Chapter 5. Series solutions of ODEs. Special functions
C.O.S. Sorzano
Biomedical Engineering
September 19, 2014
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Outline
1
Series solutions of ODEs. Special functions
Power series methods
Legendre’s equation. Legendre polynomials
Extended power series method: Frobenius method
Bessel’s equation. Bessel functions Jν (x )
Bessel’s equation. Bessel functions Yν (x )
5. Series solutions of ODEs. Special functions
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References
E. Kreyszig. Advanced Engineering Mathematics. John Wiley & sons. Chapter 5.
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Outline
1
Series solutions of ODEs. Special functions
Power series methods
Legendre’s equation. Legendre polynomials
Extended power series method: Frobenius method
Bessel’s equation. Bessel functions Jν (x )
Bessel’s equation. Bessel functions Yν (x )
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Power series methods
Power series methods
This is the standard method to solve linear ODEs with variable coefficients. A
power series is an infinite series of the form
∞
X
am (x − x0 )m = a0 + a1 (x − x0 ) + a2 (x − x0 )2 + ...
m=0
Taylor Series: f (x0 ) =
∞
P
m=0
f (m) (x0 )
m! (x
− x0 )m
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Power series methods
Example
1
1
1 2
x + x 3 + x 4 + ...
2!
3!
4!
1
1
1
e x = e + e(x − 1) + e (x − 1)2 + e (x − 1)3 + e (x − 1)4 + ...
2!
3!
4!
ex = 1 + x +
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Power series methods
Example
y0 − y = 0
Solution:
We look for a solution of the form
y=
∞
X
am x m = a0 + a1 x + a2 x 2 + ...
m=0
y0 =
∞
X
mam x m−1 = a1 + 2a2 x + 3a3 x 2 + ...
m=1
And susbtitute it in the ODE
a1 + 2a2 x + 3a3 x 2 + ... − a0 + a1 x + a2 x 2 + ... = 0
(a1 − a0 ) + (2a2 − a1 ) + (3a3 − a2 ) + ... = 0
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Power series methods
Example (continued)
(a1 − a0 ) + (2a2 − a1 ) + (3a3 − a2 ) + ... = 0

a1 − a0 = 0 ⇒ a1 = a0



2a2 − a1 = 0 ⇒ a2 = 12 a1 = 12 a0
⇒
1
3a − a2 = 0 ⇒ a3 = 13 a2 = 3·2
a0


 3
...
And in general
ak =
1
1
a0 = a0
k(k − 1)...2
k!
So the solution of the ODE is
y = a0 (1 + x +
1 2
1
x + x 3 + ...) = a0 e x
2!
3!
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Power series methods
Power series methods
y 00 + p(x )y 0 + q(x )y = 0
This method is applied to linear ODEs with variable coefficients because the
coefficients, p and q, can also be substituted by a power series.
Example: A special case of Legendre equation
It occurs in problems with spherical symmetry
(1 − x 2 )y 00 − 2xy 0 + 2y = 0
Solution:
Let’s look for a solution of the form
y = a0 + a1 x + a2 x 2 + a3 x 3 + a4 x 4 + a5 x 5 + a6 x 6 + ...
y 0 = a1 + 2a2 x + 3a3 x 2 + 4a4 x 3 + 5a5 x 4 + 6a6 x 5 + ...
y 00 = 2a2 + 6a3 x + 12a4 x 2 + 20a5 x 3 + 30a6 x 4 + ...
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Power series methods
Example (continued)
(1 − x 2 )y 00 − 2xy 0 + 2y = 0
We now compute each one of the terms appearing in the ODE
y 00
−x y
−2xy 0
2y 0
2 00
=
=
=
=
2a2
2a0
+6a3 x
−2a1 x
+2a1 x
+12a4 x 2
−2a2 x 2
−4a2 x 2
+2a2 x 2
+20a5 x 3
−6a3 x 3
−6a3 x 3
+2a3 x 3
+30a6 x 4
−12a4 x 4
−8a4 x 4
+2a4 x 4
+...
−20a5 x 5
−10a5 x 5
+2a5 x 5
−30a6 x 6
−12a6 x 6
+2a6 x 6
+...
+...
+...
And solve for the coefficients of each power
m=0:
m=1:
m=2:
m=3:
m=4:
...
2a2 + 2a0 = 0
6a3 = 0
12a4 − 4a2 = 0
20a5 − 10a3 = 0
30a6 − 18a4 = 0
⇒
⇒
⇒
⇒
⇒
a2
a3
a4
a5
a6
= −a0
=0
= 13 a2 = − 13 a0
=0
= 35 a4 = 53 − 13 a0 = − 15 a0
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Power series methods
Example (continued)
The general solution of the equation is
1
1
y = a1 x + a0 (1 − x 2 − x 4 − x 6 − ...
3
5
!
∞
X
1
y = a1 x + a0 1 −
x 2m
2m
−
1
m=1
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Power series methods
Convergence
Operations (derivatives and integrals) with a power series are valid within its
region of convergence
S = lim
n→∞
n
X
am (x − x0 )m =
m=0
∞
X
am (x − x0 )m
m=0
The region of convergence is a property of the am coefficients and its radius can
be determined with
1
R=
1
lim |am | m
m→∞
or
R=
1
lim aam+1
m
m→∞
The series is valid in the interval
(x0 − R, x0 + R)
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Power series methods
Example
∞
P
x
e =
1
1−x
1 m
m! x
m=0
∞
P
=
∞
P
xm
⇒
⇒
m=0
m!x m
m=0
⇒
1 am+1 (m+1)!
1
⇒ R=
am = 1 = m+1
m!
am+1 1 ⇒ R=
am = 1 = 1
am+1 (m+1)! am = m! = m + 1 ⇒ R =
1
lim
m→∞
1
m+1
1
lim 1
=
=
1
0
1
1
=1
=∞
m→∞
1
lim m+1
=
1
∞
=0
m→∞
Existence of power series solutions
Given the ODE
y 00 + p(x )y 0 + q(x )y = r (x )
with p, q, and r analytic at x0 , then every solution of the ODE is analytic at x0
and can be represented by a power series in terms of (x − x0 ) with a radius of
convergence R > 0.
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Power series methods
Derivative of a power series
If the power series
y (x ) =
∞
X
am (x − x0 )m
m=0
converges in |x − x0 | < R (with R > 0), then
y 0 (x ) =
∞
X
mam (x − x0 )m−1
m=1
also converges at least in the region |x − x0 | < R.
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Power series methods
Addition of two power series
If the power series
y1 (x ) =
∞
X
am (x − x0 )m
m=0
converges in |x − x0 | < R1 and
y2 (x ) =
∞
X
bm (x − x0 )m
m=0
converges in |x − x0 | < R2 , then
(y1 + y2 )(x ) =
∞
X
(am + bm )(x − x0 )m
m=0
converges at least in |x − x0 | < min(R1 , R2 ).
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Power series methods
Multiplication of two power series
Given the two power series
y1 (x ) =
∞
X
am (x − x0 )m
y2 (x ) =
m=0
∞
X
bl (x − x0 )l
l=0
its multiplication is given
=
∞
P
=
m=0
∞
P
y1 (x )y2 (x )
am (x − x0 )
m
∞
P
l
bl (x − x0 )
∞l=0
P
l
m
am (x − x0 )
bl (x − x0 )
m=0
∞ P
∞
P
l=0
am bl (x − x0 )m+l
m=0 l=0
∞
m
P
P
=
al bm−l (x − x0 )m
=
m=0
l=0
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Power series methods
Vanishing of all coefficients
If a power series has a positive radius of convergence, R, and a sum that is 0
throughout its interval, then each coefficient of the series must be 0. That is, if
∞
X
am (x − x0 )m = 0
∀x ∈ R such that |x − x0 | < R,
m=0
then
∀m
am = 0
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Exercises
Exercises
From Kreyszig (10th ed.), Chapter 5, Section 1:
5.1.7
5.1.20
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Outline
1
Series solutions of ODEs. Special functions
Power series methods
Legendre’s equation. Legendre polynomials
Extended power series method: Frobenius method
Bessel’s equation. Bessel functions Jν (x )
Bessel’s equation. Bessel functions Yν (x )
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Legendre’s equation
Legendre’s equation
(1 − x 2 )y 00 − 2xy 0 + n(n + 1)y = 0
It appears in physical problems with spherical symmetry. We transform it into
y 00 −
2x
n(n + 1)
y0 +
y =0
2
1−x
1 − x2
The coefficients p and q are analytic at x = 0 (but they are not at x = ±1).
Then, we can use a power series around 0 as a solution
∞
P
y=
am x m
m=0
∞
P
y0 =
y 00 =
mam x m−1
m=1
∞
P
m(m − 1)am x m−2
m=2
Let’s call k = n(n + 1)
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Legendre’s equation
Legendre’s equation (continued)
Subtituting into the ODE
2
(1 − x )
∞
X
!
m(m − 1)am x
m−2
− 2x
m=2
∞
X
m=2
∞
X
m=0
m(m − 1)am x m−2 −
∞
X
!
mam x
m−1
+k
m=1
∞
X
m(m − 1)am x m −
m=2
(m + 2)(m + 1)am+2 x m −
(2a2 + ka0 ) + (6a3 + (k − 2)a1 )x +
∞
X
m(m − 1)am x m −
m=2
∞
X
!
am x
m
=0
m=0
2mam x m +
m=1
∞
X
∞
X
∞
X
m=1
∞
X
kam x m = 0
m=0
2mam x m +
∞
X
kam x m = 0
m=0
((m + 2)(m + 1)am+2 − (m2 + m − k)am )x m = 0
m=2
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Legendre’s equation
Legendre’s equation (continued)
(2a2 + ka0 ) + (6a3 + (k − 2)a1 )x +
∞
X
((m + 2)(m + 1)am+2 − (m2 + m − k)am )x m = 0
m=2
m=0:
m=1:
m≥2:
a4
2a2 + ka0 = 0
a2 = − k2 a0 = − n(n+1)
a0
2!
6a3 + (k − 2)a1 = 0
a3 = − k−2
a1 = − n(n+1)−2
a1 = − (n−1)(n+2)
a1
6
3!
3!
(m + 2)(m + 1)am+2 − (m2 + m − k)am
2
+m−n(n+1)
m2 +m−k
am+2 = (m+2)(m+1)
am = m(m+2)(m+1)
am = − (n−m)(n+m+1)
am
(m+2)(m+1)
(n−2)(n+3)
(n−3)(n+4)
= −
a
a
=
−
a3 2
5
4·3
5·4
n(n+1) (n−3)(n+4)
=
− (n−2)(n+3)
−
a
=
−
a3 − (n−1)(n+2)
a1
0
4·3
2!
5·4
3!
= (n−2)n(n+1)(n+3)
a0
= (n−3)(n−1)(n+2)(n+4)
a1
4!
5!
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Legendre’s equation
Legendre’s equation (continued)
There actually two independent solutions and the general solution in (−1, 1) is a
linear combination of the two:
y = a0 y1 + a1 y2
y1
=
=
y2
=
=
a0 + a2 x 2 + a4 x 4 + ...
1−
n(n + 1) 2 (n − 2)n(n + 1)(n + 3) 4
x +
x − ...
2!
4!
a1 x + a3 x 3 + a5 x 5 + ...
x−
(n − 1)(n + 2) 3 (n − 3)(n − 1)(n + 2)(n + 4) 5
x +
x − ...
3!
5!
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Legendre’s equation
Polynomial solutions
Consider the recursion
am+2 = −
(n − m)(n + m + 1)
am
(m + 2)(m + 1)
If n is a positive integer, for m = n we have
an+2 = −
(n − n)(n + n + 1)
an = 0
(n + 2)(n + 1)
and from this point on
0 = an+2 = an+4 = an+6 = ...
If n is even, y1 is a polynomial of degree n.
If n is odd, y2 is a polynomial of degree n.
These polynomials will be referred to as Pn (x ).
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Legendre’s equation
Polynomial solutions (continued)
We choose the highest coefficient (of degree n) to be
an =
(2n)!
2n (n!)2
This choice will make Pn (1) = 1. Now we need to go back to calculate a0 using
am+2 = −
am = −
am−2 = −
(n − m)(n + m + 1)
am
(m + 2)(m + 1)
(m + 2)(m + 1)
am+2
(n − m)(n + m + 1)
m(m − 1)
am
(n − m + 2)(n + m − 1)
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Legendre’s equation
Polynomial solutions (continued)
an−2
n(n−1)
n(n−1)
= − (n−n+2)(n+n−1)
an = − 2(2n−1)
(2n)!
2n (n!)2
n(n−1)[(2n)(2n−1)(2n−2)!
= − 2(2n−1)2
n [n(n−1)!][n(n−1)(n−2)!]
2 n)
(n−1)[(
n
= − n (2n−1)(2n−2)!]
(n−1)(n−2)!]
(2n−1)2 [n(n−1)!][
n
2
(2n−2)!
= − 2n (n−1)!(n−2)!
=
an−4
(2n−2·1)!
(−1)1 2n (1!)(n−1)!(n−2·1)!
(n−2)((n−2)−1)
= − (n−(n−2)+2)(n+(n−2)−1)
an−2 = ...
(2n−4)!
= − 2n 2!(n−2)!(n−4)!
=
(2n−2·2)!
(−1)2 2n (2!)(n−2)!(n−2·2)!
and, in general,
an−2m = (−1)m
(2n − 2m)!
− m)!(n − 2m)!
2n m!(n
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Legendre’s equation
Polynomial solutions (continued)
an−2m = (−1)m
(2n − 2m)!
2n m!(n − m)!(n − 2m)!
The polynomial is finally
Pn (x ) =
M
X
(−1)m
m=0
(2n − 2m)!
xm
− m)!(n − 2m)!
2n m!(n
M = n2 if n is even
M = n−1
2 if n is odd.
Legendre’s polynomials are
orthogonal in [−1, 1]
Z1
Pi (x )Pj (x )dx =
2
δij
2i + 1
−1
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Legendre’s equation
Polynomial solutions (continued)
Fortunately, they can be calculated recursively (Bonnet’s recursion formula)
P0 (x ) = 1
P1 (x ) = x
(n + 1)Pn+1 (x ) = (2n + 1)xPn (x ) − nPn−1 (x )
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Exercises
Exercises
From Kreyszig (10th ed.), Chapter 5, Section 2:
5.2.2
5. Series solutions of ODEs. Special functions
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Exercises
Exercises
5. Series solutions of ODEs. Special functions
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Outline
1
Series solutions of ODEs. Special functions
Power series methods
Legendre’s equation. Legendre polynomials
Extended power series method: Frobenius method
Bessel’s equation. Bessel functions Jν (x )
Bessel’s equation. Bessel functions Yν (x )
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Frobenius method
Frobenius method
Let b(x ) and c(x ) be analytic functions at x = 0. Then the ODE
y 00 +
b(x ) 0 c(x )
y + 2 y =0
x
x
has at least one solution that can be represented in the form
y = xr
∞
X
am x m
(a0 6= 0)
m=0
where r may be any (real or complex) number (and r is chosen so that a0 6= 0).
The ODE also has a second solution that may be similar to the previous one (with
different r and coefficients) or may contain a logarithmic term.
Bessel’s equation
y 00 +
1 0 x 2 − ν2
y +
y =0
x
x2
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Frobenius method
Indicial equation
If b and c are not polynomials, let us expand them as
b=
∞
X
bm x
m
m=0
∞
X
cm x m
m=0
Let us calculate also the derivatives of the solution
y
y0
= xr
=
∞
P
m=0
∞
P
(m + r )am x m+r −1 = x r −1
m=0
r −1
y
00
am x m = x r (a0 + a1 x + ...)
∞
P
(m + r )am x m
m=0
= x (ra0 + (r − 1)a1 + ...)
∞
∞
P
P
=
(m + r )(m + r − 1)am x m+r −2 = x r −2
(m + r )(m + r − 1)am x m
m=0
= x r −2 (r (r − 1)a0 + (r + 1)ra1 + ...)
m=0
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Frobenius method
Indicial equation (continued)
Let us rewrite the ODE as
x 2 y 00 + xby 0 + cy = 0
And substitute the solution
x r (r (r − 1)a0 + ...) + (b0 + ...)x r (ra0 + ...) + (c0 + ...)x r (a0 + ...) = 0
Consider the coefficient of x r
(r (r − 1) + b0 r + c0 )a0 = 0
Since a0 6= 0
r (r − 1) + b0 r + c0 = 0
This is the indicial equation and provides the r of one of the solutions, and
determines the form of the other solution.
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Frobenius method
Indicial equation (continued)
r (r − 1) + b0 r + c0 = 0
One of the elements of the basis is
y1 = x r1 (a0 + a1 x + ...)
The other is
Distinct roots (including complexes) not differing by an integer 1,2,...
y2 = x r2 (A0 + A1 x + ...)
A double root:
y2 = y1 log(x ) + x r1 (A1 x + ...)
Roots differing by an integer 1,2,... (r1 > r2 )
y2 = ky1 log(x ) + x r2 (A0 + A1 x + ...)
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Frobenius method
Example: Euler-Cauchy equation
x 2 y 00 + b0 xy 0 + c0 y = 0
Solution:
The indicial equation
r (r − 1) + b0 r + c0 = 0 ⇒ r1 , r2
If r1 6= r2
If r1 = r2
y1 = x r1
y2 = x r2
y1 = x r1
y2 = x r1 log(x )
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Frobenius method
Example
x (x − 1)y 00 + (3x − 1)y 0 + y = 0
Solution:
We rewrite it as
y 00 +
3x − 1 0
1
y +
y =0
x (x − 1)
x (x − 1)
y 00 +
3x −1
x −1
x
y0 +
x
x −1
x2
y =0
−1
x
The functions b = 3x
x −1 and c = x −1 are analytic around x = 0, so we can apply
Frobenius method. Actually, we do not need the expansion of b and c
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Frobenius method
Example (continued)
Let us substitute the Frobenius solution into the ODE
x (x − 1)y 00 + (3x − 1)y 0 + y = 0
x 2 y 00 − xy 00 + 3xy 0 − y 0 + y = 0
x
2
x
r −2
∞
X
!
(m + r )(m + r − 1)am x
m
−x
+3x
x
∞
X
m=0
∞
X
!
m
(m + r )(m + r − 1)am x
m=0
m=0
r −1
x
r −2
!
(m + r )am x
m
−
x
r −1
∞
X
!
(m + r )am x
m=0
5. Series solutions of ODEs. Special functions
m
+
x
r
∞
X
!
am x
m
=0
m=0
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Frobenius method
Example (continued)
This can be rewritten as
xr
∞
X
(m + r )(m + r − 1)am x m − x r −1
m=0
(m + r )(m + r − 1)am x m
m=0
∞
+3x r
∞
X
X
∞
(m + r )am x m − x r −1
m=0
The smallest power is x
X
(m + r )am x m + x r
m=0
r −1
∞
X
am x m = 0
m=0
and its coefficient gives the indicial equation
−r (r − 1)a0 − ra0 = 0
−r 2 = 0
So, we have a double root at r = 0.
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Frobenius method
Example (continued)
First solution:
We substitute r = 0 in the equation
xr
∞
X
(m + r )(m + r − 1)am x m − x r −1
m=0
∞
X
(m + r )am x m − x r −1
m=0
That is
(m + r )(m + r − 1)am x m
m=0
∞
+3x r
∞
X
X
m=0
∞
X
∞
X
m=0
∞
X
am x m = 0
m=0
m(m − 1)am x m − x −1
m=0
+3
(m + r )am x m + x r
∞
X
m(m + 1)am x m
m=0
mam x m − x −1
∞
X
m=0
mam x m +
∞
X
am x m = 0
m=0
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Frobenius method
Example (continued)
First solution: (continued)
∞
X
m(m − 1)am x m − x −1
m=0
+3
m(m + 1)am x m
m=0
∞
X
m=0
∞
X
∞
X
mam x m − x −1
∞
X
mam x m +
m=0
(m(m − 1) + 3m + 1)am x m −
m=0
∞
X
am x m = 0
m=0
∞
X
(m(m + 1) − m)am x m−1 = 0
m=0
∞
X
(m2 + 2m + 1)am x m −
m=0
∞
X
m=0
∞
X
m2 am x m−1 = 0
m=0
(m2 + 2m + 1)am x m −
∞
X
0
(m0 + 1)2 am0 +1 x m = 0
m0 =−1
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Frobenius method
Example (continued)
First solution: (continued)
∞
X
(m2 + 2m + 1)am x m −
∞
X
0
(m0 + 1)2 am0 +1 x m = 0
m0 =−1
m=0
∞
X
(m2 + 2m + 1)am x m −
m=0
∞
X
∞
X
(m + 1)2 am+1 x m = 0
m=0
[(m2 + 2m + 1)am − (m + 1)2 am+1 ]x m = 0
m=0
∞
X
(m + 1)2 (am − am+1 )x m = 0
m=0
(m + 1)2 (am − am+1 ) = 0
am+1 = am
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Frobenius method
Example (continued)
First solution: (continued)
am+1 = am
Hence
a0 = a1 = a2 = ...
We may choose a0 = 1. The first solution is
y1 =
∞
X
m=0
xm =
1
1−x
|x | < 1
5. Series solutions of ODEs. Special functions
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Frobenius method
Example (continued)
Second solution:
We apply reduction of order
y 00 + py 0 + qy = 0
U=
3x −1
x −1
1 −
e
y12
R
pdx
Z
u=
Udx
x
y 0 + x −1
y =0
x
x2
Z
Z 3x − 1
2
1
1
−
dx = −
+
dx = −2 log |x −1|−log |x | = log
x (x − 1)
x −1 x
x (x − 1)2
y 00 +
U
1
=
(
Z
u=
1
1−x
2
)
e
log
1
x (x −1)2
1
= (1 − x )2 x (x −1)
2 =
1
x
1
log(x )
dx = log(x ) ⇒ y2 = uy1 =
x
1−x
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Frobenius method
Example (continued)
The general solution of
x (x − 1)y 00 + (3x − 1)y 0 + y = 0
is
y = c1
log(x )
1
+ c2
1−x
1−x
5. Series solutions of ODEs. Special functions
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Frobenius method
Example
(x 2 − x )y 00 − xy 0 + y = 0
Solution
Substituting the Frobenius solution in the ODE we get
!
!
∞
∞
X
X
(x 2 − x ) x r −2
(m + r )(m + r − 1)am x m − x x r −1
(m + r )am x m
m=0
m=0
+ xr
∞
X
!
am x m
=0
m=0
xr
∞
X
(m + r )(m + r − 1)am x m − x r −1
m=0
∞
X
(m + r )(m + r − 1)am x m
m=0
−x r
∞
X
m=0
(m + r )am x m + x r
∞
X
am x m = 0
m=0
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Frobenius method
Example (continued)
x
r
∞
X
m
(m + r )(m + r − 1)am x − x
r −1
m=0
∞
X
(m + r )(m + r − 1)am x m
m=0
−x
r
∞
X
m
(m + r )am x + x
r
m=0
∞
X
am x m = 0
m=0
The smallest power is r − 1 whose coefficient is
−r (r − 1)a0 = 0 ⇒ r1 = 1, r2 = 0
We have two roots, whose difference is an integer.
5. Series solutions of ODEs. Special functions
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Frobenius method
Example (continued)
First solution: Substitute r = 1 in
xr
∞
X
(m + r )(m + r − 1)am x m − x r −1
∞
X
m=0
m=0
−x r
∞
X
(m + r )am x m + x r
m=0
x
∞
X
(m + 1)mam x m −
m=0
∞
X
(m + 1)mam x m+1 −
∞
X
∞
X
(m + 1)mam x m −
m=0
m=0
∞
∞
m=0
X
m=1
(m + 1)am x m + x
m=0
∞
X
∞
(m + 1)mam x m+1 −
am x m = 0
m=0
(m + 1)mam x m − x
m=0
∞
X
∞
X
m=0
X
(m + r )(m + r − 1)am x m
(m + 1)mam x m −
X
(m + 1)am x m+1 +
∞
X
am x m = 0
m=0
∞
X
am x m+1 = 0
m=0
(m + 1)am x m+1 +
m=0
5. Series solutions of ODEs. Special functions
∞
X
am x m+1 = 0
m=0
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Frobenius method
Example (continued)
First solution: (continued)
∞
X
(m + 1)mam x m+1 −
m=0
∞
X
(m + 1)mam x m −
m=1
∞
X
(m + 1)am x m+1 +
m=0
(m + 1)mam x m+1 −
∞
X
∞
X
am x m+1 = 0
m=0
(m0 + 2)(m0 + 1)am0 +1 x m
0
+1
m0 =0
m=0
−
∞
X
m=0
∞
X
∞
X
(m + 1)am x m+1 +
∞
X
am x m+1 = 0
m=0
([(m + 1)m − (m + 1) + 1]am − (m + 2)(m + 1)am+1 ) x m+1 = 0
m=0
∞
X
m2 am − (m + 2)(m + 1)am+1 x m+1 = 0
m=0
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Frobenius method
Example (continued)
First solution: (continued)
∞
X
m2 am − (m + 2)(m + 1)am+1 x m+1 = 0
m=0
m2 am − (m + 2)(m + 1)am+1 = 0
am+1 =
m2
am
(m + 2)(m + 1)
If we choose a0 = 1, then
a1 =
02
a0 = 0 = a2 = a3 = ...
(0 + 2)(0 + 1)
So
y1 = x r1 a0 = x 1 · 1 = x
5. Series solutions of ODEs. Special functions
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Frobenius method
Example (continued)
Second solution: Let’s apply a reduction of order:
y2 = uy1 = ux
y20 = u + u 0 x
y200 = u 0 + u 00 + u 0 = u 00 + 2u 0
And substitute in the ODE
(x 2 − x )y 00 − xy 0 + y = 0
(x 2 − x )(u 00 + 2u 0 ) − x (u + u 0 x ) + ux = 0
(x 2 − x )u 00 + (x − 2)u 0 = 0
x −2
2
1
u 00
=− 2
=− +
0
u
x −x
x
1−x
5. Series solutions of ODEs. Special functions
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Frobenius method
Example (continued)
Second solution: (continued)
u 00
2
1
x −2
=− +
=− 2
u0
x −x
x
1−x
log(u 0 ) = −2 log |x | + log |x − 1| = log
x −1
x2
1
1
x −1
= − 2
x2
x
x
1
u = log(x ) +
x
1
y2 = uy1 = log(x ) +
x = x log(x ) − 1
x
u0 =
The general solution is
y = c1 x + c2 (x log(x ) − 1)
5. Series solutions of ODEs. Special functions
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Exercises
Exercises
5. Series solutions of ODEs. Special functions
September 19, 2014
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Exercises
Exercises
5. Series solutions of ODEs. Special functions
September 19, 2014
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Outline
1
Series solutions of ODEs. Special functions
Power series methods
Legendre’s equation. Legendre polynomials
Extended power series method: Frobenius method
Bessel’s equation. Bessel functions Jν (x )
Bessel’s equation. Bessel functions Yν (x )
5. Series solutions of ODEs. Special functions
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Bessel’s equation
Bessel’s equation
x 2 y 00 + xy 0 + (x 2 − ν 2 )y = 0
ν≥0
It appears in physical problems with cylindrical symmetry. We may transform it
into
1
x 2 − ν2
y 00 + y 0 +
y =0
x
x2
The functions b and c (see Frobenius method) are analytic at x = 0 so we can try
a Frobenius solution:
!
!
∞
∞
X
X
r −1
m
m
2
r −2
x
(m + r )(m + r − 1)am x
x
+x
x
+x
(m + r )am x
m=0
m=0
2
x
r
∞
X
m=0
!
am x
m
−ν
2
x
r
∞
X
!
am x
m
=0
m=0
5. Series solutions of ODEs. Special functions
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Bessel’s equation
Bessel’s equation
xr
∞
X
∞
X
(m + r )(m + r − 1)am x m + x r
m=0
(m + r )am x m
m=0
+x
∞
X
r +2
am x m − ν 2 x
∞
X
r
m=0
am x m = 0
m=0
The smallest power is r and its coefficient gives the indicial equation
r (r − 1)a0 + ra0 − ν 2 a0 = (r 2 − ν 2 )a0 = 0 ⇒ r1 , r2 = ±ν
Substituting r = r1 = ν in the equation above we get
xν
∞
X
(m + ν)(m + ν − 1)am x m + x ν
m=0
∞
X
(m + ν)am x m
m=0
∞
∞
+x ν+2
X
m=0
am x m − ν 2 x ν
X
am x m = 0
m=0
5. Series solutions of ODEs. Special functions
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Bessel’s equation
Bessel’s equation
∞
X
m(m + 2ν)am x m+ν +
m=0
∞
X
∞
X
am x m+ν+2 = 0
m=0
m(m + 2ν)am x m+ν +
∞
X
0
am0 −2 x m +ν = 0
m0 =2
m=0
(1 + 2ν)a1 x 1+ν +
∞
X
[m(m + 2ν)am + am−2 ]x m+ν = 0
m=2
(1 + 2ν)a1 = 0 ⇒ a1 = 0
m(m + 2ν)am + am−2 = 0 ⇒ am = −
1
am−2
m(2ν + m)
For odd m’s, a3 is a function of a1 (that is 0), a5 a function of a3 , ...
0 = a1 = a3 = a5 = ...
5. Series solutions of ODEs. Special functions
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Bessel’s equation
Bessel’s equation
am = −
1
am−2
m(2ν + m)
For even m’s, we can write m = 2k
a2k = −
1
1
a2k−2 = − 2
a2k−2
2k(2ν + 2k)
2 k(ν + k)
That is
a2 = −
1
1
a4 = − 2
a2 = − 2
2 2(ν + 2)
2 2(ν + 2)
k = 1, 2, ...
1
a0
+ 1)
22 (ν
1
− 2
a0
2 (ν + 1)
=
24 2!(ν
1
a0
+ 1)(ν + 2)
In general,
a2k =
(−1)m
a0
22k k!(ν + 1)(ν + 2)...(ν + k)
5. Series solutions of ODEs. Special functions
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Bessel’s equation
Bessel’s equation: Bessel’s functions of first kind Jν (x )
The first solution of Bessel’s equation is
y1 = x ν
∞
X
k=0
(−1)k
a0 x 2k
22k k!(ν + 1)(ν + 2)...(ν + k)
If ν ∈ Z, ν = n let us choose
a0 =
Then
y1 = x n
∞
X
k=0
1
2n n!
(−1)k
1 2k
x
2k
n
2 k!(n + 1)(n + 2)...(n + k) 2 n!
y1 = J n = x n
∞
X
k=0
(−1)k
x 2k
22k+n k!(n + k)!
This Bessel’s function of the first kind and order n.
5. Series solutions of ODEs. Special functions
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Bessel’s equation
Bessel’s functions of first kind Jν (x )
For large x , they fulfill
r
Jn (x ) ≈
2
nπ π cos x −
−
πx
2
4
5. Series solutions of ODEs. Special functions
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Bessel’s equation
The Gamma function
Let us define the Gamma function as
Z∞
Γ(x + 1) =
e −t t x dt
0
Integrating by parts (u = t x , dv = e −t dt) we get
Γ(x + 1) =
∞
−e −t t x 0
Z∞
+
xt
x −1 −t
e
Z∞
dt = 0 + x
0
Additionally
t x −1 e −t dt = x Γ(x )
0
Z∞
Γ(1) =
e −t dt = 1
0
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Bessel’s equation
The Gamma function
Γ(x + 1)
Γ(1)
Γ(2)
Γ(3)
Γ(4)
...
Γ(n + 1)
=
=
=
=
=
x Γ(x )
1
1Γ(1) = 1 · 1
2Γ(2) = 2 · 1
3Γ(3) = 3 · 2 · 1
=
n!
Another interesting result
√
1
Γ
= π
2
5. Series solutions of ODEs. Special functions
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Bessel’s equation
Bessel’s equation: Bessel’s functions of first kind Jν (x ) (continued)
The first solution of Bessel’s equation is
y1 = x ν
∞
X
k=0
(−1)k
a0 x 2k
22k k!(ν + 1)(ν + 2)...(ν + k)
If ν ∈
/ Z, let us choose
a0 =
Then
y1 = x ν
∞
X
k=0
1
2ν Γ(ν + 1)
(−1)k
1
x 2k
22k k!(ν + 1)(ν + 2)...(ν + k) 2ν Γ(ν + 1)
y1 = J ν = x ν
∞
X
k=0
(−1)k
x 2k
22k+ν k!Γ(ν + k + 1)
5. Series solutions of ODEs. Special functions
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Bessel’s equation
Bessel’s equation: Bessel’s functions of first kind Jν (x ) (continued)
y1 = J ν = x ν
∞
X
k=0
An interesting result is that
r
J 12 (x ) =
(−1)k
x 2k
+ k + 1)
22k+ν k!Γ(ν
2
sin(x ) J− 12 (x ) =
πx
r
2
cos(x )
πx
The general solution if ν ∈
/ Z is
y = c1 Jν + c2 J−ν
where
J−ν = x −ν
∞
X
k=0
(−1)k
x 2k
+ k + 1)
22k−ν k!Γ(−ν
5. Series solutions of ODEs. Special functions
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Bessel’s equation
Bessel’s equation: Bessel’s functions of first kind Jν (x ) (continued)
y1 = J ν = x ν
∞
X
k=0
An interesting result is that
r
J 12 (x ) =
(−1)k
x 2k
+ k + 1)
22k+ν k!Γ(ν
2
sin(x ) J− 12 (x ) =
πx
r
2
cos(x )
πx
The general solution if ν ∈
/ Z is
y = c1 Jν + c2 J−ν
where
J−ν = x −ν
∞
X
k=0
(−1)k
x 2k
+ k + 1)
22k−ν k!Γ(−ν
5. Series solutions of ODEs. Special functions
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Bessel’s equation
Bessel’s equation: Bessel’s functions of first kind Jν (x ) (continued)
For ν ∈ Z there is problem because
J−n = (−1)n Jn
that is, there is a linear dependence between the two solutions that will be solved
by Bessel’s functions of second kind.
Some useful properties
Derivatives
(x ν Jν (x ))0 = x ν Jν−1 (x )
(x −ν Jν (x ))0 = −x −ν Jν+1 (x )
Recursion
Jν−1 (x ) + Jν+1 (x ) =
2ν
Jν (x )
x
Jν−1 (x ) − Jν+1 (x ) = 2Jν0 (x )
5. Series solutions of ODEs. Special functions
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Exercises
Exercises
From Kreyszig (10th ed.), Chapter 5, Section 5:
5.4.3
5.4.6
5. Series solutions of ODEs. Special functions
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Outline
1
Series solutions of ODEs. Special functions
Power series methods
Legendre’s equation. Legendre polynomials
Extended power series method: Frobenius method
Bessel’s equation. Bessel functions Jν (x )
Bessel’s equation. Bessel functions Yν (x )
5. Series solutions of ODEs. Special functions
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Bessel’s equation
Bessel’s equation
Let’s look for a second solution of Bessel’s equation in the case of ν ∈ Z. For
simplicity we will start with n = 0. The ODE is then
x 2 y 00 + xy 0 + x 2 y = 0
xy 00 + y 0 + xy = 0
We know from the previous section that the indicial equation has a double root at
r = 0. The first solution is
y1 = J0
The second solution according to Frobenius method must be of the form
y2 = y1 log(x ) + x r
∞
X
Am x m
m=1
y2 = J0 log(x ) +
∞
X
Am x m
m=1
5. Series solutions of ODEs. Special functions
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Bessel’s equation
Bessel’s equation
y2 = J0 log(x ) +
∞
X
Am x m
m=1
y20 = J00 log(x ) + J0 x −1 +
∞
X
mAm x m−1
m=1
y200 = J000 log(x ) + 2J00 x −1 − J0 x −2 +
∞
X
m(m − 1)Am x m−2
m=1
We now substitute in the ODE
xy 00 + y 0 + xy = 0
5. Series solutions of ODEs. Special functions
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Bessel’s equation
Bessel’s equation
x
J000
2J00 x −1
log(x ) +
− J0 x
−2
+
∞
X
!
m(m − 1)Am x
m−2
m=1
+
J00
log(x ) + J0 x
−1
+
∞
X
!
mAm x
m−1
+x
J0 log(x ) +
m=1
00
(xJ(
+(
J00 (
+ xJ0 ) log(x ) + 2J00 +
0 (
(
((
∞
X
2J00 +
m=1
!
Am x
m
=0
m=1
m(m − 1)Am x m−1 +
m2 Am x m−1 +
∞
X
mAm x m−1 +
m=1
m=1
∞
X
∞
X
∞
X
∞
X
Am x m+1 = 0
m=1
Am x m+1 = 0
m=1
5. Series solutions of ODEs. Special functions
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Bessel’s equation
Bessel’s equation
We know that
Jn = x n
∞
X
k=0
(−1)k
x 2k
+ k)!
22k+n k!(n
In particular
J0 =
∞
X
(−1)m
m=0
J00 =
∞
X
(−1)m 2m
m=1
22m (m!)2
22m (m!)2
x 2m−1 =
∞
X
m=1
x 2m
(−1)m
22m−1 m!(m
− 1)!
x 2m−1
Substituting in the solution of Bessel’s equation
2
∞
X
m=1
∞
X
m=1
(−1)m
x 2m−1
2m−1
2
m!(m − 1)!
!
+
∞
X
m2 Am x m−1 +
m=1
∞
X
Am x m+1 = 0
m=1
∞
∞
m=1
m=1
X 2
X
(−1)m
x 2m−1 +
m Am x m−1 +
Am x m+1 = 0
22m−2 m!(m − 1)!
5. Series solutions of ODEs. Special functions
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Bessel’s equation
Bessel’s equation
∞
X
m=1
∞
∞
m=1
m=1
X 2
X
(−1)m
x 2m−1 +
m Am x m−1 +
Am x m+1 = 0
22m−2 m!(m − 1)!
1
−x + 2 x 3 − ... + (A1 + 4A2 x + ...) + (A1 x 2 + A2 x 3 + ...) = 0
2 2!
The only term in x 0 comes from the second series, so it must be
A1 = 0
Let’s consider now the even terms. There is none in the first series. In the second
and third series the corresponding terms are
x m−1 = x 2s
x m+1 = x 2s
⇒ m = 2s + 1
⇒ m = 2s − 1
⇒ (2s + 1)A2s+1 x 2s
⇒ A2s−1 x 2s
5. Series solutions of ODEs. Special functions
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Bessel’s equation
Bessel’s equation
Their sum gives
(2s + 1)A2s+1 + A2s−1 = 0 ⇒ A2s+1 = −
1
A2s−1
2s + 1
Consequently, if A1 = 0, so are A3 , A5 , ... Let’s go back to
∞
X
m=1
−x +
∞
∞
m=1
m=1
X
X 2
(−1)m
m−1
2m−1
m
A
x
+
Am x m+1 = 0
+
x
m
22m−2 m!(m − 1)!
1 3
x − ... + (4A2 x + 16A4 x 3 + ...) + (A2 x 3 + A4 x 5 + ...) = 0
22 2!
For the power x 1 , we have
−1 + 4A2 = 0 ⇒ A2 =
1
4
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Bessel’s equation
Bessel’s equation
∞
X
m=1
∞
∞
m=1
m=1
X 2
X
(−1)m
x 2m−1 +
m Am x m−1 +
Am x m+1 = 0
22m−2 m!(m − 1)!
For the rest of powers x 2s+1 (s = 1, 2, ...) we have
1st series
2nd series
3rd series
2m − 1 = 2s + 1 ⇒ m = s + 1
m − 1 = 2s + 1 ⇒ m = 2s + 2
m + 1 = 2s + 1 ⇒ m = 2s
s+1
(−1)s+1
= 22s(−1)
(s+1)!s!
22(s+1)−2 (s+1)!((s+1)−1)!
(2s + 2)2 A2s+2
A2s
(−1)s+1
+ (2s + 2)2 A2s+2 + A2s = 0
+ 1)!s!
22s (s
A2s+2 = −
1
(−1)s+1
A
−
2s
(2s + 2)2
22s (s + 1)!s!(2s + 2)2
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Bessel’s equation
Bessel’s equation
After some manipulation we obtain the general term
(−1)s−1
1
1 1
A2s = 2s
+
+
...
+
1
+
2 (s!)2
2 3
s
Let us call
hs = 1 +
1 1
1
+ + ... +
2 3
s
Finally, the second solution is
y2 = J0 log(x ) +
∞
X
(−1)s−1 hs
s=1
22s (s!)2
x 2s
y2 is independent of y1 = J0 , so both functions together are a basis of solutions.
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Bessel’s equation
Bessel’s equation
It is customary to use a different basis
Y0
=
=
2
π (y
2
2
π
+ (γ − log(2))J0 )
∞
P
J0 (x ) log x2 + γ +
s=1
(−1)s−1 hs 2s
22s (s!)2 x
where γ is Euler’s constant. The set {J0 , Y0 } is also a basis. In general
Yν (x ) =
Jν (x ) cos(νπ) − J−ν (x )
sin νπ
and the general solution of Bessel’s equation
x 2 y 00 + xy 0 + (x 2 − ν 2 )y = 0
ν≥0
is
y = c1 Jν + c2 Yν
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Bessel’s equation
Bessel’s functions of the second kind
For large x , they fulfill
r
Yn (x ) ≈
2
nπ π sin x −
−
πx
2
4
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Bessel’s equation
Recurrence formulas
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Modified Bessel’s equation
Modified Bessel’s equation
Bessel’s equation
x 2 y 00 + xy 0 + (x 2 − ν 2 )y = 0
The general
Modified Bessel’s equation x 2 y 00 + xy 0 − (x 2 + ν 2 )y = 0
solution of the Modified Bessel’s equation is of the form
y = c1 Iν + c2 Kν
where Iν is the modified Bessel function of first kind:
Iν (x ) = i −ν Jν (ix ) = x ν
∞
X
k=0
compare it to
Jν = x ν
∞
X
k=0
1
22k+ν k!Γ(k
+ ν + 1)
x 2k
(−1)k
x 2k
+ k + 1)
22k+ν k!Γ(ν
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Modified Bessel’s equation
Modified Bessel’s equation (continued)
and Kν is the modified Bessel function of second kind:
Kν (x ) =
π I−ν(x ) − Iν (x )
2
sin νπ
compare it to
Yν (x ) =
Jν (x ) cos(νπ) − J−ν (x )
sin νπ
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Modified Bessel’s equation
Modified Bessel’s function of first kind
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Modified Bessel’s equation
Modified Bessel’s function of second kind
5. Series solutions of ODEs. Special functions
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Modified Bessel’s equation
Blobs
B(r ) =
 q


1−


r
r0
2
m q
Im
α
1−
r
r0
2
Im (α)
0
5. Series solutions of ODEs. Special functions
0 ≤ r ≤ r0
r > r0
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Exercises
Exercises
From Kreyszig (10th ed.), Chapter 5, Section 5:
5.5.1
5.5.2
5. Series solutions of ODEs. Special functions
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Outline
1
Series solutions of ODEs. Special functions
Power series methods
Legendre’s equation. Legendre polynomials
Extended power series method: Frobenius method
Bessel’s equation. Bessel functions Jν (x )
Bessel’s equation. Bessel functions Yν (x )
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