Past exam of the methods of applied functional analysis course of the University of Oxford / 2010 / A / 3

The real Lax-Milgram theorem states: let $H$ be a Hilbert space and let $a:H\times H\to \mathbb{R}$ be a bounded bilinear form such that $a(v,v)\ge \alpha \Vert v{\Vert }_H^2$ for some $\alpha >0$ and every $v\in H$. For every bounded linear functional $\ell \in H^{\ast }$ there exists a unique $u\in H$ satisfyingSymmetry of $a$ is not required. The complex version replaces bilinearity by sesquilinearity and the lower bound by $\mathrm{Re}a(v,v)\ge \alpha \Vert v{\Vert }_H^2$.

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.

A weak solution belongs to the zero-boundary Sobolev space $H_0^1(\Omega )$ and satisfiesThe boundedness of $b$ and the Cauchy-Schwarz inequality make the drift term integrable. This weak formulation moves one derivative from the Laplacian onto the test function, while retaining the first-order drift term.

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.