10 12 14

⫺2

⫺1 t

Section 6.4. Problem 4

5.

6. The subsidiary equation is

where 3 comes from The solution is Y⫽ 3⫹ e^ⴚs

(s⫹ 2)²⫹ 1. y

r

(s²⫹ 4s ⫹ 5)Y ⫽ 3 ⫹ e^ⴚs y⫽ μ

1>2 sin (2t) 0 ⬍ t ⬍p sin (2t) p_{⬍ t ⱕ 2}p 1>2 sin (2t) t ⬎ 2p

The inverse transform (the solution of the initial value problem) is

7.

8. The subsidiary equation is

In terms of partial fractions, its solution is

Its inverse transform is

Without the -term, the solution is and approaches

a harmonic oscillation fairly soon. With the -term the first half-wave has a maximum amplitude of about 5, but from about or 10 on its graph practically coincides with the graph of that harmonic oscillation (whose maximum amplitude is This is physically understandable, since the system has damping that eventually consumes the additional energy due to the -term.

9.

10. The subsidiary equation is

Its solution is

The inverse transform of Y is

This solution is zero from 0 to and then increases rapidly. Its first negative half- wave has a smaller maximum amplitude (about 0.1) than the continuation as a harmonic oscillation with maximum amplitude of about 0.15.

1 2p

⫺ 0.1u(t ⫺p) 3⫺4e^ⴚ2t⫹2p⫹ 3e^ⴚ3t⫹3p⫺ cos t ⫺ sin t4.

y⫽ u(t ⫺¹2p) 3e^ⴚ2t⫹p⫺ e^ⴚ3t⫹3p>24 Y⫽a 1

s⫹ 2⫺ 1

s⫹ 3b e^ⴚps>2⫺a⫺ 0.4

s⫹ 2⫹ 0.3

s⫹ 3 ⫹ 0.1(s⫹ 1) s²⫹ 1 b e^ⴚps. (s²⫹ 5s ⫹ 6)Y ⫽ e^ⴚps>2⫺ s

s²⫹ 1 e^ⴚps.

⫹ 1>5( ⫺ cos (t) ⫹ 3 sin (t)) e^ⴚt

y(t)⫽ 1>5 e^tu(2⫺ t) ⫹ 1>5 e^4ⴚtu (t⫺ 2) (cos (t ⫺ 2) ⫺ 3 sin (t ⫺ 2)) d

110).

t⫽ 8 d

⫺3 cos t ⫹ sin t ⫺ 2e^ⴚ2t⫹ 6e^ⴚt d

y⫽ ⫺2e^ⴚ2t⫹ 6e^ⴚt⫺ 3 cos t ⫹ sin t ⫹ 10u(t ⫺ 1)3e^ⴚt⫹1⫺ e^ⴚ2(tⴚ1)4.

Y⫽ ⫺2

s⫹ 2 ⫹ 6

s⫹ 1⫺ 3s⫺ 1

s²⫹ 1 ⫹ 10 a 1

s⫹ 1⫺ 1

s⫹ 2b e^ⴚs. (s²⫹ 3s ⫹ 2)Y ⫽ s ⫺ 1 ⫹ 3 ⫹ 10

s²⫹ 1 ⫹ 10e^ⴚs. y⫽ 3>5 e^ⴚt⫹ 1>2 u (t ⫺ 1>4) e^ⴚ2tⴙ1>2sin (1>2t ⫺ 1>8)

y⫽ 3e^ⴚ2t sin t⫹ u(t ⫺ 1)e^ⴚ2(tⴚ1) sin (t⫺ 1).

0 2 0.15 0.1

⫺0.1 0.05

⫺0.05 4 6 8 10

t

Section 6.4. Problem 10

11.

12. The subsidiary equation is

Its solution is

Its inverse transform (the solution of the IVP) is

This is essentially a straight line, sharply deformed between pand about 8.

y⫽ 5t ⫺ 2 ⫺ 50u(t ⫺p)e^ⴚt⫹psin 2t.

Y⫽ ⫺s²⫹ 2s³⫺ 25 ⫹ 100s²e^ⴚps s²((s⫹ 1)²⫹ 4) . (s²⫹ 2s ⫹ 5)Y ⫽ 1 ⫺ 2s ⫹ 25

s² ⫺ 100e^ⴚps.

⫹ 1>2(1 ⫺ 2e^1⫺t⫹ e^ⴚ2t⫹2)u(t⫺ 1).

y(t)⫽ e^ⴚt⫺ e^ⴚ2t⫹ u(t ⫺ 2) (e^2⫺t⫺ e^ⴚ2t⫹4)

14. TEAM PROJECT. (a) If is piecewise continuous on an interval of length p, then its Laplace transform exists, and we can write the integral from zero to infinity as the series of integrals over successive periods:

If we substitute in the second integral, in the third integral, in the nth integral, then the new limits in every integral are 0 and p. Since

etc., we thus obtain

The factors that do not depend on can be taken out from under the integral signs;

this gives

l( f )⫽31 ⫹ e^ⴚsp⫹ e^ⴚ2sp⫹ Á4冮₀^{peⴚstf(t) dt.}

t

l( f )⫽ 冮₀^{peⴚstf(t) dt⫹} 冮₀^{peⴚs(t⫹p)f(t) dt⫹} 冮₀^{peⴚs(t⫹2p)f(t) dt⫹ Á .}

f(t⫹ p) ⫽ f(t), f(t⫹ 2p) ⫽ f(t), Á,

t⫽ t ⫹ (n ⫺ 1)p

Á, t⫽ t ⫹ 2p

t⫽ t ⫹ p

l( f )⫽ 冮₀^{⬁eⴚstf(t) dt⫽} 冮₀^{peⴚstf dt⫹} 冮_p^{2peⴚstf dt⫹} 冮2p^3p

e^ⴚstf dt⫹ Á . f(t)

40

10

0 2 4 6 8 10

⫺10 20 30

t

Section 6.4. Problem 12

The series in brackets is a geometric series whose sum is The theorem now follows.

(b) From (11) we obtain

Using and integrating by parts or noting

that the integral is the imaginary part of the integral

we obtain the result.

(c) From (11) we obtain the following equation by using from 0 to and from :

This gives the result.

(d) The sawtooth wave has the representation

Integration by parts gives

and thus from (11) we obtain the result

SECTION 6.5. Convolution. Integral Equations, page 232

Purpose. To find the inverse h(t) of a product of transforms whose inverses are known.

Main Content, Important Concepts Convolution f * g, its properties Convolution theorem

Application to ODEs and integral equations

H(s)⫽ F(s)G(s)

(s⬎ 0).

l( f )⫽ k

ps²⫺ ke^ⴚps s(1⫺ e^ⴚps)

⫽ ⫺p

s e^ⴚsp⫺ 1

s² (e^ⴚsp⫺ 1)

冮₀^{peⴚstt dt⫽ ⫺st eⴚst`p}₀^{⫹ 1s}冮₀^{peⴚst dt}

f(t)⫽ k

p t if 0 ⬍ t ⬍ p, f(t⫹ p) ⫽ f(t).

⫽ v

s²⫹ v²

cosh (ps>2v) sinh (ps>2v). v

s²⫹ v²

1⫹ e^ps>v

e^ps>v⫺ 1⫽ v s²⫹ v²

e^ⴚps>2v⫹ e^ps>2v e^ps>2v⫺ e^ⴚps>2v p_{>v to 2}p_>v

⫺sin vt

p_>v sin vt

冮₀^{p>ve(ⴚs⫹iv)t dt⫽} ⫺s ⫹ iv¹ e^(ⴚs⫹iv)t`

p_>v

0 ⫽⫺s ⫺ iv

s²⫹ v² (⫺e^ⴚsp>v⫺ 1) 1⫺ e^ⴚ2ps>v⫽ (1 ⫹ e^ⴚps>v)(1⫺ e^ⴚps>v)

l( f )⫽ 1

1⫺ e^ⴚ2ps>v冮₀^{p>veⴚst sin vt dt.}

1>(1 ⫺ e^ⴚps).

3 Á 4

Comment on Occurrence

In an ODE the transform of the right side is known from Step 1. Solving the subsidiary equation algebraically for causes the transform to be multiplied by the reciprocal of the factor of on the left (the transfer function ; see Sec. 6.2). This calls for the convolution theorem, unless one sees some other way or shortcut.

Very Short Courses. This section can be omitted.

Problem Set 6.5

This set concerns integrations needed to obtain convolutions (Probs. 1–7), the use of the latter in solving a certain class of integrations (Probs. 8–14), the effect of varying a parameter in an integral equation (CAS Experiment 15), a problem (Prob. 16) on general properties of convolution.

Problems 17–26 show how to obtain inverse transforms by evaluating integrals that define convolutions.

SOLUTIONS TO PROBLEM SET 6.5, page 237 1.

2.

3.

4. We obtain

5.

6.

7.

8. The integral equation can be written

By the convolution theorem it has the transform

The solution is Its inverse transform is 9. y⫹ 1 * y ⫽ 2; Y ⫽s⫹ 1² ; y⫽ 2e^ⴚt

y⫽ sin 2t.

Y⫽ 2>(s²⫹ 4).

Y⫹ 4Y>s²⫽ 2>s². y(t)⫹ 4y(t) * t ⫽ 2t.

te^ⴚt⫺ te^ⴚ2t.

e^at *e^bt⫽ 冮₀^{teateb(tⴚt) dt⫽ ebt}冮₀^{te(aⴚb)t dt⫽ eata⫺ e}⫺ b^bt

sin vt v

⫽ 1

2 ct cos vt⫹ sin vt⫺ sin (⫺vt)

2v d ⫽ 1

2t cos vt⫹ 1 2v sin vt.

冮₀^tcos vt cos (vt⫺ vt) dt ⫽1

2冮₀^t3cos vt ⫹ cos (2vt ⫺ vt)4 dv

1

2(e^t⫹ e^ⴚt)⫽ sinht

1 * sin vt⫽ 冮₀^{tsin vt dt⫽ ⫺cos vtv `t}0⫽1⫺ cos vt v

⫺t

Q(s) Y(s)

R(s) Y(s)

r(t) R(s)

10. In terms of convolution the given integral equation is

By the convolution theorem its transform is

Multiplication by gives

The inverse transform (the solution of the integral equation) is

11.

12. The integral equation can be written

This implies, by the convolution theorem, that its transform is

The solution is

Hence its inverse transform gives the answer This result can easily be checked by substitution into the given equation and integration.

14. hence

The answer is

16. Team Project. (a) Setting we have and p runs

from t to 0; thus

⫽ 冮₀^{t g(p)f(t}⫺ p) dp ⫽ g * f.

f * g ⫽ 冮₀^{tf(t)g(t⫺ t) dt ⫽} 冮_t^{0g(p)f(t⫺ p)(⫺dp)}

t⫽ t ⫺ p, dt ⫽ ⫺dp, t⫺ t ⫽ p,

y⫽ 1 ⫹ cosh t.

Y⫽ 2s²⫺ 1 s(s²⫺ 1) ⫽1

s ⫹ s

s²⫺ 1. Y a1 ⫺ 1

s²b ⫽ 2 s ⫺ 1

s³,

y(t)⫽ t ⫹ 1.

Y⫽ s²⫺ 1 s²⫹ s ⫺ 1a1

s²⫹ 1

s⫺ 1b⫽ 1 s² ⫹ 1

s.

Y⫹ s

s²⫺ 1 Y⫽ 1 s² ⫹ 1

s⫺ 1. y(t)⫹ y(t) * cosh t ⫽ t ⫹ e^t. Y⫽s^2s⫺ 1; y⫽ cosht.

y⫽ 12 sin (t12).

Y(s²⫹ 2) ⫽ 2; thus Y ⫽ 2 s²⫹ 2. s²⫹ 4

Y⫺ 2Y>(s²⫹ 4) ⫽ 2>(s²⫹ 4).

y(t)⫺ y(t) * sin 2t ⫽ sin 2t.

(b) Interchanging the order of integration and noting that we integrate over the shaded triangle in the figure, we obtain

⫽ f * (g * v).

⫽ 冮₀^tf(t)冮₀^{tⴚtg(t⫺ t ⫺ p)v(p) dp dt}

⫽ 冮₀^tv(p)冮₀^tⴚpf(t)g(t⫺ p ⫺ t) dt dp ( f * g) * v ⫽ v * ( f * g)

(c) This is a simple consequence of the additivity of the integral.

(d) Let Then for some between 0

and k. Now let Then and so that the formula follows.

(e) has the solution

etc.

18.

20.

21.

22. if and 0 if

t⬍ a

t⬎ a u(t⫺ a) * e^2t⫽ 冮_a^{t e2(tⴚt) dt⫽ e2t}冮_a^{t eⴚ2t dt⫽ 12 (e2(tⴚa)⫺ 1)}

⫺w^t ⫹^{sinh (wt)}w² .

9 * e^ⴚ3t⫽ 9冮₀^{t eⴚ3t dt⫽ 3 ⫺ 3eⴚ3t}

e^at * e^at⫽ 冮₀^{t aatea(tⴚt) dt⫽ eat}冮₀^{t dt⫽ teat}

Y⫽ 1

va v

s²⫹ v²b l(r) ⫹ y(0) s

s²⫹ v²⫹ y

r

v v s²⫹ v² s²Y⫺ sy(0) ⫺ y

r

f_k(t⫺ t~) : d(t), t~ : 0

k : 0.

t~

( f_k *f )(t)⫽ 冮₀^k1k f(t⫺ t) dt ⫽ f(t⫺ t~) t⬎ k.

t 0

p = t – = t – p t

0 p

Section 6.5. Team Project 16(b)

24. Using formula (11) in App. 3.1, convert the product in the integrand to a sum and integrate, obtaining

SECTION 6.6. Differentiation and Integration of Transforms. ODEs with Variable Coefficients, page 238

Purpose. To show that, roughly, differentiation and integration of transforms (not of functions, as before!) correspond to multiplication and division, respectively, of functions by t, with application to the derivation of further transforms and to the solution of Laguerre’s differential equation.

Comment on Application to Variable-Coefficient Equations

This possibility is rather limited; our Example 3 is perhaps the best elementary example of practical interest.

Very Short Courses. This section can be omitted.

Problem Set 6.6

Problems 2–11 concern single or twofold applications of differentiation with respect to s, as the only method wanted for solving these problems, whereas in Probs. 14–20 the student has first to select a suitable one among several possible methods suggested.

Problem 13 is an invitation to study Laguerre polynomials in somewhat more detail, in particular, to compare the locations of the extrema depending on the parameter n.

SOLUTIONS TO PROBLEM SET 6.6, page 241 2.

3.

4. We have

Differentiation and simplification gives the answer

⫽ s²⫹ 2s (s²⫹ 2s ⫹ 2)². F

r

s²⫹ 2s ⫹ 2b

r

⫽ ⫺s²⫹ 2s ⫹ 2 ⫺ (s ⫹ 1)(2s ⫹ 2) (s²⫹ 2s ⫹ 2)²

l(e^ⴚt cos t)⫽ s⫹ 1

(s⫹ 1)²⫹ 1⫽ s⫹ 1 s²⫹ 2s ⫹ 2.

1

4(s⫹ 2)^ⴚ2 F(s)⫽ ⫺3 a 4

s²⫺ 16b

r

⫽ 24s

(s²⫺ 16)²

⫽ 10 sin t ⫺ 2 sin 5t.

⫽ 3(sin t ⫺ sin 5t) ⫹ 2(sin t ⫹ sin 5t)

⫽ 24[⫺¹4 sin (⫺5t ⫹ 4t) ⫹¹6 sin (6t⫺ 5t)]`

t

0

24冮₀^t3⫺cos (5t ⫺ 4t) ⫹ cos (t ⫺ 5t ⫹ 5t)4 dt 48 (sin t) * (sin 5t) ⫽ 48冮₀^tsin t sin 5(t⫺ t) dt.