Past exam of the theoretical mechanics course of the University of Oxford / 2010 (B6, DMAB 2645)

Original solutions to the paper released at the source request. The complete editable transcription is retained in the private examination archive.

Let $\mathbf{x}(\mathbf{X},t)$ be the particle flow map from the fixed reference volume $V_0$ to the material volume, and let $J=\det (\partial \mathbf{x}/\partial \mathbf{X})$. Change variables and differentiate on the fixed domain:Use the two permitted identities, $DJ/Dt=J\nabla \cdot \mathbf{u}$ and $Df/Dt=f_t+\mathbf{u}\cdot \nabla f$. Changing back givesThis is the Reynolds transport theorem: both the value carried by each particle and the changing volume element contribute.

The traction $\mathbf{t}(\mathbf{n})$ is force per unit area across an oriented surface with unit normal $\mathbf{n}$. The stress component ${\sigma }_{ij}$ is its $i$th component on a surface normal to the $j$th coordinate axis. Cauchy's stress theorem states that this dependence is linear:For the boundary of a material volume, take the outward normal and the traction exerted by the surrounding fluid. Newton's law and the divergence theorem giveConstant density and incompressibility make the transport-theorem integrand on the left $\rho D{u}_i/Dt$. The volume is arbitrary, soSubstitute the Newtonian constitutive law. With constant viscosity,Hence

For a steady unidirectional field $\mathbf{u}=w\mathbf{k}$, incompressibility gives ${\partial }_{z}w=0$. Thus $w=w(x,y)$ and the convective acceleration vanishes. The transverse momentum equations give $p_x=p_y=0$, so $p=p(z)$. The axial equation isIts left side is independent of $z$ and its right side independent of $x,y$, so both equal a constant $G$. No slip against the translating wall gives $w=U$ on the boundary. Thus the asserted Poisson problem follows.

For the circular pipe, the axisymmetric solution isIt satisfies the boundary value and ${\Delta }_{\perp }{r}^2=4$. Integrating over the cross-section givesTherefore the laboratory-frame volume flux vanishes precisely whenFor a pressure decrease in the positive axial direction, this requires upstream wall translation.

Balance streamwise convection of order one with transverse viscous diffusion $R{e}^{-1}{\partial }_y^{2}u$. This requires a thickness $R{e}^{-1/2}$. Continuity then makes the normal velocity of that same small order. SetThe normal momentum equation makes the leading pressure independent of $Y$. It equals the outer pressure, whose inviscid tangential momentum balance gives $-p_x=U_s{U}_s^{\prime }$. The leading tangential equation and continuity are consequentlyThe divergence-free field $(u,V)$ has a streamfunction with $u={\Psi }_Y$, $V=-{\Psi }_x$. Substituting yieldsThe stationary impermeable wall has $u=V=0$, so ${\Psi }_Y(x,0)=0$ and $\Psi (x,0)$ is constant, chosen to be zero. Matching to the outer slip velocity gives ${\Psi }_Y(x,\infty )=U_s(x)$.

For the proposed form,Matching the outer velocity requires $\alpha -\beta =m$. The convective and outer-pressure terms have order $x^{2m-1}$, while $u_{YY}$ has order $x^{\alpha -3\beta }$. Equality of exponents givesThe two terms containing $\eta f^{\prime }{f}^{\prime \prime }$ cancel in $u{u}_x+V{u}_Y$, leaving $x^{2m-1}[m(f^{\prime }{)}^2-\alpha f{f}^{\prime \prime }]$. Since $U_s{U}_s^{\prime }=m{A}^2{x}^{2m-1}$, the reduced equation isThe wall and matching conditions become $\boxed{f(0)=f^{\prime }(0)=0,f^{\prime }(\infty )=A}$. This establishes a consistent similarity reduction; existence of an actual profile is a separate boundary-value question for a particular choice of parameters.

At these parameters the equation becomes $g^{\prime \prime }+g^2-4=0$, where $g=f^{\prime }$. Multiply by $g^{\prime }$ and integrate:Outer matching gives $g\to -2$ and $g^{\prime }\to 0$, so $C=16/3$. No slip gives $g(0)=0$. At that point the first integral fixes $g^{\prime }(0)=\pm 4\sqrt{2/3}$, so there are exactly two smooth initial-value solutions of the second-order equation satisfying the matching condition. Put ${\eta }_0=\mathrm{arctanh}\sqrt{2/3}$. Translation invariance of the supplied profile gives them explicitly:The first decreases monotonically from zero to $-2$. The second first rises to $4$ at ${\eta }_0$, crosses zero again at $2{\eta }_0$, and then tends to $-2$. Integrating either profile from zero recovers $f$ with $f(0)=0$.

One would not expect both branches to be equally stable. The rising branch contains flow opposite to its negative outer stream and interior inflection points, typical ingredients for shear instability. The monotone branch has $g^{\prime \prime }=4-g^2>0$ at finite positive $\eta$ and no interior inflection point, so the classical inviscid inflection-point mechanism is absent there. This motivates expecting the reversed-flow branch to be unstable and the monotone branch to be the stable candidate. A definitive viscous temporal-stability result requires perturbing the time-dependent equations; the steady first integral alone is not a stability proof.

Introduce $\mathcal{R}=\rho UL/\mu$. Balancing the streamwise pressure gradient $\Pi /L$ with cross-layer viscosity $\mu U/({\delta }^2{L}^2)$ givesAfter division by that viscous scale, the streamwise momentum equation isThe transverse equation, divided by its pressure-gradient scale, isContinuity scales exactly to ${\hat{u}}_{\hat{x}}+{\hat{w}}_{\hat{z}}=0$. With $\delta \ll 1$, ${\delta }^2\mathcal{R}\ll 1$ and $S=O(1)$, the leading equations are

No slip and impermeability give dimensional $u=-U$, $w=0$ on the upward-moving wall, so $\boxed{\hat{u}=-1,\hat{w}=0}$ there. For the interface normal, $h_x=\delta {\hat{h}}_{\hat{x}}$, and hencein its $x,z$ components. The relevant stresses scale asMultiplying by the inward normal gives the leading tractionIts leading normal component vanishes only if $\hat{p}=0$ at the interface; the tangential component then requires ${\hat{u}}_{\hat{z}}=0$.

The free surface is material. Applying the material derivative to $z-h(x,t)=0$ gives $w=h_t+u{h}_x$, which in the scaled variables becomes

Suppress hats for this calculation. Transverse pressure uniformity and zero surface pressure give $p=0$ throughout the film. Thus $u_{zz}=-S$. Integrating twice, using $u_z(h)=0$ and $u(0)=-1$, givesThe flux is $q={\int }_0^{h}udz=-h+S{h}^3/3$. Integrate continuity across the layer and apply the wall and kinematic conditions to obtain $h_t+q_x=0$. Restoring hats yieldsThe cubic flux contains downward gravity-driven transport and upward wall entrainment.

The material derivative $\boxed{D/Dt={\partial }_t+\mathbf{u}\cdot \nabla }$ measures change while following a moving gas parcel. The four equations respectively express mass conservation, momentum conservation, internal-energy balance, and the ideal gas law. The first accounts for density change under expansion or compression. The second is Newton's law with pressure as the only force. In the third the specific internal energy is $c_{v}T$ and pressure work $-p\nabla \cdot \mathbf{u}$ heats compression and cools expansion. The fourth relates thermodynamic pressure, density and temperature; $R$ is the gas constant per unit mass.

Continuity gives $D\log \rho /Dt=-\nabla \cdot \mathbf{u}$. Using $p=\rho RT$, the temperature equation gives $D\log T/Dt=-(R/c_v)\nabla \cdot \mathbf{u}$. ThereforeConsequentlyEach smooth parcel evolves isentropically under these adiabatic inviscid equations; the argument does not cover entropy-producing shocks.

Conservation of the parcel's entropy makes $p/{\rho }^{\gamma }$ constant along its trajectory. Its initial value is $p_0/{\rho }_0^{\gamma }$, soThis is a relation following the specified parcel; parcels with different initial entropy can have different constants.

Linearization about the uniform resting state givesFor irrotational motion write ${\mathbf{u}}^{\prime }=\nabla \varphi$. The momentum equation integrates to $p^{\prime }=-{\rho }_0{\varphi }_t$ after choosing the time-dependent potential gauge. Substituting into the pressure equation yieldsThe constant is the adiabatic sound speed, also given by $(\partial p/\partial \rho {)}_S^{1/2}$ at the base state.

Rigid impermeable walls require zero normal fluid velocity, so the potential satisfies a homogeneous Neumann condition. In the cylinder,It is also single-valued and periodic in $\theta$, and regular on the axis. No tangential no-slip boundary condition is imposed on this inviscid acoustic potential.

Use $\varphi =F(r)\Theta (\theta )Z(z)e^{-i\omega t}$. Periodicity gives $\Theta =\cos m\theta$ or $\sin m\theta$, $m\ge 0$, with only the cosine branch for $m=0$. The end-wall conditions give $Z=\cos (\ell \pi z/h)$, $\ell \ge 0$. With radial separation constant ${\kappa }^2$, the remaining equation isFor $\kappa >0$, boundedness on the axis selects $F=J_m(\kappa r)$. The cylindrical wall then requires $J_m^{\prime }(\kappa a)=0$. Thus the modes are real and imaginary parts ofHere take the positive derivative zeros and include the separate constant radial mode $m=0$, $k=0$. For zero radial separation constant, regularity gives $r^m$ when $m>0$, which fails the wall condition, and a constant for $m=0$. This avoids counting a formal derivative zero at the origin that would make $J_m(0r)$ identically zero. The completely constant mode has zero frequency and no velocity; the remaining modes are acoustic oscillations. The self-adjoint Neumann Laplacian has these separated modes as a complete system, so no additional nonzero natural frequencies occur.

Incompressibility and $\mathbf{u}=\nabla \varphi$ give $\boxed{{\nabla }^2\varphi =0}$. The permitted vector identity, with zero vorticity, gives $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)$. Substituting into Euler's equation yieldsThe bracket is spatially constant in a connected fluid region, soAdding a function of time to the potential changes only $F(t)$ and leaves the velocity unchanged.

Each perturbation potential is harmonic in its own half-space:Write $D_1={\partial }_t+(U/2){\partial }_x$ and $D_2={\partial }_t-(U/2){\partial }_x$. Linearizing the material-interface condition at $z=0$ givesBernoulli's equation at the displaced interface gives $p_i^{\prime }=-{\rho }_i(D_i{\tilde{\varphi }}_i+g\eta )$. Using the supplied pressure jump therefore yieldsThe potentials decay toward their respective infinities. The hydrostatic displacement terms are retained in this dynamic condition.

Put $q=|k|>0$ and ${\omega }_1=\omega -kU/2$, ${\omega }_2=\omega +kU/2$. Harmonicity and decay selectThe kinematic conditions give $B_1=i{\omega }_{1}A/q$ and $B_2=-i{\omega }_{2}A/q$. Inserting them into the dynamic condition and cancelling the nonzero interface amplitude givesExpansion proves

Let $\Sigma ={\rho }_1+{\rho }_2$, $\Delta ={\rho }_2-{\rho }_1>0$. Completing the square in the dispersion relation givesThe coefficients are real. There is an exponentially growing temporal mode exactly when the right side is negative, because one root then has positive imaginary part in $e^{-i\omega t}$. Rearrangement givesEquality is the marginal threshold rather than exponential growth.

For positive surface tension the threshold is even in $k$, tends to infinity as $k\to 0$ and as $|k|\to \infty$, and has its two equal minima atMinimizing $\Delta g/q+\gamma q$ for $q>0$ givesThus the initial state has some exponentially growing mode exactly when $|U|>U_{\min }$. If $D={\rho }_1{\rho }_2{U}^2/\Sigma$, the unstable band isGravity suppresses long wavelengths and surface tension suppresses short ones; intermediate wavelengths become unstable when the shear is large enough.

For positive depth put $c=\sqrt{gh}$. The equations becomeAdding or subtracting twice the second equation to the first givesThus the two characteristic speeds are $\boxed{dx/dt=u\pm c}$ and the corresponding conserved Riemann invariants are $\boxed{Q_{\pm }=u\pm 2c}$. At a dry front $c=0$ the two speeds coincide; the positive-depth formulas are applied up to that limit.

Before a signal traveling left at speed $c_0$ can reach the wall, the portion $-a<x<-c_{0}t$ remains at rest with $c=c_0$, $h=h_0$. In the rarefaction fan, right-running characteristics bring the unchanged invariant $Q_{+}=2c_0$. The left-running family originates at the removed dam and has speed $u-c=x/t$. Solving these two relations givesAt the dry front $c=0$, so its speed and location are $2c_0$ and $2c_{0}t$, consistent with the stated constant-speed condition. Therefore, for $0<t<a/c_0$,The velocity is zero in the untouched region and $2(c_0+x/t)/3$ in the fan. Ahead of the front there is no fluid, so $h=c=0$ and a fluid velocity need not be assigned. In particular the wall remains at $u=0,c=c_0$ until time $a/c_0$.

Inside the fan, $u+c=(4c_0+x/t)/3$. Hence a right-running characteristic satisfiesMultiplication by the integrating factor $t^{-1/3}$ givesIntegration yields $x_{+}=2c_{0}t+C{t}^{1/3}$. Requiring $x_{+}(\tau )=-c_0\tau$ determines $C=-3c_0{\tau }^{2/3}$, soThe invariant $Q_{+}=2c_0$ stays constant along it until it enters a region affected by new boundary data.

The head of the rarefaction reaches the wall at $(-a,t_0)$ with $t_0=a/c_0$. The right-running characteristic through that event is obtained by setting $\tau =t_0$ in the preceding formula:It bounds the region reached by the first wall-induced change in $Q_{+}$. To its right the characteristic information remains the original dam-break information, soBehind it, the impermeable wall must keep $u(-a,t)=0$. Continuing the original fan there would instead give $2(c_0-a/t)/3>0$ for $t>t_0$. Reflected characteristic data therefore change $Q_{+}$ and invalidate the simple fan formula. The shaded region in the characteristic diagram represents this changed domain of dependence; it includes $0<x<x_b(t)$ once that interval becomes nonempty. Its first positive extent occurs when $t/t_0>(3/2{)}^{3/2}$.