Nonlinear full-approximation multigrid — Examples

Direct multiplication gives $A{u}_{\ast }=(2,-1)$ and $(u_{\ast ,1}^3,u_{\ast ,2}^3)=(1,0)$, so $f_h=F_h(u_{\ast })=(3,-1)$. The restricted iterate is $I{u}_{\ast }=1\mathrm{/}2$, while $R{f}_h=3-1=2$.

The Galerkin coarse value at the restricted iterate is $F_H(1\mathrm{/}2)=1+1\mathrm{/}4=5\mathrm{/}4$. Thus [F1] gives $\tau (u_{\ast })=F_H(I{u}_{\ast })-R{F}_h(u_{\ast })=5\mathrm{/}4-2=-3\mathrm{/}4$ and the tau-corrected coarse right side is $R{f}_h+\tau (u_{\ast })=2-3\mathrm{/}4=5\mathrm{/}4$.

Since $F_H(1\mathrm{/}2)=5\mathrm{/}4$, the exact selected coarse solution is $v=I{u}_{\ast }=1\mathrm{/}2$. Therefore $P(v-I{u}_{\ast })=0$ and the fine correction leaves $u_{\ast }$ unchanged; the pre/post residual smoothers also leave it fixed because $F_h(u_{\ast })=f_h$. This is the instance of [F2] with the actual tau value verified above.

Set $u_{\ast }=(1,0{)}^{\mathsf{T}}$. Then $f_h=A{u}_{\ast }=(2,-1{)}^{\mathsf{T}}$, so $u_{\ast }$ is an exact fine root. Its restriction is $I{u}_{\ast }=1\mathrm{/}2$ and the restricted right side is $R{f}_h=2-1=1$.

With tau omitted, the coarse equation is $3v=R{f}_h=1$, whose exact solution is $v=1\mathrm{/}3$. The fine correction is $u_{\ast }+P(v-I{u}_{\ast })=(1,0{)}^{\mathsf{T}}+(1,1{)}^{\mathsf{T}}(1\mathrm{/}3-1\mathrm{/}2)=(5\mathrm{/}6,-1\mathrm{/}6{)}^{\mathsf{T}},$ which is not $u_{\ast }$. Thus the false claim fails for this SPD fine system and this specified rediscretized coarse operator.

For the same inputs, the correct tau is $F_H(I{u}_{\ast })-R{f}_h=3\mathrm{/}2-1=1\mathrm{/}2$, so the tau-corrected coarse right side is $3\mathrm{/}2$ and its selected root is $v=1\mathrm{/}2=I{u}_{\ast }$. The resulting correction is zero, in accord with [F2].

Let $P(v)=(0,v{)}^{\mathsf{T}}$. For any pre-smoothed $(x,y)$, [F1] gives $g_H=F_H(y)+R(-F_{\mu }(x,y))=y-y=0.$ The exact coarse solution is $v=0$, so prolonging $v-I(x,y)=-y$ sets the second fine coordinate to zero. The following post-sweep leaves it zero. Hence the actual cycle map is $E_{\mu }(x,y)=(s_{\mu }(s_{\mu }(x)),0)\mathrm{.}$

Write $a=1-\mu \mathrm{/}2$ and $r=r_{\mu }\le 0.1$. For $\mathrm{\mid }x\mathrm{\mid }\le r$, $\mathrm{\mid }{s}_{\mu }(x)\mathrm{\mid }\le r+\frac{1}{2}{r}^3\le 1.005r,\mathrm{\mid }{s}_{\mu }^{\mathrm{\prime }}(x)\mathrm{\mid }\le 1.015,\mathrm{\mid }{s}_{\mu }^{\mathrm{\prime }}(s_{\mu }(x))\mathrm{\mid }\le 1.02,$ and $\mathrm{\mid }{s}_{\mu }^{\mathrm{\prime }\mathrm{\prime }}(x)\mathrm{\mid }\le 3r$, $\mathrm{\mid }{s}_{\mu }^{\mathrm{\prime }\mathrm{\prime }}(s_{\mu }(x))\mathrm{\mid }\le 3.015r$. By [F4], $(s_{\mu }\circ s_{\mu }{)}^{\mathrm{\prime }\mathrm{\prime }}=s_{\mu }^{\mathrm{\prime }\mathrm{\prime }}(s_{\mu })(s_{\mu }^{\mathrm{\prime }}{)}^2+s_{\mu }^{\mathrm{\prime }}(s_{\mu })s_{\mu }^{\mathrm{\prime }\mathrm{\prime }}$, so these bounds give $\mathrm{\mid }(s_{\mu }\circ s_{\mu }{)}^{\mathrm{\prime }\mathrm{\prime }}(x)\mathrm{\mid }<7r$. The derivative matrix of $E_{\mu }$ is $\mathrm{diag}((s_{\mu }\circ s_{\mu }{)}^{\mathrm{\prime }}(x),0)$, so [F3] gives $\mathrm{\parallel }D{E}_{\mu }(z)-D{E}_{\mu }(z^{\mathrm{\prime }}){\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le 7r\mathrm{\parallel }z-z^{\mathrm{\prime }}{\mathrm{\parallel }}_2$ on the ball.

Since $F_{\mu }(x,y)=0$ implies $y=0$ and $x(\mu +x^2)=0$, for $\mu >0$ the unique fine root is $(0,0)$. Also $s_{\mu }^{\mathrm{\prime }}(0)=1-\mu \mathrm{/}2$, so differentiating the two sweeps gives $D{E}_{\mu }(0,0)=\left(\begin{array}{cc}(1-\mu \mathrm{/}2{)}^2& 0\\ 0& 0\end{array}\right),q_{\mu }=(1-\mu \mathrm{/}2{)}^2\mathrm{.}$

With $r^2=\mu \mathrm{/}100$, $q_{\mu }+\frac{(7r)r}{2}=(1-\mu \mathrm{/}2{)}^2+\frac{7\mu }{200}=1-\frac{193}{200}\mu +\frac{1}{4}{\mu }^2\le 1-\frac{143}{200}\mu <1.$ Thus [F2] applies on this ball. Its certified radius $r_{\mu }=\sqrt{\mu \mathrm{/}100}$ tends to zero as $\mu \downarrow 0$; at $\mu =0$, $q_0=1$ and the strict derivative hypothesis fails. For $\mu <0$, the scalar equation $\mu x+x^3=0$ has the additional roots $x=\pm \sqrt{-\mu }$, which is the pitchfork bifurcation.

Direct multiplication gives $AP=(1,1{)}^{\mathsf{T}}$ and $A_H=P^{\mathsf{T}}AP=2$. Write $m=Iu$ and $r=f-Au$. The FAS coarse target is $A_{H}m+Rr=2m+r_1+r_2\mathrm{.}$ The exact coarse solution is therefore $v=m+(r_1+r_2)\mathrm{/}2$.

Since $A_H^{-1}Rr=(r_1+r_2)\mathrm{/}2$, the coarse difference is $v-Iu=A_H^{-1}R(f-Au)$ and the fine update is $u+P(v-Iu)=u+P{A}_H^{-1}{P}^{\mathsf{T}}(f-Au),$ the standard Galerkin correction. With matching residual-correction smoothers, [F2] identifies the full V/W iterate as well.