7.2 Review of Matrices
C H A P T E R
6
Many practical engineering problems involve mechanical or electrical systems acted on by discontinuous or impulsive forcing terms. For such problems the methods described in Chapter 3 are often rather awkward to use. Another method that is especially well suited to these problems, although useful much more generally, is based on the Laplace transform. In this chapter we describe how this important method works, emphasizing problems typical of those arising in engineering applications.
6.1 Definition of the Laplace Transform
Among the tools that are very useful for solving linear differential equations are integral
transforms. An integral transform is a relation of the form β
F (s) = K (s, t) f (t) dt, (1)
α
where K (s, t) is a given function, called the kernel of the transformation, and the limits of integration α and β are also given. It is possible that α = −∞ or β = ∞, or both. The relation (1) transforms the function f into another function F , which is called the transform of f . The general idea in using an integral transform to solve a differential equation is as follows: Use the relation (1) to transform a problem for an unknown function f into a simpler problem for F , then solve this simpler problem to find F , and finally recover the desired function f from its transform F . This last step is known as “inverting the transform.”
293
Chapter 6. The Laplace Transform
There are several integral transforms that are useful in applied mathematics, but in this chapter we consider only the Laplace1 transform. This transform is defined in the following way. Let f (t) be given for t ≥ 0, and suppose that f satisfies certain conditions to be stated a little later. Then the Laplace transform of f , which we will denote by L{ f (t)} or by F(s), is defined by the equation
∞
L{ f (t)} = F(s) = e−st f (t) dt.
(2)
0
The Laplace transform makes use of the kernel K (s, t) = e−st . Since the solutions of linear differential equations with constant coefficients are based on the exponential function, the Laplace transform is particularly useful for such equations.
Since the Laplace transform is defined by an integral over the range from zero to infinity, it is useful to review some basic facts about such integrals. In the first place, an integral over an unbounded interval is called an and is defined as a limit of integrals over finite intervals; thus
∞
Af (t) dt = lim f (t) dt, (3)
a
A→∞ a where A is a positive real number. If the integral from a to A exists for each A > a, and if the limit as A → ∞ exists, then the improper integral is said to converge to that limiting value. Otherwise the integral is said to or to fail to exist. The following examples illustrate both possibilities.
Let f (t) = ect , t ≥ 0, where c is a real nonzero constant. Then E X A M P L E
A
1
∞
A
ect
ect dt = lim ect dt = lim
0
A→∞ 0
A→∞ c
0
=
1
lim
(ecA − 1).
A→∞ c
It follows that the improper integral converges if c < 0, and diverges if c > 0. If c = 0, the integrand f (t) is the constant function with value 1, and the integral again diverges.
Let f (t) = 1/t, t ≥ 1. Then E X A M P L E ∞
A
2
dt =
dt
lim
= lim ln A.
1
t
A→∞ 1
t
A→∞
Since lim ln A = ∞, the improper integral diverges.
A→∞
1The Laplace transform is named for the eminent French mathematician P. S. Laplace, who studied the relation (2) in 1782. However, the techniques described in this chapter were not developed until a century or more later. They are due mainly to Oliver Heaviside (1850 –1925), an innovative but unconventional English electrical engineer, who made significant contributions to the development and application of electromagnetic theory.
6.1Definition of the Laplace Transform Let f (t) = t−p, t ≥ 1, where p is a real constant and p = 1; the case p = 1 was E X A M P L E considered in Example 2. Then
3
∞
A
1
t−p dt = lim t−p dt = lim (A1−p − 1).
1
A→∞ 1
A→∞ 1 − p
∞
As A → ∞, A1−p → 0 if p > 1, but A1−p → ∞ if p < 1. Hence t−p dt con 1
verges for p > 1, but (incorporating the result of Example 2) diverges for p ≤ 1. These
∞
results are analogous to those for the infinite series n−p.
n=1
∞
Before discussing the possible existence of f (t) dt, it is helpful to define certain a terms. A function f is said to be piecewise continuous on an interval α ≤ t ≤ β if the interval can be partitioned by a finite number of points α = t < t < · · · < t = β so
0
1
n that
1.
f is continuous on each open subinterval t< t < t .
i −1
i
2.
f approaches a finite limit as the endpoints of each subinterval are approached from within the subinterval.
In other words, f is piecewise continuous on α ≤ t ≤ β if it is continuous there except for a finite number of jump discontinuities. If f is piecewise continuous on α ≤ t ≤ β
for every β > α, then f is said to be piecewise continuous on t ≥ α. An example of a piecewise continuous function is shown in Figure 6.1.1.
If f is piecewise continuous on the interval a ≤ t ≤ A, then it can be shown that
A
Af (t) dt exists. Hence, if f is piecewise continuous for t ≥ a, then f (t) dt exists a
a for each A > a. However, piecewise continuity is not enough to ensure convergence
∞
of the improper integral f (t) dt, as the preceding examples show.
a If f cannot be integrated easily in terms of elementary functions, the definition of
∞
convergence of f (t) dt may be difficult to apply. Frequently, the most convenient a way to test the convergence or divergence of an improper integral is by the following comparison theorem, which is analogous to a similar theorem for infinite series.
y
α
t
t
1
t2
β
FIGURE 6.1.1
A piecewise continuous function.
Chapter 6. The Laplace Transform
Theorem 6.1.1
If f is piecewise continuous for t ≥ a, if | f (t)| ≤ g(t) when t ≥ M for some positive
∞
∞
constant M, and if g(t) dt converges, then f (t) dt also converges. On
M
a
∞
the other hand, if f (t) ≥ g(t) ≥ 0 for t ≥ M, and if g(t) dt diverges, then
M ∞ f(t)dt alsodiverges.
a The proof of this result from the calculus will not be given here. It is made plausible,
∞
∞
however, by comparing the areas represented by g(t) dt and | f (t)| dt. The
M
M functions most useful for comparison purposes are ect and t−p, which were considered in Examples 1, 2, and 3.
We now return to a consideration of the Laplace transform L{ f (t)} or F(s), which is defined by Eq. (2) whenever this improper integral converges. In general, the parameter s may be complex, but for our discussion we need consider only real values of s. The foregoing discussion of integrals indicates that the Laplace transform F of a function f exists if f satisfies certain conditions, such as those stated in the following theorem.
Theorem 6.1.2
Suppose that
1.
f is piecewise continuous on the interval 0 ≤ t ≤ A for any positive A.
2.
| f (t)| ≤ K eat when t ≥ M. In this inequality K , a, and M are real constants, K
and M necessarily positive.
Then the Laplace transform L{ f (t)} = F(s), defined by Eq. (2), exists for s > a.
To establish this theorem it is necessary to show only that the integral in Eq. (2)
converges for s > a. Splitting the improper integral into two parts, we have
∞
M
∞
e−st f (t) dt = e−st f (t) dt + e−st f (t) dt.
(4)
0
0
M
The first integral on the right side of Eq. (4) exists by hypothesis (1) of the theorem; hence the existence of F (s) depends on the convergence of the second integral. By hypothesis (2) we have, for t ≥ M, |e−st f (t)| ≤ K e−steat = K e(a−s)t,
∞
and thus, by Theorem 6.1.1, F (s) exists provided that e(a−s)t dt converges. Refer M ring to Example 1 with c replaced by a − s, we see that this latter integral converges when a − s < 0, which establishes Theorem 6.1.2.
Unless the contrary is specifically stated, in this chapter we deal only with functions satisfying the conditions of Theorem 6.1.2. Such functions are described as piecewise continuous, and of exponential order as t → ∞. The Laplace transforms of some important elementary functions are given in the following examples.
6.1Definition of the Laplace Transform Let f (t) = 1, t ≥ 0. Then E X A M P L E ∞
4
L{1} = e−st dt = 1 ,s > 0.
0
s
Let f (t) = eat , t ≥ 0. Then E X A M P L E ∞
∞
5
L{eat} = e−st eat dt = e−(s−a)t dt
0
0
= 1 ,s > a.
s − a Let f (t) = sin at, t ≥ 0. Then E X A M P L E ∞
6
L{sin at} = F(s) = e−st sin at dt,s > 0.
0
Since
AF (s) = lim e−st sin at dt,
A→∞ 0 upon integrating by parts we obtain
A
AF (s) = lim −e−st cos at − se−st cos at dt
A→∞
a
a
0
0
∞
= 1 − se−st cos at dt.
aa 0 A second integration by parts then yields ∞ F (s) = 1 − s2 e−st sin at dtaa2 0 = 1 − s2 F(s).
a
a2
Hence, solving for F (s), we have F (s) = a
,
s > 0.
s2 + a2
Now let us suppose that f and f are two functions whose Laplace transforms exist
1
2
for s > a and s > a , respectively. Then, for s greater than the maximum of a and a ,
1
2
1
2
∞
L{c f (t) + c f (t)} = e−st [c f (t) + c f (t)] dt 1 1
2 2
1 1
2 2
0 ∞
∞
= ce−st f (t) dt + ce−st f (t) dt;
1
1
2
2
0
0
Chapter 6. The Laplace Transform hence
L{c f (t) + c f (t)} = c L{ f (t)} + c L{ f (t)}.
(5)
1 1
2 2
1
1
2
2
Equation (5) is a statement of the fact that the Laplace transform is a linear operator.
This property is of paramount importance, and we make frequent use of it later.
PROBLEMS
In each of Problems 1 through 4 sketch the graph of the given function. In each case determine whether f is continuous, piecewise continuous, or neither on the interval 0 ≤ t ≤ 3.
1.
2.
3.
4.
5. Find the Laplace transform of each of the following functions:
In each of Problems 21 through 24 determine whether the given integral converges or diverges.
∞
∞
21.
22.
0
0
∞
∞
23.
24.
1
0
true under less restrictive conditions, such as those of Theorem 6.1.2.
integral
∞
(i)
0
6.2Solution of Initial Value Problems
√
2
2
2
27. Consider the Laplace transform of t p, where p > −1.
(a) Referring to Problem 26, show that
∞
∞
1
0
(c) Show that
∞
√
It is possible to show that
∞
;
0
2
hence
(d) Show that
√
6.2 Solution of Initial Value Problems
In this section we show how the Laplace transform can be used to solve initial value problems for linear differential equations with constant coefficients. The usefulness of the Laplace transform in this connection rests primarily on the fact that the transform Chapter 6. The Laplace Transform of f is related in a simple way to the transform of f . The relationship is expressed in the following theorem.
Theorem 6.2.1 Suppose that f is continuous and f is piecewise continuous on any interval 0 ≤ t ≤ A. Suppose further that there exist constants K , a, and M such that | f (t)| ≤ K eat for t ≥ M. Then L{ f (t)} exists for s > a, and moreover L{ f (t)} = sL{ f (t)} − f (0).
(1)
To prove this theorem we consider the integral
Ae−st f (t) dt.
0
If f has points of discontinuity in the interval 0 ≤ t ≤ A, let them be denoted by t , t , . . . , t . Then we can write this integral as
1
2
n
A
t
t
A
1
2
e−st f (t) dt = e−st f (t) dt + e−st f (t) dt + · · · + e−st f (t) dt.
0
0
t
t
1
n
Integrating each term on the right by parts yields
A
t
t
A
1
2
e−st f (t) dt = e−st f (t) + e−st f (t) + · · · + e−st f (t)
0
0
t
t
1
n
t
t
A
1
2
+ se−st f (t) dt + e−st f (t) dt + · · · + e−st f (t) dt .
0
t
t
1
n
Since f is continuous, the contributions of the integrated terms at t , t , . . . , t cancel.
1
2
n
Combining the integrals gives
A
Ae−st f (t) dt = e−s A f (A) − f (0) + se−st f (t) dt.
0
0
As A → ∞, e−s A f (A) → 0 whenever s > a. Hence, for s > a, L{ f (t)} = sL{ f (t)} − f (0), which establishes the theorem.
If f and f satisfy the same conditions that are imposed on f and f , respectively, in Theorem 6.2.1, then it follows that the Laplace transform of f also exists for s > a
and is given by L{ f (t)} = s2L{ f (t)} − s f (0) − f (0).
(2)
Indeed, provided the function f and its derivatives satisfy suitable conditions, an expression for the transform of the nth derivative f (n) can be derived by successive applications of this theorem. The result is given in the following corollary.
Corollary 6.2.2 Suppose that the functions f, f , . . . , f (n−1) are continuous, and that f (n) is piecewise continuous on any interval 0 ≤ t ≤ A. Suppose further that there exist constants K , a, and M such that | f (t)| ≤ K eat , | f (t)| ≤ K eat , . . . , | f (n−1)(t)| ≤ K eat for t ≥ M.
Then L{ f (n)(t)} exists for s > a and is given by L{ f (n)(t)} = snL{ f (t)} − sn−1 f (0) − · · · − s f (n−2)(0) − f (n−1)(0).
(3)
We now show how the Laplace transform can be used to solve initial value problems.
It is most useful for problems involving nonhomogeneous differential equations, as we will demonstrate in later sections of this chapter. However, we begin by looking at some homogeneous equations, which are a bit simpler. For example, consider the differential equation y − y − 2y = 0 (4)
and the initial conditions y(0) = 1,y(0) = 0.
(5)
This problem is easily solved by the methods of Section 3.1. The characteristic equation is r 2 − r − 2 = (r − 2)(r + 1) = 0, (6)
and consequently the general solution of Eq. (4) is y = c e−t + c e2t .
(7)
1
2
To satisfy the initial conditions (5) we must have c + c = 1 and −c + 2c = 0;
1
2
1
2
hence c = 2 and c = 1 , so that the solution of the initial value problem (4) and (5) is
1
3
2
3
y = φ(t) = 2 e−t + 1 e2t .
(8)
3
3
Now let us solve the same problem by using the Laplace transform. To do this we must assume that the problem has a solution y = φ(t), which with its first two derivatives satisfies the conditions of Corollary 6.2.2. Then, taking the Laplace transform of the differential equation (4), we obtain L{y} − L{y} − 2L{y} = 0, (9)
where we have used the linearity of the transform to write the transform of a sum as the sum of the separate transforms. Upon using the corollary to express L{y} and L{y} in terms of L{y}, we find that Eq. (9) becomes s2L{y} − sy(0) − y(0) − [sL{y} − y(0)] − 2L{y} = 0, or (s2 − s − 2)Y (s) + (1 − s)y(0) − y(0) = 0, (10)
where Y (s) = L{y}. Substituting for y(0) and y(0) in Eq. (10) from the initial conditions (5), and then solving for Y (s), we obtain Y (s) = s − 1
=
s − 1 .
(11)
s2 − s − 2 (s − 2)(s + 1)
We have thus obtained an expression for the Laplace transform Y (s) of the solution y = φ(t) of the given initial value problem. To determine the function φ we must find the function whose Laplace transform is Y (s), as given by Eq. (11).
Chapter 6. The Laplace Transform
This can be done most easily by expanding the right side of Eq. (11) in partial fractions. Thus we write Y (s) = s − 1 = a + b = a(s + 1) + b(s − 2),
(
(12) s − 2)(s + 1)s − 2 s + 1 (s − 2)(s + 1) where the coefficients a and b are to be determined. By equating numerators of the second and fourth members of Eq. (12), we obtain s − 1 = a(s + 1) + b(s − 2), an equation that must hold for all s. In particular, if we set s = 2, then it follows that a = 1 . Similarly, if we set s = −1, then we find that b = 2 . By substituting these
3
3
values for a and b, respectively, we have Y (s) = 1/3 + 2/3 .
(13)
s − 2 s + 1
Finally, if we use the result of Example 5 of Section 6.1, it follows that 1 e2t has the
3
transform 1 (s − 2)−1; similarly, 2 e−t has the transform 2 (s + 1)−1. Hence, by the
3
3
3
linearity of the Laplace transform, y = φ(t) = 1 e2t + 2 e−t
3
3
has the transform (13) and is therefore the solution of the initial value Of course, this is the same solution that we obtained earlier.
The same procedure can be applied to the general second order linear equation with constant coefficients, ay + by + cy = f (t).
(14)
Assuming that the solution y = φ(t) satisfies the conditions of Corollary 6.2.2 for n = 2, we can take the transform of Eq. (14) and thereby obtain a[s2Y (s) − sy(0) − y(0)] + b[sY (s) − y(0)] + cY (s) = F(s), (15)
where F (s) is the transform of f (t). By solving Eq. (15) for Y (s) we find that (as + b)y(0) + ay(0)Y (s) =
+
F (s).
(16)
as2 + bs + cas2 + bs + c
The problem is then solved, provided that we can find the function y = φ(t) whose transform is Y (s).
Even at this early stage of our discussion we can point out some of the essential features of the transform method. In the first place, the transform Y (s) of the unknown function y = φ(t) is found by solving an algebraic equation rather than a differential
equation, Eq. (10) rather than Eq. (4), or in general Eq. (15) rather than Eq. (14). This is the key to the usefulness of Laplace transforms for solving linear, constant coefficient, ordinary differential equations—the problem is reduced from a differential equation to an algebraic one. Next, the solution satisfying given initial conditions is automatically found, so that the task of determining appropriate values for the arbitrary constants in the general solution does not arise. Further, as indicated in Eq. (15), nonhomogeneous equations are handled in exactly the same way as homogeneous ones; it is not necessary to solve the corresponding homogeneous equation first. Finally, the method can be applied in the same way to higher order equations, as long as we assume that the solution satisfies the conditions of the corollary for the appropriate value of n.
6.2Solution of Initial Value Problems
Observe that the polynomial as2 + bs + c in the denominator on the right side of Eq. (16) is precisely the characteristic polynomial associated with Eq. (14). Since the use of a partial fraction expansion of Y (s) to determine φ(t) requires us to factor this polynomial, the use of Laplace transforms does not avoid the necessity of finding roots of the characteristic equation. For equations of higher than second order this may be a difficult algebraic problem, particularly if the roots are irrational or complex.
The main difficulty that occurs in solving initial value problems by the transform technique lies in the problem of determining the function y = φ(t) corresponding to the transform Y (s). This problem is known as the inversion problem for the Laplace transform; φ(t) is called the inverse transform corresponding to Y (s), and the process of finding φ(t) from Y (s) is known as inverting the transform. We also use the notation L−1{Y (s)} to denote the inverse transform of Y (s). There is a general formula for the inverse Laplace transform, but its use requires a knowledge of the theory of functions of a complex variable, and we do not consider it in this book. However, it is still possible to develop many important properties of the Laplace transform, and to solve many interesting problems, without the use of complex variables.
In solving the initial value problem (4), (5) we did not consider the question of whether there may be functions other than the one given by Eq. (8) that also have the transform (13). In fact, it can be shown that if f is a continuous function with the Laplace transform F , then there is no other continuous function having the same transform. In other words, there is essentially a one-to-one correspondence between functions and their Laplace transforms. This fact suggests the compilation of a table, such as Table 6.2.1, giving the transforms of functions frequently encountered, and vice versa. The entries in the second column of Table 6.2.1 are the transforms of those in the first column. Perhaps more important, the functions in the first column are the inverse transforms of those in the second column. Thus, for example, if the transform of the solution of a differential equation is known, the solution itself can often be found merely by looking it up in the table. Some of the entries in Table 6.2.1 have been used as examples, or appear as problems in Section 6.1, while others will be developed later in the chapter. The third column of the table indicates where the derivation of the given transforms may be found. While Table 6.2.1 is sufficient for the examples and problems in this book, much larger tables are also available (see the list of references at the end of the chapter). Transforms and inverse transforms can also be readily obtained electronically by using a computer algebra system.
Frequently, a Laplace transform F (s) is expressible as a sum of several terms, F (s) = F (s) + F (s) + · · · + F (s).
(17)
1
2
n
Suppose that f (t) = L−1{F (s)}, . . . , f (t) = L−1{F (s)}. Then the function
1
1
n
n
f (t) = f (t) + · · · + f (t)
1
n
has the Laplace transform F (s). By the uniqueness property stated previously there is no other continuous function f having the same transform. Thus L−1{F(s)} = L−1{F (s)} + · · · + L−1{F (s)};
(18)
1
n
that is, the inverse Laplace transform is also a linear operator.
In many problems it is convenient to make use of this property by decomposing a given transform into a sum of functions whose inverse transforms are already known or can be found in the table. Partial fraction expansions are particularly useful in this Chapter 6. The Laplace TransformTABLE 6.2.1 Elementary Laplace Transforms f (t) = L−1{F(s)} F(s) = L{ f (t)}
Notes
1
1. 1
,s > 0 Sec. 6.1; Ex. 4
s
1
2. eat,
s > a Sec. 6.1; Ex. 5
s − an!
3. tn; n = positive integer ,s > 0 Sec. 6.1; Prob. 27
sn+1
(p + 1) 4. t p, p > −1 ,s > 0 Sec. 6.1; Prob. 27
s p+1
a 5. sin at,s > 0 Sec. 6.1; Ex. 6
s2 + a2
s
6. cos at,s > 0 Sec. 6.1; Prob. 6
s2 + a2
a 7. sinh at,s > |a| Sec. 6.1; Prob. 8
s2 − a2
s
8. cosh at,s > |a| Sec. 6.1; Prob. 7
s2 − a2
b
9. eat sin bt,
s > a Sec. 6.1; Prob. 13
(s − a)2 + b2 s − a 10. eat cos bt,
s > a Sec. 6.1; Prob. 14
(s − a)2 + b2 n!
11. tneat , n = positive integer ,
s > a Sec. 6.1; Prob. 18
(s − a)n+1 12. u (t)
e−cs ,s > 0 Sec. 6.3
c
s
13. u (t) f (t − c)e−cs F(s) Sec. 6.3
c 14. ect f (t)F(s − c) Sec. 6.3
s
15. f (ct) 1 F,c > 0 Sec. 6.3; Prob. 19
c
c
t
16.
f (t − τ)g(τ ) dτF(s)G(s) Sec. 6.6
0
17. δ(t − c)
e−cs Sec. 6.5
18. f (n)(t)sn F(s) − sn−1 f (0) − · · · − f (n−1)(0) Sec. 6.2
19. (−t)n f (t)F(n)(s) Sec. 6.2; Prob. 28
6.2Solution of Initial Value Problems connection, and a general result covering many cases is given in Problem 38. Other useful properties of Laplace transforms are derived later in this chapter.
As further illustrations of the technique of solving initial value problems by means of the Laplace transform and partial fraction expansions, consider the following examples.
Find the solution of the differential equation E X A M P L E
1
y + y = sin 2t,
(19)
satisfying the initial conditions y(0) = 2,y(0) = 1.
(20)
We assume that this initial value problem has a solution y = φ(t), which with its first two derivatives satisfies the conditions of Corollary 6.2.2. Then, taking the Laplace transform of the differential equation, we have s2Y (s) − sy(0) − y(0) + Y (s) = 2/(s2 + 4), where the transform of sin 2t has been obtained from line 5 of Table 6.2.1. Substituting for y(0) and y(0) from the initial conditions and solving for Y (s), we obtain Y (s) = 2s3 + s2 + 8s + 6 .
(21) (s2 + 1)(s2 + 4)
Using partial fractions we can write Y (s) in the form (as + b)(s2 + 4) + (cs + d)(s2 + 1)Y (s) = as + b + cs + d = .
(22)
s2 + 1 s2 + 4 (s2 + 1)(s2 + 4)
By expanding the numerator on the right side of Eq. (22) and equating it to the numerator in Eq. (21) we find that 2s3 + s2 + 8s + 6 = (a + c)s3 + (b + d)s2 + (4a + c)s + (4b + d) for all s. Then, comparing coefficients of like powers of s, we have a + c = 2,b + d = 1, 4a + c = 8, 4b + d = 6.
Consequently, a = 2, c = 0, b = 5 , and d = − 2 , from which it follows that
3
3
Y (s) =
2s + 5/3 − 2/3 .
(23)
s2 + 1 s2 + 1 s2 + 4
From lines 5 and 6 of Table 6.2.1, the solution of the given initial value problem is y = φ(t) = 2 cos t + 5 sin t − 1 sin 2t.
(24)
3
3
Find the solution of the initial value problem E X A M P L E
2
yiv − y = 0, (25) y(0) = 0,y(0) = 1,y(0) = 0,y(0) = 0.
(26)
Chapter 6. The Laplace Transform
In this problem we need to assume that the solution y = φ(t) satisfies the conditions of Corollary 6.2.2 for n = 4. The Laplace transform of the differential equation (25) is s4Y (s) − s3 y(0) − s2 y(0) − sy(0) − y(0) − Y (s) = 0.
Then, using the initial conditions (26) and solving for Y (s), we have Y (s) =
s2
.
(27)
s4 − 1
A partial fraction expansion of Y (s) is Y (s) = as + b + cs + d ,s2 − 1 s2 + 1 and it follows that (as + b)(s2 + 1) + (cs + d)(s2 − 1) = s2 (28)
for all s. By setting s = 1 and s = −1, respectively, in Eq. (28) we obtain the pair of equations 2(a + b) = 1, 2(−a + b) = 1, and therefore a = 0 and b = 1 . If we set s = 0 in Eq. (28), then b − d = 0, so d = 1 .
2
2
Finally, equating the coefficients of the cubic terms on each side of Eq. (28), we find that a + c = 0, so c = 0. Thus Y (s) = 1/2 + 1/2 , (29)
s2 − 1 s2 + 1 and from lines 7 and 5 of Table 6.2.1 the solution of the initial value (26) is y = φ(t) = sinh t + sin t .
(30)
2
The most important elementary applications of the Laplace transform are in the study of mechanical vibrations and in the analysis of electric circuits; the governing equations were derived in Section 3.8. A vibrating spring–mass system has the equation of motion
d2u
m
+ γ du + ku = F(t), (31)
dt2
dt
where m is the mass, γ the damping coefficient, k the spring constant, and F(t)
the applied external force. The equation describing an electric circuit containing an inductance L, a resistance R, and a capacitance C (an LRC circuit) is d2 Qd Q
L + R + 1 Q = E(t), (32)
dt2
dt
C where Q(t) is the charge on the capacitor and E(t) is the applied voltage. In terms of the current I (t) = d Q(t)/dt we can differentiate Eq. (32) and write d2 Id I
L + R + 1 I = d E (t).
(33)
dt2
dt
C
dt
Suitable initial conditions on u, Q, or I must also be prescribed.
6.2Solution of Initial Value Problems
We have noted previously in Section 3.8 that Eq. (31) for the spring–mass system and Eq. (32) or (33) for the electric circuit are identical mathematically, differing only in the interpretation of the constants and variables appearing in them. There are other physical problems that also lead to the same differential equation. Thus, once the mathematical problem is solved, its solution can be interpreted in terms of whichever corresponding physical problem is of immediate interest.
In the problem lists following this and other sections in this chapter are numerous initial value problems for second order linear differential equations with constant coefficients. Many can be interpreted as models of particular physical systems, but usually we do not point this out explicitly.
PROBLEMS
In each of Problems 1 through 10 find the inverse Laplace transform of the given function.
3
4
1.
2.
2
3.
4.
5.
6.
7.
8.
9.
10.
Chapter 6. The Laplace Transform
27. The Laplace transforms of certain functions can be found conveniently from their Taylor series expansions.
∞
,
1
,
(b) Let
0
Section 5.8)
∞
.
0
Assuming that the following Laplace transforms can be computed term by term, verify that
0
and
√
0
Problems 28 through 36 are concerned with differentiation of the Laplace transform.
28. Let
∞
0
35. Consider Bessel’s equation of order zero 6.2Solution of Initial Value Problems
permissible to take the inverse transform term by term, show that
∞
0
0
0
0
36. For each of the following initial value problems use the results of Problem 28 to find the
37. Suppose that
tg(t) = f (τ ) dτ.
0
If G(s) and F(s) are the Laplace transforms of g(t) and f (t), respectively, show that G(s) = F(s)/s.
38. In this problem we show how a general partial fraction expansion can be used to calculate many inverse Laplace transforms. Suppose that F(s) = P(s)/Q(s), where Q(s) is a polynomial of degree n with distinct zeros r , . . . , r and P(s) is a
1
n
polynomial of degree less than n. In this case it is possible to show that P(s)/Q(s) has a partial fraction expansion of the form P(s)
A
A
=
1
+ · · · +
n
,
(i) Q(s)s − rs − r
1
n
where the coefficients A , . . . , A must be determined.
1
n (a) Show that
(ii)
k
k
k
Hint: One way to do this is to multiply Eq. (i) by s − r and then to take the limit as s → r .
k
k
6
Many practical engineering problems involve mechanical or electrical systems acted on by discontinuous or impulsive forcing terms. For such problems the methods described in Chapter 3 are often rather awkward to use. Another method that is especially well suited to these problems, although useful much more generally, is based on the Laplace transform. In this chapter we describe how this important method works, emphasizing problems typical of those arising in engineering applications.
6.1 Definition of the Laplace Transform
Among the tools that are very useful for solving linear differential equations are integral
transforms. An integral transform is a relation of the form β
F (s) = K (s, t) f (t) dt, (1)
α
where K (s, t) is a given function, called the kernel of the transformation, and the limits of integration α and β are also given. It is possible that α = −∞ or β = ∞, or both. The relation (1) transforms the function f into another function F , which is called the transform of f . The general idea in using an integral transform to solve a differential equation is as follows: Use the relation (1) to transform a problem for an unknown function f into a simpler problem for F , then solve this simpler problem to find F , and finally recover the desired function f from its transform F . This last step is known as “inverting the transform.”
293
Chapter 6. The Laplace Transform
There are several integral transforms that are useful in applied mathematics, but in this chapter we consider only the Laplace1 transform. This transform is defined in the following way. Let f (t) be given for t ≥ 0, and suppose that f satisfies certain conditions to be stated a little later. Then the Laplace transform of f , which we will denote by L{ f (t)} or by F(s), is defined by the equation
∞
L{ f (t)} = F(s) = e−st f (t) dt.
(2)
0
The Laplace transform makes use of the kernel K (s, t) = e−st . Since the solutions of linear differential equations with constant coefficients are based on the exponential function, the Laplace transform is particularly useful for such equations.
Since the Laplace transform is defined by an integral over the range from zero to infinity, it is useful to review some basic facts about such integrals. In the first place, an integral over an unbounded interval is called an and is defined as a limit of integrals over finite intervals; thus
∞
Af (t) dt = lim f (t) dt, (3)
a
A→∞ a where A is a positive real number. If the integral from a to A exists for each A > a, and if the limit as A → ∞ exists, then the improper integral is said to converge to that limiting value. Otherwise the integral is said to or to fail to exist. The following examples illustrate both possibilities.
Let f (t) = ect , t ≥ 0, where c is a real nonzero constant. Then E X A M P L E
A
1
∞
A
ect
ect dt = lim ect dt = lim
0
A→∞ 0
A→∞ c
0
=
1
lim
(ecA − 1).
A→∞ c
It follows that the improper integral converges if c < 0, and diverges if c > 0. If c = 0, the integrand f (t) is the constant function with value 1, and the integral again diverges.
Let f (t) = 1/t, t ≥ 1. Then E X A M P L E ∞
A
2
dt =
dt
lim
= lim ln A.
1
t
A→∞ 1
t
A→∞
Since lim ln A = ∞, the improper integral diverges.
A→∞
1The Laplace transform is named for the eminent French mathematician P. S. Laplace, who studied the relation (2) in 1782. However, the techniques described in this chapter were not developed until a century or more later. They are due mainly to Oliver Heaviside (1850 –1925), an innovative but unconventional English electrical engineer, who made significant contributions to the development and application of electromagnetic theory.
6.1Definition of the Laplace Transform Let f (t) = t−p, t ≥ 1, where p is a real constant and p = 1; the case p = 1 was E X A M P L E considered in Example 2. Then
3
∞
A
1
t−p dt = lim t−p dt = lim (A1−p − 1).
1
A→∞ 1
A→∞ 1 − p
∞
As A → ∞, A1−p → 0 if p > 1, but A1−p → ∞ if p < 1. Hence t−p dt con 1
verges for p > 1, but (incorporating the result of Example 2) diverges for p ≤ 1. These
∞
results are analogous to those for the infinite series n−p.
n=1
∞
Before discussing the possible existence of f (t) dt, it is helpful to define certain a terms. A function f is said to be piecewise continuous on an interval α ≤ t ≤ β if the interval can be partitioned by a finite number of points α = t < t < · · · < t = β so
0
1
n that
1.
f is continuous on each open subinterval t< t < t .
i −1
i
2.
f approaches a finite limit as the endpoints of each subinterval are approached from within the subinterval.
In other words, f is piecewise continuous on α ≤ t ≤ β if it is continuous there except for a finite number of jump discontinuities. If f is piecewise continuous on α ≤ t ≤ β
for every β > α, then f is said to be piecewise continuous on t ≥ α. An example of a piecewise continuous function is shown in Figure 6.1.1.
If f is piecewise continuous on the interval a ≤ t ≤ A, then it can be shown that
A
Af (t) dt exists. Hence, if f is piecewise continuous for t ≥ a, then f (t) dt exists a
a for each A > a. However, piecewise continuity is not enough to ensure convergence
∞
of the improper integral f (t) dt, as the preceding examples show.
a If f cannot be integrated easily in terms of elementary functions, the definition of
∞
convergence of f (t) dt may be difficult to apply. Frequently, the most convenient a way to test the convergence or divergence of an improper integral is by the following comparison theorem, which is analogous to a similar theorem for infinite series.
y
α
t
t
1
t2
β
FIGURE 6.1.1
A piecewise continuous function.
Chapter 6. The Laplace Transform
Theorem 6.1.1
If f is piecewise continuous for t ≥ a, if | f (t)| ≤ g(t) when t ≥ M for some positive
∞
∞
constant M, and if g(t) dt converges, then f (t) dt also converges. On
M
a
∞
the other hand, if f (t) ≥ g(t) ≥ 0 for t ≥ M, and if g(t) dt diverges, then
M ∞ f(t)dt alsodiverges.
a The proof of this result from the calculus will not be given here. It is made plausible,
∞
∞
however, by comparing the areas represented by g(t) dt and | f (t)| dt. The
M
M functions most useful for comparison purposes are ect and t−p, which were considered in Examples 1, 2, and 3.
We now return to a consideration of the Laplace transform L{ f (t)} or F(s), which is defined by Eq. (2) whenever this improper integral converges. In general, the parameter s may be complex, but for our discussion we need consider only real values of s. The foregoing discussion of integrals indicates that the Laplace transform F of a function f exists if f satisfies certain conditions, such as those stated in the following theorem.
Theorem 6.1.2
Suppose that
1.
f is piecewise continuous on the interval 0 ≤ t ≤ A for any positive A.
2.
| f (t)| ≤ K eat when t ≥ M. In this inequality K , a, and M are real constants, K
and M necessarily positive.
Then the Laplace transform L{ f (t)} = F(s), defined by Eq. (2), exists for s > a.
To establish this theorem it is necessary to show only that the integral in Eq. (2)
converges for s > a. Splitting the improper integral into two parts, we have
∞
M
∞
e−st f (t) dt = e−st f (t) dt + e−st f (t) dt.
(4)
0
0
M
The first integral on the right side of Eq. (4) exists by hypothesis (1) of the theorem; hence the existence of F (s) depends on the convergence of the second integral. By hypothesis (2) we have, for t ≥ M, |e−st f (t)| ≤ K e−steat = K e(a−s)t,
∞
and thus, by Theorem 6.1.1, F (s) exists provided that e(a−s)t dt converges. Refer M ring to Example 1 with c replaced by a − s, we see that this latter integral converges when a − s < 0, which establishes Theorem 6.1.2.
Unless the contrary is specifically stated, in this chapter we deal only with functions satisfying the conditions of Theorem 6.1.2. Such functions are described as piecewise continuous, and of exponential order as t → ∞. The Laplace transforms of some important elementary functions are given in the following examples.
6.1Definition of the Laplace Transform Let f (t) = 1, t ≥ 0. Then E X A M P L E ∞
4
L{1} = e−st dt = 1 ,s > 0.
0
s
Let f (t) = eat , t ≥ 0. Then E X A M P L E ∞
∞
5
L{eat} = e−st eat dt = e−(s−a)t dt
0
0
= 1 ,s > a.
s − a Let f (t) = sin at, t ≥ 0. Then E X A M P L E ∞
6
L{sin at} = F(s) = e−st sin at dt,s > 0.
0
Since
AF (s) = lim e−st sin at dt,
A→∞ 0 upon integrating by parts we obtain
A
AF (s) = lim −e−st cos at − se−st cos at dt
A→∞
a
a
0
0
∞
= 1 − se−st cos at dt.
aa 0 A second integration by parts then yields ∞ F (s) = 1 − s2 e−st sin at dtaa2 0 = 1 − s2 F(s).
a
a2
Hence, solving for F (s), we have F (s) = a
,
s > 0.
s2 + a2
Now let us suppose that f and f are two functions whose Laplace transforms exist
1
2
for s > a and s > a , respectively. Then, for s greater than the maximum of a and a ,
1
2
1
2
∞
L{c f (t) + c f (t)} = e−st [c f (t) + c f (t)] dt 1 1
2 2
1 1
2 2
0 ∞
∞
= ce−st f (t) dt + ce−st f (t) dt;
1
1
2
2
0
0
Chapter 6. The Laplace Transform hence
L{c f (t) + c f (t)} = c L{ f (t)} + c L{ f (t)}.
(5)
1 1
2 2
1
1
2
2
Equation (5) is a statement of the fact that the Laplace transform is a linear operator.
This property is of paramount importance, and we make frequent use of it later.
PROBLEMS
In each of Problems 1 through 4 sketch the graph of the given function. In each case determine whether f is continuous, piecewise continuous, or neither on the interval 0 ≤ t ≤ 3.
1.
2.
3.
4.
5. Find the Laplace transform of each of the following functions:
In each of Problems 21 through 24 determine whether the given integral converges or diverges.
∞
∞
21.
22.
0
0
∞
∞
23.
24.
1
0
true under less restrictive conditions, such as those of Theorem 6.1.2.
integral
∞
(i)
0
6.2Solution of Initial Value Problems
√
2
2
2
27. Consider the Laplace transform of t p, where p > −1.
(a) Referring to Problem 26, show that
∞
∞
1
0
(c) Show that
∞
√
It is possible to show that
∞
;
0
2
hence
(d) Show that
√
6.2 Solution of Initial Value Problems
In this section we show how the Laplace transform can be used to solve initial value problems for linear differential equations with constant coefficients. The usefulness of the Laplace transform in this connection rests primarily on the fact that the transform Chapter 6. The Laplace Transform of f is related in a simple way to the transform of f . The relationship is expressed in the following theorem.
Theorem 6.2.1 Suppose that f is continuous and f is piecewise continuous on any interval 0 ≤ t ≤ A. Suppose further that there exist constants K , a, and M such that | f (t)| ≤ K eat for t ≥ M. Then L{ f (t)} exists for s > a, and moreover L{ f (t)} = sL{ f (t)} − f (0).
(1)
To prove this theorem we consider the integral
Ae−st f (t) dt.
0
If f has points of discontinuity in the interval 0 ≤ t ≤ A, let them be denoted by t , t , . . . , t . Then we can write this integral as
1
2
n
A
t
t
A
1
2
e−st f (t) dt = e−st f (t) dt + e−st f (t) dt + · · · + e−st f (t) dt.
0
0
t
t
1
n
Integrating each term on the right by parts yields
A
t
t
A
1
2
e−st f (t) dt = e−st f (t) + e−st f (t) + · · · + e−st f (t)
0
0
t
t
1
n
t
t
A
1
2
+ se−st f (t) dt + e−st f (t) dt + · · · + e−st f (t) dt .
0
t
t
1
n
Since f is continuous, the contributions of the integrated terms at t , t , . . . , t cancel.
1
2
n
Combining the integrals gives
A
Ae−st f (t) dt = e−s A f (A) − f (0) + se−st f (t) dt.
0
0
As A → ∞, e−s A f (A) → 0 whenever s > a. Hence, for s > a, L{ f (t)} = sL{ f (t)} − f (0), which establishes the theorem.
If f and f satisfy the same conditions that are imposed on f and f , respectively, in Theorem 6.2.1, then it follows that the Laplace transform of f also exists for s > a
and is given by L{ f (t)} = s2L{ f (t)} − s f (0) − f (0).
(2)
Indeed, provided the function f and its derivatives satisfy suitable conditions, an expression for the transform of the nth derivative f (n) can be derived by successive applications of this theorem. The result is given in the following corollary.
Corollary 6.2.2 Suppose that the functions f, f , . . . , f (n−1) are continuous, and that f (n) is piecewise continuous on any interval 0 ≤ t ≤ A. Suppose further that there exist constants K , a, and M such that | f (t)| ≤ K eat , | f (t)| ≤ K eat , . . . , | f (n−1)(t)| ≤ K eat for t ≥ M.
Then L{ f (n)(t)} exists for s > a and is given by L{ f (n)(t)} = snL{ f (t)} − sn−1 f (0) − · · · − s f (n−2)(0) − f (n−1)(0).
(3)
We now show how the Laplace transform can be used to solve initial value problems.
It is most useful for problems involving nonhomogeneous differential equations, as we will demonstrate in later sections of this chapter. However, we begin by looking at some homogeneous equations, which are a bit simpler. For example, consider the differential equation y − y − 2y = 0 (4)
and the initial conditions y(0) = 1,y(0) = 0.
(5)
This problem is easily solved by the methods of Section 3.1. The characteristic equation is r 2 − r − 2 = (r − 2)(r + 1) = 0, (6)
and consequently the general solution of Eq. (4) is y = c e−t + c e2t .
(7)
1
2
To satisfy the initial conditions (5) we must have c + c = 1 and −c + 2c = 0;
1
2
1
2
hence c = 2 and c = 1 , so that the solution of the initial value problem (4) and (5) is
1
3
2
3
y = φ(t) = 2 e−t + 1 e2t .
(8)
3
3
Now let us solve the same problem by using the Laplace transform. To do this we must assume that the problem has a solution y = φ(t), which with its first two derivatives satisfies the conditions of Corollary 6.2.2. Then, taking the Laplace transform of the differential equation (4), we obtain L{y} − L{y} − 2L{y} = 0, (9)
where we have used the linearity of the transform to write the transform of a sum as the sum of the separate transforms. Upon using the corollary to express L{y} and L{y} in terms of L{y}, we find that Eq. (9) becomes s2L{y} − sy(0) − y(0) − [sL{y} − y(0)] − 2L{y} = 0, or (s2 − s − 2)Y (s) + (1 − s)y(0) − y(0) = 0, (10)
where Y (s) = L{y}. Substituting for y(0) and y(0) in Eq. (10) from the initial conditions (5), and then solving for Y (s), we obtain Y (s) = s − 1
=
s − 1 .
(11)
s2 − s − 2 (s − 2)(s + 1)
We have thus obtained an expression for the Laplace transform Y (s) of the solution y = φ(t) of the given initial value problem. To determine the function φ we must find the function whose Laplace transform is Y (s), as given by Eq. (11).
Chapter 6. The Laplace Transform
This can be done most easily by expanding the right side of Eq. (11) in partial fractions. Thus we write Y (s) = s − 1 = a + b = a(s + 1) + b(s − 2),
(
(12) s − 2)(s + 1)s − 2 s + 1 (s − 2)(s + 1) where the coefficients a and b are to be determined. By equating numerators of the second and fourth members of Eq. (12), we obtain s − 1 = a(s + 1) + b(s − 2), an equation that must hold for all s. In particular, if we set s = 2, then it follows that a = 1 . Similarly, if we set s = −1, then we find that b = 2 . By substituting these
3
3
values for a and b, respectively, we have Y (s) = 1/3 + 2/3 .
(13)
s − 2 s + 1
Finally, if we use the result of Example 5 of Section 6.1, it follows that 1 e2t has the
3
transform 1 (s − 2)−1; similarly, 2 e−t has the transform 2 (s + 1)−1. Hence, by the
3
3
3
linearity of the Laplace transform, y = φ(t) = 1 e2t + 2 e−t
3
3
has the transform (13) and is therefore the solution of the initial value Of course, this is the same solution that we obtained earlier.
The same procedure can be applied to the general second order linear equation with constant coefficients, ay + by + cy = f (t).
(14)
Assuming that the solution y = φ(t) satisfies the conditions of Corollary 6.2.2 for n = 2, we can take the transform of Eq. (14) and thereby obtain a[s2Y (s) − sy(0) − y(0)] + b[sY (s) − y(0)] + cY (s) = F(s), (15)
where F (s) is the transform of f (t). By solving Eq. (15) for Y (s) we find that (as + b)y(0) + ay(0)Y (s) =
+
F (s).
(16)
as2 + bs + cas2 + bs + c
The problem is then solved, provided that we can find the function y = φ(t) whose transform is Y (s).
Even at this early stage of our discussion we can point out some of the essential features of the transform method. In the first place, the transform Y (s) of the unknown function y = φ(t) is found by solving an algebraic equation rather than a differential
equation, Eq. (10) rather than Eq. (4), or in general Eq. (15) rather than Eq. (14). This is the key to the usefulness of Laplace transforms for solving linear, constant coefficient, ordinary differential equations—the problem is reduced from a differential equation to an algebraic one. Next, the solution satisfying given initial conditions is automatically found, so that the task of determining appropriate values for the arbitrary constants in the general solution does not arise. Further, as indicated in Eq. (15), nonhomogeneous equations are handled in exactly the same way as homogeneous ones; it is not necessary to solve the corresponding homogeneous equation first. Finally, the method can be applied in the same way to higher order equations, as long as we assume that the solution satisfies the conditions of the corollary for the appropriate value of n.
6.2Solution of Initial Value Problems
Observe that the polynomial as2 + bs + c in the denominator on the right side of Eq. (16) is precisely the characteristic polynomial associated with Eq. (14). Since the use of a partial fraction expansion of Y (s) to determine φ(t) requires us to factor this polynomial, the use of Laplace transforms does not avoid the necessity of finding roots of the characteristic equation. For equations of higher than second order this may be a difficult algebraic problem, particularly if the roots are irrational or complex.
The main difficulty that occurs in solving initial value problems by the transform technique lies in the problem of determining the function y = φ(t) corresponding to the transform Y (s). This problem is known as the inversion problem for the Laplace transform; φ(t) is called the inverse transform corresponding to Y (s), and the process of finding φ(t) from Y (s) is known as inverting the transform. We also use the notation L−1{Y (s)} to denote the inverse transform of Y (s). There is a general formula for the inverse Laplace transform, but its use requires a knowledge of the theory of functions of a complex variable, and we do not consider it in this book. However, it is still possible to develop many important properties of the Laplace transform, and to solve many interesting problems, without the use of complex variables.
In solving the initial value problem (4), (5) we did not consider the question of whether there may be functions other than the one given by Eq. (8) that also have the transform (13). In fact, it can be shown that if f is a continuous function with the Laplace transform F , then there is no other continuous function having the same transform. In other words, there is essentially a one-to-one correspondence between functions and their Laplace transforms. This fact suggests the compilation of a table, such as Table 6.2.1, giving the transforms of functions frequently encountered, and vice versa. The entries in the second column of Table 6.2.1 are the transforms of those in the first column. Perhaps more important, the functions in the first column are the inverse transforms of those in the second column. Thus, for example, if the transform of the solution of a differential equation is known, the solution itself can often be found merely by looking it up in the table. Some of the entries in Table 6.2.1 have been used as examples, or appear as problems in Section 6.1, while others will be developed later in the chapter. The third column of the table indicates where the derivation of the given transforms may be found. While Table 6.2.1 is sufficient for the examples and problems in this book, much larger tables are also available (see the list of references at the end of the chapter). Transforms and inverse transforms can also be readily obtained electronically by using a computer algebra system.
Frequently, a Laplace transform F (s) is expressible as a sum of several terms, F (s) = F (s) + F (s) + · · · + F (s).
(17)
1
2
n
Suppose that f (t) = L−1{F (s)}, . . . , f (t) = L−1{F (s)}. Then the function
1
1
n
n
f (t) = f (t) + · · · + f (t)
1
n
has the Laplace transform F (s). By the uniqueness property stated previously there is no other continuous function f having the same transform. Thus L−1{F(s)} = L−1{F (s)} + · · · + L−1{F (s)};
(18)
1
n
that is, the inverse Laplace transform is also a linear operator.
In many problems it is convenient to make use of this property by decomposing a given transform into a sum of functions whose inverse transforms are already known or can be found in the table. Partial fraction expansions are particularly useful in this Chapter 6. The Laplace TransformTABLE 6.2.1 Elementary Laplace Transforms f (t) = L−1{F(s)} F(s) = L{ f (t)}
Notes
1
1. 1
,s > 0 Sec. 6.1; Ex. 4
s
1
2. eat,
s > a Sec. 6.1; Ex. 5
s − an!
3. tn; n = positive integer ,s > 0 Sec. 6.1; Prob. 27
sn+1
(p + 1) 4. t p, p > −1 ,s > 0 Sec. 6.1; Prob. 27
s p+1
a 5. sin at,s > 0 Sec. 6.1; Ex. 6
s2 + a2
s
6. cos at,s > 0 Sec. 6.1; Prob. 6
s2 + a2
a 7. sinh at,s > |a| Sec. 6.1; Prob. 8
s2 − a2
s
8. cosh at,s > |a| Sec. 6.1; Prob. 7
s2 − a2
b
9. eat sin bt,
s > a Sec. 6.1; Prob. 13
(s − a)2 + b2 s − a 10. eat cos bt,
s > a Sec. 6.1; Prob. 14
(s − a)2 + b2 n!
11. tneat , n = positive integer ,
s > a Sec. 6.1; Prob. 18
(s − a)n+1 12. u (t)
e−cs ,s > 0 Sec. 6.3
c
s
13. u (t) f (t − c)e−cs F(s) Sec. 6.3
c 14. ect f (t)F(s − c) Sec. 6.3
s
15. f (ct) 1 F,c > 0 Sec. 6.3; Prob. 19
c
c
t
16.
f (t − τ)g(τ ) dτF(s)G(s) Sec. 6.6
0
17. δ(t − c)
e−cs Sec. 6.5
18. f (n)(t)sn F(s) − sn−1 f (0) − · · · − f (n−1)(0) Sec. 6.2
19. (−t)n f (t)F(n)(s) Sec. 6.2; Prob. 28
6.2Solution of Initial Value Problems connection, and a general result covering many cases is given in Problem 38. Other useful properties of Laplace transforms are derived later in this chapter.
As further illustrations of the technique of solving initial value problems by means of the Laplace transform and partial fraction expansions, consider the following examples.
Find the solution of the differential equation E X A M P L E
1
y + y = sin 2t,
(19)
satisfying the initial conditions y(0) = 2,y(0) = 1.
(20)
We assume that this initial value problem has a solution y = φ(t), which with its first two derivatives satisfies the conditions of Corollary 6.2.2. Then, taking the Laplace transform of the differential equation, we have s2Y (s) − sy(0) − y(0) + Y (s) = 2/(s2 + 4), where the transform of sin 2t has been obtained from line 5 of Table 6.2.1. Substituting for y(0) and y(0) from the initial conditions and solving for Y (s), we obtain Y (s) = 2s3 + s2 + 8s + 6 .
(21) (s2 + 1)(s2 + 4)
Using partial fractions we can write Y (s) in the form (as + b)(s2 + 4) + (cs + d)(s2 + 1)Y (s) = as + b + cs + d = .
(22)
s2 + 1 s2 + 4 (s2 + 1)(s2 + 4)
By expanding the numerator on the right side of Eq. (22) and equating it to the numerator in Eq. (21) we find that 2s3 + s2 + 8s + 6 = (a + c)s3 + (b + d)s2 + (4a + c)s + (4b + d) for all s. Then, comparing coefficients of like powers of s, we have a + c = 2,b + d = 1, 4a + c = 8, 4b + d = 6.
Consequently, a = 2, c = 0, b = 5 , and d = − 2 , from which it follows that
3
3
Y (s) =
2s + 5/3 − 2/3 .
(23)
s2 + 1 s2 + 1 s2 + 4
From lines 5 and 6 of Table 6.2.1, the solution of the given initial value problem is y = φ(t) = 2 cos t + 5 sin t − 1 sin 2t.
(24)
3
3
Find the solution of the initial value problem E X A M P L E
2
yiv − y = 0, (25) y(0) = 0,y(0) = 1,y(0) = 0,y(0) = 0.
(26)
Chapter 6. The Laplace Transform
In this problem we need to assume that the solution y = φ(t) satisfies the conditions of Corollary 6.2.2 for n = 4. The Laplace transform of the differential equation (25) is s4Y (s) − s3 y(0) − s2 y(0) − sy(0) − y(0) − Y (s) = 0.
Then, using the initial conditions (26) and solving for Y (s), we have Y (s) =
s2
.
(27)
s4 − 1
A partial fraction expansion of Y (s) is Y (s) = as + b + cs + d ,s2 − 1 s2 + 1 and it follows that (as + b)(s2 + 1) + (cs + d)(s2 − 1) = s2 (28)
for all s. By setting s = 1 and s = −1, respectively, in Eq. (28) we obtain the pair of equations 2(a + b) = 1, 2(−a + b) = 1, and therefore a = 0 and b = 1 . If we set s = 0 in Eq. (28), then b − d = 0, so d = 1 .
2
2
Finally, equating the coefficients of the cubic terms on each side of Eq. (28), we find that a + c = 0, so c = 0. Thus Y (s) = 1/2 + 1/2 , (29)
s2 − 1 s2 + 1 and from lines 7 and 5 of Table 6.2.1 the solution of the initial value (26) is y = φ(t) = sinh t + sin t .
(30)
2
The most important elementary applications of the Laplace transform are in the study of mechanical vibrations and in the analysis of electric circuits; the governing equations were derived in Section 3.8. A vibrating spring–mass system has the equation of motion
d2u
m
+ γ du + ku = F(t), (31)
dt2
dt
where m is the mass, γ the damping coefficient, k the spring constant, and F(t)
the applied external force. The equation describing an electric circuit containing an inductance L, a resistance R, and a capacitance C (an LRC circuit) is d2 Qd Q
L + R + 1 Q = E(t), (32)
dt2
dt
C where Q(t) is the charge on the capacitor and E(t) is the applied voltage. In terms of the current I (t) = d Q(t)/dt we can differentiate Eq. (32) and write d2 Id I
L + R + 1 I = d E (t).
(33)
dt2
dt
C
dt
Suitable initial conditions on u, Q, or I must also be prescribed.
6.2Solution of Initial Value Problems
We have noted previously in Section 3.8 that Eq. (31) for the spring–mass system and Eq. (32) or (33) for the electric circuit are identical mathematically, differing only in the interpretation of the constants and variables appearing in them. There are other physical problems that also lead to the same differential equation. Thus, once the mathematical problem is solved, its solution can be interpreted in terms of whichever corresponding physical problem is of immediate interest.
In the problem lists following this and other sections in this chapter are numerous initial value problems for second order linear differential equations with constant coefficients. Many can be interpreted as models of particular physical systems, but usually we do not point this out explicitly.
PROBLEMS
In each of Problems 1 through 10 find the inverse Laplace transform of the given function.
3
4
1.
2.
2
3.
4.
5.
6.
7.
8.
9.
10.
Chapter 6. The Laplace Transform
27. The Laplace transforms of certain functions can be found conveniently from their Taylor series expansions.
∞
,
1
,
(b) Let
0
Section 5.8)
∞
.
0
Assuming that the following Laplace transforms can be computed term by term, verify that
0
and
√
0
Problems 28 through 36 are concerned with differentiation of the Laplace transform.
28. Let
∞
0
35. Consider Bessel’s equation of order zero 6.2Solution of Initial Value Problems
permissible to take the inverse transform term by term, show that
∞
0
0
0
0
36. For each of the following initial value problems use the results of Problem 28 to find the
37. Suppose that
tg(t) = f (τ ) dτ.
0
If G(s) and F(s) are the Laplace transforms of g(t) and f (t), respectively, show that G(s) = F(s)/s.
38. In this problem we show how a general partial fraction expansion can be used to calculate many inverse Laplace transforms. Suppose that F(s) = P(s)/Q(s), where Q(s) is a polynomial of degree n with distinct zeros r , . . . , r and P(s) is a
1
n
polynomial of degree less than n. In this case it is possible to show that P(s)/Q(s) has a partial fraction expansion of the form P(s)
A
A
=
1
+ · · · +
n
,
(i) Q(s)s − rs − r
1
n
where the coefficients A , . . . , A must be determined.
1
n (a) Show that
(ii)
k
k
k
Hint: One way to do this is to multiply Eq. (i) by s − r and then to take the limit as s → r .
k
k