Linear numerical boundary-value problems — Examples

Differentiating $u(x)=x(1-x)\mathrm{/}2$ gives $u^{\mathrm{\prime }\mathrm{\prime }}(x)=-1$, so $-u^{\mathrm{\prime }\mathrm{\prime }}=1$; also $u(0)=u(1)=0$.

Let $u_j=u(x_j)$. Since $u$ is quadratic, direct substitution gives $-u_{j-1}+2u_j-u_{j+1}=h^2$ for every interior $j$; its two endpoint values are zero. Thus the samples satisfy the system in [F1].

By [F2] the system has a unique solution, so the algorithm's output equals $(u_1,\dots ,u_N)$ and its nodal error is exactly zero for every $N\ge 1$.

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.

The row-sum expression is nonnegative and equals zero only when every matrix entry is zero; $\mathrm{\parallel }\lambda B{\mathrm{\parallel }}_{\mathrm{r}\mathrm{o}\mathrm{w},\mathrm{\infty }}=\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{r}\mathrm{o}\mathrm{w},\mathrm{\infty }}$; and [F3] applied entrywise gives $\mathrm{\parallel }B+C{\mathrm{\parallel }}_{\mathrm{r}\mathrm{o}\mathrm{w},\mathrm{\infty }}\le \mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{r}\mathrm{o}\mathrm{w},\mathrm{\infty }}+\mathrm{\parallel }C{\mathrm{\parallel }}_{\mathrm{r}\mathrm{o}\mathrm{w},\mathrm{\infty }}$. Hence it is a matrix norm.

At $N=3$, $h=1\mathrm{/}4$ and [F1] gives $A_h=16T$ with $T=\left(\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 2\end{array}\right)$. Direct multiplication gives $A_h^{-1}=\frac{1}{64}\left(\begin{array}{ccc}3& 2& 1\\ 2& 4& 2\\ 1& 2& 3\end{array}\right)$, since the displayed integer matrix multiplied by $T$ is $4I_3$.

Its value equals the induced infinity operator norm ${\sup }_{\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1}\mathrm{\parallel }Bz{\mathrm{\parallel }}_{\mathrm{\infty }}$: for every such $z$, each row obeys $\mathrm{\mid }(Bz{)}_i\mathrm{\mid }\le {\sum }_j\mathrm{\mid }{b}_{ij}\mathrm{\mid }$, while a row $i_0$ attaining the finite maximum is attained by the vector $z_j=\mathrm{sgn}(b_{i_{0}j})$, with $\mathrm{sgn}(0)=0$.

By [F2], $\mathrm{\parallel }{A}_h^{-1}z{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1\mathrm{/}8$ whenever $\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$. Taking the supremum and using step 2.1 shows $\mathrm{\parallel }{A}_h^{-1}{\mathrm{\parallel }}_{\mathrm{r}\mathrm{o}\mathrm{w},\mathrm{\infty }}\le 1\mathrm{/}8$ for every $N$.

The three absolute row sums of this inverse are $6\mathrm{/}64$, $8\mathrm{/}64$, and $6\mathrm{/}64$, so its row-sum norm is $8\mathrm{/}64=1\mathrm{/}8$. Along with the uniform bound in step 3.1, this proves the stated example and shows its constant is attained.