Past exam of the theoretical mechanics course of the University of Oxford / 2009 / B / 5

Split $I$ into the two tails and the interval $|k-k_{\ast }|<\delta$. On each tail ${\psi }^{\prime }$ has a fixed nonzero sign, so changing variable to $s=\psi (k)$ produces a nonstationary Fourier integral. Under the usual decay and regularity needed for integration by parts, these pieces are $O(t^{-1})$.

Near the nondegenerate stationary point, make the signed local changeIt has $dk/dz=1$ at zero and makes the phase exactly $\psi (k_{\ast })+{\psi }^{\prime \prime }(k_{\ast })z^2/2$. The transformed amplitude tends to $f(k_{\ast })$. Scale $z$ by $t^{-1/2}$ and use the Fresnel integral to obtainThe $O(t^{-1})$ tails are smaller than this leading stationary contribution. If ${\psi }^{\prime \prime }(k_{\ast })<0$, complex conjugation of the Gaussian phase changes the phase correction to $-\pi /4$ and replaces the second derivative by its absolute value:When the leading coefficient is nonzero this may be written as an asymptotic equivalence. For full rigor, continuity of an arbitrary amplitude alone is not sufficient for the stated nonstationary $O(t^{-1})$ rate, nor does it ensure convergence of an infinite integral. Piecewise differentiability with integrable derivative, bounded variation, or an equivalent localization/decay hypothesis supplies the usual missing assumption. The exponentially decaying beam amplitude below has the required piecewise regularity.

Use $\hat{\eta }(k,t)=\int \eta (x,t)e^{-ikx}dx$ and the inverse factor $1/(2\pi )$. Fourier transformation gives ${\hat{\eta }}_{tt}+{\lambda }^2{k}^4\hat{\eta }=0$. The supplied Lorentzian transform and zero initial velocity therefore implyAlong $x=Vt$, split the cosine into two exponentials. Their phases are ${\psi }_{\pm }=Vk\pm \lambda k^2$, with stationary points $k_{\pm }=\mp V/(2\lambda )$, phase values $\mp V^2/(4\lambda )$ and second derivatives $\pm 2\lambda$. Both amplitudes equal $a{e}^{-ϵ|V|/(2\lambda )}$. The two stationary-phase contributions are conjugate, givingThus the finite-width pulse retains a ray-dependent attenuation. In the further regime $ϵ|V|/(2\lambda )\ll 1$, it may be replaced by one, recoveringThe large-time ray limit means $V=x/t$ is fixed. Merely making both $x$ and $t$ large does not justify dropping the attenuation if $|x|/t$ becomes large compared with $\lambda /ϵ$. The leading oscillation decays as $t^{-1/2}$ because each saddle samples a Fourier band of width $t^{-1/2}$.