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The general solution of the first-order linear differential equation is $y=1+C{e}^x$. The initial condition $y(0)=1$ gives $C=0$, so the solution of this initial value problem is the constant function $y(x)=1$. Thus $\boxed{y(1)=1}$.

For the zero-data initial value problem, choose a causal Green function satisfying $L_{x}G(x,\xi )=\delta (x-\xi )$ and $G=0$ for $x<\xi$. Continuity at $x=\xi$ prevents a derivative of the Dirac delta distribution, while integration across the source givesFor $x>\xi$, part (i) gives the shifted homogeneous solutions. The zero value at the join removes the cosine term, and the unit derivative jump fixes the sine coefficient. Therefore the causal Green function of a damped oscillator isEquivalently, multiply the first expression by the Heaviside step function $H(x-\xi )$. The required solution isIts initial value and first derivative vanish: the interval has zero length at $x=0$ and the kernel vanishes at its upper limit. Differentiating twice shows that the upper-limit contribution to $y^{\prime \prime }$ is precisely $g(x)$, while the remaining homogeneous kernel terms cancel in $Ly$.

Use $x_n=nh$, $h=1/N$, and $y_n$ for the numerical approximation in the scalar initial value problem $y^{\prime }=f(x,y)$, $y(0)=y_0$, on $[0,1]$.

The familiar result follows if the force laws have Galilean invariance; the three abstract Newton's laws of motion do not guarantee that invariance. Isaac Newton's Corollary V correctly identifies a central feature of ordinary Newtonian mechanics, but its general physical scope requires assumptions about interactions as well as about the relation between force and acceleration.^[ref]

Take a constant velocity $\mathbf{U}$ and a Galilean transformationThen ${\ddot{\mathbf{x}}}_i^{\prime }={\ddot{\mathbf{x}}}_i$, ${\mathbf{x}}_i^{\prime }-{\mathbf{x}}_j^{\prime }={\mathbf{x}}_i-{\mathbf{x}}_j$, and ${\dot{\mathbf{x}}}_i^{\prime }-{\dot{\mathbf{x}}}_j^{\prime }={\dot{\mathbf{x}}}_i-{\dot{\mathbf{x}}}_j$. Any interaction depending only on these relative quantities, with unchanged parameters, supplies the same forces for the transformed initial data. Newton's second law consequently gives the same relative accelerations; uniqueness of the resulting initial value problem supplies the same relative histories. An inverse-square central interaction is a standard example. Newton's third law helps establish the common centre of mass result, but the boost argument principally needs invariant interactions and the second law.

Here is a counterexample to deduction from the three laws alone. Give two equal masses a preferred inertial frame and write $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$, $\mathbf{V}=({\dot{\mathbf{x}}}_1+{\dot{\mathbf{x}}}_2)/2$. For fixed $k>0$ and speed scale $u_{\ast }>0$, specifyThe forces are equal, opposite and central, the centre of mass is inertial, and each body obeys $m{\ddot{\mathbf{x}}}_i={\mathbf{F}}_i$. An elementary body with no force is inertial. Nevertheless,Boosting an isolated pair from $\mathbf{V}=0$ to $\mathbf{V}=\mathbf{U}$ changes its internal oscillation frequency. All three motion laws hold, but the proposed universal boost equivalence fails. The extra preferred structure in the force law is what makes the failure possible.

A passive coordinate transformation of this same model remains possible: transform the preferred frame's description too. That mathematical freedom does not establish an active dynamical symmetry which changes the physical pair's velocity relative to the preferred structure while holding the laws' background data fixed. Confusing those operations would hide the counterexample.

Finally, a change to rotating coordinates introduces centrifugal forces and Coriolis forces; their dependence on position or velocity can change internal experiments. Constant translational velocity has no analogous effect when the interactions have Galilean invariance. Corollary V expresses that physical symmetry, with the interaction assumption made explicit, rather than a consequence of action–reaction alone.

Ordinary component differences at neighboring points depend on how the local bases vary. The covariant derivative corrects for that variation:Each upper tensor index adds a connection term and each lower index subtracts one. A scalar has ${\nabla }_{a}f={\partial }_{a}f$, while the derivative of a tensor transforms as a tensor with one additional covariant index. The Christoffel symbols themselves are not tensor components; their inhomogeneous transformation cancels that of the ordinary derivative.

For the Levi-Civita connection,These express metric compatibility and zero torsion. Along $x^a(\lambda )$, parallel transport is the initial value problemIt preserves inner products for the Levi-Civita connection. Parallel transport of the tangent itself gives the affinely parametrized geodesic equation, ${\ddot{x}}^a+{\Gamma }_{bc}^a{\dot{x}}^b{\dot{x}}^c=0$.

Transport generally depends on the path. On a round sphere, transport a tangent vector around the triangle bounded by two meridians $90$ degrees apart and the intervening equatorial arc. All three sides are geodesics; the enclosed spherical area is $\pi a^2/2$. The returning vector is rotated by $90$ degrees, with sign set by traversal orientation. Locally this holonomy is controlled by curvature, even though the connection can be made zero at a single point. A vector can therefore have changing coordinate components while remaining parallel, and constant components do not generally mean parallel transport.

This construction compares directions at separate points through a chosen connection, rather than assuming that a curved manifold supplies a path-independent comparison.^[ref]

A pathline records the motion of one identified fluid element: keep its starting position $\mathbf{a}$ and starting time $t_0$ fixed, and let physical time $t$ advance. In a velocity field $\mathbf{u}(\mathbf{x},t)$ its Lagrangian trajectory $\mathbf{X}(t;t_0,\mathbf{a})$ solves the initial value problemA streamline instead freezes the time argument of the velocity field; a streakline fixes the observation time and varies the release time of many fluid elements. A pathline is useful for answering where a particular parcel goes and how long it takes, rather than merely which way the fluid is moving now. ^[ref]

For a concrete analytical model, take spatial coordinates $x,y$ in metres, time $t\ge 0$ in seconds, and the spatially uniform but time-dependent velocity fieldIts numerical components in these units are $(1,t)$. A fluid element starting at the origin at $t_0=0$ has the pathlineAt $t=1\mathrm{s}$ it is at $(1,0.5)\mathrm{m}$; at $t=2\mathrm{s}$ it is at $(2,2)\mathrm{m}$. The parabola records earlier, shallower velocity directions as well as the final direction. Differentiating gives $(\dot{x},\dot{y})=(U,at)$ and acceleration $(0,a)$, so this curved pathline needs no spatial variation of the velocity field. More generally the chain rule along a pathline gives $df(\mathbf{X}(t),t)/dt={\partial }_{t}f+\mathbf{u}\cdot \nabla f$, the material derivative of a transported scalar field.

A motivating tracer demonstration appears in S. J. Kline's film notes: water passes through a contracting channel, and hydrogen bubbles mark a small element at a wire at time $t_0$. The illustrated element is shown again $4.7\mathrm{s}$ later, with its accumulated pathline superposed. Following that same marked element across successive film frames records a history; the entire bubble pattern in one frame has a different meaning. ^[ref]

For a modern version of the readout, use sparse small markers in a transparent water channel, illuminate the viewed plane, and keep an optical camera fixed with a ruler in that plane. Record timestamps and track the same isolated marker between frames; successive positions estimate its pathline and $[\mathbf{X}(t+\Delta t)-\mathbf{X}(t)]/\Delta t$ estimates its velocity. A long exposure records a pathline segment but loses the timing within it, so its total length divided by exposure time gives only mean speed. These are proposed measurement steps based on Jean Hertzberg's teaching guidance, rather than additional measurements from Kline's film. Markers leaving the viewed plane, overlapping tracks and perspective can break particle identification; repeat with a thinner viewed slice and a lower marker concentration. ^[ref]

Let $U\subset \mathbb{R}\times {\mathbb{R}}^d$ be an open set, $d<\infty$, and $f:U\to {\mathbb{R}}^d$ a continuous function. Through every $(t_0,y_0)\in U$ there is a local solution of a differential equation: a continuously differentiable function $y:I\to {\mathbb{R}}^d$ on an open interval containing $t_0$, satisfying the displayed initial value problem pointwise. No Lipschitz condition is assumed.^[ref]

Continuity of a proposed rate law therefore guarantees at least one trajectory for a short time. For a physical model, this is a first consistency check: can the stated rule and initial condition actually be followed? Prediction also needs uniqueness. The delayed solutions of a square-root differential equation below show how the same rule and initial condition can leave a departure time undetermined. The Picard-Lindelöf theorem supplies a sufficient additional hypothesis for uniqueness; failure of that hypothesis alone does not prove nonuniqueness.

For the initial value problem $y^{\prime }=2\sqrt{|y|}$, $y(0)=0$, a forward solution of a differential equation can stay zero until any $c\ge 0$, then follow $y=(t-c{)}^2$. The delayed solutions of a square-root differential equation verify this freedom exactly.

Local existence gives neither a global solution of a differential equation nor a bound on all future values. The continuous function $f(t,y)=y^2$ already permits finite-time blowup; the finite-dimensional continuation criterion explains the additional issue of extending a trajectory within its domain.

Proof outline: choose a compact set $K=[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$. By the extreme value theorem, $M={\max }_K\Vert f\Vert <\infty$. Take $0<h\le a$ with $Mh\le r$. Forward Euler method polygons remain in the closed ball and satisfy $\Vert y_n(t)-y_n(s)\Vert \le M|t-s|$. The Arzelà-Ascoli theorem yields a convergent subsequence with uniform convergence. Uniform continuity of $f$ on $K$ makes their integral equation defect tend uniformly to zero, so the limit satisfiesThe fundamental theorem of calculus gives a continuously differentiable function. Repeat backward and join at $t_0$, where both derivatives equal $f(t_0,y_0)$.^[ref]

The compact set here uses a finite-dimensional vector space; bounded closed balls in an arbitrary Banach space need not be compact. When $M=0$, the constant solution of a differential equation $y=y_0$ works on the entire time interval in $K$. In general the proof selects a subsequence, without asserting that all polygons converge to one unique trajectory.

A periodic solution is a solution of a differential equation which is a periodic function: for some $T>0$, $x(t+T)=x(t)$ at every time. This demands repetition of the state, rather than merely repetition of the coefficients in the differential equation. A nonconstant periodic solution of an autonomous differential equation traces a periodic orbit; constant equilibrium points are also periodic functions, so they must be excluded when asserting absence of periodic orbits.

For an explicit example, consider the initial value problemIts coefficient has period $T=2\pi$. Taking a logarithmic derivative and then applying integration givesDirect differentiation verifies both the differential equation and the initial condition. After one coefficient period,Thus the repeating coefficient generally produces a repeating shape with a changed amplitude. With $b=1$, the three cases $a=-0.2,0,+0.2$ have one-period factors approximately $0.2846,1,3.5136$, respectively. Only $a=0$ gives a periodic solution with this initial condition. For $a<0$ the positive solution of a differential equation tends to zero, and for $a>0$ it tends to infinity; neither can have any positive period. The zero solution of a differential equation, obtained with a different initial condition, is periodic for every $a$.

The exponential function separates two roles:For $b=1$, even the decaying case $a=-0.2$ initially rises, because $y^{\prime }(0)=0.8>0$. The growing case $a=0.2$ has intervals of decline, because $a+\cos t$ becomes negative. This is the distinction between the instantaneous logarithmic derivative $y^{\prime }/y$ and the Floquet growth rate measured over many periods.

More generally, for a real continuous periodic function $q$ of period $T$, the scalar linear differential equation $y^{\prime }=q(t)y$ hasA nonzero solution of a differential equation is $T$-periodic exactly when $m=1$, or equivalently when the coefficient has zero integral over a period. This scalar calculation is the simplest example of a Floquet multiplier.^[ref]

Picard iteration replaces a tentative trajectory by the accumulated integral of its proposed rate. For an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, define the Picard map and its starting function byHere $y_0(t)$ is the constant starting trajectory; the same symbol without its argument denotes the initial state. A fixed point of $\mathcal{P}$ solves this integral equation, and the fundamental theorem of calculus gives the original ordinary differential equation.

For $y^{\prime }=y$, $y(0)=1$, successive integrals giveAt $t=1$, the first five values are $1,2,2.5,2.66666667,2.70833333$, while $e\approx 2.71828183$. The fourth polynomial misses by about $0.00994850$. On $0\le t\le 1$, every omitted term of the exponential series is nonnegative and increasing, so the largest error occurs at $t=1$. The tail tends to zero, giving uniform convergence to the exponential function on this interval.

The beauty of Picard iteration is that each proposed trajectory supplies a better rate history for the next one. For this example, one integration adds exactly the next term of the exponential series. In general, the integrals may be difficult to compute, but the construction still explains local existence and uniqueness.

The short contraction mapping argument needs two independent controls. Choose a time-state cylinder $[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$ on whichThe second bound is the uniform local Lipschitz condition. Set $J=[t_0-h,t_0+h]$ with $0<h\le a$, $Mh\le r$ and $Lh<1$. In the function space of continuous functions $J\to {\mathbb{R}}^d$, use the supremum norm $\Vert u{\Vert }_{\infty }={\sup }_{t\in J}\Vert u(t)\Vert$. On its closed ball $\Vert u-y_0{\Vert }_{\infty }\le r$, direct integral estimates giveThus the map preserves the closed ball and is a contraction mapping there. This closed ball is a complete metric space, so the contraction mapping theorem gives convergence to exactly one fixed point. Any other local solution of a differential equation stays in the same ball after shortening its interval, and hence agrees there.

For $y^{\prime }=y$, take $r=1$, so $|y|\le 2$ in the state ball and $M=2$, $L=1$. Choosing $h=1/2$ gives $Mh=1$ and $Lh=1/2$, with the actual solution satisfying $|e^{1/2}-1|\approx 0.64872127<1$. The plotted interval $[0,1]$ is longer than this simple guarantee: the explicitly computed exponential series converges there anyway. The condition $Lh<1$ is sufficient for this one-step contraction mapping argument, rather than a necessary lifespan bound.

Let $U\subset \mathbb{R}\times {\mathbb{R}}^d$ be an open set and $f:U\to {\mathbb{R}}^d$ a continuous function. Suppose $f$ is locally Lipschitz in the state $y$, with a Lipschitz constant uniform on a sufficiently small time-state neighborhood of each point. Then each $(t_0,y_0)\in U$ determines a unique local solution of a differential equation to the displayed initial value problem, on an interval around $t_0$. This is the local theorem in Ordinary Differential Equations and Dynamical Systems by Gerald Teschl.

For example, $y^{\prime }=y$, $y(0)=1$, has the solution of a differential equation $y(t)=e^t$: its derivative is itself and its starting value is $1$. The Lipschitz constant is $L=1$, since $|u-v|=|f(t,u)-f(t,v)|$. Thus uniqueness follows even before computing $y(1)=e\approx 2.71828183$. Picard iteration below constructs this exponential function one polynomial at a time.

The Lipschitz condition has real content. For the continuous function involving a square root,each waiting time $c\ge 0$ gives a different forward solution of a differential equation:The constant solution of a differential equation $y(t)=0$ is another choice. At $t=2$, delays $c=0,1,2$ give $y=4,1,0$ respectively. Before the join, both sides of the ordinary differential equation are zero; afterwards, $y_c^{\prime }=2(t-c)=2\sqrt{y_c}$. Both one-sided derivatives at $c$ are zero, so the join is valid. All these trajectories can also be extended by zero to $t<0$. The failed Lipschitz condition is visible in the quotientFor $\epsilon =10^{-2},10^{-4},10^{-6}$ this quotient is $20,200,2000$. Continuity still allows local existence by the Peano existence theorem; it cannot choose a waiting time. MIT's 18.034 Lecture 2, Example 2.3 develops the same waiting-time mechanism for a different ordinary differential equation, $y^{\prime }=3y^{2/3}$.

The useful promise is that an initial condition and a sufficiently regular local rule select one trajectory. It justifies comparing models and following trajectories without an arbitrary choice at each instant. A locally Lipschitz continuous rule need not keep that trajectory bounded: $y^{\prime }=1+y^2$ meets the theorem's hypotheses everywhere but has finite-time blowup. Longer existence is governed by the finite-dimensional continuation criterion.

Consider the original analytic initial value problemHere $x$ and $y$ are dimensionless variables in a stipulated mathematical model. The trajectories below are exact solutions of a differential equation, with no measured data or empirical parameter fit. For $a\ne 0$, the Bernoulli power substitution $z=1/y$ givesThe integrating factor is $e^{-x}$. Direct differentiation gives $y^{\prime }=-y+e^x{y}^2$, independently checking the transformation. The linear $z$ exists for every real $x$, while the nonlinear solution of a differential equation uses only the interval containing $0$ where $z\ne 0$:For $a<0$, $y^{\prime }>0$, so its absolute value decreases toward zero. The lost zero solution of a differential equation is recovered by substitution into the original equation. Since $f(x,y)=-y+e^x{y}^2$ is locally Lipschitz in $y$ everywhere, the Picard-Lindelöf theorem ensures uniqueness through every finite point, including $y=0$. Thus a nonzero trajectory never meets or crosses zero at a finite $x$.

For $a=1/2$, the exact solution of a differential equation is $y=e^{-x}/(2-x)$ on $(-\infty ,2)$. It first decreases, reaches a local minimum $y(1)=e^{-1}$, then increases without bound. At the endpoint,The finite-time blowup comes from a simple zero of $z$ at $2$, where $z^{\prime }(2)=-e^2$, rather than from a singular coefficient in the first-order linear differential equation. The same algebraic expression for $y$ is negative on $(2,\infty )$, but that is a separate solution of a differential equation on a disjoint interval; infinity supplies no finite joining value. For $a=-1$, $y=-e^{-x}/(1+x)$ on $(-1,\infty )$, giving a distinct decaying branch whose finite blowup is instead in the past.

The direction field also explains the positive turning point. Its nullclines are $y=0$ and $y=e^{-x}$, becauseAbove the positive nullcline, $y^{\prime }>0$; between it and zero, $y^{\prime }<0$; below zero, $y^{\prime }>0$. The moving nullcline $y=e^{-x}$ is not a solution of a differential equation: its derivative is $-e^{-x}$, whereas the prescribed slope there is zero. The $a=1/2$ trajectory crosses it at $x=1$, exactly its local minimum.

A Runge-Kutta method advances an initial value problem $y^{\prime }=f(t,y)$, $y(t_0)=y_0$, by sampling its vector field at several trial states within one time step. These samples are Runge-Kutta stages; a weighted combination of their slopes produces the next state. The construction improves on the single tangent line used by the Euler method while requiring evaluations of $f$, rather than explicit higher derivatives of the solution. It applies to scalar equations and systems of ordinary differential equations alike. The coefficient choices distinguish methods such as Heun's method, the classical fourth-order Runge-Kutta method and implicit Runge-Kutta methods.

For $s$ stages, write $A=(a_{ij})$ for the stage matrix, $c_i$ for relative time nodes and $b_i$ for output weights. A time step of length $h$ computesHere $k_i$ is a slope, with units of state per time; $h{k}_i$ is a state increment. Another common convention writes $K_i=h{k}_i$, so its stage states are $Y_i=y_n+{\sum }_j{a}_{ij}{K}_j$ and its output is $y_n+{\sum }_i{b}_i{K}_i$. These are equivalent conventions: including $h$ both inside $K_i$ and again in the output would count the time step twice. The textbook formulation uses the increment convention.^[ref]

Usually the Butcher tableau obeys $c_i={\sum }_j{a}_{ij}$, called internal consistency. Stage $i$ then combines slopes over an effective time $c_{i}h$. The second-order conditions with independent Runge-Kutta abscissae explain what changes when time nodes are specified independently. A Runge-Kutta stage is generally an auxiliary approximation, not a separate accepted solution value at its node. Several stages can even have the same time node but different state arguments, as in the classical fourth-order Runge-Kutta method.

The useful idea is to approximate the integral $y(t_n+h)-y(t_n)={\int }_{t_n}^{t_n+h}f(t,y(t))dt$ without knowing the intermediate exact trajectory. Trial states provide places to sample $f$, and Butcher order conditions make the resulting combination agree with the exact Taylor expansion. More stages alone do not guarantee a higher order of a numerical method: their coefficients must satisfy the appropriate conditions. Embedded Runge-Kutta pairs reuse these same samples for a second output and a practical local error estimator.

A second-order ordinary differential equation involves a second derivative as its highest derivative. In explicit form $y^{\prime \prime }=f(x,y,y^{\prime })$, introducing $v=y^{\prime }$ converts it into the first-order system $y^{\prime }=v$, $v^{\prime }=f(x,y,v)$. Under local regularity and Lipschitz continuity in the dependent variables, an initial value problem is specified by both $y(x_0)$ and $y^{\prime }(x_0)$. The linear case is a second-order linear differential equation.

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.

For the closed SIR model with $\gamma >0$, the infectious compartment's fraction approaches $i_{\infty }=0$. One proof outline uses the bounded, increasing removed compartment's fraction: since $r(t)-r_0=\gamma {\int }_0^{t}i(u)du$, the total time-integrated $i$ is finite. The bounded derivative $di/dt$ from the SIR model equations prevents indefinitely recurring spikes, so $i(t)$ tends to zero.

For $s_0>0$, the final size relation for an epidemic follows by dividing the SIR model equations for $s$ and $r$ while $i>0$, giving $ds/dr=-(\beta /\gamma )s$. Integration then yieldsThis also follows from the SIR first integral and conservation of population. For the stipulated initial value problem, solveThe bisection method givesAbout $5.88\%$ of the original population remains in the susceptible compartment. The $94.12\%$ is the final removed compartment's fraction, including the initially $1\%$ in the infectious compartment. The newly acquired infection during the modelled period accounts for $s_0-s_{\infty }\approx 0.93120263520$ of the total population, not $1-s_{\infty }$. Dividing by $s_0$ instead gives the proportion of initially susceptible compartment members who acquire infection; its denominator must be stated.

At the SIR infection peak, the flow into the infectious compartment only balances its outflow; it does not stop. Further infection explains why the final susceptible compartment's fraction is well below $1/3$. Also $s_{\infty }=s_0{e}^{-R_0(r_{\infty }-r_0)}>0$ for $s_0>0$ and finite $R_0$: the closed SIR model does not require everybody to acquire infection before its infectious compartment empties in the long-time limit.

If $i_0>0$ and $s_0>\gamma /\beta$, the infectious compartment initially grows. The SIR model equations give its unique later maximum whenBefore this SIR infection peak, $\beta s-\gamma >0$; afterward it is negative because the susceptible compartment keeps shrinking. Substituting $s_{\ast }$ into the SIR first integral givesFor the stipulated initial value problem,Thus about $30.38\%$ of the modelled population belongs to the infectious compartment simultaneously at the SIR infection peak. This is a peak of the infectious compartment, not of incidence, whose flow is $\beta si$. If $s_0\le \gamma /\beta$, the maximum is already $i_0$ at the start and the interior SIR infection peak formula does not apply.

The numerical solution of a differential equation puts the SIR infection peak at about day $26.63258$ with $i\approx 0.303812682386$, agreeing with the SIR first integral to within $10^{-11}$. The accompanying generator checks nonnegative fractions and conservation of population without clipping or rescaling. For steps $0.4$, $0.2$ and $0.1$ day, the largest differences between successive numerical trajectories at shared times are approximately $1.086\times 10^{-7}$ and $6.985\times 10^{-9}$. Their ratio is $15.55$, close to the factor $16$ expected when a fourth-order Runge-Kutta method's step is halved.

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.

When a dispersion relation factors as $\omega P(\omega ,k)=0$, dividing by $\omega$ discards a possible time-independent Fourier mode. This can remove a necessary stationary solution from an initial value problem. For instance, a constant-coefficient temporal equation $T^{(3)}+ia{T}^{\prime \prime }+b{T}^{\prime }=0$ has a constant solution as well as the two exponential solutions from its quadratic factor. Three initial conditions generally require all three modes. Cancellation is valid only when a nonzero frequency has already been required.

A streakline at observation time $T$ is the set of present positions of all fluid elements that passed through one fixed release point $\mathbf{a}$ during a specified history of injection. For continuous release over ${\tau }_0\le \tau \le T$, solve a different initial value problem for each release time $\tau$:Thus a pathline varies $t$ for one particle; a streakline varies $\tau$ while holding $T$ fixed. Interrupted release gives only the corresponding subsets of the streakline. A moving nozzle requires a time-dependent release position instead, so the standard fixed-point definition should not silently be applied to it. Continuous release makes streaklines easy to observe and useful for showing where material from one source has reached. ^[ref]

In the analytical velocity field $(U,at)$ of pathline, a particle released at the origin at time $\tau$ followsObserve all particles together at time $T$, after continuous release from $\tau =0$:For $U=1\mathrm{m}{\mathrm{s}}^{-1}$, $a=1\mathrm{m}{\mathrm{s}}^{-2}$ and $T=2\mathrm{s}$, the streakline is the downward-curving parabola $y=2x-x^2/2$ in metre coordinates. Releases at $\tau =0,1,2\mathrm{s}$ occupy $(2,2)$, $(1,1.5)$ and $(0,0)\mathrm{m}$, respectively. At $x=1\mathrm{m}$, the streakline lies above the pathline $y=x^2/2$ and below the origin's streamline $y=2x$. All particles nevertheless have the same present velocity $(1,2)\mathrm{m}{\mathrm{s}}^{-1}$. The streakline slope $dy/dx=a\tau /U$ generally differs from the present velocity slope $aT/U$; they agree at the nozzle. The oldest particle is the downstream endpoint, which explains why the pathline and streakline meet there without being the same curve.

Each panel fixes one observation time. Coloured dots are the present positions of particles with the indicated release times; dashed pathlines show those particles' earlier histories. The black streakline connects present positions of the continuously released family, including release times between the sampled dots. After $T=1\mathrm{s}$, particles already present keep moving and new ones enter at the origin, so the later streakline is not a frozen record of the earlier one. Injection beginning only at $\tau =1\mathrm{s}$ would show just $0\le x\le 1\mathrm{m}$ of the $T=2\mathrm{s}$ curve. Both panels are diffusion-free analytical models, not experimental photographs.

Kline's contracting-channel demonstration switches from marking an element to continuous bubble production at a marking wire: the resulting trace is a streakline. The apparatus therefore changes the curve being displayed by changing the marking history, even with the same channel. A wire produces a family of traces; conceptually each fixed emitting location supplies its own streakline. Individual bubbles are markers rather than perfect fluid elements: Jean Hertzberg's teaching notes describe hydrogen production by electrolysis at a thin cathode wire and warn that larger bubbles have more buoyancy. ^{[ref] [ref]}

A dye version uses a small fixed injection port upstream in a transparent water channel, a dilute premix of dye with the channel water, illumination and a fixed optical camera. Start continuous injection at a recorded time ${\tau }_0$ and take a short exposure at $T$: the visible filament approximates a streakline, not the accumulated pathline of one particle. Repeat with the same port and imaging geometry after changing the flow with time; compare the filament against local directions estimated from short displacements of sparse markers at $T$. Agreement in a steady flow and possible disagreement when the flow changes are the predictions being tested; no measured curves are claimed here. The protocol follows Hertzberg's advice to use small ports, little injected volume and a dilute premix with matched density and viscosity. Check for nozzle disturbance by reducing injection rate, for buoyancy by watching the premix in still water, and for diffusion by comparing filament widths at different ages. Keep exposure short enough to suppress motion blur. Dye concentration approximates a passive scalar only when it has negligible feedback on the velocity field; finite port width, diffusion and out-of-plane motion limit how literally a visible filament can be read as a geometrical streakline. ^[ref]

For an inviscid flow of constant mass density $\rho$ and depth $h$, impose excess pressure $p_1\cos (k(x-ct))$ with $k>0$ on a free surface. Neglect surface tension. A travelling solution of the linearized free-surface boundary conditions isThe velocity potential satisfies the Laplace equation and the bed no-penetration boundary condition. Its surface time derivative is $-A{c}^{2}k\coth (kh)\cos (k(x-ct))$, so the dynamic condition gives the displayed signed response. The nonnegative amplitude is $|A|$. It is finite except atthe squared phase speed of a free surface gravity wave at that wavenumber. Equality produces resonance. This describes a travelling forced component; free surface gravity waves must be added to meet general initial conditions. At exact resonance, a solution of the initial value problem generally grows with time rather than having a bounded periodic amplitude.