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A reaction's stoichiometric vector outer product multiplied by its reaction propensity function is its contribution to the infinitesimal state-increment covariance. Summing over channels gives $B$. Its normalized form $D$ divides by the two mean copy numbers. The Fokker-Planck equation writes the associated second-derivative term with $B/2$; the covariance evolution and normalized stationary Lyapunov equation instead use $B$ and $D$ themselves. A simultaneous unit loss of two species contributes positive cross noise even when their long-time stationary covariance is negative.

For the one-dimensional Itô equation $dX=b(X)dt+\sigma (X)dW$, the infinitesimal generator on smooth test functions is $L=b{\partial }_x+{\sigma }^2{\partial }_{xx}/2$. The Kolmogorov backward equation is $u_t=Lu$, while its formal adjoint produces the Fokker-Planck equation.

The diffusion matrix is the infinitesimal second jump moment, $B_{ij}(x)={\sum }_r{r}_i{r}_{j}W(x,r)$. It is symmetric and positive semidefinite and multiplies the second-derivative term of the Fokker-Planck equation.

The drift vector is the expected infinitesimal displacement per unit time. For jumps $r$ with rates $W(x,r)$, it is $A_i(x)={\sum }_r{r}_{i}W(x,r)$. It is the first jump moment entering the Fokker-Planck equation.

Let $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$ be the unnormalized conformation tensor, $G_{ij}={\partial }_j{u}_i$ the velocity gradient, and $M/\zeta$ a symmetric positive hydrodynamic mobility matrix independent of the individual connector. The connector Fokker-Planck equation has drift $G\mathbf{R}-(2H/\zeta )M\mathbf{R}$ and diffusion matrix $(2k_{B}T/\zeta )M$. Integrating its second moment twice by parts gives the displayed upper-convected derivative, $C^{\nabla }=D_{t}C-GC-C{G}^T$. If $M$ is a polynomial in $C$, the two matrices commute and the relaxation term simplifies to $-(4H/\zeta )M(C-k_{B}TI/H)$. No Gaussian distribution assumption is required for this moment closure.

With local thermal noise matched to spatially varying drag, the overdamped probability density obeys the displayed diffusion equation. Its equivalent Itô diffusion is $dX=D^{\prime }(X)dt+\sqrt{2D(X)}dW$, since the two terms in the Fokker-Planck equation combine to ${\partial }_x(D{P}_x)$. Thus $d⟨X⟩/dt=⟨D^{\prime }(X)⟩$, even with no applied force. The drift points toward greater diffusivity, but a uniform density has zero flux: the drift is balanced by the spatial variation of the noise. Holding the force-noise amplitude fixed instead would not produce this isothermal Fickian equation.

For isotropic turning in three dimensions, the tip of a marked unit vector $\mathbf{u}$ wanders over the unit sphere. Its probability density function $f$ obeys the orientational Fokker-Planck equationHere ${\nabla }_{\mathbf{u}}^2$ is the Laplace-Beltrami operator on the unit sphere. The factor two follows from the sphere surface derivative identities: ${\nabla }_{\mathbf{u}}^2{u}_i=-2u_i$, so each mean vector component decays at rate $2D_r$. This is a different orientation space from the circle of planar rotational diffusion, whose corresponding rate is $D_r$. For the same illustrative $D_r=0.25{\mathrm{s}}^{-1}$ and $t=2\mathrm{s}$, this three-dimensional autocorrelation is $e^{-1}\simeq 0.368$.

One unit vector records a pointing direction, not every aspect of a rigid body's rigid body orientation: rotation about that vector remains unspecified. Also, $C_1$ requires an identifiable arrow. An unmarked ellipsoid's long axis alone does not supply one, and projecting a three-dimensional trajectory onto a camera plane does not automatically produce the planar model.

Kinetic theory describes many-particle systems through a distribution on phase space, retaining both position and velocity. In the number-density convention, $f(t,\mathbf{x},\mathbf{v})d^{3}x{d}^{3}v$ is the expected number of particles in a small position–velocity region. For identical particles of mass $m$, its integrals give the number density $n$ and bulk velocity $\mathbf{u}$:This $f$ is not normalized to one: its integral over the whole phase space gives the total expected particle number. Transport moves particles along trajectories, while collision or diffusion terms change the distribution. The Boltzmann equation, Hamiltonian Liouville equation and Fokker-Planck equation describe different physical assumptions, rather than interchangeable collision laws.

An ideal gas is a motivating example: the pressure on a wall comes from microscopic momentum transfers, even when the bulk velocity is zero. The Maxwell-Boltzmann velocity distribution separates this random molecular motion from bulk motion; see ideal gas pressure from wall collisions. The attractive feature of kinetic theory is that a distribution can describe two opposing populations where one average velocity hides almost everything. It also tells us when averaging into a continuum approximation loses essential information: compare the intermolecular mean free path with the apparatus scale using the Knudsen number. ^[ref]

The Kramers-Moyal expansion Taylor-expands the gain terms of a jump-process master equation in the jump size. Keeping its first two terms yields a Fokker-Planck equation whose drift and diffusion coefficients are the first two infinitesimal jump moments.

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.

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.

Put $\tau =m/\zeta$. The causal Green function of the displacement equation is $G(t)={\zeta }^{-1}(1-e^{-t/\tau })$ for $t>0$ and zero for $t<0$. Its derivative has the jump needed to give $m{G}^{\prime \prime }+\zeta G^{\prime }=\delta (t)$. For the specified zero position and velocity,For Gaussian white noise these are stochastic convolutions; their first two moments follow from the stated covariance even without assuming Gaussian higher moments. In particular $⟨x⟩=⟨\dot{x}⟩=0$, andThe required equipartition theorem limit therefore fixes the fluctuation-dissipation relation for a Langevin particle:Here $k_B$ denotes the printed Boltzmann constant $k$. Set $D=k_{B}T/\zeta$, the Stokes–Einstein relation. Squaring the displacement kernel givesConsequently ${\lim }_{t\to \infty }⟨x⟩=0$ and ${\lim }_{t\to \infty }⟨x^2⟩=+\infty$, with $⟨x^2⟩=2Dt-3D\tau +o(1)$ for positive temperature. There is no finite equilibrium variance of position in an unconfined fluid. The inertial Brownian displacement with zero initial velocity has short-time variance $2D{t}^3/(3{\tau }^2)+O(t^4)$, rather than the ballistic law associated with an initially thermalized velocity.

For constant $D$, the diffusion equation has fundamental density $P(x,t)=(4\pi Dt{)}^{-1/2}{e}^{-x^2/(4Dt)}$, whose mean is zero and variance is $2Dt$. It describes the large-time, overdamped limit of the Langevin dynamics. Under the usual Gaussian-force assumption, the finite-time position density is Gaussian with the variance above and obeys a time-dependent diffusion equation with coefficient $D(1-e^{-t/\tau }{)}^2$. It does not obey the constant-$D$ equation through the initial inertial transient. The given covariance alone determines the moments, not a Gaussian law; the density statement uses the standard Gaussian thermal-noise model.

Now write $\zeta (x)={\zeta }_0+\epsilon {\zeta }_0^{\prime }x$ and $x=x_0+\epsilon x_1+O({\epsilon }^2)$, with ${\tau }_0=m/{\zeta }_0$ and $D_0=k_{B}T/{\zeta }_0$. For a fixed realization of the additive force, the first-order equation isIf the force amplitude is instead adjusted to local thermal equilibrium, its first-order noise term has zero mean in the inertial Itô formulation and does not change the following first-order mean calculation. The mixed unperturbed moment isLet $M(t)=⟨x_1(t)⟩$. Averaging and solving the scalar forced relaxation equation, with $M(0)=M^{\prime }(0)=0$, givesThus the long-time first-order coefficient and local mean drift areLiterally $\lim M=-\infty$ for ${\zeta }_0^{\prime }>0$, $+\infty$ for ${\zeta }_0^{\prime }<0$, and zero for ${\zeta }_0^{\prime }=0$; the useful finite quantity is $\lim M^{\prime }$. Because the viscosity gradient is only a local expansion, the physical drift result applies in the intermediate window ${\tau }_0\ll t\ll L_{\zeta }^2/D_0$, where $L_{\zeta }={\zeta }_0/(\epsilon |{\zeta }_0^{\prime }|)$. Extrapolating the linear drag law to arbitrarily large displacements could make the drag negative and would invalidate the model.

The isothermal diffusion with position-dependent drag interpretation uses locally matched thermal noise and $D(x)=k_{B}T/\zeta (x)$. Integrating the printed Fickian diffusion equation against $x$, with vanishing boundary fluxes, givesSince $D^{\prime }(x)=-\epsilon k_{B}T{\zeta }_0^{\prime }/{\zeta }_0^2+O({\epsilon }^2)$, this reproduces the same drift after the inertial transient. Equivalently the overdamped Itô diffusion has drift $D^{\prime }$ and noise amplitude $\sqrt{2D}$. Omitting that drift would give $P_t=(DP{)}_{xx}$, which is a different Fokker-Planck equation and not the stated Fickian equation.

Noise conventions cannot be mixed silently. If the force covariance is held at the spatially constant value $F_0=2{\zeta }_0{k}_{B}T$ while drag varies, the local overdamped diffusion coefficient is instead $\tilde{D}=F_0/(2{\zeta }^2)$ and its Itô drift is ${\tilde{D}}^{\prime }/2$. Its density equation is $P_t={\partial }_x[\tilde{D}{P}_x+({\tilde{D}}^{\prime }/2)P]$, not the isothermal Fickian equation. This still gives the same displayed first-order mean drift at $x=0$, but has different spatial diffusion and equilibrium properties. The stated Fickian interpretation therefore requires local thermal-noise matching.

Physically, thermally driven excursions travel farther on the lower-drag side. Their displacement is correlated with velocity, so averaging the position-dependent friction leaves a net drift toward greater diffusivity, without an applied force. This is not a violation of equilibrium: in the isothermal Fickian model the drift and spatially varying noise cancel in the probability flux for a uniform density. It is a local diffusion-induced shift of the centroid of an initially localized particle distribution.

Neglecting cell rotational inertia, the light and viscous torque sum to zero. Therefore the angular velocity is $\mathbf{\Omega }=\mathbf{\omega }/2+(\mathbf{p}\times \mathbf{L})/\alpha$. A material orientation evolves by $\dot{\mathbf{p}}=\mathbf{\Omega }\times \mathbf{p}$. The vector triple-product identity givesThe last term is the projection of $\mathbf{L}/\alpha$ onto the tangent plane to the unit sphere. In particular, $\mathbf{p}\cdot \dot{\mathbf{p}}=0$, as required to preserve the unit orientation. It aligns the swimmer towards the light; the vorticity rotates that orientation.

For rotational diffusion coefficient $D_r$, the phototactic orientation Fokker-Planck equation balances deterministic orientation drift with diffusion on the sphere. Its density is normalized with respect to solid angle, not $d\theta d\varphi$. When $\epsilon =0$, there is no azimuthal drift and the zero-current condition is $f_{0,\theta }=-\lambda \sin \theta f_0$. Thus $f_0=\eta e^{\lambda \cos \theta }$, and normalization givesThe limiting density at $\lambda =0$ is $1/(4\pi )$.

To justify the angular dependence of the first correction, define the operatorsThe steady Fokker-Planck equation is $\mathcal{A}f=-\epsilon \mathcal{R}f$. Its regular expansion $f=f_0+\epsilon f^{(1)}+O({\epsilon }^2)$ therefore givesThe coefficients of $\mathcal{A}$ are independent of $\varphi$, so each azimuthal Fourier harmonic is preserved. The forcing has only the sine harmonic with azimuthal number one. There can be no extra unforced harmonic in a smooth normalized first correction: writing a homogeneous solution as $f_{0}h$ gives $\mathcal{A}(f_{0}h)={\nabla }_p\cdot (f_0{\nabla }_{p}h)$. Multiplication by $h$ and integration by parts on the sphere gives $\int f_0|{\nabla }_{p}h{|}^{2}d\Omega =0$, hence $h$ is constant. Normalization $\int f^{(1)}d\Omega =0$ removes that multiple of $f_0$. It follows thatNo explicit calculation of $g$ is needed. Its dependence is on $\theta$ and the parameter $\lambda$, and its behavior at the poles must make the density smooth.

In the stated spherical coordinates, $\mathbf{p}=\cos \theta {\mathbf{e}}_x+\sin \theta \cos \varphi {\mathbf{e}}_y+\sin \theta \sin \varphi {\mathbf{e}}_z$. Azimuthal integration makes the leading $y,z$ components and the first-order $x,y$ corrections vanish. The surviving coefficients areThus the small-vorticity phototactic orientation response isBoth coefficients depend only on $\lambda$. In particular, $K_0$ is the Langevin function of $\lambda$ and is positive for positive light strength; its integral representation above suffices.

Write the diffusion process as $d{X}_t=b(X_t)dt+\sigma (X_t)d{B}_t$, with $a=\sigma {\sigma }^{\mathsf{T}}$. Its diffusion generator isThe Itô formula says that $\phi (X_t)-\phi (X_0)-{\int }_0^{t}L\phi (X_s)ds$ is a local martingale, and is a true martingale when the stopped terms are integrable. Taking expectations gives Dynkin formula for a diffusion. This is the bridge from infinitesimal stochastic dynamics to differential equations.

For the transition semigroup $P_{t}g(x)={\mathbb{E}}_{x}g(X_t)$, the Markov property gives $P_{t+s}=P_t{P}_s$. On a smooth generator domain where differentiation is valid, differentiating the semigroup identity yields the Kolmogorov backward equation ${\partial }_t{P}_{t}g=L(P_{t}g)$, with initial value $g$. If a transition density $p(t,x,y)$ exists, integration against smooth compactly supported tests and integration by parts give the Fokker-Planck equationThe backward equation acts on the starting point, while this forward equation describes the evolving probability density.

Here is a full representation proof with a source and potential. Suppose $X$ is nonexplosive, $V\ge 0$ is bounded, $f,g$ are bounded, and $u$ is a bounded classical solution on a finite slab ofFor a process started from $x$ at time $t$, put $D_s=\exp (-{\int }_t^{s}V(X_r)dr)$. The Itô product rule givesTherefore $D_{s}u(s,X_s)+{\int }_t^s{D}_{r}f(r,X_r)dr$ is a local martingale. It is bounded in absolute value by $\Vert u{\Vert }_{\infty }+(T-t)\Vert f{\Vert }_{\infty }$, so it is a true martingale, without requiring a global derivative bound. Its endpoint expectations give the parabolic Feynman-Kac formula with a sourceThis proves uniqueness in the bounded classical class. Setting $V=f=0$ recovers the backward heat or diffusion representation; the minus sign in the killing term produces the negative exponential.

For an elliptic problem, stop at the first exit $\tau$ from a bounded regular domain $D$. Suppose ${\mathbb{E}}_x\tau <\infty$, $u\in C^2(D)\cap C(\bar{D})$ is bounded and solves $(L-V)u=-f$ with boundary data $g$. The same calculation, stopped at $t\wedge \tau$, gives a martingale dominated by $\Vert u{\Vert }_{\infty }+\Vert f{\Vert }_{\infty }\tau$. Continuity puts $X_{\tau }$ on the boundary. Taking the limit and using dominated convergence gives the elliptic Feynman-Kac formulaFor $V=f=0$, harmonic functions for $L$ give stopped martingales and are averages of their exit-boundary values. For Brownian motion $L=\frac{1}{2}\Delta$, this is the probabilistic solution of the Dirichlet problem. Suitable smooth coefficients and uniform ellipticity give familiar settings where the required diffusion, exit-time integrability and classical solutions exist. The representation itself follows from the stated regularity and integrability conditions, rather than claiming classical solvability for arbitrary coefficients.

Interpret the multiplicative Gaussian white noise as the limit of a smooth short-correlation-time velocity field, hence in the Stratonovich integral convention. Put ${\partial }_i=\partial /\partial B_i$. For a realization, differentiating the field-space Dirac delta gives its field-space continuity equationThe velocity-gradient covariance has zero trace variance: contracting $T_{mn}^{ij}$ with $m=i,n=j$ gives $3-(9+3)/4=0$. Thus the gradient is trace free, consistently with the incompressible ideal induction equation.

A noise impulse in component ${\sigma }_{jn}$ changes ${\tilde{B}}_j$ by ${\tilde{B}}_n$ times that impulse. Its instantaneous retarded action on the Dirac delta is thereforeOnly earlier times affect the present field. In the white-noise limit the integral over this causal interval has ${\int }_0^t\delta (t-t^{\prime })d{t}^{\prime }=1/2$, corresponding to the symmetric short-time regularization. Applying the Furutsu–Novikov formula and then averaging givesSubstitute into the averaged field-space continuity equation:The differentiated $B_n$ term vanishes because ${\sum }_j{T}_{mj}^{ij}=0$, and the remaining contraction isConsequently the closed vector Fokker-Planck equation for Gaussian white-noise magnetic stretching isThis is a positive diffusion operator: the displayed matrix has eigenvalues $B^2$ on the two transverse directions and $B^2/2$ along $\mathbf{B}$. Replacing the endpoint half-weight by one would double the diffusion and fail to reproduce the later rate.

Differentiate the product of Dirac delta functions defining $\tilde{P}$, regarding $B$ and $\sigma$ as independent density arguments. The random continuity equation isFor $0<s<t$, the causal functional derivatives follow from the explicit strain solution and $\tilde{B}(t)=B_0\exp [{\int }_0^t\tilde{\sigma }(u)du]$:Consequently,In the Furutsu–Novikov formula, the covariance $\kappa \delta (t-s)$ samples the upper endpoint of the causal time integral. Symmetric regularization of the white noise gives half the mass there. The $B$ response vanishes as $s\uparrow t$, while the strain response tends to one, so $\mathbb{E}[f(t)\tilde{P}(t)]=-(\kappa /2){\partial }_{\sigma }P$. Averaging the continuity equation yields the Fokker-Planck equationThere is no direct diffusion term in $B$: its equation contains the time integral of colored noise, whereas the strain equation receives Gaussian white noise. The same result follows from the Itô diffusion with drift $(\sigma B,-\sigma /\tau )$ and diffusion vector $(0,\sqrt{\kappa })$.

Multiply the Fokker-Planck equation by $B^n$ and integrate over $B>0$ and all $\sigma$, assuming the required boundary terms vanish. Integration by parts givesThe mixed moment is not determined by $\mathbb{E}[B^n]$: the current strain and the accumulated magnetic field are correlated. Zero mean strain does not imply $\mathbb{E}[\sigma B^n]=0$, so this is an unclosed moment equation.

Keep the strain dependence by defining $P_n(\sigma ,t)={\int }_0^{\infty }{B}^{n}P(B,\sigma ,t)dB$. The magnetic drift term satisfiesThe strain derivatives commute with the $B$ integral. HenceThis weighted density obeys a tilted Ornstein-Uhlenbeck Fokker-Planck equation. Its integral over $\sigma$ is the desired moment, while retaining $\sigma$ makes the evolution closed.

With $x=\cos \theta$ and $D_r=0$, the Fokker-Planck equation becomes the probability-transport equationIts characteristic obeys $\dot{x}=(1-x^2)/B$. Writing $s=t/B$ and $q=\tanh s$ givesConservation of orientation probability pushes the initially uniform density forward by this Jacobian. ThusSince $q=(1-\alpha )/(1+\alpha )$ with $\alpha =e^{-2t/B}$, the deterministic alignment of an initially isotropic orientation distribution isAt $t=0$, $\alpha =1$ and $f=1/(4\pi )$; the characteristic Jacobian preserves $2\pi {\int }_{-1}^{1}fdx=1$ at every finite time. Its reciprocal is quadratic in $x$, as in the suggested substitution.