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Use the Papkovich–Neuber representation in the conventionIt gives $\nabla \cdot \mathbf{u}=0$ and $\mu {\nabla }^2\mathbf{u}=-\mu \nabla (\nabla \cdot \mathbf{\Phi })=\nabla p$. Additive pressure constants are immaterial.

Let $q=\mathbf{x}\cdot \mathbf{Ex}$, with $\mathbf{E}$ symmetric and traceless. The harmonic vector $\mathbf{Ex}/r^3$ is a derivative of $1/r$. The scalar $q/r^5$ is a decaying degree-two harmonic because $q$ is a harmonic homogeneous quadratic. These Papkovich potentials for a strained sphere have precisely the strain symmetry and decay required for the disturbance, so trySubstitution givesAt $r=a$, require ${\mathbf{u}}^{\prime }=-\mathbf{Ex}$. The independent vector structures give $B=a^5$ and $A=-5a^3/3$. The disturbance for a sphere in a uniform straining Stokes flow isIt decays at infinity and satisfies no slip exactly at the sphere. Its leading far field is a stresslet, of velocity scale $E{a}^3/r^2$.

A force-free sphere in a linear background translates with ${\mathbf{U}}_0$, by Faxén's first law, and a couple-free sphere rotates with the background rigid-body rotation $\mathbf{\Omega }$, by Faxén's rotational law. Uniform translation and solid-body rotation can then be matched without disturbance. The remaining relative boundary velocity is exactly $-\mathbf{Ex}$. The Linearity of Stokes flow and Uniqueness of Stokes flow therefore give the same disturbance. The pure strain produces no resultant force or couple, by its symmetric degree-two angular structure.

Let $\mathbf{e}=\mathbf{X}/R$, $s=\mathbf{e}\cdot \mathbf{Be}$, and $A_{\parallel }=\mathbf{A}\cdot \mathbf{e}$. Applied forces and couples below are forces and torques on the solid; by force balance their values are also the strengths exerted by that solid on the fluid. This fixes the signs in Faxén's first law and Faxén's rotational law. To hold the passive sphere fixed, set its translation and angular velocity to zero in those laws. The incident two-mode tensorial squirmer flow givesOnly the stresslet part has vorticity:Consequently, for fixed $\mathbf{A},\mathbf{B}$ as $a/R\to 0$,The leading force is the radial stresslet force. Its next correction, $O(\mu a^4|\mathbf{A}|/R^3)$, comes from the incident potential dipole. The next $\mathbf{B}$ correction, $O(\mu a^5\Vert \mathbf{B}\Vert /R^4)$, contains both the swimmer's $r^{-4}$ field and the Laplacian term in Faxén's first law. If the relevant coefficient vanishes, these next terms must be retained instead of calling the vanished term a nonzero leading approximation. Further exchanges in the method of reflections for Stokes flow are smaller still.

The passive sphere reflects primarily a Stokeslet with strength ${\mathbf{F}}_h$. At the swimmer, its velocity is $(\mathbf{I}+\mathbf{ee}){\mathbf{F}}_h/(8\pi \mu R)$. The unchanged surface slip velocity adds the free swimming velocity to the incident-flow contribution in Faxén's first law. ThusThe leading change is $O(a^3\Vert \mathbf{B}\Vert /R^3)$; its next correction is the displayed $O(a^4|\mathbf{A}|/R^4)$ term. Finite-radius Faxén's first law corrections, the passive sphere's reflected stresslet, and its rotlet contribute at order $a^5\Vert \mathbf{B}\Vert /R^5$ to translation.

For a Stokeslet at the origin,The curl of the first term is $\mathbf{F}\times \mathbf{x}/r^3$; that of the second is also $\mathbf{F}\times \mathbf{x}/r^3$, before multiplying by $1/(8\pi \mu )$. Hence ${\mathbf{\omega }}_F=\mathbf{F}\times \mathbf{x}/(4\pi \mu r^3)$. At the swimmer use displacement $-\mathbf{X}$. The leading $\mathbf{B}$ force is parallel to $\mathbf{X}$, so its Stokeslet has zero vorticity there. This eliminates the putative $O(a^3\Vert \mathbf{B}\Vert /R^4)$ rotation. The potential dipole itself has no vorticity, but the holding force it induces is not radial. Applying Faxén's rotational law to its reflected Stokeslet yieldsThe passive sphere's rotlet and reflected stresslet can affect that last order, so they cannot restore the larger missing rotation. When $\mathbf{A}\times \mathbf{X}=0$, the boxed coefficient is zero and the higher-order terms determine any rotation.

Write $r=|\mathbf{x}|$ and take $\mu$ to be the dynamic viscosity. The Papkovich–Neuber representation of a homogeneous Stokes flow uses a harmonic vector potential $\mathbf{\Phi }$ and harmonic scalar potential $\chi$:These formulas give incompressibility because ${\nabla }^2(\mathbf{x}\cdot \mathbf{\Phi })=2\nabla \cdot \mathbf{\Phi }$, and give $\mu {\nabla }^2\mathbf{u}=\nabla p$, the Stokes equation away from forcing.

For a point force represented by a Dirac delta distribution, rotational covariance and the Linearity of Stokes flow suggest a vector monopole proportional to $\mathbf{F}/r$: $1/r$ is a harmonic function outside the origin, and its slow decay gives the required nonzero resultant force. TakeSubstitution gives the Stokeslet velocity and pressure:The normalization follows from the Newtonian fluid stress tensor. Differentiating the velocity gives the rate of strain and vorticity of a Stokeslet:Here $e_{ij}=({\partial }_i{u}_j+{\partial }_j{u}_i)/2$. Thus ${\sigma }_{ij}=-p{\delta }_{ij}+2\mu e_{ij}=-3(\mathbf{F}\cdot \mathbf{x})x_i{x}_j/(4\pi r^5)$. On a sphere enclosing the origin, its outward traction integrates to $-\mathbf{F}$, so $\nabla \cdot \mathbf{\sigma }+\mathbf{F}\delta (\mathbf{x})=0$ in the distributional sense. All displayed fields away from the point force have $r>0$.

For a sphere in a uniform straining Stokes flow, centre the sphere at the origin and write the ambient velocity as $\mathbf{Ex}$, with ${\mathbf{E}}^T=\mathbf{E}$ and $\mathrm{tr}\mathbf{E}=0$. Faxén's first law and Faxén's rotational law give zero translation and rotation. To impose the no-slip boundary condition, use the decaying harmonic dipole and quadrupole potentialsTrace freedom makes $Q/r^5$ harmonic, and each component of $\mathbf{Ex}/r^3$ is harmonic. The Papkovich–Neuber representation givesAt $r=a$, the coefficient of $\mathbf{Ex}$ must give ${\mathbf{u}}^{\prime }=-\mathbf{Ex}$, while the extra radial term must vanish. Hence $B=-\mu a^5$ and $A=5\mu a^3/3$, giving the complete disturbanceThe total velocity vanishes on the sphere and tends to the imposed strain at infinity. This field has no Stokeslet or rotlet contribution, consistent with a force-free, torque-free sphere. Its leading far field is the radial stresslet $-5a^{3}Q\mathbf{x}/(2r^5)$.

Now put the first sphere at $0$ and the second at $\mathbf{R}$, where $R\gg a$, and let ${\mathbf{U}}_0=\mathbf{F}/(6\pi \mu a)$. The first sphere's leading disturbance is its Stokeslet. The rate-of-strain tensor and vorticity incident on the second sphere areThe second sphere translates with the local flow and rotates with half its local vorticity. Its force-free and torque-free conditions therefore remove any reflected force or torque monopole. The incident strain produces the stresslet just obtained. Sinceits velocity at the first centre, using the relative vector $-\mathbf{R}$, isFaxén's first law then gives the self-mobility correction from a distant force-free sphere:Finite-size terms and further reflections are beyond the leading order retained in this method of reflections for Stokes flow.

Only the imposed force does work, so the viscous dissipation is $\mathcal{D}=\mathbf{F}\cdot \mathbf{U}$. With $\zeta =6\pi \mu a$, define ${\epsilon }_R=a/R$ and $\gamma =15a^4/(4R^6)$. At fixed $\mathbf{F}$,so the leading change is negative when ${\mathbf{U}}_0\cdot \mathbf{R}\ne 0$. The direction is held fixed in taking ${\epsilon }_R\to 0$.

The fixed-force comparison of minimum viscous dissipation distinguishes the prescribed force from prescribed boundary velocity. Indeed,and henceAt the actual reduced velocity, adding the second sphere increases the dissipation relative to an isolated translating sphere at that same velocity, exactly as required by the Minimum-dissipation theorem for Stokes flow. One can extend the two-sphere velocity rigidly into the second sphere, adding zero strain and producing an admissible trial field in the domain without it. The reflected angular velocity of the first sphere contributes only a higher-order squared rotational power to this comparison.

Finally hold the second sphere's angular velocity at zero. Faxén's rotational law requires the applied coupleThe resulting rotlet adds at the first centreThis is the additional rotational constraint correction to sphere mobility. Including the strain reflection, the total leading change relative to the isolated velocity isFor motion parallel to the line of centres the constraint adds nothing at this order; for transverse motion it supplies an extra resisting correction.

To hold sphere 2 fixed against the incident angular velocity from part ii, its applied couple must generate the bare rotationThe resulting rotlet advects the force-free sphere 1. Since the field is harmonic, Faxén's first law gives

The returned rotlet has velocity $O(\Omega a^6/R^5)$ and rate of strain $O(\Omega a^6/R^6)$ at sphere 1. That strain induces a stresslet of strength $O(\mu \Omega a^9/R^6)$, whose velocity at sphere 2 is $O(\Omega a^9/R^8)$. This method of reflections for Stokes flow gives the stated order of the next correction to ${\mathbf{U}}_2$.

Write $\mathbf{\Omega }={\mathbf{G}}_1/(8\pi \mu a^3)$. The incident rotlet of sphere 1 at sphere 2 isand it is harmonic away from sphere 1. Since sphere 2 is force-free, Faxén's first law therefore givesThe vorticity of the rotlet isBecause ${\mathbf{G}}_2=0$, Faxén's rotational law gives

The symmetric rate-of-strain tensor of the incident rotlet at sphere 2 isAfter translation and rotation have matched the uniform and antisymmetric parts of the incident flow, the leading perturbation from sphere 2 is the stresslet part of the supplied straining-sphere solution:Part b gives its vorticity asAt the centre of sphere 1, $\mathbf{x}=-\mathbf{R}$, soApplying Faxén's rotational law to sphere 1 produces half this ambient vorticity and proves

To the requested order, sphere 1 has velocity $\mathbf{V}$ and creates the translating-sphere field from part a. At the centre of sphere 2, a vector distance $\mathbf{R}$ away, this incident field isSphere 2 is force free, so Faxén's first law givesThe source-dipole term is harmonic, whileThereforeThis is the two-sphere Rotne--Prager mobility through order $a^3/R^3$.

The incident strain at sphere 2 is $O(Va/R^2)$. A force-free sphere in this strain creates a stresslet of size $O(\mu a^{3}Va/R^2)=O(\mu V{a}^4/R^2)$, whose velocity back at sphere 1 is $O(V{a}^4/R^4)$. HenceThe nearly uniform returned flow merely advects sphere 1 and does not change its fixed Stokeslet strength, because its applied force remains fixed. The next scattered disturbance is therefore generated by the returned velocity gradient, of order $V{a}^4/R^5$. It induces a stresslet of size $O(\mu V{a}^7/R^5)$ at sphere 1 and hence velocity $O(V{a}^7/R^7)$ at sphere 2. This method of reflections for Stokes flow explains both the absence of an $O(V{a}^5/R^5)$ term and the next order $O(V{a}^7/R^7)$.

For a smooth three-dimensional biharmonic function regular throughout a ball, its surface average has exactly this form. Taylor expansion and rotationally invariant spherical moments turn each even term into a multiple of an iterated Laplacian; terms from the fourth derivative onward vanish when ${\Delta }^{2}f=0$. This is the mean-value identity that converts the reciprocal surface average into Faxén's first law.

For a two-mode tensorial squirmer flow of radius $a$ and a passive equal sphere held at $X=Re$, with $a/R\ll 1$, let $s=e\cdot Be$. Faxén's first law gives holding force $F_h=-9\pi \mu a^{3}se/R^2+O(\mu a^4|A|/R^3+\mu a^5\Vert B\Vert /R^4)$. Its reflected Stokeslet changes swimming velocity by $-9a^{3}se/(4R^3)$. Since this leading force is radial, its vorticity at the swimmer vanishes. The next holding force, generated by the potential dipole, yields rotation $a^{4}A\times X/(4R^6)$. Terms from the passive sphere's finite radius, reflected stresslet and rotlet are needed at higher order. These formulas assume fixed slip and no externally applied swimmer force or torque.