1 Week 8 3. Applications of the LT to ODEs Theorem 1: If the Laplace transforms of f(t), f ’ (t),...
11
1 Week 8 3. Applications of the LT to ODEs Theorem 1: If the Laplace transforms of f(t), f’ (t), and f’’ (t) exist for some s, then Alternative notation: ), 0 ( )] ( [ )] ( [ f t f s t f L L ). 0 ( ) 0 ( )] ( [ )] ( [ 2 f sf t f s t f L L ), 0 ( ) ( )] ( [ f s F s t f L ). 0 ( ) 0 ( ) ( )] ( [ 2 f f s s F s t f L
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Transcript of 1 Week 8 3. Applications of the LT to ODEs Theorem 1: If the Laplace transforms of f(t), f ’ (t),...
1
Week 8
3. Applications of the LT to ODEs
Theorem 1:
If the Laplace transforms of f(t), f’ (t), and f’’ (t) exist for some s, then
Alternative notation:
),0()]([)]([ ftfstf LL
).0()0()]([)]([ 2 fsftfstf LL
),0()()]([ fsFstf L
).0()0()()]([ 2 ffssFstf L
2
Consider a linear ODE with constant coefficients. It can be solved by the LT as follows:
LT (step 1)
ODE for y(t)
Algebraic equation for Y(s)
Y(s) = ...
y(t) = ...
solve (step 2)
inverse LT (step 3)
3
A quick review of partial fractions:
Consider
where P1 and P2 are polys. in s and the degree of P1 is
strictly smaller than that of P2. Assume also that P2 is factorised.
,)(
)()(
2
1
sP
sPsY
Then...
4
mn gfssdcssbsas
sPsY
))(())((
)()( 22
1
as
A
nn
bs
B
bs
B
bs
B
)(...
)( 221
dcss
DCs
2
.)(
...)( 222
222
11m
mm
dcss
GsF
dcss
GsF
dcss
GsF
(unrepeated linear factor)
(repeated linear factor)
(unrepeated irreducible quadratic factor)
(repeated irreducible quadratic factor)
5
Example 1:
),(52 tryyy
,1if0
,10if1
,0if0
)(
t
t
t
tr
(2)
(1)
Solution:
Step 0: Observe that
.2)0(,1)0( yy
where
and
).1u()u()( tttr
Step 1: Take the LT of (1)...
6
Step 2:
)],1[u()][u(][5][2][ ttyyy LLLLL
,e1
5)]0([2)0()0(2
ssYysYyysYs
s
hence,
,e1
5)1(222
ssYsYsYs
s
hence,
11 2
,)52(
e
)52(
1
52
4222
ssssssss
sY
s
.TermTermTerm 321
Step 3: The inverse LT.
7
,2sin2cose]Term[2
1
5
1
5
12
1 ttt
L
.)1u()1(2sin)1(2cose]Term[2
1)1(5
1
5
13
1 ttttL
,2sin2cose]Term[2
31
1 ttt L
Example 2:
Using partial fractions, simplify
.)4(
12
3
ss
s
8
4. Inversion of Laplace transformation using complex integrals
Theorem 2:
Let F(s) be the Laplace transform of f(t). Then
where γ is such that the straight line (γ – i∞, γ + i∞) is located to the right of all singular points of F(s).
,de)(2
1)(
i
i
st ssFi
tf
Question: how do we find L–1[ F(s)] if F(s) isn’t in the Table?Answer: using the following theorem.
(3)
Comment:
If F(s) decays as s → ∞ or grows slower than exponentially,
integral (3) vanishes for all t < 0.
9
,]),(res[2d)(1
N
nnC
ssFissF
Brief review of integration in complex plane:
Integrals over a closed, positively oriented contour C in a complex plane can be calculated using residues.
Let a function F(s) be analytic inside C except N points s = sn where it has poles (but not branch points, etc.).
where res[F(s), sn] are the residues of F(s) at s = sn.
Then
For example,
)...(,)(
)(res),(,
)(res 2 aGa
as
sGaGa
as
sG
i.e., when traversed, interior is on the left
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Example 4:
Using Theorem 2, find
].1[1L
Example 3:
Using Theorem 2, find
].[],)[( 2111 sas LL
The answer: the integral in the definition of the inverse LT diverges – hence, this transform doesn’t exist.
.d
d
)!1(
1,
)(
)(res 1
1
asm
m
m s
G
ma
as
sG
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Comment:
If an inverse transform, F(s), doesn’t decay as s → ∞, the corresponding f(t) isn’t a well-behaved function.
