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A bilinear square finite element uses polynomials of degree at most one separately in each coordinate and takes values at the four corners as degrees of freedom. On the unit square its nodal shape functions are $(1-\xi )(1-\eta )$, $\xi (1-\eta )$, $\xi \eta$, and $(1-\xi )\eta$, in anticlockwise corner order. Each is a tensor product of one-dimensional linear shape functions. A function in this space need not be affine, since $\xi \eta$ has total degree two. On a physical square of side $h$, its Laplacian stiffness matrix has diagonal entries $2/3$, adjacent-corner entries $-1/6$ and opposite-corner entries $-1/3$. Its Robin boundary condition edge matrix has the two-node block $(h/6)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ for a unit boundary coefficient.

A breathing mode is a collective normal mode in which a structure expands and contracts radially while preserving its overall symmetry. For a uniform scaling pattern, the displacement of point $i$ is proportional to its position relative to the center of mass, as displayed above. In a central spring framework this pattern is a normal mode only when applying the stiffness matrix returns a multiple of the same pattern with the appropriate mass matrix; radial motion alone does not establish an eigenvector.

For point masses joined by unstretched central springs obeying Hooke's law, let $n_{ij}$ be the equilibrium unit vector along an edge and $u_i$ the small displacement of its endpoint. The leading extension is $n_{ij}\cdot (u_j-u_i)$, so its quadratic elastic potential energy is $k_{ij}[n_{ij}\cdot (u_j-u_i){]}^2/2$. Consequently the block stiffness matrix has off-diagonal blocks $K_{ij}=-k_{ij}{n}_{ij}{n}_{ij}^T$ and diagonal blocks $K_{ii}={\sum }_j{k}_{ij}{n}_{ij}{n}_{ij}^T$, with absent edges contributing zero. It is symmetric and positive semidefinite. The normal modes solve $Kq={\omega }^{2}Mq$, where $M$ is the mass matrix. Rigid translations and infinitesimal rotations are rigid-motion zero modes of a spring framework. A prestressed spring can add transverse quadratic terms, so the unstretched assumption matters.

Coupled oscillators influence one another's motion through shared forces or other interactions. Two pendulums connected by a light spring provide a visible example: moving either bob changes the force on the other. For equal masses and lengths, the symmetric motion leaves the connecting spring unstretched, while the antisymmetric motion stretches it and has a higher frequency.

For a conservative linear system, the equations of motion take the form $M\ddot{\mathbf{q}}+K\mathbf{q}=0$, where $M$ is the mass matrix and $K$ the elastic stiffness matrix. A normal mode has a fixed displacement pattern $\mathbf{a}$ satisfying $K\mathbf{a}={\omega }^{2}M\mathbf{a}$. Expanding the initial motion in normal modes separates the evolution. Displacing just one pendulum generally excites several normal modes, producing beats and visible transfer of energy. A driving force excites only the modes onto which its spatial pattern has a nonzero projection.

The damped quadratic eigenvalue problem finds exponential motions of a linear mechanical model with mass matrix $M$, damping matrix $C$ and stiffness matrix $K$. Substituting $u(t)=r{e}^{\lambda t}$ into $M\ddot{u}+C\dot{u}+Ku=0$ givesThe dependence on $\lambda$ is quadratic, so solving the undamped problem $Kr={\omega }^{2}Mr$ alone does not find the damped decay rates and mode shapes.^[ref]

A useful linearization sets $z=(r,\lambda r{)}^T$ and rewrites the problem asThis generalized eigenvalue problem has twice the displacement dimension. With invertible $M$, it becomes an ordinary first-order state eigenproblem by multiplication by the block mass inverse. Singular $M$ requires more care and cannot be handled by silently using $M^{-1}$.

For an original one-coordinate example in consistent units, take $M=1$, $C=2$ and $K=5$. Then ${\lambda }^2+2\lambda +5=0$, giving $\lambda =-1\pm 2i$. Real free motions have the formThe decay rate is $1$ and damped oscillation frequency is $2$, while the undamped natural frequency is $\sqrt{5}$. Increasing $C$ to6 instead gives real roots $-1,-5$, showing that the quadratic problem also describes nonoscillatory decay.

For real symmetric positive-definite $M,K$ and positive-semidefinite $C$, an oscillatory eigenpair satisfies a useful energy check. Multiply by the conjugate transpose $r^{\ast }$ and write $m=r^{\ast }Mr>0$, $c=r^{\ast }Cr\ge 0$, $k=r^{\ast }Kr>0$. The imaginary part of $m{\lambda }^2+c\lambda +k=0$ gives $\mathrm{Re}\lambda =-c/(2m)\le 0$ when $\mathrm{Im}\lambda \ne 0$. In a multi-coordinate system, complex eigenvectors can encode relative phase differences between coordinates; nonclassical damping generally prevents replacing them by the undamped real mode shapes.

A damping matrix $C$ maps generalized velocity to the velocity-dependent resisting-force term in a linear structural dynamics model:Here $M$ is the mass matrix and $K$ is the stiffness matrix. The sign convention places $C\dot{u}$ on the left, so the actual resisting force is $-C\dot{u}$. For pure translational coordinates, entries have units $\mathrm{N}\mathrm{s}{\mathrm{m}}^{-1}$; mixed rotational and translational coordinates require their corresponding generalized units.

Consider an original two-coordinate model with three viscous dampers: one from each coordinate to a fixed support, and a third between them. With positive coefficients $c_1,c_2,c_{12}$,For $\dot{u}=(v_1,v_2{)}^T$, the removed mechanical power isThis makes $C$ a positive semidefinite matrix. In particular, the connecting damper dissipates nothing when both coordinates move at the same speed. Choosing $c_1=c_2=10$ and $c_{12}=5$ in newton-seconds per metre gives $C=\left(\begin{array}{cc}15& -5\\ -5& 15\end{array}\right)$. Speeds $(1,0)\mathrm{m}{\mathrm{s}}^{-1}$ dissipate15watts, while $(1,1)\mathrm{m}{\mathrm{s}}^{-1}$ dissipate20watts.

For symmetric constant $M,K$, multiplying the equation by ${\dot{u}}^T$ gives an energy balance:The matrix therefore encodes where dissipation occurs and which relative motions activate it.

Rayleigh damping approximates $C$ by a weighted combination of the mass and stiffness matrices. OpenSees exposes separate coefficients for mass and different stiffness states, and cautions that this model can give incorrect results in some nonlinear concentrated-plasticity analyses.^[ref] A convenient matrix prescription must still represent the intended dissipation mechanism.

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.

For a uniform piecewise-linear Galerkin heat discretization, the consistent mass matrix is $M=(h/6)\mathrm{tridiag}(1,4,1)$ and the stiffness matrix is $K=h^{-1}\mathrm{tridiag}(-1,2,-1)$. The generalized eigenvalues are $6(1-\cos \theta )/[h^2(2+\cos \theta )]$. Applying forward Euler method requires $k\lambda \le 2$, giving the displayed uniform bound for $\mu =k/h^2$. For $J$ intervals with Dirichlet endpoints the exact finite-grid upper bound is $(2-\cos (\pi /J))/[3(1+\cos (\pi /J))]$, which approaches $1/6$. Replacing the consistent mass matrix by a lumped one changes the stability restriction.

For equal rod lengths $l$, upper mass $m$ and lower mass $\alpha m>0$, use horizontal displacements $x_1=l{\theta }_1$, $x_2=l({\theta }_1+{\theta }_2)$ at first order. The quadratic kinetic energy is $m{\dot{x}}_1^2/2+\alpha m{\dot{x}}_2^2/2$, while the stiffness matrix isThe generalized eigenvalue problem for small oscillations is $Ka={\omega }^{2}Ma$, not $Ka=m{\omega }^{2}a$. In particular the lower Euler-Lagrange equation is ${\ddot{x}}_2=-(g/l)(x_2-x_1)$, independent of the lower mass factor after division by that mass. This reproduces the normal modes of a double pendulum and distinguishes the mechanical mass matrix from its stiffness matrix.

Two equal masses linked in a line by three identical springs, each stretched to twice its natural length, have longitudinal stiffness matrix $k\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)$ and transverse matrix half as large. Equal displacements of both masses select the symmetric modes. Equal longitudinal and transverse release amplitudes produce the title trajectory. It is not closed after one transverse period because the two mode frequencies have irrational ratio $\sqrt{2}$. Higher-order geometric terms are neglected in this small-displacement solution.

Nonclassical damping occurs when a linear mechanical system's damping matrix cannot be diagonalized together with its mass matrix and stiffness matrix by an undamped normal mode transformation. Consequently the damping terms couple the modal equations even though their inertia and stiffness terms are separated. This can matter when a localized dissipative element is attached to a structure whose elastic and inertial responses are distributed differently.^[ref]

For $M\ddot{u}+C\dot{u}+Ku=0$, take a complete mass-normalized undamped mode matrix $\Phi$ with ${\Phi }^{T}M\Phi =I$ and ${\Phi }^{T}K\Phi ={\Omega }^2$. Then $u=\Phi q$ givesWith distinct undamped natural frequencies, a nonzero off-diagonal $D_{ij}$ shows modal damping coupling. If a frequency is repeated, first allow a rotation of the modes within its repeated-frequency subspace: an off-diagonal entry in one arbitrary basis is not by itself proof of nonclassical damping. For symmetric $M,K,C$ with $M$ positive definite, the basis-independent condition for classical decoupling isFailure of this condition establishes nonclassical damping under those assumptions.^[ref]

For an original example in consistent normalized units, letThe damping acts only on the first physical coordinate. The undamped mode vectors $(1,1)/\sqrt{2}$ and $(1,-1)/\sqrt{2}$ giveThus an initial modal state $q=(1,0{)}^T$, $\dot{q}=(1,0{)}^T$ has ${\ddot{q}}_2=-0.2$. Dropping the off-diagonal damping terms would incorrectly predict zero initial acceleration in the second modal coordinate. Nevertheless ${\dot{u}}^{T}C\dot{u}=0.4{\dot{u}}_1^2\ge 0$: passive dissipation does not require modal decoupling.

The damped quadratic eigenvalue problem follows from $u(t)=r{e}^{\lambda t}$:In oscillatory cases its complex vectors can encode different phases at different physical coordinates. One can solve the coupled equations directly or use an appropriate damped-mode method; assigning separate scalar decay equations to the undamped modes is an additional approximation, not a consequence of positive dissipation.

Connect every pair among $n$ moving scalar coordinates and one fixed coordinate by identical springs of stiffness $k$. The coupling pattern is the complete graph $K_{n+1}$ with one vertex fixed. Its potential energy is $k{\sum }_{i<j}(X_i-X_j{)}^2/2$. The moving-coordinate stiffness matrix is the principal submatrix displayed in the header. For equal masses $m$, the uniform normal mode has squared angular frequency $k/m$, while every sum-zero vector has squared angular frequency $(n+1)k/m$. There are $n-1$ such degenerate modes. This follows directly from the Laplacian spectrum of a complete graph and its rank-one matrix form. Constant gravitational forces shift the equilibrium without changing the stiffness.

For a uniform rigid beam of mass $m$ and two identical vertical end springs of stiffness $k$, small vertical end displacements have kinetic energy $m({\dot{x}}_1^2+{\dot{x}}_1{\dot{x}}_2+{\dot{x}}_2^2)/6$ and displacement-dependent potential energy $k(x_1^2+x_2^2)/2$. The mass and stiffness matrices are $\mathsf{M}=(m/6)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ and $\mathsf{K}=k1$. Their generalized eigenvalue problem gives symmetric translation $(1,1)$ and antisymmetric rocking $(1,-1)$ at the displayed angular frequencies. The centre coordinate contains only the symmetric mode. Neglect spring inertia, damping and finite-angle geometric corrections.

For two rods of length $a$ and mass $m$, one hinged to a fixed point and the other to its free end, use the absolute rod angles from the downward vertical. The small-oscillation mass matrix is $m{a}^2\left(\begin{array}{cc}4/3& 1/2\\ 1/2& 1/3\end{array}\right)$ and the stiffness matrix is $(mga/2)\mathrm{diag}(3,1)$. Their generalized eigenvalue equation gives $7{\kappa }^2-42\kappa +27=0$ for $\kappa =a{\omega }^2/g$. The low-frequency mode has angles of the same sign and the high-frequency mode has opposite signs.

Define the mass matrix $M_{ij}={\int }_0^1{\varphi }_j{\varphi }_i$ and stiffness matrix $K_{ij}={\int }_0^1{\varphi }_j^{\prime }{\varphi }_i^{\prime }$ for $1\le i,j\le N$. With coefficient vectors $U^m$, the step system isThe matrices are symmetric, and $V^T\mathsf{A}V=\Vert v_h{\Vert }_2^2+\Delta t(\Vert v_h^{\prime }{\Vert }_2^2+\Vert v_h{\Vert }_2^2)>0$ for nonzero $V$, proving positive definiteness.

For $h=1/2$, the first basis function is the full tent at $1/2$, and the last is the half tent ending at $1$. Integration givesSince $\Delta t=2/2=1$, the requested matrix in this normalization isHere the matrix symbol $M$ is distinct from the printed integer number of time steps.

Write $u_h^m={\sum }_i{U}_i^m{\phi }_i$ and define the mass matrix and reaction-diffusion stiffness matrixThen each step isBoth component matrices are symmetric, so $A$ is symmetric. For a nonzero coefficient vector $v$, its finite element function $v_h$ is nonzero andThus $A$ is also positive definite and every update has a unique solution. Because nonneighboring hats have disjoint interior supports, the matrix is tridiagonal in nodal order.

Work in the usual bounded Lipschitz, polygonally triangulated setting in which Sobolev traces and conforming meshes are defined. Choose an $H^1$ lift $G$ of the smooth boundary data. Multiplying the Poisson equation by $v\in H_0^1(\Omega )$ and integrating by parts gives the weak formulationThe test function has zero trace, so the boundary flux term vanishes. This formulation only needs first weak derivatives.

Let $V_h$ be the conforming finite element space of continuous functions affine on each triangle, and $V_h^0$ its zero-boundary subspace. Choose a nodal interpolation $G_h$ of the boundary lift and seek $u_h\in G_h+V_h^0$ satisfying the same identity for every $v_h\in V_h^0$. General smooth boundary data need not be exactly representable by boundary piecewise linear functions; this is the usual discrete boundary approximation.

With nodal shape functions ${\phi }_j$, interior coefficients $U_j$ satisfyHere $I$ and $B$ index interior and boundary vertices. Finite element assembly sums the triangle contributions.

Take a translated lower triangle with local vertices $(0,0),(h,0),(h,h)$. Its local functions are $N_1=1-x/h$, $N_2=(x-y)/h$, $N_3=y/h$. Their gradients are constant and the area is $h^2/2$, giving the stiffness matrixTranslations preserve this matrix, while rotations or a different local vertex order permute its entries. The coupling across the hypotenuse is zero.

Each interior grid vertex belongs to six triangles. Summation gives diagonal value 4 and coupling $-1$ to each of its four horizontal and vertical neighbors. Both positive-diagonal edge couplings vanish. Thus, before prescribed boundary values are removed, an interior equation has the five-point stencilThere is no $h^{-2}$ factor in the assembled variational matrix; that factor belongs to a differently normalized finite-difference equation.

If $h=1/N$ and $m=N-1$, lexicographic ordering makes the Dirichlet-reduced matrix block tridiagonal with diagonal blocks $\mathrm{tridiag}(-1,4,-1)$ and neighboring blocks $-I$. A row with all four neighboring nodes unknown has five nonzeros: 4 once and $-1$ four times. Rows adjacent to the boundary have fewer unknown-neighbor entries; their exact count is $1+1_{i>1}+1_{i<m}+1_{j>1}+1_{j<m}$. Removed boundary contributions appear on the right, rather than as new nonzeros.

The first-order Sobolev space is $H^1(\Omega )=\{v\in L^2(\Omega ):{\partial }_{1}v,{\partial }_{2}v\in L^2(\Omega )\}$, with derivatives understood as weak derivatives and norm squared ${\int }_{\Omega }(v^2+|\nabla v{|}^2)$.

The printed normal boundary condition must be distinguished from a conormal condition. Take $K$ symmetric positive definite, as required for the symmetric matrix requested later. Integration by parts givesThe natural boundary term contains the conormal derivative $\mathbf{n}\cdot K\nabla u$, not the ordinary normal derivative ${\partial }_{n}u$. On a smooth boundary, writing $\mathbf{t}$ for its unit tangent, the literal condition ${\partial }_{n}u=0$ leaves the term $b{\partial }_{\mathbf{t}}u$, where $b=\mathbf{n}\cdot K\mathbf{t}$. Its weak form therefore retains a boundary pairing:Here the tangential derivative of the Sobolev trace can be interpreted in $H^{-1/2}$ and paired against $H^{1/2}$ on a smooth boundary. This form is generally nonsymmetric; the requested usual symmetric energy construction does not follow from the literal boundary condition.

For an explicit failure of the claimed mean-zero compatibility, take the unit disk, $K=\mathrm{diag}(2,1)$ and $u=(x^2-y^2)[1-(x^2+y^2)/2]$. It has ${\partial }_{n}u=0$ on the circle, butThus replacing the printed normal condition by a zero conormal flux is not harmless.

For the intended zero-conormal derivative problem, the requested variational and finite element conclusions are as follows. Seek $u\in H^1(\Omega )$ withPutting $v=1$ gives ${\int }_{\Omega }f=0$. On a connected domain this condition permits a solution unique modulo constants; the Poincare-Wirtinger inequality gives coercivity after fixing the mean.

The energy is $J(v)=\frac{1}{2}a(v,v)-\ell (v)$. A minimizer satisfies $\frac{d}{dt}J(u+tv){|}_{t=0}=a(u,v)-\ell (v)=0$. Conversely, a weak solution satisfiesso it is a minimizer. Testing by $u$ gives the minimum valueThis value is independent of the arbitrary additive constant when $\int f=0$. If the compatibility fails, constant shifts make $J$ unbounded below.

Write $u_h={\sum }_j{c}_j{\varphi }_j$. The Galerkin method givesSymmetry of $K$ gives $A^T=A$. For $v_h={\sum }_j{z}_j{\varphi }_j$, $z^{T}Az={\int }_{\Omega }\nabla v_h\cdot K\nabla v_h\ge 0$. Equality forces $v_h$ constant by positive definiteness and connectedness. Since the basis sums to one, the coefficient vector $1=(1,\dots ,1{)}^T$ represents that constant, givingThus the stiffness matrix is symmetric positive semidefinite and singular; the displayed compatibility condition is necessary and sufficient for the finite system. Positive definiteness of $K$ controls every gradient direction, preventing additional zero-energy directions and supplying ellipticity. A degenerate coefficient can introduce further kernel modes, while an indefinite one destroys the minimizing-energy argument.