Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 75 (Biological Physics)

The homogeneous fixed point must satisfy $v=u^2$. The singular point $u=v=0$ is outside the domain of the reaction term, so substitution in the other equation gives the unique regular spatially homogeneous equilibriumThis is the basally forced quadratic activator-inhibitor model. The linearization of its reaction terms gives the Jacobian matrixPut $a=(r-1)/(r+1)$, so $-1<a<1$. Its trace and determinant are $a-q$ and $q$, and its eigenvalues areSince $q>0$, both eigenvalues have negative real part exactly when $q>a$. HenceFor $r\le 1$ the equilibrium is always strictly linearly stable. For $r>1$, equality $q=a$ is an oscillatory threshold with ${\lambda }_{\pm }=\pm i\sqrt{a}$; the nonlinear qualification at this threshold is proved below.

Now insert a perturbation proportional to $e^{\lambda t+ikx}$. The diffusion terms subtract $k^2\mathrm{diag}(1,p)$ from $J$. With $s=k^2$ the mode trace and determinant areFor a strictly stable homogeneous state, $q>a$ makes $\tau (s)<0$ for all $s\ge 0$. A nonzero Fourier mode can therefore grow only if $\Delta (s)<0$, in which case its two real eigenvalues have opposite signs. The determinant parabola has its minimum at $s_{\ast }=(pa-q)/(2p)$, and that minimum is $q-(pa-q{)}^2/(4p)$. The two-species Turing criterion is consequently derived here asThese force $a>0$, hence $r>1$. Equivalently, solving $pa-q>2\sqrt{pq}$ for $\sqrt{p}$ givesThus the inhibitor in a reaction-diffusion system must diffuse sufficiently fast. Inside this region, the growing wavenumbers obeyAt onset, the minimum touches zero and the two determinant roots coalesce. ThereforeHere $p$ is evaluated on the onset surface. These are the Turing threshold of a basally forced quadratic activator-inhibitor model and its continuum critical wavelength. On a finite or domain with periodic boundary conditions an admissible nonzero spatial mode must additionally fall in the stated band. Equality in the diffusion criterion is marginal onset, while $q=a$ is the distinct homogeneous oscillatory threshold.

For completeness, the weak focus at the Hopf threshold of a basal activator-inhibitor model resolves the nonlinear homogeneous stability when $q=a>0$. Let $h=(1+a)/2$, $\beta =(1-a)/2$, $\omega =\sqrt{a}$, and set $u=(1+r)(1+x)$, $v=(1+r{)}^2(1+y)$, $X=x$, $Y=(hy-ax)/\omega$. Expanding the rational reaction term to cubic order giveswhereWrite $X=\varrho \cos \theta$, $Y=\varrho \sin \theta$ and evaluate these homogeneous polynomials on the unit circle. If $A=\cos \theta F_2+\sin \theta G_2$, $B=\cos \theta F_3+\sin \theta G_3$, and $C=\cos \theta G_2-\sin \theta F_2$, division of the radial and angular equations givesThe angular integral of $A$ is zero because it changes sign under $\theta \mapsto \theta +\pi$. Iterating its quadratic term once over a full cycle contributes the square of this zero integral, so the first nonzero return map term is cubic. Using $⟨{\cos }^4\theta ⟩=3/8$, $⟨{\cos }^2\theta {\sin }^2\theta ⟩=1/8$ and the corresponding sixth moments gives $⟨B⟩=a/4$ and $⟨AC⟩/\omega =(a+1)/8$. The Poincaré map isBecause $0<a<1$, sufficiently small positive radii decrease on every return. The equilibrium at $q=a$ is therefore nonlinearly asymptotically stable, with weak rather than exponential attraction, despite its neutral linear eigenvalues. The strict inequalities above describe the ordinary diffusion-driven instability from a strictly linearly stable homogeneous state.

Write the reaction term as $f(u;r)=u(1-u)(u-r)$ and assume $D>0$. At $r=1/2$, multiply the stationary travelling-wave reduction of a reaction-diffusion system $D{U}^{\prime \prime }+f(U;1/2)=0$ by $U^{\prime }$ and integrate once. Since $U^{\prime }=0$ at both asymptotic phases and ${\int }_0^{U}f(s;1/2)ds=-U^2(1-U{)}^2/4$, the integration constant is zero:The boundary orientation selects $U^{\prime }=-U(1-U)/\sqrt{2D}$. Separating variables gives the logistic front of a bistable cubic equation, with arbitrary center $q_0$:The arbitrary translation is expected from translation invariance.

To develop the requested perturbation theory, let $r=1/2+\epsilon a$ with signed $a=\pm 1$, introduce slow time $T=\epsilon t$, and put $z=(x-q(t))/\sqrt{D}$. Seekwhere $U_0(z)=(1+e^{z/\sqrt{2}}{)}^{-1}$. The explicit $T$ derivative of $\epsilon U_1$ is second order. Since ${\partial }_{r}f=-u(1-u)$, the first-order equation isDifferentiation of $U_0^{\prime \prime }+f(U_0;1/2)=0$ gives $L{U}_0^{\prime }=0$. The self-adjoint differential operator $L$ has this decaying translation mode, and $U_1\to 0$ at both ends because the phases remain exactly $1$ and $0$. Multiplying the first-order equation by $U_0^{\prime }$ and integrating by parts makes its left side zero. Therefore the translation-mode solvability for a bistable front requiresBoth integrals can be evaluated without a special-function formula. The decreasing profile takes $U$ from $1$ to $0$ and has $U_0^{\prime }=-U(1-U)/\sqrt{2}$, soandThe resulting speed isThis derives the velocity by the Fredholm solvability condition, rather than substituting a guessed speed.

With this value, the forcing actually cancels pointwise: $-c_1{U}_0^{\prime }+a{U}_0(1-U_0)=0$. Thus $L{U}_1=0$. The decaying kernel is only the translation direction: an independent solution is $U_0^{\prime }{\int }^z(U_0^{\prime }{)}^{-2}d\zeta$, which grows at at least one infinity. Fixing the center by $U_1(0)=0$ consequently gives $\boxed{U_1=0}$. In this particular bistable cubic reaction-diffusion equation the absence of shape deformation holds to all orders. Indeed $U_0^{\prime \prime }+f(U_0;r)=-(r-1/2)U_0(1-U_0)$, and the exact moving-frame equation has left side $-(\dot{q}/\sqrt{D})U_0^{\prime }$. Thereforeis the exact front for every $0<r<1$. For $r<1/2$, $c>0$ and the $u=1$ phase advances to the right; for $r>1/2$ the $u=0$ phase advances to the left. The small-bias speed is consistent with the prescribed $\dot{q}/\sqrt{D}=O(\epsilon )$ scaling.

Independence of the segment vectors makes the end-to-end distribution an $N$-fold convolution. Use the characteristic function convention $\hat{\psi }(\mathbf{k})=\int e^{i\mathbf{k}\cdot \mathbf{r}}\psi (\mathbf{r})d^{3}r$. For a three-dimensional isotropic freely jointed chain, the radial delta function fixes $|\mathbf{r}|=b$, and angular integration givesThe characteristic function of a sum of independent variables therefore gives $\hat{\Phi }(\mathbf{k},N)=[\sin (kb)/(kb){]}^N$. In the central large-$N$ scaling, $k=O((b\sqrt{N}{)}^{-1})$, soInverting this Gaussian Fourier transform yieldsThis Gaussian limit of the end-to-end distribution of a freely jointed chain is normalized with respect to $d^{3}R$, has $⟨\mathbf{R}⟩=0$ and $⟨R^2⟩=N{b}^2$, and is valid on the coil scale $R=O(b\sqrt{N})$. If a density of the scalar distance $R$ is wanted instead, it is $4\pi R^2\Phi$. The Gaussian approximation does not describe the nearly fully stretched tail or the exact finite-chain cutoff at $R=Nb$.

For the scattering calculation, radial single-segment distributions together with the given independent-segment assumption imply rotational invariance of the whole chain. Therefore each pair separation $\mathbf{S}={\mathbf{R}}_n-{\mathbf{R}}_m$ has a uniform direction conditional on its magnitude $S$. The same angular average givesAveraging its magnitude and then summing the pairs proves the required expression for the polymer scattering function:The ratio has value one at zero separation, including every diagonal term $m=n$.

The printed gyration formula leaves $m$ unsummed. The intended radius of gyration must use a double pair sum, equivalently the mean square distance from the chain's center of mass. To establish that equivalence, put ${\mathbf{R}}_{\mathrm{cm}}=N^{-1}{\sum }_n{\mathbf{R}}_n$ and expand both sides ofThus the consistent definition is $R_g^2=(2N^2{)}^{-1}{\sum }_{m,n}⟨|{\mathbf{R}}_m-{\mathbf{R}}_n{|}^2⟩$. Expanding $\sin z/z=1-z^2/6+O(z^4)$ givesTherefore the general small-$k$ result isThis Guinier expansion of a polymer scattering function requires isotropy and finite pair moments, but not Gaussian distributions.

For a Gaussian chain, each pair difference across $\ell =|n-m|$ links has variance $⟨|{\mathbf{R}}_n-{\mathbf{R}}_m{|}^2⟩=b^2\ell$ and characteristic function $e^{-k^2{b}^2\ell /6}$. Using exactly the $N$ scattering sites in the printed double sum, the finite-chain expression isThe closed expression is understood by continuity at $z=1$, where it equals $N$. The factor $N-\ell$ counts pairs with $n=m+\ell$, with the prefactor two supplying their reversed order. In this finite-site convention,This also holds for a freely jointed chain because its link covariances give the same mean square pair distances.

To obtain the complete long-chain scaling function, set $x=N{k}^2{b}^2/6\simeq k^2{R}_g^2$ and replace contour sums by integrals, keeping $x$ fixed as $N\to \infty$. With $a=k^2{b}^2/6$,The requested Debye scattering function for a Gaussian chain is consequentlyexactly for the continuous Gaussian contour convention $R_g^2=N{b}^2/6$, and asymptotically for the long discrete chain. Its small-$x$ expansion is $\mathcal{D}(x)=1-x/3+x^2/12+\cdots$, agreeing with the general gyration result, and $\mathcal{D}(x)\sim 2/x$ at large $x$. In the continuum coil range $1\ll x\ll N$ this gives $g\sim 12/(k^2{b}^2)$. At still larger $kb$, the discrete expression instead tends to the self-scattering value one.

There is a harmless finite-size convention in the source: $N$ links have $N+1$ endpoints, whereas its scattering sum counts $N$ sites. The formulas above retain its scattering normalization; if all endpoints are counted, replace the site count by $N+1$ in the finite formulas. The large-$N$ Gaussian distribution and Debye scaling function are unchanged to leading order.