The Bessel Function of Order Zero
J0 is the only solution of Bessel's equation analytic at the origin; with Bessel's integral and the bound |J0(x)| <= 1.
1 Statement
Definition 1.1 (Bessel function).
Theorem 1.2 (Bessel’s equation of order zero).
- (a)
- \(J_0\) is entire and satisfies \begin{equation}\tag{1.2} z f''(z) + f'(z) + z f(z) = 0 \qquad (z \in \C ). \end{equation}
- (b)
- If \(f\) is analytic on a disc about \(0\) and satisfies (1.2) there, then \(f = f(0)\, J_0\).
- (c)
- For every \(z \in \C \), \(J_0(z) = \dfrac {1}{\pi } \displaystyle \int _0^\pi \cos (z \sin \theta ) \dd \theta \).
- (d)
- \(\abs {J_0(x)} \leq 1\) for every real \(x\).
2 Setting
We use the following facts about power series [1, ch. 5]. A power series is analytic inside its disc of convergence and may be differentiated term by term there. If two power series agree on a disc about \(0\), they have the same coefficients.
3 Idea of the proof
Substituting a general power series into (1.2) gives a two-step recursion \(n^2 a_n = -a_{n-2}\), with \(a_1 = 0\). That recursion has exactly one solution for each choice of \(a_0\), and \(a_0 = 1\) gives the coefficients of \(J_0\).
4 Proof
Lemma 4.1 (The recursion). Let \(f(z) = \sum _{n \geq 0} a_n z^n\) converge on a disc about \(0\). Then \(f\) satisfies (1.2) on that disc if and only if
Proof. Differentiating term by term and collecting the coefficient of \(z^{n-1}\),
So \(z f'' + f' + z f = a_1 + \sum _{n \geq 2} (n^2 a_n + a_{n-2}) z^{n-1}\). This vanishes identically if and only if every coefficient is zero.
Proof of (a) and (b). Solve the recursion of Lemma 4.1. From \(a_1 = 0\) and \(a_{2m+1} = -a_{2m-1}/(2m+1)^2\), every odd coefficient is zero. For the even ones, \(a_{2m} = -a_{2m-2} / (4m^2)\), so by induction
For (b): \(f\) has a power series at \(0\), which must satisfy the recursion, so \(f = a_0 J_0\) with \(a_0 = f(0)\).
For (a): the coefficients of \(J_0\) satisfy the recursion with \(a_0 = 1\), so it remains to see that (1.1) converges everywhere. For \(z \neq 0\), the ratio of consecutive terms is \(\abs {z}^2 / (4(m+1)^2) \to 0\), so by the ratio test the radius of convergence is infinite. Then \(J_0\) is entire, and Lemma 4.1 applies on every disc.
Lemma 4.2 (Wallis). For \(m \geq 0\), \(\displaystyle \int _0^\pi \sin ^{2m}\theta \dd \theta = \pi \, \frac {(2m)!}{4^m (m!)^2}\).
Proof. Let \(I_m\) be the integral, so \(I_0 = \pi \). For \(m \geq 1\), integrate by parts with \(u = \sin ^{2m-1}\theta \) and \(\dd v = \sin \theta \dd \theta \). The boundary terms vanish, so
that is, \(I_m = \frac {2m-1}{2m} I_{m-1}\). The right-hand side of the lemma satisfies the same recursion, because
and both sides equal \(\pi \) when \(m = 0\).
Proof of (c). Fix \(z\). The series \(\cos (z\sin \theta ) = \sum _m (-1)^m z^{2m} \sin ^{2m}\theta / (2m)!\) converges uniformly in \(\theta \in [0, \pi ]\), since its \(m\)th term is at most \(\abs {z}^{2m} / (2m)!\). So we may integrate term by term and apply Lemma 4.2:
Proof of (d). For real \(x\) the integrand in (c) is a real number in \([-1, 1]\), so \(\abs {J_0(x)} \leq \frac 1\pi \int _0^\pi 1 \dd \theta = 1\).
5 Remarks
Remark 5.1 (An energy proof of (d)). Part (d) also follows from the equation alone. For real \(x > 0\), put \(E = J_0^2 + (J_0')^2\). By (1.2), \(J_0'' = -J_0 - J_0'/x\), so
Hence \(E(x) \leq E(0) = 1\), so \(\abs {J_0(x)} \leq 1\). Since \(J_0\) is even, the same holds for \(x < 0\). Physically, the term \(f'/x\) acts as friction, so the oscillation loses energy and its amplitude decays.
Remark 5.2. Away from \(0\), (1.2) is a second-order linear equation, so it has a two-dimensional space of solutions on any disc not containing \(0\). By (b), only the multiples of \(J_0\) extend analytically to \(0\). Every independent second solution is singular at the origin; the standard choice is the Bessel function \(Y_0\) of the second kind, which has a logarithmic singularity there.