Coexistence stability in a four-member hypercycle with error tail through center manifold analysis

Abstract

Establishing the conditions allowing for the stable coexistence in hypercycles has been a subject of intensive research in the past decades. Deterministic, time-continuous models have indicated that, under appropriate parameter values, hypercycles are bistable systems, having two asymptotically stable attractors governing coexistence and extinction of all hypercycle members. The nature of the coexistence attractor is largely determined by the size of the hypercycle. For instance, for two-member hypercycles the coexistence attractor is a stable node. For larger dimensions more complex dynamics appear. Numerical results on so-called elementary hypercycles with $n=3$ and $n=4$ species revealed, respectively, coexistence via strongly and weakly damped oscillations. Stability conditions for these cases have been provided by linear stability and Lyapunov functions. Typically, linear stability analysis of four-member hypercycles indicates two purely imaginary eigenvalues and two negative real eigenvalues. For this case, stability cannot be fully characterized by linearizing near the fixed point. In this letter, we determine the stability of a non-elementary four-member hypercycle which considers exponential and hyperbolic replication terms under mutation giving place to an error tail. Since Lyapunov functions are not available for this case, we use the center manifold theory to rigorously show that the system has a stable coexistence fixed point. Our results also show that this fixed point cannot undergo a Hopf bifurcation, as supported by numerical simulations previously reported.

Appendices

Appendix 1: Proof of Proposition 1

Suppose we have an equilibrium point $x^{*}\in S$ (in particular all components are greater or equal than 0) such that $x^{*}_{j}=0$ for a fixed $0\le j <n$. Then since $\dot{x}_{j+1}=x_{j+1}\left( a(Q-1)+Qx_{j}-\sum _{k=1}^{n}{x_{k}x_{k-1}}\right)$ we have either $x^{*}_{j+1}=0$ or

$$\begin{aligned} 0= & {} a(Q-1)+Qx^{*}_{j}-\sum _{k=1}^{n}{x^{*}_{k}x^{*}_{k-1}}=a(Q-1)\nonumber \\&-\sum _{k=1}^{n}{x^{*}_{k}x^{*}_{k-1}}. \end{aligned}$$

This last case would imply

$$\begin{aligned} \sum _{k=1}^{n}{x^{*}_{k}x^{*}_{k-1}}=a(Q-1)<0 \end{aligned}$$

since $a>0$ and $Q<1$, leading to a contradiction because the left-hand side should be positive or zero. Therefore, $x^{*}_{j}=0$ implies $x^{*}_{j+1}=0$ for any $0 \le j < n$ (recall we identify $x_{0} \equiv x_{n}$) and so if there is some $0\le j <n$ such that $x^{*}_{j}=0$ we must have $x^{*}=\vec {0}$. It is clear then that any fixed point $x^{*} \in S$ different from the origin must satisfy

$$\begin{aligned} a(Q-1)+Qx^{*}_{j}-\sum _{k=1}^{n}{x^{*}_{k}x^{*}_{k-1}}=0, \quad \forall \; 1\le j \le n,\nonumber \\ \end{aligned}$$

(19)

which can be rewritten as

$$\begin{aligned} x^{*}_{j}=\frac{\sum _{k=1}^{n}{x^{*}_{k}x^{*}_{k-1}}-a(Q-1)}{Q}, \quad \forall \; 1\le j \le n.\nonumber \\ \end{aligned}$$

(20)

From (20) we see that all components must be equal and so (19) becomes the quadratic equation

$$\begin{aligned} a(Q-1)+Qx^{*}_{j}-n(x^{*}_{j})^{2}=0, \end{aligned}$$

(21)

which has the same two solutions for all $j\in \{1,\dots ,n\}$:

$$\begin{aligned} x^{*}_{\pm ,1}=\dfrac{Q\pm \sqrt{Q^{2}-4na(1-Q)}}{2n}. \end{aligned}$$

It is easy to verify that if the discriminant in (21) is not negative (that is $Q^{2}/(1-Q) \ge 4na$), then such fixed points are contained in S because $0<x^{*}_{\pm ,1}< 1/n$ and so $\sum _{j=1}^{n}{x^{*}_{\pm ,1}}=nx^{*}_{\pm ,1}<1$. If $Q^{2}/(1-Q) < 4na$, then the components are complex and the only fixed point that remains in S is the origin.

Appendix 2: Computation of G(y)

By the definition of F in (4), the j-th component of the vector field G defined in (10) can be written as

$$\begin{aligned} \begin{array}{rl} G_j(y)&{}=\left( C^{-1}f\left( C\,y+x_{+}^{*}\right) \right) _j\\ &{}=\sum \limits _{k=1}^{4} C^{-1}_{jk} \left( \sum \limits _{l=1}^{4}C_{kl}y_l+x^{*}_{+,1}\right) F_{k}\\ &{}=\sum \limits _{k,l=1}^{4} C^{-1}_{jk} C_{kl} y_l F_{k}+ \sum \limits _{k=1}^{4} C^{-1}_{jk} x^{*}_{+,1}F_{k}, \end{array} \end{aligned}$$

(22)

where $F_k$ stands for $F_k(C\,y+x_{+}^{*})$, $x_{+}^{*}=(x^{*}_{+,1},x^{*}_{+,1},x^{*}_{+,1},x^{*}_{+,1})$, the notation $A_{jk}$ denotes the entry of a matrix A located in the jth row and k-th column, C is defined in (9) and

$$\begin{aligned} C^{-1}=\displaystyle \frac{1}{4}\left( \begin{array}{cccc}2&{}0&{}-2&{}0\\ 0&{}2&{}0&{}-2\\ 1&{}-1&{}1&{}-1\\ 1&{}1&{}1&{}1\end{array}\right) . \end{aligned}$$

Next, we reduce the expression of the components of $F(C\,y+x_{+}^{*})$ as:

$$\begin{aligned} \begin{array}{rcl} F_{j}(C\,y+x_{+}^{*})&{}=&{}a(Q-1)+Q\,(C\,y+x_{+}^{*})_{j-1}\\ &{} &{}-\sum \limits _{k=1}^{4}(C\,y+x_{+}^{*})_{k}(C\,y+x_{+}^{*})_{k-1}\\ &{}=&{}a(Q-1)+Q\,(C\,y)_{j-1}+Q\,x^{*}_{+,1}\\ &{} &{}-\sum \limits _{k=1}^{4}(C\,y)_{k}(C\,y)_{k-1}\\ &{} &{}-\sum \limits _{k=1}^{4} x^{*}_{+,1}((Cy)_{k}+(Cy)_{k-1}) -4\,{x^{*}_{+,1}}^2\\ &{}=&{}Q\,(C\,y)_{j-1} -\sum \limits _{k=1}^{4}(C\,y)_{k}(C\,y)_{k-1}\\ &{} &{}-\sum \limits _{k=1}^{4} x^{*}_{+,1}((C\,y)_{k}+(C\,y)_{k-1}), \end{array} \end{aligned}$$

(23)

subscripts j and k indicate components of vectors and in the last step we have used Eq. (6), and $(C\,y)_{0}:=(C\,y)_{4}$. Note that the last expression in the expansion of $F_j$ only contains one term that depends on j, $Q\,(C\,y)_{j-1}$, which can be computed directly from the definition of C in (9). The other two terms are the same for every $j\in \{1,2,3,4\}$ and can be further simplified:

• The linear term $-\sum \limits _{k=1}^{4} x^{*}_{+,1}((C\,y)_{k}+(C\,y)_{k-1})$, thanks to the cyclic structure of the variables, satisfies $\sum \limits _{k=1}^{4} ((C\,y)_{k}+(C\,y)_{k-1})=2\sum \limits _{k=1}^{4} (C\,y)_{k}$. Moreover, since $\sum \limits _{k=1}^{4} C_{km}=0$ for $m=1,2,3$ and $\sum \limits _{k=1}^{4} C_{k4}=4$, then

  $$\begin{aligned} \sum \limits _{k=1}^{4} (C\,y)_{k}=\sum \limits _{k=1}^{4} \sum \limits _{m=1}^{4} C_{km} y_m= \sum \limits _{m=1}^{4} y_m \sum \limits _{k=1}^{4} C_{km}=4\,y_4. \end{aligned}$$

  We conclude that

  $$\begin{aligned} -\sum \limits _{k=1}^{4} x^{*}_{+,1}((C\,y)_{k}+(C\,y)_{k-1})=-8\,x^{*}_{+,1}\,y_4. \end{aligned}$$

• For the quadratic term $-\sum \limits _{k=1}^{4}(C\,y)_{k}(C\,y)_{k-1}$, if we call $C_k$ the k-th row of matrix C, then it is easy to prove that

  $$\begin{aligned} \sum \limits _{k=1}^4 (C y)_{k} (C y)_{k-1}=y^{\top }\, M\, y,\end{aligned}$$

  where $M:=\sum \limits _{k=1}^4 C_{k}^{\top } C_{k-1}$, assuming that $C_0:=C_4$. A straightforward computation gives $M=\left( \begin{array}{cccc} 0&{}-2&{}0&{}0\\ 2&{}0&{}0&{}0\\ 0&{}0&{}-4&{}0\\ 0&{}0&{}0&{}4\\ \end{array}\right) ,$ and so

  $$\begin{aligned} -\sum \limits _{k=1}^4 (C y)_{k} (C y)_{k-1}=4\,y_3^2-4\,y_4^2. \end{aligned}$$

Gathering the simplified expressions for the three terms from (23), we get:

$$\begin{aligned} F_{j}(C\,y+x_{+}^{*})=Q\,(C\,y)_{j-1}-8\,x^{*}_{+,1}\,y_4+4\,y_3^{2}-4\,y_4^{2}. \end{aligned}$$

We split this expression into $F_{j}(C\,y+x_{+}^{*})=:F_{j}[1]+F[2]$, for each $j=1,2,3,4$, according to the degree (notice that the quadratic term does not depend on j), where

$$\begin{aligned} \begin{array}{rl} F_{j}[1]&{}=Q\,(C\,y)_{j-1}-8\,x^{*}_{+,1}\,y_4,\\ F[2]&{}=4\,y_3^{2}-4\,y_4^{2}. \end{array} \end{aligned}$$

We plug the above expression for $F_{j}$ into the expression of $G_j$ obtained in (22):

$$\begin{aligned} G_j= & {} \sum \limits _{k,l=1}^{4} C^{-1}_{jk} C_{kl} y_l (F_{k}[1]+F[2])\\&+\sum \limits _{k=1}^{4} C^{-1}_{jk} x^{*}_{+,1}(F_{k}[1]+F[2]). \end{aligned}$$

The terms of order 1, 2 and 3 in y are then:

$$\begin{aligned} \begin{array}{rcl} G_j[1]&{}=&{}\sum \limits _{k=1}^{4} C^{-1}_{jk} x^{*}_{+,1}\, F_{k}[1],\\ G_j[2]&{}=&{}\sum \limits _{k,l=1}^{4} C^{-1}_{jk} C_{kl} y_l \,F_{k}[1]\\ &{} &{}+\sum \limits _{k=1}^{4} C^{-1}_{jk} x^{*}_{+,1}\,F[2],\\ G_j[3]&{}=&{}\sum \limits _{k,l=1}^{4} C^{-1}_{jk} C_{kl} y_l\, F[2], \end{array} \end{aligned}$$

which, after substitution of $F_{k}[1]$ and F[2], become:

$$\begin{aligned} \begin{array}{rcl} G_j[1]&{}=&{}\sum \limits _{k=1}^{4} C^{-1}_{jk} x^{*}_{+,1}\, (Q\,(C\,y)_{k-1}-8\,x^{*}_{+,1}\,y_4),\\ G_j[2]&{}=&{}\sum \limits _{k,l=1}^{4} C^{-1}_{jk} C_{kl} y_l \,(Q\,(C\,y)_{k-1}-8\,x^{*}_{+,1}\,y_4)\\ &{} &{}+ x^{*}_{+,1}(4\,y_3^{2}-4\,y_4^{2}) \sum \limits _{k=1}^{4} C^{-1}_{jk},\\ G_j[3]&{}=&{}(4\,y_3^{2}-4\,y_4^{2})\,\sum \limits _{k,l=1}^{4} C^{-1}_{jk} C_{kl} y_l. \end{array} \end{aligned}$$

(24)

We develop each term separately, just substituting the entries of $C^{-1}$ and C, respectively, in the following two steps:

$$\begin{aligned} G[1]= & {} x^{*}_{+,1}\, \left( \begin{array}{c} Q\frac{1}{2} ((Cy)_{4}-(Cy)_{2})\\ Q\frac{1}{2} ((Cy)_{1}-(Cy)_{3})\\ Q\frac{1}{4} ((Cy)_{4}-(Cy)_{1}+(Cy)_{2}-(Cy)_{3})\\ Q\frac{1}{4} ((Cy)_{4}+(Cy)_{1}+(Cy)_{2}+(Cy)_{3})-8\,x^{*}_{+,1}\,y_4\\ \end{array}\right) \\= & {} Q\,x^{*}_{+,1}\, \left( \begin{array}{c} -y_2\\ y_1\\ -y_3\\ y_4\\ \end{array}\right) +\left( \begin{array}{c} 0\\ 0\\ 0\\ -8\,{x^{*}_{+,1}}^{2}\,y_4\\ \end{array}\right) . \end{aligned}$$

This gives the linear term provided in (11), just recalling the definitions of $\kappa =Q\,x^{*}_{+,1}$ and $c=-(Q\,x^{*}_{+,1}-8\,{x^{*}_{+,1}}^{2})=-Q\,x^{*}_{+,1}+2\,Q\,x^{*}_{+,1}+2\,a(Q-1)=\kappa -2\,a\,(1-Q)$. The quadratic terms are given by the two terms of $G_j[2]$ in (24), for $j=1,2,3,4$. On the one hand, one term only influences the fourth component:

$$\begin{aligned}&x^{*}_{+,1}(4\,y_3^{2}-4\,y_4^{2}) \sum \limits _{k=1}^{4} C^{-1}_{jk}\\&\quad = \left\{ \begin{array}{l} 0 \quad \text{ if } j=1,2,3;\\ x^{*}_{+,1}(4\,y_3^{2}-4\,y_4^{2})\quad \text{ if } j=4. \end{array}\right. \end{aligned}$$

On the other hand, the term $\sum \limits _{k,l=1}^{4} C^{-1}_{jk} C_{kl} y_l \,(Q\,(C\,y)_{k-1}-8\,x^{*}_{+,1}\,y_4)$ requires more detailed computations. For $j=1$,

$$\begin{aligned}&\sum \limits _{k,l=1}^{4} C^{-1}_{1k} C_{kl} y_l \,(Q\,(C\,y)_{k-1}-8\,x^{*}_{+,1}\,y_4)\\&\quad =\sum \limits _{k=1,3}(-1)^{\frac{k-1}{2}}\frac{1}{2}\sum \limits _{l=1}^{4} C_{kl} y_l \,(Q\,(C\,y)_{k-1}{-}8\,x^{*}_{+,1}\,y_4)\\&\quad =\frac{1}{2}(y_1+y_3+y_4)(Q(-y_2-y_3+y_4)-8 x^{*}_{+,1}y_4)\\&\qquad -\frac{1}{2}(-y_1+y_3+y_4)(Q(y_2-y_3+y_4)-8 x^{*}_{+,1}y_4)\\&\quad =Q(-y_1y_3+y_1y_4-y_2y_3-y_2y_4)-8 x^{*}_{+,1}y_1 y_4, \end{aligned}$$

which coincides with the first component of G[2] in (11). Similarly, for $j=2$:

$$\begin{aligned}&\sum \limits _{k,l=1}^{4} C^{-1}_{2k} C_{kl} y_l \,(Q\,(C\,y)_{k-1}-8\,x^{*}_{+,1}\,y_4)\\&\quad =\sum \limits _{k=2,4} (-1)^{\frac{k-2}{2}}\frac{1}{2} \sum \limits _{l=1}^{4} C_{kl} y_l \,(Q\,(C\,y)_{k-1}-8\,x^{*}_{+,1}\,y_4)\\\\&\quad =\frac{1}{2}(y_2-y_3+y_4)(Q(y_1+y_3+y_4)-8 x^{*}_{+,1}y_4)\\\\&\qquad -\frac{1}{2}(-y_2-y_3+y_4)(Q(-y_1+y_3+y_4)-8 x^{*}_{+,1}y_4)\\\\&\quad =Q(-y_1y_3+y_1y_4+y_2y_3+y_2y_4)-8 x^{*}_{+,1}y_2 y_4, \end{aligned}$$

which coincides with the second component of G[2] in (11). For $j=3$:

$$\begin{aligned}&\sum \limits _{k,l=1}^{4} C^{-1}_{3k} C_{kl} y_l \,(Q\,(C\,y)_{k-1}-8\,x^{*}_{+,1}\,y_4)\\\\&\quad =\sum \limits _{k=1}^{4} (-1)^{k-1}\frac{1}{4} \sum \limits _{l=1}^{4} C_{kl} y_l \,(Q\,(C\,y)_{k-1}-8\,x^{*}_{+,1}\,y_4)\\\\&\quad =\frac{1}{4}(y_1+y_3+y_4)(Q(-y_2-y_3+y_4)-8 x^{*}_{+,1}y_4)\\\\&\qquad -\frac{1}{4}(y_2-y_3+y_4)(Q(y_1+y_3+y_4)-8 x^{*}_{+,1}y_4)\\\\&\qquad +\frac{1}{4}(-y_1+y_3+y_4)(Q(y_2-y_3+y_4)-8 x^{*}_{+,1}y_4)\\\\&\qquad -\frac{1}{4}(-y_2-y_3+y_4)(Q(-y_1+y_3+y_4)-8 x^{*}_{+,1}y_4)\\\\&\quad =Q(-y_1y_2)-8 x^{*}_{+,1}y_3 y_4, \end{aligned}$$

which coincides with the third component of G[2] in (11). For $j=4$:

$$\begin{aligned}&\sum \limits _{k,l=1}^{4} C^{-1}_{4k} C_{kl} y_l \,(Q\,(C\,y)_{k-1}-8\,x^{*}_{+,1}\,y_4)\\\\&\quad =\sum \limits _{k=1}^{4} \frac{1}{4} \sum \limits _{l=1}^{4} C_{kl} y_l \,(Q\,(C\,y)_{k-1}-8\,x^{*}_{+,1}\,y_4)\\\\&\quad =\frac{1}{4}(y_1+y_3+y_4)(Q(-y_2-y_3+y_4)-8 x^{*}_{+,1}y_4)\\\\&\qquad +\frac{1}{4}(y_2-y_3+y_4)(Q(y_1+y_3+y_4)-8 x^{*}_{+,1}y_4)\\\\&\qquad +\frac{1}{4}(-y_1+y_3+y_4)(Q(y_2-y_3+y_4)-8 x^{*}_{+,1}y_4)\\\\&\qquad +\frac{1}{4}(-y_2-y_3+y_4)(Q(-y_1+y_3+y_4)-8 x^{*}_{+,1}y_4)\\\\&\quad =Q(-y_3^2+y_4^2)-8 x^{*}_{+,1}y_4^2, \end{aligned}$$

which, after adding the expression $x^{*}_{+,1}(4\,y_3^{2}-4\,y_4^{2})$ obtained above, coincides with the last component of G[2] in (11).

Finally, similar computations lead to:

$$\begin{aligned} G_1[3]= & {} (4\,y_3^{2}-4\,y_4^{2})\,\sum \limits _{k=1,3}(-1)^{\frac{k-1}{2}}\frac{1}{2}\sum \limits _{l=1}^{4} C_{kl} y_l\\= & {} (4\,y_3^{2}-4\,y_4^{2})\left( \frac{1}{2} (y_1+y_3+y_4)\right. \\&\left. -\frac{1}{2} (-y_1+y_3+y_4)\right) \\= & {} (4\,y_3^{2}-4\,y_4^{2})\,y_1;\\ G_2[3]= & {} (4\,y_3^{2}-4\,y_4^{2})\,\sum \limits _{k=2,4}(-1)^{\frac{k-2}{2}}\frac{1}{2}\sum \limits _{l=1}^{4} C_{kl} y_l\\= & {} (4\,y_3^{2}-4\,y_4^{2})\left( \frac{1}{2} (y_2-y_3+y_4)\right. \\&\left. -\frac{1}{2} (-y_2-y_3+y_4)\right) \\= & {} (4\,y_3^{2}-4\,y_4^{2})\,y_2;\\ G_3[3]= & {} (4\,y_3^{2}-4\,y_4^{2})\,\sum \limits _{k=1}^{4}(-1)^{{k-1}}\frac{1}{4}\sum \limits _{l=1}^{4} C_{kl} y_l\\= & {} (4\,y_3^{2}-4\,y_4^{2})\,y_3;\\ G_4[3]= & {} (4\,y_3^{2}-4\,y_4^{2})\,\sum \limits _{k=1}^{4} \frac{1}{4} \sum \limits _{l=1}^{4} C_{kl} y_l\\= & {} (4\,y_3^{2}-4\,y_4^{2})\,y_4, \end{aligned}$$

as in (11) (notice that, in the last two steps, for the sake of the presentation, we do not display the full expansion in terms of the $C_{kl}$ elements). This completes the computation to obtain the expressions in (11) from (10).