Galerkin orthogonality ($a(u-u_h,v_h)=0$)

Let $V_h\subset V$ be a conforming finite element space. Suppose the exact solution satisfies $a(u,v)=\ell (v)$ for every $v\in V$ and its Galerkin method approximation satisfies $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$, using the same bilinear form and linear functional. Subtracting the two identities gives Galerkin orthogonality:This identity itself does not require symmetry. Its interpretation as an orthogonal projection in an energy norm does require a bounded symmetric coercive bilinear form: then $a$ is an inner product, $e=u-u_h$ is perpendicular to every available discrete correction, andConsequently $u_h$ is the unique closest point in $V_h$ in the energy norm. This Pythagorean theorem in a function space explains the beauty of the Galerkin method: integrated local equations compute a global best approximation. The Céa lemma supplies the corresponding estimate in a chosen reference norm.

For the finite element solution of the one-dimensional Poisson equation, $a(w,v)={\int }_0^1{w}^{\prime }{v}^{\prime }dx$ and $\Vert w{\Vert }_a=\Vert w^{\prime }{\Vert }_{L^2(0,1)}$. Testing the error against the $i$th piecewise-linear hat function on a uniform mesh givesIn this model $e$ vanishes at every node, so each of these integrals is individually zero, although $e$ is positive inside every element. Galerkin orthogonality controls the derivative-weighted balance; it does not say that the pointwise error vanishes.

The exact identity assumes conformity and exact evaluation of the continuous forms, including any numerical integration that is used. If assembly instead uses $a_h$ and ${\ell }_h$, subtraction gives the consistency defectThus inexact numerical integration need not preserve the projection. In the Petrov-Galerkin method the error is tested against a different test space, and for a nonsymmetric bilinear form the displayed identity need not describe perpendicularity in any energy norm.

For a continuous coercive bilinear form with continuity constant $M$ and coercivity constant $\alpha$, a conforming finite-element solution satisfiesFor a symmetric problem measured in its energy norm, the constant is one.

For any $v_h\in V_h$, Galerkin orthogonality gives $a(e,e)=a(e,u-v_h)$, where $e=u-u_h$. Coercivity and continuity yield $\alpha \Vert e{\Vert }_V^2\le M\Vert e{\Vert }_V\Vert u-v_h{\Vert }_V$. Divide by $\Vert e{\Vert }_V$ when it is nonzero and take the infimum. In the symmetric case, the Cauchy-Schwarz inequality in the energy norm gives the constant one.

For a symmetric elliptic variational problem with Galerkin orthogonality, let $e=u-u_h$ and solve the dual problem $a(v,z)=(e,v{)}_{L^2}$ for every $v\in H_0^1(\Omega )$. If elliptic regularity gives $\Vert z{\Vert }_{H^2}\le C\Vert e{\Vert }_{L^2}$, then a finite element interpolation estimate givesThus $\Vert e{\Vert }_{L^2}\le Ch\Vert e{\Vert }_{H^1}$. Combining this with the Céa lemma and an $O(h)$ energy norm error gives an $O(h^2)$ L2 norm error. The additional order depends on the dual regularity assumption, which can fail on unsuitable domains.