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MATHEMATICS-I

LAPLACE TRANSFORMS

I YEAR B.Tech

By

Y. Prabhaker Reddy

Asst. Professor of Mathematics

Guru Nanak Engineering College

Ibrahimpatnam, Hyderabad.

SYLLABUS OF MATHEMATICS-I (AS PER JNTU HYD)

Name of the Unit Name of the Topic

Unit-I Sequences and Series

1.1 Basic definition of sequences and series 1.2 Convergence and divergence. 1.3 Ratio test 1.4 Comparison test 1.5 Integral test 1.6 Cauchy’s root test 1.7 Raabe’s test 1.8 Absolute and conditional convergence

Unit-II Functions of single variable

2.1 Rolle’s theorem 2.2 Lagrange’s Mean value theorem 2.3 Cauchy’s Mean value theorem 2.4 Generalized mean value theorems 2.5 Functions of several variables 2.6 Functional dependence, Jacobian 2.7 Maxima and minima of function of two variables

Unit-III Application of single variables

3.1 Radius , centre and Circle of curvature 3.2 Evolutes and Envelopes 3.3 Curve Tracing-Cartesian Co-ordinates 3.4 Curve Tracing-Polar Co-ordinates 3.5 Curve Tracing-Parametric Curves

Unit-IV Integration and its applications

4.1 Riemann Sum 4.3 Integral representation for lengths 4.4 Integral representation for Areas 4.5 Integral representation for Volumes 4.6 Surface areas in Cartesian and Polar co-ordinates 4.7 Multiple integrals-double and triple 4.8 Change of order of integration 4.9 Change of variable

Unit-V Differential equations of first order and their applications

5.1 Overview of differential equations 5.2 Exact and non exact differential equations 5.3 Linear differential equations 5.4 Bernoulli D.E 5.5 Newton’s Law of cooling 5.6 Law of Natural growth and decay 5.7 Orthogonal trajectories and applications

Unit-VI Higher order Linear D.E and their applications

6.1 Linear D.E of second and higher order with constant coefficients 6.2 R.H.S term of the form exp(ax) 6.3 R.H.S term of the form sin ax and cos ax 6.4 R.H.S term of the form exp(ax) v(x) 6.5 R.H.S term of the form exp(ax) v(x) 6.6 Method of variation of parameters 6.7 Applications on bending of beams, Electrical circuits and simple harmonic motion

Unit-VII Laplace Transformations

7.1 LT of standard functions 7.2 Inverse LT – first shifting property 7.3 Transformations of derivatives and integrals 7.4 Unit step function, Second shifting theorem 7.5 Convolution theorem-periodic function 7.6 Differentiation and integration of transforms 7.7 Application of laplace transforms to ODE

Unit-VIII Vector Calculus

8.1 Gradient, Divergence, curl 8.2 Laplacian and second order operators 8.3 Line, surface , volume integrals 8.4 Green’s Theorem and applications 8.5 Gauss Divergence Theorem and applications 8.6 Stoke’s Theorem and applications

LAPLACE TRANSFORMATION

INTRODUCTION Laplace Transformations were introduced by Pierre Simmon Marquis De Laplace (1749-1827), a French Mathematician known as a Newton of French. Laplace Transformations is a powerful Technique; it replaces operations of calculus by operations of Algebra.

Suppose an Ordinary (or) Partial Differential Equation together with Initial conditions is reduced to a problem of solving an Algebraic Equation.

Definition of Laplace Transformation: Let be a given function defined for all , then

the Laplace Transformation of is defined as

Here, is called Laplace Transform Operator. The function is known as determining function, depends on. The new function which is to be determined (i.e. F ) is called generating function, depends on.

Here

NOTE: Here Question will be in and Answer will be in.

Laplace Transformation is useful since  Particular Solution is obtained without first determining the general solution

 Non-Homogeneous Equations are solved without obtaining the complementary Integral

 Solutions of Mechanical (or) Electrical problems involving discontinuous force functions

(R.H.S function ) (or) Periodic functions other than and are obtained easily.

 The Laplace Transformation is a very powerful technique, that it replaces operations of

calculus by operations of algebra. For e.g. With the application of L.T to an Initial value problem, consisting of an Ordinary( or Partial ) differential equation (O.D.E) together with Initial conditions is reduced to a problem of solving an algebraic equation ( with any given Initial conditions automatically taken care )

APPLICATIONS

Laplace Transformation is very useful in obtaining solution of Linear D.E’s, both Ordinary and Partial, Solution of system of simultaneous D.E’s, Solutions of Integral equations, solutions of Linear Difference equations and in the evaluation of definite Integral.

Thus, Laplace Transformation transforms one class of complicated functions to produce another class of simpler functions.

ADVANTAGES

 With the application of Linear Transformation, Particular solution of D.E is obtained directly without the necessity of first determining general solution and then obtaining the particular solution (by substitution of Initial Conditions).  L.T solves non-homogeneous D.E without the necessity of first solving the corresponding homogeneous D.E.  L.T is applicable not only to continuous functions but also to piece-wise continuous functions, complicated periodic functions, step functions, Impulse functions.  L.T of various functions are readily available.

The symbol ‘ ’ denotes the L.T operator, when it operated on a function , it transforms into a function of complex variable. We say the operator transforms the function in the domain (usually called time domain) into the function in the domain (usually called complex frequency domain or simply the frequency domain)

Because the Upper limit in the Integral is Infinite,

the domain of Integration is Infinite. Thus the

Integral is an example of an Improper Integral.

The Laplace Transformation of is said to exist if the Integral Converges for some values of , Otherwise it does not exist.

Domain (time domain)

Domain (frequency domain)

Original Equation (In terms of t)

Laplace Transform

Final Equation (In terms of s)

The Laplace Transformation of , where is a non-negative Real number. Sol: We know that

Put

As

Problems

1. Find the Laplace Transformation of Sol: We know that

2. Find the Laplace Transformation of Sol: We know that

SECTIONALLY CONTINUOUS CURVES (Or) PIECE-WISE CONTINUOUS

A function is said to be Sectionally Continuous (or) Piece-wise Continuous in any Interval , if it is continuous and has finite Left and Right limits in any Sub-Interval of. In the above case Laplace Transformation holds good.

FUNCTIONS OF EXPONENTIAL ORDER

A function is said to be exponential of order as if is a finite value. Example: Verify is an exponential order (or) not? Sol:

, a finite value.

is an exponential order. Sufficient conditions for the Existence of Laplace Transformation The Laplace Transformation of exists i.e. The Improper Integral of Converges (finite value) when the following conditions are satisfied. 1) is a piece-wise continuous 2) is an exponential of order.

PROPERTIES OF LAPLACE TRANSFORMATION

LINEAR PROPERTY

Statement: If , then Proof: Given that and L.H.S: =R.H.S

FIRST SHIFTING PROPERTY (or) FIRST TRANSLATION PROPERTY

Statement: If then Proof: We know that

(Here we have taken exponential quantity as negative, but not positive, because as )

Put

Hence, If then

whenever we want to evaluate , first evaluate which is equal to and then evaluate , which will be obtained simply, by substituting in place o f in

Problems

1) Find the Laplace Transformation of

Sol: We know that and

Now,

2) Find the Laplace Transformation of

Sol: We know that and Now,

3) Find Sol: We know that I But II

From I &II, Equating the corresponding coefficients, we get , and

LAPLACE TRANSFORMATION OF DERIVATIVES

Statement: If , then

Proof: We know that

Let us consider

In view of this, Now, put

Since,

Generalizing this, we get finally

Problem: Find the by using derivatives method. we know that Given that

DIVISION BY ‘t’ METHOD (or) Laplace Integrals

Statement: If , then Proof: Let us consider Now, Integrate on both sides w.r.t by taking the Limits from then

Since and are Independent variables, by Interchanging the order of Integration,

Problems

Find the

Sol: Here

We know that

v

INVERSE LAPLACE TRANSFORMATION

Definition: If , then is known as Inverse Laplace Transformation of and it is denoted by , where is known as Inverse Laplace Transform operator and is such that.

Inverse Elementary Transformations of Some Elementary Functions

Problems based on Partial Fractions

A fraction of the form in which both powers and are positive numbers

is called rational algebraic function.

When the degree of the Numerator is Lower than the degree of Denominator, then the fraction is called as Proper Fraction.

To Resolve Proper Fractions into Partial Fractions, we first factorize the denominator into real factors. These will be either Linear (or) Quadratic and some factors may be repeated.

From the definitions of Algebra, a Proper fraction can be resolved into sum of Partial fractions.

S.No Factor of the Denominator Corresponding Partial Fractions 1.

Non-Repeated Linear Factor Ex: , [ occurs only one time] 2.

Repeated Linear Factor, repeated ‘r’ times Ex: 3.

Non-repeated Quadratic Expression Ex:

, Here atleast one of

Repeated Quadratic Expression, repeated ‘r’ times Ex:

SHIFTING PROPERTY OF INVERSE LAPLACE TRANSFORMATION

We know that

FORMULAS

If then, If and then, In general, , provided If then, If then, If then,

CONVOLUTION THEOREM

(A Differential Equation can be converted into Inverse Laplace Transformation) (In this the denominator should contain atleast two terms) Convolution is used to find Inverse Laplace transforms in solving Differential Equations and Integral Equations.

Statement: Suppose two Laplace Transformations and are given. Let and are their Inverse Laplace Transformations respectively i.e.

Then, Where is called Convolution. (Or) Falting of. Proof: Let

Now

The above Integration is within the region lying below the line, and above. (Here equation of is ) Let is taken on line and is taken on line, with as Origin.

M

APPLICATIONS OF D.E’s BY USING LAPLACE AND INVERSE LAPLACE TRANSFORMATIONS

Laplace Transform Method of solving Differential Equations yields particular solutions without necessity of first finding General solution and elimination of arbitrary constants.

Suppose the given D.Eq is of the form I is a Linear D.Eq of order 2 with constants a, b. Case 1: Suppose in Equation I, we assume a,b are constants and the boundary conditions are . We Know that and (Or) Here and is Second derivative and Procedure: Apply Laplace Transformation to equation ( I ) i.e.

Now, apply Inverse Laplace Transformation

i.e.

By solving this, we get the required answer. Case 2: If a, b are not constants (i.e. D.E with Variable Co-efficient) Let the D.E is of the form II Here are some functions of , with Initial conditions We know that

Now,

And, Apply Laplace Transformation on both sides to ( II ), we get

Substituting the boundary conditions in equation II and get the values of III Required solution is obtained by taking Inverse Laplace Transformation for equation III.

Problem

Solve by the method of Transformations, the equation and and Sol: Given I Apply Laplace Transformation on both sides, we get

II Now substitute boundary conditions Immediately before solving in equation II, we get

Some Formulas

If then Now, If , then

If then Now, If , then

Generalization: If then

Now, If , then

If then Now, If , then Generalization : If then Now, If , then If then

Now, If , then

If then Now, If , and , then Generalization: If then Now, If , and , then Similarly, If then Now, If , and , then

If and Then,

Now, If , then

Problems

 Find Sol: Here, if we observe, the denominator is in the product of factors form. So we can use partial fractions method. Or we can use the following method. We know that, If then Now, If , then I Let us consider

Substituting in I, we get

 Find

Sol: Here if we observe, it is possible to express denominator as product of partial fractions. So we can use partial fractions method also. But the method will be lengthy. So, we go for another method, by which we can solve the problem easily. Now, We know that If then Now, If , then I Let us consider

From I, we have

(Or)

If Numerator is and denominator is term, then always use model.