chapter 4 the laplace transform background: improper ...etgen/ch4-part1-sp20notes.pdf · def. if...
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Chapter 4
The Laplace Transform
Background: Improper Integrals
Recall: Definite Integral: a, b real
numbers, a ≤ b; f continuous on
[a, b]
∫ b
af(x) dx
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Improper integrals:
Type I – Infinite interval of integration
∫ ∞a
f(x) dx,∫ b
−∞ f(x) dx,∫ ∞−∞ f(x) dx
Type II –∫ ba f(x) dx f unbounded on
[a, b]
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Integrals of the form∫ ∞a
f(x) dx
Def.∫ ∞a
f(x) dx = limb→∞
∫ b
af(x) dx
a b
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∫ ∞a
f(x) dx converges if
limb→∞
∫ b
af(x) dx = L
exists, in which case
∫ ∞a
f(x) dx = L
∫ ∞a
f(x) dx diverges if
limb→∞
∫ b
af(x) dx
does not exist.
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∫ ∞1
1
x2dx
∫ ∞1
1√x
dx
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Let f(s, x) be a function of x and s.
Then
∫ b
af(s, x) dx
is a function of s. That is,
∫ b
af(s, x) dx = F (s)
Example: Set f(x, s) = 3x2s + 4xs2
∫ 2
1
(
3x2s + 4xs2)
dx =
6
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I. Definition of the Laplace Trans-
form
Definition: Let f be continuous func-
tion on [0,∞). The Laplace trans-
form of f , denoted L[f(x)], or F (s),
is given by
L[f(x)] = F (s) =∫ ∞0
e−sxf(x) dx.
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L[f(x)] = F (s) =∫ ∞0
e−sxf(x) dx.
The domain of F is the set of real
numbers s for which the improper in-
tegral converges.
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Applications of the Laplace Trans-
form
A method for solving initial-value prob-
lems for first and second order linear
differential equations with constant co-
efficients.
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Laplace transforms of elementary func-
tions
Examples:
1. f(x) ≡ c, constant, on [0,∞):
L[1] =∫ ∞0
e−sx c dx = limb→∞
∫ b
0e−sxc dx
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2. f(x) = eαx on [0,∞):
L[eαx] =∫ ∞0
e−sx eαx dx
= limb→∞
∫ b
0e−sx eαx dx =
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3. f(x) = cos βx on [0,∞):
L[cos βx] =∫ ∞0
e−sx cos βx dx
= limb→∞
∫ b
0e−sx cos βx dx
Recall, from Calculus II,
∫
e−sx cos βx dx =
e−sx(−s cos βx − β sin βx)
s2 + β2
Therefore
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∫ b
0e−sx cos βx dx
=e−sx(−s cos βx − β sin βx)
s2 + β2
]b
0
=e−sb(−s cos βb − β sin βb)
s2 + β2+
s
s2 + β2
If s > 0, then
=1
esb
(−s cos βb − β sin βb)
s2 + β2+
s
s2 + β2
Therefore,
L[cos βx] =∫ ∞0
e−sx cos βx dx =s
s2 + β2
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What functions have Laplace Trans-
forms ?
Definition : f is of exponential order
λ if there exists a positive number M
and a nonnegative number A such
that
|f(x)| ≤ Meλx on [A,∞).
Examples:
(a) Bounded functions, e.g., sin x, cos x
exponential order 013
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(b) Powers of x: f(x) = xk.
exp. order λ for any λ > 0
(c) Exponential fcns: f(x) = eax.
exp. order λ for any λ ≥ a
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(d) “Complex exponentials”:
f(x) = eax cos bx, eax sin bx, etc.
exp. order λ for any λ ≥ a
(e) f(x) = ex2is not of exponential
order.
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Theorem: Let f be a continuous
function on [0,∞). If f is of ex-
ponential order λ, then the Laplace
transform L[f(x)] = F (s) exists for
s > λ.
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f(x) F (s) = L[f(x)]
cc
s, s > 0
eαx 1
s − α, s > α
cos βxs
s2 + β2, s > 0
sin βxβ
s2 + β2, s > 0
eαx cos βxs − α
(s − α)2 + β2, s > α
eαx sin βxβ
(s − α)2 + β2, s > α
xn, n = 1,2, . . .n!
sn+1, s > 0
xn eαx, n = 1,2, . . .n!
(s − α)n+1, s > α
x cos βxs2 − β2
(s2 + β2)2, s > 0
x sin βx2βs
(s2 + β2)2, s > 0
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II. Properties of the Laplace Trans-
form
I. L is a linear operator.
That is:
L[f1(x) + f2(x)] = L[f1(x)] + L[f2(x)]
L[cf(x)] = cL[f(x)].
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Proof:
L[f1(x) + f2(x)]
=∫ ∞0
e−sx[f1(x) + f2(x)] dx
=∫ ∞0
[
e−sxf1(x) + e−sxf2(x)]
dx
=∫ ∞0
e−sxf1(x) dx +∫ ∞0
e−sxf2(x) dx
= L[f1(x)] + L[f2(x)]
L[cf(x)] =∫ ∞0
e−sx[cf(x)] dx
= c∫ ∞0
e−sxf(x) dx
= cL[f(x)]
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II. Laplace transform of derivatives.
• If f is continuously differentiable
and of exponential order λ, then L[f ′(x)]
exists for s > λ and
L[f ′(x)] = sL[f(x)] − f(0).
Proof:
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Example: L[sin βx]
sinβx =−1
β(cosβx)′
L[sinβx] =−1
βL[(cos βx)′]
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• If f is twice continuously differen-
tiable with f and f ′ of exponential
order λ, then L[f ′′(x)] exists for
s > λ and
L[f ′′(x)] = s2L[f(x)] − sf(0) − f ′(0).
Proof:
L[f ′′(x)] = L[
(f ′(x))′]
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• In general, if f, f ′, · · · , f(n−1) are of
exponential order λ, then L[f(n)(x)]
exists for s > λ and
L[f(n)(x)] = snL[f(x)] − sn−1f(0)−
sn−2f ′(0) − · · · − f(n−1)(0).
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III. L[xnf(x)] = (−1)ndnF
dsn
If L[f(x)] = F (s), then
L[xf(x)] = − dF
ds,
L[x2f(x)] =d2F
ds2
and, in general,
L[xnf(x)] = (−1)ndnF
dsn.
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Proof: We need:
If F (s) =∫ b
af(x, s) dx, then
dF
ds=
∫ b
a
∂f
∂sdx
F (s) =∫ ∞0
e−sxf(x) dx
dF
ds=
∫ ∞0
∂
∂s
(
e−sxf(x))
ds
=∫ ∞0
−xe−sxf(x) dx
= −∫ ∞0
e−sxxf(x) dx = −L[xf(x)]
Therefore, L[xf(x)] = −dF
ds
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Example: See p. 5.
F (s) =∫ 2
1
(
3x2s + 4xs2)
dx = 7s + 6s2
dF
ds(7s + 6s2) = 7 + 12s
∫ 2
1
∂
∂s
(
3x2s + 4xs2)
dx =
∫ 2
1
(
3x2 + 8xs)
dx =[
x3+4x2s]2
1= 7+12s
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Examples:
1. L[1] =1
s, s > 0
L[x] = L[x · 1] = − d
ds
(
1
s
)
=1
s2
L[x2] = L[x2 · 1] =d2
ds2
(
1
s
)
=2
s3
In general, L[xn] =n!
sn+1
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2. L[e2x] =1
s − 2, s > 2
L[xe2x] = − d
ds
(
1
s − 2
)
=1
(s − 2)2
L[x2e2x] =d2
ds2
(
1
s − 2
)
=2
(s − 2)3
——–
L[xne2x] =n!
(s − 2)n+1
In general: L[xneαx] =n!
(s − α)n+1
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IV. ”Translation”
If L[f(x)] = F (s), then
L[erxf(x)] = F (s − r).
Proof:
F (s) =∫ ∞0
e−sxf(x) dx
F (s − r) =∫ ∞0
e−(s−r)xf(x) dx
=∫ ∞0
e−sx+rxf(x) dx
=∫ ∞0
e−sxerxf(x) dx
= L[erxf(x)]
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Examples:
1. L[cos βx] = F (s) =s
s2 + β2
L[eαx cos βx] = F (s−α) =s − α
(s − α)2 + β2
2. L[sin βx] = F (s) =β
s2 + β2
L[eαx sin βx] = F (s−α) =β
(s − α)2 + β2
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Finding Laplace Transforms
Examples:
1. Find the Laplace transform of
f(x) = 3 + 4e3x − 2 cos 2x.
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2. Find the Laplace transform of
f(x) = 5x2 − 2e−3x sin 2x + 5xe4x
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3. Find the Laplace transform of the
solution of the initial-value problem:
y′ − 2y = 4x; y(0) = 3.
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4. Find the Laplace transform of the
solution of the initial-value problem:
y′′ − 2y′ + 5y = 2x + e−x;
y(0) = −2, y′(0) = 3.
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Ans:
Y (s) =
s2 + 2s + 2
s2(s + 1)(s2 − 2s + 5)+
7 − 2s
s2 − 2s + 5
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f(x) F (s) = L[f(x)]
cc
s, s > 0
eαx 1
s − α, s > α
cos βxs
s2 + β2, s > 0
sin βxβ
s2 + β2, s > 0
eαx cos βxs − α
(s − α)2 + β2, s > α
eαx sin βxβ
(s − α)2 + β2, s > α
xn, n = 1,2, . . .n!
sn+1, s > 0
xn eαx, n = 1,2, . . .n!
(s − α)n+1, s > α
x cos βxs2 − β2
(s2 + β2)2, s > 0
x sin βx2βs
(s2 + β2)2, s > 0
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Section 4.3. Inverse Laplace trans-
forms
Theorem: If f and g are continuous
functions on [0,∞), and if L[f(x)] =
L[g(x)], then f ≡ g; that is f(x) =
g(x) for all x ∈ [0,∞). (L is a one-
to-one operator.)
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Def. If F (s) is a given transform
and if the function f , continuous on
[0,∞), has the property that
L[f(x)] = F (s),
then f is called the inverse Laplace
transform of F (s), and is denoted by
f(x) = L−1[F (s)].
The operator L−1 is called the inverse
operator of L.
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The operator L−1 is linear; that is
L−1[F (s)+G(s)] = L−1[F (s)]+L−1[G(s)]
L−1[c F (s)] = cL−1[F (s)]
Finding inverse Laplace transforms:
Crucial step: Partial fraction de-
composition!
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Examples:
1. Given the initial-value problem
y′ + 2y = 3ex, y(0) = 4.
(a) Find the Laplace transform of the
solution.
(b) Find the solution by finding the in-
verse transform of the result in (a).
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(a)
y(x) = ex + 3e−2x
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2. F (s) =2s − 1
s4 + 4s2
Find L−1[F (s)] = f(x).
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Answer:
f(x) =1
2− 1
4x − 1
2cos 2x +
1
8sin 2x
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3. F (s) =s2 − 3s − 1
(s − 2)2(s + 4)
Find L−1[F (s)] = f(x).
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Answer:
f(x) =1
4e2x − 1
2xe2x +
3
4e−4x
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4. F (s) =2s − 1
(s + 1)(s2 − 4s + 13)
Find L−1[F (s)] = f(x).
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Answer:
f(x) = −1
6e−x+
1
6e2x cos 3x+
1
2e2x sin 3x
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5. Given the initial-value problem
y′′−y′−2y = sin 2x, y(0) = 1, y′(0) = 2.
(a) Find the Laplace transform of the
solution.
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(b) Find the solution by finding the in-
verse transform of the result in (a).
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Answer:
y =13
12e2x− 2
15e−x+
1
20cos 2x− 3
20sin 2x.
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6. Let y = y(x) be the solution of
the initial-value problem:
y′′−2y′−8y = 0, y(0) = α, y′(0) = 4.
Find α such that y = y(x) satisfies
limx→∞ y(x) = 0.
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