 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-2/13a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Let

u (x) = J_{0} (αx), v (x) = J_{0} (β x) .

(x u^{'})^{'} + α^{2} xu = 0, (x v^{'})^{'} + β^{2} xv = 0.

Multiply the first by

v

, the second by

u

, subtract, and integrate. The left side becomes a boundary term:

[x (v u^{'} - u v^{'})]_{0}^{1} + (α^{2} - β^{2}) \int_{0}^{1} uvx d x = 0.

Regularity at zero makes the lower boundary term vanish, while

u^{'} (1) = α J_{0}^{'} (α), v^{'} (1) = β J_{0}^{'} (β) .

Therefore

\int_{0}^{1} J_{0} (αx) J_{0} (β x) x d x = \frac{β J _{0} ( α ) J _{0}^{'} ( β ) - α J _{0} ( β ) J _{0}^{'} ( α )}{α ^{2} - β ^{2}} .

α = γ_{k}

and

β = γ_{ℓ}

with

k \neq = ℓ

, both endpoint values of

J_{0}

vanish, so the integral is zero. This is the weighted orthogonality of distinct eigenfunctions in a Sturm-Liouville problem. For the norm, hold

α = γ_{k}

and let

β \to γ_{k}

. Since

J_{0} (β) = J_{0}^{'} (γ_{k}) (β - γ_{k}) + o (β - γ_{k}),

the quotient tends to

\int_{0}^{1} J_{0} (γ_{k} x)^{2} x d x = \frac{1}{2} J_{0}^{'} (γ_{k})^{2} .

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