Spaces:
Running
Running
File size: 3,365 Bytes
83913eb | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 | #!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Assuming constant D
Original PDE, after multiplied by T (Normalize by final time)
u_t = D laplace u + ρ u(1-u)
where t = [0 1], D = D T, ρ = ρ T
For radial symmetry solution , u(r,θ)=u(r), any derivative w.r.t angle is 0
see wiki https://en.wikipedia.org/wiki/Laplace_operator
In polar coordinate , laplace u(r,θ) = u_rr + 1/r u_r
In spherical coordinate , laplace u(r,θ,φ) = u_rr + 2/r u_r
https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates
In polar coordinate , grad u = u_r
In spherical coordinate , grad u = u_r
At r=R, Neuman boundary condition u_r = 0
At r=0, symmetric at origin, u_r = 0
Therefore, the whole pde is 1d pde on domain [0,R]:
2D u_t = D (u_xx + 1/x u_x) + ρ u (1-u), neuman bc
3D u_t = D (u_xx + 2/x u_x) + ρ u (1-u), neuman bc
In PINN, we use a diffused domain, x=[0,1]
(uϕ)_t = D div ( ϕ grad u) + ρ ϕ u(1-u)
= D grad ϕ . grad u + D ϕ laplace u + ρ ϕ u(1-u)
We have
2D u_t = D ( ϕ_x u_x + ϕ u_xx + ϕ 1/x u_x ) + ρ ϕ u (1-u),
3D u_t = D ( ϕ_x u_x + ϕ u_xx + ϕ 2/x u_x ) + ρ ϕ u (1-u),
neuman bc at x=0 (for radial symmetry)
choose neuman bc at x=1 (can be arbitrary)
based on the manual
https://py-pde.readthedocs.io/en/latest/getting_started.html
this package use finite difference. The grid points are placed at cell center
By default of pde.solve, ScipySolver with an automatic, adaptive time step provided by scipy is used.
"""
import numpy as np
import matplotlib.pyplot as plt
from pde import PDE, CartesianGrid, MemoryStorage, ScalarField, plot_kymograph, UnitGrid
Dim = 2# dimesion
R = 0.5 # circle domain
N = 50 # grid size
T = 300 # final time
xi,h = np.linspace(0,0.5,N,retstep=True)
# time normalized coefficients
D = T*0.13e-4
rho = T*0.025
#%% this does not use diffused domain
# based on the example
#https://py-pde.readthedocs.io/en/latest/examples_gallery/pde_heterogeneous_diffusion.html#sphx-glr-examples-gallery-pde-heterogeneous-diffusion-py
eq = PDE({"u": f"{D}*laplace(u) + {Dim-1}*{D}*d_dx(u)/x + {rho}*u * (1-u)"},bc = {'derivative': 0})
grid = CartesianGrid([[0, R]], N)
u0 = ScalarField.from_expression(grid, "0.1*exp(-1000*x**2)")
# solve the equation and store the trajectory
storage = MemoryStorage()
eq.solve(u0, t_range=1, tracker=storage.tracker(0.1)) #interface to sample the solution
#%% plot_kymograph(storage)
x = np.array(storage.grid.coordinate_arrays)
y = np.array(storage.data)
plt.plot(x.T,y.T)
#%% this part use diffused domain
phi = "(0.5+0.5* tanh((0.5-x)/0.01))"
pde = f"{D}*dot(gradient({phi}),gradient(u)) + {D}*{phi}*laplace(u) + {Dim-1}*{D}*{phi}*d_dx(u)/x + {phi}*{rho}*u*(1-u)"
eq = PDE({"u": pde},bc = {'derivative': 0})
# the witdith of the diffused domain is approximated 0.02
# need to choose h such that arond 5 grid point resolve the interface
grid = CartesianGrid([[0, 0.7]], 200)
u0 = ScalarField.from_expression(grid, "0.1*exp(-1000*x**2)")
# solve the equation and store the trajectory
storage = MemoryStorage()
eq.solve(u0, t_range=1, tracker=storage.tracker(0.1))
# plot_kymograph(storage)
x = np.array(storage.grid.coordinate_arrays)
y = np.array(storage.data)
plt.plot(x.T,y.T)
plt.xlim([0,R]) |