PH 116C, Solution set 4
1. Read Boas, Ch. 12, Sec. 11-16, 21
2. Boas, p. 651, problem 13.7-14 [graded]
The most general bounded solution to the Laplace equation in spherical coordinates
was given in class, which we reproduce here,
u(r, θ, φ) =
∞ X
`
X
(A`m r` + B`m r−`−1 )Y`m (θ, φ) .
`=0 m=−`
Since the boundary conditions for this problem are azimuthally symmetric, it follows
that only the m = 0 term survives in the sum above, and the Ylm reduces to a Pl .
Hence, for this problem, we may write
u(r, θ) =
∞
X
(A` r` + B` r−`−1 )P` (cos θ) .
(1)
`=0
Note that the volume of interest lies in the region 1 ≤ r ≤ 2, which excludes both
r = 0 and r = ∞. Hence, both the R(r) = r` and R(r) = r−`−1 solutions must be
kept in eq. (1), as neither diverges in the region 1 ≤ r ≤ 2. The coefficients A` and B`
are determined by the boundary conditions, u(r = 1, θ) and u(r = 2, θ),
u(1, θ) =
∞
X
(A` + B` )P` (cos θ) ,
`=0
u(r, θ) =
∞
X
(A` 2` + B` 2−`−1 )P` (cos θ) .
`=0
Thus, we can project out the unknown coefficients above,
Z
2` + 1 1
A` + B` =
u(1, θ)P` (cos θ) d cos θ ,
2
−1
Z
2` + 1 1
`
−`−1
2 A` + 2
B` =
u(2, θ)P` (cos θ) d cos θ .
2
−1
(2)
(3)
In this problem, the boundary conditions on the surface of the inner and outer spherical
shells are given respectively by
(
100,
for 0 < cos θ < 1 ,
u(1, θ, φ) = 0,
u(2, θ, φ) =
0,
for −1 < cos θ < 0 .
Hence, eqs. (2) and (3) yield
A` + B` = 0 ,
`
2 A` + 2
−`−1
(4)
Z
B` = 50(2` + 1)
P` (cos θ) d cos θ .
0
1
1
(5)
It immediately follows that B` = −A` . Inserting this back into eq. (5), we end up with
Z
50(2` + 1) 1
A` = −B` = `
P` (cos θ) d cos θ .
(6)
2 − 2−`−1 0
We evaluate this integral using the results of problem 12.13–3 on p. 615 of Boas,


1,
for ` = 0 ,





Z 1

0,
for ` = 2, 4, 6, . . . ,
(7)
P` (cos θ) d cos θ =

0

(`−1)/2

(−1)
(` − 2)!!



for ` = 1, 3, 5, . . . .
 2(`+1)/2 1 (` + 1)! ,
2
Note that the integral in the case of ` = 0 is trivial, since P0 (cos θ) = 1.
Inserting the results of eqs. (6) and (7) into eq. (1) yields,
∞
X
1
(−1)(`−1)/2 (2` + 1)(` − 2)!!
1
`
r − `+1 P` (cos θ) .
u(r, θ) = 100 1 −
+50
r
r
(2(3`+1)/2 − 2−(`+1)/2 ) 12 (` + 1) !
`=1
odd `
Evaluating the first three terms of the series yields
7 3
3
−2
−4
−1
(r − r )P3 (cos θ) + . . . .
u(r, θ) = 100 (1 − r )P0 (cos θ) + (r − r )P1 (cos θ) −
7
127
3. Boas,
586, problem 12.11-5 Substituting the generalized power series y(x) =
P∞ p. n+s
, we have
n=0 an x
2x
∞
X
(n + s)(n + s − 1)an x
n+s−2
+
n=0
∞
X
(n + s)an x
n+s−1
+2
n=0
∞
X
an xn+s = 0 .
n=0
The indicial equation is the coefficient of a0 xs−2 corresponding to the n = 0 term of
the sum,
2s(s − 1) + s = 0 =⇒ 2s2 − s = 0 =⇒ s = 0, 21 .
The recursion relation is obtained from the coefficient of xn+s+1 , for n = 1, 2, 3, . . .,
which yields
2(n + s)(n + s − 1)an + (n + s)an + 2an−1 = 0
For s = 0 we have
n(2n − 1)an + 2an−1 ,
for n = 1, 2, 3, . . . ,
which yields
an =
−2
4
(−2)n
an−1 =
an−2 . . . =
a0
n(2n − 1)
n(n − 1)(2n − 1)(2n − 3)
n!(2n − 1)!!
2
Note that
(2n)! = 2n(2n − 1)(2n − 2)(2n − 3) . . . = 2n (2n − 1)!!n! .
Hence, we can write:
an =
(−4)n
a0 .
(2n)!
Inserting these coefficients back in the series gives
y(x) =
∞
X
n
an x = a0
1
2
(2n)!
n=0
n=0
For s =
∞
X
(−4x)n
a0 = a0
√
∞
X
(−1)n (2 x)2n
(2n)!
n=0
= a0 cos(2x1/2 ).
(8)
we have
n(2n + 1)a0n + 2a0n−1 = 0 , for n = 1, 2, 3, . . . ,
which yields
a0n =
−2
4
(−2)n
(−4)n 0
a0n−1 =
a0n−2 = . . . =
a00 =
a .
n(2n + 1)
n(n − 1)(2n + 1)(2n − 1)
n!(2n + 1)!!
(2n + 1)! 0
Inserting these coefficients back in the series gives
y(x) =
∞
X
1
a0n xn+ 2
=
n=0
a00
∞
1
X
(−4)n xn+ 2
n=0
(2n + 1)!
=
1 0
a
2 0
√
∞
X
(−1)n (2 x)2n+1
(2n + 1)!
n=0
= 12 a00 sin(2x1/2 ).
(9)
The general solution is then given by the linear combination of (8) and (9):
y = A cos(2x1/2 ) + B sin(2x1/2 )
4. Boas, p. 590, problem 12.12-4
We’re using the series form of the Bessel function to show that
d
J0 (x) = −J1 (x)
dx
(10)
The Bessel function Jp (x) has the following series representation [cf. eq. (12.9) on
p. 590 of Boas]:
∞
x 2n+p
X
(−1)n
.
(11)
Jp (x) =
Γ(n + 1)Γ(n + 1 + p) 2
n=0
Setting p = 0 and taking the derivative yields:
∞
∞
x 2n X
x 2n−1
d
d X
(−1)n
(−1)n n
J0 (x) =
=
,
dx
dx n=0 Γ(n + 1)Γ(n + 1) 2
Γ(n
+
1)Γ(n
+
1)
2
n=1
3
after noting that the n = 0 term does not contribute to the sum above. Since the
sums above converge uniformly for all x, it is permissible to differentiate the series
representation of J0 (x) term by term. Relabeling n → n + 1, the new summation
index n will now run from n = 0 to infinity, which yields
∞
∞
x 2n+1
X
X
d
(−1)n+1 (n + 1) x 2n+1
(−1)n
J0 (x) =
=−
= −J1 (x) .
dx
Γ(n
+
2)Γ(n
+
2)
2
Γ(n
+
1)Γ(n
+
2)
2
n=0
n=0
In the obtaining the final sum, we used Γ(n + 2) = (n + 1)Γ(n + 1), and recognized
the resulting expression for J1 (x) [cf. eq. (11) with p = 1].
5. Boas, p. 591, problem 12.13-6
Show from
cos(πp)Jp (x) − J−p (x)
,
sin(πp)
Np (x) =
(12)
that
N(2n+1)/2 (x) = (−1)n+1 J−(2n+1)/2 (x) .
For p = 12 (2n + 1) we have cos πp = 0 and sin πp = (−1)n . Plugging those values back
in eq. (12), it follows that
N(2n+1)/2 (x) = −(−1)n J−(2n+1)/2 (x) .
6. Boas, p. 593, problem 12.15-8
Using Jp−1 (x) − Jp+1 (x) = 2Jp0 (x) show that
Z ∞
Z ∞
Z
J1 (x)dx =
J3 dx = . . . =
0
and
Z
0
∞
Z
J0 (x)dx =
0
Z
J2n+1 (x)dx ,
0
∞
Z
J2 (x)dx = . . . =
0
and prove that
∞
∞
J2n (x)dx ,
0
∞
Jn (x)dx = 1
for all integral n .
0
For any odd p = 2n + 1, using the recursion formula given in eq. (15.4) on p. 592 of
Boas, it follows that
∞
Z ∞
Z ∞
Z ∞
0
J2n (x)dx − 2J2n+1 (x)
J2n+2 (x)dx =
J2n (x) − 2J2n+1 (x) dx =
0
0
Z
=
0
0
∞
J2n (x)dx ,
(13)
0
4
where we have used that fact that limx→0 J2n+1 (x) = 0 and limx→∞ J2n+1 (x) = 0 for
any non-negative integer n. The x → 0 limit follows from the series representation of
the Bessel function given in eq. (11), and the x → ∞ limit follows
√ from the observation
that the amplitude of the Bessel function graph falls off as 1/ x as x → ∞. The latter
behavior is also apparent from the asymptotic forms given in the table on p. 604 of
Boas. Since eq. (13) is true for any non-negative integer n, it immediately follows that
Z ∞
Z ∞
Z ∞
J0 (x)dx =
J2 (x)dx = . . . =
J2n (x)dx ,
0
0
0
as required.
For any even p = 2n > 0, the a similar analysis applies,
Z
Z ∞
Z ∞
0
(J2n−1 (x) − 2J2n (x))dx =
J2n+1 (x)dx =
0
0
0
Z
∞
∞
J2n−1 (x)dx − 2J2n (x)
∞
J2n−1 (x)dx ,
=
0
(14)
0
for any non-negative integer n. Hence,
Z ∞
Z ∞
Z
J1 (x)dx =
J3 dx = . . . =
0
0
∞
J2n+1 (x)dx ,
0
as required.
To complete the problem, we only have to compute two integrals. First, we use eq. (10)
to obtain
∞
Z ∞
Z ∞
d
J1 (x)dx = −
J0 (x)dx = −J0 (x) = 1 .
dx
0
0
0
In obtaining this result, we used the fact that limx→0 J0 (x) = 1 [a fact that is evident
from the √
series representation of J0 (x) given in eq. (11)], and limx→∞ J0 (x) = 0, due
to the 1/ x behavior as x → ∞. From the quoted problem 7, we have:
Z ∞
J0 (x)dx = 1 .
0
Combining all the results above, we have successfully verified that
Z ∞
Jn (x)dx = 1 ,
(15)
0
for all non-negative integers n. Since J−n (x) = (−1)n Jn (x) for any integer n
[cf. eq. (13.2) on p. 590 of Boas], we conclude that eq. (15) actually holds for any
integer n.
7. [Graded] In class we found that the dimensionless version of the equation of a lengthening pendulum is
d2 θ
dθ
z 2 + 2 + θ = 0.
(16)
dz
dz
5
I asserted that if
x = 2z 1/2 , y = z 1/2 θ,
then this transforms into Bessel’s equation with p = 1.
We show this by computing dθ/dz and d2 θ/dz 2 in terms of x and x−derivatives, and
plugging in to Eq. 16, as follows:
dx
2
= z −1/2 = , θ = y/z 1/2 = 2y/x
dz
x
so
Then
dθ
dθ dx
2 dθ
2 d 2y
4
4
=
=
=
= 2 y 0 − 3 y.
dz
dx dz
x dx
x dx x
x
x
4
d2 θ
2 d 4 dy
8
24
24
− 3 y = 3 y 00 − 4 y 0 + 5 y,
=
2
2
dz
x dx x dx x
x
x
x
with primes denoting x−derivatives. Plugging these in to Eq. 16, we get:
x 2 8
24 0 24
4 0
4
y
00
y
−
y
+
y
+
2
y
−
y
+
2
0=
2
x3
x4
x5
x2
x3
x
2
y
y
y0
0=
y 00 + 2 2 − 2 3 + 2 .
x
x
x
x
or
0 = y 00 + x−1 y 0 + (1 − 1/x2 )y,
the desired result.
8. Boas, p. 594, problem 12.16-13 (second part only, i.e. just prove that 16.2 is a solution
to 16.1)
We have
y = xa Z(bxc ),
and we want to show that when substituted into the equation
1 − 2a 0
a2 − p 2 c 2
00
c−1 2
y + (bcx ) +
y = 0,
y +
x
x2
we find a Bessel equation. We can follow the method used in class: Let
z = bxc
u = Zp (z)
and
y = xa u.
Then
dz
= bcxc−1 ,
dx
du
du dz
=
= bcxc−1 u0 ,
dx
dz dx
6
where u0 ≡ du/dz.
dy
= axa−1 u + xa (bcxc−1 u0 ) = axa−1 u + bcxa+c−1 u0
dx
d2 y
= a(a − 1)xa−2 u + axa−1 (bcxc−1 u0 ) + (a + c − 1)bcxa+c−2 u0 + bcxa+c−1 bcxc−1 u00
dx2
Now it’s time to plug these in and pray for a lot of cancellations. Let’s do this by
collecting first the terms that will be coefficients of u00 . There is just one, which is
bcxa+c−1 bcxc−1 u00 = b2 c2 xa+2c−2 u00 .
Now let’s look at the terms multiplying u0 . There are two terms from the y 00 , and one
more from the y 0 term. They add up to:
1 − 2a
a+c−1 0
a−1
c−1
a+c−2
(bcx
u ) + ax (bcx ) + (a + c − 1)bcx
u0
x
= (1 − 2a)(bcxa+c−2 u0 ) + abcxa+c−2 + (a + c − 1)bcxa+c−2 u0
= bc − 2abc + abc + abc − bc2 − bc xa+c−2 u0
= bc2 xa+c−2 u0
Finally, we look at the terms multiplying u. There is one from the y 00 term, one from
the y 0 term, and one from the y term:
a2 − p 2 c 2 a
1 − 2a a−1
a−2
c−1 2
ax
+ (bcx ) +
x u
a(a − 1)x
+
x
x2
= (bcxc−1 )2 xa + a(a − 1) + a(1 − 2a) + (a2 − p2 c2 ) xa−2 u
= (bcxc−1 )2 xa − p2 c2 xa−2 u
Thus we have:
b2 c2 xa+2c−2 u00 + bc2 xa+c−2 u0 + (bcxc−1 )2 xa − p2 c2 xa−2 u.
If we divide by b2 c2 xa+2c−2 we get
u00 + (1/b)x−c u0 + (1 − (p2 /b2 )x−2c )u.
But (1/b)x−c = 1/z so this reads
u00 + z −1 u0 + (1 − p2 /z 2 )u,
which is Bessel’s equation.
7