Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 355 (Biological Physics)

On the upper half-filament, write $\mathbf{r}(s)=(h(s),s)$ with $0\le s\le L$. To first order in slope, resistive-force theory supplies the uniform transverse load $f_x={\zeta }_{\perp }U$, and Euler--Bernoulli balance givesMidpoint symmetry and clamping impose $h(0)=h^{\prime }(0)=0$; the free-end force and moment conditions are $h^{\prime \prime \prime }(L)=h^{\prime \prime }(L)=0$. With $H=h/L$ and $\xi =s/L$,Four integrations giveIn particular, $H(1)=\mathrm{Sp}/8$ and $H^{\prime }(1)=\mathrm{Sp}/6$. The small-slope approximation first fails when the end slope is order one, giving the estimate $\mathrm{Sp}\sim 6$; parametrically, breakdown occurs at Sperm number of order one.

Choose tangent and normal vectorsso ${\hat{\mathbf{t}}}_s=-\kappa \hat{\mathbf{n}}$, ${\hat{\mathbf{n}}}_s=\kappa \hat{\mathbf{t}}$, and $\kappa ={\theta }_s$. Thenwhile the drag components are $f_n=-{\zeta }_{\perp }U\cos \theta$ and $f_t=({\zeta }_{\perp }U/2)\sin \theta$.

LetSince $N=A{\kappa }_s$, one has $\mathcal{N}=K^{\prime }$. Force balance becomesAt the free end, $\mathcal{T}(1)=\mathcal{N}(1)=K(1)=0$; at the held midpoint, position and orientation impose $\theta (0)=0$ in this convention. Reversing the normal reverses the corresponding signs without changing the shape.

Far from the midpoint at $\mathrm{Sp}\gg 1$, the filament aligns with the flow: $\theta =\pi /2$, $K=\mathcal{N}=0$. The tangential equation and free-end condition give the outer tension

Near $\xi =0$, the outer tension is $\mathrm{Sp}/2$ to leading order. The normal equation is dominated by bending and tension,because the direct normal drag is smaller by ${\mathrm{Sp}}^{-1/2}$ in the turning layer. Hence the decaying inner curvature iswhere matching and the midpoint angle determine $C$. The elastohydrodynamic boundary layer of a filament therefore has scaled arclength

For homogeneous fields the gradient terms vanish, leavingThusThe homogeneous Jacobian is triangular,with strictly negative eigenvalues $-\delta$ and $-\beta$. The fixed point is therefore linearly stable to homogeneous perturbations.

To first order about the fixed point,For a mode proportional to $e^{\sigma t+ikx}$, the linear matrix isIts trace is always negative, so instability occurs exactly when its determinant is negative. Writing $q=k^2$,At onset this quadratic touches zero, requiringThe repeated positive root isand therefore

A local excess of cells produces extra attractant. The resulting concentration gradient drives chemotaxis toward that excess, which further raises cell density and attractant production. This positive feedback causes aggregation when $\chi$ is large enough. Cell and chemical diffusion suppress short wavelengths, while death and chemical degradation suppress very long and homogeneous perturbations. Their competition selects the finite $k_c$, producing a chemotactic pattern-forming instability even though the homogeneous kinetic system is stable.

For an inextensible planar filament, ${\mathbf{r}}_s=(\cos \theta ,\sin \theta )$. The bending energy and compressive-force potential areThe straight state has $E=FL$. Free pivoting means zero end moment,The endpoints sharing the $x$ axis impose ${\int }_0^L\sin \theta ds=0$; near the straight state this removes the constant angular mode. The Neumann eigenfunctions therefore give

For $\theta =a_1\cos (\pi s/L)$, expansion through fourth order giveswhereStationarity givesThe two signs of $a_1$ are the symmetry-related buckled states of a supercritical pitchfork bifurcation.

The projected length is $L_{\parallel }={\int }_0^L\cos \theta ds$. HenceOn the buckled branch,

The quadratic energy isEach mode is a centered Gaussian, soThe $n=1$ variance diverges as $F\uparrow F_c$ because the first bending mode becomes soft; Gaussian theory then fails and the quartic term controls the fluctuations.

At zero force,With the common persistence length convention $L_p=A/(k_{B}T)$,Thermal bending therefore shortens the mean projected length even without compression. Under the strictly planar tangent-correlation convention $L_p^{(2D)}=2A/(k_{B}T)$, the same result is $L/(6L_p^{(2D)})$.

In the single-mode approximation writeFor the reference energy $U_0=L{g}_1{a}_1^2$,Up to an irrelevant constant from subtracting $⟨U_0{⟩}_0$, the variational free energy isMinimization givesso the positive optimum is

The compression estimate through quartic order isUsing $f=F/F_c$ and $L_p=A/(k_{B}T)$,This is finite and smooth at $f=1$, unlike the zero-temperature branch point. The width of the thermal rounding of a buckling transition is controlled by the dimensionless flexibility $L/L_p$ and vanishes for an increasingly stiff or cold filament.