Homogenization 2006

September 10

Part 1.

The eikonal equation,
||grad{u}||_2 = c(x,y), ...(1)
was solved with boundary condition u(0,0) = 0 for c(x,y) being:
Case 1. discrete values, 1 and alpha, randomly, and
Case 2. uniformly distributed values in [1,alpha].
The solutions were compared to the solution of
||grad{u}||_{P8} = C, ...(2)
where C is the harmonic mean of c(x,y). Number of grid point used were 51, 81, 111 and 151.

Left: Case 1, Right: Case 2. alpha = 10.

Left: error between solutions of (1) and (2) for case 1, Right: error for case (2)

Part 2.

The same test as with Part 1, except with boundary condition u = |x| + |y| on the edge of the domain.

Left: Case 1, Right: Case 2. alpha = 1.5.

Left: error between solutions of (1) and (2) for case 1, Right: error for case (2)

Summary of research during summer 2006

PDF file: report (.tex file in .txt format)
Remark: Homogenization of the Eikonal equation yeilds the same equation with a different norm!

Some numerically computed optimal path plots:

March 19

Part 1

Equation to solve: c(x/eps, y/eps) (|u_x| + |u_y|) = 1 for eps small.
Let c(.,.) be a checkerboard pattern of 1 and alpha > 1. For computation, let alpha = 2. When homogenized, equation becomes C (|u_x| + |u_y|) = 1 where C = 2 / (1 + 1/alpha), the harmonic mean of c(.,.).
Solved numerically via iteration forward Euler: u_t = c (|u_x| + |u_y|) - 1.
Used |u_x| = (max{u_(i+1,j), u_(i-1,j)} - u_(ij))/h, and similarly for |u_y|.

    Matlab Codes:
    For constant C:   homo1.m
    For checkerboard c(.,.):   nonhomo1.m

Part 2

Equation to solve: c(x/eps, y/eps)|u_x| + |u_y| = 1 for eps small.
Homogenization: C|u_x| + |u_y| = 1, where C = (3*alpha+1)/(3+alpha).
Remark: this is the same homogenization result for c(x/eps, y/eps)u_xx + u_yy = 1.

    Matlab Codes:
    For constant C:   homo2.m
    For checkerboard c(.,.):   nonhomo2.m