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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.

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.

A coarse-grid correction improves a fine-grid approximation by solving a smaller equation for the part of its approximation error representable on a coarse mesh. Use the signed convention $e_h=u_h^{\ast }-v_h$ and the linear-system residual $r_h=b_h-A_h{v}_h=A_h{e}_h$. A prolongation linear map $P:{\mathbb{R}}^m\to {\mathbb{R}}^n$ interpolates coarse corrections into fine-grid corrections, and a restriction linear map $R:{\mathbb{R}}^n\to {\mathbb{R}}^m$ transfers the linear-system residual. Solve $A_H{z}_H=R{r}_h$ exactly or approximately and update $v_h\leftarrow v_h+P{z}_H$. The plus sign follows from defining the approximation error as exact minus approximate. It is the correction equation's right-hand side $R{r}_h$, rather than merely restricted solution values, that determines the update.

For the multigrid method model $A_h=h^{-2}\mathrm{tridiag}(-1,2,-1)$, take $n=2m+1$, so $H=2h$ and the coarse nodes coincide with every second fine node. Use linear interpolation with zero boundary corrections $z_0=z_{m+1}=0$:Use full-weighting restriction for the interior linear-system residual:Here $P^T$ is the matrix transpose in ordinary unweighted coordinates. With these specific transfers, the Galerkin coarse-grid operator isThus it equals the rediscretized Poisson equation operator on this constant-coefficient one-dimensional mesh. The equality is special to the stated model and transfers. In particular, dropping $h^{-2}$ from the fine matrix while using $H^{-2}$ on the coarse mesh changes the correction unless the right-hand sides and transfers are rescaled consistently.

Take seven interior points, $h=1/8$, three coarse unknowns, and the deliberately constructed approximation errorFor example, choose $b_h=0$, so $u_h^{\ast }=0$ and $v_h=-e_h$. Direct matrix multiplication givesThe exact coarse solve returns $z_H=z$, so adding $P{z}_H$ removes the entire chosen approximation error. A broad triangular error can have a sharply localized linear-system residual: the latter measures discrete curvature, not error height. This is why the two should not be confused. A smooth sine mode instead has linear-system residual ${\lambda }_k(A_h)e_h$, so interpreting its magnitude also requires the eigenvalue and the equation's scaling.

The algebra explains the exact cancellation beyond this example. Assume $P$ has linearly independent columns and choose an invertible Galerkin coarse-grid operator $A_H=R{A}_{h}P$. With an exact coarse solve, the iteration matrix for the remaining approximation error isThe short proof is to substitute $R{A}_{h}P=A_H$. Thus every vector in the column space of $P$ is eliminated, and the corrected linear-system residual restricts to zero. These are exact projection identities, not a statement that every visually smooth vector belongs to that column space. If $A_h$ is a symmetric matrix and a positive-definite matrix, and $R=c{P}^T$ with $c>0$, then $A_H$ is also a positive-definite matrix. Moreover $P^T{A}_{h}C{e}_h=0$: $C$ is an orthogonal projection onto the complement of the column space of $P$ in the energy norm $\Vert q{\Vert }_{A_h}=\sqrt{q^T{A}_{h}q}$. Equivalently, the added correction is the best approximation to $e_h$ from that column space in this energy norm. Without these symmetry and transfer assumptions, the projection need not be orthogonal.^[ref]

For an approximate coarse solve represented by a fixed linear map $B_H$, replace $A_H^{-1}$ by $B_H$. ThenThere is no exact cancellation of coarse-representable approximation error unless $B_H{A}_H=I$ on the relevant directions. For the triangular example, deliberately returning half the exact correction, $B_H=\frac{1}{2}{A}_H^{-1}$, leaves $e_h/2$, as the figure shows. This controlled mathematical comparison illustrates an incomplete coarse solve; it is not a measured performance result or a prescription for a geometric multigrid V-cycle. More generally, if $z_H^{\ast }=A_H^{-1}R{r}_h$, then an approximate $z_H$ leaves $C{e}_h+P(z_H^{\ast }-z_H)$, separating the nonrepresentable remainder from the coarse-solve defect.

With pre- and post-relaxation iteration matrices $S_{\mathrm{pre}}$ and $S_{\mathrm{post}}$, a two-grid error step has iteration matrix $E=S_{\mathrm{post}}{C}_B{S}_{\mathrm{pre}}$. A geometric multigrid V-cycle supplies $B_H$ by one recursive coarse cycle from zero, with a direct solve only at the bottom. Effective smoothing, a coarse column space that approximates the surviving approximation error, and a sufficiently accurate coarse correction must work together. Coarsening alone supplies no convergence guarantee, and correcting an arbitrary fine-grid vector exactly would require a coarse column space large enough to contain it.

A finite element mesh partitions a domain into simple regions called elements, such as intervals, triangles or tetrahedra. In a conforming partition, distinct element interiors are disjoint, their closures cover the domain closure, and any nonempty intersection is a common face, edge or vertex of both elements. Chen's author notes give this compatibility condition for simplicial meshes.^[ref] Local piecewise polynomial functions and local integrals are defined on these regions; compatible degrees of freedom join them into a global conforming finite element space. The assembled stiffness matrix and mass matrix then inherit the local overlap pattern.

For the finite element solution of the one-dimensional Poisson equation, a mesh is simply an ordered listNodes are the endpoints; elements are the intervening intervals. Four equal elements have five nodes but only three unknown nodal values after imposing the zero Dirichlet boundary conditions. A nonuniform partition such as $0,1/8,1/4,1/2,1$ has element lengths $1/8,1/8,1/4,1/2$ and $h=1/2$, so counting elements alone does not measure the largest unresolved scale.

Refining a finite element mesh replaces some or all elements by smaller ones. For continuous piecewise linear functions in one dimension, retaining all old nodes gives nested conforming finite element spaces $V_h\subset V_{h/2}$ when every interval is bisected. The old approximation remains available, and the new interior nodes allow additional changes of slope. With a symmetric coercive bilinear form and exact integration, best approximation from Galerkin orthogonality means that the energy norm error cannot increase under such nested refinement.

For the finite element solution of the one-dimensional Poisson equation, uniform refinement from four to eight elements increases the unknown count from three to seven. The exact formulas give maximum errors $0.0078125$ and $0.001953125$, and energy norm errors approximately $0.0721688$ and $0.0360844$. Refinement near a region of strong curvature can spend unknowns where the linear interpolation error matters most; convergence in a general problem still needs approximation and regularity assumptions such as those in the finite element interpolation estimate.

The finite element method replaces an infinite-dimensional weak formulation by finitely many unknowns attached to a finite element mesh. Simple local functions, often piecewise polynomial functions, join to form a global approximation. The attraction is that each finite element can be integrated separately, while the assembled equations enforce a global balance. A piecewise-linear hat function only meets its immediate neighbors, so a large problem can have a sparse matrix rather than a dense one. Gilbert Strang's MIT lecture develops this connection from the one-dimensional weak formulation through local hats to the stiffness matrix and load integrals.^[ref]

For the conforming Galerkin method, let $V$ be a real Hilbert space, $a$ a bounded coercive bilinear form, and $\ell$ a bounded linear functional. Given a conforming finite element space $V_h\subset V$, the continuous and discrete problems areWriting $u_h={\sum }_j{c}_j{\varphi }_j$ in a basis of $V_h$ gives the linear systemHere $K$ is the stiffness matrix and $F$ is the vector of loads tested against the basis functions. The Galerkin method uses the same trial and test space in this construction. Finite element methods also use the Petrov-Galerkin method, with different trial and test spaces; testing against the trial space is therefore a property of this construction, rather than a definition of every finite element method.

The finite element solution of the one-dimensional Poisson equation below gives every entry explicitly. Refining the finite element mesh then improves the approximation by adding local degrees of freedom. For a symmetric coercive bilinear form, Galerkin orthogonality makes the discrete solution the best approximation in the energy norm; the geometry of orthogonal projection explains why this local construction yields a globally meaningful answer.

Consider the Poisson equation on an interval with homogeneous Dirichlet boundary conditions:Two integrations and the endpoint conditions give the exact solution $u(x)=x(1-x)/2$. To approximate it, multiply by a test function $v$ that vanishes at both endpoints. Integration by parts givesThe boundary term is zero. Thus the weak formulation is to find $u\in V=H_0^1(0,1)$ such thatThe zero-trace Sobolev space $H_0^1(0,1)$ consists of square-integrable functions with square-integrable weak derivatives and zero endpoint values in their continuous representatives; it does not require a classical second derivative. This is why continuous piecewise linear functions are admissible despite corners. The Poincaré inequality makes $a$ a coercive bilinear form on $V$. Strang derives this one-dimensional weak formulation, the piecewise-linear hat functions, and their load and derivative-product integrals in the cited lecture; its endpoint example uses a free left endpoint, while both endpoints here are fixed.^[ref]

Take the uniform finite element mesh $x_j=jh$, $j=0,\dots ,n$, with $h=1/n$. The continuous $P_1$ trial space isHere $P_1$ means polynomials of degree at most one on each interval. This is a conforming finite element space in $V$, with $n-1$ interior piecewise-linear hat functions ${\varphi }_i$ as a basis. Their endpoint values enforce the Dirichlet boundary conditions automatically. Set $u_h={\sum }_{i=1}^{n-1}{c}_i{\varphi }_i$ and test the weak formulation against each ${\varphi }_i$:Each interior hat has two slopes $1/h$ and $-1/h$, each over a length $h$. Its squared derivative therefore integrates to $2/h$. Adjacent hats have opposite slopes on their one shared interval, giving $-1/h$. All other derivative products vanish. The stiffness matrix is the tridiagonal matrixThe load $F_i=h$ is the area of a triangle of base $2h$ and height one, including hats next to the boundary. Boundary hats are omitted because the boundary values are prescribed. The piecewise-linear hat function article explains how the same entries arise by adding local two-node contributions.

With four mesh intervals, $h=1/4$, there are three unknowns:For example, the first row gives $8(3/32)-4(1/8)=1/4$ and the middle gives $-4(3/32)+8(1/8)-4(3/32)=1/4$. The coefficients are the nodal values $u_h(x_i)=c_i$. At $x=3/8$, linear interpolation gives $u_h(3/8)=7/64=0.109375$, whereas $u(3/8)=15/128=0.1171875$: exactness at the nodes still leaves a visible error between them.

In this model the Galerkin method solution equals the nodal approximation $I_{h}u$ obtained by linear interpolation on any finite element mesh. Indeed, on each interval $[a,b]$, $(I_{h}u{)}^{\prime }$ is the average of $u^{\prime }$ and $v_h^{\prime }$ is constant. Therefore ${\int }_a^b(u^{\prime }-(I_{h}u{)}^{\prime })v_h^{\prime }dx=0$. Summing proves Galerkin orthogonality for $I_{h}u$, and uniqueness gives $u_h=I_{h}u$. This argument relies on the constant diffusion coefficient, conformity and exact integration; nodal exactness is not a general promise for finite element methods with other operators or integration rules.

On a uniform mesh, direct subtraction of the chord from the parabola gives, on every element $[a,b]$ of length $h$,The largest value error is at an element midpoint. Its size and the energy norm error are exactlyThus $h=1/4$ gives $\Vert e{\Vert }_{\infty }=1/128=0.0078125$ and $\Vert e{\Vert }_a=1/(4\sqrt{12})\approx 0.0721688$. With eight elements, $h=1/8$, these become $1/512=0.001953125$ and $1/(8\sqrt{12})\approx 0.0360844$. Halving the mesh spacing quarters the maximum value error but only halves the energy norm error: values and slopes converge at different rates. The figure shows the small value error at its own scale, where the four-to-one reduction becomes clear.

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.

Multiply by a test function vanishing at the endpoints and apply integration by parts. The positive operator is $-u^{\prime \prime }+xu$, so the forcing changes sign. The weak formulation of this forced Airy boundary-value problem is: find $u\in V=H_0^1(0,1)$ such thatHere $V$ is the zero-boundary Sobolev space and can be equipped with $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_{L^2}$. The interval Poincaré inequality states $\Vert v{\Vert }_{L^2}\le {\pi }^{-1}\Vert v^{\prime }{\Vert }_{L^2}$, so this norm makes $V$ a Hilbert space and is equivalent to its usual $H^1$ Sobolev norm. The form is symmetric and satisfiesThe Lax-Milgram theorem says that a bounded coercive bilinear form on a Hilbert space and a bounded linear functional determine exactly one weak solution. All its hypotheses have just been verified. Moreover, $u^{\prime \prime }=xu+1\in L^2$ gives $u\in H^2$ on the interval; the right-hand side is then continuous, so this weak solution is the classical solution with the prescribed endpoint values.

Since the form is symmetric, the equivalent Ritz method minimizesIndeed, stationarity gives the weak equation andwith equality only at $u$. This proves the minimization equivalence, including the sign of the linear term.

Restricting to a conforming finite element space $V_h\subset V$ gives a unique discrete solution and Galerkin orthogonality, $a(u-u_h,v_h)=0$ for every $v_h\in V_h$. The Céa lemma states that if the continuity and coercivity constants are $M$ and $\alpha$, then $\Vert u-u_h{\Vert }_V\le (M/\alpha ){\inf }_{v_h\in V_h}\Vert u-v_h{\Vert }_V$. For this symmetric problem the exact best-approximation constant in the energy norm is one. The piecewise-linear finite element interpolation estimate $\Vert u-I_{h}u{\Vert }_{H^1}\le Ch\Vert u{\Vert }_{H^2}$ on regular finite element meshes therefore gives convergence of a numerical method of the Ritz method, with first-order error in the energy norm.

The finite element method replaces an infinite-dimensional variational problem by a finite-dimensional one, usually using functions that are polynomial on small mesh elements. Ritz method and Galerkin method describe how the discrete equations are selected; neither intrinsically requires piecewise polynomials, though those spaces make assembly local and sparse.

For a concrete realization of the two-point problem, choose homogeneous Dirichlet boundary conditions at both endpoints of $(0,1)$. The differential expression alone is not a complete boundary value problem, so this boundary choice is an explicit assumption. Let $p\in W^{1,\infty }$ with $p\ge p_0>0$, $q\in L^{\infty }$ with $q\ge 0$, and $f\in L^2$. Use the zero-boundary Sobolev space $V=H_0^1(0,1)$ with norm $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_2$. By the Poincaré inequality this is equivalent to its full first-order Sobolev norm. Multiplying by a test function and applying integration by parts givesThe endpoint terms vanish by the essential boundary condition. The form is a symmetric bilinear form, bounded byand it is a coercive bilinear form:The load is a bounded linear functional. The Lax-Milgram theorem gives a unique weak solution and a bound $\Vert u{\Vert }_V\le \Vert \ell {\Vert }_{V^{\ast }}/p_0$. Under the stated coefficient regularity it solves the differential equation in the usual weak sense.

The Ritz method minimizes the energyover $V$, or over a conforming finite-dimensional subspace $V_h\subset V$. Its first variation is $a(v,w)-\ell (w)$; symmetry and coercivity make the functional strictly convex. In fact, if $u$ is the weak solution,Thus the energy minimizer is exactly the weak solution. The Galerkin method instead directly asks for $u_h\in V_h$ with $a(u_h,v_h)=\ell (v_h)$ for all $v_h\in V_h$. For this symmetric coercive problem, Ritz-Galerkin equivalence for a symmetric coercive form shows that the discrete minimizer and Galerkin solution coincide. For a nonsymmetric problem, a Galerkin formulation can still apply but the same quadratic minimization generally cannot represent its full bilinear form; appropriate coercivity or inf-sup conditions must then justify the chosen spaces.

Subtract the continuous and discrete weak equations to obtain Galerkin orthogonality $a(u-u_h,v_h)=0$. Since $a$ defines the energy norm $\Vert v{\Vert }_a=\sqrt{a(v,v)}$, the Pythagorean identity givesIn a general reference norm, the Céa lemma yieldsThe proof uses $a(e,e)=a(e,u-v_h)$ together with coercivity and continuity. Dense approximation spaces therefore imply convergence; the estimate reduces a numerical error problem to an approximation problem.

For implementation, partition the interval by nodes $0=x_0<\cdots <x_N=1$. Choose continuous piecewise-linear functions and the interior piecewise-linear hat functions ${\varphi }_j$ as a basis, with the two boundary coefficients prescribed. Write $u_h={\sum }_j{U}_j{\varphi }_j$. The coefficients satisfyThe stiffness matrix is symmetric positive definite: $U^{T}KU=a(u_h,u_h)>0$ for nonzero interior coefficients. The local supports make it tridiagonal in this one-dimensional linear-element case.

On an element of length $h_e$, for constant element coefficients $p_e,q_e$, the two-node matrix and constant-load vector areFor variable coefficients, integrate $p,q,f$ against the same shape functions rather than silently replacing them by constants. Element contributions are added at shared nodes, and boundary values are eliminated or lifted into the right side. A sparse direct solve or a suitable iterative method gives the nodal coefficients. Numerical quadrature should preserve the required accuracy and, with positive coefficients and weights, the positive energy structure.

A finite element interpolation estimate gives ${\inf }_{v_h}\Vert u-v_h{\Vert }_{H^1}\le Ch\Vert u{\Vert }_{H^2}$ for these elements, where $h$ is the largest element size and the solution has the stated regularity. The Céa lemma then gives first-order energy error. If the dual elliptic problem has $H^2$ regularity, the Aubin–Nitsche duality argument gives second-order $L^2$ error: for $e=u-u_h$, solve $a(v,z)=(e,v{)}_{L^2}$, then use $a(e,z)=a(e,z-I_{h}z)$ to gain one further factor of $h$. Higher-order elements improve rates when the solution is sufficiently smooth.

Nonzero Dirichlet data are handled by a boundary lifting and an affine trial space. Neumann boundary conditions enter naturally through the integrated boundary term; Robin terms modify both the form and the load. With pure Neumann conditions and $q=0$, constants lie in the kernel: existence needs the load/flux compatibility condition, uniqueness needs a mean constraint, and the preceding Dirichlet coercivity argument cannot simply be reused. Conforming approximation, boundary conditions and coercivity are the ingredients connecting the finite-element construction to a justified Ritz or Galerkin solution.

Put $h=1/(N+1)$ and $x_i=ih$. The interior piecewise-linear hat functions areThey are a basis of the continuous piecewise-affine functions vanishing at both endpoints. The support of a function for each hat only overlaps those of neighbouring hats, so the stiffness matrix is a tridiagonal matrix. Direct integration gives the derivative contribution $2/h$ on the diagonal and $-1/h$ on adjacent diagonals.

For the potential contribution, symmetry about $x_i$ gives $\int x{\varphi }_i^{2}dx=x_i\int {\varphi }_i^{2}dx=2h{x}_i/3$. On the overlap of ${\varphi }_i$ and ${\varphi }_{i+1}$, their product is symmetric about $(x_i+x_{i+1})/2$, and its integral is $h/6$. This is the affine-weighted hat mass matrix calculation. Consequently,and all other entries vanish. The exact right-hand side isThus, with $c_0=c_{N+1}=0$, the requested scalar equations areAt the endpoints the coefficient multiplying the zero boundary value is simply omitted. Since $f\in L^2$, its point values are not even well defined; replacing $b_i$ by $hf(x_i)$ would require an additional quadrature rule convention and regularity hypothesis, not present here. The exact Ritz method uses the displayed integrals. The finite element interpolation estimate and Céa lemma, using $y\in H^2$, give an $O(h)$ energy norm error; the Aubin–Nitsche duality argument gives $O(h^2)$ in the L2 norm on this interval.

Choose a basis ${\varphi }_1,\dots ,{\varphi }_N$ of a conforming finite element space $V_N\subset H_0^1(0,1)$ and write $y_N={\sum }_j{c}_j{\varphi }_j$. Testing the variational problem with each basis function givesThese are also the equations $\partial J(y_N)/\partial c_i=0$ for the Ritz method. The stiffness matrix is symmetric and a positive-definite matrix: for a nonzero coefficient vector $c$, linear independence gives $v_N=\sum c_j{\varphi }_j\ne 0$, andThus there is a unique coefficient vector.

Subtracting the exact and discrete variational problems proves Galerkin orthogonality, $a(y-y_N,v_N)=0$ for every $v_N\in V_N$. In the energy norm $\Vert v{\Vert }_a=\sqrt{a(v,v)}$, the Pythagorean identity yieldsTherefore the Ritz method is the best approximation in the energy norm. The Céa lemma also gives $\Vert y-y_N{\Vert }_V\le (1+{\pi }^{-2}){\inf }_{v_N\in V_N}\Vert y-v_N{\Vert }_V$, so any dense sequence of conforming trial spaces converges.

The starting point of a conforming finite element method is a weak formulation in a Hilbert space $X$. For example, a homogeneous Dirichlet boundary condition on a bounded Lipschitz domain gives $X=H_0^1(\Omega )$. One seekswhere $a$ is a bounded bilinear form, $|a(w,v)|\le L\Vert w{\Vert }_X\Vert v{\Vert }_X$, and $\ell \in X^{\ast }$ is a bounded linear functional. A conforming finite element space $X_h\subset X$ is finite dimensional and is typically built from functions that are polynomial on each element of a finite element mesh. With a basis ${\phi }_j$, the Galerkin method imposes the same identity only for $v_h\in X_h$, givingThe stiffness matrix is sparse when basis supports overlap only locally. Local element integrals assemble its entries; the mass matrix represents an $L^2$ term. This separates the choice of trial functions from the variational principle determining their coefficients.

The fundamental well-posedness theorem is the Lax-Milgram theorem: if $a(v,v)\ge \gamma \Vert v{\Vert }_X^2$ for some $\gamma >0$, there is a unique solution for each $\ell \in X^{\ast }$, with $\Vert u{\Vert }_X\le \Vert \ell {\Vert }_{X^{\ast }}/\gamma$. The same theorem on $X_h$ gives a unique discrete solution. The constants must be independent of the mesh for uniform stability of a numerical method. Boundary conditions are built into the trial space when they are essential constraints; flux conditions arise through boundary terms in integration by parts and are natural constraints in the weak formulation.

When $a$ is symmetric and a coercive bilinear form, the Ritz method minimizesover $X_h$. Its first variation is exactly the Galerkin method, so Ritz minimization and Galerkin testing coincide for symmetric coercive problems. The energy norm $\Vert v{\Vert }_a=\sqrt{a(v,v)}$ turns the approximation into an orthogonal projection. Indeed, subtracting the exact and discrete weak equations gives Galerkin orthogonality, $a(u-u_h,v_h)=0$. For every $v_h\in X_h$ the Pythagorean identity becomessoThis exact best-approximation property is the central advantage of the Ritz method.

Symmetry is unnecessary for the Galerkin method. For any bounded, coercive form, the Céa lemma givesTo prove it, put $e=u-u_h$. Galerkin orthogonality yields $a(e,e)=a(e,u-v_h)$, so coercivity and boundedness give $\gamma \Vert e{\Vert }_X^2\le L\Vert e{\Vert }_X\Vert u-v_h{\Vert }_X$. Divide when $e\ne 0$ and take the infimum. Thus approximation capability plus uniform coercivity gives convergence. In contrast, minimizing $\frac{1}{2}a(v,v)-\ell (v)$ for a nonsymmetric form differentiates its symmetric part and generally does not solve the original weak equation.

A concrete symmetric example is the one-dimensional reaction-diffusion problem from Question 5. Continuous piecewise-linear hat functions give the diffusion stiffness matrix $S$ with diagonal $2/h$ and neighbouring entries $-1/h$, and the mass matrix $M$ with diagonal $2h/3$ and neighbouring entries $h/6$. Therefore $K=S+M$ and $F_i=i{h}^2$. Its positive-definite matrix is the coordinate form of the continuous energy. The exact smooth solution $u(x)=x-\sinh (x)/\sinh 1$ makes this an explicit test of the method, not just an abstract existence result.

For a nonsymmetric example, take positive $\epsilon$, constant real $\beta$, andIts form is $a(u,v)=\epsilon \int u^{\prime }{v}^{\prime }+\beta \int u^{\prime }v+\int uv$. Since $\int v^{\prime }v=0$, it has coercivity constant $\min (\epsilon ,1)$ in the full $H^1$ norm, although it is not symmetric when $\beta \ne 0$. The Galerkin method remains well posed. On the uniform hat basis, $K=\epsilon S+\beta C+M$, where $C_{i,i+1}=1/2$, $C_{i,i-1}=-1/2$ and $C_{ii}=0$. The skew-symmetric matrix $C$ cancels from $c^{T}Kc$, proving nonsingularity. A Ritz energy would omit precisely this advection contribution. Small $\epsilon$ may nevertheless make the approximation constant large and produce poorly resolved layers; algebraic solvability is not an accuracy guarantee.

Quantitative convergence uses a finite element interpolation estimate. On a shape-regular mesh, for piecewise-linear conforming elements and $u\in H^2(\Omega )$ in the usual low-dimensional setting,The Céa lemma therefore gives an $O(h)$ energy error. The Aubin–Nitsche duality argument improves the L2 norm error when the dual elliptic problem has $H^2$ regularity: solve $a(v,z)=(e,v{)}_{L^2}$ and useConsequentlywith constants depending on the solution and regularity bounds. The second estimate is conditional on dual regularity; corners or insufficient data regularity can reduce the rate.

The same Galerkin method handles evolution equations. For the homogeneous heat equation, the semidiscrete finite element heat equation is $M\dot{c}+Sc=0$. Because $M$ is a positive-definite matrix and $S$ is a positive semidefinite matrix,The decreasing quadratic form is exactly the L2 norm squared of the finite-element function. This separates spatial stability of a numerical method from the subsequent time integrator, just as in the Schrödinger equation example. Conformity, coercivity and approximation estimates together explain convergence.