This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 1
ζ ζ ζ 4d 4d 2 Y
dY
+ 2ζ 3
2
dζ
dζ 2 Y
+a
dζ 2 1
ζ −b 1
ζ ζ2 2ζ 3 − b 1
ζ dY
+c
dζ ζ2 1
ζ dY
+c
dζ Y = 0.
1
ζ Y = 0. b) For an ordinary point we need that the following two functions are analytic (ﬁnite) about
ζ = 0,
1
cζ
1
1
1
1
2
a
− 2b
and
.
1
1
ζ
ζ
ζ
ζ
aζ
a ζ ζ4
c) For a regular singular point we require the following two functions are analytic (ﬁnite)
about ζ = 0,
1
cζ
2
1
1
1
ζ
.
a
− 2b
and
1
1
ζ
ζ
ζ
ζ
aζ
a ζ ζ2
d) In this case a(x) = x4 , b(x) = 0, c(x) = α2 so the transformed equation is,
ζ2 d2 Y
dY
+ 2ζ
+ α2 ζ 2 Y = 0.
2
dζ
dζ e) ζ = 0 is a regular singular point.
f) Try a Frobenius expansion,
∞ an ζ n+r , Y=
n=0
∞ ζ 2Y =
ζ dY
=
dζ
2 ζ2 dY
=
dζ 2 ∞ an ζ n+r+2 =
n=0
∞ an−2 ζ n+r ,
n=2 (n + r)an ζ n+r ,
n=0
∞ (n + r)(n + r − 1)an ζ n+r .
n=0 We sub this into our equation,
∞ r(r + 1)a0 xr + (r + 1)(r + 2)a1 xr+1 + (n + r)(n + r + 1)an + α2 an−2 ζ n+r = 0
n=2 4 For the ﬁrst term to be nonzero we need r = 0 or r = −1. These are the roots to the
indicial equation. For r = 0 we then need that a1 = 0 whereas for r = −1 we do not
require that condition. The recurrence relation is,
an = − α2
an−2 .
(n + r)(n + r + 1) To solve the recurrence relation in general we get
α2
a0 ,
(r + 2)(r + 3)
α2
α4
a4 = −
a2 =
a0 .
(r + 4)(r + 5)
(r + 2)(r + 3)(r + 4)(r + 5)
a2 = − For r = 0 we have that a1 and all the odd coeﬃcients are zero and therefore
y 1 = a0 1 − α2 2 α4 4
x + x + ··· .
3!
5! For r =...
View Full
Document
 Spring '14

Click to edit the document details