Stable weighted Airy Volterra connection

[Convergence of the Volterra series] By [F2] the solution is the sum of the ordered iterates of the kernel $\pi h^{2\mathrm{/}3}[A_h(\zeta )B_h(v)-B_h(\zeta )A_h(v)]R(v)$; by [F3] each iterate gains a factor bounded by $Ch\int \mathrm{\mid }R(v)\mathrm{\mid }\mathrm{/}\sqrt{\mathrm{\mid }v\mathrm{\mid }+h^{2\mathrm{/}3}}dv={\epsilon }_h$ and a factorial denominator, so the series converges and defines the unique solution with the stated terminal data.

[Error bounds] Subtracting the leading term from the series and bounding the remainder by the exponential series give $\mathrm{\mid }w-A_h\mathrm{\mid }\le C_A{e}_h(e^{{\epsilon }_h}-1)$. To differentiate the integral equation, the boundary term vanishes because $K_h(z,z)=0$; the remaining integral contains ${\mathrm{\partial }}_z{K}_h$, which obeys the derivative estimate in [F3]. Insert the value bound from step 1.1 under this integral: ordered integration yields $\mathrm{\mid }{w}^{\mathrm{\prime }}-A_h^{\mathrm{\prime }}\mathrm{\mid }\le C_A{d}_h(e^{{\epsilon }_h}-1)$. Finally, ${\int }_{-b}^c(\mathrm{\mid }v\mathrm{\mid }+h^{2\mathrm{/}3}{)}^{-1\mathrm{/}2}dv\le 2(\sqrt{b}+\sqrt{c})$, so boundedness of $R$ implies ${\epsilon }_h\le C^{\mathrm{\prime }}h$. This proves (b).

[Growing branch and independence] The solution with data $(B_h(0),B_h^{\mathrm{\prime }}(0))$ satisfies $w_g(z)=B_h(z)-{\int }_0^z{K}_h(z,v)R(v)w_g(v)dv$. On $z\ge 0$ use ordered forward integrals and on $z\le 0$ ordered backward integrals; the growing kernel bounds in [F3] give $\mathrm{\mid }{w}_g-B_h\mathrm{\mid }\le C_A{g}_h(e^{{\epsilon }_h}-1)$ and $\mathrm{\mid }{w}_g^{\mathrm{\prime }}-B_h^{\mathrm{\prime }}\mathrm{\mid }\le C_A{\tilde{d}}_h(e^{{\epsilon }_h}-1)$ on the full interval, by the same factorial estimates. At zero the decaying solution obeys $w(0)=A_h(0)+O(h)$ and $w^{\mathrm{\prime }}(0)=A_h^{\mathrm{\prime }}(0)+O(h^{1\mathrm{/}3})$, while the growing data are exact. Their Wronskian is consequently $h^{-2\mathrm{/}3}[{\pi }^{-1}+O(h)]$, which is nonzero for small $h$ and constant because both solve the same equation without a first-derivative term. Thus the branches are independent.

[Connection] Write $u=aw$, using the Liouville identity $a=\mathrm{\mid }\zeta {\mathrm{\mid }}^{1\mathrm{/}4}\mathrm{\mid }q{\mathrm{\mid }}^{-1\mathrm{/}4}$ away from zero. The phase $2\mathrm{\mid }\zeta {\mathrm{\mid }}^{3\mathrm{/}2}\mathrm{/}(3h)$ equals the action integral divided by $h$. Normalize the decaying branch by $2\sqrt{\pi }{h}^{-1\mathrm{/}6}$ and the growing branch by $\sqrt{\pi }{h}^{-1\mathrm{/}6}$. The positive and negative Airy asymptotics of [F1] then give respectively the forbidden amplitudes $q^{-1\mathrm{/}4}{e}^{-\int \sqrt{q}\mathrm{/}h}$ and $q^{-1\mathrm{/}4}{e}^{+\int \sqrt{q}\mathrm{/}h}$, and the allowed amplitudes $2(-q{)}^{-1\mathrm{/}4}\sin (\int \sqrt{-q}\mathrm{/}h+\pi \mathrm{/}4)$ and $(-q{)}^{-1\mathrm{/}4}\cos (\int \sqrt{-q}\mathrm{/}h+\pi \mathrm{/}4)$. Steps 2.1 and 3.1 add errors bounded by the respective envelopes times $C{\epsilon }_h$, and the Airy asymptotic remainders add $Ch\mathrm{/}\mathrm{\mid }\zeta {\mathrm{\mid }}^{3\mathrm{/}2}$ on fixed matching sections. For slopes apply $u^{\mathrm{\prime }}=a^{\mathrm{\prime }}w+a{\zeta }^{\mathrm{\prime }}{w}^{\mathrm{\prime }}$ and the derivative bounds; no differentiation of an unspecified asymptotic remainder is needed. Reflection $x\mapsto -x$ handles a turning point with $q^{\mathrm{\prime }}<0$. This proves (c).