Growth of Phytoplankton

Introduction

Phytoplankton Dynamical System

In the paper of Bernard - Gouze [1] one analyse a model of phytoplankton growth based on the dynamical system

where x1 means the substrate, x2 is the phytoplankton biomass and x3 is the intracellular nutrient per biomass, with the physical domain  The previous dynΣamical system is non cooperative and has the equilibrium point  We introduce the phytoplankton vector field of components  and the maximal field line x = x(t, x0), t ∈ I, which satisfies the initial condition x(t0, x0) = x0. In order to find bounds for substrate, biomass, and intracellular nutrient per biomass, we use the techniques of optimization developed in our papers [2-6].

Bounds for Phytoplankton Substrate

We use the following problem: ftnd max f (x1, x2, x3) = x1 with the restriction x=x(t,x0). We set the critical point condition . In this case  It follows the relation  The convenient solution (critical point)  must be

The sufficient condition Hess  reduces to 

Since at the critical point we have  the condition goes to  and the convenient condition is  . Theorem 2.1. Suppose that on a evolution line (field line) it exists a point x¯ at which we have

Then the phytoplankton substrate has an upper bound at this point.

Bounds for Phytoplankton Biomass

Let us use the problem: ftnd max subject to x=x(t,x0).

Growth of phytoplankton: The critical point condition is it follows the relation The convenient solution (critical point) The sufficient condition Hess

Since at the critical point we have , the sufficient condition leads to 

Theorem: Suppose that on a evolution line (field line) it exists a point at which we have

then the x2(t) component of the corresponding field line has an upper bound at this point.

In the direct alternative, we build the composite function g(x(t, x0)). The condition

reduces to  the convenient solution is  The condition  becomes  Replacing we find x2 > 0, x1 − 2 < 0. The same result is obtained as in the previous method.

Bounds for Intracellular Nutrient Per Biomass

Now the helping problem is: compute max  with the restriction 

The critical point condition is  Since Δh = ∇h = (0,0,1), it follows the relation  The critical point  The sufficient condition Hess  reduces to 

Since at the critical point we have  the condition goes to

Theorem: Suppose that on a evolution line (field line) it exists a point x at which we have  Then the intracellular nutrient per biomass has an upper bound at this point.

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