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For constant number density $n$, constant temperature, a homogeneous incompressible linear velocity field and decaying configuration tails, defineThis free energy is a relative entropy times $n{k}_{B}T$. Differentiate it using the configuration Fokker-Planck probability current. One integration by parts gives $\dot{\mathcal{F}}=n\int \psi (H\mathbf{R}+k_{B}T\nabla \log \psi )\cdot \mathbf{V}d\mathbf{R}$. Substituting the probability current velocity gives the displayed balance: the diffusive affine contribution is $-n{k}_{B}T\mathrm{tr}K=0$. The polymer stress tensor ${\sigma }_p=n(HC-k_{B}TI)$ has mechanical power ${\sigma }_p:K=nHC:K$. Its instantaneous power need not be nonnegative, because it includes changes of stored free energy. The nonnegative quantity is $\mathcal{D}$, and equality of stress power and dissipation requires a stationary free energy.

Represent a dilute polymer by two beads connected by a spring obeying Hooke's law, with stiffness $H>0$ and bead drag coefficient $\zeta >0$. Neglect bead inertia and hydrodynamic interactions. For component-first velocity gradient $K_{ij}={\partial }_j{u}_i$, the configuration Fokker-Planck probability current isThe factor two comes from relative motion of both beads. At rest its normalized probability density is ${\psi }_{\mathrm{eq}}=(H/(2\pi k_{B}T){)}^{3/2}{e}^{-H{R}^2/(2k_{B}T)}$. The conformation tensor $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$ has dimensions of length squared; $(H/(k_{B}T))C$ is its dimensionless version. The Hookean dumbbell in planar extension is a particular imposed flow in this model.

The probability-density equation for the Ornstein-Uhlenbeck process with drift $-v$ and diffusion coefficient $\sqrt{2}$. The unit-covariance Gaussian density is stationary because its Fokker-Planck probability current vanishes. This normalization makes mean velocity decay at rate one and energy excess decay at rate two.

Let $p_T=n{k}_{B}T$, $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$, and $M=I+\alpha \sigma$, so the single-bead mobility is $M/\zeta$. Assume an incompressible solvent, constant $n,H,T,\zeta$, negligible bead inertia, and a symmetric positive-definite $M$. The last requirement makes the resistance invertible and 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 component-first gradient. Then $\mathbf{u}({\mathbf{r}}_2)-\mathbf{u}({\mathbf{r}}_1)\simeq G\mathbf{R}$. The spring forces on the two beads are ${\mathbf{F}}_1=H\mathbf{R}$ and ${\mathbf{F}}_2=-H\mathbf{R}$. Their overdamped probability currents areHere the Brownian term written as $-k_{B}T\nabla \psi$ is a force weighted by probability density. Its corresponding conditional osmotic force is $-k_{B}T\nabla \log \psi$; treating the given term as an unweighted deterministic force per bead would not produce the stated Fokker–Planck equation.

With midpoint $\mathbf{x}=({\mathbf{r}}_1+{\mathbf{r}}_2)/2$, the derivatives are ${\nabla }_{r_1}=\frac{1}{2}{\nabla }_x-{\nabla }_R$ and ${\nabla }_{r_2}=\frac{1}{2}{\nabla }_x+{\nabla }_R$. Subtracting the bead currents gives the configuration currentThe midpoint current is $\mathbf{u}\psi$ after neglecting the specified spatial Brownian diffusion. Configuration probability conservation therefore givesSince $G\mathbf{R}$ 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 diffusivity comes from relative motion of both beads.

The second moment $C$ is the unnormalized conformation tensor. Now multiply by $R_i{R}_j$ and integrate over configuration space. Assume sufficiently decaying tails so every boundary term at infinite separation vanishes. The affine current gives $GC+C{G}^T$; spring drift gives $-(2H/\zeta )(MC+CM)$; two integrations by parts of diffusion give $(4k_{B}T/\zeta )M$. HenceBecause $\sigma =nHC-p_{T}I$, $M$ is a polynomial in $C$ and commutes with it. ThusThis is the Hookean dumbbell conformation equation with anisotropic mobility, an exact second-moment closure within the stated model; no Gaussian approximation to $\psi$ was needed.

The upper-convected derivative of the identity is $I^{\nabla }=-(G+G^T)$. Applying it to the polymer stress therefore givesConsequently the coefficients and the closed stress equation arewhere ${\sigma }^{\nabla }=D_t\sigma -\sigma K-K^T\sigma$ in the printed convention. Here $\lambda$ is a polymer relaxation time, unrelated to the inverse-length parameter of question c; $\mu$ is the 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 rate removes the affine stretching and rotation of material connectors. If $\dot{F}=GF$, then a tensor transported purely as $C=F{C}_0{F}^T$ has $C^{\nabla }=0$. Under a time-dependent rigid rotation $Q$, the 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 subtraction is fixed by zero polymer extra stress at thermal equilibrium. With no flow and positive mobility, the stationary distribution isso $\boxed{{\sigma }_{\mathrm{eq}}=0}$. The constitutive equation correspondingly has a zero-stress quiescent state and a Newtonian low-rate limit $\sigma \simeq \mu (K+K^T)$. Without subtracting exactly $p_{T}I$, the spring contribution would retain an equilibrium isotropic stress and the equation would acquire corresponding reference terms. Because mobility itself depends on $\sigma$, that reference cannot simply be shifted while retaining the same model. The separate fluid pressure enforcing incompressibility is not included in this polymer extra stress.

The claimed equality does not hold for the printed joint distribution. The original PDF specifies a centered Maxwell-Boltzmann distribution of $\mathbf{v}=\dot{\mathbf{R}}$ independent of $\mathbf{R}$, withSince the double average includes number density $n$, evaluating the Gaussian integrals givesThe cross term vanishes because the conditional mean of $\mathbf{v}$ is zero. In particular, set $K=0$ and take the equilibrium configuration probability density ${\psi }_{\mathrm{eq}}\propto e^{-H{R}^2/(2k_{B}T)}$. The left side is $3n\zeta k_{B}T/m>0$, while $nHC:K=0$. This is a counterexample even at thermal equilibrium, so no rearrangement of the earlier conformation tensor equation can prove the printed statement.

The discrepancy is thermal bead drag and heat-bath power: gross viscous loss of microscopic thermal motion is replenished by the thermal bath. It is not identical to mechanical polymer stress power. Moreover, $\dot{\mathbf{R}}$ in the earlier configuration continuity equation is a probability current velocity, including a distribution-dependent osmotic drift; it cannot simultaneously be identified with an independently sampled centered Maxwellian velocity.

A consistent overdamped result is the free-energy balance of a Hookean dumbbell. Define $\mathbf{q}=H\mathbf{R}+k_{B}T{\nabla }_{\mathbf{R}}\log \psi$ andDifferentiate the relative entropy and use the configuration Fokker-Planck probability current $\psi \mathbf{V}$. After integration by parts,The affine osmotic term is $n{k}_{B}T\int \nabla \psi \cdot K\mathbf{R}d\mathbf{R}=-n{k}_{B}T\mathrm{tr}K=0$. HenceThe polymer stress tensor ${\sigma }_p=n(HC-k_{B}TI)$ has ${\sigma }_p:K=nHC:K$ in an incompressible flow. It stores or releases free energy as well as dissipating it, so the total stress-work term in part a need not itself be a purely dissipative quantity for this viscoelastic material. Stress power equals $\mathcal{D}$ only for stationary free energy, and $\mathcal{D}$ uses the probability-current slip. Neither condition turns the printed independent-Maxwellian average into that expression.

Differentiate the material derivative of the passive scalar component by component. With ${\sigma }_{ij}={\partial }_j{u}_i$,The chain rule along a Lagrangian trajectory $\dot{X}=u(X,t)$ therefore gives the covector evolution of a material scalar gradientThe transpose matters: a gradient is transported as a covector, rather than as a material separation vector. When $u_3=0$, the third row of $\sigma$ vanishes, andLet $k_h=(k_1,k_2{)}^T$ and setFor $c>0$, the matrix exponential gives the exact horizontal solutionThe two projections are onto the eigenspaces with eigenvalues $-c$ and $+c$. If $p_{-}\ne 0$, write $p_{-}=k_0(\cos \varphi ,\sin \varphi {)}^T$ with $k_0=|p_{-}|>0$. ThenThe stated growing asymptotic requires this nonzero projection. An initial gradient entirely in the other eigenspace decays, and when $c=0$ the horizontal gradient is constant.

For constant vertical shear, write $h=({\sigma }_{13},{\sigma }_{23}{)}^T$. Integrating the third component using the exact horizontal solution yieldsConsequently, if $D=h\cdot (\cos \varphi ,\sin \varphi )$ is nonzero, $k_3\sim -k_{0}D{e}^{ct}/c$, andIf $D=0$, exponential growth of $k_3$ is absent, but the ratio still tends to zero on the growing horizontal branch. Thus exponential vertical growth is generic, not unconditional. The scalar-gradient inclination in a steady planar strain is the tangent of the angle between the scalar's normal and the horizontal plane. Equivalently, it is the cotangent of the inclination of an isoconcentration surface to that plane. Large $\alpha$ corresponds to nearly horizontal scalar surfaces; small $\alpha$ corresponds to nearly vertical ones.

For the probability calculation, take $\Gamma ,\Lambda >0$. The independent random variables $a,b$ with normal distributions of variance ${\Gamma }^2$ make $c$ distributed according to the Rayleigh distribution, withThe advertised Gaussian-over-Rayleigh inclination distribution also needs $D$ to be independent of $(a,b)$; its Gaussian marginal alone does not imply this. Under that intended assumption, for $A>0$,The cumulative distribution function is zero for $A\le 0$. Differentiating givesand the probability density function is zero for negative $A$. Its integral is one because the cumulative distribution function tends to one at infinity. Independence is a real hypothesis: $D=(\Lambda /\Gamma )a$ has exactly the stated Gaussian marginal but makes $\alpha \le \Lambda /\Gamma$ almost surely, contradicting the unbounded support of this density. A sufficient physical model is isotropic Gaussian vertical shear, independent of $a,b$, whose projection in any horizontal direction has variance ${\Lambda }^2$.

If the strain varies slowly, the horizontal gradient must align with the current growing direction, and the vertical-to-horizontal ratio must adjust before that strain changes. Both adjustment times are of order $c^{-1}$. Thus the local condition is $c{\tau }_{\sigma }\gg 1$, and the typical condition for the specified statistics isThis is an adiabatic approximation, not a uniform statement about arbitrarily weak instantaneous strains: samples with $c{\tau }_{\sigma }\lesssim 1$ retain memory of earlier strain. The instantaneous eigenvector must likewise turn slowly compared with $c$.

For rapidly fluctuating vertical shear, insert the large-time horizontal solution into its stochastic differential equation:The linear combination $W=\cos \varphi W^{(1)}+\sin \varphi W^{(2)}$ is a Brownian motion: it has independent Gaussian increments and increment variance $({\cos }^2\varphi +{\sin }^2\varphi )dt=dt$. For the signed inclination $\beta =k_3{e}^{-ct}/k_0$, the Itô product rule introduces no extra cross term because the exponential factor is deterministic. It follows thatThis is an Ornstein-Uhlenbeck process, explaining why white-noise vertical shear gives an Ornstein-Uhlenbeck inclination. Its Fokker-Planck equation isIn a stationary state the Fokker-Planck probability current $j=-cBp-(g^2/2)p^{\prime }$ is constant. A nonzero constant current cannot give an integrable, nonnegative density at both ends of the real line, so $j=0$. Integrating $p^{\prime }/p=-2cB/g^2$ and normalizing with the Gaussian integral givesThe stationary variance is $g^2/(2c)$. More explicitly the explicit Ornstein-Uhlenbeck solution has mean $\beta (0)e^{-ct}$ and variance $g^2(1-e^{-2ct})/(2c)$, confirming convergence to this density. Here $c>0$ and $g\ne 0$ are needed for the stated smooth stationary density; at $c=0$ there is no confinement, while $g=0$ gives a point mass at zero.

For isotropic initial statistics, isotropic noise preserves $P=P(B,t)$. Isotropy of the noise alone does not turn an arbitrary anisotropic initial vector distribution into an isotropic one; alternatively take the angularly averaged magnitude distribution. The radial gradient satisfies ${\partial }_{j}P=P_B{B}_j/B$, so the diffusion-vector contraction in the preceding equation is $B{B}_i{P}_B/2$. Its divergence givesNow $F=4\pi B^{2}P$ is the magnitude probability density function, normalized by ${\int }_0^{\infty }FdB=1$. Sincewe obtain the radial magnetic-field Fokker-Planck equationThe outward Fokker-Planck probability current is $J=-D(B^2{F}_B-2BF)$. Zero probability flux at zero and infinity preserves normalization. For positive initial magnitude the solution below remains on $B>0$; an atom initially at $B=0$ instead remains an absorbing zero-field component.

Put $q(x)=\sigma (x{)}^{2}p(x)$. The explicit density formula and local integrability assumption giveHence its Fokker-Planck probability current $bp-q^{\prime }/2$ is zero. The stationary adjoint equation is $L^{\ast }p=-(bp{)}^{\prime }+q^{\prime \prime }/2=0$. We prove the required integral identity with cutoffs, avoiding an unstated boundary condition on derivatives of $u$.

Choose smooth $0\le {\chi }_R\le 1$, equal to one on $[-R,R]$, supported in $[-2R,2R]$, with $|{\chi }_R^{\prime }|\le C_1/R$ and $|{\chi }_R^{\prime \prime }|\le C_2/R^2$. Since $pLu=(q{u}_x{)}^{\prime }/2$, integrate the backward equation in time and integrate by parts twice in space:These integrations have compact support, as permitted in the question. Bounded coefficients and normalized $p$ giveThe right side is bounded in absolute value bywhich tends to zero. Dominated convergence on the left therefore provesThis is the cutoff proof of invariance for a zero-flux diffusion density. In particular, if an independent initial state has density $p$, part (a) and Fubini give a constant expectation of this test function at every time. This consequence alone is sufficient for the final coupling argument.

For the normalized standard Gaussian density,The two terms cancel, so the Ornstein-Uhlenbeck Fokker-Planck equation has stationary density for a Fokker-Planck equationEquivalently its Fokker-Planck probability current $-\nabla \gamma -v\gamma$ vanishes identically. The normalization follows from the Gaussian integral in each coordinate.

Here $\Gamma >0$ is a kinetic coefficient, and the additive Gaussian white noise must be centered. A rigorous starting point is a finite spatial discretization or finite set of modes of the coarse-grained field. For each field coordinate, the Fokker-Planck probability current has the continuum notationThe equilibrium Boltzmann distribution ${\mathcal{P}}_{\mathrm{eq}}\propto e^{-\beta F}$ has $\delta {\mathcal{P}}_{\mathrm{eq}}/\delta \varphi =-\beta (\delta F/\delta \varphi ){\mathcal{P}}_{\mathrm{eq}}$. Detailed balance for this time-even field requires zero configuration-space current. ThereforeThis is the Model A fluctuation-dissipation relation. Conversely, that noise strength makes the equilibrium current vanish and gives reversible relaxational dynamics when the equilibrium measure is normalizable.

Alternatively, the Onsager--Machlup path probability for Model A dynamics compares the squared required noise histories $\dot{\varphi }+\Gamma \delta F/\delta \varphi$ and $-\dot{\varphi }+\Gamma \delta F/\delta \varphi$. Their difference gives $\log ({\mathbb{P}}_F/{\mathbb{P}}_B)=-2\Gamma \Delta F/{\sigma }^2$, using the functional chain rule. The midpoint convention and time-reversal invariance of a path Jacobian justify cancellation of the common Jacobian. Comparing with the conditional path ratio gives the same result. The spatial white-noise continuum is a coarse-grained idealization: retain a microscopic cutoff when pointwise fluctuations or ultraviolet divergences matter.

With $A_i=a_{ij}{x}_j$ and constant $B_{ij}=b_{ij}$, the Fokker-Planck probability current becomesIn the stationary state we impose the stronger zero-current condition $J_i=0$, corresponding to detailed balance.

For a sufficiently regular Itô diffusion with $D=\sigma {\sigma }^T/2$, a density $\pi$ is stationary if and only if it solves the stationary Fokker-Planck equationwith integrable Fokker-Planck probability current and boundary conditions that make its outward flux vanish. Here $(\nabla \cdot (D\pi ){)}_i={\sum }_j{\partial }_j(D_{ij}\pi )$.

For the displayed parametrization, assumewhere $Z<\infty$, that $D(x)$ is symmetric positive semidefinite, and that $Q(x)$ is antisymmetric. Substituting $\nabla \pi =-\pi \nabla H$ and the stated $\Gamma$ into the probability current cancels all $D$ terms. The remaining divergence isbecause the second derivatives are symmetric in $i,j$ whereas $Q_{ij}=-Q_{ji}$. Thus these conditions, together with the boundary and regularity assumptions, imply stationarity. More generally, the divergence equation above is the exact necessary and sufficient condition; within this construction, $\pi \propto e^{-H}$ and antisymmetric $Q$ are the standard way to satisfy it.

This is a birth-death process with birth and death ratesIts transition diagram has the two arrowsThe gain into state $n$ comes from a death in state $n+1$ or a birth in state $n-1$, while the loss is the sum of both rates out of state $n$. The birth-death master equation is consequently

Multiplying by $n$, summing over the nonnegative integers, and shifting the summation indices shows that each birth contributes $+1$ and each death contributes $-1$. Thus the first-moment equation of a birth-death process givesAt a stationary state, use the variance identity $⟨n^2⟩=⟨n{⟩}^2+(\Delta n{)}^2$. Solving the resulting quadratic equation givesThe minus root is inadmissible whenever it is negative, namely when $(\Delta n{)}^2<\gamma /\alpha$. Equality would give zero mean, which is also incompatible with a positive immigration rate $\gamma$, so for $\gamma >0$ the minus branch requires $(\Delta n{)}^2>\gamma /\alpha$ even to be a possible mean.

For a continuum approximation, write $b(n)=\gamma +\beta n$ and $d(n)=\alpha n(n-1)$. Applying the Kramers-Moyal expansion to the two gain terms and retaining derivatives through second order gives the Fokker-Planck equationwhere the negative drift and infinitesimal jump variance areThis truncation requires the typical population and the scale on which $P$ varies to be much larger than the unit jump size, with the rates varying smoothly across that scale.

The positive zero of $g$ isso in particular $n_{\ast }\sim \beta /\alpha$ as $\alpha /\beta \to 0$. Put $x=n-n_{\ast }$. The linear noise approximation useswhereand, because $b(n_{\ast })=d(n_{\ast })$,In a stationary state with zero Fokker-Planck probability current,Its normalized solution is the normal distributionIt follows that

These estimates agree with the exact stationary first-moment relation on its plus branch to leading order: inserting $(\Delta n{)}^2\sim \beta /\alpha$ gives $⟨n⟩\sim \beta /\alpha$. Moreover,Thus the stationary mass lies far from the boundary $n=0$, is narrow relative to its mean, yet changes across many lattice sites. These are precisely the large-population and slow-variation conditions needed for the diffusion approximation.

The two-dimensional Fokker-Planck equation iswith Fokker-Planck probability currentThe point initial condition iswhere $\delta$ is the Dirac delta function. The reflecting boundary condition for a diffusion iswith $\mathbf{n}$ the outward unit normal. This zero-flux condition conserves the integral of the probability density function over the square.

SetThe stated curl-free condition says that $C_j=({\kappa }^{-1}{)}_{ji}{a}_i$ is a gradient, $C=\nabla \Psi$. DefineThen $a=-D\nabla V$, and the Fokker-Planck probability current becomesThis is the potential condition for a Fokker--Planck equation.

Farther than a few edge lengths from the root, replace the radial coordinate $r$ by a physical coordinate $x=ar$ and smooth the master equation. It has the same constant-coefficient Fokker-Planck equation as in (b). Under the conventional first-time-derivative truncation,The radial drift $v=-C=a(z-2)/(z\tau )$ is outward. For $z=2$ the tree is the infinite line, its radial distance is an unbiased reflected walk, and $C=0$.

The root cannot absorb probability or allow negative radial distance. Write the equation as a continuity equation, using the Fokker-Planck probability currentThe appropriate continuum boundary condition is reflecting, or zero flux:It is not generally ${\partial }_{x}P(0,t)=0$: that simpler condition is correct only when $C=0$. With the finite-bias fixed-step Gaussian convention from (b), use $D=D_{\mathrm{fixed}}=2a^2(z-1)/(z^2\tau )$ instead; the drift sign and zero-current boundary condition are unchanged. A continuum boundary condition represents the microscopic root region after coarse-graining, rather than reproducing each discrete root probability at individual step times.

Where a probability density $\psi$ is positive, its Fokker-Planck probability current can be written $\mathbf{J}=\psi \mathbf{V}$, giving a continuity equation ${\partial }_t\psi +\nabla \cdot (\psi \mathbf{V})=0$. This velocity transports probability. It is not a differentiable instantaneous velocity of an overdamped Brownian motion path. In the Hookean dumbbell model, its osmotic term is $-(2k_{B}T/\zeta )\nabla \log \psi$.

A reflecting boundary confines a diffusion without losing probability. Its forward condition is zero outward Fokker-Planck probability current, $J\cdot n=0$, and its backward condition is a zero conormal derivative.

A probability density left unchanged by the Fokker-Planck equation ${\partial }_{t}f=L^{\ast }f$. Equivalently its Fokker-Planck probability current has zero divergence; the current itself need not vanish. For the Ornstein-Uhlenbeck Fokker-Planck equation, the normalized standard Gaussian density has zero current and is stationary.

In the long-time growing horizontal-gradient branch, independent vertical shear noises combine as $\cos \varphi d{W}_1+\sin \varphi d{W}_2=dW$. The signed inclination $\beta =k_3{e}^{-ct}/k_0$ satisfies the displayed Ornstein-Uhlenbeck stochastic differential equation. For $c>0$ its stationary density is $\sqrt{c/(\pi g^2)}{e}^{-c{B}^2/g^2}$, with variance $g^2/(2c)$. The Fokker-Planck probability current vanishes at infinity and gives this Gaussian by one integration. At $c=0$ and nonzero noise there is no normalizable stationary density.