For $u>0$, the second-iterate equation $H^2(u)=u$ is equivalent to $u+H(u)=2\log r$. Write $L=\log r$ and let paired points be $L-d,L+d$. The condition $H(L-d)=L+d$ becomes $\tanh (d/2)=d/L$. At $r=9$, $L=2\log 3>2$, so strict concavity of the positive hyperbolic tangent gives one nonzero positive $d$, as well as $d=0$. Here it is exactly $d=\log 3$.

Thus the four nonnegative fixed points of $H^2$ areZero has periodic-orbit multiplier $81$ and the central point has periodic-orbit multiplier $(1-2\log 3{)}^2>1$, so both are unstable. The outer two points are exchanged by $H$, since $H(\log 3)=3\log 3$ and $H(3\log 3)=\log 3$. Their common second-iterate periodic-orbit multiplier isIts magnitude is below one, so both outer fixed points of $H^2$ are stable. This verifies the crossings and stability in the supplied graph rather than reading stability from crossing direction alone.

Apply the same vertical-then-horizontal cobweb construction to $H(u)=9u{e}^{-u}$. Since no numerical value of the large initial biological population is prescribed, an exact five-generation list is $u_1$, $H(u_1)$, $H^2(u_1)$, $H^3(u_1)$, $H^4(u_1)$. For the plotted example $u_1=5$, these values areThe initial slump is followed by rapid growth and an alternating overshoot about $\log 9\simeq 2.197225$, which is now unstable because its periodic-orbit multiplier $1-\log 9<-1$. A still larger initial biological population can remain extremely small for more than these four updates; the example illustrates the requested construction without imposing an unstated universal numerical starting value.

A large initial biological population maps to a very small positive biological population because $u{e}^{-u}\to 0$ at large $u$: the initial slump is followed by recovery. Cobwebbing alternates a vertical step from $(u_n,u_n)$ to $(u_n,H(u_n))$ with a horizontal step to the diagonal at $(u_{n+1},u_{n+1})$. The figure uses the explicit illustrative choice $u_1=5$.

For $r=2$, every positive orbit subsequently lies below the maximum $H(1)=2/e<1$. On $[0,2/e]$ the map is increasing, with the unique positive fixed point $\log 2$. Below that point $H(u)>u$ and $H(u)<\log 2$; above it $H(u)<u$ and $H(u)>\log 2$. Therefore the trapped iterates approach the fixed point monotonically from whichever side they enter. In particularThe extinction initial value remains zero, but exact extinction is never reached from a positive finite biological population in this deterministic model.

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 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.

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.

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.

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.

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.

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 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.

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.

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.

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.

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.

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.

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.

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 $H\ge 0$.

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.

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 $\boxed{\text{no nonnegative front for }0<c<2\eta }$.