Weak formulation

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Weak formulations have been an important means in the analysis of mathematical equations, since they permit the transfer of concepts of linear algebra to fields like partial differential equations. The main feature of weak formulations is that an equation is not required to hold absolutely any more (and this is not even well defined), but only with respect to certain test vectors or test functions.

We introduce them by a few examples and present the main theorem for the solution, the Lax-Milgram theorem.

1 General concept
2 Example 1: linear system of equations
3 Example 2: Poisson's equation
4 The Lax-Milgram theorem
- 4.1 Application to example 1
- 4.2 Application to Example 2
5 Reference
6 See also
7 External link

[edit] General concept

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

A u = f

where $A:V\to V'$ and $f\in V'$ .

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

[A u](v) = f (v)

Here, we call $v$ a test vector or test function.

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

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

by defining the bilinear form

a (u, v): = [A u](v).

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

[edit] Example 1: linear system of equations

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

A u = f

is: find $u\in V$ such that for all $v\in V$ the following equation holds:

(A u, v) = (f, v).

Since it is sufficient in $\mathbb R^n$ to test with basis vectors, we get

$(Au,e_i) = (f,e_i) \quad i=1,\ldots,n$ .

Actually, expanding $u=\sum_{j=1}^n u_je_j$ , we obtain the matrix form of the equation

$\mathbf A \mathbf u = \mathbf f,$

where $a i j = (A e j, e i)$ and $f i = (f, e i)$ .

The bilinear form associated to this weak formulation is

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

[edit] Example 2: Poisson's equation

Our aim is to solve Poisson's equation

- Δ u = f

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

$(u,v) = \int_\Omega uv\,dx$

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

$- \int_\Omega \Delta u v \,dx = \int_\Omega fv \,dx$ .

We can make the left side of this equation more symmetric by integration by parts using Green's formula:

$\int_\Omega \nabla u \cdot \nabla v \,dx = \int_\Omega f v \,dx.$

This is, what is usually called the weak formulation of Poisson's equation; what's missing is the space $V$ . Well, this a bit tricky and way 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 $H^1_0(\Omega)$ of functions with weak derivatives in $L 2 (Ω)$ and with zero boundary conditions, which fulfills this purpose.

We obtain the generic form by assigning

$a(u,v) = \int_\Omega \nabla u \cdot \nabla v \,dx$

and

$f(v) = \int_\Omega f v \,dx.$

[edit] 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 $V$ be a Hilbert space and $a (.,.)$ a bilinear form on $V$ , which is

bounded: $|a(u,v)| \le C \|u\| \|v\|$ and
elliptic: $a(u,u) \ge c \|u\|^2.$

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

a (u, v) = f (v)

and it holds

$\|u\| \le \frac1c \|f\|_{V'}.$

[edit] Application to example 1

Here, application of the Lax-Milgram theorem is definitely overkill, but we still can use it and give this problem the same structure as the others have.

Boundedness: all bilinear forms on $\mathbb R^n$ are bounded. In particular, we have

$|a(u,v)| \le \|A\|\,\|u\|\,\|v\|$
Ellipticity: this actually means that the real parts of the eigenvalues of $A$ are not smaller than $c$ . Since this implies in particular that no eigenvalue is zero, the system is solvable.

Additionally, we get the estimate

$\|u\| \le \frac1c \|f\|$ ,

where $c$ is the minimal real part of an eigenvalue of $A$ .

[edit] Application to Example 2

Here, as we mentioned above, we choose $V = H^1_0(\Omega)$ with the norm

$\|v\|_V := \|\nabla v\|,$

Therefore, for any $f\in [H^1_0(\Omega)]'$ , there is a unique solution $u\in V$ of Poisson's equation and we have the estimate

$\|\nabla u\| \le \|f\|_{[H^1_0(\Omega)]'}.$

[edit] Reference

P. D. Lax, A.N. Milgram, Parabolic equations, Contributions to the theory of partial differential equations (L. Bers, S. Bochner, F. John, eds.), Annals of mathematics studies, vol. 33, Princeton University Press, 1954, 167-190