Question

state and prove existence theorem of laplace transformation​

Answer 1

Answer:

2.1 Existence of Laplace transform

We give sufficient condition for the existence of LT. We need the concept of piecewise

continuous function.

Definition 1. (Piecewise continuous function) A function f is piecewise contin-

uous on the interval [a, b] if

(i) The interval [a, b] can be broken into a finite number of subintervals a = t0 <

t1 < t2 < · · · < tn = b, such that f is continuous in each subinterval (ti

, ti+1), for

i = 0, 1, 2, · · · , n − 1

(ii) The function f has jump discontinuity at ti, thus

lim

t→t

+

i

f(t)

< ∞, i = 0, 1, 2, · · · , n − 1;

lim

t→t

−

i

f(t)

< ∞, i = 1, 2, 3, · · · , n.

Note: A function is piecewise continuous on [0,∞) if it is piecewise continuous in

[0, A] for all A > 0

Example 6. The function defined by

f(t) =

t

2

, 0 ≤ t ≤ 1,

3 − t, 1 < t ≤ 2,

t + 1, 2 < t ≤ 3,