Advanced Engineering Mathematics
Ch. 5 Series Solutions of ODEs.
Special Functions

In this chapter, The three main topics are Legendre polynomials, Bessel funcitons, and
eigenfunction expansions.

Legendre’s ODE and Legendre polynomials are obtained by the power series method.

Bessel’s ODE and Bessel functions are obtained by the Frobenius method, an extension of
the power series method.

Eigenfunctions expansions are infinite series obtained by the Sturm-Liouville theory.
Advanced Engineering Mathematics
5.1 Power Series Method
∞
 Power Series : ∑ am ( x − x0 ) =a0 + a1 ( x − x0 ) + a2 ( x − x0 ) + 
m
2
m=0
• Coefficients : a0 , a1 , a2 , 
• Center : x0
∞
• Power Series in powers of x : ∑ am x m =a0 + a1 x + a2 x 2 + 
m=0

Ex 1 Maclaurin series
∞
1
= ∑ x m =1 + x + x 2 + 
1 − x m=0
∞
xm
x 2 x3
=1 + x + + + 
2! 3!
m=0 m !
ex = ∑
−1) x 2 m
(
x2 x4
cos x = ∑
= 1 − + − +
2! 4!
( 2m ) !
m=0
m
∞
−1) x 2 m +1
(
x3 x5
sin x = ∑
= x − + − +
3! 5!
m = 0 ( 2 m + 1) !
∞
m
( x < 1, geometric series )
Advanced Engineering Mathematics
5.1 Power Series Method
 Idea of the Power Series Method
0
ODE y ''+ p ( x ) y '+ q ( x ) y =
• We represent p(x) and q(x) by power series in powers of x.
• We assume a solution in the form of a power series with unknown coefficients
∞
y =∑ am x m =a0 + a1 x + a2 x 2 + a3 x 3 + 
m=0
• Series obtained by termwise differentiation
∞
y' =
a1 + 2a2 x + 3a3 x 2 + 
∑ mam x m−1 =
m =1
y=
''
∞
∑ m ( m − 1) a x = 2a + 3 ⋅ 2a x + 
m=2
• Insert the series into the ODE
m −1
m
2
3
Advanced Engineering Mathematics
5.1 Power Series Method
 Ex.2 Solve the following ODE by power series.
y '− y =
0
∞
y =∑ am x m =a0 + a1 x + a2 x 2 + a3 x 3 + 
∞
differential
y' =
a1 + 2a2 x + 3a3 x 2 + 
∑ mam x m−1 =
m=0
Insert the series into the ODE
∴ a1 = a 0 , a2 =
General solution :
m =1
0
( a + 2a x + 3a x + ) − ( a + a x + a x + ) =
2
1
2
a
a
a
a1 a0
a
= , a3 = 2 = 0 , a4 = 3 = 0 ,
2 2!
3 3!
4 4!
y = a0 + a0 x +
2
3
a0 2 a0 3
x + x +  = a0 e x
2!
3!
0

1
2
Advanced Engineering Mathematics
5.1 Power Series Method
 Theory of the Power Series Method
∞
∑ a ( x − x ) =a + a ( x − x ) + a ( x − x ) + 
m=0
2
m
m
0
0
1
0
2
0
• nth Partial Sum : sn ( x ) = a0 + a1 ( x − x0 ) + a2 ( x − x0 )2 +  + an ( x − x0 )n
• Remainder : Rn ( x ) = an +1 ( x − x0 )n +1 + an + 2 ( x − x0 )n + 2 +
• Convergent : The sequence of partial sums converges.
• Divergent : The sequence of partial sums diverges.
Advanced Engineering Mathematics
5.1 Power Series Method
 Convergence Interval. Radius of Convergence
There are three cases with respect to the convergence of the power series
Case 1. (useless) The series always converges at the center.
Case 2. (usual) If there are further values of x for which the series converges, these values form
an interval, called the convergence interval.
If the convergence interval is finite, it has the midpoint x0 .
The series converges for all x such that x − x0 < R and diverges for all x such that x − x0 > R .
Radius of convergence of the series : R
=
R
1
or
=
R
m a
lim m
m →∞
1
lim
m →∞
am+1
am
Case 3. (best) The convergence interval may sometimes be infinite, that is, the series converges
for all x.
Advanced Engineering Mathematics
5.1 Power Series Method
 Theorem 1 Existence of Power Series Solutions
r ( x ) are analytic at x = x0 , then every solution is analytic
If p, q and r in y ''+ p ( x ) y '+ q ( x ) y =
and can thus be represented by a power series in powers of x − x0 with radius of convergence R>0.
 Operations on Power Series
• A power series may be differentiated term by term.(Termwise Differentiation)
• Two power series may be added term by term.(Termwise Addition)
• Two power series may be multiplied term by term.(Termwise Multiplication)
• If a power series has a positive radius of convergence and a sum that is identically zero
throughout its interval of convergence, then each coefficient of the series must be
zero.(Vanishing of All Coefficients)
Advanced Engineering Mathematics
5.2 Legendre’s Equation. Legendre
Polynomials P ( x )
n
 Legendre’s equation : (1 − x 2 ) y ''− 2 xy '+ n ( n + 1) y =
0
∞
Substituting y = ∑ am x m and its derivatives
m=0
∞
(1 − x ) ∑ m ( m - 1) a x
2
=
m 2
m−2
∞
− 2 x ∑ ma x
m −1
∞
0
+ n ( n + 1) ∑ am x m =
m
m
=
m 1=
m 0
The first expression as two separate series.
∞
∞
=
m 2
=
m 2
∞
∞
0
∑ m ( m - 1) am x m−2 − ∑ m ( m - 1) am x m −∑ mam x m + ∑ n ( n + 1) am x m =
=
m 1=
m 0
Set m-2 = s in the first series and simply write s instead of m in the other three series.
∞
∞
∞
∞
0
∑ ( s + 2 )( s + 1) a x − ∑ s ( s - 1) a x −∑ sa x +∑ n ( n + 1) a x =
s
=s 0
s+2
=s 2
s
s
s
s
=s 1 =s 0
s
s
Advanced Engineering Mathematics
5.2 Legendre’s Equation. Legendre
Polynomials
( n − s )( n + s + 1)
( s + 2 )( s + 1)
−
as
0,1,)
Recurrence relation : as + 2 =
(s =
n ( n + 1)
( n − 1)( n + 2 ) a
−
−
a2 =
a0
a3 =
1
2!
3!
( n − 3)( n + 4 ) a
( n − 2 )( n + 3) a
−
−
a4 =
a5 =
2
3
4⋅3
5⋅ 4
( n − 3)( n − 1)( n + 2 )( n + 4 ) a
( n − 2 ) n ( n + 1)( n + 3) a
= =
0
1
4!
5!
General solution :=
y ( x ) a0 y1 ( x ) + a1 y2 ( x )
n ( n + 1) 2 ( n − 2 ) n ( n + 1)( n + 3) 4
x +
x − +
2!
4!
( n − 1)( n + 2 ) x3 + ( n − 3)( n − 1)( n + 2 )( n + 4 ) x5 − + 
y2 ( x ) = x −
3!
5!
y1 ( x ) = 1 −
Advanced Engineering Mathematics
5.2 Legendre’s Equation. Legendre
Polynomials
 Legendre Polynomials
n
We choose the coefficient an of the highest power x as

1

n!
( 2n ) ! 
=
an =
1 ⋅ 3 ⋅ 5 ( 2n − 1)
2
2n ( n !)

n a positive integer
n≠0
We solved for as in terms of as + 2
as =−
( s + 2 )( s + 1) a
( n − s )( n + s + 1) s + 2
( −1)
When n − 2m ≥ 0, an − 2 m =
m
( s ≤ n − 2)
( 2n − 2m ) !
2 m !( n − m ) !( n − 2m ) !
n
Legendre Polynomial of degree n :
M
( 2n − 2m ) !
m
Pn ( x ) =
x n−2m
( −1) n
∑
2 m !( n − m ) !( n − 2m ) !
m=0
n
n −1 

or
M =

2
2 

Advanced Engineering Mathematics
5.2 Legendre’s Equation. Legendre
Polynomials
 Examples of the Legendre polynomial
P0 ( x ) 1,=
P0 ( x ) x,
1
1
P2 ( x ) =
P3 ( x ) =
3 x 2 − 1) ,
5 x3 − 3x ) ,
(
(
2
2
1
1
P5 ( x =
P4 ( x=
) ( 63x5 − 70 x3 + 15 x )
) ( 35 x 4 − 30 x 2 + 3)
8
8
< Legendre polynomials>
Advanced Engineering Mathematics
5.3 Extended Power Series Method :
Frobenius Method
 Theorem 1 Frobenius Method
Let b(x) and c(x) be any functions that are analytic at x = 0. Then ODE y ''+
has at least one solution that can be represented in the form
∞
y ( x=
) x r ∑ am x m= x r ( a0 + a1 x + a2 x 2 + )
m=0
where the exponent r may be any number.
( a0 ≠ 0 )
b ( x)
c ( x)
y '+ 2 y =
0
x
x
Advanced Engineering Mathematics
5.3 Extended Power Series Method :
Frobenius Method
 Indicial Equation, Indicating the Form of Solutions
Multiply x
b ( x)
c ( x)
y ''+
y '+ 2 y =
0
x
x
2
x 2 y ''+ xb ( x ) y '+ c ( x ) y =
0
• We expand b(x) and c(x) in power series ( b ( x ) =b0 + b1 x + b2 x 2 +  ,
c ( x ) =c0 + c1 x + c2 x 2 + )
∞
• We differentiate y ( x =
) x r ∑ am x m= x r ( a0 + a1 x + a2 x 2 + ) term by term
m=0
x r  r ( r − 1) a0 +  + ( b0 + b1 x + ) x r [ ra0 + ] + ( c0 + c1 x + ) x r ( a0 + a1 x + ) =
0
• The equation corresponding to the power x r is  r ( r − 1) + b0 r + c0  a0 =
0.
0
• Indicial Equation : r ( r − 1) + b0 r + c0 =
Advanced Engineering Mathematics
5.3 Extended Power Series Method :
Frobenius Method
 Theorem 2 Frobenius Method. Basis of Solutions. Three Cases
Suppose that the ODE satisfies the assumptions in Theorem 1. Let r1 and r2 be the roots of the
indicial equation. Then we have the following three cases.
Case 1. Distinct Roots Not Differing by an Integer. A basis is
y1 ( x=
) x r1 ( a0 + a1 x + a2 x 2 + ) and y2 ( x=) x r2 ( A0 + A1 x + A2 x 2 + )
Case 2. Double Root. A basis is
y1 ( x =
) x r ( a0 + a1 x + a2 x 2 + ) and y2 ( x =) y1 ( x ) ln x + x r ( A1 x + A2 x 2 + )
Case 3. Roots Differing by an Integer. A basis is
y1 ( x=
) x r1 ( a0 + a1 x + a2 x 2 + ) and y2 ( x=) ky1 ( x ) ln x + x r2 ( A0 + A1 x + A2 x 2 + )
where the roots are so denoted that r1 − r2 > 0.
Advanced Engineering Mathematics
5.3 Extended Power Series Method :
Frobenius Method
 Ex.2 Illustration of Case 2(Double Root)
Solve the ODE x ( x − 1) y ''+ ( 3 x − 1) y '+ y =
0
∞
Inserting y ( x =
) x r ∑ am x m= x r ( a0 + a1 x + a2 x 2 + ) and its derivatives into the ODE
m=0
∞
∞
∑ ( m + r )( m + r − 1) am x m+r − ∑ ( m + r )( m + r − 1) am x m+r −1
m 0=
m 0
=
∞
∞
∞
0
+ 3∑ ( m + r ) am x m + r − ∑ ( m + r ) am x m + r −1 + ∑ am x m + r =
m 0=
m 0 =
m 0
=
Coefficients of the smallest power x r −1 :  −r ( r − 1) − r  a0 = 0
⇒
∴ r= 0
First Solution.
0
Coefficients of the power x s : s ( s − 1) as − ( s + 1) sas +1 + 3sas − ( s + 1) as +1 + a=
s
Choose a0 = 1.
∞
∴ y1 ( x ) = ∑ x m =
m=0
1
1− x
( x < 1)
⇒
as +=
as
1
Advanced Engineering Mathematics
5.3 Extended Power Series Method :
Frobenius Method
Second Solution. Apply the method of reduction of order
− ∫ pdx =
−∫
3x − 1
1
 2
dx =
−∫ 
+  dx =
−2 ln ( x − 1) − ln x
x ( x − 1)
 x −1 x 
u'= y e ∫
−2 − pdx
1
( x − 1) = 1 ,
=
2
( x − 1) x x
2
u = ln x,
∴ y2 = uy1 =
ln x
x −1
Advanced Engineering Mathematics
5.4 Bessel’s Equation. Bessel Functions J ( x ).
ν
 Bessel ’s equation : x 2 y ''+ xy '+ ( x 2 −ν 2 ) y =
0
Apply the Frobenius method
∞
substitute the series y = ∑ am x m + r with undetermined coefficients and its derivatives
m=0
∞
∞
=
m 0
=
m 0
∞
∞
∑ ( m + r )( m + r − 1) am x m+r + ∑ ( m + r ) am x m+r + ∑ am x m+r +2 −ν 2 ∑ am x m+r =0
=
r ( r − 1) a0 + ra0 −ν 2 a0 0
=
m 0=
m 0
=
( s 0)
1) ra1 + ( r + 1) a1 −ν 2 a1 0 =
( r +=
( s 1)
2
0 =
( s + r )( s + r − 1) as + ( s + r ) as + as=
( s 2, 3, )
− 2 − ν as
0
Indicial equation : ( r +ν )( r −ν ) =
Advanced Engineering Mathematics
5.4 Bessel’s Equation. Bessel Functions
 Coefficient Recursion for r= r1= ν
ν + 1) a1 0
( 2=
( s + 2ν ) sas + as −2 =0
a1 0
⇒
=
⇒
a3 =a5 =
 =0
0 ⇒
For s = 2m, ( 2m + 2ν ) 2ma2 m + a2 m − 2 =
a2 = −
1
22 (ν + 1)
1
a2 m =
− 2
a2 m − 2 ,
2 m ( m +ν )
a0
1
1
a4 =
− 2
a2 =
a0
2 2 (ν + 2 )
24 2!(ν + 1)(ν + 2 )
( −1)
=
∴ a2 m
=
a0 ,
m 1, 2, 
22 m m !(ν + 1)(ν + 2 ) (ν + m )
m
1, 2, 
m=
.
Advanced Engineering Mathematics
5.4 Bessel’s Equation. Bessel Functions
• Bessel Functions J n ( x ) for Integer ν = n
1
2 n!
Choose a0 = n=
a2 m
( −1)
=
,
m 1, 2, 
2m+n
2
m !( n + m ) !
m
2m
Bessel function of the first kind of order n : J n ( x ) = x ∑ 2 m(+ −n 1) x
m !( n + m ) !
m=0 2
m
∞
n
• Bessel Function J ν ( x ) for any ν ≥ 0 . Gamma Function.
Gamma function
: Γ (ν )
=
∞
ν
∫ e t dt ( ν > 0 )
−t
−1
0
Basic functional relation : Γ ( ν + 1=
n 0 ,1,)
) νΓ ( ν ) ( In general Γ ( n + 1=) n!, =
1
( −1)
Choose a0 = ν =
,
a2 m =
m 1, 2, 
2 m +ν
2 Γ (ν + 1)
2
m !Γ (ν + m + 1)
m
−1 x 2 m
Bessel function of the first kind of order ν : Jν ( x ) = x ∑ 2 m +ν ( )
m !Γ (ν + m + 1)
m=0 2
ν
∞
m
.
Advanced Engineering Mathematics
5.4 Bessel’s Equation. Bessel Functions
.
 Discovery of Properties From Series
 Theorem 3 Derivatives, Recursions
The derivative of Jν ( x ) with respect to x can be expressed by Jν −1 ( x ) and Jν +1 ( x ) by the
formulas
 xν Jν ( x )  ' = xν Jν −1 ( x )
 x −ν Jν ( x )  ' = − x −ν Jν +1 ( x )
Furthermore, Jν ( x ) and its derivative satisfy the recurrence relations
2ν
Jν −1 ( x ) + Jν +1 ( x ) =
Jν ( x )
x
Jν −1 ( x ) − Jν +1 ( x ) =
2 Jν ' ( x )
Advanced Engineering Mathematics
5.4 Bessel’s Equation. Bessel Functions
.
 Theorem 4 Elementary Jν for Half-Integer Order ν
1
2
3
2
5
2
Bessel functions Jν of orders ± , ± , ± ,  are elementary; they can be expressed by
finitely many cosines and sines and powers of x. In particular,
J 1 ( x)
=
2
2
sin x,
J− 1 ( x)
=
2
πx
2
cos x
πx
Advanced Engineering Mathematics
5.5 Bessel Functions Y ( x ) . General Solutions
ν
 n = 0 : Bessel Function of the Second Kind Y0 ( x )
Bessel’s equation : xy ''+ y '+ xy =
0
First solution : J 0 ( x )
∞
Second solution=
: y2 ( x ) J 0 ( x ) ln x + ∑ Am x m
m =1
Substitute y2 and its derivatives into ODE
∞
∞
∞
2 J 0 '+ ∑ m ( m − 1) Am x m −1 + ∑ mAm x m −1 + ∑ Am x m +1 =
0
=
m 1
=
m 1=
m 1
−1) x 2 m −1
(
J 0 ' ( x ) ∑ 2 m −1
m !( m − 1) !
=
m 12
∞
∞
∞
−1) x 2 m −1
(=
2
m −1
+ ∑ m Am x + ∑ Am x m +1 0
∑
2 m −1
m=
!( m − 1) ! m 1 =m 1
=
m 12
m
∞
m
( −1) 1 + 1 + 1 +  + 1  = ( −1) hm
A1 = A3 =  = 0, A2 m = 2 m

2 
m  22 m ( m !)2
2 ( m !)  2 3
m −1
m −1
( m = 1, 2,)
( −1) hm x 2 m J x ln x + 1 x 2 − 3 x 4 + 11 x 6 − + 
∴
=
y2 ( x ) J 0 ( x ) ln x + ∑ 2 m =
0( )
2
4
128
13824
m =1 2
( m !)
∞
m −1
Advanced Engineering Mathematics
5.5 Bessel Functions Y ( x ) . General Solutions
ν
 Bessel function of the second kind of order zero or Neumann’s function of order zero
m −1
2
 x
 ∞ ( −1) hm 2 m 
=
Y0 ( x )
x 
 J ( x )  ln + γ  + ∑ 2 m
2
π  0
 2
 m =1 2 ( m !)

 Bessel function of the second kind of order ν or Neumann’s function of order ν
=
Yν ( x )
1
 Jν ( x ) cosνπ − J −ν ( x ) 
sinνπ 
n
−n
 x
 x ∞ ( −1) ( hm + hm + n ) 2 m x n −1 ( n − m − 1) ! 2 m
lim
ln
Y=
x
Y
x
J
x
γ
x
x
=
+
+
−
(
)
(
)
(
)
∑ 22m+n m!( m + n )!
∑
ν
n


ν →n
π n
π m 0 22 m − n m !
 2
 π m 0=
=
m −1
2
 Theorem 1 General Solution of Bessel’s Equation
A general solution of Bessel’s equation for all values of ν (and x > 0 ) is
=
y ( x ) C1 Jν ( x ) + C2Yν ( x )