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Let $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ be the material derivative, and write $h$ for specific enthalpy. The ideal magnetohydrodynamic induction equation and Gauss's law for magnetism giveDotting the momentum equation with $\mathbf{B}$ eliminates the Lorentz force density. Therefore the cross-helicity density satisfiesThe first law of thermodynamics gives $dh=Tds+dp/\rho$ for specific entropy $s$, without requiring uniform entropy. ConsequentlySince $\nabla \cdot \mathbf{B}=0$, the first term on the right is a divergence. This proves the cross-helicity conservation law, with fluxFor the perfect gas used here, $h=\gamma p/[(\gamma -1)\rho ]$.

Put $u_j={\dot{p}}_j+{\partial }_i{W}_{ij}$. The noises required by a joint trajectory are $f_j=u_j+\Gamma {\mu }_j$ and $N_{ij}=W_{ij}+M{\partial }_i{\mu }_j$. Their independent Gaussian white noise weights give the joint path probability of an order parameter and its flux:For the physically reversed path, $p$ is time-even and the transport flux $W$ is time-odd. Therefore $u\mapsto -u$, $W\mapsto -W$, andThe common factor includes the two noise normalizations and the midpoint Jacobian determinant. Its calculation must include both relaxation channels. On a finite spatial grid, let $\mathsf{D}$ and $\mathsf{G}$ represent the divergence and gradient, with $\Delta =\mathsf{DG}$ the discrete Laplacian, and let ${\mathcal{H}}_n=\partial \mu /\partial p$ be the Hessian matrix of $F$ at the midpoint of time step $n$. Differentiating the two required noises with respect to the next configuration and the interval flux gives the Jacobian matrixUsing the Schur complement, its Jacobian determinant, apart from the path-independent power of $\Delta t$, isReversal visits the same midpoint configurations in reverse order, so the product of these factors is unchanged. The joint time-reversal invariance of a path Jacobian therefore involves the complete relaxation operator $\Gamma I-M\Delta$.

Taking the action difference givesRepeated component indices are summed. The second squared norm sums over both flux indices. This is mixed conserved and nonconserved order-parameter dynamics; reversing the order parameter history while leaving the flux unreversed would give the wrong probability ratio.

Writing $a_k{x}_k=\mathbf{a}\cdot \mathbf{x}$ and differentiating givesIts symmetric second-rank tensor part and antisymmetric second-rank tensor part areTaking the trace makes the two symmetric terms cancel, so the divergence isContracting the antisymmetric part with the Levi-Civita symbol gives the curl

For a velocity potential, $u=\nabla \varphi$. Writing $\theta =x-t$ givesIts divergence isThus this potential flow is generally compressible.

Since $\nabla \varphi ={\mathbf{e}}_{\varphi }/r$, the vector field has cylindrical componentsThe divergence in cylindrical coordinates is thereforeDirect use of the curl in cylindrical coordinates givesandThus, definingwe obtain

In the homogeneous background, $\mathbf{v}=0$ and the physical velocity is the Hubble flowThe comoving coordinate $\mathbf{x}$ is held fixed by the time derivative in the fluid equation, soMoreover, spatial homogeneity gives $\nabla \bar{P}=0$. The Euler equation therefore reduces toTaking the divergence in three spatial dimensions givesConsistency with the Poisson equation,then requiresThis is the pressureless, zero-cosmological-constant Raychaudhuri equation, obtained by the Newtonian fluid derivation of the Raychaudhuri equation.

Each field has only an azimuthal component and is independent of $r$ and $\varphi$. The curl in spherical coordinates therefore reduces toand its $\varphi$ component vanishes. The half-angle identitiesshow in both cases that

Apply part (a) with $g(r)=r^{-3}$. Thenaway from the origin, explicitly confirming that each resulting curl has zero divergence.

The divergence isThus the flow is incompressible.

Choose the polar axis along $\mathbf{d}$ and write $d=|\mathbf{d}|$. Sincethe velocity components areTheir divergence isThus the flow is incompressible for $r\ne 0$.

The magnetic field is $B=\nabla \times A$. The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$, so the curl of the curl identity givesA gauge transformation $A\mapsto A+\nabla \chi$ does not change $B$. We may therefore choose the Coulomb gauge $\nabla \cdot A=0$ by solving the Poisson equation ${\nabla }^2\chi =-\nabla \cdot A$. It follows component by component in a Cartesian coordinate system thatThe free-space Green function of the Laplacian then givesFor a localized steady current density, conservation of electric charge gives ${\nabla }^{\prime }\cdot J=0$; integration by parts confirms directly that this integral has zero divergence and hence obeys the Coulomb gauge.

Let the displayed integrand be $F$. Since it has no explicit dependence on $u$ or $v$, the Euler-Lagrange equations for two fields areIts derivatives areThusUsing the Laplacian, gradient, and divergence, these combine into the static Navier-Cauchy equation

For the displacement gradient tensor specified in the question,while the divergence is $\nabla \cdot \mathbf{u}=u_x+v_y$. Hence the isotropic linear-elastic energy density isExpanding the square and collecting terms yields

In the Cartesian basis, writing $F=(F_1,F_2,F_3)$,defines the divergence, while the curl isEquivalently,

In a magnetostatic equilibrium, inertia is absent and all forces balance. A force-free magnetic field is the limiting case in which the Lorentz force density vanishes:Thus the current density is parallel to the magnetic field, so for some scalar force-free parameter $\alpha (\mathbf{r})$,Using Ampère's law in magnetostatics givesTaking the divergence and using both the divergence of a curl is zero and $\nabla \cdot \mathbf{B}=0$ yieldssoThe force-free parameter is therefore constant along every magnetic field line.

Write the flux asAfter expanding the divergence, the principal symbol of a partial differential equation along a candidate solution $u$ is determined byOnly the symmetric part $A_s=(A+A^T)/2$ contributes to $A^{ik}{D}_{ik}u$. The problem is elliptic along $u$ when ${\xi }^T{A}_s\xi \ge 0$ for every $\xi$, and it is strictly elliptic where this quantity is positive for every nonzero $\xi$. It is uniformly elliptic on a set when constants $0<\lambda \le \Lambda <\infty$, independent of the point, satisfyThese definitions separate pointwise positive definiteness from a quantitative lower and upper bound; a degenerate elliptic operator may lose strict ellipticity at some jets.

Taking the divergence of the linear Euler equation and defining the peculiar-velocity divergence $\theta =\nabla \cdot \mathbf{v}$ givesThe assumption of negligible vorticity ensures that this longitudinal variable captures the velocity perturbation relevant to scalar density growth.

The Jacobian determinant of a differentiable characteristic flow map measures its change of volume. The variational equation and the Liouville formula for a fundamental matrix give the displayed formula for a complete smooth field $b$. For kinetic advection with $b=(v,F)$, its divergence is ${\nabla }_v\cdot F$. A zero divergence preserves phase-space Lebesgue measure.

In an orthonormal tetrad, $J^{\hat{0}}=\int f{d}^{3}p$ is the photon number density and $J^{\hat{i}}=\int f{e}^i{d}^{3}p$ is its number flux. Integrating the Boltzmann equation gives ${\nabla }_{\mu }{J}^{\mu }=\int C[f]d^{3}p/E$ when momentum-boundary terms vanish. Collisionless propagation and photon-number-conserving scattering give zero divergence. Photon emission, photon absorption, electron-positron annihilation in cosmology, bremsstrahlung and double Compton scattering generally do not. This is a number current, not an electric charge current: photons are neutral.

For a smooth divergence-free field on ${\mathbb{R}}^3$, the differential form $F={\iota }_{\mathbf{B}}(d{x}^1\wedge d{x}^2\wedge d{x}^3)$ is closed. The radial homotopy formula gives a global magnetic vector potential $\mathbf{A}(\mathbf{x})={\int }_0^{1}s\mathbf{B}(s\mathbf{x})\times \mathbf{x}ds$ with curl $\mathbf{B}$. For constant $\mathbf{B}$, this reduces to $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{x}$.