Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 1 / 6E (Mathematical Biology)

The nullclines are $x=f(y)$ and $y=g(x)$, both decreasing curves in the positive quadrant. To their left/right the horizontal velocity is respectively positive/negative; below/above the second nullcline the vertical velocity is positive/negative. Bounded positive synthesis functions give an inward-pointing sufficiently large rectangle, while the axes point into the quadrant. More directly, put $H(x)=f(g(x))-x$: $H(0)>0$, and $H(x)<0$ for large enough $x$. Continuity therefore gives a positive equilibrium.

At an equilibrium, the Jacobian matrix is $J=\left(\begin{array}{cc}-1& f^{\prime }(y_{\ast })\\ g^{\prime }(x_{\ast })& -1\end{array}\right)$, with eigenvalues $-1\pm \sqrt{f^{\prime }(y_{\ast })g^{\prime }(x_{\ast })}$. Its logarithmic-slope product isbecause $f(y_{\ast })=x_{\ast }$ and $g(x_{\ast })=y_{\ast }$. Thus a product exceeding one makes this equilibrium a saddle and gives $H^{\prime }(x_{\ast })>0$. Immediately on its left $H<0$, and immediately on its right $H>0$. Since $H(0)>0$ and $H$ is eventually negative, there must be other roots on both sides, crossing in the attracting direction. For simple roots those have $H^{\prime }<0$, hence two negative Jacobian eigenvalues: the saddle separates at least two stable steady-state alternatives. The divergence is $-2$, so periodic attractors cannot replace these equilibria in a bounded planar region. Nongeneric flat crossings or intervals of equilibria require a nonlinear stability interpretation; the strict saddle inequality alone does not guarantee that every other root is hyperbolic. This is the usual mutual repression stability criterion behind bistability.

The conclusion also has a nonlinear trapping justification. Choose $l<u$ on the left of the saddle with $H(l)>0>H(u)$; such endpoints exist by the signs just proved. The rectangle $[l,u]\times [g(u),g(l)]$ is forward invariant: on the vertical edges, $f(y)-l\ge f(g(l))-l>0$ and $f(y)-u\le f(g(u))-u<0$, while the horizontal-edge velocities point inward or are tangent. The same construction on the right of the saddle gives a second disjoint trapping rectangle. Each contains a nonempty compact attracting invariant set, obtained by intersecting its nested forward images; strict vertical inwardness also rules out invariant boundary trajectories. If a downward crossing is isolated, endpoints can be taken arbitrarily close to it, proving nonlinear stability, and the absence of periodic orbits makes nearby trajectories converge to that equilibrium. Intervals of equilibria instead give attracting alternatives consisting of families, so bounded monotone functions without nondegeneracy do not necessarily produce two isolated hyperbolic attractors.

At a positive equilibrium of the bounded Hill model, the mutual repression stability criterion becomesThus $mn>1$ is necessary. Because the same $\lambda$ is imposed in both synthesis functions, it is not sufficient for arbitrary unequal positive exponents. The additional equilibrium constraint isHere is an exact criterion for all the printed exponent choices, including nonintegers. If $m>1$ and $n>1$, multistability is possible: for large $\lambda$ the intermediate equilibrium satisfies $x\sim {\lambda }^{(m-1)/(mn-1)}$, $y\sim {\lambda }^{(n-1)/(mn-1)}$, so both increase without bound and $K\to mn>1$. These asymptotics follow by balancing $x{y}^m\sim \lambda$ and $y{x}^n\sim \lambda$; the limiting two algebraic equations have a nonsingular log-Jacobian when $mn\ne 1$, so a nearby exact equilibrium exists for large $\lambda$. If both exponents are at most one, $K<1$ everywhere and multistability is impossible.

For the remaining case, interchange $x,y$ if necessary so $0<m\le 1<n$. For each $x>0$, let $Y(x)>0$ be the unique solution ofwhich exists because the left side is strictly decreasing; for $m=1$ the right side always exceeds one. DefineThen the exact answer for this remaining range isThis is a one-variable test depending only on $m,n$, with no unknown $\lambda$. To prove it, parameterize all equilibria by $\lambda (x)=x(1+Y(x{)}^m)$. Implicit differentiation givesThe denominator is positive, and $\lambda (x)$ goes from zero to infinity as $x$ goes from zero to infinity. A negative-slope portion therefore forces two folds and an interval of three equilibria, with two attracting branches. If $K\le 1$ everywhere, $\lambda$ is strictly increasing apart from possible isolated stationary points, so there is only one equilibrium for each $\lambda$. This is the common-strength Hill repression criterion.

For illustration and a closed test on the boundary $m=1$, $Y=x/(1+x^n-x)$ andobtained at $x^n=(n+1)/(n-1)$. In particular $(m,n)=(1,2)$ has $mn>1$ but maximum $3\sqrt{3}/8<1$, so it cannot be multistable. For positive integer exponents the allowed choices simplify to both exponents at least two, or one equal to one and the other at least four. For arbitrary real exponents the preceding exact criterion is needed; replacing it by $mn>1$ would discard the common-strength constraint.

The unbounded power laws behave differently. Combining their equilibrium equations gives $x^{1-mn}={\lambda }^{1-m}$. If $mn\ne 1$, there is exactly one positive equilibrium,Its slope product is exactly $mn$, so it is attracting if $mn<1$ and a saddle if $mn>1$, but there are never two isolated attracting equilibria. If $mn=1$, equilibria exist only when ${\lambda }^{1-m}=1$; when this holds there is a whole curve of equilibria, with a zero tangent eigenvalue and transverse eigenvalue $-2$. This includes $m=n=1$, $xy=\lambda$, for every $\lambda$. A neutrally degenerate continuum is not bistability between isolated attractors. Thus these power laws do not generate ordinary multistability; boundedness was essential to the outer-root argument in part (a).