 Solution

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.