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 Definition 1.10.
\[\begin{split}
dE^{\mathsf{zee}} (\mbf; \vbf)
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ - \mu_0 \int_{\Omega} \Ms \ \left( \mbf + \epsilon \vbf \right) \cdot \Hbf^{\mathsf{zee}} \dxbf + \mu_0 \int_{\Omega} \Ms \ \mbf \cdot \Hbf^{\mathsf{zee}} \dxbf \right] \\
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ \cancel{- \mu_0 \int_{\Omega} \Ms \ \mbf \cdot \Hbf^{\mathsf{zee}} \dxbf} - \epsilon \mu_0 \int_{\Omega} \Ms \ \vbf \cdot \Hbf^{\mathsf{zee}} \dxbf + \cancel{\mu_0 \int_{\Omega} \Ms \ \mbf \cdot \Hbf^{\mathsf{zee}} \dxbf} \right] \\
&= - \mu_0 \int_{\Omega} \Ms \, \Hbf^{\mathsf{zee}} \cdot \vbf \, \dxbf
\end{split}\]
\[\begin{split}
dE^{\mathsf{ex}} (\mbf; \vbf)
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ \int_{\Omega} A \left( \nabla \left( \mbf + \epsilon \vbf \right) \right)^2 \dxbf - \int_{\Omega} A \left( \nabla \mbf \right)^2 \dxbf \right] \\
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ \int_{\Omega} A \left( \nabla \mbf + \epsilon \nabla \vbf \right)^2 \dxbf - \int_{\Omega} A \left( \nabla \mbf \right)^2 \dxbf \right] \\
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ \int_{\Omega} A \left( \cancel{\left( \nabla \mbf \right)^2} + \epsilon^2 \left( \nabla \vbf \right)^2 + 2 \epsilon \sum_i \nabla m_i \cdot \nabla v_i \right) \dxbf - \cancel{\int_{\Omega} A \left( \nabla \mbf \right)^2 \dxbf} \right] \\
&= \lim_{\epsilon \to 0} \left[ \int_{\Omega} A \left( \epsilon \left( \nabla \vbf \right)^2 + 2 \sum_i \nabla m_i \cdot \nabla v_i \right) \dxbf \right] \\
&= 2 \sum_i \int_{\Omega} A \nabla m_i \cdot \nabla v_i \dxbf
\end{split}\]
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
\[\begin{split}
\int_{\Omega} A \nabla m_i \cdot \nabla v_i \dxbf
&= \int_{\Omega} \nabla \cdot (v_i A \nabla m_i) \dxbf - \int_{\Omega} \nabla \cdot (A \nabla m_i) v_i \dxbf \\
&= \int_{\partial\Omega} v_i A \frac{\partial m_i}{\partial \bf{n}} \dsbf - \int_{\Omega} \nabla \cdot (A \nabla m_i) v_i \dxbf
\end{split}\]
And, by summing for \(i=1,2,3\) we finally get
\[
dE^{\mathsf{ex}} (\mbf; \vbf) = - 2 \int_{\Omega} \left[ \nabla \cdot (A \nabla \mbf) \right] \cdot \vbf \, \dxbf + 2 \int_{\partial\Omega} A \frac{\partial \mbf}{\partial \bf{n}} \cdot \vbf \, \dsbf
\]
\[\begin{split}
dE^{\mathsf{dem}} (\mbf; \vbf)
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ - \frac{\mu_0}{2} \int_{\Omega} \Ms \left( \mbf + \epsilon \vbf \right) \cdot \Hbf^{\mathsf{dem}} \left( \mbf + \epsilon \vbf \right) \dxbf + \frac{\mu_0}{2} \int_{\Omega} \Ms \mbf \cdot \Hbf^{\mathsf{dem}} (\mbf) \dxbf \right] \\
&= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ - \frac{\mu_0}{2} \int_{\Omega} \Ms \, \mbf \cdot \Hbf^{\mathsf{dem}} \left( \mbf + \epsilon \vbf \right) \dxbf - \epsilon \frac{\mu_0}{2} \int_{\Omega} \Ms \, \vbf \cdot \Hbf^{\mathsf{dem}} \left( \mbf + \epsilon \vbf \right) \dxbf + \frac{\mu_0}{2} \int_{\Omega} \Ms \, \mbf \cdot \Hbf^{\mathsf{dem}} (\mbf) \dxbf \right].
\end{split}\]
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,
\[
dE^{\mathsf{dem}} (\mbf; \vbf) = - \frac{\mu_0}{2} \int_{\Omega} \Ms \, \Hbf^{\mathsf{dem}} (\mbf) \cdot \vbf \, \dxbf .
\]
For the uniaxial anisotropy, let us first calculate
\[\begin{split}
\left( \left( \mbf + \epsilon \vbf \right) \cdot \ebf_{\mathsf{u}} \right)^2
&= \left(\mbf \cdot \ebf_{\mathsf{u}} + \epsilon \vbf \cdot \ebf_{\mathsf{u}} \right)^2 \\
&= \left( \mbf \cdot \ebf_{\mathsf{u}} \right)^2 + 2 \epsilon \left(\mbf \cdot \ebf_{\mathsf{u}} \right) \left( \vbf \cdot \ebf_{\mathsf{u}} \right) + \epsilon^2 \left( \vbf \cdot \ebf_{\mathsf{u}} \right)^2
\end{split}\]
and
\[\begin{split}
\left( \left( \mbf + \epsilon \vbf \right) \cdot \ebf_{\mathsf{u}} \right)^4
&= \left[ \left( \left( \mbf + \epsilon \vbf \right) \cdot \ebf_{\mathsf{u}} \right)^2 \right]^2 \\
&= \left[ \left( \mbf \cdot \ebf_{\mathsf{u}} \right)^2 + 2 \epsilon \left(\mbf \cdot \ebf_{\mathsf{u}} \right) \left( \vbf \cdot \ebf_{\mathsf{u}} \right) + \epsilon^2 \left( \vbf \cdot \ebf_{\mathsf{u}} \right)^2 \right]^2 \\
&= \left( \mbf \cdot \ebf_{\mathsf{u}} \right)^4
+ 4 \epsilon \left(\mbf \cdot \ebf_{\mathsf{u}} \right)^3 \left( \vbf \cdot \ebf_{\mathsf{u}} \right)
+ 6 \epsilon^2 \left(\mbf \cdot \ebf_{\mathsf{u}} \right)^2 \left( \vbf \cdot \ebf_{\mathsf{u}} \right)^2
+ 4 \epsilon^3 \left(\mbf \cdot \ebf_{\mathsf{u}} \right) \left( \vbf \cdot \ebf_{\mathsf{u}} \right)^3
+ \epsilon^4 \left( \vbf \cdot \ebf_{\mathsf{u}} \right)^4
\end{split}\]
Then,
\[\begin{split}
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
\end{split}\]
For the cubic anisotropy, (ignoring all monomials where \(\epsilon\) has a higher power than 1)
\[\begin{split}
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
\end{split}\]
Following the definition from Definition 1.10, 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:
\[\begin{split}
\langle E'(\mbf), \vbf \rangle_{\Mcal, \Rbb}
&= dE^{\mathsf{ex}} (\mbf; \vbf) + dE^{\mathsf{dem}} (\mbf; \vbf) \\
&= - 2 \int_{\Omega} \left[ \nabla \cdot (A \nabla \mbf) \right] \cdot \vbf \, \dxbf
+ 2 \int_{\partial\Omega} A \frac{\partial \mbf}{\partial \bf{n}} \cdot \vbf \, \dsbf
- \frac{\mu_0}{2} \int_{\Omega} \Ms \, \Hbf^{\mathsf{dem}} (\mbf) \cdot \vbf \, \dxbf \\
&= - \int_{\Omega} \left[ 2 \nabla \cdot (A \nabla \mbf) + \frac{\mu_0}{2} \Ms \, \Hbf^{\mathsf{dem}} (\mbf) \right] \cdot \vbf \, \dxbf
+ 2 \int_{\partial\Omega} A \frac{\partial \mbf}{\partial \bf{n}} \cdot \vbf \, \dsbf
\end{split}\]
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 Riesz representation theorem, 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
\[
E'_{\partial\Omega} (\mbf) = 2 A \frac{\partial \mbf}{\partial \bf{n}}
\]