test_ibm_polynomials Namespace Reference

Functions
	poly_D (xi)
	poly_N_c (xi)
	poly_N_nb (xi)
	poly_N_bc (xi)
	verify_1d_operator ()

Function Documentation

poly_D()

test_ibm_polynomials.poly_D

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xi

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Definition at line 3 of file test_ibm_polynomials.py.

Referenced by verify_1d_operator().

poly_N_c()

test_ibm_polynomials.poly_N_c

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xi

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Definition at line 6 of file test_ibm_polynomials.py.

Referenced by verify_1d_operator().

poly_N_nb()

test_ibm_polynomials.poly_N_nb

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xi

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Definition at line 9 of file test_ibm_polynomials.py.

Referenced by verify_1d_operator().

poly_N_bc()

test_ibm_polynomials.poly_N_bc

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xi

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Definition at line 12 of file test_ibm_polynomials.py.

verify_1d_operator()

test_ibm_polynomials.verify_1d_operator

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Verify 1D IBM Operator consistency with second derivative.

Setup:
- Domain: x in [0, 1]
- Grid: N cells, dx = 1/N
- Boundary: Located at x_b = (i_ghost + xi) * dx
- Function: u(x) = (x - x_b)^2
  u(x_b) = 0
  u'(x) = 2(x - x_b)
  u''(x) = 2

Discrete Operator L(u) should equal d^2u/dx^2 = 2.
L(u)_c = (A_c * u_c + A_nb * u_nb + B_rhs) / D_scale

Wait, in our code:
A_c_mod = D*A_c + K*A_g
A_nb_mod = D*A_nb + M*A_g
B_mod = D*B + B_fac

Standard Stencil (Diffusion):
A_c = -2/dx^2, A_nb = 1/dx^2, A_g = 1/dx^2
B = 0

Modified Stencil:
A_c' = D*(-2) + N_c*(1)
A_nb' = D*(1) + N_nb*(1)  <- (Here neighbor is fluid, ghost is on other side)
So if ghost is West:
  A_E' = D*(1) + N_E*(1) = D + 0 (since N_E applies to ghost dir)
  Wait, N_nb in code applies to the FLUID neighbor in the interpolation stencil?
  Table 1 says "N_{nb,d} applies to the fluid neighbor used in interpolation".
  Interpolation uses: Ghost(at -1), Center(0), Neighbor(1).
  So N_nb applies to Neighbor at +1.
  
  So:
  A_E' = D * A_E + N_nb * A_W (Cross term? No.)
  Let's look at code:
  modify_stencil_ibm_kernel:
  A_c += A_nb * K  (A_nb is the coeff OF the ghost direction in standard stencil)
  
  Example: Ghost is West (index 1 in loop).
  orig_AW is the coefficient for u_W in standard stencil.
  K = N_c * R (R=1 for 1D)
  M = 0
  X = N_nb * R
  
  Update:
  A_C += orig_AW * K = -2*D + 1*N_c
  
  Neighbor updates:
  mod_AW += orig_AW * (D*M - 1) = 1 * (0 - 1) = -1. 
  So A_W becomes 1 + (-1) = 0. Correct.
  
  mod_AE += orig_AE * X = 1 * N_nb = N_nb.
  So A_E becomes D*1 + N_nb.
  
  So Discrete Eq:
  (A_C' * u_C + A_E' * u_E) * (1/dx^2) / D_scale
  = [ (-2D + N_c)*u_C + (D + N_nb)*u_E ] / D
  
  RHS correction:
  B_val for West ghost:
  Code says B = 0 if is_ghost is true? 
  Wait, in `compute_ibm_geometry_kernel`:
  if (is_ghost[kp]) {
      K = N_c * R
      X = N_nb * R
      B = 0
  }
  Where is u_bc handled?
  Ah, `compute_ibm_geometry_kernel` assumes B_val is 0 in the code shown in `src/cfd_solver_ibm.cu`?
  Let's check lines 220-230.
  
  Line 226: ibm_data.B_val[...] = 0.0f;
  
  This looks suspicious! The text says N_bc * u_bc.
  The code seems to set B_val = 0.
  
  Let's verify if u_bc is handled elsewhere or if this is the Bug.
  If u_bc = 0 (Homogeneous Dirichlet), then B=0 is correct.
  Our verification case uses u=0 at boundary, so B=0 is fine for now.
  
Test for u_bc = 0.

Definition at line 15 of file test_ibm_polynomials.py.

References poly_D(), poly_N_c(), poly_N_nb(), and verify_1d_operator().

Referenced by verify_1d_operator().