The Laplace Transform - Armstrongmath.armstrong.edu/faculty/hollis/classes/DE/book3411/...185 Chapter 7 The Laplace Transform In this chapter we will explore a method for solving linear

185

Chapter 7

The Laplace Transform

In this chapter we will explore a method for solving linear differential equationswith constant coefficients that is widely used in electrical engineering. It involvesthe transformation of an initial-value problem into an algebraic equation, whichis easily solved, and then the inverse transformation back to the solution of theoriginal problem, thereby bypassing the need to solve for arbitrary constants inthe general solution. The technique is especially well suited for finding gener-alized solutions to systems driven by impulses or by discontinuous or periodicforcing functions.

7.1 Definition and Basic Properties

Given a function f(t) defined for t ≥ 0, its Laplace transform F (s) is definedas

F (s) =∫ ∞

0e−stf(t)dt. (1)

Notice that the variable s appears as a parameter in an improper integral. Wesay that the Laplace transform exists if this improper integral converges forall sufficiently large s. The notational convention used here is common, thoughnot universal: The uppercase version of the function’s name denotes its Laplacetransform, and s is used for the transform’s independent variable. Before wego further, let’s illustrate this definition and notation with a couple of simpleexamples.

• Example 1 Consider a constant function f(t) = c, t ≥ 0. Its Laplace transformis

F (s) =∫ ∞

0e−stc dt =

c

−se−st

∣∣∣∞t=0

= −cs

(limt→∞

e−st − e0)

= −cs

(0− 1) =c

s,

provided that s > 0. Note that the improper integral diverges when s ≤ 0.

• Example 2 Let f(t) = eat, t ≥ 0. Its Laplace transform is

F (s) =∫ ∞

0e−steatdt =

∫ ∞0

e−(s−a)tdt

= − 1s− a

(limt→∞

e−(s−a)t − e0)

= − 1s− a

(0− 1) =1

s− a,

provided that s > a. Note that the improper integral diverges when s ≤ a. �

186 Chapter 7. The Laplace Transform

The Laplace transform F (s) of a function f(t) is the result of applying alinear operator to f . Denoting this linear operator by L, we can write

Lf = F, or L[f ](s) = F (s).

where F is given by (1). Using this notation, the result of Example 2, for in-stance, is that

L[eat ](s) =1

s− a, s > a.

It is easy to check that the operator L is linear :

L[cf ](s) =∫ ∞

0e−stcf(t)dt = c

∫ ∞0

e−stf(t)dt = c F (s)

L[f + g](s) =∫ ∞

0e−st(f(t) + g(t))dt

=∫ ∞

0e−stf(t)dt+

∫ ∞0

e−stg(t)dt = F (s) +G(s).

Indeed, the linearity of L is a simple consequence of the linearity properties ofintegration.

Our study of Laplace transforms is primarily for the purpose of solving initial-value problems

y′′ + py′ + qy = f, y(0) = y0, y′(0) = v0, (2)

where p and q are constants. Later (starting in Section 3) we will consider gen-eralized solutions that can arise when the nonhomogeneous term f is piecewisecontinuous on [0,∞). A function f is piecewise continuous on [0,∞) if

(i) f has at most finitely many discontinuities in any bounded interval [0, T ],(ii) lim

t→0+f(t) exists, and

(iii) at every number a > 0 both limt→a−

f(t) and limt→a+

f(t) exist.

Note that condition (iii) is automatically true at any point where f is continuous,and it says that all of f ’s discontinuities in (0,∞) are simple “jump” discontinu-ities. Also, piecewise continuity guarantees integrability on any bounded interval[0, T ].

We also need a condition that will guarantee the existence of a function’sLaplace transform. We say that a function f is exponentially bounded on[0,∞) if there are constants M and r such that

|f(t)| ≤Mert for all t ≥ 0.

7.1. Definition and Basic Properties 187

If f is piecewise continuous, then∫ T

0 e−stf(t) dt will exist for all T > 0 and alls. If f is also exponentially bounded, then∫ T

0e−st|f(t)| dt ≤

∫ T

0Me(r−s)t dt =

M

r − s

(e(r−s)T − 1

);

thus∫ T

0 e−st|f(t)| dt converges as T →∞, provided that s > r. This means thatthe Laplace transform of |f | exists, which implies that the Laplace transform off exists.

Throughout the rest of this chapter we will be concerned almost exclusivelywith functions defined on [0,∞). The phrases “piecewise continuous” and “expo-nentially bounded” should always be understood to mean “piecewise continuouson [0,∞)” and “exponentially bounded on [0,∞).”

We also emphasize that throughout this chapter we will deal only with ex-ponentially bounded functions, since only those have Laplace transforms. Also,in this section and the next, all of our examples will involve ordinary elemen-tary functions—polynomials, sinusoidal functions, exponential functions, etc.—that are clearly continuous everywhere. Not until Section 3 will we begin toconsider differential equations with nonhomogeneous terms that are piecewise-continuous.

The First Differentiation Theorem

The reason why Laplace transforms are useful in solving differential equationsis embodied in the following theorem, which (together with the corollary thatfollows) we will refer to as the first differentiation theorem.

THEOREM 1 Suppose that y is continuous and exponentially bounded andthat y′ is piecewise continuous. Then L[y′] exists, and

L[y′](s) = s Y (s)− y(0). J

Proof. First, for T > 0 we use integration by parts to compute∫ T

0e−sty′(t)dt = e−sty(t)

∣∣∣Tt=0−∫ T

0(−s)e−sty(t)dt

= e−sT y(T )− y(0) + s

∫ T

0e−sty(t)dt.

This computation is valid because of the piecewise continuity of y′ and the con-tinuity of y (at 0 and T ). Now, since y is assumed to be exponentially bounded,it follows that, for s sufficiently large,

limT→∞

e−sT y(T ) = 0, and Y (s) = limT→∞

∫ T

0e−sty(t)dt exists.


Therefore, L[y′] exists, and

L[y′](s) = −y(0) + s Y (s).

By repeated application of Theorem 1, we arrive at the following corollary.

COROLLARY 1 Suppose that y and y′ are continuous and exponentiallybounded and that y′′ is piecewise continuous. Then L[y′′] exists, and

L[y′′](s) = s2Y (s)− sy(0)− y′(0).

Moreover, if y, y′, . . . , y(n−1) are continuous and exponentially bounded, and ify(n) is piecewise continuous, then L[y(n)] exists, and

L[y(n)](s) = snY (s)− sn−1y(0)− sn−2y′(0)− · · · − y(n−1)(0). J

Because of the properties stated in Theorem 1 and Corollary 1, the Laplacetransform is particularly well suited for solving linear initial-value problemswith constant coefficients. The following are two examples that indicate thebasic idea.

• Example 3 Consider the first-order initial-value problem

y′ + 2y = 4, y(0) = 1,

and let Y denote the transform of the solution y. By applying the operator Lto both sides of the differential equation and using the result of Example 1 onthe right side, we find that

s Y (s)− y(0) + 2Y (s) =4s.

Now we use the given initial value and solve for Y (s):

(s+ 2)Y (s) =4s

+ 1; so Y (s) =s+ 4

s(s+ 2).

Now our job is to find the function y(t) that has this transform. A partialfraction expansion (consult your calculus book) reveals that

Y (s) =2s− 1s+ 2

.

The results of Examples 1 and 2 now tell us that Y (s) is the transform of

y(t) = 2− e−2t.

• Example 4 Consider the initial-value problem

y′′ + 3y′ + 2y = 0, y(0) = 1, y′(0) = 0,


and let Y denote the transform of the solution y. By applying the operator Lto both sides of the differential equation, we find that

s2Y (s)− s y(0)− y′(0) + 3(s Y (s)− y(0)) + 2Y (s) = 0.

Using the given initial values and rearranging, this becomes

(s2 + 3s+ 2)Y (s)− s− 3 = 0.

Now we solve for Y (s), finding

Y (s) =s+ 3

s2 + 3s+ 2=

s+ 3(s+ 1)(s+ 2)

.

Now we look for the function y(t) that has this transform. A partial fractionexpansion reveals that

Y (s) =2

s+ 1− 1s+ 2

.

From the result of Example 2 above, we see that this is the transform of

y(t) = 2e−t − e−2t. �

The General Procedure

Examples 3 and 4 each illustrate a general procedure for solving initial valueproblems with the help of Laplace transforms:

(1) Transform each side of the differential equation, using the given initialvalues.

(2) Solve the resulting algebraic equation for the transform Y (s) of the solu-tion y(t).

(3) Find the solution y by identifying the transform Y (s) with known trans-forms.

The final step amounts to finding the inverse transform of Y (s). An importantquestion to ask is whether the operator L is actually invertible. In other words,given a transform Y (s), is there a unique function y(t), t ≥ 0, such that Ly = Y ?The answer to this is no, unless we require the function y to be continuous.However, this is not a difficulty in the context of solving differential equations,since solutions will be continuous.

Let’s now consider what happens in general when we apply the Laplacetransform technique to a linear second-order initial-value problem with constantcoefficients. Suppose that the differential equation is

P (D)y = f, with P (D) = D2 + pD + q I ,

and we have initial conditions

y(0) = y0, y′(0) = v0.


The Laplace transform of the nonhomogeneous term is simply Lf = F . Trans-forming the left side produces

LP (D)y = LD2y + pLD y + qLy= s2Y (s)− s y0 − v0 + p(s Y (s)− y0) + q Y (s)= P (s)Y (s)− (s y0 + v0 + p y0).

The transformed initial-value problem therefore becomes

P (s)Y (s)− (s y0 + v0 + p y0) = F (s),

and solving for Y (s) gives us

Y (s) =F (s)P (s)

+γ(s)P (s)

, (3)

where γ(s) = s y0 + v0 + p y0. The term F (s)/P (s) in (3) is the transform of therest solution of P (D)y = f , and the term γ(s)/P (s) in (3) is the transform of thesolution of the homogeneous equation P (D)y = 0 that satisfies the given initialconditions. The same statements are true for all linear differential equationswith constant coefficients, regardless of order.

Partial Fraction Expansions

Steps 1 and 2 of the general procedure just described are easy, provided that thetransform of the nonhomogeneous term is known. The challenging part of solvingany problem in this way lies in step 3. Extensive tables of known transforms areavailable to help in this task. As in Examples 3 and 4, partial fraction expansionis often a useful tool for expressing Y (s) is terms of known transforms.

Computer algebra systems such as Mathematica and Maple have the abilityto compute partial fraction expansions as well as all of the “standard” Laplacetransforms and inverse Laplace transforms. (You can even compute partial frac-tion expansions with the expand() function on the TI-89 and Voyage calcula-tors.) Nevertheless, it is beneficial to have some skill in efficiently computingpartial fraction expansions “by hand.”

A rational function N(s)P (s) with degree(N) < degree(P ) may be expanded into a sum

of rational functions with denominators corresponding to the linear and irreduciblequadratic factors of P (s) and with numerators that are either constant or linear. Asimple linear factor s − a of P (s) results in a single simple term of the form A

s−a , andrepeated linear factors (s− a)k can give rise to terms

A1

s− a+

A2

(s− a)2+ · · ·+ Ak

(s− a)k.


A simple irreducible quadratic factor s2 + bs + c of P (s) results in a term of the formBs+Cs2+bs+c , and repeated quadratic factors (s2 + bs+ c)k can give rise to terms

B1s+ C1

s2 + bs+ c+

B2s+ C2

(s2 + bs+ c)2+ · · ·+ Bks+ Ck

(s2 + bs+ c)k.

The number of undetermined constants needed in any partial fraction expansion is equalto the degree of the denominator P (s).

• Example 5 For any numerator N(s) with degree less than 10, there are constantsA1, A2, A3, B,C1, D1, C2, D2, E, and F such that

N(s)s3(s+ 1)(s2 + 1)2(s2 + s+ 1)

is equal to

A1

s+A2

s2+A3

s3+

B

s+ 1+C1s+D1

s2 + 1+C2s+D2

(s2 + 1)2+

Es+ F

s2 + s+ 1. �

The problem of computing the constants in a partial fraction expansion ultimatelyconsists of solving a system of linear equations. There are various ways of assembling asuitable system of equations, one of which is to clear fractions, expand products, andequate coefficients. This approach is interesting from a theoretical point of view butusually leads to a system of equations that requires considerable effort to solve. On theother hand, one can usually assemble a much simpler set of equations with the help ofa few simple algebraic manipulations and evaluations at wisely chosen values of s. Thefollowing examples illustrate these techniques.

• Example 6 Let’s find the partial fraction expansion of

s2

(s+ 1)(s+ 2)(s+ 3),

which is typical of cases involving denominators with distinct linear factors. We seekconstants A1, A2, and A3 such that

s2

(s+ 1)(s+ 2)(s+ 3)=

A1

s+ 1+

A2

s+ 2+

A3

s+ 3.

Multiplying both sides by s+ 1 gives

s2

(s+ 2)(s+ 3)= A1 +A2

s+ 1s+ 2

+A3s+ 1s+ 3

,

into which we substitute s = −1, obtaining

12

= A1.

Multiplying both sides instead by s+ 2 gives

s2

(s+ 1)(s+ 3)= A1

s+ 2s+ 1

+A2 +A3s+ 2s+ 3

,



−4 = A2.


s2

(s+ 1)(s+ 2)= A1

s+ 3s+ 1

+A2s+ 3s+ 2

+A3,


92

= A3.

Thus our partial fraction expansion is

s2

(s+ 1)(s+ 2)(s+ 3)=

1/2s+ 1

+−4s+ 2

+9/2s+ 3

.

Now, looking back over the preceding calculations, it is evident that an elegant shortcutamounts to evaluating the left-hand side at each root of the denominator after removingthe corresponding factor from the denominator; that is,

A1 =s2

(s+ 1)/////////(s+ 2)(s+ 3)

∣∣∣s=−1

=12,

A2 =s2

(s+ 1)(s+ 2)/////////(s+ 3)

∣∣∣s=−2

= −4,

A3 =s2

(s+ 1)(s+ 2)(s+ 3)/////////

∣∣∣s=−3

= −92. �

• Example 7 Let’s find the partial fraction expansion ofs

(s+ 4)3,

which is typical of cases involving denominators with repeated linear factors. We seekconstants A1, A2, and A3 such that

s2

(s+ 4)3=

A1

s+ 4+

A2

(s+ 4)2+

A3

(s+ 4)3.

We can find A3 as in the preceding example. Multiplying both sides by (s+ 4)3 gives

s2 = A1 (s+ 4)2 +A2 (s+ 4) +A3,


16 = A3.

To find A1, we’ll multiply both sides by s+ 4, obtaining

s2

(s+ 4)2= A1 +

A2

s+ 4+

A3

(s+ 4)2.


Now letting s→∞ gives

1 = A1.

So far we haves2

(s+ 4)3=

1s+ 4

+A2

(s+ 4)2+

16(s+ 4)3

.

At this point we can simply evaluate both sides at s = −3 (so that s+ 4 = 1) and solvefor A2:

9 = 1 +A2 + 16; so A2 = −8.

Thus our expansion is

s2

(s+ 4)3=

1s+ 4

− 8(s+ 4)2

+16

(s+ 4)3. �


s2

(s+ 2)(s2 + s+ 1),

in which the factor s2 + s + 1 is said to irreducible, since it has no real zeros. We seekconstants A, B, and C such that

s2

(s+ 2)(s2 + s+ 1)=

A

s+ 2+

Bs+ C

s2 + s+ 1.


s2

s2 + s+ 1= A+

(Bs+ C)(s+ 2)s2 + s+ 1

,


43

= A.

Now, letting s→∞ in the same equation gives

1 = A+B; so B = 1−A = −13.

With only C left to find, we can simply evaluate at s = 0, which gives

0 =A

2+ C; so C = −A

2= −2

3.

Thus our expansion is

s2

(s+ 2)(s2 + s+ 1)=

13

(4

s+ 2− s+ 2s2 + s+ 1

). �


5s3

(s2 + 4)(s2 + 2s+ 2),


whose denominator consists of two nonrepeated, irreducible quadratics. We seek con-stants A, B, C, and D such that

5s3

(s2 + 4)(s2 + 2s+ 2)=As+B

s2 + 4+

Cs+D

s2 + 2s+ 2.

Let’s first evaluate each side at s = 0, which gives us a simple equation involving onlyB and D:

0 = B/4 +D/2, i.e., B = −2D.

Next, let’s multiply both sides by s and let s→∞. That gives us

5 = A+ C.

Now we’ll multiply both sides by s2 + 4, obtaining

5s3

s2 + 2s+ 2= As+B +

(Cs+D)(s2 + 4)s2 + 2s+ 2

,

into which we’ll substitute s = 2i:−40i

−4 + 4i+ 2= 2Ai+B + 0.

Simplifying this gives

B + 2Ai = 4 (−2 + i),

from which we conclude that

A = 2 and B = −8.

Now returning to the relationships 5 = A+ C and B = −2D, we obtain

C = 3 and D = 4.

Finally, the desired expansion is

5s3

(s2 + 4)(s2 + 2s+ 2)=

2(s− 4)s2 + 4

+3s+ 4

s2 + 2s+ 2. �


s2(1− s)(s2 + 2s+ 2)2

,

whose denominator consists of a repeated irreducible quadratic. We seek constants A,B, C, and D such that

s2(1− s)(s2 + 2s+ 2)2

=As+B

s2 + 2s+ 2+

Cs+D

(s2 + 2s+ 2)2.

As in the previous example, we begin by evaluating both sides at s = 0 and then bymultiplying both sides by s and letting s→∞, which give

0 = B/2 +D/4 and − 1 = A+ 0.


Next, let’s clear fractions:

s2(1− s) = (As+B)(s2 + 2s+ 2) + Cs+D.

The roots of s2 + 2s+ 2 are −1± i, so we’ll evaluate both sides at s = −1 + i:

(−1 + i)2(2− i) = C (−1 + i) +D.

Simplifying this gives

D − C + Ci = −2− 4i,

from which we conclude that

C = −4 and D = −6.

Now returning to the relationship B/2 + D/4 = 0, we find that B = 3, and since wehave already that A = −1, our expansion is

s2(1− s)(s2 + 2s+ 2)2

=−s+ 3

s2 + 2s+ 2− 2(s+ 3)

(s2 + 2s+ 2)2. �

Problems

In Problems 1 through 4, use the results of Examples 1 and 2 to write down the Laplacetransform of the given function. Express the result as a single quotient.

1. f(t) = 3− 5et 2. f(t) = 2e−t + 3e−2t

3. f(t) = cosh bt 4. f(t) = sinh bt

In Problems 5 through 10, find the Laplace transform Y (s) of the solution of the giveninitial-value problem.

5. y′ + 3y = e−t, y(0) = 0 6. y′′ + y = 0, y(0) = 1, y′(0) = −1

7. y′′ + y′ + y = 0, y(0) = 0, y′(0) = 1 8. y′′ + 3y′ + 2y = 0, y(0) = 1, y′(0) = 0

9. y′′ − 4y = e−t, y(0) = 1, y′(0) = 1 10. y′′′ − y = 1, y(0) = y′(0) = y′′(0) = 0

In Problems 11 through 14, suppose that f has the Laplace transform F (s), and writedown by inspection the transform Y (s) of the differential equation’s rest solution.

11. y′′ + y′ + y = f 12. y′′ + ω2y = f 13. y′′′ + y = f 14. y′′′ − y′ + 2y = f

15. Show that y = teat satisfies

y′′ − 2ay′ + a2y = 0, y(0) = 0, y′(0) = 1,

and use that fact to find the Laplace transform of teat.

16. (a) Use the fact that y = cosωt is the solution of

y′′ + ω2y = 0, y(0) = 1, y′(0) = 0,

to find the Laplace transform of cosωt.(b) Similarly find the Laplace transform of sinωt.


17. (a) Use the fact that y = t sinωt satisfies

y′′ + ω2y = 2ω cosωt, y(0) = 0, y′(0) = 0,

to find the Laplace transform of t sinωt. Make use of the known transform ofcosωt from Problem 16.

(b) Similarly find the Laplace transform of t cosωt.

Assuming that degree(N) < degree(P ), write down the form of the partial fractionexpansion of each of the rational functions in Problems 18 through 21.

18.N(s)

s (s− 1)219.

N(s)s2(s2 + 2s+ 2)

20.N(s)

(s2 + 1)(s2 + 4)221.

N(s)s4 − 1

Use the method described at the end of Example 6 to find the partial fraction expansionof the rational functions in Problems 22 through 24.

22.s

(s− 1)(s− 2)23.

12s (s+ 1)(s− 3)

24.2s

(s− 1)(s+ 1)

Use the method of Example 7 to help find the partial fraction expansion of the rationalfunction in each of Problems 25 through 27.

25.s

(s− 1)326.

s2

(s+ 2)327.

s2

(s+ 3)4

Use the method of Example 8 to find the partial fraction expansion of the rationalfunctions in Problems 28 through 30.

28.2s

(s+ 1)(s2 + 1)29.

5s2

(s+ 2)(s2 + 1)30.

s2 + 4(s+ 1)(s2 + 2s+ 2)

Use the method of Example 9 find the partial fraction expansion of the rational functionsin Problems 31 and 32.

31.8

(s2 + 1)(s2 + 4s+ 5)32.

8s2

(s2 + 4s+ 13)(s2 + 4s+ 5)

Use the method of Example 10 find the partial fraction expansion of the rational func-tions in Problems 33 through 35.

33.s2

(s2 + 4s+ 5)234.

s2

(s2 + 2s+ 5)235.

s3

(s2 + 2s+ 5)2

36. Find general formulas for the partial fraction expansions of

(a)s2

(s2 + a2)2(b)

s3

(s2 + a2)2

Solve Problems 37 through 43 as in Examples 3 and 4.

37. y′ + y = 1, y(0) = 0 38. y′ − y = e−t, y(0) = 0

39. y′′ − y = 0, y(0) = 0, y′(0) = 1 40. y′′ − 4y = 8, y(0) = y′(0) = 0

41. y′′ + 4y′ + 3y = 6, y(0) = y′(0) = 0 42. y′′ + 3y′ + 2y = e−3t, y(0) = y′(0) = 0

7.2. More Transforms and Further Properties 197

43. y′′′ + 2y′′ − y′ − 2y = 6, y(0) = y′(0) = y′′(0) = 0

44. Suppose that L[f(t)](s) = F (s). Prove the following “scaling theorems.”

(a) L[cf(ct)] = F (s/c) b) L[f(t/c)] = c F (cs)

45. Use a comparison test to show that if the improper integral in (1) converges for somes = s0, then it converges for all s > s0.

46. (a) Argue that f(t) = et2

does not have a Laplace transform (i.e., that the definingimproper integral is divergent for all s).

(b) Argue that f(t) = tt does not have a Laplace transform. (Hint : tt = et ln t.)

47. Let 0 ≤ a < b. Assume that the fundamental theorem of calculus,∫ b

a

f ′(t) dt = f(b)− f(a),

is true when f ′ is continuous on [a, b]. Prove that the same is true if f is continuousand f ′ is piecewise continuous on [a, b].

7.2 More Transforms and Further Properties

In this section we will find Laplace transforms for more of the elementary func-tions that commonly arise as solutions of linear differential equations with con-stant coefficients. We will also derive additional properties of Laplace transformsthat will help us in finding both transforms and inverse transforms.

First, let’s recall the results of Examples 1 and 2 in Section 1:

L[c ](s) =c

s, s > 0; L[eat ](s) =

1s− a

, s > a. (1, 2)

Note that the first of these is actually a consequence of the second (with a = 0)and linearity. Also, the computation that produced the second of these remainsvalid if a is a complex constant, provided that s > Re(a). That is,

L[e(α+iβ)t ](s) =1

s− α+ iβ=

s− α− iβ(s− α)2 + β2

, s > α.

So by the Euler-DeMoivre formula we have

L[eαt(cosβt+ i sinβt)](s) =s− α− iβ

(s− α)2 + β2, s > α.

By equating real and imaginary parts we arrive at

L[eαt cosβt](s) =s− α

(s− α)2 + β2, s > α, (3)

L[eαt sinβt](s) =β

(s− α)2 + β2, s > α, (4)


and, in particular,

L[cosβt] =s

s2 + β2, s > 0, L[sinβt] =

β

s2 + β2, s > 0. (5, 6)

The First Shift Theorem

Notice that the transforms of eαt cosβt and eαt sinβt in (3) and (4) can beviewed as shifted transforms of cosβt and sinβt obtained by replacing s withs− α. It turns out that the same thing happens whenever any function is mul-tiplied by eαt. To see this, suppose that f(t) has a known transform F (s) andconsider the transform of eαtf(t):

L[eαtf(t)](s) =∫ ∞

0e−steαtf(t)dt =

∫ ∞0

e−(s−α)tf(t)dt = F (s− α).

Thus we have what we will call the first shift theorem:

L[eαtf(t)](s) = F (s− α). J

• Example 1 Let’s find the inverse transform of

F (s) =s

s2 + 4s+ 13.

Since the denominator is an irreducible quadratic, the key is to complete thesquare and then use the first shift theorem. Completing the square shows thatthe denominator is (s+2)2+9. Now in order take advantage of the shift theorem,we must express the numerator in terms of s+ 2 as well :

F (s) =s+ 2− 2

(s+ 2)2 + 9=

s+ 2(s+ 2)2 + 9

− 2(s+ 2)2 + 9

.

The first term on the right is the transform of e−2t cos 3t by (3), and the secondis the transform of −2

3 e−2t sin 3t by (4). Therefore,

f(t) = e−2t(

cos 3t− 23

sin 3t). �

• Example 2 Suppose that we wish to find the rest solution of the equation

y′′ + 2y′ + 10y = 10.

Transforming each side of the equation (and using y(0) = y′(0) = 0) gives us

(s2 + 2s+ 10)Y (s) =10s.

So the transform of the solution is

Y (s) =10

s(s2 + 2s+ 10)=

1s− s+ 2s2 + 2s+ 10

.


The denominator in the second term is (s+ 1)2 + 9. To take advantage of this,we need the numerator in terms of s+ 1. So we write

Y (s) =1s− s+ 1

(s+ 1)2 + 9− 1

(s+ 1)2 + 9

and observe (by (1), (3), and (4)) that Y (s) is the Laplace transform of

y = 1− e−t cos 3t− 13e−t sin 3t. �

The Second Differentiation Theorem

The first shift theorem describes the transform of the product eαtf(t) in termsof the transform of f . A similar result describing the transform of the producttnf(t) would also be useful. To that end, let us consider a function f(t) withknown transform F (s):∫ ∞

0e−stf(t)dt = F (s), s > s0.

Differentiating with respect to s, we obtain∗∫ ∞0

e−st(−tf(t))dt =d

dsF (s), s > s0.

Since the left side here is the transform of −tf(t), we have found we shall callthe second differentiation theorem:

L[ tf(t)](s) = − dds

F (s). J

Repeated application of this result easily produces

L[ tnf(t)](s) = (−1)ndn

dsnF (s), n = 1, 2, 3, . . . .

The second differentiation theorem is useful for the derivation of a numberof transforms. For instance, with the constant function f(t) = 1, t ≥ 0, we findthat L[t ](s) = − d

ds

(s−1)

; that is,

L[t ](s) =1s2, s > 0.

Repeating the same procedure produces the transforms of t2, t3, and so on:

L[tn ](s) =n!sn+1

, s > 0. (7)

* Here we have made the questionable move of differentiating under the integral sign; however,it can be justified in this situation by a theorem from advanced calculus.


Combining (7) with the first shift theorem yields

L[tneat ](s) =n!

(s− a)n+1, s > a. (8)

The second differentiation theorem also produces

L[t cosβt] =s2 − β2

(s2 + β2)2, s > 0, (9)

L[t sinβt] =2βs

(s2 + β2)2, s > 0. (10)

To put (9) into a form more useful for finding inverse transforms, we note that

1s2 + β2

− s2 − β2

(s2 + β2)2=

2β2

(s2 + β2)2.

Now we combine (6) with (9) to obtain

L[

sinβtβ− t cosβt

]=

2β2

(s2 + β2)2, s > 0. (11)

The following two examples illustrate the use of (8) and (11) in the context ofsolving initial-value problems.


y′′ + 4y′ + 4y = 0, y(0) = 1, y′(0) = 1.

Transforming each side of the equation gives us

s2Y (s)− s− 1 + 4(Y (s)− 1) + 4Y (s) = 0,

which simplifies to

(s2 + 4s+ 4)Y (s) = s+ 5.


Y (s) =s+ 5

s2 + 4s+ 4=s+ 2 + 3(s+ 2)2

=1

s+ 2+

3(s+ 2)2

.

Therefore, because of (2) and (8), we see that Y is the transform of

y = e−2t + 3te−2t = (1 + 3t)e−2t,

which is easily verified as the solution of the initial-value problem. �


y′′ + 4y = sin 2t, y(0) = 1, y′(0) = 0.


Transforming each side of the equation gives us

s2Y (s)− s+ 4Y (s) =2

s2 + 4.


Y (s) =2

s2+4+ s

s2 + 4=

2(s2 + 4)2

+s

s2 + 4.

By (5) and (11), we see that

y =14

(12

sin 2t− t cos 2t)

+ cos 2t . �

Problems

In Problems 1 through 3, find the inverse transform (a) by partial fraction expansion and(b) by completing the square and using the first shift theorem with the basic transformsof cosh and sinh: L[sinh bt] = b

s2−b2 and L[cosh bt] = ss2−b2 .

1.2

s2 + 2s2.

2ss2 + 4s+ 3

3.4s

s2 − 6s+ 5

Find the inverse transform of each expression in Problems 4 through 15.

4.s

s2 + 4s+ 55.

s+ 1s2 + 6s+ 13

6.s+ 8

s2 + 10s+ 347.

2ss2 + 2s+ 5

8.s

(s+ 2)29.

2s2

(s+ 1)310.

s

(s+ 3)311.

s+ 1(s− 1)3

12.12s

(s2 + 4)213.

16(s2 + 4)2

14.4s2

(s2 + 4)215.

s3

(s2 + 4)2

Use formulas (10) and (11) and the first shift theorem to find the inverse transform ofeach expression in Problems 16 through 18.

16.16

(s2 + 2s+ 5)217.

s

(s2 + 2s+ 5)218.

2s+ 6(s2 + 2s+ 2)2

In Problems 19 through 21, use the Laplace transform to solve the initial-value problem.

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

20. y′′ + 6y′ + 13y = 0, y(0) = 0, y′(0) = 2

21. y′′ + 4y′ + 4y = 0, y(0) = 0, y′(0) = 1

In Problems 22 through 27, use the Laplace transform to find the rest solution of thedifferential equation.

22. y′′ + 4y′ + 3y = 4e−t 23. y′′ + 6y′ + 13y = 169t 24. y′′ + 4y′ + 4y = e−2t

25. y′′ + 4y = 6 sin t 26. y′′ + 9y = 9 sin 3t 27. y′′ + y = cos t


28. This purpose of this problem is to derive the Laplace transform of√t.

(a) Use the definition of the Laplace transform and integration by parts to show that

L[√t ](s) =

12s

∫ ∞0

e−st√tdt .

(b) Use the substitution st = x2 to arrive at

L[√t ](s) =

k

s3/2where k =

∫ ∞0

e−x2dx .

(c) After observing that

k2 =∫ ∞

0

e−x2dx

∫ ∞0

e−y2dy =

∫ ∞0

∫ ∞0

e−(x2+y2)dx dy ,

use polar coordinates to show that k2 = π/4. Finally conclude that

L[√t ](s) =

√π

2s3/2.

29. Let f be continuous on [0,∞) with Laplace transform F (s). Apply the first differ-entiation theorem to the antiderivative

∫ t0f(τ)dτ to obtain the first integration

theorem:

L[ ∫ t

0

f(u)du](s) =

F (s)s

.

30. In this problem we will derive the second integration theorem:

L[ 1tf(t)

](s) =

∫ ∞s

F (σ)dσ.

(a) Suppose that f has the transform F (s) and that 1t f(t) has the transform Φ(s).

Apply the second differentiation theorem to show that Φ′(s) = −F (s). Hence Φis some antiderivative of −F .

(b) By the result of Problem 36, we know that Φ(s) → 0 as s → ∞. Show that∫∞sF (σ)dσ has that property, as well as being an antiderivative of −F (s), and

therefore must be the desired transform Φ(s).

In Problems 31 through 34, use the second integration theorem (see Problem 30) to findthe Laplace transform of the given function.

31.1− eat

t32.

1− cosβtt

33.sinβtt

34.sinh btt

35. Use the second integration theorem (see Problem 30) and the result of Problem 28to find the Laplace transform of f(t) = 1/

√t .

36. Asymptotic Behavior of the Laplace Transform. Suppose that f(t) is bounded onevery bounded interval [a, b], a ≥ 0. Suppose further that the Laplace transformL[f ](s) = F (s) exists for all s > s0, where s0 ≥ 0. By the following sequence ofsteps, prove that

lims→∞

F (s) = 0.


(a) Argue that e−s1tf(t)→ 0 as t→∞ for any s1 > s0.

(b) Argue that because of (a) there is a time T1 > 0 such that |e−s1tf(t)| < 1 for allt ≥ T1, and consequently |f(t)| < es1t for all t ≥ T1.

(c) Use the result of (b) to obtain

|F (s)| <∫ T1

0

e−st|f(t)|dt+∫ ∞T1

e−stes1tdt.

(d) Use the result of (c) and the fact that f is bounded on [0, T1] to obtain

|F (s)| < M

s+

1s− s1

for all s > s1,

where M is a bound on |f(t)| for all 0 ≤ t ≤ T1.

(e) Finally, conclude that F (s) → 0 as s → ∞. Also observe the stronger fact thatsF (s) is bounded for large s.

37. Combine the result of Problem 36 with the first differentiation theorem to show thatif f and f ′ have Laplace transforms and f is continuous, then

lims→∞

sF (s) = f(0).

38. The error function erf and the complementary error function erfc are

erf(t) =2√π

∫ t

0

e−x2dx and erfc(t) = 1− erf(t) =

2√π

∫ ∞t

e−x2dx

(a) Show that y = e−t2/4 satisfies the initial-value problem

y′ +12t y = 0, y(0) = 1.

Then show that the transform Y of y satisfies

s Y (s)− 1− 12Y ′(s) = 0.

(b) Use the appropriate integrating factor (see Section 2.1) to solve for Y (s) subjectto the condition Y (s)→ 0 as s→∞ (cf. Problem 36). Conclude that

L[e−t2/4](s) =

√π es

2erfc(s).

(c) Use the first integration theorem (see Problem 29) and the result of part (b) toshow that

L [ erf(t/2)] (s) =1ses

2erfc(s).

(d) Finally, use the “scaling theorem” L[f(t/k)] (s) = k F (ks) (see Problem 43 inSection 7.1) to show that, for k > 0,

L[e−t2/(4k2)](s) = k

√π ek

2s2erfc(ks) and L[

erf( t

2k

)](s) =

k

sek

2s2erfc(ks).


In Problems 39 and 40, use the first integration theorem (see Problem 29) to help infinding the solution of the given integro-differential initial-value problem.

39. y′ +∫ t

0

y(τ) dτ = t, y(0) = 0 40. y′′ +∫ t

0

y(τ) dτ = 0, y(0) = 3, y′(0) = 0

7.3 Heaviside Functions and Piecewise-Defined Inputs

The Laplace transform is particularly useful for solving initial-value problems inwhich the differential equation has a piecewise-defined nonhomogeneous term.In order to express such piecewise-defined functions in a convenient manner, wewill use the Heaviside unit step function, which is defined by

h(t) ={ 0, if t < 0;

1, if t ≥ 0. (1)

This function may be thought of as a “switch” that “turns on” at time t = 0.A switch that turns on at time t = c may be obtained by a simple shift:

h(t− c) ={ 0, if t < c;

1, if t ≥ c. (2)

Figures 1a–c show the graphs of h(t), h(t− 3), and h(t− 3)− h(t− 5).

2 4 6 8

1

2 4 6 8

1

2 4 6 8

1

Figure 1a Figure 1b Figure 1c

For further illustration, Figure 2 shows the graphs of

(a) h(t) + h(t− 2),

(b) h(t)− 2h(t− 1) + h(t− 2),

(c) h(t) + h(t− 1) + h(t− 2)− 3h(t− 3).

2 4 6 8

1

2

2 4

–1

1

2 4

1

2

3


7.3. Heaviside Functions and Piecewise-Defined Inputs 205

• Example 1 Let’s look closely at the function

f(t) = t h(t) + (2− 2t)h(t− 1) + (t− 2)h(t− 2)

with the goal of expressing it in “piecewise form.” When t < 0, all three ofthe Heaviside functions are “off,” and so f(t) = 0. When 0 ≤ t < 1, onlythe Heaviside function h(t) in the first term is “on,” and so f(t) = t. When1 ≤ t < 2, both Heaviside functions in the first two terms are “on,” and sof(t) = t+ (2− 2t) = 2− t. When t ≥ 2, allthree Heaviside functions are “on,” and sof(t) = t + (2 − 2t) + (t − 2) = 0. Thus, fhas the piecewise form

f(t) =

0, t < 0;t, 0 ≤ t < 1;2− t, 1 ≤ t < 2;0, 2 ≤ t.

Its graph is shown on the right. �

!1 1 2 3 4

1

A switch that is “on” during a given time interval a ≤ t < b is especiallyuseful and can be constructed as

h(t− a)− h(t− b) =

{0, if t < a;1, if a ≤ t < b;0, if b ≤ t.

(3)

The product of any function g(t) with h(t−a)−h(t−b) is a function that agreeswith g(t) when a ≤ t < b and is zero elsewhere. For instance, Figure 3 showsthe graphs of

(a) t (h(t)− h(t− 1)),

(b) (sin t) (h(t)− h(t− π)),

(c) (2− t)(h(t− 1)− h(t− 3)).

!1 1 2

1

"2

" 3 "2

1

!1 1 2 3 4 5

!1

1


When a piecewise-defined function consists of multiple nonzero pieces, it maybe viewed as a sum of functions of this type. This provides a convenient way ofexpressing the function in terms of Heaviside functions, as is illustrated in thenext example.


• Example 2 Consider the function

f(t) =

0, t < 0;t2, 0 ≤ t < 1;(t− 1)2, 1 ≤ t < 2;0, 2 ≤ t.

We can express f in terms of Heaviside func-tions as follows. We begin by thinking of f 1 2

1

as the sum of simpler functions f1 and f2, where

f1(t) ={t2, 0 ≤ t < 1;0, elsewhere;

and f2(t) ={

(t− 1)2, 1 ≤ t < 2;0, elsewhere.

These functions can be expressed as

f1(t) = t2(h(t)− h(t− 1)) and f2(t) = (t− 1)2(h(t− 1)− h(t− 2)).

So we have

f(t) = t2(h(t)− h(t− 1)) + (t− 1)2(h(t− 1)− h(t− 2)),

which we rearrange and write as

f(t) = t2 h(t) +((t− 1)2 − t2

)h(t− 1)− (t− 1)2 h(t− 2). �

Another way of obtaining an expression of a piecewise-defined function interms of Heaviside functions is often convenient as well. Suppose that f is givenin piecewise form as

f(t) =

0, t < 0;f0(t), 0 ≤ t < t1 ;f1(t), t1 ≤ t < t2 ;f2(t), t2 ≤ t < t3 ;

......

The representation of f in terms of Heaviside functions is

f(t) = f0(t)h(t) + (f1(t)− f0(t))h(t− t1) + (f2(t)− f1(t))h(t− t2) + · · · ,

the form of the typical term being (fnew − fprev)h(t − c). This is illustrated inthe next example.

• Example 3 The function

f(t) =

0, t < 0;t, 0 ≤ t < 1;1, 1 ≤ t < 2;3− t, 2 ≤ t < 3;0 3 ≤ t; !1 1 2 3 4

1


may be written in terms of Heaviside functions as

f(t) = t h(t) + (1− t)h(t− 1) + (3− t− 1)h(t− 2) + (0− (3− t))h(t− 3),

which after simplification becomes

f(t) = t h(t) + (1− t)h(t− 1) + (2− t)h(t− 2) + (t− 3)h(t− 3). �

The Second Shift Theorem

Given a function f(t) with Laplace transform F (s), consider the function

h(t− c)f(t− c)

where c ≥ 0. The graph of this function is the same as that of f shifted tothe right by c units and zeroed out for t < c. This is illustrated in Figure 5.

Figure 5

The Laplace transform of h(t− c)f(t− c) is computed as follows:

L[h(t− c)f(t− c)](s) =∫ ∞

0e−sth(t− c)f(t− c)dt

=∫ ∞ce−stf(t− c)dt =

∫ ∞0e−s(t+c)f(t)dt = e−csF (s).

This computation gives us the second shift theorem:

L[h(t− c)f(t− c)](s) = e−csF (s). J

As a simple consequence (with f(t − c) = 1), we obtain the transform of anysimple Heaviside function:

L[h(t− c)](s) =e−cs

s. (4)

Note that with c = 0 this gives L[h(t)](s) = 1/s, which we have already seen asthe transform of the function f(t) = 1, t ≥ 0.


• Example 4 Suppose that we want to find the inverse transform of

F (s) =1− 2e−s + e−2s

s.

By splitting this into three terms,

F (s) =1s− 2

e−s

s+e−2s

s,

we recognize that

f(t) = h(t)− 2h(t− 1) + h(t− 2).

This is the function shown in Figure 2b. �

• Example 5 Suppose that we want to find the inverse transform of

F (s) =1− 2e−s + e−2s

s2.

After splitting this into three terms and recalling that 1/s2 = L[t](s), the secondshift theorem tells us that

f(t) = t h(t)− 2(t− 1)h(t− 1) + (t− 2)h(t− 2).

This is the function in Example 1. �

• Example 6 Let’s find the inverse transform of

F (s) =π (1 + e−s)s2 + π2

.

After splitting this into two terms and recalling that π/(s2 +π2) = L[sinπt](s),the second shift theorem tells us that

f(t) = h(t) sinπt+ h(t− 1) sin(π(t− 1)).

Since sin(π(t− 1)) = − sinπt, this becomes

f(t) = (h(t)− h(t− 1)) sinπt ={ sinπt, 0 ≤ t < 1,

0, 1 ≤ t. �

As we saw in Examples 4–6, the second shift theorem makes it very straight-forward to write down the inverse transform of the product of e−cs and a knowntransform F (s). However, a slightly different form of the second shift theoremis often more convenient for the purpose of computing transforms. If we letg(t) = f(t − c), then f(t) = g(t + c) and so F (s) = L[g(t + c)](s). Thus thesecond shift theorem is equivalent to

L[h(t− c)g(t)](s) = e−csL[g(t+ c)](s). (5)


• Example 7 Consider the function

f(t) = t h(t)− t h(t− 1),

whose graph is shown in Figure 3a. To find the transform of f we apply thesecond shift theorem in the form of (5) as follows:

F (s) = L[t h(t)](s)− L[t h(t− 1)](s)= L[t](s)− e−sL[t+ 1](s)

=1s2− e−s

(1s2

+1s

)=

1− e−s(1 + s)s2

. �

• Example 8 Let’s find the transform of f(t) = eth(t) + (e−t− et)h(t− 1). Using(5), we have

F (s) = L[eth(t)](s) + L[(e−t − et)h(t− 1)](s)

=1

s− 1+ e−sL[e−(t+1) − et+1 ](s)

=1

s− 1+ e−s

(e−1

s+ 1− e

s− 1

)=

1− e1−s

s− 1+e−1−s

s+ 1. �

Initial Value Problems

Our main purpose here, of course, is to use all of this to help us solve initial-valueproblems with piecewise-defined inputs. We will illustrate with the followingexample.

• Example 9 Consider an undamped spring-mass system with a natural frequencyof ω

2π = 1 cycle per unit time, so that ω = 2π. Suppose that the mass is initiallyat rest and the platform to which the spring is attached undergoes motiondescribed by the function seen in Figure 2b. Thus we wish to find the restsolution of the equation

y′′ + 4π2y = 4π2(h(t)− 2h(t− 1) + h(t− 2)).

Formally taking Laplace transforms, we find

Y (s) =1− 2e−s + e−2s

s

4π2

(s2 + 4π2).

A partial fraction expansion shows that

Y (s) =(1− 2e−s + e−2s

) ( 1s− s

s2 + 4π2

).

The second factor is the transform of 1− cos 2πt; therefore

y(t) = (1− cos 2πt)h(t)−2(1− cos 2π(t−1))h(t−1)+(1− cos 2π(t−2))h(t−2)


by the second shift theorem. Since cos 2πt =cos 2π(t − k) for all integers k and all t, thesolution has the simple piecewise form

y(t) =

0, t < 0;1− cos 2πt, 0 ≤ t < 1;cos 2πt− 1, 1 ≤ t < 2;0, 2 ≤ t.

The graphs of this solution and the forcingfunction are seen in Figure 6. �

1 2

–1

1

Figure 6

We point out that in Example 9 we applied the first differentiation theoremto transform each side of the differential equation without taking care to checkthe hypotheses of that theorem. However, it is straightforward to verify a pos-teriori that the computation was justified. It is clear that y is continuous andexponentially bounded, and easy computations reveal that y′ is also continuousand exponentially bounded and that y′′ is piecewise continuous.

The solution in Example 9 is not a solution is the normal sense, because y′′

fails to exist at the points where f is discontinuous. Such a solution is called ageneralized solution.

Definition. Let p and q be constants, and let f be piecewise continuous on[0,∞). A generalized solution of

y′′ + py′ + qy = f

is a continuous function y on [0,∞) such that (i) y′ is continuous and (ii) thedifferential equation is satisfied at each point where f is continuous.

The following theorem on generalized solutions justifies the computation inExample 9 and similar problems. We omit the proof.

THEOREM 1 Let p and q be constants, and let f be piecewise continuous on[0,∞). Then for any numbers y0 and v0, the initial-value problem

y′′ + py′ + qy = f, y(0) = y0, y′(0) = v0,

has a unique generalized solution y on [0,∞). Moreover, y′′ is piecewise contin-uous, and if f is exponentially bounded, then so are y, y′, and y′′.

Problems

In Problems 1–3, write the given function f in piecewise form. Then sketch the graph.

1. f(t) = h(t) + h(t− 1) + h(t− 2)− h(t− 3)− h(t− 4)

2. f(t) = 4t(1− t)(h(t)− h(t− 1)

)3. f(t) = t h(t) + (1− t)h(t− 1)− h(t− 2)


In Problems 4–6, graph the given function without writing its piecewise form.

4. f(t) = 3h(t)− h(t− 1)− h(t− 2)− h(t− 3)

5. f(t) = h(t) + (1− t)h(t− 1) + (t− 2)h(t− 2)

6. f(t) = h(t)− 2h(t− 1) + h(t− 2) + h(t− 3)− 2h(t− 4) + h(t− 5)

In Problems 7–9, express the given function in terms of Heaviside functions.

7. 8. 9.

1 2 3 4

–1

1

1 2 3 4

1

1 2 3 4

1

2

Let bxc denote the “floor” (or greatest integer) function; that is,

bxc = n, where n ≤ x < n+ 1 with n an integer.

Write each of the functions in Problems 10 through 13 in terms of Heaviside functions.Also sketch the graph.

10. f(t) ={

(−1)btc, 0 ≤ t < 30, elsewhere

11. f(t) ={t− btc , 0 ≤ t < 30, elsewhere

12. f(t) ={t− bt+ .5c , 0 ≤ t < 20, elsewhere

13. f(t) ={

3− btc , 0 ≤ t < 30, elsewhere

Problems n = 14, . . . , 26: Find the Laplace transform of the function in Problem n−13.

In Problems 27 through 32, find the inverse Laplace transform and sketch its graph.

27.1− 2e−2s + e−3s

s28.

1− 2e−2s + e−3s

s229.

s+ e−2s − 2se−3s − s e−4s

s2

30.1− e−s + e−2s

s2 + 4π231.

4 (1− 2e−πs + e−2πs)s(s2 + 4)

32.2− 3e−πs + e−2πs

s2 + 1

In Problems 33 and 34, solve the initial-value problem; then graph the solution and thenonhomogeneous term on the right side of the differential equation.

33. y′ − y = −e h(t− 1), y(0) = 1 34. y′ + y = (t+ 1)(1− h(t− 2)), y(0) = 0

In Problems 35 and 36, solve the initial-value problem; then graph the solution and theforcing function.

35. y′′ + π2y = π2(h(t− 2)− h(t− 4)

), y(0) = 1, y′(0) = 0

36. y′′ + y = t h(t)− t h(t− 2π), y(0) = 0, y′(0) = 0

In Problems 37–39, obtain the transform of the given function by direct application ofthe definition of the transform, and then check by finding the transform with the secondshift theorem.

37. f(t) = h(t− 1)− h(t− 2) 38. f(t) = h(t)− 2h(t− 1) + h(t− 2)


39. f(t) = et(h(t)− h(t− 1)

)40. A piecewise constant function that is right-continuous for all t, zero on (−∞, t0),

and discontinuous at t0, t1, t2, . . . can be expressed as

f(t) = j0h(t− t0) + j1h(t− t1) + j2h(t− t2) + · · ·

where each ji is a constant. Show that ji = limt→t+

i

f(t)− limt→t−

i

f(t) . (Thus ji the

“jump” of f(t) at ti.) Show further that

F (s) =j0e

t0s + j1et1s + j2e

t2s + · · ·s

.

Using these results, redo Problems 9 and 22.

7.4 Periodic Inputs

A function f defined on [0,∞) is periodic if there is a number p > 0 such that

f(t+ p) = f(t) for all t ≥ 0.

The least such p is the function’s period . Such a function may be regarded asthe periodic extension of the function

f0(t) = f(t)(h(t)− h(t− p)),

which agrees with f on [0, p) and is zero elsewhere. We will see that the Laplacetransform of the periodic extension f of f0 can be easily expressed in terms ofthe transform of f0.

Since f(t+p) = f(t) for all t ≥ 0, it follows that f(t) = f(t−p) for all t ≥ p.Thus we can express f0 as

f0(t) = h(t)f(t)− h(t− p)f(t− p),

which allows use of the second shift theorem, producing the transform

F0(s) = F (s)− e−psF (s),

where F = Lf as usual. Now we solve for F (s), finding

F (s) =F0(s)

1− e−ps. (1)

Thus the transform of the periodic extension of f0 is simply the transform of f0

divided by 1− e−ps. The transform F0 = Lf0 in the numerator can be obtainedeither by expressing f0 in terms of Heaviside functions and applying the secondshift theorem or by evaluating the definite integral

F0(s) =∫ p

0e−stf(s)ds. (2)

7.4. Periodic Inputs 213


f(t) = (−1)btc, t ≥ 0

is the periodic extension, with period 2, of

f0(t) = h(t)− 2h(t− 1) + h(t− 2).

Graphs of f and f0 are shown below.

The transform of f0 is

F0(s) =1− 2e−s + e−2s

s;

therefore, the transform of f is∗

F (s) =1− 2e−s + e−2s

s(1− e−2s)=

1− e−s

1 + e−s1s

=tanh(s/2)

s. �


f(t) = | sin t |, t ≥ 0

can be viewed as the periodic extension of

f0(t) = sin t (h(t)− h(t− π)) = h(t) sin t+ h(t− π) sin(t− π)

with period π. These functions are plotted below.

The Laplace transform of f0 is

F0(s) = (1 + e−πs)1

s2 + 1;

* Recall that tanh x = ex−e−x

ex+e−x = 1−e−2x

1+e−2x .


consequently, the transform of f is

F (s) =1 + e−πs

1− e−πs1

s2 + 1=

coth(πs/2)s2 + 1

. �

Inverse transforms

The key to finding the inverse transform of a given transform with 1− e−ps asa factor in its denominator is the geometric series:

11− x

= 1 + x+ x2 + x3 + · · · , if |x| < 1.

Since 0 < e−ps < 1 for all s > 0 (and p > 0), we have the expansion

11− e−ps

= 1 + e−ps + e−2ps + e−3ps + · · · , s > 0.

Although the transform of any periodic function can be expressed in the form(1), a factor of 1− e−ps in the denominator does not guarantee that the inversetransform will be periodic, as we shall see in the first of the following examples.

• Example 3 The transform

F (s) =1

s (1− e−s)can be written as the series

F (s) =1s

+e−s

s+e−2s

s+e−3s

s+ · · · .

Taking inverse transforms term by term, we find that

f(t) = h(t) + h(t− 1) + h(t− 2) + h(t− 3) + · · · ,

which is easily recognizable as the floor function f(t) = btc , t ≥ 0. �


F (s) =1− e−s

s (1− e−2s)=

1s (1 + e−s)

can be written as the series

F (s) =1s− e−s

s+e−2s

s− e−3s

s+ · · · .

Taking inverse transforms term by term, we find that

f(t) = h(t)− h(t− 1) + h(t− 2)− h(t− 3) + · · · ,

which can be expressed compactly as f(t) =(1 + (−1)btc

)/2, t ≥ 0. �



F (s) =1

(s2 + 1)(1− e−πs)can be written as

F (s) =1

s2 + 1(1 + e−πs + e−2πs + e−3πs + · · · ) .

Since the first factor is the transform of sin t, we apply successive shifts to obtain

f(t) = h(t) sin t+ h(t− π) sin(t− π) + h(t− 2π) sin(t− 2π) + · · · .

Since sin(t− kπ) = (−1)k sin t for any integer k, we can write this function as

f(t) = (h(t)− h(t− π) + h(t− 2π)− h(t− 3π) + · · · ) sin t,

which is the same as

f(t) =12

( sin t+ | sin t |). �

• Example 6 Let f be as in Example 1; that is, let f be the periodic extension of

f0(t) = h(t)− 2h(t− 1) + h(t− 2)

with period 2. Let’s find the rest solution of the equation

y′′ + 4π2y = 4π2f(t).

The transform of f is

F (s) =1− 2e−s + e−2s

s (1− e−2s);

so the transform of the solution is

Y (s) =1− 2e−s + e−2s

1− e−2s

4π2

s (s2 + 4π2).

The first factor has the series expansion

1− 2e−s + e−2s

1− e−2s= (1− 2e−s + e−2s)(1 + e−2s + e−4s + e−6s + · · · )

= 1− 2e−s + 2e−2s − 2e−3s + · · · ,

and the second factor has the partial fraction expansion

4π2

s(s2 + 4π2)=

1s− s

s2 + 4π2.

So the transform can be written as

Y (s) = (1− 2e−s + 2e−2s − 2e−3s + · · · )(

1s− s

s2 + 4π2

).


Recognizing the second factor as the transform of 1− cos 2πt, we apply succes-sive shifts, obtaining the inverse transform

y(t) = (1− cos 2πt)h(t)− 2(1− cos 2π(t− 1))h(t− 1)+ 2(1− cos 2π(t− 2))h(t− 2)− 2(1− cos 2π(t− 3))h(t− 3) + · · · .

Since cos 2π(t− k) = cos 2πt for any integer k, we finally arrive at the solution

y(t) = (1− cos 2πt)(h(t)− 2h(t− 1) + 2h(t− 3)− · · · )= (−1)btc(1− cos 2πt).

Notice that this is precisely the periodic extension with period 2 of the solutionin Example 8 in Section 7.3. �

Problems

The functions in Problems 1 through 3 are the same as those in Problems 7 through 9in Section 7.3. For each of them, find the Laplace transform of the periodic extensionwith period p = 4.

1. 2. 3.

1 2 3 4

–1

1

1 2 3 4

1

1 2 3 4

1

2

4. Let f0(t) = h(t)− h(t− 1)− h(t− 2) + h(t− 3).(a) Sketch the graph of the periodic extension of f0 with period p = 3 and find its

Laplace transform.(b) Sketch the graph of the periodic extension of f0 with period p = 4 and find its

Laplace transform.

5. Let f(t) = t − btc , t ≥ 0. Graph f , describe it as a periodic extension of somerepresentative function f0, and then find its Laplace transform.

6. Let f(t) = (t − btc)2, t ≥ 0. Graph f , describe it as a periodic extension of somerepresentative function f0, and then find its Laplace transform.

In Problems 7 through 10, find the inverse Laplace transform and sketch its graph.

7.s

(s2 + 1)(1− e−πs)8.

1− e−2πs

(s2 + 1)(1− e−4πs)

9.1− e−s

s2(1− e−2s)10.

1− e−s

s2(1 + e−2s)


In Problems 11 and 12, find the rest solution of the equation and plot its graph. Notethat each equation has a forcing function similar to t− btc from Problem 5.

11. y′′ + 4π2y = 4π2(1− (t− btc)

)12. y′′ + π2y = π3

(t− btc

)13. Let f be the periodic extension of f0(t) = h(t) − 2h(t − 1) + h(t − 2) with period

p = 4. Find the rest solution of the equation

y′′ + 4π2y = 4π2f(t)

and sketch its graph along with the graph of f .

A function f defined on [0,∞) is said to be antiperiodic with “antiperiod” k iff(t + k) = −f(t) for all t ≥ 0. For example, sin t is antiperiodic with antiperiod π,and (−1)btc is antiperiodic with antiperiod 1. An antiperiodic function with antiperiodk can be described as the antiperiodic extension of

f0(t) = f(t)(h(t)− h(t− k)),

which vanishes outside the interval [0, k]. Since f(t+ k) = −f(t) for all t ≥ 0, it followsthat f0(t) can be written as

f0(t) = f(t)h(t) + h(t− k)f(t− k).

In Problems 14 through 16, assume that f is the antiperiodic extension of f0 withantiperiod k.

14. Show that the transform of f can be expressed in terms of the transform of f0 as

F (s) =F0(s)

1 + e−ks=

11 + e−ks

∫ k

0

e−stf(t)dt .

15. Let φ(t) = |f(t)| for all t ≥ 0. To electrical engineers, |f(t)| is known as the full-waverectification of f(t).

(a) Show that φ is periodic with period k and thus

Φ(s) =1

1− e−ks

∫ k

0

e−st |f(t)|dt.

(b) Conclude that if f(t) ≥ 0 for 0 ≤ t < k, then the following relationship holdsbetween the Laplace transforms of f(t) and φ(t) = |f(t)|:

Φ(s) =1 + e−ks

1− e−ksF (s) = coth(ks/2)F (s).

16. The function g(t) = 12

(f(t) + |f(t)|

)is the half-wave rectification of f(t).

(a) Sketch the graph of the half-wave rectification of sin 2πt.

(b) Show that if f(t) ≥ 0 for 0 ≤ t < k, then the following relationship holds betweenthe Laplace transforms of f(t) and g(t) = 1

2

(f(t) + |f(t)|

):

G(s) =1

1− e−2ks

∫ k

0

e−stf(t)dt =F (s)

1− e−ks.


17. Use the result of Problem 14 to rework Example 1.

18. Use the result of Problem 15 to rework Example 2.

19. Use the result of Problem 15 in a backward fashion to obtain the result of Example1 from the transform L[h(t)] = 1/s.

In Problems 20 and 21, sketch a graph of the given function, and use the appropriateresult from either Problem 15 or 16, along with the first shift theorem, to find its Laplacetransform.

20. g(t) = e−t | sinπt | 21. g(t) = 12 e−t( sinπt+ | sinπt |

).

7.5 Impulses and the Dirac Distribution

Consider a spring-mass system being acted upon by some external force. ByNewton’s second law, force is the rate of change in momentum:

F (t) =dµ

dt,

where µ = mv. Thus, as illustrated in Fig-ure 1, a force acting on the mass over atime interval [t0, t1] results in a change inmomentum given by

∆µ = µ(t1)− µ(t0) =∫ t1

t0

F (t)dt.

Force

Momentum

Figure 1A force acting on the mass over a verysmall time interval [ t, t+ ∆t] might result, for instance, from stiking the masswith a hammer. It will be useful to imagine an approximation to such a forcein the form of an impulse—an idealized force concentrated at a single point t0in time and resulting in an instantaneous change in momentum. Our goal hereis to develop a model for such a force.

Consider a unit change in momentum caused by a force F (t) whose value isa constant b between time t = 0 and t = ∆t and zero at all other times; that is,

F (t) = (h(t)− h(t−∆t)) b and ∆µ =∫ ∆t

0b dt = 1.

This arrangement is always possible no matter how short the time interval is.In fact, b and ∆t are related by b∆t = 1; therefore, b = 1/∆t, and so

F (t) =h(t)− h(t−∆t)

∆t. (1)

The limit of such forces as ∆t → 0 defines a unit impulse at time t = 0and may be thought of a generalized derivative of the Heaviside function h(t).

7.5. Impulses and the Dirac Distribution 219

Clearly this will not be a function in the usual sense, but we will use notationthat seems to suggest that it is. To capture the limiting behavior of F as ∆t→ 0,we define the symbol δ(t), with units of time−1 in the current context, by thefollowing properties:

(i) δ(t) = 0 for all t 6= 0; (ii)∫ ∞−∞

δ(t)dt = 1.

We emphasize that δ(t) is not a function of t in the usual sense, because (i) and(ii) are contradictory in the realm of ordinary calculus. Instead, δ(t) is a pecu-liar invention that serves as a “limit” for a sequence of functions whose valuesconverge to zero except at a single point, but whose integrals do not converge tozero. It is an example of a class of “generalized functions” called distributions.This particular distribution δ(t) is known as the Dirac distribution, or theDirac delta function, after the great physicist Paul Dirac.

Let g(t) be a continuous function on (−∞,∞). Since δ(t) represents the“limit” as ∆t→ 0 of the function F (t) in (1), we define∫ ∞−∞

g(t)δ(t)dt = lim∆t→0

∫ ∞−∞

g(t)h(t)− h(t−∆t)

∆tdt = lim

∆t→0

1∆t

∫ ∆t

0g(t)dt.

This last quantity is the limit as ∆t→ 0 of the average value of g(t) over [0,∆t].Since g is continuous, this limit must be g(0). Therefore, it is consistent with(i) and (ii) above to define

(iii)∫ ∞−∞

g(t)δ(t)dt = g(0) for any continuous g on (−∞,∞).

A unit impulse at time t = c is represented by δ(t − c), the unit impulse att = 0 shifted to occur at t = c. A simple change of variables applied to (iii) gives

(iv)∫ ∞−∞

g(t)δ(t− c)dt = g(c) for any continuous g on (−∞,∞).

These last two properties put us in a position to state the Laplace transformof δ(t) and δ(t− c). Applying (iii) and (iv) with g(t) = e−st, we find that

L[δ(t)](s) = 1 and L[δ(t− c)](s) = e−cs. (2)

Initial-Value Problems

Impulses can cause discontinuities in the derivatives of the solution of a differ-ential equation, or in the solution itself in the case of a first-order equation.Therefore, initial values should be understood in the sense of left-sided lim-its. For example, in a second-order initial-value problem, the initial conditionsy(0) = y0 and y′(0) = v0 are understood to mean

y(0−) = limt→0−

y(t) = y0 and y′(0−) = limt→0−

y′(t) = v0. (3)


The rest solution of a second-order equation is understood to be the solution yfor which y(t) = 0 for all t ≤ 0, which implies that each of the statements in (3)holds with y0 = v0 = 0.

• Example 1 Suppose that an undamped spring-mass system with with mass m,stiffness k, and natural angular frequency ω =

√k/m is initially at rest. At time

t = 0, the mass is struck with a hammer, imparting an instantaneous transferof momentum ∆µ. The subsequent motion may be modeled with the equation

my′′ + k y = ∆µ · δ(t) .

(Note that all terms have units of force.) With zero initial values, the trans-formed equation is

ms2Y (s) + k Y (s) = ∆µ · 1 .

Thus the transform of the rest solution is

Y (s) =∆µ

ms2 + k=

∆µ/ms2 + ω2

,

and consequently the solution is

y(t) =∆µmω

h(t) sinωt .

For t ≥ 0, this is the same as the solution of my′′ + k y = 0 with y(0) = 0and initial velocity y′(0) = ∆µ/m. Also note that, if viewed on (−∞,∞), thesolution is continuous, but its derivative has a jump discontinuity at t = 0. �

• Example 2 Suppose that an undamped spring-mass system with mass m, stiff-ness k, and natural angular frequency ω =

√k/m is initially in motion with

position y(0) = 0 and velocity y′(0) = v0. At a later time t = t1, the mass isstruck with a hammer, imparting an instantaneous transfer of momentum ∆µ.The motion of the mass may be modeled by

my′′ + k y = ∆µ · δ(t− t1), y(0) = 0, y′(0) = v0.

The transformed equation is

m (s2Y (s)− v0) + k Y (s) = ∆µ e−t1s;

so the transform of the solution is

Y (s) =∆µ e−t1s +mv0

ms2 + k=(

∆µ e−t1s

mω+v0

ω

)ω

s2 + ω2.

Since the second factor is the transform of sinωt, we apply the second shifttheorem to obtain

y(t) =∆µmω

h(t− t1) sin (ω(t− t1)) +v0

ωh(t) sinωt.

7.5. Impulses and the Dirac Distribution 221

Note that if ∆µ = mv0 and t1 = (2k + 1)πω for some integer k ≥ 0, then fort ≥ t1 we have

y(t) =v0

ωsin(ω t− (2k + 1)π) +

v0

ωsinωt =

v0

ω(− sin(ωt) + sinωt) = 0,

and so the impulse instantaneously stops the motion. �

Circuits. Let Q(t) be the charge on the capacitor in a simple RC, LC, or RLCcircuit. (See Sections 2.2.3 and 5.1.) If current flows for some period of time[t0, t1], then there is a resulting change in the capacitor’s charge given by

∆Q = Q(t1)−Q(t0) =∫ t1

t0

Q′(t)dt.

Since Q′(t) is precisely the current I(t), this becomes

∆Q =∫ t1

t0

I(t)dt.

So, just as an instantaneous unit change in the momentum of a mass can bedescribed with the Dirac distribution, so can an instantaneous unit change inthe charge on a capacitor. A (nearly) instantaneous change in charge mightoccur due to a quick discharge across a spark gap or a quick charge from a briefspike in source voltage.

• Example 3 Consider an RC circuit with R = C = 1 and an initial charge ofQ0 = 0 on the capacitor. A voltage source produces large spikes at t = 0 andt = 1 and is zero elsewhere. With y representing the charge, the model is

y′ + y = a0δ(t) + a1δ(t− 1), y(0) = 0,

where a0 and a1 are the changes in charge that occur due the voltage spikes.The transform of the solution is

Y (s) = (a0 + a1e−s)

1s+ 1

.

Thus the solution is

y = a0e−th(t) + a1e

−(t−1)h(t− 1) = (a0h(t) + a1e h(t− 1))e−t. �

Problems

Evaluate each integral in Problems 1 through 6.

1.∫ ∞−∞

(t2 + 3) δ(t− 2)dt 2.∫ 1

−1

(t2 + 3) δ(t)dt 3.∫ ∞−∞

δ(t− π) cos t dt

4.∫ π

−πδ(t− π

6

)sin t dt 5.

∫ 1

−1

δ(t)1 + t2

dt 6.∫ ∞−∞

δ(t+ 1)e−t dt


In Problems 7 through 12, find and graph the rest solution.

7. y′′ = δ(t)− δ(t− 1) 8. y′′ = δ(t)− 2δ(t− 1) + δ(t− 2)

9. y′′ + y = δ(t− 1) 10. y′′ + y = δ(t) + δ(t− π)

11. y′′ + 3y′ + 2y = δ(t)− δ(t− ln 3) 12. y′′ + 2y′ + y = δ(t− 1)− δ(t− 2)

In Problems 13-15, find and graph the solution satisfying y(0) = 1.

13. y′ = δ(t−1) 14. y′ + y = δ(t−ln 2)−δ(t−ln 3) 15. y′ + y =−δ(t−ln 2)

16. Consider the rest solution of y′ − y = δ(t)− a δ(t− 1). Find a so that the solution iszero for all t ≥ 1.

17. Show that if p and q are constants, then the rest solution of y′′ + p y′ + q y = δ(t)and the solution of y′′ + p y′ + q y = 0, y(0) = 0, y′(0) = 1, have the same Laplacetransform; hence the solutions agree for all t ≥ 0.

18. (a) Using Laplace transforms, find the solution of y′ = δ(t), y(0−) = 0. What inter-pretation of δ(t) does this imply?

(b) Let a > 0 and find the solution of y′ = 1a

(h(t)− h(t− a)

), y(0−) = 0 in terms of

a. Sketch the graph of the forcing function and the solution for a = 1, 0.5, and0.1. What happens as a→ 0+?

19. (a) Let a > 0, and find the rest solution of y′′+ y = 1a

(h(t)− h(t− a)

)in terms of a.

(b) Show that if t ≥ a, then y(t) = 1a

(cos(t− a)− cos t

).

(c) Argue that if t > 0, then y(t)→ sin t as a→ 0+.(d) Compare with the rest solution of y′′ + y = δ(t).

20. Find an impulse-driven system whose rest solution is as in the following figure.

–1 1 2 3 4 5 6

–1

1

7.6 Convolution

In this section we will study integral equations of the form

y(t) +∫ t

0g(t− x)y(x)dx = f(t), (1)

where f and g are given functions of t. An equation of this form is called a linearVolterra integral equation of convolution type. General linear Volterra integralequations are of the form

α(t)y(t) +∫ t

ag(t, x)y(x)dx = f(t).

7.6. Convolution 223

When g(t, x) can be expressed as a function of t − x as in (1), the integralinvolved is called a convolution integral, which defines the convolution of thetwo functions g and y, sometimes denoted by g ∗ y. That is,

(g ∗ y)(t) =∫ t

0g(t− x)y(x)dx,

and so (1) can be written in the abbreviated form

y + g ∗ y = f.

Integral equations (and integro-differential equations) arise in applications wherebehavior is influenced by past history as well as the state at time t. Convolutionintegrals, in particular, arise when the contribution at time t of the history attime x < t depends upon the elapsed time t− x.

• Example 1 The convolution of t and t2 is

t ∗ t2 =∫ t

0(t− x)x2 dx =

(t

3x3 − 1

4x4

)]x=t

x=0

=112t4 .

The convolution of et and e−t is

et ∗ e−t =∫ t

0et−xe−x dx =

∫ t

0ete−2x dx = −1

2ete−2x

∣∣∣x=t

x=0

=12et (1− e−2t) = sinh t. �

It is clear from the linearity properties of integration that for a given g theoperator K defined by Ky = g ∗ y is linear; that is,

g ∗ (c y) = c (g ∗ y) and g ∗ (u+ v) = g ∗ u+ g ∗ v.

It is also true that convolution is commutative; that is, g∗y = y∗g. To see this,simply make a change of variable τ = t− x in the integral as follows:

g ∗y =∫ t

0g(t−x)y(x)dx =

∫ 0

tg(τ)y(t− τ)(−dτ) =

∫ t

0y(t− τ)g(τ)dτ = y ∗ g.

The Convolution Theorem

With an eye toward solving (1), we would like to know how to express theLaplace transform of g ∗ y in terms of the individual transforms of g and y. Theanswer is provided by the convolution theorem:

L[ g ∗ y ](s) = G(s)Y (s), J

where the transform exists for all s such that both transforms G and Y exist.


To establish this result, first note that by the definition of the transform,

L[g ∗ y ](s) =∫ ∞

0e−st

(∫ t

0g(t− x)y(x)dx

)dt =

∫ ∞0

∫ t

0e−stg(t− x)y(x)dx dt.

The planar region over which we are integrating is {(t, x) | 0 ≤ x ≤ t <∞}; sointerchanging the order of integration produces

L[g ∗ y ](s) =∫ ∞

0

∫ ∞xe−stg(t− x)y(x) dt dx =

∫ ∞0y(x)

∫ ∞xe−stg(t− x) dt dx.

This step can be justified by the absolute convergence of the improper integralsinvolved, which follows from the existence of the Laplace transforms of g and y.Now the change of variable τ = t− x in the inner integral gives us

L[g ∗ y ](s) =∫ ∞

0y(x)

∫ ∞0e−s(τ+x)g(τ) dτ dx

=∫ ∞

0e−sxy(x)

∫ ∞0e−sτg(τ) dτ dx

=∫ ∞

0e−sxy(x)G(s) dx = G(s)

∫ ∞0e−sxy(x) dx = G(s)Y (s).

The next two examples illustrate how the convolution theorem can be usedto help calculate certain convolution integrals.

• Example 2 Consider the convolution

cos t ∗ sin t =∫ t

0cos(t− x) sinx dx.

According to the convolution theorem,

L[cos t ∗ sin t](s) =(

s

s2 + 1

)(1

s2 + 1

)=

s

(s2 + 1)2.

By (10) in Section 7.2, we recognize therefore that∫ t

0cos(t− x) sinx dx =

t

2sin t. �

• Example 3 Consider the convolution

√t ∗√t =

∫ t

0

√t− x

√x dx.

According to the convolution theorem and Problem 28 in Section 7.2,

L[√t ∗√t](s) =

( √π

2s3/2

)2

=π

4s3.


Therefore we conclude that∫ t

0

√t− x

√x dx =

π

8t2. �

Integral Equations

We will now look at a few examples to indicate how the convolution theoremcan be used to solve linear Volterra integral equations of convolution type.

• Example 4 Consider the integral equation

y(t) +∫ t

0y(x)dx = 1.

Viewing the integral as 1 ∗ y, we see that the transform of the solution satisfies

Y (s) +1sY (s) =

1s.

Solving for Y (s) yields Y (s) = 1/(s+ 1), and so y(t) = e−t. Note that differen-tiating this integral equation results in y′+y = 0 and that the integral equationimplies the initial value y(0) = 1. Thus it is no surprise that y = e−t. �


y(t) +∫ t

0(t− x)y(x)dx = 1.

Viewing the integral as t ∗ y, we see that the transform of the solution satisfies

Y (s) +1s2Y (s) =

1s.

Solving for Y (s) produces Y (s) = s/(s2 + 1), and consequently y(t) = cos t. �


y(t) + 3∫ t

0sin(t− x)y(x)dx = 4.

Viewing the integral as (sin t) ∗ y, we see that Y (s) satisfies

Y (s) +3

s2 + 1Y (s) =

4s.

Solving for Y (s) produces

Y (s) = 4s2 + 1

s (s2 + 4)=

3s

+3s

s2 + 4.

Therefore, the solution is y(t) = 1 + 3 cos 2t. �


The Green’s Function Revisited

In Section 6.2, it was shown that, given any two linearly independent solutionsu and v of the homogeneous equation y′′ + p y′ + q y = 0, the rest solution of

y′′ + p y′ + q y = f

could be obtained by variation of constants and expressed as

y(t) =∫ t

0G0(t, x)f(x)dx,

where G0(t, x) is the Green’s function

G0(t, x) =u(x)v(t) + u(t)v(x)u(x)v′(x)− u′(x)v(x)

corresponding to the operator P (D) = D2+pD+qI . Laplace transform methodsand the convolution theorem provide another approach to finding the Green’sfunction for an operator P (D) with constant coefficients.

Suppose that P (D) has constant coefficients and that f(t) is a given functionwith Laplace transform F (s). Recall that the Laplace transform of the restsolution of P (D)y = f is

Y (s) =1

P (s)F (s).

Thus, by the convolution theorem,

y =∫ t

0G0(t− x)f(x)dx, where LG0 =

1P (s)

.

Thus we can find the Green’s function for P (D) by finding the inverse Laplacetransform of 1/P (s). An interesting implication is that this type of Green’sfunction G0—for operators with constant coefficients—is always a function ofjust one variable.

• Example 7 The transform of the rest solution of y′′ + y = f is

Y (s) =1

s2 + 1F (s).

Thus, LG0 = 1s2+1

. Therefore G0(t) = sin t, and by the convolution theorem,

y =∫ t

0sin(t− x)f(x)dx. �

• Example 8 Consider the third-order operator P (D) = D3 + 2D2 − D − 2I ,which factors as (D+ I )(D− I )(D+ 2I ). The corresponding Green’s function


G0 is the inverse transform of

1P (s)

=1

(s+ 1)(s− 1)(s+ 2)=

16

(1

s− 1− 3s+ 1

+2

s+ 2

).

Therefore,

G0(t) =16(et − 3e−t + 2e−2t

),

and the rest solution of y′′′ + 2y′′ − y′ − 2y = f can be represented by

y =16

∫ t

0

(et−x − 3e−(t−x) + 2e−2(t−x)

)f(x) dx. �

It is interesting to note that, since the Green’s function G0(t) correspondingto the operator P (D) has the Laplace transform 1/P (s), it is actually the restsolution of the equation

P (D)y = δ(t),

where δ(t) is the Dirac distribution. Consequently, for any f with a Laplacetransform, the rest solution of P (D)y = f(t) is the convolution of f with therest solution of P (D)y = δ(t). Therefore, having found the rest solution ofP (D)y = δ(t), one has, in a sense, also found the rest solution of P (D)y = f(t)for any f .

Problems

In Problems 1 through 4, use the definition (and commutativity) to compute the con-volution.

1.√t ∗ t 2. t ∗ t 3. 1 ∗ cos t 4. t ∗ e−t

In Problems 5 through 8, evaluate the convolution by means of the convolution theorem.

5. sin t ∗ sin 2t 6. et ∗ et 7. et ∗ cos t 8. tm∗ tn, for integers m,n > 0

Solve the given integral equation in each of Problems 9 through 14.

9. y + e−t ∗ y = 1 10. et ∗ y = e−t 11. y + t ∗ y = t

12. y + t ∗ y = t2 13. y − t ∗ y = h(t)− h(t− 1) 14. y − t ∗ y = |1− t |

Solve the given integro-differential equation in Problems 15 through 17.

15. y′ + e−2t ∗ y = h(t), y(0) = 0 16. y′ + y − 2t ∗ y = 0, y(0) = 5

17. y′ − (cos t) ∗ y = h(t), y(0) = 0

In Problems 18 and 19, use the results of Problems 15 and 16 in Section 7.4 to help insolving the given integral equation.

18. (cos t) ∗ y = | sin t | 19. 2(cos t) ∗ y = sin t+ | sin t |


Suppose that v and v have Laplace transforms and that u(0) = u0, u′(0) = u1, v(0) =

v0, v′(0) = v1. Use the convolution theorem and the first differentiation theorem to

formally verify the identities in Problems 20 through 22.

20. u ∗ v′ − u′∗ v = u0v − v0u21. (u ∗ v)′ = u ∗ v′ + v0u = u′∗ v + u0v

22. (u ∗ v)′′ = u ∗ v′′ + v1u+ v0u′ = u′′∗ v + u1v + u0v

′

23. Conclude from Problem 21 that y = t ∗ eat satisfies y′ = ay + t. Also, the definitionof convolution implies that y(0) = 0. Use this to compute t ∗ eat.

24. Conclude from Problem 22 that y = g ∗ sin at satisfies y′′ + a2y = ag. Also concludefrom Problem 21 and the definition of convolution that y′(0) = y(0) = 0. Use thisto compute cos bt ∗ sin at.

25. Let y be the rest solution of y′′ + p y′ + q y = f, where p and q are constants withq 6= 0, and let u be the solution of u′′ + p u′ = 0, u(0) = 0, u′(0) = 1.

(a) Using the results of Problems 20 through 22, show that y satisfies the linearVolterra integral equation of convolution type

y + q u ∗ y = u ∗ f.

(b) Express the problem y′′ + y′ + y = et, y(0) = y′(0) = 0, as a linear Volterraintegral equation of convolution type.

26. Suppose that f and g have Laplace transforms and that P is a polynomial withconstant coefficients. Show that u is the rest solution of P (D)y = f , if and only ifg ∗ u is the rest solution of P (D)y = g ∗ f.

In Problems 27 through 30, find the Green’s function G0(t) of the given operator, andwrite down the rest solution of P (D)y = f .

27. P (D)y = y′′ − y 28. P (D)y = y′′ + 2y′ + 2y

29. P (D)y = y′′ + 3y′ + 2y 30. P (D)y = y′ + y

In Problems 31 through 33, find the Green’s function G0(t) of the given operator, andwrite down the solution of T y = f, y(0) = 0.

31. T y = y′ + 2e−3t ∗ y 32. T y = y′ + e−2t ∗ y 33. T y = y′ + 3 cos t ∗ y

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The Laplace Transform - Armstrongmath.armstrong.edu/faculty/hollis/classes/DE/book3411/...185 Chapter 7 The Laplace Transform In this chapter we will explore a method for solving linear