LAPLACE TRANSFORMS

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LAPLACE TRANSFORMS
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LAPLACE TRANSFORMS
Given a real-valued function f(x) defined for all
x ? 0, we define its Laplace transform as
where p is a real number.
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Thus the Laplace transform of f(x) is a function
F(p) of p and is defined only for those p for
which the improper integral
exists.
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From definition we easily show that if f(x)1,
then
for all p gt 0
Also if f(x)x,
for all p gt 0
If f(x)eax,
for p gt a
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Let f(x)sin ax. We know
Thus for all pgt0,
exists and equals
i.e.
for p gt 0
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The Laplace transform is a linear transformation.
Thus
Hence
for all p gt 0
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Definition of inverse Laplace transform
Given a function F(p), we say its inverse Laplace
transform is L-1F(p) f(x) if
Lf(x) F(p). Thus
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Lerchs theorem (on the uniqueness of the inverse
Laplace transform)
If
and also if
then
at all points x except where either one of them
is discontinous.
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Sufficient conditions for the existence of
Laplace transform of a function.
Definition Let a, b be a finite interval. A
real-valued function f(x) is said to be piecewise
continuous on a, b if it is continuous at all
points of a, b except a finite number
and at these (finite number of) points of
discontinuity, both the left hand and right hand
limits exist as finite numbers.
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Example
Let
then f(x) is continuous at all points of 1,3
except at
x2.
LH limit of f(x) at x 2 is 2. RH limit of f(x)
at x 2 is 1. Hence f(x) is piecewise continuous
on 1,3.
Example
Let f(x) x greatest integer x.
Then f(x) is piecewise continuous on 0,4.
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Example
Let f(x) . Then f(x)
is not piecewise continuous on 0,2.
Definition A real valued function f(x) is said to
be piecewise continuous for all x 0 if it is
piecewise continuous on every finite interval
0,b.
Definition A real valued function f(x) is said
to be of exponential order for x 0 if there
exist constants M, c such that
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Thus f(x) sin 2x is of exponential order as we
can take M 1, c 0.
f(x) x2 is of exponential order as we can take
M 2, c 1.
Fact Let f(x) be piecewise continuous and be of
exponential order on x 0.
Then Lf(x) exists for all p gt c and
and hence ? 0 as p ? ? .
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A counterexample Let f(x)
Then f(x) is not piecewise continuous on x
0. But yet its Laplace transform exists for all p
gt 0.
Put pxt. Assume pgt0
for all p gt 0
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Properties of Laplace transform
Theorem 1 Let f(x) be continuous for all x 0.
Suppose f ?(x) exists, is piecewise continuous
and is of exponential order on x 0. Then Lf
?(x)(p) p Lf(x) f(0)
Proof
Integrating by parts (and noting that f ? is of
exponential order implies f is of exponential
order) we get
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as

• Corollary
• If f, f ? are continuous and if f ? is piecewise
  continuous and is of exponential order on x 0,
  then

Lf ?(p) p2 Lf(x)- p f(0) - f ?(0)
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2) If f, f , f ,, f(n-1) are continuous and
if f(n) is piecewise continuous and is of
exponential order on x 0, then
Application 1) Let f(x) x. Hence f(x) is
continuous and f ? (x) 1 is of exponential
order. Thus Lf ? (x) pLf (x) f (0)
gives
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2) Let f(x)xn Thus Lf(n)(x)
pnLf(x)-pn-1f(0)--f(n-1)(0) i.e
Ln!pnLxn or n! L1 pnLxn Hence
Lxn n!/pn1 (p gt 0)
3) Let f(x)sin ax Hence f(x) is continuous and
f? (x) a cos ax is continuous and is of
exponential order on x 0.
Thus Lf ? (x) pLf (x) f (0) or
La cos ax pLsin ax pa/(p2a2) ?
Lcos ax p/(p2a2) p gt 0
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Theorem 2 Let f be pw continuous and of
exponential order on x 0. Let a 0 be a
constant. Then
Proof
Hence g ?(x) f(x) is pw continuous and is of
exponential order. Thus
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or
Corollary If f is pw continuous and of
exponential order then
Application
1)
?
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2)
?
?
?
p gt 1
End of Lecture