Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 355 (Biological Physics)

Write the reaction terms asA nonzero homogeneous equilibrium satisfies $v_{\ast }=\beta u_{\ast }$ and $1+u_{\ast }-\gamma v_{\ast }=0$, henceFor physically positive populations it exists exactly when

The reaction Jacobian matrix at this equilibrium isIts trace and determinant areThe equilibrium is therefore stable to spatially uniform perturbations when

For a spatial Fourier mode of wavenumber $k$, put $q=k^2$. The linearized reaction-diffusion system has matrixIts trace is smaller than $\mathrm{tr}J$, whileThe two-species diffusion-driven instability criterion says that this upward-opening quadratic becomes negative for some $q>0$ precisely whenCombining all conditions, a Turing instability may occur in the region

At onset the discriminant vanishes, and the double root isThe threshold relation givesso the critical wavenumber isIf $d=1$, uniform stability requires $\beta >1$ whereas diffusion-driven instability requires $d>\beta$. These inequalities are incompatible, so the Turing region vanishes.

Take $z$ upward along the straight rod. The part above height $z$ has weight $\lambda g(L-z)$, so, with tensile force positive, the internal tension is compressive:For a small transverse displacement $X(z)$, the quadratic bending and gravitational energies areThe Euler-Lagrange equation for this functional iswhich is exactly

Clamping at the bottom fixes displacement and slope:At the free upper end, bending moment and transverse force vanish:Since $\sigma (L)=0$, the four boundary conditions are

Set $u=X^{\prime }$. Integrating the field equation once and using the free-end shear condition givesor, with $s=L-z$,Introduce the dimensionless similarity coordinateand write $u={\eta }^{1/3}F(\eta )$. Direct substitution reduces the equation toThis is the Bessel differential equation of order $1/3$, so

The free-moment condition is $u^{\prime }(L)=0$. As $\eta \to 0$,The second term has nonzero limiting $z$ derivative, so the free-end condition forces $b=0$. The free-shear condition $u^{\prime \prime }(L)=0$ then follows from the differential equation. At the clamp, $u(0)=0$, givingLet $j_{-1/3,1}$ be the smallest positive zero of this Bessel function. The first self-buckling threshold isor equivalently

At large separation, the finite-thickness van der Waals combination has the expansionThus the attraction governed by the Hamaker constant decays asand the screened electrostatic term decays exponentially on the Debye–Hückel screening length. The Helfrich repulsion, however, decays only asConsequently the total interaction approaches its unbound value zero from above as $d\to \infty$.

A finite bound minimum must have nonpositive energy to beat the state at infinity. Because the large-$d$ interaction is positive, such a minimum cannot move continuously to infinity while remaining globally stable. At the transition it instead becomes degenerate with the $d=\infty$ state at a finite spacing and then loses global stability. The equilibrium spacing therefore jumps from finite $d_{\ast }$ to infinity, making this a discontinuous, first-order unbinding transition.

For a dilute stack, the membrane number per unit normal length is $1/d=\varphi /\delta$. Dividing the fluctuation repulsion per area by the repeat distance gives its free-energy density

The short-range electrostatic repulsion and long-range van der Waals attraction enter at second-virial order. In general, for an effective pair energy $U_{A_H}(\Gamma )$ over relative configurations $\Gamma$, the second virial coefficient has the Mayer-integral formwith a fixed normalization by the microscopic membrane thickness making $B_2$ dimensionless here. The mean-field contribution is quadratic in membrane concentration. The required two-power free energy can therefore be writtenChanges of microscopic normalization merely rescale $B_2$ by a positive constant and do not affect the transition.

Write the pair energy asAt $A_H=0$ the interaction is repulsive, so the Mayer integrand and hence $B_2$ are positive. Increasing $A_H$ strengthens attraction andFor sufficiently strong attraction, negative configurations dominate and $B_2<0$. Continuity therefore gives a critical $A_H^{\ast }$ with $B_2(A_H^{\ast })=0$. The Taylor theorem giveswhere $r=-B_2^{\prime }(A_H^{\ast })>0$.

Write the free energy asWhen $A_H<A_H^{\ast }$, $B_2>0$ and the global minimum on $\varphi \ge 0$ is ${\varphi }_{\ast }=0$, corresponding to an unbound stack. When $A_H>A_H^{\ast }$, $B_2<0$, and minimization givesUsing $B_2\sim -r(A_H-A_H^{\ast })$ yieldsSince $d_{\ast }=\delta /{\varphi }_{\ast }$,so the continuous membrane unbinding transition has exponent