Definitions and Examples

DEFINITION 10.2.1 (Piece-wise Continuous Function)  
  1. A function $ f(t)$ is said to be a piece-wise continuous function on a closed interval $ [a, b] \subset {\mathbb{R}}$ , if there exists finite number of points $ a = t_0 < t_1 < t_2 < \cdots < t_N = b$ such that $ \; f(t)$ is continuous in each of the intervals $ (t_{i-1}, \; t_i)$ for $ 1 \leq i \leq N$ and has finite limits as $ t$ approaches the end points, see the Figure 10.1.
  2. A function $ f(t)$ is said to be a piece-wise continuous function for $ t \geq 0$ , if $ f(t)$ is a piece-wise continuous function on every closed interval $ [a, b] \subset [0, \infty).$ For example, see Figure 10.1.

Figure 10.1: Piecewise Continuous Function
\includegraphics[scale=0.5]{piecewise.eps}

DEFINITION 10.2.2 (Laplace Transform)   Let $ f: [0, \infty) \longrightarrow {\mathbb{R}}$ and $ s \in {\mathbb
{R}}.$ Then $ F(s),$ for $ s \in {\mathbb{R}}$ is called the LAPLACE TRANSFORM of $ f(t),$ and is defined by

$\displaystyle {\mathcal L}(f(t)) = F(s) = \int_0^\infty f(t) e^{-s t} dt$

whenever the integral exists.
(Recall that $ \int\limits_{0}^\infty g(t) dt $ exists if $ \lim\limits_{b \longrightarrow \infty} \int\limits_0^b g(t) d(t)$ exists and we define $ \int\limits_{0}^\infty g(t) dt =
\lim\limits_{b \longrightarrow \infty} \int\limits_0^b g(t) d(t)$ .)

Remark 10.2.3  
  1. Let $ f(t)$ be an EXPONENTIALLY BOUNDED function, i.e.,

    $\displaystyle \vert f(t) \vert \leq M e^{{\alpha}t}\;\; {\mbox{ for all }}
\;\...
... real numbers }} \; {\alpha}\; {\mbox{ and }} \; M \; {\mbox{ with }} \; M > 0.$

    Then the Laplace transform of $ f$ exists.
  2. Suppose $ F(s)$ exists for some function $ f$ . Then by definition, $ \lim\limits_{b {\longrightarrow}\infty}
\int_0^{\mathbf b}f(t) e^{-s t} dt$ exists. Now, one can use the theory of improper integrals to conclude that

    $\displaystyle \lim\limits_{s {\longrightarrow}\infty} F(s) = 0.$

    Hence, a function $ F(s)$ satisfying

    $\displaystyle \lim\limits_{s {\longrightarrow}\infty} F(s) \; {\mbox{ does not exist or}} \; \lim\limits_{s {\longrightarrow}\infty} F(s) \ne 0,$

    cannot be a Laplace transform of a function $ f$ .

DEFINITION 10.2.4 (Inverse Laplace Transform)   Let $ {\mathcal L}(f(t)) = F(s)$ . That is, $ F(s)$ is the Laplace transform of the function $ f(t).$ Then $ f(t)$ is called the inverse Laplace transform of $ F(s)$ . In that case, we write $ f(t) = {\mathcal L}^{-1}(F(s)).$



Subsections
A K Lal 2007-09-12