A second-order differential equation can still fail to control a function. The operator may contain second derivatives, yet only differentiate in some directions. Elliptic operators are the class of second-order operators whose principal part sees every direction, so boundary value problems behave like equilibrium problems rather than transport problems. The central question is: when does $Lu=f$ determine an unknown from boundary data, support energy estimates, and improve the regularity of solutions? The answer lies in the quadratic form attached to the highest-order coefficients. [example: A Second Derivative That Misses a Direction] Let $U=(0,1)^2\subsetneq\mathbb R^2$ and define $Lu=-\partial_{x_1}^2u$. For \begin{align*} u(x_1,x_2)&=\sin(2\pi x_2), \end{align*} the function does not depend on $x_1$, so \begin{align*} \partial_{x_1}u(x_1,x_2)&=0,\\ \partial_{x_1}^2u(x_1,x_2)&=0,\\ Lu(x_1,x_2)&=-\partial_{x_1}^2u(x_1,x_2)=0 \end{align*} for every $(x_1,x_2)\in U$. At the same time, \begin{align*} \partial_{x_2}u(x_1,x_2)&=2\pi\cos(2\pi x_2), \end{align*} so the solution can vary in the $x_2$ direction while the operator does not detect that variation. The principal matrix for the second-order part is \begin{align*} A&=\begin{pmatrix}1&0\\0&0\end{pmatrix}. \end{align*} For $\xi=(\xi_1,\xi_2)\in\mathbb R^2$, \begin{align*} A\xi&=\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}\xi_1\\ \xi_2\end{pmatrix} =\begin{pmatrix}\xi_1\\0\end{pmatrix},\\ \xi^\top A\xi&=\begin{pmatrix}\xi_1&\xi_2\end{pmatrix}\begin{pmatrix}\xi_1\\0\end{pmatrix} =\xi_1^2+\xi_2\cdot 0 =\xi_1^2. \end{align*} If $\xi=(0,\xi_2)$ with $\xi_2\ne0$, then $\xi\ne0$ but \begin{align*} \xi^\top A\xi&=0^2=0. \end{align*} Thus the formula contains a second derivative, but its principal quadratic form loses the whole $x_2$ direction. [/example] The example shows why ellipticity is a condition on the principal part, not on the visual appearance of the formula. Before a boundary value problem can be estimated, the second-order coefficients must be tested against arbitrary directions. The following definition names the basic pointwise positivity condition. [definition: Elliptic Matrix Field] Let $U\subsetneq\mathbb R^n$ be open. A measurable matrix field $A:U\to\mathbb R^{n\times n}$ is elliptic if for $\mathcal L^n$-a.e. $x\in U$ and every $\xi\in\mathbb R^n\setminus\{0\}$, \begin{align*} \sum_{i,j=1}^n A_{ij}(x)\xi_i\xi_j&>0. \end{align*} [/definition] Pointwise positivity prevents a fixed direction from disappearing at almost every fixed point. Estimates over a domain need a quantitative lower bound that does not collapse from point to point, so the primary definition records the uniform version. ## Definition Pointwise ellipticity still allows the quadratic form to become arbitrarily small as $x$ varies. That is too weak for estimates, because an energy bound must use one constant across the whole region being tested. Uniform ellipticity isolates the quantitative condition that prevents the operator from becoming nearly degenerate on smaller and smaller sets. [definition: Uniformly Elliptic Matrix Field] Let $U\subsetneq\mathbb R^n$ be open. A measurable matrix field $A:U\to\mathbb R^{n\times n}$ is uniformly elliptic on $U$ if there exists $\theta>0$ such that for $\mathcal L^n$-a.e. $x\in U$ and every $\xi\in\mathbb R^n$, \begin{align*} \sum_{i,j=1}^n A_{ij}(x)\xi_i\xi_j&\ge \theta |\xi|^2. \end{align*} [/definition] The constant $\theta$ is the ellipticity constant. When $A$ is also bounded, the principal quadratic form is comparable with the Euclidean form $|\xi|^2$, which is why the Laplacian becomes the model operator. ## Operator Forms ### Divergence and Nondivergence Forms Weak solutions are built from [integration by parts](/theorems/2098), so the coefficients must be arranged in a form compatible with test functions and weak derivatives. This motivates the divergence form operator. [definition: Divergence Form Elliptic Operator] Let $U\subsetneq\mathbb R^n$ be open. Let $a_{ij},b_i,c\in L^\infty(U)$ for $1\le i,j\le n$, and assume $A=(a_{ij})$ is uniformly elliptic on $U$. The divergence form elliptic operator with these coefficients is the distribution-valued operator \begin{align*} L:H_0^1(U)&\to H^{-1}(U) \end{align*} defined by \begin{align*} Lu&=-\sum_{i,j=1}^n\partial_{x_i}\left(a_{ij}\partial_{x_j}u\right)+\sum_{i=1}^n b_i\partial_{x_i}u+c u \end{align*} in $\mathcal D'(U)$. [/definition] Divergence form is adapted to [integration by parts](/theorems/210) because derivatives can be transferred onto test functions. In maximum principles, pointwise comparison, and classical regularity estimates, however, the equation is read at a point and the coefficients act directly on the Hessian. That viewpoint cannot be recovered from integration by parts when coefficients are merely rough, so it has to be recorded as a separate form of operator. [definition: Nondivergence Form Elliptic Operator] Let $U\subsetneq\mathbb R^n$ be open. Let $a_{ij},b_i,c\in L^\infty(U)$ for $1\le i,j\le n$, and assume $A=(a_{ij})$ is uniformly elliptic on $U$. The nondivergence form elliptic operator with these coefficients is the operator \begin{align*} L:C^2(U)&\to L^1_{\mathrm{loc}}(U) \end{align*} defined $\mathcal L^n$-a.e. by \begin{align*} Lu&=-\sum_{i,j=1}^n a_{ij}\partial_{x_i}\partial_{x_j}u+\sum_{i=1}^n b_i\partial_{x_i}u+c u. \end{align*} [/definition] The two forms agree for constant coefficients. For variable coefficients they support different notions of solution, so a page about elliptic operators must keep both available. ### The Laplacian as Model [example: The Laplacian] Let $L=-\Delta$ on an [open set](/page/Open%20Set) $U\subsetneq\mathbb R^n$. In nondivergence form this means the second-order coefficient matrix is the identity matrix, so \begin{align*} a_{ij}&=\delta_{ij},\\ Lu&=-\sum_{i,j=1}^n \delta_{ij}\partial_{x_i}\partial_{x_j}u =-\sum_{i=1}^n \partial_{x_i}\partial_{x_i}u =-\sum_{i=1}^n\partial_{x_i}^2u. \end{align*} For every $\xi=(\xi_1,\dots,\xi_n)\in\mathbb R^n$, the principal quadratic form is \begin{align*} \sum_{i,j=1}^n\delta_{ij}\xi_i\xi_j &=\sum_{i=1}^n \delta_{ii}\xi_i\xi_i+\sum_{\substack{1\le i,j\le n\\ i\ne j}}\delta_{ij}\xi_i\xi_j\\ &=\sum_{i=1}^n \xi_i^2+\sum_{\substack{1\le i,j\le n\\ i\ne j}}0\cdot \xi_i\xi_j\\ &=\sum_{i=1}^n \xi_i^2\\ &=|\xi|^2. \end{align*} Thus \begin{align*} \sum_{i,j=1}^n\delta_{ij}\xi_i\xi_j&\ge 1\cdot |\xi|^2 \end{align*} for every $\xi\in\mathbb R^n$, so $-\Delta$ is uniformly elliptic with ellipticity constant $1$. In divergence form the principal part is \begin{align*} -\sum_{i,j=1}^n\partial_{x_i}\left(\delta_{ij}\partial_{x_j}u\right) &=-\sum_{i=1}^n\partial_{x_i}\left(\partial_{x_i}u\right) =-\sum_{i=1}^n\partial_{x_i}^2u. \end{align*} With no first-order or zero-order terms, the associated [bilinear form](/page/Bilinear%20Form) satisfies, for $u\in H_0^1(U)$, \begin{align*} B[u,u] &=\int_U\sum_{i,j=1}^n\delta_{ij}\partial_{x_j}u\,\partial_{x_i}u\,d\mathcal L^n\\ &=\int_U\sum_{i=1}^n \partial_{x_i}u\,\partial_{x_i}u\,d\mathcal L^n\\ &=\int_U\sum_{i=1}^n |\partial_{x_i}u|^2\,d\mathcal L^n\\ &=\int_U|\nabla u|^2\,d\mathcal L^n. \end{align*} Thus the matrix test and the variational test say the same thing: the identity principal matrix charges every coordinate direction, and the energy records exactly the full squared gradient. [/example] The Laplacian supplies the Euclidean benchmark. General elliptic operators ask which parts of harmonic function theory survive after replacing $|\xi|^2$ by a uniformly comparable variable quadratic form. ## Principal Symbol and Directional Control To understand which part of an operator controls small-scale oscillation, one freezes the coefficients and keeps only the highest-order terms. First-order drift and zero-order terms can affect size and transport, but they do not determine which infinitesimal directions are penalised by the second derivatives. The resulting quadratic expression is the local directional test that detects ellipticity, degeneracy, and anisotropy. [definition: Principal Symbol] Let $U\subsetneq\mathbb R^n$ be open and let \begin{align*} Lu&=-\sum_{i,j=1}^n a_{ij}(x)\partial_{x_i}\partial_{x_j}u+\sum_{i=1}^n b_i(x)\partial_{x_i}u+c(x)u \end{align*} be a second-order nondivergence form operator. Choose measurable representatives of the coefficients $a_{ij}$ when they are defined only up to $\mathcal L^n$-a.e. equality. The principal symbol of $L$ is the measurable function \begin{align*} p_L:U\times\mathbb R^n&\to\mathbb R \\ (x,\xi)&\mapsto \sum_{i,j=1}^n a_{ij}(x)\xi_i\xi_j. \end{align*} [/definition] Changing the representatives changes $p_L$ only on a subset of $U\times\mathbb R^n$ whose $x$-projection has $\mathcal L^n$-measure zero. For this reason, symbol inequalities for rough coefficients are understood for $\mathcal L^n$-a.e. $x$, just as matrix ellipticity is. The same inequality should be recognisable whether we speak about the coefficient matrix or the symbol. Otherwise the symbol language could appear to impose a different condition from uniform ellipticity, creating two incompatible tests for the same directional control. The theorem below resolves this possible mismatch by verifying that the a.e. matrix bound and the a.e. symbol bound are exactly the same requirement. [quotetheorem:4945] This is not an additional hypothesis hidden inside the symbol. The theorem lets the page move freely between coefficient language and symbol language: estimates can be stated in terms of the matrix field, while intuition about directions and high-frequency behaviour can be read from $p_L(x,\xi)$. The a.e. qualification is also important, because rough coefficients are only meant to control the operator outside null exceptional sets. This characterisation makes degeneracy visible. If the symbol can vanish away from $\xi=0$, the operator has directions along which high-frequency oscillations are not penalised. [example: Degeneracy Along a Line] Let $U=(-1,1)^2$ and \begin{align*} Lu&=-\partial_{x_1}^2u-x_1^2\partial_{x_2}^2u. \end{align*} The second-order coefficient matrix is \begin{align*} A(x)&=\begin{pmatrix}1&0\\0&x_1^2\end{pmatrix}, \end{align*} because the coefficient of $\partial_{x_1}^2u$ is $1$, the coefficient of $\partial_{x_2}^2u$ is $x_1^2$, and there are no mixed second derivatives. For $\xi=(\xi_1,\xi_2)\in\mathbb R^2$, \begin{align*} A(x)\xi &=\begin{pmatrix}1&0\\0&x_1^2\end{pmatrix} \begin{pmatrix}\xi_1\\ \xi_2\end{pmatrix} \\ &=\begin{pmatrix}\xi_1\\ x_1^2\xi_2\end{pmatrix}, \end{align*} and therefore \begin{align*} \xi^\top A(x)\xi &=\begin{pmatrix}\xi_1&\xi_2\end{pmatrix} \begin{pmatrix}\xi_1\\ x_1^2\xi_2\end{pmatrix} \\ &=\xi_1^2+x_1^2\xi_2^2. \end{align*} At points with $x_1=0$, this becomes \begin{align*} \xi^\top A(x)\xi &=\xi_1^2+0^2\xi_2^2\\ &=\xi_1^2. \end{align*} In particular, for $\xi=(0,1)$, \begin{align*} \xi^\top A(x)\xi &=0^2+0^2\cdot 1^2\\ &=0, \end{align*} so the principal quadratic form does not charge the $x_2$ direction along the line $\{x_1=0\}$. The stronger obstruction is that no uniform ellipticity constant can exist. Let $\theta>0$ and again choose $\xi=(0,1)$. Then \begin{align*} |\xi|^2&=0^2+1^2=1,\\ \xi^\top A(x)\xi &=0^2+x_1^2\cdot 1^2\\ &=x_1^2. \end{align*} If $|x_1|<\min\{\sqrt{\theta},1\}$, then $x\in U$ for $x_2\in(-1,1)$ and \begin{align*} \xi^\top A(x)\xi &=x_1^2\\ &<\theta\\ &=\theta|\xi|^2. \end{align*} The set \begin{align*} \{(x_1,x_2)\in(-1,1)^2: |x_1|<\min\{\sqrt{\theta},1\}\} \end{align*} has positive two-dimensional [Lebesgue measure](/page/Lebesgue%20Measure), so the required inequality fails on a positive-measure subset of $U$. Thus the operator is not uniformly elliptic on $U$: its second-order part becomes arbitrarily weak in the $x_2$ direction as $x_1$ approaches $0$. [/example] Degenerate operators are important, but they belong to a different theory. Uniform ellipticity is the hypothesis that keeps the second-order part comparable to the Laplacian at every point and every scale. ## Weak Formulation and Energy ### Variational Encoding Divergence form operators are designed to act on functions with weak derivatives. Instead of requiring $Lu$ pointwise, we ask the equation to hold after testing against $v\in H_0^1(U)$. [definition: Bilinear Form of a Divergence Form Operator] Let $U\subsetneq\mathbb R^n$ be open. Let $a_{ij},b_i,c\in L^\infty(U)$ for $1\le i,j\le n$, and assume $A=(a_{ij})$ is uniformly elliptic on $U$. Let \begin{align*} Lu&=-\sum_{i,j=1}^n\partial_{x_i}(a_{ij}\partial_{x_j}u)+\sum_{i=1}^n b_i\partial_{x_i}u+c u. \end{align*} The associated bilinear form is the map \begin{align*} B:H_0^1(U)\times H_0^1(U)&\to\mathbb R \end{align*} defined by \begin{align*} B[u,v]&=\int_U\sum_{i,j=1}^n a_{ij}\partial_{x_j}u\partial_{x_i}v\,d\mathcal L^n+\int_U\sum_{i=1}^n b_i\partial_{x_i}u\,v\,d\mathcal L^n+\int_U c u v\,d\mathcal L^n. \end{align*} [/definition] To use $B$ as a Hilbert-space object, we need it to be continuous with respect to the Sobolev norm. This is the basic estimate that makes weak formulations stable under limits. [quotetheorem:4948] This boundedness result says that the weak operator is stable under Sobolev convergence: small changes in $u$ and $v$ in the $H^1(U)$ norm produce controlled changes in the value of $B[u,v]$. That is the analytic reason the variational formulation can be handled inside Hilbert-space functional analysis rather than by pointwise differentiability. Continuity alone does not state the PDE; it only tells us that the left-hand side is meaningful. For rough functions, the expression $Lu$ may not exist pointwise, and boundary values are imposed through the choice of test space rather than by evaluating $u$ on $\partial U$. This problem requires a definition that encodes the equation as an identity against every admissible [test function](/page/Test%20Function). [definition: Weak Solution of a Dirichlet Problem] Let $U\subsetneq\mathbb R^n$ be open, let $B:H_0^1(U)\times H_0^1(U)\to\mathbb R$ be the bilinear form associated to a divergence form elliptic operator, and let $F:H_0^1(U)\to\mathbb R$ be a continuous linear functional, so $F\in H^{-1}(U)$. A function $u\in H_0^1(U)$ is a weak solution of $Lu=F$ with homogeneous Dirichlet boundary condition if \begin{align*} B[u,v]&=F(v) \end{align*} for every $v\in H_0^1(U)$. [/definition] ### Energy Control A weak formulation can still be underdetermined unless the energy controls the unknown. If $B[u,u]$ can be small while $\|u\|$ is large, then the equation may fail to distinguish different candidates or may not respond continuously to the data. This obstruction motivates the Hilbert-space lower bound tying the quadratic energy to the norm of the solution. [definition: Coercive Bilinear Form] Let $H$ be a real [Hilbert space](/page/Hilbert%20Space). A bilinear form $B:H\times H\to\mathbb R$ is coercive if there exists $\alpha>0$ such that \begin{align*} B[u,u]&\ge \alpha\|u\|_H^2 \end{align*} for every $u\in H$. [/definition] Boundedness and coercivity address complementary obstructions: boundedness makes testing against $v$ continuous, while coercivity prevents loss of control in the solution direction. With only one of these properties, a variational equation can be meaningful but still fail to have a stable unique solution. The Hilbert-space result below is the bridge from those two estimates to solvability for every continuous linear datum. [quotetheorem:4946] The theorem turns an elliptic PDE into a solvable variational equation. The standard computation for the Poisson equation shows how the coercive estimate looks in practice. [example: Energy Estimate for the Poisson Equation] Let $U\subsetneq\mathbb R^n$ be bounded and open, let $f\in L^2(U)$, and suppose $u\in H_0^1(U)$ is a weak solution of \begin{align*} -\Delta u&=f. \end{align*} For the Poisson equation, the weak formulation is \begin{align*} \int_U \nabla u\cdot\nabla v\,d\mathcal L^n&=\int_U f v\,d\mathcal L^n \end{align*} for every $v\in H_0^1(U)$. Since $u\in H_0^1(U)$, we may choose $v=u$, giving \begin{align*} \int_U \nabla u\cdot\nabla u\,d\mathcal L^n&=\int_U f u\,d\mathcal L^n. \end{align*} Writing $\nabla u=(\partial_{x_1}u,\dots,\partial_{x_n}u)$, \begin{align*} \nabla u\cdot\nabla u &=\sum_{i=1}^n \partial_{x_i}u\,\partial_{x_i}u\\ &=\sum_{i=1}^n |\partial_{x_i}u|^2\\ &=|\nabla u|^2, \end{align*} so \begin{align*} \|\nabla u\|_{L^2(U)}^2 &=\int_U |\nabla u|^2\,d\mathcal L^n\\ &=\int_U f u\,d\mathcal L^n. \end{align*} By *Cauchy-Schwarz* in $L^2(U)$, \begin{align*} \int_U f u\,d\mathcal L^n &\le \left|\int_U f u\,d\mathcal L^n\right|\\ &\le \|f\|_{L^2(U)}\|u\|_{L^2(U)}. \end{align*} By *Poincare's inequality* for $H_0^1(U)$, there is a constant $C_P>0$ depending on $U$ such that \begin{align*} \|u\|_{L^2(U)}&\le C_P\|\nabla u\|_{L^2(U)}. \end{align*} Combining the last three displays gives \begin{align*} \|\nabla u\|_{L^2(U)}^2 &\le \|f\|_{L^2(U)}\|u\|_{L^2(U)}\\ &\le C_P\|f\|_{L^2(U)}\|\nabla u\|_{L^2(U)}. \end{align*} If $\|\nabla u\|_{L^2(U)}=0$, then the desired estimate is immediate. If $\|\nabla u\|_{L^2(U)}>0$, dividing by $\|\nabla u\|_{L^2(U)}$ yields \begin{align*} \|\nabla u\|_{L^2(U)}&\le C_P\|f\|_{L^2(U)}. \end{align*} Thus the forcing controls the Dirichlet energy of the solution before any explicit formula for $u$ is used. [/example] This estimate is the prototype for a priori bounds: control the solution before constructing it, then use compactness or Hilbert-space theory to obtain it. ## Boundary Value Problems and Maximum Principles Elliptic equations model equilibrium. For suitable signs of the lower-order terms, this means an interior maximum cannot be created without boundary support. [definition: Classical Solution of an Elliptic Equation] Let $U\subsetneq\mathbb R^n$ be open, let $f\in C(U)$, and let \begin{align*} L:C^2(U)&\to C(U) \end{align*} be a second-order elliptic operator with continuous coefficients. A function $u\in C^2(U)$ is a classical solution of $Lu=f$ in $U$ if \begin{align*} Lu(x)&=f(x) \end{align*} for every $x\in U$. [/definition] Classical solutions permit pointwise reasoning at interior extrema. With the convention \begin{align*} Lu&=-\sum_{i,j=1}^n a_{ij}\partial_{x_i}\partial_{x_j}u+\sum_{i=1}^n b_i\partial_{x_i}u+c u, \end{align*} the maximum principle pairs the inequality $Lu\le0$ with a nonnegative zero-order coefficient. At a positive interior maximum the second-order part contributes with the same sign as $c u$, so the operator cannot be negative there unless the maximum is forced by the boundary. The comparison question is whether ellipticity turns this local obstruction to interior peaks into global control by boundary values. If a new positive peak could appear inside the domain, Dirichlet data would no longer determine the sign or size of a solution. [quotetheorem:100] The sign of the zero-order term is part of the [comparison principle](/theorems/4870), not a cosmetic convention. For the displayed convention, a negative coefficient makes $c u$ negative at a positive interior maximum, so it can offset the nonnegative contribution from the second-order part. The boundary estimate can then fail even for a very smooth one-dimensional equation. [example: Failure When $c<0$] Let $U=(0,\pi)$ and define \begin{align*} Lu&=-u''-u. \end{align*} Here the zero-order coefficient is $c=-1<0$. For $u(x)=\sin x$, the first two derivatives are \begin{align*} u'(x)&=\cos x,\\ u''(x)&=-\sin x. \end{align*} Substituting into the operator gives, for every $x\in(0,\pi)$, \begin{align*} Lu(x) &=-u''(x)-u(x)\\ &=-(-\sin x)-\sin x\\ &=\sin x-\sin x\\ &=0. \end{align*} At the boundary points, \begin{align*} u(0)&=\sin 0=0,\\ u(\pi)&=\sin \pi=0, \end{align*} so $u=0$ on $\partial U=\{0,\pi\}$. However, for every $x\in(0,\pi)$, \begin{align*} 0&<x<\pi \end{align*} implies \begin{align*} \sin x&>0, \end{align*} and hence \begin{align*} u(x)&>0. \end{align*} Thus a nonzero positive interior solution can satisfy both $Lu=0$ in $U$ and zero boundary data when the zero-order coefficient is negative. [/example] Boundary value problems also require a choice of what is prescribed on the boundary. The most direct choice is the value of the solution itself. [definition: Dirichlet Boundary Condition] Let $U\subsetneq\mathbb R^n$ be a bounded Lipschitz domain, and let \begin{align*} \operatorname{Tr}:H^1(U)&\to H^{1/2}(\partial U) \end{align*} be the trace map. A Dirichlet boundary condition for $u\in H^1(U)$ with boundary datum $g\in H^{1/2}(\partial U)$ is the condition \begin{align*} \operatorname{Tr}u&=g \end{align*} in $H^{1/2}(\partial U)$. [/definition] Dirichlet data prescribe values, but variational problems naturally expose another boundary quantity after integration by parts. Boundary terms contain the normal component of the coefficient-weighted gradient, so fixing only the trace of $u$ is not the only natural way to close the problem. The corresponding condition prescribes the outward flux through the boundary. [definition: Neumann Boundary Condition] Let $U\subsetneq\mathbb R^n$ be a bounded $C^1$ domain, let $A=(a_{ij})\in C(\overline U;\mathbb R^{n\times n})$, and let $\nu:\partial U\to\mathbb R^n$ be the outward unit normal. The conormal derivative map associated to $A$ is \begin{align*} \partial_{\nu_A}:C^1(\overline U)&\to C(\partial U) \end{align*} defined by \begin{align*} \partial_{\nu_A}u&=\sum_{i,j=1}^n a_{ij}\partial_{x_j}u\,\nu_i. \end{align*} A Neumann boundary condition for $u\in C^1(\overline U)$ with boundary datum $g\in C(\partial U)$ is the condition \begin{align*} \partial_{\nu_A}u&=g \end{align*} on $\partial U$. [/definition] Dirichlet problems usually fix constants, while pure Neumann problems often determine the solution only up to an additive constant. This distinction reappears in spectral theory and variational methods. ## Regularity and Interior Smoothing A weak solution begins with limited differentiability, and the equation may contain only distributional derivatives. Even so, ellipticity often suppresses interior roughness because every direction is controlled by the principal part. The local issue is to separate smoothing forced by the equation from singularities caused by boundary geometry or boundary data. [definition: Interior Regularity] Let $U\subsetneq\mathbb R^n$ be open. For each open set $V\subsetneq U$, let $X(V)$ and $Y(V)$ be function spaces on $V$. An elliptic equation has interior $X$-to-$Y$ regularity if every weak or classical solution $u$ of $Lu=f$ satisfying $f\in X(V)$ on an open set $V\subsetneq U$ belongs to $Y(W)$ for every open set $W$ with $\overline W\subsetneq V$. [/definition] The first model theorem is needed to show the smoothing effect for the Poisson equation away from the boundary. It is the Sobolev version of elliptic regularity: $L^2$ forcing gives local second weak derivatives. [quotetheorem:4871] Sobolev estimates measure derivatives in an integral sense, but they do not by themselves give pointwise control of second derivatives. For classical solutions, roughness in the coefficients or forcing can still obstruct a clean Hessian estimate. The theorem below uses Hölder continuity as the scale in which the elliptic equation can recover pointwise second-order regularity without losing track of the data's modulus of continuity. [quotetheorem:4947] Regularity up to the boundary is harder because the boundary can create singularities. The next example separates interior smoothing from global smoothness. [example: A Corner Can Destroy Global $H^2$ Regularity] Let the reentrant corner have opening angle $\omega$ with $\pi<\omega<2\pi$, and use polar coordinates $(r,\theta)$ centered at that corner, with $0<\theta<\omega$. Set \begin{align*} \alpha&=\frac{\pi}{\omega}. \end{align*} Then $0<\alpha<1$. The model corner mode \begin{align*} v(r,\theta)&=r^\alpha\sin(\alpha\theta) \end{align*} vanishes on the two sides of the corner, since \begin{align*} v(r,0)&=r^\alpha\sin 0=0,\\ v(r,\omega)&=r^\alpha\sin(\alpha\omega) =r^\alpha\sin\pi =0. \end{align*} Using the polar formula \begin{align*} \Delta v&=\partial_r^2v+\frac1r\partial_rv+\frac1{r^2}\partial_\theta^2v, \end{align*} we compute \begin{align*} \partial_r v&=\alpha r^{\alpha-1}\sin(\alpha\theta),\\ \partial_r^2 v&=\alpha(\alpha-1)r^{\alpha-2}\sin(\alpha\theta),\\ \partial_\theta v&=\alpha r^\alpha\cos(\alpha\theta),\\ \partial_\theta^2v&=-\alpha^2 r^\alpha\sin(\alpha\theta). \end{align*} Therefore \begin{align*} \Delta v &=\alpha(\alpha-1)r^{\alpha-2}\sin(\alpha\theta) +\frac1r\alpha r^{\alpha-1}\sin(\alpha\theta) +\frac1{r^2}\left(-\alpha^2r^\alpha\sin(\alpha\theta)\right)\\ &=\left(\alpha(\alpha-1)+\alpha-\alpha^2\right)r^{\alpha-2}\sin(\alpha\theta)\\ &=\left(\alpha^2-\alpha+\alpha-\alpha^2\right)r^{\alpha-2}\sin(\alpha\theta)\\ &=0. \end{align*} Thus the corner mode is harmonic away from the vertex. Its second radial derivative has size \begin{align*} |\partial_r^2v|^2 &=\alpha^2(\alpha-1)^2r^{2\alpha-4}\sin^2(\alpha\theta). \end{align*} On any angular subinterval where $\sin(\alpha\theta)$ is bounded below by a positive constant, the local square integral near $r=0$ contains a constant multiple of \begin{align*} \int_0^\varepsilon r^{2\alpha-4}\,r\,dr &=\int_0^\varepsilon r^{2\alpha-3}\,dr. \end{align*} Since $\alpha<1$, we have \begin{align*} 2\alpha-3&<-1, \end{align*} so this integral diverges at $0$. Hence this corner mode is not locally $H^2$ at the reentrant corner. For the Dirichlet problem \begin{align*} -\Delta u&=1, \qquad u=0\text{ on }\partial U, \end{align*} the forcing is smooth, but the local expansion of $u$ near a reentrant corner can contain a nonzero multiple of this harmonic corner mode. When that coefficient is nonzero, the computation above shows that $u$ cannot belong to $H^2(U)$ globally, even though elliptic regularity still gives $H^2$ regularity on compact subsets away from the corner. [/example] Ellipticity improves solutions locally, but global regularity depends on the boundary geometry and compatibility of the boundary data. ## Green Functions and Fundamental Solutions Another way to solve elliptic equations is to understand the response to a point source. This leads from differential operators to distributions and integral kernels. [definition: Fundamental Solution] Let $L:\mathcal D'(\mathbb R^n)\to\mathcal D'(\mathbb R^n)$ be the distributional extension of a linear differential operator on $\mathbb R^n$. A distribution $\Gamma\in\mathcal D'(\mathbb R^n)$ is a fundamental solution for $L$ if \begin{align*} L\Gamma&=\delta_0 \end{align*} in $\mathcal D'(\mathbb R^n)$. [/definition] For the Laplacian, the fundamental solution is explicit. It gives the basic singularity that appears throughout potential theory. [example: Fundamental Solution of the Laplacian] For $n\ge3$, write $\sigma_{n-1}=\mathcal H^{n-1}(S^{n-1})$ and set \begin{align*} \Gamma(x)&=\frac{1}{(n-2)\sigma_{n-1}}|x|^{2-n}, \qquad x\ne0. \end{align*} If $r=|x|$ and $\gamma(r)=\frac{1}{(n-2)\sigma_{n-1}}r^{2-n}$, then \begin{align*} \gamma'(r)&=\frac{1}{(n-2)\sigma_{n-1}}(2-n)r^{1-n} =-\frac{1}{\sigma_{n-1}}r^{1-n},\\ \gamma''(r)&=-\frac{1}{\sigma_{n-1}}(1-n)r^{-n} =\frac{n-1}{\sigma_{n-1}}r^{-n}. \end{align*} For a radial function in $\mathbb R^n$, \begin{align*} \Delta \Gamma(x) &=\gamma''(r)+\frac{n-1}{r}\gamma'(r)\\ &=\frac{n-1}{\sigma_{n-1}}r^{-n} +\frac{n-1}{r}\left(-\frac{1}{\sigma_{n-1}}r^{1-n}\right)\\ &=\frac{n-1}{\sigma_{n-1}}r^{-n} -\frac{n-1}{\sigma_{n-1}}r^{-n}\\ &=0 \end{align*} for $x\ne0$. The normalization is chosen so that the outward flux of $-\nabla\Gamma$ through $\partial B_\varepsilon(0)$ equals $1$: \begin{align*} -\int_{\partial B_\varepsilon(0)}\partial_\nu\Gamma\,d\mathcal H^{n-1} &=-\int_{\partial B_\varepsilon(0)}\gamma'(\varepsilon)\,d\mathcal H^{n-1}\\ &=-\left(-\frac{1}{\sigma_{n-1}}\varepsilon^{1-n}\right) \mathcal H^{n-1}(\partial B_\varepsilon(0))\\ &=\frac{1}{\sigma_{n-1}}\varepsilon^{1-n} \left(\sigma_{n-1}\varepsilon^{n-1}\right)\\ &=1. \end{align*} Thus $\Gamma$ is harmonic away from the origin and has exactly one unit of distributional mass at the origin, so $-\Delta\Gamma=\delta_0$ in $\mathcal D'(\mathbb R^n)$. For $n=2$, set \begin{align*} \Gamma(x)&=\frac{1}{2\pi}\log\frac{1}{|x|} =-\frac{1}{2\pi}\log |x|, \qquad x\ne0. \end{align*} With $r=|x|$ and $\gamma(r)=-\frac{1}{2\pi}\log r$, \begin{align*} \gamma'(r)&=-\frac{1}{2\pi r},\\ \gamma''(r)&=\frac{1}{2\pi r^2}. \end{align*} The two-dimensional radial Laplacian gives \begin{align*} \Delta\Gamma(x) &=\gamma''(r)+\frac1r\gamma'(r)\\ &=\frac{1}{2\pi r^2} +\frac1r\left(-\frac{1}{2\pi r}\right)\\ &=\frac{1}{2\pi r^2}-\frac{1}{2\pi r^2}\\ &=0 \end{align*} for $x\ne0$, and the flux normalization is \begin{align*} -\int_{\partial B_\varepsilon(0)}\partial_\nu\Gamma\,d\mathcal H^1 &=-\int_{\partial B_\varepsilon(0)}\gamma'(\varepsilon)\,d\mathcal H^1\\ &=-\left(-\frac{1}{2\pi\varepsilon}\right)(2\pi\varepsilon)\\ &=1. \end{align*} Hence the logarithmic kernel also satisfies $-\Delta\Gamma=\delta_0$ in the distributional sense. Now let $f\in C_c^\infty(\mathbb R^n)$ and define \begin{align*} u(x)&=\int_{\mathbb R^n}\Gamma(x-y)f(y)\,d\mathcal L^n(y). \end{align*} The singularity of $\Gamma(x-y)$ at $y=x$ is locally integrable, and $f$ has compact support, so the integral is well-defined. Formally using $-\Delta_x\Gamma(x-y)=\delta_y(x)$ gives \begin{align*} -\Delta u(x) &=\int_{\mathbb R^n}\left(-\Delta_x\Gamma(x-y)\right)f(y)\,d\mathcal L^n(y)\\ &=\int_{\mathbb R^n}\delta_y(x)f(y)\,d\mathcal L^n(y)\\ &=f(x). \end{align*} Thus convolution with the fundamental solution is the model whole-space solution formula for the Poisson equation. [/example] A whole-space fundamental solution does not know about boundary conditions. On a bounded domain, the cleanest first definition is the model Dirichlet Green function for the Laplacian on a smooth domain: it records a point source in the interior and a zero boundary trace in the ordinary Sobolev trace sense, without pretending that the same minimal definition covers every rough elliptic operator. [definition: Green Function for the Dirichlet Problem] Let $U\subsetneq\mathbb R^n$ be a bounded $C^\infty$ domain. A Dirichlet Green function for $-\Delta$ on $U$ is a kernel \begin{align*} G:(U\times U)\setminus\{(x,x):x\in U\}&\to\mathbb R \end{align*} such that for each $y\in U$, the section $G_y:x\mapsto G(x,y)$ belongs to $L^1_{\mathrm{loc}}(U)$, defines the [regular distribution](/page/Regular%20Distribution) $T_{G_y}\in\mathcal D'(U)$, satisfies $G_y\in H^1(V)$ for every open set $V$ with $\overline V\subsetneq U\setminus\{y\}$, satisfies $\chi G_y\in H^1(U)$ and $\operatorname{Tr}(\chi G_y)=0$ on $\partial U$ for every $\chi\in C^\infty(\overline U)$ with $\chi=0$ on a neighbourhood of $y$, and satisfies \begin{align*} -\Delta T_{G_y}&=\delta_y \end{align*} in $\mathcal D'(U)$. [/definition] The representation theorem is needed to turn the kernel into a solution formula. It shows how source data are accumulated across the whole domain. [quotetheorem:42] For fixed $x\in U$, the theorem treats $G(x,\cdot)$ as the kernel in the $y$ variable: its distributional identity is $-\Delta_yG(x,\cdot)=\delta_x$. The formula separates the interior forcing contribution from the boundary contribution, so nonzero Dirichlet data enter through the normal derivative term on $\partial U$ rather than being hidden in the volume integral. Green functions reveal the nonlocal character of solving an elliptic equation: changing the source in one part of a connected domain can change the solution everywhere. ## Spectral Viewpoint On a bounded domain, an elliptic operator with boundary conditions often behaves like an infinite-dimensional positive matrix. Its eigenfunctions are the natural modes of the domain. [definition: Dirichlet Eigenvalue Problem] Let $U\subsetneq\mathbb R^n$ be bounded and open, and let $L:\mathcal D(L)\subsetneq L^2(U)\to L^2(U)$ be an elliptic operator with homogeneous Dirichlet boundary condition. The Dirichlet eigenvalue problem is to find $\lambda\in\mathbb R$ and nonzero $u\in\mathcal D(L)$ such that \begin{align*} Lu&=\lambda u \end{align*} in $L^2(U)$. [/definition] The definition by itself only poses a question; it does not say whether enough eigenpairs exist to describe arbitrary data. In infinite dimensions, eigenvectors need not span the space, and spectral values can accumulate or fail to appear as isolated modes. For the Dirichlet Laplacian on a bounded domain, compactness restores a discrete Fourier-like basis of modes. [quotetheorem:533] The theorem is the spectral analogue of finite-dimensional diagonalisation: symmetry makes the modes orthogonal, ellipticity gives positivity, and compactness of the resolvent turns the infinite-dimensional problem into a countable sequence of eigenmodes. The one-dimensional case gives the familiar model for this spectral decomposition. [example: Eigenfunctions on an Interval] On $U=(0,\pi)$, consider \begin{align*} -u''&=\lambda u, \qquad u(0)=u(\pi)=0. \end{align*} We compute the nonzero solutions satisfying both boundary conditions. If $\lambda=0$, then $u''=0$, so \begin{align*} u(x)&=Ax+B. \end{align*} The boundary conditions give \begin{align*} u(0)&=B=0,\\ u(\pi)&=A\pi+B=A\pi=0, \end{align*} hence $A=B=0$, so no nonzero eigenfunction occurs when $\lambda=0$. If $\lambda<0$, write $\lambda=-\mu^2$ with $\mu>0$. Then \begin{align*} -u''&=-\mu^2u \end{align*} is equivalent to \begin{align*} u''&=\mu^2u. \end{align*} The solutions are \begin{align*} u(x)&=Ae^{\mu x}+Be^{-\mu x}. \end{align*} The condition $u(0)=0$ gives \begin{align*} A+B&=0, \end{align*} so $B=-A$. Then \begin{align*} u(\pi) &=Ae^{\mu\pi}-Ae^{-\mu\pi}\\ &=A\left(e^{\mu\pi}-e^{-\mu\pi}\right). \end{align*} Since $\mu>0$, we have $e^{\mu\pi}-e^{-\mu\pi}\ne0$, and therefore $A=0$. Thus $B=0$, so no nonzero eigenfunction occurs when $\lambda<0$. If $\lambda>0$, write $\lambda=\mu^2$ with $\mu>0$. Then \begin{align*} -u''&=\mu^2u \end{align*} is equivalent to \begin{align*} u''+\mu^2u&=0. \end{align*} The solutions are \begin{align*} u(x)&=A\cos(\mu x)+B\sin(\mu x). \end{align*} The condition $u(0)=0$ gives \begin{align*} A\cos 0+B\sin 0&=0,\\ A&=0. \end{align*} Thus \begin{align*} u(x)&=B\sin(\mu x). \end{align*} The condition $u(\pi)=0$ gives \begin{align*} B\sin(\mu\pi)&=0. \end{align*} For a nonzero eigenfunction we need $B\ne0$, so \begin{align*} \sin(\mu\pi)&=0. \end{align*} This holds exactly when \begin{align*} \mu\pi&=k\pi \end{align*} for some $k\in\mathbb N$, hence \begin{align*} \mu&=k. \end{align*} Therefore \begin{align*} \lambda&=\mu^2=k^2, \qquad u(x)=B\sin(kx). \end{align*} Taking $B=1$, the eigenpairs are \begin{align*} \phi_k(x)&=\sin(kx),\\ \lambda_k&=k^2, \end{align*} for $k\in\mathbb N$. Indeed, \begin{align*} \phi_k'(x)&=k\cos(kx),\\ \phi_k''(x)&=-k^2\sin(kx), \end{align*} so \begin{align*} -\phi_k''(x) &=-\left(-k^2\sin(kx)\right)\\ &=k^2\sin(kx)\\ &=\lambda_k\phi_k(x), \end{align*} and the boundary values are \begin{align*} \phi_k(0)&=\sin 0=0,\\ \phi_k(\pi)&=\sin(k\pi)=0. \end{align*} Thus each additional oscillation raises the eigenvalue from $k^2$ to $(k+1)^2$, so higher modes carry larger second-derivative energy. [/example] The spectral viewpoint links elliptic theory with compact operators, self-adjointness, heat flow, and geometry. ## Beyond and Connected Topics The closest continuation is [Second Order Elliptic Equations](/page/Second%20Order%20Elliptic%20Equations), where the existence, uniqueness, maximum principle, and regularity theory are developed for boundary value problems. That page takes the structural definitions here and turns them into the main PDE toolkit: comparison, weak solvability, and estimates with boundary conditions. The weak formulation depends on [Sobolev Space](/page/Sobolev%20Space), especially $H_0^1(U)$, weak derivatives, trace theory, and compact embeddings. This is the functional setting in which a divergence form equation can be read as an identity in $H^{-1}(U)$ rather than as a pointwise formula. The variational side connects elliptic operators to [Calculus of Variations](/page/Calculus%20of%20Variations), where elliptic equations often arise as Euler-Lagrange equations for energy functionals. In that setting, uniform ellipticity is the analytic shadow of convexity in the gradient variable. The kernel viewpoint connects to [Distribution](/page/Distribution), [Fourier Transform](/page/Fourier%20Transform), and potential theory, since fundamental solutions are distributions and constant-coefficient estimates are often Fourier multiplier estimates. This viewpoint is especially useful when the source is singular or when the operator is studied through its symbol. The spectral side connects to compact operators and [self-adjoint operators](/page/Self-Adjoint%20Operators). On manifolds, elliptic operators such as the Laplace-Beltrami operator connect PDE with geometry and topology. ## References [Second Order Elliptic Equations](/page/Second%20Order%20Elliptic%20Equations). [Sobolev Space](/page/Sobolev%20Space). [Calculus of Variations](/page/Calculus%20of%20Variations). [Distribution](/page/Distribution). [Fourier Transform](/page/Fourier%20Transform). Evans, *Partial Differential Equations* (2010). Gilbarg and Trudinger, *Elliptic Partial Differential Equations of Second Order* (2001). Folland, *Introduction to Partial Differential Equations* (1995).