A two-node semilinear diffusion FAS cycle has an explicit local rate

The continuous model is $-u^{\mathrm{\prime }\mathrm{\prime }}+u^3=0$ on $(0,3)$ with zero endpoint values, and this proof concerns the specified finite-difference residual. The vectors $(1,1{)}^{\mathsf{T}}$ and $(1,-1{)}^{\mathsf{T}}$ are eigenvectors of $A$ with eigenvalues $1$ and $3$, so $z^{\mathsf{T}}Az\ge \mathrm{\parallel }z{\mathrm{\parallel }}_2^2$. For real $s,t$, $(s-t)(s^3-t^3)=(s-t{)}^2(s^2+st+t^2)\ge 0$, since $s^2+st+t^2=((2s+t{)}^2+3t^2)\mathrm{/}4$. Therefore, for $a,b\in {\mathbb{R}}^2$, $⟨a-b,F(a)-F(b)⟩=(a-b{)}^{\mathsf{T}}A(a-b)+{\sum }_{i=1}^2(a_i-b_i)(a_i^3-b_i^3)\ge \mathrm{\parallel }a-b{\mathrm{\parallel }}_2^2$. Since $F(0)=0$, any other root $a$ would give $0=⟨a,F(a)-F(0)⟩\ge \mathrm{\parallel }a{\mathrm{\parallel }}_2^2$, so $u_{\ast }=0$ is the unique root. Direct differentiation gives $DF(0)=A$.

Direct multiplication gives $AP=P$, $P^{\mathsf{T}}AP=2$, and hence $F_H(v)=P^{\mathsf{T}}F(Pv)=2v+2v^3\mathrm{.}$ Writing $m=Iu=(u_1+u_2)\mathrm{/}2$, [F1] gives the coarse target $g(u)=F_H(m)-P^{\mathsf{T}}F(u)=2m^3-u_1^3-u_2^3\mathrm{.}$ The scalar map $\varphi (v)=2v+2v^3$ has ${\varphi }^{\mathrm{\prime }}(v)=2+6v^2\ge 2$ and limits $\pm \mathrm{\infty }$ at $v\to \pm \mathrm{\infty }$, so for every $g$ it has one exact root. An exact direct solver is $v(g)=\frac{2}{\sqrt{3}}\sinh \text{\hspace{0.17em}⁣}(\frac{1}{3}\mathrm{arsinh}\text{\hspace{0.17em}⁣}\left(\frac{3\sqrt{3}}{4}g\right)),$ because $\sinh (3t)=3\sinh (t)+4{\sinh }^3(t)$; thus $\varphi (v(g))=g$. The displayed formula is a composition of smooth real functions, so the selected branch, correction map $C$, and cycle map $E$ are $C^2$ on an open neighborhood of the ball. At $g=0$ the branch has $v=0$ and ${\varphi }^{\mathrm{\prime }}(0)=2$.

The pre/post map has derivative at zero $B_s=DS(0)=I_2-\frac{1}{4}A=\left(\begin{array}{cc}1\mathrm{/}2& 1\mathrm{/}4\\ 1\mathrm{/}4& 1\mathrm{/}2\end{array}\right)\mathrm{.}$ The coarse-branch equation has right side $g(u)$, a cubic with $Dg(0)=0$. Differentiating $\varphi (v(u))=g(u)$ at zero gives $2Dv(0)=Dg(0)=0$. The derivative of the coarse correction is $Q=I_2-PI=I_2-\frac{1}{2}\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right),$ the orthogonal projection onto $(1,-1)$. The symmetric vector $(1,1)$ is an eigenvector of $B_s$ with eigenvalue $3\mathrm{/}4$, and $(1,-1)$ has eigenvalue $1\mathrm{/}4$. Thus for $E=S\circ C\circ S$, [F2] gives $DE(0)=B_{s}Q{B}_s=\frac{1}{16}Q,q=\mathrm{\parallel }DE(0){\mathrm{\parallel }}_2=\frac{1}{16}\mathrm{.}$

It remains to bound the derivative variation. Set $r=1\mathrm{/}100$ and $m(u)=(u_1+u_2)\mathrm{/}2$. On $\mathrm{\parallel }u{\mathrm{\parallel }}_2\le r$, $\mathrm{\mid }m\mathrm{\mid }\le r\mathrm{/}\sqrt{2},\mathrm{\mid }g(u)\mathrm{\mid }<2r^3,\mathrm{\parallel }Dg(u){\mathrm{\parallel }}_2<6r^2,\mathrm{\parallel }{D}^{2}g(u){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}<11r\mathrm{.}$ The first inequality is Cauchy--Schwarz. The second follows from $g=2m^3-u_1^3-u_2^3$ and $\mathrm{\mid }{u}_1{\mathrm{\mid }}^3+\mathrm{\mid }{u}_2{\mathrm{\mid }}^3\le \mathrm{\parallel }u{\mathrm{\parallel }}_2^3$. Differentiating gives $Dg=6m^{2}Dm-3(u_1^2,u_2^2)$ and $D^{2}g=12m(Dm\otimes Dm)-\mathrm{diag}(6u_1,6u_2)$, which give the last two bounds. If $\varphi (v(u))=g(u)$, then $\mathrm{\mid }v\mathrm{\mid }\le \mathrm{\mid }g\mathrm{\mid }\mathrm{/}2<r^3$, $\mathrm{\parallel }Dv\mathrm{\parallel }\le \mathrm{\parallel }Dg\mathrm{\parallel }\mathrm{/}2<3r^2$, and twice differentiating yields $\mathrm{\parallel }{D}^{2}v\mathrm{\parallel }<\frac{1}{2}(11r+12r^3(3r^2{)}^2)=5.5r+54r^7\mathrm{.}$ Writing $C(u)=Qu+Pv(u)$, it follows that $\mathrm{\parallel }DC\mathrm{\parallel }<1.001$ and $\mathrm{\parallel }{D}^{2}C\mathrm{\parallel }<0.08$ on this ball.

Since $\mathrm{\parallel }{B}_s{\mathrm{\parallel }}_2=3\mathrm{/}4$, $\mathrm{\parallel }S(u){\mathrm{\parallel }}_2\le 3r\mathrm{/}4+r^3\mathrm{/}4<r$ for $u$ in the radius-$r$ ball. Orthogonality of $Qu$ and $Pv(u)$ gives $\mathrm{\parallel }C(S(u)){\mathrm{\parallel }}_2^2\le r^2+2r^6<4r^2$. On the radius-$2r$ ball, $\mathrm{\parallel }DS{\mathrm{\parallel }}_2<0.751$ and $\mathrm{\parallel }{D}^{2}S{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 0.03$; on the radius-$r$ ball, $\mathrm{\parallel }{D}^{2}S{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 0.015$. The formula [F5] for $E=S\circ C\circ S$ therefore gives $\mathrm{\parallel }{D}^{2}E(u){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 0.03(1.001\cdot 0.751{)}^2+0.751(0.08\cdot {0.751}^2+1.001\cdot 0.015)<0.07<1.$ For each unit $w\in {\mathbb{R}}^2$, apply [F4] on the open radius-$r$ ball to the map $u\mapsto DE(u)w$; the Hessian bound gives derivative norm less than $1$. By continuity the resulting Lipschitz bound extends to the closed ball, so $\mathrm{\parallel }DE(x)-DE(y){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \mathrm{\parallel }x-y{\mathrm{\parallel }}_2$ there.

The pre/post map $S$ fixes zero because $F(0)=0$. At $u=0$, [F1] gives restricted value $Iu=0$ and coarse target $F_H(0)=0$; by step 1.2, the exact selected coarse root is $v=0$, so the correction is zero and $E(0)=0$. By steps 1.3 and 2.1, $q=1\mathrm{/}16$, $L=1$, and $r=1\mathrm{/}100$, hence $q+Lr\mathrm{/}2=\frac{1}{16}+\frac{1}{200}=\frac{27}{400}<1.$ Applying [F3] proves invariance of the ball and the stated local error rate. This single fixed mesh example makes no mesh-independent claim.