Past exam of the numerical solution of differential equations course of the University of Oxford / 2011 / B / 5

The five-point equations for the Poisson equation arewith prescribed boundary values. If $L_h\varphi \le 0$, an interior maximal value is at most the average of its four neighbours. Since no neighbour exceeds the maximum, equality forces all four to equal it. Repeating along adjacent nodes propagates that value to the boundary. Thus the discrete maximum principle gives $\max \varphi \le {\max }_{\partial {\Omega }_h}\varphi$, in particular the usual positive-maximum bound with zero included.

For an $O(h^2)$ error estimate assume the exact solution has bounded continuous fourth partial derivatives up to the boundary. The local truncation error $\tau =L_{h}u-f$ then satisfies $\Vert \tau {\Vert }_{\infty }\le C{h}^2$ by the Taylor theorem. Put $e=u{|}_{{\Omega }_h}-U$, so $L_{h}e=\tau$ and $e=0$ on boundary nodes. The Discrete Poisson barrier on a squaresatisfies $L_{h}B=1$ exactly, $B\ge 0$ on the grid boundary, and $B\le 1/8$. Set $\epsilon =\Vert \tau {\Vert }_{\infty }$. Then $L_h(\pm e-\epsilon B)\le 0$, with nonpositive boundary values. Applying the discrete maximum principle twice givesFor nonzero smooth Dirichlet boundary data $g$, impose $U=g$ at the boundary nodes. The error still vanishes there, so the same argument and order apply, under the same fourth-derivative regularity of the exact solution. Smooth boundary values alone do not remove every possible corner-regularity issue.

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.