Euclidean Surface Measure, Divergence, and Green Identities: Examples

The tangent columns are $(e_i,{\mathrm{\partial }}_{i}h)$, so $D{X}^{T}DX=I+Dh(Dh{)}^T$ and F1 gives density $\sqrt{1+\mathrm{\mid }Dh{\mathrm{\mid }}^2}$. The vector $(-Dh,1)$ has zero dot product with each tangent column, length $\sqrt{1+\mathrm{\mid }Dh{\mathrm{\mid }}^2}$, and positive last component. It points out of $z<h(y)$, as moving in its direction increases $z-h(y)$ to first order by $1+\mathrm{\mid }Dh{\mathrm{\mid }}^2>0$. Dividing by its length and multiplying by the density proves $\nu dS=(-Dh,1)dy$.

For affine h, $Dh=a$, so $dS=\sqrt{1+\mathrm{\mid }a{\mathrm{\mid }}^2}dy$ and $\nu =(-a,1)\mathrm{/}\sqrt{1+\mathrm{\mid }a{\mathrm{\mid }}^2}$. In the stated instance $a=(2,-1)$, giving $D{X}^{T}DX=\left(\begin{array}{cc}5& -2\\ -2& 2\end{array}\right)$, determinant $10-4=6$, and $\nu =(-2,1,1)\mathrm{/}\sqrt{6}$. Integration over the unit square gives area $\sqrt{6}\cdot 1=\sqrt{6}$ and flux $\int (e_3\cdot \nu )\sqrt{6}dy=\int 1dy=1$. For the upper unit hemisphere, $h(y)=\sqrt{1-\mathrm{\mid }y{\mathrm{\mid }}^2}$ on $\mathrm{\mid }y\mathrm{\mid }<1$. Here $Dh=-y\mathrm{/}h$, so $\sqrt{1+\mathrm{\mid }Dh{\mathrm{\mid }}^2}=1\mathrm{/}h$ and $\nu =(y,h)$; its last component is positive and it is the radial outward unit vector. The equator is outside this graph. Rotated full-sphere graph charts cover it in an atlas of the full sphere, but no graph chart contained in the closed upper hemisphere covers an equator point.

The sphere is a C1 boundary: near any point one nonzero component of $x-a$ lets its equation be solved as a smooth square-root graph, with the ball on the inner side. Its outward normal is $\nu =(x-a)\mathrm{/}R$. For $F=x-a$, each ${\mathrm{\partial }}_i{F}_i=1$, so $\mathrm{div}F=n$, and on the boundary $F\cdot \nu =R$. F1 yields $n\mathrm{\mid }{B}_R(a)\mathrm{\mid }=R\mathrm{\mid }\mathrm{\partial }{B}_R(a)\mathrm{\mid }$; dividing by R and using F2 gives both stated area formulas. The ball has positive volume since it contains a cube of positive side length. In dimension three a spherical chart $X(\theta ,\varphi )=(\cos \theta \sin \varphi ,\sin \theta \sin \varphi ,\cos \varphi )$, $0<\theta <2\pi$, $0<\varphi <\pi$, has tangent squared lengths ${\sin }^2\varphi$ and 1 and zero cross inner product, hence density $\sin \varphi$. Rotated charts cover its omitted meridian and poles.

For a constant vector b all partial derivatives vanish. Applying F1 gives ${\int }_{\mathrm{\partial }{B}_R(a)}b\cdot \nu dS={\int }_{B_R(a)}0dx=0$. For the radial field and $r=\mathrm{\mid }x-a\mathrm{\mid }>0$, ${\mathrm{\partial }}_i{F}_i=f(r)+f^{\mathrm{\prime }}(r)(x_i-a_i{)}^2\mathrm{/}r$. Summing gives $\mathrm{div}F=nf(r)+r{f}^{\mathrm{\prime }}(r)$. The stipulated C1 extension supplies the value at the centre and the hypotheses of F1; no assertion about a singular f at zero is needed. On the sphere the flux density is the constant $f(R)R$, proving the claimed flux.

For the explicit polynomial field $F=\mathrm{\mid }x-a{\mathrm{\mid }}^2(x-a)$ the extension is automatic. Its divergence is $(n+2)\mathrm{\mid }x-a{\mathrm{\mid }}^2$ also at the centre by direct differentiation. Thus its flux is $R^3\mathrm{\mid }\mathrm{\partial }{B}_R(a)\mathrm{\mid }$, and F1 gives ${\int }_{B_R(a)}\mathrm{\mid }x-a{\mathrm{\mid }}^{2}dx=R^3\mathrm{\mid }\mathrm{\partial }{B}_R(a)\mathrm{\mid }\mathrm{/}(n+2)$. In particular at R=1, a=0 this moment is $n\mathrm{\mid }{B}_1\mathrm{\mid }\mathrm{/}(n+2)$ by F2.

For a box ${\prod }_i(a_i,b_i)$, its 2n faces are $x_i=a_i$ and $x_i=b_i$, with remaining coordinates in their closed intervals. Each is a compact subset of an affine regular hypersurface, with area element the ordinary product measure and outward normal $-e_i$ or $e_i$. Put every face boundary (at least two endpoint coordinates) into E. On each face it is a finite union of parameter-coordinate hyperplanes and is null: a bounded hyperplane strip has arbitrarily small volume by giving the fixed coordinate an arbitrarily short interval. Off E exactly one coordinate is an endpoint and the domain is locally on one side of that plane. Thus all three boxes satisfy the specified finite-face conditions.

The common face $S=\{0\}\times [0,1{]}^{n-1}$ has normal $e_1$ for $Q_{-}$ and $-e_1$ for $Q_{+}$. The same continuous F has the same trace on both sides, so the sum of its flux densities is $F\cdot e_1+F\cdot (-e_1)=0$ pointwise off E. F1 applies to both restrictions of F. Their volume integrals add to the integral on Q because the omitted plane S is volume-null by the strip argument in step 1.1. Their exposed face integrals add to the faces of Q, while the two integrals over S cancel. This proves the asserted identity and all required gluing data.

For $F(x)=x$, the divergence is n and each small box has volume one. On $Q_{-}$ the face $x_1=-1$ contributes 1 and $x_1=0$ contributes 0; on $Q_{+}$ the face $x_1=1$ contributes 1 and $x_1=0$ contributes 0. For each $i\ge 2$ the face $x_i=1$ contributes 1 on each box and $x_i=0$ contributes 0. Each box therefore has flux $1+(n-1)=n$, and Q has flux $2+2(n-1)=2n$, equal to its volume integral. For the further constant field $F=e_1$ the shared fluxes are explicitly 1 and -1, exhibiting nonzero cancellation.

The ball is bounded with smooth boundary, and F is a polynomial C1 field on its closure. Directly $\mathrm{div}F={\sum }_{i=1}^{n}1=n$, so its integral is $n\mathrm{\mid }{B}_R\mathrm{\mid }$. This is positive: $(-R\mathrm{/}(2\sqrt{n}),R\mathrm{/}(2\sqrt{n}){)}^n\subset B_R$ has positive product volume.

The inward unit normal is $-x\mathrm{/}R$ on $\mathrm{\mid }x\mathrm{\mid }=R$. Consequently $F\cdot {\nu }_{\mathrm{i}\mathrm{n}}=-\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}R=-R$ and its flux is $-R\mathrm{\mid }\mathrm{\partial }{B}_R\mathrm{\mid }=-n\mathrm{\mid }{B}_R\mathrm{\mid }$ by F1. Step 1.1 proves this differs from the positive divergence integral, although every domain and field regularity hypothesis holds. It is exactly the orientation hypothesis that fails.

Present Q by its 2n closed coordinate faces with normals $\pm e_i$ and the set E where at least two coordinates are endpoints. As in the explicit box verification F1, the parameter boundaries are finite unions of bounded coordinate hyperplanes and have measure zero, and every boundary point outside E has a single planar one-sided neighborhood. This verifies the finite piecewise hypotheses.

Let p have at least two endpoint coordinates i and j. Approach p through relative interiors of the i-face, moving every other endpoint coordinate slightly into (0,1); also approach through relative interiors of the j-face. The outward normals along these two sequences are the distinct constants ${\epsilon }_i{e}_i$ and ${\epsilon }_j{e}_j$, with ${\epsilon }_k=-1$ at endpoint 0 and $+1$ at endpoint 1. If a C1 graph chart as in F2 existed at p, its normal would be the continuous function $(-Dh,1)\mathrm{/}\sqrt{1+\mathrm{\mid }Dh{\mathrm{\mid }}^2}$, transformed by the chart rotation and given the unique outward sign. At neighboring planar points this normal must be the corresponding coordinate normal, since orthogonality to the plane and the outward side uniquely determine it. Continuity at p would force those two distinct constant vectors to have the same limit, a contradiction. This proves failure at every edge and vertex, including the origin.

Write ${\mathrm{\Omega }}^{\ast }=\mathrm{\Omega }\setminus {\bar{B}}_r(a)$. Its two boundary parts are separated by a positive distance, so their original graph neighborhoods can be shrunk to exclude the other part. The old outward normals are unchanged. At the new sphere the side belonging to ${\mathrm{\Omega }}^{\ast }$ is $\mathrm{\mid }x-a\mathrm{\mid }>r$, so its outward direction points into the deleted ball and its normal is $-(x-a)\mathrm{/}r$. A finite subdivision of its compact C1 boundary into chart faces gives the presentation F1: choose finitely many small closed graph-coordinate boxes covering the boundary, remove previous box interiors in their order, and include their boundaries in E. Each such boundary is a Lipschitz image of parameter-box sides, hence surface-null in overlapping charts; the transition maps are C1 with bounded derivatives on these compact boxes. F2 then applies. For an explicit instance take concentric balls ${\mathrm{\Omega }}^{\ast }=B_R(0)\setminus {\bar{B}}_r(0)$ and $F=x$, $0<r<R$. The outer flux is $R\mathrm{\mid }\mathrm{\partial }{B}_R\mathrm{\mid }=n\mathrm{\mid }{B}_R\mathrm{\mid }$ and the inner flux is $-r\mathrm{\mid }\mathrm{\partial }{B}_r\mathrm{\mid }=-n\mathrm{\mid }{B}_r\mathrm{\mid }$ by F3, giving $n(\mathrm{\mid }{B}_R\mathrm{\mid }-\mathrm{\mid }{B}_r\mathrm{\mid })={\int }_{{\mathrm{\Omega }}^{\ast }}\mathrm{div}F$.

For K the caps are the closed d-dimensional balls in the planes $t=a_0,b_0$, with normals $-e_t,+e_t$. For d at least two, cover the unit sphere by the 2d closed patches on which a chosen signed coordinate has maximal absolute value. Such a patch is parametrized, after coordinate permutation, by $\omega (y)=(y,\pm 1)\mathrm{/}\sqrt{1+\mathrm{\mid }y{\mathrm{\mid }}^2}$ for $y\in [-1,1{]}^{d-1}$, with the corresponding sign in the other coordinates immaterial to its image. The map is regular on an open neighborhood of this cube. Lateral faces are $X(y,t)=(R(t)\omega (y),t)$ on the closed cube times $[a_0,b_0]$. Because R is strictly positive, these are regular compact hypersurface patches. Their overlaps lie on parameter-box boundaries. Put these seams and both rims into E. Parameter-box boundaries are null; C1 transition maps on compact subpatches are Lipschitz, so they preserve these null sets (cover by cubes and multiply their volumes by a fixed Lipschitz bound to the parameter dimension). On a cap its rim is null: enclosing it in annuli of thickness epsilon gives volume tending to zero by ball scaling from F3. For d=1 the caps are intervals and the lateral faces are the two straight segments $(\pm R(t),t)$; E consists of their four endpoints. Off E every face is smooth and K is on the stated one side. This verifies all of F1.

On the lateral face $g(x,t)=\mathrm{\mid }x\mathrm{\mid }-R(t)$ vanishes and K is $g<0$; $\mathrm{\nabla }g=(\omega ,-c)$ is nonzero and points outward. Thus $\nu =(\omega ,-c)\mathrm{/}\sqrt{1+c^2}$. In the parametrization of step 1.2 the tangential y columns are $(RD\omega ,0)$ and the time column is $(c\omega ,1)$. Their cross inner products vanish because $\omega \cdot D\omega =0$. The Gram determinant is therefore $(1+c^2)R^{2d-2}\det (D{\omega }^{T}D\omega )$, giving $dS=\sqrt{1+c^2}R(t{)}^{d-1}d\sigma (\omega )dt$ by F3. For d=1 each of the two rays has arclength density $\sqrt{1+c^2}dt$, the same formula with counting measure on $S^0$. F2 now gives the complete cap-plus-side identity for every C1 field on the closure of K.

As a direct calculation take the constant space-time field $F=e_t$, of divergence zero. Put $v_d=\mathrm{\mid }{B}_1^d\mathrm{\mid }$ (so $v_1=2$). The two cap fluxes sum to $v_d(R(b_0{)}^d-R(a_0{)}^d)$. The lateral flux, by step 2.1 and $\mathrm{\mid }{S}^{d-1}\mathrm{\mid }=d{v}_d$ from F3 for d at least two (and the two rays for d=1), is $-cd{v}_d{\int }_{a_0}^{b_0}R(t{)}^{d-1}dt=-v_d(R(b_0{)}^d-R(a_0{)}^d)$, using $\frac{d}{dt}R(t{)}^d=dcR(t{)}^{d-1}$. This equality also holds when c=0 because both expressions vanish; no division by c is required. The total flux is exactly zero.

For a bottom tip let $R(t)=c(t-a_0)$ with c>0, and truncate at $a_0+\delta$. Steps 1.2–2.1 give F2 on the truncated region. If $\mathrm{\mid }F\mathrm{\mid }\le M$ and $\mathrm{\mid }\mathrm{div}F\mathrm{\mid }\le L$ on the full closure, the artificial cap flux is at most $M{v}_d(c\delta {)}^d$. F4, applied to the bounded measurable indicator times the bounded divergence on the bounded cylinder, bounds the omitted volume integral by $L{v}_d{c}^d{\delta }^{d+1}\mathrm{/}(d+1)$. The omitted lateral flux is bounded by $M\sqrt{1+c^2}d{v}_d{\int }_0^{\delta }(cs{)}^{d-1}ds=M\sqrt{1+c^2}{v}_d{c}^{d-1}{\delta }^d$, including d=1 via its two rays. Each error tends to zero. The unchanged top cap, the lateral improper integral (absolutely convergent by the same estimate), and the full volume integral therefore satisfy the limiting identity. For a top tip substitute $s=b_0-t$ and replace c by $\mathrm{\mid }c\mathrm{\mid }$ in all three bounds; the artificial cap orientation changes but its absolute bound does not. Thus no regular chart at the apex is assumed.