The Laplace Transform

CHAPTER 4 The Laplace Transform

4.1 Introduction

The Laplace transform provides an effective method of solving initial-value problems for linear differential equations with constant coefficients. However, the usefulness of Laplace transforms is by no means restricted to this class of problems. Some understanding of the basic theory is an essential part of the mathematical background of engineers, scientists and mathematicians.

The Laplace transform is deﬁned in terms of an integral over the interval [0, ∞). In- tegrals over an inﬁnite interval are called improper integrals, a topic studied in Calculus II.

DEFINITION Let f be a continuous function on [0, ∞). The Laplace transform of f, denoted by L[(f(x)], or by F (s), is the function given by

∞ L[f(x)] = F (s)=Z e−sxf(x) dx. (1) 0 The domain of F is the set of all real numbers s for which the improper integral converges.

In more advanced treatments of the Laplace transform the parameter s assumes complex values, but the restriction to real values is sufficient for our purposes here. Note that L transforms a function f = f(x) into a function F = F (s) of the parameter s. The continuity assumption on f will hold throughout the first three sections. It is made for convenience in presenting the basic properties of L and for applying the Laplace transform method to solving initial-value problems. In the last two sections of this chapter we extend the definition of L to a larger class of functions, the piecewise continuous functions on [0, ∞). There we will apply L to the problem of solving nonhomogeneous equations in which the nonhomogeneous term is piecewise continuous. This will involve some extension of our concepts of differential equation and solution.

As indicated above, the primary application of Laplace transforms of interest to us is solving linear diﬀerential equations with constant coeﬃcients. Referring to our work in Chapter 3, the functions which arise naturally in the treatment of these equations are:

p(x)erx,p(x) cos βx, p(x) sin βx, p(x)erx cos βx, p(x)erx sin βx where p is a polynomial.

We begin by calculating the Laplace transforms of some simple cases of these functions.

115 Example 1. Let f(x)=1· e0x ≡ 1on[0, ∞). By the Deﬁnition,

∞ b L[1] = Z e−sx · 1 dx = lim Z e−sx dx 0 b→∞ 0

− b − " e sx # e sb 1 −1 1 = lim = lim + = lim + . b→∞ −s b→∞ −s s b→∞ sesb s 0

Now, lim −1/sesb exists if and only if s>0, and in this case b→∞ −1 lim =0. b→∞ sesb Thus, 1 L[1] = ,s>0. s

Example 2. Let f(x)=erx on [0, ∞). Then,

∞ b L[erx]=Z e−sx · erx dx = lim Z e−(s−r)x dx 0 b→∞ 0

b  −(s−r)x  " −(s−r)b # e e 1 = lim = lim + . b→∞  −(s − r)  b→∞ −(s − r) s − r 0 The limit exists (and has the value 0) if and only if s − r>0. Therefore 1 L[erx]= , s>r. s − r Note that if r = 0, then we have the result in Example 1.

Example 3. Let f(x) = cos βx on [0, ∞). Then,

∞ b L[cos βx]=Z e−sx · cos βx dx = lim Z e−sx cos βx dx 0 b→∞ 0

− b e sx[−s cos βx − β sin βx] = lim . b→∞ s2 + β2 0 (Note the integral was calculated using integration by parts; also, it is a standard entry in a table of integrals.)

Now, 1 s cos βb + β sin βb s L[cos βx]=− lim · + . b→∞ e−sb s2 + β2 s2 + β2

Since [s cos βb+ β sin βb]/(s2 + β2) is bounded, the limit exists (and has the value 0) if and only if s>0. Therefore, s L[cos βx]= ,s>0. s2 + β2

116 The following table gives a basic list of the Laplace transforms of functions that we will encounter in this chapter. While the entries in the table can be verified using the Definition, some of the integrations involved are complicated. The properties of the Laplace transform presented in the next section provide a more efficient way to obtain many of the entries in the table. Handbooks of mathematical functions, for example the CRC Standard Mathematical Tables, give extensive tables of Laplace transforms.

Table of Laplace Transforms

f(x) F (s)=L[f(x)]

1 1 ,s>0 s 1 eαx , s>α s − α s cos βx ,s>0 s2 + β2 β sin βx ,s>0 s2 + β2 s − α eαx cos βx , s>α (s − α)2 + β2 β eαx sin βx , s>α (s − α)2 + β2 n! xn,n=1, 2,... ,s>0 sn+1 n! xn erx,n=1, 2,... , s>r (s − r)n+1 s2 − β2 x cos βx ,s>0 (s2 + β2)2 2βs x sin βx ,s>0 (s2 + β2)2

Exercises 4.1

Use the deﬁnition of the Laplace transform to ﬁnd the Laplace transform of the given function.

1. f(x)=x.

2. f(x)=x2.

117 3. f(x) = sin x.

4. f(x)=xerx.

5. f(x) = sinh x.

6. f(x) = cosh x. eax 7. Use the fact that Z eax cos bx dx = [a cos bx + b sin bx] to find L[erx cos βx] a2 + b2 by the definition. eax 8. Use the fact that Z eax sin bx dx = [a sin bx − b cos bx] to find L[erx sin βx] a2 + b2 by the definition.

2π Z e−sx sin xdx 9. Show that L[sin x]= 0 . 1 − e−2πs 10. Let f be a continuous function on [0, ∞) and suppose that f is periodic with period p. That is f is continuous and f(x + p)=f(x),p>0, for all x. Show

that p Z e−sxf(x) dx L[f(x)] = 0 . 1 − e−ps

118 4.2 Basic Properties of the Laplace Transform

In the preceding section we defined the Laplace transform and calculated the Laplace transforms of some of the functions that occur in solving linear differential equations with constant coefficients. In this section we consider the basic question of the existence of the Laplace transform of a function f, and we develop the properties of the Laplace transform that will be used in solving initial value problems.

To motivate the material in this section, consider the diﬀerential equation

y00 + ay0 + by = f(x) (2) where a and b are constants and f is a continuous function on [0, ∞). If we assume that y = y(x) is a solution of (1) and formally apply L, we obtain

00 0 L y (x)+ay (x)+by(x) = L[f(x)]. (3) The right-hand side of this equation suggests the basic question of the existence of L[f(x)]. That is, for what functions f does L[f] exist?

DEFINITION A function f, continuous on [0, ∞), is said to be of exponential order λ, λ a real number, if there exists a positive number M and a nonnegative number A such that | f(x)|≤Meλx on [A, ∞). Example 1. (a) If f is a bounded function on [0, ∞) [for example, f(x) = cos βx or f(x)=sinβx], then f is of exponential order 0. f bounded implies that there exists a positive number M such that | f(x)|≤M for all x ∈ [0, ∞). Therefore,

| f(x)|≤M = Me0x on [0, ∞).

[Note: if f(x) = cos βx or f(x) = sin βx, then we could take M = 1.]

(b) Let f(x)=x on [0, ∞). For any positive number λ, x lim =0 x→∞ eλx by L’Hˆopital’srule. Therefore, there exists a nonnegative number A such that x ≤ 1on[A, ∞). eλx This implies that x ≤ eλx =1· eλx on [A, ∞) and f(x)=x is of exponential order λ. The same argument can be used to show that f(x)=xα,α any real number, is of exponential order λ for any positive number λ. In general, if p = p(x) is a polynomial, then p is of exponential order λ for any positive number λ.

119 (c) If f(x)=erx, then f is of exponential order λ for any λ ≥ r.

(d) Consider the function f(x)=ex2 .If f is of exponential order λ for some λ, then there exists a positive number M and a nonnegative number A such that

2 2 ex ≤ Meλx on [A, ∞) which implies e−λxex ≤ M on [A, ∞).

But, 2 lim e−λxex = lim ex(x−λ) = ∞, x→∞ x→∞ a contradiction. Thus f(x)=ex2 is not of exponential order λ for any positive number λ.

Our ﬁrst property of L is a suﬃcient condition for L[f(x)] to exist. We shall omit the proof. THEOREM 1. Let f be a continuous function on [0, ∞). If f is of exponential order λ, then the Laplace transform L[f(x)] = F (s) exists for s>λ.

We now turn to the left-hand side of equation (2) where we have L applied to the linear combination y00(x)+ay0(x)+by(x). THEOREM 2. The operator L is a linear operator. That is, if g and h are continuous functions on [0, ∞), and if each of L[g(x)] and L[h(x)] exists for s>λ, then L[g(x)+h(x)] and L[cg(x)],c constant, each exist for s>λ, and

L[g(x)+h(x)] = L[g(x)] + L[h(x)]

L[cg(x)] = cL[g(x)].

Proof: ∞ b L[g(x)+h(x)] = Z e−sx[g(x)+h(x)] dx = lim Z e−sx[g(x)+h(x)] dx 0 b→∞ 0 b b = lim Z e−sxg(x) dx + Z e−sxh(x) dx b→∞ 0 0

b b = lim Z e−sxg(x) dx + lim Z e−sxh(x) dx b→∞ 0 b→∞ 0 ∞ ∞ = Z e−sxg(x) dx + Z e−sxh(x) dx 0 0 = L[g(x)] + L[h(x)]

The proof that L[cg(x)] = cL[g(x)] for any constant c is left as an exercise.

120 COROLLARY Let g1(x),g2(x), ..., gn(x) be continuous functions on [0, ∞). If L[g1(x)], L[g2(x)], ...,L[gn(x)] all exist for s>λ, and if c1,c2,...,cn are real numbers, then

L [c1g1(x)+c2g2(x)+···+ cngn(x)] exists for s>λ and

L [c1g1(x)+c2g2(x)+···+ cngn(x)] = c1L[g1(x)] + c2L[g2(x)] + ···+ cnL[gn(x)].

According to the Corollary, if L[y00], L[y0], and L[y] all exist for s>λ for some λ, then L[y00 + ay0 + by]=L[y00]+aL[y0]+bL[y].

The next property gives a relationship between the Laplace transform of the derivative of a function and the Laplace transform of the function itself.

THEOREM 3. Let g be a continuously diﬀerentiable function on [0, ∞). If g is of exponential order λ, then L[g0(x)] exists for s>λ and

L[g0(x)] = sL[g(x)] − g(0).

Proof: By the deﬁnition of L, we have

∞ b L[g0(x)] = Z e−sxg0(x) dx = lim Z e−sxg0(x) dx. 0 b→∞ 0 Now, using integration by parts,

b b Z −sx 0 −sx b Z −sx e g (x) dx = e g(x) 0 + s e g(x) dx 0 0 b = e−sbg(b) − g(0) + s Z e−sxg(x) dx 0 Therefore,

b L[g0(x)] = lim e−sbg(b) − g(0) + s Z e−sxg(x) dx b→∞ 0 b = lim e−sbg(b) − g(0) + s lim Z e−sxg(x) dx b→∞ b→∞ 0 = lim e−sbg(b) − g(0) + sL[g(x)] b→∞ provided lim e−sbg(b) exists. b→∞ Since g is of exponential order λ, there exist constants M and A such that

| g(x)|≤Meλx

121 on [A, ∞). Therefore,

−sb −sb λb −(s−λ)b e g(b) ≤ e Me = Me for all b>A, and it follows that

lim e−sbg(b) = 0 for s>λ. b→∞

We can now conclude that

L[g0(x)] = sL[g(x)] − g(0).

COROLLARY Let g be a function which is n-times continuously diﬀerentiable on [0, ∞). If each of the functions g, g0,...,g(n−1) is of exponential order λ, then L[g(n)] exists for s>λ and

L[g(n)(x)] = snL[g(x)] − sn−1g(0) − sn−2g0(0) −···−g(n−1)(0).

In particular, if g00 is continuous on [0, ∞), and if g and g0 are of exponential order λ, then L[g00(x)] exists for s>λ and

L[g00(x)] = s2L[g(x)] − sg(0) − g0(0).

The proof of the Corollary follows by induction.

Application to Initial-Value Problems

Consider the ﬁrst order initial-value problem

y0 + ay = f(x); y(0) = α where α is a real number and the function f = f(x) has Laplace transform F = F (s).

Let y = y(x) be the solution of the problem. Applying the properties of L established above, we have

0 L y (x)+ay(x) = L[f(x)]

L[y0(x)] + aL[y(x)] = F (s)(by linearity)

sL[y(x)] − y(0) + a L[y(x)] = F (s)(by Theorem 3)

(s + a)L[y(x)] − y(0) = F (s)

Applying the initial condition y(0) = α, and solving for L[y(x)] = Y (s), we get F (s)+α Y (s)= . s + a

122 Next consider the second order initial-value problem

y00 + ay0 + by = f(x); y(0) = α, y0(0) = β. where α and β are real numbers, and f is a function with Laplace transform L[f(x)] = F (s).

Let y = y(x) be the solution of this problem. Then

00 0 L y (x)+ay (x)+by(x) = L[f(x)] (equation (2))

L[y00(x)] + aL[y0(x)] + bL[y(x)] = F (s)(by linearity) s2L[y(x)] − sy(0) − y0(0) + a {sL[y(x)] − y(0)} + bL[y(x)] = F (s)(by the Corollary to Theorem 3)

(s2 + as + b)L[y(x)] − sy(0) − y0(0) − ay(0) = F (s)

Applying the initial conditions y(0) = α, y0(0) = β and solving for L[y(x)] = Y (s), we get F (s)+αs + β + aα Y (s)= . s2 + as + b

Implicit in these derivations is the assumption that the Laplace transform of y and its derivatives exist. Assuming that this is the case, the importance of these results is that it gives us the Laplace transform of the solution of an initial-value problem directly. The question now is: Knowing Y (s), what is y(x)? This is the topic of the next section, “inverting” the Laplace transform.

Example 2. Find the Laplace transform L[y(x)] = Y (s) of the solution of the initial-value problem y0 − 2y =2e−3x; y(0) = −2. SOLUTION If y = y(x) is the solution, then 2 L[y0(x) − 2y(x)] = L[2e−3x]=2L[e−3x]= s +3 2 L[y0(x)] − 2L[y(x)] = s +3 2 sL[y(x)] − y(0) − 2L[y(x)] = s +3 2 (s − 2)L[y(x)] + 2 = s +3 2 (s − 2)L[y(x)] = − 2 s +3 Therefore, 2 2 L[y(x)] = Y (s)= − . (s − 2)(s +3) s − 2

123 Example 3. Find the Laplace transform L[y(x)] = Y (s) of the solution of the initial-value problem y00 +4y =3e2x; y(0) = 5,y0(0) = −2. SOLUTION If y = y(x) is the solution, then 3 L[y00(x)+4y(x)] = L[3e2x]=3L[e2x]= s − 2 3 L[y00(x)] + 4L[y(x)] = s − 2 3 s2L[y(x)] − sy(0) − y0(0) + 4L[y(x)] = s − 2 3 (s2 +4)L[y(x)] − 5s − (−2) = s − 2 3 (s2 +4)L[y(x)] = +5s − 2 s − 2 Therefore, 3 5s − 2 L[y(x)] = Y (s)= + . (s − 2)(s2 +4) s2 +4

In the next section we will see how to go from Y (s)toy(x).

Additional Applications and Properties

Although the main use of Theorem 3 and its Corollary is in solving initial-value problems, the results can also be used to determine entries in a table of Laplace transforms. For example, if f is a continuously diﬀerentiable function and L[f(x)] is known, then L[f 0(x)] can be determined: L[f 0(x)] = sL[f(x)] − f(0).

Example 4. In Example 3, Section 4.1, we showed that s L[cos βx]= . s2 + β2 We could use essentially the same calculations to obtain L[sin βx], but recall that the integrations involved are a little “messy.”

Here is a simpler way to obtain L[sin βx]. Since [cos βx]0 = −β sin βx, we can write

s2 L[−β sin βx]=L[(cos βx)0]=sL[cos βx] − cos(0) = − 1 s2 + β2 −β2 −βL[sin βx]= . s2 + β2 Therefore, β L[sin βx]= . s2 + β2

124 Here are two more properties of the Laplace transform that are useful in determining entries in a table of transforms.

THEOREM 4. Let f be a continuous function on [0, ∞), and be of exponential order λ. Then F (s)=L[f(x)] has derivatives of all orders, and for s>λ,

dnF (s) =(−1)nL [xnf(x)] ,n=1, 2, ... . dsn

Justification By the definition of L[f(x)] = F (s), we have ∞ F (s)=Z e−sxf(x) dx. 0 Differentiation of this equation with respect to s can be justified and yields dF ∞ d ∞ = Z [e−sxf(x)] dx = Z e−sx[−xf(x)] dx ds 0 ds 0 = L[−xf(x)] = −L[xf(x)] dF Therefore, L[xf(x)] = − . ds Differentiating a second time, we have

d2F d dF ∞ d Z −sx − 2 = = e [ xf(x)] dx ds ds ds 0 ds ∞ = Z e−sx[x2f(x)] dx = L[x2f(x)] 0 The result stated in the theorem follows by mathematical induction. 1 Example 5. From the table at the end of Section 4.1, L[erx]= , s>r. Therefore, s − r d 1 1 L[xerx]=−F 0(s)=− = , s>r, ds s − r (s − r)2

d2 1 2 L[x2erx]=F 00(s)= = , s>r, ds2 s − r (s − r)3 d3 1 6 L[x3erx]=−F 000(s)=− = , s>r, ds3 s − r (s − r)4 and so on.

The ﬁnal property we’ll consider is called the translation property of L.

THEOREM 5. If f is a continuous function on [0, ∞), and if L[f(x)] = F (s) exists for s>λ, then for any real number r,

L[erxf(x)] = F (s − r) for s>λ+ r.

125 Proof: From the deﬁnition of the Laplace transform, ∞ ∞ F (s − r)=Z e−(s−r)xf(x) dx = Z e−sxerxf(x) dx 0 0 = L[erxf(x)].

Example 6.

(a) Since L[x]=1/s2,s>0, we have 1 L[xerx]=F (s − r)= , s>r (c.f. Example 5) (s − r)2 s (b) Since L[cos βx]= ,s>0, we have s2 + β2 s − r L[erx cos βx]=F (s−r)= ,s>r (as indicated in the Table, Section 4.1). (s − r)2 + β2

Summary of the Properties of L

1. If f is continuous on [0, ∞), and of exponential order λ, then L[f(x)] = F (s) exists for s>λ. (Theorem 1)

2. L is a linear operator (Theorem 2): L[f + g]=L[f]+L[g]; L[cf]=cL[f].

3. If f is continuously diﬀerentiable on [0, ∞), and of exponential order λ, then L[f 0(x)] exists for s>λ and

L[f 0(x)] = sL[f(x)] − f(0).

If f is twice continuously diﬀerentiable on [0, ∞), and of exponential order λ, then L[f 00(x)] exists for s>λ and

L[f 00(x)] = s2L[f(x)] − sf(0) − f 0(0).

And so on. (Theorem 3)

4. If f is continuous on [0, ∞), and of exponential order λ, then F (s)=L[f(x)] has derivatives of all orders, and for s>λ, dnF (s) =(−1)nL [xnf(x)] ,n=1, 2, ... . (Theorem 4) dsn

5. If f is continuous on [0, ∞), and if L[f(x)] = F (s) exists for s>λ, then for any real number r,

L[erxf(x)] = F (s − r) for s>λ+ r. (Theorem 5)

126 Exercises 4.2

Use the properties of the Laplace transform and the Table to ﬁnd L[f].

1. f(x)=3− 2x + x2.

2. f(x)=2e−x − 3 sin 4x.

3. f(x)=3+4e3x − 2 cos 2x.

4. f(x)=2xe−2x +5e2x cos 3x.

5. f(x)=5x2 − 2e−3x sin 2x.

6. f(x)=2− 3x +4x2e2x.

7. f(x)=x sin x +2x cos 2x. s 8. Show that L[cosh βx]= . s2 − β2 β 9. Show that L[sinh βx]= . s2 − β2 10. Find L[3 cosh 2x − 5 sinh 3x].

11. Find L[e2x sinh x + e−x cosh 3x].

12. Find L[x cosh 2x]. 1 13. Show that L[xerx]= by: (s − r)2

(a) Using Property 4. (b) Using Property 5.

2βs 14. Use Property 4 to show that L[x sin βx]= . (s2 + β2)2 Find the Laplace transform Y (s) of the solution of the given initial-value problem.

15. y0 − 2y =0; y(0 = 1.

16. y0 − 2y = x; y(0) = 1.

17. y0 +4y =2e2x − 3 sin 3x; y(0) = −3.

18. y00 +2y0 − 8y =0; y(0) = 4,y0(0) = −2.

19. y00 +6y0 +9y =0; y(0) = 0,y0(0) = 2.

20. y00 − 2y0 +5y =2x + e−x; y(0) = −2,y0(0) = 0.

127 21. y00 − 2y0 − 15y =3+4e−3x; y(0) = 1,y0(0) = −3.

22. y00 − 4y0 +4y =5e2x; y(0) = −3,y0(0) = 2.

Here is another property of the Laplace transform. x 23. Let f be a continuous function on [0, ∞) and assume that both f and Z f(t) dt 0 are of exponential order λ on [0, ∞). Show that if F (s)=L[f(x)], then

x 1 L Z f(t) dt = F (s). 0 s

x 24. Find L Z sin 2tdt by using: 0 (a) The property given in Exercise 23. (b) By ﬁrst calculating the integral and then taking the Laplace transform of the result.

128 4.3 Inverse Laplace Transforms and Initial-Value Problems

In Section 4.2 we saw that the Laplace transform of the solution y = y(x) of the initial-value problem y00 + ay0 + by = f(x); y(0) = α, y0(0) = β is given by F (s)+αs + β + aα L[y(x)] = Y (s)= s2 + as + b where F (s)=L[f(x)] is the Laplace transform of f.

Now that we know L[y(x)], the obvious question is: What is y(x)? The general problem of ﬁnding a function with a given Laplace transform is called the inversion problem. The inversion problem and its application to solving initial-value problems is the topic of this section.

If f is continuous on [0, ∞), and if the Laplace transform, L[f(x)] = F (s) exists for s>λ, then the function F is uniquely determined by f; that is, the operator L is itself a function. Our ﬁrst result states that L is a one-to-one function. A proof of this result is beyond the scope of this introductory treatment.

THEOREM 1. If f and g are continuous functions on [0, ∞), and if L[f(x)] = L[g(x)], then f ≡ g; that is f(x)=g(x) for all x ∈ [0, ∞).

The following deﬁnition gives the terminology and notation used in treating the inversion problem.

DEFINITION If F (s) is a given transform and if the function f, continuous on [0, ∞), has the property that L[f(x)] = F (s), then f is called the inverse Laplace transform of F (s), and is denoted by f(x)=L−1[F (s)]. The operator L−1 is called the inverse operator of L.

There is a general formula for the inverse operator L−1 corresponding to (1), Section 4.1, but use of the formula requires a knowledge of complex-valued functions, a topic which is treated in more advanced courses.

The relationship between L and L−1 is given by the following equations:

L−1 {L[f(x)]} = f(x)

− L L 1[F (s)] = F (s) for all functions f, continuous on [0, ∞), such that L[f(x)] = F (s).

For convenience here, we reproduce the table of Laplace transforms given at the end of Section 4.1.

129 Table of Laplace Transforms

f(x) F (s)=L[f(x)]

A simple way to interpret Theorem 1 is that the table can be read either from left to right or from right to left. That is, the table is simultaneously a table of Laplace transforms and of inverse Laplace transforms. 1 Example 1. (a) If L[f(x)] = F (s)= , then f(x)=e4x. s − 4 s (b) If L[f(x)] = F (s)= , then f(x) = cos 3x. s2 +9 s +2 s +2 s − (−2) (c) If L[f(x)] = F (s)= = = , then f(x)= s2 +4s +13 (s +2)2 +9 [s − (−2)]2 +9 e−2x cos 3x.

The properties of the Laplace transform operator L can be used to derive corresponding properties of its inverse operator L−1. For our purposes, the most important property is that of linearity.

THEOREM 2. The operator L−1 is linear; that is

L−1[F (s)+G(s)] = L−1[F (s)] + L−1[G(s)], and

130 L−1[cF (s)] = cL−1[F (s)],cany constant.

The proof is left as an exercise.

Example 2. Find L−1[F (s)] if 7 6 F (s)= − . s +2 s2 +4

SOLUTION Since L−1 is a linear operator,

7 6 L−1[F (s)] = L−1 − s +3 s2 +4 1 2 =7L−1 − 3L−1 s +3 s2 +4 Now, reading the table from right to left, we see that 1 2 L−1 = e−3x and L−1 = sin 2x. s +3 s2 +4 Therefore, 7 6 L−1 − =7e−3x − 3 sin 2x. s +3 s2 +4

The translation property of L (Theorem 5, Section 4.2) is also useful in ﬁnding inverse transforms. The analog of Theorem 5 for inverse transforms is:

THEOREM 3. If f is continuous on [0, ∞), and if L[f(x)] = F (s) exists for s>λ, then, for any real number r,

L−1[F (s − r)] = erxf(x).

The following examples illustrate the kinds of manipulations that typically occur in calculating inverse Laplace transforms. The basic strategy is to try to re-write a given expression F (s) as sum of terms which appear in the table.

Example 3. Find L−1[F (s)] if 4 1 F (s)= + . (s − 3)2 s2 − 2s +10

SOLUTION 1 1 L−1[F (s)] = 4L−1 + L−1 (s − 3)2 s2 − 2s +10 From the table, 1 L−1 = xe3x. (s − 3)2

131 1 To put in a form in the table, we complete the square in the denominator s2 − 2s +10 and “adjust” the numerator: 1 1 1 1 3 = = = . s2 − 2s +10 s2 − 2s +1+9 (s − 1)2 +9 3 (s − 1)2 +9

From the table,

1 1 3 L−1 = L−1 = 1 ex sin 3x. s2 − 2s +10 3 (s − 1)2 +9 3

Putting the two results together, we have

4 1 L−1 + =4xe3x + 1 ex sin 3x. (s − 3)2 s2 − 2s +10 3

Example 4. Find L−1[F (s)] if 2s +1 F (s)= . s2 − 2s − 8

SOLUTION By factoring the denominator, we can write 2s +1 2s +1 F (s)= = . s2 − 2s − 8 (s + 2)(s − 4)

Now, by partial fraction decomposition,

2s +1 3 1 = 2 + 2 . (s + 2)(s − 4) s − 4 s +2

Therefore 2s +1 3 1 1 1 L−1 = L−1 + L−1 = 3 e4x + 1 e−2x. s2 − 2s − 8 2 s − 4 2 s +2 2 2

Example 5. Find L−1[F (s)] if 2s +4 F (s)= . (s − 2)(s2 − 4s +8)

SOLUTION The quadratic factor in the denominator cannot be factored into linear factors.

By partial fraction decomposition 2s +4 2 −2s +6 F (s)= = + . (s − 2)(s2 − 4s +8) s − 2 s2 − 4s +8

Next, we complete the square in the denominator of the second term: 2 −2s +6 2 −2s +6 2 −2s +6 + = + = + . s − 2 s2 − 4s +8 s − 2 s2 − 4s +4+4 s − 2 (s − 2)2 +4

132 Finally, we “adjust” the numerator of the second term so that we can use the Table: 2 −2s +6 2 −2(s − 2) + 2 2 −2(s − 2) 2 + = + = + + . s − 2 (s − 2)2 +4 s − 2 (s − 2)2 +4 s − 2 (s − 2)2 +4 (s − 2)2 +4

Now 2s +4 1 s − 2 2 L−1 = L−1 2 − 2 + (s − 2)(s2 − 4s +8) s − 2 (s − 2)2 +4 (s − 2)2 +4 1 s − 2 2 =2L−1 − 2 L−1 + L−1 s − 2 (s − 2)2 +4 (s − 2)2 +4

=2e2x − 2 e2x cos 2x + e2x sin 2x.

Solution of Initial-Value Problems

Here we complete the application of Laplace transforms to the solution of initial-value problems. For our ﬁrst example, we ﬁnish Example 2, Section 4.2. Example 6. Find the solution of the initial-value problem y0 − 2y =2e−3x; y(0) = −2.

SOLUTION From the Example, if y = y(x) is the solution, then 2 L[y0(x) − 2y(x)] = L[2e−3x]=2L[e−3x]= s +3 2 L[y0(x)] − 2L[y(x)] = s +3 2 sL[y(x)] − y(0) − 2L[y(x)] = s +3 2 (s − 2)L[y(x)] + 2 = s +3 2 (s − 2)L[y(x)] = − 2 s +3 Therefore, 2 2 L[y(x)] = Y (s)= − . (s − 2)(s +3) s − 2

Now, by partial fraction decomposition 2 2/5 2/5 = − . (s − 2)(s +3) s − 2 s +3 Therefore, 2/5 2/5 2 −8/5 2/5 Y (s)= − − = − s − 2 s +3 s − 2 s − 2 s +3 and −8/5 2/5 y(x)=L−1 − = − 8 e2x − 2 e−3x. s − 2 s +3 5 5

133 Next, we ﬁnish Example 3 of Section 4.2.

Example 7. Find the solution of the initial-value problem

y00 +4y =3e2x; y(0) = 5,y0(0) = −2.

SOLUTION From the Example, if y = y(x) is the solution, then 3 L[y00(x)+4y(x)] = L[3e2x]=3L[e2x]= s − 2 3 L[y00(x)] + 4L[y(x)] = s − 2 3 s2L[y(x)] − sy(0) − y0(0) + 4L[y(x)] = s − 2 3 (s2 +4)L[y(x)] − 5s − (−2) = s − 2 3 (s2 +4)L[y(x)] = +5s − 2 s − 2 Therefore, 3 5s − 2 L[y(x)] = Y (s)= + . (s − 2)(s2 +4) s2 +4

Now, by partial fraction decomposition,

3 3 3 s + 3 = 8 − 8 4 . (s − 2)(s2 +4) s − 2 s2 +4

Therefore,

3 3 s + 3 5s − 2 3 1 37 s 11 2 Y (s)= 8 − 8 4 + = + − s − 2 s2 +4 s2 +4 8 s − 2 8 s2 +4 8 s2 +4 and 3 1 37 s 11 2 y(x)=L−1[Y (s)] = L−1 + L−1 − L−1 8 s − 2 8 s2 +4 8 s2 +4 3 2x 37 − 11 = 8 e + 8 cos 2x 8 sin 2x.

Example 8. Find the solution of the initial-value problem

y00 − 5y0 +6y = x − 1; y(0) = 0,y0(0) = 1.

134 SOLUTION If y = y(x) is the solution, then 1 1 L[y00(x) − 5y0(x)+6y(x)] = L[x − 1] = L[x] −L[1] = − s2 s 1 − s L[y00(x)] − 5L[y0(x)] + 6L[y(x)] = s2 1 − s s2L[y(x)] − sy(0) − y0(0) − 5 {sL[y(x)] − y(0)} +6L[y(x)] = s2 1 − s (s2 − 5s +6)L[y(x)] − 1= s2 1 − s (s2 − 5s +6)L[y(x)] = +1. s2 Therefore, 1 − s 1 1 − s 1 L[y(x)] = Y (s)= + = + . s2(s2 − 5s +6) s2 − 5s +6 s2(s − 2)(s − 3) (s − 2)(s − 3)

Now, by partial fraction decomposition,

1 − s 1 1 1 1 1 1 2 1 = − + + − s2(s − 2)(s − 3) 36 s 6 s2 4 s − 2 9 s − 3 and 1 −1 1 = + . (s − 2)(s − 3) s − 2 s − 3 Therefore, 1 1 1 1 3 1 7 1 Y (s)=− + − + 36 s 6 s2 4 s − 2 9 s − 3 and − 1 1 − 3 2x 7 3x y(x)= 36 + 6 x 4 e + 9 e .

In many applications of diﬀerential equations it is not required to determine the solutions explicitly. Instead what is needed is information about the solutions. Often such information can be obtained by analyzing their Laplace transforms. The next example illustrates this type of application.

Example 9. Consider the diﬀerential equation

y00 − y0 − 6y =2e−x on [0, ∞) together with the condition y(0) = −1. From Chapter 3, we know that the general solution of the diﬀerential equation has the form

−2x 3x −x y(x)=C1e + C2e + Ae (∗) where C1,C2 are arbitrary constants and A is a constant which can be determined

135 The question we want to examine is: Can we choose a value for α = y0(0) so that solution of the resulting initial value-problem

y00 − y0 − 6y =2e−x; y(0) = −1,y0(0) = α has limit 0 as x →∞? Since e−2x → 0 and e−x → 0asx →∞, we want to choose α so that the coeﬃcient of the e3x term is 0.

If y = y(x) is the solution of the initial-value problem, then 2 L[y00(x) − y0(x) − 6y(x)] = L[2e−x]= s +1 2 L[y00(x)] −L[y0(x)] − 6L[y(x)] = s +1 2 s2L[y(x)] − sy(0) − y0(0) −{sL[y(x)] − y(0)}−6L[y(x)] = s +1 2 (s2 − s − 6)L[y(x)] + s − α − 1= s +1 2 (s2 − s − 6)L[y(x)] = +1+α − s. s +1 Therefore, 2 1+α − s L[y(x)] = Y (s)= + (s + 1)(s2 − s − 6) s2 − s − 6 2 1+α − s = + . (s + 1)(s + 2)(s − 3) (s + 2)(s − 3)

Now, by partial fraction decomposition,

2 A B C − 1 2 1 = + + = 2 + 5 + 10 (s + 1)(s + 2)(s − 3) s +1 s +2 s − 3 s +1 s +2 s − 3 and 1+α − s D E − α+3 α−2 = + = 5 + 5 . (s + 2)(s − 3) s − 2 s − 3 s +2 s − 3 Combining these results, we have

α +1 1 2α − 3 1 1 1 Y (s)=− + − . 5 s +2 10 s − 3 2 s +1

Clearly, if 2α − 3 = 0, that is, if α =3/2, then the e3x term in (∗) is eliminated. The resulting solution is: − 1 −2x − 1 −x y(x)= 2 e 2e . This solution has initial values: y(0) = −1,y0(0) = 3, and lim y(x)=0. 2 x→∞

136 Exercises 4.3

Find L−1[F (s)]

6 1. F (s)= . s +7 1 2. F (s)= . 2s +2 1 3. F (s)= . s2 +25 4 3 4. F (s)= − . s s − 4 s +4 5. F (s)= . s2 +8s +17 4 6. F (s)= . s2 − 6s +13 s +4 7. F (s)= . s2 +4s +8 2 5 1 8. F (s)= − + . s s2 s3 2 s 9. F (s)= − . (s +2)2 s2 − 2s +2 s +3 10. F (s)= . s2 − 2s +9 Use partial fraction decomposition to ﬁnd the inverse Laplace transform. 1 11. F (s)= . (s + 1)(s2 +1) s +3 12. F (s)= . s2 − s − 2 1 13. F (s)= . s(s2 +4) 1 14. F (s)= . s2 − 1 s2 − 3s − 1 15. F (s)= . s3 + s2 − 2s s 16. F (s)= . s4 − 1 4 17. F (s)= . s2(s − 1)(s − 2) Find the solution of the initial-value problem.

137 18. y0 − 4y =0; y(0) = −2.

19. y0 +2y = ex; y(0) = 1.

20. y0 − 3y = e2x; y(0) = 2.

21. y0 + y = sin x; y(0) = 1.

22. y00 +4y =0; y(0) = 2,y0(0) = 2.

23. y00 − 2y0 +2y =0; y(0) = 0,y0(0) = 1.

24. y00 − y = sin x; y(0) = 1,y0(0) = 1.

25. y00 − y = ex; y(0) = 1,y0(0) = 0.

26. y00 − y0 − 2y = sin 2x; y(0) = 1,y0(0) = 1.

00 0 x 7 0 9 27. y +2y + y = x + e ; y(0) = − 4 ,y(0) = 4 . 28. y00 +2y0 + y =4e−x; y(0) = 2,y0(0) = −1.

29. y00 +3y0 +2y =6ex; y(0) = 2,y0(0) = −1.

30. y00 − 2y0 +5y =3e−2x; y(0) = 1,y0(0) = 1.

31. Given the initial-value problem

y00 − y0 − 6y =2e−x; y(0) = α, y0(0) = −1. (See Example 9)

What value should be assigned to α so that the resulting solution will have limit 0 as x →∞?

32. What initial conditions should be assigned with the diﬀerential equation

y00 + y = e−x

so that lim y(x) = 0 where y = y(x) is the solution. x→∞ 33. Consider the diﬀerential equation: y00 + y0 − 2y = 3 sin 2x together with the initial value y(0) = 2. For what value(s) of β = y0(0) will the resulting solution(s) be bounded?

34. Consider the diﬀerential equation: y00 +3y0 +2y = x together with the initial value y0(0) = −2. For what value(s) of α = y(0) will the resulting solution(s) be bounded?

The Laplace transform method applies to initial-value problems in which the initial values are speciﬁed at x = 0. Actually, the method can be applied when the initial

138 conditions are speciﬁed at some point a =6 0. All that is required is a simple change of independent variable; a translation. For example, the initial-value problem

d2y dy − 3 +2y = x; y(1) = 0,y0(1) = 1 dx2 dx is transformed into d2y dy − 3 +2y = t +1,y(0) = 0,y0(0) = 1 dt2 dt by setting t = x−1. With this transformation, we have: t = 0 when x =1;x = t+1; and dy dy dt dy dy = = · 1= dx dt dx dt dt d2y d dy d dy dt d dy d2y = = = · 1= . dx2 dx dx dt dx dx dt dt dt2

35. Find the solution of the initial-value problem: y00 − 3y0 +2y = x; y(1) = 0,y0(1) = 1 by ﬁrst solving the transformed problem.

36. Use the Laplace transform method to solve the initial-value problem

y0 − 2y =1+x; y(2) = −1.

139 4.4 Applications to Discontinuous Functions

In the preceding sections we assumed that the functions under consideration were continuous on [0, ∞). In this section we consider Laplace transforms of certain types of discontinuous functions that occur in various applications.

A particular example of the type of discontinuous function that we will be considering in this section is the unit step function, or Heaviside function u. This function is deﬁned on (−∞, ∞)by ( 0 x<0 u(x)= (1) 1 x ≥ 0. The graph of u is

y 3

x -10 -5 5 10 -1

It is clear that u is continuous on (−∞, ∞) except at x =0.At x = 0, lim u(x) x→0 does not exist. However, note that the left-hand and right-hand limits of u at x =0 do exist: lim u(x) = 0 and lim u(x)=1. x→0− x→0+

Recall from calculus that a function f is continuous at a point c if and only if lim f(x)=f(c), and lim f(x) exists if and only if the left-hand and right-hand limits of x→c x→c f at c both exist and are equal. DEFINITION 1. (Jump Discontinuity) Let the function f = f(x) be deﬁned on an interval I and continuous except at a point c ∈ I, c not an endpoint of I. If the left-hand and right-hand limits of f at c both exist but are not equal, then f is said to have a jump (or ﬁnite ) discontinuity at c. Example 1. (a) The unit step function u is continuous on (−∞, ∞) except at x =0 where it has a jump discontinuity.

(b) Let f be deﬁned on [−5, 5] by

 x +5 −5 ≤ x<−2  1  − −2 ≤ x<0 f(x)= x 2x 0 ≤ x ≤ 2    −(x − 5) 2

140 The graph of f is

4 3

1 x -5 -2 2 5 -1

Clearly f is discontinuous at x = −2,x=0, and x = 2. Since

1 lim f(x) = 3 and lim f(x)= 2, x→−2− x→−2+

and lim f(x) = 4 and lim f(x)=3, x→2− x→2+ f has jump discontinuities at x = −2 and x =2.

The discontinuity at x = 0 is not a jump discontinuity because lim f(x) does not x→0− exist (f(x) →∞as x → 0−).

 x 0 ≤ x<1   3 x =1 g(x)= 21≤ x<5  −(x−5)  e x ≥ 5 The graph of g is

x 1 5 -1

This function has jump discontinuities at x = 1 and x = 5. Note that the value of g at x = 1 is independent of the left-hand and right-hand limits of g at x =1.

DEFINITION 2. (Piecewise Continuous) A function f deﬁned on an interval I is piecewise continuous on I if it is continuous on I except for at most a ﬁnite number of

points c1,c2,...,cn of I at which it has a jump discontinuity.

141 Remark A continuous function is also piecewise continuous.

Referring to Example 1, the unit step function u is piecewise continuous on (−∞, ∞); the function g in (c) is piecewise continuous on [0, ∞). the function f in (b) is not piecewise continuous on [−5, 5] because the discontinuity at x = 0 is not a jump discontinuity.

In this section we want to apply Laplace transform methods to piecewise continuous functions. Before we can calculate the Laplace transform of a piecewise function we have to know how to integrate such a function.

Suppose that f is piecewise continuous on [a, b] with jump discontinuities at c1 < c2 < ···

b c1 c2 b Z f(x) dx = Z f(x) dx + Z f(x) dx + ···+ Z f(x) dx. a a c1 cn Example 2. Let f be deﬁned on [0, 3] by

( x 0 ≤ x ≤ 1 f(x)= 4 − (x − 1)2 1

The graph of f is

1 x 1 2 3 -1

We have

3 1 3 Z f(x) dx = Z xdx+ Z (4 − (x − 1)2) dx 0 0 1 x2 1 (x − 1)3 3 35 = + 4x − = . 2 0 3 1 6

The integration of piecewise continuous functions can be extended to include (improper) integrals on an inﬁnite interval.

142 Example 3. Consider the function g of Example 1: ∞ 1 5 ∞ Z g(x) dx = Z xdx+ Z 2 dx + Z e−(x−5) dx 0 0 1 5 x2 1 b = +[2x]5 + lim Z e−(x−5) dx 1 →∞ 2 0 b 5 1 b = + 8 + lim h−e−(x−5)i 2 b→∞ 5 17 19 = +[0+1]= . 2 2

The following theorem is an extension of Theorem 1, Section 4.2. THEOREM 1. If the function f is piecewise continuous on [0, ∞), and of exponential order λ, then the Laplace transform L[f(x)] exists for s>λ.

Let c be a real number. The translation of the unit step function u by c is the function uc = u(x − c) deﬁned on (−∞, ∞)by ( 0 x

The graph of uc for c>0is

x c

-1

Example 4. The Laplace transform of uc,c>0is e−cs L[u(x − c)] = ,s>0. (3) s Proof: By deﬁnition ∞ L[u(x − c)] = Z e−sxu(x − c) dx 0 c ∞ = Z e−sx · 0 dx + Z e−sx · 1 dx 0 c b e−sx b = lim Z e−sx dx = lim b→∞ c b→∞ −s c e−sb e−sc e−cs = lim + = ,s>0. b→∞ −s s s

143 Note that if c = 0, then u0 = u(x − 0) = u(x) ≡ 1 on [0, ∞), and L[u(x)] = L[1] = 1/s, s > 0 as we saw in Section 4.1.

Example 5. (a) Let f be the function deﬁned on [0, ∞)by

( 10≤ x<5 f(x)= 25≤ x<∞. y

3 2 1 x 1 3 5 7 -1

Then

∞ 5 ∞ L[f(x)] = Z e−sxf(x) dx = Z e−sx · 1 dx + Z e−sx · 2 dx 0 0 5 −e−sx 5 b = + lim Z 2e−sx dx →∞ s 0 b 0 −e−5s 1 2e−sx b = + + lim →∞ s s b −s 5 −e−5s 1 2e−sb 2e−5s = + + lim + s s b→∞ −s s −e−5s 1 2e−5s 1 e−5s = + + = + ,s>0. s s s s s

(b) We calculate L[g(x)] where g is the function given in Example 1(c).

∞ 1 5 ∞ L[g(x)] = Z e−sxg(x) dx = Z e−sx · xdx+ Z e−sx · 2 dx + Z e−sxe−(x−5) dx 0 0 1 5 −xe−sx e−sx 1 −2e−sx 5 b − Z −[x(s+1)−5] = 2 + + lim e dx s s 0 s 1 b→∞ 0 ∞ −e−s −e−s 1 −2e−5s 2e−s "−e−[x(s+1)−5] # = + + + + + lim s s2 s2 s s b→∞ s +1 5 e−s e−s 1 2e−5s 2e−s e−5s = − − + − + + s s2 s2 s s s +1 1 e−s e−s 2e−5s e−5s = + − − + . s2 s s2 s s +1

144 The unit step function and its translations can be used to obtain translations of arbitrary functions. For example, if f is deﬁned on [0, ∞) and c>0, then the function

( 0 x

y y

x x c c

THEOREM 2. Let f be deﬁned on [0, ∞) and suppose L[f(x)] = F (s) exists for s>λ.If c>0, then L[fc(x)] exists for s>λ and is given by

−cs L[fc(x)] = L[f(x − c)u(x − c)] = e F (s).

Proof: By the deﬁnition,

∞ Z −sx L[fc(x)] = L[f(x − c)u(x − c)] = e f(x − c)u(x − c) dx 0 b = lim Z e−sxf(x − c) dx. b→∞ c

Now let t = x − c. Then

x = t + c, dx = dt, and t = 0 when x = c.

With this change of variable,

b b−c lim Z e−sxf(x − c) dx = lim Z e−s(t+c)f(t) dt b→∞ c b→∞ 0 b−c = e−cs lim Z e−stf(t) dt b→∞ 0 ∞ = e−cs Z e−stf(t) dt = e−csF (s) 0 since b − c →∞ as b →∞.

145 This theorem can be expressed in an equivalent manner using the inverse Laplace transform.

COROLLARY If L−1[F (s)] = f(x) and c>0, then

−1 −cs L [e F (s)] = f(x − c)u(x − c)=fc(x).

We now illustrate how to use translations of the unit step function and Theorem 2 to calculate Laplace transforms of piecewise continuous functions.

Example 6. We calculate L[f(x)] where f is the function given in Example 4:

( 10≤ x<5 f(x)= 25≤ x<∞.

Since f has a jump discontinuity at x = 5, we’ll write f in terms of u(x − 5). Deﬁne f1(x)by ( 10≤ x<5 f1(x)= =1− u(x − 5) 05≤ x<∞ and f2(x)by ( 00≤ x<5 f2(x)= =2u(x − 5). 25≤ x<∞ Then

f(x)=f1(x)+f2(x)=1− u(x − 5)+2u(x − 5) = 1 + u(x − 5) and 1 e−5s L[f(x)] = L[1 + u(x − 5)] = L[1] + L[u(x − 5)] = + ,s>0. s s

( x2 0 ≤ x<2 Example 7. Let h(x)= . Calculate L[h(x)]. 32≤ x<∞.

SOLUTION The graph of h is

4 3 2 1 x 1 2 3 -1

2 2 Let h1(x)=x − x u(x − 2) and h2(x)=3u(x − 2). Then

2 2 h(x)=h1(x)+h2(x)=x − x u(x − 2)+3u(x − 2)

146 and 2 e−2s L[h(x)] = L[x2] −L[x2 u(x − 2)] + 3L[u(x − 2)] = −L[x2 u(x − 2)] + 3 . s3 s Before we can apply Theorem 2 to calculate L[x2 u(x − 2)] we must have the coeﬃcient x2 of u(x − 2) expressed as a function of (x − 2). Since

x2 =[(x − 2) + 2]2 =(x − 2)2 +4(x − 2) + 4 we have

L[x2 u(x − 2)] = L (x − 2)2u(x − 2) + 4(x − 2)u(x − 2) + 4u(x − 2)

= L[(x − 2)2u(x − 2)] + 4L[(x − 2)u(x − 2)] + 4L[u(x − 2)] 2 1 1 = e−2s +4e−2s +4e−2s . s3 s2 s It now follows that 2 2 1 1 e−2s 2 2e−2s 4e−2s e−2s L[h(x)] = − e−2s +4e−2s +4e−2s +3 = − − − . s3 s3 s2 s s s3 s3 s2 s

Example 8. Let  20≤ x<2  g(x)= x 2 ≤ x<4 .  4 x ≥ 4

The graph of g is

4 3 2 1 x 2 4

Set

g1(x)=2− 2 u(x − 2)

g2(x)=xu(x − 2) − xu(x − 4) = [(x − 2) + 2)]u(x − 2) − [(x − 4) + 4]u(x − 4)

=(x − 2)u(x − 2) + 2u(x − 2) − (x − 4)u(x − 4) − 4u(x − 4)

g3(x)=4u(x − 4).

Then

g(x)=g1(x)+g2(x)+g3(x)=2+(x − 2)u(x − 2) − (x − 4)u(x − 4).

147 Therefore, 2 1 1 L[g(x)] = + e−2s − e−4s s s2 s2

The Corollary to Theorem 2 is used to calculate inverse Laplace transforms.

Example 9. Suppose 1 e−4s F (s)= + . s2 s − 2 Then 1 e−4s 1 e−4s L−1 + = L−1 + L−1 s2 s − 2 s2 s − 2 From the Table of Laplace transforms L−1 1/s2 = x. To calculate L−1 e−4s/(s − 2), let F (s)=1/(s − 2). Then L−1[F (s)] = e2x. Therefore, by the Corollary to Theorem 2,

e−4s L−1 = e2(x−4)u(x − 4). s − 2 It now follows that ( x 0 ≤ x<4 L−1[F (s)] = x + e2(x−4)u(x − 4) = . x + e2(x−4) 4 ≤ x<∞

Example 10. Suppose e−s e−πs F (s)= + . (s +1)2 s2 +4 Then e−s e−πs L−1[F (s)] = L−1 + L−1 . (s +1)2 s2 +4 −1 −s 2 2 To calculate L [e /(s +1) ], let F1(s)=1/(s +1) . Then, from the Table of Laplace −1 −x transforms, L [F1(s)] = xe . Therefore, e−s L−1 =(x − 1)e−(x−1) u(x − 1). (s +1)2

To calculate L−1[e−πs/(s2 + 4)], let 1 1 2 F (s)= = . 2 s2 +4 2 s2 +4 −1 1 From the Table of Laplace transforms, L [F2(s)] = 2 sin 2x. Therefore, e−πs L−1 = 1 sin 2(x − π) u(x − π)= 1 sin 2xu(x − π). (sin 2(x − π) = sin 2x) s2 +4 2 2 Finally, L−1 − −(x−1) − 1 − [F (s)] = (x 1)e u(x 1) + 2 sin 2xu(x π)  00≤ x<1  =  (x − 1)e−(x−1) 1 ≤ x<π (x − 1)e−(x−1) + 1 sin 2xx≥ π  2

148 Exercises 4.4

Use the deﬁnition of the Laplace transform to ﬁnd L[f].

 x 0 ≤ x<1  1. f(x)= x − 11≤ x<2 .   0 x ≥ 2

( −10≤ x<1 2. f(x)= . x − 1 x ≥ 1

( 00≤ x<5 3. f(x)= . 2 x ≥ 5

( 2x 0 ≤ x<3 4. f(x)= . 1 x ≥ 3

 10≤ x<2  5. f(x)= x − 22≤ x<4 .  −(x−4)  e x ≥ 4

( 10≤ x<1 6. f(x)= −11≤ x<2 and extended periodically on [2, ∞). See Exercises 4.1, Problem 10.

( 2x 0 ≤ x<2 7. f(x)= 22≤ x<4 and extended periodically on [4, ∞). (Sketch the graph of f.)

8. f(x)=x − n on [n, n +1),n=0, 1, 2, .... (Sketch the graph of f.)

Express the function f in terms of the unit step function and its translations. Then ﬁnd L[f].

( x2 0 ≤ x<3 9. f(x)= . 3xx≥ 3

 sin x 0 ≤ x<π  10. f(x)= sin 2xπ≤ x<2π . Sketch the graph of f.   sin 3xx≥ 2π

149  00≤ x<π  11. f(x)= 1+cosxπ≤ x<2π .   2 cos xx≥ 2π

 x 0 ≤ x<2  12. f(x)= 02≤ x<4 .  2  (x − 4) x ≥ 4

 x 0 ≤ x<1    2 − x 1 ≤ x<3 13. f(x)= . Sketch the graph of f. x − 43≤ x<4    0 x ≥ 4 Calculate the inverse Laplace transform of F (s).

e−s 2e−3s 6e−4s 14. F (s)= + − . s s s 1+e−πs 15. F (s)= . s2 +1 1 e−2s 16. F (s)= − . s +1 s +1 s +(s − 1)e−πs 17. F (s)= . s2 +1 2 e−2s 2e−2s 4e−3s 18. F (s)= + + − . s2 s2 s3 s e−2s 19. F (s)= . s(s +1)

1 − e−2s 20. F (s)= . s2 + π2

150 4.5 Initial-Value Problems with Piecewise Continuous Nonhomogeneous Terms

In this section we illustrate the application of Laplace transforms to the solution of nonhomogeneous initial-value problems in which the forcing function f is piecewise continuous.

Example 1. Find the solution of the initial-value problem

y0 +2y = f(x); y(0) = 4

( x 0 ≤ x<3 where f(x)= 5 x ≥ 3.

SOLUTION The function f has a jump discontinuity at x = 3. Therefore, the ﬁrst step is to express f in terms of u3(x)=u(x − 3). You should verify that f(x)=x−xu(x−3)+5u(x−3) = x−[(x−3)+3]u(x−3)+5u(x−3) = x−(x−3)u(x−3)+2u(x−3).

Now, taking the Laplace transform of the equation, we get 1 e−3s e−3s L[y0 +2y]=sL[y] − y(0) + 2L[y]= − +2 s2 s2 s Applying the initial condition and solving for L[y], we ﬁnd that 1 e−3s e−3s 4 L[y]=Y (s)= + +2 + s2(s +2) s2(s +2) s(s +2) s +2

4s2 +1 (2s +1)e−3s = + . s2(s +2) s2(s +2)

By partial fraction decomposition 4s2 +1 1 1 1 1 17 1 = − + + s2(s +2) 4 s 2 s2 4 s +2 (2s +1)e−3s 3 e−3s 1 e−3s 3 e−3s = + − s2(s +2) 4 s 2 s2 4 s +2

Therefore, 1 1 1 1 17 1 3 e−3s 1 e−3s 3 e−3s Y (s)=− + + + + − 4 s 2 s2 4 s +2 4 s 2 s2 4 s +2 and

L−1 − 1 1 17 −2x 3 − 1 − − − 3 −2(x−3) − y(x)= [Y (s)] = 4 + 2 x + 4 e + 4 u(x 3) + 2 (x 3) u(x 3) 4 e u(x 3)

( − 1 + 1 x + 17e−2x 0 ≤ x<3 = 4 2 4 − 17 −2x − 3 −2(x−3) ≥ x 1+ 4 e 4 e x 3

151 The graph of y is

x 3

Example 2. Find the solution of the initial-value problem

y00 +4y = f(x); y(0) = 1,y0(0) = 0

( 10≤ x<π/2 where f(x)= −2 x ≥ π/2.

SOLUTION The function f has a jump discontinuity at x = π/2. Therefore, we express f in terms of u(x − π/2): f(x)=1− 3u(x − π/2).

Applying the Laplace transform to the equation and using the initial conditions, we get 1 3e−πs/2 L[y00 +4y]=s2L[y] − sy(0) − y0(0) + 4L[y]= − s s 1 3e−πs/2 (s2 +4)L[y] − s = − . s s Solving for L[y]=Y (s), we have 1 1 s Y (s)= − 3e−πs/2 + . s(s2 +4) s(s2 +4) s2 +4

By partial fraction decomposition, 1 1 1 1 s = − . s(s2 +4) 4 s 4 s2 +4 Therefore, 1 1 1 s 3 e−πs/2 3 s s Y (s)= − − + e−πs/2 + 4 s 4 s2 +4 4 s 4 s2 +4 s2 +4 and

L−1 1 − 1 − 3 − 3 − − y(x)= [Y (s)] = 4 4 cos 2x 4 u(x π/2) + 4 cos 2(x π/2)u(x π/2) + cos 2x  1 3 ≤ 4 + 4 cos 2x 0 x<π/2 =  − 1 x ≥ π/2.  2

152 The graph of y is

x Π 2 -0.5

-1

Exercises 4.5

Solve the initial-value problem.

1. y0 − 2y = f(x); y(0) = 2 where

( 10≤ x<1 f(x)= . 2 x ≥ 1

2. y0 +3y = f(x); y(0) = 1 where

( sin x 0 ≤ x<π f(x)= . 0 x ≥ π

3. y00 + y = f(x); y(0) = 0,y0(0) = 1 where

( 10≤ x<1 f(x)= . 0 x ≥ 1

4. y00 +4y = f(x); y(0) = 1,y0(0) = 0 where

( 00≤ x<2 f(x)= . 1 x ≥ 2

5. y00 +2y0 + y = f(x); y(0) = 0,y0(0) = 0 where

( 10≤ x<2 f(x)= . x − 1 x ≥ 2

6. y00 +4y = f(x); y(0) = 0,y0(0) = 0 where

( sin x 0 ≤ x<2π f(x)= . 0 x ≥ 2π

153 7. y00 − 4y0 +3y = f(x); y(0) = 0,y0(0) = 0 where

( −10≤ x<1 f(x)= . 1 x ≥ 1

8. y00 − 3y0 +2y = f(x); y(0) = 0,y0(0) = 0 where

( 00≤ x<2 f(x)= . 2x − 4 x ≥ 2

9. y00 +2y0 + y = f(x); y(0) = 3,y0(0) = −1 where

( ex 0 ≤ x<1 f(x)= . ex − 1 x ≥ 1

10. y00 − 4y0 +4y = f(x); y(0) = 1,y0(0) = −1 where

( e2x 0 ≤ x<2 f(x)= . −e2x x ≥ 2

154