The Backward Error Compensation Method for Level Set Equation Wayne Lawton and Jia Shuo Department of Mathematics National University of Singapore

Level Set Method (Osher and Sethian 1988) The interface is represented as a zero level set of a Lipschitz continuous functionThe interface is represented as a zero level set of a Lipschitz continuous function φ(x,t) The evolution equation ofThe evolution equation of φ(x,t) under a velocity field u: grad Φ t + u · grad φ= 0

Applications –Multiphase flows –Stefan problem –Kinetic crystal growth –Image processing and computer vision Advantages –Naturally handle topological changes and complex geometries of the interfaces –Simple formulae for unit normal and curvature n = grad φ κ = div (gradφ/|gradφ|)

Conventional Numerical Schemes Schemes for hyperbolic conservation laws –Spatial: Essential Non-Oscillatory (ENO) –Temporal: Total Variation Diminishing Runge- Kutta (TVD-RK)

Backward Error Compensation (Dupont and Liu 2003) Consider the ODE: y’= f(t,y) Advance it one step from t n to t n+1 by forward Euler method: y 1 n+1 = y n + Δt f n Solve the ODE backward from t n+1 to t n y 2 n = y 1 n+1 - Δt f n+1 If no numerical errors, y n = y 2 n. Let e = y n - y 2 n

Assume that the errors in the forward and backward process are the same. Let y 3 n = y n + e/2 = y n + (y n - y 2 n )/2 Starting with y 3 n to remove the principal components of the error at t n+1, solve the ODE forward again. y n+1 = y 3 n + Δt f(t n, y 3 n )

Backward Error Compensation y n+1 = y n + Δt(f n+1 + f(t n, y 3 n ) – f n )/2 2 nd order modified Euler scheme y n+1 = y n + Δt(f n+1 + f n )/2 (f n+1 + f(t n, y 3 n ) – f n ) - (f n+1 + f n ) =O(Δt 2 ) The forward Euler with backward error compensation is 2 nd order accuracy

Theorem: The backward error compensation algorithm can improve the order of accuracy of kth order Taylor method y n+1 = y n + Δt f n + Δt 2 f’ n /2 + …+ Δt k f (k-1) n /k! by one if k is an odd positive number.

Consider 1-D level set equation Φ t + uφ x = 0 1 st order upwind scheme for φ x (φ x ) i = (φ i – φ i-1 )/Δx, if u i > 0 (φ i+1 – φ i )/Δx, if u i < 0 Assume u =1, the 1 st order upwind scheme can be written as Φ n+1 i = (1-λ) φ n i + λφ n i-1, where λ= Δt/Δx.

Applying the backward error compensation, Φ n+1 i = (λ 2 /2+λ 3 /2) φ n i-2 + (λ/2+2 λ 2 -3λ 3 )φ n i-1 + (1-5λ 2 /2+3λ 3 /2)φ n i + (-λ/2+λ 2 -λ 3 /2)φ n i+1 (1) The local truncation error is O(Δx 3 ) It not only improves the temporal order of accuracy by one, but also improves the spatial order by one.

Stable Condition Theorem: If 0<λ ≤ 1.5 and u n+1 is defined by (1) with periodic boundary condition, then ||u n+1 || 2 ≤ ||u n || 2. Proof: by expansion in Fourier series. Remark: Backward error compensation with center difference creates a stable scheme for 0<λ ≤ 3 1/2.

Reinitialization Keep Φ as a distance function at each time step PDE approach Φ τ = sgn(Φ 0 )(1-|gradΦ|) Φ(x,0) = Φ 0 (x) ENO and TVD-RK

For the grid x near the interface, we do a Taylor expansion to find an accurate approximation of the orthogonal projection y of x on the interface. y = x + rp, where p = grad φ/|grad φ| Φ(x) + |grad φ |r + (p T H(φ)p) r 2 = φ(y) = 0, where H is the Hessian matrix of φ. Then r is the distance from x to the interface. x y p

Accuracy Check Move a unit circle Φ 0 =(x 2 + y 2 ) 1/2 -1 with a constant velocity (1,1) in a 4x4 periodic box. 64x64128x128256x256order error9.32e e-45.51e-52.04

Numerical Results Rotational Velocity Field Rotate a slotted disk in a 100x100 square with the velocity field u(x,y) = pi(50-y)/314; v(x,y) = pi(x-50)/314. We compute for t=628 on a 200x200 grid.

T = 0T = 628

Applications in Multiphase Flows Solve Navier-Stokes equation using projection method Solve the level set equation using backward error compensation Reinitialization by solving the quadratic equation near the interface

ρ w / ρ A = 100; μ w / μ A = 10 Δx=Δy=0.005; Δt=0.001 Numerical Examples

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