ORDINARY DIFFERENTIAL EQUATION (ODE) LAPLACE TRANSFORM.
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ORDINARY ORDINARY DIFFERENTIAL DIFFERENTIAL EQUATION (ODE) EQUATION (ODE) LAPLACE LAPLACE TRANSFORM TRANSFORM
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Transcript of ORDINARY DIFFERENTIAL EQUATION (ODE) LAPLACE TRANSFORM.
ORDINARY ORDINARY DIFFERENTIAL DIFFERENTIAL
EQUATION (ODE) EQUATION (ODE)
LAPLACE LAPLACE TRANSFORMTRANSFORM
1.1 Definition of the Laplace Transform
Let be a function defined on . The Laplace transform of is
defined to be the function given by the integral
The domain of the transform is taken to be all values of s for which the integral exists. The Laplace transform of is denoted both by
And the alternate notation .
)(tf ),0( )(tf
)(sF
0
)()( dttfesF st
)(sF)(tf )(sF
fL
Laplace transform
)(tf
fLsF )(
dttfesF st )()(0
Note: It should be understood that the integral that defines the Laplace Transform is an improper integral, defined by
bst
b
st dttfedttfesF00
)(lim)()(
Example 1: Simplest Laplace transform
Find the Laplace transform of the constant function , where .
1)( tf 0tSolution: From the definition of the Laplace transform, we have
bst
b
st dtedteL00
lim1.1
s
e
s
sb
b
bt
tb s
e st1
limlim
0
s
1
Example 2: Laplace transform of Exponential Functions
Find the Laplace transform of , where k is any constant.
ktetf )(
ks 1
Solution: From the definition of the Laplace transform, we have
btks
b
ktstkt dtedteeeL0
)(
0
lim.
1lim1
lim )(
0
)(
bks
b
bt
tb
eksks
e tks
for ks
t
)(tf
L
s
s
t
L
1)( tf
)(tf
)(sF
ssF
1)(
)(sF
kssF
1
)(
k
ktetf )(
Theorem 1: Laplace transform, a linear transformation
Let f and g be functions whose Laplace transform exists on a common domain. The Laplace transform satisfies the two properties of being a linear transformation:
gLfLgfL fcLcfL
where c is an arbitrary constant.
Example 3: Find the Laplace transform teL 235
tt eLLeL 22 3535
teLL 2315
21
31
5ss
)2()5(2
sss 0s
Theorem 2: Laplace transform of Derivative
Let f is a continuous function whose derivative is piecewise continuous
on , and if both functions are of exponential order , then both
and exists for , moreover,
'f
),0[
1.2 Properties of the Laplace Transform
te
)0(' ffsLfL
'fL
)1( nf
s
Theorem 3: Laplace transform of Higher Derivatives
Let the functions f, f’, f’’…. are continuous on and is
piecewise continuous and if these functions are of exponential order ,
then the Laplace transform of all these functions exist for , moreover,
fL
),0[ )1( nf )(nf
te
s
)0().....0(')0( )1(21 nnnnn ffsfsfLsfL
Theorem 3: Translation Property
If the Laplace transform exists for , then
for .
fLsF )( s
)0(' ffsLfL
as
Example 4: Translation Property
Find bteL at cosSolution: From Table of Laplace Transform we know that
Using Translation Property, we have
22
)(cosbs
ssFbtL
22)(
)(cosbas
asasFbteL at
Theorem 4: Laplace Transform
If is a piecewise continuous function on and of exponential
order then for ,
)('' tftL
)(tf ) ,0[
te s
)()1( sds
FdftL n
nnn
Example 5: Find bttL cos
Solution: Since
we have
22
)(cosbs
ssFbtL
222
22
22 )(cos
bs
bs
bs
sdsd
bttL
222
22
22222 )(
1
)(
)2(
bs
bs
bsbs
ss
Example 6: Combining the Rules
Find the transform
deLtt sin0
2Solution: Starting from the inside, we compute the following Laplace
transform in succession using rules from Table 2.
STEP 1:
STEP 2:
STEP 3:
STEP 4:
1
1sin
2 s
L
222 )1(
2
1
1sin
s
s
sdsd
L
22220 )1(
2
)1(
21 sin
ss
ss
dL
220
2
)1)2(
2 sin
sdeL t
1.3 Inverse Laplace TransformPURPOSE
To introduce the inverse Laplace transform and show how it can
be found by using tables with the help of some algebraic manipulations,
such as the partial fraction decomposition.
)(1 sFL
DEFINITION
The inverse Laplace transform of a function is the unique continuous
Function f that satisfies
which is denoted by . In case all functions f that satisfy
are discontinuous on select any piecewise
continuous function that satisfies this condition to be .
)(sF
)(sFfL
)()(1 tfsFL
)(sFfL ) ,0(
)(1 sFL
111
s
L
Common inverse Laplace Transform1. 5.
2. 6.
• 7.
4.
nn ts
nL
11 !
ateas
L
11
btbs
bL sin22
1
btbs
sL cos22
1
btbs
bL sinh22
1
btbs
sL cosh22
1
Example 7: Simple Inverses
Find the inverse Laplace Transform of the following functions. (see slide 12)
(a) Solution:
(b) Solution:
(c) Solution:
46
)(s
sF
16
4)(
2 s
sF
313
1 !3t
sL
ts
L 4sin4
422
1
106
1)(
2
sssF
1)3(
1
104
12
12
1
sL
ssL
te t sin3
Theorem 5: Linearity of the Inverse Laplace Transform
If two inverse Laplace transform and exists, then
(i)
(ii)
These conditions state that the inverse Laplace Transform is a linear transformation
FL 1 GL 1
GLFLGFL 111
FcLcFL 11
Example 8: Preliminary Algebraic Manipulation
Find the inverse transform of
Solution:
5
31
22
1
ssL
5
55
31
12
5
31
22
112
1
sL
sL
ssL tet 5sin
53
2
RULES FOR PARTIAL FRACTION DECOMPOSITION If P(s) and Q(s) are polynomials in s, and if the degree of P(s) is less than
the degree of Q(s), then exists and can be found by first
writing as its partial fraction decomposition.
)(/)(1 sQsPL
)(/)( sQsP
= terms of the partial fraction decomposition
•Linear Factor: For each factor ( ) in the denominator of Q(s), there corresponds a term of the form
in the partial fraction decomposition.
•Power of Linear Factor: For each power of a linear factor in the denominator of Q(s), there corresponds a term of the form
in the partial fraction decomposition.
)()(sQsP
bas
basA
nbas )(
nn
bas
A
bas
Abas
A
)(......
)( 221
3. Quadratic Factor: For each factor in the denominator of Q(s), there corresponds a term of the form
in the partial fraction decomposition.
•Power of Quadratic Factor: For each power of a quadratic factor in the denominator of Q(s), there corresponds a term of the form
in the partial fraction decomposition.
)( 2 cbsas
cbsas
BAs
2
ncbsas )( 2
nnn
cbsas
BsA
cbsas
BsAcbsas
BsA
)(......
)()( 22211
Example 9: Two Linear Factors
Find
Solution: Since the denominator contains two linear factors, we write
Hence
)1(11
ssL
)1()1(
1)1(1
ssBsA
sB
sA
ss
0 BsAs1A
111
)1(1
ssss
1B
111
)1(1 11
ssL
ssL
111 11
sL
sL te1
Example 10: One Linear and One Quadratic Factor
Find
Solution: Since the denominator contains one linear factors and one
Quadratic factor, we try the decomposition
Hence
)1(
12
1
ssL
)1(
)
)1(
)()1(
1)1(
12
22
2
2
22
ss
csBsAAs
ss
cBsssA
s
cBssA
ss
1A 0c 1 1 BBA
1
1
)1(
12
12
1
s
ss
Lss
L
1
12
11
s
sL
sL tcos1
Example 11: Power of Linear Factor
Find
Solution: Since the denominator contains one linear factors and one
Quadratic factor, we try the decomposition
3
21
)1(s
sL
3
2
323
2
)1(
)1()1(
)1()1(1)1(
s
csBsA
s
C
s
BsA
s
s
cBBsAAsAscsBsA 2)1()1( 22
1A 2 02 BBA 021 0 ccBA1c
323
2
)1(
1
)1(
21
1
)1(
ssss
s
31
211
3
21
)1(
221
)1(
21
1
)1( sL
sL
sL
s
sL ttt ettee 2
21
2
Example 12: Completing the Square
Find
Solution: Since the discriminant of the quadratic expression is negative, it cannot be factored into real linear factors. Hence, we complete
the square we get
Hence,
52
12
1
ssL
4)1(41252 222 sssss
4)1(
221
4)1(
1
52
12
12
12
1
sL
sL
ssL
te t 2sin21
PURPOSE
To show how the initial-value problem
can be solved by transforming the differential equation into an algebraic
equation in , which can then be solved by using simple
algebra. We then show how the inverse Laplace Transform
can be found, obtaining the solution .
1.4 Initial- Value Problems
)(''' tfcybyay
0)0( yy
')0(' 0yy
)()(1 tysYL
yLsY )(
)(ty
DEFINITION: STEPS IN Solving Differential Equations
STEP 1: Take the Laplace Transform of each of the given differential equation, obtaining an algebraic equation in the transform of the solution . The initial conditions for the problem will be included in the algebraic equations.
STEP 2: Solve the algebraic equation obtained in Step 1 for the transform of the unknown solution.
STEP 3: Find the inverse transform to find the solution .
yL
yL
)(ty
Transform problem:
Solution of the transformed problem:
Solution of the original problem:
1)1( 2 yLs
Typical problem:
0'' yy
0)0( y
1)0(' y
1
12
s
yL
tty sin)(
Solve the transformed
problem
(Direct)
solution
Example 13: First-Order Equation
Solve the initial-value problem
Solution: Since the differential equation is an identity between two functions of t, their transform are also equal. Hence
By the linearity of the Laplace Transform, we have
Using the derivative property and
substituting the initial condition into the above equation,
we find
teyy 3'
1)0( y
teLyyL 3'
teLyLyL 3'
)0(' yysLyL
1)0( y
1
131
s
yLysL
Solving for gives
Hence the inverse gives:
1
131
s
yLysL
yL
)3)(1(
2
ss
syL
12
11
13
ss
ssyL
)3)(1(1
)3(1
sss
)3(2
1)1(2
1)3(
1sss
ttt eLeLeL 33
21
21
ttt eeety 33
21
21
)( tt ee 3
21
)3)(1()1()3(
)3()1(
sssBsA
sB
sA
0 BA 13 BA
1)(3 AA
21A
21B
Example 14: Use of the Laplace Transform
Solve the initial-value problem
Solution: Taking the Laplace transform of both sides
we obtain
tyy sin4'' 0)0( y1)0(' y
tLyyL sin4''
tLyLyL sin4''
Using the derivative property: )0(' yysLyL
)0(')0('' 2 ysyyLsyL
1
1sin4)0(')0(
22
s
tLyLysyyLs
1
1sin4)0(')0(
22
s
tLyLysyyLs
Substituting the initial conditions and and using the
more suggestive notation , we obtain
Solving for gives
yLsY )(
1
1)(41)(
22
s
sYsYs
0)0( y 1)0(' y
)(sY
1
11)4)((
22
s
ssY1
21
1
1)4)(( 2
2
22
s
s
sssY
)4)(1(
2)( 22
2
ss
ssY
Taking partial fraction of )4)(1(
2)( 22
2
ss
ssY
)4()1( 22
s
DCs
s
BAs
)1)(()4)(( 22 sDCssBAs
2)4()4()()( 223 sDBsCAsDBsCA
1DB0CA 04 CA 24 DB
solve
31B
32D
0CA
tLtLss
sY 2sin32
sin31
)4(3
2
)1(3
1)(
22
Hence the solution is ttty 2sin32
sin31
)(
p4=Plot[((1/3)Sin[t]+(2/3)Sin[2t]),{t,0,4*Pi}] by Mathematica
2 4 6 8 10 12
-0.75
-0.5
-0.25
0.25
0.5
0.75 ttty 2sin32
sin31
)(
Example 15: Differential equation with Damping
Solve the initial-value problem
Solution: Taking the Laplace transform of both sides
teyyy 323'' 0)0( y1)0(' y
s
yLyyLysyyLs1
2)0(3)0(')0(2
Solving for give yL
s
yLyLyLs1
2312
Substituting the initial conditions
)1)(2)(3(
4
)23)(3(
42
sss
s
sss
syL
)1()2()3()1)(2)(3(4
sC
sB
sA
ssss
)1()2()3()1)(2)(3(4
sC
sB
sA
ssss
21A
Finding the undetermined coefficients gives
2B23C
)1(2
3)2(
2)3(2
1
sss
yL
ttt eeety 23
221
)( 23
ttt eLeLeL 23
221 23
Hence we have
p5=Plot[((1/2)Exp[-3t]-(2)Exp[-2t]+(3/2)Exp[-t]),{t,0,5}]
1 2 3 4 5
0.05
0.1
0.15
0.2
0.25
0.3
ttt eeety 23
221
)( 23
