t = 0 min, valve a is closed and valve b is opened, …facstaff.cbu.edu/wschrein/media/m231...

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.


(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.


(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.


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.”


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.


(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).


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).


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.


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.


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.


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.


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.


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.


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.


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.


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:


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).


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).

⇤


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).

⇤


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)

⇤


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).


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.


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.


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)


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.


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


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.


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.


(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

#=


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.

t = 0 min, valve a is closed and valve b is opened, …facstaff.cbu.edu/wschrein/media/m231...

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