Airy envelopes and asymptotic coefficients

[Definition and ODE] By [F2], $B=E+S$ is smooth and satisfies $B^{\mathrm{\prime }\mathrm{\prime }}=zB$ since the two one-sided forcing terms cancel; analytic continuation gives an entire solution.

[Positive asymptotics] For $Z>0$, substitute $t=\sqrt{Z}+Z^{-1\mathrm{/}4}u$ in $E(Z)$. Its exponent is $2Z^{3\mathrm{/}2}\mathrm{/}3-u^2-u^3\mathrm{/}(3Z^{3\mathrm{/}4})$, with $u\ge -Z^{3\mathrm{/}4}$. On this domain the real exponent after removal of $2Z^{3\mathrm{/}2}\mathrm{/}3$ is at most $-2u^2\mathrm{/}3$. Split at $\mathrm{\mid }u\mathrm{\mid }=Z^{3\mathrm{/}4}\mathrm{/}4$: the outer contribution is exponentially small; on the central part expand the cubic exponential through its linear term. That linear term is odd and integrates to zero on the symmetric central interval; the remainder is bounded by $C{Z}^{-3\mathrm{/}2}\mathrm{\mid }u{\mathrm{\mid }}^6{e}^{-c{u}^2}$ after multiplication by the Gaussian. Thus $E(Z)={\pi }^{-1\mathrm{/}2}{Z}^{-1\mathrm{/}4}{e}^{2Z^{3\mathrm{/}2}\mathrm{/}3}(1+O(Z^{-3\mathrm{/}2}))$. For $E^{\mathrm{\prime }}$ the extra amplitude $t=\sqrt{Z}+Z^{-1\mathrm{/}4}u$ gives the same relative bound: its extra linear term is odd, and its first nonzero cross term is also of relative order $Z^{-3\mathrm{/}2}$. The nonstationary sine integral at positive $Z$ is $O(Z^{-1})$, and its first derivative is polynomially bounded, by integration by parts after scaling $t=\sqrt{Z}s$ with phase derivative $1+s^2$. Both are negligible relative to the exponential terms. This proves the positive formulas in (b).

[Negative asymptotics and endpoint cancellation] Substituting $t=u\mathrm{/}Z$ gives $E(-Z)=1\mathrm{/}(\pi Z)+O(Z^{-4})$, since $\mathrm{\mid }{e}^{-u^3\mathrm{/}(3Z^3)}-1\mathrm{\mid }\le u^3\mathrm{/}(3Z^3)$ and the moments against $e^{-u}$ are finite. In the sine integral use $t=\sqrt{Z}s$, $\lambda =Z^{3\mathrm{/}2}$, and the phase $s^3\mathrm{/}3-s$. Near the endpoint $s=0$, integration by parts supplies $-1\mathrm{/}(\pi Z)$; the next endpoint terms start at $O(Z^{-4})$ because the second derivative of the phase vanishes there. Near $s=1$, the compact Morse-coordinate/Fourier-moment estimate proved in [F1] gives the stationary contribution ${\pi }^{-1\mathrm{/}2}{Z}^{-1\mathrm{/}4}\cos (2Z^{3\mathrm{/}2}\mathrm{/}3+\pi \mathrm{/}4)$ with absolute error $O(Z^{-7\mathrm{/}4})$. On the remaining region repeated nonstationary integration gives a smaller error. The two endpoint contributions cancel in $B(-Z)=E(-Z)+S(-Z)$. Apply the same argument directly to the first differentiated integrals (amplitude $t$), not by differentiating an unspecified remainder: the endpoint terms cancel again, and the stationary term is ${\pi }^{-1\mathrm{/}2}{Z}^{1\mathrm{/}4}\sin (2Z^{3\mathrm{/}2}\mathrm{/}3+\pi \mathrm{/}4)$ with absolute error $O(Z^{-5\mathrm{/}4})$. This gives the negative formulas in (b).

[Wronskian and envelopes] $A{B}^{\mathrm{\prime }}-A^{\mathrm{\prime }}B$ has derivative $A{B}^{\mathrm{\prime }\mathrm{\prime }}-A^{\mathrm{\prime }\mathrm{\prime }}B=0$ by step 1.1 and [F1]. Taking $Z\to +\mathrm{\infty }$ in step 2.1 and [F1] gives the constant ${\pi }^{-1}$. The upper envelopes follow from steps 2.1 and 2.2 outside a compact interval and continuity inside it. For the lower envelope, [F1] gives $A>0$ and $A^{\mathrm{\prime }}<0$ for large positive $z$. In fact $A$ has no zero on $[0,\mathrm{\infty })$: if there were a last zero $z_0$, positivity on $(z_0,\mathrm{\infty })$ would give $A^{\mathrm{\prime }\mathrm{\prime }}=zA\ge 0$ there, so $A^{\mathrm{\prime }}$ would be increasing with limit zero and hence nonpositive, contradicting $A(z_0)=0<A(z)$ for $z>z_0$. A last zero exists if any does, since zeros are isolated and positivity already holds beyond a fixed compact interval. The positive asymptotics and the positive minimum of $A(z)(1+z{)}^{1\mathrm{/}4}{e}^{2z^{3\mathrm{/}2}\mathrm{/}3}$ on that compact interval now yield the global lower envelope. This proves (a)-(c).