Unit - 4
Laplace transform
Differential equations with boundary values can be solved by using Laplace transform without finding general solutions and the values of the arbitrary constants.
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 which 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.
Applications of LT:
Laplace transform is very useful in obtaining solution of linear differential equations, both ordinary and partial, solution of system of simultaneous differential equations, solution of integral equations, solution of linear difference equations and in the evaluation of definite integrals.
Advantages of LT:
1. With the application of Laplace transform, particular solution of differential equation (D.E.) is obtained directly without the necessity of first determining general solution and then obtaining the particular solution (by substitution of initial conditions).
2. L.T. Solves non-homogeneous D.E. Without the necessity of first solving the corresponding homogeneous D.E.
3. L.T. Is applicable not only to continuous functions but also to piecewise continuous functions, complicated periodic functions, step functions and impulse functions.
Conditions for the existence of Laplace transforms-
The Laplace transform of f(t) exists for s>a, if
1. f(t) is a discontinuous 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-
LT of standard functions-
1.
Proof: by the definition of Laplace transform-
So that-
2.
Proof: by the definition of Laplace transform-
Let
So that-
We know that-
Then-
3.
Proof: by the definition of Laplace transform-
4.
Proof: here we know that,
So that-
5.
Proof: here we know that,
So that-
6.
Proof: here we know that,
So that-
7.
Proof: here we know that,
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 that, by definition,
Change of scale property-
Example: Find-
Given that,
Sol.
By using change of scale property-
Existence theorem-
The Laplace transform of f(t) exists for s>a if –
- F(t) is continuous
- is finite.
The above conditions are not necessary but sufficient.
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-
L[t2 f(t)] = (-1)2 d2/ds2 [F(s)]
L[t3 f(t)] = (-1)3 d3/ds3 [F(s)]
And
Example: Find the Laplace transform of .
Sol. Here-
Now-
Example: Find the Laplace transform of the function f(t) = using the theorem of transform of derivative.
Sol:
f (0) = 0, f ‘(0) = 0, f’’ (t ) = 2
By theorem
So that
Example: Find the Laplace transform of the function f(t) = t.cos at using the theorem of transform of derivative.
Sol:
f’ = −at · sin at + cos at
f’’ = −a(at cos at + sin at ) − a sin at
f’’ = − t cos at − 2a sin at
f (0) = 0, f’(0) = 1
Using theorem for 2nd derivative
Rearranging
Example: Find-
Sol:
The integrand is –
We know that
Using division by ‘t’
Using theorem on LT of integral:
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.
9. 10.
11. 12.
13. 14.
15. 16.
17. 18.
General properties of inverse Laplace theorem:
- Linearity property:
Let L{f (t )} = F (s) and L {g(t )} = G(s) then
Where are any two constants
Proof:
From the Linearity property for L.T., we
Know that
Taking inverse L.T. On either side, we get
Since f(t) = g(t) = we have
2. First shift or translation theorem
If then
Proof: From the first translation property on L.T., we have
Then
3. Change of scale property:
Proof:
From change of scale property for L.T.
Take a = 1/k then
Or
Thus
Example: (Linearity property): Find the inverse Laplace transform of the following:
Sol:
Example: (Linearity property): Find the inverse Laplace transform of the following:
Sol:
Example: (Change of scale property): Find
Sol:
We know that
Rewriting
with a = 4
Where
Thus applying change of scale property (with a = 3)
Example: (First shift theorem): Find the inverse LT of
Sol:
Example: (First shift theorem): Find the inverse LT of
Sol:
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 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-
Inverse transform of logarithmic & inverse trigonometric functions
We can find the inverse LT of logarithmic and inverse trigonometric function functions as below:
Example: Find the inverse LT of-
Sol:
So that
Since
Thus
Example: Find the inverse LT of
Sol:
Here
Now
By using inverse Laplace theorem of derivatives,
Thus
Example: Evaluate the following-
Sol:
Here
We know that
Thus applying inverse LT of integrals, we get
Convolution theorem
Convolution is used to find inverse Laplace transforms in solving differential equations and integral equations.
If
Then
Proof:
We have
Where the double integral is taken over the infinite region in the first quadrant lying between the lines x = 0 and x = t.
If we change the order of integration, the above integrals changes to-
Hence proved.
Example: Apply convolution theorem to evaluate-
Sol.
Since
By convolution theorem, we get-
Because- f(u) = cos au, g(t-u) = cos b(t – u)
Example: Obtain
Sol:
By using convolution theorem:
Step by step procedure to solve a linear differential equation by using Laplace transform-
1. Take Laplace transform of both sides of the given differential equation.
2. Transpose the terms with 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-
References:
- E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
- P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
- S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
- W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
- N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
- B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.
- T. Veerarajan, “Engineering Mathematics”, Tata Mcgraw-Hill, New Delhi, 2010