A Fast Stokes Solver Isuru Fernando University of Illinois, Urbana - - PowerPoint PPT Presentation

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A Fast Stokes Solver Isuru Fernando University of Illinois, Urbana - - PowerPoint PPT Presentation

Dec 06, 2022 328 likes •567 views

A Fast Stokes Solver Isuru Fernando University of Illinois, Urbana Champaign idf2@illinois.edu December 6, 2017 Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 1 / 18 Stokes PDEs 2 u p + f = 0 u = 0 where u

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SLIDE 1

A Fast Stokes Solver

Isuru Fernando

University of Illinois, Urbana Champaign idf2@illinois.edu

December 6, 2017

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 1 / 18

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SLIDE 2

Stokes PDEs

µ∇2u − ∇p + f = 0 ∇ · u = 0 where u is the velocity vector, µ is the viscosity, p is the pressure and f is the force.

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 2 / 18

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SLIDE 3

Stokes PDEs: 3D case

µ ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2

  • − ∂p

∂x + fx = 0 µ ∂2v ∂x2 + ∂2v ∂y2 + ∂2v ∂z2

  • − ∂p

∂y + fy = 0 µ ∂2w ∂x2 + ∂2w ∂y2 + ∂2w ∂z2

  • − ∂p

∂z + fz = 0 ∂u ∂x + ∂v ∂y + ∂w ∂z = 0

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 3 / 18

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SLIDE 4

Green’s Function solution: The Stokeslet

u(r) = F · J(r) p(r) = F · r 4π|r|3 where r =

  • x

y z

  • J(r) =

1 8πµ I |r| + rr |r|3

  • Isuru Fernando (UIUC)

A Fast Stokes solver December 6, 2017 4 / 18

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SLIDE 5

Green’s Function solution: The Stokeslet

    u v w p     = 1 8πµ|r|3     |r|2 + x2 xy xz xy |r|2 + y2 yz xz yz |r|2 + z2 2µx 2µy 2µz       fx fy fz  

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 5 / 18

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SLIDE 6

Current implementation

pytential uses 9 kernel functions. (Reuse 3 kernels) Drawbacks of this method are, Takes more storage than necessary Redundant computations

Derivative relationships in Taylor series Common Subexpression Elimination

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 6 / 18

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SLIDE 7

Revisiting local expansion

Let x be the targets and y be the sources and f be the charge/force f (y)φ(x − y) ≈

  • |p|≤k

Dp

x f (y)φ(x − y)

  • x=c

p! (x − c)p Here

f (y)Dp

x φ(x−y)

  • y=c

p!

depends on source and center, while (x − c)p depends on target and center

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 7 / 18

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SLIDE 8

Local expansion for Stokes

In the stokes equation case u is a linear combination of 3 functions φ1, φ2, φ3 and there are 3 charges f1, f2, f3 u(x − y) =

3

  • i=1

fi(y)φi(x − y) ≈

  • |p|≤k

Dp

x

3

i=1 fi(y)φi(x − y)

  • x=c

p! (x − c)p

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 8 / 18

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SLIDE 9

Revisiting multipole expansion

Let x be the targets and y be the sources and f be the charge/force f (y)φ(x − y) ≈ f (y)

  • |p|≤k

Dp

x φ(x − y)

  • y=c

p! (y − c)p

  • r equivalently

f (y)φ(x − y) ≈

  • |p|≤k

Dp

x φ(x − y)

  • y=c

p! f (y)(y − c)p Here

Dp

x φ(x−y)

  • y=c

p!

depends on target and center, while f (y)(y − c)p depends on source and center

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 9 / 18

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SLIDE 10

Multipole expansion for Stokes

In the stokes equation case conside the first velocity component u which is a linear combination of 3 functions φ1, φ2, φ3 and there are 3 charges f1, f2, f3 u(x−y) =

3

  • i=1

fi(y)φi(x−y) ≈

3

  • i=1

 fi(y)

  • |p|≤k

Dp

x φi(x − y)

  • y=c

p! (y − c)p   u(x − y) ≈

  • |p|≤k

3

  • i=1

Dp

x φi(x − y)

  • y=c

p! fi(y)(y − c)p

  • Needs 12 multipoles

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 10 / 18

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SLIDE 11

Multipole to local translation for Stokes

Let φ1,1, φ1,2, φ1,3, φ2,1, . . . , φ3,3 be the 12 multipole kernels. Then, u = φ1,1 + φ1,2 + φ1,3 v = φ2,1 + φ2,2 + φ2,3 w = φ3,1 + φ3,2 + φ3,3 p = φ4,1 + φ4,2 + φ4,3

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 11 / 18

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SLIDE 12

Stokes PDE Derivatives

µ ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2

  • − ∂p

∂x + fx = 0 (1) µ ∂2v ∂x2 + ∂2v ∂y2 + ∂2v ∂z2

  • − ∂p

∂y + fy = 0 (2) µ ∂2w ∂x2 + ∂2w ∂y2 + ∂2w ∂z2

  • − ∂p

∂z + fz = 0 (3) ∂u ∂x + ∂v ∂y + ∂w ∂z = 0 (4)

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 12 / 18

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SLIDE 13

Stokes PDE Derivatives

µ ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2

  • − ∂p

∂x + fx = 0 (1) µ ∂2v ∂x2 + ∂2v ∂y2 + ∂2v ∂z2

  • − ∂p

∂y + fy = 0 (2) µ ∂2w ∂x2 + ∂2w ∂y2 + ∂2w ∂z2

  • − ∂p

∂z + fz = 0 (3) ∂u ∂x + ∂v ∂y + ∂w ∂z = 0 (4)

∂(1) ∂x + ∂(2) ∂y + ∂(3) ∂z

gives us ∂2p ∂x2 + ∂2p ∂y2 + ∂2p ∂z2 = 0 (5) Pressure term satisfies the Laplace equation !

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 12 / 18

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SLIDE 14

Laplace Derivative Reduction

∂2p ∂x2 + ∂2p ∂y2 + ∂2p ∂z2 = 0 ∂a+b+2p ∂xa+2yb + ∂a+b+2p ∂xayb+2 + ∂a+b+2p ∂xaybz2 = 0 ∂a+b+2p ∂xaybz2 = −∂a+b+2p ∂xa+2yb − ∂a+b+2p ∂xayb+2

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 13 / 18

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SLIDE 15

Laplace Derivative Reduction

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 14 / 18

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SLIDE 16

Stokes Derivative Reduction

µ ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2

  • − ∂p

∂x + fx = 0 (1) ∂a+b+2u ∂xaybz2 = 1 µ ∂a+b+1p ∂xa+1yb − ∂a+b+2u ∂xa+2yb − ∂a+b+2u ∂xayb+2

  • (a)

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 15 / 18

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SLIDE 17

Stokes Derivative Reduction

µ ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2

  • − ∂p

∂x + fx = 0 (1) ∂a+b+2u ∂xaybz2 = 1 µ ∂a+b+1p ∂xa+1yb − ∂a+b+2u ∂xa+2yb − ∂a+b+2u ∂xayb+2

  • (a)

µ ∂2v ∂x2 + ∂2v ∂y2 + ∂2v ∂z2

  • − ∂p

∂y + fy = 0 (2) ∂a+b+2v ∂xaybz2 = 1 µ ∂a+b+1p ∂xayb+1 − ∂a+b+2v ∂xa+2yb − ∂a+b+2v ∂xayb+2

  • (b)

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 15 / 18

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SLIDE 18

Stokes Derivative Reduction

µ ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2

  • − ∂p

∂x + fx = 0 (1) ∂a+b+2u ∂xaybz2 = 1 µ ∂a+b+1p ∂xa+1yb − ∂a+b+2u ∂xa+2yb − ∂a+b+2u ∂xayb+2

  • (a)

µ ∂2v ∂x2 + ∂2v ∂y2 + ∂2v ∂z2

  • − ∂p

∂y + fy = 0 (2) ∂a+b+2v ∂xaybz2 = 1 µ ∂a+b+1p ∂xayb+1 − ∂a+b+2v ∂xa+2yb − ∂a+b+2v ∂xayb+2

  • (b)

∂u ∂x + ∂v ∂y + ∂w ∂z = 0 (4) ∂a+b+1w ∂xaybz = −∂a+b+1u ∂xa+1yb − ∂a+b+1v ∂xayb+1 (c)

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 15 / 18

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SLIDE 19

Stokes Derivative Reduction

∂a+b+2u ∂xaybz2 = 1 µ ∂a+b+1p ∂xa+1yb − ∂a+b+2u ∂xa+2yb − ∂a+b+2u ∂xayb+2

  • (a)

∂a+b+2v ∂xaybz2 = 1 µ ∂a+b+1p ∂xayb+1 − ∂a+b+2v ∂xa+2yb − ∂a+b+2v ∂xayb+2

  • (b)

∂a+b+1w ∂xaybz = −∂a+b+1u ∂xa+1yb − ∂a+b+1v ∂xayb+1 (c) ∂a+b+2p ∂xaybz2 = −∂a+b+2p ∂xa+2yb − ∂a+b+2p ∂xayb+2 (d)

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 16 / 18

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SLIDE 20

Stokeslet Kernel Derivative Reduction

    u v w p     = 1 8πµ|r|3     |r|2 + x2 xy xz xy |r|2 + y2 yz xz yz |r|2 + z2 2µx 2µy 2µz       fx fy fz  

  • u

v w p

  • satisfies the Stokes PDEs

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 17 / 18

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SLIDE 21

Stokeslet Kernel Derivative Reduction

    u v w p     = 1 8πµ|r|3     |r|2 + x2 xy xz xy |r|2 + y2 yz xz yz |r|2 + z2 2µx 2µy 2µz       fx fy fz  

  • u

v w p

  • satisfies the Stokes PDEs

    u0 v0 w0 p0     = 1 8πµ|r|3     |r|2 + x2 xy xz 2µx    

  • u0

v0 w0 p0

  • satisfies the Stokes PDEs

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 17 / 18

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SLIDE 22

Questions?

Isuru Fernando (UIUC) A Fast Stokes solver December 6, 2017 18 / 18