Multiplying a Function by tⁿ

The Laplace transform of the product of a function f(t) with t can be found by differentiating the Laplace transform of f(t). If F(s) = {f(t)} and if we assume that interchanging of differentiation and integration is possible, then

that is,

Similarly,

The preceding two cases suggest the following general result for {tⁿf(t)}.

EXAMPLE 1 Using Theorem 4.4.1

If we wish to find the Laplace transforms and all we need do is take the negative of the derivative with respect to s of , found in Example 1, followed by the negative of the derivative with respect to s of

To find transforms of functions tⁿe^at we can use either the first translation theorem or Theorem 4.4.1. For example,

Theorem 4.3.1: {te^3t} = = =

Theorem 4.4.1:

EXAMPLE 2 An Initial-Value Problem

Theorem 4.4.1 is sometimes useful in solving initial-value problems involving differential equations with variable coefficients that are monomials For example, if a linear second-order differential equation contains the term then from Theorem 4.4.1 with its Laplace transform is

(2)

The effect of the last term in (2) is to change a second-order DE in t to a first-order DE in s. The next example illustrates the procedure.

EXAMPLE 3 An Initial-Value Problem

Solve

SOLUTION

Using (2) above and (7) of Section 4.2 the Laplace transform of the differential equation is

Simplifying, we obtain a linear first-order differential equation

or

Integrating with respect to s and solving for Y(s) then gives

The inverse Laplace transform is then

Applying the given initial condition to the last expression gives

The solution of the IVP is then ≡

Convolution

If functions f and g are piecewise continuous on the interval then the convolution of f and g, denoted by the symbol f * g, is a function defined by the integral

(3)

Because we are integrating in (3) with respect to the variable τ (the lowercase Greek letter tau), the convolution f * g is a function of t. To emphasize this fact, (3) is also written (f * g)(t).

EXAMPLE 4 Convolution of Two Functions

Properties of the Convolution

As the notation f * g suggests, the convolution (3) is often interpreted as a generalized product of two functions f and g. Indeed, convolution has some properties that are similar to ordinary multiplication. If f, g, and h are piecewise continuous functions on then

The commutative law f * g = g * f simply states that two integrals are equal. To prove this, we use the change of variables in (3),

f * g = g * f.

But one has to be careful. For example, if the dot denotes multiplication and * convolution, then but f *

Convolution Theorem

We have seen that if f and g are both piecewise continuous for t ≥ 0, then the Laplace transform of a sum f + g is the sum of the individual Laplace transforms. While it is not true that the Laplace transform of the product fg is the product of the Laplace transforms, we see in the next theorem—called the convolution theorem—that the Laplace transform of the generalized product f * g is the product of the Laplace transforms of f and g.

PROOF:

Let

Proceeding formally, we have

Holding τ fixed, we let t = τ + β, dt = dβ, so that

In the tτ-plane we are integrating over the shaded region in FIGURE 4.4.1. Since f and g are piecewise continuous on [0, ∞) and of exponential order, it is possible to interchange the order of integration:

{f * g}. ≡

Theorem 4.4.2 shows that we can find the Laplace transform of the convolution f * g of two functions without actually evaluating the definite integral as we did in (4). The next example illustrates the idea.

EXAMPLE 5 Using Theorem 4.4.2

Inverse Form of Theorem 4.4.2

The convolution theorem is sometimes useful in finding the inverse Laplace transforms of the product of two Laplace transforms. From Theorem 4.4.2 we have

(5)

Many of the results in the table of Laplace transforms in Appendix C can be derived using (5). For instance, in the next example we obtain entry 25 of the table:

{sin kt − kt cos kt} = (6)

EXAMPLE 6 Inverse Transform as a Convolution

Transform of an Integral

When g(t) = 1 and {g(t)} = G(s) = 1/s, the convolution theorem implies that the Laplace transform of the integral of f is

.(8)

The inverse form of (8),

,(9)

can be used in lieu of partial fractions when sⁿ is a factor of the denominator and f(t) = ⁻¹{F(s)} is easy to integrate. For example, we know for f(t) = sin t that F(s) = 1/(s² + 1), and so by (9)

and so on.

Volterra Integral Equation

The convolution theorem and the result in (8) are useful in solving other types of equations in which an unknown function appears under an integral sign. In the next example, we solve a Volterra integral equation for f(t),

(10)

The functions g(t) and h(t) are known. Notice that the integral in (10) has the convolution form (3) with the symbol h playing the part of g.

EXAMPLE 7 An Integral Equation

Series Circuits

In a single-loop or series circuit, Kirchhoff’s second law states that the sum of the voltage drops across an inductor, resistor, and capacitor is equal to the voltage impressed on the circuit. Now the voltage drops across an inductor, resistor, and capacitor are, respectively,

where is the current and is the charge on the capacitor. The constants L, R, and C are called, in turn, the inductance, resistance, and capacitance. Hence, in an LR-series circuit, such as that shown in FIGURE 4.4.2(a), the sum of the voltage drops across the inductance and the resistor is the differential equation

(11)

for the current The sum of the voltage drops across the resistor and the capacitor for the RC-series circuit in Figure 4.4.2(b) is

(12)

Because and are related by , we can write Substituting the last expression in (12) yields an integral equation for the current

(13)

For the LC-series circuit shown in Figure 4.4.2(c), the sum of the voltage drops gives an equation that involves both an integral and the derivative of

(14)

Finally, for the LRC-series circuit in Figure 4.4.2(d), we have

(15)

Equations (14) and (15) are called integrodifferential equations. In view of the transform (8), we can use the Laplace transform to solve equations such as (11), (13), (14), and (15).

EXAMPLE 8 An Integrodifferential Equation

Determine the current i(t) in a single-loop LRC-circuit when L = 0.1 h, R = 2 Ω, C = 0.1 F, i(0) = 0, and the impressed voltage is

E(t) = 120t − 120t (t − 1).

SOLUTION

Using the given data, equation (15) becomes

Now by (8), { i(τ) dτ} = I(s)/s, where I(s) = {i(t)}. Thus the Laplace transform of the integrodifferential equation is

Multiplying this equation by 10s, using s² + 20s + 100 = (s + 10)², and then solving for I(s) gives

By partial fractions,

From the inverse form of the second translation theorem, (15) of Section 4.3, we finally obtain

Written as a piecewise-defined function, the current is

(16)

Using the last form of the solution and a CAS, we graph i(t) on each of the two intervals and then combine the graphs. Note in FIGURE 4.4.3 that even though the input E(t) is discontinuous, the output or response i(t) is a continuous function. ≡

Periodic Function

If a periodic function f has period T, T > 0, then f(t + T) = f(t). The Laplace transform of a periodic function can be obtained by integration over one period.

PROOF:

Write the Laplace transform of f as two integrals:

When we let t = u + T, the last integral becomes

Therefore

Solving the equation in the last line for {f(t)} proves the theorem. ≡

EXAMPLE 9 Transform of a Periodic Function

Find the Laplace transform of the periodic function shown in FIGURE 4.4.4.

SOLUTION

The function in the figure is called a square wave and has period For can be defined by

and outside the interval by Now from Theorem 4.4.3,

(17)

≡

EXAMPLE 10 A Periodic Impressed Voltage

Determine the current in a single loop LR-series circuit when and the impressed voltage is the square-wave function given in Figure 4.4.4 with period

SOLUTION

The differential equation for the current is given in (11). If we use the result in (17) of the preceding example with then the Laplace transform of the DE is

(18)

To find the inverse Laplace transform of the last function, we first make use of geometric series. With the identification x = e^−s, s > 0, the geometric series

From

we can then rewrite (18) as

By applying the form of the second translation theorem to each term of both series we obtain

or, equivalently,

.

To interpret the solution, let us assume for the sake of illustration that R = 1, L = 1, and 0 ≤ t < 4. In this case

.

in other words,

(19)

The graph of i(t) for 0 ≤ t < 4, given in FIGURE 4.4.5, was obtained with the help of a CAS. ≡

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

4.4.1 Derivatives of Transforms

In Problems 1–8, use Theorem 4.4.1 to evaluate the given Laplace transform.

1. {te^−10t}
2. {t³e^t}
3. {t cos 2t}
4. {t sinh 3t}
5. {t sinh t}
6. {t² cos t}
7. {te^2t sin 6t}
8. {te^−3t cos 3t}

In Problems 9–14, use the Laplace transform to solve the given initial-value problem. Use the table of Laplace transforms in Appendix C as needed.

9. y′ + y = t sin t, y(0) = 0
10. y′ − y = te^t sin t, y(0) = 0
11. y″ + 9y = cos 3t, y(0) = 2, y′(0) = 5
12. y″ + y = sin t, y(0) = 1, y′(0)= −1
13. y″ + 16y = f(t), y(0) = 0, y′(0) = 1, where
    

    

14. y″ + y = f(t), y(0) = 1, y′(0) = 0, where
    

    

In Problems 15 and 16, use a graphing utility to graph the indicated solution.

15. y(t) of Problem 13 for 0 ≤ t < 2π
16. y(t) of Problem 14 for 0 ≤ t < 3π

In Problems 17–20, use Theorem 4.4.1 to reduce the given differential to a linear first-order DE in the transformed function Solve the first-order DE and then find Use the table of Laplace transforms in Appendix C as needed. Note: Your solution may contain an arbitrary constant.

In Problems 21 and 22, proceed as in Problems 17–20 to solve the given differential equation. But in each of these problems, you will have to use Theorem 4.2.3 on to eliminate a term containing a constant c of integration.

4.4.2 Transforms of Integrals

In Problems 23–26, proceed as in Example 4 and find the convolution f * g of the given functions. After integrating find the Laplace transform of f * g.

In Problems 27–38, proceed as in Example 5 and find the Laplace transform of f * g using Theorem 4.4.2. Do not evaluate the convolution integral before transforming.

27. {1 * t³}
28. {t² * te^t}
29. {e^−t * e^t cos t}
30. {e^2t * sin t}
31. 
32. 
33. 
34. 
35. 
36. 
37. 
38. 

In Problems 39–42, use (9) to evaluate the given inverse transform.

39. ⁻¹
40. ⁻¹
41. ⁻¹
42. ⁻¹

In Problems 43 and 44, use (3) and (5) to solve the given initial-value problem. Let

43. 
44. 
45. The table in Appendix C does not contain an entry for
    

    

    (a) Use (5) along with the result in (6) to evaluate this inverse transform. Use a CAS as an aid in evaluating the convolution integral.

    (b) Reexamine your answer to part (a). Could you have obtained the result in a different manner?

46. Use the Laplace transform and the result of Problem 45 to solve the initial-value problem
    

    y″ + y = sin t + t sin t, y(0) = 0, y′(0) = 0.

    Use a graphing utility to graph the solution.

In Problems 47–58, use the Laplace transform to solve the given integral equation or integrodifferential equation.

47. 
48. 
49. 
50. 
51. 
52. 
53. 
54. 
55. 
56. 
57. 
58. 
59. Solve equation (13) subject to with
    

    

    Use a graphing utility to graph the solution for

60. Solve equation (14) subject to i(0) = 0 with
    

    

    

    Use a graphing utility to graph the solution for

In Problems 61 and 62, solve equation (15) subject to i(0) = 0 with L, R, C, and E(t) as given. Use a graphing utility to graph the solution for 0 ≤ t ≤ 3.

61. L = 0.1 h, R = 3 Ω, C = 0.05 F,
    

    E(t) = 100 [(t − 1)− (t − 2)]

62. L = 0.005 h, R = 1 Ω, C = 0.02 F,
    

    E(t) = 100 [t − (t − 1) (t − 1)]

63. The Laplace transform exists, but without finding it solve the initial-value problem
    

    

64. Solve the integral equation
    

    

4.4.3 Transform of a Periodic Function

In Problems 65–70, use Theorem 4.4.3 to find the Laplace transform of the given periodic function.

In Problems 71 and 72, solve equation (11) subject to i(0) = 0 with E(t) as given. Use a graphing utility to graph the solution for 0 ≤ t ≤ 4 in the case when L = 1 and R = 1.

71. E(t) is the meander function in Problem 65 with amplitude 1 and a = 1.
72. E(t) is the sawtooth function in Problem 67 with amplitude 1 and b = 1.

In Problems 73 and 74, solve the model for a driven spring/mass system with damping

where the driving function f is as specified. Use a graphing utility to graph x(t) for the indicated values of t.

73. m = , β = 1, k = 5, f is the meander function in Problem 65 with amplitude 10, and a = π, 0 ≤ t ≤ 2π.
74. m = 1, β = 2, k = 1, f is the square wave in Figure 4.4.4 with amplitude 5, and a = π, 0 ≤ t ≤ 4π.

Discussion Problems

75. Show how to use the Laplace transform to find the numerical value of the improper integral
76. In Problem 63 we were able to solve an initial-value problem without knowing the Laplace transform In this problem you are asked to find the actual transformed function by solving another initial-value problem.
    

    (a) If then show that y is a solution of the initial-value problem

    

    (b) Find by using the Laplace transform to solve the problem in part (a). [Hint: First find Y(0) by rereading the material on the error function in Section 2.3. Then in the solution of the resulting linear first-order DE in Y(s) integrate on the interval [0, s]. It also helps to use a dummy variable of integration.]

In Problems 77 and 78, use Theorem 4.4.1 with ,

where to find the inverse Laplace transform of the given function.

In Problems 79 and 80, use the known result

where to find the Laplace transform of the given function. The symbols a and k are positive constants.

Computer Lab Assignments

81. In this problem you are led through the commands in Mathematica that enable you to obtain the symbolic Laplace transform of a differential equation and the solution of the initial-value problem by finding the inverse transform. In Mathematica the Laplace transform of a function y(t) is obtained using LaplaceTransform [y[t], t, s]. In line two of the syntax, we replace LaplaceTransform [y[t], t, s] by the symbol Y. (If you do not have Mathematica, then adapt the given procedure by finding the corresponding syntax for the CAS you have on hand.)
    

    Consider the initial-value problem

    y″ + 6y′ + 9y = t sin t, y(0) = 2, y′(0) = –1.

    Precisely reproduce and then, in turn, execute each line in the given sequence of commands. Either copy the output by hand or print out the results.

    diffequat = y″[t] + 6y′[t] + 9y[t] == t Sin[t]

    transformdeq = LaplaceTransform [diffequat, t, s]/.

    {y[0] − > 2, y′[0] − > −1,

    LaplaceTransform [y[t], t, s] − > Y}

    soln = Solve[transformdeq, Y] // Flatten

    Y = Y/. soln

    InverseLaplaceTransform[Y, s, t]

82. Appropriately modify the procedure of Problem 81 to find a solution of
    

    y‴ + 3y′ − 4y = 0, y(0) = 0, y′(0) = 0, y″(0) = 1.

83. The charge q(t) on a capacitor in an LC-series circuit is given by
    

    + q = 1− 4 (t− π)+ 6 (t− 3π), q(0) = 0, q′(0) = 0.

    Appropriately modify the procedure of Problem 81 to find q(t). Graph your solution.