EXAMPLE 1 Laplace Transform

Let f(t) = 1 when t 0. Find F(s).

Solution. From (1) we obtain by integration

Such an integral is called an improper integral and, by definition, is evaluated according to the rule

Hence our convenient notation means

We shall use this notation throughout this chapter.

EXAMPLE 2 Laplace Transform (e^at) of the Exponential Function) e^at

Let f(t) = e^at when t 0, where a is a constant. Find (f).

Solution. Again by (1),

hence, when s − a > 0,

THEOREM 1 Linearity of the Laplace Transform

The Laplace transform is a linear operation; that is, for any functions f(t) and g(t) whose transforms exist and any constants a and b the transform of af(t) + bg(t) exists, and

PROOF

This is true because integration is a linear operation so that (1) gives

EXAMPLE 3 Application of Theorem 1: Hyperbolic Functions

Find the transforms of cosh at and sinh at.

Solution. Since cosh and sinh , we obtain from Example 2 and Theorem 1

EXAMPLE 4 Cosine and Sine

Derive the formulas

Solution. We write L_e = (cos ωt) and L_s = (sin ωt). Integrating by parts and noting that the integral-free parts give no contribution from the upper limit ∞, we obtain

By substituting L_s into the formula for L_e on the right and then by substituting L_e into the formula for L_s on the right, we obtain

THEOREM 2 First Shifting Theorem, s-Shifting

If f (t) has the transform F(s) (where s > k for some k), then e^atf(t) has the transform F(s − a) (where s − a > k). In formulas,

or, if we take the inverse on both sides,

PROOF

We obtain F(s − a) by replacing s with s − a in the integral in (1), so that

If F(s) exists (i.e., is finite) for s greater than some k, then our first integral exists for s − a > k. Now take the inverse on both sides of this formula to obtain the second formula in the theorem. (CAUTION! −a in F(s − a) but + a in e^atf(t).)

EXAMPLE 5 s-Shifting: Damped Vibrations. Completing the Square

From Example 4 and the first shifting theorem we immediately obtain formulas 11 and 12 in Table 6.1,

For instance, use these formulas to find the inverse of the transform

Solution. Applying the inverse transform, using its linearity (Prob. 24), and completing the square, we obtain

We now see that the inverse of the right side is the damped vibration (Fig. 114)

THEOREM 3 Existence Theorem for Laplace Transforms

If f(t) is defined and piecewise continuous on every finite interval on the semi-axis t 0 and satisfies (2) for all t 0 and some constants M and k, then the Laplace transform (f) exists for all s > k.

PROOF

Since f(t) is piecewise continuous, e^−stf(t) is integrable over any finite interval on the t-axis. From (2), assuming that s > k (to be needed for the existence of the last of the following integrals), we obtain the proof of the existence of (f) from

PROBLEM SET 6.1

1–16 LAPLACE TRANSFORMS

Find the transform. Show the details of your work. Assume that a, b, ω, θ are constants.

1. 3t + 12
2. (a − bt)²
3. cos πt
4. cos² ωt
5. e^2t sinh t
6. e^−t sinh 4t
7. sin (ωt + θ)
8. 1.5 sin (3t − π/2)
9. 
10. 
11. 
12. 
13. 
14. 
15. 
16. 

17–24 SOME THEORY

• 17. Table 6.1. Convert this table to a table for finding inverse transforms (with obvious changes, e.g., ⁻¹(1/sⁿ) = tⁿ⁻¹/(n − 1), etc).
• 18. Using (f) in Prob. 10, find (f₁) where f₁(t) = 0 if t 2 and f₁(t) = 1 if t > 2.
• 19. Table 6.1. Derive formula 6 from formulas 9 and 10.
• 20. Nonexistence. Show that does not satisfy a condition of the form (2).
• 21. Nonexistence. Give simple examples of functions (defined for all t 0) that have no Laplace transform.
• 22. Existence. Show that . [Use (30) in App. 3.1.] Conclude from this that the conditions in Theorem 3 are sufficient but not necessary for the existence of a Laplace transform.
• 23. Change of scale. If (f(t)) = F(s) and c is any positive constant, show that (f(ct)) = F(s/c)/c (Hint: Use (1).) Use this to obtain (cos ωt) from (cos t).
• 24. Inverse transform. Prove that ⁻¹ is linear. Hint: Use the fact that is linear.

25–32 INVERSE LAPLACE TRANSFORMS

Given F(s) = (f), find f(t). a, b, L, n are constants. Show the details of your work.

• 25.
• 26.
• 27.
• 28.
• 29.
• 30.
• 31.
• 32.

33–45 APPLICATION OF s-SHIFTING

In Probs. 33–36 find the transform. In Probs. 37–45 find the inverse transform. Show the details of your work.

• 33. t²e^−3t
• 34. ke^−at cos ωt
• 35. 0.5e^−4.5t sin 2πt
• 36. sinh t cos t
• 37.
• 38.
• 39.
• 40.
• 41.
• 42.
• 43.
• 44.
• 45.

THEOREM 1 Laplace Transform of Derivatives

The transforms of the first and second derivatives of f(t) satisfy

Formula (1) holds if f(t) is continuous for all t 0 and satisfies the growth restriction (2) in Sec. 6.1 and f′(t) is piecewise continuous on every finite interval on the semi-axis t 0. Similarly, (2) holds if f and f′ are continuous for all t 0 and satisfy the growth restriction and f″ is piecewise continuous on every finite interval on the semi-axis t 0.

PROOF

We prove (1) first under the additional assumption that f′ is continuous. Then, by the definition and integration by parts,

Since f satisfies (2) in Sec. 6.1, the integrated part on the right is zero at the upper limit when s > k, and at the lower limit it contributes −f(0). The last integral is (f). It exists for because of Theorem 3 in Sec. 6.1. Hence (f′) exists when s > k and (1) holds.

If f′ is merely piecewise continuous, the proof is similar. In this case the interval of integration of f′ must be broken up into parts such that f′ is continuous in each such part.

The proof of (2) now follows by applying (1) to f″ and then substituting (1), that is

THEOREM 2 Laplace Transform of the Derivative f⁽ⁿ⁾ of Any Order

Let f, f′, …, f⁽ⁿ⁻¹⁾ be continuous for all t 0 and satisfy the growth restriction (2) in Sec. 6.1. Furthermore, let f⁽ⁿ⁾ be piecewise continuous on every finite interval on the semi-axis t 0. Then the transform of f⁽ⁿ⁾ satisfies

EXAMPLE 1 Transform of a Resonance Term (Sec. 2.8)

Let f(t) = t sin ωt. Then f(0) = 0, f′(t) = sin ωt + ωt cos ωt, f′(0) = 0, f″ = 2ω cos ωt − ω²t sin ωt. Hence by (2),

EXAMPLE 2 Formulas 7 and 8 in Table 6.1, Sec. 6.1

This is a third derivation of (cos ωt) and (sin ωt); cf. Example 4 in Sec. 6.1. Let f(t) = cos ωt. Then f(0) = 1, f′(0) = 0, f″(t) = −ω² cos ωt. From this and (2) we obtain

Similarly, let g = sin ωt. Then g(0) = 0, g′ = ω cos ωt. From this and (1) we obtain

THEOREM 3 Laplace Transform of Integral

Let F(s) denote the transform of a function f(t) which is piecewise continuous for t 0 and satisfies a growth restriction (2) , Sec.6.1. Then, for s > 0, s > k, and t > 0,

PROOF

Denote the integral in (4) by g(t). Since f(t) is piecewise continuous, g(t) is continuous, and (2), Sec. 6.1, gives

This shows that g(t) also satisfies a growth restriction. Also, g′(t) = f(t), except at points at which f(t) is discontinuous. Hence g′(t) is piecewise continuous on each finite interval and, by Theorem 1, since g(0) = 0 (the integral from 0 to 0 is zero)

Division by s and interchange of the left and right sides gives the first formula in (4), from which the second follows by taking the inverse transform on both sides.

EXAMPLE 3 Application of Theorem 3: Formulas 19 and 20 in the Table of Sec. 6.9

Using Theorem 3, find the inverse of .

Solution. From Table 6.1 in Sec. 6.1 and the integration in (4) (second formula with the sides interchanged) we obtain

This is formula 19 in Sec. 6.9. Integrating this result again and using (4) as before, we obtain formula 20 in Sec. 6.9:

It is typical that results such as these can be found in several ways. In this example, try partial fraction reduction.

EXAMPLE 4 Initial Value Problem: The Basic Laplace Steps

Solve

Solution. Step 1. From (2) and Table 6.1 we get the subsidiary equation [with Y = (y)]

Step 2. The transfer function is Q = 1/(s² − 1), and (7) becomes

Simplification of the first fraction and an expansion of the last fraction gives

Step 3. From this expression for Y and Table 6.1 we obtain the solution

The diagram in Fig. 116 summarizes our approach.

EXAMPLE 5 Comparison with the Usual Method

Solve the initial value problem

Solution. From (1) and (2) we see that the subsidiary equation is

The solution is

Hence by the first shifting theorem and the formulas for cos and sin in Table 6.1 we obtain

This agrees with Example 2, Case (III) in Sec. 2.4. The work was less.

EXAMPLE 6 Shifted Data Problems

This means initial value problems with initial conditions given at some t = t₀ > 0 instead of t = 0. For such a problem set , so that t = t₀ gives and the Laplace transform can be applied. For instance, solve

Solution. We have and we set . Then the problem is

where . Using (2) and Table 6.1 and denoting the transform of by , we see that the subsidiary equation of the “shifted” initial value problem is

Solving this algebraically for , we obtain

The inverse of the first two terms can be seen from Example 3 (with ω = 1), and the last two terms give cos and sin,

Now , so that the answer (the solution) is

PROBLEM SET 6.2

1–11 INITIAL VALUE PROBLEMS (IVPS)

Solve the IVPs by the Laplace transform. If necessary, use partial fraction expansion as in Example 4 of the text. Show all details.

1. y′ + 5.2y = 19.4 sin 2t, y(0) = 0
2. y′ + 2y = 0, y(0) = 1.5
3. y″ − y′ − 6y = 0, y(0) = 11, y′(0) = 28
4. y″ + 9y = 10e^−t, y(0) = 0, y′(0) = 0
5. 
6. y″ − 6y′ + 5y = 29 cos 2t, y(0) = 3.2, y′(0) = 6.2
7. y″ + 7y′ + 12y = 21e^3t, y(0) = 3.5, y′(0) = −10
8. y″ − 4y′ + 4y = 0, y(0) = 8.1, y′(0) = 3.9
9. y″ − 4y′ + 3y = 6t − 8, y(0) = 0, y′(0) = 0
10. y″ + 0.04y = 0.02t², y(0) = −25, y′(0) = 0
11. y″ + 3y′ + 2.25y = 9t³ + 64, y(0) = 1, y′(0) = 31.5

12–15 SHIFTED DATA PROBLEMS

Solve the shifted data IVPs by the Laplace transform. Show the details.

• 12. y″ − 2y′ − 3y = 0, y(4) = −3, y′(4) = −17
• 13. y′ − 6y = 0, y(−1) = 4
• 14. y″ + 2y′ + 5y = 50t − 100, y(2) = −4, y′(2) = 14
• 15. y″ + 3y′ − 4y = 6e^2t−3, y(1.5) = 4, y′(1.5) = 5

16–21 OBTAINING TRANSFORMS BY DIFFERENTIATION

Using (1) or (2), find (f) if f(t) equals:

• 16. t cos 4t
• 17. te^−at
• 18. cos²2t
• 19. sin² ωt
• 20. sin⁴ t. Use Prob. 19.
• 21. cosh² t
• 22. PROJECT. Further Results by Differentiation. Proceeding as in Example 1, obtain

  

  and from this and Example 1: (b) formula 21, (c) 22, (d) 23 in Sec. 6.9,

  

23–29 INVERSE TRANSFORMS BY INTEGRATION

Using Theorem 3, find f(t) if (F) equals:

• 23.
• 24.
• 25.
• 26.
• 27.
• 28.
• 29.
• 30. PROJECT. Comments on Sec. 6.2. (a) Give reasons why Theorems 1 and 2 are more important than Theorem 3.

  (b) Extend Theorem 1 by showing that if f(t) is continuous, except for an ordinary discontinuity (finite jump) at some t = a (>0), the other conditions remaining as in Theorem 1, then (see Fig. 117)

  

  (c) Verify (1*) for f(t) = e^−t if 0 < t < 1 and 0 if t > 1.

  (d) Compare the Laplace transform of solving ODEs with the method in Chap. 2. Give examples of your own to illustrate the advantages of the present method (to the extent we have seen them so far).

  

  Fig. 117. Formula (1*)

THEOREM 1 Second Shifting Theorem; Time Shifting

If f(t) has the transform F(s), then the “shifted function”

has the transform e^−as F(s). That is, if {f(t)} = F(s), then

Or, if we take the inverse on both sides, we can write

PROOF

We prove Theorem 1. In (4), on the right, we use the definition of the Laplace transform, writing τ for t (to have t available later). Then, taking e^−as inside the integral, we have

Substituting τ + a = t, thus τ = t − a, dτ = dt in the integral (CAUTION, the lower limit changes!), we obtain

To make the right side into a Laplace transform, we must have an integral from 0 to ∞, not from a to ∞. But this is easy. We multiply the integrand by u(t − a). Then for t from 0 to a the integrand is 0, and we can write, with as in (3),

(Do you now see why u(t − a) appears?) This integral is the left side of (4), the Laplace transform of in (3). This completes the proof.

EXAMPLE 1 Application of Theorem 1. Use of Unit Step Functions

Write the following function using unit step functions and find its transform.

Solution. Step 1. In terms of unit step functions,

Indeed, 2(1 − u(t − 1)) gives f(t) for 0 < t < 1, and so on.

Step 2. To apply Theorem 1, we must write each term in f(t) in the form f(t − a)u(t − a). Thus, 2(1 − u(t − 1)) remains as it is and gives the transform 2(1 − e^−s)/s. Then

Together,

If the conversion of f(t) to f(t − a) is inconvenient, replace it by

(4**) follows from (4) by writing f(t − a) = g(t), hence f(t) = g(t + a) and then again writing f for g. Thus,

as before. Similarly for . Finally, by (4**),

EXAMPLE 2 Application of Both Shifting Theorems. Inverse Transform

Find the inverse transform f(t) of

Solution. Without the exponential functions in the numerator the three terms of F(s) would have the inverses (sin πt)/π, (sin πt)/π, and te^−2t because 1/s² has the inverse t, so that 1/(s + 2)² has the inverse te^−2t by the first shifting theorem in Sec. 6.1. Hence by the second shifting theorem (t-shifting),

Now sin(πt − π) = −sin πt and sin(πt − 2π) = sin πt, so that the first and second terms cancel each other when t > 2. Hence we obtain f(t) = 0 if 0 < t < 1, −(sin πt)/π if 1 < t < 2, 0 if 2 < t < 3, and (t − 3)e^−2(t−3) if t < 3. See Fig. 123.

EXAMPLE 3 Response of an RC-Circuit to a Single Rectangular Wave

Find the current i(t) in the RC-circuit in Fig. 124 if a single rectangular wave with voltage V₀ is applied. The circuit is assumed to be quiescent before the wave is applied.

Solution. The input is V₀[u(t − a) − u(t − b)]. Hence the circuit is modeled by the integro-differential equation (see Sec. 2.9 and Fig. 124)

Fig. 124. RC-circuit, electromotive force v(t), and current in Example 3

Using Theorem 3 in Sec. 6.2 and formula (1) in this section, we obtain the subsidiary equation

Solving this equation algebraically for I(s), we get

the last expression being obtained from Table 6.1 in Sec. 6.1. Hence Theorem 1 yields the solution (Fig. 124)

that is, i(t) = 0 if t < a, and

where K₁ = V₀e^a/(RC)/R and K₂ = V₀e^b/(RC)/R.

EXAMPLE 4 Response of an RLC-Circuit to a Sinusoidal Input Acting Over a Time Interval

Find the response (the current) of the RLC-circuit in Fig. 125, where E(t) is sinusoidal, acting for a short time interval only, say,

and current and charge are initially zero.

Solution. The electromotive force E(t) can be represented by (100 sin 400t)(1 − u(t − 2π)). Hence the model for the current i(t) in the circuit is the integro-differential equation (see Sec. 2.9)

From Theorems 2 and 3 in Sec. 6.2 we obtain the subsidiary equation for I(s) = (i)

Solving it algebraically and noting that s² + 110s + 1000 = (s + 10)(s + 10)(s + 100), we obtain

For the first term in the parentheses (…) times the factor in front of them we use the partial fraction expansion

Now determine A, B, D, K by your favorite method or by a CAS or as follows. Multiplication by the common denominator gives

We set s = −10 and −100 and then equate the sums of the s³ and s² terms to zero, obtaining (all values rounded)

Since K = 258.66 = 0.6467 · 400, we thus obtain for the first term I₁ in I = I₁ − I₂

From Table 6.1 in Sec. 6.1 we see that its inverse is

This is the current i(t) when 0 < t < 2π. It agrees for 0 < t < 2π with that in Example 1 of Sec. 2.9 (except for notation), which concerned the same RLC-circuit. Its graph in Fig. 63 in Sec. 2.9 shows that the exponential terms decrease very rapidly. Note that the present amount of work was substantially less.

The second term I₁ of I differs from the first term by the factor e^−2πs. Since cos 400(t − 2π) = cos 400t and sin 400(t − 2π) = sin 400t, the second shifting theorem (Theorem 1) gives the inverse i₂(t) = 0 if 0 < t < 2π, and for > 2π it gives

Hence in i(t) the cosine and sine terms cancel, and the current for t > 2π is

It goes to zero very rapidly, practically within 0.5 sec.

PROBLEM SET 6.3

1. Report on Shifting Theorems. Explain and compare the different roles of the two shifting theorems, using your own formulations and simple examples. Give no proofs.

2–11 SECOND SHIFTING THEOREM, UNIT STEP FUNCTION

Sketch or graph the given function, which is assumed to be zero outside the given interval. Represent it, using unit step functions. Find its transform. Show the details of your work.

• 2. t (0 < t < 2)
• 3. t − 2(t > 2)
• 4. cos 4t(0 < t < π)
• 5. e^t(0 < t < π/2)
• 6. sin πt(2 < t < 4)
• 7. e^−πt (2 < t < 4)
• 8. t²(1 < t < 2)
• 9.
• 10. sinh t (0 < t < 2)
• 11. sin t(π/2 < t < π)

12–17 INVERSE TRANSFORMS BY THE 2ND SHIFTING THEOREM

Find and sketch or graph f(t) if (f) equals

• 12. e^−3s/(s − 1)³
• 13. 6(1 − e^−πs)/(s² + 9)
• 14. 4(e^−2s − 2e^−5s)/s
• 15. e^−3s/s⁴
• 16. 2(e^−s − e^−3s)/(s² − 4)
• 17. (1 + e^−2π(s+1))(s + 1)/((s + 1)² + 1)

18–27 IVPs, SOME WITH DISCONTINUOUS INPUT

Using the Laplace transform and showing the details, solve

• 18. 9y″ − 6y′ + y = 0, y(0) = 3, y′(0) = 1
• 19. y″ + 6y′ + 8y = e^−3t − e^−5t, y(0) = 0, y′(0) = 0
• 20. y″ + 10y′ + 24y = 144t², y(0) = 19/12, y′(0) = −5
• 21. y″ + 9y = 8 sin t if 0 < t < π and 0 if t > π; y(0) = 0, y′(0) = 4
• 22. y″ + 3y′ + 2y = 4t if 0 < t < 1 and 8 if t > 1; y(0) = 0, y′(0) = 0
• 23. y″ + y′ − 2y = 3 sin t − cos t if 0 < t < 2π and 3 sin 2t − cos 2t if t > 2π; y(0) = 1, y′(0) = 0
• 24. y″ + 3y′ + 2y = 1 if 0 < t < t and 0 if t > 1; y(0) = 0, y′(0) = 0
• 25. y″ + y = t if 0 < t < 1 and 0 if t > 1; y(0) = 0, y′(0) = 0
• 26. Shifted data. y″ + 2y′ + 5y = 10 sin t if 0 < t < 2π and 0 if t > 2π; y(π) = 1, y′(π) = 2e^−π − 2
• 27. Shifted data. y″ + 4y = 8t² if 0 < t < 5 and 0 if t > 5; y(1) = 1 + cos 2, y′(1) = 4 − 2 sin 2

28–40 MODELS OF ELECTRIC CIRCUITS

28–30 RC-CIRCUIT

Using the Laplace transform and showing the details, find the current in the circuit i(t) in Fig. 126, assuming i(0) = 0 and:

• 28. R = 1 kΩ (= 1000 Ω), L = 1 H, ν = 0 if 0 < t < π, and 40 sin t V if t > π
• 29. R = 25 Ω, L = 0.1 H, ν = 490 e^−5t V if 0 < t < 1 and 0 if t > 1
• 30. R = 10 Ω, L = 0.5 H, ν = 200t V if 0 < t < 2 and 0 if t > 2

  

  Fig. 126. Problems 28–30

  

  Fig. 127. Problem 31

• 31. Discharge in RC-circuit. Using the Laplace transform, find the charge q(t) on the capacitor of capacitance C in Fig. 127 if the capacitor is charged so that its potential is V₀ and the switch is closed at t = 0.

32–34 RL-CIRCUIT

Using the Laplace transform and showing the details, find the current i(t) in the circuit in Fig. 128 with R = 10 Ω and C = 10⁻² F, where the current at t = 0 is assumed to be zero, and:

• 32. ν = 0 if t < 4 and 1.4 · 10⁶e^−3t V if t > 4
• 33. ν = 0 if t < 2 and 100(t − 2) V if t > 2
• 34. ν(t) = 100 V if 0.5 < t < 0.6 and 0 otherwise. Why does i(t) have jumps?

  

  Fig. 128. Problems 32–34

35–37 LC-CIRCUIT

Using the Laplace transform and showing the details, find the current i(t) in the circuit in Fig. 129, assuming zero initial current and charge on the capacitor and:

• 35. L = 1 H, C = 10⁻² F, ν = −9900 cos t V if π < t < 3π and 0 otherwise
• 36. 0 < t < 1 and 0 if t > 1
• 37. L = 0.5 H, C = 0.05 F, ν = 78 sin t V if 0 < t < π and 0 if t > π

  

  Fig. 129. Problems 35–37

38–40 RLC-CIRCUIT

Using the Laplace transform and showing the details, find the current i(t) in the circuit in Fig. 130, assuming zero initial current and charge and:

• 38. R = 4 Ω, L = 1 H, C = 0.05 F, ν = 34e^−t V if 0 < t < 4 and 0 if t > 4
• 39. R = 2 Ω, L = 1 H, C = 0.5 F, ν(t) = 1 k V if 0 < t < 2 and 0 if t > 2

  

  Fig. 130. Problems 38–40

• 40. R = 2 Ω, L = 1 H, C = 0.1 F, ν = 255 sin t V if 0 < t < 2π and 0 if t > 2π

  

  Fig. 131. Current in Problem 40

EXAMPLE 1 Mass–Spring System Under a Square Wave

Determine the response of the damped mass–spring system (see Sec. 2.8) under a square wave, modeled by (see Fig. 133)

Solution. From (1) and (2) in Sec. 6.2 and (2) and (4) in this section we obtain the subsidiary equation

Using the notation F(s) and partial fractions, we obtain

From Table 6.1 in Sec. 6.1, we see that the inverse is

Therefore, by Theorem 1 in Sec. 6.3 (t-shifting) we obtain the square-wave response shown in Fig. 133,

EXAMPLE 2 Hammerblow Response of a Mass–Spring System

Find the response of the system in Example 1 with the square wave replaced by a unit impulse at time t = 1.

Solution. We now have the ODE and the subsidiary equation

Solving algebraically gives

By Theorem 1 the inverse is

y(t) is shown in Fig. 134. Can you imagine how Fig. 133 approaches Fig. 134 as the wave becomes shorter and shorter, the area of the rectangle remaining 1?

EXAMPLE 3 Four-Terminal RLC-Network

Find the output voltage response in Fig. 135 if R = 20 Ω, L = 1 H, C = 10⁻⁴ F, the input is δ(t)(a unit impulse at time t = 0), and current and charge are zero at time t = 0.

Solution. To understand what is going on, note that the network is an RLC-circuit to which two wires at A and B are attached for recording the voltage v(t) on the capacitor. Recalling from Sec. 2.9 that current i(t) and charge q(t) are related by i = q′ = dq/dt, we obtain the model

From (1) and (2) in Sec. 6.2 and (5) in this section we obtain the subsidiary equation for Q(s) = (q)

By the first shifting theorem in Sec. 6.1 we obtain from Q damped oscillations for q and v; rounding 9900 ≈ 99.50², we get (Fig. 135)

EXAMPLE 4 Unrepeated Complex Factors. Damped Forced Vibrations

Solve the initial value problem for a damped mass–spring system acted upon by a sinusoidal force for some time interval (Fig. 136),

Solution. From Table 6.1, (1), (2) in Sec. 6.2, and the second shifting theorem in Sec. 6.3, we obtain the subsidiary equation

We collect the Y-terms, (s² + 2s + 2)Y, take to the right, and solve,

For the last fraction we get from Table 6.1 and the first shifting theorem

In the first fraction in (6) we have unrepeated complex roots, hence a partial fraction representation

Multiplication by the common denominator gives

We determine A, B, M, N. Equating the coefficients of each power of s on both sides gives the four equations

We can solve this, for instance, obtaining M = −A from (a), then A = B from (c), then N = −3A from (b), and finally A = −2 from (d). Hence a = −2, B = −2, N = 6, and the first fraction in (6) has the representation

The sum of this inverse and (7) is the solution of the problem for 0 < t < π, namely (the sines cancel),

In the second fraction in (6), taken with the minus sign, we have the factor e^−πs so that from (8) and the second shifting theorem (Sec. 6.3) we get the inverse transform of this fraction for t > 0 in the form

The sum of this and (9) is the solution for t > π,

Figure 136 shows (9) (for 0 < t < π) and (10) (for t > π), a beginning vibration, which goes to zero rapidly because of the damping and the absence of a driving force after t = π.

PROBLEM SET 6.4

1. CAS PROJECT. Effect of Damping. Consider a vibrating system of your choice modeled by

   

   (a) Using graphs of the solution, describe the effect of continuously decreasing the damping to 0, keeping k constant.

   (b) What happens if c is kept constant and k is continuously increased, starting from 0?

   (c) Extend your results to a system with two δ-functions on the right, acting at different times.

2. CAS EXPERIMENT. Limit of a Rectangular Wave. Effects of Impulse.

   (a) In Example 1 in the text, take a rectangular wave of area 1 from 1 to 1 + k. Graph the responses for a sequence of values of k approaching zero, illustrating that for smaller and smaller k those curves approach the curve shown in Fig. 134. Hint: If your CAS gives no solution for the differential equation, involving k, take specific k's from the beginning.

   (b) Experiment on the response of the ODE in Example 1 (or of another ODE of your choice) to an impulse δ(t − a) for various systematically chosen a (> 0); choose initial conditions y(0) ≠ 0, y′(0) = 0. Also consider the solution if no impulse is applied. Is there a dependence of the response on a? On b if you choose bδ(t − a)? Would −δ(t − ã) with ã > a annihilate the effect of δ(t − a)? Can you think of other questions that one could consider experimentally by inspecting graphs?

3–12 EFFECT OF DELTA (IMPULSE) ON VIBRATING SYSTEMS

Find and graph or sketch the solution of the IVP. Show the details.

• 3. y″ + 4y = δ(t − π), y(0) = 8, y′ (0) = 0
• 4. y″ + 16y = 4δ(t − 3π), y(0) = 2, y′(0) = 0
• 5. y″ + y = δ(t − π) − δ(t − 2π), y(0) = 0, y′(0) = 1
• 6. y″ + 4y′ + 5y = δ(t − 1), y(0) = 0, y′(0) = 3
• 7.
• 8. y″ + 3y′ + 2y = 10(sin t + δ(t − 1)), y(0) = 1, y′(0) = −1
• 9. y″ + 4y′ + 5y = [1 − u(t − 10)]e^t − e¹⁰δ(t − 10)), y(0) = 0, y′(0) = 1
• 10.
• 11. y″ + 5y′ + 6y = u(t − 1) + δ(t − 2), y(0) = 0, y′(0) = 1
• 12. y″ + 2y′ + 5y = 25t − 100δ(t − π), y(0) = −2, y′(0) = 5
• 13. PROJECT. Heaviside Formulas. (a) Show that for a simple root a and fraction A/(s − a) in F(s)/G(s) we have the Heaviside formula

  

  (b) Similarly, show that for a root a of order m and fractions in

  

  we have the Heaviside formulas for the first coefficient

  

  and for the other coefficients

  

• 14. TEAM PROJECT. Laplace Transform of Periodic Functions

  (a) Theorem. The Laplace transform of a piecewise continuous function f(t) with period p is

  

  Prove this theorem. Hint: Write .

  Set t = (n − 1)p in the nth integral. Take out e^−(n−1)p from under the integral sign. Use the sum formula for the geometric series.

  (b) Half-wave rectifier. Using (11), show that the half-wave rectification of sin ωt in Fig. 137 has the Laplace transform

  

  (A half-wave rectifier clips the negative portions of the curve. A full-wave rectifier converts them to positive; see Fig. 138.)

  (c) Full-wave rectifier. Show that the Laplace transform of the full-wave rectification of sin ωt is

  

  

  Fig. 137. Half-wave rectification

  

  Fig. 138. Full-wave rectification

  (d) Saw-tooth wave. Find the Laplace transform of the saw-tooth wave in Fig. 139.

  

  Fig. 139. Saw-tooth wave

• 15. Staircase function. Find the Laplace transform of the staircase function in Fig. 140 by noting that it is the difference of kt/p and the function in 14(d).

  

  Fig. 140. Staircase function

THEOREM 1 Convolution Theorem

If two functions f and g satisfy the assumption in the existence theorem in Sec. 6.1, so that their transforms F and G exist, the product H = FG is the transform of h given by (1) . (Proof after Example 2.)

EXAMPLE 1 Convolution

Let H(s) = 1/[(s − a)s]. Find h(t).

Solution. 1/(s − a) has the inverse f(t = e^at, and 1/s has the inverse g(t) = 1. With f(τ) = e^aτ and g(t − τ) ≡ 1 we thus obtain from (1) the answer

To check, calculate

EXAMPLE 2 Convolution

Let H(s) = 1/(s² + ω²)². Find h(t).

Solution. The inverse of 1/(s² + ω² is (sin ωt)/ω. Hence from (1) and the first formula in (11) in App. 3.1 we obtain

in agreement with formula 21 in the table in Sec. 6.9.

PROOF

We prove the Convolution Theorem 1. CAUTION! Note which ones are the variables of integration! We can denote them as we want, for instance, by τ and p, and write

We now set t = p + τ, where τ is at first constant. Then p = t − τ, and t varies from τ to ∞. Thus

τ in F and t in G vary independently. Hence we can insert the G-integral into the F-integral. Cancellation of e^−sτ and e^sτ then gives

Here we integrate for fixed τ over t from τ to ∞ and then over τ from 0 to ∞. This is the blue region in Fig. 141. Under the assumption on f and g the order of integration can be reversed (see Ref. [A5] for a proof using uniform convergence). We then integrate first over τ from 0 to t and then over t from 0 to ∞, that is,

This completes the proof.

EXAMPLE 3 Unusual Properties of Convolution

f * 1 ≠ f in general. For instance,

(f * f)(t) 0 may not hold. For instance, Example 2 with ω = 1 gives

EXAMPLE 4 Repeated Complex Factors. Resonance

In an undamped mass–spring system, resonance occurs if the frequency of the driving force equals the natural frequency of the system. Then the model is (see Sec. 2.8)

where is the spring constant, and m is the mass of the body attached to the spring. We assume y(0) = 0 and y′(0) = 0, for simplicity. Then the subsidiary equation is

This is a transform as in Example 2 with ω = ω₀ and multiplied by Kω₀. Hence from Example 2 we can see directly that the solution of our problem is

We see that the first term grows without bound. Clearly, in the case of resonance such a term must occur. (See also a similar kind of solution in Fig. 55 in Sec. 2.8.)

EXAMPLE 5 Response of a Damped Vibrating System to a Single Square Wave

Using convolution, determine the response of the damped mass–spring system modeled by

This system with an input (a driving force) that acts for some time only (Fig. 143) has been solved by partial fraction reduction in Sec. 6.4 (Example 1).

Solution by Convolution. The transfer function and its inverse are

Hence the convolution integral (3) is (except for the limits of integration)

Now comes an important point in handling convolution. r(τ) = 1 if 1 < τ < 2 only. Hence if t < 1, the integral is zero. If 1 < t < 2, we have to integrate from τ = 1 (not 0) to t. This gives (with the first two terms from the upper limit)

If t > 2, we have to integrate from τ = 1 to 2 (not to t). This gives

Figure 143 shows the input (the square wave) and the interesting output, which is zero from 0 to 1, then increases, reaches a maximum (near 2.6) after the input has become zero (why?), and finally decreases to zero in a monotone fashion.

EXAMPLE 6 A Volterra Integral Equation of the Second Kind

Solve the Volterra integral equation of the second kind³

Solution. From (1) we see that the given equation can be written as a convolution, y − y* sin t = t. Writing Y = (y) and applying the convolution theorem, we obtain

The solution is

Check the result by a CAS or by substitution and repeated integration by parts (which will need patience).

EXAMPLE 7 Another Volterra Integral Equation of the Second Kind

Solve the Volterra integral equation

Solution. By (1) we can write y − (1 + t) * y = 1 − sinh t. Writing Y = (y), we obtain by using the convolution theorem and then taking common denominators

(s² − s − 1)/s cancels on both sides, so that solving for Y simply gives

PROBLEM SET 6.5

1–7 CONVOLUTIONS BY INTEGRATION

Find:

1. 1 * 1
2. 1 * sin ωt
3. e^t * e^−t
4. (cos ωt) * (cos ωt
5. (sin ωt) * (cos ωt)
6. e^at * e^bt(a ≠ b)
7. t * e^t

8–14 INTEGRAL EQUATIONS

Solve by the Laplace transform, showing the details:

• 8.
• 9.
• 10.
• 11.
• 12.
• 13.
• 14.
• 15. CAS EXPERIMENT. Variation of a Parameter.

  (a) Replace 2 in Prob. 13 by a parameter k and investigate graphically how the solution curve changes if you vary k, in particular near k = −2.

  (b) Make similar experiments with an integral equation of your choice whose solution is oscillating.

• 16. TEAM PROJECT. Properties of Convolution. Prove:

  (a) Commutativity, f * g = g * f

  (b) Associativity, (f * g) * ν = f * (g * ν)

  (c) Distributivity, f * (g₁ + g₂) = f * g₁ + f * g₂

  (d) Dirac's delta. Derive the sifting formula (4) in Sec. 6.4 by using f_k with a = 0 [(1), Sec. 6.4] and applying the mean value theorem for integrals.

  (e) Unspecified driving force. Show that forced vibrations governed by

  

  with ω ≠ 0 and an unspecified driving force r(t) can be written in convolution form,

  

17–26 INVERSE TRANSFORMS BY CONVOLUTION

Showing details, find f(t) if (f) equals:

• 17.
• 18.
• 19.
• 20.
• 21.
• 22.
• 23.
• 24.
• 25.
• 26. Partial Fractions. Solve Probs. 17, 21, and 23 by partial fraction reduction.

EXAMPLE 1 Differentiation of Transforms. Formulas 21–23 in Sec. 6.9

We shall derive the following three formulas.

Solution. From (1) and formula 8 (with ω = β) in Table 6.1 of Sec. 6.1 we obtain by differentiation (CAUTION! Chain rule!)

Dividing by 2β and using the linearity of , we obtain (3).

Formulas (2) and (4) are obtained as follows. From (1) and formula 7 (with ω = β) in Table 6.1 we find

From this and formula 8 (with ω = β) in Table 6.1 we have

On the right we now take the common denominator. Then we see that for the plus sign the numerator becomes s² − β² + s² + β² = 2s², so that (4) follows by division by 2. Similarly, for the minus sign the numerator takes the form s² − β² − s² − β² = −2β², and we obtain (2). This agrees with Example 2 in Sec. 6.5.

EXAMPLE 2 Differentiation and Integration of Transforms

Find the inverse transform of .

Solution. Denote the given transform by F(s). Its derivative is

Taking the inverse transform and using (1), we obtain

Hence the inverse f(t) of F(s) is f(t) = 2(1 − cos ωt)/t. This agrees with formula 42 in Sec. 6.9.

Alternatively, if we let

From this and (6) we get, in agreement with the answer just obtained,

the minus occurring since s is the lower limit of integration.

In a similar way we obtain formula 43 in Sec. 6.9,

EXAMPLE 3 Laguerre's Equation. Laguerre Polynomials

Laguerre's ODE is

We determine a solution of (9) with n = 0, 1, 2, …. From (7)–(9) we get the subsidiary equation

Simplification gives

Separating variables, using partial fractions, integrating (with the constant of integration taken to be zero), and taking exponentials, we get

We write l_n = ⁻¹(Y) and prove Rodrigues's formula

These are polynomials because the exponential terms cancel if we perform the indicated differentiations. They are called Laguerre polynomials and are usually denoted by L_n (see Problem Set 5.7, but we continue to reserve capital letters for transforms). We prove (10). By Table 6.1 and the first shifting theorem (s-shifting),

because the derivatives up to the order n − 1 are zero at 0. Now make another shift and divide by n! to get [see (10) and then (10*)]

PROBLEM SET 6.6

1. REVIEW REPORT. Differentiation and Integration of Functions and Transforms. Make a draft of these four operations from memory. Then compare your draft with the text and write a 2- to 3-page report on these operations and their significance in applications.

2–11 TRANSFORMS BY DIFFERENTIATION

Showing the details of your work, find (f) if f(t) equals:

• 2. 3t sinh 4t
• 3.
• 4. te^−t cos t
• 5. t cos ωt
• 6. t² sin 3t
• 7. t² cosh 2t
• 8. te^−kt sin t
• 9.
• 10. tⁿe^kt
• 11.
• 12. CAS PROJECT. Laguerre Polynomials. (a) Write a CAS program for finding l_n(t) in explicit form from (10). Apply it to calculate l₀, …, l₁₀. Verify that l₀, …, l₁₀ satisfy Laguerre's differential equation (9).

  (b) Show that

  

  and calculate l₀, …, l₁₀ from this formula.

  (c) Calculate l₀, …, l₁₀ recursively from l₀ = 1, l₁ = 1 − t by

  

  (d) A generating function (definition in Problem Set 5.2) for the Laguerre polynomials is

  

  Obtain l₀, …, l₁₀ from the corresponding partial sum of this power series in x and compare the l_n with those in (a), (b), or (c).

• 13. CAS EXPERIMENT. Laguerre Polynomials. Experiment with the graphs of l₀, …, l₁₀, finding out empirically how the first maximum, first minimum, … is moving with respect to its location as a function of n. Write a short report on this.

14–20 INVERSE TRANSFORMS

Using differentiation, integration, s-shifting, or convolution, and showing the details, find f(t) if (f) equals:

• 14.
• 15.
• 16.
• 17.
• 18.
• 19.
• 20.

EXAMPLE 1 Mixing Problem Involving Two Tanks

Tank T₁ in Fig. 144 initially contains 100 gal of pure water. Tank T₂ initially contains 100 gal of water in which 150 lb of salt are dissolved. The inflow into T₁ is 2 gal/min from T₂ and 6 gal/min containing 6 lb of salf from the outside. The inflow into T₂ is 8 gal/min from T₁. The outflow from T₂ is 2 + 6 = 8 gal/min, as shown in the figure. The mixtures are kept uniform by stirring. Find and plot the salt contents y₁(t) and y₂(t) in T₁ and T₂ respectively.

Solution. The model is obtained in the form of two equations

for the two tanks (see Sec. 4.1). Thus,

The initial conditions are y₁(0) = 0, y₂(0) = 150. From this we see that the subsidiary system (2) is

We solve this algebraically for Y₁ and Y₂ by elimination (or by Cramer's rule in Sec. 7.7), and we write the solutions in terms of partial fractions,

By taking the inverse transform we arrive at the solution

Figure 144 shows the interesting plot of these functions. Can you give physical explanations for their main features? Why do they have the limit 100? Why is y₂ not monotone, whereas y₁ is? Why is y₁ from some time on suddenly larger than y₂? Etc.

EXAMPLE 2 Electrical Network

Find the currents i₁(t) and i₂(t) in the network in Fig. 145 with L and R measured in terms of the usual units (see Sec. 2.9), ν(t) = 100 volts if 0 t 0.5 sec and 0 thereafter, and i(0) = 0, i′(0) = 0.

Solution. The model of the network is obtained from Kirchhoff's Voltage Law as in Sec. 2.9. For the lower circuit we obtain

Fig. 145. Electrical network in Example 2

and for the upper

Division by 0.8 and ordering gives for the lower circuit

and for the upper

With i₁(0) = 0, i₂(0) = 0 we obtain from (1) in Sec. 6.2 and the second shifting theorem the subsidiary system

Solving algebraically for I₁ and I₂ gives

The right sides, without the factor 1 − e^−s/2, have the partial fraction expansions

and

respectively. The inverse transform of this gives the solution for ,

According to the second shifting theorem the solution for , that is,

Can you explain physically why both currents eventually go to zero, and why i₁(t) has a sharp cusp whereas i₂(t) has a continuous tangent direction at ?

EXAMPLE 3 Model of Two Masses on Springs (Fig. 146)

The mechanical system in Fig. 146 consists of two bodies of mass 1 on three springs of the same spring constant k and of negligibly small masses of the springs. Also damping is assumed to be practically zero. Then the model of the physical system is the system of ODEs

Here y₁ and y₂ are the displacements of the bodies from their positions of static equilibrium. These ODEs follow from Newton's second law, Mass × Acceleration = Force, as in Sec. 2.4 for a single body. We again regard downward forces as positive and upward as negative. On the upper body, −ky₁ is the force of the upper spring and k(y₂ − y₁) that of the middle spring, y₂ − y₁ being the net change in spring length—think this over before going on. On the lower body, −k(y₂ − y₁) is the force of the middle spring and −ky₂ that of the lower spring.

We shall determine the solution corresponding to the initial conditions y₁(0) = 1, y₂(0) = 1, , . Let Y₁ = (y₁) and Y₂ = (y₂). Then from (2) in Sec. 6.2 and the initial conditions we obtain the subsidiary system

This system of linear algebraic equations in the unknowns Y₁ and Y₂ may be written

Elimination (or Cramer's rule in Sec. 7.7) yields the solution, which we can expand in terms of partial fractions,

Hence the solution of our initial value problem is (Fig. 147)

We see that the motion of each mass is harmonic (the system is undamped!), being the superposition of a “slow” oscillation and a “rapid” oscillation.

PROBLEM SET 6.7

1. TEAM PROJECT. Comparison of Methods for Linear Systems of ODEs

   (a) Models. Solve the models in Examples 1 and 2 of Sec. 4.1 by Laplace transforms and compare the amount of work with that in Sec. 4.1. Show the details of your work.

   (b) Homogeneous Systems. Solve the systems (8), (11)–(13) in Sec. 4.3 by Laplace transforms. Show the details.

   (c) Nonhomogeneous System. Solve the system (3) in Sec. 4.6 by Laplace transforms. Show the details.

2–15 SYSTEMS OF ODES

Using the Laplace transform and showing the details of your work, solve the IVP:

• 2. y′₁ + y₂ = 0, y₁ + y′₂ = 2 cos t, y₁(0) = 1, y₂(0) = 0
• 3. y′₁ = −y₁ + 4y₂, y′₂ = 3y₁ − 2y₂, y₁(0) = 3, y₂(0) = 4
• 4. y′₁ = 4y₂ − 8 cos 4t, y′₂ = −3y₁ − 9 sin 4t, y₁(0) = 0, y₂(0) = 3
• 5. y′₁ = y₂ + 1 − u(t − 1), y′₂ = −y₁ + 1 − u(t − 1), y₁(0) = 0, y₂(0) = 0
• 6. y′₁ = 5y₁ + y₂, y′₂ = y₁ + 5y₂, y₁(0) = 1, y₂(0) = −3
• 7. y′₁ = 2y₁ − 4y₂ + u(t − 1)e^t,

  y′₂ = y₁ − 3y₂ + u(t − 1)e^t, y₁(0) = 3, y₂(0) = 0

• 8. y′₁ = −2y₁ + 3y₂, y′₂ = 4y₁ − y₂, y₁(0) = 4, y₂(0) = 3
• 9. y′₁ = 4y₁ + y₂, y′₂ = −y₁ + 2y₂, y₁(0) = 3, y₂(0) = 1
• 10. y′₁ = −y₂, y′₂ = −y₁ + 2[1 − u(t − 2π)] cos t, y₁(0) = 1, y₂(0) = 0
• 11. y″₁ = y₁ + 3y₂, y″₂ = 4y₁ − 4e^t, y₁(0) = 2, y′₁(0) = 3, y₂(0) = 1, y′₂(0) = 2
• 12. y″₁ = −2y₁ + 2y₂, y″₂ = 2y₁ − 5y₂, y₁(0) = 1, y′₁(0) = 0, y₂(0) = 3, y′₂(0) = 0
• 13. y″₁ + y₂ = −101 sin 10t, y″₂ + y₁ = 101 sin 10t, y₁(0) = 0, y′₁(0) = 6, y₂(0) = 8, y′₂(0) = −6
• 14.
• 15.

FURTHER APPLICATIONS

• 16. Forced vibrations of two masses. Solve the model in Example 3 with k = 4 and initial conditions y₁(0) = 1, y′₁(0) = 1, y₂(0) = 1, y′₂ = −1 under the assumption that the force 11 sin t is acting on the first body and the force −11 sin t on the second. Graph the two curves on common axes and explain the motion physically.
• 17. CAS Experiment. Effect of Initial Conditions. In Prob. 16, vary the initial conditions systematically, describe and explain the graphs physically. The great variety of curves will surprise you. Are they always periodic? Can you find empirical laws for the changes in terms of continuous changes of those conditions?
• 18. Mixing problem. What will happen in Example 1 if you double all flows (in particular, an increase to containing 12 lb of salt from the outside), leaving the size of the tanks and the initial conditions as before? First guess, then calculate. Can you relate the new solution to the old one?
• 19. Electrical network. Using Laplace transforms, find the currents i₁(t) and i₂(t) in Fig. 148, where ν(t) = 390 cos t and i₁(0) = 0, i₂(0) = 0. How soon will the currents practically reach their steady state?

  

  Fig. 148. Electrical network and currents in Problem 19

• 20. Single cosine wave. Solve Prob. 19 when the EMF (electromotive force) is acting from 0 to 2π only. Can you do this just by looking at Prob. 19, practically without calculation?

The main purpose of Laplace transforms is the solution of differential equations and systems of such equations, as well as corresponding initial value problems. The Laplace transform F(s) = (f) of a function f(t) is defined by

This definition is motivated by the property that the differentiation of f with respect to t corresponds to the multiplication of the transform F by s; more precisely,

etc. Hence by taking the transform of a given differential equation

and writing (y) = Y(s) we obtain the subsidiary equation

Here, in obtaining the transform (r) we can get help from the small table in Sec. 6.1 or the larger table in Sec. 6.9. This is the first step. In the second step we solve the subsidiary equation algebraically for Y(s). In the third step we determine the inverse transform y(t) = ⁻¹(Y), that is, the solution of the problem. This is generally the hardest step, and in it we may again use one of those two tables. Y(s) will often be a rational function, so that we can obtain the inverse ⁻¹(Y) by partial fraction reduction (Sec. 6.4) if we see no simpler way.

The Laplace method avoids the determination of a general solution of the homogeneous ODE, and we also need not determine values of arbitrary constants in a general solution from initial conditions; instead, we can insert the latter directly into (4). Two further facts account for the practical importance of the Laplace transform. First, it has some basic properties and resulting techniques that simplify the determination of transforms and inverses. The most important of these properties are listed in Sec. 6.8, together with references to the corresponding sections. More on the use of unit step functions and Dirac's delta can be found in Secs. 6.3 and 6.4, and more on convolution in Sec. 6.5. Second, due to these properties, the present method is particularly suitable for handling right sides given by different expressions over different intervals of time, for instance, when is a square wave or an impulse or of a form such as r(t) = cos t if 0 t 4π and 0 elsewhere.

The application of the Laplace transform to systems of ODEs is shown in Sec. 6.7. (The application to PDEs follows in Sec. 12.12.)