Linear traction from shrinking local contact balance

Fix $(t,x,n)$. Since $x\in {\mathrm{\Omega }}_t$ and ${\mathrm{\Omega }}_t$ is open, choose a ball about $x$ contained in ${\mathrm{\Omega }}_t$; for sufficiently small $h$ it contains the pillboxes with $0<\epsilon \le 1$. Choose a square $A_h$ of side $h$ in the plane through $x$ perpendicular to $n$ and let $K$ be the closed square prism of thickness $\epsilon h$ around it. Round this prism, or any tetrahedron constructed below, by taking the open distance-$\eta$ parallel body of the sharp convex polytope, with $\eta =\delta h$. Its boundary consists of face-parallel pieces and circular edge and spherical vertex pieces that meet tangentially, hence is $C^1$. At fixed shape and $h$, each face-parallel area tends to the original face area as $\delta \downarrow 0$; the total area of the edge pieces is $O(\delta h^2)$ and of vertex pieces is $O({\delta }^2{h}^2)$. The added volume is contained in prisms of thickness $O(\delta h)$ along faces, cylinders of radius $O(\delta h)$ along edges, and balls of radius $O(\delta h)$ at vertices, so it is $O(\delta h^3)$. Choose $h$ and then $\delta$ small enough that the rounded domains have closure in the chosen ball; they are admissible in the balance hypothesis. On face-parallel pieces the normals are the original face normals and the traction integrands converge uniformly by continuity in $y$; the bounded traction times the vanishing curved-piece area tends to zero. The bounded body-force integrand times the vanishing volume difference also tends to zero. The finite, chart-independent surface integral from [F4] makes each face integral unambiguous. Thus the assumed balance passes to each sharp-polytope limit as $\delta \downarrow 0$ at fixed $h$ (and fixed $\epsilon$ for the pillbox).

Now let $\epsilon \downarrow 0$ at fixed $h$. The side area is $O(\epsilon h^2)$ and the volume is $O(\epsilon h^3)$, so their resultants vanish by local boundedness. The two caps converge to the same square with opposite outward normals, giving ${\int }_{A_h}(\tau (t,y,n)+\tau (t,y,-n))d{S}_y=0$. The cap measure is the finite chart-independent face measure from [F4]. Divide by $\mathrm{\mid }{A}_h\mathrm{\mid }=h^2$ and let $h\downarrow 0$; continuity in $y$ yields $\tau (t,x,n)+\tau (t,x,-n)=0$.

First take $n\in S^2$ with all coordinates nonzero. For $i=1,2,3$ put $v_i=x+(h\mathrm{/}{n}_i)e_i$, and take the tetrahedron with vertices $x,v_1,v_2,v_3$. Its oblique face lies in $n\cdot (y-x)=h$ and has outward normal $n$ because the tetrahedron lies on the side $n\cdot (y-x)\le h$. Its diameter is $O(h)$ and its oblique face area $A$ satisfies $A\asymp h^2$ for this fixed $n$. The other face opposite $v_i$ has outward normal $-\mathrm{sgn}(n_i)e_i$; orthogonal projection of the oblique face onto that coordinate plane is exactly this face, so its area is $\mathrm{\mid }{n}_i\mathrm{\mid }A$. Equivalently the oriented area vector of the oblique face is $An$, whose $i$th component is its signed projected area. Round this tetrahedron by the construction in step 1.1 and pass to the sharp limit. Divide the balance by $A$ and let $h\downarrow 0$: the bounded volume resultant is $O(h^3)\mathrm{/}A\to 0$, while continuity makes the mean traction on each face converge to its value at $x$. The face integrals are well-defined by [F3,F4]. Hence $\tau (t,x,n)+{\sum }_{i=1}^3\mathrm{\mid }{n}_i\mathrm{\mid }\tau (t,x,-\mathrm{sgn}(n_i)e_i)=0$. Using step 2.1 gives $\tau (t,x,n)={\sum }_i{n}_i\tau (t,x,e_i)$.

If a coordinate of $n$ is zero, approximate $n$ by unit vectors with nonzero coordinates and use continuity in $n$ to retain the same formula. Define the $i$th column of $\sigma (t,x)$ to be $\tau (t,x,e_i)$. Then the formula gives $\tau (t,x,n)=\sigma (t,x)n$ for every unit $n$, and these three column values force uniqueness.