Inviscid Damping and Enhanced Dissipation of the Boundary Layer for 2D Navier–Stokes Linearized Around Couette Flow in a Channel

Abstract

We study the 2D Navier–Stokes equations linearized around the Couette flow $(y,0)^t$ in the periodic channel ${\mathbb {T}} \times [-1,1]$ with no-slip boundary conditions in the vanishing viscosity $\nu \rightarrow 0$ limit. We split the vorticity evolution into the free evolution (without a boundary) and a boundary corrector that is exponentially localized to at most an $O(\nu ^{1/3})$ boundary layer. If the initial vorticity perturbation is supported away from the boundary, we show inviscid damping of both the velocity and the vorticity associated to the boundary layer. For example, our $L^2_t L^1_y$ estimate of the boundary layer vorticity is independent of $\nu$, provided the initial data is $H^1$. For $L^2$ data, the loss is only logarithmic in $\nu$. Note both such estimates are false for the vorticity in the interior. To the authors’ knowledge, this inviscid decay of the boundary layer vorticity seems to be a new observation not previously isolated in the literature. Both velocity and vorticity satisfy the expected $O(\exp (-\delta \nu ^{1/3}\alpha ^{2/3}t))$ enhanced dissipation in addition to the inviscid damping. Similar, but slightly weaker, results are obtained also for $H^1$ data that is against the boundary initially. For $L^2$ data against the boundary, we at least obtain the boundary layer localization and enhanced dissipation.

Appendices

Technical Lemmas

1.1 The free evolution

Let us briefly recall the properties of the initial value problem

$$\begin{aligned}&\partial _t \omega + y \partial _x \omega = \nu \Delta \omega , \end{aligned}$$

(A.1)

$$\begin{aligned}&\omega (0) = \omega _{in}, \end{aligned}$$

(A.2)

$$\begin{aligned}&u = \nabla ^{\perp }\Delta ^{-1}\omega . \end{aligned}$$

(A.3)

Via Fourier transform in both variables $(x,y) \mapsto (\alpha ,\eta )$ we derive (see [7] for a modern treatment),

$$\begin{aligned} {\widehat{\omega }}(t,\alpha ,\eta )&= {\widehat{\omega }}_{in}(\alpha ,\eta + \alpha t) \exp \left( -\nu \alpha ^2 t - \nu \int _0^t\left| \eta + \alpha (t-\tau ) \right| ^2 d\tau \right) . \end{aligned}$$

(A.4)

The following lemma quantifies the enhanced dissipation and inviscid damping.

Lemma A.1

There holds the pointwise-in-time enhanced dissipation estimate

$$\begin{aligned} \left| \left| {\widehat{\omega }}(t,\alpha ,\cdot ) \right| \right| _{L^2_y} \lesssim e^{-\alpha ^2 \nu t - \frac{1}{12} \nu \alpha ^2 t^3} \left| \left| {\widehat{\omega }}_{in}(\alpha ,\cdot ) \right| \right| _{L^2_y}, \end{aligned}$$

(A.5)

there holds the time-averaged inviscid damping and enhanced dissipation

$$\begin{aligned} \left| \left| e^{\nu \alpha ^2 t+ \frac{1}{12}\nu \alpha ^2 t^3} {\widehat{u}}(t,\alpha ,\cdot ) \right| \right| _{L^2_t L^2_y} \lesssim \frac{1}{\left| \alpha \right| ^{1/2}}\left| \left| {\widehat{\omega }}_{in}(\alpha ,\cdot ) \right| \right| _{L^2_y}, \end{aligned}$$

(A.6)

and, more generally, the following $L^2_t H^{-s}_y$ estimate for $s > 1/2$:

$$\begin{aligned} \left| \left| e^{\nu \alpha ^2 t+ \frac{1}{12}\nu \alpha ^{2} t^3} {\widehat{\omega }}(t,\alpha ,\cdot ) \right| \right| _{L^2_t H^{-s}_y} \lesssim \frac{1}{\left| \alpha \right| ^{1/2}} \left| \left| {\widehat{\omega }}_{in}(\alpha ,\cdot ) \right| \right| _{L^2}. \end{aligned}$$

(A.7)

Proof

Estimate (A.5) is immediate from the formula (A.4). Estimate (A.6) follows from (A.7). Finally, we observe by (A.4),

$$\begin{aligned} \int _0^\infty \int _{-\infty }^\infty e^{2\alpha ^2 \nu t + \frac{1}{6}\nu \alpha ^2 t^3} \left\langle \eta \right\rangle ^{-2s} \left| {\widehat{\omega }}(t,\alpha ,\eta ) \right| ^2 d\eta&\lesssim \int _0^\infty \int _{-\infty }^\infty \left\langle \eta \right\rangle ^{-2s} \left| {\widehat{\omega }}_{in}(\alpha ,\eta + \alpha t) \right| ^2 d\eta dt \\&\lesssim \int _{-\infty }^\infty \left( \int _0^\infty \left\langle \eta - \alpha t \right\rangle ^{-2s} dt\right) \left| {\widehat{\omega }}_{in}(\alpha ,\eta ) \right| ^2 d\eta \\&\lesssim \frac{1}{\left| \alpha \right| } \left| \left| \omega _{in} \right| \right| _{L^2}^2, \end{aligned}$$

which completes the proof of (A.7). $\quad \square$

1.2 Airy functions estimates

Recall the definition of the homogeneous solutions of the Orr–Sommerfeld problem in terms of Airy functions (2.13). Standard asymptotics for the Airy functions gives the following

Lemma A.2

The homogeneous solutions $H_-(z)$ and $H_+(z)$ satisfy the following for $z=z_r+\delta i \subset \left\{ z \in {\mathbb {C}}: 0< z_i < K + \frac{1}{100}\left| z_r \right| \right\}$,

$$\begin{aligned} |H_-(z)|\lesssim _K&\frac{1}{\langle z\rangle ^{1/4}}e^{-\frac{{2}}{3}Re(e^{i\pi /4}z^{3/2})}, \quad z_r\ge 0; \end{aligned}$$

(A.8a)

$$\begin{aligned} |H_-(z)|\lesssim _K&\frac{1}{\langle z\rangle ^{1/4}}e^{\frac{{2}}{3}Re(e^{i\pi /4}z^{3/2})}, \quad z_r\le 0;\end{aligned}$$

(A.8b)

$$\begin{aligned} |H_+(z)|\lesssim _K&\frac{1}{\langle z\rangle ^{1/4}}e^{\frac{{2}}{3}Re(e^{i\pi /4}z^{3/2})},\quad z_r\ge 0;\end{aligned}$$

(A.8c)

$$\begin{aligned} |H_+(z)|\lesssim _K&\frac{1}{\langle z\rangle ^{1/4}}e^{-\frac{{2}}{3}Re(e^{i\pi /4}z^{3/2})}, \quad z_r\le 0. \end{aligned}$$

(A.8d)

If $z\in \{z\in {\mathbb {C}}|z_i<0\}$, then the solutions $H_-(z)$ and $H_+(z)$ satisfy the following

$$\begin{aligned} |H_-(z)|\lesssim _K&\frac{1}{\langle z\rangle ^{1/4}}e^{-\frac{{2}}{3}Re(e^{i\pi /4}z^{3/2})}, \quad z_r\in {\mathbb {R}};\end{aligned}$$

(A.9a)

$$\begin{aligned} |H_+(z)|\lesssim _K&\frac{1}{\langle z\rangle ^{1/4}}e^{\frac{{2}}{3}Re(e^{i\pi /4}z^{3/2})}, \quad z_r\in {\mathbb {R}}. \end{aligned}$$

(A.9b)

Proof

Recall the definition of the $H_\pm$ in (2.13). For $\left| z \right| \lesssim 1$, (A.8) follows since $\left| H_{\pm } \right| \lesssim 1$ on bounded sets. For $\left| z \right|$ larger, we apply the asymptotic expansion for $|z|\rightarrow \infty$, $|\mathrm {ph} z|<\pi -\zeta$, for any $\zeta >0$ (see e.g. [42]):

$$\begin{aligned} Ai(z)=\frac{1}{2z^{1/4}\sqrt{\pi }}e^{-2z^{3/2}/3}(1+O(|z|^{-3/2})). \end{aligned}$$

Note that for the second and fourth inequalities, one must be careful to treat the branch cut (1.31) correctly. For the estimates (A.9), the proof is the same. One only needs to be careful with the fact that now the phase is $\mathrm {ph}(z)\in (-\pi ,0)$ since $z_r<0$. $\quad \square$

In addition to the standard Airy functions, we also require the integrated version used by Romanov [46],

$$\begin{aligned} A_0(z)=\int _{ze^{i \pi /6}}^{i0+\infty } Ai(t)dt, \end{aligned}$$

(A.10)

where for definiteness we take the contour to be the straight line connecting $ze^{i\pi /6}$ to 0 and then the ray connecting 0 to $i0 + \infty$. The following property of $A_0$ proved by Romanov [46] is crucial.

Lemma A.3

(Lemma 2 and following remark, [46]). The function $A_0(z)$ has no zeros in the sector $-\pi \le \mathrm {ph} (z)\le 2\pi /3$ and in the half-plane $\{z| z_i \le \delta _0\}$ for some universal constant $\delta _0>0$. Moreover, the following quantity a is strictly positive

$$\begin{aligned} a=a(\delta )=-\max _{ z_i\le \delta }Re \frac{A_0'(z)}{A_0(z)}>0, \quad \delta \in [0,\delta _0). \end{aligned}$$

(A.11)

We also have the following lower bound of $A_0$, which also follows from [(3.4) [46]]; see also [48].

Lemma A.4

(Integrated Airy function asymptotics) . The integrated Airy function (A.10) has the following lower bounds for $z=z_r+i z_i \subset \left\{ z \in {\mathbb {C}}: 0< z_i< K + \frac{1}{100}\left| z_r \right| \right\} \cup \left\{ z \in {\mathbb {C}}: z_i < 0 \right\}$,

$$\begin{aligned} |A_0(z)| > rsim _K&\frac{1}{\langle z\rangle ^{3/4}}e^{\frac{2}{3}Re(z^{3/2}e^{i\pi /4})},\quad z_r\le 0; \end{aligned}$$

(A.12a)

$$\begin{aligned} |A_0(z)| > rsim _K&\frac{1}{\langle z\rangle ^{3/4}}e^{-\frac{2}{3}Re(z^{3/2}e^{i\pi /4})},\quad z_r\ge 0. \end{aligned}$$

(A.12b)

1.3 Power-gain lemmas

The following lemma is used often in our paper. Similar estimates are used in e.g. [24] and the references therein, however, here we use a different range of complex parameters, so we include a sketch for the sake of completeness.

Lemma A.5

Case 1: Consider the spectral parameter c lying on the vertical line $\Gamma _\alpha :=\{c=c_r-(\alpha \nu +\delta \epsilon )i, c_r\in {\mathbb {R}}\}$. From (2.13a), the Langer variables R, Z have the explicit form

$$\begin{aligned} R=\frac{r-c_r}{\epsilon }+\delta i=:R_r+\delta i,\quad Z=\frac{z-c_r}{\epsilon }+\delta i=:Z_r+\delta i. \end{aligned}$$

For $\delta$ small enough, the following estimates hold (recall also (1.31))

$$\begin{aligned} \int _0^{Z_r}&\frac{1}{\langle R\rangle ^{1/4}}e^{\frac{2}{3}Re((R_r+\delta i)^{3/2}e^{i\pi /4})}dR_r\lesssim \frac{1}{\langle Z\rangle ^{3/4}}e^{\frac{2}{3}Re ((Z_r+\delta i)^{3/2}e^{\pi i/4})},\quad Z_r\ge 0. \end{aligned}$$

(A.13a)

$$\begin{aligned} \int _{Z_r}^0&\frac{1}{\langle R\rangle ^{1/4}}e^{\frac{2}{3}Re((R_r+\delta i)^{3/2}e^{i\pi /4})}dR_r\lesssim \frac{1}{\langle Z\rangle ^{3/4}}e^{\frac{2}{3}Re ((Z_r+\delta i)^{3/2}e^{\pi i/4})},\quad Z_r\le 0, \end{aligned}$$

(A.13b)

$$\begin{aligned} \int _{Z_r}^\infty&\frac{1}{\langle R\rangle ^{1/4}}e^{-\frac{2}{3}Re((R_r+\delta i)^{3/2}e^{i\pi /4})}dR_r\lesssim \frac{1}{\langle Z\rangle ^{3/4}}e^{-\frac{2}{3}Re ((Z_r+\delta i)^{3/2}e^{\pi i/4})},\quad Z_r\ge 0, \end{aligned}$$

(A.13c)

$$\begin{aligned} \int _{-\infty }^{Z_r}&\frac{1}{\langle R\rangle ^{1/4}}e^{-\frac{2}{3}Re((R_r+\delta i)^{3/2}e^{i\pi /4})}dR_r\lesssim \frac{1}{\langle Z\rangle ^{3/4}}e^{-\frac{2}{3}Re ((Z_r+\delta i)^{3/2}e^{\pi i/4})},\quad Z_r\le 0. \end{aligned}$$

(A.13d)

Case 2: In the high mode case, i.e., $\alpha ^2\nu \ge C_0$, we consider the spectral parameter c lying on the vertical line $\Gamma _\alpha :=\{c=c_r-(1-\kappa )\alpha \nu i, c_r\in {\mathbb {R}}\}$. From (2.13a), the Langer variables R, Z have the explicit form

$$\begin{aligned} R=\frac{r-c_r}{\epsilon }-\kappa \alpha ^{4/3}\nu ^{2/3} i,\quad Z=\frac{z-c_r}{\epsilon }-\kappa \alpha ^{4/3}\nu ^{2/3} i. \end{aligned}$$

The following estimates hold (recall also (1.31))

$$\begin{aligned} \int _0^{Z_r}&\frac{1}{\langle R\rangle ^{1/4}}e^{\frac{2}{3}Re((R_r-\kappa \alpha ^{4/3}\nu ^{2/3} i)^{3/2}e^{i\pi /4})}dR_r\lesssim \frac{1}{\langle Z\rangle ^{3/4}}e^{\frac{2}{3}Re ((Z_r-\kappa \alpha ^{4/3}\nu ^{2/3} i)^{3/2}e^{\pi i/4})},\quad Z_r\ge 0. \end{aligned}$$

(A.14a)

$$\begin{aligned} \int _{Z_r}^0&\frac{1}{\langle R\rangle ^{1/4}}e^{-\frac{2}{3}Re((R_r-\kappa \alpha ^{4/3}\nu ^{2/3} i)^{3/2}e^{i\pi /4})}dR_r\lesssim \frac{1}{\langle Z\rangle ^{3/4}}e^{-\frac{2}{3}Re ((Z_r-\kappa \alpha ^{4/3}\nu ^{2/3} i)^{3/2}e^{\pi i/4})},\quad Z_r\le 0,\end{aligned}$$

(A.14b)

$$\begin{aligned} \int _{Z_r}^\infty&\frac{1}{\langle R\rangle ^{1/4}}e^{-\frac{2}{3}Re((R_r-\kappa \alpha ^{4/3}\nu ^{2/3} i)^{3/2}e^{i\pi /4})}dR_r\lesssim \frac{1}{\langle Z\rangle ^{3/4}}e^{-\frac{2}{3}Re ((Z_r-\kappa \alpha ^{4/3}\nu ^{2/3} i)^{3/2}e^{\pi i/4})},\quad Z_r\ge 0,\end{aligned}$$

(A.14c)

$$\begin{aligned} \int _{-\infty }^{Z_r}&\frac{1}{\langle R\rangle ^{1/4}}e^{\frac{2}{3}Re((R_r-\kappa \alpha ^{4/3}\nu ^{2/3} i)^{3/2}e^{i\pi /4})}dR_r\lesssim \frac{1}{\langle Z\rangle ^{3/4}}e^{\frac{2}{3}Re ((Z_r-\kappa \alpha ^{4/3}\nu ^{2/3} i)^{3/2}e^{\pi i/4})},\quad Z_r\le 0. \end{aligned}$$

(A.14d)

Proof

Proof of (A.13):

The inequality is immediate for $\left| Z_r \right| \le 4$, hence, consider next the case $\left| Z_r \right| \ge 4$. In this case, the following holds on the integration interval:

$$\begin{aligned} e^{\frac{2}{3}Re((R_r+\delta i)^{3/2}e^{\pi i/4})}\le e^{\frac{2}{3}Re((Z_r+\delta i)^{1/2}(R_r+\delta i)e^{i\pi /4})},\quad Z_r\ge R_r\ge 2; \end{aligned}$$

(A.15)

this follows by e.g. differentiation. Next, we decompose the integral in (A.13a) into two parts:

$$\begin{aligned} \int _0^{Z_r}\frac{1}{\langle R\rangle ^{1/4}}e^{\frac{2}{3}Re((R_r+\delta i)^{3/2}e^{i\pi /4})}dR_r&=\left( \int _0^{Z_r-\sqrt{Z_r}}+\int _{Z_r-\sqrt{Z_r}}^{Z_r}\right) \frac{1}{\langle R\rangle ^{1/4}}e^{\frac{2}{3}Re((R_r+\delta i)^{3/2}e^{i\pi /4})}dR_r \nonumber \\&=:T_1+T_2. \end{aligned}$$

(A.16)

First consider $T_2$. Combining (A.15) and the following relation on the integration interval,

$$\begin{aligned} \langle R\rangle \ge (1+|Z_r-\sqrt{Z_r}|^2+\delta ^2)^{1/2}\ge (1+\frac{1}{4}|Z|^2)^{1/2}\ge \frac{1}{2}\langle Z\rangle , \end{aligned}$$

together with $\delta \le \frac{1}{100}\le \frac{1}{400}Z_r$, we have

$$\begin{aligned} T_2&\lesssim \int _{Z_r-\sqrt{Z_r}}^{Z_r}\frac{1}{\langle Z\rangle ^{1/4}}e^{\frac{2}{3}Re((Z_r+\delta i)^{1/2}(R_r+\delta i)e^{i\pi /4})}dR_r \lesssim \frac{1}{\langle Z\rangle ^{1/4}}\frac{1}{Re((Z_r+\delta i)^{1/2}e^{i\pi /4})}e^{\frac{2}{3}Re((Z_r+\delta i)^{3/2}e^{i\pi /4})}\nonumber \\&\lesssim \frac{1}{\langle Z\rangle ^{3/4}}e^{\frac{2}{3}Re((Z_r+\delta i)^{3/2}e^{i\pi /4})}. \end{aligned}$$

(A.17)

This completes the treatment of the $T_2$ term. The $T_1$ term in (A.16), using (A.15) and the fact that $\delta \le \frac{1}{100}\le \frac{1}{400} \left| Z_r \right|$ is easier and is hence omitted for the sake of brevity. Hence (A.13a) follows.

The remaining inequalities (A.13a)–(A.13d) hold by analogous arguments once the suitable analogue of (A.15) is similarly deduced via differentiation. We omit the details for the sake of brevity.

Proof of (A.14):

In this case, $\Gamma _\alpha :=\{c=c_r-(1-\kappa )\alpha \nu i,\quad c_r\in {\mathbb {R}}\}$. We prove the following monotonicity inequalities adapted for use in high frequencies:

$$\begin{aligned} e^{\frac{2}{3}Re((Y_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)^{3/2}e^{i\pi /4})}&\le e^{\frac{2}{3}Re((W_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)^{3/2}e^{i\pi /4})},\quad W_r\ge Y_r\ge 0;\\ e^{\frac{2}{3}Re((Y_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)^{3/2}e^{i\pi /4})}&\le e^{\frac{2}{3}Re((W_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)^{1/2}(Y_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)e^{i\pi /4})},\quad W_r\ge Y_r\ge 0;\\ e^{-\frac{2}{3}Re((Y_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)^{3/2}e^{i\pi /4})}&\le e^{-\frac{2}{3}Re((W_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)^{3/2}e^{i\pi /4})},\quad W_r\le Y_r\le 0. \\ e^{-\frac{2}{3}Re((Y_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)^{3/2}e^{i\pi /4})}&\le e^{-\frac{2}{3}Re((W_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)^{1/2}(Y_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)e^{i\pi /4})},\quad W_r\le Y_r\le 0. \end{aligned}$$

Consider the case $W_r\le Y_r\le 0$ i.e., the third and fourth inequalities. First,

$$\begin{aligned}&-\frac{d}{dw}Re((w-\kappa \alpha ^{4/3}\nu ^{2/3}i)^{1/2}(Y_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)e^{i\pi /4})\\&\quad =-\frac{1}{2}Re((w-\kappa \alpha ^{4/3}\nu ^{2/3}i)^{-1/2}(Y_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)e^{i\pi /4}). \end{aligned}$$

Since $W_r\le Y_r$, $\mathrm {ph}(W_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)\le \mathrm {ph}(Y_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)$, so that the derivative is negative and the fourth inequality follows. The third inequality follows by a similar argument. For the $W_r\ge Y_r\ge 0$ case, we have

$$\begin{aligned}&\frac{d}{dw}Re((w-\kappa \alpha ^{4/3}\nu ^{2/3}i)^{1/2}(Y_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)e^{i\pi /4})\\&\quad =\frac{1}{2}Re((w-\kappa \alpha ^{4/3}\nu ^{2/3}i)^{-1/2}(Y_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)e^{i\pi /4}). \end{aligned}$$

Since $W_r\ge Y_r\ge 0$, we have $0\le -\frac{1}{2}\mathrm {ph}(W_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)\le -\frac{1}{2}\mathrm {ph}(Y_r-\kappa \alpha ^{4/3}\nu ^{2/3}i)$ and hence the derivative is positive. This completes the proof of the second inequality; the first inequality follows similarly. Combining these inequalities with the argument used to prove (A.13) yields the inequalities (A.14). $\quad \square$

By a similar argument, one can prove the following lemma (used in Sect. 6).

Lemma A.6

Consider the spectral parameter c lying on the half-lines as in Sect. 6,

$$\begin{aligned} \Gamma ^+&=\{c| c_i+\alpha \nu +\delta \alpha ^{-1/3}\nu ^{1/3}=-\tan (\theta )(c_r-200)_+, \quad c_r\in [200,\infty )\};\\ \Gamma ^-&=\{c|c_i+\alpha \nu +\delta \alpha ^{-1/3}\nu ^{1/3}=-\tan (\theta )(-c_r-200)_+, \quad c_r\in (-\infty ,-200]\}. \end{aligned}$$

The Langer variables are

$$\begin{aligned} R=\frac{r-c_r}{\epsilon }-\frac{c_i+\alpha \nu }{\epsilon }i =R_r+iR_i,\quad Z=\frac{z-c_r}{\epsilon }-\frac{c_i+\alpha \nu }{\epsilon }i=Z_r+iZ_i. \end{aligned}$$

Then the following estimates are satisfied (recall also (1.31))

$$\begin{aligned} \int _{\frac{-1-c_r}{\epsilon }}^{Z_r}&\frac{1}{\langle R\rangle ^{1/4}}e^{\frac{2}{3}Re((R_r+ iR_i)^{3/2}e^{i\pi /4})}dR_r\lesssim \frac{1}{\langle Z\rangle ^{3/4}}e^{\frac{2}{3}Re ((Z_r+ iZ_i)^{3/2}e^{\pi i/4})},\quad Z_r\ge \max \left\{ 0,\frac{-1-c_r}{\epsilon }\right\} ; \end{aligned}$$

(A.18a)

$$\begin{aligned} \int _{Z_r}^{\frac{1-c_r}{\epsilon }}&\frac{1}{\langle R\rangle ^{1/4}}e^{\frac{2}{3}Re((R_r+iR_i )^{3/2}e^{i\pi /4})}dR_r\lesssim \frac{1}{\langle Z\rangle ^{3/4}}e^{\frac{2}{3}Re ((Z_r+ iZ_i)^{3/2}e^{\pi i/4})},\quad Z_r\le \min \left\{ 0,\frac{1-c_r}{\epsilon }\right\} ; \end{aligned}$$

(A.18b)

$$\begin{aligned} \int _{Z_r}^\infty&\frac{1}{\langle R\rangle ^{1/4}}e^{-\frac{2}{3}Re((R_r+iR_i)^{3/2}e^{i\pi /4})}dR_r\lesssim \frac{1}{\langle Z\rangle ^{3/4}}e^{-\frac{2}{3} Re((Z_r+iZ_i)^{3/2}e^{i\pi /4})},\quad Z_r\ge 0; \end{aligned}$$

(A.18c)

$$\begin{aligned} \int _{-\infty }^{Z_r}&\frac{1}{\langle R\rangle ^{1/4}}e^{-\frac{2}{3}Re((R_r+iR_i)^{3/2}e^{i\pi /4})}dR_r\lesssim \frac{1}{\langle Z\rangle ^{3/4}}e^{-\frac{2}{3}Re((Z_r+iZ_i)^{3/2}e^{i\pi /4})},\quad Z_r\le 0. \end{aligned}$$

(A.18d)

1.4 Detailed Green’s function

The following denotes the full expression of the boundary resolvent ${\mathcal {R}}_b$ directly as an integral operator on ${\widehat{\omega }}_{in}$; see 2.34:

$$\begin{aligned} \omega _b&=-\int _{\Gamma _\alpha }\frac{ 2\epsilon ^{-2}}{\pi D(\alpha ,c)}\left( \int _{-1}^1 e^{-\alpha w} H_+(W) dw\right) \nonumber \\&\quad \times \,\int _{-1}^1{\bigg [\int _{z}^1 e^{\alpha r}H_-(R)dr}\bigg ] H_+(Z)H_-(Y) {\widehat{\omega }}_{in}(\alpha ,z) dzdc\nonumber \\&\quad -\int _{\Gamma _\alpha }\frac{ 2\epsilon ^{-2}}{\pi D(\alpha ,c)}\left( \int _{-1}^1 e^{-\alpha w} H_+(W) dw\right) \nonumber \\&\quad \times \,\int _{-1}^1 {\bigg [\int _{-1}^z e^{\alpha r} H_+(R)dr\bigg ]} H_-(Z)H_-(Y) {\widehat{\omega }}_{in}(\alpha ,z)dzdc\nonumber \\&\quad -\int _{\Gamma _\alpha }\frac{ 2\epsilon ^{-2}}{\pi D(\alpha ,c)}\left( -\int _{-1}^1 e^{\alpha w} H_+(W) dw\right) \nonumber \\&\quad \times \,\int _{-1}^1 {\bigg [\int _{z}^1e^{-\alpha r}H_-(R)dr\bigg ]} H_+(Z)H_-(Y) {\widehat{\omega }}_{in}(\alpha ,z)dzdc\nonumber \\&\quad -\int _{\Gamma _\alpha }\frac{ 2\epsilon ^{-2}}{\pi D(\alpha ,c)}\left( -\int _{-1}^1 e^{\alpha w} H_+(W) dw\right) \nonumber \\&\quad \times \,\int _{-1}^1 {\bigg [\int _{-1}^ze^{-\alpha r}H_+(R)dr \bigg ]} H_-(Z)H_-(Y) {\widehat{\omega }}_{in}(\alpha ,z)dzdc\nonumber \\&\quad -\int _{\Gamma _\alpha }\frac{ 2\epsilon ^{-2}}{\pi D(\alpha ,c)}\left( -\int _{-1}^1 e^{-\alpha w}H_-(W) dw\right) \nonumber \\&\quad \times \,\int _{-1}^1{\bigg [\int _{z}^1 e^{\alpha r}H_-(R)dr}\bigg ] H_+(Z)H_+(Y){\widehat{\omega }}_{in}(\alpha ,z) dzdc\nonumber \\&\quad -\int _{\Gamma _\alpha }\frac{ 2\epsilon ^{-2}}{\pi D(\alpha ,c)}\left( -\int _{-1}^1 e^{-\alpha w}H_-(W) dw\right) \nonumber \\&\quad \times \,\int _{-1}^1 {\bigg [\int _{-1}^z e^{\alpha r} H_+(R)dr\bigg ]} H_-(Z)H_+(Y) {\widehat{\omega }}_{in}(\alpha ,z) dzdc\nonumber \\&\quad -\int _{\Gamma _\alpha }\frac{ 2\epsilon ^{-2}}{\pi D(\alpha ,c)}\left( \int _{-1}^1 e^{\alpha w}H_-(W) dw\right) \nonumber \\&\quad \times \,\int _{-1}^1 {\bigg [\int _{z}^1e^{-\alpha r}H_-(R)dr\bigg ]} H_+(Z)H_+(Y) {\widehat{\omega }}_{in}(\alpha ,z) dzdc\nonumber \\&\quad -\int _{\Gamma _\alpha }\frac{ 2\epsilon ^{-2}}{\pi D(\alpha ,c)}\left( \int _{-1}^1 e^{\alpha w}H_-(W) dw\right) \nonumber \\&\quad \times \,\int _{-1}^1 {\bigg [\int _{-1}^ze^{-\alpha r}H_+(R)dr \bigg ]} H_-(Z)H_+(Y) {\widehat{\omega }}_{in}(\alpha ,z) dzdc\nonumber \\&=:\sum _{j=1}^8\int _{-1}^{1}K_j(y,z) {\widehat{\omega }}_{in}(\alpha ,z)dz. \end{aligned}$$

(A.19)

Estimates on the Evans Function

1.1 Evans function estimates

The main goal of this section is to prove the following lemma, which in term implies Lemma 2.1.

Lemma B.1

There exists a universal $\nu _0$ such that for $\nu \in (0,\nu _0]$, the Evans function $D(\alpha ,c)$ is non-zero except in the region

$$\begin{aligned} \left\{ z \in {\mathbb {C}}\big | \ z_i < -\alpha \nu - \delta \alpha ^{-1/3} \nu ^{1/3} \right\} , \end{aligned}$$

(B.1)

for $\delta <\delta _0$ (a sufficiently small universal constant).

Moreover, the following bounds are satisfied.

• (i) For any $c\in \{c\in {\mathbb {C}}|c=c_r-i\alpha \nu -i\delta \alpha ^{-1/3}\nu ^{1/3},\quad \forall c_r\in (-\infty ,\infty )\}$:

  $$\begin{aligned} |D(\alpha , c)| > rsim \bigg |\int _{-1}^1 e^{-\alpha w}H_-(W)dw\bigg |\bigg |\int _{-1}^1 e^{\alpha w}H_+(W)dw\bigg |. \end{aligned}$$

  (B.2)

  Moreover, there holds

  $$\begin{aligned} \left| \int _{-1}^1e^{\alpha w}H_+(W)dw\right| > rsim \frac{\epsilon e^\alpha }{\langle \alpha \epsilon \rangle }\left| A_0\left( \frac{-1+ c_r}{\epsilon }+\delta i\right) \right| , \end{aligned}$$

  (B.3)
  $$\begin{aligned} \left| \int _{-1}^1e^{-\alpha w}H_-(W)dw\right| > rsim \frac{\epsilon e^\alpha }{\langle \alpha \epsilon \rangle }\left| A_0\left( \frac{-1-c_r}{\epsilon }+i\delta \right) \right| . \end{aligned}$$

  (B.4)

• (ii) For $\alpha ^2\nu \ge C_0$ and $c\in \{c\in {\mathbb {C}}|c=c_r-i(1-\kappa )\alpha \nu ,\quad \forall c_r\in (-\infty ,\infty )\}$:

  $$\begin{aligned} |D(\alpha , c)| > rsim \bigg |\int _{-1}^1 e^{-\alpha w}H_-(W)dw\bigg |\bigg |\int _{-1}^1 e^{\alpha w}H_+(W)dw\bigg |. \end{aligned}$$

  (B.5)

  Moreover, there holds

  $$\begin{aligned} \left| \int _{-1}^1e^{\alpha w}H_+(W)dw\right| > rsim {\epsilon e^\alpha }\left| A_0\left( \frac{-1+ c_r}{\epsilon }+\delta i\right) \right| , \end{aligned}$$

  (B.6)
  $$\begin{aligned} \left| \int _{-1}^1e^{-\alpha w}H_-(W)dw\right| > rsim \epsilon e^\alpha \left| A_0\left( \frac{-1-c_r}{\epsilon }+i\delta \right) \right| . \end{aligned}$$

  (B.7)

Proof

Consider case (i) first. First, define the variables $d,C_j^\star$

$$\begin{aligned} d=&-ic+\alpha \nu , \end{aligned}$$

(B.8a)

$$\begin{aligned} C_1^\star&:=\frac{-1-id}{\epsilon }=\frac{-1-c_r}{\epsilon }+\delta i,\end{aligned}$$

(B.8b)

$$\begin{aligned} C_2^\star&:=\frac{-1-i{\overline{d}}}{\epsilon }=\frac{-1+c_r}{\epsilon }+\delta i. \end{aligned}$$

(B.8c)

By definition (2.13a) for the spectral parameter c on the vertical line $c_i\equiv -\alpha \nu -\delta \epsilon$ we get

$$\begin{aligned} Z:=\frac{z-id}{\epsilon }=\frac{z-c_r}{\epsilon }+\delta i. \end{aligned}$$

(B.9)

Define also the function u(z, t)

$$\begin{aligned} u(z,t)=\frac{A_0(z+t)}{A_0(z)}=\exp \bigg (\int _0^t \frac{A_0'(z+s)}{A_0(z+s)}ds\bigg ). \end{aligned}$$

(B.10)

Step 1: Rephrasing the lower bound (B.2). First, we show that the lower bound of the Evans function $|D(\alpha ,c)|$ (B.2) is follows from the follwing:

$$\begin{aligned} e^{2\alpha }&\frac{1}{1+q}\left| 1-e^{-2\alpha }u\left( C_j^\star ,\frac{2}{\epsilon }\right) \right| \ge \frac{1}{1+q}\left| 1-e^{2\alpha }u\left( C_j^\star ,\frac{2}{\epsilon }\right) \right| \nonumber \\&+\alpha \epsilon \int _0^{2/\epsilon }(e^{2\alpha -\alpha \epsilon t}+e^{\alpha \epsilon t})\left| u\left( C_j^\star ,t\right) \right| dt+\frac{q}{1+q}(1+e^{2\alpha -2a/\epsilon }) \end{aligned}$$

(B.11)

for some fixed universal constant $q>0$. To this end, we first note that the lower bound (B.2) is implied by the relations:

$$\begin{aligned} \frac{1}{1+q}\bigg |\int _{-1}^1 e^{-\alpha w}Ai\left( e^{i\pi /6}\frac{w-id}{\epsilon }\right) dw\bigg |\ge \bigg |\int _{-1}^1 e^{\alpha w}Ai\left( e^{i\pi /6}\frac{w-id}{\epsilon }\right) dw\bigg |;\end{aligned}$$

(B.12)

$$\begin{aligned} \frac{1}{1+q}\bigg |\int _{-1}^1 e^{\alpha w}Ai\left( e^{i5\pi /6}\frac{w-id}{\epsilon }\right) dw\bigg |\ge \bigg |\int _{-1}^1 e^{-\alpha w}Ai\left( e^{i5\pi /6}\frac{w-id}{\epsilon }\right) dw\bigg |. \end{aligned}$$

(B.13)

Secondly, by substitution ($x=-1+\epsilon t$ in (B.12) and $x=1-\epsilon t$ in (B.13)) and the fact that $A_i(w)=\overline{Ai({\overline{w}})}$, (B.12) and (B.13) are equivalent to

$$\begin{aligned}&\frac{1}{1+q}e^{2\alpha }\bigg |\int _0^{2/\epsilon }e^{-\alpha \epsilon t} Ai\left( e^{i\pi /6}\left( \frac{-1-id}{\epsilon }+t\right) \right) dt\bigg |\nonumber \\&\quad \ge \bigg |\int _0^{2/\epsilon }e^{\alpha \epsilon t} Ai\left( e^{i\pi /6}\left( \frac{-1-id}{\epsilon }+t\right) \right) dt\bigg |\end{aligned}$$

(B.14)

$$\begin{aligned}&\frac{1}{1+q}e^{2\alpha }\bigg |\int _0^{2/\epsilon }e^{-\alpha \epsilon t} Ai\left( e^{i\pi /6}\left( \frac{-1-i{\overline{d}}}{\epsilon }+t\right) \right) dt\bigg |\nonumber \\&\quad \ge \bigg |\int _0^{2/\epsilon }e^{\alpha \epsilon t} Ai\left( e^{i\pi /6}\left( \frac{-1-i{\overline{d}}}{\epsilon }+t\right) \right) dt\bigg |, \end{aligned}$$

(B.15)

which in turn hold provided the following is satisfied:

$$\begin{aligned}&e^{2\alpha }\frac{1}{1+q}\bigg |\int _0^{2/\epsilon }e^{-\alpha \epsilon t} Ai\left( e^{i\pi /6}(C_j^\star +t)\right) dt\bigg |\nonumber \\&\quad \ge \bigg |\int _0^{2/\epsilon }e^{\alpha \epsilon t} Ai(e^{i\pi /6}(C_j^\star +t))dt\bigg |,\quad j=1,2, \end{aligned}$$

(B.16)

where the quantities $C_j^\star$ (B.8a) take values in the domain

$$\begin{aligned} Re(C_j^\star )&\le \max \left\{ \frac{-1- c_r}{\epsilon },\frac{-1+ c_r}{\epsilon }\right\} ,\quad Im(C_j^\star )=\frac{- c_i-\alpha \nu }{\epsilon }= \delta ,\quad {\alpha \ge 0}. \end{aligned}$$

(B.17)

According to Lemma A.3, $|A_0(C_j^\star )|$ is non-zero in this domain. Integrating both integrals in the inequality (B.16) by parts and then dividing by $|A_0(C_j^\star )|$ yield that

$$\begin{aligned}&\frac{e^{2\alpha }}{1+q}\left| \int _0^{2/\epsilon }(\alpha \epsilon )e^{-\alpha \epsilon t} \frac{A_0(C_j^\star +t)}{A_0(C_j^\star )}dt+\frac{e^{-2\alpha }A_0(C_j^\star +\frac{2}{\epsilon })}{A_0(C_j^\star )}-1\right| \nonumber \\&\quad \ge \left| -\int _0^{2/\epsilon } \alpha \epsilon e^{\alpha \epsilon t} \frac{A_0(C_j^\star +t)}{A_0(C_j^\star )}dt+\frac{e^{2\alpha }A_0(C_j^\star +\frac{2}{\epsilon })}{A_0(C_j^\star )}-1\right| ,\quad j=1,2. \end{aligned}$$

(B.18)

Combining it with the definition of u (B.10), we obtain the result (B.11).

Step 2: Proof of the inequality (B.11). Recalling from [46], we obtain

$$\begin{aligned} |u(C_j^\star ,t)|=\bigg |\exp \bigg (\int _0^t Re\frac{A_0'(C_j^\star +s)}{A_0(C_j^\star +s)}ds\bigg )\bigg |\le e^{-at}, \quad \forall t\in \left[ 0,\frac{2}{\epsilon }\right] ,\quad j=1,2,\nonumber \\ \end{aligned}$$

(B.19)

where a is defined in (A.11). Next, we square both sides of (B.11) (note both sides are positive), use $u(C_j^\star ,2\epsilon ^{-1})=-e^{-2a/\epsilon }$ and the upper bound (B.19), to obtain that the following implies (B.11),

$$\begin{aligned} e^{2\alpha }(1+e^{-2\alpha -2a/\epsilon })\ge & {} 1+e^{2\alpha -2a/\epsilon }+(1+q)\alpha \epsilon \int _0^{2/\epsilon }(e^{2\alpha -\alpha \epsilon t}+e^{\alpha \epsilon t})e^{-at}dt\nonumber \\&+ q(1+e^{2\alpha -2a/\epsilon }). \end{aligned}$$

(B.20)

Now substituting $t=(1+s)/\epsilon$ and dividing both sides by $\alpha e^{\alpha -a/\epsilon }$ yields the equivalent inequality

$$\begin{aligned} e^{a/\epsilon }\frac{a}{\epsilon }\int _0^1\frac{\sinh \alpha s}{\sinh \alpha }\sinh \frac{as}{\epsilon }ds\ge 1+\frac{q}{4(1+q)}\left( \frac{e^\alpha }{\sinh \alpha }+\frac{e^{2a/\epsilon }}{e^\alpha \sinh \alpha }\right) . \end{aligned}$$

(B.21)

The left hand side is calculated through integration by parts as

$$\begin{aligned} e^{a/\epsilon }\frac{a}{\epsilon }\int _0^1\frac{\sinh \alpha s}{\sinh \alpha }\sinh \frac{as}{\epsilon }ds= \frac{e^{2a/\epsilon }+1}{2}\left( \frac{1-\frac{\alpha \epsilon }{a} \frac{\tanh (a/\epsilon )}{\tanh (\alpha )}}{1-(\alpha \epsilon /a)^2}\right) =:F\left( \alpha ,\frac{a}{\epsilon }\right) . \end{aligned}$$

(B.22)

As a result, the following lower bound of the function F yields the inequality (B.11)

$$\begin{aligned} F\left( \alpha ,\frac{a}{\epsilon }\right) >1+\frac{q}{4(1+q)}\left( \frac{e^\alpha }{\sinh \alpha }+\frac{e^{2a/\epsilon }}{e^\alpha \sinh \alpha }\right) . \end{aligned}$$

(B.23)

The remaining part is devoted to proving this lower bound. Recall some properties of the function F from [46]. The function $F(\alpha ,\beta )$ is decreasing in terms of $\alpha$ and is increasing in terms of $\beta$. Next we distinguish between two regimes: K sufficiently large such that $Ka\ge 100$ where a is defined in (A.11), we define

$$\begin{aligned}&(1)\text { Low modes: } \alpha \epsilon \le Ka; \end{aligned}$$

(B.24)

$$\begin{aligned}&(2)\text { High modes: } \alpha \epsilon \ge Ka. \end{aligned}$$

(B.25)

In the low mode case (1), to derive the lower bound, we consider the minimum of the function $F(\alpha ,\beta )$ in the domain (here $B_K$ is a constant chosen sufficiently large relative to K)

$$\begin{aligned} D:=\{(\alpha ,\beta )|\beta \ge B_K, K\beta \ge \alpha , \alpha \ge 1\}. \end{aligned}$$

A figure of the region D can be found in Fig. 2. Due to monotonicity, the minimum of $F(\alpha ,\beta )$ is achieved on the half-line $\alpha =K\beta , \beta \ge B_K$. On this line segment, explicit calculation yields

$$\begin{aligned} \min _{(\alpha ,\beta )\in D}F(\alpha ,\beta )\ge & {} \min _{\beta \ge B_K}F(K\beta ,\beta )=\min _{\beta \ge B_K}\frac{e^{2\beta }+1}{2}\frac{1-K\frac{\tanh (\beta )}{\tanh (K\beta )}}{1-K^2}\nonumber \\\ge & {} \min _{\beta \ge B_K}{ \frac{e^{2\beta }}{2(K+1)}-\frac{1}{2(K-1)}}. \end{aligned}$$

(B.26)

Choosing q sufficiently small relative to $K^{-1}$ and $\nu$ sufficiently small relative to a and $K^{-1}$, (B.26) yields (B.23).

Fig. 2

Domain of interest

For the high mode case (2), we estimate the function F as follows:

$$\begin{aligned} F(\alpha ,a/\epsilon )&=\frac{e^{2a/\epsilon }+1}{2}\frac{\frac{\alpha \epsilon }{a}\frac{\tanh (a/\epsilon )}{\tanh (\alpha )}-1}{(\alpha \epsilon /a)^2-1}\ge \frac{e^{2a\alpha ^{1/3}\nu ^{-1/3}}+1}{2}\frac{K\frac{\tanh (a\alpha ^{1/3}\nu ^{-1/3})}{\tanh \alpha }-1}{\alpha ^{4/3}\nu ^{2/3}/a^2-1}\nonumber \\&\ge \frac{K/2}{\alpha ^{4/3}}\frac{e^{\alpha ^{1/3}}+1}{2}, \end{aligned}$$

(B.27)

which implies (B.23).

Step 3: Proof of inequalities (B.3) and (B.4). Recall (B.8a). By arguments similar to that used to prove (B.2), a suitable lower bound b as follows yields (B.3) and (B.4)

$$\begin{aligned} e^{\alpha }\epsilon \left| \int _0^{2/\epsilon } e^{-\alpha \epsilon t}H_-(C_j^\star +t)dt\right| \ge b>0. \end{aligned}$$

(B.28)

Recalling the definition of $A_0$(A.10) and u (B.10), then an integration by parts yields that the following implies (B.28):

$$\begin{aligned} e^{\alpha }\epsilon \left| u\bigg (C_j^\star ,\frac{2}{\epsilon }\bigg )e^{-2\alpha }-1\right| \ge \epsilon e^\alpha \left| \int _0^{2/\epsilon }e^{-\alpha \epsilon t}\alpha \epsilon \left| u(C_j^\star , t)\right| dt\right| +\frac{b}{|A_0(C_j^\star )|}. \end{aligned}$$

(B.29)

Using the upper bound (B.19) and $u(C_j^\star ,2\epsilon ^{-1})=-e^{-2a/\epsilon }$, we see that (B.3) and (B.4) hold if the following is satisfied:

$$\begin{aligned} e^\alpha (1-e^{-2\alpha -2a/\epsilon })\ge e^\alpha \int _0^{2/\epsilon } e^{-\alpha \epsilon t}\alpha \epsilon e^{-at}dt+\frac{b}{|A_0(C_j^\star )|\epsilon }. \end{aligned}$$

(B.30)

A calculation shows that (B.30) holds if

$$\begin{aligned} \bigg (\frac{a}{\alpha \epsilon +a}\bigg )(e^{\alpha }-e^{-\alpha -2a/\epsilon })\ge \frac{b}{|A_0(C_j^\star )|\epsilon }. \end{aligned}$$

(B.31)

To prove the inequality (B.31), we distinguish between the high modes and low modes (B.24) again. If $\alpha$ is small, i.e., $\alpha \epsilon \le Ka$, the inequality (B.31) is satisfied if b is small

$$\begin{aligned} b\le \frac{e^\alpha |A_0(C_j^\star )|\epsilon }{(K+1)2}. \end{aligned}$$

(B.32)

For the high modes $\alpha \epsilon \ge Ka\ge 100$, the inequality (B.31) holds if

$$\begin{aligned} b\le \frac{e^\alpha a|A_0(C_j^\star )|\epsilon }{2(\alpha \epsilon +a)}\approx \frac{e^\alpha |A_0(C_j^\star )|\epsilon }{\alpha \epsilon }. \end{aligned}$$

Combining the estimates in different regimes, we obtain that (B.3) and (B.4) hold as long as the lower bound b in (B.28) is smaller than

$$\begin{aligned} b\lesssim \frac{e^\alpha |A_0(C_j^\star )|\epsilon }{\langle \alpha \epsilon \rangle }. \end{aligned}$$

Combining it with the asymptotic expansion of $A_0$ in Lemma A.4 yields the estimate (2.18). This completes the proof part (i) of the lemma.

Step 4: Part (ii)—Proof of the inequalities (B.5), (B.6) and (B.7) in the high mode case. Let us comment on the proof of (B.5), (B.6), and (B.7). Here we use the observation [(3.5), [46]]: for sufficiently large $R>0$, and for all z in $G_R:=\{z||z|\ge R,-13\pi /12\le \mathrm {ph} z\le \pi /12\}$, the following inequality holds for some universal constant $B > 0$,

$$\begin{aligned} Re\frac{A_0'(z)}{A_0(z)}\le -R^{1/2}\cos \frac{7\pi }{24}+\frac{B}{R}. \end{aligned}$$

(B.33)

Following the estimate (B.33), we have that the Langer variables in this case satisfy

$$\begin{aligned} C_j^\star =\frac{-1\pm c_r}{\epsilon }-\kappa \frac{\alpha \nu }{\epsilon }i,\quad |C_j^\star | > rsim \alpha ^{4/3}\nu ^{2/3}. \end{aligned}$$

Now from (B.33) and the definition of a in (A.11) gives $a > rsim \kappa ^{1/2}\alpha ^{2/3}\nu ^{1/3}=\kappa ^{1/2}\alpha \epsilon > rsim 1$. Therefore, by choosing K sufficiently large we have $\alpha \epsilon \le K(\kappa ) a$. This reduces to case (1) in the previous steps. The inequalities (B.5), (B.6) (B.7) with implicit constants depending on $\kappa$ follow by the same arguments as above. Note that $\langle \alpha \epsilon \rangle$ will not appear. $\quad \square$

1.2 Evans function estimate in the connection region

Recall the contours $\Gamma _t^{\pm }$, $\Gamma _E$, and $\Gamma _j^{\pm }$. On the contour, it is clear that we may write $c_i$ as a function of $c_r$; denote this function $c_i = \Gamma (c_r)$.

Lemma B.2

(Connection region Evans function). There exists a universal $\nu _0$ such that for $\nu \in (0,\nu _0]$, the Evans function $D(\alpha ,c)$ is non-zero except in the region

$$\begin{aligned} \left\{ z \in {\mathbb {C}}\big | \Gamma (z_r) > z_i \right\} . \end{aligned}$$

(B.34)

Moreover, on $\Gamma _t^{\pm }, \Gamma _j^{\pm }$ the lower bounds (B.2), (B.3), and (B.4) all hold.

Proof

The proof is similar to Lemma B.1 but some changes are required because $c_i$ is no longer constant and the $C_j^\star$’s, defined as

$$\begin{aligned}&C_1^\star =\frac{-1-id}{\epsilon }=\frac{-1- c_r}{\epsilon }+\frac{-c_i-\alpha \nu }{\epsilon } i, \end{aligned}$$

(A.35)

$$\begin{aligned}&C_2^\star =\frac{-1-i{\overline{d}}}{\epsilon }=\frac{-1+ c_r }{\epsilon }+\frac{-c_i-\alpha \nu }{\epsilon } i, \end{aligned}$$

(A.36)

are no longer in the region specified in Lemma A.3. As above, one must bound a from below on the contour:

$$\begin{aligned} -\max _{c \in \Gamma _2^{\pm }\cup \Gamma _t^\pm } Re\frac{A_0'(C_j^\star +s)}{A_0(C_j^\star +s)}>0,\quad \forall s\in [0,{2}{\epsilon }^{-1}]. \end{aligned}$$

(A.37)

For this, we use again (B.33). Since the angle between the region $\Gamma _2^+\cup \Gamma _t^+$ ($\Gamma _2^-\cup \Gamma _t^-$) and the positive imaginary (negative) axis are small, the argument $C_j^\star +s$ in the equation (A.37) is inside the domain $G_R$. This completes the proof of (A.37); the rest of the argument follows similarly to Lemma B.1. $\quad \square$