Modelling of Biological Decontamination of a Water Resource in Natural Environment and Related Feedback Strategies

Abstract

We show how to combine numerical schemes and calibration of systems of o.d.e. to provide efficient feedback strategies for the biological decontamination of water resources. For natural resources, we retain to introduce any bacteria in the resource and treat it aside preserving a constant volume of the resource at any time. The feedback strategies are derived from the minimal time synthesis of the system of o.d.e.

Appendix 1: Links Between PDE and ODE Models

In this appendix we start from the transport-diffusion equation (2.5) and justify the simplified equations (2.1c) and (2.1d). Let us recall Eq.  (2.5):

$$\begin{aligned} \displaystyle \frac{\partial {S_{L}}}{\partial t } + \mathbf{U}\cdot \nabla S_{L}-\nu _{S}\Delta S_{L}= 0. \end{aligned}$$

(5.1)

We define the averaged pollutant concentration over the domain:

$$\begin{aligned} \overline{S_{L}}(t) = \displaystyle \frac{1}{|\Omega |}\int _{\Omega } S_{L}(t,x,y)\,dxdy. \end{aligned}$$

(5.2)

Integrating Eq. (5.1) over the whole domain $\Omega$, we have:

$$\begin{aligned} \displaystyle \frac{d {\overline{S_{L}}}}{d t }+\displaystyle \frac{1}{|\Omega |}\int _{\Omega } \mathbf{U}(t,x,y)\cdot \nabla S_{L}(t,x,y)\,dxdy -\displaystyle \frac{\nu _{S}}{|\Omega |}\int _{\Omega } \Delta S_{L}(t,x,y)\,dxdy = 0. \end{aligned}$$

Thanks to the free divergence of $\mathbf{U}$ and to boundary conditions (2.8), (2.9) and (2.10), we obtain:

$$\begin{aligned} \displaystyle \frac{d {\overline{S_{L}}}}{d t }-\displaystyle \frac{1}{|\Omega |}\int _{\Gamma _{in}}\displaystyle \frac{Q\,S_{L}}{|\Gamma _{in}|}\, ds+\displaystyle \frac{1}{|\Omega |}\int _{\Gamma _{out}}\displaystyle \frac{Q\,S_{L}}{|\Gamma _{out}|}\,ds-\displaystyle \frac{\nu _{S}}{|\Omega |}\int _{\Gamma _{in}} \displaystyle \frac{\partial {S_{L}}}{\partial n }\,ds=0, \end{aligned}$$

where n classically denotes the outward-pointing vector normal to $\Omega$ on $\partial \Omega$.

We now replace the boundary condition (2.8a) on $\Gamma _{in}$ by the following Robin condition²:

$$\begin{aligned} -\nu _{S} \displaystyle \frac{\partial {S_{L}}}{\partial n }=\displaystyle \frac{Q}{|\Gamma _{in}|}(S_{L}-S_{in}), \end{aligned}$$

(5.3)

and obtain:

$$\begin{aligned} \displaystyle \frac{d {\overline{S_{L}}}}{d t }-\displaystyle \frac{Q}{|\Omega |}S_{in}+\displaystyle \frac{Q}{|\Omega |\,|\Gamma _{out}|}\int _{\Gamma _{out}}S_{L}\,ds=0. \end{aligned}$$

Now, assuming that the average concentration close to $\Gamma _{out}$ is the same as the average concentration in the whole domain (in particular this is true if one assumes that $S_{L}$ does not depend on space), then we have:

$$\begin{aligned} \displaystyle \frac{d {\overline{S_{L}}}}{d t }=\displaystyle \frac{Q}{|\Omega |}(S_{in}-\overline{S_{L}}), \end{aligned}$$

(5.4)

which is exactly the one zone model.

Now, if we consider (as it can be inferred from Fig. 2) that we have two zones in the lake (the blue one $\Omega _{1}$ and the red one $\Omega _{2}$) separated by an interface $\Gamma _{1/2}$ on which we have $\mathbf{U}\cdot n_{1}=0$ where $n_{1}$ is the outward-pointing vector normal to $\Omega _{1}$ on $\Gamma _{1/2}$, we can restart from Eq. (5.1) but integrate it over the active (blue) zone $\Omega _{1}$ only, and obtain after the same kind of computation:

$$\begin{aligned} \displaystyle \frac{d {S_{1}}}{d t }-\displaystyle \frac{Q}{|\Omega _{1}|}S_{in}+\displaystyle \frac{Q}{|\Omega _{1}|\,|\Gamma _{out}|}\int _{\Gamma _{out}}S_{L}\, ds-\displaystyle \frac{\nu _{S}}{|\Omega _{1}|}\int _{\Gamma _{1/2}} \displaystyle \frac{\partial {S_{L}}}{\partial n_1 }\,ds=0, \end{aligned}$$

where $S_{1}$ is the averaged pollutant concentration in the active zone $\Omega _{1}$:

$$\begin{aligned} S_{1}(t) = \displaystyle \frac{1}{|\Omega _{1}|}\int _{\Omega _{1}} S_{L}(t,x,y)\,dxdy. \end{aligned}$$

(5.5)

Assuming again than the average concentration close to $\Gamma _{out}$ is the same as the average concentration in $\Omega _{1}$, we have:

$$\begin{aligned} \displaystyle \frac{d {S_{1}}}{d t } = \displaystyle \frac{Q}{|\Omega _{1}|}(S_{in}-S_{1}) +\displaystyle \frac{\nu _{S}}{|\Omega _{1}|}\int _{\Gamma _{1/2}} \displaystyle \frac{\partial {S_{L}}}{\partial n_1 }\,ds, \end{aligned}$$

(5.6)

Considering now the dead zone $\Omega _{2}$ with the outward-pointing vector normal $n_{2}=-n_{1}$ on $\Gamma _{1/2}$, the same computations lead to:

$$\begin{aligned} \displaystyle \frac{d {S_{2}}}{d t } = \displaystyle \frac{\nu _{S}}{|\Omega _{2}|}\int _{\Gamma _{2/1}} \displaystyle \frac{\partial {S_{L}}}{\partial n_2 }\,ds, \end{aligned}$$

(5.7)

with

$$\begin{aligned} S_{2}(t) = \displaystyle \frac{1}{|\Omega _{2}|}\int _{\Omega _{2}} S_{L}(t,x,y)\,dxdy. \end{aligned}$$

(5.8)

We now consider that the integral of the normal derivative of $S_{L}$ along the interface $\Gamma _{1/2}$ scales like $S_{2}-S_{1}$, that is there exists a positive constant K such that

$$\begin{aligned} \int _{\Gamma _{1/2}} \displaystyle \frac{\partial {S_{L}}}{\partial n_1 }\,ds=K\,(S_{2}-S_{1}). \end{aligned}$$

(5.9)

Finally, considering Eqs. (5.6) and (5.7) in which the integral is replaced by the above approximation, we indeed obtain the two ODEs (2.1c) and (2.1d).