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For the series-fed parallel LC voltage response, use phasors with time dependence $e^{i\omega t}$. The parallel branch has impedance $iX$, where $X=\omega L/(1-{\omega }^{2}LC)$. ThereforeAt the ideal parallel resonance ${\omega }_0=1/\sqrt{LC}$ its input impedance diverges. The input-current phase changes sign on crossing this pole; expressions there are interpreted by limits.

The capacitor voltage equals the parallel-branch voltage. Define $q=R(1-{\omega }^{2}LC)/(\omega L)$. ThenAt low frequency $H\sim i\omega L/R$; at high frequency $H\sim -i/(\omega RC)$; at resonance $H=1$. Thus it is a band-pass voltage filter or resonant frequency selector. Its phase decreases continuously from $+\pi /2$ through zero to $-\pi /2$. Half-power points satisfy $|q|=1$ and

For the supplied component values and $\omega =2\pi (50){\mathrm{s}}^{-1}$, $X=421.745\Omega$ and $V_{C,\mathrm{pk}}=99.3046\mathrm{V}$. The element peak currents areInductor and capacitor currents are opposite phasors; their magnitudes subtract here rather than add. The resistor carries the input current, withThese results assume ideal components and sinusoidal steady state.

In sinusoidal steady state, complex impedance is the voltage phasor divided by the current phasor. Its modulus is the amplitude ratio and its argument is the voltage lead. With the $e^{i\omega t}$ convention, $I=dq/dt=i\omega CV$, so $Z_C=1/(i\omega C)$.

The PDF labels $V_1$ across $R_1$, not across the complete driven network. If $I$ is the current through $R_1$, then $V_1=I{R}_1$ and $I=Y_p{V}_2$, whereThe resistor-referenced parallel-resonator response isWriting $b(\omega )=\omega C-1/(\omega L)$,The maximum occurs at ${\omega }_0=1/\sqrt{LC}$, where $|H|=R/R_1$ and the phase is zero. Below resonance the phase is positive, tending to $+\pi /2$; above it the phase is negative, tending to $-\pi /2$. The amplitude tends to zero at both extremes: the inductor shorts the parallel network at low frequency, and the capacitor does so at high frequency. At resonance their reactive currents cancel, leaving only the resistor current.

If instead $V_{\mathrm{in}}$ denotes the voltage across the complete circuit, its different transfer ratio is $V_2/V_{\mathrm{in}}=1/[1+R_1/R+i{R}_{1}b]$, with peak $R/(R+R_1)$. This distinction follows directly from the printed voltage endpoints.

For sinusoidal steady state with convention $e^{i\omega t}$, complex impedance is the voltage phasor divided by the current phasor. Its modulus is the amplitude ratio and its argument is the voltage-current phase difference. Since $V_L=LdI/dt$,The original circuit is a series RLC circuit with output across the inductor. Its transfer function isFor the original capacitance, the outputs are $10.15\mathrm{mV}$ at $9\mathrm{MHz}$ and $4.65\mathrm{mV}$ at $10\mathrm{MHz}$, a ratio $0.458$. The current resonance is $f_0=1/(2\pi \sqrt{LC})=8.967\mathrm{MHz}$, so the circuit favors the first transmitter.

The resistor voltage is in phase with the current; the inductor voltage leads it by $\pi /2$ and the capacitor voltage lags by $\pi /2$. Large reactive voltages therefore nearly cancel in their phasor sum. A large inductor voltage is consistent with a small supply voltage: it is inductor voltage magnification, not power gain in a passive circuit.

Reducing the capacitance raises $f_0$ to $9.452\mathrm{MHz}$, between the two transmitters. Direct evaluation gives $7.02\mathrm{mV}$ and $7.22\mathrm{mV}$, respectively. Their similar detunings give approximately equal amplitudes; they are not exactly equal, partly because the numerator of $H$ also depends on frequency.

The two-tone output of a linear resonant circuit, for coherent monochromatic sources of equal input amplitude and arbitrary fixed input phases, isWith $A\simeq B=V_o$, this is approximatelyThus a $9.5\mathrm{MHz}$ rapid oscillation has a beat envelope repeating in magnitude at $1\mathrm{MHz}$. The signed envelope cosine has frequency $0.5\mathrm{MHz}$. Unequal amplitudes leave a small nonzero envelope minimum; independently drifting transmitter phases make the beating drift. No new frequency is generated by this ideal linear circuit.

The moment of inertia about a specified axis is $I=\int s^{2}dm$, where $s$ is perpendicular distance from the axis. The parallel axis theorem states $I=I_{\mathrm{cm}}+m{d}^2$ for two parallel axes separated by $d$, one passing through the centre of mass of a body of mass $m$.

It is easiest to calculate the eccentric disc by subtraction. Let $m_f$ be the mass of the unperforated disc and $m_h$ the removed mass. Uniform thickness and density give $m_h=m_f/4$ and $m_f-m_h=M$, hence $m_f=4M/3$ and $m_h=M/3$. The hole is tangent to the rim, so its centre, and therefore the shaft axis, lies $R/2$ from the original disc centre. About the shaft axis,The shaft contributes $I_{\mathrm{shaft}}=M(R/2{)}^2/2=M{R}^2/8$. ThusThe motor's torque is $P/\omega$ at nonzero angular velocity, while the resisting torque is $-\gamma \omega$. The rotational equation of motion and its equivalent energy balance areAt the steady state, input power equals dissipated power, givingFor the acceleration, use $u={\omega }^2$. Then $\dot{u}+(2\gamma /I)u=2P/I$ with $u(0)=0$, soThe approach to ${\omega }_0$ takes infinite time. This constant-power rotational acceleration idealizes the start: $P/\omega$ diverges at rest, although the energy balance has a well-defined limiting solution. A real motor has finite starting torque.

Here $\mathbf{H}$ is magnetic field intensity, $\mathbf{J}$ free electric current density, and $\mathbf{D}$ electric displacement. In vacuum, $\mathbf{B}={\mu }_0\mathbf{H}$ and $\mathbf{D}={ϵ}_0\mathbf{E}$. Integrating the equation gives Ampère's circuital law with displacement current; in a steady state the displacement term vanishes.

Take $z$ upward and define $V={\varphi }_{\mathrm{upper}}-{\varphi }_{\mathrm{lower}}=V_0\sin \omega t$. Neglect fringing and use the low-frequency thin-capacitor approximation. ThenInside the gap, a circular Ampère loop encloses only displacement current:Because $\hat{\mathbf{z}}\times \hat{\mathbf{\varphi }}=-\hat{\mathbf{r}}$, the Poynting vector isIt points inward while the electrical energy is increasing and outward while it decreases. The net power entering the curved side isThis is the leading quasistatic energy balance. The time-averaged field energies areFor the axisymmetric high-frequency capacitor field when $\omega a/c$ is not small, induction and radial propagation prevent a spatially uniform $E_z$. In an ideal axially uniform axisymmetric interior model, Faraday's and Ampère's equations giveThe solution regular at the axis and normalized to the central voltage is approximatelyAt $\omega =\pi c/a$, the radial profile is $J_0(\pi r/a)$: its first zero is at $r\simeq 0.765a$, and the edge has opposite phase, $J_0(\pi )\simeq -0.304$. This is an illustrative thin-disc interior profile; the exact driven high-frequency field also depends on the feed and edge/radiation boundary conditions. Continuing the uniform quasistatic field would be inconsistent.

For flux-driven convective layer cooling, use $E=C{\theta }^{4/3}$ with the same unspecified coefficient as in (b). The steady heat balance givesThe corresponding boundary-layer thickness isSet $\theta ={\theta }_{\ast }+ϵ$ and expand the heat loss to first order. Since $E^{\prime }({\theta }_{\ast })=(4/3)C{\theta }_{\ast }^{1/3}=4F/(3{\theta }_{\ast })$,This is the near-equilibrium relaxation time. The full equation is nonlinear, so an arbitrary large initial temperature departure need not decay as one exponential. Positive heat loss derivative makes the steady state stable. In particular, $\tau$ scales with $d{\delta }_{\ast }$, rather than $d^2$, because convection shortens the effective heat-transport distance to the boundary layer.

A persistent barrier to whole-mantle circulation adds an internal thermal bottleneck and permits the deep mantle to retain a larger temperature excess. In a single well-mixed layer there is one interior temperature and one surface heat-loss law. The coupled thermal histories of layered mantle convection arise because separately mixed layers can have different temperatures, exchanging heat through an interfacial boundary layer as well as losing heat at the surface. Their thermal histories then have coupled relaxation modes rather than the single mode in (c).

For example, let ${\theta }_l,{\theta }_u$ be lower- and upper-layer temperature excesses, $H_l,H_u$ their heat capacities per area, andWith no internal heating the budgets are $H_l{\dot{\theta }}_l=F-J$ and $H_u{\dot{\theta }}_u=J-E$. At steady state,Thus a fixed basal flux still leaves the surface eventually with $E=F$, but the lower layer is hotter than it would be in the single-temperature model. For a given deep temperature instead, the extra interfacial drop reduces outward heat transfer. This favours delayed loss of deep primordial heat and distinct upper- and lower-mantle cooling histories. A numerical change in cooling time cannot be obtained from layer count alone: interface transport, layer thicknesses, temperature-dependent viscosity, internal heating, and whether material can cross the interface matter. Merely replacing $d$ by $d/2$ in (c) misses the added interface and is not a model of two-layer cooling.

Linearizing at the positive steady state gives $n^{\prime }=f^{\prime }(N^{\ast })n$, so a small displacement decays as $n(t)=n(0)e^{-t/\tau }$. The recovery time of a stable equilibrium is thereforeAs $\delta \uparrow r$ with $r,b$ fixed, $N^{\ast }\to 0$ and $\tau \to \infty$. Thus the equilibrium population approaches extinction and recovers increasingly slowly after small perturbations. This is critical slowing down. Exactly at the threshold the leading equation is $N^{\prime }\sim -rb{N}^3$, which gives algebraic decay $N(t)\sim (2rbt{)}^{-1/2}$ instead of exponential recovery.

Write $f(N)=N[r/(1+b{N}^2)-\delta ]$. A steady state is a nonnegative zero of $f$. If $r<\delta$, the bracket is strictly negative for every $N\ge 0$, so only $N=0$ exists. If $r>\delta$, there are exactly two nonnegative steady states,For an autonomous scalar ordinary differential equation, a steady state with $f^{\prime }<0$ is locally asymptotically stable, while one with $f^{\prime }>0$ is unstable. HereAt the positive steady state, use $1+b(N^{\ast }{)}^2=r/\delta$ to obtainTherefore zero is stable for $r<\delta$ and unstable for $r>\delta$, whereas the positive steady state is stable whenever it exists. At the omitted equality $r=\delta$, $f(N)=-\delta b{N}^3/(1+b{N}^2)$; zero is nonhyperbolic but still attracts every nonnegative initial state.

The nonnegative bifurcation diagram has the branch $N=0$ for every $r>0$, stable for $r<\delta$ and unstable for $r>\delta$. At $r=\delta$ the stable positive branch emerges:The figure in (iv) uses solid curves for stable steady states and a dashed curve for the unstable zero branch. At the meeting point the zero steady state is nonhyperbolic and attracts from the nonnegative side. If the equation is formally extended to negative $N$, there is also the symmetric branch $-N^{\ast }$, giving a supercritical pitchfork bifurcation; its negative half is omitted in the biological diagram. This qualification avoids calling the single nonnegative square-root branch a full pitchfork by itself.

For $r>s$, $a>0$. When $0<a\le 1/4$, the roots of ${\mu }^2-\mu +a$ are real and lie strictly between zero and one. When $a>1/4$, they are a complex-conjugate pair of modulus $\sqrt{a}$. Thus the positive steady state is linearly asymptotically stable exactly when $0<a<1$, and unstable when $a>1$. Since $a=2s-2s^2/r$ increases strictly with $r$, the crossing occurs atAt the crossing,Every nonzero real linearized perturbation has period six. This establishes the intended linear stability threshold and linearized period six.

The nonlinear period-six claim in the printed paper does not follow, and is false as a local bifurcating-branch claim for this map. One can check this directly rather than infer a nonlinear period from the eigenvalues. At $r=r_c$ normalize $N_t=N^{\ast }(1+u)$, $N_{t-1}=N^{\ast }(1+v)$. The exact planar map isIts cubic Taylor expansion isHere is a concrete check of the sixfold return. Put $z=(u,v{)}^T$, $Q=u^2-uv+v^2$, andComposing the displayed cubic expansion six times cancels all quadratic terms and givesIn particularThe quartic coefficient is strictly negative. Hence, sufficiently near the equilibrium, every nonzero point has smaller $Q$ after six iterations: the critical nonlinear map has no nearby nontrivial period-six orbit, and the equilibrium itself is locally attracting despite its neutral linearization.

One can also rule out a branch of small period-six orbits on either side of the crossing. Normalize about the positive equilibrium at each nearby $r$, and put $\eta =a-1$. The derivative of the sixth power of $\left(\begin{array}{cc}1& -a\\ 1& 0\end{array}\right)$ at $a=1$ is $3I+K$. Smooth dependence on $r$ therefore givesThe leading matrix has determinant $(3\eta +\alpha Q{)}^2+3(\eta +\beta Q{)}^2$. Both terms could vanish with $Q>0$ only if $\alpha =3\beta$, butThus its norm on nonzero $z$ is bounded below by a positive constant times $(|\eta |+\Vert z{\Vert }^2)\Vert z\Vert$, and the smaller remainder cannot cancel it. There is no nearby nonzero solution of $F_{\eta }^6(z)=z$. The eigenvalues satisfy the nonresonance conditions through order four, their moduli cross outward with increasing $r$, and the cubic radial coefficient is negative. The standard Neimark–Sacker bifurcation theorem therefore gives a supercritical invariant closed curve. Its rotation is near the linear resonance $1/6$; it is not a local branch of exact period-six orbits.

Finally, the discrete population map does not preserve nonnegativity for arbitrary large lagged populations: $r/(1+b{N}_{t-1}^2)-\delta$ can be negative. The positive-equilibrium stability conclusions above are local and do not assume a global positivity property that the formula lacks.

Let $f(X,Y)=X[r/(1+b{Y}^2)-\delta ]$. At the positive steady state, $1+b(N^{\ast }{)}^2=r/s$, where $s=\delta +1$. The two partial derivatives areThe first-order Taylor expansion with $N_t=N^{\ast }+n_t$ therefore givesEquivalently, the linearization on $(n_t,n_{t-1})$ has matrix $\left(\begin{array}{cc}1& -a\\ 1& 0\end{array}\right)$ and characteristic polynomial ${\mu }^2-\mu +a$.

At a constant steady state $N_t=N$, the discrete equation reduces toWrite $s=\delta +1>1$. For $r<s$, no nonzero real root is possible, so zero is the only real steady state. For $r>s$ the real roots areThus there is exactly one strictly positive steady state, $N^{\ast }=\sqrt{(r-s)/(bs)}$. There are three real steady states in total, two of them nonnegative. At $r=s$ the only one is zero. This distinction keeps the positive-population statement separate from the count over all real numbers.

In the steady state, conservation of probability requires $J_x=0$, so $J=j_0$ is constant. The sign is positive because increasing gap reaches the incorporation boundary, whereas the deterministic drift points towards smaller gap. ThusAn integrating factor $e^{\beta Fx}$ givesSubstitution makes the check particularly transparent:so ${\partial }_{t}p=-J_x=0$. The constant offset is necessary for nonzero transport; a pure Boltzmann distribution would have $J=0$ and could not describe a steady succession of addition events. The $F=0$ limit is taken after applying the boundary conditions, avoiding the apparent singularity of the displayed parametrization.

A uniform steady state satisfies $f(u^{\ast },v^{\ast })=g(u^{\ast },v^{\ast })=0$. Set $u=u^{\ast }+\tilde{u}$, $v=v^{\ast }+\tilde{v}$ and keep only first-order terms in the perturbations. Since the base state is constant, its Laplacian vanishes. Taylor expansion givesThe zero-flux boundary conditions linearize to homogeneous Neumann conditions, ${\partial }_n\tilde{u}={\partial }_n\tilde{v}=0$. The off-diagonal entries of the Jacobian describe the cross-effects of the two chemicals, while the diagonal entries describe local self-response.

Let $J_X={\lim }_{x\downarrow 0}X(x)X^{\prime }(x)>0$. The spatial ordinary differential equation gives $(X{X}^{\prime }{)}^{\prime }=-\lambda X$, so its leading outlet behavior isAgain this is a square-root drainage boundary. For the separable aquifer drawdown, the dimensional drainage flux isEquivalently, with ${\tau }_d=\varphi /(K\lambda )$, it is $Q_0(t^{\ast })[1+(t-t^{\ast })/{\tau }_d{]}^{-2}$ for an exact separated profile. Integrating the spatial ordinary differential equation gives $J_X=\lambda {\int }_0^{L}Xdx$. This verifies the global mass conservation checkThus the late drainage flux follows an inverse-square power law, while stored water decays inversely with time.

The aquifer drainage hydrograph starts at zero, rises linearly in the early rainfall similarity solution, and approaches the total recharge $RL$ if rainfall lasts long enough for a finite-length steady state. Indeed, steady mass conservation gives $q_x=R$ and $q(L)=0$, hence $Q_0=RL$. At $t^{\ast }$, the outlet flux stays continuous for a regular inherited flux profile, then recedes as water drains; the eventual decay is proportional to $(t-t^{\ast }+\text{constant}{)}^{-2}$. An arbitrary transition requires the full time-dependent partial differential equation, so matching an early rainfall formula directly to a separated drought formula is only a schematic approximation.

The curve illustrates these regimes and a possible $t^{\ast }$. Its numerical shape and the relative duration of the regimes are schematic; the early rainfall and late drought formulas apply in their respective asymptotic ranges.

The null space consists of concentration patterns on which transport produces no change. Here it is the familyEvery box has the same concentration. Ocean circulation may continue, but inflow and outflow of the chemical tracer balance in each box. With $\mathbf{b}=0$ these are exactly the steady states; physically admissible members have $\alpha \ge 0$. With nonzero $\mathbf{b}$, adding $\alpha 1$ to any particular steady state leaves its transport balance unchanged, but a null space vector alone need not balance the sources and sinks.

To identify the tracer inventory, apply conserved inventory from a left null vector. Directly summing each column givesConsequently, for the supplied ordinary differential equations,For $\mathbf{b}=0$, $M$ is a conserved quantity. The connected box model equalizes concentrations, approaching the unique member of the null space with the initial inventory:The arithmetic mean is appropriate here because the supplied matrix has zero column sums. In a general box model the conserved quantity is $w^T\mathbf{c}$, where $w$ is a positive left null space vector; for physical concentrations expressed per unit volume, these weights represent the box volumes up to a common factor. One must establish the weights from the equations, rather than infer equal volumes from a schematic drawing.

Nonnegative initial concentrations remain nonnegative under transport alone. Indeed $A+2I$ is a nonnegative matrix, sohas nonnegative entries for $t\ge 0$: it is a positive matrix semigroup. Its column sums are one by $1^{T}A=0$. To see why equalization occurs, set $d=\mathbf{c}-(M/4)1$ for $\mathbf{b}=0$; then $1^{T}d=0$ andThe right side vanishes only for constant $d$, which in the zero-sum vector subspace means $d=0$. Its strict negative bound on the unit sphere in that vector subspace gives decay to zero. Thus the matrix exponential redistributes inventory without creating negative concentrations; arbitrary prescribed sinks can still exhaust the tracer if they are allowed to act indefinitely.

A steady state ${\mathbf{c}}_{\ast }$ satisfies $0=A{\mathbf{c}}_{\ast }-\mathbf{b}$, henceThe sign is positive on the right after rearranging the supplied equation. A solution exists exactly when $\mathbf{b}$ lies in the column space of $A$, which above was found to be the zero-sum vector subspace. Equivalently, multiply by the left null space vector $1^T$:This condition is necessary and sufficient for an algebraic steady state, since both the column space and that zero-sum vector subspace have dimension $3$. It expresses global conservation of mass: in the closed box model, total tracer production must balance total tracer removal. With the paper's convention $-\mathbf{b}$, a positive $b_i$ is a sink and a negative $b_i$ is a source. If the sum is nonzero, $dM/dt=-{\sum }_i{b}_i$ and the inventory changes linearly in time, preventing a steady state.

When the condition holds, all algebraic steady states form the affine solution space of a linear equation ${\mathbf{c}}_p+\alpha 1$, for any particular solution $A{\mathbf{c}}_p=\mathbf{b}$. The initial inventory selects one member through $1^T{\mathbf{c}}_{\ast }=1^T\mathbf{c}(0)$. For a physically admissible solution, its four concentrations must additionally be nonnegative; the compatibility condition alone imposes no such inequalities.

A P-body, or processing body, is a cytoplasmic assembly of RNA and proteins associated with nontranslating messenger RNA and factors involved in decapping and RNA decay. Its molecular composition and exchange with the surrounding cytoplasm distinguish it from a membrane-bounded container or a fixed pile of cellular waste.

The 2003 yeast study localized decay-related factors and RNA degradation intermediates to these foci, and altered RNA-turnover steps to test their functional relationship to processing.^[ref] This connects cellular location to a biochemical pathway; it does not establish that every RNA must pass through a visible focus to be degraded.

A 2005 study tested P-body assembly more directly. RNA digestion disrupted enriched P-body preparations. In cells, inhibition of translation initiation increased P-bodies, whereas inhibition of elongation trapped messages in polysomes and promoted P-body dissociation.^[ref] The opposite outcomes show why saying only that translation was inhibited leaves an important mechanistic distinction unresolved.

An original pool model explains why a larger focus need not mean faster degradation. Let $N$ count RNA molecules in a processing pool, with entry rate $J$ and first-order removal rate $kN$:With stipulated $J=6$ molecules per minute and $k=0.3\mathrm{mi}{\mathrm{n}}^{-1}$, the steady state contains 20 molecules. Lowering $k$ to $0.1\mathrm{mi}{\mathrm{n}}^{-1}$ increases that pool to 60, while the steady removal flux remains $kN=6$ molecules per minute. Even if removal represents degradation, pool size and throughput are different measurements. This is a teaching model, not measured P-body kinetics; actual pools can also exchange RNA through other routes.

Fluorescence microscopy measures labeled components under a stated imaging procedure. Interpreting intensity as RNA number needs an additional response model, and counting spots need not measure either molecular number or reaction flux. Localization, composition, assembly and turnover consequently provide complementary tests of what a P-body is doing.

A signalling output returns to its original steady state after a persistent change of input. This concerns the final output, not absence of an initial response or invariance of the internal state. Robust perfect adaptation additionally requires that the return property persist over a range of parameters.