Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 61 / 1 / d / Solution

For asymptotic analysis it is useful to separate the spectral density from the observation time. For a fixed horizon $T\ge t$, $H(k,T)$ already has this property. If the boundary traces extend to all positive times with suitable growth, their one-sided Laplace transforms give the infinite-horizon versionFor bounded traces these integrals converge in $D_{+}$; boundary values are taken by an Abel limit where needed. The finite-horizon formula permits this replacement: the tail from $t$ to infinity contributes an analytic, decaying integrand with zero integral around $\partial D_{+}$. The resulting spectral functions are independent of $t$.

They can be made particularly explicit on the two contour pieces. On the real axis and upper curve, respectively,The second identity follows because $k^3+k$ is real on $C$: the remaining roots are $\overline{k}$ and $-2\mathrm{Re}k$. Therefore rewrite the solution asThis isolates all large-time dependence in the exponential, allowing the stationary phase method or the method of steepest descent. Along $x/t=\xi$, the phase is $i[k\xi -k-k^3]$, and its stationary points satisfy $3k^2=\xi -1$. Thus the real saddles coalesce at $\xi =1$; any poles arising from the boundary Laplace transforms also have to be retained during contour deformation.

The source specifies smooth boundary traces, not their large-time decay or growth. Accordingly a universal numerical long-time profile cannot be inferred. The infinite-horizon formula requires the stated additional transform hypothesis; for more general data the finite-horizon density $H(k,T)$ remains valid, or an appropriately shifted Laplace contour must be used. This qualification does not leave an unknown boundary trace in the representation.