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A bilinear form $B:X\times X\to \mathbb{R}$ on a real normed vector space is bounded when one constant $0\le C<\infty$ satisfiesThis is an upper bound on the magnitude of every pairing. It says nothing about the sign of $B(v,v)$ and does not require a symmetric bilinear form. The chosen norm is part of the hypothesis. Boundedness is equivalent to joint continuity, as shown by the boundedness criterion for a bilinear form.^[ref]

For example, on ${\mathbb{R}}^2$ with the Euclidean norm, the bilinear formsboth satisfy the upper bound with $C=1$, by the Cauchy-Schwarz inequality. Yet their quadratic forms are $B_{-}(v,v)=x^2-y^2$ and $B_0(v,v)=x^2$: the former takes negative values and the latter vanishes at the nonzero vector $(0,1)$. Thus neither is a coercive bilinear form. The latter is also a degenerate bilinear form, whereas $B_{-}$ is nondegenerate. Boundedness controls size, not positivity or loss of directions.

Every bilinear form on a real or complex finite-dimensional vector space equipped with any norm is bounded. In coordinates, write $B(u,v)=u^{T}Av$ using its matrix of a bilinear form. The Cauchy-Schwarz inequality gives $|u^{T}Av|\le \Vert A{\Vert }_2\Vert u{\Vert }_2\Vert v{\Vert }_2$, with $\Vert A{\Vert }_2$ the spectral norm; finite-dimensional equivalence of norms transfers this to the chosen norm. The conclusion concerns each fixed norm and fixed bilinear form: it does not supply a single constant for an arbitrary family of matrices or changing dimensions.

On an infinite-dimensional vector space, the upper bound is an additional condition. Equip the finitely supported sequences $c_{00}$ with the norm inherited from the l2 sequence space and setEach pairing is a finite sum, so this is a well-defined bilinear form. For the basis vector $e_n$, however, $\Vert e_n{\Vert }_2=1$ and $B(e_n,e_n)=n$; no finite $C$ works. It is even a coercive bilinear form with $\alpha =1$, since $\sum n{v}_n^2\ge \sum v_n^2$. Consequently coercivity alone does not imply boundedness on a normed vector space. This example uses $c_{00}$, which is not a Hilbert space; it does not define this pairing on all of the l2 sequence space.

A bounded bilinear form defines a bounded linear operator $T:X\to X^{\prime }$ by $(Tu)(v)=B(u,v)$, where $X^{\prime }$ is the continuous dual space. Its operator norm is the smallest admissible $C$:This converts an upper bound for pairings into an upper bound for a linear map taking an unknown vector to a linear functional.

The clamped problem uses the weak solution identity $\int \Delta u\Delta v=\int fv$ for every $v$ in the clamped second-order Sobolev space. For $f\in L^2(U)$, the clamped Hessian identity and the Poincaré inequality make this a bounded coercive bilinear form. The Lax-Milgram theorem gives a unique weak solution without a mean-zero condition on $f$.

A real bilinear form $B$ on a normed vector space $X$ is coercive when there is one constant $\alpha >0$ such thatThe lower bound uniformly controls every direction in the chosen norm. Unlike the upper bound defining a bounded bilinear form, it forces positive diagonal values at every nonzero vector. No symmetry is assumed. For complex Hilbert spaces, the usual analogue uses a sesquilinear form and $\mathrm{Re}B(v,v)\ge \alpha \Vert v{\Vert }^2$; it is not the same assertion for a complex bilinear form.

A concrete bounded and coercive example on ${\mathbb{R}}^2$, with the Euclidean norm, isIts quadratic form isThe best lower constant is $\alpha =2$, attained at $(1,0)$. Nevertheless, $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$, so coercivity does not imply a symmetric bilinear form. The symmetric part of a matrix $S=(A+A^T)/2=\mathrm{diag}(2,3)$ determines $q$, while the skew-symmetric matrix part changes mixed pairings without changing $q$. This is the symmetric and alternating parts of a bilinear form decomposition.^[ref]

For the upper constant, direct multiplication gives $A^{T}A=\left(\begin{array}{cc}5& -1\\ -1& 10\end{array}\right)$. Its eigenvalues are $(15\pm \sqrt{29})/2$, so the optimal upper constant, the spectral norm, is $C=\sqrt{(15+\sqrt{29})/2}$. The two constants answer different questions: how large a mixed pairing can be, and how small a self-pairing can be relative to squared norm.

The plotted quadratic forms show the mechanism: $x^2-y^2$ descends in one direction, $x^2$ leaves one direction uncontrolled, and $2x^2+3y^2$ controls both. These are self-pairings; their level sets cannot reveal the skew-symmetric matrix part. Calling $q$ a quadratic energy here refers to this diagnostic quantity, not to a claim that a nonsymmetric equation comes from minimizing an energy functional.

On a real finite-dimensional vector space with any norm, strict positivity $B(v,v)>0$ for every nonzero $v$ does imply coercivity. Indeed the quadratic form is continuous, and its positive minimum on the compact unit sphere supplies $\alpha$. If $B$ is also a symmetric bilinear form, this is positive definiteness. For general $B$, it is the symmetric and alternating parts of a bilinear form decomposition that isolates the positive part. Switching between equivalent norms preserves coercivity, but usually changes its constant.

The infinite-dimensional vector space distinction is explicit on the real l2 sequence space:The Cauchy-Schwarz inequality gives boundedness with $C=1$, and $B(v,v)>0$ for every nonzero vector. But $B(e_n,e_n)=1/n$ with $\Vert e_n{\Vert }_2=1$, so no positive $\alpha$ works. On the span of $e_1,\dots ,e_N$, the best lower constant is $1/N$: each finite restriction is coercive, but the control disappears as $N$ grows. Strict positivity is therefore weaker than a uniform lower bound.

The practical payoff is a stability estimate for a weak formulation. Assume $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is both a bounded bilinear form and a coercive bilinear form, and the load $\ell \in H^{\prime }$ is a bounded linear functional. The Lax-Milgram theorem gives a unique $u\in H$ satisfying $B(u,v)=\ell (v)$ for all $v\in H$, including nonsymmetric $B$.^[ref] The dual norm $\Vert \ell {\Vert }_{H^{\prime }}={\sup }_{v\ne 0}|\ell (v)|/\Vert v\Vert$ measures load size. Testing with $v=u$ gives the short energy estimateFor $u=0$ the bound holds directly; otherwise divide by $\Vert u\Vert$. Subtracting the equations for two loads and using the same argument gives $\Vert u_1-u_2\Vert \le \Vert {\ell }_1-{\ell }_2{\Vert }_{H^{\prime }}/\alpha$. Thus a positive lower constant limits the response to load errors. The upper bound and completeness of the Hilbert space enter the existence theorem; the estimate itself uses the lower bound and the load's dual norm.

For the displayed matrix $A$, take $\ell (v)=f^{T}v$. Then $B(u,v)=\ell (v)$ for every $v$ means $A^{T}u=f$, not $Au=f$, with this ordering of the bilinear form. Since $\det A=7$,For $f=t(1,0{)}^T$, the exact response is $u=t(3,-1{)}^T/7$ and $\Vert u{\Vert }_2=\sqrt{10}|t|/7<|t|/2$. For $f=t(0,1{)}^T$, it is $u=t(1,2{)}^T/7$, with norm $\sqrt{5}|t|/7$. The actual maximal amplification is the operator norm $\Vert (A^T{)}^{-1}{\Vert }_2=\sqrt{2/(15-\sqrt{29})}<1/2$: the coercivity estimate is valid but need not be sharp for a nonsymmetric bilinear form.

On the zero-boundary Sobolev space $H_0^1(0,1)$ with norm $\Vert u^{\prime }{\Vert }_{L^2}$, assume measurable coefficients $0<p_0\le p\le p_1$ and $0\le q\le q_1$. The Poincaré inequality and Cauchy-Schwarz inequality give $|a(u,v)|\le (p_1+q_1/{\pi }^2)\Vert u^{\prime }{\Vert }_2\Vert v^{\prime }{\Vert }_2$, while nonnegativity of $q$ gives $a(u,u)\ge p_0\Vert u^{\prime }{\Vert }_2^2$. Thus the form is a bounded bilinear form and a coercive bilinear form. For a source in $H^{-1}$, the Lax-Milgram theorem yields a unique weak solution and every conforming Galerkin method approximation. No derivative of the bounded coefficient $p$ is needed for this weak formulation.

On a bounded open set, suppose $b\in L^{\infty }$ has weak divergence in $L_{\mathrm{loc}}^1$ with $\mathrm{div}b\le 0$. For $\phi \in C_0^{\infty }$, testing the divergence identity with ${\phi }^2/2$ gives $\int (b\cdot \nabla \phi )\phi =-\frac{1}{2}\int (\mathrm{div}b){\phi }^2\ge 0$. The left side is continuous in the zero-boundary Sobolev space $H_0^1$, by the Poincaré inequality and boundedness of $b$. Density therefore extends its nonnegativity to every $u\in H_0^1$. One need not assume that $(\mathrm{div}b)u^2$ is integrable. It follows that $a(u,v)=\int \nabla u\cdot \nabla v+\int (b\cdot \nabla u)v$ is a coercive bilinear form in the gradient norm, giving unique weak Dirichlet solvability by the Lax-Milgram theorem.

Let $U\subset {\mathbb{R}}^d$ be bounded, $A\in {\mathbb{R}}^d$ satisfy $\Vert A{\Vert }_2<2$, and $f\in L^2(U)$. The bilinear formon $H_0^1(U)$ is bounded. The Cauchy-Schwarz inequality and $2ab\le a^2+b^2$ giveso it is a coercive bilinear form. The Lax-Milgram theorem therefore gives a unique weak solution of $-\Delta v+v+A\cdot \nabla v=f$ with zero Dirichlet boundary condition.

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.

The zero-boundary Korn inequality controls the full displacement gradient tensor by the infinitesimal strain tensor for $v\in H_0^1(\Omega ;{\mathbb{R}}^d)$. On a bounded Lipschitz domain, combining it with the Poincaré inequality makes a positive linear elasticity energy a coercive bilinear form in the first-order Sobolev norm. The zero Dirichlet boundary condition removes the otherwise uncontrolled rigid displacements.

For this zero-boundary version one can prove an exact identity. For a smooth compactly supported vector field, two applications of integration by parts giveExpanding the symmetric gradient and using density in the zero-boundary Sobolev space therefore yieldsIn three dimensions this also gives the stronger trace-free versionHere the matrix-valued L2 norm uses the Frobenius inner product. This explains why the trace-free form printed in the examination suffices for the usual Korn inequality estimate.

The theorem supplies solvability for a coercive weak formulation.

A bounded bilinear form $a$ on a real Hilbert space $H$ that is a coercive bilinear form represents every bounded linear functional uniquely: for each $\ell \in H^{\prime }$, there is a unique $u\in H$ with $a(u,v)=\ell (v)$ for every $v\in H$.

For a proof, the Riesz representation theorem gives a bounded linear operator $T$ and a vector $g$ with $a(u,v)=⟨Tu,v⟩$ and $\ell (v)=⟨g,v⟩$. Coercivity and the Cauchy-Schwarz inequality give $\Vert Tu\Vert \ge \alpha \Vert u\Vert$, so $T$ is injective and its image is closed: convergence of $T{u}_n$ makes $u_n$ a Cauchy sequence, whose limit supplies the image limit. If $v$ belongs to the orthogonal complement of the image, then $a(u,v)=0$ for every $u$; setting $u=v$ and using coercivity forces $v=0$. Thus the image is also dense and equals $H$. There is consequently a unique $u$ with $Tu=g$, and $\Vert u\Vert \le \Vert \ell \Vert /\alpha$. The complex version uses a bounded sesquilinear form with $\mathrm{Re}a(u,u)\ge \alpha \Vert u{\Vert }^2$.

For $L=-d^2/d{x}^2+x$ on $(0,1)$, with homogeneous conditions $v(0)=0$, $v^{\prime }(1)=0$, integration by parts gives $⟨v,Lv⟩={\int }_0^1(v^{\prime 2}+x{v}^2)dx>0$ for nonzero $v$. The affine energy $J[u]=\frac{1}{2}{\int }_0^1(u^{\prime 2}+x{u}^2)dx$, minimized over $H^1$ functions with $u(0)=a$, yields $Lu=0$ and the natural boundary condition $u^{\prime }(1)=0$. The inequality $\Vert v{\Vert }_2^2\le \frac{1}{2}\Vert v^{\prime }{\Vert }_2^2$ on $v(0)=0$ proves that the form is a coercive bilinear form; Lax-Milgram theorem supplies a unique critical point, and $J[u+v]-J[u]=\frac{1}{2}\int (v^{\prime 2}+x{v}^2)$ proves that it is the unique minimum.

Consider $a(v,v)={\int }_b^c[|v^{\prime }{|}^2+V{v}^2]$ on real $H^1$ functions with $v(c)=0$, where $V_{-}=\max (-V,0)$. The pointwise estimate in the Endpoint Poincaré inequality gives ${\int }_b^c{V}_{-}{v}^2\le \theta \Vert v^{\prime }{\Vert }_2^2$ with $\theta ={\int }_b^c{V}_{-}(x)(c-x)dx$. If $\theta <1$, then $a(v,v)\ge (1-\theta )\Vert v^{\prime }{\Vert }_2^2\ge (1-\theta )[1+(c-b{)}^2/2{]}^{-1}\Vert v{\Vert }_{H^1}^2$. Thus the form is a coercive bilinear form even though the potential can change sign. For $V(x)=x$ on $(-1,1)$, $\theta ={\int }_{-1}^0(-x)(1-x)dx=5/6$, yielding the explicit constant $1/18$.

A real bilinear form is a coercive bilinear form on $V$ when $a(v,v)\ge \alpha \Vert v{\Vert }_V^2$ for all $v\in V$ and some fixed $\alpha >0$. It is a bounded bilinear form, equivalently continuous, when $|a(v,w)|\le M\Vert v{\Vert }_V\Vert w{\Vert }_V$ for one finite $M$. A continuous linear functional satisfies $|\ell (v)|\le C_{\ell }\Vert v{\Vert }_V$.

The Galerkin method seeks $u_h\in S$ with $a(u_h,v_h)=\ell (v_h)$ for all $v_h\in S$. Subtraction from the continuous equations gives Galerkin orthogonality: $a(u-u_h,v_h)=0$ for every $v_h\in S$. Under the stated continuity and coercivity assumptions, Céa's lemma saysIndeed, with $e=u-u_h$ and any $v_h\in S$, Galerkin orthogonality eliminates $v_h-u_h$:If $e=0$ the claim is immediate; otherwise cancel $\Vert e{\Vert }_V$ and take the infimum. No symmetry of $a$ was used. These statements define an approximation when it exists; a finite-dimensional $S$ gives existence and uniqueness by the Lax-Milgram theorem, as does a closed vector subspace.

For the interval example write $L=c-b$ and $R=\max (|b|,|c|)$. The one-dimensional Sobolev representative of $v\in V$ satisfiesby the Cauchy-Schwarz inequality. The representation follows by taking the absolutely continuous function primitive of the weak derivative; its difference from $v$ has zero weak derivative, hence is constant, and $v(c)=0$ fixes that constant. Integrating the estimate gives the Endpoint Poincaré inequalityThus the nonnegative potential $x^2$ givesFor continuity, $|x|\le R$, the Cauchy-Schwarz inequality for the two integrals, and then its two-dimensional version giveSimilarly $\boxed{|\ell (v)|\le \Vert f{\Vert }_2\Vert v{\Vert }_2\le \Vert f{\Vert }_2\Vert v{\Vert }_V}$, proving continuity of the linear functional.

For the last displayed integral take $\Omega =(-1,1)$ and retain the endpoint condition $v(1)=0$, the domain-consistent interpretation of the changed formula. The changed form is still coercive. A negative potential on part of the interval does not by itself destroy coercivity. The sharper pointwise estimate above givesDiscarding only the positive part of the potential yieldsThe proved Endpoint Poincaré inequality with $L=2$ gives $\Vert v{\Vert }_V^2\le 3\Vert v^{\prime }{\Vert }_2^2$, and thereforeThis is a concrete application of negative-potential domination by endpoint control. The constant is sufficient, without claiming to be optimal.

For a sufficiently regular solution, put $w=-\Delta u$, with $\Delta$ the Laplacian. Then $-\Delta u-w=0$ and $u-\Delta w=u+{\Delta }^{2}u=f$. The zero Dirichlet boundary conditions on $u$ and $\Delta u$ become zero Dirichlet boundary conditions on $u$ and $w$. Conversely, eliminating $w$ from these two equations recovers the fourth-order equation with the biharmonic operator. For the weak solution we will construct, these interior identities hold as distributional identities; zero Sobolev traces are imposed on $u$ and $w$, without assuming classical fourth derivatives of $u$.

Let $V=H_0^1(\Omega )\times H_0^1(\Omega )$, the Cartesian product of two zero-boundary Sobolev spaces, with normThis is a Hilbert space. Applying integration by parts separately to the two equations gives the weak formulationEquivalently, for $U=(u,w)$ and $T=(v,z)$, define the bilinear form and linear functionalWe seek $U\in V$ with $b(U,T)=F(T)$ for every $T\in V$.

The skew-coupled elliptic system has a useful bounded bilinear form estimate with constant one. Indeed, write the integrand as the dot product of $(\nabla p,\nabla q,-q,p)$ and $(\nabla v,\nabla z,v,z)$. The Cauchy-Schwarz inequality givesFor equal arguments, the two zeroth-order coupling terms cancel. If $K$ is the constant in the supplied Poincaré inequality, thenThus $b$ is a coercive bilinear form, although it is not a symmetric bilinear form. The Lax-Milgram theorem applies and yieldsIn particular, $\Vert (u,w){\Vert }_V\le (1+K)\Vert f{\Vert }_0$. Testing with compactly supported smooth test functions identifies $-\Delta u=w$ and $-\Delta w=f-u$ as distributional identities. This gives the asserted relation to the original boundary value problem at the available regularity.

Take $V_h={\mathcal{U}}_h\times {\mathcal{W}}_h\subset V$. The Galerkin method asks for $U_h=(u_h,w_h)\in V_h$ such thatSince a finite-dimensional subspace is closed, $V_h$ is itself a Hilbert space. The bounded bilinear form and coercive bilinear form estimates in part (a) restrict to $V_h$ with the same constants. The Lax-Milgram theorem therefore gives exactly one $U_h$ for every $h$.

Put $E=U-U_h=(e_u,e_w)$. Subtracting the continuous and discrete weak formulations gives Galerkin orthogonality:For any $Y_h=(v_h,z_h)\in V_h$, the difference $Y_h-U_h$ also belongs to $V_h$, soThe last step uses the constant-one Cauchy-Schwarz inequality estimate from part (a). Bounding the four terms separately by a sum would obscure the precise constant requested here.

The two minimization problems over ${\mathcal{U}}_h$ and ${\mathcal{W}}_h$ are independent. Each has a best approximation because its vector subspace is finite-dimensional. Consequently, ifthen the preceding inequality gives exactlyBoth errors belong to the zero-boundary Sobolev space. Applying the Poincaré inequality componentwise now yieldsWhen $U=U_h$ the desired estimate is automatic; otherwise divide by the nonzero norm. ThusThe constant $C=1+K$ is independent of $h$. This is the Céa lemma estimate for this particular skew-coupled elliptic system.

Use the Sobolev space with a partial Dirichlet conditionwhere $T$ is the Sobolev trace operator. The restricted Sobolev trace operator is continuous into $L^2({\Gamma }_D)$; its kernel is consequently a closed vector subspace of $H^1(\Omega )$. Thus $V$, with the first-order Sobolev norm, is a Hilbert space. The endpoint convention for ${\Gamma }_D$ and ${\Gamma }_N$ does not affect this definition: finitely many boundary points have zero boundary length. It would be inappropriate to impose pointwise boundary values on arbitrary $H^1$ functions in two dimensions.

For a smooth solution, Green's first identity givesThe boundary term vanishes because $Tv=0$ on ${\Gamma }_D$ and the normal derivative is zero on ${\Gamma }_N$. This suggests the weak formulationHere $a$ is exactly the first-order Sobolev inner product. In particular the Cauchy-Schwarz inequality givesThus $a$ is a bounded bilinear form and a coercive bilinear form, both with constant one. The Lax-Milgram theorem proves existence and uniqueness, together withThe positive reaction term already controls the entire Sobolev norm, so no Poincaré inequality is needed here. The zero Neumann boundary condition is the natural part of the weak formulation; it is not an additional restriction on the test space. For this weak solution, $-\Delta u+u=f$ holds as a distributional identity, and the zero normal derivative on ${\Gamma }_N$ is understood through the corresponding Green's first identity.