Dynamics of a Producer–Grazer Model Incorporating the Effects of Phosphorus Loading on Grazer’s Growth

Abstract

Phosphorus is an essential element for all life forms, and it is also a limiting nutrient in many aquatic ecosystems. To keep track of the mismatch between the grazer’s phosphorus requirement and producer phosphorus content, stoichiometric models have been developed to explicitly incorporate food quality and food quantity. Most stoichiometric models have suggested that the grazer dynamics heavily depends on the producer phosphorus content when the producer has insufficient nutrient content [low phosphorus (P):carbon (C) ratio]. However, recent laboratory experiments have shown that the grazer dynamics are also affected by excess producer nutrient content (extremely high P:C ratio). This phenomenon is known as the “stoichiometric knife edge.” While the Peace et al. (Bull Math Biol 76(9):2175–2197, 2014) model has captured this phenomenon, it does not explicitly track P loading of the aquatic environment. Here, we extend the Peace et al. (2014) model by mechanistically deriving and tracking P loading in order to investigate the growth response of the grazer to the producer of varying P:C ratios. We analyze the dynamics of the system such as boundedness and positivity of the solutions, existence and stability conditions of boundary equilibria. Bifurcation diagram and simulations show that our model behaves qualitatively similar to the Peace et al. (2014) model. The model shows that the fate of the grazer population can be very sensitive to P loading. Furthermore, the structure of our model can easily be extended to incorporate seasonal P loading.

Appendices

Appendix A. Proof of Theorem 3.1

Proof

We consider a solution $S(t) \equiv (x(t), y(t), Q(t),P_f(t))$ of System (3) with initial condition in $\Omega$. Hence, $0< x(0), 0< y(0) , q< Q(0)< \hat{Q}, 0< P_f(0), Q(0) x(0)+\theta y(0)+ P_f(0) < P_{f_{\mathrm{in}}}$. Assume that there exists a time $t_1 > 0$ such that S(t) touches or crosses a boundary of $\bar{\Omega }$ (closure of $\Omega$) for the first time, and then $(x(t), y(t), Q(t),P_f(t))\in \Omega$ for $0 \le t<t_1$. We will have several cases to consider.

Case 1$x(t_1) = 0$. Let $\overline{f}=f'(0)=\lim _{x\rightarrow 0} \frac{f(x)}{x}$ and $\overline{y}=\max _{t \in [0, t_1]} y(t) < \frac{P_{f_{\mathrm{in}}}}{\theta }$. Then for every $t \in [0,t_1]$, we have

$$\begin{aligned} x'&=bx \min \left\{ 1-\frac{x}{K},1-\frac{q}{Q} \right\} -\min \left\{ f(x),\frac{\hat{f}\theta }{Q} \right\} y-\rho x\\&\ge -f(x)y-\rho x \ge -\bar{f}\bar{y}x -\rho x = -(\bar{f} \bar{y}+ \rho )x\equiv \alpha _1 x, \end{aligned}$$

where $\alpha _1$ is a constant. Then, $x(t)\ge x(0)e^{\alpha _1 t} > 0$ for $t \in [0, t_1]$, which implies that $x(t_1)\ge x(0)e^{\alpha _1 t_1} > 0$, a contradiction. This proves that $S(t_1)$ does not reach this boundary.

Case 2$y(t_1) = 0$. Then, for every $t \in [0, t_1]$, we have

$$\begin{aligned} y'&=\min \left\{ \hat{e}f(x),\frac{Q}{\theta }f(x),\hat{e}\hat{f}\frac{\theta }{Q}\right\} y-d y-\rho y,\\&\ge -d y-\rho y=-(d+\rho )y \equiv \alpha _2 y, \end{aligned}$$

where $\alpha _2$ is a constant. This implies that $y(t_1) \ge y(0)e^{\alpha _2 t_1} > 0$, a contradiction. This proves that $S(t_1)$ does not reach this boundary.

Case 3$Q(t_1)=q$. Then, for every $t \in [0, t_1]$, we have

$$\begin{aligned} Q'&=\upsilon (P_f,Q)-b\min \left\{ Q\left( 1-\frac{x}{K}\right) ,Q-q\right\} \\&\ge -b\min \left\{ Q\left( 1-\frac{x}{K}\right) ,Q-q\right\} \\&\ge -b(Q-q). \end{aligned}$$

This implies that $Q(t_1) \ge q + (Q(0) - q)e^{-bt_1} > q$, a contradiction. This proves that $S(t_1)$ cannot cross this boundary.

Case 4$Q(t_1) = \hat{Q}$. Since $\upsilon (P_f(t_1),Q(t_1)) = 0$

$$\begin{aligned} \frac{dQ}{dt}\Bigr |_{\begin{array}{c} t_1 \end{array}}=-b\min \Big \{ Q(t_1)\Big (1-\frac{x(t_1)}{K}\Big ),Q(t_1)-q\Big \} < 0 . \end{aligned}$$

Thus, $S(t_1)$ cannot cross this boundary.

Case 5$P_f(t_1)=0$. Since $\upsilon (P_f(t_1),Q(t_1)) = 0$ and $\rho (P_{f_{\mathrm{in}}}-P_f(t_1))=\rho P_{f_{\mathrm{in}}}$

$$\begin{aligned} \frac{\mathrm{d}P_f}{\mathrm{d}t}\bigg |_{\begin{array}{c} t_1 \end{array}}&=\rho P_{f_{\mathrm{in}}}+\theta d y(t_1)+ \min \left\{ f(x(t_1)),\frac{\hat{f}\theta }{Q(t_1)}\right\} \\&\quad y(t_1) \left( Q(t_1)-\min \left\{ \hat{e},\frac{Q(t_1)}{\theta }\right\} \theta \right) >0. \end{aligned}$$

Thus, $S(t_1)$ cannot cross this boundary.

Case 6$P_f(t_1)+Q(t_1)x(t_1)+\theta y(t_1)= P_{f_{\mathrm{in}}}$. Let $z(t)=P_{f_{\mathrm{in}}}-P_f(t)-Q(t)x(t)-\theta y(t)$, then $z(t_1)=0$ and $z(t)>0$ for $0 \le t < t_1$. Then for every $t \in [0,t_1]$, we have

$$\begin{aligned} z'&=-P_f'-Q'x-Q x'-\theta y'\\&=-\rho (P_{f_{\mathrm{in}}}-P_f -Qx-\theta y)\\&=-\rho z, \end{aligned}$$

where $\rho$ is a constant. This implies that $z(t_1)= e^{-\rho t_1} z(0) > 0$, a contradiction. This proves that $S(t_1)$ cannot cross this boundary. $\square$

Appendix B. Proof of Theorem 3.2

Proof

At $E_0=(0,0,Q_0,P_{f_{\mathrm{in}}} )$, the Jacobian matrix is

$$\begin{aligned} \begin{array}{lc} \ J(E_0)= &{} \left( \begin{array}{@{}cccc@{}} b\left( 1-\frac{q}{Q_0}\right) -\rho &{} 0 &{} 0 &{} 0 \\ 0 &{} -d-\rho &{} 0 &{} 0 \\ 0 &{} 0 &{} \frac{\mathrm{d} \upsilon }{\mathrm{d}Q}\bigg |_{\begin{array}{c} E_0 \end{array}}-b &{} \frac{\mathrm{d} \upsilon }{ \mathrm{d}P_f}\bigg |_{\begin{array}{c} E_0 \end{array}} \\ -\upsilon (Q,P_f)\bigg |_{\begin{array}{c} E_0 \end{array}}&{} \theta d &{} 0 &{} -\rho \end{array}\right) \\ \end{array} \end{aligned}$$

where $\frac{\mathrm{d} \upsilon }{ \mathrm{d}Q}\bigg |_{\begin{array}{c} E_0 \end{array}}-b=-\frac{\hat{c} P_{f_{\mathrm{in}}}}{ (\hat{a}+ P_{f_{\mathrm{in}}}) (\hat{Q} -q)}-b<0$. The eigenvalues of $J(E_0)$ are $b(1-\frac{q}{Q_0})-\rho$, $-d-\rho$, $\frac{\mathrm{d} \upsilon }{ \mathrm{d}Q}\bigg |_{\begin{array}{c} E_0 \end{array}}-b$, and $-\rho$.

If $b(1-\frac{q}{Q_0})>\rho$, then the distinct real eigenvalues are of opposite signs. So $E_0$ is unstable in the form of a saddle. $\square$

Appendix C. Proof of Theorem 3.3

Proof

Assume that $\min \left\{ \hat{e}f(x_1),\frac{Q_1}{\theta }f(x_1),\hat{e}\hat{f}\frac{\theta }{Q_1}\right\} < d+\rho$. To prove that $E_1$ is locally asymptotically stable, we consider two cases ( $1 - \frac{x}{K} < 1 -\frac{q}{Q}$ and $1 - \frac{x}{K} > 1 -\frac{q}{Q}$) where we look at the linearized system and use the Routh–Hurwitz criterion.

Case 1 $1 - \frac{x}{K} < 1 -\frac{q}{Q}$

Here, $E_1=(x_1,0,Q_1,P_{f_1})= \Big (\frac{K(b-\rho )}{b},0,\frac{\hat{c}P_{f_1}\hat{Q}}{(\hat{a}+P_{f_1})(\hat{Q}-q) \rho +\hat{c}P_{f_1}},P_{f_1}\Big )$ by Sect. 3.2 and the Jacobian takes the following form:

$$\begin{aligned} \begin{array}{lc} \ J(E_1)= &{} \left( \begin{array}{@{}cccc@{}} \rho -b &{}F_y(E_1) &{} 0 &{} 0 \\ 0 &{} G(E_1) &{} 0 &{} 0 \\ H_x(E_1)&{} 0 &{} \frac{\mathrm{d} \upsilon }{ \mathrm{d} Q}\bigg |_{\begin{array}{c} E_1 \end{array}}-\rho &{} \frac{\mathrm{d} \upsilon }{ \mathrm{d} P_f}\bigg |_{\begin{array}{c} E_1 \end{array}}\\ -\upsilon (P_f,Q)\bigg |_{\begin{array}{c} E_1 \end{array}}&I_y(E_1)&-\frac{K(b-\rho )}{b} \frac{\mathrm{d} \upsilon }{\mathrm{d} Q}\bigg |_{\begin{array}{c} E_1 \end{array}}&-\rho -\frac{K(b-\rho )}{b} \frac{\mathrm{d} \upsilon }{\mathrm{d} P_f}\bigg |_{\begin{array}{c} E_1 \end{array}} \end{array}\right) \\ \end{array} \end{aligned}$$

where $G(E_1)=\min \Bigg \{ \hat{e}f(x_1),\frac{Q_1}{\theta }f(x_1),\hat{e}\hat{f}\frac{\theta }{Q_1}\Bigg \} -d-\rho$.

Let $\rho <b$, $G(E_1)<0$, $\alpha _1=\frac{\mathrm{d} \upsilon }{ \mathrm{d}Q}\bigg |_{\begin{array}{c} E_1 \end{array}}=-\frac{\hat{c} P_f}{ (\hat{a}+ P_f) (\hat{Q} -q)}\bigg |_{\begin{array}{c} E_1 \end{array}}<0$, and $\alpha _2=\frac{\mathrm{d} \upsilon }{ \mathrm{d}P_f}\bigg |_{\begin{array}{c} E_1 \end{array}}=\frac{\hat{c}\hat{a}(\hat{Q}-Q)}{(\hat{a}+ P_f) ^{2} (\hat{Q}-q )}\bigg |_{\begin{array}{c} E_1 \end{array}}>0$. The eigenvalues of $J(E_1)$ are $\rho -b$, $G(E_1)$, $-\rho$, and $-\frac{K(b-\rho )}{b} \alpha _2+\alpha _1-\rho$, which are all negative. Thus, $E_1$ is locally asymptotically stable for this case.

Case 2 $1 - \frac{x}{K} > 1 -\frac{q}{Q}$

Here, $E_1=(x_1,0,Q_1,P_{f_1})= \Big (\frac{(b-\rho )\left( P_{f_{\mathrm{in}}}\left( (q-\hat{Q})(q \rho -\hat{c})b-\hat{Q}\hat{c}\rho \right) -(\hat{Q}-q)\hat{a}bq\rho \right) }{((q-\hat{Q})(q \rho -\hat{c})b-\hat{Q}\hat{c}\rho )qb},0,\frac{bq}{b-\rho },-\frac{(q-\hat{Q})\hat{a}\rho b q}{(q-\hat{Q})(q \rho -\hat{c})b-\hat{Q} \hat{c} \rho }\Big )$ by Sect. 3.2 and the Jacobian takes the following form:

$$\begin{aligned} \begin{array}{lc} \ J(E_1)= &{} \left( \begin{array}{@{}cccc@{}} 0&{} F_y(E_1) &{} F_Q(E_1)&{} 0 \\ 0 &{} G(E_1) &{} 0 &{} 0 \\ 0 &{} 0 &{} \frac{\mathrm{d} \upsilon }{\mathrm{d} Q}\bigg |_{\begin{array}{c} E_1 \end{array}}-b &{} \frac{\mathrm{d} \upsilon }{\mathrm{d} P_f}\bigg |_{\begin{array}{c} E_1 \end{array}}\\ -\upsilon (P_f,Q)\bigg |_{\begin{array}{c} E_1 \end{array}} &{} I_y(E_1) &{}I_Q(E_1) &{} I_{P_f}(E_1) \end{array}\right) \\ \end{array} \end{aligned}$$

where

$$\begin{aligned} F_Q(E_1)=&\frac{bqx_1}{Q_1^2},\\ G(E_1)=&\min \left\{ \hat{e}f(x_1),\frac{Q_1}{\theta }f(x_1),\hat{e}\hat{f}\frac{\theta }{Q_1}\right\} -d-\rho ,\\ I_Q(E_1)=&-x_1\frac{\mathrm{d} \upsilon }{ \mathrm{d} Q}\bigg |_{\begin{array}{c} E_1 \end{array}}=x_1\frac{\hat{c} P_{f_1}}{ (\hat{a}+ P_{f_1}) (\hat{Q} -q)} , \end{aligned}$$

and

$$\begin{aligned} I_{P_f}(E_1)=-\rho -x_1\frac{\mathrm{d} \upsilon }{\mathrm{d} P_f}\bigg |_{\begin{array}{c} E_1 \end{array}}=-\rho -x_1\frac{\hat{c}\hat{a}(\hat{Q}-Q_1)}{(\hat{a}+ P_{f_1} ) ^{2} (\hat{Q}-q )}. \end{aligned}$$

Let $\alpha =\frac{\hat{c}}{(\hat{a}+P_{f_1})(\hat{Q}-q)}>0$. Then, the Jacobian simplifies down to

$$\begin{aligned} \begin{array}{lc} \ J(E_1)= &{} \left( \begin{array}{@{}cccc@{}} 0&{} F_y(E_1) &{} F_Q(E_1)&{} 0 \\ 0 &{} G(E_1) &{} 0 &{} 0 \\ 0 &{} 0 &{} -\alpha P_{f_1} -b &{} \frac{(\hat{Q}-Q_1)\hat{a}\alpha }{\hat{a}+P_{f_1}} \\ -\alpha P_{f_1} (\hat{Q}-Q_1) &{} I_y (E_1) &{}x_1 P_{f_1} \alpha &{} -\rho -x_1\frac{(\hat{Q}-Q_1)\hat{a}\alpha }{\hat{a}+P_{f_1}} \end{array}\right) \\ \end{array} \end{aligned}$$

One of the eigenvalues of $J(E_1)$ is $G(E_1)=\min \left\{ \hat{e}f(x_1),\frac{Q_1}{\theta }f(x_1),\hat{e}\hat{f}\frac{\theta }{Q_1}\right\} -d-\rho$ and the remaining three eigenvalues are given by the eigenvalues of the matrix $J'(E_1)$ written as:

$$\begin{aligned} \begin{array}{lc} \ J'(E_1)= &{} \left( \begin{array}{@{}cccc@{}} 0&{} F_Q(E_1)&{} 0 \\ 0 &{} -\alpha P_{f_1} -b &{} \frac{(\hat{Q}-Q_1)\hat{a}\alpha }{\hat{a}+P_{f_1}} \\ -\alpha P_{f_1} (\hat{Q}-Q_1) &{}x_1 P_{f_1} \alpha &{} -\rho -x_1\frac{(\hat{Q}-Q_1)\hat{a}\alpha }{\hat{a}+P_{f_1}} \end{array}\right) \\ \end{array} \end{aligned}$$

The characteristic equation of the above matrix is given by :

$$\begin{aligned} \lambda ^3+A_1 \lambda ^2+A_2 \lambda +A_3= 0, \end{aligned}$$

where

$$\begin{aligned} A_1&=-tr(J'(E_1))=\frac{-Q_1 \hat{a}\alpha x_1+\hat{Q}\hat{a}\alpha x_1+P_{f_1}^2 \alpha +P_{f_1}\hat{a}\alpha +P_{f_1}b +P_{f_1}\rho +\hat{a}b+\hat{a}\rho }{\hat{a}+P_{f_1}},\\ A_2&=\frac{-Q_1\hat{a}\alpha bx_1+\hat{Q}\hat{a}\alpha b x_1+P_{f_1}^2\alpha \rho +P_{f_1}\hat{a}\alpha \rho +P_{f_1} b\rho +\hat{a}b\rho }{\hat{a}+P_{f_1}},\\ A_3&=-\det (J'(E_1))=\frac{P_{f_1}Q_1^2\hat{a}\alpha ^2 b q x_1 -2P_{f_1}Q_1\hat{Q}\hat{a}\alpha ^2bqx_1 +P\hat{Q}^2\hat{a}\alpha ^2bqx_1}{Q_1^2(\hat{a}+P_{f_1})}. \end{aligned}$$

Now by Routh–Hurwitz criterion, $E_1$ is locally asymptotically stable for this case whenever the following conditions are satisfied:

$$\begin{aligned} {\left\{ \begin{array}{ll} A_1A_2>A_3, \\ A_i>0, \quad i=1,2,3 \quad \hbox {and}\\ \min \left\{ \hat{e}f(x_1),\frac{Q_1}{\theta }f(x_1),\hat{e}\hat{f}\frac{\theta }{Q_1}\right\} -d-\rho <0. \end{array}\right. } \end{aligned}$$

(C.1)

$\square$