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For a solenoidal vector field $J$ with compact support and suitable integrability, apply the divergence theorem to $x_i{x}_{j}J$. It gives zero boundary term and hence the displayed antisymmetry. Equivalently, for any constant vector $C$, the vector triple product identity yields $\int (C\cdot r)JdV=\frac{1}{2}C\times \int (J\times r)dV$. A piecewise smooth field must have no normal jump at its support boundary; otherwise its weak divergence contains a surface contribution and the argument fails.

On a bounded open set, suppose $b\in L^{\infty }$ has weak divergence in $L_{\mathrm{loc}}^1$ with $\mathrm{div}b\le 0$. For $\phi \in C_0^{\infty }$, testing the divergence identity with ${\phi }^2/2$ gives $\int (b\cdot \nabla \phi )\phi =-\frac{1}{2}\int (\mathrm{div}b){\phi }^2\ge 0$. The left side is continuous in the zero-boundary Sobolev space $H_0^1$, by the Poincaré inequality and boundedness of $b$. Density therefore extends its nonnegativity to every $u\in H_0^1$. One need not assume that $(\mathrm{div}b)u^2$ is integrable. It follows that $a(u,v)=\int \nabla u\cdot \nabla v+\int (b\cdot \nabla u)v$ is a coercive bilinear form in the gradient norm, giving unique weak Dirichlet solvability by the Lax-Milgram theorem.

Introduce coordinates $s=x+y$, $t=x-y$, so $dxdy=\frac{1}{2}dsdt$. For a test function $\phi (x,y)=\Phi (s,t)$,almost everywhere. HenceThe first derivative integrates to zero in $s$, and the second to zero in $t$, using compact support. Thus $\boxed{\mathrm{div}V=0}$ as a weak divergence. Geometrically, the jump across $x+y=0$ is $(1,-1)$, perpendicular to its normal $(1,1)$; there is no distributional flux concentrated on the interface.

Give $H_0^1$ the equivalent Hilbert space norm $\Vert u{\Vert }_H=\Vert \nabla u{\Vert }_2$, using the Poincaré inequality. DefineThe Cauchy-Schwarz inequality gives $|a(u,v)|\le (1+C_P\Vert b{\Vert }_{\infty })\Vert u{\Vert }_H\Vert v{\Vert }_H$ and $|\ell (v)|\le C_P\Vert f{\Vert }_2\Vert v{\Vert }_H$.

For $\phi \in C_0^{\infty }(\Omega )$, the definition of weak divergence, applied to ${\phi }^2/2$, givesThe left side is continuous in the $H_0^1$ norm. Approximate any $u\in H_0^1$ by test functions and pass to the limit on the left to obtain $\int (b\cdot \nabla u)u\ge 0$. This density argument is essential: local integrability of $\mathrm{div}b$ alone does not justify treating $(\mathrm{div}b)u^2$ as integrable for arbitrary $u\in H_0^1$.

Consequently $a(u,u)\ge \Vert u{\Vert }_H^2$, so $a$ is a coercive bilinear form. The Lax-Milgram theorem supplies a unique weak solution with $\boxed{\Vert \nabla u{\Vert }_2\le C_P\Vert f{\Vert }_2}$. The reusable estimate is coercivity from a drift with nonpositive weak divergence.