A duck swims at a constant velocity $(-V, 0)$, where $V>0$, on the surface of infinitely deep water. Surface tension can be neglected, and the dispersion relation for the linear surface water waves (relative to fluid at rest) is $\omega^{2}=g|\mathbf{k}|$. Show that the wavevector $\mathbf{k}$ of a plane harmonic wave that is steady in the duck's frame, i.e. of the form

$$\operatorname{Re}\left[A e^{i\left(k_{1} x^{\prime}+k_{2} y\right)}\right]$$

where $x^{\prime}=x+V t$ and $y$ are horizontal coordinates relative to the duck, satisfies

$$\left(k_{1}, k_{2}\right)=\frac{g}{V^{2}} \sqrt{p^{2}+1}(1, p)$$

where $\hat{\mathbf{k}}=(\cos \phi, \sin \phi)$ and $p=\tan \phi$. [You may assume that $|\phi|<\pi / 2 .$ ]

Assume that the wave pattern behind the duck can be regarded as a Fourier superposition of such steady waves, i.e., the surface elevation $\eta$ at $\left(x^{\prime}, y\right)=R(\cos \theta, \sin \theta)$ has the form

$$\eta=\operatorname{Re} \int_{-\infty}^{\infty} A(p) e^{i \lambda h(p ; \theta)} \mathrm{d} p \quad \text { for }|\theta|<\frac{1}{2} \pi$$

where

$$\lambda=\frac{g R}{V^{2}}, \quad h(p ; \theta)=\sqrt{p^{2}+1}(\cos \theta+p \sin \theta)$$

Show that, in the limit $\lambda \rightarrow \infty$ at fixed $\theta$ with $0<\theta<\cot ^{-1}(2 \sqrt{2})$,

$$\eta \sim \sqrt{\frac{2 \pi}{\lambda}} \operatorname{Re}\left\{\frac{A\left(p_{+}\right)}{\sqrt{h_{p p}\left(p_{+} ; \theta\right)}} e^{i\left(\lambda h\left(p_{+} ; \theta\right)+\frac{1}{4} \pi\right)}+\frac{A\left(p_{-}\right)}{\sqrt{-h_{p p}\left(p_{-} ; \theta\right)}} e^{i\left(\lambda h\left(p_{-} ; \theta\right)-\frac{1}{4} \pi\right)}\right\},$$

where

$$p_{\pm}=-\frac{1}{4} \cot \theta \pm \frac{1}{4} \sqrt{\cot ^{2} \theta-8}$$

and $h_{p p}$ denotes $\partial^{2} h / \partial p^{2}$. Briefly interpret this result in terms of what is seen.

Without doing detailed calculations, briefly explain what is seen as $\lambda \rightarrow \infty$ at fixed $\theta$ with $\cot ^{-1}(2 \sqrt{2})<\theta<\pi / 2$. Very briefly comment on the case $\theta=\cot ^{-1}(2 \sqrt{2})$.

[Hint: You may find the following results useful.

$$\begin{gathered} h_{p}=\left\{p \cos \theta+\left(2 p^{2}+1\right) \sin \theta\right\}\left(p^{2}+1\right)^{-1 / 2} \\ \left.h_{p p}=(\cos \theta+4 p \sin \theta)\left(p^{2}+1\right)^{-1 / 2}-\left\{p \cos \theta+\left(2 p^{2}+1\right) \sin \theta\right\} p\left(p^{2}+1\right)^{-3 / 2} \cdot\right] \end{gathered}$$