Skip to main content

Bessel function second-derivative identity

Short answer

Using the identity 2J_n' = J_{n-1} - J_{n+1}, differentiate to obtain 2J_n'' = J_{n-1}' - J_{n+1}', then substitute the same identity for J_{n-1}' and J_{n+1}'. This gives 4J_n'' = J_{n-2} - 2J_n + J_{n+2}.

Show that 4J_n''(x) = J_(n−2)(x) − 2J_n(x) + J_(n+2)(x).

Use the standard relation obtained from the series of Jₙ:

2Jₙ′(x) = Jₙ₋₁(x) − Jₙ₊₁(x) ... (1)

This holds for every order n, so we can apply it again to each term.

Differentiate (1)

2Jₙ″ = Jₙ₋₁′ − Jₙ₊₁′

Apply (1) with n replaced by n − 1 and by n + 1:

2Jₙ₋₁′ = Jₙ₋₂ − Jₙ
2Jₙ₊₁′ = Jₙ − Jₙ₊₂

Combine

Multiply 2Jₙ″ = Jₙ₋₁′ − Jₙ₊₁′ by 2:

4Jₙ″ = 2Jₙ₋₁′ − 2Jₙ₊₁′
     = (Jₙ₋₂ − Jₙ) − (Jₙ − Jₙ₊₂)
     = Jₙ₋₂ − 2Jₙ + Jₙ₊₂

Hence 4Jₙ″(x) = Jₙ₋₂(x) − 2Jₙ(x) + Jₙ₊₂(x).

Proof of relation (1) (for completeness)

From the two standard results xJₙ′ = nJₙ − xJₙ₊₁ and xJₙ′ = −nJₙ + xJₙ₋₁ (obtained from d/dx[x⁻ⁿJₙ] = −x⁻ⁿJₙ₊₁ and d/dx[xⁿJₙ] = xⁿJₙ₋₁), adding gives 2xJₙ′ = x(Jₙ₋₁ − Jₙ₊₁).

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics chapter 4 questions →