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The sedimentation coefficient $s$ is radial drift velocity divided by local centrifugal acceleration. It has units of time and combines particle properties with those of the surrounding liquid. In a dilute, deterministic mathematical model with negligible particle inertia, an effective displaced volume $V$, particle mass $m$, local liquid density ${\rho }_f$ and linear friction coefficient $f>0$ giveThe subtraction represents buoyancy. Equal mass does not imply equal $s$: changing shape changes drag force, and changing the medium changes both buoyancy and viscosity.^[ref]

For an isolated sphere of radius $b$ and uniform density ${\rho }_p$, Stokes's law supplies $f=6\pi \eta b$, where $\eta$ is dynamic viscosity, and $m=(4\pi /3)b^3{\rho }_p$. ThereforeThis original force-balance check predicts fourfold $s$ on doubling $b$ with everything else fixed, and zero drift when ${\rho }_p={\rho }_f$. It applies to dilute isolated spheres in Stokes flow, not arbitrary folded proteins or crowded suspensions. If $s$ and $\omega$ remain constant in a uniform medium, the ordinary differential equation $dr/dt=s{\omega }^{2}r$ has the exponential function solution $r(t)=r(0)e^{s{\omega }^{2}t}$. Arrival at an outer boundary therefore depends on starting position as well as $s$; a separation has no universal size cutoff.

A first-order ordinary differential equation is separable if its explicit form factors as $y^{\prime }=a(t)b(y)$. On an interval where $b(y)\ne 0$, write $dy/b(y)=a(t)dt$ and take antiderivatives to obtain an implicit solution:An initial condition determines $C$ and selects a branch and domain. The useful simplification is that the two integrals each involve only one variable, even when the original equation is nonlinear.

For Newton's law of cooling, division by $y-T_a$ gives $dy/(y-T_a)=-kdt$. To keep the logarithm dimensionless, integrate relative to the initial excess temperature:When $y_0\ne T_a$, continuity preserves the sign of the excess temperature until any potential zero, giving $y-T_a=(y_0-T_a)e^{-k(t-t_0)}$. This formula never reaches zero at finite time. Taking its derivative verifies the ordinary differential equation, and substitution at $t_0$ verifies the initial condition.

Always check what division removed. Each zero $b(y_{\ast })=0$ gives a constant solution of a differential equation $y=y_{\ast }$. Cooling therefore also has the equilibrium point $y=T_a$, omitted by division but included in the final explicit formula. For the nonunique initial value problem $y^{\prime }=\sqrt{|y|}$, division by $\sqrt{|y|}$ only describes the moving pieces: a valid solution may remain zero for an arbitrary time before joining one. Separation is a method for finding solutions, not a uniqueness theorem.

A solution to a partial differential equation that factors into functions of separate variables, for example $u(x,t)=X(x)T(t)$. For the heat equation $u_t=D{u}_{xx}$, substitution yields $T^{\prime }/(DT)=X^{\prime \prime }/X$, so both sides are a common constant wherever division is valid. The resulting ordinary differential equations must be solved together with the original boundary conditions. Individual separated solutions generate more general solutions by superposition when the equation and boundary data are linear.

A solution of a partial differential equation expressed as a product of functions of separate groups of variables. Substitution reduces many linear partial differential equations to ordinary differential equations; superposition then combines the resulting normal modes.

Signaling adaptation is a decline of a response toward its earlier level while the initiating signal remains present. Specify which response is measured and its baseline: complete signaling adaptation means eventual return to that baseline; partial signaling adaptation leaves a lasting displacement. The continuing signal distinguishes signaling adaptation from simply switching off the stimulus. Signaling adaptation describes behavior, while signaling feedback describes a causal arrangement, so neither implies the other.

By reducing a persistent response to background ligand concentration, signaling adaptation can make a biological cell sensitive to changes rather than only to absolute levels. Resetting the response helps leave room for another change over a useful range of ligand concentrations. This is valuable in chemotaxis: swimming bacteria encounter changing chemical conditions, and their signal transduction influences whether a run continues or a tumble occurs. Signaling adaptation helps separate a change encountered during motion from a lasting background. This usefulness does not imply exact return or equal sensitivity at every possible concentration. ^{[ref] [ref]}

Here is a solvable incoherent feedforward loop toy that exhibits complete signaling adaptation without signaling feedback. The input $s(t)$ jumps from $0$ to $1$ at $t=0$ and remains $1$, a Heaviside step function. Let $x$ be a fast activating branch, $a$ a slower inhibitory branch, and $r=x-a$ a readout. The activities $s,x,a,r$ are dimensionless variables; physical time is in seconds. Choose time constants ${\tau }_x=1$ second and ${\tau }_a=5$ seconds, with initial conditions $x(0)=a(0)=0$:Both branches read $s$ directly. Neither branch depends on $r$, and $a$ does not change $x$: the subtraction only defines the readout. Thus the slow opposing influence is feedforward, with no closed feedback control loop.

For $t\ge 0$, solving the two ordinary differential equations givesWriting $t$ for the numerical time in seconds, these are $x=1-e^{-t}$, $a=1-e^{-t/5}$ and $r=e^{-t/5}-e^{-t}$. Initially $r=0$, then $x$ rises faster than $a$, producing a pulse. At long times $x$ and $a$ both approach $1$, so $r$ returns to $0$ despite sustained $s=1$. For this upward step $e^{-t/5}\ge e^{-t}$, hence $r(t)\ge 0$. The striking feature of this signaling adaptation toy is that both branches remain active while the readout becomes quiet: the continuing background and the response to its arrival are separable.

Find the pulse maximum by differentiating the exponential functions. With numerical time in seconds, the derivative isThe stationary condition is $e^{-4t/5}=1/5$, so the natural logarithm givesBefore this time the derivative is positive; afterwards it is negative. Together with $r(0)=0$ and ${\lim }_{t\to \infty }r(t)=0$, this proves that it is the unique pulse maximum. As another numerical check, at $t=10$ seconds $x\simeq 0.999955$, $a\simeq 0.864665$ and $r\simeq 0.135290$: the slow branch is still catching up.

The subtraction $r=x-a$ is a mathematical readout, not a claim that an actual chemical concentration equals this difference for arbitrary inputs. Nonnegativity was established only for the specified upward step; other input histories can give a negative signed readout. Complete signaling adaptation here also relies on equal eventual branch gains: changing the readout to $r=x-0.9a$ leaves $r_{\ast }=0.1$ for $s=1$. This toy illustrates cancellation and different time constants; it does not establish molecular implementation or robustness to arbitrary parameter changes. Ma and colleagues' 2009 study separately analyzed biochemical mathematical models with negative feedback and incoherent feedforward loops, identifying additional conditions for signaling adaptation in those models. ^[ref]

A historical connection to feedback control comes from Yi, Huang, Simon and Doyle's 2000 analysis of a bacterial chemotaxis model. They identified integral action in its regulation of receptor activity: accumulated deviations alter a slower adjustment, which in turn changes the measured activity. At a stationary state, an adjustment that integrates a nonzero error cannot remain constant, explaining how zero error can follow. Their conclusion depended on stated biochemical assumptions and stable behavior; it was not a claim that every biological cell uses that mechanism. A transient pulse alone, as the incoherent feedforward loop toy shows, cannot identify the underlying signaling feedback architecture. ^[ref]

Signaling feedback returns the effect of a downstream event to an earlier stage of signal transduction, closing a loop analogous to feedback control. Trace the causal arrows: if an earlier relay activates a later relay and the later relay changes the earlier relay's activity, the return path closes the loop. Negative feedback opposes the initiating change; positive feedback reinforces it. Signaling feedback is a possible property of signal transduction, not a requirement of every pathway.

For a schematic example, let relay $A$ activate relay $B$. If $B$ inhibits $A$, increased $A$ eventually produces an opposing influence: negative feedback. If $B$ activates $A$, increased $A$ produces a reinforcing influence: positive feedback. These signs describe effects on activity, rather than the direction in which molecules travel. A closed loop containing two inhibitions can have positive feedback, because inhibiting an inhibitor reinforces the original change. The sign belongs to the whole loop.

Negative feedback can restrain excessive activation and help preserve a usable range of responses. Positive feedback can support a decisive transition: Alberts and colleagues describe an enzyme whose product increases its own activity. Such reinforcement can contribute to switchlike behavior, but a positive feedback arrow alone does not prove bistability. A different structure, an incoherent feedforward loop, sends opposing influences forward from the same input; inhibition there is not a downstream return. ^{[ref] [ref]}

A worked mathematical model shows why negative feedback does not automatically produce complete signaling adaptation. Let $s$ be a fixed normalized input, $r$ a normalized response, $g\ge 0$ a dimensionless feedback gain, and $\tau >0$ a time constant. Suppose response removal contributes $-r$ and a return proportional to the response contributes $-gr$. The ordinary differential equation isSetting the derivative to zero gives the equilibrium point $r_{\ast }=s/(1+g)$. A displacement $\delta r=r-r_{\ast }$ obeys $\tau d(\delta r)/dt=-(1+g)\delta r$, so its decaying exponential function establishes asymptotic stability. For a unit input switched on at $t=0$, initial condition $r(0)=0$, $g=2$ and $\tau =1$ second,The negative feedback reduces the eventual response from $1$ to $1/3$, but this response rises monotonically and never returns to its zero baseline. Attenuation and complete signaling adaptation are different properties. This deliberately simple mathematical model specifies no molecular pathway; integral action provides a contrasting way to remove a lasting offset in a stable feedback control system. ^[ref]

A similarity ansatz assumes a profile invariant under a prescribed rescaling, commonly in the displayed power-law form. Substitution into a partial differential equation fixes compatible exponents and can reduce it to an ordinary differential equation for $F$. Boundary conditions and conserved quantities must also be compatible with the scaling; the ansatz alone does not establish existence of a solution.

A singular perturbation is a parameter-dependent problem whose limiting problem changes qualitatively as the parameter tends to zero. In an ordinary differential equation, a small coefficient of the highest derivative can reduce the order of the limiting differential equation, preventing it from satisfying all the original boundary conditions. Different length or time scales often require different asymptotic expansions.^[ref]

Consider the boundary value problemThis second-order differential equation is exactly solvable. Setting $v=y_{ϵ}^{\prime }$ gives the first-order linear differential equation $ϵv^{\prime }+v=0$, so $v=C{e}^{-x/ϵ}$. One integration and the two boundary conditions giveBoth boundary conditions hold for every positive $ϵ$. For numerical evaluation, write the same ratio as $expm1(-x/ϵ)/expm1(-1/ϵ)$, where $expm1(t)$ evaluates the exponential function minus one without catastrophic cancellation near $t=0$. Keep the denominator for finite $ϵ$: at $ϵ=0.2$ it is approximately $0.993262$, rather than exactly one.

Setting $ϵ=0$ gives $y_0^{\prime }=0$. The outer solution selected by the right boundary condition is therefore $y_0=1$, which fails the left boundary condition. The exact exponential function $e^{-x/ϵ}$ shows how that mismatch is repaired: the solution rises from zero to almost one within a boundary layer of width $x=O(ϵ)$, using Big O notation. For each fixed $x>0$, $y_{ϵ}(x)\to 1$, while $y_{ϵ}(0)=0$ always. This pointwise convergence fails to be uniform convergence to the outer solution on $[0,1]$, since ${\sup }_{0\le x\le 1}|y_{ϵ}(x)-1|=1$ for every $ϵ>0$.

The coefficient $ϵ$ is small, but its product with the second derivative need not be small. Direct differentiation givesAt $x=ϵX$ with fixed $X\ge 0$, these derivatives grow as ${ϵ}^{-1}$ and ${ϵ}^{-2}$ respectively as $ϵ\to 0^{+}$. Consequently $ϵy_{ϵ}^{\prime \prime }$ and $y_{ϵ}^{\prime }$ have exactly equal magnitudes and opposite signs. Even away from the boundary layer, their magnitudes remain equal in this example; both become exponentially small at each fixed positive $x$. The outer expansion neglects an exponentially small transient there, rather than establishing that one derivative term is smaller than the other everywhere.

To predict the width without knowing the exact solution, suppose an order-one change occurs over distance $\delta$. The derivative scales are $y^{\prime }=O({\delta }^{-1})$ and $y^{\prime \prime }=O({\delta }^{-2})$. The dominant balance $ϵ/{\delta }^2\sim 1/\delta$ gives $\delta \sim ϵ$. The resulting inner variable $X=x/ϵ$ makes the boundary layer visible at a fixed scale. This is the useful feature of singular perturbation: an apparently disappearing term identifies the missing structure when one measures its derivatives on the appropriate scale.

For frequency-dependent transmission, the closed SIR model is a system of ordinary differential equations for the fractions $s,i,r$:Here $\beta >0$ is the coefficient of frequency-dependent transmission, and $\gamma >0$ is the recovery rate, interpreted more generally as the removal rate. Both have units of inverse time. With a constant recovery rate, the associated individual time in the infectious compartment has an exponential distribution with mean $1/\gamma$; the SIR model does not require every person to spend exactly that long there. The fractions themselves have no units.^[ref]

For counts $S=Ns$, $I=Ni$, $R=Nr$, the same SIR model equations readThe factor $1/N$ keeps the same $\beta$ in both descriptions. Dropping it while keeping these units would fail dimensional analysis. Some existing SIR model variants use the unnormalized mass-action infection convention instead, including the SIR susceptible-recovered identity; their coefficient is $b=\beta /N$ in the notation used here.

Adding the SIR model equations gives $d(s+i+r)/dt=0$, a short proof of conservation of population. Nonnegative starting fractions stay nonnegative: the outflow from either the susceptible compartment or the infectious compartment vanishes when that group is empty, while $dr/dt\ge 0$. Thus the susceptible compartment can only shrink and the removed compartment can only grow.

For a concrete initial value problem, stipulateAt the start, the flow into the infectious compartment is $0.3\times 0.99\times 0.01=0.00297$ of the total per day, and the flow into the removed compartment is $0.1\times 0.01=0.00100$ of the total per day. Hence $di/dt=0.00197$ per day. If $N=1000$, these correspond to model rates of $2.97$ and $1.00$ people per day, and net growth $1.97$ people per day. A rate need not be an integer count of events on any particular day.

For this SIR model, the basic reproduction number isIt describes one introduced member of the infectious compartment in an otherwise wholly susceptible compartment. The instantaneous reproduction number at the actual starting state is $R_0{s}_0=2.97$, obtained by freezing the current conditions and the current susceptible compartment's fraction. More generally it is $R_{0}s(t)$ in this SIR model. The basic reproduction number is a ratio of two rates, so it has no units; $r_0$ is a different quantity, the initial removed compartment's fraction.

While $i>0$, its instantaneous relative growth isInitially this equals $0.3\times 0.99-0.1=0.197{\mathrm{day}}^{-1}$. Freezing $s\approx s_0$ gives the early exponential growth approximation $i(t)\approx 0.01e^{0.197t}$ with $t$ measured in days. It must fail as the susceptible compartment shrinks. An epidemic invasion threshold occurs at $R_0{s}_0=1$; if $i_0=0$, the deterministic SIR model stays at $i=0$ even above that threshold.

A classical solution is a sufficiently differentiable function that satisfies a differential equation and any prescribed initial or boundary conditions. A weak solution instead satisfies an appropriate integral or distributional identity in a specified function space. For a smooth locally Lipschitz continuous vector field, an ordinary differential equation has a unique local solution for each initial state; this does not by itself ensure global existence or boundedness.

A stationary solution of an evolution differential equation is independent of time. Substituting ${\partial }_{t}u=0$ leaves the spatial boundary value problem that determines its profile. For an ordinary differential equation this is an equilibrium; for a partial differential equation it may vary in space. A stationary profile need not have zero flux, although zero flux is forced in a one-dimensional conserved system with reflecting endpoints.

The steady-state tracking error iswhen this finite limit exists. A finite steady-state tracking error does not require the reference signal, the plant output, or the control input to become constant: all can grow while their difference settles. If the tracking error grows without bound or keeps oscillating, report that behavior rather than assigning a finite steady-state tracking error. Eventual accuracy and asymptotic stability are different questions.

For the first-order low-pass filter $\tau \dot{y}+y=Ku+d$ introduced in reference signal, assume exact measurement, constant control disturbance $d$, and no actuator saturation or feedback delay. Under proportional control $u=k_p(r-y)$ with $k_p>0$, a constant reference signal $r=R$ givesThe decaying homogeneous solution has rate $(1+K{k}_p)/\tau >0$, so this equilibrium point is approached from any finite initial condition. A nonzero tracking error is needed to maintain the control input that balances the model's tendency to relax toward $y=d$. This differs from the accumulating controlled plant $\dot{x}=u+w$, for which proportional control already tracks a constant reference signal exactly when $w=0$. The presence or absence of a step offset depends on both the controlled plant and the controller.

For a ramp reference signal $r=vt$, the same proportional control givesIts long-time particular solution grows with slope $v/(1+K{k}_p)$ when $v\ne 0$. Thus the controlled plant and homogeneous closed-loop response are stable, but the tracking error has no finite steady-state tracking error. A changing goal requires a different performance test from a fixed destination.

Now use proportional-integral control $u=k_{p}e+k_{i}z$, $\dot{z}=e$, with $k_i>0$. For the same controlled plant, differentiate the error equation on intervals where the inputs are smooth:The characteristic polynomial $\tau {\lambda }^2+(1+K{k}_p)\lambda +K{k}_i$ has both roots of a polynomial in the open left half-plane: real roots have negative sum and positive product, and a nonreal complex conjugate pair has negative real part. This proves decay of the homogeneous error for this model. It does not prove stability of an arbitrary controlled plant controlled by positive gains. At a reference jump, use the original ordinary differential equations to set the new initial conditions, rather than treating the reference as smooth at that jump.

With a constant reference signal and constant control disturbance, proportional-integral control therefore gives $e_{\mathrm{ss}}=0$. At its equilibrium point, $y_{\ast }=R$, $u_{\ast }=(R-d)/K$, and $z_{\ast }=(R-d)/(K{k}_i)$. The stored integral maintains the required command while the current tracking error vanishes. For a ramp reference signal, however, the forcing above is $v$, andThe tracking error settles even though the accumulated integral and the required control input grow. This finite offset is the error needed to keep increasing that command. For $r=a{t}^2/2$ with $a\ne 0$, the forcing is $\tau a+at$ and even this proportional-integral control leaves a tracking error growing with slope $a/(K{k}_i)$. Integral action is therefore a precise mechanism for selected input classes, not a promise of perfect tracking for every reference signal. ^[ref]

Here are completely explicit synthetic examples with $y(0)=z(0)=0$, $d=0$, and no sensor bias. Choose $\tau =2\mathrm{s}$, $K=1{}^{\circ }\mathrm{C}$ per percentage point, $k_p=1$ percentage point per degree, and $k_i=0.5$ percentage points per degree-second. Thus $K{k}_p=1$ is dimensionless, and $K{k}_i=0.5{\mathrm{s}}^{-1}$. In the following formulas, $t$ denotes the numerical time in seconds and $y,e,r$ denote numerical temperature rises in degrees. For the unit-step reference signal $r=1$, proportional control and proportional-integral control respectively giveFor the ramp reference signal $r=0.2t$, the corresponding solutions areThe control inputs here are numerical heater-command percentage points. Substitution into $2\dot{y}+y=u$ and into the controller equations checks these solutions without invoking an unverified limiting rule. The proportional control step offset is $0.5$ degrees; the proportional-integral control step offset is zero. The proportional control ramp error grows; the proportional-integral control ramp offset is $0.4$ degrees, exactly $v/(K{k}_i)$.

The conclusions have specific limits. With constant sensor bias $b$, stable proportional-integral control for a feasible constant reference signal drives the measured tracking error to zero but leaves the physical discrepancy ${\epsilon }_{\mathrm{ss}}=b$. Under proportional control, the measured offset instead becomes $(R-b-d)/(1+K{k}_p)$. If actuator saturation prevents the required control input, zero step offset is not achievable by these formulas. A changing control disturbance contributes $-\dot{d}$ to the error equation, while feedback delay can change stability; neither is covered by the constant-disturbance result. These restrictions are what make steady-state tracking error useful: it identifies the input, measurement convention, and model for which an accuracy claim is actually justified.

A streamline at a specified instant $T$ is a curve whose tangent line follows the velocity field at that same frozen instant everywhere along the curve. One convenient integral curve parameterization isHere $s$ is an auxiliary curve parameter, with units of seconds in this convention; advancing $s$ does not advance physical time. Any regular reparameterization preserves the tangent geometry. Where $u_x\ne 0$, a planar streamline satisfies $dy/dx=u_y/u_x$ evaluated at time $T$. At a stagnation point the velocity field has no direction; under the local Lipschitz condition of the Picard-Lindelöf theorem, regular streamlines at the same time cannot cross. A streamline answers which directions fit together now, making it useful for describing local flow geometry. ^[ref]

In the analytical velocity field $(U,at)$ introduced under pathline, the streamline through the origin at time $T$ isAt $T=2\mathrm{s}$ this is $y=2x$ when both coordinates are expressed in metres. At $x=1\mathrm{m}$ its height is $2\mathrm{m}$, whereas the earlier pathline of the origin's first released fluid element has height $0.5\mathrm{m}$. Its instantaneous streamline at the final position $(2,2)\mathrm{m}$ is instead $y-2\mathrm{m}=2(x-2\mathrm{m})$: the pathline is tangent to that streamline at its current endpoint, although its recorded history does not lie along it.

The green streamline uses the velocity field at $T=2\mathrm{s}$ throughout; the orange pathline uses times $0\le t\le 2\mathrm{s}$; the purple streakline uses different release times at the single observation time $T=2\mathrm{s}$. The grey arrows all show the instantaneous velocity direction at $T$. These are exact analytical curves with chosen units, not measured dye shapes.

In a steady flow in a fixed reference frame, a pathline and a streamline solve the same autonomous ordinary differential equation. Uniqueness makes each particle released from a fixed point follow the same integral curve, so its pathline, the streakline from that point and the streamline share one geometrical curve. Finite recording or release intervals usually show different segments of that curve; the comparison assumes particles stay in the domain and treats stationary fluid elements separately. Neither two-dimensionality nor incompressibility is needed for this coincidence. Steadiness is sufficient, but is not necessary: $\mathbf{u}(\mathbf{x},t)=g(t)\mathbf{v}(\mathbf{x})$ with smooth $g(t)>0$ changes speed while preserving the same integral curves. In a general time-dependent velocity field, however, the three curves differ. Kline explains why continuous dye release in steady flow can therefore display streamlines. ^[ref]

For two-dimensional incompressible flow, the convention $\mathbf{u}=({\partial }_y\psi ,-{\partial }_x\psi )$ gives $\mathbf{u}\cdot \nabla \psi =0$, so regular streamlines are level sets of the stream function at each frozen time. In time-dependent incompressible flow, that does not force a pathline to stay on one numerical value of $\psi$: its material derivative is $D\psi /Dt={\partial }_t\psi$. The distinction is the difference between a geometric snapshot and a parcel's history.

For a fluid occupying $z<0$, match an outer stagnation point flow $u=x,w=-z$ to a flat impermeable interface with dimensionless shear stress $\beta x$. With boundary layer thickness $\delta ={\mathrm{Re}}^{-1/2}$, inner variable $\zeta =z/\delta$ and $\gamma =\delta \beta$, the leading boundary-layer equation admits $u=x{f}^{\prime }(\zeta )$ and $w=-\delta f(\zeta )$. Incompressibility is automatic, and tangential momentum conservation gives the displayed ordinary differential equation, with $f(0)=0$, $f^{\prime \prime }(0)=\gamma$ and $f^{\prime }\to 1$ at $\zeta \to -\infty$. The sign of this far-field endpoint is fixed by the fluid domain. This is a prescribed-traction interface; it does not impose a no-slip boundary condition.

Neglecting the one in a strongly repressed Hill equation gives $\dot{Y}=\beta K^n{Y}^{-n}-\alpha Y$. The substitution $z=Y^{n+1}$ makes this a linear ordinary differential equation. It yields $Y_{\mathrm{st}}=(\beta K^n/\alpha {)}^{1/(n+1)}$ and half-response time $-\log (1-2^{-(n+1)})/[(n+1)\alpha ]$. The singular initial slope is an artifact of the approximation, whose range is not uniform at arbitrarily large $n$ and fixed production/degradation ratio.

For smooth bounded vector fields, the support of a Stratonovich diffusion in uniform path topology is the closure of solutions to the controlled ordinary differential equation with the same drift and controls in the Cameron-Martin space of Wiener measure. The support of enhanced Brownian motion and the universal limit theorem prove this through continuity of the enhanced solution map. For an equation written in Itô form, the controlled drift is the Itô drift minus $\frac{1}{2}{\sum }_{i}D{V}_i{V}_i$.

A synaptic potential is a postsynaptic change in membrane potential caused by activity at a synapse. Unlike the regenerative action potential, a passive synaptic potential is graded: its size depends on input strength, the initial membrane potential and the receiving cell membrane's properties. Local voltage changes can decay before reaching the soma; their timing and overlap help determine whether an action potential follows. See Ion Channels and the Electrical Properties of Membranes.

Depolarization means that the interior becomes less negative relative to the exterior; hyperpolarization means that it becomes more negative. Excitation and inhibition instead describe whether an input promotes or opposes postsynaptic action potential generation. An inhibitory postsynaptic potential can be depolarizing, and shunting inhibition can occur with no voltage change at rest. The receptor, permeating ions, ionic gradients and reversal potential matter. See Purves and colleagues: excitation, inhibition and depolarizing inhibition.

A worked passive model treats the receiving membrane as one compartment with capacitance $C>0$, leak conductance $g_L>0$, and activated synaptic conductances $g_e,g_i\ge 0$:All electric currents are inward-positive. Each term $g(E-V)$ vanishes at its reversal potential $E$; $\mathrm{nS}\times \mathrm{mV}=\mathrm{pA}$. With constant parameters, the ordinary differential equation hasHere $G$ is total conductance, $1/G$ is effective electrical resistance, $V_{\infty }$ is the steady state and $\tau$ the time constant. Without imposed electric current, $V_{\infty }$ is a conductance-weighted average of the reversal potentials. This approximation omits spatial structure, changing ionic gradients and active voltage-gated ion channels; it calculates passive voltage responses, not actual action potential waveforms. See 3.1 Synapses for the mechanism behind the terms; the inward-positive convention here reverses that source's synaptic-current sign.

Take the same illustrative leak and capacitance as under shunting inhibition, start at $V(0)=-65\mathrm{mV}$ and set $I_{\mathrm{ext}}=0$. Opening $g_e=2\mathrm{nS}$ with $E_e=0\mathrm{mV}$ gives $V_{\infty }=-650/12\simeq -54.167\mathrm{mV}$ and $\tau \simeq 16.667\mathrm{ms}$. Its initial inward electric current is $2(0+65)=130\mathrm{pA}$. Adding $g_i=20\mathrm{nS}$ with $E_i=-65\mathrm{mV}$ changes $V_{\infty }$ to $(-650-1300)/32=-60.9375\mathrm{mV}$ and $\tau =6.25\mathrm{ms}$. This is shunting inhibition of the same excitatory drive.

For comparison, $g_i=20\mathrm{nS}$ with $E_i=-75\mathrm{mV}$ alone gives $V_{\infty }\simeq -71.667\mathrm{mV}$: hyperpolarization. With $E_i=-60\mathrm{mV}$ instead, it alone gives $V_{\infty }\simeq -61.667\mathrm{mV}$: depolarization. Yet adding this depolarizing conductance to the $g_e=2\mathrm{nS}$ excitatory drive lowers that drive's $V_{\infty }$ from $-54.167$ to $(-650-1200)/32=-57.8125\mathrm{mV}$. At an illustrative firing threshold $V=-50\mathrm{mV}$, its electric current is $20(-60+50)=-200\mathrm{pA}$, opposing threshold approach. The combined trace initially rises faster than the excitatory drive alone, before settling below it. Thus the sign of a voltage change at rest alone does not determine excitation or inhibition; a claim about firing must specify input strength, timing and the neuron's state.

A system of ordinary differential equations specifies derivatives of several unknown functions of one independent variable. A vector-valued form ${\mathbf{y}}^{\prime }=\mathbf{f}(t,\mathbf{y})$ bundles its scalar equations. A scalar second-order ordinary differential equation $y^{\prime \prime }=F(t,y,y^{\prime })$ becomes the first-order system $(y_1^{\prime },y_2^{\prime })=(y_2,F(t,y_1,y_2))$ by introducing position and velocity variables. Conversely the first equation identifies $y_2=y_1^{\prime }$ and the second recovers the original equation. Its initial condition must specify both variables.

Compare the displayed ordinary differential equation with $u^{\prime }=1+u^2$, $u(0)=1$, whose solution is $u=\tan (x+\pi /4)$. Their difference satisfies $w^{\prime }=(y+u)w+x^2$, $w(0)=0$. The integrating factor formula makes $w\ge 0$ on the common forward existence interval. Since $u$ diverges at $\pi /4$, the increasing solution $y$ has a maximal forward endpoint $0<x_{\ast }\le \pi /4$. If $y$ remained bounded at this finite endpoint, local existence with the smooth right side would continue it, a contradiction. Hence $y\to +\infty$. The Picard-Lindelöf theorem guarantees local existence, not existence through such a finite-time blow-up. On $|x|\le 1$, $|y|\le 10$, a bound $|1+x^2+y^2|\le 102$ and distance nine to the nearer horizontal box edge give the explicit guaranteed interval $|x|\le 3/34$.

Temporal summation is overlap of responses to inputs arriving at different times. It can occur at one repeatedly activated synapse or across several synapses; it is not defined by reuse of a particular contact. For prescribed electric current inputs in a passive compartment, residual depolarization from earlier pulses changes the voltage reached by later pulses. The membrane time constant sets how rapidly that residual decays, while the duration of synaptic conductance and spatial filtering by a passive dendritic cable can add other timescales.

For the diagram, write $v=V-E_L$, choose inward-positive imposed electric current $I$, and use the ordinary differential equationA rectangular pulse of amplitude $I_0=300\mathrm{pA}$ and duration $d=5\mathrm{ms}$, starting from rest, ends withAfterwards its contribution decays as $A{e}^{-(t-d)/{\tau }_m}$. The superposition principle therefore gives, at the end of the third pulse with onset spacing $\Delta \ge d$,For $\Delta =5\mathrm{ms}$ this is $15.829\mathrm{mV}$, so the peak membrane potential is $-49.171\mathrm{mV}$. For $\Delta =30\mathrm{ms}$ it is $8.447\mathrm{mV}$, giving $-56.553\mathrm{mV}$. Both protocols inject $3I_{0}d=4.5\mathrm{pC}$ of electric charge; equal charge does not imply equal peak voltage because the leak drains charge while it is delivered.

In the adjacent-pulse case, the illustrative neuronal firing threshold $-50\mathrm{mV}$ is crossed at $10+20\log 2\approx 23.863\mathrm{ms}$. A leaky integrate-and-fire model would emit its first event there and reset, so its later trajectory would differ from the passive diagram. The widely spaced train never reaches that threshold. Thus changing input timing alone can change whether an event occurs. A real neuron's neuronal firing threshold is also state-dependent, and synaptic conductance pulses are not prescribed electric current pulses.^[ref]

Measured responses to repeated presynaptic action potentials may also change through synaptic plasticity. A larger second response can therefore reflect a stronger second input as well as residual postsynaptic voltage. To isolate passive temporal summation, compare known injected electric current waveforms with their single-pulse responses under stable recording conditions; to test actual synapses, measure whether input efficacy changes. These are proposed discriminating comparisons, not additional controls claimed for the slice experiment.^[ref]