Lax-Milgram theorem

The theorem supplies solvability for a coercive weak formulation.

A bounded bilinear form $a$ on a real Hilbert space $H$ that is a coercive bilinear form represents every bounded linear functional uniquely: for each $\ell \in H^{\prime }$, there is a unique $u\in H$ with $a(u,v)=\ell (v)$ for every $v\in H$.

For a proof, the Riesz representation theorem gives a bounded linear operator $T$ and a vector $g$ with $a(u,v)=⟨Tu,v⟩$ and $\ell (v)=⟨g,v⟩$. Coercivity and the Cauchy-Schwarz inequality give $\Vert Tu\Vert \ge \alpha \Vert u\Vert$, so $T$ is injective and its image is closed: convergence of $T{u}_n$ makes $u_n$ a Cauchy sequence, whose limit supplies the image limit. If $v$ belongs to the orthogonal complement of the image, then $a(u,v)=0$ for every $u$; setting $u=v$ and using coercivity forces $v=0$. Thus the image is also dense and equals $H$. There is consequently a unique $u$ with $Tu=g$, and $\Vert u\Vert \le \Vert \ell \Vert /\alpha$. The complex version uses a bounded sesquilinear form with $\mathrm{Re}a(u,u)\ge \alpha \Vert u{\Vert }^2$.

A coercive zero-Dirichlet weak form gives a bounded solution map from $L^2$ into $H_0^1$. Composing with the compact Sobolev embedding into $L^2$ gives the compact inverse $K$. Transposing the bilinear form produces its Hilbert-space adjoint inverse $K^{\ast }$. The Fredholm compatibility condition for a shifted equation is orthogonality of $f$ to the homogeneous adjoint solutions. Indeed $(Kf,w)=(f,w)/\lambda$ when $K^{\ast }w=w/\lambda$, and the zero spectral parameter is excluded at resonance by coercivity.

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.

On the zero-boundary Sobolev space $H_0^1(0,1)$ with norm $\Vert u^{\prime }{\Vert }_{L^2}$, assume measurable coefficients $0<p_0\le p\le p_1$ and $0\le q\le q_1$. The Poincaré inequality and Cauchy-Schwarz inequality give $|a(u,v)|\le (p_1+q_1/{\pi }^2)\Vert u^{\prime }{\Vert }_2\Vert v^{\prime }{\Vert }_2$, while nonnegativity of $q$ gives $a(u,u)\ge p_0\Vert u^{\prime }{\Vert }_2^2$. Thus the form is a bounded bilinear form and a coercive bilinear form. For a source in $H^{-1}$, the Lax-Milgram theorem yields a unique weak solution and every conforming Galerkin method approximation. No derivative of the bounded coefficient $p$ is needed for this weak formulation.

For $H=H_0^1(U)\times H^1(U)$, the coupled formsatisfiesThe Poincaré inequality for $u$ then controls both $\Vert u{\Vert }_2$ and $\Vert w{\Vert }_2$, so the form is coercive on $H$. The Lax-Milgram theorem therefore gives a unique weak solution. The first component has a Dirichlet boundary condition, while the second component's Neumann boundary condition is the natural boundary condition in the weak formulation.

On $H=\{v\in H^1(0,1):v(0)=0\}$, the formis bounded and coercive. The Hardy inequality on an interval controls its singular term, while the one-sided Poincaré inequality controls the first-order term. Thus the Lax-Milgram theorem gives a unique weak solution for every functional $v\mapsto {\int }_0^1(f/x)(v/x)$ with $f/x\in L^2(0,1)$.

Let a positive-measure part ${\Gamma }_D$ of the boundary of a bounded connected domain carry a homogeneous Dirichlet boundary condition, and let the complementary part carry a homogeneous Neumann boundary condition. Onthe weak Poisson problem isThe Poincare inequality with a partial Dirichlet boundary makes the left side coercive, so the Lax-Milgram theorem gives a unique weak solution.

For a bounded open $U\subset {\mathbb{R}}^n$, $m^2>0$, and $f\in L^2(U)$, there is a unique $u\in H_0^1(U)$ such thatfor every $v\in H_0^1(U)$.

The solution operator $(-\Delta +m^2{)}^{-1}:L^2(U)\to L^2(U)$ is compact on every bounded open set because it maps boundedly into $H_0^1(U)$ and the Rellich-Kondrashov compactness theorem for H01 embeds that space compactly into $L^2(U)$.

Let $U\subset {\mathbb{R}}^d$ be bounded, $A\in {\mathbb{R}}^d$ satisfy $\Vert A{\Vert }_2<2$, and $f\in L^2(U)$. The bilinear formon $H_0^1(U)$ is bounded. The Cauchy-Schwarz inequality and $2ab\le a^2+b^2$ giveso it is a coercive bilinear form. The Lax-Milgram theorem therefore gives a unique weak solution of $-\Delta v+v+A\cdot \nabla v=f$ with zero Dirichlet boundary condition.