A milk protein is a protein present in milk, including caseins and whey proteins. Different fractions have different aggregation and processing properties, so total protein concentration alone does not specify the behavior of the mixture.^[ref]

A milk fat globule is a lipid-rich droplet surrounded by an interfacial membrane derived from the secreting cell. Dispersion of many droplets contributes to the physical properties of milk. Fat content and droplet structure are different measurements.^[ref]

The fat-enriched fraction of milk.

The fraction remaining after substantial fat removal from milk.

The liquid fraction after curd separation; it retains lactose, soluble proteins and many dissolved salts.

For a compact Lie group that may be disconnected, fix a maximal torus $T$ and define the normalizer quotient $N_G(T)/T$. It is finite. To prove this, identify $T$ with its Lie algebra modulo the kernel lattice of its exponential map. A torus automorphism preserves this lattice. Its differential is therefore an integral invertible matrix in a lattice basis. A connected subgroup of the normalizer induces a continuous family of such matrices, hence acts trivially on $T$.

The identity component $N_G(T{)}^0$ thus centralizes $T$ and lies in $G^0$. The maximal-torus centralizer theorem for the connected compact group gives $N_G(T{)}^0=T$. The quotient is consequently discrete and compact, hence finite. Its action on $T$ need not be faithful when $G$ is disconnected: for a finite group, $T$ is trivial and $N_G(T)/T=G$. This is why the quotient $N_G(T)/C_G(T)$ must not silently replace $N_G(T)/T$ in this convention.

A period-$k$ orbit of a Poincaré map corresponds to a closed trajectory, hence a periodic orbit of the underlying flow, meeting the chosen section at the $k$ distinct successive return points. Its time period is the sum of the $k$ return times, not necessarily $k$ times a single common time. Stability of the return-map orbit describes transverse stability of that closed trajectory; the flow itself also has the neutral time-translation direction.

Follow a perturbation through one complete circuit. The chain rule givesAfter $m$ circuits the linear perturbation is $M^m\delta x_0$. With distinct eigenvalues, the eigenvectors span over the complex numbers, and each component is multiplied by ${\lambda }_j^m$. Thus the periodic orbit is linearly asymptotically stable exactly whenThese are the orbit's discrete Floquet multipliers. Any modulus greater than one causes instability; a unit modulus prevents decay in the linear system, and nonlinear stability then needs further analysis. The finitely many intermediate derivative factors do not affect convergence to zero between complete circuits.

A periodic orbit of least period $k$ is a list of distinct points $x_0,\dots ,x_{k-1}$ satisfying $f(x_i)=x_{i+1}$ for $i<k-1$ and $f(x_{k-1})=x_0$. Equivalently $f^k(x_0)=x_0$, with no smaller positive iterate fixing $x_0$. For $k=1$ this is a fixed point.

The explicit travelling wave speed increases with $q$:Its two tails change differently. Near the leading empty state, $U\sim a^{-2/q}{e}^{-kz}$ with $k=\sqrt{2/(q+2)}$ decreasing as $q$ grows: that tail lengthens. Near the filled state, $1-U\sim (2a/q)e^{bz}$, with $b=q/\sqrt{2(q+2)}$ increasing: that tail becomes steeper. Thus the change is not a uniform sharpening of the whole profile. For example the half-height slope is $-\frac{k}{2}(1-2^{-q/2})$.

For meaningful comparison, the figure aligns all fronts at $U(0)=1/2$ by choosing $a=2^{q/2}-1$. It displays the speed increase and the unequal changes of the two tails. Since $f^{\prime }(0)=1$ for every $q$, the minimum admissible speed remains two; the explicit family has $c(q)>2$ for $q>0$. The invariant-triangle proof above extends with boundary derivative $U^{q+1}\ge 0$.

Set $Y=U^{q/2}$ and $k=2b/q$. The proposed profile with $a,b>0$ givesSubstituting into $U^{\prime \prime }+c{U}^{\prime }+U(1-U^q)=0$ and dividing in the interior by $U(1-Y)$ yieldsEquating the constant and $Y$ coefficients givesHence the exact generalized Fisher front isThe free positive $a$ translates the front. It connects one on the left to zero on the right and is strictly decreasing. At $q=1$, its speed is $5/\sqrt{6}$.

The positive quadrant is invariant. The non-axis nullclines are $u+v=1$ and $u=1/\beta$. Below $u+v=1$, $u$ increases; above it, $u$ decreases. To the left of $u=1/\beta$, $v$ increases; to the right, it decreases.

For $\beta >1$, the stable manifold of the interior saddle separates trajectories tending to $(1,0)$ from those for which $u\to 0$ and $v\to \infty$. Exactly on that manifold, the trajectory tends to the saddle. For the illustrated $r=1,\beta =2$, the separating manifold is exactly $v=u$, since the vector field preserves that diagonal.

For $\beta <1$, every strictly positive initial state ultimately has $u\to 0$ and $v\to \infty$. Indeed $\dot{u}\le ru(1-u)$ implies $\mathrm{limsup}u\le 1$, so eventually $1-\beta u$ is bounded below by a positive constant. Then $v$ grows exponentially, forcing $u$ down. At $\beta =1$, a positive $v$ eventually drives $u$ below one; thereafter $v$ increases. A finite positive limit for $v$ would force $u\to 1$, inconsistent with $\dot{u}\to -rv<0$, so the same exclusion outcome holds.

There are no interior periodic orbits: the Bendixson-Dulac criterion with multiplier $1/(uv)$ givesThe phase portraits in the preceding figure show the threshold geometry and the absence of a bounded attracting coexistence state. Initial states on the axes retain the corresponding one-species dynamics.

The equilibrium points areThe third is biologically interior only for $\beta >1$, and coincides with $(1,0)$ at $\beta =1$. The Jacobian matrix isAt $(0,0)$ the eigenvalues are $r,1$, so it is a repelling node. At $(1,0)$ they are $-r,1-\beta$, giving a stable node for $\beta >1$ and a saddle for $\beta <1$.

At the interior equilibrium,so coexistence is a saddle. The formal equilibrium with negative $v$ for $\beta <1$ is not a population state; there its trace is negative and determinant positive.

At $\beta =1$ the boundary equilibrium is nonhyperbolic. To determine its behavior into the positive quadrant, put $x=u-1$. A local center manifold has $x=-v+v^2/r+O(v^3)$, so $\dot{v}=v^2+O(v^3)>0$ for small positive $v$. Thus it is unstable to introduction of species $v$, though attracting along the $v=0$ axis. The boundary/interior branches undergo a transcritical bifurcation at this parameter.

The term $ru$ is unrestricted growth of species $u$, and $-r{u}^2$ represents its density-dependent self-limitation, with scaled carrying capacity one. The interaction $-ruv$ suppresses $u$ in the presence of species $v$.

Species $v$ has intrinsic growth $v$ and suppression $-\beta uv$ by $u$. Thus both interspecific effects are harmful: this is a competition model, with no self-limitation term for $v$. Its unbounded growth when $u$ disappears is a consequence of that modeling choice. The parameter $r$ is the growth-time ratio, while $\beta$ measures the effect of $u$ on $v$.

The two positive fixed points merge whenHere $f^{\prime }(N_c)=1$, $f^{\prime \prime }(N_c)\ne 0$, and increasing $S$ decreases $f$, so this is a saddle-node bifurcation of the map. For $S<S_c$, the cobweb descends to zero below $N_{-}$, rises towards $N_{+}$ between the positive fixed points, and descends towards $N_{+}$ above it. Monotonicity of $f$ prevents a trajectory from overshooting across the attracting fixed point. Thus $N_{-}$ is an extinction threshold.

For $S>S_c$, $f(N)<N$ for every $N>0$. The iterates decrease to a nonnegative fixed point, necessarily zero. At $S=S_c$, initial states below $N_c$ tend to zero and those above tend to $N_c$. The sterile-release population threshold therefore combines bistability below the critical release with extinction above it.

Write $f(N)=\alpha N^2/(\beta N^2+S)$. Its nonnegative fixed points are $N=0$ and, when ${\alpha }^2\ge 4\beta S$,Indeed a positive fixed point satisfies $\beta N^2-\alpha N+S=0$. The linear stability multiplier isAt a positive fixed point, $\beta N^2+S=\alpha N$, henceFor strict discriminant positivity, $N_{+}$ is attracting and $N_{-}$ repelling; the extinction state is superstable. At equality the positive point has multiplier one. Sinceit attracts from above and repels from below at equality. This is a semistable fixed point, whose one-sided behavior requires the nonlinear calculation.

One explicit future itinerary is obtained by concatenating all binary words, ordered first by length and then lexicographically:Extend it arbitrarily to negative indices, for example by zeros. Every finite binary block occurs in its forward itinerary, and indeed occurs arbitrarily far out as a subword of longer words. Its shift orbit is therefore dense in the full two-symbol shift space. It is not eventually periodic: arbitrarily long runs of zeros and ones occur, together with all other finite words.

The full shift has sensitive dependence because sequences agreeing on an arbitrarily large initial window can differ just beyond it, and shifting brings that disagreement into view. Thus this itinerary supplies a dense aperiodic orbit in the chaotic symbolic system assumed by the question. A merely nonrepeating sequence would not be enough: restricted aperiodic sequences can have much simpler, nonchaotic orbit closures. Nor does symbolic description alone assign a numerical Lyapunov exponent without information about the underlying map and its coding.

Choose $\boxed{\dots 00000.11111\dots }$, with zeros on the entire left tail and ones on the entire right tail. Forward shifts converge in the symbolic metric to the all-one fixed point, while backward shifts converge to the all-zero fixed point. Its two limiting invariant states differ, so it is a heteroclinic orbit. Reversing the two tails reverses these asymptotic roles.

The sequence $\boxed{\dots 000010000.00000\dots }$ has exactly one $1$ and otherwise zero entries. Under forward shifts that exceptional symbol moves out of every fixed observation window; the same happens under backward shifts in the other direction. Thus both temporal limits are the all-zero fixed sequence, while the orbit itself is nonconstant: it is homoclinic to that fixed point. The bi-infinite coding specifies a chosen full orbit even when the original interval map is not invertible.

Use the primitive repeating word $\boxed{\dots 001001.001001\dots }$. One shift changes the sequence and three shifts return it, so the least period is three. A word such as $000$ repeated would instead have least period one; primitivity is what distinguishes a true three-cycle from a periodic notation for a fixed point.