Past exam of the theoretical mechanics course of the University of Oxford / 2007 / B

For a material volume, mass conservation is $d{\int }_{V(t)}\rho dV/dt=0$. The transport theorem and arbitrariness of that volume give $\boxed{{\rho }_t+\nabla \cdot (\rho u)=0}$. Balance of linear momentum includes the pressure traction and body force:Apply transport, the divergence theorem and the mass equation to obtainFor small perturbations of the uniform resting barotropic state, write $\rho ={\rho }_0+{\rho }^{\prime }$, $p=p_0+c_0^2{\rho }^{\prime }+\cdots$. Linearized mass and momentum areTaking the curl shows that initially vanishing vorticity remains zero, so the acoustic velocity has a potential $u=\nabla \varphi$. Integrating momentum, using the additive freedom of a velocity potential to remove the spatially constant integration term, gives ${\rho }^{\prime }=-({\rho }_0/c_0^2){\varphi }_t$. Continuity now yields $\boxed{{\varphi }_{tt}=c_0^2{\nabla }^2\varphi }$.

For spherical symmetry, set $\varphi =e^{-i\omega t}F(r)$ and $k=\omega /c_0$. The radial equation is $(r^2{F}^{\prime }{)}^{\prime }+k^2{r}^{2}F=0$, or $(rF{)}^{\prime \prime }+k^2(rF)=0$. HenceRigid boundaries require radial velocity $F^{\prime }=0$ at both radii. A form automatically satisfying the condition at $a$ isAt $b$, the condition $b(rF{)}^{\prime }-(rF)=0$ becomesTherefore the nonzero oscillatory eigenfrequencies obeyThe undivided sine-cosine form avoids artificial issues from dividing by a cosine. The formal $k=0$ root is the spatially constant zero-frequency potential additive freedom of a velocity potential, not a propagating acoustic oscillation. This is the radial acoustic spectrum between rigid concentric spheres.

For a vector $a$ described in axes rotating at constant angular velocity $\Omega$, the rotating-axes theorem is $(da/dt{)}_{\mathrm{inertial}}=(da/dt{)}_{\mathrm{rotating}}+\Omega \times a$. Inertial particle velocity is $u+\Omega \times x$. Applying the theorem once more gives inertial acceleration $Du/Dt+2\Omega \times u+\Omega \times (\Omega \times x)$. Thus the rotating momentum equation isIf $g=-\nabla \Phi$ and density is constant, the centrifugal term can be absorbed into the reduced pressure in a rotating fluid:For small disturbances with $\Omega =\Omega e_z$, linearize to $u_t-2\Omega v=-P_x/\rho$, $v_t+2\Omega u=-P_y/\rho$, $w_t=-P_z/\rho$, together with incompressibility. Put $\zeta =v_x-u_y$. Taking the vertical curl gives ${\zeta }_t=2\Omega w_z$. Taking divergence gives ${\nabla }^{2}P=2\rho \Omega \zeta$. Applying the Laplacian to the vertical momentum equation therefore gives ${\partial }_t{\nabla }^{2}w=-2\Omega {\zeta }_z$. Another time derivative yieldsAll three velocity components are retained, even for fields independent of $y$; rotation produces the transverse component.

Set $m=\pi /h$ and seek $w=a{e}^{-i\omega t}W(x)\sin mz$. The rigid-wall conditions are then automatic and the scalar equation givesFor $0<|\omega |<2|\Omega |$, define $q=m\sqrt{4{\Omega }^2/{\omega }^2-1}$. The two possible oscillatory factors are distinguished by energy propagation. With horizontal wavenumber $k$, the dispersion relation is ${\omega }^2=4{\Omega }^2{m}^2/(k^2+m^2)$ and $c_{g,x}=-\omega k/(k^2+m^2)$. An outgoing wave in $x>0$ must have $\omega k<0$, givingFor $|\omega |>2|\Omega |$, define $\kappa =m\sqrt{1-4{\Omega }^2/{\omega }^2}$. Reject the exponentially growing branch, leaving the evanescent responseThe real part is the physical velocity. The radiation or bounded-decay condition is needed in addition to the wavemaker boundary value to pick a unique half-channel response. Steady forcing ($\omega =0$) is a singular limit: with nonzero rotation a steady geostrophic balance and impermeable top/bottom require $w=0$, incompatible with a nonzero imposed sine profile. At the cutoff equality, the bounded outgoing polarization also degenerates. These exceptional cases cannot be obtained by blindly substituting into the two strict-inequality formulas. The propagating sign is the radiation sign for an inertial wave.

Long-wave geometry makes vertical acceleration small, so pressure is approximately hydrostatic, $p=p_{\mathrm{atm}}+\rho g(h-z)$. The leading horizontal velocity is nearly depth-independent. Integrating incompressibility through the depth and using the impermeable bed and kinematic free-surface condition gives $\boxed{h_t+(hu{)}_x=0}$. The horizontal Euler equation then gives $\boxed{u_t+u{u}_x+g{h}_x=0}$.

Linearizing about a uniform resting layer gives wave speed $\boxed{c=\sqrt{gh}}$. Since $h=c^2/g$, the nonlinear equations can be written $c_t+u{c}_x+(c/2)u_x=0$ and $u_t+u{u}_x+2c{c}_x=0$. Combining them verifiesLet $c_0=\sqrt{g{h}_0}$ and $V\ge 0$ be the receding wall's speed, so its position is $x=-Vt$ and its fluid velocity condition is $u=-V$. The outward-moving rarefaction has the other invariant fixed at its resting value, $u-2c=-2c_0$. In its varying part, self-similarity gives $x/t=u+c$. ThereforeThe wall-adjacent state has $u=-V$ and $c=c_w=c_0-V/2$. Its right-going characteristic curve speed is $c_0-3V/2$, whereas the undisturbed head travels at $c_0$. The complete solution in the fluid $x>-Vt$ isBoth edges match continuously and the wall velocity condition holds. The assumption $V^2<4g{h}_0$ means $V<2c_0$, giving positive wall depth and placing the fan tail to the right of the wall by speed $c_w>0$. The depth everywhere is $h=c^2/g$. This is a shallow-water rarefaction from a receding wall.

Work in the rest frame of the planar shock, with nonzero mass flux $j=\rho u$. Conservation of mass gives $[j]=0$; the tangential and normal components of balance of linear momentum give $[jv]=0$ and $[p+ju]=0$. Thus $[v]=0$. For an ideal gas, specific internal energy is $e=c_{v}T=p/[(\gamma -1)\rho ]$, and specific enthalpy is $e+p/\rho =\gamma p/[(\gamma -1)\rho ]$. Total energy flux is $j[(u^2+v^2)/2+e+p/\rho ]$. Its continuity, with $v$ already continuous, givesPut $c^2=\gamma p/\rho$ and $M=u/c$. Since $p=ju/(\gamma M^2)$, the normal component of balance of linear momentum says $u[1+1/(\gamma M^2)]$ is constant. HenceThe energy equation says $u^2[1/2+1/((\gamma -1)M^2)]$ is constant, givingSet $a=M_{-}^2$, $b=M_{+}^2$, square the first relation and equate to the second. After clearing positive denominators the equation factors asThe first factor is the unchanged state. For a genuine shock the second factor vanishes, yieldingThe physical selection principle is entropy increase in the direction of flow. For $u_{\pm }>0$, the upstream side is $x<0$ and the downstream side is $x>0$. A compressive entropy-admissible shock has $\boxed{M_{-}>1,M_{+}<1}$; interpreting the printed “ahead” and “behind” spatial labels must respect these explicit side assignments. The expansion-shock branch would violate the second law.

Substitution gives the velocity ratioSince $v$ is unchanged while the normal component decreases, $|v|/u$ increases. Thus an oblique flow bends towards the shock plane, equivalently away from its normal. If $v=0$ there is no angular bending; the flow stays normal and simply slows. The tangential continuity, not a change in the tangential speed, causes the deflection.