Asymptotic Stability of Precessing Domain Walls for the Landau–Lifshitz–Gilbert Equation in a Nanowire with Dzyaloshinskii–Moriya Interaction

Abstract

We consider a ferromagnetic nanowire and we focus on an asymptotic regime where the Dzyaloshinskii-Moriya interaction is taken into account. First we prove a dimension reduction result via $\Gamma$-convergence that determines a limit functional E defined for maps $m:\mathbb {R}\rightarrow \mathbb {S}^2$ in the direction $e_1$ of the nanowire. The energy functional E is invariant under translations in $e_1$ and rotations about the axis $e_1$. We fully classify the critical points of finite energy E when a transition between $-e_1$ and $e_1$ is imposed; these transition layers are called (static) domain walls. The evolution of a domain wall by the Landau–Lifshitz–Gilbert equation associated to E under the effect of an applied magnetic field $h(t)e_1$ depending on the time variable t gives rise to the so-called precessing domain wall. Our main result proves the asymptotic stability of precessing domain walls for small h in $L^\infty ([0, +\infty ))$ and small $H^1(\mathbb {R})$ perturbations of the static domain wall, up to a gauge which is intrinsic to invariances of the functional E.

Appendices

Appendix A: Toolbox

We compute for $a,b,c\in \mathbb {R}^3$, $g= (y,\phi ) \in G$, $m, {\tilde{m}}:\mathbb {R}\rightarrow \mathbb {S}^2$, $m=(m_1, m_2, m_3)$:

$$\begin{aligned}{} & {} a \wedge ( b \wedge c) = (a\cdot c) b - (a \cdot b)c, \quad e_1\wedge a=(a\cdot e_2) e_3-(a\cdot e_3) e_2, \\{} & {} R_\phi a\wedge R_\phi b=R_\phi (a\wedge b), \quad \partial _\phi (R_\phi a)=e_1\wedge (R_\phi a)=R_\phi (e_1\wedge a), \, \,\\{} & {} g.m\wedge g.{\tilde{m}}=g.(m\wedge {\tilde{m}}),\, \, \partial _y (g.m) = -g. \partial _x m, \, \, \partial _\phi (g.m) = e_1 \wedge g.m=g.(e_1 \wedge m),\\{} & {} \partial _x m \cdot (e_1\wedge m)=m_2\partial _x m_3-m_3\partial _x m_2, \quad |e_1\wedge m|^2=1-m_1^2. \end{aligned}$$

Appendix B: Some Properties of the Sets $\mathscr {H}^1$ and $H^1(\mathbb {R}, \mathbb {S}^2)$

We recall that for $m = (m_1,m_2,m_3) \in \mathbb {R}^3$, we denote $m' = (m_2,m_3) \in \mathbb {R}^2$ the last two coordinates.

For $s \geqslant 1$, we defined for every $m:\mathbb {R}\rightarrow \mathbb {R}^3$,

$$\begin{aligned} \Vert m \Vert _{\mathscr {H}^s} : = \Vert m_2 \Vert _{L^2} + \Vert m_3 \Vert _{L^2} + \Vert m \Vert _{\dot{H}^s}, \end{aligned}$$

where $\dot{H}^s$ was defined in (1.11). We start by proving the following interpolation inequality¹³$\mathscr {H}^s(\mathbb {R}, \mathbb {S}^2)\subset \dot{H}^1(\mathbb {R}, \mathbb {S}^2)$ for $s\in [1,2]$. We also prove the behaviour at $\pm \infty$ of maps in $\mathscr {H}^s(\mathbb {R}, \mathbb {S}^2)$.

Lemma B.1

Let $s \in [1,2]$. There exists $C>0$ such that for any $m: \mathbb {R} \rightarrow \mathbb {S}^2$ with $\Vert m \Vert _{\mathscr {H}^s} <+\infty$, then $\partial _x m \in L^2$ and

$$\begin{aligned} \Vert \partial _x m \Vert _{L^2} \leqslant C \Vert m \Vert _{\mathscr {H}^s} (1+ \Vert m \Vert _{\mathscr {H}^s}^2 ); \end{aligned}$$

(B.1)

in particular, $E(m)<\infty$, and m admits limits belonging to $\{ (\pm 1,0,0) \}$ as $x \rightarrow \pm \infty$.

Proof

We divide the proof in two steps:

Step 1: $s=1$. In this case, the desired inequality follows by the definition of $\mathscr {H}^1$. Note that

$$\begin{aligned} |\partial _x m\cdot (e_1\wedge m)|\leqslant |\partial _x m'|\cdot |m'|\leqslant \frac{1}{2}(|\partial _x m'|^2+ |m'|^2). \end{aligned}$$

We infer that $E(m)\leqslant C \Vert m \Vert ^2_{\mathscr {H}^1}$.

Step 2: $s \in (1,2]$. First observe that $m_2 \in H^s=L^2\cap \dot{H}^s$ and by interpolation, we have

$$\begin{aligned} \Vert \partial _x m_2 \Vert _{L^2} \lesssim \Vert m_2 \Vert _{L^2} + \Vert m_2 \Vert _{\dot{H}^s} \lesssim \Vert m \Vert _{\mathscr {H}^s}, \end{aligned}$$

and similarly $m_3 \in H^s$ and $\Vert \partial _x m_3 \Vert _{L^2} \lesssim \Vert m \Vert _{\mathscr {H}^s}$. In particular, $|m_2|, |m_3| \in H^1$; as m takes its values in $\mathbb {S}^2$, this means that $\sqrt{1-m_1^2} = |(m_2,m_3)| \in H^1$ and

$$\begin{aligned} \Vert \sqrt{1-m_1^2} \Vert _{H^1} \lesssim \Vert (m_2,m_3) \Vert _{H^1} \lesssim \Vert m \Vert _{\mathscr {H}^s}. \end{aligned}$$

(B.2)

In particular, the functions $m_2, m_3, |m_1|$ are continuous and have the following limits

$$\begin{aligned} m_2(\pm \infty )=m_3(\pm \infty )=0 \quad \text {and} \quad |m_1|(\pm \infty )=1. \end{aligned}$$

We now consider

$$\begin{aligned} A : = \{ x \in \mathbb {R}: |m_1(x)| < \frac{1}{2} \} \end{aligned}$$

which is an open bounded set. First, we estimate the length |A| of A:

$$\begin{aligned} \Vert m \Vert _{\mathscr {H}^s}^2 \geqslant \int _{\mathbb {R}} (1-m_1^2) \, dx\geqslant \int _A \frac{3}{4}\, dx \geqslant \frac{3}{4} |A|. \end{aligned}$$

(B.3)

Next, we estimate the $L^2$ norm of $\partial _x m_1$. On the set $A^c = \mathbb {R}\setminus A$, one has

$$\begin{aligned} |\partial _x m_1| \leqslant 2 \frac{|m_1 \partial _x m_1|}{\sqrt{1-m_1^2}} = 2 \big | \partial _x \sqrt{1-m_1^2}\big |\quad \text {a.e. in } A^c. \end{aligned}$$

Therefore, by (B.2),

$$\begin{aligned} \int _{A^c} |\partial _x m_1|^2 \, dx\leqslant 4 \Vert \partial _x \sqrt{1-m_1^2} \Vert _{L^2}^2 \lesssim \Vert m \Vert _{\mathscr {H}^s}^2. \end{aligned}$$

It remains to estimate the $L^2$ norm of $\partial _x m_1$ on A. As A is open and bounded, we write

$$\begin{aligned} A = \bigcup _{k\in \mathscr {K}} I_k, \quad I_k = (a_k^-, a_k^+), \quad |m_1(a_k^\pm )|=\frac{1}{2}, \end{aligned}$$

where $\{I_k\}_{k\in \mathscr {K}}$ is a (at most) countable family of disjoint open bounded intervals. We have

where denotes the average on the bounded interval $I_k$.

Estimating $J_2$. Let $\mathscr {J} = \big \{ k \in \mathscr {K}\, : \, m_1(a_k^-) \ne m_1(a_k^+) \big \}$. Then $|m_1(a_k^-) - m_1(a_k^+)|=1$ if $k \in \mathscr {J}$ and $m_1(a_k^-) - m_1(a_k^+) =0$ if $k \notin \mathscr {J}$. Thus,

To estimate the length $|I_k|$ of $I_k$ for $k\in \mathscr {J}$, as $m_1(a_k^-)$ and $m_1(a_k^+)$ have different sign, the continuity of $m_1$ implies the existence of $b_k \in (a_k^-, a_k^+)$ such that $m_1(b_k) =0$. Denoting by $f = \sqrt{1-m_1^2} \in H^1$ (by (B.2)), the Cauchy-Schwarz inequality implies

$$\begin{aligned} 1- \frac{\sqrt{3}}{2} = | f(a_k^-) - f(b_k)|\leqslant \int _{a_k^-}^{b_k} |\partial _x f| \, dx\leqslant |b_k - a_k^-|^{1/2} \left( \int _{a_k^-}^{b_k} |\partial _x f|^2 \, dx\right) ^{1/2}. \end{aligned}$$

Arguing similarly on $[b_k,a_k^+]$, we get after squaring, for $c = 7/2 - 2\sqrt{3} >0$:

$$\begin{aligned} c \leqslant (|b_k - a_k^-| + |b_k - a_k^+|) \int _{I_k} |\partial _x f|^2\, dx = |I_k| \int _{I_k} |\partial _x f|^2\, dx . \end{aligned}$$

Therefore, by (B.2),

$$\begin{aligned} J_2=\sum _{k \in \mathscr {J}} \frac{1}{|I_k|} \leqslant \frac{1}{c} \sum _{k \in \mathscr {J}} \int _{I_k} |\partial _x f|^2 \, dx\lesssim \int _{\mathbb {R}} |\partial _x f|^2 \, dx\lesssim \Vert m \Vert _{\mathscr {H}^s}^2. \end{aligned}$$

Estimating $J_1$. We consider separately the cases $s \in (1,2)$ and $s=2$.

Case 1. Assume that $s \in (1,2)$. We have by Jensen’s inequality:

where we used (B.3). As $2+4(s-1) \leqslant 6$, this completes the case when $s <2$.

Case 2. Assume that $s=2$. As before, we have by Jensen’s inequality

where we used (B.3). The bound (B.1) follows.

Finally, $m_1$ is continuous (as $\partial _x m_1 \in L^2$) and we saw that $|m_1| \rightarrow 1$ as $x \rightarrow +\infty$. As $\{ \pm 1 \}$ is discrete, we infer that $m_1 \rightarrow a^+$ as $x \rightarrow +\infty$ with $a^+ \in \{ \pm 1 \}$. The same argument shows that $m_1 \rightarrow a^-$ as $x \rightarrow -\infty$ with $a^- \in \{ \pm 1 \}$. $\quad \square$

In the next lemma, we prove the reverse inequality with respect to Lemma B.1 relating E(m) and $\Vert m\Vert _{\mathscr {H}^1}$. In other words, the seminorm $\Vert \cdot \Vert _{\mathscr {H}^1}$ fully characterises the set of finite energy configurations $E(m)<\infty$.

Lemma B.2

If $\gamma ^2<1$ then there exists a constant $C_\gamma >0$ such that

$$\begin{aligned} E(m)\geqslant C_\gamma \Vert m\Vert ^2_{\mathscr {H}^1} \quad \text {for every } m: \mathbb {R} \rightarrow \mathbb {S}^2. \end{aligned}$$

Proof

Let $m=(m_1, m_2, m_3):\mathbb {R}\rightarrow \mathbb {S}^2$ and $m'=(m_2, m_3)$. As $|\gamma |<1$, we can choose $a,b\in (-1,1)$ such that $ab=\gamma$. Then

$$\begin{aligned} a^2|\partial _x m'|^2+2\gamma \partial _x m\cdot (e_1 \wedge m)+b^2 |m'|^2=(a\partial _x m_{2}-b m_{3})^2+(a\partial _x m_{3}+b m_{2})^2. \end{aligned}$$

Therefore,

$$\begin{aligned} E(m)\geqslant \frac{1}{2} \int _\mathbb {R}(\partial _x m_1)^2+(1-a^2) |\partial _x m'|^2 +(1-b^2) |m'|^2\, dx\geqslant C_\gamma \Vert m\Vert ^2_{\mathscr {H}^1}. \end{aligned}$$

(If one chooses $a^2=b^2=|\gamma |$, then $C_\gamma$ can be taken equal to $\frac{1-|\gamma |}{2}$). $\quad \square$

In the following, we study more closely the norms $\Vert \cdot \Vert _{\mathscr {H}^1}$ and $\Vert \cdot \Vert _{H^1}$ for maps with target $\mathbb {S}^2$. First, note that Lemma B.2 implies that two maps $m, {\tilde{m}}\in \mathscr {H}^1(\mathbb {R}, \mathbb {S}^2)$ may have different limits at $\pm \infty$, so $m-{\tilde{m}}$ might not belong to $L^2$ (e.g., ${\tilde{m}}=-m$). However, if m and ${\tilde{m}}$ have the same limits at $\pm \infty$, then we show that $m-{\tilde{m}}\in L^2$. Next we prove an important stability property for non-constant maps: if m and ${\tilde{m}}$ are close to each other in $\mathscr {H}^1$ and they are not constant, then they have the same limits at $\pm \infty$ and $m-{\tilde{m}}$ is small in $L^2$.

Proposition B.3

• 1) Let $m, {\tilde{m}}\in \mathscr {H}^1(\mathbb {R}, \mathbb {S}^2)$ with the same limits at infinity, i.e., $m(\pm \infty ) = {\tilde{m}}(\pm \infty )$. Then $m-{\tilde{m}} \in H^1$.

• 2) Let $m\in \mathscr {H}^1(\mathbb {R}, \mathbb {S}^2)$ such that $m_1$ is not a constant function. Then for all $\varepsilon >0$, there exists $\delta = \delta (m)>0$ such that if ${\tilde{m}}\in \mathscr {H}^1(\mathbb {R}, \mathbb {S}^2)$ satisfies $\Vert m - {\tilde{m}} \Vert _{\mathscr {H}^1} \leqslant \delta$, then

  $$\begin{aligned} \Vert m - {\tilde{m}} \Vert _{H^1} \leqslant \varepsilon ; \end{aligned}$$

  (B.4)

  in particular, $m(\pm \infty ) = {\tilde{m}}(\pm \infty )$.

Remark B.4

The case of a constant component $m_1$ is different. For example, observe that if $m=e_1$, then ${\tilde{m}}=-e_1$ has the property that $\Vert m-{\tilde{m}}\Vert _{\mathscr {H}^1}=0$, but $m-{\tilde{m}}\notin L^2$.

Proof

We divide the proof in several steps:

Step 1. We prove 1). Consider $u = m - {\tilde{m}}$. Since $m, {\tilde{m}}\in \mathscr {H}^1(\mathbb {R}, \mathbb {S}^2)$, then $\Vert u \Vert _{\mathscr {H}^1} < +\infty$, so that $u_2,u_3 \in H^1$ and $\partial _x u_1 \in L^2$. It remains to prove that $u_1\in L^2$. As $m(\pm \infty ) = {\tilde{m}}(\pm \infty )$, we deduce that $u_1=m_1-{\tilde{m}}_1$ (which is continuous) tends to 0 at $\pm \infty$. By Lemma B.1, we have $(m_2,m_3), ({\tilde{m}}_2, {\tilde{m}}_3)\rightarrow 0$ as $x\rightarrow \pm \infty$; combined with $m(\pm \infty ) = {\tilde{m}}(\pm \infty )$, we deduce $\lim _{x\rightarrow \pm \infty } |m_1 + {\tilde{m}}_1|=2|m(\pm \infty )|=2$. Thus, there exists $R>0$ such that

$$\begin{aligned} |m_1(x) - {\tilde{m}}_1(x)| \leqslant 1/2 \quad \text {and} \quad |m_1(x) + {\tilde{m}}_1(x) | \geqslant 1 \quad \text { if } \, |x| \geqslant R. \end{aligned}$$

(B.5)

We estimate the $L^2$ norm of $u_1$ in the set $\big \{|x| \geqslant R\big \}$ and then in the set $\big \{|x| \leqslant R\big \}$. (B.5) implies

$$\begin{aligned} \forall |x| \geqslant R, \quad u_1(x)^2 \leqslant \frac{1}{2} |m_1(x) - {\tilde{m}}_1(x) | \leqslant \frac{1}{2} |m_1(x)^2 - {\tilde{m}}_1(x)^2|, \end{aligned}$$

yielding

$$\begin{aligned}&\int _{\mathbb {R} \setminus [-R,R]} u_1(x)^2 dx \leqslant \frac{1}{2} \int _{\mathbb {R} \setminus [-R,R]} |m_1(x)^2 - {\tilde{m}}_1(x)^2| dx \\&\quad \leqslant \frac{1}{2} \int _{\mathbb {R}} |m_2^2 +m_3^2 - {\tilde{m}}_2^2 -{\tilde{m}}_3^2|\, dx < \infty \end{aligned}$$

because $m, {\tilde{m}}$ are $\mathbb {S}^2$-valued maps and $(m_2,m_3), ({\tilde{m}}_2, {\tilde{m}}_3)\in L^2$. The estimate of the $L^2$ norm of $u_1$ in the set $\{|x| \leqslant R \}$ is easy: as $|u_1| \leqslant |m_1| + |{\tilde{m}}_1| \leqslant 2$, it follows

$$\begin{aligned} \int _{-R}^R u_1(x)^2 dx \leqslant 8 R < +\infty . \end{aligned}$$

Hence $u_1 \in L^2(\mathbb {R})$.

Step 2. We prove 2). First in the following claim, we prove smallness of the $L^\infty$ norm of $u = m - {\tilde{m}}$ provided that $\Vert m - {\tilde{m}} \Vert _{\mathscr {H}^1}$ is small.

Claim B.5

Let $m\in \mathscr {H}^1(\mathbb {R}, \mathbb {S}^2)$ with $m_1$ non-constant and let $u^n$ be a sequence such that $m + u^n: \mathbb {R} \rightarrow \mathbb {S}^2$ and $\Vert u^n \Vert _{\mathscr {H}^1} \rightarrow 0$ as $n\rightarrow \infty$. Then $\Vert u^n \Vert _{L^\infty } \rightarrow 0$ as $n\rightarrow \infty$.

Proof of Claim

The assumption $\Vert u^n \Vert _{\mathscr {H}^1} \rightarrow 0$ as $n\rightarrow \infty$ implies $\Vert u^n_2 \Vert _{H^1}, \Vert u^n_3 \Vert _{H^1} \rightarrow 0$; thus, the embedding $H^1\hookrightarrow L^\infty$ yields $u^n_2$, $u^n_3 \rightarrow 0$ in $L^\infty$ as $n\rightarrow \infty$. It remains to show that $u_1^n \rightarrow 0$ in $L^\infty$. For this, observe that the constraints $m, m + u^n \in \mathbb {S}^2$ yield

$$\begin{aligned} 2m \cdot u^n + |u^n|^2 =0 \quad \text {in} \quad \mathbb {R}. \end{aligned}$$

Combined with $u^n_2,u^n_3 \rightarrow 0$ in $L^\infty$ as $n\rightarrow \infty$, since $|m|=1$ in $\mathbb {R}$, we deduce

$$\begin{aligned} 2m_1 u^n_1 + (u^n_1)^2 \rightarrow 0 \quad \text {in } L^\infty \quad \text {as }n\rightarrow \infty . \end{aligned}$$

(B.6)

Recall that $\Vert u^n_1 \Vert _{L^\infty }\leqslant \Vert u^n \Vert _{L^\infty } \leqslant 2$. Up to a subsequence, we can assume that $u^n_1(0)\rightarrow c$ for some limit $c\in [-2,2]$.

Step 1. We claim that $c=0$ and for any $R >0$, $u_1^n\rightarrow 0$ in $L^\infty ([-R,R])$. Indeed, it suffices to consider large R, and as $m_1$ is not constant in $\mathbb {R}$, we can assume that $m_1$ is not constant in $[-R,R]$. For such R and any $x \in [-R,R]$,

$$\begin{aligned} |u^n_1(x)-c|\leqslant |u^n_1(0)-c|+\sqrt{R}\Vert \partial _x u^n_1\Vert _{L^2(-R,R)}. \end{aligned}$$

As $\partial _x u^n_1 \rightarrow 0$ in $L^2$, we deduce that $u_1^n\rightarrow c$ in $L^\infty ([-R,R])$. Passing to the limit in (B.6), it yields $2 m_1 c + c^2 =0$ in $[-R,R]$. If $c\ne 0$, then we would obtain $m_1 = -c/2$ is constant in $[-R,R]$ which contradicts our assumption. Thus, $c=0$, i.e., $u_1^n\rightarrow 0$ in $L^\infty ([-R,R])$ as $n\rightarrow \infty$.

Step 2. Convergence on all $\mathbb {R}$ of $u^n_1 \rightarrow 0$ in $L^\infty$.

Assume for the sake of contradiction that there is $\delta \in (0, \frac{1}{2})$, a sequence of points $x_n \in \mathbb {R}$ and a subsequence still denoted $u^n$ for simplicity of notations, such that $|u^n_1(x_n)| \geqslant \delta$ for every n.

Up to choosing a further subsequence, we can also assume $x_n\rightarrow x_\infty$ with $x_\infty \in [-\infty , \infty ]$. First observe that $x_\infty \in \{\pm \infty \}$. Indeed, if $x_\infty \in \mathbb {R}$, then choosing $R>0$ sufficiently large, we would have that $x_\infty \in (-R,R)$ and $u_1^n\rightarrow 0$ in $L^\infty ([-R,R])$ which contradicts the fact that $|u^n_1(x_n)| \geqslant \delta >0$ and $x_n\rightarrow x_\infty \in (-R,R)$. Therefore, $x_0=\pm \infty$: we will assume $x_0=+\infty$ (the other case follows similarly).

By Lemma B.2, we know that $m_1$ has limits belonging to $\{\pm 1\}$ as $x\rightarrow \pm \infty$; fix a large R, such that $|m_1|\geqslant \frac{1}{2}$ outside $(-R,R)$. As $x_n \rightarrow +\infty$ and $u_1^n\rightarrow 0$ in $L^\infty ([-R,R])$, there exists $N>0$ such that for every $n\geqslant N$, $x_n \geqslant 2R$ and $|u^n_1|\leqslant \frac{\delta }{2}$ in $[-R,R]$. As $|u^n_1(x_n)| \geqslant \delta$ and $x_n\geqslant 2R$, by continuity of $u^n_1$, there exists $y_n> R$ such that $|u^n_1(y_n)|=\delta$ for every $n\geqslant N$. Using $|m_1(y_n)|\geqslant \frac{1}{2}$ (because $y_n\notin [-R,R]$), we deduce that for $n\geqslant N$:

$$\begin{aligned} |2m_1(y_n) u^n_1(y_n) + (u^n_1)^2(y_n)| \geqslant 2 |m_1(y_n)| \delta -\delta ^2\geqslant \delta (1-\delta )\geqslant \frac{\delta }{2}. \end{aligned}$$

This contradicts (B.6).

This argument shows that every subsequence of $u^n_1$ has a further subsequence converging to the unique limit 0 in $L^\infty$; as $L^\infty$ is a Hausdorff space, we conclude that $u^n_1 \rightarrow 0$ in $L^\infty$. $\quad \square$

End of proof of point 2). We now prove smallness of the $H^1$ norm of u as stated in (B.4). By Lemma B.2, there exists $R \geqslant 1$ such that $|m_1| \geqslant 1/2$ for every $|x| \geqslant R$. Let $\varepsilon \in (0,1)$. The above claim implies the existence of $\delta =\delta (m) >0$ such that if $m+u: \mathbb {R} \rightarrow \mathbb {S}^2$ and $\Vert u \Vert _{\mathscr {H}^1} \leqslant \delta$, then $\Vert u \Vert _{L^\infty } \leqslant \frac{\varepsilon }{2\sqrt{R}}<\frac{1}{2}$. As $m+u$ and m are $\mathbb {S}^2$-valued maps, we have

$$\begin{aligned} 2m_1 u_1 =- u_1^2 + v, \quad v: = -2(m_2u_2 + m_3 u_3) - u_2^2 - u_3^2. \end{aligned}$$

Squaring the above equality, as $|m_1| \geqslant 1/2$ outside $(-R,R)$ and $|u_1|\leqslant |u| \leqslant \frac{1}{2}$, we get

$$\begin{aligned} u_1^2 \leqslant (2m_1 u_1)^2 \leqslant 2 (u_1^4 + v^2) \leqslant \frac{1}{2} u_1^2 + 2 v^2, \quad \text {for } |x| \geqslant R \end{aligned}$$

and so $u_1^2 \leqslant 4 v^2$ outside $(-R,R)$. We now integrate on $|x| \geqslant R$; as $|m|=1$ and $|u|^2\leqslant |u|\leqslant \frac{1}{2}$ in $\mathbb {R}$ yielding $|v|\leqslant 3(|u_2|+|u_3|)$ in $\mathbb {R}$, we obtain for some universal constant $C>1$:

$$\begin{aligned} \int _{|x| \geqslant R} u_1^2 \leqslant 4 \Vert v \Vert _{L^2}^2 \lesssim \Vert (u_2,u_3) \Vert _{L^2}^2 \leqslant (C-1) \Vert u \Vert _{\mathscr {H}^1}^2. \end{aligned}$$

Combined with

$$\begin{aligned} \int _{-R}^R u_1^2 \, dx \leqslant 2 R \Vert u_1 \Vert _{L^\infty }^2 \leqslant \frac{\varepsilon ^2}{2}, \end{aligned}$$

we get

$$\begin{aligned} \Vert u \Vert _{H^1}^2 = \Vert u_1 \Vert _{L^2}^2 + \Vert u \Vert _{\dot{H}^1}^2 \leqslant \frac{\varepsilon ^2}{2} + C \Vert u \Vert _{\mathscr {H}^1}^2 \leqslant \frac{\varepsilon ^2}{2} + C \delta ^2. \end{aligned}$$

Up to lowering $\delta >0$ further, we can also assume that $C \delta ^2 \leqslant \varepsilon ^2/2$, and so $\Vert u \Vert _{H^1} \leqslant \varepsilon$.

$\square$

As mentioned in Remark B.4, the case of a constant first coordinate should be considered separately, and for this, we have to take into account the following (mirror) symmetry in the first component

$$\begin{aligned} m^c := (-m_1,m_2,m_3). \end{aligned}$$

(B.7)

The following result holds for continuous maps m not necessarily belonging to $\mathscr {H}^1$.

Lemma B.6

Let $m=(m_1, m_2, m_3): \mathbb {R} \rightarrow \mathbb {S}^2$ be a continuous map with $m_1=c$ constant in $\mathbb {R}$.

• (1) If $c\ne 0$, then there exist $\delta ,C>0$ (depending on¹⁴c) such that if ${\tilde{m}}: \mathbb {R} \rightarrow \mathbb {S}^2$ satisfies $\Vert m - {\tilde{m}} \Vert _{\mathscr {H}^1} \leqslant \delta$, then

  $$\begin{aligned} \min ( \Vert m - {\tilde{m}} \Vert _{H^1}, \Vert m^c - {\tilde{m}} \Vert _{H^1}) \leqslant C \Vert m - {\tilde{m}} \Vert _{\mathscr {H}^1}. \end{aligned}$$

• (2) If $c=0$, then there exists a universal $C>0$ such that for every ${\tilde{m}}: \mathbb {R} \rightarrow \mathbb {S}^2$,

  $$\begin{aligned} \Vert m - {\tilde{m}} \Vert _{H^1} \leqslant C(\Vert m - {\tilde{m}} \Vert _{\mathscr {H}^1}+\Vert (m - {\tilde{m}})' \Vert ^{1/2}_{L^1}), \end{aligned}$$

  where we recall the notation $(m - {\tilde{m}})'=(m_2 - {\tilde{m}}_2, m_3 - {\tilde{m}}_3)$.

Proof

Denote $u:= {\tilde{m}}-m$. As in the proof of Proposition B.3, since m and ${\tilde{m}}$ are $\mathbb {S}^2$-valued, we get

$$\begin{aligned} -c^2\leqslant 2c u_1 + u_1^2 = v, \quad v: = -2 (m_2u_2 + m_3 u_3) - u_2^2 - u_3^2, \quad \text {in } \mathbb {R}\end{aligned}$$

so that

$$\begin{aligned} u_1 \in \bigg \{ - c \pm \sqrt{c^2 + v} \bigg \} \quad \text {in } \mathbb {R}. \end{aligned}$$

Case 1. $c=0$. In this case, $|(m_2,m_3)|=1$ (note that $m\notin \mathscr {H}^1$) and $|u_1| = \sqrt{v}$; then there is a constant $C>0$ (independent of m) such that

$$\begin{aligned} \Vert u_1\Vert ^2_{L^2}=\Vert v\Vert _{L^1}\leqslant C( \Vert (u_2,u_3) \Vert ^2_{L^2}+ \Vert (u_2,u_3) \Vert _{L^1}), \end{aligned}$$

which leads to conclusion 2).

Case 2. $c\ne 0$. As $|m|=1$ and $|u|^2\leqslant 2|u|\leqslant 4$ in $\mathbb {R}$, we obtain that

$$\begin{aligned} \Vert v \Vert _{L^2} \lesssim \Vert (u_2,u_3) \Vert _{L^2} \leqslant M \Vert u \Vert _{\mathscr {H}^1}\end{aligned}$$

(B.8)

for some universal constant $M>0$ (not depending on m). Also, as $\partial _x v=2c\partial _x u_1+ 2u_1 \partial _x u_1$ and $|u_1|\leqslant 2$, we deduce that $\Vert v \Vert _{\dot{H}^1}\lesssim \Vert u \Vert _{\mathscr {H}^1}$. By the Sobolev embedding $H^1\hookrightarrow L^\infty$, up to possibly increase the above constant M, it follows

$$\begin{aligned} \Vert v \Vert _{L^\infty }\leqslant M \Vert u \Vert _{\mathscr {H}^1}. \end{aligned}$$

Choosing $\delta \leqslant c^2/(2M)$, if $\Vert u \Vert _{\mathscr {H}^1} \leqslant \delta$, then the above estimate yields $\Vert v \Vert _{L^\infty } \leqslant c^2/2$, and in particular, $v+c^2\geqslant c^2/2>0$ in $\mathbb {R}$. As $u_1$ is a continuous function belonging to $\bigg \{ - c \pm \sqrt{c^2 + v} \bigg \}$, then there exists a sign $\sigma \in \{ \pm 1 \}$ such that

$$\begin{aligned} u_1 = -c + \sigma \sqrt{c^2 +v} = |c| (\sigma - {{\,\mathrm{\text {sgn}}\,}}(c)) - \sigma |c| \left( 1 - \sqrt{1+\frac{v}{c^2}} \right) \, \, \text { in } \mathbb {R}. \end{aligned}$$

Note that

$$\begin{aligned} \left| 1 - \sqrt{1+\frac{v}{c^2}} \right| \leqslant \frac{|v|}{c^2} \, \, \text { in } \mathbb {R}. \end{aligned}$$

Subcase i). $\sigma = {{\,\mathrm{\text {sgn}}\,}}(c)$. In this case, $|u_1| \leqslant \left| \frac{v}{c}\right|$ in $\mathbb {R}$ and we get by (B.8):

$$\begin{aligned}\Vert u_1 \Vert _{L^2} \leqslant \frac{M}{c} \Vert u \Vert _{\mathscr {H}^1}, \end{aligned}$$

and so

$$\begin{aligned} \Vert m-{\tilde{m}} \Vert _{H^1} = \Vert u \Vert _{H^1} \leqslant \left( 1+ \frac{M}{c} \right) \Vert u \Vert _{\mathscr {H}^1}. \end{aligned}$$

Subcase ii). $\sigma = -{{\,\mathrm{\text {sgn}}\,}}(c)$. In this case,

$$\begin{aligned} |u_1 +2c | \leqslant \left| \frac{v}{c}\right| , \end{aligned}$$

and similarly, we conclude

$$\begin{aligned}{} & {} \Vert m^c-{\tilde{m}} \Vert _{H^1} = \Vert (2c+u_1,u_2,u_3) \Vert _{H^1} \leqslant \left( 1+ \frac{M}{c} \right) \Vert u \Vert _{\mathscr {H}^1}. \end{aligned}$$

$\square$

We go back to magnetisations m with non constant first component. Actually, the smallness (B.4) in Proposition B.3 can be quantified to a linear bound under an additional nondegeneracy assumption on m

Lemma B.7

Let $m\in \mathscr {H}^1(\mathbb {R}, \mathbb {S}^2)$ be such that $m_1$ is non-constant and satisfies the nondegeneracy condition for some $c>0$:

$$\begin{aligned} |m_1| + |\partial _x m_1| \geqslant c, \quad \text {a.e. in } \, \mathbb {R}. \end{aligned}$$

Then there exist $\delta ,C>0$ (depending on m¹⁵) such that if ${\tilde{m}}: \mathbb {R} \rightarrow \mathbb {S}^2$ satisfies $\Vert m - {\tilde{m}} \Vert _{\mathscr {H}^1} \leqslant \delta$, then

$$\begin{aligned} \Vert m - {\tilde{m}} \Vert _{H^1} \leqslant C \Vert m - {\tilde{m}} \Vert _{\mathscr {H}^1}. \end{aligned}$$

We emphasise that the nondegeneracy assumption is verified for $w_*$¹⁶, and this is our purpose in Theorem 1.4.

Proof

Denote $u = {\tilde{m}} - m$ as before. If $u_1=0$, then we are done. Otherwise, we may assume that $\Vert u_1\Vert _{L^2}>0$. In the following, $M>0$ is a universal constant that can change from line to line. We fix $\varepsilon >0$ that is given later (it will be defined below on (B.9)). By Proposition B.3, there exists $\delta \in (0,1)$ so that $\Vert u\Vert _{\mathscr {H}^1}\leqslant \delta$ implies $\Vert u \Vert _{H^1} \leqslant \varepsilon$. By the constraint $|m|=|{\tilde{m}}|=1$, it follows

$$\begin{aligned} -m_1 u_1=\frac{1}{2} |u|^2 +m_2 u_2+m_3 u_3 \quad \text { in } \, \mathbb {R}\end{aligned}$$

yielding

$$\begin{aligned} \Vert m_1 u_1\Vert _{L^2} \leqslant \frac{1}{2} (\Vert u_1\Vert _{L^4}^2 + \Vert u_2\Vert _{L^4}^2+\Vert u_3\Vert _{L^4}^2)+ \Vert m_2u_2\Vert _{L^2} + \Vert m_3u_3\Vert _{L^2}. \end{aligned}$$

Using the Gagliardo-Nirenberg inequality $\Vert f\Vert _{L^4}\lesssim \Vert \partial _x f\Vert ^{1/4}_{L^2} \Vert f\Vert ^{3/4}_{L^2}$ and the Hölder inequality $\Vert m_ju_j\Vert _{L^2}\leqslant \Vert u_j\Vert _{L^4}\Vert m_j\Vert _{L^4}$ for $j=2, 3$, we deduce that

$$\begin{aligned} \Vert m_1 u_1\Vert _{L^2}&\leqslant \frac{1}{2}\Vert u_1\Vert _{L^\infty }\Vert u_1\Vert _{L^2}+M(\Vert m\Vert _{\mathscr {H}^1}\Vert u\Vert _{\mathscr {H}^1}+\Vert u\Vert ^2_{\mathscr {H}^1})\\&\leqslant \frac{1}{2} \varepsilon \Vert u_1\Vert _{L^2}+M(\Vert m\Vert _{\mathscr {H}^1}+1)\Vert u\Vert _{\mathscr {H}^1}. \end{aligned}$$

(We used $\Vert u\Vert ^2_{\mathscr {H}^1}\leqslant \Vert u\Vert _{\mathscr {H}^1}\leqslant \delta <1$ and the Sobolev embedding $\Vert u_1\Vert _{L^\infty } \leqslant \Vert u \Vert _{H^1} \leqslant \varepsilon$). Combined with the non-degeneracy condition and using again the Gagliardo-Nirenberg and Hölder inequality, we obtain

$$\begin{aligned} c \int _{\mathbb {R}} u_1^2 \, dx&\leqslant \int _{\mathbb {R}} (|m_1| + |\partial _x m_1|) u_1^2 \, dx \leqslant \Vert m_1 u_1\Vert _{L^2} \Vert u_1\Vert _{L^2}+\Vert \partial _x m_1\Vert _{L^2} \Vert u_1\Vert ^2_{L^4}\\&\leqslant \frac{1}{2} \varepsilon \Vert u_1\Vert ^2_{L^2}+M(\Vert m\Vert _{\mathscr {H}^1}+1)\Vert u\Vert _{\mathscr {H}^1} \Vert u_1\Vert _{L^2} +M \Vert m\Vert _{\mathscr {H}^1}\Vert u\Vert ^{1/2}_{\mathscr {H}^1} \Vert u_1\Vert ^{3/2}_{L^2}. \end{aligned}$$

We choose

$$\begin{aligned} \varepsilon := c. \end{aligned}$$

(B.9)

Taking $s> 0$ given¹⁷ by $s^2= \Vert u_1\Vert _{L^2}(\leqslant \Vert u_1\Vert _{H^1}\leqslant \varepsilon <\infty )$ and denoting $a:=M(\Vert m\Vert _{\mathscr {H}^1}+1)>0$, after dividing by $s^2$, we obtain

$$\begin{aligned} \frac{c}{2} s^2-a\Vert u\Vert ^{1/2}_{\mathscr {H}^1}s-a\Vert u\Vert _{\mathscr {H}^1}\leqslant 0. \end{aligned}$$

The discriminant of the above quadratic form in s is positive of order $O(\Vert u\Vert _{\mathscr {H}^1} (\Vert m \Vert _{\mathscr {H}^1}^2 + c^2))$, so both roots are real numbers of order $\Vert u\Vert ^{1/2}_{\mathscr {H}^1}$, and in particular $s=\Vert u_1\Vert ^{1/2}_{L^2}$ is of order $\Vert u\Vert ^{1/2}_{\mathscr {H}^1}$ which yields the conclusion. $\quad \square$

Remark B.8

If ${\tilde{m}}$ and m are not close in $\mathscr {H}^1$, then one can not expect any bound of the form

$$\begin{aligned} \Vert m - {\tilde{m}} \Vert _{H^1} \leqslant F(\Vert m -{\tilde{m}} \Vert _{\mathscr {H}^1}) \end{aligned}$$

for any function $F: \mathbb {R}_+ \rightarrow \mathbb {R}_+$ which takes finite values. To see this, let us give an example of a family of magnetisations m, ${\tilde{m}}$, such that $\Vert m -{\tilde{m}} \Vert _{\mathscr {H}^1}$ remains bounded, but $\Vert m - {\tilde{m}} \Vert _{H^1}$ is unbounded. For this, let $\theta$ be a smooth function in $\mathbb {R}$ such that $\theta (x) = 0$ for $x < -1$, $\theta (x) = \pi$ for $x > 1$ and $\theta$ is increasing on $[-1,1]$. Consider ${\tilde{m}} = (1,0,0)$ constant and m such that $m_1$ has two transitions from $-1$ to 1 (given by $\cos \theta$) separated at a distance of order $R>2$: for example, we can choose such $m_1$ to be

$$\begin{aligned} m_1(x) = {\left\{ \begin{array}{ll} \cos \theta (x+R) \quad \text {for} \quad x \leqslant 0, \\ \cos \theta (-x+R) \quad \text {for} \quad x \geqslant 0. \end{array}\right. } \end{aligned}$$

Let $m_2 = \sqrt{1-m_1^2} \in H^1$ and $m_3 =0$. Then $m \in \mathscr {H}^1$ and one sees that $\Vert m - {\tilde{m}} \Vert _{\mathscr {H}^1}$ is constant for large R, but of course $\Vert m - {\tilde{m}} \Vert _{L^2}=O(\sqrt{R}) \rightarrow +\infty$ as $R \rightarrow +\infty$.

Appendix C: Coercivity of a Schrödinger Operator

Lemma C.1

Let $L = - \Delta +V$ where $V \in L^\infty (\mathbb {R}^N)$ has the property that there exist $R>0$ and $c >0$ such that $V(x) \geqslant c$ for every $|x| \geqslant R$. Assume that there exists $\phi \in H^1(\mathbb {R}^N)$ such that$L \phi = 0$ in the sense of distributions and $\phi >0$ in $\mathbb {R}^N$. Then $\ker L=\mathbb {R}\phi$ and there exists $\lambda >0$ (small) such that for every $v \in H^1(\mathbb {R}^N)$, $0\leqslant (Lv,v)_{H^{-1}, H^1} \leqslant \frac{1}{\lambda }\Vert v\Vert ^2_{H^1}$,

$$\begin{aligned} (Lv,v)_{H^{-1}, H^1} \geqslant 4 \lambda \Vert v \Vert _{H^1}^2 - \frac{1}{\lambda } (v, \phi )^2_{L^2} \end{aligned}$$

(C.1)

and for every $v \in H^2(\mathbb {R}^N)$,

$$\begin{aligned} \Vert Lv\Vert ^2_{L^2} \geqslant 4 \lambda \Vert v \Vert _{H^2}^2 - \frac{1}{\lambda } (v, \phi )^2_{L^2}. \end{aligned}$$

(C.2)

Proof

Note that if $\phi \in H^1(\mathbb {R}^N)$ is a solution of $L\phi =0$, as $V\in L^\infty (\mathbb {R}^N)$, then $\phi \in H^2(\mathbb {R}^N)$. Moreover, standard elliptic regularity implies that $\phi \in W^{2,p}_{loc}(\mathbb {R}^N)$ for every $p<\infty$, in particular, $\phi \in C^1(\mathbb {R}^N)$. Therefore, the condition $\phi >0$ in $\mathbb {R}^N$ makes sense pointwise in $\mathbb {R}^N$. Also, we will assume without loss of generality that $\Vert \phi \Vert _{L^2}=1$.

Step 1. For every $v\in C^\infty _c(\mathbb {R}^N)$, set $w:=\frac{v}{\phi }\in C_c^1(\mathbb {R}^N)$ with compact support. Integrating by parts, we obtain:

$$\begin{aligned} \int _{\mathbb {R}^N} Lv\cdot v \, dx&= \int _{\mathbb {R}^N} \bigg (|\nabla v|^2 +V(x) v^2\bigg ) \, dx=\int _{\mathbb {R}^N} \bigg (|\nabla (\phi w)|^2 +V(x) \phi ^2 w^2\bigg )\, dx\\&=\int _{\mathbb {R}^N} \bigg ( \phi ^2 |\nabla w|^2+w^2 |\nabla \phi |^2 +\frac{1}{2} \nabla (\phi ^2) \cdot \nabla (w^2) +V(x) \phi ^2 w^2\bigg )\, dx\\&{=} \int _{\mathbb {R}^N} \bigg (\phi ^2 |\nabla w|^2 +w^2 |\nabla \phi |^2 +V(x) \phi ^2 w^2\bigg )\, dx -\frac{1}{2} (\Delta (\phi ^2), w^2) \\&=\int _{\mathbb {R}^N} \phi ^2 | \nabla w|^2 \, dx+ (L\phi , w^2 \phi )=\int _{\mathbb {R}^N} \phi ^2 | \nabla w|^2 \, dx \geqslant 0, \end{aligned}$$

because $L\phi =0$ and $\Delta (\phi ^2)=2\phi \Delta \phi +2|\nabla \phi |^2\in H^{-1}$ on any open bounded set. By density, for every $v\in H^1(\mathbb {R}^N)$, there exists a sequence $v_n\in C^\infty _c(\mathbb {R}^N)$ such that $v_n\rightarrow v$ and $\nabla v_n\rightarrow \nabla v$ in $L^2$ and a.e. in $\mathbb {R}^N$. In particular, $\nabla \big (\frac{v_n}{\phi } \big )\rightarrow \nabla \big (\frac{v}{\phi } \big )$ a.e. in $\mathbb {R}^N$. Since $V\in L^\infty (\mathbb {R}^N)$, it follows by Fatou’s lemma:

$$\begin{aligned} (Lv,v)=\lim _n (L v_n,v_n)=\liminf _n \int _{\mathbb {R}^N} \phi ^2 \big | \nabla \big (\frac{v_n}{\phi } \big )\big |^2 \, dx\geqslant \int _{\mathbb {R}^N} \phi ^2 \big | \nabla \big (\frac{v}{\phi } \big )\big |^2 \, dx\geqslant 0. \end{aligned}$$

As a consequence, if $v\in H^1$ belongs to $\ker L$, since $\phi >0$ in $\mathbb {R}^N$, then $v\in \mathbb {R}\phi$. Also, as $V\in L^\infty$, we conclude that $0\leqslant (Lv,v) \leqslant \max \big (1, \Vert V\Vert _{L^\infty }\big )\Vert v\Vert ^2_{H^1}$ for every $v\in H^1$.

Step 2. Let

$$\begin{aligned} a = \inf \{ (Lv,v) : v \in H^1, \Vert v \Vert _{L^2} =1, (v ,\phi )_{L^2} =0 \}. \end{aligned}$$

By Step 1, we know that $a\geqslant 0$. The aim is to prove that $a > 0$. For that, we consider a minimising sequence $(v_n)_n$ in $H^1$. As for $v \in H^1$,

$$\begin{aligned} (L|v|,|v|) = \int _{\mathbb {R}^N} \bigg (|\nabla v|^2 +V(x) v^2\bigg ) \, dx=(Lv,v), \end{aligned}$$

we can furthermore assume that $v_n \geqslant 0$ for all n. As for some $C>0$, $V \geqslant -C$ in $\mathbb {R}^N$, we infer that for any $v \in H^1$,

$$\begin{aligned} \Vert v \Vert _{H^1}^2 \leqslant (Lv,v) + (C+1) \Vert v \Vert _{L^2}^2, \end{aligned}$$

so that $(v_n)_n$ is bounded in $H^1$. Up to extracting a subsequence, we can assume that for some ${\underline{v}} \in H^1$, $v_n \rightharpoonup {\underline{v}}$ in $H^1$, $v_n \rightarrow {\underline{v}}$ in $L^2$ locally on compact sets and a.e. in $\mathbb {R}^N$ and $(Lv_n, v_n)\rightarrow a$ as $n\rightarrow \infty$. In particular, ${\underline{v}} \geqslant 0$. Now as $(v_n,\phi )_{L^2} =0$, since $v_n \rightharpoonup {\underline{v}}$ in $L^2$ and $\phi \in L^2$, we deduce $({\underline{v}},\phi )_{L^2}=0$, and as $\phi >0$ and ${\underline{v}}\geqslant 0$, we conclude ${\underline{v}}=0$ a.e. in $\mathbb {R}^N$.

Now we argue by contradiction and assume that $a=0$, that is $(Lv_n,v_n) \rightarrow 0$. As $V+(C+c) {\mathbb {1}_{|x| \leqslant R}} \geqslant c$, we get from strong $L^2$ convergence in B(0, R) of $v_n$ to ${\underline{v}}$, we see that

$$\begin{aligned} 0 \leqslant \int _{m R} \left( |\nabla v_n|^2 + c |v_n|^2 \right) dx&\leqslant \int _{\mathbb {R}} |\nabla v_n|^2 + ( V + (C+c) {\mathbb {1}_{|x| \leqslant R}}) |v_n|^2 dx \\&\leqslant (Lv_n,v_n) + (C+c) \int _{|x| \leqslant R} |v_n|^2 dx \rightarrow 0. \end{aligned}$$

This implies that $\Vert v_n \Vert _{H^1} \rightarrow 0$, and the convergence $v_n \rightarrow {\underline{v}} =0$ is strong in $H^1$. But $\Vert v_n \Vert _{L^2} =1$, so this strong convergence also implies $\Vert {\underline{v}} \Vert _{L^2} =1$, a contradiction. Hence $a >0$.

Step 3. We now prove inequality (C.1). By Step 2 and homogeneity, we get that if $v \in H^1$ and $(v,\phi )_{L^2}=0$, then

$$\begin{aligned} (Lv,v) \geqslant a \Vert v \Vert _{L^2}^2. \end{aligned}$$

Then, still assuming that $(v,\phi )=0$, since $V \geqslant -C$ in $\mathbb {R}^N$, there holds for $b:=\frac{a}{a+C+1}\in (0,1)$:

$$\begin{aligned} (Lv,v)&= b \int _{\mathbb {R}^N} |\nabla v|^2 + (1-b) (Lv,v) + b \int _{\mathbb {R}^N} V v^2 \nonumber \\&\geqslant b \int _{\mathbb {R}^N} |\nabla v|^2 + \int _{\mathbb {R}^N} v^2 (a(1-b) + b V) \geqslant b \Vert v \Vert _{H^1}^2, \end{aligned}$$

(C.3)

because $a(1-b) + b V \geqslant a(1-b) -b C = b$.

Let $v \in H^1$ (no longer assuming the orthogonality condition). We define the $L^2$ orthogonal decomposition $w:=v- (v, \phi )_{L^2} \phi$, so that $(w,\phi )_{L^2} =0$. As $L\phi =0$ and L is self-adjoint, we compute using (C.3):

$$\begin{aligned} (Lv,v)&=(Lw,w)+2(v,\phi ) (L\phi , w)+(v,\phi )^2 (L\phi ,\phi )\\&= (Lw,w) \geqslant b \Vert w \Vert _{H^1}^2 = b \Vert v - (v, \phi ) \phi \Vert _{H^1}^2 \\&\geqslant b (\Vert v \Vert _{H^1} - |(v,\phi )|\Vert \phi \Vert _{H^1})^2 \geqslant b \left( \frac{1}{2} \Vert v \Vert _{H^1}^2 - |(v,\phi )|^2\Vert \phi \Vert _{H^1}^2 \right) . \end{aligned}$$

where we used $(x-y)^2 \geqslant \frac{1}{2} x^2 - y^2$. The desired inequality follows for $\lambda := \min \big (\frac{1}{b\Vert \phi \Vert ^2_{H^1}}, \frac{b}{8}\big )$.

Step 4. We finally prove estimate (C.2), with a similar strategy as (C.1). By Step 2, we know that for every $v\in H^2(\mathbb {R}^N)$ with $(v, \phi )_{L^2}=0$,

$$\begin{aligned} \Vert Lv\Vert _{L^2} \Vert v\Vert _{L^2}\geqslant (Lv,v)\geqslant a\Vert v\Vert _{L^2}^2 \end{aligned}$$

yielding $\Vert Lv\Vert _{L^2}\geqslant a \Vert v\Vert _{L^2}$ with $a>0$.

Let ${\tilde{b}} = \frac{a}{1+2a+2\Vert V \Vert _{L^\infty }} >0$, then, as in Step 3, we compute for $v\in H^2(\mathbb {R}^N)$ such that $(v, \phi )_{L^2}=0$:

$$\begin{aligned} \Vert Lv\Vert ^2_{L^2}&= 2 {\tilde{b}} \Vert \Delta v \Vert _{L^2}^2 + (1-2{\tilde{b}}) \Vert Lv \Vert _{L^2}^2 + 4 {\tilde{b}} (\Delta v,Vv) + 2 {\tilde{b}} \Vert V v \Vert _{L^2}^2 \\&\geqslant {\tilde{b}} \Vert \Delta v \Vert _{L^2}^2 + a (1-2{\tilde{b}}) \Vert v \Vert _{L^2}^2 - 2 {\tilde{b}} \Vert V v \Vert _{L^2}^2 \\&\geqslant {\tilde{b}} \Vert \Delta v \Vert _{L^2}^2 + (a (1-2{\tilde{b}}) -2\Vert V \Vert _{L^\infty } {\tilde{b}}) \Vert v \Vert _{L^2}^2 \\&\geqslant {\tilde{b}} (\Vert \Delta v \Vert _{L^2}^2 + \Vert v \Vert _{L^2}^2). \end{aligned}$$

because $(a (1-2{\tilde{b}}) -2\Vert V \Vert _{L^\infty } {\tilde{b}}) = {\tilde{b}}$. As $\Vert \Delta v \Vert _{L^2} + \Vert v \Vert _{L^2}$ controls the $\Vert v \Vert _{H^2}$ norm, we can conclude that for some $b'>0$,

$$\begin{aligned} \Vert Lv \Vert _{L^2} \geqslant b' \Vert v \Vert _{H^2}. \end{aligned}$$

For general $v \in H^2$ (without assuming that $(v, \phi )_{L^2}=0$), we set $w:=v- (v, \phi ) \phi$. Then $w\in H^2$ (as $\phi \in H^2$) with $(w, \phi )_{L^2}=0$ and since $Lv=Lw$, the desired inequality (C.2) follows as in Step 3. $\quad \square$

Appendix D: BV Curves to $\mathbb {S}^2$

We claimed in the sketch of proof of Theorem 4.1 that any $\mathbb {S}^2$ valued map in $H^1([-A,A])$ can not have an image which is dense in any cap. Performing a change a chart, it suffices prove that a map in $H^1([0,1], [0,1]^2)$ does not have a dense image (in $[0,1]^2$). We actually provide a short quantitative proof in the slightly more general setting of BV maps.

Lemma D.1

Let $m \in BV([0,1],[0,1]^2)$. Then m([0, 1]) is not dense in $[0,1]^2$.

More precisely, there exists a square Q of length $\frac{1}{5(\Vert m \Vert _{BV}+1)}$ such that $Q \cap m([0,1]) = \varnothing$.

Proof

Let $n \geqslant 2$ be an integer to be fixed later, and consider a partition of $[0,1[^2$ into $n^2$ squares $(Q_j)_{1 \leqslant j \leqslant n^2}$ of length 1/n, of the form $[\alpha /n, (\alpha +1) /n[ \times [\beta /n, (\beta +1) /n[$ for $0 \leqslant \alpha , \beta \leqslant n-1$.

For all $1 \leqslant j \leqslant n^2$, denote $Q_j'$ the smaller square of length 1/(3n) which has the same center as $Q_j$: if $Q_j = [\alpha /n, (\alpha +1) /n[ \times [\beta /n, (\beta +1) /n[$ then $Q_j' = [(\alpha +1/3) /n, (\alpha +2/3) /n[ \times [(\beta +1/3) /n, (\beta +2/3) /n[$.

Assume that for all $1 \leqslant j \leqslant n^2$, $Q_j' \cap m([0,1]) \ne \varnothing$, and consider $y_j \in [0,1]$ such that $m(y_j) \in Q_j' \cap m([0,1])$, and $(x_j)_j$ is the reordering of the $(y_j)_j$ that is $0 \leqslant x_1< x_2< \cdots < x_{n^2} \leqslant 1$.

By construction, for any $1 \leqslant j <k \leqslant n^2$, the distance between $Q_j'$ and $Q_k'$ is at least 2/(3n) so that $d(m(x_j), m(x_{j+1})) \geqslant 2/(3n)$. Therefore, by definition of the BV norm,

$$\begin{aligned} \Vert m \Vert _{BV} \geqslant \sum _{j=1}^{n^2-1} d(m(x_{j}), m(x_{j+1})) \geqslant \sum _{j=1}^{n^2-1} \frac{2}{3n} \geqslant \frac{2}{3} \left( n - \frac{1}{n} \right) \geqslant \frac{2n-1}{3}. \end{aligned}$$

(Recall $n \geqslant 2$). As a consequence, $\Vert m \Vert _{BV} \geqslant 1$, and for any integer $n > (3 \Vert m \Vert _{BV}+ 1)/2$, there exists j such that the $Q_j'$ such that $Q_j' \cap m([0,1]) = \varnothing$. The conclusion follows by choosing the integer $n_0$ in the interval $((3 \Vert m \Vert _{BV}+ 1)/2, 3 (\Vert m \Vert _{BV}+ 1)/2]$ (notice that $n_0 \geqslant 3$). Then $Q_j'$ has length

$$\begin{aligned}{} & {} \frac{1}{3n_0} \geqslant \frac{2}{9(\Vert m \Vert _{BV}+ 1)} \geqslant \frac{1}{5(\Vert m \Vert _{BV}+1)}. \end{aligned}$$

$\square$