Céa and Aubin–Nitsche error estimates — Examples

For $x=(x_1,x_2)$, $Ax=(x_1-x_2,x_1+x_2)$, so $\mathrm{\parallel }Ax{\mathrm{\parallel }}_2^2=2\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$ and $\mathrm{\mid }a(x,y)\mathrm{\mid }\le \sqrt{2}\mathrm{\parallel }x{\mathrm{\parallel }}_2\mathrm{\parallel }y{\mathrm{\parallel }}_2$. Also $a(x,x)=x_1^2+x_2^2$, so the coercivity constant is $\alpha =1$. Finally, $a(e_1,e_2)=1$ and $a(e_2,e_1)=-1$, so the form is not symmetric.

Since $u=e_2$ and $F(v)=a(e_2,v)$, the continuous variational equation holds for every $v\in {\mathbb{R}}^2$. Write $u_h=c{e}_1$. Testing the discrete equation with $e_1$ gives $ca(e_1,e_1)=a(e_2,e_1)$, hence $c=-1$ because $a(e_1,e_1)=1$ and $a(e_2,e_1)=-1$. Thus $u_h=-e_1$.

For $v_h=t{e}_1$, $\mathrm{\parallel }u-v_h{\mathrm{\parallel }}_2^2=\mathrm{\parallel }{e}_2-t{e}_1{\mathrm{\parallel }}_2^2=1+t^2$, whose infimum is $1$ at $t=0$. Meanwhile $u-u_h=e_1+e_2$, so $\mathrm{\parallel }u-u_h{\mathrm{\parallel }}_2=\sqrt{2}$. Therefore the ratio is $\sqrt{2}=M\mathrm{/}\alpha$, and the constant in [F1] cannot be decreased in general.

The function $\varphi$ is absolutely continuous, vanishes at $0$ and $1$, and has derivative $2$ on $(0,1\mathrm{/}2)$ and $-2$ on $(1\mathrm{/}2,1)$. Its derivative is in $L^2$, so [F2] gives $\varphi \in H_0^1(0,1)$.

Put $p(x)=\frac{1}{2}x(1-x)$. It is absolutely continuous with zero endpoint values, $p^{\mathrm{\prime }}=\frac{1}{2}-x\in L^2$, and $p^{\mathrm{\prime }\mathrm{\prime }}=-1$; [F2] gives $p\in H_0^1(0,1)$.

Poincar'e constant for this interval. Let $v\in H_0^1(0,1;\mathbb{K})$ and $\delta >0$. By [F7], choose $\phi \in C_c^{\mathrm{\infty }}(0,1;\mathbb{K})$ with $\mathrm{\parallel }v-\phi {\mathrm{\parallel }}_{H^1}<\delta$. For $0<x\le 1$, apply the real second fundamental theorem to the real and imaginary parts of $\phi$ on $[0,x]$; their derivatives are continuous and hence Riemann integrable by [F8]. The Riemann--Lebesgue equality in [F8] and the definition of the integral over a set give $\phi (x)=({\phi }^{\mathrm{\prime }},{\mathbf{1}}_{[0,x]}{)}_{L^2(0,1)}$, since $\phi (0)=0$. By [F9], $\mathrm{\parallel }{\mathbf{1}}_{[0,x]}{\mathrm{\parallel }}_2=\sqrt{x}$, so [F4] gives $\mathrm{\mid }\phi (x)\mathrm{\mid }\le \sqrt{x}\mathrm{\parallel }{\phi }^{\mathrm{\prime }}{\mathrm{\parallel }}_2\le \mathrm{\parallel }{\phi }^{\mathrm{\prime }}{\mathrm{\parallel }}_2$; this also holds at $x=0$. Integrating the squared bound over $(0,1)$ and using ${\lambda }_1((0,1))=1$ and [F10] yields $\mathrm{\parallel }\phi {\mathrm{\parallel }}_2\le \mathrm{\parallel }{\phi }^{\mathrm{\prime }}{\mathrm{\parallel }}_2$. Therefore $\mathrm{\parallel }v{\mathrm{\parallel }}_2\le \mathrm{\parallel }\phi {\mathrm{\parallel }}_2+\delta \le \mathrm{\parallel }{\phi }^{\mathrm{\prime }}{\mathrm{\parallel }}_2+\delta \le \mathrm{\parallel }{v}^{\mathrm{\prime }}{\mathrm{\parallel }}_2+2\delta$. Since $\delta$ is arbitrary, $\mathrm{\parallel }v{\mathrm{\parallel }}_2\le \mathrm{\parallel }{v}^{\mathrm{\prime }}{\mathrm{\parallel }}_2$ for every $v\in H_0^1(0,1;\mathbb{K})$, so the required Poincar'e constant is $C_P=1$.

For $v\in C_c^{\mathrm{\infty }}(0,1)$, apply [F5] to $p^{\mathrm{\prime }}$ with test $\bar{v}$. Since $p^{\mathrm{\prime }\mathrm{\prime }}=-1$, it gives $a(p,v)={\int }_0^1{p}^{\mathrm{\prime }}\overline{v^{\mathrm{\prime }}}dx=-{\int }_0^1{p}^{\mathrm{\prime }\mathrm{\prime }}\bar{v}dx={\int }_0^1\bar{v}dx$. Both sides are continuous in the $H^1$ norm by [F4]. Since $H_0^1$ is the $W^{1,2}$-closure of these tests, the identity extends to all $v\in H_0^1(0,1)$ by [F7]. By [F6], the datum $v\mapsto {\int }_0^1\bar{v}$ lies in $H^{-1}(0,1)$. The interval $I$ is open and nonempty, is bounded in direction $e=1$ between $a=0$ and $b=1$, and step 1.3 supplies $C_P=1$; [A1] supplies the required axioms. Thus [F3] applies and makes $p$ the unique weak solution.

Direct integration gives $a(\varphi ,\varphi )={\int }_0^{1\mathrm{/}2}4dx+{\int }_{1\mathrm{/}2}^{1}4dx=4,{\int }_0^1\varphi (x)dx=\frac{1}{2}\mathrm{.}$ For $u_h=c\varphi$, the Galerkin equation tested with $\varphi$ is $4c={\int }_0^1\varphi =\frac{1}{2}$, hence $c=\frac{1}{8}$ and $u_h=\frac{1}{8}\varphi$.

On the first half interval, $(p-u_h{)}^{\mathrm{\prime }}=\frac{1}{4}-x$; on the second, $(p-u_h{)}^{\mathrm{\prime }}=\frac{3}{4}-x$. Therefore $\mathrm{\parallel }p-u_h{\mathrm{\parallel }}_a^2={\int }_0^{1\mathrm{/}2}(\frac{1}{4}-x{)}^{2}dx+{\int }_{1\mathrm{/}2}^1(\frac{3}{4}-x{)}^{2}dx=\frac{1}{96}+\frac{1}{96}=\frac{1}{48}\mathrm{.}$ By [F1], with the Hilbert and coercivity hypotheses verified in [F3] and the Hermitian property in [F4], this computed Galerkin solution is the energy-best approximation.

Put $\alpha =3\pi \mathrm{/}2$ and $\lambda =2\mathrm{/}3$. The domain is bounded Lipschitz: at the origin, the rotation $s=(x_1+x_2)\mathrm{/}\sqrt{2}$, $t=(x_2-x_1)\mathrm{/}\sqrt{2}$ writes it locally as $t>-\mathrm{\mid }s\mathrm{\mid }$; every other boundary point lies on a line segment or at a right-angle corner, where a rotation gives a Lipschitz graph. Define $\chi (r)=1-\sigma (4r-1)$, so $\chi =1$ for $r\le 1\mathrm{/}4$ and $\chi =0$ for $r\ge 1\mathrm{/}2$, and set $\mathrm{\Phi }(r,\theta )=r^{\lambda }\sin (\lambda \theta )$ and $u=\chi \mathrm{\Phi }$. Polar differentiation gives ${\mathrm{\partial }}_{rr}\mathrm{\Phi }=\lambda (\lambda -1)r^{\lambda -2}\sin (\lambda \theta )$ and $\mathrm{\Delta }\mathrm{\Phi }=0$. Also $\mathrm{\mid }u\mathrm{\mid }\le C{r}^{2\mathrm{/}3}$ and $\mathrm{\mid }\mathrm{\nabla }u\mathrm{\mid }\le C{r}^{-1\mathrm{/}3}$ near the corner. By [F5], ${\int }_{r<1\mathrm{/}4}(\mathrm{\mid }u{\mathrm{\mid }}^2+\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^2)dx\le C{\int }_0^{1\mathrm{/}4}(r^{4\mathrm{/}3}+r^{-2\mathrm{/}3})rdr<\mathrm{\infty }$; on the remaining support $u$ is smooth with bounded derivatives by [F3, F4]. Thus $u$ and its classical first derivatives are in $L^2(\mathrm{\Omega })$.

Let ${ϵ}_n=1\mathrm{/}(100(n+1))$, ${\eta }_n(r)=\sigma ((r-{ϵ}_n)\mathrm{/}{ϵ}_n)$, and ${\kappa }_n(\theta )=\sigma ((\theta -{ϵ}_n)\mathrm{/}{ϵ}_n)\sigma ((\alpha -\theta -{ϵ}_n)\mathrm{/}{ϵ}_n)$. Then $u_n={\eta }_n{\kappa }_{n}u$ belongs to $C_c^{\mathrm{\infty }}(\mathrm{\Omega })$: it vanishes for $r\le {ϵ}_n$, for $r\ge 1\mathrm{/}2$, and within angular distance ${ϵ}_n$ of either boundary ray. The cutoff derivatives satisfy $\mathrm{\mid }\mathrm{\nabla }{\eta }_n\mathrm{\mid }\le C\mathrm{/}{ϵ}_n$ and $\mathrm{\mid }\mathrm{\nabla }{\kappa }_n\mathrm{\mid }\le C\mathrm{/}({ϵ}_{n}r)$ on their transition regions by [F3, F4].

The radial cutoff region contributes $O({ϵ}_n^{10\mathrm{/}3})$ to the squared function error and $O({ϵ}_n^{4\mathrm{/}3})$ to the squared gradient error; also ${\int }_{{ϵ}_n<r<2{ϵ}_n}\mathrm{\mid }u\mathrm{\nabla }{\eta }_n{\mathrm{\mid }}^2=O({ϵ}_n^{4\mathrm{/}3})$. In the angular strips of width $2{ϵ}_n$, $\mathrm{\mid }u\mathrm{\mid }\le C{r}^{2\mathrm{/}3}\min (\theta ,\alpha -\theta )$ and $\mathrm{\mid }\mathrm{\nabla }u\mathrm{\mid }\le C{r}^{-1\mathrm{/}3}$; hence $\mathrm{\mid }u\mathrm{\nabla }{\kappa }_n\mathrm{\mid }\le C{r}^{-1\mathrm{/}3}$. Polar integration [F5] bounds the squared function and gradient errors on those strips by $O({ϵ}_n)$. These estimates give $u_n\to u$ in $H^1(\mathrm{\Omega })$, so [F1] gives $u\in H_0^1(\mathrm{\Omega })$.

Define $f=-\mathrm{\Delta }u$ on $\mathrm{\Omega }$. Since $\chi =1$ and $\mathrm{\Phi }$ is harmonic for $r<1\mathrm{/}4$, while the derivatives of $\chi$ are supported in the compact annulus $1\mathrm{/}4\le r\le 1\mathrm{/}2$, $f$ is bounded and supported in that annulus. Thus $f\in L^2(\mathrm{\Omega })$.

On $0<r<1\mathrm{/}4$, the classical radial second derivative is a unit-direction contraction of the Cartesian Hessian. If $u\in H^2(\mathrm{\Omega })$, [F10] identifies its weak Hessian with the classical Hessian on $\mathrm{\Omega }$, and $\mathrm{\mid }{e}_r^{\mathsf{T}}{D}^{2}u{e}_r\mathrm{\mid }\le \mathrm{\mid }{D}^{2}u{\mathrm{\mid }}_{\mathrm{F}}$ makes the radial derivative square-integrable. But [F5] gives ${\int }_{0<r<1\mathrm{/}4}\mathrm{\mid }{\mathrm{\partial }}_{rr}u{\mathrm{\mid }}^{2}dx=\mathrm{\mid }\lambda (\lambda -1){\mathrm{\mid }}^2({\int }_0^{3\pi \mathrm{/}2}{\sin }^2(\lambda \theta )d\theta )\left({\int }_0^{1\mathrm{/}4}{r}^{2\lambda -3}dr\right)=\mathrm{\infty },$ since the angular integral is $3\pi \mathrm{/}4$ and $2\lambda -3=-5\mathrm{/}3<-1$. This contradicts square-integrability of the Hessian, so $u\mathrm{\notin }{H}^2(\mathrm{\Omega })$.

Set $F(v)={\int }_{\mathrm{\Omega }}f\bar{v}$. By [F6] and $\mathrm{\parallel }v{\mathrm{\parallel }}_{L^2}\le \mathrm{\parallel }v{\mathrm{\parallel }}_{H^1}$, $F$ is bounded and conjugate-linear, so $F\in H^{-1}(\mathrm{\Omega })$ by [F9]. For every test $v\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$, support is away from $\mathrm{\partial }\mathrm{\Omega }$ and classical integration by parts gives ${\int }_{\mathrm{\Omega }}\mathrm{\nabla }u\cdot \overline{\mathrm{\nabla }v}=F(v)$. [F10] gives the weak-derivative interpretation, and both sides are continuous in the $H^1$ norm by [F6]; density [F1] extends the identity to all $v\in H_0^1(\mathrm{\Omega })$. Thus $u$ is a zero-Dirichlet weak solution by [F8].

Let $\tilde{u}\in H_0^1(\mathrm{\Omega })$ be any other zero-Dirichlet weak solution for this $f$. By [F8], subtracting the two variational identities and testing with $w=u-\tilde{u}$ gives $\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_{L^2}^2=0$ by [F6]. Since $\mathrm{\Omega }\subset \{-1<x_1<1\}$, [F7] gives $\mathrm{\parallel }w{\mathrm{\parallel }}_{L^2}\le 2\mathrm{\parallel }\mathrm{\nabla }w{\mathrm{\parallel }}_{L^2}=0$. Thus $u=\tilde{u}$ as a Sobolev class, so an $H^2$ weak solution would force $u\in H^2(\mathrm{\Omega })$.

Step 4.1 shows that no other weak solution can lie in $H^2$, while step 2.3 excludes $u$ itself. Thus this $L^2$ forcing on a bounded Lipschitz domain has no $H^2$ zero-Dirichlet weak solution, refuting the universal statement. The Aubin–Nitsche estimate remains conditional on its actual adjoint regularity and approximation hypotheses [F11].