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Elliptic boundary value problem
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<h2 id="the-main-example">The main example</h2> <p>In two dimensions, let x , y {\displaystyle x,y} be the coordinates. We will use the <a href="/facts/Notation_for_differentiation/PTCbwS0Y">subscript notation</a> u x , u x x {\displaystyle u_{x},u_{xx}} for the first and second <a href="/facts/Partial_derivative/JWHsBkAu">partial derivatives</a> of u {\displaystyle u} with respect to x {\displaystyle x} , and a similar notation for y {\displaystyle y} . We define the gradient ∇ u = ( u x , u y ) {\displaystyle \nabla u=(u_{x},u_{y})} , the <a href="/facts/Laplace_operator/kov9u3PT">Laplace operator</a> Δ u = u x x + u y y {\displaystyle \Delta u=u_{xx}+u_{yy}} and the divergence ∇ ⋅ ( u , v ) = u x + v y {\displaystyle \nabla \cdot (u,v)=u_{x}+v_{y}} . Note from the definitions that Δ u = ∇ ⋅ ( ∇ u ) {\displaystyle \Delta u=\nabla \cdot (\nabla u)} . </p><p>The main example for boundary value problems is the Laplace operator, </p> Δ u = f in Ω , {\displaystyle \Delta u=f{\text{ in }}\Omega ,} u = 0 on ∂ Ω ; {\displaystyle u=0{\text{ on }}\partial \Omega ;} <p>where Ω {\displaystyle \Omega } is a region in the plane and ∂ Ω {\displaystyle \partial \Omega } is the boundary of that region. The function f {\displaystyle f} is known data and the solution u {\displaystyle u} is what must be computed. </p><p>The solution u {\displaystyle u} can be interpreted as the stationary or limit distribution of heat in a metal plate shaped like Ω {\displaystyle \Omega } with its boundary ∂ Ω {\displaystyle \partial \Omega } kept at zero degrees. The function f {\displaystyle f} represents the intensity of heat generation at each point in the plate. After waiting for a long time, the temperature distribution in the metal plate will approach u {\displaystyle u} . </p> <h3>Second-order linear problems</h3> <p>In general, a boundary-value problem (BVP) consists of a <a href="/facts/Partial_differential_equation/DyPsBlv7">partial differential equation</a> (PDE) subject to a <a href="/facts/Boundary_condition/QEfmUqmP">boundary condition</a>. For now, <a href="/facts/Partial_differential_equation/DyPsBlv7">second-order PDEs</a> subject to a <a href="/facts/Dirichlet_boundary_condition/m9wsgG89">Dirichlet boundary condition</a> will be considered. </p><p>Let U {\displaystyle U} be an <a href="/facts/Open_set/QfTCIqMu">open</a>, <a href="/facts/Bounded_set/yCldjtoy">bounded</a> subset of R n {\displaystyle \mathbb {R} ^{n}} , denote its boundary as ∂ U {\displaystyle \partial U} and the variables as x = ( x 1 , . . . , x n ) {\displaystyle x=(x_{1},...,x_{n})} . Representing the PDE as a partial <a href="/facts/Differential_operator/Nr15J0IE">differential operator</a> L {\displaystyle L} acting on an unknown function u = u ( x ) {\displaystyle u=u(x)} of x ∈ U {\displaystyle x\in U} results in a BVP of the form { L u = f in U u = 0 on ∂ U , {\displaystyle \left\{{\begin{aligned}Lu&=f&&{\text{in }}U\\u&=0&&{\text{on }}\partial U,\end{aligned}}\right.} where f : U → R {\displaystyle f:U\rightarrow \mathbb {R} } is a given function f = f ( x ) {\displaystyle f=f(x)} and u : U ∪ ∂ U → R {\displaystyle u:U\cup \partial U\rightarrow \mathbb {R} } and the operator L {\displaystyle L} is either of the form: L u ( x ) = − ∑ i , j = 1 n ( a i j ( x ) u x i ) x j + ∑ i = 1 n b i ( x ) u x i ( x ) + c ( x ) u ( x ) , {\displaystyle Lu(x)=-\sum _{i,j=1}^{n}(a_{ij}(x)u_{x_{i}})_{x_{j}}+\sum _{i=1}^{n}b_{i}(x)u_{x_{i}}(x)+c(x)u(x),} or L u ( x ) = − ∑ i , j = 1 n a i j ( x ) u x i x j + ∑ i = 1 n b ~ i ( x ) u x i ( x ) + c ( x ) u ( x ) , {\displaystyle Lu(x)=-\sum _{i,j=1}^{n}a_{ij}(x)u_{x_{i}x_{j}}+\sum _{i=1}^{n}{\tilde {b}}_{i}(x)u_{x_{i}}(x)+c(x)u(x),} for given coeficient functions a i j ( x ) , b i ( x ) , c ( x ) {\displaystyle a_{ij}(x),b_{i}(x),c(x)} . </p><p>The PDE L u = f {\displaystyle Lu=f} is said to be in divergence form in case of the former and in nondivergence form in case of the latter. If the functions a i j {\displaystyle a_{ij}} are <a href="/facts/Continuously_differentiable/jQqTgLk1">continuously differentiable</a> then both cases are equivalent for b ~ i ( x ) = b i ( x ) + ∑ j a i j , x j ( x ) . {\displaystyle {\tilde {b}}_{i}(x)=b_{i}(x)+\sum _{j}a_{ij,x_{j}}(x).} In matrix notation, we can let a ( x ) {\displaystyle a(x)} be an n × n {\displaystyle n\times n} matrix valued function of x {\displaystyle x} and b ( x ) {\displaystyle b(x)} be a n {\displaystyle n} -dimensional column vector-valued function of x {\displaystyle x} , and then we may write (the divergence form as) L u = − ∇ ⋅ ( a ∇ u ) + b T ∇ u + c u {\displaystyle Lu=-\nabla \cdot (a\nabla u)+b^{T}\nabla u+cu} One may assume, without loss of generality, that the matrix a {\displaystyle a} is symmetric (that is, for all i , j , x {\displaystyle i,j,x} , a i j ( x ) = a j i ( x ) {\displaystyle a_{ij}(x)=a_{ji}(x)} . We make that assumption in the rest of this article. </p><p>We say that the operator L {\displaystyle L} is <i>elliptic</i> if, for some constant α > 0 {\displaystyle \alpha >0} , any of the following equivalent conditions hold: </p> <ol><li> λ min ( a ( x ) ) > α ∀ x {\displaystyle \lambda _{\min }(a(x))>\alpha \;\;\;\forall x} (see <a href="/facts/Eigenvalue/8TjEoT8u">eigenvalue</a>).</li> <li> u T a ( x ) u > α u T u ∀ u ∈ R n {\displaystyle u^{T}a(x)u>\alpha u^{T}u\;\;\;\forall u\in \mathbb {R} ^{n}} .</li> <li> ∑ i , j = 1 n a i j u i u j > α ∑ i = 1 n u i 2 ∀ u ∈ R n {\displaystyle \sum _{i,j=1}^{n}a_{ij}u_{i}u_{j}>\alpha \sum _{i=1}^{n}u_{i}^{2}\;\;\;\forall u\in \mathbb {R} ^{n}} .</li></ol> <p>If the second-order partial differential operator L {\displaystyle L} is elliptic, then the associated BVP is called an elliptic boundary-value problem. </p> <h3>Boundary conditions</h3> <p>The above BVP is a particular example of a <a href="/facts/Dirichlet_problem/uHErgNCb">Dirichlet problem</a>. The <a href="/facts/Neumann_problem/0e0YBW4G">Neumann problem</a> is </p> L u = f in Ω {\displaystyle Lu=f{\text{ in }}\Omega } and u ν = g on ∂ Ω {\displaystyle u_{\nu }=g{\text{ on }}\partial \Omega } <p>where u ν {\displaystyle u_{\nu }} is the derivative of u {\displaystyle u} in the direction of the outwards pointing normal of ∂ Ω {\displaystyle \partial \Omega } . In general, if B {\displaystyle B} is any <a href="/facts/Trace_operator/7uQ4kSE4">trace operator</a>, one can construct the boundary value problem </p> L u = f in Ω {\displaystyle Lu=f{\text{ in }}\Omega } and B u = g on ∂ Ω {\displaystyle Bu=g{\text{ on }}\partial \Omega } . <p>In the rest of this article, we assume that L {\displaystyle L} is elliptic and that the boundary condition is the Dirichlet condition u = 0 on ∂ Ω {\displaystyle u=0{\text{ on }}\partial \Omega } . </p> <h2 id="sobolev-spaces">Sobolev spaces</h2> <p>The analysis of elliptic boundary value problems requires some fairly sophisticated tools of <a href="/facts/Functional_analysis/MIp233MT">functional analysis</a>. We require the space H 1 ( Ω ) {\displaystyle H^{1}(\Omega )} , the <a href="/facts/Sobolev_space/34qxWCyr">Sobolev space</a> of "once-differentiable" functions on Ω {\displaystyle \Omega } , such that both the function u {\displaystyle u} and its partial derivatives u x i {\displaystyle u_{x_{i}}} , i = 1 , … , n {\displaystyle i=1,\dots ,n} are all <a href="/facts/Square_integrable/IASblcJJ">square integrable</a>. That is: H 1 ( Ω ) = { u ∈ L 2 ( Ω ) , u x j ∈ L 2 ( Ω ) , 1 ≤ i ≤ n } . {\displaystyle H^{1}(\Omega )=\left\{u\in L^{2}(\Omega ),\;\;u_{x_{j}}\in L^{2}(\Omega ),\;\;1\leq i\leq n\right\}.} There is a subtlety here in that the partial derivatives must be defined "in the weak sense" (see the article on Sobolev spaces for details.) The space H 1 {\displaystyle H^{1}} is a <a href="/facts/Hilbert_space/aDFMh4lV">Hilbert space</a>, which accounts for much of the ease with which these problems are analyzed. </p><p>Unless otherwise noted, all derivatives in this article are to be interpreted in the weak, Sobolev sense. We use the term "strong derivative" to refer to the classical derivative of calculus. We also specify that the spaces C k {\displaystyle C^{k}} , k = 0 , 1 , … {\displaystyle k=0,1,\dots } consist of functions that are k {\displaystyle k} times strongly differentiable, and that the k {\displaystyle k} th derivative is continuous. </p> <h2 id="weak-or-variational-formulation">Weak or variational formulation</h2> <p>The first step to cast the boundary value problem as in the language of Sobolev spaces is to rephrase it in its weak form. Consider the Laplace problem Δ u = f {\displaystyle \Delta u=f} . Multiply each side of the equation by a "test function" φ {\displaystyle \varphi } and <a href="/facts/Integrate_by_parts/aV6JgBNd">integrate by parts</a> using <a href="/facts/Green%2527s_theorem/WBDk654A">Green's theorem</a> to obtain </p> − ∫ Ω ∇ u ⋅ ∇ φ + ∫ ∂ Ω u ν φ = ∫ Ω f φ {\displaystyle -\int _{\Omega }\nabla u\cdot \nabla \varphi +\int _{\partial \Omega }u_{\nu }\varphi =\int _{\Omega }f\varphi } . <p>We will be solving the Dirichlet problem, so that u = 0 on ∂ Ω {\displaystyle u=0{\text{ on }}\partial \Omega } . For technical reasons, it is useful to assume that φ {\displaystyle \varphi } is taken from the same space of functions as u {\displaystyle u} is so we also assume that φ = 0 on ∂ Ω {\displaystyle \varphi =0{\text{ on }}\partial \Omega } . This gets rid of the ∫ ∂ Ω {\displaystyle \int _{\partial \Omega }} term, yielding </p> A ( u , φ ) = F ( φ ) {\displaystyle A(u,\varphi )=F(\varphi )} (*) <p>where </p> A ( u , φ ) = ∫ Ω ∇ u ⋅ ∇ φ {\displaystyle A(u,\varphi )=\int _{\Omega }\nabla u\cdot \nabla \varphi } and F ( φ ) = − ∫ Ω f φ {\displaystyle F(\varphi )=-\int _{\Omega }f\varphi } . <p>If L {\displaystyle L} is a general elliptic operator, the same reasoning leads to the bilinear form </p> A ( u , φ ) = ∫ Ω ∇ u T a ∇ φ − ∫ Ω b T ∇ u φ − ∫ Ω c u φ {\displaystyle A(u,\varphi )=\int _{\Omega }\nabla u^{T}a\nabla \varphi -\int _{\Omega }b^{T}\nabla u\varphi -\int _{\Omega }cu\varphi } . <p>We do not discuss the Neumann problem but note that it is analyzed in a similar way. </p> <h3>Continuous and coercive bilinear forms</h3> <p>The map A ( u , φ ) {\displaystyle A(u,\varphi )} is defined on the Sobolev space H 0 1 ⊂ H 1 {\displaystyle H_{0}^{1}\subset H^{1}} of functions which are once differentiable and zero on the boundary ∂ Ω {\displaystyle \partial \Omega } , provided we impose some conditions on a , b , c {\displaystyle a,b,c} and Ω {\displaystyle \Omega } . There are many possible choices, but for the purpose of this article, we will assume that </p> <ol><li> a i j ( x ) {\displaystyle a_{ij}(x)} is <a href="/facts/Continuously_differentiable/jQqTgLk1">continuously differentiable</a> on Ω ¯ {\displaystyle {\bar {\Omega }}} for i , j = 1 , … , n , {\displaystyle i,j=1,\dots ,n,} </li> <li> b i ( x ) {\displaystyle b_{i}(x)} is continuous on Ω ¯ {\displaystyle {\bar {\Omega }}} for i = 1 , … , n , {\displaystyle i=1,\dots ,n,} </li> <li> c ( x ) {\displaystyle c(x)} is continuous on Ω ¯ {\displaystyle {\bar {\Omega }}} and</li> <li> Ω {\displaystyle \Omega } is bounded.</li></ol> <p>The reader may verify that the map A ( u , φ ) {\displaystyle A(u,\varphi )} is furthermore <a href="/facts/Bilinear_map/LH1lHVeP">bilinear</a> and <a href="/facts/Continuous_function/sKbl02pB">continuous</a>, and that the map F ( φ ) {\displaystyle F(\varphi )} is <a href="/facts/Linear/fChtg3KC">linear</a> in φ {\displaystyle \varphi } , and continuous if (for instance) f {\displaystyle f} is square integrable. </p><p>We say that the map A {\displaystyle A} is <a href="/facts/Coercive_function/6Yb5rG1q">coercive</a> if there is an α > 0 {\displaystyle \alpha >0} for all u , φ ∈ H 0 1 ( Ω ) {\displaystyle u,\varphi \in H_{0}^{1}(\Omega )} , </p> A ( u , φ ) ≥ α ∫ Ω ∇ u ⋅ ∇ φ . {\displaystyle A(u,\varphi )\geq \alpha \int _{\Omega }\nabla u\cdot \nabla \varphi .} <p>This is trivially true for the Laplacian (with α = 1 {\displaystyle \alpha =1} ) and is also true for an elliptic operator if we assume b = 0 {\displaystyle b=0} and c ≤ 0 {\displaystyle c\leq 0} . (Recall that u T a u > α u T u {\displaystyle u^{T}au>\alpha u^{T}u} when L {\displaystyle L} is elliptic.) </p> <h3>Existence and uniqueness of the weak solution</h3> <p>One may show, via the <a href="/facts/Lax%25E2%2580%2593Milgram_lemma/hXd2jSoW">Lax–Milgram lemma</a>, that whenever A ( u , φ ) {\displaystyle A(u,\varphi )} is coercive and F ( φ ) {\displaystyle F(\varphi )} is continuous, then there exists a unique solution u ∈ H 0 1 ( Ω ) {\displaystyle u\in H_{0}^{1}(\Omega )} to the weak problem (*). </p><p>If further A ( u , φ ) {\displaystyle A(u,\varphi )} is symmetric (i.e., b = 0 {\displaystyle b=0} ), one can show the same result using the <a href="/facts/Riesz_representation_theorem/9CjKDUII">Riesz representation theorem</a> instead. </p><p>This relies on the fact that A ( u , φ ) {\displaystyle A(u,\varphi )} forms an inner product on H 0 1 ( Ω ) {\displaystyle H_{0}^{1}(\Omega )} , which itself depends on <a href="/facts/Poincar%25C3%25A9%2527s_inequality/7W5a0H9G">Poincaré's inequality</a>. </p> <h2 id="strong-solutions">Strong solutions</h2> <p>We have shown that there is a u ∈ H 0 1 ( Ω ) {\displaystyle u\in H_{0}^{1}(\Omega )} which solves the weak system, but we do not know if this u {\displaystyle u} solves the strong system </p> L u = f in Ω , {\displaystyle Lu=f{\text{ in }}\Omega ,} u = 0 on ∂ Ω , {\displaystyle u=0{\text{ on }}\partial \Omega ,} <p>Even more vexing is that we are not even sure that u {\displaystyle u} is twice differentiable, rendering the expressions u x i x j {\displaystyle u_{x_{i}x_{j}}} in L u {\displaystyle Lu} apparently meaningless. There are many ways to remedy the situation, the main one being regularity. </p> <h3>Regularity</h3> <p>A regularity theorem for a linear elliptic boundary value problem of the second order takes the form </p><p>Theorem <i>If (some condition), then the solution u {\displaystyle u} is in H 2 ( Ω ) {\displaystyle H^{2}(\Omega )} , the space of "twice differentiable" functions whose second derivatives are square integrable.</i> </p><p>There is no known simple condition necessary and sufficient for the conclusion of the theorem to hold, but the following conditions are known to be sufficient: </p> <ol><li>The boundary of Ω {\displaystyle \Omega } is C 2 {\displaystyle C^{2}} , or</li> <li> Ω {\displaystyle \Omega } is convex.</li></ol> <p>It may be tempting to infer that if ∂ Ω {\displaystyle \partial \Omega } is piecewise C 2 {\displaystyle C^{2}} then u {\displaystyle u} is indeed in H 2 {\displaystyle H^{2}} , but that is unfortunately false. </p> <h3>Almost everywhere solutions</h3> <p>In the case that u ∈ H 2 ( Ω ) {\displaystyle u\in H^{2}(\Omega )} then the second derivatives of u {\displaystyle u} are defined <a href="/facts/Almost_everywhere/1quRFtJE">almost everywhere</a>, and in that case L u = f {\displaystyle Lu=f} almost everywhere. </p> <h3>Strong solutions</h3> <p>One may further prove that if the boundary of Ω ⊂ R n {\displaystyle \Omega \subset \mathbb {R} ^{n}} is a <a href="/facts/Smooth_manifold/aSnbXKcV">smooth manifold</a> and f {\displaystyle f} is infinitely differentiable in the strong sense, then u {\displaystyle u} is also infinitely differentiable in the strong sense. In this case, L u = f {\displaystyle Lu=f} with the strong definition of the derivative. </p><p>The proof of this relies upon an improved regularity theorem that says that if ∂ Ω {\displaystyle \partial \Omega } is C k {\displaystyle C^{k}} and f ∈ H k − 2 ( Ω ) {\displaystyle f\in H^{k-2}(\Omega )} , k ≥ 2 {\displaystyle k\geq 2} , then u ∈ H k ( Ω ) {\displaystyle u\in H^{k}(\Omega )} , together with a <a href="/facts/Sobolev_imbedding_theorem/jxJvCHML">Sobolev imbedding theorem</a> saying that functions in H k ( Ω ) {\displaystyle H^{k}(\Omega )} are also in C m ( Ω ¯ ) {\displaystyle C^{m}({\bar {\Omega }})} whenever 0 ≤ m < k − n / 2 {\displaystyle 0\leq m<k-n/2} . </p> <h2 id="numerical-solutions">Numerical solutions</h2> <p>While in exceptional circumstances, it is possible to solve elliptic problems explicitly, in general it is an impossible task. The natural solution is to approximate the elliptic problem with a simpler one and to solve this simpler problem on a computer. </p><p>Because of the good properties we have enumerated (as well as many we have not), there are extremely efficient numerical solvers for linear elliptic boundary value problems (see <a href="/facts/Finite_element_method/eUcfP1vn">finite element method</a>, <a href="/facts/Finite_difference_method/EoG81Qn9">finite difference method</a> and <a href="/facts/Spectral_method/25VTd7oc">spectral method</a> for examples.) </p> <h2 id="eigenvalues-and-eigensolutions">Eigenvalues and eigensolutions</h2> <p>Another Sobolev imbedding theorem states that the inclusion H 1 ⊂ L 2 {\displaystyle H^{1}\subset L^{2}} is a compact linear map. Equipped with the <a href="/facts/Spectral_theorem/Vqc04B21">spectral theorem</a> for compact linear operators, one obtains the following result. </p><p>Theorem <i>Assume that A ( u , φ ) {\displaystyle A(u,\varphi )} is coercive, continuous and symmetric. The map S : f → u {\displaystyle S:f\rightarrow u} from L 2 ( Ω ) {\displaystyle L^{2}(\Omega )} to L 2 ( Ω ) {\displaystyle L^{2}(\Omega )} is a compact linear map. It has a <a href="/facts/Basis_(linear_algebra)/89IPoN6c">basis</a> of <a href="/facts/Eigenvector/8TjEoT8u">eigenvectors</a> u 1 , u 2 , ⋯ ∈ H 1 ( Ω ) {\displaystyle u_{1},u_{2},\dots \in H^{1}(\Omega )} and matching <a href="/facts/Eigenvalue/8TjEoT8u">eigenvalues</a> λ 1 , λ 2 , ⋯ ∈ R {\displaystyle \lambda _{1},\lambda _{2},\dots \in \mathbb {R} } such that</i> </p> <ol><li> S u k = λ k u k , k = 1 , 2 , … , {\displaystyle Su_{k}=\lambda _{k}u_{k},k=1,2,\dots ,} </li> <li> λ k → 0 {\displaystyle \lambda _{k}\rightarrow 0} <i>as</i> k → ∞ {\displaystyle k\rightarrow \infty } ,</li> <li> λ k ≩ 0 ∀ k {\displaystyle \lambda _{k}\gneqq 0\;\;\forall k} ,</li> <li> ∫ Ω u j u k = 0 {\displaystyle \int _{\Omega }u_{j}u_{k}=0} <i>whenever</i> j ≠ k {\displaystyle j\neq k} <i>and</i></li> <li> ∫ Ω u j u j = 1 {\displaystyle \int _{\Omega }u_{j}u_{j}=1} <i>for all</i> j = 1 , 2 , … . {\displaystyle j=1,2,\dots \,.} </li></ol> <h3>Series solutions and the importance of eigensolutions</h3> <p>If one has computed the eigenvalues and eigenvectors, then one may find the "explicit" solution of L u = f {\displaystyle Lu=f} , </p> u = ∑ k = 1 ∞ u ^ ( k ) u k {\displaystyle u=\sum _{k=1}^{\infty }{\hat {u}}(k)u_{k}} <p>via the formula </p> u ^ ( k ) = λ k f ^ ( k ) , k = 1 , 2 , … {\displaystyle {\hat {u}}(k)=\lambda _{k}{\hat {f}}(k),\;\;k=1,2,\dots } <p>where </p> f ^ ( k ) = ∫ Ω f ( x ) u k ( x ) d x . {\displaystyle {\hat {f}}(k)=\int _{\Omega }f(x)u_{k}(x)\,dx.} <p>(See <a href="/facts/Fourier_series/s3fsxPAT">Fourier series</a>.) </p><p>The series converges in L 2 {\displaystyle L^{2}} . Implemented on a computer using numerical approximations, this is known as the <a href="/facts/Spectral_method/25VTd7oc">spectral method</a>. </p> <h3>An example</h3> <p>Consider the problem </p> u − u x x − u y y = f ( x , y ) = x y {\displaystyle u-u_{xx}-u_{yy}=f(x,y)=xy} on ( 0 , 1 ) × ( 0 , 1 ) , {\displaystyle (0,1)\times (0,1),} u ( x , 0 ) = u ( x , 1 ) = u ( 0 , y ) = u ( 1 , y ) = 0 ∀ ( x , y ) ∈ ( 0 , 1 ) × ( 0 , 1 ) {\displaystyle u(x,0)=u(x,1)=u(0,y)=u(1,y)=0\;\;\forall (x,y)\in (0,1)\times (0,1)} (Dirichlet conditions). <p>The reader may verify that the eigenvectors are exactly </p> u j k ( x , y ) = sin ( π j x ) sin ( π k y ) {\displaystyle u_{jk}(x,y)=\sin(\pi jx)\sin(\pi ky)} , j , k ∈ N {\displaystyle j,k\in \mathbb {N} } <p>with eigenvalues </p> λ j k = 1 1 + π 2 j 2 + π 2 k 2 . {\displaystyle \lambda _{jk}={1 \over 1+\pi ^{2}j^{2}+\pi ^{2}k^{2}}.} <p>The Fourier coefficients of g ( x ) = x {\displaystyle g(x)=x} can be looked up in a table, getting g ^ ( n ) = ( − 1 ) n + 1 π n {\displaystyle {\hat {g}}(n)={(-1)^{n+1} \over \pi n}} . Therefore, </p> f ^ ( j , k ) = ( − 1 ) j + k + 1 π 2 j k {\displaystyle {\hat {f}}(j,k)={(-1)^{j+k+1} \over \pi ^{2}jk}} <p>yielding the solution </p> u ( x , y ) = ∑ j , k = 1 ∞ ( − 1 ) j + k + 1 π 2 j k ( 1 + π 2 j 2 + π 2 k 2 ) sin ( π j x ) sin ( π k y ) . {\displaystyle u(x,y)=\sum _{j,k=1}^{\infty }{(-1)^{j+k+1} \over \pi ^{2}jk(1+\pi ^{2}j^{2}+\pi ^{2}k^{2})}\sin(\pi jx)\sin(\pi ky).} <h2 id="maximum-principle">Maximum principle</h2> <p>There are many variants of the maximum principle. We give a simple one. </p><p>Theorem. <i>(Weak maximum principle.) Let u ∈ C 2 ( Ω ) ∩ C 1 ( Ω ¯ ) {\displaystyle u\in C^{2}(\Omega )\cap C^{1}({\bar {\Omega }})} , and assume that c ( x ) = 0 ∀ x ∈ Ω {\displaystyle c(x)=0\;\forall x\in \Omega } . Say that L u ≤ 0 {\displaystyle Lu\leq 0} in Ω {\displaystyle \Omega } . Then max x ∈ Ω ¯ u ( x ) = max x ∈ ∂ Ω u ( x ) {\displaystyle \max _{x\in {\bar {\Omega }}}u(x)=\max _{x\in \partial \Omega }u(x)} . In other words, the maximum is attained on the boundary.</i> </p><p>A strong maximum principle would conclude that u ( x ) ≨ max y ∈ ∂ Ω u ( y ) {\displaystyle u(x)\lneqq \max _{y\in \partial \Omega }u(y)} for all x ∈ Ω {\displaystyle x\in \Omega } unless u {\displaystyle u} is constant. </p> <h2 id="notes">Notes</h2> <ul><li><a href="/facts/Lawrence_C._Evans/0XluMrne">Evans, Lawrence C.</a> (2010). <a href="https://math24.wordpress.com/wp-content/uploads/2013/02/partial-differential-equations-by-evans.pdf"><i>Partial differential equations</i></a> (PDF). <a href="/facts/Graduate_Studies_in_Mathematics/APstozeU">Graduate Studies in Mathematics</a>. Vol. 19 (Second edition of 1998 original ed.). Providence, RI: <a href="/facts/American_Mathematical_Society/fr5gotCt">American Mathematical Society</a>. <a href="/facts/Doi_(identifier)/muM9Etpq">doi</a>:<a href="https://doi.org/10.1090%2Fgsm%2F019">10.1090/gsm/019</a>. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 978-0-8218-4974-3. <a href="/facts/MR_(identifier)/uP137L11">MR</a> <a href="https://mathscinet.ams.org/mathscinet-getitem?mr=2597943">2597943</a>.</li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>Swetz, Faauvel, Bekken, "Learn from the Masters", 1997, MAA ISBN 0-88385-703-0, pp.128-9 <a href="/wiki/ISBN_(identifier)" target="_blank">/wiki/ISBN_(identifier)</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> </ol>