method of applied math
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Lecture 1 Sujin Khomrutai – 1 / 23
Method of Applied MathLecture 7: Laplace Transform
Sujin Khomrutai, Ph.D.
Shift in t
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 2 / 23
Definition. Let a be a positive constant. The function
f(t− a)H(t− a) =
{
0 t < a
f(t− a) t ≥ a
is called the shifting of f(t) by a.
EX. (t− 1)2H(t− 1)
et−3H(t− 3)
sin(2t− 2π)H(t− π) = sin(2(t− π))H(t− π)
Shift in t
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 3 / 23
Theorem. Let F (s) = L[f(t)] and a a positive constant. Then
L[f(t− a)H(t− a)] = e−asF (s).
Thus
L−1[e−asF (s)] = f(t− a)H(t− a).
Proof. By definition,
L[f(t− a)H(t− a)] =
∫
∞
0
e−stf(t− a)H(t− a)dt
=
∫
∞
a
e−stf(t− a)dt.
Shift in t
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 4 / 23
Use the change of variable x = t− a, the integral becomes
∫
∞
a
e−stf(t− a)dt =
∫
∞
0
e−s(x+a)f(x)dx = e−saF (s).
So
L[f(t− a)H(t− a)] = e−asF (s).
Example 1
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 5 / 23
EX. Find the Laplace transform for each of the following functions
1. f(t) = (t− 3)2H(t− 3)
2. g(t) =
{
0 0 < t < 1
et−1 t ≥ 1
3. h(t) =
{
0 0 < t < 4
2t− 8 t ≥ 4
Shift in t
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 6 / 23
Note. A function of the form
f(t) =
{
f1(t) 0 < t < a
f2(t) t ≥ a
is equal to
f(t) = f1(t)(1−H(t− a)) + f2(t)H(t− a)
= f1(t)− f1(t)H(t− a) + f2(t)H(t− a)
EX.
f(t) =
{
t− 5 0 < t < 2
cos t t ≥ 2= (t−5)(1−H(t−2))+(cos t)H(t−2)
Multiplication with Heaviside Function
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 7 / 23
Theorem.
L[f(t)H(t− a)] = e−asL[f(t + a)].
Proof. Let g(t) = f(t+ a). Then g(t− a) = f(t) hence
f(t)H(t− a) = g(t− a)H(t− a).
By the previous theorem,
L[f(t)H(t− a)] = L[g(t− a)H(t− a)] = e−asG(s)
Since G(s) = L[g(t)] = L[f(t + a)], the desired identity is true.
Example 2
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 8 / 23
EX. Find the Laplace transform for each of the following functions
1. f(t) = tH(t− 3)
2. g(t) = e−tH(t− 1)
3. h(t) =
{
1 0 < t < 7
cos t t ≥ 7
Example 3
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 9 / 23
EX. Find the inverse Laplace transform for each of the followingfunctions
1. F (s) =e−3s
s− 5
2. G(s) =e−5s
s2
3. (∗) P (s) =e−2s
s(s2 + 1)
Example 4
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 10 / 23
EX. Solve the IVP:
y′′ + 4y = f(t), y(0) = 1, y′(0) = 0.
where
f(t) =
{
0 if 0 < t < 4
3 if t ≥ 4
ANS.
y(t) = cos 2t+3
4(1− cos(2t− 8))H(t− 4)
Impulse functions
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 11 / 23
Definition Let a ≥ 0. The unit impulse function at a orDirac delta function at a is
δ(t− a) = limε→0
1
ε[H(t− a)−H(t− a− ε)].
Note. It can be proved that the following identity holds
∫
∞
0
f(t)δ(t− a) dt = f(a),
for any continuous function f .
Laplace Transform of δ(t− a)
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 12 / 23
Theorem For each a ≥ 0, we have
L [δ(t− a)] = e−as.
Thus
L−1[
e−as]
= δ(t− a).
In particular,
L[δ(t)] = 1, L−1[1] = δ(t).
Example 5
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 13 / 23
EX. Find the Laplace transform for each of the following functions
1. f(t) = 2δ(t)2. g(t) = δ(t− 1) + 3δ(t− 5)3. h(t) = 3δ(t− 4)− δ(t− 8)
Example 6
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 14 / 23
EX. Find the inverse Laplace transform for each of the followingfunctions
1. F (s) = 72. G(s) = 3e−4s
3. P (s) = 2− e−s + 3e−2s
Example 7
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 15 / 23
EX. Solve the IVP
y′′ + 9y = δ(t− 1), y(0) = y′(0) = 0
ANS.
y(t) =1
3sin(3t− 3)H(t− 1)
Question 1
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 16 / 23
Q. Find the Laplace transform of
1. f(t) = t2H(t− 3)
2. g(t) = 2δ(t)− 5δ(t− 9)
Question 2
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 17 / 23
Q. Find the inverse Laplace transform of
1. F (s) =e−9s
s2 + 1
2. G(s) =e2s + 5
e3s
Convolution
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 18 / 23
Q. Given two Laplace transforms
F (s) = L[f(t)], G(s) = L[g(t)],
does there exists a function/formula having the Laplace transform
F (s)G(s)?
Definition. The convolution of two function f(t), g(t) is thefunction, denoted f ∗ g, given by
(f ∗ g)(t) =
∫
t
0
f(t− τ)g(τ)dτ.
Example 8
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 19 / 23
EX. Find the following convolutions
1. 1 ∗ 1
2. 1 ∗ et
3. t ∗ t
Convolution
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 20 / 23
Theorem.
L[(f ∗ g)(t)] = F (s)G(s)
so
L−1[F (s)G(s)] = (f ∗ g)(t) =
∫
t
0
f(t− τ)g(τ)dτ.
Proof. We have
L[(f ∗ g)(t)] =
∫
∞
0
e−st
∫
t
0
f(t− τ)g(τ)dτdt = F (s)G(s).
Example 9
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 21 / 23
EX. Find the Laplace transform for each of the following functions
1. f(t) = (t ∗ sin t)
2. g(t) =
∫
t
0
(t− τ)eτdτ
3. h(t) =
∫
t
0
et−τ cos 2τdτ
Example 10
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 22 / 23
EX. For each of the following functions, express the inverseLaplace transform in terms of the convolution
1. F (s) =s
(s2 + 4)(s2 + 1)
2. G(s) =1
s(s2 + 1)
3. P (s) =1
s4(s− 5)
Example 11
Prop 5: t-shifting
EX 1.
EX 2.
EX 3.
EX 4.Impulse and Diracdelta
Laplace δ(t − a)
EX 5.
EX 6.
EX 7.
Q 1.
Q 2.
Convol.
EX 8.
Thm. Conv
EX 9.
EX 10.
EX 11.
Lecture 1 Sujin Khomrutai – 23 / 23
EX. Using the convolution, find the solution formula for each ofthe following IVPs
1. y′′ − 5y′ + 6y = f(t), y(0) = y′(0) = 0
2. y′′ + 10y′ + 24y = f(t), y(0) = 1, y′(0) = 0
3. y′′ − 4y′ − 5y = f(t), y(0) = 2, y′(0) = 1