The Lax-Milgram Theorem is a fundamental result in functional analysis, particularly significant in the study of partial differential equations (PDEs). It provides conditions under which a bilinear form on a Hilbert space can be "inverted" to guarantee the existence and uniqueness of a solution to a certain linear equation.

Example

Let $f \in L^2(\Omega), c \in L^{\infty}(\Omega)$. Assume that $c \geq 0$. Then the problem: Find $u \in V=H_0^1(\Omega)$ such that

$$ \forall v \in V, \quad \int_{\Omega}(\nabla u \cdot \nabla v+c u v) d x=\int_{\Omega} f v d x $$

has one and only one solution.

Analysis:

Let:

• $\Omega \subset R ^n$ be a bounded open set (assumed to have suitable regularity),
• $V=H_0^1(\Omega)$ : the Sobolev space of functions vanishing on the boundary in the sense of traces,
• $f \in L^2(\Omega)$,
• $c \in L^{\infty}(\Omega)$ with $c(x) \geq 0$ a.e.

We consider the bilinear form:

$$ a(u, v):=\int_{\Omega} \nabla u \cdot \nabla v d x+\int_{\Omega} c u v d x $$

and the linear form:

$$ L(v):=\int_{\Omega} f v d x $$

We want to find $u \in V$ such that

$$ a(u, v)=L(v), \quad \forall v \in V . $$

Step 1: Show $a(\cdot, \cdot)$ is bounded (continuous)

Since $\nabla u, \nabla v \in L^2(\Omega)^n$ and $c \in L^{\infty}(\Omega)$, we have:

$$ |a(u, v)| \leq\|\nabla u\|{L^2}\|\nabla v\|{L^2}+\|c\|{L^{\infty}}\|u\|{L^2}\|v\|_{L^2} . $$