A first-order Neumann closure limits the global rate

For $u(x)=-x^2\mathrm{/}2+(c+1)x$, one has $u^{\mathrm{\prime }\mathrm{\prime }}=-1$, $u(0)=0$, and $u^{\mathrm{\prime }}(1)=-(1)+(c+1)=c$, so $u$ solves the stated mixed-boundary BVP.

Put $d_j=U_j-U_{j-1}$. Each interior equation is equivalent to $d_{j+1}=d_j-h^2$, while the boundary closure sets $d_{N+1}=hc$. Starting from this last difference and applying $d_j=d_{j+1}+h^2$ for $j=N,\dots ,1$ determines every $d_j$; then $U_0=0$ and $U_j=U_{j-1}+d_j$ determine a unique discrete vector.

The exact quadratic samples satisfy each centered interior equation exactly. Their final backward difference is $({\bar{U}}_{N+1}-{\bar{U}}_N)\mathrm{/}h=u^{\mathrm{\prime }}(1)-h{u}^{\mathrm{\prime }\mathrm{\prime }}(1)\mathrm{/}2=c+h\mathrm{/}2$, so the boundary difference error for $e_j:=U_j-{\bar{U}}_j$ is $e_{N+1}-e_N=-h^2\mathrm{/}2$.

Subtracting the identical interior equations gives $2e_j-e_{j-1}-e_{j+1}=0$, hence the increments $e_j-e_{j-1}$ are constant. Since $e_0=0$ and the final increment is $-h^2\mathrm{/}2$ by step 2.1, every increment equals $-h^2\mathrm{/}2$, and therefore $e_j=-j{h}^2\mathrm{/}2=-h{x}_j\mathrm{/}2$ for $0\le j\le N+1$.

Thus ${\max }_{0\le j\le N+1}\mathrm{\mid }{e}_j\mathrm{\mid }=h\mathrm{/}2$. This is not $O(h^2)$ as $N\to \mathrm{\infty }$: such a bound would imply $1\mathrm{/}(2h)\le C$ for all $h=1\mathrm{/}(N+1)$, whereas $1\mathrm{/}(2h)=(N+1)\mathrm{/}2$ is unbounded. The first-order boundary closure has reduced the global rate to first order despite the interior stencil being exact on the quadratic solution.