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For a smooth bounded domain and a bounded continuous scalar function $f$, this problem has a weak solution. Let $S$ be the zero-Dirichlet inverse of $-\Delta$, defined by the Lax-Milgram theorem. The Poincaré inequality gives $\Vert Sh{\Vert }_2\le C_P^2\Vert h{\Vert }_2$. The Rellich-Kondrachov compactness theorem makes $S:L^2\to L^2$ compact. The Nemytskii operator $N_f$ is continuous and has bounded image, so $T=S\circ N_f$ is a nonlinear compact operator mapping a sufficiently large closed $L^2$ ball into itself. The Schauder fixed-point theorem gives a fixed point, necessarily in $H_0^1$, and its defining identity is the desired weak formulation. This proves existence; uniqueness requires additional assumptions.

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 $f$.

This is the closure of the compactly supported test functions $C_c^{\infty }(U)$ in the $H^2(U)$ norm. On a smooth bounded domain it consists precisely of the $H^2$ functions with zero value and zero normal derivative traces. Tangential first derivatives then vanish as well. The clamped Hessian identity makes the Laplacian norm equivalent to the full $H^2$ norm on this space, enabling a Lax-Milgram theorem treatment of the clamped biharmonic problem.

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.

For $H=H_0^1(U)\times H^1(U)$, the coupled formsatisfiesThe Poincaré inequality for $u$ then controls both $\Vert u{\Vert }_2$ and $\Vert w{\Vert }_2$, so the form is coercive on $H$. The Lax-Milgram theorem therefore gives a unique weak solution. The first component has a Dirichlet boundary condition, while the second component's Neumann boundary condition is the natural boundary condition in the 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 smooth uniformly elliptic differential expression with symmetric Dirichlet bilinear form has an unbounded self-adjoint operator realization $A_D$ on $L^2(U)$, whose domain is $H^2(U)\cap H_0^1(U)$ on a smooth bounded domain. If the form is strictly positive, the Lax-Milgram theorem and elliptic regularity make $A_D^{-1}$ a bounded map from $L^2$ to $H^2\cap H_0^1$. Its action on $L^2$ is compact by the Rellich-Kondrachov compactness theorem and self-adjoint by symmetry of the form. The spectral theorem for compact Hermitian operators therefore gives an orthonormal basis of smooth Dirichlet eigenfunctions with positive eigenvalues tending to infinity.

As the jump-length weight tends to infinity with the gradient weight fixed, the limiting problem minimizes $\Vert u-g{\Vert }_2^2+\alpha \Vert \nabla u{\Vert }_2^2$ on $H^1$. It has a unique minimizer by the Lax-Milgram theorem and strict convexity. Without prescribed boundary values, its weak equation is $u-\alpha \Delta u=g$ with natural Neumann boundary conditions.

For a fixed sufficiently regular image edge set, the Mumford–Shah functional is a strictly convex quadratic energy on the Sobolev space of its complement. A first variation gives $u-\alpha \Delta u=g$ within each region and separate one-sided Neumann boundary conditions on free edges. A jump need not vanish across an edge. The weak solution is unique by the Lax-Milgram theorem; this does not prove uniqueness when the edge set is also optimized.

On the unit interval, the weak form is $a(u,v)={\int }_0^{1}fv$, with $a(u,v)={\int }_0^1(u^{\prime }{v}^{\prime }+xuv)$ on the zero-boundary Sobolev space. The nonnegative potential makes this form coercive in the derivative norm, and the Poincaré inequality makes that norm equivalent to the Sobolev norm. The Lax-Milgram theorem gives a unique solution. A conforming Ritz method minimizes the corresponding strictly convex quadratic energy and produces a symmetric positive-definite tridiagonal matrix for piecewise-linear hats.

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.

For a symmetric uniformly elliptic divergence-form operator with zero conormal flux and zero reaction, Lax-Milgram theorem constructs the shifted inverse $K:L^2\to H^1$. The compact Rellich-Kondrachov theorem embedding makes it compact on $L^2$, and symmetry makes it self-adjoint. If the homogeneous kernel consists of constants, then $\ker (I-K)$ is the constant line and $K1=1$. The Fredholm alternative makes solvability equivalent to $Kf$ being orthogonal to constants, or $⟨Kf,1⟩=⟨f,K1⟩=\int f=0$. The weak solution is unique up to an additive constant. For a disconnected domain the identical argument gives one compatibility condition per connected component.

The Sobolev space and norm areIn one dimension these functions have continuous representatives and well-defined endpoint traces. integration by parts gives the weak problem: find $u\in H^1(a,b)$ such that $a(u,v)=\ell (v)$ for all $v\in H^1(a,b)$, whereIndeed the integration boundary term is $-u^{\prime }(b)v(b)+u^{\prime }(a)v(a)$; substituting $u^{\prime }(a)=u(a)-A$ and $u^{\prime }(b)=B-u(b)$ gives the two positive boundary terms on the left.

If $0\notin [a,b]$, put $m={\min }_{[a,b]}{x}^2>0$. Then $a(v,v)\ge \Vert v^{\prime }{\Vert }_2^2+m\Vert v{\Vert }_2^2\ge \min (1,m)\Vert v{\Vert }_{H^1}^2$. If $0\in [a,b]$, the supplied trace inequality givesThus the form is coercive in both cases. The trace theorem and boundedness of $x^2$ give continuity of $a$ and $\ell$, and the Lax-Milgram theorem gives existence and uniqueness.

The Sobolev space $H^1(0,1)$ consists of square-integrable functions whose weak first derivative is square-integrable, with norm $\Vert v{\Vert }_{H^1}^2={\int }_0^1(v^2+v^{\prime 2})$. Its one-dimensional representatives have well-defined endpoint traces.

Multiplying by $v\in H^1$ and integrating by parts givesThe two Robin conditions give $u^{\prime }(0)=u(0)$ and $u^{\prime }(1)=-u(1)$, so defineA weak solution is $u\in H^1(0,1)$ satisfying $a(u,v)=\ell (v)$ for every $v\in H^1(0,1)$. The boundary conditions are natural; no endpoint derivative is imposed in the trial space.

Since $\cos x\ge \cos 1>0$ on this interval,This proves coercivity on the full space. Trace continuity and Cauchy-Schwarz give continuity of the form, so the Lax-Milgram theorem also ensures a unique weak solution.

The Sobolev space $H^1(b,c)$ consists of $L^2$ functions with a weak first derivative in $L^2$, with normIn one dimension these functions have absolutely continuous representatives and well-defined endpoint Sobolev traces. PutThis is a closed subspace, hence a Hilbert space. Both trial and test functions belong to $V$; one does not impose a derivative boundary condition in this space.

Integrate the differential equation against $v\in V$:The printed condition uses the ordinary derivative, $u^{\prime }(c)=-u(c)$, so the weak form isSeek $u\in V$ with $a(u,v)=\ell (v)$ for all $v\in V$. The coefficient $e^c$ in the endpoint term must be retained: replacing the ordinary derivative with the conormal flux would change the boundary problem.

Let $L=c-b$. The Poincaré inequality on $V$ can be stated as $\Vert v{\Vert }_{L^2}\le L\Vert v^{\prime }{\Vert }_{L^2}$. Also $|v(c)|\le \sqrt{L}\Vert v^{\prime }{\Vert }_{L^2}$. Consequentlyproving coercivity. By Cauchy-Schwarz inequality and the trace bound,Thus continuity holds with $M=e^c(1+L)$. The load is continuous if $f\in L^2$, and the Lax-Milgram theorem gives a unique weak solution.

Choose a finite-dimensional conforming subspace $V_h\subset V$, such as continuous piecewise linear functions vanishing at $b$, and solve $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$. Subtracting the weak equations gives Galerkin orthogonality $a(u-u_h,v_h)=0$.

Céa's lemma states, for a continuous coercive bilinear form with constants $M,\alpha$ on a Hilbert space,To prove it, put $e=u-u_h$ and use orthogonality to write $a(e,e)=a(e,u-v_h)$ for any $v_h\in V_h$. Coercivity and continuity give $\alpha \Vert e{\Vert }^2\le M\Vert e\Vert \Vert u-v_h\Vert$. If $e=0$ the claim is immediate; otherwise cancel its norm and take the infimum. Any approximation theorem making the infimum tend to zero under mesh refinement therefore gives strong convergence $u_h\to u$ in $H^1$.

Finally, $a$ is symmetric and positive definite on $V$. It is an inner product, so $\Vert v{\Vert }_a=\sqrt{a(v,v)}$ is an energy norm on all of $V$: homogeneity and definiteness are immediate, and Cauchy-Schwarz inequality for this inner product gives the triangle inequality. No smaller space is required. In this norm the continuity and coercivity constants are one. More precisely, Galerkin orthogonality givessoThe finite element solution is the exact best energy-norm approximation.

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.