Green s Functions, Boundary Integral Equations and Rotational Symmetry

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1 Green s Functions, Boundary Integral Equations and Rotational Symmetry...or, How to Construct a Fast Solver for Stokes Equation Saibal De Advisor: Shravan Veerapaneni University of Michigan, Ann Arbor February 2, 2018 Saibal De (U of M) Axis-symmetric BIE February 2, / 42

2 Why bother?! Particulate flow omnipresent in many physical/biological processes Porous media (soil) Suspensions (proteins in cell membranes) Sedimentation Low Reynold s number = Stokes equation Saibal De (U of M) Axis-symmetric BIE February 2, / 42

3 Outline 1 Green s Functions and Boundary Integral Equations 2 Rotational Invariance and Fourier Series 3 Application to Stokes flow 4 Numerical Aspects 5 Some Results Saibal De (U of M) Axis-symmetric BIE February 2, / 42

4 Green s Functions and Boundary Integral Equations Saibal De (U of M) Axis-symmetric BIE February 2, / 42

5 Fundamental principles of electrostatics Given charge distribution, find out potential function Two main ideas: Coulomb s law Superposition u(x) = 1 Q 4π x x Saibal De (U of M) Axis-symmetric BIE February 2, / 42

6 An example: potential due to charged spherical conductor Charge Q is dumped into a spherical conductor of radius R Can only have one type of charge (positive or negative) Charge will be distributed over the surface Charge distribution will be uniform Uniform surface charge density of σ(x ) = Q 4πR 2 Potential given by u(x) = 1 σ(x ) 4π Γ x x dγ(x ) Some symmetry arguments u(x) = 1 Q 4π x for x > R Saibal De (U of M) Axis-symmetric BIE February 2, / 42

7 A More Mathematical Point of View Rewrite u(x) = Γ [ ] 1 1 4π x x σ(x ) dγ(x ) Characteristics: A linear operator Takes information about σ on surface of ball Constructs value of u in exterior of ball The kernel G(x, x ) = 1 1 4π x x is the Green s function for 3D Laplace equation Saibal De (U of M) Axis-symmetric BIE February 2, / 42

8 Green s functions Can be defined for any linear partial differential equation, Lu = f G solves: LG(x, x ) = δ(x x ) Linearity allows us to construct more solutions: u(x) = G(x, x )σ(x ) dν(x ) Saibal De (U of M) Axis-symmetric BIE February 2, / 42

9 Layer potentials Arbitrary closed, simple surface Γ Single layer potential u(x) = [Sσ](x) := G(x, x )σ(x ) dγ(x ), Γ x Γ Double layer potential u(x) = [Dµ](x) := Γ G(x, x ) ν(x ) µ(x ) dγ(x ), x Γ Saibal De (U of M) Axis-symmetric BIE February 2, / 42

10 Extending layer potentials up to boundary Theorem Γ of class C 2, σ C(Γ) = Sσ can be extended continuously everywhere, in particular to Γ Theorem Γ of class C 2, µ C(Γ) = Dµ can be extended continuously to Γ, with limit [Dµ] ± (x) = 1 µ(x) + [Dµ](x) for x Γ 2 where [Dµ] ± (x) = lim [Dµ](x ± hν(x)), x h 0+ Γ Saibal De (U of M) Axis-symmetric BIE February 2, / 42

11 Boundary integral equations Boundary value problem Lu = 0 in domain, u = f on Γ Single layer Double layer interior [Sσ](x) = f(x) + 1 µ(x) + [Dµ](x) = f(x) 2 Double layer exterior 1 µ(x) + [Dµ](x) = f(x) 2 Saibal De (U of M) Axis-symmetric BIE February 2, / 42

12 What got swept under the rug... Is integral operator u = L σ onto? Why is it enough to have sources only on the surface? Saibal De (U of M) Axis-symmetric BIE February 2, / 42

13 References BIEs and Laplace equation Kress, Linear Integral Equations, Saibal De (U of M) Axis-symmetric BIE February 2, / 42

14 Rotational Invariance and Fourier Series Saibal De (U of M) Axis-symmetric BIE February 2, / 42

15 Exploiting rotational symmetry Saibal De (U of M) Axis-symmetric BIE February 2, / 42

16 Fourier decomposition Single layer equations Γ G(x, x )σ(x ) dγ(x ) = f(x) Multiply by 1 2π e inθ and integrate w.r.t. θ: γ [ π π ] G(x, x )e in(θ θ ) dθ σ n (r, z )r dγ(r, z ) = f n (r, z) Saibal De (U of M) Axis-symmetric BIE February 2, / 42

17 Rotationally invariant kernel Need: G(x, x ) = G(θ θ, r, z, r, z ) Single layer BIE reduces to: 2π G n (r, z, r, z )σ n (r, z )r dγ(r, z ) = f n (r, z) γ Similar formulation for other layer potentials Saibal De (U of M) Axis-symmetric BIE February 2, / 42

18 Nyström discretization Integral equation 2π G n (r, z, r, z )σ n (r, z )r dγ(r, z ) = f n (r, z) γ Pick quadrature rule {(r i, z i ), w i } I i=1 on γ Discretize I 2π G n (r i, z i, r j, z j )σ n (r j, z j )r j γ (r j, z j )w j f n (r i, z i ) j=1 A linear system with unknown σ n (r j, z j ) Saibal De (U of M) Axis-symmetric BIE February 2, / 42

19 What does G n look like? Let Then Simplify G n = G n = G(x, x ) = 1 1 4π x x 1 π e int 32π 3 π r 2 + r 2 2rr cos t + (z z ) dt 2 1 π cos nt 32π 3 π r 2 + r 2 2rr cos t + (z z ) dt 2 Write in terms of special functions, G n = 2 Q rr n 1 (χ), χ = r2 + r 2 + (z z ) 2 2 2rr Saibal De (U of M) Axis-symmetric BIE February 2, / 42

20 What are all the Q stuff? Legendre differential equation: Two solutions: Maximal P λ (χ) Minimal Q λ (χ) Recurrence relations: (1 χ 2 )u 2χu + λ(λ + 1)u = 0 Q n 1/2 (χ) = 4 n 1 2n 1 χq n 3/2(χ) 2n 3 2n 1 Q n 5/2(χ) Saibal De (U of M) Axis-symmetric BIE February 2, / 42

21 References Reduction to Fourier modes Young, Hao and Martinsson, A high-order Nystrom discretization scheme for boundary integralequations defined on rotationally symmetric surfaces, Relation to Legendre Q functions Cohl and Tohline, A Compact Cylindrical Greens Function Expansion for the Solution of Potential Problems, Saibal De (U of M) Axis-symmetric BIE February 2, / 42

22 Application to Stokes flow Saibal De (U of M) Axis-symmetric BIE February 2, / 42

23 A linearization of Navier-Stokes Incompressible, isotropic with Galilean invariant stress: t u + (u )u = 1 p + µ 2 u + g ρ 0 ρ 0 u = 0 Drop the acceleration and advection term, no external force: p + µ 2 u = 0 u = 0 Saibal De (U of M) Axis-symmetric BIE February 2, / 42

24 The Stokeslet Green s function for Stokes equation: G(x, x ) = 1 [ I 8πµ x x + J(x, ] x ) x x 3 where Single layer BIE J(x, x ) = (x x )(x x ) G(x, x )σ(x ) dγ(x ) = f(x) Γ Saibal De (U of M) Axis-symmetric BIE February 2, / 42

25 Making J rotationally invariant Define cos φ sin φ 0 U(φ) = sin φ cos φ U(φ) rotates a vector about z-axis by angle φ Define J(x, x ) = U(θ)J(x, x )U(θ ) = [U(θ)(x x )][U(θ )(x x )] Saibal De (U of M) Axis-symmetric BIE February 2, / 42

26 Rotationally invariant Stokes BIE Single layer equation, Rotationally invariant kernel Γ G(x, x )σ(x ) dγ(x ) = f(x) G(x, x ) = U(θ) G(x, x )U(θ ) Transformed equation G(x, x )[U(θ )σ(x )] dγ(x ) = [U(θ)f(x)] Γ Saibal De (U of M) Axis-symmetric BIE February 2, / 42

27 References Stokes equation and related BIEs Pozrikidis, Boundary integral and singularity methods for linearized viscous flow, 1992 Saibal De (U of M) Axis-symmetric BIE February 2, / 42

28 Numerical Aspects Saibal De (U of M) Axis-symmetric BIE February 2, / 42

29 Algorithm for single body Set up quadrature on γ; generate grid points on Γ Compute kernel matrices associated to G n Apply rigid motion, apply transformation to f Compute Fourier coefficients for f Solve the linear systems G n σ n = f n Compute σ, undo transformation and rigid motion to get σ Evaluate field u = Sσ Saibal De (U of M) Axis-symmetric BIE February 2, / 42

30 Issue: Selecting appropriate quadrature The kernel G n are singular, need to adapt Nyström Solution: Use specially designed quadratures Kapur-Rokhlin Alpert Kolm-Rokhlin (Gauss) Saibal De (U of M) Axis-symmetric BIE February 2, / 42

31 Issue: Instability of Q recurrences Recurrence relations for Q n 1/2 (χ) is unstable for all χ > 1! Solution 1: use the recursion formula differently Miller s algorithm Solution 2: don t use recursion for all values of χ For χ (say), use recursion For χ > , use FFT to evaluate integral Saibal De (U of M) Axis-symmetric BIE February 2, / 42

32 Issue: Multi-body simulations Formally, N b L ij σ j = f i j=1 After discretization, L 11 L L 1Nb σ 1 f 1 L 21 L L 2Nb σ = f 2. System is too big! L Nb 1 L Nb 2... L Nb N b σ Nb f Nb Saibal De (U of M) Axis-symmetric BIE February 2, / 42

33 Issue: Multi-body simulations Solution: Split L into two parts: L = D + B Rewrite (D + B)σ = f = (I + D 1 B)σ = D 1 f Use iterative solvers: σ k+1 = D 1 f D 1 Bσ k...or, more sophisticated GMRES Saibal De (U of M) Axis-symmetric BIE February 2, / 42

34 Issue: Particle-particle interactions Evaluating Bσ k is expensive Solution: Use FMM to reduce complexity for far off particles Saibal De (U of M) Axis-symmetric BIE February 2, / 42

35 References Singular quadrature on curves Hao, Barnett, Martinsson and Young, High-order accurate Nystrom discretization of integral equations with weakly singular kernels on smooth curves in the plane, 2011 Fast Multipole Methods Saibal De (U of M) Axis-symmetric BIE February 2, / 42

36 Some Results Saibal De (U of M) Axis-symmetric BIE February 2, / 42

37 Error for single body simulations Saibal De (U of M) Axis-symmetric BIE February 2, / 42

38 Run-time for single body simulation Saibal De (U of M) Axis-symmetric BIE February 2, / 42

39 Example flow 1: interior domain Saibal De (U of M) Axis-symmetric BIE February 2, / 42

40 Example flow 2: exterior domain Saibal De (U of M) Axis-symmetric BIE February 2, / 42

41 Next Steps Move beyond fixed bodies Resolve collision between particles Saibal De (U of M) Axis-symmetric BIE February 2, / 42

42 Thanks for Listening! Questions? Saibal De (U of M) Axis-symmetric BIE February 2, / 42

lim = F F = F x x + F y y + F z

lim = F F = F x x + F y y + F z Physics 361 Summary of Results from Lecture Physics 361 Derivatives of Scalar and Vector Fields The gradient of a scalar field f( r) is given by g = f. coordinates f g = ê x x + ê f y y + ê f z z Expressed