Download! - Edhole.com
Download
Report
Transcript Download! - Edhole.com
Chapter 6
Series Solutions of Linear Equations
1
NCCU
Wireless Comm. Lab.
Outline
Using power series to solve a differential equation. First, we
should decide the point we choose to be the expanding point that
is ordinary or not.
If the point is not an ordinary point, decide it a regular or irregular
singular point, then use the Frobenius’ series to solve the problem.
Introduce the Bessel equation and the Legendre’s equation.
2
NCCU
Wireless Comm. Lab.
Introduction
In applications, higher order linear equations with variable
coefficients are just as important as, if not more important than,
differential equations with constant coefficient.
Considering a equation y xy 0, it does not possess elementary
solutions. But we can find two linear independent solutions of
y xy 0 by using the series expansion.
3
NCCU
Wireless Comm. Lab.
6.1 Solutions About Ordinary Points
In section 4.7, without understanding that the most higher-order
ordinary equations with variable coefficients cannot be solved in
terms of elementary functions.
The usual strategy for solving differential equations of this sort is
to assume a solution in the form of an infinite series and proceed
in a manner similar to the method of undetermined coefficients.
4
NCCU
Wireless Comm. Lab.
6.1.1 Review of Power Series
Definition
A power series in x a is an infinite series of the form
cn ( x a ) n
c0 c1 ( x a ) c2 ( x a )
2
n 0
Such a series is also said to be a power series centered at a.
For example, the power series
c ( x 1)
n
n
is centered at a =1.
n 0
Convergence
A power series
c ( x a)
n 0
n
n
is convergent at a specified value of x if its
sequence of partial sums SN ( x) converges- that is,
N
limS x lim c
N
N
N n 0
n
( x a ) n constant.
If the limit does not exist at x, the series is said to be divergent.
5
NCCU
Wireless Comm. Lab.
Interval of Convergence
Every power series has an interval of convergence. The interval of
convergence is the set of all real number x for which the series converges.
Note: We will use the ratio test to see the series is convergence or divergence for
x x the region of R , where R will be defined immediately
and R is the radius of convergence.
Radius of Convergence
As we mentioned that the R is assigned to be an interval boundary to check
the series for its convergence property and R is also called the radius of
convergence.
What’s the meaning for the value R?
It means a distance from the point x to the nearest singular point.(see in Theorem 6.1)
Bringing a concept, the singular point possess between convergent and
divergent region.
6
NCCU
Wireless Comm. Lab.
If R 0, then a power series
diverges for x a R.
c ( x a) converges for
n
n 0
n
x a R and
For example, if the series converges for x=a or for all x, then R is equal to
0 or .
Recall that x a R
is equivalent to a R x a R.
Note: A power series may or may not converge at the endpoints a-R or a+R of this
interval.
Absolute Convergence
Within its interval of convergence a power series converges absolutely.
c
n 0
n
( x a ) n converges, where a R x a R.
7
NCCU
Wireless Comm. Lab.
Ratio test
Convergence of a power series
the ratio test.
c ( x a)
n 0
n
n
can often be determined by
Suppose that cn 0 ,n and that
lim
n
cn 1 ( x a)n 1
cn 1
x
a
L.
lim
cn ( x a)n
c
n
n
If L<1 the series converges absolutely, if L>1 the series diverges, and if L=1
the test is inconclusive.
8
NCCU
Wireless Comm. Lab.
Example:
A power series
( x 5)n
, the ratio test gives
n
n 0 4 ( n 1)
( x 5) n 1
( n 1)
1
4 n 1 ( n 2)
x
5
x 5 ( L );
n
lim
lim
( x 5)
4
n
n 4( n 2)
n
4 ( n 1)
1
The series converges absolutely for x 5 1 (L<1), we get 1 x 9.
4
The series diverges for L>1, that is x 9 or x 1.
Test for the convergence of the boundary for x=1 or 9.
n 0
( 4) n
4 n ( n 1)
n 0
( 1) n
for x 1, it will converge.
( n 1)
(4) n
1
for x 9, it will diverge.
n
n 1 4 ( n 1)
n 1 ( n 1)
So the int erval of convergence of the series is [1, 9).
9
NCCU
Wireless Comm. Lab.
A Power Series Defines a Function
A power series defines a function f ( x) cn ( x a) whose domain is the
n 0
interval of convergence of the series. If the radius of the convergence is
R>0, then f is continuous, differential, and integrable on the interval
(a R, a R). Thus, f ( x ) and f ( x)dx can be found by term-by-term
differentiation and integration.
n
n
If y cn x is a power series in x, then the first two derivatives are
n 0
Since y c0 c1 x c2 x
cn x
So y c1 2c2 x
n 1
2
ncn x
y 2c2 3 2c3 x
cn x n
n
n 0
ncn x n 1
n 1
n (n 1)cn x
n2
n(n 1)cn x n 2 .
n2
It will be useful to substitute y, y, and y into the 2nd differential equation.
10
NCCU
Wireless Comm. Lab.
Identity Property
If c ( x a ) 0,
n
n 0
n
R 0 for all numbers x in the interval of convergence,
then cn 0 n.
Analytic at a Point
A function f is analytic at a point a if it can be represented by a power
series in x-a with a positive or infinite radius of convergence.
For example,
x2
e 1 x
2!
x
x2
cos x 1
2!
xn
n!
n 0
x2n
( 1)
2n !
n
For analytic, the function is continuous, differential ,
xn
n!
and integrable in the interval of convergence.
( 1) n x 2 n
2n !
n 0
x3
( 1) n 2 x 2 n 1
( 1) n 2 x 2 n 1
sin x x
for x .
3!
(2n 1)!
(2
n
1)!
n 0
These Taylor series centered at 0, called Maclaurin series,
show that e x , cos x, and sin x are analytic at x 0.
11
NCCU
Wireless Comm. Lab.
Arithmetic of Power Series
Power series can be combined through the operations of addition,
multiplication, and division.
Example:
Using the series expansion to describe the sin x cos x
( 1) n 2 x 2 n 1 x 2 m
sin x cos x
(2n 1)! m 0 2m !
n 0
x3
(x
3!
( 1) n 2 x 2 n 1
(2n 1)!
x2
) (1
2!
x2n
2n !
)
x3 x5
x7
x
x, x .
3
20 252
It is obvious that the power series sin x cos x converges on the same interval.
12
NCCU
Wireless Comm. Lab.
Shifting the Summation Index
It is important to combine two or more summations with different index, so
it may need to shift the summation index. You may see the rule by the
following example.
Example 1:Adding
Two Power
Series
Write n ( n 1)cn x n 2 cn x n 1
n 2
( If y
n 0
c
n
n 0
x n is a Taylor series at centered point 0, the subject
is liking to combine the y xy .)
From the original subject ,
start from x1
start from x 0
n(n 1)c
n 2
with n 3
start from x1
n
x
n 2
cn x
n 0
start from x1
with n 0
n 1
2c2 x n ( n 1)cn x
0
n 3
13
n 2
cn x n 1
n 0
NCCU
Wireless Comm. Lab.
Considering the term of
n(n 1)c x
n 2
n
n 3
n(n 1)c x
n 2
n
n 3
cn x n 1
n 0
let k n - 3
let k n
cn x
n 1
n 0
(k 3)(k 2)ck 3 x
k 0
k 1
ck x k 1
k 0
[( k 3)( k 2)ck 3 ck ] x k 1
k 0
n( n 1)cn x
n 2
n 2
cn x
n 0
n 1
2c2 [(k 3)(k 2)ck 3 ck ]x k 1
k 0
Note : For the shifting summation approach ,
Step 1.You may figure out the minimum order of each component , then take the terms
out from the summation.
Step 2.Substitute a new index to the old one for preparing to combine the summation.
Step 3.Create a new summation for the original subject.
14
NCCU
Wireless Comm. Lab.
6.1.2 Power Series Solutions
Definition 6.1 Ordinary and Singular Points
A point x0 is said to be an ordinary point of the differential equation
a2 ( x) y a1 ( x) y a0 ( x) y 0, if both P(x) and Q(x) in the standard form
y P( x ) y Q ( x ) y 0 are analytic at x0 . A point that is not an ordinary
point is said to be a singular point of the equation.
For analytic, it means that a2 ( x ) 0 in y
a1 ( x )
a ( x)
y 0
y 0.
a2 ( x )
a2 ( x )
Considering the differential equation y cos x y e x y 0 and y ln x y e x y 0.
–
y cos x y e x y 0
y ln x y e x y 0
For y cos x y e x y 0
For y ln x y e x y 0
P( x ) cos x, Q ( x ) e x
P( x) ln x, Q( x) e x
x 0 is the ordinary point,
x 0 is a singular point,
because P( x ) and Q ( x ) are analytic.
because P ( x ) is not analytic.
15
NCCU
Wireless Comm. Lab.
Polynomial Coefficients
A polynomial is analytic at any value x, and a rational function
is analytic except at points where its denominator is zero.
Thus if a 2 ( x), a1 ( x), ,and a 0 ( x) are polynomials with no common factors,
then both rational function P(x)=
a ( x)
a1 ( x)
and Q(x)= 0
are analytic except
a2 ( x)
a2 ( x)
where a 2 ( x) 0.
For x x0 is an ordinary point of a2 ( x) y a1 ( x) y a0 ( x) y 0, if a2 ( x0 ) 0,
whereas x x0 is a singular point of a2 ( x) y a1 ( x) y a0 ( x) y 0, if a2 ( x0 ) 0.
16
NCCU
Wireless Comm. Lab.
Example : If the equation is ( x 2 1)( x 3) 2 y ( x 1) y ( x 1)( x 3) y 0
Consider for the ordinary and singular points.
The original equation can be rewritten to y
( x 1)
( x 1)( x 3)
y
y 0.
( x 2 1)( x 3) 2
( x 2 1)( x 3) 2
1.For a2 ( x) ( x 2 1)( x 3) 2 , let a2 ( x) 0 to calculate the possible point that make the function
not analytic, so x 1, -3.
( x 1)
1
( x 2 1)( x 3) 2 ( x 1)( x 3)
( x 1)( x 3)
1
Q( x) 2
( x 1)( x 3) 2 x 1
3.Combining 1 and 2.
2.P ( x)
So x 1, -3 are the singular point for this function.
Note : x 1,-3 are also called the regular singular point
in Definition 6.2.
17
NCCU
Wireless Comm. Lab.
Theorem 6.1 Existence of Power Series Solutions
If
x x0 is an ordinary point of the differential equation
a2 ( x) y a1 ( x) y a0 ( x) y 0, we can always find two linearly independent
solutions in the form of a power series centered at x0 -that is,
y
c
n
(x x0 ) n . A series solution converges at least on some interval
n 0
defined by x x0 R, whereas R is the distance from x0
singular point.
to the closest
A solution of the form y cn ( x x0 )n is said to be a solution about the ordinary point
n 0
x0 .
18
NCCU
Wireless Comm. Lab.
Example 2. Power Series Solutions
Solve y xy 0.
There are no finite singular points, Theorem 6.1 guarantees
two power series solutions centered at 0, convergent for x .
Substituting y cn x and y cn n(n -1) x n-2 into the
n
n 0
n 2
differential equation.
y xy cn n(n -1) x
n 2
let k n-3
for the sec ond term
and let k n
for the third term
n-2
cn x
n 0
n 1
2c2 x ck 3 (k 3)(k 2) x
0
2c2 x cnn (n -1) x
0
k 0
n-2
n 3
k 1
cn x n 1
n 0
ck x k 1
k 0
2c2 (k 3)(k 2)ck 3 ck x k 1 0
k 0
19
NCCU
Wireless Comm. Lab.
x,
x satisfies 2c2 ( k 3)( k 2)ck 3 ck x k 1 0.
k 0
We should let 2c2 0 and ( k 3)( k 2)ck 3 ck 0, where k 0,1, 2,
ck
, k 0,1, 2
( k 3)( k 2)
Substitute the value of k into the recursive term,
So we get c2 0 and ck 3
.
.
c0
c
0
23
6
c1
c
k 1, c4
1
43
12
c2
0
k 2, c5
0
5 4
20
c3
c0
1
c
k 3, c6
0
65
30
6 180
c4
1
c1
c
k 4, c7
1
76
42 12
504
k 0, c3
k 5, c8
c5
1
c
2 0
8 7
56 20
20
NCCU
Wireless Comm. Lab.
So we reconstruct the term
( k 3)( k 2)c
k 3
ck x k 1
k 0
to be the form of
x3
x6
c0 1
6
180
x4
x7
c1 x
12 504
We substitue it into the origin series y
c
n
xn.
n 0
x3
x6
So, we get y c0 1
6
180
x4
x7
c1 x
12 504
.
Another description of y c0 y1 ( x ) c1 y2 ( x ),
( 1) n
we get y1 ( x ) 1
x 3n 2
(3n 1)3n
n 1 2 3
y2 ( x ) x
n 1
3 4
( 1) n
x 3n 1
(3n )(3n 1)
21
NCCU
Wireless Comm. Lab.
Example 3. Power Series Solution
Solve ( x2 1) y xy y 0.
There shows singular points at x i, and so the power series solution
centered at 0 will converge at least for x 1, where 1 is the distance
in the complex plane from the origin to x i.
Substituing y
cn x
n 0
n
, y ncn x n 1 , and
n 1
y cn n(n -1) x n-2 into the differential equation.
n2
22
NCCU
Wireless Comm. Lab.
( x 1) y xy y
2
n2
n2
(2c2 c0 ) x (6c3 c1 c1 ) x
0
let k n -2
for the fourth term
let k n
for another summations
1
2c2 c0 6c3 x
1
2c2 c0 6c3 x
1
cnn(n -1) x cnn(n -1) x
n
n-2
ncn x
n
n 1
cn n(n -1) x cn n(n -1) x
n2
n4
n
n-2
n
c
x
n
n 0
ncn x
n
n2
cn x n
n2
k
c
k
(
k
-1)
x
c
(
k
2)
(
k
1)
x
kc
x
c
x
k
k
k 2
k
k
k 2
k 2
k (k -1) k 1 c
k 2
k
k
k
k 2
k 2
(k 2)(k 1)ck 2 x k 0
For the equality of the equation to 0 x, x 1.
We have 2c2 c0 0, 6c3 0, and k (k -1) k 1 ck ( k 2)( k 1)ck 2 0, k=2,3,4, .
c0 2c2 , c3 0, ck 2 -
(k 1)(k 1)
k -1
ck ck , k 2.3.4. .
(k 2)(k 1)
k 2
23
NCCU
Wireless Comm. Lab.
By iteration of k , we get
1
11
1
k 2, c4 - c2 - c0 - c0
4
42
8
2
k 3, c5 - c3 0
5
3
1 1 1
k 4, c6 - c4 - c0 c0
6
2 8 16
k 5, c7 0
So we rewritten the power series y cn x n to y c0 (1
n 0
1 2 1 4
x x
2
8
) c1 x c0 y1 ( x) c1 y2 ( x).
1 2
1 3 5 (2n 3) 2 n
Thus y1 ( x) 1 x ( 1) n 1
x
n
2
2
n
!
n2
y2 ( x ) x
, where x 1.
24
NCCU
Wireless Comm. Lab.
Example 4. Three-Term Recurrence Relation
If we seek a power series solution y c x for the differential equation
n
n 0
y (1 x) y 0.
cn n(n 1) x
y (1 x ) y
n 2
n 2
(2c2 c0 ) x
0
c
n 3
n
n ( n 1) x
2c2 c0
c
k 1
k 2
k n-2
cn x cn x n 1
n
n 0
n 2
n 0
cn x cn x n 1
n
n 0
k n
k n 1
( k 2)( k 1) ck ck 1 x k 0
For the equality x ,
We get c2
n 1
Let
n
x .
1
ck ck 1
c0 and ck 2
k 1.2.3.
2
( k 2)( k 1)
25
.
NCCU
Wireless Comm. Lab.
Iterating the value k into the recurrence relation
1
c0 ,
2
c c0
c
c
k 1, c3 1
1 0
3 2
6
6
1
c0 c1
c2 c1
c
1
2
k 2, c4
1
c0
43
12
12 24
c c2
c
c
k 3, c5 3
1 0
5 4
120 30
with the condition c2
y
c
n 0
c0 (1
n
xn
1 2 1 3
1 4
1 5
x x
x
x
2
6
24
30
26
) c1 ( x
1 3
1 4
1
x
x
x5
6
12
120
).
NCCU
Wireless Comm. Lab.
Nonpolynomial Coefficients
The next example illustrates how to find a power series solution about the
ordinary point x0 0 of a differential equation when its coefficients are
not polynomials.
Example 5. ODE with Nonpolynomial Coefficients
Solve y (cos x) y 0.
We see that x 0 is an ordinary point of the equation,
because we know cos x is analytic at x 0.
Substituting y cn x , y n (n -1)cn x n 2 , and
n
n 0
n2
( 1) n x 2 n
cos x
into the equation.
2n !
n0
27
NCCU
Wireless Comm. Lab.
(1) k x 2 k
n
y (cos x) y n(n -1)cn x
c
x
n
2k ! n 0
n2
k 0
1 2 1 4
1 6
2
3
2c2 6c3 x 12c4 x 20c5 x (1 x
x
x )(c0 c1 x c2 x 2
2
24
720
1
1
2c2 c0 (6c3 c1 ) x (12c4 c2 c0 ) x 2 (20c5 c3 c1 ) x 3 0.
2
2
For the equality x, x .
n2
1
1
We get 2c2 c0 0, 6c3 c1 0, 12c4 c2 c0 0, 20c5 c3 c1 0,
2
2
1
1
1
1
So c2 c0 , c3 c1 , c4 c0 , c5 c1 , .
2
6
12
30
1
1
1
1
y cn x n c0 (1 x 2 x 4 ) c1 ( x x 3 x 5 ).
2
12
6
30
n 0
28
)
.
NCCU
Wireless Comm. Lab.
Solution Curves
n
y
c
x
n
The approximate graph of a power series solution
can be
n 0
obtained in several ways. We can always resort to graphing the terms in the
sequence of partial sums of the series—in other words, the graphs of the
N
polynomials
S N x cn x n .
n 0
For a large value of N,
lim S N x lim
N is l arg e
N is l arg e
N
n
c
x
n
y.
n 0
N
n
By this, S N x cn x will give us some information about the
n 0
behavior of y(x) near the ordinary point.
29
NCCU
Wireless Comm. Lab.
Remarks
Even though we can generate as many terms as desired in
series solution y c x either through the use of a
recurrence relation or, as in Example 4, by multiplication, it
may not be possible to deduce any general term for the
coefficients cn . We may have to settle, as we did in Example 4
and 5, for just writing out the first few terms of the series.
n
n 0
n
30
NCCU
Wireless Comm. Lab.
6.2 Solutions About Singular Points
The two differential equations y xy 0 and x y y 0 are similar
only in that they are both examples of simple linear second-order
equations with variable coefficients.
2
We saw in the preceding section that since x=0 is an ordinary
point of the first equation, there is no problem in finding two
linear independent power series solutions centered at that point.
In the contrast, because x=0 is a singular point(which is defined
in Definition 6.1) of the second ODE, finding two infinite series
solutions of the equation about that point becomes a more
difficult task.
31
NCCU
Wireless Comm. Lab.
Regular and Irregular Singular Points
A singular point at x x0 of a linear differential equation
a2 ( x ) y a1 ( x ) y a0 ( x ) y 0
is further classified as either regular or irregular. The classification again
depends on the functions P and Q in the standard form
y P( x ) y Q ( x ) y 0.
Note : Where P ( x )
a1 ( x )
a ( x)
and Q ( x ) 0
.
a2 ( x )
a2 ( x )
Definition 6.2 Regular and Irregular Singular Points
A singular point x0 is said to be a regular singular point of the differential equation
a2 ( x) y a1 ( x) y a0 ( x) y 0, if the function p ( x) P ( x) ( x - x0 ) and q( x) Q( x) ( x - x0 ) 2
are both analytic at x0 . A singular point that is not regular is said to be an
irregular singular point of the equation.
32
NCCU
Wireless Comm. Lab.
If x x0 is a regular singular point of the differential equation y P( x) y Q( x) 0,
then multiplying ( x - x0 ) 2 in both side,
( x - x0 ) 2 y ( x - x0 ) 2 P( x) y ( x - x0 ) 2 Q( x) y 0 ( x - x0 ) 2 y ( x - x0 ) p( x) y q( x) y 0.
For regular singular point p( x) P( x) ( x - x0 ) and q( x) Q( x) ( x - x0 ) 2
are both analytic at x0 .
So ( x - x0 ) 2 y ( x - x0 ) p ( x) y q ( x) y 0, where p and q are analytic at the point x x0 .
Example 1. Classification of Singular Points
It should be clear that x 2 and x -2 are singular points of
( x 2 - 4) 2 y 3( x 2) y 5 y 0.
Comparing with the general form of y P ( x ) y Q ( x ) y 0,
it is clearly that P ( x )
3( x 2)
3
5
and
Q
(
x
)
.
( x 2 - 4) 2
( x 2) 2 ( x 2)
( x 2 - 4) 2
By the equivalent form ( x - x0 ) 2 y ( x - x0 ) p ( x ) y q ( x ) y 0,
where p ( x ) P ( x ) ( x - x0 ) and q ( x ) Q ( x ) ( x - x0 ) 2 .
33
NCCU
Wireless Comm. Lab.
Checking x 2 and x -2 is the regular or irregular singular point.
3
5
2
and
q
(
x
)
(
x
2)
Q
(
x
)
.
( x 2) 2
( x 2) 2
From p ( x ) and q ( x ) are analytic at x 2, so x 2 is a regular singular point.
First if x 2, p ( x ) ( x - 2) P ( x)
3
5
and q ( x) ( x 2) 2 Q ( x)
.
( x 2)( x 2)
( x 2) 2
Although q ( x ) is analytic at x -2, p ( x ) is not analytic at x -2.
Second if x -2, p ( x ) ( x 2) P ( x)
Combining these conditions , x -2 is an irregular singular point.
34
NCCU
Wireless Comm. Lab.
Theorem 6.2 Frobenius’ Theorem
If x x0 is a regular singular point of the differential equation
a2 ( x) y a1 ( x) y a0 ( x) y 0, then there exists at least one solution of the
form
r
n
y ( x x0 ) cn ( x x0 ) cn ( x x0 ) n r ,
n 0
n 0
where the number r is a constant to be determined. The series will
converge at least on some interval 0 x x0 R.
If we consider a differential equation that has a regular singular point,
then we can substitute y
c ( x x ) n r into the DE like the approach
n 0
n
0
we did before by using the power series to solve with the ordinary point.
35
NCCU
Wireless Comm. Lab.
Example 2. Two series solutions
Because x 0 is a regular singular point of the differential equation
3 xy y - y 0,
we try to find a solution of the form y cn x n r .
n 0
From y c0 x r c1 x1 r
cn x n r
y c0 rx r 1 c1 (1 r ) x r
y c0 r (r 1) x
r 2
,
cn (n r ) x
c1 (1 r )rx
r 1
n r 1
(n r )cn x n r 1
n 0
cn (n r )(n r 1) x
nr 2
cn (n r )(n r 1) x n r 2 ,
n 0
now we substitute y , y , and y into the equation.
3 xy y - y 3cn (n r )(n r 1) x
n 0
3r (r 1) r c0 x
r 1
let k n
for 4th term
(n r )cn x
n 0
3cn (n r )(n r 1) x
n r 1
n 1
let k n -1
for 2nd and 3rd term
n r 1
r (3r - 2)c0 x
r 1
n r 1
cn x n r
n 0
( n r ) cn x
n 1
n r 1
cn x n r
n 0
(k r 1)(3k 3r 1)ck 1 ck x k r 0
k 0
36
NCCU
Wireless Comm. Lab.
This implies r (3r - 2)c0 0 and (k r 1)(3k 3r 1)ck 1 ck 0, k 0,1, 2,
.
If we let c0 0, we will see the all terms in recursive term that will equal to zero, it ' s a trivial solution.
So, we choose r (3r - 2) 0 and ck 1
ck
, k 0,1, 2,
(k r 1)(3k 3r 1)
ck
, k 0,1, 2,
(k 1)(3k 1)
ck
2
r2 , ck 1
, k 0,1, 2,
3
(k 1)(3k 5)
From (1) we get
If r1 0, ck 1
(1)
.
(2)
From (2) we get
c1 c0
c1
c0
8
c
c3 0
168
c2
c2
cn
c0
n !1 4 (3n 2)
.
c0
5
c0
80
c
c3 0
2640
cn
c0
.
n ! 5 8 11 (3n 2)
37
NCCU
Wireless Comm. Lab.
From each set contains the same coefficient c0 , so we omit the term.
So we find two independent solutions on the entired x - axis.
1
n
y1 ( x) x 1
x
n
!
1
4
(3
n
2)
n 1
0
1
y2 ( x) x 1
xn .
n 1 n ! 5 8 11 (3n 2)
By the ratio test it can be demonstrated that both y1 ( x) and y2 ( x) converge
2
3
for all finite values of x that is, x .
Hence, by the superposition principle, y c1 y1 ( x) c2 y2 ( x) is another solution
of this differential equation. On any interval not containing the origin, this linear
combinition represents the general solution of the differential equation.
38
NCCU
Wireless Comm. Lab.
Indicial Equation
Equation r (3r 2) 0 is called the indicial equation of the previous example,
and the value r1 0 and r2 2 are called the indicial roots, or exponents, of
3
the singularity x=0.
In general, after substituting y cn x
into the given
n 0
differential equation and simplifying, the indicial equation is a quadratic
equation in r that results from equating the total coefficient of the lowest
power of x to zero.
nr
We solve for the two values of r and substitute these value
ck
into a recurrence relation such as ck 1
.
(k r 1)(3k 3r 1)
By Theorem 6.2, there is at least one solution of the assumed
series form that can be found.
39
NCCU
Wireless Comm. Lab.
Example for Indicial Equation
From y P( x ) y Q ( x ) y 0, if we multiply x 2 on both side,
we get x 2 y x xP ( x ) y x 2Q ( x ) y 0.
As the previous concept , we know that the p( x ) xP( x ) and
q( x ) x 2Q ( x ) are both analytic.
So, we let p( x ) a0 a1 x a2 x 2
and q( x ) b0 b1 x b2 x 2 .
Substituting y cn x n r , p( x ), and q( x ) into the differential equation,
n 0
(n r)(n r 1)cn x
n 0
nr
(n r ) a0 a1 x a2 x cn x
2
n 0
nr
b0 b1 x b2 x 2 cn x n r 0.
n 0
As we find the indicial equation, we will take from the lowest order of n like x r ,
so we let n 0, we get the indicial equation r ( r -1) ra0 b0 0.
40
NCCU
Wireless Comm. Lab.
Three Cases
Case I:
If r1 and r2 are ristinct and do not differ by an integer,
then there exist two linearly independent solutions y1 ( x )
and y2 ( x ) of equation a2 ( x ) y a1 ( x ) y a0 ( x ) y 0 of
the form y
c
n 0
Case II:
n
x n r . This is the case of Example 2.
If r1 r2 N , where N is a positive integer, then there exist two linearly
independent solutions of equation a2 ( x ) y a1 ( x ) y a0 ( x ) y 0 of the form
y1 ( x )
c
n 0
n
x n r1 , c0 0
y2 ( x ) Cy1 ( x ) ln x bn x n r2 , b0 0,
n 0
where C is a constant that could be zero.
Note : Why y2 ( x ) is equal to Cy1 ( x ) ln x bn x n r2 and how to obtain it ?
n 0
The solution y2 ( x ) can be obtained by y 2 ( x ) y1 ( x )
41
exp( P ( x )dx )
y12 ( x )
dx .
NCCU
Wireless Comm. Lab.
From y P ( x ) y Q ( x ) y 0, we get
y1 P ( x ) y1 Q ( x ) y1 0 y2 y1 P ( x ) y1 Q ( x ) y1 0 (1)
y2 P ( x ) y2 Q ( x ) y2 0 y1 y2 P ( x ) y2 Q ( x ) y 2 0 (2)
By (2) - (1),
y1 y2 y2 y1 P ( x ) y1 y2 y2 y1 0
( y1 y2 y2 y1 ) P ( x ) y1 y2 y2 y1 0 " First order linear differential equation "
y1 y2 y2 y1 c exp( P ( x ) dx )
c exp( P ( x ) dx )
y1 y2 y2 y1
y12
y12
c exp( P ( x ) dx )
y2
d(
)
y1
y12
y2
y1
c exp( P ( x ) dx )
so y2 y1
y12
, let c 1,
exp( P ( x ) dx )
y12
. So you can calculate y 2 ( x ) by using the formula.
42
NCCU
Wireless Comm. Lab.
Case III: If r1 r2 then there always exists two linearly
independent solutions of a2 ( x) y a1 ( x) y a0 ( x) y 0 of the form
y1 ( x )
c
n 0
n
x n r1 , c0 0
y2 ( x ) Cy1 ( x ) ln x bn x n r2 ,
n 1
The solution y2 ( x ) can also be obtained by y 2 ( x ) y1 ( x )
exp( P ( x )dx )
y12 ( x )
dx .
Note : We will see how the formula of y 2 ( x ) works by the following example.
43
NCCU
Wireless Comm. Lab.
Example 5.
Find the general solution of xy y 0.
You can use y1 ( x ) of Example 4. and find y 2 ( x ) by substituting y1 ( x) into
y2 ( x ) y1 ( x )
exp( P ( x ) dx )
y12 ( x )
dx .
1 2
1 3
1
x
x
x4
2
12
144
Note : There is wrong in Example 4. for y1 ( x ).
y1 ( x ) x
y2 ( x ) y1 ( x )
y1 ( x )
exp( 0dx )
dx
2
1 2
1 3
1
x
x
x
x4
2
12
144
dx
1
7
19
1
y1 ( x ) 2
x
5 4
7 5
x
x 12
72
3
x-x
x
x
12
12
dx
7
19
1
y1 ( x ) ln x
x
x2
12
144
x
7
19
1
y2 ( x ) y1 ( x ) ln x y1
x
x2
.
144
x 12
On the interval (0, ), the general solution is y c1 y1 ( x ) c2 y2 ( x ).
44
NCCU
Wireless Comm. Lab.
Remark
When the difference of indicial roots r1 r2 is a positive integer
(r1 r2 ), it sometimes pays to iterate the recurrence relation using
the smaller root r2 first.
Since r is the root of a quadratic equation, it could be complex.
Here we do not concern this case.
If x=0 is an irregular singular point, we may not be able to find
any solution of the form y cn x n r .
n 0
45
NCCU
Wireless Comm. Lab.
6.3 Two Special Equations
The two differential equations
x 2 y xy ( x 2 2 ) y 0 (1)
and
(1- x 2 ) y - 2 xy n(n 1) y 0 (2)
occur frequently in advanced studies in applied mathematics,
physics, and engineering.
They are called Bessel’s equation and Legendre’s equation,
respectively.
In solving (1) we shall assume 0, whereas in (2) we shall
consider only the case when n is a nonnegative integer.
46
NCCU
Wireless Comm. Lab.
Solution of Bessel’s Equation
Substituting
y cn x n r
n 0
(n r )(n r 1)c
n
n 0
x
nr
into the Bessel’s equation,
( n r )cn x
n 0
nr
cn x
nr 2
n 0
2
c
n 0
n
x nr
c0 ( r 2 r r 2 ) x r c1 r ( r 1) ( r 1) 2 x r 1
k 2 r
k 0
2
r 2 ck 2 ck x k r 2 0
2
Taking r 2 2 0, ( r1 , r2 - ) and k 2 r r 2 ck 2 ck 0,
ck
we get r r1 , c1 0, and ck 2
, k 0,1, 2, .
( k 2)( k 2 2 )
If c1 0, it is easy to say ck 0, k 1, 3, 5, 7, .
Let k 2 2n, n 1, 2, 3,
, so c2 n
47
cn 2
.
22 n(n )
NCCU
Wireless Comm. Lab.
c0
2 2 1 (1 )
c0
c2
c4 2
4
2 2 (2 ) 2 1 2 (1 )(2 )
c0
c4
c6 2
6
2 4 (4 ) 2 1 2 3 (1 )(2 )(3 )
Thus c2
( 1) n c0
c2 n 2 n
, n 1, 2, 3, .
2 n !(1 )(2 ) ( n )
First we define the gamma function (1 ) ( ),
(1 1) (1 )(1 ) (1 ) ( ) (1 ) ( 1)( 1)
(1 )!
So, we multiply (1 ) at the numerator and denominator,
( 1) n c0 (1 )
c2 n 2 n
, n 1, 2, 3, .
2 n !(1 )(1 )(2 ) ( n )
From (1 )(1 )(2 ) ( n ) !(1 )(2 ) ( n ) (1 n ),
c2 n
( 1) n c0 (1 )
( 1) n c0 (1 )
2n
.
2 n ! !(1 )(2 ) ( n ) 2 2 n n !(1 n )
48
NCCU
Wireless Comm. Lab.
So, y1 x
c0 c2 x
r1
x
n 0
2
x
( 1) n c0 (1 )
x2n
2n
2 n !(1 n )
c
2n
n 0
x 2n
( 1) n c0 (1 )2 x
n !(1 n ) 2
n 0
Let a1 c0 (1 )2 , then y1 a1
n 0
( 1) n
x
n ! (1 n ) 2
2 n
2 n
a1 J ( x ).
2 n
( 1) n
x
We denoted that J ( x )
.
n 0 n ! (1 n ) 2
If we take r2 into the equation, we will get
y2 a2
n 0
( 1) n
x
n ! (1 n ) 2
2 n
a2 J ( x ).
( 1) n
x
We denoted that J ( x )
n 0 n ! (1 n ) 2
Thus y y1 ( x ) y2 ( x ) a1 J ( x ) a2 J ( x ).
2 n
Note : J ( x ) and J - ( x ) are called Bessel functions of the first kind
of the order and - , repectively.
49
NCCU
Wireless Comm. Lab.
Example 1. General Solution: Not an Integer
Put out atteneion on this differential equation
1
1
) y 0 on (0, ), we can see that .
4
2
Then the general solution is y c1 J 1 ( x ) c2 J 1 ( x ).
x 2 y xy ( x 2
2
2
First we talk about the properties of Bessel functions
of order m, m 0,1, 2,
.
(i ) J m ( x) ( 1) m J m ( x) (ii) J m ( x) ( 1) m J m ( x)
0, m 0
(iii ) J m (0)
1, m 1
(iv) lim Ym ( x)
x 0
Note : J m ( x) and J m ( x) are linear dependent , where m 0,1, 2,
50
.
NCCU
Wireless Comm. Lab.
Bessel Functions of the Second Kinds
cos J ( x ) - J - ( x)
,
sin
J ( x ) and Y ( x) can be the linearly independent solutions of
If integer, by linear mapping Y ( x)
x 2 y xy ( x 2 - 2 ) y 0, so y c1 J ( x ) c2 J ( x ) or y c1 J ( x ) c2Y ( x ).
If integer, according to L ' Hospital ' s rule that lim Y ( x) exists.
m
By linear mapping Ym ( x) lim Y ( x), Ym ( x) and J m ( x) are the linearly
m
independent solutions of x 2 y xy ( x 2 - m 2 ) y 0, so y c1 J ( x) c2Y ( x).
Note : We drop the solution of y c1 J ( x) c2 J ( x),
because J ( x) and J ( x) are linear dependent from the previous page.
So, for any value of the general solution of x 2 y xy ( x 2 2 ) y 0
on (0, ) can be modeled by y c1 J ( x ) c2Y ( x ).
Note : Y ( x ) is called the Bessel function of the second kind of order .
51
NCCU
Wireless Comm. Lab.
Following plot will show us the first and second kind of Bessel
function.
52
NCCU
Wireless Comm. Lab.
Example 2. General Solution: an Integer
Considering x 2 y xy ( x 2 - 9) y 0 on (0, ),
we see 2 9 and 3, so the general solution is
y c1 J 3 ( x ) c2Y3 ( x ).
Parametric Bessel Equation
For the parametic equation x 2 y xy ( 2 x 2 - 2 ) y 0,
the general solution y c1 J ( x) c2Y ( x) or
y c1 J ( x) c2 J ( x), where integer.
Note : We will prove it at the next page.
53
NCCU
Wireless Comm. Lab.
From x 2 y xy ( 2 x 2 - 2 ) y 0, let z= x,
dy dy dz
dy
then
, y =
,
dx dz dx
dz
2
d 2 y d dy dz
d
y
2
.
2
2
dx
dz dz dx
dz
Substituting into the DE, we get
2
d
y
dy
2 2
2 2
2
x
x
(
x
)y 0
2
dz
dz
2
dy
2 d y
2
2
z
z
(
z
)y 0
2
dz
dz
So, y c1 J ( z ) c2Y ( z ) or y c1 J ( z ) c2 J ( z ).
Then y c1 J ( x ) c2Y ( x ) or y c1 J ( x ) c2 J ( x ).
54
NCCU
Wireless Comm. Lab.
Example 4. Derivation Using the Series Definition
Derive the formula xJ ( x ) J ( x ) xJ 1 ( x ).
( 1) n
x
From J ( x )
n 0 n ! (1 n ) 2
n
(
1)
(2n ) x
xJ ( x )
n 0 n ! (1 n ) 2
( 1) n
x
n 0 n ! (1 n ) 2
2 n
2 n
,
2 n
( 1) n n
x
2
n 0 n ! (1 n ) 2
( 1) n n
x
J ( x ) x
(
n
1)!
(1
n
)
2
n 1
( 1) k ( k 1) x
J ( x ) x
k 0 k ! (2 k ) 2
55
2 n
2 n 1
2 k 1
, let k n - 1
J ( x ) xJ 1 ( x ).
NCCU
Wireless Comm. Lab.
From the example 4. xJ ( x ) J ( x ) xJ 1 ( x ), we divide
x at both side, J ( x )
J ( x )
x
x
J ( x ) J 1 ( x ).
J ( x ) J 1 ( x ), multiply x - at both side.
x - J ( x ) x - 1 J ( x ) x - J 1 ( x )
d
x - J ( x )
x - J 1 ( x ).
dx
So, J 0 ( x ) J 1 ( x ) and Y0 Y1 ( x ).
Spherical Bessel Functions
When the order is half an odd integer, that is,
1
3
5
, , , , the Bessel function of first kind J ( x )
2
2
2
can be expressed in terms of the elementary functions sin x,
cos x, and powers of x. Such Bessel functions are called
Spherical Bessel functions.
56
NCCU
Wireless Comm. Lab.
Example 5. Spherical Bessel Function with
1
2
.
Find an alternative expression for J 1 ( x ).
2
( 1) n
x
1
n ! (1 n
) 2
2
1
With
, we get J 1 ( x )
2
2
By (1 ) ( ) and (
1
)
2
1
)
2
3
n 1 : (1
)
2
n 0 : (1
n n : (1 n
n 0
1
2
3 1
3!
3
2 2
2
1
(2 n - 1)!
)
2
2 2 n 1 n !
.
, we find
.
2n
( 1) n
x
So, J 1 ( x )
(2 n - 1)!
2
n 0 n !
2
2 n 1
2
n!
1
1
From sin x x x3
x5
3!
5!
2
1
2
so we get J 1 ( x )
2n
1
2
n 0
2
x
n 0
( 1) n
x 2 n 1.
(2 n 1)!
( 1) n
x 2 n 1 ,
(2 n 1)!
2
sin x.
x
57
NCCU
Wireless Comm. Lab.
Now we bring out an extra concept that is not mentioned in
textbook, that is, the Modified Bessel Equation.
x 2 y xy ( x 2 u 2 ) y 0 is called the Modified Bessel Equation.
It changes from x 2 y xy (i 2 x 2 u 2 ) y 0. We get the general
solution y c1 J u (ix ) c2Yu (ix ), where u 0,1, 2, 3,
y c1 J u (ix ) c2 J u (ix ), where u 0,1, 2,
( 1) n
ix
And J u (ix )
n 0 n ! (1 n u ) 2
( 1) n i 2 n
x
i
n 0 n ! (1 n u ) 2
u
2 n u
, or
.
( 1) n i 2 ni u x
n 0 n ! (1 n u ) 2
2 n u
2 n u
i u I u ( x ).
2 n u
( 1) n i 2 n
x
Where I u ( x )
is called the
n
!
(1
n
u
)
2
n 0
" Modified Bessel function of 1st kind ".
Following the similar approach, you can get J - u (ix ) i u I u ( x ).
58
NCCU
Wireless Comm. Lab.
From the equation x 2 y xy ( x 2 u 2 ) y 0, the solution is
y c1 J u (ix ) c2 J u (ix ), where u m, m 0,1, 2,
.
We modify it to y A1 I u ( x) A2 I u ( x).
y c1 J u (ix ) c2Yu (ix ), where u m, m 0,1, 2,
.
We modify it to y A1 I m ( x ) A2 K m ( x),
I u ( x) I u ( x)
where K u ( x)
, K m ( x) limK u ( x).
2
sin u
u m
K u ( x) is called the " Modified Bessel function of 2nd kind ".
59
NCCU
Wireless Comm. Lab.
Example.
Solve x 2 y xy ( 4 x 2 6) y 0.
x 2 y xy [(2i ) 2 x 2 (
6 )2 ] y 0
So, the general solution
y c1 I
6
(2 x ) c2 K
6
(2 x ).
Solution of Legendre’s Equation
Since x 0 is the ordinary point of the equation (1- x 2 ) y - 2 xy ( 1) y 0,
let y cn x n substitute into the equation and combine the summation,
n 0
(1- x 2 ) y - 2 xy ( 1) y ( 1)c0 2c2 ( 1)( 2)c1 6c3 x
(n 2)(n 1)c
n 2
n 2
( n )( n 1)cn x n 0.
60
NCCU
Wireless Comm. Lab.
From the previous equation , ( 1)c0 2c2 0
( 1)( 2)c1 6c3 0
( n 2)( n 1) cn 2 ( n )( n 1) cn 0
or c2
( 1)c0
2!
( 1)( 2)c1
c3
3!
( n )( n 1)cn
cn 2
, n 2, 3, 4,
( n 2)( n 1)
Take n 2, 3, 4,
into cn 2 , we get
( 2)( 3)c2
43
( 3)( 4) c3
c5
5 4
( 4)( 5)c4
c6
65
( 5)( 6)c5
c7
76
c4
.
( 2) ( 1)( 3) c0
4!
( 3)( 1)( 2)( 4) c1
5!
( 4)( 2) ( 1)( 3)( 5) c0
6!
( 5)( 3)( 1)( 2)( 4)( 6) c1
7!
61
NCCU
Wireless Comm. Lab.
( 1) 2 ( 2) ( 1)( 3) 4
y c0 1
x
x
2!
4!
( 1)( 2) 3 ( 3)( 1)( 2)( 4)
c1 x
x
3!
5!
c0U ( x ) c1V ( x), where
x5
( 2) ( 1)( 3) 4
x
2!
4!
( 1)( 2) 3 ( 3)( 1)( 2)( 4) 5
V ( x ) x
x
x
3!
5!
U ( x ) 1
( 1)
x2
(i ) If n, n 0,1, 2, 3,
.
" Note : Here n means a nonnegative integer."
U ( x ) and V ( x) will diverge at x 1, it violate the physic phenomenon.
So, n, n 0,1, 2, 3,
, the original equation becomes
(1 - x 2 ) y - 2 xy n(n 1) y 0, n 0,1, 2, 3,
, the solution is
y c0U n ( x) c1Vn ( x)
62
NCCU
Wireless Comm. Lab.
(ii )When n 0, 2, 4,
, U n ( x) has a finite terms, but Vn ( x) has an infinite terms.
n 1, 3, 5,
, U n ( x) has an infinite terms, but Vn ( x ) has a finite terms.
So, we recombine the U n ( x ) and Vn ( x ), put the same term together.
Note : Separate the infinite terms and the finite terms.
U n ( x )
U (1) , n 0, 2, 4,
n
Define Pn ( x)
,
Vn ( x) , n 1,3,5,
Vn (1)
where Pn ( x) is the "first kind of n - th order Legendre ' s equation ".
Vn ( x) U n (1), n 0, 2, 4,
Qn ( x)
U n ( x)Vn (1), n 1,3,5,
,
where Qn ( x ) is the "second kind of n - th order Legendre ' s equation ".
63
NCCU
Wireless Comm. Lab.
1
1
(3 x 2 1), and P3 ( x ) (5 x 3 3 x ).
2
2
1
1
1
1 x
Q0 ( x ) x x 3 x 5 ln(
)
3
5
2
1 x
1
1
x
1 x
Q1 ( x ) x ( x x 3 x 5 ) ln(
) 1
3
5
2
1 x
3x 2 1
1 3 1 5
3
3x 2 1 1 x
3
Q2 ( x )
(x x x ) x
ln(
) x.
2
3
5
2
4
1 x
2
We get P0 ( x ) 1, P1 ( x ) x, P2 ( x )
The general solution can be expressed by y APn ( x ) BQn ( x ).
x 1, Qn ( x ) will diverge, B 0.
odd function, n 1, 3,5,
So, the solution is y APn ( x ), where Pn ( x )
even function, n 0, 2, 4,
.
We will show the plot of Pn ( x ) at next page.
64
NCCU
Wireless Comm. Lab.
Example from P305. Q19
For (1 - x 2 ) y - 2 xy 2 y 0, x 1.
We get n 1, so y AP1 ( x ) BQ1 ( x ), where
P1 ( x ) x and Q1 ( x )
x
1 x
ln(
) 1.
2
1 x
Properties of Pn ( x )
(i ) Pn ( x ) ( 1) n Pn ( x )
(ii ) Pn (1) 1
(iii ) Pn ( 1) ( 1) n
(iv ) Pn (0) 0, n 0dd
( v ) Pn (0) 0, n even.
65
NCCU
Wireless Comm. Lab.
Conclusion
Here we point out two type of these special function, it will be
categorized to represent with a special solution, that we will
find out in the mathematics handbook.
66
NCCU
Wireless Comm. Lab.