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Write $N=S+I$. Inthe term $N$ represents production of healthy new plants by both healthy and diseased plants at unit per-capita rate. The term $-NS$ is density-dependent loss of healthy plants through competition, and $-\beta IS$ is loss through mass-action infection. Inthe term $-NI$ is the same density-dependent loss for diseased plants, while $+\beta IS$ is the matching gain when healthy plants become diseased. Thus $\beta$ is the transmission-rate coefficient per healthy-diseased encounter.

Adding the equations cancels the infection terms and gives the logistic differential equationThe disease therefore changes how the population is divided between healthy and diseased plants but has no effect on the total-population dynamics: infected plants reproduce at the same rate, all offspring enter the healthy class, and infection causes no additional death in this model.

Using $S=N-I$ reduces the system to the logistic differential equationSeparation of variables and the initial values giveDifferentiation yields the incidence

Write the scalar field energy as $E=T+V$, whereUnder the Derrick scaling ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$, a change of variables givesA finite-energy solution of the Euler-Lagrange equation is stationary under this admissible variation. The Derrick virial identity is therefore

The static field equation is ${\varphi }^{\prime \prime }=U^{\prime }(\varphi )$, so integration givesThe polynomial before $C$ is nonnegative and vanishes, so requiring the minimum to be zero fixes $C=0$. Hence the vacuum manifold iswhich has three elements. A finite-energy scalar-field kink can join only adjacent vacua: a solution cannot cross the intermediate vacuum at finite $x$ because its first integral would have $\varphi ={\varphi }^{\prime }=0$ there and the Picard-Lindelöf theorem would make it constant. There are therefore four oriented topological sectors,comprising two increasing kinks and their two antikinks. Symmetry under $\varphi \mapsto -\varphi$ and spatial reflection generates all four from one profile.

For the $0\to 1$ sector, completing the square gives the Bogomolny boundEquality holds for the Bogomolny equationWith $y={\varphi }^2$, this becomes the logistic differential equation $y^{\prime }=2y(1-y)$. Translation invariance supplies an arbitrary center $x_0$, and the explicit kink in a phi-six model isIt tends to $0$ and $1$ at the two spatial ends and saturates the bound. Its mass is consequently

For the kink in a phi-six model, choose the sector from $\varphi =0$ to $\varphi =2$ and the increasing sign of the Bogomolny equation:For $y={\varphi }^2$ this becomes the logistic differential equationAfter translating the centre to $x_0$, its solution isIt tends to $0$ as $x\to -\infty$ and to $2$ as $x\to +\infty$. Spatial reflection and $\varphi \mapsto -\varphi$ generate the other kink and antikink sectors.

Because the first-order equation saturates the Bogomolny bound, its mass is

Two populations grow at the same per-capita rate and each loses members at a rate proportional to their product. On the invariant diagonal $x=y=z$, the logistic differential equation gives $z(t)=\lambda /[\mu +(\lambda -\mu )e^{-\lambda t}]$ for $z(0)=1$ and positive rates. The positive symmetric equilibrium $(\lambda /\mu ,\lambda /\mu )$ is stable along that diagonal but unstable transversely: $(x-y{)}^{\prime }=\lambda (x-y)$, while the Jacobian matrix has eigenvalues $-\lambda ,+\lambda$. Thus an exactly symmetric deterministic solution can relax toward coexistence while a small imbalance grows. A corresponding density-dependent Markov jump process has a controlled fluid limit on each fixed finite interval, rather than uniform control for all time.