Actuator saturation means the applied control input reaches a device limit even if the controller requests a larger or smaller command. In the heater interpretation, the applied command is bounded between $0$ and $100\%$. If integral action continues accumulating a persistent tracking error beyond that limit, its stored integral can grow without producing the requested physical correction, delaying recovery when the goal changes. The unconstrained steady-state tracking error formulas require that the necessary command can actually be applied. ^[ref]

A feedback gain sets the strength of a controller's response to its observations. For proportional control, $u=ke$ means that a tracking error $e=0.1$ produces a control input $u=0.2$ when $k=2$.

Quantum controllability concerns the transformations reachable under specified quantum control inputs. Unitary operator controllability concerns entire propagators; density operator controllability concerns conjugation on every spectral orbit; pure-state controllability concerns normalized state vectors or their physical rays. These notions need not coincide.

For $T>0$ and the constant matrices over the real numbers in the Kalman rank criterion, the finite-time Controllability Gramian accumulates the influence of all command times. Its column space equals that of the controllability matrix, and $W_T$ is a symmetric positive-definite matrix exactly when the controllability matrix has rank of a matrix $n$. If so, put $\delta =x_1-e^{AT}{x}_0$. Among square-integrable functions steering $x_0$ to $x_1$ at $T$, the commandminimizes ${\int }_0^T\Vert u(t){\Vert }^{2}dt$, with value ${\delta }^T{W}_T^{-1}\delta$. Here $T$ in a matrix superscript denotes transpose. This cost is a squared input norm, not necessarily physical energy, and the optimizing command need not satisfy an amplitude bound. ^[ref]

For the double integrator, $e^{As}B=(s,1{)}^T$, so direct integration givesFor a rest-to-rest distance $d$, $u_{\ast }(t)=6d/T^2-12dt/T^3$ and the minimum squared input norm is $12d^2/T^3$. Compare the two constant commands in the double integrator example, whose cost is $16d^2/T^3$: both attain the target, but the smooth command costs less. At $T=2$, $d=1$, however, $u_{\ast }$ peaks at $3/2$, so it violates $|u|\le 1$ while the two constant commands remain admissible. Controllability, optimal control, and constrained reachability in control theory answer different questions.

For a finite-dimensional continuous-time linear time-invariant system $\dot{x}=Ax+Bu$ on ${\mathbb{R}}^n$, with constant matrices $A,B$ over the real numbers, no state constraints, and every piecewise continuous function $u:[0,T]\to {\mathbb{R}}^m$ allowed as a control input without an amplitude bound, controllability in any fixed $T>0$ is equivalent to the controllability matrix having rank of a matrix $n$. The same equivalence holds with all square-integrable functions as inputs. A stronger useful statement iswhere $e^{AT}$ is a matrix exponential and $\mathrm{im}\mathcal{C}$ is the column space. Thus a rank of a matrix $r<n$ leaves only an $r$-dimensional affine subspace reachable at that time. ^[ref]

The Kalman rank criterion allows arbitrarily large commands as $T$ shrinks. It does not assert controllability with bounded control inputs, nor test a general nonlinear dynamical system just by counting its inputs. For the double integrator, $B=(0,1{)}^T$ and $AB=(1,0{)}^T$ give two independent directions, despite there being only one control input.

For the linear time-invariant system $\dot{x}=Ax+Bu$, with $A\in {\mathbb{R}}^{n\times n}$ and $B\in {\mathbb{R}}^{n\times m}$, the controllability matrix has $n$ rows and $nm$ columns. The columns of $B$ are directions influenced directly by the control inputs; $AB,A^{2}B,\dots$ account for how the dynamics carry their influence into further directions. The column space of the controllability matrix is the zero-state reachable set for any $T>0$ when inputs are unrestricted. ^[ref]

An internal variable records a material state not determined by instantaneous deformation and temperature alone. Accumulated plastic strain, a relaxing internal strain and an evolving damage parameter are examples. Two specimens at the same overall strain and temperature can therefore carry different stresses because their loading histories leave different internal states.

For an isothermal scalar illustration, let the free energy per unit volume be $\psi (\epsilon ,\alpha )=E(\epsilon -\alpha {)}^2/2$. Set $\sigma ={\partial }_{\epsilon }\psi =E(\epsilon -\alpha )$ and $\dot{\alpha }=\sigma /\eta$ with $\eta >0$. Then the mechanical dissipation is $\sigma \dot{\epsilon }-\dot{\psi }={\sigma }^2/\eta \ge 0$. The hidden strain $\alpha$ evolves even when the total strain is held fixed, so the stress relaxes. This shows why an internal variable needs both an energetic role and an evolution law in a complete constitutive equation.

For the material-first nominal stress tensor, the displayed local conservation of energy law balances mechanical work, heat supply and outgoing heat flux. Integrating over a fixed reference material volume and using the divergence theorem gives its integral form. The full total-energy balance includes kinetic energy and external mechanical work; momentum conservation reduces it to this internal-energy balance.

The Clausius-Duhem inequality is a local expression of the Second law of thermodynamics for a simple continuum whose thermodynamic entropy flux is heat flux divided by positive absolute temperature. With specific entropy $s$, outward heat flux $\mathbf{q}$ and heat supply $r$ per unit mass, it readsCombining this with the local internal energy balance and writing the specific Helmholtz free energy as $\psi =e-Ts$ givesHere $D$ is the rate-of-strain tensor and the dot denotes a material derivative.^[ref]

The inequality constrains constitutive equations, including both mechanical dissipation and thermal conduction. For Fourier's law $\mathbf{q}=-k\nabla T$ with $k\ge 0$, the thermal contribution is $k|\nabla T{|}^2/T\ge 0$. This direct sign check explains why passive thermal conduction carries heat down a temperature gradient. More general continua can require a different thermodynamic entropy flux, so the stated assumption matters.

Thermoelasticity couples elasticity and temperature through a specific Helmholtz free energy $\psi (F,\theta )$. Without constraints its reversible laws are $P_{Ii}={\rho }_0{\psi }_{,F_{iI}}$ and $\eta =-{\psi }_{,\theta }$. Mixed derivatives give the same coupling tensor in the thermal stress response and the strain dependence of specific entropy, with opposite signs.

Let an interface move at signed speed $c$ in a reference unit normal direction $n$, and set $[a]=a^{+}-a^{-}$ with $n$ pointing from minus to plus. For continuous reference density and no surface sources, distributional momentum conservation, conservation of energy and the Lagrangian entropy inequality giveFor a level-set interface $\varphi =0$, ${\varphi }_t=-c|\nabla \varphi |$, the singular part of ${\partial }_{t}a$ is $-c[a]{\delta }_S$ and that of $\mathrm{Div}b$ is $n\cdot [b]{\delta }_S$, proving these formulas. Using $V=-c$ reverses the speed terms. Density jumps require jumps of the complete conserved densities.

State needed to resume a suspended execution stream, including saved processor registers and its position in a call stack. An operating-system kernel can keep parts in a per-task record and parts on a kernel call stack. An execution context identifies and preserves that stream's state; it is not a duplicate of all the computer memory reachable through its address space.

// Target: computer-science.bigb

An omnivore feeds on both primary producers and consumers. Its mixed diet explains why a trophic level need not be a whole number or a fixed label throughout an organism's life.

A decomposer breaks down dead organic material and absorbs products of that breakdown; many bacteria and fungi perform this role. Decomposers connect detritus from different trophic levels to material recycling, so placing all decomposers at one final step of every food chain is misleading. See Clark, Douglas and Choi, Biology 2e, 46.1: grazing and detrital food webs.

For a smooth stationary wire $z=x^2/(4a)$ and release from rest at height $h$, the normal reaction does no work. The energy equation is $[1+x^2/(4a^2)]{\dot{x}}^2/2+g{x}^2/(4a)=gh$. Integrating the travel time between turning points gives the displayed period. Its small-amplitude expansion is $T=2\pi \sqrt{2a/g}[1+h/(4a)+O((h/a{)}^2)]$. The mass cancels because both gravitational and inertial forces are proportional to it.

A constitutive equation is frame indifferent when a superposed rigid spatial rotation changes its outputs by their tensor transformation rules. For an elastic Cauchy stress tensor, this requires $\sigma (QF)=Q\sigma (F)Q^T$. It differs from material isotropy, which rotates reference material directions.

The closure of the displayed connected spiral adds the unit circle. Its union with the closed unit disk is therefore connected but not path-connected. A path from the spiral to the disk would have a first contact time. Before contact its reciprocal radial offset defines a continuous parameter tending to infinity. The intermediate value theorem then forces sequences of path points with parameters integer and half-integer, approaching opposite circle points. This contradicts continuity at the first contact. Thus the spiral and disk form separate path-connected components.

A sequence of connected subsets with each consecutive pair intersecting has connected union. If the union admitted a separation, each connected subset would lie entirely on one side. Consecutive intersections force consecutive subsets onto the same side, hence all subsets onto one side, contradicting that both sides of the separation are nonempty.

Consider the graph of a function $x\mapsto \sin (1/x)$, using the sine of the multiplicative inverse of $x$, together with a vertical closed interval:Give $T$ the subspace topology from the Euclidean plane. The symbol $\times$ denotes a Cartesian product. The right endpoint $(1,\sin 1)\approx (1,0.841471)$ belongs to $G$. The continuous map $x\mapsto (x,\sin (1/x))$ sends the connected real interval $(0,1]$ onto $G$. Hence $G$ is connected, and its closure $\overline{G}=T$ is connected by closure of a connected set. Every height in $[-1,1]$ is approached as $x$ tends to zero; for example,For $n=1$, these inputs are $p_1=2/(5\pi )\approx 0.127324$ and $q_1=2/(7\pi )\approx 0.090946$. For any other height $y\in [-1,1]$, choose $\theta \ge 1$ with $\sin \theta =y$; then the sequence $(1/(\theta +2\pi n),y)$ lies in $G$ and has limit of a sequence $(0,y)$.

Nevertheless, $T$ has exactly two path components, $G$ and $L$. Within $G$, join inputs $a,b\in (0,1]$ by the continuous path $t\mapsto ((1-t)a+tb,\sin (1/((1-t)a+tb)))$. Within $L$, use $t\mapsto (0,(1-t)c+td)$ for heights $c,d\in [-1,1]$.

The obstruction to a continuous path between $L$ and $G$ is persistent oscillation, not a visible gap. Proof outline: suppose such a path exists, directed from $L$ to $G$, and take its last time in $L$. Immediately afterwards its first coordinate $x(t)>0$ approaches zero and its second coordinate is $\sin (1/x(t))$. The intermediate value theorem forces the first coordinate to pass through inputs of both displayed types arbitrarily close to that departure time. The second coordinate therefore takes both values $1$ and $-1$ arbitrarily close to that time, contradicting continuity. ^[ref]

The sampled graph of a function illustrates the formulas; it does not prove connectedness or the failure of path connectedness. Taking a closure preserves connectedness, but this example shows that it can destroy path connectedness.

For the conjugate base ${\mathrm{A}}^{-}$ accepting a proton from water,This base dissociation constant uses the same water thermodynamic activity approximation and standard state conventions as the acid dissociation constant. Multiplying the two relations cancels the partner thermodynamic activities:Here $K_w$ is the aqueous ion product associated with self-ionization of water. At fixed temperature, a larger acid dissociation constant gives a smaller base dissociation constant for its own conjugate base; this reciprocal rule compares matching conjugate acid-base pairs. A weak acid does not necessarily have a conjugate base that reacts almost completely with water. For example, stipulated $K_a=10^{-5}$ and $K_w=10^{-14}$ give $K_b=10^{-9}$.^[ref]

The value of $K_w$ changes with temperature; $10^{-14}$ is the familiar approximate value near $25{}^{\circ }\mathrm{C}$, not a universal constant. If water is retained explicitly in both full equilibrium constants, their product instead equals $a_{{\mathrm{H}}_3{\mathrm{O}}^{+}}{a}_{{\mathrm{OH}}^{-}}/a_{{\mathrm{H}}_2\mathrm{O}}^2$, the full equilibrium constant for $2{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{OH}}^{-}$. The acid dissociation constant, base dissociation constant and self-ionization of water convention must agree.