Marine Reserve Design with Ocean Currents and Multiple Objectives

Abstract

There is a growing tendency to consider marine reserves as a management and conservation tool. We investigate a spatial bio-economic model to determine fishery profits under conservation efforts. Rather than imposing a marine reserve on our model, we ask, “When and where should a marine reserve be implemented ?” For one-dimensional habitat, we determine conditions under which marine reserves emerge as a part of the optimal policy. Depending upon the size of the habitat, the optimal strategy is either to avoid fishing or to fish at maximum rate. The effect of ocean currents is analyzed through numerical simulations. We find that in the presence of strong currents, a marine reserve may become ineffective. If the currents are at low rate, a marine reserve could emerge as a variable management tool.

Appendices

Appendix A: Proof of Theorem 2.1

To this end, we consider the control set

$$F=\left\{ E\in L_{\infty }\left( \left] 0,L\right[ \right) :0\leq E\leq E_{\max }\right\} $$

By standard elliptic theory, for every $E\in L_{\infty }$, problem (M) has a unique positive solution u $\in W^{2,p}\left ( \left ] 0,L\right [ \right )$, ∀p>1. Moreover, there exists a positive constant M ^∗ = M ^∗(L,r,k,μ)>0 such that $\left \Vert u\right \Vert _{W^{2,p}}\leq M^{\ast };$ see for instance [31] and [53]. Since J(E)≤C, where C is a constant, we can choose a maximizing sequence $\left \{ E_{n}\right \} \subset F$ such that

$$J\left( E_{n}\right) \rightarrow {\sup J(E).} $$

For every $E_{n}\in L_{\infty }$, problem (M) has a unique positive solution u _n $\in W^{2,p}\left ( \left ] 0,L\right [ \right )$, ∀p>1. Using u _n as a test function in (M), we obtain

$$\left\Vert u_{n}\right\Vert_{{H_{0}^{1}}}\leq C, $$

for some positive constant C. Since ${H_{0}^{1}}\subset L^{2}$, with compact injection, then after passing to a subsequence, there exists u ^∗ such that $u_{n}\rightarrow$ u ^∗ strongly in $L^{2}\left ( \left ] 0,L \right [ \right )$. Note that $\left \Vert u_{n}\right \Vert _{W^{2,p}}\leq M^{\ast }$ and from the compact injection $W^{2,p}\left ( \left ] 0,L\right [ \right ) \subset C^{0}\left ( \left [ 0,L\right ] \right )$ for p large enough, it follows that $\left \Vert u^{\ast }\right \Vert _{\infty }$ ≤M ^∗. Similarly, since E _n is uniformly bounded in L ², then there exists a subsequence such that $E_{n}\rightharpoonup$ E ^∗, weakly. We need to prove that

$$J\left( E^{\ast }\right) ={\sup J(E)}. $$

The quantity $J\left ( E_{n}\right )$ can be written

$$\begin{array}{@{}rcl@{}} J\left( E_{n}\right) &=&{\int\limits_{0}^{L}}pE_{n}(x)u_{n}(x)dx+Q{\int\limits_{0}^{L}}u_{n}(x)dx \\ &=&{\int\limits_{0}^{L}}pE_{n}(x)u_{n}(x)dx-{\int\limits_{0}^{L}}pE_{n}(x)u^{ \ast }(x)dx\\ &&+{\int\limits_{0}^{L}}pE_{n}(x)u^{\ast }(x)dx+Q{\int\limits_{0}^{L}}u_{n}(x)dx, \end{array} $$

hence

$$\begin{array}{@{}rcl@{}} &&\left\vert J\left( E_{n}\right) -J\left( E^{\ast }\right) \right\vert \leq \left( pE_{\max }+Q\right) {\int\limits_{0}^{L}}\\ &&\left\vert u_{n}-u^{\ast }\right\vert dx+M^{\ast }{\int\limits_{0}^{L}}\left\vert E_{n}-E^{\ast }\right\vert dx. \end{array} $$

We deduce that

$$J\left( E^{\ast }\right) =\sup J(E). $$

Appendix B: Proof of Lemma 2.1

We follow [29], by contradiction, assuming, that there exist an interval $[a ; b]\subset [ 0; L]$ such as

$$ \lambda_{2}\left( x\right) =-p\text{ \ \ \ on }[a;b]. $$

(7.2)

This implies that

$$\frac{d\lambda_{2}}{dx}=0. $$

From the second equation of $\left ( S\right )$, we obtain

$$-\lambda_{1}\left( x\right) +kp=0. $$

which gives that

$$ \lambda_{1}\left( x\right) =-kp. $$

(7.3)

Then

$$ \frac{d\lambda_{1}\left( x\right) }{dx}=0. $$

(7.4)

Hence,

$$\left\{ \begin{array}{l} \frac{d\lambda_{1}\left( x\right) }{dx}=0\text{ \ \ \ on }[a;b], \\ \frac{d\lambda_{2}\left( x\right) }{dx}=0\text{ \ \ \ on }[a;b]. \end{array} \right. $$

It means that $\left ( \lambda _{1},\lambda _{2}\right )$ is at equilibrium $\left ( -kp,-p\right )$ on [a;b].

Consequently,

$$\lambda_{2}\left( x\right) =-p,\text{ \ \ for all }x\in [ 0;L]\text{,} $$

this constitutes a contradiction with the transversality condition.

We conclude that for each x∈[0;L], $\lambda _{2}\left ( x\right ) \neq -p$.

Appendix C: Proof of Lemma 3.1

• (i) Transversality conditions implies that $\frac {d\lambda _{2}^{{}}}{dx} \left ( 0\right ) =-\lambda _{1}\left ( 0\right )$. This gives that

  $$\left\{ \begin{array}{c} s_{1}c_{1}+s_{2}c_{2}=-\lambda_{1}\left( 0\right) , \\ c_{1}+c_{2}=\frac{pE_{\max }+Q}{\mu +E_{\max }}. \end{array} \right. $$

It follows that

$$\left\{ \begin{array}{l} c_{1}=-\frac{\lambda_{1}\left( 0\right) +s_{2}\left( \frac{pE_{\max }+Q}{ \mu +E_{\max }}\right) }{s_{1}-s_{2}}, \\ c_{2}=\frac{\lambda_{1}\left( 0\right) +s_{1}\left( \frac{pE_{\max }+Q}{\mu +E_{\max }}\right) }{s_{1}-s_{2}}. \end{array} \text{ }\right. $$

Therefore, the function

$$\begin{array}{@{}rcl@{}} \lambda_{2}\left( x\right) =&-&\frac{\lambda_{1}\left( 0\right) +s_{2}\left( \frac{pE_{\max }+Q}{\mu +E_{\max }}\right) }{s_{1}-s_{2}}e^{s_{1}x}\\&+&\frac{ \lambda_{1}\left( 0\right) +s_{1}\left( \frac{pE_{\max }+Q}{\mu +E_{\max }} \right) }{s_{1}-s_{2}}e^{s_{2}x}\\&-&\left( \frac{pE_{\max }+Q}{\mu +E_{\max }} \right) , \end{array} $$

satisfies

$$\lambda_{2}\left( 0\right) =0. $$

Now, it is clear that the transversality condition

$$\lambda_{2}\left( L\right) =0, $$

will be satisfied if and only if

$$\lambda_{1}(0)\,=\,{\lambda_{1}^{0}}:=\!\frac{\frac{pE_{\max }+Q}{\mu +E_{\max }}}{ e^{s_{2}L}-e^{s_{1}L}}\left[ s_{1}\left( 1\,-\,e^{s_{2}L}\right) \,+\,s_{2}\left( \!-1\,+\,e^{s_{1}L}\right)\! \right] . $$

Note that

$$\lambda_{1}(0)=-\frac{pE_{\max }+Q}{\mu +E_{\max }}\left[ \alpha s_{1}+\beta s_{2}\right] , $$

with

$$0<\alpha =\frac{(e^{s_{2}T}-1)}{e^{s_{2}T}-e^{s_{1}T}}<1\text{ and }0<\beta = \frac{(1-e^{s_{1}T})}{e^{s_{2}T}-e^{s_{1}T}}<1, $$

and

$$\alpha +\beta =1. $$

Hence,

$$-s_{2}\left( \frac{p+E_{\max }Q}{\mu +E_{\max }}\right) <\lambda_{1}\left( 0\right) <-s_{1}\left( \frac{pE_{\max }+Q}{\mu +E_{\max }}\right) . $$

Let

$$\begin{array}{@{}rcl@{}} \lambda_{1}\left( x\right) &=&{\lambda_{1}^{0}}-\left( \mu +E_{\max }\right)\left( \frac{c_{1}}{s_{1}}e^{s_{1}x}+\frac{c_{2}}{s_{2}}e^{s_{2}x}-Qx\right)\\ &&+\left( \mu +E_{\max }\right) \left( \frac{c_{1}}{s_{1}}+\frac{c_{2}}{s_{2}} \right) , \end{array} $$

then $\left ( \lambda _{1},\lambda _{2}\right )$ is the desired solution.

• (ii) The stable manifold is a line through P _λ with slope $\frac {1}{ s_{2}}$ and unstable manifold is a line through P _λ with slope $\frac {1}{s_{1}}.\$The intercept of stable manifold with λ ₁-axis is the point $-s_{1}\left ( \frac {pE_{\max }+Q}{\mu +E_{\max }}\right )$ . The intercept of unstable manifold with λ ₁-axis is the point $-s_{2}\left ( \frac {pE_{\max }+Q}{\mu +E_{\max }}\right ) .\$The region that lies above the stable and unstable manifold is positively invariant. The lowest point of this region is the equilibrium point $P_{\lambda }=\left ( \lambda _{1}^{\ast },\lambda _{2}^{\ast }\right )$. It follows that when $-s_{2}\left ( \frac {pE_{\max }+Q}{\mu +E_{\max }}\right )$ ≤ λ ₁(0)$\leq -s_{1}\left ( \frac {pE_{\max }+Q}{\mu +E_{\max }}\right ),$ then $\lambda _{2}(x)\geq \lambda _{2}^{\ast }$, for all $x\in \left [ 0,L\right ]$. In order to establish the existence of a minimum, we write λ ₂ as a function of λ ₁, λ ₂ = f(λ ₁). We do not know the function f, but we can deduce its variation using the adjoint system. For the variation, we have

  $$\frac{d\lambda_{2}}{d\lambda_{1}}=\frac{k\lambda_{2}+\lambda_{1}}{ pE_{\max }+Q+\lambda_{2}\left( \mu +E_{\max }\right) }, $$

  it vanishes on the line

  $$\lambda_{2}=\frac{-1}{k}\lambda_{1} $$

  It is clear that$\frac {d\lambda _{2}}{d\lambda _{1}}>0$ if $\lambda _{2}> \frac {-1}{k}\lambda _{1}$ and $\frac {d\lambda _{2}}{d\lambda _{1}}<0$ if $\lambda _{2}<\frac {-1}{k}\lambda _{1}$. Moreover,

  $$\frac{d^{2}\lambda_{2}}{d{\lambda_{1}^{2}}}=\frac{\left\{ k(k\lambda_{2}+\lambda_{1})\left( pE_{\max }+Q+\lambda_{2}\left( \mu +E_{\max }\right) \right) +\left( pE_{\max }+Q+\lambda_{2}\left( \mu +E_{\max }\right) \right)^{2}-\left( \mu +E_{\max }\right) \left( k\lambda_{2}+\lambda_{1}\right)^{2}\right\} }{\left( pE_{\max }+Q+\lambda_{2}\left( \mu +E_{\max }\right) \right)^{2}}, $$

  when k λ ₂ + λ ₁=0, then

  $$\frac{d^{2}\lambda_{2}}{d{\lambda_{1}^{2}}}=1>0, $$

  and the function is convex in a neighborhood of the minimum.

• (iii) From the relation

  $$\begin{array}{@{}rcl@{}} \lambda_{2}\left( x\right) =&-&\frac{\lambda_{1}\left( 0\right) +s_{2}\left( \frac{pE_{\max }+Q}{\mu +E_{\max }}\right) }{s_{1}-s_{2}}e^{s_{1}x}\\&+&\frac{ \lambda_{1}\left( 0\right) +s_{1}\left( \frac{pE_{\max }+Q}{\mu +E_{\max }} \right) }{s_{1}-s_{2}}e^{s_{2}x}\\&-&\frac{pE_{\max }+Q}{\mu +E_{\max }}, \end{array} $$

  it follows that

  $$\begin{array}{@{}rcl@{}} \frac{\lambda_{2}\left( x\right) }{dx}=&-&\frac{\lambda_{1}\left( 0\right) +s_{2}\left( \frac{pE_{\max }+Q}{\mu +E_{\max }}\right) }{s_{1}-s_{2}} s_{1}e^{s_{1}x}\\&+&\frac{\lambda_{1}\left( 0\right) +s_{1}\left( \frac{ pE_{\max }+Q}{\mu +E_{\max }}\right) }{s_{1}-s_{2}}s_{2}e^{s_{2}x}. \end{array} $$

The minimum is achieved when

$$\frac{\lambda_{2}\left( x\right) }{dx}=0. $$

This implies that

$$e^{\left( s_{2}-s_{1}\right) x}=\left( \frac{\lambda_{1}\left( 0\right) +s_{2}\left( \frac{pE_{\max }+Q}{\mu +E_{\max }}\right) }{\lambda_{1}\left( 0\right) +s_{1}\left( \frac{pE_{\max }+Q}{\mu +E_{\max }}\right) }\right) \frac{s_{1}}{s_{2}}, $$

and

$$x^{\ast }=\frac{1}{\left( s_{2}-s_{1}\right) }\ln \left( \frac{\lambda_{1}\left( 0\right) +s_{2}\left( \frac{pE_{\max }+Q}{\mu +E_{\max }}\right) }{\lambda_{1}\left( 0\right) +s_{1}\left( \frac{pE_{\max }+Q}{\mu +E_{\max } }\right) }\times \frac{s_{1}}{s_{2}}\right) , $$

and the minimal value is

$$\lambda_{2}\left( x^{\ast }\right) \,=\,e^{s_{2}x^{\ast }}\!\!\left( \! \frac{\lambda_{1}\left( 0\right) \,+\,s_{1}\left( \frac{pE_{\max }+Q}{\mu +E_{\max }}\right) }{s_{1}}\right) \!-\left( \frac{pE_{\max }+Q}{\mu +E_{\max }}\right) . $$

Appendix D: Proof of Corollary 4.1

Since the equilibrium $P_{\lambda }=\left ( \lambda _{1}^{\ast },\lambda _{2}^{\ast }\right )$ is a saddle point, then $\lambda _{2}^{\min }>\lambda _{2}^{\ast }$ . From the expression of $\lambda _{2}^{\ast }$, it is clear that $\lambda _{2}^{\ast }$ ≥−p. Hence $\lambda _{2}^{\min }>-p$.

Appendix E: Proof of Corollary 4.2

Let $\left ( \lambda _{1},\lambda _{2}\right )$ be a solution of(S) satisfying $\lambda _{2}\left ( 0\right ) =\lambda _{2}\left ( L\right ) =0$ and $\left ( \omega _{1},\omega _{2}\right )$ be a solution of (S) with k=0, and satisfying $\omega _{2}\left ( 0\right ) =\omega _{2}\left ( L\right ) =0$. We obtain that

$$\frac{d\lambda_{2}}{dx}\leq \frac{d\omega_{2}}{dx}. $$

Then by comparison principle, we deduce that

$$\lambda_{2}(x)\leq \omega_{2}(x)\text{ for all }x\in \left[ 0,L\right] . $$

From theorem 3 and inequality 34, in [29], it follows that there exists a real value L _{c r i} such that if L>L _{c r i} , the solution corresponding to k=0, hits the switching line ω ₂=−p for some $x\in \left [ 0,\frac {L}{2}\right ]$. This implies that $\lambda _{2}^{\min }<-p$.