Past exam of the topics in applied mathematics course of the University of Oxford / 2007 / A

The iteration of a map is $F(u)=u/(2+u^2)$. For every $u>0$, $0<F(u)<u/2$, so $0<u_n\le 2^{-n}{u}_0$ and $\boxed{u_n\to 0}$. A cobweb alternates vertical steps to the graph and horizontal steps to the diagonal, approaching the origin monotonically.

More precisely $u_{n+1}/u_n\to 1/2$. The product $2^n{u}_n=u_0{\prod }_{j<n}(1+u_j^2/2{)}^{-1}$ has a finite strictly positive limit because ${\sum }_j{u}_j^2<\infty$. Thus $\boxed{u_n\sim C(u_0)2^{-n}}$ for a positive constant depending on the initial biological population.

A positive fixed point requires $\beta +u_{\ast }^{\gamma }=\alpha$, so it exists exactly when $\boxed{\alpha >\beta }$ and is $u_{\ast }=(\alpha -\beta {)}^{1/\gamma }$. Differentiating the map at it givesFor $1<\gamma \le 2$ and $0<\beta /\alpha <1$, this fixed-point multiplier is strictly between $-1$ and $1$. Hence the positive fixed point is linearly asymptotically stable, including $\gamma =2$; positivity of $\beta$ makes the lower bound strict.

The fixed-point multiplier decreases continuously with $\gamma$ and first reaches the unit circle at $-1$. Solving $1-{\gamma }_{\ast }(1-\beta /\alpha )=-1$ givesIt is stable for $1<\gamma <{\gamma }_{\ast }$ and unstable for $\gamma >{\gamma }_{\ast }$. At the critical value linear stability is inconclusive. The crossing at $-1$ is the local signature of a period-doubling bifurcation, rather than a crossing through $+1$.

At low density the per-generation growth factor is approximately $\alpha /\beta$. The power $\gamma$ controls how sharply crowding reduces recruitment or survival when resources become limiting; at high density $F(u)\sim \alpha u^{1-\gamma }$. Larger $\gamma$ therefore means a stronger overcompensation in a population model.

The flip threshold says that sufficiently strong density feedback can destabilize a constant positive biological population and create alternating generations, under the usual nondegeneracy conditions. A local loss of fixed-point stability alone does not establish chaos, nor does it prove every ecological system with stronger crowding must follow the same route.

A fixed point of $F$ is also a fixed point of $F\circ F$. By the chain rule,If the original fixed-point multiplier has modulus greater than one, its square also exceeds one, so the same fixed point is unstable for the twice-iterated map. New stable fixed points of the twice-iterated map, if present, must be different points; they can represent a genuine two-cycle of the original map.

Law of mass action assigns net substrate gain $k_{0}s$ to autocatalytic replication, substrate-enzyme binding flux $k_{1}se$, dissociation flux $k_{-1}c$ and product-formation flux $k_{2}c$. ThereforeComplex breakdown to product restores free enzyme but does not restore substrate. Product concentration, if required, satisfies $\dot{p}=k_{2}c$.

Adding the last two equations gives $d(e+c)/dt=0$. The initial conditions therefore imply $\boxed{e+c=e_0}$, or $e=e_0-c$. Substitute this in the first two balances:The enzyme inventory is conserved even while substrate can grow autocatalytically.

Use $s=s_{0}u$, $c=e_{0}v$ and slow time $\boxed{\tau =k_1{e}_{0}t}$. Dividing the equations by their respective concentration and time scales givesThus $u^{\prime }=\eta u-u+uv+(K-\lambda )v$ and $ϵv^{\prime }=u-uv-Kv$. Notice that $K>\lambda >0$ and the physical enzyme fraction satisfies $0\le v\le 1$. The initial values are $u(0)=1$, $v(0)=0$ for this particular normalization; the threshold analysis compares different initial substrate abundances using a fixed reference scale.

At equilibrium $v=u/(u+K)$ and adding $u^{\prime }+ϵv^{\prime }$ gives $0=\eta u-\lambda v$. The positive equilibrium isThis condition also gives $\eta <1$. The nullclines are $v=u/(u+K)$ and $v=(1-\eta )u/(u+K-\lambda )$. The second lies above the first for $0<u<u_e$ and below it for $u>u_e$. Below the first nullcline $v^{\prime }>0$; above it $v^{\prime }<0$. Below the second nullcline $u^{\prime }<0$; above it $u^{\prime }>0$.

At the origin the Jacobian has trace $\eta -1-K/ϵ<0$ and determinant $(\lambda -\eta K)/ϵ>0$. Its off-diagonal entries are positive, making the eigenvalues real, so this is an attracting node. At the positive equilibrium the slow reduced derivative is $\eta -\lambda K/(u_e+K{)}^2=\eta v_e>0$, while the fast derivative is $-(u_e+K)/ϵ<0$; equivalently the full Jacobian determinant is negative. Thus the positive point is a saddle equilibrium. Its stable manifold separates extinction from unbounded substrate growth.

After an initial layer of duration $O(ϵ)$ in slow time, $v\simeq u/(u+K)$ while $u$ changes only slightly. The reduced substrate equation isBelow $u_e$ it decreases to zero; above $u_e$ it increases without bound. The exact threshold for initial $v=0$ is the intersection of the saddle equilibrium's stable manifold with that axis. For small $ϵ$ it isIn dimensional initial-substrate units the leading threshold is $s_0{u}_{\ast }\simeq k_2{e}_0/k_0-(k_{-1}+k_2)/k_1$. Comparing the actual initial value $u(0)=1$ to this threshold determines its fate; the reference scale can be held fixed when varying the initial concentration.

The initial sequestration produces an $O(ϵ)$ correction if desired. Write the stable manifold as $u=u_e+ϵ\varphi (v)+O({ϵ}^2)$. The invariance equation at leading order gives ${\varphi }^{\prime }(v)=F(u_e,v)/G(u_e,v)=-(1-\eta )$ and $\varphi (v_e)=0$. Hencefor fixed parameters away from degenerate limits. The positive correction reflects the substrate consumed in building complexes from the initially enzyme-free-complex state.

The physical strip $u\ge 0$, $0\le v\le 1$ is invariant, and the off-diagonal derivatives are nonnegative there, so ordered initial substrate values remain ordered. There are only the attracting origin and the saddle equilibrium. With Dulac function $1/u$, the divergence of the weighted vector field is $-(K-\lambda )v/u^2-(1+K/u)/ϵ<0$, excluding periodic orbits in its interior. The saddle equilibrium separatrix therefore gives the unique threshold on the initial axis: below it solutions tend to extinction, at it they tend to the saddle equilibrium, and above it they escape to infinity. Growth is at most exponential, so this is $u\to \infty$ as time tends to infinity, not a finite-time blowup. This is autocatalytic enzyme clearance with a saddle threshold.

The travelling coordinate has $A_t=-c{a}^{\prime }$, $A_{xx}=a^{\prime \prime }$. Therefore the profile equation isThe wave moves right with speed $c$. It joins a populated equilibrium behind to an empty equilibrium ahead.

The phase-plane equations are $a^{\prime }=b$, $b^{\prime }=-cb-f(a)$. Nonnegative equilibria require $b=0$ and $f(a)=0$, hence only $\boxed{(0,0),(\eta ,0)}$. At the origin $f^{\prime }(0)={\eta }^2$, so the linear eigenvalues solveFor $0<c<2\eta$ the roots are complex with negative real part, giving a stable focus. For $c>2\eta$ both are negative real numbers, giving a stable node. Equality is the repeated-root threshold and is not covered by either strict classification.

A nonconstant trajectory approaching a stable focus winds around it infinitely often. Its $a$ coordinate therefore changes sign repeatedly near zero. Since the biological population profile must remain nonnegative, such a trajectory cannot represent the wave. The only nonnegative trajectory avoiding that winding at the origin would be the equilibrium itself, which cannot join the populated state. Thus $\boxed{\text{no nonnegative front for }0<c<2\eta }$.

For $0<a<\eta$, $f(a)=a({\eta }^2-a^2)/(1+a^2)>0$. On the horizontal edge $b=0$,The vector field points downward into the proposed trapping region.

On $b=-\eta a$ with $a>0$,Thus $\boxed{db/da<-\eta }$. Since $a^{\prime }=b<0$, comparing slopes alone can reverse one's intuition about crossing direction. More directly, for $H=b+\eta a$ the inward test isso the sloping edge also points into $H\ge 0$.

The top edge points down by (a), the sloping edge points upward in the inward normal sense by (b), and on $a=\eta$ the velocity $a^{\prime }=b\le 0$ points left. At $a=0$ the region contains only the equilibrium $(0,0)$. Thus the closed triangle is forward invariant.

The given unstable saddle equilibrium branch enters its interior from $(\eta ,0)$ and stays in this compact region. Along it $a^{\prime }=b\le 0$, so $a$ decreases and has a limit. An interior limiting orbit with $b\ne 0$ would keep changing $a$, while an interior point with $b=0$, $0<a<\eta$ has $b^{\prime }<0$ and is not an equilibrium. Hence the only possible forward limit is $(0,0)$. Ordinary-differential-equation uniqueness prevents reaching and leaving an equilibrium in finite profile coordinate. The whole orbit therefore joins $(\eta ,0)$ at $y=-\infty$ to $(0,0)$ at $y=+\infty$ with $a\ge 0$ and $a^{\prime }=b\le 0$.

Consequently $\boxed{\text{a nonnegative monotone front exists for every }c>2\eta }$. This proves the requested strict-speed result without claiming an unproved conclusion at the repeated-root threshold. It is a trapping triangle for a saturating monostable invasion front.

At a homogeneous equilibrium, add the two reaction equations: $f+g=1-b=0$, so $b=1$. Then $f=0$ gives $a=3/4$. Thus $\boxed{(a_{\ast },b_{\ast })=(3/4,1)}$.

The reaction Jacobian isIts trace is $-1/2$ and determinant is $1$, so both eigenvalues have negative real parts. Explicitly they are $(-1\pm i\sqrt{15})/4$. Therefore the homogeneous reaction equilibrium is a stable focus in the absence of diffusion.

The cosine modes satisfy both zero-flux endpoint conditions and ${\partial }_x^2\cos (k_{n}x)=-k_n^2\cos (k_{n}x)$. Linearizing the reaction terms gives $f_a\delta a+f_b\delta b$ and $g_a\delta a+g_b\delta b$. Cancelling the common mode factor yieldsThe derivatives are exactly the entries of the matrix computed in (ii). The zero mode reproduces the stable homogeneous perturbation; positive integers give the possible spatial instabilities.

Put $q=k_n^2$. The mode matrix has trace $-1/2-(1+D)q<0$, so a growing eigenvalue occurs exactly when its determinant is negative:This upward quadratic has two positive roots only if $D<1/2$ and $D^2-5D+1/4>0$. Within $0<D<1$ this isFor such $D$, defineThe unstable modes satisfy $\boxed{\sqrt{q_{-}}<k_n<\sqrt{q_{+}}}$. These are strict inequalities: at either endpoint the leading eigenvalue is zero rather than positive.

If $D\ge 1/2$, both nonconstant coefficients of $\Delta (q)$ are nonnegative, so it is positive for $q\ge 0$. If $D_c<D<1/2$, its discriminant is negative, so it is again strictly positive. Therefore $\boxed{D>D_c⟹\text{no diffusion-driven instability}}$.

At $D=D_c$ the double root can only give a neutral zero eigenvalue if it coincides with an allowed spatial mode; it still gives no positive growth rate. This is the finite-wavenumber onset of a Turing instability, not an oscillatory crossing since the mode trace stays negative.

Small $L$ makes every nonzero $k_n=n\pi /L$ large, placing them above the finite instability band, where diffusion dominates reaction. The homogeneous zero mode is already stable.

For $0<D<D_c$, the first possible band entry as $L$ increases is $n=1$ crossing $k_1=\sqrt{q_{+}}$. Hence the critical length isAt equality the mode is neutral; it is unstable just above this value, while $\pi /\sqrt{q_{+}}<L<\pi /\sqrt{q_{-}}$. Thus strictly positive growth has infimum $L_c$, with no attained smallest length if “unstable” means $\mathrm{Re}\lambda >0$. The conventional threshold answer is $L_c$. Larger modes cannot enter earlier because their first-entry lengths are $n{L}_c$.