Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 353 / 3

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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