Liouville type results for local minimizers of the micromagnetic energy

Abstract

We study local minimizers of the micromagnetic energy in small ferromagnetic 3d convex particles for which we justify the Stoner–Wohlfarth approximation: given a uniformly convex shape $\Omega \subset \varvec{\mathbf {R}}^3$, there exist $\delta _c$>0 and $C > 0$ such that for $0 < \delta \le \delta _c$ any local minimizer $\mathbf {m}$ of the micromagnetic energy in the particle $\delta \Omega$ satisfies $\Vert \nabla \mathbf {m} \Vert _{L^2} \leqslant C \delta ^2$. In the case of ellipsoidal particles we strengthen this result by proving that, for $\delta$ small enough, local minimizers are exactly spatially uniform. This last result extends W.F. Brown’s fundamental theorem for fine 3d ferromagnetic particles Brown (J Appl Phys 39:463–488, 1968), Di Fratta et al. (Physica B 407(9):1368–1371, 2011) which states the same result but only for global minimizers. As a by-product of the method that we use, we establish a new Liouville type result for locally minimizing $p$-harmonic maps with values into a closed subset of a Hilbert space. Namely, we establish that in a smooth uniformly convex domain of $\mathbf {R}^d$ any local minimizer of the $p$-Dirichlet energy ($p > 1$, $p \ne d$) is constant.

Appendix A (proof of Proposition 5)

Let $\Omega \subset \mathbf {R}^d$ be a bounded convex smooth open set with diameter $\delta > 0$ and assume that $0 \in \partial \Omega$. We consider a real valued function $f \in C^{\infty } ( \bar{\Omega } )$, (the result for $f \in H^1 (\Omega )$ is obtained by density of $C^{\infty } ( \bar{\Omega })$ in $H^1 (\Omega )$ and by continuity of the trace mapping $f \in H^1 (\Omega ) \mapsto f_{| \partial \Omega } \in L^2 (\partial \Omega )$). We have to estimate the quantity

$$\begin{aligned} I (f) :=\int \limits _{\Omega } \int \limits _{\partial \Omega } | f (x) - f (y) |^2 \left( \mathbf {n} (y) \cdot y \right) d\mathcal {H}^{d - 1} (y) \, dx. \end{aligned}$$

For $y \in \partial \Omega$, we define the following weighted mean value of $f$ along the segment $(0, 1) y$:

$$\begin{aligned} \langle f \rangle _y^{} : = \frac{d + 1}{2} \int \limits _0^1 r^{\frac{d - 1}{2}} f (ry) \, dr. \end{aligned}$$

We then decompose $f (y)$ as $\langle f \rangle _y^{} + [f (y) - \langle f \rangle _y^{}]$ to get $I (f) \le 2 (| \Omega | I_1 (f)_{} + I_2 (f))$, with

$$\begin{aligned} I_1 (f) :=\int \limits _{\partial \Omega } | f (y) - \langle f \rangle _y^{} |^2 \left( \mathbf {n} (y) \cdot y \right) d \mathcal {H}^{d - 1} (y),\end{aligned}$$

(72)

$$\begin{aligned} I_2 (f) :=\int \limits _{\Omega } \int \limits _{\partial \Omega } | f (x) - \langle f \rangle _y^{} |^2 \left( \mathbf {n} (y) \cdot y \right) d \mathcal {H}^{d - 1} (y) \, dx. \end{aligned}$$

(73)

We start by estimating $I_1 (f)$. Let us fix $y \in \partial \Omega$, we have,

$$\begin{aligned} f (y) - \langle f \rangle _y^{}&= \frac{(d + 1)}{2} \int \limits _0^1 r^{\frac{d - 1}{2}} (f (y) - f (ry)) d r\end{aligned}$$

(74)

$$\begin{aligned}&= \frac{(d + 1)}{2} \int \limits _0^1 r^{\frac{d - 1}{2}} (1 - r) \int \limits _0^1 y \cdot \nabla f ((r + (1 - r) s) y) \,ds \, dr \end{aligned}$$

(75)

Using the change of variable $s = (t - r) / (1 - r)$ and then Fubini, we obtain,

$$\begin{aligned} f (y) - \langle f \rangle _y^{} = \frac{(d + 1)}{2} \int \limits _0^1 r^{\frac{d - 1}{2}} \int \limits _r^1 y \cdot \nabla f (ty) \, dt \, dr = \int \limits _0^1 y \cdot \nabla f (ty) t^{\frac{d + 1}{2}} d t. \end{aligned}$$

Squaring and using the Jensen inequality, we get:

$$\begin{aligned} | f (y) - \langle f \rangle _y^{} |^2 \, \le \, | y |^2 \int \limits _0^1 | \nabla f (ty) |^2 t^{d + 1} d t \, \le \, \delta ^2 \int \limits _0^1 | \nabla f (ty) |^2 t^{d - 1} d t. \end{aligned}$$

Then, we multiply by $\left( y \cdot \mathbf {n} (y) \right)$ and integrate in $y \in \partial \Omega$. Using the change of variable $z = \psi (y, t) :=ty$, which maps $\partial \Omega \times (0, 1)$ onto $\Omega \setminus \{ 0 \}$, we get

$$\begin{aligned} I_1 (f) \, \le \, \delta ^2_{} \int \limits _{\Omega } | \nabla f |^2 (z) \frac{[t (z)]^{d - 1} \left( y (z) \cdot \mathbf {n} (y (z)) \right) }{J_{\psi } (\psi ^{- 1} (z))} \, dz \,\, = \,\, \delta ^2 \int \limits _{\Omega } | \nabla f |^2 (z) d z , \end{aligned}$$

(76)

with the notation, $\psi ^{- 1} (z) = : (y (z), t (z))$ and $J_{\psi } (y, t) = \sqrt{\det D \psi ^T \cdot D \psi } (y, t)$. Indeed, introducing the orthogonal decomposition $\mathbf {R}^d = T_y \partial \Omega \oplus \mathbf {R} \mathbf {n} (y) \simeq T_y \partial \Omega \oplus \mathbf {R}$, we compute the Jacobian matrix of $\psi$ in these spaces:

$$\begin{aligned} D \psi _{} (y, t) \, = \, \left( \begin{array}{cc} t \mathrm{Id }_{T_y \partial \Omega } &{} \left( y - \left( y \cdot \mathbf {n} (y) \right) \mathbf {n} (y) \right) \\ 0 &{} \left( y \cdot \mathbf {n} (y) \right) \end{array}\right) . \end{aligned}$$

The Jacobian determinant of $\psi$ is $J_{\psi } (y, s) = t^{d - 1} \left( y \cdot \mathbf {n} (y) \right)$.

Now we bound $I_2 (f)$. We first use the definition of $\langle f \rangle _y^{}$ and the Cauchy–Schwarz inequality to get for every $(x, y) \in \Omega \times \partial \Omega$:

$$\begin{aligned} | f (x) - \langle f \rangle _y^{} |^2 \, \le \, \frac{(d + 1)^2}{4} \int \limits _0^1 | f (x) - f (ry) |^2 r^{d - 1} d r \end{aligned}$$

Integrating in $y \in \partial \Omega$, and using the change of variable $z = \psi (y, r)$ as above, we obtain (after integration in $x \in \Omega$):

$$\begin{aligned} I_2 (f) \, \le \, \frac{(d + 1)^2}{4} \int \limits _{\Omega \times \Omega } | f (x) - f (z) |^2 \, dx\, dz \, \le \, (d + 1)^2 | \Omega |^{^{}} C_{P^{}}^2 \Vert \nabla f \Vert _{L^2}^2 . \end{aligned}$$

(77)

Inequality (5) follows from (76) and (77) with

$$\begin{aligned} \frac{C_P'}{\delta } = \sqrt{2 \left[ 1 + (d + 1)^2 \left( \frac{C_P}{\delta } \right) ^2 \right] } \, \le \, \sqrt{2} \left( 1 + \frac{(d + 1) C_P}{\delta } \right) . \end{aligned}$$

Since $C_P / \delta \le 1 / \pi$, we have $C_P' / \delta \le \sqrt{2} (1 + (d + 1) / \pi )$ as claimed.