Past exam of the numerical solution of differential equations course of the University of Oxford / 2010 / B / 5

The coefficient signs alone do not imply the assertion as printed. Two interior nodes can have rows $L_h{\psi }_1={\psi }_2-{\psi }_1$ and $L_h{\psi }_2={\psi }_1-{\psi }_2$, with a separate boundary node. Taking interior values one and boundary value zero satisfies every stated coefficient inequality and $L_h\psi =0$, but its maximum is not on the boundary. Connectedness of the geometric domain does not ensure connectedness of this stencil graph.

The usual extra assumption is that every interior node has a path of positive stencil coefficients to a prescribed boundary node. Under this assumption let $M\ge 0$ be the global maximum and suppose an interior node $p$ attains it. ThenAll summands are nonpositive, so equality forces ${\psi }_r=M$ at every neighbor with $c_{p,r}>0$. Propagate along a path to the boundary to find a boundary node with value $M$. This proves the discrete maximum principle under boundary-reaching stencil connectivity, and exhibits why the literal hypotheses need that qualification.

A diameter-only bound also needs the normalization of a Laplacian stencil, beyond its coefficient signs. For example, a single interior row $L_h{e}_p=\epsilon (e_a-e_p)$ with $e_a=0$ allows $|L_h{e}_p|=1$ and $|e_p|=1/\epsilon$ on a domain of fixed diameter. Thus arbitrary rescaling of the printed coefficients rules out a universal diameter-only estimate.

For the intended normalized Laplacian stencil, assume $L_{h}1=0$ and $L_h\varphi =4$ for ${\varphi }_p=|x_p-x_0{|}^2$, with $x_0\in \Omega$, as holds for the standard two-dimensional central stencil. Let $D=\mathrm{diam}\Omega$. Then $0\le {\varphi }_p\le D^2$. Use the truncation error convention $T_p=L_h{u}_p-f_p$, so $L_h{e}_p=T_p$; reversing the convention does not change the absolute-value argument.

The two functions ${\psi }_p^{\pm }=\pm e_p+(T/4){\varphi }_p$ satisfy $L_h{\psi }^{\pm }\ge -T+T=0$. On the prescribed boundary $e_a=0$, so their maximum boundary values are at most $T{D}^2/4$. Apply the discrete maximum principle from part (a), with its boundary-connectivity assumption. Since $\varphi \ge 0$,This is the standard quadratic-barrier error estimate for a discrete Laplacian. The calculation identifies exactly the additional normalization needed for the bound intended by the question.

The displayed estimate requires a nonnegative barrier. As printed, a counterexample is obtained with two interior nodes, each having row $L_h{v}_p=v_a-v_p$ to one boundary node. Set $e_a=0$, $e_p=1$, and ${\varphi }_a=-1$, ${\varphi }_p=-2$. Both interior rows have $|T_p|=1$ and $L_h{\varphi }_p=1$. Put one node in each of $J_1,J_2$ and take $T=S=d_1=d_2=1$. Then the proposed right-hand side is $-1$ while $E=1$.

For the intended use, suppose ${\varphi }_p\ge 0$ and the stencil has the maximum principle established in part (a). Set $C=\max \{T/d_1,S/d_2\}$. On either subregion,At boundary nodes these functions equal $C{\varphi }_a$. Hencewhich proves the intended discrete elliptic barrier estimateFor a normalized stencil with $L_{h}1=0$ and a sign-changing barrier, shifting it by its minimum gives the valid general replacement $E\le C({\max }_a{\varphi }_a-{\min }_p{\varphi }_p)$. The supplied barrier in part (d) is positive, so the printed estimate applies there without a sign problem.

The given stencil has nonnegative neighbor weights, zero row sums and a path from every unknown node to the Dirichlet boundary. Thus the discrete maximum principle applies, including at the reflected Neumann edge. The barrier is positive throughout the square.

At an ordinary interior node, central differences differentiate this quadratic exactly, giving $L_h\varphi =2+2=4$. At $x=1$, the reflected horizontal part isand the vertical part is two. Thus $L_h\varphi =4+4/h$ there. Part (c) therefore permitsOn the Dirichlet boundary, the largest barrier value is $4+1/4=17/4$. ConsequentlyThe $O(h)$ truncation error at the Neumann edge is divided by the barrier response $O(h^{-1})$, producing a contribution of order $h^2$. Thus its lower local order does not reduce this global maximum-norm error bound.