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Replace $L_h$ by $\boxed{{\tilde{L}}_h=L_h+I}$. At a positive interior maximum $v>0$, the neighbour average is no larger than $v$, so $({\tilde{L}}_h\varphi {)}_{jk}\ge v>0$. Thus ${\tilde{L}}_h\varphi \le 0$ cannot have such an interior maximum. The positive-maximum form of the discrete maximum principle therefore remains valid:With $e=u{|}_{{\Omega }_h}-U$ the error equation is ${\tilde{L}}_{h}e=\tau$, where $\tau =O(h^2)$ under the preceding regularity assumptions. Since ${\tilde{L}}_{h}B=1+B\ge 1$ for the same Discrete Poisson barrier on a square, the comparison functions $\pm e-\Vert \tau {\Vert }_{\infty }B$ proveThere is no loss of uniqueness: apply the positive-maximum argument to both signs of a homogeneous discrete solution. The continuous weak formulation also has the coercive bilinear form ${\int }_{\Omega }(|\nabla v{|}^2+v^2)$, so the added positive reaction term strengthens its positive energy.

The Lax-Milgram theorem gives, for a bounded coercive bilinear form on a real Hilbert space and every bounded linear functional, a unique vector representing that functional through the form, with the inverse bound supplied by coercivity. The Fredholm alternative for a compact operator $K$ says that $I+\mu K$ is invertible if its kernel is zero; in general its range is closed and consists exactly of vectors orthogonal to $\ker (I+\mu K{)}^{\ast }$ in the Hilbert-space case.

Uniform ellipticity gives $\sum a_{ij}{\xi }_i{\xi }_j\ge \theta |\xi {|}^2$ for some $\theta >0$. Bounded coefficients make $B_{\lambda }$ bounded on $H^1$, and for $\lambda <-\gamma$,The forcing $v\mapsto \int fv$ is bounded on $H^1$. The Lax-Milgram theorem therefore gives a unique weak solution for every $\lambda <-\gamma$.

When $c=0$, choose the shift ${\lambda }_0=-1$ and let $Kf$ be the unique solution of $B[Kf,v]+\int (Kf)v=\int fv$. Thus $K=(L+I{)}^{-1}$ in the weak sense. The equation $Lu=f$ is equivalent to $u=K(f+u)$, orConversely a solution in $L^2$ of this operator equation lies in $H^1$ because it equals $K(f+u)$, and satisfies the variational equation. The assumed boundedness $K:L^2\to H^1$, followed by the compact Rellich embedding $H^1\hookrightarrow L^2$, proves compactness of $K$ as an $L^2$ operator.

By the supplied kernel assumption, $\ker (I-K)$ consists of constants. Also $K1=1$. Since $K$ is a self-adjoint operator, the Fredholm alternative condition is $Kf\perp 1$, andTherefore a weak solution exists exactly when the mean of $f$ is zero, and solutions differ by a constant. On a disconnected domain the actual kernel would consist of constants on each component and the condition would apply separately on each; the question's given kernel assumption specifies the single-constant case.

For a general coefficient matrix, the stated variational definition encodes zero conormal flux $(aDu)\cdot n=0$. It agrees with ordinary zero normal derivative only under additional conditions on $a$. The argument above uses the weak formulation explicitly defined in the paper, so does not require that stronger boundary identification.

For a symmetric bounded bilinear form $B$ that is strictly positive on nonzero vectors, and a bounded linear functional $\ell$, set $I(v)=B(v,v)-2\ell (v)$. The first variation is $DI(v)[w]=2(B(v,w)-\ell (w))$. A weak solution $u$ of $B(u,w)=\ell (w)$ satisfies $I(u+w)-I(u)=B(w,w)$, so it is the unique minimizer. Conversely, any minimizer solves the weak equation. A coercive bilinear form gives existence by the Lax-Milgram theorem; strict positivity alone does not.

The Ritz method minimizes a quadratic energy over a finite-dimensional trial space. For basis functions ${\varphi }_j$ and bilinear form $a$, its linear system is $Ka=F$ with $K_{jk}=a({\varphi }_k,{\varphi }_j)$ and $F_j=⟨f,{\varphi }_j⟩$.

For a bounded symmetric coercive bilinear form on a Hilbert space, a solution of $a(u,w)=\ell (w)$ minimizes $J(v)=a(v,v)/2-\ell (v)$. The displayed identity proves strict minimality and uniqueness. Restricting to a conforming finite element space makes the discrete Ritz method equivalent to the Galerkin method. Galerkin orthogonality gives exact best approximation in the energy norm; the Céa lemma translates it into a reference-norm estimate.

A skew-coupled elliptic system has zeroth-order couplings with opposite signs. For the displayed example with homogeneous Dirichlet boundary conditions, its weak formulation on $V=H_0^1(\Omega )\times H_0^1(\Omega )$ has the bilinear formThe coupling cancels for equal arguments, so $b((p,q),(p,q))=\Vert \nabla p{\Vert }_0^2+\Vert \nabla q{\Vert }_0^2$. On a domain satisfying the Poincaré inequality, this is a coercive bilinear form. With the product first-order Sobolev norm, the Cauchy-Schwarz inequality applied to $(\nabla p,\nabla q,-q,p)$ and $(\nabla v,\nabla z,v,z)$ gives $|b(P,T)|\le \Vert P{\Vert }_V\Vert T{\Vert }_V$. The Lax-Milgram theorem therefore applies even though the bilinear form is not symmetric. Eliminating $w=-\Delta u$ recovers ${\Delta }^{2}u+u=f$, involving the biharmonic operator.

The stiffness matrix represents a variational bilinear form $a$ in a finite element basis. A symmetric coercive bilinear form gives a positive-definite matrix. Local support of the piecewise-linear hat functions produces a sparse matrix, allowing efficient numerical solution.

On l2 sequence space ${\ell }^2$, let $(Lu{)}_n=u_n/n$ and $f_n=1/n$. This bounded self-adjoint operator has $⟨Lu,u⟩>0$ for nonzero $u$, but the equation $Lu=f$ requires $u_n=1$, which is not in ${\ell }^2$. Taking the first $N$ coordinates equal to one makes $I(u)=-{\sum }_{n=1}^{N}1/n\to -\infty$. Thus strict positivity in isolation does not give existence in that Hilbert space; a coercive bilinear form and a continuous forcing functional in its norm are essential hypotheses for the usual minimization proof.

A continuous symmetric coercive bilinear form $a$ on a real Hilbert space and a continuous linear functional $\ell$ give a unique minimum of $I$. The parallelogram identity makes any minimizing sequence Cauchy in the energy norm; coercivity and completeness give its limit. Differentiation gives $a(u,w)=\ell (w)$, and $I(v)-I(u)=a(v-u,v-u)$ proves uniqueness. A nonsymmetric operator instead differentiates to its symmetric part. Mere pointwise positivity without a lower bound in the chosen norm need not yield a minimizing vector.

A variational problem seeks a stationary point or minimizer of a functional on a specified function space. For a symmetric coercive bilinear form $a$ and bounded linear functional $\ell$ on a real Hilbert space, minimizing $J(v)=a(v,v)/2-\ell (v)$ is equivalent to the weak formulation $a(u,v)=\ell (v)$ for every test vector. The Lax-Milgram theorem gives a unique solution, and $J(u+w)-J(u)=a(w,w)/2$ proves unique minimization. Specifying the admissible space and its boundary constraints is part of the problem.

On a bounded Lipschitz domain, the Poisson equation $-\Delta u=f$ with Robin boundary condition ${\partial }_{n}u+\beta u=0$ has the displayed weak formulation in $H^1(\Omega )$ when $\beta$ is a bounded boundary coefficient. Boundary integrals use Sobolev traces. It follows from Green's first identity, requiring only first weak derivatives of $u$. If $\beta$ is uniformly positive, testing the difference of two solutions against itself proves uniqueness: its gradient and its boundary Sobolev trace vanish. The Poincare inequality with a boundary trace also makes this a coercive bilinear form.

Let a positive-measure part ${\Gamma }_D$ of the boundary of a bounded connected domain carry a homogeneous Dirichlet boundary condition, and let the complementary part carry a homogeneous Neumann boundary condition. Onthe weak Poisson problem isThe Poincare inequality with a partial Dirichlet boundary makes the left side coercive, so the Lax-Milgram theorem gives a unique weak solution.