**Application of Laplace transforms and Laplace transforms of some useful functions**

Objective

To study the method of solving differential equations by using Laplace transform.

To study Laplace transforms of unit step function, unit impulse function and periodic functions.

Modules

Module I- Application to Differential equations

Module II- Unit step function

Module III- Unit impulse function

Module IV- Laplace transform of periodic functions

**Module I- Application to Differential equations**

Laplace transforms can be used to solve ordinary as well as partial differential equations. We shall apply this method to solve only ordinary linear differential equations with constant coefficients. The advantage of this method is that it yields the particular solution directly without the necessity of first finding the general solution and then evaluating the arbitrary constants.

Consider the following linear differential equations with constant coefficients:

where *k1, k2,…, kn are constants, withthe n* initial conditions

.

This equation is known as initial value problem.

**Steps to solve the differential equation**

- Take Laplace transforms on both sides using the notations and .
- Use the given initial conditions.
- Simplify the algebraic equation in the form.
- Take inverse Laplace transforms on both sides to obtain the desired solution.

Example 1.Solve the equation

, where at t = 0.

Solution.

The given equation is

.

Taking the Laplace transform on both sides, we obtain

Using the given conditions, the equation reduces to

or.

Therefore

, by resolving into partial fractions

Taking the inverse Laplace transforms on both sides, we get

i.e.,

i.e.,, which is the desired solution.

Note.Boundary value problem can also be solved by Laplace transform method. The following example illustrates the method.

Example 2. Solve the boundary value problem

with .

Solution.

Let, a constant.

Then taking Laplace transforms on both sides, we get

.

Taking inverse Laplace transform on both sides, we get

i.e.,.

Now from the given condition , we have

i.e.,.

Hence the particular solution is

.

**Solution of simultaneous linear equations with constant coefficients**

Consider the first order simultaneous linear equations with constant coefficients as follows:

where x and y are two dependent variable on the independent variable t and all ai’s and bi’s are constant terms. To solve this, the following steps may be used:

- Take Laplace transforms of both the equations using the notations

- Use the given initial conditions and obtain two equations in and .
- Solve for and so as to obtain

and

- Take inverse Laplace transforms and obtain the solution x(t) and y(t).

Example 3. Solve given x(0) = 1, y(0) = 0.

Solution.

Taking Laplace transforms of both equations, we get

and.

Solving for we get

.

Taking inverse Laplace transform, we get

.

Now from the given second equation, we have

.

Integrating both sides we obtain

, where c is the constant of integration.

From the given condition y(0) = 0, we obtain

.

Thus.

Hence the required solution of the given simultaneous equations is

, .

**Module II- Unit step function**

Definition

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

, where

When a = 0, we have

The product is defined as

**Laplace transform of unit step function**

By definition, we have

In particular,.

For ab, a difference of unit step functions may be defined as

.

**Second shifting theorem**

If , then

By definition, we have

, where t – a = p

.

Remark: From above we have and if a = 0, we have .

Example 4. Find the Laplace transform of .

Solution.

Express as a function of t – 2, we have

Therefore

Comparing with , we get a = 2 and .

Therefore.

Therefore

.

Example 5. Find the inverse Laplace transform of .

Solution.

By second shifting theorem, we have

.

Let, then .

Therefore

.

**Module III- Unit impulse function**

Impulse is considered as a force of very large magnitude applied for just an instant and the function representing the impulse is called Dirac-delta function. Mathematically it is considered as the limiting form of the function,

, as .

As , the function tends to be infinite at t = a and zero elsewhere, with the characteristic property that its integral across t = a is unity.

Now the integral

represents unit impulse at t = a. Hence the limiting form of as is expressed as unit impulse function denoted by .

Thus the unit impulse function is defined as follows:

such that .

**Laplace transform of unit impulse function**

We can write

.

Now

.

Hence

.

In particular if a = 0, and so .

**Filtering property of unit impulse function**

If f(t) is continuous and integrable in , then

**Relation between u(t - a) and **

We have

.

Example 6. A beam has its ends clamped at *x = 0 and x = l. A concentrated load W* acts vertically downwards at the point x = l/2. Find the resulting deflection.

Solution.

The differential equation for the given beam problem can be taken as

,

where y = y(x) represents the deflection with the boundary conditions

Taking Laplace transform on both sides we get

Using the given conditions and taking and , we obtain

.

Taking inverse Laplace transform on both sides, we obtain

i.e.,.

Using the given boundary conditions, we obtain

and.

Solving these equations, we obtain

and .

Hence the resulting deflection is

.

**Module IV- Laplace transform of periodic functions**

If f(t) is a periodic function with period k, then

By definition, we have

Put t = u, t = u + k, t = u + 2k, … in the successive integrals, we get

Since *f(t) is periodic we have f(u) = f(u + k) = f(u + 2k*) = …

Therefore we get

Example 7. Find the Laplace transform of the following rectangular wave function given by

and f(t + 2c) = f(t).

Solution.

The given function periodic with period k = 2c.

Therefore

.

Summary

In the session we have discussed the method of finding solution of linear ordinary differential equations by using the Laplace transform method. Also we have discussed Laplace transforms of unit step function, unit impulse function and periodic function.

**Assignment questions**

- Solve using Laplace transform

- Solve the simultaneous equations

- State and prove second shifting theorem.
- Find the Laplace transform of the function.
- Find the inverse Laplace transform of .
- Find the Laplace transform of the triangular wave function of period 2c given by

.

Reference

- The Laplace Transform, Shaum Outline Series, Shaum Publishing Company, New York.
- Advanced Engineering Mathematics by E. Kreyszig, John Wylie & Sons, New York (1999).

Quiz

- A solution of is

- The value of is

a.

b.

c.

- The value of is

a.

b.1/t

c.

- If , then

- The relation between unit step function and unit impulse function is

Answers

- a2.b3.c4.a5.b

FAQs

1. Solve using Laplace transform

, where at t = 0.

Answer.

Taking the Laplace transform on both sides, we get

.

Using the given conditions it reduces to

.

i.e.,

, Resolving into partial fractions

Taking inverse Laplace transforms on both sides, we get

.

2. The currents i1 and i2 in mesh are given by the differential equations

.

Find the currents *i1 and i2 by Laplace transform if i1 = i2 = 0 when t* = 0.

Answer.

Let L{i1} = L1 and L{i2} = L2.

Taking Laplace transform of both equations, we get

and

and by using the initial conditions.

Solving for L1 and L2 we obtain

and .

Taking inverse Laplace transforms, we obtain

and .

3. Find .

Answer.

Let

The function also can be written as

which is known as stair case function.

4. Find the Laplace transform of the triangular wave function

and f(t + 2a) = f(t).

Answer.

The given function is a periodic function with period 2a.

Now

.

Glossary

Function: It is an assignment *f from a set A into another set B; the set A is called domain of f* and the set of all function values is called the range of f.

Laplace transform:Let f(t) be a function of t defined for all . Then the Laplace transform of t denoted by L{f(t)} is defined by

,

provided the integral exists. Here s is a parameter which may be real or complex.

Laplace transform of derivative: If f(t) and its first (n - 1) derivatives be continuous, then

.

Periodic function: A function f(t) is said to be periodic with period T if f(t + T) = f(t), for all t.

Partial fraction:Suppose that is a proper rational function and is a product of polynomials. Thencan be expressed as sum of simpler rational functions, each of which is called a partial fraction. This process is called decomposition ofinto partial fractions.

The decomposition depends on the nature of the factors of.

If. i.e., a product of non-repeating linear functions, then

, where are constants.

Ifi.e., some factors repeating, then

, where are constants.

If some factors are quadratic, but non-repeating, then corresponding to these factors, the partial fraction is in the form.