Faculty of Engineering
Department of Mechanical Engineering
[Mathematics (3)]
Prepared by
Dr. Mohamed Ramadan ZeenEldeen
Dr. Karima Mohamed Oraby
Academic Year
2021-2022
Table of contents
Chapter one (Functions of Several Variables)….……………1
Chapter Two (Multiple Integral)………………………..………..66
Chapter Three (Infinite series)………………………..………..111
Chapter Four (Ordinary Differential Equations)……………..173
Chapter five (Laplace Transform ………………………………219
Chapter Six (Partial Differential Equations in Cartesian
Coordinates)……………………………………………………….244
References.……………………………………………………......276
Chapter One : Function of Several Variables
Chapter one
Function of Several variabels
Introduction
Through our previous study of the function in one variable x
Which is in the form y = f ( x ) Where x is the independent variable
and y is the dependent variables varies depending on the variable x
In this section we define functions of more than one independent variable
and discuss ways to graph them. Real-valued functions of several
independent real variables are defined similarly to functions in the singlevariable case. Points in the domain are ordered pairs (triples, quadruples,
n-tuples) of real numbers, and values in the range are real numbers as we
have worked with all along.
Definition: Suppose D is a set of n-tuples of real numbers
( x1 , x2 ,... , xn ) . A real-valued function f on D is a rule that assigns a
unique (single) real number
w = f ( x1 , x2 ,... , xn )
to each element in D. The set D is the function's domain. The set of wvalues taken on by f is the function's range. The symbol w is the
dependent variable of f, and f is said to be a function of the n
independent variables x1 to x n . We also call the x j ' s the function's
input variables and call w the function's output variable.
Functions of Two Variables
If 𝑓 is a function of two independent variables, we usually call the
independent variables x and y and the dependent variable z and we
picture the domain of 𝑓 as a region in the 𝑥𝑦 −plane (Figure .1).
If f
is a function of three independent variables, we call the
independent variables
x , y , and z
, and the dependent variable
w , and we picture the domain as a region in space.
1
Chapter One : Function of Several Variables
In applications, we tend to use letters that remind us of what the
variables stand for. To say that the volume of a right circular cylinder is
a function of its radius and height, we might write V = f (r , h) . To be
more specific, we might replace the notation f ( r , h) by the formula
that calculates the value of 𝑉 from the values of 𝑟 and ℎ, and write
V = r 2 h . In either case, r and h would be the independent
variables and 𝑉 the dependent variable of the function.
Figure .1.
Example ( 1 )
The equation of the sphere in the space which have unit radius and its
center is the origin has the form
x2 + y2 + z2 = 1
So
z = 1− x2 − y2
If we take the positive sigin , then
z = 1− x2 − y2
Represented the distance between any point on the upper surface of the
ball and the horizontal plane 𝑥𝑜𝑦
2
Example ( 3 ): Describe the domain of the function z = y − x
Solution: Since f is defined only where y − x 0 , the domain
is the closed, unbounded region shown in Figure 2. The parabola
2
2
Chapter One : Function of Several Variables
y = x 2 is the boundary of the domain. The points above the parabola
make up the domain's interior.
Figure .2
Example ( 3): These are functions of two variables. Note the restriction
that may apply to their domain in order to obtain a real value for the
dependent variable
z
3
Chapter One : Function of Several Variables
Example ( 4): Find the domain and range of the function
z = f ( x, y ) = 9 − y 2 − x 2
Solution:
f
The domain of
D=
(x, y ) : 9 − y
2
is
− x 2 0 = ( x, y ) : x 2 + y 2 9
which is the disk with center (0 ,0 ) and radius 3 (see Figure 3).
Figure .3
The range of
f is
Z : Z =
Since
Z
9 − y2 − x2 , ( x, y ) D
Z 0
is a positive square root,
9 − y2 − x2 9
. Also
9 − y2 − x2 3
So the range is
Z : 0 Z
3
= 0 , 3
( 2 ) The Limit of function in two variables
4
Chapter One : Function of Several Variables
If the values of f(x, y) lie arbitrarily close to a fixed real number M for
all points (x, y) sufficiently close to a point (a, b ) , we say that
f
approaches the limit M as (x, y) approaches (a, b ) . This is similar to the
informal definition for the limit of a function of a single variable. Notice,
however, that if (a, b ) lies in the interior of j's domain, (x, y) can approach
(a, b ) from any direction. For the limit to exist, the same limiting value
must be obtained whatever direction of approach is taken. We illustrate
this issue in several examples following the definition
Definition:
We say that the function z = f ( x, y ) approaches the limit M as
(x, y )
approaches (a, b ) and write
lim
( x , y ) → ( a ,b )
f ( x , y) = M
If for every 0 , there exists a corresponding number 0 for all
(x, y ) in the domain of f such that :
f ( x, y ) − M
i.e.,
0 ( ) 0
Such that
f ( x , y) − M
Whenever
x − a ,
0
y − b
i.e.,
( x − a ) 2 + ( y − b) 2
In this case we write
lim
( x , y ) → ( a ,b )
Or
f ( x , y) = M
f ( x, y ) → M
as ( x, y ) → (a, b )
5
Chapter One : Function of Several Variables
Remark(1): From the above definition of the limit of the function, it is
clear that there is no requirement on how the independent variables
( x, y ) approach the to any point (a, b ) that there is no requirement on
the track for the pair ( x, y ) could be to get to the point (a, b ) , so the
path can be straight lines like
y − b = m( x − a )
Or any curve , like y = ( x ) such that , when x → a then y → b , for
example the parabola
( y − b) 2 = m ( x − a )
This means that ,
1- when we take any path , if the results of the limit depends on m ,
then the limit does not exists
2- If the results of the limit is the same for any number of paths , then
the limit exists
Figure .4
Remark(2):
The limit does not exists if : the successive limits.
6
Chapter One : Function of Several Variables
lim ( lim f ( x, y ) ) lim ( lim f ( x , y ))
x →a
y −b
y →b
x→ a
But if we have
lim ( lim f ( x, y ) ) = lim ( lim f ( x , y ))
x →a
y −b
y →b
x→a
Then the limit
lim
( x → y ) → ( a ,b )
f ( x , y)
It can exist (not a requirement) and can not exist, and these are called
successive limits.
As with single-variable functions, the limit of the sum of two functions is
the sum of their limits (when they both exist), with similar results for the
limits of the differences, constant multiples, products, quotients, powers,
and roots.
Theorem [1]: Properties of Limits of Functions of Two variables
The following rules hold if M 1 , M 2 and K are real numbers and
lim
( x , y ) → (a , b)
f ( x, y ) = M 1 ,
lim
( x , y ) → (a , b)
g ( x, y ) = M 2 .
Then:
(I)
(II)
(III)
lim
( x , y ) → (a , b)
lim
f ( x , y ) g ( x , y ) = M 1 M 2
( x , y ) → (a , b)
f ( x , y ) . g ( x , y )= M 1 M 2
f ( x , y) M 1
=
( x , y ) → (a , b)
g
(
x
y
)
M2
;M2 0
lim
Example (4)
Determine the limit of the following function exixts or not ?
x y
f ( x , y) = x 2 + y 2
0
; ( x , y ) ( 0 , 0)
; ( x, y ) = ( 0 , 0)
7
Chapter One : Function of Several Variables
The successive limits.
Solution:
lim
lim
x→0 y →0
xy
2
x + y2
= 0
lim
lim
y →0 x→0
xy
2
x + y2
= 0
Since the two limits equal , so we still can not rule on the presence of
the limit or not
Then , by using the path y = mx :
xy
x (m x)
= lim 2
lim 2
2
2 2
x →0
x→0
x
+
y
x
+
m
x
y →0
m x2
= lim 2
x → 0 x (1 + m 2 )
=
m
1+ m2
Since the limit depends on m , then the limit does not exists
Example (5):
Prove that the following limits does not exist
(I)
x − 3y
( x , y ) → ( 0 , 0 ) 3 x − y
(II)
x3 − y3
lim 3
3
x→0 x + y
y→0
(III)
2x − y
lim 2
2
x→0 x + y
y→0
lim
Solution:
(I)
The successive limits.
x − 3y
lim lim
x→0
y → 0 3x − y
x
= lim
x→0 3 x
x − 3y
lim lim
y→0
x → 0 3x − y
−3y
= lim
= 3
y→0 − y
8
1
=
3
Chapter One : Function of Several Variables
Then it is not exists
x3 − y3
lim lim 3
3
x→0
y →0 x + y
(II)
= 1
− y3
x3 − y3
= lim 3
lim lim 3
y→0
x→0 x + y3
y →0 y
Then it is not exists
2x − y
lim lim 2
2
x→0
y →0 x + y
(III)
x3
= lim 3
x→0 x
= − 1
2x
= lim 2 =
x→0 x
−y
2x − y
= lim 2 = −
lim lim 2
2
y→0
x→0 x + y y →0 y
Then it is not exists
Example (6): Show that the limit
Solution:
lim
( x , y ) → (0 , 0)
2x y 2
2
4 is not exists .
x +y
2
By using the path x = my
2x y 2
lim
( x , y ) → (0 , 0) x 2 + y 4
The limit depends on m
2my 4
= lim
y → 0 ( m 2 + 1) y 4
2m
=
1+ m 2
Then it is not exists
Example (7):
Show that the limit
lim
( x , y ) → (0 , 0
f ( x , y ) is not exists , where
x2y
f (x , y ) = x 4 + y 2
0
if
if
Solution:
9
( x , y ) (0, 0)
(x , y ) = (0, 0)
Chapter One : Function of Several Variables
y = mx 2
By using the path
x2 y
= lim 4
x →0
x + y2
y→ 0
x 2 (m x 2 )
= lim 4
2
4
x→0
x +m x
m x4
m
=
= lim 4
2
2
x →0
x (1 + m ) 1 + m
The limit depends on m
Then it is not exists
Example (8):
Discuss the following limit
y − x 1+ x
( x , y ) → ( 0 , 0) y + x 1 + y
lim
Solution:
y − x 1+ x
lim lim f ( x , y ) = lim lim
x→0
y
+
x
1
+
y
y
→
0
x
→
0
y
→
0
− x 1+ x
= − lim (1 + x) = − 1
= lim
.
1 x → 0
x→0 x
but
y − x 1+ x
lim lim f ( x , y ) = lim lim
y →0
x→0
y → 0 x → 0 y + x 1 + y
y 1
=1
= lim
y → 0 y 1+ 0
lim lim f ( x , y ) lim lim f ( x , y )
x →0
y →0
y →0 x →0
The limit does not exists.
Example (9):
(I)
Evaluate the following limit
(
x sin x 2 + y 2
lim
x →0
x2 + y2
y →0
(
)
)
tan −1 (3x y + 1)
−1
sin ( xy − 1)
( ii) ( x , ylim
) → ( 2 , 1)
10
Chapter One : Function of Several Variables
Solution:
By using the path y = mx
(I)
(
x sin ( x 2 + m 2 x 2 )
sin x 2 (1+ m 2 )
lim
= lim x
2
2 2
x →0
x +m x
x 2 (1+ m 2 )
x →0
By using the properties of limits
(
sin x 2 (1 + m 2 )
= lim ( x ) lim
2
2
x →0
x →0
x (1 + m )
Solution:
The first:
lim
Prove that
) = 0 (1) = 0
( II ) By direct subsitiution, we find
tan −1 3( 2) (1) + 1 tan −1 7
tan −1 7
=
=
= tan −1 7
−1
−1
2
sin (2 (1) − 1)
sin 1
2
Example (10):
)
( x , y ) → (1 , 2 )
ﺗﻌوﯾض ﻣﺑﺎﺷر ﻗﺑل اي
ﺣﺎﺟﺔ
( x 2 + 2 y) = 5
We have two methods
From definition
f ( x , y) − 5 = x 2 + 2 y − 5
Assuming the existence of a small number 1 so that
x −1
; and
y − 2
then
− x −1 ; and − y − 2
1 − x + 1 ; and
1 − 2 + 2 x 2
So
2
2 − y + 2
+ 2 + 1 ; and 4 − 2 2 y 4 + 2
− 5 − 4 + 2 x 2 + 2 y − 5 4 + 2 5
f ( x , y) − 5 5
if we take , so = 5
11
Chapter One : Function of Several Variables
0 ( ) =
5
f ( x , y ) −5
Where
At
x −1
then
y − 2
and
lim
( x , y ) → (1 , 2 )
f ( x , y) = 5
The second method: By using Theorem 1
lim
( x , y ) → (1 , 2 )
(x
2
)
+ 2y =
(x )+
2
lim
( x , y ) → (1 , 2 )
( )
lim
( x , y ) → (1 , 2 )
(2 y )
= lim x 2 + lim (2 y ) = 1 + 4 = 5
x →1
y→ 2
Exercises on limits
1- Prove that by using the definition of limit
(3 x y ) = 6
lim
(I)
( x , y ) → (1, 2)
(II)
(III)
lim
( x , y ) → ( 2 , 1)
lim
(x
( x , y ) → (1 , −2)
2
)
+ y2 = 5
(3x − 4 y ) = 11
2- Prove that
(I)
(II)
lim
( x , y ) → ( 0 , 0)
sin −1 (xy − 2) 1
−1
=
tan (3xy − 6) 3
x + y −1
= 0
lim
( x , y ) → ( 0 , 1)
x − 1− y
, x0 , y1
3- Consider the function
1
y + x sin
f ( x , y) =
y
0
; y0
;y = 0
Find the following limits
lim
f ( x , y)
(I)
( x , y ) → ( 0 , 0)
(II)
lim lim f ( x , y )
y →0 x →0
12
Chapter One : Function of Several Variables
lim f ( x , y )
(III) lim
x →0 y →0
3- 4- Prove that the following limit does not exists
2x − 5 y
lim
x→0
5x − 2 y
y →0
lim
5- Prove that the following limit
( x , y ) → (0 , 0)
f ( x , y)
and the successive limit exists for the function
x2 − y2
x y
f ( x, y ) = x2 + y2
0
( x , y ) ( 0, 0)
( x , y ) = ( 0, 0)
6- Prove that the successive limit exists for the function f ( x, y ) but tle
limit
lim
( x , y ) → (0 , 0)
f ( x , y ) not exists
(I)
x− y
f ( x , y ) =
x
+
y
(II)
x2 y2
f ( x , y ) = 4
4
2 2
x
+
y
−
x
y
(III)
x3 + y3
f ( x , y) = x − y
0
(IV)
,
x y
,
x=y
x2 + y2
f ( x , y) = x 2 − y 2
0
,
x y
,
x=y
7- Find each of the following limits at the corresponding point
(I)
f ( x , y) =
x2 + y2 − 2
(II)
tan −1 x
f ( x , y ) =
x
+
y
13
at
(1 , 2)
at
(1 , 1)
Chapter One : Function of Several Variables
(III)
f ( x , y) = e
−
1
x
at
log y
(1 , 1)
1
f ( x , y ) =
at
(9 , 9)
x
−
y
8- Find each of the following limits , or explain that the limit does not
exists
(IV)
(I)
lim
( x , y ) → ( −1 , 3)
(II)
sin ( x + y )
( x , y ) → (0 , 0)
x
+
y
(III)
xy
( x , y ) → (0 , 0) x 2 + y 2
(IV)
(V)
2 x y + y 2
x2
lim
lim
lim
( x , y ) → (0 , 0)
(y
2
(
log x 2 + y 2
))
3x 2 − 2 xy
lim
( x , y ) → ( 3 , 1) x 2 + xy + y 2
Continuity
As with functions of a single variable, continuity is defined in
terms of limits.
Continuity of function of two variables
Definition: A function z = f (x, y ) is continuous at the point (a, b ) in its
domain if
lim
( x , y ) → ( a , b)
f (x , y ) = f (a , b ) ,
(1) - z = f (x, y ) is defined at (a, b )
f (x , y ) exists
(2) - ( x , y )lim
→ ( a , b)
(3) -
lim
( x , y ) → ( a , b)
f (x , y ) = f (a , b )
14
Chapter One : Function of Several Variables
A function is continuous if it is continuous at every point of its domain.
As with the definition of limit, the definition of continuity applies at
boundary points. As well as interior points of the domain of f. The only
requirement is that each point ( x, y ) near (a, b ) be in the domain of
z = f ( x, y ) .
A consequence of Theorem 1 is that algebraic combinations of
continuous functions are continuous at every point at which all the
functions involved are defined . This means that sums, differences,
constant multiples, products, quotients, and powers of continuous
functions are continuous where defined In particular, polynomials and
rational functions of two variables are continuous at every point at which
they are defined .
Remark:
If the function z = f ( x, y ) is continuous at the point (a, b )
, then the function f ( x, b ) is continuous at the point x = a . Also the
f (a, y ) is continuous at the point y = b . But The converse is not true
i.e., the function z = f ( x, y ) may be continuous at every variables
separately but not continuous at the point (a, b )
Exampels on continuity
Example(11): Let
2 xy
f ( x , y) = x 2 + y 2
0
( x , y ) (0 , 0)
( x , y ) = ( 0 , 0)
We have
lim f ( x , 0) = 0 = f ( 0 , 0)
x→0
lim f ( 0 , y ) = 0 = f ( 0 , 0)
y→ 0
15
Chapter One : Function of Several Variables
the function z = f ( x, y ) is continuous at every variables separately but
not continuous at the point (0,0 ) since the limit not exists
Exampels on continuity
Example ( 12 ):
Test the continuity of the following function at the origin
x2 y
f (x , y ) = x 2 + y 4
0
; (x , y ) (0, 0)
;x =y =0
Solution
f ( 0 , 0) = 0
→ (1)
The limits
x2 y
lim f ( x , y ) = lim 2
x→0
x→0
x + y4
y→0
y→0
2
By using the path x = my
m 2 y 4 (y )
m 2y 5
= lim
= lim 4
2
2
4
2
x →0
(my ) + y x → 0 y (1 + m )
m 2y
= lim 2 = 0
y →0
m +1
→ (2)
From ( 1 ) and ( 2 ) we find that
The function is continuous at (0,0 ) .
Example ( 13 ):
Discus the continuity of the function at origin
sin xy
g ( x , y) = x y
2
( x, y ) (0,0)
( x, y ) = ( 0 , 0 )
16
Chapter One : Function of Several Variables
Solution:
g ( 0 , 0) = 2
→ (1)
The limit
sin (x y )
lim g ( x , y ) = lim
x→0
x→0
x
y
y→0
y→0
By using the path y = mx
(
sin mx 2
= lim
2
x→0
mx
From
) = 1
→ (2)
)2( ، )1(
we find that
lim g ( x , y ) g (0 , 0)
x→0
y→0
The function is not continuous.
Example ( 14 ) :
Discus the continuity of the function at origin
x3 − y3
f ( x , y) = x 3 + y 3
0
; ( x , y ) (0, 0)
; ( x , y ) = (0 , 0)
Solution:
f ( 0 , 0) = 0
→ (1)
The limit
x3 − y3
lim f ( x , y ) = lim 3
3
x→0
x→0
x
+
y
y→0
y→0
By using the path
y = mx
17
Chapter One : Function of Several Variables
x3 − m3 x3
x 3 (1 − m 3 )
lim f ( x , y ) = lim 3
=
lim
3 3 x→0 3
3
x→0
x→0
x
+
m
x
x
(
1
+
m
)
y→0
1 − m3
=
3
1
+
m
→ (2)
From ( 1 ) and ( 2 ) we find that the limit depends on
m
the function is discontinuous.
Example ( 15):
Prove that the following function is continuous at the point (0,0 )
xy
f ( x , y) = x 2 + y 2
0
; ( x , y ) (0, 0)
; ( x , y ) = (0 , 0)
Solution
f ( 0 , 0) = 0
→ (1)
The limit
x y
lim f ( x , y ) = lim
2
2
x→0
x→0
x +y
y→0
y→0
By using the path y = mx
m x2
lim f ( x , y ) = lim
2
2 2
x→0
x→0
y→0
x +m x
m
=
lim (x )
1 + m2 x → 0
2
= lim m x
2
x
→
0
x 1+ m
=0
From ( 1 ) and ( 2 ) we find that
lim f ( x , y) = f (0 , 0)
x →0
y→ 0
The function is continuous at (0,0 ) .
18
→ (2)
Chapter One : Function of Several Variables
Exercises on continuity
1- Discuss the continuity of the function at the origin
x2 − y2
f ( x , y) = x 2 + y 2
0
; ( x , y ) (0, 0)
; ( x , y ) = (0 , 0)
2- Discuss the continuity of the function at (1,2 )
x 2 + 2 y
f ( x , y) =
0
; ( x , y ) (1 , 2)
; ( x , y ) = (1 , 2)
3- Discuss the continuity of the function at the origin
x2
f ( x , y) = x 2 + y 2
0
; ( x , y ) (0 , 0)
; ( x , y ) = (0 , 0)
4- Discuss the continuity of the function at the origin
(I)
5 sin xy
xy
f ( x , y) =
5
(II)
2 tan xy
f ( x , y ) = xy
2
(III)
1
f ( x , y) = x 2 + y 2
0
(IV)
x4 − y4
f ( x , y) = x 4 + y 4
0
; ( x , y ) (0 , 0)
; ( x , y ) = (0 , 0)
; ( x , y ) (0 , 0)
; ( x , y ) = (0 , 0)
; ( x , y ) (0 , 0)
; ( x , y ) = (0 , 0)
; ( x , y ) (0 , 0)
; ( x , y ) = (0 , 0)
19
Chapter One : Function of Several Variables
(V)
x3 y3
f ( x , y) = x 2 + y 2
0
; ( x , y ) (0 , 0)
; ( x , y ) = (0 , 0)
5- Discuss the continuity of the function at (1,0 )
( x − 1) y
f ( x , y ) = ( x −1) 2 + y 2
0
; ( x , y)
(1 , 0)
; ( x , y ) = (1 , 0)
6- Discuss the continuity of the function at (0,2 )
x ( y − 2)
h ( x , y ) = x 2 + ( y − 2) 2
0
(0 , 2)
; (x , y ) =
(0 , 2)
; ( x , y)
Partial Derivatives of a Function of Two Variables
We define the partial derivative of z = f (x, y ) with respect to x at the
point
( x 0 , y 0 ) as the ordinary derivative of f (x , y 0 ) with respect
to x at the point x = x 0 To distinguish partial derivatives from ordinary
derivatives we use the symbol a rather than the d previously used. In the
definition, h represents a real number, positive or negative.
If we have the function z = f ( x, y ) , then the partial differentiation
for this function with respect to the variable (for example) x is defined
as the normal differential of function, as it is a function of the
independent variable x and the rest of the variables is fixed amounts "
y fixed ". And it symbolized to the partial derivative of z = f (x, y ) with
respect to x
fx =
z f
=
= f x ( x , y)
x
x
20
Chapter One : Function of Several Variables
The function z = f (x, y ) is differentiable partially with respect to
the limit hlim
→0
x
if
f ( x + h , y) − f ( x , y)
exists .
h
Similarly the function z = f (x, y ) is differentiable partially with
respect to
y
if the limit klim
→0
f ( x , y + k ) − f ( x , y)
exists .
k
Also the partial derivatives for z = f (x, y ) at the point (a, b ) is defined
by
f
f ( a , b)
=
= f x ( a , b)
x
x (a , b)
f
y
f ( a , b)
=
= f y ( a , b)
y
(a , b)
i.e.,
f
f ( a + h , b) − f ( a , b)
= lim
h
→
0
h
x (a , b)
f
y
f ( a , b + k ) − f ( a , b)
= lim
k→ 0
k
(a , b)
Example ( 16 ):
Let
f ( x , y ) = e x sin y
By using the definition, find
f x (1,1)
and
f y (1,1)
Solution:
f ( a + h , b) − f ( a , b)
h
f ( 1 + h ,1) − f (1 , 1)
e1 + h sin 1 − e sin 1
f x (1 , 1) = lim
= lim
h→ 0
h→ 0
h
h
h
e −1
= e sin 1 lim
h→ 0
h
f
x
( a , b ) = hlim
→ 0
By using L’Hopital Theorem
= e sin 1 lim e h = e sin 1
h→0
21
Chapter One : Function of Several Variables
similarly
f
x
f x
f ( a , b + k ) − f ( a ,b )
k
f ( 1 , 1 + k ) − f (1 , 1)
e sin (1 + k ) − e sin 1
(1 , 1) = lim
= lim
k →0
k →0
k
k
sin ( 1 + k ) − sin1
= e lim
k →0
k
( a , b ) = klim
→0
By using L’Hopital Theorem
= e lim ( cos ( 1 + k ) ) = e cos 1
h→0
Example (17):
Let
x 2 − xy
f ( x , y) = x + y
0
Find f x (0,0 )
and
; ( x , y ) ( 0 , 0)
; ( x , y ) = ( 0 , 0)
f y (0,0 )
Solution:
f ( a + h , b) − f ( a , b)
h
f ( h , 0) − f ( 0 , 0)
f x (0 , 0) = lim
h→ 0
h
h2 − h (0)
− 0
h+0
= lim h = 1
= lim
h→ 0 h
h→ 0
h
f x ( 0 , 0) = 1
f
x
( a , b ) = hlim
→ 0
similarly
22
Chapter One : Function of Several Variables
f
y
( a , b ) = klim
→ 0
f ( a , b + k ) − f ( a , b)
k
0 − 0( k) − 0
f ( 0 , k ) − f ( 0 , 0)
0+ k
f y (0 , 0) = lim
= lim
k → 0
k → 0
k
k
0
= lim
= 0 f y ( 0 , 0) = 0
k → 0 k
Example ( 18 ):
Show that the function
x2 y
f ( x , y) = x 4 + y 2
0
; x2 + y2 0
; ( x , y ) = ( 0 , 0)
Has derivative of the first order at the point
connected at the point. (0,0 )
(0,0) while it is not
Solution;
f
x
f ( 0 + h , 0) − f ( 0 , 0)
h
0 − 0
= lim
=0
h→ 0
h
( 0 , 0) = hlim
→ 0
Similarly
f y ( 0 , 0) = lim
h→0
f ( 0 , 0 + h) − f ( 0 , 0)
0−0
= lim
=0
h→0
h
h
The partial derivatives of the first order exist at the point (0,0 ) .
But by studying the continuity , we find . f (0,0) = 0
The limit :
x2 y
lim f ( x , y ) = lim 4
x→0
x→0
x + y2
y→0
y→0
2
by using the path y = mx
23
→ (1)
Chapter One : Function of Several Variables
m
x2 ( m x2 )
m x4
lim f ( x , y ) = lim 4
=
lim
=
2
x→0
x → 0 x + m2 x4
x → 0 x 4 (1 m 2 )
1+ m
y→0
From ( 1 ) and ( 2 ) we find that the limit depends on
(2)
m
the function is discontinuous at (0,0 )
From the above example it is clear that there is a difference between
ordinary derivatives and partial derivatives in its relationship with the
continuity . In a single-variable function, we see that the continuity is
condition of the presence of the first derivative and if the first derivative
present then the function must be connected to the function.
As in the case of multivariate functions relationship between
communication and partial derivatives will become clear from the
following theories:
Theorem: If the partial derivative f x exists in the neighborhood of the
point (a, b ) and the derivative f y (a, b ) exists . Then for any point
(a + h, b + k ) in the neighborhood of the point (a, b ) we have
f (a + h, b + k ) − f (a, b ) = hf x (a + h, b + k ) + kf y (a, b + g )
Where 0 1 , and g is a function on k tends to zero with k
Proof:
But f (a + h, b + k ) − f (a, b ) = f (a + h, b + k ) − f (a, b + k ) + f (a, b + k ) − f (a, b )
f (a + h, b + k ) − f (a, b + k ) = hf x (a + h, b + k )
→ (1)
From Lagrangian mean theorem 0 1 and since f y (a, b ) exist , we
have
lim
k →0
f ( a ,b + k) − f ( a ,b )
= f y ( a , b)
k
Then
f (a, b + k ) − f (a, b ) = k ( f y (a, b ) + g )
Where k → 0 when g → 0
From (1) and (2), the teorem is proved
24
→ (2)
Chapter One : Function of Several Variables
Differentiability
If y = f (x) is differentiable at x = x0 , then the change in the
value of f that results from changing x from x 0 to x0 + x is given by
an equation of the form
y = f ( x0 ) x + x
in which 1 → 0 as x → 0 .
For functions of two variables, the analogous property becomes the
definition of differentiability. The following Theorem tells us when to
expect the property to hold.
Theorem3: Suppose that the first partial derivatives of z = f (x, y ) are
defined throughout an open region R containing the point ( x0 , y 0 ) and
that f x and f y are continuous at ( x0 , y 0 ) .Then the change
Z = f ( x0 + x, y0 + y ) − f ( x0 , y0 )
in the value of
f that results from moving from (x0 ,
y 0 ) to another
point (x0 + x , y 0 + y ) in R satisfies an equation of the form
Z = f x ( x0 , y0 )x + f y ( x0 , y0 )y + 1x + 2 y
in which each of
1 , 2 → 0 as both
(x, y ) → (0,0)
Definition: A function z = f (x, y ) is differentiable at ( x0 , y 0 ) if
f x ( x0 , y0 ) and f y ( x0 , y0 ) exist and Z satisfies an equation of the
form
Z = f x ( x0 , y0 )x + f y ( x0 , y0 )y + 1x + 2 y
25
Chapter One : Function of Several Variables
in which each of
1 , 2 → 0 as both
(x, y ) → (0,0) . We call f
differentiable if it is differentiable at every point in its domain, and say
that its graph is a smooth surface.
Because of this definition, an immediate corollary of Theorem 3 is that
a function is differentiable at ( x0 , y 0 ) if its first partial derivatives are
continuous there.
Corollary of Theorem 3: If the partial derivatives
f x and f y of the
function z = f (x, y ) are continuous throughout an open region R, then
f is differentiable at every point of R.
If z = f (x, y ) is differentiable, then the definition of differentiability
assures that Z = f ( x0 + x, y0 + y ) − f ( x0 , y0 ) approaches 0 as
(x, y ) → (0,0) . This tells us that a function of two variables is
continuous at every point where it is differentiable.
Theorem 4 ( Dlfferentiabillty Implies Continuity )
If a function z = f (x, y ) is differentiable at ( x0 , y 0 ) , then I is continuous
at ( x0 , y 0 ) .
Second-Order Partial Derivatives
When we differentiate a function z = f ( x, y ) twice, we produce its
second-order derivatives. These derivatives are usually denoted by
(I)
x
f 2 f
=
= fxx
2
x
x
(II)
y
f 2 f
=
= f xyy
2
y y
(III)
y
f
2 f
=
= fyx
x y x
(IV)
x
f
2 f
y
= x y = f x y
26
Chapter One : Function of Several Variables
The second – order derivatives can be defined at any point (a, b )
in the form :
(V)
f x x ( a , b) = lim
(VI)
f x y ( a , b) = lim
(VII)
f y y ( a , b) = lim
h →0
h →0
f x ( a + h , b) − f x ( a , b)
h
f y ( a + h , b ) − f y ( a , b)
h
f y ( a , b + k ) − f y ( a , b)
k →0
k
f x ( a , b + k ) − f x ( a , b)
(VIII) f y x ( a , b) = lim
k →0
k
If this limits exist.
Similarly , the derivative of any order can be defined , for example:
3 f
3 f
=
= f xxy
x x y
x2 y
Example (19 ):
Let
x2 − y2
x
y
f ( x , y) =
x2 + y2
0
; ( x , y ) (0 , 0)
; ( x , y ) = (0 , 0)
By using the definition of derivatives, find the first and second order
derivatives at (0,0 )
Solution:
f ( a + h , b) − f ( a , b)
h→0
h
f ( h , 0 ) − f ( 0 , 0)
f x ( 0 , 0) = lim
h→0
h
2
h −0
− 0
h(0) 2
h
+
0
0
= lim
= lim
=0
h→0
h→0 h
h
fx (0,0) = 0
f x ( a , b) = lim
27
Chapter One : Function of Several Variables
Similarly
f ( a , b + k ) − f ( a , b)
k →0
k
f ( 0 , k ) − f ( 0 , 0)
f y ( 0 , 0) = lim
k →0
k
0− k 2
−0
(0) (k )
2
0
+
k
= lim
h→0
h
f y ( 0,0) = 0
f y ( a , b) = lim
The second order derivatives
f ( a + h , b) − f x ( a , b)
f x x ( a , b) = lim x
h→0
h
f ( h , 0) − f x ( 0 , 0)
f xx ( 0 , 0) = lim x
h→0
h
= lim
h→0
f x (h,0)
h
→ (1)
x 3 y − xy 3
f ( x , y) = 2
x + y2
Since
( x 2 + y 2 ) (3x 2 y − y 3 ) − ( x 3 y − xy 3 ) (2 x)
fx =
(x2 + y 2 )2
fx
(h
(h , 0) =
2
)
+ 0 3h 2 (0) − 0 − 2h h3 (0) − h (0)
(h 2 + 0) 2
f x (h , 0) = 0
Substitution in ( 1 )
f xx (0 ,0) = lim
h→ 0
0
= lim (0) = 0
h h→ 0
f xx (0 ,0) = 0
28
Chapter One : Function of Several Variables
f x ( a , b + k ) − f x ( 0 , 0)
k →0
k
f ( 0 , k ) − f x ( 0 , 0)
(0 ,0) = lim x
k→0
k
f (0 , k )
= lim x
k→0
k
f xy (a , b) = lim
f xy
(0 + k ) 3 ( 0) − k − (0 − 0) (0)
2
3
(0 + k 2 ) 2
k
lim =
k →0
− k5
4
−k
lim = k = lim
= lim (−1)
k →0
k →0
k →0
k
k
f x y (0 , 0) = − 1
Similarly , we can find
Example ( 20 )
f xy (0,0), f yy (0,0)
Find the first and the second order derivatives of
f (x , y ) = x y +e
3
the function
x y2
Solution: The first step is to calculate both first partial derivatives.
f
(x , y ) = x 3y
+e x
y
2
By differentiate with respect to x
f x = 3x y + y e
2
→ (1)
2 x y2
By differentiate with respect to y
f y = x 3 (1) + 2 x y e x y
→ (2)
2
Now we find both partial derivatives of each first partial:
By differentiate ( 1 ) with respect to x
f xx = 6 xy + y 2 y 2 e xy = 6 xy + y 4 e xy
2
By differentiate ( 1 ) with respect to y
29
2
Chapter One : Function of Several Variables
y
f
2 f
x
= y x = f yx
= 3 x 2 (1) + y 2 2 xy e xy
= 3 x 2 + 2 xy 3 e xy
2
2
+ 2 y e xy
+ 2 y e xy
2
By differentiate ( 2 ) with respect to x
x
f
2 f
=
= f yx
y
x
y
= 3 x 2 + 2 xy 3 e xy
2
+ 2 y e xy
2
By differentiate ( 2 ) with respect to y
f yy = (0) + 2 x y 2 xy e xy
= 2 x e xy
Example (21 )
Prove that
Let
2
u ( x , y , z ) = (x 2 + y 2 + z 2 )
−
Solution:
u ( x , y , z ) = (x + y + z
2
2
2
)
−
1
2
By differentiate with respect to x
(
1 2
x + y2 + z2
2
(
+ (1)e xy
2 xy 2 + 1
u xx + u yy + u zz = 0
ux = −
2
= −x x2 + y 2 + z 2
)
−3 / 2
(2 x )
)
−3 / 2
By differentiate again with respect to x
30
1
2
2
2
Chapter One : Function of Several Variables
− 3
u xx = − x x 2 + y 2 + z 2
2
(
(
= 3x 2 x 2 + y 2 + z 2
)
−5 / 2
) (2 x) + (x
−5 / 2
(
2
+ y2 + z2
) (− 1)
−3 / 2
)
−3 / 2
− x2 + y2 + z2
→ (1)
Similarly , by differentiate twice with respect to y
(
u yy = 3 y 2 x 2 + y 2 + z 2
)
−5 / 2
(
(
− x2 + y2 + z2
2 2
2
2
Also , u zz = 3 y x + y + z
)
−5 / 2
)
−3 / 2
→ (2)
− (x 2 + y 2 + z 2 )
−3 / 2
From ( 1 ) , ( 2 ) and ( 3 ) by collection we get
u xx + u yy + u zz = 0
Example ( 22): Find the first and second derivatives of
z = x tan ( x y )
Solution:
z
= xy sec 2 (xy ) + tan(xy )
x
z
zy =
= x 2 sec 2 (xy )
y
zx =
→ (1)
→ (2)
By differentiate ( 1 ) with respect to x
z xx = 2 xy 2 sec 2 (xy ) tan (xy ) + 2 y sec 2 (xy )
By differentiate ( 1 ) with respect to y
z yx = 2 x 2 y sec 2 (xy ) tan (xy ) + 2 x sec 2 (xy )
By differentiate ( 2 ) with respect to x
z xy = 2 xy 2 sec 2 ( xy ) tan ( xy ) + 2 y sec 2 (xy )
By differentiate ( 2 ) with respect to y
31
→ (3)
Chapter One : Function of Several Variables
z yy = 2 x 3 sec 2 ( xy ) tan ( xy )
Remark:
We notice that in the above example ,
z xy = z yx
. Does this exists everywhere or there exist a condition to
have f xy = f yx , the following theorem gives the conditions
Theorem (1):
If both f x , f y are differentiable at the point (a, b ) in the domain D of
the function f . Then
f xy (a, b ) = f
yx
(a, b )
Theorem (2)
If the function f ( x, y ) is defined in the domain D , and f xy , f yx exists
and continuous at the point (a, b ) in D . Then
f xy (a, b ) = f yx (a, b )
Example (23) : Let
x2 y2
f ( x, y ) = x 2 + y 2
0
Prove that :
; (x, y ) (0,0 )
; (x, y ) = (0,0 )
f xy (0,0 ) = f yx (0,0 )
Solution:
Since ,
f (h,0 ) − f (0,0 )
f x (0,0 ) = lim
=0
h →0
h
Also ,
f y (0,0 ) = 0
32
Chapter One : Function of Several Variables
fx
(x, y ) = (x
f y ( x, y ) =
(x
2
)
(x
+ y 2 2 xy 2 − x 2 y 2 2 x
2x 4 y
2
+ y2
2
+ y2
)
2
=
(x
2 xy 4
2
+ y2
)
2
)
2
f (0, k ) − f x (0,0 )
f yx (0,0 ) = lim x
=0
k →0
k
f y (h,0 ) − f y (0,0 )
= 0
f xy (0,0 ) = lim
h →0
h
f xy (0,0 ) = f yx (0,0 )
Exercises on partial derivatives
1. Find for the following function f x (0,0 ), f y (0,0 )
x 2 − xy
f ( x, y ) = x + y
0
; ( x, y ) (0,0 )
; (x, y ) = (0,0 )
2- proved that the following functions are not continuous at the origin,
despite of the exists of partial derivatives of the first order in the
domain of the function and at the original point .
(I)
x3 + y3
f ( x, y ) = x − y
0
(II)
x2 y
f ( x, y ) = x 4 + y 2
0
;x y
;x = y
; ( x, y ) (0,0 )
; ( x, y ) = (0,0 )
3-Let
xy tan( y, x )
f ( x, y ) =
0
; (x, y ) (0,0)
; (x, y ) = (0,0)
Prove that : x f x + y f y = 2 f
4- Discuss the continuity and differentiability of the functions
33
Chapter One : Function of Several Variables
xy
2
f ( x, y ) = x + y 2
0
(I)
if
x2 + y2 0
if
x= y=0
xy 2
f ( x, y ) = x 2 + y 2
0
(II)
; ( x, y ) 0
; (x, y ) = (0,0)
1
f (x, y ) = y sin
if x 0
x
5- Find the partial derivative of first and second order of the following
functions:
(III)
(I)
f ( x, y ) = x tan −1 ( xy )
(II)
f (x, y ) = e xy + log( x + y )
(III)
x
f (x, y ) = log
x+ y
(IV)
f ( x, y, z ) = tan −1 ( xyz )
(V)
f ( x, y ) = x y
(VI)
f ( x, y, z ) = sin x + z y
(VII)
h(x, y ) = x 2 cos( y ) + y 2 sin (x )
(VIII)
h(x, y ) = y cos −1 (x + y )
(IX)
k ( x, y ) = x tan −1 ( y cos( x ))
(X)
f ( x, y, z ) = x yz
(XI)
f ( x, y, z ) = e x log x 2 + y 2 + z 2
(
)
2
(
)
6- Prove that the function : z = log ( x − a ) + ( y − b )
2z 2z
+
= 0 except at
relation
x 2 y 2
2
(x, y ) = (a, b )
y
y
7- Prove that the function : z = x cos + tan
x
x
2
2
satisfy the relation x z xx + 2 xyz xy + y z yy = 0
34
2
satisfy the
Chapter One : Function of Several Variables
8- Prove that the function u = e cos( y ) + e sin ( z )
x
y
Prove that : u xy = u yx , u xz = u zx , u yz = u zy
( )
2
9- Let the function z = f x y be differentiable function . Prove that :
xz x = 2 yz y
10- Find the value of the first partial derivatives of each function at the
corresponding point :
(I)
f ( x, y ) = e x log ( y )
; (0, e )
(II)
g ( x, y ) =
x
x+ y
; (1,2)
(III)
h(x, y ) = e − x sin (x + 2 y )
; 0,
4
The Chain Rule
For functions of more than one variable, the Chain Rule has several
versions, each of them giving a rule for differentiating a composite
function. The first version
(Theorem 2) deals with the case where z = f ( x, y ) and each of the
variables x and y is, in turn, a function of a variable t . This means that
z = f ( x, y ) is indirectly a function of t , z = f ( g (t ), h(t ) ) , and the Chain
Rule gives a formula for differentiating as a function of t . We assume
that f is differentiable . Recall that this is the case when f x and f y are
continuous .
Theorem 2: Suppose that z = f ( x, y ) is a differentiable function of x
and y , where x = g (t ) ; y = h(t ) are both differentiable functions of t .
Then z is a differentiable function of t and
dz
z dx
z dy
=
+
dt
x dt
y dt
35
Chapter One : Function of Several Variables
Since
dz =
z
z
dx +
dy
x
y
dx =
And
→ (1)
dx
dt
dt
, dy =
dy
dt
dt
but z is a differentiable function
dz =
z dx
z dy
dt +
dt
x dt
y dt
→ (2)
Also
dz =
dz
dt
dt
→ (3)
Then we have
dz
dt
=
xy
Example (24): If z = e
Find
dz
dt
at t =
2
z dx
z dy
+
x dt
y dt
; x = t cos(t )
; y = t sin (t )
2
Solution:
z = e xy
2
2
z
= y 2 e xy
x
,
2
z
= 2 xye xy
y
x = t cos(t )
dx
= −t sin (t ) + cos(t )
dt
dy
y = t sin (t )
= t cos(t ) + sin (t )
dt
dz
z dx
z dy
=
+
dt
x dt
y dt
From (1),(2) and (3) we have
36
→ (1)
→ (2)
→ (3)
Chapter One : Function of Several Variables
(
)
(
)
2
2
dz
= y 2 e xy (− t sin (t ) + cos(t )) + xe xy (t cos(t ) + sin (t ))
dt
= e xy
at t =
2
2
(y cos(t ) − t sin (t )) + xt cos(t ) + sin (t )
2
x=0
;y =
2
2
0 2
dz
3
2
=e
0 − + 0 = −
dt
2
8
4
We now consider the situation where z = f ( x, y ) but each of x and y
is a function of two variables u and v : where
x = g (u , v ) ; y = h(u , v ) . Then z = f ( x, y ) is indirectly a function of
u ,and
Suppose that z = f ( x, y ) is a differentiable function
of x and y , where x = g (u , v ) ; y = h(u , v ) are differentiable functions
of u and v Then
Theorem 3: Suppose that z = f ( x, y ) is a differentiable function of x
and y , where x = y(u, v ) , y = h(u, v ) are both differentiable functions of
u and v .
Then z is a differentiable function of of u and v , then
z
z
=
u
x
z
z
=
v
x
x
z
+
u
y
x
z
+
v
y
y
u
y
v
Since
x
x
du +
dv
u
v
y
y
dy =
du +
dv
u
v
dx =
→ (4)
But z = f ( x, y ) is a differentiable function of x and y ,so z is a
differentiable function of of u and v
37
Chapter One : Function of Several Variables
dz =
But
dz =
z
z
du + dv
u
v
→ (5)
z
z
dx + dy . Then
x
y
dz =
z x
x
z y
y
du +
dv +
du +
dv
x u
v
y u
v
z x
z y
=
x u + y u
z x
z y
du +
x v + y v
dv
Comparing with (5) , we have
z
z x
z
=
+
u
x u
y
z
z x
z
=
+
v
x v
y
y
u
y
v
3
3
Example (25): If z = x − xy + y where
x = r cos( )
Find :
Solution:
y = r sin ( )
;
z
z
,
r
z
= 3x 2 − y
x
,
z
= −x + 3y 2
y
x
= cos( )
r
x = r cos( )
x = − r sin ( )
y
= sin ( )
r
y = r sin ( )
y = r cos( )
z
z x
z y
=
+
r
x r
y r
From (1),(2) and (3) we have
38
→ (1)
→ (2)
→ (3)
Chapter One : Function of Several Variables
z
= 3 x 2 − y (cos ) + 3 y 2 − x (sin )
r
= 3r 3 cos 3 + sin 3 − 2r sin cos
(
)
(
)
(
)
z
z x
z y
=
+
x
y
From (1),(2) and (3) we have
z
= 3 x 2 − y (− r sin ) + 3 y 2 − x (r cos )
= 3r 3 sin cos 2 + cos sin 2 + r 2 sin 2 − cos 2
(
)
(
)
)
(
(
)
Example (26): If v = v(x, y ) where x = r cos( ) ; y = r sin ( )
2
v
1 v
v
v
Prove that: + = + 2
r
x
r
y
2
2
2
Solution:
x
x
x
x
= (1) cos( ) =
=
r
r
r
r
x = r cos( )
x = − r sin ( ) = − y x = − y
y
y
y
y
= (1) sin ( ) =
=
r
r
r
r
y = r sin ( )
y
y
= r cos( ) = x
= x
v
v x
v y
=
+
r
x r
y r
By (1) and (2)
v
r
=
x v
+
r x
y v
r y
Multiply by r
r
v
v
v
= x
+ y
r
x
y
39
→ (1)
→ (2)
Chapter One : Function of Several Variables
2
v v
v
2 v
2 v
+ 2 xy
r
=x
+ y
x y
r
x
y
2
2
→ (3)
2
Similarly
v
=
v x
x
+
v y
y
By (1) and (2)
v
v
v
= −y
+ x
x
y
2
v v
v
2 v
2 v
− 2 xy
= y
+ x
x y
x
y
2
2
→ (4)
From (3) and ( 4) by addition
v
v
2
2 v
2
2 v
r +
= x + y + x + y
r
x
y
2
2
(
2
2
)
(
)
2
v 2 v 2
= x + y +
x
y
x 2 + y 2 = r 2 cos 2 ( ) + r 2 sin 2 ( )
(
2
2
)
→ (5)
= r 2 (1) = r 2
2
From (5) by dividing on r we have
v
v
+
y
x
2
2
1 v
v
=
+ 2
r
r
2
Example (27): If z = f (x, y ) where
2
x = ue v ; y = ue − v
(
2
2
2
Prove that: u z uu − uz u + z vv = 2 x z xx + y z yy
Solution:
x = ue v
x
x
= ev =
u
u
x = ue v = x
v
40
x
x
=
u
u
x
= x
v
→ (1)
)
Chapter One : Function of Several Variables
y
y
−v
=
e
=
u
u
y = ue −v
y = −ue −v = − y
v
z z x z y
zu =
=
+
u x u y u
y y
=
u u
y
= −y
v
→ (2)
By (1) and (2)
x
y
zu = z x + z y
u
u
Multiply by u
u z u = x z x + y z y
→ (3)
By differentiate (3) with respect to u
(xz x + yz y )
u
(xz x + yz y ) x + (xz x + yz y ) y
+ zu =
x
u y
u
uz uu + z u (1) =
uz uu
from (1), (2)
x
y
uz uu + z u = (xz xx + z x + yz yx ) + (xz yx + yz yy + z y )
u
u
Multiply by u
u 2 z uu + uz u = x 2 z xx + xz x + xyz yx + xyz xy + y 2 z yy + yz y
Similarly
zv =
z
z x
z y
=
+
v
x v
y v
From (1) and (2)
z v = xz x − yz y
→ (5)
41
→ (4)
Chapter One : Function of Several Variables
By differentiate (5) with respect to v
(xz x − yz y )
v
x
y
z vv = (xz x − yz y ) + (xz x − yz y )
x
v y
v
z vv = (xz xx + z x − yz yx )(x ) + (xz xy − yz yy − z y )(− y )
z vv =
z vv = x 2 z xx + xz x − xyz yx − xyz xy + y 2 z yy + yz y
By adding (4) and (6), we have
u 2 z uu + uz u + z vv = 2 x 2 z xx + 2 xz x + 2 y 2 z yy + 2 yz y
(
)
u 2 z uu + uz u + z vv = 2 x 2 z xx + y 2 z yy + 2(xz x + yz y )
2
2
2
By using (3) u z uu − uz u + z vv = 2(x z xx + y z yy )
Example (28): If
z = f ( x, y ) where
x=
Prove that :
( II )
(I)
(
1
log u 2 + v 2
2
)
v
; y = tan−1
u
(z x )2 + (z y )2 = (u 2 + v 2 )(zu )2 + (zzv )2
(
)
z xx + z yy = u 2 + v 2 z uu + z vv
Solution:
42
→ (6)
Chapter One : Function of Several Variables
v
v
y = tan −1 tan( y ) =
u
u
v
sin ( y ) =
, cos( y ) =
u2 + v2
(
x = log u + v
2
2
)
1
2
= log
u
u2 + v2
u2 + v2
v
v
e x =
x = log
sin ( y )
sin ( y )
x =
u
u
x = log
e x =
cos( y )
cos( y )
u = e x cos( y )
v = e x sin ( y )
u
u
= e x cos( y ) = u ,
= − e x sin ( y ) = −v
x
y
v
v
x
x
= e sin ( y ) = v ,
= e cos( y ) = u
x
y
→ (1)
z
z u
z v
=
+
x
u x
v x
z
z
z
=u
+v
x
u
v
→ (2)
Squaring the equation (2)
z z
z
2 z
2 z
=u
+v
+ 2uv
u v
x
u
v
2
2
2
→ (3)
similarly
z
z u
z v
=
+
y
u y
v y
z
z
z
= −v
+u
y
u
v
→ (4)
Squaring the equation (4)
2
z
z z
z
2 z
= v 2
+u
− 2uv
u v
u
v
y
2
2
By adding (3) and (5) we have
43
→ (5)
Chapter One : Function of Several Variables
z
x
2
z
+
y
2
(
= u
(
= u
2
2
+v
+v
2
2
z
u
)
2
(
+ u
+v
2
2
z
v
)
2
2
z 2
z
+
v
u
)
From (2) by differentiate with respect to
x
z
z
+v
u
x u
v
(uz u + vzv ) u + (uz u + vzv ) v
=
u
x v
x
= (uz uu + z u + vzvu )(u ) + (uz uv + vzvv + z v )(v )
z xx =
z xx
z xx = u 2 z uu + uz u + uvzvu + uvzuv + v 2 z vv + vz v
From (4) by differentiate with respect to
→ (6)
y
(− vzu + uz v )
y
(− vzu + uz v ) u + (− vzu + uz v ) v
=
u
y v
y
= (− vzuu + uz vu + z v )(− v ) + (− vzuv − z u + uz vv )(u )
z yy =
z yy
z yy = v 2 z uu − uvzvu − vzv − uvzuv − uz u + u 2 z vv
→ (7)
By adding (6) and (7) we have
(
= (u
)
)(z
(
)
z xx + z yy = u 2 + v 2 z uu + u 2 + v 2 z vv
z xx + z yy
2
+ v2
uu
+ z vv )
Exercise on The Chain Rule
(1) Prove that
Laplace's equation u xx + u yy = 0 Transformed by the
substitution x = r cos( )
To
u rr +
; y = r sin ( )
1
1
u +
ur = 0
2
r
r
(2) Prove that the substitution x = r cos( )
44
; y = r sin ( )
Chapter One : Function of Several Variables
Transform Cauchy –Riemann equation u y = −v x , u x = v y to
vr = −
1
1
u , u r = v
r
r
y
−y
(3) If f = f (u , v ) where u = xe , v = xe , Prove that :
(
x 2 f xx − xf x + f yy = 2 u 2 f uu + v 2 f vv
)
(4) Let x = u cos − v sin , y = u sin + v cos
Prove that , if V = v( x, y ) then
V V
V V
+
=
+
u v
x y
2
2
(5) If z = f (x y ) where
Prove that :
2
2
2
f is differentiable function .
z
z
x
= 2 y
y
x
u
u
(6) Let z = f ( x, y ) and x = e cos t , y = e sin t .
2
2z 2z
2z
− 2u z
+
=e
+
Prove that :
2
x 2
y 2
t 2
u
u
−v
v
−u
(7) Let z = f ( x, y ) and x = e + e , y = e + e
Prove that :
z uu − 2 z uv + z vv = x 2 z xx − 2 xyz xy + y 2 z yy + xz x + yz y
(8) If u = ( x + at ) + ( x − at )
2
2u
u
2
= a
Prove that :
2
t
x 2
45
Chapter One : Function of Several Variables
2v 2v
+
(9) proved that the amount
does not change as a result
x 2
y 2
of the transfer of the original axes x, y Parallel to their position the
point (a, b )
(10) If z = f ( x, y ) + g (u ) where u = xy .
Prove that : w = xz x − yz y does not depend on the function g , then
find
w
f = xye x − y
when
(11) Let V be a function on two variables x, y , and both x, y is
a function on two variables u, v satisfy xu = y v , xv = − y u . Prove that
(
)
: Vuu + Vvv = xu + xv (Vxx + V yy )
2
2
2
2
2
(12) Let u = y log ( y + v ) − r where r = x + y .
Prove that : u xx + u yy =
1
y+r
(13) Prove that : If we change the variables
(
x, y in the equation
)
z xx + 2 xy 2 z x + 2 y − y 3 z y + x 2 y 2 = 0
1
, then we can get the same
v
by y
By the variables u, v , where x = uv, y =
equation by replacing
u
by
x
and
v
(14) If the function f ( x, y ) changed to the function (u, v ) by the
substitutions x = u cosh v, y = u sinh v , then prove that
f xx − f yy = uu −
46
1
1
vv + u
2
u
u
Chapter One : Function of Several Variables
Directional Derivatives in the Plane
We know that if f ( x, y ) is differentiable, then the rate at which f
changes with respect to t along a differentiable curve
x = g ( t ) , y = h ( t ) is
d f
f dx f dy
=
+
dt
x dt
y dt
At any point p0 ( x0 , y0 ) = p0 ( g ( t0 ) , h ( t0 ) ) , h(to)), this equation
gives the rate of change of f with respect to increasing t and
therefore depends, among other things, on the direction of motion
along the curve. If the curve is a straight line and t is the arc length
parameter along the line measured from p0 in the direction of a given
unit vector u , then df dt is the rate of change of f with respect to
distance in its domain in the direction of u . By varying u , we find the
rates at which f changes with respect to distance as we move through
p0 in different directions. We now define this idea more precisely.
Suppose that the function f ( x, y ) is defined throughout a region R in
the
xy -plane, that p0 ( x0 , y0 ) is a point in R , and that u = u1i + u2 j is
a unit vector. Then the equations
x = x0 + su1 , y = y0 + su2
parameterize the line through p0 parallel to u . If the parameter s
measures arc length from p0 in the direction of u , we find the rate of
change of f at p0 in the direction of u by calculating df ds at p0
Figure ( 5).
47
Chapter One : Function of Several Variables
Figure .5 The rate of change of
f in the direction of
u
which f changes along this line at
at a point
p0
is the rate at
p0
Definition: The derivative of f at p0 ( x0 , y0 ) in the direction of the
unit vector u = u1i + u2 j is the number
d f
ds
f ( x0 + su1 , y0 + su2 ) − f ( x0 , y0 )
= lim
s →0
s
u , p0
( 1)
Provided the limit exists.
The directional derivative defined by Equation (1) is also denoted by
(Du f ) p
0
"The derivative of f at
The partial derivatives
p0
in the direction of
f x ( x0 , y0 ) and f y ( x0 , y0 )
u"
are the directional
derivatives of f at p 0 in the i and j directions. This observation can be seen
by comparing Equation (1) to the definitions of the two partial derivatives given
before.
Example(29): Using the definition, find the derivative of f (x, y ) = x 2 + xy
at p0 (1, 2 ) in the direction of the unit vector u = (1
2 )i + (1 2 ) j
Solution:
d f
ds
f ( x0 + su1 , y0 + su2 ) − f ( x0 , y0 )
= lim
s →0
s
u , p0
f (1 + s
= lim
s →0
(1 +
= lim
s →0
1
1
,2 + s
) − f (1,2)
2
2
s
2s
s2
3s
s2
+
)+(2 +
+
)−3
2
2
2
2
s
5s
+ s2
5
5
2
= lim
= lim
+ s=
s →0
s →0
s
2
2
The rate of change of f ( x, y ) = x + xy at p0 (1, 2 ) in the direction
2
u
is 5
2
For a physical interpretation of the directional derivative, suppose that T = f ( x, y )
is the temperature at each point (x, y) over a region in the plane. Then
48
f ( x0 , y0 ) is
Chapter One : Function of Several Variables
the temperature at the point p0 ( x0 , y0 ) and (Du f
)p
0
is the instantaneous rate of
change of the temperature at p 0 stepping of f in the direction u.
Calculation and Gradients
We now develop an efficient formula to calculate the directional derivative for a
differentiable function f . We begin with the line
x = x0 + su1 , y = y0 + su 2
through p0 ( x0 , y0 ) , parameterized with the arc length parameter s increasing in
the direction of the unit vector
d f
d s
u ,
p0
u = u1i + u2 j .Then by the Chain Rule we find
f
dx
f
=
+
y
x p0 ds
f
=
x
u1
p0
f
f
=
i +
y
x
p0
dy
p0 ds
f
+
y
u2
p0
j • u1 i + u 2 j
p0
(3)
Equation (3) says that the derivative of a differentiable function f in the direction of
u at p0 is the dot product of u with the special vector called the gradient of f at
p0 .
Definition: The gradient vector (gradient) of f ( x, y ) at a point p0 ( x0 , y0 ) is the
vector
f =
f
f
i+
j
x
y
obtained by evaluating the partial derivatives of f at p 0 .
Theorem: The Directional Derivative Is a Dot Product
If f ( x, y ) differentiable in an open region containing p0 ( x0 , y0 ) , then
d f
ds
= (f
u , p0
)p
0
• u,
the dot product of the gradient f at
p0
and
u .
Find the derivative of f ( x, y ) = xe y + cos( xy )
at the point ( 2,0 ) in the direction of the unit vector v = 3i − 4 j
Solution: The direction of v is the unit vector obtained by dividing v by its
Example(29):
length:
49
Chapter One : Function of Several Variables
u=
v v 3
4
= = i− j
v 5 5
5
The partial derivatives of f are everywhere continuous and at (2, 0) are given by
f x (2,0) = ( e y − y sin ( xy )) ( 2 , 0 ) = e 0 − 0 =1
f y (2,0) = ( x e y − x sin ( xy )) ( 2 , 0 ) = 2 e 0 − 2 0 = 2
The gradient of f at (2, 0) is
f |( 2 , 0 ) = f x ( 2 , 0 ) i + f y ( 2 , 0 ) j = i + 2 j
Figure ( ). The derivative of f at (2, 0) in the direction of v is therefore
(Du f )|( 2, 0 ) = f
3 4
3 8
|( 2 , 0 ) u = ( i + 2 j ) ( i − j ) = − = − 1
5 5
5 5
Figure .6 Picture f
as a vector in the domain of f
.
Tangent Planes and Differentials
In this section we define the tangent plane at a point on a smooth surface in space.
Then we show how to calculate an equation of the tangent plane from the partial
derivatives of the function defining the surface. This idea is similar to the definition of
the tangent line at a point on a curve in the coordinate plane for single-variable
functions . We then study the total differential and linearization of functions of
several variables.
Tangent Planes and Normal Lines
If
r = g ( t ) i + h( t ) j + k (t )k
is a smooth curve on the level surface
50
Chapter One : Function of Several Variables
f ( x, y , z ) = c
of a differentiable function f , then f ( g (t ), h(t ), k (t ) ) = c .
Differentiating both sides of this equation with respect to t leads to
d
d
f ( g (t ), h(t ), k (t ) ) =
(c )
dt
dt
f dg
f dh
f dk
+
+
=0
x dt
y dt
z dt
f
dg
f
f
dh
dk
x i + y j + z k
dt i + dt j + dt k
=0
f
dr
dt
At every point along the curve, f is orthogonal to the curve's velocity vector.
Now let us restrict our attention to the curves that pass through p 0 Figure ( ). All
the velocity vectors at p 0 are orthogonal to f at Po, so the curves' tangent lines all
lie in the plane through p 0 normal to f . We now define this plane.
Figure .7
The gradient f is orthogonal to the velocity vector of every smooth curve in
the surface through p 0 . The velocity vectors at Po therefore lie in a common
plane, which we call the tangent plane at p 0 .
Definition: The tangent plane at the point p0 ( x0 , y0 , z0 ) on the level surfaces
f
( x, y , z ) = c of the differentiable function f is the plain through p0
normal to t a point is the vector
f | p0
The normal line of the surfaces at p 0 is the line through p 0 parallel to f | p .
0
the tangent plane and normal line have the following equations:
Tangent plane to f ( x, y, z ) = c at p0 ( x0 , y0 , z0 )
f x ( p0 )(x − x0 ) + f y ( p0 ) ( y − y0 ) + f z ( p0 ) ( z − z0 ) = 0
Normal line to f ( x, y , z ) = c
at the point p0 ( x0 , y0 , z0 )
51
Chapter One : Function of Several Variables
x = x0 + f x ( p0 ) t ,
Example(30):
y = y0 + f y ( p0 ) t ,
z = z0 + f z ( p0 ) t
Find the tangent plane and normal line of the surface
f ( x, y , z ) = x 2 + y 2 + z − 9 = 0
at the point p0 ( 1,2,4 ) .
Solution: The surface is shown in Figure 8. The tangent plane is the plane through
p 0 perpendicular to the gradient of f at p 0 . The gradient is
f | p0 = ( 2 xi + 2 yj + k ) (1, 2, 4 ) = 2i + 4 j + k
The tangent plane is therefore the plane
2 (x − 1) + 4 ( y − 2) + ( z − 4) = 0
or
2 x + 4 y + z = 14
The line normal to the surface at p 0 is
x = 1 + 2t ,
Figure .8
y = 2+4 t ,
z =4+ t
The tangent plane and normal line to this surface at p 0
To find an equation for the plane tangent to a smooth surface z = f (x, y ) at a point
p0 ( x0 , y0 , z0 ) where z0 = p0 ( x0 , y0 ) we first observe that the equation z = f (x, y ) is
equivalent to f ( x, y ) − z = 0 . The surface z = f (x, y ) is therefore the zero level surface
of the function F ( x, y, z ) = f (x, y ) − z . The partial derivatives of F are
( f ( x, y ) − z ) = f x − 0 = f x
x
( f ( x, y ) − z ) = f y − 0 = f y
Fy =
y
Fx =
52
Chapter One : Function of Several Variables
Fz =
z
( f ( x, y ) − z ) = 0 −1= − 1
The formula
Fx ( p0 )(x − x0 ) + Fy ( p0 ) ( y − y0 ) + Fz ( p0 ) ( z − z0 ) = 0
for the plane tangent to the level surface at Po therefore reduces to
f x ( x0 , y0 )( x − x0 ) + f y ( x0 , y0 ) ( y − y0 ) +
- ( z − z0 ) = 0
Example(31):
( 4)
Find the plane tangent to the surface
z = x cos y − y e x
at the point p0 ( 0,0,0 ) .
Solution: We calculate the partial derivatives of
f ( x, y ) = x cos y + y e x
and use Equation (4):
f x (0,0) = ( cos y − y e x ) ( 0, 0 ) = 1 − 0 1 = 1
f y (0,0) = ( − x sin y − e x ) ( 0, 0 ) = 0 −1 = − 1
The tangent plane is therefore the plane
1( x − 0 ) − 1( y − 0) − ( z − 0) = 0
or
x− y−z=0
The Harmonic Function
Definition: The function f (x, y ) is said to be harmonic function if it
satisfies Laplace's equation i.e.,
f xx + f yy = 0
everywhere on D .
This is usually written as
where
Example(32):
2 =
2 f = 0
2
2
+
x 2
y 2
The following functions are harmonic
x
• The function f ( x, y ) = e sin y
• The function f ( x, y ) = cos x sin ( − y )
• The function
f ( x, y ) = ln ( x 2 + y 2 ) define on R 2 − (0,0)
• The function
f ( x, y ) = ( x 2 + y 2 )1− n 2 define on R 2 − (0,0) for n 2
53
Chapter One : Function of Several Variables
Theorem: If u = u ( x, y ), v = v( x, y ) two functions on the variables x, y ,
satisfying the Cauchy–Riemann equations
u
v
=
,
x
y
And assume that
v
u
= −
x
y
u yx , u xy , v yx , v xy are continuous functions . Then
u, v are harmonic unction.
Proof:
u
v
=
x
y
differentiating with respect to
x
2u
v
=
x 2 x y
2u
2v
2 =
xy
x
u
v
=−
y
x
differentiating with respect to
→ (1)
y
2u
v
=−
2
y x
y
2u
2v
=
−
yx
y 2
→ (2)
By adding (1) and(2)
2u 2u
2v
2v
+
=
−
xy yx
x 2
y 2
From continuity condition
2u 2u
2 + 2 =0
x
y
u
is harmonic
Similarly , for proving that
v
is harmonic
54
Chapter One : Function of Several Variables
v
u
= −
x
y
differentiating with respect to
x
2v
u
=
−
x y
x 2
2v
2u
2 =−
xy
x
v u
=
y x
→ (3)
differentiating with respect to
y
2v
u
=
2
y x
y
2v
2u
2 =
yx
y
→ (4)
By adding (3) and(4)
2 v 2 v 2u
2u
+
=
−
x 2 y 2 yx xy
From continuity condition
2v 2v
+
=0
x 2 y 2
v is harmonic
Theorem:
If f = f (u , v ) is a harmonic function in the variables u, v and
u = u ( x, y ), v = v( x, y ) satisfying the Cauchy–Riemann equations . Then
f = f ( x, y ) is harmonic in the variables x, y
Proof:
Since
f = f (u , v )
is a harmonic function in the variables u, v
55
Chapter One : Function of Several Variables
2 f 2 f
2 + 2 =0
u
v
f f u f v
=
+
x u x v x
→ (1)
differentiating with respect to
x
2 f
f u f v
=
+
2
x u x x v x
x
f 2 u u f f 2 v v f
=
+
+
+
u x 2 x x u v x 2 x x v
f 2 u u f u f v
=
+
+
u x 2 x u u x v u x
+
f 2 v v f u f v
+
+
v x 2 x u v x v v x
2 f 2 f u
2 f u v 2 f v
f 2 u f 2 v
=
+
2
+
+
+
uv x x v 2 x
u x 2 v x 2
x 2
u 2 x
2
2
From Cauchy–Riemann equations.
u v v
u
= , =−
x y x
y
2 f 2 f u
2 f u u 2 f
=
+
−2
uv x y v 2
x 2
u 2 x
2
2
u
f 2 u f 2 u
+
−
→ ( 2)
u x 2 v xy
y
Similarly
2 f 2 f
= 2
y 2
u
2
u
2 f u u 2 f u
f 2 u f 2 u
+ 2
+
+
→ (3) By
−
uv x y v 2 x
u x 2 v yx
y
2
adding (2) and (3)
2
2
2 f 2 f u u 2 f 2 f
+
= +
+
x 2 y 2 x y u 2 v 2
By using (1)
2 f 2 f
+
=0
x 2 y 2
56
Chapter One : Function of Several Variables
Homogeneous function
Definition: A homogeneous function of two variables x and y is a realvalued function f ( x, y ) that satisfies the condition
f (x, y ) = n f ( x, y ) for some constant
n and all real numbers .
The constant n is called the degree of homogeneity.
Example(33):
The function
(x , y ) = x
f
2
y
sin
−1
x
y 4
y
log
+
x
x
y
Is a homogeneous function of degree three , since
4
4
x
y y
f (x, y ) = (x ) 2 y 2 sin −1
+
log
y
x
x
(
)
= xy sin
3
2
−1
4
x
y
3 y
log
+
y
x
x
2
x
y4
−1 y
= xy sin +
log
y
x
x
f (x, y ) = 3 f ( x, y )
3
Euler Theorem for homogeneous function
Theorem: Let
x, y . Then
f ( x, y )
be a homogeneous function of degree n in the variables
f
f
+ y
= nf
x
y
(1)
x
(2)
x2
2
2 f
2 f
f
2
+
2
xy
+
y
= n(n − 1) f
2
2
xy
x
y
Proof:
Let f ( x, y ) be a homogeneous function of degree n in the variables
x, y
57
Chapter One : Function of Several Variables
f (x, y ) = n f ( x, y )
Put u = x, v = y , where
→ (1)
u
v
= x,
=
f (u , v ) = n f ( x, y )
Differentiate both sides w.r.to
f (u , v ) = n n −1 f ( x, y )
f u
f v
+
= n n −1 f ( x, y )
u
v
f
f
x
+ y
= n n −1 f ( x, y )
u
v
At = 1 we have u = x, v = y
x
f
f
+y
= nf (x, y )
x
y
→ (2)
From (1) differentiate both sides w.r. to
x
2 f
f
2 f
f
x
+
+ y
= n
2
x
yx
x
x
x
2 f
2 f
f
(
)
+
y
=
n
−
1
yx
x
x 2
Multiply by
x2
x
2 f
2 f
f
(
)
+
xy
=
n
−
1
x
yx
x
x 2
→ (3)
From (2) differentiate both sides w.r. to y
58
Chapter One : Function of Several Variables
2 f
2 f
f
f
(
)
+y
+
=
n
xy
y
y
y 2
x
2 f
2 f
f
x
+y
= (n − 1)
2
xy
y
y
Multiply by
xy
y
2 f
2 f
f
+ y2
= (n − 1) y
2
xy
y
y
→ (4)
By adding (3) and (4)
2
f
2 f
2 f
f
2 f
x
+ 2 xy
+y
= (n − 1) x
+y
2
2
xy
y
x
y
x
2
From the first part of the theorem , we have
2
2 f
2 f
2 f
x
+ 2 xy
+y
= n(n − 1) f
xy
x 2
y 2
2
If
Example(34):
6
x
x
y y
f ( x, y ) = x 2 y 3 sin −1 +
tan −1 + x 5 log
x
x
y
y
Prove that :
(I)
(II)
x
x
2
f
f
+ y
= 5f
x
y
2
2 f
2 f
f
2
+ 2 xy
+ y
= 20 f
2
2
xy
x
y
Solution: By testing the homogeneous of the function
(
f (x, y ) = x
2
2
6 6
x
x
y y
y sin +
tan −1 + 5 x 5 log
x
x
y
y
)(
3
3
)
−1
6
x
x
y
5 y
f (x, y ) = x y sin +
tan −1 + 5 x 5 log
x
x
y
y
f (x, y ) = 5 f (x, y )
5
2
3
−1
59
Chapter One : Function of Several Variables
The function is homogeneous of degree 5
Then by using Euler’s theorem , we have
(I)
x
f
f
+ y
= 5 f
x
y
Also
x2
(II)
2
2 f
2 f
2 f
+
2
xy
+
y
= 5(5 − 1) f = 20 f
xy
x 2
y 2
Example(35): If u = log v where
degree n in the variables x, y .
Prove that :
x
Solution:
Since
variables x, y .
v
is homogeneous function of
u
u
+ y
= n
x
y
v
is homogeneous function of degree
n in the
From Euler’s theorem
v
v
+ y
= nv
x
y
u = log v
→ (1)
x
u
u v
u
1 v
=
=
x
v x
x
v x
Multiply by
x
x
u x v
=
x v x
→ (2)
Also
u
u v
u
1 v
=
=
y
v y
y
v y
Multiply by
y
u y v
=
y v y
y
→ (3)
60
Chapter One : Function of Several Variables
By adding (2) and (3)
x
u
u
1 v
v
+ y
=
x
+ y
x
y
v x
y
From (1) , we have
x
u
u 1
+y
= (nv ) = n
x
y v
Exercises on Homogeneous functions
n
x− y
prove that
(1) If w =
x+ y
( a) x
w
w
+ y
=0
x
y
( b ) x2
2
2w
2w
2 w
+
2
xy
+
y
=0
xy
x 2
y 2
x
2
2
(2) ) If f (x, y ) = x − y g prove that
y
(
)
(a) xf x + yf y = 2 f
( b)
x 2 f xx + 2 xyf xy + y 2 f yy = 2 f
y
y
( 3) Prove that the function z = x cos + y tan
x
x
Satisfying the relation
x 2 z xx + 2 xyz xy + y 2 z yy = 0
y
prove that:
x
4 2
−1
(4) If z = x y sin
(I)
xz x + yz y = 6 z
(II)
x 2 z xx + 2 xyz xy + y 2 z yy = 30 z
61
Chapter One : Function of Several Variables
3
3
(5) If f (x, y ) = 3 x + y Prove that:
x
f
f
+ y
= f
x
y
y
Prove that:
x
If z = x
(6)
x 2 z xx + 2 xyz xy + y 2 z yy = 0
( 7) Let z = f ( x, y ) where x, y are two homogeneous functions of
degree P on the variables u, v .
Prove that :
u 2 f uu + 2ufzuv + v 2 f vv + uf u + vf v =
(
P 2 x 2 f xx + 2 xyf xy + y 2 f yy + xf x + yf y
)
(8) If f = f (x1 , x 2 , x3 ) is homogeneous function of degree n .
And
u=
x1
x
, v = 2 , w = x3
x3
x3
Prove that :
w
f
= n f
w
Taylor series in several variables
The Taylor series may also be generalized to functions of more than one
variable . For example, for a function z = f ( x, y ) that depends on two
variables, x and y, the Taylor series to second order about the
point (a, b) is
f ( x, y ) = f ( a , b ) + ( x − a ) f x ( a , b ) + ( y − b ) f y ( a , b ) +
1
( x − a ) 2 f xx (a, b) + 2( x − a )( y − b) f xy + ( y − b) 2 f yy (a, b) +
2!
62
Chapter One : Function of Several Variables
Find the Taylor series generated by
Example(36):
f ( x, y ) = ( x + y 2 + 1)
1
2
at ( a, b) = (1,0) .
Solution: We first need to computes all the necessary partial
derivatives:
1
−
1
f x = ( x + y 2 + 1) 2
2
f y = y ( x + y + 1)
2
−
1
2
3
f xx
−
1
= − ( x + y 2 + 1) 2
4
f yy = ( x + y + 1)
2
−
1
2
− y ( x + y + 1)
2
2
−
3
2
3
f xy
−
y
= f yx= − ( x + y 2 + 1) 2
2
Evaluating these derivatives at the origin gives the Taylor coefficients
f (1,0) = 2
f x (1,0) =
2
4
f y (1,0) = 0
f xx (1,0) = −
f yy (1,0) =
1
8 2
1
2
f xy (1,0) = f yx (1, 0 ) = 0
Substituting these values in to the general formula
63
Chapter One : Function of Several Variables
f ( x, y ) = f ( a , b ) + ( x − a ) f x ( a , b ) + ( y − b ) f y ( a , b ) +
1
( x − a ) 2 f xx ( a, b) + 2( x − a )( y − b) f xy + ( y − b) 2 f yy ( a, b) +
2!
produces
1
2
−1
1
f (1,0) = 2 + ( x − 1) + ( y − 0) 0 + ( x − 1) 2
+ 2( x − 1)( y − 0) 0 + ( y − 0) 2 +
4
8 2
2
2!
= 2 +
2
1 2 −1
( x − 1) +
y +
( x − 1) 2 + .
4
2 2
16 2
Example(37):
Find the Taylor series generated by
f ( x, y ) = e x log( y + 1) at (a, b) = (0,0) .
Solution: We first need to computes all the necessary partial
derivatives:
f x = e x log( y + 1)
ex
fy =
1+ y
f xx = e x log( y + 1)
f yy
ex
= −
(1 + y ) 2
f xy = f yx
ex
=
(1 + y ) 2
Evaluating these derivatives at the origin gives the Taylor coefficients
f (0,0) = 0
f x (0,0) = 0
f y (0,0) = 1
64
Chapter One : Function of Several Variables
f xx (0,0) = 0
f yy (0,0) = − 1
f xy (0,0) = f yx (0, 0 ) = 1
Substituting these values in to the general formula
f ( x, y ) = f ( a , b ) + ( x − a ) f x ( a , b ) + ( y − b ) f y ( a , b ) +
1
( x − a ) 2 f xx (a, b) + 2( x − a )( y − b) f xy + ( y − b) 2 f yy (a, b) +
2!
produces
f (0,0) = 0 + ( x − 0)0 + ( y − 0)1 +
= y+x y −
1
( x − 0) 2 0 + 2( x − 0)( y − 0) 1 + ( y − 0) 2 (−1) +
2!
y2
+ .
2
Since log( y + 1) is analytic in | y | < 1, we have
y2
e log( y + 1) = y + x y −
+
2
x
65
Chapter Two :
Multiple Integral
Chapter 2
MULTIPLE INTEGRAL
In this chapter we consider the integral of a function of two variables f ( x, y ) over a
region in the plane and the integral of a function of three variables f ( x, y, z ) over a
region in space. These multiple integrals are defined to be the limit of approximating
Riemann sums, much like the single-variable integrals. We illustrate several
applications of multiple integrals, including calculations of volumes, areas in the plane,
moments, and centers of mass.
Double and Iterated Integrals over Rectangles
In Calculus1 we defined the definite integral of a continuous function f (x) over an
interval [a, b] as a limit of Riemann sums. In this section we extend this idea to define
the double integral of a continuous function of two variables f ( x, y ) over a bounded
rectangle R in the plane. In both cases the integrals are limits of approximating
Riemann sums. The Riemann sums for the integral of a single-variable function f (x)
are obtained by partitioning a finite interval into thin subintervals, multiplying the
width of each subinterval by the value of f at a point C K inside that subinterval,
and then adding together all the products. A similar method of partitioning,
multiplying, and summing is used to construct double integrals.
Double Integrals
We begin our investigation of double integrals by considering the simplest
type of planar region, a rectangle. We consider a function f ( x, y ) defined
c y d
on a rectangular region R , R : a x b ,
We subdivide R into small rectangles using a network of lines parallel to
the x- and y-axes (Figure 1). The lines divide R into n rectangular pieces,
where the number of such pieces n gets large as the width and height of
each piece gets small. These rectangles form a partition of R . A small
rectangular piece of width x and height y has area A = x y .
If we number the small pieces partitioning R in some order, then their
areas are given by
numbers A1 , A2 , , An where
rectangle.
66
AK is the area of the k th small
Chapter Two :
Multiple Integral
Figure ( 1) Rectangular grid partitioning the region R into small rectangles of area
AK = x
K
yK
To form a Riemann sum over R , we choose a point ( x k , y k ) in the
k th small rectangle, multiply the value of
f
at that point by the area
AK , and add together the products:
Sn =
n
k =1
f ( x k , y k ) Ak
Depending on how we pick ( x k , y k ) in the k th small rectangle, we may
get different values for S n
We are interested in what happens to these Riemann sums as the
widths and heights of all the small rectangles in the partition of R
approach zero. The norm of a partition P ,written P , is the largest width
or height of any rectangle in the partition. If
P = 0.1 then all the
rectangles in the partition of R have width at most 0.1 and height at
most 0.1.
Sometimes the Riemann sums converge as the norm of P goes to zero,
written P → 0 . The resulting limit is then written as
n
lim
P →0
k =1
f ( xk , y k ) Ak
As P → 0 and the rectangles get narrow and short, their number n
increases, so we can also write this limit as
67
Chapter Two :
n
lim
n →
k =1
Multiple Integral
f ( xk , y k ) Ak
with the understanding that P → 0 , and hence Ak → 0 , as n → .
There are many choices involved in a limit of this kind. The collection of small
rectangles is determined by the grid of vertical and horizontal lines that determine a
rectangular partition of R. In each of the resulting small rectangles there is a choice of
an arbitrary point
( x k , y k ) at which f is evaluated. These choices together determine a single Riemann
sum. To form a limit, we repeat the whole process again and again, choosing partitions
whose rectangle widths and heights both go to zero and whose number goes to
infinity.
When a limit of the sums S n exists, giving the same limiting value no
matter what choices are made, then the function f is said to be
integrable and the limit is called the double integral of f over R ,
written as
f ( x, y)dA
or
f ( x, y)dx dy
R
R
It can be shown that if f ( x, y ) is a continuous function throughout R ,
then f is integrable, as in the single-variable case. Many discontinuous
functions are also integrable, including functions that are discontinuous
only on a finite number of points or smooth curves.
We leave the proof of these facts to a more advanced text.
Double Integrals as Volumes
When f ( x, y ) is a positive function over a rectangular region R in the xy
-plane, we may interpret the double integral of f over R as the
volume of the 3-dimensional solid region over the xy -plane bounded
below by R and above by the surface Z = f ( x, y ) Figure (2)
68
Chapter Two :
Multiple Integral
Figure (2) Approximating solids with rectangular boxes leads us to define the
volumes of more general solids as double integrals. The volume of the solid
shown here is the double integral of f ( x, y ) over the base region R .
in the sum S n = f ( x k , y k ) Ak is the
Each term f ( xk , y k ) Ak
volume of a vertical rectangular box that approximates the volume
of the portion of the solid that stands directly above the base AK . The
sum S n thus approximates what we want to call the total volume of the
solid. We define this volume to be
Volume =
lim
Sn =
n→
f ( x, y)dA where
Ak → 0 , as n → .
R
Fubini's Theorem for calculating Double Integrals
Suppose that we wish to calculate the volume under the plane
z = 4 − x − y over the rectangular region R = 0 x 2 , 0 y 1
in the
xy -plane. If we apply the method of slicing , with slices perpendicular to
the x-axis Figure.3, then the volume is
x=2
A( x ) dx (1)
x =0
where A(x) is the cross-sectional area at x . For each value of x , we
may calculate A(x) as the integral
69
Chapter Two :
Multiple Integral
y =1
A( x) =
(4 − x −
y ) dy
(2)
y =0
which is the area under the curve z = 4 − x − y in the plane of the crosssection at x . In calculating A(x) , x is held fixed and the integration
takes place with respect to y. Combining Equations (1) and (2), we see
that the volume of the entire solid is
y =1
Volume = A( x)dx = ( 4 − x − y ) dy dx
x =0 y =0
x =0
x=2
x=2
y =1
x=2
x =2
7
y2
(
− x )dx
4
y
−
xy
−
dx
=
=
2
2
y =0
x =0
x =0
2
7
x2
x
−
=
= 5
2
2
0
(3)
If we just wanted to write a formula for the volume, without carrying out
any of the integrations, we could write
2 1
Volume =
(4 − x − y ) dy dx
0 0
The expression on the right, called an iterated or repeated integral, says
that the volume is obtained by integrating 4 − x − y with respect to y
from y = 0 to y =1 , holding x fixed, and then integrating the resulting
expression in x with respect to x from x = 0 to x = 2. The limits of
integration 0 and 1 are associated with y , so they are placed on the
integral closest to dy . The other limits of integration, 0 and 2, are
associated with the variable x , so they are placed on the outside integral
symbol that is paired with dx .
What would have happened if we had calculated the volume by slicing
with planes perpendicular to the y -axis (Figure 4). As a function of y ,
the typical cross-sectional area is
70
Chapter Two :
Multiple Integral
x=2
x =2
x2
A( y ) = ( 4 − x − y )dx = 4 x −
− yx
= 6 − 2y
2
x =0
x =0
(4)
The volume of the entire solid is therefore
y =1
Volume =
y =1
A ( y )dy = (6 − 2 y ) d
y =0
y = 6y − y
y =0
1
2
=5
0
in agreement with our earlier calculation.
Again, we may give a formula for the volume as an iterated integral
by writing
1 2
Volume =
(4 − x −
y ) dx dy
0 0
The expression on the right says we can find the volume by integrating
4 − x − y with respect to x from x = 0 to x = 2 as in Equation (4) and
integrating the result with respect to y from y = 0 to y = 1. In this.
iterated integral, the order of integration is first x and then y , the
reverse of the order in Equation (3).
What do these two volume calculations with iterated integrals have
to do with the double integral
Fubini's Theorem: If
f ( x, y ) is continuous throughout the rectangular
region R : a x b ,
c y d , then
d b
b d
c a
a c
f ( x, y)dA = f ( x. y)dxdy = f ( x, y)dydx
R
Fubini's Theorem says that double integrals over rectangles can be
calculated as iterated integrals. Thus, we can evaluate a double integral
by integrating with respect to one variable at a time.
71
Chapter Two :
Multiple Integral
Fubini's Theorem also says that we may calculate the double integral by
Figure (4) To obtain. The
Figure (3) To obtain the cross-sectional
area A(x ) , we hold x fixed and
cross-sectional area A(y), we hold y
fixed and integrate with respect to x.
integrate with respect to y .
integrating in either order, a genuine convenience. When we calculate a
volume by slicing, we may use either planes perpendicular to the x-axis or
planes perpendicular to the y-axis.
Example (1): Calculate
f ( x, y ) dA
for
R
f ( x, y ) = 100 − 6 x 2 y and
R:
0 x 2,
− 1 y 1
Solution: Figure 5 displays the volume beneath the surface. By Fubini's
Theorem,
R
1 2
1
f ( x, y ) dA = (100 − 6 x y ) dxdy = 100 x − 2 x 3 y
2
−1 0
1
x=2
x =0
−1
= ( 200 − 16 y ) dy = 200 y − 8 y
−1
1
2
= 400
−1
Reversing the order of integration gives the same answer:
72
dy
Chapter Two :
2 1
Multiple Integral
2
f ( x, y ) dA = (100 − 6 x y ) dydx = 100 y − 3 x 2 y 2
R
2
0 −1
2
y =1
y = −1
dx
0
2
= (100 − 3 x ) − (−100 − 3 x ) dx = 200 dx = 400
2
2
0
0
Figure (5) The double integral
f ( x, y) dA . gives the volume under
R
this surface over the rectangular region R
Example (2): Find the volume of the region bounded above by the elliptical
paraboloid
Z = 10 + x 2 + 3 y 2 and below by the rectangle
R:
0 x 2,
− 1 y 1
Solution: The surface and volume are shown in Figure 6. The volume is
given by the double integral
73
Chapter Two :
Multiple Integral
Figure (6) : The double integral gives the volume under this surface aver the
rectangular region R
V
=
(10
+ x 2 + 3 y 2 ) dA
R
1
2
=
(10
=
10 y
0
+ x 2 + 3 y 2 ) dydx
0
y =2
1
+ x
2
y + y
3
dx
y =0
0
1
=
( 20
+ 2 x 2 + 8) dx
0
1
2
86
= 20 x +
x3 + 8 x
=
3
3
0
Example (3): Calculate
y sin( xy) dA where
R =
[1, 2 ] [ 0 , ]
R
Solution 1: If we first integrate with respect to x , we get
R
2
y sin( xy) dA =
y sin( xy) dx dy
0 1
=
− cos( xy)
x = 2
x = 1
dy
0
=
(− cos 2 y + cos y ) dy
0
= −
1
2
74
sin 2 y + sin y
0
= 0
Chapter Two :
Multiple Integral
For a function f that takes on
both positive and negative
values,
f ( x, y) dA is a
R
difference of volumes: V1 − V2 ,
where V1 is the volume above
R and below the graph of f
and V 2 is the volume below R
and above the graph. The fact
that the integral in Example ( 3) is
0 means that these two volumes
and are equal.
Figure (7)
Solution 2: If we reverse the order of integration, we get
2
y sin( xy) dA = y sin( xy) dy dx
R
1 0
To evaluate the inner integral we use integration by parts , so
1
y cos( xy)
0 y sin( xy) dy = − x 0 − x 0 cos( xy) dy
cos x
1
=−
+ 2 sin ( xy) 0
x
x
cos x sin x
=−
+
x
x2
If we now integrate the first term by parts with u = −1 x and dv = cos x dx , we
get du = − dx x 2 , v = sin x , and
−
cosx
x
dx = −
sin x
sin x
−
dx
x
x2
Therefore
(−
cos x
x
+
sin x
sin x
)dx = −
2
x
x
And so
75
Chapter Two :
2
1 0
Multiple Integral
sin x
y sin( xy) dy dx = −
x
1
sin 2
=−
+ sin = 0
2
2
Example (4):
f ( x, y ) =
Let
y 2 − x2
( x2 + y 2 )2
1
Calculate (i) f ( x, y ) dx dy
0 0
1
(ii) f ( x, y ) dy dx
0 0
1
1
Solution: since
y2 − x2
0 ( x 2 + y 2 ) 2 dx =
1
1
2x 2
−
0 x 2 + y 2 ( x 2 + y 2 ) 2 dx
1
1
=
0
1
1
x2
dx − 2 2
dx
x2 + y2
+ y 2 )2
0 (x
(1)
we use integration by parts , so
x2
1
x
1
dx
( x 2 + y 2 ) 2 dx = − 2 ( x 2 + y 2 ) + 2 ( x 2 + y 2 )
Substitution in (1) we get
1
0
1
y2 − x2
x
1
dx
=
=
2
2 2
2
2
2
(x + y )
x + y 0 1+ y
1
f
(
x
,
y
)
dx
dy =
0 0
1
1
dy
1+ y
2
= tan −1 y
1
0
=
0
Similarly ,
1
1
1 y2 − x2
0 0 f ( x, y ) dy dx = 0 0 x 2 + y 2 dy dx
1
1
=−
dx
1+ x
2
= − tan −1 x
0
1
0
=−
4
We notice that , in this example
1 1
1 1
f ( x, y ) dx dy
0 0
0 0
76
f ( x, y ) dy dx
4
Chapter Two :
Multiple Integral
This is because the function f ( x, y ) is discontinuous on the region R bounded by
the rectangle 0 y 1
Example ( 5 ): Calculate
,0 x 1
(x sin y − ye ) dx dy
x
where
D
the region
D
bounded by the curves x = 1
, x = −1
,y=0
,y=
2
Solution 1: If we first integrate with respect to x and then with respect to y , we
get
Figure (8)
I = ( x sin y − y e x ) dx dy =
0 −1
1
2
2
=
1
x2
x
0 2 sin y − y e dy
−1
2
2
( − e y + e −1 y ) dy = ( e −1 − e ) y dy
0
0
= (e
−1
y2 2
2
− e )
=
( e −1 − e )
8
2 0
Solution 2: If we first integrate with respect to y and then x , we get
2
x
I = ( x sin y − y e ) dy dx =
−1 0
1
1
= (x −
−1
2
8
y2 x 2
−
x
cos
y
−
e dx
2
0
−1
1
1
e ) dx =
x
x2
2 x
2
−
e
=
( e −1 − e )
2
8
8
−1
77
Chapter Two :
Multiple Integral
Double Integrals over General Regions
For single integrals, the region over which we integrate is always an interval. But for
double integrals, we want to be able to integrate a function f not just over
rectangles but also over regions D of more general shape, such as the one illustrated
in Figure 9. We suppose that D is a bounded region, which means that D can be
enclosed in a rectangular region R as in figure 10. Then we define a new function F
with domain R by
f ( x, y )
F ( x , y) =
0
if ( x , y ) is in D
if ( x, y ) is in R but not in D
Figure (9)
(1)
Figure (10)
If the double integral of F exists over R , then we define the double integral
of over D by
Definition:
f ( x, y )dA = F ( x, y )dA
D
(2)
R
Where F is given by Equation (1)
In the case where we can still interpret
f ( x, y )dA
as the volume of
D
the solid that lies above D and under the surface z = f ( x, y ) (the graph
of f ). You can see that this is reasonable by comparing the graphs of f
78
Chapter Two :
Multiple Integral
and in Figures 11 and 12 and remembering that
F ( x, y)dA is the
R
volume under the graph of F .
Figure (12)
Figure ( 11)
Figure 12 also shows that F is likely to have discontinuities at the
boundary points of D . Nonetheless, if f is continuous on D and the
boundary curve of D is “well behaved” , then it can be shown that
F ( x, y)dA exists and therefore f ( x, y )dA exists. In particular, this
R
D
is the case for the following types of regions.
A plane region is said to be of type I if it lies between the graphs of
two continuous function of x , that is ,
D = ( x, y ) | a x b ,
g1 ( x ) y g 2 ( x )
Where g1 ( x) and g 2 ( x) are continuous on a, b . Some examples of type
I regions are shown in Figure 13
Figure (13)
Some type I regions
79
Chapter Two :
Multiple Integral
In order to evaluate f ( x, y )dA when D is a region of type I, we
D
choose a rectangle R = a, b c , d that contains D , as in Figure 6, and
we let F be the function given by Equation 1 ; that is, F agrees with f
on D and F is 0 outside D . Then, by Fubini’s Theorem,
f ( x, y ) dA = F ( x, y ) dA =
D
R
b d
F ( x, y )dydx
a c
Figure (14)
F ( x, y ) = 0
Observe that
if
y g1 ( x )
or
y g 2 ( x)
because
( x, y)
then lies outside D Therefore
d
g2 ( x)
g2 ( x)
c
g1 ( x )
g1 ( x )
F ( x , y ) dy = F ( x, y) dy = f ( x, y) dy
Because F ( x, y ) = f ( x, y ) when g1 ( x) y g 2 ( x) . Thus, we have the
following formula that enables us to evaluate the double integral as an
iterated integral.
Theorem: If
f
is continuous on a type I region D such that
D = ( x , y ) | a x b , g1 ( x ) y g 2 ( x )
b g2 ( x)
Then
f ( x, y )dA = a g ( x)f ( x, y)dydx
D
(3)
1
The integral on the right side of (3) is an iterated integral that is similar to
the ones we considered in the preceding section, except that in the inner
80
Chapter Two :
Multiple Integral
integral we regard as being constant not only in
f ( x, y )
but alo in the limit
of integration , g1 ( x) and g 2 ( x)
We also consider plane regions of type II , which can be expressed as
D = ( x , y ) | c y d , h1 ( y ) x h2 ( y )
(4)
Where h1 and h2 are continuous. Two such regions are illustrated in
Figure 7. Using the same methods that were used in establishing (3), we
can show that
d h2 ( y )
f ( x, y )dA = c h ( y )f ( x, y)dxdy
D
(5)
1
where D is a type II region given by Equation 4.
Figure (15) Some type II regions
Properties of Double Integrals
Like single integrals, double integrals of continuous functions have
algebraic properties that are useful in computations and applications.
If ( x, y ) and ( x, y ) are continuous on the bounded region D , then the
following properties hold
(a) Sum and difference
( x, y) ( x, y) ds = ( x, y) ds ( x, y) ds
D
D
(b) Constant multiple
81
D
Chapter Two :
Multiple Integral
a ( x, y )ds = a ( x, y)ds
D
D
(c) Additivity If D is the union of two non overlapping regions
D1 and D2 except perhaps on their boundaries
f ( x, y)ds = f ( x, y)ds + f ( x, y)ds
D
Example( 6): Calculate
D1
D2
( x + 2 y ) dA
where D is the region
D
2
2
bounded by the parabolas y = 2 x and y = 1 + x
2
2
2
Solution: The parabolas intersect when 2 x = 1 + x , that is, x = 1 , so
x = 1 . We note that the region D , sketched in Figure 16, is a type I
region but not a type II region and we can write.
D = ( x , y ) | − 1 x 1, 2 x 2 y 1 + x 2
Figure (16)
Since the lower boundary is y = 2 x and the upper boundary is y = 1 + x , Equation
2
2
(3) gives
82
Chapter Two :
Multiple Integral
1 1+ x 2
( x + 2 y ) dA = ( x + 2 y ) dy dx
−1 2 x 2
D
1
= xy + y 2
y =1+ x 2
y =2 x 2
dx
−1
1
= ( −3 x 4 − x 3 + 2 x 2 + x + 1) dx
−1
1
x5 x 4
x3 x 2
32
= −3
−
+2
+
+ x =
5
4
3
2
−1 15
NOTE ● When we set up a double integral as in Example 1, it is essential to draw a
diagram. Often it is helpful to draw a vertical arrow as in Figure 8. Then the limits of
integration for the inner integral can be read from the diagram as follows: The
arrow starts at the lower boundary y = g1 ( x) , which gives the lower limit in the
integral, and the arrow ends at the upper boundary y = g 2 ( x) , which gives the
upper limit of integration. For a type II region the arrow is drawn horizontally from
the left boundary to the right boundary.
Example( 7): Evaluate
( x
2
+ y 2 ) dA where D is the region
D
bounded by the curves
y = 0 , y = x2
,x = 0
,x =1
Solution: first we must sketch the region D bounded by
y = 0 , y = x2
,x = 0
, x = 1 , see Figure 17
Figure (17)
83
Chapter Two :
2
2
( x + y )dA =
D
Multiple Integral
0
x2
2
y
2
2
(
x
+
y
)
dy
dx
=
x
y
+
dx
0
0
3 0
1 x2
1
3
1
x5 x7
x6
26
=(x +
) dx =
+
=
3
21 0 105
5
0
1
4
Example(8): Calculate
D
sin x
dx dy
x
where D is the triangle in
the xy-plane bounded by the x − axis , the line y = x
and the line x = 1
Solution: The region of integration is shown in Figure 18. If we
integrate first with respect to y and then with respect to x, we find
Figure (18)
1
x
0
0
(
y=x
sin x
sin x
dy ) dx = y
dx = sin x dx
x
x
y =0
0
0
1
1
= − cos1 + 1 0.46
If we reverse the order of integration and attempt to calculate
1
1
0
y
sin x
dx dy
x
84
Chapter Two :
we run into a problem because
Multiple Integral
((sin x)
x )dx cannot be expressed in
terms of elementary functions (there is no simple antiderivative).
There is no general rule for predicting which order of integration will be
the good one in circumstances like these. If the order you first choose
doesn't work, try the other. Sometimes neither order will work, and then
we need to use numerical approximations.
Example(9): Calculate
e
y/x
dy dx where D is the triangle in the
D
𝑥𝑦 −plane bounded by the lines 𝑦 = 𝑥, 𝑦 = 2𝑥, the lines x = 1 , x = 2
Solution: The region of integration is shown in Figure 19. If we integrate
first with respect to y and then with respect to x, we find
Figure (19)
e
2 2x
y/x
dy dx =
D
1
e
x
2
y
x
2
=
2x
y
dy dx = x e x
x
1
dx
2
x
2
( x e − x e ) dx = ( e − e ) x dx
1
1
2
x2
3
= ( e2 − e )
( e2 − e )
=
2
2
1
85
Chapter Two :
Example(10): Calculate
( xy)
Multiple Integral
where D is the region
dA
D
2
2
bounded by the parabolas y = x and y = x
4
4
Solution : The parabolas intersect when x = x , that is, x − x = 0 , so
x = 1 , 0 . We note that the region D , sketched in Figure 20
Figure (20-b)
Figure (20-a)
Case (1): If we integrate first with respect to
1
( xy)dA =
D
0
x
y , from Figure (20-a)
1
1
(
xy
)
dy
dx
=
x y2
2
20
x
x
x
dx
2
1
1
1
1 x3 x6
1
2
5
= ( x − x ) dx =
−
=
20
2 3
6 0 12
Case (2): If we integrate first with respect to
1
( xy)dA =
D
0
y
1
x , from Figure (20-b)
1
(
xy
)
dx
dy
=
y x2
2
20
y
y
y
dy
2
1
1
1
1 y3 y6
1
2
5
= ( y − y )dy =
−
=
20
2 3
6 0 12
1 1
Example ( 11 ) : Evaluate the iterated integral
sin ( y
0 x
.
86
2
) dy dx
Chapter Two :
Multiple Integral
Solution. If we try to evaluate the integral as it stands, we are faced with
the task of first evaluating sin( y 2 ) dy . But it’s impossible to do so in finite
terms since sin( y 2 ) dy is not an elementary function. So we must Change
the order of integration. This is accomplished by first expressing the given
iterated integral as a double integral. Using (3) backward, we have
1 1
sin ( y
2
) dy dx =
f ( x, y ) dA
D
0 x
Where
D = ( x , y ) | 0 x 1,
x
y 1
We sketch this region D in Figure (21-a ). Then from Figure (21-b) we see that an
alternative description of D is.
D = ( x , y ) | 0 y 1, 0 x y
This enables us to use (5) to express the double integral as an iterated integral in the
reverse order:
1 1
sin ( y
2
)dy dx =
f ( x, y )dA
D
0 x
1 y
=
sin ( y
1
2
) dx dy = x sin ( y )
0 0
1
=
0
x= y
0
2
dy
x =0
1
1
y sin ( y 2 ) dy = − cos ( y 2 )
2
0
1
= ( 1 − cos1 )
2
Figure (21-a)
Figure (21-b)
87
Chapter Two :
Multiple Integral
Property 3 can be used to evaluate double integrals over regions that
are neither type I nor type II but can be expressed as a union of regions
of type I or type II.
Example (12 ) Evaluate the integral
xy dx dy . where
D is
D
x - axis , the parabolas y = x 2 and the line
the region bounded by
y = 2− x
.
Solution: We sketch this region D in Figure (22).
Figure ( 22)
1 x = 2− y
xy dx dy =
0 x=
D
1
xy dx dy =
0
y
y
x2
2
2− y
y
dy
1
1
= 1 (2 − y ) 2 − ydy = 1 y y 2 − 5 y + 4dy
2
2
0
0
1
1
1
1 y4 5y3
7
3
2
(
y
−
5
y
+
4
y
)
dy
=
−
+ 2y2 =
=
20
2 4
3
0 24
If we want to Change the order of integration , then the region D is
divided into two regions D1 , D2 , Figure (23)
88 (23)
Figure
Chapter Two :
Multiple Integral
xy dx dy = xy dy dx + xy dy dx
D
D1
D2
2 2− x
1 x2
=
=
xydy dx
0
0
1
x
y2
2
0
=
=
1
2
+
x y dy dx
1
x2
0
2
dx +
5
dx +
0
1
1
x
0
1
2
x
y2
2
2− x
dx
0
2
(x
3
− 4 x 2 + 4 x ) dx
1
1
1
5
7
+
(
)=
12
2 12
24
Example( 13) Evaluate the integral
xy dx dy . where
D is the
D
region in the first quadrant bounded by y - axis , the line y = x and
2
2
the circle x + y = 1
Solution: We sketch this region D in Figure (24).
Figure(24-b)
Figure (24-a)
From Figure (17-a) , we have
1
2 y=
xy dx dy =
D
0
1
1− x 2
xy dy dx =
y=x
2
0
89
x
y2
2
x
1− x 2
dx
Chapter Two :
Multiple Integral
1
1
=
2
x (1 − x)
2
2
−x
2
0
1
2
1
dx =
2
(x − 2x
3
) dx
0
1
1 x2
1
−
x4
=
2 2
2
0
2
1
16
=
If we want to Change the order of integration , then the region D is divided into
two regions
D1 , D2 ,
Figure (24-b)
xy dx dy = xy dx dy + xy dx dy
D
D1
D2
1
2
y
xydx dy
=
0
1− y 2
1
+
x y dx dy
1
0
0
2
1
2
=
0
y
2
x
2
y
0
1
dy +
1
y
2
x
2
1− y 2
0
dy
2
1
=
1
2
2
y 3 dy +
0
1
2
1
(y −
y 3 ) dx
1
2
=
1
1
1
+
=
32
32
16
Example ( 14 ): Find the volume of the tetrahedron bounded by the planes
x = 2 y,
x = 0,
x + 2 y + z = 2 and
z=0
Solution:
In a question such as this, it’s wise to draw two diagrams: one of the three
dimensional solid and another of the plane region D over which it lies.
Figure 13 shows the tetrahedron T bounded by the coordinate planes
x = 0,
z = 0 , , the vertical plane x = 2 y , and the plane
x + 2 y + z = 2 . Since the plane x + 2 y + z = 2 intersects the – x y
plane (whose equation is z = 0 ) in the line x + 2 y = 2 , we see that T
lies above the triangular region D in the x y -plane bounded by the lines
90
Chapter Two :
, x = 2 y,
x + 2 y = 2 and
Multiple Integral
x = 0 . (See Figure 25.) The plane
z = 2 − x − 2 y , so the required
volume lies under the graph of the function z = 2 − x − 2 y and above
x + 2 y + z = 2 can
be written as
D = ( x , y ) | 0 x 1,
x
2
y 1− x
2
Figure ( 25)
V =
D
1
1− x 2
Z d x d y = ( 2 − x − 2 y ) dy dx
0 x
2
1
= 2 y − xy − y
2
0
Example( 15 ):
x = 0,
y =1− x
y= x
1
2
dx =
2
(x
2
− 2 x + 1) dx =
0
Calculate the volume bounded by
y = 0,
x + y + z = 1,
Solution: We sketch this region D in Figure (26).
91(26)
Figure
z = 0.
1
3
Chapter Two :
V =
R
Multiple Integral
1− y
Z d x d y = (1 − x − y ) dx dy
0
0
1
1− y
x2
= (1 − y ) x −
2 0
0
1
1
1
1
2
(
1
−
y
)
dy
=
0 2
6
dy =
Area by Double Integration
In this section we show how to use double integrals to calculate the
areas of bounded regions in the plane, and to find the average value of a
function of two variables.
Areas of Bounded Regions in the Plane
If we take f ( x, y ) = 1 in the definition of the double integral over a region
R in the preceding section, the Riemann sums reduce to
Sn =
n
k =1
f ( xk , yk ) Ak =
n
A
K =1
K
This is simply the sum of the areas of the small rectangles in the
partition of R , and approximates what we would like to call the area of
R . As the norm of a partition of R approaches zero, the height and
width of all rectangles in the partition approach zero, and the coverage
of R becomes increasingly complete . We define the area of R to be
the limit
n
lim
P →0
A
k
k =1
=
d
A
R
Definition: The area of a closed, bounded plane region R is
A =
d
R
92
A
Chapter Two :
Multiple Integral
As with the other definitions in this chapter, the definition here applies to a
greater variety of regions than does the earlier single-variable definition of area, but
it agrees with the earlier definition on regions to which they both apply.
To evaluate the integral in the definition of area, we integrate the
constant function f ( x, y ) = 1 over R .
Example( 16 ) Find the area of the region R bounded by y = x and the
2
parabola y = x in the first quadrant.
Solution: We sketch the region (Figure 27 ), noting where the two curves
intersect at the origin and
(1,1) , and calculate the area as
Figure ( 27)
x
A = dy dx = d y dx =
R
0
x 2
1
1
y
x
x2
dx
0
1
x2
x3
1
=( x− x )d x=
−
=
3 0
6
2
0
1
2
Example( 17 ) Find the area of the region R bounded by y = x and the
2
parabola y = 2 − x
Solution: We sketch the region (Figure 28 ), noting where the two
curves intersect at (1,1), (−2,2) , and calculate the area as
93
Chapter Two :
Multiple Integral
Figure ( 28) )
2− x
A = dy dx = d y dx =
R
− 2
x
2
1
1
2− x
y x
2
dx
−2
1
x2
x3
9
= ( 2 − x − x ) d x = 2 x −
−
=
2
3 −2
2
−2
1
2
On the other hand, if we reverse the order of integration, then the
region R will divided into two region
Double Integral in Polar Coordinates
Integrals are sometimes easier to evaluate if we change to polar
coordinates. This section shows how to accomplish the change and how
to evaluate integrals over regions whose boundaries are given by polar
equations
When we defined the double integral of a function over a region R in
the xy -plane, we began by cutting R into rectangles whose sides were
parallel to the coordinate axes. These were the natural shapes to use
because their sides have either constant x -values or constant y values.
Suppose that we want to evaluate a double integral , where is one of
94
Chapter Two :
Multiple Integral
the regions shown in Figure 29. In either case the description of in terms
of rectangular coordinates is rather complicated but is easily
described using polar coordinates
In polar coordinates, the natural shape is a ''polar rectangle" whose
sides have constant r - and -values.
Figure ( 29 )
Recall that the polar coordinates , ( r , ) of a point are related to the
rectangular coordinates, ( x, y ) by the equations
x = r cos ,
r 2 = x2 + y 2 ,
y = r sin
Although we have defined the double integral in terms of ordinary
rectangles, it can be shown that, for continuous functions , we always
obtain the same answer using polar coordinates
Change to Polar Coordinates in a Double Integral
Theorem: If
f
is continuous on a polar rectangle R given by ,
0a r b ,
, where ,
0 − 2
, then
b
f ( x, y ) d A = f ( r cos , r sin ) r dr d
R
(2)
a
The formula in (2) says that we convert from rectangular to polar
coordinates in a double integral by writing x = r cos and
95
Chapter Two :
Multiple Integral
y = r cos , using the appropriate limits of integration for r - and
, and replacing dA by r dr d .
Be careful not to forget the additional factor r the right side of
Formula 2. A classical method for remembering this is shown in Figure 5,
where the “infinitesimal” polar rectangle can be thought of as an
Example( 18 ): Evaluate ,
(3x + 4 y
2
) d A where R is the region in
R
2
2
2
2
the upper half-plane bounded by the circles x + y = 1 and x + y = 4
Solution: The region can be described as
R = ( x , y ) | y 0 , 1 x 2 + y 2 4
It is the half-ring shown in Figure 1(b), and in polar coordinates it is given
by , 1 r 2 , 0 . Therefore, by Formula 2,
(3x + 4 y
2
2
) d A = ( 3r cos + 4 r 2 sin 2 ) r dr d
R
0 1
= r 3 cos + r 4 sin 2
2
1
d
0
= ( 7 cos + 15 sin 2 ) d
0
15
15
15
= 7 sin +
−
sin 2 =
2
4
2
0
Example( 19 ): Find the volume of the solid bounded by the plane
z = 0 and the paraboloid z = 1 − x − y
2
2
Solution: If we put z = 0 in the equation of the paraboloid, we get
1 = x 2 + y 2 . This means that the plane intersects the paraboloid in the
2
2
circle 1 = x + y , so the solid lies under the paraboloid and above the
2
2
circular disk D given by x + y 1 [see Figures 30]. In polar coordinates
D is given by , 0 r 1 , 0 2 .
96
Chapter Two :
Multiple Integral
Figure ( 30)
Since
1 − x 2 − y 2 =1 − r 2 ,
V =
the volume is
2
2
(1 − x − y ) dA =
D
2 1
(1 − r
2
) r dr d
0 0
2
1
r2
r4
= d ( r − r ) dr = 2
−
=
4 0
2
2
0
0
1
3
If we had used rectangular coordinates instead of polar coordinates,
then we would have obtained
V =
( 1 − x
1
2
− y ) dA =
2
−1 −
D
1− x 2
(1 − x
2
− y 2 ) dy dx
1− x 2
which is not easy to evaluate because it involves finding the following
integrals:
1 − x 2 dx ,
Example ( 20 ): Evaluate ,
x
2
e
1 − x 2 dx ,
x2 + y2
(1 − x
2
)
3
2
dx
d y dx where R is the
R
semicircular region bounded by the x-axis and the curve y = 1 − x 2
Solution: In Cartesian coordinates, the integral in question is a nonx +y
elementary integral and there is no direct way to integrate e
with respect to either x or y . Yet this integral and others like it are
important in mathematics--in statistics, for example--and we need
2
97
2
Chapter Two :
Multiple Integral
to find a way to evaluate it. Polar coordinates save the day. Substituting
r 2 = x2 + y 2 ,
x = r cos ,
and replacing dx dy
by
r dr d
y = r sin
enables us to evaluate the integral as
Figure ( 31 ) The semicircular region
e
x2 + y 2
d y dx =
R
1
e
r2
0 0
=
0
0 r 1, 0
1
2
1
r dr d = e r d
2
0
0
1
( e −1 ) d =
( e −1 )
2
2
If f (r , ) is the constant function whose value is 1 , then the integral
of f over R is the area of R .
Area in Polar Coordinates
The area of a closed and bounded region R in the polar coordinate
plane is
A = r dr d
R
This formula for area is consistent with all earlier formulas, although we
do not prove this fact.
2
Example ( 21 ): Find the area enclosed by the lemniscate r = 4 cos 2
98
Chapter Two :
Multiple Integral
Solution: We graph the lemniscate to determine the limits of
integration (Figure 32 ) and see from the symmetry of the region that
the tota1 area is 4 times the first-quadrant portion.
Figure ( 32 ) To integrate over the shaded region , we run
4 cos 2 and from 0 to
4 cos 2
4
A =4
0
r dr d = 4
0
=4
4
from 0 to
4
r
r=
4 cos 2
2
2
0
4
r
d
r =0
2 cos 2 d = 4 sin 2 0 4
=4
0
Substitutions in Multiple Integrals
The goal of this section is to introduce you to the ideas involved in
coordinate transformations. You will see how to evaluate multiple
integrals by substitution in order to replace complicated integrals by ones
that are easier to evaluate. Substitutions accomplish this by simplifying
the integrand, the limits of integration, or both. A thorough discussion of
multi variable transformations and substitutions, and the Jacobian, is best
left to a more advanced course following a study of linear algebra.
99
Chapter Two :
Multiple Integral
Substitutions in Double Integrals
Suppose that a region G in the uv -plane is transformed one-to-one into
the region R in the xy -plane by equations of the form
x = g( u , v ) ,
y =h(u,v)
as suggested in Figure 33. We call R the image of G under the
transformation, and G the preimage of R . Any function f ( x, y )
defined on R can be thought of as a function f ( g (u , v), h(u , v)) defined
on G as well. How is the integral of f ( x, y ) over R related to the integral
of f ( g (u , v), h(u , v)) over G ?
The answer is: If g , h and f have continuous partial derivatives and
J ( u , v ) (to be discussed in a moment) is zero only at isolated points, if
at all, then
f ( x, y ) dx dy = f ( g (u, v), h(u, v)) J ( u, v )
R
du dv
(1)
G
Cartesian
uv -plane
Cartesian xy -plane
Figure (33)
The factor J ( u , v ) , whose absolute value appears in Equation (I), is the
Jacobian of the coordinate transformation, named after German
mathematician Carl Jacobi. It measures how much the transformation is
expanding or contracting the area around a point in G as G is
transformed into R .
100
Chapter Two :
Multiple Integral
Definition: The Jacobian determinant or Jacobian of the coordinate
transformation x = g ( u , v ) , y = h ( u , v ) is ,
x
u
J (u ,v) =
y
u
x
v x y y x
=
−
y u v u v
v
( 2)
The Jacobian can also denoted by
J (u ,v)=
Example ( 1): Evaluate ,
( x, y)
(u ,v)
y − x d y dx where
D is the region bounded
D
by the lines
y = x +1
, y = x −3
,y=
−1
7
x+
3
3
,y=
−1
x+5
3
Solution: In Cartesian coordinates, the region of integral in question is
D
y = x +1
, y = x −3
By using the substitution
,y=
−1
7
x+
3
3
u = y−x
,y=
,v = y +
−1
x+5
3
1
x
3
The region D changes to the region D in Figure ( 34 ), since
y = x + 1,
y =
y = x−3
gives u = 1, u = −3 ,
−1
7
−1
x+ , y =
x+5
3
3
3
v =
gives
7
, v = 5.
3
To apply Equation (I), we need to find the corresponding uv -region and
the Jacobian of the transformation. To find them, we first solve Equations
(4) for x and y in terms of u and v . From those equations it is easy to
see that
x=
−3
3
u+ v
4
4
x
u
J (u ,v) =
y
u
,y =
x
−3
v
4
=
y
1
4
v
101
3
3
1
3
u+ v
4
4
4 = −3
4
4
J =
3
4
Chapter Two :
Multiple Integral
Figure ( 34 )
1
3
3 3
−3
u + v dudv
4
4 4
( y − x) dx dy = 4 u + 4 v −
D
D
5 1
3
3
= u du dv = u du dv = -8
4 D
4 7 −3
3
Example ( 2 ): Evaluate ,
e
( y − x ) /( x + y )
dx dy where D is the
D
region in xy - plane bounded by the lines
y = 0 ,x = 0,x + y = 2 ,x + y = 4
Solution: We sketch the region D of integration in the xy -plane and identify
its boundaries Figure ( 35 )
Figure ( 35-a)
By using the substitution
Figure ( 35-b )
u = y − x,
v= y+x
The region D changes to the region D , since
102
(4)
Chapter Two :
y = 0 gives
Multiple Integral
u = −v ,
u = − x, v = x i.e.,
x = 0 gives u = y , v = y i.e.,
u=v,
v=2 ,
x + y = 2 gives
x + y = 4 gives v = 4 .
To apply Equation (I), we need to find the corresponding
uv -region
and the
Jacobian of the transformation. To find them, we first solve Equations (4) for
y in terms of u and v . From those equations it is easy to see that
x=
1
(v − u ),
2
x
u
J (u ,v) =
y
u
I=
e
y =
x
1
v − 2
=
y
1
2
v
( y − x ) /( x + y )
1
=
2
4
v
e
1
2 = −1 J =1
1
2
2
2
D
u
v
du dv
2 −v
1
=
2
and
1
(v + u )
2
dx dy =
D
x
u
1 v
e d u dv
2
v
u
v
v
e
dv
2
−v
4
4
1
1
1
=
( e − ) v d v = 3 ( e − )
2
e 2
e
Example ( 3 ): Evaluate , I =
xy dx dy
where D is the region in xy -
D
plane bounded by the lines
y2 = x
, y 2 = 2x
x2 = y
, x2 = 2y
Solution: We sketch the region D of integration in the xy -plane and
identify its boundaries Figure ( 36) .
By using the substitution
y2
u=
,
x
x2
v=
y
(5)
The region D changes to the region D , which is the square
u = 1, u = 2, v = 1, v = 2
103
Chapter Two :
Multiple Integral
Figure ( 36-a)
Figure ( 36-b )
, we first solve Equations (5) for x and y in terms of u and v . From
those equations it is easy to see that
x = ( u v2 )
J (u ,v) =
1
, y = ( u2 v )
3
1 2 2 −2 3
v (u v )
3
( x, y )
=
(u , v) 3 u v (u 2 v )( u 2 v ) −2 3
2
1
3
−2
2
2
u v(u v ) 3 1
1
3
=
−
J
=
1 2 2 −2 3
2
3
u (u v )
3
We can find the Jacobian from the relation
( x, y )
1
=
(u , v )
(u , v )
( x, y )
u
(u ,v)
x
=
v
( x, y )
x
u
y2
−
y
x2
=
v
2x
y
y
\
J =
1
3
2y
−x
x
2
= 3 J =
y2
, uv = xy
Then the integration gives
2
2
3
1
1
I = uv du dv = uv du )
dv =
4
3
3 11
D
104
1
3
Chapter Two :
Note: When we change from
( r , )
( x, y )
Multiple Integral
coordinate to the polar coordinate
x = r cos
, the relation is
, y = r sin
The Jacobian is
x
(x , y)
r
J =
=
y
( r , )
r
x
cos
=
y
sin
− r sin
r cos
= r
So we have ,
D
2 ( )
f ( x, y ) dx dy = F ( r , ) r dr d
= 1 ( )
=
Example ( 4 ): Evaluate ,
e
−( x2 + y 2 )
dx dy
where D is the
D
region in xy - plane bounded by the circle
x2 + y2 = a2
Solution: The region D of integration in the xy -plane is a circle
x2 + y2 = a2
By using the substitution
x = r cos
The value of the Jacobian
, y = r sin
( x, y )
= r
( r , )
The region D changes to the region D , which is the rectangle
,0 2
0r a
Figure ( 37 )
Then the integration
105
Chapter Two :
e
I =
−r 2
D
=
Multiple Integral
2
2
a −r 2
−1
− r2
r dr d = e r dr d =
e
2 0
0 0
− 1 − a2
(e
−1)
2
2
a
d
0
d = ( e − a −1)
2
0
(
)
2
2
Example ( 5 ): Evaluate , I = log x + y + 1 dx dy where D is
D
x2 + y2 = a2
the region in xy - plane bounded by the circle
Solution: The region D of integration in the xy -plane is a circle
x 2 + y 2 = a 2 . By using the substitution
x = r cos
, y = r sin
The region D changes to the region D , which is the rectangle
,0 2
0ra
I =
log (x
2
)
+ y 2 + 1 dx dy = log ( r 2 + 1) r dr d
D
D
2
a
= r log ( r 2 + 1) d r d
00
1
=
2
2
( ( r
2
a
)
+ 1) log ( r 2 + 1 − ( r 2 + 1)
0
d
0
1
=
( a 2 + 1) log ( a 2 + 1) − ( a 2 + 1) + 1
2
= ( a 2 + 1) log ( a 2 + 1) − 1 +
Exercises
106
2
d
0
Chapter Two :
1-
Multiple Integral
Evaluate the iterated integral.
2 4
(a)
4 4
2 xy dy dx
(b)
1 0
e
0
y ) dx dy
1 0
2
ln 2 ln 5
(c)
x
(2 +
2 x+ y
dy dx
(d)
2
y sin x dx dy
−1 0
1
2- Evaluate the double integral. over the given region
(a)
( 6 y
2
− 2 x ) dA ,
R : 0 x 1, 0 y 2
R
(a)
xy 3
R x 2 + 1 dA ,
R : 0 x 1, 0 y 2
(a)
xy cos y dA
R : −1 x 1 , 0 y
,
R
3- Find the volume of the region bounded above by the plane
z= y
2 and below by the rectangle
R : 0 x 4, 0y 2
4- Find the volume of the region bounded above by the paraboloid
z = x 2 + y 2 and below by the square R : − 1 x 1, − 1 y 1
5- Find the volume of the region bounded above by the elliptical
Paraboloid z = 16 − x − y
2
2
and below by the square
R : 0 x 2, 0 y 2
6- Write an iterated integral for
f ( x, y) dA over the described region
R
a) vertical cross-sections, (b) horizontal cross-sections.
(I)
R bounded by y = 0, y = x and
107
x= 9
R using (
Chapter Two :
(I)
R bounded by
(I)
R bounded by y = x 2 and
y = 0, y = 1 , x = 0
7- Finding Limits of Integration
Multiple Integral
and
y = ln x
y = x+2
f ( x, y) dx dy
where D is the region bounded
D
by the curves
(I)
x = 2 , x = 3 , y = −1, y = 5
(II)
y = 0, y = 1 − x 2
(III)
x2 + y2 = a2
(IV)
y=
(V)
y =0 ,y =a
2
1+ x2
, y = x2
,y = x
, y = x − 2a
8- Sketch the region of integration and evaluate the integral.
x
(a)
sin x
x sin y dy dx
y dy dx
(b)
0
0 0
ln 8 ln y
(c)
e
1
x+ y
0
1 y2
dx dy
(d)
0
3y
2
e xy dx dy
0 0
9- Sketch the region of integration, reverse the order of integration, and evaluate
the integral.
(a)
sin y
dy dx
y
0 x
2
2 4− x
(c)
0
0
xe2 y
dy dx
4− y
3
(b)
1
x
e
0
y 3
3
dx dy
1 1
(d)
x
2
e xy dx dy
0 y
10- Sketch 1he region bounded by 1he given lines and curves. Then express the
region's area as an iterated double integral and evaluate 1he integral
108
Chapter Two :
Multiple Integral
x
( a ) The lines y = 2, y = 1 − x and the curve y = e
2
2
( a ) The parabolas x = y − 1 and x = 2 y − 2
2
( a ) The parabola x = y − y and the line y = − x
11 - Change the Cartesian integral into an equivalent polar integral. Then evaluate
the polar integral.
1− y 2
1
(a)
(x
0
+ y 2 ) dx dy
2
0
(ln 2 ) 2 − y 2
ln 2
(b)
0
(c)
dx dy
1
dy dx
( x + y 2 )2
1
x2 + y2
0
2 x− x2
2
e
2
0
12- Sketch the region of integration and convert each polar integral or sum of
integrals to a Cartesian integral or sum of integrals. Do not evaluate the integrals.
2 1
(a)
r
0
0
2
(b)
sin cos dr d
3
csc
r
6
r
0
cos dr d
1
4 2 sec
(c)
2
sin 2 dr d
5
0
13- Use the transformation
x + 2y = , x − y =v
to evaluate the
integral
2
32 − 2 y
( x + 2 y) e
0
( y − x)
dx dy
y
by lust writing it as an integral over a region G in the uv-plane
109
Chapter Two :
14- Use the transformation
2
integral
x = u + (1 2 ) v , y = v
to evaluate the
( y +4)
2
y
0
Multiple Integral
3
( 2 x − y ) e ( 2 x − y ) dx dy
2
y 2
by lust writing it as an integral over a region G in the uv-plane
2
2
15- Use the transformation x = u − v , y = 2uv
1 2
integral
1− x
0
to evaluate the
x 2 + y 2 dy dx
0
by lust writing it as an integral over a region G in the uv-plane
110
Chapter Three : Infinite Sequences and Series
Chapter 3
Infinite Series
If we try to add the terms of an infinite sequence
an n=1
we get an expression of the form
a1 + a 2 + + a n + ....
An infinite series is the sum of an infinite sequence of numbers. The goal of
this section is to understand the meaning of such an infinite sum and to
develop methods to calculate it. Since there are infinitely many terms to add in
an infinite series, we can not just keep adding to see what comes out. Instead
we look at the result of summing the first n terms of the sequence and
stopping. The sum of the first n terms
Sn =
n
a
i =1
i
= a1 + a2 + + an
is an ordinary finite sum and can be calculated by normal addition. It is called
the n th partial sum. As n gets larger, we expect the partial sums to get
closer and closer to a limiting value in the same sense that the terms of a
sequence approach a limit. For example, to assign meaning to an expression
like
1+
1
1
1
1
+
+
+
+ ...
2
4
8
16
we add the terms one at a time from the beginning and look for a pattern in
how these partial sums grow
111
Chapter Three : Infinite Sequences and Series
Indeed there is a pattern. The partial sums form a sequence whose n th term is
Sn = 2 −
1
2 n−1
This sequence of partial sums converges to 2 because
lim 2
n →
say "the sum of the infinite
1+
1
n −1
=0
. We
1
1
1
1
+
+
+
+ ... is 2 ’’
2
4
8
16
Is the sum of any finite number of terms in this series equal to 2? No. Can we
actually add an infinite number of terms one by one? No. But we can still define
their sum by defining it to be the limit of the sequence of partial sums as
n → , in this case 2 .
Our knowledge of sequences and limits enables us to break away from the
confines of finite sums.
Definition: Given a sequence of numbers an an expression of the form
a1 + a 2 + a3 + + a n + =
112
a
n =1
n
Chapter Three : Infinite Sequences and Series
is an infinite series. The number
sequence
S n
a n . is the n th term of the series. The
defined by
S 1 = a1
S 2 = a1 + a 2
S n = a1 + a 2 + + a n
is the sequence of partial sums of the series, the number s n . being the n th
partial sum. If the sequence of partial sums converges to a limit L, we say that
the series converges and that its sum is L. In this case, we also write
a1 + a2 + a3 + + an + = an = L
n=1
If the sequence of partial sums of the series does not converge, we say that the
series diverges.
Example ( 1 )
Show that the "telescoping" series
1
n(n + 2)
n =1
Solution:
an =
is convergence and find its sums
1
A
B
=
+
n(n + 2)
n
n+2
By using partial fractions , we have
an =
A =
1
−1
,B =
2
2
1 1
1
−
2 n
n + 2
Then
113
Chapter Three : Infinite Sequences and Series
S n = a1 + a2 + a3 + + an
=
1
1
1 1
1 1
1
1
)
−
(
+
+
)
−
(
+
)
−
(
+
)
−
1
(
n+2
n
3 5
2 4
3
2
=
1
1
1
1+ −
2 n + 2
2
=
1
1 3
−
2
2 n + 2
And whereas
lim S
n→
1
n(n + 2)
a
n =1
, an =
n
a
2)
n =1
Solution:
an =
lim
n→
1 3
1
3
−
=
2 2
n + 2
4
3
4
Prove that the following series are divergent
1)
=
is converges and its sum is
n =1
Example ( 2 )
n
n +1 −
, a n = Log (1 +
n
n
1
)
n
(1)
n +1 −
n
S n = a1 + a 2 + a3 + + a n
= ( 2 − 1) + ( 3 −
Sn =
2 ) + ( 4 − 3) + + ( n + 1 −
n + 1 −1
lim S n = lim ( n + 1 − 1) =
n →
a
n =1
n
n →
divergent
(2)
114
n)
Chapter Three : Infinite Sequences and Series
1
n +1
) = Log (
) = Log ( n + 1) − Log ( n)
n
n
S n = a1 + a 2 + a3 + + a n
a n = Log (1 +
= ( Log 2 − Log1) + ( Log 3 − Log 2) + + ( Log ( n + 10 − Logn)
S n = Log ( n + 1)
lim S n = lim Log ( n + 1) = Log =
n →
n →
a
n =1
n
divergent
Special kinds of series
Given a series, we want to know whether it converges or not. In this section
and the next two, we study series with nonnegative terms. Such a series
converges if its sequence of partial sums is bounded. If we establish that a
given series does converge, we generally do not have a formula available for
its sum, so we investigate methods to approximate the sum instead.
1. Harmonic series
The series on the form
1
1
1
1
1+ + + + + =
2
3
4
n
1
n
n =1
Is a divergent series , since
S1 = 1
1
2
1 1
S3 = 1 + +
2 3
1 1 1
S4 = 1 + + +
2 3 4
S2 = 1 +
1 1
+
2 4
1
1 1
1 1
S4 1 + + ( + ) = 1 + +
2
4 4
2 2
S3 1 +
Similarly, and in general
115
Chapter Three : Infinite Sequences and Series
1 1 1
1 1 1 1
1
1
S n = 1 + + ( + ) + ( + + + ) + ( + + ) +
2 3 4
5 6 7 8
9
16
1 1 1
1 1 1 1
1
1
1+ + ( + ) + ( + + + ) + ( + + ) +
2 4 4 8 8 8 8 16
16
1 1 1
1 + + + +
2 2 2
lim S n =
n →
therefore, the harmonic series diverges.
2- Geometric Series
Geometric series are series of the form
a + ar + ar
2
+ + ar
n −1
+ =
ar
n −1
n =1
in which a and r are fixed real numbers and a o . The series can also be
written as
1+
ar
n =0
n
. The ratio r can be positive, as in
1
1
1
+
+ ... + ( ) n −1 + ... ,
2
4
2
r = 1
2
, a =1
or negative, as in
1−
1
1
1
+ − ... + ( − ) n −1 + ...,
3
9
3
r =− 1
3
, a =1
If r =1 , the n th partial sum of the geometric series is
S n = a + a (1 ) + a (1 ) 2 + + a (1 ) n−1 + = n a
and the series diverges because
lim S
n →
n
= , depending on the sign of a .
If r = −1 , the series diverges because the nth partial sums alternate between a
and O.
116
Chapter Three : Infinite Sequences and Series
If r 1 , we can determine the convergence or divergence of the series in
the following way:
S n = a + ar + ar 2 + ar n −1
rSn = ar + ar 2 + ar 3 + + ar n −1 + ar n
S n − rSn = a − ar n
a − ar n
a
ar n
Sn =
=
−
1− r
1− r
1− r
a
ar n
lim S n = lim (
−
)
1− r
1− r
n→
n→
=
a
ar n
− lim
1− r
1− r
n→
We have two cases
(I) If r 1 r
lim
n →
(II) If
n
→ , as n →
ar n
= −
1− r
r 1
The series divergent
r n → 0 , as n →
ar n
lim
=0
n → 1 − r
The series convergent
Theorem ( 1):
2
n−1
If r 1 , the geometric series a + ar + ar + + ar + converges to
a
1− r
ar
n =1
n −1
=
a
first term
=
,
1− r
1 − common ratio
If r 1 , The series divergent
117
r 1
Chapter Three : Infinite Sequences and Series
We have determined when a geometric series converges or diverges, and to
what value. Often we can determine that a series converges without knowing
the value to which it converges, as we will see in the next several sections. The
a
formula
for the sum of a geometric series applies only when the
1− r
summation index begins with n = 1 in the expression
ar
n −1
n =1
( or with the index n = 0 if we write the series as
ar
n =0
n
)
Example ( 3 ): Find the sum of the geometric series
5−
10
20
40
+
−
+
3
9
27
Solution: The first term is a = 5 and the common ratio is r = −
r =
2
1
3
5−
2
. Since
3
the series is convergent and its sum is
10
20
40
5
5
+
−
+ =
=
=3
2
5
3
9
27
1 − (−
)
3
3
Example ( 4 ): Is the series
2
2n
31− n
n =1
convergent or divergent?
n −1
Solution: Let’s rewrite the n th term of the series in the form a r
:
2
2n
n =1
1− n
3
4n
4
= n −1 = 4( ) n −1
3
n =1 3
n =`
We recognize this series as a geometric series with a = 4 and the common ratio
4
3
4
3
is r = . Since r = 1 the series is diverges
118
Chapter Three : Infinite Sequences and Series
(3) P-
Series
P- series are series of the form
1
1
1
1
=
+
+
+
p
p
p
p
n
1
2
3
n=1
Where P is constant ,
1) P- series is convergent if p 1
2) P- series is divergent if p 1
Example ( 5 ):
Determine which series are convergent and which are divergent of the
following series
(a)
n =1
(c )
n =1
n +1
n
(b)
n =1
3
n2
(d)
3
10
n =1
n
1
n
Solution:
n +1
=
n
1
1
) = 1 +
n
n=1
n=1
n=1
n=1 n
1
since the harmonic series
is divergent, also
n =1 n
(a)
(1 +
n +1
is divergent
n
n =1
1 Is divergent , so the series
n =1
( b)
n =1
3
=
10 n
1
3(10 )
n =1
1
1 , The series is convergent
10
3
10
=
=
1
3
1−
10
Is a geometric series and a = 3, r =
and its sum is
n
a
1− r
119
Chapter Three : Infinite Sequences and Series
(c)
n =1
3
1
=
3
2
n2
n =1 n
This is a P series and p = 2 1 , so it is converges
(d)
n =1
1
n
=
n =1
1
n1 / 2
1
This is a P series and p =
1 , so it is diverges
2
The nth-Term Test for a Divergent Series:
One reason that a series may fail to converge is that its terms don't become
small.
Example ( 6 ): The series
n =1
n +1 2 3 4
n +1
= + + + ... +
+ ...
n
1 2 3
n
diverges because the partial sums eventually outgrow every reassigned
number. Each term is greater than
n ".
Theorem 1: If the series
a
n =1
n
1 , so the sum of n terms is greater than
an = 0
convergent , then lim
n→
Proof:
Let S n =
Since
n
a
a
i =1
n
i
= a1 + a2 + + an , then a n = S n − S n −1
is convergent, the sequence { S n } is convergent.
lim S n = L , since
n →
n − 1 → as n → , we
S n −1 = L . Therefore
also have nlim
→
lim an = lim ( Sn − Sn −1 ) = lim Sn − lim Sn −1
n →
n →
n →
n →
= L −L= 0
120
Chapter Three : Infinite Sequences and Series
an = 0 ,
NOTE : The converse of Theorem 1 is not true in general. If lim
n→
we cannot conclude that
a
is convergent. Observe that for the harmonic
n
1 n we have an = 1 n → 0
7 that 1 n is divergent.
as n → , but we showed in Example
series
The Test for Divergence: If
lim an does not exist or if lim an 0 ,
n →
n→
then the series
a
n =1
n
is divergent.
The Test for Divergence follows from Theorem 1 because, if the series is
not divergent, then it is convergent, and so
lim an = 0 .
n→
n2
Example (7 ): Show that the series
diverges.
2
n =1 5n + 2
n2
1
lim
a
=
lim
=
lim
n
2
Solution: n →
n → 5n + 2
n→
5+ 2
=
n2
1
0
5
So the series diverges by the Test for Divergence.
. Example (8) Show that the series
1
2 3 4
+ + + + diverges
2 3 4 5
Solution:
1
2
3
4
+
+
+
+
2
3
4
5
liman =
n →
=
n =1
n
n +1
n
lim n + 1 = lim
n →
n →
=1 0
the series diverges
121
an =
1
1+
1
n
n
n +1
Chapter Three : Infinite Sequences and Series
an 0 , we know that is divergent. If we find
NOTE 2 : If we find that nlim
→
an = 0 , we know nothing about the convergence or divergence
that lim
n→
an = 0 , the series
.Remember the warning in Note 1: If lim
n→
a
n =1
n
might
converge or it might diverge.
Combining Series
Whenever we have two convergent series, we can add them term by term,
subtract them term by term, or multiply them by constants to make new
convergent series.
Theorem:
If
a
n =1
n
b
converges to a and
(1) Sum Rule :
converges to b , then
n
n =1
n =1
n =1
n =1
(an + bn ) = an + bn = a + b
n =1
n =1
n =1
(2) Difference Rule: (an − bn ) = an − bn = a − b
(3) Constant Multiple Rule:
n =1
n =1
K an = K an = Ka
Example (9): Find the sum of the series (
n =1
Solution: The series
, so
1
n =k +1
n = k +1
2n
=
Also, we find that
1
3
1
+ n ).
n(n + 1) 2
a = 1 and r = 1
is
a
geometric
series
with
2
2
2
n
1
2
1−
1
2
=1
1
n( n + 1) = 1
n =1
122
Chapter Three : Infinite Sequences and Series
So, by Theorem 8, the given series is convergent and
3
1
1
1
(
+
)
=
3
+
n
n( n + 1) 2 n
n =1
n =1 n ( n + 1)
n =1 2
3 1 + 1 = 4
NOTE 3 A finite number of terms doesn’t affect the convergence or
_
divergence of a series. For instance, suppose that we were able to show that
n
the series
n=4
n =1
n
+1
3
is convergent. Since
n
1
2
3
n
=
+
+
+ 3
3
9
28
n +1 2
+1
n=4 n
it follows that the entire series
nn
3
n =1
+ 1 is convergent.
Similarly, if it is known that the series
a
n = N +1
an =
n =1
N
an +
n =1
converges, then the full series
n
a
n = N +1
n
is also convergent
The Integral and Comparison Tests
In general, it is difficult to find the exact sum of a series. We were able to
accomplish this for geometric series and the series
1
n(n + 1)
because in
each of those cases we could find a simple formula for the partial sum . But
usually it is not easy to compute . Therefore, in this section and the next we
develop tests that enable us to determine whether a series is convergent or
divergent without explicitly finding its sum. In some cases, however, our
methods will enable us to find good estimates of the sum.
123
Chapter Three : Infinite Sequences and Series
In this section we deal only with series with positive terms, so the partial sums
are increasing. In view of the Monotonic Sequence Theorem, to decide whether
a series is convergent or divergent, we need to determine whether the partial
sums are bounded or not
The Integral Test
The Integral Test: Suppose f is a continuous, positive, decreasing function
( if for all x 1 , x i x j : f (x i ) f (x j )
) on 1 , ) and let a n = f (n) .
Then the series
a
n =1
n
is convergent if and only if the improper integral
f ( x)dx
is convergent. In other words:
1
(a) If
f ( x)dx
is convergent, then
a
n =1
1
(b) If
f ( x)dx is divergent, then
a
n =1
1
n
n
is convergent.
is divergent.
_
Example (10 ): Show that the p -series
1
n
n =1
p
=
1
1
1
1
+ p + p + + p +
p
1
2
3
n
( p is a real constant ) Converges if p 1 , and diverges if p 1
If p 1 , then f ( x) =
Solution:
1
is a positive decreasing function
xp
of x. Since
1
dx
=
xp
a
x1− p
−p
1 x dx = alim
→ 1 − p
1
=
1
1
lim ( p -1 - 1) , p 1
1 − p a → a
124
Chapter Three : Infinite Sequences and Series
Remember that
p 1
if
, then p − 1 0 alim
→
1
a p −1
=0
the series converges by the Integral Test. We emphasize that the sum
1
. The series converges, but we don't know the
1− p
of the p-series is not
value it converges to.
If p 1 , then 1 − p 0 and
dx
1
p = x − p dx =
lim ( a1-p - 1) =
x
1 − p a →
1
1
The series diverges by the Integral Test.
If p = 1 , we have
h
h
dx
dx
h
p =
= ln x 1 = ln h − ln 1
x
x
1
1
h
lim
h → 1
dx
= ln − 0 =
x
The series diverges
Example (11 ): Which of the following series converge, and which diverge?
n
(a) 2
n =1 n + 1
1
(b)
n =1 n ln n
(c)
ne
−n 2
n =1
Solution:
( a ) For the series
The function
n
n =1
f ( x)=
x
x +1
2
n
+1
2
is positive, continuous, and decreasing
for x 1 , and
125
Chapter Three : Infinite Sequences and Series
h
f ( x )dx = l im
h →
1
1
(
)
x
1
2
dx
=
lim
ln
n
+1
2
h
→
x +1
2
h
1
1
ln 2
2
=−
=
The series
n
n =1
n
is divergent
+1
2
( b ) For the series
f ( x)=
The function
1
n ln n
n =2
1
x ln x
is positive, continuous, and decreasing
for x 2 , and
2
2
f ( x ) dx =
1
dx = ln ln x 2
x ln x
= ln ln − ln ln 2
=
The series
1
n ln n
n =2
is divergent
( c ) For the series
The function
for x 1 , and
ne
−n 2
n =1
f ( x ) = xe − x
2
is positive, continuous, and decreasing
126
Chapter Three : Infinite Sequences and Series
an = ne − n = f ( n)
2
f ( x ) = xe− x
2
2
1
f ( x ) dx = xe
dx = − e − x
2
1
1
1
1
= − e − + e −1
2
2
1
=
2e
1
The series
ne
−n 2
− x2
is convergent
n =1
The Comparison Tests
We have seen how to determine the convergence of geometric series, p-series,
and a few others. We can test the convergence of many more series by
comparing their terms to those of a series whose convergence is known.
The Comparison Test : Suppose that
a
n
and
b
n
are series with positive
terms
.( a )
If
b
n
is convergent and 0 an bn for all
n,
then
a
n
is also
convergent.
b
(b) If
n
is divergent and
an bn 0 for all n , then
a
n
is also
divergent.
In using the Comparison Test we must, of course, have some known
series
b
n
for the purpose of comparison. Most of the time we use either a ,
p-series [ converges if p 1 and diverges if p 1 ] or a geometric series [
converges
if r 1 and diverges if r 1 ].
Proof: In Part (a), Let S n be the partial sums of
127
a
n
and
Chapter Three : Infinite Sequences and Series
Let
tn
be the partial sums of
And since 0 an bn for all
b
n
is convergent , then
n→
n
In Part (b ), since
Where
a
n
and
lim t
n →
n
n
exists
is also convergent.
an bn 0 for all n and Sn rn
lim r
n→
n
n , so 0 S n t n
a
lim S exists , then
b
n
= , then
lim S
n→
n
=
divergent
1
Example ( 12 ): Test the series
for convergence or divergence
n =1 n!
Solution: Observe that
n ! 2 n−1
1
n!
1
Since the series
2
1
1
n −1
n! 2
1
2
1
1
n! 2
n −1
n −1
n−1
is geometric series with r =
1
1
2
1
Then it is convergent , so
is convergent.
n =1 n!
Example (13 ): Test the series
1
1
1
+
+
+
log 2 log 3 log 4
for convergence or divergence
128
Chapter Three : Infinite Sequences and Series
Solution: Since
1
1
1
1
+
+
+=
log 2
log 3
log 4
n = 2 log n
Observe that
n Log n
1
1
n
log n
n=2
1
1
log n n = 2 n
1
Since the series is harmonic series and it is divergent , so
n
1
log n is
n =2
divergent.
The terms of the series being tested must be smaller than those of a
convergent series or larger than those of a divergent series. If the terms are
larger than the terms of a convergent series or smaller than those of a
divergent series, then the Comparison Test doesn’t apply. Consider, for
instance, the series
n =1
1
2
−1
n
The inequality
1
1
2n − 1
2n
is useless as far as the Comparison Test is concerned because
is convergent and
b = 12
n
n
an bn .Nonetheless, we have the feeling that 1 n
2 −1
ought to be convergent because it is very similar to the convergent geometric
series . In such cases the following test can be used.
129
Chapter Three : Infinite Sequences and Series
The Limit Comparison Test: Suppose that
lim
terms. If
n →
a
n
and
b
n
are series with positive
an
=k
bn
Where k is a finite number and k 0 , then either both series converge or
both diverge.
Proof:
Since lim
n →
an
= c , then for any
bn
0 , there exists N
such that for all n N , we have
an
−k
bn
k −
an
k +
bn
( k − ) an ( k + )bn
If we choose
( 1 ) If
b
n
such that k − is positive , then
is convergent , then from an (k + )bn we have
a
is divergent , then from an (k − )bn we have
a
n
is
n
is
convergent .
( 2 ) If
b
n
divergent .
Hint: The comparison test is preferred to use if the n- term of the
series is a fraction all of the numerator and denominator is a
polynomial , in this case we choose bn as the difference between
the orders of numerator and denominator .
130
Chapter Three : Infinite Sequences and Series
The following table is useful to choose
b
n
an
bn
Log ( n)
n
n!
2 n−1
1
sin
n
1
n
1
sin
n
2
1
sin
n
1
sin −1
n
3
1
tan −1
n
1
n
2
1
n
1
n
1
n
3
Example ( 14 ): Which of the following series converge, and which diverge?
(a)
n2 − 1
n4 + 1
( b )
n+3
3n 2 + 8n
Solution:
We apply the Limit Comparison Test to each series.
n2 −1
= an
(a) 4
n +1
For large n , we expect
an
2
to behave like n
n4
=1
n2
since the leading terms dominate for large n , we let
b
n
=
1
n2
This is p - series , and p = 2 1 , so it is convergent
131
Chapter Three : Infinite Sequences and Series
an
n2 −1 2
n4 − n2
= lim 4
n = lim
bn
n4 +1
n → n + 1
n →
lim
n →
1
n2 = 1
= lim
1
n →
1+ 4
n
1−
a ,b
the two series
n
n
(0, )
are similarly
n2 −1
= a n is convergent
So 4
n +1
(b)
n+3
= an
2
+ 8n
3n
b
We let
n
=
1
n
This is p - series , and p = 1 , so it is divergent
lim
n →
an
=
bn
n
2
lim 3n
n →
=
1
3
2
+ 3n
=
+ 8n
3
n
8
3+
n
1+
lim
n →
(0, )
the two series are similarly , So
n+3
= an is divergent
2
+ 8n
3n
Example ( 15 ):
Which of the following series converge, and which diverge?
2n 3 − 3n 2
4
3
n =1 7 n + 100n + 7
(a)
(c )
sin
n =1
3
1
n
(b)
n =1
(d)
n =1
132
2n 2 + 4
n7 / 2 + n
2
tan −1 3
n
2n + 3
Chapter Three : Infinite Sequences and Series
Solution :
2n 3 − 3n 2
( a ) For
4
3
n =1 7 n + 100n + 7
we choose
lim
n →
b
=
n
1
which is divergent
n
an
2n 3 − 3n 2
n
= lim
4
3
bn
+ 100 n + 7 1
n → 7 n
=
2
7
2n 3 − 3n 2
7n 4 + 100n3 + 7 is divergent
The n the series
( b ) For
n =1
we choose
(0, )
2n 2 + 4
=
n7 / 2 + n
b
=
n
a
n
1
which is divergent
n3 / 2
This is p - series , and p = 3 2 1 , so it is convergent
lim
n→
an
=
bn
the two series
2n 2 + 4 n 3 / 2
n7 / 2 + n
1
lim
n→
a ,b
n
2n 2 + 4
=
So 7 / 2
+n
n=1 n
a
( c ) For
sin
we choose
b
n =1
n
n
3
n
are similarly
is convergent
1
n
=
1
n
3
133
=2
(0, )
Chapter Three : Infinite Sequences and Series
This is p - series , and p = 3 1 , so it is convergent
1
1
sin
sin
n =
n
lim
1
1
n →
n
n3
3
3
lim
n →
an
= lim
bn
n →
=1
So
sin
n=1
3
1
is convergent
n
( d ) For
n =1
we choose
(0, )
2
tan −1 3
n =
2n + 3
b
n
=
a
n
2
1
n3 = 2
n
n4
This is p - series , and p = 4 1 , so it is convergent
lim
n →
an
= lim
bn
n →
= lim
n →
2
tan −1 3
n n
1
2n + 3
n3
n
2n +
2
tan −1 3
n
2
lim
2
3 n →
3
n
2
tan −1 3
1
n
=
2 lim
2
2
n →
3
n
=1
(0, )
So
n =1
2
tan −1
3
n
2n +
3
is convergent
134
Chapter Three : Infinite Sequences and Series
The Ratio and Root Tests
The Ratio Test measures the rate of growth (or decline) of a series by
examining the ratio
a n +1
a n . For a geometric series
ar
, this rate is a
n
n +1
n +1
constant ( ( ar ) / ( ar ) = r ) , and the series converges if and only if its
ratio is less than 1 in absolute value. The Ratio Test is a powerful rule
extending that result.
The Ratio Test:
Let
a
n
be a series with positive terms and suppose that :
lim
n →
a n +1
=l
an
Then ( a ) the series convergent if l 1
( b ) the series divergent if l 1
( c ) the test is inconclusive if l = 1
1
For l = 1 , the two series
n =1 n
Note:
and
1
n
n =1
2
show that some other test for convergence must be used when l = 1
For :
1
n
n =1
:
1
For : 2 :
n =1 n
a n +1
1 /(n + 1)
n
=
=
→1
an
1/ n
n +1
a n +1 1 /(n 2 + 1)
n2
=
= 2
→1
an
1/ n 2
n +1
In both cases, l = 1 , yet the first series diverges, whereas the second converges.
The Ratio Test is often effective when the terms of a series contain factorials of
expressions involving n or expressions raised to a power involving n
Example (16 ):
Investigate the convergence of the following series:
135
Chapter Three : Infinite Sequences and Series
a)
n =1
n!
nn
d)
b)
(n !)2
(2n ) !
c)
e
e)
n
n
Solution: ( a ) For
an =
n2
en
n4
e n2
n!
n
n =1
n
(n + 1)!
(n + 1)n+1
(n + 1)! n n = (n + 1)(n!) n n
(n + 1)n+1 n ! (n + 1)n (n + 1) n !
n!
an +1 =
nn
an +1
=
an
n
=
n +1
lim
n →
an +1
=
an
n
n +1
lim
n
n →
= e −1 =
1
1
e
The series is convergent
( b ) For
(n !)2
(2n ) !
136
−n
=
lim
n →
n
1
1 +
n
−1
Chapter Three : Infinite Sequences and Series
(n!)2 a
(( n + 1)!)2
=
n +1
(2n )!
(2n + 2 )!
2
a
(n + 1)! ( 2n)
n +1 =
(2n + 2 )! (n!)2
an
(n + 1)2 (n!)2
( 2n)!
=
(2n + 2 )(2n + 2 )( 2n)! (n!)2
an =
n 2 + 2n + 1
=
4n 2 + 6n + 2
lim
n →
lim
n + 2n + 1
=
4n 2 + 6n + 2
n →
2
1
+
n
n2
6
2
4+
+
n
n2
1+
2
a n +1
=
an
lim
n →
1
1
4
=
The series is convergent
( c ) For
n
e
an =
an +1
an
n
n
n +1
an +1 =
n
e
e n +1
n +1
en
=
e en
n
=
lim
n →
n +1
en
an +1
an
=
1
e
=
The series is convergent
( d ) For
n2
en
137
lim
n →
1
1
e
n +1
n
Chapter Three : Infinite Sequences and Series
an
=
lim
n →
n2
2n
a n +1 =
a n +1
an
1
=
2
( n + 1) 2
2 n +1
n + 1
n
lim
n →
2
1
1
2
=
The series is convergent
n4
e
( c ) For
an =
n2
n4
en
an +1 =
2
an +1
( n + 1) 4
en
=
2
an
n4
e n + 2 n +1
( n + 1) 4
e ( n +1)
2
2
lim
n→
an +1
=
an
lim
n→
=
( n + 1) 4
n 4 e 2 n +1
( n + 1) 4
=
4
2 n +1
n e
By using L’ Hopital Rule
lim
n→
lim
n →
an +1
=
an
a n +1
=
an
=
lim n (
lim e
n →
lim
lim
n →
= lim
n →
4
n→
n →
=
4(n + 1) 3
=
2 n +1
2 n +1 3
2e
+ 4e
n
2 n +1
)
12( n + 1) 2
8n 3 + 12n 2 + 2( 2n 4 + 4n 3 )e 2 n +1
(
)
12( n + 1) 2
e 2 n +1 16n 3 + 12n 2 + 4n 4
24( n + 1)
2 n +1
2
e
72n + 24n + 48n 3 + 8n 4
24
e 2 n +1 144n + 24 + 144n 2 + 32n 3 + 2e 2 n +1 (72n 2 + 24n + 48n 3 + 8n 4 )
24
=
= 0 1
(
The series is convergent
138
)
Chapter Three : Infinite Sequences and Series
Example (17 ):
Investigate the convergence of the following series:
I)
2n
n =1 n!
III)
II)
n5 8n+1
3n
n=1
IV)
n 25
n
n =1 3
n
n=1
n!
5n
100
Solution: We apply the Ratio Test to each series.
(I)
2n
n =1 n!
an
2n
2 n +1
=
a n +1 =
n!
( n + 1)!
lim
n →
2 n +1
n!
n
lim
2
n → ( n + 1)!
2
= lim
= 0 1
n → n + 1
a n +1
=
an
The series is convergent
(II) For
n =1
n 5 8n+1
3n
n 5 8 n +1
( n + 1) 5 8 n + 2
an =
a n +1 =
3n
3 n +1
( n + 1) 5 8 n + 2
a n +1
3n
lim
= lim
n +1
5
n +1
an
3
n 8
n →
n →
5
8( n + 1)
= lim
3n 5
n →
8
=
1
3
The series is divergent
(III)
For
n =1
n 25
3n
139
Chapter Three : Infinite Sequences and Series
an =
lim
n →
n 25
( n + 1) 25
a
=
n +1
3n
3 n +1
( n + 1) 25
a n +1
3n
= lim
an
3 n +1
n 25
n →
1
=
1
3
The series is convergent
(IV)
For
n =1
an =
lim
n →
n!
n100 5 n
n!
( n + 1)!
a n +1 =
n
n 5
( n + 1)100 5 n +1
100
( n + 1) n!
a n +1
n100 5 n
= lim
=
100
an
n!
5 n +1
n → ( n + 1)
The series is divergent
The Root Test:
It is sometimes known as the Cauchy root test or Cauchy's radical test. For
a series .
The convergence tests we have so far for
a
n
, work best when the formula
for a n , is relatively simple. However, consider the series with the terms
an
n n
= 2
1 n
2
;
n odd
; n even
To investigate convergence we write out several terms of the series:
a
n =1
n
1
1
3
1
5
1
7
+ 2 + 3 + 4 + 5 + 6 + 7 +
2
2
2
2
2
2
2
1
1
3
1
5
1
7
=
+
+
+
+
+
+
+
2
4
8
16
32
64
128
=
140
Chapter Three : Infinite Sequences and Series
Clearly, this is not a geometric series. The n th - term approaches zero as
n → , so the n th -Term Test does not tell us if the series diverges. The
Integral Test does not look promising. The Ratio Test produces
1
a n +1
2n
=
n +1
an
2
;
n
;
odd
n
even
As n → , the ratio is alternately small and large and has no limit. However,
we will see that the following test establishes that the series converges.
The Root Test:
Let
a
n
be a series with positive terms and suppose that :
lim
n
n →
an = l
Then ( a ) the series convergent if l 1
( b ) the series divergent if l 1
( c ) the test is inconclusive if l = 1
a
Example ( 18 ): Does
an
n
converge or not ?
n n
= 2
1 n
2
;
n odd
; n even
Solution: We apply the Root Test finding that
n
an
n n
2
=
1
2
; n odd
; n even
141
Chapter Three : Infinite Sequences and Series
1
2
Therefore
Since
n
n
an
n
2
n a
n → 1 as n → , we have lim
n =
n→
n
1
2
by the Sandwich Theorem. The limit is less than 1 , so the series converges
by the Root Test .
Example (19):
Which of the following series converge, and which diverge?
n2
(a) n
n =1 2
(b)
n =1
2n
n3
Solution: We apply the Root Test to each series
n2
( a ) For n
n =1 2
2
lim
n
n →
an =
lim
n →
n n
1
=
1
2
2
The series is convergent
( b ) For
lim
n
n →
n =1
2n
n3
an =
→
lim
n
n →
2n
=
n3
lim
n →
2
n
3
n
2
1
13
The series is divergent
Example (20 ): Which of the following series converge, and which diverge?
(a)
n =1
1
nn
(b)
n
n =1 n + 1
Solution:
142
n2
Chapter Three : Infinite Sequences and Series
1
nn
( a ) For
n =1
1
an =
n
1
n an =
n
n
lim n a n = lim
n →
n →
1
1
=
= 0 1
n
The series is convergent
n
( b ) For
n=1 n + 1
an
n
n2
n
=
n +1
an =
lim
n
n →
n
n2
n
n +1
an =
n
=
n +1
n
Lnlim
n +1
→
n +1
= lim
n
n →
= e −1 =
n2
−n
=
lim
n→
n
n
n
1
1 +
n
−1
1
1
e
The series is convergent
Example ( 21 ): Prove that the following series is convergent .
a2 n =
1
3n
a2 n−1 =
,
Solution:
2 n a2 n =
2n
1
3
n
1
=
3
n
2n
=
143
1
3
1
3n+1
Chapter Three : Infinite Sequences and Series
1
2 n−1 a2 n−1 = 2 n−1
3
n +1
1
=
3
1
lim 2 n−1 a2 n−1 = lim
n→
n→ 3
1
lim n an =
1
3
n→
n +1
2 n −1
n +1
2 n −1
1
2
1
1
= =
3
3
The series is convergent
Alternating Series, Absolute and Conditional
Convergence:
A series in which the terms are alternately positive and negative is an
alternating series.
Here are three examples:
1−
1
1 1 1
(−1) n +1
+ − + ++
+
2
3 4 5
n
→ (1)
1
1 1
(−1) n 4
− 2 +1−
+ − + +
+
4 8
2
2n
→ (2)
1 − 2 + 3 − 4 + 5 − + ( −1) n+1 n +
→ (3)
We see from these examples that the nth term of an alternating series is of
the form
(− 1)
n =1
n +1
an
, an 0
Series (1), called the alternating harmonic series, converges, as we will see in a
moment . Series (2), a geometric series with ratio r = -1/2, converges to -2/[1 +
(1/2)] = -4/3.
Series (3) diverges because the n th term does not approach zero.
144
Chapter Three : Infinite Sequences and Series
We prove the convergence of the alternating harmonic series by applying the
Alternating Series Test. The Test is for convergence of an alternating series and
cannot be used to conclude that such a series diverges.
The Alternating Series Test (Leibniz's Test)
Theorem: The series
(− 1)
n =1
n +1
an
an 0
,
converges if the following conditions are satisfied:
( 1 ) The positive
(2)
lim a
n →
n
an ' s are (eventually) non increasing: an+1 an , n N
=0
Proof:
If n is an even integer, say n = 2m , then the sum of the first n term is
S 2 m = a1 − a2 + a3 − a4 + a5 − a6 + + a2 m−1 − a2 m
= (a1 − a2 ) + (a3 − a4 ) + + (a2 m−1 − a2 m ) 0
I.e.,
S 2 n = a1 − (a 2 − a3 ) − (a 4 − a5 ) − − (a 2 m − 2 − a 2 m −1 ) − a 2 m
The first equality shows that S 2 m is the sum of m nonnegative terms since
each term in parentheses is positive or zero. Hence , S 2 m+ 2 S 2 m , and the
sequence S 2 m is non-decreasing. The second equality shows that S 2 m a1 .
Since S 2 m is non-decreasing and bounded from above, it has a limit, say
lim S
m→
2m
If n is an odd integer, say n = 2m +1
=L
→ (1)
then the sum of the first n terms is
S 2 n+1 = S 2 n + a2 n+1
145
Chapter Three : Infinite Sequences and Series
an → 0
lim S
2n
= L
S 2 n+1 = S 2 n + a2 n+1 → L = L + 0
→ (2)
Combining the results of Equations (1) and (2) gives
lim S
n →
n
=L
Rather than directly verifying the definition an+1 an , n N , a second
way to show that the Sequence
an is non-increasing is to define a
f ( n ) = a n . That is, the values of
differentiable function f ( x ) satisfying
f ( x ) match the values of the sequence at every positive integer n. If
f ( x ) 0 for all x greater than or equal to some positive integer N, then
f ( x ) is non-increasing for x N . It follows that f ( n ) f (n + 1 ) ), or
an an +1 for n N .
Example (22):
Which of the following series converge, and which diverge?
1
sin −1
n
( a ) (− 1)n
2n + 1
n=1
Solution:
(b)
( a ) lim
n →
(− 1)n+1
n=1
n
1
sin −1
n =0
a n = lim
2n + 1
n →
Also
1
sin −1
n
an =
2n + 1
an+1 an
an+1
1
sin −1
n +1
=
2n + 3
146
Chapter Three : Infinite Sequences and Series
so the series is convergent by the Alternating Series Test
( b ) an =
1
1
an+1 =
n
n +1
Where
1
1
n +1 n
an+1 an
0
→ (1)
lim a
n →
= lim
n
n →
1
=0
n
→ (2)
so the series is convergent by the Alternating Series Test
Absolute and Conditional Convergence
Given any series
a
n =1
n
, we can consider the corresponding series
an = a1 +
n =1
a2 + a3 +
whose terms are the absolute values of the terms of the original series.
Definition: A series
a
n =1
n
is called absolutely convergent if the series of
absolute values
a
n =1
Notice that if
a
n =1
n
n
is convergent.
is a series with positive terms, then an = an and so absolute
convergence is the same as convergence.
Example( 23 ): The series
(−1) n−1
1
1
1
=
1
−
+
−
+
n2
2 2 32 4 2
n =1
147
Chapter Three : Infinite Sequences and Series
is absolutely convergent because
n =1
( −1) n −1
=
n2
1
n
n =1
= 1+
2
1
1
1
+
+
+
22 32
42
is a convergent P -series ( P = 2 1 ).
Example (24 ):
We know that the alternating harmonic series
n =1
( −1) n −1
1
1
1
=1 −
+
−
+
n
2
3
4
is convergent (see Example 1), but it is not absolutely convergent because the
corresponding series of absolute values is
n =1
( −1) n −1
=
n
1
n
= 1+
n =1
1
1
1
+ +
+
2
3
4
which is the harmonic series ( P -series with P = 1 ) and is therefore divergent.
Example 24 shows that it is possible for a series to be convergent but not
absolutely convergent. However, the following theorem shows that absolute
convergence implies convergence.
Absolute convergence is important for two reasons. First, we have good
tests for convergence of series of positive terms. Second, if a series converges
absolutely, then it converges, as we now prove.
Theorem 1. If a series
a
n
is absolutely convergent, then it is convergent.
To see why Theorem 1 is true, observe that the inequality
0 an + an
148
2 an
Chapter Three : Infinite Sequences and Series
is true because an
convergent, then
is either an or − a n . If
an is convergent, so
Therefore, by the Comparison Test,
(a
a = (a
n
n
n
a
2
n
is absolutely
a n is convergent.
+ an ) is convergent. Then
+ an ) − an
is the difference of two convergent series and is therefore convergent.
Example ( 25 ): Determine whether the series is convergent or divergent
n =1
cos n cos1
cos 2 cos 3
=
+
+
+
2
2
n
1
22
32
.Solution: This series has both positive and negative terms, but it is not
alternating. (The first term is positive, the next three are negative, and
the following three are positive. The signs change irregularly ) . We can
apply the Comparison Test to the series of absolute values
n =1
cos n
cos n
=
n2
n2
n =1
Since cos n 1 for all n , we have
cos n
n2
We know that
1 / n
2
1
n2
is convergent ( P -series with P = 2 ) and therefore is
convergent by the Comparison Test. Thus, the given series
(cos n) / n
absolutely convergent and therefore convergent by Theorem 1.
Definition: A series that converges but does not converge absolutely
converges conditionally.
149
2
is
Chapter Three : Infinite Sequences and Series
To determine whether the series
a
n =1
n
is absolutely convergent or
Conditional convergence , we study the series
a
n =1
n
, ten we have the
cases:
Divergent
convergent
We study the
series
then
is absolutely
convergent
If it convergent
The series is
Conditional
convergence
Example (26): Test the convergence of the series
(− 1)n+1
n =1
n +1
Solution:
Studying the series
a
n=1
Studying the series
n
1
n=1 2n − 1
=
n=1
By using integral test
(− 1)n+1
n +1
It is an alternating series, so
150
Chapter Three : Infinite Sequences and Series
1
dx
1
= ln (2 x − 1)1
2x − 1 2
1- lim a n = lim
n →
=
2- an +1 an
it is divergent
n=1
(− 1)n+1
n →
1
=0
2n − 1
the series convergent
is conditional convergent
n +1
Example ( 27 ): Test the absolutely and conditional convergent of the
series
(− 1)n+1
ln (n )
2
n=2
n
Solution:
Studying the series
Studying the series
1
2
n=2 n ln n
an =
n=2
ln (n )
I =
1
2
n=2
By using integral test
(− 1)n+1
n
It is an alternating series, so
dx
x ln 2 x
1- lim = lim
n →
put
ln x = u
n →
2- an =
dx
= du
x
1
=0
n ln 2 n
1
n ln 2 n
du 1
1
I = 2 = − =
u
u 2 ln 2
ln 2
an+1 =
the series is convergent
an+1 an
1
(n + 1) ln 2 (n + 1)
the series is convergent
151
Chapter Three : Infinite Sequences and Series
(− 1)n+1
ln (n )
n=1
2
n
is absolutely convergent
Example ( 28 ): Test the absolutely and conditional convergent of the
series
(− 1)n−1 2n
n =1
n2
Solution:
Studying the series
Studying the series
2n
an = 2
n=1
n=1 n
(− 1)n−1 2n
n=1
n2
By using ratio test
a n +1
2 n +1 n 2
=
n
lim
lim
2
2
n → a n
n → (n + 1)
2n
=
lim
2
n → n
n
= 2 lim
n → n + 1
= 2 1
2
By L’Hopital
= lim
n →
the series is divergent
2 n ln 2
=
2n
2 n (ln 2)
=
lim
2
n →
2
the series is divergent
The series
(− 1)n−1 2n
n=1
n2
is divergent
Example ( 29 ): Test the convergence or divergence of the series
sin 1 sin 2 sin 3 sin 4
− 3 / 2 + 3 / 2 − 3 / 2 + → (1)
13 / 2
2
3
4
Solution:
The absolute series
152
Chapter Three : Infinite Sequences and Series
sin 1 sin 2 sin 3 sin 4
+ 3 / 2 + 3 / 2 + 3 / 2 + → (2)
13 / 2
2
3
4
and whereas
sin n
1
3/ 2
3/ 2
n
n
1
n
the series
3/ 2
is convergent
The series( 2 ) is convergent .
Power Series:
A power series is a series of the form
c (x − a )
n =0
n
n
x is a variable and the
of the series. For each fixed x ,
Where
= c0 + c1 ( x − a) +c2 ( x − a) 2 +
c0 , c1 , c2 , are constants called the coefficients
is called a power series in ( x − a ) or a power series centered at
series about a .
a or a power
. A power series may converge for some values of and diverge for other values
of x . The sum of the series is a function
If a = 0 we get the power series
c
n =0
n
x n = c0 + c1 x +c2 x 2 +
We will see that a power series defines a function f (x) on a certain interval
where it converges. Moreover, this function will be shown to be continuous
and differentiable over the interior of that interval.
153
Chapter Three : Infinite Sequences and Series
To study the power series we use the ratio test to find the interval of
convergent
l = lim
n →
a n +1 x n +1
a n +1
=
x
lim
a
an x n
n →
n
By putting
a
1
= lim n +1
R
an
n →
We have
l =
x
R
Then the power series
a
n
xn :
(I) convergence if x R
(II) Divergent if x R
(III) The series may be convergent or divergent at x = R i.e., − R x R
Theorem: For a given power series c n ( x − a) n there are only three
possibilities:
(i) The series converges only when x = a .
(ii) The series converges for all x .
(iii) There is a positive number such that the series converges if x − a R
and diverges if
x−a R.
The number R in case (iii) is called the radius of convergence of the
power series. By convention, the radius of convergence is R = 0 in case (i)
and R = in case (ii).
The interval of convergence of a power series is the interval that consists of
all values of x for which the series converges.
154
Chapter Three : Infinite Sequences and Series
In case (i) the interval consists of just a single point a . In case (ii) the
interval is ( − , ) .In case (iii) note that the inequality
x − a R can
be rewritten as a − R x a + R . When x is an endpoint of the
interval, that is, x = a R , anything can happen -the series might converge
at one or both endpoints or it might diverge at both endpoints. Thus, in case
(iii) there are four possibilities for the interval of convergence:
( a − R , a + R ) ( a − R, a + R ] [ a − R, a + R ) [ a − R , a + R ]
Example ( 30):
of the series
(I)
Find the radius of convergence and interval of convergence
(n!) x
n
n =0
xn
n!
n =0
x
x2
xn
(III) 1 +
+ 2
+ n
2 +1
2 +2
2 +n
2
3
x
x
xn
(IV) 1 − x + 2 − 3 + + (−1) n n +
2
3
n
(II)
Solution:
(I)
(n!) x
a
1
( n + 1)!
= lim n +1 = lim
R
an
n!
n →
n →
n
n=0
= lim
n →
(n + 1)(n!)
=
( n!)
R = 0
The series is convergent only at
(II)
n =0
x=0
xn
n!
155
Chapter Three : Infinite Sequences and Series
an =
1
=
R
=
1
1
a n +1 =
n!
( n + 1)!
lim
n →
a n +1
=
an
n1
lim ( n + 1)!
n →
n!
1
1
lim (n + 1)n! = lim n + 1 =
n →
=0
n →
1
=0 R=
R
The interval of convergence is − x .
i.e., the series convergence for all
(III)
x
x
x2
xn
xn
1+
+
+ n
=
2 + 1 22 + 2
2 + n n =0 2 n + n
an =
1
1
a n +1 = n +1
2 +n
2
+ ( n + 1)
n
a n +1
2n + n
= n +1
an
2
+ ( n + 1)
1
=
R
lim
n →
a n +1
=
an
lim
n →
2n + n
2 n +1 + ( n + 1)
By using L’Hopital
1
=
R
lim
n →
2 n Log 2 + 1
2 n +1 Log 2 + 1
1
2 n ( Log 2) 2
1
1
= lim n +1
= lim =
2
R
2
( Log 2)
n → 2
n → 2
R = 2
The interval of convergence is − 2 x 2 .
(IV)
(−1) n
n =1
xn
nn
156
Chapter Three : Infinite Sequences and Series
an
( −1) n
( −1) n +1
=
a n +1 =
nn
( n + 1) n +1
a n +1
( −1) n n
=
an
( n + 1) n +1
1
=
R
=
lim
n →
n →
lim
−n
lim
n +1
n
lim
n
1
1 +
n
n →
lim
n
n →
=
=
n
n +1
n →
=
a n +1
an
( −1) n n
( n + 1) n +1
1
n +1
1
lim
n → n + 1
−1
1
= 0
1
= 0 R =
R
The interval of convergence is − x . i.e., the series
convergence for all x
Example ( 31 ):
Find the radius of convergence and interval of convergence of the
following series
(I)
xn
n =1 n( n + 1)
(II)
( n!) 2 n
x
n =0 ( 2n)!
Solution:
xn
n =1 n( n + 1)
1
1
an =
a n +1 =
n( n + 1)
( n + 1)(n + 2)
a
n( n + 1)
n
n +1 =
=
an
( n + 1)(n + 2)
n+2
(I)
a
1
n
= lim n +1 = lim
= lim
R
an
n →
n → n + 2
n →
R =1
157
1
2
1+
n
=1
Chapter Three : Infinite Sequences and Series
The interval of convergence of the series is
−1 x 1
At x = 1 the series becomes
1
1
=
2
n=1 n( n + 1)
n=1 n + n
By using the comparisons test
1
b = n
n
2
an
n2
lim
= lim 2
= 1 (0, )
n → bn
n → n + n
a ,b
n
n
are similar , and since
1
b = n
n
2
p
- series p = 2 1
It is convergent
So the series
1
1
=
is convergent at x = 1
2
n=1 n( n + 1)
n=1 n + n
At x = −1 the series becomes
(−1) n
n=1 n( n + 1)
This is an alternating series
lim a n =
n →
1
1
lim n(n + 1) =
n →
n( n + 1) ( n + 1)(n + 2)
1
1
n( n + 1)
( n + 1)(n + 2)
a n a n +1
The series convergent
The interval of convergence of the series is
158
=0
Chapter Three : Infinite Sequences and Series
−1 x 1
(II)
n =0
( n!) 2
xn
( 2n)!
2
a n +1
( 2n!) 2 ( n + 1)!
n 2 + 2n + 1
=
=
an
( n!) 2 ( 2n + 2)!
4n 2 + 6n + 2
a n +1
1
n 2 + 2n + 1
1
= lim
= lim
=
2
R
an
4
+ 6n + 2
n →
n → 4n
R = 4
The interval of convergence of the series is − 4 x 4
Example (32 ):
Find the radius of convergence and interval of convergence of
( x − 3) 2 n
n 2 5n
n =1
the following series
Solution:
( x − 3) 2 n
( x − 3) 2 n + 2
an =
a n +1 =
n 2 5n
( n + 1) 2 5 n +1
a
1
( x − 3) 2 n + 2
n 2 5n
= lim n +1 = lim
2 n +1
R
an
( x − 3) 2 n
n →
n → ( n + 1) 5
= lim
n →
( x − 3) 2
1
5
x−3
−
−
5
5 x−3
5 +3
5 +3 x
5 +3
At the boundary point
1- At x = 5 + 3
(5) n
1
=
2 n
2
n=1 n 5
n=1 n
p
- series p = 2 1
159
Chapter Three : Infinite Sequences and Series
It is convergent
2- At x = − 5 + 3
(− 5 ) 2 n
1
=
2 n
2
n 5
n =1
n =1 n
It is convergent .
p - series p = 2 1
The interval of convergence of the series is
− 5 +3 x
5 +3
Representations of Functions as Power Series
This section shows how functions that are infinitely differentiable
generate power series called Taylor series.
In many cases, these series can provide useful polynomial approximations of
the generating functions. Because they are used routinely by mathematicians
and scientists, Taylor series are considered one of the most important topics
of this chapter.
If
f (x) is a continuous function in
x and also
f ( x), f ( x), f n ( x) exists
and continuous on the closed interval a x b , then we can expand the
function f (x) by using
•
Taylor Series:
Definition: Let
f be a function with derivatives of all orders throughout
some interval containing
generated by
f ( x) =
k =0
•
a
as an interior point. Then the Taylor series
f at x = a is:
f ( k ) (a)
f (a)
f ( x)
f n−1 ( x)
k
2
( x − a) = f (a) +
( x − a) +
( x − a) + +
( x − a) n−1 + ...
k!
1!
2!
3!
Maclaurin Series
If a = 0 in Taylor Series then we have Maclaurin Series for the function
f (x) in the form.
160
Chapter Three : Infinite Sequences and Series
f (0)
f ( x) 2
f n −1 ( x) n −1
f ( x) = f (0) +
( x) +
( x) + +
( x) + ...
1!
2!
3!
Example ( 33 ):
Find the Taylor series generated by f ( x) =
1
at a = 2 .
x
Where, if any-where, does the series converge to 1 x ?
Solution: We need to find f ( 2) , f ( 2) , f ( 2), Taking derivatives we
f ( x) = x −1 , f ( x) = − x −2 , f ( x ) = 2 ! x −3 , , f ( n ) ( x) = (−1) n n ! x − ( n+1)
get
So that
f (2) = 2
−1
1
1 f ( x )
1
f ( n ) ( x) (−1) n
−2
= , f (2) = − 2 = − 2 ,
= 3 , ,
= n+1
2
2!
n!
2
2
2
The Taylor series is
f (2)
f (2)
f n−1 (2)
2
f ( x) = f (2) +
( x − 2) +
( x − 2) + +
( x − 2) n−1 + ...
1!
2!
3!
n
1 ( x − 2) ( x − 2) 2
2 ( x − 2)
= − 2 +
+
+
(
−
1
)
+
2
2
23
2 n+1
Similarly, we can find Maclaurin series for some famous function, for example
x x2 x3
xn
xn
1) e = 1 + +
+
++
+ =
1! 2! 3!
n!
n = 0 n!
x
x2 x4
x 2 n−2
x 2n−2
n −1
n −1
2) cos x = 1 −
+
+ + (−1)
+ = (−1)
2! 4!
(2n − 2)!
(2n − 2)!
n =1
x x3 x5
x 2 n −1
x 2 n −1
n −1
n −1
3) sin x = −
+
+ + (−1)
+ = (−1)
1! 3! 5!
(2n − 1)!
(2n − 1)!
n =1
x2 x4 x6
x 2n−2
x 2n−2
4) cosh x = 1 +
+
+
++
+ =
2! 4! 6!
(2n − 2)!
n =1 ( 2n − 2)!
x x3 x5 x7
x 2 n −1
x 2 n −1
5) sinh x = +
+
+
++
+ =
1! 3! 5! 7!
(2n − 1)!
n =1 ( 2n − 1)!
161
Chapter Three : Infinite Sequences and Series
The Binomial series
In this section we introduce the binomial series for estimating powers and
roots of binomial expressions (1 + x ) . We also show how series can be used
to evaluate non-elementary integrals and limits that lead to indeterminate
forms, and we provide a derivation of the Taylor series for f ( x) = tan −1 x . This
section concludes with a reference table of frequently used series.
n
The Binomial Series for Powers and Roots
n
The Taylor series generated by f ( x) = (1 + x ) , when m is constant, is
n
n(n − 1) 2
n(n − 1)(n − 2) 3
x+
x +
x +
1!
2!
3!
n(n − 1)(n − 2) (n − k + 1) k
+
x +
k!
(1 + x) n = 1 +
This series, called the binomial series, converges absolutely for x 1 . To
derive the series, we first list the function and its derivatives:
f ( x ) = (1 + x ) n
f ( x ) = n (1 + x ) n −1
f ( x ) = n ( n − 1 )(1 − x )
n−2
f
(k )
( x ) = n ( n − 1)( n − 2) ( n − k + 1) x ( n − k )
We then evaluate these at x = 0 and substitute into the Taylor series formula
to obtain Series (I).
If n is an integer greater than or equal to zero, the series stops after ( n + 1 )
terms because the coefficients from k = n + 1
It = m + I on are zero. If n is not a positive integer or zero, the series is infinite
and converges for x 1 .
162
Chapter Three : Infinite Sequences and Series
n
Our derivation of the binomial series shows only that it is generated by (1 + x )
x 1 . The derivation does not show that the series
and converges for
converges to (1 + x ) .
n
It does, but we omit the proof.
The Binomial Series
n k
x ,
−
1
x
1
,
(
1
+
x
)
=
1
+
For
k =1 k
n
n n(n − 1) (n − 2) (n − k + 1)
where we define =
k!
k
Example ( 34 ):
If n = − 1
− 1
= − 1 ,
1
− 2 − 1( − 2 )
=
=1,
1
2
!
− 1 − 1(−2) (−3) (−1 − k + 1)
k!
=
= (−1) k ( ) = ( − 1 ) k
k!
k!
k
With these coefficient values and with x replaced by -x , the binomial series
formula gives the familiar geometric series
3
1
−1
= (1 + x) = 1 + (−1) k x k = 1 − x + x 2 − x 3 + x 4 + + (−1) k x k +
1+ x
k =1
Similarly , we can find
1
= 1 + x + x 2 + x 3 + x 4 + + x n−1 +
;−1 x 1
1− x
2) (1 + x 2 ) −1 = 1 − x 2 + x 4 − x 6 + x 8 +
;−1 x 1
1)
163
Chapter Three : Infinite Sequences and Series
f ( x) =
Example ( 35 ): Express
1
( x − 2)( x − 5)
as a power series and find
radius of convergent
Solution:
From partial fraction
f ( x) =
1 1
1
−
3 x −5
x−2
1
−1 x
=
x−5
5 0 5
1
−1 x
=
x−2
2 0 2
n
n
→ (1)
= −
0
= −
0
xn
5n +1
; x 5
xn
2 n +1
; x 2
The intersection of the intervals of convergence is x 2 , so
f ( x) =
1 1
1 n
n+1 − n+1 x
3 0 2
5
;x 2
Application of Power series
1- Evaluating Nonelementary Integrals :
Taylor series can be used to express nonelementary integrals in terms of series.
2
Integrals like sin ( x) dx
arise in the study of the diffraction of light.
f ( x) =
c
n =1
n
xn
n
f ( x ) = cn x dx = cn
n =1
n=1
x2
x3
= c0 + c1
+ c2
+ c3
2
3
164
x n+1
n +1
x4
+ ; x R
4
Chapter Three : Infinite Sequences and Series
e− x
Example ( 36 ): Express
2
as a power series and find
1 − e− x
dx .
approximate value of
2
x
0
2
1
Solution:
From Maclurian series
x
x2
x3
e =1+
+
+
+
;− x
1!
2!
3!
x2
x4
x6
−x2
e
=1−
+
−
+
;− x
1!
2!
3!
2
1 − ex
1
x2
x4
x6
=
1
−
+
−
+
x2
x2
1!
2!
3!
x
x2
x4
x6
x8
+
−
+
2!
3!
4!
5!
=1−
Since the series is convergent in the interval of integration
1
1 − e−x
x3
x5
x7
2 dx = x −
+
−
+
x
3 2! 5 3! 7 4! 0
0
1
1
1
1
= 1−
+
−
+
−
3 2! 5 3! 7 4! 9 5!
= 1 − 0.16667 + 0.0333 − 0.00595 + 0.00692 − 0.00013
2
1
0.862
Also we can differentiate term by term
If we have
f ( x) =
c
n =1
n
xn
d
n
c
x
=
ncn x n −1
n
dx n =1
n =1
= c1 + 2c2 x + 2c3 x 2 + 4c4 x 3 + ; x R
f ( x ) =
165
Chapter Three : Infinite Sequences and Series
Example ( 37 ):
Express
f ( x) = tan −1 x
as a power series in
x
Solution:
(
)
d
1
tan −1 x =
dx
1 + x2
From binomial series
1
= (1 + x 2 ) −1 = 1 − x 2 + x 4 − x 6 +
2
1+ x
d
tan −1 x = 1 − x 2 + + x 4 − x 6 + ;−1 x 1
dx
(
)
Integration term by term we have
tan −1 x = x −
x3 x5 x 7
+
−
++ C
3
5
7
Since, at x = 0 we have tan −1 x = 0 , so C = 0
tan −1 x = x −
x3
x5
x7
+
−
+ ;−1 x 1
3
5
7
Example (38 ): Prove that
x3 x5 x 7
1+ x
ln
+
+
+ ;−1 x 1
= 2 x +
3
5
7
1− x
Solution:
d 1+ x d
ln
= ln(1 + x) − ln(1 − x)
dx 1 − x dx
1
1
+
1+ x 1− x
= 1 − x + x 2 − x3 + + 1 + x + x 2 + x3 +
=
(
(
= 2 1 + x2 + x4 + x6
) (
+ )
166
)
Chapter Three : Infinite Sequences and Series
Integration term by term we have
x3
x5
x7
1+ x
ln
=
2
x
+
+
+
++ C
3
5
7
1− x
1+ x
Since, at x = 0 we have ln
= 0 , so C = 0
1− x
x3
x5
x7
1+ x
ln
+
+
+
= 2 x +
3
5
7
1− x
Example ( 39 ): Find the summation of the series
x+
x3
x5
x7
x 2 n −1
+
+
++
+
3
5
7
2n − 1
Solution: From the ratio test , this series is convergent when
x 1.
Let this summation equal F (x) . i.e.,
F ( x) = x +
x3
x5
x7
x 2 n −1
+
+
++
+
3
5
7
2n − 1
Differentiate term by term , we have
F ( x ) = 1 + x 2 + x 4 + + x 2 n − 2 +
=
x
F ( x) =
1
1 − x2
dx
1− x
2
; x 1
= tan −1 x
x
0
= tanh −1 x
0
=
1
1+ x
ln
2
1− x
So , we have
x+
x3
x5
x7
x 2 n −1
1 1+ x
+
+
++
+ = ln
3
5
7
2n − 1
2 1− x
; x 1
167
Chapter Three : Infinite Sequences and Series
2- Evaluating Indeterminate Forms
We can sometimes evaluate indeterminate forms by expressing the functions
involved as Taylor series.
Example ( 40 ): find the following limit by using the power series
lim
x →0
1 − cos x
x2
Solution:
x2
+
2!
1 − cos x
1
=
−
2
x
2!
cos x = 1 −
x4
−
4!
x2
+
4!
x6
+
6!
x4
−
6!
By taking the limit for both sides when x → 0
lim
x →0
1 − cos x
1
=
2
2
x
General Exercises
1. Test the convergent or divergent of the following series:
3n 2 + 3n
4
n=1 n + 2
1)
3)
n
n=1
5)
n=1
4)
n − ln n
2
(n + 1)(n − 3)
6)
3/ 2
1
sin 3
7) n=1 n n
cos2 n
n
9) n=1 5
1
11) n=1 n
13)
n =1
n=1
n
n + 2n
1
n+3
3
tan −1 n
3
n=1 n + 3
n + sin n
2
8) n=1 3n + 6n
2n
10) n=1 3n + 1
12)
n
n +1
n=1
3n + 6n − 1
(n 2 + 1)1 / 2
2
2)
14)
2n
n=1
168
+1
2n + 1
3
+ 5n
n
n=1
n
3
2
Chapter Three : Infinite Sequences and Series
n+ n
3
15) n=1 4n − 2
2n + 1
3/ 2
17) n=1 n (3n + 1)
19)
3
n =1
5
n
n
n
n
3
2
21) n=1 n
5+ n
23)
n=1 1 + n
ln n
2
+1
16)
n
18)
n
20)
n =1
n=1
n +1
n+2
n e
3 −n 2
n =1
(n!) 2
22) n=1 (2n)!
n2
n3 + 3n
25) n
2
n=1
n
24)
n=1 n + 1
n 2 2n
26)
n!
n=1
n 2 5n−1
27) 2n−1
n=1 3
(n + 3)(n + 7)
29)
2n
n=1
n10
31) n
n =1 2
5n−1
28) 3 n−1
n=1 n 4
(n + 5)!
30) 2
2
n=1 n n!2
nn
32) 3
n =1 n n!
23n−1
34) n+6
n =1 8
n3 4n
33)
n!
n =1
n!
35)
n =1 ( 2n)!
n!+3n
n=1 ( n + 1)!
(n + 2)!−n!
39)
2n
n=1
1
41)
n =1 4n − 1
n
43)
3
n=1 ( n + 1)
ln n
45)
n =1 n
37)
n 2 n!
n=1 (n + 3)!
n!
38) n2
n =1 3
(n + 3)!−(n + 1)!
40)
2n (n + 2)!
n=1
n +1
42) 2
n=1 n + 2n − 2
1
44) n
n =1 e
1
46)
n =1 n ln n
1
48)
2
n=1 n ln(ln ln n)
36)
e − n+1
n +1
n=1
1
1
1
1
49) 1 + 3 + 3 + 3 + 3 +
2
3
4
n
1
1
1
1
50)
+
+
+
+
2 ln 2 3 ln 3 4 ln 4
(1 + n) ln(n + 1)
47)
169
Chapter Three : Infinite Sequences and Series
1
1
1
1
+
+
+
+
2
5
10
1 + n2
3n n!
102 n
52)
53) n
n
n=1 ( 2n − 1)!
n =1
51)
2 n + ln n n
55) n
n
n =2 2 ln n
cos n + n
57)
n2 + 1
n=1
sin n 3
54)
n2
n =1
tan −1 n + cos n
n2 + 1
n=1
cos n
58)
n
n =1 n ln n
56)
59)
n =1
1
n(n + 1)
( 2 ) Prove that the following series is convergent and find its sums.
1
1
1
+
+
+
1 2
23
3 4
1
1
1
(II)
+
+
+
1 3
35
57
(I)
2)
Prove that the general term in the series tends to zero nevertheless the
series is divergent
(
n +1 −
n
n =1
)
3) Test the absolutely and conditional convergent of the series
1)
(−1)
n+1
n=1
cos2 n
2n
2 − n2
3) (−1)
(n − 3) 2
n=1
n
ln n n
n2
n=1
1
cos
7) (−1) n 3/ 2n
n
n=1
1
9) (−1) n n sin
n
n=1
5)
(−1) n
11) (−1) n−1
n=1
13) (−1) n−1
n=1
2)
cos n
3
n
n=1
n3 2n
4) (−1)
n!
n=1
1
6) (−1) n−1
2
n(ln n )
n=2
n−1
8)
(−1)n
n=1
2n
en
1
10) (−1) n+1 1 +
1
n(n + 1)
n
2n
12) (−1) n−1 2
n
n=1
n
2
n +1
14) (−1) n−1
1
15) (−1) n
n ln n
n=2
−n
n=1
n=1
2n
n3
n3
16) (−1) 2 4 / 3
(n + 1)
n=1
n
170
Chapter Three : Infinite Sequences and Series
17) (−1) n
n=1
23 n
32 n
18) (−1) n
n=1
n3
2 n−1
4) Find the radius of convergence and interval of convergence of the
following series
1)
n =1
xn
n
xn
2) (−1) 3
n
n =1
n
x
4) n
n =1 3
n
x2n
3) (−1)
n+3
n =1
n
x n−1
6) (−1)
(ln n) 2n−1
n=1
xn
5)
n+ 2
n=1 n 5
n
( x − 2) 2 n
n 9n
n =1
7)
9)
n 2 ( x − 3) n
10) n!( x + 5) 2 n+1
n=1
11)
n =1
n=1
n =1
n +1
( x + 4)
n 2 3n
13) (−2) n
( x + 3)
n
12)
n =1
n+1
(3 x − 12) 2 n
15)
n
n =1
x n −1
17)
n
n =1 n 4
nx n
19) n
n =1 2 (3n − 1)
21)
n 2 ( x + 4) n
n!
n =1
8)
n xn
n =1
(2 x − 4) n
n3 / 2 3n
3n ( x − 2) 2 n+1
n!
n =1
14)
nn n
x
16)
n
!
n =1
18)
(−1)
n =1
n −1
x 2 n −1
(2n − 1)!
xn
20) 2 n
n =1 n 3
x2n
22) 2 n
n
n =1 2 ( n!)
(n!) 2 n
23)
x
n =1 ( 2n)!
5) Prove that the series
171
Chapter Three : Infinite Sequences and Series
1
( x + n)( x + n − 1)
n =1
Is convergent for all
x
except at
x = 0,−1,−2, and find its
sum.
6) Find the radius of convergence and interval of convergence of the
following series
(1 + x 2 ) −1 = 1 − x 2 + x 4 − x 6 +
Then find the expansion of
tan −1 x
. Then prove that
1 1 1
=1− + − +
4
3 5 7
7) Expand the function in the integral in power series , then find the value of
integration
1
1)
sin x dx
2
1/ 2
2)
0
1
3)
3
2
x cos x dx
0
1
5)
0
1/ 2
7)
16 − x dx
2
−x
e
dx
2
0
1 − e− x
x 2 dx
0
1
9)
2
x
e
dx
0
1/ 2
4)
2
(1 + x )
0
1
2 1/ 3
dx
tan −1 x
6)
dx
x
0
1
1 − cos x
8)
dx
x
0
1
sin x
10)
dx
x
0
11)
172
Chapter Four
Differential Equations (DEs)
4.1 Introduction
This is an introduction to ordinary differential equations. We
describe the main ideas to solve certain differential equations, such us
first order scalar equations, second order linear equations, and systems of
linear equations. We use power series methods to solve variable
coefficients second order linear equations. We introduce Laplace
transform methods to find solutions to constant coefficients equations with
generalized source functions. We provide a brief introduction to boundary
value problems.
4.2 Overview of Differential Equations. A differential equation is an
equation, where the unknown is a function and both the function and its
derivatives may appear in the equation. Differential equations are essential for a
mathematical description of nature they lie at the core of many physical theories.
For example, let us just mention Newton's and Lagrange's equations for classical
mechanics, Maxwell's equations for classical electromagnetism, Schrödinger's
equation for quantum mechanics, and Einstein's equation for the general theory of
gravitation.
Example:
Newton's law is the differential equation
173
Where 𝑚 is the particle is mass,
𝑑2𝑥
𝑑𝑡 2
is the particle acceleration, and 𝑓 is the force
acting on the particle, the unknown is 𝑥(𝑡) -the position of the particle in space at
the time t. As we see above, the force may depend on time, on the particle position
in space, and on the particle velocity.
4.3 Classification of differential equations:
There are many types of differential equations, and a wide variety of solution
techniques, even for equations of the same type. We will introduce some
terminology that helps in classification of equations and, by extension, selection of
the solution techniques.
An ordinary differential equation, or ODE, is an equation that depends on one or more
derivatives of functions of a single variable.
A partial differential equation, or PDE, is an equation that depends on one or more
partial derivatives of functions of several variables. In many cases, PDE are solved by
reducing to multiple ODE.
The order of a differential equation is the order of the highest derivative of any
unknown function in the equation.
The degree of the DE is the highest power of the higher derivatives term in the
equations.
A differential equation is linear if any linear combination of solutions of the equation
is also a solution of the equation. A differential equation that is not linear is said to be
nonlinear.
Nonlinear equations are, in general, very difficult to solve, so in many cases one
approximates a nonlinear equation by a linear equation, called a linearization, that is
more readily solved.
174
EXAMPLE:
Classify the following differential equation :
ex
dy
3y x2 y :
dx
o Separable and not linear.
o Linear and not separable.
o Both separable and linear.
o Neither separable nor linear.
Of course the answer here is both separable and linear. (Check the answer).
EXAMPLE:
175
Classify the following differential equation :
w
dw
3t 10 :
dt
o Separable and not linear.
o Linear and not separable.
o Both separable and linear.
o Neither separable nor linear.
(Separable and not linear).
Note: the equation that can be written in the form
dy
p(t ) y q(t ) is linear.
dt
EXAMPLE:
Classify the following differential equation as linear or nonlinear. Give the order and the
5
degree x
d3y
dy
2 y 0 .
3
dx
dx
Defintion: An ordinary differential equation (ODE) is an equation that contains one or
several derivatives of an unknown function, which we usually call y(x) (or sometimes
y(t) if the independent variable is time t). The equation may also contain y itself, known
functions of x (or t), and constants. For example:
are ordinary differential equations (ODEs). The term ordinary distinguishes them from
partial differential equations (PDEs), which involve partial derivatives of an unknown
function of two or more variables. For instance, a PDE with unknown function u of two
variables x and y is
2u 2u
0
x 2 y 2
PDEs have important engineering applications, but they are more complicated than
ODEs. An ODE is said to be of order n if the nth derivative of the unknown function y is
176
the
highest derivative of y in the equation. The concept of order gives a useful
classification into ODEs of first order, second order, and so on. Thus, (1) is of first order,
(2) of second
Order, and (3) of third order.
Verification of Solution
c
x
Verify that y (c : an arbitrary constant) is a solution of the ODE xy y for all x 0
. Indeed, differentiate y
c
c
c
to get y 2 . Multiply this by x, obtaining xy
;
x
x
x
thus, xy y the given ODE
Solution by Calculus. Solution Curves
The ODE y cos x can be solved directly by integration on both sides. Indeed, using
calculus, we obtain y cos xdx sin x c , where c is an arbitrary constant. This is a
family of solutions. Each value of c gives one of these curves.
We see that each ODE in these examples has a solution that contains an arbitrary
constant c. Such a solution containing an arbitrary constant c is called a general
solution of the ODE.
(We shall see that c is sometimes not completely arbitrary but must be restricted to some
interval to avoid complex expressions in the solution.)
We shall develop methods that will give general solutions uniquely (perhaps except for
notation). Hence we shall say the general solution of a given ODE (instead of a general
solution). In most cases, general solutions exist, and every solution not containing an
arbitrary constant is obtained as a particular solution by assigning a suitable value to c.
Exceptions to these rules occur but are of minor interest in applications.
In most cases the unique solution of a given problem, hence a particular solution, is
obtained from a general solution by an initial condition y x0 y 0 with given values x0
177
and y 0 , that is used to determine a value of the arbitrary constant c. Geometrically this
condition means that the solution curve should pass through the point in the xy-plane. An
ODE, together with an initial condition, is called an initial value problem. Thus, if the
ODE is explicit, the initial value problem is of the form y f x, y , yx0 y 0
178
Separable Ordinary Differential Equations
179
EXAMPLE: Heating an Office Building (Newton’s Law of Cooling)
180
Suppose that in winter the daytime temperature in a certain office building is maintained at 70°F. The
heating is shut off at 10 P.M. and turned on again at 6 A.M. On a certain day the temperature inside
the building at 2 A.M. was found to be 65°F. The outside temperature was 50°F at 10 P.M. and had
dropped to 40°F by 6 A.M. What was the temperature inside the building when the heat was turned
on at 6 A.M.?
Physical information. Experiments show that the time rate of change of the temperature T of a body
B (which conducts heat well, for example, as a copper ball does) is proportional to the difference
between T and the temperature of the surrounding medium (Newton’s law of cooling).
Solution. Step 1 . Setting up a model. Let T(t) be the temperature inside the building and TA
the outside temperature (assumed to be constant in Newton’s law). Then by Newton’s law,
Such experimental laws are derived under idealized assumptions that rarely hold exactly.
However, even if a model seems to fit the reality only poorly (as in the present case), it may
still give valuable qualitative information.
To see how good a model is, the engineer will collect experimental data and compare
them with calculations from the model.
Step 2 . General solution. We cannot solve the equation because we do not know TA just
that it varied between 50°F and 40°F, so we follow the Golden R ule: If you cannot solve
your problem, try to solve a simpler one. We solve this equation with the unknown
function TA replaced with the average of the two known values, or 45°F. For physical
reasons we may expect that this will give us a reasonable approximate value of T in the
building at 6 A.M. For constant (or any other constant value) the ODE is separable.
Separation, integration, and taking exponents gives the general solution
181
Step 3 . Particular solution. We choose 10 P.M. to be t=0. Then the given initial
condition is T(0)=70 and yields a particular solution, call it Tp. By substitution,
Step 4 . Determination of k. We use T(4)=65 where t=4 is 2 A.M. Solving algebraically
for k and inserting k into Tp (t) gives
Step 5 . A nswer and interpretation. 6 A.M. is t=8 (namely, 8 hours after 10 P.M.), and
Hence the temperature in the building dropped 9°F, a result that looks reasonable.
EXAMPLE: Leaking Tank. Outflow of Water Through a Hole (Torricelli’s Law)
This is another prototype engineering problem that leads to an ODE. It concerns the
outflow of water from a cylindrical tank with a hole at the bottom (Fig.).
You are asked to find the height of the water in the tank at any time if the tank has diameter
2 m, the hole has diameter 1 cm, and the initial height of the water when the hole is opened
is 2.25 m. When will the tank be empty?
Physical information. Under the influence of gravity the outflowing water has velocity
Where h(t) is the height of the water above the hole at time t, and g=980 cm/sec2 =32.17
ft/ sec2 is the acceleration of gravity at the surface of the earth.
Solution. Step 1 . Setting up the model. To get an equation, we relate the decrease in
water level h(t) to the outflow. The volume of the outflow during a short time t is
182
must equal the change h of the volume of the water in the tank. Now
183
PROBLEMS:
1. Constant of integration. Why is it important to introduce the constant of
integration immediately when you integrate?
184
185
186
Exact Ordinary Differential Equations
187
188
189
190
191
192
EXAMPLE: Solve
SOLUTION:
or
x y' + y + 4 = 0
(y + 4) dx + x dy = 0
M = y+4 ;
N = x
Check the exactness by
M
=1=
y
Solve for
N
x
Exact Differential
u = Mdx + k(y)
u = x y + 4x + k(y)
But
u
= N(x,y)
y
or
x + k'(y) = x
we have
k'(y) = 0
or
k(y) = c*
Thus, we have the solution u = c or
x y + 4x + c* = c
or
x y + 4x = c'
#
2
EXAMPLE:
(1 - sin x tan y) dx + (cos x sec y) dy = 0
SOLUTION:
M = 1-sin x tan y
2
N = cos x sec y
193
2
M
N
= -sin x sec y =
x
y
Exact Differential
The solution is
u = Mdx + k(y)
= x + cos x tan y + k(y)
u
= N(x,y)
y
or
2
2
cos x sec y + k(y)' = cos x sec y
k'(y) = 0
or
k(y) = c*
The solution u = c becomes
x + cos x tan y = c'
#
Linear Differential Equations
Definitions
th
An n –order differential equation is linear if it can be written in the form
y(n)+ an-1(x) y(n-1)+ ... + a1(x) + ao(x) y = f(x)
Hence, a first-order linear equation has the form
y'+ p(x) y = r(x)
e.g.,
2x
st
1 – order linear
y' - y = e
y'' + a(x) y' + b(x) y = f(x)
nd
2 –order linear
Homogeneous Differential Equations
If the function f(x) = 0 [or r(x) = 0], then the above linear differential
equation is said to be homogeneous; otherwise, it is said to be
nonhomogeneous.
194
e.g.
y' - y = 0
homogeneous
2x
y' - y = e
nonhomogeneous
Solution of the First-Order Linear Differential Equations
Homogeneous Equation
The solution of the linear homogeneous equation
y' + p(x) y = 0
Can be obtained by separation of variables
dy/y= - p(x) dx
y(x) = c e
- p(x)dx
or
Nonhomogeneous Equations
The nonhomogeneous equation
y' + p(x) y = r(x)
can be written in the following form
[p(x) y - r(x)] dx + dy = 0
which is of the form
P(x) dx + Q(x) dy = 0
with
P(x) = p(x) y - r(x)
Q(x) = 1
Since
p(x) 0
the above equation is not exact differential. However, we have the
integrating factor
195
F x e
P x dx
for the differential equation. Multiply the differential equation by the
integrating factor, we have
[y' + p(x) y]
e
p(x)dx
= ye
p ( x ) dx
yp( x)e
p ( x ) dx
r(x) e
p(x)dx
According to chain rule, the left hand side of the above equation is the
derivative of y e
p(x)dx
, we have
y e
p(x)dx
= r(x) e
p(x)dx
Integrating both sides of the above equation with respect to x, we have
ye
or
p(x)dx
y =
e
= r ( x) e
p(x)dx
p(x)dx
r ( x) e
dx +
p(x)dx
c
dx +c
Alternative Solution Procedure：
i.
Rearrange the equation in the standard form: y' + p(x) y = r(x)
ii.
Derive the integrating factor e
iii.
Multiply both sides of the given equation by this factor
iv.
Integrate both sides of the resulting equation. Note that the
integral of the left is always just y times the integrating factor.
v.
Solve the integrated equation for y.
EXAMPLE:
Solve the IVP: y' = y + x
y' - y = x
2
p(x)dx
=
e
-dx
,
y' + p(x) y = r(x)
thus, the integrating factor is
e
2
p(x) dx
–x
=e
196
y(0) = 1
Multiply both sides of the differential equation according to the alternative
procedure, we have
e
–x
2
(y' - y) = x e
–x
or
2
( y e )' = x e
–x
–x
Integrating both sides, we have
–x
ye
=
x
x
2
2
Thus,
y = c e - (x + 2 x + 2)
Since
y(0) = 1
x
–x
2
e x dx + c = c - ( x + 2 x + 2 ) e
--- general solution
c = 3
2
Thus y = 3 e - ( x + 2 x + 2 )--- particular solution for y(0) = 1
3
EXAMPLE:
y'= x -2 x y
y(1) = 1
SOLUTION:
y' + 2 xy = x
3
y' + p(x) y = r(x)
- The integrating factor is
e
p(x)dx
= e
Multiply both sides by e
2x dx
= e
x2
x2
and integrating, we have
ex2 ( y' + 2 x y ) = x ex2
3
or
e
e
or
x2
x2
y
y =
= x e
3
2
3 x
x e
x2
dx + c
y =
197
e
- x2
x 2 x 2 -1
e
c
2
Since y(1) = 1
(integration by parts)
1 1
ce1 , we have c = e
2
y=……..(Complete)
thus,
EXERCISES
I. Solve the following Differential Equations:
1.
y' + 2 x y = - 6 x
2.
y dx - dy = x y dx + x dy
3.
(x + y ) y' + (2xy + 1) = 0, y(2) = -2
4.
2 x y' = 10 x y + y
5
y' = y
6
( 4xy + 6y ) + ( 2x + 6xy ) y' = 0
7
x y' - 3xy = - 2y
8
( 4y - x ) y'= y
9
x y' - 2 y = x e
10
2 x y y' + ( x - 1 ) y = x e
11
2
y' = x e-x - 3 x y
12
( x + y + 1 ) y' + x y = 0
13
( y + tan ( x + y ) ) y' + y = 0
14
x y' = y + 5 x y + 4 x
15
( 3 x e + 2 y ) dx + ( x e + x ) dy = 0
16
( x + y - 2 ) y' + ( x - y ) = 0
2
2
2
2
3
5
2
2
2
5/3
3
x
2
2
x
3
2
with y(0) = -1
2
2
2
y
2
2
198
y
II.
2
2
17
(1 + x ) dy + x y dx =
1+x
dx
18
(2x + 3y - 5) y' + (x + 2y - 3) = 0
Under what conditions is the following differential equation exact?
2
y
3
y
3
y
(c x y e + 2 cos y ) + (x e y + x e + k x sin y ) y' = 0
Solve the exact equation.
199
SECOND-ORDER DIFFERENTIAL EQUATIONS
Second-order linear differential equation has the following basic equation :
d2y
dy
p
(
x
)
q( x) y r ( x)
dx
dx 2
or in alternative notation,
y p( x) y q ( x ) y r ( x )
where p(x), q(x), and r(x) are continuous functions.
If r(x) = 0 for all x, then, the equation
d2y
dy
p( x) q( x) y 0 is said to be homogeneous.
2
dx
dx
If r ( x) 0 for all x, then, the equation
d2y
dy
p( x) q( x) y r ( x) is said to be nonhomogeneous.
2
dx
dx
Examples of second-order homogeneous differential equations :
d 2 y dy
2y 0
dx 2 dx
y e x y 0
Examples of second-order nonhomogeneous differential equations :
200
d2y
2 dy
x
xy e x
2
dx
dx
y y 3 y sin x
Solving Second-Order Linear Homogeneous Differential Equations With Constant
Coefficients
Two continuous functions f and g are said to be linearly dependent if one is a constant
multiple of the other. If neither is a constant multiple of the other, then they are called
linearly independent.
Examples:
f(x) = sin x and g(x) = 3 sin x
f(x) = x
and
g ( x) x 2
=>
linearly dependent
=>
linearly independent
The following theorem is central to the study of second-order linear differential
equations:
Theorem :
If y1 y1 ( x) and y2 y2 ( x) are linearly independent solutions of the homogeneous
equation
d2y
dy
p
(
x
)
q( x) y 0
dx
dx 2
(1)
then
y ( x ) c1 y1 ( x ) c2 y2 ( x )
( 2)
201
is the general solution of (1) in the sense that every solution of (1) can be obtained from
(2) by choosing appropriate values for the arbitrary constants c1 and c 2 ; conversely, (2)
is a solution of (1) for all choices of c1 and c 2 .
At this level, we restrict our attention to second-order linear homogeneous differential
equations with constant coefficients only, i.e.
d2y
dy
p qy 0
2
dx
dx
(basic form of equation)
where p and q are constants.
or we can rewrite :
d2y
dy
a 2 b cy 0
dx
dx
(basic form)
where a, b, and c are constants.
Replacing
d2y
dy
with m 2 ,
with m1 , and y with m 0 will result
2
dx
dx
=> is called “auxiliary quadratic equation”
or “auxiliary equation”.
Thus, the general solution of the 2nd – order linear differential equation
a
d2y
dy
b cy 0 where a, b, and c are constants
2
dx
dx
depends on the roots of the auxiliary quadratic equation
am2 bm c 0
i) if b 2 4ac 0
such that
(2 distinct real roots m1 and m2 )
then y c1e m x c2e m x
1
2
202
am 2 bm c 0
ii) if b 2 4ac 0
(1 real repeated root m1 m2 ( m) )
then y c1e mx c2 xe mx
iii) if b 2 4ac 0
(2 complex roots m1 i and m2 i )
then y ex (c1 cos x c2 sin x)
Note :
Each of these solutions contains two arbitrary constants c1 and c 2 since solution
of 2nd – order differential involves 2 integrations.
EXAMPLE: Solve the following differential equations:
1.
Sol.
2
d2y
dy
5 3y 0
2
dx
dx
Auxiliary eqn. : 2m 2 5m 3 0
(2m 1)(m 3) 0
m
1
or m 3
2
the general solution, y c1e
1
x
2
c 2 e 3 x
203
2.
d2y
dy
4
4y 0
dx
dx 2
Auxiliary eqn. : m2 4m 4 0
(m 2)(m 2) 0
m2
the general solution, y c1e 2 x c2 xe 2 x
3.
d2y
dy
2
4y 0
dx
dx 2
Auxiliary eqn. : m2 2m 4 0
2 4 16
2
m 1 3i
m
a 1,
b 3
the general solution, y e x (c1 cos 3x c2 sin 3x )
Solving Second-Order Linear Nonhomogeneous Differential Equations With Constant
Coefficients
Now, we consider 2nd – order differential equation of the form
d2y
dy
p
qy r ( x)
dx
dx 2
204
where p and q are constants and r(x) is a continuous function.
Theorem :
The general solution of y py qy r (x) is
y( x) c1 y1 ( x) c2 y2 ( x) y p ( x)
where c1 y1 ( x) c2 y2 ( x) is the general solution of the homogeneous equation
y py qy 0
and y p (x) is any solution of y py qy r (x) .
In short, what is meant from the above theorem is that, the general solution for 2nd –
order differential equation of the form y py qy r (x) is
y = Complementary equation (C.E.) + Particular solution (P.S.)
where the C.E. is the solution of the differential equation of the form
y py qy 0
(homogeneous)
and the P.S. is any solution of the complete differential equation. It is determined by
trial solution or initial guessing based on the form of r(x).
Equation
Initial guess for
yp
y py qy k
yp A
y py qy keax
y p Ae ax
y py qy a0 a1x ...an x n
y p A0 A1 x ... An x n
205
y py qy a1 cos bx a2 sin bx y p A1 cosbx A2 sin bx
Modification rule : If any term in the initial guess is a solution of the C.E. , then the
correct form for y p is obtained by multiplying the initial guess by the smallest positive
integer power of x required so that no term is a solution of the C.E.
In summary, a P.S. of an equation of the form y py qy ke ax can be obtained as follows :
ax
Step 1. Start with y p Ae as an initial guess.
Step 2. Determine if the initial guess is a solution of the C.E.
y py qy 0 .
Step 3. If the initial guess is not a solution of the C.E. , then
y p Ae ax is the correct form of a P.S.
Step 4. If the initial guess is a solution of the C.E., then multiply it by
the smallest positive integer power of x required to produce a function that is not a solution of the C.E. . This
2 ax
ax
will yield either y p Axe or y p Ax e .
EXAMPLES: Solve the following differential equations:
206
d2y
dy
6
5y 3
dx
dx 2
1.
y C. E P. S .
C. E .
Auxiliary eqn. : m2 6m 5 0
(m 5)(m 1) 0
m 5 or m 1
the general solution for C. E , y c1e5 x c2 e x
P. S .
r ( x) 3
the initial guess : y p A
u sin g y A
dy
d2y
0,
0
dx
dx 2
substitute these values int o the original eqn.
d2y
dy
6
5y 3
dx
dx 2
0 6(0) 5 A 3
A
3
5
the general solution for P. S ., y p
Thus, the general solution is
y c1e5 x c2 e x
3
5
207
3
5
d2y
dy
2 2 2 5 y cos x 5 sin x
dx
dx
2.
y C. E P. S .
C. E .
Auxiliary eqn. : 2m2 2m 5 0
2 36
4
1 3
m i
2 2
m
the gen. sol. for C. E , y
1
x
3
e 2 (c1 cos
3
x c2 sin x )
2
2
P. S .
r ( x ) cos x 5 sin x
the initial guess : y p A1 cos x A2 sin x
u sin g y A1 cos x A2 sin x
dy
A1 sin x A2 cos x
dx
d2y
A1 cos x A2 sin x
dx 2
208
substitute these values int o the original eqn.
d2y
dy
2 2 2 5 y cos x 5 sin x
dx
dx
2( A1 cos x A2 sin x ) 2( A1 sin x A2 cos x )
5( A1 cos x A2 sin x ) cos x 5 sin x
cos x (3 A1 2 A2 ) sin x ( 2 A1 3 A2 ) cos x 5 sin x
compare coefficients of sin x and cos x:
3 A1 2 A2 1
2 A1 3 A2 5
solve for A1 & A2 yields
A1 1,
A2 1
the general solution for P. S ., y p cos x sin x
Thus, the general solution is
ye
1
x
2
3
3
(c1 cos x c2 sin x ) cos x sin x
2
2
Example 3
Solve
2
d2y
dy
dy
3 3.125 y sin x , y (0) 5,
( x 0) 3
2
dx
dx
dx
Solution
The homogeneous equation is given by
(2 D 2 3D 3.125 ) y 0
The characteristic equation is
2r 2 3r 3.125 0
209
The roots of the characteristic equation are
3 32 4 2 3.125
r
2 2
3 9 25
4
3 16
4
3 4i
4
0.75 i
Therefore the homogeneous part of the solution is given by
y H e 0.75 x ( K 1 cos x K 2 sin x)
The particular part of the solution is of the form
y P A sin x B cos x
d2
d
2 2 A sin x B cos x 3 A sin x B cos x 3.125( A sin x B cos x ) sin x
dx
dx
2
d
A cos x B sin x 3( A cos x B sin x) 3.125( A sin x B cos x) sin x
dx
2( A sin x B cos x) 3( A cos x B sin x) 3.125 ( A sin x B cos x) sin x
(1.125 A 3B ) sin x (1.125 B 3 A) cos x sin x
Equating coefficients of sin x and cos x on both sides, we get
1.125 A 3B 1
1.125 B 3 A 0
Solving the above two simultaneous linear equations we get
210
A 0.109589
B 0.292237
Hence
y P 0.109589 sin x 0.292237 cos x
The complete solution is given by
y e 0.75 x ( K 1 cos x K 2 sin x) (0.109589 sin x 0.292237 cos x)
To find K1 and K 2 we use the initial conditions
y (0) 5,
dy
( x 0) 3
dx
From y (0) 5 we get
5 e 0.75( 0) ( K 1 cos(0) K 2 sin( 0)) (0.109589 sin( 0) 0.292237 cos(0))
5 K 1 0.292237
K 1 5.292237
dy
0.75e 0.75 x ( K 1 cos x K 2 sin x) e 0.75 x ( K 1 sin x K 2 cos x)
dx
0.109589 cos x 0.292237 sin x
From
dy
( x 0) 3,
dx
we get
3 0.75e 0.75( 0) ( K1 cos(0) K 2 sin( 0)) e 0.75( 0) ( K1 sin( 0) K 2 cos(0))
0.109589 cos(0) 0.292237 sin( 0)
3 0.75 K 1 K 2 0.109589
211
3 0.75(5.292237 ) K 2 0.109589
K 2 6.859588
The complete solution is
y e 0.75 x (5.292237 cos x 6.859588 sin x) 0.109589 sin x 0.292237 cos x
Example 4
Solve
2
d2y
dy
dy
6 3.125 y cos( x) , y (0) 5,
( x 0) 3
2
dx
dx
dx
Solution
The homogeneous part of the equations is given by
(2 D 2 6 D 3.125 ) y 0
The characteristic equation is given by
2r 2 6r 3.125 0
r
6 6 2 4(2)(3.125)
2(2)
6 36 25
4
6 11
4
1.5 0.829156
0.670844 ,2.329156
Therefore, the homogeneous solution y H is given by
212
y H K1e 0.670845 x K 2 e 2.329156 x
The particular part of the solution is of the form
y P A sin x B cos x
Substituting the particular part of the solution in the differential equation,
d2
d
( A sin x B cos x) 6 ( A sin x B cos x)
2
dx
dx
3.125( A sin x B cos x) cos x
2
d
( A cos x B sin x) 6( A cos x B sin x)
dx
3.125( A sin x B cos x) cos x
2
2( A sin x B cos x) 6( A cos x B sin x)
3.125( A sin x B cos x) cos x
(1.125 A 6 B) sin x (1.125 B 6 A) cos x cos x
Equating coefficients of cos x and sin x we get
1.125B 6 A 1
1.125 A 6 B 0
The solution to the above two simultaneous linear equations are
A 0.161006
B 0.0301887
Hence the particular part of the solution is
y P 0.161006 sin x 0.0301887 cos x
Therefore the complete solution is
y yH yP
y ( K1e 0.670845 x K 2 e 2.329156 x ) 0.161006 sin x 0.0301887 cos x
Constants K1 and K 2 can be determined using initial conditions. From y (0) 5 ,
213
y (0) K 1 K 2 0.0301887 5
K1 K 2 5 0.0301887 4.969811
Now
dy
0.670845 K 1e ( 0.670845) x 2.329156 K 2 e ( 2.329156) x
dx
0.161006 cos x 0.0301887 sin x
From
dy
( x 0) 3
dx
0.670845 K 1 2.329156 K 2 0.161006 3
0.670845 K 1 2.329156 K 2 3 0.161006
0.670845 K 1 2.329156 K 2 2.838994
We have two linear equations with two unknowns
K 1 K 2 4.969811
0.670845 K 1 2.329156 K 2 2.838994
Solving the above two simultaneous linear equations, we get
K 1 8.692253
K 2 3.722442
The complete solution is
y (8.692253e 0.670845 x 3.722442e 2.329156 x )
0.161006sin x 0.0301887 cos x.
Sample Case 3
When
X e ax sin bx
or e ax cos bx ,
214
the particular part of the solution is of the form
e ax ( A sin bx B cos bx) ,
we can get A and B by substituting
y e ax ( A sin bx B cos bx)
in the left hand side of differential equation and equating coefficients.
Example 5
Solve
2
d2y
dy
dy
5 3.125 y e x sin x , y (0) 5,
( x 0) 3
2
dx
dx
dx
Solution
The homogeneous equation is given by
(2 D 2 5D 3.125 ) y 0
The characteristic equation is given by
2r 2 5r 3.125 0
r
5 5 2 4(2)(3.125)
2(2)
5 25 25
4
50
4
1.25,1.25
Since roots are repeated, the homogeneous solution y H is given by
y H ( K 1 K 2 x)e ( 1.25) x
215
The particular part of the solution is of the form
y P e x ( A sin x B cos x)
Substituting the particular part of the solution in the ordinary differential equation
2
2
d 2 x
d
{e ( A sin x B cos x)} 5 {e x ( A sin x B cos x)}
2
dx
dx
x
3.125{e ( A sin x B cos x)} e x sin x
d
{e x ( A sin x B cos x) e x ( A cos x B sin x)}
dx
5{e x ( A sin x B cos x) e x ( A cos x B sin x)} 3.125e x ( A sin x B cos x) e x sin x
2{e x ( A sin x B cos x) e x ( A cos x B sin x) e x ( A cos x B sin x) e x ( A sin x B cos x)}
5{e x ( A sin x B cos x) e x ( A cos x B sin x)} 3.125e x ( A sin x B cos x) e x sin x
1.875e x ( A sin x B cos x) e x ( A cos x B sin x) e x sin x
1.875( A sin x B cos x) ( A cos x B sin x) sin x
(1.875 A B) sin x ( A 1.875 B) cos x sin x
Equating coefficients of cos x and sin x on both sides we get
A 1.875 B 0
1.875 A B 1
Solving the above two simultaneous linear equations we get
A 0.415224 and
B 0.221453
Hence,
y P e x (0.415224 sin x 0.221453 cos x)
Therefore complete solution is given by
y yH yP
216
y ( K 1 xK 2 )e 1.25 x e x (0.415224 sin x 0.221453 cos x)
Constants K1 and K 2 can be determined using initial conditions,
From y (0) 5, we get
K 1 0.221453 5
K 1 5.221453
Now
dy
1.25K 1e 1.25 x 1.25K 2 xe 1.25 x K 2 e 1.25 x
dx
e x (0.415224 cos x 0.221453 sin x) e x (0.415224 sin x 0.221453 cos x)
From
dy
(0) 3, we get
dx
1.25K1e 1.25( 0) 1.25K 2 (0)e 1.25( 0) K 2 e 1.25( 0)
e 0 (0.415224 cos(0) 0.221453sin(0)) e 0 (0.415224 sin(0) 0.221453 cos(0) 3
1.25 K 1 K 2 0.221453 0.415224 3
1.25 K 1 K 2 3.193771
1.25(5.221453 ) K 2 3.193771
K 2 9.720582
Substituting
K 1 5.221453 and
K 2 9.720582
in the solution, we get
y (5.221453 9.720582 x)e 1.25 x e x (0.415224 sin x 0.221453 cos x)
217
The forms of the particular part of the solution for different right hand sides of ordinary
differential equations are given below
y P x
X
a0 a1 x a 2 x 2
b0 b1 x b2 x 2
e ax
Ae ax
sin(bx)
A sin(bx) B cos(bx)
e ax sin(bx)
e ax A sin(bx) B cos(bx)
cos(bx)
A sin(bx) B cos(bx)
e ax cos(bx)
e ax A sin(bx) B cos(bx)
218
Chapter five
Laplace Transforms
If y f (x) is defined at all positive values of x , the Laplace transform denoted by Y (s )
is given by
Y ( s) L{ f ( x)} e sx f ( x)dx
where s is a
0
parameter, which can be a real or complex number. We can get back f (x) by taking the
inverse Laplace transform of Y (s ) .
L1{Y ( s )} f ( x)
Laplace transforms are very useful in solving differential equations. They give the
solution directly without the necessity of evaluating arbitrary constants separately.
The following are Laplace transforms of some elementary functions
L(1)
1
s
L( x n )
L(e ax )
n!
s n 1
, where n 0,1,2,3....
1
sa
L(sin ax)
a
s a2
L(cos ax)
s
s a2
2
2
219
L(sinh ax)
a
s a2
L(cosh ax)
s
s a2
2
2
The following are the inverse Laplace transforms of some common functions
1
L1 1
s
1 ax
L1
e
sa
x n 1
1
, where n 1,2,3......
L n
s n 1!
1
1
L1
n
s a
e ax x n 1
n 1!
1
L1 2
2
s a
1
sin ax
a
s
L1 2
cos ax
2
s a
1
L1 2
2
s a
1
sinh ax
a
s
L1 2
cosh at
2
s a
1 ax
1
e sin bx
L1
2
2
s a b b
sa
e ax cos bx
L1
2
2
s a b
220
s
L1
s2 a2
2
1
x sin ax
2a
221
Properties of Laplace transforms
222
Linear property
If a, b, c are constants and f ( x), g ( x), and h(x) are functions of x then
L[af ( x) bg ( x) ch( x)] aL( f ( x)) bL( g ( x)) cL (h( x))
Shifting property
If
L{ f ( x )} Y ( s )
then
L{e at f ( x)} Y ( s a )
Using shifting property we get
L e ax x n
n!
L e ax sin bx
L e ax cos bx
b
s a 2 b 2
sa
s a 2 b 2
L e ax sinh bx
L e ax cosh bx
, n0
s a n1
b
s a 2 b 2
sa
s a 2 b 2
Scaling property
If
L{ f ( x )} Y ( s )
223
then
L{ f (ax)}
1 s
Y
a a
Laplace transforms of derivatives
If the first n derivatives of f (x) are continuous then
L{ f ( x)} e sx f n ( x)dx
n
0
Using integration by parts we get
e
0
sx
e sx f n 1 ( x) ( s )e sx f n 2 ( x)
f ( x)dx
2 sx
n 3
n 1
n 1 sx
( s ) e f ( x) ...... (1) ( s ) e f ( x) 0
n
(1) n ( s ) n e sx f ( x)dx
0
f
n 1
(0) sf
n 2
(0) s f
2
n 3
(0) ............. s
n 1
f (0) s
n
e
sx
f ( x )dx
0
s nY ( s ) f n 1 (0) sf n 2 (0) s 2 f n 3 (0) ........ s n 1 f (0)
Laplace transform technique to solve ordinary differential equations
The following are steps to solve ordinary differential equations using the Laplace
transform method
(A) Take the Laplace transform of both sides of ordinary differential equations.
(B) Express Y (s ) as a function of s .
(C) Take the inverse Laplace transform on both sides to get the solution.
Let us solve Examples 1 through 4 using the Laplace transform method.
Example
Solve
224
3
dy
2 y e x , y ( 0) 5
dx
Solution
Taking the Laplace transform of both sides, we get
dy
L 3 2 y L e x
dx
3[ sY ( s) y (0)] 2Y ( s)
1
s 1
Using the initial condition, y (0) 5 we get
3[ sY ( s) 5] 2Y ( s)
1
s 1
(3s 2)Y ( s)
1
15
s 1
(3s 2)Y ( s)
15s 16
s 1
Y ( s)
15s 16
( s 1)(3s 2)
Writing the expression for Y (s ) in terms of partial fractions
15s 16
A
B
( s 1)(3s 2) s 1 3s 2
15s 16
3 As 2 A Bs B
( s 1)(3s 2)
( s 1)(3s 2)
15 s 16 3 As 2 A Bs B
Equating coefficients of s 1 and s 0 gives
3 A B 15
225
2 A B 16
The solution to the above two simultaneous linear equations is
A 1
B 18
Y (s)
1
18
s 1 3s 2
1
6
s 1 s 0.666667
Taking the inverse Laplace transform on both sides
6
1
1
L1{Y ( s)} L1
L
s 1
s 0.666667
Since
1
at
L1
e
sa
The solution is given by
y ( x) e x 6e 0.666667 x
Example
Solve
2
dy
3 y e 1.5 x , y (0) 5
dx
Solution
Taking the Laplace transform of both sides, we get
dy
L 2 3 y L e 1.5 x
dx
226
2[ sY ( s) y (0)] 3Y ( s)
1
s 1.5
Using the initial condition y (0) 5 , we get
2[ sY ( s) 5] 3Y ( s)
1
s 1.5
(2s 3)Y ( s)
1
10
s 1.5
(2s 3)Y ( s)
10s 16
s 1.5
10s 16
( s 1.5)(2s 3)
Y ( s)
10s 16
2( s 1.5)(s 1.5)
10s 16
2( s 1.5) 2
5s 8
( s 1.5) 2
Writing the expression for Y (s ) in terms of partial fractions
5s 8
A
B
2
s 1.5 ( s 1.5) 2
( s 1.5)
5s 8
As 1.5 A B
2
( s 1.5)
( s 1.5) 2
5s 8 As 1.5 A B
Equating coefficients of s 1 and s 0 gives
A5
1 .5 A B 8
227
The solution to the above two simultaneous linear equations is
A5
B 0 .5
Y ( s)
5
0.5
s 1.5 ( s 1.5) 2
Taking the inverse Laplace transform on both sides
5 1 0.5
L1{Y ( s)} L1
L
2
s 1.5
( s 1.5)
Since
1
1
1
ax
L1
e and L
2
sa
( s a)
xe ax
The solution is given by
y ( x) 5e 1.5 x 0.5 xe 1.5 x
Example
Solve
2
d2y
dy
dy
( x 0) 3
3 3.125 y sin x , y (0) 5,
2
dx
dx
dx
Solution
Taking the Laplace transform of both sides
d2y
dy
L 2 2 3 3.125 y Lsin x
dx
dx
228
and knowing
d2y
dy
L 2 s 2Y s sy 0 x 0
dx
dx
dy
L sY s y 0
dx
1
s 1
L(sin x)
2
we get
dy
1
2 s 2Y ( s) sy (0) ( x 0) 3sY ( s) y (0) 3.125Y ( s) 2
dx
s 1
2 s 2Y ( s) 5s 3 3sY ( s) 5 3.125Y ( s)
s2s 3 3.125Y (s) 10s 21
1
s 1
2
1
s 1
2
s2s 3 3.125Y (s)
1
10s 21
s 1
2s
22 10s 3 10s 21s 2
( s 2 1)
2
3s 3.125 Y ( s)
Y ( s)
2
10s 3 21s 2 10s 22
s 2 1 2s 2 3s 3.125
Writing the expression for Y (s ) in terms of partial fractions
As B
Cs D
10s 3 21s 2 10s 22
2
2
2s 2 3s 3.125
s 1
s 1 2s 2 3s 3.125
As 3 As Bs 2 B 2Cs 3 3Cs 2 3.125Cs 2 Ds 2 3Ds 3.125D
2 s 2 3s 3.125 s 2 1
10s 21s 10s 22
s 2 1 2 s 2 3s 3.125
3
2
229
A 2C s 3 B 3C 2 D s 2 A 3.125C 3D s B 3.125D
s
2
1 2 s 2 3s 3.125
10s 21s 10s 22
s 2 1 2 s 2 3s 3.125
3
2
Equating terms of s 3 , s 2 , s 1 and s 0 gives
A 2C 10
B 3C 2 D 21
A 3.125 C 3D 10
B 3.125 D 22
The solution to the above four simultaneous linear equations is
A 10 .584474
B 21 .657534
C 0.292237
D 0.109589
Hence
Y ( s)
2s
2
10.584474s 21.657534 0.292237s 0.109589
2s 2 3s 3.125
s2 1
3s 3.125 2{( s 2 1.5s 0.5625 ) 1} 2{( s 0.75) 2 1}
Y ( s)
10.584474( s 0.75) 13.719179 0.292237s 0.109589
2{(s 0.75) 2 1}
s2 1
5.292237( s 0.75)
6.859589
0.292237s 0.109589
2
2
{(s 0.75) 1}
{(s 0.75) 1}
( s 2 1)
( s 2 1)
Taking the inverse Laplace transform of both sides
230
5.292237( s 0.75)
6.859589
L1
L1{Y ( s )} L1
2
2
{(s 0.75) 1}
{(s 0.75) 1
0.292237s
1 0.109589
L1
L 2
2
s 1
s 1
s 0.75
1
6.859589L1
L1{Y ( s )} 5.292237L1
2
2
{(s 0.75) 1}
{(s 0.75) 1
1
s
1
0.292237L1 2
0.109589L 2
s 1
s 1
Since
sa
e ax cos bx
L1
2
2
s a b
b
e ax sin bx
L1
2
2
s a b
1
L1 2
2
s a
sin ax
s
L1 2
cos ax
2
s a
The complete solution is
y ( x) 5.292237e 0.75 x cos x 6.8595859e 0.75 x sin x
0.292237 cos x 0.109589 sin x
e 0.75 x 5.292237 cos x 6.859589 sin x 0.292237 cos x 0.109589 sin x
Example
Solve
2
d2y
dy
dy
( x 0) 3
6 3.125 y cos x , y (0) 5,
2
dx
dx
dx
231
Solution
Taking the Laplace transform of both sides
d2y
dy
L 2 2 6 3.125 y Lcos x
dx
dx
and knowing
d2y
dy
L 2 s 2Y s sy 0 x 0
dx
dx
dy
L sY s y 0
dx
L(cos x)
s
s 1
2
we get
dy
s
2 s 2Y ( s) sy (0) ( x 0) 6sY ( s) y (0) 3.125Y ( s) 2
dx
s 1
2 s 2Y ( s) 5s 3 6sY ( s) 5 3.125Y ( s)
s
s 1
s(2s 6) 3.125Y (s)
s
10s 36
s 1
2s
36 10s 3 11s 36s 2
s2 1
2
6s 3.125 Y ( s)
Y ( s)
2
2
10s 3 36s 2 11s 36
s 2 1 2s 2 6s 3.125
Writing the expression for Y (s ) in terms of partial fractions
As B
Cs D
10s 3 36s 2 11s 36
2
2
2s 2 6s 3.125
s 1
s 1 2s 2 6s 3.125
232
As 3 As Bs 2 B 2Cs 3 6Cs 2 3.125Cs 2 Ds 2 6 Ds 3.125D
2 s 2 6 s 3.125 s 2 1
10s 36s 11s 36
s 2 1 2 s 2 6 s 3.125
3
2
A 2C s 3 B 6C 2 D s 2 A 3.125C 6 D s B 3.125D
s
2
1 2 s 2 6 s 3.125
10s 36s 11s 36
s 2 1 2 s 2 6 s 3.125
3
2
Equating terms of s 3 , s 2 , s 1 and s 0 gives
A 2C 10
B 6C 2 D 36
A 3.125 C 6 D 11
B 3.125 D 36
The solution to the above four simultaneous linear equations is
A 9.939622
B 35 .496855
C 0.0301886
D 0.161006
Then
Y ( s)
2s
2
9.939622s 35.496855 0.0301886s 0.161006
2s 2 6s 3.125
s2 1
6s 3.125 2{( s 2 3s 2.25) 0.6875} 2{( s 1.5) 2 0.829156 2 }
Y ( s)
9.939622( s 1.5) 20.587422 0.0301886s 0.161006
2{(s 1.5) 2 0.8291562 }
s2 1
233
4.969811( s 1.5)
10.293711
2
2
{(s 1.5) 0.829156 } {(s 1.5) 2 0.8291562 }
0.0301886s 0.161006
s2 1
s2 1
Taking the inverse Laplace transform on both sides
4.969811( s 1.5)
10.293711
L1
L1{Y ( s )} L1
2
2
2
2
{(s 1.5) 0.829156 }
{(s 1.5) 0.829156
0.0301886s
1 0.161006
L1
L 2
2
s 1
s 1
( s 1.5)
1
4.969811L1
10.293711L1
2
2
2
2
( s 1.5) 0.829156
( s 1.5) 0.829156
s
1
0.161006L1 2
0.0301886L1 2
( s 1)
( s 1)
Since
sa
e ax cosh bx
L1
2
2
s a b
1 ax
1
e sinh bx
L1
2
2
s
a
b
b
1
L1 2
2
s a
1
sin ax
a
s
L1 2
cos ax
2
s a
The complete solution is
234
10 .293711 1.5 x
e
sinh( 0.829156 x)
0.829156
0.0301886 cos x 0.161006 sin x
y ( x) 4.969811 e 1.5 x cosh(0.829156 x)
e 0.829156 x e 0.829156 x
e 1.5 x 4.969811
2
0.030188 cos x 0.161006 sin x
e 0.829156 x e 0.829156 x
12.414685
2
e 1.5 x 8.692248e 0.829156 x 3.722437e 0.829156 x 0.0301886 cos x
0.161006 sin x
Example
Solve
2
d2y
dy
dy
( x 0) 3
5 3.125 y e x sin x , y (0) 5,
2
dx
dx
dx
Solution
Taking the Laplace transform of both sides
d2y
dy
L 2 2 5 3.125 y L e x sin x
dx
dx
knowing
d2y
dy
L 2 s 2Y s sy 0 x 0
dx
dx
dy
L sY s y 0
dx
L(e x sin x)
1
( s 1) 2 1
we get
235
dy
1
2 s 2Y ( s ) sy (0) ( x 0) 5sY ( s ) y (0) 3.125Y ( s )
dx
( s 1) 2 1
1
2 s 2Y ( s ) 5s 3 5sY ( s ) 5 3.125Y ( s )
( s 1) 2 1
s2s 5 3.125Y (s) 10s 31
1
( s 1) 2 1
s(2s 5) 3.125Y (s)
1
10s 31
( s 1) 2 1
2s
63 10s 3 82s 51s 2
s 2 2s 2
2
5s 3.125 Y ( s)
Y ( s)
10s 3 51s 2 82s 63
s 2 2s 2 2s 2 5s 3.125
Writing the expression for Y (s ) in terms of partial fractions
As B
Cs D
10s 3 51s 2 82s 63
2s 2 5s 3.125 s 2 2s 2 s 2 2s 2 2s 2 5s 3.125
2Cs 3 5Cs 2 3.125Cs 2 Ds 2 5 Ds 3.125D As 3 2 As 2 2 As Bs 2 2 Bs 2 B
2s 2 5s 3.125 s 2 2 s 2
10s 51s 82s 63
s 2 s 2 2 s 2 5s 3.125
3
2
2
2C As 3 5C 2 D 2 A B s 2 3.125C 5D 2 A 2 B s 3.125D 2 B
s
2
2 s 2 2 s 2 5s 3.125
10s 3 51s 2 82s 63
s 2 2 s 2 2 s 2 5s 3.125
Equating terms of s 3 , s 2 , s 1 and s 0 gives four simultaneous linear equations
2C A 10
5C 2 D 2 A B 51
236
3.125 C 5 D 2 A 2 B 82
3.125 D 2 B 63
The solution to the above four simultaneous linear equations is
A 10 .442906
B 32 .494809
C 0.221453
D 0.636678
Then
Y ( s)
2s
2
10.442906s 32.494809 0.221453s 0.636678
2s 2 5s 3.125
s 2 2s 2
5s 3.125 2{( s 2 2.5s 1.5625 )} 2( s 1.25) 2
Y ( s)
10.442906( s 1.25) 19.441176 0.221453( s 1) 0.415225
2( s 1.25) 2
( s 1) 2 1
5.221453( s 1.25) 9.720588 0.221453( s 1) 0.415225
( s 1.25) 2
( s 1.25) 2
( s 1) 2 1
( s 1) 2 1
Taking the inverse Laplace transform on both sides
9.720588
5.221453
L1
L1{Y ( s )} L1
2
( s 1.25)
( s 1.25)
0.221453( s 1)
0.415225
L1
L1
2
2
(
s
1
)
1
(
s
1
)
1
1
1
9.720588L1
5.221453L1
2
( s 1.25)
( s 1.25)
( s 1)
1
0.415225L1
0.221453L1
2
2
(
s
1
)
1
(
s
1
)
1
Since
237
sa
e ax cos bx
L1
2
2
s a b
b
e ax sin bx
L1
2
2
s a b
1
ax
L1
e
sa
1
L1
n
( s a)
e ax x n1
(n 1)!
The complete solution is
y ( x) 5.221453e 1.25 x 9.720588e 1.25 x x 0.221453e x cos x
0.415225e x sin x
e 1.25 x 5.221453 9.720588 x e x (0.221453 cos x 0.415225 sin x)
EXERCISES
238
239
240
241
242
243
Chapter Six
Partial Differential Equations in Cartesian
Coordinates
An equation involving partial derivatives of an unknown function of two or more
independent variables is called a partial differential equation.
Compared with ordinary differential equations, far more problems in physical sciences
lead to partial differential equations. In fact, most of mathematical physics deals with
partial differential equations.
In general, the totality of solutions of a partial differential equation is very large.
However, a unique solution of a partial differential equation corresponding to a given
physical problem can usually be obtained by the use of either boundary and/or initial
conditions. In practice, the boundary conditions frequently serve as a guide in choosing
a particular form of the solution, which satisfies the partial differential equation as well
as the boundary conditions.
The field of partial differential equation is very wide. We are going to focus our
attention on the equations that arise most often in physics, namely
244
Wave Equations and Separation of Variables
245
246
247
248
249
250
251
252
253
254
255
256
257
Diffusion Equations
258
259
260
261
262
Laplace Equations
263
264
265
266
267
268
269
270
MAPLE Example
271
272
273
274
275
References
1. James Stewart, Calculus, Early Transcendental, Sixth Edition, Mc Master
University, 2008.
2. H. Anton , I. C. Bivens and S. Davis, calculus, 10th-edition, 2017
3. G. Nagy, Ordinary Differential Equations, Michigan State University, 2019.
4. K. Tang, Mathematical Methods for Engineers and Scientists 3, Springer-Verlag
Berlin Heidelberg, 2007.
5. Miroslav Lovric´, Functions of several variables, Department of Mathematics and
Statistics McMaster University, 2011.
6. Erwin Kreyszig, Herbert Kreyszig and Edward Normination, Advanced
Engineering Mathematics, Copyright © 2011, 2006, 1999 by John Wiley & Sons,
Inc. All rights reserved.
7. George Articolo, Partial differential equations and boundary value problems,
Copyright © 2009, Elsevier Inc. All rights reserved.
8. Jeffrey R. Chasnov, Differential equations for Engineers, Copyright © 2019 by
Jeffrey Robert Chasnov .
9. Robert C. Rogers, the Calculus of Several Variables, September 29, 2011.
10.Paul Dawkins, Calculus of several variables, Lamar University
(2010).
7. Internet sources for the following keywords:
- Functions of several variables
- Partial derivatives
- Limits and continuity for functions of several variables
- Ordinary differential Equations
- Partial differential equations.
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