# Weak formulation

(Redirected from Lax–Milgram theorem)

Weak formulations are important tools for the analysis of mathematical equations that permit the transfer of concepts of linear algebra to solve problems in other fields such as partial differential equations. In a weak formulation, an equation is no longer required to hold absolutely (and this is not even well defined) and has instead weak solutions only with respect to certain "test vectors" or "test functions". This is equivalent to formulating the problem to require a solution in the sense of a distribution.[citation needed]

We introduce weak formulations by a few examples and present the main theorem for the solution, the Lax–Milgram theorem.

## General concept

Let ${\displaystyle V}$ be a Banach space. We want to find the solution ${\displaystyle u\in V}$ of the equation

${\displaystyle Au=f}$,

where ${\displaystyle A:V\to V'}$ and ${\displaystyle f\in V'}$, with ${\displaystyle V'}$ being the dual of ${\displaystyle V}$.

Calculus of variations tells us that this is equivalent to finding ${\displaystyle u\in V}$ such that for all ${\displaystyle v\in V}$ holds:

${\displaystyle [Au](v)=f(v)}$.

Here, we call ${\displaystyle v}$ a test vector or test function.

We bring this into the generic form of a weak formulation, namely, find ${\displaystyle u\in V}$ such that

${\displaystyle a(u,v)=f(v)\quad \forall v\in V,}$

by defining the bilinear form

${\displaystyle a(u,v):=[Au](v).}$

Since this is very abstract, let us follow this by some examples.

## Example 1: linear system of equations

Now, let ${\displaystyle V=\mathbb {R} ^{n}}$ and ${\displaystyle A:V\to V}$ be a linear mapping. Then, the weak formulation of the equation

${\displaystyle Au=f}$

involves finding ${\displaystyle u\in V}$ such that for all ${\displaystyle v\in V}$ the following equation holds:

${\displaystyle \langle Au,v\rangle =\langle f,v\rangle ,\,}$

where ${\displaystyle \langle \cdot ,\cdot \rangle }$ denotes an inner product.

Since ${\displaystyle A}$ is a linear mapping, it is sufficient to test with basis vectors, and we get

${\displaystyle \langle Au,e_{i}\rangle =\langle f,e_{i}\rangle \quad i=1,\ldots ,n.\,}$

Actually, expanding ${\displaystyle u=\sum _{j=1}^{n}u_{j}e_{j}}$, we obtain the matrix form of the equation

${\displaystyle \mathbf {A} \mathbf {u} =\mathbf {f} ,}$

where ${\displaystyle a_{ij}=\langle Ae_{j},e_{i}\rangle }$ and ${\displaystyle f_{i}=\langle f,e_{i}\rangle }$.

The bilinear form associated to this weak formulation is

${\displaystyle a(u,v)=\mathbf {v} ^{T}\mathbf {A} \mathbf {u} .}$

## Example 2: Poisson's equation

Our aim is to solve Poisson's equation

${\displaystyle -\nabla ^{2}u=f,\,}$

on a domain ${\displaystyle \Omega \subset \mathbb {R} ^{d}}$ with ${\displaystyle u=0}$ on its boundary, and we want to specify the solution space ${\displaystyle V}$ later. We will use the ${\displaystyle L^{2}}$-scalar product

${\displaystyle \langle u,v\rangle =\int _{\Omega }uv\,dx}$

to derive our weak formulation. Then, testing with differentiable functions ${\displaystyle v}$, we get

${\displaystyle -\int _{\Omega }(\nabla ^{2}u)v\,dx=\int _{\Omega }fv\,dx.}$

We can make the left side of this equation more symmetric by integration by parts using Green's identity and assuming that ${\displaystyle v=0}$ on ${\displaystyle \partial \Omega }$:

${\displaystyle \int _{\Omega }\nabla u\cdot \nabla v\,dx=\int _{\Omega }fv\,dx.}$

This is what is usually called the weak formulation of Poisson's equation; what's missing is the space ${\displaystyle V}$, which is beyond the scope of this article. The space must allow us to write down this equation. Therefore, we should require that the derivatives of functions in this space are square integrable. Now, there is actually the Sobolev space ${\displaystyle H_{0}^{1}(\Omega )}$ of functions with weak derivatives in ${\displaystyle L^{2}(\Omega )}$ and with zero boundary conditions, which fulfills this purpose.

We obtain the generic form by assigning

${\displaystyle a(u,v)=\int _{\Omega }\nabla u\cdot \nabla v\,dx}$

and

${\displaystyle f(v)=\int _{\Omega }fv\,dx.}$

## The Lax–Milgram theorem

This is a formulation of the Lax–Milgram theorem which relies on properties of the symmetric part of the bilinear form. It is not the most general form.

Let ${\displaystyle V}$ be a Hilbert space and ${\displaystyle a(\cdot ,\cdot )}$ a bilinear form on ${\displaystyle V}$, which is

1. bounded: ${\displaystyle |a(u,v)|\leq C\|u\|\|v\|}$ and
2. coercive: ${\displaystyle a(u,u)\geq c\|u\|^{2}.}$

Then, for any ${\displaystyle f\in V'}$, there is a unique solution ${\displaystyle u\in V}$ to the equation

${\displaystyle a(u,v)=f(v)}$

and it holds

${\displaystyle \|u\|\leq {\frac {1}{c}}\|f\|_{V'}.}$

### Application to example 1

Here, application of the Lax–Milgram theorem is definitely a stronger result than is needed, but we still can use it and give this problem the same structure as the others have.

• Boundedness: all bilinear forms on ${\displaystyle \mathbb {R} ^{n}}$ are bounded. In particular, we have
${\displaystyle |a(u,v)|\leq \|A\|\,\|u\|\,\|v\|\,}$
• Coercivity: this actually means that the real parts of the eigenvalues of ${\displaystyle A}$ are not smaller than ${\displaystyle c}$. Since this implies in particular that no eigenvalue is zero, the system is solvable.

${\displaystyle \|u\|\leq {\frac {1}{c}}\|f\|,\,}$

where ${\displaystyle c}$ is the minimal real part of an eigenvalue of ${\displaystyle A}$.

### Application to example 2

Here, as we mentioned above, we choose ${\displaystyle V=H_{0}^{1}(\Omega )}$ with the norm

${\displaystyle \|v\|_{V}:=\|\nabla v\|,}$

where the norm on the right is the ${\displaystyle L^{2}}$-norm on ${\displaystyle \Omega }$ (this provides a true norm on ${\displaystyle V}$ by the Poincaré inequality). But, we see that ${\displaystyle |a(u,u)|=\|\nabla u\|^{2}}$ and by the Cauchy–Schwarz inequality, ${\displaystyle |a(u,v)|\leq \|\nabla u\|\,\|\nabla v\|}$.

Therefore, for any ${\displaystyle f\in [H_{0}^{1}(\Omega )]'}$, there is a unique solution ${\displaystyle u\in V}$ of Poisson's equation and we have the estimate

${\displaystyle \|\nabla u\|\leq \|f\|_{[H_{0}^{1}(\Omega )]'}.}$