Accelerated finite elements schemes for parabolic stochastic partial differential equations

Abstract

For a class of finite elements approximations for linear stochastic parabolic PDEs it is proved that one can accelerate the rate of convergence by Richardson extrapolation. More precisely, by taking appropriate mixtures of finite elements approximations one can accelerate the convergence to any given speed provided the coefficients, the initial and free data are sufficiently smooth.

Appendix

The aim of this section is to prove that the example described in Sect. 6.2 satisfies Assumptions 2.4 and 2.5.

For $k=1,\ldots , d$, let $e_k\in {\mathbb {Z}}^d$ denote the k-th unit vector of ${\mathbb {R}}^d$; then ${\mathbb {G}}={\mathbb {Z}}^d$ and

$$\begin{aligned} \Gamma =\left\{ \sum _{k=1}^d \epsilon _k e_k \; : \; \epsilon _k\in \{-1,0,1\} \, \text{ for } k=1, \ldots , d \right\} . \end{aligned}$$

For fixed $k=1, \ldots , d$ (resp. $k\ne l \in \{1, \ldots , d\}$) let

$$\begin{aligned} {{\mathcal {I}}}(k)=\{1, \ldots , d\} \setminus \{k\}, \quad \text{ resp. } {{\mathcal {I}}}(k,l)=\{1, \ldots , d\} \setminus \{k,l\}. \end{aligned}$$

(7.1)

Note that in the particular case $d=1$, the functions $\psi$ gives rise to the classical linear finite elements. Then for $\mathbf{i} \in {{\mathbb {Z}}}^d$, we have for $k=0, 1, \ldots , d$:

$$\begin{aligned} R_\mathbf{i}:= (\psi _\mathbf{i},\psi )= {d\atopwithdelims ()k} \Big (\frac{1}{6}\Big )^k \Big ( \frac{2}{3}\Big )^{d-k} \quad \text{ if } \; \sum _{l=1}^d |i_l|=k. \end{aligned}$$

(7.2)

Furthermore, given $k=0, 1, \ldots , d$, there are $2^k$ elements $\mathbf{i}\in {\mathbb {Z}}^d$ such that $\sum _{l=1}^d |i_l|=k$. Therefore, we deduce

$$\begin{aligned} \sum _{\mathbf{i}\in {\mathbb {Z}}^d} (\psi _\mathbf{i},\psi )= \sum _{k=0}^d 2^k {d\atopwithdelims ()k} \Big (\frac{1}{6}\Big )^k \Big ( \frac{2}{3}\Big )^{d-k} = \sum _{k=0}^d {d\atopwithdelims ()k} \Big (\frac{2}{6}\Big )^k \Big ( \frac{2}{3}\Big )^{d-k}=1, \end{aligned}$$

which yields the first compatibility conditon in (2.20).

We at first check that Assumption 2.4 holds true, that is

$$\begin{aligned} \delta \sum _{\mathbf{i}\in {\mathbb {Z}}^d} U_\mathbf{i}^2 = \delta |U|^2_{\ell _2({\mathbb {Z}}^d)} \le \Big |\sum _{\mathbf{i}\in {\mathbb {Z}}^d} U_\mathbf{i} \psi _\mathbf{i}\Big |_{L^2}^2 =\sum _{\mathbf{i}, \mathbf{j}\in {\mathbb {Z}}^d} R_{\mathbf{i}-\mathbf{j}} U_\mathbf{i} U_\mathbf{j}, \quad U\in \ell _2({\mathbb {Z}}^d). \end{aligned}$$

for some $\delta >0$. For $U\in \ell _2({{\mathbb {Z}}}^d)$ and $k=1, \ldots , d,$ let $T_k U=U_{e_k}$, where $e_k$ denotes the k-th vector of the canonical basis.

For $U\in \ell _2({\mathbb {Z}}^d)$ we have

$$\begin{aligned} \Big | \sum _\mathbf{i} U_\mathbf{i} \psi _\mathbf{i}\Big |_{L^2}^2&= \sum _\mathbf{i\in {\mathbb {Z}}^d} \int _{[0,1]^d} \left[ U_\mathbf{i} \prod _{j=1}^d ( 1-{x_j}) + \sum _{k=1}^d (T_k U)_\mathbf{i}\; x_k \prod _{j\in {{\mathcal {I}}}(k)}(1-x_j) \right. \\&\left. \quad + \sum _{1\le k_1<k_2\le d} (T_{k_1} \circ T_{k_2} U)_\mathbf{i}\; x_{k_1} x_{k_2}\right. \\&\left. \quad \times \prod _{j\in {{\mathcal {I}}}(k_1,k_2)}(1-x_j) +\cdots + (T_1 \circ T_2 \cdots \circ T_d U)_\mathbf{i} \prod _{k=1}^d x_k\right] ^{2} dx, \end{aligned}$$

Given $\alpha \in (0,1)$ if we let

$$\begin{aligned} I(\alpha )= & {} \int _0^\alpha (1-x)^2 dx = \alpha - \alpha ^2 + \frac{\alpha ^3}{3}, \\ J(\alpha )= & {} \int _0^\alpha x(1-x) dx = \frac{\alpha ^2}{2}-\frac{\alpha ^3}{3}, \; K(\alpha )= \int _0^\alpha x^2 dx = \frac{\alpha ^3}{3} , \end{aligned}$$

restricting the above integral on the set $[0,\alpha ]^d$, expanding the square and using the Cauchy Schwarz inequality we deduce the existence of some constants $C(\gamma _1,\gamma _2, \gamma _3)$ defined for $\gamma _i\in \{0,1, \ldots , d\}$ such that

$$\begin{aligned} \Big | \sum _\mathbf{i} U_\mathbf{i} \psi _\mathbf{i}\Big |_{L^2}^2 \ge&\sum _\mathbf{i} |U_\mathbf{i}|^2 \Big [ I(\alpha )^d + {d\atopwithdelims ()1} K(\alpha )\, I(\alpha )^{d-1} \\&+ {d\atopwithdelims ()2} K(\alpha )^2\, I(\alpha )^{d-2} + \cdots + K(\alpha )^d\Big ] \\&-\, 2 \Big ( \sum _\mathbf{i} |U_\mathbf{i}|^2 \Big ) \sum _{\gamma _1+\gamma _2+\gamma _3=d, \gamma _2+\gamma _3\ge 1} C(\gamma _1,\gamma _2, \gamma _3)\; I(\alpha )^{\gamma _1}\; J(\alpha )^{\gamma _2} \; K(\alpha )^{\gamma _3}\\ \ge&\, |U|_{\ell _2({{\mathbb {Z}}}^d)}^2 \Big ( \alpha ^d - \sum _{l=d+1}^{3d} C_l \alpha ^l \Big ), \end{aligned}$$

where $C_l$ are some positive constants. Choosing $\alpha$ small enough, we have $\big | \sum _\mathbf{i} U_\mathbf{i} \psi _\mathbf{i}\big |_{L^2}^2 \ge \frac{\alpha ^d}{2} |U|_{\ell _2({{\mathbb {Z}}}^d)}^2$, which implies the invertibility Assumption 2.4.

We now prove that the compatibility Assumption 2.5 holds true. For $l=1, \ldots , d$, $n=0, \ldots , d-1$:

$$\begin{aligned} (D_l \psi _\mathbf{i} , D_l\psi )&= - \, 2^{d-1-n} \, \Big ( \frac{1}{6}\Big )^n \, \Big ( \frac{1}{3}\Big )^{d-1-n} \; \text{ for } |i_l|=1, \; \sum _{r \ne l, 1\le n\le d} |i_r|=n, \end{aligned}$$

(7.3)

$$\begin{aligned} (D_l \psi _\mathbf{i} , D_l \psi )&= + \, 2^{d-n} \, \Big (\frac{1}{6}\Big )^n \, \Big ( \frac{1}{3}\Big )^{d-1-n} \; \text{ for } |i_l|=0, \; \sum _{r\ne l, 1\le r\le d} |i_r|=n. \end{aligned}$$

(7.4)

For $n=1, \ldots , d-1$ and $k_1<k_2<\cdots < k_n$ with $k_r\in {{\mathcal {I}}}(l)$, where ${{\mathcal {I}}}(l)$ is defined in (7.1), let

$$\begin{aligned} \Gamma _l(k_1, \ldots , k_n)&=\Big \{ \sum _{r=1}^n \epsilon _{k_r} e_{k_r} \; : \; \epsilon _{k_r}\in \{ -1, 1\}, r=1, \ldots , n \Big \} ,\\ \Gamma _l(l;k_1, \ldots , k_n)&=\Big \{ \epsilon _l e_l + \sum _{r=1}^n \epsilon _{k_r} e_{k_r} \; : \; \epsilon _l \in \{-1,1\} \; \text{ and } \epsilon _{k_r}\in \{ -1, 1\}, r=1, \ldots , n\Big \} . \end{aligned}$$

Then $| \Gamma _l (k_1, \ldots , k_n)|=2^n$ while $| \Gamma _l (l;k_1, \ldots , k_n)|=2^{n+1}$. For $l=1, \ldots , d$, the identities (7.3) and (7.4) imply

$$\begin{aligned} \sum _{\lambda \in \Gamma }&(D_l \psi , D_l \psi _\lambda ) = \left[ (D_l\psi , D_l\psi ) + \sum _{\epsilon _l\in \{-1,+1\}} (D_l \psi , D_l \psi _{\epsilon _l e_l})\right] \\&\quad + \sum _{n=1}^{d-1} \sum _{k_1<k_2< \cdots <k_n, k_j\in {{\mathcal {I}}}(l)} \left[ \sum _{\lambda \in \Gamma _l(k_1,\ldots , k_n)} (D_l\psi , D_l\psi _\lambda ) + \sum _{\lambda \in \Gamma _l(l; k_1, \ldots , k_n)} (D_l\psi , D_l\psi _\lambda ) \right] \\&= \Big [ 2^d \Big ( \frac{1}{3}\Big )^{d-1} -2\times 2^{d-1} \Big ( \frac{1}{3}\Big )^{d-1}\Big ] \\&\quad +\sum _{n=1}^{d-1} 2^n {d-1 \atopwithdelims ()n} \Big [ 2^{d-n} \Big ( \frac{1}{6} \Big )^n \Big ( \frac{1}{3}\Big )^{d-1-n} -2\times 2^{d-1-n} \Big (\frac{1}{6}\Big )^n \Big ( \frac{1}{3}\Big )^{d-1-n}\Big ] =0 . \end{aligned}$$

This proves the second identity in (2.20) when $i=j$. Furthermore, (7.3) implies

$$\begin{aligned} \sum _{\lambda \in \Gamma } R^{ll}_\lambda \lambda _l \lambda _l&= - \sum _{\lambda \in \Gamma } (D_l \psi , D_l \psi _\lambda ) \lambda _l \lambda _l = - \sum _{\epsilon _l\in \{-1,1\}} \big (D_l \psi , D_l \psi _{\epsilon _l e_l} \big ) \\&\quad - \sum _{n=1}^{d-1} \sum _{k_1<k_2< \cdots <k_n, k_j\in {{\mathcal {I}}}(l)} \sum _{\lambda \in \Gamma _l(l;k_1, \ldots , k_n)} (D_l \psi , D_l \psi _\lambda )\\&= 2\times 2^{d-1} \Big (\frac{1}{3}\Big )^{d-1} + \sum _{d=1}^n {d-1 \atopwithdelims ()n} 2^{n+1} \times 2^{d-1-n} \Big ( \frac{1}{6}\Big )^n \Big ( \frac{1}{3}\Big )^{d-1-n} \\&= 2 \sum _{n=0}^{d-1} {d-1 \atopwithdelims ()n} \Big (\frac{2}{6}\Big )^n \Big (\frac{2}{3}\Big )^{d-1-n} =2,\quad l=1,\ldots ,d. \end{aligned}$$

Furthermore, given $k\ne l \in \{1, \ldots , d\}$,

$$\begin{aligned} \sum _{\lambda \in \Gamma } R^{ll}_\lambda \lambda _k\lambda _k&= - \sum _{\lambda \in \Gamma } (D_l \psi , D_l \psi _\lambda ) \lambda _k \lambda _l=0. \end{aligned}$$

Indeed, for $n=1, \ldots , d-1$, $k_1<\cdots < k_n$ where $k_r\in {{\mathcal {I}}}(l)$ and at least one of the indices $k_r$ is equal to k for $r=1, \ldots ,n$, given $\lambda \in \Gamma _l(k_1, \ldots , k_n)$ we have using (7.3) and (7.4)

$$\begin{aligned} \sum _{\epsilon _l\in \{-1,1\}} (D_l\psi ,D_l\psi _{\epsilon _le_l+\lambda }) \lambda _k \lambda _l = - 2^{d-1-n} \Big (\frac{1}{6}\Big )^n \Big (\frac{1}{3}\Big )^{d-1-n} \times (-1+1)=0. \end{aligned}$$

This proves the second identity in (2.22) when both derivatives agree.

Also note that for $k\ne l \in \{1, \ldots , d\}$ we have $\sum _{\lambda \in \Gamma } R^{kl}_\lambda =0$. Indeed, for $\lambda$ as above

$$\begin{aligned} (D_k \psi , D_l \psi _\lambda )&+ \sum _{\epsilon _l\in \{-1,1\}} (D_k\psi , D_l\psi _{\epsilon _le_l+\lambda }) \\&= 2^{d-n} \Big (\frac{1}{6}\Big )^n \Big (\frac{1}{3}\Big )^{d-1-n} -2\times 2^{d-1-n} \Big (\frac{1}{6}\Big )^n \Big (\frac{1}{3}\Big )^{d-1-n} =0, \end{aligned}$$

while $R^{kl}_\lambda =0$ for other choices of $\lambda \in \Gamma$.

We now study the case of mixed derivatives. Given $k\ne l \in \{1, \ldots , d\}$ recall that ${{\mathcal {I}}}(k,l)=\{ 1, \ldots , d\} \setminus \{k,l\}$. Then for $k\ne l \in \{1, \ldots , d\}$ and $\mathbf{i} \in {\mathbb {Z}}^d$ we have for $n=0, \ldots , d-2$

$$\begin{aligned} (D_k \psi _\mathbf{i}, D_l \psi )&=\,0 \; \text{ if } |i_k \, i_l|\ne 1, \end{aligned}$$

(7.5)

$$\begin{aligned} (D_k \psi _\mathbf{i}, D_l \psi )&= - \,\Big (\frac{1}{2}\Big )^2 \; \Big ( \frac{1}{6}\Big )^n \; \Big (\frac{2}{3}\Big )^{d-n-2} \; \text{ if } \; i_k\, i_l=1, \quad \sum _{r\in {{\mathcal {I}}}(k,l)} |i_r|=n, \end{aligned}$$

(7.6)

$$\begin{aligned} (D_k \psi _\mathbf{i}, D_l \psi )&= +\, \Big (\frac{1}{2}\Big )^2 \; \Big ( \frac{1}{6}\Big )^n \; \Big (\frac{2}{3}\Big )^{d-n-2} \; \text{ if } \; i_k \, i_l=- 1, \quad \sum _{r\in {{\mathcal {I}}}(k,l)} |i_r|=n. \end{aligned}$$

(7.7)

For $n=1, \ldots , d-2$ and $k_1<\cdots <k_n$ with $k_r\in {{\mathcal {I}}}(k,l)$ for $r=1,\ldots , n$, set

$$\begin{aligned} \Gamma _{k,l}(k_1, \ldots , k_n)=\Big \{ \sum _{r=1}^n \epsilon _{k_r} e_{k_r} \; : \; \epsilon _r\in \{-1,1\} \Big \}. \end{aligned}$$

For $n=0$ there is no such family of indices $k_1<\cdots <k_n$ and we let $\Gamma _{k,l}(\emptyset )=\{0\}$. Thus for $n=0, \ldots , d-2$, $| \Lambda _{k,l}(k_1,\ldots , k_n)|=2^n$. Using the identities (7.5)–(7.7) we deduce

$$\begin{aligned} \sum _{\lambda \in \Gamma }&(D_k \psi , D_l \psi _\lambda ) = \sum _{n=0}^{d-2} \; \sum _{k_1<k_2<\cdots <k_n, k_r\in {{\mathcal {I}}}(k,l)} \sum _{\lambda \in \Gamma _{k,l}(k_1,\ldots , k_n)} \big [ (D_k\psi , D_l\psi _{e_k+e_l+\lambda }) \nonumber \\&\quad + (D_k\psi , D_l\psi _{e_k-e_l+\lambda })+ (D_k\psi , D_l\psi _{-e_k+e_l+ \lambda }) + (D_k\psi , D_l\psi _{-e_k-e_l+\lambda }) \big ] \nonumber \\ =&\sum _{n=0}^{d-2} {d-2 \atopwithdelims ()n} \, 2^n \, \Big [ - \Big ( \frac{1}{2}\Big )^2 \, \Big (\frac{1}{6}\Big )^n \, \Big (\frac{2}{3}\Big )^{d-2-n} + \Big ( \frac{1}{2}\Big )^2 \, \Big (\frac{1}{6}\Big )^n \, \Big (\frac{2}{3}\Big )^{d-2-n} \nonumber \\&\quad + \Big ( \frac{1}{2}\Big )^2 \, \Big (\frac{1}{6}\Big )^n \, \Big (\frac{2}{3}\Big )^{d-2-n} - \Big ( \frac{1}{2}\Big )^2 \, \Big (\frac{1}{6}\Big )^n \, \Big (\frac{2}{3}\Big )^{d-2-n}\Big ] = 0,\quad k\ne l. \end{aligned}$$

(7.8)

This completes the proof of the second identity in (2.20) when $i\ne j$, and hence (2.20) holds true. Furthermore, the identities (7.6) and (7.7) imply for $i\ne j \in \{1, \ldots , d\}$ and $\{i,j\}=\{k,l\}$

$$\begin{aligned} \sum _{\lambda \in \Gamma }&(D_k \psi , D_l \psi _\lambda )\lambda _k \lambda _l = \sum _{n=0}^{d-2} \; \sum _{k_1<k_2<\cdots <k_n, k_r\in {{\mathcal {I}}}(k,l)} \sum _{\lambda \in \Gamma _{k,l}(k_1,\ldots , k_n)} \big [ (D_k\psi , D_l\psi _{e_k+e_l+\lambda }) \\&\qquad -\, (D_k\psi , D_l\psi _{e_k-e_l+\lambda }) - (D_k\psi , D_l\psi _{-e_k+e_l+ \lambda }) + (D_k\psi , D_l\psi _{-e_k-e_l+\lambda }) \big ] \\&\quad = -\, 4 \, \Big ( \frac{1}{2}\Big )^2 \sum _{n=0}^{d-2} {d-2 \atopwithdelims ()n} \, 2^n \, \Big (\frac{1}{6}\Big )^n \, \Big (\frac{2}{3}\Big )^{d-2-n}\\&\quad =- \sum _{n=0}^{d-2} {d-2 \atopwithdelims ()n} \, \Big (\frac{2}{6}\Big )^n \Big ( \frac{2}{3}\Big )^{d-2-n}= -1. \end{aligned}$$

Equation (7.5) proves that $(D_k \psi , D_l \psi _\lambda )=0$ if $|\lambda _k \lambda _l|\ne 1$. Hence using (7.8) we deduce that for any $r=1, \ldots , d$,

$$\begin{aligned} \sum _{\lambda \in \Gamma } (D_k \psi , D_l \psi _\lambda )\lambda _r \lambda _r =0. \end{aligned}$$

Let $r\in {{\mathcal {I}}}(k,l)$ and for $n=1, \ldots , d-3$, let $k_1<\cdots < k_n$ be such that $k_j\in \{1, \ldots , d\} \setminus \{k,l,r\}$ and $\lambda = \sum _{j=1}^n \epsilon _{k_j} e_{k_j}$ for $\epsilon _{k_j}\in \{-1,1\}$, $j=1, \ldots , n$. Then for any choice of $\epsilon _k$ and $\epsilon _l$ in $\{-1,1\}$ the equalities (7.6) and (7.7) imply that

$$\begin{aligned} (D_k\psi , D_l \psi _{\lambda + \epsilon _k e_k + \epsilon _l e_l + e_r} ) = (D_k\psi , D_l \psi _{\lambda + \epsilon _k e_k + \epsilon _l e_l - e_r} ) . \end{aligned}$$

This clearly yields that for $r\in {{\mathcal {I}}}(k,l)$ we have

$$\begin{aligned} \sum _{\lambda \in \Gamma } (D_k \psi , D_l \psi _\lambda )\lambda _k \lambda _r = \sum _{\lambda \in \Gamma } (D_k \psi , D_l \psi _\lambda )\lambda _l \lambda _r =0. \end{aligned}$$

Finally, given $n=2, \ldots , d$ and $k_1< \cdots <k_n$ where the terms $k_j\in {{\mathcal {I}}}(k,l)$, then given any choice of $\epsilon _k$ and $\epsilon _l$ in $\{-1,1\}$, the value of $(D_k \psi , D_l \psi _{\epsilon _k e_k + \epsilon _l e_l + \lambda })$ does not depend on the value of $\lambda \in \Gamma _{k,l}(k_1,\ldots , k_n)$. Therefore, if we fix $r_1\ne r_2$ in the set ${{\mathcal {I}}}(k,l)$, for fixed n there are as many choices of indices $k_1<\cdots <k_n$ such that $\epsilon _{r_1} \epsilon _{r_2}=1$ that of such indices such that $\epsilon _{r_1} \epsilon _{r_2}=-1$. This proves

$$\begin{aligned} \sum _{\lambda \in \Gamma } (D_k\psi , D_l\psi _\lambda ) \lambda _{r_1} \lambda _{r_2}=0, \end{aligned}$$

which completes the proof of the first identity in (2.22) for mixed derivatives; hence (2.22) holds true.

We now check the compatibility condition (2.21). Fix $j\in \{1, \ldots , d\}$; then

$$\begin{aligned} (D_j\psi , \psi )= & {} 2^{d-1} \Big ( \prod _{k\ne j} \int _0^1 (1-x_k)^2 dx_k\Big ) \nonumber \\&\quad \times \Big [ \int _0^1 (-1) (1-x_j) dx_j + \int _{-1}^0 (1+x_j) dx_j\Big ] =0, \end{aligned}$$

(7.9)

while

$$\begin{aligned} (D_j \psi , \psi _{e_j})&=2^{d-1} \Big ( \prod _{k\ne j} \int _0^1 (1-x_k)^2 dx_k\Big ) \int _0^1 (-1) \big (1+(x_j-1)\big ) dx_j = -\, \frac{1}{2} \Big (\frac{2}{3}\Big )^{d-1}, \nonumber \\ (D_j \psi , \psi _{-e_j})&=2^{d-1} \Big ( \prod _{k\ne j} \int _0^1 (1-x_k)^2 dx_k\Big ) \int _{-1}^0 \big (1- (x_j+1)\big ) dx_j = \frac{1}{2} \Big (\frac{2}{3}\Big )^{d-1}. \end{aligned}$$

(7.10)

For $n=1, \ldots , d-1$ and $k_1<\cdots <k_n$ where the indexes $k_r$, $r=1, \ldots , n$ are different from j we have for any $\lambda \in \Gamma _j(k_1,\ldots , k_n)$

$$\begin{aligned} (D_j\psi , \psi _{\lambda }) =&\, 2^{d-(n+1)} \Big (\prod _{k\in \Gamma \setminus \{j,k_1, \ldots , k_n\}} \int _0^1 (1-x_k)^2 dx_k\Big ) \times \Big ( \prod _{r=1}^n \int _0^1 x_{k_r} (1-x_{k_r}) dx_{k_r} \Big ) \nonumber \\&\times \, \Big [ \int _0^1 (-1) (1-x_j) dx_j + \int _{-1}^0 (1+x_j) dx_j\Big ] = 0, \end{aligned}$$

(7.11)

while

$$\begin{aligned} (D_j\psi , \psi _{e_j+\lambda }) =&\, 2^{d-(n+1)} \Big (\prod _{k\in \Gamma \setminus \{j,k_1, \ldots , k_n\}} \int _0^1 (1-x_k)^2 dx_k\Big )\nonumber \\&\times \Big ( \prod _{r=1}^n \int _0^1 x_{k_r} (1-x_{k_r}) dx_{k_r} \Big ) \nonumber \\&\times \int _0^1 (-1) \big ( 1+ (x_j-1)\big ) dx_j = -\frac{1}{2} \; \Big ( \frac{2}{3}\Big )^{d-(n+1)} \; \Big ( \frac{1}{6}\Big )^n , \end{aligned}$$

(7.12)

and

$$\begin{aligned} (D_j\psi , \psi _{-e_l+\lambda }) =&\, 2^{d-(n+1)} \Big (\prod _{k\in \Gamma \setminus \{j,k_1,\ldots , k_n\}} \int _0^1 (1-x_k)^2 dx_k\Big )\nonumber \\&\times \Big ( \prod _{r=1}^n \int _0^1 x_{k_r} (1-x_{k_r}) dx_{k_r} \Big ) \nonumber \\&\times \int _{-1}^0 \big ( 1- (x_j+1)\big ) dx_j = \frac{1}{2} \; \Big ( \frac{2}{3}\Big )^{d-(n+1)} \; \Big ( \frac{1}{6}\Big )^n . \end{aligned}$$

(7.13)

Note that the number of terms $(D_j\psi , \psi _{\epsilon _l e_l+\lambda })$ with $\epsilon _l=-1$ or $\epsilon _l =+1$ is equal to ${d-1 \atopwithdelims ()n} 2^n$. Therefore, the identities (7.9)–(7.13) imply that for any $j=1, \ldots , d$ we have

$$\begin{aligned}&\sum _{\lambda \in \Gamma } \lambda _j R^j_\lambda =- \sum _{\lambda \in \Gamma } (D_j \psi , \psi _\lambda )\; \lambda _j = \frac{1}{2} \Big (\frac{2}{3}\Big )^{d-1} - \frac{1}{2} \Big (\frac{2}{3}\Big )^{d-1} (-1) \end{aligned}$$

(7.14)

$$\begin{aligned}&\qquad + \frac{1}{2} \sum _{n=1}^{d-1} {d-1 \atopwithdelims ()n} 2^n \Big ( \frac{2}{3}\Big )^{d-1-n} \Big ( \frac{1}{6}\Big )^n - \frac{1}{2}\sum _{n=0}^{d-1} {d-1 \atopwithdelims ()n} 2^n \Big ( \frac{2}{3}\Big )^{d-1-n} \Big ( \frac{1}{6}\Big )^n \times (-1) \nonumber \\&\quad = \sum _{n=0}^{d-1} {d-1 \atopwithdelims ()n} \Big ( \frac{2}{6}\Big )^n \Big (\frac{2}{3}\Big )^{d-1-n} = 1. \end{aligned}$$

(7.15)

This proves (2.21) when $i=k$.

Let $k\ne j \in \{1,\ldots , d\}$ and given $n=1,\ldots , d-1$ let $k_1<\cdots < k_n$ be indices that belong to ${{\mathcal {I}}}(j)$ such that one of the indices $k_r, r=1,\ldots , n$ is equal to k. Given any $\lambda \in \Gamma _j(k_1,\ldots , k_n)$ we deduce that

$$\begin{aligned} (D_j \psi , \psi _{e_l+\lambda }) \lambda _k + (D_j \psi , \psi _{-e_l+\lambda }) \lambda _k =0. \end{aligned}$$

This completes the proof of the identity (2.21).

In order to complete the proof of the validity of Assumption 2.5, it remains to check that the identities in (2.23) hold true. Recall that for $\lambda \in \Gamma$ and $i,j,k,l\in \{1,\ldots ,d\}$ we have

$$\begin{aligned} Q_\lambda ^{ij,kl}=\int _{{\mathbb {R}}^d} z_k z_l D_j\psi _\lambda (z) D_i^*\psi (z) dz =-\int _{{\mathbb {R}}^d} z_k z_l D_j\psi _\lambda (z) D_i \psi (z) dz. \end{aligned}$$

For $p=1,\ldots , 4$, $n=1,\ldots , d-p$ and $i_1,\ldots , i_p \in \{1, \ldots ,d\}$ with $i_1,\ldots , i_p$ pairwise different let

$$\begin{aligned}&{\mathcal {I}}_n(i_1,\ldots , i_p):=\Big \{ \sum _{\alpha =1}^n \epsilon _\alpha e_{k_\alpha } ; \epsilon _\alpha \in \{-1,+1\}, \; 1\le k_1< \cdots <k_n\le d, \\&\quad k_\alpha \not \in \{i_1,\ldots , i_p\} \, \text{ for } \alpha =1,\ldots , n\Big \}, \end{aligned}$$

and ${\mathcal {I}}_0(i_1,\ldots , i_p)=\{0\}$.

First suppose that $i=j$.

First let $k=l=i$; then for $n=0,\ldots , d-1$ and $\mu \in {\mathcal {I}}_n(i)$ we have

$$\begin{aligned} Q_{\mu }^{ii,ii} + Q_{\mu + e_i}^{ii,ii} + Q_{\mu - e_i}^{ii,ii}=0. \end{aligned}$$

Let $k=l$ with $k\ne i$; then then for $n=0,\ldots , d-1$ and $\mu \in {\mathcal {I}}_n(i)$ we have

$$\begin{aligned} Q_{\mu }^{ii,kk} + Q_{\mu + e_i}^{ii,kk} + Q_{\mu -e_i}^{ii,kk}=0. \end{aligned}$$

Let $l=i$ and $k\ne i$; then for $n=0,\ldots , d-2$, $\epsilon \in \{-1,+1\}$ and $\mu \in {\mathcal {I}}_n(i,k)$ we have

$$\begin{aligned} Q_{\mu + \epsilon e_i + e_k}^{ii,ki} + Q_{\mu + \epsilon e_i - e_k}^{ii,ki}=0. \end{aligned}$$

A similar result holds for $k=i$ and $l\ne i$. Furthermore, $Q_\lambda ^{ii,ki}=0$ if $\lambda$ is not equal to $\mu + \epsilon e_i + e_k$ or $\mu + \epsilon e_i - e_k$ with $\mu \in {\mathcal {I}}_n(i,k)$ for some n.

Let $k\ne l$ with $k\ne i$ and $l\ne i$; then for $n=0,\ldots , d-2$, $\epsilon \in \{-1,+1\}$ and $\mu \in {\mathcal {I}}_n(k,l)$ we have

$$\begin{aligned} Q_{\mu + \epsilon e_k + e_l}^{ii,kl} + Q_{\mu + \epsilon e_k - e_l}^{ii,kl}=0, \end{aligned}$$

while $Q_\lambda ^{ii,kl}=0$ if $\lambda$ is not equal to $\mu + \epsilon e_i + e_k$ or $\mu + \epsilon e_i - e_k$ with $\mu \in {\mathcal {I}}_n(i,k)$ for some n.

We now suppose that $i\ne j$.

First suppose that $k=i$ and $l=j$; then for $n=0,\ldots , d-1$ and $\mu \in {\mathcal {I}}_n(i)$ we have

$$\begin{aligned} Q_{\mu }^{ij,ij} + Q_{\mu + e_j}^{ij,ij} + Q_{\mu -e_j}^{ij,ij}=0. \end{aligned}$$

Let $k=l=i$; then for $n=0,\ldots , d-2$, $\epsilon \in \{-1+1\}$ and $\mu \in {\mathcal {I}}_n(i,j)$ we have

$$\begin{aligned} Q_{\mu + \epsilon e_i +e_j}^{ij,ii} + Q_{\mu + \epsilon e_i -e_j}^{ij,ii}=0, \end{aligned}$$

while $Q_\lambda ^{ij,ii}=0$ is $\lambda$ is not equal to $\mu +\epsilon e_i +e_j$ or $\mu + \epsilon e_i-e_j$ where $\mu \in {\mathcal {I}}_n(i,j)$ for some n. A similar result holds exchanging i and j for $k=l=j$.

Let $k=l$ with $k\not \in \{i,j\}$ and $l\not \in \{i,j\}$; then for $n=0,\ldots , d-2$, $\epsilon \in \{-1,+1\}$ and $\mu \in {\mathcal {I}}_n(i,j)$ we have

$$\begin{aligned} Q_{\mu + \epsilon e_i +e_j}^{ij,kk} + Q_{\mu + \epsilon e_i -e_j}^{ij,kk}=0, \end{aligned}$$

while $Q_\lambda ^{ij,kk}=0$ if $\lambda$ is not equal to $\mu +\epsilon e_i +e_j$ or $\mu + \epsilon e_i-e_j$ with $\mu \in {\mathcal {I}}_n(i,j)$ for some n.

Let $l=i$ and $k\not \in \{i,j\}$; then for $n=0,\ldots , d-2$, $\epsilon \in \{-1+1\}$ and $\mu \in {\mathcal {I}}_n(i,k)$ we have

$$\begin{aligned} Q_{\mu + \epsilon e_i +e_k}^{ij,ki} + Q_{\mu + \epsilon e_i -e_k}^{ij,ki}=0, \end{aligned}$$

while $Q_\lambda ^{ij,ki}=0$ is $\lambda$ is not equal to $\mu +\epsilon e_i +e_k$ or $\mu + \epsilon e_i-e_k$ where $\mu \in {\mathcal {I}}_n(i,k)$ for some n. A similar result holds exchanging i and j for $k=l=j$.

Finally, let $k\ne l$ with $k\not \in \{i,j\}$ and $l\not \in \{i,j\}$; then for $n=0,\ldots , d-4$, $\epsilon _i, \epsilon _j, \epsilon _k \in \{-1+1\}$ and $\mu \in {\mathcal {I}}_n(i,j,k,l)$ we have

$$\begin{aligned} Q_{\mu + \epsilon _i e_i +\epsilon _j e_j + \epsilon _k e_k +e_l}^{ij,kl} + Q_{\mu + \epsilon _i e_i +\epsilon _j e_j + \epsilon _k e_k -e_l}^{ij,kl}=0, \end{aligned}$$

while $Q_\lambda ^{ij,kl}=0$ is $\lambda$ is not equal to $\mu + \epsilon _i e_i +\epsilon _j e_j + \epsilon _k e_k +e_l$ or $\mu + \epsilon _i e_i +\epsilon _j e_j + \epsilon _k e_k -e_l$ where $\mu \in {\mathcal {I}}_n(i,j,k,l)$ for some n. These computations complete the proof of the first identity in (2.23). Recall that for $i,k\in \{1,\ldots , d\}$ and $\lambda \in \Gamma$ we let

$$\begin{aligned} {\tilde{Q}}_\lambda ^{i,k}:=\int _{{\mathbb {R}}^d} z_k D_i\psi _\lambda (z) \psi (z) dz. \end{aligned}$$

Let $k=i$; for $n=0,\ldots , d-1$ and $\mu \in {\mathcal {I}}_n(i)$ we have

$$\begin{aligned} {\tilde{Q}}^{i,i}_\mu + {\tilde{Q}}^{i,i}_{\mu + e_i} + {\tilde{Q}}^{i,i}_{\mu -e_i}=0. \end{aligned}$$

Let $k\ne i$; for $n=0,\ldots , d-2$, $\epsilon \in \{-1,0, +1\}$ and $\mu \in {\mathcal {I}}_n(i,k)$ we have

$$\begin{aligned} {\tilde{Q}}_{\mu + \epsilon e_i + e_k}^{i,k} + {\tilde{Q}}_{\mu + \epsilon e_i - e_k}^{i,k}=0 \end{aligned}$$

while ${\tilde{Q}}_\lambda ^{i,k}=0$ if $\lambda$ is not equal to $\mu +\epsilon e_i +e_k$ or $\mu + \epsilon e_i-e_k$ where $\mu \in {\mathcal {I}}_n(i,k)$ for some n. This completes the proof of the second identity in (2.23); therefore Assumption 2.5 is satisfied for these finite elements. This completes the verification of the validity of Assumptions 2.4–2.5 for the function $\psi$ defined by (6.2).