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A carrying capacity is an abundance scale at which a specified environment's net population growth vanishes. In the logistic growth equation, positive unharvested populations approach $K$. It depends on the species, resources and environment rather than being an immutable universal population limit.

The useful distinction is between a growth rate and a sustainable abundance. In a logistic differential equation, $r$ controls how quickly a population responds, while the carrying capacity $K$ sets the positive equilibrium point. Increasing $r$ with $K$ fixed accelerates approach without raising that equilibrium. Increasing $K$ changes the destination. The logistic differential equation diagrams show populations approaching the same carrying capacity from below and above.

In ecological use, the carrying capacity describes what a particular environment can support under specified conditions. Nutrient supply, available space, and accumulated waste can change those conditions; it is not a permanent property of a species alone. A population temporarily above a carrying capacity is possible. In the simplest constant-$K$ logistic growth equation, its net growth becomes negative; real population dynamics can include delays, seasonal variation, and changing resources that the basic mathematical model omits. ^[ref].

For $a>0$, the positive amplitude ordinary differential equation $R^{\prime }=aR-R^3$ has solution $R(t)=\sqrt{a}[1+(a/R_0^2-1)e^{-2at}{]}^{-1/2}$. This follows from the logistic differential equation for $R^2$. Every $R_0>0$ approaches $\sqrt{a}$ as a monotone function, and the positive distance to this limit decreases; an oscillating complex wave carrying this amplitude has no ordinary monotone ordering.

The Forward Euler method applied to the logistic differential equation $y^{\prime }=ry(1-ay)$ becomes the logistic map after $u=a(\lambda -1)y/\lambda$, $\lambda =1+r\Delta t$. The positive equilibrium point of the differential equation has perturbations decaying as $e^{-rt}$. Its discrete fixed-point multiplier is $1-r\Delta t=2-\lambda$: for $r\Delta t>2$, the discretization has an artificial oscillatory instability, leading initially to alternation on successive steps. This contrasts with the positive perturbation multiplier $e^{-r\Delta t}$ of the exact time-step map.

A growing linear mode can make a formally small derivative nonlinearity order one at large distance. After the crossover scaling, the equation $X{H}^{\prime \prime }-H^{\prime }+(H^{\prime }{)}^2=0$ reduces to a logistic differential equation in $\log X$ for $H^{\prime }$. Matching $H\sim X^2/2$ at small $X$ gives $H^{\prime }=X/(1+X)$ and the displayed primitive with $H(0)=0$. It changes from quadratic growth to linear growth without a finite turning singularity.

The logistic differential equation is the nonlinear ordinary differential equationAs a mathematical model of population dynamics, it treats abundance $N$ as a continuous nonnegative quantity. The constant $r$ has units of inverse time, while the carrying capacity $K$ has the same units as $N$. The useful idea is to combine growth resembling exponential growth at low abundance with a slowing approach to a finite abundance. The three trajectories below already show an important distinction: starting above the carrying capacity gives a decreasing curve, whereas starting below it gives an increasing one.

For $N>0$, dividing the derivative by abundance gives the instantaneous per-capita rate:It decreases linearly with $N$: it approaches $r$ at low abundance, is $r/2$ at $N=K/2$, vanishes at the carrying capacity, and is negative above it. This per-capita rate differs from the total rate $N^{\prime }$, which also counts how many individuals contribute. It is an instantaneous rate, rather than a fixed percentage increase over a whole day. The linear per-capita assumption and the resulting qualitative picture are developed in MIT 18.03SC's logistic-model notes.

Writing $f(N)=rN(1-N/K)$, the equilibrium points satisfy $f(N)=0$, so the constant solutions of a differential equation are $N(t)=0$ and $N(t)=K$. The phase line points upward for $0<N<K$ and downward for $N>K$. Also $f^{\prime }(0)=r>0$ and $f^{\prime }(K)=-r<0$: the criterion for fixed point stability for an autonomous differential equation makes zero an unstable equilibrium and $K$ an asymptotically stable equilibrium point. Every positive initial condition approaches $K$ as $t\to \infty$; $N_0=0$ stays exactly zero. Instability of zero means that a small positive population grows away from it, not that a zero population spontaneously appears.

For every $N_0>0$, the unique solution of a differential equation for all $t\ge 0$ isThe denominator is positive on this domain, including when $N_0>K$. For $0<N_0<K$, this is a translated and scaled logistic function; its extension to all real times is a sigmoid function. For $N_0>K$ the coefficient of the exponential function is negative, and the future curve instead decreases toward $K$. No nonconstant positive solution of a differential equation crosses $K$ or reaches it at finite time. A short derivation uses the dimensionless variable $z=K/N-1$: the chain rule gives $z^{\prime }=-rz$, so exponential decay yields $z(t)=(K/N_0-1)e^{-rt}$. The substitution divides by $N$, so the zero equilibrium point must be included separately.

The total growth rate is a downward-opening quadratic function of abundance. Completing the square gives its maximum without solving the differential equation:The balance is beautiful: more individuals contribute to growth, but each contributes less, and the strongest total increase occurs halfway to the carrying capacity. This is a maximum over possible abundances. A trajectory starting above $K/2$ never passes through that abundance in its future.

The chain rule connects this maximum to the shape of the time trajectory through its second derivative:For $0<N_0<K/2$, the trajectory initially has convexity and then concavity, with its one future inflection point atFor $N_0=K/2$, this point lies at the initial time, with concavity for $t>0$. For $K/2<N_0<K$, the inflection point belongs to the past extension, so the future trajectory has concavity throughout. Above $K$, the decreasing trajectory has convexity and no inflection point. The two constant solutions of a differential equation have no change of concavity. Thus an entire logistic function has an S shape, but its observed future segment need not show both bends. See also ^[ref].

For the plotted example with $K=1000$, $r=0.4{\mathrm{day}}^{-1}$ and $N_0=100$, time $t$ measured in days givesThe initial total rate is $N^{\prime }(0)=0.4\cdot 100\cdot 0.9=36$ individuals per day, and the initial per-capita rate is $0.36{\mathrm{day}}^{-1}$. The fastest total increase is $rK/4=100$ individuals per day, reached at $t_{1/2}=\ln 9/0.4\approx 5.49306$ days. At that time the per-capita rate has already fallen to $0.2{\mathrm{day}}^{-1}$.

These parameters are illustrative choices, rather than an empirical fit. The equation specifies deterministic evolution under a phenomenological assumption: it summarizes how net growth depends on abundance without separately modelling every process that affects it. The carrying capacity is a parameter for a specified setting, not a universal constant for all populations. In his 1838 paper "Notice sur la loi que la population suit dans son accroissement", Correspondance mathématique et physique 10, pp. 113–121, Pierre-François Verhulst proposed $p^{\prime }=mp-n{p}^2$, equivalent to this equation with $r=m$ and $K=m/n$. He also noted that different assumed growth-retarding terms could describe the observations similarly while giving very different limiting populations. This is an early illustration of why a good-looking curve alone does not establish a unique long-term prediction.^[ref]

The logistic differential equation describes continuous time. The logistic map instead advances in discrete steps and can exhibit a period-doubling cascade and Devaney chaos; the monotone positive trajectories here cannot. Their connection through the Forward Euler method is explained under Euler discretization of logistic growth, where a large step can introduce behaviour absent from the original differential equation.

The logistic growth equation applies the logistic differential equation to population dynamics. For positive growth rate $r$ and carrying capacity $K$, an initial condition $0<N(0)<K$ gives $N(t)=K/[1+b{e}^{-rt}]$, where $b=K/N(0)-1$. Its per-capita growth falls linearly with abundance, and a trajectory starting below $K/2$ reaches an inflection point when $N=K/2$. The exact solutions of a differential equation and their diagrams are developed under logistic differential equation. A translated and scaled logistic function has the same form. In nonlinear regression for biological size, an analogous saturation curve need not imply that circumference itself obeys a mechanistic population law.

A nonlinear ordinary differential equation depends nonlinearly on the unknown function or its derivatives: it cannot be written in the form defining a linear differential equation. For example, $y^{\prime }=y^2$ contains a square of the unknown function, and $y^{\prime }=y(1-y)$ is a logistic differential equation. A coefficient such as $t^2$ depending only on the independent variable does not make an ordinary differential equation nonlinear. This distinction concerns dependence on the unknown, rather than the shape of its function graph.

For a concrete initial value problem, consider $y^{\prime }=y^2$, $y(0)=y_0>0$. Its solution of a differential equation isTaking its derivative gives $y^{\prime }=y_0^2/(1-y_{0}t{)}^2=y^2$, and evaluating at zero checks the initial condition. The stated interval contains zero and ends at finite-time blowup: $y$ tends to positive infinity as $t$ approaches $1/y_0$ from below. Increasing the positive initial condition brings finite-time blowup closer. This is a simple mathematical example with dimensionless $t$ and $y$, rather than a calibrated physical model.

Choose $y_0=1$ and $y_0=2$, giving $y_1=1/(1-t)$ and $y_2=2/(1-2t)$. At $t=1/4$ they have values $4/3$ and $4$, so their sum $s=y_1+y_2$ has value $16/3$. On the common interval $0\le t\le 0.4$, both are valid solutions of a differential equation, but their sum fails the same ordinary differential equation:Already at $t=0$, $s=3$ and its derivative is $1+4=5$, whereas $y^{\prime }=y^2$ would require $9$. At $t=1/4$, the defect $s^{\prime }-s^2$ is $-32/3$. The cross term explains the failure of the superposition principle, even though each individual solution of a differential equation is a smooth function.

The first panel shows $y_1$, $y_2$ and their invalid sum only up to $t=0.4$, before either finite-time blowup. The second measures the defect in the ordinary differential equation, including the computed value at $t=1/4$. For a homogeneous linear differential equation, the corresponding defect of a sum is zero. Nonlinear ordinary differential equations can describe state-dependent growth such as the logistic differential equation, whose negative quadratic term limits growth instead of accelerating it without bound. Recognizing the distinction tells us when we may assemble solutions of a differential equation from simpler pieces and when we must account for their interaction. See MIT's worked counterexample to nonlinear superposition.

For the logistic differential equation, separate variables to obtain $\log |y/(1-ay)|=rt+C$ away from the constant solutions. Applying the positive initial condition givesThe formula also includes the constant solution $y_0=1/a$. Below $1/a$ positive solutions increase; above $1/a$ they decrease. They approach $1/a$ as $t\to \infty$. Equivalently, linearizing $F(y)=ry(1-ay)$ gives $F^{\prime }(0)=r>0$ and $F^{\prime }(1/a)=-r<0$. Thus zero is unstable and $1/a$ is asymptotically stable.

The Forward Euler method gives $y_{n+1}=(1+r\delta t)y_n-ra\delta t{y}_n^2$. Set $\lambda =1+r\delta t$ and $u_n=a(\lambda -1)y_n/\lambda$; multiplying the recurrence by this scaling factor gives the logistic mapIts fixed points are $u=0$ and $u_{\ast }=1-1/\lambda$. The fixed-point multipliers are $F^{\prime }(0)=\lambda$ and $F^{\prime }(u_{\ast })=2-\lambda$. Since $\lambda >1$, zero is always unstable. The positive fixed point is stable for $1<\lambda <3$ and unstable for $\lambda >3$. For $\lambda >3$, a small displacement obeys ${ϵ}_{n+1}\simeq (2-\lambda ){ϵ}_n$: its sign alternates and its amplitude grows, giving an oscillatory instability whose period is $2\delta t$.

This is an artifact of Euler discretization of logistic growth. Exact small perturbations decay by the positive factor $e^{-r\delta t}$ per time step; a one-dimensional autonomous differential equation cannot have a nonconstant periodic orbit, since its solutions move monotonically between equilibrium points. The artificial instability occurs when $r\delta t>2$, beyond the stable step range of this discretization.

For $\lambda =3.2$, remove the fixed point factors from $F(F(u))-u$ to obtain ${\lambda }^2{u}^2-\lambda (\lambda +1)u+\lambda +1=0$. The resulting logistic map two-cycle isThese points map to each other. Their periodic-orbit multiplier is $F^{\prime }(u_{-})F^{\prime }(u_{+})=4+2\lambda -{\lambda }^2=0.16$, so the two-cycle attracts nearby iterates. The cobweb shows a small initial departure from $u_{\ast }=0.6875$ growing with alternating sign and then settling into the finite rectangle between $u_{-}$ and $u_{+}$. The time trace shows the corresponding finite-amplitude oscillation.

Putting $\epsilon =0$ gives the local regular perturbation result $y_0=(x^2+1)/2$. It cannot describe the whole half-line: its derivative grows like $x$, making $\epsilon y^{\prime }$ order one at $x=O({\epsilon }^{-1})$.

The useful exact change of variables isDirect substitution turns the equation into two autonomous differential equations,The initial values here are $z(0)=w(0)=\epsilon$. Near the origin, linearizing this system gives eigenvalues $1$ and $-1-\epsilon$. Equivalently, neglecting only the derivative-square nonlinearity in the original equation gives the two modes $x^2$ and $x^{-\epsilon }$. Their matching coefficients areThus the growing derivative has $w\sim Cx$. Keeping these unexpanded linear-mode coefficients is convenient for matching; it does not assert that the linearized solution is the exact nonlinear one.

At the distinguished limit $X=Cx=O(1)$, set $R=\log X$. To leading order the derivative equation becomes $w_R=w(1-w)$, the logistic differential equation. Matching $w\sim e^R$ as $R\to -\infty$ selectsThen $z_R+z=w$, and the growing-mode matching $z\sim X/2$ givesThis proves the logistic crossover in a weakly nonlinear Euler equation. In particular the nonlinear-region leading solution is $y\sim H(Cx)/(\epsilon C)$, or ${\epsilon }^{-2}H(\epsilon x)$ with the constants taken at their lowest order.

A composite asymptotic expansion retaining the small-distance stable mode and satisfying the initial value exactly isSince $H(X)=X^2/2+O(X^3)$, this matches $A{x}^2+B{x}^{-\epsilon }$ when $Cx\ll 1$ and hence $(x^2+1)/2$ at fixed $x$ as $\epsilon \to 0$. Its derivative at one differs from the prescribed value only by $O(\epsilon )$, within leading-order accuracy. For $Cx\gg 1$ it gives $y\sim x/\epsilon$. Thus the composite covers the initial region, the nonlinear transition and the arbitrarily large-$x$ regime; uniform leading accuracy is understood relative to the growing solution, not as an absolute error independent of unbounded $x$.

There is no later hidden growth regime. The exact positive system keeps $z,w$ in $[0,1+\epsilon {]}^2$, because its vector field points inward at every boundary. Its positive equilibrium point is $z=w=1+\epsilon$, with Jacobian matrix having trace $-2-3\epsilon$ and determinant $1+\epsilon$. The origin has no attracting direction inside the positive quadrant, while the divergence $-\epsilon -2w<0$ excludes cycles by the Bendixson-Dulac theorem. The Poincaré-Bendixson theorem therefore gives convergence to the positive stable equilibrium. At fixed small positive $\epsilon$, the exact far-field slope is $y/x\to (1+\epsilon )/\epsilon$, whose difference from $1/\epsilon$ is only relative $O(\epsilon )$, consistent with the global leading composite.

For $r>0$, the polar coordinates satisfyA positive constant radius must therefore be one. The nonconstant periodic solutions arewith least period $2\pi$ and counterclockwise orientation. The origin is also a stationary solution, hence periodic in the degenerate sense of a constant function; it is a point rather than a nontrivial closed orbit. There are no other constant-radius solutions.

For $r_0>0$, put $v=r^2$. Then $\dot{v}=2v(1-v)$, a logistic differential equation. Solving this separable differential equation givesThe denominator is positive for every forward time. When $0<r_0<1$, the radius increases monotonically to one; when $r_0>1$, it decreases monotonically to one; and when $r_0=1$ it stays there. Thus every nonzero orbit approaches the unit circle while continuing to rotate. More precisely, its distance from the phase-matched circular solution tends to zero; its Cartesian coordinates do not converge to a single point. For $r_0=0$, the solution stays at the origin.

The origin is the only equilibrium point. Its Jacobian matrix iswhose eigenvalues are $1\pm i$. It is an unstable focus. The unit circle is an attracting limit cycle, rather than a curve of equilibria: on it the vector field remains nonzero. Linearizing the radial equation at $r=1$ gives $\dot{\delta r}=-2\delta r+O(\delta r^2)$, confirming radial attraction. All other radii are strictly monotone, so there are no additional nontrivial periodic orbits.

At the unscaled state $(x,y)$, write the Lévy kernel of a Markov jump process on jump displacements $z$ asThe destination-kernel convention translates each Dirac mass by $(x,y)$. The total holding rate is the mass of this kernel. At either zero population the corresponding death rates vanish, so no negative states are reached. Births in the total population are bounded by a Yule process with per-individual rate $\lambda$, while deaths only decrease it; this also gives nonexplosion.

Put $Z^N=(X/N,Y/N)$. This is a density-dependent Markov jump process, with jumps of size $1/N$ and rates of order $N$. Its drift gives the deterministic approximationUniqueness preserves the diagonal $x=y=z$. For positive $\lambda ,\mu$, the resulting logistic differential equation has solutionIf $\lambda <\mu$, it decreases from one to $\lambda /\mu$; if $\lambda >\mu$, it increases to that same equilibrium; equality gives the constant solution one. In particular positive rates do not give extinction on this exactly symmetric deterministic trajectory. Transverse perturbations behave differently: subtracting the equations gives $(x-y{)}^{\prime }=\lambda (x-y)$, so the positive symmetric equilibrium has an unstable difference direction. The symmetric competition model distinguishes diagonal logistic relaxation from this symmetry-breaking instability.

For an explicit error estimate, use the martingale decomposition of a density-dependent jump process:Fix $t_0,\delta >0$ and choose $R>{\sup }_{t\le t_0}z(t)+\delta$. Stop at the first exit ${\sigma }_R$ from the box $[0,R{]}^2$. On that box, $F$ has Lipschitz constant $L_R=\lambda +2\mu R$ in the supremum norm. Each coordinate of $M^N$ has jumps bounded by $1/N$ and predictable quadratic variation at mostIndeed a coordinate's positive and negative jump rates are $N\lambda x$ and $N\mu xy$, and the squared jump size is $1/N^2$. Independent jump clocks give zero cross-bracket.

The Bernstein bound for bounded-jump martingales gives, for each coordinate,The Gronwall inequality applied up to ${\sigma }_R$ givesAn exit from the box already entails an error larger than $\delta$, by the choice of $R$, so the stopped estimate controls the event of any error on the full interval. Take $\eta =\delta e^{-L_R{t}_0}$ and use the union bound on the two coordinates:For fixed time horizon and tolerance this is exponentially small in $N$, proving a fluid limit. The transverse instability explains why this finite-horizon estimate should not be extrapolated to arbitrarily long times without changing its constants.

Write $c=(1-a-b)/a$. The admissible equilibrium points are $O=(0,0)$, $E=(1-a,0)$ when $a\le 1$, and $P=(b,c)$ when $a+b\le 1$. The Jacobian matrix isAt $O$ the eigenvalues are $(1-a)/a$ and $-1$; at $E$ they are $-(1-a)/a$ and $(1-a-b)/b$. At a positive coexistence equilibrium point,It is therefore asymptotically stable. It is a node for $b^2>4a(1-a-b)$, a focus for the reverse inequality, and has repeated real eigenvalues at equality. The node–focus change does not alter the attracting equilibrium point's topology.

There are three open physical parameter regions:

The axes are invariant. On $x=0$, $y$ decays exponentially; on $y=0$, prey obey the logistic differential equation. Positivity follows because each right-hand side has its population as a factor. Also $x$ is bounded by logistic comparison, while $W=x+by$ satisfies $\dot{W}=(1-a)x/a-x^2/a-by$, so all forward trajectories are bounded. The Dulac function $1/(xy)$ has divergence $-1/(ay)<0$ in the interior, excluding periodic orbits by the Bendixson-Dulac criterion.

In the extinction region, $\dot{x}\le -(a-1)x/a$, and then $\dot{y}/y=x/b-1$ is eventually negative. In the prey-only region, the upper bound $\mathrm{limsup}x\le 1-a<b$ implies predator decay and convergence of interior prey to $1-a$. In the coexistence region the Lyapunov functionhas $\dot{\mathcal{V}}=-(x-b{)}^2/a\le 0$. Its positive-quadrant sublevel sets are compact, and the only invariant subset with $x=b$ also has $y=c$. The LaSalle invariance principle proves convergence of every interior trajectory to $P$. These facts justify the complete qualitative phase portraits, not just their local arrows.

At $a=1$, $O$ and $E$ meet in a transcritical bifurcation; positive prey decay algebraically. At $a+b=1$ with $a<1$, $E$ and $P$ exchange stability. On its centre manifold, $x=b-ay+O(y^2)$ gives $\dot{y}=-a{y}^2/b+O(y^3)$, so the physically positive side is still attracting. There is no interior Hopf bifurcation since the coexistence trace never vanishes.

A Markov jump process relative to $({\mathcal{F}}_t)$ is an adapted càdlàg process with piecewise-constant paths and finitely many jumps on each bounded time interval, whose conditional future law given ${\mathcal{F}}_t$ depends only on its current state. For a time-homogeneous process on a countable state space, write $q(x,y)\ge 0$ for its transition rates and $q(x)={\sum }_{y\ne x}q(x,y)<\infty$. In state $x$, the holding time is exponential of rate $q(x)$ and the next state is $y$ with probability $q(x,y)/q(x)$; a zero rate makes the state absorbing. The generator is $\mathcal{A}f(x)={\sum }_{y\ne x}q(x,y)(f(y)-f(x))$. Nonexplosion is included here so that the process is defined at every finite time.

Here is a finite-jump, bounded-rate form of the Kurtz fluid limit theorem. For each $N$, let $Z^N$ be a chain on a countable state space $E_N\subseteq {\mathbb{R}}^d$ with jumps $\ell /N$ at rates $N{\beta }_{\ell }(z)$, with $\ell$ in a fixed finite subset $\mathcal{J}\subset {\mathbb{R}}^d$. Assume the nonnegative ${\beta }_{\ell }$ are uniformly bounded, admissible jumps remain in the state space, andis globally Lipschitz with constant $L$. Suppose $Z_0^N\to z_0$ in probability, and let $z$ solve $\dot{z}=F(z)$, $z(0)=z_0$. Then for every finite $T$,Bounded total jump rates ensure nonexplosion; the globally Lipschitz drift gives a unique global ODE solution. These explicit hypotheses suffice for the population example, without needing a more general local version of the theorem.

To prove the result, compensate the jumps of this density-dependent Markov jump process. The drift and the coordinate martingales satisfyHere $⟨M^{N,i}⟩$ is predictable quadratic variation, not the sum of realized squared jumps. The compensation identity follows because each possible increment $\ell /N$ has intensity $N{\beta }_{\ell }$, and its squared coordinate increment contributes ${\ell }_i^2{\beta }_{\ell }/N$ per unit time. Bounded rates and jump sizes make these martingales square-integrable.

We use the Doob L2 maximal inequality in the explicit form $\mathbb{E}{\sup }_{s\le T}|M_s{|}^2\le 4\mathbb{E}|M_T{|}^2$ for a scalar square-integrable martingale starting at zero. Applying it coordinatewise and setting $K={\sup }_z{\sum }_{\ell }|\ell {|}^2{\beta }_{\ell }(z)<\infty$ givesSubtract the ODE integral equation. Lipschitz continuity and Gronwall's inequality yieldThe first term tends to zero in probability by hypothesis, and the second by its maximal second-moment estimate. More quantitatively, for $\delta >0$,This proves the stated uniform finite-horizon fluid limit.

For the cells, division replaces one cell by two and increases population by one. At population $k\le N$, the total rate is the sum of the $k$ identical cell rates, namelyThe state $N$ is absorbing. Set $Z_t^N={\xi }_t/N$; its jump is $1/N$ with rate $N\beta (z)$, where $\beta (z)=z(1-z)$ on $[0,1]$. Extending this function by zero outside $[0,1]$ makes it bounded, nonnegative and globally Lipschitz, with Lipschitz constant one. Starting from $Z_0^N=p$, or from $\lfloor pN\rfloor /N$ when integer rounding is necessary, the limiting ODE is the logistic differential equationSeparation of variables, or differentiating $z/(1-z)$, gives its solution. The logistic cell-division process therefore satisfiesHere $K\le 1/4$, so the maximal martingale second moment is at most $T/N$. With exact initial density $p$, the preceding pathwise bound even gives $\mathbb{E}{\sup }_{s\le T}|Z_s^N-z(s){|}^2\le T{e}^{2T}/N$. With integer rounding, the bound $\mathbb{E}{\sup }_{s\le T}|Z_s^N-z(s){|}^2\le 2e^{2T}(T/N+1/N^2)$ follows from $|Z_0^N-p|\le 1/N$ and $(a+b{)}^2\le 2a^2+2b^2$. Thus the deterministic logistic curve approximates the density with fluctuations of order $N^{-1/2}$ on fixed time intervals. The convergence statement is on finite horizons; it does not interchange the limits $t\to \infty$ and $N\to \infty$.

Use the complete system in the PDF: the TeX omits $\dot{y}=y(4x-1)/8$. Both coordinate axes are invariant, and the equilibrium points are $(0,0)$, $(1,0)$ and $(1/4,3/4)$. The Jacobian matrix isAt $(0,0)$ its eigenvalues are $1,-1/8$, and at $(1,0)$ they are $-1,3/8$: both points are saddle equilibria. At the coexistence point,Thus the interior point is a stable spiral. The nontrivial nullclines are $x+y=1$ and $x=1/4$: $x$ increases below the first and decreases above it, while $y$ increases to the right of the second and decreases to the left. The rotation near coexistence is counterclockwise.

A local linearization by itself does not prove that every interior solution has the same limit. For this logistic predator-prey model, let $x_{\ast }=1/4,y_{\ast }=3/4$ and use the Lyapunov functionThe inequality $u-1-\log u\ge 0$ makes $V$ nonnegative, with equality only at coexistence. Its sublevel sets are compact inside the positive quadrant, because $V$ diverges both at the axes and at infinity. Direct differentiation cancels the predator-prey cross terms and givesOn $\{\dot{V}=0\}$ one has $x=x_{\ast }$, and remaining there requires $\dot{x}=-x_{\ast }(y-y_{\ast })=0$. The largest invariant subset is therefore the coexistence point. The LaSalle invariance principle provesOn $x=0$, $y$ decays as $e^{-t/8}$ toward the origin. On $y=0$, the logistic differential equation takes every $x(0)>0$ to one. The origin itself stays fixed. These boundary trajectories are the exceptions to the interior convergence statement.

For $r>0$, differentiate $r^2=x^2+y^2$ and use the angular derivative $(x\dot{y}-y\dot{x})/r^2$. The polar coordinates satisfyHence the rotation is clockwise. The origin is an unstable equilibrium, while $r=1$ is a limit cycle of period $2\pi$. Writing $u=r^2$ gives the logistic differential equation $\dot{u}=2u(1-u)$. For $r(t_0)=r_0>0$,All nonzero solutions tend to the unit-circle attracting limit cycle as $t\to \infty$, spiralling outward for $r_0<1$ and inward for $r_0>1$. For $0<r_0<1$ the solution exists for every real time and approaches the origin as $t\to -\infty$; for $r_0=1$ it remains periodic in both time directions. For $r_0>1$, the denominator first vanishes atThis finite-time blow-up of an ordinary differential equation occurs backward in time. Such exterior solutions have no $t\to -\infty$ behavior: their maximal time interval is $(t_{\ast },\infty )$. The zero solution exists for all time and has no defined polar angle.

Put $a=\mu -k^2=Q^2>0$. The shared travelling phase again cancels advection in the nonlinear Ginzburg-Landau equation; the remaining amplitude ordinary differential equation isFor $R>0$, $Y=R^2$ obeys the logistic differential equation $Y^{\prime }=2Y(a-Y)$. Equivalently, $Z=1/Y$ solves $Z^{\prime }=-2aZ+2$, whence $Z(t)=a^{-1}+(R_0^{-2}-a^{-1})e^{-2at}$. The nonconstant cubic amplitude saturation solution isIt takes the specified $R_0$ at $t=0$. The denominator of the cubic amplitude saturation solution remains positive for all future time: if $R_0<Q$ its bracket exceeds one, and if $R_0>Q$ it lies between $Q^2/R_0^2$ and one. Thus the positive solution is global forward in time. $R_0=Q$ would give the excluded constant solution; $R_0=0$ would remain at zero rather than relax to the positive wave.