A weak formulation moves derivatives from an unknown solution onto test functions. It lowers regularity requirements, incorporates natural boundary conditions, and exposes coercivity or monotonicity needed for existence.
Poisson equation
For \[ -\Delta u=f\quad\text{in }\Omega,\qquad u=0\quad\text{on }\partial\Omega, \] multiply by \(v\in C_c^\infty(\Omega)\) and integrate by parts: \[ \int_\Omega\nabla u\cdot\nabla v\,\mathrm{d}x =\int_\Omega fv\,\mathrm{d}x. \]
Definition. A weak solution is \(u\in H_0^1(\Omega)\) satisfying \[ a(u,v)=F(v)\quad\text{for every }v\in H_0^1(\Omega), \] where \[ a(u,v)=\int_\Omega\nabla u\cdot\nabla v,\qquad F(v)=\int_\Omega fv. \]
Lax--Milgram well-posedness
Theorem (Lax--Milgram). Let \(V\) be a Hilbert space, \(a:V\times V\to\mathbb R\) bounded and coercive: \[ |a(u,v)|\le M\|u\|_V\|v\|_V,\qquad a(v,v)\ge\alpha\|v\|_V^2. \] For every \(F\in V^*\), there is a unique \(u\in V\) with \[ a(u,v)=F(v)\quad(v\in V), \] and \[ \|u\|_V\le\alpha^{-1}\|F\|_{V^*}. \]
Proof. Riesz representation gives a bounded operator \(A:V\to V\) and vector \(f\in V\) such that \[ a(u,v)=\langle Au,v\rangle,\qquad F(v)=\langle f,v\rangle. \] Coercivity implies \(\|Au\|\|u\|\ge\langle Au,u\rangle\ge\alpha\|u\|^2\), so \(A\) is injective and bounded below. Its range is closed. If \(w\perp\operatorname{Ran}A\), then \(\langle Aw,w\rangle=0\), forcing \(w=0\); the range is dense and hence all of \(V\). Thus \(Au=f\) has a unique solution and the estimate follows.
Neumann boundary conditions
For \[ -\Delta u=f,\qquad \partial_nu=g\text{ on }\partial\Omega, \] the weak form on \(H^1(\Omega)\) is \[ \int_\Omega\nabla u\cdot\nabla v =\int_\Omega fv+\int_{\partial\Omega}gv. \] Constants lie in the kernel, so compatibility \[ \int_\Omega f+\int_{\partial\Omega}g=0 \] is necessary. Restricting to zero-mean functions restores coercivity and uniqueness modulo constants.
Energy minimization
Proposition. If \(a\) is symmetric and coercive, the weak solution uniquely minimizes \[ J(v)=\frac12a(v,v)-F(v) \] over \(V\).
Proof. If \(u\) solves the weak equation, \[ J(u+w)-J(u)=\frac12a(w,w)\ge\frac\alpha2\|w\|^2. \] Conversely, first variation of a minimizer in every direction gives the weak equation.
Distributional and weak solutions
Every \(H_0^1\) weak solution is a distributional solution because the identity holds for compactly supported smooth tests. The converse requires enough integrability to make \(u\) belong to the energy space. Boundary conditions are encoded through traces or boundary terms, not visible in a purely interior distributional equation.