Module-1

Laplace Transform

1.1 Laplace transform, Existence theorem

Let f(t) be any function of t defined for all positive values of t. Then the Laplace transform of the function f(t) is defined as-

Provided that the integral exists, here ‘s’ is the parameter that could be real or complex.

The inverse of the Laplace transform can be defined as below-

Here

f(t) is called the inverse Laplace transform of

L is called the Laplace transformation operator.

Conditions for the existence of Laplace transform-

The Laplace transform of f(t) exists for s>a, if

1. f(t) is a continuous function.

2. is finite.

Important formulae-

1.

2.

3.

4.

5.

6.

7.

Example-1: Find the Laplace transform of the following functions-

1.                            2.

Sol. 1.

Here

So that we can write it as-

Now-

2.  Since 

Or            

Now-

Example-2: Find the Laplace transform of (1 + cos 2t)

Sol.

So that-

Properties and theorems of LT

1. Linearity property-

Let a and b be any two constants and , any two functions of t, then-

Proof:

Hence proved.

2. First shifting property (Theorem)- If

Proof: By definition-

Let (s – a) = r

Hence proved.

We can find the following results with the help of the above theorem-

1.

2.

4.

5.

6.

7.

Here s>a in each case.

Example-1: Find the Laplace transform of t sin at.

Sol. Here-

Example-2: Find the Laplace transform of

Sol. Here-

So that-

As we know that-  

So that-

Hence-

Example-3: Find the Laplace transform of the following function-

Sol. The given function f(t) can be written as-

So, by definition,

Existence theorem-

The Laplace transform of f(t) exists for s>a if –

1. F(t) is continuous
2. is finite.

The above conditions are not necessary but sufficient.

1.2 Laplace transforms of derivatives and integrals

Laplace transform of the derivative of f(t)-

Here

Proof: by the definition of Laplace transform-

On integrating by parts, we get-

Since
 

Then-

So that-

Laplace transform of integral of f(t) -

Proof: Suppose

We know that-

So that-

Putting the values of and , we get-

Laplace transform of the function multiplied by t

If , then-

Proof:                   

Differentiate w.r.t. x, we get-

Similarly-

And

Example-4: Find the Laplace transform of .

Sol. Here-

Now-

1.3 Initial and final value theorems

We can solve the initial value problems directly without first finding a general solution by using the Laplace transform method for solving initial value problems.

Example: solve the initial value problem-

Sol.

Take the Laplace transform of the given differential equation, we get-

Using the initial conditions, we get-

On taking Laplace inverse of Y(s), we get-

Example: using Laplace transform find the solution of the following initial value problem-

y’’ + 25y  = 10 cos 5t, y(0) = 2, y’(0) = 0

Sol.

Here we have-

y’’ + 25y  = 10 cos 5t, y(0) = 2, y’(0) = 0

On taking Laplace transform of the given differential equation and using initial conditions, we get-

Example: using Laplace transform find the solution of the following initial value problem-

y’ + y =

Sol. On taking the Laplace transform and using the initial conditions, we get-

Thus

On breaking into partial fractions, we get-

Example: Solve the following initial boundary value problem-

Sol.

Taking Laplace to transform with respect to ‘t’ and using-

We get-

Or

It is a linear ordinary differential equation and its solution is given by-

When x = 0,

Hence C =

So that-

Now taking Laplace inverse of , we get-

1.4 Unit step function, Dirac- delta function

Unit step function

The unit step function u(t – a) is defined as-

Laplace transform of unit functions-

Example-1: Express the function given below in terms of a unit step function and find it's Laplace transform as well-

Sol. Here we are given-

So that-

Example-2: Find the Laplace transform of the following function by using unit step function-

Sol.

Since

Dirac-delta function-

The impulse function is also known as the Dirac-delta function.

Impulse- When a large force acts for a short time, then the product of the force and the time is called impulse.

The unit impulse function is the limiting function.

= 0, otherwise

The unit impulse function can be defined as-

And

Laplace transform of unit impulse function-

We know that-

Mean value theorem-  

As , then we get- 

When then we have

Example-1: Evaluate-

1.

Sol.1. As we know that- 

So that-

2. As we know that-

Example-2:

Sol.

1.5 Laplace transform of a periodic function

Suppose the function f(t) be periodic with period , then-

Similarly-

Now-

Put t = z + in the second integral of the above equation and t = z + 2 in the third integral and so on.

We get-

F(t) is periodic with period we can write-

Here is a geometric progression with a common ratio , and we know that the sum of infinite terms in G.P. Is given by
 

Then-

1.6 Inverse Laplace transform, Convolution theorem

Inverse Laplace transforms-

The inverse of the Laplace transform can be defined as below-

Here

f(t) is called the inverse Laplace transform of

L is called the Laplace transformation operator.

Important formulae-

1. 2.

3. 4.

5. 6.

7. 8.

Example: Find the inverse Laplace transform of the following functions-

1.

2.

Sol.

1.

2.

Example: Find the inverse Laplace transform of-

Sol.

Multiplication by ‘s’ -

Example: Find the inverse Laplace transform of-

Sol.

Division by s-

Example: Find the inverse Laplace transform of-

Sol.

Inverse Laplace transform of derivative-

Example: Find

Sol.

Inverse Laplace transform by using partial fraction

We can find the inverse Laplace transform by using the partial fractions method described below-

Example: Find the Laplace inverse of-

Sol.

We will convert the function into partial fractions-

Example: Find the inverse transform of-

Sol.

First, we will convert it into partial fractions-

Inverse Laplace transform by convolution theorem-

According to the convolution theorem-

Example: Find

Sol.

Therefore by the convolution theorem-

1.7 Application to solve simple linear and simultaneous differential equations

Step by step procedure to solve a linear differential equation by using Laplace transform-

1. Take the Laplace transform of both sides of the given differential equation.

2. Transpose the terms with a negative sign to the right.

3. Divide by the coefficient of , getting as a known function of s.

4. Resolve the function of s into partial fractions and take the inverse transform of both sides.

We will get y as a function of t. Which is the required solution.

Example-1: Use Laplace transform method to solve the following equation-

Sol. Here we have-

Take Laplace transform of both sides, we get-

It becomes-

(

So that-

Now breaking it into partial fractions-

We get the following results on inversion-

Example-2: Use Laplace transform method to solve the following equation-

Sol.

Here, taking the Laplace transform of both sides, we get

It becomes-

On inversion, we get-

Example-3: Use Laplace transform method to solve the following equation-

Sol. Here we have-

Taking Laplace transform of both sides, we get-

We get on putting given values-

On inversion, we get-

Example-4: Find the solution of the initial value problem by using Laplace transform-

Sol. Here we have-

Taking Laplace transform, we get-

Putting the given values, we get-

On inversion, we get-

4

Now-

The solution of simultaneous differential equations by using Laplace transform-

Example: solve the following differential equation by using Laplace transform-

Here D = d/dt and

Sol.

Here we have-

Now multiply (1) by D+1 and (2) by D – 1 we get-

Now subtract (4) from (3), we get-

By taking Laplace to transform we get-

Put the value of in (1) we get-

By taking Laplace to transform we get-

Which is the required answer.

Reference Books

1. B.S. Grewal: Higher Engineering Mathematics; Khanna Publishers, New Delhi.

2. B.V. Ramana: Higher Engineering Mathematics; Tata McGraw- Hill Publishing Company Limited, New Delhi.

3. Peter V.O’ Neil. Advanced Engineering Mathematics, Thomas ( Cengage) Learning.

4. Kenneth H. Rosem: Discrete Mathematics its Application, with Combinatorics and Graph Theory; Tata McGraw- Hill Publishing Company Limited, New Delhi

5. K.D. Joshi: Foundation of Discrete Mathematics; New Age International (P) Limited,

Publisher, New Delhi.