Lecture 21 Laplace transforms Lecture 21 Laplace transforms revisited You have seen Laplace transforms

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• Lecture 21

Laplace transforms revisited

You have seen Laplace transforms in an earlier course, e.g., AMATH 250 or MATH 228, so we

shall not dwell on the basics. The most important background material is listed in Section 3.1

of the AMATH 351 Course Notes, which we also include below.

Given a real- or complex-valued function y(t), t ≥ 0, the Laplace transform (LT) of y, to be denoted by L[y] = Y (s), is defined by

Y (s) =

0

e−sty(t)dt (1)

for all complex values of s such that the above improper integral converges. (That being said,

in most applications we shall be considering only real values of s.) Note that the LT is a linear

operator:

L[c1f + c2g] = c1L[f ] + c2L[g], (2)

provided that each transform exists.

Example 1: f(t) = eat. Then

F (s) = L[eat] = ∫

0

e−sueat dt (3)

=

0

e(a−s)t dt

= lim b→∞

∫ b

0

e(a−s)t dt

= lim b→∞

[

1

a− se (a−s)t

]t=b

t=0

= 1

a− s limb→∞ [

e(a−s)b − 1 ]

= 1

s− a , Re(s) > Re(a).

Note: From this point onward, we shall understand the following meaning of the improper

integral, ∫

0

h(t) dt = lim b→∞

∫ b

0

h(t) dt , (4)

200

• and omit writing “limb→∞”.

Example 2: f(t) = tn, n = 0, 1, 2, · · · .

L[eat] = ∫

0

e−sutn dt (5)

= n!

sn+1 , Re(s) > 0.

Definition: If f(t) is piecewise continuous on each interval [0, b] for b > 0 and there is a

constant α such that f(t) = O(eαt) as t → +∞, then f(t) is said to be of exponential order α as t → +∞. In this case, F (s) = L[f ] exists for Re(s) > α.

Note: The mathematical statement,

f(t) = O(g(t)) as t → a , (6)

(where a could be ±∞), means that there exists a constant K ≥ 0 such that

|f(t)| ≤ K|g(t)| for all t in some neighbourhood of a . (7)

A list of LTs for functions commonly encountered in Calculus and applications

along with some basic properties of LTs is presented in the AMATH 351 Course

Notes by J. Wainwright on Page 233.

Differentiation formulae:

1. If f is continuous and f ′ is piecewise continous on any interval [0, b], b > 0, and f is of

exponential order α as t → +∞, then

L[f ′] = sL[f ]− f(0), for Re(s) > α. (8)

201

• Proof:

L[f ′] = ∫

0

e−stf ′(t) dt

= e−stf(t) ∣

0 + s

0

e−stf(t) dt (integration by parts)

= −f(0) + sL[f ] . (9)

2. Replacing f by f ′ yields the result

L[f ′′] = s2L[f ]− sf(0)− f ′(0). (10)

Shift theorems:

1. First shift theorem: If F (s) = L[f ] exists for Re(s) > α ≥ 0, then

L[ectf ] = F (s− c) for Re(s− c) > α, (11)

where c is a constant.

Proof: Let g(t) = ectf(t) for t ≥ 0. Then

G(s) = L[g] = ∫

0

e−stectf(t) dt

=

0

e−(s−c)tf(t) dt

= F (s− c) . (12)

2. Second shift theorem: If F (s) = L[f ] exists for Re(s) > α ≥ 0, and c is a positive constant, then

L[H(t− c)f(t− c)] = e−csF (s) for Re(s) > α, (13)

where H is the Heaviside step function, defined by

H(t) =

0, if t < 0

1, if t ≥ 0 (14)

202

• Proof: Let g(t) = H(t−c)f(t−c), which implies that g(t) = 0 for t < c and g(t) = f(t−c) for t ≥ c. Then

G(s) = L[g] = ∫

c

e−stf(t− c) dt

=

0

e−s(v+c)f(v) dv (change of variable v = t− c =⇒ t = v + c · · · )

= e−sc ∫

0

e−svf(v) dv

= e−scF (s) . (15)

Solving linear second order DEs with LTs

Let’s review quickly how solutions to linear second order differential equations with constant

coefficients can be obtained from LTs.

Consider the problem

y′′ + 3y′ + 2y = et, y(0) = 1, y′(0) = 2. (16)

Of course, we could solve this problem using the “traditional method:”

1. Find linearly independent solutions u1 and u2 to the associated homogeneous DE and

then construct its general solution yh = C1u1 + C2u2,

2. Find the particular solution yp of the inhomogeneous DE,

3. Construct the general solution y = yh + yp and then solve for C1 and C1 to fit the initial

conditions.

Taking LTs of both sides of the DE:

L[y′′ + 3y′ + 2y] = L[et] (17)

The RHS is simply 1

s− 1. By linearity of the LT, the LHS becomes

L[y′′] + 3L[y′] + 2L[y] = [s2Y (s)− sy(0)− y′(0)] + 3[sY (s)− y(0)] + 2Y (s) (18)

= (s2 + 3s+ 2)Y (s)− s− 5

203

• Thus Eq. (17) becomes

(s+ 1)(s+ 2)Y (s)− (s+ 5) = 1 s− 1 , (19)

which can be rearranged to give

Y (s) = s+ 5

(s+ 1)(s+ 2) +

1

(s− 1)(s+ 1)(s+ 2) . (20)

The following partial fraction decompositions can be performed:

s+ 5

(s+ 1)(s+ 2) =

4

s+ 1 − 3

s+ 2 (21)

and 1

(s− 1)(s+ 1)(s+ 2) = 1

6 · 1 s− 1 −

1

2 · 1 s + 1

+ 1

3 · 1 s+ 2

, (22)

to give

Y (s) = 7

2 · 1 s+ 1

− 8 3 · 1 s+ 2

+ 1

6 · 1 s− 1 . (23)

Taking inverse LTs gives

y(t) = 7

2 e−t − 8

3 e−2t +

1

6 et . (24)

At this point, one may well wonder whether it was worth all of the effort to solve this problem

using Laplace transforms. Perhaps not, for this specific problem, but LTs do reveal some general

mathematical properties that are not evident using the more classical techniques, as we’ll see

below.

With the above comments in mind, we now consider the more general second order IVP as a

motivation for further studies of the LT:

y′′ + py′ + qy = u(t), y(0) = y0, y ′(0) = v0. (25)

Taking LTs of both sides yields, after a little algebra,

Y (s)[s2 + ps + q]− (s+ p)y0 − v0 = U(s), (26)

where U(s) is the LT of u(t). We may rearrange this equation to solve for Y (s):

Y (s) = U(s)

s2 + ps+ q +

(s+ p)y0 + v0 s2 + ps+ q

. (27)

204

• Once p and q are prescribed, the second term on the RHS is quite straightforward to invert,

involving either exponentials, trigonometric functions or both. In fact, the reader should rec-

ognize that inverting the second term will yield the solution of the associated homogeneous DE

with the original boundary conditions:

y′′ + py′ + qy = 0, y(0) = y0, y ′(0) = v0. (28)

To see this, simply set u(t) = 0 in Eq. (25). Therefore, we can express the LT in Eq. (27) as

Y (s) = Yp(s) + Yh(s), (29)

where Yh = L[yh], yh being the solution of Eq. (28).

205

• Lecture 22

Laplace transforms (cont’d)

We continue with our discussion from the previous lecture, namely, more general second order

IVP as a motivation for further studies of the LT:

y′′ + py′ + qy = u(t), y(0) = y0, y ′(0) = v0. (30)

Taking LTs of both sides yields, after a little algebra,

Y (s)[s2 + ps + q]− (s+ p)y0 − v0 = U(s), (31)

where U(s) is the LT of u(t). We may rearrange this equation to solve for Y (s):

Y (s) = U(s)

s2 + ps+ q +

(s+ p)y0 + v0 s2 + ps+ q

. (32)

The LT in Eq. (27) may be expressed as

Y (s) = Yp(s) + Yh(s), (33)

where Yh = L[yh], yh being the solution of Eq. (28).

Let us now focus on the LT involving the transform U(s) and write it as

Yp(s) = G(s)U(s), where G(s) = 1

s2 + ps+ q . (34)

For reasons that will become clear below, G(s) is known as the transfer function associated

with the linear differential operator,

y′′ + py′ + qy , (35)

that comprises our original problem. If we now apply the inverse LT to (34), we obtain

yp(t) = L−1[G(s)U(s)]. (36)

206

• If we knew the functional form of u(t), hence G(s), we might well be able to invert the above

LT to obtain yp. But one may well wish to ask, “How is the RHS, i.e., the inverse transform of

G(s)U(s), related, if at all, to the inverse transforms L−1[G(s)] and L−1[U(s)]?” Is it, by any chance, the product of the inverse transforms L−1[G(s)] and L−1[U(s)]? The answer, unfortu- nately, is “no.”

Example: Given

f(t) = 1 and g(t) = t. (37)

Then

F (s) = 1

s and G(s) =

1

s2 . (38)

Let us now write G(s) as

G(s) = 1

s · 1 s . (39)

Then

L−1[G(s)] = L−1[1 s · 1 s ] = t. (40)

But

L−1[1 s ] = 1. (41)

Therefore, we see that

t = L−1[ 1 s2 ] 6= L−1[1

s ] · L−1[1

s ] = 1 (42)

In general, suppose that F (s), G(s) and H(s) are LTs of, respectively, f(t), g(t) and h(t). And

suppose further that

H(s) = F (s)G(s). (43)

The question is,

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