A reentrant-corner mode separates fixed-degree h error from hp error

With $\lambda =2\mathrm{/}3$, polar differentiation gives $\mathrm{\mid }u\mathrm{\mid }=r^{\lambda }\mathrm{\mid }\sin (\lambda \theta )\mathrm{\mid }$ and $\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^2={\lambda }^2{r}^{2\lambda -2}$. The polygon is bounded, and the area element is $rdrd\theta$, so $u\in L^2(\mathrm{\Omega })$ and $\mathrm{\nabla }u\in L^2(\mathrm{\Omega })$ because ${\int }_0^R{r}^{2\lambda +1}dr$ and ${\int }_0^R{r}^{2\lambda -1}dr$ are finite at zero. The origin is on the boundary and $u$ is smooth in $\mathrm{\Omega }$, so these classical derivatives are its weak derivatives; hence $u\in H^1(\mathrm{\Omega })$.

In the polar orthonormal frame the Hessian entries are $u_{rr}=\lambda (\lambda -1)r^{\lambda -2}\sin (\lambda \theta )$, $u_{r\theta }^{\mathrm{o}\mathrm{r}\mathrm{t}\mathrm{h}}=\lambda (\lambda -1)r^{\lambda -2}\cos (\lambda \theta )$, and $u_{\theta \theta }^{\mathrm{o}\mathrm{r}\mathrm{t}\mathrm{h}}=\lambda (1-\lambda )r^{\lambda -2}\sin (\lambda \theta )$. Counting both symmetric off-diagonal entries gives $\mathrm{\mid }{D}^{2}u{\mathrm{\mid }}^2=2{\lambda }^2(1-\lambda {)}^2{r}^{2\lambda -4}$. Its area integral contains ${\int }_0^1{r}^{2\lambda -3}dr={\int }_0^1{r}^{-5\mathrm{/}3}dr=\mathrm{\infty }$. If $u\in H^2$, its weak Hessian agrees with its classical Hessian on every punctured sub-sector, contradicting this divergent integral. Thus $u\mathrm{\notin }{H}^2(\mathrm{\Omega })$.

On the sector, $u=\mathrm{Im}{z}^{2\mathrm{/}3}$ for the branch $0<\arg z<3\pi \mathrm{/}2$. Repeated differentiation gives ${\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }=j}\mathrm{\mid }{D}^{\alpha }u\mathrm{\mid }\le C^{j}j!r^{2\mathrm{/}3-j}$. Set ${\beta }_0=1\mathrm{/}2$ at the origin and ${\beta }_i=0$ at every other polygon corner. The other-corner factors in ${\mathrm{\Phi }}_{\beta +j-2}$ are bounded above by $C^j$ on the bounded polygon; the origin factor changes the radial power to $r^{1\mathrm{/}2+j-2+2\mathrm{/}3-j}=r^{-5\mathrm{/}6}$. Hence the squared weighted integral near the origin is bounded by $C^{2j}(j!{)}^2{\int }_0^R{r}^{-5\mathrm{/}3}rdr$, which is finite. Away from the origin the mode has a local analytic branch at every point of the closure, so all remaining weighted integrals are bounded by another $C^{2j}(j!{)}^2$ factor. Since $j!\mathrm{/}(j-2)!=j(j-1)\le 2^j$, the bound is at most $(2C{)}^j(j-2)!$; taking $d_u=\max (1,2C)$ and $C_u=d_u^2$ proves the exact $B_{\beta }^2$ factorial bound in [F3]. The pointwise supplier [F4] then gives the continuous representative.

Define $c_p={\inf }_{q\in {\mathbb{P}}_p(\widehat{T})}\mathrm{\parallel }\mathrm{\nabla }(u-q){\mathrm{\parallel }}_{L^2(\widehat{T})}$. The gradients of ${\mathbb{P}}_p(\widehat{T})$ form a finite-dimensional, hence closed, subspace of $L^2(\widehat{T};{\mathbb{R}}^2)$ by [F2]. If $c_p=0$, then $\mathrm{\nabla }u$ lies in the closure of that subspace, hence equals $\mathrm{\nabla }q$ almost everywhere for some $q\in {\mathbb{P}}_p(\widehat{T})$. Both gradients are continuous on the punctured open triangle. If they differed at one point, continuity would give a nonempty open rectangle on which they differ, contradicting their almost-everywhere equality. Thus they agree throughout that open set; differentiating gives $D^{2}u=D^{2}q$. Since the nondegenerate triangle $\widehat{T}$ has a vertex at the origin, its interior contains a sector of positive angle near $0$, and the same radial Hessian integral computed in step 2.1 diverges on that sector. This contradicts the bounded polynomial Hessian and proves $c_p>0$.

For any $q_h\in {\mathbb{P}}_p(T_h)$ set $\widehat{q}(y)=h^{-2\mathrm{/}3}{q}_h(hy)\in {\mathbb{P}}_p(\widehat{T})$. Homogeneity $u(hy)=h^{2\mathrm{/}3}u(y)$ and the change of variables $x=hy$ give $\mathrm{\parallel }{\mathrm{\nabla }}_x(u-q_h){\mathrm{\parallel }}_{L^2(T_h)}=h^{2\mathrm{/}3}\mathrm{\parallel }{\mathrm{\nabla }}_y(u-\widehat{q}){\mathrm{\parallel }}_{L^2(\widehat{T})}\ge c_p{h}^{2\mathrm{/}3}$. Taking the infimum proves the asserted local lower bound. If $T_h$ is one mesh cell of degree $p$, restricting any global finite-element error to that cell gives the same lower bound.

The mode vanishes on the two edges $\theta =0$ and $\theta =3\pi \mathrm{/}2$, so it satisfies precisely the boundary condition on ${\mathrm{\Gamma }}_D$. Step 2.2 verifies $u\in B_{\beta }^2(\mathrm{\Omega })$ for the stated weights. Applying [F3] to any mesh and uniform degree sequence satisfying its geometric grading, cell-count, and degree hypotheses proves the global hp upper estimate on this same polygon. Together with step 4.1 this is the stated fixed-degree local h lower bound versus global hp upper bound. No outer-edge zero condition is used.