 Solution

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.