Stray field computation by inverted finite elements: a new method in micromagnetic simulations

Abstract

In this paper, we propose a new method for computing the stray-field and the corresponding energy for a given magnetization configuration. Our approach is based on the use of inverted finite elements and does not need any truncation. After analyzing the problem in an appropriate functional framework, we describe the method and we prove its convergence. We then display some computational results which demonstrate its efficiency and confirm its full potential.

Appendix A: Solving the problem in the case of an inhomogeneously magnetized ball (numerical example 2)

The resolution of the system in the case of a ball $B_{r_0}$ and ${\varvec{M}}$ given by (57) can be done by means of a decomposition on spherical harmonics $(Y_\ell ^m)_{\ell \geqslant 0, -\ell \leqslant m \leqslant \ell }$ (which is orthonormal with respect to the inner product in $L^2({\mathbb S}^2)$). We have outside the ball $B_{r_0}$:

$$ \Delta u = 0\,\, \text{ in }\,\, {\mathbb R}^3 \backslash \overline{B}_{r_0}. $$

Developing u on the basis of spherical harmonics gives (see, e.g., [35]):

$$ u( {\varvec{x}}) = \sum _{\ell = 0}^{+\infty } \sum _{m=-\ell }^\ell A_\ell ^m \left( \frac{r}{{r_0}}\right) ^{-\ell -1} Y_\ell ^m( \varphi , \theta ). $$

where $(A_\ell ^m)_{\ell \geqslant 0, -\ell \leqslant m \leqslant \ell }$ is a sequence of complex coefficients. On the other hand, we have in the interior of the ball

$$ \Delta u = \textrm{div}\, {\varvec{M}}= 2 \frac{\cos \theta }{r} = \frac{\alpha }{r} Y_1^0(\varphi , \theta ) \text{ in } \overline{B}_{r_0}, \; \; \text{ with } \alpha = 4 \sqrt{\frac{\pi }{3}}. $$

Fig. 8

(Example 3) The computed energy of an homogeneously magnetized unit cube

Writing

$$ u ( {\varvec{x}}) = \sum _{\ell = 0}^{+\infty } \sum _{m=-\ell }^\ell u_\ell ^m(r) Y_\ell ^m ( \varphi , \theta ), \text{ in } \overline{B}_{r_0}, $$

gives:

$$ \frac{1}{r^2} \frac{d}{dr } (r^2 \frac{d u_\ell ^m }{dr }) (r) - \frac{\ell (\ell + 1) }{r^2} u_\ell ^m(r) = \frac{\alpha }{r} \delta _{\ell , 1} \delta _{m, 0} \text{ for } \text{ all } \ell \geqslant 0 \text{ and } -\ell \leqslant m \leqslant \ell , $$

where $\delta _{i, j}$, $i \in {\mathbb N}$, $j \in {\mathbb N}$, denotes the usual Kronecker delta. The solutions of these equations are of the form

$$ u_\ell ^m (r) = B_\ell ^m \left( \frac{r}{{r_0}}\right) ^{\ell } + C_\ell ^m \left( \frac{r}{{r_0}}\right) ^{-\ell -1} + \frac{\alpha }{3} r \ln ( \frac{r}{{r_0}}) \delta _{\ell , 1} \delta _{m, 0}, $$

where $B_\ell ^m$ and $C_\ell ^m$ are constants. Since $u \in W^1_0({\mathbb R}^3)$, we deduce that $u_{B_{r_0}} \in H^1(B_{r_0})$. Necessarily $C_\ell ^m = 0$ for all $\ell \geqslant 0$ and $|m| \leqslant \ell$. Since $[u]_{\partial \Omega } = 0$, we deduce that $B_\ell ^m = A_\ell ^m$ for all $\ell \geqslant 0$ and $|m| \leqslant \ell$. In addition,

$$ \left[ \frac{\partial u}{\partial r} \right] _{\partial \Omega } = - {\varvec{M}}\cdot {\varvec{n}}= - {\varvec{M}}\cdot {\varvec{e}}_r = 0. $$

Thus, for all $\ell \geqslant 0$ and $|m| \leqslant \ell$, we have

$$ \ell \frac{B^m_\ell }{{r_0}} + \frac{ \alpha }{3} \delta _{\ell , 1} \delta _{m, 0} = -\frac{ \ell + 1 }{{r_0}} A^m_\ell . $$

Thus, $A_l^m = B_l^m=0$ for $(\ell , m) \ne (1, 0)$ and

$$ A_1^0 = B_1^0 = - \frac{\alpha {r_0}}{9}. $$

Thus, if $| {\varvec{x}}| \geqslant {r_0}$ then

$$ u( {\varvec{x}}) = A_1^0 \left( \frac{{r_0}}{r}\right) ^{2} Y_1^0 = - \frac{{r_0}^3}{9 r^2} \alpha Y_1^0 = - \frac{2 {r_0}^3 \cos \theta }{9 r^2} = - \frac{2 {r_0}^3 z }{9 r^3}. $$

and

$$ \Vert \nabla u\Vert ^2_{L^2({\mathbb R}^3 \backslash \overline{B}_{r_0})} = \frac{32 \pi }{243}{r_0}^3. $$

If $| {\varvec{x}}| \leqslant {r_0}$ then

$$ u ( {\varvec{x}}) = (B_1^0 \frac{r}{{r_0}} +\frac{\alpha }{3} r \ln ( \frac{r}{{r_0}})) Y_1^0 = ( - \frac{r}{9} + \frac{r}{3} \ln ( \frac{r}{{r_0}})) \alpha Y_1^0= \frac{2 r \cos \theta }{9} ( - 1 + 3 \ln ( \frac{r}{{r_0}})). $$

Thus,

$$ u = \frac{2 z }{9} ( - 1 + 3 \ln ( \frac{r}{{r_0}})), $$

and

$$ \Vert \nabla u\Vert ^2_{L^2( \overline{B}_{r_0})} =\frac{64 \pi }{243}{r_0}^3. $$

Thus, the energy of the corresponding stray field is

$$ {\mathscr {E}}_{sf} (u) = \frac{1}{2} \int _{{r_0}^3} |\nabla u|^2 dx = \frac{16}{81} \pi {r_0}^3. $$