Math 257/316 Assignment 3 Solutions
1. The steady-state temperature distribution y(x) along a wire 0 ≤ x ≤ 1, cooled by its
surroundings, and held fixed at zero temperature at its left endpoint, solves:
d
p(x)y 0 − y = 0,
dx
y(0) = 0,
where p(x) ≥ 0 represents its (spatially dependent) thermal conductivity. If the
conductivity degenerates at the left endpoint as p(x) = xs for some 0 ≤ s < 1, how
does the temperature behave near that point? (I.e. just give the form of the the first
term in a series solution about x0 = 0 (for x > 0).) Does the problem have a solution
if s = 1? s > 1?
Using the product rule, and then dividing through by p(x) we find the ODE reads
y 00 +
p0 (x) 0
1
y −
y = 0,
p(x)
p(x)
and for p(x) = xs , this is
1
s 0
y − s y = 0,
x
x
If s = 0, this just y 00 = y, so the general solution is c1 ex + c2 e−x , and the condition
y(0) = 0 implies c2 = −c1 , so y(x) = c1 (ex − e−x ) ≈ 2c1 x as x → 0+. For s > 0,
the coefficients xs and − x1s are singular at x = 0, so x = 0 is a singular point. Note
however, that
hsi
1
2
α := lim x
= s and β := lim x − s = lim −x2−s = 0
x→0+
x→0+
x→0+
x
x
y 00 +
(assuming s < 2) are finite. Strictly speaking, x2−s is not analytic (unless s = 0 or
s = 1), so x = 0 is not a regular singular point, but we may still model the benaviour
of solutions on those of the corresponding Euler equation, with indicial equation
0 = r(r − 1) + αr + β = r2 + (s − 1)r = r[r − (1 − s)]
whose roots are r = 0 and r = 1 − s. These two roots lead to solutions behaving like
y2 (x) = x1−s + b2 x2−s + · · ·
y1 (x) = 1 + a1 x + · · · ,
1
as x → 0+, and the general solution is y(x) = ay1 (x)+by2 (x). The condition y(0) = 0
implies a = 0, and so
y(x) ≈ bx1−s for x near 0,
for some constant b. If s > 1, y2 (x) is unbounded as x → 0+ and so we need
a = b = 0 and the only solution is y ≡ 0. Note that if s = 1, the indicial equation
has double root r = 0 and we find
y1 (x) = 1 + a1 x + · · · ,
y2 (x) = ln(x) + a1 x ln(x) + b1 x + · · ·
as x → 0+, and so the only solution y(x) = ay1 (x) + by2 (x) with y(0) = 0 has
a = b = 0 and so y ≡ 0.
2. Find (the first three non-zero terms of) a series solution (about x0 = 0) of this initial
value problem for a spherical Bessel equation:
x2
d2 y
dy
5
+ 2x
+ (x2 − )y = 0,
2
dx
dx
16
y(0) = 0,
lim x3/4 y 0 (x) = 1.
x→0+
Dividing through by the lead coefficient reveals
y 00 +
Since p(x) = x2 and q(x) = 1 −
Since, as x → 0,
2 0
5
y + (1 −
)y = 0.
x
16x2
5
16x2
are singular at x = 0, x = 0 is a singular point.
5
5
→
16
16
are finite, this is a regular singular point. The indicial equation,
xp(x) = 2 → 2,
x2 q(x) = x2 −
0 = r(r − 1) + 2r + (−5/16) = r2 + r − 5/16 = (r + 5/4)(r − 1/4),
has roots r1 = 1/4 and r2 = −5/4. Thus we will find Frobenius series solutions (for
x > 0) of the form
y2 (x) = x−5/4 + b1 x−1/4 + b2 x3/4 + · · · ,
y1 (x) = x1/4 + a1 x5/4 + a2 x9/4 + · · · ,
and the general solution will be
y(x) = c1 y1 (x) + c2 y2 (x) = c2 x−5/4 + c2 b1 x−1/4 + c1 x1/4 + c2 b2 x3/4 + c1 a1 x5/4 + · · · .
The requirement y(0) = 0 implies that c2 = 0 (other wise y(x) is unbounded as
x → 0+ !). The other requirement then implies
5
3/4 0
3/4 1
−3/4
1/4
1 = lim x y (x) = lim x
c1 x
+ c1 a1 x + · · ·
x→0+
x→0+
4
4
1
4
1
= lim c1 + c1 a1 x + · = c1 .
x→0+ 4
5
4
2
So c1 = 4. Now let’s find the (first few terms of the) series
we do not
P for y1 (notice
n+1/4 (remember,
have to bother finding y2 !). As usual, differentiate y = ∞
a
x
n=0 n
we have already chosen r = 1/4) and substitute into our ODE to find
0 = x2
∞
X
∞
n=0
=
∞
X
∞
∞
X
X
1
7
3
1
1
3
1
5 X
(n + )(n − )an xn− 4 + 2x
an xr+ 4
(n + )an xn− 4 + x2
an xn+ 4 −
4
4
4
16
n=0
1
1
3
(n + )(n − )an xn+ 4 +
4
4
n=0
∞
X
∞
X
n=0
∞
X
1
1
2(n + )an xn+ 4 +
4
n=0
n=0
∞
X
9
an xn+ 4
n=0
∞
X
∞
X
1
5
−
an xr+ 4
16
n=0
∞
X
1
1
1
1
5
1
1
3
(n + )(n − )an xn+ 4 +
2(n + )an xn+ 4 +
an−2 xn+ 4 −
an xr+ 4
4
4
4
16
n=0
n=0
n=2
n=0
5
1 3 1
5
1
10
5
a0 x1/4 +
a1 x5/4
= − · + −
· +
−
4 4 2 16
4 5
4
16
∞ X
1
1
3
1
5
+
(n + )(n − ) + 2(n + ) −
an + an−2 xn+ 4
4
4
4
16
n=2
∞
X
1
39
3
1/4
5/4
= [0] a0 x +
a1 x +
n(n + )an + an−2 xn+ 4 .
16
2
=
n=2
Notice that the coefficient of the first term is already zero – that’s because we already
chose r = 1/4 to be a root of the indicial equation ! – and so a0 is free. As indicated
above, we’ll simply choose a0 = 1. The second term shows us that a1 = 0. And then
the recurrence relation is
an−2
an = −
, n = 2, 3, 4, . . . .
n(n + 32 )
So
1
a2
1
a0
=− ,
a3 = 0,
a4 = − =
,...,
7
7
22
7 · 22
1
so y1 (x) = x1/4 − 17 x9/4 + 7·22
x17/4 + · · · , and
a2 = −
4
2
y(x) = 4y1 (x) = 4x1/4 − x9/4 +
x17/4 + · · · .
7
7 · 11
3. In the notes, one solution of the Bessel equation of order zero
x2 y 00 + xy 0 + x2 y = 0
is found to be
∞
X
1 2
1 4
1
(−1)m 2m
6
J0 (x) = 1 − 2 x + 2 2 x − 2 2 2 x + · · · =
x .
2
2 4
2 4 6
22m (m!)2
m=0
3
Find a second (linearly independent) solution of the form
y2 (x) = J0 (x) ln(x) + ỹ(x)
as follows:
(a) show that for y2 to solve the Bessel equation, ỹ must solve the ode
x2 ỹ 00 + xỹ 0 + x2 ỹ + 2xJ00 (x) = 0;
Since
y20 = J00 ln(x) + J0 (x)
1
+ ỹ 0 ,
x
y200 = J000 ln(x) + 2J00 (x)
1
1
− J0 (x) 2 + ỹ 00 ,
x
x
we require
0 = x2 y200 + xy20 + x2 y2 = [x2 J000 + xJ00 + x2 J0 ] ln(x) + x2 ỹ 00 + xỹ 0 + x2 ỹ + 2xJ00 (x),
and since J0 itself solves the ODE, the term with the ln(x) disappears, leaving
0 = x2 ỹ 00 + xỹ 0 + x2 ỹ + 2xJ00 (x).
(b) find the first three non-zero terms of a series solution ỹ(x) =
ode.
P∞
n=1 bn x
n
of this
After observing
1
1
1
J00 (x) = − x + 2 x3 − 2 2 x5 + · · · ,
2
2 4
2 4 6
as usual, we differentiate the power series and plug in to the ODE to find:
∞
∞
∞
X
X
X
1
1 3
1
2
n−2
n−1
2
n
5
0=x
n(n − 1)bn x
+x
nbn x
+x
bn x + 2x − x + 2 x − 2 2 x + · · ·
2
2 4
2 4 6
n=1
n=1
n=1
∞
∞
∞
X
X
X
1
1
=
n(n − 1)bn xn +
nbn xn +
bn xn+2 + −x2 + x4 − 6 x4 + · · ·
8
2 3
n=1
n=1
n=1
∞
∞
∞
X
X
X
1 4
1 4
n
n
n
2
=
n(n − 1)bn x +
nbn x +
bn−2 x + −x + x − 6 x + · · ·
8
2 3
n=1
n=1
n=3
1
= b1 x + (4b2 − 1)x2 + (9b3 + b1 )x3 + (16b4 + b2 + )x4 + (25b5 + b3 )x5
8
1
4
+ (36b6 + b4 − 6 )x + · · ·
2 3
4
and so:
b1 = 0,
b5 = −
1
b2 = ,
4
b3
= 0,
25
b1
1
1
3
= 0,
b4 = − (b2 + ) = − 7 ,
9
16
8
2
1
1
11
b6 = − (b4 − 6 ) = 9 3 .
36
2 3
2 ·3
b3 = −
Thus
1
3
11
ỹ = x2 − 7 x4 + 8 3 x6 + · · · .
4
2
2 ·3
So putting it all together:
1
1
1
3
y2 (x) = J0 (x) ln(x) + ỹ = ln(x) − x2 ln(x) + x2 − 6 x4 ln(x) − 7 x4 + · · · .
4
4
2
2
5