1

Contents

Introduction 2

Chapter 1. Laplace Transform 3

1.1 Definition of the Laplace transform ……………………..…….…… 4

1.1.1 Examples of Laplace transform …………………………..… 4

1.2 Existence of Laplace transform …………………………………....….. 6

1.3 Basic Properties of Laplace transform …………………………….... 8

1.4 Laplace transform of the first and second derivative ……….…. 11

1.5 Definition of Heaviside function ………………………………….……. 12

1.6 Table of Laplace transforms ……………………………………………… 13

Chapter 2. Inverse Laplace Transform 14

2.1 Definition: Inverse Laplace transform ……………….….………..… 15

2.1.1 Lerch's Theorem ……………………………………..……….….. 15

2.1.2 Heaviside expansion formula …………………………….….. 15

2.2 Examples of Inverse Transform ……………………………………….. 17

Chapter 3. Applications of Laplace Transform 21

3.1 Introduction…………………………………………………………….……. 22

3.2 Example (Distinct real roots, but on matches term ) ………. 23

3.3 Example (Differential equations with discontinuous

Forcing function)………………………………………………………………..….. 25

Conclusion 28

References 29

2

INTRODUCTION

The Laplace transform is an integral transform method which is

particularly useful in solving linear ordinary differential equations. It finds

very wide applications in various areas of physics, electrical engineering,

control engineering, optics, mathematics and signal processing.

The Laplace transform can be interpreted as a transformation from the time

domain where inputs and outputs are functions of time to the frequency

domain where inputs and outputs are function of complex angular

frequency.

We will focus on the case of real variables and give some applications of

Laplace transform to solve different kinds of ordinary differential equations.

3

Chapter 1

Laplace Transform

4

𝟏. 𝟏 Definition of the Laplace transform In This chapter we will give :

• Definition of Laplace transform,

• Compute Laplace transform by definition, including piecewise continuous

functions.

Definition: Given a function 𝑓(𝑡), 𝑡 ≥ 0 , its Laplace transform 𝐹(𝑠) =

ℒ{𝑓(𝑡)} is defined as

𝑭(𝒔) = 𝓛{ 𝒇(𝒕) } = ∫ 𝒆−𝒔𝒕 ∞

𝟎𝒇(𝒕)𝒅𝒕 = 𝐥𝐢𝐦

𝑨→∞∫ 𝒆−𝒔𝒕

𝑨

𝟎𝒇(𝒕)𝒅𝒕

We say that the transform converges if the limit exists, and diverges if not.

Next, we will give some examples on computing the Laplace transform of

usual functions by using the definition.

1.1.1 Examples of Laplace Transform

𝑬xample 1.

Let 𝑓(𝑡) = 1 for 𝑡 ≥ 0, then

𝐹(𝑠) = ℒ{𝑓(𝑡)} = 𝑙𝑖𝑚𝐴→∞

∫ 𝑒−𝑠𝑡

𝐴

0

. 1 𝑑𝑡 = 𝑙𝑖𝑚𝐴→∞

−1

𝑠𝑒−𝑠𝑡 |

𝐴0

= 𝑙𝑖𝑚𝐴→∞

−1

𝑠[𝑒−𝑠𝐴 − 1] =

1

𝑠 . (𝑠 ≥ 0)

5

Example 2. Let 𝑓(𝑡) = 𝑒𝑎𝑡,

then 𝐹(𝑠) = ℒ{𝑓(𝑡)} = lim𝐴→∞

∫ 𝑒−𝑠𝑡 𝐴

0𝑒𝑎𝑡𝑑𝑡 = lim

𝐴→∞∫ 𝑒−(𝑠−𝑎)

𝐴

0𝑑𝑡 =

lim𝐴→∞

−1

𝑠−𝑎𝑒−(𝑠−𝑎)𝑡 |

𝐴0

= lim𝐴→∞

−1

𝑠−𝑎(𝑒−(𝑠−𝑎)𝐴 − 1) =

1

(𝑠−𝑎) , (𝑠 > 𝑎).

𝑬xample 3.

Let 𝑓(𝑡) = 𝑡𝑛, for 𝑛 ≥ 1 integer.

𝐹(𝑠) = lim𝐴→∞

∫ 𝑒−𝑠𝑡 𝑡𝑛𝐴

0

𝑑𝑡 = lim𝐴→∞

{𝑡𝑛𝑒−𝑠𝑡

−𝑠 |

𝐴0

− ∫𝑛𝑡𝑛−1𝑒−𝑠𝑡

−𝑠

𝐴

0

𝑑𝑡}

= 0 +𝑛

𝑠 lim𝐴→∞

∫ 𝑒−𝑠𝑡 𝑡𝑛−1𝐴

0𝑑𝑡 =

𝑛

𝑠ℒ{𝑡𝑛−1} .

So we get a recursive relation

𝓛{𝒕𝒏} = 𝒏

𝒔 𝓛{𝒕𝒏−𝟏}, ∀𝒏 ≥ 1,

which means

ℒ{𝑡𝑛−1} =𝑛−1

𝑠ℒ{𝑡𝑛−2}, ℒ{𝑡𝑛−2} =

𝑛−2

𝑠ℒ{𝑡𝑛−3}, …

By induction, we get

ℒ{𝑡𝑛} =𝑛

𝑠ℒ{𝑡𝑛−1} =

𝑛

𝑠

(n−1)

sℒ{𝑡𝑛−2} =

𝑛

𝑠

(n−1)

s

(𝑛−2)

𝑠ℒ{𝑡𝑛−3}

= ⋯ =𝑛

𝑠

(𝑛 − 1)

𝑠

(𝑛 − 2)

𝑠…

1

𝑠ℒ{1} =

𝑛!

𝑠𝑛

1

𝑠=

𝑛!

𝑠𝑛+1, (𝑠 > 0)

6

𝟏. 𝟐 Existence of Laplace transform

We will try to give a sufficient condition for the existence of Laplace

transform. For this, we need to recall the concept of piecewise continuous

function.

Definition 1. (Piecewise continuous function)

A function 𝑓 is piecewise continuous on the internal [𝑎. 𝑏] if

(i) The internal [𝑎. 𝑏] can be broken into a finite number of

subintervals 𝑎 = 𝑡0 < 𝑡1 < 𝑡2 < ⋯ < 𝑡𝑛 = 𝑏,

Such that 𝑓 is continuous in each subinterval

(𝑡𝑖 , 𝑡𝑖+1), for 𝑖 = 1,2,3, … . , 𝑛 − 1

(ii) The function 𝑓 has jump discontinuity at 𝑡𝑖 ,thus

│ lim𝑡→𝑡𝑖

+𝑓(𝑡) │ < ∞, 𝑖 = 0,1,2, … , 𝑛 − 1

│ lim𝑡→𝑡𝑖

−𝑓(𝑡) │ < ∞, 𝑖 = 0,1,2, … , 𝑛.

Note: A function is piecewise continuous on [0 , ∞) if it is piecewise

continuous in [0, 𝐴] for all 𝐴 > 0

Example : The function defined by

𝑓(𝑡) = {1

2 − 𝑡, 0 ≤ 𝑡 < 2,

𝑡 + 1, 2 ≤ 𝑡 ≤ 3,

is NOT piecewise continuous on [0,3]

Definition 2. (Exponential order ) A function 𝑓 is said to be of exponential

order if there exist constants 𝑀 and 𝛼 such that

7

│f(t)│≤ 𝑀𝑒𝛼𝑡 for sufficiently large 𝑡.

Consequence: Any polynomial is of exponential order. This is clear from the

fact that

𝑒𝑎𝑡 = ∑𝑡𝑛𝑎𝑛

𝑛!∞𝑛=0 ≥

𝑡𝑛𝑎𝑛

𝑛! ⟹ 𝑡𝑛 ≤

𝑛!

𝑎𝑛𝑒𝑎𝑡

But, for example, 𝑓(𝑡) = et2 is not of exponential order.

𝑺𝒖𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒄𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒆𝒙𝒊𝒔𝒕𝒆𝒏𝒄𝒆 𝒐𝒇 𝑳𝒂𝒑𝒍𝒂𝒄𝒆 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎:

Let 𝑓 be a piecewise continuous function in [0 , ∞) and is of exponential

order. Then Laplace transform 𝐹 (𝑠) of 𝑓 exists for 𝑠 > 𝛼, where 𝛼 is a real

number that depends on 𝑓 .

𝑷𝒓𝒐𝒐𝒇: Since 𝑓 is of exponential order, there exists 𝐴, 𝑀 and 𝑐 such that

│𝑓(𝑡)│ ≤ 𝑀𝑒𝑐𝑡 for 𝑡 ≥ 𝐴.

Now we write 𝐼 = ∫ 𝑓(𝑡)∞

0𝑒−𝑠𝑡𝑑𝑡 = 𝐼1 + 𝐼2,

Where 𝐼1 = ∫ 𝑓(𝑡)𝐴

0𝑒−𝑠𝑡𝑑𝑡 and 𝐼2 = ∫ 𝑓(𝑡)

∞

𝐴𝑒−𝑠𝑡𝑑𝑡 .

Since 𝑓 is piecewise continuous, 𝐼1 exists. For the second integral 𝐼2 , we

note that for 𝑡 ≥ 𝐴

│𝑒−𝑠𝑡𝑓(𝑡)│ ≤ 𝑀𝑒−(𝑠−𝑐)𝑡 . Thus

∫ │𝑓(𝑡)∞

𝐴𝑒−𝑠𝑡│𝑑𝑡 ≤ 𝑀 ∫ 𝑒−(𝑠−𝑐)𝑑𝑡 ≤ 𝑀 ∫ 𝑒−(𝑠−𝑐)𝑑𝑡

∞

0

∞

𝐴 =

𝑀

𝑠−𝑐 , 𝑠 > 𝑐.

Since the integral in 𝐼2 converges absolutely for

𝐼2 converges for 𝑠 > 𝑐 Thus, both 𝐼1 and 𝐼2 exist and

hence 𝐼 exists for 𝑠 > 𝑐.

8

1.3 Basic properties of Laplace transform

𝑻𝒉𝒆𝒐𝒓𝒆𝒎 𝟏. (𝑼𝒏𝒊𝒒𝒖𝒆𝒏𝒆𝒔𝒔 𝒐𝒇 𝑳𝒂𝒑𝒍𝒂𝒄𝒆 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎)

Let 𝑓 (𝑡) and 𝑔(𝑡) be two functions such that 𝐹(𝑠) = 𝐺 (𝑠) for all 𝑠 > 𝑘.

Then 𝑓(𝑡) = 𝑔(𝑡) at all 𝑡 where both are continuous.

Theorem 2. (Linearity) Suppose that 𝐹1(𝑠) = ℒ(𝑓1(𝑡)) exists for 𝑠 > 𝑎1 and

𝐹2(𝑠) = ℒ(𝑓2(𝑡)) exists for 𝑠 > 𝑎2 .

Then ℒ(𝑐1𝑓1(𝑡) + 𝑐2𝑓2(𝑡)) = 𝑐1𝐹1(𝑠) + 𝑐2𝐹2(𝑠),

for 𝑠 > 𝑎 , 𝑤ℎ𝑒𝑟𝑒 𝑎 = 𝑚𝑎𝑥{𝑎1, 𝑎1}.

The proof follows from the linearity of the integral.

𝑬xample 4. Find the Laplace transform of 𝒔𝒊𝒏 (𝒂𝒕) and 𝒄𝒐𝒔 (𝒂𝒕).

Method 1. We can compute by definition, with integration-by-parts, twice. (

this method is long.)

Method 2. Use the Euler's formula

𝑒𝑖𝑎𝑡 = cos 𝑎𝑡 + 𝑖 sin 𝑎𝑡, ⟹ ℒ{𝑒𝑖𝑎𝑡} = ℒ{cos 𝑎𝑡} + 𝑖ℒ{sin 𝑎𝑡}.

By Example 2 we have

ℒ{𝑒𝑖𝑎𝑡} =1

𝑠−𝑖𝑎=

1(𝑠+𝑖𝑎)

(𝑠−𝑖𝑎)(𝑠+𝑖𝑎)=

𝑠+𝑖𝑎

𝑠2+𝑎2 =𝑠

𝑠2+𝑎2 + 𝑖𝑎

𝑠2+𝑎2 .

Comparing the real and imaginary parts, we get

𝓛{𝐜𝐨𝐬 𝒂𝒕} =𝒔

𝒔𝟐 + 𝒂𝟐, 𝓛{𝐬𝐢𝐧 𝒂𝒕} =

𝒂

𝒔𝟐 + 𝒂𝟐, (𝒔 > 0).

9

Remark: Now we will use ∫ ∞

0 instead of lim

𝐴→∞ ∫

𝐴

0 , without causing

confusion.

For piecewise continuous functions, Laplace transform can be computed by

integrating each integral and add up at the end.

𝑬xample 5. Find the Laplace transform of

𝑓(𝑡) = {1, 0 ≤ 𝑡 < 2

𝑡 − 2, 2 ≤ 𝑡.

We do this by definition :

𝐹(𝑠) = ∫ 𝑒−𝑠𝑡

∞

0

𝑓(𝑡)𝑑𝑡 = ∫ 𝑒−𝑠𝑡 𝑑𝑡 +2

0

∫ (𝑡 − 2 )𝑒−𝑠𝑡 𝑑𝑡∞

2

= 1

−𝑠𝑒−𝑠𝑡 |

2𝑡 = 0

+ (𝑡 − 2)1

−𝑠𝑒−𝑠𝑡 |

∞𝑡 = 2

− ∫1

−𝑠

𝐴

2𝑒−𝑠𝑡 𝑑𝑡

=1

−𝑠(𝑒−2𝑠 − 1) + (0 − 0) +

1

𝑠

1

−𝑠𝑒−𝑠𝑡│

∞𝑡 = 2

= 1

−𝑠(𝑒−2𝑠 − 1) +

1

𝑠2𝑒−2𝑠

Example : Consider 𝑓(𝑡) = cosh(𝑤𝑡).

𝑇ℎ𝑒𝑛 𝓛(cosh(wt)) =1

2(𝓛(𝑒𝑤𝑡) + 𝓛(𝑒−𝑤𝑡))

=1

2(

1

s − w+

1

s + w) =

s

s2 − w2.

Example : Consider 𝑓(𝑡) = sinh(𝑤𝑡) . Proceeding as above, we find

𝐹(𝑠) =𝑤

(s2 − w2)

𝑻𝒉𝒆𝒐𝒓𝒆𝒎 𝟑. (𝑭𝒊𝒓𝒔𝒕 𝒔𝒉𝒊𝒇𝒕𝒊𝒏𝒈 𝒕𝒉𝒆𝒐𝒓𝒆𝒎)

𝐼𝑓 ℒ(𝑓(𝑡)) = 𝐹(𝑠), Then 𝓛(𝒆𝒂𝒕𝒇(𝒕)) = 𝑭(𝒔 − 𝒂),

11

𝑷𝒓𝒐𝒐𝒇: Suppose ℒ(𝑓(𝑡)) = 𝐹(𝑠) holds for 𝑠 > 𝑘. Now

ℒ( 𝑒𝑎𝑡𝑓(𝑡) ) = ∫ 𝑒−𝑠𝑡 𝑒𝑎𝑡 ∞

0𝑓(𝑡)𝑑𝑡 = ∫ 𝑒−(𝑠−𝑎)𝑡

∞

0𝑓(𝑡)𝑑𝑡

= 𝐹(𝑠 − 𝑎), 𝑠 − 𝑎 > 𝑘.

Example : Consider 𝑓(𝑡) = 𝑒−5𝑡cos (4𝑡) . Since

ℒ(cos(𝑎𝑡)) =𝑠

𝑠2+16 ⟹ ℒ( 𝑒−5𝑡 cos(4𝑡)) =

𝑠+5

(𝑠+5)2+16

Proposition.

𝐼𝑓 ℒ(𝑓(𝑡)) = 𝐹(𝑠), 𝑡ℎ𝑒𝑛 𝐹(𝑠) → 0 𝑎𝑠 𝑠 → ∞ .

Proof : We prove this for a piecewise continuous function which is of

exponential order. But the result is valid for any function for which Laplace

transform exists. Now

𝐼 = ∫ 𝑒−𝑠𝑡 𝑓(𝑡)∞

0𝑑𝑡 ⟹ │𝐼│ ≤ ∫ 𝑒−𝑠𝑡 │𝑓(𝑡)│

∞

0𝑑𝑡.

Now since the function is of exponential order, there exists 𝑀, 𝛼 and 𝐴 such

that │𝑓(𝑡)│ ≤ 𝑀1𝑒𝛼𝑡 for 𝑡 ≥ 𝐴. Also, since the function is piecewise

continuous in [0, 𝐴], we must have │𝑓(𝑡)│ ≤ 𝑀2𝑒𝛽𝑡 for 0 ≤ 𝑡 ≤ 𝐴

except possibly at some finite number of points where 𝑓(𝑡) is not defined.

Now we take 𝑀 = max{𝑀1, 𝑀2}

and 𝛾 = max{𝛼, 𝛽 } . Then we have

│𝐹(𝑠)│ = │𝐼│ ≤ ∫ 𝑒−𝑠𝑡

∞

0

│𝑓(𝑡)│𝑑𝑡 ≤ 𝑀 ∫ 𝑒−(𝑠−𝛾)𝑡 𝑑𝑡 =

𝑀

𝑠 − 𝛾

∞

0

, 𝑠 > 𝛾.

Thus, 𝐹(𝑠) → 0 𝑎𝑠 𝑠 → ∞ .

𝑪𝒐𝒎𝒎𝒆𝒏𝒕: Any function 𝐹(𝑠) without this behavior can not be Laplace

transform of a certain function.

11

For example,

𝑠

(𝑠−1) , sin(s),

𝑠2

(1+𝑠2) are not Laplace transform of any function.

1.4 Laplace transform of the first and second derivative

Let 𝐹(𝑠) = ℒ(𝑓(𝑡)) = ∫ 𝑒−𝑠𝑡𝑓(𝑡)𝑑𝑡∞

0

Then ℒ(𝑓′(𝑡)) = ∫ 𝑒−𝑠𝑡𝑓′(𝑡)𝑑𝑡∞

0

To evaluate the integral, we use integration by Parts:

Let 𝑢 = 𝑒−𝑠𝑡 𝑑𝑣 = 𝑓′(𝑡)𝑑𝑡

𝑑𝑢 = −𝑠𝑒−𝑠𝑡𝑑𝑡 𝑣 = 𝑓(𝑡)

ℒ(𝑓′(𝑡)) = 𝑒−𝑠𝑡𝑓(𝑡)│∞0

+ 𝑠 ∫ 𝑒−𝑠𝑡𝑓(𝑡)𝑑𝑡∞

0

= 0 − 𝑓(0) + 𝑠𝐹(𝑠)

Then

𝓛{𝒇′(𝒕)} = 𝒔𝑭(𝒔) − 𝒇(𝟎)

For the second derivative: ℒ(𝑓′′(𝑡)) = ∫ 𝑒−𝑠𝑡𝑓′′(𝑡)𝑑𝑡∞

0

𝑢 = 𝑒−𝑠𝑡 𝑑𝑣 = 𝑓′′(𝑡)𝑑𝑡

𝑑𝑢 = −𝑠𝑒−𝑠𝑡𝑑𝑡 𝑣 = 𝑓′(𝑡)

ℒ(𝑓′′(𝑡)) = 𝑒−𝑠𝑡𝑓′(𝑡)│∞0

+ 𝑠 ∫ 𝑒−𝑠𝑡𝑓′(𝑡)𝑑𝑡∞

0

= −𝑓′(0) + 𝑠[𝑠𝐹(𝑠) − 𝑓(0)]

Then

𝓛{𝒇′′(𝒕)} = 𝒔𝟐𝑭(𝒔) − 𝒔𝒇(𝟎) − 𝒇′(𝟎)

12

1.5 Definition of the Heaviside function Definition: The Heaviside function or unit step function is defined for real

numbers by

𝐻𝑎(𝑡) = 𝐻(𝑡 − 𝑎) = 𝑢𝑎(𝑡) = {1 ; 𝑡 ≥ 𝑎0 ; 𝑡 < 𝑎

Laplace transform of Heaviside function :

We have

Proof

Assume 𝑎 ≥ 0, by direct Laplace transform

𝐼 = ℒ(𝐻(𝑡 − 𝑎)) = ∫ 𝐻(𝑡 − 𝑎)∞

0

𝑒−𝑠𝑡𝑑𝑡

= ∫ (1)𝑒−𝑠𝑡𝑑𝑡∞

𝑎

Because 𝐻(𝑡 − 𝑎) = 0 for 0 ≤ 𝑡 < 𝑎

𝐼 = ∫ (1)𝑒−𝑠(𝑥+𝑎)𝑑𝑥∞

𝑎 Change of variable 𝑡 = 𝑥 + 𝑎

= 𝑒−𝑎𝑠 ∫ (1)𝑒−𝑠𝑥𝑑𝑥∞

0, Constant 𝑒−𝑎𝑠 moves outside integral

= 𝑒−𝑎𝑠( 1

𝑠 ) Apply ℒ(1) =

1

𝑠

𝓛( 𝑯(𝒕 − 𝒂) ) = 𝓛( 𝑢𝑎(𝑡) ) =𝒆−𝒂𝒔

𝒔

y

t

a

1

13

1.6 Table of Laplace transforms

𝑓(𝑡) for 𝑡 ≥ 0 𝐹(𝑠) = ℒ(𝑓) = ∫ 𝑒−𝑠𝑡𝑓(𝑡)𝑑𝑡∞

0

1 1

𝑠

𝑒𝑎𝑡 1

𝑠 − 𝑎

𝑡𝑛 𝑛!

𝑠𝑛+1, 𝑛 𝑖𝑠 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟

𝑠𝑖𝑛(𝑏𝑡) 𝑏

𝑠2 + 𝑏2

𝑐𝑜𝑠(𝑏𝑡) 𝑠

𝑠2 + 𝑏2

𝑠𝑖𝑛ℎ(𝑏𝑡) 𝑏

𝑠2 − 𝑏2

𝑐𝑜𝑠ℎ(𝑏𝑡) 𝑠

𝑠2 − 𝑏2

𝑓′(𝑡) 𝑠ℒ(𝑓) − 𝑓(0)

𝑓′′(𝑡) 𝑠2ℒ(𝑓) − 𝑠𝑓(0) − 𝑓′(0)

𝑒𝑎𝑡𝑓(𝑡) ℒ(𝑓)(𝑠 − 𝑎)

𝐻𝑎(𝑡) = 𝑢𝑎(𝑡) = 𝑢(𝑡 − 𝑎) = {0 ; 𝑡 < 𝑎1 ; 𝑡 ≥ 𝑎

𝑒−𝑎𝑠

𝑠

𝑢(𝑡 − 𝑎)𝑓(𝑡 − 𝑎) 𝑒−𝑎𝑠ℒ(𝑓)(𝑠)

𝑢(𝑡 − 𝑎)𝑔(𝑡) 𝑒−𝑎𝑠ℒ(𝑔(𝑡 − 𝑎))(𝑠)

14

Chapter 2

Inverse Laplace Transform

15

2.1 Definition: Inverse of Laplace transform

𝓛−𝟏 (𝐹(𝑠)) = 𝑓(𝑡) 𝐢𝐟 𝐹(𝑠) = ℒ(𝑓(𝑡) )

Proposition: (Linearity)

ℒ−1(𝛼𝐹 + 𝛽𝐺) = 𝛼ℒ−1(𝐹) + 𝛽ℒ−1(𝐺)

Proof:

This follows from the linearity of ℒ and holds in the domain common

to F and G. ℒ(𝛼𝑓 + 𝛽𝑔) = 𝛼ℒ(𝑓) + 𝛽ℒ(𝑔)

Then ℒ−1(𝛼ℒ(𝑓) + 𝛽ℒ(𝑔)) = 𝛼𝑓 + 𝛽𝑔

So ℒ−1(𝛼𝐹 + 𝛽𝐺) = 𝛼ℒ−1(𝐹) + 𝛽ℒ−1(𝐺)

2.1.1 (Lerch's Theorem : Uniqueness of the Inverse Laplace

transform) Suppose that 𝑓(𝑡) and 𝑔(𝑡) are continuous on [0, ∞) and of exponential

order 𝛼. If ℒ{𝑓(𝑡)} = ℒ{𝑔(𝑡)} for all 𝑠 > 𝛼,

Then 𝑓(𝑡) = 𝑔(𝑡) for all 𝑡 ≥ 0.

2.1.2 Heaviside expansion formula. Let 𝑃 , 𝑄 𝜖 ℂ[𝑋] two polynomials, where the degree of 𝑄 is greater than the

degree of 𝑃.

(i) If all complex zeroes 𝑎1, 𝑎2, … , 𝑎𝑛 of Q(𝑝) are simple, then

ℒ−1 (𝑃(𝑝)

𝑄(𝑝)) = ∑

𝑃(𝑎𝑗)

𝑄′(𝑎𝑗)𝑒𝑎𝑗𝑡 .𝑛

𝑗=1 (1)

(ii) If the different zeroes 𝑎1, 𝑎2, … , 𝑎𝑛 of Q(𝑝) have the

multiplicities 𝑚1, 𝑚2, … , 𝑚𝑛 respectively, we denote

𝐹𝑗(𝑝) =(𝑝 − 𝑎𝑗)

𝑚𝑗𝑃(𝑝)

𝑄(𝑝).

Then ℒ−1 (𝑃(𝑝)

𝑄(𝑝)) = ∑ 𝑒𝑎𝑗𝑡𝑛

𝑗=1 ∑𝐹𝑗

(𝑘)

𝑘!(𝑚𝑗−1−𝑘)!

𝑚𝑗−1𝑘=0 (2)

16

Proof:

Without hurting the generality, we can suppose that Q(s) is monic .

Therefore 𝑄(𝑝) = (𝑝 − 𝑎1)(𝑝 − 𝑎2) … (𝑝 − 𝑎𝑛)

For =1,2,...,n , denoting

𝑄(𝑝) = (𝑝 − 𝑎𝑗)𝑄𝑗(𝑝),

One has 𝑄𝑗(𝑎𝑗) ≠ 0.

We have a partial fraction expansion of the form

𝑃(𝑝)

𝑄(𝑝)=

𝐶1

𝑝−𝑎1+

𝐶2

𝑝−𝑎2+ ⋯ +

𝐶𝑛

𝑝−𝑎𝑛 (3)

With constants 𝐶𝑗 . According to the linearity and the formula 1 of the parent

entry, one gets

ℒ−1 (𝑃(𝑝)

𝑄(𝑝)) = ∑

𝑃(𝑎𝑗)

𝑄′(𝑎𝑗)

𝑛𝑗=1 𝑒𝑎𝑗𝑡 . (4)

For determining the constants 𝐶𝑗 , multiply (3) by −𝑎𝑗 .

It yields 𝑃(𝑝)

𝑄𝑗(𝑝)= 𝐶𝑗 + (𝑝 − 𝑎𝑗) ∑

𝑐𝑗

𝑝−𝑎𝑖𝑖≠𝑗

Setting to this identity 𝑝 = 𝑎𝑗 gives the value

𝐶𝑗 =𝑃(𝑝)

𝑄𝑗(𝑎𝑗) (5)

But sine

𝑄′(𝑝) = [(𝑝 − 𝑎𝑗)𝑄𝑗(𝑝)]′

= 𝑄𝑗(𝑝) + (𝑝 − 𝑎𝑗)𝑄𝑗′(𝑝)

We see that

𝑄𝑗′(𝑝) = 𝑄𝑗(𝑎𝑗);

17

Thus the equation (5) may be written

𝐶𝑗 =𝑃(𝑎𝑗)

𝑄′(𝑎𝑗) (6)

2.2 Examples of Inverse Transform As an example, from the Laplace transforms table, we see that

ℒ (sin(6𝑡)) =6

𝑠2+36 .

Written in the inverse transform notation

ℒ−1 (6

𝑠2 + 36) = sin(6𝑡)

Recall that ℒ(𝑡𝑛) =𝑛!

𝑠𝑛+1 . Hence, for example,

ℒ−1 (7!

𝑠8) = 𝑡7. Here, by examining the power 𝑠8 we saw that n=7. Now

consider ℒ−1 (5

𝑠11) . Here n+1=11. Hence n=10. Now, we need to make

the numerator to be 10! .

ℒ−1 (5

𝑠11) = 5ℒ−1 (

1

𝑠11) =

5

10!ℒ−1 (

10!

𝑠11) =

5

10!𝑡10.

More Examples of Inverse Laplace transform

Consider ℒ−1 (7𝑠+15

𝑠2+2) . The form of the denominator, 𝑠2 + 𝑘2, is that of the

Laplace transforms of sin, 𝑐𝑜𝑠 .

ℒ−1 (7𝑠 + 15

𝑠2 + 2) = ℒ−1 (

7𝑠

𝑠2 + 2) + ℒ−1 (

15

𝑠2 + 2)

= 7ℒ−1 (𝑠

𝑠2+2) + 15ℒ−1 (

1

𝑠2+2) = 7 𝑐𝑜𝑠(√2𝑡) +

15

√2𝑠𝑖𝑛 (√2𝑡)

18

Partial Fractions Consider the rational expression

3𝑠 + 5

𝑠2 − 3𝑠 − 10=

3𝑠 + 5

(𝑠 − 5)(𝑠 + 2)

The denominator is factored, and the degree of the numerator is at least

one less than that of the denominator, in fact , it is exactly one less than the

degree of the denominator.

We can, therefore, put the rational expression in partial fractions. This

means for constants

A and B, we have the decomposition

3𝑠 + 5

(𝑠 − 5)(𝑠 + 2)=

𝐴

(𝑠 − 5)+

𝐵

(𝑠 + 2)

To determine A and B, first clear the denominators:

3𝑠 + 5

(𝑠 − 5)(𝑠 + 2)(𝑠 − 5)(𝑠 + 2)

=𝐴

(𝑠 − 5)(𝑠 − 5)(𝑠 + 2) +

𝐵

(𝑠 + 2)(𝑠 − 5)(𝑠 + 2).

Thus we have the polynomial equality:

3𝑠 + 5 = 𝐴(𝑠 + 2) + 𝐵(𝑠 − 5) = (𝐴 + 𝐵)𝑠 + 2𝐴 − 5𝐵.

By comparing the coefficients of s and constant coefficients, we get two

equations in A and B.

𝐴 + 𝐵 = 3

2𝐴 − 5𝐵 = 5

19

We can solve for A and B by using Cramer's rule and get

𝐴 =20

7 and 𝐵 =

1

7 .

We can now determine the inverse transform

ℒ−1 (3𝑠 + 5

(𝑠 − 5)(𝑠 + 2)) = ℒ−1 (

𝐴

𝑠 − 5+

𝐵

𝑠 + 2)

= 𝐴ℒ−1 (1

𝑠 − 5) + 𝐵ℒ−1 (

1

𝑠 + 2) =

20

7𝑒5𝑡 +

1

7𝑒−2𝑡 .

This could also have been directly determined by using a formula from the

table of Laplace transform.

Partial Fractions: More Examples

To find ℒ−1 (3𝑠+4

(𝑠−2)(𝑠2+7))

Put 3𝑠+4

(𝑠−2)(𝑠2+7) in partial fractions.

Since 𝑠2 + 7 is a quadratic, when it is put in partial fractions,

its numerator must be a general polynomial of degree one .

3𝑠+4

(𝑠−2)(𝑠2+7)=

𝐴

𝑠−2+

𝐵𝑠+𝐷

(𝑠2+7)

3𝑠 + 4

(𝑠 − 2)(𝑠2 + 7)(𝑠 − 2)(𝑠2 + 7)

=𝐴

𝑠−2(𝑠 − 2)(𝑠2 + 7)+

𝐵𝑠+𝐷

(𝑠2+7)(𝑠 − 2)(𝑠2 + 7)

21

Hence, we have the equality of polynomials:

3𝑠 + 4 = 𝐴(𝑠2 + 7) + ( 𝐵𝑠 + 𝐷)(𝑠 − 2)

= (𝐴 + 𝐵)𝑠2 + (−2𝐵 + 𝐷)𝑠 + 7𝐴 − 2𝐷

Comparing the coefficient of 𝑠2, s , and constant coefficients,

𝐴 + 𝐵 = 0 ,−2𝐵 + 𝐷 = 3 , and 7𝐴 − 2𝐷 = 4.

From the first equation , we get that 𝐵 = −𝐴. Sub in the second, to get

2𝐴 + 𝐷 = 3 (1)

7𝐴 − 2𝐷 = 4 (2)

Then, 𝐴 =det (

3 14 −2

)

det (2 17 −2

)=

−10

−11, and D=

𝑑𝑒𝑡(2 37 4

)

𝑑𝑒𝑡(2 17 −2

)=

−13

−11.

Hence , 𝐴 =10

11, 𝐵 = −𝐴 = −

10

11, 𝐷 =

13

11.

21

Chapter 3

Application of the Laplace

transform

22

3.1 Introduction

Differential equations, whether ordinary or partial, describe the ways

certain quantities of interest vary over time. These equations are generally

coupled with initial conditions at time 𝑡 = 0 and boundary conditions.

Laplace transform is a powerful technique to solve differential equations. It

transforms an IVP in ODE to algebraic equations. The solution of the

algebraic equations is than back-transformed to the original problem.

Steps of the method:

• Take Laplace transform of both sides of the differential equation and we

will get an algebraic equation for 𝑌(𝑠) = ℒ(𝑦(𝑡)) .

• Solve this equation to get Y(s).

• Take inverse Laplace transform to get 𝑦(𝑡) = ℒ−1{𝑌}.

23

3.2 Example 1. (Ordinary differential equation with distinct real

roots, but one matches term) Solve the initial value problem by Laplace transform,

𝑦′′ − 2𝑦′ − 3𝑦 = 𝑒3𝑡 , y(0)=0, y'(0)=1.

Solution

Take Laplace transform on both sides of the equation,

we get ℒ{𝑦′′} − 2ℒ{𝑦′} − 3ℒ{𝑦} = ℒ{𝑒3𝑡}

⟹ 𝑠2𝑌(𝑠) − 1 − 2(𝑠𝑌(𝑠)) − 3𝑌(𝑠) =1

𝑠−3

⟹ (𝑠2 − 2𝑠 − 3)𝑌(𝑠) = 1 +1

𝑠 − 3=

𝑠 − 2

𝑠 − 3

⟹ 𝑌(𝑠) =𝑠−2

(𝑠−3)(𝑠2−2𝑠−3)=

𝑠−2

(𝑠−3)2(𝑠+1)

We use partial fraction decomposition

𝑌(𝑠) =𝑠 − 2

(𝑠 − 3)2(𝑠 + 1)=

𝐴

𝑠 + 1+

𝐵

𝑠 − 3+

𝐶

(𝑠 − 3)2

Compare the numerators:

𝑠 − 2 = 𝐴(𝑠 − 3)2 + 𝐵(𝑠 + 1)(𝑠 − 3) + 𝐶(𝑠 + 1)

Set 𝑠 = −1, we get 𝐴 =−3

16 .

Set 𝑠 = 3, we get =1

4 .

Set 𝑠 = 0, we get 𝐵 =3

16 .

So

𝑌(𝑠) =−3

16

1

𝑠 + 1+

3

16

1

𝑠 − 3+

1

4

1

(𝑠 − 3)2

𝑦(𝑡) = ℒ−1{𝑌(𝑠)} =−3

16𝑒−𝑡 +

3

16𝑒3𝑡 +

1

4𝑡𝑒3𝑡

24

Compare this to the method of undetermined coefficient:

General solution of the equation should be 𝑦 = 𝑦𝐻 + 𝑦𝑝 , where 𝑦𝐻 is the

general solution to the homogeneous equation and 𝑦𝑝 is a particular

solution. The characteristic equation is 𝑟2 − 2𝑟 − 3 = (𝑟 + 1)(𝑟 − 3) = 0,

has two roots

𝑟1 = −1, 𝑟2 = 3 and 𝑦𝐻 = 𝑐1𝑒−𝑡 + 𝑐2𝑒3𝑡 .

Since 3 is a root, so the form of the Particular solution is

𝑦𝑝= 𝐴𝑡𝑒3𝑡. This discussion concludes that the solution should be of the form

𝑦 = 𝑐1𝑒−𝑡 + 𝑐2𝑒3𝑡 + 𝐴𝑡𝑒3𝑡

For some constants 𝑐1, 𝑐2, 𝐴. This fits well with our result.

25

3.3 Example 2: Differential equations with discontinuous forcing

functions We now use Laplace transform to solve an initial value problem with

discontinuous force, and then describe behavior of solutions to try to make

physical sense of them.

We will start with the following problem:

Solve the following initial value problem

𝑦′′ + 𝑦′ + 𝑦 = 𝑓(𝑡), where 𝑓(𝑡) = { 0 ; 0 ≤ 𝑡 < 2

1 ; 𝑡 ≥ 2

𝑦(0) = 1, 𝑦′(0) = 0.

Solution

𝐿𝑒𝑡 ℒ(𝑦(𝑡)) = 𝑌(𝑠) , thus ℒ(𝑦′) = 𝑠𝑌 − 1 and ℒ(𝑦′′) = 𝑠2𝑌 − 𝑠.

Also, we have ℒ{𝑓(𝑡)} = ℒ{𝑢2(𝑡)} = 𝑒−2𝑠 1

𝑠 .

Then

𝑠2𝑌 − 𝑠 + 𝑠𝑌 − 1 + 𝑌 = 𝑒−2𝑠 1

𝑠,

⟹ (𝑠2 + 𝑠 + 1)𝑌(𝑠) =𝑒−2𝑠

𝑠+ (𝑠 + 1)

Which gives

𝑌(𝑠) =𝑒−2𝑠

𝑠(𝑠2 + 𝑠 + 1)+

𝑠 + 1

𝑠2 + 𝑠 + 1

Now we need to find the inverse Laplace transform for Y(s), for this we have

to do partial fraction decomposition first. It has the form

1

𝑠(𝑠2 + 𝑠 + 1)=

𝐴

𝑠+

𝐵𝑠 + 𝐶

𝑠2 + 𝑠 + 1

26

Compare the numerators of both sides; we get

1 = 𝐴(𝑠2 + 𝑠 + 1) + (𝐵𝑠 + 𝐶). 𝑠

Set 𝑠 = 0, we get 𝐴 = 1.

Compare 𝑠2-term: 0 = 𝐴 + 𝐵, so 𝐵 = −𝐴 = −1.

Compare s-term: 0 = 𝐴 + 𝐶, so 𝐶 = −𝐴 = −1.

So

𝑌(𝑠) = 𝑒−2𝑠 (1

𝑠−

𝑠 + 1

𝑠2 + 𝑠 + 1) +

𝑠 + 1

𝑠2 + 𝑠 + 1

We need to rewrite the fraction as:

𝑠 + 1

𝑠2 + 𝑠 + 1=

𝑠 + 1

(𝑠 +12)2 + (

√32 )2

=

(𝑠 +12) +

1

√3.√32

(𝑠 +12)2 + (

√32 )2

So

ℒ−1 {𝑠 + 1

𝑠2 + 𝑠 + 1} = 𝑒−

12

𝑡 (cos (√3

2𝑡) +

1

√3sin (

√3

2𝑡)).

We deduce that

𝑦(𝑡) = 𝑢2(𝑡) [1 − 𝑒−12

(𝑡−2)(cos (

√3

2(𝑡 − 2)) +

1

√3sin (

√3

2(𝑡 − 2)))]

+ 𝑒−12

𝑡 (cos (√3

2𝑡) +

1

√3sin (

√3

2𝑡))

We can make the following remarks

27

• We see that the solution of the homogeneous equation is

𝑒−12

𝑡 (𝑐1𝑐𝑜𝑠√3

2𝑡 + 𝑐2𝑠𝑖𝑛

√3

2𝑡)

and these terms appear in the solution.

• Actually the solution consists of two parts: the forced response and the

homogeneous solution.

• Furthermore, the function 𝑓 has a discontinuity at 𝑡 = 2,

and we see a jump in the solution also for 𝑡 = 2, as in the presence

of term 𝑢2(𝑡).

28

Conclusion

In this project we presented some applications of Laplace transform to

solve differential equations. Laplace transform is also a very effective

mathematical tool to simplify very complex problems in the area of stability

and control. With the ease of application of Laplace transform in a variety of

scientific applications, many research software have made it possible to

simulate the Laplace transformable equations directly, which has made a

good advancement in the research field.

29

References

[1] web.mit.edu/jorloff/www/18.03.../laplaceuniqueness.pdf

[2] www2.fiu.edu/~aladrog/InverseLaplace.pdf

[3] home.iitk.ac.in/~sghorai/TEACHING/.../ode17.pdf

[4] math.stackexchange.com/.../uniqueness-of-the-laplace-transform.

[5] https://ccrma.stanford.edu/.../Existence_Laplace_Transform...

[6] www.sosmath.com/diffeq/laplace/basic/basic.htm

[7] www2.fiu.edu/~aladrog/IntroLaplaceTransform.pdf

[8] mathworld.wolfram.com/LerchsTheorem.html

[9] acstaff.cbu.edu/wschrein/media/.../M231L114.

[10] ww2.fiu.edu/~aladrog/PropLaplaceTransform.pdf

[11] Fo]rier.eng.hmc.edu/.../Laplace_Transform/node5.html

[12] en.wikipedia.org/wiki/Inverse_Laplace_transform

[13] www.utdallas.edu/.../Inverse_Laplace_Transforms.pdf

[14] archive.nathangrigg.net/.../laplace-with-heaviside function- Nathan [pdf]

[15] tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

[16] jdebug.org/practical-applications-of-laplace-transform/.pdf