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On a shape-regular mesh in dimensions at most three, nodal piecewise-linear interpolation of $u\in H^2(\Omega )$ satisfieswhere $h$ is the largest element diameter and $C$ is independent of $h$. For homogeneous Dirichlet boundary conditions, $I_{h}u$ belongs to the conforming finite element space. Scaling each element to a reference element gives the estimate; shape regularity bounds the scaling constants, while subtraction of an affine function leaves an error controlled by the second derivatives.

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.

Solutions to all three questions of the finite element methods for partial differential equations course of the University of Oxford Part C examination, sat on 6 June 2023. Source: original released PDF, bundle pages 325–328; paper pages 1–4. The original numbering is retained; question statements are not reproduced.

All function spaces and bilinear forms below are real. Write $(p,q{)}_0={\int }_{\Omega }pq$ for the L2 inner product, and $\Vert p{\Vert }_1^2=\Vert p{\Vert }_0^2+\Vert \nabla p{\Vert }_0^2$ for the first-order Sobolev norm; the notation applies componentwise to vector fields. We use the Poincaré inequality explicitly allowed by the paper. On an arbitrary unbounded Lipschitz domain that inequality would need an additional assumption. The convergence assertions use the usual full conforming finite element spaces, or vector subspaces containing the indicated interpolants; merely being a vector subspace of the stated piecewise polynomial functions is insufficient. Part 1(d) explains the intended meaning of positivity for its nonsymmetric matrix.

Use the continuous piecewise-linear nodal interpolants $I_{h}u$ and $I_{h}w$ on the shape-regular mesh. Since $u,w\in H^2(\Omega )$ in two dimensions, the Sobolev embedding theorem supplies continuous representatives, so the nodal values are well-defined. Their zero Sobolev traces imply zero boundary nodal values. Hence $I_{h}u\in {\mathcal{U}}_h$ and $I_{h}w\in {\mathcal{W}}_h$ for the usual full conforming finite element spaces.

The permitted finite element interpolation estimates give constants $C_u,C_w$ determined by the shape regularity and $\Omega$, independent of $h$, such thatSubstitute these choices in the best approximation bound from part (b):For fixed $u,w$, this constant is independent of $h$. A shape-regular mesh suffices; no uniformity of the individual element sizes is needed, since $h$ is their maximum diameter.

The inclusion of these interpolants is an essential assumption. If the wording “consist of” were read as permitting arbitrary smaller vector subspaces, taking ${\mathcal{U}}_h={\mathcal{W}}_h=\{0\}$ for every $h$ would prevent the claimed convergence for a nonzero solution. The full conforming finite element space interpretation is also needed for the dimension assertion in part (d).

Let $x_1,\dots ,x_N$ be the interior nodes, and let ${\varphi }_1,\dots ,{\varphi }_N$ be the nodal basis of the common conforming finite element space, with ${\varphi }_j(x_i)={\delta }_{ij}$ and zero boundary values. These piecewise-linear hat functions give the expansionsTesting the first equation with ${\varphi }_i$ and then the second equation with ${\varphi }_i$ gives the linear systemHere $S$ is the diffusion stiffness matrix and $M$ the mass matrix. Their entries are symmetric in $i,j$, so both are symmetric matrices.

For $c\in {\mathbb{R}}^N$, put $v_c={\sum }_j{c}_j{\varphi }_j$. The two quadratic forms areIf $c\ne 0$, the linear independence of the basis ensures $v_c\ne 0$, so $c^{\mathsf{T}}Mc>0$. If $c^{\mathsf{T}}Sc=0$, the Poincaré inequality and zero Dirichlet boundary condition force $v_c=0$, contradicting $c\ne 0$. Thus both $S$ and $M$ are positive-definite matrices.

For $X=(p,q)\in {\mathbb{R}}^N\times {\mathbb{R}}^N$, symmetry of $M$ cancels the cross terms:ThereforeThis is the positivity of a quadratic form intended by the last sentence of the question. For $N>0$, $A$ is a nonsymmetric matrix, since $M\ne 0$. Under the convention that “positive-definite matrix” includes symmetry, that last sentence needs this qualification: the symmetric part of a matrix is positive definite. The quadratic lower bound for a nonsymmetric matrix still implies that $A$ is an invertible matrix: $AX=0$ would give $X^{\mathsf{T}}AX=0$, hence $X=0$.

For a conforming triangular finite element mesh, takeThis is the quadratic conforming finite element space. Across a common edge of a polygon, the restrictions from neighboring triangles are univariate quadratic polynomials. Their agreement at the two endpoints and the midpoint makes their difference a degree-at-most-two polynomial with three distinct roots of a polynomial, so it vanishes on the entire common edge of a polygon. Thus matching the local finite element degrees of freedom produces the required continuity. Similarly, setting the three nodal values on every boundary edge carrying the Dirichlet boundary condition to zero makes its entire restriction vanish.

Continuity across element interfaces means the piecewise gradients give the global weak gradient, with no jump contribution, so $V_h\subset V$. There are finitely many local finite element degrees of freedom, hence $V_h$ is finite-dimensional. The Galerkin method seeksThe bounded bilinear form and coercive bilinear form estimates from part (b) remain valid on $V_h$. Since a finite-dimensional subspace is closed, the Lax-Milgram theorem gives exactly one $u_h$. Alternatively, the stiffness matrix plus the mass matrix is a positive-definite matrix in any basis of $V_h$. The same existence and uniqueness conclusion holds for any finite-dimensional vector subspace of this full space.

Put $e=u-u_h\in V$. The Aubin–Nitsche duality argument uses the dual weak formulationBy symmetry of $a$ and the Lax-Milgram theorem, this has a unique solution. Taking $v=z$ and using the Cauchy-Schwarz inequality showsWe must justify $H^2$ elliptic regularity for this dual problem from the result actually supplied. That result concerns $\Delta$, whereas the dual equation is $-\Delta z+z=e$. Put $g=z-e\in L^2(\Omega )$. The given result produces $w\in H^2(\Omega )\cap V$ with $\Delta w=g$, zero normal derivative on ${\Gamma }_N$, andBy Green's first identity, for every $v\in V$,Set $v=w-z$. Then $\nabla (w-z)=0$. On the connected triangle, a Sobolev function with zero weak gradient is constant, and its zero Sobolev trace on ${\Gamma }_D$ forces that constant to be zero. Hence $z=w$, provingThis proves the needed reaction-diffusion regularity from Laplacian regularity without assuming the dual solution already has second derivatives in the L2 space.

Only $H^2$ regularity is available for $z$. Use its continuous piecewise-linear interpolant $L_{h}z$, which belongs to the quadratic conforming finite element space $V_h$ because ${\mathcal{P}}_1(K)\subset {\mathcal{P}}_2(K)$. It has the same zero Dirichlet boundary condition and satisfies the finite element interpolation estimateThus no $H^3$ assumption on the dual solution is necessary. Taking $v=e$ in the dual weak formulation and using Galerkin orthogonality givesIf $\Vert e{\Vert }_0=0$, the desired bound holds. Otherwise divide by $\Vert e{\Vert }_0$ and use part (d):All constants depend only on the fixed triangle and shape regularity. The acute-angle assumption enters through the supplied mixed-boundary elliptic regularity result.

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.

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.

Consider a bounded Lipschitz domain $\Omega$ and homogeneous Dirichlet boundary conditions forAssume $A$ is a bounded real symmetric matrix field with ${\xi }^{T}A(x)\xi \ge a_0|\xi {|}^2$ uniformly, $c\ge 0$ is bounded, and $f$ defines a bounded linear functional on $V=H_0^1(\Omega )$, the zero-boundary Sobolev space. The weak formulation isThe Cauchy-Schwarz inequality gives continuity $|a(u,v)|\le M\Vert u{\Vert }_V\Vert v{\Vert }_V$. Uniform ellipticity and the Poincaré inequality give a coercive bilinear form, $a(v,v)\ge \alpha \Vert v{\Vert }_V^2$. The Lax-Milgram theorem therefore supplies a unique weak solution, with $\Vert u{\Vert }_V\le \Vert \ell {\Vert }_{V^{\prime }}/\alpha$. Its proof represents $a$ by a bounded linear operator $T$ using the Riesz representation theorem; coercivity gives an injective operator with closed image, and a zero orthogonal complement makes that image all of $V$. This yields existence, uniqueness and the displayed estimate.

For a conforming finite element space $V_h\subseteq V$, the Galerkin method finds $u_h\in V_h$ such that $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$. With a basis ${\varphi }_1,\dots ,{\varphi }_N$, this is the linear system $KU=F$, where the stiffness matrix is $K_{ij}=a({\varphi }_j,{\varphi }_i)$ and $F_i=\ell ({\varphi }_i)$. Its positive-definite matrix property proves unique discrete solvability. Local support gives a sparse matrix through elementwise assembly.

For the symmetric problem, the Ritz method minimizes $J(v)=a(v,v)/2-\ell (v)$ over $V_h$. Differentiating in every direction $v_h$ gives precisely the Galerkin method equations. Conversely, if $u_h$ solves them, $J(u_h+w_h)-J(u_h)=a(w_h,w_h)/2$, proving it is the unique minimum. The same argument over $V$ gives $J(v)-J(u)=\Vert v-u{\Vert }_a^2/2$, where $\Vert v{\Vert }_a=\sqrt{a(v,v)}$ is the energy norm. Thus the Ritz method chooses the best trial function in that norm.

Subtracting the continuous and discrete equations gives Galerkin orthogonality, $a(u-u_h,v_h)=0$. For arbitrary $v_h\in V_h$ and $e=u-u_h$, it follows that $a(e,e)=a(e,u-v_h)$. Coercivity and continuity prove the Céa lemma:In the symmetric energy norm, the Cauchy-Schwarz inequality improves the constant to one. Equivalently, Galerkin orthogonality gives the Pythagorean identity $\Vert u-v_h{\Vert }_a^2=\Vert u-u_h{\Vert }_a^2+\Vert u_h-v_h{\Vert }_a^2$. These estimates establish stability and show that approximation properties of the trial space determine convergence.

For continuous piecewise-linear elements on a shape-regular mesh in dimensions at most three, a finite element interpolation estimate gives $\Vert u-I_{h}u{\Vert }_{H^1}\le Ch\Vert u{\Vert }_{H^2}$ when $u\in H^2(\Omega )$. Combining it with the Céa lemma yields $O(h)$ energy norm error. More generally, degree-$p$ elements give $O(h^p)$ error in $H^1$ under the corresponding regularity and approximation assumptions. Mere mesh refinement cannot supply an order whose required solution regularity is absent.

An additional order in the L2 norm follows under a dual elliptic regularity assumption. Solve $a(v,z)=(e,v{)}_{L^2}$ and assume $\Vert z{\Vert }_{H^2}\le C\Vert e{\Vert }_{L^2}$. The Aubin–Nitsche duality argument uses Galerkin orthogonality to obtainTherefore $\Vert e{\Vert }_{L^2}\le Ch\Vert e{\Vert }_{H^1}$, giving $O(h^2)$ for the piecewise-linear case with $u\in H^2$. This improvement explicitly requires regularity of the dual problem.

As a concrete example, solve $-u^{\prime \prime }=1$ on $(0,1)$ with zero endpoint values. On a uniform mesh of spacing $h$, the piecewise-linear hat functions give $K_{jj}=2/h$, $K_{j,j\pm 1}=-1/h$ and $F_j=h$. The discrete equations are $2U_j-U_{j-1}-U_{j+1}=h^2$, whose solution is $U_j=x_j(1-x_j)/2$. The exact solution is $u(x)=x(1-x)/2$, so $u_h$ is its piecewise-linear nodal interpolant. On each interval $[x_j,x_{j+1}]$,Integrating the squared error and squared derivative error gives the explicit rateswhere the $H_0^1$ norm here is $\Vert v^{\prime }{\Vert }_{L^2}$.

For nonsymmetric coercive problems, the Galerkin method, Galerkin orthogonality and the Céa lemma still apply, but the symmetric quadratic minimization interpretation of the Ritz method generally does not. Nonconforming trial spaces, inexact integration and noncoercive equations need additional arguments beyond the conforming coercive theory proved here.

A piecewise-linear hat function is a continuous piecewise linear function that selects one node of a finite element mesh: it is one there and zero at every other node. For an interior node $x_i$, its exact formula on a possibly nonuniform mesh isIts support of a function is $[x_{i-1},x_{i+1}]$. It rises on the left interval and falls on the right, producing its hat-shaped graph. The interior hat functions form a basis of the continuous $P_1$ conforming finite element space with homogeneous Dirichlet boundary conditions. Chen's author notes define this nodal basis and illustrate it in one and two dimensions.^[ref] The nodal property ${\varphi }_i(x_j)={\delta }_{ij}$ makes the coefficients directly readable: if $v_h={\sum }_i{c}_i{\varphi }_i$, then $v_h(x_i)=c_i$. Unlike the global Lagrange cardinal polynomials, each of these functions is confined to two neighboring elements.