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

Let $t(n)$ denote traction: the force per unit area exerted across an oriented internal surface with unit normal $n$. Define ${\sigma }_{ij}$ as component $i$ of the traction on a coordinate plane whose normal is $e_j$. Balance of linear momentum on a shrinking tetrahedron makes the surface-force terms, of order area, dominate its inertia and body forces, of order volume. Resolving its oblique face into the three coordinate projected areas gives Cauchy's traction formulaOpposite faces have opposite tractions, as follows independently from a shrinking pillbox. This fixes both the index and normal conventions.

For a material volume $V(t)$, the Reynolds transport theorem says $d{\int }_{V(t)}fdV/dt={\int }_{V(t)}[f_t+\nabla \cdot (fu)]dV$. Apply it componentwise to $f=\rho u_i$ and use mass conservation. With no body force in this calculation, balance of linear momentum and the divergence theorem giveArbitrariness of the material volume yields $\boxed{\rho D{u}_i/Dt={\partial }_j{\sigma }_{ij}}$, with $D/Dt={\partial }_t+u_j{\partial }_j$.

For constant dynamic viscosity and incompressible velocity, ${\partial }_j{u}_j=0$, soThus $\boxed{\rho (u_t+u\cdot \nabla u)=-\nabla p+\mu {\nabla }^{2}u}$.

In fully developed steady pipe flow $u=w(x,y)e_z$, the convective acceleration vanishes. The transverse equations give $p_x=p_y=0$ and the axial equation gives $p_z=\mu (w_{xx}+w_{yy})$. One side depends only on $z$ and the other only on $(x,y)$, so both equal a constant. HenceIf $n$ is outward from the fluid cross-section, the wall-on-fluid axial shear per length is $\mu \oint {\partial }_{n}wds=\mu {\int }_A\Delta wdA=A{p}_z$. The fluid-on-pipe force is its negative:For forward pressure-driven flow $p_z<0$, this force points along the flow. Its value depends on cross-sectional area and pressure drop, even though the detailed velocity profile depends on the shape.

Let the thin transverse length be $R{e}^{-1/2}$ times the longitudinal scale and write $Y=R{e}^{1/2}y$, $V=R{e}^{1/2}v$. Continuity then remains $u_x+V_Y=0$. Transverse momentum gives $p_Y=0$ to leading order; the quiescent exterior has constant pressure, hence $p_x=0$. Axial viscous diffusion in $x$ is smaller by $R{e}^{-1}$, leavingSymmetry about the jet centreline requires $V=0$ and $u_Y=0$ at $Y=0$. At large positive $Y$, require $u\to 0$ and $u_Y\to 0$. The transverse velocity can approach a finite nonzero scaled entrainment velocity; imposing $V\to 0$ as well would rule out the jet. The physical transverse velocity remains smaller by $R{e}^{-1/2}$ and matches an entraining outer flow.

Using continuity, the momentum equation is ${\partial }_x(u^2)+{\partial }_Y(uV-u_Y)=0$. Integrating across the jet, the boundary terms vanish because $u,u_Y\to 0$ while $V$ is finite. ThusApart from the dimensional density and velocity/length factors, this is the conserved axial momentum flux supplied by the nozzle. Entrainment increases volume flux without adding axial momentum from the resting exterior.

For $u=x^{\alpha }{f}^{\prime }(\eta )$ with $\eta =Y/x^{\beta }$, use the stream function $\psi =x^{\alpha +\beta }f(\eta )$ so that $u={\psi }_Y$ and $V=-{\psi }_x$. Momentum conservation requires $2\alpha +\beta =0$. Inertia and diffusion scale as $x^{2\alpha -1}$ and $x^{\alpha -2\beta }$, so $\alpha +2\beta =1$. Therefore $\boxed{\alpha =-1/3,\beta =2/3}$. In particularSubstitution reduces momentum to $3f^{(3)}+f{f}^{\prime \prime }+(f^{\prime }{)}^2=0$, with $f(0)=f^{\prime \prime }(0)=0$ and $f^{\prime }(\infty )=0$. Integrating first gives $3f^{\prime \prime }+f{f}^{\prime }=0$, the integration constant vanishing at the centreline. A second integration gives $3f^{\prime }+f^2/2=C$. Parameterize its positive constant as $C=18A^2$ and solve the resulting Riccati equation with $f(0)=0$:The momentum integral fixes its scale, since ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^{4}zdz=4/3$:The upper-half entrainment is $V\to -2A{x}^{-2/3}$; the negative side has the opposite sign, drawing ambient fluid inward. This is the Bickley jet, valid in the slender far-downstream laminar regime, rather than in the immediate nozzle region.

Scale lateral distance with $L$, transverse distance with $h$, lateral velocity with $U$, and the normal velocity with $Uh/L$. Balancing lateral pressure force with transverse viscous stress gives pressure scale $\mu UL/h^2$. The ratio of inertia to that stress is $U{h}^2/(\nu L)$, while lateral viscous derivatives are smaller by $(h/L{)}^2$. The two stated smallness conditions therefore reduce the lateral equations to $\mu u_{zz}=p_x$, $\mu v_{zz}=p_y$. The normal component of balance of linear momentum gives $p_z=0$ at leading order; continuity remains $u_x+v_y+w_z=0$. The no-slip, impermeable rigid plates require $u=v=w=0$ at both plate surfaces.

Integrating with those boundary values givesDepth-integrated continuity has no normal boundary flux, so $\nabla \cdot \bar{u}=0$ and $\boxed{{\nabla }^{2}p=0}$ away from the injection point.

Neglecting surface tension, the fluid free boundary has $p=p_{\mathrm{ext}}$, and its normal velocity is the fluid velocity: $\dot{R}={\bar{u}}_r(R)$. Any constant capillary pressure for a circular interface can instead be included in that spatially uniform boundary pressure. Conservation of injected volume gives $2\pi rh{\bar{u}}_r=Q$. HenceThe logarithmic pressure singularity represents the ideal point source, which in a physical cell is replaced by a finite inlet. The formula is an exact radial solution of the depth-averaged model for either sign of $Q$, while $R^2>0$.

Its robustness differs sharply between the signs. With viscous fluid inside a much less viscous exterior, injection displaces the less viscous phase and is stabilizing; suction draws the exterior into the viscous region and is prone to fingering. One can see this directly without relying on a slogan. Write the boundary as $R+a_n\cos n\theta$ and its regular pressure perturbation as $b_n{r}^n\cos n\theta$. Linearizing the constant-pressure condition gives $\delta p(R)=-a_n{p}_0^{\prime }(R)=a_{n}Q/(2\pi hKR)$. The perturbation radial velocity is then $-n{a}_{n}Q/(2\pi h{R}^2)$, and evaluation of the base velocity at the displaced boundary adds $-a_{n}Q/(2\pi h{R}^2)$. ThusModes decay for injection and grow for suction. Accordingly the suction circle is a mathematically possible symmetric solution but generally an unstable prediction, failing before its formal collapse in a perturbed experiment. This is Stability of an interior Hele-Shaw blob under injection and suction.

Two-dimensional incompressibility permits $u={\psi }_y$, $v=-{\psi }_x$. Taking the curl of the Stokes momentum equation eliminates pressure and gives ${\nabla }^2(v_x-u_y)=0$. Since $v_x-u_y=-{\nabla }^2\psi$, the stream function obeys $\boxed{{\nabla }^4\psi =0}$.

Scale radius by $a$, velocity by $U$ and stream function by $Ua$. In polar coordinates $u_r=r^{-1}{\psi }_{\theta }$, $u_{\theta }=-{\psi }_r$. No penetration makes $\psi$ constant on the cylinder; choose that constant zero. No slip also requires ${\psi }_r=0$ at $r=1$. The uniform stream has $\psi =r\sin \theta$, specifying the far-field condition.

The boundary forcing has angular mode $\sin \theta$, so its corresponding mode must have $\psi =f(r)\sin \theta$. Define $L_1=d^2/d{r}^2+r^{-1}d/dr-r^{-2}$. Then ${\nabla }^4\psi =(L_1^{2}f)\sin \theta$, while $L_1{r}^s=(s^2-1)r^{s-2}$. Thus the indicial equation of $L_1^2$ is $(s^2-1)[(s-2{)}^2-1]=(s-3)(s-1{)}^2(s+1)=0$, givingThe repeated exponent one supplies the logarithmic term. The condition $f^{\prime }\to 1$ first forces $A=B=0$ and $C=1$. Then $f(1)=0$ forces $D=-1$, but $f^{\prime }(1)=C-D=2\ne 0$. No additional Fourier mode can repair the missing boundary condition in this angular mode. Therefore the unbounded-domain Stokes problem has no solution with both no slip and a nonzero uniform far stream.

This is Stokes' paradox. It identifies a nonuniform approximation, rather than the absence of physical flow. For any small but nonzero Reynolds number, convective inertia becomes important sufficiently far away. An inner Stokes region must be matched to an outer Oseen or Navier-Stokes region that retains that inertia. Alternatively, finite external boundaries introduce a different well-posed physical problem; they cannot be sent to infinity while keeping the naive Stokes solution unchanged.