Incorporating carbon dioxide into a stoichiometric producer–grazer model

Abstract

Climate change models predict increases in atmospheric carbon dioxide concentration. As ecosystems equilibrate with the atmosphere, stimulation of photosynthesis is expected to occur. However, growth limitation due to soil nutrients may potentially limit sequestration of carbon. Additionally, changes in producer nutritional quality may cause a decline in grazer populations. Here we extend the WKL model to allow for consideration of the impacts of elevated atmospheric carbon dioxide concentration on producer–grazer dynamics. We do so by explicitly tracking the free carbon in the medium and allowing the producer’s growth rate to be limited by available carbon instead of light. This model is analyzed using primarily local bifurcation analysis. Overall, these analyses show that carbon sequestration due to increased atmospheric carbon dioxide can be limited by insufficient available phosphorus. Furthermore, increased atmospheric carbon dioxide will cause decreased stoichiometric quality of producers where available phosphorus is limiting.

Appendices

Appendix A

This appendix contains the proof of Theorem 1.

Proof

Let $X(t) \equiv (x(t),y(t),p(t))$ be a solution of Eqs. (6)–(8) with initial conditions in $\Omega$. Then, $0< x(0)< \text {min}\{T_P/q, h(T_C)\}, 0< y(0), 0< p(0), p(0) + \theta y(0) < T_P,$ and $x(0) + y(0) < T_C$. Assume for the sake of contradiction that there is time $t_1 > 0$ such that X(t) touches or crosses the boundary of ${\bar{\Omega }}$ (closure of $\Omega$) for the first time. Therefore, $(x(t),y(t),p(t)) \in \Omega$ for $0 \le t < t_1$. We now consider cases for which part of the boundary of ${\bar{\Omega }}$ $X(t_1)$ lies on.

Case 1 $x(t_1) = 0$ but $p(t_1) \ne 0$. Since $(x(t),y(t),p(t)) \in \Omega$ for $0 \le t < t_1$, then $0 < p(t)$ and $p(t) + \theta y(t) \le T_P$ for $0 \le t \le t_1$. Also, all parameters are assumed to be positive. Therefore, $y(t) \le T_P/\theta - p(t)/\theta < T_P/\theta$ for $0 \le t \le t_1$. Let $p_1 = \text {min}\{p(t): t \in [0,t_1]\} > 0$, $x_1 = \text {max}\{x(t): t \in [0,t_1] \} > 0$, and $y_1 = \text {max}\{y(t): t \in [0,t_1] \} > 0$. Both $x_1$ and $y_1$ are guaranteed to exist since the initial conditions are in $\Omega$ and thus are positive. Then for $0 \le t \le t_1$, we have (using Eq. 6)

$$\begin{aligned} \frac{\text {d}x}{\text {d}t}&= rx\left( 1-\frac{x}{\text {min}\{p/q, h(T_C-x-y)\}}\right) -f(x)y-l_xx \\&\ge rx\left( 1-\frac{\text {min}\{T_P/q, h(T_C)\}}{\text {min}\{p_1/q, h(T_C-x_1-y_1)\}}\right) - f^\prime (0)(T/\theta )x - l_xx\\&=\left[ r\left( 1-\frac{\text {min}\{T_P/q, h(T_C)\}}{\text {min}\{p_1/q, h(T_C-x_1-y_1)\}}\right) - f^\prime (0)(T/\theta ) - l_x\right] x \equiv \mu x, \end{aligned}$$

where $\mu$ is a constant. Then, $x(t) \ge x(0)e^{\mu t}$ for $0 \le t \le t_1$, which implies $x(t_1) \ge x(0)e^{\mu t_1} > 0$, which is a contradiction.

Case 2 $x(t_1) = \text {min}\{T_P/q, h(T_C)\}$. Since $(x(t),y(t),p(t)) \in \Omega$ for $0 \le t < t_1$, we know $0 \le y(t)$ and $p(t) + \theta y(t) \le T_P$ for $0 \le t \le t_1$. Therefore, $p(t) \le T_P - \theta y(t) \le T_P$ for $0 \le t \le t_1$. Clearly $T_C - x(t) - y(t) \le T_C$ for $x(t), y(t) \ge 0$. Since h(C) is non-decreasing, then $h(T_C - x(t) - y(t)) \le h(T_C)$ for $x(t), y(t) \ge 0$. Hence, for $0 \le t \le t_1$, we have

$$\begin{aligned} \frac{\text {d}x}{\text {d}t}&= rx\left( 1-\frac{x}{\text {min}\{p/q, h(T_C-x-y)\}}\right) - f(x)y - l_xx \\&\le rx\left( 1-\frac{x}{\text {min}\{T_P/q, h(T_C)\}}\right) . \end{aligned}$$

Note that the right hand side is logistic growth in x with the carrying capacity given by $\text {min}\{T_P/q, h(T_C)\}$. The standard comparison argument yields that $x(t) < \text {min}\{T_P/q, h(T_C)\}$ for all $0 \le t \le t_1$, a contradiction.

Case 3 $y(t_1) = 0$. For $0 \le t \le t_1$, we have from Eq. (7)

$$\begin{aligned} \frac{\text {d}y}{\text {d}t}&= {\hat{e}} \text {min} \left\{ 1, \frac{p/x}{\theta }\right\} f(x)y - {\hat{d}}y - l_yy \\&= \left( {\hat{e}} \text {min} \left\{ 1, \frac{p/x}{\theta }\right\} f(x) - {\hat{d}} - l_y\right) y \ge -({\hat{d}}+l_y)y \end{aligned}$$

since all parameters are assumed to be positive; $f(0) = 0, f'(x) > 0, f''(x) \le 0$ for $x \ge 0$; and $(x(t),y(t),p(t)) \in \Omega$ for $0 \le t < t_1$. Hence, $y(t) \ge y(0)e^{-({\hat{d}}+l_y)t} > 0$ for $0 \le t \le t_1$, a contradiction.

Case 4 $p(t_1) = 0$. Since $(x(t),y(t),p(t)) \in \Omega$ for $0 \le t < t_1$, we know $p(t) + \theta y(t) \le T_P$ for $0 \le t \le t_1$. Therefore, $T_P-p(t)-\theta y(t) \ge 0$ for $0 \le t \le t_1$, and since we assume that in general $g(0) = 0$ and $g'(P) > 0$ for $P \ge 0$, then $g(T_P - p(t) - \theta y(t)) \ge 0$ for $0 \le t \le t_1$.

Also, since $0 \le p(t)$ and $p(t) + \theta y(t) \le T_P$ for $0 \le t \le t_1$, then $y(t) \le T_P/\theta - p(t)/\theta \le T_P/\theta$ for $0 \le t \le t_1$, and therefore $-T_P/\theta \le -y(t)$.

Thus, for $0 \le t \le t_1$, we have from Eq. (8)

$$\begin{aligned} \frac{\text {d}p}{\text {d}t}&= g(T_P - p - \theta y)x - \frac{p}{x}f(x)y - dp \ge - \frac{p}{x}f(x)y - dp \\&\ge [-f^\prime (0)(T_P/\theta ) - d]p \equiv \nu p, \end{aligned}$$

where $\nu$ is a constant. Thus, $p(t) \ge p(0) e^{\nu t} > 0$ for $0 \le t \le t_1$, a contradiction.

Case 5 $p(t_1) + \theta y(t_1) = T_P$. Let $z(t) = T_P - p(t) - \theta y(t)$. Since $p(t_1) + \theta y(t_1) = T_P$, then $z(t_1) = 0$. Also, since $t_1$ is assumed to be the first time that X(t) touches or crosses the boundary of ${\bar{\Omega }}$, then $p(t) + \theta y(t) < T_P$ for $0 \le t < t_1$ and thus $z(t) > 0$ for $0 \le t < t_1$. Then for $0 \le t \le t_1$, we have

$$\begin{aligned} \frac{\text {d}z}{\text {d}t}&=\frac{\text {d}}{\text {d}t}(T_P - p - \theta y) = \frac{\text {d}}{\text {d}t}T_P -\frac{\text {d}p}{\text {d}t} - \theta \frac{\text {d}y}{\text {d}t} =-\frac{\text {d}p}{\text {d}t} - \theta \frac{\text {d}y}{\text {d}t} \\&= -\left( g(T_P - p - \theta y)x - \frac{p}{x}f(x)y - dp \right) - \theta \left( {\hat{e}} \text {min} \left\{ 1, \frac{p/x}{\theta }\right\} f(x)y - {\hat{d}}y - l_yy \right) \\&= -g(T_P-p-\theta y)x + \frac{p}{x}f(x)y + dp - \theta {\hat{e}} \text {min} \left\{ 1, \frac{p/x}{\theta }\right\} f(x)y + \theta {\hat{d}} y + \theta l_y y \\&\ge -g(T_P-p-\theta y)x + dp + \theta {\hat{d}} y + \theta l_y y \ge -g(z)x + dp + ({\hat{d}} + l_y)\theta y\\&\ge -g^\prime (0)z\text {min}\{T_P/q, h(T_C)\} + \text {min}\{d,{\hat{d}} + l_y\}(T_P-z) \\&=\text {min}\{d,{\hat{d}} + l_y\}T_P - [g^\prime (0)\text {min}\{T_P/q, h(T_C)\} + \text {min}\{d, {\hat{d}} + l_y\}]z \equiv {\tilde{\mu }} - {\tilde{v}}z, \end{aligned}$$

where ${\tilde{u}} > 0$ and ${\tilde{v}} > 0$ are constant. Thus $z(t) \ge e^{-{\tilde{v}}t} z(0) > 0$ for $0 \le t \le t_1$, a contradiction.

Case 6 $x(t_1) + y(t_1) = T_C$. Consider the limit of $\text {d}x/\text {d}t$ as $t \rightarrow t_1$:

$$\begin{aligned} \lim _{t \rightarrow t_1} \frac{\text {d}x}{\text {d}t}&= \lim _{t \rightarrow t_1}\left[ rx\left( 1-\frac{x}{\text {min} \{p/q, h(T_C-x-y) \}} \right) - f(x)y - l_xx\right] \\&= \lim _{t \rightarrow t_1} rx\left( 1-\frac{x}{\text {min} \{p/q, h(T_C-x-y) \}} \right) - \lim _{t \rightarrow t_1} f(x)y - \lim _{t \rightarrow t_1} l_xx \\&= rx(t_1) - \lim _{t \rightarrow t_1}\frac{rx^2}{\text {min}\{p/q, h(T_C-x-y) \}} - f(x(t_1))y(t_1) - l_xx(t_1). \end{aligned}$$

Since $h(0) = 0$, then the limit as $t \rightarrow t_1$ of $\text {min}\{p/q, h(T_C-x-y) \}$ is 0. However, $x(t_1) > 0$. Therefore, as $t \rightarrow t_1$, the limit of $\text {d}x/\text {d}t$ is $-\infty$. Clearly this implies that $\text {d}y/\text {d}t \rightarrow -\infty$ as $t \rightarrow t_1$. But then as $t \rightarrow t_1$, $x(t) + y(t) \rightarrow -\infty$, a contradiction. $\square$

Appendix B

This appendix contains all the necessary information to find the equilibria assuming Holling type I functional responses for f and g. In addition, we assume $\alpha = 0$ and $\rho (C) = 1$.

Substituting $f(x) = cx$, $g(P) = {\hat{c}}P$ and $h(C) = \gamma C$ into Eqs. (6)–(8), we get the following system:

$$\begin{aligned} \frac{\text {d}x}{\text {d}t}&= rx\left( 1-\frac{x}{\text {min} \{p/q, \gamma (T_C-x-y) \}} \right) - cxy - l_xx, \\ \frac{\text {d}y}{\text {d}t}&= {\hat{e}} \text {min} \left\{ 1, \frac{p/x}{\theta }\right\} cxy - {\hat{d}}y - l_yy, \\ \frac{\text {d}p}{\text {d}t}&= {\hat{c}}(T_P- p - \theta y)x - \frac{p}{x}cxy - dp. \end{aligned}$$

Equilibria satisfy

$$\begin{aligned} 0&= {\bar{x}}\left( r\left( 1-\frac{{\bar{x}}}{\text {min} \{{\bar{p}}/q, \gamma (T_C-{\bar{x}}-{\bar{y}}) \}} \right) - c{\bar{y}} - l_x\right) \equiv {\bar{x}}F({\bar{x}},{\bar{y}},{\bar{p}}), \\ 0&= {\bar{y}}\left( {\hat{e}} \text {min} \left\{ 1, \frac{{\bar{p}}/{\bar{x}}}{\theta }\right\} c{\bar{x}} - {\hat{d}} - l_y\right) \equiv {\bar{y}}G({\bar{x}},{\bar{y}},{\bar{p}}), \\ 0&= {\hat{c}}(T_P- {\bar{p}} - \theta {\bar{y}}){\bar{x}} - c{\bar{p}}{\bar{y}} - d{\bar{p}} \equiv H({\bar{x}},{\bar{y}},{\bar{p}}). \end{aligned}$$

Clearly the trivial extinction equilibrium $E_0 = (0,0,0)$ is a possible solution of the above. We can also explicitly find the forms of the grazer extinction equilibria. The grazer extinction equilibria always take the form $({\bar{x}},0,{\bar{p}})$, where ${\bar{x}}$ and ${\bar{p}}$ are decided by what is limiting the producer and other parameter-based conditions.

When the producer is phosphorus limited at equilibrium (${\bar{p}}/q \le \gamma (T_C-{\bar{x}})$), a grazer extinction equilibrium is given by $({\bar{x}}, 0,{\bar{p}})$ where

$$\begin{aligned} {\bar{x}}&= \frac{{\hat{c}}T_P(r-l_x)-dqr}{{\hat{c}}qr}, \\ {\bar{p}}&= \frac{{\hat{c}}T_P(r-l_x)-dqr}{{\hat{c}}(r-l_x)}. \end{aligned}$$

We assume all parameters are positive. For ${\bar{x}} \ge 0$, we need ${\hat{c}}T_P(r-l_x) - dqr \ge 0$. Then, for ${\bar{p}} \ge 0$, we need $r - l_x >0$.

When the producer is carbon limited at equilibrium ($\gamma (T_C-{\bar{x}}) \le {\bar{p}}/q$), we have $({\bar{x}}, 0,{\bar{p}})$ where

$$\begin{aligned} {\bar{x}}&= \frac{\gamma T_C(r-l_x)}{r+\gamma r - \gamma l_x}, \\ {\bar{p}}&= \frac{{\hat{c}} \gamma T_CT_P(r-l_x)}{{\hat{c}}\gamma T_C(r-l_x) + d(r + \gamma r - \gamma l_x)}. \end{aligned}$$

For this equilibrium to be biologically feasible and not equal to $E_0$, we require either $r - l_x > 0$ and $r + \gamma r - \gamma l_x > 0$, or $r - l_x < 0$ and $r + \gamma r - \gamma l_x < 0$.

To determine parameter-based conditions for which grazer extinction equilibria are present, we substitute the equilibria into ${\bar{p}}/q = \gamma (T_C - {\bar{x}})$. With some rearrangement, we find that for both the phosphorus and carbon limited equilibria, ${\bar{p}}/q = \gamma (T_C - {\bar{x}})$ when

$$\begin{aligned} {\hat{c}} \gamma T_P(r-l_x)^2 + ({\hat{c}}T_P - {\hat{c}}q \gamma T_C - dq\gamma )r(r-l_x) - dqr^2 = 0. \end{aligned}$$

A grazer extinction equilibrium takes the phosphorus limited form when ${\bar{p}}/q \le \gamma (T_C - {\bar{x}})$, or equivalently,

$$\begin{aligned} \frac{{\hat{c}} \gamma T_P(r-l_x)^2 + ({\hat{c}}T_P - {\hat{c}}q\gamma T_C - dq\gamma )r(r-l_x) - dqr^2}{{\hat{c}}qr(r-l_x)} \le 0. \end{aligned}$$

A grazer extinction equilibrium takes the carbon limited form when ${\bar{p}}/q \ge \gamma (T_C - {\bar{x}})$, that is,

$$\begin{aligned} \frac{{\hat{c}}\gamma T_P(r - l_x)^2 +({\hat{c}}T_P - {\hat{c}}q\gamma T_C - dq\gamma )r(r-l_x) - dqr^2}{({\hat{c}}q\gamma T_C(r-l_x) + dq(r + \gamma r - \gamma l_x))(r+\gamma r - \gamma l_x)} \ge 0. \end{aligned}$$

There may also be coexistence equilibria, which would satisfy

$$\begin{aligned} 0&= r\left( 1-\frac{{\bar{x}}}{\text {min} \{{\bar{p}}/q, \gamma (T_C-{\bar{x}}-{\bar{y}}) \}} \right) - c{\bar{y}} - l_x = F({\bar{x}},{\bar{y}},{\bar{p}}), \\ 0&= {\hat{e}} \text {min} \left\{ 1, \frac{{\bar{p}}/{\bar{x}}}{\theta }\right\} c{\bar{x}} - {\hat{d}} - l_y = G({\bar{x}},{\bar{y}},{\bar{p}}), \\ 0&= {\hat{c}}(T_P- {\bar{p}} - \theta {\bar{y}}){\bar{x}} - c{\bar{p}}{\bar{y}} - d{\bar{p}} = H({\bar{x}},{\bar{y}},{\bar{p}}). \end{aligned}$$

Appendix C

This appendix contains all the necessary information to find the equilibria assuming Holling type II functional responses for f and g. In addition, we assume $\alpha = 0$ and $\rho (C) = 1$.

Using $f(x) = cx/(a+x)$, $g(P) = {\hat{c}}P/({\hat{a}} + P)$, and $h(C) = \gamma C$,

$$\begin{aligned} \frac{\text {d}x}{\text {d}t}&= rx\left( 1-\frac{x}{\text {min} \{p/q, \gamma (T_C-x-y) \}} \right) - \frac{cx}{a+x}y - l_xx, \\ \frac{\text {d}y}{\text {d}t}&= {\hat{e}} \text {min} \left\{ 1, \frac{p/x}{\theta }\right\} \frac{cx}{a+x}y - {\hat{d}}y - l_yy, \\ \frac{\text {d}p}{\text {d}t}&= \frac{{\hat{c}}(T_P - p - \theta y)}{{\hat{a}} + T_P - p - \theta y}x - \frac{p}{x}\frac{cx}{a+x}y - dp. \end{aligned}$$

Equilibria satisfy

$$\begin{aligned} 0&= {\bar{x}}\left( r\left( 1-\frac{{\bar{x}}}{\text {min} \{{\bar{p}}/q, \gamma (T_C-{\bar{x}}-{\bar{y}}) \}} \right) - \frac{c}{a+{\bar{x}}}{\bar{y}} - l_x\right) \equiv {\bar{x}}F({\bar{x}},{\bar{y}},{\bar{p}}), \\ 0&= {\bar{y}}\left( {\hat{e}} \text {min} \left\{ 1, \frac{{\bar{p}}/{\bar{x}}}{\theta }\right\} \frac{c{\bar{x}}}{a+{\bar{x}}}- {\hat{d}} - l_y\right) \equiv {\bar{y}}G({\bar{x}},{\bar{y}},{\bar{p}}), \\ 0&= \frac{{\hat{c}}(T_P - {\bar{p}} - \theta {\bar{y}})}{{\hat{a}} + T_P - {\bar{p}} - \theta {\bar{y}}}{\bar{x}} -\frac{c{\bar{p}}}{a+{\bar{x}}}{\bar{y}} - d{\bar{p}} \equiv H({\bar{x}},{\bar{y}}, {\bar{p}}). \end{aligned}$$

Clearly the trivial extinction equilibrium $E_0 = (0,0,0)$ is a possible solution of this system. We can explicitly find the forms of the grazer extinction equilibria, dependent upon what is limiting the producer and other parameter-based conditions.

When the producer is phosphorus limited at equilibrium (${\bar{p}}/q \le \gamma (T_C-{\bar{x}}$)), the grazer extinction equilibrium is given by $({\bar{x}},0,{\bar{p}})$ where

$$\begin{aligned} {\bar{x}}&= \frac{{\bar{p}}}{q}\left( 1-\frac{l_x}{r}\right) = \frac{{\hat{c}}T_P(r-l_x)^2 - dqr({\hat{a}} + T_P)(r-l_x)}{{\hat{c}}qr(r-l_x) - dq^2r^2}, \\ {\bar{y}}&= 0, \\ {\bar{p}}&= \frac{{\hat{c}}T_P(r-l_x) - dqr({\hat{a}} + T_P)}{{\hat{c}}(r-l_x) - dqr}. \end{aligned}$$

Consider the equation for ${\bar{x}}$ that includes ${\bar{p}}$. Since we require ${\bar{x}}, {\bar{y}}, {\bar{p}} \ge 0$ for the equilibrium to be biologically feasible, then we require $r - l_x \ge 0$, since otherwise ${\bar{x}}$ is negative for positive ${\bar{p}}$. The equation for ${\bar{p}}$ would also yield additional conditions for non-negativity, which are not included here.

When the producer is carbon limited at equilibrium ($\gamma (T_C-{\bar{x}}) \le {\bar{p}}/q$), we have multiple possible grazer extinction equilibria, given by $({\bar{x}}, 0, {\bar{p}})$ where

$$\begin{aligned} {\bar{x}}&= \frac{\gamma T_C(r-l_x)}{r + \gamma r - \gamma l_x}, \\ {\bar{y}}&= 0, \\ {\bar{p}}&= \frac{{\hat{c}}\gamma T_C(r-l_x)+d({\hat{a}}+T_P)(r+\gamma r - \gamma l_x)}{2d(r + \gamma r - \gamma l_x)} \\&\quad \pm \, \frac{\sqrt{({\hat{c}} \gamma T_C(r-l_x)+d({\hat{a}}+T_P)(r + \gamma r- \gamma l_x))^2-4{\hat{c}}d\gamma T_CT_P(r-l_x)(r+\gamma r - \gamma l_x)}}{2d(r + \gamma r- \gamma l_x)}. \end{aligned}$$

For ${\bar{x}} \ge 0$, we need either $r - l_x \ge 0$ and $r + \gamma r - \gamma l_x > 0$, or $r - l_x \le 0$ and $r + \gamma r - \gamma l_x < 0$. Note that $r - l_x > 0$ and $r + \gamma r - \gamma l_x < 0$ is not possible for positive parameters. Thus, the equilibrium is not biologically feasible if $r - l_x < 0$ and $r + \gamma r - \gamma l_x > 0$. As in the phosphorus limited case, the equation for ${\bar{p}}$ would also yield conditions for feasibility of the equilibria.

To determine conditions for which grazer equilibria are present dependent on parameter values, we substitute the grazer extinction equilibria found above into ${\bar{p}}/q = \gamma (T_C-{\bar{x}})$. Using the values of ${\bar{x}}$ and ${\bar{p}}$ for the phosphorus limited equilibrium, we require

$$\begin{aligned}&{\hat{c}}rT_P(r-l_x) - dqr^2({\hat{a}} + T_P) - \gamma T_C({\hat{c}}qr(r-l_x) - dq^2r^2) \\&\quad +\, {\hat{c}}\gamma T_P (r-l_x)^2 - dqr\gamma ({\hat{a}} + T_P)(r-l_x) = 0, \end{aligned}$$

for ${\bar{p}}/q = \gamma (T_C - {\bar{x}})$, and for phosphorus limited producer growth at equilibrium (${\bar{p}}/q \le \gamma (T_C-{\bar{x}}))$, we require

$$\begin{aligned} \frac{{\hat{c}}rT_P(r-l_x) - dqr^2({\hat{a}} + T_P) - \gamma T_C({\hat{c}}qr(r-l_x) - dq^2r^2) + {\hat{c}}\gamma T_P (r-l_x)^2 - dqr\gamma ({\hat{a}} + T_P)(r-l_x)}{{\hat{c}}qr(r-l_x) - dq^2r^2}&\le 0. \end{aligned}$$

Using the values of ${\bar{x}}$ and ${\bar{p}}$ for the carbon limited equilibrium, we require

$$\begin{aligned} 0&= {\hat{c}}\gamma T_C(r-l_x)+d({\hat{a}}+T_P)(r+\gamma r - \gamma l_x) - 2dqr\gamma T_C \\&\quad \pm \sqrt{({\hat{c}} \gamma T_C(r-l_x)+d({\hat{a}}+T_P)(r + \gamma r- \gamma l_x))^2-4{\hat{c}}d\gamma T_CT_P(r-l_x)(r+\gamma r - \gamma l_x)} \end{aligned}$$

for ${\bar{p}}/q = \gamma (T_C - {\bar{x}})$, and for carbon limited producer growth at equilibrium ($\gamma (T_C - {\bar{x}}) \le {\bar{p}}/q$), we require

$$\begin{aligned} 0&\le \frac{{\hat{c}}\gamma T_C(r-l_x)+d({\hat{a}}+T_P)(r+\gamma r - \gamma l_x)- 2dqr\gamma T_C}{2d(r + \gamma r - \gamma l_x)} \\&\quad \pm \, \frac{\sqrt{({\hat{c}} \gamma T_C(r-l_x)+d({\hat{a}}+T_P)(r + \gamma r- \gamma l_x))^2-4{\hat{c}}d\gamma T_CT_P(r-l_x)(r+\gamma r - \gamma l_x)}}{2d(r + \gamma r- \gamma l_x)}. \\ \end{aligned}$$

There may also be coexistence equilibria, which would satisfy

$$\begin{aligned} 0&= r\left( 1-\frac{{\bar{x}}}{\text {min} \{{\bar{p}}/q, \gamma (T_C-{\bar{x}}-{\bar{y}}) \}} \right) - \frac{c}{a+{\bar{x}}}{\bar{y}} - l_x = F({\bar{x}},{\bar{y}},{\bar{p}}), \\ 0&= {\hat{e}} \text {min} \left\{ 1, \frac{{\bar{p}}/{\bar{x}}}{\theta }\right\} \frac{c{\bar{x}}}{a+{\bar{x}}}- {\hat{d}} - l_y = G({\bar{x}},{\bar{y}},{\bar{p}}), \\ 0&= \frac{{\hat{c}}(T_P - {\bar{p}} - \theta {\bar{y}})}{{\hat{a}} + T_P - {\bar{p}} - \theta {\bar{y}}}{\bar{x}} -\frac{c{\bar{p}}}{a+{\bar{x}}}{\bar{y}} - d{\bar{p}} = H({\bar{x}},{\bar{y}}, {\bar{p}}). \end{aligned}$$

Appendix D

This appendix contains the proof of Theorem 4.

Proof

Let $u = x/p$, then

$$\begin{aligned} \frac{\text {d}u}{\text {d}t}&= \frac{\text {d}}{\text {d}t}\frac{x}{p} = \frac{(\text {d}x/\text {d}t)p - x(\text {d}p/\text {d}t)}{p^2} = \frac{\text {d}x/\text {d}t}{p} - x\frac{\text {d}p/\text {d}t}{p^2} \\&= \frac{rx}{p}\left( 1-\frac{x}{\text {min} \{p/q, h(T_C-x-y) \}} \right) - \frac{f(x)y}{p} - \frac{l_xx}{p} \\&\quad - \, \frac{x^2}{p^2}g(T_P - p - \theta y) + \frac{f(x)y}{p} +d\frac{x}{p} \\&= \frac{rx}{p}\left( 1-\frac{x}{\text {min} \{p/q, h(T_C-x-y) \}} \right) - \frac{l_xx}{p} - \frac{x^2}{p^2}g(T_P - p - \theta y) +d\frac{x}{p} \\&= ru\left( 1-\frac{x}{\text {min} \{p/q, h(T_C-x-y) \}} \right) - l_xu - u^2g(T_P - p - \theta y) +du. \\ \end{aligned}$$

Since $g(0) = 0$ and $g^\prime (P) > 0$ for $P \ge 0$, then $- u^2g(T_P-p-\theta y) \le 0$. Note that $\text {min} \{p/q, h(T_C-x-y) \} \le p/q$ if and only if

$$\begin{aligned} - \frac{1}{\text {min} \{p/q, h(T_C-x-y) \}} \le -\frac{1}{p/q}. \end{aligned}$$

Thus

$$\begin{aligned} \frac{\text {d}u}{\text {d}t}&\le ru\left( 1-\frac{x}{p/q} \right) - l_xu +du=ru(1-qu) - l_xu +du = ru(1-qu) + (d-l_x)u\\&\le ru(1-qu) + ru\frac{d-l_x}{r} = ru\left( 1 - qu + \frac{d-l_x}{r} \right) . \end{aligned}$$

Hence, $u \le \text {min}\{x(0)/p(0), [1 + (d-l_x)/r]/q \} \equiv m$. Consider the equation for $(\text {d}p/\text {d}t)$:

$$\begin{aligned} \frac{\text {d}p}{\text {d}t}= & {} g(T_P - p - \theta y)x - \frac{p}{x}f(x)y - dp \le g(T_P)x - dp \le g(T_P)mp - dp \\= & {} (g(T_P)m - d)p. \end{aligned}$$

Since $d > mg(T_P)$ implies $g(T_P)m - d < 0$, then $\text {d}p/\text {d}t < 0$ and thus $p \rightarrow 0$ as $t \rightarrow \infty$.

Now, consider the equation for $(\text {d}x/\text {d}t)$:

$$\begin{aligned} \frac{\text {d}x}{\text {d}t}&= rx\left( 1-\frac{x}{\text {min} \{p/q, h(T_C-x-y) \}} \right) - f(x)y - l_xx \\&\le rx(1-(qx/p)). \end{aligned}$$

Hence $\text {lim sup}_{t \rightarrow \infty } x(t) \le p/q$ and $x \rightarrow 0$ as $t \rightarrow \infty$. Given the dependence of y on x, this implies that $y \rightarrow 0$ as $t \rightarrow \infty$. Therefore, the extinction steady state $E_0 = (0,0,0)$ is globally asymptotically stable if $d > mg(T_P)$, where $m = \text {min}\{x(0)/p(0),[1+(d-l_x)/r]/q\}$. $\square$

Appendix E

This appendix contains all the necessary calculations for the stability results in Sect. 3.3 using Holling type I functional responses for f and g.

For $f(x) = cx$, $g(P) = {\hat{c}}P$, and $h(C) = \gamma C$, using F, G, H as defined in “Appendix B”,

$$\begin{aligned} \frac{\partial F}{\partial x}&= {\left\{ \begin{array}{ll} -\frac{r}{p/q}, &{} p/q \le \gamma (T_C-x-y), \\ -\frac{r(T_C-y)}{\gamma (T_C-x-y)^2}, &{} p/q> \gamma (T_C-x-y), \end{array}\right. } \\ \frac{\partial F}{\partial y}&= {\left\{ \begin{array}{ll} -c, &{} p/q \le \gamma (T_C-x-y), \\ -\frac{rx}{\gamma (T_C-x-y)^2}-c, &{} p/q> \gamma (T_C-x-y), \end{array}\right. } \\ \frac{\partial F}{\partial p}&= {\left\{ \begin{array}{ll} \frac{rx}{p^2/q}, &{} p/q \le \gamma (T_C-x-y), \\ 0, &{} p/q> \gamma (T_C-x-y), \end{array}\right. } \\ \frac{\partial G}{\partial x}&= {\left\{ \begin{array}{ll} 0, &{} p/x \le \theta , \\ c{\hat{e}}, &{} p/x> \theta , \end{array}\right. }&\frac{\partial H}{\partial x}&= {\hat{c}}(T_P-p-\theta y), \\ \frac{\partial G}{\partial y}&= 0,&\frac{\partial H}{\partial y}&= -{\hat{c}}\theta x - cp, \\ \frac{\partial G}{\partial p}&= {\left\{ \begin{array}{ll} \frac{c{\hat{e}}}{\theta }, &{} p/x \le \theta , \\ 0, &{} p/x > \theta , \end{array}\right. }&\frac{\partial H}{\partial p}&= -{\hat{c}}x-cy-d. \end{aligned}$$

We compute the products/sums we need for the Jacobian:

$$\begin{aligned} xF_x&= {\left\{ \begin{array}{ll} -\frac{rx}{p/q}, &{} p/q \le \gamma (T_C-x-y), \\ -\frac{rx(T_C-y)}{\gamma (T_C-x-y)^2}, &{} p/q> \gamma (T_C-x-y), \end{array}\right. } \\ F + xF_x&= {\left\{ \begin{array}{ll} r\left( 1-\frac{2x}{p/q}\right) -cy-l_x, &{} p/q \le \gamma (T_C-x-y), \\ r\left( 1-\frac{x(2T_C-x-2y)}{\gamma (T_C-x-y)^2}\right) -cy-l_x, &{} p/q> \gamma (T_C-x-y), \end{array}\right. } \\ xF_y&= {\left\{ \begin{array}{ll} -cx, &{} p/q \le \gamma (T_C-x-y), \\ -\frac{rx^2}{\gamma (T_C-x-y)^2}-cx, &{} p/q> \gamma (T_C-x-y), \end{array}\right. } \\ xF_p&= {\left\{ \begin{array}{ll} \frac{rx^2}{p^2/q}, &{} p/q \le \gamma (T_C-x-y), \\ 0, &{} p/q> \gamma (T_C-x-y), \end{array}\right. } \\ yG_x&= {\left\{ \begin{array}{ll} 0, &{} p/x \le \theta , \\ c{\hat{e}}y, &{} p/x> \theta , \end{array}\right. } \\ G+yG_y&= G = {\hat{e}}\text {min}\left\{ 1, \frac{p/x}{\theta } \right\} cx - {\hat{d}} - l_y, \\ yG_p&= {\left\{ \begin{array}{ll} \frac{c{\hat{e}}y}{\theta }, &{} p/x \le \theta , \\ 0, &{} p/x > \theta . \end{array}\right. } \\ \end{aligned}$$

Case 1 Producer nutrient limited and grazer quality limited

The corresponding Jacobian matrix with $E_P$ from Theorem 2 substituted in is

$$\begin{aligned} A_1 = \begin{bmatrix} a_{11} &{} a_{12} &{} a_{13} \\ a_{21} &{} a_{22} &{} a_{23} \\ a_{31} &{} a_{32} &{} a_{33} \end{bmatrix}, \end{aligned}$$

where

$$\begin{aligned} a_{11}&= l_x - r, \\ a_{12}&= -\frac{c{\hat{c}}T_P(r-l_x)-cdqr}{{\hat{c}}qr}, \\ a_{13}&= \frac{(r-l_x)^2}{qr}, \\ a_{21}&= 0, a_{22} = \frac{c{\hat{c}}{\hat{e}}T_P(r-l_x)-cd{\hat{e}}qr}{{\hat{c}}\theta (r-l_x)} - {\hat{d}} - l_y,\\ a_{23}&= 0, \\ a_{31}&= \frac{dqr}{r-l_x}, \\ a_{32}&= -\frac{{\hat{c}}T_P\theta (r-l_x)-dqr\theta }{qr} - \frac{c{\hat{c}}T_P(r-l_x)-cdqr}{{\hat{c}}(r-l_x)}, \\ a_{33}&= -\frac{{\hat{c}}T_P(r-l_x)}{qr}. \end{aligned}$$

The three eigenvalues for $A_1$ are (see “Appendix G” for proof)

$$\begin{aligned} \lambda _1&= \frac{c{\hat{c}}{\hat{e}}T_P(r-l_x)-cd{\hat{e}}qr}{{\hat{c}}\theta (r-l_x)} - {\hat{d}} - l_y, \\ \lambda _2&= -\frac{1}{2}\sqrt{\left( l_x-r + \frac{{\hat{c}}T_P(r-l_x)}{qr}\right) ^2 + 4d(r-l_x)} + \frac{1}{2}\left( l_x-r - \frac{{\hat{c}}T_P(r-l_x)}{qr}\right) , \\ \lambda _3&= \frac{1}{2}\sqrt{\left( l_x-r + \frac{{\hat{c}}T_P(r-l_x)}{qr}\right) ^2 + 4d(r-l_x)} + \frac{1}{2}\left( l_x-r - \frac{{\hat{c}}T_P(r-l_x)}{qr}\right) . \end{aligned}$$

Case 2 Producer nutrient limited and grazer quantity limited

The corresponding Jacobian matrix with $E_P$ from Theorem 2 substituted in is

$$\begin{aligned} A_2= \begin{bmatrix} a_{11} &{} a_{12} &{} a_{13} \\ a_{21} &{} a_{22} &{} a_{23} \\ a_{31} &{} a_{32} &{} a_{33} \end{bmatrix}, \end{aligned}$$

where all entries except $a_{22}$ are the same as in $A_1$ (Case 1 above), and

$$\begin{aligned} a_{22}&= \frac{c{\hat{c}}{\hat{e}}T_P(r-l_x)-cd{\hat{e}}qr}{{\hat{c}}qr} - {\hat{d}} - l_y. \end{aligned}$$

As in Case 1, we can explicitly find the eigenvalues for $A_2$:

$$\begin{aligned} \lambda _1&= \frac{c{\hat{c}}{\hat{e}}T_P(r-l_x)-cd{\hat{e}}qr}{{\hat{c}}qr} - {\hat{d}} - l_y, \\ \lambda _2&= -\frac{1}{2}\sqrt{\left( l_x-r + \frac{{\hat{c}}T_P(r-l_x)}{qr}\right) ^2 + 4d(r-l_x)} + \frac{1}{2}\left( l_x-r - \frac{{\hat{c}}T_P(r-l_x)}{qr}\right) , \\ \lambda _3&= \frac{1}{2}\sqrt{\left( l_x-r + \frac{{\hat{c}}T_P(r-l_x)}{qr}\right) ^2 + 4d(r-l_x)} + \frac{1}{2}\left( l_x-r - \frac{{\hat{c}}T_P(r-l_x)}{qr}\right) . \end{aligned}$$

Case 3 Producer carbon limited and grazer quality limited

The Jacobian with $E_C$ from Theorem 2 substituted in takes the form

$$\begin{aligned} A_3 = \begin{bmatrix} a_{11} &{} a_{12} &{} a_{13} \\ a_{21} &{} a_{22} &{} a_{23} \\ a_{31} &{} a_{32} &{} a_{33} \end{bmatrix}, \end{aligned}$$

where

$$\begin{aligned} a_{11}&= -\frac{(r-l_x)(r+\gamma r - \gamma l_x)}{r}, \\ a_{12}&= -\frac{\gamma (r-l_x)^2}{r}- \frac{c\gamma T_C(r-l_x)}{r+\gamma r - \gamma l_x}, \quad a_{13} = 0, \\ a_{21}&= 0, a_{22} =\frac{c{\hat{c}}{\hat{e}} \gamma T_CT_P(r-l_x)}{{\hat{c}}\gamma \theta T_C(r-l_x) + d\theta (r+\gamma r - \gamma l_x)} - {\hat{d}} - l_y, a_{23} = 0, \\ a_{31}&= \frac{{\hat{c}}dT_P(r+\gamma r - \gamma l_x)}{{\hat{c}}\gamma T_C(r-l_x) + d(r+\gamma r - \gamma l_x)}, \\ a_{32}&= -\frac{{\hat{c}}\theta \gamma T_C(r-l_x)}{r+\gamma r - \gamma l_x} - \frac{c{\hat{c}}\gamma T_CT_P(r-l_x)}{{\hat{c}}\gamma T_C(r-l_x) + d(r+\gamma r - \gamma l_x)}, \\ a_{33}&= -\frac{{\hat{c}}\gamma T_C(r-l_x)}{r+\gamma r - \gamma l_x}-d. \end{aligned}$$

A 3 x 3 matrix with zeros in these specific entries has its eigenvalues along the main diagonal, since it can be decomposed into a 2 x 2 upper triangular matrix and the eigenvalue given by $a_{22}$. Thus, in this case

$$\begin{aligned} \lambda _1&= -\frac{(r-l_x)(r+\gamma r - \gamma l_x)}{r}, \\ \lambda _2&= \frac{c{\hat{c}}{\hat{e}} \gamma T_CT_P(r-l_x)}{{\hat{c}}\gamma \theta T_C(r-l_x) + d\theta (r+\gamma r - \gamma l_x)} - {\hat{d}} - l_y, \\ \lambda _3&= -\frac{{\hat{c}}\gamma T_C(r-l_x)}{r+\gamma r - \gamma l_x}-d. \end{aligned}$$

Consider $r + \gamma r - \gamma l_x$ and $r - l_x$. We see that for $r, \gamma , l_x > 0$,

$$\begin{aligned} r + \gamma r - \gamma l_x< 0 \Rightarrow r + \gamma (r - l_x)< 0 \Rightarrow r-l_x< -\frac{r}{\gamma } \Rightarrow r - l_x < 0. \end{aligned}$$

Also, since all parameters are assumed to be positive,

$$\begin{aligned} r - l_x> 0 \Rightarrow \gamma (r-l_x)>0 \Rightarrow r + \gamma (r-l_x)> 0 \Rightarrow r +\gamma r - \gamma l_x > 0. \end{aligned}$$

For stability of the grazer extinction equilibrium, we need $\lambda _2 < 0$, which requires

$$\begin{aligned} \frac{c{\hat{c}}{\hat{e}} \gamma T_CT_P(r-l_x)}{{\hat{c}}\gamma \theta T_C(r-l_x) + d\theta (r+\gamma r - \gamma l_x)} < {\hat{d}} + l_y \end{aligned}$$

and either $r + \gamma r - \gamma l_x < 0$ (in which case $r-l_x < 0$) or $r-l_x > 0$ (in which case $r + \gamma r - \gamma l_x > 0$). Then, the grazer extinction equilibrium is a stable node since $\lambda _1, \lambda _3 < 0$ when $r + \gamma r - \gamma l_x$ and $r - l_x$ have the same sign.

For either $r + \gamma r - \gamma l_x < 0$ or $r-l_x > 0$, and

$$\begin{aligned} \frac{c{\hat{c}}{\hat{e}} \gamma T_CT_P(r-l_x)}{{\hat{c}}\gamma \theta T_C(r-l_x) + d\theta (r+\gamma r - \gamma l_x)} > {\hat{d}} + l_y, \end{aligned}$$

the grazer extinction equilibrium is a saddle with a two-dimensional stable manifold and a one-dimensional unstable manifold.

Now, for $r - l_x < 0$ and $r + \gamma r - \gamma l_x > 0$ we have a few cases. $\lambda _1$ is guaranteed to be positive, but the signs of $\lambda _2$ and $\lambda _3$ depend on additional conditions.

For $\lambda _2 < 0$, we need either ${\hat{c}}\gamma \theta T_C(r-l_x) + d\theta (r+\gamma r - \gamma l_x) > 0$ or ${\hat{c}}\gamma \theta T_C(r-l_x) + d\theta (r+\gamma r - \gamma l_x) < 0$ and

$$\begin{aligned} \frac{c{\hat{c}}{\hat{e}} \gamma T_CT_P(r-l_x)}{{\hat{c}}\gamma \theta T_C(r-l_x) + d\theta (r+\gamma r - \gamma l_x))} < {\hat{d}} + l_y . \end{aligned}$$

For $\lambda _3 < 0$, we need

$$\begin{aligned} -\frac{{\hat{c}}\gamma T_C(r-l_x)}{r+\gamma r - \gamma l_x}< d. \end{aligned}$$

Hence, for $r-l_x < 0$ and $r+\gamma r - \gamma l_x > 0$, we can have either a saddle with a one- or two-dimensional unstable manifold, or an unstable node. However, when $r-l_x < 0$ and $r+\gamma r - \gamma l_x > 0$, the equilibrium is not biologically feasible.

Case 4 Producer carbon limited and grazer quantity limited

The Jacobian with $E_C$ from Theorem 2 substituted in is as follows:

$$\begin{aligned} A_4 = \begin{bmatrix} a_{11} &{} a_{12} &{} a_{13} \\ a_{21} &{} a_{22} &{} a_{23} \\ a_{31} &{} a_{32} &{} a_{33} \end{bmatrix}, \end{aligned}$$

where all entries are the same as $A_3$ (Case 3), except $a_{22}$ and

$$\begin{aligned} a_{22}&= \frac{c{\hat{e}}\gamma T_C(r-l_x)}{r+\gamma r - \gamma l_x} - {\hat{d}} - l_y. \end{aligned}$$

As in the previous case, this matrix has its eigenvalues along the main diagonal. Thus, we have

$$\begin{aligned} \lambda _1&= -\frac{(r-l_x)(r+\gamma r - \gamma l_x)}{r}, \\ \lambda _2&= \frac{c{\hat{e}}\gamma T_C(r-l_x)}{r+\gamma r - \gamma l_x} - {\hat{d}} - l_y, \\ \lambda _3&= -\frac{{\hat{c}}\gamma T_C(r-l_x)}{r+\gamma r - \gamma l_x}-d. \end{aligned}$$

If $r + \gamma r - \gamma l_x < 0$, then we know $r-l_x < 0$. Therefore, if $r + \gamma r - \gamma l_x < 0$, then $\lambda _1, \lambda _3 < 0$ and we also need for stability

$$\begin{aligned} \frac{c{\hat{e}}\gamma T_C(r-l_x)}{r+\gamma r - \gamma l_x} < {\hat{d}} + l_y. \end{aligned}$$

If $r-l_x > 0$, then we know $r + \gamma r - \gamma l_x > 0$ and therefore $\lambda _1, \lambda _3 < 0$. For $\lambda _2$ to be less than 0 and therefore for stability we also require

$$\begin{aligned} \frac{c{\hat{e}}\gamma T_C(r-l_x)}{r+\gamma r - \gamma l_x} < {\hat{d}} + l_y. \end{aligned}$$

Hence, the grazer extinction equilibrium is a stable node for

$$\begin{aligned} \frac{c{\hat{e}}\gamma T_C(r-l_x)}{r+\gamma r - \gamma l_x} < {\hat{d}} + l_y \end{aligned}$$

and either $r + \gamma r - \gamma l_x < 0$ or $r-l_x > 0$.

For either $r + \gamma r - \gamma l_x < 0$ or $r-l_x > 0$, and

$$\begin{aligned} \frac{c{\hat{e}}\gamma T_C(r-l_x)}{r+\gamma r - \gamma l_x} > {\hat{d}} + l_y, \end{aligned}$$

the grazer extinction equilibrium is a saddle with a two-dimensional stable manifold and a one-dimensional unstable manifold.

For $r-l_x < 0$ and $r + \gamma r - \gamma l_x > 0$, we have $\lambda _1 > 0$ and $\lambda _2 < 0$. Therefore, we have a saddle. If $\lambda _3 < 0$, the saddle has a two-dimensional stable manifold and a one-dimensional unstable manifold; if $\lambda _3 > 0$, it has a one-dimensional stable manifold and a two-dimensional unstable manifold. Note that for $r-l_x < 0$ and $r+\gamma r - \gamma l_x > 0$, the equilibrium is not biologically feasible.

Appendix F

This appendix contains all the necessary calculations for the stability results in Sect. 3.3 using Holling type II functional responses for f and g.

For $f(x) = cx/(a+x)$, $g(P) = {\hat{c}}P/({\hat{a}} + P)$, and $h(C) = \gamma C$, using F, G, H as defined in “Appendix C”,

$$\begin{aligned} \frac{\partial F}{\partial x}&= {\left\{ \begin{array}{ll} -\frac{r}{p/q}+\frac{cy}{(a+x)^2}, &{} p/q \le \gamma (T_C-x-y), \\ -\frac{r(T_C-y)}{\gamma (T_C-x-y)^2} + \frac{cy}{(a+x)^2}, &{} p/q> \gamma (T_C-x-y), \end{array}\right. } \\ \frac{\partial F}{\partial y}&= {\left\{ \begin{array}{ll} -\frac{c}{a+x}, &{} p/q \le \gamma (T_C-x-y), \\ -\frac{rx}{\gamma (T_C-x-y)^2}-\frac{c}{a+x}, &{} p/q> \gamma (T_C-x-y), \end{array}\right. } \\ \frac{\partial F}{\partial p}&= {\left\{ \begin{array}{ll} \frac{rx}{p^2/q}, &{} p/q \le \gamma (T_C-x-y), \\ 0, &{} p/q> \gamma (T_C-x-y), \end{array}\right. }\\ \frac{\partial G}{\partial x}&= {\left\{ \begin{array}{ll} -\frac{c {\hat{e}} p}{\theta (a+x)^2}, &{} p/x \le \theta , \\ \frac{ac{\hat{e}}}{(a+x)^2}, &{} p/x> \theta , \end{array}\right. } \\ \frac{\partial G}{\partial y}&= 0, \\ \frac{\partial G}{\partial p}&= {\left\{ \begin{array}{ll} \frac{c{\hat{e}}}{\theta (a+x)}, &{} p/x \le \theta , \\ 0, &{} p/x > \theta , \end{array}\right. } \\ \frac{\partial H}{\partial x}&= \frac{{\hat{c}}(T_P-p-\theta y)}{{\hat{a}} + T_P-p-\theta y} + \frac{cpy}{(a+x)^2}, \\ \frac{\partial H}{\partial y}&= -\frac{{\hat{a}}{\hat{c}}\theta x}{({\hat{a}}+T_P-p-\theta y)^2} - \frac{cp}{a+x},\\ \frac{\partial H}{\partial p}&= -\frac{{\hat{a}}{\hat{c}}x}{({\hat{a}}+T_P-p-\theta y)^2}-\frac{cy}{a+x}-d. \\ \end{aligned}$$

The necessary products and sums for the Jacobian are

$$\begin{aligned} xF_x&= {\left\{ \begin{array}{ll} -\frac{rx}{p/q}+\frac{cxy}{(a+x)^2}, &{} p/q \le \gamma (T_C-x-y), \\ -\frac{rx(T_C-y)}{\gamma (T_C-x-y)^2} + \frac{cxy}{(a+x)^2}, &{} p/q> \gamma (T_C-x-y), \end{array}\right. } \\ F + xF_x&= {\left\{ \begin{array}{ll} r\left( 1-\frac{2x}{p/q}\right) - \frac{acy}{(a+x)^2} - l_x, &{} p/q \le \gamma (T_C-x-y), \\ r\left( 1-\frac{x(2T_C - x - 2y)}{\gamma (T_C-x-y)^2}\right) -\frac{acy}{(a+x)^2}-l_x, &{} p/q> \gamma (T_C-x-y), \end{array}\right. } \\ xF_y&= {\left\{ \begin{array}{ll} -\frac{cx}{a+x}, &{} p/q \le \gamma (T_C-x-y), \\ -\frac{rx^2}{\gamma (T_C-x-y)^2}-\frac{cx}{a+x}, &{} p/q> \gamma (T_C-x-y), \end{array}\right. } \\ xF_p&= {\left\{ \begin{array}{ll} \frac{rx^2}{p^2/q}, &{} p/q \le \gamma (T_C-x-y), \\ 0, &{} p/q> \gamma (T_C-x-y), \end{array}\right. } \\ yG_x&= {\left\{ \begin{array}{ll} -\frac{c {\hat{e}} p y}{\theta (a+x)^2}, &{} p/x \le \theta , \\ \frac{ac{\hat{e}}y}{(a+x)^2}, &{} p/x> \theta , \end{array}\right. } \\ G+yG_y&= G = {\hat{e}} \text {min} \left\{ 1, \frac{p/x}{\theta }\right\} \frac{cx}{a+x} - {\hat{d}} - l_y, \\ yG_p&= {\left\{ \begin{array}{ll} \frac{c{\hat{e}}y}{\theta (a+x)}, &{} p/x \le \theta , \\ 0, &{} p/x > \theta . \end{array}\right. } \\ \end{aligned}$$

Case 1 Producer nutrient limited and grazer quality limited

The corresponding Jacobian matrix with $E_P$ from Theorem 3 substituted in is

$$\begin{aligned} B_1 = \begin{bmatrix} b_{11} &{} b_{12} &{} b_{13} \\ b_{21} &{} b_{22} &{} b_{23} \\ b_{31} &{} b_{32} &{} b_{33} \end{bmatrix}, \end{aligned}$$

where for $E_P = ({\bar{x}}, 0, {\bar{p}})$ (Theorem 3)

$$\begin{aligned} b_{11}&= l_x - r, \\ b_{12}&= -\frac{c{\bar{p}}(r-l_x)}{arq + {\bar{p}}(r-l_x)}, \\ b_{13}&= \frac{(r-l_x)^2}{qr},\\ b_{21}&= 0, \\ b_{22}&= \frac{c{\hat{e}}qr{\bar{p}}}{\theta (aqr+{\bar{p}}(r-l_x))} -{\hat{d}} - l_y, \\ b_{23}&= 0, \\ b_{31}&= \frac{dqr}{r-l_x}\\ b_{32}&=-\frac{{\hat{a}}{\hat{c}}\theta {\bar{p}}(r-l_x)}{qr({\hat{a}} + T_P - {\bar{p}})^2} - \frac{cqr{\bar{p}}}{aqr + {\bar{p}}(r-l_x)},\\ b_{33}&= -\frac{{\hat{a}}{\hat{c}}{\bar{p}}(r-l_x)}{qr({\hat{a}}+T_P-{\bar{p}})^2}-d. \end{aligned}$$

As in Holling type I Case 1, we can explicitly determine the eigenvalues:

$$\begin{aligned} \lambda _1&= \frac{c{\hat{e}}qr{\bar{p}}}{\theta (aqr + {\bar{p}}(r-l_x))} - {\hat{d}} - l_y, \\ \lambda _2&= -\frac{1}{2}\sqrt{\left( l_x - r + \frac{{\hat{a}}{\hat{c}}{\bar{p}}(r-l_x)}{qr({\hat{a}}+T_P-{\bar{p}})^2}+d\right) ^2 + 4d(r-l_x)} \\&\quad + \frac{1}{2}\left( l_x - r - \frac{{\hat{a}}{\hat{c}}{\bar{p}}(r-l_x)}{qr({\hat{a}}+T_P-{\bar{p}})^2}-d \right) , \\ \lambda _3&=\frac{1}{2}\sqrt{\left( l_x - r + \frac{{\hat{a}}{\hat{c}}{\bar{p}}(r-l_x)}{qr({\hat{a}}+T_P-{\bar{p}})^2}+d\right) ^2 + 4d(r-l_x)} \\&\quad + \frac{1}{2}\left( l_x - r - \frac{{\hat{a}}{\hat{c}}{\bar{p}}(r-l_x)}{qr({\hat{a}}+T_P-{\bar{p}})^2}-d \right) . \end{aligned}$$

Again, the eigenvalues are not particularly illuminating. Also similar to the Holling type I case, stability does not depend on $T_C$ or $\gamma$. Given the producer is nutrient limited at this equilibrium, it is reasonable that the carbon-dependent carrying capacity has no impact on its stability.

Case 2 Producer nutrient limited and grazer quantity limited

The corresponding Jacobian with $E_P$ from Theorem 3 substituted in is

$$\begin{aligned} B_2 = \begin{bmatrix} b_{11} &{} b_{12} &{} b_{13} \\ b_{21} &{} b_{22} &{} b_{23} \\ b_{31} &{} b_{32} &{} b_{33} \end{bmatrix}, \end{aligned}$$

where all entries are the same as $B_1$ (Case 1) except $b_{22}$, which is

$$\begin{aligned} b_{22}&= \frac{c{\hat{e}}{\bar{p}}(r-l_x)}{aqr+{\bar{p}}(r-l_x)}-{\hat{d}}-l_y, \end{aligned}$$

for $E_P = ({\bar{x}}, 0, {\bar{p}})$ (Theorem 3).

The eigenvalues are

$$\begin{aligned} \lambda _1&= \frac{c{\hat{e}}{\bar{p}}(r-l_x)}{aqr + {\bar{p}}(r-l_x)} - {\hat{d}} - l_y, \\ \lambda _2&= -\frac{1}{2}\sqrt{\left( l_x - r + \frac{{\hat{a}}{\hat{c}}{\bar{p}}(r-l_x)}{qr({\hat{a}}+T_P-{\bar{p}})^2}+d\right) ^2 + 4d(r-l_x)} \\&\quad + \, \frac{1}{2}\left( l_x - r - \frac{{\hat{a}}{\hat{c}}{\bar{p}}(r-l_x)}{qr({\hat{a}}+T_P-{\bar{p}})^2}-d \right) , \\ \lambda _3&=\frac{1}{2}\sqrt{\left( l_x - r + \frac{{\hat{a}}{\hat{c}}{\bar{p}}(r-l_x)}{qr({\hat{a}}+T_P-{\bar{p}})^2}+d\right) ^2 + 4d(r-l_x)}\\&\quad +\, \frac{1}{2}\left( l_x - r - \frac{{\hat{a}}{\hat{c}}{\bar{p}}(r-l_x)}{qr({\hat{a}}+T_P-{\bar{p}})^2}-d \right) . \end{aligned}$$

The eigenvalues do not provide any stability conclusions in this case. As in Case 2 for Holling type I functional responses, stability does not depend on $T_C$, $\gamma$, or $\theta$. The producer is nutrient limited at equilibrium, so $T_C$ and $\gamma$ should not impact stability of the equilibrium. Note that since the grazer is quantity limited at equilibrium, we would also expect the stability of the equilibrium to not depend on the grazer’s P:C ratio, $\theta$.

Case 3 Producer carbon limited and grazer quality limited

Here we use $E_{C+}$ or $E_{C-}$ from Theorem 3 and compute

$$\begin{aligned} B_3 = \begin{bmatrix} b_{11} &{} b_{12} &{} b_{13} \\ b_{21} &{} b_{22} &{} b_{23} \\ b_{31} &{} b_{32} &{} b_{33} \end{bmatrix}, \end{aligned}$$

where

$$\begin{aligned} b_{11}&= -\frac{(r-l_x)(r+\gamma r - \gamma l_x)}{r}, \\ b_{12}&= -\frac{\gamma (r-l_x)^2}{r} - \frac{c \gamma T_C(r-l_x)}{a(r+\gamma r - \gamma l_x) + \gamma T_C(r-l_x)}, \\ b_{13}&= 0, \\ b_{21}&= 0,\\ b_{22}&= \frac{c{\hat{e}}(r+\gamma r - \gamma l_x){\bar{p}}}{a\theta (r+\gamma r - \gamma l_x) + \gamma \theta T_C(r-l_x)} - {\hat{d}} - l_y, \\ b_{23}&= 0,\\ b_{31}&= \frac{{\hat{c}}(T_P-{\bar{p}})}{{\hat{a}} + T_P-{\bar{p}}}, \\ b_{32}&= -\frac{{\hat{a}}{\hat{c}}\gamma \theta T_C(r-l_x)}{(r + \gamma r - \gamma l_x)({\hat{a}} + T_P - {\bar{p}})^2} - \frac{c(r + \gamma r - \gamma l_x){\bar{p}}}{a(r + \gamma r - \gamma l_x) + \gamma T_C(r-l_x)}, \\ b_{33}&= -\frac{{\hat{a}}{\hat{c}} \gamma T_C(r-l_x)}{(r + \gamma r - \gamma l_x)({\hat{a}}+T_P-{\bar{p}})^2}-d, \end{aligned}$$

where ${\bar{x}}$ and ${\bar{p}}$ correspond to the possible carbon-limited equilibrium values from Theorem 3.

The corresponding eigenvalues are along the main diagonal, as in the corresponding Holling type I case:

$$\begin{aligned} \lambda _1&= -\frac{(r-l_x)(r+\gamma r - \gamma l_x)}{r}, \\ \lambda _2&= \frac{c{\hat{e}}(r+\gamma r - \gamma l_x){\bar{p}}}{a\theta (r+\gamma r - \gamma l_x) + \gamma \theta T_C(r-l_x)} - {\hat{d}} - l_y, \\ \lambda _3&= -\frac{{\hat{a}}{\hat{c}} \gamma T_C(r-l_x)}{(r + \gamma r - \gamma l_x)({\hat{a}}+T_P-{\bar{p}})^2}-d. \end{aligned}$$

For these eigenvalues, we consider stability cases using $r- l_x$ and $r + \gamma r - \gamma l_x$. As in the Holling type I case, we recognize that $r + \gamma r - \gamma l_x < 0$ implies $r - l_x < 0$, and $r - l_x > 0$ implies $r + \gamma r - \gamma l_x > 0$, when all parameters are assumed to be positive. Therefore we consider three cases.

If $r + \gamma r - \gamma l_x < 0$ (and therefore $r - l_x < 0$), then $\lambda _1 < 0$ and $\lambda _3 < 0$. The sign of $\lambda _2$ depends on the sign of ${\bar{p}}$. Looking at the forms of the equilibria in Sect. 3.2, we observe that for $r + \gamma r - \gamma l_x < 0$ and $r - l_x < 0$, both equilibria have a positive ${\bar{p}}$. Hence, the equilibrium is a stable node for $r + \gamma r - \gamma l_x < 0$ and

$$\begin{aligned} \frac{c{\hat{e}}(r+\gamma r - \gamma l_x){\bar{p}}}{a\theta (r+\gamma r - \gamma l_x) + \gamma \theta T_C(r-l_x)} < {\hat{d}} + l_y. \end{aligned}$$

If $r + \gamma r - \gamma l_x < 0$ but the additional condition is not satisfied, then the equilibrium is a saddle with a two-dimensional stable manifold and a one-dimensional unstable manifold.

If $r - l_x > 0$ (and therefore $r + \gamma r - \gamma l_x > 0$), then $\lambda _1 < 0$ and $\lambda _3 < 0$. Since ${\bar{p}} > 0$ for $r - l_x > 0$ and both forms of ${\bar{p}}$, then the equilibrium is a stable node for $r - l_x > 0$ and

$$\begin{aligned} \frac{c{\hat{e}}(r+\gamma r - \gamma l_x){\bar{p}}}{a\theta (r+\gamma r - \gamma l_x) + \gamma \theta T_C(r-l_x)} < {\hat{d}} + l_y. \end{aligned}$$

If $r - l_x > 0$ but $\lambda _2 > 0$ because the above condition is not satisfied, then the equilibrium is a saddle with a two-dimensional stable manifold and a one-dimensional unstable manifold.

Lastly, if $r - l_x < 0$ and $r + \gamma r - \gamma l_x > 0$, then $\lambda _1 > 0$. The signs of $\lambda _2$ and $\lambda _3$ depend on additional conditions. For the equilibrium to be a saddle with a two-dimensional stable manifold and a one-dimensional unstable manifold when $r - l_x < 0$ and $r + \gamma r - \gamma l_x > 0$, we require

$$\begin{aligned} \frac{c{\hat{e}}(r+\gamma r - \gamma l_x){\bar{p}}}{a\theta (r+\gamma r - \gamma l_x) + \gamma \theta T_C(r-l_x)}&< {\hat{d}} + l_y, \\ -\frac{{\hat{a}}{\hat{c}} \gamma T_C(r-l_x)}{(r + \gamma r - \gamma l_x)({\hat{a}}+T_P-{\bar{p}})^2}&< d. \end{aligned}$$

Note that here the sign of ${\bar{p}}$ is not determined by the signs of $r - l_x$ and $r + \gamma r - \gamma l_x$ alone. If either one of these additional conditions is not satisfied (i.e., $\lambda _2 > 0$ or $\lambda _3 > 0$) and the other is satisfied, then the equilibrium is a saddle with a one-dimensional stable manifold and a two-dimensional unstable manifold. If both of these additional conditions are not satisfied, then the equilibrium is an unstable node. However, in this case, the equilibrium is only biologically feasible when the signs of $r - l_x$ and $r + \gamma r - \gamma l_x$ are the same.

Case 4 Producer carbon limited and grazer quantity limited

The Jacobian with $E_{C+}$ or $E_{C-}$ from Theorem 3 substituted in takes the form

$$\begin{aligned} B_4 = \begin{bmatrix} b_{11} &{} b_{12} &{} b_{13} \\ b_{21} &{} b_{22} &{} b_{23} \\ b_{31} &{} b_{32} &{} b_{33} \end{bmatrix}, \end{aligned}$$

where all entries are the same as in $B_3$, except

$$\begin{aligned} b_{22}&= \frac{c{\hat{e}}\gamma T_C(r-l_x)}{a(r + \gamma r - \gamma l_x) + \gamma T_C (r-l_x)} - {\hat{d}} - l_y. \end{aligned}$$

Once again, the eigenvalues are along the main diagonal, using ${\bar{p}}$ from $E_{C+}$ or $E_{C-}$ from Theorem 3:

$$\begin{aligned} \lambda _1&= -\frac{(r-l_x)(r+\gamma r - \gamma l_x)}{r}, \\ \lambda _2&= \frac{c{\hat{e}}\gamma T_C(r-l_x)}{a(r + \gamma r - \gamma l_x) + \gamma T_C (r-l_x)} - {\hat{d}} - l_y, \\ \lambda _3&= -\frac{{\hat{a}}{\hat{c}} \gamma T_C(r-l_x)}{(r + \gamma r - \gamma l_x)({\hat{a}}+T_P-{\bar{p}})^2}-d. \end{aligned}$$

We consider cases based on $r - l_x$ and $r + \gamma r - \gamma l_x$. Note that we cannot have $r - l_x > 0$ and $r + \gamma r - \gamma l_x < 0$ for positive $r, \gamma$ and $l_x$.

If $r - l_x < 0$ and $r + \gamma r - \gamma l_x < 0$, then $\lambda _1 < 0$ and $\lambda _3 < 0$. Therefore, the equilibrium is a stable node if $r + \gamma r - \gamma l_x < 0$ and

$$\begin{aligned} \frac{c{\hat{e}}\gamma T_C(r-l_x)}{a(r + \gamma r - \gamma l_x) + \gamma T_C (r-l_x)} < {\hat{d}} + l_y. \end{aligned}$$

If this additional condition is not satisfied (i.e., $\lambda _2 > 0$), then the equilibrium is a saddle with a two-dimensional stable manifold and a one-dimensional unstable manifold.

Similarly, if $r - l_x > 0$ and $r + \gamma r - \gamma l_x > 0$, then $\lambda _1 < 0$ and $\lambda _3 < 0$. Hence, the equilibrium is a stable node if $r - l_x > 0$ and

$$\begin{aligned} \frac{c{\hat{e}}\gamma T_C(r-l_x)}{a(r + \gamma r - \gamma l_x) + \gamma T_C (r-l_x)} < {\hat{d}} + l_y. \end{aligned}$$

If the above condition is not satisfied, then the equilibrium is a saddle with a two-dimensional stable manifold and a one-dimensional unstable manifold.

If $r - l_x < 0$ and $r + \gamma r - \gamma l_x > 0$, then $\lambda _1 > 0$. The equilibrium is a saddle with a two-dimensional stable manifold and a one-dimensional unstable manifold if

$$\begin{aligned} \frac{c{\hat{e}}\gamma T_C(r-l_x)}{a(r + \gamma r - \gamma l_x) + \gamma T_C (r-l_x)}&< {\hat{d}} + l_y, \\ -\frac{{\hat{a}}{\hat{c}} \gamma T_C(r-l_x)}{(r + \gamma r - \gamma l_x)({\hat{a}}+T_P-{\bar{p}})^2}&<d. \end{aligned}$$

If either of these conditions is not met, the equilibrium is a saddle with a one-dimensional stable manifold and a two-dimensional unstable manifold; if both conditions are not met, the equilibrium is an unstable node. Note that if $r - l_x < 0$ and $r + \gamma r - \gamma l_x > 0$, then the equilibrium is not biologically feasible.

Appendix G

This appendix proves the form of the eigenvalues for the nutrient limited grazer extinction equilibria.

Claim: For a matrix of the form

$$\begin{aligned} A = \begin{bmatrix} a &{} b &{} c \\ 0 &{} d &{} 0 \\ f &{} g &{} h \end{bmatrix} \end{aligned}$$

where all entries are real numbers, the eigenvalues are

$$\begin{aligned} \lambda _1&= d, \\ \lambda _2&= \frac{1}{2}\left( -\sqrt{(a-h)^2 + 4cf} + a + h\right) , \\ \lambda _3&= \frac{1}{2}\left( \sqrt{(a-h)^2 + 4cf} + a + h\right) . \end{aligned}$$

Proof

We know an eigenvalue $\lambda$ of matrix A satisfies $det(A - \lambda I) = 0$. Hence, we solve this equation for $\lambda$.

We see that

$$\begin{aligned} det(A - \lambda I) = det\left( \begin{bmatrix} a - \lambda &{} b &{} c \\ 0 &{} d - \lambda &{} 0 \\ f &{} g &{} h - \lambda \end{bmatrix} \right) . \end{aligned}$$

Using a cofactor expansion along the second row, we see

$$\begin{aligned} det(A - \lambda I)&= (d-\lambda )det\left( \begin{bmatrix} a - \lambda &{} c \\ f &{} h - \lambda \end{bmatrix} \right) \\&= (d-\lambda )((a-\lambda )(h-\lambda ) - cf) \\&= (d-\lambda )(ah - (a+h)\lambda + \lambda ^2 - cf) \\&= (d-\lambda )(\lambda ^2 - (a+h)\lambda + (ah-cf)). \end{aligned}$$

We can find solutions to the following using the quadratic formula:

$$\begin{aligned} (d-\lambda )(\lambda ^2 - (a+h)\lambda + (ah-cf)) = 0. \end{aligned}$$

The resulting solutions are

$$\begin{aligned} \lambda _1&= d, \\ \lambda _2&= \frac{1}{2}\left( \sqrt{(a+h)^2 - 4(ah-cf)} + a + h\right) , \\ \lambda _3&= \frac{1}{2}\left( -\sqrt{(a+h)^2 - 4(ah-cf)} + a + h\right) . \end{aligned}$$

Clearly

$$\begin{aligned} (a+h)^2 - 4(ah-cf)= & {} a^2 + 2ah + h^2 - 4ah + 4cf \\= & {} a^2 - 2ah+ h^2 + 4cf = (a-h)^2 + 4cf, \end{aligned}$$

and this proves the claim. $\square$