Download ODE Homework 6

ODE Homework 6
5.2. Series Solutions Near an Ordinary Point, Part I
1. Seek power series solution of the equation
y 00 + k 2 x2 y = 0,
k a constant
about the the point x0 = 0. Find the recurrence relation; also
find the first four terms in each of two solutions y1 and y2 (unless
the series terminates sooner). By evaluating the Wronskian
W (y1 , y2 )(x0 ), show that y1 and y2 form a fundamental set of
solutions. If possible, find the general terms in each solutions.
[§5.2 #4]
∞
P
Sol. Let y =
an xn . Then
n=0
y 00 =
∞
X
n(n − 1)an xn−2 =
n=2
∞
X
(n + 2)(n + 1)an+2 xn
n=0
Substituting into the equation we have
∞
∞
X
X
n
2 2
(n + 2)(n + 1)an+2 x + k x
an x n = 0
n=0
n=0
Before proceeding, write
2 2
k x
∞
X
n
an x =
n=0
∞
X
k 2 an−2 xn
n=2
It follows that
∞
X
2a2 + 6a3 x +
[(n + 2)(n + 1)an+2 + k 2 an−2 ]xn = 0
n=2
We obtain a2 = a3 = 0 and the recurrence relation
an+2 = −
k2
an−2 , n = 2, 3, · · ·
(n + 2)(n + 1)
According to the recurrence relation we have that a4m+2 =
a4m+3 = 0, m = 0, 1, 2, · · · , and
k2
k2
k4
a0 , a8 = −
a4 =
a0 , · · ·
3·4
7·8
3·4·7·8
k2
k2
k4
a5 = −
a1 , a9 = −
a5 =
a1 , · · ·
4·5
8·9
4·5·8·9
a4 = −
By setting a0 = 1, a1 = 0 and a0 = 0, a1 = 1 respectively, we
get
k2 4
k4 8
k 6 12
x +
x −
x + ···
12
672
88704
k2 5
k4 9
k6
y2 (x) = x − x +
x −
x13 + · · ·
20
1440
224640
1 0 = 1, which shows
It is easy to see that W (y1 , y2 )(0) = 0 1 that y1 , y2 form a fundamental set of solutions. Moreover,
y1 (x) = 1 −
k2
(−1)m k 2m
a4m−4 = · · · =
a0
4m(4m − 1)
3 · 4 · 7 · 8 · · · (4m − 1) · 4m
k2
(−1)m k 2m
=−
a4m−3 = · · · =
a1
4m(4m + 1)
4 · 5 · 8 · 9 · · · (4m − 3)(4m + 1)
a4m = −
a4m+1
so
∞
X
(−1)m k 2m
y1 (x) = 1 +
x4m
3 · 4 · 7 · 8 · · · (4m − 1) · 4m
m=1
y2 (x) = x +
∞
X
(−1)m k 2m
x4m+1
4
·
5
·
8
·
9
·
·
·
(4m
−
3)(4m
+
1)
m=1
In fact, y1 and y2 can be written as
∞
∞
X
X
(−1)m k 2m Γ 54 4m+1
(−1)m k 2m Γ 34 4m
x , y2 (x) = x+
x
y1 (x) = 1+
4m m!Γ 4m+3
4m m!Γ 4m+5
2
m=1 2
m=1
4
4
2. Seek power series solution of the equation
(1 + x2 )y 00 − 4xy 0 + 6y = 0
about the the point x0 = 0. Find the recurrence relation; also
find the first four terms in each of two solutions y1 and y2 (unless
the series terminates sooner). By evaluating the Wronskian
W (y1 , y2 )(x0 ), show that y1 and y2 form a fundamental set of
solutions. If possible, find the general terms in each solutions.
[§5.2 #9]
Sol. Let y =
∞
P
an xn . Then
n=0
y0 =
00
y =
∞
X
nan xn−1 =
n=1
∞
X
∞
X
(n + 1)an+1 xn
n=0
n(n − 1)an x
n−2
n=2
=
∞
X
(n + 2)(n + 1)an+2 xn
n=0
Substituting into the equation we have
2
(1+x )
∞
X
n
(n+2)(n+1)an+2 x −4x
n=0
∞
X
n
(n+1)an+1 x +6
n=0
∞
X
an x n = 0
n=0
Before proceeding, write
x
2
∞
X
n
(n + 2)(n + 1)an+2 x =
n=0
∞
X
n(n − 1)an xn
n=2
∞
∞
X
X
nan xn
(n + 1)an+1 xn =
x
n=1
n=0
It follows that
∞
X
2(3a0 +a2 )+2(a1 +3a3 )x+ [(n+2)(n+1)an+2 +n(n−1)an −4nan +6an ]xn = 0
n=2
We obtain a2 = −3a0 , a3 = − 13 a1 and the recurrence relation
an+2 = −
n(n + 1) − 4n + 6
(n − 2)(n − 3)
an = −
an , n = 2, 3, · · ·
(n + 2)(n + 1)
(n + 1)(n + 2)
Observe that for n = 2, 3 we get a4 = a5 = 0, and thus
an = 0, ∀ n ≥ 4. Hence the general solution for the equation is
of the form
1
y(x) = a0 + a1 x − 3a0 x2 − a1 x3
3
Letting a0 = 1, a1 = 0, and a0 = 0, a1 = 1 we get respectively
y1 (x) = 1 − 3x2 ,
1
y2 (x) = x − x3
3
The Wronskian
3
1 − 3x2 x − x3
W (y1 , y2 )(x) = −6x
1 − x2
= (x2 + 1)2
Hence W (y1 , y2 )(0) = 1 which shows that y1 , y2 form a fundamental set of solutions.
3. Seek power series solution of the equation
2y 00 + (x + 1)y 0 + 3y = 0
about the the point x0 = 2. Find the recurrence relation; also
find the first four terms in each of two solutions y1 and y2 (unless
the series terminates sooner). By evaluating the Wronskian
W (y1 , y2 )(x0 ), show that y1 and y2 form a fundamental set of
solutions. If possible, find the general terms in each solutions.
[§5.2 #14]
∞
P
Sol. Let y =
an (x − 2)n . Then
n=0
0
y =
y 00 =
∞
X
n−1
nan (x − 2)
n=1
∞
X
∞
X
=
(n + 1)an+1 (x − 2)n
n=0
n(n − 1)an (x − 2)n−2 =
∞
X
(n + 2)(n + 1)an+2 (x − 2)n
n=0
n=2
Substituting into the equation we have
∞
∞
∞
X
X
X
an (x−2)n = 0
(n+1)an+1 (x−2)n +3
(n+2)(n+1)an+2 (x−2)n +(x+1)
2
n=0
n=0
n=0
Before proceeding, write
∞
X
(x + 1)
(n + 1)an+1 (x − 2)n
n=0
∞
X
= (x − 2 + 3)
(n + 1)an+1 (x − 2)n
n=0
∞
∞
X
X
n+1
=
(n + 1)an+1 (x − 2)
+3
(n + 1)an+1 (x − 2)n
=
n=0
∞
X
n=0
nan (x − 2)n + 3
n=1
∞
X
(n + 1)an+1 (x − 2)n
n=0
It follows that
∞
X
4a2 +3a1 +3a0 + [2(n+2)(n+1)an+2 +nan +3(n+1)an+1 +3an ](x−2)n = 0
n=1
We obtain the recurrence relation
an+2 = −
3
n+3
an+1 −
an , n = 0, 1, 2 · · ·
2(n + 2)
2(n + 2)(n + 1)
According to the recurrence relation, we have
3
a2 = − a1 −
4
1
a3 = − a2 −
2
3
a4 = − a3 −
8
..
.
3
a0
4
1
1
3
a1 = a1 + a0
3
24
8
5
9
1
a2 = a1 + a0
24
64
64
By setting a0 = 1, a1 = 0 and a0 = 0, a1 = 1 respectively, we
get
3
1
3
y1 (x) = 1 − (x − 2)2 + (x − 2)3 + (x − 2)4 + · · ·
4
8
64
3
1
9
2
y2 (x) = (x − 2) − (x − 2) + (x − 2)3 + (x − 2)4 + · · ·
4
24
64
1 0 = 1, which shows
It is easy to see that W (y1 , y2 )(2) = 0 1 that y1 , y2 form a fundamental set of solutions.
4. (a) By making the change of variable x − 1 = t and assuming
that y has a Taylor series in powers of t, find two series
solutions of
y 00 + (x − 1)2 y 0 + (x2 − 1)y = 0
in powers of x − 1.
(b) Show that you obtain the same result by assuming that y
has a Taylor series in powers of x − 1 and also expressing
the coefficient x2 − 1 in powers of x − 1.
[§5.2 #19]
Sol.
(a) Let x − 1 = t, then x = t + 1. Let z(t) = y(t + 1), then the
equation can be transformed to
z 00 + t2 z 0 + (t2 + 2t)z = 0
Let z =
∞
P
an tn . Then
n=0
0
∞
X
00
n=1
∞
X
z =
z =
n−1
nan t
∞
X
=
(n + 1)an+1 tn
n=0
n−2
n(n − 1)an t
=
n=2
∞
X
(n + 2)(n + 1)an+2 tn
n=0
Substituting into the equation we have
∞
∞
∞
X
X
X
(n+1)an+1 tn +(t2 +2t)
an tn = 0
(n+2)(n+1)an+2 tn +t2
n=0
n=0
n=0
Before proceeding, write
t
2
∞
X
n
(n + 1)an+1 t =
(t + 2t)
(n − 1)an−1 tn
n=2
n=0
2
∞
X
∞
X
n
an t =
n=0
∞
X
n=2
n
an−2 t +
∞
X
2an−1 tn
n=1
It follows that
∞
X
2a2 +2(a0 +3a3 )t+ [(n+2)(n+1)an+2 +(n−1)an−1 +an−2 +2an−1 ]tn = 0
n=2
We obtain a2 = 0, a3 = − 13 a0 and the recurrence relation
an+2 = −
1
1
an−1 −
an−2 , n = 2, 3 · · ·
n+2
(n + 2)(n + 1)
According to the recurrence relation, we have
1
a4 = − a1 −
4
1
a5 = − a2 −
5
1
a6 = − a3 −
6
1
a7 = − a4 −
7
..
.
1
a0
12
1
1
a1 = − a1
20
20
1
1
a2 = a0
30
18
1
1
5
a3 = a1 +
a0
42
28
252
By setting a0 = 1, a1 = 0 and a0 = 0, a1 = 1 respectively,
we get
1
1
1
5 7
z1 (t) = 1 − t3 − t4 + t6 +
t + ···
3
12
18
252
1
1
1
z2 (t) = t − t4 − t5 + t7 + · · ·
4
20
28
or
1
1
1
5
y1 (x) = 1 − (x − 1)3 − (x − 1)4 + (x − 1)6 +
(x − 1)7 + · · ·
3
12
18
252
1
1
1
y2 (x) = (x − 1) − (x − 1)4 − (x − 1)5 + (x − 1)7 + · · ·
4
20
28
(b) Since the Taylor series for x2 − 1 about x = 1 is
x2 − 1 = 2(x − 1) + (x − 1)2
Then the equation becomes
y 00 + (x − 1)2 y 0 + [2(x − 1) + (x − 1)2 ]y = 0
which is identical to the transformed equation with t =
x − 1.
5. The Hermite Equation. The equation
y 00 − 2xy 0 + λy = 0,
−∞ < x < ∞,
where λ is a constant, is known as the Hermite equation. It is
an important equation in mathematical physics.
(a) Find the first four terms in each of two solutions about x =
0 and show that they form a fundamental set of solutions.
(b) Observe that if λ is a nonnegative even integer, then one
or the other of the series solutions terminates and becomes a polynomial. Find the polynomial solutions for
λ = 0, 2, 4, 6, 8, and 10. Note that each polynomial is determined only up to a multiplicative constant.
(c) The Hermite polynomial Hn (x) is defined as the polynomial solution of the Hermite equation with λ = 2n for
which the coefficient of xn is 2n . Find H0 (x), · · · , H5 (x).
[§5.2 #21]
Sol.
(a) Let y =
∞
P
an xn . Then
n=0
0
y =
y 00 =
∞
X
nan x
n=1
∞
X
n−1
=
∞
X
(n + 1)an+1 xn
n=0
n(n − 1)an xn−2 =
n=2
∞
X
(n + 2)(n + 1)an+2 xn
n=0
Substituting into the equation we have
∞
∞
∞
X
X
X
n
n
(n+2)(n+1)an+2 x −2x
(n+1)an+1 x +λ
an x n = 0
n=0
n=0
n=0
Before proceeding, write
2x
∞
X
n
(n + 1)an+1 x =
n=0
∞
X
2nan xn
n=1
It follows that
∞
X
[(n + 2)(n + 1)an+2 − 2nan + λan ]xn = 0
2a2 + λa0 +
n=1
We obtain the recurrence relation
an+2 =
2n − λ
an , n = 0, 1, 2, · · ·
(n + 2)(n + 1)
Thus
λ
2·2−λ
λ(λ − 2 · 2)
a2 = − a0 , a 4 =
a2 =
a0 , · · ·
2
3·4
2·3·4
2·1−λ
2·3−λ
(λ − 2 · 1)(λ − 2 · 3)
a3 =
a1 , a 5 =
a3 =
a1 , · · ·
2·3
4·5
2·3·4·5
More precisely, we have
a2m =
a2m+1
2(2m − 2) − λ
λ(λ − 4) · · · (λ − 4m + 4)
a2m−2 = · · · = (−1)m
a0
2m(2m − 1)
(2m)!
2(2m − 1) − λ
(λ − 2)(λ − 6) · · · (λ − 4m + 2)
=
a2m−1 = · · · = (−1)m
a1
2m(2m + 1)
(2m + 1)!
By setting a0 = 1, a1 = 0 and a0 = 0, a1 = 1 respectively,
we get
λ 2 λ(λ − 4) 4 λ(λ − 4)(λ − 8) 6
x +
x −
x + ···
2!
4!
6!
∞
X
λ(λ − 4) · · · (λ − 4m + 4) 2m
=1+
(−1)m
x
(2m)!
m=1
y1 (x) = 1 −
λ − 2 3 (λ − 2)(λ − 6) 5 (λ − 2)(λ − 6)(λ − 10) 7
x +
x −
x + ···
3!
5!
7!
∞
X
(λ − 2)(λ − 6) · · · (λ − 4m + 2) 2m+1
x
=x+
(−1)m
(2m + 1)!
m=1
1 0 = 1, which
It is easy to see that W (y1 , y2 )(0) = 0 1 shows that y1 , y2 form a fundamental set of solutions.
(b) If λ = 2k with k = 0, 1, 2, · · · ,, then the recurrence relation
becomes
2n − 2k
an+2 =
an , n = 0, 1, 2, · · ·
(n + 2)(n + 1)
y2 (x) = x −
which implies that ak+2 = 0 and hence ak+2` = 0, ` =
0, 1, 2, · · · . More precisely, if k = 2j, then y1 (x) becomes a
polynomial with degree at most k and if k = 2j + 1, then
y2 (x) becomes a polynomial with degree at most k. In particular, we obtain the polynomial solutions corresponding
to λ = 0, 2, 4, 6, 8, 10 are
λ = 0 ⇒ y1 (x) = 1,
λ = 2 ⇒ y2 (x) = x
λ = 4 ⇒ y1 (x) = 1 − 2x2 ,
λ = 6 ⇒ y2 (x) = x − 32 x3
4 4
2
λ = 8 ⇒ y1 (x) = 1 − 4x + 3 x λ = 10 ⇒ y2 (x) = x − 43 x3 +
(c) If λ = 2n. By letting a0 = a1 = 1, we have that
2n(2n − 4) · · · (2n − 4m + 4)
a0
(2m)!
(2n − 2)(2n − 6) · · · (2n − 4m + 2)
= · · · = (−1)m
a1
(2m + 1)!
a2m = · · · = (−1)m
a2m+1
for m = 1, 2, · · · , b n2 c. It follows that the coefficient of xn ,
in y1 and y2 , is
(
4` `!
(−1)` (2`)!
for n = 2`
an =
4` `!
(−1)` (2`+1)!
for n = 2` + 1
4 5
x
15
Hence, we have that
(
2n · (−1)k ` (2`)!
y1 (x)
for n = 2`
y (x) = (−1)k ` (2`)!
`!
4` `! 1
Hn (x) =
y2 (x) for n = 2` + 1
2n · (−1)` (2`+1)!
y2 (x) = (−1)k ` 2·(2`+1)!
`!
4` `!
Therefore, the Hermite polynomials H0 (x), · · · , H5 (x) are
H0 (x) = 1,
H2 (x) = 4x2 − 2,
H4 (x) = 16x4 − 48x2 + 12,
H1 (x) = 2x
H3 (x) = 8x3 − 12x
H5 (x) = 32x5 − 160x3 + 120x
5.3. Series Solutions Near an Ordinary Point, Part II
6. The Chebyshev Equation. The Chebyshev differential equation is
(1 − x2 )y 00 − xy 0 + α2 y = 0,
where α is a constant.
(a) Determine two solutions in powers of x for |x| < 1 and
show that they form a fundamental set of solutions.
(b) Show that if α is a nonnegative integer n, then there is a
polynomial solution of degree n. These polynomials, when
properly normalized, are called the Chebyshev polynomials. They are very useful in problems that require a polynomial approximation to a function defined on −1 ≤ x ≤ 1.
(c) Find a polynomial solution for each of the cases α = n =
0, 1, 2, 3.
[§5.3 #10]
Sol.
∞
P
(a) Let y =
an xn . Then
n=0
0
∞
X
00
n=1
∞
X
y =
y =
nan x
n−1
∞
X
=
(n + 1)an+1 xn
n=0
n(n − 1)an x
n=2
n−2
=
∞
X
(n + 2)(n + 1)an+2 xn
n=0
Substituting into the equation we have
2
(1−x )
∞
X
n=0
n
(n+2)(n+1)an+2 x −x
∞
X
n=0
n
(n+1)an+1 x +α
2
∞
X
n=0
an x n = 0
It follows that
∞
X
(2a2 +α a0 )+(6a3 +(α −1)a1 )x+ [(n+2)(n+1)an+2 +(α2 −n2 )an ]xn = 0
2
2
n=2
We obtain the recurrence relation
n2 − α2
an+2 =
an , n = 0, 1, 2, · · ·
(n + 2)(n + 1)
According to the recurrence relation we have that
α2
22 − α2
α2 (22 − α2 )
a0 , a 4 =
a2 = −
a0 , · · ·
2!
4·3
4!
32 − α2
(32 − α2 )(12 − α2 )
12 − α2
a1 , a 5 =
a5 =
a1 , · · ·
a3 =
3!
5·4
5!
That is,
a2 = −
α2 (22 − α2 ) · · · ((2m − 2)2 − α2 )
a2m = −
a0 , m = 1, 2, · · ·
(2m)!
(12 − α2 )(32 − α2 ) · · · ((2m − 1)2 − α2 )
a2m+1 =
a1 , m = 1, 2, · · ·
(2m + 1)!
Therefore we get two linearly independent solutions by setting (a0 , a1 ) = (1, 0) and (a0 , a1 ) = (0, 1) respectively
y1 (x) = 1 −
∞
X
α2 (22 − α2 ) · · · ((2m − 2)2 − α2 ) 2m
x
(2m)!
m=1
∞
X
(12 − α2 )(32 − α2 ) · · · ((2m − 1)2 − α2 ) 2m+1
y2 (x) = x +
x
(2m + 1)!
m=1
(b) If α = n a nonnegative integer. For n = 2k, then
k
X
α2 (22 − α2 ) · · · ((2m − 2)2 − α2 ) 2m
y1 (x) = Pn (x) = 1−
x
(2m)!
m=1
is a polynomial with degree 2k = n. If n = 2k + 1, then
k
X
(12 − α2 )(32 − α2 ) · · · ((2m − 1)2 − α2 ) 2m+1
x
y2 (x) = x+
(2m + 1)!
m=1
is a polynomial with degree 2k + 1 = n.
(c) The polynomial solutions corresponding to α = 0, 1, 2, 3
are
α = 0 ⇒ P0 (x) = 1,
α = 1 ⇒ P1 (x) = x
2
α = 2 ⇒ P2 (x) = 1 − 2x , α = 3 ⇒ P3 (x) = x − 43 x3
7. The Legendre Equation. The Legendre equation is
(1 − x2 )y 00 − 2xy 0 + α(α + 1)y = 0.
The point x = 0 is an ordinary point of the equation, and the
distance from the origin to the nearest zero of P (x) = 1 − x2
is 1. Hence the radius of convergence of series solutions about
x = 0 is at leat 1. Also notice that we need to consider only
α > −1 because if α ≤ −1, then the substituting α = −(1 + γ),
where γ ≥ 0, leads to the Legendre equation (1 − x2 )y 00 − 2xy 0 +
γ(γ + 1)y = 0.
Show that two solutions of the Legendre equation for |x| < 1
are
α(α + 1) 2 α(α − 2)(α + 1)(α + 3) 4
y1 (x) = 1 −
x +
x
2!
4!
∞
X
α · · · (α − 2m + 2)(α + 1) · · · (α + 2m − 1) 2m
(−1)m
+
x
(2m)!
m=3
(α − 1)(α + 2) 3 (α − 1)(α − 3)(α + 2)(α + 4) 5
x +
x
3!
5!
∞
X
(α − 1) · · · (α − 2m + 1)(α + 2) · · · (α + 2m) 2m+1
+
(−1)m
x
(2m + 1)!
m=3
y2 (x) = x −
[§5.3 #22]
∞
P
Proof. Let y =
an xn . Then
n=0
y0 =
00
y =
∞
X
nan xn−1 =
n=1
∞
X
∞
X
(n + 1)an+1 xn
n=0
n(n − 1)an x
n−2
n=2
=
∞
X
(n + 2)(n + 1)an+2 xn
n=0
Substituting into the equation we have
∞
∞
∞
X
X
X
(1−x2 )
(n+2)(n+1)an+2 xn −2x
(n+1)an+1 xn +α(α+1)
an x n = 0
n=0
n=0
n=0
It follows that
(2a2 + α(α + 1)a0 ) + (6a3 + (α − 1)(α + 2)a1 )x
∞
X
+
[(n + 2)(n + 1)an+2 + (α − n)(α + n + 1)an ]xn = 0
n=2
We obtain the recurrence relation
(α − n)(α + n + 1)
an+2 = −
an , n = 0, 1, 2, · · ·
(n + 2)(n + 1)
According to the recurrence relation we have that
α(α − 2)(α + 1)(α + 3)
α(α + 1)
a0 , a 4 = −
a0 , · · ·
2!
4!
(α − 1)(α + 2)
(α − 1)(α − 3)(α + 2)(α + 4)
a3 = −
a1 , a 5 =
a1 , · · ·
3!
5!
That is,
a2 = −
α · · · (α − 2m + 2)(a + 1) · · · (α + 2m − 1)
a0 , m = 1, 2, · · ·
(2m)!
(α − 1) · · · (α − 2m + 1)(a + 2) · · · (α + 2m)
a1 , m = 1, 2, · · ·
= (−1)m
(2m + 1)!
a2m = (−1)m
a2m+1
Therefore we get two linearly independent solutions by setting
(a0 , a1 ) = (1, 0) and (a0 , a1 ) = (0, 1) respectively
α(α + 1) 2 α(α − 2)(α + 1)(α + 3) 4
x +
x
2!
4!
∞
X
α · · · (α − 2m + 2)(α + 1) · · · (α + 2m − 1) 2m
+
(−1)m
x
(2m)!
m=3
y1 (x) = 1 −
(α − 1)(α + 2) 3 (α − 1)(α − 3)(α + 2)(α + 4) 5
x +
x
3!
5!
∞
X
(α − 1) · · · (α − 2m + 1)(α + 2) · · · (α + 2m) 2m+1
(−1)m
x
+
(2m + 1)!
m=3
y2 (x) = x −
5.4. Euler Equations; Regular Singular Points
8. Find all singular points of the equation
2
x2 (1 − x2 )y 00 + y 0 + 4y = 0
x
and determine whether each one is regular or irregular.
[§5.4 #19]
Sol. Rewrite the equation as the form
y 00 +
x3 (1
2
4
y0 + 2
y=0
2
−x )
x (1 − x2 )
2
and let p(x) = x3 (1−x
2 ) , q(x) =
are x = 0, ±1. At x = 0
4
.
x2 (1−x2 )
The singular points
2
x→0
− x2 )
which does not exist, so x = 0 is an irregular singular point. At
x = −1,
−2
= −1
lim (x + 1)p(x) = lim 3
x→−1
x→−1 x (x − 1)
−4(x + 1)
lim (x + 1)2 q(x) = lim 2
=0
x→−1
x→−1 x (x − 1)
so x = −1 is an regular singular point. At x = 1,
−2
= −1
lim (x − 1)p(x) = lim 3
x→1
x→−1 x (x + 1)
4(x − 1)
lim (x − 1)2 q(x) = lim 2
=0
x→1
x→−1 x (x + 1)
so x = 1 is an regular singular point.
lim xp(x) =
x2 (1
9. Find all singular points of the equation
x2 y 00 + xy 0 + (x2 − ν 2 )y = 0
Bessel equation
and determine whether each one is regular or irregular.
[§5.4 #22]
Sol. Rewrite the equation as the form
1
x2 − ν 2
y=0
y 00 + y 0 +
x
x2
and let p(x) = x1 , q(x) =
x = 0. At x = 0,
x2 −ν 2
.
x2
The only singular point is
lim xp(x) = lim 1 = 1
x→0
x→0
2
lim x q(x) = lim (x2 − ν 2 ) = −ν 2
x→0
x→0
so x = 0 is an regular singular point.
10. Find all singular points of the equation
y 00 + ln |x|y 0 + 3xy = 0
and determine whether each one is regular or irregular.
[§5.4 #29]
Sol. Since ln |x| is singular at x = 0, so the only singular point
is x = 0. Furthermore, since ln |x| is not analytic at x = 0, so
does the function x ln |x|. Hence x = 0 is an irregular singular
point.
11. Find all singular points of the equation
x2 y 00 + 2(ex − 1)y 0 + (e−x cos x)y = 0
and determine whether each one is regular or irregular.
[§5.4 #30]
Sol. Rewrite the equation as the form
2(ex − 1) 0 e−x cos x
y 00 +
y +
y=0
x2
x2
x
−x
x
. The only singular point
and let p(x) = 2(ex2−1) , q(x) = e xcos
2
is x = 0. At x = 0,
2(ex − 1)
2ex
lim xp(x) = lim
= lim
=2
x→0
x→0
x→0 1
x
lim x2 q(x) = lim e−x cos x = 1
x→0
x→0
so x = 0 is an regular singular point.
12. Find all singular points of the equation
x2 y 00 − (3 sin x)y 0 + (1 + x2 )y = 0
and determine whether each one is regular or irregular.
[§5.4 #32]
Sol. Rewrite the equation as the form
3 sin x 0 1 + x2
y +
y=0
x2
x2
2
x
, q(x) = 1+x
. The only singular point is
and let p(x) = − 3 sin
x2
x2
x = 0. At x = 0,
3 sin x
3 cos x
lim xp(x) = − lim
= − lim
= −3
x→0
x→0
x→0
x
1
lim x2 q(x) = lim (1 + x2 ) = 1
y 00 −
x→0
x→0
so x = 0 is an regular singular point.
13. Consider the Euler equation x2 y 00 + αxy 0 + βy = 0. Find conditions on α and β so that:
(a) All solutions approach zero as x → 0.
(b) All solutions are bounded as x → 0.
(c) All solutions approach zero as x → ∞.
(d) All solutions are bounded as x → ∞.
(e) All solutions are bounded both as x → 0 and as x → ∞.
[§5.4 #39]
Sol. Note that the general solution of the equation in any interval containing the origin is determined by the roots r1 and
r2 of the equation
F (r) = r(ρ − 1) + αr + β = r2 + (α − 1)r + β
that is, r1 + r2 = 1 − α, r1 r2 = β, more precisely
p
−(α − 1) ± (α − 1)2 − 4β
r1 , r2 =
2
And

if r1 , r2 ∈ R, r1 6= r2
 c1 |x|r1 + c2 |x|r2 ,
r1
r1
c
|x|
+
c
|x|
ln
|x|,
y(x) =
1
2
if r1 , r2 ∈ R, r1 = r2

λ
λ
c1 |x| cos µ ln |x| + c2 |x| sin µ ln |x| , if r1 , r2 = λ ± iµ
, µ = 4β − (α − 1)2 .
where λ = − α−1
2
(a) For solutions to approach zero as x → 0, we need that
r1 , r2 > 0 ⇒ α < 1, β > 0,
if r1 , r2 ∈ R, r1 6= r2
r1 = r2 > 0 ⇒ α < 1, β > 0, if r1 , r2 ∈ R, r1 = r2
λ > 0 ⇒ α < 1,
if r1 , r2 = λ ± iµ
Thus, if α < 1 and β > 0, the all solutions approach zero
as x → 0.
(b) For solutions are bounded as x → 0, we need that
r1 , r2 ≥ 0 ⇒ α < 1, β ≥ 0,
if r1 , r2 ∈ R, r1 6= r2
r1 = r2 > 0 ⇒ α < 1, β > 0,
if r1 , r2 ∈ R, r1 = r2
λ > 0 or λ = 0, µ 6= 0 ⇒ α < 1 or α = 1, β > 0, if r1 , r2 = λ ± iµ
Thus, if α < 1 and β ≥ 0 or α = 1 and β > 0, the all
solutions are bounded as x → 0.
(c) For solutions to approach zero as x → ∞, we need that
r1 , r2 < 0 ⇒ α > 1, β > 0,
if r1 , r2 ∈ R, r1 6= r2
r1 = r2 < 0 ⇒ α > 1, β > 0, if r1 , r2 ∈ R, r1 = r2
λ < 0 ⇒ α > 1,
if r1 , r2 = λ ± iµ
Thus, if α > 1 and β > 0, the all solutions approach zero
as x → ∞.
(d) For solutions are bounded as x → ∞, we need that
r1 , r2 ≤ 0 ⇒ α > 1, β ≥ 0,
if r1 , r2 ∈ R, r1 6= r2
r1 = r2 < 0 ⇒ α > 1, β > 0,
if r1 , r2 ∈ R, r1 = r2
λ < 0 or λ = 0, µ 6= 0 ⇒ α > 1 or α = 1, β > 0, if r1 , r2 = λ ± iµ
Thus, if α > 1 and β ≥ 0 or α = 1 and β > 0, the all
solutions are bounded x → ∞.
(e) For solutions are bounded both as x → 0 and as x → ∞,
we need that
impossible,
if r1 , r2 ∈ R, r1 6= r2
impossible,
if r1 , r2 ∈ R, r1 = r2
λ = 0, µ 6= 0 ⇒ α = 1, β > 0, if r1 , r2 = λ ± iµ
Thus, if α = 1 and β > 0 and , the all solutions are bounded
both as x → 0 and as x → ∞.
14. Using the method of reduction of order, show that if r1 is a
repeated root of
r(r − 1) + αr + β = 0,
then xr1 and xr1 ln x are solutions of x2 y 00 + αxy 0 + βy = 0 for
x > 0. [§5.4 #40]
Proof. Assume that y = v(x)xr1 , then
y 0 = xr1 v 0 +r1 xr1 −1 v, y 00 = xr1 v 00 +2r1 xr1 −1 v 0 +r1 (r1 −1)xr1 −2 v
Substituting into the equation, we get
xr1 +2 v 00 + [2r1 + α]xr1 +1 v 0 + [r1 (r1 − 1) + αr1 + β]xr1 v = 0
If r1 is a repeated root of r(r − 1) + αr + β = 0, then
r1 (r1 − 1) + αr1 + β = 0, 2r1 + α − 1 = 0
Hence v satisfies the equation
xr1 +2 v 00 + xr1 +1 v 0 = xr1 +1 [xv 00 + v 0 ] = 0
and this immediately implies that v(x) = ln x, since x > 0.
Therefore the second solution for the Euler equation is xr1 ln x.
15. Show that the point x = 0 is a regular singular point of the
equation
2x2 y 00 + 3xy 0 − (1 + x)y = 0
∞
P
Try to find solutions of the form
an xn . Show that (except
n=0
for constant multiplies) there are no nonzero solutions of this
form in our problem. Thus, in this case, the general solution
can not be found in this manner. This is typical of equations
with singular points.
[§5.4 #42]
Proof. Rewrite the equation as the form
1+x
3
y 00 + y 0 −
y=0
2x
2x2
3
, q(x) = − 1+x
. The only singular point is
and let p(x) = 2x
2x2
x = 0. At x = 0,
3
3
lim xp(x) = lim =
x→0
x→0 2
2
1
1
+x
=−
lim x2 q(x) = − lim
x→0
x→0
2
2
∞
P
so x = 0 is a regular singular point. Let y =
an xn . Substin=0
tuting into the equation we have
∞
∞
∞
X
X
X
n
2
n
an x n = 0
(n+1)an+1 x −(1+x)
(n+2)(n+1)an+2 x +3x
2x
n=0
n=0
n=0
It follows that
−a0 + (2a1 − a0 )x +
∞
X
[2n(n − 1)an + 3nan − an − an−1 ]xn = 0
n=2
We obtain a0 = 0, a1 = 12 a0 = 0 and the recurrence relation
1
an−1 , n = 2, 3, · · ·
(2n − 1)(n + 1)
Thus we can conclude that an = 0, ∀ n = 0, 1, 2, · · · . Hence
y(x) = 0 is the only solution that can be obtained.
an =
5.5. Series Solutions Near a Regular Singular Point, Part I
16. (a) Show that the equation
1
x2 y 00 + xy 0 + x2 −
y=0
9
has a regular singular point at x = 0.
(b) Determine the indicial equation, the recurrence relation,
and the roots of the indicial equation.
(c) Find the series solution (x > 0) corresponding to the largest
root.
(d) If the roots are unequal and do not differ by an integer,
find the series solution corresponding to the smaller root
also.
[§5.5 #2]
Sol.
(a) Rewrite the equation as the form
1
9x2 − 1
y 00 + y 0 +
y=0
x
9x2
9x2 −1
.
9x2
and let p(x) = x1 , q(x) =
x = 0. At x = 0,
The only singular point is
lim xp(x) = lim 1 = 1
x→0
x→0
9x2 − 1
1
=−
x→0
x→0
9
9
so x = 0 is a regular singular point.
∞
P
(b) Let y =
an xn+r . Then
lim x2 q(x) = lim
n=0
0
y =
∞
X
(n+r)an x
n+r−1
∞
X
(n+r)(n+r−1)an xn+r−2
,y =
00
n=2
n=0
Substituting into the equation we have
∞
∞
∞
X
X
X
n+r
n+r
2 1
(n+r)(n+r−1)an x + (n+r)an x + x −
an xn+r = 0
9 n=0
n=0
n=0
It follows that
1 r+1
1 r a0 x + (r + 1)r + (r + 1) −
a1 x
r(r − 1) + r −
9
9
∞ h
i
X
1
+
(n + r)(n + r − 1) + (n + r) −
an + an−2 xn+r = 0
9
n=2
We obtain the indicial equation
r2 −
1
=0
9
with roots r = ± 13 . For either value of r it is necessary to
take a1 = 0 in order that the coefficient of xr+1 be zero.
Hence we get the recurrence relation
an+2 = −
1
an−2 , n = 2, 3, · · ·
(n + r)2 − 19
(c) For r = 13 , we have
an+2 = −
n+
1
1 2
3
−
1
9
an−2 =
−an−2
, n = 2, 3, · · ·
n n + 23
According to the recurrence relation and a1 = 0, we have
that a2m+1 = 0, m = 0, 1, 2, · · · . For n = 2m, we have that
−a2m−2
a2m−4
=
a2m = 2
1
4
2 m m+ 3
2 m(m − 1) m + 13 m − 1 + 23
(−1)m a0
= · · · = 2m
2 m! m + 13 · · · 1 + 31
By setting a0 = 1, we get
∞
X
1
y1 (x) = x 3 1 +
(−1)m
2m
x
22m m! m + 13 · · · 1 + 13
m=1
∞
X
(−1)m Γ 34
1
2m
= x3 1 +
x
22m m!Γ 3m+4
3
m=1
(d) Since r2 = − 13 6= r1 = 13 and r1 − r2 = 32 is not an integer,
we can calculate a second series solution corresponding to
r = − 31 . For r = − 31 , we have
an+2 = −
1
n−
1 2
3
−
1
9
an−2 =
−an−2
, n = 2, 3, · · ·
n n − 23
According to the recurrence relation and a1 = 0, we have
that a2m+1 = 0, m = 0, 1, 2, · · · . For n = 2m, we have that
−a2m−2
a2m−4
=
a2m = 2
1
4
2 m m− 3
2 m(m − 1) m − 13 m − 1 − 23
(−1)m a0
= · · · = 2m
2 m! m − 13 · · · 1 − 13
By setting a0 = 1, we get
∞
X
1
y1 (x) = x 3 1 +
(−1)m
2m
x
22m m! m − 13 · · · 1 − 13
m=1
∞
X
(−1)m Γ 23
− 13
2m
=x
1+
x
22m m!Γ 3m+2
3
m=1
17. The Legendre equation of order α is
(1 − x2 )y 00 − 2xy 0 + α(α + 1)y = 0.
It was shown that x = ±1 are regular singular points.
(a) Determine the indicial equation and its roots for the point
x = 1.
(b) Find a series solution in powers of x − 1 for x − 1 > 0.
Hint: Write 1 + x = 2 + (x − 1) and x = 1 + (x − 1).
Alternatively, make the change of variable x − 1 = t and
determine a series solutions in powers of t.
[§5.5 #11]
Sol.
(a) By letting x − 1 = t and let u(t) = y(t + 1), the Legendre
equation can be transformed to
(t2 + 2t)u00 (t) + 2(t + 1)u0 (t) − α(α + 1)u(t) = 0
Note that
2(t + 1)
2(t + 1)
= lim
=1
t→0
t→0 t + 2
t(t + 2)
α(α + 1)t
α(α + 1)
= − lim
=0
− lim t2 ·
t→0
t→0
t(t + 2)
t+2
lim t ·
so t = 0 is a regular singular point for above equation. Let
∞
P
u=
an tn+r . Then
n=0
u0 =
∞
∞
X
X
(n+r)(n+r−1)an tn+r−2
(n+r)an tn+r−1 , u00 =
n=2
n=0
Substituting into the equation we have
∞
∞
X
X
(n + r)(n + r − 1)an tn+r +
2(n + r)(n + r − 1)an tn+r−1
n=0
n=0
+
∞
X
n+r
2(n + r)an t
n=0
+
∞
X
n+r−1
2(n + r)an t
−
n=0
∞
X
α(α + 1)an xn+r = 0
n=0
It follows that
[2r(r − 1) + 2r]a0 tr−1
∞
X
+
2(n + r + 1)2 an+1 + [(n + r)(n + r + 1) − α(α + 1)]an tn+r = 0
n=0
We obtain the indicial equation 2r2 = 0 with repeated
roots r = 0.
(b) Since r = 0, the recurrence relation becomes
an+1 =
α(α + 1) − n(n + 1)
an , n = 0, 1, 2, · · ·
2(n + 1)2
According to the recurrence relation, we have that
α(α + 1) − (n − 1)n
an−1
2n2
[α(α + 1) − (n − 1)n][α(α + 1) − (n − 2)(n − 1)]
an−2
=
2
22 n(n − 1)
[α(α + 1) − n(n − 1)] · · · [α(α + 1) − 2 · 1] · α(α + 1)
= ··· =
a0
2
2n n!
an =
By setting a0 = 0 and replacing t = x − 1, we get the series
solution for the Legendre equation is
∞
X
[α(α + 1) − n(n − 1)] · · · [α(α + 1) − 2 · 1] · α(α + 1)
y1 (x) =
(x−1)n
2
n
2 n!
n=0
Note that since r = 0 is repeated root for the indicial
equation, there will be only one series solution of the form
∞
P
y=
an (x − 1)n+r .
n=0
18. The Chebyshev equation of order α is
(1 − x2 )y 00 − xy 0 + α2 y = 0,
where α is a constant.
(a) Show that x = 1 and x = −1 are regular singular points
and find the exponents at each of these singularities.
(b) Find two solutions about x = 1.
[§5.5 #12]
Sol.
(a) Rewrite the equation of the form
y 00 −
α2
x
0
y
+
y=0
1 − x2
1 − x2
2
x
α
and let p(x) = − 1−x
2 , q(x) = 1−x2 . There are two singular
points which are x = ±1. At x = 1,
x
1
=
x→1
x→1 x + 1
2
α(x
−
1)
q0 = lim (x − 1)2 q(x) = lim
=0
x→1
x→1 (x + 1)
p0 = lim (x − 1)p(x) = lim
At x = −1,
1
x
=
x→−1
x→−1 x − 1
2
α(x + 1)
q0 = lim (x + 1)2 q(x) = lim
=0
x→−1
x→−1 (x − 1)
p0 = lim (x + 1)p(x) = lim
Hence x = ±1 are regular singular points. Note that the
indicial equation is given by
1
=0
r(r − 1) + p0 r + q0 = r r −
2
with distinct real roots r = 0, r = 12 .
(b) Set t = x − 1 and u(t) = y(t + 1), the equation can be
transformed as
(t2 + 2t)u00 + (t + 1)u0 − α2 u = 0
By a. we know that t = 0 is a regular singular point for
∞
P
above equation. Let u =
an tn+r . Substituting into the
n=0
equation we have
∞
X
n+r
(n + r)(n + r − 1)an t
+
∞
X
2(n + r)(n + r − 1)an tn+r−1
n=0
n=0
+
∞
X
n=0
n+r
(n + r)an t
+
∞
X
n=0
n+r−1
(n + r)an t
−
∞
X
α2 an xn+r = 0
n=0
That is,
∞
∞
∞
X
X
1 n+r−1 X 2 n+r
2
n+r
(n+r) an t +
2(n+r) n+r− an t
−
α an t
=0
2
n=0
n=0
n=0
It follows that
h 1 i r−1
a0 t
2r r −
2
∞
X
1
+
2(n + r + 1) n + r +
an+1 + [(n + r)2 − α2 ]an tn+r = 0
2
n=0
we obtain r(2r − 1)a0 = 0 and the recurrence relation
an+1 =
α2 − (n + r)2
an , n = 0, 1, 2, · · ·
2(n + r + 1) n + r + 21
Assuming that a0 6= 0, then for r1 = 12 , according to the
recurrence relation,
4α2 − (2n − 1)2
[4α2 − (2n − 1)2 ][4α2 − (2n − 3)2 ]
an−1 =
an−2
4n(2n + 1)
42 n(n − 1)(2n + 1)(2n − 1)
[4α2 − (2n − 1)2 ][4α2 − (2n − 3)2 ] · · · [4α2 − 12 ]
a0
= ··· =
4n n!(2n + 1)(2n − 1) · · · 5 · 3
[4α2 − (2n − 1)2 ][4α2 − (2n − 3)2 ] · · · [4α2 − 12 ]
=
4n n! (2n+1)!
2n n!
an =
=
[4α2 − (2n − 1)2 ][4α2 − (2n − 3)2 ] · · · [4α2 − 12 ]
, n = 1, 2, · · ·
2n (2n + 1)!
On the other hand, for r = 0, according the recurrence
relation,
α2 − (n − 1)2
[α2 − (n − 1)2 ][α2 − (n − 2)2 ]
an−1 =
an−2
n(2n − 1)
n(n − 1)(2n − 1)(2n − 3)
[α2 − (n − 1)2 ][α2 − (n − 2)2 ] · · · [α2 − 12 ] · α2
= ··· =
a0
n!(2n − 1)(2n − 3) · · · 3 · 1
[α2 − (n − 1)2 ][α2 − (n − 2)2 ] · · · [α2 − 12 ] · α2
=
(2n−1)!
n! 2n−1
(n−1)!
an =
=
2n−1 [α2 − (n − 1)2 ][α2 − (n − 2)2 ] · · · [α2 − 12 ] · α2
, n = 1, 2 · · ·
n(2n − 1)!
By letting a0 = 1 and replacing t = x − 1, we get two
linearly independent solutions of the equation
y1 (x) =
√
∞
X
[4α2 − (2n − 1)2 ][4α2 − (2n − 3)2 ] · · · [4α2 − 12 ]
n
x−1 1+
(x − 1)
2n (2n + 1)!
n=1
y2 (x) = 1 +
∞
X
2n−1 [α2 − (n − 1)2 ][α2 − (n − 2)2 ] · · · [α2 − 12 ] · α2
n=1
n(2n − 1)!
19. The Bessel equation of order zero is
x2 y 00 + xy 0 + x2 y = 0.
(a) Show that x = 0 is a regular singular point.
(b) Show that the roots of the indicial equation are r1 = r2 = 0.
(x − 1)n
(c) Show that one solution for x > 0 is
J0 (x) = 1 +
∞
X
(−1)n x2n
n=1
22n (n!)2
.
(d) Show that the series for J0 (x) converges for all x. The
function J0 is known as the Bessel function of the first
kind of order zero.
[§5.5 #14]
Proof.
(a) Rewrite the equation of the form
1
y 00 + y 0 + y = 0
x
and let p(x) = x1 , q(x) = 1. The only singular points is
x = 0. At x = 0,
p0 = lim xp(x) = lim 1 = 1
x→0
x→0
2
q0 = lim x q(x) = lim x2 = 0
x→0
x→1
Hence x = 0 is a regular singular point.
∞
P
(b) Let y =
an xn+r . Then
n=0
0
y =
∞
X
(n+r)an x
n=0
n+r−1
∞
X
(n+r)(n+r−1)an xn+r−2
,y =
00
n=2
Substituting into the equation we have
∞
∞
∞
X
X
X
n+r
n+r
(n+r)(n+r−1)an x + (n+r)an x +
an xn+r+2 = 0
n=0
n=0
n=0
It follows that
∞
X
[r(r−1)+r]a0 xr +[r(r+1)+(r+1)]a1 xr+1 + [(n+r)2 an +an−2 ]xn+r = 0
n=2
The indicial equation is r2 = 0 with repeated real roots
r1 = r2 = 0.
(c) For r = 0, it is necessary to take a1 = 0 in order that
the coefficient of xr+1 be zero. Note that the recurrence
relation is of the form
an−2
an = − 2 , n = 2, 3, · · ·
n
The fact that a1 = 0 implies that a2m+1 = 0, ∀ m =
0, 1, 2, · · · . And
a2m = −
a2m−2
a2m−4
(−1)m a0
=
=
·
·
·
=
22 m2
24 m2 (m − 1)2
22m (m!)2
By setting a0 = 1, we obtain one solution for x > 0 of the
equation
J0 (x) = 1 +
∞
X
(−1)n x2n
n=1
22n (n!)2
(−1)n
,
22n (n!)2
n = 1, 2, · · · . Then
(−1)n+1 a 22n+2 ((n+1)!)2 1
n+1 = lim
lim =0
= lim n
(−1)
2
n→∞
n→∞
an
22n (n!)2 n→∞ 4(n + 1)
(d) Let an =
which shows that the radius of convergence of J0 (x) is infinite, that is, J0 (x) converges for all x.
20. the Bessel equation of order one is
x2 y 00 + xy 0 + (x2 − 1)y = 0.
(a) Show that x = 0 is a regular singular point.
(b) Show that the roots of the indicial equation are r1 = 1 and
r2 = −1.
(c) Show that one solution for x > 0 is
∞
x X (−1)n x2n
J1 (x) =
.
2 n=0 (n + 1)!n!22n
(d) Show that the series for J1 (x) converges for all x. The
function J1 is known as the Bessel function of the first
kind of order one.
(e) Show that it is impossible to determine a second solution
of the form
∞
X
−1
x
bn x n ,
x > 0.
n=0
[§5.5 #16]
Proof.
(a) Rewrite the equation of the form
1
x2 − 1
y 00 + y 0 +
y=0
x
x2
2
and let p(x) = x1 , q(x) = x x−1
2 . The only singular points is
x = 0. At x = 0,
p0 = lim xp(x) = lim 1 = 1
x→0
x→0
2
q0 = lim x q(x) = lim (x2 − 1) = −1
x→0
x→1
Hence x = 0 is a regular singular point.
∞
P
(b) Let y =
an xn+r . Then
0
y =
n=0
∞
X
(n+r)an x
n+r−1
∞
X
,y =
(n+r)(n+r−1)an xn+r−2
00
n=0
n=2
Substituting into the equation we have
∞
∞
∞
∞
X
X
X
X
n+r
n+r
n+r+2
(n+r)(n+r−1)an x + (n+r)an x +
an x
−
an xn+r = 0
n=0
n=0
∞
X
n=0
[(n + r)2 − 1]an xn+r +
∞
X
n=0
an−2 xn+r = 0
n=2
n=0
It follows that
2
r
2
(r −1)a0 x +[(r+1) −1]a1 x
r+1
∞
X
[(n+r)2 −1]an +an−2 xn+r = 0
+
n=2
2
The indicial equation is r − 1 = 0 with distinct real roots
r1 = 1, r2 = −1.
(c) For either r = ±1, it is necessary to take a1 = 0 in order
that the coefficient of xr+1 be zero. Note that the recurrence relation is of the form
an−2
an = −
, n = 2, 3, · · ·
(n + r)2 − 1
The fact that a1 = 0 implies that a2m+1 = 0, ∀ m =
0, 1, 2, · · · . If r = 1, then
a2m−2
a2m−2
a2m−4
a2m = −
=−
= 2 2
2
(2m + 1) − 1
4m(m + 1)
4 m (m + 1)(m − 1)
a2m−6
=− 3
4 m(m + 1) · (m − 1)m · (m − 2)(m − 1)
(−1)m a0
= ··· = m
4 m!(m + 1)!
By setting a0 = 12 , we obtain one solution for x > 0 of the
equation
∞
x X (−1)n x2n
J1 (x) =
.
2 n=0 (n + 1)!n!22n
(−1)n
.,
(n+1)!n!22n
n = 0, 1, · · · . Then
(−1)n+1
a 2n+2
1
(n+2)!(n+1)!2
n+1 = lim
lim =0
= lim n
(−1)
n→∞
n→∞
n→∞
an
4(n + 2)(n + 1)
(n+1)!n!22n (d) Let an =
which shows that the radius of convergence of J0 (x) is infinite, that is, J1 (x) converges for all x.
(e) For r = −1 the recurrence relation becomes
an−2
an = −
, n = 2, 3, · · ·
(n − 1)2 − 1
If n = 2 the coefficient of a2 zero and we cannot calculate
a2 . Consequently it is not possible to find a series solution
∞
P
of the form x−1
bn x n .
n=0