Chapter 6 Series Solutions of Linear Equations

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?????????????,????????????? ??????????
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Chapter 6 Series Solutions of Linear Equations
? DE ? linear ? coefficients ? polynomials ?
?? DE ? solutions ? polynomial ???
(? Cauchy-Euler Method ?? Taylor Series ?????)
???? power series centered at x0
Power series ????? (Sec. 6-1)
x0 is a non-singular point (Sec. 6-2)
??
regular singular point (Sec. 6-3)
x0 is a singular point
irregular singular point
examples (Sec. 6-4)
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Section 6-1 Reviews of Power Series
6-1-1 ??
1. Power series
2. Convergence
exists
????Ratio test (test for convergence)
L lt 1 converge
L gt 1 diverge
L 1 ???
3. Radius of Convergence R
L lt 1 if x - x0 lt R
L gt 1 if x - x0 gt R
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Example 1 (text page 238)
For the Power series
for 1 lt x lt 5
Interval of convergence (1, 5)
However, since when x 1, the power series
becomes
which is also convergent, the interval of
convergence is modified as
Interval of convergence 1, 5)
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6-1-2 Maclaurin Series (Taylor Series)
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Maclaurin Series (Taylor Series) Interval of Convergence
(-8, 8)
(-8, 8)
(-8, 8)
(-1, 1
(-1, 1)
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Example 2 (text page 240)
Find a power series representation of exsinx
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Section 6-2 Solutions about Ordinary Points
Suppose that the solution is
6-2-1 ??????
(1) Linear
(2) x0 is not a singular point
(3) It is better that a0(x), a1(x), ., an(x),
g(x) are all polynomials.
(or can be expressed by Taylor series)
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6-2-2 ????
Step 1 ? ?? (x0 ??? ordinary
point)
?? singular point ??? ordinary point
Step 2 ?? (???? (x - x0)k )
Step 3 ??
Step 4 ????,? cn ????????
Step 5 Obtained independent solutions and
general solution
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6-2-3 ??
Example 5 (text page 246)
Set
since P(x) 0 and Q(x) x are analytic at 0
Step 1
set k n 1
set k n -2
Step 2 ??
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Step 3
2c2 0
Step 4
c2 0
?
recurrence relation
c0, c1 ????
k 1
k 2
k 3
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????,??? cn ???????? (? c0 ? c1 ??)
k 4
k 5
k 6
k 7
k 8
k 9

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Step 5
y1
y2
??? ? ??
ratio test
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Example 6 (text page 248)
(analytic at x 0)
Radius of convergence?
Step 1
k n
k n
k n - 2
Step 2
k n
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Step 3
k 0
k 1
Step 4
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c0, c1 ????

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Step 5
y2
y1
??? ? ??
x lt 1 (Why?)
ratio test
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Example 8 (text page 250)
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6-2-4 ??
1. Analytic at x0 If a function can be
expressed as a power series and the radius of
convergence of the power series is nonzero ?????
f(x) ? x0 ???analytic ?? (1) f(x0) should be
neither ? nor -? (2) f(m)(x0) should be neither ?
nor -? m 1, 2, 3, .
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2. Ordinary Point and Singular Point
? For the 2nd order linear DE
Definition 6.1
x0 is an ordinary point of the 2nd order linear
DE if both P(x) and Q(x) are analytic at x0
Otherwise, x0 is a singular point .
Theorem 6.1 If x0 is an ordinary point of
the 2nd order linear DE, then we can find two
linearly independent solutions in the form of a
power series centered at x0 , i.e.,
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? For the kth order linear DE
Extension of Definition 6.2.1
x0 is an ordinary point of the kth order linear
DE if P0(x), P1(x), P2(x), . , Pk-1 (x), are
analytic at x0
Otherwise, x0 is a singular point .
Extension of Theorem 6.2.1 If x0 is an
ordinary point of the nth order linear DE, then
we can find n linearly independent solutions in
the form of a power series centered at x0 ,
i.e.,
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6-2-5 Interval of Convergence ?????
????? ??
???
????? (???,????? ????????????????????)
?? R ? x0 ???? singular point ??? Singular point
can be a complex number , see Example 6
?????????? convergence
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6-2-6 ??
(1) ?? nonhomogeneous ???.. (2) ?????????????
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6-2-7 ????????
(1) ????????? (a) convergence, (b) radius of
convergence, (c) analytic at x0, (d)
singular point, (e) ordinary point (2) ????
Taylor series (? page 349) (3) Index ????????
(a) ???? xk ???,(b) ?????????? (c) Index
??????? (4) nth order linear DE ?? n ? linearly
independent ? (5) ????? interval of convergence
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Section 6-3 Solutions about Singular Points
????
6-3-1 ??????
(1) Linear
(2) (x -x0)Pn-1 (x), (x -x0)2Pn-2 (x), . , (x
-x0)n-1P1(x), (x -x0)nP0(x) are analytic at
x0
(?? Section 6-2 ?? Pn-1 (x), Pn-2 (x), . ,
P1(x), P0(x) are analytic at x0)
(3) It is better that P0(x), P1(x), ., Pn-1(x)
are all polynomials.
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6-3-2 ??
Singular Points ????
? If x0 is a singular point but (x -x0)Pn-1 (x),
(x -x0)2Pn-2 (x), . , (x -x0)n-1P1(x), (x
-x0)nP0(x) are analytic at x0
x0 regular singular point
? If (x -x0)Pn-1 (x), (x -x0)2Pn-2 (x), . ,
(x -x0)n-1P1(x), (x -x0)nP0(x) are not analytic
at x0
x0 irregular singular point
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Example 1 (text page 253)
x 2 is a
point
x -2 is a
point
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6-3-3 ??
?????
????
Theorem 6.3.1 Frobenius Theorem
? x0 ? linear DE ????? regular singular point ???
linear DE ???????
???
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Process
Step 1 ? ??
Step 2 Power ??
Step 3 ??
Step 4 ?? r
Step 5 ????,? cn ????????
Step 6 ? Step 4 ??? r ?? Step 5 ?????
independent solutions ? general solution
Step 7 (???)
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(Step 7)
? (1) r ??? ? (2) r ?????????,?? Step 6 ???????
independent ?
?
?????? y2(x)
(?? Section 6-3 ? Examples 4, 5)
? r ?????????,?? Step 6 ?????? independent
?, ????????
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6-3-4 ??
Example 2 (text page 255)
Step 1 ? ??
Step 2 Power ??
n k - 1
n k
n k
k n 1
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Step 3 ??
Step 4 ?? r
Step 5
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Step 6
? r 0
? r 2/3


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Solution of Example 2 (??????????)
Set c0 0
(????? x ???)
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Examples 4, 5 (text pages 258, 259)
Step 1 ? ??
Step 2 ??
n k -1
n k
Step 3 ??
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Step 4
Step 5
Step 6
? r 1
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Step 6
? r 0
k 1 ????
??,???? Step 3,?
(k 1, r 0 ??)
c0 ???? 0, c1 ?????
.
???????,???
m n - 1
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???????? y2(x) ?? y1(x), ????? Sec. 4-2reduction
of order ?????
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long division ???
?????? lnx ?
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6-3-5 ???????
??
(1)
(x)
(x -2)
(x -1)
(x2)
(x3)
(x4)
(x5)
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6-3-6 Indicial Equation
2nd order case
If x0 is a regular singular point
where
?? p(x) ? q(x) ?? analytic
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? y(x), y'(x), y''(x), p(x), q(x) ??
?? (x ?x0) r ? coefficient ?
indicial equation
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? linear DE ? 2nd order ?,r ???
??
??
a0 p(x0)
b0 q(x0)
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For the 2nd order case
two roots r1, r2
(Case 1) r1 ? r2 and r1, r2 are real, r2 - r1
? integer
?????? ??
(Case 2) r1 ? r2 and r1, r2 are real, r2 - r1
integer
???????? ??
??????
?????
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(Case 3) r1 r2 ?
C ???? 0
???
(Case 4) r1 ? r2 and r1, r2 are complex
??????
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6-3-7 Indicial Equation for Higher Order Case
(??)
? linear DE ? nth order ?
??? r ???
??
??
k 0, 1, 2, ., n -1
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6-3-8 ?????????
(1) Index ?????????? (????? power
?????,??????? k) ??? page 374 ? Step 2,?????
xkr 1 ?? xkr (2) ? x 0 ? regular singular
point, ? x0 0 ?? (3) ??? ck ? ck-1 (? ck-1 ?
ck) ? recursive relation ????????? cn
??????? (???????? 0) (4) ????? 0 ??? (?
page 380) (5) ?????? y2(x) ? y1(x) ????? (? pages
380, 381)
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(6)?????????? ?????????? ????? xr (7)
Interval of solution ?????, ? interval
?????singular point, ??? regular singular
point (8) ?????
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Section 6-4 Special Functions
(?????????)
Special cases of Sections 6-2 and 6-3
? Bessels equation of order v
Solution
1st kind Bessel function
2nd kind Bessel function
? Legendres equation of order n
One of the solution Legendre polynomials
(See page 409)
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????
? Gamma function

• ? modified Bessel equation of order v
• modified Bessel equation of the 1st kind
• modified Bessel equation of the 2nd kind
• Bessel ??????
• ?

?c1Iv( x) c2Kv(x)
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6.4.1 Bessels Equation
6.4.1.1 Solving for Bessels equation of order v
Steps 13 ?
??
?????? (See text pages 262, 263) ??
two roots v and -v
Step 4
Step 5
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Step 6
? r v
? r -v
?? c1 0, c3 c5 c7 c9 .. 0
when r v
when r -v
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6.4.1.2 Gamma function a generalization of n!
properties of Gamma function
(1)
when n is a positive integer
(2)
???? Appendix 1
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(3)
when n is a negative integer or n 0
(4)
?(4)
?(x)
?(3)
?(2)
?(1)
x
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6.4.1.1 ?? Solving for Bessel function
when r v
Set
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??,? r -v
set
Two independent solutions of the Bessels equation
??
When r v
When r -v
?? Bessel functions of the first kind of order v
and -v
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6.4.1.3 Bessel function of the second kind
??,?? roots ??? 2v
(1) ? 2v ?????,Bessels equation ????
c1Jv(x) c2J-v(x) (????? c1Jv(x)
c2Yv(x))
(2) ? 2v ???,? v m 1/2 (m ?????) ?,Bessels
equation ???? c1Jv(x) c2J-v(x) (????? c1Jv(x)
c2Yv(x)) (3) ? 2v ???,? v ??????, Bessels
equation ??? c1Jv(x) c2Yv(x)
Yv(x) Bessel function of the second kind of
order v (???)
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Yv(x) Bessel function of the second kind of
order v
? m ????, Ym(x) ???
? LHopitals rule ??
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6.4.1.4 Bessel function of the 1st kind (order m
????)???
(1) J0(0) 1, Jm(0) 0 for m ? 0
(2) Zero crossing ???,?? m ??????? (? Table 6.4.1)
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(3)
(4) (5) (6)
when m is an integer
when m is an integer
? Example 6, text pages 268, 269
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6.4.1.5 Bessel function of the 2nd kind (order m
????)???
(1)
(2) Zero crossing ???,?? m ???????
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6.4.1.6 Bessels equation ???
?c1Jv(x) c2Yv(x)
?c1Jv(? x) c2Yv(? x)
(A)
Proof Set t ?x
Similarly,
??
? t ??? Bessel equation
y c1Jv(t) c2Yv(t) c1Jv(? x) c2Yv(? x)
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(B) modified Bessel equation of order v
?c1Iv( x) c2Kv(x)
??
??? modified Bessel function of the first kind of
order v
??? modified Bessel function of the second kind
of order v
? v ????,??? limit
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(C)
?
??????,??????????????
Example 4 (text page 266)
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6.4.1.7 Spherical Bessel Functions
Jv(x) ?
?,??? spherical Bessel functions
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6.4.2 Legendres Equation
6.4.2.1 Legendres Equation
??,?? (???text pages 270, 271)
Two linearly independent solutions are
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(a) When n is not an integer, both the two
solutions have infinite number of terms. (b)
When n is an even integer, y1(x) has finite
number of terms. In y1(x), the coefficient
of xk is zero when k gt n. (c) When n is an odd
integer, y2(x) has finite number of terms.
In y2(x), the coefficient of xk is zero when k gt
n.
y1(x) when n is an even integer and y2(x) when n
is an odd integer are called the Legendre
polynomials
(denoted by Pn(x)).
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???
(? Pn(1) ???? 1)
? y1(x)
? y2(x)
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Legendre polynomials
Interval x ? -1, 1
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6.4.2.2 Properties of Legendre Polynomials
(1)
even / odd symmetry
(2)
(3)
when n is odd
(4)
when n is even
(5)
recursive relation
Rodrigues formula
(6)
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(7)
If m ? n
orthogonality property
(8) ???? x ? -1, 1 ??? continuous ??? f(x)
?????
??
?? orthogonality property
??
Orthogonality property ?? Legendre polynomials
??????
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6.4.2.3 ??????? orthogonal polynomial
? Chebychev polynomials
???? filter design ??
They are the solutions of
? Hermite polynomials
?????????????
They are the solutions of
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6.4.3 Section 6-4 ???????
(1) ????,????,??,?????? ?????? (2) ???
Gamma function
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Review of Chapter 6
?????? Linear DE,? coefficients ??? polynomials
? ? Pm(x) ? x x0 ?? analytic x0 ? ordinary
point
????
? ? Pm(x) ? x x0 ??? analytic ?? (x - x0 )n
-mPm(x) ? x x0 ?? analytic x0 ? regular
singular point
????
??,?????
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Exercise for practice
Sec. 6-1 4, 9, 12, 24, 30, 32, 34 Sec. 6-2
2, 10, 13, 18, 20, 22, 23, 26, 27 Sec. 6-3
4, 9, 13, 16, 22, 24, 26, 28, 29, 31, 33, 36
Review 6 6, 7, 10, 14, 19, 20