Weak formulation

Let ${\displaystyle V}$ be a Banach space, let ${\displaystyle V'}$ be the dual space of ${\displaystyle V}$, let ${\displaystyle A\colon V\to V'}$, and let ${\displaystyle f\in V'}$. A vector ${\displaystyle u\in V}$ is a solution of the equation

$${\displaystyle Au=f}$$

if and only if for all ${\displaystyle v\in V}$,

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

Here, ${\displaystyle v}$ is called a test vector (in general) or a test function (if ${\displaystyle V}$ is a function space).

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

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

by defining the bilinear form

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

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

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

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 to specify the solution space ${\displaystyle V}$ later, one can use the ${\displaystyle L^{2}}$-scalar product

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

to derive the weak formulation. Then, testing with differentiable functions ${\displaystyle v}$ yields

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

The left side of this equation can be made 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. Functions in the solution space ${\displaystyle V}$ must be zero on the boundary, and have square-integrable derivatives. The appropriate space to satisfy these requirements is the Sobolev space ${\displaystyle H_{0}^{1}(\Omega )}$ of functions with weak derivatives in ${\displaystyle L^{2}(\Omega )}$ and with zero boundary conditions, so ${\displaystyle V=H_{0}^{1}(\Omega )}$.

The generic form is obtained by assigning

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

and

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

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

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

and it holds

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

Application to example 1

Here, application of the Lax–Milgram theorem is a stronger result than is needed.

• 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.

Additionally, this yields the estimate

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

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

Application to example 2

Here, 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

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

• Babuška–Lax–Milgram theorem
• Lions–Lax–Milgram theorem

• Lax, Peter D.; Milgram, Arthur N. (1954), "Parabolic equations", Contributions to the theory of partial differential equations, Annals of Mathematics Studies, vol. 33, Princeton, N. J.: Princeton University Press, pp. 167–190, doi:10.1515/9781400882182-010, ISBN 9781400882182, MR 0067317, Zbl 0058.08703

• MathWorld page on Lax–Milgram theorem