Existence, Stability and Slow Dynamics of Spikes in a 1D Minimal Keller–Segel Model with Logistic Growth

Abstract

We analyze the existence, linear stability, and slow dynamics of localized 1D spike patterns for a Keller–Segel model of chemotaxis that includes the effect of logistic growth of the cellular population. Our analysis of localized patterns for this two-component reaction–diffusion (RD) model is based, not on the usual limit of a large chemotactic drift coefficient, but instead on the singular limit of an asymptotically small diffusivity $d_2=\epsilon ^2\ll 1$ of the chemoattractant concentration field. In the limit $d_2\ll 1$, steady-state and quasi-equilibrium 1D multi-spike patterns are constructed asymptotically. To determine the linear stability of steady-state N-spike patterns, we analyze the spectral properties associated with both the “large” ${{\mathcal {O}}}(1)$ and the “small” o(1) eigenvalues associated with the linearization of the Keller–Segel model. By analyzing a nonlocal eigenvalue problem characterizing the large eigenvalues, it is shown that N-spike equilibria can be destabilized by a zero-eigenvalue crossing leading to a competition instability if the cellular diffusion rate $d_1$ exceeds a threshold, or from a Hopf bifurcation if a relaxation time constant $\tau$ is too large. In addition, a matrix eigenvalue problem that governs the stability properties of an N-spike steady-state with respect to the small eigenvalues is derived. From an analysis of this matrix problem, an explicit range of $d_1$ where the N-spike steady-state is stable to the small eigenvalues is identified. Finally, for quasi-equilibrium spike patterns that are stable on an ${{\mathcal {O}}}(1)$ time-scale, we derive a differential algebraic system (DAE) governing the slow dynamics of a collection of localized spikes. Unexpectedly, our analysis of the KS model with logistic growth in the singular limit $d_2\ll 1$ is rather closely related to the analysis of spike patterns for the Gierer–Meinhardt RD system.

Appendices

Solvability of the Outer Problem: Turing Instability of the Base State

In this appendix, we relate the solvability of the outer problem (2.22) to Turing bifurcation points in the parameter $d_1$ for the spatially uniform base state $u=v=0$ of (1.2). This analysis will motivate Remark 2.1.

On an interval of length L, with homogeneous Neumann conditions for u and v, we linearize (1.2) around $u=v=0$ by setting $u=e^{\lambda t + ikx}\Phi$ and $v=e^{\lambda t + ikx}N$, where $k={m \pi /L}$ with $m=1,2,\ldots$. We readily obtain that

$$\begin{aligned} \left( \begin{array}{cc} -d_1 k^2 + \mu \bar{u} - \lambda &{} 0\\ 1 &{} -\epsilon ^2 k^2 -1 - \lambda \\ \end{array} \right) \left( \begin{array}{c} \Phi \\ N \\ \end{array} \right) = {{\textbf{0}}} \,, \end{aligned}$$

(A.1)

which has a nontrivial solution if and only if $\lambda =-1-\epsilon ^2 k^2$ or $\lambda =-d_1k^2+\mu \bar{u}$. As such, with $k={m \pi /L}$, there is a zero-eigenvalue crossing associated with the spatially uniform state $u=v=0$ at the critical values

$$\begin{aligned} d_1 = \frac{\mu \bar{u} L^2}{m^2 \pi ^2} , \quad m=1,2,\ldots . \end{aligned}$$

(A.2)

This base-state is linearly stable on a domain of length L when $d_1>{\mu \bar{u} L^2/\pi ^2}$. Setting $L=2$, consistent with (1.2), we conclude that (A.2) coincides precisely with the “resonant” values of $d_1$ in (2.23) for the outer problem.

However, in our construction of N-spike steady-state patterns for (1.2), the spatially uniform base state approximates the outer solution $w_o$ only on intervals of length 2/N. Upon setting $L={2/N}$ in (A.2), this observation suggests that the outer solution for an N-spike steady-state should be linearly stable when $d_1> {4\mu \bar{u} /(N^2 \pi ^2)}$. This latter threshold also has the alternative interpretation that it is the smallest value of $d_1$ for which the outer solution $w_o$ is always positive in $|x|<1$. In particular, for an N-spike steady-state, it is easy to verify that this positivity condition for $w_o$ holds when $d_1> d_{1pN}:= {{\bar{u}} \, \mu /\lambda _1}$, where $\lambda _1:={N^2\pi ^2/4}$ is the first non-zero eigenvalue of the negative Neumann Laplacian -${d^2/\textrm{d}x^2}$ on $(-1/N,1/N)$. We remark that for quasi-equilibrium patterns with unequally spaced spikes, this positivity threshold must be modified to (2.36).

Next, we verify that the outer problem (2.22) is solvable for an N-spike steady-state pattern when $d_1=d_{1Tm}$, where $d_{1Tm}$ is one of the “resonant” values in (2.23) with $m=1,\ldots ,N-1$. For the steady-state problem, where $v_{\max k}=v_{\max 0}$ and where $x_k=x_{k}^{0}$, with $x_{k}^{0}$ as given in (2.29), (2.22) is solvable at $d_1=d_{1Tm}$ if and only if

$$\begin{aligned} \int _{-1}^{1} w_{oh} {{\mathcal {L}}}_0 w_o \, \textrm{d}x= & {} \frac{2{\bar{\chi }} \epsilon }{3} v_{\max 0}^3 \sum _{k=1}^{N} \int _{-1}^{1} w_{oh}(x)\, \delta (x-x_k^{0}) \, \textrm{d}x \nonumber \\= & {} \frac{2{\bar{\chi }} \epsilon }{3} v_{\max 0}^3 \sum _{k=1}^{N} \cos \left( \frac{m\pi }{2}\frac{(2k-1)}{N}\right) =0 . \end{aligned}$$

(A.3)

The trigonometric sum in (A.3) can be readily evaluated for $m=1,\ldots ,N-1$ with the result

$$\begin{aligned} \sum _{k=1}^{N} \cos \left( \frac{m\pi }{2}\frac{(2k-1)}{N}\right) = \frac{\sin (m\pi )}{2\sin \left( {m\pi /N}\right) }=0 . \end{aligned}$$

(A.4)

As a consequence, (2.22) is solvable for an N-spike steady-state even when $d_1={{\mathcal {T}}}_e$ (see (2.35)).

Finally, we remark that when $d_1={4\mu \bar{u}/(m^2 \pi ^2)}$, for some $m=1,\ldots ,N-1$, a solution (non-unique) to (2.22) for an N-spike steady-state can be represented as $u_o\sim w_o=\frac{2}{3}{\bar{\chi }}\epsilon v_{\max 0}^3\sum _{k=1}^N G_{m}(x;x_k^{0})$. Here, with the operator ${{\mathcal {L}}}_0$ of (2.22), the modified Green’s function $G_{m}(x;\xi )$ satisfies

$$\begin{aligned} {{\mathcal {L}}}_0 G_m = \delta (x-\xi ) - w_{oh}(\xi ) w_{oh}(x) , \quad |x|\le 1 ; \qquad G_{mx}(\pm 1;\xi )=0. \end{aligned}$$

(A.5)

Although $G_m$ can be found analytically, for simplicity we have restricted our analysis only to when $d_1\in {{\mathcal {T}}}_e$.

Calculation of ${\mathcal {G}}_{\lambda }$ and ${\mathcal {P}}$

In this appendix, we show how to determine the matrix spectrum of ${{\mathcal {G}}}_{\lambda }$, as defined in (3.16) of Sect. 3. Moreover, we calculate ${{\mathcal {P}}}$, as defined in (4.27) of Sect. 4. To do so, we introduce an auxiliary problem for $y=y(x)$, given by

$$\begin{aligned} \frac{d_1}{\mu }y^{\prime \prime }+ {{{\hat{u}}}} y=0\,, \quad -1<x<1\,; \quad y^{\prime }(\pm 1)=0\,; \quad [y]_j=0\,, \quad \Big [\frac{d_1}{\mu }y^{\prime } \Big ]_j=b_j\,, \end{aligned}$$

(B.1)

for $j=1,\ldots ,N$, where $[y]_j:=y(x_j^+)-y(x_j^-)$ and $x_j=x_j^{0}$ is given by (2.29). Here ${\hat{u}}:=\bar{u}-{\tau \lambda _0/\mu }$. This problem is solvable when $d_1\ne {4\mu {\hat{u}}/(m^2\pi ^2)}$ for $m=1,2,\ldots$. When $\tau =0$, (B.1) is always solvable when $d_1\in {{\mathcal {T}}}_e$.

With the exception of this restricted set for $d_1$, the solution to (B.1) can be represented in terms of the Green’s function $G_{\lambda }(x;x_k)$, satisfying (3.13), as $y=\sum _{k=1}^N b_k G_{\lambda }(x;x_k)$. Upon defining ${\varvec{y}}:=\left( y_{1},\ldots ,y_N\right) ^T$, $\varvec{\langle y^{\prime }\rangle }:=\left( y^{\prime }_{1},\ldots , y^{\prime }_{N}\right) ^T$ and ${\varvec{b}}:=\left( b_1,\ldots ,b_N\right) ^T$, where $y_j=y(x_j)$ and $\langle y^{\prime }\rangle _j= \big (y^{\prime }(x_j^{+})+y^{\prime }(x_j^{-})\big )/2$, we identify the eigenvalue-dependent Green’s matrix ${{\mathcal {G}}}_{\lambda }$ of (3.16) and ${{\mathcal {P}}}$ of (4.27) as

$$\begin{aligned} {\varvec{y}}={\mathcal {G}}_{\lambda }{\varvec{b}}\,, \qquad \varvec{ \langle y^{\prime }\rangle } ={\mathcal {P}}{\varvec{b}}\,. \end{aligned}$$

(B.2)

Next, we show how to represent ${{\mathcal {G}}}_{\lambda }$ and ${{\mathcal {P}}}$ in terms of tridiagonal matrices. By solving (B.1) on each subinterval, and enforcing the continuity conditions $[y]_j=0$ for $j=1,\ldots ,N$, we get

$$\begin{aligned} y=\left\{ \begin{array}{ll} y_1\frac{\cos [\theta _\lambda (1+x)]}{\cos [\theta _\lambda (1+x_1)]}\,,&{}-1<x<x_1\,,\\ y_j\frac{\sin [\theta _\lambda (x_{j+1}-x)]}{\sin [\theta _\lambda (x_{j+1}-x_j)]}+ y_{j+1}\frac{\sin [\theta _\lambda (x-x_j)]}{\sin [\theta _\lambda (x_{j+1}-x_j)]}\,, &{}x_j<x<x_{j+1}\,,\quad j=1,\ldots ,N-1\,,\\ y_{N}\frac{\cos [\theta _\lambda (1-x)]}{\cos [\theta _\lambda (1-x_{N})]}\,,&{}x_{N}<x<1\,. \end{array} \right. \end{aligned}$$

(B.3)

Then, upon satisfying the jump conditions in (B.1) we can write ${\varvec{b}}$ as

$$\begin{aligned} {\varvec{b}} = \frac{d_1\theta _{\lambda }}{\mu } {{\mathcal {D}}} {\varvec{y}} , \qquad \text{ where } \quad \theta _{\lambda }:= \sqrt{\frac{\mu }{d_1}\left( \bar{u}-\frac{\tau \lambda _0}{\mu }\right) } . \end{aligned}$$

(B.4)

Here, for $d_1\ne {4\mu {\hat{u}}/(m^2\pi ^2)}$ with $m=1,2,\ldots$, ${{\mathcal {D}}}$ is the invertible tridiagonal matrix defined by

$$\begin{aligned} {\mathcal {D}}=\left( \begin{array}{ccccccc} d &{} f&{}0&{}\cdots &{}0 &{}0&{} 0\\ f &{} e&{}f&{}\cdots &{}0 &{}0&{} 0\\ 0 &{} f&{}e&{}\ddots &{}0 &{}0&{} 0\\ \vdots &{} \vdots &{}\ddots &{}\ddots &{}\ddots &{}\vdots &{} \vdots \\ 0 &{} 0&{}0&{}\ddots &{}e &{}f&{} 0\\ 0 &{} 0&{}0&{}\cdots &{}f &{}e&{} f\\ 0 &{} 0&{}0&{}\cdots &{}0 &{}f&{} d\\ \end{array} \right) \,. \end{aligned}$$

(B.5)

The matrix entries of ${{\mathcal {D}}}$, for which the identity $d=f+e$ holds, are

$$\begin{aligned} \textrm{d}\equiv \tan (\theta _\lambda /N)-\cot (2\theta _\lambda /N)\,,\quad e\equiv -2\cot (2\theta _\lambda /N)\,,\quad f\equiv \csc (2\theta _\lambda /N)\,. \end{aligned}$$

(B.6)

By combining (B.4) with the first equation in (B.2), we conclude for $d_1\ne {4\mu {\hat{u}}/(m^2\pi ^2)}$ for $m=1,2,\ldots$ that

$$\begin{aligned} {{\mathcal {G}}}_{\lambda } = \sqrt{\frac{\mu }{d_1{\hat{\mu }}}} {{\mathcal {D}}}^{-1}, \quad \text{ with } \quad {\hat{u}}=\bar{u} - \frac{\tau \lambda _0}{\mu } . \end{aligned}$$

(B.7)

When $\tau =0$, we remark that (B.7) holds when $d_1\in {{\mathcal {T}}}_e$.

Since ${{\mathcal {D}}}$ is a tridiagonal matrix with a constant row sum, its eigenpairs $\kappa _j$ and $\textbf{q}_j$ for $j=1,\ldots ,N$ can be calculated explicitly (see Iron et al. 2001), with the following result:

Proposition B.1

The eigenvalues $\kappa _j$ and the normalized eigenvectors of ${\mathcal {D}}$ are

$$\begin{aligned} \left\{ \begin{array}{ll} \kappa _1=e+2f\,;\quad \kappa _j=e+2f\cos \left( \pi (j-1)/N\right) \,, &{}j=2,\ldots ,N,\\ {\varvec{q}}_1=\frac{1}{\sqrt{N}}(1,\ldots ,1)^T\,,\quad q_{l,j}=\sqrt{\frac{2}{N}}\cos \left( \frac{\pi (j-1)}{N}(l-\frac{1}{2})\right) \,, \quad &{}j=2,\ldots ,N\,,\,\, l=1,\ldots ,N, \end{array} \right. \end{aligned}$$

where $\varvec{q_j}=(q_{1,j},\ldots ,q_{N,j})^T$ and d, e and f are given by (B.6). By using (B.7), the eigenvalues $\sigma _j$ of ${{\mathcal {G}}}_{\lambda }$ when $d_1\ne {4\mu {\hat{u}}/(m^2\pi ^2)}$ for $m=1,2,\ldots$, are

$$\begin{aligned} \sigma _j = \sqrt{\frac{\mu }{d_1{\hat{u}}}} \left[ e + 2f \cos \left( \frac{\pi (j-1)}{N}\right) \right] ^{-1}, \qquad j=1,\ldots ,N . \end{aligned}$$

(B.8)

By setting $\lambda _0=0$, we use (B.7) and Proposition B.1 to calculate $a_g$, as defined in (2.30). For $d_1\in {{\mathcal {T}}}_e$, we get

$$\begin{aligned}{} & {} a_g = \sum _{k=1}^N G\big (x_j^0;x_k^0\big )= \sqrt{\frac{\mu }{d_1\bar{u}}} {{\mathcal {D}}}^{-1} \sqrt{N} {{\varvec{q}}}_1 = \sqrt{\frac{\mu }{d_1\bar{u}}}\frac{1}{(e+2f)}= \frac{1}{2} \sqrt{\frac{\mu }{d_1\bar{u}}} \cot \left( \frac{\theta }{N}\right) , \nonumber \\{} & {} \quad \theta = \sqrt{\frac{\mu \bar{u}}{d_1}}. \end{aligned}$$

(B.9)

To determine ${{\mathcal {P}}}$ when $\lambda _0=0$, we use (B.3) to write $\varvec{\langle y^{\prime }\rangle }$ in terms of ${\varvec{y}}$ as $\varvec{\langle y^{\prime }\rangle }=-\left( {\theta /2}\right) \csc \Big ({2\theta /N}\Big ){\mathcal {C}}^T{\varvec{y}}$, where $\theta =\sqrt{\mu \bar{u}/d_1}$ and ${{\mathcal {C}}}$ is the tridiagonal matrix defined by

$$\begin{aligned} {\mathcal {C}}:=\left( \begin{array}{ccccccc} 1 &{} 1&{}0&{}\cdots &{}0 &{}0&{} 0\\ -1 &{} 0&{}1&{}\cdots &{}0 &{}0&{} 0\\ 0 &{} -1&{}0&{}\ddots &{}0 &{}0&{} 0\\ \vdots &{} \vdots &{}\ddots &{}\ddots &{}\ddots &{}\vdots &{} \vdots \\ 0 &{} 0&{}0&{}\ddots &{}0 &{}1&{} 0\\ 0 &{} 0&{}0&{}\cdots &{}-1 &{}0&{} 1\\ 0 &{} 0&{}0&{}\cdots &{}0 &{}-1&{} -1\\ \end{array} \right) . \end{aligned}$$

(B.10)

By combining the second equation in (B.2) with this result, we conclude for $d_1\in {{\mathcal {T}}}_e$ that

$$\begin{aligned} {\mathcal {P}}=-\frac{\mu }{2d_1}\csc \left( \frac{2\theta }{N}\right) \mathcal C^T{\mathcal {D}}^{-1}\,. \end{aligned}$$

(B.11)

Proof of Theorem 3.1

For convenience, we drop the overbars in (3.24) to rewrite the NLEP as

$$\begin{aligned}&\Psi _{0 z z}+ U_0\Psi _0-\alpha U_0\frac{\int _{-\infty }^\infty { U_0}^2\Psi _0 \, \textrm{d}z }{\int _{-\infty }^\infty { U_0}^2 \, \textrm{d}z}=\Lambda \,, \quad -\infty<z<+\infty \,; \nonumber \\&\quad \Psi _0\, \text{ bounded } \text{ as } \,\,\, |z|\rightarrow \infty \,. \end{aligned}$$

(C.1)

Here $U_0=2\textrm{sech}^2 z$, $\Lambda :=\delta ^2(\lambda _0+1)$ with $\delta :={2/({\bar{\chi }} v_{\max 0})}$. It is well-known (Kolokolnikov et al. 2009) that the homoclinic solution to $w_{zz}- w+{w}^3=0$ on $-\infty<z<\infty$ with $w(0)>0$, $w^{\prime }(0)=0$ and $w\rightarrow 0$ as $|z|\rightarrow \infty$ is $w=\sqrt{2}\textrm{sech}(y)$. Therefore, we have $U_0=w^2$ and the NLEP (C.1) becomes

$$\begin{aligned}&\Psi _{0zz}+ w^2\Psi _0- \alpha w^2\frac{\int _{-\infty }^\infty { w}^4\Psi _0 \, \textrm{d}z }{\int _{-\infty }^\infty { w}^4 \, \textrm{d}z}=\Lambda \Psi _0\,, \quad -\infty<z<+\infty ; \nonumber \\&\quad \Psi _0 \text{ bounded } \text{ as }\,\,\, |z|\rightarrow \infty \,. \end{aligned}$$

(C.2)

There is a standard approach (Wei 1999) to study (C.2). Firstly, we focus on the following local eigenvalue problem:

$$\begin{aligned} \Psi _{0zz}+ w^2\Psi _0=\lambda \Psi _0\,, \quad -\infty<z<\infty \,; \qquad \Psi _0 \,\,\, \text{ bounded } \text{ as } \,\,\, |z|\rightarrow \infty \,. \end{aligned}$$

(C.3)

As shown in Kolokolnikov et al. (2009), the principal eigenvalue of (C.3) is $\lambda =1$ and the corresponding eigenfunction is $\Psi _0=w$. Next, we transform (C.2) into a form more amenable for analysis. To this end, we observe from the ODE $w^{\prime \prime }-w+w^3=0$ that $w^2$ satisfies

$$\begin{aligned} (w^2)_{zz}- 4w^2+3w^4=0\,, \quad -\infty< z<+\infty \,; \qquad w\rightarrow 0 \quad \text{ as } \quad |z|\rightarrow \infty \,. \end{aligned}$$

(C.4)

Therefore, upon multiplying the $\Psi _0$-equation in (C.2) by $w^2$ and integrating it over $(-\infty ,\infty )$ by parts, we get

$$\begin{aligned} \int _{-\infty }^\infty (w^2)_{zz}\Psi _0\, \textrm{d}z+\int _{-\infty }^\infty w^4\Psi _0\, \textrm{d}z- \alpha \int _{-\infty }^\infty w^4\Psi _0\, \textrm{d}z=\Lambda \int _{-\infty }^\infty w^2\Psi _0\, \textrm{d}z. \end{aligned}$$

(C.5)

Next, upon substituting (C.4) into (C.5), we obtain

$$\begin{aligned} (4-\Lambda )\int _{-\infty }^\infty w^2\Psi _0\, \textrm{d}z=(2+\alpha )\int _{-\infty }^\infty w^4\Psi _0\, \textrm{d}z. \end{aligned}$$

(C.6)

Then, by using (C.6), we transform the NLEP in (C.2) into the following form, as written in (3.41):

$$\begin{aligned} \Psi _{0zz}+w^2\Psi _0- \kappa \frac{\int _{-\infty }^\infty w^2\Psi _0\, \textrm{d}z}{\int _{-\infty }^\infty w^4 \, \textrm{d}z}\, w^2=\Lambda \Psi _0\,, \qquad \kappa :=\frac{\alpha (4-\Lambda )}{(2+\alpha )}\,, \end{aligned}$$

(C.7)

Next, we test (C.7) against the conjugate $\Psi _0^*$ and by integrating the resulting expression by parts we get

$$\begin{aligned} \int _{-\infty }^\infty \vert \Psi _{0z}\vert ^2\, \textrm{d}z-\int _{-\infty }^\infty w^2\vert \Psi _0\vert ^2\, \textrm{d}z+\Lambda \int _{-\infty }^\infty \vert \Psi _0\vert ^2\, \textrm{d}z =-\frac{\alpha (4-\Lambda )\big \vert \int _{-\infty }^\infty w^2\Psi _0\, \textrm{d}z \big \vert ^2}{(2+\alpha )\int _{-\infty }^\infty w^4\, \textrm{d}z}. \end{aligned}$$

(C.8)

We first claim that $\Lambda$ is real-valued when $\alpha$ is real-valued. To show this, the imaginary part of (C.8) yields

$$\begin{aligned} \text{ Im }(\Lambda )\int _{-\infty }^\infty \vert \Psi _0\vert ^2\, \textrm{d}z =\frac{\alpha \, \text{ Im }(\Lambda )}{2+\alpha }\frac{\big \vert \int _{-\infty }^\infty w^2\Psi _0\, \textrm{d}z\big \vert ^2}{\int _{-\infty }^\infty w^4\, \textrm{d}z}. \end{aligned}$$

(C.9)

Then, upon invoking the Cauchy-Schwartz inequality, we obtain

$$\begin{aligned} \frac{\alpha }{2+\alpha }\frac{\big \vert \int _{-\infty }^\infty w^2\Psi _0\, \textrm{d}z \big \vert ^2}{\int _{-\infty }^\infty \, w^4\textrm{d}z}\le \frac{\alpha }{2+\alpha }{\int _{-\infty }^\infty \vert \Psi _0\vert ^2\, \textrm{d}z}. \end{aligned}$$

(C.10)

Upon substituting this inequality into (C.9), we conclude that $\text{ Im }(\Lambda )=0$. This completes the proof of our claim. It immediately follows that (C.8) is also real-valued when $\alpha$ is real-valued.

The next step is to study the sign of $\Lambda$ in (C.7). We claim that

$$\begin{aligned} \int _{-\infty }^\infty \vert \Psi _{0z}\vert ^2\, \textrm{d}z- \int _{-\infty }^\infty w^2\vert \Psi _0\vert ^2\, \textrm{d}z\ge - \frac{\big \vert \int _{-\infty }^\infty w^2\Psi _0\, \textrm{d}z\big \vert ^2 }{\int _{-\infty }^\infty w^2\, \textrm{d}z}. \end{aligned}$$

Similarly as the proof of Lemma 5 in Kolokolnikov et al. (2009), this claim is established if we can equivalently show that the real eigenvalues $\upsilon$ of the following NLEP are non-positive:

$$\begin{aligned} \Delta \Psi _0+w^2\Psi _0-w^2\frac{\int _{-\infty }^\infty w^2\Psi _0 \textrm{d}z}{\int _{-\infty }^\infty w^2 \textrm{d}z}=\upsilon \Psi _0. \end{aligned}$$

(C.11)

We first observe that if $\Psi _0\equiv 1$, then $\upsilon =0.$ Next, we observe that (C.11) is equivalent to solving

$$\begin{aligned} (L_{0}-\upsilon )\Psi _0=w^2\,, \qquad \int _{-\infty }^\infty w^2\Psi _0\, \textrm{d}z= \int _{-\infty }^\infty w^2\, \textrm{d}z=4. \end{aligned}$$

(C.12)

As such, we define $\Xi$ as

$$\begin{aligned} \Xi (\upsilon ):= \int _{-\infty }^\infty w^2(L_{0}-\upsilon )^{-1}w^2 \, \textrm{d}z-4. \end{aligned}$$

Since the operator is self-adjoint and $L_0 (1)=w^2$, we obtain that $\Xi (0)=0.$ By differentiating in $\Xi$ we get

$$\begin{aligned} \Xi ^{\prime }(\upsilon )=\int _{-\infty }^\infty w^2(L_{0}-\upsilon )^{-2}w^2 \, \textrm{d}z =\int _{-\infty }^\infty [(L_{0}-\upsilon )^{-1}w^2]^2\, \textrm{d}z>0. \end{aligned}$$

Noting that $L_0$ admits a single positive eigenvalue at $\upsilon =1,$ it follows that $\Xi$ has a single pole at $\upsilon =1$ and that there are no other poles for $\upsilon >0.$ On the other hand, as $\upsilon \rightarrow +\infty ,$ we have

$$\begin{aligned} \Xi (\upsilon )\sim -\frac{1}{\upsilon }\int _{-\infty }^\infty w^4\, \textrm{d}z\rightarrow 0^{-}. \end{aligned}$$

To summarize, $\Xi (\upsilon )$ has a vertical asymptote at $\upsilon =1$; $\Xi (0) = 0$, $\Xi \rightarrow 0^{-}$ as $\upsilon \rightarrow \infty$ and $\Xi$ is increasing for $\upsilon >0$. It follows that $\Xi (\upsilon )\not =0$ for all $\upsilon >0$, which proves our claim.

Next, from (C.8), we conclude that when $\Lambda \ge {4/({\bar{\chi }}^{2}v_{\max 0}^{2})}$ we have

$$\begin{aligned} -1+\Lambda \frac{\int _{-\infty }^\infty w^2\, \textrm{d}z}{\int _{-\infty }^\infty w^4\, \textrm{d}z} \le -\frac{\alpha (4-\Lambda )}{2+\alpha }\frac{\int _{-\infty }^\infty w^2 \, \textrm{d}z}{\int _{-\infty }^\infty w^4\, \textrm{d}z}. \end{aligned}$$

(C.13)

By using the identity $4\int _{-\infty }^\infty w^2\, \textrm{d}z=3\int _{-\infty }^\infty w^4\,\textrm{d}z$, (C.13) implies that $\alpha \le 1-{3\Lambda /4}$. By observing that the condition $\Lambda \ge {4/({\bar{\chi }}^{2}v_{\max 0}^{2})}$ holds when $\lambda _0<0$, we conclude that $\lambda _0<0$ when

$$\begin{aligned} \alpha \le 1-3{\bar{\chi }}^{-2}v_{\max 0}^{-2}\,. \end{aligned}$$

(C.14)

Similarly as in Wei (1999), we find when $\alpha =1$, $\Psi _0\equiv 1$ is an eigenfunction such that (C.2) admits the zero eigenvalue. If $\alpha >1,$ we claim there exists a positive real eigenvalue of (C.2). In fact, assume that some $\Lambda$ satisfies $\Lambda \ge 0$. Then, one obtains that (C.2) can be written as the equivalent form

$$\begin{aligned} \Psi _0=\alpha \frac{\int _{-\infty }^\infty w^4\Psi _0\, \textrm{d}z}{\int _{-\infty }^\infty w^4 \,\textrm{d}z} (L_0-\Lambda )^{-1}w^2\,, \quad \text{ where } \quad L_0\Psi _0=\Psi _{0zz}+ w^2\Psi _0\,, \end{aligned}$$

and where $\alpha$ satisfies $\int _{-\infty }^\infty w^4\, \textrm{d}z=\alpha \int _{-\infty }^\infty \left[ (L_0-\Lambda )^{-1}w^2\right] w^4\, \textrm{d}z$. Then, we define $R(\Lambda )$ as

$$\begin{aligned} R(\Lambda ):= \int _{-\infty }^\infty w^4\, \textrm{d}z-\alpha \int _{-\infty }^\infty \left[ (L_0-\Lambda )^{-1}w^2\right] w^4\, \textrm{d}z. \end{aligned}$$

Since $R(0)=(1-\alpha )\int _{-\infty }^\infty w^4\, \textrm{d}z<0$ and $R(\Lambda )\rightarrow +\infty$ as $\Lambda \rightarrow 1^{-},$ we conclude that there exists a positive $\Lambda \in (0,1)$ such that $R(\Lambda )=0.$ This finishes the proof of our claim.

By comparing this result and (C.14), it follows that there is still a gap region between $1-3{\bar{\chi }}^{-2}v_{\max 0}^{-2}$ and 1. To eliminate this gap, and obtain a refined prediction of the threshold $\alpha _c$, we shall rewrite the solution to (C.7) in terms of the hypergeometric function and perform a detailed asymptotic expansion of it similar to that in Wei and Winter (2002).

To do so, we first recall the definition and some properties of generalized hypergeometric functions (Slater 1966). The generalized hypergeometric functions ${}_pF_q(a_1,\cdots ,a_p;b_1,\cdots ,b_q;z)$ are defined by the following series:

$$\begin{aligned} {}_pF_{q}(a_1,\cdots ,a_p;b_1,\cdots ,b_q;z)=1+\frac{a_1\cdots a_p}{b_1\cdots b_q}\frac{z}{1!}+\frac{(a_1+1)\cdots (a_p+1)}{(b_1+1)\cdots (b_q+1)}\frac{z^2}{2!}+\cdots \,. \end{aligned}$$

(C.15)

Their derivatives satisfy a recursion formula, given by

$$\begin{aligned}&\frac{\textrm{d}}{\textrm{d}z}{}_pF_{q}(a_1,\cdots ,a_p;b_1,\cdots ,b_q;z)\nonumber \\&\quad = \frac{\Pi _{i=1}^Pa_i}{\Pi _{i=1}^q b_i}{}_pF_{q}(a_1+1,\cdots ,a_p+1;b_1+1,\cdots , b_q+1;z)\,. \end{aligned}$$

(C.16)

In addition, the relationship between

${}_{p+1}F_{q+1}(a_1,\cdots ,a_p;b_1,\cdots ,b_q;z)$ and ${}_pF_{q}(a_1,\cdots ,a_p;b_1,\cdots ,b_q;z)$ is

$$\begin{aligned}&{}_{p+1}F_{q+1}(a_1,\cdots ,a_p,a_{p+1};b_1,\cdots ,b_q,b_{q+1};z)\nonumber \\&\quad =\frac{\Gamma (b_{q+1})}{\Gamma (a_{p+1})\Gamma (b_{q+1}-a_{p+1})} \nonumber \\&\qquad \int _{0}^1 t^{a_{p+1}-1}(1-t)^{b_{q+1}-a_{p+1}-1}{}_{p} F_{q}(a_1,\cdots ,a_p;b_1,\cdots ,b_q;tz)\, \textrm{d}t, \end{aligned}$$

(C.17)

where $\Gamma$ is the Gamma function $\Gamma (z):=\int _0^{\infty }t^{z-1}e^{-t} \textrm{d}t$. In particular, when $p=2$ and $q=1$, (C.15) becomes the ordinary hypergeometric function, which satisfies

$$\begin{aligned} {}_2F_1(a_1,a_2;b_1;1)=\frac{\Gamma (b_1)\Gamma (b_1-a_2-a_1)}{\Gamma (b_1-a_1)\Gamma (b_1-a_2)}\,, \qquad b_1>a_1+a_2\,. \end{aligned}$$

(C.18)

In addition, for $\vert z\vert <1$, ${}_2F_1(a_1;b_1,b_2;z)$ has the following recursion formula:

$$\begin{aligned} {}_2F_1(a_1,a_2;b_1;z)=(1-z)^{b_1-a_2-a_1}{}_2 F_1(b_1-a_1;b_1-a_2,b_1;z)\,, \qquad b_1<a_1+a_2\,. \end{aligned}$$

(C.19)

With this preliminary background, we return to the NLEP (C.7) and use generalized hypergeometric functions to calculate the critical value of $\alpha$, labeled by $\alpha _c$, for which $\lambda _0=0$ is an eigenvalue. This implies that $\Lambda =\delta ^2$ in (C.7). By defining ${\bar{z}}:=2z$, (C.7) can be written when $\lambda _0=0$ and $\Lambda =\delta ^2$ as

$$\begin{aligned} \Psi _{0{\bar{z}}{\bar{z}}}+\frac{w^2}{4}\Psi _0- \frac{\bar{\kappa }}{4}\frac{\int _{-\infty }^\infty w^2\Psi _0 \, d{\bar{z}}}{\int _{-\infty }^\infty w^2 \, d{\bar{z}}}w^2=\frac{\delta ^2}{4}\Psi _0\,, \qquad \bar{\kappa } := \frac{\alpha (4-\Lambda )}{2+\alpha } \frac{\int _{-\infty }^{\infty } w^2 \, \textrm{d}\bar{z}}{\int _{-\infty }^{\infty } w^4 \, \textrm{d}\bar{z}}\,. \end{aligned}$$

(C.20)

To use the standard results in Wei and Winter (2002), we define ${\bar{w}}:=\frac{3}{2}\textrm{sech}^2({\bar{z}}/2)$ and $\delta _1:=\delta /2$, so that (C.20) becomes

$$\begin{aligned} \Psi _{0{\bar{z}}{\bar{z}}}+\frac{{\bar{w}}}{3}\Psi _0-\frac{\bar{\kappa }}{3} \frac{\int _{-\infty }^\infty {\bar{w}}\Psi _0 \, d{\bar{z}}}{\int _{-\infty }^\infty {\bar{w}} \, d{\bar{z}}}{\bar{w}}=\delta _1^2\Psi _0\,. \end{aligned}$$

(C.21)

Next, as was shown in Wei and Winter (2002), (C.21) can be transformed into a local problem with an integral constraint:

$$\begin{aligned} \Psi _{0{\bar{z}}{\bar{z}}}+\frac{{\bar{w}}}{3}\Psi _0=\delta _1^2\Psi _0+{\bar{w}}\,, \qquad \int _{0}^\infty {\bar{w}}\Psi _0\, d{\bar{z}}=\frac{3}{\bar{\kappa }}\int _0^\infty {\bar{w}} \, d{\bar{z}}\,. \end{aligned}$$

(C.22)

Upon defining G by $\Psi _0={\bar{w}}^{\delta _1}G$, we substitute this relation into (C.22) to obtain

$$\begin{aligned} G_{{\bar{z}}{\bar{z}}}-2\delta _1\frac{{\bar{w}}_{{\bar{z}}}}{{\bar{w}}}G_{{\bar{z}}}+ \left[ \frac{1}{3}-\frac{\delta _1}{3}(1+2\delta _1)\right] {\bar{w}} G= {\bar{w}}^{1-\delta _1}\,. \end{aligned}$$

(C.23)

We next define ${\tilde{z}}:={2{\bar{w}}/3}$ and rewrite (C.23) as

$$\begin{aligned} {\tilde{z}}(1-{\tilde{z}})G_{{\tilde{z}}{\tilde{z}}}+\left[ c-(a+b+1){\tilde{z}}\right] G_{{\tilde{z}}} -abG=\left( \frac{3}{2}\right) ^{1-\delta _1}{\tilde{z}}^{-\delta _1}\,, \end{aligned}$$

(C.24)

where we have labeled a, b, and c by $a=\delta _1+1$, $b=\delta _1-{1/2}$ and $c=1+2\delta _1$.

With this reformulation, we now solve (C.24) in terms of hypergeometric functions. To begin, we recall from Kolokolnikov et al. (2009) that the two linear independent solutions to the homogeneous counterpart of (C.24) are

$$\begin{aligned} {}_2F_1(a,b;c;{\tilde{z}})\,, \qquad {\tilde{z}}^{1-c}{}_2F_1(a-c+1,b-c+1;2-c;{\tilde{z}})\,. \end{aligned}$$

(C.25)

As such, we need only find a particular solution, labeled by $G_1$, of (C.24). To do so, we write $G_1$ in the form $G_1({\tilde{z}})={\tilde{z}}^{i}\sum _{k=0}^{\infty } c_{k}{\tilde{z}}^k$, where the constants i and $c_k$ need to be determined. Upon substituting this infinite series into (C.24), we solve the resulting recursion equations for i and $c_k$ to get

$$\begin{aligned} G_1=\left( \frac{3}{2}\right) ^{1-\delta _1}(1-\delta _1^2)^{-1}\tilde{z}^{1-\delta _1} {}_3F_2\bigg (1,\frac{1}{2},2;2-\delta _1,2+\delta _1;{\tilde{z}}\bigg )\,. \end{aligned}$$

(C.26)

It is verify that $\Psi _0={\bar{w}}^{\delta _1}G_1\rightarrow 0$ as ${\bar{z}}\rightarrow +\infty .$ However, we must have $\Psi _{0{\bar{z}}}(0)=0$ since $\Psi _0$ is even. To enforce this condition, we write $\Psi _0$ as a linear combination of $G_1$ and the first homogeneous solution $G_2$ in (C.25) as

$$\begin{aligned} \Psi _0={\bar{w}}^{\delta _1}(G_1+AG_2)\,, \qquad \text{ where } \quad G_2:={}_2F_1\Big (\delta _1+1,\delta _1-\frac{1}{2};2\delta _1+1;{\tilde{z}}\Big )\,, \end{aligned}$$

(C.27)

where the constant A will be determined below. To determine A, we apply (C.16) on (C.26) to get

$$\begin{aligned} \frac{dG_1}{d{\tilde{z}}}= \left( \frac{3}{2}\right) ^{1-\delta _1} (1-\delta _1^2)^{-1}{}_3 F_2\bigg (2,\frac{3}{2},3;3-\delta _1,3+\delta _1;{\tilde{z}}\bigg )\,. \end{aligned}$$

By using (C.17), together with (C.18) and (C.19), we further calculate for $\vert {\tilde{z}}\vert \rightarrow 1^-$, that

$$\begin{aligned} \frac{dG_1}{d{\tilde{z}}}\sim \left( \frac{3}{2}\right) ^{1-\delta _1} \frac{(1-{\tilde{z}})^{-1/2}}{4}{}_2F_1\left( 1,\frac{3}{2};3;1\right) \sim \left( \frac{3}{2}\right) ^{1-\delta _1} (1-{\tilde{z}})^{-1/2}\,. \end{aligned}$$

(C.28)

Similarly, from (C.16), (C.18) and (C.19), we get that the asymptotic behavior of $G_2$ in (C.27) as $\vert {\tilde{z}}\vert \rightarrow 1^-$ is

$$\begin{aligned} \frac{dG_2}{d{\tilde{z}}}\sim \frac{(1+\delta _1)(\delta _1-\frac{1}{2})}{2\delta _1+1}(1-{\tilde{z}})^{-\frac{1}{2}}\frac{\Gamma (2\delta _2+2)\Gamma (\frac{1}{2})}{\Gamma (2\delta _1+1)\Gamma (\delta _1+\frac{1}{2})}. \end{aligned}$$

(C.29)

Upon combining (C.28) and (C.29), we conclude that $\Psi _{0{\bar{z}}}(0)=0$ holds when

$$\begin{aligned} A=\left( \frac{3}{2}\right) ^{1-\delta _1}\frac{\Gamma (1+\delta _1)\Gamma \big (\frac{1}{2}+\delta _1\big )}{\big (\frac{1}{2}-\delta _1\big ) \Gamma (1+2\delta _1)\Gamma \big (\frac{1}{2}\big )}\,. \end{aligned}$$

(C.30)

This gives us an explicit form for $\Psi _0$ in (C.27).

Next, we focus on the integral constraint in (C.22). To begin, we calculate for $\delta _1\ll 1$ that

$$\begin{aligned} \int _0^\infty {\bar{w}}^{1+\delta _1}G_1\, d{\bar{z}}=&-\frac{3}{2}\int _0^1 \bar{w}^{1+\delta _1}\frac{G_1}{{\bar{w}}_{{\bar{z}}}}\, \textrm{d}{\tilde{z}} = \bigg (\frac{3}{2}\bigg )^2(1-\delta _1^2)^{-1} \frac{\Gamma (2)\Gamma \big (\frac{1}{2}\big )}{\Gamma \big (\frac{5}{2}\big )} {}_4\nonumber \\&\quad F_3\Big (1,\frac{1}{2},2,2;2-\delta _1,2+\delta _1,\frac{5}{2};1\Big )\nonumber \\ \sim&3(1-\delta _1^2)^{-1}\,, \end{aligned}$$

(C.31)

and

$$\begin{aligned} \int _0^\infty {\bar{w}}^{1+\delta _1}G_2 \, d{\bar{z}}=&-\frac{3}{2}\int _0^1 \bar{w}^{1+\delta _1}\frac{G_2}{{\bar{w}}_{{\bar{z}}}}\, \textrm{d}{\tilde{z}}\nonumber \\ =&\bigg (\frac{3}{2}\bigg )^{1+\delta _1}(1-\delta _1^2)^{-1} \frac{\Gamma (1+\gamma _1)\Gamma \big (\frac{1}{2}\big )}{\Gamma \big (\frac{3}{2}\big )}{}_3\nonumber \\&F_2\Big (1+\delta _1,\delta _1-\frac{1}{2}, 1+\delta _1;2\delta _1+1,\frac{3}{2}+\delta _1;1\Big )\,. \end{aligned}$$

(C.32)

Moreover, we calculate that

$$\begin{aligned} \int _0^\infty {\bar{w}} \, d{\bar{z}}=-\frac{3}{2}\int _0^1 \frac{{\bar{w}}}{{\bar{w}}_{{\bar{z}}}}\, \textrm{d}{\tilde{z}}=\frac{3}{2}\int _0^{1}\frac{1}{\sqrt{1-{\tilde{z}}}}\, \textrm{d}{\tilde{z}}=3\,. \end{aligned}$$

(C.33)

Upon collecting (C.31), (C.32) and (C.33), we use the constraint in (C.22), with A as in (C.30), to obtain

$$\begin{aligned}&(1-\delta _1^2)^{-1}{}_4F_3\Big (1,\frac{1}{2},2,2;2-\delta _1,2+\delta _1, \frac{5}{2};1\Big )\nonumber \\&\qquad + \frac{A}{3}\Big (\frac{3}{2}\Big )^{1+\delta _1} \frac{\Gamma (1+\delta _1)\Gamma \big (\frac{1}{2}\big )}{\Gamma \big (\frac{3}{2}+\delta _1\big )}{}_3F_2\Big (1+\delta _1,\delta _1 -\frac{1}{2},1+\delta _1;2\delta _1+1,\frac{3}{2}+\delta _1;1\Big ) =\frac{3}{\bar{\kappa }} \,. \end{aligned}$$

(C.34)

As a partial verification of our computation, if we let $\delta _1=0$ then (C.34) yields that $\bar{\kappa }=1$. This agrees precisely with our leading order threshold $\alpha _{c}\sim 1.$ To seek a refined approximation of this threshold, as obtained by the next order term of $\alpha _c$, we expand (C.34) up to $O(\delta _1).$ To do so, we use the standard result in Bühring (1987) to find

$$\begin{aligned}&{}_3F_2\Big (1+\delta _1,\delta _1-\frac{1}{2},1+\delta _1;2\delta _1+1, \frac{3}{2}+\delta _1;1\Big ) \nonumber \\&\quad = \frac{\Gamma (b_1)\Gamma (b_2)}{\Gamma (a_3)\Gamma (a_1+1) \Gamma (a_2+1)}{}_3F_2\Big (\delta _1,\frac{1}{2},1;2+\delta _1,\frac{1}{2}+ \delta _1;1\Big )\,, \end{aligned}$$

(C.35)

where $a_1=1+\delta _1$, $a_2=\delta _1-{1/2}$, $a_3=1+\delta _1$, $b_1=2\delta _1+1$, and $b_2=\delta _1+{3/2}$.

Next, we expand

$$\begin{aligned} \frac{\Gamma (b_1)\Gamma (b_2)}{\Gamma (a_3)\Gamma (a_1+r)\Gamma (a_2+r)}&= \frac{\Gamma (1+2\delta _1)\Gamma (\frac{3}{2}+\delta _1)}{\Gamma (1+\delta _1)\Gamma (2+\delta _1)\Gamma (\frac{1}{2}+\delta _1)} = \frac{\frac{1}{2}+\delta _1}{1+\delta _1}+{{\mathcal {O}}}(\delta _1^2)\,, \end{aligned}$$

(C.36a)

$$\begin{aligned} {}_3F_2\Big (\delta _1,\frac{1}{2},1;2+\delta _1,\frac{1}{2}+\delta _1;1\Big )&= 1+\delta _1+{{\mathcal {O}}}(\delta _1^2)\,. \end{aligned}$$

(C.36b)

Upon substituting (C.36) into (C.35), we conclude that

$$\begin{aligned}&{}_3F_2\Big (1+\delta _1,\delta _1-\frac{1}{2},1+\delta _1;2\delta _1+1, \frac{3}{2}+\delta _1;1\Big )\nonumber \\&\quad =\frac{1}{2} \left[ 1+2\delta _1+{{\mathcal {O}}}(\delta _1^2)\right] =\frac{1}{2}+\delta _1 + {{\mathcal {O}}}(\delta _1^2)\,. \end{aligned}$$

(C.37)

Then, by using the identity ${\Gamma ^2(1+\delta _1)/\Gamma (1+2\delta _1)}=1+{{\mathcal {O}}}(\delta _1^2)$. we substitute (C.37) into (C.34), and recall that $\delta _1={\delta /2}$ where $\delta ={2/(v_{\max 0}{\bar{\chi }})}$. This yields

$$\begin{aligned} {{\bar{\kappa }}}=1-\delta _1+{{\mathcal {O}}}(\delta _1^2)=1-\frac{\delta }{2}+ {{\mathcal {O}}}(\delta ^2) = 1-\frac{1}{{\bar{\chi }} v_{\max 0}} + {{\mathcal {O}}}(v_{\max 0}^{-2}) \,. \end{aligned}$$

(C.38)

Finally, by relating $\bar{\kappa }$ to $\alpha$ using (C.20), and noting the identity $4\int _{-\infty }^\infty w^2\, \textrm{d}\bar{z}=3\int _{-\infty }^\infty w^4\, \textrm{d}\bar{z}$, we conclude that (C.38) provides the following refined threshold at which $\lambda _0=0$, which completes the proof of Theorem 3.1:

$$\begin{aligned} \alpha _{c}\sim 1-\frac{3}{2{\bar{\chi }} v_{\max 0}}\,. \end{aligned}$$

(C.39)

Computation of Partial Derivatives for Quasi-Equilibria

In this appendix, we derive an approximation for ${d v_{\max j}/ds_j}$ from our quasi-equilibrium construction, and we calculate some related partial derivatives that are needed in our analysis. From (2.28), $v_{\max j}$ and $C_j$ satisfy

$$\begin{aligned} v_{\max j}^2=\frac{2 C_{j}}{{\bar{\chi }} } e^{{\bar{\chi }} v_{\max j}} - \frac{2 s_j}{{\bar{\chi }}}+s^2_j\,, \qquad C_je^{{\bar{\chi }} s_j}=s_j \,. \end{aligned}$$

(D.1)

Upon differentiating the equation for $v_{\max j}$ with respect to $s_j$, and labeling $v_{\max j}^{\prime }:= {d v_{\max j}/d s_j}$, we get

$$\begin{aligned} 2v_{\max k}v_{\max j}^{\prime }=(2e^{-{\bar{\chi }} s_j}-2{\bar{\chi }} s_je^{-{\bar{\chi }} s_j}) \frac{e^{{\bar{\chi }} v_{\max j}}}{{\bar{\chi }}}+2s_je^{-{\bar{\chi }} s_j}e^{{\bar{\chi }} v_{\max j}} v_{\max j}^{\prime }-\frac{2}{{\bar{\chi }}}+2s_j \,, \end{aligned}$$

(D.2)

We solve for $v_{\max j}^{\prime }$ in (D.2), while eliminating $C_j$ in (D.1). After some algebra we obtain

$$\begin{aligned} v_{\max j}^{\prime }=\frac{ \left( {v_{\max j}^2/s_j} - \bar{\chi } v_{\max j}^2 +\bar{\chi } s_j^2 -s_j\right) }{2 v_{\max j} - \bar{\chi } v_{\max j}^2 - 2 s_j + \bar{\chi } s_j^2}\,. \end{aligned}$$

(D.3)

Since $s_j={{\mathcal {O}}}(\epsilon |\log \epsilon |^3)$ and $v_{\max j}={{\mathcal {O}}}(|\log \epsilon |)$, we obtain upon retaining only the first term in the numerator and the first two terms in the denominator that for $\epsilon \rightarrow 0$

$$\begin{aligned} v_{\max j}^{\prime }=\frac{dv_{\max j}}{d s_j} \sim -\frac{\zeta _{\max j}}{\bar{\chi } s_j} \,, \qquad \zeta _{\max j}:= \left( 1 - \frac{2}{\bar{\chi } v_{\max j}}\right) ^{-1}\,. \end{aligned}$$

(D.4)

This result (D.4) is needed in (3.34) for analyzing the Jacobian of the quasi-equilibrium construction.

In a similar way, by taking the partial derivative of $v_{\max k}$ with respect to the location $x_i$ of the $i{\text{ th }}$ spike in the quasi-equilibrium pattern, we readily derive the following result for $\epsilon \rightarrow 0$ that is needed in (5.32) and (5.34):

$$\begin{aligned} \partial _{x_i} v_{\max k} \sim -\frac{\zeta _{\max k}}{\bar{\chi } s_k}\partial _{x_i} s_{k} \,, \qquad \zeta _{\max k}= \left( 1 - \frac{2}{\bar{\chi } v_{\max k}}\right) ^{-1}\,. \end{aligned}$$

(D.5)

Calculation of ${\mathcal {G}}_g$ and ${\mathcal {P}}_g$

In this appendix, for $d_1\in {{\mathcal {T}}}_e$, we calculate the matrix spectrum of ${{\mathcal {G}}}_g$, as given in (4.27), as well as the matrix ${{\mathcal {P}}}_g$ that was defined in (4.24). To do so, for $d_1\in {{\mathcal {T}}}_e$, we introduce the auxiliary BVP

$$\begin{aligned} \frac{d_1}{\mu }y^{\prime \prime } +{\bar{u}} y=0\,, \quad 1<x<1\,; \quad y^{\prime }(\pm 1)=0\,;\quad \Big [\frac{d_1}{\mu }y\Big ]_j=b_j\,,\quad \Big [\frac{d_1}{\mu }y^{\prime }\Big ]_j=0\,, \end{aligned}$$

(E.1)

for $j=1,\ldots ,N$. Here $[y]_j:=y(x_j^+)-y(x_j^-)$ with $x_j=x_j^0$ as given by (2.29). The solution to (E.1) is $y=\sum _{k=1}^N b_k g(x;x_k)$, where the dipole Green’s function $g(x;x_k)$ satisfies (4.22). Upon defining ${\varvec{y}}^{\prime }:=\left( y^{\prime }_{1},\ldots ,y^{\prime }_{N}\right) ^T$, $\varvec{\langle y\rangle }:=\left( \langle y\rangle _1,\ldots , \langle y\rangle _N\right) ^T$, and ${\varvec{b}}:=\left( b_1,\ldots ,b_N \right) ^T$, where $y^{\prime }_j=y^{\prime }(x_j)$ and $\langle y\rangle _j=\big (y(x_j^+)+y(x_j^-)\big )/2$, we conclude that

$$\begin{aligned} {\varvec{y}}^{\prime }={\mathcal {G}}_g{\varvec{b}}\,, \qquad \langle {\varvec{y}}\rangle ={\mathcal {P}}_g{\varvec{b}}\,. \end{aligned}$$

(E.2)

The inverses of ${\mathcal {G}}_g$ and ${\mathcal {P}}_g$ exist and are tridiagonal when $d_1\in {{\mathcal {T}}}_e$. To show this, we solve (E.1) on each subinterval where we impose the continuity conditions on $y^{\prime }$ across $x_j$. This yields that

$$\begin{aligned} y=\left\{ \begin{array}{ll} -\frac{y_1^{\prime }}{\theta }\frac{\cos [\theta (1+x)]}{\sin [\theta (1+x_1)]}\,, &{}-1<x<x_1\,,\\ \frac{y_j^{\prime }}{\theta }\frac{\cos [\theta (x_{j+1}-x)]}{\sin [\theta (x_{j+1}-x_j)]}-\frac{y_{j+1}^{\prime }}{\theta } \frac{\cos [\theta (x-x_j)]}{\sin [\theta (x_{j+1}-x_j)]}\,,&{}x_j<x<x_{j+1}\,, \quad j=1,\ldots ,N-1\,,\\ \frac{y_{N}^{\prime }}{\theta }\frac{\cos [\theta (1-x)]}{\sin [\theta (1-x_{N})]} \,,&{}x_{N}<x<1\,, \end{array} \right. \end{aligned}$$

(E.3)

where $\theta =\sqrt{\mu {\bar{u}}/d_1}.$ By using (E.3), we satisfy the jump conditions in (E.1) to get

$$\begin{aligned} {\mathcal {D}}_g{\varvec{y}}^{\prime }=\frac{\mu \theta }{d_1}{\varvec{b}}\,, \qquad {\mathcal {G}}_g=\frac{\mu \theta }{d_1} {\mathcal {D}}_g^{-1}\,. \end{aligned}$$

(E.4)

Here, for $d_1\in {{\mathcal {T}}}_e$, ${\mathcal {D}}_g$ is the invertible tridiagonal matrix defined by

$$\begin{aligned} {\mathcal {D}}_g=\left( \begin{array}{ccccccc} d_g &{} f_g&{}0&{}\cdots &{}0 &{}0&{} 0\\ f_g &{} e_g&{}f_g&{}\cdots &{}0 &{}0&{} 0\\ 0 &{} f_g&{}e_g&{}\ddots &{}0 &{}0&{} 0\\ \vdots &{} \vdots &{}\ddots &{}\ddots &{}\ddots &{}\vdots &{} \vdots \\ 0 &{} 0&{}0&{}\ddots &{}e_g &{}f_g&{} 0\\ 0 &{} 0&{}0&{}\cdots &{}f_g &{}e_g&{} f_g\\ 0 &{} 0&{}0&{}\cdots &{}0 &{}f_g&{} d_g\\ \end{array} \right) \,. \end{aligned}$$

(E.5)

where $d_g=\cot \left( {2\theta /N}\right) +\cot \left( {\theta /N}\right)$, $e_g=2\cot \left( {2\theta /N}\right)$ and $f_g=-\csc \left( {2\theta /N}\right)$, for which the identity $d_g=e_g-f_g$ holds. When $d_1\in {{\mathcal {T}}}_e$ (see (2.35)), i.e. ${2\theta /N}<\pi$, we see that $e_g$, $d_g$ and $f_g$ are well-defined.

Similarly, we rewrite $\varvec{\langle y\rangle }$ in terms of ${\varvec{y}}^{\prime }$ as $\varvec{\langle y\rangle }=- (2\theta )^{-1}\csc \left( {2\theta /N}\right) {\mathcal {C}}{\varvec{y}}^{\prime }$. where ${{\mathcal {C}}}$ was defined in (B.10). By combining the second equation in (E.2) with this result we obtain for $d_1\in {{\mathcal {T}}}_e$ that

$$\begin{aligned} {\mathcal {P}}_g=-\frac{\mu }{2d_1}\csc \left( \frac{2\theta }{N}\right) {\mathcal {C}}{\mathcal {D}}_g^{-1}\,. \end{aligned}$$

(E.6)

The matrix spectrum of the tridiagonal matrix ${{\mathcal {D}}}_g$, labeled by ${\mathcal {D}}_g {\varvec{v}}=\xi \varvec{\nu }$ where $\varvec{\nu }=(\nu _1,\ldots ,\nu _N)^T$, is readily calculated as in Iron et al. (2001) and the result is summarized in Proposition 4.2.

Finally, when $\lambda _0=0$, we establish a key identity

$$\begin{aligned} {{\mathcal {P}}}^T = - {{\mathcal {P}}}_g \,, \end{aligned}$$

(E.7)

which relates (4.27) for ${{\mathcal {P}}}$ when $\lambda _0=0$ to (4.24). One way to derive this identity is to observe from (B.6) that when $\lambda _0=0$, we have $e_g = - e$, $f_g = -f$, and $d_g=-e+f$. By using these expressions in (E.5) a direct matrix multiplication yields the identity ${{\mathcal {C}}} {{\mathcal {D}}}_g = - {{\mathcal {D}}} {{\mathcal {C}}}$, where ${{\mathcal {D}}}$ and ${{\mathcal {C}}}$ are defined in (B.5) and (B.10), respectively. The result (E.7) follows by comparing (E.6) and (B.11), and noting that ${{\mathcal {D}}}$ and ${{\mathcal {D}}}_g$ are symmetric.

Diagonalization of the Matrix ${{\mathcal {M}}}$ for the Small Eigenvalues

In this appendix, when $d_1\in {{\mathcal {T}}}_e$, we show how to diagonalize the matrix ${{\mathcal {M}}}$ in (4.29) to obtain the result given in Proposition 4.3 for the small eigenvalues. From (4.29), the matrix for the small eigenvalues is

$$\begin{aligned}&{\mathcal {M}}=\frac{2{\bar{\chi }}}{3}v_{\max 0}^3{\mathcal {G}}_g- \frac{2v_{\max 0}^2\zeta _0}{a_g}{\mathcal {P}}\left( I+\frac{3\zeta _0}{{\bar{\chi }} a_gv_{\max 0}} {\mathcal {G}}\right) ^{-1}\nonumber \\&\quad {\mathcal {P}}_g+\frac{s_0 {\bar{u}}\mu }{\epsilon d_1} I\,, \qquad \zeta _0:=\left( 1-\frac{2}{\bar{\chi }v_{\max 0}}\right) ^{-1}\,. \end{aligned}$$

(F.1)

We begin by focusing on the middle term in ${{\mathcal {M}}}$. We first introduce the matrix decomposition of ${{\mathcal {D}}}$ by ${{\mathcal {D}}} = {{\mathcal {Q}}} {{\mathcal {K}}} {{\mathcal {Q}}}^T$, where ${{\mathcal {K}}}=\text{ diag }(\kappa _1,\ldots ,\kappa _N)$ and ${{\mathcal {Q}}}$ is the orthogonal matrix formed from the eigenvectors ${\varvec{q}}_j$ in Proposition B.1 when $\tau =0$. For $\tau =0$, the eigenvalues $\kappa _j$ of ${{\mathcal {D}}}$ are related to the eigenvalues $\xi _j$ of ${{\mathcal {D}}}_g$ by

$$\begin{aligned} \kappa _1= & {} 2\tan \left( {\theta /N}\right) , \quad \kappa _j=-\xi _j= -2\cot \left( {2\theta /N}\right) \nonumber \\{} & {} + 2 \csc \left( {2\theta /N}\right) \cos \left( {\pi (j-1)/N}\right) , \quad j=2,\ldots ,N. \end{aligned}$$

(F.2)

By using (B.7) with $\lambda _0=0$, we obtain that ${{\mathcal {G}}} = \sqrt{\frac{\mu }{d_1\bar{u}}} {{\mathcal {Q}}} {\mathcal {K}}^{-1} {{\mathcal {Q}}}^T$, which yields

$$\begin{aligned} {{\mathcal {P}}} \left( I+\frac{3\zeta _0}{{\bar{\chi }} a_gv_{\max 0}}\mathcal G\right) ^{-1} {{\mathcal {P}}}_g = {{\mathcal {P}}} {{\mathcal {Q}}} \left( I+\frac{3\zeta _0}{{\bar{\chi }} a_gv_{\max 0}} \sqrt{\frac{\mu }{{\bar{u}} d_1}}{{\mathcal {K}}}^{-1} \right) ^{-1}Q^T {{\mathcal {P}}}_g\,. \end{aligned}$$

(F.3)

Next, we use (E.6) and (B.11) to conclude that ${{\mathcal {P}}}{{\mathcal {D}}}=\left( {{\mathcal {P}}}_g{{\mathcal {D}}}_g\right) ^T$ so that

$$\begin{aligned} {{\mathcal {P}}} = \left( {{\mathcal {P}}}_g{{\mathcal {D}}}_g\right) ^T {{\mathcal {D}}}^{-1} =\left( {{\mathcal {P}}}_g{{\mathcal {D}}}_g\right) ^T {{\mathcal {Q}}} {{\mathcal {K}}}^{-1} {{\mathcal {Q}}}^T \,. \end{aligned}$$

(F.4)

By combining (F.4) and (F.3), and using ${{\mathcal {Q}}}{{\mathcal {Q}}}^T=I$, we get

$$\begin{aligned} {{\mathcal {P}}} \left( I+\frac{3\zeta _0}{{\bar{\chi }} a_gv_{\max 0}}\mathcal G\right) ^{-1} {{\mathcal {P}}}_g = {{\mathcal {R}}} {{\mathcal {D}}_g}^{-1}\,, \qquad \text{ where } \quad {{\mathcal {R}}} := \left( {{\mathcal {P}}}_g{{\mathcal {D}}}_g\right) ^T {{\mathcal {Q}}} {{\mathcal {H}}} {{\mathcal {Q}}}^T \left( {{\mathcal {P}}}_g{{\mathcal {D}}}_g\right) \,. \end{aligned}$$

(F.5)

Here ${{\mathcal {R}}}$ is defined in terms of a diagonal matrix ${{\mathcal {H}}}$ given by

$$\begin{aligned} {{\mathcal {H}}} := \left( \frac{3\zeta _0}{{\bar{\chi }} a_gv_{\max 0}} \sqrt{\frac{\mu }{{\bar{u}} d_1}} I + {{\mathcal {K}}}\right) ^{-1}= \text{ diag }(h_1,\ldots ,h_N) \,. \end{aligned}$$

(F.6)

Therefore, by using ${{\mathcal {G}}}_g=\frac{\mu \theta }{d_1}{{\mathcal {D}}}_g^{-1}$ from (E.4), together with (F.5), we can write (F.1) as

$$\begin{aligned} {\mathcal {M}}=\left( \frac{2{\bar{\chi }}}{3}v_{\max 0}^3 \frac{\mu \theta }{d_1} I + \frac{s_0 {\bar{u}}\mu }{\epsilon d_1} {{\mathcal {D}}_g} - \frac{2v_{\max 0}^2\zeta _0}{a_g} {{\mathcal {R}}} \right) {{\mathcal {D}}}_g^{-1} \,. \end{aligned}$$

(F.7)

Next, we must focus on analyzing the matrix ${{\mathcal {R}}}$ defined by (F.5). By using (E.6), we obtain

$$\begin{aligned} \left( {{\mathcal {P}}}_g {{\mathcal {D}}}_g\right) ^T = -\frac{\mu }{2d_1} \csc \left( \frac{2\theta }{N}\right) {\mathcal {C}}\,, \end{aligned}$$

where ${{\mathcal {C}}}$ is given in (B.10). In this way, it is convenient to write ${{\mathcal {R}}}$ as

$$\begin{aligned} {{\mathcal {R}}} = \frac{\mu ^2}{4d_1^2} \csc ^{2}\left( \frac{2\theta }{N}\right) {{\mathcal {C}}}^T {{\mathcal {Q}}} {{\mathcal {H}}} {{\mathcal {Q}}}^T {{\mathcal {C}}} = \frac{\mu ^2}{4d_1^2} \csc ^{2}\left( \frac{2\theta }{N}\right) {{\mathcal {Q}}}_g {{\mathcal {Q}}}_g^T {{\mathcal {C}}}^T {{\mathcal {Q}}} {{\mathcal {H}}} {{\mathcal {Q}}}^T {{\mathcal {C}}} {{\mathcal {Q}}}_g {{\mathcal {Q}}}_g^T \,, \end{aligned}$$

where ${{\mathcal {Q}}}_g$ are the normalized eigenvectors of ${{\mathcal {D}}}_g$ (see Proposition 4.2), arising in the matrix decomposition

$$\begin{aligned} {{\mathcal {D}}}_g = {{\mathcal {Q}}}_g {{\mathcal {K}}}_g {{\mathcal {Q}}}_g^T \,, \qquad {{\mathcal {K}}}_g =\text{ diag }(\xi _1,\ldots ,\xi _N)\,, \end{aligned}$$

(F.8)

where $\xi _j$ are the eigenvalues of ${{\mathcal {D}}}_g$ as given in Proposition 4.2. In this way, we can write ${{\mathcal {R}}}$ as

$$\begin{aligned} {{\mathcal {R}}} = {{\mathcal {Q}}}_g \Sigma {{\mathcal {Q}}}_g^T \,, \qquad \text{ where } \quad \Sigma := \frac{\mu ^2}{4d_1^2} \csc ^{2}\left( \frac{2\theta }{N}\right) {{\mathcal {S}}} {{\mathcal {H}}} {{\mathcal {S}}}^T \,, \quad {{\mathcal {S}}}:={{\mathcal {Q}}}_g^{T} {{\mathcal {C}}}^T {{\mathcal {Q}}} \,. \end{aligned}$$

(F.9)

The key step in the analysis is the calculation of $\Sigma$ in (F.9) using the explicit forms for the matrices ${{\mathcal {Q}}}_g$, ${{\mathcal {C}}}$, and ${{\mathcal {Q}}}$, as was done in section 4.2 of Iron et al. (2001). This calculation in Iron et al. (2001) showed that $\Sigma$ is a diagonal matrix given by

$$\begin{aligned} \Sigma&= \text{ diag }(\omega _1,\ldots , \omega _N) \,, \qquad \text{ where } \qquad \omega _j := \frac{\mu ^2}{d_1^2}\csc ^2\left( \frac{2\theta }{N}\right) \nonumber \\&\quad \sin ^2\left( \frac{(j-1)}{N}\pi \right) h_j \,, \quad j=1,\ldots ,N \,. \end{aligned}$$

(F.10)

Here $h_j$, for $j=1,\ldots ,N$, are the diagonal entries of ${{\mathcal {H}}}$ that can be identified from (F.6).

Upon substituting (F.9) and ${{\mathcal {D}}}_g^{-1}= {{\mathcal {Q}}}_g {{\mathcal {K}}}_g^{-1} {{\mathcal {Q}}}_g^T$ into (F.7), and recalling (4.28), we obtain that the matrix eigenvalue problem for the small eigenvalues reduces to

$$\begin{aligned} \lambda {\varvec{c}}\sim -\epsilon ^3\beta _0{\mathcal {M}}{\varvec{c}}\,, \qquad \text{ where } \quad {\mathcal {M}}={{\mathcal {Q}}}_g \left( a {{\mathcal {K}}}_g^{-1} + b I - \frac{2v_{\max 0}^2\zeta _0}{a_g} \Sigma {{\mathcal {K}}}_g^{-1} \right) {{\mathcal {Q}}}_g^{T} \,. \end{aligned}$$

(F.11)

This key result shows that ${{\mathcal {M}}}$ is diagonalizable by the eigenspace ${{\mathcal {Q}}}_g$ of the Green’s dipole matrix. In (F.11),

$$\begin{aligned} a:= \frac{2 {\bar{\chi }} }{3} v_{\max 0}^3 \left( \frac{\mu \theta }{d_1}\right) \,, \qquad b: = \frac{s_0{\bar{u}} \mu }{\epsilon d_1} = \frac{2{\bar{\chi }}}{3} v_{\max 0}^3 \left( \frac{a_g{{\bar{u}}}\mu }{d_1}\right) \,, \end{aligned}$$

(F.12)

where we have used the result $s_0\sim {2\bar{\chi }a_gv_{\max 0}^3\epsilon /3}$ from (2.30) to simplify b.

Finally, by introducing $\varvec{{\tilde{c}}}={\mathcal {Q}}_{g}^T {\varvec{c}}$ in (F.11), we readily obtain from (F.12) that the small eigenvalues are given explicitly as in (4.33) of Proposition 4.3. The constants $\omega _j$, as given in (4.34), are obtained from (F.10) by using the diagonal entries of ${{\mathcal {H}}}$ that can be identified from (F.6) and (F.2).

Bifurcation Point for the Emergence of Asymmetric Steady-States

In this appendix we verify that the simultaneous zero-eigenvalue crossing threshold for the small eigenvalues, as given in (4.40), coincides with the bifurcation point at which asymmetric steady-state solution branches bifurcate from the symmetric steady-state branches constructed in Sect. 2.

To do so, we proceed in a similar way as in Ward and Wei (2002) by constructing a steady-state solution of (1.2) on a canonical domain $|x|\le \ell$, with $u_x=v_x=0$ at $x=\pm \ell$ and with a spike centered at $x=0$. On this domain, the leading-order outer solution $u_{o\ell }(x)$ satisfies (see (2.22))

$$\begin{aligned} {{\mathcal {L}}}_{0\ell } u_{o\ell } := \frac{d_1}{\mu }u_{o\ell xx}+{\bar{u}} u_{o\ell }=\frac{2{\bar{\chi }}\epsilon }{3 } v_{\max \ell }^3 \, \delta (x) \,, \quad |x|\le \ell \,; \qquad u_{o\ell x}(\pm \ell )=0 \,, \end{aligned}$$

(G.1)

where, in analogy with (2.33), $v_{\max \ell }$ satisfies the dominant balance

$$\begin{aligned} \frac{1}{2}v_{\max \ell }^2 \sim \frac{s_{\ell }}{\bar{\chi }} e^{\bar{\chi }v_{\max \ell }} , \qquad \text{ with } \quad s_{\ell }=u_{o\ell }(0) . \end{aligned}$$

(G.2)

To solve (G.1) we let $G_{\ell }(x)$ be the Green’s function satisfying ${{\mathcal {L}}}_{0\ell } G_{\ell }=\delta (x)$, with $G_{\ell x}(\pm \ell )=0$. For $\theta \ne {m\pi /\ell }$ with $m=1,2,\ldots$, where $\theta =\sqrt{{\mu \bar{u}/d_1}}$, we obtain that

$$\begin{aligned} u_{o\ell }(x)=\frac{2{\bar{\chi }}}{3}\epsilon v_{\max \ell }^3 \, G_{\ell }(x) , \qquad \text{ where } \quad G_{\ell }(x)=\frac{\mu \cos \left[ \theta (\ell -|x|)\right] }{2\theta d_1 \sin (\theta \ell )}. \end{aligned}$$

(G.3)

By evaluating (G.3) at $x=0$ we can calculate $s_{\ell }$, which is needed in (G.2) for determining $v_{\max \ell }$. In this way, we obtain after some algebra that at $x=\ell$

$$\begin{aligned} u_{o\ell }(\ell ) = c {{\mathcal {B}}}(\ell ) , \qquad \text{ where } \quad {{\mathcal {B}}}(\ell ):= \frac{v_{\max \ell }^3}{\sin (\theta \ell )} , \quad c: = \frac{\epsilon \bar{\chi }}{3\bar{u}} \sqrt{\frac{\mu \bar{u}}{d_1}} . \end{aligned}$$

(G.4a)

Here $v_{\max \ell }$ as a function of $\ell$ satisfies the nonlinear algebraic equation

$$\begin{aligned} v_{\max \ell } e^{\bar{\chi } v_{\max \ell }} \cot (\theta \ell ) = {\bar{\chi }/(2c)}. \end{aligned}$$

(G.4b)

As similar to the analysis in Ward and Wei (2002) for the GM model, the construction of asymmetric steady-state patterns for (1.2) relies on determining $\ell _1$ and $\ell _2$ for which ${{\mathcal {B}}}(\ell _1)={\mathcal B}(\ell _2)$. As a result, we have $u_{o\ell }(\ell _1)=u_{o\ell }(\ell _2)$, which allows for the construction of a $C^{1}$ global solution on $|x|\le 1$ with $M_1$ and $M_2$ small and large spikes, respectively, when the length constraint $\ell _1 M_1 + \ell _2 M_2=1$ is satisfied (cf. Ward and Wei 2002).

The bifurcation point along the steady-state symmetric branch where such asymmetric equilibria emerge is determined by setting ${{\mathcal {B}}}^{\prime }(\ell )=0$ with $\ell ={1/N}$. From (G.4a) and the logarithmic derivative of (G.4b) we get

$$\begin{aligned} {{\mathcal {B}}}^{\prime }(\ell )= & {} \frac{v_{\max \ell }^2}{\sin (\theta \ell )} \left[ 3 v_{\max \ell }^{\prime } - \theta v_{\max \ell } \cot (\theta \ell )\right] ,\nonumber \\{} & {} \quad \qquad v_{\max \ell }^{\prime }\left( 1 + \frac{1}{\bar{\chi } v_{\max \ell }}\right) = \frac{\theta }{\bar{\chi } \sin (\theta \ell )\cos (\theta \ell )} . \end{aligned}$$

(G.5)

Upon combining these two equations we conclude that

$$\begin{aligned} {{\mathcal {B}}}^{\prime }(\ell ) = \frac{\theta \, v_{\max \ell }^3}{ \sin ^2(\theta \ell )\cos (\theta \ell )} \left[ \frac{3}{1+ \bar{\chi } v_{\max \ell }} - \cos ^2(\theta \ell )\right] . \end{aligned}$$

(G.6)

By setting ${{\mathcal {B}}}^{\prime }(\ell )=0$ with $\ell ={1/N}$, and using the double-angle formula for $\cos ^{2}(\theta \ell )$, we readily obtain that the threshold value of $\theta$ is

$$\begin{aligned} \cos \left( \frac{2\theta }{N}\right) = \frac{1-a_1}{1+a_1} , \qquad \text{ where } \quad a_1 = \frac{1}{3} \left( \bar{\chi } v_{\max } -2\right) . \end{aligned}$$

(G.7)

This threshold agrees precisely with the zero-eigenvalue crossing result (4.40) for the small eigenvalues.

Computation of $\beta _0$ and $\beta _j$

In this appendix, we show how to obtain the estimate (5.26) for $\beta _j$, where $\beta _j$ was defined in (5.17) of Sect. 5. For simplicity, in the analysis below we will drop the subscript j in $V_{0 j}$, $v_{\max j}$, $C_j$, $s_j$, and $v_{\max j}$.

We begin by recalling from (2.3) that the leading order steady state v-equation for the spike profile is

$$\begin{aligned}&V_0^{\prime \prime }-V_0+C e^{{\bar{\chi }} V_0}=0\,, \qquad -\infty<y<+\infty \,;\nonumber \\&\quad \qquad V_0(0)=v_{\max } \,, \quad V_0(\infty )=s \,, \end{aligned}$$

(H.1)

where $v_{\max }^2= 2C e^{{\bar{\chi }} v_{\max }}-2 s+ s^2$ and $C=se^{-{\bar{\chi }} s}$.

From the results in Proposition 2.1 for the sub-inner region, we conclude that there exists a positive constant $y_0={{\mathcal {O}}}\left( {1/v_{\max }}\right) \ll 1$ such that

$$\begin{aligned}{} & {} V_0 \sim v_{\max }+\frac{1}{{\bar{\chi }}} \log \left[ \textrm{sech}^2\left( \frac{v_{\max }{\bar{\chi }} y}{2} \right) \right] , \quad 0<y<y_0 ; \nonumber \\{} & {} \qquad U_0 \sim \frac{{\bar{\chi }} }{2} v_{\max }^2\textrm{sech}^2\Big (\frac{v_{\max }{\bar{\chi }} y}{2} \Big ) , \quad 0<y<y_0 . \end{aligned}$$

The decay behavior of $U_0$ and $V_0$ is obtained by noting that $V_0^{\prime \prime }-V_0+\bar{\chi }s V_0\approx 0$ for $y>y_0$. Since $s\ll 1$, this yields $V_{0}^{\prime \prime }-V_0\approx 0$. With this observation, and by enforcing continuity across $y=y_0$, we estimate

$$\begin{aligned}&V_0\sim \left\{ \begin{array}{ll} v_{\max }+\frac{1}{{\bar{\chi }}}\log \left[ \textrm{sech}^2 \left( \frac{v_{\max }{\bar{\chi }} y}{2}\right) \right] \,,&{}y<y_0\,,\\ v_{\max }e^{-(y-y_0)}+\frac{1}{{\bar{\chi }}}\log \left[ \textrm{sech}^2\Big (\frac{v_{\max }\bar{\chi }y_0}{2}\Big )\right] \,,&{}y>y_0\,, \end{array} \right. \,, \nonumber \\&\quad \qquad V_{0}^{\prime }\sim \left\{ \begin{array}{ll} -v_{\max }\tanh (\frac{v_{\max }{\bar{\chi }} y}{2})\,,&{}y<y_0\,,\\ -v_{\max }e^{-(y-y_0)}\,,&{}y>y_0\,. \end{array} \right. \end{aligned}$$

(H.2)

Moreover, since $U_0=Ce^{{\bar{\chi }} V_0},$ we obtain in a similar way that

$$\begin{aligned} U_0\sim \left\{ \begin{array}{ll} \frac{{\bar{\chi }} }{2}v_{\max }^2\textrm{sech}^2\Big (\frac{v_{\max }{\bar{\chi }} y}{2}\Big ) \,,&{}y<y_0\,,\\ Ce^{{\bar{\chi }} v_{\max }e^{-(y-y_0)}}\Big (\textrm{sech}^2\Big (\frac{v_{\max }{\bar{\chi }} y_0}{2}\Big )\Big )\, ,&{}y>y_0\,. \end{array} \right. \end{aligned}$$

(H.3)

By using (H.3) we calculate that

$$\begin{aligned} \int _0^y \frac{1}{U_0}\, \textrm{d}\xi \sim \left\{ \begin{array}{ll} \frac{2}{{\bar{\chi }} v_{\max }^2}\Big (\frac{y}{2}+ \frac{\sinh (v_{\max }{\bar{\chi }} y)}{2{\bar{\chi }} v_{\max }}\Big )\,, &{}y<y_0\,,\\ \frac{2}{{\bar{\chi }} v_{\max }^2}\Big (\frac{y_0}{2}+\frac{\sinh (v_{\max }{\bar{\chi }} y_0)}{2{\bar{\chi }} v_{\max }}\Big )+\frac{1}{s}(y-y_0)\,, &{}y>y_0\,. \end{array} \right. \end{aligned}$$

Then, upon multiplying by $U_0$, we obtain

$$\begin{aligned}&U_0\int _0^y \frac{1}{U_0}\, \textrm{d}\xi \nonumber \\&\sim \left\{ \begin{array}{ll} \frac{y}{2} \textrm{sech}^2(\frac{v_{\max }{\bar{\chi }} y}{2})+\frac{1}{2v_{\max }{\bar{\chi }}}\tanh (\frac{v_{\max }{\bar{\chi }} y}{2})\textrm{sech}(\frac{v_{\max }{\bar{\chi }} y}{2})\,, &{}y<y_0\,,\\ (y-y_0)+\frac{2C}{{\bar{\chi }} v_{\max }^2}e^{{\bar{\chi }} v_{\max }e^{-(y-y_0)}}\textrm{sech}^2\Big (\frac{v_{\max }{\bar{\chi }} y_0}{2}\Big )\Big (\frac{y_0}{2}+\frac{\sinh (v_{\max }{\bar{\chi }} y_0)}{2{\bar{\chi }} v_{\max }}\Big )\,, &{}y>y_0\,. \end{array} \right. \end{aligned}$$

(H.4)

By multiplying (H.4) with $V_0^{\prime }$ from (H.2) and integrating, we observe that the dominant contribution to the integrand arises from multiplying the $y-y_0$ term in (H.4) with the $-v_{\max }e^{-(y-y_0)}$ term in (H.2). In this way,

$$\begin{aligned} \int _0^\infty U_{0}V_{0}^{\prime }\Big (\int _0^y\frac{1}{U_{0}}\, \textrm{d}\xi \Big )\, \textrm{d}y\sim -v_{\max }\int _{y_0}^{+\infty } e^{-(y-y_0)} (y-y_0)\, \textrm{d}y\sim -v_{\max }\,. \end{aligned}$$

In a similar way, we estimate that $\int _{0}^{+\infty }\left( V_{0}^{\prime }\right) ^2\, \textrm{d}y\sim v^2_{\max } \int _{y_0}^{+\infty } e^{-2(y-y_0)}\, \textrm{d}y\sim {v_{\max }^2/2} \,.$ We conclude from (5.17) that $\beta _j\sim {2/v_{\max }}$, as was claimed in (5.26).

Next, we recall from (4.28) in our analysis of the small eigenvalues that $\beta _0=-{\int _0^\infty yV_0^{\prime }\, \textrm{d}y/\int _0^\infty \left( V_{0}^{\prime } \right) ^2 \, \textrm{d}y}$. By using (H.4) and (H.2), we can readily verify that

$$\begin{aligned} \int _0^\infty yV_0^{\prime }\, \textrm{d}y\sim \int _0^\infty U_{0}V_{0}^{\prime } \Big (\int _0^y\frac{1}{U_{0}}\, \textrm{d}\xi \Big ) \, \textrm{d}y, \end{aligned}$$

which establishes that $\beta _j\sim \beta _0$ when evaluated at the steady-state solution.

The Equivalence Between Some Matrices

In this appendix, we show the relationship between the matrices

$$\begin{aligned}&\nabla {\mathcal {G}}:=(\partial _{x_j} G(x_j^0;x_k^0))_{N\times N}\,, \qquad (\nabla {\mathcal {G}})^T:=(\partial _{x_k}G(x_j^0;x_k^0))_{N\times N}\,,\nonumber \\&\qquad \nabla ^2 {\mathcal {G}}:=(\partial _{x_j}\partial _{x_k} G(x_j^0;x_k^0))_{N\times N}\,, \end{aligned}$$

(I.1)

used in the linearization of the DAE system and the matrices ${{\mathcal {P}}}$, ${{\mathcal {P}}}_g$, and ${{\mathcal {G}}}_g$, as defined in (4.27), (4.24), and (4.27), respectively, that were used in Sect. 4 in our analysis of the small eigenvalues. Recall that the diagonal entries in the matrices in (I.1) were defined in (5.27) in terms of the regular part R of the Green’s function (see (5.20)).

We first show that $\nabla G={{\mathcal {P}}}$. To establish this, we use the decomposition (5.20) to obtain

$$\begin{aligned} G_{x}(x;x_k) = \left\{ \begin{array}{ll} \frac{\mu }{2d_1} + R_x(x;x_k) ,&{}x>x_k,\\ -\frac{\mu }{2d_1} + R_x(x;x_k) ,&{}x<x_k.\\ \end{array} \right. \end{aligned}$$

(I.2)

As such, we identify that the average across the $k{\text{ th }}$ spike is simply $\langle G_x\rangle _k={\left( G_x(x_k^+;x_k)+G_x(x_k^-;x_k)\right) /2}=R_x(x_k;x_k)$. By comparing (I.1) and (4.27), and recalling (5.27) for $j=k$, we conclude that $\nabla G={{\mathcal {P}}}$.

Next, we show that $\left( \nabla {\mathcal {G}}\right) ^T=-{{\mathcal {P}}}_g$. We first differentiate the BVP (2.24) for $G(x;x_k)$ with respect to $x_k$ to get

$$\begin{aligned} \frac{d_1}{\mu } \left( \partial _{x_k} G(x;x_k)\right) _{xx} + \bar{u} \left( \partial _{x_k} G(x;x_k) \right) = - \delta ^{\prime }(x-x_k); \qquad \partial _{x} \left( \partial _{x_k} G(x;x_k) \right) \vert _{x=\pm 1}=0 . \end{aligned}$$

By comparing this result with the BVP (4.22) satisfied by the dipole Green’s function, we conclude that

$$\begin{aligned} \partial _{x_k} G(x;x_k) = - g(x;x_k), \qquad -1<x<1 , \end{aligned}$$

(I.3)

so that for $j\ne k$ we have $\partial _{x_k} G(x_j;x_k) = - g(x_j;x_k)$. It follows that the off-diagonal entries in $\left( \nabla {{\mathcal {G}}}\right) ^T$ and ${{\mathcal {P}}}_g$ are identical. For the diagonal entries, where $j=k$, we use (I.3) and the decomposition (5.20) to obtain

$$\begin{aligned} g(x;x_k) = \left\{ \begin{array}{ll} \partial _{x_k}\left( \frac{\mu }{d_1}(x-x_k) + R(x;x_k)\right) = -\frac{\mu }{d_1} - \partial _{x_k} R(x;x_k) \,,&{}x>x_k\,,\\ \partial _{x_k}\left( -\frac{\mu }{d_1}(x-x_k) + R(x;x_k)\right) = \frac{\mu }{d_1} - \partial _{x_k} R(x;x_k) \,,&{}x< x_k\,,\\ \end{array} \right. \end{aligned}$$

(I.4)

Upon defining $\langle g\rangle _k=\frac{1}{2} \left( g(x_{k}^{+};x_k) +g(x_{k}^{-};x_k)\right)$, we conclude from (I.4) and the reciprocity $R(x;y)=R(y;x)$ of the Green’s function that $\langle g\rangle _k=-\partial _{x_k}R(x;x_k)\vert _{x=x_k}= -\partial _{x_k}R(x_k;x)\vert _{x=x_k}$. This implies that the diagonal entries of ${\mathcal {P}}_g$ in (4.24) are the same as those of $(\nabla {\mathcal {G}})^T$ in (I.1). It follows that $\left( \nabla {\mathcal {G}}\right) ^T={{\mathcal {P}}}^T = -{{\mathcal {P}}}_g$. We remark that the relation ${{\mathcal {P}}}^T = -{{\mathcal {P}}}_g$ was also derived using an alternative approach in (E.7) at the end of Appendix E.

Our next identity is to establish that $\nabla ^2 {\mathcal {G}}=-{{\mathcal {G}}}_g$. The equivalence between the off-diagonal entries in these matrices, where $j\ne k$, is established by setting $x=x_j$ in (I.3) and differentiating in $x_j$ to obtain

$$\begin{aligned} \partial _{x_j} \left[ \partial _{x_k} G(x_j;x_k)\right] = - \partial _{x_j} g(x_j;x_k) = - \partial _{x} g(x;x_k)\vert _{x=x_j} . \end{aligned}$$

Next, we differentiate (I.4) with respect to x and upon evaluating at $x=x_k$, we compare the resulting expression with (5.27) to obtain that

$$\begin{aligned} g_{x}(x:x_k)\vert _{x=x_k} = -\frac{\partial }{\partial x}\vert _{x=x_k} \frac{\partial }{\partial y} \vert _{y=x_k} R(x,y) = -\partial ^2_{x_k} G(x_j;x_k) , \quad j=k. \end{aligned}$$

We conclude that the diagonal entries in $\nabla ^2 {\mathcal {G}}$ and $-{{\mathcal {G}}}_g$ are also identical. It follows that $\nabla ^2 {\mathcal {G}}=-{{\mathcal {G}}}_g$.

Finally, we calculate $R_{xx}(x;x_j)\vert _{x=x_j}$ as needed in (5.41). By using the decomposition (5.20) we write (I.2) as

$$\begin{aligned} G_x(x;x_j) = -\frac{\mu }{2d_1} + \frac{\mu }{d_1} H(x-x_j) + R_x(x;x_j), \end{aligned}$$

where H(z) is the Heavyside function. Therefore, $G_{xx}(x;x_j) = \frac{\mu }{d_1} \delta (x-x_j) + R_{xx}(x;x_j)$ on $|x|<1$. Upon substituting this expression into the BVP (2.24) for G, we conclude that $R_{xx}(x;x_j) = \frac{\bar{u}\mu }{d_1} G(x;x_j)$, so that

$$\begin{aligned} R_{xx}(x_j;x_j) = \frac{\bar{u}\mu }{d_1} G(x_j;x_j) . \end{aligned}$$

(I.5)