Results on the Spectral Stability of Standing Wave Solutions of the Soler Model in 1-D

Abstract

We study the spectral stability of the nonlinear Dirac operator in dimension $1+1$, restricting our attention to nonlinearities of the form $f(\left\langle \psi ,\beta \psi \right\rangle _{\mathbb {C}^2}) \beta$. We obtain bounds on eigenvalues for the linearized operator around standing wave solutions of the form $e^{-i\omega t} \phi _0$. For the case of power nonlinearities $f(s)= s |s|^{p-1}$, $p>0$, we obtain a range of frequencies $\omega$ such that the linearized operator has no unstable eigenvalues on the axes of the complex plane. As a crucial part of the proofs, we obtain a detailed description of the spectra of the self-adjoint blocks in the linearized operator. In particular, we show that the condition $\left\langle \phi _0,\beta \phi _0 \right\rangle _{\mathbb {C}^2} > 0$ characterizes groundstates analogously to the Schrödinger case.

Appendices

Appendix A. ODE arguments

In this appendix, we give the proofs of Propositions 2.1 (basic properties of the groundstates) and 2.5 because some intermediate steps are useful in both proofs. However, we stress that the proof of Proposition 2.5 requires spectral properties of $L_2$ that have been established in Proposition 2.2 and Lemma 2.3.

Proof of Proposition 2.1

Equation (3) satisfied by $\phi _0(\omega ) = (v_\omega , u_\omega )^{\textsf{T}}$ can be written as

$$\begin{aligned} {\left\{ \begin{array}{ll} v_\omega '= - (M_\omega +\omega ) u_\omega \\ u_\omega '= -(M_\omega -\omega ) v_\omega , \end{array}\right. } \end{aligned}$$

(50)

with $M_\omega = m - f\!\left( v_\omega ^2 - u_\omega ^2\right)$ as defined in (9). Therefore, we assume $v_\omega (0)\geqslant 0$ since $-\phi _0(\omega )$ is a solution if and only if $\phi _0(\omega )$ is a solution. We define $F(s) := \int _0^s f(t) \,\textrm{d}t$ on $(0,+\infty )$ and ${\tilde{F}}\in {\mathcal {C}}^0([0,+\infty ))\cap {\mathcal {C}}^1((0,+\infty ))$ by ${\tilde{F}}(0)=0$ and ${\tilde{F}}(s) := F(s)/s$ on $(0,+\infty )$. The continuity at the origin derives from $f'>0$ on $(0,+\infty )$ by Assumption 1.1, as it gives

$$\begin{aligned} 0<F(s) = \int _0^s f(t) \,\textrm{d}t < s f(s) \quad \text { on } (0,+\infty ), \end{aligned}$$

(51)

where we used the assumption $f(0)=0$ for the positivity of F.

Existence, positivity and regularity. The existence is established in [2, Lemma 3.2]. In its notation, we have $g(s) = m - f(s)$ and $G(s) = m s - F(s)$, and have to check that

$$\begin{aligned} \exists \, s^*_\omega>0, \, \forall \, s \in (0, s^*_\omega ), \, \frac{G(s)}{s} > \omega = \frac{G(s^*_\omega )}{s^*_\omega } \ne g(s^*_\omega ). \end{aligned}$$

First, $G(s)/s>g(s)$ on $(0,+\infty )$ by (51)—hence checking already, for any $s^*_\omega >0$, the non-equality—, which implies $\lim _{s\searrow 0} G(s)/s \geqslant g(0) = m > \omega$. Second, denoting $f^{-1}$ the inverse of $f:[0,+\infty )\rightarrow [0,\lim _{+\infty } f)$, $\lim _\infty f \geqslant m$ from Assumption 1.1 gives

$$\begin{aligned} \forall \, \epsilon>0, \, \forall \, s> f^{-1}(m-\epsilon ), \quad {\tilde{F}}(s) \geqslant \frac{1}{s} \int _{f^{-1}(m-\epsilon )}^s f(t) \,\textrm{d}t > \frac{s-f^{-1}(m-\epsilon )}{s} (m-\epsilon ). \end{aligned}$$

Thus $\lim _{+\infty } {\tilde{F}} \geqslant m$, and $\lim _{s\rightarrow +\infty } G(s)/s \leqslant 0 < \omega$. Third, $s^2 {\tilde{F}}'(s)= s f(s) - F(s)>0$ on $(0,+\infty )$ by (51). Summarizing, $G(s)/s = m - {\tilde{F}}(s)$ is strictly decreasing on $(0,+\infty )$ with $\omega$ is in its image set, hence there is a unique $s^*_\omega >0$ satisfying the properties. The existence of a non-zero solution $(v_\omega , u_\omega ) \in H^1(\mathbb {R},\mathbb {R}^2)$ to (3) with $v_\omega$ is even and $u_\omega$ odd is therefore proved by [2, Lemma 3.2]. Moreover, since $u_\omega (0)=0$ and $v_\omega (0)\geqslant 0$, we can assume that $v_\omega (0)>0$ as, otherwise, $\phi _0(\omega )$ is the trivial solution by the Cauchy theory.

   To prove (i), notice that (50) is the Hamiltonian system $h_\omega (u_\omega ,v_\omega )=0$ associated to

$$\begin{aligned} h_\omega (u,v) := \frac{1}{2} \left( \omega \left( v^2 + u^2\right) - m \left( v^2 - u^2\right) + F\circ \left( v^2 - u^2\right) \right) . \end{aligned}$$

On one hand, since $f>{\tilde{F}}$ on $(0,+\infty )$ by (51) and $v_\omega (0)>0$, (50) implies

$$\begin{aligned} u_\omega '(0) = (\omega - m + f\circ v_\omega ^2(0) ) v_\omega (0)> \left( \omega - m + {\tilde{F}}\circ v_\omega ^2(0)\right) v_\omega (0) = (m - \omega )v_\omega (0) > 0. \end{aligned}$$

On another hand, it also gives $u_\omega (x) \ne 0$ if $x \ne 0$ since, otherwise, there would be $x >0$ (by oddity of u) such that $(v_\omega (x), u_\omega (x))= (v_\omega (0), u_\omega (0))$ and the solution would be periodic (in x), contradicting $\phi _0(\omega ) \in L^2(\mathbb {R})$. Hence, we conclude that $u_\omega >0$ on $(0,+\infty )$, since $u_\omega$ is continuous, non-zero on $(0,+\infty )$ with $u_\omega '(0)>0$ and $u_\omega (0)=0$.

   Moreover, $d_\omega :=v_\omega ^2 - u_\omega ^2>0$ on $\mathbb {R}$. Indeed, $d_\omega (0)=v_\omega (0)^2>0$ and, if $d_\omega (x)=0$ for some $x\ne 0$, then $0=h_\omega (u_\omega ,v_\omega )(x) = \frac{\omega }{2} (v_\omega (x)^2 + u_\omega (x)^2)$ hence $v_\omega (x)= u_\omega (x) = 0$ contradicting $u_\omega \ne 0$ on $\mathbb {R}{\setminus }\{0\}$ just obtained. Finally, $v_\omega> u_\omega >0$ on $(0,+\infty )$ because $|v_\omega |> u_\omega >0$ on $(0,+\infty )$, since $d_\omega >0$ on $\mathbb {R}$, with $v_\omega$ continuous and $v_\omega (0)>0$.

Since $(v_\omega , u_\omega )\in H^1(\mathbb {R})\subset {\mathcal {C}}^0(\mathbb {R})$ and $f\in {\mathcal {C}}^0(\mathbb {R})$ by Assumption 1.1, (50) gives $(v_\omega , u_\omega )\in {\mathcal {C}}^1(\mathbb {R})$. Moreover, $f\circ d_\omega \in {\mathcal {C}}^1(\mathbb {R})$ since $d_\omega >0$ on $\mathbb {R}$ and $f\in {\mathcal {C}}^1(\mathbb {R}{\setminus }\{0\})$ by Assumption 1.1, hence (50) actually gives $(v_\omega , u_\omega )\in {\mathcal {C}}^2(\mathbb {R})$.

Uniqueness and continuity in $\omega$ of the initial condition. We have

$$\begin{aligned} h_\omega (v_\omega (0),0)=0 \quad \Leftrightarrow \quad {\tilde{F}}(v_\omega ^2(0)) = m - \omega . \end{aligned}$$

Moreover, we already prove that ${\tilde{F}}\in {\mathcal {C}}^0([0,+\infty ))\cap {\mathcal {C}}^1((0,+\infty ))$ is strictly increasing on $[0,+\infty )$ with $\lim _{+\infty } {\tilde{F}} \geqslant m$. Thus, it is bijective from $[0,+\infty )$ to $[0,\lim _{+\infty } {\tilde{F}})$ and has a continuous inverse with $v_\omega (0)^2 = {\tilde{F}}^{-1}(m-\omega )>0$. By the Cauchy theory, this implies the uniqueness of $\phi _0(\omega )$.

Pointwise decay of $\phi _0(\omega ,\cdot )$ as $\omega \rightarrow m$. We start by noticing that $h_\omega (u_\omega ,v_\omega ) = 0$ gives

$$\begin{aligned} 0< \omega (v_\omega ^2 + u_\omega ^2) < m (v_\omega ^2 - u_\omega ^2) = m d_\omega \quad \text { on } \mathbb {R}, \end{aligned}$$

(52)

due to $F >0$ on $(0,+\infty )$, $v_\omega ^2(0)> 0 = u_\omega ^2(0)$, and the parities of $v_\omega$ and $u_\omega$. Moreover, $\left| \left| d_\omega \right| \right| _{L^\infty } = d_\omega (0)$, because $(d_\omega )' = - 4 \omega u_\omega v_\omega$ by (50) and $v_\omega>u_\omega >0$ on $(0,+\infty )$, hence

$$\begin{aligned} \frac{\omega }{m} \left| \left| \phi _0(\omega , \cdot ) \right| \right| ^2_{L^\infty } = \frac{\omega }{m} \left| \left| v_\omega \right| \right| ^2_{L^\infty } \leqslant \frac{\omega }{m} \left| \left| v_\omega ^2+u_\omega ^2 \right| \right| _{L^\infty } \leqslant \left| \left| d_\omega \right| \right| _{L^\infty } = d_\omega (0) = {\tilde{F}}^{-1}(m-\omega ), \end{aligned}$$

for $\omega \in (0,m)$. Since ${\tilde{F}}^{-1}(0) = 0$, this establishes half of (iv).

Uniform exponential decay of $\phi _0(\omega ,\cdot )$. In order to prove differentiability, it is important to obtain a pointwise upper bound uniform in $\omega$, for $\omega$ bounded away from m and 0. Thus, for fixed $\epsilon > 0$, we assume that $m^2-\omega ^2 \geqslant \epsilon ^2 > 0$. In view of (52), it is sufficient to bound $d_\omega$, and by symmetry we assume that $x \geqslant 0$. We have

$$\begin{aligned} (d_\omega )' = - 4 \omega u_\omega v_\omega = -2\omega \left( (v_\omega ^2 + u_\omega ^2)^2 - d_\omega ^2 \right) ^{1/2} = - 2\left( \left( m d_\omega - F(d_\omega ) \right) ^ 2 - \omega ^2 d_\omega ^2 \right) ^{1/2}, \end{aligned}$$

using $h(v_\omega , u_\omega ) = 0$ for the last equality. Since $\lim _{s\searrow 0} (m - {\tilde{F}}(s)) \geqslant m > \omega$ (see the proof of existence and remember that $m - {\tilde{F}}(s)=G(s)/s$), we define $s_\epsilon$ such that

$$\begin{aligned} \left( \left( m s - F(s) \right) ^ 2 - \omega ^2 s^2 \right) ^{1/2} = s \left( \left( m - {\tilde{F}}(s) \right) ^ 2 - \omega ^2 \right) ^{1/2} \geqslant \frac{\epsilon }{2} s, \quad \forall \, s \in [0,s_\epsilon ] . \end{aligned}$$

Since $d_\omega$ is strictly decreasing on $[0,+\infty )$ hence bijective on it, we define $x^*(\omega ,\epsilon )$ by $x^*(\omega ,\epsilon ) = 0$ if $d_\omega (0) \leqslant s_\epsilon$ and $x^*(\omega ,\epsilon ) = d_\omega ^{-1}(s_\epsilon )$ otherwise, and we have

$$\begin{aligned} (d_\omega )' (x)\leqslant - \epsilon d_\omega (x), \quad \forall \, x \geqslant x^*(\omega ,\epsilon ). \end{aligned}$$

Integrating it, yields (ii) since it gives

$$\begin{aligned} d_\omega (x) \leqslant d_\omega (x^*(\omega ,\epsilon )) \exp {\left( - \epsilon (x - x^*(\omega ,\epsilon )) \right) } \leqslant s_\epsilon \exp {\left( - \epsilon (x - x^*(\omega ,\epsilon )) \right) }. \end{aligned}$$

Decays of Q. By the definition of Q, the positivity of $d_\omega$, the one of $f'$ on $(0,+\infty )$ from Assumption 1.1, and (52), we bound

$$\begin{aligned} \left| \left| Q_\omega (x) \right| \right| _{\mathbb {C}^2 \mapsto \mathbb {C}^2} = (v_\omega ^2(x) + u_\omega ^2(x)) f'(d_\omega (x)) \leqslant \frac{m}{\omega } d_\omega (x) f'(d_\omega (x)). \end{aligned}$$

It yields (iii), the exponential decay in x, due to the exponential decay of $v_\omega$ and $u_\omega$, and to $\lim _{s\rightarrow 0^+} s f'(s) = 0$ from Assumption 1.1. It also gives the statement on Q in (iv):

$$\begin{aligned} \lim _{\omega \rightarrow m} \sup _{x \in \mathbb {R}}\left| \left| Q_\omega (x) \right| \right| _{\mathbb {C}^2 \mapsto \mathbb {C}^2} \leqslant \lim _{\omega \rightarrow m} \frac{m}{\omega } \sup _{s \in (0, v_\omega ^2(0))} s f'(s) = 0. \end{aligned}$$

$\square$

Proof of Proposition 2.5

We freely use the notation introduced in the previous proof. We remind the reader that the following proof of differentiability of $\omega \mapsto \phi _0(\omega )$ is needed because we do not assume continuity of $f'$ at 0.

Uniform continuity in $\omega$. We show that $\omega \mapsto \phi _0(\omega )$ is continuous with values in ${\mathcal {C}}^0(\mathbb {R})$. We rewrite the nonlinear equation (3) as an initial value problem on $[0, +\infty )$ of the form

$$\begin{aligned} \partial _x \phi _0(\omega ) = B(\omega , \phi _0(\omega )) \phi _0(\omega ), \quad \phi _0(\omega , 0) = \left( {\tilde{F}}^{-1}(m-\omega ) , 0 \right) ^{\textsf{T}}, \end{aligned}$$

with

$$\begin{aligned} B(\omega , \phi ) = i \omega \sigma _2 - (m - f(\left\langle \phi ,\sigma _3 \phi \right\rangle _{\mathbb {C}^2}))\sigma _1 . \end{aligned}$$

Here, $B(\omega , \phi )$ is of class ${\mathcal {C}}^1$ in $\omega$ and $\phi$ (as long as $\phi _1^2 - \phi _2^2$ is bounded away from zero, which is the case on any bounded interval), and thus $\omega \mapsto \phi _0(\omega ,x)$ is continuous in $\omega$, uniformly for x in bounded intervals.

   We now use the exponential decay from the previous proof. The pointwise continuity implies that $x^*(\omega , \epsilon )$ is bounded for $\omega \in [\epsilon , \sqrt{m^2-\epsilon ^2}]$, since otherwise there would be $\omega ^*$ such that $d_{\omega ^*}(x) > s_\epsilon$ for all $x \geqslant 0$. Thus, there exist $r_\epsilon$ and $A_\epsilon$ such that

$$\begin{aligned} \left| \left| \phi _0(\omega , x) \right| \right| _{\mathbb {C}^2} \leqslant A_\epsilon \exp (-\epsilon |x|)\, \quad \text { for all } \quad \omega \in [\epsilon , \sqrt{m^2-\epsilon ^2}] \text { and } x \in \mathbb {R}. \end{aligned}$$

The restriction that $\omega \geqslant \epsilon$ comes from (52). Combined with the uniform continuity on bounded intervals, this shows that $\omega \mapsto \phi _0(\omega )$ is continuous with values in ${\mathcal {C}}^0(\mathbb {R})$.

Differentiability. We fix $\epsilon > 0$ and $\omega \in [\epsilon , \sqrt{m^2-\epsilon ^2}]$. Assuming $\alpha \in [\epsilon , \sqrt{m^2-\epsilon ^2}]$, substracting the groundstate equations for $\omega$ and $\alpha$ gives

$$\begin{aligned} 0=&{} \left( D_m - \frac{\omega + \alpha }{2}{{\,{{\,\mathrm{\mathbbm {1}}\,}}\,}}\right) (\phi _{0}(\omega ) - \phi _{0}(\alpha ) ) -(\omega -\alpha )\frac{\phi _{0}(\omega ) + \phi _{0}(\alpha ) }{2} \\{}&{}\qquad - (f\circ d_\omega ) \sigma _3 \phi _{0}(\omega ) + (f\circ d_\alpha ) \sigma _3 \phi _{0}(\alpha ). \end{aligned}$$

For fixed $x \in \mathbb {R}$, we apply the mean value theorem to f in order to find ${{\tilde{d}}}(x)$, between $d_\alpha (x)$ and $d_\omega (x)$, such that

$$\begin{aligned}{} & {} (f\circ d_\omega ) \sigma _3 \phi _{0}(\omega ) - (f\circ d_\alpha ) \sigma _3 \phi _{0}(\alpha )\\{} & {} \quad = f'({{\tilde{d}}})(d_\omega - d_\alpha )\sigma _3 \frac{\phi _{0}(\omega ) + \phi _{0}(\alpha ) }{2} + \frac{f\circ d_\omega + f\circ d_\alpha }{2}\sigma _3(\phi _{0}(\omega ) - \phi _{0}(\alpha )) \\{} & {} \quad = 2 {\widetilde{Q}}_\alpha (\phi _{0}(\omega ) - \phi _{0}(\alpha )) + \frac{f\circ d_\omega + f\circ d_\alpha }{2}\sigma _3(\phi _{0}(\omega ) - \phi _{0}(\alpha )), \end{aligned}$$

where we have defined

$$\begin{aligned} {\widetilde{Q}}_\alpha := f'({{\tilde{d}}}) \left( \sigma _3 \frac{\phi _{0}(\omega ) + \phi _{0}(\alpha ) }{2} \right) \left( \sigma _3 \frac{\phi _{0}(\omega ) + \phi _{0}(\alpha ) }{2} \right) ^{\textsf{T}}. \end{aligned}$$

Inserting this in the previous identity, we obtain

$$\begin{aligned} (\omega -\alpha )\frac{\phi _{0}(\omega ) + \phi _{0}(\alpha ) }{2}&= \left( D_m - \frac{\omega + \alpha }{2}{{\,\mathrm{\mathbbm {1}}\,}}- \frac{f\circ d_\omega + f\circ d_\alpha }{2} \sigma _3 - 2 {\widetilde{Q}}_\alpha \right) (\phi _{0}(\omega ) - \phi _{0}(\alpha ))\\&=: {\tilde{L}}_{2, \alpha } (\phi _{0}(\omega ) - \phi _{0}(\alpha )). \end{aligned}$$

We claim that ${\tilde{L}}_{2, \alpha }$ converges to $L_2(\omega )$, in ${\mathcal {B}}(H^1(\mathbb {R}), L^2(\mathbb {R}))$, as $\alpha$ tends to $\omega$. For the first two terms, this is clear. For the third term, we use

$$\begin{aligned} \left| \left| (f\circ d_\omega - f\circ d_\alpha ) \Psi \right| \right| _{L^2} \leqslant \left| \left| f\circ d_\omega - f\circ d_\alpha \right| \right| _{L^\infty } \left| \left| \Psi \right| \right| _{L^2} \end{aligned}$$

and the uniform continuity from the previous proof. For the last term, we use the uniform convergence on bounded intervals (where $d_\omega$ and $d_\alpha$ are bounded away from zero). In order to treat large values of x, we factorize out ${\tilde{d}}$ in the definition of ${\widetilde{Q}}_\alpha$:

$$\begin{aligned} {\widetilde{Q}}_\alpha = {\tilde{d}} f'({\tilde{d}}) \left( \sigma _3 \frac{\phi _{0}(\omega ) + \phi _{0}(\alpha ) }{2 \sqrt{{\tilde{d}}}} \right) \left( \sigma _3 \frac{\phi _{0}(\omega ) + \phi _{0}(\alpha ) }{2 \sqrt{{\tilde{d}}}} \right) ^{\textsf{T}}, \end{aligned}$$

where the factors in parenthesis are bounded in view of (52) and their pre-factor ${\tilde{d}} f'({\tilde{d}})$ vanishes at infinity since $\lim _{s\rightarrow 0^+} s f'(s) = 0$ by Assumption 1.1.

   The convergence in ${\mathcal {B}}(H^1(\mathbb {R}), L^2(\mathbb {R}))$ implies norm resolvent convergence and, in particular, convergence of the spectrum. Recall that for each fixed $\omega$, $L_2(\omega )$ has zero as an isolated simple eigenvalue with an odd eigenfunction. Therefore, for sufficiently small $\left| \alpha -\omega \right|$,

$$\begin{aligned} {\tilde{L}}_{2, \alpha } : H^{1,\textrm{even}}(\mathbb {R}) \mapsto L^{2,\textrm{even}}(\mathbb {R}) \end{aligned}$$

is invertible and we conclude that

$$\begin{aligned} \frac{\phi _0(\omega )- \phi _0(\alpha )}{\omega -\alpha } = ({\tilde{L}}_{2, \alpha })^{-1} \frac{\phi _{0}(\omega ) + \phi _{0}(\alpha ) }{2}. \end{aligned}$$

This shows the differentiability of $\omega \mapsto \phi _0(\omega )$ as a function with values in $H^{1}(\mathbb {R},\mathbb {C}^2)$, with $\partial _\omega \phi _0(\omega ) = \left( L_2( \omega )\right) ^{-1} \phi _0(\omega )$. This function is continuous in $\omega$ because $\left( L_2( \omega )\right) ^{-1}$ is bounded on $L^{2,\textrm{even}}(\mathbb {R})$. Taking the inner product with $\phi _0(\omega )$ concludes the proof. $\square$

Appendix B. Proof of Lemma 4.4

We start from (18), where we introduce the parameter $\theta :=\frac{\mu \left| \left| \left| Q \right| \right| \right| }{4(t+\omega )}$ for shortness, and study the function

$$\begin{aligned} g_\theta (\alpha ,\eta ) := (1-\alpha )\eta ^2 + (1+\alpha ) - \frac{4\eta \theta }{1+\alpha }. \end{aligned}$$

Since we do not know the value of $\eta >0$, we need to study the question: Given $\xi \geqslant 0$, for which $\theta >0$ the inequality

$$\begin{aligned} \inf _{\eta >0}\max _{\alpha \in [0,1]} g_\theta (\alpha ,\eta ) \geqslant \xi \end{aligned}$$

is verified? Notice that

$$\begin{aligned} \partial _\alpha g_\theta (\alpha ,\eta ) = -\eta ^2 + \frac{4 \theta }{(1+\alpha )^2}\eta + 1 \end{aligned}$$

is a decreasing function of $\alpha >0$ for any $\eta >0$. Thus,

$$\begin{aligned} \partial _\alpha g_\theta (\alpha ,\eta ) \geqslant \partial _\alpha g_\theta (1,\eta ) = -\eta ^2 + \theta \eta + 1. \end{aligned}$$

Defining $\eta _\star$ as the positive number such that $-\eta _\star ^2 +\theta \eta _\star +1=0$, i.e.,

$$\begin{aligned} \eta _\star = \frac{ \theta + \sqrt{\theta ^2 + 4}}{2} > 1, \end{aligned}$$

we have for all $\eta \in (0,\eta _\star ]$ that

$$\begin{aligned} \partial _\alpha g_\theta (\alpha ,\eta ) \geqslant -\eta ^2 + \theta \eta + 1 \geqslant -\eta _\star ^2 + \theta \eta _\star + 1 = 0. \end{aligned}$$

Thus, for $\eta \in (0,\eta _\star ]$, $\alpha \mapsto g_\theta (\alpha ,\eta )$ is an increasing function on [0, 1] and its maximum is $g_\theta (1,\eta ) = 2(1 - \eta \theta )$.

   Similarly, for any $\alpha >0$,

$$\begin{aligned} \partial _\alpha g_\theta (\alpha ,\eta ) \leqslant \partial _\alpha g_\theta (0,\eta ) = -\eta ^2 + 4 \theta \eta + 1. \end{aligned}$$

Defining $\eta _\circ$ as the positive number such that $-\eta _\circ ^2 +4\theta \eta _\circ +1=0$, i.e., $\eta _\circ :=2\theta + \sqrt{4\theta ^2+1}$, we have for all $\eta \geqslant \eta _\circ$ that

$$\begin{aligned} \partial _\alpha g_\theta (\alpha ,\eta ) \leqslant -\eta ^2 + 4\theta \eta + 1 \leqslant -\eta _\circ ^2 + 4\theta \eta _\circ + 1 = 0. \end{aligned}$$

Thus, for $\eta \geqslant \eta _\circ$, $\alpha \mapsto g_\theta (\alpha ,\eta )$ is a decreasing function on [0, 1] and its maximum is $g_\theta (0,\eta ) = \eta ^2 - 4\theta \eta + 1$.

   Finally, for $\eta \in (\eta _\star ,\eta _\circ )$, $\alpha \mapsto g_\theta (\alpha ,\eta )$ is increasing then decreasing on [0, 1], and reaches its maximum at $\alpha _+:=\sqrt{\frac{4\eta \theta }{\eta ^2-1}} - 1$ with value

$$\begin{aligned} h(\eta ) := g_\theta (\alpha _+,\eta ) = 2\left( \eta ^2-2\sqrt{\theta }\sqrt{\eta (\eta ^2-1)}\right) . \end{aligned}$$

Now, since $\eta \mapsto g_\theta (0,\eta ) = \eta ^2 - 4\theta \eta + 1\geqslant 2$ on $(\eta _\circ ,+\infty )$, $\eta \mapsto g_\theta (1,\eta ) = 2(1 - \eta \theta )$ is decreasing ($\theta >0$), $h(\eta _\star ) = 2(1 - \eta _\star \theta )$ by construction, and (using the identity $\eta _\star ^2= \theta \eta _\star + 1$)

$$\begin{aligned} h'(\eta _\star ) = 2\left( 2\eta _\star - \sqrt{\theta }\frac{3\eta _\star ^2-1}{\sqrt{\eta _\star (\eta _\star ^2-1)}}\right) = -2\theta <0, \end{aligned}$$

we conclude that

$$\begin{aligned} \inf _{\eta >0}\max _{\alpha \in [0,1]} g_\theta (\alpha ,\eta ) = \inf _{\eta _\star< \eta < \eta _\circ } h(\eta ) = h(\eta _\theta ) = 2\eta _\theta ^2\frac{3-\eta _\theta ^2}{3\eta _\theta ^2 -1}, \end{aligned}$$

where $\eta _\theta$ is defined as the unique real number in $(1,+\infty )$ such that

$$\begin{aligned} 4\frac{\eta _\theta ^3(\eta _\theta ^2-1)}{(3\eta _\theta ^2-1)^2} = \theta . \end{aligned}$$

(53)

Note that the l.h.s. of (53) being a strictly increasing function on $(1,+\infty )$—hence one-to-one from $(1,+\infty )$ to $(0,+\infty )\ni \theta$—, it gives that (53) has a unique solution $\eta _\theta$ in $(1,+\infty )$ and, on another hand that $\eta _\star<\eta _\theta <\eta _\circ$. Indeed, recalling that $1<\eta _\star <\eta _\circ$ and the equations they respectively solve, we have

$$\begin{aligned} \eta _\star <\eta _\theta&\Leftrightarrow \theta> 4\frac{\eta _\star ^3(\eta _\star ^2-1)}{(3\eta _\star ^2-1)^2} = \frac{4\eta _\star ^4}{(3\eta _\star ^2-1)^2}\theta \Leftrightarrow \eta _\star ^2>1 \end{aligned}$$

and

$$\begin{aligned} \eta _\theta<\eta _\circ&\Leftrightarrow \theta< 4\frac{\eta _\circ ^3(\eta _\circ ^2-1)}{(3\eta _\circ ^2-1)^2} = \frac{16\eta _\circ ^4}{(3\eta _\circ ^2-1)^2}\theta \Leftrightarrow 3\eta _\circ ^2-1 < 4\eta _\circ ^2. \end{aligned}$$

Moreover, as a by-product, the problem has a solution only for $\xi <2$, since

$$\begin{aligned} \inf _{\eta >0}\max _{\alpha \in [0,1]} g_\theta (\alpha ,\eta ) = \inf _{\eta _\star< \eta< \eta _\circ } h(\eta ) = h(\eta _\star ) = 2(1 - \eta _\star \theta ) < 2. \end{aligned}$$

Summarizing, we have obtained $\eta _\theta >1$ defined by (53) for which

$$\begin{aligned} \inf _{\eta >0} \max _{\alpha \in [0,1]} g_\theta (\alpha ,\eta ) \geqslant \xi \quad \Leftrightarrow \quad h(\eta _\theta ) \geqslant \xi . \end{aligned}$$

Now, keeping in mind that $\xi \in [0,2)$, that $\eta \mapsto 2\eta ^2\frac{3-\eta ^2}{3\eta ^2 -1}$ is strictly decreasing and that $\eta \mapsto 4\frac{\eta _\theta ^3(\eta _\theta ^2-1)}{(3\eta _\theta ^2-1)^2}$ is stritly increasing, we have

$$\begin{aligned} \inf _{\eta >0} \max _{\alpha \in [0,1]} g_\theta (\alpha ,\eta ) \geqslant \xi \Leftrightarrow \quad h(\eta _\theta ) \geqslant \xi \quad&\Leftrightarrow \quad \eta _\theta \leqslant \frac{\sqrt{6-3\xi + \sqrt{9(2-\xi )^2+8\xi }}}{2}\\&\Leftrightarrow \quad \frac{\mu \left| \left| \left| Q \right| \right| \right| }{4(t+\omega )}=:\theta \leqslant \theta _+(\xi ), \end{aligned}$$

where $\theta _+$ is defined in (23), that is, $\theta _+:[0,2)\rightarrow (0,3\sqrt{3}/8]$ with

$$\begin{aligned} \theta _+(\xi ):= 2\frac{\left( 2 - 3\xi + \sqrt{9(2-\xi )^2+8\xi }\right) \left( 6-3\xi + \sqrt{9(2-\xi )^2+8\xi }\right) ^{\frac{3}{2}}}{\left( 14 - 9 \xi + 3 \sqrt{9(2-\xi )^2+8\xi }\right) ^2}. \end{aligned}$$

$\square$

Appendix C. Details on Some Computations

1.1 C.1. $L^\infty$-norms of terms involving ${\tilde{u}}$ and ${\tilde{v}}$.

We give here the details on computing several $L^\infty$-norms that we need in Sect. 7.

Expansion of $\left| \left| ({\tilde{v}}^2-{\tilde{u}}^2)^p \right| \right| _\infty$. We have

$$\begin{aligned} 0 < ({\tilde{v}}^2-{\tilde{u}}^2)^p&= (p+1)(m-\omega )\frac{1-\tanh ^2}{1-\nu \tanh ^2} = (p+1)\frac{m^2-\omega ^2}{m-\omega +2\omega \cosh ^2} \nonumber \\&\leqslant (p+1)(m-\omega ) =\left| \left| ({\tilde{v}}^2-{\tilde{u}}^2)^p \right| \right| _\infty = ({\tilde{v}}^2-{\tilde{u}}^2)^p(0) \nonumber \\&= \frac{p+1}{2m}\kappa ^2 + \frac{p+1}{8m^3}\kappa ^4 + O\!\left( \kappa ^6\right) . \end{aligned}$$

(54)

Expansion of $\left| \left| ({\tilde{v}}^2-{\tilde{u}}^2)^{p-1}{\tilde{u}}{\tilde{v}} \right| \right| _\infty$. Defining $h_3(y):=\frac{y\left( 1-y^2\right) }{\left( 1-\nu y^2\right) ^2}$, we have

$$\begin{aligned} ({\tilde{v}}^2-{\tilde{u}}^2)^{p-1}{\tilde{u}}{\tilde{v}} = (p+1) \sqrt{\nu } (m-\omega ) h_3(\tanh ). \end{aligned}$$

Since $h_3$, defined on $(-1,1)$, attains at $\pm \sqrt{y_3}$, with $y_3:=\frac{ - 3(1-\nu ) + \sqrt{9(1-\nu )^2+4\nu }}{2\nu }$, its extrema $\pm \frac{\sqrt{y_3}(1-y_3)}{(1-\nu y_3)^2}$, we have the non-relativistic expansion

$$\begin{aligned} \left| \left| ({\tilde{v}}^2-{\tilde{u}}^2)^{p-1}{\tilde{u}}{\tilde{v}} \right| \right| _\infty&= (p+1) \sqrt{\nu } (m-\omega ) \left| \left| h_3 \right| \right| _\infty \nonumber \\&=(p+1)(m-\omega )\frac{\sqrt{y_3}(1-y_3)}{(1-\nu y_3)^2} = \frac{p+1}{6\sqrt{3}m^2}\kappa ^3\!\left( 1+O\!\left( \kappa ^2\right) \right) . \end{aligned}$$

(55)

Expansion of $\left| \left| ({\tilde{v}}^2-{\tilde{u}}^2)^{p-1}{\tilde{u}}^2 \right| \right| _\infty$. Since on [0, 1), $h_1(y):=\frac{1-y}{1-\nu y}\frac{\nu y}{1-\nu y}$ is nonnegative with a maximum $\frac{\nu }{4(1-\nu )} = \frac{m-\omega }{8\omega }$ at $\frac{1}{2-\nu }$, we have

$$\begin{aligned}{} & {} 0 \leqslant (v^2-u^2)^{p-1}u^2 = (p+1)(m-\omega )h_1\left( \tanh ^2\right) \nonumber \\{} & {} \quad \leqslant (p+1)\frac{(m-\omega )^2}{8\omega } = \left| \left| ({\tilde{v}}^2-{\tilde{u}}^2)^{p-1}{\tilde{u}}^2 \right| \right| _\infty = \frac{p+1}{32 m^3}\kappa ^4\!\left( 1+O\!\left( \kappa ^2\right) \right) . \end{aligned}$$

(56)

Expansion of $\left| \left| ({\tilde{v}}^2-{\tilde{u}}^2)^p - \frac{(p+1)\kappa ^2}{2m\cosh ^2} \right| \right| _\infty$. Since $h_2(y):= \frac{1}{m-\omega +2\omega y} - \frac{1}{2m y}$ is positive on $(0,+\infty )$ with a maximum $\frac{1}{m}\frac{\sqrt{m}-\sqrt{\omega }}{\sqrt{m}+\sqrt{\omega }}$ at $\frac{1}{2}\left( 1+\sqrt{\frac{m}{\omega }}\right)$, we have

$$\begin{aligned} 0< & {} ({\tilde{v}}^2-{\tilde{u}}^2)^p - \frac{(p+1)\kappa ^2}{2m\cosh ^2} = (p+1)\kappa ^2h_2\left( \cosh ^2\right) \nonumber \\\leqslant & {} (p+1) \frac{\kappa ^2}{m}\frac{\sqrt{m}-\sqrt{\omega }}{\sqrt{m}+\sqrt{\omega }} = \left| \left| ({\tilde{v}}^2-{\tilde{u}}^2)^p - \frac{(p+1)\kappa ^2}{2m\cosh ^2} \right| \right| _\infty \nonumber \\= & {} (p+1)\frac{\kappa ^4}{8 m^3}\!\left( 1+O\!\left( \kappa ^2\right) \right) . \end{aligned}$$

(57)

1.2 C.2. Bounds on derivatives of $h \in F_k$

For fixed ${\mathfrak {s}}$ defined in (34), we denote by

$$\begin{aligned} p_j(x) = P_{{\mathfrak {s}}}^{{\mathfrak {s}}+1-j}(\tanh (p\kappa x)), \quad j = 1, \ldots , \lceil {\mathfrak {s}} \rceil , \end{aligned}$$

the eigenfunctions of the Schrödinger operator

$$\begin{aligned} -\frac{1}{2}\partial _x^2 + p^2 \kappa ^2 V_{\text {PT}}({\mathfrak {s}}, p\kappa x) = -\frac{1}{2}\partial _x^2 - p^2 \kappa ^2 \frac{{\mathfrak {s}}({\mathfrak {s}}+1)}{2\cosh ^2(p\kappa x)}. \end{aligned}$$

They satisfy the eigenvalue equation

$$\begin{aligned} - p_j'' (x) = p^2 \kappa ^2 \left( \frac{{\mathfrak {s}}({\mathfrak {s}}+1)}{\cosh ^2(p\kappa x)} - ({\mathfrak {s}}+1-j)^2 \right) p_j(x). \end{aligned}$$

We want to bound the first and second derivatives of $h \in {{\,\textrm{span}\,}}\{p_1, \ldots , p_k\}$. First,

$$\begin{aligned} {} \sup _{h \in {{\,\text {span}\,}}\{p_1, \ldots , p_k\} {\setminus }\{0\}} \frac{\left| \left| h' \right| \right| }{\left| \left| h \right| \right| }&{} =\max _{j \in \{1, \cdots , k\}} \frac{\left| \left| p_j' \right| \right| }{\left| \left| p_j \right| \right| }\quad \text { and} \\\sup _{h \in {{\,\text {span}\,}}\{p_1, \ldots , p_k\} {\setminus }\{0\}} \frac{\left| \left| h'' \right| \right| }{\left| \left| h \right| \right| }&{} =\max _{j \in \{1, \cdots , k\}} \frac{\left| \left| p_j'' \right| \right| }{\left| \left| p_j \right| \right| }. \end{aligned}$$

For the first bound, we multiply the eigenvalue equation by $p_j$, integrate by parts in the first term and bound $\cosh ^{-2}\leqslant 1$ in order to obtain

$$\begin{aligned} \left| \left| p_j' \right| \right| ^2 = p^2 \kappa ^2 \left( \left\langle p_j, \frac{{\mathfrak {s}}({\mathfrak {s}}+1)}{\cosh ^2(p\kappa \cdot )}p_j \right\rangle - ({\mathfrak {s}}+1-j)^2\left| \left| p_j \right| \right| ^2 \right) \leqslant p^2 \kappa ^2 {\mathfrak {s}}({\mathfrak {s}}+1)\left| \left| p_j \right| \right| ^2. \end{aligned}$$

For the second bound, we take the norm on both sides of the eigenvalue equation and, since $0<{\mathfrak {s}}+1-j\leqslant {\mathfrak {s}}$ and $0<({\mathfrak {s}}+1-j)^2\leqslant {\mathfrak {s}}^2<{\mathfrak {s}}({\mathfrak {s}}+1)$ for $j\in \llbracket 1, \lceil {\mathfrak {s}} \rceil \rrbracket$, we obtain for all $j\in \llbracket 1, \lceil {\mathfrak {s}} \rceil \rrbracket$ that

$$\begin{aligned} \frac{\left| \left| p_j'' \right| \right| }{\left| \left| p_j \right| \right| }\leqslant & {} p^2 \kappa ^2 \left| \left| \frac{{\mathfrak {s}}({\mathfrak {s}}+1)}{\cosh ^2(p\kappa \cdot )} - ({\mathfrak {s}}+1-j)^2 \right| \right| _\infty \\= & {} p^2 \kappa ^2 \max \left\{ {\mathfrak {s}}({\mathfrak {s}}+1) - ({\mathfrak {s}}+1-j)^2, ({\mathfrak {s}}+1-j)^2 \right\} \leqslant p^2 \kappa ^2 {\mathfrak {s}}({\mathfrak {s}}+1). \end{aligned}$$