= Solution
Let $p_T=nk_BT$, $C=\langle\mathbf R\mathbf R^T\rangle$, and $M=I+\alpha\sigma$, so the single-bead <Hydrodynamic mobility matrix>[mobility] is $M/\zeta$. Assume an <Incompressible flow>[incompressible solvent], constant $n,H,T,\zeta$, negligible <Inertia>[bead inertia], and a symmetric <Positive-definite matrix>[positive-definite] $M$. The last requirement makes the <Hydrodynamic resistance matrix>[resistance] invertible and <Diffusion>[Brownian diffusion] physically admissible. At a given position and time, $M$ depends on the moment $C$ but not on the individual configuration $\mathbf R$.
Use the source's <gradient> convention $K_{ij}=\partial_i u_j$, and write $G=K^T$ for the <Velocity gradient>[component-first gradient]. Then $\mathbf u(\mathbf r_2)-\mathbf u(\mathbf r_1)\simeq G\mathbf R$. The <Hooke's law>[spring forces] on the two beads are $\mathbf F_1=H\mathbf R$ and $\mathbf F_2=-H\mathbf R$. Their overdamped <Fokker-Planck probability current>[probability currents] are
$$
\mathbf j_i=\mathbf u_i\psi+\frac M\zeta
\left(\mathbf F_i\psi-k_BT\nabla_{\mathbf r_i}\psi\right).
$$
Here the <Brownian motion>[Brownian term] written as $-k_BT\nabla\psi$ is a <force> weighted by <probability density>. Its corresponding conditional <Osmotic pressure>[osmotic force] is $-k_BT\nabla\log\psi$; treating the given term as an unweighted deterministic <force> per bead would not produce the stated <Fokker-Planck equation>[Fokker–Planck equation].
With midpoint $\mathbf x=(\mathbf r_1+\mathbf r_2)/2$, the derivatives are $\nabla_{r_1}=\tfrac12\nabla_x-\nabla_R$ and $\nabla_{r_2}=\tfrac12\nabla_x+\nabla_R$. Subtracting the bead currents gives the <Fokker-Planck probability current>[configuration current]
$$
\boxed{\mathbf j_R=G\mathbf R\psi-
\frac{2H}{\zeta}M\mathbf R\psi-
\frac{2k_BT}{\zeta}M\nabla_R\psi.}
$$
The <Fokker-Planck probability current>[midpoint current] is $\mathbf u\psi$ after neglecting the specified spatial <Diffusion>[Brownian diffusion]. <Continuity equation>[Configuration probability conservation] therefore gives
$$
\boxed{\partial_t\psi+\mathbf u\cdot\nabla_x\psi+
\nabla_R\cdot\left(G\mathbf R\psi-\frac{2H}{\zeta}M\mathbf R\psi\right)
=\frac{2k_BT}{\zeta}\nabla_R\cdot(M\nabla_R\psi).}
$$
Since $G\mathbf R$ has components $R_j\partial_j u_i$, this agrees with the source's $\mathbf R\cdot\nabla\mathbf u$. The factor two in both spring relaxation and configuration <Diffusion coefficient>[diffusivity] comes from relative motion of both beads.
The <second moment> $C$ is the unnormalized <conformation tensor>. Now multiply by $R_iR_j$ and integrate over <configuration space>. Assume sufficiently decaying tails so every boundary term at infinite separation vanishes. The affine current gives $GC+CG^T$; spring drift gives $-(2H/\zeta)(MC+CM)$; two <integrations by parts> of <diffusion> give $(4k_BT/\zeta)M$. Hence
$$
D_tC-GC-CG^T=-\frac{2H}{\zeta}(MC+CM)+\frac{4k_BT}{\zeta}M,
\qquad D_t=\partial_t+\mathbf u\cdot\nabla_x.
$$
Because $\sigma=nHC-p_TI$, $M$ is a <polynomial> in $C$ and <Commuting operators>[commutes] with it. Thus
$$
C^{\nabla}\equiv D_tC-GC-CG^T
=-\frac{4H}{\zeta}M\left(C-\frac{k_BT}{H}I\right).
$$
This is the <Hookean dumbbell conformation equation with anisotropic mobility>, an exact <Moment closure>[second-moment closure] within the stated model; no <Gaussian distribution>[Gaussian approximation] to $\psi$ was needed.
The <upper-convected derivative> of the <Identity matrix>[identity] is $I^{\nabla}=-(G+G^T)$. Applying it to the <Stress>[polymer stress] therefore gives
$$
\sigma^{\nabla}=-\frac{4H}{\zeta}M\sigma+p_T(G+G^T).
$$
Consequently the coefficients and the closed <Constitutive equation>[stress equation] are
$$
\boxed{\lambda=\frac{\zeta}{4H},\qquad
\mu=\lambda p_T=\frac{n k_BT\zeta}{4H},\qquad
\lambda\sigma^{\nabla}+\sigma+\alpha\sigma^2
=\mu(K+K^T),}
$$
where $\sigma^{\nabla}=D_t\sigma-\sigma K-K^T\sigma$ in the printed convention. Here $\lambda$ is a <polymer relaxation time>, unrelated to the inverse-length parameter of question <1/c>; $\mu$ is the <Viscosity>[polymer viscosity] scale. The coefficient $\alpha$ has inverse-stress units. In the <dimensional mobility parameter in the Giesekus model>, the dimensionless factor is $b=\alpha p_T$, not the dimensional $\alpha$ printed in this problem.
The <Upper-convected derivative>[upper-convected rate] removes the affine <Conformation tensor>[stretching] and <Rotation>[rotation of material connectors]. If $\dot F=GF$, then a <tensor> transported purely as $C=FC_0F^T$ has $C^{\nabla}=0$. Under a time-dependent <Rotation>[rigid rotation] $Q$, the <Objective time derivative>[rate transforms] as $Q C^{\nabla}Q^T$: this is <material frame indifference>, not an ordinary componentwise time derivative. The resulting equation is the <Giesekus model> in these dimensional conventions.
The <Isotropic tensor>[isotropic subtraction] is fixed by \b[zero <Stress>[polymer extra stress] at <thermal equilibrium>]. With no flow and <Hydrodynamic mobility matrix>[positive mobility], the <stationary distribution> is
$$
\psi_{\rm eq}=\left(\frac{H}{2\pi k_BT}\right)^{3/2}
\exp\left(-\frac{H R^2}{2k_BT}\right),\qquad
C_{\rm eq}=\frac{k_BT}{H}I,
$$
so $\boxed{\sigma_{\rm eq}=0}$. The <constitutive equation> correspondingly has a zero-stress quiescent state and a <Newtonian fluid>[Newtonian low-rate limit] $\sigma\simeq\mu(K+K^T)$. Without subtracting exactly $p_TI$, the spring contribution would retain an equilibrium <Isotropic tensor>[isotropic stress] and the equation would acquire corresponding reference terms. Because <Hydrodynamic mobility matrix>[mobility] itself depends on $\sigma$, that reference cannot simply be shifted while retaining the same model. The separate <Pressure>[fluid pressure] enforcing <Incompressible flow>[incompressibility] is not included in this <Stress>[polymer extra stress].
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