Power Series Solutions of Differential Equations, Lecture notes of Differential Equations

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Chapter 6

Chapter 6 Series Solutions of Linear

Equations

Second-order differential equation: y” + P(x)y’ + Q(x)y = 0依P(x)及Q(x)是否為解析而 分類為 (1) Ordinary Point: 若P及Q兩者均解析，則y(x)有兩個獨立解析解 (2) Singular Point: 其它條件下，則y(x)可能有二個、一個或無解析解，但仍是保有 兩個獨立解

6.1 Solutions About Ordinary Points

Review of Power Series Definition of a Power Series:

c (^) n x a n n

=

∞

0

a series centered at x = a

Convergence For a specific value of x a power series is a series of constants. If the series equals a finite real constant for the given x, then the series is said to converge at x. Otherwise it is said to diverge at x. Interval of Convergence Radius of Convergence A power series converges for | x - a | < R where R is the radius of convergence. (1) R = 0: A singe point of convergence (2) R = finite (3) R = infinite: Converging for all x. Convergence at an Endpoint The power series converges at x = R + a or x = R - a Absolute Convergence

c (^) n x a n n

=

∞

0

is convergent.

Finding the Interval of Convergence Ratio Test of a series

n

n

n

n n

c x a c x a

L

→∞

lim (^ ) =

1

1

indicates that the series converges absolutely for L < 1. Therefor, the radius of convergence is

R c n c

n n

lim→∞ + 1

A Power Series Defines a Function(Analytic Function).

f x c (^) n x a n n

=

∞

0 If f(x) has a radius of convergence R > 0, then f(x) is continuous, differentiable, and integrable on the interval of convergence.

f x nc (^) n x a n n

=

∞

1

∞

=

n 0

n 1 n (^) n 1 f(x)dx C c (x a)

lim lim n

n n n

c c

x x →∞ n

→∞

it is true for all x. This indicates that the radius of convergence R is infinite.

Ordinary and Singular Points Suppose the linear second-order differential equation a 2 (x)y" + a 1 (x)y' + a 0 (x)y = 0 is put into the form y" + P(x)y' + Q(x)y = 0. A point x = x 0 is said to be an ordinary point of the equation if both P(x) and Q(x) are analytic at x 0 ; that is, both P(x) and Q(x) have a power series in (x - x 0 ) with a positive radius of convergence. A point that is not an ordinary point is said to be a singular point of the equation.

Example: xy" + (sinx)y = 0 has an ordinary point at x = 0. y" + (lnx)y = 0 has a singular point at x = 0.

Existence of Solutions If x = x 0 is an ordinary point of the equation, we can always find two linearly independent analytic solutions in the form of power series centered at x 0 :

y c (^) n x x n n

=

∞

∑ (^0 )

0 A series solution will converge at least for x − x 0 c 2 1 c 0 2

= , and c 3 = 0.

c n n n + = −^ − cn (^2) +

for n ≥ 2.

Iteration of the last formula gives

c 4 1 c 2 4

c 5 = 0 c 6 3 c 4 6

c 7 = 0 c 8 5 c 6 8

We conclude that for k ≥ 2 c (^) 2k-1 = 0 and

2 k^ (^ )^ k^14 (^ )^ k^1 c 0

2 4 6 ( 2 k)

c 1 13 5 (^2 k^3 ) 4 6 8 ( 2 k)

c 1 1 3 5 (^2 k^3 ) ⋅ ⋅ ⋅

K

K

K

K .

The solution is

x c x 2 4 6 ( 2 k)

( 1 )^135 (^2 k^3 ) 2

y c 1 x 1 k 2

0 2 k^12 k + ⎥

∞

=

− K

K

Example 2: Solve y" + (cosx)y = 0 Solution: The cosine function can be expanded as a series:

cos ( ) ( )!

x x n

n n n

=

∞

2

0 With the assumption of y as a series of x:

y c xn n n

=

∞

0

we can find

n n c x x m n n c x n

m m n n m n

=

∞

=

∞

=

∞

∑ 1 2 ∑^1 2 ∑^0

2

2

0 0 i.e.,

( 2 ) ( 6 ) ( 12 1 ) ( ) 2

c 2 + c 0 + c 3 + c 1 x + c 4 + c 2 − c 0 x^2 + c 5 + c 3 − c 1 x^3 + K= 0

Hence, with c 0 and c 1 being arbitrary constants, all other coefficients can be represented by them:

c 2 1 c 0 2

c 3 1 c 1 6

c 4 1 c 2 c 0 c 0 12

c 5 1 c 3 c 1 c 1 20

The two roots of the r equation gives two independent recursive relationships: (1) r = 0, c c n (^) n n

• =^ n (^1) ( + 1 )( 3 + 1 ) which gives c c n+ (^) n n

0 ( 1 )!1 4 7 K ( 3 1 )

The solution form is

⎟⎟⎠

∞

=

n 0

1 0 xn^1 (n 1 )! 1 4 7 ( 3 n 1 )

y c 1 1 K (2) r = 2/3, c c n (^) n n

• =^ n (^1) ( 3 + 5 )( + 1 ) which gives c c n+ (^) n n

0 ( 1 )! 5 8 11 K( 3 5 ) The solution form is

⎟⎟⎠

∞

=

n 0

2 / 3 0 n 5 / 3 (^2 0) (n 1 )! 5 8 11 ( 3 n 5 )x

c y c x K

The general solution is

⎟⎟⎠

⎞ ⎜⎜⎝

⎛

• ⋅ ⋅ ⋅ ⋅ + ⎟⎟+ + ⎠

⎞ ⎜⎜⎝

⎛

• ⋅ ⋅ ⋅ ⋅ +

= +∑ ∑

∞

∞ +

n 0

2 2 /^30 n^5 /^3 n 0

1 n^1 x (n 1 )! 5811 ( 3 n 5 )

x c x c (n 1 )! 14 7 ( 3 n 1 )

y c 1 1 K K

Transforming the regular singular point into an ordinary point If x = x 0 is a regular singular point, then the transformation y = (x-x 0 ) ru(x) renders the equation for u(x) to be ordinary at the point x = x 0.

Example: Solve the equation

y 0 4 x

y 1 1 x

y" 1 2 ⎟⎠ =

+ +⎛^ −

Solution: x = 0 is a regular singular point for the equation. We will try to find some r and use the transformation form y = xru(x) to get an ordinary equation for u(x). The equation for u is

u 0 x

u 1 r^1 /^4 x

u" 2 r^1 2

2 ⎟⎟^ = ⎠

The condition under which the equation becomes ordinary at x = 0 and u(x) has two analytic solutions is 2r + 1 = 0 and r^2 - 1/4 = 0 Therefore, r = -1/2 and the equation is u" + u = 0 The solution for u(x) is u = c 1 cosx + c 2 sinx and the solution y(x) is

y c^ x x

c x x

= 1 cos^ +^2 sin

Indicial Equation The equation of r is called the indicial equation of the problem. For a second-order differential equation, it is a quadratic equation in r that results from equating the total coefficient of the lowest power of x to zero.

For the case in which x = 0 is a regular singular point, The differential equation y" + P(x)y' + Q(x)y = 0 has the following forms of coefficient functions:

P x p x

( ) = 0 + p 1 + p x 2 +L

Q x q x

q x

( ) = 02 + 1 + q 2 + q x 3 +L

A special case: If q 0 = 0, we can conclude that y has at least one analytic solution. We try the solution form y c xn n^ r n

=

0

, and the indicial equation is r(r - 1)c 0 + p 0 rc 0 = 0. The

roots are r = 0 or r = 1 - p 0. The root r = 0 guarantees the solution must be analytic. The sufficient condition that the equation at a regular singular point has at least one analytic solution is that xQ(x) is analytic. If we set y = xru(x) and hope that u(x) is an analytic function at x = 0, then r( r− 1 )xr^ −^2 u(x)+ 2 rxr−^1 u(x)+xru"(x)+P(x) (rx r−^1 u(x)+xru(x)) +Q(x)xru(x)= 0 That is,

Q(x) u(x) 0 x

rP(x) x

P(x))u(x) r(r^1 ) x

u" (x) (^2 r 2 ⎟⎠ =

+ + +⎛^ − + +

The condition that u(x) has at least one analytic solution is r(r - 1) + rp 0 + q 0 = 0 The condition that u(x) has two analytic soltions is r(r - 1) + rp 0 + q 0 = 0 and 2r + p 0 = 0 and rp 1 + q 1 = 0 That is, the indicial equation must be r(r - 1) + rp 0 + q 0 = 0. Now we classify the equation according to the property of the two roots.

Case I Roots Not differing by an Integer If r 1 and r 2 are distinct and do not differ by an integer, then there exist two linearly independent solutions of the form

y c xn n^ r n

1 0

= +^1

=

∞

y b xn n^ r n

2 0

= +^2

=

∞

where c 0 ≠ 0 and b 0 ≠ 0. Note that if the difference of r 1 and r 2 is an integer, the two solution forms may merge into a single one.

Example: Solve 2xy" + (1 + x)y' + y = 0 Solution: Assume

The reason why we choose the larger root for y 1 (x) is to make sure c 0 ≠ 0. If we choose the smaller one, then we may find that c 0 = c 1 = c 2 = .... = cN-1 = 0, i.e., the solution form for the smaller one merges into the one for the larger root. Then the independent solution for the smaller root can be found from the formula

(^2 1) y 1 (x) 2

exp P(x)dx y (x) y(x)

We first calculate the term involving with the exponential term

− = − ⎛^ + + +

∫ ∫ x p p x dx

exp P(x)dx exp p

0 12 L

= ⎛^ − + +^2 +L

0 1 2 p 2 x exp p lnx px^1

= −^ ⎛^ +^2 +L

1 2

p (^) p x 2

x 0 exp px^1

Therefore, exp ( − (^) ∫ P(x)dx) =x−p^0 ( 1 +A 1 x+A 2 x^2 +L)

and tis yields ( ) ( )

dx x 1 ax a x

y y x^1 Ax A x 2 2 1 2

2 r

p 1 2 2 2 1 1

0

L

L

where we take c 0 = 1. The indicial equation is r(r - 1) + rp 0 + q 0 = (r - r 1 )(r - r 1 + N) = 0 then we have p 0 = N - 2r 1 + 1 Hence, y 2 becomes as

y 2 = y (^1) ∫ x−^1 −N ( 1 +C 1 x+C 2 x^2 +L)dx

For the case N > 0, the above formula gives

dx x

C

x

x

y y C N

1 N 1 2 1 N

= ⎛^ + + +

+ L

− − + L N 1

Cx N

C y lnx y(x) x N^1 N^1 N 1 1

=

∞

C N y x x ∑b xn n^ r

n

1 0

( ) ln^2

where CN may be zero.

Example 1: Solve xy" + (x - 6)y' - 3y = 0. Solution: We try the solution form:

y c xn n^ r n

=

∞

0 and the equation leads to

c (^) n n r n r x n^ r c n r x c n r x c x n

n

n r n

n

n r n

n

n r n

=

∞ (^) +

=

∞ (^) + −

=

∞ (^) +

=

∞

∑ 1 1 ∑ 6 ∑ 3 ∑^0

0 0

1 0 0

i.e.,

c (^) n n r n r x n^ r c n r x c n r x c x n

n n^ r n

n n^ r n

n n^ r n

=−

∞ (^) +

=

∞

=−

∞ (^) +

=

∞

∑ 1 +^ +^ +^ +^ ∑ +^ −^ ∑ +^ +^ −^ ∑ =

1 0

1 1 0

For n = -1, we can obtain the indicial equation is r(r - 1) - 6r = 0 and the two roots are r = 0 and r = 7. The iteration formula is (n + r + 1)(n + r - 6)cn+1 + (n + r - 3)cn = 0, n = 0, 1, 2, ... Note: 由small r 先試。因為相對於small r可能沒有Forbenius form solution，若事實上 是有的，則可找到；若果真沒有，則會收斂到larger r，故是較佳的解題策略。 For the case of r = 0, the iteration is (n + 1)(n - 6)c (^) n+1 +(n - 3)cn = 0 That is,

c 1 1 c 0 2

c 2 1 c 1 c 0 5

c 3 1 c 2 c 0 12

c 4 = c 5 = c 6 = 0 c 7 is arbitrary (for n = 6) and for n ≥ 7 ,

c n^ c n (^) n n

• = −^ − n (^1) + −

i.e.,

c n n n n

n = −^ ⋅^ ⋅ ⋅^ ⋅^ − − ⋅ ⋅ ⋅ ⋅

1 K

K

The general solution is

⎟⎟⎠

= ⎛^ − + −

∞

=

n 8

0 2 3 7 7 n^1 xn (n 7 )! 8 910 n

x c x (^1 )^456 (n^4 ) 120

x^1 10

x^1 2

y c 1 1 K

K

Example 2: Solve xy" + 3y' -y = 0. Solution: We try the solution form:

y c xn n^ r n

=

∞

0

and the equation leads to

c (^) n n r n r x n^ r c n r x c x n

n n^ r n

n n^ r n

=−

∞

=−

∞ (^) +

=

∞

∑ 1 +^ +^ +^ +^ ∑ +^ +^ −^ ∑ =

1

1 1 0

For n = -1, we find the indicial equation: r(r - 1) + 3r = 0 and the roots are r = 0 and r = -2. For the case of r = -2, the iteration formula is c (^) n+1(n - 1)(n + 1) = c (^) n i.e., c 1 (-1)(1) = c 0 c 2 (0)(2) = c 1 This indicates both c 1 and c 0 are zero, and c 2 is arbitrary. For n ≥ 3 ,

y 2 = y (^1) ∫ x−^1 ( 1 +C 1 x+C 2 x^2 +L)dx

⎟ ⎠

= + ⎛^ +^2 +L

1 1 1 2 C 2 x y lnx y Cx^1

=

∞

y x x ∑b xn n^ r

n

1 1

( ) ln^1

Therefore, there always exist two linearly independent solutions:

y c xn n^ r n

1 0

= +^1

=

∞

∑ ,^ c^0 ≠^0

y y x x b xn n^ r n

2 1 1

= + +^1

=

∞

( ) ln ∑

Example: Solve xy" + y' - 4y = 0 Solution: We try the solution form

y c xn n^ r n

=

∞

0 and the equation leads to

c (^) n n r n r x n^ r c n r x c x n

n n^ r n

n n^ r n

=

∞ (^) + −

=

∞ (^) +

=

∞

∑ 1 1 ∑ 4 ∑^0

0

1 0 0 i.e.,

c (^) n n r n r x n^ r c n r x c x n

n

n r n

n

n r n

=−

∞

=−

∞ (^) +

=

∞

∑ 1 +^ +^ +^ +^ ∑ +^ +^ −^ ∑ =

1

1 1 0

For n = -1, we can find the indicial equation r(r - 1) + r = 0 and the two roots are r = 0 with multiplicity 2. The iteration formula is

c c n (^) n

• =^ n (^1) + 2

i.e.,

c c n n

n = (^4 ) ( !) The solution is

y c x n

n n

n

(^1 0 ) 0

=^4

=

∞

The second linearly independent solution is obtained by

− 2 2 3

2 1 1

P(x)dx 2 1 x 9

x 1 4 x 4 x^16

dx y(x) dx y (x)

y y (x) e K

We consider the factor

⎟ ⎠

⎟ = − ⎛^ + + +

⎛ (^) + + (^2) + (^3) +K −^22 x (^3) K 9

x 1 2 4 x 4 x^16 9

1 4 x 4 x^16

K K⎟ + K

⎟ − ⎛^ + + +

• ⎛^ +^2 +^3 +^22 x^33 9

x 4 4 x 4 x^16 9

34 x 4 x^16

x x 2 x^3 K

and the integration becomes

= ⎛^ − + − x + dx 9

8 40 x^1472 x

y y(x)^12 2 1 K

⎟ ⎠

= ⎛^ − +^2 −^3 +K

1 27 x y (x)lnx 8 x 20 x^1472

The general solution is y = C 1 y 1 (x) + C 2 y 2 (x) Note:本題可用Laplace Transform求解為 對原方程式取Laplace Transform得

(s Y(s) sA B) sY(s) A 4 Y(s) 0 ds

− d^2 − − + − − = where y(0) = A, y’(0) = B

Y 0 s

s 4 ds

dY 2 =

∞ = +

∞

n 0 n^1

n n 0 n!s

c^4 s

exp^4 s

Y (s) c^1 where c is an arbitrary constant.

Inverse Laplace Transform得 ∑

∞

n 0 2

n n (n!)

y(x) c^4 x

故僅能找到一獨立解。另一解因無Laplace Transform而無法尋得。

6.3 Two Special Equation

Bessel's Equation: x^2 y" + xy' + (x^2 - ν^2 ) = 0 where ν ≥ 0.

Legendre's Equation: (1 - x^2 )y" - 2xy" + n(n + 1)y = 0 where n is a nonnegative integer. Solution of Bessel's Equation Since x = 0 is a regular singular point for Bessel's equation, we assume that the solution form is

y c xn n^ r n

=

∞

0

and then

x y xy x y c (^) n n r n r x n^ r c n r x c x c x n

n n^ r n

n n^ r n

n n^ r n

2 2 2 0 0

2 2

2 0

" + ' + ( − ) = ( + )( + − 1 ) + + ( + ) + −

∞ (^) +

∞ − + =

∞ (^) +

∞

When n = 0, c 0 r(r - 1) + c 0 r - ν^2 c 0 = 0, i.e., the indicial equation ( c 0 ≠ 0 ) r 2 - ν^2 = 0 When n = 1, c 1 (r + 1)r + c 1 (r + 1) - ν^2 c 1 = 0 i.e.,

Two special cases are Γ(1) = 1 and

⎟=^ π ⎠

Therefore, we have Γ(n+1) = n! for a positive integer n. For -1 < x < 0, Gamma function is defined as

Γ ( x) Γ(^ x ) x

= +^1.

The same formula may then be used to the open interval (-2,-1), then to the open interval (-3,-2), and so on. The graph of Gamma function thus extended is shown in the following graph.

(2) If ν is half an odd positive integer, then Jν(x) and J-ν(x) are linearly independent. If fact, Jν(x) is an elementary function if the order ν is half an odd integer. For example,

J x x 1 2/ (^ )^ =^2 sinx π

and J x x − 1 2/ (^ )^ =^2 cosx π

From the above results, we conclude that y = c 1 Jν(x) + c^2 J^ - ν(x)^ for^ ν _^ integer.

The Graph of Bessel's functions

Example: Solve x y^2 xy x 2 1 y 4

Solution: Since ν = 1/2, the general solution is y = c 1 J (^) 1/2(x) + c 2 J (^) -1/2(x)

Bessel Functions of the Second Kind If ν _ integer, the function defined by the linear combination Y (^) ν x νπJ^ ν^ x^ J^ ν x νπ

( ) cos^ (^ )^ (^ ) sin

= −^ −

and the function Jν(x) are linearly independent. Thus another form of the general solution is y = c 1 Jν(x) + c 2 Yν(x). As ν → m, where m is an integer, Ym(m) exists Y (^) m x Y x m

( ) = lim ( ) ν→ ν and is linearly independent of Jm(x). Hence, for any value of ν, the general solution can be written as y = c 1 Jν(x) + c 2 Yν(x) Yν(x) is sometimes called Neumann's function or the Bessel's function of the second kind of order ν.

The Graph of Neumann's functions

Example: Solve x^2 y" + xy' + (x^2 - 9)y = 0 Solution: Since ν = 3, the general solution is y = c 1 J 3 (x) + c 2 Y 3 (x)

Parametric Bessel Equation x^2 y" + xy' + (λ^2 x^2 - ν^2 )y = 0 The general solution is y = c 1 Jν(λx) + c 2 Yν(λx) As we may se, this equation appears in the solution of Laplace's equation in polar coordinates.

Properties of Bessel functions of order m, where m = 0, 1, 2, .... (i) J (^) -m(x) = (-1) m^ J (^) m(x)

Solution of Legendre's Equation Since x = 0 is an ordinary point of Legendre's equation, we assume that the solution form is

y c xk k k

=

∞

0

Therefore, the equation leads to

( 1 2 ) " 2 ( 1 ) 2 ( 2 )( 1 ) ( 1 ) 2 ( 1 ) 0 2 1 0

− − + + = (^) + + + − − − + +

∞

∞

∞

∞

x y xy n n y ∑ c k k k x k ∑ c k k x ∑ c kx n n ∑c x

k

k k k

k k k

k k k When k = 0,

2c 2 + n(n + 1)c 0 = 0, i.e., c 2 n n^1 c^0 2

= − (^ + )

When k = 1,

6c 3 - 2c 1 + n(n + 1)c 1 = 0, i.e., c 3 n^1 n^2 c^1 6

= − (^ −^ )(^ + )

When k ≥ 2,

c n^ k^ n^ k^ c k (^) k k

• = −^ −^ +^ + k (^2) + +

Thus, for at least |x| < 1, we obtain two linearly independent power series solutions

⎢⎣

= ⎡^ − +^2 + − + +^4

(^1 04)! x x (n^2 )n(n^1 )(n^3 ) 2!

y (x) c 1 n(n^1 )

− −^ −^ +^ +^ +^ + 2

n 4 n 2 n n 1 n 3 n (^5) x 6

6 K

and

2 1 3 x^5 5!

x (n^3 )(n^1 )(n^2 )(n^4 ) 3!

y c x (n^1 )(n^2 ) + − − + + ⎢⎣

= ⎡^ − − +

− −^ −^ −^ +^ +^ +^ + 2

n 5 n 3 n 1 n 2 n 4 n (^6) x 7

7 K

Notice that if n is an even integer, the first series terminates with xn, whereas y 2 (x) is an ifinite series. Similarly when n is an odd integer, y 1 (x) is an infinite series while y 2 (x)

terminates with xn. That is, when n is a nonnegative integer, we obtain an nth-degree polynomial solution of Legendre's equation. In particular, we choose the specific values of c 0 and c 1 : c 0 = 1^ for n = 0 c n n

n 0

1 2 1 3 5^1

= − ⋅^ ⋅^ −

( ) /^ K(^ )

K

for n = 2, 4, 6, ...

and c 1 = 1 for n = 1 c n n

n 1

= − ⋅^ ⋅

( )/ K

K

for n = 3, 5, 7, ...

Legendre Polynomials The specific nth-degree polynomial solutions are called Legendre polynomials and are denoted by Pn(x). They are P 0 (x) = 1 P 1 (x) = x

( 3 x 1 ) 2

P (x)^12 2 =^ − 2 (^5 x^3 x) P (x)^13 3 = −

( 35 x 30 x 3 ) 8

P (x)^142 4 =^ − + 8 (^63 x^70 x^15 x) P (x)^153 5 = − +

Properties (i) Pn(-x) = (-1) nPn(x) (ii) Pn(1) = 1 (iii) Pn(-1) = (-1) n (iv) Pn(0) = 0, n = 1, 3, 5, ... (v) Pn′ ( ) 0 = 0 , n = 0, 2, 4, ....

Graph of the Legendre Polynomials

Recurrence Relation Using binomial series, we can formally show that. ( ) (^) ∑

∞

=

− − + = n 0

n n

2 1 /^2 1 2 xt t P (x)t.

The function on the left is called the generating function for Legendrepolynomials. Differentiating both sides with respect to t gives ( ) (^) ∑

∞

=

− + −^ − = −

n 1

1 2 xt t^2 3 /^2 (x t) nPn(x)tn^1

so that after multiplying by 1 - 2xt + t 2 we have

( ) ( )∑

∞

=

− + −^ − = − + −

n 1

1 2 xt t^2 1 /^2 (x t) 1 2 xt t^2 nPn(x)tn^1