MATH 308 Differential Equationsjdkim/math308spring2020/math308week7.pdfChapter 6 The Laplace Transform Section 6.1 Deﬁnition of the Laplace Transform In this Chapter, we will study

JD Kim Texas A&M University Math 308 Differential Equations Spring 2020, week 7

MATH 308 Differential EquationsSpring 2020, WEEK 7

JoungDong Kim

Week 7: Section 6.1, 6.2, 6.3 Definition of the Laplace Transform, Solution of Initial Value Prob-lems, Step Functions

Chapter 6 The Laplace Transform

Section 6.1 Definition of the Laplace Transform

In this Chapter, we will study a new technique for solving linear ODE: Laplace Transform.Comparing with what we’ve learned in Chapter 2 and 3, this method is more advanced and requiremore skills of integral and algebra. Nevertheless, this method is more solid in the sense that we don’trely on the guess of the solutions. In addition, it can handle a much wider class of ODE than thosemethods we learned in the previous Chapters.

Improper Integrals

The Improper integral is defined as

∫

∞

a

f(t)dt := limA→∞

∫ A

a

f(t)dt,

where A is positive real number.If the limit exists, we say the improper integral converges to that limit, otherwise, it is said to

diverge.

Ex.1) Consider the improper integral,∫

∞

0

ectdt.

1



∞

1

1

tdt.


∞

1

t−pdt.

2


Piecewise Continuous

Definition 0.1 (Piecewise Continuous). A function f is said to be piecewise continuouson an interval α ≤ t ≤ β if the interval can be partitioned by a finite number of points α = t0 <

t1 < · · · < tn = β so that

(a) f is continuous on each open subinterval ti−1 < t < ti.

(b) f has a finite limit as the endpoints of each subinterval are approached from within thesubinterval.

Ex.4) Sketch the graph of

f(x) =

x, 0 ≤ x ≤ 1

3− x, 1 < x ≤ 2

1, 2 < x ≤ 3

Determine whether f is continous, piecewise continuous, or neither.

Definition 0.2 (Exponential Order). A function f is said to be of exponential order if|f(t)| ≤ Keat for t ≥ M , where K, a, and M are real constants.

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The Laplace Transform

Definition 0.3 (The Laplace Transform). Let f(t) is defined on [0,∞). The Laplace trans-form of f(t), denoted by L{f(t)} or by F (s), is defined by the improper integral

L{f(t)} = F (s) :=

∫

∞

0

e−stf(t)dt.

Ex.5) Let f(t) = 1, t ≥ 0. What is L{f(t)}?

Ex.6) Let f(t) = eat, t ≥ 0. What is L{f(t)}?

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Ex.7) Find the Laplace transform of f(t) = sin(at).

5


Theorem 0.4 (Existence of the Laplace Transform). Suppose that

(a) f is piecewise continuous on the interval [0, A] for any positive number A.

(b) |f(t)| ≤ Keat when t ≥ M . In this inequality, K, a and M are real constants, K and M

necessarily positive.

Then the Laplace Transform L{f(t)} = F (s) is well defined for s > a.

Note: The main point of Theorem 0.4 is that the Laplace transform is well defined for piecewisecontinuous functions. These functions may have finite many discontinuity. Recall that in theprevious Chapters, we only consider continuous functions. This technique can handle the ODEwith discontinuous right hand side g(t).

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Ex.8) Find the Laplace transform of f(t) =

{

t, 0 ≤ t < 1

0, 1 ≤ t < ∞.

Ex.9) Let

f(t) =

1, 0 ≤ t < 1

k, t = 1

0, t > 1

,

where k is a constant (typically large). What is L{f(t)}?

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Theorem 0.5 (Linearity of the Laplace Transform). Suppose that f1 and f2 are two functionswhose Laplace transforms exist. Then

L{c1f1(t) + c2f2(t)} = c1L{f1(t)}+ c2L{f2(t)},

where c1 and c2 are constants.

Ex.10) Find the Laplace transform of f(t) = 6 + 2e3t + 4 sin(5t).

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Section 6.2 Solution of Initial Value Problems

In this section, we will use couple of examples to show the general mechanism of using Laplacetransform to solve initial value problems for linear differential equations with constant coefficients.

Theorem 0.6 (Lifting Property of Laplace transform). Suppose that f is continuous andf ′ is piecewise continuous on any interval 0 ≤ t ≤ A. Suppose further that there exists constantsK, a and M such that |f(t)| ≤ Keat for t ≥ M , then L{f ′(t)} exists for s > a, and moreover

L{f ′(t)} = sL{f(t)} − f(0).

If we apply this on f ′′, we obtain the following identity for s > a:

L{f ′′(t)} = s2L{f(t)} − sf(0)− f ′(0).

Theorem 0.7. Suppose that the functions f, f ′, · · · , f (n−1) are continuous and that f (n) is piece-wise continuous on any interval 0 ≤ t ≤ A. Suppose further that there exists constants K, a andM such that |f(t)| ≤ Keat, |f ′(t)| ≤ Keat, · · · , |f (n−1)(t)| ≤ Keat for t > M . Then L{f (n)(t)}exists for s > a and is given by

L{f (n)(t)} = snL{f(t)} − s(n−1)f(0)− s(n−2)f ′(0)− · · · − sf (n−2)(0)− f (n−1)(0).

9


Definition 0.8 (Inverse Laplace Transform). Suppose that f(t) has a Laplace transform givenby L{f(t)} = F (s). Then f(t) is called the inverse Laplace transform of F (s) which we denoteby L−1{F (s)} = f(t).

Note. Since the Laplace transform is a linear operator, the inverse Laplace transform is also alinear operator. That is,

L−1{c1F1(s) + c2F2(s)} = c1L−1{F1(s)}+ c2L

−1{F2(s)},

where c1 and c2 are constants.

Ex.11) Find the inverse Laplace transform of each function.

(a) F (s) =4

(s− 1)3

(b) F (s) =2s− 3

s2 − 4

Note. You can find the Laplace Transform Table at the end of this note. This table makes itsimpler to find inverse Laplace transforms.

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Ex.12) Find the inverse Laplace transform of

F (s) =8s2 − 4s+ 12

s3 + 4s

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Ex.13) Use the Laplace transform to solve the initial value problem

y′′ − y′ − 2y = 0, y(0) = 1, y′(0) = 0.

12



y′′ + y = 0, y(0) = 2, y′(0) = 1.

13



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

14



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

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Summary

In general, if we have an initial value problem;

a0y(n) + a1y

(n−1) + · · ·+ any = g(t),

with initial conditions;

y(0) = b0, y′(0) = b1, · · · , y(n−1)(0) = bn−1.

The method of Laplace transform can be summarized as following three steps:

(a) Apply the Laplace transform on the equation. Using Theorem 0.7, we obtain an algebraicequation for Y (s) = L{y(t)}.

(b) Solve for Y (s).

(c) Solve for y(t) = L−1{Y (s)}.

First two steps are relatively easy and fast. The main difficulty is the last step. In fact, we alsohave an explicit expression of the inverse Laplace transform. However, this is beyond the materialof this course. What we can do is to use algebra tool to split/write Y (s) as a sum of several termswhich are in the Laplace transform table. Then we can check the table to find the inverse Laplacetransform for each term and finally we have y(t) as the sum of them.

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Section 6.3 Step functions

Unit step function

Definition 0.9 (Unit step function or Heaviside function). The unit step function orHeaviside function is defined by

uc(t) =

{

0, t < c

1, t ≥ c.

Since the Laplace transform involves values of t ∈ [0,∞), we are only interested in values of c ≥ 0.The graph of y = uc(t) is shown below. Note that the function can also be scaled or have anegative step.

Ex.17) Sketch the graph ofh(t) = uπ(t)− u2π(t).

Ex.18) Sketch the graph off(t) = (t− 3)u2(t)− (t− 2)u3(t)

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General discussion on piecewise continuous function

• f(t) = m(ua(t)− ub(t)), 0 < a < b.

• h(t) = g(t)(ua(t)− ub(t)), 0 < a < b.

18


Ex.19) Sketch the graph of

f(t) =

1, 0 ≤ t < 1

−1, 1 ≤ t < 2

2, 2 ≤ t < 3

1, t ≥ 3

Express f(t) in terms of the unit step function uc(t).

Ex.20) Consider the function

f(t) =

5, 0 ≤ t < 2

1, 2 ≤ t < 3

−1, t ≥ 3

.

Sketch the graph of the function and express f(t) in terms of the unit step functions uc(t).

19


Ex.21) Sketch the graph of the given function and express f(t) in terms of unit step functions uc(t).

f(t) =

t, 0 ≤ t < 2

2, 2 ≤ t < 5

7− t, 5 ≤ t < 7

0, t ≥ 7

.

Ex.22) Find the Laplace transform of uc(t).

20


For a given function f defined for t ≥ 0, we will often consider the related function g defined by

g(t) = uc(t)f(t− c) =

{

0, t < c

f(t− c), t ≥ c(1)

which represents a translation of f a distance of c units in the positive t direction.

Theorem 0.10. If F (s) = L{f(t)} exists for s ≥ a ≥ 0, and if c is a positive constant, then

L{uc(t)f(t− c)} = e−csF (s), s > a.

Conversely, if f(t) = L−1{F (s)}, then

uc(t)f(t− c) = L−1{e−csF (s)}.

21


Ex.23) Find the Laplace transform of the given function.

(a) f(t) =

{

0, t < 1

t2 − 2t+ 2, t ≥ 1

(b) f(t) =

{

0, 0 ≤ t < 2

t2 + t, t ≥ 2.

22


Ex.24) Find the inverse Laplace transform of each function

(a) F (s) =1− e−2s

s2

(b) F (s) =e−2s

s2 + s− 2

23


Theorem 0.11. If F (s) = L{f(t)} exists for s > a ≥ 0 and if c is a constant, then

L{ectf(t)} = F (s− c), s > a + c.

Conversely, if f(t) = L−1{F (s)}, then

ectf(t) = L−1{F (s− c)}.


G(s) =1

s2 − 4s+ 5.

24



G(s) =(s− 2)e−s

s2 − 4s+ 3

25


Table 1: Laplace Transform Table

f(t) = L−1{F (s)} F (s) = L{f(t)}

1. 11

s, s > 0

2. eat1

s− a, s > a

3. tn, n = positive integern!

sn+1, s > 0

4. tp, p > −1Γ(p+ 1)

Sp+1, s > 0

5. sin ata

s2 + a2, s > 0

6. cos ats

s2 + a2, s > 0

7. sinh ata

s2 − a2, s > |a|

8. cosh ats

s2 − a2, s > |a|

9. eat sin btb

(s− a)2 + b2, s > a

10. eat cos bts− a

(s− a)2 + b2, s > a

11. tneat, n = positive integern!

(s− a)n+1, s > a

12. uc(t)e−cs

s, s > 0

13. uc(t)f(t− c) e−csF (s)

14. ectf(t) F (s− c)

15. f(ct)1

cF(s

c

)

, c > 0

16.∫ t

0f(t− τ)g(τ)dτ F (s)G(s)

17. δ(t− c) e−cs

18. f (n)(t) snF (s)− sn−1F (0)− · · · − f (n−1)(0)

19. (−t)nf(t) F (n)(s)

Note. Laplace transform and inverse Laplace transform are linear operator.

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MATH 308 Differential Equationsjdkim/math308spring2020/math308week7.pdfChapter 6 The Laplace Transform Section 6.1 Deﬁnition of the Laplace Transform In this Chapter, we will study