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Article Contents

# Existence of multi-peak solutions to the Schnakenberg model with heterogeneity on metric graphs

• * Corresponding author
The first author was supported by JSPS KAKENHI Grant Number 20J12212. The second author was supported by JSPS KAKENHI Grant Numbers 17H01092, 19K03587
• In this paper, we study the existence of spiky stationary solutions of the Schnakenberg model with heterogeneity on compact metric graphs. These solutions are constructed by using the Liapunov–Schmidt reduction method and taking the same strategy as that in [14,11]. First, we give the abstract theorem on the existence of multi-peak solutions for general compact metric graphs under several assumptions for the associated Green's function. In particular, we reveal that how locations of concentration points and amplitudes of spiky solutions are determined by the interaction of the heterogeneity with the geometry of the compact metric graph, represented by Green's function. Second, we apply our abstract theorem to the $Y$-shaped metric graph and the $H$-shaped metric graph in non-heterogeneity case. In particular, we show the precise effect of the geometry of those compact graphs to the locations of concentration points for these concrete graphs, respectively.

Mathematics Subject Classification: Primary 35B25, 35R02; Secondary 35K57, 35Q92.

 Citation:

• Figure 1.  $Y$-shaped metric graph

Figure 2.  A one-peak solution and a two-peak solution on the interval $(-1, 1)$. $L$ is a length of $(-1,1)$

Figure 3.  A concentration point $t_0$ of a one-peak solution on the $Y$-shaped graph. By Theorem 2.2, we have $A+l_2+l_3 = L/2$

Figure 4.  Concentration points $t_1^0$, $t_2^0$ of a two-peak solution on the $Y$-shaped graph. Case A: By Theorem 2.3, it holds that $l_1 = l_2$ and $A_1+A_2+l_3 = L/2$. Moreover, we also have $A_1 = A_2$. Case B: A distance between $t_1^0$ and $t_2^0$ is $L/2$ and $A+l_2+l_3 = L/4$ holds

Figure 5.  $H$-shaped metric graph

Figure 6.  A concentration point of a one-peak solution on the $H$-shaped graph. If $t^0\in e_3$, by Theorem 3.2, then we have $A_1+l_1+l_2=L/2$ and $A_2+l_4+l_5=L/2$

Figure 7.  Concentration points of a two-peak solution on the $H$-shaped graph (CaseA, Case C, and Case D). Case A: Using Theorem 3.3, we obtain $l_1 = l_2$. Case C: $A_1+A_2+l_2 = L/2$ and $l_1 = l_3+l_4+l_5$ is required. Then, we also have $A_1 = A_2$. Case D: A distance between $t_1^0$ and $t_2^0$ is $L/2$ and $A_1+l_1+l_2 = A_2+l_4+l_5 = L/4$ holds

Figure 8.  Concentration points of a two-peak solution on the $H$-shaped graph (Case B). The point $B\in [0,l_3]$ divides $[0,l_3]$ internally in the ratio $l_5:l_2$. We have $A_1+B = A_2+(l_3-B)$

Figure 9.  $\hat{\mathcal{G}}$ is an arbitrary metric and $\mathcal{G}$ is defined by $\mathcal{G}: = \hat{\mathcal{G}} \cup \{e\}$. If a edge $e$ is sufficiently long, then can we construct a one-peak solution which concentrates near $t^0 = l_e-L/2 \in e = [0,l_e]$?

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