= Solution
{c}
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 $L_h\phi=2+2=4$. At $x=1$, the reflected horizontal part is
$$
\frac{2[\phi(1-h,y)-\phi(1,y)]}{h^2}=\frac4h+2,
$$
and the vertical part is two. Thus $L_h\phi=4+4/h$ there. Part (c) therefore permits
$$
C=\max\left\{\frac{K_1h^2}4,\frac{K_2h}{4+4/h}\right\}
=\frac{h^2}4\max\left\{K_1,\frac{K_2}{1+h}\right\}
\leq\frac{h^2}4\max\{K_1,K_2\}.
$$
On the Dirichlet boundary, the largest barrier value is $4+1/4=17/4$. Consequently
$$
\boxed{E\leq\frac{17}{16}\max\{K_1,K_2\}h^2,\qquad K_3=\frac{17}{16}\max\{K_1,K_2\}.}
$$
The $O(h)$ <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.
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