where the operator $L$ is given by

$$L = \begin{pmatrix} \alpha L_1 & -L_2 \\ L_1 & \alpha L_2 \end{pmatrix}$$

This operator is not self-adjoint. Yet, it is a relatively compact perturbation of the operators

$$\begin{pmatrix} \alpha L_1 & -L_1 \\ L_1 & \alpha L_1 \end{pmatrix} = L_1 M \quad \text{and} \quad \begin{pmatrix} \alpha L_2 & -L_2 \\ L_2 & \alpha L_2 \end{pmatrix} = L_2 M,$$

where $M$ is a matrix with constant coefficients

$$M := \begin{pmatrix} \alpha & -1 \\ 1 & \alpha \end{pmatrix}.$$

This matrix has two complex eigenvalues:

• $\alpha + i$, with related eigenvector $(1, i)$,

• $\alpha - i$, with related eigenvector $(1, -i)$.

We know that the potential in both $L_1$ and $L_2$ has a limit at $\pm\infty$, which is $1 + h_0$. Thus, the essential spectrum of $L_1$ and $L_2$ is given by $\sigma_{ess}(L_1) = \sigma_{ess}(L_2) = [1 + h_0, \infty)$, and we obtain straightforwardly the essential spectrum of $L_1 M$ and $L_2 M$, which in turn is also the essential spectrum of $L$: $(\alpha + i)\cdot[1 + h_0, \infty) \cup (\alpha - i)\cdot[1 + h_0, \infty)$. A similar analysis leads to the same result for the adjoint of this operator.

In particular, this essential spectrum is completely embedded in $\{z \in \mathbb{C} | \text{Re } z > 0\}$. Thus, in order to study the linear stability, one should study the remaining eigenvalues of $L$, or rather of $L^\top$, and in particular the eigenvalues embedded in $\{z \in \mathbb{C} | \text{Re } z \le 0\}$. The kernel of $L^\top$ can be easily computed thanks to the kernel of $L_1$ and $L_2$ given in Proposition 3.1:

$$\ker L^\top = \mathbb{C} \begin{pmatrix} \alpha \\ 1 \end{pmatrix} \partial_x \theta_{h_0} + \mathbb{C} \begin{pmatrix} -1 \\ \alpha \end{pmatrix} \sin \theta_{h_0}.$$

These two functions are related to the invariance by translation and rotation around $e_1$, as already explained previously, and can therefore be put aside by some modulation. In view of the results found above on the dynamics in the nonlinear setting, one would expect at least another eigenvalue with negative real part, and more probably two ($L$ is real, so any eigenvalue $\lambda \in \mathbb{C} \setminus \mathbb{R}$ would give rise to a second eigenvalue $\bar{\lambda}$). However, since the operator $L$ is not self-adjoint, this analysis is much more difficult and it is still an open problem to the best of the author's knowledge. A positive answer to this question would still give a more precise hint of the linear and nonlinear dynamics, as it would roughly say that most solutions to (LLG) close to a stationary solution would evolve into a different state, except maybe on a manifold of dimension 2. This would be better than the result in Theorem 1.10, which states this behaviour only on a subset in form of a cone near the stationary solution.

## ACKNOWLEDGEMENTS

The author acknowledges partial support by the ANR project MOSICOF ANR-21-CE40-0004. The author expresses gratitude to Raphaël Côte for pointing out the existence of these structures and for engaging in several insightful discussions. While the subject looked poor on first sight, it actually revealed several very interesting behaviors of the dynamics of the Landau-Lifshitz-Gilbert equation on a nanowire, completing our previous study. The author extends appreciation to Gilles Carbou for generously sharing his code, which served as the foundation of the code of the present article. Special thanks are also extended to Quentin Chauleur for his meticulous review of the initial version of this article.

## A. RAYLEIGH QUOTIENTS AND EIGENFUNCTIONS

In this appendix, we recall a particular case of the Rayleigh quotients, where the supremum over linear subspace of codimension fixed can be explicitly thanks to the eigenfunctions of the previous eigenvalues.
A. RAYLEIGH QUOTIENTS AND EIGENFUNCTIONS

In this appendix, we recall a particular case of the Rayleigh quotients, where the supremum over linear subspace of codimension fixed can be explicit thanks to the eigenfunctions of the previous eigenvalues.

**Lemma A.1.** Let $\lambda_1 < \lambda_2$ be the first two eigenvalues of a self-adjoint bounded-from-below operator $(\mathrm{Dom}(L), L)$ on an Hilbert space $H$, associated with the quadratic form $Q(u) := \langle Lu, u \rangle_H$. Assume that these eigenvalues are simple with respecting normalized eigenvectors $\psi_1, \psi_2 \in \mathrm{Dom}(L) \subset \mathrm{Dom}(Q)$ and that the essential spectrum is included in $[\lambda_\infty, \infty)$ where $\lambda_\infty > \lambda_2$. Then

$$ \inf_{u \in \mathrm{span}(\psi_1, \psi_2)^\perp \cap \mathrm{Dom}(Q) \setminus \{0\}} \frac{\langle Lu, u \rangle_H}{\|u\|_H^2} =: a = \mu_3(L) := \sup_{\phi_1, \phi_2 \in \mathrm{span}(\phi_1, \phi_2)^\perp \cap \mathrm{Dom}(Q) \setminus \{0\}} \inf \frac{\langle Lu, u \rangle_H}{\|u\|_H^2}, $$

where $\mu_3(L)$ is the third Rayleigh quotient of $L$.

*Proof.* By definition of $\mu_3(L)$, we obviously have $a \le \mu_3(L)$. By contradiction, assume that $a < \mu_3(L)$. By definition of $a$, there exists a non-trivial vector $\psi_3 \in \mathrm{Dom}(Q) \setminus \{0\}$ such that $\langle \psi_3, \psi_1 \rangle_H = \langle \psi_3, \psi_2 \rangle_H = 0$ and $Q(\psi_3) < \lambda \|\psi_3\|_H^2$ for any $\lambda \in (a, \mu_3(L))$ to be fixed later. We can assume that $\|\psi_3\|_H = 1$. Let $\mu_2(L)$ be the second Rayleigh quotient

$$ \mu_2(L) := \sup_{\phi_1} \inf_{u \in \mathrm{span}(\phi_1)^\perp \cap \mathrm{Dom}(Q) \setminus \{0\}} \frac{\langle Lu, u \rangle_H}{\|u\|_H^2}. $$

It is known that $\mu_2(L) = \lambda_2$ (see [3]). Moreover, since $\lambda_2$ is simple and since the bottom of the essential spectrum is larger than $\lambda_\infty > \lambda_2$, we also know that $\mu_2(L) < \mu_3(L)$. Then, we can assume furthermore $\lambda \ge \mu_2(L) = \lambda_2$.

Now, let $\phi_1, \phi_2 \in H$. We know that $\mathrm{span}(\psi_1, \psi_2, \psi_3)$ is a vector space with dimension 3 which is included in $\mathrm{Dom}(Q) \subset H$, whereas $\mathrm{span}(\phi_1, \phi_2)^\perp$ is a vector space with codimension (at least) 2 in $H$. Therefore, there exists a non-trivial vector $u \in \mathrm{span}(\psi_1, \psi_2, \psi_3) \cap \mathrm{span}(\phi_1, \phi_2)^\perp$, and this $u$ is obviously in $\mathrm{Dom}(Q)$. We can decompose $u$ in the basis $(\psi_1, \psi_2, \psi_3)$:

$$ u = \langle u, \psi_1 \rangle_H \psi_1 + \langle u, \psi_2 \rangle_H \psi_2 + \langle u, \psi_3 \rangle_H \psi_3. $$

In particular, we can compute $Q(u)$ with this expansion. Indeed, since $\psi_1$ and $\psi_2$ are eigenvectors, and as $(\psi_1, \psi_2, \psi_3)$ are orthogonal to each other in $H$, all the cross terms vanish:

$$ \begin{aligned} Q(u) &= \langle u, \psi_1 \rangle_H^2 Q(\psi_1) + \langle u, \psi_2 \rangle_H^2 Q(\psi_2) + \langle u, \psi_3 \rangle_H^2 Q(\psi_3) \\ &= \underbrace{\lambda_1 \langle u, \psi_1 \rangle_H^2}_{< \lambda_2 \le \lambda} + \underbrace{\lambda_2 \langle u, \psi_2 \rangle_H^2}_{\le \lambda} + \underbrace{\langle u, \psi_3 \rangle_H^2 Q(\psi_3)}_{< \lambda} \\ &\le \lambda (\langle u, \psi_1 \rangle_H^2 + \langle u, \psi_2 \rangle_H^2 + \langle u, \psi_3 \rangle_H^2) = \lambda \|u\|_H^2. \end{aligned} $$

Therefore, $\inf_{u \in \mathrm{span}(\phi_1, \phi_2)^\perp \cap \mathrm{Dom}(Q) \setminus \{0\}} \frac{\langle Lu, u \rangle_H}{\|u\|_H^2} \le \lambda$. Since this is true for any $\phi_1, \phi_2 \in H$, we get $\mu_3(L) \le \lambda < \mu_3(L)$, and thus the contradiction. $\square$

B. PROOF OF THE MODULATION LEMMA

In this section, we prove Lemma 3.11 which decomposes the magnetization with a nice gauge satisfying two orthogonal equalities. In brief, this result is a consequence of the implicit function theorem. We start with defining and studying the appropriate function: let

$$ \begin{gathered} \mathcal{F} : (L^\infty + L^2) \times G \to \mathbb{R}^2 \\ (m, g) \mapsto \left( \int m \cdot g \cdot \partial_x w_{h_0} \, dx \quad \int m \cdot (e_1 \wedge g \cdot w_{h_0}) \, dx \right). \end{gathered} $$

This integrals are well-defined since both $\partial_x w_{h_0}$ and $e_1 \wedge w_{h_0}$ converge exponentially to 0 at $\pm\infty$ (they are also of class $W^{k,1}$ for all $k \in \mathbb{N}$). We begin by some obvious properties on this function.
**Lemma A.1.** Let $\lambda_1 < \lambda_2$ be the first two eigenvalues of a self-adjoint bounded-from-below operator (Dom(L), L) on an Hilbert space H, associated with the quadratic form $Q(u) := \langle Lu, u \rangle_H$. Assume that these eigenvalues are simple with respecting normalized eigenvectors $\psi_1, \psi_2 \in \text{Dom}(L) \subset \text{Dom}(Q)$ and that the essential spectrum is included in $[\lambda_\infty, \infty)$ where $\lambda_\infty > \lambda_2$. Then

$$ \inf_{u \in \text{span}(\psi_1, \psi_2)^\perp \cap \text{Dom}(Q) \setminus \{0\}} \frac{\langle Lu, u \rangle_H}{\|u\|_H^2} =: a = \mu_3(L) := \sup_{\phi_1, \phi_2 \ u \in \text{span}(\phi_1, \phi_2)^\perp \cap \text{Dom}(Q) \setminus \{0\}} \inf \frac{\langle Lu, u \rangle_H}{\|u\|_H^2}, $$

where $\mu_3(L)$ is the third Rayleigh quotient of L.

*Proof.* By definition of $\mu_3(L)$, we obviously have $a \le \mu_3(L)$. By contradiction, assume that $a < \mu_3(L)$. By definition of a, there exists a non-trivial vector $\psi_3 \in \text{Dom}(Q) \setminus \{0\}$ such that $\langle \psi_3, \psi_1 \rangle_H = \langle \psi_3, \psi_2 \rangle_H = 0$ and $Q(\psi_3) < \lambda \|\psi_3\|_H^2$ for any $\lambda \in (a, \mu_3(L))$ to be fixed later. We can assume that $\|\psi_3\|_H = 1$. Let $\mu_2(L)$ be the second Rayleigh quotient

$$ \mu_2(L) := \sup_{\phi_1} \inf_{u \in \text{span}(\phi_1)^\perp \cap \text{Dom}(Q) \setminus \{0\}} \frac{\langle Lu, u \rangle_H}{\|u\|_H^2}. $$

It is known that $\mu_2(L) = \lambda_2$ (see [4]). Moreover, since $\lambda_2$ is simple and since the bottom of the essential spectrum is larger than $\lambda_\infty > \lambda_2$, we also know that $\mu_2(L) < \mu_3(L)$. Then, we can assume furthermore $\lambda \ge \mu_2(L) = \lambda_2$.

Now, let $\phi_1, \phi_2 \in H$. We know that $\text{span}(\psi_1, \psi_2, \psi_3)$ is a vector space with dimension 3 which is included in $\text{Dom}(Q) \subset H$, whereas $\text{span}(\phi_1, \phi_2)^\perp$ is a vector space with codimension (at least) 2 in H. Therefore, there exists a non-trivial vector $u \in \text{span}(\psi_1, \psi_2, \psi_3) \cap \text{span}(\phi_1, \phi_2)^\perp$, and this $u$ is obviously in $\text{Dom}(Q)$. We can decompose $u$ in the basis $(\psi_1, \psi_2, \psi_3)$:

$$ u = \langle u, \psi_1 \rangle_H \psi_1 + \langle u, \psi_2 \rangle_H \psi_2 + \langle u, \psi_3 \rangle_H \psi_3. $$

In particular, we can compute $Q(u)$ with this expansion. Indeed, since $\psi_1$ and $\psi_2$ are eigenvectors, and as $(\psi_1, \psi_2, \psi_3)$ are orthogonal to each other in H, all the cross terms vanish:

$$ \begin{aligned} Q(u) &= \langle u, \psi_1 \rangle_H^2 Q(\psi_1) + \langle u, \psi_2 \rangle_H^2 Q(\psi_2) + \langle u, \psi_3 \rangle_H^2 Q(\psi_3) \\ &= \underbrace{\lambda_1}_{< \lambda_2 \le \lambda} \langle u, \psi_1 \rangle_H^2 + \underbrace{\lambda_2}_{\le \lambda} \langle u, \psi_2 \rangle_H^2 + \langle u, \psi_3 \rangle_H^2 \underbrace{Q(\psi_3)}_{< \lambda} \\ &\le \lambda (\langle u, \psi_1 \rangle_H^2 + \langle u, \psi_2 \rangle_H^2 + \langle u, \psi_3 \rangle_H^2) = \lambda \|u\|_H^2. \end{aligned} $$

Therefore, $\inf_{u \in \text{span}(\phi_1, \phi_2)^\perp \cap \text{Dom}(Q) \setminus \{0\}} \frac{\langle Lu, u \rangle_H}{\|u\|_H^2} \le \lambda$. Since this is true for any $\phi_1, \phi_2 \in H$, we get $\mu_3(L) \le \lambda < \mu_3(L)$, and thus the contradiction. $\square$

## B. PROOF OF THE MODULATION LEMMA

In this section, we prove Lemma 3.11 which decomposes the magnetization with a nice gauge satisfying two orthogonal equalities. In brief, this result is a consequence of the implicit function theorem. We start with defining and studying the appropriate function: let

$$ \begin{gathered} \mathcal{F} : (L^\infty)^3 \times G \to \mathbb{R}^2 \\ (m, g) \mapsto \left( \int m \cdot g \cdot \partial_x w_{h_0} \, dx \quad \int m \cdot (e_1 \wedge g \cdot w_{h_0}) \, dx \right). \end{gathered} $$

This integrals are well-defined since both $\partial_x w_{h_0}$ and $e_1 \wedge w_{h_0}$ converge exponentially to 0 at $\pm\infty$ (they are also of class $W^{k,1}$ for all $k \in \mathbb{N}$). We begin by some obvious properties on this function.

**Lemma B.1.** The function $\mathcal{F}$ is linear in the first variable and of class $C^\infty$. Moreover, there also holds

$$ \mathcal{F}(m, g) = \left( \int (-g) \cdot m \cdot \partial_x w_{h_0} \, dx \quad \int (-g) \cdot m \cdot (e_1 \wedge w_{h_0}) \, dx \right) = \left( \int ((-g) \cdot m - w_{h_0}) \cdot \partial_x w_{h_0} \, dx \quad \int ((-g) \cdot m - w_{h_0}) \cdot (e_1 \wedge w_{h_0}) \, dx \right). $$
**Lemma B.1.** The function $\mathfrak{F}$ is linear in the first variable and of class $\mathcal{C}^\infty$. Moreover, there also holds

$$ \mathfrak{F}(m, g) = \left( \int (-g) \cdot m \cdot \partial_x w_{h_0} \, dx \right) = \left( \int ((-g) \cdot m - w_{h_0}) \cdot \partial_x w_{h_0} \, dx \right). $$

Then, we show that $\partial_g \mathfrak{F}(w_{h_0}, (0, 0))$ is invertible.

**Lemma B.2.** Writing $g = (y, \phi)$, there holds $\mathfrak{F}(w_{h_0}, (0, 0)) = 0$ and

$$ \partial_y \mathfrak{F}(w_{h_0}, (0, 0)) = \left( -\int |\partial_x w_{h_0}|^2 dx \quad , \quad \partial_\phi \mathfrak{F}(w_{h_0}, (0, 0)) = \left( \begin{array}{c} 0 \\ -\int |e_1 \wedge w_{h_0}|^2 dx \end{array} \right) \right). $$

Therefore, $\partial_g \mathfrak{F}(w_{h_0}, (0, 0))$ is invertible.

*Proof.* Since $\partial_y((-g) \cdot m) = (-g) \cdot \partial_x m$ and $\partial_\phi((-g) \cdot m) = (-g) \cdot (e_1 \wedge m)$, we get

$$ \partial_y \mathfrak{F}(w_{h_0}, (0, 0)) = \left( \begin{array}{c} -\int |\partial_x w_{h_0}|^2 dx \\ -\int \partial_x w_{h_0} \cdot (e_1 \wedge w_{h_0}) \end{array} \right), \quad \partial_\phi \mathfrak{F}(w_{h_0}, (0, 0)) = \left( \begin{array}{c} -\int \partial_x w_{h_0} \cdot (e_1 \wedge w_{h_0}) \\ -\int |e_1 \wedge w_{h_0}|^2 dx \end{array} \right). $$

On the other hand, we know that $\partial_x w_{h_0} = \partial_x \theta_{h_0} n_{h_0}$ and $e_1 \wedge w_{h_0} = \sin \theta_{h_0} e_3$, so that $\partial_x w_{h_0} \cdot (e_1 \wedge w_{h_0}) = 0$ on $\mathbb{R}$.
Last, we know that $\int |\partial_x w_{h_0}|^2 dx > 0$ and $\int |e_1 \wedge w_{h_0}|^2 dx > 0$, which leads to the conclusion. $\square$

Last before proving Lemma 3.11, we show that $g.f$ is close to $f$ when $g$ is close to $(0, 0)$.

**Lemma B.3.** There exists $C > 0$ such that, for any $g \in G$, there holds

$$ \|g \cdot w_{h_0} - w_{h_0}\|_{H^1} \leq C|g|. $$

*Proof.* First, we know that, for any $f \in \mathcal{H}^2$ and $y \in \mathbb{R}$, there holds

$$ \|\tau_y \cdot f - f\|_{H^1} \leq C|y|\|\partial_x f\|_{H^1}. $$

Moreover, using the expression of $w_{h_0}$ in terms of $\theta_{h_0}$, there holds

$$ R_{\phi} \cdot w_{h_0} - w_{h_0} = \sin \theta_{h_0} \begin{pmatrix} 0 \\ \cos \phi - 1 \\ \sin \phi \end{pmatrix}, $$

so that

$$ \|R_{\phi} \cdot w_{h_0} - w_{h_0}\|_{H^1} \leq C|\phi|. $$

Therefore,

$$ \begin{aligned} \|g \cdot w_{h_0} - w_{h_0}\|_{H^1} &\leq \|\tau_y R_{\phi} \cdot w_{h_0} - R_{\phi} \cdot w_{h_0}\|_{H^1} + \|R_{\phi} \cdot w_{h_0} - w_{h_0}\|_{H^1} \\ &\leq C|y|\|w_{h_0}\|_{H^1} + C|\phi| \leq C|g|. \end{aligned} \quad \square $$

*Proof of Lemma 3.11.* By Lemma B.2, the implicit function theorem gives some $\tilde{\delta}_1 > 0$ such that, for all $w \in L^\infty$ satisfying

$$ \|m - w_{h_0}\|_{L^\infty} < \tilde{\delta}_1, $$

there exists a unique $g \in G$ satisfying $|g| < C\|m - w_{h_0}\|_{L^\infty}$ for some $C > 0$ such that

$$ \int m \cdot g \cdot \partial_x w_{h_0} \, dx = \int m \cdot (e_1 \wedge g \cdot w_{h_0}) \, dx = 0. $$

Then, Lemma B.1 implies that $\eta := (-g) \cdot m - w_{h_0}$ satisfies

$$ \int \eta \cdot \partial_x w_{h_0} \, dx = \int \eta \cdot (e_1 \wedge w_{h_0}) \, dx = 0. $$
Then, we show that $\partial_g \mathfrak{F}(w_{h_0}, (0,0))$ is invertible.

**Lemma B.2.** Writing $g = (y, \phi)$, there holds $\mathfrak{F}(w_{h_0}, (0,0)) = 0$ and

$$ \partial_y \mathfrak{F}(w_{h_0}, (0,0)) = \left( -\int |\partial_x w_{h_0}|^2 dx \quad , \quad \partial_\phi \mathfrak{F}(w_{h_0}, (0,0)) = \left( \begin{array}{c} 0 \\ -\int |e_1 \wedge w_{h_0}|^2 dx \end{array} \right) \right). $$

Therefore, $\partial_g \mathfrak{F}(w_{h_0}, (0,0))$ is invertible.

*Proof.* Since $\partial_y((-g).m) = (-g).\partial_x m$ and $\partial_\phi((-g).m) = (-g).(e_1 \wedge m)$, we get

$$ \partial_y \mathfrak{F}(w_{h_0}, (0,0)) = \left( \begin{array}{c} -\int |\partial_x w_{h_0}|^2 dx \\ -\int \partial_x w_{h_0} \cdot (e_1 \wedge w_{h_0}) \end{array} \right), \quad \partial_\phi \mathfrak{F}(w_{h_0}, (0,0)) = \left( \begin{array}{c} -\int \partial_x w_{h_0} \cdot (e_1 \wedge w_{h_0}) \\ -\int |e_1 \wedge w_{h_0}|^2 dx \end{array} \right). $$

On the other hand, we know that $\partial_x w_{h_0} = \partial_x \theta_{h_0} n_{h_0}$ and $e_1 \wedge w_{h_0} = \sin \theta_{h_0} e_3$, so that $\partial_x w_{h_0} \cdot (e_1 \wedge w_{h_0}) \equiv 0$ on $\mathbb{R}$.
Last, we know that $\int |\partial_x w_{h_0}|^2 dx > 0$ and $\int |e_1 \wedge w_{h_0}|^2 dx > 0$, which leads to the conclusion. $\square$

Last before proving Lemma 3.11, we show that $g.f$ is close to $f$ when $g$ is close to $(0,0)$.

**Lemma B.3.** There exists $C > 0$ such that, for any $g \in G$, there holds

$$ \|g \cdot w_{h_0} - w_{h_0}\|_{H^1} \leq C|g|. $$

*Proof.* First, we know that, for any $f \in \mathcal{H}^2$ and $y \in \mathbb{R}$, there holds

$$ \|\tau_y \cdot f - f\|_{H^1} \leq C|y|\|\partial_x f\|_{H^1}. $$

Moreover, using the expression of $w_{h_0}$ in terms of $\theta_{h_0}$, there holds

$$ R_{\phi} \cdot w_{h_0} - w_{h_0} = \sin \theta_{h_0} \begin{pmatrix} 0 \\ \cos \phi - 1 \\ \sin \phi \end{pmatrix}, $$

so that

$$ \|R_{\phi} \cdot w_{h_0} - w_{h_0}\|_{H^1} \leq C|\phi|. $$

Therefore,

$$ \begin{aligned} \|g \cdot w_{h_0} - w_{h_0}\|_{H^1} &\le \|\tau_y R_{\phi} \cdot w_{h_0} - R_{\phi} \cdot w_{h_0}\|_{H^1} + \|R_{\phi} \cdot w_{h_0} - w_{h_0}\|_{H^1} \\ &\le C|y|\|w_{h_0}\|_{H^1} + C|\phi| \le C|g|. \end{aligned} \quad \square $$

*Proof of Lemma 3.11.* By Lemma B.2, the implicit function theorem gives some $\tilde{\delta}_1 > 0$ and $C > 0$ such that, for all $m \in L^\infty$ satisfying

$$ \|m - w_{h_0}\|_{L^\infty} < \tilde{\delta}_1, $$

there exists a unique $g \in G$ satisfying $|g| \le C\|m - w_{h_0}\|_{L^\infty}$ such that

$$ \int m \cdot g \cdot \partial_x w_{h_0} \, dx = \int m \cdot (e_1 \wedge g \cdot w_{h_0}) \, dx = 0. $$

Then, Lemma B.1 implies that $\eta := (-g).m - w_{h_0}$ satisfies

$$ \int \eta \cdot \partial_x w_{h_0} \, dx = \int \eta \cdot (e_1 \wedge w_{h_0}) \, dx = 0. $$

Now, we use the embedding of $H^1$ into $L^\infty$, that is the existence of $C_S > 0$ such that for all $f \in H^1$,

$$ \|f\|_{L^\infty} \le C_S \|f\|_{H^1}. $$
Now, we use the embedding of $H^1$ into $L^\infty$, that is the existence of $C_S > 0$ such that for all $f \in H^1$,

$$ \|f\|_{L^\infty} < C_S \|f\|_{H^1}. $$

Therefore, a similar conclusion can be reached if $m - w_{h_0} \in H^1$ satisfies (3.12) for $\delta_1 := \frac{\delta_1}{C_S}$. Last, in such a case, we can estimate $\|\eta\|_{H^1}$ as follows:

$$ \begin{aligned} \|\eta\|_{H^1} &= \|m - g.w_{h_0}\|_{H^1} \\ &\le \|m - w_{h_0}\|_{H^1} + \|g.w_{h_0} - w_{h_0}\|_{H^1} \\ &\le \|m - w_{h_0}\|_{H^1} + C|g| \\ &\le \|m - w_{h_0}\|_{H^1} + C\|m - w_{h_0}\|_{L^\infty} \\ &\le (1 + CC_S)\|m - w_{h_0}\|_{H^1}, \end{aligned} $$

where we used Lemma B.3 from the second to the third line. □

## REFERENCES

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[5] R. Côte and R. Ignat. Asymptotic stability of precessing domain walls for the Landau-Lifshitz-Gilbert equation in a nanowire with Dzyaloshinskii-Moriya interaction. *Comm. Math. Phys.*, 401(3):2901–2957, 2023.

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Therefore, a similar conclusion can be reached if $m - w_{h_0} \in H^1$ satisfies (3.12) for $\delta_1 := \frac{\delta_1}{C_S}$. Last, in such a case, we can estimate $\|\eta\|_{H^1}$ as follows:

$$ \begin{aligned} \|\eta\|_{H^1} &= \|m - g.w_{h_0}\|_{H^1} \\ &\le \|m - w_{h_0}\|_{H^1} + \|g.w_{h_0} - w_{h_0}\|_{H^1} \\ &\le \|m - w_{h_0}\|_{H^1} + C|g| \\ &\le \|m - w_{h_0}\|_{H^1} + C\|m - w_{h_0}\|_{L^\infty} \\ &\le (1 + CC_S)\|m - w_{h_0}\|_{H^1}, \end{aligned} $$

where we used Lemma B.3 from the second to the third line. □

## REFERENCES

[1] F. Alouges and A. Soyeur. On global weak solutions for Landau-Lifshitz equations: existence and nonuniqueness. *Nonlinear Anal.*, 18(11):1071-1084, 1992.

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