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Lecture 1 — Model problems and the Ritz–Galerkin method

This lecture introduces the variational point of view that will be used throughout the course. We start from a model elliptic problem, derive its weak formulation, and then restrict the problem to a finite-dimensional space. The main results are the Lax–Milgram lemma, Galerkin orthogonality, and Céa’s lemma.

1. A model problem

Let Ω⊂Rd\Omega \subset \mathbb{R}^d be a bounded domain with sufficiently regular boundary ∂Ω\partial\Omega. We consider the Poisson problem with homogeneous Dirichlet boundary conditions:

Uniformly refined mesh for the model Poisson problem

Figure 1:Mesh generated by the course test suite.

−Δu=fin Ω,u=0on ∂Ω.-\Delta u = f \quad \text{in } \Omega, \qquad u = 0 \quad \text{on } \partial\Omega.

Here ff is the given source term and uu is the unknown solution. This is a useful model problem because it contains the main ingredients of elliptic finite element methods while remaining simple enough to analyze explicitly.

There are two broad ways to obtain a numerical approximation:

The second point of view is particularly useful on non-rectangular domains and when the data or the solution are not smooth enough for the strong equation to be interpreted pointwise.

2. The weak formulation

Let vv be a smooth test function that vanishes on the boundary. Multiplying (1) by vv, integrating over Ω\Omega, and integrating by parts gives

∫Ω∇u⋅∇v dx=∫Ωfv dx.\int_\Omega \nabla u \cdot \nabla v\,\mathrm{d}x = \int_\Omega f v\,\mathrm{d}x.

The boundary term vanishes because both uu and vv have zero trace. The natural space for this problem is

V=H01(Ω),V = H_0^1(\Omega),

and the weak problem is:

Find u∈V such that a(u,v)=⟨f,v⟩∀v∈V,\text{Find } u \in V \text{ such that }\quad a(u,v) = \langle f,v\rangle \quad \forall v \in V,

where

a(u,v):=∫Ω∇u⋅∇v dx.a(u,v) := \int_\Omega \nabla u \cdot \nabla v\,\mathrm{d}x.

The right-hand side is written as a duality pairing because, in general, the data may belong to V′V', the dual of VV, rather than to L2(Ω)L^2(\Omega).

3. Well-posedness and the Lax–Milgram lemma

The abstract form of the problem is determined by a bilinear form a:V×V→Ra:V\times V\to\mathbb{R} and a functional f∈V′f\in V'. Two properties are central:

For the Poisson problem, boundedness follows from the Cauchy–Schwarz inequality:

∣a(u,v)∣≤∥∇u∥L2(Ω)∥∇v∥L2(Ω).\lvert a(u,v)\rvert \leq \lVert\nabla u\rVert_{L^2(\Omega)} \lVert\nabla v\rVert_{L^2(\Omega)}.

On H01(Ω)H_0^1(\Omega), Poincaré’s inequality makes the seminorm ∥∇v∥L2(Ω)\lVert\nabla v\rVert_{L^2(\Omega)} equivalent to the H1H^1-norm. Hence the form is also coercive, and the weak Poisson problem has a unique solution.

4. The Ritz–Galerkin approximation

Let Vh⊂VV_h\subset V be a finite-dimensional conforming space. The Galerkin approximation is obtained by restricting the weak problem to VhV_h:

Find uh∈Vh such that a(uh,vh)=⟨f,vh⟩∀vh∈Vh.\text{Find } u_h\in V_h \text{ such that }\quad a(u_h,v_h)=\langle f,v_h\rangle \quad \forall v_h\in V_h.

Because the same boundedness and coercivity estimates hold on the subspace VhV_h, the discrete problem is also uniquely solvable.

Choose a basis {φ1,…,φN}\{\varphi_1,\ldots,\varphi_N\} of VhV_h and write

uh=∑j=1NUjφj.u_h = \sum_{j=1}^N U_j\varphi_j.

Testing (9) with each basis function gives the linear system

∑j=1Na(φj,φi)Uj=⟨f,φi⟩,i=1,…,N.\sum_{j=1}^N a(\varphi_j,\varphi_i)U_j = \langle f,\varphi_i\rangle, \qquad i=1,\ldots,N.

Thus the stiffness matrix and load vector are

Aij=a(φj,φi),Fi=⟨f,φi⟩,AU=F.A_{ij}=a(\varphi_j,\varphi_i), \qquad F_i=\langle f,\varphi_i\rangle, \qquad A\mathbf{U}=\mathbf{F}.

When aa is symmetric, as for the Poisson problem, AA is symmetric positive definite. This is the algebraic form of the Ritz–Galerkin method.

The Ritz interpretation

For a symmetric bilinear form, define the energy functional

J(w):=12a(w,w)−⟨f,w⟩.J(w) := \frac{1}{2}a(w,w)-\langle f,w\rangle.

The solution of the continuous problem is the unique minimizer of JJ over VV, while uhu_h is the unique minimizer over VhV_h:

u=arg min⁡w∈VJ(w),uh=arg min⁡wh∈VhJ(wh).u = \operatorname*{arg\,min}_{w\in V}J(w), \qquad u_h = \operatorname*{arg\,min}_{w_h\in V_h}J(w_h).

The term Ritz emphasizes this minimization property; Galerkin emphasizes the orthogonality condition that characterizes the discrete solution.

5. Galerkin orthogonality

The exact solution also satisfies the weak equation for every vh∈Vhv_h\in V_h, since Vh⊂VV_h\subset V. Subtracting the discrete equation gives

a(u−uh,vh)=0∀vh∈Vh.a(u-u_h,v_h)=0 \quad \forall v_h\in V_h.

This is Galerkin orthogonality: the error u−uhu-u_h is orthogonal to the discrete space with respect to the bilinear form aa. It is the key identity behind the basic a priori error estimate.

6. Céa’s lemma

The estimate separates the analysis into two parts:

  1. M/αM/\alpha measures the stability of the variational problem;

  2. the infimum measures the best approximation available in VhV_h.

To see the mechanism, take any vh∈Vhv_h\in V_h. By coercivity and Galerkin orthogonality,

α∥u−uh∥V2≤a(u−uh,u−uh)=a(u−uh,u−vh)+a(u−uh,vh−uh)=a(u−uh,u−vh)≤M∥u−uh∥V∥u−vh∥V.\begin{aligned} \alpha\lVert u-u_h\rVert_V^2 &\leq a(u-u_h,u-u_h)\\ &=a(u-u_h,u-v_h)+a(u-u_h,v_h-u_h)\\ &=a(u-u_h,u-v_h)\\ &\leq M\lVert u-u_h\rVert_V\lVert u-v_h\rVert_V. \end{aligned}

After cancelling the error norm and taking the infimum over vhv_h, we obtain Céa’s estimate.

For finite elements, this result reduces the error analysis to an approximation question: how well can the exact solution be approximated by the chosen finite element space? The following lectures will address the construction of these spaces and the estimates needed to answer that question.

Summary