(7.4) Operational Properties II

129 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^𝑛𝑓(𝑡)} = (−1)^𝑛 𝑑^𝑛

𝑑𝑠^𝑛𝐹(𝑠).

EXAMPLE 1 Evaluate

(𝑎) ℒ{𝑡𝑓(𝑡)} (𝑏) ℒ{𝑡²𝑓(𝑡)} (𝑐) ℒ{𝑡𝑒^3𝑡} Solution:

(𝑎) ℒ{𝑡𝑓(𝑡)} = (−1)¹ 𝑑

𝑑𝑠𝐹(𝑠) = − 𝑑

𝑑𝑠𝐹(𝑠) (𝑏) ℒ{𝑡²𝑓(𝑡)} = (−1)² 𝑑²

𝑑𝑠²𝐹(𝑠) = 𝑑² 𝑑𝑠²𝐹(𝑠) (𝑐) ℒ{𝑡𝑒^3𝑡}

By Section 7.3,

ℒ{𝑡𝑒^3𝑡} = ℒ{𝑡}|_{𝑠→𝑠−3} = 1 𝑠²|

𝑠→𝑠−3= 1 (𝑠 − 3)² By Section 7.4,

ℒ{𝑡𝑒^3𝑡} = (−1)¹ 𝑑

𝑑𝑠ℒ{𝑒^3𝑡} = − 𝑑 𝑑𝑠( 1

𝑠 − 3) = −(0) − (1)

(𝑠 − 3)² = 1 (𝑠 − 3)².

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} = ^𝑘

𝑠²+𝑘², and 𝑛 = 1, Theorem 7.4.1 gives ℒ{𝑡 sin 𝑘𝑡} = − 𝑑

𝑑𝑠ℒ{sin 𝑘𝑡} = − 𝑑

𝑑𝑠( 𝑘

𝑠²+ 𝑘²) = 2𝑘𝑠 (𝑠²+ 𝑘²)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

𝑓 ∗ 𝑔 = ∫ 𝑓(𝜏)𝑔(𝑡 − 𝜏)𝑑𝜏^𝑡

0

… … … (1) = ∫ 𝑔(𝜏)𝑓(𝑡 − 𝜏)𝑑𝜏^𝑡

0

= 𝑔 ∗ 𝑓

and is called the convolution of 𝑓 and 𝑔. The convolution 𝑓 ∗ 𝑔 is a function of 𝑡. For example,

130

𝑒^𝑡 ∗ sin 𝑡 = ∫ 𝑒^{𝑡 𝜏}sin(𝑡 − 𝜏) 𝑑𝜏

0

=1

2(− sin 𝑡 − cos 𝑡 + 𝑒^𝑡).

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:

ℒ{𝑓(𝑡) ∗ 𝑔(𝑡)} = ℒ{𝑓(𝑡)}. ℒ{𝑔(𝑡)} = ℒ {∫ 𝑔(𝜏)𝑓(𝑡 − 𝜏)𝑑𝜏^𝑡

0

} = ℒ {∫ 𝑓(𝜏)𝑔(𝑡 − 𝜏)𝑑𝜏^𝑡

0

} = ℒ{𝑔(𝑡) ∗ 𝑓(𝑡)}

EXAMPLE 3 (Transform of Convolution) Evaluate

ℒ {∫ 𝑒^{𝑡 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:

ℒ {∫ 𝑒^𝜏sin(𝑡 − 𝜏) 𝑑𝜏

𝑡 0

} = ℒ{𝑒^𝑡}. ℒ{sin 𝑡} = 1

𝑠 − 1. 1

𝑠²+ 1= 1

(𝑠 − 1)(𝑠²+ 1).

Remark 2: (Inverse Form of Theorem 7.4.2)

ℒ

⁻¹

{

𝐹

(

𝑠

)

. 𝐺

(

𝑠

)} =

𝑓 ∗ 𝑔. … … … (2)

TRANSFORM OF AN INTEGRAL:

When 𝑔(𝑡) = 1 and ℒ{𝑔(𝑡)} = 𝐺(𝑠) =¹_𝑠, the convolution theorem implies that the Laplace transform of the integral of 𝑓 is

ℒ {∫ 𝑓(𝜏)𝑑𝜏

𝑡 0

} = ℒ{𝑓(𝑡) ∗ 1} = ℒ{𝑓(𝑡)}. ℒ{1} = 𝐹(𝑠).1

𝑠 =𝐹(𝑠)

𝑠 . … … … (3) The inverse form of (3),

∫ 𝑓(𝜏)𝑑𝜏^𝑡

0

=

ℒ

⁻¹{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),

131

𝑓(𝑡) = 𝑔(𝑡) + ∫ 𝑓(𝜏)ℎ(𝑡 − 𝜏)𝑑𝜏^𝑡

0

. … … … (5)

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:

𝑓(𝑡) = 3𝑡²− 𝑒^−𝑡− ∫ 𝑓(𝜏)𝑒^{𝑡 𝑡−𝜏}𝑑𝜏

0

for 𝑓(𝑡).

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)} = 𝐹(𝑠) and ℒ{𝑒^𝑡} = ¹

𝑠−1:

𝐹(𝑠) = 3. 2

𝑠³− 1

𝑠 + 1− 𝐹(𝑠). 1 𝑠 − 1.

After solving the last equation for 𝐹(𝑠) and carrying out the partial fraction decomposition, we find 𝐹(𝑠) = 6

𝑠³− 6 𝑠⁴+1

𝑠 − 2 𝑠 + 1. The inverse transform then gives

𝑓(𝑡) = 3

ℒ

⁻¹{2!

𝑠³}

− ℒ

⁻¹{3!

𝑠⁴}

+ ℒ

⁻¹{1

𝑠}

− 2ℒ

⁻¹{ 1

𝑠 + 1}

= 3𝑡

²

− 𝑡

³

+ 1 − 2𝑒

^−𝑡

.

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)} = 1

1 − 𝑒^−𝑠𝑇∫ 𝑒^−𝑠𝑇𝑓(𝑡)𝑑𝑡

𝑇 0

.

Exercises 7.4: Pages 289-290

(𝟐) Use Theorem 7.4.1 to evaluate the given Laplace transform.

ℒ{𝑡³𝑒^𝑡}

.

Solution:

With 𝑓(𝑡) = 𝑒^𝑡, 𝐹(𝑠) = ℒ{𝑓(𝑡)} = ℒ{𝑒^𝑡} = ¹

𝑠−1, and 𝑛 = 3, Theorem 7.4.1 gives

132

ℒ{𝑡³𝑒^𝑡} = (−1)³ 𝑑³

𝑑𝑠³ℒ{𝑒^𝑡} = − 𝑑³ 𝑑𝑠³( 1

𝑠 − 1) = − (− 6

(𝑠 − 1)⁴) = 6 (𝑠 − 1)⁴.

---

(𝟐𝟐) Use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the integral before transforming.

ℒ

{

𝑒^2𝑡∗ 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

𝑠 − 2. 1

𝑠²+ 1= 1

(𝑠 − 2)(𝑠²+ 1).

---

(𝟒𝟏) Use the Laplace transform to solve the given integral equation or integrodifferential equation.

𝑓

(

𝑡

)

+

∫

𝑓

(

𝜏

)

𝑑𝜏

𝑡 0

= 1

.

Solution:

ℒ{𝑓(𝑡)} + ℒ {∫ 𝑓(𝜏) 𝑑𝜏

𝑡 0

} = ℒ{1}, 𝐹(𝑠) +𝐹(𝑠)

𝑠 =1 𝑠, 𝐹(𝑠) [1 +1

𝑠] =1 𝑠, (𝑠 + 1

𝑠 ) 𝐹(𝑠) = 1 𝑠, 𝐹(𝑠) = 1

𝑠 + 1. The inverse transform then gives

𝑓(𝑡) =

ℒ

⁻¹{ 1

𝑠 + 1}

= 𝑒

^−𝑡

.

133 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 ∗ 𝑒^𝑡cos 𝑡

}.

---

In the following problems, use the Laplace transform to solve the given integral equation or integrodifferential equation.

(

𝟑𝟗

)

𝑓

(

𝑡

)

= 𝑡𝑒^𝑡+

∫

𝜏𝑓

(

𝑡 − 𝜏

)

𝑑𝜏

𝑡 0

.

(

𝟒𝟑

)

𝑓

(

𝑡

)

= 1 + 𝑡 −8

3

∫ (

𝜏 − 𝑡

)

³𝑓

(

𝜏

)

𝑑𝜏

𝑡 0

.