hp finite elements and spectral elements — Examples

For $d=1$, the Rodrigues polynomial in [F1] is ${\varphi }_n(t)=(-1{)}^n{D}^n(t^n(1-t{)}^n)\mathrm{/}n!$. Its value at $t=1$ is $1$: writing $s=1-t$, the first nonzero term of $t^n(1-t{)}^n=(1-s{)}^n{s}^n$ is $s^n$, and $D_t^n=(-1{)}^n{D}_s^n$. Define $K_{p-1}(t)={\sum }_{n=0}^{p-1}(2n+1){\varphi }_n(t)$. Orthogonality, the norm in [F1], and ${\varphi }_n(1)=1$ give $K_{p-1}(1)={\sum }_{n=0}^{p-1}(2n+1)=p^2$, ${\int }_0^1{K}_{p-1}=1$, and $\mathrm{\parallel }{K}_{p-1}-1{\mathrm{\parallel }}_2^2=p^2-1$.

Put $q_p=(K_{p-1}-1)\mathrm{/}2$. Then $\mathrm{deg}{q}_p\le p-1$, ${\int }_0^1{q}_p=0$, $q_p(1)=(p^2-1)\mathrm{/}2$, and ${\int }_0^1{q}_p^2=(p^2-1)\mathrm{/}4$. Define $v_p(x)=1+{\int }_0^{x+1}{q}_p(t)dt$ on $(-1,0)$ and $v_p(x)=0$ on $(0,1)$. Its cell restrictions have degree at most $p$, its jump is $1$, its left derivative at the interface is $q_p(1)$, and its right derivative is $0$. Therefore the broken derivative energy is $(p^2-1)\mathrm{/}4$ and the interface flux term is $-2(q_p(1)\mathrm{/}2)(1)=-(p^2-1)\mathrm{/}2$. Substituting into the stated form yields $a_h(v_p,v_p)=\eta -(p^2-1)\mathrm{/}4$, which is negative whenever $p^2>4\eta +1$. Meanwhile ${\sum }_K\mathrm{\parallel }{v}_p^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2+[v_p{]}^2=(p^2-1)\mathrm{/}4+1>0$. This disproves coercivity for every fixed $\eta$ once $p$ is sufficiently large.

Since $P_2^{\mathrm{\prime }}(x)=3x$, its interior GLL node is $0$, and the two endpoints complete the nodes. The cardinal polynomials are ${\mathrm{\ell }}_{-1}=x(x-1)\mathrm{/}2$, ${\mathrm{\ell }}_0=1-x^2$, and ${\mathrm{\ell }}_1=x(x+1)\mathrm{/}2$. Integrating them gives weights $1\mathrm{/}3,4\mathrm{/}3,1\mathrm{/}3$.

Legendre orthogonality [F3] gives the exact diagonal entries $\int P_0^2=2$, $\int P_1^2=2\mathrm{/}3$, and $\int P_2^2=2\mathrm{/}5$. The nodes and weights in step 1.1 give $Q_2(P_0^2)=2$, $Q_2(P_1^2)=2\mathrm{/}3$, and $Q_2(P_2^2)=1$. Every cross product has degree at most three, so exactness through degree $2p-1=3$ makes the off-diagonal entries zero. The top correction is $1-2\mathrm{/}5=3\mathrm{/}5$, which agrees with [F2] at $p=2$ and top coefficient $1$.

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.

For a multi-index $\nu =(n,j)$, $D^{\nu }u=0$ when $j>0$, while ${\mathrm{\partial }}_x^{n}u(x,y)=n!(4-x{)}^{-n-1}$. Since $4-x\ge 3$ on the square, $\mathrm{\mid }{D}^{\nu }u\mathrm{\mid }\le \frac{1}{3}\mathrm{\mid }\nu \mathrm{\mid }!2^{-\mathrm{\mid }\nu \mathrm{\mid }}$ in every case. Thus the hypothesis of [F1] holds with $M=1\mathrm{/}3$, $R=2$, and $q=3\mathrm{/}4$.

The form $a$ is the $H^1$ inner product. By [F3], $F(w)=a(u,w)$ is bounded, and $u$ is the exact variational solution by its definition. Applying [F2] with the constants established in step 1.1 gives $\mathrm{\parallel }u-u_p{\mathrm{\parallel }}_{H^1}\le C(1\mathrm{/}3)(3\mathrm{/}4{)}^p$.