Past exam of the theoretical mechanics course of the University of Oxford / 2011 / B

Let $R(t)$ rotate vectors from the rotating frame to the inertial frame. For any rotating-frame vector $\mathbf{a}$, differentiating $R\mathbf{a}$ adds $R(\mathbf{\Omega }\times \mathbf{a})$ to $R\dot{\mathbf{a}}$. Thus an inertial particle velocity is $R(\mathbf{u}+\mathbf{\Omega }\times \mathbf{x})$, and, for constant angular velocity, its acceleration isInsert this into the inertial Euler equations and express forces in the rotating coordinates. This givesThe additional terms are the Coriolis acceleration and the centripetal term associated with frame rotation. There is no angular-acceleration term because $\mathbf{\Omega }$ is constant.

Write the conservative body force as $\mathbf{g}=-\nabla {\Phi }_g$. The supplied vector identity gives $\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{x})=-\frac{1}{2}\nabla |\mathbf{\Omega }\times \mathbf{x}{|}^2$. For constant density, defineMoving the centrifugal and conservative terms to the right then gives ${\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}+2\mathbf{\Omega }\times \mathbf{u}=-{\rho }^{-1}\nabla P$. Only the gradient of this reduced pressure matters, so an additive spatial constant is immaterial.

Linearize about rest, with the equilibrium forces already absorbed in the reduced pressure. The quadratic convective acceleration drops out. Since $\mathbf{k}\times (u\mathbf{i}+v\mathbf{j}+w\mathbf{k})=-v\mathbf{i}+u\mathbf{j}$, the component equations areThe scan prints $v_w$ in the continuity equation; the derivative required by incompressibility is $v_y$.

To eliminate pressure, put $\zeta =v_x-u_y$ and ${\nabla }_h^2={\partial }_x^2+{\partial }_y^2$. Taking the vertical component of the curl of momentum givesTaking horizontal divergence of momentum and substituting continuity givesDifferentiate this in $z$ and use $P_z/\rho =-w_t$:One more time derivative, followed by the vorticity relation, yields the Poincaré equation for inertial wavesThis derivation retains all three velocity components, including the transverse component coupled by rotation.

Write $\omega$ for the nonzero forcing frequency, to distinguish it from the vertical velocity $w$. Put $k_z=\pi /h$ and $Z(z)=\cos (\pi (2z+h)/(2h))=-\sin (k_{z}z)$. Then $Z(0)=Z(h)=0$ and $Z^{\prime \prime }=-k_z^{2}Z$, so it meets the rigid-wall conditions. Substituting $w=\mathrm{Re}(f(x)e^{-i\omega t}Z(z))$ into the scalar wave equation givesorThe radiation sign for an inertial wave must exclude any disturbance arriving from $x=+\infty$. This means an radiation condition for propagating waves, or bounded decay for an evanescent wave.

For $0<|\omega |<2|\Omega |$, put $q=k_z\sqrt{4{\Omega }^2/{\omega }^2-1}$. The solutions are oscillatory. The inertial-wave dispersion relation is ${\omega }^2=4{\Omega }^2{k}_z^2/(k_x^2+k_z^2)$; its horizontal group velocity is $c_{g,x}=-\omega k_x/(k_x^2+k_z^2)$. Positive energy transport into the channel therefore requires $k_x=-\mathrm{sgn}(\omega )q$. The solution isFor a positive frequency, the phase travels towards negative $x$ while the group velocity travels towards positive $x$. Choosing the outgoing sign from the phase velocity alone would give the wrong result.

For $|\omega |>2|\Omega |$, set $\kappa =k_z\sqrt{1-4{\Omega }^2/{\omega }^2}$. Excluding the exponentially growing solution givesThis disturbance decays away from the wavemaker and does not propagate horizontally. At the cutoff the ODE degenerates to $f^{\prime \prime }=0$; zero frequency is not a nonzero forced harmonic mode with this vertical shape.

There is a second printed inconsistency: the stated velocity $u\mathbf{i}+w\mathbf{k}$, if taken literally with no $y$ dependence and no transverse velocity, cannot support a nontrivial rotating inertial wave. Its transverse momentum equation would give $2\Omega u=0$, then continuity and the rigid walls would force $w=0$. The scalar solutions above are the intended inertial-wave solutions when two-dimensional means dependence on $x,z$ while retaining the transverse component $v$. For a mode, $-i\omega v+2\Omega u=0$ explicitly supplies the generally nonzero transverse velocity. Thus the scalar equation can be solved as requested, but the full vector field needs this correction to the printed two-component description.

Take the nondegenerate stationary point in the interior of the integration interval. Choose a smooth cutoff equal to one near $k_{\star }$. Away from its support, ${\psi }^{\prime }$ is bounded away from zero. Integration by parts using $e^{it\psi }=(it{\psi }^{\prime }{)}^{-1}{\partial }_k{e}^{it\psi }$ makes that nonstationary contribution $O(t^{-1})$, under the usual smooth-phase assumptions.

Near the stationary point use the signed coordinateSince ${\psi }^{\prime \prime }(k_{\star })>0$, this is locally invertible, with $ds/dk=\sqrt{{\psi }^{\prime \prime }(k_{\star })}$ at the point. The local integral becomesScale $s=\sqrt{2/t}r$. The supplied Fresnel integral gives ${\int }_{-\infty }^{\infty }{e}^{i{r}^2}dr=e^{i\pi /4}\sqrt{\pi }$. Approximating the smooth amplitude by its value at zero, with a cutoff and a nonstationary estimate for the tails, therefore givesIf $f(k_{\star })\ne 0$, this is the requested asymptotic equivalence. The phase shift $\pi /4$ comes from the positive quadratic curvature. A stationary point at an integration endpoint instead needs a one-sided formula.

The printed auxiliary estimate asserting an $O({\gamma }^{-1})$ Fourier integral for every continuous amplitude is too strong without extra regularity. The argument here uses integration by parts only for the actual smooth, cutoff nonstationary amplitudes, where that estimate is justified; continuity alone generally guarantees convergence to zero, not this rate.

For a negative second derivative, the corresponding local normal form is $\psi =\psi (k_{\star })-s^2/2$. The oscillatory Gaussian integral is the complex conjugate of the positive-curvature one, so the phase shift changes sign. ThusWhen the leading amplitude is nonzero, the equality may again be written as an asymptotic equivalence. Both signs are summarized by the factor $e^{i\pi \mathrm{sgn}({\psi }^{\prime \prime })/4}$.

Irrotationality permits a velocity potential with $\mathbf{u}=({\varphi }_x,{\varphi }_z)$. Incompressibility gives Laplace's equation in the moving domain $z<\eta (x,t)$. Neglecting surface tension and taking the free-surface pressure as constant atmospheric pressure, the full surface conditions areThe kinematic condition makes the free surface material; the dynamic condition is Bernoulli's equation on that surface. A time-dependent additive gauge in $\varphi$ has absorbed the spatially uniform Bernoulli constant. Deep below the surface, impose $\nabla \varphi \to 0$ as $z\to -\infty$, together with suitable horizontal decay. Initially $\eta (x,0)={\eta }_0(x)$ and the fluid is at rest, so $\nabla \varphi (x,z,0)=0$; its initial constant may be chosen as zero.

Linearize about a flat quiescent surface. The product ${\varphi }_x{\eta }_x$ and the kinetic-energy term are quadratic, and the boundary conditions can be evaluated at $z=0$ to first order. Thus ${\eta }_t={\varphi }_z$ and ${\varphi }_t+g\eta =0$ there. Fourier transformation in $x$ changes ${\partial }_x^2$ into multiplication by $-k^2$, givingThe transform uses $e^{-ikx}$ in its forward definition; the inverse transform therefore uses $e^{ikx}/(2\pi )$. The zero-wavenumber potential has an irrelevant spatially constant gauge freedom.

For $k\ne 0$, the solution decaying at depth is $\hat{\varphi }(k,z,t)=C(k,t)e^{|k|z}$. The surface conditions give ${\hat{\eta }}_t=|k|C$ and $C_t=-g\hat{\eta }$, henceInitial rest implies ${\hat{\eta }}_t(k,0)=0$. Define the transform of the initial elevation by ${\hat{\eta }}_0(k)={\int }_{\mathbb{R}}{\eta }_0(x)e^{-ikx}dx$. Solving this oscillator equation and inverting the transform givesThe formula also includes the continuous zero-frequency limit. For real even initial data, the transform is real and even, ensuring a real even surface profile at every time.

For wavenumber selection by a moving observer of deep-water waves, an observer with nonzero velocity $V$ follows $x=Vt$ up to a constant initial displacement. Splitting the cosine into exponentials gives phase functions ${\psi }_{\pm }(k)=kV\pm \sqrt{g|k|}$. Their stationary points satisfyso the selected wavenumbers areThe two signs arise in the conjugate branches needed for a real wave. Equivalently the observer selects wave packets whose group velocity equals $V$. Stationary-phase contributions decay as $t^{-1/2}$, whereas smooth nonstationary contributions decay more rapidly, so these are the dominant wavenumbers when the initial spectrum is nonzero there.

For real even data, one may make this explicit. Set $k_{\ast }=g/(4V^2)$. The stationary curvature has magnitude $2|V{|}^3/g$, yieldingwhen its leading coefficient is nonzero. If that spectral value vanishes, the leading stationary term vanishes too. At $V=0$ there is no finite nonzero stationary wavenumber selected by this condition.

A weak solution is a field $\mathbf{u}$ for which the composed densities $\mathbf{P}(x,t,\mathbf{u})$ and fluxes $\mathbf{Q}(x,t,\mathbf{u})$ are locally integrable andfor every smooth compactly supported vector test function $\mathbf{\phi }$ in the spacetime domain. This is the distributional form of the conservation law, obtained by moving derivatives onto the test function. If tests are allowed to meet an initial boundary $t=0$, the specified initial density contributes the additional term $\int \mathbf{P}(x,0,{\mathbf{u}}_0)\cdot \mathbf{\phi }(x,0)dx$ in the usual initial-value formulation.

If the fields are continuously differentiable and obey the pointwise conservation law, integration by parts against a compactly supported test function gives the weak identity, with no boundary terms. Conversely, integrate the weak identity by parts for a continuously differentiable field:for every test function. The fundamental lemma of the calculus of variations, applied to each component, makes the continuous integrand vanish at every point. Hence the classical and weak formulations are equivalent for continuously differentiable fields.

Write the shock as $x=s(t)$ and let the plus trace be on $x>s(t)$. Locally its discontinuous part can be represented with the Heaviside factor $H(x-s(t))$. Distributional differentiation gives ${\partial }_{t}H=-s^{\prime }(t)\delta (x-s(t))$ and ${\partial }_{x}H=\delta (x-s(t))$. The singular part of the conservation law is thereforeIt must vanish, so the Rankine-Hugoniot conditions areThe smooth parts of the fields still satisfy the classical conservation law on each side. The jump condition ensures that the moving discontinuity creates no extra source of the conserved quantities.

Put $v=u-V$, the velocity in the shock frame. The mass jump condition gives a common mass flux $j=\rho v$. Subtracting $V$ times the mass balance from the momentum balance givesFor energy, transform the conserved energy to the shock frame by subtracting $V$ times momentum and adding $V^2/2$ times mass. The resulting energy flux isFor a genuine shock, $j\ne 0$; dividing its continuous energy flux by the same nonzero $j$ on both sides givesThe second term is the ideal-gas specific enthalpy. A contact discontinuity has zero relative mass flux and does not in general require equal specific enthalpies; the division here is appropriate for the shock specified in the question.

Let the undisturbed state ahead have $({\rho }_0,p_0,u_0)=({\rho }_0,p_0,0)$, and the uniform compressed state behind have velocity $u_1=U$. The piston boundary fixes that downstream velocity. Mass and momentum jumps giveThus ${\rho }_1/{\rho }_0=V/(V-U)$. Equality of specific stagnation enthalpy in the shock frame isSubstitute $p_1=p_0+{\rho }_{0}UV$ and $1/{\rho }_1=(V-U)/({\rho }_{0}V)$, then multiply by $2(\gamma -1)V$. The resulting difference factors as $U[-2V^2+(\gamma +1)UV+2\gamma p_0/{\rho }_0]=0$. Since the piston moves into the gas with $U>0$, division givesThis determines the constant speed of the compression shock ahead of the piston.

For a piston-driven normal shock with $U>0$, $c_0>0$ and $\gamma >1$, the quadratic has one positive and one negative root. The forward-moving shock usesSince the square root is greater than $(\gamma +1)U$,Therefore there is always a forward root with shock speed greater than the gas and piston speed under the stated compression assumptions. This also makes the density ratio $V/(V-U)$ positive and greater than one. Evaluating the quadratic at $V=c_0$ gives a negative value, so the positive root additionally has $V>c_0$: it is supersonic relative to the undisturbed gas. The negative root does not describe the shock driven into the region $x>0$.