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Consider a nonlinear diffusion equation with uniform supply, $\varphi h_t=D_m(h^m{h}_x{)}_x+R$, on $x>0$, with an absorbing boundary $h(0,t)=0$ and initially $h=0$. Far from the boundary, $h=Rt/\varphi$. Balancing the time derivative, supply, and diffusion givesThe similarity solution satisfiesThe outward boundary volume flux per unit width is $Q=R\ell c_m$, where $c_m={\lim }_{\eta \downarrow 0}{F}^m{F}^{\prime }$. Integrating the equation givesThus the growing region of depleted storage fixes the discharge prefactor, and $Q\propto t^{(m+1)/2}$.

A time-dependent response can be represented by its Fourier transform in angular frequency. With $\hat{f}(\omega )=\int f(t)e^{i\omega t}dt$, a time derivative transforms to $-i\omega \hat{f}$ when boundary terms vanish. A time-translation-invariant response convolution becomes multiplication by a frequency-dependent response coefficient. This is why a dispersive electrical conductivity is naturally specified in the frequency domain. The inverse-transform modes use $e^{-i\omega t}$.

The time derivative of a generalized coordinate. It is an independent velocity argument of a mechanical Lagrangian, but need not have the units or direction of a Cartesian velocity: for an angular coordinate it is an angular velocity. A regular Legendre transform replaces these variables with the corresponding canonical momenta.

The velocity of a fluid element is the time derivative of its position with its Lagrangian coordinates $X,Y$ held fixed. Direct differentiation givesReplacing the Lagrangian coordinates by the current Eulerian coordinates therefore gives the velocity fieldThe distinction between following a fluid element and differentiating at fixed Eulerian coordinates is what connects the given deformation map to this velocity field.

Yes, for genuine precession of a nonspherical disc. Let its unit normal be $\mathbf{n}=(\sin \alpha \cos \omega t,\sin \alpha \sin \omega t,\cos \alpha )$. Its trace-free mass quadrupole moment has the axisymmetric formThe $xz,yz$ entries vary at $\omega$, and $xx-yy,xy$ at $2\omega$. Their nonzero second and third time derivatives give an oscillating far-field metric perturbation and positive average radiated power. For $\alpha =\pi /2$ the $\omega$ terms vanish but the $2\omega$ terms remain. The precise condition is $\omega \ne 0$ and $\sin \alpha \ne 0$: the literal inequality $\alpha \ne 0$ also allows $\alpha =\pi$, where the normal is fixed and no such gravitational radiation occurs. For a uniform thin disc of mass $M_d$ and radius $b$, $Q_0=-M_d{b}^2/4$, confirming that it is not zero.

Let $z$ measure height above the substrate, and write $\mu =\rho \nu$. In the lubrication approximation, the horizontal scale $\ell$ is much larger than the typical height $H_0$. Vertical momentum balance is hydrostatic to leading order, while the horizontal balance retains the dominant vertical shear:The no-slip boundary condition gives $u(0)=0$; negligible air viscosity and constant surface tension give $u_z(h)=0$. Integrating twice and then through the depth yieldsThe free-surface kinematic boundary condition and incompressibility imply $h_t+q_x=0$. ThereforeThe cross-sectional area $A=\int hdx$ is conserved when the endpoint volume flux vanishes.

The relevant small groups follow from the same balances. Put ${ϵ}_h=H_0/\ell$ and $U_{\ast }=g{H}_0^3/(3\nu \ell )$. Small ${ϵ}_h$ suppresses horizontal viscous derivatives and makes vertical viscous corrections to hydrostatic pressure smaller by $O({ϵ}_h^2)$. The ratio of horizontal inertia to vertical viscous stress isThus a small full-length Reynolds number is sufficient but not necessary: thin-layer geometry reduces the inertial ratio. The unsteady viscous ratio $H_0^2/(\nu t_{\ast })$ must likewise be small; on the spreading timescale $t_{\ast }=\ell /U_{\ast }$ it equals the displayed group. If surface tension is retained, its pressure-gradient correction relative to gravity isThe Bond number here uses the horizontal scale, because the capillary pressure is $O(\gamma H_0/{\ell }^2)$ whereas the hydrostatic pressure is $O(\rho g{H}_0)$. We also neglect air shear and evaporation and assume a Newtonian fluid with constant kinematic viscosity.

To find the similarity solution, put $K=g/(3\nu )$. Conservation of $A$ gives $h\sim A/\ell$, and balancing the time derivative against the nonlinear diffusion gives ${\ell }^5\sim K{A}^{3}t$. A normalized ansatz isSubstitution gives $(f^3{f}^{\prime }{)}^{\prime }=-(f+\xi f^{\prime })/5$. For the symmetric zero-flux profile, one integration gives $f^3{f}^{\prime }=-\xi f/5$, and on its positive support a second integration givesThus the solution is a compactly supported Barenblatt solution of the porous medium equation. Let $\ell (t)$ denote its half-width. In dimensional variables,Here the positive-part notation sets the height to zero outside $|x-x_c|<\ell$. The area normalization fixes the coefficient, not merely the exponent. With $x-x_c=\ell \sin \theta$,Writing the remaining integral as $I$, the final widths areThe height falls like $t^{-1/5}$ and the width grows like $t^{1/5}$. The volume flux tends to zero at the front, even though the profile slope diverges; this gives a valid compactly supported weak solution of the reduced equation. The diverging edge slope means the lubrication approximation is not uniform at a microscopic contact line. Surface-tension or contact-line physics supplies an inner regularization without changing the gravity-dominated bulk similarity law in its validity regime. This is an exact source-type solution for $t>0$; an arbitrary finite initial drop approaches an appropriate late-time profile rather than being exactly represented by this singular initial profile.

The initial Fourier transform is $\hat{\zeta }(k,0)=2b{\zeta }_0/(1+b^2{k}^2)$. Initial rest gives zero time derivative, so each wave mode evolves by $\cos (\sqrt{g|k|}t)$. Fourier inversion therefore givesOn $x=Vt$, the product of cosines produces phases ${\phi }_{\pm }(k)=Vk\pm \sqrt{gk}$ with prefactor $b{\zeta }_0/\pi$. Only the minus phase has a stationary point:Expanding quadratically there and evaluating the Gaussian oscillatory integral gives the stationary-phase factor $\sqrt{2\pi /(t{\phi }_{-}^{\prime \prime }(k_{\ast }))}$ and phase shift $+\pi /4$. The other phase has no stationary point. Using $q=\sqrt{k}$ makes both endpoint phases smooth with nonzero endpoint derivative and amplitude vanishing at zero, so endpoint contributions are smaller. ThusThe selected wavenumber has group velocity $V$, explaining why it determines the long-time signal on this ray.

Assume $B$ is continuously differentiable. Apply the chain rule to each component with $z_j=t{x}_j$:Contracting the first identity with $x_j$ provesHere $\nabla$ differentiates with respect to $x$ at fixed $t$, while the time derivative holds $x$ fixed. The equality follows directly even at $t=0$; no division by $t$ is needed. It holds componentwise for the vector field, so it is not restricted to scalar functions.

Substitute the separated solution $y=e^{-\alpha t}F(t)\sin (nx)$ into the damped wave equation. Its time derivatives givewhereas $y_{xx}=-n^2{e}^{-\alpha t}F\sin (nx)$. ThusThe last two conditions follow from $y(x,0)=0$ and $y_t(x,0)=\sin (nx)$. For $0<\alpha <n$, let $\nu =\sqrt{n^2-{\alpha }^2}>0$. The simple harmonic motion equation has solution of a differential equationThis is an underdamped normal mode: its oscillations have the decaying factor $e^{-\alpha t}$.

For positive linear drag, $v_x$ decays exponentially toward zero. The vertical velocity decreases monotonically toward $-mg/\alpha$, with an initially steeper gradient $-g-(\alpha /m)v_0\sin \theta$ and positive second time derivative for an upward launch. It crosses zero atThe right panel shows the two components and the formal terminal velocity. To identify the end of the physical flight, integrate the vertical velocity once:The first positive root of $y(t)=0$ is ground contact. The terminal velocity need not be reached before that event; dashed portions in the sketch are mathematical extensions without a ground surface.

For ordinary conservative classical mechanics, the Lagrangian is $L(q,\dot{q},t)=T-V$, where $T$ is the kinetic energy and $V$ the potential energy. More generally it is the function whose action $\int Ldt$ satisfies the principle of stationary action. The canonical momenta and Hamiltonian areFor a regular Lagrangian, the first equations can be inverted to express the generalized velocities in terms of $q,p,t$; this defines the Legendre transform on phase space.

Along a solution of the Euler-Lagrange equations, ${\dot{p}}_i=\partial L/\partial q_i$. Taking a total time derivative givesThus an autonomous Lagrangian hasFor the usual time-independent mechanical coordinates and a velocity-independent potential energy, the kinetic energy is quadratic in the generalized velocities, so ${\sum }_i{p}_i{\dot{q}}_i=2T$ and $H=T+V$. The conservation proof itself uses the Euler-Lagrange equations, rather than assuming this special expression for energy.

For electric potential difference and electric current varying as sinusoids, choose the phasor conventionThe electrical impedance isIts modulus gives the voltage-to-current amplitude ratio, and its complex argument gives the voltage-minus-current phase difference. For ideal components,These follow from Ohm's law $V=RI$, the capacitor law $I=CdV/dt$, and the inductor law $V=LdI/dt$. In phasor notation a time derivative becomes multiplication by $i\omega$, turning the linear ordinary differential equations of the electrical circuit into algebraic equations. Electrical impedances add in a series circuit; their reciprocals, the electrical admittances, add in a parallel circuit. This computes both amplitudes and phase differences in the steady response to sinusoids.

Definitions and pressure convention. $R(t)$ is the scale factor, $t$ cosmic time and dots time derivatives. $G$ is the gravitational constant, $c$ the speed of light, $\rho$ the mass-equivalent density of the material fluid and $k=+1,0,-1$ the spatial-curvature sign in this normalization. $\Lambda$ here has units of inverse time squared, rather than inverse length squared. Comparison of the acceleration equation with $\ddot{R}/R=\Lambda /3-4\pi G(\rho +3P/c^2)/3$ fixesThis $ϵ$ is a pressure parameter, unrelated to Question 1's radiative efficiency. pressureless matter has $P=0$ and $ϵ=0$. When solving the generalized model below, $ϵ$ is assumed constant.

Evidence for a positive vacuum component. Standardized Type Ia supernovae measure the luminosity-distance–redshift relation. Their corrected apparent magnitudes at cosmological distances imply greater distances than a decelerating matter-dominated model of comparable parameters, favoring late accelerated expansion. Light-curve standardization, interstellar extinction, selection bias and possible population evolution must be controlled rather than treating every supernova as an identical candle.^[ref][ref]

The cosmic microwave background and baryon acoustic oscillation measurements constrain geometry and matter density in combination. Near spatial flatness together with a subcritical matter density leaves room for an additional smooth component; cosmic ages and structure growth supply consistency tests. A positive cosmological constant fits these observations within the standard framework. They do not establish that every allowed dark-energy theory must be a strictly constant vacuum energy; the inference depends on the cosmological and gravitational model.

pressureless matter evolution. Set $k=0$ and $ϵ=0$. Since $\dot{H}=\ddot{R}/R-H^2$, eliminating $\rho$ givesFor $H=b\coth (at)$, $\dot{H}=-ab{\mathrm{csch}}^2(at)$ and ${\coth }^{2}x=1+{\mathrm{csch}}^{2}x$. ThereforeOn the expanding Big-Bang branch with positive $\Lambda$, chooseThe origin $t=0$ is the Big Bang; an arbitrary time translation would shift this origin. Integrating gives $R\propto {\sinh }^{2/3}(at)$. At early times $H\sim 2/(3t)$, while at late times $H\to \sqrt{\Lambda /3}$.

Age and the density fractions. Dividing the flat constraint by $H^2$, with ${\rho }_{\mathrm{crit}}=3H^2/(8\pi G)$, givesThe solution implies $\sqrt{{\Omega }_{\Lambda }}=\tanh (at)$, soHere $H_0=2.3331\times 10^{-18}{\mathrm{s}}^{-1}$, using the printed megaparsec conversion. Taking one year as $365.25$ days givesFor the same $H_0$, an Einstein-de Sitter universe hasA cosmic model must be older than its oldest objects, with additional time allowed for their formation. Old stellar populations are dated through stellar-evolution fitting, particularly the globular cluster main-sequence turnoff, with uncertainties from distances, composition and input stellar physics. A pre-examination globular-cluster age constraint on cosmology obtained a $95\%$ lower cosmic-age limit of $11\mathrm{Gyr}$ after allowing for formation time. This conflicts with the $9.05\mathrm{Gyr}$ model but is compatible with the $13.09\mathrm{Gyr}$ calculation.^[ref] Individual stellar age estimates are not exact clocks, and the numerical $13.09\mathrm{Gyr}$ is the result for the specified $H_0$ and matter fraction, not a claim to be a contemporary best-fit age.

Constant nonzero pressure parameter. Eliminating $\rho$ before setting the pressure to zero instead givesHenceFor constant $ϵ>-3$, the positive-density expanding Big-Bang branch isEquivalently use $t^{\prime }=(3+ϵ)t/3$ in the pressureless matter solution. The scale factor obeys $R\propto {\sinh }^{2/(3+ϵ)}[(3+ϵ)H_{\Lambda }t/2]$, and the early limit is $H\simeq 2/[(3+ϵ)t]$. The age of a flat barotropic matter-vacuum universe with the same present density fractions and $H_0$ isThusThe small positive pressure makes the early universe expand with a smaller age at the same late expansion parameters.

The acceleration transition. Normalize the present scale factor to one. Constant $ϵ$ and the continuity equation give $\rho ={\rho }_0(1+z{)}^{3+ϵ}$. Dividing the acceleration equation by $H_0^2$ givesAt the barotropic matter-vacuum acceleration transition its right-hand side vanishes. For the parameters here, $1+ϵ>0$, so the past transition isAt greater redshift the material term dominates and expansion decelerates; at smaller redshift vacuum energy dominates and expansion accelerates. The pressure factor $1+ϵ$ belongs in the numerator of the gravitating material term; omitting it would confuse this model with pressureless matter.

With constant rest mass, $\mathbf{p}=m\gamma \mathbf{v}$ and the chain rule gives $\dot{\gamma }={\gamma }^3(\mathbf{v}\cdot \mathbf{a})/c^2$. Hence the laboratory relativistic force isUse the vector triple product identity $\mathbf{v}\times (\mathbf{v}\times \mathbf{a})=\mathbf{v}(\mathbf{v}\cdot \mathbf{a})-v^2\mathbf{a}$, and ${\gamma }^3(1-v^2/c^2)=\gamma$, to obtainThe relativistic energy is $E=m\gamma c^2$, whose time derivative is $m{\gamma }^3\mathbf{v}\cdot \mathbf{a}$. Taking the dot product of the first expression for $\mathbf{F}$ with $\mathbf{v}$ gives the same quantity, because $\gamma +{\gamma }^3{v}^2/c^2={\gamma }^3$. Thus $dE/dt=\mathbf{F}\cdot \mathbf{v}$. Since $d/d\tau =\gamma d/dt$, the four-force isIts tensor contraction with the four-velocity vanishes: $F^{\alpha }{U}_{\alpha }=-{\gamma }^2\mathbf{F}\cdot \mathbf{v}+{\gamma }^2\mathbf{F}\cdot \mathbf{v}=0$. This orthogonality also follows by differentiating $p^{\alpha }{p}_{\alpha }=-m^2{c}^2$. Resolving the laboratory acceleration relative to $\mathbf{v}$ gives ${\mathbf{F}}_{\parallel }=m{\gamma }^3{\mathbf{a}}_{\parallel }$ and ${\mathbf{F}}_{\perp }=m\gamma {\mathbf{a}}_{\perp }$.

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.

The proof is a single time derivative. Apply the product rule to each dot product and use Newton's second law:Integrating from $0$ to $T$ gives the endpoint identity; the stated sublinear condition removes the endpoint term only in the infinite-time limit. Equivalently, for the scalar inertia $I=\frac{1}{2}{\sum }_i{m}_i|{\mathbf{r}}_i{|}^2$ used in the pairwise moment-of-inertia identity, $\dot{I}=G$ and $\ddot{I}=2K+{\sum }_i{\mathbf{r}}_i\cdot {\mathbf{F}}_i$. Specifying this factor of one half avoids confusion with another common inertia convention.

In On a Mechanical Theorem applicable to Heat by R. Clausius (1870, pp. 123-124), the author defines stationary motion through bounded positions and velocities, includes planetary orbits and elastic vibrations, and names the average negative half of the force-position sum the “virial”. That historical convention differs from the auxiliary $G(t)$ defined above. His proof on pp. 126-127 keeps an endpoint term, then explains why it vanishes over a complete period or becomes small in a long bounded irregular motion. The same paper separates internal and external forces and includes uniform boundary pressure on pp. 125-126: confinement was already part of the formulation, rather than a reason to omit external stresses.^[ref]