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Use the well-mixed natural ventilation model for a distributed floor source. Let the building height be $H_b$, its volume be $\mathcal{V}$, its temperature excess be $\theta$, and the exterior mass density be $\rho$. Let ${\alpha }_T$ be the coefficient of thermal expansion, so reduced gravity is $g^{\prime }=g{\alpha }_T\theta$. Convert the total heat input $Q$ to the temperature-volume flux $\mathcal{H}=Q/(\rho c_p)$, where $c_p$ is specific heat capacity. If the quoted heat flux is per floor area, first multiply it by that area.

For effective opening area for pressure-driven ventilation, define the equal-opening hydraulic coefficient $K_A$ such that the signed throughflow $q$ obeys $q|q|=K_A\Pi$, with $\Pi$ the total driving pressure divided by mass density. In the effective-area convention of question 6, each opening has discharge $A\sqrt{\delta p/\rho }$, so $K_A=A^2/2$. The usual physical-area orifice law instead gives $K_A=C_d^2{A}^2$; that prefactor does not affect the stability conclusions.

Set $W=\Delta P/\rho >0$ and $G_T=g{\alpha }_T{H}_b$. Positive $q$ denotes entry at the floor and exit at the roof. hydrostatic pressure and wind combine to giveBecause the inflowing air is at exterior temperature in either direction, the heat budget isThe well-mixed ventilation temperature balance at steady state is thereforeAll roots must satisfy $\theta >0$ and their appropriate flow-direction inequality; squaring a pressure relation without imposing those conditions could add spurious branches.

The assisting case has a unique positive solution and it is stable. For opposing wind, define ${\theta }_w=W/G_T$. There is always one solution with $\theta >{\theta }_w$, a buoyancy-dominated upward flow. Below ${\theta }_w$, the heat-removal curve is $R(\theta )=\sqrt{K_A}\theta \sqrt{W-G_T\theta }$. It is zero at both endpoints, and its maximum occurs at $\theta =2{\theta }_w/3$:For $0<\mathcal{H}<R_{\max }$ there are two downward-flow roots as well as the upward-flow root. Equivalently, setting $y=G_T\theta /W$ and $\epsilon =G_T\mathcal{H}/(\sqrt{K_A}{W}^{3/2})$, the equilibrium curve isFor linear stability analysis of the opposing-wind ventilation bistability, linearize the heat budget: the eigenvalue is $-R^{\prime }(\theta )/\mathcal{V}$. The cooler downward root has $0<\theta <2{\theta }_w/3$ and $R^{\prime }>0$, so it is stable. The warmer downward root has $2{\theta }_w/3<\theta <{\theta }_w$ and $R^{\prime }<0$, so it is unstable and separates the two basins. The upward root has $R^{\prime }>0$ and is stable. At $\mathcal{H}=R_{\max }$ the two downward roots merge at a saddle-node bifurcation; above that value only the upward equilibrium remains.

For ventilation switching under wind and heating changes, increasing wind pressure or decreasing heat input both reduce $\epsilon$, so both can create the pair of wind-driven equilibria. They nevertheless have different physical and transient effects. On the hot upward branch, implicit differentiation of $\mathcal{H}=\sqrt{K_A}\theta \sqrt{G_T\theta -W}$ shows that increasing $W$ increases $\theta$, while decreasing $\mathcal{H}$ decreases $\theta$. Both reduce the upward ventilation rate, but only a wind change shifts ${\theta }_w$.

Under slow parameter variation, a state follows its current stable branch while that branch persists; the upward branch does not disappear when wind increases or heating decreases. A sufficiently large abrupt wind increase can raise ${\theta }_w$ beyond the current temperature and place that state below the new unstable threshold, causing flow reversal and cooling towards the wind-dominated equilibrium. A jump insufficient to cross its basin boundary returns to the upward equilibrium.

At fixed $W$, reducing a still-positive heat input cannot force an initially upward-flow state through ${\theta }_w$: at that boundary the temperature balance has $\dot{\theta }=\mathcal{H}/\mathcal{V}>0$. Thus heat reduction alone does not cause that reversal in this model. Conversely, decreasing wind or increasing heat from the stable downward branch eventually destroys it at the fold and forces a switch to the upward branch, giving history dependence. Wind strengthening and heat reduction have the same effect on the steady dimensionless control parameter, but not the same temperature change or switching dynamics. These stability statements use quasi-steady opening flow and a single well-mixed temperature; other stratification or airflow-inertia models require their own stability analysis.

To find the threefold phase-locked equilibria, write $C=R{e}^{i\theta }$ with $R>0$; separating real and imaginary parts givesA steady state satisfiessoWriting $y=R_0^2$ and $D=1+4\mu -4{\omega }^2$, the amplitude branches areThere are two distinct positive roots for $\mu >{\omega }^2-1/4$, except at $(\mu ,\omega )=(0,0)$, where $y_{-}=0$ and only the upper root is nonzero. Indeed their sum is positive in this range and their product is ${\mu }^2+{\omega }^2$. Below the fold condition there are none. At equality, there is one positive repeated root $y={\omega }^2+1/4$, the saddle-node bifurcation limit.

For each positive amplitude, the sine and cosine determine $3{\theta }_0$ modulo $2\pi$, giving three phases separated by $2\pi /3$. Thus the source's “two states” means two amplitude branches modulo the threefold spatial symmetry. Generically there are six nonzero complex equilibria, three on each branch, not literally two.

The polar Jacobian matrix at an equilibrium isOn the lower branch, $\det J=-3y_{-}\sqrt{D}<0$, so it is a saddle equilibrium and unstable. On the upper branch, $\det J=3y_{+}\sqrt{D}>0$ andbecause existence implies $\mu >-1/4$. Therefore every upper-branch equilibrium is asymptotically stable, and every lower-branch equilibrium is a saddle equilibrium, away from the degenerate endpoints. The eigenvalues are unchanged by the smooth polar coordinate transformation at $R_0>0$. The origin, not covered by those coordinates, has eigenvalues $\mu \pm i\omega$ and is stable for $\mu <0$ and unstable for $\mu >0$.

Write the deterministic drift of the Adler phase equation as $f(\theta )=\omega -ϵ\sin \theta$. Its minimum is $\omega -ϵ$ and its maximum is $\omega +ϵ$. For $0<\omega <ϵ$, the zero condition has exactly two solutions in the specified interval:The linearization of a dynamical system at an equilibrium point gives $\delta \dot{\theta }=f^{\prime }({\theta }_{\ast })\delta \theta$, with $f^{\prime }=-ϵ\cos \theta$. Defining $\kappa =\sqrt{{ϵ}^2-{\omega }^2}$ gives $f^{\prime }({\theta }_s)=-\kappa <0$, so ${\theta }_s$ is stable; $f^{\prime }({\theta }_u)=+\kappa >0$, so ${\theta }_u$ is unstable. For $\omega >ϵ$, $f$ is positive everywhere and there are no equilibrium points: the phase runs continuously. This is the distinction between phase locking and running phase dynamics.

With mobility scaled to one, the effective force is $-V^{\prime }$. Therefore the tilted washboard potential isFor $\omega <ϵ$, its alternating local minima and maxima trap noise-free trajectories in wells. Minima coincide with ${\theta }_s+2\pi j$, and maxima with ${\theta }_u+2\pi j$. For $\omega >ϵ$, $V^{\prime }=ϵ\sin \theta -\omega <0$ everywhere: there are no wells and the particle slides down the tilt. The potential is defined on the unwrapped phase and obeys $V(\theta +2\pi )=V(\theta )-2\pi \omega$; it is not a single-valued periodic equilibrium potential on the circle.

At the transition $\omega =ϵ$, the two equilibrium points merge at $\theta =\pi /2$ in a saddle-node bifurcation. There $f\simeq ϵ(\theta -\pi /2{)}^2/2$, so the point is attracting from the left and repelling from the right; a zero linear derivative alone does not establish stable trapping.

There are axis equilibria $E_{\pm }=(\pm \sqrt{\mu },0)$ when $\mu \ge 0$, and $E_a=(a,a^2-\mu )$ for every $\mu$. At an axis equilibrium $(x_{\ast },0)$ the eigenvalues of the Jacobian matrix are $2x_{\ast }$ and $a-x_{\ast }$. Thus a negative axis root is a stable node when $a<x_{\ast }$ and a saddle when $a>x_{\ast }$. A positive axis root is an unstable node when $a>x_{\ast }$ and a saddle when $a<x_{\ast }$. Equalities give a zero eigenvalue.

At $E_a$,It is a saddle for $\mu <a^2$. For $\mu >a^2$, it is stable if $a<0$, unstable if $a>0$, with real node eigenvalues for $a^2<\mu \le 2a^2$ and a focus for $\mu >2a^2$. At $\mu =2a^2$ the node-focus transition is not a bifurcation of hyperbolicity when $a\ne 0$. For $a=0$, $E_a$ has imaginary eigenvalues when $\mu >0$ and the linearized system is a centre. For $a\ne 0$, the candidate local bifurcation values are $\boxed{\mu =0and\mu =a^2}$. When $a=0$ these zero-eigenvalue events coincide at $\mu =0$, while the off-axis equilibrium also has zero-real-part imaginary eigenvalues for every $\mu >0$.

The centre manifold theorem reduces local dynamics near zero-real-part eigenvalues to an invariant manifold tangent to their eigenspace; transverse directions with nonzero real parts retain their stable or unstable character. For a varying parameter, adjoin $\dot{\mu }=0$ and use an extended centre manifold for a parameter.

Take $a=1$. At $\mu =0$ near the origin, $y=0$ is an exact extended centre manifold. Its reduced equation isThis is a saddle-node bifurcation: no nearby axis equilibria for $\mu <0$, one at zero, and two for $\mu >0$. The transverse eigenvalue is positive, so the negative root is a saddle and the positive root an unstable node for small positive $\mu$.

At $\mu =1$, put $u=x-1$, $v=y$, $\nu =\mu -1$. Then $\dot{u}=2u+u^2-v-\nu$, $\dot{v}=-uv$, $\dot{\nu }=0$. Write the extended centre manifold as $u=h(v,\nu )$. Its invariance equation is $h_v(-hv)=2h+h^2-v-\nu$. Matching degrees givesHence the reduced dynamics areThe exact equilibrium branches are $v=0$ and $v=-\nu$, crossing at zero and exchanging stability in the centre direction. This is a transcritical bifurcation; the transverse eigenvalue remains positive. For $\nu <0$ the axis branch is an unstable node and the off-axis branch a saddle; for $\nu >0$ their roles reverse locally.

When $a=0$, the two parameter values merge and the origin at $\mu =0$ has a double zero eigenvalue with nonzero nilpotent linear part. The centre manifold is then two-dimensional, so the preceding one-dimensional classifications cannot simply be reused. Also the off-axis branch has imaginary eigenvalues for every $\mu >0$, with identically zero trace: this is not a generic Hopf bifurcation obtained by varying $\mu$. These degeneracies already follow without computing a new normal form.

The fixed point zero has multiplier $F^{\prime }(0)=\mu$, so it is attracting for $|\mu |<1$ and repelling for $|\mu |>1$. Other real fixed points satisfyTheir multiplier is $M=1+bx+2a{x}^2$, and they are attracting exactly when $-2<bx+2a{x}^2<0$. At equality $M=\pm 1$ the linear test is inconclusive. For $b\ne 0$ two branches meet at $x_s=-b/(2a)$, ${\mu }_s=1+b^2/(4a)$, a generic saddle-node bifurcation. At $\mu =1$, the nonzero branch crossing zero gives a transcritical bifurcation when $b\ne 0$.

The three sketches use $(a,b)=(1,0),(1,1/2),(-1,-1/2)$ respectively; case (iii) therefore illustrates the subcritical-flip choice.