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Let $t=T(\mathbf{x})$ be a smooth wavefront and suppose the first jump in time derivatives of displacement field is $[{\partial }_t^k\mathbf{u}]=a^{[k]}\mathbf{n}$, with $k\ge 2$. Substitution of $\mathbf{u}=a^{[k]}\mathbf{n}(t-T{)}_{+}^k/k!+\cdots$ into the elastic equation first gives the eikonal equation $|\nabla T|=1/\alpha$ and longitudinal polarization $\mathbf{n}=\alpha \nabla T$. Projecting the next-order compatibility condition onto $\mathbf{n}$ gives $2\rho \alpha \mathbf{n}\cdot \nabla a^{[k]}+a^{[k]}\nabla \cdot (\rho \alpha \mathbf{n})=0$, hence the displayed conservation law. Coefficients and wavefront must be sufficiently differentiable for this classical transport equation; its local form fails at a caustic.

Put $B_j=[m(x)(x-\bar{x}{)}^j{]}_0^L$, $M_j=\int m(x)(x-\bar{x}{)}^{j}dx$, $A=\int \alpha (t)dt$, and $C=B_0(B_2-M_1)-B_1(B_1-M_0)$. For quadratic bending $h_0=-\alpha (t)(x-\bar{x}{)}^2$, the linear recoil correction in elongated-body theory with a stable post-stroke relaxation gives the displayed angle. It follows by setting time derivatives to zero in the integrated lateral and angular momentum equations. The simpler $2UA$ is exact in the zero-body-mass limit, or when $B_1=M_0$, but is not a generic finite-body-mass identity. For $L=1$, $\bar{x}=1/2$, and $m(x)=m_{\ast }{x}^2$, it becomes $2UA/(1+2M_b/m_{\ast })$. Body deformation inertia affects derivative forcing during the stroke, not this final integral relation when the deformation and its time derivative return to zero.

The squared L2 norm of a probability density function derivative measures roughness and appears in optimal smoothing constants. For $r=2$, its directional derivative along a smooth compactly supported perturbation $a$ is $2\int f^{\prime \prime }{a}^{\prime \prime }$. When integration by parts is valid twice, this equals $2\int f^{(4)}a$. Thus regular root-$n$ estimation requires substantially more regularity than merely estimating a smooth probability density function.

A section of the square of the canonical bundle, locally $q=\varphi (z)d{z}^2$; its coefficient changes by the square of the coordinate derivative. A holomorphic quadratic differential has holomorphic coefficient, and its flat metric of a quadratic differential is $|q{|}^{1/2}$.

Let ${\delta }_t=\sqrt{p_t}-\sqrt{p}=(t/2)g\sqrt{p}+r_t$, with $\Vert r_t{\Vert }_2=o(|t|)$. Since $p_t-p=2\sqrt{p}{\delta }_t+{\delta }_t^2$, the Cauchy-Schwarz inequality gives $\Vert p_t-p-tpg{\Vert }_1\le 2\Vert r_t{\Vert }_2+\Vert {\delta }_t{\Vert }_2^2=o(|t|)$. Thus differentiability in quadratic mean suffices to differentiate bounded density integrals, even when the density itself has no useful pointwise derivative.

A quadratic spline with simple knots is a continuously differentiable function that is a polynomial of degree at most two between successive knots. Its first derivative is continuous and piecewise linear, while its second derivative is piecewise constant and can jump at knots. Constant tails impose vanishing first derivatives at the outermost knots. Such a spline arises from first-derivative roughness penalization when observations are interval averages, as in the quadratic smoothing spline for interval averages.

A quadrature rule approximates an integral by a finite weighted sum of sampled function values:The quadrature nodes $x_i$ specify where to sample, and the quadrature weights $w_i$ specify how much each sample contributes. The density $\rho (x)$ specifies the target integral, whereas the discrete quadrature weights belong to the approximation. Ordinary numerical integration has $\rho =1$. Gaussian quadrature and the midpoint quadrature rule choose interior quadrature nodes; the trapezoidal rule includes endpoints.^[ref]

For ordinary numerical integration, exactness on the constant function $1$ requires ${\sum }_i{w}_i=b-a$; exactness on $x$ additionally requires ${\sum }_i{w}_i{x}_i=(b^2-a^2)/2$. These two checks make the quadrature rule exact for every affine function. For example, both the midpoint quadrature rule and trapezoidal rule give $6$ for ${\int }_0^2(2x+1)dx$. Testing progressively higher polynomial degrees defines the degree of precision of a quadrature rule. Exactness describes a class of integrands, rather than promising exact results for every function.

Composite midpoint quadrature rules and composite trapezoidal rules apply a simple quadrature rule on every panel of a partition of an interval. This buys accuracy by shrinking the panels while keeping the elementary formula fixed. For continuous functions sampled exactly, their quadrature errors tend to zero as the largest panel width tends to zero. The familiar second-order bounds require stronger assumptions about derivatives. The signed approximation error used here is $Q(f)-I_{\rho }(f)$; some sources define quadrature error with the opposite sign, so the convention must accompany any formula.

For a fixed Hilbert space, a bounded self-adjoint Hamiltonian operator with a finite step $H_i\to H_f$ at $t=0$ does not deliver an instantaneous rotation of the quantum state. Integrating the Schrödinger equation across a short interval givesFor a normalized quantum state and $\Vert H(t)\Vert \le M$ in operator norm, the difference has norm at most $2ϵM/\hslash$, which tends to zero. The time derivative can jump even though the quantum state is continuous. Here $\hslash$ is the reduced Planck constant. An instantaneous replacement of the eigenbasis changes coordinates, not the vector being represented.

For a real switch occupying $0\le t\le \tau$, remove any real scalar contribution $a(t)I$ that produces only a global phase, and write $|\tilde{\psi }(t)⟩=e^{i{\int }_0^{t}a(s)ds/\hslash }|\psi (t)⟩$. The same integral argument gives the sufficient bounded-model estimateThus $\epsilon \ll 1$ controls the sudden approximation up to a global phase. For any normalized final energy eigenstate, its probability amplitude differs from the instantaneous-switch value by at most $\epsilon$, and its Born rule probability differs by at most $2\epsilon$. These are sufficient error bounds, not necessary criteria. For the fixed-gap two-level sudden quench, any switching path $H(t)=-\Delta \mathbf{n}(t)\cdot \mathbf{\sigma }/2$ with a unit vector $\mathbf{n}(t)$ has operator norm $\Delta /2$, giving $\epsilon \le \Delta \tau /(2\hslash )$.

This estimate is deliberately for bounded Hamiltonian operators. An unbounded energy spectrum does not supply a finite global $M$; relevant quantum states, operator domains and excited energy eigenvalues require separate control. The adiabatic theorem also requires hypotheses about the path and its spectral gaps; there is no universal switch-duration criterion covering all spectra. The cold-atom collapse and revival after a lattice quench illustrates how a single switch can be sudden for one degree of freedom and slow for another.

The quantum variational principle estimates the lowest energy by choosing a trial wavefunction and computing its expectation value, without solving the Schrödinger equation exactly. Minimizing over a manageable family of trial wavefunctions gives an upper bound; enlarging that family can improve the estimate.

Let $H$ be a self-adjoint operator bounded below, and put $E_{\ast }=\inf \sigma (H)$. For $\Vert \psi \Vert =1$ in its operator domain $D(H)$,Trial states may also lie in the form domain $Q(H)=D((H-bI{)}^{1/2})$, where $b\le E_{\ast }$. Then use the quadratic form of a positive quantum Hamiltonian shifted by $b$:This extends the expectation value beyond $D(H)$ without requiring $H\psi$ to exist. Equality for a normalized state holds exactly in the energy eigenspace at $E_{\ast }$, if that eigenvalue exists. ^[ref].

A short proof makes the direction of the bound memorable. If $H$ has a complete orthonormal basis of energy eigenstates ${\varphi }_n$, writeThe expectation value is a weighted average of allowed energies:Equality requires every occupied energy eigenstate to have energy $E_{\ast }$. ^[ref]. With continuous spectrum, the spectral theorem replaces the sum by an integral against the scalar spectral measure; the same nonnegative-difference argument works.

For an unnormalized trial wavefunction, divide by its squared norm: $R_H[\psi ]=q_H[\psi ]/\Vert \psi {\Vert }^2$ is the form version of the Rayleigh quotient. The quantum variational principle supplies an upper bound on $E_{\ast }$; it supplies neither a lower bound on $E_{\ast }$ nor, by itself, a guarantee that a bound state exists.

To see the method at work, choose the real Gaussian wave packetThe Gaussian integral gives wavefunction normalization and the two useful position averages:Thus its position probability density is a normal distribution with standard deviation $1/\sqrt{2a}$: large $a$ means narrow confinement. In units $\hslash =m=1$, integration by parts and ${\psi }_a^{\prime }=-ax{\psi }_a$ give the kinetic energyThese smooth, rapidly decaying trial wavefunctions belong to the operator domains of both polynomial Quantum Hamiltonians below. Narrowing the Gaussian wave packet increases its kinetic energy; widening it increases the potential energy in either confining potential.

For the quantum harmonic oscillator, take $\hslash =m=\omega =1$:The trial values at $a=1/2,1,2$ are respectively $0.625,0.5,0.625$. The derivative $E_{\mathrm{ho}}^{\prime }(a)=(1-a^{-2})/4$ changes sign at $a=1$, where kinetic energy and potential energy are both $1/4$. Here the trial family contains an exact ground state:The zero-point energy remains positive even at the best width. For comparison, removing the potential leaves $E(a)=a/4\to 0$ as $a\to 0$: the free Quantum Hamiltonian on the line has spectral bottom zero but no normalizable zero-energy energy eigenstate. A sequence of ever broader trial wavefunctions need not approach an actual bound state.

For a pure quartic oscillator, keep $\hslash =m=1$ and set the quartic coefficient to one:The trial values at $a=1,2,3$ are respectively $1,0.6875,0.833333\dots$. Setting the derivative to zero givesThe sign of this derivative changes from negative to positive at the optimum, giving the smallest energy within this Gaussian wave packet family:At this width, $⟨T⟩\approx 0.454280$ and $⟨V⟩\approx 0.227140$, so $2⟨T⟩=4⟨V⟩$, as expected from the power-law quantum virial theorem for a scale-optimized trial wavefunction. The number $0.681420$ is an upper estimate of the ground-state energy, not an asserted exact energy eigenvalue. Indeed,depends on $x$ for every $a>0$, so no member of this trial family is an energy eigenstate. Nevertheless, a single width parameter already turns a difficult Schrödinger equation into a useful, explicit energy estimate.

The local and mean lensing convergence obey $2\kappa =2\bar{\kappa }+r{\bar{\kappa }}^{\prime }$. Hence the radial derivative of the lens mapping is $1-{\alpha }^{\prime }(r)=1-2\kappa +\bar{\kappa }$. Its zeros are radial gravitational-lensing critical curves, where radial stretching diverges and images merge on crossing the mapped gravitational-lensing caustic.

For the radial Laplace operator $L=-r^{-2}(r^2{u}^{\prime }{)}^{\prime }$ on $0<r<R$, require a finite value at the origin and $u(R)=0$. The map $u\mapsto v=ru$ preserves the norm from $L^2((0,R),r^{2}dr)$ to $L^2(0,R)$ and sends $L$ to $-d^2/d{r}^2$. The regular endpoint condition becomes $v(0)=0$. The self-adjoint realization therefore corresponds to the ordinary Dirichlet boundary conditions for the one-dimensional second derivative. Its eigenvalues and eigenfunctions are displayed above, with $u_n(0)=n\pi /R$ understood by a limit. Their squared weighted norm is $R/2$. Completeness and orthogonality follow from the Fourier sine series on $(0,R)$.

Let $-1<s<1$ and let $S_s$ be the outward-oriented boundary of $x^2+y^2+z^2\le 1$, $z\le s$. For a continuous function $h:[-1,s]\to \mathbb{R}$, the vector field $\mathbf{F}=h(z)(x,y,z{)}^T$ has outward fluxOn the curved unit sphere, the outward unit normal is $(x,y,z{)}^T$, so the normal component is $h(z)$. A horizontal spherical band has area $2\pi dz$, producing the first term. On the top disk, the outward unit normal is $(0,0,1{)}^T$, the normal component is $sh(s)$, and the area is $\pi (1-s^2)$, producing the second term.

For $h=f^{\prime }$ with a continuous first derivative, the fundamental theorem of calculus givesThis direct surface integral proof needs no second derivative of $f$. For $h=1$, it reduces to $\pi (2+3s-s^3)$; the divergence theorem identifies this with three times the enclosed volume.

In any finite-dimensional Euclidean space, this extension is a smooth function and all its partial derivatives at the origin vanish. Away from zero, repeated differentiation gives finite sums of coordinate polynomials divided by powers of $|x|$, multiplied by $e^{-1/|x{|}^2}$. Every such term is bounded in absolute value by $C|x{|}^{-N}{e}^{-1/|x{|}^2}$ near zero. Since Gaussian decay dominates every power, this bound tends to zero faster than any prescribed positive power of $|x|$. Induction using difference quotients then proves smoothness and zero derivatives of every order at zero.

The origin is nevertheless a strict global minimum, because $F(x)>0$ for $x\ne 0$. All its Taylor series coefficients vanish, so the function is not real analytic at zero. Adding a constant preserves the stationary minimum but changes the zeroth-order value. An expression written only for $x\ne 0$ has no stationary point at the excluded origin until an extension is specified.

Radiant flux is the power carried by electromagnetic radiation: radiation energy transferred per unit time, measured in watts, $\mathrm{W}=\mathrm{J}{\mathrm{s}}^{-1}$. Specify what is counted: all radiation emitted by a source, radiation within one beam, or radiation intercepted by a photodetector. Writing $Q(t)$ for the cumulative energy crossing the specified boundary givesThe first expression is an instantaneous derivative; the second is an average over the stated time interval. If $0.020\mathrm{J}$ arrives during $0.010\mathrm{s}$, the mean radiant flux is $2\mathrm{W}$. The photodetector's absorbed power equals this incident radiant flux only if it absorbs all of the incident radiation.

Keeping radiant flux fixed while widening a beam reduces its irradiance, the incident power per receiving area. Radiance additionally resolves the propagation direction and has units $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$; an optical camera can collect different radiant fluxes from sources with the same radiance but different apparent solid angles. Luminous flux, measured in lumens, weights the optical spectrum by the human eye's standardized response instead of counting energy equally at every wavelength. Equal radiant fluxes need not give equal luminous fluxes.^[ref]

Radiant power also differs from the optical power of an optical lens, which is reciprocal focal length in dioptres. Likewise, astronomical radiative flux is power per area. Stating the units prevents these shared words from hiding different physical quantities.