Laplace Transform

Introduction

In many problems, a function is transformed to another function through a relation of the type:

where is a known function. Here, is called integral transform of . Thus, an

integral transform sends a given function into another function . This transformation of

into provides a method to tackle a problem more readily. In some cases, it affords solutions to otherwise difficult problems. In view of this, the integral transforms find numerous applications in engineering problems. Laplace transform is a particular case of integral transform

(where is defined on and ). As we will see in the following, application of Laplace transform reduces a linear differential equation with constant coefficients to an algebraic equation, which can be solved by algebraic methods. Thus, it provides a powerful tool to solve differential equations.

It is important to note here that there is some sort of analogy with what we had learnt during the study of logarithms in school. That is, to multiply two numbers, we first calculate their logarithms, add them and then use the table of antilogarithm to get back the original product. In a

similar way, we first transform the problem that was posed as a function of to a problem in

, make some calculations and then use the table of inverse Laplace transform to get the solution of the actual problem.

In this chapter, we shall see same properties of Laplace transform and its applications in solving differential equations.

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Definitions and Examples DEFINITION 10.2.1 (Piece-wise Continuous Function)

1. A function is said to be a piece-wise continuous function on a closed interval

, if there exists finite number of points such

that is continuous in each of the intervals for and has finite limits as approaches the end points, see the Figure 10.1.

2. A function is said to be a piece-wise continuous function for , if is a

piece-wise continuous function on every closed interval For example, see Figure 10.1.

Figure 10.1: Piecewise Continuous Function

DEFINITION 10.2.2 (Laplace Transform) Let and Then for

is called the LAPLACE TRANSFORM of and is defined by

whenever the integral exists.

(Recall that exists if exists and we define

.) Remark 10.2.3

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1. Let be an EXPONENTIALLY BOUNDED function, i.e.,

Then the Laplace transform of exists.

2. Suppose exists for some function . Then by definition, exists. Now, one can use the theory of improper integrals to conclude that

Hence, a function satisfying

cannot be a Laplace transform of a function .

DEFINITION 10.2.4 (Inverse Laplace Transform) Let . That is, is the

Laplace transform of the function Then is called the inverse Laplace transform of

. In that case, we write

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Examples EXAMPLE 10.2.5

1. Find where

Solution:

Note that if then

Thus,

In the remaining part of this chapter, whenever the improper integral is calculated, we will not explicitly write the limiting process. However, the students are advised to provide the details.

2. Find the Laplace transform of where Solution: Integration by parts gives

3.

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4. Find the Laplace transform of a positive integer.

Solution: Substituting we get

5.

6. Find the Laplace transform of Solution: We have

7.

8. Compute the Laplace transform of Solution:

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9.

Note that the limits exist only when Hence,

10. 11. Similarly, one can show that

12. Find the Laplace transform of

Solution: Note that is not a bounded function near (why!). We will still show that

the Laplace transform of exists.

13.

Recall that for calculating the integral one needs to consider the double integral

14. 15. It turns out that

16.

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17. Thus, for

We now put the above discussed examples in tabular form as they constantly appear in applications of Laplace transform to differential equations.

Table 10.1: Laplace transform of some Elementary Functions

1

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Properties of Laplace Transform LEMMA 10.3.1 (Linearity of Laplace Transform)

1. Let . Then

2.

3. If and , then

The above lemma is immediate from the definition of Laplace transform and the linearity of the definite integral.

EXAMPLE 10.3.2

1. Find the Laplace transform of

Solution: Thus

2. Similarly,

3. Find the inverse Laplace transform of . Solution:

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4.

Thus, the inverse Laplace transform of is

THEOREM 10.3.3 (Scaling by ) Let be a piecewise continuous function with Laplace

transform Then for Proof.

By definition and the substitution we get

EXERCISE 10.3.4

1. Find the Laplace transform of

where and are arbitrary constants.

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Figure 10.2:

2. Find the Laplace transform of the function given by the graphs in Figure 10.2.

3. If , find .

The next theorem relates the Laplace transform of the function with that of .

THEOREM 10.3.5 (Laplace Transform of Differentiable Functions) Let for be a

differentiable function with the derivative, being continuous. Suppose that there exist

constants and such that for all If then

(10.3.1)

Proof.

Note that the condition for all implies that

So, by definition,

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We can extend the above result for derivative of a function , if

exist and is continuous for . In this case, a repeated use of Theorem 10.3.5, gives the following corollary.

COROLLARY 10.3.6 Let be a function with If

exist and is continuous for then

(10.3.2)

In particular, for , we have

(10.3.3)

COROLLARY 10.3.7 Let be a piecewise continuous function for . Also, let

. Then

EXAMPLE 10.3.8

1. Find the inverse Laplace transform of

Solution: We know that Then and therefore,

2. Find the Laplace transform of

Solution: Note that and Also,

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Now, using Theorem 10.3.5, we get

LEMMA 10.3.9 (Laplace Transform of ) Let be a piecewise continuous function

with If the function is differentiable, then

Proof.

By definition, The result is obtained by differentiating both sides with respect to .

Suppose we know the Laplace transform of a and we wish to find the Laplace transform of

the function Suppose that exists. Then writing gives

Thus, for some real number . As , we get

Hence, we have the following corollary.

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COROLLARY 10.3.10 Let and Then

EXAMPLE 10.3.11

1. Find

Solution: We know Hence

2. Find the function such that

Solution: We know and

By lemma 10.3.9, we know that Suppose Then

Therefore,

Thus we get

LEMMA 10.3.12 (Laplace Transform of an Integral) If then

Equivalently, Proof. By definition,

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We don't go into the details of the proof of the change in the order of integration. We assume that the order of the integrations can be changed and therefore

Thus,

EXAMPLE 10.3.13

1. Find

Solution: We know Hence

2. Find Solution: By Lemma 10.3.12

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3. Find the function such that

Solution: We know So,

LEMMA 10.3.14 ( -Shifting) Let Then for Proof.

EXAMPLE 10.3.15

1. Find

Solution: We know Hence

2. Find

Solution: By -Shifting, if then . Here, and

Hence,

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Inverse Transforms of Rational Functions

Let be a rational function of . We give a few examples to explain the methods for

calculating the inverse Laplace transform of

EXAMPLE 10.3.16

1. DENOMINATOR OF MATHEND000# HAS DISTINCT REAL ROOTS:

Solution: Thus,

2. DENOMINATOR OF MATHEND000# HAS DISTINCT COMPLEX ROOTS:

Solution: Thus,

3. DENOMINATOR OF MATHEND000# HAS REPEATED REAL ROOTS:

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Solution: Here,

Solving for and , we get

Thus,

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Transform of Unit Step Function DEFINITION 10.3.17 (Unit Step Function) The Unit-Step function is defined by

EXAMPLE 10.3.18

Figure 10.3: Graphs of and

LEMMA 10.3.19 ( -Shifting) Let Define by

Then and

Proof. Let . Then and so,

If , then and Since the functions and

take the same value for all we have Thus,

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EXAMPLE 10.3.20 Find

Solution: Let with Since

Hence, by Lemma 10.3.19

EXAMPLE 10.3.21 Find , where Solution: Note that

Thus,

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Limiting Theorems

The following two theorems give us the behaviour of the function when and when .

THEOREM 10.4.1 (First Limit Theorem) Suppose exists. Then

Proof. We know Therefore

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EXAMPLE 10.4.2

1. For let Determine such that Solution: Theorem 10.4.1 implies

Thus,

2. If find Solution: Theorem 10.4.1 implies

On similar lines, one has the following theorem. But this theorem is valid only when is bounded as approaches infinity.

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THEOREM 10.4.3 (Second Limit Theorem) Suppose exists. Then

provided that converges to a finite limit as tends to 0 .

Proof.

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EXAMPLE 10.4.4 If find Solution: From Theorem 10.4.3, we have

We now generalise the lemma on Laplace transform of an integral as convolution theorem.

DEFINITION 10.4.5 (Convolution of Functions) Let and be two smooth functions. The

convolution, is a function defined by

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Check that

1.

2. If then

THEOREM 10.4.6 (Convolution Theorem) If and then

Remark 10.4.7 Let for all Then we know that Thus, the Convolution Theorem 10.4.6 reduces to the Integral

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Application to Differential Equations Consider the following example.

EXAMPLE 10.5.1 Solve the following Initial Value Problem:

Solution: Let Then

and the initial conditions imply

Hence,

(10.5.1)

Now, if we know that is a rational function of then we can compute from by using the method of PARTIAL FRACTIONS (see Subsection 10.3.1). EXAMPLE 10.5.2

1. Solve the IVP

with and

Solution: Note that . Thus,

Taking Laplace transform of the above equation, we get

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Which gives

Hence,

Remark 10.5.3 Even though is a DISCONTINUOUS function at the solution and

are continuous functions of , as exists. In general, the following is always true:

Let be a solution of Then both and are continuous functions of time. EXAMPLE 10.5.4

1. Consider the IVP with and Find

Solution: Applying Laplace transform, we have

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Using initial conditions, the above equation reduces to

This equation after simplification can be rewritten as

Therefore, From Example 10.4.2.1, we see that and hence

2. Show that is a solution of

where

Solution: Here, Hence,

3. Show that is a solution of

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Solution: Here, Hence,

4. Solve the following IVP.

Solution: Taking Laplace transform of both sides and using Theorem 10.3.5, we get

Solving for we get

So,

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Transform of the Unit-Impulse Function Consider the following example.

EXAMPLE 10.6.1 Find the Laplace transform, , of

Solution: Note that By linearity of the Laplace transform, we get

Remark 10.6.2

1. Observe that in Example 10.6.1, if we allow to approach 0 , we obtain a new function,

say That is, let

This new function is zero everywhere except at the origin. At origin, this function tends to infinity. In other words, the graph of the function appears as a line of infinite height at

the origin. This new function, , is called the UNIT-IMPULSE FUNCTION (or Dirac's delta function).

2. We can also write

3. In the strict mathematical sense does not exist. Hence, mathematically

speaking, does not represent a function. 4. However, note that

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5. Also, observe that Now, if we take the limit of both sides, as approaches zero (apply L'Hospital's rule), we get

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