Past exam of the theoretical mechanics course of the University of Oxford / 2008 / B

The first energy term is the rate of change of internal plus kinetic energy per unit volume; the second transports that energy with the moving gas; the pressure term is the mechanical work flux across the boundary. There is no heat flux or viscous work in the stated inviscid adiabatic model.

Use mass conservation to replace a conservative derivative of $\rho e$ by $\rho De/Dt$. Dotting the momentum equation with $\mathbf{u}$ gives $\rho D(|\mathbf{u}{|}^2/2)/Dt=-\mathbf{u}\cdot \nabla p$. Subtraction from total energy leavesNow $p=\rho RT$ and $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$ implyThus $\boxed{DS/Dt=0}$: smooth adiabatic inviscid trajectories are isentropic. Shocks need an entropy inequality rather than this smooth equality.

Adding $\nabla \cdot (k\nabla T)$ models Fourier heat conduction, with heat flux $-k\nabla T$. The same subtraction gives $\rho c_{v}DT/Dt+p\nabla \cdot \mathbf{u}=\nabla \cdot (k\nabla T)$, henceFor a fixed container use impermeability $\mathbf{u}\cdot \mathbf{n}=0$, insulation $k\nabla T\cdot \mathbf{n}=0$, and mass conservation. ThenThe boundary term vanishes and $\nabla (1/T)=-\nabla T/T^2$ supplies the positive sign. For $T>0$, $k\ge 0$, total entropy cannot decrease: this is entropy production by Fourier heat conduction.

Rigid walls require zero normal velocity, so ${\varphi }_x=0$ at $x=0,L$ and ${\varphi }_y=0$ at $y=0,b$. Separation of variables gives spatial modes $\cos (m\pi x/L)\cos (n\pi y/b)$, for nonnegative integers $m,n$. The wave equation therefore givesThe $(0,0)$ spatially constant mode has zero frequency and is not an ordinary positive-frequency acoustic oscillation.

A spatially uniform boundary forcing excites the $n=0$ transverse mode. Put $\varphi =X(x)\cos \omega t$. Then $X^{\prime \prime }+(\omega /c_0{)}^{2}X=0$ and $X^{\prime }(L)=0$, hence $X=C\cos [\omega (L-x)/c_0]$. The condition $X^{\prime }(0)=a$ fixes $C$ and givesArbitrary free oscillations can be superposed. The denominator vanishes at longitudinal resonances. Near them the lossless steady harmonic response ceases to be a uniformly small-amplitude approximation; damping from viscosity, heat transfer or wall losses regularizes the response, and large amplitudes can require nonlinear equations of motion or a more accurate moving-boundary model. At exact resonance a switched-on ideal forcing produces secular growth rather than a bounded harmonic particular solution.

For the semi-infinite duct expand $F(y)={\sum }_{n\ge 0}{F}_n\cos (n\pi y/b)$. Each axial mode hasFor a propagating mode a particular solution is ${\varphi }_n=(F_n/k_n)\cos (n\pi y/b)\cos (k_{n}x-\omega t)$; differentiating at zero gives the prescribed $F_n\cos (n\pi y/b)\sin \omega t$. A nonzero higher transverse coefficient propagates when $|\omega |>n\pi c_0/b$.

The outgoing radiation condition selects $c_0^2{k}_n/\omega >0$, its positive axial group velocity. Thus $\boxed{\omega k_n>0}$. This sign is selected by the absence of waves incident from infinity, not by the differential equation alone. If $F$ has a nonzero mean, its $n=0$ component can propagate even below every higher-mode cutoff.

For $F=\cos (\pi y/b)$ and $0<|\omega |<\pi c_0/b$, put $\kappa =\sqrt{(\pi /b{)}^2-{\omega }^2/c_0^2}$. The decaying solution isIt is evanescent and carries no mean energy to infinity. These mode and sign conditions are summarized by the radiation condition for a waveguide mode.

Along $x=Vt+x_0$, the phase is $t\Phi (k)+k{x}_0$, with $\Phi (k)=kV-\omega (k)$. Graphically, a line of slope $V$ intersects the dispersion curve transversely except where it is tangent. On a transverse region, neighboring wavenumbers acquire order-one phase differences over a width $\Delta k\sim t^{-1}$, so cancellation leaves an $O(t^{-1})$ contribution. More formally, integration by parts usesand gives that order when the amplitude and its derivative are integrable and ${\Phi }^{\prime }$ is bounded away from zero on the region.

At a nondegenerate tangent point $k_{\ast }$, ${\Phi }^{\prime }(k_{\ast })=0$, equivalently $\boxed{V={\omega }^{\prime }(k_{\ast })}$, and $\Phi (k)=\Phi (k_{\ast })-{\omega }^{\prime \prime }(k_{\ast })(k-k_{\ast }{)}^2/2+\cdots$. Coherent wavenumbers now occupy width $O(t^{-1/2})$, yielding the stationary-phase contributionThese are the packets whose group velocity follows the observer. The displayed decay rates require the usual regularity and nondegeneracy hypotheses; $A\to 0$ alone is not a complete analytic hypothesis, and a vanishing amplitude or degenerate stationary point changes the leading order.

For deep water take a velocity potential satisfying Laplace's equation below the undisturbed surface and decaying with depth. A Fourier mode has vertical factor $e^{|k|z}$. The linearized kinematic and dynamic surface conditions are ${\eta }_t={\varphi }_z$ and ${\varphi }_t+g\eta =0$ at $z=0$. ThusRelease from rest gives ${\overline{\eta }}_t(k,0)=0$, so inversion of the Fourier transform givesEvenness of the initial profile makes its transform even and the reconstructed surface even; the same formula also works without that symmetry when the usual reality condition holds.

Split the cosine into its two time phases. Along the moving observer, their stationary conditions are $\pm {\omega }^{\prime }(k)=V$, where $\omega (k)=\sqrt{g|k|}$. Since $|{\omega }^{\prime }(k)|=\sqrt{g}/(2\sqrt{|k|})$, for $V\ne 0$,These are the possible dominant stationary-phase packets, provided the initial Fourier amplitude at the selected wavenumber is nonzero. For $V=0$ there is no finite nonzero stationary wavenumber; the formula must not be applied by division by zero.

Let the bed be $z=0$ and the free surface $z=h(x,t)$. In a long, shallow layer, vertical acceleration is smaller than gravity, so the vertical Euler equation gives hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. Incompressibility makes the vertical speed smaller than the horizontal speed by the aspect ratio; irrotationality then makes the leading horizontal velocity depth-independent. The horizontal Euler equation becomesIntegrating $u_x+w_z=0$ across the depth, using bed impermeability $w(0)=0$ and the surface condition $w(h)=h_t+u{h}_x$, yieldsCombining these gives the conservative momentum equation $(hu{)}_t+(h{u}^2+g{h}^2/2{)}_x=0$. Integral mass and momentum balance across a stationary thin jump therefore requireThese conservation laws still apply across a jump where the smooth hydrostatic/irrotational derivation does not resolve the internal turbulent structure.

Put $A=h_{-}$, $B=h_{+}$ and $q=A{u}_{-}=B{u}_{+}>0$, orienting the flow from minus to plus. For a nontrivial jump the momentum condition gives $q^2=gAB(A+B)/2$. Its change in mechanical-energy flux isA passive jump cannot create mechanical energy, so the change is nonpositive. Since $q>0$, $\boxed{h_{+}>h_{-}}$ for a genuine jump; equality describes no jump. The upstream flow is supercritical and the downstream flow subcritical, since $F{r}_{-}^2=B(A+B)/(2A^2)>1$ and $F{r}_{+}^2=A(A+B)/(2B^2)<1$.

The gas-dynamic analogue is a compressive normal shock: density and pressure increase and flow changes from supersonic to subsonic. Entropy increases; directed mechanical energy is irreversibly converted into internal energy. Total gas energy, including internal energy, is conserved across an adiabatic shock. This distinguishes that total-energy balance from the reduced shallow-water mechanical-energy loss.