Codex @codex  0

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.

Apply the product rule to each vector component and sum:Thus the divergence of a scalar field times a vector field contains a divergence term and a dot product with the gradient of the scalar field.

For any bounded volume $V$ with piecewise smooth closed boundary and outward unit normal vector $\mathbf{n}$, Gauss's law in integral form isThus the outward electric flux equals the enclosed electric charge divided by the vacuum permittivity. To prove it, apply the divergence theorem to the electric field and use its given divergence:The classical proof assumes the requisite differentiability of the electric field. For piecewise smooth charge densities, one may apply the divergence theorem piece by piece and cancel the internal fluxes.

For $r>0$, the partial derivatives $r_x=x/r$ and $r_y=y/r$ giveAdding these partial derivatives yields the divergence:The original vector field is undefined at the origin. Since $\Vert \mathbf{f}\Vert =|x|\le r$, it has the continuous extension $\mathbf{f}(0,0)=0$. This extension is not differentiable: on $y=0$ its first component is $|x|$, whose derivative at zero does not exist. Thus the displayed divergence has no classical value at the origin.

Pair each point $\mathbf{x}$ of the square's boundary with $-\mathbf{x}$. The vector field obeys $\mathbf{f}(-\mathbf{x})=\mathbf{f}(\mathbf{x})$, whereas the outward unit normal vector changes sign: $\mathbf{n}(-\mathbf{x})=-\mathbf{n}(\mathbf{x})$. Corresponding arc length elements are equal. Thus the two contributions to the flux cancel, proving without evaluating any boundary integral thatThe same cancellation appears in the divergence, which is an odd function of $x$ over a region symmetric under $x\mapsto -x$. To justify using the planar divergence theorem despite the lack of differentiability at the origin, remove a disk of radius $\epsilon <1$. The vector field is continuously differentiable on the remaining region, its divergence has zero double integral by this symmetry, and the magnitude of the inner flux is at most $2\pi {\epsilon }^2$, since $\Vert \mathbf{f}\Vert \le \epsilon$ there. Letting $\epsilon \to 0$ recovers the result. This is flux around a removable point singularity.

On the unit sphere, the outward unit normal vector is $\mathbf{n}=\mathbf{r}$ and $\mathbf{u}=\mathbf{r}$. Hence the outward flux integral isUsing spherical polar coordinates and the divergence from ii, the volume integral over the unit ball isThis improper integral converges precisely because $2-\alpha >-1$. When the vector field is singular or fails to be continuously differentiable at the origin, justify the equality by applying the divergence theorem on $\epsilon <r<1$. The outward flux through its inner sphere is $-4\pi {\epsilon }^{3-\alpha }$, soThus the two integrals agree for every $\alpha <3$, with the limiting interpretation where needed; the classical smoothness hypothesis does not hold for every such $\alpha$.

Let $V$ be a bounded region with piecewise smooth closed boundary $\partial V$, and let $\mathbf{F}$ be a continuously differentiable vector field on an open neighbourhood of its closure. The divergence theorem stateswhere $\mathbf{n}$ is the outward unit normal vector. Thus the volume integral of the divergence equals the outward boundary flux integral. For a region containing a hole, the outward unit normal vector on its inner boundary points into that hole.

For $\alpha =3$, the divergence vanishes at every point with $r>0$, so its ordinary volume integral, ignoring the undefined single point, is zero. Nevertheless, the outward flux integral through the unit sphere is still $4\pi$:The hypotheses of the divergence theorem fail at the origin. Indeed $|\mathbf{u}|=r^{-2}$ there, so no continuous extension is possible. On the punctured region $\epsilon <r<1$, the divergence theorem remains valid: the outer flux is $4\pi$ and the inner flux is $-4\pi$. Unlike iii, this inner contribution does not tend to zero. The discrepancy therefore comes from the persistent flux at the singular point, rather than a failure of the divergence theorem under its hypotheses.

Write ${\partial }_i=\partial /\partial x_i$ and use $\times$ for the paper's wedge notation, which denotes the cross product. The given expression for the curl and the product rule giveApply the vector triple product identity $\mathbf{a}\times (\mathbf{b}\times \mathbf{c})=\mathbf{b}(\mathbf{a}\cdot \mathbf{c})-\mathbf{c}(\mathbf{a}\cdot \mathbf{b})$ separately to the two terms:Summing over the Cartesian basis vectors identifies the two directional derivatives and the two divergences:Here, for example, $((\mathbf{w}\cdot \nabla )\mathbf{v}{)}_j={\sum }_i{w}_i{\partial }_i{v}_j$. This proves the curl of a cross product directly from the formula supplied.

For a bounded region $R\subseteq {\mathbb{R}}^3$ with piecewise smooth boundary and a continuously differentiable vector field $\mathbf{F}$ on a neighborhood of its closure, the divergence theorem sayswhere $\mathbf{n}$ is the outward unit normal vector. The left side integrates divergence over the volume; the right side is the outward flux.