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For $u_{tt}=\Delta u+\alpha u$ with a homogeneous Dirichlet boundary condition, let ${\lambda }_1>0$ be the first Dirichlet Laplacian eigenvalue. On a bounded one-dimensional interval, finite-energy displacement remains bounded for all time and all initial data exactly when $\alpha <{\lambda }_1$. The conserved energy $\frac{1}{2}(\Vert u_t{\Vert }_2^2+\Vert \nabla u{\Vert }_2^2-\alpha \Vert u{\Vert }_2^2)$ controls the gradient and, by the Poincaré inequality and one-dimensional integration, the maximum norm. At equality a mode can grow linearly; above it a mode can grow exponentially. Bounded oscillations need not have a long-time limit.

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 $f$.

For $u\in H_0^2(U)$, ${\sum }_{i,j}\Vert {\partial }_{ij}u{\Vert }_2^2=\Vert \Delta u{\Vert }_2^2$. Two integrations by parts prove this for compactly supported test functions, and $H^2$ density extends it to the clamped second-order Sobolev space. Together with the Poincare-Wirtinger inequality applied to the mean-zero first derivatives and the zero-boundary Poincaré inequality, it controls the full $H^2$ norm by $\Vert \Delta u{\Vert }_2$. The boundary conditions are essential to this exact identity on bounded domains.

A perturbation preserving both values and slopes has $\eta =\eta \prime =0$ at the endpoints. Apply the Poincaré inequality to $\eta$ and then to $\eta \prime$. The resulting inequality makes a quadratic second variation proportional to $\int ({\eta }^2-(\eta \prime \prime {)}^2)$ nonpositive. If a functional is quadratic, its exact expansion about a stationary function then proves a global maximum.

For fixed-endpoint variations, a second variation is coercively positive if the displayed estimate holds for one constant $c>0$ and every admissible variation. The endpoint Poincaré inequality makes this equivalent to a positive lower bound in the $W^{1,2}$ norm. It is stronger than strict positivity on each nonzero variation: $q(\phi )=\int {\phi }^2$ is strictly positive, but high-frequency variations make its ratio to $\int {\phi }^{\prime 2}$ tend to zero.

For $q=\int (p{\phi }^{\prime 2}+2r\phi {\phi }^{\prime }+s{\phi }^2)$ with continuous coefficients, coercive positivity implies $p\ge c$. Indeed, if $p-c<0$ on an interval, multiply a smooth function supported there by $\sin (nx)$. Its negative highest-derivative contribution has order $n^2$, while the mixed term has order at most $n$ and the remaining term is bounded. This contradicts the displayed inequality for large $n$. In particular coercive positivity implies the Strengthened Legendre condition, whereas mere strict positivity does not.

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.

For suitable convex integral functionals, ordered boundary values of two minimizers imply the same ordering in the domain. Affine minimizers can therefore serve as upper and lower barriers.

For a nonnegative potential $V(x)\to \infty$ as $|x|\to \infty$, the confining-potential energy space isA local Poincaré inequality controls the missing $L^2$ term, so this energy norm makes ${\Sigma }_V$ a Hilbert space.

For the angular endpoint conditions $\theta (0)=\theta (1)=0$, take $U=(EI/L){\int }_0^1[{\theta }^{\prime 2}/2-{\pi }^2\lambda (1-\cos \theta )]d\xi$. The Poincaré inequality and $1-\cos t\le t^2/2$ give $U\ge 0$ for $0\le \lambda \le 1$. At $\lambda =1$ a nonzero continuous angle still has strictly positive energy because $1-\cos t<t^2/2$ for $t\ne 0$. Thus the straight state remains the unique global minimum at the critical parameter even though the quadratic variation vanishes on the first sine mode.

Along that mode the energy is ${\pi }^{2}EI{A}^4/(64L)+O(A^6)$ at criticality. For $\lambda >1$ a sufficiently small first-mode variation lowers energy, so the straight state loses local minimality. Near the first threshold, expansion of the Euler strut equation gives $A_0(A_0^2-8{\lambda }_1)=0$ when $\lambda =1+ϵ{\lambda }_1$ and $\theta \sim \sqrt{ϵ}{A}_0\sin \pi \xi$. The resulting supercritical pitchfork bifurcation is stabilized by the same positive quartic term. A zero second variation is therefore an inconclusive quadratic test, not proof of failure of minimization.

The direct method takes a minimizing sequence, extracts a subsequence using compactness, and uses lower semicontinuity to show that its limit attains the infimum.

For planar Gaussian free field theory, this is the completion of real test functions $C_c^{\infty }(D)$ in the norm induced by the Dirichlet inner product, $(f,g{)}_{\nabla }=(2\pi {)}^{-1}{\int }_D\nabla f\cdot \nabla g$. On a general unbounded domain this convention must be distinguished from completion in the inhomogeneous $H^1$ norm. A conformal map identifies its energy norm with that on the unit disc by conformal invariance.

On a domain $U\subseteq {\mathbb{R}}^n$, a common normalization of the Dirichlet inner product isThe Poincaré inequality makes its induced seminorm a norm on $H_0^1(U)$.

An elliptic boundary value problem couples an elliptic differential equation in a domain to boundary conditions. Its weak formulation is often an equation for a bounded bilinear form on a Sobolev space.

For the one-dimensional Sobolev representative with $v(c)=0$, the fundamental theorem of calculus gives $v(x)=-{\int }_x^c{v}^{\prime }(t)dt$. The Cauchy-Schwarz inequality then gives $|v(x){|}^2\le (c-x)\Vert v^{\prime }{\Vert }_2^2$. Integrating proves the displayed, generally nonoptimal Poincaré inequality. In particular $\Vert v{\Vert }_{H^1}^2\le [1+(c-b{)}^2/2]\Vert v^{\prime }{\Vert }_2^2$. Only one endpoint must be fixed, which is useful for mixed boundary conditions and for controlling a negative potential.

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.

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.

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.