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Solutions to Bessel function problems
Question 1. The Bessel function J is defined by the power series
J(x) =
∑^ ∞
n=
)n x^2 n n!^2
Prove that J is twice differentiable and that J satisfies the differential equa- tion d dx
x
dJ dx
Answer Using Example 5.3.10 of ECA, we know J is differentiable on R since it is of the form (^) ∞ ∑
n=
an n!
xn
with a 2 n = (−1)
n 4 nn! and^ a^2 n+1^ = 0, and^ {an}^ is certainly bounded. Differentiat- ing term by term, we have
J′(x) =
∑^ ∞
n=
)n 2 nx^2 n−^1 n!^2
Clearly J′(x) is also differentiable, again by Example 5.3.10 of ECA. Now xJ′(x) is given by
xJ′(x) =
∑^ ∞
n=
)n 2 x^2 n n!(n − 1)!
and differentiating term by term, we have
d dx
(xJ′(x)) =
∑^ ∞
n=
)n 4 nx^2 n−^1 n!(n − 1)!
∑^ ∞
n=
)n x^2 n+ n!^2 = −xJ(x)
which was to be shown.
Question 2. Using the differential equation, show that
d dx
J′(x)^2 + J(x)^2
x
J′(x)^2
Use this to prove carefully that both J′^ and J are bounded for x ≥ 0. What happens for x ≤ 0?
Answer The differential equation can be rewritten as
xJ′′(x) + J′(x) + xJ(x) = 0
So using the chain rule, the derivative of J′^2 + J^2 is
d dx
J′(x)^2 + J(x)^2
= 2J′(x)J′′(x) + 2J(x)J′(x)
= 2J′(x)
− (^1) x J′(x) − J(x)
= −
x
J′(x)^2
Because we can write
J′(x) = −^12 x
∑^ ∞
n=
)n x^2 n n!(n + 1)!
J′(x)/x is continuous. Thus so is J′(x)^2 /x^2 , and thus so is J′(x)^2 /x^2 · x = J′(x)^2 /x. So we can integrate using the fundamental theorem of calculus to say
J′(x)^2 + J(x)^2 = J′(0)^2 + J(0)^2 +
∫ (^) x
0
t
J′(t)^2 dt
∫ (^) x
0
t
J′(t)^2 dt
≤ 1
since (^2) t J′(t)^2 ≥ 0 for t ≥ 0. We conclude that |J(x)| ≤ 1 and |J′(x)| ≤ 1 for all x ≥ 0. The same is true for x ≤ 0, since J is an even function and J′^ is an odd function.
and this limit exists, we must have
lim x→∞ J′(x) = 0
(otherwise the improper integral would be divergent). Now by the differential equation, we have
J′′(x) +
x
J′(x) + J(x) = 0
which implies that lim x→∞ J′′(x) = L
However, we have
J′(x) =
∫ (^) x
0
J′′(t) dt
and since J′(x) → 0, we must have J′′(x) → 0 as well for the improper integral to converge. So it is impossible for J not to have a local minimum if J(x) < 0 for all x > x 1. Thus J has a local minimum at some point x 2 , and for x > x 2 and sufficiently close to x 2 we have J′(x) > 0. Now from the formula
d dx
(xJ′(x)) = −xJ(x)
we know that since J is negative, xJ′(x) is increasing for x > x 1 , by assump- tion that J(x) < 0 for x > x 1. Choose some r > x 2 such that J′(r) > 0. Then for x ≥ r, xJ′(x) ≥ rJ′(r). Thus by the fundamental theorem of calculus,
J(x) = J(x 2 ) +
∫ (^) x
x 2
J′(t) dt ≥ J(x 2 ) +
∫ (^) x
x 2
rJ′(r) t
dt = J(x 2 ) + rJ′(r) ln
x x 2
Since as x → ∞, ln x → ∞, J(x) is eventually positive for x sufficiently large, contradicting our assumption that J had no more zeroes after x 1. So in fact J does have another zero. If on the other hand J(x 1 ) = 0 and J′(x 1 ) > 0, then we could simply repeat the previous argument with the function −J(x) which satisfies the same differential equation, so again we can find another zero larger than x 1. (Technically we should worry that J′(x 1 ) = 0, but we skip that part.)
