chap6 series bessel legendre
TRANSCRIPT
NCCU Wireless Comm. Lab.1
Chapter 6Series Solutions of Linear Equations
NCCU Wireless Comm. Lab.2
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.
NCCU Wireless Comm. Lab.3
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 it does not possess elementary solutions. But we can find two linear independent solutions of
by using the series expansion.
0,y xy
0y xy
NCCU Wireless Comm. Lab.4
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.
NCCU Wireless Comm. Lab.5
6.1.1 Review of Power Series Definition
A power series in is an infinite series of the form
Such a series is also said to be a power series centered at a.
For example, the power series is centered at a =1.
ConvergenceA power series is convergent at a specified value of x if its
sequence of partial sums converges- that is,
If the limit does not exist at x, the series is said to be divergent.
x a2
0 1 20
( ) ( ) ( )nn
n
c c x a c x a c x a
0
( 1)nn
n
c x
0
( )nn
n
c x a
( )NS x
0
( ) constant.lim limN
nN n
N N n
S x c x a
NCCU Wireless Comm. Lab.6
Interval of ConvergenceEvery 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
Radius of ConvergenceAs 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.
,
.
x x the region of R where R will be defined immediately
and R is the radius of convergence
NCCU Wireless Comm. Lab.7
If then a power series converges for and diverges for
For example, if the series converges for x=a or for all x, then R is equal to 0 or
Recall that is equivalent to
Note: A power series may or may not converge at the endpoints a-R or a+R of this interval.
Absolute ConvergenceWithin its interval of convergence a power series converges absolutely.
.
x a R .a R x a R
0,R 0
( )nn
n
c x a
x a R .x a R
0
( ) , .nn
n
c x a converges where a R x a R
NCCU Wireless Comm. Lab.8
Ratio testConvergence of a power series can often be determined by
the ratio test.
Suppose that
If L<1 the series converges absolutely, if L>1 the series diverges, and if L=1 the test is inconclusive.
0
( )nn
n
c x a
11 1
0 ,
( ) .( )lim lim
n
nn n
nn nn n
c n and that
c x a cx a Lc x a c
NCCU Wireless Comm. Lab.9
Example: A power series the ratio test gives
The series converges absolutely for (L<1), we get
The series diverges for L>1, that is
Test for the convergence of the boundary for x=1 or 9.
0
( 5) ,4 ( 1)
n
nn
xn
1
1
( 5)( 1) 14 ( 2) 5 5 ( );
( 5) 4( 2) 44 ( 1)
lim lim
n
n
nn n
n
xnn x x L
x nn
1 5 14
x 1 9.x
9 1.x or x
0 0
1 1
( 4) ( 1) 1, .4 ( 1) ( 1)
(4) 1 9, .4 ( 1) ( 1) int [1,9).
n n
nn n
n
nn n
for x it will convergen n
for x it will divergen n
So the erval of convergence of the series is
NCCU Wireless Comm. Lab.10
A Power Series Defines a FunctionA power series defines a function whose domain is the 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 Thus, and can be found by term-by-term
differentiation and integration.
If is a power series in x, then the first two derivatives are
It will be useful to substitute into the differential equation.
0
( ) ( )nn
n
f x c x a
( , ).a R a R ( )f x ( )f x dx
0
nn
n
y c x
2
0 1 20
1 11 2
1
2 22 3
2
2
2 3 2 ( 1) ( 1) .
n nn n
n
n nn n
n
n nn n
n
Since y c c x c x c x c x
So y c c x nc x nc x
y c c x n n c x n n c x
, , y y and y 2nd
NCCU Wireless Comm. Lab.11
Identity Property
Analytic at a PointA 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,
0
( ) 0, 0 interval ,
0 .
nn
n
n
If c x a R for all numbers x in the of convergence
then c n
2
0
2 2 2
0
3 2 2 1 2 2 1
0
1 2! ! !
( 1)cos 1 ( 1)2! 2 ! 2 !
( 1) ( 1)sin . 3! (2 1)! (2 1)!
n nx
n
n n nn
n
n n n n
n
x x xe xn n
x x xxn n
x x xx x for xn n
These Taylor series centere
0, ,
, cos , sin 0.x
d at called Maclaurin series
show that e x and x are analytic at x
, , , integrable interval .
For analytic the function is continuous differentialand in the of convergence
NCCU Wireless Comm. Lab.12
Arithmetic of Power SeriesPower series can be combined through the operations of addition,
multiplication, and division.
Example:
2 2 1 2
0 0
3 2 2 1 2 2
3 5 7
Using expansion sin cos
( 1)sin cos(2 1)! 2 !
( 1)( ) (1 )3! (2 1)! 2! 2 !
, .3 20 252
n n m
n m
n n n
the series to describe the x x
x xx xn m
x x x xxn n
x x xx x x
It is obvious that
sin cos interval.the power series x x converges on the same
NCCU Wireless Comm. Lab.13
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 2 1
2 0
0
( 1)
( point 0,
.)
,
n nn n
n n
nn
n
Write n n c x c x
If y c x is a Taylor series at centered the subject
is liking to combine the y xy
From the original subject
1 1
0 1
3 0
start from x start from x
start from x start from x with n with n
2 1 0 2 12
2 0 3 0
( 1) 2 ( 1)
n n n nn n n n
n n n n
n n c x c x c x n n c x c x
NCCU Wireless Comm. Lab.14
2 1
3 0
( 1)
- 3
( 1)
n nn n
n n
nn
Considering the term of n n c x c x
let k n let k n
n n c x
2 1 1 13
3 0 0 0
13
0
2 1 12 3
2 0 0
( 3)( 2)
[( 3)( 2) ]
( 1) 2 [( 3)( 2) ]
: ,
n k kn k k
n n k k
kk k
k
n n kn n k k
n n k
c x k k c x c x
k k c c x
n n c x c x c k k c c x
Note For the shifting summation approachSt
1. minimum , . 2. a
ep You may figure out the order of each component then take the termsout from the summation
Step Substitute new index to the old one for preparing to combin . 3. .
e the summationStep Create a new summation for the original subject
NCCU Wireless Comm. Lab.15
6.1.2 Power Series Solutions Definition 6.1 Ordinary and Singular Points
A point is said to be an ordinary point of the differential equation if both P(x) and Q(x) in the standard form are analytic at A point that is not an ordinary
point is said to be a singular point of the equation.
Considering the differential equation and
–
0x2 1 0( ) ( ) ( ) 0,a x y a x y a x y
( ) ( ) 0y P x y Q x y 0.x
1 02
2 2
( ) ( )For analytic, it means that ( ) 0 in 0.( ) ( )
a x a xa x y y ya x a x
cos 0xy x y e y ln 0.xy x y e y
cos 0 ln 0
cos 0 ln 0
( ) cos , ( )
x x
x x
x
y x y e y y x y e y
For y x y e y For y x y e y
P x x Q x e
( ) ln , ( )0 point, 0 singular point,
( ) ( ) . ( )
xP x x Q x ex is the ordinary x is abecause P x and Q x are analytic because P x is not analyti
.c
NCCU Wireless Comm. Lab.16
Polynomial Coefficients
2 1 0
, points denominator .
if a ( ), a ( ), ,and a ( ) are polynomials with no common factors,
then
A polynomial is analytic at any value x and a rational functionis analytic except at where its is zeroThus x x x
01
2 2
2
0 2 1 0 2 0
0
a ( )a ( ) both rational function P(x)= and Q(x)= are analytic except ( ) ( )
where a ( ) 0.
point ( ) ( ) ( ) 0, ( ) 0, singular point
xxa x a x
x
For x x is an ordinary of a x y a x y a x y if a xwhereas x x is a
2 1 0 2 0 ( ) ( ) ( ) 0, ( ) 0.of a x y a x y a x y if a x
NCCU Wireless Comm. Lab.17
2 2
2
: ( 1)( 3) ( 1) ( 1)( 3) 0 singular points.
( 1) ( 1)(
Example If the equation is x x y x y x x yConsider for the ordinary and
xThe original equation can be rewritten to yx x
2 2 2
2 22 2
( 1)( 3) 0.3) ( 1)( 3)
1. ( ) ( 1)( 3) , ( ) 0 point , 1, -3.
2.
x xy yx x
For a x x x let a x to calculate the possible that make the functionnot analytic so x
P
2 2
2 2
( 1) 1( )( 1)( 3) ( 1)( 3)( 1)( 3) 1 ( )
( 1)( 3) 1 3. 1 2. 1, -3 singular point .
: 1,-3
xxx x x xx xQ x
x x xCombining andSo x are the for this function
Note x
singular point 6.2.
are also called the regularin Definition
NCCU Wireless Comm. Lab.18
Theorem 6.1 Existence of Power Series Solutions If is an ordinary point of the differential equation we can always find two linearly independent
solutions in the form of a power series centered at -that is, A series solution converges at least on some interval defined by whereas R is the distance from to the closest singular point.
A solution of the form is said to be a solution about the ordinary point
0x x
2 1 0( ) ( ) ( ) 0,a x y a x y a x y
0x
00
( ) .nn
n
y c x x
0 ,x x R 0x
00
( )nn
n
y c x x
0.x
NCCU Wireless Comm. Lab.19
Example 2. Power Series SolutionsSolve
0.y xy
-2
0 2
singular points, 6.1 0, .
( -1) into
n nn n
n n
There are no finite Theorem guaranteestwo power series solutions centered at convergent for x
Substituting y c x and y c n n x the
differential equatio
-2 1 0 -2 12
2 0 3 0
0 1 12 3
0 0
2
-3 sec
.
( -1) 2 ( -1)
2 ( 3)( 2)
2 ( 3)( 2)
n n n nn n n n
n n n n
k kk k
k k
let k nfor the ond term
and let k nfor the third term
n
y xy c n n x c x c x c n n x c x
c x c k k x c x
c k k
13
00k
k kk
c c x
NCCU Wireless Comm. Lab.20
12 3
0
2 3
2 3
, 2 ( 3)( 2) 0.
2 0 ( 3)( 2) 0, 0,1,2, .
0 , 0,1,2 .( 3)( 2)
into
kk k
k
k k
kk
x x satisfies c k k c c x
We should let c and k k c c where kcSo we get c and c k
k kSubstitute the value of k the recur
0 03
1 14
25
3 0 06
4 1 17
5 28
,
0, 2 3 6
1, 4 3 12
02, 05 4 20
13, 6 5 30 6 180
14, 7 6 42 12 504
15, 08 7 56 20
sive termc ck c
c ck c
ck c
c c ck c
c c ck c
c ck c
NCCU Wireless Comm. Lab.21
13
0
3 6 4 7
0 1
0
3 6
0
( 3)( 2)
16 180 12 504
into .
, 16 180
kk k
k
nn
n
So we reconstruct the term k k c c x
x x x xto be the form of c c x
We substitue it the origin series y c x
x xSo we get y c
4 7
1
0 1 1 2
3 21
1
3 12
1
.12 504
( ) ( ),
( 1) ( ) 12 3 (3 1)3
( 1) ( )3 4 (3 )(3 1)
nn
n
nn
n
x xc x
Another description of y c y x c y x
we get y x xn n
y x x xn n
NCCU Wireless Comm. Lab.22
Example 3. Power Series SolutionSolve 2( 1) 0.x y xy y
singular points , 0 1, 1 distance
.
There shows at x i and so the power series solutioncentered at will converge at least for x where is the
in the complex plane from the origin to x i
Substituing y c
1
10
-2
2
, ,
( -1) into .
nn
n
nn
n
nn
n
y nc x andx
y c n n x the differential equation
NCCU Wireless Comm. Lab.23
2
1
0 12 0 3 1 1
2
-2
2 2 0
-2
2 4 2
( 1)
(2 ) (6 )
( -1) ( -1)
( -1) ( -1)
nn
n
nn
n
le
let k nfor another summations
n n nn n n
n n n
n n nn n n
n n n
x y xy y nc x
c c x c c c x nc x
c n n x c n n x c x
c n n x c n n x c x
-2
12 0 3
2
12 0 3 2
22 2 2
2
2 6 ( 2)
2 6 ( -1) 1 ( 2)( 1) 0
0 , 1.
( -1) ( 1)t k n
for the fourth termk
kk
k k
k k kk k k
k k k
k
k
c c c x k k kc x
c c c x k k k c k k c
For the equality of the equation to x x
We have
c k x c k x c x
x
2 0 3 2
0 2 3 2
2 0, 6 0, ( -1) 1 ( 2)( 1) 0, k=2,3,4, .
( 1)( 1) -12 , 0, - - , 2.3.4. .( 2)( 1) 2
k k
k k k
c c c and k k k c k k c
k k kc c c c c c kk k k
NCCU Wireless Comm. Lab.24
4 2 0 0
5 3
6 4 0 0
7
2 40 1 0 1
0
, 1 1 1 12, - - -4 4 2 823, - 053 1 1 14, - - 6 2 8 16
5, 0
1 1 (1 ) ( )2 8
nn
n
By iteration of k we get
k c c c c
k c c
k c c c c
k c
So we rewritten the power series y c x to y c x x c x c y x
1 2
2 1 21
2
2
( ).
1 1 3 5 (2 3) ( ) 1 ( 1)2 2 !
( ) , 1.
n nn
n
c y x
nThus y x x xn
y x x where x
NCCU Wireless Comm. Lab.25
Example 4. Three-Term Recurrence Relation If we seek a power series solution for the differential equation
0
nn
n
y c x
(1 ) 0.y x y
2 1
2 0 0
0 2 12 0
3 1 0
(1 ) ( 1)
(2 ) ( 1)
n n nn n n
n n n
n n nn n n
n n n
y x y c n n x c x c x
c c x c n n x c x c x
Let k
2 0 2 11
12 0 2
- 2 1
2 ( 2)( 1) 0
, .1 1.2.3. .2 ( 2)( 1)
kk k k
k
k kk
n k n k n
c c c k k c c x
For the equality x x
c cWe get c c and c kk k
NCCU Wireless Comm. Lab.26
2 0
1 0 013
0 12 1 1
4 0
3 2 015
0
2 30
into 1 ,2
1, 3 2 6 6
1122,
4 3 12 12 24
3, 5 4 120 30
1 1 1(12 6 24
nn
n
Iterating the value k the recurrence relation
with the condition c c
c c cck c
c cc c ck c c
c c cck c
y c x
c x x
4 5 3 4 51
1 1 1 1) ( ).30 6 12 120
x x c x x x x
NCCU Wireless Comm. Lab.27
Nonpolynomial CoefficientsThe next example illustrates how to find a power series solution about the
ordinary point of a differential equation when its coefficients are not polynomials.
Example 5. ODE with Nonpolynomial CoefficientsSolve
0 0x
(cos ) 0.y x y
2
0 2
2
0
0 point , cos 0.
, ( -1) ,
( 1)cos into .2 !
n nn n
n n
n n
n
We see that x is an ordinary of the equationbecause we know x is analytic at x
Substituting y c x y n n c x and
xx the equationn
NCCU Wireless Comm. Lab.28
22
2 0 0
2 3 2 4 6 22 3 4 5 0 1 2
2 32 0 3 1 4 2 0 5 3 1
( 1)(cos ) ( -1)2 !1 1 12 6 12 20 (1 )( )2 24 720
1 12 (6 ) (12 ) (20 ) 0.2 2
, .
k kn n
n nn k n
xy x y n n c x c xk
c c x c x c x x x x c c x c x
c c c c x c c c x c c c x
For the equality x x
2 0 3 1 4 2 0 5 3 1
2 0 3 1 4 0 5 1
2 4 3 50 1
0
1 1 2 0, 6 0, 12 0, 20 0, .2 2
1 1 1 1 , , , , .2 6 12 30
1 1 1 1(1 ) ( ).2 12 6 30
nn
n
We get c c c c c c c c c c
So c c c c c c c c
y c x c x x c x x x
NCCU Wireless Comm. Lab.29
Solution CurvesThe approximate graph of a power series solution can be 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 polynomials
For a large value of N, By this, will give us some information about the behavior of y(x) near the ordinary point.
0
nn
n
y c x
0
.N
nN n
n
S x c x
arg arg 0
lim lim .N
nN nN is l e N is l e n
S x c x y
0
Nn
N nn
S x c x
NCCU Wireless Comm. Lab.30
RemarksEven though we can generate as many terms as desired in series solution 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 We may have to settle, as we did in Example
4 and 5, for just writing out the first few terms of the series.
0
nn
n
y c x
.nc
NCCU Wireless Comm. Lab.31
6.2 Solutions About Singular Points The two differential equations are similar
only in that they are both examples of simple linear second-order equations with variable coefficients.
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.
20 and x 0y xy y y
NCCU Wireless Comm. Lab.32
Regular and Irregular Singular Points
Definition 6.2 Regular and Irregular Singular Points
0
2 1 0
singular point ( ) ( ) ( ) 0
.
A at x x of a linear differential equationa x y a x y a x y
is further classified as either regular or irregular The classification againdepends on the
1 0
2 2
standard ( ) ( ) 0.
( ) ( ): ( ) ( ) .( ) ( )
functions P and Q in the formy P x y Q x y
a x a xNote Where P x and Q xa x a x
0
22 1 0 0 0
0
singular point singular point
( ) ( ) ( ) 0, ( ) ( ) ( - ) ( ) ( ) ( - ) . sing
A x is said to be a regular of the differential equation
a x y a x y a x y if the function p x P x x x and q x Q x x xare both analytic at x A
ular point
singular point .that is not regular is said to be an
irregular of the equation
NCCU Wireless Comm. Lab.33
Example 1. Classification of Singular Points
0
20
2 2 2 20 0 0 0 0
singular point ( ) ( ) 0,
( - ) ,
( - ) ( - ) ( ) ( - ) ( ) 0 ( - ) ( - ) ( ) ( ) 0.
If x x is a regular of the differential equation y P x y Q x
then multiplying x x in both side
x x y x x P x y x x Q x y x x y x x p x y q x y
For reg
2
0 0
0
20 0 0
singular point ( ) ( ) ( - ) ( ) ( ) ( - ) .
( - ) ( - ) ( ) ( ) 0, point .
ular p x P x x x and q x Q x x xare both analytic at x
So x x y x x p x y q x y where p and q are analytic at the x x
2 2
2
2 -2 singular points
( - 4) 3( 2) 5 0. ( ) ( ) 0,
3( 2) ( )( - 4
It should be clear that x and x are of
x y x y yComparing with the general form of y P x y Q x y
xit is clearly that P xx
2 2 2 2
20 0
20 0
3 5 ( ) .) ( 2) ( 2) ( - 4)
( - ) ( - ) ( ) ( ) 0,
( ) ( ) ( - ) ( ) ( ) ( - ) .
and Q xx x x
By the equivalent form x x y x x p x y q x y
where p x P x x x and q x Q x x x
NCCU Wireless Comm. Lab.34
22 2
2 -2 singular point.3 5 2, ( ) ( - 2) ( ) ( ) ( - 2) ( ) .
( 2) ( 2) ( ) ( ) 2, 2 singular po
Checking x and x is the regular or irregular
First if x p x x P x and q x x Q xx x
From p x and q x are analytic at x so x is a regular
22
int.
3 5 -2, ( ) ( 2) ( ) ( ) ( 2) ( ) .( 2)( 2) ( 2)
( ) -2, ( ) -2. , -2 singular poi
Second if x p x x P x and q x x Q xx x x
Although q x is analytic at x p x is not analytic at xCombining these conditions x is an irregular
nt.
NCCU Wireless Comm. Lab.35
Theorem 6.2 Frobenius’ Theorem If is a regular singular point of the differential equation then there exists at least one solution of the form
where the number r is a constant to be determined. The series will converge at least on some interval
If we consider a differential equation that has a regular singular point, then we can substitute into the DE like the
approach
we did before by using the power series to solve with the ordinary point.
0x x
2 1 0( ) ( ) ( ) 0,a x y a x y a x y
0 0 00 0
( ) ( ) ( ) ,r n n rn n
n n
y x x c x x c x x
00 .x x R
00
( )n rn
n
y c x x
NCCU Wireless Comm. Lab.36
Example 2. Two series solutions
0
10 1
10 1
0 singular point 3 - 0,
.
,
(
n rn
n
r r n rn
r
Because x is a regular of the differential equationxy y y
we try to find a solution of the form y c x
From y c x c x c x
y c rx c
1 1
0
2 1 2 20 1
0
1 ) ( ) ( )
( 1) (1 ) ( )( 1) ( )( 1) ,
, , into .
3 - 3 ( )( 1
r n r n rn n
n
r r n r n rn n
n
n
r x c n r x n r c x
y c r r x c r rx c n r n r x c n r n r x
now we substitute y y and y the equation
xy y y c n r n r
1 1
0 0 0
1 1 10
1 1 0
-1 2 3
10
4
) ( )
3 ( 1) 3 ( )( 1) ( )
(3 - 2)nd rd
th
n r n r n rn n
n n n
r n r n r n rn n n
n n n
let k nfor and term
r
let k nfor term
x n r c x c x
r r r c x c n r n r x n r c x c x
r r c x
10
( 1)(3 3 1) 0k rk k
k
k r k r c c x
NCCU Wireless Comm. Lab.37
0 1
0
(3 - 2) 0 ( 1)(3 3 1) 0, 0,1, 2, . 0, , ' .
, (3 - 2) 0
k kThis implies r r c and k r k r c c kIf we let c we will see the all terms in recursive term that will equal to zero it s a trivial solution
So we choose r r a
1
1 1
2 1
, 0,1, 2, .( 1)(3 3 1)
0, , 0,1, 2, (1) ( 1)(3 1)
2 , , 0,1,2, . (2)3 ( 1)(3 5)
(1)
kk
kk
kk
cnd c kk r k r
cIf r c kk k
cr c kk k
From we get
01 0 1
0 02 2
03
(2)
5
8 80
168
From we getcc c c
c cc c
cc
03
0 0
2640
! 1 4 (3 2) ! 5n n
cc
c cc cn n n
.
8 11 (3 2)n
NCCU Wireless Comm. Lab.38
0
01
1
23
21
, . - .
1( ) 1! 1 4 (3 2)
1( ) 1! 5 8 11 (3 2)
n
n
n
n
From each set contains the same coefficient c so we omit the termSo we find two independent solutions on the entired x axis
y x x xn n
y x x xn n
1 2
.
( ) ( )
By the ratio test it can be demonstrated that both y x and y x convergefor all finite values of x
1 1 2 2
, .
, superposition , ( ) ( ) . interval ,
that is x
Hence by the principle y c y x c y x is another solutionof this differential equation On any not containing the origin this linearcombinition represents t
.he general solution of the differential equation
NCCU Wireless Comm. Lab.39
Indicial Equation Equation is called the indicial equation of the previous example,
and the value are called the indicial roots, or exponents, of the singularity x=0.
In general, after substituting into the given 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.
We solve for the two values of r and substitute these value into a recurrence relation such as
By Theorem 6.2, there is at least one solution of the assumed series form that can be found.
(3 2) 0r r 1 2
20 3
r and r
0
n rn
n
y c x
1 .( 1)(3 3 1)
kk
cck r k r
NCCU Wireless Comm. Lab.40
Example for Indicial Equation
2
2 2
2
0
( ) ( ) 0, ,
( ) ( ) 0.
, ( ) ( )
( ) ( ) .
, ( )
From y P x y Q x y if we multiply x on both side
we get x y x xP x y x Q x y
As the previous concept we know that the p x xP x and
q x x Q x are both analytic
So we let p x a a
2 21 2 0 1 2
0
2 20 1 2 0 1 2
0 0 0
( ) .
, ( ), ( ) into ,
( )( 1) ( ) 0.
n rn
n
n r n r n rn n n
n n n
x a x and q x b b x b x
Substituting y c x p x and q x the differential equation
n r n r c x n r a a x a x c x b b x b x c x
As w
0 0
, , 0, ( -1) 0.
re find the indicial equation we will take from the lowest order of n like xso we let n we get the indicial equation r r ra b
NCCU Wireless Comm. Lab.41
Three Cases Case I:
Case II:
1 2
1
2 2 1 0
0
integer, ( ) ( ) ( ) ( ) ( ) 0
. n rn
n
If r and r are ristinct and do not differ by anthen there exist two linearly independent solutions y xand y x of equation a x y a x y a x y of
the form y c x This is
2.the case of Example
1
1 2
2 1 0
1 00
, integer, ( ) ( ) ( ) 0
( ) , 0
n rn
n
If r r N where N is a positive then there exist two linearlyindependent solutions of equation a x y a x y a x y of the form
y x c x c
2
2
2 1 00
2 10
2
( ) ( ) ln , 0,
constant .
: ( ) ( ) ln ?
( )
n rn
n
n rn
n
y x Cy x x b x b
where C is a that could be zero
Note Why y x is equal to Cy x x b x and how to obtain it
The solution y x can be obtained by y
2 1 21
exp( ( ) )( ) ( ) dx .
( )
P x dxx y x
y x
NCCU Wireless Comm. Lab.42
1 1 1 2 1 1 1
2 2 2 1 2 2 2
1 2 2 1 1 2 2 1
1 2 2 1
( ) ( ) 0,
( ) ( ) 0 ( ) ( ) 0 (1)
( ) ( ) 0 ( ) ( ) 0 (2)
(2) - (1),
( ) 0
( ) (
From y P x y Q x y we get
y P x y Q x y y y P x y Q x y
y P x y Q x y y y P x y Q x y
By
y y y y P x y y y y
y y y y P
1 2 2 1
1 2 2 1
1 2 2 12 21 1
22
1 1
22
1 1
2 1 21
) 0 " "
exp( ( ) )
exp( ( ) )
exp( ( ) )( )
exp( ( ) ) , 1,
exp( ( ) )
x y y y y First order linear differential equation
y y y y c P x dx
c P x dxy y y yy y
c P x dxydy y
c P x dxy let cy y
P x dxso y y
y
2 . ( ) using .So you can calculate y x by the formula
NCCU Wireless Comm. Lab.43
Case III: If then there always exists two linearly independent solutions of of the form
1 2r r
2 1 0( ) ( ) ( ) 0a x y a x y a x y
1
2
1 00
2 11
2 2 1 21
2
( ) , 0
( ) ( ) ln ,
exp( ( ) ) ( ) ( ) ( ) dx .
( )
: ( )
n rn
n
n rn
n
y x c x c
y x Cy x x b x
P x dxThe solution y x can also be obtained by y x y x
y x
Note We will see how the formula of y x works by the following exa
.mple
NCCU Wireless Comm. Lab.44
Example 5. Find the general solution of 0.xy y
1 2 1
2 1 21
2 3 41
1
2 1
( ) 4. ( ) ( ) into
exp( ( ) )( ) ( ) .
( )1 1 1( )2 12 144
: 4. ( ).
exp( 0( ) ( )
You can use y x of Example and find y x by substituting y x
P x dxy x y x dx
y x
y x x x x x
Note There is wrong in Example for y x
dy x y x
22 3 4
1 1 23 4 5
21
22 1 1
)
1 1 12 12 144
1 1 7 19( ) ( )5 7 12 72-
12 121 7 19( ) ln
12 1441 7 19( ) ( ) ln .
12 144 interval (0, )
xdx
x x x x
dxy x y x x dxx xx x x x
y x x x xx
y x y x x y x xx
On the
1 1 2 2, ( ) ( ).the general solution is y c y x c y x
NCCU Wireless Comm. Lab.45
Remark When the difference of indicial roots is a positive integer it sometimes pays to iterate the recurrence relation using
the smaller root 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
1 2r r
1 2( ),r r
2r
0
.n rn
n
y c x
NCCU Wireless Comm. Lab.46
6.3 Two Special Equations The two differential equations
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 whereas in (2) we shall consider only the case when n is a nonnegative integer.
2 2 2
2
( ) 0 (1)
(1- ) - 2 ( 1) 0 (2)
x y xy x y
and x y xy n n y
0,
NCCU Wireless Comm. Lab.47
Solution of Bessel’s Equation Substituting into the Bessel’s equation,
0
n rn
n
y c x
2 2
0 0 0 0
2 2 2 10 1
2 2 22
0
22 2 21 2 2
( )( 1) ( )
( ) ( 1) ( 1)
2 0
0,( , - ) 2 0
n r n r n r n rn n n n
n n n n
r r
k rk k
k
k k
n r n r c x n r c x c x c x
c r r r x c r r r x
k r r c c x
Taking r r r and k r r c c
1 1 2
1
22 2
,
, 0, , 0,1,2, .( 2)( 2 2 )
0, 0, 1,3,5,7, .
2 2 , 1,2,3, , .2 ( )
kk
k
nn
cwe get r r c and c kk k
If c it is easy to say c kcLet k n n so c
n n
NCCU Wireless Comm. Lab.48
02 2
024 2 4
046 2 6
02 2
2 1 (1 )
2 2 (2 ) 2 1 2 (1 )(2 )
2 4 (4 ) 2 1 2 3 (1 )(2 )(3 )
( 1) , 1, 2,3, .2 !(1 )(2 ) ( )
n
n n
cThus c
ccc
ccc
cc nn n
First we defin
02 2
(1 ) ( ),(1 1) (1 ) (1 ) (1 ) ( ) (1 ) ( 1) ( 1)
(1 )!, (1 ) denominator,
( 1) (1 ) , 2 ! (1 )(1 )(2 ) ( )
n
n n
e the gamma function
So we multiply at the numerator and
cc nn n
0 02 2 2
1, 2,3, .
(1 )(1 )(2 ) ( ) !(1 )(2 ) ( ) (1 ),
( 1) (1 ) ( 1) (1 ) .2 ! !(1 )(2 ) ( ) 2 ! (1 )
n n
n n n
From n n n
c ccn n n n
NCCU Wireless Comm. Lab.49
1 2 21 0 2 2
0
220 0
20 0
2
1 0 1 1 10
,
( 1) (1 ) ( 1) (1 )22 ! (1 ) ! (1 ) 2
( 1) (1 )2 , ( ).! (1 ) 2
r nn
n
nn nn
nn n
nn
n
So y x c c x x c x
c c xx xn n n n
xLet a c then y a a J xn n
We denoted th
2
0
2
2
2 2 20
2
0
( 1) ( ) .! (1 ) 2
into ,
( 1) ( ).! (1 ) 2
( 1) ( )! (1 ) 2
nn
n
nn
n
nn
n
xat J xn n
If we take r the equation we will get
xy a a J xn n
xWe denoted that J xn n
Thus y y
1 2 1 2
-
( ) ( ) ( ) ( ).
: ( ) ( ) - , .
x y x a J x a J x
Note J x and J x are called Bessel functions of the first kindof the order and repectively
NCCU Wireless Comm. Lab.50
Example 1. General Solution: Not an Integer
0
, 0,1,2, .
( ) ( ) ( 1) ( ) ( ) ( ) ( 1) ( )0, 0
( ) (0) ( ) lim ( )1, 1
: ( ) ( )
m mm m m m
m mx
m m
First we talk about the properties of Bessel functionsof order m m
i J x J x ii J x J xm
iii J iv Y xm
Note J x and J x are li
, 0,1,2, .near dependent where m
12
2 2
1 2 12
1 1( ) 0 (0, ), .4 2
( ) ( ).
Put out atteneion on this differential equation
x y xy x y on we can see that
Then the general solution is y c J x c J x
NCCU Wireless Comm. Lab.51
Bessel Functions of the Second Kinds -
2 2 21 2 1 2
cos ( ) - ( ) integer, ( ) ,sin
( ) ( )
( - ) 0, ( ) ( ) ( ) ( ).
integer,
J x J xIf by linear mapping Y x
J x and Y x can be the linearly independent solutions of
x y xy x y so y c J x c J x or y c J x c Y x
If
2 2 21 2
' ' lim ( ) .
( ) lim ( ), ( ) ( )
( - ) 0, ( ) ( ).:
m
m m mm
according to L Hospital s rule that Y x exists
By linear mapping Y x Y x Y x and J x are the linearly
independent solutions of x y xy x m y so y c J x c Y xNote W
1 2
2 2 2
( ) ( ), ( ) ( ) .
, ( ) 0 (0, ) model
e drop the solution of y c J x c J xbecause J x and J x are linear dependent from the previous page
So for any value of the general solution of x y xy x yon can be
1 2ed ( ) ( ).
: ( ) second .
by y c J x c Y x
Note Y x is called the Bessel function of the kind of order
NCCU Wireless Comm. Lab.52
Following plot will show us the first and second kind of Bessel function.
NCCU Wireless Comm. Lab.53
Example 2. General Solution: an Integer
Parametric Bessel Equation
2 2
2
1 3 2 3
( - 9) 0 (0, ),
9 3, ( ) ( ).
Considering x y xy x y on
we see and so the general solution isy c J x c Y x
2 2 2 2
1 2
1 2
( - ) 0, ( ) ( )
( ) ( ), integer.
: .
For the parametic equation x y xy x ythe general solution y c J x c Y x ory c J x c J x where
Note We will prove it at the next page
NCCU Wireless Comm. Lab.54
2 2 2 2
2 22
2 2
22 2 2 2 2
2
22 2 2
2
1
( - ) 0, let z= x,dy dythen , y = ,dx
dy d y .
into the DE,
( - ) 0
( - ) 0
,
From x y xy x ydy dzdz dx dz
d y d dzdx dz dz dx dzSubstituting we get
d y dyx x x ydz dz
d y dyz z z ydz dz
So y c J
2 1 2
1 2 1 2
( ) ( ) ( ) ( ). ( ) ( ) ( ) ( ).
z c Y z or y c J z c J zThen y c J x c Y x or y c J x c J x
NCCU Wireless Comm. Lab.55
Example 4. Derivation Using the Series Definition1
2
0
2
0
2 2
0 0
( ) ( ) ( ).
( 1) ( ) ,! (1 ) 2
( 1) (2 )( )! (1 ) 2
( 1) ( 1)2! (1 ) 2 ! (1 ) 2
nn
n
nn
n
n nn n
n n
Derive the formula xJ x J x xJ x
xFrom J xn n
n xxJ xn n
x n xn n n n
2 1
1
2 1
10
( 1)( ) , -1( 1)! (1 ) 2
( 1) ( 1)( ) ( ) ( ).! (2 ) 2
nn
n
kk
k
n xJ x x let k nn n
k xJ x x J x xJ xk k
NCCU Wireless Comm. Lab.56
Spherical Bessel Functions
1
1
-1
- - 1 - - -1
4. ( ) ( ) ( ),
, ( ) ( ) ( ).
( ) ( ) ( ), .
( ) ( ) ( ) ( )
From the example xJ x J x xJ x we divide
x at both side J x J x J xx
J x J x J x multiply x at both sidex
dx J x x J x x J x x J x x Jdx
1
0 1 0 1
( ).
, ( ) ( ) ( ).
x
So J x J x and Y Y x
integer, , 1 3 5, , , , ( )2 2 2
expressed sin ,cos , .
When the order is half an odd that is
the Bessel function of first kind J x
can be in terms of the elementary functions xx and powers of x Such Bessel
.
functions are calledSpherical Bessel functions
NCCU Wireless Comm. Lab.57
Example 5. Spherical Bessel Function with 1 .2
12
122
102
3
expression ( ).
1 ( 1) , ( ) .12 2! (1 )2
1 (1 ) ( ) ( ) , 2
1 10 : (1 )2 23 3 1 3!1: (1 )2 2 2 2
1 (2: (1 )2
nn
n
Find an alternative for J x
xWith we get J xn n
By and we find
n
n
n n n
2 1
122 2 1
10 02
2 1
3 5 2 1
0
12
-1)! .2 !
( 1) 2 ( 1), ( ) .(2 -1)! 2 (2 1)!!2 !
1 1 ( 1) sin - ,3! 5! (2 1)!
2 ( ) sin .
n
nn nn
n nn
nn
n
nn
xSo J x xn x nnn
From x x x x xn
so we get J x xx
NCCU Wireless Comm. Lab.58
Now we bring out an extra concept that is not mentioned in textbook, that is, the Modified Bessel Equation. 2 2 2
2 2 2 2
1 2
1 2
( ) 0 .
( ) 0. ( ) ( ), 0,1,2,3, , ( ) ( ),
u u
u u
x y xy x u y is called the Modified Bessel Equation
It changes from x y xy i x u y We get the generalsolution y c J ix c Y ix where u ory c J ix c J ix where u
2 22
0 0
22
0
22
0
0,1,2, .
( 1) ( 1) ( )! (1 ) 2 ! (1 ) 2
( 1) ( ). ! (1 ) 2
( 1) ( ) ! (1 ) 2
n u n un n n u
un n
n un nu u
un
n un n
un
ix i i xAnd J ixn n u n n u
i xi i I xn n u
i xWhere I x is called then n u
-
" 1 ".
, ( ) ( ).uu u
Modified Bessel function of st kind
Following the similar approach you can get J ix i I x
NCCU Wireless Comm. Lab.59
2 2 2
1 2
1 2
1 2
1
( ) 0, ( ) ( ), , 0,1, 2, .
modify ( ) ( ).
( ) ( ), , 0,1, 2, . modify (
u u
u u
u u
m
From the equation x y xy x u y the solution isy c J ix c J ix where u m mWe it to y A I x A I x
y c J ix c Y ix where u m mWe it to y A I x
2) ( ),( ) ( ) ( ) , ( ) ( ).
2 sin( ) " 2 ".
limm
u uu m u
u m
u
A K xI x I xwhere K x K x K x
uK x is called the Modified Bessel function of nd kind
NCCU Wireless Comm. Lab.60
Example.
Solution of Legendre’s Equation
2 2
2 2 2 2
1 26 6
x ( 4 6) 0.
[(2 ) ( 6) ] 0,
(2 ) (2 ).
Solve y xy x y
x y xy i x ySo the general solutiony c I x c K x
2
0
20 2 1 3
0 point (1- ) - 2 ( 1) 0,
into ,
(1- ) - 2 ( 1) ( 1) 2 ( 1)( 2) 6
(
nn
n
Since x is the ordinary of the equation x y xy y
let y c x substitute the equation and combine the summation
x y xy y c c c c x
n
22
2)( 1) ( )( 1) 0.nn n
n
n c n n c x
NCCU Wireless Comm. Lab.61
0 2
1 3
2
02
13
2
, ( 1) 2 0 ( 1)( 2) 6 0 ( 2)( 1) ( )( 1) 0
( 1) 2!
( 1)( 2) 3!
( )(
n n
n
From the previous equation c cc c
n n c n n ccor c
cc
n nc
2
2 04
3 15
4 06
7
1) , n 2,3,4, .( 2)( 1)
n 2,3,4, into , ( 2)( 3) ( 2) ( 1)( 3)
4 3 4!( 3)( 4) ( 3)( 1)( 2)( 4)
5 4 5!( 4)( 5) ( 4)( 2) ( 1)( 3)( 5)
6 5 6!(
n
n
cn n
Take c we getc cc
c cc
c cc
c
5 15)( 6) ( 5)( 3)( 1)( 2)( 4)( 6)7 6 7!
c c
NCCU Wireless Comm. Lab.62
2 40
3 51
0 1
2 4
3
( 1) ( 2) ( 1)( 3)12! 4!
( 1)( 2) ( 3)( 1)( 2)( 4)3! 5!
( ) ( ), ( 1) ( 2) ( 1)( 3)( ) 1
2! 4!( 1)( 2) ( 3)( 1)( 2)(( )
3!
y c x x
c x x x
c U x c V x where
U x x x
V x x x
54) .5!
( ) , 0,1,2,3, " : integer."( ) ( ) 1, ., , 0,1,2,3, ,
(1
x
i If n n Note Here n means a nonnegativeU x and V x will diverge at x it violate the physic phenomenonSo n n the original equation becomes
2
0 1
- ) - 2 ( 1) 0, 0,1,2,3, , ( ) ( )n n
x y xy n n y n the solution isy c U x cV x
NCCU Wireless Comm. Lab.63
( ) 0, 2,4, , ( ) , ( ) infinit . 1,3,5, , ( ) infinite , ( ) .
, ( ) ( ),
n n
n n
n n
ii When n U x has a finite terms but V x has an e termsn U x has an terms but V x has a finite terms
So we recombine the U x and V x
.: infinite .
( ) , 0, 2, 4,(1)
( ) ,( ) , 1,3,5,(1)
( ) " - '
n
nn
n
n
n
put the same term togetherNote Separate the terms and the finite terms
U x nU
Define P xV x nV
where P x is the first kind of n th order Legendre s e
".
( ) (1), 0, 2,4, ( ) ,
( ) (1), 1,3,5,
( ) "second - ' ".
n nn
n n
n
quation
V x U nQ x
U x V n
where Q x is the kind of n th order Legendre s equation
NCCU Wireless Comm. Lab.64
2 30 1 2 3
3 50
3 51
2 23 5
2
1 1 ( ) 1, ( ) , ( ) (3 1), ( ) (5 3 ).2 2
1 1 1 1( ) ln( )3 5 2 1
1 1 1( ) ( ) ln( ) 13 5 2 1
3 1 1 1 3 3 1 1 3( ) ( ) ln( ) .2 3 5 2 4 1 2
We get P x P x x P x x and P x x x
xQ x x x xx
x xQ x x x x xx
x x xQ x x x x x xx
The general sol
expressed ( ) ( ).1, ( ) , 0.
, 1,3,5,, ( ), ( ) .
, 0,2,4,
( )
n n
n
n n
n
ution can be by y AP x BQ xx Q x will diverge B
odd function nSo the solution is y AP x where P x
even function n
We will show the plot of P x at
.next page
NCCU Wireless Comm. Lab.65
Example from P305. Q19
Properties of
( )nP x
( ) ( ) ( 1) ( ) ( ) (1) 1
( ) ( 1) ( 1) ( ) (0) 0, 0
( ) (0) 0, .
nn n n
nn n
n
i P x P x ii P
iii P iv P n dd
v P n even
2
1 1
1 1
(1- ) - 2 2 0, 1.
1, ( ) ( ),1( ) ( ) ln( ) 1.
2 1
For x y xy y x
We get n so y AP x BQ x wherex xP x x and Q x
x
NCCU Wireless Comm. Lab.66
ConclusionHere 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.