Tensor FEM for Spectral Fractional Diffusion

Abstract

We design and analyze several finite element methods (FEMs) applied to the Caffarelli–Silvestre extension that localizes the fractional powers of symmetric, coercive, linear elliptic operators in bounded domains with Dirichlet boundary conditions. We consider open, bounded, polytopal but not necessarily convex domains \(\varOmega \subset {\mathbb {R}}^d\) with \(d=1,2\). For the solution to the Caffarelli–Silvestre extension, we establish analytic regularity with respect to the extended variable \(y\in (0,\infty )\). Specifically, the solution belongs to countably normed, power-exponentially weighted Bochner spaces of analytic functions with respect to y, taking values in corner-weighted Kondrat’ev-type Sobolev spaces in \(\varOmega \). In \(\varOmega \subset {\mathbb {R}}^2\), we discretize with continuous, piecewise linear, Lagrangian FEM (\(P_1\)-FEM) with mesh refinement near corners and prove that the first-order convergence rate is attained for compatible data \(f\in \mathbb {H}^{1-s}(\varOmega )\) with \(0<s<1\) denoting the fractional power. We also prove that tensorization of a \(P_1\)-FEM in \(\varOmega \) with a suitable hp-FEM in the extended variable achieves log-linear complexity with respect to \({\mathscr {N}}_\varOmega \), the number of degrees of freedom in the domain \(\varOmega \). In addition, we propose a novel, sparse tensor product FEM based on a multilevel \(P_1\)-FEM in \(\varOmega \) and on a \(P_1\)-FEM on radical-geometric meshes in the extended variable. We prove that this approach also achieves log-linear complexity with respect to \({\mathscr {N}}_\varOmega \). Finally, under the stronger assumption that the data be analytic in \(\overline{\varOmega }\), and without compatibility at\(\partial \varOmega \), we establish exponential rates of convergence ofhp-FEM for spectral fractional diffusion operators in energy norm. This is achieved by a combined tensor product hp-FEM for the Caffarelli–Silvestre extension in the truncated cylinder with anisotropic geometric meshes that are refined toward \(\partial \varOmega \). We also report numerical experiments for model problems which confirm the theoretical results. We indicate several extensions and generalizations of the proposed methods to other problem classes and to other boundary conditions on \(\partial \varOmega \).

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Correspondence to Enrique Otárola.

Additional information

This paper is dedicated to Professor Ivo Babuška on the occasion of his 90th birthday.

The results in this paper were obtained when the authors met at the MFO Oberwolfach during the WS 1711 in March 2017. The research of RHN was supported in part by NSF Grants DMS-1109325 and DMS-1411808. The research of EO was supported in part by CONICYT through Project FONDECYT 3160201. The research of AJS was supported in part by NSF Grant DMS-1418784. The research of JMM was supported by the Austrian Science Fund (FWF) Project F 65. The research of CS was supported by the Swiss National Science Foundation (SNSF) under Grant No. 159940.

Communicated by Philippe G. CIARLET.

Appendices

A Proof of Lemma 11

We will only show (5.53), (5.54) as the estimates (5.55), (5.56) are proved using similar arguments; see, for instance, the proof of [6, Theorem 8]. We distinguish between the first element \(I_1\), the terminal element \(I_M\), and the remaining ones. We write \(h_i = |I_i|\). We simplify the exposition by assuming \(X = {{\mathbb {R}}}\). It is convenient to define, for each interval \(I_i\), \(i=2,\ldots ,M\), the quantity \(C_i\) by

$$\begin{aligned} C_i^2:= \sum _{\ell =1}^\infty (2 K_u)^{-\ell } \frac{1}{\ell !^2} \Vert u^{(\ell )}\Vert ^2_{L^2(\omega _{\alpha + 2\ell -2\beta ,\gamma }, I_i)}. \end{aligned}$$
(A.1)

We observe that, since \(u \in {{\mathscr {B}}}_{\beta ,\gamma }^1(C_u,K_u)\),

$$\begin{aligned} \sum _{i=2}^M C_i^2 \le 2 C_u^2, \end{aligned}$$
(A.2)

where we recall that the space \({{\mathscr {B}}}_{\beta ,\gamma }^1(C_u,K_u)\) corresponds to a class of analytic functions and is defined as in (5.51). We begin the proof with an auxiliary result about linear interpolation on the reference element.

Lemma 15

(linear interpolant) Let X be a Hilbert space, \({\widehat{K}} = (0,1)\), and let \(\widetilde{\pi }_1\) be the linear interpolant in the points 1 / 2, 1. Let \(\alpha > -1\) and \(\delta \le 1\). Then, for \(u \in C((0,1];X)\) and provided the terms on the right-hand side are finite, we have

$$\begin{aligned} \int _{{\widehat{K}}} y^\alpha \Vert u - \widetilde{\pi }_1 u\Vert _X^2\, \, {\mathrm{d}}y&\lesssim \int _{{\widehat{K}}} y^{\alpha +2\delta } \Vert u^\prime \Vert _X^2\, \, {\mathrm{d}}y , \end{aligned}$$
(A.3)
$$\begin{aligned} \int _{{\widehat{K}}} y^\alpha \Vert (u - \widetilde{\pi }_1 u)^\prime \Vert _X^2\, \, {\mathrm{d}}y&\lesssim \int _{{\widehat{K}}} y^{\alpha +2\delta } \Vert u^{\prime \prime }\Vert _X^2\, \, {\mathrm{d}}y, \end{aligned}$$
(A.4)

where the hidden constant is independent of u.

Proof

For notational simplicity, we will prove the lemma only for the case \(X = {{\mathbb {R}}}\).

We begin with the proof of (A.3). Since \((u - \widetilde{\pi }_1 u)(1) = 0\), we have, for \(y \in {\widehat{K}}\),

$$\begin{aligned} (u - \widetilde{\pi }_1 u)(y) = \int _1^y (u - \widetilde{\pi }_1 u)^\prime (t)\,\, {\mathrm{d}}t, \end{aligned}$$

so that

$$\begin{aligned} \int _0^1 y^\alpha |u - \widetilde{\pi }_1 u|^2\, \, {\mathrm{d}}y\le & {} 2 \int _0^1 y^\alpha \left| \int _{y}^1 |u^\prime (t)|\,\, {\mathrm{d}}t\right| ^2 \, {\mathrm{d}}y\\&\quad +\, 2 \int _0^1 y^\alpha \left| \int _{y}^1 |(\widetilde{\pi }_1 u)^\prime (t)|\,\, {\mathrm{d}}t\right| ^2 \, {\mathrm{d}}y. \end{aligned}$$

From Hardy’s inequality (e.g., [25, Chapter 2, Theorem 3.1]), we infer

$$\begin{aligned} \int _0^1 y^\alpha \left| \int _{y}^1 |u^\prime (t)|\,\, {\mathrm{d}}t\right| ^2\, \, {\mathrm{d}}y \le 4(\alpha +1)^{-2} \int _0^1 y^{\alpha +2} |u^\prime (y)|^2\, \, {\mathrm{d}}y. \end{aligned}$$

From \( (\widetilde{\pi }_1 u)^\prime = 2\int _{1/2}^1 u^\prime (t)\,\, {\mathrm{d}}t \), we obtain \( | (\widetilde{\pi }_1 u)^\prime |^2 \le C \int _{1/2}^1 t^{\alpha +2\delta } |u^\prime (t)|^2\,\, {\mathrm{d}}t \) and therefore, in view of \(\alpha > -1\), the estimate

$$\begin{aligned} \int _0^1 y^\alpha | (\widetilde{\pi }_1 u)'|^2\, \, {\mathrm{d}}y \lesssim \int _0^1 y^{\alpha +2} |u^\prime (y)|^2\, \, {\mathrm{d}}y. \end{aligned}$$

This concludes the proof of (A.3) for the case \(\delta = 1\). Since the integration range is \(y \in \widehat{K} = (0,1)\), we may replace \(y^{\alpha +2}\) by \(y^{\alpha +2 \delta }\).

Next, we prove (A.4) and assume that the right-hand side of (A.4) is finite. Again, it suffices to consider the limiting case \(\delta = 1\) and use Hardy’s inequality. We mention in passing that (A.5) actually shows that \(u \in C^1((0,1]; X)\). We write

$$\begin{aligned} (u - \widetilde{\pi }_1 u)^\prime (y)&= u^\prime (y) - 2 \int _{1/2}^1 u^\prime (t)\, \, {\mathrm{d}}t = 2\int _{1/2}^1 \left( u^\prime (y) - u^\prime (t) \right) \,\, {\mathrm{d}}t \nonumber \\&= 2\int _{1/2}^1 \int _{t}^y u^{\prime \prime }(\tau )\,\, {\mathrm{d}}\tau \,\, {\mathrm{d}}t . \end{aligned}$$
(A.5)

Therefore,

$$\begin{aligned} \int _{0}^1 y^\alpha |(u -&\widetilde{\pi }_1 u)^\prime (y)|^2\,\, {\mathrm{d}}y = 4 \int _{0}^1 y^\alpha \left| \int _{1/2}^1 \int _{t}^y u^{\prime \prime }(\tau )\,\, {\mathrm{d}}\tau \, \, {\mathrm{d}}t \right| ^2\, \, {\mathrm{d}}y \\&\le 2 \int _{1/2}^1 \int _{0}^1 y^\alpha \left| \int _{t}^y |u^{\prime \prime }(\tau )|\,\, {\mathrm{d}}\tau \right| ^2 \, {\mathrm{d}}y \, {\mathrm{d}}t\\&\lesssim \int _{1/2}^1 \int _{0}^1 y^\alpha \left[ \left| \int _{y}^1 |u^{\prime \prime }(\tau )|\,\, {\mathrm{d}}\tau \right| ^2 + \left| \int _{t}^1 |u^{\prime \prime }(\tau )|\,\, {\mathrm{d}}\tau \right| ^2 \right] \, {\mathrm{d}}y \, {\mathrm{d}}t\\&\lesssim \int _{0}^1 y^{\alpha +2} |u^{\prime \prime }(y)|^2\,\, {\mathrm{d}}y + \int _{1/2}^1 y^{\alpha +2} |u^{\prime \prime }(y)|^2\, \, {\mathrm{d}}y, \end{aligned}$$

where in the last step we applied Hardy’s inequality. Lemma 15 is thus proved. \(\square \)

With Lemma 15, we can estimate the error on the first element \(I_1\) as follows: Scaling the estimate (A.3) gives

$$\begin{aligned} \Vert u - \pi ^{\mathbf {r}}_{y} u\Vert _{L^2(\omega _{\alpha ,0},I_1)} \le C h_1^{\beta } \Vert u^\prime \Vert _{L^2(\omega _{\alpha +2-2\beta ,0},I_1)}, \end{aligned}$$
(A.6)

and the assumption allows us to insert the weight \(e^{\gamma y}\) on both sides of (A.6).

We now bound the error contributions from elements away from the origin, i.e., on the elements \(I_i\), \(i=2,\ldots ,M\). These elements satisfy \(h_i \sigma /(1-\sigma )= {{\mathrm{dist}}}(I_i,0) \). For \(I_i = (y_{i-1}, y_i)\), the pull-back \({\widehat{u}}_i: = u|_{I_i} \circ F_{I_i}\) satisfies

$$\begin{aligned}&\Vert {\widehat{u}}_i^{(\ell +1)}\Vert ^2_{L^2(-1,1)} =(h_i/2)^{-1+2(\ell +1)} \Vert u^{(\ell +1)}\Vert ^2_{L^2(I_i)} \\&\quad \le (h_i/2)^{-1+2(\ell +1)} e^{-\gamma y_{i-1}} \max _{y \in I_i} y^{-\alpha -2(\ell +1)+2\beta } \Vert u^{(\ell +1)}\Vert ^2_{L^2(\omega _{\alpha +2(\ell +1)-2\beta ,\gamma },I_i)} \\&\quad \lesssim e^{-\gamma y_{i-1}} h_i^{-1 + 2 (\ell +1)} h_i^{-\alpha - 2 (\ell +1)+2\beta } (2 \sigma /(1-\sigma ))^{-2(\ell +1)} C_i^2 (2 K_u)^{2(\ell +1)} (\ell +1)!^2, \end{aligned}$$

where in the last step we have used (A.1). The operator \({\widehat{\varPi }}_r\) given by Lemma 10 then yields the existence of a \(b>0\) that depends solely on \(K_u\) and \(\sigma \) for which

$$\begin{aligned} \Vert {\widehat{u}} - {\widehat{\varPi }}_{{\mathbf {r}}_i}{\widehat{u}}\Vert _{L^2(-1,1)} \lesssim C_i e^{-\gamma y_{i-1}} e^{-b {\mathbf {r}}_i} h_i^{-(1+\alpha )/2 + \beta }. \end{aligned}$$

Scaling back to \(I_i\) and using again \(h_i \sim {\text {*}}{dist}(I_i,0)\) yields

$$\begin{aligned} \Vert u - \pi ^{{\mathbf {r}}}_{y} u\Vert ^2_{L^2(\omega _{\alpha ,\gamma },I_i)} \le C h_i^{2\beta } C_i^2 e^{-2 b {\mathbf {r}}_i}. \end{aligned}$$

Summation over i and taking the slope of the linear degree vector sufficiently large (see, for instance, the proof of [6, Theorem 8] for details) give

for suitable \(b' > 0\). Combining this with (A.6) gives the desired (5.53).

It remains to prove (5.54). We begin with a preparatory result.

Lemma 16

(exponential decay) Let X be a Hilbert space, and let \(\delta \in {{\mathbb {R}}}\), \(\gamma > 0\), . Then, the following holds for in items (i), (ii) and for in items (iii), (iv) with implied constants depending solely on \(\delta \), \(\gamma \), and :

  1. (i)

    If \(\lim _{y \rightarrow \infty } u(y) = 0\) and , then

    (A.7)
  2. (ii)

    If then \(\lim _{y \rightarrow \infty } u(y) = 0.\)

  3. (iii)

    If \(\lim _{y \rightarrow \infty } u^{(j)}(y)=0\) for \(j=0\), 1 and , then

    (A.8)
  4. (iv)

    If , then \(\lim _{y \rightarrow \infty } u(y) = \lim _{y\rightarrow \infty } u^\prime (y) = 0\).

Proof

We will only prove items (i) and (ii) as the remaining two are proved by similar arguments.

We begin the proof with the following observation: There is a constant \(c>0\) (that depends only on \(\delta \), , and \(\gamma \)) such that for

(A.9)

For \(\delta \ge 0\), this is immediate. For \(\delta < 0\), one integrates by parts once to discover that the leading-order asymptotics (as ) of the integral is .

We now proceed with the proof of (A.7): Since \(\gamma > 0\), we can write

and (A.7) follows from (A.9). The assertion of item (ii) follows by a similar argument, starting from \(u(y) = u(\eta ) + \int _{\eta }^y u^\prime (t)\,\, {\mathrm{d}}t\), squaring, multiplying by \(\exp (-\gamma \eta )\), and integrating in \(\eta \) from y to \(\infty \). \(\square \)

To prove (5.54) we have to estimate . Lemma 16, (i) shows

(A.10)

Since is obtained from \(\pi ^{\mathbf {r}}_{y}\) by a correction on the terminal element \(I_M\), the desired (5.54) follows easily from (A.10), if we recall that .

B Analysis of the Decoupling Eigenvalue Problem

Lemma 17

(weighted Poincaré) Let and \(\alpha \in (-1,1)\). Then, for with there holds

(B.1)

Proof

From we get . Hence, for ,

which finishes the proof. \(\square \)

Lemma 18

(eigenvalue upper bound) Let and \(\alpha \in (-1,1)\). Assume that \((v,\mu )\) satisfy

(B.2)

Then, .

Proof

We compute, using Lemma 17

which finishes the proof. \(\square \)

We also need lower bounds for eigenvalues.

Lemma 19

(eigenvalue lower bound) Let \(\alpha > -1\). Let \({\mathscr {G}}^M\) be an arbitrary mesh on with the property that for all elements \(I_i\), \(i=2,\ldots ,M\), not abutting \(y=0\) there holds \(|I_i| \le C_{geo} {{\mathrm{dist}}}(I_i,0)\). Let be a subspace of the space of piecewise polynomials of degree q on \({\mathscr {G}}^M\). Then, with \(h_{min}\) denoting the smallest element size,

(B.3)

where the hidden constant depends solely on \(C_{geo}\) and \(\alpha \).

Proof

We emphasize that the condition \(h_i \le C_{geo} {\text {*}}{dist}(I_i,0)\) is satisfied for all meshes where neighboring elements have comparable size. We also remark that (slightly) sharper estimates (in the dependence on the polynomial degree q) are possible on geometric meshes with linear degree vector. We write \(h_i = |I_i|\). We will use the polynomial inverse estimate (B.6). For the first element \(I_1 = (0,y_1)\), we calculate for \(v \in V_h\) and its pull-back \(\widehat{v}:= v|_{I_1} \circ F_{I_1}\)

$$\begin{aligned} \Vert v^\prime \Vert ^2_{L^2(y^\alpha ,{\widehat{K}})}&= (h_1/2)^{\alpha +1-2} \int _{-1}^1 (1+y)^\alpha |\widehat{v}^\prime (y)|^2\,\, {\mathrm{d}}y \nonumber \\&{\mathop {\lesssim }\limits ^{(B.6)}} h_1^{\alpha +1-2} q^4 \int _{-1}^1 (1+y)^\alpha |\widehat{v}(y)|^2\,\, {\mathrm{d}}y \sim h_1^{-2} q^4 \Vert v\Vert ^2_{L^2(y^\alpha ,I_1)}. \end{aligned}$$
(B.4)

For the remaining elements \(I_i\), we exploit that the assumption \(h_i \le C_{geo} {{\mathrm{dist}}}(I_i,0)\) ensures that the weight is slowly varying within them, i.e.,

$$\begin{aligned} \max _{y \in I_i} y^\alpha \le (1+C_{geo})^{|\alpha |} \min _{y \in I_i} y^\alpha , \qquad i=2,\ldots ,M. \end{aligned}$$

Hence, the polynomial inverse estimate (B.6) (with \(\alpha =\beta = 0\) there) yields by scaling arguments

$$\begin{aligned} \Vert v^\prime \Vert _{L^2(y^\alpha ,I_i)} \le C h_i^{-1} q^2 \Vert v\Vert _{L^2(y^\alpha ,I_i)}. \end{aligned}$$
(B.5)

Combining (B.4), (B.5) yields the result. \(\square \)

Lemma 20

(polynomial inverse estimate) Let \({\widehat{K}} = (-1,1)\). For \(\alpha \), \(\beta > -1\), there is \(C_{\alpha ,\beta }>0\) such that for all \(q \in {{\mathbb {N}}}_0\) and all \(w \in {{\mathbb {P}}}_q({\widehat{K}})\):

$$\begin{aligned} \int _{-1}^1 (1+y)^\alpha (1-y)^\beta |w^\prime (y)|^2\, {\mathrm{d}}y&\le C_{\alpha ,\beta } q^4 \int _{-1}^1 (1+y)^\alpha (1-y)^\beta |w(y)|^2\, {\mathrm{d}}y. \end{aligned}$$
(B.6)

Proof

Step 1: We assert (B.6) for \(\alpha = \beta \). From [10, Chap. III, Props. 6.1, 6.3], we get for \(w \in {{\mathbb {P}}}_q({\widehat{K}})\)

$$\begin{aligned} \int _{-1}^1 (1-y^2)^\alpha |w^\prime (y)|^2\, {\mathrm{d}}y&{\mathop {\lesssim }\limits ^{[{10, Chap.~{III}, Prop.~{6.1}}]}} q^2 \int _{-1}^1 (1-y^2)^{\alpha +1} |w^\prime (y)|^2\, {\mathrm{d}}y \\&{\mathop {\lesssim }\limits ^{[10, Chap.~{III}, Prop.~{6.3}]}} q^4 \int _{-1}^1 (1-y^2)^{\alpha } |w(y)|^2\, {\mathrm{d}}y. \end{aligned}$$

Simple scaling arguments imply for arbitrary (fixed) finite intervals (ab) and \(\alpha ' > -1\) for all \(w \in {{\mathbb {P}}}_q(a,b)\):

$$\begin{aligned} \int _{a}^b (y - a)^{\alpha '} (b-y)^{\alpha '} |w^\prime (y)|^2\, {\mathrm{d}}y \lesssim q^4 \int _{a}^b (y - a)^{\alpha '} (b-y)^{\alpha '} |w(y)|^2\, {\mathrm{d}}y . \end{aligned}$$
(B.7)

Step 2: We show (B.6) for \((\alpha ,\beta ) = (\alpha ,0)\). (By symmetry, this also shows the case \((\alpha ,\beta ) = (0,\beta )\)). Since (B.7) implies for \(w \in {{\mathbb {P}}}_q({\widehat{K}})\)

$$\begin{aligned} \int _{0}^1 (1+y)^\alpha |w^\prime (y)|^2\, {\mathrm{d}}y \lesssim \int _{0}^1 |w^\prime (y)|^2\, {\mathrm{d}}y \lesssim q^4 \int _{0}^1 |w(y)|^2\, {\mathrm{d}}y, \end{aligned}$$

it suffices to prove the bound \(\int _{-1}^0 (1+y)^\alpha |w^\prime (y)|^2\, {\mathrm{d}}y \lesssim q^4 \int _{-1}^1 (1+y)^\alpha |w(y)|^2\, {\mathrm{d}}y\). To that end, define \(\widetilde{w}(y):= (1-y) w(y) \in {{\mathbb {P}}}_{q+1}({\widehat{K}})\) and note \(\widetilde{w}^\prime (y) = (1-y) w^\prime (y) - w(y)\). Then,

$$\begin{aligned}&\int _{-1}^0(1+y)^\alpha |w^\prime (y)|^2\, {\mathrm{d}}y \le \int _{-1}^0(1+y)^\alpha |w^\prime (y) (1-y)|^2\, {\mathrm{d}}y \\&\quad \le 2 \int _{-1}^0(1+y)^\alpha \left( |\widetilde{w}^\prime (y)|^2 + |w(y)|^2\right) \, {\mathrm{d}}y \\&\quad {\mathop {\lesssim }\limits ^{(B.7)}} (q+1)^4 \int _{-1}^1 (1+y)^\alpha \left( (1-y)^\alpha |\widetilde{w}(y)|^2 + |w(y)|^2\right) \, {\mathrm{d}}y \\&\quad \lesssim (q+1)^4 \int _{-1}^1 (1+y)^\alpha \left( (1-y)^\alpha (1-y)^2 |w(y)|^2 + |w(y)|^2\right) \, {\mathrm{d}}y. \end{aligned}$$

Since \(\alpha > -1\), we have \((1- y)^{2+\alpha } \lesssim 1\), which allows us to conclude the proof of the case \(\beta = 0\) in (B.6).

Step 3: For arbitrary \(\alpha \), \(\beta > -1\), we use (scaled versions of) Step 2:

$$\begin{aligned}&\int _{-1}^1 (1+y)^\alpha (1-y)^\beta |w^\prime (y)|^2\, {\mathrm{d}}y \lesssim \int _{-1}^0 (1+y)^\alpha |w^\prime (y)|^2\, {\mathrm{d}}y + \int _{0}^1 (1-y)^\beta |w^\prime (y)|^2\, {\mathrm{d}}y\\&\quad {\mathop {\lesssim }\limits ^{\text {Step 2}}} q^4 \int _{-1}^0 (1+y)^\alpha |w(y)|^2\, {\mathrm{d}}y + q^4 \int _{0}^1 (1-y)^\beta |w(y)|^2\, {\mathrm{d}}y\\&\quad \lesssim q^4 \int _{-1}^1 (1+y)^\alpha (1-y)^\beta |w(y)|^2\, {\mathrm{d}}y. \end{aligned}$$

This concludes the proof. \(\square \)

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Banjai, L., Melenk, J.M., Nochetto, R.H. et al. Tensor FEM for Spectral Fractional Diffusion. Found Comput Math 19, 901–962 (2019). https://doi.org/10.1007/s10208-018-9402-3

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Keywords

  • Fractional diffusion
  • Nonlocal operators
  • Weighted Sobolev spaces
  • Regularity estimates
  • Finite elements
  • Anisotropic hp-refinement
  • Corner refinement
  • Sparse grids
  • Exponential convergence

Mathematics Subject Classification

  • 26A33
  • 65N12
  • 65N30