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Set $y={\sum }_{j\ge 0}{a}_j{x}^j$ and equate coefficients in the power series. The constant coefficient gives $a_2=0$. For $m\ge 1$, the coefficient of $x^m$ givesThe square bracket is $(m+1)(m+2)$, so the recurrence relation is $a_{j+3}=a_j$ for $j\ge 0$. The initial conditions give $a_0=1$, $a_1=0$, $a_2=0$. Hence

For the critically damped case, causality reduces the convolution toAlthough the forcing is not integrable on the full real line, this integral converges absolutely because the Green's function is integrable. Thus the calculation is valid directly; the corresponding Fourier transform can be interpreted as a tempered distribution.

Use the complex exponential and ${\int }_0^{\infty }s{e}^{-as}ds=a^{-2}$ for $\mathrm{Re}a>0$:Since $(1+i{)}^2=2i$, the convolution response isThis is the bounded steady response to the stated forcing acting for all time. Without an initial condition or the convolution selection, the general solution additionally contains $(C_1+C_{2}t)e^{-\kappa t}$. No switch-on at $t=0$ was specified, so zero initial displacement and velocity should not be imposed on this all-time forcing.

Project the initial condition onto the orthogonal spatial eigenfunctions. Their squared norms in the ordinary inner product areThe latter follows from the integrals of ${\cos }^2(n\pi x)$ and ${\sin }^2(n\pi x)$ being $1/2$, with the sine-cosine cross term zero. ThereforeFor square-integrable data the expansion converges to the data in the L2 norm at $t=0$; for $t>0$ the decaying high modes are smoothed. Initial data with zero $e^x$ projection have no growing term.

Take the causal solution, extending the input and output by zero before the device starts at $t=0$. The two zero initial conditions ensure that this extension adds no impulsive terms to the first two distributional derivatives. The Fourier transform of a derivative is multiplication by $i\omega$, soFactoring the quadratic and taking partial fraction decomposition yieldsThe inverse Fourier transform follows directly by integrating decaying exponentials:where $H$ is the Heaviside step function. This damped-oscillator impulse response has $R(0^{+})=0$ and $R^{\prime }(0^{+})=1$, as required for a unit impulse.

For $b^2<4mk$, put ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2}$. Complex-conjugate roots give the real solutionThis is underdamping: oscillation at ${\omega }_d$ persists within a decaying exponential envelope. damping reduces the oscillation frequency below ${\omega }_0$. Constants in all three cases are determined by two initial conditions.

Equilibrium and undamped motion. Resolve forces along the incline, taking positive displacement down the slope. At equilibrium, Hooke's law balances the downhill component of weight:If $x$ is measured from that equilibrium position, Newton's second law becomesThus the motion is a harmonic oscillator withGravity shifts the equilibrium extension but drops out of this oscillation frequency because the downhill force component is constant.

Viscous damping. For a damped harmonic oscillator, put $\beta =b/(2m)$. Substitution of $z=A{e}^{-pt}$ in the supplied linear ordinary differential equation givesIn the underdamped regime $\beta <{\omega }_0$, define ${\omega }_d=\sqrt{{\omega }_0^2-{\beta }^2}$. The two characteristic roots are $p=\beta \pm i{\omega }_d$, and the general real displacement isHere $C,D$, or equivalently $A_{\mathrm{env}}$ and $\varphi$, are fixed by the initial conditions. For release from rest at $x_0$, the condition $\dot{x}(0)=0$ requires $D=\beta x_0/{\omega }_d$, not $D=0$:The curve oscillates about zero between exponentially shrinking envelopes $\pm A_{\mathrm{env}}{e}^{-\beta t}$, where $A_{\mathrm{env}}=|x_0|\sqrt{1+(\beta /{\omega }_d{)}^2}$ for this release. The upper panel illustrates that shape; its damping strength is chosen to make the envelope visible.

Frequency change. Viscous damping reduces the angular frequency and lengthens the oscillation period:For light damping, ${\omega }_d\simeq {\omega }_0[1-b^2/(8mk)]$, so the frequency shift is second order in $b$, even though the envelope decay rate is first order in $b$.

Will the block reach the stop? The given parameters giveFor $x_0=-0.260\mathrm{m}$, differentiating the release-from-rest solution givesIt is positive throughout the first half-cycle. The first positive turning point therefore occurs at $t_{+}=\pi /{\omega }_d=0.993460\mathrm{s}$. The turning points of an underdamped release from rest give its unobstructed height:Since $x_{+}>0.250\mathrm{m}$ and the displacement increases continuously up to that point,Solving $x(t)=0.250\mathrm{m}$ gives first contact at approximately $0.91117\mathrm{s}$. The plotted dashed continuation is the hypothetical free motion after that time; the post-impact motion is not determined by the supplied damping model.

Write $\gamma =b/(2m)$ and $\nu =\sqrt{{\omega }^2-{\gamma }^2}$, where $\omega =\sqrt{k/m}$ is the undamped angular frequency. The linear drag gives the vector damped harmonic oscillator equationFor the original circular initial conditions, ${\mathbf{r}}_0=(c,c)$ and ${\mathbf{v}}_0=\omega (-c,c)$, the exact underdamped solution isUsing $|{\mathbf{r}}_0{|}^2=2c^2$, $|{\mathbf{v}}_0{|}^2=2{\omega }^2{c}^2$ and ${\mathbf{r}}_0\cdot {\mathbf{v}}_0=0$ givesThis exact potential energy is useful because it satisfies the prescribed initial conditions, including $\dot{U}(0)=k{\mathbf{r}}_0\cdot {\mathbf{v}}_0=0$. In the stated weak-damping limit $\gamma \ll \omega$, $\nu \simeq \omega$ and the square bracket differs from one by a small oscillatory term. The leading approximation is thereforeThe radius is approximately $\sqrt{2}c{e}^{-bt/(2m)}$, so the trajectory is approximately a shrinking logarithmic spiral. The exactly exponential result would require slightly adjusted initial velocity; with the specified undamped circular velocity, the small modulation above is real. This is the initially circular damped planar oscillator.

Set $s=r_0+\mu t$. The proposed velocity and its derivative areThe remaining term in the linear ordinary differential equation isso the sum is exactly $g$. Also $v(0)=g(r_0-r_0)/(4\mu )=0$, establishing both the equation and its initial condition:The integrating factor $(r_0+\mu t{)}^3$ explains where the expression comes from:The result has units of velocity. Its continuous $\mu \to 0$ limit is $gt$, despite the apparent $1/\mu$ in the written expression.

At the ridge axis, $T(z,0)=T_1$ for $0<z<a$ in the intended cooling model. Upwelling and magmatic heat supply continually bring hot material into the newly formed plate. The abrupt meeting of this hot interior with the cold upper boundary is an idealized corner discontinuity, resolved diffusively for $t>0$. The printed series fails this initial condition, as shown in the parent Solution.

For a mode of positive frequency, differentiation of the causal convolution givesThe first boundary term vanishes because $\sin 0=0$; the second supplies $f(t)$ because $\cos 0=1$. Both initial values vanish. This verifies the forced-oscillator equation and its initial conditions, including a step source in the piecewise or distributional sense.

The spatial Dirac delta function reduces the excitation toFor an event at $t_s\ge 0$, put $\tau =t-t_s$. Integrating the oscillator kernel from $t_s$ to $t$ gives the normal-mode response to a step seismic moment:Non-unit mass normalization introduces the same factor $1/N_{\alpha }$ as above. Complex modes must be combined with their conjugate partners to produce a real displacement. The formal $\omega \to 0$ limit is $a=C{\tau }^{2}H(\tau )/2$; for rigid-body modes here, $C=0$ because their strain is zero.

The step produces a static component plus continuing oscillations in this lossless model. It does not predict physical damping. A point source is appropriate when rupture dimensions are small compared with the wavelengths being analysed and rupture duration is short compared with their periods. Long-period observations of a relatively small earthquake can meet these conditions; high-frequency finite-fault directivity generally cannot. A finite source-time function replaces the ideal instantaneous step when its duration matters.

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.

P. J. E. Peebles made the connection between microscopic processes and cosmological observables quantitative. His 1968 paper treated cosmological recombination as a kinetic problem. Capturing an electron directly into the hydrogen ground state releases a photon that can ionize another atom. Net neutralization depends on escape of resonance radiation through cosmological redshift and on two-photon spontaneous emission from the $2s$ state. An equilibrium Saha equation alone does not give that history.^[ref]

This matters for an acoustic prediction because the hydrogen ionization fraction fixes the supply of free electrons for Thomson scattering. In physical units the scattering rate per photon is ${\Gamma }_T=n_e{\sigma }_{T}c$. The cosmological visibility function describes the distribution of last-scattering times. Thus atomic physics helps determine when the photon-baryon acoustic oscillator is read out, as well as how rapidly tight coupling fails.

In 1970 P. J. E. Peebles and J. T. Yu followed adiabatic cosmological perturbations by numerically integrating the photon collision equation through cosmological recombination. Their calculation connected acoustic evolution and dissipation to residual matter structure and microwave brightness fluctuations. It did not assume that a perfect photon-baryon fluid remained valid throughout last scattering. Their illustrative cosmological models and normalization were historical choices, rather than the present Lambda-CDM model; their predicted brightness variance versus angular resolution should not be relabelled as a modern measured $C_{\ell }$ spectrum.^[ref]

The continuing idea is a forward prediction: specify initial conditions, evolve the coupled matter and radiation, project onto the sky, and compare the result with an instrument's readout. Coherent initial phases are needed for acoustic variance peaks explains why a collection of oscillating modes does not automatically produce a peaked power spectrum. The BOOMERANG acoustic-peak measurement provides a later observational test of this physical programme.

Use dimensionless $x$ and $t$, the periodic domain $[0,2\pi ]$, and the same initial conditionWith zero PDE source term, compare the heat equation $u_t=u_{xx}$ and the transport equation $u_t+u_x=0$. Their exact solutions areEach cosine satisfies ${\partial }_x^2\cos (nx)=-n^2\cos (nx)$, so the heat equation reduces its amplitude according to $A_n^{\prime }=-n^2{A}_n$. The superposition principle combines the constant and two Fourier modes. The transport equation shifts the argument of the entire profile. Both formulas meet the initial condition and matching periodic boundary conditions exactly.^[ref]

At $t=1$, the heat amplitudes are approximately $0.368$ and $0.0000494$: the small ripples disappear much sooner than the broad variation. The transport equation retains amplitudes $1$ and $0.4$, shifting the profile right by one unit with wraparound. Both preserve spatial average $1$, but only the heat equation approaches the uniform profile. This gives diffusion its usefulness for smoothing uneven temperatures, while the transport equation models a profile carried by a flow. The signed profile can represent a scaled temperature deviation.

To restore units, let physical position be $X=\ell x$ and time be $T=\tau t$. For $U_T=D{U}_{XX}$, the dimensionless coefficient is $D\tau /{\ell }^2$, with $D$ measured in length squared per time. For $U_T+c{U}_X=0$, it is $c\tau /\ell$, with $c$ measured in length per time. Both plotted coefficients equal one when $D={\ell }^2/\tau$ and $c=\ell /\tau$. Thus the example compares specified models on the same scales, rather than identifying diffusivity with velocity.

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]

Physical cosmology connects the universe's history to the measurable behaviour of its contents. The Friedmann equations describe cosmic expansion; atomic physics, radiation pressure and gravity determine what that expanding universe looks like. A particularly revealing example is the route from cosmological recombination through the photon-baryon fluid to the Cosmic microwave background acoustic peaks. The universe supplies the initial conditions and the laboratory; the prediction must still pass through a model of the telescope before it can be compared with observations.

The practical reward is that one sky contains several different tests. The Cosmic microwave background power spectrum carries information about the sound horizon, the cosmological baryon density and cosmic geometry. These are related quantities with different observational signatures. Angular sound-horizon degeneracy explains why measuring a characteristic angle alone cannot count the universe's baryons.

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.

Proportional control chooses a control input proportional to the current tracking error. For the ideal controlled plant and constant reference signal $r$, its ordinary differential equation is $\dot{x}=k(r-x)+w$. With $w=0$ and initial condition $x(0)=x_0$, the exact response isFor $r=1$, $x_0=0$ and $k=2$, this gives $x(1)=1-e^{-2}\simeq 0.864665$ and $x(2)=1-e^{-4}\simeq 0.981684$. To verify the formula, substitute it into the ordinary differential equation and check the initial condition; its decaying exponential function explains the approach to the goal.

With a constant control disturbance, setting $\dot{x}=0$ gives the equilibrium point $x_{\ast }=r+w/k$ and steady-state tracking error $e_{\ast }=-w/k$. For $w=0.2$, doubling the feedback gain from $k=2$ to $k=4$ reduces the offset from $0.1$ to $0.05$. This improvement is a statement about the ideal model: it neglects feedback delay and measurement noise, so it does not establish that arbitrarily large feedback gains are safe. ^[ref]

In the first diagram's switched control disturbance experiment, $k=2$ and $w=0.2$ from $t=2$ give, for $t\ge 2$,Thus $x(4)\simeq 1.097833$ and $x(8)\simeq 1.099999$. At the equilibrium point, the controller supplies $u=-0.2$ to cancel the control disturbance. Proportional control needs a nonzero tracking error to produce that persistent correction.

Proportional-integral control is the proportional and integral part of a PID controller, with the derivative term omitted. Proportional-integral control combines proportional control with integral action. For the ideal controlled plant, a constant reference signal, exact measurement and a constant control disturbance, the ordinary differential equations giveThe characteristic polynomial is ${\lambda }^2+k_p\lambda +k_i$. Positive $k_p$ and $k_i$ imply asymptotic stability of the equilibrium point: real roots of a polynomial have negative sum and positive product, hence are both negative; a nonreal complex conjugate pair has real part $-k_p/2<0$. This is a short stability check for this ideal controlled plant, rather than a rule for every dynamical system.

At the equilibrium point, $e_{\ast }=0$, $x_{\ast }=r$, and $z_{\ast }=-w/k_i$. For $w=0.2$ and $k_i=1$, the stored value $z_{\ast }=-0.2$ supplies $u_{\ast }=-0.2$ even though the tracking error is zero. This separates the eventual accuracy from the transient response: integral action cancels the lasting offset but can make the plant output pass beyond the goal. The authors discuss practical limits on integral action in Åström and Murray, Feedback Systems, Chapter 11.

For $k_p=2$, $k_i=1$, $x(0)=z(0)=0$ and $w=0$, the characteristic polynomial is $(\lambda +1{)}^2$. The initial conditions are $e(0)=1$ and $\dot{e}(0)=-2$, givingThe tracking error changes sign at $t=1$. Since $\dot{x}=(2-t)e^{-t}$, the largest plant output occurs at $t=2$ and is $1+e^{-2}\simeq 1.135335$, about $13.53\%$ above the reference signal.

For a separate comparison, start the control disturbance $w=0.2$ at $t=4$. For $t\ge 4$, write $s=t-4$. The exact responses areThe proportional control response tends to $1.1$, while the proportional-integral control response returns to $1$. At $t=8$, the values are approximately $1.099966$ and $1.017001$. The added term in the proportional-integral control response eventually vanishes, exposing how memory removes a constant control disturbance's lasting effect.

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.

Release the upper rod at ${\theta }_1(0)=\alpha =0.06$ radians while keeping the lower rod vertical, ${\theta }_2(0)=0$, with both angular velocities zero. The initial condition is the sum $\mathbf{\theta }(0)=(\alpha /2){\mathbf{a}}_{-}+(\alpha /2){\mathbf{a}}_{+}$. The solution from linearization is thereforeAt release the lower angle vanishes by cancellation between the two normal modes. Their different angular frequencies then destroy that cancellation. Differentiating twice gives ${\ddot{\theta }}_1(0)=-2g\alpha /\ell$ and ${\ddot{\theta }}_2(0)=2g\alpha /\ell$, agreeing with the small-angle limit of the exact double pendulum equations of motion.

The quadratic mechanical energy at release is $E_2=mg\ell {\alpha }^2$. Each normal mode contributes half, because $(4+2\sqrt{2})(2-\sqrt{2})=(4-2\sqrt{2})(2+\sqrt{2})=4$. Equal coordinate amplitudes need not mean equal energies in general; here the modal inertias and squared angular frequencies happen to compensate. Releasing instead along $\mathbf{\theta }(0)=\alpha {\mathbf{a}}_{-}$ or $\alpha {\mathbf{a}}_{+}$, with zero angular velocities, excites just one normal mode.