Past exam of the topics in applied mathematics course of the University of Oxford / 2008 / A

Write $P(N)=f(N,0)$. Its three positive zeros of a function are $1/2,1,2$, and expansion and differentiation giveThus its local minimum and maximum occur respectively atwith valuesThe function graph approaches the value $1$ as $N\downarrow 0$, decreases through its first zero to the minimum, increases through its second zero to the maximum, and then decreases through $2$ to $-\infty$. This fixes the shape and all axis crossings of the cubic polynomial.

Changing $a$ translates the function graph vertically without moving either stationary point. For $a=1/2$ both extrema are above the horizontal axis: their heights are approximately $0.421109$ and $0.764076$. For $a=-1/2$ both are below it, at approximately $-0.578891$ and $-0.235924$. The function graph for $a=1/2$ therefore has only its final positive-axis crossing, whereas the function graph for $a=-1/2$ has only its first positive-axis crossing. The original plots mark the extrema and also show the equilibrium point branches used below.

Put $F(N)=f(N,1/2)$. A cobweb plot alternates a vertical move to the function graph $y=F(N)$ with a horizontal move to the diagonal $y=N$. Beginning at $N_0=3/2$, the first iterates areThe diagram shows the ensuing approach to $1/2$ with alternating deviations. More precisely, if $e_t=N_t-1/2$, Taylor expansion at the fixed point givesFor sufficiently small nonzero deviations the sign reverses each step and their magnitudes contract. The local approach is an alternating, geometrically damped oscillation, with asymptotic contraction factor $3/4$ per iterate.

First establish a uniform upper bound. Direct subtraction givesbecause the quadratic polynomial has positive leading coefficient and discriminant $49-56=-7$. The assumed positivity of $F(N_0)$ therefore places $N_1$ in $(0,3/2)$.

This interval is forward invariant. On $[0,1/2]$ and on $[1,3/2]$, the three factors in $P(N)$ have nonnegative product, hence $F(N)\ge 1/2$. On $[1/2,1]$ the product is nonpositive, butso $F(N)\ge 1/8>0$. Together with the upper bound this proves $F((0,3/2))\subset (0,3/2)$. Induction now givesThe hypothesis on the first image is essential for initial values outside that invariant interval.

For an autonomous scalar differential equation, linear stability of a simple equilibrium point $N_e$ is determined by the derivative $f_N(N_e,a)=P^{\prime }(N_e)$: a negative derivative attracts nearby solutions and a positive derivative gives repulsion in a dynamical system of nearby solutions.

Here $f(0,a)=1+a\ge 1/2>0$ throughout the prescribed mathematical model parameter range, while $f(N,a)\to -\infty$ as $N\to \infty$. If there is only one equilibrium point, the function graph must cross the axis on one of its decreasing branches, outside the interval between its two extrema. The crossing is simple and $P^{\prime }(N_e)<0$. Thus the unique equilibrium point is linearly asymptotically stable. At a saddle-node bifurcation there are two distinct equilibrium points, one of them double, so that case does not fall under the one-equilibrium hypothesis.

With three distinct positive roots of an equation, the negative-leading cubic polynomial has the sign patternEquivalently its derivative is negative at the first root of an equation, positive at the middle root of an equation, and negative at the final root of an equation. The phase line arrows therefore point toward the two outer equilibrium points and away from the middle one:The middle equilibrium point separates the two basin of attraction in the bistability mathematical model parameter interval.

Yes: there is a quasistatic hysteresis loop. Equilibrium points lie on the mathematical model parameter curve $a=-P(N)$. Its two saddle-node bifurcations occur at the stationary points computed in part (a), at mathematical model parameter valuesBetween these values there are two attracting branches separated by a repelling branch. Starting below $a_{\mathrm{down}}$ and increasing $a$ slowly, a solution follows the lower attracting branch. This branch persists through the bistability interval and ends only at $a_{\mathrm{up}}$, where it meets the repelling branch in a saddle-node bifurcation. The solution then moves to the upper attracting branch.

If the direction of mathematical model parameter change is reversed, the upper branch persists until $a_{\mathrm{down}}$, where a different saddle-node bifurcation forces the downward transition. The two transition values differ, and both lie inside the allowed range $[-1/2,1/2]$. Thus the biological population selected at an intermediate mathematical model parameter value depends on its prior branch. This is the equilibrium point hysteresis shown in the branch diagram; finite-rate passage can additionally delay the jumps near the saddle-node bifurcations.

The term $rIS$ is a mass-action incidence rate: each infective individual makes potentially transmitting contacts with susceptible individuals, and the total rate is proportional to the product of the two compartment sizes. An infection transfers one animal from $S$ to $I$, accounting for opposite signs in their equations. The per-capita mortality rate of an infective individual is $a$, giving transfer $aI$ from $I$ to $R$.

There are no births, recoveries or additional deaths in this model. Adding the three equations gives $\dot{S}+\dot{I}+\dot{R}=0$, so the total initial biological population is conserved when the dead compartment is included. Before quarantine, $S$ decreases whenever $I,S>0$; after effective isolation no further transmission is permitted.

Write $S_0=S_{initial}$ and $I_0=I_{initial}$. Until quarantine and while $I,S>0$, division of the equations givesIntegration from the initial state yields the first integralLet $S_{\ast }$ be the susceptible count at the first time $I=I_{\ast }$. It is the root of an equation reached on the decreasing-$S$ outbreak trajectory ofFor a strictly rising threshold crossing this is the larger admissible root of an equation, with $a/r<S_{\ast }<S_0$; if the threshold equals the peak, $S_{\ast }=a/r$. The assumption that quarantine is reached selects the physically relevant branch and excludes a merely hypothetical declining-branch crossing after an earlier one.

Since there is no incubation period here, all removals from $S$ have already become infection cases by quarantine. No further susceptible individuals are infected after isolation. The requested number among the initially susceptible individuals is therefore

The delayed influx $rI(t-\tau )S(t-\tau )$ represents animals exposed one incubation period earlier who enter the infective individual compartment now. They die at the instantaneous rate $aI(t)$ once in that compartment. The displayed bookkeeping removes an animal from $S$ at the same delayed time that it enters $I$, preserving $S+I+R$.

For this bookkeeping, $S$ must include the exposed animals which have not yet completed incubation period. If $S$ were instead defined as strictly unexposed susceptible individuals, its loss term would be the current exposure rate and a separate latent compartment would be needed. The calculations here follow the printed delay differential equation. A full initial-value problem for a delay differential equation also needs a history on $[-\tau ,0]$; the following small-delay expansion of an epidemic transfer rate concerns a smooth compatible history and the regular regime assumed by the question.

In the first-order Taylor expansion correction it suffices to use the zeroth-order derivatives. They giveThe delayed product is consequentlyMultiplying by $r$ and substituting into the two transfer equations producesThe opposite signs of $f$ and $g$ are also required by conservation law of $S+I+R$: this correction changes the timing of the transfer between $S$ and $I$, while leaving the instantaneous death term unchanged.

At the retained order define $K=1+\tau r(I-S)$. Neglecting terms of order $a\tau$ in the incidence correction givesDividing and expanding consistently,Thus the susceptible count $S_{\ast }$ at the first threshold crossing obeys the same retained-order equation as before:It is important to distinguish completed case transfers at quarantine from all infections already initiated. Let the quarantine time be $T$. Isolation prevents exposures after $T$, but animals exposed during $(T-\tau ,T)$ still finish their incubation period during $(T,T+\tau )$. Their number, in the printed bookkeeping, isThe infection cases transferred by time $T$ among the initial $S$ class number $S_0-S_{\ast }$. Adding those pending infection cases gives the eventual number who catch the infectious disease:This adds only infections caused by exposures before isolation; it does not allow secondary transmission after quarantine. If one asks instead only for the transfers completed by the instant of quarantine, the answer is $S_0-S_{\ast }$. The distinction is necessary because the model keeps incubating animals in its $S$ compartment.

An unbiased random walk of bacteria leads, on the continuum scale, to diffusion with coefficient $D$, represented by $D{B}_{XX}$. The cell proliferation rate $\nu GB$ is proportional both to the available biological growth factor concentration and to the bacterial density. Consumption of biological growth factor at rate $\mu GB$ gives its negative reaction term.

There is no diffusion term for $G$ because the biological growth factor is attached to the substrate rather than freely transported. There is also no chemotactic drift in this model: the specified motion is random, while the biological growth factor changes cell proliferation rather than directional movement. Both reaction terms vanish if either bacteria or biological growth factor are absent.

Choose a positive reference biological growth factor concentration $G_{ref}$, for example its concentration ahead of the travelling wave. SetwhereThe chain rule then transforms the equations intoEach displayed coefficient is one with the chosen scales. Hence the dimensionless system isThe free choice of $G_{ref}$ expresses the freedom to normalize the as-yet unspecified concentration scale.

Take profiles $b=b(y)$ and $g=g(y)$ with $y=x-ct$, and use a prime for differentiation with respect to $y$. The chain rule gives $b_t=-c{b}^{\prime }$, $b_{xx}=b^{\prime \prime }$ and $g_t=-c{g}^{\prime }$. ThereforeThese are coupled ordinary differential equations for a right-moving travelling wave with speed $c>0$. A first-order formulation, using $v=b^{\prime }$, is $b^{\prime }=v$, $v^{\prime }=-cv-gb$, $g^{\prime }=gb/c$. In particular a positive profile has $g^{\prime }>0$, consistent with depleted substrate behind the travelling front and fresh biological growth factor ahead.

Substitute $gb=c{g}^{\prime }$ into the bacterial profile equation. This giveswhich integrates once toAt the asymptotic constant states the profile derivative tends to zero. The state ahead of the travelling wave, where $b\to 0$ and $g\to 1$, therefore gives $C=c$. The state behind it, where $g\to 0$ and $b\to b_0$, gives $C=c{b}_0$. Since $c>0$,This first integral expresses the matching between biomass produced and biological growth factor consumed after the chosen nondimensionalization.

At the leading edge $b\to 0$ and $g\to 1$. The linearized bacterial equation iswith characteristic exponentsFor $0<c<2$ they form a nonreal complex conjugate pair. A nonzero profile approaching the travelling front then has an exponentially decaying oscillatory leading term, and repeatedly changes sign. This is incompatible with the assumed positive bacterial density. The leading edge must therefore have real characteristic exponents, requiringAt $c=2$ the repeated characteristic exponent is $-1$ and polynomial factors in the asymptotic profile may occur. The argument supplies a necessary lower bound; it is not an independent existence proof for every speed satisfying it.

The first integral and $b_0=1$ yield the linear identity $b^{\prime }+cb=c(1-g)$. Multiply by the integrating factor $e^{cy}$ and integrate from $-\infty$. Since the bounded limit $b(-\infty )=1$ makes $e^{cy}b(y)\to 0$ there,Positivity of $g$ makes the last integral nonnegative, and strictly positive at any finite point of a strictly positive travelling wave. Thuswith equality approached only at the far-behind boundary. This bound uses the first integral rather than an unsupported assumption that the bacterial profile is monotone.

For the modified strain the travelling wave equations areIntegrating the second across the entire travelling wave gives ${\int }_{\mathbb{R}}gbdy=c$, since the biological growth factor profile still goes from $0$ to $1$. Integrating the first, using the constant asymptotic states and vanishing derivative at infinity, givesConsequentlyThe subtracted integral is strictly positive because the profiles are positive and $\eta >0$. It is finite: $b$ is bounded by continuity and its finite endpoint limits, and $\int gb=c$. The additional density-dependent inhibition therefore reduces the bacterial density left behind; positivity of $b_1$ is part of the assumed positive-wave solution.

The term $\alpha {\nabla }^{2}q$ is cell diffusion arising from random movement. Chemotaxis up the concentration gradient gives the cell flux $\beta q\nabla c$, whose negative divergence is the transport contribution $-\beta \nabla \cdot (q\nabla c)$. The reaction term $q(1-q)$ gives growth governed by the logistic differential equation with dimensionless carrying capacity one.

The chemoattractant itself diffuses, is produced by cells at rate $\kappa \eta q$, and decays at rate $\kappa c$. The homogeneous Neumann boundary conditions suppress the normal diffusive flux of each quantity. They also suppress the normal chemotactic cell flux because the normal derivative of $c$ vanishes. Thus the rectangular boundary is impermeable, and the spatially integrated cell count changes through cell proliferation rather than transport across that boundary.

For spatially homogeneous perturbations write $q=1+u$ and $c=\eta +v$. The linearized reaction equations areThis triangular matrix has eigenvalues $-1$ and $-\kappa$, both negative. Hence the homogeneous steady state is linearly asymptotically stable against homogeneous perturbations. Even when $\kappa =1$, where the two eigenvalues coincide and a polynomial factor may occur in the matrix exponential, all such perturbations still decay.

For the cosine mode in the question, the Laplacian acts by multiplication by $-s$, whereThese modes obey the Neumann boundary conditions. In the chemotactic term, put $q=1+u$, $c=\eta +v$. Then $\nabla c=\nabla v$, and $q\nabla c=\nabla v+u\nabla v$; the second product is a quadratic polynomial in perturbation amplitude. Thus the linearized equations areSubstituting the modal time dependence givesSeparating this matrix into its transport part and the reaction Jacobian matrix gives the required form, with the same $J$ as above.

For the modal matrix $M(s)$, $s\ge 0$, its trace is always negative:Its determinant isThe proposed inequality is exactly $\beta \kappa \eta <\alpha \kappa +1$. It makes the coefficient of $s$ positive, while the leading and constant coefficients are already positive. Therefore $D(s)>0$ for every $s\ge 0$, independent of which discrete wave numbers the domain allows.

A real two-by-two matrix with negative trace and positive determinant has both eigenvalues with negative real part: if they are real they have positive product and negative sum, and if nonreal their common real part is half the trace. Every mode therefore decays, proving linear stability for all domain sizes under the stated sufficient condition.

Here $L_1=1$ and $L_2=1/2$, so a perturbation varying in both coordinates has $n,m\ge 1$ andWith the given unit coefficients the determinant becomes $D(s)=(1+s{)}^2-\beta s$. The trace remains negative, so instability occurs precisely when $D(s)<0$, orThe right-hand side is strictly increasing for $s>1$, since its derivative is $1-s^{-2}>0$. Its smallest allowed value is attained by the mode $(n,m)=(1,1)$, giving the onset thresholdAt this value the first mixed mode has a zero eigenvalue; it becomes unstable for $\beta >{\beta }_{\mathrm{crit}}$. Thus the critical value is the infimum of the strictly unstable range, rather than a strictly unstable point itself. Its spatial factor is $\cos (\pi x)\cos (2\pi y)$.