Power series methods of solving ordinary differential equations 511
Now try the following exercise
Exercise 197 Further problems on Bessel’s
equation and Bessel’s functions
1. Determine the power series solution of Bessel’s equation: x 2 d 2 y
dx 2 + x
d y
dx
+ (x 2 −v 2 )y = 0
when v = 2, up to and including the term in x 6 .
y = Ax 2
1 −
x
2
12
+
x
4
384
− · · ·
2. Find the power series solution of the
Bessel function: x 2 y + x y +
x 2 − v 2
y = 0
in terms of the Bessel function J 3 (x) when
v = 3. Give the answer up to and including the
term in x 7 .
⎡
⎢
⎢
⎣
y = AJ 3 (x) =
x
2
3
1
4
−
x 2
2 2 5
+
x 4
2 5 6
− · · ·
⎤
⎥
⎥
⎦
3. Evaluate the Bessel functions J 0 (x) and J 1 (x)
when x = 1, correct to 3 decimal places.
[J 0 (x) = 0.765, J 1 (x) = 0.440]
52.7 Legendre’s equation and
Legendre polynomials
Another important differential equation in physics
and engineering applications is Legendre’s equation
of the form: (1 − x 2 )
d
2 y
dx 2 − 2x
d y
dx
+ k(k + 1)y = 0 or
(1 − x 2 )y − 2x y + k(k + 1)y = 0 where k is a real
constant.
Problem 12. Determine the general power series
solution of Legendre’s equation.
To solve Legendre’s equation
(1 − x 2 )y − 2x y + k(k + 1)y = 0 using the Frobenius
method:
(i) Let a trial solution be of the form
y = x
c
a 0 + a 1 x + a 2 x
2
+ a 3 x
3
+· · · + a r x
r
+ · · ·
(43)
where a 0 = 0,
i.e. y = a 0 x
c
+ a 1 x
c+1
+ a 2 x
c+2
+ a 3 x
c+3
+ · · · + a r x
c+r
+ · · ·
(44)
(ii) Differentiating equation (44) gives:
y
= a 0 cx
c−1
+ a 1 (c + 1)x
c
+ a 2 (c + 2)x
c+1
+ · · ·
+ a r (c + r)x
c+r−1
+ · · ·
and y
= a 0 c(c − 1)x
c−2
+ a 1 c(c + 1)x
c−1
+ a 2 (c + 1)(c + 2)x
c
+ · · ·
+ a r (c + r − 1)(c + r)x
c+r−2
+ · · ·
(iii) Substituting y, y and y into each term of the
given equation:
1 − x 2
y − 2x y + k(k + 1)y = 0 gives:
a 0 c(c − 1)x
c−2
+ a 1 c(c + 1)x
c−1
+ a 2 (c + 1)(c + 2)x
c
+ · · ·
+ a r (c + r − 1)(c + r)x
c+r−2
+ · · ·
− a 0 c(c − 1)x
c
− a 1 c(c + 1)x
c+1
− a 2 (c + 1)(c + 2)x
c+2
− · · ·
− a r (c + r − 1)(c + r)x
c+r
− · · · − 2a 0 cx
c
− 2a 1 (c + 1)x
c+1
− 2a 2 (c + 2)x
c+2
− · · ·
− 2a r (c + r)x
c+r
− · · · + k
2 a 0 x
c
+ k
2 a 1 x
c+1
+ k
2 a 2 x
c+2
+ · · · + k
2 a r x
c+r
+ · · · + ka 0 x
c
+ ka 1 x
c+1
+ · · ·
+ ka r x
c+r
+ · · · = 0
(45)
(iv) The indicial equation is obtained by equating the
coefficient of the lowest power of x (i.e. x c−2 ) to
zero. Hence, a 0 c(c − 1) = 0 from which, c = 0 or
c = 1 since a 0 = 0.
For the term in x c−1 , i.e. a 1 c(c + 1) = 0 With
c = 1, a 1 = 0; however, when c = 0, a 1 is indeterminate, since any value of a 1 combined with
the zero value of c would make the product zero.
For the term in x c+r ,
a r+2 (c + r + 1)(c + r + 2) −a r (c + r − 1)
(c + r) − 2a r (c + r) + k 2 a r + ka r = 0
Now try the following exercise
Exercise 197 Further problems on Bessel’s
equation and Bessel’s functions
1. Determine the power series solution of Bessel’s equation: x 2 d 2 y
dx 2 + x
d y
dx
+ (x 2 −v 2 )y = 0
when v = 2, up to and including the term in x 6 .
y = Ax 2
1 −
x
2
12
+
x
4
384
− · · ·
2. Find the power series solution of the
Bessel function: x 2 y + x y +
x 2 − v 2
y = 0
in terms of the Bessel function J 3 (x) when
v = 3. Give the answer up to and including the
term in x 7 .
⎡
⎢
⎢
⎣
y = AJ 3 (x) =
x
2
3
1
4
−
x 2
2 2 5
+
x 4
2 5 6
− · · ·
⎤
⎥
⎥
⎦
3. Evaluate the Bessel functions J 0 (x) and J 1 (x)
when x = 1, correct to 3 decimal places.
[J 0 (x) = 0.765, J 1 (x) = 0.440]
52.7 Legendre’s equation and
Legendre polynomials
Another important differential equation in physics
and engineering applications is Legendre’s equation
of the form: (1 − x 2 )
d
2 y
dx 2 − 2x
d y
dx
+ k(k + 1)y = 0 or
(1 − x 2 )y − 2x y + k(k + 1)y = 0 where k is a real
constant.
Problem 12. Determine the general power series
solution of Legendre’s equation.
To solve Legendre’s equation
(1 − x 2 )y − 2x y + k(k + 1)y = 0 using the Frobenius
method:
(i) Let a trial solution be of the form
y = x
c
a 0 + a 1 x + a 2 x
2
+ a 3 x
3
+· · · + a r x
r
+ · · ·
(43)
where a 0 = 0,
i.e. y = a 0 x
c
+ a 1 x
c+1
+ a 2 x
c+2
+ a 3 x
c+3
+ · · · + a r x
c+r
+ · · ·
(44)
(ii) Differentiating equation (44) gives:
y
= a 0 cx
c−1
+ a 1 (c + 1)x
c
+ a 2 (c + 2)x
c+1
+ · · ·
+ a r (c + r)x
c+r−1
+ · · ·
and y
= a 0 c(c − 1)x
c−2
+ a 1 c(c + 1)x
c−1
+ a 2 (c + 1)(c + 2)x
c
+ · · ·
+ a r (c + r − 1)(c + r)x
c+r−2
+ · · ·
(iii) Substituting y, y and y into each term of the
given equation:
1 − x 2
y − 2x y + k(k + 1)y = 0 gives:
a 0 c(c − 1)x
c−2
+ a 1 c(c + 1)x
c−1
+ a 2 (c + 1)(c + 2)x
c
+ · · ·
+ a r (c + r − 1)(c + r)x
c+r−2
+ · · ·
− a 0 c(c − 1)x
c
− a 1 c(c + 1)x
c+1
− a 2 (c + 1)(c + 2)x
c+2
− · · ·
− a r (c + r − 1)(c + r)x
c+r
− · · · − 2a 0 cx
c
− 2a 1 (c + 1)x
c+1
− 2a 2 (c + 2)x
c+2
− · · ·
− 2a r (c + r)x
c+r
− · · · + k
2 a 0 x
c
+ k
2 a 1 x
c+1
+ k
2 a 2 x
c+2
+ · · · + k
2 a r x
c+r
+ · · · + ka 0 x
c
+ ka 1 x
c+1
+ · · ·
+ ka r x
c+r
+ · · · = 0
(45)
(iv) The indicial equation is obtained by equating the
coefficient of the lowest power of x (i.e. x c−2 ) to
zero. Hence, a 0 c(c − 1) = 0 from which, c = 0 or
c = 1 since a 0 = 0.
For the term in x c−1 , i.e. a 1 c(c + 1) = 0 With
c = 1, a 1 = 0; however, when c = 0, a 1 is indeterminate, since any value of a 1 combined with
the zero value of c would make the product zero.
For the term in x c+r ,
a r+2 (c + r + 1)(c + r + 2) −a r (c + r − 1)
(c + r) − 2a r (c + r) + k 2 a r + ka r = 0