Lecture-IV
L APLACE T RANSFORMS
G. Ramesh
21 Sep 2015
Lecture-IV
O UTLINE
1
L ECTURE -IV
Lecture-IV
L APLACE TRANSFORM OF D ERIVATIVES
Let f : [0, ∞) → R be a continuous function with exponential
order α. Assume that f 0 is piecewise continuous. Then
• L(f 0 ) exists and
• L(f 0 ) = sL(f ) − f (0).
Lecture-IV
L APLACE TRANSFORM OF D ERIVATIVES
Let f : [0, ∞) → R be a continuous function with exponential
order α. Assume that f 0 is piecewise continuous. Then
• L(f 0 ) exists and
• L(f 0 ) = sL(f ) − f (0).
Example
1
Let f (t) = sin2 (t). Find L(f )?
Lecture-IV
L APLACE TRANSFORM OF D ERIVATIVES
Let f : [0, ∞) → R be a continuous function with exponential
order α. Assume that f 0 is piecewise continuous. Then
• L(f 0 ) exists and
• L(f 0 ) = sL(f ) − f (0).
Example
1
2
Let f (t) = sin2 (t). Find L(f )?
2
Ans: L(f ) =
.
2
s(s + 4)
Find L(t sin(wt)), t ∈ [0, ∞)?.
Lecture-IV
L APLACE TRANSFORM OF D ERIVATIVES
Let f : [0, ∞) → R be a continuous function with exponential
order α. Assume that f 0 is piecewise continuous. Then
• L(f 0 ) exists and
• L(f 0 ) = sL(f ) − f (0).
Example
1
2
Let f (t) = sin2 (t). Find L(f )?
2
Ans: L(f ) =
.
2
s(s + 4)
Find L(t sin(wt)), t ∈ [0, ∞)?.
2ws
Ans: L(f ) = 2
.
(s + w 2 )2
Lecture-IV
GENERALIZATION
T HEOREM
00
Let f , f 0 , f , · · · , f n−1 be continuous on [0, ∞) and
f j (j = 1, 2, . . . , n) be of exponential order α. Then f n is
piecewise continuous and
L(f n ) = sn L(f ) − sn−1 f (0) − sn−2 f 0 (0)−, . . . , −f n−1 (0).
Lecture-IV
GENERALIZATION
T HEOREM
00
Let f , f 0 , f , · · · , f n−1 be continuous on [0, ∞) and
f j (j = 1, 2, . . . , n) be of exponential order α. Then f n is
piecewise continuous and
L(f n ) = sn L(f ) − sn−1 f (0) − sn−2 f 0 (0)−, . . . , −f n−1 (0).
Example
Solve the IVP: y 0 + 4y = et , y (0) = 2.
Lecture-IV
GENERALIZATION
T HEOREM
00
Let f , f 0 , f , · · · , f n−1 be continuous on [0, ∞) and
f j (j = 1, 2, . . . , n) be of exponential order α. Then f n is
piecewise continuous and
L(f n ) = sn L(f ) − sn−1 f (0) − sn−2 f 0 (0)−, . . . , −f n−1 (0).
Example
Solve the IVP: y 0 + 4y = et , y (0) = 2.
9
1 1
1 • L(y ) =
+
5 s+4
5 s−1
Lecture-IV
GENERALIZATION
T HEOREM
00
Let f , f 0 , f , · · · , f n−1 be continuous on [0, ∞) and
f j (j = 1, 2, . . . , n) be of exponential order α. Then f n is
piecewise continuous and
L(f n ) = sn L(f ) − sn−1 f (0) − sn−2 f 0 (0)−, . . . , −f n−1 (0).
Example
Solve the IVP: y 0 + 4y = et , y (0) = 2.
9
1 1
1 • L(y ) =
+
5 s+4
5 s−1
9 −4t 1 t
• y (t) = e
+ e.
5
5
Lecture-IV
L APLACE TRANSFORMS FOR INTEGRALS
T HEOREM
Let f : [0, ∞) → R be a piecewise continuous function with
exponential order α. Then
Z t
F(s)
L
, (s > 0, s > α)
f (u)du =
s
0
Hence
L−1
F(s)
s
Z
t
f (u)du).
=
0
Lecture-IV
L APLACE TRANSFORMS FOR INTEGRALS
T HEOREM
Let f : [0, ∞) → R be a piecewise continuous function with
exponential order α. Then
Z t
F(s)
L
, (s > 0, s > α)
f (u)du =
s
0
Hence
L−1
F(s)
s
Z
0
EXAMPLE -1
Find
L−1
1
2
s(s + w 2 )
t
f (u)du).
=
Lecture-IV
L APLACE TRANSFORMS FOR INTEGRALS
T HEOREM
Let f : [0, ∞) → R be a piecewise continuous function with
exponential order α. Then
Z t
F(s)
L
, (s > 0, s > α)
f (u)du =
s
0
Hence
L−1
F(s)
s
Z
t
f (u)du).
=
0
EXAMPLE -1
1
Find
s(s2 + w 2 )
1 − cos(wt)
Answer: f (t) =
.
w2
L−1
Lecture-IV
EXAMPLE -2
Find
L−1
1
2
2
s (s + w 2 )
Lecture-IV
EXAMPLE -2
1
Find
s2 (s2 + w 2 )
t − sin(wt)
w
.
Answer: f (t) =
w2
L−1
Lecture-IV
D IFFERENTIATION OF L APLACE TRANSFORMS
T HEOREM
Let f : [0, ∞) → R be piecewise continuous and has
exponential order α. Then
d
ds (F(s))
= (−1)(tf (t)).
d
Hence L−1 (tf (t)) = (−1) ds
(F(s)).
E XAMPLES
s2 − w 2
• L t cos(wt) = 2
.
s + w2
2ws
• L t sin(wt) = 2
.
s + w2
Lecture-IV
E XERCISES
Find the inverse Laplace transforms of the following:
s+a
• L−1 log( s+b
)
2 +a2
• L−1 log( ss2 +b
2)
Lecture-IV
I NTEGRATION OF L APLACE TRANSFORM
Let f : [0, ∞) → R be a piecewise continuous function with
f (t)
exponenetial order α. Assume that lim
exists. Then
t→0+ t
Z ∞
f (t)
F (u)du = L
(s > α).
t
s
E XAMPLES
R∞
sin t
1 L
= s
t
1
dx
x 2 +1
=
π
2
− tan−1 (s) = tan−1 ( 1s ), s > 0
Lecture-IV
I NTEGRATION OF L APLACE TRANSFORM
Let f : [0, ∞) → R be a piecewise continuous function with
f (t)
exponenetial order α. Assume that lim
exists. Then
t→0+ t
Z ∞
f (t)
F (u)du = L
(s > α).
t
s
E XAMPLES
R∞
sin t
1 L
= s
t
sinh wt
2 L
=
t
1
dx
x 2 +1
=
π
2
− tan−1 (s) = tan−1 ( 1s ), s > 0
Lecture-IV
I NTEGRATION OF L APLACE TRANSFORM
Let f : [0, ∞) → R be a piecewise continuous function with
f (t)
exponenetial order α. Assume that lim
exists. Then
t→0+ t
Z ∞
f (t)
F (u)du = L
(s > α).
t
s
E XAMPLES
R∞
sin t
1 L
= s x 21+1 dx = π2 − tan−1 (s) = tan−1 ( 1s ), s > 0
t
sinh wt
2 L
= 12 ln s+w
s−w (s > w).
t
Lecture-IV
C ONVOLUTION PRODUCT
Let f , g : [0, ∞) → R be two functions. Then the convolution of f
and g is defined by
Z
(f ∗ g)(t) =
f (τ )g(t − τ )dτ
0
if the above integral exists.
t
Lecture-IV
C ONVOLUTION PRODUCT
Let f , g : [0, ∞) → R be two functions. Then the convolution of f
and g is defined by
Z
t
(f ∗ g)(t) =
f (τ )g(t − τ )dτ
0
if the above integral exists.
If f , g are piecewise continuous, the the above integral exists.
EXAMPLE
Let f (t) = et and g(t) = t. Then
(f ∗ g)(t) = et − t − 1.
Lecture-IV
L APLACE TRANSFORM
If f , g : [0, ∞) → R are piecewise continuous with exponential
order α, then
L(f ∗ g) = L(f ).L(g)(s > α)
Lecture-IV
L APLACE TRANSFORM
If f , g : [0, ∞) → R are piecewise continuous with exponential
order α, then
L(f ∗ g) = L(f ).L(g)(s > α)
1
1
• L(eat ∗ ebt ) = s−a
s−b (s > a, s > b)
• L−1
1
1
s−a s−b
= eat ∗ ebt .
Find L−1
1 1
s2 s−1
1
Find L−1
1
(s+1)2
2
3
Solve the integral equation
Z
t
y (τ ) sin(t − τ )dτ.
y (t) = t +
0
Lecture-IV
PARTIAL F RACTIONS
Useful in finding the inverse Laplace transform when it is
difficult to recognize that a given function is a Laplace transform
of a known function.
Lecture-IV
PARTIAL F RACTIONS
Useful in finding the inverse Laplace transform when it is
difficult to recognize that a given function is a Laplace transform
of a known function.
E XAMPLE
F(s) =
1
; (s > 3)
(s − 2)(s − 3)
Lecture-IV
PARTIAL F RACTIONS
Useful in finding the inverse Laplace transform when it is
difficult to recognize that a given function is a Laplace transform
of a known function.
E XAMPLE
1
; (s > 3)
(s − 2)(s − 3)
B
A
+
(A, B are constants)
Write F(s) =
s−2 s−3
F(s) =
Lecture-IV
PARTIAL F RACTIONS
Useful in finding the inverse Laplace transform when it is
difficult to recognize that a given function is a Laplace transform
of a known function.
E XAMPLE
1
; (s > 3)
(s − 2)(s − 3)
B
A
+
(A, B are constants)
Write F(s) =
s−2 s−3
Substituting s = 2, we get A = −1.
F(s) =
Lecture-IV
PARTIAL F RACTIONS
Useful in finding the inverse Laplace transform when it is
difficult to recognize that a given function is a Laplace transform
of a known function.
E XAMPLE
1
; (s > 3)
(s − 2)(s − 3)
B
A
+
(A, B are constants)
Write F(s) =
s−2 s−3
Substituting s = 2, we get A = −1.
s = 3 gives B = 1.
1
1
Hence F(s) =
−
.
s−3 s−2
F(s) =
Lecture-IV
L INEAR FACTORS
Let F(s) =
P(s)
, where P and Q are polynomials in s such
Q(s)
that
1
degree of P is less than or equals to the degree of Q
2
P and Q have no common factors.
Lecture-IV
L INEAR FACTORS
Let F(s) =
P(s)
, where P and Q are polynomials in s such
Q(s)
that
1
degree of P is less than or equals to the degree of Q
2
P and Q have no common factors.
Then
1
For each factor of the form as + b of Q(s), there
A
corresponds a partial fraction of the form
(A is a
as + b
contsant)
Lecture-IV
L INEAR FACTORS
Let F(s) =
P(s)
, where P and Q are polynomials in s such
Q(s)
that
1
degree of P is less than or equals to the degree of Q
2
P and Q have no common factors.
Then
1
2
For each factor of the form as + b of Q(s), there
A
corresponds a partial fraction of the form
(A is a
as + b
contsant)
For each repeated linear factor of the form (as + b)n of
Q(s), there corresponds a partial fraction of the form
A1
A2
An
+
+ ··· +
(A0 s are contsants)
as + b (as + b)2
(as + b)n i
Lecture-IV
Q UADRATIC FACTORS
1
For each factor of the form as2 + bs + c of Q(s), there
As + B
corresponds a partial fraction of the form 2
as + bs + c
(A, B are contsants)
Lecture-IV
Q UADRATIC FACTORS
1
2
For each factor of the form as2 + bs + c of Q(s), there
As + B
corresponds a partial fraction of the form 2
as + bs + c
(A, B are contsants)
For each repeated factor of the form (as2 + bs + c)n of
Q(s), there corresponds a partial fraction of the form
A1 s + B1
A2 s + B2
An s + Bn
+
+ ··· +
2
2
2
as + bs + c (as + bs + c)
(as2 + bs + c)n
(Ai , Bi0 s are contsants)
Lecture-IV
E XAMPLES
Find L−1
s+1 ;s>1
s2 (s − 1)
Lecture-IV
E XAMPLES
s+1 ;s>1
s2 (s − 1)
C
s+1
As + B
• 2
+
=
2
s−1
s (s − 1)
s
Find L−1
Lecture-IV
E XAMPLES
s+1 ;s>1
s2 (s − 1)
C
s+1
As + B
• 2
+
=
2
s−1
s (s − 1)
s
• substituting s = 0, 1, 2, we get B = −1, C = 2, A = 1
Find L−1
Lecture-IV
E XAMPLES
s+1 ;s>1
s2 (s − 1)
C
s+1
As + B
• 2
+
=
2
s−1
s (s − 1)
s
• substituting s = 0, 1, 2, we get B = −1, C = 2, A = 1
s+1 • L−1 2
= 1 − t + 2et .
s (s − 1)
Find L−1
Lecture-IV
E XAMPLES
s+1 ;s>1
s2 (s − 1)
C
s+1
As + B
• 2
+
=
2
s−1
s (s − 1)
s
• substituting s = 0, 1, 2, we get B = −1, C = 2, A = 1
s+1 • L−1 2
= 1 − t + 2et .
s (s − 1)
Find L−1
Find L−1
2s2
;s>1
(s2 + 1)(s − 1)2
Lecture-IV
E XAMPLES
s+1 ;s>1
s2 (s − 1)
C
s+1
As + B
• 2
+
=
2
s−1
s (s − 1)
s
• substituting s = 0, 1, 2, we get B = −1, C = 2, A = 1
s+1 • L−1 2
= 1 − t + 2et .
s (s − 1)
Find L−1
2s2
;s>1
(s2 + 1)(s − 1)2
Solution: − cos(t) + et + tet .
Find L−1
Lecture-IV
Solve the IVP:
y 00 + 3y 0 + 2y = t + 1, y (0) = 1, y 0 (0) = 0.
Lecture-IV
Solve the IVP:
y 00 + 3y 0 + 2y = t + 1, y (0) = 1, y 0 (0) = 0.
• L(y ) =
s3 + 3s2 + s + 1
; s>0
s2 (s + 1)(s + 2)
Lecture-IV
Solve the IVP:
y 00 + 3y 0 + 2y = t + 1, y (0) = 1, y 0 (0) = 0.
s3 + 3s2 + s + 1
; s>0
s2 (s + 1)(s + 2)
s3 + 3s2 + s + 1
As + b
C
D
• 2
=
+
+
2
s+1 s+2
s (s + 1)(s + 2)
s
• L(y ) =
Lecture-IV
Solve the IVP:
y 00 + 3y 0 + 2y = t + 1, y (0) = 1, y 0 (0) = 0.
s3 + 3s2 + s + 1
; s>0
s2 (s + 1)(s + 2)
s3 + 3s2 + s + 1
As + b
C
D
• 2
=
+
+
2
s+1 s+2
s (s + 1)(s + 2)
s
• L(y ) =
1
−3
• A = −1
4 , B = 2 , C = 2, D = 4
Lecture-IV
Solve the IVP:
y 00 + 3y 0 + 2y = t + 1, y (0) = 1, y 0 (0) = 0.
s3 + 3s2 + s + 1
; s>0
s2 (s + 1)(s + 2)
s3 + 3s2 + s + 1
As + b
C
D
• 2
=
+
+
2
s+1 s+2
s (s + 1)(s + 2)
s
• L(y ) =
1
−3
• A = −1
4 , B = 2 , C = 2, D = 4
• y (t) = 2t − 14 + 2et − 34 e−2t .
Lecture-IV
P ERIODIC F UNCTIONS
A function f : R → R is called periodic with period L if
f (x) = f (x + L) for all x ∈ R
Lecture-IV
P ERIODIC F UNCTIONS
A function f : R → R is called periodic with period L if
f (x) = f (x + L) for all x ∈ R
E XAMPLES
1 the functions cos(x), sin(x) are periodic with period 2π
2
the functions tan(x) and cot(x) are periodic with period π
3
A constant function is periodic with any period
4
f (x) = x n , x ∈ R (n ∈ N) is not periodic
5
the functions ex , cosh(x) are not periodic
Lecture-IV
• If L is a period for a periodic function, then nL is a period
for each n ∈ Z
Lecture-IV
• If L is a period for a periodic function, then nL is a period
for each n ∈ Z
• the smallest period is called the fundamental period for the
function.
• 2π is the fundamental period for cos(x) and sin(x)
• a constant function has no fundamental period.
Lecture-IV
• If L is a period for a periodic function, then nL is a period
for each n ∈ Z
• the smallest period is called the fundamental period for the
function.
• 2π is the fundamental period for cos(x) and sin(x)
• a constant function has no fundamental period.
T HEOREM
Let f be a periodic function with period L and L(f ) = F(s).
Then
Z L
1
F(s) =
e−st f (t)dt.
1 − esL 0
Lecture-IV
THANK YOU