6.1 The Laplace transform

6.1 The Laplace transform

The Laplace transform

is an efficient method to solve certain ODE or PDE problems.

The transform turns an ODE into an algebraic equation.

We solve the algebraic equation and apply the inverse transform to get our desired solution.

The related Fourier transform, requires more understanding of complex numbers.

We write L 

f (t)

= F (s) for the Laplace transform of f (t):

L 

f (t)

= F (s) ^def =

Z _∞

0

e ^−st f (t) dt.

Often, t is time and s is frequency.

Some simple transforms: f (t) = 1 L{f (t)} = L{1} =

Z _∞

0

1 e ^−st dt =

 e ^−st

−s

 _∞

t=0

= lim

B →∞

 e ^−st

−s

 B t=0

= lim

B →∞

 e ^−sB

−s − 1

−s



= 1 s The limit (the improper integral) only exists if s > 0, so

L{1} =

 ₁

s if s > 0

DNE if s ≤ 0

Some simple transforms: f (t) = e ^−at L 

e ^−at

=

Z _∞

0

e ^−st e ^−at dt

=

Z _∞

0

e ^−(s+a)t dt

=

"

e ^−(s+a)t

−(s + a)

# _∞

t=0

= 1

s + a . The limit only exists if s + a > 0. So

L{e ^−at } =

 ₁

s+a if s + a > 0

DNE if s + a ≤ 0

Some simple transforms: f (t) = t L{t} =

Z _∞

0

e ^−st t dt

=

 −te ^−st s

 _∞

t=0

+ 1 s

Z _∞

0

e ^−st dt

= 0 + 1 s

 e ^−st

−s

 _∞

t=0

= 1 s ² . Again, the limit only exists if s > 0. So

L{t} =

 ₁

s ² if s > 0

DNE if s ≤ 0

Some simple transforms: Heaviside unit step funnction u(t) This function is

u(t) =

( 0 if t < 0, 1 if t ≥ 0.

Suppose a ≥ 0 is some constant.

Then u(t − a) is the function that is 0 for t < a and 1 for t ≥ a.

Let us find the Laplace transform of u(t − a), L 

u(t − a)

=

Z _∞

0

e ^−st u(t − a) dt

=

Z _∞

a

e ^−st dt

=

 e ^−st

−s

 _∞

t=a

= e ^−as

s ,

where of course s > 0

A short table of Laplace transforms f (t) L 

f (t)

f (t) L 

f (t)

C ^C _s sin(ωt) _{s 2 +ω} ^ω ₂

t _s ¹ ₂ cos(ωt) _{s 2 +ω} ^s ₂

t ² _s ² ₃ sinh(ωt) _{s 2 −ω} ^ω ₂

t ⁿ _{s n+1} ^n! cosh(ωt) _{s 2 −ω} ^s ₂

e ^−at _s+a ¹ u(t − a) ^{e −as} _s

Linearity of Laplace transform

Theorem (Linearity of the Laplace transform)

Suppose that A, B, and C are constants. Then L 

Af (t) + Bg (t)

= A L 

f (t)

+ B L 

g (t) , and in particular

L 

Cf (t)

= C L 

f (t)

.

Functions that do not have Laplace transforms The function ¹ _t , because the integral diverges for all s.

The issue is the function ¹ _t “blows up” too fast as t → 0 ⁺ . Similarly, e ^{t 2} does not have a Laplace transform.

The issue is the function e ^{t 2} grows too fast as t → ∞.

The function f (t) = tan(t) blows up as t → (2n + 1) ^π ₂ for

n = 0, 1, 2, . . . so the Laplace transform based on the Riemann integral doesn’t exist.

However, it is possible to define the Laplace transform of f (t) = tan(t) by using the Cauchy Principal Value integral.

So when does the Laplace transform of a given function f (t) exist?

It is convenient to restrict our attention to a class of functions

for which the existence of the Laplace transform is guaranteed.

Functions of exponential order

A function f (t) is of exponential order as t goes to infinity if

|f (t)| ≤ Me ^ct ,

for some constants M and c, for large t (say t > t ₀ for some t ₀ ).

One way to check this condition is to try to compute

t lim →∞

|f (t)|

e ^ct .

If this limit exists and is finite, f (t) is of exponential order.

Bounded functions like sin(t) and cos(t) are of exponential order.

Any polynomial is of exponential order.

The exponential function e ^at is of exponential order.

So are cosh(at) and sinh(at).

Existence and Uniqueness of the Laplace transform

Theorem (Existence)

Let f (t) be continuous and of exponential order for a certain constant c. Then F (s) = L 

f (t)

is defined for all s > c.

Theorem (Uniqueness)

Let f (t) and g (t) be continuous and of exponential order.

Suppose that there exists a constant C , such that F (s) = G (s) for

all s > C . Then f (t) = g (t) for all t ≥ 0.

Uniqueness and piecewise continuous functions f (t), g (t) Piecewise continuous means a function is continuous except perhaps at a discrete set of points.

A version of the uniqueness theorem for piecewise continuous functions allows concluding that f (t) = g (t) at points of

continuity, but not at discontinuities.

For example, the unit step function is sometimes defined using

u(0) = ¹ ₂ . This new step function, however, has the exact same

Laplace transform as the one we defined earlier where u(0) = 1.

The inverse transform

The Laplace transform allows us to convert an ODE into an algebraic equation.

We solve the algebraic equation in the frequency domain, and then want to get back to the time domain.

Given a function F (s), we wish to find a function f (t) such that L 

f (t)

= F (s).

Suppose F (s) = L 

f (t)

for some function f (t). Define the inverse Laplace transform as

L ⁻¹ 

F (s) def

= f (t).

There is an integral formula for the inverse, but it requires complex numbers and path integrals. For us, computing the inverse by

means of a table will suffice.

Example: inverse transform of 1/(s + 1)

Take F (s) = _s+1 ¹ . Find the inverse Laplace transform.

We look at the table and see the line

e ^−at 1

s + a which means

L 

e ^−at

= 1

s + a but which also means

e ^−at = L ⁻¹

 1

s + a



Since we wish to find L ⁻¹ n

1 s+1

o , we let a = 1 to obtain

L ⁻¹

 1

s + 1



= e ^−t .

Linearity of the inverse Laplace transform

As the Laplace transform is linear, the inverse Laplace transform is also linear. That is,

L ⁻¹ 

AF (s) + BG (s)

= A L ⁻¹ 

F (s)

+ B L ⁻¹ 

G (s) .

Linearity can be used in computing inverse Laplace transforms, as

shown below.

Example: Inverse Laplace transform of (s ² + s + 1)/(s ³ + s) This F (s) does not appear in the table.

We use the method of partial fractions to write F as a sum, each term of the sum appearing in the table.

We factor the denominator as s(s ² + 1) and write s ² + s + 1

s ³ + s = A

s + Bs + C s ² + 1 .

Putting the right-hand side over a common denominator and equating the numerators A(s ² + 1) + s(Bs + C ) = s ² + s + 1.

Expanding and equating coefficients we obtain A + B = 1, C = 1, A = 1, so B = 0 and

F (s) = s ² + s + 1

s ³ + s = 1

s + 1

s ² + 1 . By linearity of the inverse Laplace transform we get

L ⁻¹

 s ² + s + 1 s ³ + s



= L ⁻¹

 1 s



+ L ⁻¹

 1

s ² + 1



= 1 + sin t.

Shifting property

L 

e ^−at f (t)

= F (s + a), where F (s) is the Laplace transform of f (t).

The reverse version

L ⁻¹ 

F (s + a)

= e ^−at f (t), of the shifting property is most often used.

Givenor example, to compute L ⁻¹ n

1

quadratic in s

o ,

we complete the square, write the quadratic as (s + a) ² + b

and then use the shifting property.

Example: L ⁻¹ n

1 s ² +4s+8

o

First we complete the square to make the denominator (s + 2) ² + 4.

Next we find

L ⁻¹

 1

s ² + 4



= 1

2 sin(2t).

Putting it all together with the shifting property, we find L ⁻¹

 1

s ² + 4s + 8



= L ⁻¹

 1

(s + 2) ² + 4



= 1

2 e ^−2t sin(2t).

Example: L ⁻¹ n

1

s ² +4s+13

o in detail First we complete the square:

s ² + 4s

+ 13 = (s + 2) ² − 4

+ 13 = (s + 2) ² + 9. Now L ⁻¹

 1

(s + 2) ² + 9



= L ⁻¹ {F (s + 2)} where F (s) = 1

s ² + 9 . By the shifting property,

L ⁻¹ {F (s + 2)} = e ^−2t f (t) where f (t) = L ⁻¹

 1

s ² + 9

 . Next we find (from the table)

L ⁻¹

 1

s ² + 9



= 1

3 sin(3t) ≡ f (t).

Putting it all together, we find L ⁻¹

 1

s ² + 4s + 13



= L ⁻¹

 1

(s + 2) ² + 9



= e ^−2t

 1

3 sin(3t)



.

Example: L ⁻¹ n

1

s ² +4s+13

o briefly

L ⁻¹

 1

s ² + 4s + 13



= L ⁻¹

( 1

(s + 2) ² − 4

+ 13 )

= L ⁻¹

 1

(s + 2) ² + 9



(complete the square)

= e ^−2t L ⁻¹

 1

s ² + 9



(by shifting property)

= e ^−2t

 1

3 sin(3t)



(by table)