Elliptic finite-difference methods — Examples

For the exact samples $u_i=x_i^2$, $D_h^{+}{u}_0=(h^2-0)\mathrm{/}h=h$, whereas $u^{\mathrm{\prime }}(0)=0$. Thus the imposed Neumann row has residual exactly $h$.

Define $V_i=x_i^2+h(1-x_i)$. Then $V_0=h$, $V_1=h$, so $D_h^{+}{V}_0=0$; also $V_N=1$. Since the second difference of the quadratic part is $2$ and the second difference of the affine part is $0$, $-{\delta }_h^2{V}_i=-2$ at every interior node. Hence $V$ satisfies [A1].

If two grid solutions existed, their difference $d$ would satisfy ${\delta }_h^2{d}_i=0$ for $1\le i<N$, $D_h^{+}{d}_0=0$, and $d_N=0$. The recurrence says $d_i-d_{i-1}$ is constant in $i$; the forward boundary row makes that constant $0$, and $d_N=0$ then makes every $d_i=0$. Therefore the solution in step 1.2 is unique.

The error is $V_i-u_i=h(1-x_i)$. Because $0\le x_i\le 1$, its maximum absolute value is $h$, attained at $i=0$. The boundary residual and global error are both first order in $h$.

Set $x_i=ih$ and ${\varphi }_i=x_i(1-x_i)\mathrm{/}2$ for $0\le i\le N$. Then ${\varphi }_0={\varphi }_N=0$, $-{\delta }_h^2{\varphi }_i=1$, so $A_N\varphi =\mathbf{1}$, and $0\le {\varphi }_i\le 1\mathrm{/}8$ because $x(1-x)\le 1\mathrm{/}4$ on $[0,1]$.

If $w_0=w_N=0$ and $A_{N}w\ge 0$ at every interior index, a negative interior minimum $w_k<0$ would give $(A_{N}w{)}_k=(2w_k-w_{k-1}-w_{k+1})\mathrm{/}{h}^2\le 0$. Equality is then forced, so both neighboring values equal $w_k$; repeating this along the finite grid reaches a zero endpoint, a contradiction. Therefore $w_i\ge 0$ everywhere.

The kernel of $A_N$ is zero: if $A_{N}v=0$ and $v$ has a positive interior maximum, the equation forces both neighboring values to equal that maximum, which propagates to a zero endpoint; applying the same argument to $-v$ rules out a negative minimum. Thus $v=0$, and the square finite matrix is invertible.

Let $g$ be any right-hand side, let $v=A_N^{-1}g$, and put $M=\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}$. If $M=0$, then $v=0$. If $M>0$, $A_N(M\varphi +v)=M\mathbf{1}+g\ge 0$ and $A_N(M\varphi -v)=M\mathbf{1}-g\ge 0$; step 1.2 makes both grid functions nonnegative. Hence $\mathrm{\mid }{v}_i\mathrm{\mid }\le M{\varphi }_i\le M\mathrm{/}8$ for each $i$, proving $\mathrm{\parallel }{A}_N^{-1}g{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}8$.

At $x=x_i$ and $h=h_x$, the two midpoint coefficients are $a_{+}=1+(x+h\mathrm{/}2{)}^2$ and $a_{-}=1+(x-h\mathrm{/}2{)}^2$, while the one-sided differences of $x^2$ are $D_{+}u=2x+h$ and $D_{-}u=2x-h$. Therefore $D_h^{\mathrm{f}\mathrm{l}\mathrm{u}\mathrm{x}}u=(a_{+}(2x+h)-a_{-}(2x-h))\mathrm{/}h=2+6x^2+h^2\mathrm{/}2$.

The centered differences are ${\delta }_{x,h}^{2}u=2$ and $D_{x,h}^{0}u=2x$, while $a(x,y)=1+x^2$ and $a_x(x,y)=2x$. Hence $D_h^{\mathrm{e}\mathrm{x}\mathrm{p}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{e}\mathrm{d}}u=2(1+x^2)+(2x)(2x)=2+6x^2$, so subtracting from step 1.1 gives exactly $h_x^2\mathrm{/}2$.

For every interior node, $D_x^{+}{D}_x^{-}{x}_i^2=2$ and $D_y^{+}{D}_y^{-}(y_j^2)=2$; the other coordinate second differences vanish. Thus $(A_h^1{u}^{\ast }{)}_{ij}=-(2+2)=-4=f(x_i,y_j)$.

The sampled values $u_{ij}^{\ast }$ also satisfy the exact Dirichlet boundary rows, so they solve the full discrete system.

Any two discrete solutions have a difference in $X_{h,0}$ in the kernel of $A_h^1$. By [F2] that kernel is zero, so the solution is unique and equals $u^{\ast }$.