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For two electrostatic potentials satisfying the Poisson equation $\Delta {\varphi }_i=-{\rho }_i/{\epsilon }_0$, Green's second identity yields the displayed reciprocity formula. The product rule for divergence applied to ${\varphi }_1\nabla {\varphi }_2-{\varphi }_2\nabla {\varphi }_1$ cancels the cross-gradient terms. The normal points outward from the integration region; for an exterior region this means inward toward the excluded object on its inner boundary. Boundary terms may be removed only under suitable decay or boundary conditions.

This is a Gaussian thermostat, with generator and angular accelerationThe canonical fields $X$ and $V$ preserve Liouville volume of a surface geodesic flow. To check this directly, use the canonical coframe of a surface unit tangent bundle $(\alpha ,\beta ,\omega )$, with dual frame $(X,H,V)$ andFor the positive volume $\mu =\alpha \wedge \beta \wedge \omega$, contraction gives ${\iota }_X\mu =\beta \wedge \omega$ and ${\iota }_V\mu =\alpha \wedge \beta$. Both two-forms are closed by the displayed structure equations, so the Cartan formula for the Lie derivative gives ${\mathcal{L}}_X\mu ={\mathcal{L}}_V\mu =0$.

The product rule for divergence now yieldsThe vertical vector field of a surface unit tangent bundle fixes the base point and satisfies $V(v)=iv$, $V(iv)=-v$. Since $E$ depends only on the base point,This is the divergence of a Gaussian thermostat. Reversing the overall sign of the chosen volume form does not change its divergence, so the answer is also valid if Liouville volume is written as the oppositely oriented contact volume.

Combining the electric-field equations gives the Poisson equation ${\nabla }^2{\varphi }_i=-{\rho }_i/{\epsilon }_0$. Apply the product rule for divergence toThe two cross terms in its divergence cancel, leavingThis electrostatic reciprocity identity follows from the divergence theorem, and is exactly Green's second identity for these potentials. Integrating and rearranging givesThe boundary orientation is outward from the chosen region. For smooth charge densities this follows directly from classical differentiation. For isolated point charges, one may excise small spheres and take their radii to zero, provided the other potential is regular at each charge.

Use suffix notation, with repeated indices summed. Contracting the two Levi-Civita symbols givesApplying the product rule to $h=A_i{B}_i$ and substituting the two evolution equations yieldsIn the second line we renamed dummy indices in $A_i{B}_j{\partial }_j{u}_i$ and then combined product derivatives. HenceNo zero-divergence assumption was used in this local identity.

When both fields are solenoidal vector fields, the product rule for divergence rewrites this as a conservation law:Because the volume is fixed, differentiation under its integral and the divergence theorem giveunder the two tangency conditions. Thus the volume integral is conserved. This is helicity conservation by tangent boundary conditions; when $\mathbf{B}=\nabla \times \mathbf{A}$, the density is the magnetic helicity density.

For the given axisymmetric field, the curl in spherical coordinates givesThere are no $\varphi$ derivatives and $A_r=0$. ConsequentlyThe volume form in spherical polar coordinates is $r^2\sin \theta drd\theta d\varphi$, soHere $B_r=0$ at $r=a$, as required for the magnetic-field tangency condition in the conservation argument; tangency of $\mathbf{u}$ is a separate hypothesis.

The formal adjoint of $\mathcal{L}=\Delta -k^2$ is itself. With the bilinear convention used in Green second identity, take any complex-valued $v$ satisfying $\mathcal{L}v=0$. The product rule for divergence gives the familyEquivalently, the differential one-formis a closed differential form. Arbitrary linear combinations of adjoint solutions give further members of this family. One can also write it as $\nabla \cdot [v^2\nabla (u/v)]=0$ wherever $v\ne 0$, but the first formula has no division and remains valid at zeros of $v$. For a sesquilinear convention replace $v$ by $\overline{v}$; no such conjugation is built into the bilinear identity used below.

Let $a$ be any constant vector and take $u=\varphi a$. The product rule for divergence gives $\nabla \cdot u=a\cdot \nabla \varphi$. The divergence theorem on a bounded region with piecewise smooth boundary, oriented outward, therefore yieldsSince $a$ is arbitrary, equality of all components provesHere $\varphi$ is continuously differentiable on a neighbourhood of the closed region.

For the side of this right circular cone, the parameter tangents are $x_r=(\cos \theta ,\sin \theta ,\sqrt{3})$ and $x_{\theta }=(-r\sin \theta ,r\cos \theta ,0)$. Their cross product in the outward order isThe sign is outward because the solid right circular cone lies at smaller cylindrical radius for fixed height. Reversing the parameter order reverses the oriented surface element.

To check the integral identity, the closed boundary must include the top Euclidean disk $z=1$, radius $1/\sqrt{3}$, as well as the curved side. For $\varphi =z^2$, horizontal components cancel on integrating $\theta$. The side contribution isOn the top Euclidean disk $\varphi =1$ and $d\mathbf{S}=e_{z}dA$, so its contribution is $(\pi /3)e_z$. The total is $(\pi /6)e_z$. Independently, the cross-section of the solid at height $z$ has area $\pi z^2/3$, and $\nabla \varphi =2z{e}_z$, givingThus the two sides agree. The curved side alone is not a closed surface and does not satisfy this volume identity. The apex has zero area; alternatively one can truncate at height $\epsilon$ and let $\epsilon \downarrow 0$, with the extra boundary contribution vanishing.

For an arbitrary constant vector $\mathbf{c}$, the product rule for divergence givesThe divergence theorem therefore impliesSince this holds for every $\mathbf{c}$,

For $\varphi =x+y$, $\nabla \varphi =(1,1,0)$. The ball of radius $a$ has volume $4\pi a^3/3$, soOn the sphere, use $x=a\sin \theta \cos \phi$, $y=a\sin \theta \sin \phi$ and the stated outward oriented surface element. The first component of the surface integral isand symmetry gives the same second component and a zero third component. The two sides agree.

The product rule for divergence gives $\nabla \cdot (r^2\mathbf{r})=\nabla (r^2)\cdot \mathbf{r}+r^2\nabla \cdot \mathbf{r}$. Since $\nabla \cdot \mathbf{r}=3$, the result is

The product rule for divergence, $\nabla \cdot \mathbf{r}=3$, and the result in (a) give, for $r>0$,This is a pointwise identity away from the origin, where the radial unit vector is undefined.