= Solution
Use the deterministic <Hookean dumbbell model>, with a <zero-rest-length spring> and negligible connector drag. For $\mathbf R=\mathbf r_1-\mathbf r_2$, its elastic <forces> on the two beads are
$$
\mathbf F_1=-H\mathbf R,\qquad \mathbf F_2=+H\mathbf R.
$$
Here $H>0$ is the <spring constant>. No independent thermal forcing from <Brownian motion> or bead <inertia> is included. The unperturbed <linear velocity field> is $\mathbf L\mathbf r$, with $\operatorname{tr}\mathbf L=0$ by <incompressibility>.
The fluid <velocity> used in each bead's drag law is the imposed flow plus the disturbance from the \i[other] bead, with its own singular field omitted. At leading order in $a/R$, the evenness of the <Oseen tensor>, $\mathsf G(-\mathbf R)=\mathsf G(\mathbf R)$, gives
$$
\begin{aligned}
\dot{\mathbf r}_1
&=\mathbf L\mathbf r_1+\mathsf G(\mathbf R)\mathbf F_2
+\frac{\mathbf F_1}{\zeta},\\
\dot{\mathbf r}_2
&=\mathbf L\mathbf r_2+\mathsf G(\mathbf R)\mathbf F_1
+\frac{\mathbf F_2}{\zeta},
\qquad \boxed{\zeta=6\pi\mu a}.
\end{aligned}
$$
These are the pair <hydrodynamic mobility matrix> relations. Their signs follow from balancing each spring <force> with $-\zeta(\dot{\mathbf r}_i-\mathbf u_{{\rm ext},i})$, the fluid drag on that bead.
Adding the equations cancels both the direct spring slips and the cross-induced <velocities>, since $\mathbf F_1+\mathbf F_2=0$. For the midpoint $\mathbf r_c=(\mathbf r_1+\mathbf r_2)/2$,
$$
\boxed{\dot{\mathbf r}_c=\mathbf L\mathbf r_c.}
$$
The midpoint therefore follows the undisturbed <linear velocity field>. Subtracting the equations instead gives the <hydrodynamic interaction of a Hookean dumbbell>:
$$
\begin{aligned}
\dot{\mathbf R}
&=\mathbf L\mathbf R+
\left(\zeta^{-1}I-\mathsf G(\mathbf R)\right)
(\mathbf F_1-\mathbf F_2)\\
&=\boxed{\mathbf L\mathbf R-\frac{2H}{\zeta}
\left[I-\frac{\zeta}{8\pi\mu R}
(I+\widehat{\mathbf R}\widehat{\mathbf R}^T)\right]\mathbf R}.
\end{aligned}
$$
The dimensionless interaction matrix is
$$
\zeta\mathsf G(\mathbf R)
=\frac{3a}{4R}(I+\widehat{\mathbf R}\widehat{\mathbf R}^T).
$$
Its transverse and longitudinal <eigenvalues> are $3a/(4R)$ and $3a/(2R)$, respectively. Since the spring <force> is longitudinal, the separation equation can also be written
$$
\boxed{\dot{\mathbf R}
=\mathbf L\mathbf R-\frac{2H}{\zeta}
\left(1-\frac{3a}{2R}\right)\mathbf R.}
$$
The relative correction to the free-draining spring relaxation rate is therefore $3a/(2R)$, explicitly proportional to $a/R$. Entrainment by the other bead reduces their relative approach for given opposing spring <forces>.
This is a far-field <hydrodynamic interaction>, valid for $R\gg a$; it is not a near-contact law and omits higher reflections and lubrication. The printed target also fixes the spring's rest length implicitly. For an ordinary <spring> obeying <Hooke's law> with nonzero rest length $R_0$, the elastic <force> would be $-H(R-R_0)\widehat{\mathbf R}$, and every spring term above would acquire the factor $1-R_0/R$. The displayed result uses the <zero-rest-length spring> convention appropriate to the ideal <Hookean dumbbell model>.
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