Subdiffusion in a nonconvex polygon

Transcription

1 Subdiffusion in a nonconvex polygon Kim Ngan Le and William McLean The University of New South Wales Bishnu Lamichhane University of Newcastle Monash Workshop on Numerical PDEs, February 2016

2 Outline Time-fractional diffusion Poisson equation on a nonconvex polygon Error bounds for fractional diffusion Numerical examples Smoothing property of diffusion equations

3 Time-fractional diffusion Let 0 < α < 1 and seek u = u(x, t) satisfying t u 1 α t K 2 u = 0 for x Ω and t > 0, together with the initial condition u(x, 0) = u 0 (x) and homogeneous Dirichlet boundary conditions, u(x, t) = 0 for x Ω and t > 0. Riemann Liouville fractional derivative in time, t 1 α v(x, t) = t t 0 (t s) α 1 v(x, s) ds. Γ(α) Recover classical (Brownian) diffusion in limit α 1.

4 Spatial discretisation by finite elements Assume that Ω is a 2D polygon. Triangulate Ω and let V h H0 1 (Ω) denote the corresponding space of continuous, piecewise-linear finite element functions that vanish on Ω. Variational solution u : [0, ) H0 1 (Ω) satisfies t u, v + a( 1 α t u, v) = 0 for all v H 1 0 (Ω), with u(0) = u 0, where a(u, v) = K Ω u v dx. Finite element solution u h : [0, ) V h satisfies t u h, χ + a( 1 α t u h, χ) = 0 for all χ V h, with u h (0) = u 0h u 0.

5 Method of lines Let N = dim V h = number of degrees of freedom, U(t) = vector of nodal values of u h (t), M = mass matrix, S = stiffness matrix. Obtain a system of integrodifferential equations in R N : M t U + S 1 α t U = 0. Reduces to usual method of lines for the heat equation in the limiting case when α 1.

6 Convex case Notation: Av = K 2 v : H 1 0 (Ω) H 1 (Ω), and P h : L 2 (Ω) V h denotes the orthoprojector. Theorem (McLean & Thomée, 2010) Let u 0h = P h u 0. If u 0 L 2 (Ω), then u h (t) u(t) Ct α h 2 u 0, for t > 0. If u 0 D(A) = H 2 (Ω) H0 1 (Ω), then u h (t) u(t) Ch 2 Au 0, for t 0. Classical case α = 1 proved by Bramble, Schatz, Thomée & Wahlbin (SINUM, 1977).

7 Error analysis relies on H 2 -regularity of the Poisson problem, K 2 u = f (x) for x Ω, with u(x) = 0 for x Ω. If Ω is convex and f L 2 (Ω), then the (variational) solution u belongs to H 2 (Ω), and u 2 C f. Here, we use the abbreviation v m = v H m (Ω). But H 2 -regularity breaks down if Ω is a polygon with a re-entrant corner.

8 Poisson equation on a nonconvex polygon Suppose only one re-entrant corner at the origin with angle π/β for 1/2 < β < 1. u = 0 π/β 2 u = f

9 Polar coordinates Separating variables, we find that the functions u n ± = r ±nβ sin(nβθ), n {1, 2, 3,...}, satisfy 2 u n ± = 0 for 0 < r < and 0 < θ < π/β, with u n ± = 0 for 0 < r < and θ = 0 or π/β. Let η = η(r) be a C cutoff function equal to 1 for small r. Then, ηu n + H0 1 (Ω) but ηun / H0 1 (Ω) for all n 1. Also, ηu n + H 2 (Ω) for all n 2 but ηu 1 + / H2 (Ω).

10 Singular behaviour Recall A = K 2 and put R = { Av : v H0 1 (Ω) H 2 (Ω) }, N = { v L 2 (Ω) : Av = 0 in Ω, v = 0 on Ω }. Lemma L 2 (Ω) = R N and dim N = 1. Theorem There exists q N (depending only on Ω and η) such that if f L 2 (Ω) then the variational solution u H0 1 (Ω) of the Poisson problem, K 2 u = f (x) for x Ω, with u(x) = 0 for x Ω, satisfies u q, f ηu C f.

11 Local mesh refinement Consider a family of shape-regular triangulations T h. For each element T h, let For some γ 1, assume h = diameter of, r = distance from 0 to, h = max T h h. chr 1 1/γ h Chr 1 1/γ whenever h γ r 1, and ch γ h Ch γ whenever r < h γ. Elements increase in size from h γ near 0 to h when r c > 0.

12 Generalised polygon with γ = 3/2

13 Approximation property Given v C(Ω) let Π h v V h denote the nodal interpolant to v. Theorem If f L 2 (Ω), then the solution of the Poisson problem satisfies K 2 u = f (x) for x Ω, with u(x) = 0 for x Ω, u Π h u Chɛ(h, γ) f and u Π h u H 1 0 (Ω) Cɛ(h, γ) f, where h γβ / γ 1 β, 1 γ < 1/β, ɛ(h, γ) = h log(1 + h 1 ), γ = 1/β, h/ β γ 1, γ > 1/β.

14 Finite element error for the Poisson problem Corollary u h u Cɛ(h, γ) 2 f and u h u 1 Cɛ(h, γ) f. Proof. Error bound in H0 1 (Ω) follows from quasi-optimality, u h u 1 C min v V h v u 1. Error bound in L 2 (Ω) follows by usual duality argument (Nitsche trick). So for a quasi-uniform mesh (γ = 1) we have u h u Ch 2β f and u h u 1 Ch β f.

15 Error bounds for fractional diffusion Let P h : L 2 (Ω) V h denote the orthoprojector and R h : H 1 0 (Ω) V h the Ritz projector. Theorem For general initial data u 0 L 2 (Ω) and all t > 0, and u h (t) u(t) u 0h P h u 0 + Ct α ɛ(h, γ) 2 u 0 u h (t) u(t) 1 Ct α u 0h P h u 0 + C(t α + t 2α )ɛ(h, γ) u 0. For smoother initial data such that A 1+δ u 0 L 2 (Ω) and for 0 t T, u h (t) u(t) u h0 R h u 0 + Cδ 1 ɛ(h, γ) 2 A 1+δ u 0.

16 Integral representation of u(t) The proof relies on the Laplace transform, Since we have and so û(z) = 0 e zt u(t) dt. t u + 1 α t Au = 0 zû(z) u 0 + z 1 α Aû = 0 (z α I + A)û(z) = z α 1 u 0. Inversion formula: for a suitable contour Γ in the complex plane, u(t) = 1 e zt û(z) dz, t > 0. 2πi Γ

17 Integral representation of u h (t) Similarly, where A h : V h V h is defined by Thus, t u h + 1 α t A h u = 0 A h v, w = a(v, w) for v, w V h. (z α I + A h )û h (z) = z α 1 u 0h and u h (t) = 1 e zt û h (z) dz, t > 0, 2πi Γ leading to a representation of the error, u h (t) u(t) = 1 2πi Γ e zt z α 1[ (z α I +A h ) 1 u 0h (z α I +A) 1 u 0 ] dz.

18 Numerical examples We choose Ω to be the domain 0 < r < 1 and 0 < θ < π/β for β = 3/2. In the first example we solve an (inhomogeneous) fractional diffusion equation with exact solution ( t α ) u = 1 + r β (1 r) sin(βθ), Γ(α + 1) with α = 1/2. Time discretisation uses a quadrature approximation to the Laplace inversion formula.

19 Errors u h (t) u(t) at t = Quasiuniform (γ =1) Locally refined (γ =1/β) Error in L 2 (Ω) Degrees of Freedom

20 In the second example we impose mixed boundary conditions and choose the initial data to be the first eigenfunction of A = K 2, u 0 = J β/2 (ωr) sin( 1 2 βθ), where ω is the smallest positive zero of J β/2. For triangulations T h with γ = 2/β = 3 we observe the following convergence rates at t = 1. α = 1/4 α = 3/4 h N error rate error rate e e e e e e e e e e

21 Error u h (t) u(t) as a function of t, for α = 1/ Error in L 2 (Ω) 10-5 h =2 5 h =2 6 h = t

22 Smoothing property of diffusion equations Let φ 1, φ 2, φ 3,... denote the orthonormal Let (φ n, λ n ) denote the nth eigenpair of A = K 2 in Ω: Thus, φ n H0 1 (Ω) with Aφ n = λ n φ n in Ω, with φ n = 0 on Ω. a(φ n, φ m ) = λ n δ mn and φ n, φ m = δ mn. Expand the solution of the fractional diffusion equation in a generalised Fourier series u(t) = u n (t)φ n. n=1 Modes u n (t) are damped as n, for any fixed t > 0.

23 Mittag Leffler function Each mode satisfies a fractional relaxation equation, Using the identity t u n + λ n 1 α t u n = 0, with u n (0) = u(0), φ n. β t t α Γ(α + 1) = we find u n (t) = u n (0)E α ( λ n t α ) where E α (z) = p=0 t α β Γ(α β + 1) z p Γ(1 + pα).

24 Series representation Thus, Note that u(t) = u n (0)E α ( λ n t α )φ n. n=1 E α ( λ n t α ) E 1 ( λ n t) = e λnt as α 1, giving the usual representation for the solution of the heat equation.