15 Introduction to Laplace Transforms

Suppose f (x) is defined on [0,∞). Then the Laplace Trans-

form of f (x) is

L(f (x)) = F (s) =∫ ∞0e−sxf (x)dx, provided the integral con-

verges.

Note:∫ ∞0e−sxf (x)dx = limR→∞

∫ R0e−sxf (x)dx.

Note: If the integral converges for some s = s0, then it con-

verges for all s ≥ s0.

We view the Laplace Transform as an operator; it maps func-

tions into other functions.

103

ex: L(eax) =∫ ∞0e−sxeaxdx =

∫ ∞0e−(s−a)xdx

= limR→∞∫ R0e−(s−a)xdx

= limR→∞e−(s−a)x

−(s−a) |R0

= limR→∞e−(s−a)R−1−(s−a)

= limR→∞[−1

(s − a)e(s−a)R+

1

s− a].

If s > a, limR→∞ = 1s−a.

Hence L(eax) = 1s−a, s > a.

Note: Letting a = 0 above we get

L(1) = 1/s for s > 0.

104

An important property of the Laplace Transform is that it is

linear. We have

L(af (x) + bg(x)) =∫ ∞0e−sx(af (x) + bg(x))dx

= a∫ ∞0e−sxf (x)dx + b

∫ ∞0e−sxg(x)dx

= aL(f (x)) + bL(g(x)), provided the

integrals exist.

Suppose f (x) is complex valued, i.e., f (x) = u(x) + iv(x).

Then

L(f ) = L(u + iv) = L(u) + iL(v) = U(s) + iV (s).

Hence

Re(L(f )) = U(s) = L(u(x)) = L(Ref (x))

and

Im(L(f )) = V (s) = L(v(x)) = L(Imf (x)).

105

ex: Let f (x) = eibx = cos bx+ i sin bx.

L(eibx) =1

s− ib.

Note: This involves taking limits in the complex plane.

1

s− ib =1

s− ib ·s + ib

s + ib=

s

s2 + b2+

ib

s2 + b2.

Thus

L(cos bx) =s

s2 + b2, L(sin bx) =

b

s2 + b2.

Note: These formulas could also have been obtained directly

by Integration by Parts (twice).

106

ex: Find L(11 + 5e4t − 6 sin 2t).

Solution: 11s + 5

s−4 −12s2+4

.

Question: Which functions have Laplace Transforms?

In general, functions that grow too fast will not have them.

Hence some continuous functions do not have them.

ex: f (x) = ex2.∫ ∞0e−sxex

2dx→∞ ∀s > 0.

This follows since for large x and any fixed s,ex

2

esx≥ M and∫ ∞

0Mdx diverges.

We say f (t) is of exponential order α if f (t) is continuous

on 0 ≤ t < ∞ and there exist positive constants T and M

such that |f (t)| ≤Meαt for all t ≥ T .

Notation: Σα = the set of functions of exponential order α.

107

Some examples.

1. ex2

is not in Σα for any α.

2. e2x cos 3x ∈ Σ2 since |e2x cos 3x| ≤ 1e2x.

3. xn ∈ Σ1. This follows since

ex = 1 + x + x2/2! + . . .+ xn/n! + . . .

Hence xn

n! ≤ ex and thus xn ≤ n!ex.

We need a few more definitions. A function f (t), defined on

[a,b], has a jump discontinuity at t0 ∈ (a, b) if f (t) is

discontinuous at t0 and the one sided limits

limt→t−0f (t)

and

limt→t+0f (t)

exist as finite numbers.

108

A function f (t) is said to be piecewise continuous on [a,b]

if f (t) is continuous at every point in [a,b] except possibly at

a finite number of points where it has a jump discontinuity. It

is piecewise continuous on [0,∞) if it is piecewise continuous

on [0,N] for all N > 0.

Theorem 15.1 If f (t) is piecewise continuous on [0,∞)

and of exponential order α, then L(f ) exists for s > α.

Note: If L(f ) exists, it is unique.

Theorem 15.2 If L(f ) = L(g) and f (x) and g(x) are both

continuous, then f (x) = g(x).

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If L(f (x)) = F (s), then f (x) = L−1(F (s)) is called the in-

verse Laplace Transform of F (s).

Note: By Theorem 15.2 the inverse transform is unique if we

restrict our attention to continuous functions.

ex:

L−1(2

s2 + 4) = sin 2x.

A very important property of the inverse laplace transform is

that it is linear, i.e.,

L−1(c1F (s) + c2G(s)) = c1L−1F (s) + c2L

−1G(s).

This property is used over and over again when using Laplace

Transforms to solve d.e.’s.

ex:

L−1(4

s2 + 4) = 2L−1(

2

s2 + 4) = 2 sin 2x.

110

We now illustrate the use of Laplace Transforms to solve a

simple first order d.e.

ex:

y′+ y = 3e2x, y(0) = 0.

Begin by taking the L.T. of both sides, assuming L(y) and

L(y′) exist. Then

L(y′+ y) = L(3e2x).

L(y′) + L(y) =

3

s− 2(23)

(24)

Question: What is L(y′) in terms of L(y)?

111

Solution: We need to use Integration by Parts.

L(y′) =

∫ ∞0e−sxy

′(x)dx.

Let

u = e−sx

du = −se−sx

v = y(x)

dv = y′(x)dx

Hence

L(y′) = e−sxy(x)|∞0 + s

∫ ∞0e−sxy(x)dx

= limR→∞e−sRy(R)− y(0) + sL(y(x))

We assume y(x) ∈ Σα for some α. Then

limR→∞y(R)

esR≤ limR→∞

MeαR

esR= 0

if s > α.

Thus

L(y′) = sL(y)− y(0).

112

Returning to (23), since y(0) = 0, we have

sL(y) + L(y) =3

s− 2.

Hence, letting Y (s) = L(y)

Y (s)(s + 1) =3

s− 2.

Thus

Y (s) =3

(s + 1)(s− 2)=−1

s + 1+

1

s− 2.

Review Partial Fractions

Since L−1 is linear we take the inverse Laplace Transform of

both sides and get

y(x) = L−1(1

s− 2− 1

s + 1)

= L−1(1

s− 2)− L−1(

1

s + 1)

= e2x − e−x.

113

Before using Laplace Transforms to solve second order d.e.’s,

we need to develop a few more properties.

Theorem 15.3 Let f (x) be continuous on [0,∞], f′(x)

piecewise continuous on [0,∞] and f (x), f′(x) ∈ Σα. Then

L(f′(x)) exists for s > α and

L(f′) = sL(f (x))− f (0).

To solve 2nd order d.e.’s we need an extension of the previous

result.

L(f′′(x)) = sL(f

′(x))− f ′(0)

= s[sL(f (x))− f (0)]− f ′(0)

= s2L(f (x))− sf (0)− f ′(0).

114

Theorem 15.4 Let f (x) ∈ Σα and

G(x) =∫ x0f (t)dt. Then

G(x) ∈ Σα and L(G(x)) =F (s)

s.

Proof:

F (s) = L(f (x)) = L(G′(x)) = sL(G(x)) −G(0).

Since G(0) = 0,

L(G(x)) =F (s)

s.

ex: Recall L(1) =1

s. Also x =

∫ x0

1dt. Hence

L(x) = L(∫ x0

1dt) =F (s)

s=

1/s

s=

1

s2.

Theorem 15.5 (Shifting Theorem)

If f ∈ Σα then L(e−axf (x)) = F (s + a)

if s > α− a.

Proof: Let u = s + a.

L(e−axf (x)) =∫ ∞0e−sxe−axf (x)dx

=∫ ∞0e−(s+a)xf (x)dx

=∫ ∞0e−uxf (x)dx

= F (u)

= F (s + a)2

115

ex: Find L(e−ax cos bx).

Since L(cos bx) =s

s2 + b2we have

L(e−ax cos bx) =s + a

(s + a)2 + b2.

ex: Find L(eaxx).

Since L(x) = 1s2 we have L(eaxx) = 1

(s−a)2 .

Question: L(?) = F′(s)?

Solution:

F′(s) =

d

ds(∫ ∞0e−sxf (x)dx).

If f (x) ∈ Σα, we may differentiate under the integral sign.

Thus

F′(s) =

∫ ∞0e−sx(−xf (x))dx = L(−xf (x)).

Similarly,

F′′(s) = L(x2f (x))

F′′′

(s) = L(−x3f (x)).

In general,

F (n)(s) = L((−x)nf (x)).

116

ex: L(cos bx) =s

s2 + b2. HenceL(x cos bx) = −L(−x cos bx) =

− d

ds(

s

s2 + b2) =

s2 − b2

(s2 + b2)2.

ex: L(eax) = 1s−a. Hence

L(xeax) = − d

ds(

1

s− a) =1

(s− a)2.

L(x2eax) =d2

ds2(

1

s− a) =2

(s− a)3.

In general we get

L(xneax) =n!

(s− a)n+1.

Note: If a = 0, L(xn) =n!

sn+1.

Theorem 15.6 If f (x) ∈ Σα, then

lims→∞F (s) = 0.

Note: As a result, functions such as 1, s, sin s . . . can not be

Laplace Transforms.

117

We’re almost ready to solve 2nd order d.e.’s using Laplace

Transforms. To do this we will need to use the linearity of

the inverse transform.

ex: Find L−1(5

(s + 2)4+

3

2s2 + 8s + 10+

3s + 2

s2 + 2s + 10).

Since L(tn) =n!

sn+1, by letting n = 3 and using the shifting

formula we get L(e−2tt3) =6

(s + 2)4. Thus

L−1(5

(s + 2)4) =

5e−2tt3

6.

Next we note that

3

2s2 + 8s + 10=

3

2(s2 + 4s + 5)=

3

2((s + 2)2 + 12).

Thus

L−1(3

2s2 + 8s + 10) =

3e−2t sin t

2.

118

Finally,

3s + 2

s2 + 2s+ 10=

3(s + 1)

(s + 1)2 + 32− 1

(s + 1)2 + 32

= 3s + 1

(s + 1)2 + 32− 1

3

3

((s + 1)2 + 32).

Thus

L−1(3s + 2

s2 + 2s + 10) = 3e−t cos 3t − e−t sin 3t

3.

So

L−1(5

(s + 2)4+

3

2s2 + 8s + 10+

3s + 2

s2 + 2s + 10)

=5e−2tt3

6+

3e−2t sin t

2+ 3e−t cos 3t − e−t sin 3t

3.

119

16 Solving 2nd Order D.E.’s with Laplace Trans-

forms

ex: Solve

y′′

+ 4y′+ 3y = 0, y(0) = 3, y

′(0) = 1.

Take the L. T. of both sides.

L(y′′

+ 4y′+ 3y) = 0.

Recall

L(y′) = sL(y)− y(0)

L(y′′) = s2L(y)− sy(0)− y′(0)

Thus, by the linearity of the L. T.,

s2Y (s)− 3s− 1 + 4[sY (s)− 3] + 3Y (s) = 0.

Hence

(s2 + 4s+ 3)Y (s) = 3s + 13.

Thus

Y (s) =3s + 13

s2 + 4s + 3

=3s + 13

(s + 1)(s + 3)

=5

s + 1− 2

s + 3.

Hence

y(x) = 5L−1(1

s + 1)− 2L−1(

1

s + 3) = 5e−x − 2e−3x.

120

ex: Solve

y′′

+ 2y′+ 5y = 0, y(0) = 2, y

′(0) = −4.

Take the Laplace Transform of both sides and use linearity.

L(y′′) + 2L(y

′) + 5L(y) = 0.

Hence

s2Y (s)− sy(0)− y′(0) + 2[sY (s)− y(0)] + 5Y (s) = 0.

We get

Y (s)[s2 + 2s + 5] = 2s.

Now

Y (s) =2s

s2 + 2s + 5

=2s

(s + 1)2 + 4

=2(s + 1)

(s + 1)2 + 4− 2

(s + 1)2 + 4.

Thus

y(x) = 2e−x cos 2x− e−x sin 2x.

121

ex: Solve

y′′

+ y = ex sin 2x, y(0) = 1, y′(0) = 3.

After taking the L.T. of both sides and doing a bit of algebra

we get

Y (s)[s2 + 1]− s− 3 =2

(s− 1)2 + 4.

Hence

Y (s) =s + 3

s2 + 1+

2

(s2 + 1)[(s− 1)2 + 4].

We now use partial fractions to express

2

(s2 + 1)[(s− 1)2 + 4]=As +B

s2 + 1+

Cs +D

(s− 1)2 + 4.

We get

A = 1/5, B = 2/5, C = −1/5, D = 0.

Hence Y (s) =s

s2 + 1+

3

s2 + 1+

s

5(s2 + 1)+

2

5(s2 + 1)−

s

5[(s− 1)2 + 4].

Thus Y (s) =6s

5(s2 + 1)+

17

5(s2 + 1)− s− 1

5[(s− 1)2 + 4]− 1

5[(s− 1)2 + 4].

Finally we get

y(x) =6 cos x

5+

17 sin x

5− ex cos 2x

5− ex sin 2x

10.

122

Let’s talk about some of the examples in the text. Recall that

to solve a 2nd order d.e. using L.T.’s we must be given initial

conditions at zero. If they are given elsewhere we essentially

“shift the coordinate axis” to force the initial conditions to be

at zero. We first look at a simple example.

ex:

y′(t) = 2t, y(1) = 3.

We get y(t) = t2 + C. Of course

y(1) = 3⇒ C = 2.

Hence

y(t) = t2 + 2.

Suppose the only technique we have for solving this problem

involves having an initial condition at the origin.

123

Let x = t− 1 or t = x + 1. We then must solve

y′(x) = 2(x + 1), y(0) = 3.

We get y(x) = x2 + 2x + C and

y(0) = 3⇒ C = 3.

Hence y(x) = x2 + 2x + 3.

In terms of t we get

y(t) = (t− 1)2 + 2(t− 1) + 3

= t2 − 2t + 1 + 2t− 2 + 3

= t2 + 2

as before.

Now let’s look at Ex 3 on p. 383.

124

ex:

w′′(t)− 2w

′(t) + 5w(t) = −8eπ−t, w(π) = 2, w

′(π) = 12.

We attack this problem the same way.

Let x = t− π or t = x + π. We can now solve

w′′(x)− 2w

′(x) + 5w(x) = −8e−x, w(0) = 2, w

′(0) = 12.

The solution is

w(x) = 3ex cos 2x+ 4ex sin 2x− e−x.

In terms of t the solution is

w(t) = 3et−π cos 2(t− π) + 4et−π sin 2(t− π) − e−(t−π).

or

w(t) = 3et−π cos 2t+ 4et−π sin 2t − e−(t−π).

125

Before looking at Example 4 (p. 384) we first recall a few facts.

1. L((−t)nf (t)) = F (n)(s).

2. If f (t) ∈ Σα then lims→∞F (s) = 0.

3. L(y′) = sL(y)− y(0).

4. L(y′′) = s2L(y)− sy(0)− y′(0).

ex: Solve

y′′

+ 2ty′ − 4y = 1, y(0) = y

′(0) = 0.

Note: The above equation does not have constant coefficients.

We have no technique as yet to solve this.

126

We begin by taking the L. T. of both sides and use the linearity

of the L. T.

L(y′′) = s2Y (s).

L(ty′) = = − d

dsL(y

′)

= − d

ds[sY (s)]

= −sY ′(s)− Y (s).

Also

L(1) =1

s.

After some algebra we get

Y′(s) + (

3

s− s

2)Y (s) =

1

2s2.

This is a first order equation with an integrating factor and

can be solved.

Y (s) =1

s3+ C

es2/4

s3.

However if Y (s) is the L. T. of a function of exponential order,

we must have lims→∞Y (s) = 0. This means that

C = 0. Hence Y (s) = 1s3 and the solution is

y(t) =t2

2.

127