Bilinear Optimal Control for a Nonlinear Parabolic Equation Involving Nonlocal-in-Time Term

Abstract

We study a bilinear optimal control problem for an evolution equation with a nonlinear term that depends on both the state and its time integral. First, we establish existence and uniqueness results for this evolution equation. Then, we derive weak maximum principle results to improve the regularity of the state equation. We proceed by formulating an optimal control problem aimed at steering the system’s state to a desired final state. Finally, we demonstrate that this optimal control problem admits a solution and derive the first- and second-order optimality conditions.

1. Introduction

We denote by $\mathrm{\Omega }$ a bounded open set of ${\mathbb{R}}^d,d=2,3$ with a Lipschitz boundary $\mathrm{\Gamma }$. Also, let $\omega$ be an open subset of $\mathrm{\Omega }$. For any $T>0$, we set $Q=\mathrm{\Omega }\times (0,T)$, $\mathrm{\Sigma }=\mathrm{\Gamma }\times (0,T)$ and ${\omega }_T=\omega \times (0,T)$. Then, we turn our attention to the optimal control problem:

$$\begin{array}{c}\hfill {\displaystyle minJ(y,v),}\end{array}$$

(1)

subject to the constraints that y solves the nonlinear and nonlocal-in-time parabolic

$$\left\{\begin{array}{ccccc}\hfill {\displaystyle \frac{\partial y}{\partial t}-\Delta y+\beta \left({\int }_0^{t}g\left(s\right)y(x,s)ds\right)y}& =\hfill & {\displaystyle v{\chi }_{\omega }y}\hfill & \mathrm{in}\hfill & Q=\mathrm{\Omega }\times (0,T),\hfill \\ \hfill {\displaystyle \frac{\partial y}{\partial \nu }}& =\hfill & 0\hfill & \mathrm{on}\hfill & \mathrm{\Sigma }=\mathrm{\Gamma }\times (0,T),\hfill \\ \hfill {\displaystyle y(·,0)}& =\hfill & {\displaystyle y^0}\hfill & \mathrm{in}\hfill & \mathrm{\Omega },\hfill \end{array}\right.$$

(2)

where

$$\begin{array}{c}\hfill {\displaystyle J(y,v)=\frac{1}{2}\parallel y(·,T)-y_d{\parallel }_{L^2\left(\mathrm{\Omega }\right)}^2+\frac{N}{2}{\parallel v\parallel }_{L^2\left({\omega }_T\right)}^2,}\end{array}$$

(3)

with $N>0$ and $y_d\in L^{\infty }\left(\mathrm{\Omega }\right)$ are given, and the set of admissible controls

$${\mathcal{U}}_{ad}=\left\{v\in L^{\infty }\left({\omega }_T\right):v_a\le v\le v_b\right\},$$

(4)

with $v_b,v_b\in \mathbb{R},v_b>v_a$. The function $y^0\in L^{\infty }\left(\mathrm{\Omega }\right)$, $g\in L^{\infty }(0,T)$, $\beta \in {\mathcal{C}}^2\left(\mathbb{R}\right)\cap L^{\infty }\left(\mathbb{R}\right)$ and ${\chi }_{\omega }$ denote the characteristic function of the control set $\omega$. We assume that there exists $g_{\infty }>0$ such that

$${\parallel g\parallel }_{L^{\infty }(0,T)}\le g_{\infty }$$

(5)

and the function $\beta \in C^2\left(\mathbb{R}\right)$, and there exists $M>0$ such that

$$1\le \beta \left(s\right)\le M\mathrm{for}\mathrm{all}s\in \mathbb{R}$$

(6)

and

$$|{\beta }^{\prime }\left(s\right)|+|{\beta }^{″}\left(s\right)|\le M\mathrm{for}\mathrm{all}s\in \mathbb{R},$$

(7)

Equation (2) can be viewed as a model for population dynamics, where $y(x,t)$ represents the population density at a specific location x and time t. The real function ${\displaystyle {\int }_0^{t}g\left(s\right)y(x,s)ds}$ represents at any given time t and, at the location x, the aggregated history of a population’s density weighted by factors such as pathogen exposure, stress or other time-dependent influences. This dynamic approach allows a wide range of biological processes to be captured, including pathogen dynamics, resource competition and adaptive immunity or resistance. The real function ${\displaystyle \beta \left({\int }_0^{t}g\left(s\right)y(x,s)ds\right)}$ models at any given time t and at the location x the mortality rate as a response to cumulative environmental, physiological or population-related factors. As a rate function, $\beta$ is positive and naturally bounded. We refer to [1], where the proof that $\beta$ can be lower bounded, i.e., $1\le \beta \left(s\right),s\in \mathbb{R}$ is given. Mathematically, the assumption (6)–(7) on $\beta$ ensure the well-posedness of the system, the boundedness of solutions, and the characterization of the local optimal control.

In the literature, a system similar to (2), but with homogeneous Dirichlet boundary conditions and a second member for the state equation living in $L^2\left(Q\right)$, has been studied. In [1], the author proved the existence and uniqueness of the weak solution and analyzed its asymptotic behavior. The establishment of the existence of a weak solution relies on the use of a fixed-point argument and Lebesgue’s dominated convergence theorem. For uniqueness, the standard approach of considering two solutions satisfying the same system and then proving that they are equal by using a subtraction technique that leads to a differential inequality, which forces the difference between the two solutions to be zero, has been applied. Another similar nonlocal-in-time parabolic problem, but with a different nonlinearity and homogeneous Dirichlet boundary conditions, has been studied in [2]. The nonlocal term involves an integral over the entire time interval, making the problem non-evolutionary in the traditional sense because the solution at any time depends on future values. The author proved the existence of a weak solution as well as its uniqueness. For the existence, the author built a weakly continuous map and applied Tikhonov’s fixed-point theorem. Using the standard method, the author proved the uniqueness of the solution under the conditions that the derivative of the nonlocal-in-time term is bounded, and the time interval T is sufficiently small. From the same author, Ref. [3] presents a similar system but with a slightly different nonlocal-in-time nonlinear term. The difference lies in the fact that the integral over the entire time interval is weighted. The author proved the existence of a weak solution without the use of any continuity properties of the solution in time. He has two existence results, one where the nonlinearity is a continuous bounded function and the other where the nonlinear term is a non-negative function that is not necessarily bounded. The demonstrations involve the use of estimates and weak convergences to show that a sequence of approximate solutions converges to a weak solution.

Optimal control of nonlinear partial differential Equations (PDEs) has been a subject of intensive works and applications in various fields. We refer, for instance, to [4,5,6,7,8,9,10,11] and the references therein. It is well established that the first-order optimality condition is both necessary and sufficient for studying optimal control problems associated with linear partial differential equations. However, in the case of nonlinear partial differential equations, the cost function is no longer convex. As a result, while the first-order optimality condition remains necessary, it is not sufficient. To address this, the second-order optimality condition is required, which involves analyzing the second derivative of the cost function with respect to the control. This ensures the local uniqueness of the optimal control. In this paper, we study an optimal control problem for a time-nonlocal parabolic equation. We start by establishing some regularity results for both the state equation and the cost function. Then, we prove that our optimal control problem as the local optimal solution that we characterize by means of the necessary and sufficient first-order and second-order optimality conditions.

The rest of this paper is organized as follows. Section 2 is devoted to the general framework and preliminary results. In Section 3, we establish a maximum principle and study the existence and uniqueness of a generalized nonlocal-in-time problem. Then, we deduce similar properties for system (2). In Section 4, we study the Lipschitz continuity and the differentiability of the solution to (2), then we prove existence of a local optimal control to problem (1) and characterize it using first- and second-order optimality conditions. Some concluding remarks are also given in Section 5.

2. General Framework and Preliminary Results

We give the general setting and some results that will be useful for the study of the control problem (1)–(2).

We recall that the Sobolev Space $H^1\left(\mathrm{\Omega }\right))$ is defined as

$$H^1\left(\mathrm{\Omega }\right))=\left\{{\displaystyle \rho \in L^2\left(\mathrm{\Omega }\right),\frac{\partial \rho }{\partial x_i}\in L^2\left(\mathrm{\Omega }\right),i=1,\cdots ,d}\right\}.$$

Equipped with the norm

$${\parallel \rho \parallel }_{H^1\left(\mathrm{\Omega }\right)}={\left({\int }_{\mathrm{\Omega }}{\rho }^2\left(x\right)dx+\sum _{i=1}^d{\int }_{\mathrm{\Omega }}{\left({\displaystyle \frac{\partial \rho }{\partial x_i}}\left(x\right)\right)}^{2}dx\right)}^{1/2},$$

the space $H^1\left(\mathrm{\Omega }\right))$ is a Hilbert space. We denote by ${\left(H^1\left(\mathrm{\Omega }\right)\right)}^{\prime }$ the dual of $H^1\left(\mathrm{\Omega }\right))$. Next, we introduce the space

$$\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))=\left\{{\displaystyle \rho \in L^2(0,T;H^1\left(\mathrm{\Omega }\right));\frac{\partial \rho }{\partial t}\in L^2\left(0,T;{\left(H^1\left(\mathrm{\Omega }\right)\right)}^{\prime }\right)}\right\}.$$

Endowed with the norm

$${\displaystyle {\parallel \psi \parallel }_{\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))}={\left({\parallel \psi \parallel }_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}^2+{∥\frac{\partial \psi }{\partial t}∥}_{L^2(0,T;{\left(H^1\left(\mathrm{\Omega }\right)\right)}^{\prime })}^2\right)}^{1/2},}$$

(8)

the space $\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ is a Hilbert space. Furthermore, we have from ([12], Theorem 1.1, p. 102) that the following embedding is continuous:

$$\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))\hookrightarrow C([0,T];L^2\left(\mathrm{\Omega }\right)).$$

(9)

Also, using ([13] Theorem 16.1, p. 110) and ([14], Theorem 5.1, p. 58), we obtain the following compact embedding:

$$\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))\hookrightarrow L^2(0,T;H^{\frac{1}{2}}\left(\mathrm{\Omega }\right)).$$

(10)

The following results are essential for proving the existence of a solution to the nonlinear Equation (2), as well as the existence of a solution to the optimization problem discussed in Section 4.

Lemma 1.

Let the sequence $\left(y_n\right)$ be such that

$$\parallel y_n{\parallel }_{\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))}\le C_2$$

(11)

for some constant $C_2>0$ independent of $n.$ Then, there exists $y\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ such that

$$y_n\rightharpoonup y\mathit{weakly}\mathit{in}\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right)),$$

(12)

$$y_n\rightharpoonup y\mathit{weakly}\mathit{in}{L}^2(0,T;H^1\left(\mathrm{\Omega }\right)),$$

(13)

$$y_n\to y\mathit{strongly}\mathit{in}{L}^2\left(Q\right).$$

(14)

Proof. 

We then deduce from (11) that there exists $y\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ such that

$$y_n\rightharpoonup y\mathrm{weakly}\mathrm{in}\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right)).$$

Hence, from the compact embedding (10), we obtain that

$$y_n\to y\mathrm{strongly}\mathrm{in}{L}^2(0,T;H^{\frac{1}{2}}\left(\mathrm{\Omega }\right)),$$

from which, thanks to the continuous embedding of $L^2(0,T;H^{\frac{1}{2}}\left(\mathrm{\Omega }\right))$ into $L^2\left(Q\right)$, we have that

$$y_n\to y\mathrm{strongly}\mathrm{in}{L}^2\left(Q\right).$$

□

From now on, we adopt the following notation for any appropriate function $\varphi$:

$$\begin{array}{ccc}\hfill h_{\varphi }^g(x,t)& =\hfill & {\displaystyle {\int }_0^{t}g\left(s\right)\varphi (x,s)ds\mathrm{for}\mathrm{a}.\mathrm{e}.(x,t)\in Q,}\hfill \\ \hfill h_{\varphi }^{\vartheta }(x,t)& =\hfill & {\displaystyle {\int }_{T-t}^T\vartheta (x,s)\varphi (x,s)ds\mathrm{for}\mathrm{a}.\mathrm{e}.(x,t)\in Q,}\hfill \end{array}$$

(15)

where $g\in L^{\infty }(0,T)$ and $\vartheta \in L^{\infty }\left(Q\right)$. We assume that there exists ${\vartheta }_{\infty }>0$ such that

$${\parallel \vartheta \parallel }_{L^{\infty }\left(Q\right)}\le {\vartheta }_{\infty }.$$

(16)

Lemma 2.

Let $g\in L^{\infty }(0,T)$ and $\vartheta \in L^{\infty }\left(Q\right)$ be such that (5) and (16) hold. Also, let $\varphi \in L^{\infty }\left(Q\right).$ Then,

$$\parallel h_{\varphi }^g{\parallel }_{L^{\infty }\left(Q\right)}\le g_{\infty }T{\parallel \varphi \parallel }_{L^{\infty }\left(Q\right)}$$

(17)

and

$$\parallel h_{\varphi }^{\vartheta }{\parallel }_{L^{\infty }\left(Q\right)}\le {\vartheta }_{\infty }T{\parallel \varphi \parallel }_{L^{\infty }\left(Q\right)},$$

(18)

where $h_{\varphi }^g$ and $h_{\varphi }^{\vartheta }$ are defined as in (15).

Proof. 

Using (5), (15), the Cauchy–Schwarz inequality and the fact that $\varphi \in L^{\infty }\left(Q\right)$, we easily have (17) and (18). □

Lemma 3.

Let the functions $g\in L^2(0,T),$$\beta \in {\mathcal{C}}^2\left(\mathbb{R}\right)$ and $\vartheta \in L^{\infty }\left(Q\right)$ be such that (5), (6), (7) and (16) hold. Also, let $\varphi \in L^2\left(Q\right)$ and $\left(z_n\right)$ be sequence of $L^2\left(Q\right)$ such that

$$z_n\to z\mathit{strongly}\mathit{in}{L}^2\left(Q\right).$$

(19)

Then, we have the following estimation

$$\parallel h_{\varphi }^g{\parallel }_{L^2\left(Q\right)}\le g_{\infty }T{\parallel \varphi \parallel }_{L^2\left(Q\right)}$$

(20)

and

$$\parallel h_{\varphi }^{\vartheta }{\parallel }_{L^2\left(Q\right)}\le {\vartheta }_{\infty }T{\parallel \varphi \parallel }_{L^2\left(Q\right)},$$

(21)

where $h_{\varphi }^g$ and $h_{\varphi }^{\vartheta }$ are defined as in (15). Moreover, as $n\to \infty ,$

$$h_{z_n}^g\to h_z^g\mathit{strongly}\mathit{in}{L}^2\left(Q\right),$$

(22)

$$h_{z_n}^{\vartheta }\to h_z^{\vartheta }\mathit{strongly}\mathit{in}{L}^2\left(Q\right),$$

(23)

$${\beta }^{\left(l\right)}\left(h_{z_n}^g\right)\to {\beta }^{\left(l\right)}\left(h_z^g\right)\mathit{strongly}\mathit{in}{L}^2\left(Q\right),k=0,1,2,$$

(24)

where ${\beta }^{\left(l\right)}$ stands for the derivative of order $l=0,1,2$ of the function β with ${\beta }^{\left(0\right)}=\beta$.

Proof. 

Using (15) and the Cauchy–Schwarz inequality, we have that

$$\begin{array}{ccc}\hfill \parallel h_{\varphi }^g{\parallel }_{L^2\left(Q\right)}^2& =\hfill & {\displaystyle {\int }_0^T{\int }_{\mathrm{\Omega }}{\left({\int }_0^{t}g\left(s\right)\varphi (x,s)ds\right)}^{2}dxdt}\hfill \\ & \le \hfill & T^2{g}_{\infty }^2{\parallel \varphi \parallel }_{L^2\left(Q\right)}^2,\hfill \end{array}$$

from which we deduce (20). Proceeding exactly as above while using (16), we deduce (21).

From the linearity of the integral, (20) and (21), we can write

$$\begin{array}{ccc}\hfill \parallel h_{z_n}^g-h_z^g{\parallel }_{L^2\left(Q\right)}^2& =\hfill & {\displaystyle {\int }_0^T{\int }_{\mathrm{\Omega }}{\left({\int }_0^{t}g\left(s\right)\left(z_n(x,s)-z(x,s)\right)ds\right)}^{2}dxdt}\hfill \\ & \le \hfill & T^2{g}_{\infty }^2{\parallel z_n-z\parallel }_{L^2\left(Q\right)}^2\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \parallel h_{z_n}^{\vartheta }-h_z^{\vartheta }{\parallel }_{L^2\left(Q\right)}^2& =\hfill & {\displaystyle {\int }_0^T{\int }_{\mathrm{\Omega }}{\left({\int }_{T-t}^T\vartheta (x,s)\left(z_n(x,s)-z(x,s)\right)ds\right)}^{2}dxdt}\hfill \\ & \le \hfill & T^2{\vartheta }_{\infty }^2{\parallel z_n-z\parallel }_{L^2\left(Q\right)}^2.\hfill \end{array}$$

Passing to the limit when $n\to \infty$ in these two inequalities while using (19), we obtain (22) and (23). To prove (24), we use the Lebesgue theorem. Indeed, in view of (22), we can extract a subsequence of ${\displaystyle {\left(h_{z_n}^g\right)}_n}$ still denoted ${\displaystyle {\left(h_{z_n}^g\right)}_n}$ such that

$$h_{z_n}^g(x,t)⟶h_z^g(x,t)\mathrm{for}\mathrm{a}.\mathrm{e}.(x,t)\in Q,\mathrm{as}n\to +\infty ,$$

and since $\beta \in {\mathcal{C}}^2\left(\mathbb{R}\right)$, we have that

$${\beta }^{\left(l\right)}\left(h_{z_n}^g\right)(x,t)⟶{\beta }^{\left(l\right)}\left(h_z^g\right)(x,t)\mathrm{for}\mathrm{a}.\mathrm{e}.(x,t)\in Q,\mathrm{as}n\to +\infty .$$

Therefore, using (6) and (7), we have that there exists $M>0$ such that

$$\left|{\beta }^{\left(l\right)}\left(h_{z_n}^g\right)(x,t)\right|\le M\mathrm{for}\mathrm{a}.\mathrm{e}.(x,t)\in Q.$$

Since Q is bounded domain, it follows from the Lebesgue dominated convergence theorem that

$${\beta }^{\left(l\right)}\left(h_{z_n}^g\right)⟶{\beta }^{\left(l\right)}\left(h_z^g\right)\mathrm{strongly}\mathrm{in}{L}^2\left(Q\right),l=0,1,2\mathrm{as}n\to +\infty .$$

□

3. Analysis of the State Equation

In this section, we establish maximum principle and study the existence and uniqueness of a generalized nonlocal-in-time problem. Actually, the study of this auxiliary system allow us obtain similar properties for system (2) which also includes related systems such as its adjoint model. More precisely, we consider the following system:

$$\left\{\begin{array}{ccccc}\hfill {\displaystyle \frac{\partial y}{\partial t}-\Delta y+b_0{h}_y^g+g\left(t\right)h_y^{\vartheta }+\mu \beta \left(h_y^g\right)y+a_{0}y}& =\hfill & {\displaystyle v{\chi }_{\omega }y+f}\hfill & \mathrm{in}\hfill & Q,\hfill \\ \hfill {\displaystyle \frac{\partial y}{\partial \nu }}& =\hfill & 0\hfill & \mathrm{on}\hfill & \mathrm{\Sigma },\hfill \\ \hfill {\displaystyle y(·,0)}& =\hfill & {\displaystyle y^0}\hfill & \mathrm{in}\hfill & \mathrm{\Omega },\hfill \end{array}\right.$$

(25)

where $\mu \ge 0$, $v{\chi }_{\omega }\in L^{\infty }\left(Q\right)$, $f\in L^2\left(Q\right)$ and $y^0\in L^2\left(\mathrm{\Omega }\right)$. The functions $g\in L^{\infty }(0,T)$, $b_0\in L^{\infty }\left(Q\right),$$\vartheta \in L^{\infty }\left(Q\right)$ and $a_0\in L^{\infty }\left(Q\right)$. The function $h_y^{\vartheta }$ and $h_y^g$ are given by notation (15). We assume that (16) holds and that there exist $b_{\infty }>0$ and $a_{\infty }>0$ such that

$$\parallel b_0{\parallel }_{L^{\infty }\left(Q\right)}\le b_{\infty },$$

(26)

and

$$\parallel a_0{\parallel }_{L^{\infty }\left(Q\right)}\le a_{\infty }.$$

(27)

Before going further, we give the definition of a weak solution to system (25).

Definition 1.

Assume that $\beta \in {\mathcal{C}}^2\left(\mathbb{R}\right)$ satisfies (6)–(7). Let $\mu \ge 0$, $v\in L^{\infty }\left({\omega }_T\right)$, $g\in L^{\infty }\left(Q\right),$$f\in L^2\left(Q\right)$, $b_0\in L^{\infty }\left(Q\right),$$\vartheta \in L^{\infty }\left(Q\right),$$a_0\in L^{\infty }\left(Q\right)$, and $y^0\in L^2\left(\mathrm{\Omega }\right)$. We say that a function $y\in L^2(0,T;H^1\left(\mathrm{\Omega }\right))$ is a weak solution to (25) if

$$\begin{array}{ccc}{\displaystyle {\int }_{\mathrm{\Omega }}{y}^0\varphi (·,0)\mathrm{d}x+{\int }_{Q}f\varphi dxdt}\hfill & =\hfill & {\displaystyle -{\int }_0^T{〈\frac{\partial \varphi }{\partial t},y〉}_{{\left(H^1\left(\mathrm{\Omega }\right)\right)}^{\prime },H^1\left(\mathrm{\Omega }\right)}dt+{\int }_Q{a}_{0}y\varphi dxdt}\hfill \\ \hfill & +\hfill & {\displaystyle {\int }_Q\nabla \varphi ·\nabla ydxdt+\mu {\int }_Q\beta \left(h_y^g\right)y\varphi dxdt-{\int }_{{\omega }_T}vy\varphi dxdt}\hfill \\ \hfill & +\hfill & {\displaystyle {\int }_Q{b}_0{h}_y^g\varphi dxdt+{\int }_{Q}g\left(t\right)h_y^{\vartheta }\varphi dxdt,}\hfill \end{array}$$

(28)

holds for every $\varphi \in H\left(Q\right):=\{\phi \in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))\mathrm{such}\mathrm{that}\varphi (·,T)=0\}$.

3.1. Maximum Principle

In this subsection, we will prove some estimates in the space $L^{\infty }\left(Q\right)$. One necessary result is the one directly obtained from the Gagliardo–Niremberg Theorem (see, e.g., [15,16]) which we recall here.

Lemma 4.

Let $y\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$. Then, $y\in L^{\mu }\left(Q\right)$, with $\mu ={\displaystyle \frac{2(d+2)}{d}}$. In addition, there exists a constant $C=C\left(d\right)$ such that the following estimate

$${\int }_Q{\left|y\right|}^{\mu }dxdt\le {C\parallel y\parallel }_{L^{\infty }(0,T;L^2\left(\mathrm{\Omega }\right))}^{{\displaystyle \frac{4}{d}}}{|y\parallel }_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}^2$$

(29)

holds.

Another preliminary result in proving the maximum principle is the following theorem.

Theorem 1

([17], Lemma 4.1.1). We consider ρ a non-negative and non-increasing function on $[k_0,+\infty )$ such that

$$\rho \left(m\right)\le {\left({\displaystyle \frac{C}{m-k}}\right)}^{\theta }{\left(\rho \left(k\right)\right)}^{\eta },\forall m>k>k_0,$$

(30)

where C, $\theta , \eta$ are positive constants with $\eta >1$. Then, $\rho \left(m\right)=0$ for all $m\ge k_0+\varpi$, where $\varpi :=C{2}^{{\displaystyle \frac{\eta }{\eta -1}}}{\left(\rho \left(k_0\right)\right)}^{{\displaystyle \frac{\eta -1}{\theta }}}$.

Theorem 2.

Let $\mu \ge 0,$$v\in L^{\infty }\left({\omega }_T\right)$, $f\in L^{\infty }\left(Q\right)$ and $y^0\in L^{\infty }\left(\mathrm{\Omega }\right).$ Also, let $g\in L^{\infty }(0,T)$, $b_0\in L^{\infty }\left(Q\right),$$\vartheta \in L^{\infty }\left(Q\right),$$a_0\in L^{\infty }\left(Q\right)$ and $\beta \in {\mathcal{C}}^2\left(\mathbb{R}\right)$ be such that (5), (26), (16), (27), and (6) hold. Assume that $y\in \mathbb{W}(0,T,H^1\left(\mathrm{\Omega }\right))$ is a weak solution to (25). Then, we have that $y\in L^{\infty }\left(Q\right)$. Moreover, there exists $C(d,T,\mathrm{\Omega },a_{\infty },b_{\infty },\parallel v{\parallel }_{L^{\infty }\left({\omega }_T\right)},{\vartheta }_{\infty },g_{\infty })>0$ such that

$${\parallel y\parallel }_{L^{\infty }\left(Q\right)}\le C\left({\parallel f\parallel }_{L^{\infty }\left(Q\right)}+{\parallel y^0\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}\right).$$

(31)

Proof. 

First we need to make the following change for variable: $p=e^{-rt}y$ where $y\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ is a solution to (25) and $r>0$. It follows that $p\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ is a weak solution to

$$\left\{{\displaystyle \begin{array}{c}\hfill {\displaystyle \frac{\partial p}{\partial t}-\Delta p=-(r+a_0)p-\mu \beta \left(h_y^g\right)p}\\ \hfill {\displaystyle -e^{-rt}{b}_0{h}_{e^{r(·)}p}^g-e^{-rt}g\left(t\right)h_{e^{r(·)}p}^{\vartheta }}\\ \hfill {\displaystyle +v{\chi }_{\omega }p+e^{-rt}f}& \mathrm{in}\hfill & Q,\hfill \\ \hfill \frac{\partial p}{\partial \nu }=0& \mathrm{on}& \mathrm{\Sigma },\\ \hfill p(·,0)=y^0& \mathrm{in}& \mathrm{\Omega },\end{array}}\right.$$

(32)

where in view of (15), ${\displaystyle h_{e^{r(·)}p}^g(x,t)={\int }_0^t{e}^{rs}g\left(s\right)p(x,s)ds}$ and ${\displaystyle h_{e^{r(·)}p}^{\vartheta }(x,t)={\int }_{T-t}^T{e}^{rs}{b}_0(x,s)p(x,s)ds,}$ for a.e. $(x,t)\in Q$.

Next, we introduce a real number $\kappa$ such that $\kappa >\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}$. We easily deduce that for almost every $t\in (0,T)$, $p\left(t\right)-\kappa \in H^1\left(\mathrm{\Omega }\right)$, and then we set $p^{+}\left(t\right):=max\{p\left(t\right)-\kappa ,0\}$ which also resides in $H^1\left(\mathrm{\Omega }\right)$. Multiplying the first equation of system (25) by $\phi \in H^1\left(\mathrm{\Omega }\right)$ and then integrating by parts over $\mathrm{\Omega }$, we obtain the following:

$$\begin{array}{c}\hfill {\displaystyle {\int }_{\mathrm{\Omega }}{e}^{-rt}f\left(t\right)\phi \left(t\right)dx={〈\frac{\partial p}{\partial t}\left(t\right),\phi \left(t\right)〉}_{{\left(H^1\left(\mathrm{\Omega }\right)\right)}^{\prime },H^1\left(\mathrm{\Omega }\right)}}\\ \hfill {\displaystyle +{\int }_{\mathrm{\Omega }}\nabla p\left(t\right)·\nabla \phi \left(t\right)dx+{\int }_{\mathrm{\Omega }}(a_0+r)p\left(t\right)\phi \left(t\right)dx}\\ \hfill {\displaystyle +\mu {\int }_{\mathrm{\Omega }}\beta \left(h_y^g\right)\left(t\right)p\left(t\right)\phi \left(t\right)dx-{\int }_{\omega }v\left(t\right)p\left(t\right)\phi \left(t\right)dx}\\ \hfill {\displaystyle +{\int }_{\mathrm{\Omega }}{e}^{-rt}{b}_0{\int }_0^t{e}^{rs}g\left(s\right)p(x,s)ds\phi dx}\\ \hfill {\displaystyle +{\int }_{\mathrm{\Omega }}{e}^{-rt}g\left(t\right){\int }_{T-t}^T{e}^{rs}\vartheta (x,s)p(x,s)ds\phi dx.}\end{array}$$

(33)

Hence,

$$\begin{array}{ccc}\hfill {\displaystyle {\int }_{\mathrm{\Omega }}{e}^{-rt}f\left(t\right)\phi \left(t\right)dx}& =\hfill & {\displaystyle {〈\frac{\partial }{\partial t}(p-\kappa )\left(t\right),\phi \left(t\right)〉}_{{\left(H^1\left(\mathrm{\Omega }\right)\right)}^{\prime },H^1\left(\mathrm{\Omega }\right)}+{\int }_{\mathrm{\Omega }}\nabla (p-\kappa )\left(t\right)·\nabla \phi \left(t\right)dx}\hfill \\ & +\hfill & {\displaystyle \mu {\int }_{\mathrm{\Omega }}\beta \left(h_y^g\right)\left(t\right)(p-\kappa )\left(t\right)\phi dx+\kappa \mu {\int }_{\mathrm{\Omega }}\beta \left(h_y^g\right)\left(t\right)\phi \left(t\right)dx}\hfill \\ & +\hfill & {\displaystyle {\int }_{\mathrm{\Omega }}(a_0+r)(p-\kappa )\left(t\right)\phi \left(t\right)dx+\kappa {\int }_{\mathrm{\Omega }}{e}^{-rt}g\left(t\right){\int }_{T-t}^T{e}^{rs}\vartheta (x,s)ds\phi dx}\hfill \\ & +\hfill & {\displaystyle {\int }_{\mathrm{\Omega }}{e}^{-rt}{b}_0\left[{\int }_0^t{e}^{rs}g\left(s\right)(p-\kappa )(x,s)ds\right]\phi \left(t\right)dx}\hfill \\ & +\hfill & {\displaystyle \kappa {\int }_{\mathrm{\Omega }}{e}^{-rt}{b}_0\left[{\int }_0^t{e}^{rs}g\left(s\right)ds\right]\phi \left(t\right)dx+\kappa {\int }_{\mathrm{\Omega }}(a_0+r)\phi \left(t\right)dx}\hfill \\ & -\hfill & {\displaystyle {\int }_{\omega }v\left(t\right)(p-\kappa )\left(t\right)\phi \left(t\right)dx-\kappa {\int }_{\omega }v\left(t\right)\phi \left(t\right)dx}\hfill \\ & +\hfill & {\displaystyle {\int }_{\mathrm{\Omega }}{e}^{-rt}{b}_0{\int }_{T-t}^T{e}^{rs}g\left(s\right)(p-\kappa )(x,s)ds\phi dx.}\hfill \end{array}$$

Taking $\phi =p^{+}$ in this latter identity yields

$$\begin{array}{ccc}\hfill {\displaystyle {\int }_{\mathrm{\Omega }}{e}^{-rt}f\left(t\right)p^{+}\left(t\right)dx}& =\hfill & {\displaystyle \frac{1}{2}\frac{d}{dt}{\parallel p^{+}\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^2+\mu {\int }_{\mathrm{\Omega }}\beta \left(h_y^g\right)\left(t\right){\left(p^{+}\left(t\right)\right)}^{2}dx}\hfill \\ & +\hfill & {\displaystyle \mu \kappa {\int }_{\mathrm{\Omega }}\beta \left(h_y^g\right)\left(t\right)p^{+}\left(t\right)dx+{\int }_{\mathrm{\Omega }}|\nabla p^{+}{\left(t\right)|}^{2}dx+r{\parallel p^{+}\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^2}\hfill \\ & +\hfill & {\displaystyle {\int }_{\mathrm{\Omega }}{a}_0{\left(p^{+}\left(t\right)\right)}^{2}dx+{\int }_{\mathrm{\Omega }}{e}^{-rt}{b}_0\left[{\int }_0^t{e}^{rs}g\left(s\right)(p-\kappa )(x,s)ds\right]p^{+}\left(t\right)dx}\hfill \\ & +\hfill & {\displaystyle \kappa {\int }_{\mathrm{\Omega }}{e}^{-rt}{b}_0\left[{\int }_0^t{e}^{rs}g\left(s\right)ds\right]p^{+}\left(t\right)dx+\kappa {\int }_{\mathrm{\Omega }}(a_0+r)p^{+}\left(t\right)dx}\hfill \\ & -\hfill & {\displaystyle {\int }_{\omega }v\left(t\right){\left(p^{+}\left(t\right)\right)}^{2}dx-\kappa {\int }_{\omega }v\left(t\right)p^{+}\left(t\right)dx}\hfill \\ & +\hfill & {\displaystyle {\int }_{\mathrm{\Omega }}{e}^{-rt}{b}_0{\int }_{T-t}^T{e}^{rs}g\left(s\right)(p-\kappa )(x,s)ds{p}^{+}\left(t\right)dx}\hfill \\ & +\hfill & {\displaystyle \kappa {\int }_{\mathrm{\Omega }}{e}^{-rt}g\left(t\right){\int }_{T-t}^T{e}^{rs}\vartheta (x,s)ds{p}^{+}\left(t\right)dx,}\hfill \end{array}$$

and from (5), (6), (26), and (27) we obtain that for a.e. $t\in (0,T)$,

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2}\frac{d}{dt}\parallel p^{+}{\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^2+{\int }_{\mathrm{\Omega }}|\nabla p^{+}{\left(t\right)|}^{2}dx+r{\parallel p^{+}\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^2\le }\\ \hfill {\displaystyle a_{\infty }\parallel p^{+}{\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^2+b_{\infty }{g}_{\infty }{T}^{1/2}{\parallel p^{+}\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}{\left({\int }_{\mathrm{\Omega }}{\int }_0^t{\left(p^{+}\left(s\right)\right)}^{2}dsdx\right)}^{1/2}}\\ \hfill {\displaystyle +{\vartheta }_{\infty }{g}_{\infty }{T}^{1/2}\parallel p^{+}{\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}{\left({\int }_{\mathrm{\Omega }}{\int }_{T-t}^T{\left(p^{+}\left(s\right)\right)}^{2}dsdx\right)}^{1/2}+{\int }_{\mathrm{\Omega }}\left|f\left(t\right)\right||p^{+}\left(t\right)|dx}\\ \hfill {\displaystyle +a_{\infty }\kappa {\int }_{\mathrm{\Omega }}{p}^{+}\left(t\right)dx+{\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)}\parallel p^{+}{\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^2+{\kappa \parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)}{\int }_{\mathrm{\Omega }}\left|p^{+}\left(t\right)\right|dx}\\ \hfill {\displaystyle +\kappa ({\vartheta }_{\infty }+b_{\infty })g_{\infty }T{\int }_{\mathrm{\Omega }}{p}^{+}\left(t\right)dx-r\kappa {\int }_{\mathrm{\Omega }}{p}^{+}\left(t\right)dx.}\end{array}$$

(34)

Now, observing that $p^{+}\left(0\right):=max(y\left(0\right)-k,0)=0$ and integrating (34) on $(0,s)$, with $s\in [0,T]$ and then using Cauchy–Schwarz’s inequality, we obtain that

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2}\parallel p^{+}{\left(s\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^2+{\int }_0^s{\int }_{\mathrm{\Omega }}|\nabla p^{+}{\left(t\right)|}^{2}dxdt+r{\int }_0^s{\parallel p^{+}\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^{2}dt\le }\\ \hfill {\displaystyle -r\kappa {\int }_0^s{\int }_{\mathrm{\Omega }}{p}^{+}\left(t\right)dxdt+b_{\infty }{g}_{\infty }T{\int }_0^s{\parallel p^{+}\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^{2}dt}\\ \hfill {\displaystyle +{\vartheta }_{\infty }{g}_{\infty }T{\int }_0^s\parallel p^{+}{\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^{2}dt+a_{\infty }{\int }_0^s{\parallel p^{+}\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^{2}dt+}\\ \hfill {\displaystyle {\int }_0^s{\int }_{\mathrm{\Omega }}\left|f\left(t\right)\right||p^{+}{\left(t\right)|dxdt+\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)}{\int }_0^s{\parallel p^{+}\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^{2}dt+}\\ \hfill {\displaystyle {\kappa \parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)}{\int }_0^s{\int }_{\mathrm{\Omega }}\left|p^{+}\left(t\right)\right|dxdt+\kappa ({\vartheta }_{\infty }+b_{\infty })g_{\infty }T{\int }_0^s{\int }_{\mathrm{\Omega }}{p}^{+}\left(t\right)dxdt}\\ \hfill {\displaystyle +a_{\infty }\kappa {\int }_0^s{\int }_{\mathrm{\Omega }}{p}^{+}\left(t\right)dxdt+}\end{array}$$

and then choosing $r=1+({\vartheta }_{\infty }+b_{\infty })g_{\infty }T+a_{\infty }+{\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)}$, we finally obtain

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2}\parallel p^{+}{\left(s\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^2+{\int }_0^s\parallel p^{+}{\left(t\right)\parallel }_{H^1\left(\mathrm{\Omega }\right)}^{2}dt\le {\int }_Q\left|f\right||p^{+}|dxdt.}\end{array}$$

(35)

Hence,

$${\displaystyle \underset{t\in [0,T]}{sup}\parallel p^{+}{\left(t\right)\parallel }_{L^2\left(\mathrm{\Omega }\right)}^2\le 2{\int }_Q\left|f\right||p^{+}|dxdt,}$$

(36)

$${\displaystyle \parallel p^{+}{\parallel }_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}^2\le {\int }_Q\left|f\right||p^{+}|dxdt,}$$

(37)

and it follows from (36) that

$$\parallel p^{+}{\parallel }_{L^{\infty }(0,T;L^2\left(\mathrm{\Omega }\right))}^2\le 2{\int }_Q\left|f\right||p^{+}|dxdt.$$

(38)

Now, using Lemma 4, we obtain that

$$\begin{array}{c}{\displaystyle {\left({\int }_Q{\left|p^{+}\right|}^{\frac{2(d+2)}{d}}dxdt\right)}^{\frac{d}{d+2}}\le C\left(d\right){\left[\parallel p^{+}{\parallel }_{L^{\infty }(0,T;L^2\left(\mathrm{\Omega }\right))}^{\frac{4}{d}}{\parallel p^{+}\parallel }_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}^2\right]}^{\frac{d}{d+2}}}\hfill \end{array}$$

(39)

for some $C\left(d\right)>0$. Observing that

$${\displaystyle {\left[\parallel p^{+}{\parallel }_{L^{\infty }(0,T;L^2\left(\mathrm{\Omega }\right))}^{\frac{4}{d}}{\parallel p^{+}\parallel }_{L^2((0,T;H^1\left(\mathrm{\Omega }\right))}^2\right]}^{\frac{d}{d+2}}=\parallel p^{+}{\parallel }_{L^{\infty }(0,T;L^2\left(\mathrm{\Omega }\right))}^{\frac{4}{d+2}}{\parallel p^{+}\parallel }_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}^{\frac{2d}{d+2}}}$$

and using the Young’s inequality with $p={\displaystyle \frac{d+2}{2}}$ and $q={\displaystyle \frac{d+2}{d}}$ and then (39), we have that

$$\begin{array}{c}{\displaystyle {\left({\int }_Q{\left|p^{+}\right|}^{\frac{2(d+2)}{d}}dxdt\right)}^{\frac{d}{d+2}}\le C\left(d\right)\left[\parallel p^{+}{\parallel }_{L^{\infty }(0,T;L^2\left(\mathrm{\Omega }\right))}^2+{\parallel p^{+}\parallel }_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}^2\right].}\hfill \end{array}$$

(40)

Therefore, it follows from (37) and (38) that

$$\begin{array}{c}{\displaystyle {\left({\int }_Q{\left|p^{+}\right|}^{\frac{2(d+2)}{d}}dxdt\right)}^{\frac{d}{d+2}}\le 3C\left(d\right){\int }_Q\left|f\right||p^{+}|dxdt.}\hfill \end{array}$$

(41)

We set $B_{\kappa }:=\{(x,t)\in Q,p(x,t)>\kappa \}$. Then, $B_{\kappa }$ is a measurable set. Moreover, we deduce from (41) that

$${\displaystyle {\left({\int }_{B_{\kappa }}{\left|p^{+}\right|}^{\frac{2(d+2)}{d}}dxdt\right)}^{\frac{d}{d+2}}\le \left(3C\left(d\right){\int }_{B_{\kappa }}\left|f\right||p^{+}|dxdt\right)}$$

(42)

Using Hölder’s inequality with $p={\displaystyle \frac{2(d+2)}{d+4}}$ and $q={\displaystyle \frac{2(d+2)}{d}}$ in the right hand side of (42), we obtain

$${\displaystyle {\int }_{B_{\kappa }}{\left|p^{+}\right|}^{\frac{2(d+2)}{d}}dxdt\le {\left(3C\left(d\right)\right)}^{\frac{d+2}{d}}{\left({\int }_{B_{\kappa }}{\left|f\right|}^{\frac{2(d+2)}{d+4}}dxdt\right)}^{\frac{d+4}{2d}}{\left({\int }_{B_{\kappa }}{\left|p^{+}\right|}^{\frac{2(d+2)}{d}}dxdt\right)}^{\frac{1}{2}},}$$

from which we deduce that

$$\begin{array}{ccc}\hfill {\displaystyle {\left({\int }_{B_{\kappa }}{\left|p^{+}\right|}^{\frac{2(d+2)}{d}}dxdt\right)}^{\frac{1}{2}}}& \le \hfill & {\displaystyle {\left(3C\left(d\right)\right)}^{\frac{d+2}{d}}{\left({\int }_{B_{\kappa }}{\left|f\right|}^{\frac{2(d+2)}{d+4}}dxdt\right)}^{\frac{d+4}{2d}}}\hfill \\ \hfill & \le \hfill & {\left(3C\left(d\right)\right)}^{{\displaystyle \frac{d+2}{d}}}{\parallel f|}_{L^{\infty }\left(Q\right)}^{{\displaystyle \frac{d+2}{d}}}{\left|B_{\kappa }\right|}^{{\displaystyle \frac{d+4}{2d}}},\hfill \end{array}$$

(43)

where $|B_{\kappa }|$ denotes the measure of the set $B_{\kappa }$.

Let $m>\kappa$, then $B_m\subseteq B_{\kappa }$. Consequently, for a.e. $(x,t)\in B_m$, we have $p(x,t)-k>m-\kappa$ and then, $p-\kappa =p^{+}$.

Thus,

$$\begin{array}{ccc}{\displaystyle {\int }_{B_{\kappa }}{|p^{+}|}^{\frac{2(d+2)}{d}}dxdt}\hfill & \ge \hfill & {\displaystyle {\int }_{B_m}|p^{+}{|}^{\frac{2(d+2)}{d}}dxdt={\int }_{B_m}{|y-k|}^{\frac{2(d+2)}{d}}dxdt}\hfill \\ \hfill & \ge \hfill & {\displaystyle {\int }_{B_m}{(m-k)}^{\frac{2(d+2)}{d}}dxdt={(m-k)}^{\frac{2(d+2)}{d}}\left|B_m\right|.}\hfill \end{array}$$

(44)

Then, we obtain that

$$\begin{array}{ccc}|B_m{|}^{{\displaystyle \frac{1}{2}}}\hfill & \le \hfill & {\displaystyle \frac{1}{{(m-k)}^{\frac{(n+2)}{n}}}{\left({\int }_{B_{\kappa }}{\left|p^{+}\right|}^{\frac{2(d+2)}{d}}dxdt\right)}^{\frac{1}{2}}}\hfill \\ & \le \hfill & {\displaystyle {\left(\frac{3C\left(d\right)}{m-k}\right)}^{\frac{d+2}{d}}{\parallel f\parallel }_{L^{\infty }\left(Q\right)}^{\frac{d+2}{d}}{\left|B_{\kappa }\right|}^{\frac{d+4}{2d}}}\hfill \end{array}$$

Hence,

$$\begin{array}{ccc}|B_m|\hfill & \le \hfill & {\displaystyle {\left(\frac{3C\left(d\right)}{m-k}\right)}^{\frac{2(d+2)}{d}}{\parallel f\parallel }_{L^{\infty }\left(Q\right)}^{\frac{2(d+2)}{d}}{\left|B_{\kappa }\right|}^{\frac{d+4}{d}}.}\hfill \end{array}$$

Observing that $B_m\subset B_{\kappa }$ for any $m>k$, we have that $|B_m|\le |B_{\kappa }|$. Hence, applying Theorem 1 with $\phi \left(k\right)=|B_{\kappa }|,$$k_0={\parallel y^0\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}$ and $\varpi ={C(d,T,\mathrm{\Omega })\parallel f\parallel }_{L^{\infty }\left(Q\right)}$, we deduce that $|B_m|=0$ where $m=C(d,T,\mathrm{\Omega })\left({\parallel f\parallel }_{L^{\infty }\left(Q\right)}+{\parallel y^0\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}\right)$. Therefore, we have

$$p(x,t)\le C(d,T,\mathrm{\Omega })\left({\parallel f\parallel }_{L^{\infty }\left(Q\right)}+{\parallel y^0\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}\right)\forall (x,t)\in Q.$$

We prove by setting $p^{-}=min(p+k,0)$ and ${\tilde{B}}_{\kappa }=\{(x,t)\in Q:p(x,t)<-k\}$ that the set ${\tilde{B}}_m$, where $p<-C(d,T,\mathrm{\Omega })\left({\parallel f\parallel }_{L^{\infty }\left(Q\right)}+{\parallel y^0\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}\right)$, is a measure zero set. Consequently,

$${\parallel p\parallel }_{L^{\infty }\left(Q\right)}\le C(d,T,\mathrm{\Omega })\left({\parallel f\parallel }_{L^{\infty }\left(Q\right)}+{\parallel y^0\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}\right).$$

Now, since $p=e^{-(a_{\infty }+({\vartheta }_{\infty }+b_{\infty })g_{\infty }T+\parallel v{\parallel }_{L^{\infty }\left({\omega }_T\right)}+1)t}y$, we deduce that

$${\parallel y\parallel }_{L^{\infty }\left(Q\right)}\le C(d,T,\mathrm{\Omega })e^{(a_{\infty }+({\vartheta }_{\infty }+b_{\infty })g_{\infty }T+\parallel v{\parallel }_{L^{\infty }\left({\omega }_T\right)}+1)T}\left({\parallel f\parallel }_{L^{\infty }\left(Q\right)}+{\parallel y^0\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}\right).$$

□

From this Theorem based on the generalized model (25), we can derive similar properties for system (2).

Corollary 1.

Let $v\in L^{\infty }\left({\omega }_T\right)$ and $y^0\in L^{\infty }\left(\mathrm{\Omega }\right)$. Also, let $g\in L^{\infty }(0,T)$ and $\beta \in {\mathcal{C}}^2\left(\mathbb{R}\right)$ be such that (5) and (6) hold. Assume that system (2) has a weak solution $y\in \mathbb{W}(0,T,H^1\left(\mathrm{\Omega }\right))$. Then, $y\in L^{\infty }\left(Q\right)$ and there exists a constant $C(d,T,\mathrm{\Omega })>0$ such that

$${\parallel y\parallel }_{L^{\infty }\left(Q\right)}\le C(d,T,\mathrm{\Omega })e^{(\parallel v{\parallel }_{L^{\infty }\left(Q\right)}+1)T}{\parallel y^0\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}.$$

(45)

Proof. 

Proceeding as in Theorem 2 but with $a_0\equiv 0$, $\vartheta \equiv 0$, $b_0\equiv 0$ and $f\equiv 0$, we easily prove that $y\in L^{\infty }\left(Q\right)$ and that estimate (45) holds. □

In what following, we need the following result.

Lemma 5.

Let ${\left({\psi }_n\right)}_n$ be sequence of $L^{\infty }\left(Q\right)$. Assume that there exists ${\psi }_{\infty }>0$ and $\psi \in L^2\left(Q\right)$ such that

$$\parallel {\psi }_n{\parallel }_{L^{\infty }\left(Q\right)}\le {\psi }_{\infty }$$

and

$${\psi }_n\to \psi \mathit{strongly}\mathit{in}{L}^2\left(Q\right),\mathrm{as}n\to +\infty .$$

Then, for any $\zeta \in L^2\left(Q\right)$,

$${\psi }_n\zeta \to \psi \zeta \mathit{strongly}\mathit{in}{L}^2\left(Q\right),\mathrm{as}n\to +\infty .$$

(46)

Proof. 

Since ${\psi }_n\to \psi \mathrm{strongly}\mathrm{in}{L}^2\left(Q\right),asn\to +\infty ,$ we can extract a subsequence of ${\displaystyle {\left({\psi }_n\right)}_n}$ still denoted ${\left({\psi }_n\right)}_n$ such that

$${\psi }_n(x,t)⟶\psi (x,t)\mathrm{for}\mathrm{a}.\mathrm{e}.(x,t)\in Q,\mathrm{as}n\to +\infty .$$

Therefore, for any $\zeta \in L^2\left(Q\right)$, we obtain that

$${\psi }_n(x,t)\zeta (x,t)⟶\psi (x,t)\zeta (x,t)\mathrm{for}\mathrm{a}.\mathrm{e}.(x,t)\in Q,\mathrm{as}n\to +\infty$$

and as there exists ${\psi }_{\infty }>0$ such that

$$\left|\zeta (x,t){\psi }_n(x,t)\right|\le {\psi }_{\infty }\left|\zeta (x,t)\right|\mathrm{for}\mathrm{a}.\mathrm{e}.(x,t)\in Q.$$

Relation (46) follows from the Lebesgue’s dominated convergence theorem. □

3.2. Existence and Uniqueness Results

Theorem 3.

Let $\mu \ge 0$, $v\in L^{\infty }\left({\omega }_T\right)$, $f\in L^{\infty }\left(Q\right)$ and $y^0\in L^{\infty }\left(\mathrm{\Omega }\right)$. Also, let $g\in L^{\infty }(0,T)$, $b_0\in L^{\infty }\left(Q\right),$$\vartheta \in L^{\infty }\left(Q\right),$$a_0\in L^{\infty }\left(Q\right)$, and $\beta \in {\mathcal{C}}^2\left(\mathbb{R}\right)$ be such that (5), (26), (16), (27), (6) and (7) hold. Then, there exists a unique weak solution $y\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ to (25) in the sense of Definition 1.

Moreover, there exist $C\left(M,g_{\infty },b_{\infty },a_{\infty },{\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},\mathrm{\Gamma },T\right)>0$ and $C(a_{\infty },b_{\infty },g_{\infty },\parallel v{\parallel }_{L^{\infty }\left({\omega }_T\right)},T)>0$ depending continuously on ${\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)}$ such that

$${\displaystyle {\parallel y\parallel }_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}\le C(a_{\infty },{\vartheta }_{\infty },b_{\infty },g_{\infty }{,\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},T)\left({\parallel f\parallel }_{L^2\left(Q\right)}+{\parallel y^0\parallel }_{L^2\left(\mathrm{\Omega }\right)}\right),}$$

(47)

and

$${\parallel y\parallel }_{\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))}\le C\left(M,g_{\infty },b_{\infty },a_{\infty },{\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},{\vartheta }_{\infty },T\right)\left({\parallel f\parallel }_{L^2\left(Q\right)}+{\parallel y^0\parallel }_{L^2\left(\mathrm{\Omega }\right)}\right).$$

(48)

Proof. 

We proceed in three steps.

Step 1. We prove the existence results and estimations (48).

We use the Schauder’s fixed-point theorem. So, Let B be a non-empty, closed and convex subset of $L^2\left(Q\right)$ given by

$$B:=\left\{z\in L^2\left(Q\right),{∥z∥}_{L^2\left(Q\right)}\le C_1\right\},$$

(49)

where

$$C_1:=C(a_{\infty },b_{\infty },g_{\infty }{,\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},T)\left({∥f∥}_{L^2\left(Q\right)}+{∥y^0∥}_{L^2\left(\mathrm{\Omega }\right)}\right).$$

(50)

Let $z\in B$. In view of Theorem A1, we have constructed the following nonlinear map

$${\displaystyle \mathbb{F}:B\to B}$$

where $\mathbb{F}\left(z\right)=y\left(z\right)$. Here, $y\left(z\right)\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ is the weak solution to (A1):

$$\left\{{\displaystyle \begin{array}{ccccc}\hfill {\displaystyle \frac{\partial y}{\partial t}-\Delta y+b_0{h}_y^g+g\left(t\right)h_y^{\vartheta }+\mu \beta \left(h_z\right)y+a_{0}y}& =\hfill & {\displaystyle v{\chi }_{\omega }y+f}\hfill & \mathrm{in}\hfill & Q,\hfill \\ \hfill \frac{\partial y}{\partial \nu }& =\hfill & 0\hfill & \mathrm{on}\hfill & \mathrm{\Sigma },\hfill \\ \hfill y(·,0)& =\hfill & y^0\hfill & \mathrm{in}\hfill & \mathrm{\Omega },\hfill \end{array}}\right.$$

Proving that $\mathbb{F}$ has a fixed point will be enough to prove that (25) has a solution in $L^2\left(Q\right)$. To this end, we use the Schauder’s fixed-point theorem.

Let $z\in B$. Since $\mathbb{F}\left(z\right)=y\left(z\right)$, where $y\left(z\right)$ is the unique weak solution to (A1), using Corollary A1, we have that there exist $C(a_{\infty },b_{\infty },g_{\infty },\parallel v{\parallel }_{L^{\infty }\left({\omega }_T\right)},T)>0$ and $C\left(M,g_{\infty },b_{\infty },a_{\infty },{\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},\mathrm{\Gamma },T\right)>0$ such that

$${\displaystyle {\parallel y\parallel }_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}\le C(a_{\infty },{\vartheta }_{\infty },b_{\infty },g_{\infty }{,\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},T)\left({\parallel f\parallel }_{L^2\left(Q\right)}+{\parallel y^0\parallel }_{L^2\left(\mathrm{\Omega }\right)}\right),}$$

(51)

and

$${\parallel y\parallel }_{\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))}\le C\left(M,\mu ,g_{\infty },{\vartheta }_{\infty },b_{\infty },a_{\infty },{\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},T\right)\left({\parallel f\parallel }_{L^2\left(Q\right)}+{\parallel y^0\parallel }_{L^2\left(\mathrm{\Omega }\right)}\right).$$

(52)

From (51), we deduce that $\mathbb{F}\left(B\right)\subset B$, and from (52), we have that $\mathbb{F}\left(B\right)$ is relatively compact in B because $\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ is relatively compact in $L^2\left(Q\right)$. It remains to be proven that $\mathbb{F}$ is continuous on B to obtain that $\mathbb{F}$ a fixed point. So, let $z_k$ be a sequence of B such that

$$z_k⟶z\mathrm{strongly}\mathrm{in}{L}^2\left(Q\right),\mathrm{as}k\to +\infty .$$

(53)

Let $y_k=\mathbb{F}\left(z_k\right)$, the solution to (A1) with $z=z_k$:

$$\left\{{\displaystyle \begin{array}{ccccc}\hfill {\displaystyle \frac{\partial y_k}{\partial t}-\Delta y_k+b_0{h}_{y_k}^g+g\left(t\right)h_{y_k}^{\vartheta }+\mu \beta \left(h_{z_k}^g\right)y_k+a_0{y}_k}& =\hfill & {\displaystyle v{\chi }_{\omega }{y}_k+f}\hfill & \mathrm{in}\hfill & Q,\hfill \\ \hfill \frac{\partial y_k}{\partial \nu }& =\hfill & 0\hfill & \mathrm{on}\hfill & \mathrm{\Sigma },\hfill \\ \hfill y_k(·,0)& =\hfill & y^0\hfill & \mathrm{in}\hfill & \mathrm{\Omega },\hfill \end{array}}\right.$$

Then, from Corollary A1, we have that $y_k$ verifies

$$\begin{array}{c}\hfill {\displaystyle {\int }_{\mathrm{\Omega }}{y}^0\varphi (·,0)\mathrm{d}x=-{\int }_0^T{〈\frac{\partial \varphi }{\partial t},y_k〉}_{{\left(H^1\left(\mathrm{\Omega }\right)\right)}^{\prime },H^1\left(\mathrm{\Omega }\right)}dt-{\int }_{{\omega }_T}v{y}_k\varphi dxdt}\\ \hfill {\displaystyle +{\int }_Q\nabla \varphi ·\nabla y_{k}dxdt+\mu {\int }_Q\beta \left(h_{z_k}^g\right)y_k\varphi dxdt+{\int }_Q{a}_0{y}_k\varphi dxdt}\\ \hfill {\displaystyle +{\int }_Q{b}_0{h}_{y_k}^g\varphi dxdt+{\int }_{Q}g\left(t\right)h_{y_k}^{\vartheta }\varphi dxdt-{\int }_{Q}f\varphi dxdt,}\end{array}$$

(54)

and there exist positive constants $C_1=C(a_{\infty },{\vartheta }_{\infty },b_{\infty },g_{\infty }{,\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},T)$ and $C_2=C\left(M,\mu ,g_{\infty },b_{\infty },{\vartheta }_{\infty },a_{\infty },{\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)}T\right)$ such that

$${\displaystyle \parallel y_k{\parallel }_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}\le C_1\left({\parallel f\parallel }_{L^2\left(Q\right)}+{\parallel y^0\parallel }_{L^2\left(\mathrm{\Omega }\right)}\right),}$$

(55)

and

$$\parallel y_k{\parallel }_{\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))}\le C_2\left({\parallel f\parallel }_{L^2\left(Q\right)}+{\parallel y^0\parallel }_{L^2\left(\mathrm{\Omega }\right)}\right).$$

(56)

From (56), we have that there exists $y\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ and a subsequence of ${\left(y_k\right)}_k$ still denoted by ${\left(y_k\right)}_k$ such that

$$y_k\rightharpoonup y\mathrm{weakly}\mathrm{in}{L}^2(0,T;H^1\left(\mathrm{\Omega }\right))\mathrm{as}k\to +\infty ,$$

(57)

and

$${\displaystyle \frac{\partial y_k}{\partial t}\rightharpoonup \frac{\partial y}{\partial t}\mathrm{weakly}\mathrm{in}{L}^2\left(0,T;{\left(H^1\left(\mathrm{\Omega }\right)\right)}^{\prime }\right)\mathrm{as}k\to +\infty .}$$

(58)

Using Lemma 1 with $y_n=y_k$, we deduce from (56) that

$$y_k\to y\mathrm{strongly}\mathrm{in}{L}^2\left(Q\right)\mathrm{as}k\to +\infty .$$

(59)

Thanks to Lemma 3, we have from (22) and (23) that

$$h_{z_k}^g\to h_z^g\mathrm{strongly}\mathrm{in}{L}^2\left(Q\right)\mathrm{as}k\to +\infty ,$$

(60)

$$h_{y_k}^g\to h_y^g\mathrm{strongly}\mathrm{in}{L}^2\left(Q\right)\mathrm{as}k\to +\infty$$

(61)

and

$$h_{y_k}^{\vartheta }\to h_y^{\vartheta }\mathrm{strongly}\mathrm{in}{L}^2\left(Q\right)\mathrm{as}k\to +\infty .$$

(62)

Applying Lemma 3 again, we have from (24) with $l=0$ that

$$\beta \left(h_{z_k}^g\right)\to \beta \left(h_z^g\right)\mathrm{strongly}\mathrm{in}{L}^2\left(Q\right)\mathrm{as}k\to +\infty .$$

(63)

From (6), (7) and Lemma 5, we obtain for any $\varphi \in H\left(Q\right)$ that

$$\varphi \beta \left(h_{z_k}^g\right)⟶\varphi \beta \left(h_z^g\right)\mathrm{strongly}\mathrm{in}{L}^2\left(Q\right),\mathrm{as}k\to +\infty .$$

(64)

Passing to the limit in (54) when $k\to +\infty$ while using (5), (16), (27), (26), (57), (59), (61), (62), (63), and (64), we obtain that

$$\begin{array}{ccc}{\displaystyle {\int }_{\mathrm{\Omega }}{y}^0\varphi (·,0)\mathrm{d}x+{\int }_{Q}f\varphi dxdt}\hfill & =\hfill & {\displaystyle -{\int }_0^T{〈\frac{\partial \varphi }{\partial t},y〉}_{{\left(H^1\left(\mathrm{\Omega }\right)\right)}^{\prime },H^1\left(\mathrm{\Omega }\right)}dt-{\int }_{{\omega }_T}vy\varphi dxdt}\hfill \\ \hfill & +\hfill & {\displaystyle {\int }_Q\nabla \varphi ·\nabla ydxdt+\mu {\int }_Q\beta \left(h_z^g\right)y\varphi dxdt+{\int }_Q{a}_{0}y\varphi dxdt}\hfill \\ \hfill & +\hfill & {\displaystyle {\int }_Q{b}_0{h}_y^g\varphi dxdt+{\int }_{Q}g\left(t\right)h_y^{\vartheta }\varphi dxdt.}\hfill \end{array}$$

(65)

Consequently, y is a weak solution to (A1). The uniqueness of the solution to (A1) allows us to conclude that $\mathbb{F}\left(z_k\right)=y_k$ converges strongly to $\mathbb{F}\left(z\right)=y$ in $L^2\left(Q\right)$. This means that $\mathbb{F}$ is continuous on B.

Step 2. We prove estimate (48)

If we set $z=e^{-rt}y$ with $r=a_{\infty }+({\vartheta }_{\infty }+b_{\infty })g_{\infty }T+{\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)}+1,$ where y is a solution to (25), we obtain that $z\in W(0,T;H^1\left(\mathrm{\Omega }\right))$ is a solution to

$$\left\{\begin{array}{c}\hfill {\displaystyle \frac{\partial z}{\partial t}-\Delta z+\mu \beta \left(h_y^g\right)z=-e^{-rt}{b}_0{h}_{e^{r(·)}z}^g-e^{-rt}g\left(t\right)h_{e^{r(·)}z}^{\vartheta }}\\ \hfill {\displaystyle -(a_0+r)z+v{\chi }_{\omega }z+e^{-rt}f}& \mathrm{in}\hfill & Q,\hfill \\ \hfill {\displaystyle \frac{\partial z}{\partial \nu }=0}& \mathrm{on}\hfill & \mathrm{\Sigma },\hfill \\ \hfill {\displaystyle z(·,0)=y^0}& \mathrm{in}\hfill & \mathrm{\Omega },\hfill \end{array}\right.$$

(66)

where in view of notation (15), ${\displaystyle h_{e^{r(·)}z}^g(x,t)={\int }_0^t{e}^{rs}g\left(s\right)z(x,s)ds}$ and ${\displaystyle h_{e^{r(·)}z}^{\vartheta }(x,t)={\int }_{T-t}^T{e}^{rs}\vartheta (x,s)z(x,s)ds,}$ for a.e. $(x,t)\in Q$.

Then, proceeding exactly as for the computation of (A11) and (A13), we prove that there there exist $C(a_{\infty },{\vartheta }_{\infty },b_{\infty },g_{\infty },\parallel v{\parallel }_{L^{\infty }\left({\omega }_T\right)},T)>0$ and $C\left(M,\mu ,g_{\infty },b_{\infty },{\vartheta }_{\infty },a_{\infty },{\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},T\right)>0$ such that

$${\displaystyle {\parallel y\parallel }_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}\le C(a_{\infty },{\vartheta }_{\infty },b_{\infty },g_{\infty }{,\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},T)\left({\displaystyle {\parallel f\parallel }_{L^2\left(Q\right)}+{\parallel y^0\parallel }_{L^2\left(\mathrm{\Omega }\right)}}\right).}$$

$$\begin{array}{c}{\parallel y\parallel }_{\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))}\le C\left(M,\mu ,g_{\infty },b_{\infty },{\vartheta }_{\infty },a_{\infty },{\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},T\right)\left({\parallel f\parallel }_{L^2\left(Q\right)}+{\parallel y^0\parallel }_{L^2\left(\mathrm{\Omega }\right)}\right),\hfill \end{array}$$

Step 3. We show the uniqueness of the solution.

Let $y_1,y_2$ be two solutions to (25) with the same source term f and initial value $y^0$. If we set $z=e^{-rt}(y_1-y_2$) with $r>0$, we obtain that $z\in W(0,T;H^1\left(\mathrm{\Omega }\right))$ is a solution to

$$\left\{\begin{array}{ccccc}\hfill {\displaystyle \frac{\partial z}{\partial t}-\Delta z+\eta (y_1,y_2,z)+(\beta \left(h_{y_1}^g\right)+a_0+r)z}& =\hfill & {\displaystyle v{\chi }_{\omega }z}\hfill & \mathrm{in}\hfill & Q,\hfill \\ \hfill {\displaystyle \frac{\partial z}{\partial \nu }}& =\hfill & {\displaystyle 0}\hfill & \mathrm{on}\hfill & \mathrm{\Sigma },\hfill \\ \hfill {\displaystyle z(·,0)}& =\hfill & {\displaystyle 0}\hfill & \mathrm{in}\hfill & \mathrm{\Omega },\hfill \end{array}\right.$$

(67)

where

$$\eta (y_1,y_2,z)=\mu \beta \left(h_{y_1}^g\right)z+\mu e^{-rt}\left[\beta \left(h_{y_1}^g\right)-\beta \left(h_{y_2}^g\right)\right]y_2+e^{-rt}{b}_0{h}_{e^{r_1(·)}z}^g+e^{-rt}g\left(t\right)h_{e^{r(·)}z}^{\vartheta }.$$

If we multiply the first equation of (67) by z and integrate over Q, we obtain

$$\begin{array}{c}{\displaystyle \frac{1}{2}{∥z(·,T)∥}_{L^2\left(\mathrm{\Omega }\right)}^2+{\parallel \nabla z\parallel }_{L^2\left(Q\right)}^2+{\int }_Q\beta \left(h_{y_1}^g\right)z^{2}dxdt+r{\parallel z\parallel }_{L^2\left(Q\right)}^2=}\hfill \\ {\displaystyle -\mu {\int }_Q{e}^{-rt}\left[\beta \left(h_{y_1}^g\right)-\beta \left(h_{y_2}^g\right)\right]y_{2}zdxdt-{\int }_Q{e}^{-rt}{b}_0\left({\int }_0^t{e}^{r_{1}s}g\left(s\right)z(x,s)ds\right)zdxdt-}\hfill \\ {\displaystyle {\int }_Q{e}^{-rt}g\left(t\right)\left({\int }_{T-t}^T{e}^{r_{1}s}\vartheta (x,s)z(x,s)ds\right)zdxdt-{\int }_Q{a}_0{z}^{2}dxdt+{\int }_{{\omega }_T}vs.z^{2}dxdt}\hfill \end{array}$$

which in view of (6) and the Mean Value Theorem implies that

$$\begin{array}{ccc}\hfill {\displaystyle {\parallel \nabla z\parallel }_{L^2\left(Q\right)}^2+r{\parallel z\parallel }_{L^2\left(Q\right)}^2}& \le \hfill & {\displaystyle M\mu {\int }_Q\left({\int }_0^t\left|g\left(s\right)\right|\left|z(x,s)\right|ds\right)\left|y_2\right|\left|z\right|dxdt+a_{\infty }{\parallel z\parallel }_{L^2\left(Q\right)}^2}\hfill \\ & +\hfill & {\displaystyle b_{\infty }{\int }_Q\left({\int }_0^t\left|g\left(s\right)\right|\left|z(x,s)\right|ds\right)\left|z\right|dxdt+{\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)}{\parallel z\parallel }_{L^2\left(Q\right)}^2}\hfill \\ & +\hfill & {\displaystyle g_{\infty }{\int }_Q\left({\int }_{T-t}^T\left|\vartheta (x,s)\right|\left|z(x,s)\right|ds\right)\left|z\right|dxdt.}\hfill \end{array}$$

Hence, thanks to Theorem 2, (20) and (21), we deduce that

$$\begin{array}{ccc}\hfill {\displaystyle {\parallel \nabla z\parallel }_{L^2\left(Q\right)}^2+r{\parallel z\parallel }_{L^2\left(Q\right)}^2}& \le \hfill & {\displaystyle M\mu g_{\infty }TC\left({\parallel f\parallel }_{L^{\infty }\left({\omega }_T\right)}+{\parallel y^0\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}\right){\parallel z\parallel }_{L^2\left(Q\right)}^2}\hfill \\ & +\hfill & (b_{\infty }{g}_{\infty }T+{\vartheta }_{\infty }{g}_{\infty }T+a_{\infty }+{\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)}{)\parallel z\parallel }_{L^2\left(Q\right)}^2,\hfill \end{array}$$

where $C=C(d,T,\mathrm{\Omega },a_{\infty },b_{\infty },g_{\infty })>0$.

Choosing ${\displaystyle r\ge M\mu g_{\infty }TC\left({\parallel f\parallel }_{L^{\infty }\left({\omega }_T\right)}+{\parallel y^0\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}\right)+(b_{\infty }{g}_{\infty }T+{\vartheta }_{\infty }{g}_{\infty }T+a_{\infty }+{\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)})+1,}$

$${\displaystyle {\parallel \nabla z\parallel }_{L^2\left(Q\right)}^2+{\parallel z\parallel }_{L^2\left(Q\right)}^2\le 0,}$$

from which we deduce that $z=0$ in Q. This means that $y_1=y_2\mathrm{in}Q.$ □

Corollary 2.

Let $v\in L^{\infty }\left({\omega }_T\right)$ and $y^0\in L^{\infty }\left(\mathrm{\Omega }\right)$. Also, let $g\in L^{\infty }(0,T)$ and $\beta \in {\mathcal{C}}^2\left(\mathbb{R}\right)$ be such that (5), (6) and (7) hold. Then, there exists a unique weak solution $y\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ to (2) provided that

$$\begin{array}{ccc}{\displaystyle {\int }_{\mathrm{\Omega }}{y}^0\varphi (·,0)\mathrm{d}x}\hfill & =\hfill & {\displaystyle -{\int }_0^T{〈\frac{\partial \varphi }{\partial t}\left(t\right),y\left(t\right)〉}_{{\left(H^1\left(\mathrm{\Omega }\right)\right)}^{\prime },H^1\left(\mathrm{\Omega }\right)}dt}\hfill \\ \hfill & +\hfill & {\displaystyle {\int }_Q\nabla \varphi ·\nabla ydxdt+{\int }_Q\beta \left(h_y^g\right)y\varphi dxdt-{\int }_{{\omega }_T}vy\varphi dxdt,}\hfill \end{array}$$

(68)

holds for every $\varphi \in H\left(Q\right)=\{\phi \in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))\mathrm{such}\mathrm{that}\varphi (·,T)=0\}$.

Moreover, there exist positive constants $C(g_{\infty },\parallel v{\parallel }_{L^{\infty }\left({\omega }_T\right)},T)$ and $C\left(M,g_{\infty },{\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},T\right)$ depending continuously on ${\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)}$ such that

$${\displaystyle {\parallel y\parallel }_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}\le C(g_{\infty }{,\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},T)\parallel y^0{\parallel }_{L^2\left(\mathrm{\Omega }\right)}}$$

(69)

and

$${\parallel y\parallel }_{\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))}\le C\left(M,g_{\infty },{\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},T\right){\parallel y^0\parallel }_{L^2\left(\mathrm{\Omega }\right)}.$$

(70)

Proof. 

Proceeding as in Theorem 3 while taking $\mu =1$, $f\equiv 0$, $\vartheta \equiv 0,$ $a_0\equiv 0$ and $b_0\equiv 0$, we obtain the existence of a weak solution to (2) as well as its uniqueness and the estimates (69) and (70). □

Lemma 6.

Let $g\in L^{\infty }(0,T)$, $v\in L^{\infty }\left({\omega }_T\right)$, $y^0\in L^{\infty }\left(\mathrm{\Omega }\right)$ and $\beta \in {\mathcal{C}}^2\left(\mathbb{R}\right)$ be such that (6) and (7) hold. Also, let $y=y\left(v\right)\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))\cap L^{\infty }\left(Q\right)$ be the solution of (2). Then, there exist $M>0$ and $C(d,T,\mathrm{\Omega })>0$ such that

$$\parallel y{\beta }^{\prime }\left(h_y^g\right){\parallel }_{L^{\infty }\left(Q\right)}\le MC\left({d,T,\mathrm{\Omega },\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)}\right){\parallel y^0\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},$$

(71)

and

$$\parallel \beta \left(h_y^g\right){\parallel }_{L^{\infty }\left(Q\right)}\le M.$$

(72)

Proof. 

Since $y=y\left(v\right)\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))\cap L^{\infty }\left(Q\right)$ is solution of (2), using (6), (7) and (45), we deduce (71) and (72) □

We now consider the system:

$$\left\{\begin{array}{c}\hfill {\displaystyle -\frac{\partial q}{\partial t}-\Delta q+\beta \left(h_y^g\right)q=v{\chi }_{\omega }q}\\ \hfill {\displaystyle -g\left(t\right)\left({\int }_t^{T}y(x,\tau ){\beta }^{\prime }\left(h_y^g\right)(x,\tau )q(x,\tau )d\tau \right)}& \mathrm{in}\hfill & Q,\hfill \\ \hfill {\displaystyle \frac{\partial q}{\partial \nu }=0}& \mathrm{in}\hfill & \mathrm{\Sigma },\hfill \\ \hfill q(v;·,T)=y(v;·,T)-y_d& \mathrm{in}\hfill & \mathrm{\Omega }.\hfill \end{array}\right.$$

(73)

Proposition 1.

Let $v\in L^{\infty }\left({\omega }_T\right)$ and $y_d\in L^{\infty }\left(\mathrm{\Omega }\right)$ and $y=y\left(v\right)\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))\cap L^{\infty }\left(Q\right)$ be solution to (2). Also, let $g\in L^{\infty }(0,T)$ and $\beta \in {\mathcal{C}}^2\left(\mathbb{R}\right)$ be such that (5), (6) and (7) hold. Then, (73) has a unique weak solution $q\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))\cap L^{\infty }\left(Q\right)$. Moreover, there exist $C=C(M,d,T,\mathrm{\Omega },\parallel v{\parallel }_{L^{\infty }\left({\omega }_T\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty })>0$ such that

$${\parallel q\parallel }_{\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))}\le C\left(\parallel y^0{\parallel }_{L^2\left(\mathrm{\Omega }\right)}+{\parallel y_d\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}\right)$$

(74)

and

$${\parallel q\parallel }_{L^{\infty }\left(Q\right)}\le C\left(\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}+{\parallel y_d\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}\right).$$

(75)

Proof. 

If we make the change for variable $t\mapsto T-t$ in (73), we obtain that $\tilde{q}(x,t)=q(x,T-t)$ is a solution to

$$\left\{\begin{array}{c}\hfill {\displaystyle \frac{\partial \tilde{q}}{\partial t}-\Delta \tilde{q}+\beta \left({\tilde{h}}_y^g\right)\tilde{q}=\tilde{v}{\chi }_{\omega }\tilde{q}}\\ \hfill -g(T-t)\left({\int }_{t-T}^{T}y(x,\tau ){\beta }^{\prime }\left(h_y^g\right)(x,\tau )q(x,\tau )d\tau \right)& =\hfill & 0\hfill & \mathrm{in}\hfill & Q,\hfill \\ \hfill {\displaystyle \frac{\partial \tilde{q}}{\partial \nu }=0}& \mathrm{in}\hfill & \mathrm{\Sigma },\hfill \\ \hfill \tilde{q}(v;·,0)=y(v;·,0)-y_d& \mathrm{in}\hfill & \mathrm{\Omega },\hfill \end{array}\right.$$

(76)

where ${\displaystyle {\tilde{h}}_y^g(x,t)={\int }_0^{T-t}g\left(s\right)y(x,s)ds}$, for almost every $(x,t)\in Q$, $\tilde{v}(x,t)=v(x,T-t)$. Since $y=y\left(v\right)\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))\cap L^{\infty }\left(Q\right)$ is a solution of (2), we have on the one hand that $y(v;·,0)-y_d=y^0-y_d\in L^{\infty }\left(\mathrm{\Omega }\right)$, and on the other hand, using (71) and (72), we have that $\vartheta =y{\beta }^{\prime }\left(h_y^g\right)\in L^{\infty }\left(Q\right)$ and $a_0=\beta \left(h_y^g\right)\in L^{\infty }\left(Q\right)$. Therefore, $\tilde{q}(x,t)=q(x,T-t)$ is a solution to (A1) with $\mu =0$, $b_0=0,$$f=0$, $a_0=\beta \left(h_y^g\right)$ and $\vartheta =y{\beta }^{\prime }\left(h_y^g\right)$. Thanks to Corollary A1 and Theorem 2, we deduce that there exists a unique weak solution $\tilde{q}\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))\cap L^{\infty }\left(Q\right)$ to (76). Moreover, using (71) and (72), the continuous embedding of $L^{\infty }\left(\mathrm{\Omega }\right)$ into $L^2\left(\mathrm{\Omega }\right)$, the continuous embedding of $L^{\infty }\left(Q\right)$ into $L^2\left(Q\right)$, (A16) and (31) with $\mu =0$, $b_0=0,$$f=0$, $a_0=\beta \left(h_y^g\right),$$\vartheta =y{\beta }^{\prime }\left(h_y^g\right)$, we have that there exists $C=C(M,d,T,\mathrm{\Omega },\parallel v{\parallel }_{L^{\infty }\left({\omega }_T\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty })>0$ such that

$${\displaystyle \parallel \tilde{q}{\parallel }_{\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))}\le C_1\left(\parallel y_d{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}+{\parallel y^0\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}\right).}$$

Using (31), (71) and (72) again with $\mu =0$, $b_0=0,$$f=0$, $a_0=\beta \left(h_y^g\right),$$\vartheta =y{\beta }^{\prime }\left(h_y^g\right)$, then (70), the embedding $\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ into $C([0,T];L^2\left(\mathrm{\Omega }\right))$, (71) and (72), we deduce that there exists $C=C(d,g_{\infty },M,T,\mathrm{\Omega },\parallel v{\parallel }_{L^{\infty }\left({\omega }_T\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)})>0$ such that

$$\parallel \tilde{q}{\parallel }_{L^{\infty }\left(Q\right)}\le C\left(\parallel y_d{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}+{\parallel y^0\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}\right).$$

Since $\tilde{q}(x,t)=q(x,T-t)$, we deduce (74) and (75). □

4. Resolution of an Optimal Control Problem

From Corollaries 1 and 2, we know that the weak solution y of system (2) resides in $\mathbb{X}=\mathbb{W}(0,T,H^1\left(\mathrm{\Omega }\right))\cap L^{\infty }\left(Q\right).$ We introduce the following space:

$$\mathbb{Y}=\left\{{\displaystyle \varphi \in \mathbb{X}:\frac{\partial \varphi }{\partial t}-\Delta \varphi \in L^{\infty }\left(Q\right)}\right\},$$

(77)

and we observe that endowed with the graph norm

$${\parallel \varphi \parallel }_{\mathbb{Y}}={\parallel \varphi \parallel }_{\mathbb{X}}+{∥{\displaystyle \frac{\partial \varphi }{\partial t}}-\Delta \varphi ∥}_{L^{\infty }\left(Q\right)},\forall \varphi \in \mathbb{Y},$$

$\mathbb{Y}$ is a Banach space.

We also set the mapping $\mathbb{S}:\mathbb{Y}\times L^{\infty }\left({\omega }_T\right)\to L^{\infty }\left(Q\right)\times L^{\infty }\left(\mathrm{\Omega }\right)$ given by

$$(y,v)\mapsto \mathbb{S}(y,v):=\left({\displaystyle \frac{\partial y}{\partial t}-\Delta y+\beta \left({\int }_0^{t}g\left(s\right)y(x,s)ds\right)y-v{\chi }_{\omega }y,y\left(0\right)-y^0}\right),$$

(78)

which is well defined, and the state equation as well as the initial data of system (2) can be viewed as

$$\mathbb{S}(y,v)=0.$$

Moreover, taking into account the control-to-state mapping

$$\begin{array}{ccccc}\hfill \mathcal{S}& \hfill :& L^{\infty }\left({\omega }_T\right)& \to \hfill & \mathbb{X}\hfill \\ & & v& \mapsto \hfill & y\left(v\right),\hfill \end{array}$$

we can rewrite the optimal control problem (1) as:

$$\underset{v\in {\mathcal{U}}_{ad}}{inf}J(\mathcal{S}\left(v\right),v),$$

(79)

where

$$J\left(\mathcal{S}\right(v),v)=J\left(y\right(v),v).$$

(80)

4.1. Sensitivity Analysis

Lemma 7.

Let $\beta \in {\mathcal{C}}^2\left(\mathbb{R}\right)$ be such that (6)–(7) hold and $y=y\left(v\right)\in \mathbb{X}$ be the solution of (2). Then, the mapping $\mathbb{S}$ is of class ${\mathcal{C}}^2$. Furthermore, the control-to-state mapping $\mathcal{S}$ is also of class ${\mathcal{C}}^2$.

Proof. 

It is clear that the second component of $\mathbb{S}$ is of class ${\mathcal{C}}^{\infty }$ with respect to y and v. Since $\beta \in {\mathcal{C}}^2\left(\mathbb{R}\right),$ the first component of $\mathbb{S}$ is also of class ${\mathcal{C}}^2$. Therefore, we can conclude that $\mathbb{S}$ is of class ${\mathcal{C}}^2$. Moreover,

$${\displaystyle \frac{\partial }{\partial y}\mathbb{S}(y,v)\phi =\left(\frac{\partial \phi }{\partial t}-\Delta \phi +y{\beta }^{\prime }\left(h_y^g\right)h_{\phi }^g+\beta \left(h_y^g\right)\phi -v{\chi }_{\omega }\phi ,\phi \left(0\right)\right).}$$

Let ${\phi }^0\in L^{\infty }\left(\mathrm{\Omega }\right)$ and $f\in L^{\infty }\left(Q\right)$. We consider the linear problem

$$\left\{\begin{array}{ccccc}\hfill {\displaystyle \frac{\partial \phi }{\partial t}-\Delta \phi +y{\beta }^{\prime }\left(h_y^g\right)h_{\phi }^g+\beta \left(h_y^g\right)\phi }& =\hfill & f+v{\chi }_{\omega }\phi \hfill & \mathrm{in}\hfill & Q,\hfill \\ \hfill {\displaystyle \frac{\partial \phi }{\partial \nu }}& =\hfill & 0\hfill & \mathrm{in}\hfill & \mathrm{\Sigma },\hfill \\ \hfill \phi (·,0)& =\hfill & {\phi }^0\hfill & \mathrm{in}\hfill & \mathrm{\Omega }.\hfill \end{array}\right.$$

(81)

Let $b_0=y{\beta }^{\prime }\left(h_y^g\right)$ and $a_0=\beta \left(h_y^g\right)$. Then, from (71), (72), $b_0\in L^{\infty }\left(Q\right),$$a_0\in L^{\infty }\left(Q\right)$ and the system (81) is the same as (25) with $\mu =0,$$\vartheta =0$, $y^0={\phi }^0$. Consequently, we have from Theorems 2 and 3 that there exists a unique $\phi =\phi ({\phi }^0,f)\in \mathbb{X}$ which depends continuously on ${\phi }^0\in L^{\infty }\left(\mathrm{\Omega }\right)$ and on $f\in L^{\infty }\left(Q\right).$ Since $y\in L^{\infty }\left(Q\right)$ and $\beta \in {\mathcal{C}}^2\left(\mathbb{R}\right)$ satisfies (6) and (7), we have ${\displaystyle \frac{\partial \phi }{\partial t}-\Delta \phi =f+v{\chi }_{\omega }\phi -y{\beta }^{\prime }\left(h_y^g\right)h_{\phi }^g-\beta \left(h_y^g\right)\phi \in L^{\infty }\left(Q\right).}$ Therefore, $\phi \in \mathbb{Y}$. Hence, ${\displaystyle \frac{\partial }{\partial y}\mathbb{S}(y,v)}$ defines an isomorphism from $\mathbb{Y}$ to $L^{\infty }\left(Q\right)\times L^{\infty }\left(\mathrm{\Omega }\right)$. Using the Implicit Function Theorem, we deduce that $\mathbb{S}(y,v)=0$ implicitly defines the control-to-state operator $\mathcal{S}:v\mapsto y\left(v\right)$ which is itself of class ${\mathcal{C}}^2$. □

Proposition 2.

Let $v,w,\pi \in L^{\infty }\left({\omega }_T\right)$. Under the assumptions of Lemma 7, the first derivative of the control-to-state operator ${\mathcal{S}}^{\prime }\left(v\right)$ is given by ${\mathcal{S}}^{\prime }\left(v\right)w=y_w\left(v\right)$, where $y_w\left(v\right)\in \mathbb{X}$ is the unique weak solution to

$$\left\{\begin{array}{c}\hfill {\displaystyle \frac{\partial y_w\left(v\right)}{\partial t}-\Delta y_w\left(v\right)=-\beta \left(h_y^g\right)y_w\left(v\right)}\\ \hfill +(v{y}_w\left(v\right)+wy\left(v\right)){\chi }_{\omega }-y\left(v\right){\beta }^{\prime }\left(h_y^g\right)h_{y_w\left(v\right)}^g& in\hfill & Q,\hfill \\ \hfill {\displaystyle \frac{\partial y_w\left(v\right)}{\partial \nu }=0}& in\hfill & \mathrm{\Sigma },\hfill \\ \hfill y_w(v;·,0)=0& in\hfill & \mathrm{\Omega },\hfill \end{array}\right.$$

(82)

 where according to (15) ${\displaystyle h_y^g(x,t)={\int }_0^{t}g\left(s\right)y(x,s)ds}$ for almost every $(x,t)\in Q$. The second derivative ${\mathcal{S}}^{″}\left(v\right)$ is given by ${\mathcal{S}}^{″}\left(v\right)(w,\pi )=y_{w\pi }\left(v\right)$, where $y_{w\pi }\left(v\right)\in \mathbb{X}$ is the unique weak solution to

$$\left\{\begin{array}{c}\hfill {\displaystyle \frac{\partial }{\partial t}{y}_{w\pi }\left(v\right)-\Delta y_{w\pi }\left(v\right)=-\beta \left(h_y^g\right)y_{w\pi }\left(v\right)}\\ \hfill -y\left(v\right){\beta }^{\prime }\left(h_y^g\right)h_{y_{w\pi }\left(v\right)}^g+v{\chi }_{\omega }{y}_{w\pi }\left(v\right)\\ \hfill +\pi {\chi }_{\omega }{y}_w\left(v\right)+w{\chi }_{\omega }{y}_{\pi }\left(v\right)+\mathcal{F}& in\hfill & Q,\hfill \\ \hfill {\displaystyle \frac{\partial }{\partial \nu }{y}_{w\pi }\left(v\right)=0}& in\hfill & \mathrm{\Sigma },\hfill \\ \hfill y_{w\pi }(v;·,0)=0& in\hfill & \mathrm{\Omega },\hfill \end{array}\right.$$

(83)

where

$$\begin{array}{ccc}\hfill \mathcal{F}& =\hfill & {\displaystyle -{\beta }^{\prime }\left(h_y^g\right)y_w{h}_{y_{\pi }}^g-{\beta }^{\prime }\left(h_y^g\right)y_{\pi }{h}_{y_w}^g-{\beta }^{″}\left(h_y^g\right)y\left(v\right)h_{y_w}^g{h}_{y_{\pi }}^g,}\hfill \end{array}$$

(84)

and $y_w={\mathcal{S}}^{\prime }\left(v\right)w,y_{\pi }={\mathcal{S}}^{\prime }\left(v\right)\pi$ are respective solutions to (82) with $w=w$ and $w=\pi$. Moreover, for every $v,w\in L^{\infty }\left(\mathrm{\Omega }\right)$, the linear mapping $w\mapsto y_w={\mathcal{S}}^{\prime }\left(v\right)w$ can be extended to a linear continuous mapping on $L^2\left({\omega }_T\right)$ and there exists $C=C(M,d,T,\mathrm{\Omega },\parallel v{\parallel }_{L^{\infty }\left({\omega }_T\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty })>0$ such that

$$\parallel y_w{\parallel }_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}\le C{\parallel w\parallel }_{L^2\left({\omega }_T\right)}$$

(85)

and

$$\parallel y_w{\parallel }_{\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))}\le C{\parallel w\parallel }_{L^2\left({\omega }_T\right)}.$$

(86)

Proof. 

Let $v,w\in L^{\infty }\left({\omega }_T\right)$. We have from Lemma 7 that the control-to-state mapping $\mathcal{S}:L^{\infty }\left({\omega }_T\right)\to \mathbb{X},v\mapsto y$ is of class ${\mathcal{C}}^2$. Therefore, ${\mathcal{S}}^{\prime }\left(v\right)w=y_w\left(v\right)$ and ${\mathcal{S}}^{″}\left(v\right)(w,\pi )=y_{w\pi }\left(v\right)$ exist. Using (6) and (7), we obtain after some computations that $y_w\left(v\right)$ is a solution to (82) and $y_{w\pi }\left(v\right)$ is a solution to (83).

Applying Theorems 2 and 3 to $y_w$ solution to the system (82) with $f=wy\left(v\right)$, $b_0=y{\beta }^{\prime }\left(h_y^g\right),$$a_0=\beta \left(h_y^g\right)$, $\vartheta =0$ and $\mu =0$, we deduce that $y_w\in \mathbb{X}$. Moreover, using (31), (47), (48), (71) and (72), we obtain that there exists $C=C(M,d,T,\mathrm{\Omega },\parallel v{\parallel }_{L^{\infty }\left({\omega }_T\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty })>0$ such that

$$\begin{array}{ccc}\hfill {\displaystyle \parallel y_w{\parallel }_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}}& \le \hfill & {C\parallel w\parallel }_{L^2\left({\omega }_T\right)},\hfill \\ \hfill {\displaystyle \parallel y_w{\parallel }_{\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))}}& \le \hfill & {C\parallel w\parallel }_{L^2\left({\omega }_T\right)}\hfill \end{array}$$

and

$$\parallel y_w{\parallel }_{L^{\infty }\left(Q\right)}\le {C(M,d,T,\mathrm{\Omega },\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty }{)\parallel w\parallel }_{L^{\infty }\left({\omega }_T\right)}.$$

(87)

Proceeding as above, we also have that the $y_{\pi }$ solution to the system (82) with $w=\pi$ belongs to $\mathbb{X}$ and satisfies the same estimation as $y_w$. Since $y_w$ and $y_{\pi }$ belong to $L^{\infty }\left(Q\right)$, we have from Lemma 2 that $h_{y_{\pi }}^g$ and $h_{y_w}^g$ belong also to $L^{\infty }\left(Q\right)$. Consequently, using (6)–(7), the fact that the solution y to (2) belongs to $\mathbb{X}$, we obtain that $\mathcal{F}$ defined in (84) belongs to $L^{\infty }\left(Q\right)$. Therefore, applying Theorem 3 and Theorem 2 to $y_{w\pi }$ solution to (83) with $f=\mathcal{F}+\pi y_w\left(v\right)+w{y}_{\pi }\left(v\right)){\chi }_{\omega }\in L^{\infty }\left(Q\right)$, $\mu =0,$$\vartheta =0$, $b_0=y{\beta }^{\prime }\left(h_y^g\right)\in L^{\infty }\left(Q\right)$ and $a_0=\beta \left(h_y^g\right)\in L^{\infty }\left(Q\right),$ we deduce that the problem (83) has also a unique weak solution $y_{w\pi }\in \mathbb{X}.$ □

Proposition 3.

Let $v\in L^{\infty }\left({\omega }_T\right)$ and $y^0\in L^{\infty }\left(\mathrm{\Omega }\right)$. Let $g\in L^{\infty }(0,T)$ and $\beta \in {\mathcal{C}}^2\left(\mathbb{R}\right)$ be such that (5), (6) and (7) hold. Also, let $y=y\left(v\right)\in \mathbb{X}$ and $q=q\left(v\right)\in \mathbb{X}$ be the unique weak solution to (2) and to (73), respectively. Then, the mappings $v\mapsto \mathcal{S}\left(v\right)=y\left(v\right)$ and $v\mapsto q\left(v\right)$ are Lipschitz continuous functions from $L^2\left({\omega }_T\right)$ onto $L^2(0,T;H^1\left(\mathrm{\Omega }\right))$. More precisely, for all $v_1,v_2\in L^{\infty }\left({\omega }_T\right)$, there is More precisely, for all $v_1,v_2\in L^{\infty }\left({\omega }_T\right)$, there is a constant $C_1=C(M,d,T,\mathrm{\Omega },\parallel v_2{\parallel }_{L^{\infty }\left(Q\right)},\parallel v_1{\parallel }_{L^{\infty }\left(Q\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty })>0$ and $C_2=C\left(d,T,\mathrm{\Omega },\parallel v_2{\parallel }_{L^{\infty }\left({\omega }_T\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty },M,\parallel v_1{\parallel }_{L^{\infty }\left({\omega }_T\right)},{\parallel y_d\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}\right)>0$ such that the following estimates hold

$$\parallel \mathcal{S}\left(v_1\right)-\mathcal{S}\left(v_2\right){\parallel }_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}\le C_1{\parallel v_1-v_2\parallel }_{L^2\left({\omega }_T\right)}$$

(88)

and

$$\parallel q\left(v_1\right)-q\left(v_2\right){\parallel }_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}\le C_2{\parallel v_1-v_2\parallel }_{L^2\left({\omega }_T\right)}.$$

(89)

Proof. 

Let $v_1,v_2\in L^{\infty }\left({\omega }_T\right)$ and $r>0$. Let $p=e^{-rt}(y_1-y_2)$ where $y_1=y\left(v_1\right)$ and $y_2=y\left(v_2\right)$ are solutions of (2) with $v=v_1$ and $v=v_2,$ respectively. Then, $p\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ is a weak solution to

$$\left\{\begin{array}{c}\hfill {\displaystyle \frac{\partial p}{\partial t}+rp-\Delta p+\beta \left(h_{y_1}^g\right)p=v_1{\chi }_{\omega }p}\\ \hfill {\displaystyle +{\mathrm{e}}^{-rt}\left[(v_1-v_2){\chi }_{\omega }{y}_2-\left(\beta \left(h_{y_1}^g\right)-\beta \left(h_{y_2}^g\right)\right)y_2\right]}& \mathrm{in}\hfill & Q,\hfill \\ \hfill {\displaystyle \frac{\partial p}{\partial \nu }=0}& \mathrm{on}\hfill & \mathrm{\Sigma },\hfill \\ \hfill {\displaystyle p(·,0)=0}& \mathrm{in}\hfill & \mathrm{\Omega }.\hfill \end{array}\right.$$

(90)

We multiply the first equation in (90) by p and intergate by parts over Q, which gives us

$$\begin{array}{c}{\displaystyle \frac{1}{2}{∥p(·,T)∥}_{L^2\left(\mathrm{\Omega }\right)}^2+r{∥p∥}_{L^2\left(Q\right)}^2+{∥\nabla p∥}_{L^2\left(Q\right)}^2+{\int }_Q\beta \left(h_{y_1}^g\right)p^{2}dxdt}\hfill \\ {\displaystyle \le {\int }_{{\omega }_T}{v}_1{p}^{2}dxdt+{\int }_{{\omega }_T}{\mathrm{e}}^{-rt}\left|(v_1-v_2)\right|\left|y_2\right|\left|p\right|dxdt}\hfill \\ {\displaystyle +{\int }_Q{\mathrm{e}}^{-rt}\left|\beta \left(h_{y_1}^g\right)-\beta \left(h_{y_2}^g\right)\right|\left|y_2\right|\left|p\right|dxdt,}\hfill \end{array}$$

which in view of (6) and (45), we obtain that

$${\displaystyle \begin{array}{c}\hfill {\displaystyle \frac{1}{2}{∥p(·,T)∥}_{L^2\left(\mathrm{\Omega }\right)}^2+r{∥p∥}_{L^2\left(Q\right)}^2+{∥\nabla p∥}_{L^2\left(Q\right)}^2\le \frac{1}{2}{\parallel p\parallel }_{L^2\left(Q\right)}^2}\\ \hfill +C(d,T,\mathrm{\Omega },\parallel v_2{\parallel }_{L^{\infty }\left(Q\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)})\parallel v_1-v_2{\parallel }_{L^2\left({\omega }_T\right)}^2\\ \hfill +\parallel v_1{\parallel }_{L^{\infty }\left({\omega }_T\right)}{\parallel p\parallel }_{L^2\left(Q\right)}^2\\ \hfill +C(M,d,T,\mathrm{\Omega },\parallel v_2{\parallel }_{L^{\infty }\left(Q\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty }{,T)\parallel p\parallel }_{L^2\left(Q\right)}^2.\end{array}}$$

Choosing $r=C(M,d,T,\mathrm{\Omega },\parallel v_2{\parallel }_{L^{\infty }\left(Q\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty },T)+\parallel v_1{\parallel }_{L^{\infty }\left(\omega T\right)}+1$ in this latter inequality, ${\displaystyle \frac{1}{2}{∥p(·,T)∥}_{L^2\left(\mathrm{\Omega }\right)}^2+\frac{1}{2}{∥p∥}_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}^2\le C(d,T,\mathrm{\Omega },\parallel v_2{\parallel }_{L^{\infty }\left(Q\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)})\parallel v_1-v_2{\parallel }_{L^2\left({\omega }_T\right)}^2.}$ Since $p=e^{-rt}(y_1-y_1)$ we deduce there exists $C_1=C(M,d,T,\mathrm{\Omega },\parallel v_2{\parallel }_{L^{\infty }\left(Q\right)},\parallel v_1{\parallel }_{L^{\infty }\left(Q\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\Omega \right)},g_{\infty })>0$ such that (88) holds that

$$\begin{array}{c}\hfill {\displaystyle {∥y(v_1;·,T)-y(v_2;·,T)∥}_{L^2\left(\mathrm{\Omega }\right)}={∥y_1(·,T)-y_2(·,T)∥}_{L^2\left(\mathrm{\Omega }\right)}\le C_1{\parallel v_1-v_2\parallel }_{L^2\left({\omega }_T\right)}^2.}\end{array}$$

(91)

Next, let $v_1,v_2\in L^{\infty }\left({\omega }_T\right)$ and $r>0$. Let $z=e^{rt}(q_1-q_2)$ where $q_1=q\left(v_1\right)$ and $q_2=q\left(v_2\right)$ are solutions of (73) with $v=v_1$ and $v=v_2$, respectively. Then, $z\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ is a weak solution to

$$\left\{\begin{array}{ccccc}\hfill {\displaystyle -\frac{\partial z}{\partial t}-\Delta z+rz+\beta \left(h_{y_1}^g\right)z}& =\hfill & v_1{\chi }_{\omega }z+e^{rt}(v_1-v_2){\chi }_{\omega }{q}_2+{\eta }_q\hfill & \mathrm{in}\hfill & Q,\hfill \\ \hfill {\displaystyle \frac{\partial z}{\partial \nu }}& =\hfill & 0\hfill & \mathrm{in}\hfill & \mathrm{\Sigma },\hfill \\ \hfill z(·,T)& =\hfill & e^{rT}(y(v_1;·,T)-y(v_2;·,T))\hfill & \mathrm{in}\hfill & \mathrm{\Omega }\hfill \end{array}\right.$$

(92)

where

$$\begin{array}{ccc}\hfill {\eta }_q& =\hfill & {\displaystyle -e^{rt}g\left(t\right){\int }_t^T(y_1-y_2){\beta }^{\prime }\left(h_{y_1}^g\right)q_{2}d\tau -e^{rt}g\left(t\right){\int }_t^T{y}_1{e}^{-r\tau }{\beta }^{\prime }\left(h_{y_1}^g\right)zd\tau }\hfill \\ & -\hfill & {\displaystyle e^{rt}g\left(t\right){\int }_t^T{y}_2{q}_2\left({\beta }^{\prime }\left(h_{y_1}^g\right)-{\beta }^{\prime }\left(h_{y_2}^g\right)\right)d\tau -e^{rt}{q}_2(\beta \left(h_{y_1}^g\right)-\beta \left(h_{y_2}^g\right)).}\hfill \end{array}$$

Using the linearity of the integral, the notation (5), (6), (7), (15), (20) of Lemma 3, the Cauchy–Schwarz inequality and the fact that $y_1,y_2,q_1,q_2\in L^{\infty }\left(Q\right)$, we have that

$$\begin{array}{ccc}\hfill {\displaystyle {\int }_Q{\eta }_{q}zdxdt}& \le \hfill & {\displaystyle e^{rT}M{g}_{\infty }\parallel q_2{\parallel }_{L^{\infty }\left(Q\right)}T\parallel y_1-y_2{\parallel }_{L^2\left(Q\right)}{\parallel z\parallel }_{L^2\left(Q\right)}}\hfill \\ & +\hfill & {\displaystyle g_{\infty }M\parallel y_1{\parallel }_{L^{\infty }\left(Q\right)}T{\parallel z\parallel }_{L^2\left(Q\right)}^2}\hfill \\ & +\hfill & {\displaystyle M\parallel q_2{\parallel }_{L^{\infty }\left(Q\right)}{e}^{rT}{g}_{\infty }T\parallel y_1-y_2{\parallel }_{L^2\left(Q\right)}{\parallel z\parallel }_{L^2\left(Q\right)}}\hfill \\ & +\hfill & {\displaystyle g_{\infty }^2{e}^{rT}\parallel q_2{\parallel }_{L^{\infty }\left(Q\right)}\parallel y_2{\parallel }_{L^{\infty }\left(Q\right)}M{T}^2\parallel y_1-y_2{\parallel }_{L^2\left(Q\right)}{\parallel z\parallel }_{L^2\left(Q\right)},}\hfill \end{array}$$

which in view of Young’s inequality implies that

$$\begin{array}{c}{\displaystyle {\int }_Q{\eta }_{q}zdxdt\le \frac{1}{2}{e}^{2rT}{M}^2{g}_{\infty }^2\parallel q_2{\parallel }_{L^{\infty }\left(Q\right)}^2\left(1+T^2+T^2{g}_{\infty }^2{\parallel y_2\parallel }_{L^{\infty }\left(Q\right)}^2\right){\parallel y_1-y_2\parallel }_{L^2\left(Q\right)}^2}\hfill \\ +{\parallel z\parallel }_{L^2\left(Q\right)}^2\left[{\displaystyle \frac{3}{2}}+g_{\infty }M{\parallel y_1\parallel }_{L^{\infty }\left(Q\right)}T\right]\hfill \end{array}$$

(93)

If we multiply the first equation in (92) by z and integrate by parts over Q, we obtain

$$\begin{array}{c}{\displaystyle \frac{1}{2}{∥z(·,0)∥}_{L^2\left(\mathrm{\Omega }\right)}^2+r{∥z∥}_{L^2\left(Q\right)}^2+{∥\nabla z∥}_{L^2\left(Q\right)}^2+{\int }_Q\beta \left(h_{y_1}^g\right)z^{2}dxdt\le }\hfill \\ {\displaystyle \parallel v_1{\parallel }_{L^{\infty }\left({\omega }_T\right)}{\parallel z\parallel }_{L^2\left(Q\right)}^2+\frac{1}{2}{e}^{2rT}\parallel q_2{\parallel }_{L^{\infty }\left(Q\right)}^2\parallel v_1-v_2{\parallel }_{L^2\left({\omega }_T\right)}+\frac{1}{2}{\parallel z\parallel }_{L^2\left(Q\right)}^2+}\hfill \\ {\displaystyle {\int }_Q|{\eta }_q\left|\right|z|dxdt+\frac{1}{2}{∥e^{rT}(y(v_1;·,T)-y(v_2;·,T))∥}_{L^2\left(\mathrm{\Omega }\right)}^2.}\hfill \end{array}$$

Choosing $r=3+g_{\infty }M\parallel y_1{\parallel }_{L^{\infty }\left(Q\right)}T+{\parallel v_1\parallel }_{L^{\infty }\left({\omega }_T\right)}$ in this latter inequality and using (6) and (93), we deduce that

$$\begin{array}{c}\hfill {\displaystyle {∥z∥}_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}^2\le \frac{1}{2}{e}^{2rT}\parallel q_2{\parallel }_{L^{\infty }\left(Q\right)}^2{\parallel v_1-v_2\parallel }_{L^2\left({\omega }_T\right)}}\\ \hfill {\displaystyle +\frac{e^{2rT}}{2}{\parallel y_1-y_2\parallel }_{L^2\left(Q\right)}^2\left[1+M^2{g}_{\infty }^2\parallel q_2{\parallel }_{L^{\infty }\left(Q\right)}^2(1+T+\parallel y_2{\parallel }_{L^{\infty }\left(Q\right)}^2)\right]}\\ \hfill {\displaystyle +\frac{1}{2}{∥e^{rT}(y(v_1;·,T)-y(v_2;·,T))∥}_{L^2\left(\mathrm{\Omega }\right)}^2,}\end{array}$$

which in view of (45), (75), (88) and (91) implies that

$${\displaystyle {∥z∥}_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}^2\le C_2{\parallel v_1-v_2\parallel }_{L^2\left({\omega }_T\right)}^2,}$$

for some $C_2=C(d,T,\mathrm{\Omega },\parallel v_2{\parallel }_{L^{\infty }\left(Q\right)},\parallel v_1{\parallel }_{L^{\infty }\left(Q\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},M,g_{\infty },\parallel y_d{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)})>0$.

Since $z=e^{rt}(q_1-q_2)$, we have that

$${\displaystyle {∥q_1-q_2∥}_{L^2(0,T;H^1\left(\mathrm{\Omega }\right))}^2\le C{\parallel v_1-v_2\parallel }_{L^2\left({\omega }_T\right)}^2.}$$

□

4.2. Optimization Problem

We introduce the following notion of local solutions.

Definition 2.

We say that $v\in {\mathcal{U}}_{ad}$ is an $L^p$-local solution of (1) if there exists $\epsilon >0$ such that $J\left(\mathcal{S}\right(v),v)\le J\left(\mathcal{S}\right(w),w)$ for every $w\in {\mathcal{U}}_{ad}\cap B_{\epsilon }^p\left(v\right)$ where $B_{\epsilon }^p\left(v\right)=\{w\in L^p{\left(\mathrm{\Omega }\right):\parallel w-v\parallel }_{L^p\left(Q\right)}\le \epsilon \}$. We say that v is a strict local minimum of (1) if the above inequality is strict whenever $v\ne w$.

Theorem 4.

Let $N>0$, $v\in {\mathcal{U}}_{ad},$$y^0\in L^{\infty }\left(\mathrm{\Omega }\right)$ and $y^d\in L^{\infty }\left(\mathrm{\Omega }\right)$. Then, there exists at least a solution $v\in {\mathcal{U}}_{ad}$ of the optimal control problem (2) and (79).

Proof. 

Since $J\left(\mathcal{S}\right(v),v)\ge 0$ for all $v\in {\mathcal{U}}_{ad}$ we can have a minimizing sequence ${\left\{(\mathcal{S}\left(v_n\right),v_n)\right\}}_n\subset \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))\times {\mathcal{U}}_{ad}$ such that

$$\underset{n\to \infty }{lim}J(\mathcal{S}\left(v_n\right),v_n)\to \underset{w\in {\mathcal{U}}_{ad}}{min}J(\mathcal{S}\left(w\right),w).$$

From the structure of cost function J given by (3), there exists a constant $C>0$ independent of n such that

$$\parallel v_n{\parallel }_{L^2\left({\omega }_T\right)}\le C.$$

(94)

Since $y_n=\mathcal{S}\left(v_n\right)$ is the solution of (2) associated with the control $v_n$, we know that $y_n$ satisfies

$$\begin{array}{c}{\displaystyle -{\int }_0^T{〈\frac{\partial \varphi }{\partial t},y_n〉}_{{\left(H^1\left(\mathrm{\Omega }\right)\right)}^{\prime },H^1\left(\mathrm{\Omega }\right)}dt+{\int }_Q\nabla \varphi \nabla y_{n}dxdt+}\hfill \\ {\displaystyle {\int }_Q\beta \left(h_{y_n}^g\right)y_n\varphi dxdt={\int }_{\mathrm{\Omega }}{y}^0\varphi (·,0)\mathrm{d}x+{\int }_{{\omega }_T}{v}_n{y}_n\varphi dxdt}\hfill \\ \mathrm{for}\mathrm{every}\varphi \in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right)),\varphi (·,T)=0\mathrm{in}\mathrm{\Omega }\hfill \end{array}$$

(95)

and in view of (70) in Corollary 2, there exists a constant $C>0$ independent of n such that

$$\parallel y_n{\parallel }_{\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))}\le C.$$

(96)

Using, on the one hand, the boundedness of ${\mathcal{U}}_{ad}$ in $L^{\infty }\left({\omega }_T\right)$ and (94), we have that there exist $v\in L^{\infty }\left({\omega }_T\right)$ such that

$$v_n\rightharpoonup v\mathrm{weakly}\mathrm{in}{L}^2\left({\omega }_T\right),$$

(97)

$$v_n\rightharpoonup v\mathrm{weak}-\mathrm{star}\mathrm{in}{L}^{\infty }\left({\omega }_T\right),$$

(98)

and on the other hand, using (96) and Lemma 1, we deduce the existence of $y\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ such that

$$y_n\rightharpoonup y\mathrm{weakly}\mathrm{in}\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right)),$$

(99)

$$y_n\rightharpoonup y\mathrm{weakly}\mathrm{in}{L}^2(0,T;H^1\left(\mathrm{\Omega }\right))$$

(100)

and

$$y_n\to y\mathrm{strongly}\mathrm{in}{L}^2\left(Q\right).$$

(101)

Thanks to (22) and (101),

$$h_{y_n}^g\to h_y^g\mathrm{strongly}\mathrm{in}{L}^2\left(Q\right)\mathrm{as}n\to +\infty .$$

(102)

From (6) and Lemma 5, we have that for any $\varphi \in L^2\left(Q\right)$,

$$\varphi \beta \left(h_{y_n}^g\right)⟶\varphi \beta \left(h_y^g\right)\mathrm{strongly}\mathrm{in}{L}^2\left(Q\right),\mathrm{as}n\to +\infty .$$

(103)

Passing to the limit in (95), while using the convergences (98), (100), (101) and (103), we obtain that

$$\begin{array}{c}\hfill {\displaystyle {\int }_{\mathrm{\Omega }}{y}^0\varphi (·,0)\mathrm{d}x+{\int }_{{\omega }_T}vy\varphi dxdt=-{\int }_0^T{〈\frac{\partial \varphi }{\partial t},y〉}_{{\left(H^1\left(\mathrm{\Omega }\right)\right)}^{\prime },H^1\left(\mathrm{\Omega }\right)}dt}\\ \hfill {\displaystyle +{\int }_Q\nabla \varphi \nabla ydxdt+{\int }_Q\beta \left(h_y^g\right)y,\varphi dxdt,}\\ \hfill \mathrm{for}\mathrm{every}\varphi \in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right)),\varphi (·,T)=0\mathrm{in}\mathrm{\Omega }.\end{array}$$

(104)

Thanks to Equation (68), $y=y\left(v\right)$ is a weak solution of (2). In addition, since ${\mathcal{U}}_{ad}$ is a closed convex subset of $L^{\infty }\left({\omega }_T\right)$, we have that $v\in {\mathcal{U}}_{ad}.$

Using (97) and (101) and the lower semi-continuity of the cost functional J, it is deduced that

$$J(\mathcal{S}\left(v\right),v)\le \underset{n\to \infty }{lim\; inf}J(\mathcal{S}\left(v_n\right),v_n)=\underset{w\in {\mathcal{U}}_{ad}}{inf}J(\mathcal{S}\left(w\right),w).$$

□

Proposition 4

(Twice Fréchet differentiability of J). Let $y=y\left(v\right)\in \mathbb{X}$ be the solution of (2) and set $\mathcal{J}\left(v\right)=J\left(\mathcal{S}\right(v),v),$ where $J\left(\mathcal{S}\right(v),v)$ is given by (80). Then, following the hypothesis of Lemma 7, the functional $\mathcal{J}:L^{\infty }\left({\omega }_T\right)\to \mathbb{R}$ is twice continuously Fréchet-differentiable, and for every $\pi ,v,w\in L^{\infty }\left({\omega }_T\right)$, we have

$${\mathcal{J}}^{\prime }\left(v\right)w={\int }_{{\omega }_T}\left(yq+Nv\right)wdxdt$$

(105)

and

$$\begin{array}{ccc}{\mathcal{J}}^{″}\left(v\right)[w,\pi ]\hfill & =\hfill & {\displaystyle {\int }_{{\omega }_T}(\pi y_w+w{y}_{\pi })qdxdt+{\int }_Q\mathcal{F}qdxdt}\hfill \\ \hfill & +\hfill & {\displaystyle {\int }_{\mathrm{\Omega }}{y}_w(v;x,T)y_{\pi }(v;x,T)dx+N{\int }_{{\omega }_T}w\pi dxdt.}\hfill \end{array}$$

(106)

where $y_w={\mathcal{S}}^{\prime }\left(v\right)w\in \mathbb{X},y_{\pi }={\mathcal{S}}^{\prime }\left(u\right)\pi \in \mathbb{X}$ are solutions to (82) with $v=v$ and $v=w$, respectively. $\mathcal{F}\in L^{\infty }\left(Q\right)$ is defined in (84), and $q\in \mathbb{X}$ is the unique weak solution to the adjoint Equation (73).

Proof. 

From Lemma 7, we deduce that $\mathcal{J}$ is twice continuously Fréchet-differentiable.

Let $\pi ,v,w\in L^{\infty }\left({\omega }_T\right)$. After some computations, we obtain

$${\mathcal{J}}^{\prime }\left(v\right)w={\int }_{\mathrm{\Omega }}{y}_w(v;x,T)(y(v;x,T)-y_d)dx+N{\int }_{{\omega }_T}vwdxdt.$$

(107)

If we multiply the first equation in (82) by $\varrho \in C^{\infty }\left(\overline{Q}\right)$ such that ${\displaystyle \frac{\partial \varrho }{\partial \nu }}=0$ on $\mathrm{\Sigma }$ and integrate by parts over Q, we obtain

$$\begin{array}{c}\hfill {\displaystyle {\int }_Q\left(-\frac{\partial p}{\partial t}-\Delta p+\beta \left(h_y^g\right)p-vp{\chi }_{\omega }\right)y_w\left(v\right)dxdt=}\\ \hfill {\displaystyle -{\int }_{Q}y\left(v\right){\beta }^{\prime }\left(h_y^g\right)h_{y_w\left(v\right)}^{g}pdxdt+{\int }_{{\omega }_T}wy\left(v\right)pdxdt}\\ \hfill {\displaystyle -{\int }_{\mathrm{\Omega }}{y}_w(v;x,T)p(x,T)dx.}\end{array}$$

(108)

Hence, using the fact that ${\displaystyle h_{y_w\left(v\right)}^g(x,t)={\int }_0^{t}g\left(s\right)y_w\left(v\right)(x,s)ds}$, we have

$$\begin{array}{c}\hfill {\displaystyle {\int }_{Q}y\left(v\right){\beta }^{\prime }\left(h_y^g\right)h_{y_w\left(v\right)}^{g}pdxdt}\\ \hfill {\displaystyle ={\int }_{\mathrm{\Omega }}{\int }_0^{T}p(x,t)y\left(v\right)(x,t){\beta }^{\prime }\left(h_y^g\right)(x,t)\left({\int }_0^{t}g\left(s\right)y_w\left(v\right)(x,s)ds\right)dtdx}\\ \hfill {\displaystyle ={\int }_{\mathrm{\Omega }}{\int }_0^T{y}_w\left(v\right)(x,s)g\left(s\right)\left({\int }_s^{T}p(x,t)y\left(v\right)(x,t){\beta }^{\prime }\left(h_y^g\right)(x,t)dt\right)dsdx}\\ \hfill {\displaystyle ={\int }_Q{y}_w\left(v\right)(x,s)\left[g\left(s\right)\left({\int }_s^{T}p(x,t)y\left(v\right)(x,t){\beta }^{\prime }\left(h_y^g\right)(x,t)dt\right)\right]dsdx.}\end{array}$$

Therefore, (108) can be rewritten as

$$\begin{array}{c}\hfill {\displaystyle {\int }_Q\left(-\frac{\partial p}{\partial t}-\Delta p+\beta \left(h_y^g\right)p-v{\chi }_{\omega }p\right)y_w\left(v\right)dxdt}\\ \hfill {\displaystyle +{\int }_Q\left(\left[g\left(t\right)\left({\int }_t^{T}p(x,s)y\left(v\right)(x,s){\beta }^{\prime }\left(h_y^g\right)(x,s)ds\right)\right]\right)y_w\left(v\right)dxdt=}\\ \hfill {\displaystyle {\int }_{{\omega }_T}wy\left(v\right)pdxdt-{\int }_{\mathrm{\Omega }}{y}_w(v;x,T)p(x,T)dx.}\end{array}$$

(109)

So, if p verifies

$$\left\{\begin{array}{c}\hfill {\displaystyle -\frac{\partial p}{\partial t}-\Delta p=-\beta \left(h_y^g\right)p}\\ \hfill {\displaystyle -g\left(t\right)\left({\int }_t^{T}y(x,s){\beta }^{\prime }\left(h_y^g\right)(x,s)p(x,s)ds\right)+v{\chi }_{\omega }p}& \mathrm{in}\hfill & Q,\hfill \\ \hfill {\displaystyle \frac{\partial p}{\partial \nu }=0}& \mathrm{in}\hfill & \mathrm{\Sigma },\hfill \\ \hfill p(v;·,T)=y(v;·,T)-y_d& \mathrm{in}\hfill & \mathrm{\Omega },\hfill \end{array}\right.$$

where $y(v;·,T)-y_d\in L^{\infty }\left(\mathrm{\Omega }\right)$, then, in view of Proposition 1 and the uniqueness of solution to (73), $p=q\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))\cap L^{\infty }\left(Q\right)$. Hence, (109) becomes

$${\int }_{\mathrm{\Omega }}{y}_w(v;x,T)(y(v;x,T)-y_d)dx={\int }_{{\omega }_T}yqwdxdt.$$

(110)

Combining (107) and (110), we obtain (105).

Next, after some calculations, we prove that

$$\begin{array}{c}{\displaystyle {\mathcal{J}}^{″}\left(v\right)[w,\pi ]={\int }_{\mathrm{\Omega }}{y}_{w\pi }(v;x,T)(y(v;x,T)-y_d)dx}\hfill \\ {\displaystyle +{\int }_{\mathrm{\Omega }}{y}_w(v;x,T)y_{\pi }(v;x,T)dx+N{\int }_{{\omega }_T}w\pi dxdt,}\hfill \end{array}$$

(111)

where $y_{w\pi }\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$, the solution to (83). Multiplying the first equation of (73) by the $y_{w\pi }$ solution to (83) and integrating by parts over Q, we obtain

$$\begin{array}{c}{\displaystyle {\int }_{\mathrm{\Omega }}{y}_{w\pi }(v;x,T)(y(v;x,T)-y_d)dx={\int }_{{\omega }_T}(\pi y_w+w{y}_{\pi })qdxdt+{\int }_Q\mathcal{F}qdxdt.}\hfill \end{array}$$

(112)

Thus, if we combine (111) and (112), we deduce (106). □

Theorem 5.

Let $v\in {\mathcal{U}}_{ad}$ be an $L^{\infty }$-local minimum for (1). Then,

$${\mathcal{J}}^{\prime }\left(v\right)(w-v)\ge 0\mathit{for}\mathit{every}w\in {\mathcal{U}}_{ad},$$

(113)

equivalently

$${\int }_{{\omega }_T}(yq+Nv)(w-v)\mathrm{d}x\ge 0\mathit{for}\mathit{every}w\in {\mathcal{U}}_{ad},$$

(114)

where q is the unique weak solution to (73).

Proof. 

Let $w\in {\mathcal{U}}_{ad}$ be arbitrary and v be an $L^{\infty }$-local minimum. Since ${\mathcal{U}}_{ad}$ is convex, we have that $v+\lambda (w-v)\in {\mathcal{U}}_{ad}$ for all $\lambda \in (0,1].$ Then, for all $w\in {\mathcal{U}}_{ad}$,

$${\displaystyle 0\le \underset{\lambda \downarrow 0}{lim}\frac{\mathcal{J}(v+\lambda ((w-v))-\mathcal{J}(v)}{\lambda }={\mathcal{J}}^{\prime }\left(v\right)(w-v),}$$

which in view of (105) implies that

$${\displaystyle {\int }_{{\omega }_T}(yq+Nv)((w-v)\mathrm{d}x\ge 0\mathrm{for}\mathrm{every}w\in {\mathcal{U}}_{ad}.}$$

□

Lemma 8.

Assume that Lemma 7 holds. Let $v,w,\pi \in L^{\infty }\left({\omega }_T\right)$, ${\mathcal{J}}^{\prime }\left(v\right)w$ and ${\mathcal{J}}^{″}\left(v\right)[w,\pi ]$ be given by (105) and (106), respectively. Then, the linear mapping $w\mapsto {\mathcal{J}}^{\prime }\left(v\right)w$ can be extended to a linear continuous mapping ${\mathcal{J}}^{\prime }\left(v\right):L^2\left({\omega }_T\right)\to \mathbb{R}$ and the bilinear mapping $(w,\pi )\mapsto {\mathcal{J}}^{″}\left(v\right)[w,\pi ]$ can be extended to a bilinear continuous mapping on ${\mathcal{J}}^{″}\left(v\right):L^2\left({\omega }_T\right)\times L^2\left({\omega }_T\right)\to \mathbb{R}$ given by (106).

Proof. 

Let $v\in L^{\infty }\left({\omega }_T\right)$ and $w\in L^2\left({\omega }_T\right)$. From (105), we have that

$${\mathcal{J}}^{\prime }\left(v\right)w={\int }_{{\omega }_T}\left(yq+Nv\right)wdxdt,$$

where $y=y\left(v\right)$ and $q=q\left(v\right)$ are the solution of (2) and (73), respectively.

Using the Cauchy–Schwarz inequality, (45) and (75), we have that

$$|{\mathcal{J}}^{\prime }{\left(v\right)w|\le C\parallel w\parallel }_{L^2\left({\omega }_T\right)}$$

for some $C=C(g_{\infty },M,T,d,\mathrm{\Omega },\parallel v{\parallel }_{L^{\infty }\left({\omega }_T\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},\parallel y_d{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)})>0.$ Thus, the mapping $w\mapsto {\mathcal{J}}^{\prime }\left(v\right)w$ is linear and continuous on $L^2\left({\omega }_T\right)$.

From (84),

$${\displaystyle \mathcal{F}=-{\beta }^{\prime }\left(h_y^g\right)y_w{h}_{y_{\pi }}^g-{\beta }^{\prime }\left(h_y^g\right)y_{\pi }{h}_{y_w}^g-{\beta }^{″}\left(h_y^g\right)y{h}_{y_w}^g{h}_{y_{\pi }}^g,}$$

where $y=y\left(v\right)\in \mathbb{X}$, $y_{\pi }\in \mathbb{X}$ and $y_w\in \mathbb{X}$ are solutions to (2), (82) and (83), respectively.

Using (7) and the Cauchy–Schwarz inequality, (17) of Lemma 2, we deduce that

$$\begin{array}{c}\hfill {\displaystyle {\int }_Q\left|\mathcal{F}\right|dxdt\le M\parallel y_w{\parallel }_{L^2\left(Q\right)}\parallel h_{y_{\pi }}^g{\parallel }_{L^{2}y\left(Q\right)}+M\parallel y_{\pi }{\parallel }_{L^2\left(Q\right)}{\parallel h_{y_w}^g\parallel }_{L^2\left(Q\right)}}\\ \hfill +{M\parallel y\parallel }_{L^{\infty }\left(Q\right)}\parallel h_{y_w}^g{\parallel }_{L^2\left(Q\right)}{\parallel h_{y_{\pi }}^g\parallel }_{L^2\left(Q\right)},\end{array}$$

which in view of (20), (45) and (85), and of Lemma 3 implies that there exist two positive constants $C_2={C(d,T,\mathrm{\Omega },\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},)>0$ and $C_1={C(M,d,T,\mathrm{\Omega },\parallel v\parallel }_{L^{\infty }\left({\omega }_T\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty },)>0$ such that

$$\begin{array}{ccc}\hfill {\displaystyle {\int }_Q\left|\mathcal{F}\right|dxdt}& \le \hfill & {\displaystyle M{C}_1^2{\parallel w\parallel }_{L^2\left({\omega }_T\right)}{g}_{\infty }{T\parallel \pi \parallel }_{L^2\left({\omega }_T\right)}+M{C}_1^2{\parallel \pi \parallel }_{L^2\left({\omega }_T\right)}{g}_{\infty }T{\parallel w\parallel }_{L^2\left({\omega }_T\right)}}\hfill \\ & +\hfill & M{g}_{\infty }^2{T}^2{C}_1^2{C}_2{\parallel w\parallel }_{L^2\left({\omega }_T\right)}{\parallel \pi \parallel }_{L^2\left({\omega }_T\right)}.\hfill \end{array}$$

Hence, we have that

$${\displaystyle {\parallel \mathcal{F}\parallel }_{L^1\left(Q\right)}\le {C\parallel w\parallel }_{L^2\left({\omega }_T\right)}{\parallel \pi \parallel }_{L^2\left({\omega }_T\right)},}$$

(115)

for some $C=C(M,d,T,\mathrm{\Omega },\parallel v{\parallel }_{L^{\infty }\left({\omega }_T\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty },)>0.$

Applying the Cauchy–Schwartz inequality to (106), we obtain

$$\begin{array}{ccc}{\displaystyle |{\mathcal{J}}^{″}\left(v\right)[w,\pi ]|}\hfill & \le \hfill & {\displaystyle \left[{\parallel \pi \parallel }_{L^2\left({\omega }_T\right)}\parallel y_w{\parallel }_{L^2\left(Q\right)}+{\parallel w\parallel }_{L^2\left({\omega }_T\right)}{\parallel y_{\pi }\parallel }_{L^2\left(Q\right)}\right]{\parallel q\parallel }_{L^{\infty }\left(Q\right)}}\hfill \\ & +\hfill & {\parallel \mathcal{F}\parallel }_{L^1\left(Q\right)}{\parallel q\parallel }_{L^{\infty }\left(Q\right)}+\parallel y_w{\parallel }_{L^2\left(Q\right)}{\parallel y_{\pi }\parallel }_{L^2\left(Q\right)}\hfill \\ \hfill & +\hfill & {N\parallel \pi \parallel }_{L^2\left({\omega }_T\right)}{\parallel w\parallel }_{L^2\left({\omega }_T\right)},\hfill \end{array}$$

(116)

which in view of (75), (85) and (115) implies that

$$|{\mathcal{J}}^{″}{\left(v\right)[w,\pi ]|\le C\parallel w\parallel }_{L^2\left({\omega }_T\right)}{\parallel \pi \parallel }_{L^2\left({\omega }_T\right)}$$

(117)

for some constant there exists $C=C(M,d,T,\mathrm{\Omega },\parallel v{\parallel }_{L^{\infty }\left({\omega }_T\right)},\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty },)>0.$ □

Since the cost functional $\mathcal{J}$ given by (80) is non-convex, the first-order optimality conditions given in Theorem 5 are necessary but not sufficient for optimality. Sufficient second-order conditions are required. To this end, we first recall the definition of the feasible direction and the critical cone that can be found [18,19].

Definition 3.

1. 

The set of feasible directions is the set $A\left(v\right)$ defined by

$$A\left(v\right)=\{\pi \in L^{\infty }\left({\omega }_T\right):\pi =\lambda (w-v)\mathit{for}\mathit{some}\lambda >0\mathit{and}w\in {\mathcal{U}}_{ad}\}.$$

(118)

2. 

The critical cone is the set $\widehat{C}\left(v\right)$ defined by

$$\widehat{C}\left(v\right)={\mathit{cl}}_{L^2\left({\omega }_T\right)}\left(A\left(v\right)\right)\cap \{w\in L^2\left({\omega }_T\right):{\mathcal{J}}^{\prime }\left(v\right)w=0\}.$$

(119)

where ${\mathit{cl}}_{L^2\left({\omega }_T\right)}\left(A\left(v\right)\right)$ denotes the closure in $L^2\left({\omega }_T\right)$ of the admissible set $A\left(v\right)$.

Next, proceeding as in ([18] (Page 18)) (see also [19] (page 273)) we prove the following result that gives a characterization of the critical cone $\widehat{C}\left(v\right)$ defined in (119).

Proposition 5.

Let $v\in {\mathcal{U}}_{ad}$ and w be defined for a.e $(x,t)\in {\omega }_T$ as follows:

$$\left\{\begin{array}{ccc}{\displaystyle w(x,t)\ge 0}\hfill & if\hfill & v(x,t)=v_a,\hfill \\ {\displaystyle w(x,t)\le 0}\hfill & if\hfill & v(x,t)=v_b,\hfill \\ w(x,t)=0\hfill & if\hfill & Nv(x,t)+q(x,t)\ne 0,\hfill \end{array}\right.$$

(120)

where q is a solution to (73). The critical cone $\widehat{C}\left(v\right)$ defined in (119) can be rewritten as

$$\widehat{C}\left(v\right)=\{w\in L^2\left({\omega }_T\right):w\mathit{fulfills}\left(120\right)\}.$$

(121)

Proceeding as in ([20] (p. 246)) or [21,22], we prove the following result.

Proposition 6

(Necessary second-order optimality conditions). Let $v\in {\mathcal{U}}_{ad}$ be a $L^{\infty }$-local solution of system (79). Then, ${\mathcal{J}}^{″}\left(v\right)w^2\ge 0$, for all $w\in \widehat{C}\left(v\right)$.

Theorem 6.

Let $v\in {\mathcal{U}}_{ad}$ be a control satisfying the first-order optimality conditions (113). Then, the following hold:

1. 

The functional $\mathcal{J}:L^{\infty }\left({\omega }_T\right)\to \mathbb{R}$ is of class ${\mathcal{C}}^2$. Furthermore, if we denote by $\mathcal{L}(L^2\left({\omega }_T\right),\mathbb{R})$ and by $\mathcal{B}(L^2\left({\omega }_T\right),\mathbb{R})$ the spaces of linear continuous functional on $L^2\left({\omega }_T\right)$ and bilinear continuous functional on $L^2\left({\omega }_T\right)\times L^2\left({\omega }_T\right),$ then for every $v\in {\mathcal{U}}_{ad}$, the following continuous extensions exist:

$${\mathcal{J}}^{\prime }\left(v\right)\in \mathcal{L}(L^2\left({\omega }_T\right),\mathbb{R})\mathit{and}{\mathcal{J}}^{″}\left(v\right)\in \mathcal{B}(L^2\left({\omega }_T\right),\mathbb{R}).$$

(122)

2. 

For any sequence ${\left\{(v_n,w_n)\right\}}_{n=1}^{\infty }\subset {\mathcal{U}}_{ad}\times L^2\left({\omega }_T\right)$ such that

$$v_n\to v.\mathit{strongly}\mathit{in}{L}^2\left({\omega }_T\right),$$

(123)

$$w_n\rightharpoonup w\mathit{weakly}\mathit{in}{L}^2\left({\omega }_T\right),$$

(124)

as $n\to \infty$,

$${\displaystyle {\mathcal{J}}^{\prime }\left(v\right)w=\underset{n\to \infty }{lim}{\mathcal{J}}^{\prime }\left(v_n\right)w_n}$$

(125)

and

$${\displaystyle {\mathcal{J}}^{″}\left(v\right)[w,w]\le \underset{n\to \infty }{lim\; inf}{\mathcal{J}}^{″}\left(v_n\right)[w_n,w_n].}$$

(126)

In addition, if $w=0$, then

$$N\underset{n\to \infty }{lim\; inf}{\parallel w_n\parallel }_{L^2\left({\omega }_T\right)}^2\le \underset{n\to \infty }{lim\; inf}{\mathcal{J}}^{″}\left(v_n\right)[w_n,w_n].$$

(127)

Proof. 

We have that $v\in {\mathcal{U}}_{ad}$ is a control satisfying the first-order optimality conditions (113). Thus, from Proposition 4 and Lemma 8, we have (122).

To prove (125), (126) and (127), we proceed in two steps.

Step 1. We prove (125).

Let $v\in {\mathcal{U}}_{ad}$ and the sequence ${\left\{(v_n,w_n)\right\}}_{n=1}^{\infty }\subset {\mathcal{U}}_{ad}\times L^2\left({\omega }_T\right)$ be such that (123) and (124) hold. Then, we have that there exist $C_4>0$ and $C_5>0$ independent of n such that

$$\begin{array}{ccc}\hfill \parallel v_n{\parallel }_{L^{\infty }\left({\omega }_T\right)}& \le \hfill & C_4,\hfill \\ \hfill \parallel w_n{\parallel }_{L^2\left({\omega }_T\right)}& \le \hfill & C_5.\hfill \end{array}$$

(128)

If we denote by $y_n=y\left(v_n\right)$ and $y=y\left(v\right)$, the solutions of (2) with $v=v_n$ and $v=v$, respectively, and by $q_n=q\left(v_n\right)$ and $q=q\left(v\right),$ the solutions to (73) with $v=v_n$ and $v=v$, respectively, then from (45), (70), (74) and (75), we have that $y,y_n,q$ and $q_n$ belong to $\mathbb{X}$. Moreover, from (45), (85), (75) and (87) we obtain

$$\parallel y_n{\parallel }_{L^{\infty }\left({\omega }_T\right)}\le C(d,T,\mathrm{\Omega },C_4,\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)}),$$

(129)

and

$$\parallel q_n{\parallel }_{L^{\infty }\left({\omega }_T\right)}\le C(d,T,\mathrm{\Omega },C_4,\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},\parallel y_d{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty },M),$$

(130)

for some $C(d,T,\mathrm{\Omega },C_4,\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)})>0$ and $C(d,T,\mathrm{\Omega },C_4,\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},\parallel y_d{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty },M)>0$. Using the Lipschitz continuity of the mapping $v\mapsto y\left(v\right)$ and $v\mapsto q\left(v\right)$ given in Proposition 3, we have that

$$y_n\to y\mathrm{strongly}\mathrm{in}{L}^2(0,T;H^1\left(\mathrm{\Omega }\right))\mathrm{as}n\to \infty$$

(131)

and

$$q_n\to q\mathrm{strongly}\mathrm{in}{L}^2(0,T;H^1\left(\mathrm{\Omega }\right))\mathrm{as}n\to \infty .$$

(132)

Thanks to (130) and (132), we have that

$$q_n\rightharpoonup q\mathrm{weak}-\mathrm{star}\mathrm{in}{L}^{\infty }\left(Q\right),\mathrm{as}n\to \infty .$$

(133)

On the other hand, in view of (129) and (130), the sequences $\left(y_n{q}_n\right)$ are bounded in $L^{\infty }\left(Q\right)$ and thus in $L^2\left(Q\right)$ since Q is bounded. Therefore, we can prove using (129), (130), (131) and (132) that

$$y_n{q}_n\to yq\mathrm{strongly}\mathrm{in}{L}^2\left(Q\right)\mathrm{as}n\to \infty .$$

(134)

Observing that

$${\mathcal{J}}^{\prime }\left(v_n\right)w_n={\int }_{{\omega }_T}\left(y_n{q}_n+N{v}_n\right)w_{n}dxdt,$$

we have using (123), (124) and (134) that

$$\underset{n\to \infty }{lim}{\mathcal{J}}^{\prime }\left(v_n\right)w_n={\int }_{{\omega }_T}\left(yq+Nv\right)wdxdt={\mathcal{J}}^{\prime }\left(v\right)w.$$

Step 2. We show (126) and (127).

Let $z_n=y_w\left(v_n\right)$ and $z=y_w\left(v\right)$ be the solutions to (82) with $(v,w)=(v_n,w_n)$ and $(v,w)=(v,w)$, respectively. Then, $z_n,z\in \mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ and from (86) and (128), we have that there exist $C_4>0$ and $C_5>0$ independent of n such that

$$\parallel z_n{\parallel }_{\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))}\le C(d,T,\mathrm{\Omega },C_4,\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty },M),$$

(135)

for some $C(d,T,\mathrm{\Omega },C_4,\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty },M)>0$ independent of n. Thanks to (20) of Lemma 3,

$$\parallel h_z^g{\parallel }_{L^2\left(Q\right)}\le g_{\infty }T{\parallel z\parallel }_{L^2\left(Q\right)},$$

(136)

and

$$\parallel h_{z_n}^g{\parallel }_{L^2\left(Q\right)}\le C(d,T,\mathrm{\Omega },C_4,\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty },M),$$

(137)

for some $C(d,T,\mathrm{\Omega },C_4,\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},g_{\infty },M)>0$.

In addition, in view of (135), the compact embedding of $\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ into $L^2\left(Q\right)$ and the continuous embedding of $\mathbb{W}(0,T;H^1\left(\mathrm{\Omega }\right))$ into $\mathcal{C}\left([0,T];L^2\left(\mathrm{\Omega }\right)\right)$, we deduce that there exists $z\in L^2\left(Q\right)$ such that

$$z_n\to z\mathrm{strongly}\mathrm{in}{L}^2\left(Q\right)\mathrm{as}n\to \infty$$

(138)

and

$$z_n(·,T)\rightharpoonup z(·,T)\mathrm{weakly}\mathrm{in}{L}^2\left(\mathrm{\Omega }\right)\mathrm{as}n\to \infty .$$

(139)

Thanks to (22) and (138) of Lemma 3,

$$h_{z_n}^g\to h_z^g\mathrm{strongly}\mathrm{in}{L}^2\left(Q\right),\mathrm{as}n\to \infty .$$

(140)

Now, from (84) and (106),

$$\begin{array}{c}{\displaystyle {\mathcal{J}}^{″}\left(v\right)[w,w]=2{\int }_{{\omega }_T}wzqdxdt+{\int }_{\mathrm{\Omega }}z{(x,T)}^{2}dx+N{\int }_{{\omega }_T}{w}^{2}dxdt}\hfill \\ {\displaystyle -2{\int }_Q{\beta }^{\prime }\left(h_y^g\right)z{h}_z^{g}qdxdt-{\int }_Q{\beta }^{″}\left(h_y^g\right)y\left(v\right){\left(h_z^g\right)}^{2}qdxdt.}\hfill \end{array}$$

Thus,

$$\begin{array}{c}{\displaystyle {\mathcal{J}}^{″}\left(v_n\right)[w_n,w_n]=2{\int }_{{\omega }_T}{w}_n{z}_n{q}_{n}dxdt+{\int }_{\mathrm{\Omega }}{z}_n{(x,T)}^{2}dx+N{\int }_{{\omega }_T}{w}_n^{2}dxdt}\hfill \\ {\displaystyle -2{\int }_Q{\beta }^{\prime }\left(h_{y_n}^g\right)z_n{h}_{z_n}^g{q}_{n}dxdt-{\int }_Q{\beta }^{″}\left(h_{y_n}^g\right)y_n{\left(h_{z_n}^g\right)}^2{q}_{n}dxdt.}\hfill \end{array}$$

(141)

where we recall $z_n=y_w\left(v_n\right)$ and $z=y_w\left(v\right)$ as the solutions to (82) with $(v,w)=(v_n,w_n)$ and $(v,w)=(v,w)$, respectively. Since $\beta \in {\mathcal{C}}^2$ satisfies (6) and (7), $v\in {\mathcal{U}}_{ad}$ and the sequence ${\left\{(v_n,w_n)\right\}}_{n=1}^{\infty }\subset {\mathcal{U}}_{ad}\times L^2\left({\omega }_T\right)$ satisfies (123) and (124), we have from (7), (129), (130), (135) and (136) that

$${\displaystyle {∥{\beta }^{\prime }\left(h_{y_n}^g\right)q_n{z}_n∥}_{L^2\left(Q\right)}dxdt\le M\parallel q_n{\parallel }_{L^{\infty }\left(Q\right)}{\parallel z_n\parallel }_{L^2\left(Q\right)}\le C}$$

and

$${\displaystyle {∥{\beta }^{″}\left(h_{y_n}^g\right)q_n{y}_n{h}_{z_n}^g∥}_{L^2\left(Q\right)}dxdt\le M\parallel q_n{\parallel }_{L^{\infty }\left(Q\right)}\parallel y_n{\parallel }_{L^{\infty }\left(Q\right)}{\parallel h_{z_n}^g\parallel }_{L^{\infty }\left(Q\right)}\le C}$$

where there exists $C=C(d,T,\mathrm{\Omega },C_4,\parallel y^0{\parallel }_{L^{\infty }\left(\mathrm{\Omega }\right)},\parallel y_d{\parallel }_{L^{\infty }\left(Q\right)},g_{\infty },M)>0.$ Consequently, for any $\varphi \in L^2\left(Q\right),$

$$\begin{array}{ccc}{\displaystyle \left|{\int }_Q{\beta }^{\prime }\left(h_{y_n}^g\right)q_n{z}_n\varphi dxdt-{\int }_Q{\beta }^{\prime }\left(h_y^g\right)qz\varphi dxdt\right|}\hfill & \le \hfill & {\displaystyle {\int }_Q\left|{\beta }^{\prime }\left(h_{y_n}^g\right)\varphi q_n(z_n-z)\right|dxdt}\hfill \\ & +\hfill & {\displaystyle {\int }_Q\left|{\beta }^{\prime }\left(h_{y_n}^g\right)\varphi (q_{n}z-qz)\right|dxdt}\hfill \\ & +\hfill & {\displaystyle {\int }_Q\left|\left({\beta }^{\prime }\left(h_{y_n}^g\right)\varphi -{\beta }^{\prime }\left(h_y^g\right)\varphi \right)qz\right|dxdt}\hfill \end{array}$$

and

$$\begin{array}{c}{\displaystyle \left|{\int }_Q{\beta }^{″}\left(h_{y_n}^g\right)q_n{y}_n{h}_{z_n}^g\varphi dxdt-{\int }_Q{\beta }^{″}\left(h_y^g\right)qy{h}_z^g\varphi dxdt\right|\le }\hfill \\ {\displaystyle {\int }_Q\left|{\beta }^{″}\left(h_{y_n}^g\right)y_n{q}_n\varphi \left(h_{z_n}^g-h_z^g\right)\right|dxdt}\hfill \\ {\displaystyle +{\int }_Q\left|{\beta }^{″}\left(h_{y_n}^g\right)h_z^g{q}_n(y_n\varphi -y\varphi )\right|dxdt}\hfill \\ {\displaystyle +{\int }_Q\left|{\beta }^{″}\left(h_{y_n}^g\right)h_z^{g}y(q_n\varphi -q\varphi )\right|dxdt}\hfill \\ {\displaystyle +{\int }_Q\left|h_z^{g}yq\left({\beta }^{″}\left(h_{y_n}^g\right)\varphi -{\beta }^{″}\left(h_y^g\varphi \right)\right)\right|dxdt,}\hfill \end{array}$$

which in view of the Cauchy—Schwarz inequality, (7), (129), (130) and (135) implies that

$$\begin{array}{c}{\displaystyle \left|{\int }_Q{\beta }^{\prime }\left(h_{y_n}^g\right)q_n{z}_n\varphi dxdt-{\int }_Q{\beta }^{\prime }\left(h_y^g\right)qz\varphi dxdt\right|\le }\hfill \\ {\displaystyle M\parallel q_n{\parallel }_{L^{\infty }\left(Q\right)}{\parallel \varphi \parallel }_{L^2\left(Q\right)}{\parallel z_n-z\parallel }_{L^2\left(Q\right)}}\hfill \\ +{M\parallel z\parallel }_{L^2\left(Q\right)}{\parallel q_n\varphi -q\varphi \parallel }_{L^2\left(Q\right)}\hfill \\ +{\parallel q\parallel }_{L^{\infty }\left(Q\right)}{\parallel z\parallel }_{L^2\left(Q\right)}{\parallel {\beta }^{\prime }\left(h_{y_n}^g\right)\varphi -{\beta }^{\prime }\left(h_y^g\right)\varphi \parallel }_{L^2\left(Q\right)}\hfill \end{array}$$

(142)

and

$$\begin{array}{c}{\displaystyle \left|{\int }_Q{\beta }^{″}\left(h_{y_n}^g\right)q_n{y}_n{h}_{z_n}^g\varphi dxdt-{\int }_Q{\beta }^{″}\left(h_y^g\right)qy{h}_z^g\varphi dxdt\right|\le }\hfill \\ M\parallel y_n{\parallel }_{L^{\infty }\left(Q\right)}\parallel q_n{\parallel }_{L^{\infty }\left(Q\right)}{\parallel \varphi \parallel }_{L^2\left(Q\right)}{\parallel h_{z_n}^g-h_z^g\parallel }_{L^2\left(Q\right)}\hfill \\ +M\parallel q_n{\parallel }_{L^{\infty }\left(Q\right)}\parallel h_z^g{\parallel }_{L^2\left(Q\right)}{\parallel y_n\varphi -y\varphi \parallel }_{L^2\left(Q\right)}\hfill \\ +M\parallel h_z^g{\parallel }_{L^2\left(Q\right)}{\parallel q_n\varphi -q\varphi \parallel }_{L^2\left(Q\right)}\hfill \\ {\displaystyle +{\parallel y\parallel }_{L^{\infty }\left(Q\right)}{\parallel q\parallel }_{L^{\infty }\left(Q\right)}\parallel {\beta }^{″}\left(h_{y_n}^g\right)\varphi -{\beta }^{″}(h_y^g{\varphi \parallel }_{L^2\left(Q\right)}{\parallel h_z^g\parallel }_{L^2\left(Q\right)}.}\hfill \end{array}$$

(143)

Thanks to (7), (24), (129), (130), (131), (132) and Lemma 5, for any $\varphi \in L^2\left(Q\right)$,

$$\begin{array}{ccc}\hfill q_n\varphi & \to \hfill & q\varphi \mathrm{strongly}\mathrm{in}{L}^2\left(Q\right),\hfill \\ \hfill y_n\varphi & \to \hfill & y\varphi \mathrm{strongly}\mathrm{in}{L}^2\left(Q\right),\hfill \\ \hfill {\beta }^{″}\left(h_{y_n}^g\right)\varphi & \to \hfill & {\beta }^{″}(h_y^g\varphi \mathrm{strongly}\mathrm{in}{L}^2\left(Q\right),\hfill \\ \hfill {\beta }^{\prime }\left(h_{y_n}^g\right)\varphi & \to \hfill & {\beta }^{\prime }\left(h_y^g\right)\varphi \mathrm{strongly}\mathrm{in}{L}^2\left(Q\right),\hfill \end{array}$$

as $n\to +\infty$. Therefore, passing to limit in (142) and in (143) while using the above convergence, (129), (130), (138) and (140), we deuce that

$${\int }_Q{\beta }^{\prime }\left(h_{y_n}^g\right)q_n{z}_n\varphi dxdt\to {\int }_Q{\beta }^{\prime }\left(h_y^g\right)qz\varphi dxdt,\forall \varphi \in L^2\left(Q\right)$$

and

$${\int }_Q{\beta }^{″}\left(h_{y_n}^g\right)q_n{y}_n{h}_{z_n}^g\varphi dxdt\to {\int }_Q{\beta }^{″}\left(h_y^g\right)qy{h}_z^g\varphi dxdt,\forall \varphi \in L^2\left(Q\right).$$

Thus

$${\beta }^{\prime }\left(h_{y_n}^g\right)q_n{z}_n\rightharpoonup {\beta }^{\prime }\left(h_y^g\right)qz\mathrm{weakly}\mathrm{in}{L}^2\left(Q\right)$$

(144)

and

$${\beta }^{″}\left(h_{y_n}^g\right)q_n{y}_n{h}_{z_n}^g\rightharpoonup {\beta }^{″}\left(h_y^g\right)qy{h}_z^g\mathrm{weakly}\mathrm{in}{L}^2\left(Q\right),$$

(145)

as $n\to \infty .$ Since $q_n{z}_n$ is bounded in $L^2\left(Q\right)$ because (130) and (135),

$${\int }_Q{(q_n{z}_n-qz)}^{2}dxdt\le 2{\parallel q_n\parallel }_{L^{\infty }\left(Q\right)}{\int }_Q{(z_n-z)}^{2}dxdt+2{\int }_Q{z}^2{(q_n-q)}^{2}dxdt.$$

Passing to the limi in this latter inequality while using (130), (133), (138) and the fact that $z^2\in L^1\left(Q\right)$, we deduce that

$$\underset{n\to \infty }{lim}{\int }_Q{(q_n{z}_n-qz)}^{2}dxdt=0.$$

This means that

$$q_n{z}_n\to qz\mathrm{strongly}\mathrm{in}{L}^2\left(Q\right).$$

(146)

Then, using (141), (124), (144), (145), (138), (139) and (146), we obtain that

$$\begin{array}{c}\hfill {\displaystyle \underset{n\to \infty }{lim\; inf}{\mathcal{J}}^{″}\left(v_n\right)[w_n,w_n]=\underset{n\to \infty }{lim}2{\int }_{{\omega }_T}{w}_n{z}_n{q}_{n}dxdt}\\ \hfill {\displaystyle +\underset{n\to \infty }{lim\; inf}\left({\int }_Q{z}_n{(·,T)}^{2}dxdt+N{\int }_{{\omega }_T}{w}_n^{2}dxdt\right)}\\ \hfill {\displaystyle +\underset{n\to \infty }{lim}\left({\displaystyle -2{\int }_Q{\beta }^{\prime }\left(h_{y_n}^g\right)z_n{h}_{z_n}^g{q}_{n}dxdt-{\int }_Q{\beta }^{″}\left(h_{y_n}^g\right)y_n{\left(h_{z_n}^g\right)}^2{q}_{n}dxdt}\right)}\\ \hfill {\displaystyle \ge 2{\int }_{{\omega }_T}wzqdxdt+{\int }_Q{z}^2(0,T)dxdt+N{\int }_{{\omega }_T}{w}^{2}dxdt}\\ \hfill {\displaystyle -2{\int }_Q{\beta }^{\prime }\left(h_y^g\right)z{h}_z^{g}qdxdt-{\int }_Q{\beta }^{″}\left(h_y^g\right)y{\left(h_z^g\right)}^{2}qdxdt}\\ \hfill ={\mathcal{J}}^{″}\left(v\right)[w,w].\end{array}$$

Finally, if $w=0$, then $z=y_w$, being the solution of (105), we have $z=0$. Hence, passing to the lower limit in (141) when $n\to \infty$, we obtain (127). □

We have the following result giving sufficient second-order conditions for locally optimal solutions.

Theorem 7.

Let $v\in {\mathcal{U}}_{ad}$ be a control satisfying the first-order optimality conditions (113) and

$${\mathcal{J}}^{″}\left(v\right)[w,w]>0\forall w\in C\left(v\right)\setminus \left\{0\right\}.$$

(147)

Then, there exist $\gamma >0$ and $\eta >0$ such that the quadratic growth condition holds

$$\mathcal{J}\left(w\right)\ge {\mathcal{J}\left(v\right)+\eta \parallel w-v\parallel }_{L^2\left({\omega }_T\right)}^2,\forall w\in {\mathcal{U}}_{ad}\cap B_{\gamma }^2\left(v\right).$$

(148)

Therefore, v is locally optimal in the sense of $L^2\left({\omega }_T\right)$.

Proof. 

Relation (148) is obtained by proceeding as ([19] Theorem 2.3, page 265) while using Proposition 6. □

5. Concluding Remarks

In this paper, we studied the optimal control of a nonlinear parabolic equation involving a nonlocal-in-time term. We proved some regularity results of systems and the existence of the optimal solution to our control problem. Since the system is nonlinear and thus the cost function non-convex, we obtain local uniqueness by means of the second optimality conditions.