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CHAPTER 7
Laplace Transforms
1. Introduction: A Mixing Problem
Example. Initially, 30 kg of salt are dissolved in 1000 L of water in a tank.The tank has two input valves, A and B, and one exit valve C. At time t = 0,valve A is opened, delivering 6 L/min of a brine solution contining 0.4 kg of saltper liter. At t = 0 min, valve A is closed and valve B is opened, delivering 6L/min of brine at a concentration of 0.2 kg/L.The exit valve C, which emptiesthe tank at 6 L/min, maintains the contents of the tank at a constant volume.Assuming the solution is kept well-stirred, determine the amount of salt in thetank at all times t > 0.
This gives us an input rate of
g(t) =
(2.4, 0 < t < 10
1.2, t > 10.
128
1. INTRODUCTION: A MIXING PROBLEM 129
Then x(t) changes at a rate
d
dtx(t) = g(t)� 3x(t)
500or
dx
dt+
3
500x = g(t)
with ICx(0) = 30.
With the methods we have so far, noting that g(t) is constant on each of thetwo intervals it is defined for, we have two equations that need to be solvedby undetermined coe�cients. But there is another method, using the Laplacetransform. This is an alternate method of solving linear IVP’s of all orders.The idea:
We transform an IVP in the variable t (the t-domain) by integration into analgebraic equation in the variable s (the s-domain), solve that equation usingalgebra, and then translate that answer back to the original t-domain. Thefollowing chart compares the two methods.
130 7. LAPLACE TRANSFORMS
For now, the solution can be written as
x(t) = 400� 370e�3t/500 � 200 ·(
0, t 10
[1� e�3(t�10)/500], t � 10.
A graph of the solution follows:
2. DEFINITION OF THE LAPLACE TRANSFORM 131
2. Definition of the Laplace Transform
Definition (1 — The Laplace Transform). Let f(t) be a function whosedomain includes all of t � 0. The Laplace transform of f is the function Fdefined by the improper integral
F (s) :=
Z 1
0e�stf(t) dt.
The domain of F is the set of s-values for which the improper integral exists.
Notation.f(t)
L7�! F (s)
L{f} = F or L{f}(s) = F (s)
or a bit sloppy, but useful,
L{f(t)} = F (s).Example.
(1)
L{1} =
Z 1
0e�st · 1 dt = lim
N!1
h� e�st
s
iN
0= lim
N!1
h� e�sN
s+
1
s
i=
1
s, s > 0.
(2)
L{eat} =
Z 1
0e�steat dt =
Z 1
0e(a�s)t dt =
limN!1
he(a�s)t
a� s
iN
0= lim
N!1
he(a�s)N
a� s� 1
a� s
i=
1
s� a, s > a.
132 7. LAPLACE TRANSFORMS
(3)
L{sin bt} =
Z 1
0e�st sin bt dt =
sin bt e�st
+&
b cos bt �e�st
s�&
�b2 sin bt+ e�st
s2h� e�st
ssin bt� be�st
s2cos bt
i10� b2
s2
Z 1
0e�st sin bt dt =)
⇣1 +
b2
s2
⌘L{sin bt} =
limN!1
h� e�sN
ssin bN � be�sN
s2cos bN +
b
s2
i=
b
s2, s > 0 =)
L{sin bt} =b
s2· s2
s2 + b2, s > 0 =)
L{sin bt} =b
s2 + b2, s > 0.
2. DEFINITION OF THE LAPLACE TRANSFORM 133
(4) Let f(t) =
8><>:
2, 0 < t < 5
0, 5 < t < 10
e4t, t > 10
.
L{f(t)} =
Z 1
0e�stf(t) dt =
Z 5
0e�st · 2 dt +
Z 10
5e�st · 0 dt +
Z 1
10e�ste4t dt =
2
Z 5
0e�st dt + lim
N!1
Z N
10e(4�s)t dt =
2h� e�st
s
i5
0+ lim
N!1
he(4�s)t
4� s
iN
10=
2h1
s� e�5s
s
i+ lim
N!1
he(4�s)N
4� s� e10(4�s)
4� s
i=
2
s� 2e�5s
s+
e�10(s�4)
s� 4, s > 4.
Theorem (1 — Linearity of the Laplace Transform). Let f , f1 and f2 befunctions whose Laplace transforms exist for s > ↵ and let c be a constant.Then, for s > ↵,
L{f1 + f2} = L{f1} + L{f2},L{cf} = cL{f}.
The proof here follows easily from the same properties for integrals.
134 7. LAPLACE TRANSFORMS
Example.
L{11 + 5e4t � 6 sin 2t} =
11L{1} + 5L{e4t}� 6L{sin 2t} =
11 · 1
s+ 5 · 1
s� 4� 6 · 2
s2 + 4=
11
s+
5
s� 4� 12
s2 + 4, s > 4.
Definition. A function f(t) is said to have a jump discontinuity at t0 2(a, b) if f(t) is discontinuous at t0, but the one-sided limits
limt!t�0
f(t) and limt!t+0
f(t)
exist as finite numbers.
Definition (2 — Piecewise Continuous Functions).
A function f(t) is piecewise continuous on a finite interval [a, b] if it is contin-uous at each point of [a, b] except for at most a finite number of jump discon-tinuities.
A function f(t) is piecewise continuous on the infinite interval [0,1] if it ispiecewise continuous on every finite interval [0, T ].
There are functions for which the integral for the Laplace transform does not
converge. Examples are f(t) =1
t, which grows too fast near 0, and g(t) = et2,
which increases too rapidly as t!1.
However, roughly speaking, the Laplace transform of a piecewise continuousfunction exists, provided that the function does not grow “faster than an expo-nential.”
2. DEFINITION OF THE LAPLACE TRANSFORM 135
Definition (3 — Exponential Order). A function f(t) is said to be ofexponential order ↵ as t ! 1 if there are positive constants T and M suchthat
|f(t)| Me↵t for all t > T.
Notation. If f(t) is of exponential order ↵, we write
f(t) = O(e↵t) (as t!1).
O(e↵t) is read as “Big Oh of e↵t.”Example.
(1) f(t) = e5t sin 2t.|f(t)| = |e5t sin 2t| e5t =)
f(t) = O(e5t)
f(t) is exponential order 5.
136 7. LAPLACE TRANSFORMS
(2) f(t) = tn, t � 0.
et = 1 + t +t2
2!+
t3
3!+ · · · +
tn
n!+ · · · =)
et >tn
n!=) |tn| = tn < n!et =)
f(t) = O(et)
f(t) is exponential order 1.
Theorem (Long-Term Behavior of Functions of Exponential Order). Iff(t) = O(e↵t) as t!1, then for s > ↵,
limt!1
e�stf(t) = 0.
Proof.
|f(t)| Me↵t =) �Me↵t f(t) Me↵t =)�Me(↵�s)t e�stf(t) Me(↵�s)t t!1
=)0 lim
t!1e�stf(t) 0 =)
limt!1
e�stf(t) = 0.
⇤
Corollary (Long-Term Behavior of Laplace Transforms). If a functionf(t) is bounded and integrable on every finite interval [0, T ] and is of ex-ponential order ↵, then
lims!1
L{f}(s) = 0.
3. PROPERTIES OF THE LAPLACE TRANSFORM 137
Theorem (2 — Conditions for Existence of the Transform). If f(t) ispiecewise continuous on [0,1) and is of exponential order ↵, then L{f}(s)exists for s > ↵.
3. Properties of the Laplace Transform
Theorem (3 — Translation in s). If the Laplace transform L{f}(s) =F (s) exists for s > a, then
L{eatf(t)}(s) = F (s� a)
for s > ↵ + a.
Proof.
L{eatf(t)}(s) =
Z 1
0e�steatf(t) dt =
Z 1
0e�(s�a)tf(t) dt = F (s� a).
⇤
Example. We know
L{sin bt}(s) = F (s) =b
s2 + b2.
Then, by translation in s,
L{eat sin bt}(s) = F (s� a) =b
(s� a)2 + b2.
Theorem (4 — Laplace Transform of the Derivative). Let f(t) be di↵er-entiable on [0,1) and f 0(t) be piecewise continuous on [0,1), with bothof exponential order ↵. Then, for s > ↵,
L{f 0(t)}(s) = sL{f(t)}(s)� f(0).
138 7. LAPLACE TRANSFORMS
Example. L{sin bt}(s) =b
s2 + b2, s > 0. Find L{cos bt}(s).
|sin bt| 1 = 1e0t =) sin bt = O(e0t) =) ↵ = 0.
L{cos bt}(s) =1
bL{b cos bt}(s) =
1
b
hsL{sin bt}(s)� sin 0
i=
1
b· s · b
s2 + b2=
s
s2 + b2, s > 0.
The following theorems are needed for higher-order linear DE’s.
Theorem (Laplace Transform of the Second Derivative). Let f(t) and f 0(t)be continuous on [0,1) and let f 00(t) be piecewise continuous on [0,1),with all these functions of exponential order ↵. Then, for s > ↵,
L{f 00(t)}(s) = s2L{f(t)}(s)� sf(0)� f 0(0).
Theorem (5 — Laplace Transform of Higher-Order Derivatives).
Let f(t), f 0(t), . . . , f (n�1)(t) be continuous on [0,1) and let f (n)(t) bepiecewise continuous on [0,1), with all these functions of exponential or-der ↵. Then, for s > ↵,
L{f (n)(t)}(s) = snL{f(t)}(s)� s(n�1)f(0)� s(n�2)f 0(0)� · · ·� f (n�1)(0).
3. PROPERTIES OF THE LAPLACE TRANSFORM 139
Theorem (Transform of an Integral). If f(t) is piecewise continuous on[0,1) and of exponential order ↵, then
L
(Z t
0f(⌧ ) d⌧
)(s) =
1
sL{f(t)}(s)
for s > ↵.
Proof.
Let g(t) =
Z t
0f(⌧ ) d⌧ . Note that g(0) = 0 and g0(t) = f(t). Then
L{f(t)}(s) = L{g0(t)}(s) = sL
(Z t
0f(⌧ ) d⌧
)� 0 =)
L
(Z t
0f(⌧ ) d⌧
)=
1
sL{f(t)}(s)
⇤
Example.
L
(Z t
0e4⌧ d⌧
)=
1
sL{e4t} =
1
s· 1
s� 4=
1
s(s� 4).
140 7. LAPLACE TRANSFORMS
The following theorem shows that if F (s) is the Laplace transform of f(t),then F 0(s) is also a Laplace transform of some function of t. In fact,
F 0(s) = L{�tf(t)}(s).
More generally:
Theorem (Derivatives of the Laplace Transform). Suppose a function f(t)is piecewise continuous on [0,1) and of exponential order ↵. Let F (s) =L{f(t)}(s). Then
L{tnf(t)}(s) = (�1)ndnF
dsn(s).
Example.
L{e2t}(s) =1
s� 2.
d
ds
⇣ 1
s� 2
⌘=
d
ds
⇥(s� 2)�1
⇤= �(s� 2)�2.
d2
ds2
⇣ 1
s� 2
⌘= 2(s� 2)�3;
d3
ds3
⇣ 1
s� 2
⌘= �6(s� 2)�4 = � 6
(s� 2)4.
Thus
L{t3e2t}(s) = (�1)3⇣� 6
(s� 2)4
⌘=
6
(s� 2)4.
4. INVERSE LAPALCE TRANSFORM 141
Thus we have that multiplying a function by t corresponds to the negative ofthe derivative of the transform. Similarly, the following states that dividing fby t corresponds to an integral of its transform.
Theorem (Integral of a Transform). Suppose a function f(t) is piecewisecontinuous on [0,1) and of exponential order ↵. Furthermore, suppose
limt!0+
f(t)
texists and is finite. Then
Lnf(t)
t
o(s) =
Z 1
sF (u) du
where F (s) = L{f}(s).
4. Inverse Lapalce Transform
Definition (4 — Inverse Laplace Transform). An inverse Laplace transformof a function F (s) is the unique function f(t) that is continuous on [0,1) andsatisfies L{f(t)}(s) = F (s).
In case all functions that satisfy L{f(t)}(s) = F (s) are discontinuous on [0,1),we select a piecewise continuous function that satisfies L{f(t)}(s) = F (s).
In both cases, we writeL�1{F} = f.
Theorem (Theorem 7 — Linearity of the Inverse Transform). Assume thatL�1{F}, L�1{F1}, and L�1{F2} exist and are continuous on [0,1) andlet c be any constant. Then
L�1{F1 + F2} = L�1{F1} + L�1{F2}, and
L�1{cF} = cL�1{F}.In other words, L�1 is a linear operator.
142 7. LAPLACE TRANSFORMS
IfL{f(t)}(s) = F (s),
then alsoL�1{F (s)}(t) = f(t).
f(t)L�! �
L�1F (s)
Example.
L�1n� 1
s+
1
s2+
2
s + 1
o= �L�1
n1
s
o+ L�1
n 1
s2
o+ 2L�1
n 1
s + 1
o= �1 + t + 2e�t.
Problem (Page 374 # 6). Find L�1n 3
(2s + 5)3
o.
Solution.
L�1n 3
(2s + 5)3
o
=3
8L�1
n 1
(s + 5/2)3
o
=3
16t2e�5t/2.
⇤
5. SOLVING INITIAL VALUE PROBLEMS 143
Problem (Page 375 # 24). Find {L}�1n 7s2 � 51s + 84
(s� 1)(s2 � 4s + 13)
o.
Solution.
Using partial fractions,
7s2 � 51s + 84
(s� 1)(s2 � 4s + 13)=
5
s� 1+
2s� 19
s2 � 4s + 13
= 51
s� 1+
2s� 19
(s� 2)2 + 9
= 51
s� 1+ 2
s� 2
(s� 2)2 + 9� 5
3
(s� 2)2 + 9.
Then
L�1n 7s2 � 51s + 84
(s� 1)(s2 � 4s + 13)
o
= 5L�1n 1
s� 1
o+ 2L�1
n s� 2
(s� 2)2 + 9
o� 5L�1
n 3
(s� 2)2 + 9
o
= 5et + 2e2t cos 3t� 5e2t sin 2t.
⇤
5. Solving Initial Value Problems
Solving IVP’s – The Method of Laplace Transforms
(1) Take the Laplace transform of both sides of the DE.
(2) Use the initial conditions and properties of the transform to express thealgebraic equation obtained in Step 1 in terms of the transform of the solution.
(3) Solve the equation in Step 2 for the transform of the solution.
(4) Determine the solution by finding a continuous (or piecewise continuous)function with a transform the same as the one obtained in Step 3.
144 7. LAPLACE TRANSFORMS
Theorem (Lerch’s Theorem — Uniqueness of Inverse Laplace Transforms).Suppose f(t) and g(t) are continuous on [0,1) and of exponential order↵. If L{f(t)}(s) = L{g(t)}(s) for all s > ↵, then f(t) = g(t) for all t � 0.
Example. Solve y0 � y = e5t, y(0) = 0.
L{y0 � 5y}(s) = L{e5t}(s) =)L{y0}(s)� 5L{y}(s) = L{e5t}(s) =)
sL{y}(s)� y(0)� 5L{y}(s) = L{e5t}(s) =)
(s� 5)L{y}(s) =1
s� 5=)
L{y}(s) =1
(s� 5)2=)
y(t) = L�1n 1
(s� 5)2
o=)
y(t) = te5t.
5. SOLVING INITIAL VALUE PROBLEMS 145
Example. Solve y0 + y = t, y(0) = 1.
L{y0 + y}(s) = L{t}(s) =) L{y0}(s) + L{y}(s) = L{t}(s) =)sL{y}(s)� y(0) + L{y}(s) = L{t}(s) =)
sL{y}(s)� 1 + L{y}(s) =1
s2=)
(s + 1)L{y}(s) =1
s2+ 1 =
s2 + 1
s2=)
L{y}(s) =s2 + 1
s2(s + 1).
————————————————————————————-
s2 + 1
s2(s + 1)=
As + B
s2+
C
s + 1=
As2 + As + Bs + B + Cs2
s2(s + 1)=)
8><>:
A + C = 1
A + B = 0
B = 1
=) A = �1 =) C = 2.
————————————————————————————-Then
L{y}(s) = �1
s+
1
s2+
2
s + 1=)
y(t) = �L�1n1
s
o+ L�1
n 1
s2
o+ 2L�1
n 1
s + 1
o= �1 + t + 2e�t
is the solution.
146 7. LAPLACE TRANSFORMS
Example. Solve y0(t) + y(t) = sin t, y(0) = 1.
L{y0(t)}(s) + L{y(t)}(s) = L{sin t}(s) =)
sL{y(t)}(s)� y(0) + L{y(t)}(s) =1
s2 + 1=)
(s + 1)L{y(t)}(s)� 1 =1
s2 + 1=)
(s + 1)L{y(t)}(s) =1
s2 + 1+ 1 =
s2 + 2
s2 + 1=)
L{y(t)}(s) =s2 + 2
(s2 + 1)(s + 1).
————————————————————————————-
s2 + 2
(s2 + 1)(s + 1)=
As + B
s2 + 1+
C
s + 1=
As2 + As + Bs + B + Cs2 + C
(s2 + 1)(s + 1)=)
8><>:
A + C = 1
A + B = 0
B + C = 2
=) A = �1
2, B =
1
2, C =
3
2.
————————————————————————————-Then
L{y(t)}(s) = �1
2· s
s2 + 1+
1
2· 1
s2 + 1+
3
2· 1
s + 1=)
y(t) = �1
2L�1
n s
s2 + 1
o+
1
2L�1
n 1
s2 + 1
o+
3
2L�1
n 1
s + 1
o=)
y(t) = �1
2cos t +
1
2sin t +
3
2e�t
is the solution.
5. SOLVING INITIAL VALUE PROBLEMS 147
Example.
(1) Solve y00 + 4y = 0, x(0) = y0(0) = 2.
L{y00}(s) + 4L{y}(s) = L{0}(s) =)s2L{y}(s)� sx(0)� x0(0) + 4L{y}(s) = 0 =)
(s2 + 4)L{y}(s)� 2s� 2 = 0 =)
L{y}(s) =2s + 2
s2 + 4= 2 · s
s2 + 4+
2
s2 + 4=)
y(t) = 2L�1n s
s2 + 4
o+ L�1
n 2
s2 + 4
o=)
y(t) = 2 cos 2t + sin 2t
is the solution.
148 7. LAPLACE TRANSFORMS
Example. Solve y00 � 2y0 + 5y = �8e�t, y(0) = 2, y0(0) = 12.
L{y00}(s)� 2L{y0}(s) + 5L{y}(s) = �8L{e�t}(s) =)
s2L{y}(s)� sy(0)� y0(0)� 2⇥sL{y}(s)� y(0)
⇤+ 5L{y}(s) =
�8
s + 1=)
s2L{y}(s)� 2s� 12� 2sL{y}(s) + 4 + 5L{y}(s) =�8
s + 1=)
(s2 � 2s + 5)L{y}(s) = 2s + 8� 8
s + 1=
2s2 + 10s
s + 1=)
L{y}(s) =2s2 + 10s
(s2 � 2s + 5)(s + 1)=)
————————————————————————————-
2s2 + 10s
(s2 � 2s + 5)(s + 1)=
As + B
s2 � 2s + 5+
C
s + 1=
As2 + As + Bs + B + Cs2 � 2Cs + 5C
(s2 � 2s + 5)(s + 1)=)
8><>:
(1)A + C = 2
(2)A + B � 2C = 10
(3)B + 5C = 0
=)
8>>><>>>:
A + C = 2
[(2)� (3)]A� 7C = 10
����������8C = �8
=)
8><>:
C = �1
A = 3
B = 5
————————————————————————————-
L{y}(s) =3s + 5
s2 � 2s + 5� 1
s + 1=
3s� 1
(s� 1)2 + 4+ 4
2
(s� 1)2 + 4� 1
s + 1=)
y(t) = 3L�1n s� 1
(s� 1)2 + 4
o+ 4L�1
n 2
(s� 1)2 + 4
o� L�1
n 1
s + 1
o=)
y(t) = 3et cos 2t + 4et sin 2t� e�t
is the solution.
5. SOLVING INITIAL VALUE PROBLEMS 149
Example. Solve x00 + x = t2 + 2, x(0) = 1, x0(0) = �1.
L{x00}(s) + L{x}(s) = L{t2}(s) + 2L{1}(s) =)
s2L{x}(s)� sx(0)� x0(0) + L{x}(s) =2
s3+
2
s=)
s2L{x}(s)� s + 1 + L{x}(s) =2s2 + 2
s3=)
(s2 + 1)L{x}(s) = s� 1 +2s2 + 2
s3=
s4 � s3 + 2s2 + 2
s3=)
L{x}(s) =s4 � s3 + 2s2 + 2
s3(s2 + 1)=)
————————————————————————————-
s4 � s3 + 2s2 + 2
s3(s2 + 1)=
A
s+
B
s2+
C
s3+
Ds + E
s2 + 1=
As4 + As2 + Bs3 + Bs + Cs2 + C + Ds4 + Es3
s3(s2 + 1)=)
8>>>>>><>>>>>>:
A + D = 1
B + E = �1
A + C = 2
B = 0 =) E = �1
C = 2 =) A = 0 =) D = 1————————————————————————————-
L{x} =2
s3+
s
s2 + 1� 1
s2 + 1=)
x(t) = L�1n 2
s3
o+ L�1
n s
s2 + 1
o� L�1
n 1
s2 + 1
o=)
x(t) = t2 + cos t� sin t
is the solution.
150 7. LAPLACE TRANSFORMS
6. Transforms of Discontinuous and Periodic Functions
Definition (5 — Unit Step Function). The unit step (Heaviside)
function u(t) is defined by
u(t) =
(0, t < 0
1, t > 0.
Note that this function is undefined at 0. Also, if the function jumps att = a,
u(t� a) =
(0, t < a
1, t > a,
and if the height of the jump is a constant M ,
Mu(t� a) =
(0, t < a
M, t > a.
6. TRANSFORMS OF DISCONTINUOUS AND PERIODIC FUNCTIONS 151
Definition (Rectangular Window Function). The rectangular window
function ⇧a,b(t) is defined by
⇧a,b(t) = u(t� a)� u(t� b) =
8><>:
0, t < a
1, a < t < b
0, t > b
.
We can also use this function to choose a window for a function:
152 7. LAPLACE TRANSFORMS
Any piecewise continuous functioncan be expressed in terms of window andstep functions.
Example.
f(t) =
8><>:
t, 0 < t < 1
t2, 1 < t < 3
3 ln(t), t > 3
can be written as
f(t) = t⇧0,1(t) + t2⇧1,3(t) + 3 ln(t)u(t� 3)
= tu(t) + (t2 � t)u(t� 1) + (3 ln t� t2)u(t� 3).
6. TRANSFORMS OF DISCONTINUOUS AND PERIODIC FUNCTIONS 153
Example.
For a � 0,
L{u(t� a)} =
Z 1
0e�stu(t� a) dt =
Z 1
ae�st dt =
limN!1
h� e�st
s
iN
a= lim
N!1
he�sN
�s� e�as
�s
i=
e�as
s, s > 0.
Also,
L�1ne�as
s
o(t) = u(t� a) and
L�1n
⇧a,b(t)o
(s) = L�1nu(t� a)� u(t� b)
o=
e�sa � e�sb
s, 0 < a < b.
Just as the shift in s describes the e↵ect on the Laplace transform of multi-plying a function by eat, the next theorem illustrates an analogousd e↵ect ofmultiplying the Lapace transform of a function by e�as.
Theorem (8 — Translation in t). Let F (s) = L{f}(s) exist for s > ↵ �0. If a is a positive constant, then
L{f(t� a)u(t� a)}(s) = e�asF (s),
and, conversely, an inverse Laplace transform of e�asF (s) is given by
L�1ne�asF (s)
o(t) = f(t� a)u(t� a).
Proof.
L{f(t� a)u(t� a)} =
Z 1
0e�stf(t� a)u(t� a) dt =
Z 1
ae�stf(t� a) dt =
v = t� a, dv = dtZ 1
0e�s(v+a)f(v) dv = e�as
Z 1
0e�svf(v) dv = e�asF (s).
⇤
154 7. LAPLACE TRANSFORMS
Corollary.
L{g(t)u(t� a)}(s) = e�asL{g(t + a)}(s).Proof.
Identify g(t) with f(t� a), so f(t) = g(t + a). ⇤
Example. Find L{(sin t)u(t� ⇡)}(s).
Let g(t) = sin t and a = ⇡. Then
g(t + a) = g(t + ⇡) = sin(t + ⇡) = � sin t.
Thus,
L{g(t + a)}(s) = �L{sin t(s)} = � 1
s2 + 1=)
L{(sin t)u(t� ⇡)}(s) = �e�⇡s 1
s2 + 1.
Problem (Page 394 # 14). Find L�1n e�3s
s2 + 9
o(t).
Solution.
We have F (s) =1
s2 + 9, so f(t) = L�1
nF (s)
o(t) =
sin 3t
3. Thus
L�1n e�3s
s2 + 9
o(t) = f(t� 3)u(t� 3) =
sin(3t� 9)
3u(t� 3).
⇤
6. TRANSFORMS OF DISCONTINUOUS AND PERIODIC FUNCTIONS 155
Example (Abrupt Change). We begin with a filled 500 L tank with 2 kg/Lof salt. There are two input valves. Valve A flows at 12 L/min with 4 kg/L ofsalt. Valve B flows at 12 L/min with 6 kg/L of salt. There is one output valve,Valve C, which also flows at 12 L/min.
The inflow starts with Valve A for 10 minutes, switches to valve B for the next10, then back to A for the duration. Valve C is open continuously during thistime.
Find the amount of salt in the tank at any time.
Let t = time in minutes.
Let x(t) = kg of salt in tank at time t.
dx
dt= 12
(4 for 0 t < 10, t > 20
6 10 < t < 20
)� 12
x(t)
500
x0 +3
125x = 12
⇥4 + 2u(t� 10)� 2u(t� 20)
⇤
156 7. LAPLACE TRANSFORMS
We need to solve
x0 +3
125x = 48 + 24u(t� 10)� 24(t� 20), x(0) = 1000.
L{x0}(s) +3
125L{x}(s) = 48L{1}(s) + 24L{u(t� 10)}(s)
� 24L{u(t� 20)}(s) =)
sL{x}(s)� x(0) +3
125L{x}(s) = 48L{1}(s) + 24L{u(t� 10)}(s)
� 24L{u(t� 20)}(s) =)⇣s +
3
125
⌘L{x}(s)� 1000 = 48 · 1
s+ 24 · e�10s
s� 24 · e�20s
s=)
L{x}(s) = 10001
s + 3125
+ 481
s(s + 3125)
+ 24e�10s 1
s(s + 3125)
� 24e�20s 1
s(s + 3125)
=)
x(t) = 1000L�1n 1
s + 3125
o(t) + 48L�1
(1
s⇣s + 3
125
⌘)
(t)
+ 24L�1
(e�10s 1
s⇣s + 3
125
⌘)
(t)� 24L�1
(e�20s 1
s⇣s + 3
125
⌘)
(t).
6. TRANSFORMS OF DISCONTINUOUS AND PERIODIC FUNCTIONS 157
The following is needed in three di↵erent terms.————————————————————————————-
1
s⇣s + 3
125
⌘ =A
s+
B
s + 3125
=As + 3
125A + Bs
s⇣s + 3
125
⌘ =)
(A + B = 03
125A = 1=) A =
125
3and B = �125
3————————————————————————————-
This gives us
L�1
(1
s⇣s + 3
125
⌘)
(t) = L�1
(1253
s�
1253
s + 3125
)(t) =
125
3L�1
(1
s
)(t)� 125
3L�1
(1
s + 3125
)(t) =
125
3� 125
3e�
3125t.
Recalling
x(t) = 1000L�1n 1
s + 3125
o(t) + 48L�1
(1
s⇣s + 3
125
⌘)
(t)
+ 24L�1
(e�10s 1
s⇣s + 3
125
⌘)
(t)� 24L�1
(e�20s 1
s⇣s + 3
125
⌘)
(t),
the solution is
x(t) = 1000e�3
125t+48⇣125
3� 125
3e�
3125t
⌘+24
⇣125
3� 125
3e�
3125(t�10)
⌘u(t�10)
� 24⇣125
3� 125
3e�
3125(t�20)
⌘u(t� 20) =)
x(t) = 1000h2�e�
3125t+
⇣1�e�
3125(t�10)
⌘u(t�10)�
⇣1�e�
3125(t�20)
⌘u(t�20)
i.
158 7. LAPLACE TRANSFORMS
Maple. See abrupt-Heaviside.mw or abrupt-Heaviside.pdf.
Definition (Periodic Function). A function f(t) is said to be periodic ofperiod T (6= 0) if
f(t + T ) = f(t)
for all t in the domain of f .
Example.
f(t) =
(1, 0 < t < 1
�1, 1 < t < 2, and f(t) has period 2.
A notation for the windowed version of a periodic function is
fT (t) = f(t)⇧0,T (t) = f(t)⇥u(t)� u(t� T )
⇤=
(f(t), 0 < t < T
0, otherwise.
6. TRANSFORMS OF DISCONTINUOUS AND PERIODIC FUNCTIONS 159
The Laplace transform of fT (t) is given by
FT (s) =
Z 1
0e�stfT (t) dt =
Z T
0e�stf(t) dt.
Theorem (9 — Transform of Periodic Function). If f has period T andis piecewise continuous on [0, T ], then the Laplace transforms
F (s) =
Z 1
0e�stf(t) dt and FT (s) =
Z T
0e�stf(t) dt
are related by
FT (s) = F (s)⇥1� e�sT
⇤or F (s) =
FT (s)
1� e�sT.
Problem (Page 395 # 38). Solve
y00 + 5y0 + 6y = tu(t� 2), y(0) = 0, y0(0) = 1.Solution.
With Y (s) = L{y(t)}(s),
L{y00(t)}(s) + 5L{y0(t)}(s) + 6L{y(t)}(s) = L{tu(t� 2)}(s) =)
s2Y (s)� sy(0)� y0(0) + 5⇥sY (s)� y(0)
⇤+ 6Y (s)
L{(t� 2)u(t� 2)} + 2L{u(t� 2)} =)
s2Y (s)� y0(0) + 5⇥sY (s)� 1
⇤+ 6Y (s) =
e�2s
s2+ 2
e�2s
s=)
(s2 + 5s + 6)Y (s) = 1 +e�2s(2s + 1)
s2=)
Y (s) =1
(s + 2)(s + 3)+
e�2s(2s + 1)
s2(s + 2)(s + 3)=) (using partial fractions)
160 7. LAPLACE TRANSFORMS
Y (s) =1
s + 2� 1
s + 3+ e�2s
1
6s2+
7
36s� 3
4(s + 2)+
5
9(s + 3)
�=)
y(t) = L�1
(1
s + 2� 1
s + 3+ e�2s
1
6s2+
7
36s� 3
4(s + 2)+
5
9(s + 3)
�)=)
y(t) = e�2t � e�3t +
7
36+
t� 2
6� 3e�2(t�2)
4+
5e�3(t�2)
9
�u(t� 2).
⇤
8. Impulses and the Dirac Delta Function
To deal with violent forces of short duration, we introduce the Dirac DeltaFunction.
Definition. The Dirac Delta function �(t) is characterized the followingtwo properties:
(⇤) �(t) =
(0, t 6= 0
1, t = 0,
and
(⇤⇤)Z 1
�1f(t)�(t) dt = f(0)
for any function f(t) that is continuous on an open interval containing t = 0.Note.
(1) By shifting the argument of �(t), we have
(⇤0) �(t� a) =
(0, t 6= a
1, t = a,
and
(⇤⇤0)Z 1
�1f(t)�(t� a) dt = f(a)
for any function f(t) that is continuous on an open interval containing t = a.
8. IMPULSES AND THE DIRAC DELTA FUNCTION 161
(2) �(t � a) is not a function in the usual sense, but is called a generalizedfunction or distribution.
(3)
Z 1
�1�(t� a) = 1. This is a special case of (⇤⇤0) by taking f(t) = 1.
(4)
Z t
�1�(t� a) dt =
(0, t < a
1, t > a= u(t� a).
(5) Di↵erentiating the above with respect to t, and using the FundamentalTheorem of Calculus,
�(t� a) = u0(t� a).Example.
For a > 0,
L{�(t� a)} =
Z 1
0e�st�(t� a) dt =
e�st �(t� a)+&
�se�st � u(t� a)he�stu(t� a)
i10
+ s
Z 1
0e�stu(t� a) dt =
limT!1
he�sTu(T � a)
i+ s · e�sa
s= e�sa, s > 0.
Maple. See abrupt-Heaviside.mw or abrupt-Heaviside.pdf.
162 7. LAPLACE TRANSFORMS
Example. A rancher has 1600 longhorn cattle, with a per capita growthrate of 8% in the herd. The rancher decides to get rid of this herd in two yearstime by selling the same number every 6 months starting 6 months from now.How many should be sold at a time?
Let p(t) = # of cattle at time t (in years).
The input or growth rate is .08p.
Let N = # sold each 6 months.
Then the total number sold at time t is
n(t) = Nu⇣t� 1
2
⌘+ Nu(t� 1) + Nu
⇣t� 3
2
⌘+ Nu(t� 2).
Then the output or selling rate is
n0(t) = N�⇣t� 1
2
⌘+ N�(t� 1) + N�
⇣t� 3
2
⌘+ N�(t� 2).
The DE that then models this isdp
dt= .08p�N�
⇣t� 1
2
⌘�N�(t� 1)�N�
⇣t� 3
2
⌘�N�(t� 2).
Then our IVP is
8. IMPULSES AND THE DIRAC DELTA FUNCTION 163
p0�.08p = �N�⇣t�1
2
⌘�N�(t�1)+N�
⇣t�3
2
⌘�N�(t�2). p(0) = 1600 =)
L{p0}(s)� L{.08p}(s) =
�NhL
n�⇣t�1
2
⌘o(s)+L
n�⇣t�1
⌘o(s)+L
n�⇣t�3
2
⌘o(s)+L
n�⇣t�2
⌘o(s)
i
=) sL{p}(s)� p(0)� .08L{p}(s) =
�NhL
n�⇣t�1
2
⌘o(s)+L
n�⇣t�1
⌘o(s)+L
n�⇣t�3
2
⌘o(s)+L
n�⇣t�2
⌘o(s)
i
=) (s� .08)L{p}(s)� 1600 = �Nhe�
12s + e�s + e�
32s + e�2s
i=)
L{p}(s) =1600
s� .08�N
h e�12s
s� .08+
e�s
s� .08+
e�32s
s� .08+
e�2s
s� .08
i=)
p(t) = 1600L�1n 1
s� .08
o(t)�N
hL�1
ne�
12s 1
s� .08
o(t)
+L�1ne�s 1
s� .08
o(t)+L�1
ne�
32s 1
s� .08
o(t)+L�1
ne�2s 1
s� .08
o(t)
i=)
p(t) = 1600e.08t �Nhe.08
�t�1
2
�u⇣t� 1
2
⌘+ e.08
�t�1
�u⇣t� 1
⌘+
e.08�t�3
2
�u⇣t� 3
2
⌘+ e.08
�t�2
�u⇣t� 2
⌘iTo find N , we use the fact that p(2+) = 0.
N =1600e.16
e.12 + e.08 + e.04 + 1⇡ 442.
Maple. See abrupt-Dirac.mw or abrupt-Dirac.pdf.
164 7. LAPLACE TRANSFORMS
Example. An underground storage tank contains 1000 gal of gasoline. Attime t = 0, a pump is turned on and gas is pumped into the tank at 50 gal/min.The pump turns o↵ automatically when the 6000 gal capacity of the tank isreached. A leak in the tank develops at t = 4 minutes with a model for thegallons leaked being g(t) = t2 � 8t + 17, t � 4.
(a) Express the gallons leaked as a step function.
g(t) =
(0, 0 t < 4
t2 � 8t + 17, t > 4
= (t2 � 8t + 17)u(t� 4).
(b) Express the number of gallons in the tank in terms of t.
Let x(t) = the number of gallons in the tank at time t.
x(t) = 1000 + 50t� (t2 � 8t + 17)u(t� 4), 0 t < 4, t > 4.
(c) Will the pump shut o↵ automatically?
For t > 4,x(t) = 983 + 58t� t2.
x0(t) = 58� 2t = 0 () t = 29.
x00(t) = �2 =) x(29) = 1824 is the maximum number of gal.
Thus the pump never turns o↵.
(d) When does the tank become empty?
983 + 58t� t2 = 0 ()t = 29 ± 4
p114 ()
t ⇡ 71.7 or t ⇡ �13.7.
The tank will empty in 71.7 minutes.
8. IMPULSES AND THE DIRAC DELTA FUNCTION 165
(e) Model the rate of change of gallons in the tank.
We have
g(t) = (t2 � 8t + 17)u(t� 4), 0 t <1, t 6= 4 =)g0(t) = (2t� 8)u(t� 4) + (t2 � 8t + 17)�(t� 4)
dx
dt= 50�
h(2t� 8)u(t� 4) + (t2 � 8t + 17)�(t� 4)
i, x(0) = 1000
x0 = 50� (2t� 8)u(t� 4)� (t2 � 8t + 17)�(t� 4), x(0) = 1000
(f) For what times is the equation in part (e) valid?
The equation is valid until the tank overflows or becomes empty.
(g) Find a formula for the number of gallons of gasoline in the tank.
L{(t2 � 8t + 17)�(t� 4)}(s) =Z 1
0e�st(t2 � 8t + 17)�(t� 4) dt =
limN!1
"Z N
0e�st(t2 � 8t + 17)�(t� 4) dt
#=
u = e�st(t2 � 8t + 17) dv = �(t� 4) dt
du = (�se�st(t2 � 8t + 17) + e�st(2t� 8)) dt v = u(t� 4)
limN!1
"✓e�st(t2 � 8t + 17)u(t� 4)
◆����N
0
+ s
Z N
0e�st(t2 � 8t + 17)u(t� 4) dt�
Z T
0e�st(2t� 8)u(t� 4) dt
#=
166 7. LAPLACE TRANSFORMS
limN!1
"0+s
Z N
0e�st(t2�8t+17)u(t�4) dt�
Z N
0e�st(2t�8)u(t�4) dt
#=
limN!1
"s
Z N
4e�st(t2 � 8t + 17) dt�
Z N
4e�st(2t� 8) dt
#=
✓e�4s +
2e�4s
s2
◆� 2e�4s
s2= e�4s.
Maple. See Dirac integral.mw or Dirac integral.pdf.
Then
x0 = 50� (2t� 8)u(t� 4)� (t2 � 8t + 17)�(t� 4), x(0) = 1000 =)
L{x0}(s) = L{50}(s)�L{(2t�8)u(t�4)}(s)�L{(t2�8t+17)�(t�4)}(s) =)
sL{x}(s)� x(0) = 50L{1}(s)� 2L{(t� 4)u(t� 4)}(s)� e�4s =)
sL{x}(s)� 1000 =50
s� 2e�4sL{t}(s)| {z }
shift in t
�e�4s =)
sL{x}(s) = 1000 +50
s� 2e�4s 1
s2� e�4s =)
L{x}(s) =1000
s+
50
s2� 2e�4s
s3� e�4s
s=)
x = 1000L�1n1
s
o(t) + 50L�1
n 1
s2
o(t)� L�1
ne�4s 2
s3
o(t)� L�1
ne�4s
s
o(t) =)
x = 1000� 50t� (t� 4)2u(t� 4)� u(t� 4) =)x(t) = 1000 + 50t� (t2 � 8t + 17)u(t� 4)
is the number of gallons in the tank after t minutes until the tank is empty.