Dynamics of Phytoplankton Communities Under Photoinhibition

Abstract

We analyzed a model of phytoplankton competition for light in a well-mixed water column. The model, proposed by Gerla et al. (Oikos 120:519–527, 2011), assumed inhibition of photosynthesis at high irradiance (photoinhibition). We described the global behavior through mathematical analyses, providing a general solution to the multi-species competition for light with photoinhibition. We classified outcomes of 2- and 3-species competitions as examples, and evaluated feasibility of the theoretical predictions using empirical relationships between photosynthetic production and irradiance. Numerical simulations with published p–I curves indicate that photoinhibition may often lead to strong Allee effects and competitive facilitation among species. Hence, our results suggest that photoinhibition may play a major role in organizing phytoplankton communities.

Appendix

1.1 A.1 Basic Properties

We will demonstrate the shape of function

where p(I) satisfies p(0)=0, p(I)∈(0,∞) for 0<I≤I _in, dp/dI<∞, dp/dI>0 for 0≤I<I _opt, and dp/dI<0 for I _opt<I≤I _in. We claim that g(0)=0 and g(I _in)=p(I _in), and there is a unique point $\hat{x}$ such that $d g(\hat{x})/dx = 0$, and dg/dx>0 for $x\in[0,\hat{x})$; dg/dx<0 for $x \in(\hat{x}, I_{\mathrm{in}}]$.

The proof of g(0)=0 is straightforward:

because $\int_{0}^{I_{\mathrm{in}}} \frac{p(I)}{I}\,dI$ is bounded and lim _x→0(lnI _in−lnx)=∞. Likewise, g(I _in)=p(I _in) can be proved by

Next, we prove that g(x) is a unimodal function with a peak at $\hat{x} \in(0, I_{\mathrm{opt}})$. The derivative of g(x) is

Let

Because p(0)=0 from the assumption, we have

Therefore, g ₁(0)<g ₂(0) and g ₁(I _in)=g ₂(I _in)=0.

The derivatives of g ₁(x) and g ₂(x) are:

(12)

(13)

Because p′(x)<0 for x∈(I _opt,I _in], we have $|g_{1}^{\prime}(I_{\mathrm{in}}^{-})|>|g_{2}^{\prime}(I_{\mathrm{in}}^{-})|$. Since g ₁(I _in)=g ₂(I _in), it follows that g ₁(x)>g ₂(x) for x near I _in. By the intermediate value theorem, there exists a point $\hat{x} \in(0,I_{\mathrm{in}})$ such that $g_{1}(\hat{x})=g_{2}(\hat{x})$.

From $g_{1}(\hat{x})=g_{2}(\hat{x})$, g ₁(I _in)=g ₂(I _in), there exists a point $\bar{x} \in(\hat{x},I_{\mathrm{in}})$ such that $g_{1}^{\prime}(\bar{x})=g_{2}^{\prime}(\bar{x})$ according to Rolle’s theorem. From (12) and (13), we have $p^{\prime}(\bar{x})=0$. From assumption of p(x), it follows that $\bar{x}=I_{\mathrm{opt}}$. Hence, $\hat{x}<I_{\mathrm{opt}}<I_{\mathrm{in}}$.

Suppose there exists another point $\hat{x}^{\prime}\in(0,I_{\mathrm{in}})$ with $g_{1}(\hat{x}^{\prime})=g_{2}(\hat{x}^{\prime})$. If $\hat{x}^{\prime}\in(\hat{x},I_{\mathrm{opt}}]$, then by Rolle’s theorem there exists a point $s\in(\hat{x},\hat{x}^{\prime})$ such that p′(s)=0. It contradicts to that p′(I)>0 for 0≤I<I _opt. If $\hat{x}^{\prime}\in(I_{\mathrm{opt}},I_{\mathrm{in}})$, then by Rolle’s theorem there exists a point s∈(I _opt,I _in) such that p′(s)=0. It contradict to that p′(I)<0 for 0≤I _opt<I<I _in. Thus, $\hat{x}$ is the only point with $g_{1}(\hat{x})=g_{2}(\hat{x})$.

Next, we prove the following theorem. In the following, we let L(t)=I _out(t) for convenience. From (11),

(14)

Theorem 2

The solutions of (10) are positive and bounded.

Proof

Assume there exists t ₁>0 and some i∈{1,2,…,n} such that

By reversing time, let τ=−t and we consider backward behavior of the solution of (10) with initial data x _i (0)=0,x _j (0)=x _j (t ₁)>0. It follows that x _i (τ)=0 for all τ<0. By the uniqueness of ordinary differential equations, we have x _i (−t ₁)=0, a contradiction. Thus, the solutions are positive if the initial condition is in Ω.

To prove the boundedness of solution, we consider the differential inequalities

where

with $g_{i}(L_{\max,i})=\max_{s\in[0,I_{0}]}g_{i}(s)$. Let y(t) be the solution of

$$y'=\bigl[G_i\bigl(I_0 \exp(-k_iy)\bigr)-d_i\bigr] y. $$

Then lim _t→∞ y(t)=y _i , where y _i satisfies G _i (I ₀exp(−k _i y _i ))−d _i =0. Hence, for given small ϵ, x _i (t)≤y _i +ϵ for all t large. Hence, x _i (t) is bounded for all time t for i=1,2,…,n. □

The next theorem says that if the light intensity is weak enough, then some species die out.

Theorem 3

If λ _i >I ₀, then lim _t→∞ x _i (t)=0.

Proof

Since L(t)≤I ₀ for all t≥0, we have L(t)<λ _i for all t≥0. From g _i (L(t))≤G _i (L(t)),

$$\frac{x_i'}{x_i}=g_i\bigl(L(t)\bigr)-d_i\leq G_i\bigl(L(t)\bigr)-d_i\leq G_i(I_0)-d_i<0, $$

then x _i (t)→0 as t→∞. □

1.2 A.2 Local Stability of Equilibria

For simplicity, let

$$f_i\bigl(x_1(t),x_2(t), \ldots,x_n(t)\bigr):=\bigl[g_i\bigl(L(t) \bigr)-d_i\bigr]x_i(t). $$

We denote the Jacobian of (10) at an equilibrium E is $J(E)=[m_{ij}]\in\mathbb{R}^{n\times n}$, where

For the equilibrium E ₀, the Jacobian at E ₀ is

$$J(E_0) = \left [ \begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} g_1(I_0)-d_1 & 0 & 0 & \ldots&0 \\ 0 & g_2(I_0)-d_2 &0 &\ldots&0 \\ \vdots\\ 0 & 0 &\ldots & 0 & g_n(I_0)-d_n \\ \end{array} \right ]. $$

Obviously the eigenvalues of J(E ₀) are g _i (I ₀)−d _i , for i=1,2,…,n.

If I ₀∉S, then g _i (I ₀)−d _i <0 for all i and E ₀ is locally asymptotically stable.

If I ₀∈S, then there exists some i such that I ₀∈(λ _i ,μ _i ). Hence, g _i (I ₀)−d _i >0, and E ₀ is unstable. If there exists some j such that I ₀∉(λ _j ,μ _j ), then J(E ₀) has negative eigenvalues and E ₀ is saddle with the local stable manifold

$$W_\mathrm{loc}^s(E_0)= \Biggl\{\sum _{i=1}^n c_ie_i: c_i=0 \mbox{ except some }j \mbox{ with } I_0\notin( \lambda_j,\mu_j). \Biggr\}, $$

where e _i is the eigenvector corresponding to the eigenvalue of (g _i (I ₀)−d _i ). Hence, the dimension of W ^s(E ₀), the stable manifold of E ₀, is at most (n−1) and

$$W^s(E_0)\subset\bigl\{(x_1, \ldots,x_n): x_i=0 \mbox{ if } I_0\in( \lambda_{i},\mu_{i})\bigr\}. $$

Therefore, W ^s(E ₀)∩Ω=∅.

For the equilibria $E_{\lambda_{r}}$, the Jacobian evaluated at $E_{\lambda_{r}}$ is

$$J(E_{\lambda_r}) = \left [ \begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad }c@{\quad}c@{\quad}c} m_{11} & 0 &\ldots&\ldots& \ldots&\ldots&\ldots& \ldots&0 \\ 0 & m_{22} &0 &\ldots&\ldots& \ldots&\ldots&\ldots&0 \\ \vdots\\ 0 &\ldots &0 &m_{r-1,r-1} &0 &\ldots&\ldots&\ldots &0\\ m_{r1} &\ldots&\ldots &m_{r,r-1} &m_{r,r} &m_{r,r+1} &\ldots&\ldots &m_{rn}\\ 0 &\ldots&\ldots&\ldots&0 &m_{r+1,r+1} &0 &\ldots&0\\ \vdots\\ 0 &\ldots&\ldots&\ldots&\ldots&\ldots&\ldots& 0 & m_{nn} \\ \end{array} \right ], $$

where

The eigenvalues of $J(E_{\lambda_{r}})$ are $-k_{r}x_{\lambda_{r}}g_{r}'(\lambda_{r})\lambda_{r}<0$, and g _j (λ _r )−d _j , for all j≠r.

When λ _r is the endpoint of a component of S, we have that g _j (λ _r )−d _j <0 for j≠r. Hence, $E_{\lambda_{r}}$ is locally asymptotically stable.

If λ _r is not the endpoint of a component of S, then there exists j such that λ _r ∈(λ _j ,μ _j ). The eigenvector corresponding to the negative eigenvalue $-k_{r}x_{\lambda_{r}}g_{r}'(\lambda_{r})\lambda_{r}$ is

$$v_r=(0,\ldots,0,x_r,0,\ldots,0),\quad x_r>0 \mbox{ in the }r\mbox{th component}, $$

and the eigenvectors corresponding to the negative eigenvalues g _i (λ _r )−d _i are

$$v_i=(0,\ldots,0,k_r,0,\ldots,0,-k_i,0, \ldots,0), $$

where k _r , k _j >0 are in the ith , and rth component of vector v _i , respectively. We know that $E_{\lambda_{r}}$ is saddle and the local stable manifold is

$$W_\mathrm{loc}^s(E_{\lambda_r})= \Biggl\{E_{\lambda_r}+\sum _{i=1}^nc_i v_i:\ c_i=0 \mbox{ except } c_r \mbox{ and some } j\mbox{ with } \lambda_r\notin(\lambda_j,\mu_j) \Biggr\}. $$

Hence, the stable manifold of $E_{\lambda_{r}}$ is

$$W^s(E_{\lambda_r})\subset\bigl\{(x_1, \ldots,x_n): x_i=0 \mbox{ if } \lambda_r\in( \lambda_{i},\mu_{i})\bigr\}, $$

and $W^{s}(E_{\lambda_{r}})\cap\varOmega=\emptyset$.

For the equilibria $E_{\mu_{r}}$, the Jacobian evaluated at $E_{\mu_{r}}$ is $J(E_{\mu_{r}})$, the structure is similar to $J(E_{\lambda_{r}})$ and

The eigenvalues are $-k_{r}x_{\mu_{r}}g_{r}'(\mu_{r})\mu_{r}>0$, an g _j (μ _r )−d _j , for all j≠r.

For the case μ _r is the endpoint of a component of S, then g _j (μ _r )−d _j <0 for all j≠r. Hence, $E_{\mu_{r}}$ is saddle with one dimensional unstable manifold

$$W^u(E_{\mu_r})= \bigl\{E_{\mu_r}+sv_r: v_r=(0,\ldots,0,x_r,0,\ldots,0), x_r>0 \mbox{ in the }r\mbox{th component} \bigr\}. $$

The eigenvectors corresponding to the negative eigenvalues g _j (λ _r )−d _j for all j≠r are

$$v_j=(0,\ldots,0,k_r,0,\ldots,0,-k_j,0, \ldots,0), $$

where k _r , k _j >0 are in the jth, and rth components of the vector v _j , respectively. $E_{\mu_{r}}$ is saddle which stable manifold $W^{s}(E_{\mu_{r}})$ is tangent to

$$W_\mathrm{loc}^s(E_{\mu_r})= \Biggl\{E_{\mu_r}+\sum _{j=1}^{n}c_jv_j: c_r=0, c_j\neq0, \mbox{ for }j\neq r \Biggr\}, $$

at $E_{\mu_{r}}$. Thus, the (n−1)-dimensional stable manifold of $E_{\mu_{r}}$ satisfies

$$W^s(E_{\mu_r})\subset\bigl\{(x_1,x_2, \ldots,x_n), x_i>0 \mbox{ for all } i\bigr\}, $$

and $W^{s}(E_{\mu_{r}})\cap\varOmega\neq\emptyset$.

For the case μ _r is not the endpoint of a component of S, then there exists j such that μ _r ∈(λ _j ,μ _j ) and g _j (λ _r )−d _j >0. If there exists some i such that μ _r ∈(λ _i ,μ _i ), then $E_{\mu_{r}}$ is saddle which stable manifold $W^{s}(E_{\mu_{r}})$ is tangent to

$$W_\mathrm{loc}^s(E_{\mu_r})= \Biggl\{E_{\mu_r}+\sum _{j=1}^nc_jv_j: c_j=0 \mbox{ except } c_i \mbox{ with } \mu_r\notin(\lambda_i,\mu_i) \Biggr\}, $$

at $E_{\mu_{r}}$. Hence, the stable manifold of $E_{\mu_{r}}$ is

$$W^s(E_{\mu_r})\subset\bigl\{(x_1, \ldots,x_n): x_i=0 \mbox{ if } \mu_r\in( \lambda_{i},\mu_{i})\bigr\}, $$

and $W^{s}(E_{\mu_{r}})\cap\varOmega=\emptyset$.

1.3 A.3 The Proof of Theorem 1

To prove Theorem 1, we need the following three lemmas. We note that the following proofs are similar to those in Butler and Wolkowicz (1985). We present them for the sake of completeness of the paper.

Lemma 4

If lim _t→∞ x _i (t)>0, then lim _t→∞ L(t)=λ _i or μ _i and lim _t→∞ x _j (t)=0 for j≠i.

Proof

Since lim _t→∞ x _i (t) exists and it is positive, and $|x_{i}''(t)|$ is bounded, then $x_{i}'(t)$ converges to 0 as t goes to infinity, i.e.,

$$g_i\bigl(L(t)\bigr)-d_i\to0 \quad\mbox{as } t\to \infty. $$

Therefore, lim _t→∞ L(t)=λ _i or μ _i .

For j≠i, we prove lim _t→∞ x _j (t)=0 by contradiction. First, we assume lim _t→∞ x _j (t)>0, then by the similar argument as above, we obtain that lim _t→∞ L(t)=λ _j or μ _j , a contradiction. Thus, lim _t→∞ x _j (t) does not exist and lim sup _t→∞ x _j (t)>0. Then there exists a subsequence t _m increases to infinity as m goes to infinity, and x _j (t _m ) converges to lim sup _t→∞ x _j (t) and $x_{j}'(t_{m})=0$. Hence, g _j (L(t _m ))−d _j =0 and L(t _m )=λ _j or μ _j , a contradiction to lim _t→∞ L(t)=λ _i or μ _i . Thus, we have lim _t→∞ x _j (t)=0. □

Lemma 5

If lim _t→∞ L(t)=γ, then γ∈{I ₀, the endpoints of a component of S}.

• 1. If γ=I ₀, then I ₀∉S, and lim _t→∞ x _i (t)=0 for all i.

• 2. If γ=λ _i or μ _i , the endpoint of a component of S, then γ<I ₀ and $\lim_{t\to\infty}x_{i}(t)=x_{\lambda_{i}}$ or $x_{\mu_{i}}$ and lim _t→∞ x _j (t)=0 for j≠i.

Proof

We prove by contradiction. If not, then γ∉{I ₀, the endpoints of a component of S}. There are two possibilities: γ∈S and γ∉S.

If γ∈S, from the assumption lim _t→∞ L(t)=γ, then for ϵ>0 there exists some i and T _ϵ >0 such that L(t)⊂(λ _i ,μ _i ) for t≥T _ϵ . It follows that $x_{i}'(t)\geq0$ for t≥T _ϵ , and from the fact x _i (t) is bounded above, then lim _t→∞ x _i (t)>0. By Lemma 4, we have that γ=lim _t→∞ L(t)=λ _i or μ _i . Since γ is not endpoints of S, there exists j≠i such that γ∈(λ _j ,μ _j ). It follows that L(t)⊂(λ _j ,μ _j ) for all large t. By similar argument as above, we have lim _t→∞ L(t)=λ _j or μ _j , a contradiction.

If γ∉S, then for ϵ>0 there exists T _ϵ >0 s.t. L(t)∈(γ−ϵ,γ+ϵ)⊂S ^c for t≥T _ϵ . Hence $\frac{x_{i}'(t)}{x_{i}(t)}=g_{i}(L(t))-d_{i}<0$ for all i for t≥T _ϵ . Therefore, lim _t→∞ x _i (t)=0 for all i, and lim _t→∞ L(t)=I ₀, a contradiction.

• 1. Let γ=I ₀. Assume I ₀∈S and from the convergence of L(t), there exists i such that L(t)∈(λ _i ,μ _i ) for all t is large. By similar argument as above, we have lim _t→∞ x _i (t)>0 and lim _t→∞ L(t)=λ _i or μ _i , it is a contradiction. Thus, I ₀∉S.

  Now we prove lim _t→∞ x _i (t)=0 for all i by contradiction. First we assume that there exists i such that lim _t→∞ x _i (t)>0. Then lim _t→∞ L(t)=λ _i or μ _i , a contradiction. Assume lim _t→∞ x _i (t) does not exist and lim sup _t→∞ x _i (t)>0. Then there exists a sequence {t _m } increases to infinity as t goes to infinity such that $x_{i}'(t_{m})=0$ and lim _m→∞ x _i (t _m )=lim sup _t→∞ x _i (t). It follows that g _i (L(t _m ))−d _i =0 and L(t _m )=λ _i or μ _i for all m, a contradiction to γ=I ₀. Thus, lim _t→∞ x _i (t)=0 for all i.

• 2. It is clear that γ<I ₀, since L(t)≤I ₀ for all t. lim _t→∞ x _j (t)=0 for j≠i follows from the above argument. Thus, it follows that $\lim_{t\to\infty}L(t)= \lim_{t\to\infty}[I_{0} e^{-k_{i} x_{i}(t)}]=\lambda_{i}$ or μ _i , or equivalently $\lim_{t\to\infty}x_{i}(t)=x_{\lambda_{i}}$ or $x_{\mu_{i}}$. □

Lemma 6

L(t) converges as t goes to infinity.

Proof

If not, then there exist increasing sequences {t _m },{τ _m } such that

$$L'(t_m)=0, \lim_{m\to\infty}L(t_m)= \limsup_{t\to\infty}L(t):=\overline{L}, $$

$$L'(\tau_m)=0, \lim_{m\to\infty}L( \tau_m)=\liminf_{t\to\infty }L(t):=\underline{L}. $$

Note that there are some i∈{1,2,…,n} such that x _i (t) do not tend to zero. Since

$$L'(t_m)=-L(t_m) \Biggl[\sum _{i=1}^{n}k_ix_i'(t_m) \Biggr]=0, $$

for each m there are some j _m ∈{1,2,…,n} satisfies $x_{j_{m}}'(t_{m})\geq0$. There exists some j such that j _m =j for infinitely many m. For this j, we choose a subsequence of {t _m }, also named {t _m }, such that $x_{j}'(t_{m})\geq0$. It follows that L(t _m )∈[λ _j ,μ _j ] for all m and $\overline{L}\in[\lambda_{j},\mu_{j}]$. Similarly, we can find some k and a subsequence of {τ _m }, also named {τ _m }, such that L(τ _m )∈[λ _k ,μ _k ] for all m and $\underline{L}\in[\lambda_{k},\mu_{k}]$.

If $\overline{L}\in[\lambda_{j},\mu_{j}]\subset[\lambda_{p_{1}},\mu_{q_{1}}]$ and $\underline{L}\in[\lambda_{k},\mu_{k}]\subset[\lambda_{p_{2}},\mu_{q_{2}}]$ where $(\lambda_{p_{1}},\mu_{q_{1}})$ and $(\lambda_{p_{2}},\mu_{q_{2}})$ are two disjoint components of S. Then there exists an increasing sequence {s _m } with t _m <s _m <τ _m such that L′(s _m )<0 and $L(s_{m})\in(\mu_{q_{2}},\lambda_{p_{1}})\cap S^{c}$ for all m. Hence, $x_{i}'(s_{m})<0$ for all i and L′(s _m )>0, a contradiction.

Thus, [λ _j ,μ _j ] and [λ _k ,μ _k ] belong to the same set [λ _p ,μ _q ], that is, $\overline{L}$ and $\underline{L}$ belong to [λ _p ,μ _q ], where (λ _p ,μ _q ) is a component of S.

If there does not exist γ∈Γ, Γ={λ _i , μ _i :i=1,2,…,n}, s.t. $\gamma\in(\underline{L},\overline{L})$, then there exists some r s.t. L(t)∈[λ _r ,μ _r ]⊂[λ _p ,μ _q ] for all large t. Then we have $x_{r}'(t)>0$ for all large t. From the boundedness of x _r , it follows that $\lim_{t\to\infty}x_{r}(t)=x_{r}^{*}\geq0$ and lim _t→∞ L(t)=λ _r or μ _r , a contradiction.

Thus, there exists some γ∈Γ s.t. $\gamma\in(\underline{L},\overline{L})$, let γ ₁, γ ₂ be two consecutive elements of Γ such that $\gamma_{1}<\underline{L}<\gamma_{2}$. Since L(t) oscillates, there exists T ₁<T ₂ such that L(T ₁)=L(T ₂)=γ ₂, γ ₁<L(t)≤γ ₂ for t∈[T ₁,T ₂] and L′(T ₁)<0<L′(T ₂). From

$$L'(t)=-L(t) \Biggl[\sum_{i=1}^{n}k_ix_i'(t) \Biggr], $$

we have

$$\sum_{i=1}^{n}k_ix_i'(T_1)>0> \sum_{i=1}^{n}k_ix_i'(T_2). $$

We divide the above summation into two parts, one is $x_{i}'(T_{j})<0$, i.e. g _i (γ ₂)<d _i , the other is $x_{i}'(T_{j})>0$, i.e. g _i (γ ₂)>d _i . Therefore, we have

$$\sum_{g_i(\gamma_2)<d_i}k_ix_i'(T_1)+ \sum_{g_i(\gamma_2)>d_i}k_ix_i'(T_1) >\sum_{g_i(\gamma_2)<d_i}k_ix_i'(T_2)+ \sum_{g_i(\gamma_2)>d_i}k_ix_i'(T_2), $$

and

(15)

For the case g _i (γ ₂)<d _i , i.e., γ ₂∉(λ _i ,μ _i ), then L([T ₁,T ₂]) is disjoint from (λ _i ,μ _i ). Hence, $x_{i}'(t)<0$ for t∈[T ₁,T ₂] and x _i (T ₂)<x _i (T ₁). For the case g _i (γ ₂)>d _i , i.e., λ _i <γ ₂<μ _i , we have L([T ₁,T ₂])⊂(λ _i ,μ _i ). Hence $x_{i}'(t)>0$ for t∈[T ₁,T ₂] and x _i (T ₂)>x _i (T ₁). Thus,

a contradiction to (15). Hence, the theorem holds. □