Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 39C (Waves)

Substitute the monochromatic waveinto the differential equation. Since $(ik{)}^9=i{k}^9$, cancellation of the common exponential givesThus the dispersion relation is

From the dispersion relation, the phase velocity and group velocity areBoth are even functions of $k$, both attain their minimum value $U$ at $k=0$, and both increase like $|k{|}^8$. For every $k\ne 0$,The graph of $c_g$ therefore lies above the graph of $c_p$ away from their common minimum.

A wave crest moves at the phase velocity, while the envelope of a narrow wave packet moves at the group velocity. Since $c_g>c_p$ for every nonzero wavenumber, the crests move more slowly than the packet.

The Fourier modes evolve according to the dispersion relation found in part (a), soFor $V>U$, the stationary-point equation isIt has the two real solutionsDefineThe one-dimensional stationary-phase formula gives one contribution from each stationary point:Because the initial field is real, its Fourier transform has the conjugate symmetry $A(-k)=A(k{)}^{\ast }$. Hence the same result can be written

If the initial field is also an even function, then $A(k)$ is real and even. Provided $A(k_0)\ne 0$, the leading term becomesIts large zeros satisfyThereforeIf $A(k_0)=0$, the leading stationary-phase coefficient vanishes and a higher-order term determines the zeros.

For $V<U$,for every real $k$, so there is no real stationary point. The ray $x=Vt$ travels more slowly than the minimum group velocity $U$ and receives no leading stationary-phase wave packet. Under the usual smoothness and decay assumptions on $A$, the nonstationary oscillatory integral can be integrated by parts repeatedly, sofor every fixed $N$ for which the needed derivatives are integrable. In particular it is asymptotically much smaller than the $t^{-1/2}$ signal when $V>U$.