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Application of Quasi-Monte Carlo Methods to Elliptic PDEs with Random Diffusion Coefficients: A Survey of Analysis and Implementation

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Abstract

This article provides a survey of recent research efforts on the application of quasi-Monte Carlo (QMC) methods to elliptic partial differential equations (PDEs) with random diffusion coefficients. It considers and contrasts the uniform case versus the lognormal case, single-level algorithms versus multi-level algorithms, first-order QMC rules versus higher-order QMC rules, and deterministic QMC methods versus randomized QMC methods. It gives a summary of the error analysis and proof techniques in a unified view, and provides a practical guide to the software for constructing and generating QMC points tailored to the PDE problems. The analysis for the uniform case can be generalized to cover a range of affine parametric operator equations.

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Acknowledgments

We graciously acknowledge many insightful discussions and valuable comments from our collaborators Josef Dick, Mahadevan Ganesh, Thong Le Gia, Alexander Gilbert, Ivan Graham, Yoshihito Kazashi, James Nichols, Pieterjan Robbe, Robert Scheichl, Christoph Schwab and Ian Sloan. We especially thank Mahadevan Ganesh for suggesting an alternative proof strategy to improve some existing estimates. We are also grateful for the financial supports from the Australian Research Council (FT130100655 and DP150101770) and the KU Leuven research fund (OT:3E130287 and C3:3E150478).

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Correspondence to Frances Y. Kuo.

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Communicated by Albert Cohen.

Appendix: Selected Proofs

Appendix: Selected Proofs

In this section we provide the proofs for Lemmas 6.16.8. For simplicity of presentation, in the proofs we will often omit the arguments \({\pmb {x}}\) and \({\pmb {y}}\) in our notation. We start by collecting some identities and estimates that we need for the proofs.

We will make repeated use of the Leibniz product rule

$$\begin{aligned} \partial ^{\pmb {\nu }}(AB) = \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} (\partial ^{{\pmb {m}}} A)\, (\partial ^{{\pmb {\nu }}-{\pmb {m}}} B), \end{aligned}$$
(9.1)

and the identity

$$\begin{aligned} \nabla \cdot (A\, \nabla B) = A\, \varDelta B + \nabla A \cdot \nabla B. \end{aligned}$$
(9.2)

We also need the combinatorial identity

$$\begin{aligned} \sum _{\mathop {\scriptstyle {|{\pmb {m}}|=i}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} = {\textstyle {\left( {\begin{array}{c}|{\pmb {\nu }}|\\ i\end{array}}\right) }}, \end{aligned}$$
(9.3)

which follows from a simple counting argument (i.e., consider the number of ways to select i distinct balls from some baskets containing a total number of \(|{\pmb {\nu }}|\) distinct balls). The identity (9.3) is used to establish the following identities

$$\begin{aligned} \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }}\, |{\pmb {m}}|!\,|{\pmb {\nu }}-{\pmb {m}}|!&= (|{\pmb {\nu }}|+1)!, \end{aligned}$$
(9.4)
$$\begin{aligned} \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} |{\pmb {m}}|! \,(|{\pmb {\nu }}-{\pmb {m}}|+1)!&= \frac{(|{\pmb {\nu }}|+2)!}{2}, \end{aligned}$$
(9.5)
$$\begin{aligned} \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} \frac{(|{\pmb {m}}|+2)!}{2}\,\frac{(|{\pmb {\nu }}-{\pmb {m}}|+2)!}{2}&= \frac{(|{\pmb {\nu }}|+5)!}{120}. \end{aligned}$$
(9.6)

Additionally, we need the recursive estimates in the next two lemmas. The proofs can be found in [25] and [66], respectively.

Lemma 9.1

Given a sequence of non-negative numbers \({\pmb {b}}=(b_j)_{j\in {\mathbb {N}}}\), let \(({\mathbb {A}}_{\pmb {\nu }})_{{\pmb {\nu }}\in {\mathfrak {F}}}\) and \(({\mathbb {B}}_{\pmb {\nu }})_{{\pmb {\nu }}\in {\mathfrak {F}}}\) be non-negative numbers satisfying the inequality

$$\begin{aligned} {\mathbb {A}}_{\pmb {\nu }}\le \sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\,b_j\, {\mathbb {A}}_{{\pmb {\nu }}-{\pmb {e}}_j} + {\mathbb {B}}_{\pmb {\nu }}\quad \text{ for } \text{ any } {\pmb {\nu }}\in {\mathfrak {F}}\quad (\hbox {including }~{\pmb {\nu }}={\pmb {0}}). \end{aligned}$$

Then

$$\begin{aligned} {\mathbb {A}}_{\pmb {\nu }}\le \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }}\, |{\pmb {m}}|!\, {\pmb {b}}^{\pmb {m}}\,{\mathbb {B}}_{{\pmb {\nu }}-{\pmb {m}}} \quad \text{ for } \text{ all } {\pmb {\nu }}\in {\mathfrak {F}}. \end{aligned}$$

The result holds also when both inequalities are replaced by equalities.

Lemma 9.2

Given a sequence of non-negative numbers \({\pmb {\beta }}=(\beta _j)_{j\in {\mathbb {N}}}\), let \(({\mathbb {A}}_{\pmb {\nu }})_{{\pmb {\nu }}\in {\mathfrak {F}}}\) and \(({\mathbb {B}}_{\pmb {\nu }})_{{\pmb {\nu }}\in {\mathfrak {F}}}\) be non-negative numbers satisfying the inequality

$$\begin{aligned} {\mathbb {A}}_{\pmb {\nu }}\le \sum _{\mathop {\scriptstyle {{\pmb {m}}\ne {\pmb {\nu }}}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} {\pmb {\beta }}^{{\pmb {\nu }}-{\pmb {m}}} {\mathbb {A}}_{\pmb {m}}+ {\mathbb {B}}_{\pmb {\nu }}\quad \text{ for } \text{ any } {\pmb {\nu }}\in {\mathfrak {F}}\quad (\hbox {including } \ {\pmb {\nu }}={\pmb {0}}). \end{aligned}$$

Then

$$\begin{aligned} {\mathbb {A}}_{\pmb {\nu }}\le \sum _{{\pmb {k}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {k}}\end{array}}\right) }} \Lambda _{|{\pmb {k}}|}\, {\pmb {\beta }}^{\pmb {k}}\, {\mathbb {B}}_{{\pmb {\nu }}-{\pmb {k}}} \quad \text{ for } \text{ all } {\pmb {\nu }}\in {\mathfrak {F}}, \end{aligned}$$

where the sequence \((\Lambda _n)_{n\ge 0}\) is defined recursively by

$$\begin{aligned} \Lambda _0 := 1 \quad \text{ and }\quad \Lambda _n := \sum _{i=0}^{n-1} {\textstyle {\left( {\begin{array}{c}n\\ i\end{array}}\right) }} \Lambda _i \quad \text{ for } \text{ all } n\ge 1. \end{aligned}$$
(9.7)

The result holds also when both inequalities are replaced by equalities. Moreover, we have

$$\begin{aligned} \Lambda _n \le \frac{n!}{\alpha ^n} \quad \text{ for } \text{ all } n\ge 0. \end{aligned}$$
(9.8)

1.1 Proof of Lemma 6.1

This result was proved in [15]. Since the same proof strategy is used repeatedly in subsequent more complicated proofs, we include this relatively simple proof as a first illustration.

Let \(f\in V^*\) and \({\pmb {y}}\in U\). We prove this result by induction on \(|{\pmb {\nu }}|\). For \({\pmb {\nu }}={\pmb {0}}\), we take \(v = u(\cdot ,{\pmb {y}})\) in (3.5) to obtain

$$\begin{aligned} \int _D a({\pmb {x}},{\pmb {y}})\, |\nabla u({\pmb {x}},{\pmb {y}})|^2 \,{\mathrm {d}}{\pmb {x}}= \int _D f({\pmb {x}})\,u({\pmb {x}},{\pmb {y}})\,{\mathrm {d}}{\pmb {x}}, \end{aligned}$$

which leads to

$$\begin{aligned} a_{\min }\, \Vert u(\cdot ,{\pmb {y}})\Vert _V^2 \le \Vert f\Vert _{V^*} \Vert u(\cdot ,{\pmb {y}})\Vert _V \quad \implies \quad \Vert u(\cdot ,{\pmb {y}})\Vert _V \le \frac{\Vert f\Vert _{V^*}}{a_{\min }}, \end{aligned}$$

as required (see also (3.7)).

Given any multi-index \({\pmb {\nu }}\) with \(|{\pmb {\nu }}|\ge 1\), suppose that the result holds for any multi-index of order \(\le |{\pmb {\nu }}|-1\). Applying the mixed derivative operators \(\partial ^{\pmb {\nu }}\) to the variational formulation (3.5), recalling that f is independent of \({\pmb {y}}\), and using the Leibniz product rule (9.1), we obtain the identity (suppressing \({\pmb {x}}\) and \({\pmb {y}}\) in our notation)

$$\begin{aligned} \int _D \bigg ( \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} (\partial ^{{\pmb {m}}} a)\, \nabla (\partial ^{{\pmb {\nu }}-{\pmb {m}}} u) \cdot \nabla z \bigg )\,{\mathrm {d}}{\pmb {x}}= 0 \qquad \text{ for } \text{ all }\quad z\in V. \end{aligned}$$

Observe that due to the linear dependence of \(a({\pmb {x}},{\pmb {y}})\) on the parameters \({\pmb {y}}\), the partial derivative \(\partial ^{{\pmb {m}}}\) of a with respect to \({\pmb {y}}\) satisfies

$$\begin{aligned} \partial ^{{\pmb {m}}} a({\pmb {x}},{\pmb {y}}) = {\left\{ \begin{array}{ll} a({\pmb {x}},{\pmb {y}}) &{} \text{ if } {\pmb {m}}= {\pmb {0}}, \\ \psi _j({\pmb {x}}) &{} \text{ if } {\pmb {m}}= {\pmb {e}}_j, \\ 0 &{} \text{ otherwise }. \end{array}\right. } \end{aligned}$$
(9.9)

Taking \(z = \partial ^{\pmb {\nu }}u(\cdot ,{\pmb {y}})\) and separating out the \({\pmb {m}}= {\pmb {0}}\) term, we obtain

$$\begin{aligned} \int _D a\, |\nabla (\partial ^{\pmb {\nu }}u)|^2 \,{\mathrm {d}}{\pmb {x}}= - \sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\, \int _D \psi _j\,\nabla (\partial ^{{\pmb {\nu }}-{\pmb {e}}_j}u) \cdot \nabla (\partial ^{{\pmb {\nu }}} u)\,{\mathrm {d}}{\pmb {x}}, \end{aligned}$$

which yields

$$\begin{aligned} a_{\min }\, \Vert \nabla (\partial ^{\pmb {\nu }}u) \Vert _{L^2}^2&\le \sum _{j\ge 1}\nu _j\,\Vert \psi _j\Vert _{L^\infty } \Vert \nabla (\partial ^{{\pmb {\nu }}-{\pmb {e}}_j} u)\Vert _{L^2}\, \Vert \nabla (\partial ^{{\pmb {\nu }}} u)\Vert _{L^2}, \end{aligned}$$

and therefore

$$\begin{aligned} \Vert \nabla (\partial ^{\pmb {\nu }}u) \Vert _{L^2} \le \sum _{j\ge 1}\nu _j\, b_j\, \Vert \nabla (\partial ^{{\pmb {\nu }}-{\pmb {e}}_j} u)\Vert _{L^2}, \end{aligned}$$

where we used the definition of \(b_j\) in (2.3). The induction hypothesis then gives

$$\begin{aligned} \Vert \nabla (\partial ^{\pmb {\nu }}u) \Vert _{L^2}&\le \sum _{j\ge 1}\nu _j\, b_j\, |{\pmb {\nu }}-{\pmb {e}}_j|!\, {\pmb {b}}^{{\pmb {\nu }}-{\pmb {e}}_j}\,\frac{\Vert f\Vert _{V^*}}{a_{\min }} = |{\pmb {\nu }}|!\,{\pmb {b}}^{\pmb {\nu }}\,\frac{\Vert f\Vert _{V^*}}{a_{\min }}. \end{aligned}$$

This completes the proof. \(\square \)

1.2 Proof of Lemma 6.2

This result corresponds to [69, Theorem 6]. Here we take a more direct route with the proof and the bound depends on the sequence \(\overline{{\pmb {b}}}\) which is simpler than the sequence in [69] (there the sequence depends on an additional parameter \(\kappa \in (0,1]\) and other constants), at the expense of increasing the factorial factor from \(|{\pmb {\nu }}|!\) to \((|{\pmb {\nu }}|+1)!\). For simplicity we consider here the case \(f\in L^2(D)\), but the proof can be generalized to the case \(f\in H^{-1+t}(D)\) for \(t\in [0,1]\) as in [69].

Let \(f\in L^2(D)\) and \({\pmb {y}}\in U\). For \({\pmb {\nu }}={\pmb {0}}\), we apply the identity (9.2) to the strong formulation (2.1) to obtain (formally, at this stage, since we do not yet know that \(\varDelta u(\cdot ,{\pmb {y}})\in L^2(D)\))

$$\begin{aligned} - a({\pmb {x}},{\pmb {y}})\, \varDelta u({\pmb {x}},{\pmb {y}}) \ = \nabla a({\pmb {x}},{\pmb {y}})\cdot \nabla u({\pmb {x}},{\pmb {y}}) + f({\pmb {x}}), \end{aligned}$$

which leads to

$$\begin{aligned} a_{\min }\, \Vert \varDelta u(\cdot ,{\pmb {y}})\Vert _{L^2} \le \Vert \nabla a(\cdot ,{\pmb {y}})\Vert _{L^\infty }\,\Vert \nabla u(\cdot ,{\pmb {y}})\Vert _{L^2} + \Vert f\Vert _{L^2}. \end{aligned}$$

Combining this with (3.7) gives

$$\begin{aligned} \Vert \varDelta u(\cdot ,{\pmb {y}})\Vert _{L^2} \le \frac{\sup _{{\pmb {z}}\in U}\Vert \nabla a(\cdot ,{\pmb {z}})\Vert _{L^\infty }}{a_{\min }}\,\frac{\Vert f\Vert _{V^*}}{a_{\min }} + \frac{\Vert f\Vert _{L^2}}{a_{\min }} \le C\,\Vert f\Vert _{L^2}, \end{aligned}$$
(9.10)

where we could take

$$\begin{aligned} C := C_\mathrm{emb}\left( \frac{\sup _{{\pmb {z}}\in U}\Vert \nabla a(\cdot ,{\pmb {z}})\Vert _{L^\infty }}{a_{\min }^2} + \frac{1}{a_{\min }}\right) , \quad \text{ with }\quad C_\mathrm{emb} := \sup _{f\in L^2(D)} \frac{\Vert f\Vert _{V^*}}{\Vert f\Vert _{L^2}}. \end{aligned}$$

Thus, the result holds for \({\pmb {\nu }}= {\pmb {0}}\) (see also (3.8)).

For \({\pmb {\nu }}\ne {\pmb {0}}\), we apply \(\partial ^{\pmb {\nu }}\) to the strong formulation (2.1) and use the Leibniz product rule (9.1) to obtain (suppressing \({\pmb {x}}\) and \({\pmb {y}}\))

$$\begin{aligned} \nabla \cdot \bigg (\sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} (\partial ^{{\pmb {m}}} a)\, \nabla (\partial ^{{\pmb {\nu }}-{\pmb {m}}} u) \bigg ) = 0. \end{aligned}$$

Using again (9.9) and separating out the \({\pmb {m}}= {\pmb {0}}\) term yield the following identity

$$\begin{aligned} \nabla \cdot (a\nabla (\partial ^{\pmb {\nu }}u)) = - \nabla \cdot \left( \sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\,\psi _j({\pmb {x}})\, \nabla (\partial ^{{\pmb {\nu }}-{\pmb {e}}_j} u) \right) . \end{aligned}$$

Applying the identity (9.2) to both sides yields (formally)

$$\begin{aligned} a\,\varDelta (\partial ^{\pmb {\nu }}u) + \nabla a \cdot \nabla (\partial ^{\pmb {\nu }}u) = - \sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\, \left( \psi _j\, \varDelta (\partial ^{{\pmb {\nu }}-{\pmb {e}}_j} u) \, + \, \nabla \psi _j\cdot \nabla (\partial ^{{\pmb {\nu }}-{\pmb {e}}_j} u) \right) . \end{aligned}$$

In turn, we obtain

$$\begin{aligned} a_{\min } \,\Vert \varDelta (\partial ^{\pmb {\nu }}u)\Vert _{L^2}&\le \sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\, \bigg (\Vert \psi _j\Vert _{L^\infty }\, \Vert \varDelta (\partial ^{{\pmb {\nu }}-{\pmb {e}}_j} u)\Vert _{L^2} + \Vert \nabla \psi _j\Vert _{L^\infty } \, \Vert \nabla (\partial ^{{\pmb {\nu }}-{\pmb {e}}_j} u)\Vert _{L^2} \bigg ) \\&\qquad + \Vert \nabla a\Vert _{L^\infty } \,\Vert \nabla (\partial ^{\pmb {\nu }}u)\Vert _{L^2}, \end{aligned}$$

which leads to

$$\begin{aligned} \underbrace{\Vert \varDelta (\partial ^{\pmb {\nu }}u)\Vert _{L^2}}_{{\mathbb {A}}_{\pmb {\nu }}}&\le \sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\,b_j\, \underbrace{\Vert \varDelta (\partial ^{{\pmb {\nu }}-{\pmb {e}}_j} u)\Vert _{L^2}}_{{\mathbb {A}}_{{\pmb {\nu }}-{\pmb {e}}_j}} \,+\, B_{\pmb {\nu }}, \end{aligned}$$

where we used the definition of \(b_j\) in (2.3), and

$$\begin{aligned} B_{\pmb {\nu }}:= \sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\, \frac{\Vert \nabla \psi _j\Vert _{L^\infty }}{a_{\min }} \, \Vert \nabla (\partial ^{{\pmb {\nu }}-{\pmb {e}}_j} u)\Vert _{L^2} + \frac{\Vert \nabla a\Vert _{L^\infty }}{a_{\min }} \,\Vert \nabla (\partial ^{\pmb {\nu }}u)\Vert _{L^2}. \end{aligned}$$

Note that this formulation of \(B_{\pmb {\nu }}\) cannot be used as \({\mathbb {B}}_{\pmb {\nu }}\) in Lemma 9.1 because the base step \({\mathbb {A}}_{\pmb {0}}\le B_{\pmb {0}}\) does not hold. From Lemma 6.1 we can estimate

$$\begin{aligned} B_{\pmb {\nu }}&\le \sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\, \frac{\Vert \nabla \psi _j\Vert _{L^\infty }}{a_{\min }} \, |{\pmb {\nu }}-{\pmb {e}}_j|!\, {\pmb {b}}^{{\pmb {\nu }}-{\pmb {e}}_j}\, \frac{\Vert f\Vert _{V^*}}{a_{\min }} + \frac{\sup _{{\pmb {z}}\in U} \Vert \nabla a(\cdot ,{\pmb {z}})\Vert _{L^\infty }}{a_{\min }} \,|{\pmb {\nu }}|!\, {\pmb {b}}^{{\pmb {\nu }}}\, \frac{\Vert f\Vert _{V^*}}{a_{\min }} \\&\le C\,|{\pmb {\nu }}|!\, \overline{{\pmb {b}}}^{{\pmb {\nu }}}\, \Vert f\Vert _{L^2(D)} \,=:\, {\mathbb {B}}_{\pmb {\nu }}, \end{aligned}$$

where we used the definition of \(\overline{b}_j\ge b_j\) in (2.4). This definition of \({\mathbb {B}}_{\pmb {\nu }}\) ensures that the base step \({\mathbb {A}}_{\pmb {0}}\le {\mathbb {B}}_{\pmb {0}}\) does hold; see (9.10). Now we apply Lemma 9.1 to conclude that

$$\begin{aligned} \Vert \varDelta (\partial ^{\pmb {\nu }}u)\Vert _{L^2}&\le \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }}\, |{\pmb {m}}|!\, {\pmb {b}}^{\pmb {m}}\, C\,|{\pmb {\nu }}-{\pmb {m}}|!\, \overline{{\pmb {b}}}^{{\pmb {\nu }}-{\pmb {m}}}\, \Vert f\Vert _{L^2} \le C\,(|{\pmb {\nu }}|+1)!\, \overline{{\pmb {b}}}^{{\pmb {\nu }}}\, \Vert f\Vert _{L^2}, \end{aligned}$$

where we used the identity (9.4). This completes the proof. \(\square \)

1.3 Proof of Lemma 6.3

This result appeared as a technical step in the proof of [69, Theorem 7], but only first derivatives were considered there, i.e., \(\nu _j\le 1\) for all j. Here we consider general derivatives, and we make use of Lemma 6.2 instead of [69, Theorem 6] so that the sequence \(\overline{{\pmb {b}}}\) is different, the factorial factor is larger, and we restrict to \(f\in L^2(D)\).

Let \(f\in L^2(D)\), \({\pmb {y}}\in U\) and \({\pmb {\nu }}\in {\mathfrak {F}}\). Galerkin orthogonality for the FE method yields

$$\begin{aligned} {\mathscr {A}}({\pmb {y}};u(\cdot ,{\pmb {y}})-u_h(\cdot ,{\pmb {y}}), z_h)\,= \,0 \qquad \text {for all} \quad z_h \in V_h, \end{aligned}$$
(9.11)

Let \({\mathscr {I}}:V\rightarrow V\) denote the identity operator and let \({\mathscr {P}}_h = {\mathscr {P}}_h({\pmb {y}}) :V\rightarrow V_h\) denote the parametric FE projection onto \(V_h\) which is defined, for arbitrary \(w\in V\), by

$$\begin{aligned} {\mathscr {A}}({\pmb {y}}; {\mathscr {P}}_h({\pmb {y}}) w - w, z_h) = 0 \qquad \text {for all} \quad z_h\in V_h . \end{aligned}$$
(9.12)

In particular, we have \(u_h = {\mathscr {P}}_h u \in V_h\) and

$$\begin{aligned} {\mathscr {P}}_h^2({\pmb {y}}) \,\equiv \, {\mathscr {P}}_h({\pmb {y}}) \quad \text{ on }\quad V_h . \end{aligned}$$
(9.13)

Moreover, since \(\partial ^{\pmb {\nu }}u_h \in V_h\) for every \({\pmb {\nu }}\in {\mathfrak {F}}\), it follows from (9.13) that

$$\begin{aligned} ({\mathscr {I}}- {\mathscr {P}}_h({\pmb {y}}))(\partial ^{\pmb {\nu }}u_h(\cdot ,{\pmb {y}})) \,\equiv \, 0. \end{aligned}$$
(9.14)

Thus,

$$\begin{aligned} \Vert \nabla \partial ^{\pmb {\nu }}(u - u_{h})\Vert _{L^2}&= \Vert \nabla {\mathscr {P}}_h \partial ^{\pmb {\nu }}(u - u_{h}) + \nabla ({\mathscr {I}}- {\mathscr {P}}_h) \partial ^{\pmb {\nu }}(u - u_{h}) \Vert _{L^2} \nonumber \\&\le \Vert \nabla {\mathscr {P}}_h \partial ^{\pmb {\nu }}(u - u_{h})\Vert _{L^2} \,+\, \Vert \nabla ({\mathscr {I}}- {\mathscr {P}}_h) \partial ^{\pmb {\nu }}u \Vert _{L^2} . \end{aligned}$$
(9.15)

We stress here that, since the parametric FE projection \({\mathscr {P}}_h({\pmb {y}})\) depends on \({\pmb {y}}\), in general we have \(\partial ^{\pmb {\nu }}(u(\cdot ,{\pmb {y}}) - u_{h}(\cdot ,{\pmb {y}})) \ne ({\mathscr {I}}- {\mathscr {P}}_h({\pmb {y}})) (\partial ^{\pmb {\nu }}u(\cdot ,{\pmb {y}}))\); this is why we need the estimate (9.15).

Now, applying \(\partial ^{\pmb {\nu }}\) to (9.11) and recalling (9.9), we get for all \(z_h\in V_h\),

$$\begin{aligned} \int _D a\, \nabla \partial ^{{\pmb {\nu }}} (u-u_h)\cdot \nabla z_h \,{\mathrm {d}}{\pmb {x}}= - \sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\, \int _D \psi _j \nabla \partial ^{{\pmb {\nu }}-{\pmb {e}}_j}(u-u_h)\cdot \nabla z_h \,{\mathrm {d}}{\pmb {x}}. \end{aligned}$$
(9.16)

Choosing \(z_h = {\mathscr {P}}_h \partial ^{\pmb {\nu }}(u-u_h)\) and using the definition (9.12) of \({\mathscr {P}}_h\), the left-hand side of (9.16) is equal to \(\int _D a\, |\nabla {\mathscr {P}}_h \partial ^{{\pmb {\nu }}} (u-u_h)|^2 \,{\mathrm {d}}{\pmb {x}}\). Using the Cauchy–Schwarz inequality, we then obtain

$$\begin{aligned}&a_{\min }\, \Vert \nabla {\mathscr {P}}_h \partial ^{{\pmb {\nu }}} (u-u_h)\Vert _{L^2}^2 \\&\quad \le \, \sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\, \Vert \psi _j\Vert _{L^\infty } \Vert \nabla \partial ^{{\pmb {\nu }}-{\pmb {e}}_j}(u-u_h)\Vert _{L^2}\, \Vert \nabla {\mathscr {P}}_h \partial ^{{\pmb {\nu }}} (u-u_h)\Vert _{L^2}. \end{aligned}$$

Canceling one common factor from both sides, we arrive at

$$\begin{aligned} \Vert \nabla {\mathscr {P}}_h \partial ^{{\pmb {\nu }}} (u-u_h)\Vert _{L^2}&\le \sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\,b_j\, \Vert \nabla \partial ^{{\pmb {\nu }}-{\pmb {e}}_j}(u-u_h)\Vert _{L^2}, \end{aligned}$$
(9.17)

where we used the definition of \(b_j\) in (2.3). Substituting (9.17) into (9.15), we then obtain

$$\begin{aligned} \underbrace{\Vert \nabla \partial ^{\pmb {\nu }}(u - u_{h})\Vert _{L^2}}_{{\mathbb {A}}_{\pmb {\nu }}}&\le \sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\,b_j\, \underbrace{\Vert \nabla \partial ^{{\pmb {\nu }}-{\pmb {e}}_j}(u-u_h)\Vert _{L^2}}_{{\mathbb {A}}_{{\pmb {\nu }}-{\pmb {e}}_j}} \,+\, \underbrace{\Vert \nabla ({\mathscr {I}}- {\mathscr {P}}_h) \partial ^{\pmb {\nu }}u\Vert _{L^2}}_{{\mathbb {B}}_{\pmb {\nu }}} . \end{aligned}$$

Noting that \({\mathbb {A}}_{\pmb {0}}= {\mathbb {B}}_{\pmb {0}}\), we now apply Lemma 9.1 to obtain

$$\begin{aligned} \Vert \nabla \partial ^{\pmb {\nu }}(u - u_{h})\Vert _{L^2}&\le \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }}\, |{\pmb {\nu }}|!\, {\pmb {b}}^{\pmb {\nu }}\, \Vert \nabla ({\mathscr {I}}- {\mathscr {P}}_h) \partial ^{{\pmb {\nu }}-{\pmb {m}}} u\Vert _{L^2}. \end{aligned}$$

Next we use the FE estimate (3.9) that for all \({\pmb {y}}\in U\) and \(w\in H^2(D)\), we have \(\Vert \nabla ({\mathscr {I}}- {\mathscr {P}}_h) w \Vert _{L^2} \,\lesssim \, h\, \Vert \varDelta w\Vert _{L^2}\). Hence, from Lemma 6.2 we obtain

$$\begin{aligned} \Vert \nabla \partial ^{\pmb {\nu }}(u - u_{h})\Vert _{L^2}&\,\lesssim \, \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }}\, |{\pmb {\nu }}|!\, {\pmb {b}}^{\pmb {\nu }}\, h\, \Vert \varDelta (\partial ^{{\pmb {\nu }}-{\pmb {m}}} u)\Vert _{L^2} \\&\,\lesssim \, \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }}\, |{\pmb {\nu }}|!\, {\pmb {b}}^{\pmb {\nu }}\, h\, (|{\pmb {\nu }}-{\pmb {m}}|+1)!\, \overline{{\pmb {b}}}^{{\pmb {\nu }}-{\pmb {m}}}\, \Vert f\Vert _{L^2} \\&\,\lesssim \, h\, \frac{(|{\pmb {\nu }}|+2)!}{2}\,\overline{{\pmb {b}}}^{\pmb {\nu }}\, \Vert f\Vert _{L^2}, \end{aligned}$$

where we used the identity (9.5). This completes the proof.

1.4 Proof of Lemma 6.4

This result generalizes [69, Theorem 7] from first derivatives to general derivatives. The proof is based on a duality argument since G is a bounded linear functional. It makes use of Lemma 6.3, and therefore, the sequence \(\overline{{\pmb {b}}}\) is different, the factorial factor is larger, and we restrict to \(f,G\in L^2(D)\) here.

Let \(f,G\in L^2(D)\) and \({\pmb {y}}\in U\). We define \(v^G(\cdot ,{\pmb {y}})\in V\) and \(v_h^G(\cdot ,{\pmb {y}})\in V_h\) via the adjoint problems

$$\begin{aligned} {\mathscr {A}}({\pmb {y}}; w, v^G(\cdot ,{\pmb {y}})) = G( w ) \quad&\text {for all}\quad w\in V, \end{aligned}$$
(9.18)
$$\begin{aligned} {\mathscr {A}}({\pmb {y}}; w_h, v_h^G(\cdot ,{\pmb {y}})) = G ( w_h ) \;\;\;\;\,\quad&\text {for all}\quad w_h\in V_h. \end{aligned}$$
(9.19)

Due to Galerkin orthogonality (9.11) for the original problem, by choosing the test function \(w = u(\cdot ,{\pmb {y}}) - u_h(\cdot ,{\pmb {y}})\) in (9.18), we obtain

$$\begin{aligned} G(u(\cdot ,{\pmb {y}})-u_h(\cdot ,{\pmb {y}})) = {\mathscr {A}}({\pmb {y}}; u(\cdot ,{\pmb {y}})-u_h(\cdot ,{\pmb {y}}), v^G(\cdot ,{\pmb {y}}) - v_h^G(\cdot ,{\pmb {y}})). \end{aligned}$$
(9.20)

From the Leibniz product rule (9.1) and (9.9) we have for \({\pmb {\nu }}\in {\mathfrak {F}}\)

$$\begin{aligned} \partial ^{\pmb {\nu }}G(u - u_{h})&= \int _D \partial ^{\pmb {\nu }}\big (a \,\nabla (u - u_{h}) \cdot \nabla \big (v^G - v^G_h \big ) \big )\, {\mathrm {d}}{\pmb {x}}\\&= \int _D \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} (\partial ^{{\pmb {m}}} a)\, \partial ^{{\pmb {\nu }}-{\pmb {m}}} \big ( \nabla (u - u_{h}) \cdot \nabla \big (v^G - v^G_h \big ) \big )\, {\mathrm {d}}{\pmb {x}}\\&= \int _D a\, \partial ^{{\pmb {\nu }}} \big ( \nabla (u - u_{h}) \cdot \nabla \big (v^G - v^G_h \big ) \big )\, {\mathrm {d}}{\pmb {x}}\\&\quad + \sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\, \int _D \psi _j\,\partial ^{{\pmb {\nu }}-{\pmb {e}}_j} \big ( \nabla (u - u_{h}) \cdot \nabla \big (v^G - v^G_h \big ) \big )\, {\mathrm {d}}{\pmb {x}}\\&= \int _D a\, \sum _{{\pmb {k}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {k}}\end{array}}\right) }} \nabla \partial ^{\pmb {k}}(u- u_{h}) \cdot \nabla \partial ^{{\pmb {\nu }}-{\pmb {k}}} \big (v^G - v^G_h \big )\, {\mathrm {d}}{\pmb {x}}\\&\quad + \sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\, \int _D \psi _j\,\sum _{{\pmb {k}}\le {\pmb {\nu }}-{\pmb {e}}_j} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}-{\pmb {e}}_j\\ {\pmb {k}}\end{array}}\right) }} \nabla \partial ^{\pmb {k}}(u- u_{h}) \cdot \nabla \partial ^{{\pmb {\nu }}-{\pmb {e}}_j-{\pmb {k}}} \big (v^G - v^G_h \big )\, {\mathrm {d}}{\pmb {x}}. \end{aligned}$$

The Cauchy–Schwarz inequality then yields

$$\begin{aligned}&|\partial ^{\pmb {\nu }}G(u - u_{h})| \le a_{\max } \sum _{{\pmb {k}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {k}}\end{array}}\right) }} \Vert \nabla \partial ^{\pmb {k}}(u- u_{h})\Vert _{L^2} \, \Vert \nabla \partial ^{{\pmb {\nu }}-{\pmb {k}}} \big (v^G - v^G_h \big )\Vert _{L^2} \nonumber \\&\quad + \sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\, \Vert \psi _j\Vert _{L^\infty }\,\sum _{{\pmb {k}}\le {\pmb {\nu }}-{\pmb {e}}_j} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}{-}{\pmb {e}}_j\\ {\pmb {k}}\end{array}}\right) }} \Vert \nabla \partial ^{\pmb {k}}(u{-} u_{h})\Vert _{L^2}\Vert \nabla \partial ^{{\pmb {\nu }}-{\pmb {e}}_j{-}{\pmb {k}}} \big (v^G {-} v^G_h \big )\Vert _{L^2}. \end{aligned}$$
(9.21)

We see from (the proof of) Lemma 6.3 that

$$\begin{aligned} \Vert \nabla \partial ^{\pmb {k}}(u - u_{h})\Vert _{L^2}&\,\lesssim \, h\, \frac{(|{\pmb {k}}|+2)!}{2}\,\overline{{\pmb {b}}}^{\pmb {k}}\, \Vert f\Vert _{L^2}. \end{aligned}$$
(9.22)

Since the bilinear form \({\mathscr {A}}({\pmb {y}};\cdot ,\cdot )\) is symmetric and since the representer g for the linear functional G is in \(L^2(D)\), all the results hold verbatim also for the adjoint problem (9.18) and for its FE discretisation (9.19). Hence, as in (9.22), we obtain

$$\begin{aligned} \Vert \nabla \partial ^{{\pmb {\nu }}-{\pmb {k}}} (v^G - v^G_{h})\Vert _{L^2}&\,\lesssim \, h \, \frac{(|{\pmb {\nu }}-{\pmb {k}}|+2)!}{2}\, \overline{{\pmb {b}}}^{{\pmb {\nu }}-{\pmb {k}}}\, \Vert G\Vert _{L^2}. \end{aligned}$$
(9.23)

Substituting (9.22) and (9.23) into (9.21), and using \(\Vert \psi _j\Vert _{L^\infty } = a_{\min } b_j \le a_{\max } \overline{b}_j\), yields

$$\begin{aligned}&|\partial ^{\pmb {\nu }}G(u - u_{h})| \\&\quad \,\lesssim \, a_{\max }\,\Vert f\Vert _{L^2}\,\Vert G\Vert _{L^2}\, h^2\, \bigg ( \sum _{{\pmb {k}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {k}}\end{array}}\right) }}\, \frac{(|{\pmb {k}}|+2)!}{2}\,\overline{{\pmb {b}}}^{\pmb {k}}\, \frac{(|{\pmb {\nu }}-{\pmb {k}}|+2)!}{2}\,\overline{{\pmb {b}}}^{{\pmb {\nu }}-{\pmb {k}}} \\&\qquad + \sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\,\overline{b}_j\, \sum _{{\pmb {k}}\le {\pmb {\nu }}-{\pmb {e}}_j} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}-{\pmb {e}}_j\\ {\pmb {k}}\end{array}}\right) }}\,\frac{(|{\pmb {k}}|+2)!}{2}\,\overline{{\pmb {b}}}^{\pmb {k}}\, \frac{(|{\pmb {\nu }}-{\pmb {e}}_j-{\pmb {k}}|+2)!}{2}\,\overline{{\pmb {b}}}^{{\pmb {\nu }}-{\pmb {e}}_j-{\pmb {k}}} \bigg ) \\&\quad \,\lesssim \, a_{\max }\,\Vert f\Vert _{L^2}\,\Vert G\Vert _{L^2}\, h^2\, \bigg (\frac{(|{\pmb {\nu }}|+5)!}{120} +\sum _{j\in {\mathrm {supp}}({\pmb {\nu }})} \nu _j\,\frac{(|{\pmb {\nu }}-{\pmb {e}}_j|+5)!}{120} \bigg )\overline{{\pmb {b}}}^{\pmb {\nu }}\\&\quad \,\lesssim \, a_{\max }\,\Vert f\Vert _{L^2}\,\Vert G\Vert _{L^2}\, h^2\, \frac{(|{\pmb {\nu }}|+5)!}{120}\, \overline{{\pmb {b}}}^{\pmb {\nu }}, \end{aligned}$$

where we used the identity (9.6). This completes the proof. \(\square \)

1.5 Proof of Lemma 6.5

This result is [46, Theorem 14]. We include the proof here since, unlike in the uniform case where we do induction directly for the quantity \(\Vert \nabla (\partial ^{{\pmb {\nu }}}u(\cdot ,{\pmb {y}}))\Vert _{L^2}\), here we need to work with \(\Vert a^{1/2}(\cdot ,{\pmb {y}})\,\nabla (\partial ^{{\pmb {\nu }}}u(\cdot ,{\pmb {y}}))\Vert _{L^2}\), and this technical step is needed for the subsequent proof.

Let \(f\in V^*\) and \({\pmb {y}}\in U_{\pmb {b}}\). We first prove by induction on \(|{\pmb {\nu }}|\) that

$$\begin{aligned} \Vert a^{1/2}(\cdot ,{\pmb {y}})\,\nabla (\partial ^{{\pmb {\nu }}}u(\cdot ,{\pmb {y}}))\Vert _{L^2} \le \Lambda _{|{\pmb {\nu }}|}\,{\pmb {\beta }}^{\pmb {\nu }}\, \frac{\Vert f\Vert _{V^*}}{\sqrt{a_{\min }({\pmb {y}})}}, \end{aligned}$$
(9.24)

where the sequence \((\Lambda _n)_{n\ge 0}\) is defined recursively by (9.7) and satisfies (9.8).

We take \(v = u(\cdot ,{\pmb {y}})\) in the weak form (3.5) to obtain

$$\begin{aligned} \int _D a \,|\nabla u |^2\,{\mathrm {d}}{\pmb {x}}&\;\le \; \Vert f\Vert _{V^*}\,\Vert u(\cdot ,{\pmb {y}})\Vert _{V} \;\le \; \frac{\Vert f\Vert _{V^*}}{\sqrt{a_{\min }({\pmb {y}})}} \left( \int _D a \, |\nabla u |^2 \,{\mathrm {d}}{\pmb {x}}\right) ^{1/2}, \end{aligned}$$

and then cancel the common factor from both sides to obtain (9.24) for the case \({\pmb {\nu }}= {\pmb {0}}\). Given any multi-index \({\pmb {\nu }}\) with \(|{\pmb {\nu }}| = n\ge 1\), we apply \(\partial ^{\pmb {\nu }}\) to (3.5) to obtain

$$\begin{aligned} \int _D \bigg ( \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} (\partial ^{{\pmb {\nu }}-{\pmb {m}}} a) \,\nabla (\partial ^{{\pmb {m}}} u) \cdot \nabla z\bigg ) {\mathrm {d}}{\pmb {x}}= 0 \qquad \text{ for } \text{ all }\quad z\in V\;. \end{aligned}$$

Taking \(z = \partial ^{\pmb {\nu }}u(\cdot ,{\pmb {y}})\), separating out the \({\pmb {m}}= {\pmb {\nu }}\) term, dividing and multiplying by a, and using the Cauchy–Schwarz inequality, we obtain

$$\begin{aligned}&\int _D a\, |\nabla (\partial ^{{\pmb {\nu }}} u) |^2 {\mathrm {d}}{\pmb {x}}= - \sum _{\mathop {\scriptstyle {{\pmb {m}}\ne {\pmb {\nu }}}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} \int _D (\partial ^{{\pmb {\nu }}-{\pmb {m}}} a) \,\nabla (\partial ^{{\pmb {m}}} u) \cdot \nabla (\partial ^{{\pmb {\nu }}} u) \, {\mathrm {d}}{\pmb {x}}\\&\quad \;\le \; \sum _{\mathop {\scriptstyle {{\pmb {m}}\ne {\pmb {\nu }}}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} \bigg \Vert \frac{\partial ^{{\pmb {\nu }}-{\pmb {m}}} a (\cdot ,{\pmb {y}})}{a(\cdot ,{\pmb {y}})}\bigg \Vert _{L^\infty } \left( \int _D a |\nabla (\partial ^{{\pmb {m}}} u) |^2\,{\mathrm {d}}{\pmb {x}}\right) ^{1/2} \left( \int _D a |\nabla (\partial ^{{\pmb {\nu }}} u) |^2\,{\mathrm {d}}{\pmb {x}}\right) ^{1/2}. \end{aligned}$$

We observe from (2.6) that

$$\begin{aligned} \partial ^{{\pmb {\nu }}-{\pmb {m}}} a = a\prod _{j\ge 1} (\sqrt{\mu _j}\,\xi _j)^{\nu _j-m_j} \quad \text {for all} \quad {\pmb {\nu }}\ne {\pmb {m}}, \end{aligned}$$
(9.25)

and therefore

$$\begin{aligned} \bigg \Vert \frac{\partial ^{{\pmb {\nu }}-{\pmb {m}}} a (\cdot ,{\pmb {y}})}{a(\cdot ,{\pmb {y}})}\bigg \Vert _{L^\infty } = \bigg \Vert \prod _{j\ge 1} (\sqrt{\mu _j}\,\xi _j)^{\nu _j-m_j}\bigg \Vert _{L^\infty } \le {\pmb {\beta }}^{{\pmb {\nu }}-{\pmb {m}}}. \end{aligned}$$
(9.26)

Thus, we arrive at

$$\begin{aligned} \left( \int _D a |\nabla (\partial ^{{\pmb {\nu }}} u) |^2\,{\mathrm {d}}{\pmb {x}}\right) ^{1/2}&\le \sum _{\mathop {\scriptstyle {{\pmb {m}}\ne {\pmb {\nu }}}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }}\, {\pmb {\beta }}^{{\pmb {\nu }}-{\pmb {m}}} \left( \int _D a |\nabla (\partial ^{{\pmb {m}}} u)|^2\,{\mathrm {d}}{\pmb {x}}\right) ^{1/2}. \end{aligned}$$

We now use the inductive hypothesis that (9.24) holds when \(\vert {\pmb {\nu }}\vert \le n-1\) in each of the terms on the right-hand side to obtain

$$\begin{aligned} \left( \int _D a |\nabla (\partial ^{{\pmb {\nu }}} u) |^2\,{\mathrm {d}}{\pmb {x}}\right) ^{1/2}&\le \sum _{i=0}^{n-1} \sum _{\mathop {\scriptstyle {|{\pmb {m}}|=i}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }}\, {\pmb {\beta }}^{{\pmb {\nu }}-{\pmb {m}}}\, \Lambda _i\, {\pmb {\beta }}^{\pmb {m}}\, \frac{\Vert f\Vert _{V^*}}{\sqrt{a_{\min }({\pmb {y}})}} \\&= \sum _{i=0}^{n-1} {\textstyle {\left( {\begin{array}{c}n\\ i\end{array}}\right) }} \Lambda _i\, {\pmb {\beta }}^{\pmb {\nu }}\frac{\Vert f\Vert _{V^*}}{\sqrt{a_{\min }({\pmb {y}})}} = \Lambda _n\, {\pmb {\beta }}^{\pmb {\nu }}\frac{\Vert f\Vert _{V^*}}{\sqrt{a_{\min }({\pmb {y}})}}, \end{aligned}$$

where we used the identity (9.3). This completes the induction proof of (9.24).

The desired bound in the lemma is obtained by applying (9.8) on the right-hand side of (9.24), and by noting that the left-hand side of (9.24) can be bounded from below by \(\sqrt{a_{\min }({\pmb {y}})}\,\Vert \partial ^{{\pmb {\nu }}} u(\cdot ,{\pmb {y}}) \Vert _V\). The case \({\pmb {\nu }}={\pmb {0}}\) corresponds to (3.12). This completes the proof. \(\square \)

1.6 Proof of Lemma 6.6

This result was proved in [66] based on an argument similar to the proof of Lemma 6.2 in the uniform case. The tricky point of the proof is in recognizing that for the recursion to work in the lognormal case we need to multiply the expression by \(a^{-1/2}(\cdot ,{\pmb {y}})\), which is not intuitive.

Let \(f\in L^2(D)\) and \({\pmb {y}}\in U_{\overline{{\pmb {\beta }}}}\). For any multi-index \({\pmb {\nu }}\ne {\pmb {0}}\), we apply \(\partial ^{\pmb {\nu }}\) to (2.1) to obtain (formally, at this stage)

$$\begin{aligned} \nabla \cdot \partial ^{\pmb {\nu }}(a\nabla u) = \nabla \cdot \left( \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} (\partial ^{{\pmb {\nu }}-{\pmb {m}}} a)\, \nabla (\partial ^{\pmb {m}}u) \right) = 0. \end{aligned}$$

Separating out the \({\pmb {m}}= {\pmb {\nu }}\) term yields the following identity

$$\begin{aligned} g_{\pmb {\nu }}:= \nabla \cdot (a\nabla (\partial ^{\pmb {\nu }}u))&= - \nabla \cdot \Bigg (\sum _{\mathop {\scriptstyle {{\pmb {m}}\ne {\pmb {\nu }}}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} (\partial ^{{\pmb {\nu }}-{\pmb {m}}} a)\, \nabla (\partial ^{\pmb {m}}u) \Bigg ) \\&= - \sum _{\mathop {\scriptstyle {{\pmb {m}}\ne {\pmb {\nu }}}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} \nabla \cdot \left( \frac{\partial ^{{\pmb {\nu }}-{\pmb {m}}} a}{a}\, (a \nabla (\partial ^{\pmb {m}}u)) \right) \\&= - \sum _{\mathop {\scriptstyle {{\pmb {m}}\ne {\pmb {\nu }}}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} \left( \frac{\partial ^{{\pmb {\nu }}-{\pmb {m}}} a}{a}\, g_{\pmb {m}}\, + \, \nabla \left( \frac{\partial ^{{\pmb {\nu }}-{\pmb {m}}} a}{a}\right) \cdot (a \nabla (\partial ^{\pmb {m}}u)) \right) , \end{aligned}$$

where we used the identity (9.2). Due to Assumption (L2) we may multiply \(g_{\pmb {\nu }}\) by \(a^{-1/2}\) and obtain, for any \(|{\pmb {\nu }}| > 0\), the recursive bound

$$\begin{aligned} \Vert a^{-1/2}g_{\pmb {\nu }}\Vert _{L^2}&\le \sum _{\mathop {\scriptstyle {{\pmb {m}}\ne {\pmb {\nu }}}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} \bigg (\bigg \Vert \frac{\partial ^{{\pmb {\nu }}-{\pmb {m}}} a}{a}\bigg \Vert _{L^\infty }\, \Vert a^{-1/2} g_{\pmb {m}}\Vert _{L^2} \nonumber \\&\quad \quad \quad \quad \quad \quad + \; \bigg \Vert \nabla \bigg (\frac{\partial ^{{\pmb {\nu }}-{\pmb {m}}} a}{a}\bigg )\bigg \Vert _{L^\infty }\, \Vert a^{1/2} \nabla (\partial ^{\pmb {m}}u)\Vert _{L^2} \bigg ). \end{aligned}$$
(9.27)

By assumption, \(-g_{\pmb {0}}=f\in L^2(D)\), so that we obtain (by induction with respect to \(|{\pmb {\nu }}|\)) from (9.27) that \(a^{-1/2}(\cdot ,{\pmb {y}})\,g_{\pmb {\nu }}(\cdot ,{\pmb {y}})\in L^2(D)\), and hence from Assumption (L2) that \(g_{\pmb {\nu }}(\cdot ,{\pmb {y}})\in L^2(D)\) for every \({\pmb {\nu }}\in {\mathfrak {F}}\). The above formal identities therefore hold in \(L^2(D)\).

To complete the proof, it remains to bound the above \(L^2\) norm. Applying the product rule to (9.25) we obtain

$$\begin{aligned} \nabla \left( \frac{\partial ^{{\pmb {\nu }}-{\pmb {m}}} a}{a}\right) = \sum _{k\ge 1} (\nu _k-m_k)(\sqrt{\mu _k}\,\xi _k)^{\nu _k-m_k-1} (\sqrt{\mu _k}\,\nabla \xi _k) \prod _{\mathop {\scriptstyle {j\ne k}}\limits ^{\scriptstyle {j\ge 1}}} (\sqrt{\mu _j}\,\xi _j)^{\nu _j-m_j}. \end{aligned}$$

Due to the definition of \(\overline{\beta }_j\) in (2.9), this implies, in a similar manner to (9.26), that

$$\begin{aligned} \bigg \Vert \nabla \bigg (\frac{\partial ^{{\pmb {\nu }}-{\pmb {m}}} a}{a}\bigg )\bigg \Vert _{L^\infty } \le |{\pmb {\nu }}-{\pmb {m}}| \,\overline{{\pmb {\beta }}}^{{\pmb {\nu }}-{\pmb {m}}}. \end{aligned}$$
(9.28)

Substituting (9.26) and (9.28) into (9.27), we conclude that

$$\begin{aligned} \underbrace{\Vert a^{-1/2}g_{\pmb {\nu }}\Vert _{L^2}}_{{\mathbb {A}}_{\pmb {\nu }}} \le \sum _{\mathop {\scriptstyle {{\pmb {m}}\ne {\pmb {\nu }}}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }}\,{\pmb {\beta }}^{{\pmb {\nu }}-{\pmb {m}}}\, \underbrace{\Vert a^{-1/2}g_{\pmb {m}}\Vert _{L^2}}_{{\mathbb {A}}_{\pmb {m}}} \,+\, B_{\pmb {\nu }}, \end{aligned}$$

where

$$\begin{aligned} B_{\pmb {\nu }}\,:=&\, \sum _{\mathop {\scriptstyle {{\pmb {m}}\ne {\pmb {\nu }}}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }}\, |{\pmb {\nu }}-{\pmb {m}}| \,\overline{{\pmb {\beta }}}^{{\pmb {\nu }}-{\pmb {m}}}\, \Vert a^{1/2} \nabla (\partial ^{\pmb {m}}u)\Vert _{L^2} \\ \,\le&\, \sum _{\mathop {\scriptstyle {{\pmb {m}}\ne {\pmb {\nu }}}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }}\, |{\pmb {\nu }}-{\pmb {m}}| \,\overline{{\pmb {\beta }}}^{{\pmb {\nu }}-{\pmb {m}}}\, \Lambda _{|{\pmb {m}}|}\,{\pmb {\beta }}^{\pmb {m}}\, \frac{\Vert f\Vert _{V^*}}{\sqrt{a_{\min }({\pmb {y}})}} \le \overline{\Lambda }_{|{\pmb {\nu }}|}\,\overline{{\pmb {\beta }}}^{\pmb {\nu }}\frac{\Vert f\Vert _{V^*}}{\sqrt{a_{\min }({\pmb {y}})}}, \end{aligned}$$

where we used (9.24) and again the identity (9.3) to write, with \(n = |{\pmb {\nu }}|\),

$$\begin{aligned} \sum _{\mathop {\scriptstyle {{\pmb {m}}\ne {\pmb {\nu }}}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }}\, |{\pmb {\nu }}-{\pmb {m}}|\, \Lambda _{|{\pmb {m}}|} = \sum _{i=0}^{n-1} \sum _{\mathop {\scriptstyle {|{\pmb {m}}|=i}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }}\, (n-i)\, \Lambda _i = \sum _{i=0}^{n-1} {\textstyle {\left( {\begin{array}{c}n\\ i\end{array}}\right) }}\, (n-i)\, \Lambda _i \,=:\, \overline{\Lambda }_n. \end{aligned}$$

Since \({\mathbb {A}}_{\pmb {0}}= \Vert a^{-1/2} f\Vert _{L^2} \le \Vert f\Vert _{L^2}/\sqrt{a_{\min }({\pmb {y}})}\), we now define

$$\begin{aligned} {\mathbb {B}}_{\pmb {\nu }}:= C_\mathrm{emb}\, \overline{\Lambda }_{|{\pmb {\nu }}|}\,\overline{{\pmb {\beta }}}^{\pmb {\nu }}\frac{\Vert f\Vert _{L^2}}{\sqrt{a_{\min }({\pmb {y}})}}, \end{aligned}$$

so that \({\mathbb {A}}_{\pmb {0}}\le {\mathbb {B}}_{\pmb {0}}\) and \(B_{\pmb {\nu }}\le {\mathbb {B}}_{\pmb {\nu }}\) for all \({\pmb {\nu }}\). We may now apply Lemma 9.2 to obtain

$$\begin{aligned} \Vert a^{-1/2}g_{\pmb {\nu }}\Vert _{L^2}&\le \sum _{{\pmb {k}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {k}}\end{array}}\right) }}\Lambda _{|{\pmb {k}}|}\,{\pmb {\beta }}^{\pmb {k}}\, C_\mathrm{emb}\, \overline{\Lambda }_{|{\pmb {\nu }}-{\pmb {k}}|}\,\overline{{\pmb {\beta }}}^{{\pmb {\nu }}-{\pmb {k}}} \frac{\Vert f\Vert _{L^2}}{\sqrt{a_{\min }({\pmb {y}})}}. \end{aligned}$$
(9.29)

Note the extra factor \(n-i\) in the definition of \(\overline{\Lambda }_n\) compared to \(\Lambda _n\) in (9.7) so that \(\Lambda _n\le \overline{\Lambda }_n\). Using the bound in (9.8) with \(\alpha \le \ln 2\), we have

$$\begin{aligned} \overline{\Lambda }_n \le \sum _{i=0}^{n-1} {\textstyle {\left( {\begin{array}{c}n\\ i\end{array}}\right) }} (n-i)\,\frac{i!}{\alpha ^i} = \frac{n!}{\alpha ^n}\,\alpha \,\sum _{i=0}^{n-1} \frac{\alpha ^{n-i-1}}{(n-i-1)!} = \frac{n!}{\alpha ^n}\,\alpha \,\sum _{k=0}^{n-1} \frac{\alpha ^k}{k!} \le \frac{n!}{\alpha ^n}\,\alpha \, e^\alpha \le \frac{n!}{\alpha ^n}, \end{aligned}$$

where the final step is valid provided that \(\alpha \,e^\alpha \le 1\). Thus, it suffices to choose \(\alpha \le 0.567\cdots \). For convenience we take \(\alpha = 0.5\) to bound (9.29). This together with the identity (9.4) gives

$$\begin{aligned} \Vert a^{-1/2}g_{\pmb {\nu }}\Vert _{L^2} \le C_\mathrm{emb}\, (|{\pmb {\nu }}|+1)!\,2^{|{\pmb {\nu }}|}\,\overline{{\pmb {\beta }}}^{\pmb {\nu }}\,\frac{\Vert f\Vert _{L^2}}{\sqrt{a_{\min }({\pmb {y}})}}. \end{aligned}$$
(9.30)

Since \(a^{-1/2}g_{\pmb {\nu }}= a^{-1/2}\nabla \cdot (a\nabla (\partial ^{\pmb {\nu }}u)) = a^{1/2}\varDelta (\partial ^{\pmb {\nu }}u) + a^{-1/2}\,(\nabla a\cdot \nabla (\partial ^{\pmb {\nu }}u))\) by applying (9.2), we have

$$\begin{aligned} \Vert a^{1/2}\varDelta (\partial ^{\pmb {\nu }}u)\Vert _{L^2} \le \Vert a^{-1/2}g_{\pmb {\nu }}\Vert _{L^2} \,+\, \Vert a^{-1/2}\,(\nabla a\cdot \nabla (\partial ^{\pmb {\nu }}u))\Vert _{L^2}, \end{aligned}$$

which yields

$$\begin{aligned} \sqrt{a_{\min }({\pmb {y}})}\,\Vert \varDelta (\partial ^{\pmb {\nu }}u)\Vert _{L^2} \le \Vert a^{-1/2}g_{\pmb {\nu }}\Vert _{L^2} \,+\, \frac{\Vert \nabla a(\cdot ,{\pmb {y}})\Vert _{L^\infty } }{a_{\min }({\pmb {y}})} \Vert a^{1/2}\nabla (\partial ^{\pmb {\nu }}u)\Vert _{L^2}, \end{aligned}$$

and in turn

$$\begin{aligned} \Vert \varDelta (\partial ^{\pmb {\nu }}u)\Vert _{L^2} \le \frac{\Vert a^{-1/2}g_{\pmb {\nu }}\Vert _{L^2}}{\sqrt{a_{\min }({\pmb {y}})}} \,+\, \frac{\Vert \nabla a(\cdot ,{\pmb {y}})\Vert _{L^\infty } }{a_{\min }({\pmb {y}})}\, \frac{\Vert a^{1/2}\nabla (\partial ^{\pmb {\nu }}u)\Vert _{L^2}}{\sqrt{a_{\min }({\pmb {y}})}}. \end{aligned}$$
(9.31)

Substituting (9.30) and (9.24) into (9.31), and using \(\Lambda _{|{\pmb {\nu }}|}\le 2^{|{\pmb {\nu }}|}|{\pmb {\nu }}|!\) and \({\pmb {\beta }}^{\pmb {\nu }}\le \overline{{\pmb {\beta }}}^{\pmb {\nu }}\), we conclude that

$$\begin{aligned} \Vert \varDelta (\partial ^{\pmb {\nu }}u)\Vert _{L^2}&\le C_\mathrm{emb}\, \bigg ( \frac{1}{a_{\min }({\pmb {y}})} + \frac{\Vert \nabla a(\cdot ,{\pmb {y}})\Vert _{L^\infty }}{a_{\min }^2({\pmb {y}})}\bigg ) (|{\pmb {\nu }}|+1)!\,2^{|{\pmb {\nu }}|}\,\overline{{\pmb {\beta }}}^{\pmb {\nu }}\,\Vert f\Vert _{L^2}. \end{aligned}$$

This completes the proof. \(\square \)

1.7 Proof of Lemma 6.7

This result was proved in [66]. We include the proof here to provide a complete unified view of the proof techniques discussed in this survey.

Let \(f\in L^2(D)\) and \({\pmb {y}}\in U_{\overline{{\pmb {\beta }}}}\). Following (9.12)–(9.15) in the uniform case, we can write in the lognormal case

$$\begin{aligned} \Vert a^{1/2} \nabla \partial ^{\pmb {\nu }}(u - u_{h})\Vert _{L^2} \le \Vert a^{1/2} \nabla {\mathscr {P}}_h \partial ^{\pmb {\nu }}(u - u_{h})\Vert _{L^2} \,+\, \Vert a^{1/2} \nabla ({\mathscr {I}}- {\mathscr {P}}_h) \partial ^{\pmb {\nu }}u \Vert _{L^2}. \end{aligned}$$
(9.32)

Now, applying \(\partial ^{\pmb {\nu }}\) to (9.11) and separating out the \({\pmb {m}}= {\pmb {\nu }}\) term, we get for all \(z_h\in V_h\) in the lognormal case that

$$\begin{aligned} \int _D a\, \nabla \partial ^{{\pmb {\nu }}} (u-u_h)\cdot \nabla z_h \,{\mathrm {d}}{\pmb {x}}= - \sum _{\mathop {\scriptstyle {{\pmb {m}}\ne {\pmb {\nu }}}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} \int _D (\partial ^{{\pmb {\nu }}-{\pmb {m}}} a) \nabla \partial ^{{\pmb {m}}}(u-u_h)\cdot \nabla z_h \,{\mathrm {d}}{\pmb {x}}. \end{aligned}$$
(9.33)

Choosing \(z_h = {\mathscr {P}}_h \partial ^{\pmb {\nu }}(u-u_h)\) and using the definition (9.12) of \({\mathscr {P}}_h\), the left-hand side of (9.33) is equal to \(\int _D a\, |\nabla {\mathscr {P}}_h \partial ^{{\pmb {\nu }}} (u-u_h)|^2 \,{\mathrm {d}}{\pmb {x}}\). Dividing and multiplying the right-hand side of (9.33) by a, and using the Cauchy–Schwarz inequality, we then obtain

$$\begin{aligned}&\int _D a\, |\nabla {\mathscr {P}}_h \partial ^{{\pmb {\nu }}} (u-u_h)|^2 \,{\mathrm {d}}{\pmb {x}}\\&\quad \le \, \sum _{\mathop {\scriptstyle {{\pmb {m}}\ne {\pmb {\nu }}}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} \left\| \frac{\partial ^{{\pmb {\nu }}-{\pmb {m}}} a}{a}\right\| _{L^\infty } \left( \int _D a\, |\nabla \partial ^{{\pmb {m}}}(u-u_h)|^2\,{\mathrm {d}}{\pmb {x}}\right) ^{\frac{1}{2}} \left( \int _D a\, |\nabla {\mathscr {P}}_h \partial ^{{\pmb {\nu }}} (u-u_h)|^2\,{\mathrm {d}}{\pmb {x}}\right) ^{\frac{1}{2}}. \end{aligned}$$

Canceling one common factor from both sides and using (9.26), we arrive at

$$\begin{aligned} \Vert a^{1/2} \nabla {\mathscr {P}}_h \partial ^{{\pmb {\nu }}} (u-u_h)\Vert _{L^2}&\le \sum _{\mathop {\scriptstyle {{\pmb {m}}\ne {\pmb {\nu }}}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }}{\pmb {\beta }}^{{\pmb {\nu }}-{\pmb {m}}}\, \Vert a^{1/2} \nabla \partial ^{{\pmb {m}}}(u-u_h)\Vert _{L^2}. \end{aligned}$$
(9.34)

Substituting (9.34) into (9.32), we then obtain

$$\begin{aligned}&\underbrace{\Vert a^{1/2} \nabla \partial ^{\pmb {\nu }}(u - u_{h})\Vert _{L^2}}_{{\mathbb {A}}_{\pmb {\nu }}} \\&\le \sum _{\mathop {\scriptstyle {{\pmb {m}}\ne {\pmb {\nu }}}}\limits ^{\scriptstyle {{\pmb {m}}\le {\pmb {\nu }}}}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }}{\pmb {\beta }}^{{\pmb {\nu }}-{\pmb {m}}} \underbrace{\Vert a^{1/2} \nabla \partial ^{{\pmb {m}}}(u-u_h)\Vert _{L^2}}_{{\mathbb {A}}_{\pmb {m}}} \,+\, \underbrace{\Vert a^{1/2} \nabla ({\mathscr {I}}- {\mathscr {P}}_h) \partial ^{\pmb {\nu }}u \Vert _{L^2}}_{{\mathbb {B}}_{\pmb {\nu }}}. \end{aligned}$$

Note that we have \({\mathbb {A}}_{\pmb {0}}= {\mathbb {B}}_{\pmb {0}}\). Now applying Lemma 9.2 with \(\alpha =0.5\), together with (3.14), Lemma 6.6 and (9.5), we conclude that

$$\begin{aligned} \Vert a^{1/2} \nabla \partial ^{\pmb {\nu }}(u - u_{h})\Vert _{L^2} \,&\le \, \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} \Lambda _{|{\pmb {m}}|}\,{\pmb {\beta }}^{\pmb {m}}\, \Vert a^{1/2} \nabla ({\mathscr {I}}- {\mathscr {P}}_h) \partial ^{{\pmb {\nu }}-{\pmb {m}}} u \Vert _{L^2} \\&\,\lesssim \, h\, a_{\max }^{1/2}({\pmb {y}})\, \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} \Lambda _{|{\pmb {m}}|}\, {\pmb {\beta }}^{\pmb {m}}\, \Vert \varDelta (\partial ^{{\pmb {\nu }}-{\pmb {m}}} u)\Vert _{L^2} \\&\,\lesssim \, h\, T({\pmb {y}})\,a_{\max }^{1/2}({\pmb {y}})\, \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} \, |{\pmb {m}}|!\,2^{|{\pmb {m}}|}\, {\pmb {\beta }}^{\pmb {m}}\, (|{\pmb {\nu }}-{\pmb {m}}|+1)!\\&\quad \quad 2^{|{\pmb {\nu }}-{\pmb {m}}|}\,\overline{{\pmb {\beta }}}^{{\pmb {\nu }}-{\pmb {m}}}\, \Vert f\Vert _{L^2} \\&\,\lesssim \, h T({\pmb {y}})\,\, a_{\max }^{1/2}({\pmb {y}})\, \frac{(|{\pmb {\nu }}|+2)!}{2}\,2^{|{\pmb {\nu }}|}\, \overline{{\pmb {\beta }}}^{{\pmb {\nu }}}\, \Vert f\Vert _{L^2} , \end{aligned}$$

where \(T({\pmb {y}})\) is defined in (3.13). This completes the proof. \(\square \)

1.8 Proof of Lemma 6.8

This result was proved in [66]. Again we include the proof here to provide a complete unified view of the proof techniques discussed in this survey.

Let \(f,G\in L^2(D)\) and \({\pmb {y}}\in U_{\overline{{\pmb {\beta }}}}\). Following (9.18)–(9.20) in the uniform case, and using the Leibniz product rule (9.1), we have for the lognormal case that

$$\begin{aligned} \partial ^{\pmb {\nu }}G(u - u_{h}) \,&=\, \int _D \partial ^{\pmb {\nu }}\left( a \,\nabla (u - u_{h}) \cdot \nabla \left( v^G - v^G_h \right) \right) \, {\mathrm {d}}{\pmb {x}}\\&= \int _D \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} (\partial ^{{\pmb {\nu }}-{\pmb {m}}} a)\, \partial ^{\pmb {m}}\left( \nabla (u - u_{h}) \cdot \nabla \left( v^G - v^G_h \right) \right) \, {\mathrm {d}}{\pmb {x}}\\&= \int _D \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} (\partial ^{{\pmb {\nu }}-{\pmb {m}}} a)\, \sum _{{\pmb {k}}\le {\pmb {m}}} {\textstyle {\left( {\begin{array}{c}{\pmb {m}}\\ {\pmb {k}}\end{array}}\right) }} \nabla \partial ^{\pmb {k}}(u- u_{h}) \cdot \nabla \partial ^{{\pmb {m}}-{\pmb {k}}} \left( v^G - v^G_h \right) \, {\mathrm {d}}{\pmb {x}}\\&= \int _D \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} \frac{\partial ^{{\pmb {\nu }}-{\pmb {m}}} a}{a} \sum _{{\pmb {k}}\le {\pmb {m}}} {\textstyle {\left( {\begin{array}{c}{\pmb {m}}\\ {\pmb {k}}\end{array}}\right) }} \left( a^{1/2}\nabla \partial ^{\pmb {k}}(u - u_{h})\right) \nonumber \\&\quad \quad \cdot \left( a^{1/2}\nabla \partial ^{{\pmb {m}}-{\pmb {k}}} \left( v^G - v^G_h \right) \right) \, {\mathrm {d}}{\pmb {x}}. \end{aligned}$$

Using the Cauchy–Schwarz inequality and (9.26), we obtain

$$\begin{aligned}&|\partial ^{\pmb {\nu }}G(u - u_{h}) | \nonumber \\&\le \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} {\pmb {\beta }}^{{\pmb {\nu }}-{\pmb {m}}} \sum _{{\pmb {k}}\le {\pmb {m}}} {\textstyle {\left( {\begin{array}{c}{\pmb {m}}\\ {\pmb {k}}\end{array}}\right) }} \Vert a^{1/2} \nabla \partial ^{\pmb {k}}(u - u_{h})\Vert _{L^2}\, \Vert a^{1/2} \nabla \partial ^{{\pmb {m}}-{\pmb {k}}} \left( v^G - v^G_h \right) \Vert _{L^2}. \end{aligned}$$
(9.35)

We have from Lemma 6.7 that

$$\begin{aligned} \Vert a^{1/2} \nabla \partial ^{\pmb {k}}(u - u_{h})\Vert _{L^2} \,\lesssim \, h \, T({\pmb {y}})\,a_{\max }^{1/2}({\pmb {y}})\, \frac{(|{\pmb {k}}|+2)!}{2}\, 2^{|{\pmb {k}}|}\, \overline{{\pmb {\beta }}}^{{\pmb {k}}}\,\Vert f\Vert _{L^2}. \end{aligned}$$
(9.36)

Since the bilinear form \({\mathscr {A}}({\pmb {y}};\cdot ,\cdot )\) is symmetric and since the representer g for the linear functional G is in \(L^2\), all the results hold verbatim also for the adjoint problem (9.18) and for its FE discretisation (9.19). Hence, as in (9.36), we obtain

$$\begin{aligned} \Vert a^{1/2} \nabla \partial ^{{\pmb {m}}-{\pmb {k}}} (v^G - v^G_{h})\Vert _{L^2}&\lesssim h T({\pmb {y}})\,\,a_{\max }^{1/2}({\pmb {y}})\, \frac{(|{\pmb {m}}-{\pmb {k}}|+2)!}{2}\,2^{|{\pmb {m}}-{\pmb {k}}|}\,\overline{{\pmb {\beta }}}^{{\pmb {m}}-{\pmb {k}}}\,\Vert G\Vert _{L^2}. \end{aligned}$$
(9.37)

Substituting (9.36) and (9.37) into (9.35), and using the identity (9.6), we obtain

$$\begin{aligned} |\partial ^{\pmb {\nu }}G(u - u_{h}) |&\,\lesssim \, h^2\, T^2({\pmb {y}})\,a_{\max }({\pmb {y}})\, \sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} \frac{(|{\pmb {m}}|+5)!}{120}\,2^{|{\pmb {m}}|}\,\overline{{\pmb {\beta }}}^{{\pmb {\nu }}} \Vert f\Vert _{L^2} \,\Vert G\Vert _{L^2}. \end{aligned}$$

Using again (9.3), with \(n=|{\pmb {\nu }}|\) we have

$$\begin{aligned}&\sum _{{\pmb {m}}\le {\pmb {\nu }}} {\textstyle {\left( {\begin{array}{c}{\pmb {\nu }}\\ {\pmb {m}}\end{array}}\right) }} 2^{|{\pmb {m}}|}\,\frac{(|{\pmb {m}}|+5)!}{120} \\&\quad {=} \sum _{i=0}^{n} {\textstyle {\left( {\begin{array}{c}n\\ i\end{array}}\right) }} 2^{i}\,\frac{(i+5)!}{120} {=} n!\,\sum _{i=0}^{n} \frac{(i+1)(i{+}2)(i{+}3)(i+4)(i+5) 2^{i}}{120(n-i)!} \le \frac{(n+5)!}{120} 2^n e. \end{aligned}$$

This yields the required bound in the lemma. This completes the proof. \(\square \)

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Kuo, F.Y., Nuyens, D. Application of Quasi-Monte Carlo Methods to Elliptic PDEs with Random Diffusion Coefficients: A Survey of Analysis and Implementation. Found Comput Math 16, 1631–1696 (2016). https://doi.org/10.1007/s10208-016-9329-5

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