(7.4) Operational Properties II

(7.4) Operational Properties II
DERIVATIVES OF A TRANSFORM:
THEOREM 7.4.1 (Derivatives of Translations)
If 𝐹 (𝑠) = ℒ {𝑓 (𝑡 )} and 𝑛 = 1, 2, 3, … , then
ℒ {𝑡 𝑛 𝑓(𝑡 )} = (−1)𝑛
EXAMPLE 1
Evaluate
(𝑎) ℒ {𝑡𝑓(𝑡 )}
𝑑𝑛
𝐹 (𝑠 ).
𝑑𝑠 𝑛
(𝑏) ℒ {𝑡 2 𝑓 (𝑡 )}
(𝑐 ) ℒ {𝑡𝑒 3𝑡 }
Solution:
𝑑
𝑑
𝐹 (𝑠 ) = − 𝐹 (𝑠 )
𝑑𝑠
𝑑𝑠
2
𝑑
𝑑2
2
2
(𝑏) ℒ {𝑡 𝑓(𝑡 )} = (−1)
𝐹 (𝑠 ) = 2 𝐹 (𝑠 )
𝑑𝑠 2
𝑑𝑠
(𝑐 ) ℒ{𝑡𝑒 3𝑡 }
(𝑎) ℒ{𝑡𝑓(𝑡 )} = (−1)1
By Section 7.3,
ℒ{𝑡𝑒 3𝑡 } = ℒ {𝑡 }|𝑠→𝑠−3 =
By Section 7.4,
ℒ{𝑡𝑒 3𝑡 } = (−1)1
1
1
|
=
𝑠 2 𝑠→𝑠−3 (𝑠 − 3)2
(0) − (1)
𝑑
𝑑
1
1
)=−
ℒ {𝑒 3𝑡 } = − (
=
.
2
(𝑠 − 3)
(𝑠 − 3)2
𝑑𝑠
𝑑𝑠 𝑠 − 3
EXAMPLE 2 (Using Theorem 7.4.1)
Evaluate
ℒ {𝑡 sin 𝑘𝑡 }.
Solution:
With 𝑓(𝑡 ) = sin 𝑘𝑡, 𝐹 (𝑠) = ℒ {𝑓(𝑡 )} = ℒ {sin 𝑘𝑡 } =
ℒ {𝑡 sin 𝑘𝑡 } = −
𝑘
𝑠 2 +𝑘 2
, and 𝑛 = 1, Theorem 7.4.1 gives
𝑑
𝑑
𝑘
2𝑘𝑠
)
ℒ {sin 𝑘𝑡 } = − ( 2
=
.
(𝑠 2 + 𝑘 2 )2
𝑑𝑠
𝑑𝑠 𝑠 + 𝑘 2
TRANSFORMS OF INTEGRALS:
CONVOLUTION:
If functions 𝑓 and 𝑔 are piecewise continuous on the interval [0, ∞), then a special product,
denoted by 𝑓 ∗ 𝑔, is defined by the integral
𝑡
𝑓 ∗ 𝑔 = ∫ 𝑓 (𝜏)𝑔(𝑡 − 𝜏)𝑑𝜏
… … … … … … … … … … … … … (1)
0
𝑡
= ∫ 𝑔(𝜏)𝑓 (𝑡 − 𝜏)𝑑𝜏 = 𝑔 ∗ 𝑓
0
and is called the convolution of 𝑓 and 𝑔. The convolution 𝑓 ∗ 𝑔 is a function of 𝑡.
For example,
129
(7.4) Operational Properties II
𝑡
𝑡
𝑒 ∗ sin 𝑡 = ∫ 𝑒 𝜏 sin(𝑡 − 𝜏) 𝑑𝜏 =
0
1
(− sin 𝑡 − cos 𝑡 + 𝑒 𝑡 ).
2
THEOREM 7.4.2 (Convolution Theorem)
If 𝑓 (𝑡 ) and 𝑔(𝑡 ) are piecewise continuous on [0, ∞) and of exponential order, then
ℒ {𝑓 ∗ 𝑔} = ℒ {𝑓 (𝑡 )}. ℒ {𝑔(𝑡 )} = 𝐹 (𝑠)𝐺 (𝑠).
Remark 1:
𝑡
𝑡
ℒ {𝑓(𝑡 ) ∗ 𝑔(𝑡 )} = ℒ {𝑓 (𝑡 )}. ℒ {𝑔(𝑡 )} = ℒ {∫ 𝑔(𝜏)𝑓 (𝑡 − 𝜏)𝑑𝜏} = ℒ {∫ 𝑓(𝜏)𝑔(𝑡 − 𝜏)𝑑𝜏}
0
0
= ℒ {𝑔(𝑡 ) ∗ 𝑓 (𝑡 )}
EXAMPLE 3 (Transform of Convolution)
Evaluate
𝑡
ℒ {∫ 𝑒 𝜏 sin(𝑡 − 𝜏) 𝑑𝜏}.
0
Solution:
With 𝑓(𝑡 ) = 𝑒 𝑡 and 𝑔(𝑡 ) = sin 𝑡, the convolution theorem states that the Laplace transform of the
convolution of 𝑓 and 𝑔 is the product of their Laplace transforms:
𝑡
1
1
1
ℒ {∫ 𝑒 𝜏 sin(𝑡 − 𝜏) 𝑑𝜏} = ℒ {𝑒 𝑡 }. ℒ {sin 𝑡 } =
. 2
=
.
𝑠 − 1 𝑠 + 1 (𝑠 − 1)(𝑠 2 + 1)
0
Remark 2: (Inverse Form of Theorem 7.4.2)
ℒ −1 {𝐹(𝑠). 𝐺(𝑠)} = 𝑓 ∗ 𝑔. … … … … … … … … … (2)
TRANSFORM OF AN INTEGRAL:
1
When 𝑔(𝑡 ) = 1 and ℒ {𝑔(𝑡 )} = 𝐺 (𝑠) = , the convolution theorem implies that the Laplace
transform of the integral of 𝑓 is
𝑠
𝑡
1 𝐹 (𝑠 )
ℒ {∫ 𝑓 (𝜏)𝑑𝜏} = ℒ {𝑓(𝑡 ) ∗ 1} = ℒ {𝑓(𝑡 )}. ℒ {1} = 𝐹 (𝑠). =
.
𝑠
𝑠
0
… … … … … … … … … (3)
The inverse form of (3),
𝑡
∫ 𝑓 (𝜏)𝑑𝜏 = ℒ−1 {
0
𝐹 (𝑠 )
}.
𝑠
… … … … … … … … … (4)
VOLTERRA INTEGRAL EQUATION:
The convolution theorem and the result in (3) are useful in solving other types of equations in which
an unknown function appears under an integral sign. In the next example we solve a Volterra
integral equation for 𝑓 (𝑡 ),
130
(7.4) Operational Properties II
𝑡
𝑓 (𝑡 ) = 𝑔(𝑡 ) + ∫ 𝑓 (𝜏)ℎ(𝑡 − 𝜏)𝑑𝜏 . … … … … … … … … … (5)
0
The functions 𝑔(𝑡 ) and ℎ(𝑡 ) are known. Notice that the integral in (5) has the convolution form (1)
with the symbol ℎ playing the part of 𝑔.
EXAMPLE 4 (An Integral Equation )
Solve:
𝑡
𝑓 (𝑡 ) = 3𝑡 2 − 𝑒 −𝑡 − ∫ 𝑓 (𝜏)𝑒 𝑡−𝜏 𝑑𝜏
for 𝑓(𝑡 ).
0
Solution:
In the integral we identify ℎ(𝑡 − 𝜏) = 𝑒 𝑡−𝜏 so that ℎ(𝑡 ) = 𝑒 𝑡 . We take the Laplace transform of
each term; in particular, by Theorem 7.4.2 the transform of the integral is the product of ℒ {𝑓 (𝑡 )} =
1
𝐹 (𝑠) and ℒ {𝑒 𝑡 } =
:
𝑠−1
𝐹 (𝑠) = 3.
2
1
1
(
)
−
−
𝐹
𝑠
.
.
𝑠3 𝑠 + 1
𝑠−1
After solving the last equation for 𝐹 (𝑠) and carrying out the partial fraction decomposition, we find
𝐹 (𝑠 ) =
6
6 1
2
− 4+ −
.
3
𝑠
𝑠
𝑠 𝑠+1
The inverse transform then gives
−1
𝑓 (𝑡 ) = 3ℒ
2!
3!
1
1
{ 3 } − ℒ−1 { 4 } + ℒ−1 { } − 2ℒ−1 {
} = 3𝑡2 − 𝑡3 + 1 − 2𝑒−𝑡 .
𝑠
𝑠
𝑠
𝑠+1
TRANSFORM OF A PERIODIC FUNCTION:
PERIODIC FUNCTION:
If a periodic function has period 𝑇, 𝑇 > 0, then 𝑓 (𝑡 + 𝑇) = 𝑓 (𝑡 ). The next theorem shows that the
Laplace transform of a periodic function can be obtained by integration over one period.
THEOREM 7.4.3 (Transform of a Periodic Function)
If 𝑓 (𝑡 ) is piecewise continuous on [0, ∞), of exponential order, and periodic with period 𝑇, then
ℒ {𝑓(𝑡 )} =
𝑇
1
∫
𝑒 −𝑠𝑇 𝑓 (𝑡 )𝑑𝑡.
−𝑠𝑇
1−𝑒
0
Exercises 7.4: Pages 289-290
(𝟐) Use Theorem 7.4.1 to evaluate the given Laplace transform.
ℒ {𝑡 3 𝑒 𝑡 }.
Solution:
With 𝑓(𝑡 ) = 𝑒 𝑡 , 𝐹 (𝑠) = ℒ {𝑓 (𝑡 )} = ℒ {𝑒 𝑡 } =
1
, and 𝑛 = 3, Theorem 7.4.1 gives
𝑠−1
131
(7.4) Operational Properties II
ℒ {𝑡 3 𝑒 𝑡 } = (−1)3
𝑑3
𝑑3
1
6
6
𝑡}
{
(
) = − (−
)=
ℒ
𝑒
=
−
.
3
3
4
(𝑠 − 1)
(𝑠 − 1)4
𝑑𝑠
𝑑𝑠 𝑠 − 1
-----------------------------------------------------------------------------------------------------------------------------------
(𝟐𝟐) Use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the integral
before transforming.
ℒ{𝑒2𝑡 ∗ sin 𝑡}.
Solution:
With 𝑓(𝑡 ) = 𝑒 2𝑡 and 𝑔(𝑡 ) = sin 𝑡, the convolution theorem states that the Laplace transform of the
convolution of 𝑓 and 𝑔 is the product of their Laplace transforms:
ℒ {𝑒 2𝑡 ∗ sin 𝑡 } = ℒ {𝑒 2𝑡 }. ℒ {sin 𝑡 } =
1
1
1
. 2
=
.
𝑠 − 2 𝑠 + 1 (𝑠 − 2)(𝑠 2 + 1)
------------------------------------------------------------------------------------------------------------------------------(𝟒𝟏) Use the Laplace transform to solve the given integral equation or integrodifferential
equation.
𝑡
𝑓(𝑡) + ∫ 𝑓(𝜏) 𝑑𝜏 = 1.
0
Solution:
𝑡
ℒ {𝑓(𝑡 )} + ℒ {∫ 𝑓(𝜏) 𝑑𝜏} = ℒ {1},
0
𝐹 (𝑠 ) 1
= ,
𝑠
𝑠
1
1
𝐹 (𝑠) [1 + ] = ,
𝑠
𝑠
𝑠+1
1
(
) 𝐹 (𝑠 ) = ,
𝑠
𝑠
1
𝐹 (𝑠 ) =
.
𝑠+1
𝐹 (𝑠 ) +
The inverse transform then gives
−1
𝑓 (𝑡 ) = ℒ
{
1
} = 𝑒−𝑡 .
𝑠+1
132
(7.4) Operational Properties II
Exercises 7.4: Pages 289-290
(𝟑) Use Theorem 7.4.1 to evaluate the given Laplace transform.
ℒ {𝑡 cos 2𝑡 }.
-------------------------------------------------------------------------------------------------------------------------------
(𝟐𝟏) Use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the integral
before transforming.
ℒ{𝑒−𝑡 ∗ 𝑒𝑡 cos 𝑡}.
-------------------------------------------------------------------------------------------------------------------------------
In the following problems, use the Laplace transform to solve the given integral equation or
integrodifferential equation.
𝑡
(𝟑𝟗) 𝑓(𝑡) = 𝑡𝑒 + ∫ 𝜏𝑓(𝑡 − 𝜏) 𝑑𝜏.
𝑡
0
𝑡
8
(𝟒𝟑) 𝑓(𝑡) = 1 + 𝑡 − ∫ (𝜏 − 𝑡)3 𝑓(𝜏) 𝑑𝜏.
3
0
133