Past exam of the advanced fluid dynamics course of the University of Oxford / 2018 / 2

Use the same constant dynamic viscosity $\mu$ for both incompressible Stokes flows, and let ${\mathbf{f}}^{(a)}$ be body force per unit volume. Their equations and Newtonian fluid stress tensor arewhere $e_{ij}^{(a)}=({\partial }_i{u}_j^{(a)}+{\partial }_j{u}_i^{(a)})/2$ is the strain-rate tensor. Consider the cross-work fluxBy the product rule,The pressure terms vanish by incompressibility. Symmetry of the stress tensor removes antisymmetric velocity gradient contributions, leaving $2\mu e^{(1)}:e^{(2)}$ in both cross terms; they cancel. Apply the divergence theorem, with $\mathbf{n}$ outward from $V$, and rearrange:This is the Lorentz reciprocal theorem with body forces. It depends on reciprocal linear viscous response in the same fluid; arbitrary unrelated constitutive laws would not supply the cross-term cancellation.

Write the supplied velocity in the radial formAt $r=a$, $f=1$ and $g=0$, giving the no-slip boundary condition $\mathbf{u}=\mathbf{U}$. At infinity $\mathbf{u}\to 0$ and the supplied gauge-fixed pressure tends to zero. These are the laboratory-frame conditions for a sphere translating through otherwise quiescent fluid.

The exterior Stokes equations can also be checked directly:andTo compute surface traction in translating-sphere Stokes flow, let $\mathbf{n}=\mathbf{x}/a$ point from the sphere into the fluid. At the surface, $f^{\prime }(a)=-3/(2a)$, $g^{\prime }(a)=3/(2a^3)$ and $g(a)=0$. HenceSubstitute into the Newtonian fluid stress tensor. Its viscous contribution to the surface traction is $3\mu [-\mathbf{U}+(\mathbf{U}\cdot \mathbf{n})\mathbf{n}]/(2a)$, so the normal part cancels the pressure contribution:The vector is constant over the entire sphere in the stated pressure gauge. Integrating it gives the fluid's drag force on the sphere,This is Stokes's law. On an exterior fluid domain the outward domain normal at its inner boundary is $-\mathbf{n}$; that distinction will matter in the reciprocal calculation.

Let $S_a$ be the sphere surface, ${\mathbf{n}}_s$ point from the sphere into the fluid, and writeBecause the prescribed exterior body forces are unchanged, they cancel on subtraction: the disturbance is a homogeneous Stokes flow in the exterior domain. On the sphere the no-slip boundary condition isallowing an arbitrary sphere rotation ${\mathbf{\omega }}_s$.

Use as auxiliary flow the translating-sphere solution of part b, with arbitrary test velocity $\mathbf{U}$. Its surface velocity is $\mathbf{U}$ and its surface traction is $-3\mu \mathbf{U}/(2a)$. Applying the Lorentz reciprocal theorem on the exterior domain gives, after cancelling the common inner-normal minus sign,The outer surface terms vanish for decaying disturbances: velocities are at most $O(r^{-1})$, stresses $O(r^{-2})$, and the integrated cross-work is $O(r^{-1})$.

There are no body forces inside the sphere in the original ambient problem. Thus the smooth ambient stress tensor has zero divergence there, andAlso ${\int }_{S_a}\mathbf{x}dS=0$, so the rotation term drops out of the left side. Consequently the reciprocal identity reduces toSince $\mathbf{U}$ is arbitrary, the result before imposing mechanical balance isFor a freely suspended sphere with negligible inertia and no applied external force, ${\mathbf{F}}_H=0$. Its sphere velocity as a surface average of ambient Stokes flow is thereforeThe implicit force-free condition is essential. More generally, an external force ${\mathbf{F}}_{\mathrm{ext}}$ gives $\mathbf{V}=⟨{\mathbf{u}}_{\infty }{⟩}_{S_a}+{\mathbf{F}}_{\mathrm{ext}}/(6\pi \mu a)$. For example, an externally pulled sphere in the zero ambient flow has nonzero velocity, contrary to an unrestricted surface-average-only claim. No torque-free assumption is needed for this translational identity, because the auxiliary surface traction is constant.

The undisturbed ambient solution is smooth and free of body forces throughout the ball $r\le a$. Its Stokes equations implyTaking the divergence gives ${\nabla }^2{p}_{\infty }=0$, and a further Laplacian therefore givesEach velocity component obeys the biharmonic equation in the whole ball.

Expand a component of ${\mathbf{u}}_{\infty }$ at the centre and average over $\mathbf{x}=a\mathbf{n}$. The odd spherical moments vanish, whileThus the spherical mean of a biharmonic function givesThe truncation is exact, not just a small-sphere approximation: all terms containing two or more Laplacians vanish. One can avoid any Taylor-convergence issue by letting $M(r)=⟨{\mathbf{u}}_{\infty }(r\mathbf{n})⟩$. Since ${\nabla }^2{\mathbf{u}}_{\infty }$ is harmonic, its spherical mean is ${\nabla }^2{\mathbf{u}}_{\infty }(0)$, soIntegrating gives $M(r)={\mathbf{u}}_{\infty }(0)+r^2{\nabla }^2{\mathbf{u}}_{\infty }(0)/6$ for every admissible $r$.

Combining with the force-free result of part c proves Faxén's first law in the requested form:The force-free qualification carries over. Local absence of ambient body forces and regularity inside the ball are also needed; if forcing or a singularity lies inside it, the harmonic-pressure and biharmonic-mean argument does not apply unchanged.