Laplace Transform Introduction

Transcription

1 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. 1

2 Definitions and Examples DEFINITION (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 A function is said to be a piece-wise continuous function for, if is a piece-wise continuous function on every closed interval Figure For example, see Figure 10.1: Piecewise Continuous Function DEFINITION (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 ) 2

3 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 (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 3

4 Examples EXAMPLE 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. 4

5 4. Find the Laplace transform of a positive integer. Solution: Substituting we get Find the Laplace transform of Solution: We have Compute the Laplace transform of Solution: 5

6 9. Note that the limits exist only when Hence, 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 integral one needs to consider the double It turns out that 16. 6

7 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 7

8 Properties of Laplace Transform LEMMA (Linearity of Laplace Transform) 1. Let. Then If and, then The above lemma is immediate from the definition of Laplace transform and the linearity of the definite integral. EXAMPLE Find the Laplace transform of Solution: Thus 2. Similarly, 3. Find the inverse Laplace transform of. Solution: 8

9 4. Thus, the inverse Laplace transform of is THEOREM (Scaling by ) Let be a piecewise continuous function with Laplace transform Then for Proof. By definition and the substitution we get EXERCISE Find the Laplace transform of where and are arbitrary constants. 9

10 Figure 10.2: 2. Find the Laplace transform of the function given by the graphs in Figure If, find. The next theorem relates the Laplace transform of the function with that of. THEOREM (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, 10

11 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 , gives the following corollary. COROLLARY Let be a function with If exist and is continuous for then (10.3.2) In particular, for, we have (10.3.3) COROLLARY Let be a piecewise continuous function for. Also, let. Then EXAMPLE Find the inverse Laplace transform of Solution: We know that Then and therefore, 2. Find the Laplace transform of Solution: Note that and Also, 11

12 Now, using Theorem , we get LEMMA (Laplace Transform of ) Let be a piecewise continuous function with If the function is differentiable, then Proof. By definition, respect to. The result is obtained by differentiating both sides with 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. 12

13 COROLLARY Let and Then EXAMPLE Find Solution: We know Hence 2. Find the function such that Solution: We know and By lemma , we know that Suppose Then Therefore, Thus we get LEMMA (Laplace Transform of an Integral) If then Equivalently, Proof. By definition, 13

14 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 Find Solution: We know Hence 2. Find Solution: By Lemma

15 3. Find the function such that Solution: We know So, LEMMA ( -Shifting) Let Then for Proof. EXAMPLE Find Solution: We know Hence 2. Find Solution: By -Shifting, if then. Here, and Hence, 15

16 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 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: 16

17 Solution: Here, Solving for and, we get Thus, 17

18 Transform of Unit Step Function DEFINITION (Unit Step Function) The Unit-Step function is defined by EXAMPLE Figure 10.3: Graphs of and LEMMA ( -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, 18

19 height6pt width 6pt depth 0pt EXAMPLE Find Solution: Let with Since Hence, by Lemma EXAMPLE Find Solution: Note that, where Thus, 19

20 Limiting Theorems The following two theorems give us the behaviour of the function when and when. THEOREM (First Limit Theorem) Suppose exists. Then Proof. We know Therefore as height6pt width 6pt depth 0pt EXAMPLE For let Determine such that Solution: Theorem implies Thus, 2. If find Solution: Theorem implies On similar lines, one has the following theorem. But this theorem is valid only when bounded as approaches infinity. is 20

21 THEOREM (Second Limit Theorem) Suppose exists. Then provided that converges to a finite limit as tends to 0. Proof. height6pt width 6pt depth 0pt EXAMPLE If Solution: From Theorem , we have find We now generalise the lemma on Laplace transform of an integral as convolution theorem. DEFINITION (Convolution of Functions) Let and be two smooth functions. The convolution, is a function defined by 21

22 Check that If then THEOREM (Convolution Theorem) If and then Remark Let for all Then we know that Thus, the Convolution Theorem reduces to the Integral 22

23 Application to Differential Equations Consider the following example. EXAMPLE 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 ). EXAMPLE Solve the IVP with and Solution: Note that. Thus, Taking Laplace transform of the above equation, we get 23

24 Which gives Hence, Remark 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 Consider the IVP with and Find Solution: Applying Laplace transform, we have 24

25 Using initial conditions, the above equation reduces to This equation after simplification can be rewritten as Therefore, From Example , we see that and hence 2. Show that is a solution of where Solution: Here, Hence, 3. Show that is a solution of 25

26 Solution: Here, Hence, 4. Solve the following IVP. Solution: Taking Laplace transform of both sides and using Theorem , we get Solving for we get So, 26

27 Transform of the Unit-Impulse Function Consider the following example. EXAMPLE Find the Laplace transform,, of Solution: Note that By linearity of the Laplace transform, we get Remark Observe that in Example , 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, delta function)., is called the UNIT-IMPULSE FUNCTION (or Dirac's 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 27

28 5. Also, observe that Now, if we take the limit of both sides, as approaches zero (apply L'Hospital's rule), we get 28