Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 353 (Noisy Mechanics)

Stationarity makes the covariance depend only on $\tau =t-s$. For $\tau >0$,At the reversed lag, the $s>t$ formula givesand thereforeThe same identity follows directly by exchanging the two random variables in $G_{ij}(t,s)=⟨x_i(t)x_j(s)⟩$.

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Split the Fourier integral at zero and use the two stationary covariance branches:Equivalently, Fourier transforming the Multivariate Ornstein-Uhlenbeck process equation givesUnit white-noise covariance then yields the Ornstein-Uhlenbeck power spectrumso

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For $X=(x,p{)}^T$,Solving $A\sigma +\sigma A^T=b{b}^T$ givesThe canonical ensemble density proportional to $e^{-H/(k_{B}T)}$ for $H=p^2/(2m)+\lambda x^2/2$ factorizes into independent centered Gaussians. Its equipartition variances are exactly $⟨x^2⟩=k_{B}T/\lambda$ and $⟨p^2⟩=m{k}_{B}T$, while the absence of an $xp$ term gives ${\sigma }_{12}=0$.

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LetThe second column of $(A+i\omega I{)}^{-1}$ is ${\Delta }^{-1}(1/m,i\omega {)}^T$. HenceIn particular,This is the thermally broadened resonance of the damped harmonic oscillator.

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The stationary solution ofisIt follows from the Itô isometry thatThus $q$ is zero-mean colored noise with correlation time ${\nu }^{-1}$. The equation $\dot{x}=\mu (-\lambda x+q)$ is an overdamped harmonic particle of mobility $\mu$ driven by that correlated random force.

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For $X=(x,q{)}^T$,The Lyapunov equation gives ${\sigma }_{qq}=\alpha /2$, ${\sigma }_{xq}=\lambda {\sigma }_{xx}$, and $(\mu \lambda +\nu ){\sigma }_{xq}=\mu {\sigma }_{qq}$. Therefore

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Take $\nu \to \infty$ withSince $(\nu /2)e^{-\nu |\tau |}\to \delta (\tau )$,Thus $\mu q$ tends to thermal white forcing of covariance $2\mu k_{B}T\delta (\tau )$, as required by the fluctuation--dissipation relation. The equal-time position variance becomesthe equilibrium harmonic variance.

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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.

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Normalization makes the derivative of the additive $1$ in $\delta F/\delta P=V+\log P+1$ vanish. Using $\dot{P}=-\nabla \cdot J$, the no-flux boundary condition, and integration by parts givesSubstituting the gradient current,Positive definiteness makes equality possible only when $\nabla (V+\log P)=0$, equivalently $J=0$. Thus $F$ is a strict Fokker--Planck free-energy functional away from stationarity.

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Let $Z={\int }_X{e}^{-V}dx$ and $P_B=e^{-V}/Z$. ThenBy nonnegativity of Kullback-Leibler divergence,with equality exactly when $P=P_B$ almost everywhere. Hence a normalizable $P_B$ is the unique minimizer and, because $\dot{F}<0$ elsewhere, the unique steady density compatible with the boundary condition.

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The independent additive noises give diagonal diffusion $\kappa =\mathrm{diag}(b_1^2,b_2^2)$. ThusThe mixed derivatives of ${\kappa }^{-1}a$ areThey agree precisely when

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Under the potential condition, integrate ${\partial }_i\Psi =a_i/b_i^2$ to obtainSince $V=-2\Psi$, the zero-current steady density on the positive quadrant isThe equivalent coefficient $r_2{c}_2/b_2^2$ may be used for the cross term.

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If $c_1,c_2<0$, the cross term in the exponent becomes positive. Along rays with both densities large it grows quartically, while the self-limiting death terms are only negative cubics. The candidate density is therefore not normalizable and the Fokker--Planck free-energy functional is not bounded below. Deterministically, mutual nutrient enhancement eventually overwhelms each species' quadratic crowding death and drives runaway growth, potentially in finite time. The stochastic model consequently has no steady probability density and sends probability toward arbitrarily large populations. This signals failure of the idealized growth law at high density; resource depletion or stronger saturation must regularize a biological model.

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The functional derivatives areWith the stated Fourier convention,The first is conserved order-parameter dynamics; its deterministic rate and conserved-noise amplitude vanish at $q=0$. The second is nonconserved order-parameter dynamics.

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The deterministic relaxation matrix isIts right eigenvectors are the hydrodynamic modes. Writing $A_q=a+\kappa q^2$ and $B_q=\bar{a}+\bar{\kappa }{q}^2$, their decay rates obey

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For $M=\Gamma =1$, $a=\bar{a}$, and $\kappa =\bar{\kappa }$, put $s=a+\kappa q^2$. ThenAt small $q$, second-order perturbation in $c$ gives the slow conserved and fast nonconserved eigenvaluesKeeping terms through $q^2$,Stability of the long-wavelength mode requires $a^2>c^2$.

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For the slow mode choose $e_B=(1,\beta {)}^T$. The second row gives $c+s\beta ={\lambda }_B\beta$, so to the requested orderFor the fast mode choose $e_A=(\alpha ,1{)}^T$. The first row gives $({\lambda }_A-q^{2}s)\alpha =c{q}^2$, henceThus the conserved mode contains an order-one slaved nonconserved component, while the fast mode contains only an $O(q^2)$ conserved component.

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Write $({\varphi }_q(0),0{)}^T=C_B{e}_B+C_A{e}_A$. The equations $C_A=-\beta C_B$ and ${\varphi }_q(0)=C_B(1-\alpha \beta )$ giveSince $\alpha \beta =-c^2{q}^2/a^2$,Therefore $G_{\varphi }=f_1{e}^{-{\lambda }_{B}t}+f_2{e}^{-{\lambda }_{A}t}$. At $q=0$, conservation makes ${\varphi }_0$ exactly constant. Any nonzero overlap with the fast mode would change it on the finite timescale $a^{-1}$, so conservation requires $f_2\to 0$; the explicit $q^2$ factor enforces this.

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At $|\mathbf{q}|=1$, let $s=a+\kappa$. The matrix is symmetric:Its modes areBecause $(1,0{)}^T=(e_{-}+e_{+})/2$,

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The second components of $e_{-}$ and $e_{+}$ have opposite signs. Therefore

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For $|\mathbf{q}|=1$, the quadratic free-energy kernel isThe equilibrium covariance is $V{K}^{-1}$ under the Fourier normalization in the question. Sincethe steady cross-correlator isIf Fourier modes are normalized by $V^{-1/2}$, the factor $V$ is absent. The negative sign reflects the energetic preference for opposite signs when $c>0$.

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For the Poisson process,Differentiating at zero gives

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The scaled cumulant-generating function isBy Cramér theorem, its Legendre transform is stationary at $e^{{\lambda }_{\ast }}=x/\mu$ for $x\ge 0$. Thuswith $I(x)=+\infty$ for $x<0$ and the continuous convention $0\log 0=0$.

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Assume ${\mu }_2>{\mu }_1$, so the upper tail is governed by its boundary. The large deviation principle gives the exponentially equivalent estimatewhere now $I(x)=x\log (x/{\mu }_1)-x+{\mu }_1$. Increasing capacity to ${\mu }_3>{\mu }_2$ reduces this estimate by at least $10^6$ whenEquivalently,

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A Brownian motion has independent stationary Gaussian increments, so

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The increment law immediately givesThe difference quotient has second momentThus increments scale as $\sqrt{h}$, rather than $h$, and no finite derivative is suggested. In fact this heuristic is strengthened by the theorem on nowhere differentiability of Brownian motion.

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Let $Q=(\sigma {\sigma }^T{)}^{-1}$ and use its induced inner product. The autonomous Lagrangian isIts conserved Hamiltonian isOn an infinite-duration fluctuation path $H=0$, so $\Vert \dot{\varphi }\Vert =\Vert a\Vert$. Expanding the action then givesSince metric arclength satisfies $ds=\Vert d\varphi \Vert =\Vert \dot{\varphi }\Vert dt=\Vert a\Vert dt$,This is the geometric minimum action representation and is independent of the speed used to parametrize the path.

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For constant isotropic diffusion and detailed balance, the drift has gradient formfor a positive scalar mobility $M$ after absorbing temperature and diffusion constants into $V$. Deterministic relaxation follows $a$ downhill. The least-action escape trajectory is its time reverse,Its tangent is therefore parallel to $\nabla V$. Since the gradient is normal to every level set $V=\text{constant}$, the escape path from a local minimum crosses all equipotentials orthogonally.

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