Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2009 / A / 1

For $x>0$, the real and imaginary parts of the exponent divided by $x$ areThus constant-phase contours satisfy $u^3-3u{v}^2=c$. Through zero, $c=0$ gives the six rays $\theta =\pi /6+j\pi /3$. The three rays $\theta =\pi /6,5\pi /6,3\pi /2$ have decreasing real part of the exponent and are steepest descent contours; the alternating three are steepest ascent rays.

Through one, the phase is $c=1$. Its branch passing through that point can be writtenwith the cube root continuous from $t=1$. The exponent is $ix-xs$, so increasing $s$ from zero gives descent and decreasing $s$ gives ascent.

Deform the finite interval into the ray $t=e^{i\pi /6}r$, $0\le r<\infty$, followed by the reverse of the descent branch from one. The integrand is entire, so Cauchy's integral theorem permits the deformation.

For a precise connector at infinity, use $t=(a+iR{)}^{1/3}$, $0\le a\le 1$. Its length is $O(R^{-2/3})$ and the exponential has modulus $e^{-xR}$. Its contribution is therefore $O(R^{-2/3}{e}^{-xR})\to 0$. This avoids relying on an undifferentiated claim that every large circular arc decays.

A saddle point of the exponential phase occurs where $(i{t}^3{)}^{\prime }=3i{t}^2=0$. Hence the only saddle isIt is degenerate: the second phase derivative vanishes there, while the third is $6i\ne 0$. This cubic degeneracy produces the $x^{-1/3}$ scale. The other endpoint $t=1$ is not a saddle, since the phase derivative there is $3i$.

On the selected steepest descent contour put $t=e^{i\pi /6}r$, $r\ge 0$. Then $i{t}^3=-r^3$ and $dt=e^{i\pi /6}dr$. Its exact contribution isUse the Gamma function integral and its recurrence $\Gamma (4/3)=\Gamma (1/3)/3$ to obtainThere is no remainder here: the complete infinite descent ray was integrated exactly.

The endpoint branch has $t=(1+is{)}^{1/3}$ and $dt=(i/3)(1+is{)}^{-2/3}ds$. Its reversed orientation in the deformed contour givesAt $s=0$,Watson's lemma, or elementary Laplace's method with a Taylor remainder, now yieldsAdding the exact cubic-saddle contribution gives the two leading termsIf the next term is retained, it is $-2e^{ix}/(9x^2)$. The different powers reflect a degenerate saddle at zero and an ordinary endpoint at one.