Competition between phytoplankton and bacteria: exclusion and coexistence

Abstract

Resource-based competition between microorganisms species in continuous culture has been studied extensively both experimentally and theoretically, mostly for bacteria through Monod and Contois “constant yield” models, or for phytoplankton through the Droop “variable yield” models. For homogeneous populations of N bacterial species (Monod) or N phytoplanktonic species (Droop), with one limiting substrate and under constant controls, the theoretical studies indicated that competitive exclusion occurs: only one species wins the competition and displaces all the others (Armstrong and McGehee in Am Nat 115:151, 1980; Hsu and Hsu in SIAM J Appl Math 68:1600–1617, 2008). The winning species expected from theory is the one with the lowest “substrate subsistence concentration” $s^{\star }$, such that its corresponding equilibrium growth rate is equal to the dilution rate $D$. This theoretical result was validated experimentally with phytoplankton (Tilman and Sterner in Oecologia 61(2):197–200, 1984) and bacteria (Hansen and Hubell in Science 207(4438):1491–1493, 1980), and observed in a lake with microalgae (Tilman in Ecology 58(22):338–348, 1977). On the contrary for aggregating bacterial species described by a Contois model, theory predicts coexistence between several species (Grognard et al. in Discrete Contin Dyn Syst Ser B 8(1):73–93, 2007). In this paper we present a generalization of these results by studying a competition between three different types of microorganisms: planktonic (or free) bacteria (represented by a generalized Monod model), aggregating bacteria (represented by a Contois model) and free phytoplankton (represented by a Droop model). We prove that the outcome of the competition is a coexistence between several aggregating bacterial species with a free species of bacteria or phytoplankton, all the other free species being washed out. This demonstration is based mainly on the study of the substrate concentration’s evolution caused by competition; it converges towards the lowest subsistence concentration $s^{\star }$, leading to three different types of competition outcomes: (1) the best free bacteria or phytoplankton competitor excludes all other species; (2) only some aggregating bacterial species coexist in the chemostat; (3) A coexistence between the single best free species, with one or several aggregating species.

Appendix

1.1 A Step 3: case $a$: $L$ attains $s^{\star }$ in finite time

In this case

• if $s^{\star } = s^{z\star }_1$ we consider Fig. 8 where $L$ attains $s^{\star }$ after a finite time $t^L$:

  $$\begin{aligned} \forall t \ge t^L, L(t) = s^{\star } \end{aligned}$$

Fig. 8

Visual explanation of the demonstration of Lemma 10—Case a: $L$ attains $s^{\star }$ in finite time $t^L$ (Q-model). $(i)$ $q_1$ is repeatedly higher than $Q_1(s^{\star })+\theta$ $(\bullet )$. $(ii)$ Because ${\dot{q}}_1$ is upper bounded by $\rho ^m_1$, so that $q_1$ is higher than $Q_1(s^{\star })+\theta /2$ during non negligible time intervals (dashed lines represent ${\dot{q}}_1 = \rho ^m_1$). Thus $z_1$ diverges, which is a contradiction

1.1.1 Substep 3a.1: after a finite time larger than $t^L$, $q_1$ is repeatedly higher than $Q_1(s^{\star }) + \theta$

Since $\min _i(S^z_k(q_k)) \ge L$, we know that

$$\begin{aligned} \forall t>t^L, q_1(t)\ge Q_1(s^{\star }) \end{aligned}$$

As $s$ does not converge to $s^{\star }$, we also know from Lemma 6 that $q_1$ does not converge towards $Q_1(s^{\star })$:

$$\begin{aligned} \exists \theta >0, \forall t>0, \exists t^q>t, |q_1(t^q) - Q_1(s^{\star })| > \theta \end{aligned}$$

Those two facts imply that the repeated exits of $q_1(t)$ from the $\theta$-interval around $Q_1(s^{\star })$ take place above $Q_1(s^{\star })$ for any $t^q > t^L$, so that, in that case, we have $q_1(t^q)>Q_1(s^{\star }) + \theta$. In Fig. 8, such $t^q$ time instants are represented by $\bullet$.

1.1.2 Substep 3a.2: $q_1$ is higher than $Q_1(s^{\star }) + \theta /2$ during non negligible time intervals

Since the $q_1$-dynamics are upper bounded with

$$\begin{aligned} {\dot{q}}_1 \le \rho ^m_1 \end{aligned}$$

we know that every time $q_1$ is higher than $Q_1(s^{\star }) + \theta$, it has been higher than $Q_1(s^{\star }) + \theta /2$ during a time interval of minimal duration $A(\theta ) = \frac{\theta }{2 \rho ^m_1}$. On Fig. 8, ${\dot{q}}_1 = \rho ^m_1$ is represented by the dashed lines.

1.1.3 Substep 3a.3: then $z_1$ diverges, which is impossible

From time $t^L$ on, we have that $q_1\ge Q_1(s^{\star }) \Rightarrow \gamma _1(q_1) \ge D$, so that $z_1(t)$ is non decreasing. During each of the time interval where $q_1$ is higher than $Q_1(s^{\star }) + \theta /2$, the increase of $z_1$ is lower bounded by

$$\begin{aligned} {\dot{z}}_1 = \gamma _1(Q_1(s^{\star }) + \theta /2) - D = C(\theta ) > 0 \end{aligned}$$

so that every $t^q$ time we have

$$\begin{aligned} z_1(t^q)-z_1(t^q-A(\theta )) > C(\theta ) A(\theta ) \end{aligned}$$

As such increases occurs repeatedly, and as $z_1$ is non decreasing, $z_1$ diverges. This is a contradiction because $z_1$ is upper bounded [see (12)].

• if $s^{\star } = s^{x\star }_1$ then the non convergence of $s$ to $s^{\star }$, and the fact that $s \ge s^{\star }$ will cause $s$ to be non negligibly “away” from $s^{\star }$, so that $x_1$ will diverge, causing a contradiction with (11). This is exactly the same demonstration as above (in the case $s^{\star }=s^{z\star }_1$) without needing the $q_k$ study.

• if $s^{\star } = s^{y\star }$ then $s+\sum _{j=1}^{N_y} Y_j(s)$ will always be higher than $s_{in}=s^{y\star }+\sum _{j=1}^{N_y} Y_j(s^{y\star })$ without converging to $s_{in}$, which is in contradiction with (27).

1.2 B Step 3: Case $b$: $L$ never attains $s^{\star }$

In this case $L(t)$ converges towards a value ${\hat{L}} \in (0,s^{\star }]$, because it is non decreasing and bounded in $[0,s^{\star }]$, so that

$$\begin{aligned} \forall \epsilon >0, \exists t^L(\epsilon )>0, \forall t>t^L(\epsilon ), \arrowvert L(t)-{\hat{L}}\arrowvert <\epsilon \end{aligned}$$

We consider the neighborhood of ${\hat{L}}$ in Fig. 9.

Fig. 9

Visual explanation of the demonstration of Lemma 10—Case b: $L$ never attains $s^{\star }$. $(i)$ $s$ is repeatedly higher than ${\hat{L}} + \lambda \; (\bullet )$. $(ii)$ ${\dot{s}}$ is upper bounded by $D s_{in}$, so that $s$ is higher than ${\hat{L}} + \lambda /2$ during non negligible time intervals (dashed lines represent ${\dot{s}} = D s_{in}$) $(iii)$ during such a time intervals $L = \min _k(S^z_k(q_k))$ (or $\min _j(S^y_j(y_j))$) is increasing non negligibly towards $s$, so that $L$ cannot both converge towards ${\hat{L}}$ and stay lower than ${\hat{L}}$ during the whole time interval: there is a contradiction

1.2.1 Substep 3b.1: after a finite time, $s$ is repeatedly higher than ${\hat{L}} + \lambda$

Since, from the beginning of the proof of Lemma 10, we know that $s$ does not converge to any constant value, hence not to ${\hat{L}}$,

$$\begin{aligned} \exists \lambda >0, \forall t>0, \exists t^s>t, |s(t^s) - {\hat{L}}| > \lambda \end{aligned}$$

Since $L$ is increasing and converges to ${\hat{L}}$, it reaches ${\hat{L}} - \lambda$ in finite time $t^L(\lambda )$. After this finite time, $s$ is higher than ${\hat{L}} + \lambda$ on every $t^s$ time instants, which are represented by $\bullet$ in Fig. 9.

1.2.2 Substep 3b.2: $s$ is higher than ${\hat{L}} + \lambda /2$ during non negligible time intervals

Because of the boundedness of ${\dot{s}}$

$$\begin{aligned} {\dot{s}} \le D s_{in} \end{aligned}$$

every time $s$ is higher than ${\hat{L}} + \lambda$, it has been higher than ${\hat{L}} + \lambda /2$ during a non negligible time interval of minimal duration $A(\lambda ) = \frac{\lambda }{2 D s_{in}}$. On Fig. 9 the case ${\dot{s}} = D s_{in}$ is represented by dashed lines.

1.2.3 Substep 3b.3: $L = \min _k(S^z_k(q_k))$ (or $\min _j(S^y_j(y_j))$) is increasing non negligibly towards $s$, so that $L$ cannot both converge towards ${\hat{L}}$ and stay lower than ${\hat{L}}$ during the whole time interval: there is a contradiction

Like in previous proofs, we are interested in what happens during the $[t^s - A(\lambda ),t^s]$ time-interval, with $t^s -A(\lambda )>t^L(\epsilon )$ (for some $\epsilon <\lambda$). Since, during this time-interval, $s(t) > {\hat{L}} + \lambda /2$ and $L < {\hat{L}}$, we know that there exists a $k$ such that $L(t^s) = S^z_k(q_k(t^s)) < {\hat{L}}$, or a $j$ such that $L(t^s) = S^y_j(y_j(t^s)) < {\hat{L}}$.

For both this step (3b.3) we choose to first only present arguments for the case $L(t^s) = \min _k(S^z_k(q_k))$; almost similar arguments for the case $L(t^s) = \min _j(S^y_j(y_j))$ will then be briefly presented.

• if $L(t^s) = \min _k(S^z_k(q_k))$, then during the whole considered time-interval, as $S^z_k(q_k)$ was increasing, we know that

  $$\begin{aligned} {\hat{L}} - \epsilon < L \le S^z_k(q_k) \le S^z_k(q_k(t^s)) < {\hat{L}} \end{aligned}$$

  (29)

  so that $Q_k({\hat{L}}-\epsilon ) < q_k(t) < Q_k({\hat{L}})$. For the $k$ species, the dynamics of $q_k$ can then be lower bounded:

  $$\begin{aligned} {\dot{q}}_k \ge \rho _k({\hat{L}}+\lambda /2) - f_k(Q_k({\hat{L}})) \end{aligned}$$

  and then

  $$\begin{aligned} {\dot{q}}_k \ge \rho _k({\hat{L}}+\lambda /2) - \rho _k({\hat{L}}) = G_k(\lambda ) \end{aligned}$$

  positive, so that the increase of $q_k$ during the $[t^s - A(\lambda ),t^s]$ time-interval is also lower bounded:

  $$\begin{aligned} q_k(t^s) - q_k(t^s - A(\lambda )) \ge G_k(\lambda ) A(\lambda ) = H_k(\lambda ) \end{aligned}$$

  Since $Q_k = S^{z^{-1}}_k$ is locally Lipschitz with constant $K$ (because $f'_k > 0$), we have

  $$\begin{aligned} q_k(t^s)-q_k(t^s-A(\lambda ))&= Q_k(S^z_k(q_k(t^s)))-Q_k(S^z_k(q_k(t^s-A(\lambda ))))\\&< K \left[ S^z_k(q_k(t^s)) - S^z_k(q_k(t^s-A(\lambda ))) \right] \end{aligned}$$

  so that the corresponding increase of $S^z_k(q_k)$ is lower bounded with

  $$\begin{aligned} S^z_k(q_k(t^s)) - S^z_k(q_k(t^s - A(\lambda ))) \ge \frac{1}{K} H_k(\lambda ) \end{aligned}$$

  and then

  $$\begin{aligned} S^z_k(q_k(t^s - A(\lambda ))) < {\hat{L}} - \frac{1}{K} H_k(\lambda ) \end{aligned}$$

  which implies the same higher bound for $L$:

  $$\begin{aligned} L(t^s - A(\lambda )) < {\hat{L}} - \frac{1}{K} H_k(\lambda ) \end{aligned}$$

  By choosing $\epsilon < \frac{1}{K} H_k(\lambda )$, this inequality is contradictory with (29) so that Case 2 is not possible

• if $L(t^s) = \min _j(S^y_j(y_j))$, then the same arguments can be developed for the $j$ species, with a lower bound $G_j(\lambda )$ on the $y_j$ dynamics:

  $$\begin{aligned} G_j(\lambda ) = \beta _j({\hat{L}}+\lambda /2,Y_j({\hat{L}})) - \beta _j({\hat{L}},Y_j({\hat{L}})) > 0 \end{aligned}$$

  and then an increase of variable $y_j$ at least equal to $H_j(\lambda ) = G_j(\lambda ) A(\lambda )$ followed by a non negligible increase of $L$

  $$\begin{aligned} L(t^s - A(\lambda )) < {\hat{L}} - \frac{1}{K} H_j(\lambda ) \end{aligned}$$

  because $Y_j$ is locally Lipschitz. Finally a contradiction also occurs when $\epsilon < \frac{1}{K} H_j(\lambda )$:

  $$\begin{aligned} L(t^s - A(\lambda )) < {\hat{L}} - \epsilon \end{aligned}$$

1.3 C Computation of system $\Sigma$ Jacobian Matrix and eigenvalues for all the equilibria

Computation of the Jacobian Matrix of system $\Sigma$, with $s=s_{in}-\sum _{i=1}^{N_x} x_i-\sum _{j=1}^{N_y} y_j-\sum _{k=1}^{N_z} q_k z_k$.

$$\begin{aligned} \left( \begin{array}{cccc} J^{xx} &{} J^{xy} &{} J^{xz} &{} J^{xq} \\ J^{yx} &{} J^{yy} &{} J^{yz} &{} J^{yq} \\ J^{zx} &{} J^{zy} &{} J^{zz} &{} J^{zq} \\ J^{qx} &{} J^{qy} &{} J^{qz} &{} J^{qq} \end{array} \right) \end{aligned}$$

where

$$\begin{aligned}&J^{xx}_{ii} = \alpha _i(s)-D -\frac{\partial \alpha _i}{\partial s} x_i\quad \hbox {and} \quad \forall l \ne i, J^{xx}_{il} = -\frac{\partial \alpha _i}{\partial s} x_i \\&J^{xy}_{ij} = -\frac{\partial \alpha _i}{\partial s} x_i \\&J^{xz}_{ik} = -\frac{\partial \alpha _i}{\partial s} x_i q_k \\&J^{xq}_{ik} = -\frac{\partial \alpha _i}{\partial s} x_i z_k \end{aligned}$$

and

$$\begin{aligned}&J^{yx}_{ji} = -\frac{\partial \beta _j}{\partial s} y_j \\&J^{yy}_{jj} = \beta _j(s,y_j)-D + \frac{\partial \beta _j}{\partial y_j}y_j -\frac{\partial \beta _j}{\partial s} y_j \quad \hbox {and}\quad \forall l \ne j, J^{yy}_{jl} = -\frac{\partial \beta _j}{\partial s} y_j \\&J^{yz}_{jk} = -\frac{\partial \beta _j}{\partial s} y_j q_k \\&J^{yq}_{jk} = -\frac{\partial \beta _j}{\partial s} y_j z_k \end{aligned}$$

and

$$\begin{aligned}&J^{zx}_{ki} = 0 \\&J^{zy}_{kj} = 0 \\&J^{zz}_{kk} = \gamma _k(q_k)-D \quad \hbox {and} \quad \forall l \ne k, J^{zz}_{kl} = 0 \\&J^{zq}_{kk} = \frac{\partial \gamma _k}{\partial q_k} z_k \quad \hbox {and} \quad \forall l \ne k, J^{zq}_{kl} = 0 \end{aligned}$$

and

$$\begin{aligned} J^{qx}_{ki}&= -\frac{\partial \rho _k}{\partial s} \\ J^{qy}_{kj}&= -\frac{\partial \rho _k}{\partial s} \\ J^{qz}_{kl}&= -\frac{\partial \rho _k}{\partial s} q_l \\ J^{qq}_{kk}&= -\frac{\partial \rho _k}{\partial s} z_k - \frac{\partial f_k}{\partial q_k} \quad \hbox {and} \quad \forall l \ne k, J^{qq}_{kl} = -\frac{\partial \rho _k}{\partial s} z_l \end{aligned}$$

Fortunately for eigenvalue computations, at equilibria the null biomasses will simplify the matrix:

• when $x_i=0$, then the whole $i^{th}$ line gives eigenvalue $\alpha _i(s)-D$ (denoted “$x_i$-eigenvalue”) and can be deleted, as well as the $i^{th} column$;

• when $y_j=0$ then the whole $N_x+j^{th}$ line gives eigenvalue $\beta _j(s,y_j)-D$ (denoted “$y_j$-eigenvalue”) and can be deleted, as well as the $N_x+j^{th}$ corresponding column;

• when $z_k=0$ then the whole $N_x+N_y+k^{th}$ line gives eigenvalue $\gamma _k(q_k)-D$ (denoted “$z_k$-eigenvalue”) and can be deleted, as well as the $N_x+N_y+k^{th}$ column; in a second step, the whole $N_x+N_y+N_z+k^{th}$ column can also be deleted and gives eigenvalue $\frac{-\partial f_k}{\partial q_k}$ (denoted “$q_k$-eigenvalue”), as well as the $N_x+N_y+N_z+k^{th}$ line.

1.3.1 C.1 Complete washout equilibrium

With this in hand, we see that for equilibrium ${\tilde{E}}_0\;(x_i=y_j=z_k=0)$ the Jacobian matrix is triangular, so that the eigenvalues lay on the diagonal. They are:

• $\alpha _i(s_{in})-D$

• $\beta _j(s_{in},0)-D$

• $\gamma _k(Q_k(s_{in}))-D$

• $- \frac{\partial f_k}{\partial q_k} \quad \hbox {(negatives)}$

We denote $n_x,\; n_y,\; n_z$ the number of M-, C- and Q-species verifying the inequalities of Hypothesis 5, and thus having the possibility to be at equilibrium with a positive biomass, under controls $D$ and $s_{in}$. Each of these species has a positive corresponding eigenvalue on this equilibrium, so that equilibrium ${\tilde{E}}_0$ has $n_x+n_y+n_z$ positive eigenvalues, and $N_x-n_x+N_y-n_y+ 2 N_z - n_z$ negative eigenvalues.

1.3.2 C.2 M-only equilibria

For equilibrium $E^{x\star }_i$ we get all the previously cited x-, y-, z- and q-eigenvalues:

• $\alpha _l(s^{x\star }_i)-D$ whose signs are the same as $sign(s^{x\star }_i-s^{x\star }_l)$

• $\beta _j(s^{x\star }_i,0)-D$ which are positive if the $j^{th}$ species is $s^{x\star }_i$-compliant, or negative else;

• $\gamma _k(Q_k(s^{x\star }_i))-D$ whose signs are the same as $sign(s^{x\star }_i-s^{z\star }_k)$

• $\frac{-\partial f_k}{\partial q_k}$ which are all negative

and the remaining eigenvalue corresponds to the positive $x_i$-only dynamics:

$$\begin{aligned} {\dot{x}}_i=(\alpha _i(s_{in}-x_i)-D)x_i \end{aligned}$$

which yields the eigenvalue $-\frac{\partial \alpha _i}{\partial s} x_i^{\star }$ for free bacteria species $i$. Each free species with a substrate subsistence concentration $s^{x\star }_l$ or $s^{z\star }_k$ lower than $s^{x\star }_i$ gives a positive eigenvalue. Among all the $E^{x\star }_i$ equilibria, only $E^{x\star }_1$ is stable if and only if $s^{\star }=s_1^{x\star }<s_1^{z\star }$, and if all the C-species are not $s^{x\star }_1$-compliant.

1.3.3 C.3 Q-only equilibria

For Equilibrium $E^{z\star }_k$ we get all the

• $x$-eigenvalues whose signs are the sign of $sign(s^{z\star }_k-s^{x\star }_i)$;

• $y$-eigenvalues: as previously, $y$-eigenvalues are positive if the corresponding C-species is $s^{z\star }_k$-compliant and negative else;

• $z_l$-eigenvalues whose signs are the sign of $sign(s^{z\star }_k-s^{z\star }_l)$;

• $q_l$-eigenvalues for all $l \ne k$ (negative);

and the remaining eigenvalues correspond to the positive $(z_k,q_k)$-only dynamics:

$$\begin{aligned} \left\{ \begin{array}{l} {\dot{z}}_k= (\gamma _k(q_k) - D) z_k \\ {\dot{q}}_k= \rho _k(s_{in}-q_kz_k) - f_k(q_k) \end{array}\right. \end{aligned}$$

and we obtain the following resulting matrix:

$$\begin{aligned} \left( \begin{array}{cccc} 0 &{} \frac{\partial \gamma _k}{\partial q_k} z_k \\ -\frac{\partial \rho _k}{\partial s} q_k &{} -\frac{\partial \rho _k}{\partial s} z_k-\frac{\partial f_k}{\partial q_k} \end{array} \right) \end{aligned}$$

which has negative trace and positive determinant, so that its two eigenvalues are real negative. Just like before, each free species with a substrate subsistence concentration $s^{x\star }_i$ or $s^{z\star }_l$ lower than $s^{z\star }_k$ gives a positive eigenvalue. Among all the $E^{z\star }_k$ equilibria, only $E^{z\star }_1$ is stable if and only if $s^{\star }=s_1^{z\star }<s_1^{x\star }$, and if all the C-species are not $s^{z\star }_1$-compliant.

1.3.4 C.4 C-only equilibria

Now let us consider the $E^{y\star }_G$ equilibria for which all $j \in G$ (where $G$ represents a subset of $\{1,\ldots ,N_y\}$) C-species coexist in the chemostat under substrate concentration $s^{y\star }_G$, while all the free species are washed out. $s^{y\star }_G$ is defined by $s^{y\star }_G + \sum _{j \in G} Y_j(s^{y\star }_G) = s_{in}$. Note that some of the $G$ species can have a null biomass on these equilibria, as $Y_j(s^{y\star }_G)$ might be null for some $j \in G$.

This gives all the

• $x$-eigenvalues whose sign are the same as the signs of $s^{x\star }_i-s^{y\star }_G$;

• $z$-eigenvalues whose sign are the same as the signs of $s^{z\star }_k-s^{y\star }_G$;

• $q$-eigenvalues (negative).

All the $y_j$ species who are not included in $G$ give negative eigenvalues if they are not $s^{y\star }_G$-compliant, and positive eigenvalues else; their eigenvalues cannot be null because of technical Hypothesis 7. All the $y_j$ species who are included in $G$ but have a null biomass $Y_j(s^{y\star }_G)$ on the $E^{y\star }_G$ equilibrium give negative eigenvalues. Now let us study the remaining matrix $J^{yy}_G$ which is composed of all the $j \in G$ lines of $J^{yy}$, for which $Y_j(s^{y\star }_G)>0$, and thus $\beta _j(s^{y\star }_G, Y_j(s^{y\star }_G))=D$:

$$\begin{aligned} {\dot{y}}_j=\left( \beta _j(s-\sum _ly_l,y_j)-D\right) y_j \end{aligned}$$

which yields the Jacobian matrix:

$$\begin{aligned} J^{yy}_G = \left( \begin{array}{ccccc} -a_1-b_1 &{} \ldots &{} -a_1 &{} \ldots &{} -a_1\\ \vdots &{} \ddots &{} \vdots &{} &{} \vdots \\ -a_j &{} \ldots &{} -a_j-b_j &{} \ldots &{} -a_j \\ \vdots &{} &{} \vdots &{} \ddots &{} \vdots \\ -a_n &{} \ldots &{} -a_n &{} \ldots &{} -a_n-b_n \\ \end{array} \right) \end{aligned}$$

with $a_j = \frac{\partial \beta _j}{\partial s}Y_j(s^{y\star }_G)>0$ and $b_j = -\frac{\partial \beta _j}{\partial y_j}Y_j(s^{y\star }_G)>0$.

Let us show that this matrix has only real negative eigenvalues, by using the definition of an eigenvalue $\lambda =(A + B i)$, where $A \in {\mathbb {R}}$ is the real part and $B \in {\mathbb {R}}$ the imaginary part:

$$\begin{aligned} J^{yy}_G \cdot \left( \begin{array}{l} y_1 \\ \vdots \\ y_n \end{array} \right) = (A+Bi) \left( \begin{array}{l} y_1 \\ \vdots \\ y_n \end{array} \right) \end{aligned}$$

(30)

We obtain $n$ equations:

$$\begin{aligned} -b_j y_j - a_j \sum _l y_l = (A+Bi) y_j \end{aligned}$$

and thus

$$\begin{aligned} (A+Bi+b_j) y_j = - a_j \sum _l y_l \end{aligned}$$

(31)

If we have $A+Bi+b_j = 0$ for some $j$, then $B=0$ and $A=-b_j<0$ so that we have a negative eigenvalue.

Else, isolating $y_j$ yields

$$\begin{aligned} y_j=\frac{- a_j \sum _l y_l}{b_j+A+Bi} \end{aligned}$$

Summing over $j$, we obtain

$$\begin{aligned} \sum _j y_j=\sum _j\left( \frac{- a_j \sum _l y_l}{b_j+A+Bi}\right) \end{aligned}$$

Now if $\sum _j y_j=0$, since some $y_j$ must be different of $0$, (31) yields, for that $j$, that $A+Bi+b_j = 0$ so that again $B=0$ and $A=-b_j<0$.

Else, simplifying the sums of $y_l$ and $y_j$, this yields

$$\begin{aligned} 1&= \sum _j\left( \frac{- a_j}{b_j+A+Bi}\right) \\&= \sum _j\left( \frac{- a_j(b_j+A-Bi)}{(b_j+A)^2+B^2}\right) \\&= \sum _j\left( \frac{- a_j(b_j+A)}{(b_j+A)^2+B^2}\right) +i\sum _j\left( \frac{a_jB)}{(b_j+A)^2+B^2}\right) \end{aligned}$$

Since the left-hand-side is real, the imaginary part of the right-hand side must be zero, which imposes $B=0$. For the right-hand-side to be positive, at least one of the $b_j+A$ must be negative, which translates into $\min _j (b_j+A)<0$ and

$$\begin{aligned} A<-\min _j b_j<0 \end{aligned}$$

We conclude from this that all eigenvalues of this matrix are real negative.

Finally, an $E^{y\star }_G$ equilibrium is stable if and only if all the C-species not contained in $G$ are not $s^{y\star }_G$-compliant (this is equivalent to saying that $s^{y\star }_G = s^{y\star }, \hbox { with } s^{y\star }= s^{y\star }_{\{1,\ldots ,N_y\}}$), and if $s^{\star }=s^{y\star }$.

1.3.5 C.5 M-coexistive equilibria

In this section we consider equilibria $E^{(x,y)\star }_{i,G}$ where free bacteria species $x_i$ coexists with the C-species in $G$, a subset of $\{1,\ldots ,N_y\}$, under substrate concentration $s^{x\star }_i$.

We obtain here all the

• $x_l$-eigenvalues ($l \ne i$) whose signs are the signs of $s^{x\star }_i-s^{x\star }_l$;

• $z$-eigenvalues whose sign is the sign of $s^{x\star }_i-s^{z\star }_k$;

• $q$-eigenvalues (negative);

$y_j$-eigenvalues with $j$ not in $G$ are positive if $y_j$ is $s^{x\star }_i$-compliant and negative else; $y_j$-eigenvalues with $j$ in $G$ but have a null biomass $Y_j(s^{x\star }_i)$ give negative eigenvalues. For the remaining C-species, and species $x_i$, we obtain the following system:

$$\begin{aligned} \left\{ \begin{array}{l} {\dot{x}}_i = \left( \alpha _i(s_{in}-x_i-\sum _ly_l) - D\right) x_i \\ {\dot{y}}_j = \left( \beta _j(s_{in}-x_i-\sum _ly_l ,y_j) - D\right) y_j \\ \end{array}\right. \end{aligned}$$

and the Jacobian matrix:

$$\begin{aligned} \left( \begin{array}{cccccc} -a_0 &{} -a_0 &{} \ldots &{} -a_0 &{} \ldots &{} -a_0 \\ -a_1 &{} -a_1-b_1 &{} \ldots &{} -a_1 &{} \ldots &{} -a_1\\ \vdots &{}\vdots &{} \ddots &{} \vdots &{} &{} \vdots \\ -a_j &{} -a_j &{} \ldots &{} -a_j-b_j &{} \ldots &{} -a_j \\ \vdots &{} \vdots &{} &{} \vdots &{} \ddots &{} \vdots \\ -a_n &{} -a_n &{} \ldots &{} -a_n &{} \ldots &{} -a_n-b_n \\ \end{array} \right) \end{aligned}$$

with $a_0=\frac{\partial \alpha _i}{\partial s}x_i^{\star }>0,\; a_j = \frac{\partial \beta _j}{\partial s}Y_j(s^{x\star }_i)>0$ and $b_j = -\frac{\partial \beta _j}{\partial y_j}Y_j(s^{x\star }_i)>0$ (for $j\in \{1,\ldots ,n\}$). This matrix has exactly the same form has the one considered on Appendix C.4. The only difference being that the there is no “$b_0$” in the first element of the matrix. Defining a $b_0=0$, we can then conclude that all eigenvalues are real and negative because, following the development of Appendix C.4, we obtain

$$\begin{aligned} A<-\min _j b_j=0 \end{aligned}$$

Finally, only equilibrium $E^{(x,y)\star }_{1,\{1,\ldots ,N_y\}}$ can be stable if and only if $s^{\star }=s^{x\star }_1$.

1.3.6 C.6 Q-coexistive equilibria

In this section we consider equilibria $E^{(z,y)\star }_{k,G}$ where phytoplankton species $z_k$ coexists with the aggregating species in $G$, a subset of $\{1,\ldots ,N_y\}$, under substrate concentration $s^{z\star }_k$.

We obtain here all the

• $x$-eigenvalues whose signs are the signs of $s^{z\star }_k-s^{x\star }_i$;

• $z_l$-eigenvalues ($l \ne j$) whose sign are the signs of $s^{z\star }_k-s^{z\star }_l$;

• $q$-eigenvalues (negative);

$y_j$-eigenvalues with $j$ not in $G$ are positive if $y_j$ is $s^{z\star }_k$-compliant and negative else; $y_j$-eigenvalues with $j$ in $G$ but have a null biomass $Y_j(s^{z\star }_k)$ give negative eigenvalues. For the remaining C-species, and species $z_k$, we obtain the following model

$$\begin{aligned} \left\{ \begin{array}{l} {\dot{y}}_j = (\beta _j(s_{in}-\sum _ly_l-q_kz_k,y_j) - D) y_j\\ {\dot{z}}_k = (\gamma _k(q_k) - D) z_k \\ {\dot{q}}_k = \rho _k(s_{in}-\sum _ly_l-q_kz_k) - f_k(q_k) \end{array}\right. \end{aligned}$$

and, swapping the last two equations and using $f_k(q_k)=\gamma _k(q_k)q_k$, we get the Jacobian matrix:

$$\begin{aligned} \left( \begin{array}{ccccccc} -a_1-b_1 &{} \ldots &{} -a_1 &{} \ldots &{} -a_1 &{} -a_1 z_k^{\star } &{} -a_1 q_k^{\star } \\ \vdots &{} \ddots &{} \vdots &{} &{} \vdots &{} \vdots &{} \vdots \\ -a_j &{} \ldots &{} -a_j-b_j &{} \ldots &{} -a_j &{} -a_j z_k^{\star } &{} -a_j q_k^{\star } \\ \vdots &{} &{} \vdots &{} \ddots &{} \vdots &{} \vdots \\ -a_n &{} \ldots &{} -a_n &{} \ldots &{} -a_n-b_n &{} -a_n z_k^{\star } &{} -a_n q_k^{\star } \\ -a_{n+1} &{} \ldots &{} -a_{n+1} &{} \ldots &{} -a_{n+1} &{} -a_{n+1} z_k^{\star }-b_{n+1}q_k^{\star }-\gamma _k &{} -a_{n+1} q_k^{\star }\\ 0 &{} \ldots &{} 0 &{} \ldots &{} 0 &{} b_{n+1} z_k^{\star }&{}0 \end{array} \right) \end{aligned}$$

with $a_j = \frac{\partial \beta _j}{\partial s}Y_j(s^{z\star }_k)>0$ and $b_j = -\frac{\partial \beta _j}{\partial y_j}Y_j(s^{z\star }_k)>0$ for $j\in \{1,\ldots ,n\}$, with $a_{n+1}=\frac{\partial \rho _k}{\partial s}$ and $b_{n+1}=\frac{\partial \gamma _k}{\partial q_k}$. By using the definition of eigenvalue $\lambda =(A + B i)$ [see (30)] we follow a similar path to that of Appendix C.4. We show that the eigenvalues are real and negative by isolating $\sum _j y_j+q_k^{\star } z_k+z_j^{\star } q_k$ instead of $\sum _j y_j$ with $(y_1,\ldots ,y_n,q_k,z_k)$ a candidate eigenvector. One of the crucial arguments concerns the case where the summation is zero and with either some $y_j\ne 0$, which leads to identical arguments to those of Appendix C.4 or, if all $y_j= 0,q_k$ and $z_k$ are both non-zero and of opposite signs. In the latter case, the $q_k$ line of the eigenvector definition then becomes $(-b_{n+1}q_k^{\star }-\gamma _k)q_k=(A+Bi)q_k$ so that necessarily $B=0$ and $A<0$. The remainder of the proof is unchanged.

Finally, only equilibrium $E^{(z,y)\star }_{1,\{1,\ldots ,N_y\}}$ can be stable if and only if $s^{\star }=s^{z\star }_1$.

Remark 7

The same work can be done for the whole system (9), where the eigenvalues are the same, plus the $-D$ eigenvalue which arises from mass balance dynamics (10).