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A Reduced Basis Method For Fractional Diffusion Operators II

Tobias Danczul and Joachim Schöberl

Abstract

We present a novel numerical scheme to approximate the solution map s ↦ u(s) := 𝓛sf to fractional PDEs involving elliptic operators. Reinterpreting 𝓛s as an interpolation operator allows us to write u(s) as an integral including solutions to a parametrized family of local PDEs. We propose a reduced basis strategy on top of a finite element method to approximate its integrand. Unlike prior works, we deduce the choice of snapshots for the reduced basis procedure analytically. The integral is interpreted in a spectral setting to evaluate the surrogate directly. Its computation boils down to a matrix approximation L of the operator whose inverse is projected to the s-independent reduced space, where explicit diagonalization is feasible. Exponential convergence rates are proven rigorously.

A second algorithm is presented to avoid inversion of L. Instead, we directly project the matrix to the subspace, where its negative fractional power is evaluated. A numerical comparison with the predecessor highlights its competitive performance.

Acknowledgements

The authors acknowledge support from the Austrian Science Fund (FWF) through grant number F 65 and W1245.

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6 Appendix

Proof

Proof ofTheorem 2.2. It suffices to show that

K(V0,V1)2(t;u)=u02u,v(t)0.(24)

There holds

uv(t)02=u022u,v(t)0+v(t)02.

Let uk := 〈u,φk0 to deduce from Lemma 2.1

t2v(t)12=k=1t2λk2uk2(1+t2λk2)2=k=1uk21+t2λk2uk2(1+t2λk2)2=u,v(t)0v(t)02,

which proves (24) and concludes the proof.□

Proof

Proof ofTheorem 2.3. One observes that for any F ∈ 𝓥0

F,φk=R1F,φk1=λk2R1F,φk0=λk2Φk,R1F=λk2Φk,F1,

from which we conclude that (λkΦk)k=1 is a 𝓥−1-orthonormal system of eigenfunctions. Since

0=F,Φk1=λk2F,φk

for all k ∈ ℕ implies that F = 0, it is also a basis. This proves the claim.□

Proof

Proof ofTheorem 2.5. Due to

F,Φk1=λk2F,φk

and Theorem 2.3, there holds

FH1s(V1,V0)2=k=1λk22sF,Ψk12=k=1λk2sF,φk2=FHs(V0,V1)2,

proving the first equality in (8). The second one follows by means of (3). Furthermore, one observes

R1LH1s(V1,V0)F=R1k=1λk22sF,Ψk1Ψk=R1k=1λk2sF,ψkΨk=k=1λk22sF,φkR1Φk=k=1λk2sF,φkφk=LHs(V0,V1)F,

confirming the first equality in (9). The latter is a consequence of (4). The remainder follows as LHs(V0,V1)F=LHs(V0,V1)1f for F ∈ 𝓥0.□

Published Online: 2021-03-09

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