Space-Time Localized Radial Basis Function Collocation Methods for PDEs
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1 Space-Time Localized Radial Basis Function Collocation Methods for PDEs Alfa Heryudono Department of Mathematics University of Massachusetts Dartmouth ICERM Topical Workshop Providence August 27 This research is partly supported by NSF DMS Joint work with grad student: Jacob Sousa Computations are mostly done on UMassD Rapid Prototyping Server / 2
2 Dealing with Time-Dependent PDEs for RBF Methods Method of Lines RBF discretization in space + common ODE solver in time Min changes of PS/FD codes: replace differentiation matrices with RBF versions (Global, RBF-FD, RBF-PU, etc) PS/FD treatments for BCs: Strip-rows, Strip-rows move over columns, fictitious pts/ rect projection (for multiple bcs), penalty, etc Stability for linear pde case: Eigenvalue and Pseudospectra Simultaneous Space-Time RBF Boundary value collocation problem in space-time domain Time is treated as another space variable RBF-BVP solver have been studied for quite a while Less worry about choosing ODE solver based on PDE types Adaptivity, moving boundary, and BCs: same treatments as in BVP cases No need to rewrite the pde due to var trans (eg in moving boundary case) Analyzing stability is not clear (eg in moving boundary case) Might be epensive to solve (eg finding preconditioner, non-linear case) 2 / 2
3 Space-Time PS Collocation Method: D+t linear case PDE : IC : BC : u t = u (,t) [,) (,T] u(,) = f() u(,t) = g(t) Use PS or Block PS (Driscoll-Fornberg) to create differentiation matrices / 2
4 Space-Time PS Collocation Method: 2D+t, linear case ut = u + F (, y, t) IC : (, y, t) Ω (, T ] u(, y, ) = f (, y ) BC : u( Ω, t) = g ( Ω, t) kron s disease is worse in 2D + t case 5 PDE : P = symrcm(plinop); 2 L = gpuarray(linop(p,p)); -2 PL = gpuarray(plinop(p,p)); r = gpuarray(rhs(p)); 2-6 MAXITER = 3; TOL = e-4; RESTART = []; 3 [Ugpu,FLAG,RELRES,ITER,RESVEC] = nz = gmres(l,r,restart,tol,maxiter,pl); U(P) = gather(ugpu); / 2
5 Space-Time PS Collocation Method: D+t, nonlinear case Human tear film dynamics: D model: see H et al 27 h t +q = on X(t), where ( ) h 3 q(,t) = Sh 3 +βh2 Boundary conditions h(x(t),t) = h(,t) = h q(x(t),t) = X t h +Q top q(,t) = Q bot 2 8 t h(,t) Advance the solution in space-time domain: Slab by Slab (Show MATLAB) 5 / 2
6 RBF-FD Differentiation Matrices where Ω Ω s j () = n loc k= λ k φ k (), whereφ k ()isaradialbasisfunction centered at k Or in Lagrange formulation as s j () = n loc k= Ψ k ()u k, Ψ = [ Ψ () Ψ n loc () ] = [ φ () φ n loc () ][ A ], Ψ = [ Ψ () Ψn loc () ] = [ φ () φn loc () ][ A ], The matri A with entries a lk = φ k ( l ), l,k =,,n loc is local RBF interpolation matri BYODM: Bring Your Own Differentiation Matrices 6 / 2
7 Getting the space-time domain This is probably for programming on a lazy Sunday: Use Mathematica s DiscretizeRegion family commands Surprisingly, Mathematica has many built-in funky domains too This is also useful if you want to compare results with finite-element R = ImplicitRegion[-6 Sin[t] <=, {{, -, }, {t,, 5 Pi}}]; ev = DiscretizeRegion[R]; pts = MeshCoordinates[ev]; Eport["spacetimedommat", pts]; To obtained boundary points, you can use Mathematica or boundary command in MATLAB 7 / 2
8 t+d Advection Eample PDE : u t = u (,t) [X(t),) (,T] IC : u(,) = f() BC : u(,t) = g(t) solution in space-time domain IMQ-RBF: +(εr) 2 r2 = ( i ) 2 +(t t i ) 2 D t D f() = e ( 5+35y)2,g(t) = I u = f g P = symrcm(l); u(p) = L(P,P)\RHS(P); or MAXITER = 2; TOL = e-3; RESTART = []; [ML,MU] = ilu(l(p,p),struct( type, ilutp, droptol,e-6)); u(p) = gmres(l(p,p),rhs(p),restart,tol,maxiter,ml,mu); portion of system matri after applying MATLAB symrcm 8 / 2
9 t+d Advection Eample PDE : u t = u +F(,t) (,t) [X(t),) (,T] IC : u(,) = f() BC : u(,t) = g(t) solution in space-time domain IMQ-RBF: +(εr) 2 r2 = ( i ) 2 +(t t i ) 2 Get RBF-QR diffmat from Elisabeth s website f() = e ( 5+35y)2 n loc = 5,ε= D t D I u = F f g Test case Test case 2 Test case 3 Test case N slope = / 2
10 t+d Advection with Variable Speed Eample PDE : u t = a(,t)u +F(,t) (,t) [X(t),) (,T] IC : u(,) = f() BC : u(,t) = g(t) solution in space-time domain a(,t) = ep((+t)(+cos(3)) f() = e ( 5+35y)2 Bribe Varun for PHS diffmat n loc = 5,ε= D t diag(a)d F slope = -456 I u = f g Test case Test case 2 Test case 3 Test case N / 2
11 Analyzing Stability? Let s take a look at one step ( 2 levels) space-time global RBF method 2 j n n t + t 2 a j n n b t for a simple -D advection equation u PDE: t = u for [a,b) IC: u(,) = u () when t = BC: u(b,t) = g(t) at = b u() = n λ j φ(ε j )+ n λ j φ(ε j ), j= j= where { j } and { j } are centers at the old time level and new time level respectively / 2
12 2 j n n t + t 2 a j n n b t Our goal is to find the unknowns {λ j } and {λ j } This can be done by enforcing initial and boundary data and satisfying the PDE at the interior points that lead to solving system of linear equations A C B D λ λ n λ λ n = u o ( ) u o ( n ) g(t) / 2
13 A B C D λ λ n λ λ n = u o ( ) u o ( n ) g(t) The block matrices A,B,C,D are all n n matrices with elements: A ij = φ(ε i j ) B ij = φ(ε i j ) C ij = Lφ(ε i j ) D ij = Lφ(ε i j ) for all i,j =,,n and L := t The last row C and D must be slightly modified to satisfy the boundary condition at n = b 2 / 2
14 Amplification Matri and Stability Region The process of marching in time to the new time level is given by where u( ) u( n ) = G G = [ B A ][ A B C D u( ) u( n ) ] [ I and I is an n n identity matri The method is numerically stable if spectral radius ρ(g) < ], log(ε) log( t / ) IMQ, g(t) =, N = 5: to avoid blowing up the solution, the ratio of t/ vs shape parameter ε must be away from the darker alley in the the stability region, ie we want to avoid ρ(g) 3 / 2
15 Adaptivity for BVP based on Residual subsampling Initial coarse collection of nonoverlapping regular boes in R d that cover the domain Ω of interest 2 Geometric adaptation 3 Refining and Coarsening steps D initial boes Irregular geometry 75 2D initial boes 5 see 5 25 y y Driscoll & H (27) / 2
16 Rules of refining and coarsening centers Refinement strategy: converting all check points if Coarsening strategy: reactivate all RBF centers if all of its grand children have residual any of them have residual errors are greater than θ r described as into RBF centers as dots and remove errors less than θ c described as its parent a layout of quadtree With this rule, centers are located as leaves 5 / 2
17 Depth first search algorithm Pruning device to save computing pairwise distances:o(n q log(n)) instead of O(N) per query point Partial updates for lists of neighbors Embarassingly parallel neighbors search Values at are computed using local RBF interpolant of the bo whose midpoint is the parent node of the check points average distance computations Uniform nodes distribution N n q log 4 N Some non-uniform nodes distribution average distance computations N N n q log 4 N N 6 / 2
18 t+d Nonlinear Eample Burgers Equation 5 υu uu = u t, < < u(,t) = u(,t) = u(,) = sin(2π)+ 2 sin(π) where, υ = 3 MATLAB s fsolve is used to solve the nonlinear system Jacobian file is provided t 2 t t 2 t 5 N = / 2
19 Dealing with Multiple Boundary Conditions PDE : Tear film PDE in terms of h (,t) [X(t),) (,T] IC : h(,) = f() BC : h(,t) = h(x(t),t) = h h (,t) = g (t) h (X(t),t) = g 2 (X(t),t) PDE : h t = Sq,S is a constant q = nonlinear flu (,t) [X(t),) (,T] IC : h(,) = f() BC : h(,t) = h(x(t),t) = h q(,t) = g (t) q(x(t),t) = g 2 (X(t),t) h(,t) t / 2
20 t+2d Advection Eample u t = 5u +75u y +F(,y,t) (,y) [,) [,) u(,y,t) = f (,y,t) u(,,t) = f 2 (,,t) u(,y,) = g(,y) n loc = 5,ε=75 2 Error Test case Test case N /3 9 / 2
21 t+2d Wave Eample u tt = u (,y) (,) (,) u(,y,t) = at the boundary u(,y,) = g(,y) u t (,y,) = etra ghost/fictitious points for enforcing u t 2 / 2
22 On-going study or future questions Stability: Can it only be done through adaptivity? Least-Squares Space-time RBF-PU might be worth to try Adaptivity in terms of partitions Move away from points adaptivity Preconditioner? Possible GR application Application to 2D +t Human Tear Film Dynamics Enforce my grad students to finish the papers Fully open 2/3 open /3 open Closed y y y y ====================== 2 / 2
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