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

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.

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.

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.

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.

For a piston-driven normal shock with $U>0$, $c_0>0$ and $\gamma >1$, the quadratic has one positive and one negative root. The forward-moving shock usesSince the square root is greater than $(\gamma +1)U$,Therefore there is always a forward root with shock speed greater than the gas and piston speed under the stated compression assumptions. This also makes the density ratio $V/(V-U)$ positive and greater than one. Evaluating the quadratic at $V=c_0$ gives a negative value, so the positive root additionally has $V>c_0$: it is supersonic relative to the undisturbed gas. The negative root does not describe the shock driven into the region $x>0$.

Let the undisturbed state ahead have $({\rho }_0,p_0,u_0)=({\rho }_0,p_0,0)$, and the uniform compressed state behind have velocity $u_1=U$. The piston boundary fixes that downstream velocity. Mass and momentum jumps giveThus ${\rho }_1/{\rho }_0=V/(V-U)$. Equality of specific stagnation enthalpy in the shock frame isSubstitute $p_1=p_0+{\rho }_{0}UV$ and $1/{\rho }_1=(V-U)/({\rho }_{0}V)$, then multiply by $2(\gamma -1)V$. The resulting difference factors as $U[-2V^2+(\gamma +1)UV+2\gamma p_0/{\rho }_0]=0$. Since the piston moves into the gas with $U>0$, division givesThis determines the constant speed of the compression shock ahead of the piston.