为什么我对一组 ODE 的数值解是无限的?

计算科学 稳定
2021-12-03 00:12:05

我正在尝试解决以下线性偏微分方程 对于 ,的函数。其中

uxx=[iωb||+u],
b||x=iωLux,
Lu=iωb||,
ux=uxb||=b_paru=u_perpxz
=iksinαz,
L=cos2α2z2+ω2vA2(x,z).

为了解决这个问题,我正在尝试使用线条的方法。我离散化给出一组 ODE该代码适用于的情况。附图显示解决方案趋于无穷大。这是否意味着该方案在数值上不稳定?请注意,如果我更改读取的代码行zvA(x,z)=vA0

    dux_dx = -(1j * omega * b_par + 1j * k_perp * u_perp - \
                np.sin(alpha) / (2 * dz) * (np.roll(u_perp,-1) - np.roll(u_perp,1)))

    dux_dx = -(1j * omega * b_par_exact(x,z) + nabla_perp0 * u_perp_exact(x,z))

然后代码完美运行。这表明矩阵求逆导致解趋于无穷大。你知道如何解决这个问题吗?

代码:

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits import mplot3d
from scipy.integrate import solve_ivp
from scipy.linalg import solve_banded

def ux_exact(x,z):
    return ux0 * np.exp(1j * (kx * x + kz * z))

def b_par_exact(x,z):   # Note that b_par denotes b_{||}
    return b_par0 * np.exp(1j * (kx * x + kz * z))

def u_perp_exact(x,z): # Note that perp is short for perpendicular
    return u_perp0 * np.exp(1j * (kx * x + kz * z))

def calc_u_perp(x, b_par):
    # Calculate u_perp by inverting the matrix ab,
    # Note that A is nearly tridiagonal and so a computationally efficent algorithm can be used
    # See https://www.cfd-online.com/Wiki/Tridiagonal_matrix_algorithm_-_TDMA_(Thomas_algorithm)
    [ab, u, v] = calc_A_u_v(x)
    d = vector_d(b_par)

    y = solve_banded((1,1), ab, d)
    q = solve_banded((1,1), ab, u)

    vTy = v[0] * y[0] + v[-1] * y[-1]
    vTq = v[0] * q[0] + v[-1] * q[-1]
    u_perp = y - vTy / (1 + vTq) * q

    return u_perp

def calc_A_u_v(x):
    a = np.full(nz, np.cos(alpha) ** 2 / dz ** 2, dtype=complex) # Off diagonal elements
    b = -2 * a + omega ** 2 / vA0 ** 2                           # Diagonal elements
    b[0]  += b[1]
    b[-1] += a[0] ** 2 / b[1]

    ab = np.zeros((3, nz), dtype=complex)
    ab[0,:] = a
    ab[1,:] = b
    ab[2,:] = a

    u = np.zeros(nz, dtype=complex)
    v = np.zeros(nz, dtype=complex)
    u[0] = -b[1]
    u[-1] = a[0]
    v[0] = 1
    v[-1] = -a[0] / b[1]

    return [ab, u, v]

def vector_d(b_par):
    return -omega * (k_perp * b_par + \
            1j * np.sin(alpha) / (2 * dz) * (np.roll(b_par,-1) - np.roll(b_par,1)))

def dUdx(x, U):
    ux = U[0:nz]
    b_par = U[nz:]
    u_perp = calc_u_perp(x,b_par)
    # dux_dx = -(1j * omega * b_par_exact(x,z) + nabla_perp0 * u_perp_exact(x,z))
    dux_dx = -(1j * omega * b_par + 1j * k_perp * u_perp - \
                np.sin(alpha) / (2 * dz) * (np.roll(u_perp,-1) - np.roll(u_perp,1)))
    db_par_dx = -1j / omega * ( \
                np.cos(alpha) ** 2 / dz ** 2 * (np.roll(ux,-1) - 2 * ux  + np.roll(ux,1))  + \
                omega ** 2 / vA0 ** 2 * ux)
    return np.concatenate((dux_dx, db_par_dx))

Lz = 1
kx = 1
kz = np.pi / Lz

alpha = 0.25 * np.pi
k_perp = 0.5 * np.pi
vA0 = 1
omega = vA0 * np.sqrt(kx ** 2 + k_perp ** 2 + kz ** 2 - 2 * kz * k_perp * np.sin(alpha) + 0j)

nx = 256
x_min = 0
x_max = 6
dx = (x_max - x_min) / (nx - 1)
x = np.linspace(x_min, x_max, nx)

nz = 128
z_min = 0
z_max =  2 * Lz
dz = (z_max - z_min) / nz
z = np.linspace(z_min + dz / 2, z_max - dz / 2, nz)

X, Z = np.meshgrid(x, z)

nabla_perp0 = 1j * (k_perp - kz * np.sin(alpha))
nabla_par0  = 1j * kz * np.cos(alpha)
L0          = nabla_par0 ** 2 + omega ** 2 / vA0 ** 2

ux0     = 1
b_par0  = -L0 / (omega * kx) * ux0
u_perp0 = -1j * kx * nabla_perp0 / (L0 + nabla_perp0 ** 2) * ux0

ux_x_min    = ux_exact(x_min, z)
b_par_x_min = b_par_exact(x_min, z)
U_x_min = np.concatenate((ux_x_min, b_par_x_min))
sol = solve_ivp(dUdx, [x_min, x_max], U_x_min, t_eval=x)

ux = sol.y[0:nz]
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
surf = ax.plot_surface(X, Z, np.real(ux))
ax.set_xlabel('x')
ax.set_ylabel('z')
ax.set_title(r'Re($u_x$)')
plt.savefig('Figures/real_part_ux.png')

输出图: 在此处输入图像描述

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