Numerical Solutions of Geometric Partial Differential Equations. Adam Oberman McGill University

Transcription

1 Numerical Solutions of Geometric Partial Differential Equations Adam Oberman McGill University

2 Sample Equations and Schemes

3 Fully Nonlinear Pucci Equation! "#+ "#* "#) "#( "#$ "#' "#& "#% "#! "!!!"#$ " "#$! Figure 4. Surface plot of the Pucci solution, for = 3, n = 256. Plot of the midline of the solutions, increasing with = 2, 2.5, 3, 5, n = 256.

4 Mean Curvature image: Evans-Spruck Fig. 3. Surface plot: initial data, and solution at time Fig. 2. Contour plots of the.02, and.02 contours at times 0,.015,.03,.045 =

5 n Infinity Laplacian u = 1 Du 2 m i,j=1 u xi x j u xi u xj = 0 ded, open set in R, along with Dirichlet bou

6 Fractional Obstacle Problem (with) Yanghong Huang 1.2 Numerical ( Δ) α/2 u Exact ( Δ) α/2 u u (a) ( Δ) α/2 u

7 Filtered Schemes for Hamilton Jacobi with Tiago Salvador

8 Obtain High Accuracy in 1d (even if solutions not smooth)

9 Obtain 2nd order accuracy in 2d

10 General Convex Envelopes Directionally Convex Envelopes Rank 1 Convex Envelope: Laminate (scalar) quasi-convex envelope: make level sets of function convex With Yanglong Ruan

11 Microstructure in Laminates

12 Four Gradient Example

13 Four Gradient Example

14 Four D (2X2) Example

15 Numerical Solution of the Infinity Laplace Equation via solution of the absolutely minimizing Lipschitz extension problem in a discrete setting

16 The discrete Lipschitz extension problem. Definition. Given distinct x 0,..., x n in R m, and values u i = u(x i ), for i = 1,... n, the discrete Lipschitz constant at x 0, is L(u 0 ) = max n i=1 Li (u 0 ) = max n i=1 u 0 u i x 0 x i Problem. Minimize the discrete Lipschitz constant of u at x 0, (computed with respect to the points x 1,..., x n ) over the value u 0 = u(x 0 ) min u 0 L(u 0 )

17 Now solve the problem at every point on a grid.

18 n Infinity Laplacian u = 1 Du 2 m i,j=1 u xi x j u xi u xj = 0 f(x, y) = x 4/3 y 4/3, i ded, open set in R, along with Dirichlet bou

19 Metric induced by different stencils dθ } dx dθ } dx dx dθ } Figure 1. Grids for the 5, 9, and 17 point schemes, and level sets of the cones for the corresponding schemes.

20 Convergence of the scheme Theorem. Let u be a C 2 function in a neighborhood of x 0. Suppose we are given neighbors x 1,..., x n, arranged symmetrically on a grid. Let u be the solution of the discrete minimal Lipschitz extension problem computed with respect to the points x 1,..., x n, and let i, j be the indices which maximize the relaxed discrete gradient. Then u(x 0 ) = 1 (u(x 0 ) u ) + O(d + dx) d i d j Theorem (Convergence). The solution of the di erence scheme defined above converges (uniformly on compact sets) as dx, d 0 to the solution of (IL). Proof. Convergence to the solution of (IL) follows from consistency and degenerate ellipticity (monotonicity) of the scheme by [Barles- Souganidis].

21 Mean Curvature

22 Interpretations Catte-Dibos-Koepfler (1985) morphological scheme for mean curvature Kohn-Serfaty (2005) deterministic control based approach to motion by mean curvature. Ryo Takei (2007) M.S. thesis:

23 Failure of naive difference scheme Simply replace all the terms in the equation by a finite difference. Explicit in time. 1 0!1 0.5 Use exact steady solution (with straight level sets) on periodic domain !0.5! Numerical solution contracts over time to a constant !0.1 Monotone scheme converges for this example.!0.2!0.3!0.4!0.5!0.5!0.4!0.3!0.2!

24 Other schemes discretize equation written in divergence structure will get capping at local min/max of level set function. div( grad u/ grad u ). When u has local max, get nonzero divergence, even if function has straight level sets. Expect similar behavior for FEM method.

25 Scheme: part 1 of 2

26 Scheme: part 1 of 2 PDE: u t = Du div Du Du = d2 u dt 2, t = (u y, u x ) u 2 x + u 2 y

27 Scheme: part 1 of 2 PDE: u t = Du div Du Du = d2 u dt 2, t = (u y, u x ) u 2 x + u 2 y Use this interpretation to discretize spatial operator by finite differences d 2 u u(x + dx t) 2u(x) + u(x dx t) = dt2 dx 2 + O(dx 2 )

28 Scheme: part 1 of 2 PDE: u t = Du div Du Du = d2 u dt 2, t = (u y, u x ) u 2 x + u 2 y Use this interpretation to discretize spatial operator by finite differences d 2 u u(x + dx t) 2u(x) + u(x dx t) = dt2 dx 2 + O(dx 2 ) Q: How to find a monotone discretization of this operator?

29 Scheme: part 2 of 2

30 Scheme: part 2 of 2 u = median {u 1, u 2,..., u 12 } = u 2 + u 7 2 ( + t) + ( t)

31 Scheme: part 2 of 2 u = median {u 1, u 2,..., u 12 } = = u 2 + u 7 2 2( + t) + ( t) u(x + dx t) + u(x dx t) 2 + O(dw)

32 Scheme: part 2 of 2 u = median {u 1, u 2,..., u 12 } = = u 2 + u 7 2 2( + t) + ( t) u(x + dx t) + u(x dx t) 2 + O(dw) d 2 u dt 2 = 2u 2u(x) dx 2 + O(dx 2 + dw)

33 Scheme: part 2 of 2 u = median {u 1, u 2,..., u 12 } = = u 2 + u 7 2 2( + t) + ( t) u(x + dx t) + u(x dx t) 2 + O(dw) d 2 u dt 2 = 2u 2u(x) dx 2 + O(dx 2 + dw) Scheme is consistent, with additional error due to directional resolution, decreased by widening stencil.

34 Fattening image: Evans-Spruck

35 Fattening image: Evans-Spruck Fig. 3. Surface plot: initial data, and solution at time.03 =

36 Fattening image: Evans-Spruck Fig. 3. Surface plot: initial data, and solution at time Fig. 2. Contour plots of the.02, and.02 contours at times 0,.015,.03,.045 =