Stability Analysis of Spike Solutions to the Schnakenberg Model with Heterogeneity on Metric Graphs

Abstract

In this paper, we consider the linear stability of spiky stationary solutions on compact metric graphs for the Schnakenberg model with heterogeneity. The existence of spiky solutions has been shown by the author and Kurata in the work (Ishii and Kurata 2021). By studying the associated linearized eigenvalue problem, we establish the abstract theorem on the stability of the solutions for general compact metric graphs. In particular, the associated Green’s function plays an important role in calculating eigenvalues, and we reveal the several needed conditions for Green’s function on general graphs. To show the stability, we calculate two eigenvalues of order O(1) and of order o(1), respectively. The stability of eigenvalues of order O(1) is shown by using the lemma of Wei and Winter for non-local eigenvalue problem. The stability of eigenvalues of order o(1) is determined by the interaction of the heterogeneity with Green’s function. Moreover, based on the abstract theorem, we give precise stability thresholds with respect to diffusion constants for the solutions without heterogeneity function on the Y-shaped graph and the H-shaped graph. In particular, compared with the one-dimensional interval case, we obtain new phenomena on the stability of two-peak solutions by the effect of the geometry of these concrete graphs. In addition, we also present the effect of heterogeneity by using a typical example.

Appendix

In this section, we check the two assumptions (GS1) and (GS2) for Lemma 3.1 and Lemma 4.1, respectively.

Check of (GS1) and (GS2) for Lemma 3.1. Let $(t_1,t_2) \in e_1\times e_2$ (or $e_1\times e_1$ with $t_1<t_2$). Moreover, let $y,z \in (-r(\varvec{t}),r(\varvec{t}))$. We first check (GS1). By (3.1), we deduce

$$\begin{aligned} K_{ii}(y,z)=\frac{1}{2D}(|y-z|-|z|)-\frac{y^2}{2DL}, \quad K_{ij}(y,z)=-\frac{y^2}{2DL}\;(i \ne j),\qquad \end{aligned}$$

(9.49)

$K_{ij}(y,z)$ is given in (2.7). Thus, ${\hat{K}}_{ij}(u,z)=-y^2/(2DL)$. Then, we obtain $\partial _y{\hat{K}}_{ij}(y,z)=-y/(DL)$ and $\partial _z{\hat{K}}_{ij}(y,z)=0$. From (3.16) and (3.26), we find that

$$\begin{aligned} \partial _{t_1}m_{21}(\varvec{t})=\partial _{t_2}m_{12}(\varvec{t})=0 ,\quad \partial _{t_1}m_{1m}(\varvec{t})=\partial _{t_2}m_{2m}(\varvec{t})=-1/(DL), \end{aligned}$$

(9.50)

for $m=1,2$. Thus, we complete the check of (GS1). Next, we check (GS2). Put $\sigma (\varvec{t},s):=G(t_1,s)-G(t_2,s)$. In particular, $\sigma (\varvec{t},t_j)=\sigma _j(\varvec{t})$. By (3.1), we can calculate

$$\begin{aligned} \begin{aligned} D\sigma (\varvec{t},z+t_j)&= \frac{1}{2}\bigr [|t_1-(z+t_j)|-(t_1+z+t_j)\bigl ]\chi _{e_1}(t_j) \\&\quad -\frac{1}{2}\bigr [|t_2-(z+t_j)|-(t_2+z+t_j)\bigl ]\chi _{e_2}(t_j) \\&\quad -\frac{1}{2L}(t_1-l_1)^2+\frac{1}{2L}(t_2-l_2)^2+\frac{l_1^2}{2L}-\frac{l_2^2}{2L}. \end{aligned} \end{aligned}$$

(9.51)

Therefore, we arrive at

$$\begin{aligned} \begin{aligned} D(\sigma (\varvec{t},z+t_j)-\sigma _j(\varvec{t}))&= \frac{1}{2}\bigr [|t_1-(z+t_j)|-|t_1-t_j|-z \bigl ]\chi _{e_1}(t_j) \\&\quad -\frac{1}{2}\bigr [|t_2-(z+t_j)|-|t_2-t_j|-z \bigl ]\chi _{e_2}(t_j) . \end{aligned} \end{aligned}$$

(9.52)

If $(t_1,t_2) \in e_1\times e_2$, then

$$\begin{aligned} D(\sigma (\varvec{t},z+t_j)-\sigma _j(\varvec{t}))=2^{-1}(-1)^{j}\bigr [z-|z| \bigl ]. \end{aligned}$$

(9.53)

Also, (3.16) implies $(Dq_1(\varvec{t}),Dq_2(\varvec{t}))=(-2^{-1},2^{-1})$. Thus we obtain

$$\begin{aligned} G(t_1,z+t_j)-G(t_2,z+t_j)=\sigma _j(\varvec{t})+q_j(\varvec{t})z+\tau _j(z), \end{aligned}$$

(9.54)

where $D\tau _j(z):=-2^{-1}(-1)^{j}|z|$ is an even function on $(-r(\varvec{t}),r(\varvec{t}))$. On the other hand, if $(t_1,t_2) \in e_1\times e_1$ with $t_1<t_2$, then by combining (9.52) and (3.26), it holds that

$$\begin{aligned} G(t_1,z+t_j)-G(t_2,z+t_j)=\sigma _j(\varvec{t})+q_j(\varvec{t})z+\tau _j(z), \end{aligned}$$

(9.55)

where $(Dq_1(\varvec{t}),Dq_2(\varvec{t}))=(2^{-1},2^{-1})$ and $D\tau _j(z):=-2^{-1}(-1)^{j}|z|$. Hence, we finish the check of (GS2). $\square$

Check of (GS1) and (GS2) for Lemma 4.1. Since (GS1) and (GS2) for Lemma 4.1 can be checked by same argument above, we check only the case of $(t_1, t_2) \in e_1 \times e_3$. Let $y,z\in (-r(\varvec{t}),r(\varvec{t}))$. We first check (GS1). By using (4.3), we obtain

$$\begin{aligned} G(y+t_i,z+t_j)-G(t_i,z+t_j)=m_{ij}(\varvec{t})y+K_{ij}(y,z), \end{aligned}$$

(9.56)

where

$$\begin{aligned} m_{11}(\varvec{t})= & {} -\frac{1}{2D}-\frac{t_1-l_1}{DL} ,\quad m_{22}(\varvec{t})=-\frac{1}{2D}-\frac{t_2-l_3}{DL}+\frac{l_4+l_5}{DL}, \end{aligned}$$

(9.57)

$$\begin{aligned} m_{12}(\varvec{t})= & {} -\frac{t_1-l_1}{DL} ,\quad m_{21}(\varvec{t})=-\frac{t_2-l_3}{DL}+\frac{l_4+l_5}{DL}, \end{aligned}$$

(9.58)

and

$$\begin{aligned} K_{ii}(y,z)=\frac{1}{2D}(|y-z|-|z|)-\frac{y^2}{2DL},\quad K_{ij}(y,z)=-\frac{y^2}{2DL} \; (i\ne j).\qquad \end{aligned}$$

(9.59)

Then, since ${\hat{K}}_{ij}(y,z)=-y^2/(2DL)$, we have $\partial _y{\hat{K}}_{ij}(y,z)=-y/(DL)$ and $\partial _z{\hat{K}}_{ij}(y,z)=0$. Therefore, combining (9.57), (9.58), and two formulas above, we obtain

$$\begin{aligned} \partial _{t_1}m_{21}(\varvec{t})=\partial _{t_2}m_{12}(\varvec{t})=0 ,\quad \partial _{t_1}m_{1m}(\varvec{t})=\partial _{t_2}m_{2m}(\varvec{t})=-1/(DL), \end{aligned}$$

(9.60)

for $m=1,2$. Thus, we complete the check of (GS1). Next, we check (GS2). From (4.3), we deduce

$$\begin{aligned} \begin{aligned}&D\sigma (\varvec{t},z+t_j)= \frac{1}{2}\bigr [|t_1-(z+t_j)|-(t_1+z+t_j)\bigl ]\delta _{1j} \\&\quad -\frac{1}{2}\bigr [|t_2-(z+t_j)|-(t_2+z+t_j)\bigl ]\delta _{2j}-\frac{1}{2L}(t_1-l_1)^2 \\&\quad +\frac{1}{2L}(t_2-l_3)^2+\frac{l_1^2}{2L}-\frac{l_2^2}{2L} -\frac{l_4+l_5}{L}(t_2-l_3+l_1), \end{aligned} \end{aligned}$$

(9.61)

where $\sigma (\varvec{t},z+t_j):=G(t_1,z+t_j)-G(t_2,z+t_j)$. Noting that $\sigma (\varvec{t},t_j)=\sigma _j(\varvec{t})$, we can calculate

$$\begin{aligned} \begin{aligned}&D(\sigma (\varvec{t},z+t_j)-\sigma _j(\varvec{t}))= \frac{1}{2}\bigr [|t_1-(z+t_j)|-|t_1-t_j|-z \bigl ]\delta _{1j} \\&\quad -\frac{1}{2}\bigr [|t_2-(z+t_j)|-|t_2-t_j|-z \bigl ]\delta _{2j} =\frac{(-1)^{j}}{2}\bigr [z-|z| \bigl ]. \end{aligned} \end{aligned}$$

(9.62)

Since $(Dq_1(\varvec{t}),Dq_2(\varvec{t}))=(-2^{-1},2^{-1})$ follows from (9.57) and (9.58), we have

$$\begin{aligned} G(t_1,z+t_j)-G(t_2,z+t_j)=\sigma _j(\varvec{t})+q_j(\varvec{t})z+\tau _j(z), \end{aligned}$$

(9.63)

where $D\tau _j(z):=-2^{-1}(-1)^{j}|z|$ is an even function on $(-r(\varvec{t}),r(\varvec{t}))$. Thus, we finish the check of (GS2). $\square$