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Weak formulation

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Weak formulations are tools for the analysis of mathematicalequations that permit the transfer of concepts oflinear algebra to solve problems in other fields such aspartial differential equations. In a weak formulation, equations or conditions are no longer required to hold absolutely (and this is not even well defined) and has insteadweak solutions only with respect to certain "test vectors" or "test functions". In astrong formulation, the solution space is constructed such that these equations or conditions are already fulfilled.

TheLax–Milgram theorem, named afterPeter Lax andArthur Milgram who proved it in 1954, provides weak formulations for certain systems onHilbert spaces.

General concept

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LetV{\displaystyle V} be aBanach space, letV{\displaystyle V'} be thedual space ofV{\displaystyle V}, letA:VV{\displaystyle A\colon V\to V'} be alinear map, and letfV{\displaystyle f\in V'}. A vectoruV{\displaystyle u\in V} is a solution of the equation

Au=f{\displaystyle Au=f}

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

(Au)(v)=f(v).{\displaystyle (Au)(v)=f(v).}

A particular choice ofv{\displaystyle v} is called atest vector (in general) or atest function (ifV{\displaystyle V} is a function space).

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

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

by defining thebilinear form

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

Example 1: linear system of equations

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Now, letV=Rn{\displaystyle V=\mathbb {R} ^{n}} andA:VV{\displaystyle A:V\to V} be alinear mapping. Then, the weak formulation of the equation

Au=f{\displaystyle Au=f}

involves findinguV{\displaystyle u\in V} such that for allvV{\displaystyle v\in V} the following equation holds:

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

where,{\displaystyle \langle \cdot ,\cdot \rangle } denotes aninner product.

SinceA{\displaystyle A} is a linear mapping, it is sufficient to test withbasis vectors, and we get

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

Actually, expandingu=j=1nujej{\displaystyle u=\sum _{j=1}^{n}u_{j}e_{j}}, we obtain thematrix form of the equation

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

whereaij=Aej,ei{\displaystyle a_{ij}=\langle Ae_{j},e_{i}\rangle } andfi=f,ei{\displaystyle f_{i}=\langle f,e_{i}\rangle }.

The bilinear form associated to this weak formulation is

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

Example 2: Poisson's equation

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To solvePoisson's equation

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

on a domainΩRd{\displaystyle \Omega \subset \mathbb {R} ^{d}} withu=0{\displaystyle u=0} on itsboundary, and to specify the solution spaceV{\displaystyle V} later, one can use theL2{\displaystyle L^{2}}-scalar product

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

to derive the weak formulation. Then, testing withdifferentiable functionsv{\displaystyle v} yields

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

The left side of this equation can be made more symmetric byintegration by parts usingGreen's identity and assuming thatv=0{\displaystyle v=0} onΩ{\displaystyle \partial \Omega }:

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

This is what is usually called the weak formulation ofPoisson's equation.Functions in the solution spaceV{\displaystyle V} must be zero on the boundary, and have square-integrablederivatives. The appropriate space to satisfy these requirements is theSobolev spaceH01(Ω){\displaystyle H_{0}^{1}(\Omega )} of functions withweak derivatives inL2(Ω){\displaystyle L^{2}(\Omega )} and with zero boundary conditions, soV=H01(Ω){\displaystyle V=H_{0}^{1}(\Omega )}.

The generic form is obtained by assigning

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

and

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

The Lax–Milgram theorem

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

LetV{\displaystyle V} be a realHilbert space anda(,){\displaystyle a(\cdot ,\cdot )} abilinear form onV{\displaystyle V}, which is

  1. bounded:|a(u,v)|Cuv;{\displaystyle |a(u,v)|\leq C\|u\|\|v\|\,;} and
  2. coercive:a(u,u)cu2.{\displaystyle a(u,u)\geq c\|u\|^{2}\,.}

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

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

and it holds

u1cfV.{\displaystyle \|u\|\leq {\frac {1}{c}}\|f\|_{V'}\,.}

Application to example 1

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

Additionally, this yields the estimateu1cf,{\displaystyle \|u\|\leq {\frac {1}{c}}\|f\|,}wherec{\displaystyle c} is the minimal real part of an eigenvalue ofA{\displaystyle A}.

Application to example 2

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Here, chooseV=H01(Ω){\displaystyle V=H_{0}^{1}(\Omega )} with the normvV:=v,{\displaystyle \|v\|_{V}:=\|\nabla v\|,}

where the norm on the right is theL2{\displaystyle L^{2}}-norm onΩ{\displaystyle \Omega } (this provides a true norm onV{\displaystyle V} by thePoincaré inequality).But, we see that|a(u,u)|=u2{\displaystyle |a(u,u)|=\|\nabla u\|^{2}} and by theCauchy–Schwarz inequality,|a(u,v)|uv{\displaystyle |a(u,v)|\leq \|\nabla u\|\,\|\nabla v\|}.

Therefore, for anyf[H01(Ω)]{\displaystyle f\in [H_{0}^{1}(\Omega )]'}, there is a unique solutionuV{\displaystyle u\in V} ofPoisson's equation and we have the estimate

uf[H01(Ω)].{\displaystyle \|\nabla u\|\leq \|f\|_{[H_{0}^{1}(\Omega )]'}.}

See also

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References

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External links

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