Weak formulations are 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, equations or conditions are 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". In a strong formulation, the solution space is constructed such that these equations or conditions are already fulfilled. The Lax–Milgram theorem, named after Peter Lax and Arthur Milgram who proved it in 1954, provides weak formulations for certain systems on Hilbert spaces.
General concept Let V {\displaystyle V} be a Banach space, let V ′ {\displaystyle V'} be the dual space of V {\displaystyle V} , let A : V → V ′ {\displaystyle A\colon V\to V'} be a linear map, and let f ∈ V ′ {\displaystyle f\in V'} . A vector u ∈ V {\displaystyle u\in V} is a solution of the equation
A u = f {\displaystyle Au=f}
if and only if for all v ∈ V {\displaystyle v\in V} ,
( A u ) ( v ) = f ( v ) . {\displaystyle (Au)(v)=f(v).}
A particular choice of v {\displaystyle v} is called a test vector (in general) or a test function (if V {\displaystyle V} is a function space). To bring this into the generic form of a weak formulation, find u ∈ V {\displaystyle u\in V} such that
a ( u , v ) = f ( v ) ∀ v ∈ V , {\displaystyle a(u,v)=f(v)\quad \forall v\in V,}
by defining the bilinear form
a ( u , v ) := ( A u ) ( v ) . {\displaystyle a(u,v):=(Au)(v).}
Example 1: linear system of equations Now, let V = R n {\displaystyle V=\mathbb {R} ^{n}} and A : V → V {\displaystyle A:V\to V} be a linear mapping. Then, the weak formulation of the equation
A u = f {\displaystyle Au=f}
involves finding u ∈ V {\displaystyle u\in V} such that for all v ∈ V {\displaystyle v\in V} the following equation holds:
⟨ A u , v ⟩ = ⟨ f , v ⟩ , {\displaystyle \langle Au,v\rangle =\langle f,v\rangle ,}
where ⟨ ⋅ , ⋅ ⟩ {\displaystyle \langle \cdot ,\cdot \rangle } denotes an inner product. Since the inner product is bilinear, it is sufficient to test with basis vectors, and we get
⟨ A u , e i ⟩ = ⟨ f , e i ⟩ , i = 1 , … , n . {\displaystyle \langle Au,e_{i}\rangle =\langle f,e_{i}\rangle ,\quad i=1,\ldots ,n.}
Actually, expanding u = ∑ j = 1 n u j e j {\displaystyle u=\sum _{j=1}^{n}u_{j}e_{j}} , we obtain the matrix form of the equation
A u = f , {\displaystyle \mathbf {A} \mathbf {u} =\mathbf {f} ,}
where a i j = ⟨ A e j , e i ⟩ {\displaystyle a_{ij}=\langle Ae_{j},e_{i}\rangle } and f i = ⟨ f , e i ⟩ {\displaystyle f_{i}=\langle f,e_{i}\rangle } . The bilinear form associated to this weak formulation is
a ( u , v ) = v T A u . {\displaystyle a(u,v)=\mathbf {v} ^{T}\mathbf {A} \mathbf {u} .}
Example 2: Poisson's equation To solve Poisson's equation
− ∇ 2 u = f , {\displaystyle -\nabla ^{2}u=f,}
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