Physical test scenario 4 - 1D bloch type 180° domain wall problem

[AdamsMP92]

Motivation

A one-dimensional 180-degree domain wall provides an analytical reference solution for testing exchange, uniaxial anisotropy, boundary constraints, and energy minimisation.

For a sufficiently long sample, the equilibrium domain-wall profile is known analytically. The numerical equilibrium state can therefore be compared directly with the expected magnetisation profile and domain-wall width.

Physical model

Consider a one-dimensional magnetic system along the (x)-axis with exchange energy density

$$w_{\mathrm{ex}} = A \left| \frac{d\mathbf{m}}{dx} \right|^2$$

and uniaxial anisotropy along the (z)-axis,

$$w_{\mathrm{a}} = K \left(1-m_z^2\right).$$

The magnetisation is constrained to form a 180-degree wall between the two domains,

$$\mathbf{m}(0)=\mathbf{e}_z, \qquad \mathbf{m}(L)=-\mathbf{e}_z.$$

Assuming rotation in the (yz)-plane,

$$\mathbf{m}(x) = \begin{bmatrix} 0\\\ \sin\theta(x)\\\ \cos\theta(x) \end{bmatrix},$$

the equilibrium equation is

$$2A\frac{d^2\theta}{dx^2} = K\sin(2\theta).$$

For an isolated wall centred at (x_0), the analytical solution is

$$m_x(x)=0,$$ $$m_y(x) = \mathrm{sech} \left( \frac{x-x_0}{\Delta} \right),$$

and

$$m_z(x) = -\tanh \left( \frac{x-x_0}{\Delta} \right),$$

where the domain-wall width parameter is

$$\Delta = \sqrt{\frac{A}{K}}.$$

Proposed test setup

A one-dimensional mesh can be used with exchange and uniaxial anisotropy only.

import numpy as np
import discretisedfield as df
import micromagneticmodel as mm
import oommfc as oc

Ms = 1.4e6
A = 30e-12
K = 520e3

length = 100e-9
cell = 1e-9
x0 = length / 2

region = df.Region(
    p1=(0, 0, 0),
    p2=(length, cell, cell),
)

mesh = df.Mesh(
    region=region,
    cell=(cell, cell, cell),
)

system = mm.System(name="domain_wall_profile")

system.energy = (
    mm.Exchange(A=A)
    + mm.UniaxialAnisotropy(K=K, u=(0, 0, 1))
)

system.m = df.Field(
    mesh,
    nvdim=3,
    value=initial_m,
    norm=Ms,
)

The initial state should contain a continuous transition between (+\mathbf{e}_z) and (-\mathbf{e}_z). The end cells should be fixed to preserve the 180-degree wall during relaxation.

For example,

def initial_m(pos):
    x = pos[0]

    mz = 1.0 - 2.0 * x / length
    my = np.sqrt(max(0.0, 1.0 - mz**2))

    return (0.0, my, mz)

The system can then be relaxed using an energy-minimisation driver or strongly damped time evolution.

Analytical reference implementation

import numpy as np


def analytical_domain_wall(x, x0, A, K):
    """Return the analytical 180-degree domain-wall profile."""
    x = np.asarray(x, dtype=float)

    delta = np.sqrt(A / K)
    u = (x - x0) / delta

    mx = np.zeros_like(x)
    my = 1.0 / np.cosh(u)
    mz = -np.tanh(u)

    return np.column_stack((mx, my, mz))

For the proposed parameters,

delta = np.sqrt(A / K)

which gives approximately

$$\Delta \approx 7.60\,\mathrm{nm}.$$

Primary assertion

After relaxation, the normalised numerical magnetisation should be compared with the analytical profile:

m_exact = analytical_domain_wall(
    x=x_coordinates,
    x0=x0,
    A=A,
    K=K,
)

np.testing.assert_allclose(
    m_simulated,
    m_exact,
    rtol=...,
    atol=...,
)

The comparison may exclude a small number of cells at both ends because the analytical expression describes an isolated wall on an infinite domain, whereas the numerical model uses a finite interval with pinned boundary values.

A component-wise error can also be used:

error_my = np.max(
    np.abs(m_simulated[:, 1] - m_exact[:, 1])
)

error_mz = np.max(
    np.abs(m_simulated[:, 2] - m_exact[:, 2])
)

Additional checks

The following properties can be tested independently:

$$m_x(x)\approx0,$$ $$m_y(x_0)\approx1, \qquad m_z(x_0)\approx0,$$ $$m_z(x_0-\xi) \approx -m_z(x_0+\xi),$$

and

$$m_y(x_0-\xi) \approx m_y(x_0+\xi).$$

The wall width can also be extracted from the slope at its centre:

$$\left. \frac{dm_z}{dx} \right|_{x=x_0} = -\frac{1}{\Delta}.$$

Therefore,

$$\Delta_{\mathrm{sim}} = - \left[ \left. \frac{dm_z}{dx} \right|_{x=x_0} \right]^{-1}$$

can be compared with

$$\Delta = \sqrt{\frac{A}{K}}.$$

Suggested parameterisation

The test can be parametrised over several exchange and anisotropy values while keeping the mesh resolution sufficiently smaller than the wall width:

$$\Delta x \ll \sqrt{\frac{A}{K}}.$$

This checks whether the simulated wall width follows the expected scaling,

$$\Delta\propto\sqrt{A}, \qquad \Delta\propto K^{-1/2}.$$

Expected value

This test would validate:

• the exchange energy term;
• uniaxial anisotropy;
• the relative scaling of exchange and anisotropy;
• fixed boundary regions;
• the relaxation or minimisation driver;
• the spatial discretisation of magnetisation gradients;
• and recovery of an analytical equilibrium profile.

Unlike tests that only check whether relaxation converges, this scenario compares the complete spatial magnetisation profile with a known analytical solution.