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Faxén's first law gives the external force needed to translate a rigid sphere with the no-slip boundary condition, of radius $a$ with velocity $\mathbf{U}$ in an ambient Stokes flow ${\mathbf{u}}_{\infty }$. In an unbounded Newtonian fluid of dynamic viscosity $\mu$, with the ambient Stokes flow smooth and free of body forces throughout the sphere's ball,The hydrodynamic force on the sphere is $-{\mathbf{F}}_{\mathrm{ext}}$. This sign convention gives the Stokes drag law when the ambient velocity vanishes and gives a holding force opposite to a uniform ambient velocity when the sphere is fixed.

The Lorentz reciprocal theorem for Stokes flow, using a translating sphere in Stokes flow as the auxiliary solution, reduces the ambient contribution to its surface average. The spherical mean of a biharmonic function makes that average exactly ${\mathbf{u}}_{\infty }({\mathbf{x}}_c)+(a^2/6){\nabla }^2{\mathbf{u}}_{\infty }({\mathbf{x}}_c)$. The finite-size correction therefore measures spatial curvature of the ambient velocity; it vanishes in a uniform or linear Stokes flow. A force-free sphere translates with this corrected ambient velocity.

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.

For two incompressible Stokes flows $(\mathbf{u},\mathbf{\sigma })$ and $(\hat{\mathbf{u}},\hat{\mathbf{\sigma }})$ with the same viscosity and no body forces, the Lorentz reciprocal theorem for Stokes flow statesIndeed the divergence of their difference is zero: stress divergences vanish, pressure contracts with zero velocity divergence, and the two symmetric strain-rate contractions cancel.

Apply it to the decaying disturbance ${\mathbf{u}}^{\prime }=\mathbf{u}-{\mathbf{u}}_{\infty }$ around the sphere and an auxiliary translating sphere of velocity $\mathbf{V}$. On the body surface, ${\mathbf{u}}^{\prime }=\mathbf{U}+{\mathbf{\Omega }}_s\times \mathbf{x}-{\mathbf{u}}_{\infty }$. Using normals from the body into the fluid, the given auxiliary traction is $-3\mu \mathbf{V}/(2a)$; reversing both normal conventions does not change the reciprocal equality. The background traction integrates to zero because it is regular inside the sphere. The disturbance's hydrodynamic force is consequently ${\mathbf{F}}_H=-\mathbf{F}$, where $\mathbf{F}$ is the externally applied force. The rotational term integrates to zero. Thus for arbitrary $\mathbf{V}$,Therefore $\mathbf{U}=\mathbf{F}/(6\pi \mu a)+⟨{\mathbf{u}}_{\infty }{⟩}_{S_a}$.

The background Stokes equations imply ${\nabla }^2{p}_{\infty }=0$ and ${\nabla }^4{\mathbf{u}}_{\infty }=0$. The spherical mean of a biharmonic function gives exactly $⟨{\mathbf{u}}_{\infty }⟩={\mathbf{u}}_{\infty }+(a^2/6){\nabla }^2{\mathbf{u}}_{\infty }$, evaluated at the sphere center. One can obtain it from Taylor expansion and $⟨x_i{x}_j⟩=a^2{\delta }_{ij}/3$; every subsequent spherical average contains ${\nabla }^4$ and vanishes. This requires the background to be regular throughout the enclosed ball. Hence the Faxén relation isThe printed denominator $6\pi \mu$ omits $a$. The given traction integrates to $-6\pi \mu a\mathbf{V}$, fixing that factor; the printed force term also fails dimensional consistency unless its force were redefined, which the question does not do.

For a no-slip sphere in a smooth ambient Stokes flow whose body forces lie outside the sphere and are unchanged by insertion, use the translating sphere as a reciprocal auxiliary field. Its uniform surface traction selects the ambient surface average and annihilates a rigid rotation term. The ambient stress has zero net surface force because its divergence vanishes inside the ball. Thus the sphere's hydrodynamic force is $-6\pi \mu a(\mathbf{V}-⟨{\mathbf{u}}_{\infty }⟩)$; mechanical balance gives the displayed formula. The surface-average-only expression requires a force-free inertialess sphere. The spherical mean of a biharmonic function makes this average exactly $[1+(a^2/6){\nabla }^2]{\mathbf{u}}_{\infty }(0)$.