Laplace Transform
Ordinary Differential Equations
Dr. Marco A Roque Sol
12/01/2015
Dr. Marco A Roque Sol
Ordinary Differential Equations
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To deal effectively with functions having jump discontinuities, it is
very helpful to introduce a function known as the unit step
function or Heaviside function. This function will be denoted by
uc and is defined by
0 t<c
uc (t) =
1 t≥c
Dr. Marco A Roque Sol
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The step can also be negative. For instance
1 t<c
u(t) = (1 − uc (t)) =
0 t≥c
Dr. Marco A Roque Sol
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In fact, any step function can be written as a linear combination of
uc (t)’s functions. For instance consider the function

2 0≤t<1



1 1≤t<2
f (t) =
2 2≤t<3



0 3≤t
Dr. Marco A Roque Sol
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We start with the function f1 (t) = 2u0 , which agrees with f (t) on
[0, 1). To produce the jump down of one unit at t = 1, we add
−u1 (t) to f1 (t), obtaining f2 (t) = 2 − u1 (t), which agrees with
f (t) on [1, 2). The jump up of one unit at t = 2 corresponds to
adding u2 (t), which gives f3 (t) = 2u0 − u1 (t) + u2 (t). Thus we
obtain
f (t) = f3 (t) = 2u0 − u1 (t) + u2 (t)
The Laplace transform of uc for c ≥ 0 is easily determined:
Z ∞
Z ∞
e −cs
−st
L {uc } =
e uc dt =
e −st dt =
, s>0
s
0
c
Dr. Marco A Roque Sol
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For a given function f defined for t ≥ 0, we will often want to
consider the related function g defined by
0
t<c
u(t) = (1 − uc (t)) =
f (t − c) c ≤ t
which represents a translation of f a distance c in the positive t
direction.
Dr. Marco A Roque Sol
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In terms of the unit step function we can write g (t) in the
convenient form
g (t) = uc (t)f (t − c)
Theorem 6.3
If F (s) = L {f (t)} exists for 0 ≤ a < s, and if c is a positive
constant, then
L {uc (t)f (t − c)} = e −cs L {f (t)} = e −cs F (s),
Conversely, if f (t) = L −1 {F (s)}, then
L −1 {e −cs F (s)} = uc (t)f (t − c)
Dr. Marco A Roque Sol
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Example 6.10
If the function f is defined
sin(t)
0 ≤ t < π/4
f (t) =
sin(t) + cos(t − π/4) π/4 ≤ t
find L {f (t)}.
Dr. Marco A Roque Sol
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Solution
Note that f (t) = sint + g (t), where
0 ≤ t < π/4
g (t) =
cos(t − π/4) π/4 ≤ t
Thus
g (t) = uπ/4 (t)cos(t − π/4)
Dr. Marco A Roque Sol
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and
L {f (t)} = L {sin(t)} + L {uπ/4 (t)cos(t − π/4)} =
L {sin(t)} + e −πs/4 L {cos(t)}
and using the table of Laplace Transforms
L {f (t)} =
1
s
1 + se −πs/4
−πs/4
+
e
=
s2 + 1
s2 + 1
s2 + 1
Dr. Marco A Roque Sol
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Let’s consider the following theorem
Theorem 6.4
If F (s) = L {f (t)} exists for 0 ≤ a < s, and if c is a constant, then
L {e ct f (t)} = F (s − c),
s >a+c
Conversely, if f (t) = L −1 {F (s)}, then
L −1 {F (s − c)} = e ct f (t)
Dr. Marco A Roque Sol
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Example 6.11
Find the inverse transform of
G (s) =
s2
1
− 4s + 5
Solution
First of all the polynomial s 2 − 4s + 5, has complex roots. By
completing the square in the denominator, we can write
1
G (s) =
= F (s − 2)
(s − 2)2 + 1
where F (s) = (s 2 + 1)−1 . L −1 {F (s)} = sin(t). It follows from
the previous theorem that
g (t) = L −1 {G (s)} = e 2t sin(t)
Dr. Marco A Roque Sol
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Considering solving the IVP
ay 00 + by 0 + cy = g (t);
y (0) = y 0 (0) = 0
Using Laplace Transform
L {ay 00 + by 0 + cy = g (t)}
aL {y 00 } + bL {y 0 } + cL {y } = L {g (t)}
= a s 2 Y (s) − sy (0) − y 0 (0) + b [sY (s) − y (0)] + ca [Y (s)] = G (s)
Y (s) =
as 2
G (s)
;
+ bs + c
where Y (s) = L {y (t)} and g (t) is a piecewise continuous force.
Dr. Marco A Roque Sol
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Example 6.12
Solve the initial value problem
2y 00 + y 0 + 2y = g (t);
y (0) = 0; y 0 (0) = 0
where

 0 0≤t<5
1 5 ≤ t < 20 = u5 − u20
g (t) =

0 20 ≤ t
Dr. Marco A Roque Sol
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Solution
First of all the polynomial s 2 − 4s + 5, has complex roots. By
completing the square in the denominator, we can write
G (s) =
1
= F (s − 2)
(s − 2)2 + 1
where F (s) = (s 2 + 1)−1 . L −1 {F (s)} = sin(t), it follows from
the previous theorem that
g (t) = L −1 {G (s)} = e 2t sin(t)
Dr. Marco A Roque Sol
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The Laplace transform of the above equation is
Y (s) =
Y (s) =
G (s)
(e −5s − e −2s )/s
=
2s 2 + s + 2
2s 2 + s + 2
e −5s − e −2s
= e −5s − e −2s H(s) = e −5s H(s) − e −2s H(s)
2
s (2s + s + 2)
with H(s) = 1/[s(2s 2 + s + 2)]. Thus, if h(t) = L −1 {H(s)} =
L −1 {1/s[2s 2 + s + 2]}, and using the Theorem 6.3
(L −1 {e −cs F (s)} = uc (t)f (t − c) ) we have
Dr. Marco A Roque Sol
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y (t) = u5 h(t − 5) − u20 h(t − 20)
Finally, to determine h(t), we use the partial fraction expansion of
H(s)
a
bs + c
1
} = L −1 { } + L −1 { 2
}
s (2s 2 + s + 2)
s
2s + s + 2
and finding the coefficients a, b, and c, we get
h(t) = L −1 {
1/2
(−s) + (−1/2)
} + L −1 {
}=
s
2s 2 + s + 2
1/2
1
(s + 1/4) + 1/4
L −1 {
}−
L −1 {
}
s
2
(s + 1/4)2 + 15/16
h(t) = L −1 {
Dr. Marco A Roque Sol
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L
−1
1
s + 1/4
√
}+
L −1 {
2
(s + 1/4)2 + ( 15/4)2
#
√
1
15/4
√ L −1 {
√
}
15
(s + 1/4)2 + ( 15/16)2
1/2
{
}−
s
but
L −1 {
Dr. Marco A Roque Sol
1
1/2
}=
s
2
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i
√
1
s + 1/4
1 h t/4
−1
√
L {
} =
e cos( 15t/4)
2
2
(s + 1/4)2 + ( 15/4)2
"
#
√
h√
i
√
1
15/4
√
√
L −1 {
} = ( 15/15)e −t/4 sin( 15t/4)
2 15
(s + 1/4)2 + ( 15/4)2
Therefore
h(t) =
i
√
√
√
1 1 h t/4
−
e cos( 15t/4) + ( 15/15)e −t/4 sin( 15t/4)
2 2
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And the solution is given by
y (t) = u5 h(t − 5) − u20 h(t − 20)
√
1 1 h (t−5)/4
−
e
cos( 15(t − 5)/4)
2 2
i
√
√
+ ( 15/15)e −(t−5)/4 sin( 15(t − 5)/4)
y (t) = u5
√
1 1 h (t−20)/4
−
e
cos( 15(t − 20)/4)
2 2
i
√
√
+ ( 15/15)e −(t−20)/4 sin( 15(t − 20)/4)
−u20
Dr. Marco A Roque Sol
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Example 6.13
Solve the initial value problem
y 00 + 4y = g (t);
y (0) = 0; y 0 (0) = 0
where

0≤t<5
 0
t −5
(t − 5)/5 5 ≤ t ≤ 10 = u5 − u10
g (t) =
+ u10 =

5
1
10 ≤ t
1
1
u5 (t − 5) − u10 (t − 10)
5
5
Dr. Marco A Roque Sol
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Solution
1
1
1
1
G (s) = e −5s 2 − e −10s 2
5
s
5
s
Y (s) =
G (s)
(e −5s − e −10s )/(5s 2 )
1
=
= (e −5s − e −10s )H(s) =
2
2
s +4
(s + 4)
5
1 −5s
e F (s) − e −10s F (s)
5
1
1
y (t) = u5 (t)f (t − 5) − u10 (t)f (t − 10)
5
5
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where
F (s) =
1
1
a
b
cs + d
= + 2+ 2
=
2
2
5 s (s + 4)
s
s
s +4
as(s 2 + 4) + b(s 2 + 4) + cs 3 + ds 2
s 2 (s 2 + 4)
a + c = 0;
b + d = 0;
4a = 0;
4b = 1
Then, the solution is a = 0, b = 1/4, c = 0, and d = −1/4
F (s) =
1/4 1 2
+
s
8 s2 + 4
Dr. Marco A Roque Sol
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Finally, we have f (t)
1
1
f (t) = t − sin(2t)
4
8
and the solution y (t) is given by
1
1
1
y (t) = u5 (t − 5) − sin2(t − 5)
5
4
8
1
1
1
− u10 (t − 10) − sin2(t − 10)
5
4
8
Dr. Marco A Roque Sol
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Consider the Initial Value Problem
ay 00 + by 0 + cy = g (t)
where
g (t) =
0
otherwise
1/ t0 ≤ t ≤ t0 + We can solve the problem using Laplace Transform technique, but
Can we still solve the problem when → 0 ?
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Example 6.14
Solve the initial value problem
y 00 + y = I0 g (t);
y (0) = 0; y 0 (0) = 0
where
g (t) =
1
1/ 0 < t < = (1 − u )
0
≤t
Dr. Marco A Roque Sol
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Solution
G (s) =
Y (s) =
G (s)
s2 + 1
11
11
1 1
− e −s
=
(1 − e −s )
s
s
s s
I0 1
−s )
s (1 − e
s2 + 1
=
I0
I0
I0
(1 − e −s )F (s) = F (s) − e −s F (s)
where
F (s) =
1
1
s
= − 2
s(s 2 + 1)
s
s +1
Dr. Marco A Roque Sol
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and
f (t) = 1 − cos(t) =⇒ f (t − ) = 1 − cos(t − )
Therefore
I0
I0
y (t) = L −1 { F (s)} − L −1 { e −s F (s)}
I0
[1 − cost(t) − u (t) (1 − cos(t − ))]
I0
0≤t<
(1 − cos(t − ))
y (t) =
I0
(cos(t
−
)
−
cos(t))
≤t
Dr. Marco A Roque Sol
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Moreover, we have
1 − cos() 1 − cos(t) =
lim→0 ≤ lim→0 sin()
=0
1
On the other hand, for t ≥ , pause when → 0 , we have
lim→0
lim→0
cos(t − ) − cos(t)
sin(t − )
= lim→0
= sin(t)
1
Thus
y (t) = lim→0 y = I0 sin(t),
Dr. Marco A Roque Sol
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Unit Impulse
The Unit Impulsion Function or Dirac Delta Function (
https://en.wikipedia.org/wiki/Paul_Dirac ) at the point
t0 , denoted by δ(t − t0 ) is defined by
∞ t = t0
δ(t − t0 ) =
0 t 6= t0
such that if a ≤ t ≤ b, then
Z
b
f (t)δ(t − t0 ) = f (t0 )
a
In particular, from the above difinition, we have
b
Z
δ(t − t0 )dt =
a
or
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1 a ≤ t0 ≤ b
0 otherwise
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Z
∞
f (t)δ(t − t0 )dt = f (t0 )
−∞
Z
∞
δ(t − t0 )dt = 1
−∞
Now, using the definition of δ(t − t0 ), the Laplace Transform of
the Dirac Delta Function is given by
Z ∞
L {δ(t − t0 )} =
δ(t − t0 )e −st = e −st0 ; t0 > 0
0
In particular
L {δ(t)} = 1
Dr. Marco A Roque Sol
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Example 6.15
Solve the initial value problem
y 00 + y = I0 δ(t);
y (0) = 0; y 0 (0) = 0
Solution
Y (s) =
y (t) =
s2
I0
+1
I0 L −1 {1
} = I0 sin(t)
s2 + 1
Dr. Marco A Roque Sol
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Example 6.16
Solve the initial value problem
2y 00 + y 0 + 2y = δ(t − 5);
y (0) = 0; y 0 (0) = 0
Solution
Y (s) =
G (s)
e −5s
e −5s
=
=
F (s)
2s 2 + s + 2
2s 2 + s + 2
2
where
F (s) =
1
s2
+
1
2s
+1
Dr. Marco A Roque Sol
=
1
(s +
1 2
4)
+
15
16
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Therefore −1
f (t) = L {
1
1
(s +
p
e
15/16
1 2
4)
+
15
16
}=
r
−t/4
sin
15
t
4
!
4
= √ e −t/4 sin
15
√
15
t
4
!
Hence, the solution is
1
y (t) = u5 (t)f (t − 5)
2
!
√
4 −(t−5)/4
15
1
(t − 5) =
y (t) = u5 (t) √ e
sin
2
4
15
(
0
√
0≤t<5
15
1
4
−(t−5)/4
√
sin 4 (t − 5)
5≤t
2 u5 (t) 15 e
Dr. Marco A Roque Sol
Ordinary Differential Equations