# Static micromagnetics

```{contents}
:local:
```

As summed up in {cite}`abert2019`, *Static micromagnetics is the theory of stable magnetization configurations.*

The magnetization is found as minimizer of the following optimization problem:

$$

\min_{\mbf \in \Mcal} E(\mbf) \qquad \text{subject to} \qquad | \mbf (\xbf) | = 1 \quad \forall \xbf \in \Omega \qquad \text{and} \qquad | \mbf (\xbf) | = 1 \quad \forall \xbf \in \partial\Omega.
$$ (static-micromagnetics-optimization-problem)

As the minimizer $\mbf$ is necessarily a critical point of $E$, we derive the optimality conditions using the Lagrangian multiplier technique.

## Energy derivative

To proceed, we need to evaluate the derivative of the energy $E: \Mcal \to \Rbb$. As the energy is the sum of different components, we can write:

$$
dE(\mbf; \vbf) = dE^{\mathsf{zee}} (\mbf; \vbf) + dE^{\mathsf{ex}} (\mbf; \vbf) + dE^{\mathsf{dem}} (\mbf; \vbf) + dE^{\mathsf{ani}} (\mbf; \vbf).
$$

Hence, we can derive the directional derivatives for all components following {prf:ref}`def-differentiability-banach`.

### Zeeman energy

$$
dE^{\mathsf{zee}} (\mbf; \vbf)
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ - \mu_0 \int_{\Omega} \Ms \ \left( \mbf + \epsilon \vbf \right) \cdot \Hbf^{\mathsf{zee}} \dxbf + \mu_0 \int_{\Omega} \Ms \ \mbf \cdot \Hbf^{\mathsf{zee}} \dxbf \right] \\
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ \cancel{- \mu_0 \int_{\Omega} \Ms \ \mbf \cdot \Hbf^{\mathsf{zee}} \dxbf} - \epsilon \mu_0 \int_{\Omega} \Ms \ \vbf \cdot \Hbf^{\mathsf{zee}} \dxbf  + \cancel{\mu_0 \int_{\Omega} \Ms \ \mbf \cdot \Hbf^{\mathsf{zee}} \dxbf} \right] \\
&= - \mu_0 \int_{\Omega} \Ms \, \Hbf^{\mathsf{zee}} \cdot \vbf \, \dxbf
$$

### Exchange energy

$$
dE^{\mathsf{ex}} (\mbf; \vbf)
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ \int_{\Omega} A \left( \nabla \left( \mbf + \epsilon \vbf \right) \right)^2 \dxbf - \int_{\Omega} A \left( \nabla \mbf \right)^2 \dxbf \right] \\
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ \int_{\Omega} A \left( \nabla \mbf + \epsilon \nabla \vbf \right)^2 \dxbf - \int_{\Omega} A \left( \nabla \mbf \right)^2 \dxbf \right] \\
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ \int_{\Omega} A \left( \cancel{\left( \nabla \mbf \right)^2} + \epsilon^2 \left( \nabla \vbf \right)^2 + 2 \epsilon \sum_i \nabla m_i \cdot \nabla v_i \right) \dxbf - \cancel{\int_{\Omega} A \left( \nabla \mbf \right)^2 \dxbf} \right] \\
&= \lim_{\epsilon \to 0} \left[ \int_{\Omega} A \left( \epsilon \left( \nabla \vbf \right)^2 + 2 \sum_i \nabla m_i \cdot \nabla v_i \right) \dxbf \right] \\
&= 2 \sum_i \int_{\Omega} A \nabla m_i \cdot \nabla v_i \dxbf
$$

As we would like to unload all the derivatives on $\mbf$ rather than $\vbf$, we can go one step further. Integrating by parts and using the divergence theorem, we get

$$
\int_{\Omega} A \nabla m_i \cdot \nabla v_i \dxbf
&= \int_{\Omega} \nabla \cdot (v_i A \nabla m_i) \dxbf - \int_{\Omega} \nabla \cdot (A \nabla m_i) v_i \dxbf \\
&= \int_{\partial\Omega} v_i A \frac{\partial m_i}{\partial \bf{n}} \dsbf - \int_{\Omega} \nabla \cdot (A \nabla m_i) v_i \dxbf
$$

And, by summing for $i=1,2,3$ we finally get

$$
dE^{\mathsf{ex}} (\mbf; \vbf) = - 2 \int_{\Omega} \left[ \nabla \cdot (A \nabla \mbf) \right] \cdot \vbf \, \dxbf + 2 \int_{\partial\Omega} A \frac{\partial \mbf}{\partial \bf{n}} \cdot \vbf \, \dsbf
$$


### Demagnetization energy

$$
dE^{\mathsf{dem}} (\mbf; \vbf)
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ - \frac{\mu_0}{2} \int_{\Omega} \Ms \left( \mbf + \epsilon \vbf \right) \cdot \Hbf^{\mathsf{dem}} \left( \mbf + \epsilon \vbf \right) \dxbf + \frac{\mu_0}{2} \int_{\Omega} \Ms \mbf \cdot \Hbf^{\mathsf{dem}} (\mbf) \dxbf \right] \\
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ - \frac{\mu_0}{2} \int_{\Omega} \Ms \, \mbf \cdot \Hbf^{\mathsf{dem}} \left( \mbf + \epsilon \vbf \right) \dxbf - \epsilon \frac{\mu_0}{2} \int_{\Omega} \Ms \, \vbf \cdot \Hbf^{\mathsf{dem}} \left( \mbf + \epsilon \vbf \right) \dxbf + \frac{\mu_0}{2} \int_{\Omega} \Ms \, \mbf \cdot \Hbf^{\mathsf{dem}} (\mbf) \dxbf \right].
$$

As the demagnetization field $\Hbf^{\mathsf{dem}}$ is linear in the magnetization $\mbf$, we have

$$
\lim_{\epsilon \to 0} \Hbf^{\mathsf{dem}} (\mbf + \epsilon \vbf) = \Hbf^{\mathsf{dem}} (\mbf).
$$

Therefore,

$$
dE^{\mathsf{dem}} (\mbf; \vbf) = - \frac{\mu_0}{2} \int_{\Omega} \Ms \, \Hbf^{\mathsf{dem}} (\mbf) \cdot \vbf \, \dxbf .
$$


### Anisotropy energy

For the uniaxial anisotropy, let us first calculate

$$
\left( \left( \mbf + \epsilon \vbf \right) \cdot \ebf_{\mathsf{u}} \right)^2
&= \left(\mbf \cdot \ebf_{\mathsf{u}} + \epsilon \vbf \cdot \ebf_{\mathsf{u}} \right)^2 \\
&= \left( \mbf \cdot \ebf_{\mathsf{u}} \right)^2 + 2 \epsilon \left(\mbf \cdot \ebf_{\mathsf{u}} \right) \left( \vbf \cdot \ebf_{\mathsf{u}} \right) + \epsilon^2 \left( \vbf \cdot \ebf_{\mathsf{u}} \right)^2
$$

and

$$
\left( \left( \mbf + \epsilon \vbf \right) \cdot \ebf_{\mathsf{u}} \right)^4
&= \left[ \left( \left( \mbf + \epsilon \vbf \right) \cdot \ebf_{\mathsf{u}} \right)^2 \right]^2 \\
&= \left[ \left( \mbf \cdot \ebf_{\mathsf{u}} \right)^2 + 2 \epsilon \left(\mbf \cdot \ebf_{\mathsf{u}} \right) \left( \vbf \cdot \ebf_{\mathsf{u}} \right) + \epsilon^2 \left( \vbf \cdot \ebf_{\mathsf{u}} \right)^2 \right]^2 \\
&= \left( \mbf \cdot \ebf_{\mathsf{u}} \right)^4
+ 4 \epsilon \left(\mbf \cdot \ebf_{\mathsf{u}} \right)^3 \left( \vbf \cdot \ebf_{\mathsf{u}} \right)
+ 6 \epsilon^2 \left(\mbf \cdot \ebf_{\mathsf{u}} \right)^2 \left( \vbf \cdot \ebf_{\mathsf{u}} \right)^2
+ 4 \epsilon^3 \left(\mbf \cdot \ebf_{\mathsf{u}} \right) \left( \vbf \cdot \ebf_{\mathsf{u}} \right)^3
+ \epsilon^4 \left( \vbf \cdot \ebf_{\mathsf{u}} \right)^4
$$

Then,

$$
dE^{\mathsf{aniu}} (\mbf; \vbf)
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ - \int_{\Omega} \left[ K_{\mathsf{u}1} \left( \left( \mbf + \epsilon \vbf \right) \cdot \ebf_{\mathsf{u}} \right)^2 + K_{\mathsf{u}2} \left( \left( \mbf + \epsilon \vbf \right) \cdot \ebf_{\mathsf{u}} \right)^4 \right] \dxbf + \int_{\Omega} \left[ K_{\mathsf{u}1} \left( \mbf \cdot \ebf_{\mathsf{u}} \right)^2 + K_{\mathsf{u}2} \left( \mbf \cdot \ebf_{\mathsf{u}} \right)^4 \right] \dxbf \right] \\
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \int_{\Omega} \left[ \cancel{- K_{\mathsf{u}1} \left( \mbf \cdot \ebf_{\mathsf{u}} \right)^2} - \epsilon^2 K_{\mathsf{u}1} \left( \vbf \cdot \ebf_{\mathsf{u}} \right)^2 - 2 \epsilon K_{\mathsf{u}1} \left(\mbf \cdot \ebf_{\mathsf{u}} \right) \left( \vbf \cdot \ebf_{\mathsf{u}} \right)
- \cancel{K_{\mathsf{u}2} \left( \mbf \cdot \ebf_{\mathsf{u}} \right)^4}
- 4 \epsilon K_{\mathsf{u}2} \left(\mbf \cdot \ebf_{\mathsf{u}} \right)^3 \left( \vbf \cdot \ebf_{\mathsf{u}} \right)
- 6 \epsilon^2 K_{\mathsf{u}2} \left(\mbf \cdot \ebf_{\mathsf{u}} \right)^2 \left( \vbf \cdot \ebf_{\mathsf{u}} \right)^2
- 4 \epsilon^3 K_{\mathsf{u}2} \left(\mbf \cdot \ebf_{\mathsf{u}} \right) \left( \vbf \cdot \ebf_{\mathsf{u}} \right)^3
- \epsilon^4 K_{\mathsf{u}2}  \left( \vbf \cdot \ebf_{\mathsf{u}} \right)^4
+ \cancel{K_{\mathsf{u}1} \left( \mbf \cdot \ebf_{\mathsf{u}} \right)^2} + \cancel{K_{\mathsf{u}2} \left( \mbf \cdot \ebf_{\mathsf{u}} \right)^4} \right] \dxbf \\
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \int_{\Omega} \left[ - 2 \epsilon K_{\mathsf{u}1} \left(\mbf \cdot \ebf_{\mathsf{u}} \right) \left( \vbf \cdot \ebf_{\mathsf{u}} \right) - 4 \epsilon K_{\mathsf{u}2} \left(\mbf \cdot \ebf_{\mathsf{u}} \right)^3 \left( \vbf \cdot \ebf_{\mathsf{u}} \right) \right] \dxbf \\
&= - \int_{\Omega} \left[ 2 K_{\mathsf{u}1} \left(\mbf \cdot \ebf_{\mathsf{u}} \right) \, \ebf_{\mathsf{u}} + 4 K_{\mathsf{u}2} \left(\mbf \cdot \ebf_{\mathsf{u}} \right)^3 \, \ebf_{\mathsf{u}} \right] \cdot \vbf \, \dxbf
$$

For the cubic anisotropy, (ignoring all monomials where $\epsilon$ has a higher power than 1)

$$
dE^{\mathsf{anic}} (\mbf; \vbf)
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ \int_{\Omega} \left[ K_{\mathsf{c}1} \left( (m_1+\epsilon v_1)^2 (m_2+\epsilon v_2)^2 + (m_2+\epsilon v_2)^2 (m_3+\epsilon v_3)^2 + (m_3+\epsilon v_3)^2 (m_1+\epsilon v_1)^2 \right) + K_{\mathsf{c}2} (m_1+\epsilon v_1)^2 (m_2+\epsilon v_2)^2 (m_3+\epsilon v_3)^2 \right] \dxbf - \int_{\Omega} \left[ K_{\mathsf{c}1} \left( m_1^2 m_2^2 + m_2^2 m_3^2 + m_3^2 m_1^2 \right) + K_{\mathsf{c}2} m_1^2 m_2^2 m_3^2 \right] \dxbf \right] \\
%
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \int_{\Omega} \left[
K_{\mathsf{c}1} \left( (m_1^2 + 2 \epsilon m_1 v_1) (m_2^2 + 2 \epsilon m_2 v_2) + (m_2^2 + 2 \epsilon m_2 v_2) (m_3^2 + 2 \epsilon m_3 v_3) + (m_3^2 + 2 \epsilon m_3 v_3) (m_1^2 + 2 \epsilon m_1 v_1) \right)
+ K_{\mathsf{c}2} (m_1^2 + 2 \epsilon m_1 v_1) (m_2^2 + 2 \epsilon m_2 v_2) (m_3^2 + 2 \epsilon m_3 v_3)
- K_{\mathsf{c}1} \left( m_1^2 m_2^2 + m_2^2 m_3^2 + m_3^2 m_1^2 \right)
- K_{\mathsf{c}2} m_1^2 m_2^2 m_3^2
\right] \dxbf \\
%
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \int_{\Omega} \left[
K_{\mathsf{c}1} \left( m_1^2 m_2^2 + 2 \epsilon m_1^2 m_2 v_2 + 2 \epsilon m_1 m_2^2 v_1 + m_2^2 m_3^2 + 2 \epsilon m_2^2 m_3 v_3 + 2 \epsilon m_2 m_3^2 v_2 + m_1^2 m_3^2 + 2 \epsilon m_1 m_3^2 v_1 + 2 \epsilon m_1^2 m_3 v_3 \right)
+ K_{\mathsf{c}2} \left( \cancel{m_1^2 m_2^2 m_3^2} + 2 \epsilon m_1^2 m_2 m_3^2 v_2 + 2 \epsilon m_1 m_2^2 m_3^2 v_1 + 2 \epsilon m_1^2 m_2^2 m_3 v_3\right)
- K_{\mathsf{c}1} \left( m_1^2 m_2^2 + m_2^2 m_3^2 + m_3^2 m_1^2 \right)^2
- \cancel{K_{\mathsf{c}2} m_1^2 m_2^2 m_3^2}
\right] \dxbf \\
%
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \int_{\Omega} \left[
K_{\mathsf{c}1} \left( \cancel{m_1^2 m_2^2} + 2 \epsilon m_1^2 m_2 v_2 + 2 \epsilon m_1 m_2^2 v_1 + \cancel{m_2^2 m_3^2} + 2 \epsilon m_2^2 m_3 v_3 + 2 \epsilon m_2 m_3^2 v_2 + \cancel{m_1^2 m_3^2} + 2 \epsilon m_1 m_3^2 v_1 + 2 \epsilon m_1^2 m_3 v_3 \right)
+ K_{\mathsf{c}2} \left( \cancel{m_1^2 m_2^2 m_3^2} + 2 \epsilon m_1^2 m_2 m_3^2 v_2 + 2 \epsilon m_1 m_2^2 m_3^2 v_1 + 2 \epsilon m_1^2 m_2^2 m_3 v_3\right)
- K_{\mathsf{c}1} \left( \cancel{m_1^2 m_2^2} + \cancel{m_2^2 m_3^2} + \cancel{m_3^2 m_1^2} \right)^2
- \cancel{K_{\mathsf{c}2} m_1^2 m_2^2 m_3^2}
\right] \dxbf \\
%
&= 2 \int_{\Omega} \left[
K_{\mathsf{c}1} \left( m_1^2 m_2 v_2 + m_1 m_2^2 v_1 + m_2^2 m_3 v_3 + m_2 m_3^2 v_2 + m_1 m_3^2 v_1 + m_1^2 m_3 v_3 \right)
+ K_{\mathsf{c}2} \left( m_1^2 m_2 m_3^2 v_2 + m_1 m_2^2 m_3^2 v_1 + m_1^2 m_2^2 m_3 v_3 \right)
\right] \dxbf \\
%
&= 2 \int_{\Omega} \left[ K_{\mathsf{c}1}
\left( \begin{array}{c}
m_1 (m_2^2 + m_3^2) \\
m_2 (m_1^2 + m_3^2) \\
m_3 (m_1^2 + m_2^2)
\end{array} \right) \cdot \vbf
+ K_{\mathsf{c}2}
\left( \begin{array}{c}
m_1 (m_2 m_3)^2 \\
m_2 (m_1 m_3)^2 \\
m_3 (m_1 m_2)^2
\end{array} \right) \cdot \vbf
\right] \dxbf
$$


### Gateaux derivative of the energy

Following the definition from {prf:ref}`def-differentiability-banach`, we define the Gateaux derivative as an operator $E' : \Lcal (\Mcal, \Rbb)$:

$$
\langle E'(\mbf), \vbf \rangle_{\Mcal, \Rbb} := dE (\mbf; \vbf).
$$

Taking as example an energy composed of demagnetization and exchange, we get:

$$
\langle E'(\mbf), \vbf \rangle_{\Mcal, \Rbb}
&= dE^{\mathsf{ex}} (\mbf; \vbf) + dE^{\mathsf{dem}} (\mbf; \vbf) \\
&= - 2 \int_{\Omega} \left[ \nabla \cdot (A \nabla \mbf) \right] \cdot \vbf \, \dxbf
+ 2 \int_{\partial\Omega} A \frac{\partial \mbf}{\partial \bf{n}} \cdot \vbf \, \dsbf
- \frac{\mu_0}{2} \int_{\Omega} \Ms \, \Hbf^{\mathsf{dem}} (\mbf) \cdot \vbf \, \dxbf \\
&= - \int_{\Omega} \left[ 2 \nabla \cdot (A \nabla \mbf) + \frac{\mu_0}{2} \Ms \, \Hbf^{\mathsf{dem}} (\mbf) \right] \cdot \vbf \, \dxbf
+ 2 \int_{\partial\Omega} A \frac{\partial \mbf}{\partial \bf{n}} \cdot \vbf \, \dsbf
$$

In general, we can split this derivative as an operator $E'_\Omega$ acting on the internal domain and an operator $E'_{\partial\Omega}$ acting on the boundary. Depending on the type of energy considered, the operator $E'_{\partial\Omega}$ might be null.

In this case, we get

$$
\langle E'_\Omega (\mbf), \vbf \rangle_{\Mcal, \Rbb} = - \int_{\Omega} \left[ 2 \nabla \cdot (A \nabla \mbf) + \frac{\mu_0}{2} \Ms \, \Hbf^{\mathsf{dem}} (\mbf) \right] \cdot \vbf \, \dxbf
$$

and

$$
\langle E'_{\partial\Omega} (\mbf), \vbf \rangle_{\Mcal, \Rbb} = 2 \int_{\partial\Omega} A \frac{\partial \mbf}{\partial \bf{n}} \cdot \vbf \, \dsbf.
$$

By the {prf:ref}`Riesz representation theorem <thm-riesz-representation>`, we can identify the operators in $\Lcal(\Mcal, \Rbb)$ with magnetizations in $\Mcal$:

$$
E'_\Omega (\mbf) = - \left[ 2 \nabla \cdot (A \nabla \mbf) + \frac{\mu_0}{2} \Ms \, \Hbf^{\mathsf{dem}} (\mbf) \right]
$$

and

$$
E'_{\partial\Omega} (\mbf) = 2 A \frac{\partial \mbf}{\partial \bf{n}}
$$


## Discretize-then-optimize approach

In this approach, first we discretize the optimization problem {eq}`static-micromagnetics-optimization-problem` in $n$ variables using the finite element approximation and the optimize the problem in $\Rbb^n$.

### Finite element discretization

Given a grid $\Tcal$ of points $\{ \xbf_i \}_{i=1}^n$ and a finite element basis $\{ \varphi_i \}_{i=1}^n$ such that

$$
\varphi_i(\xbf_j) = \delta_{i,j}
$$





### Lagrangian approach

We proceed with the Lagrangian multiplier technique. Let

$$
\mathcal{L} (\mbf, \lambda, \mu) := E(\mbf) + \int_\Omega \lambda \left( |\mbf|^2 - 1  \right) \dxbf + \int_{\partial \Omega} \mu \left( |\mbf|^2 - 1  \right) \dsbf.
$$

with $\lambda: \Omega \to \Rbb$ and $\mu: \partial\Omega \to \Rbb$ multipliers to enforce space contraints.










## Optimize-then-discretize approach

````{important}
Work in progress. Not sure this will work.
````


If the constraints were convex in $\mbf$, we could follow Section 1.7.1 in {cite}`hinze+2008`, because $\Mcal$ is a Banach space and $dE: \Mcal \to \Rbb$ is Gateaux differentiable.