the laplace transform of step functions

The Laplace Transform of step functions (Sect. 6.3).

I Overview and notation.

I The definition of a step function.

I Piecewise discontinuous functions.

I The Laplace Transform of discontinuous functions.

I Properties of the Laplace Transform.

Overview and notation.

Overview: The Laplace Transform method can be used to solveconstant coefficients differential equations with discontinuoussource functions.

Notation:If L[f (t)] = F (s), then we denote L−1[F (s)] = f (t).

Remark: One can show that for a particular type of functions f ,that includes all functions we work with in this Section, thenotation above is well-defined.

Example

From the Laplace Transform table we know that L[eat

]=

1

s − a.

Then also holds that L−1[ 1

s − a

]= eat . C

The definition of a step function.

DefinitionA function u is called a step function at t = 0 iff holds

u(t) =

{0 for t < 0,

1 for t > 0.

Example

Graph the step function values u(t) above, and the translationsu(t − c) and u(t + c) with c > 0.

Solution:u(t)

t

1

0

u(t − c)

t

1

0 c

u(t + c)

0

1

t− c

C


Remark: Given any function values f (t) and c > 0, then f (t − c)is a right translation of f and f (t + c) is a left translation of f .

Example

f ( t ) = e

0 t

a t

1

f ( t ) f ( t ) = e

0 t

1

c

a ( t − c )f ( t )

a t

0 t

f ( t )

1

f ( t ) = u ( t ) e f ( t ) = u ( t − c ) e

0 t

1

c

a ( t − c )f ( t )



Example

f ( t ) = e

0 t

a t

1

f ( t )

f ( t ) = e

0 t

1

c

a ( t − c )f ( t )

a t

0 t

f ( t )

1

f ( t ) = u ( t ) e f ( t ) = u ( t − c ) e

0 t

1

c

a ( t − c )f ( t )



Example

f ( t ) = e

0 t

a t

1

f ( t ) f ( t ) = e

0 t

1

c

a ( t − c )f ( t )

a t

0 t

f ( t )

1

f ( t ) = u ( t ) e f ( t ) = u ( t − c ) e

0 t

1

c

a ( t − c )f ( t )



Example

f ( t ) = e

0 t

a t

1

f ( t ) f ( t ) = e

0 t

1

c

a ( t − c )f ( t )

a t

0 t

f ( t )

1

f ( t ) = u ( t ) e

f ( t ) = u ( t − c ) e

0 t

1

c

a ( t − c )f ( t )



Example

f ( t ) = e

0 t

a t

1

f ( t ) f ( t ) = e

0 t

1

c

a ( t − c )f ( t )

a t

0 t

f ( t )

1

f ( t ) = u ( t ) e f ( t ) = u ( t − c ) e

0 t

1

c

a ( t − c )f ( t )

Piecewise discontinuous functions.

Example

Graph of the function b(t) = u(t − a)− u(t − b), with 0 < a < b.

Solution: The bump function b can be graphed as follows:

u(t −a)

0 a b

1

t

u(t −b)

0 a b

1

t

t0 a b

b(t)

1

C

Piecewise discontinuous functions.

Example

Graph of the function f (t) = eat[u(t − 1)− u(t − 2)

].

Solution:

a t

t1 2

1

a t

f ( t ) = e [ u ( t −1 ) − u ( t −2 ) ]

[ u ( t −1 ) − u ( t −2 ) ]

e

y

Notation: The function values u(t − c) are denoted in thetextbook as uc(t).

The Laplace Transform of discontinuous functions.

TheoremGiven any real number c > 0, the following equation holds,

L[u(t − c)] =e−cs

s, s > 0.

Proof:

L[u(t − c)] =

∫ ∞

0

e−stu(t − c) dt =

∫ ∞

ce−st dt,

L[u(t − c)] = limN→∞

−1

s

(e−Ns − e−cs

)=

e−cs

s, s > 0.

We conclude that L[u(t − c)] =e−cs

s.




s, s > 0.

Proof:

L[u(t − c)] =

∫ ∞

0

e−stu(t − c) dt

=

∫ ∞

ce−st dt,


−1

s

(e−Ns − e−cs

)=

e−cs

s, s > 0.


s.




s, s > 0.

Proof:

L[u(t − c)] =

∫ ∞

0


∫ ∞

ce−st dt,


−1

s

(e−Ns − e−cs

)=

e−cs

s, s > 0.


s.




s, s > 0.

Proof:

L[u(t − c)] =

∫ ∞

0


∫ ∞

ce−st dt,


−1

s

(e−Ns − e−cs

)

=e−cs

s, s > 0.


s.




s, s > 0.

Proof:

L[u(t − c)] =

∫ ∞

0


∫ ∞

ce−st dt,


−1

s

(e−Ns − e−cs

)=

e−cs

s, s > 0.


s.


Example

Compute L[3u(t − 2)].

Solution: L[3u(t − 2)] = 3L[u(t − 2)] = 3e−2s

s.

We conclude: L[3u(t − 2)] =3e−2s

s. C

Example

Compute L−1[e−3s

s

].

Solution: L−1[e−3s

s

]= u(t − 3).

We conclude: L−1[e−3s

s

]= u(t − 3). C


Example


Solution: L[3u(t − 2)] = 3L[u(t − 2)]

= 3e−2s

s.


s. C

Example


s

].


s

]= u(t − 3).


s

]= u(t − 3). C


Example



s.


s. C

Example


s

].


s

]= u(t − 3).


s

]= u(t − 3). C


Example



s.


s. C

Example


s

].


s

]

= u(t − 3).


s

]= u(t − 3). C


Example



s.


s. C

Example


s

].


s

]= u(t − 3).


s

]= u(t − 3). C

Properties of the Laplace Transform.

Theorem (Translations)

If F (s) = L[f (t)] exists for s > a > 0 and c > 0, then holds

L[u(t − c)f (t − c)] = e−cs F (s), s > a.

Furthermore,

L[ect f (t)] = F (s − c), s > a + c .

Remark:

I L[translation (uf )

]= (exp)

(L[f ]

).

I L[(exp) (f )

]= translation

(L[f ]

).

Equivalent notation:

I L[u(t − c)f (t − c)] = e−cs L[f (t)],

I L[ect f (t)] = L[f ](s − c).


Example

Compute L[u(t − 2) sin(a(t − 2))

].

Solution: L[sin(at)] =a

s2 + a2, L[u(t − c)f (t − c)] = e−cs L[f (t)].

L[u(t − 2) sin(a(t − 2))

]= e−2s L[sin(at)] = e−2s a

s2 + a2.

We conclude: L[u(t − 2) sin(a(t − 2))

]= e−2s a

s2 + a2. C

Example

Compute L[e3t sin(at)

].

Solution: Recall: L[ect f (t)] = L[f ](s − c).

We conclude: L[e3t sin(at)

]=

a

(s − 3)2 + a2, with s > 3. C


Example


].


s2 + a2,

L[u(t − c)f (t − c)] = e−cs L[f (t)].

L[u(t − 2) sin(a(t − 2))


s2 + a2.


]= e−2s a

s2 + a2. C

Example


].



]=

a

(s − 3)2 + a2, with s > 3. C


Example


].



L[u(t − 2) sin(a(t − 2))


s2 + a2.


]= e−2s a

s2 + a2. C

Example


].



]=

a

(s − 3)2 + a2, with s > 3. C


Example


].



L[u(t − 2) sin(a(t − 2))

]= e−2s L[sin(at)]

= e−2s a

s2 + a2.


]= e−2s a

s2 + a2. C

Example


].



]=

a

(s − 3)2 + a2, with s > 3. C


Example


].



L[u(t − 2) sin(a(t − 2))


s2 + a2.


]= e−2s a

s2 + a2. C

Example


].



]=

a

(s − 3)2 + a2, with s > 3. C


Example

Find the Laplace transform of f (t) =

{0, t < 1,

(t2 − 2t + 2), t > 1.

Solution: Using step function notation,

f (t) = u(t − 1)(t2 − 2t + 2).

Completing the square we obtain,

t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2 = (t − 1)2 + 1.

This is a parabola t2 translated to theright by 1 and up by one. This is adiscontinuous function.

1

10

f ( t )

t


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


f (t) = u(t − 1)(t2 − 2t + 2).


t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2

= (t − 1)2 + 1.


1

10

f ( t )

t


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


f (t) = u(t − 1)(t2 − 2t + 2).


t2 − 2t + 2 = (t2 − 2t + 1)− 1 + 2 = (t − 1)2 + 1.


1

10

f ( t )

t


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.

Solution: Recall: f (t) = u(t − 1)[(t − 1)2 + 1

].

This is equivalent to

f (t) = u(t − 1) (t − 1)2 + u(t − 1).

Since L[t2] = 2/s3, and L[u(t − c)g(t − c)] = e−cs L[g(t)], then

L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)] = e−s 2

s3+ e−s 1

s.

We conclude: L[f (t)] =e−s

s3

(2 + s2

). C


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


].


f (t) = u(t − 1) (t − 1)2 + u(t − 1).

Since L[t2] = 2/s3,

and L[u(t − c)g(t − c)] = e−cs L[g(t)], then

L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)] = e−s 2

s3+ e−s 1

s.


s3

(2 + s2

). C


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


].


f (t) = u(t − 1) (t − 1)2 + u(t − 1).

Since L[t2] = 2/s3, and L[u(t − c)g(t − c)] = e−cs L[g(t)],

then

L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)] = e−s 2

s3+ e−s 1

s.


s3

(2 + s2

). C


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


].


f (t) = u(t − 1) (t − 1)2 + u(t − 1).


L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)]

= e−s 2

s3+ e−s 1

s.


s3

(2 + s2

). C


Example


{0, t < 1,

(t2 − 2t + 2), t > 1.


].


f (t) = u(t − 1) (t − 1)2 + u(t − 1).


L[f (t)] = L[u(t − 1) (t − 1)2] + L[u(t − 1)] = e−s 2

s3+ e−s 1

s.


s3

(2 + s2

). C


Remark: The inverse of the formulas in the Theorem above are:

L−1[e−cs F (s)

]= u(t − c) f (t − c),

L−1[F (s − c)

]= ect f (t).

Example

Find L−1[ e−4s

s2 + 9

].

Solution: L−1[ e−4s

s2 + 9

]=

1

3L−1

[e−4s 3

s2 + 9

].

Recall: L−1[ a

s2 + a2

]= sin(at). Then, we conclude that

L−1[ e−4s

s2 + 9

]=

1

3u(t − 4) sin

(3(t − 4)

). C



L−1[e−cs F (s)

]= u(t − c) f (t − c),

L−1[F (s − c)

]= ect f (t).

Example

Find L−1[ e−4s

s2 + 9

].


s2 + 9

]=

1

3L−1

[e−4s 3

s2 + 9

].

Recall: L−1[ a

s2 + a2

]= sin(at).

Then, we conclude that

L−1[ e−4s

s2 + 9

]=

1

3u(t − 4) sin

(3(t − 4)

). C



L−1[e−cs F (s)

]= u(t − c) f (t − c),

L−1[F (s − c)

]= ect f (t).

Example

Find L−1[ e−4s

s2 + 9

].


s2 + 9

]=

1

3L−1

[e−4s 3

s2 + 9

].

Recall: L−1[ a

s2 + a2

]= sin(at). Then, we conclude that

L−1[ e−4s

s2 + 9

]=

1

3u(t − 4) sin

(3(t − 4)

). C


Example

Find L−1[ (s − 2)

(s − 2)2 + 9

].

Solution: L−1[ s

s2 + a2

]= cos(at), L−1

[F (s − c)

]= ect f (t).

We conclude: L−1[ (s − 2)

(s − 2)2 + 9

]= e2t cos(3t). C

Example

Find L−1[ 2e−3s

s2 − 4

].

Solution: Recall: L−1[ a

s2 − a2

]= sinh(at)

and L−1[e−cs F (s)

]= u(t − c) f (t − c).


Example

Find L−1[ (s − 2)

(s − 2)2 + 9

].

Solution: L−1[ s

s2 + a2

]= cos(at),

L−1[F (s − c)

]= ect f (t).


(s − 2)2 + 9

]= e2t cos(3t). C

Example

Find L−1[ 2e−3s

s2 − 4

].


s2 − a2

]= sinh(at)


]= u(t − c) f (t − c).


Example

Find L−1[ (s − 2)

(s − 2)2 + 9

].

Solution: L−1[ s

s2 + a2

]= cos(at), L−1

[F (s − c)

]= ect f (t).


(s − 2)2 + 9

]= e2t cos(3t). C

Example

Find L−1[ 2e−3s

s2 − 4

].


s2 − a2

]= sinh(at)


]= u(t − c) f (t − c).


Example

Find L−1[ 2e−3s

s2 − 4

].

Solution: Recall:

L−1[ a

s2 − a2

]= sinh(at), L−1

[e−cs F (s)

]= u(t − c) f (t − c).

L−1[ 2e−3s

s2 − 4

]= L−1

[e−3s 2

s2 − 4

].

We conclude: L−1[ 2e−3s

s2 − 4

]= u(t − 3) sinh

(2(t − 3)

). C


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Find the roots of the denominator:

s± =1

2

[−1±

√1 + 8

]⇒

{s+ = 1,

s− = −2.

Therefore, s2 + s − 2 = (s − 1) (s + 2).

Use partial fractions to simplify the rational function:

1

s2 + s − 2=

1

(s − 1) (s + 2)=

a

(s − 1)+

b

(s + 2),

1

s2 + s − 2= a(s + 2) + b(s − 1) =

(a + b) s + (2a− b)

(s − 1) (s + 2).


Example

Find L−1[ e−2s

s2 + s − 2

].


s± =1

2

[−1±

√1 + 8

]

⇒

{s+ = 1,

s− = −2.

Therefore, s2 + s − 2 = (s − 1) (s + 2).


1

s2 + s − 2=

1

(s − 1) (s + 2)=

a

(s − 1)+

b

(s + 2),

1

s2 + s − 2= a(s + 2) + b(s − 1) =

(a + b) s + (2a− b)

(s − 1) (s + 2).


Example

Find L−1[ e−2s

s2 + s − 2

].


s± =1

2

[−1±

√1 + 8

]⇒

{s+ = 1,

s− = −2.

Therefore, s2 + s − 2 = (s − 1) (s + 2).


1

s2 + s − 2=

1

(s − 1) (s + 2)=

a

(s − 1)+

b

(s + 2),

1

s2 + s − 2= a(s + 2) + b(s − 1) =

(a + b) s + (2a− b)

(s − 1) (s + 2).


Example

Find L−1[ e−2s

s2 + s − 2

].


s± =1

2

[−1±

√1 + 8

]⇒

{s+ = 1,

s− = −2.

Therefore, s2 + s − 2 = (s − 1) (s + 2).


1

s2 + s − 2=

1

(s − 1) (s + 2)

=a

(s − 1)+

b

(s + 2),

1

s2 + s − 2= a(s + 2) + b(s − 1) =

(a + b) s + (2a− b)

(s − 1) (s + 2).


Example

Find L−1[ e−2s

s2 + s − 2

].


s± =1

2

[−1±

√1 + 8

]⇒

{s+ = 1,

s− = −2.

Therefore, s2 + s − 2 = (s − 1) (s + 2).


1

s2 + s − 2=

1

(s − 1) (s + 2)=

a

(s − 1)+

b

(s + 2),

1

s2 + s − 2= a(s + 2) + b(s − 1) =

(a + b) s + (2a− b)

(s − 1) (s + 2).


Example

Find L−1[ e−2s

s2 + s − 2

].


s± =1

2

[−1±

√1 + 8

]⇒

{s+ = 1,

s− = −2.

Therefore, s2 + s − 2 = (s − 1) (s + 2).


1

s2 + s − 2=

1

(s − 1) (s + 2)=

a

(s − 1)+

b

(s + 2),

1

s2 + s − 2= a(s + 2) + b(s − 1)

=(a + b) s + (2a− b)

(s − 1) (s + 2).


Example

Find L−1[ e−2s

s2 + s − 2

].


s± =1

2

[−1±

√1 + 8

]⇒

{s+ = 1,

s− = −2.

Therefore, s2 + s − 2 = (s − 1) (s + 2).


1

s2 + s − 2=

1

(s − 1) (s + 2)=

a

(s − 1)+

b

(s + 2),

1

s2 + s − 2= a(s + 2) + b(s − 1) =

(a + b) s + (2a− b)

(s − 1) (s + 2).


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Recall:1

s2 + s − 2=

(a + b) s + (2a− b)

(s − 1) (s + 2)

a + b = 0, 2a− b = 1, ⇒ a =1

3, b = −1

3.

L−1[ e−2s

s2 + s − 2

]=

1

3L−1

[e−2s 1

s − 1

]− 1

3L−1

[e−2s 1

s + 2

].

Recall: L−1[ 1

s − a

]= eat , L−1

[e−cs F (s)

]= u(t − c) f (t − c),

L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2) e(t−2) − 1

3u(t − 2) e−2(t−2).

Hence: L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2)

[e(t−2) − e−2(t−2)

]. C


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Recall:1

s2 + s − 2=

(a + b) s + (2a− b)

(s − 1) (s + 2)

a + b = 0,

2a− b = 1, ⇒ a =1

3, b = −1

3.

L−1[ e−2s

s2 + s − 2

]=

1

3L−1

[e−2s 1

s − 1

]− 1

3L−1

[e−2s 1

s + 2

].

Recall: L−1[ 1

s − a

]= eat , L−1

[e−cs F (s)

]= u(t − c) f (t − c),

L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2) e(t−2) − 1

3u(t − 2) e−2(t−2).


s2 + s − 2

]=

1

3u(t − 2)

[e(t−2) − e−2(t−2)

]. C


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Recall:1

s2 + s − 2=

(a + b) s + (2a− b)

(s − 1) (s + 2)

a + b = 0, 2a− b = 1,

⇒ a =1

3, b = −1

3.

L−1[ e−2s

s2 + s − 2

]=

1

3L−1

[e−2s 1

s − 1

]− 1

3L−1

[e−2s 1

s + 2

].

Recall: L−1[ 1

s − a

]= eat , L−1

[e−cs F (s)

]= u(t − c) f (t − c),

L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2) e(t−2) − 1

3u(t − 2) e−2(t−2).


s2 + s − 2

]=

1

3u(t − 2)

[e(t−2) − e−2(t−2)

]. C


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Recall:1

s2 + s − 2=

(a + b) s + (2a− b)

(s − 1) (s + 2)

a + b = 0, 2a− b = 1, ⇒ a =1

3, b = −1

3.

L−1[ e−2s

s2 + s − 2

]=

1

3L−1

[e−2s 1

s − 1

]− 1

3L−1

[e−2s 1

s + 2

].

Recall: L−1[ 1

s − a

]= eat , L−1

[e−cs F (s)

]= u(t − c) f (t − c),

L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2) e(t−2) − 1

3u(t − 2) e−2(t−2).


s2 + s − 2

]=

1

3u(t − 2)

[e(t−2) − e−2(t−2)

]. C


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Recall:1

s2 + s − 2=

(a + b) s + (2a− b)

(s − 1) (s + 2)

a + b = 0, 2a− b = 1, ⇒ a =1

3, b = −1

3.

L−1[ e−2s

s2 + s − 2

]=

1

3L−1

[e−2s 1

s − 1

]− 1

3L−1

[e−2s 1

s + 2

].

Recall: L−1[ 1

s − a

]= eat ,

L−1[e−cs F (s)

]= u(t − c) f (t − c),

L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2) e(t−2) − 1

3u(t − 2) e−2(t−2).


s2 + s − 2

]=

1

3u(t − 2)

[e(t−2) − e−2(t−2)

]. C


Example

Find L−1[ e−2s

s2 + s − 2

].

Solution: Recall:1

s2 + s − 2=

(a + b) s + (2a− b)

(s − 1) (s + 2)

a + b = 0, 2a− b = 1, ⇒ a =1

3, b = −1

3.

L−1[ e−2s

s2 + s − 2

]=

1

3L−1

[e−2s 1

s − 1

]− 1

3L−1

[e−2s 1

s + 2

].

Recall: L−1[ 1

s − a

]= eat , L−1

[e−cs F (s)

]= u(t − c) f (t − c),

L−1[ e−2s

s2 + s − 2

]=

1

3u(t − 2) e(t−2) − 1

3u(t − 2) e−2(t−2).


s2 + s − 2

]=

1

3u(t − 2)

[e(t−2) − e−2(t−2)

]. C

Equations with discontinuous sources (Sect. 6.4).

I Differential equations with discontinuous sources.I We solve the IVPs:

(a) Example 1:

y ′ + 2y = u(t − 4), y(0) = 3.

(b) Example 2:

y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).

(c) Example 3:

y ′′+y ′+5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t), t ∈ [0, π)

0, t ∈ [π,∞).

Differential equations with discontinuous sources.

Example

Use the Laplace transform to find the solution of the IVP

y ′ + 2y = u(t − 4), y(0) = 3.

Solution: Compute the Laplace transform of the whole equation,

L[y ′] + 2L[y ] = L[u(t − 4)] =e−4s

s.

From the previous Section we know that[s L[y ]−y(0)

]+2L[y ] =

e−4s

s⇒ (s +2)L[y ] = y(0)+

e−4s

s.

Introduce the initial condition, L[y ] =3

(s + 2)+ e−4s 1

s(s + 2),

Use the table: L[y ] = 3L[e−2t

]+ e−4s 1

s(s + 2).


Example


y ′ + 2y = u(t − 4), y(0) = 3.


L[y ′] + 2L[y ] = L[u(t − 4)]

=e−4s

s.


]+2L[y ] =

e−4s

s⇒ (s +2)L[y ] = y(0)+

e−4s

s.


(s + 2)+ e−4s 1

s(s + 2),


]+ e−4s 1

s(s + 2).


Example


y ′ + 2y = u(t − 4), y(0) = 3.


L[y ′] + 2L[y ] = L[u(t − 4)] =e−4s

s.


]+2L[y ] =

e−4s

s⇒ (s +2)L[y ] = y(0)+

e−4s

s.


(s + 2)+ e−4s 1

s(s + 2),


]+ e−4s 1

s(s + 2).


Example


y ′ + 2y = u(t − 4), y(0) = 3.


L[y ′] + 2L[y ] = L[u(t − 4)] =e−4s

s.


]+2L[y ] =

e−4s

s

⇒ (s +2)L[y ] = y(0)+e−4s

s.


(s + 2)+ e−4s 1

s(s + 2),


]+ e−4s 1

s(s + 2).


Example


y ′ + 2y = u(t − 4), y(0) = 3.


L[y ′] + 2L[y ] = L[u(t − 4)] =e−4s

s.


]+2L[y ] =

e−4s

s⇒ (s +2)L[y ] = y(0)+

e−4s

s.


(s + 2)+ e−4s 1

s(s + 2),


]+ e−4s 1

s(s + 2).


Example


y ′ + 2y = u(t − 4), y(0) = 3.


L[y ′] + 2L[y ] = L[u(t − 4)] =e−4s

s.


]+2L[y ] =

e−4s

s⇒ (s +2)L[y ] = y(0)+

e−4s

s.

Introduce the initial condition,

L[y ] =3

(s + 2)+ e−4s 1

s(s + 2),


]+ e−4s 1

s(s + 2).


Example


y ′ + 2y = u(t − 4), y(0) = 3.


L[y ′] + 2L[y ] = L[u(t − 4)] =e−4s

s.


]+2L[y ] =

e−4s

s⇒ (s +2)L[y ] = y(0)+

e−4s

s.


(s + 2)+ e−4s 1

s(s + 2),


]+ e−4s 1

s(s + 2).


Example


y ′ + 2y = u(t − 4), y(0) = 3.

Solution: Recall: L[y ] = 3L[e−2t

]+ e−4s 1

s(s + 2).

We need to invert the Laplace transform on the last term.Partial fractions:

1

s(s + 2)=

a

s+

b

(s + 2)=

a(s + 2) + bs

s(s + 2)=

(a + b) s + (2a)

s(s + 2)

We get, a + b = 0, 2a = 1. We obtain: a =1

2, b = −1

2. Hence,

1

s(s + 2)=

1

2

[1

s− 1

(s + 2)

].


Example


y ′ + 2y = u(t − 4), y(0) = 3.


]+ e−4s 1

s(s + 2).

We need to invert the Laplace transform on the last term.

Partial fractions:

1

s(s + 2)=

a

s+

b

(s + 2)=

a(s + 2) + bs

s(s + 2)=

(a + b) s + (2a)

s(s + 2)


2, b = −1

2. Hence,

1

s(s + 2)=

1

2

[1

s− 1

(s + 2)

].


Example


y ′ + 2y = u(t − 4), y(0) = 3.


]+ e−4s 1

s(s + 2).


1

s(s + 2)=

a

s+

b

(s + 2)=

a(s + 2) + bs

s(s + 2)=

(a + b) s + (2a)

s(s + 2)


2, b = −1

2. Hence,

1

s(s + 2)=

1

2

[1

s− 1

(s + 2)

].


Example


y ′ + 2y = u(t − 4), y(0) = 3.


]+ e−4s 1

s(s + 2).


1

s(s + 2)=

a

s+

b

(s + 2)

=a(s + 2) + bs

s(s + 2)=

(a + b) s + (2a)

s(s + 2)


2, b = −1

2. Hence,

1

s(s + 2)=

1

2

[1

s− 1

(s + 2)

].


Example


y ′ + 2y = u(t − 4), y(0) = 3.


]+ e−4s 1

s(s + 2).


1

s(s + 2)=

a

s+

b

(s + 2)=

a(s + 2) + bs

s(s + 2)

=(a + b) s + (2a)

s(s + 2)


2, b = −1

2. Hence,

1

s(s + 2)=

1

2

[1

s− 1

(s + 2)

].


Example


y ′ + 2y = u(t − 4), y(0) = 3.


]+ e−4s 1

s(s + 2).


1

s(s + 2)=

a

s+

b

(s + 2)=

a(s + 2) + bs

s(s + 2)=

(a + b) s + (2a)

s(s + 2)


2, b = −1

2. Hence,

1

s(s + 2)=

1

2

[1

s− 1

(s + 2)

].


Example


y ′ + 2y = u(t − 4), y(0) = 3.


]+ e−4s 1

s(s + 2).


1

s(s + 2)=

a

s+

b

(s + 2)=

a(s + 2) + bs

s(s + 2)=

(a + b) s + (2a)

s(s + 2)

We get, a + b = 0, 2a = 1.

We obtain: a =1

2, b = −1

2. Hence,

1

s(s + 2)=

1

2

[1

s− 1

(s + 2)

].


Example


y ′ + 2y = u(t − 4), y(0) = 3.


]+ e−4s 1

s(s + 2).


1

s(s + 2)=

a

s+

b

(s + 2)=

a(s + 2) + bs

s(s + 2)=

(a + b) s + (2a)

s(s + 2)


2, b = −1

2.

Hence,

1

s(s + 2)=

1

2

[1

s− 1

(s + 2)

].


Example


y ′ + 2y = u(t − 4), y(0) = 3.


]+ e−4s 1

s(s + 2).


1

s(s + 2)=

a

s+

b

(s + 2)=

a(s + 2) + bs

s(s + 2)=

(a + b) s + (2a)

s(s + 2)


2, b = −1

2. Hence,

1

s(s + 2)=

1

2

[1

s− 1

(s + 2)

].


Example


y ′ + 2y = u(t − 4), y(0) = 3.

Solution: Recall:1

s(s + 2)=

1

2

[1

s− 1

(s + 2)

].

The algebraic equation for L[y ] has the form,

L[y ] = 3L[e−2t

]+

1

2

[e−4s 1

s− e−4s 1

(s + 2)

].

L[y ] = 3L[e−2t

]+

1

2

(L[u(t − 4)]− L

[u(t − 4) e−2(t−4)

]).

We conclude that

y(t) = 3e−2t +1

2u(t − 4)

[1− e−2(t−4)

]. C


Example


y ′ + 2y = u(t − 4), y(0) = 3.

Solution: Recall:1

s(s + 2)=

1

2

[1

s− 1

(s + 2)

].


L[y ] = 3L[e−2t

]+

1

2

[e−4s 1

s− e−4s 1

(s + 2)

].

L[y ] = 3L[e−2t

]

+1

2

(L[u(t − 4)]− L

[u(t − 4) e−2(t−4)

]).

We conclude that

y(t) = 3e−2t +1

2u(t − 4)

[1− e−2(t−4)

]. C


Example


y ′ + 2y = u(t − 4), y(0) = 3.

Solution: Recall:1

s(s + 2)=

1

2

[1

s− 1

(s + 2)

].


L[y ] = 3L[e−2t

]+

1

2

[e−4s 1

s− e−4s 1

(s + 2)

].

L[y ] = 3L[e−2t

]+

1

2

(L[u(t − 4)]

− L[u(t − 4) e−2(t−4)

]).

We conclude that

y(t) = 3e−2t +1

2u(t − 4)

[1− e−2(t−4)

]. C


Example


y ′ + 2y = u(t − 4), y(0) = 3.

Solution: Recall:1

s(s + 2)=

1

2

[1

s− 1

(s + 2)

].


L[y ] = 3L[e−2t

]+

1

2

[e−4s 1

s− e−4s 1

(s + 2)

].

L[y ] = 3L[e−2t

]+

1

2

(L[u(t − 4)]− L

[u(t − 4) e−2(t−4)

]).

We conclude that

y(t) = 3e−2t +1

2u(t − 4)

[1− e−2(t−4)

]. C



(a) Example 1:

y ′ + 2y = u(t − 4), y(0) = 3.

(b) Example 2:

y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).

(c) Example 3:

y ′′+y ′+5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t), t ∈ [0, π)

0, t ∈ [π,∞).


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).

Solution:

Rewrite the source function usingstep functions.

b ( t )

1

0 pi t

t

u ( t )

1

0 pi

u ( t − pi )

1

0 pi t


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).

Solution: The graphs imply: b(t) = u(t)− u(t − π)

Now is simple to find L[b], since

L[b(t)] = L[u(t)]− L[u(t − π)] =1

s− e−πs

s.

So, the source is L[b(t)] =(1− e−πs

) 1

s, and the equation is

L[y ′′] + L[y ′] +5

4L[y ] =

(1− e−πs

) 1

s.


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


Now is simple to find L[b],

since

L[b(t)] = L[u(t)]− L[u(t − π)] =1

s− e−πs

s.


) 1


L[y ′′] + L[y ′] +5

4L[y ] =

(1− e−πs

) 1

s.


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).



L[b(t)] = L[u(t)]− L[u(t − π)]

=1

s− e−πs

s.


) 1


L[y ′′] + L[y ′] +5

4L[y ] =

(1− e−πs

) 1

s.


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).



L[b(t)] = L[u(t)]− L[u(t − π)] =1

s− e−πs

s.


) 1


L[y ′′] + L[y ′] +5

4L[y ] =

(1− e−πs

) 1

s.


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).



L[b(t)] = L[u(t)]− L[u(t − π)] =1

s− e−πs

s.


) 1

s,

and the equation is

L[y ′′] + L[y ′] +5

4L[y ] =

(1− e−πs

) 1

s.


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).



L[b(t)] = L[u(t)]− L[u(t − π)] =1

s− e−πs

s.


) 1


L[y ′′] + L[y ′] +5

4L[y ] =

(1− e−πs

) 1

s.


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).

Solution: So: L[y ′′] + L[y ′] +5

4L[y ] =

(1− e−πs

) 1

s.

The initial conditions imply: L[y ′′] = s2 L[y ] and L[y ′] = s L[y ].

Therefore,(s2 + s +

5

4

)L[y ] =

(1− e−πs

) 1

s.

We arrive at the expression: L[y ] =(1− e−πs

) 1

s(s2 + s + 5

4

) .


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


4L[y ] =

(1− e−πs

) 1

s.

The initial conditions imply:

L[y ′′] = s2 L[y ] and L[y ′] = s L[y ].

Therefore,(s2 + s +

5

4

)L[y ] =

(1− e−πs

) 1

s.


) 1

s(s2 + s + 5

4

) .


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


4L[y ] =

(1− e−πs

) 1

s.

The initial conditions imply: L[y ′′] = s2 L[y ]

and L[y ′] = s L[y ].

Therefore,(s2 + s +

5

4

)L[y ] =

(1− e−πs

) 1

s.


) 1

s(s2 + s + 5

4

) .


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


4L[y ] =

(1− e−πs

) 1

s.


Therefore,(s2 + s +

5

4

)L[y ] =

(1− e−πs

) 1

s.


) 1

s(s2 + s + 5

4

) .


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).

Solution: Recall: L[y ] =(1− e−πs

) 1

s(s2 + s + 5

4

) .

Denoting: H(s) =1

s(s2 + s + 5

4

) ,

we obtain, L[y ] =(1− e−πs

)H(s).

In other words: y(t) = L−1[H(s)

]− L−1

[e−πs H(s)

].


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).

Solution: Recall: y(t) = L−1[H(s)

]− L−1

[e−πs H(s)

].

Denoting: h(t) = L−1[H(s)

], the L[ ] properties imply

L−1[e−πsH(s)

]= u(t − π) h(t − π).

Therefore, the solution has the form

y(t) = h(t)− u(t − π) h(t − π).

We only need to find h(t) = L−1[ 1

s(s2 + s + 5

4

)].


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


]− L−1

[e−πs H(s)

].


],

the L[ ] properties imply

L−1[e−πsH(s)

]= u(t − π) h(t − π).


y(t) = h(t)− u(t − π) h(t − π).


s(s2 + s + 5

4

)].


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


]− L−1

[e−πs H(s)

].


], the L[ ] properties imply

L−1[e−πsH(s)

]= u(t − π) h(t − π).


y(t) = h(t)− u(t − π) h(t − π).


s(s2 + s + 5

4

)].


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).

Solution: Recall: h(t) = L−1[ 1

s(s2 + s + 5

4

)].

Partial fractions: Find the zeros of the denominator,

s± =1

2

[−1±

√1− 5

]⇒ Complex roots.

The partial fraction decomposition is:

H(s) =1(

s2 + s + 54

)s

=a

s+

(bs + c)(s2 + s + 5

4

)


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


s(s2 + s + 5

4

)].

Partial fractions:

Find the zeros of the denominator,

s± =1

2

[−1±

√1− 5

]⇒ Complex roots.


H(s) =1(

s2 + s + 54

)s

=a

s+

(bs + c)(s2 + s + 5

4

)


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


s(s2 + s + 5

4

)].


s± =1

2

[−1±

√1− 5

]⇒ Complex roots.


H(s) =1(

s2 + s + 54

)s

=a

s+

(bs + c)(s2 + s + 5

4

)


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


s(s2 + s + 5

4

)].


s± =1

2

[−1±

√1− 5

]

⇒ Complex roots.


H(s) =1(

s2 + s + 54

)s

=a

s+

(bs + c)(s2 + s + 5

4

)


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


s(s2 + s + 5

4

)].


s± =1

2

[−1±

√1− 5

]⇒ Complex roots.


H(s) =1(

s2 + s + 54

)s

=a

s+

(bs + c)(s2 + s + 5

4

)


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).

Solution: Recall: H(s) =1(

s2 + s + 54

)s

=a

s+

(bs + c)(s2 + s + 5

4

) .


1 = a(s2 + s +

5

4

)+ s (bs + c) = (a + b) s2 + (a + c) s +

5

4a.

This equation implies that a, b, and c , are solutions of

a + b = 0, a + c = 0,5

4a = 1.


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


s2 + s + 54

)s

=a

s+

(bs + c)(s2 + s + 5

4

) .


1 = a(s2 + s +

5

4

)+ s (bs + c)

= (a + b) s2 + (a + c) s +5

4a.


a + b = 0, a + c = 0,5

4a = 1.


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


s2 + s + 54

)s

=a

s+

(bs + c)(s2 + s + 5

4

) .


1 = a(s2 + s +

5

4

)+ s (bs + c) = (a + b) s2 + (a + c) s +

5

4a.


a + b = 0, a + c = 0,5

4a = 1.


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).

Solution: So: a =4

5, b = −4

5, c = −4

5.

Hence, we have found that,

H(s) =1(

s2 + s + 54

)s

=4

5

[1

s− (s + 1)(

s2 + s + 54

)]We have to compute the inverse Laplace Transform

h(t) =4

5L−1

[1

s− (s + 1)(

s2 + s + 54

)]


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).

Solution: So: a =4

5, b = −4

5, c = −4

5.


H(s) =1(

s2 + s + 54

)s

=4

5

[1

s− (s + 1)(

s2 + s + 54

)]

We have to compute the inverse Laplace Transform

h(t) =4

5L−1

[1

s− (s + 1)(

s2 + s + 54

)]


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).

Solution: So: a =4

5, b = −4

5, c = −4

5.


H(s) =1(

s2 + s + 54

)s

=4

5

[1

s− (s + 1)(

s2 + s + 54

)]We have to compute the inverse Laplace Transform

h(t) =4

5L−1

[1

s− (s + 1)(

s2 + s + 54

)]


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).

Solution: Recall: h(t) =4

5L−1

[1

s− (s + 1)(

s2 + s + 54

)].

In this case we complete the square in the denominator,

s2 + s +5

4=

[s2 + 2

(1

2

)s +

1

4

]− 1

4+

5

4=

(s +

1

2

)2+ 1.

So: h(t) =4

5L−1

[1

s− (s + 1)[(

s + 12

)2+ 1

]].

That is, h(t) =4

5L−1

[1

s

]− 4

5L−1

[ (s + 1

2

)+ 1

2[(s + 1

2

)2+ 1

]].


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


5L−1

[1

s− (s + 1)(

s2 + s + 54

)].


s2 + s +5

4=

[s2 + 2

(1

2

)s +

1

4

]− 1

4+

5

4

=(s +

1

2

)2+ 1.

So: h(t) =4

5L−1

[1

s− (s + 1)[(

s + 12

)2+ 1

]].

That is, h(t) =4

5L−1

[1

s

]− 4

5L−1

[ (s + 1

2

)+ 1

2[(s + 1

2

)2+ 1

]].


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


5L−1

[1

s− (s + 1)(

s2 + s + 54

)].


s2 + s +5

4=

[s2 + 2

(1

2

)s +

1

4

]− 1

4+

5

4=

(s +

1

2

)2+ 1.

So: h(t) =4

5L−1

[1

s− (s + 1)[(

s + 12

)2+ 1

]].

That is, h(t) =4

5L−1

[1

s

]− 4

5L−1

[ (s + 1

2

)+ 1

2[(s + 1

2

)2+ 1

]].


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


5L−1

[1

s

]− 4

5L−1

[ (s + 1

2

)+ 1

2[(s + 1

2

)2+ 1

]].

h(t) =4

5L−1

[1

s

]− 4

5L−1

[ (s + 1

2

)[(

s + 12

)2+ 1

]]− 2

5L−1

[ 1[(s + 1

2

)2+ 1

]].

Recall: L−1[F (s − c)

]= ect f (t). Hence,

h(t) =4

5

[1− e−t/2 cos(t)− 1

2e−t/2 sin(t)

].

We conclude: y(t) = h(t) + u(t − π)h(t − π). C


Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


5L−1

[1

s

]− 4

5L−1

[ (s + 1

2

)+ 1

2[(s + 1

2

)2+ 1

]].

h(t) =4

5L−1

[1

s

]− 4

5L−1

[ (s + 1

2

)[(

s + 12

)2+ 1

]]− 2

5L−1

[ 1[(s + 1

2

)2+ 1

]].


]= ect f (t).

Hence,

h(t) =4

5

[1− e−t/2 cos(t)− 1

2e−t/2 sin(t)

].



Example


y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).


5L−1

[1

s

]− 4

5L−1

[ (s + 1

2

)+ 1

2[(s + 1

2

)2+ 1

]].

h(t) =4

5L−1

[1

s

]− 4

5L−1

[ (s + 1

2

)[(

s + 12

)2+ 1

]]− 2

5L−1

[ 1[(s + 1

2

)2+ 1

]].


]= ect f (t). Hence,

h(t) =4

5

[1− e−t/2 cos(t)− 1

2e−t/2 sin(t)

].




(a) Example 1:

y ′ + 2y = u(t − 4), y(0) = 3.

(b) Example 2:

y ′′ + y ′ +5

4y = b(t),

y(0) = 0,

y ′(0) = 0,b(t) =

{1, t ∈ [0, π)

0, t ∈ [π,∞).

(c) Example 3:

y ′′+y ′+5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t), t ∈ [0, π)

0, t ∈ [π,∞).


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).

Solution:


g ( t )

0 tpi

1

y

sin ( t )

1

0 pi t

y y

0

1

u ( t ) − u ( t − pi )

pi t


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).

Solution:


g ( t )

0 tpi

1

y

sin ( t )

1

0 pi t

y

y

0

1

u ( t ) − u ( t − pi )

pi t


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).

Solution:


g ( t )

0 tpi

1

y

sin ( t )

1

0 pi t

y y

0

1

u ( t ) − u ( t − pi )

pi t


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).

Solution: The graphs imply: g(t) =[u(t)− u(t − π)

]sin(t).

Recall the identity: sin(t) = − sin(t − π). Then,

g(t) = u(t) sin(t)− u(t − π) sin(t),

g(t) = u(t) sin(t) + u(t − π) sin(t − π).

Now is simple to find L[g ], since

L[g(t)] = L[u(t) sin(t)] + L[u(t − π) sin(t − π)].


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).


]sin(t).

Recall the identity: sin(t) = − sin(t − π).

Then,






Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).


]sin(t).







Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).


]sin(t).




Now is simple to find L[g ],

since



Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).


]sin(t).







Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).

Solution: So: L[g(t)] = L[u(t) sin(t)] + L[u(t − π) sin(t − π)].

L[g(t)] =1

(s2 + 1)+ e−πs 1

(s2 + 1).

Recall the Laplace transform of the differential equation

L[y ′′] + L[y ′] +5

4L[y ] = L[g ].


Therefore,(s2 + s +

5

4

)L[y ] =

(1 + e−πs

) 1

(s2 + 1).


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).


L[g(t)] =1

(s2 + 1)+ e−πs 1

(s2 + 1).


L[y ′′] + L[y ′] +5

4L[y ] = L[g ].

The initial conditions imply:

L[y ′′] = s2 L[y ] and L[y ′] = s L[y ].

Therefore,(s2 + s +

5

4

)L[y ] =

(1 + e−πs

) 1

(s2 + 1).


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).


L[g(t)] =1

(s2 + 1)+ e−πs 1

(s2 + 1).


L[y ′′] + L[y ′] +5

4L[y ] = L[g ].

The initial conditions imply: L[y ′′] = s2 L[y ]

and L[y ′] = s L[y ].

Therefore,(s2 + s +

5

4

)L[y ] =

(1 + e−πs

) 1

(s2 + 1).


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).


L[g(t)] =1

(s2 + 1)+ e−πs 1

(s2 + 1).


L[y ′′] + L[y ′] +5

4L[y ] = L[g ].


Therefore,(s2 + s +

5

4

)L[y ] =

(1 + e−πs

) 1

(s2 + 1).


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).

Solution: Recall:(s2 + s +

5

4

)L[y ] =

(1 + e−πs

) 1

(s2 + 1).

L[y ] =(1 + e−πs

) 1(s2 + s + 5

4

)(s2 + 1)

.

Introduce the function H(s) =1(

s2 + s + 54

)(s2 + 1)

.

Then, y(t) = L−1[H(s)] + L−1[e−πs H(s)].

Differential equations with discontinuous sources.Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).

Solution: Recall: y(t) = L−1[H(s)] + L−1[e−πs H(s)], and

H(s) =1(

s2 + s + 54

)(s2 + 1)

.


s± =1

2

[−1±

√1− 5

]⇒ Complex roots.


1(s2 + s + 5

4

)(s2 + 1)

=(as + b)(

s2 + s + 54

) +(cs + d)

(s2 + 1).



y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).


H(s) =1(

s2 + s + 54

)(s2 + 1)

.

Partial fractions:

Find the zeros of the denominator,

s± =1

2

[−1±

√1− 5

]⇒ Complex roots.


1(s2 + s + 5

4

)(s2 + 1)

=(as + b)(

s2 + s + 54

) +(cs + d)

(s2 + 1).



y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).


H(s) =1(

s2 + s + 54

)(s2 + 1)

.


s± =1

2

[−1±

√1− 5

]⇒ Complex roots.


1(s2 + s + 5

4

)(s2 + 1)

=(as + b)(

s2 + s + 54

) +(cs + d)

(s2 + 1).



y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).


H(s) =1(

s2 + s + 54

)(s2 + 1)

.


s± =1

2

[−1±

√1− 5

]

⇒ Complex roots.


1(s2 + s + 5

4

)(s2 + 1)

=(as + b)(

s2 + s + 54

) +(cs + d)

(s2 + 1).



y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).


H(s) =1(

s2 + s + 54

)(s2 + 1)

.


s± =1

2

[−1±

√1− 5

]⇒ Complex roots.


1(s2 + s + 5

4

)(s2 + 1)

=(as + b)(

s2 + s + 54

) +(cs + d)

(s2 + 1).


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).

Solution: So:1(

s2 + s + 54

)(s2 + 1)

=(as + b)(

s2 + s + 54

) +(cs + d)

(s2 + 1).

Therefore, we get

1 = (as + b)(s2 + 1) + (cs + d)(s2 + s +

5

4

),

1 = (a + c) s3 + (b + c + d) s2 +(a +

5

4c + d

)s +

(b +

5

4d).

This equation implies that a, b, c , and d , are solutions of

a + c = 0, b + c + d = 0, a +5

4c + d = 0, b +

5

4d = 1.


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).

Solution: So: a =16

17, b =

12

17, c = −16

17, d =

4

17.

We have found: H(s) =4

17

[ (4s + 3)(s2 + s + 5

4

) +(−4s + 1)

(s2 + 1)

].

Complete the square in the denominator,

s2 + s +5

4=

[s2 + 2

(1

2

)s +

1

4

]− 1

4+

5

4=

(s +

1

2

)2+ 1.

H(s) =4

17

[ (4s + 3)[(s + 1

2

)2+ 1

] +(−4s + 1)

(s2 + 1)

].


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).

Solution: So: a =16

17, b =

12

17, c = −16

17, d =

4

17.


17

[ (4s + 3)(s2 + s + 5

4

) +(−4s + 1)

(s2 + 1)

].


s2 + s +5

4=

[s2 + 2

(1

2

)s +

1

4

]− 1

4+

5

4

=(s +

1

2

)2+ 1.

H(s) =4

17

[ (4s + 3)[(s + 1

2

)2+ 1

] +(−4s + 1)

(s2 + 1)

].


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).

Solution: So: a =16

17, b =

12

17, c = −16

17, d =

4

17.


17

[ (4s + 3)(s2 + s + 5

4

) +(−4s + 1)

(s2 + 1)

].


s2 + s +5

4=

[s2 + 2

(1

2

)s +

1

4

]− 1

4+

5

4=

(s +

1

2

)2+ 1.

H(s) =4

17

[ (4s + 3)[(s + 1

2

)2+ 1

] +(−4s + 1)

(s2 + 1)

].


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).

Solution: So: H(s) =4

17

[ (4s + 3)[(s + 1

2

)2+ 1

] +(−4s + 1)

(s2 + 1)

].

Rewrite the polynomial in the numerator,

(4s + 3) = 4(s +

1

2− 1

2

)+ 3 = 4

(s +

1

2

)+ 1,

H(s) =4

17

[4

(s + 1

2

)[(s + 1

2

)2+ 1

] +1[(

s + 12

)2+ 1

] − 4s

(s2 + 1)+

1

(s2 + 1)

],


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).


17

[ (4s + 3)[(s + 1

2

)2+ 1

] +(−4s + 1)

(s2 + 1)

].


(4s + 3) = 4(s +

1

2− 1

2

)+ 3

= 4(s +

1

2

)+ 1,

H(s) =4

17

[4

(s + 1

2

)[(s + 1

2

)2+ 1

] +1[(

s + 12

)2+ 1

] − 4s

(s2 + 1)+

1

(s2 + 1)

],


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).


17

[ (4s + 3)[(s + 1

2

)2+ 1

] +(−4s + 1)

(s2 + 1)

].


(4s + 3) = 4(s +

1

2− 1

2

)+ 3 = 4

(s +

1

2

)+ 1,

H(s) =4

17

[4

(s + 1

2

)[(s + 1

2

)2+ 1

] +1[(

s + 12

)2+ 1

] − 4s

(s2 + 1)+

1

(s2 + 1)

],



y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).

Solution:

H(s) =4

17

[4

(s + 1

2

)[(s + 1

2

)2+ 1

] +1[(

s + 12

)2+ 1

] − 4s

(s2 + 1)+

1

(s2 + 1)

],

Use the Laplace Transform table to get H(s) equal to

H(s) =4

17

[4L

[e−t/2 cos(t)

]+L

[e−t/2 sin(t)

]− 4L[cos(t)]+L[sin(t)]

].

H(s) = L[ 4

17

(4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)

)].


Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).

Solution: Recall:

H(s) = L[ 4

17


)].

Denote:

h(t) =4

17

[4e−t/2 cos(t) + e−t/2 sin(t)− 4 cos(t) + sin(t)

].

Then, H(s) = L[h(t)]. Recalling: L[y(t)] = H(s) + e−πs H(s),

L[y(t)] = L[h(t)] + e−πs L[h(t)].



Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).

Solution: Recall:

H(s) = L[ 4

17


)].

Denote:

h(t) =4

17


].

Then, H(s) = L[h(t)].

Recalling: L[y(t)] = H(s) + e−πs H(s),

L[y(t)] = L[h(t)] + e−πs L[h(t)].



Example


y ′′ + y ′ +5

4y = g(t),

y(0) = 0,

y ′(0) = 0,g(t) =

{sin(t) t ∈ [0, π)

0 t ∈ [π,∞).

Solution: Recall:

H(s) = L[ 4

17


)].

Denote:

h(t) =4

17


].

Then, H(s) = L[h(t)]. Recalling: L[y(t)] = H(s) + e−πs H(s),

L[y(t)] = L[h(t)] + e−πs L[h(t)].


Generalized sources (Sect. 6.5).

I The Dirac delta generalized function.

I Properties of Dirac’s delta.

I Relation between deltas and steps.

I Dirac’s delta in Physics.

I The Laplace Transform of Dirac’s delta.

I Differential equations with Dirac’s delta sources.

The Dirac delta generalized function.

DefinitionConsider the sequence of functions for n > 1,

δn(t) =

0, t < 0

n, 0 6 t 61

n

0, t >1

n.

d

0

1

2

3

2

1

1 t

3

nd

d

d

The Dirac delta generalized function is given by

limn→∞

δn(t) = δ(t), t ∈ R.

Remarks:

(a) There exist infinitely many sequences δn that define the samegeneralized function δ.

(b) For example, compare with the sequence δn in the textbook.


DefinitionConsider the sequence of functions for n > 1,

δn(t) =

0, t < 0

n, 0 6 t 61

n

0, t >1

n. d

0

1

2

3

2

1

1 t

3

nd

d

d

The Dirac delta generalized function is given by

limn→∞

δn(t) = δ(t), t ∈ R.

Remarks:

(a) There exist infinitely many sequences δn that define the samegeneralized function δ.

(b) For example, compare with the sequence δn in the textbook.


d

0

1

2

3

2

1

1 t

3

nd

d

d

R − { 0 }

0 t

delta

Remarks:

(a) The Dirac δ is a function on the domain R− {0}, andδ(t) = 0 for t ∈ R− {0}.

(b) δ at t = 0 is not defined, since δ(0) = limn→∞ n = +∞.

(c) δ is not a function on R.

Properties of Dirac’s delta.

Remark: The Dirac δ is not a function.

We define operations on Dirac’s δ as limits n →∞ of theoperation on the sequence elements δn.

Definition

δ(t − c) = limn→∞

δn(t − c),

a δ(t) + b δ(t) = limn→∞

[a δn(t) + b δn(t)

],

f (t) δ(t) = limn→∞

[f (t) δn(t)

],∫ b

aδ(t) dt = lim

n→∞

∫ b

aδn(t) dt,

L[δ] = limn→∞

L[δn].




Definition


δn(t − c),



],


[f (t) δn(t)

],

∫ b

aδ(t) dt = lim

n→∞

∫ b

aδn(t) dt,

L[δ] = limn→∞

L[δn].




Definition


δn(t − c),



],


[f (t) δn(t)

],∫ b

aδ(t) dt = lim

n→∞

∫ b

aδn(t) dt,

L[δ] = limn→∞

L[δn].


Theorem ∫ a

−aδ(t) dt = 1, a > 0.

Proof: ∫ a

−aδ(t) dt = lim

n→∞

∫ a

−aδn(t) dt = lim

n→∞

∫ 1/n

0n dt

∫ a

−aδ(t) dt = lim

n→∞

[n(t∣∣∣1/n

0

)]= lim

n→∞

[n

1

n

].

We conclude:

∫ a

−aδ(t) dt = 1.


Theorem ∫ a

−aδ(t) dt = 1, a > 0.

Proof: ∫ a

−aδ(t) dt

= limn→∞

∫ a

−aδn(t) dt = lim

n→∞

∫ 1/n

0n dt

∫ a

−aδ(t) dt = lim

n→∞

[n(t∣∣∣1/n

0

)]= lim

n→∞

[n

1

n

].

We conclude:

∫ a

−aδ(t) dt = 1.


Theorem ∫ a

−aδ(t) dt = 1, a > 0.

Proof: ∫ a

−aδ(t) dt = lim

n→∞

∫ a

−aδn(t) dt

= limn→∞

∫ 1/n

0n dt

∫ a

−aδ(t) dt = lim

n→∞

[n(t∣∣∣1/n

0

)]= lim

n→∞

[n

1

n

].

We conclude:

∫ a

−aδ(t) dt = 1.


Theorem ∫ a

−aδ(t) dt = 1, a > 0.

Proof: ∫ a

−aδ(t) dt = lim

n→∞

∫ a

−aδn(t) dt = lim

n→∞

∫ 1/n

0n dt

∫ a

−aδ(t) dt = lim

n→∞

[n(t∣∣∣1/n

0

)]= lim

n→∞

[n

1

n

].

We conclude:

∫ a

−aδ(t) dt = 1.


Theorem ∫ a

−aδ(t) dt = 1, a > 0.

Proof: ∫ a

−aδ(t) dt = lim

n→∞

∫ a

−aδn(t) dt = lim

n→∞

∫ 1/n

0n dt

∫ a

−aδ(t) dt = lim

n→∞

[n(t∣∣∣1/n

0

)]

= limn→∞

[n

1

n

].

We conclude:

∫ a

−aδ(t) dt = 1.


Theorem ∫ a

−aδ(t) dt = 1, a > 0.

Proof: ∫ a

−aδ(t) dt = lim

n→∞

∫ a

−aδn(t) dt = lim

n→∞

∫ 1/n

0n dt

∫ a

−aδ(t) dt = lim

n→∞

[n(t∣∣∣1/n

0

)]= lim

n→∞

[n

1

n

].

We conclude:

∫ a

−aδ(t) dt = 1.


TheoremIf f : R → R is continuous, t0 ∈ R and a > 0, then∫ t0+a

t0−aδ(t − t0) f (t) dt = f (t0).

Proof: Introduce the change of variable τ = t − t0,

I =

∫ t0+a

t0−aδ(t − t0) f (t) dt =

∫ a

−aδ(τ) f (τ + t0) dτ,

I = limn→∞

∫ a

−aδn(τ) f (τ + t0) dτ = lim

n→∞

∫ 1/n

0n f (τ + t0) dτ

Therefore, I = limn→∞

n

∫ 1/n

0F ′(τ + t0) dτ , where we introduced the

primitive F (t) =

∫f (t) dt, that is, f (t) = F ′(t).



t0−aδ(t − t0) f (t) dt = f (t0).


I =

∫ t0+a

t0−aδ(t − t0) f (t) dt

=

∫ a

−aδ(τ) f (τ + t0) dτ,

I = limn→∞

∫ a


n→∞

∫ 1/n

0n f (τ + t0) dτ


n

∫ 1/n


primitive F (t) =




t0−aδ(t − t0) f (t) dt = f (t0).


I =

∫ t0+a

t0−aδ(t − t0) f (t) dt =

∫ a

−aδ(τ) f (τ + t0) dτ,

I = limn→∞

∫ a


n→∞

∫ 1/n

0n f (τ + t0) dτ


n

∫ 1/n


primitive F (t) =




t0−aδ(t − t0) f (t) dt = f (t0).


I =

∫ t0+a

t0−aδ(t − t0) f (t) dt =

∫ a

−aδ(τ) f (τ + t0) dτ,

I = limn→∞

∫ a

−aδn(τ) f (τ + t0) dτ

= limn→∞

∫ 1/n

0n f (τ + t0) dτ


n

∫ 1/n


primitive F (t) =




t0−aδ(t − t0) f (t) dt = f (t0).


I =

∫ t0+a

t0−aδ(t − t0) f (t) dt =

∫ a

−aδ(τ) f (τ + t0) dτ,

I = limn→∞

∫ a


n→∞

∫ 1/n

0n f (τ + t0) dτ


n

∫ 1/n


primitive F (t) =




t0−aδ(t − t0) f (t) dt = f (t0).


I =

∫ t0+a

t0−aδ(t − t0) f (t) dt =

∫ a

−aδ(τ) f (τ + t0) dτ,

I = limn→∞

∫ a


n→∞

∫ 1/n

0n f (τ + t0) dτ


n

∫ 1/n

0F ′(τ + t0) dτ ,

where we introduced the

primitive F (t) =




t0−aδ(t − t0) f (t) dt = f (t0).


I =

∫ t0+a

t0−aδ(t − t0) f (t) dt =

∫ a

−aδ(τ) f (τ + t0) dτ,

I = limn→∞

∫ a


n→∞

∫ 1/n

0n f (τ + t0) dτ


n

∫ 1/n


primitive F (t) =




t0−aδ(t − t0) f (t) dt = f (t0).

Proof: So, I = limn→∞

n

∫ 1/n

0F ′(τ + t0) dτ , with f (t) = F ′(t).

I = limn→∞

n[F (τ + t0)

∣∣∣1/n

0

]= lim

n→∞n

[F

(t0 +

1

n

)− F (t0)

].

I = limn→∞

[F

(t0 + 1

n

)− F (t0)

]1n

= F ′(t0) = f (t0).

We conclude:

∫ t0+a

t0−aδ(t − t0) f (t) dt = f (t0).



t0−aδ(t − t0) f (t) dt = f (t0).


n

∫ 1/n

0F ′(τ + t0) dτ , with f (t) = F ′(t).

I = limn→∞

n[F (τ + t0)

∣∣∣1/n

0

]

= limn→∞

n[F

(t0 +

1

n

)− F (t0)

].

I = limn→∞

[F

(t0 + 1

n

)− F (t0)

]1n

= F ′(t0) = f (t0).

We conclude:

∫ t0+a

t0−aδ(t − t0) f (t) dt = f (t0).



t0−aδ(t − t0) f (t) dt = f (t0).


n

∫ 1/n

0F ′(τ + t0) dτ , with f (t) = F ′(t).

I = limn→∞

n[F (τ + t0)

∣∣∣1/n

0

]= lim

n→∞n

[F

(t0 +

1

n

)− F (t0)

].

I = limn→∞

[F

(t0 + 1

n

)− F (t0)

]1n

= F ′(t0) = f (t0).

We conclude:

∫ t0+a

t0−aδ(t − t0) f (t) dt = f (t0).



t0−aδ(t − t0) f (t) dt = f (t0).


n

∫ 1/n

0F ′(τ + t0) dτ , with f (t) = F ′(t).

I = limn→∞

n[F (τ + t0)

∣∣∣1/n

0

]= lim

n→∞n

[F

(t0 +

1

n

)− F (t0)

].

I = limn→∞

[F

(t0 + 1

n

)− F (t0)

]1n

= F ′(t0)

= f (t0).

We conclude:

∫ t0+a

t0−aδ(t − t0) f (t) dt = f (t0).



t0−aδ(t − t0) f (t) dt = f (t0).


n

∫ 1/n

0F ′(τ + t0) dτ , with f (t) = F ′(t).

I = limn→∞

n[F (τ + t0)

∣∣∣1/n

0

]= lim

n→∞n

[F

(t0 +

1

n

)− F (t0)

].

I = limn→∞

[F

(t0 + 1

n

)− F (t0)

]1n

= F ′(t0) = f (t0).

We conclude:

∫ t0+a

t0−aδ(t − t0) f (t) dt = f (t0).

Relation between deltas and steps.

TheoremThe sequence of functions for n > 1,

un(t) =

0, t < 0

nt, 0 6 t 61

n

1, t >1

n.

u

t0 1/3 1/2 1

3 2 1u u u

1

n

satisfies, for t ∈ (−∞, 0) ∪ (0, 1/n) ∪ (1/n,∞), both equations,

u′n(t) = δn(t), limn→∞

un(t) = u(t), t ∈ R.

Remark:

I If we generalize the notion of derivative asu′(t) = lim

n→∞δn(t), then holds u′(t) = δ(t).

I Dirac’s delta is a generalized derivative of the step function.

Dirac’s delta in Physics.

Remarks:

(a) Dirac’s delta generalized function is useful to describeimpulsive forces in mechanical systems.

(b) An impulsive force transmits a finite momentum in aninfinitely short time.

(c) For example: The momentum transmitted to a pendulumwhen hit by a hammer. Newton’s law of motion says,

m v ′(t) = F (t), with F (t) = F0 δ(t − t0).

The momentum transfer is:

∆I = lim∆t→0

mv(t)∣∣∣t0+∆t

t0−∆t= lim

∆t→0

∫ t0+∆t

t0−∆tF (t) dt = F0.

That is, ∆I = F0.


Remarks:



(c) For example: The momentum transmitted to a pendulumwhen hit by a hammer.

Newton’s law of motion says,



∆I = lim∆t→0

mv(t)∣∣∣t0+∆t

t0−∆t= lim

∆t→0

∫ t0+∆t

t0−∆tF (t) dt = F0.

That is, ∆I = F0.


Remarks:




m v ′(t) = F (t),

with F (t) = F0 δ(t − t0).


∆I = lim∆t→0

mv(t)∣∣∣t0+∆t

t0−∆t= lim

∆t→0

∫ t0+∆t

t0−∆tF (t) dt = F0.

That is, ∆I = F0.


Remarks:






∆I = lim∆t→0

mv(t)∣∣∣t0+∆t

t0−∆t= lim

∆t→0

∫ t0+∆t

t0−∆tF (t) dt = F0.

That is, ∆I = F0.


Remarks:






∆I = lim∆t→0

mv(t)∣∣∣t0+∆t

t0−∆t

= lim∆t→0

∫ t0+∆t

t0−∆tF (t) dt = F0.

That is, ∆I = F0.


Remarks:






∆I = lim∆t→0

mv(t)∣∣∣t0+∆t

t0−∆t= lim

∆t→0

∫ t0+∆t

t0−∆tF (t) dt

= F0.

That is, ∆I = F0.


Remarks:






∆I = lim∆t→0

mv(t)∣∣∣t0+∆t

t0−∆t= lim

∆t→0

∫ t0+∆t

t0−∆tF (t) dt = F0.

That is, ∆I = F0.

The Laplace Transform of Dirac’s delta.

Recall: The Laplace Transform can be generalized from functionsto δ,

as follows, L[δ(t − c)] = limn→∞

L[δn(t − c)].

TheoremL[δ(t − c)] = e−cs .

Proof:

L[δ(t − c)] = limn→∞

L[δn(t − c)], δn(t) = n[u(t)− u

(t − 1

n

)].


n(L[u(t − c)]− L

[u(t − c − 1

n

)])L[δ(t − c)] = lim

n→∞n(e−cs

s− e−(c+ 1

n)s

s

)= e−cs lim

n→∞

(1− e−sn )(

sn

) .

This is a singular limit, 00 . Use l’Hopital rule.


Recall: The Laplace Transform can be generalized from functionsto δ, as follows, L[δ(t − c)] = lim

n→∞L[δn(t − c)].


Proof:


L[δn(t − c)], δn(t) = n[u(t)− u

(t − 1

n

)].


n(L[u(t − c)]− L

[u(t − c − 1

n

)])L[δ(t − c)] = lim

n→∞n(e−cs

s− e−(c+ 1

n)s

s

)= e−cs lim

n→∞

(1− e−sn )(

sn

) .




n→∞L[δn(t − c)].


Proof:


L[δn(t − c)],

δn(t) = n[u(t)− u

(t − 1

n

)].


n(L[u(t − c)]− L

[u(t − c − 1

n

)])L[δ(t − c)] = lim

n→∞n(e−cs

s− e−(c+ 1

n)s

s

)= e−cs lim

n→∞

(1− e−sn )(

sn

) .




n→∞L[δn(t − c)].


Proof:


L[δn(t − c)], δn(t) = n[u(t)− u

(t − 1

n

)].


n(L[u(t − c)]− L

[u(t − c − 1

n

)])L[δ(t − c)] = lim

n→∞n(e−cs

s− e−(c+ 1

n)s

s

)= e−cs lim

n→∞

(1− e−sn )(

sn

) .




n→∞L[δn(t − c)].


Proof:


L[δn(t − c)], δn(t) = n[u(t)− u

(t − 1

n

)].


n(L[u(t − c)]− L

[u(t − c − 1

n

)])


n(e−cs

s− e−(c+ 1

n)s

s

)= e−cs lim

n→∞

(1− e−sn )(

sn

) .




n→∞L[δn(t − c)].


Proof:


L[δn(t − c)], δn(t) = n[u(t)− u

(t − 1

n

)].


n(L[u(t − c)]− L

[u(t − c − 1

n

)])L[δ(t − c)] = lim

n→∞n(e−cs

s− e−(c+ 1

n)s

s

)

= e−cs limn→∞

(1− e−sn )(

sn

) .




n→∞L[δn(t − c)].


Proof:


L[δn(t − c)], δn(t) = n[u(t)− u

(t − 1

n

)].


n(L[u(t − c)]− L

[u(t − c − 1

n

)])L[δ(t − c)] = lim

n→∞n(e−cs

s− e−(c+ 1

n)s

s

)= e−cs lim

n→∞

(1− e−sn )(

sn

) .




n→∞L[δn(t − c)].


Proof:


L[δn(t − c)], δn(t) = n[u(t)− u

(t − 1

n

)].


n(L[u(t − c)]− L

[u(t − c − 1

n

)])L[δ(t − c)] = lim

n→∞n(e−cs

s− e−(c+ 1

n)s

s

)= e−cs lim

n→∞

(1− e−sn )(

sn

) .

This is a singular limit, 00 .

Use l’Hopital rule.



n→∞L[δn(t − c)].


Proof:


L[δn(t − c)], δn(t) = n[u(t)− u

(t − 1

n

)].


n(L[u(t − c)]− L

[u(t − c − 1

n

)])L[δ(t − c)] = lim

n→∞n(e−cs

s− e−(c+ 1

n)s

s

)= e−cs lim

n→∞

(1− e−sn )(

sn

) .



Proof: Recall: L[δ(t − c)] = e−cs limn→∞

(1− e−sn )(

sn

) .

limn→∞

(1− e−sn )(

sn

) = limn→∞

(− sn2 e−

sn )(

− sn2

) = limn→∞

e−sn = 1.

We therefore conclude that L[δ(t − c)] = e−cs .

Remarks:

(a) This result is consistent with a previous result:∫ t0+a

t0−aδ(t − t0) f (t) dt = f (t0).

(b) L[δ(t − c)] =

∫ ∞

0δ(t − c) e−st dt = e−cs .

(c) L[δ(t − c) f (t)] =

∫ ∞

0δ(t − c) e−st f (t) dt = e−cs f (c).



(1− e−sn )(

sn

) .

limn→∞

(1− e−sn )(

sn

)

= limn→∞

(− sn2 e−

sn )(

− sn2

) = limn→∞

e−sn = 1.


Remarks:


t0−aδ(t − t0) f (t) dt = f (t0).

(b) L[δ(t − c)] =

∫ ∞


(c) L[δ(t − c) f (t)] =

∫ ∞




(1− e−sn )(

sn

) .

limn→∞

(1− e−sn )(

sn

) = limn→∞

(− sn2 e−

sn )(

− sn2

)

= limn→∞

e−sn = 1.


Remarks:


t0−aδ(t − t0) f (t) dt = f (t0).

(b) L[δ(t − c)] =

∫ ∞


(c) L[δ(t − c) f (t)] =

∫ ∞




(1− e−sn )(

sn

) .

limn→∞

(1− e−sn )(

sn

) = limn→∞

(− sn2 e−

sn )(

− sn2

) = limn→∞

e−sn

= 1.


Remarks:


t0−aδ(t − t0) f (t) dt = f (t0).

(b) L[δ(t − c)] =

∫ ∞


(c) L[δ(t − c) f (t)] =

∫ ∞




(1− e−sn )(

sn

) .

limn→∞

(1− e−sn )(

sn

) = limn→∞

(− sn2 e−

sn )(

− sn2

) = limn→∞

e−sn = 1.


Remarks:


t0−aδ(t − t0) f (t) dt = f (t0).

(b) L[δ(t − c)] =

∫ ∞


(c) L[δ(t − c) f (t)] =

∫ ∞




(1− e−sn )(

sn

) .

limn→∞

(1− e−sn )(

sn

) = limn→∞

(− sn2 e−

sn )(

− sn2

) = limn→∞

e−sn = 1.


Remarks:

(a) This result is consistent with a previous result:

∫ t0+a

t0−aδ(t − t0) f (t) dt = f (t0).

(b) L[δ(t − c)] =

∫ ∞


(c) L[δ(t − c) f (t)] =

∫ ∞




(1− e−sn )(

sn

) .

limn→∞

(1− e−sn )(

sn

) = limn→∞

(− sn2 e−

sn )(

− sn2

) = limn→∞

e−sn = 1.


Remarks:


t0−aδ(t − t0) f (t) dt = f (t0).

(b) L[δ(t − c)] =

∫ ∞


(c) L[δ(t − c) f (t)] =

∫ ∞


Differential equations with Dirac’s delta sources.

Example

Find the solution y to the initial value problem

y ′′ − y = −20 δ(t − 3), y(0) = 1, y ′(0) = 0.

Solution: Compute: L[y ′′]− L[y ] = −20L[δ(t − 3)].

L[y ′′] = s2 L[y ]−s y(0)−y ′(0) ⇒ (s2−1)L[y ]−s = −20 e−3s ,

We arrive to the equation L[y ] =s

(s2 − 1)− 20 e−3s 1

(s2 − 1),

L[y ] = L[cosh(t)]− 20L[u(t − 3) sinh(t − 3)],

We conclude: y(t) = cosh(t)− 20 u(t − 3) sinh(t − 3). C


Example


y ′′ − y = −20 δ(t − 3), y(0) = 1, y ′(0) = 0.


L[y ′′] = s2 L[y ]−s y(0)−y ′(0)

⇒ (s2−1)L[y ]−s = −20 e−3s ,


(s2 − 1)− 20 e−3s 1

(s2 − 1),




Example


y ′′ − y = −20 δ(t − 3), y(0) = 1, y ′(0) = 0.


L[y ′′] = s2 L[y ]−s y(0)−y ′(0) ⇒ (s2−1)L[y ]−s = −20 e−3s ,


(s2 − 1)− 20 e−3s 1

(s2 − 1),




Example

Find the solution to the initial value problem

y ′′ + 4y = δ(t − π)− δ(t − 2π), y(0) = 0, y ′(0) = 0.

Solution: Compute: L[y ′′] + 4L[y ] = L[δ(t − π)]− L[δ(t − 2π)],

(s2 + 4)L[y ] = e−πs − e−2πs ⇒ L[y ] =e−πs

(s2 + 4)− e−2πs

(s2 + 4),

that is, L[y ] =e−πs

2

2

(s2 + 4)− e−2πs

2

2

(s2 + 4).

Recall: e−cs L[f (t)] = L[u(t − c) f (t − c)]. Therefore,

L[y ] =1

2L

[u(t−π) sin

[2(t−π)

]]− 1

2L

[u(t−2π) sin

[2(t−2π)

]].


Example


y ′′ + 4y = δ(t − π)− δ(t − 2π), y(0) = 0, y ′(0) = 0.


(s2 + 4)L[y ] = e−πs − e−2πs

⇒ L[y ] =e−πs

(s2 + 4)− e−2πs

(s2 + 4),


2

2

(s2 + 4)− e−2πs

2

2

(s2 + 4).


L[y ] =1

2L

[u(t−π) sin

[2(t−π)

]]− 1

2L

[u(t−2π) sin

[2(t−2π)

]].


Example


y ′′ + 4y = δ(t − π)− δ(t − 2π), y(0) = 0, y ′(0) = 0.


(s2 + 4)L[y ] = e−πs − e−2πs ⇒ L[y ] =e−πs

(s2 + 4)− e−2πs

(s2 + 4),


2

2

(s2 + 4)− e−2πs

2

2

(s2 + 4).


L[y ] =1

2L

[u(t−π) sin

[2(t−π)

]]− 1

2L

[u(t−2π) sin

[2(t−2π)

]].


Example


y ′′ + 4y = δ(t − π)− δ(t − 2π), y(0) = 0, y ′(0) = 0.


(s2 + 4)L[y ] = e−πs − e−2πs ⇒ L[y ] =e−πs

(s2 + 4)− e−2πs

(s2 + 4),


2

2

(s2 + 4)− e−2πs

2

2

(s2 + 4).

Recall: e−cs L[f (t)] = L[u(t − c) f (t − c)].

Therefore,

L[y ] =1

2L

[u(t−π) sin

[2(t−π)

]]− 1

2L

[u(t−2π) sin

[2(t−2π)

]].


Example


y ′′ + 4y = δ(t − π)− δ(t − 2π), y(0) = 0, y ′(0) = 0.


(s2 + 4)L[y ] = e−πs − e−2πs ⇒ L[y ] =e−πs

(s2 + 4)− e−2πs

(s2 + 4),


2

2

(s2 + 4)− e−2πs

2

2

(s2 + 4).


L[y ] =1

2L

[u(t−π) sin

[2(t−π)

]]− 1

2L

[u(t−2π) sin

[2(t−2π)

]].


Example


y ′′ + 4y = δ(t − π)− δ(t − 2π), y(0) = 0, y ′(0) = 0.

Solution: Recall:

L[y ] =1

2L

[u(t−π) sin

[2(t−π)

]]− 1

2L

[u(t−2π) sin

[2(t−2π)

]].

This implies that,

y(t) =1

2u(t − π) sin

[2(t − π)

]− 1

2u(t − 2π) sin

[2(t − 2π)

],

We conclude: y(t) =1

2

[u(t − π)− u(t − 2π)

]sin(2t). C

the laplace transform of step functions

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