18.305 Exam 1,
October 18, 04.
Closed Book.
Problem: Consider the differential equation
y” + x 4 y : 0.
(a) Locate and classify the singular points, finite or infinite, of this differential equation. (10%)
(b) Find the WKB solutions of this equation. For what values of x are the WKB solutions good
approximations of the solution? (25%)
(c) Find the Maclaurin series solution of this equation. For what values of x are these series
convergent? (25%)
(d) Find the entire asymptotic series of the solutions which are useful when the magnitude of x is
very large. For what values are these series convergent? (25%)
(e) Find the exact solution of this equatiion. (15%)
Solutions:
(a) and (b):
The infinity is an irregular singular point of this equation.
Since x 4 is always positive, the WKB solutions are oscillatory. Putting
p : x 2 , X pdx : x 3 /3,
we have
y çWKB : x ?1 expΩçix 3 /3æ.
The rank of the irregular singular point at K is 3.
c. Let
y : > a n x n , a ?1 : a ?2 : 6 6 6 : 0.
We have
y” : > a n nΩn ? 1æx n?2
and
x 4 y : > a n x n+4 : > a n?6 x n?2 .
Thus the recurrence formula is
a n nΩn ? 1æ : ?a n?6 .
Let
n : 6m,
then
a
a <Ω5/6æ
: Ω?1æ m 2m 0
.
a 6m : ? 2 6Ωm?1æ
6 m!<Ωm + 5/6æ
6 mΩm ? 1
/6æ
Thus one of the Maclaurin solutions is
K
Ω?1æ m x 6m
y 1 : >
2m
.
6 m!<Ωm + 5/6æ
m:0
Setting
n : 6m + 1,
we have
a
a <Ω7/6æ
: Ω?1æ m 2m 1
.
a 6m+1 : ? 2 6Ωm?1æ+1
6 Ωm + 1/6æΩmæ
6 m!<Ωm + 7/6æ
Thus the second Maclaurin solution is
K
Ω?1æ m x 6m+1
.
6
m!<Ωm
+
7/6æ
m:0
These seires converge for all finite values of x.
y2 :>
2m
(d) Let
y : expΩix 3 /3æY,
then
ΩD + ix 2 æΩD + ix 2 æY + x 4 Y : 0,
or
Y” + 2ix 2 Y r + 2ixY : 0.
Let
Y : > A n x ?1?n , A ?1 : A ?2 : 6 6 6 : 0.
We have
Y” : > A n Ωn + 1æΩn + 2æx ?3?n : > A n?3 Ωn ? 2æΩn ? 1æx ?n
and
2ix 2 Y r + 2ixY : ?2i > A n nx ?n .
We get
Ωn ? 2æΩn ? 1æ
An :
A n?3 , n ; 0.
2in
Setting n : 3m, we get
3Ωm ? 2/3æΩm ? 1/3æ
<Ωm + 1/3æ<Ωm + 2/3æ
A 3m :
.
A 3Ωm?1æ : Ω 3 æ m
2im
2i
m!<Ω2/3æ<Ω1/3æ
Thus one of the asymptotic solutions is
K
3
Ω3/2iæ m <Ωm + 1/3æ<Ωm + 2/3æx ?3m?1
y 3 : e ix /3 >
.
m!
m:0
The second asymptotic solution is obtained by taking the complex conjugate of y 3 . We get
K
3
Ω3i/2æ m <Ωm + 1/3æ<Ωm + 2/3æx ?3m?1
y 4 : e ?ix /3 >
m!
m:0
These series converge for no value of x.
(e) The asymptotic forms of the Bessel functions are t ?1/2 expΩçitæ.
Thus we put
x 3 /3 : t, y : t 1/6 Z.
We have
d : 3 2/3 t 2/3 d .
dx
dt
Therefore
d2y
: 3 4/3 t 2/3 d t 2/3 d t 1/6 Z : 3 4/3 t 3/2 Ω d + 5 æΩ d + 1 æZ,
2
6t dt
6t
dt
dt
dt
dx
and the differential equation for Z is
d 2 Z + 1 dZ + Ω1 ? 1 æZ : 0.
t dt
36t 2
dt 2
The solution of the equation above is,
Z : aJ 1/6 Ωtæ + bJ ?1/6 Ωtæ.
Hence
3
3
yΩxæ : ax 1/2 J 1/6 Ω x æ + bx 1/2 J ?1/6 Ω x æ.
3
3