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In a cross-section normal to the cylinder axes, measure $y$ from the line of contact. The two circular boundaries have the parabolic approximationfor their separation. The contact point of the meniscus is at $y=w+o(w)$, so the leading cross-sectional area isThe area of the small meniscus cap is $O(w^4/a^2)$ and is lower order. At its upper end the gap has width $h(w)=w^2/a$. A tangent semicircle therefore has radius $w^2/(2a)$ and curvature

The Young–Laplace equation makes the liquid pressure, relative to the nearly uniform gas pressure,At a fixed $y$, lubrication theory gives a planar Poiseuille flow through a gap of width $h(y)$, with axial flux per unit $y$Integrating across the cusp,The continuity equation $A_t+Q_z=0$ now gives

Set $q=w^3$. Since $w^4{w}_z=\frac{1}{5}(q^{5/3}{)}_z$, this is the porous medium equationConservation of the fixed volumeand dimensional analysis give the self-similar solution $q=t^{-3/8}F(z/t^{3/8})$. One integration of the resulting ordinary differential equation yields the compactly supported Barenblatt solutionEquivalently,Its tip is $z_N=(C/B{)}^{1/2}{t}^{3/8}$. Usingin the volume constraint gives $C^2=8aV\sqrt{B}/\pi$. Therefore

For vertical cylinders at equilibrium, hydrostatic pressure gives $p-p_{\mathrm{atm}}=-\rho gz$. Balancing this with the capillary pressure $-2\gamma a/w^2$ gives the large-height profile

Vertically integrating the continuity equation and imposing a steady axisymmetric flow givesThus the inward-positive mass accretion rate isThe zero-torque condition at $R_{\mathrm{ISCO}}$ gives the standard Shakura--Sunyaev thin disk relationCombining this relation with mass conservation, or substituting it into the stated drift equation, yieldsFar outside the inner edge this reduces to $u_R\simeq -3\nu /(2R)$.

The probability density and probability current areFor a stationary state, $\psi (x,t)=\chi (x)e^{-iEt/\hslash }$, so $\rho =|\chi {|}^2$ is independent of time. The continuity equation then gives $\partial J/\partial x=0$. Hence

The nearly occluding sphere in a cylindrical tube has gap thickness $H=O(ϵa)$ and axial length $L=O(\sqrt{aH})=O(a\sqrt{ϵ})$. In the gap, lubrication theory therefore givesFar ahead of and behind the sphere, the length and width scales are both $a$, so

In the sphere frame the tube wall at $y=0$ moves with velocity $-U$ and the sphere surface at $y=h$ is stationary. The local Couette-Poiseuille flow in a thin gap isIts axial volume flux per unit circumferential width isDifferentiating the profile and eliminating $p_x$ gives the two wall stressesThe continuity equation integrated across the gap says that changes of $q$ along $x$ are balanced by circumferential flux divergence. Circumferential variations occur on scale $a$, much longer than the axial scale $a\sqrt{ϵ}$, so $q=q(\theta )$ is independent of $x$ at leading order.

Put $h=h_0(1+{\xi }^2)$ with $x=\sqrt{2a{h}_0}\xi$. The leading $O({ϵ}^{-3/2})$ pressure jump must vanish in the global balance from part ii. Equivalently, the pressure recovery condition in lubrication flow givesUsing $I_2=\pi /2$ and $I_3=3\pi /8$,

The total flux through the gap is consequentlyFar from the sphere, translation of the tube contributes $-\pi a^{2}U$, while Hagen-Poiseuille flow contributes $\pi a^4{p}_x/(8\mu )$. Equating these fluxes gives

At the next order, substitute $q=-2U{h}_0/3$ into the tube-wall shear and integrate through the gap:The balance in part ii then gives the leading pressure drop

Finally, the shear on the sphere integrates tobecause $I_1=\pi$ and $I_2=\pi /2$. Thus the narrow gap exerts no net leading axial shear force on the sphere even though its local shear is nonzero; the leading hydrodynamic force transfer to the sphere is through the large lubrication pressure acting on its gently sloping surface.

For a binary fluid mixture, the scalar compositional order parameter $\varphi (\mathbf{r})$ may be taken as the local concentration difference between its two molecular species. Its spatial integral is fixed by their separately fixed total amounts:Locally, $\varphi$ can change only through transport and therefore obeys a continuity equation. By contrast, $\mathbf{p}$ is the local polar order parameter measuring the mean tail-to-head orientation of the surfactant molecules. Individual molecules can rotate in place, so the integral of $\mathbf{p}$ need not be conserved.

The standard Shakura--Sunyaev thin disk assumes: a steady state; axial symmetry; a geometrically thin disk $H/R\ll 1$; Newtonian, nearly Keplerian circular motion outside the inner edge; subsonic radial drift $|u_R|\ll u_{\varphi }$; vertical hydrostatic equilibrium; negligible disk self-gravity; an optically thick, locally thermal spectrum; local radiative balance between viscous heating and cooling; and a local alpha-viscosity stress with a zero-torque inner boundary near the innermost stable circular orbit. These assumptions also exclude dynamically dominant winds and large-scale external torques from the standard solution.

Start from the continuity equationIntegrate vertically, use axial symmetry and steadiness, and define the surface density of a disk $\Sigma =\int \rho dz$. Thenso the inward-positive accretion rate is

The homogeneous part of the continuity equation gives $\dot{\bar{n}}+3H\bar{n}=0$. At first order,For the density contrast $\delta =\delta n/\bar{n}$, the background equation cancels the expansion terms, leaving

The electric dipole moment of a localized charge density isDifferentiate and use the continuity equation $\dot{\rho }+\nabla \cdot \mathbf{J}=0$:Localization makes the boundary integral vanish. Component by component,This is the time derivative of the electric dipole moment.

For a plane wave $A{e}^{i(kx-Et/\hslash )}$, the probability current isThe incident amplitude is one and the transmitted amplitude is $T$, soThis is the transmission probability for quantum tunnelling. In the stationary state, $\partial \rho /\partial t=0$, so the continuity equation makes the net current independent of position. The reflected current therefore supplies the remainder:

Write $\mu =\hat{\mathbf{k}}\cdot \hat{\mathbf{p}}$ and use the Legendre polynomial recurrence relation. The definition in the question is inverted byFor Newtonian-gauge potentials the scanned geodesic equation is the standard identityIts monopole and dipole are $S_0={\Phi }^{\prime }$ and $S_1=k\Psi /3$. Projecting the photon Boltzmann hierarchy onto $P_0$ and $P_1$ therefore givesThe first is the photon continuity equation; the second is the photon Euler equation, with the photon quadrupole providing the anisotropic-stress term. Indeed, with photon density contrast ${\delta }_{\gamma }=4{\Theta }_0$ and velocity divergence ${\theta }_{\gamma }=3k{\Theta }_1$,

For $\gamma =5/3$, the shock data give ${\rho }_{\mathrm{ps}}=4{\rho }_0$ and $u_{\mathrm{ps}}=3V_s/4$. Since $V_s=R/(3t)$, the prescribed homologous spherical flow isPut $\xi =r/R(t)$ and $\rho ={\rho }_{0}D(\xi )$. The spherical continuity equation becomesso $D\propto {\xi }^9$. Matching the post-shock density givesThe material acceleration isThe radial Euler equations for an inviscid fluid thus give $dp/dr=3\rho r/(16t^2)$. Integrating from the centre to the shock,Since $p_{\mathrm{ps}}={\rho }_0{R}^2/(12t^2)$,

Integrating the incompressible continuity equation $u_x+v_y=0$ across the moving gap, with $v(x,0)=0$ and $v(x,h)=-V$, gives the Reynolds lubrication equationHencefor a constant $C$, and the flux formula from part (b) givesDefine the spatial meanBecause both ends meet fluid at pressure $p_0$, the pressure recovery condition in lubrication flow isIt determinesTherefore the requested pressure gradient in a translating and squeezing finite gap is

With $\mathcal{G}=0$ and $\mathcal{T}=2\pi v_m{h}^{\prime }\Sigma$, the evolution equation reduces to the radial continuity equationor, for the mass per unit radius $M_r=2\pi r\Sigma$,The outward mass flux is $F_M=2\pi r{v}_m\Sigma$. If $\dot{M}>0$ denotes the inward accretion rate, steady mass conservation requires $F_M=-\dot{M}$. HenceThis is positive because the magnetic drift velocity satisfies $v_m<0$.

Substitution into the diffusive flux givesThe continuity equation for this drift-diffusion equation with a potential therefore reduces, after cancelling the nonzero exponential, toThe reflecting boundary conditions become $D(u^{\prime }+\beta V^{\prime }u)=0$ at both endpoints. Thus each solution of this eigenvalue problem supplies a separated solution of the original diffusion equation.

For a candidate curve $\rho$, define the velocity field by the convolutionIt is well defined with a finite first absolute moment, because $|K(x-z)|\le L(|x|+|z|)$. Equality of mixed partial derivatives givesThe Hessian matrix bound justifies moving a spatial derivative through this convolution. For smooth densities, the transport equation consequently becomes the continuity equationThis form defines the equation for singular probability measures as well. With initial probability measure ${\rho }_0$, a weak solution is a curve in $C([0,\infty ),{\mathcal{P}}_1)$ such that every $\phi \in C_c^1([0,\infty )\times {\mathbb{R}}^2)$ satisfiesUsing smooth test functions gives the same definition by approximation on compact sets. Continuity in the Wasserstein distance bounds first absolute moments on compact time intervals, so the displayed integrals exist.

For the empirical measure, the chain rule and the particle equations giveIntegrating in time, with zero terminal contribution because $\phi$ has compact support, gives precisely the weak identity and the correct initial empirical measure. Thus ${\mu }^N$ is a global weak solution. The self-interaction term vanishes since $K(0)=0$.

A finite first moment suffices for the velocity, but may not suffice for the printed potential. For example, $W(z)=|z{|}^2/2$ has all the stated properties. If $\mu$ is the law of $(S,0)$ with density $2s^{-3}$ on $s\ge 1$, then $\mathbb{E}S=2$ but $\mathbb{E}{S}^2=\infty$, and $(W\ast \mu )(x)=\infty$ for every $x$. Thus the literal definition of $V$ needs a finite second absolute moment, or another hypothesis ensuring integrability of $W$.

For the specified class ${\mathcal{P}}_1$, define the velocity fields directly by $b_{\mu }=K\ast \mu$ and $b_{\hat{\mu }}=K\ast \hat{\mu }$. Equivalently, use the renormalized interaction potentialThe bounded Hessian matrix and $\nabla W(0)=0$ implyThus $V_{\mathrm{ren}}$ is finite, is twice continuously differentiable in space, and has $\nabla V_{\mathrm{ren}}=\nabla W\ast \mu$, by dominated differentiation under the integral sign. Its rotated negative gradient is exactly $b_{\mu }$. Whenever $W\ast \mu$ is finite, the two potentials differ only by an $x$-independent constant and give the same transport equation.

We check the hypotheses for characteristic representation of a measure-valued continuity equation. For every $x,y$,For any coupling $\gamma \in \Gamma ({\mu }_t,{\hat{\mu }}_t)$, the Lipschitz condition on $K$ also givesTaking the infimum over couplings proves the uniform boundApplying the same estimate at two times of one curve proves time continuity of $b_{\mu }$, uniformly in $x$. Also $|m_1({\mu }_t)-m_1({\mu }_s)|\le d_1({\mu }_t,{\mu }_s)$, so $m_1({\mu }_t)$ is bounded on compact time intervals. Spatial partial derivatives of $b_{\mu }$ are continuous and bounded by $L$, using dominated convergence in space and weak convergence of probability measures in time for the bounded continuous kernel $DK$. All corresponding facts hold for $\hat{\mu }$.

The characteristic representation of a measure-valued continuity equation says that a continuous velocity field, continuously differentiable and globally Lipschitz in space and with locally uniform linear growth, has a global characteristic flow map $X_{\mu }(t;s,a)$ solvingFor each $f\in {\mathcal{P}}_1$, its pushforward measure is the unique weak solution in $C([0,\infty ),{\mathcal{P}}_1)$ of the associated continuity equation. The hypotheses have all been verified above, and the fields have zero divergence, so these are also the two required transport equations. Explicitly,For completeness, the Gronwall inequality givesso these pushforward measures have finite first absolute moments. Integrating $|X_{\mu }(t;0,a)-X_{\mu }(s;0,a)|$ against $f$ proves continuity in the Wasserstein distance. The chain rule proves the weak identity. For uniqueness at a fixed terminal time $t$, use the backward test function $\phi (s,x)=\psi (X_{\mu }(t;s,x))$, where $\psi$ is continuously differentiable with compact support. It satisfies ${\partial }_s\phi +b_{\mu }\cdot \nabla \phi =0$ and has compact spatial support uniformly for $0\le s\le t$, by the backward linear-growth bound. The weak identity with temporal cutoffs therefore gives $\int \psi d{\rho }_t=\int \psi (X_{\mu }(t;0,a))df(a)$, identifying every weak solution with the displayed pushforward measure.

Finally compare the two characteristic curves with the same initial point $a$. Put ${\Delta }_t=X_{\mu }(t;0,a)-X_{\hat{\mu }}(t;0,a)$. The spatial Lipschitz condition and the boxed velocity estimate givefor almost every time; the integral form gives the same conclusion at zeros of $\Delta$. The Gronwall inequality impliesThe pushforward measure $(X_{\mu }(t;0,\cdot ),X_{\hat{\mu }}(t;0,\cdot ){)}_{\#}f$ is a coupling of ${\rho }_t,{\hat{\rho }}_t$. Integrating the last bound against the probability measure $f$ provesThe same constant $L$ depends only on the uniform bound for the Hessian matrix of $W$, and is independent of the prescribed curves and their common initial data.

For a real potential, the time-dependent Schrödinger equation isMultiply it by ${\psi }^{\ast }$, combine with its conjugate multiplied by $\psi$, and note that the real-potential terms cancel. This giveswhere the probability current isThe product rule gives the divergence identity because the gradient cross terms cancel.

For $F(r)=\sin (kr)/r$, the supplied radial identity gives $\Delta F=-k^{2}F$ for $r>0$. The time factor gives $i\hslash {\psi }_t=E\psi$, so the free equation holds forThe apparent origin singularity is removable: $F(0)=k$ and $F(r)=k-k^3{r}^2/6+\cdots$, so the equation extends smoothly there as well. Since the spatial factor is real and its time dependence is a common phase, ${\psi }^{\ast }\nabla \psi =F\nabla F$ is real. Thus $\mathbf{j}=0$, while $\rho =F^2$ is independent of time. Both terms in the continuity equation vanish separately. This is a standing spherical wave; the real spatial factor carries no net radial probability current.

Use $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and restore $\hslash$ temporarily. The free Dirac equation is $(i\hslash {\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0$, with $c=1$. Its Hermitian conjugate is $-i\hslash ({\partial }_{\mu }{\psi }^{†})({\gamma }^{\mu }{)}^{†}-m{\psi }^{†}=0$. Multiplying on the right by ${\gamma }^0$ and using $({\gamma }^{\mu }{)}^{†}{\gamma }^0={\gamma }^0{\gamma }^{\mu }$ gives the adjoint equationThe Dirac adjoint and original equations give $({\partial }_{\mu }\bar{\psi }){\gamma }^{\mu }\psi =im\bar{\psi }\psi /\hslash$ and $\bar{\psi }{\gamma }^{\mu }{\partial }_{\mu }\psi =-im\bar{\psi }\psi /\hslash$. Adding gives the continuity equationThe Dirac current transforms as a four-vector; the positive density is the time component, not a Lorentz scalar. With ordinary units the number current is $c\bar{\psi }{\gamma }^{\mu }\psi$ and the electron charge current is $-e$ times it.

For a charging parallel-plate capacitor of area $A$, neglect fringing. The field in the gap is $E=Q/({ϵ}_{0}A)$, so the displacement current through the gap isAn Amperian loop around the feed wire can bound a surface cutting the wire, which has conduction current $I$, or a bulging surface passing through the dielectric gap, which has no conduction current. Including displacement current makes both surfaces give the same magnetic circulation. It is also required by charge conservation: the divergence of the Ampère-Maxwell equation and Gauss's law yields $\nabla \cdot \mathbf{J}+\partial \rho /\partial t=0$, the continuity equation.

Take the divergence of the proposed Ampère's law. Since the divergence of a curl vanishes identically, it implies $\nabla \cdot \mathbf{J}=0$. However, local charge conservation requires the continuity equationFor example, electric current entering a charging capacitor accumulates electric charge on its plates, so the charge density is time dependent. The proposed equation cannot describe that process.

From Gauss's law, $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. Thus $\nabla \cdot ({ϵ}_0{\partial }_t\mathbf{E})={\partial }_t\rho$, which supplies precisely the missing divergence. Adding the displacement current density gives the Ampère-Maxwell equation:Its divergence now yields the correct continuity equation. Through a fixed surface, the associated displacement current is $I_D={ϵ}_{0}d{\Phi }_E/dt$, where ${\Phi }_E$ is the electric flux. In a vacuum capacitor gap, this equals the conduction electric current charging the plates, even though no charges cross the gap. Charge conservation fixes the required divergence of the correction; by itself it does not exclude adding a further divergence-free vector field.