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# Optimal harvesting control for an age-dependent competing population with diffusion

## Abstract

This paper is mainly concerned with optimal harvesting policy for competing species with age dependence and diffusion. The existence and uniqueness of solutions for the system are proved by using the Banach fixed point theorem. The existence and uniqueness of optimal control are discussed by Ekeland’s variational principle. The maximum principle is obtained.

MSC: 35B10, 49K05, 65L12, 92B05.

## 1 Introduction

The optimal harvesting control of age-structured and size-structured single species have been widely studied in the literature –. The control problems of multi-species have been investigated in the literature –. The objective function represents the total harvesting, respectively, in –. The diffusion factor has not been considered in –. It is well known that the profit due to harvesting is a quite important problem for optimal harvesting. In this paper, our purpose is to consider the optimal control of the profit functional for diffusion population. Specifically, we deal with the following optimal harvesting problem:

$sup ∑ i = 1 n ∫ Q [ u i ( a , t , x ) p i u ( a , t , x ) − u i 2 ( a , t , x ) ] dadtdx$
(1.1)

for all $u=( u 1 (a,t,x), u 2 (a,t,x),…, u n (a,t,x))∈U$, where the corresponding state variable $p u =( p 1 u , p 2 u ,…, p n u )$ satisfies the state system

${ ∂ p i ∂ a + ∂ p i ∂ t − k i Δ p i = f i − μ i ( a , t , x ) p i − u i ( a , t , x ) p i ∂ p i ∂ a + ∂ p i ∂ t − k i Δ p i = − ∑ k = 1 , k ≠ i n λ i k ( a , t , x ) P k ( t , x ) p i , ( a , t , x ) ∈ Q , ∂ p i ∂ ν ( a , t , x ) = 0 , ( a , t , x ) ∈ Σ , p i ( 0 , t , x ) = ∫ 0 A β i ( a , t , x ) p i ( a , t , x ) d a , ( t , x ) ∈ Ω T , p i ( a , 0 , x ) = p i 0 ( a , x ) , ( a , x ) ∈ Ω A , P i ( t , x ) = ∫ 0 A p i ( a , t , x ) d a , ( t , x ) ∈ Ω T , i = 1 , 2 , … , n ,$
(1.2)

where $p i (a,t,x)$ represents the density of the $i$th population. $k i$ is the diffusion rate of the $i$th population. We assume that the populations have the same life expectancy $A$, $0. $μ i (a,t,x)$ is the average morality of the $i$th population, and $β i (a,t,x)$ describes the average fertility of the $i$th population. $λ i k (a,t,x)$ represents the interaction coefficients ($i,k=1,2,…,n$, $k≠i$). $Ω⊂ R N$ ($N=1,2,3$) is a bounded domain with a smooth enough boundary $∂Ω$, $Q=(0,A)×(0,T)×Ω$, $Ω T =Ω×(0,T)$, $Ω A =Ω×(0,A)$, $∑=(0,A)×(0,T)×∂Ω$, and $ν$ is the outward unit normal. The function $p i 0 (a,x)$ gives the initial density distribution of the population, and $u i (a,t,x)$ represents the harvesting effort function, which is the control variable in the model and satisfies

where $γ i j ∈ L ∞ (Q)$, $j=1,2$, $i=1,2,…,n$. The integral

$J( u 1 , u 2 ,…, u n )= ∑ i = 1 n ∫ Q [ u i ( a , t , x ) p i u ( a , t , x ) − u i 2 ( a , t , x ) ] dadtdx$
(1.3)

represents the profit due to harvesting.

Throughout this paper, we assume that:

(A1): $β i (a,t,x)∈ L ∞ (Q)$, $0⩽ β i (a,t,x)≤M$, where $M$ is a constant.

(A2): $μ i ∈ L loc ∞ ([0,A]×[0,T]× Ω ¯ )$, $μ i (a,t,x)≥ μ 0 (a,t)≥0$ a.e. in $Q$, $i=1,2,…,n$, where $μ 0 ∈ L loc ∞ ([0,A]×[0,T])$, $∫ 0 A μ 0 (a,t+a−A)da=+∞$.

(A3): $λ i k ∈ L ∞ (Q)$, $0≤ λ i k (a,t,x)≤B$, where $B$ is a positive constant ($i,k=1,2,…,n$, $k≠i$).

(A4): $f i ∈ L ∞ (Q)$, $p i 0 (a,x)∈ L ∞ ( Ω A )$, $p i 0 (a,x)≥0$, $0≤ f i (a,t,x)≤ B 0$, where $B 0$ is a constant, $i=1,2,…,n$.

The rest of this paper is organized as follows. In Section 2, we prove that under the assumptions listed above, the system has a unique non-negative solution. In Section 3, the necessary conditions of optimality for the control problems is given. In the final section, we prove the existence and uniqueness of the optimal control.

## 2 The existence and uniqueness of solution for system (1.2)

For the sake of convenience, we introduce the following definitions of the solution.

### Definition 2.1

Given the solution of system (1.2), the function $p i ∈ L 2 (Q)$, $i=1,2,…,n$, which belongs to $C( S ¯ ; L 2 (Ω))∩AC(S; L 2 (Ω))∩ L 2 (S; H 1 (Ω))∩ L loc 2 (S; H 2 (Ω))$ for almost any characteristic line $S$ of the equation

$a−t= a 0 − t 0 ,(a,t)∈(0,A)×(0,T),( a 0 , t 0 )∈{0}×(0,T)∪(0,A)×{0},$

satisfies

(2.1)

where $P i (t,⋅)= ∫ 0 A p i (a,t,⋅)da$, $i=1,2,…,n$.

Then we rewrite the characteristic line $S$ as

$S= { ( a , t ) ∈ ( 0 , A ) × ( 0 , T ) ; a − t = a 0 − t 0 } = { ( a 0 + s , t 0 + s ) ; s ∈ ( 0 , α ) } ,$

here $( a 0 +α, t 0 +α)∈{A}×(0,T)∪(0,A)×{T}$.

We introduce the following notations:

### Theorem 2.1

For any given$u=( u 1 , u 2 ,…, u n )∈U$, system (1.2) has a unique non-negative solution$p u =( p 1 u , p 2 u ,…, p n u )∈ L 2 (Q; R n )$such that

1. (i)

$0≤ p i u ≤ M 1$, $∀(a,t,x)∈Q$, $i=1,2,…,n$, where $M 1$ is a positive constant;

2. (ii)

$p u$ is continuous in $u$.

### Proof

For any given $h=( h 1 , h 2 ,…, h n )∈ L 2 (Q; R n )$, $h≥0$, we define

$H i (t,x)= ∫ 0 A h i (a,t,x)da,i=1,2,…,n.$

The given system

${ ∂ p i ∂ a + ∂ p i ∂ t − k i Δ p i = f i − μ i ( a , t , x ) p i − u i p i ∂ p i ∂ a + ∂ p i ∂ t − k i Δ p i = − ∑ k = 1 , k ≠ i n λ i k ( a , t , x ) H k ( t , x ) p i , ( a , t , x ) ∈ Q , ∂ p i ∂ ν ( a , t , x ) = 0 , ( a , t , x ) ∈ Σ , p i ( 0 , t , x ) = ∫ 0 A β i ( a , t , x ) p i ( a , t , x ) d a , ( t , x ) ∈ Ω T , p i ( a , 0 , x ) = p i 0 ( a , x ) , ( a , x ) ∈ Ω A , P i ( t , x ) = ∫ 0 A p i ( a , t , x ) d a , ( t , x ) ∈ Ω T , i = 1 , 2 , … , n ,$
(2.2)

by Theorem 4.1.3 in , we know that the above system has a unique non-negative solution,

$p h = ( p 1 h , p 2 h , … , p n h ) ∈ L 2 ( Q ; R n ) , p i h (A,t,x)=0,∀(t,x)∈ Ω T ,i=1,2,…,n.$

From the comparison principle of linear system , it follows that

where $p ¯ i (a,t,x)∈ L ∞ (Q)$, and this is the solution of the following system:

${ ∂ p i ∂ a + ∂ p i ∂ t − k i Δ p i = f i − μ i ( a , t , x ) p i , ( a , t , x ) ∈ Q , ∂ p i ∂ ν ( a , t , x ) = 0 , ( a , t , x ) ∈ Σ , p i ( 0 , t , x ) = ∫ 0 A β i ( a , t , x ) p i ( a , t , x ) d a , ( t , x ) ∈ Ω T , p i ( a , 0 , x ) = p i 0 ( a , x ) , ( a , x ) ∈ Ω A .$

For any $h k =( h 1 k , h 2 k ,…, h n k )∈ L 2 (Q; R n )$, $0≤ h i k ≤ p ¯ i$, let the corresponding state be $p k =( p 1 k , p 2 k ,…, p n k )$ ($k=1,2$), $w=( w 1 , w 2 ,…, w n ):= p 1 − p 2$. It follows from (1.2) that

${ ∂ w i ∂ a + ∂ w i ∂ t − k i Δ w i = − ( μ i + u i ) w i − ∑ k = 1 , k ≠ i n λ i k ( a , t , x ) H k 1 ( t , x ) w i ∂ w i ∂ a + ∂ w i ∂ t − k i Δ w i = − ∑ k = 1 , k ≠ i n λ i k ( a , t , x ) [ H k 1 ( t , x ) − H k 2 ( t , x ) ] p i 2 , ( a , t , x ) ∈ Q , ∂ w i ∂ ν ( a , t , x ) = 0 , ( a , t , x ) ∈ Σ , w i ( 0 , t , x ) = ∫ 0 A β i ( a , t , x ) w i ( a , t , x ) d a , w i ( a , 0 , x ) = 0 , ( a , x ) ∈ Ω A , H i j ( t , x ) = ∫ 0 A h i j ( a , t , x ) d a , ( t , x ) ∈ Ω T , i = 1 , 2 , … , n , j = 1 , 2 .$
(2.3)

Multiplying the $i$th equation in system (2.3) by $w i$ ($i=1,2,…,n$), integrating on $(0,A)×(0,t)×Ω$, and according to the Hölder inequality, we have

$∥ w i ( ⋅ , t , ⋅ ) ∥ L 2 ( Ω A ) 2 ≤ ( A M 2 + B ∥ p ¯ i ∥ L ∞ ( Q ) ) ∫ 0 t ∥ w i ( ⋅ , τ , ⋅ ) ∥ L 2 ( Ω A ) 2 d τ + ∑ k = 1 n A 2 B ∥ p ¯ i ∥ L ∞ ( Q ) ∫ 0 t ∥ h k 1 − h k 2 ∥ L 2 ( Ω A ) 2 d τ = C ˜ ∫ 0 t ( ∑ k = 1 n ∥ h k 1 ( ⋅ , τ , ⋅ ) − h k 2 ( ⋅ , τ , ⋅ ) ∥ L 2 ( Ω A ) 2 + ∥ w i ( ⋅ , τ , ⋅ ) ∥ L 2 ( Ω A ) 2 ) d τ ,$
(2.4)

where $C ˜ = max 1 ≤ i ≤ n {A M 2 +B ∥ p ¯ i ∥ L ∞ ( Q ) , A 2 B ∥ p ¯ i ∥ L ∞ ( Q ) }$. By using the Gronwall inequality, we have

$∥ w i ( ⋅ , t , ⋅ ) ∥ L 2 ( Ω A ) 2 ≤C ∑ k = 1 n ∫ 0 t ∥ h k 1 ( ⋅ , τ , ⋅ ) − h k 2 ( ⋅ , τ , ⋅ ) ∥ L 2 ( Ω A ) 2 dτ,$
(2.5)

where $C= C ˜ e C ˜ T$. Set

$I:= { h = ( h 1 , h 2 , … , h n ) ∈ L 2 ( Q ; R n ) : 0 ≤ h i ( a , t , x ) ≤ p ¯ i ( a , t , x ) , ∀ ( a , t , x ) ∈ Q } .$

Define the mapping $G:I→I$

$(Gh)(a,t,x)= p h (a,t,x),∀(a,t,x)∈Q.$

Denote the norm on $L 2 (Q; R N )$ by

$∥ h ∥ L 2 ( Q ) = ( ∑ i = 1 n ∥ h i ∥ L 2 ( Q ) 2 ) 1 2 .$

Define the following norm for $C>0$:

$∥ h i ∥ ∗ = ∫ 0 T ∥ h i ( ⋅ , t ) ∥ L 2 ( Ω A ) 2 e − 4 n C t dt, ∥ h ∥ ∗ 2 = ( ∑ i = 1 n ∥ h i ∥ ∗ 2 ) 1 2 ,i=1,2,…,n.$

Obviously the norm $∥ ⋅ ∥ L 2 ( Q )$ is equivalent to the norm $∥ ⋅ ∥ ∗$. Using (2.5), we get

$∥ G h 1 − G h 2 ∥ ∗ = ∥ p 1 − p 2 ∥ ∗ = ( ∑ i = 1 n ∥ w i ∥ ∗ 2 ) 1 2 = ( ∑ i = 1 n ∫ 0 T ∥ w i ( ⋅ , t , ⋅ ) ∥ L 2 ( Ω A ) 2 e − 4 n C t d t ) 1 2 ≤ ( ∑ i = 1 n ∫ 0 T C ∑ k = 1 n ∫ 0 t ∥ h k 1 ( ⋅ , τ , ⋅ ) − h k 2 ( ⋅ , τ , ⋅ ) ∥ L 2 ( Ω A ) 2 d τ ⋅ e − 4 n C t d t ) 1 2 = ( ∫ 0 T ∑ k = 1 n ∥ h k 1 ( ⋅ , s , ⋅ ) − h k 2 ( ⋅ , s , ⋅ ) ∥ L 2 ( Ω A ) 2 ⋅ ∫ s T n C e − 4 n C t d t d s ) 1 2 ≤ 1 2 ( ∫ 0 T ∑ k = 1 n ∥ h k 1 ( ⋅ , s , ⋅ ) − h k 2 ( ⋅ , s , ⋅ ) ∥ L 2 ( Ω A ) 2 ⋅ e − 4 n C s d s ) 1 2 = 1 2 ∥ h 1 − h 2 ∥ ∗ .$

It is obvious that $GI⊆I$ and $G$ is a contraction on $(I, ∥ ⋅ ∥ ∗ )$, so there is a unique fixed point, which is the solution of system (1.2).

Since $p ¯ i ∈ L ∞ (Q)$, let $M 1 =max{esssup| p ¯ i (a,t,x)|,1≤i≤n}$, we have $0≤ p i u ≤ M 1$, a.e. $(a,t,x)∈Q$. In the following, we study the continuity of the solution of (1.2) for the control variable $u$. Let $u k =( u 1 k , u 2 k ,…, u n k )∈U$, $k=1,2$, $y i (a,t,x)= p i u 1 (a,t,x)− p i u 2 (a,t,x)$. From (1.2), it follows that

${ ∂ y i ∂ a + ∂ y i ∂ t − k i Δ y i = − ( μ i + u i 1 ) y i ∂ y i ∂ a + ∂ y i ∂ t − k i Δ y i = − ∑ k = 1 , k ≠ i n λ i k p i u 2 ∫ 0 A y k ( a , t , x ) d a − ( u i 1 − u i 2 ) p i u 2 ∂ y i ∂ a + ∂ y i ∂ t − k i Δ y i = − ∑ k = 1 , k ≠ i n λ i k ( a , t , x ) P k u 1 ( t , x ) y i , ( a , t , x ) ∈ Q , ∂ y i ∂ ν ( a , t , x ) = 0 , ( a , t , x ) ∈ Σ , y i ( 0 , t , x ) = ∫ 0 A β i ( a , t , x ) y i ( a , t , x ) d a , ( t , x ) ∈ Ω T , y i ( a , 0 , x ) = 0 , ( a , x ) ∈ Ω A , P i u j ( t , x ) = ∫ 0 A p i u j ( a , t , x ) d a , ( t , x ) ∈ Ω T , i = 1 , 2 , … , n , j = 1 , 2 .$
(2.6)

In a similar manner as that in (2.4), we deduce that

$∥ y i ( ⋅ , τ , ⋅ ) ∥ L 2 ( Ω A ) 2 ≤ ( A M 2 + ∥ p ¯ i ∥ L ∞ ( Q ) + B ∥ p ¯ i ∥ L ∞ ( Q ) ) ∫ 0 t ∥ y i ( ⋅ , τ , ⋅ ) ∥ L 2 ( Ω A ) 2 d τ + ∥ p ¯ i ∥ L ∞ ( Q ) ∫ 0 t ∥ u i 1 − u i 2 ∥ L 2 ( Ω A ) 2 d τ + 2 ∑ k = 1 , k ≠ i n A B ∥ p ¯ k ∥ L ∞ ( Q ) ∫ 0 t ∫ Ω A y i 2 d x d a d τ .$

Let $C ¯ =A M 2 + ∥ p ¯ i ∥ L ∞ ( Q ) +B ∥ p ¯ i ∥ L ∞ ( Q ) +2AB ∑ k = 1 , k ≠ i n ∥ p ¯ k ∥ L ∞ ( Q )$, we have

$∥ y i ( ⋅ , t , ⋅ ) ∥ L 2 ( Ω A ) 2 ≤ C ¯ ( ∫ 0 t ∥ u i 1 ( ⋅ , τ , ⋅ ) − u i 2 ( ⋅ , τ , ⋅ ) ∥ L 2 ( Ω A ) 2 d τ + ∫ 0 t ∥ y i ( ⋅ , τ , ⋅ ) ∥ L 2 ( Ω A ) 2 d τ ) .$
(2.7)

Adding up (2.7) from $i=1$ to $n$ and using the Gronwall inequality, we get

$∑ i = 1 n ∥ y i ( ⋅ , t , ⋅ ) ∥ L 2 ( Ω A ) 2 ≤K ∑ i = 1 n ∫ 0 t ∥ u i 1 ( ⋅ , τ , ⋅ ) − u i 2 ( ⋅ , τ , ⋅ ) ∥ L 2 ( Ω A ) 2 dτ,$
(2.8)

where $K= C ¯ e C ¯ T$. Multiplying (2.8) by $e − n r t$ and integrating on $(0,T)$, we have

$∥ p 1 − p 2 ∥ ∗ ≤ K T ∥ u 1 − u 2 ∥ ∗ .$
(2.9)

Inequality (2.9) implies that $p u$ is continuous with respect to $u$. This completes the proof. □

## 3 The necessary conditions

Before stating our main results, we prove the following lemma, which is useful in proving our results.

### Lemma 3.1

Let$( u ∗ , p ∗ )$be an optimal pair for Problem (1.1), for any$0<ε<1$and for any$v=( v 1 , v 2 ,…, v n )∈ L 2 (Q; R n )$, $v i >0$, $u ∗ +ε(v− u ∗ )= u ε ∈U$, where$u ε =( u 1 ε , u 2 ε ,…, u n ε )$, then the following limit holds:

where$p ∗ =( p 1 u ∗ , p 2 u ∗ ,…, p n u ∗ )$, $u ∗ =( u 1 ∗ , u 2 ∗ ,…, u n ∗ )$, and$z=( z 1 , z 2 ,…, z n )$is the solution of

${ ∂ z i ∂ a + ∂ z i ∂ t − k i Δ z i = − ( μ i ( a , t , x ) + u i ∗ ) z i − ( v i − u i ∗ ) p i u ∗ ∂ z i ∂ a + ∂ z i ∂ t − k i Δ z i = − ∑ k = 1 , k ≠ i n λ i k ( a , t , x ) Z k ( t , x ) p i u ∗ ∂ z i ∂ a + ∂ z i ∂ t − k i Δ z i = − ∑ k = 1 , k ≠ i n λ i k P k u ∗ ( t , x ) z i , ( a , t , x ) ∈ Q , ∂ z i ∂ ν ( a , t , x ) = 0 , ( a , t , x ) ∈ Σ , z i ( 0 , t , x ) = ∫ 0 A β i ( a , t , x ) z i ( a , t , x ) d a , ( t , x ) ∈ Ω T , z i ( a , 0 , x ) = 0 , ( a , x ) ∈ Ω A , P i u ∗ ( t , x ) = ∫ 0 A p i u ∗ ( a , t , x ) d a , ( t , x ) ∈ Ω T , Z i ( t , x ) = ∫ 0 A z i ( a , t , x ) d a , ( t , x ) ∈ Ω T , i = 1 , 2 , … , n .$
(3.1)

### Proof

The existence and uniqueness of the solution of (3.1) follow in the same manner as by Theorem 2.1, the solution $z=( z 1 , z 2 ,…, z n )∈ L 2 (Q; R n )$, and $0≤ z i ≤L$, where $L$ is a constant. We introduce the following notations:

$z ε ( a , t , x ) = 1 ε [ p u ε ( a , t , x ) − p u ∗ ( a , t , x ) ] , ( a , t , x ) ∈ Q , x ε ( a , t , x ) = ε z ε ( a , t , x ) , ( a , t , x ) ∈ Q .$

It is straightforward that $z ε$ is the solution of

${ ∂ z i ε ∂ a + ∂ z i ε ∂ t − k i Δ z i ε = − ( μ i + u i ∗ ) z i ε − ( v i − u i ∗ ) p i u ε ∂ z i ε ∂ a + ∂ z i ε ∂ t − k i Δ z i ε = − ∑ k = 1 , k ≠ i n λ i k ( a , t , x ) Z k ε ( t , x ) p i u ε ∂ z i ε ∂ a + ∂ z i ε ∂ t − k i Δ z i ε = − ∑ k = 1 , k ≠ i n λ i k ( a , t , x ) P k u ∗ ( t , x ) z i ε , ( a , t , x ) ∈ Q , ∂ z i ε ∂ ν ( a , t , x ) = 0 , ( a , t , x ) ∈ Σ , z i ε ( 0 , t , x ) = ∫ 0 A β i ( a , t , x ) z i ε ( a , t , x ) d a , ( t , x ) ∈ Ω T , z i ( a , 0 , x ) = 0 , ( a , x ) ∈ Ω A ,$
(3.2)

where $P i u ∗ (t,x)= ∫ 0 A p i u ∗ (a,t,x)da$, $Z i ε (t,x)= ∫ 0 A z i ε (a,t,x)da$, $(t,x)∈ Ω T$.

By (2.9), we have

$∥ x ε ∥ ∗ = ∥ p u ε − p u ∗ ∥ ∗ ≤ K T ∥ ε ( v − u ∗ ) ∥ ∗ = K T ε ∥ v − u ∗ ∥ ∗ .$
(3.3)

So we can claim that $x ε →0$ in $L 2 (Q; R n )$, as $ε→ 0 +$. By (3.1) and (3.2), we see that $z ε −z=( z 1 ε − z 1 , z 2 ε − z 2 ,…, z n ε − z n )∈ L 2 (Q; R n )$ is the solution of the following equation:

${ ∂ ( z i ε − z i ) ∂ a + ∂ ( z i ε − z i ) ∂ t − k i Δ ( z i ε − z i ) = − μ i ( z i ε − z i ) − u i ∗ ( z i ε − z i ) − ( v i − u i ∗ ) ( p i u ε − p i u ∗ ) − ∑ k = 1 , k ≠ i n λ i k ( ∫ 0 A ( z k ε − z k ) d a ) p i u ε − ∑ k = 1 , k ≠ i n λ i k ( a , t , x ) Z k ( t , x ) ( p i u ε − p i u ∗ ) + ∑ k = 1 , k ≠ i n λ i k ( ∫ 0 A ( p k u ε − p k u ∗ ) d a ) z i ε − ∑ k = 1 , k ≠ i n λ i k ( ∫ 0 A p k u ∗ ( a , t , x ) d a ) ( z i ε − z i ) , ( a , t , x ) ∈ Q , ∂ ( z i ε − z i ) ∂ v ( a , t , x ) = 0 , ( a , t , x ) ∈ Σ , ( z i ε − z i ) ( 0 , t , x ) = ∫ 0 A β i ( a , t , x ) ( z i ε − z i ) d a , ( t , x ) ∈ Ω T , z i ( a , 0 , x ) = 0 , ( a , x ) ∈ Ω A , i = 1 , 2 , … , n ,$
(3.4)

where $P i u ∗ (t,x)= ∫ 0 A p i u ∗ (a,t,x)da$, $Z i (t,x)= ∫ 0 A z i (a,t,x)da$, $(t,x)∈ Ω T$.

Multiplying the first equation in (3.4) by $z i ε − z i$, integrating on $(0,A)×(0,t)×Ω$, using the Gronwall inequality, in the same manner as in (2.5), and taking $x ε →0$ in $L 2 (Q; R n )$, we deduce

Consequently, the proof is completed. By the same argument as in (2.9), the following lemma holds. □

In order to give the necessary conditions, we introduce the adjoint equations of (3.1).

### Lemma 3.2

If$q u 1$, $q u 2$are the solutions of the following system corresponding to the control$u 1$, $u 2$, respectively:

${ ∂ q j ∂ a + ∂ q j ∂ t + k j Δ q j = ( μ j + u j ∗ ) q j + u j ∗ − q j ( 0 , t , x ) β j ( a , t , x ) ∂ q j ∂ a + ∂ q j ∂ t + k j Δ q j = + ∑ k = 1 , k ≠ j n λ j k ( a , t , x ) q j P k u ∗ ( t , x ) ∂ q j ∂ a + ∂ q j ∂ t + k j Δ q j = + ∑ k = 1 , k ≠ j n ∫ 0 A λ k j ( a , t , x ) p k u ∗ q j ( a , t , x ) d a , ( a , t , x ) ∈ Q , ∂ q j ∂ ν ( a , t , x ) = 0 , ( a , t , x ) ∈ Σ , q j ( a , T , x ) = q j ( A , t , x ) = 0 , ( a , x ) ∈ Ω A , P j u ∗ ( t , x ) = ∫ 0 A p j u ∗ ( a , t , x ) d a , ( t , x ) ∈ Ω T , j = 1 , 2 , … , n .$
(3.5)

Then

$∥ q u 1 − q u 2 ∥ ∗ ≤ K T ∥ u 1 − u 2 ∥ ∗ , ∥ q u ∥ ∗ ≤ M 2 ,$

where$q u 1 =( q 1 u 1 , q 2 u 1 ,…, q n u 1 )$, $q u 2 =( q 1 u 2 , q 2 u 2 ,…, q n u 2 )$, $u 1 =( u 1 1 , u 2 1 ,…, u n 1 )$, $u 2 =( u 1 2 , u 2 2 ,…, u n 2 )$, $K ¯ >0$, $M 2$are constant and independent of$q u 1$, $q u 2$, $u 1$, $u 2$.

We are now able to state the main result in this section.

### Theorem 3.1

Suppose that (A1)-(A5) hold, if$( u ∗ , p ∗ )$is an optimal pair for Problem (1.1), and$q=( q 1 , q 2 ,…, q n )$is the solution of (3.5), which is corresponding to the control$u ∗ =( u 1 ∗ , u 2 ∗ ,…, u n ∗ )$, then we have the following necessary conditions:

$∑ i = 1 n ∫ 0 A ∫ 0 T ∫ Ω ( v i − u i ∗ ) [ ( 1 + q i ) p i u ∗ − 2 u i ∗ ] dxdtda≤0.$

### Proof

The existence and uniqueness of the solution $q=( q 1 , q 2 ,…, q n )$ to (3.5) can be proved by Theorem 2.1. For any $v i ∈ L 2 (Q)$ and any $0<ε<1$, $u i ∗ +ε( v i − u i ∗ )∈ U i$. Since $u ∗ =( u 1 ∗ , u 2 ∗ ,…, u n ∗ )$ is an optimal control for (1.1), we get

$∑ i = 1 n ∫ Q [ u i ∗ p i u ∗ − ( u i ∗ ) 2 ] dadtdx≥ ∑ i = 1 n ∫ Q [ u i ε p i u ε − u i ε 2 ] dadtdx,$

which implies

$∑ i = 1 n ∫ 0 A ∫ 0 T ∫ Ω [ u i ∗ p i u ε − p i u ∗ ε + ( v i − u i ∗ ) ( p i u ε − 2 u i ∗ − ε ( v i − u i ∗ ) ) ] dxdtda≤0.$
(3.6)

By Lemma 3.1 and inequality (3.3), passing to the limit as $ε→ 0 +$ in (3.6), we conclude

$∑ i = 1 n ∫ 0 A ∫ 0 T ∫ Ω [ u i ∗ z i + ( v i − u i ∗ ) ( p i u ∗ − 2 u i ∗ ) ] dxdtda≤0.$
(3.7)

Multiplying (3.5) by $z i$, then integrating over $Q$, we deduce that

$∑ i = 1 n ∫ 0 A ∫ 0 T ∫ Ω ( z i u i ∗ ) ( a , t , x ) d x d t d a = ∑ i = 1 n ∫ 0 A ∫ 0 T ∫ Ω q i ( v i − u i ∗ ) p i u ∗ ( a , t , x ) d x d t d a .$
(3.8)

Combining (3.7) with (3.8), we have

$∑ i = 1 n ∫ 0 A ∫ 0 T ∫ Ω ( v i − u i ∗ ) [ ( 1 + q i ) p i u ∗ − 2 u i ∗ ] dxdtda≤0.$
(3.9)

This completes the proof. □

## 4 The existence and uniqueness of optimal control

In this section, we prove the existence and uniqueness of optimal control. From , we give the following lemma.

### Lemma 4.1

Let$X$be a reflexive Banach space and let$φ:X→(−∞,+∞]$be a lower semi-continuous convex function. If$X 0$is a closed, convex, and bounded subset of$X$, then$φ$attains its infimum on$X 0$. In other words, then there is$x 0 ∈ X 0$such that

$φ( x 0 )=inf { φ ( x ) , x ∈ X 0 } .$

The main conclusion is presented as follows:

### Theorem 4.1

If$T$is small enough, then Problem (1.1) has a unique optimal control in$U$.

### Proof

For any $u=( u 1 , u 2 ,…, u n )$, $v=( v 1 , v 2 ,…, v n )$, $u i ∈ U i$, $v i ∈ U i$, $i=1,2,…,n$, we define

$H(ε)=J ( ε u + ( 1 − ε ) v ) ,$

we shall prove that $H ′ (ε)$ is strictly monotone. Indeed, denote by $( p 1 ε , p 2 ε ,…, p n ε )$ and $( p 1 ε + δ , p 2 ε + δ ,…, p n ε + δ )$ the state corresponding to controls $ε( u 1 , u 2 ,…, u n )+(1−ε)( v 1 , v 2 ,…, v n )$ and $(ε+δ)( u 1 , u 2 ,…, u n )+(1−(ε+δ))( v 1 , v 2 ,…, v n )$, respectively.

By (1.3), we have

$H ′ ( ε ) = ∑ i = 1 n lim δ → 0 1 δ [ J ( ( ε + δ ) u i + ( 1 − ε − δ ) v i ) − J ( ε u i + ( 1 − ε ) v i ) ] = lim δ → 0 ∑ i = 1 n 1 δ ∫ Q [ ( p i ε + δ − p i ε ) ( v i + ( ε + δ ) ( u i − v i ) ) + p i ε δ ( u i − v i ) ] d a d t d x − lim δ → 0 ∑ i = 1 n 1 δ ∫ Q [ δ 2 ( u i − v i ) 2 + 2 ( v i + ε ( u i − v i ) ) δ ( u i − v i ) ] d a d t d x = ∑ i = 1 n ∫ Q [ z i ε ( v i + ε ( u i − v i ) ) + p i ε ( u i − v i ) ] d a d t d x − ∑ i = 1 n ∫ Q 2 ( v i + ε ( u i − v i ) ( u i − v i ) ) d a d t d x ,$
(4.1)

where $z ε =( z 1 ε , z 2 ε ,…, z n ε )$ is the solution of Problem (3.1) corresponding to the control $v i +ε( u i − v i )$. By the same argument as given in (3.8), we have

$∫ Q z i ε ( v i + ε ( u i − v i ) ) dadtdx= ∫ Q p i ε q i ε ( u i − v i )dadtdx.$
(4.2)

Combining (4.1) and (4.2), we obtain

$H ′ (ε)= ∑ i = 1 n ∫ Q ( u i − v i ) [ p i ε ( 1 + q i ε ) − 2 ( v i + ε ( u i − v i ) ) ] dtdtdx,$
(4.3)

where $( q 1 ε , q 2 ε )$ is the solution of Problem (3.5) corresponding to the control $v 1 +ε( u 1 − v 1 )$ ($v 2 +ε( u 2 − v 2 )$).

Given $ε,ρ∈(0,1)$, and $ε≠ρ$, and the norm $∥ ⋅ ∥ L 2 ( Q )$ being equivalent to the norm $∥ ⋅ ∥ ∗$, combining Theorem 2.1 and Lemma 3.1, we have

$[ H ′ ( ε ) − H ′ ( ρ ) ] ( ε − ρ ) = ( ε − ρ ) ∑ i = 1 n ∫ Q ( u i − v i ) [ p i ε ( 1 + q i ε ) − p i ρ ( 1 + q i ρ ) ] d a d t d x − 2 ( ε − ρ ) 2 ∑ i = 1 n ∫ Q ( u i − v i ) 2 d a d t d x = ( ε − ρ ) ∑ i = 1 n ∫ Q [ ( u i − v i ) ( p i ε − p i ρ ) + ( u i − v i ) ( p i ε q i ε − p i ρ q i ρ ) ] d a d t d x − 2 ( ε − ρ ) 2 ∑ i = 1 n ∫ Q ( u i − v i ) 2 d a d t d x ≤ ( ε − ρ ) 2 K T ∑ i = 1 n ∥ u i − v i ∥ L 2 ( Q ) 2 + ( ε − ρ ) 2 M 2 K T ∑ i = 1 n ∥ u i − v i ∥ L 2 ( Q ) 2 + ( ε − ρ ) 2 M 1 K ¯ T ∑ i = 1 n ∥ u i − v i ∥ L 2 ( Q ) 2 − 2 ( ε − p ) 2 ∑ i = 1 n ∥ u i − v i ∥ L 2 ( Q ) 2 ,$

where $T$ is small enough. If $∥ u i − v i ∥ L 2 ( Q ) ≠0$, then we have

$[ H ′ ( ε ) − H ′ ( ρ ) ] (ε−ρ)≤− ( ε − ρ ) 2 ∥ u i − v i ∥ L 2 ( Q ) 2 <0.$

Hence, $H ′ (ε)$ is strictly monotone. Consequently, $J(u)$ is strictly concave in $U$.

Define the function $Φ: L 2 (Q; R N )→(−∞,+∞)$:

It is clear that $Φ$ is concave in $L 2 (Q; R N )$. It follows that the nonlinear function $Φ: L 2 (Q; R N )→$$(−∞,+∞)$ is upper weakly semi-continuous. Since $U$ is a closed, convex, and bounded subset in $L 2 (Q; R N )$ and $J(u)$ is strictly convex in $U$, by Lemma 4.1, $J(u)$ attains the unique maximum in $U$, which implies Problem (1.1) has a unique optimal control in $U$. □

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## Acknowledgements

The authors would like to thank the anonymous referees for their valuable comments on and suggestions regarding the original manuscript. The project is supported by Graduate Student Innovation of Jilin Normal University (201114) and the Department of Education of Jilin Province (2013445).

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