Laplace Tranform Notes Kerala University Semester 6

Laplace Transforms

Sumesh S. S.*
Assistant Professor
Department of Mathematics
Mar Ivanios College, Thiruvananthapuram.



Contents
1 Laplace transform 1

2 Laplace transform of some elementary functions 3

3 Transforms of Derivatives and Integrals 14

4 Unit Step Function (Heaviside Function) 25

5 Unit impulse function 30

6 Periodic Functions 31

7 Convolution 33

8 Differentiation and Integration of Transforms 38

9 Inverse Laplace Transforms 46


1 LAPLACE TRANSFORM
Rb
A relation of the form F(s) = a k(s,t) f (t) dt which transforms a given function f (t) into
another function F(s), is called an integral transform. Here K(s,t) is called the kernal of the
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1

, 2


transform and F(s) the transform of f (t). The most common integral transforms are

(i) a = 0, b = ∞, K(s,t) = e−st , (Laplaces)
1
(ii) a = −∞, b = ∞, K(s,t) = √ eist (Fourier)
2π

The idea behind any transform is that given problem can be solved more easily in the trans-
formed domain. Laplace transform reduces the problem of solving a differential equation to an
algebraic problem. It is widely used in problems where the

Definition 1.1
Let f (t) be a function defined for all t ≥ 0. The Laplace transform of f (t) is defined as
Z ∞
L [ f (t)] = e−st f (t) dt = F(s), (1)
0

provided the integral exists. Here s is a parameter real or complex. The function f (t)
whose transform is F(s) is said to be the inverse transform of F(s) and is denoted by
L −1 [F(s)]. Thus if L [ f (t)] = F(s), then f (t) = L −1 [F(s)].




EXISTENCE OF LAPLACE TRANSFORM

A function f (t) is said to be of exponential order or satisfies growth restriction if there exist
constants M and a such that | f (t)| ≤ M eat for all positive t.

Theorem 1.1 (Existence Theorem for Laplace Transforms)
If f (t) is defined and piecewise continuous on every finite interval on the semi-axis t ≥ 0
and is of exponential order for all for all t ≥ 0 and some constants M and a, then the
Laplace transform of f (t) exists for all s > a.


Proof. Since f (t) is piecewise continuous, e−st f (t) is integrable over any finite interval on the




Sumesh S S

, 3


t-axis. Also f (t) is of exponential order so that for s > a we get,
Z ∞
|L [ f (t)]| = e−st f (t) dt
0
Z ∞
≤ |e−st f (t)| dt
Z0∞
= e−st | f (t)| dt
Z0 ∞
≤ e−st Meat dt
0
Z k
= M lim e−(s−a)t dt
k→∞ 0
" #k
e−(s−a)t
= M lim
k→∞ −(s − a)
0
" #
e−(s−a)k 1
= M lim −
k→∞ −(s − a) −(s − a)
M
= .
s−a
Hence L [ f (t)] exists for s > a.

I Note that Z ∞ Z k
f (x) dx = lim f (x) dx
a k→∞ a

provided, the limit on the right side exist.



LINEARITY OF THE LAPLACE TRANSFORM

The Laplace transform is a linear operation. i.e. for any function f (t) and g(t) whose Laplace
transform exist and any constants a and b,

L [a f (t) + bg(t)] = aL [ f (t)] + bL [g(t)].

Proof. By definition,
Z ∞
L [a f (t) + bg(t)] = e−st [a f (t) + bg(t)] dt
0Z
∞ Z ∞
−st
=a e f (t) dt + b e−st g(t) dt
0 0

= aL [ f (t)] + bL [g(t)].



2 LAPLACE TRANSFORM OF SOME ELEMENTARY
FUNCTIONS

Sumesh S S

, 4


I Note that x→∞
lim e−x = 0 and lim e−ax = 0 for any a > 0.
x→∞

1
(i) L (1) = , s > 0.
s

Proof. By definition,
Z ∞
L (1) = e−st 1 dt
0
Z k
= lim e−st 1 dt
k→∞ 0
 −st k
e
= lim
k→∞ −s 0
 −sk 
e 1
= lim +
k→∞ −s s
 
1
= 0+
s
1
= , s > 0.
s

1
(ii) L (eat ) = , s > a.
s−a

Proof.
Z ∞
L [e ] =
at
e−st eat dt
Z0 ∞
= e−(s−a)t dt
0
" #k
e−(s−a)t
= lim
k→∞ −(s − a)
0
" #
e−(s−a)k 1
= lim +
k→∞ −(s − a) s−a
1
= if s > a.
s−a

n!
(iii) L (t n ) = where n is a positive integer.
sn+1
1
Proof. We prove this formula by induction. For n = 0, t n = t 0 = 1 and L (1) = . Thus
s
the result is true for n = 0. We now make the induction hypothesis that it holds for any
n!
positive integer n. i.e. L (t n ) = n+1 .
s




Sumesh S S