Part 1 - Properties of Laplace Transform

DIFFERENTIAL EQUATIONS 2.1

PROPERTIES OF LAPLACE TRANSFORMATION



Let f be a function defined for t ≥ 0. Then the integral 

e

ℒ{ f(t)} =

 st

f (t ) dt

0

is said to be the Laplace transform of f, provided the integral converges.



Transforms of Some Basic Functions. (a) ℒ{1} = (c) ℒ{eat} =

1 s

(b) ℒ{tn} = n!/sn+1, n = 1, 2, 3, …

1 sa

(e) ℒ{cos kt} =

s s  k2

(g) ℒ{cosh kt} =



2

(d) ℒ{sin kt} =

k s  k2

(f) ℒ{sinh kt} =

k s  k2

2

2

s s  k2 2

Theorem 2.1.1: Linearity Property ℒ is a linear transform. From a linear combination of functions we can write 

 e af (t )  st



 bg (t ) dt  a  e

0

0

 st



f (t )dt  b  e st g (t )dt 0

whenever both integrals converge for s > c. Hence it follows that ℒ af (t )  bg (t )  aℒ  f (t ) + bℒ g (t ) = aF(s) + bG(s)



(1)

First Translation Theorem Theorem 2.1.2: First-Shift Property If ℒ{ f(t)} = F(s) exists for s > c, and a is any real number, then ℒ{ e at f (t ) } = ℒ  f (t ) ssa = F(s  a) for s > c + a.

(2)

Proof: By definition of Laplace transform, 

ℒ{ e f (t ) } =  e e f (t )dt  at

 st at

0



e

( s  a ) t

f (t )dt  F ( s  a) .

0



The symbolism ℒ{ e at f (t ) } = ℒ  f (t ) ssa where s  s  a means that in the Laplace transform F(s) of f(t) we replace the symbol s wherever it appears by s – a.



If s is a real variable, then the graph of F(s – a) is the graph of F(s) shifted on the s-axis by the amount of |a|. If a > 0, the graph of F(s) is shifted a units to the right, whereas if a < 0, the graph is shifted |a| units to the left. See Figure 2.1.1.

Figure 2.1.1: Right shift on s-axis



Example 2.1.1: Evaluate, (a) ℒ{ e5t t 3 }

Solution: (b) ℒ{ e 5t t 3 } = ℒ t 3 

(b) ℒ{ e 2t cos 4t }

(c) ℒ{ e at sin bt }

s s 5

= (3!/s4)|s s5

From ℒ{tn} = n!/sn+1, n = 3.

= 6/(s – 5)4

(c)

ℒ{ e 2t cos 4t } = ℒ cos 4t ss ( 2) =

s s  16 ss 2

=

s2 ( s  2) 2  16

2

From ℒ{cos kt} =

s s  k2 2

(c)

= ℒ sin bt ssa

ℒ{ e at sin bt }

b s  b 2 s s  a b = s  a 2  b 2 =



2

Multiplying a function by tn The Laplace transform of the product of a function f(t) with t can be found by differentiating the Laplace transform of f(t). Assume that F(s) = ℒ{f(t)} exists, then 

d d F s   e st f (t )dt ds ds 0 



 s e 

 st



f (t ) dt

0



   e st t f (t ) dt 0

=  ℒ{tf(t)} = ℒ{ t f(t)}

linearity of ℒ

That is, ℒ{tf(t)} = 

d ℒ{f(t)} ds

Using the last result, we can find the Laplace transform of t2f(t). ℒ{t2f(t)} = ℒ{t  tf(t)} = 

d ℒ{tf(t)} ds

= 

d d [ ℒ{f(t)}] ds ds

=

d2 ℒ{f(t)} ds 2

The preceding two cases suggest the general result for ℒ{tnf(t)}. 

Theorem 2.1.3: Derivatives of the Laplace Transform Property Let F(s) = ℒ{f(t)} exists and n = 1, 2, 3, …, then n dn n d ℒ{f(t)} = (  1 ) F ( s) ds n ds n

ℒ{tnf(t)} = (1) n



(3)

Example 2.1.2: Evaluate (a) ℒ{t sin kt}

(b) ℒ{t2 sin kt}

Solutions: (a) ℒ{t sin kt}

= 

d ℒ{sin kt} ds

= 

d  k  ds  s 2  k 2 

=

(b) ℒ{t2 sin kt}

s

2ks 2

 k2

2

= 

d ℒ{t sin kt} ds

= 

d ds

 2ks   2 2  s  k 2  

2k 3  6ks2

s

2

= (1) 2

 k2



3

d2  k  ds 2  s 2  k 2 

d2 = (1) [ℒ{sin kt}] ds 2 2

k s  k2 2

differentiate with respect to s (k constant)



= ℒ{t  t sin kt}

= 

since ℒ{sin kt} =



Note: To find transforms of functions tneat we can use either Theorem 2.1.2 or Theorem 2.1.3.

For example, evaluate ℒ{ te 3t } ℒ{ te 3t } = ℒ t ss 3 =

Theorem 2.1.2:

ℒ{ te 3t } = 

Theorem 2.1.3:

1 s2

= s s 3

1 s  32

d d  1  ℒ{e3t} =  = ds ds  s  3 

1 s  32

Exercises 2.2

(1) Find F(s) as indicated. (a) ℒ{te10t}

(b) ℒ{t3 e2t}

(c) ℒ{t(et + e2t)2}

(a) ℒ{ et sin 3t }

(f) ℒ 1  et  3e 4t cos 5t

(2) Evaluate the following using Theorem 2.1.3 (Derivatives of the Laplace Transform Property). (a) ℒ{te10t}

(b) ℒ{t cos 2t}

(c) ℒ{t2 sinh t}

(d) ℒ{te2t sin 6t}

Answers 2.2 (1) Find F(s) as indicated. (b) ℒ{te10t} =

1 s  102

(c) ℒ{t(et + e2t)2} =

(b) ℒ{t3 e2t} =

1 2 1   2 2 s  2 s  3 s  42

(f) ℒ 1  et  3e 4t cos 5t =

(d) ℒ{ et sin 3t } =

s s 1 s4  3 2 s  25 s  1  25 s  42  25 2

6 s  24

3 s  12  9

(2) Evaluate the following using Theorem 2.1.3. (b) ℒ{te10t} =

1 s  102

(c) ℒ{t2 sinh t} =

(b) ℒ{t cos 2t} =

6s 2  2

s

2



1

3

s2  4

s

(d) ℒ{te2t sin 6t} =

2

4



2

12s  24

s  2

2

 36



2