Optimal Control Related to Weak Solutions of a Chemotaxis-Consumption Model

Abstract

In the present work we investigate an optimal control problem related to the following chemotaxis-consumption model in a bounded domain $\Omega \subset {\mathbb {R}}^3$:

$$\begin{aligned} \partial _t u - \Delta u = - \nabla \cdot (u \nabla v), \quad \partial _t v - \Delta v = - u^s v + f \,v\, 1_{\Omega _c}, \end{aligned}$$

with $s \ge 1$, endowed with isolated boundary conditions and initial conditions for (u, v), being u the cell density, v the chemical concentration and f the control acting in the v-equation through the bilinear term $f \,v\, 1_{\Omega _c}$, in a subdomain $\Omega _c \subset \Omega$. We address the existence of optimal control restricted to a weak solution setting, where, in particular, uniqueness of state (u, v) given a control f is not clear. Then by considering weak solutions satisfying an adequate energy inequality, we prove the existence of optimal control subject to uniformly bounded controls. Finally, we discuss the relation between the considered control problem and two other related ones, where the existence of optimal solution can not be proved.

1 Introduction

We consider an optimal control problem subject to a chemotaxis-consumption model. Let $\Omega \subset {\mathbb {R}}^3$ be a bounded domain, denoting by $\Gamma$ its boundary, and define $Q: = (0,T) \times \Omega$, for a given final time $T > 0$. Let $u=u(t,x)$ and $v=v(t,x)$ be the density of cell population and the concentration of chemical substance, respectively, defined in $(t,x) \in Q$. The aforementioned chemotaxis-consumption model is given by the PDE system

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t u - \Delta u = - \nabla \cdot (u \nabla v), \quad \partial _t v - \Delta v = - u^s v, \quad \hbox {in}\, Q,\\ \partial _\eta u |_{\Gamma } = \partial _\eta v |_{\Gamma } = 0, \quad u(0,\cdot ) = u^0, \quad v(0, \cdot ) = v^0, \quad \hbox {in}\, \Omega , \end{array}\right. \end{aligned}$$

(1.1)

where $s \ge 1$, $\partial _\eta u |_{\Gamma }$ denotes the outer normal derivative of u on the boundary and $u^0, v^0 \ge 0$ in $\Omega$ are the initial conditions.

The controlled problem consists of the chemotaxis-consumption model (1.1) with a term $f v 1_{\Omega _c}$ added to the chemical equation:

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t u - \Delta u = - \nabla \cdot (u \nabla v), \quad \partial _t v - \Delta v = - u^s v + f v 1_{\Omega _c}, \quad \hbox {in}\, Q, \\ \partial _\eta u |_{\Gamma } = \partial _\eta v |_{\Gamma } = 0, \quad u(0) = u^0, \quad v(0) = v^0, \quad \hbox {in}\, \Omega , \end{array}\right. \end{aligned}$$

(1.2)

where $\Omega _c \subset \Omega$ is the control subdomain and $f: Q_c:=(0,T)\times \Omega _c \rightarrow {\mathbb {R}}$ is the control function. This term $fv1_{\Omega _c}$ means that we are controlling the system by acting directly on the chemical equation on the subdomain $\Omega _c$, increasing (where $f\ge 0$) or decreasing (where $f\le 0$) the chemical concentration, while the control on the cell equation is exerted indirectly. The use of this bilinear term $fv1_{\Omega _c}$ as control allows to preserve the positivity of v independently of the sign of f. The control of chemotaxis systems through the direct action on the chemical equation has been considered in previous studies [2, 7, 10,11,12,13, 16, 18, 20].

To contextualize this work, we highlight the closest related previous literature, beginning with some theoretical studies on the uncontrolled problem (1.1). For the case $s = 1$, the existence of a global weak solution in smooth and convex 3D domains is proved in [22]. Moreover, these solutions become smooth after a large enough period of time and their long time behavior is studied. In [23], a parabolic-elliptic version of (1.1) is studied, leading to results on the existence, uniqueness and long time behavior of a global classical solution in n-dimensional smooth domains.

Still considering $s = 1$, some authors have also focused on the model (1.1) but coupled with a surrounding fluid (called chemotaxis-Navier–Stokes model), as a particular case of the models introduced by [25]. In [26], the existence of global weak solutions is proved in smooth and convex 3D domains. Moreover, these solutions become smooth after a large enough period of time and their long time behavior is studied. These results are extended in [14] to nonconvex but smooth domains. In [27] the author studies the asymptotic behavior of the chemotaxis-Navier–Stokes equations in 2D domains for adequate generalizations of the chemotaxis and consumption terms, proving the convergence towards constant states in the $L^{\infty }$-norm. In [29], existence of global weak solutions for the chemotaxis-Navier–Stokes equations is established in 3D smooth and convex domains, and the study of the asymptotic behavior of these solutions is carried out in [30].

All the aforementioned works are related to the chemotaxis-consumption model (1.1) with $s = 1$, i.e. the chemotaxis and consumption terms are given, respectively, by

$$\begin{aligned} \nabla \cdot (u \nabla v) \text{ and } -uv. \end{aligned}$$

(1.3)

However, from the modeling point of view, it may be useful to study models with more general terms. Indeed, different assumptions during the modeling process can lead to different terms in the equations and, therefore, it would be useful to have results for a model with more general terms, rather than for a specific case. In particular, when the potential consumption $-u^s v$ is considered, the action of the consumption term is more accentuated because, compared to the bilinear case $-u v$, the consumption is stronger when $u > 1$ and is weaker for $u < 1$. Moreover, it is also interesting to know, for instance, what is the effect of different chemotaxis and consumption terms in the properties of the solutions, such as regularity, boundedness, asymptotic behavior and so on.

In [26], already cited above, the chemotaxis and consumption terms considered had the form $\nabla \cdot (u F'(u) \chi (v) \nabla v)$ and $-F(u)h(v)$, including (1.3) as a particular case. The results were then obtained under hypotheses on the behavior of the functions $\chi (v)$, h(v), F(u) and its derivative $F'(u)$. More recently, problem (1.1) has been studied in [5] and the effects of any consumption power $s\ge 1$ in the regularity of the solutions have been addressed. In addition, the class of considered domains has been enlarged, including 2D and 3D domains which are not necessarily smooth or convex.

Another relevant topic in chemotaxis models is the existence or non-existence of blowing-up solutions. If we consider model (1.1), this question has been answered in 2D domains. In fact, considering $s = 1$ and 2D convex domains, existence and uniqueness of a global classical solution that is uniformly bounded up to infinity time is proved in [22]; and for $s \ge 1$ and more general nonconvex 2D domains, existence and uniqueness of a global strong solution that is uniformly bounded up to infinity time is proved in [5]. As far as we know, this question remains open for 3D domains. In fact, some authors investigated hypotheses that could lead to no blow-up results in n-dimensional spaces, for $n \ge 3$. They obtained advances under smallness assumptions on the chemotaxis coefficient and on $\Vert {v^0} \Vert _{L^{\infty }(\Omega )}$. On this subject, we refer the interested reader to [21] and [1], for the problem (1.1) with $s = 1$. In addition, these results are extended to other related chemotaxis models with consumption in [8, 9].

The works cited so far address the analysis of the chemotaxis-consumption model (1.1) and helps us understand how the system evolves from the given initial data. However, when it comes to PDE models which describe physical phenomena, such as the chemotaxis models, as important as the analysis of the system itself, are the studies of control problems related to them. Concerning chemotaxis models, particularly relevant are the optimal control problems. Due to the low regularity of the weak solutions and the lack of uniqueness results in 3D, some works are concentrated in 2D domains, where one usually has existence and uniqueness of strong solution to the controlled problem, which allows to prove the existence of optimal control, and to derive an optimality system, where existence and regularity of Lagrange multipliers are deduced. In this direction, we refer the reader to the works on control problems related to: the Keller–Segel model [18]; a chemorepulsion-production model [11, 13]; a Keller–Segel logistic model [2]; a chemotaxis model with indirect consumption [31]; and a chemotaxis-haptotaxis model [20].

For optimal control problems related to chemotaxis models in 3D domains the situation is more complex because, in general, we have results of existence of weak solutions, however, in many cases, there is not any result on the existence and uniqueness of global in time strong solutions. In this setting, it is usual to introduce a regularity criterion, which is a mild additional regularity hypothesis, sufficient to conclude that a weak solution satisfying this regularity is actually the unique strong solution. A didactic introduction to this kind of adaptation can be found in [4], for an optimal control problem related to the Navier–Stokes equations in 3D domains. We refer the reader to [12], where a regularity criterion is established to study an optimal control problem related to a chemorepulsion-production model in 3D domains. The drawback of using a regularity criterion is that it is not clear in general if the admissible set is nonempty. In [12], the authors show that if $\Omega _c = \Omega$, that is, if the control acts in the whole domain, and the initial chemical density $v^0$ is strictly positive and separated from zero, then the admissible set is nonempty. To do that, the idea is to define the control f a posteriori, depending on a regular pair (u, v).

Although the study of optimal control problems related to chemotaxis models is an interesting and growing topic, we still have a relative low number of studies concerning the chemotaxis-consumption model (1.1). In fact, to the best of our knowledge, we can cite two works: [10] and [16]. In [16], the authors apply a regularity criterion to a chemotaxis-consumption-Navier–Stokes model, proving the existence of optimal control and the existence of Lagrange multipliers, arriving at the optimality system. In [10], an optimal control problem subject to strong solutions of (1.2) is studied. A sharp regularity criterion is proved and used to show the existence of optimal solution and the existence and uniqueness of Lagrange multipliers associated to any local optimal solution. A generic PDE linear system is also used to study the regularity of the Lagrange multipliers. In both [16] and [10], it is only possible to prove that the admissible set is nonempty for controls acting in the whole domain $\Omega$.

In view of all exposed so far, the objective of the present work is to study an optimal control problem related to (1.1) for which we are able to prove the existence of global optimal solution in the weak solutions setting, that is, without using any regularity criterion or hypothesis over the admissible set. To achieve it, an important novelty of this work is to consider weak solutions of the controlled model (1.2) satisfying an energy inequality (see (1.6) below). Afterwards, we prove existence of global optimal solution and, to conclude, we discuss the relation between this optimal control problem and two other related ones, where the existence of optimal solution can not be proved.

We remark that the deduction of first order optimality conditions in this weak solution setting remains as an open problem. Indeed, since we only have the weak regularity, it is not possible to prove the well-posedness of the linearized problem around a local optimal solution, which is the essential hypothesis to apply a generic Lagrange multiplier method as in [10]. Also, given a control f, we do not have in general uniqueness of the state (u, v), hence it is not possible to define the “control-to-state” mapping and follow the procedure used in [16] to compute the derivative of the state with respect to the control.

1.1 Main Results

Along this paper, we impose the following hypotheses:

$$\begin{aligned}{} & {} \Omega \subset {\mathbb {R}}^3\, \hbox {is a bounded domain with boundary}\, \Gamma \, \hbox {of class}\, C^{2,1},\nonumber \\{} & {} \quad f \in L^q(Q_c), \text{ for } \text{ some } q > 5/2, \end{aligned}$$

(1.4)

$$\begin{aligned}{} & {} (u^0, v^0) \in L^p(\Omega ) \times W^{2-2/q,q}(\Omega ), \end{aligned}$$

(1.5)

with $p = 1 + \varepsilon$, for some $\varepsilon > 0$, if $s = 1$, and $p = s$, if $s > 1$.

Let us define the specific functional spaces appearing for the weak solution setting. For $s \in [1,2)$,

$$\begin{aligned} \begin{array}{rl} X_u = &{} \hspace{-4pt} \Big \{ u \ | \ u\in L^{\infty }(0,T;L^s(\Omega )) \cap L^{5s/3}(Q), \ \nabla u \in L^{5s/(3 + s)}(Q), \\ &{} \qquad \partial _t u \in L^{5s/(3 + s)}(0,T;W^{1,5s/(4s-3)}(\Omega )') \Big \}, \end{array} \end{aligned}$$

for $s \ge 2$,

$$\begin{aligned} \begin{array}{rl} X_u = &{} \Big \{ u\ | \ u \in L^{\infty }(0,T;L^s(\Omega )) \cap L^{5s/3}(Q),\ \nabla u \in L^{2}(Q), \\ &{} \qquad \partial _t u \in L^2(0,T;H^1(\Omega )') \Big \}, \end{array} \end{aligned}$$

and for $s \ge 1$,

$$\begin{aligned} \begin{array}{rl} X_v = &{} \hspace{-4pt} \Big \{ v\ | \ v \in L^{\infty }(Q), \ \nabla v \in L^{\infty }(0,T;L^2(\Omega )) \cap L^4(Q) \cap L^2(0,T;H^1(\Omega )), \\ &{} \qquad \Delta v \in L^2(Q), \ \partial _t v \in L^{5/3}(Q) \Big \}. \end{array} \end{aligned}$$

We also introduce the bounded convex set for the control

$$\begin{aligned} B_q(M) = \Big \{ f \in L^q(Q_c) \ | \ \Vert {f} \Vert _{L^q(Q_c)} \le M \Big \}. \end{aligned}$$

Definition 1.1

(Weak Solution of (1.2)) A pair (u, v) is called a weak solution of (1.2) if $u(t,x),v(t,x) \ge 0$ a.e. $(t,x) \in Q$, with

$$\begin{aligned} u \in X_u, \ v \in X_v \end{aligned}$$

and satisfying the initial conditions for (u, v), the u-equation of (1.2) and the boundary condition of u in the variational sense

$$\begin{aligned} \langle \partial _t u, \varphi \rangle + \int _\Omega \nabla u\cdot \nabla \varphi \ dx = \int _\Omega u \nabla v\cdot \nabla \varphi \ dx, \quad a.e. \ t > 0, \quad \forall \, \varphi \in W^{1,5s/(4s-3)}(\Omega ), \end{aligned}$$

the v-equation of (1.2) point-wisely a.e. $(t,x) \in Q$ (in fact, the v-equation is satisfied in $L^{5/3}(Q)$) and, since $\Delta v \in L^2(Q)$, the boundary condition of v in the sense of $H^{-1/2}(\Gamma )$. $\square$

Remark 1.2

Considering the regularity of the weak solution pair (u, v) and the regularity of the time derivatives, $u_t$ and $v_t$, we can specify the sense in which the initial conditions are attained. Indeed, we are able to conclude that (u, v) is weakly continuous from $[0,\infty )$ into $L^s(\Omega ) \times H^1(\Omega )$, if $s \in [1,2]$, and $L^2(\Omega ) \times H^1(\Omega )$, if $s \ge 2$ (see [5, Remark 2]). $\square$

The proof of existence of weak solution to the controlled problem (1.2) is based in the treatment of the uncontrolled model (1.1) given in [5] and extended to the model (1.2), which has a non-smooth control f as a coefficient. An important step in [5] is the obtaining of an energy inequality using the change of variable from (u, v) to (u, z), with $z = \sqrt{v + \alpha ^2}$, where $\alpha > 0$ is a sufficiently small but fixed real number, independently of (u, v) and f, which will be chosen in Lemmas 3.5, 3.6 and 3.8 below. Here, the energy inequality satisfied by the constructed weak solutions of (1.2) will also be written in terms of (u, z). In fact, we consider the energy

$$\begin{aligned} E(u,z)(t) = \frac{s}{4} \displaystyle {\int _{\Omega }}{g(u(t,x)) \ dx} + \frac{1}{2} \int _{\Omega }{| {\nabla z(t,x)} | _{}^2} \ dx, \end{aligned}$$

where

$$\begin{aligned} g(u) = \left\{ \begin{array}{rl} (u+1)ln(u+1) - u, &{} \text{ if } s = 1, \\ \dfrac{1}{s(s-1)} \, u^s, &{} \text{ if } s > 1. \end{array} \right. \end{aligned}$$

We have the following result of existence of weak solutions to (1.2).

Theorem 1.3

(Existence of energy inequality weak solutions) Given $f \in L^q(Q)$ ($q > 5/2$), there is a non-negative weak solution (u, v) of (1.2) satisfying the following energy inequality

$$\begin{aligned} \begin{aligned}&E(u,z)(t_2) + \beta \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{| {\nabla [u + 1]^{s/2}} | _{}^2 \ dx} \ dt + \dfrac{1}{4} \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{u^s | {\nabla z} | _{}^2 \ dx} \ dt \\&\quad \quad + \beta \Bigg ( \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{| {D^2 z} | _{}^2 \ dx} \ dt + \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{\frac{| {\nabla z} | _{}^4}{z^2} \ dx} \ dt \Bigg ) \\&\quad \le E(u,z)(t_1) + {\mathcal {K}}(\Vert {f} \Vert _{L^q(Q)}, \Vert {v_0} \Vert _{W^{2-2/q,q}(\Omega )}), \end{aligned} \end{aligned}$$

(1.6)

for a.e. $t_1,t_2 \in [0,T]$. Here, ${\mathcal {K}}(\Vert {f} \Vert _{L^q(Q)},\Vert {v_0} \Vert _{W^{2-2/q,q}(\Omega )})$ is a continuous and increasing function with respect to $\Vert {f} \Vert _{L^q(Q)}$ and $\beta > 0$ is a constant, independent of (u, v, f). Moreover, inequality (1.6) is also valid for $t_1 = 0$, with $E(u,z)(0) = E(u^0,v^0)$. Finally, in the case $s > 1$, inequality (1.6) also holds for all $t_2 \in (t_1,T]$.

Remark 1.4

The existence of weak solutions satisfying an energy inequality is commonly seen, for instance, for fluid models, and is used to prove either weak-strong uniqueness results [15] or large time behaviour [17]. In the present work, we use this “energy inequality weak solution” setting in order to prove the existence of global optimal solution. To the best of our knowledge, this is the first time that the concept of weak solution with energy inequality is applied to this purpose for chemotaxis models. $\square$

Next we introduce the minimization problem. Consider the functional

$$\begin{aligned} J: L^{5s/3}(Q) \times L^2(Q) \times L^q(Q) \longrightarrow {\mathbb {R}} \end{aligned}$$

given by

$$\begin{aligned}&J(u,v,f) : = \dfrac{3\gamma _u}{5s} \int _0^T{\Vert {u(t) - u_d(t)} \Vert _{L^{5s/3}(\Omega )}^{5s/3} \ dt} \\&\displaystyle {+} \dfrac{\gamma _v}{2} \int _0^T{\Vert {v(t) - v_d(t)} \Vert _{L^2(\Omega )}^2 \ dt} + \dfrac{\gamma _f}{q} \int _0^T{\Vert {f(t)} \Vert _{L^q(\Omega )}^q \ dt}, \end{aligned}$$

where $(u_d,v_d) \in L^{5s/3}(Q) \times L^2(Q)$ represents the desired states, $\gamma _u, \gamma _v > 0$ measure the errors in the states and $\gamma _f>0$ the cost of the control. In view of the existence result, Theorem 1.3, one could expect the following admissible sets

$$\begin{aligned} \begin{array}{rl} S_{ad}^w = &{} \hspace{-4pt} \Big \{ (u,v,f) \in X_u \times X_v \times L^q(Q) \ | \ (u,v) \text{ is } \text{ a } \\ &{} \quad \text{ weak } \text{ solution } \text{ of } \mathrm{(1.2)} \text{ with } \text{ control } f \Big \} \end{array} \end{aligned}$$

or

$$\begin{aligned} \begin{array}{rl} S_{ad}^E = &{} \hspace{-4pt} \Big \{ (u,v,f) \in X_u \times X_v \times L^q(Q) \ | \ (u,v) \text{ is } \text{ a } \text{ weak } \text{ solution } \text{ of } \mathrm{(1.2)} \\ &{} \quad \text{ with } \text{ control } f \text{ and } \text{ satisfies } \text{ the } \text{ energy } \text{ inequality } \mathrm{(1.6)} \Big \} \end{array} \end{aligned}$$

and then state the corresponding minimization problems

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle {m}in \ J(u,v,f) \\ \text{ subject } \text{ to } (u,v,f) \in S_{ad}^w, \end{array} \right. \end{aligned}$$

(1.7)

or

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle {m}in \ J(u,v,f) \\ \text{ subject } \text{ to } (u,v,f) \in S_{ad}^E. \end{array} \right. \end{aligned}$$

(1.8)

Thanks to Theorem 1.3 we have that both $S_{ad}^w$ and $S_{ad}^E$ are nonempty sets. However, we are not able to prove that problem (1.7) or (1.8) has a solution, as we will analyze in Remark 1.8 and Subsect. 4.1, respectively.

Therefore, in order to find an optimal control related to weak solutions of (1.2), we introduce the following admissible set, for each $M>0$:

$$\begin{aligned} \begin{array}{rl} S_{ad}^M = &{} \hspace{-4pt} \Big \{ (u,v,f) \in X_u \times X_v \times B_q(M) \ | \ (u,v) \text{ is } \text{ a } \text{ weak } \text{ solution } \text{ of } \\ &{} (1.2) \text{ with } \text{ control } f \text{ and } \text{ satisfies } (1.6)\, \text{ changing } \text{ the } \\ &{} \text{ constant } {\mathcal {K}}\big (\Vert {f} \Vert _{L^q(Q)}, \Vert {v_0} \Vert _{W^{2-2/q,q}(\Omega )}\big ) \text{ by } {\mathcal {K}}\big (M, \Vert {v_0} \Vert _{W^{2-2/q,q}(\Omega )}\big ) \Big \} \end{array} \end{aligned}$$

and the corresponding minimization problem

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle {m}in \ J(u,v,f) \\ \text{ subject } \text{ to } (u,v,f) \in S_{ad}^M. \end{array} \right. \end{aligned}$$

(1.9)

Again, from Theorem 1.3, one has $S_{ad}^M \ne \emptyset$. But now, we are able to prove the following result.

Theorem 1.5

(Existence of optimal control) For each $M > 0$, the optimal control problem (1.9) has at least one global optimal solution, that is, there is $(\overline{u},\overline{v},\overline{f}) \in S_{ad}^M$ such that

$$\begin{aligned} J(\overline{u},\overline{v},\overline{f}) = \min _{(u,v,f) \in S_{ad}^M} J(u,v,f). \end{aligned}$$

Remark 1.6

As it can be observed in Subsect. 4.1, to prove Theorem 1.5, it will be fundamental to have an energy structure such as inequality (1.6). In fact, considering the minimizing sequence argument used in Subsect.  4.1, the energy inequality (1.6) is the key point to guarantee that all the possible limits of the minimizing sequence are weak solutions of the controlled model. In fact, the corresponding energy estimates must be strong enough to guarantee that the possible limits of the minimizing sequence are weak solutions of the controlled model. Therefore, if the model does not admit an energy structure, as it seems to be the case in [28], for example, it is not clear how to prove the existence of optimal solution. $\square$

By construction, we have the following relation between problems (1.7) and (1.9):

$$\begin{aligned} J_{inf}^w:=\inf _{(u,v,f) \in S_{ad}^w} J(u,v,f) \le \min _{(u,v,f) \in S_{ad}^M} J(u,v,f). \end{aligned}$$

(1.10)

On the other hand, in this paper we will obtain the following relation between the minimization problems (1.8) and (1.9) for M large enough:

Theorem 1.7

If

$$\begin{aligned} M^q \ge \frac{q}{\gamma _f} \inf _{(u,v,f) \in S_{ad}^E}{J(u,v,f)}, \end{aligned}$$

we have the inequality

$$\begin{aligned} \min _{(u,v,f) \in S_{ad}^M} J(u,v,f) \le \inf _{(u,v,f) \in S_{ad}^E}{J(u,v,f)}:=J_{inf}^E. \end{aligned}$$

(1.11)

Remark 1.8

From Theorem 1.5, for each $M > 0$ there is $(u^M,v^M,f^M) \in S_{ad}^M$ such that

$$\begin{aligned} J\big (u^M,v^M,f^M\big ) = \min _{(u,v,f) \in S_{ad}^M} J(u,v,f). \end{aligned}$$

Let $M_2> M_1 > 0$. Since $S_{ad}^{M_1} \subset S_{ad}^{M_2}$ then $J(u^M,v^M,f^M)$ decreases as M increases. Therefore, since $J(u^M,v^M,f^M)$ is bounded from below, there exists $\displaystyle {\lim _{M \rightarrow \infty } \ J(u^M,v^M,f^M)}$ and, accounting for (1.10) and (1.11), one has the inequalities

$$\begin{aligned} J_{inf}^w \le \lim _{M \rightarrow \infty } \ J(u^M,v^M,f^M) \le J_{inf}^E. \end{aligned}$$

Let $(u^{\infty },v^{\infty },f^{\infty }) \in L^{5s/3}(Q) \times L^2(Q) \times L^q(Q)$ be the weak limit of a subsequence of $\{ (u^M,v^M,f^M) \}_M$. Then, the weakly lower semicontinuity of J in $L^{5s/3}(Q) \times L^2(Q) \times L^q(Q)$ leads to

$$\begin{aligned} J(u^{\infty },v^{\infty },f^{\infty }) \le \lim _{M \rightarrow \infty } \ J\big (u^M,v^M,f^M\big ). \end{aligned}$$

(1.12)

In our opinion, the proof or the refutation of the following two questions are interesting open problems:

• (1) $(u^{\infty },v^{\infty },f^{\infty }) \in S_{ad}^w$ ?

• (2) $\displaystyle {\lim _{M \rightarrow \infty }} \ J(u^M,v^M,f^M) = J_{inf}^w$ ?

In fact, if (2) were valid, then $J_{inf}^w$ could be approximated by $\displaystyle {\min _{(u,v,f) \in S_{ad}^M}} J(u,v,f)$ as $M \rightarrow \infty$. On the other hand, if (1) and (2) were valid, then $(u^{\infty },v^{\infty },f^{\infty })$ becomes an optimal solution of (1.7). Indeed, from (1) we have $J_{inf}^w \le J(u^{\infty },v^{\infty },f^{\infty })$, and from (2) and (1.12), we have $J(u^{\infty },v^{\infty },f^{\infty }) \le J_{inf}^w$. $\square$

The rest of the paper is organized as follows. In Sect. 2 we present some preliminary results. The existence of weak solutions satisfying the energy inequality (1.6) for the controlled problem (proof of Theorem 1.3) is established in Sect. 3 and, in Sect.  4 we study the optimal control problem, proving Theorems 1.5 and 1.7.

2 Preliminary Results

Lemma 2.1

Let $\Omega \subset {\mathbb {R}}^3$ be a bounded Lipschitz domain. We have

$$\begin{aligned} \Vert {v} \Vert _{L^{10/3}(\Omega )} \le C \Vert {v} \Vert _{L^2(\Omega )}^{2/5} \Vert {v} \Vert _{H^1(\Omega )}^{3/5}. \end{aligned}$$

Lemma 2.2

[3] Let B be a Banach space, let $\{ w_n \}$ be a sequence in B and $w \in B$. Either if $w_n \rightarrow w$ weakly* or weakly in B then $\{ w_n \}$ is bounded in B and $\Vert {w} \Vert _{B} \le \liminf \Vert {w_n} \Vert _{B}$.

Lemma 2.3

[24] Let X and Y be two Banach spaces such that $X \subset Y$ with a continuous injection. If $\phi \in L^{\infty }(0,T;X)$ and $\phi \in C([0,T];Y)$, then $\phi \in C_w([0,T];X)$.

Lemma 2.4

[6] Let $\Omega$ be a bounded domain of ${\mathbb {R}}^N$ such that $\Gamma$ is of class $C^2$. Let $p \in (1,3)$, $w^0 \in W^{2-2/p,p}(\Omega )$ and $h \in L^p(Q)$. Then the problem

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t w - \Delta w = h, \quad \text{ in } Q, \\ \partial _\eta w |_{\Gamma } = 0, \quad \text{ on } (0,T) \times \Gamma , \\ w(0,\cdot ) = w^0, \quad \text{ in } \Omega , \end{array}\right. \end{aligned}$$

has a unique solution

$$\begin{aligned} w \in C\big ([0,T]; W^{2-2/p,p}(\Omega )\big ) \cap L^p(0,T;W^{2,p}(\Omega )), \ \partial _t w \in L^p(Q). \end{aligned}$$

Moreover, there is a positive constant $C = C(p,T,\Omega )$ such that

$$\begin{aligned} \begin{array}{l} \Vert {w} \Vert _{C([0,T];W^{2 - 2/p,p}(\Omega ))} + \Vert {w} \Vert _{L^p(0,T;W^{2,p}(\Omega ))} + \Vert {\partial _t w} \Vert _{L^p(Q)} \\ \le C \big ( \Vert {h} \Vert _{L^p(Q)} + \Vert {w^0} \Vert _{W^{2 - 2/p,p}(\Omega )}\big ). \end{array} \end{aligned}$$

(2.1)

Lemma 2.5

[5] Let $w_1$ and $w_2$ be nonnegative real numbers. For each $s \ge 1$ we have

$$\begin{aligned} | {w_2^s - w_1^s} | _{} \le s | {w_2 + w_1} | _{}^{s-1} | {w_2 - w_1} | _{}. \end{aligned}$$

Using Lemma 2.5, we can prove the following.

Lemma 2.6

Let $p \in (1, \infty )$ and let $\{ w_n \}$ be a sequence of nonnegative functions in $L^p(Q)$ such that $w_n \rightarrow w$ in $L^p(Q)$ as $n \rightarrow \infty$. Then, for every $r \in (1,p)$, $w_n^r \rightarrow w^r$ in $L^{p/r}(Q)$ as $n \rightarrow \infty$.

Lemma 2.7

(Compactness in Bochner spaces [19]) Let X, B and Y be Banach spaces, let

$$\begin{aligned} F \subset \Big \{ f \in L^1(0,T;Y) \ | \ \partial _t f \in L^1(0,T;Y) \Big \}. \end{aligned}$$

Suppose that $X \subset B \subset Y$, with compact embedding $X \subset B$ and continuous embedding $B \subset Y$. Let the set F be bounded in $L^q(0,T;B) \cap L^1(0,T;X)$, for $1 < q \le \infty$, and $\Big \{ \partial _t f, \ \forall f \in F \Big \}$ be bounded in $L^1(0,T;Y)$. Then F is relatively compact in $L^p(0,T;B)$, for $1 \le p < q$.

3 Existence of the Controlled Problem

The existence of weak solutions of the uncontrolled problem (1.1) is proved in [5], by means of a sequence of truncated problems. To define these truncated problems, we are going to use a mollifier regularization of the control $f \in L^q(Q)$ defined via convolution, considering a sequence (see [3])

$$\begin{aligned} \begin{array}{c} f_m \in C^{\infty }_c(\overline{Q}), \quad \Vert {f_m} \Vert _{L^r(Q)} \le \Vert {f} \Vert _{L^r(Q)}, \, \text{ for } r \in [1,q], \\ f_m \rightarrow f \text{ strongly } \text{ in } L^q(Q). \end{array} \end{aligned}$$

(3.1)

Then, we prove the existence of solution of the controlled problem (1.2) satisfying, in addition, the energy inequality (1.6), using the following controlled truncated problems:

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t u_m - \Delta u_m = - \nabla \cdot (T^m(u_m) \nabla v_m), \\ \partial _t v_m - \Delta v_m = - T^m(u_m)^s v_m + {f_m} v_m 1_{\Omega _c}, \\ \partial _\eta u_m |_{\Gamma } = \partial _\eta v_m |_{\Gamma } = 0, \quad u_m(0) = u^0_m, \quad v_m(0) = v^0, \end{array} \right. \end{aligned}$$

(3.2)

for each $m \in {\mathbb {N}}$, where the truncation function $T^m \in C^2({\mathbb {R}})$ is defined by

$$\begin{aligned} T^m(u) = \left\{ \begin{array}{cl} - 1, &{} \text{ if } u \le -1, \\ C^2 \text{ extension }, &{} \text{ if } u \in (-1,0), \\ u, &{} \text{ if } u \in [0, m], \\ C^2 \text{ extension }, &{} \text{ if } u \in (m,m + 2), \\ m + 1, &{} \text{ if } u \ge m + 2, \end{array} \right. \end{aligned}$$

with $(u^0,v^0)$ satisfying (1.5) and $u^0_m \in C^{\infty }(\overline{\Omega })$ being mollifier regularizations of $u^0$ extended to ${\mathbb {R}}^N$ and having the following properties (see [5] for more details):

$$\begin{aligned} u^0_m\ge 0,\quad \int _\Omega u^0_m = \int _\Omega u^0,\quad u^0_m \rightarrow u^0 \text{ strongly } \text{ in } L^p(\Omega ). \end{aligned}$$

(3.3)

3.1 A $\varvec{L^{\infty }}$ Function Bounding $\varvec{v_m}$ from Above

In [5], where the uncontrolled model ($f \equiv 0$) is considered, a crucial step to prove the existence of a weak solution to (1.1) as a limit of solutions of the truncated models (3.2) is to get m-independent bounds for $\Vert {v_m} \Vert _{L^{\infty }(Q)}$. This fact also remains essential in our case, $f \not \equiv 0$. But, while in the case where $f \equiv 0$ this m-independent bound is obtained by straightforward calculations, it is not so obvious now by considering a control f with $f_+ \not \equiv 0$ in general. The next result will help us to build a function $w \in L^{\infty }$ bounding $v_m$ from above for all m.

Lemma 3.1

Let $\Omega$ be a bounded domain of ${\mathbb {R}}^3$ such that $\Gamma$ is of class $C^2$. Let $w^0 \in W^{2-2/q,q}(\Omega )$ and $\tilde{f} \in L^q(Q)$, for some $q > 5/2$. Then the problem

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t w - \Delta w = \tilde{f} \, w, \quad \text{ in } Q, \\ \partial _\eta w |_{\Gamma } = 0, \quad \text{ on } (0,T) \times \Gamma , \\ w(0,x) = w^0 \ \text{ in } \Omega , \end{array}\right. \end{aligned}$$

(3.4)

has a unique solution

$$\begin{aligned} w \in C\big ([0,T]; W^{2-2/q,q}(\Omega )\big ) \cap L^q(0,T;W^{2,q}(\Omega )), \ \partial _t w \in L^q(Q), \end{aligned}$$

In particular, there is a positive constant $C(\Vert \tilde{f}\Vert _{L^q(Q)}, \Vert w^0\Vert _{W^{2-2/q,q}(\Omega )})$ such that

$$\begin{aligned} \Vert {w} \Vert _{L^{\infty }(Q)} \le C\big (\Vert \tilde{f}\Vert _{L^q(Q)}, \Vert w^0\Vert _{W^{2-2/q,q}(\Omega )}\big ). \end{aligned}$$

(3.5)

In fact, inequality (3.5) will provide an estimate for $\Vert {v_m} \Vert _{L^{\infty }(Q)}$ in terms of the control f. We remark that it is the main reason why we need to assume in this work that $f \in L^q(Q)$, for some $q > 5/2$.

Proof

The key idea here is the injection $W^{2-2/q,q}(\Omega ) \subset L^{\infty }(\Omega )$, the reason why we suppose that $q > 5/2$. The proof is divided in two steps.

Step 1 (Existence and uniqueness of problem (3.4)):

For any solution w of (3.4) such that

$$\begin{aligned} w \in L^{\infty }\big (0,T;H^1(\Omega )\big ) \cap L^2\big (0,T;H^2(\Omega )\big ), \ \partial _t w \in L^2(Q), \end{aligned}$$

(3.6)

we have

$$\begin{aligned} \begin{array}{l} \displaystyle \Vert {w(t)} \Vert _{L^p(\Omega )}^p + \beta \int _0^t{\Vert {\nabla [w(r)]^{p/2}} \Vert _{L^2(\Omega )}^2 \ dr} \\ \displaystyle \le C \Vert {w^0} \Vert _{L^p(\Omega )}^p \exp \left( C p^{5/2} \int _0^t \left( \big \Vert \tilde{f}(r)\big \Vert _{L^{5/2}(\Omega )}^{5/2} + 1\right) \ dr\right) , \end{array} \end{aligned}$$

(3.7)

$a.e. \ t \in (0,T)$. In fact, we test the equation in (3.4) by $p \, w^{p-1}$ and define $\tilde{w}: = w^{p/2}$. Using (2.1), we obtain

$$\begin{aligned}&\dfrac{d}{dt} \Vert {\tilde{w}(t)} \Vert _{L^2(\Omega )}^2 + \dfrac{4 p(p - 1)}{p^2} \int _0^t{\Vert {\nabla \tilde{w}} \Vert _{L^2(\Omega )}^2 \ dr} \le p \int _{\Omega } \tilde{f} \ \tilde{w}^2 \ dx \\&\quad \le C\, p \,\big \Vert \tilde{f}\big \Vert _{L^{5/2}(\Omega )} \Vert {\tilde{w}} \Vert _{L^{10/3}(\Omega )}^2 \\&\quad \le C \,p\, \big \Vert \tilde{f}\big \Vert _{L^{5/2}(\Omega )} \Vert {\tilde{w}} \Vert _{L^2(\Omega )}^{4/5} \Vert {\tilde{w}} \Vert _{H^1(\Omega )}^{6/5} \\&\quad \le C(\delta ) \, p^{5/2} \big \Vert \tilde{f}\big \Vert _{L^{5/2}(\Omega )}^{5/2} \Vert {\tilde{w}} \Vert _{L^2(\Omega )}^2 + \delta \Vert {\tilde{w}} \Vert _{L^2(\Omega )}^2 + \delta \Vert {\nabla \tilde{w}} \Vert _{L^2(\Omega )}^2. \end{aligned}$$

Hence, choosing $\delta > 0$ small enough to absorb the last term in the right hand side and going back to the notation w we obtain

$$\begin{aligned} \dfrac{d}{dt} \Vert {w(t)} \Vert _{L^p(\Omega )}^p + \beta \int _0^t{\Vert {\nabla [w(r)]^{p/2}} \Vert _{L^2(\Omega )}^2 \ dr} \le C p^{5/2} \big ( \big \Vert \tilde{f}\big \Vert _{L^{5/2}(\Omega )}^{5/2} + 1\big ) \Vert {w(t)} \Vert _{L^p(\Omega )}^p \end{aligned}$$

and Gronwall’s Lemma leads us to (3.7).

For $\tilde{f}$ regular enough one can prove that (3.4) has a unique solution satisfying (3.6) by using Galerkin’s method, for example. But accounting for the dependence of w on $\Vert \tilde{f}\Vert _{L^{5/2}(\Omega )}^{5/2}$ given by (3.7), we are actually able to prove that (3.4) has a unique strong solution satisfying (3.6) and (3.7) under a weaker assumption on the regularity of $\tilde{f}$. It is enough that $\tilde{f} \in L^{5/2}(Q)$, for instance. The uniqueness is proved by comparing two possibly distinct solutions of (3.4) and concluding that they are in fact the same solution.

Step 2 (Proof of the $L^\infty$ estimate (3.5)):

Since $\tilde{f} \in L^q(Q)$ and $q > 5/2$, (3.7) implies that there are $\tilde{q} \in (5/2,q)$ and a positive constant $\tilde{C}(\Vert \tilde{f}\Vert _{L^q(Q)}, \Vert w^0\Vert _{W^{2-2/q,q}(\Omega )})$ such that $f w \in L^{\tilde{q}}(Q)$ with

$$\begin{aligned} \big \Vert \tilde{f} \, w\big \Vert _{L^{\tilde{q}}(Q)} \le \tilde{C}\big (\big \Vert \tilde{f}\big \Vert _{L^q(Q)}, \Vert w^0\Vert _{W^{2-2/q,q}(\Omega )}\big ). \end{aligned}$$

(3.8)

From (3.8) and (2.1) we can conclude,

$$\begin{aligned} \Vert {w} \Vert _{C([0,T];W^{2-2/\tilde{q},\tilde{q}}(\Omega ))} \le C\big (\big \Vert \tilde{f}\big \Vert _{L^q(Q)}, \Vert w^0\Vert _{W^{2-2/q,q}(\Omega )}\big ). \end{aligned}$$

But since $\tilde{q} > 5/2$, we have $C([0,T];W^{2-2/\tilde{q},\tilde{q}}(\Omega )) \subset L^{\infty }(Q)$, hence

$$\begin{aligned} \Vert {w} \Vert _{L^{\infty }(Q)} \le C\big (\big \Vert \tilde{f}\big \Vert _{L^q(Q)}, \Vert w^0\Vert _{W^{2-2/q,q}(\Omega )}\big ). \end{aligned}$$

Finally, since $w \in L^{\infty }(Q)$, we have $\tilde{f} \ w \in L^q(Q)$ and therefore we can use Lemma 2.4 to conclude that

$$\begin{aligned} w \in C\big ([0,T]; W^{2-2/q,q}(\Omega )\big ) \cap L^q\big (0,T;W^{2,q}(\Omega )\big ), \ \partial _t w \in L^q(Q). \end{aligned}$$

$\square$

3.2 Existence for the Controlled Truncated Problem and the First Uniform Estimates

Theorem 3.2

Given $f_m$ and $(u^0_m,v^0)$ satisfying (3.1), (3.3) and $v^0 \in W^{2 - 2/q,q}(\Omega )$, respectively, there is a unique solution $(u_m,v_m)$ of (3.2) with regularity

$$\begin{aligned} \begin{array}{c} u_m \in L^{\infty }(0,T;H^1(\Omega )) \cap L^2(0,T;H^2(\Omega )), \ \partial _t u_m \in L^2(0,T;L^2(\Omega )), \\ v_m \in L^{\infty }(Q) \cap L^{\infty }(0,T;H^2(\Omega )), \quad \Delta v_m \hbox { and } \partial _t v_m \in L^2(0,T;H^1(\Omega )), \end{array} \end{aligned}$$

and satisfying

$$\begin{aligned}{} & {} u_m(t,x), v_m(t,x) \ge 0, \ a.e. \ (t,x) \in Q, \end{aligned}$$

(3.9)

$$\begin{aligned}{} & {} \int _{\Omega }{u_m(t,x) \ dx} = \int _{\Omega }{u^0_m(x) \ dx}= \int _{\Omega }{u^0(x) \ dx}, \ a.e. \ t \in (0,T). \end{aligned}$$

(3.10)

Moreover, there is a positive, continuous and increasing function of $\Vert {f} \Vert _{L^q(Q)}$, ${\mathcal {K}}_1(\Vert f\Vert _{L^q(Q)}, \Vert v^0\Vert _{W^{2-2/q,q}(\Omega )})$, also independent of m, such that

$$\begin{aligned} \Vert {v_m} \Vert _{L^{\infty }(Q)}, \Vert {v_m} \Vert _{L^2(0,T; H^1(\Omega ))} \le {\mathcal {K}}_1(\Vert {f} \Vert _{L^q(Q)}, \Vert {v^0} \Vert _{W^{2-2/q,q}(\Omega )}). \end{aligned}$$

(3.11)

Proof

Concerning the proof of existence and uniqueness of solution to (3.2), the truncation $T^m(\cdot )$ simplifies the treatment of the chemotaxis and consumption terms, $- \nabla \cdot (T^m(u_m) \nabla v_m)$ and $- T^m(u_m)^s v_m$ respectively, and one can deal with the control term $f_m v_m$, likewise in the proof of existence of problem (3.4), in Lemma 3.1. The uniqueness is proved by comparing two possibly different solutions. The regularity of the solution pair $(u_m,v_m)$ and properties (3.9) and (3.10) can be proved following the ideas in [5]. Finally, we prove (3.11), beginning by the estimate in the $L^{\infty }$-norm. Using the already proved property (3.9) of $v_m$ in the $v_m$-equation of (3.2), we obtain

$$\begin{aligned} \partial _t v_m - \Delta v_m \le {(f_m)_+} \, v_m \quad a.e. \ (t,x) \in Q. \end{aligned}$$

(3.12)

On the other hand, accounting for Lemma 3.1, as well as (3.4), with $\tilde{f} = {(f_m)_+}$ and $w^0 = v^0$, we consider w satisfying

$$\begin{aligned} \partial _t w - \Delta w = {(f_m)_+} \, w \quad a.e. \ (t,x) \in Q, \end{aligned}$$

(3.13)

with $\partial _\eta w |_{\Gamma } = 0$ and $w(0,x) = v^0(x)$. Subtracting (3.13) from (3.12) we conclude that $(v_m - w)$ satisfies

$$\begin{aligned} \left\{ \begin{array}{l} \partial _t (v_m - w) - \Delta (v_m - w) \le {(f_m)_+} \, (v_m - w) \quad a.e. \ (t,x) \in Q, \\ \partial _\eta (v_m - w) |_{\Gamma } = 0, \quad (v_m - w)(0,x) = 0. \end{array} \right. \end{aligned}$$

Multiplying the above inequality by $(v_m - w)_+$ and using (2.1) to estimate the right hand side term leads us to $(v_m - w)_+(t,x) = 0$ $a.e. \ (t,x) \in Q$, that is, $v_m(t,x) \le w(t,x)$ $a.e. \ (t,x) \in Q$. Then, taking the $L^q$-norm inequality of (3.1) into account, the bound in the $L^{\infty }$-norm for $v_m$ is a consequence of the estimate for w given in (3.5), with $\tilde{f} = (f_m)_+$ and $w^0 = v^0$, from Lemma 3.1. The bound of $v_m$ in the norm of $L^2(0,T;H^1(\Omega ))$ is obtained by testing the $v_m$-equation of (3.2) by $v_m$ and conveniently estimating the term on the right hand side using Holder’s inequality, the interpolation inequality (2.1) and Young’s inequality. $\square$

3.3 Energy Inequality

Analogously to [5], we consider the variable $z_m(t,x) = \sqrt{v_m(t,x) + \alpha ^2}$ and the rewritten problem

$$\begin{aligned} \begin{array}{rl} \partial _t u_m - \Delta u_m &{} = - \nabla \cdot (T^{m}(u_m) \nabla (z_m)^2) \\ \partial _t z_m - \Delta z_m - \dfrac{| {\nabla z_m} | _{}^2}{z_m} &{} = - \dfrac{1}{2} T^{m}(u_m)^s \left( z_m - \dfrac{\alpha ^2}{z_m} \right) + \dfrac{1}{2} {f_m} \left( z_m - \dfrac{\alpha ^2}{z_m} \right) 1_{\Omega _c} \\ \partial _\eta u_m |_{\Gamma } &{} = \partial _\eta z_m |_{\Gamma } = 0 \\ u_m(0) &{} = u^0_m, \quad z_m(0) = \sqrt{v^0 + \alpha ^2}, \end{array}\nonumber \\ \end{aligned}$$

(3.14)

which is equivalent to the controlled truncated problem (3.2). From the equivalence of (3.2) and (3.14) and from the results given in Lemma 3.2, we have the following.

Corollary 3.3

Given $f_m$ and $(u^0_m,v^0)$ satisfying (3.1), (3.3) and $v^0 \in W^{2 - 2/q,q}(\Omega )$, respectively, there is a unique solution $(u_m,z_m)$ of (3.14) with regularity

$$\begin{aligned} \begin{array}{c} u_m \in L^{\infty }(0,T;H^1(\Omega )) \cap L^2(0,T;H^2(\Omega )), \ \partial _t u_m \in L^2(0,T;L^2(\Omega ))), \\ z_m \in L^{\infty }(Q) \cap L^{\infty }(0,T;H^2(\Omega )), \quad \Delta z_m \hbox { and } \partial _t z_m \in L^2(0,T;H^1(\Omega )), \end{array} \end{aligned}$$

and satisfying the m-uniform estimates

$$\begin{aligned}{} & {} u_m(t,x) \ge 0 \ \text{ and } \ z_m(t,x) \ge \alpha , \ a.e. \ (t,x) \in Q,\nonumber \\{} & {} \quad \int _{\Omega }{u_m(t,x) \ dx} = \int _{\Omega }{u^0_m(x) \ dx} = \int _{\Omega }{u^0(x) \ dx}, \ a.e. \ t \in (0,T),\nonumber \\{} & {} \quad \Vert {z_m} \Vert _{L^{\infty }(Q)}, \Vert {z_m} \Vert _{L^2(0,T;H^1(\Omega ))} \le {\mathcal {K}}_1\big (\Vert {f} \Vert _{L^q(Q)}, \Vert {v^0} \Vert _{W^{2-2/q,q}(\Omega )}\big ). \end{aligned}$$

(3.15)

Using this change of variables, we obtain an energy inequality involving the control f. In this subsection, in order to simplify the notation, we drop the m subscript and denote the solution $(u_m,z_m)$ of (3.14) by (u, z). We begin with the following lemma.

Lemma 3.4

The solution (u, z) of (3.14), satisfies the inequality

$$\begin{aligned} \begin{array}{l} \dfrac{1}{2} \dfrac{d}{dt} \Vert {\nabla z} \Vert _{L^2(\Omega )}^2 + C_1 \Bigg ( \displaystyle {\int _{\Omega }}{| {D^2 z} | _{}^2 \ dx} + \displaystyle {\int _{\Omega }}{\frac{| {\nabla z} | _{}^4}{z^2} \ dx} \Bigg ) + \dfrac{1}{2} \displaystyle {\int _{\Omega }{T^{m}(u)^s | {\nabla z} | _{}^2 \ dx}} \\ \qquad \le \dfrac{s}{2} \alpha \displaystyle {\int _{\Omega }}{T^{m}(u)^{s-1} | {\nabla z} | _{} | {\nabla T^{m}(u)} | _{} \ dx} + {\mathcal {K}}_1^2 \Vert {f} \Vert _{L^2(\Omega )}^2 \\ \qquad \qquad + C_2 \Vert {\nabla z} \Vert _{L^2(\Omega )}^2 + \dfrac{s}{4} \displaystyle {\int _{\Omega }}{T^{m}(u)^{s-1} \nabla (z^2) \cdot \nabla T^{m}(u) \ dx}. \end{array} \end{aligned}$$

Proof

The proof of this lemma is analogous to the proof of Lemma 23 of [5]. We also use property (3.1) which says that $\Vert {f_m} \Vert _{L^2(Q)} \le \Vert {f} \Vert _{L^2(Q)}$. $\square$

Now we need to prove m-independent estimates for $(u_m,z_m)$. These estimates will be obtained from an energy inequality as in [5, Subsects. 5.1 and 5.2], with the modifications due to the treatment of the term related to the control f. We remark that we will use property (3.1) to keep the dependence on the norms of the control f rather than in terms of the norms of the mollified functions $f_m$. As in [5], we have to separate the proof of the energy inequality for the cases $s = 1$, $s \in (1,2)$ and $s \ge 2$. In fact, the test function used for the u-equation is $g'(u)$, where $g''(u) = u^{s-2}$. Note that, if $s = 1$, both $g'(u)$ and $g''(u)$ are singular at $u = 0$; if $s \in (1,2)$, then only $g''(u)$ is singular at $u = 0$; and if $s \ge 2$ then neither $g'(u)$ nor $g''(u)$ have any singularity.

Next we consider the function $g_m$ defined by $g_m(r) = \displaystyle {\int _0^r}{g'_m(\theta ) \ d\theta }$, where $g_m'(\theta )$ is defined for $\theta \ge 0$ by

$$\begin{aligned} g_m'(\theta ) = \left\{ \begin{array}{rl} ln(T^m(\theta ) + 1), &{} \text{ if } s = 1, \\ \dfrac{T^m(\theta )^{s-1}}{(s-1)}, &{} \text{ if } s > 1, \end{array} \right. \end{aligned}$$

and the energy

$$\begin{aligned} E_m(u,z)(t) = \frac{s}{4} \int _{\Omega }{g_m(u(t,x)) \ dx} + \frac{1}{2} \int _{\Omega }{| {\nabla z(t,x)} | _{}^2} \ dx. \end{aligned}$$

(3.16)

Lemma 3.5

(Energy inequality for $\varvec{s = 1}$) The solution (u, z) of the problem (3.14) satisfies, for sufficiently small $\alpha > 0$,

$$\begin{aligned} \begin{aligned}&\dfrac{d}{dt} E_m(u,z)(t) + \beta \displaystyle {\int _{\Omega }}{| {\nabla [T^{m}(u) + 1]^{1/2}} | _{}^2 \ dx} + \dfrac{1}{4} \displaystyle {\int _{\Omega }}{T^{m}(u) | {\nabla z} | _{}^2 \ dx} \\&\quad + \beta \Bigg ( \displaystyle {\int _{\Omega }}{| {D^2 z} | _{}^2 \ dx} + \displaystyle {\int _{\Omega }}{\frac{| {\nabla z} | _{}^4}{z^2} \ dx} \Bigg ) \le C ({\mathcal {K}}^2) \Vert {\nabla z} \Vert _{L^2(\Omega )}^2 + {\mathcal {K}}_1^2 \Vert {f} \Vert _{L^2(\Omega )}^2. \end{aligned} \end{aligned}$$

(3.17)

Proof

We follow [5], pointing out the most relevant steps to deal with the control term and make explicit the dependence on ${\mathcal {K}}_1 = {\mathcal {K}}_1(\Vert f\Vert _{L^q(Q)}, \Vert v^0\Vert _{W^{2-2/q,q}(\Omega )})$, from Corollary 3.3. By testing the u-equation of problem (3.14) by $ln(T^{m}(u) + 1)$, we obtain

$$\begin{aligned} \frac{d}{{dt}}\int _{\Omega } {g_{m} (u)\;dx} + \int _{\Omega } {\frac{{(T^{m} )^{\prime }(u)}}{{T^{m} (u) + 1}}{\text { }}\left| {\nabla u^{2} } \right| dx} = \left( {\frac{{T^{m} (u)}}{{T^{m} (u) + 1}}\nabla (z^{2} ),\nabla T^{m} (u)} \right) . \end{aligned}$$

Since $0 \le (T^{m})'(u) \le C$, we have $((T^{m})'(u))^2 \le C (T^{m})'(u)$, and we can write

$$\begin{aligned} \int _{\Omega }{\frac{(T^{m})'(u)}{T^{m}(u) + 1} | {\nabla u} | _{}^2 \ dx} \ge C \int _{\Omega }{\frac{((T^{m})'(u))^2}{T^{m}(u) + 1} | {\nabla u} | _{}^2 \ dx} \ge C \int _{\Omega }{| {\nabla [T^{m}(u) + 1]^{1/2}} | _{}^2 \ dx}. \end{aligned}$$

Hence, using that $\dfrac{1}{(T^{m}(u) + 1)} \le \dfrac{1}{\sqrt{T^{m}(u) + 1}}$, we have

$$\begin{aligned} \begin{gathered} \frac{d}{{dt}}\int _{\Omega } {g_{m} (u)\;dx} + C\int _{\Omega } {\left| {\nabla [T^{m} (u) + 1]^{{1/2}} } \right| ^{2} dx} = 2\left( {\frac{{T^{m} (u) + 1 - 1}}{{T^{m} (u) + 1}}z\nabla z,\nabla T^{m} (u)} \right) \\ \quad = \left( {\nabla (z^{2} ),\nabla T^{m} (u)} \right) - 2\left( {z\nabla z,\frac{{\nabla T^{m} (u)}}{{T^{m} (u) + 1}}} \right) \\ \quad \le \left( {\nabla (z^{2} ),\nabla T^{m} (u)} \right) + 2\left\| z \right\| _{{L^{\infty } (\Omega )}} \left\| {\nabla z} \right\| _{{L^{2} (\Omega )}} \left\| {\nabla [T^{m} (u) + 1]^{{1/2}} } \right\| _{{L^{2} (\Omega )}} . \\ \end{gathered} \end{aligned}$$

Using Young’s inequality and (3.15), we arrive at

$$\begin{aligned} \begin{array}{rl} &{} \dfrac{d}{dt} \displaystyle {\int _{\Omega }}{g_m(u) \ dx} + C \displaystyle {\int _{\Omega }}{| {\nabla [T^{m}(u) + 1]^{1/2}} | _{}^2 \ dx}\\ &{}\quad \le (\ {\nabla (z^2)} \ , \ {\nabla T^{m}(u)} \ ) + {\mathcal {K}}_1^2 \Vert {\nabla z} \Vert _{L^2(\Omega )}^2. \end{array} \end{aligned}$$

(3.18)

If we add the inequality of Lemma 3.4, for $s = 1$, to 1/4 times (3.18), then the terms $\displaystyle {\int _{\Omega }{\nabla T^{m}(u) \cdot \nabla (z^2) \ dx}}$ cancel and we obtain

$$\begin{aligned}&\frac{d}{dt} \Bigg [ \frac{1}{4} \int _{\Omega }{g_m(u) \ dx} + \frac{1}{2} \Vert {\nabla z} \Vert _{L^2(\Omega )}^2 \Bigg ] + C \int _{\Omega }{| {\nabla [T^{m}(u) + 1]^{1/2}} | _{}^2 \ dx}\nonumber \\&\quad + \frac{1}{2} \int _{\Omega }{T^{m}(u) | {\nabla z} | _{}^2 \ dx} + C_1 \Bigg ( \int _{\Omega }{| {D^2 z} | _{}^2 \ dx} + \int _{\Omega }{\frac{| {\nabla z} | _{}^4}{z^2} \ dx} \Bigg ) \nonumber \\&\quad \le \frac{\sqrt{\alpha }}{2} \int _{\Omega }{| {\nabla z} | _{} | {\nabla T^{m}(u)} | _{} \ dx} + {\mathcal {K}}_1^2 \Vert {f} \Vert _{L^2(\Omega )}^2 + (C_2 + {\mathcal {K}}_1^2) \Vert {\nabla z} \Vert _{L^2(\Omega )}^2 \nonumber \\&\quad \le \int _{\Omega }{\alpha | {\nabla [T^{m}(u) + 1]^{1/2}} | _{} | {\sqrt{T^{m}(u) + 1}} | _{} | {\nabla z} | _{} \ dx} \nonumber \\&\quad + {\mathcal {K}}_1^2 \Vert {f} \Vert _{L^2(\Omega )}^2 + (C_2 + {\mathcal {K}}_1^2) \Vert {\nabla z} \Vert _{L^2(\Omega )}^2. \end{aligned}$$

(3.19)

We can deal with the first term in the right hand side of the inequality using Holder’s and Young’s inequality,

$$\begin{aligned}&\int _{\Omega }{\alpha | {\nabla [T^{m}(u) + 1]^{1/2}} | _{} | {\sqrt{T^{m}(u) + 1}} | _{} | {\nabla z} | _{} \ dx} \le \alpha ^2 \ C(\delta ) \int _{\Omega }{T^{m}(u) | {\nabla z} | _{}^2 \ dx} \\&\quad + \delta \Vert {\nabla [T^{m}(u) + 1]^{1/2}} \Vert _{L^2(\Omega )}^2 + \alpha ^2 \ C(\delta ) \int _{\Omega }{| {\nabla z} | _{}^2 \ dx}. \end{aligned}$$

Therefore, we can first choose $\delta > 0$ and then $\alpha > 0$ sufficiently small in order to use the terms on the left hand side of inequality (3.19) to absorb the first two terms on the right hand side of the above inequality and finally obtain the desired inequality (3.17). $\square$

Now we obtain the energy inequalities for $s \in (1,2)$ and for $s \ge 2$, respectively. Analogously to Lemma 3.5, we follow the ideas of [5], making the necessary changes in order to deal with the control term and to make explicit the dependence on the positive constant ${\mathcal {K}}_1 = {\mathcal {K}}_1(\Vert f\Vert _{L^q(Q)}, \Vert v^0\Vert _{W^{2-2/q,q}(\Omega )})$, from Corollary 3.3. Since these changes were covered in Lemma 3.5, next we will state the results, skipping their proofs.

Lemma 3.6

(Energy inequality for $\varvec{s \in (1,2)}$) The solution (u, z) of the problem (3.14) satisfies, for sufficiently small $\alpha > 0$,

$$\begin{aligned} \begin{aligned}&\dfrac{d}{dt} E_m(u,z)(t) + \beta \displaystyle {\int _{\Omega }}{| {\nabla [T^{m}(u) + 1]^{s/2}} | _{}^2 \ dx} + \dfrac{1}{4} \displaystyle {\int _{\Omega }}{T^m(u)^s | {\nabla z} | _{}^2 \ dx} \\&\quad + \beta \Bigg ( \displaystyle {\int _{\Omega }}{| {D^2 z} | _{}^2 \ dx} + \displaystyle {\int _{\Omega }}{\frac{| {\nabla z} | _{}^4}{z^2} \ dx} \Bigg ) \le C ({\mathcal {K}}_1^2) \Vert {\nabla z} \Vert _{L^2(\Omega )}^2 + {\mathcal {K}}_1^2 \Vert {f} \Vert _{L^2(\Omega )}^2. \end{aligned} \end{aligned}$$

(3.20)

Remark 3.7

In [5], in the lemma where the authors prove the energy inequality for $s \in (1,2)$, the term $\displaystyle {\int _{\Omega }}{| {\nabla [T^{m}(u) + 1/j]^{s/2}} | _{}^2 dx}$ is estimated by

$$\begin{aligned} \displaystyle {\int _{\Omega }}{| {\nabla [T^{m}(u) + 1/j]^{s/2}} | _{}^2 dx} \ge 0, \text{ for } \text{ all } j \in {\mathbb {N}}, \end{aligned}$$

but it can be estimated by

$$\begin{aligned} \displaystyle {\int _{\Omega }}{| {\nabla [T^{m}(u) + 1/j]^{s/2}} | _{}^2 dx} \ge \displaystyle {\int _{\Omega }}{| {\nabla [T^{m}(u) + 1]^{s/2}} | _{}^2 dx}, \text{ for } \text{ all } j \in {\mathbb {N}}, \end{aligned}$$

instead, yielding (3.20). $\square$

Lemma 3.8

(Energy inequality for $\varvec{s \ge 2}$) The solution (u, z) of the problem (3.14) satisfies, for sufficiently small $\alpha > 0$,

$$\begin{aligned} \begin{array}{l} \dfrac{d}{dt} E_m(u,z)(t) + \displaystyle {\int _{\Omega }}{| {\nabla [T^m(u)]^{s/2}} | _{}^2 dx} + \dfrac{1}{4} \displaystyle {\int _{\Omega }}{T^m(u)^s | {\nabla z} | _{}^2 dx} \\ \quad + \beta \Bigg ( \displaystyle {\int _{\Omega }}{| {D^2 z} | _{}^2 dx} + \displaystyle {\int _{\Omega }}{\frac{| {\nabla z} | _{}^4}{z^2} dx} \Bigg ) \le C ({\mathcal {K}}_1^2) \Vert {\nabla z} \Vert _{L^2(\Omega )}^2 + {\mathcal {K}}_1^2 \Vert {f} \Vert _{L^2(\Omega )}^2. \end{array} \end{aligned}$$

(3.21)

3.4 $\varvec{m}$-Independent Estimates and Passage to the Limit as $\varvec{m \rightarrow \infty }$

In the present subsection we go back to the notation $(u_m,z_m)$ and $(u_m,v_m)$ to the solution of (3.14) and (3.2), respectively.

3.4.1 $\varvec{m}$-Independent Estimates for $\varvec{\nabla v_m}$

We will integrate the energy inequalities (3.17), (3.20) and (3.21) with respect to t, from 0 to $T > 0$. We take into account that, because of Corollary 3.3, we have the following bounds independently of m:

$$\begin{aligned} \nabla z_m \text{ is } \text{ bounded } \text{ in } L^{2}(Q) \end{aligned}$$

and

$$\begin{aligned} 0 < \alpha \le z_m(t,x) \le {\mathcal {K}}_1, \ a.e. \ (t,x) \in Q. \end{aligned}$$

(3.22)

We also use the hypothesis (1.5) on the initial data $u^0,v^0$ to prove that the energy given in (3.16) at time $t = 0$, $E_m(u_m,z_m)(0)$, is also bounded, independently of m. Thus we conclude that

$$\begin{aligned}{} & {} \nabla z_m \text{ is } \text{ bounded } \text{ in } L^{\infty }(0,T;L^2(\Omega )) \cap L^4(Q),\\{} & {} T^m(u_m)^{s/2} \nabla z_m \text{ and } \Delta z_m \text{ are } \text{ bounded } \text{ in } L^2(Q). \end{aligned}$$

But using the fact that $z_m = \sqrt{v_m + \alpha ^2}$ and (3.22) we can conclude that

$$\begin{aligned}{} & {} \nabla v_m \text{ is } \text{ bounded } \text{ in } L^{\infty }(0,T;L^2(\Omega )) \cap L^4(Q), \end{aligned}$$

(3.23)

$$\begin{aligned}{} & {} T^m(u_m)^{s/2} \nabla v_m \text{ and } \Delta v_m \text{ are } \text{ bounded } \text{ in } L^2(Q). \end{aligned}$$

(3.24)

3.4.2 Case $\varvec{s \in [1,2)}$

First, following [5], to which we refer the reader that might be interested in more details, we prove the existence of weak solution (u, v) to (1.2). Next, to conclude the proof of Theorem 1.3, letting $z = \sqrt{v + \alpha ^2}$, we prove the energy inequality (1.6).

Existence of weak solution to (1.2):

In order to prove the existence of a weak solution (u, v) to (1.2), first we obtain m-independent estimates for the solutions $(u_m,v_m)$ of (3.2) and then we use compactness results in weak*, weak and strong topologies to pass to the limit as $m \rightarrow \infty$.

Let

$$\begin{aligned} \qquad g'(r) = \left\{ \begin{array}{cc} ln(r) &{} \text{ if } s = 1, \\ r^{s-1}/(s-1) &{} \text{ if } s \in (1,2), \end{array} \right. \forall r > 0. \end{aligned}$$

and let

$$\begin{aligned} g(r) = \int _0^r{g'(\theta ) \ d\theta } = \left\{ \begin{array}{cc} r ln(r) - (r - 1) &{} \text{ if } s = 1, \\ r^s/s(s-1) &{} \text{ if } s \in (1,2). \end{array} \right. \end{aligned}$$

Notice that $g''(r) = r^{s-2}, \ \forall r > 0$, in all cases. We test the $u_m$-equation of (3.2) by $g'(u_m + 1)$ and obtain

$$\begin{aligned}&\dfrac{d}{dt} \int _{\Omega } g(u_m + 1) \ dx + \dfrac{4}{s^2} \int _{\Omega }{| {\nabla [u_m + 1]^{s/2}} | _{}^2 \ dx} \\&\quad = \int _{\Omega }{T^m(u_m) (u_m + 1)^{s/2-1} \nabla v_m \cdot \nabla u_m\, (u_m + 1)^{s/2-1} \ dx} \\&\quad = \dfrac{2}{s} \int _{\Omega }{\dfrac{T^m(u_m)^{1-s/2}}{(u_m + 1)^{1-s/2}} T^m(u_m)^{s/2} \nabla v_m \cdot \nabla [u_m + 1]^{s/2} \ dx} \\&\quad \le \dfrac{2}{s} \Bigg ( \int _{\Omega }{T^m(u_m)^s | {\nabla v_m} | _{}^2 \ dx} \Bigg )^{1/2} \Bigg ( \int _{\Omega }{| {\nabla [u_m + 1]^{s/2}} | _{}^2 \ dx} \Bigg )^{1/2} \end{aligned}$$

and thus we have

$$\begin{aligned} \dfrac{d}{dt} \int _{\Omega }{g(u_m + 1) dx} + \dfrac{2}{s^2} \int _{\Omega }{| {\nabla [u_m + 1]^{s/2}} | _{}^2 dx} \le \dfrac{1}{4} \int _{\Omega }{T^m(u_m)^s | {\nabla v_m} | _{}^2 dx}. \end{aligned}$$

Integrating with respect to t from 0 to T we obtain

$$\begin{aligned}&\int _{\Omega }{g(u_m(T) + 1) dx} + \dfrac{2}{s^2} \int _{0}^T{\int _{\Omega }{| {\nabla [u_m + 1]^{s/2}} | _{}^2 dx} dt}\\&\quad \le \dfrac{1}{4} \int _0^T{\int _{\Omega }{T^m(u_m)^s | {\nabla v_m} | _{}^2 dx} dt} + \int _{\Omega }{g(u^0 + 1) dx}. \end{aligned}$$

Then, because of (3.11), (1.5) and the definition of g and (3.24) we conclude that

$$\begin{aligned} (u_m + 1)^{s/2} \text{ is } \text{ bounded } \text{ in } L^{\infty }(0,T;L^2(\Omega )) \cap L^2(0,T;H^1(\Omega )). \end{aligned}$$

(3.25)

Using the Sobolev inequality $H^1(\Omega ) \subset L^6(\Omega )$ and interpolation inequalities we obtain

$$\begin{aligned} (u_m)^{s/2} \text{ is } \text{ bounded } \text{ in } L^{10/3}(Q). \end{aligned}$$

The latter and (3.25) imply that

$$\begin{aligned} u_m \text{ is } \text{ bounded } \text{ in } L^{\infty }(0,T;L^s(\Omega )) \cap L^{5s/3}(Q). \end{aligned}$$

(3.26)

From (3.26) we can conclude, using the $v_m$-equation of (3.2) that

$$\begin{aligned} \partial _t v_m \text{ is } \text{ bounded } \text{ in } L^{5/3}(Q). \end{aligned}$$

Reminding that $s \in [1,2)$, if we use (3.25) and (3.26) in the relation

$$\begin{aligned} \nabla u_m = \nabla (u_m + 1) = \nabla \big ( (u_m + 1)^{s/2} \big )^{2/s} = \dfrac{2}{s} (u_m + 1)^{1 - s/2} \ \nabla (u_m + 1)^{s/2}. \end{aligned}$$

then we also have

$$\begin{aligned} u_m \text{ is } \text{ bounded } \text{ in } L^{5s/(3+s)}(0,T;W^{1,5s/(3+s)}(\Omega )). \end{aligned}$$

(3.27)

Considering the chemotaxis term of the $u_m$-equation of (3.2), we can write $T^m(u_m) \nabla v_m$ as

$$\begin{aligned} T^m(u_m) \nabla v_m = T^m(u_m)^{1-s/2} T^m(u_m)^{s/2} \nabla v_m. \end{aligned}$$

Then, we have $T^m(u_m)^{1-s/2}$ bounded in $L^{10s/(6-3s)}(Q)$, because of (3.25), and $T^m(u_m)^{s/2} \nabla v_m$ bounded in $L^2(Q)$, because of (3.24), and hence we can conclude that

$$\begin{aligned} T^m(u_m) \nabla v_m \text{ is } \text{ bounded } \text{ in } L^{5s/(3+s)}(Q). \end{aligned}$$

(3.28)

Then, if we consider the $u_m$-equation of (3.2), from (3.27) and (3.28) we conclude that

$$\begin{aligned} \partial _t u_m \text{ is } \text{ bounded } \text{ in } L^{5s/(3+s)}(0,T;(W^{1,5s/(4s-3)}(\Omega ))'). \end{aligned}$$

Now we are going to obtain compactness for $\{ u_m \}$ which is necessary in order to pass to the limit as $m \rightarrow \infty$ in the nonlinear terms of the equations of (3.2).

We observe that $W^{1,5s/(3+s)}(\Omega ) \subset L^q(\Omega )$, with continuous embedding for $q = 15s/(9-2s)$ and compact embedding for $q \in [1,15s/(9-2s))$. Then, since $s \in [1,2)$, we have $5s/3 < 15s/(9-2s)$ and therefore the embedding $W^{1,5s/(3+s)}(\Omega ) \subset L^{5s/3}(\Omega )$ is compact. Note also that $q = 5s/3 \ge 5/3 > 1$.

Now we can use Lemma 2.7 with

$$\begin{aligned} X = W^{1,5s/(3+s)}(\Omega ),\quad B = L^{5s/3}(\Omega ), \quad Y = \big ( W^{1,5s/(4s-3)}(\Omega ) \big )' \end{aligned}$$

and $q = 5s/3$, to conclude that there is a subsequence of $\{ u_m \}$ (still denoted by $\{ u_m \}$) and a limit function u such that

$$\begin{aligned} u_m \longrightarrow u \text{ weakly } \text{ in } L^{5s/(3+s)}(0,T;W^{1,5s/(3+s)}(\Omega )), \end{aligned}$$

and

$$\begin{aligned} u_m \longrightarrow u \text{ strongly } \text{ in } L^p(Q), \ \forall p \in [1,5s/3). \end{aligned}$$

(3.29)

Using the Dominated Convergence Theorem we can conclude from (3.29) that

$$\begin{aligned} T^m(u_m) \rightarrow u \text{ strongly } \text{ in } L^p(Q), \ \forall p \in [1,5s/3). \end{aligned}$$

(3.30)

It stems from the convergence (3.30) and Lemma 2.6 that

$$\begin{aligned} (T^m(u_m))^s \rightarrow u^s \text{ strongly } \text{ in } L^q(Q), \ \forall q \in [1,5/3). \end{aligned}$$

(3.31)

The convergence of $v_m$ is better. There is a subsequence of $\{ v_m \}$ (still denoted by $\{ v_m \}$) and a limit function v such that

$$\begin{aligned} \begin{array}{c} v_m \rightarrow v \text{ weakly* } \text{ in } L^{\infty }(Q) \cap L^{\infty }(0,T;H^1(\Omega )), \\ v_m \rightarrow v \text{ weakly } \text{ in } L^2(0,T;H^2(\Omega )), \\ \nabla v_m \rightarrow \nabla v \text{ weakly } \text{ in } L^4(Q), \\ \text{ and } \partial _t v_m \rightarrow \partial _t v \text{ weakly } \text{ in } L^{5/3}(Q). \end{array} \end{aligned}$$

(3.32)

Now we are going to use the weak and strong convergences obtained so far to pass to the limit as $m \rightarrow \infty$ in the equations of problem (3.2). Since passing to the limit in the linear terms is simpler, as well as in the term $f_m v_m$, because of the strong convergence of $f_m$ given in (3.1), we focus on the nonlinear terms of the equations. We are going to identify the limits of the nonlinear terms related to chemotaxis and consumption,

$$\begin{aligned} T^m(u_m) \nabla v_m \text{ and } T^m(u_m)^s v_m, \end{aligned}$$

respectively, with

$$\begin{aligned} u \nabla v \text{ and } u^s v. \end{aligned}$$

In fact, considering the chemotaxis term, because of (3.30), (3.23) and (3.32), we can conclude that

$$\begin{aligned} T^m(u_m) \nabla v_m \longrightarrow u \nabla v \text{ weakly } \text{ in } L^{20s/(5s+12)}(Q). \end{aligned}$$

Considering now the consumption term, considering (3.31) and (3.32) we conclude that

$$\begin{aligned} T^m(u_m)^s v_m \longrightarrow u^s v \text{ weakly } \text{ in } L^{5/3}(Q). \end{aligned}$$

With these identifications and all previous convergences, it is possible to pass to the limit as $m \rightarrow \infty$ in each term of the equations of (3.2).

Energy inequality (1.6):

In order to finish we must prove the energy inequality (1.6). First we obtain an integral inequality for the solution $(u_m,z_m)$ of (3.14), where we remind that $z_m = \sqrt{v_m + \alpha ^2}$, for small enough but fixed $\alpha > 0$, being $(u_m,v_m)$ the solution of (3.2). According to Lemmas (3.5) and (3.6), $(u_m,z_m)$ satisfies

$$\begin{aligned} \begin{array}{l} E_m(u_m,z_m)(t_2) + \beta \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{| {\nabla [T^{m}(u_m) + 1]^{s/2}} | _{}^2 \ dx} \ dt\\ \quad + \dfrac{1}{4} \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{T^m(u_m)^s | {\nabla z_m} | _{}^2 \ dx} \ dt \\ \quad + \beta \Bigg ( \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{| {D^2 z_m} | _{}^2 \ dx} \ dt + \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{\frac{| {\nabla z_m} | _{}^4}{z_m^2} \ dx} \ dt \Bigg ) \\ \le E_m(u_m,z_m)(t_1) + C ({\mathcal {K}}_1^2) \displaystyle {\int _{t_1}^{t_2} \Vert {\nabla z_m} \Vert _{L^2(\Omega )}^2} \ dt + {\mathcal {K}}_1^2 \displaystyle {\int _{t_1}^{t_2} \Vert {f} \Vert _{L^2(\Omega )}^2} \ dt, \end{array} \end{aligned}$$

(3.33)

where $E_m$ is given by (3.16). Next we collect some convergences that can be obtained from the m-independent bounds and the weak and strong convergences proved so far and that will be useful to pass to the limit in the inequality (3.33). Recalling that we denote $z = \sqrt{v + \alpha ^2}$, we have, in particular,

$$\begin{aligned} \begin{array}{c} (u_m + 1)^{s/2} \longrightarrow (u + 1)^{s/2} \text{ weakly* } \text{ in } L^{\infty }(0,T;L^2(\Omega )), \\ (u_m + 1)^{s/2} \longrightarrow (u + 1)^{s/2} \text{ weakly } \text{ in } L^2(0,T;H^1(\Omega )), \\ \nabla T^m(u_m)^{s/2} \longrightarrow \nabla u^{s/2} \text{ weakly } \text{ in } L^2(Q), \\ T^m(u_m)^{s/2} \nabla z_m \longrightarrow u^{s/2} \nabla z \text{ weakly } \text{ in } L^2(Q), \\ \dfrac{\nabla z_m}{\sqrt{z_m}} \longrightarrow \dfrac{\nabla z }{\sqrt{z}} \text{ weakly } \text{ in } L^4(Q), \\ D^2 z_m \longrightarrow D^2 z \text{ weakly } \text{ in } L^2(Q), \\ \nabla z_m \longrightarrow \nabla z \text{ strongly } \text{ in } L^2(Q). \end{array} \end{aligned}$$

(3.34)

Recalling that we are dealing with the case $s \in [1,2)$, let

$$\begin{aligned} E(u,z)(t) = \frac{s}{4} \displaystyle {\int _{\Omega }}{g(u(t,x)) \, dx} + \frac{1}{2} \int _{\Omega }{| {\nabla z(t,x)} | _{}^2} dx, \end{aligned}$$

where

$$\begin{aligned} g(u) = \left\{ \begin{array}{rl} (u+1)ln(u+1) - u, &{} \text{ if } s = 1, \\ \dfrac{u^s}{s(s-1)}, &{} \text{ if } s \in (1,2). \end{array} \right. \end{aligned}$$

Then the following convergence will be also necessary.

Lemma 3.9

$E_m(u_m,z_m) \longrightarrow E(u,z)$ in $L^1(0,T)$.

Proof

From (3.34) we have that $\nabla z_m \rightarrow \nabla z$ in $L^2(Q)$ which, in particular, leads us to

$$\begin{aligned} \int _{\Omega }{| {\nabla z_m(t,x)} | _{}^2} dx \longrightarrow \int _{\Omega }{| {\nabla z(t,x)} | _{}^2} dx \quad \text{ in } L^1(0,T). \end{aligned}$$

Then, it remains to prove that

$$\begin{aligned} \int _{\Omega }{g_m(u_m) \, dx} \longrightarrow \int _{\Omega }{g(u) \, dx} \quad \text{ in } L^1(0,T). \end{aligned}$$

(3.35)

We begin by rewriting $g_m(u_m) - g(u)$ as

$$\begin{aligned} g_m(u_m) - g(u) = g_m(u_m) - g_m(u) + g_m(u) - g(u). \end{aligned}$$

(3.36)

For the first difference in (3.36), $g_m(u_m) - g_m(u)$, using that $g_m'$ and $g'$ are monotone increasing functions and that $g_m'(r) \le g'(r)$ for all $r \ge 0$, we have

$$\begin{aligned} | {g_m(u_m) - g_m(u)} | _{}&= | {\int _u^{u_m}{g_m'(\theta ) \ d \theta }} | _{} \\&\le | {u_m - u} | _{} (g_m'(u_m) + g_m'(u)) \le | {u_m - u} | _{} (g'(u_m) + g'(u)). \end{aligned}$$

Then, for $s = 1$, we have

$$\begin{aligned} | {g_m(u_m) - g_m(u)} | _{} \le C | {u_m - u} | _{} (ln(u_m + 1) + ln(u + 1)) \end{aligned}$$

and, for $s \in (1,2)$, we have

$$\begin{aligned} | {g_m(u_m) - g_m(u)} | _{} \le C | {u_m - u} | _{} (| {u_m} | _{}^{s-1} + | {u} | _{}^{s-1}). \end{aligned}$$

Considering the case $s \in (1,2)$, since from (3.26) we have $(| {u_m} | _{}^{s-1} + | {u} | _{}^{s-1})$ bounded in $L^{5s/(3s-3)}(Q)$ and, from (3.29), we have $| {u_m - u} | _{} \rightarrow 0$ in $L^{5s/(2s+3)}(Q)$, we conclude that

$$\begin{aligned} g_m(u_m) - g_m(u) \longrightarrow 0 \quad \text{ in } L^1(Q). \end{aligned}$$

(3.37)

Considering now the case $s = 1$, from (3.26) we have $ln(u_m + 1) + ln(u + 1)$ bounded in $L^p(Q)$, for all $p \in [1,\infty )$. Then, analogously to the case $s \in (1,2)$, we use (3.29) and obtain (3.37) also for $s = 1$. From (3.37) we conclude, in particular, that

$$\begin{aligned} \int _{\Omega }{g_m(u_m) \ dx} - \int _{\Omega }{g_m(u) \ dx} \longrightarrow 0 \quad \text{ in } L^1(0,T). \end{aligned}$$

(3.38)

For the second difference in (3.36), $g_m(u) - g(u)$, we use the Dominated Convergence Theorem. In fact, we write

$$\begin{aligned} g_m(u) - g(u) = \int _0^u{g_m'(\theta ) - g'(\theta ) \ d \theta }. \end{aligned}$$

Using this expression one can verify that

$$\begin{aligned} g_m(u) - g(u) \longrightarrow 0, \ a.e. \ (t,x) \in Q. \end{aligned}$$

Next we note that $g_m(u) - g(u)$ is bounded by $2 g(u) \in L^1(Q)$. Therefore we conclude, by using the Dominated Convergence Theorem, that $g_m(u) - g(u) \rightarrow 0$ in $L^1(Q)$ and, in particular

$$\begin{aligned} \int _{\Omega }{g_m(u) \, dx} - \int _{\Omega }{g(u) \, dx} \longrightarrow 0 \quad \text{ in } L^1(0,T). \end{aligned}$$

(3.39)

With (3.38) and (3.39) we obtain (3.35), finishing the proof. $\square$

Lemma 3.10

For $s \ge 1$ we have $v \in C_w([0,T];H^1(\Omega ))$ and $u \in C_w([0,T];L^s(\Omega ))$.

Proof

For any $s \ge 1$, we have $v \in L^{\infty }(Q) \subset L^{5/3}(Q)$ and $\partial _t v \in L^{5/3}(Q)$, which implies that $v \in C([0,T];L^{5/3}(\Omega ))$. Since $v \in L^{\infty }(0,T;H^1(\Omega ))$, by Lemma 2.3 one has $v \in C_w([0,T];H^1(\Omega ))$. On the other hand, $u \in L^{5s/(3+s)}(0,T;W^{1,5s/(3+s)}(\Omega ))$ and $\partial _t u \in L^{5s/(3+s)}(0,T;(W^{1,5s/(4s-3)}(\Omega ))')$. Since from (3.34), $u \in L^{\infty }(0,T;L^s(\Omega ))$, we conclude, using Lemma 2.3, that $u \in C_w([0,T];L^s(\Omega ))$. $\square$

Next we pass to the limit in inequality (3.33). Because of Lemma 3.9, we conclude that, up to a subsequence,

$$\begin{aligned} E_m(u_m,z_m)(t) \longrightarrow E(u,z)(t) \quad a.e. \ t \in [0,T]. \end{aligned}$$

(3.40)

Accounting for the properties of $u_m(0) = u^0_m$ and $v_m(0) = v^0$ we conclude that (3.40) holds, in particular, for $t = 0$.

Therefore, using the convergences (3.34), the weak lower semicontinuity of the norm (Lemma 2.2) and the almost everywhere pointwise convergence (3.40), we are able to pass to the limit in (3.33) and conclude that, for a.e. $t_1,t_2 \in [0,T]$, with $t_2 > t_1$ (including $t_1 = 0$), we have

$$\begin{aligned} \begin{array}{l} E(u,z)(t_2) + \beta \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{| {\nabla [u + 1]^{s/2}} | _{}^2 dx} \, dt + \dfrac{1}{4} \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{u^s | {\nabla z} | _{}^2 dx} \, dt \\ + \beta \Bigg ( \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{| {D^2 z} | _{}^2 dx} \, dt + \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{\frac{| {\nabla z} | _{}^4}{z^2} dx} \, dt \Bigg ) \\ \le E(u,z)(t_1) + C ({\mathcal {K}}_1^2) \displaystyle {\int _{t_1}^{t_2} \Vert {\nabla z} \Vert _{L^2(\Omega )}^2} \ dt + {\mathcal {K}}_1^2 \displaystyle {\int _{t_1}^{t_2} \Vert {f} \Vert _{L^2(\Omega )}^2} \ dt. \end{array} \end{aligned}$$

Accounting for the m-independent bound for $\nabla z_m$ given in (3.15) and the strong convergence of $\nabla z_m$ to $\nabla z$ given in (3.34), we have $\Vert {\nabla z} \Vert _{L^2(Q)} \le {\mathcal {K}}_1$. And since ${\mathcal {K}}_1 = {\mathcal {K}}_1(\Vert f\Vert _{L^q(Q)},\Vert v^0\Vert _{W^{2-2/q,q}(\Omega )})$, we conclude that there is other constant ${\mathcal {K}} = {\mathcal {K}}(\Vert f\Vert _{L^q(Q)},\Vert v^0\Vert _{W^{2-2/q,q}(\Omega )})$, which is increasing and continuous with respect to $\Vert {f} \Vert _{L^q(Q)}$, such that, for a.e. $t_1,t_2 \in [0,T]$, with $t_2 > t_1$ (including $t_1 = 0$), we have

$$\begin{aligned} \begin{array}{l} E(u,z)(t_2) + \beta \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{| {\nabla [u + 1]^{s/2}} | _{}^2 dx} \, dt + \dfrac{1}{4} \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{u^s | {\nabla z} | _{}^2 dx} \, dt \\ + \beta \Bigg ( \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{| {D^2 z} | _{}^2 dx} \, dt + \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{\frac{| {\nabla z} | _{}^4}{z^2} dx} \, dt \Bigg ) \\ \le E(u,z)(t_1) + {\mathcal {K}}(\Vert f\Vert _{L^q(Q)},\Vert v^0\Vert _{W^{2-2/q,q}(\Omega )}). \end{array} \end{aligned}$$

(3.41)

To finish, we consider the case $s > 1$. For simplicity, let us write (3.41) in terms of the energy $E(u,z)(\cdot )$ and the dissipative term $D(u,z)(t_1,t_2)$, for a.e. $t_1,t_2 \in [0,T]$, with $t_2 > t_1$, as

$$\begin{aligned} E(u,z)(t_2) + D(u,z)(t_1,t_2) \le E(u,z)(t_1) + {\mathcal {K}}(\Vert f\Vert _{L^q(Q)},\Vert v^0\Vert _{W^{2-2/q,q}(\Omega )}). \end{aligned}$$

Now, let $t_2 \in (t_1, T]$ and let $\{ t_2^n \}_n$ be a sequence such that $t_2^n \rightarrow t_2$ as $n \rightarrow \infty$ and such that, for all n, we have

$$\begin{aligned} E(u,z)(t_2^n) + D(u,z)(t_1,t_2^n) \le E(u,z)(t_1) + {\mathcal {K}}(\Vert {f} \Vert _{L^q(Q)},\Vert v^0\Vert _{W^{2-2/q,q}(\Omega )}).\nonumber \\ \end{aligned}$$

(3.42)

If we take the $\liminf$ in both sides of (3.42), we obtain

$$\begin{aligned} \liminf _{n \rightarrow \infty }{E(u,z)(t_2^n)} + D(u,z)(t_1,t_2) \le E(u,z)(t_1) + {\mathcal {K}}(\Vert {f} \Vert _{L^q(Q)},\Vert v^0\Vert _{W^{2-2/q,q}(\Omega )}). \end{aligned}$$

Then, from the definition of E for $s > 1$ and Lemma 2.3, we have

$$\begin{aligned} E(u,z)(t_2) \le \liminf _{n \rightarrow \infty }{E(u,z)(t_2^n)} \end{aligned}$$

and therefore we conclude that, for $s > 1$, the energy inequality (3.41) is satisfied for a.e. $t_1 \in [0,T]$, and for all $t_2 \in (t_1,T]$.

3.4.3 Case $\varvec{s \ge 2}$

Existence of weak solution to (1.2):

The procedure for the case $s \ge 2$ is slightly different. First we note that, integrating the energy inequality (3.21) from Lemma 3.8 with respect to t, we have

$$\begin{aligned} \nabla T^m(u_m)^{s/2} \text{ is } \text{ bounded } \text{ in } L^2(Q). \end{aligned}$$

(3.43)

We also remind that we defined $g_m'(r) = T^m(r)^{s-1}/(s-1)$, for $s \ge 2$. Then we have

$$\begin{aligned} T^m(r)^s = s \int _0^r{(T^m)'(\theta ) T^m(\theta )^{s-1} d\theta } \le C s \int _0^r{T^m(\theta )^{s-1} d\theta } = C s(s - 1) g_m(r). \end{aligned}$$

Therefore it also stems from integrating the energy inequality (3.21) with respect to t that

$$\begin{aligned} T^m(u_m)^{s/2} \text{ is } \text{ bounded } \text{ in } L^{\infty }(0,T;L^2(\Omega )). \end{aligned}$$

(3.44)

From (3.44) and (3.43) we can conclude that

$$\begin{aligned} T^m(u_m)^{s/2} \text{ is } \text{ bounded } \text{ in } L^{10/3}(Q), \end{aligned}$$

that is,

$$\begin{aligned} T^m(u_m) \text{ is } \text{ bounded } \text{ in } L^{5s/3}(Q). \end{aligned}$$

(3.45)

For each fixed $m \in {\mathbb {N}}$, consider the zero measure set ${\mathcal {N}} \subset (0,T)$ such that

$$\begin{aligned} u_m(t^{*},\cdot ), v_m(t^{*},\cdot ) \in H^1(\Omega ), \ \forall t^{*} \in (0,T) \setminus {\mathcal {N}}. \end{aligned}$$

Then, for each fixed $t^{*} \in (0,T) \setminus {\mathcal {N}}$, let us consider the sets

$$\begin{aligned} \{ 0 \le u_m \le 1 \} = \Big \{ x \in \Omega \ | \ 0 \le u_m(t^{*},x) \le 1 \Big \} \end{aligned}$$

and

$$\begin{aligned} \{ u_m \ge 1 \} = \Big \{ x \in \Omega \ | \ u_m(t^{*},x) \ge 1 \Big \}. \end{aligned}$$

Now note that, since $s \ge 2$, we have

$$\begin{aligned}&\int _{\Omega } T^m(u_m(t^{*},x)^2 | {\nabla v_m(t^{*},x)} | _{}^2 dx \\&\quad \le \int _{\{ 0 \le u_m \le 1 \}}{| {\nabla v_m(t^{*},x)} | _{}^2 dx} + \int _{\{ u_m \ge 1 \}}{T^m(u_m(t^{*},x))^s | {\nabla v_m(t^{*},x)} | _{}^2 dx} \\&\quad \le \int _{\Omega }{| {\nabla v_m(t^{*},x)} | _{}^2 dx} + \int _{\Omega }{T^m(u_m(t^{*},x))^s | {\nabla v_m(t^{*},x)} | _{}^2 dx}. \end{aligned}$$

The last inequality is valid for all $t^{*} \in (0,T) {\setminus } {\mathcal {N}}$, then if we integrate in the variable t we obtain

$$\begin{aligned} \int _0^\infty {\int _{\Omega }{T^m(u_m(t,x)^2 | {\nabla v_m(t,x)} | _{}^2 dx} \, dt}&\le \int _0^\infty {\int _{\Omega }{| {\nabla v_m(t,x)} | _{}^2 dx} \,dt} \\&+ \int _0^\infty {\int _{\Omega }{T^m(u_m(t,x))^s | {\nabla v_m(t,x)} | _{}^2 dx} \, dt}. \end{aligned}$$

Therefore by (3.11) and (3.24) we can conclude that

$$\begin{aligned} T^m(u_m) \nabla v_m \text{ is } \text{ bounded } \text{ in } L^2(Q). \end{aligned}$$

(3.46)

Now we test the $u_m$-equation of problem (3.2) by $u_m$. This gives us

$$\begin{aligned}&\dfrac{1}{2} \dfrac{d}{dt} \Vert {u_m} \Vert _{L^2(\Omega )}^2 + \Vert {\nabla u_m} \Vert _{L^2(\Omega )}^2 = \int _{\Omega }{T^m(u_m) \nabla v_m \cdot \nabla u_m \ dx} \\&\quad \le \dfrac{1}{2} \int _{\Omega }{T^m(u_m)^2 | {\nabla v_m} | _{}^2 dx} + \dfrac{1}{2} \Vert {\nabla u_m} \Vert _{L^2(\Omega )}^2, \end{aligned}$$

hence we have

$$\begin{aligned} \dfrac{d}{dt} \Vert {u_m} \Vert _{L^2(\Omega )}^2 + \Vert {\nabla u_m} \Vert _{L^2(\Omega )}^2 \le \int _{\Omega }{T^m(u_m)^2 | {\nabla v_m} | _{}^2 dx}. \end{aligned}$$

Integrating with respect to t, we conclude from (3.46) that

$$\begin{aligned} u_m \text{ is } \text{ bounded } \text{ in } L^{\infty }(0,T;L^2(\Omega )) \end{aligned}$$

(3.47)

and

$$\begin{aligned} \nabla u_m \text{ is } \text{ bounded } \text{ in } L^2(Q). \end{aligned}$$

(3.48)

Then, if we consider the $u_m$-equation of (3.2), by applying (3.48) and (3.46) we conclude that

$$\begin{aligned} \partial _t u_m \text{ is } \text{ bounded } \text{ in } L^2(0,T;(H^1(\Omega ))'). \end{aligned}$$

(3.49)

Considering (3.24), (3.11) and (3.45) we conclude from the $v_m$-equation of (3.2) that

$$\begin{aligned} \partial _t v_m \text{ is } \text{ bounded } \text{ in } L^2(0,T;L^{3/2}(\Omega )). \end{aligned}$$

Now, using (3.47), (3.48) and (3.49) we can conclude that there is a subsequence of $\{ u_m \}$, still denoted by $\{ u_m \}$, and a limit function u such that

$$\begin{aligned} \begin{array}{c} u_m \longrightarrow u \text{ weakly* } \text{ in } L^{\infty }(0,T;L^2(\Omega )), \\ \nabla u_m \longrightarrow \nabla u \text{ weakly } \text{ in } L^2(Q), \\ \partial _t u_m \longrightarrow u \text{ weakly } \text{ in } L^2 \Big ( 0,\infty ;\big (H^1(\Omega ) \big )' \Big ). \end{array} \end{aligned}$$

By applying the compactness result Lemma 2.7, one has

$$\begin{aligned} u_m \longrightarrow u \text{ strongly } \text{ in } L^2(Q). \end{aligned}$$

Using the Dominated Convergence Theorem and (3.45) we can also prove that

$$\begin{aligned} T^m(u_m) \longrightarrow u \text{ strongly } \text{ in } L^p(Q), \ \forall p \in (1,5s/3), \end{aligned}$$

and using Lemma 2.6,

$$\begin{aligned} T^m(u_m)^s \longrightarrow u^s \text{ strongly } \text{ in } L^p(Q), \ \forall p \in (1,5/3). \end{aligned}$$

From the global in time estimate (3.44) we can conclude that, up to a subsequence,

$$\begin{aligned} T^m(u_m) \rightarrow u \text{ weakly* } \text{ in } L^{\infty }(0,T;L^s(\Omega )), \end{aligned}$$

hence, in particular,

$$\begin{aligned} u \in L^{\infty }(0,T;L^s(\Omega )). \end{aligned}$$

For $s \ge 2$, if we consider the functions $v_m$, we have the same m-independent estimates that we had for $s \in [1,2)$. Then we have the same convergences given in (3.32).

Following the ideas of Subsect. 3.4.2, we can identify the limits of $T^m(u_m) \nabla v_m$ and $T^m(u_m)^s v_m$ with $u \nabla v$ and $u^s v$, respectively.

Energy inequality (1.6):

One can reach it by following the reasoning used in Subsect. 3.4.2 for $s \in [1,2)$.

4 Existence of an Optimal Control

In the present section we first prove Theorem 1.5, in Subsect. 4.1, establishing the existence of solution to the minimization problem (1.9). Afterwards, we prove Theorem 1.7 in Subsect. 4.2.

4.1 Proof of Theorem 1.5

Since the functional J in (1.9) is nonnegative,

$$\begin{aligned} J_{inf}: = \displaystyle {\inf _{(u,v,f) \in S_{ad}^M}} J(u,v,f) \ge 0 \end{aligned}$$

is well defined and there is a minimizing sequence $\{ (u_n,v_n,f_n) \} \subset S_{ad}^M$ such that

$$\begin{aligned} \lim _{n \rightarrow \infty }{J(u_n,v_n,f_n)} = J_{inf}. \end{aligned}$$

Next we prove that there is $(\overline{u},\overline{v},\overline{f}) \in S_{ad}^M$, that will be defined as the limit of a subsequence of $\{ (u_n,v_n,f_n) \}_n$, such that $J(\overline{u},\overline{v},\overline{f}) = J_{inf}$.

Since $(u_n,v_n,f_n) \in S_{ad}^M$, we have

$$\begin{aligned} \left\{ \begin{array}{l} \langle \partial _t u_n, \varphi \rangle + \displaystyle {\int _\Omega } \nabla u_n\cdot \nabla \varphi \ dx = \displaystyle {\int _\Omega } u_n \nabla v_n \cdot \nabla \varphi \ dx \\ \partial _t v_n - \Delta v_n = - u_n^s v_n + f_n v_n 1_{\Omega _c}, \\ \partial _\eta u_n |_{\Gamma } = \partial _\eta v_n |_{\Gamma } = 0, \ u_n(0) = u^0, \ v_n(0) = v^0, \end{array}\right. \end{aligned}$$

(4.1)

for every $\varphi \in W^{1,5s/(4s-3)}(\Omega )$. Denoting $z_n = \sqrt{v_n + \alpha ^2}$, we have

$$\begin{aligned} \begin{array}{l} E(u_n,z_n)(t_2) + \beta \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{| {\nabla [u_n + 1]^{s/2}} | _{}^2 dx} \, dt \\ + \dfrac{1}{4} \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{u_n^s | {\nabla z_n} | _{}^2 dx} \,dt + \beta \Bigg ( \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{| {D^2 z_n} | _{}^2 dx} \, dt \\ + \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{\frac{| {\nabla z_n} | _{}^4}{z^2} dx} \, dt \Bigg ) \le E(u_n,z_n)(t_1) + {\mathcal {K}}(M,\Vert {v_0} \Vert _{W^{2-2/q,q}(\Omega )}). \end{array} \end{aligned}$$

(4.2)

Since $(u_n,v_n,f_n) \in S^M_{ad}$, we have

$$\begin{aligned} \Vert {f_n} \Vert _{L^q(Q)} \le M. \end{aligned}$$

(4.3)

Then, comparing $v_n$ with the solution $w_n$ of (3.4), with $\tilde{f} = f_n$ and $w^0 = v^0$, yields $0 \le v_n(t,x) \le w_n(t,x)$ a.e. $(t,x) \in Q$. From Lemma 3.1 and (4.3), we obtain $\Vert {v_n} \Vert _{L^{\infty }(Q)} \le \Vert {w_n} \Vert _{L^{\infty }(Q)} \le {\mathcal {K}}_1(M)$ and, in particular, we conclude that there is a constant $C(M) > 0$ such that

$$\begin{aligned} 0 < \alpha \le z_n(t,x) \le C(M), \ a.e. \ (t,x) \in Q. \end{aligned}$$

(4.4)

With the energy inequality (4.2) and the pointwise bound (4.4) at hand, we are able to get the same estimates of Subsect. 3.4 and pass to the limit as $n \rightarrow \infty$. In fact, from (4.2), we conclude that, for $s \ge 1$, we have the following bounds independently of n:

$$\begin{aligned} \begin{array}{c} \nabla z_n \text{ is } \text{ bounded } \text{ in } L^{\infty }(0,T;L^2(\Omega )) \cap L^4(Q), \\ u_n^{s/2} \nabla z_n \text{ and } D^2 z_n \text{ are } \text{ bounded } \text{ in } L^2(Q). \end{array} \end{aligned}$$

But using the fact that $z_n = \sqrt{v_n + \alpha ^2}$ and (4.4) we can conclude that

$$\begin{aligned} \begin{array}{c} \nabla v_n \text{ is } \text{ bounded } \text{ in } L^{\infty }(0,T;L^2(\Omega )) \cap L^4(Q), \\ u_n^{s/2} \nabla v_n \text{ and } \Delta v_n \text{ are } \text{ bounded } \text{ in } L^2(Q). \end{array} \end{aligned}$$

From (4.2) (and by testing the $u_n$-equation of (4.1) by $\varphi = 1$, in the case $s = 1$) we also have

$$\begin{aligned} \begin{array}{c} \nabla [u_n + 1]^{s/2} \text{ is } \text{ bounded } \text{ in } L^2(Q), \\ (u_n + 1)^{s/2} \text{ is } \text{ bounded } \text{ in } L^{\infty }(0,T;L^2(\Omega )). \end{array} \end{aligned}$$

Afterwards, using some ideas of Subsect. 3.4 we conclude that for $s \in [1,2)$ we have

$$\begin{aligned} \begin{array}{c} u_n \text{ is } \text{ bounded } \text{ in } L^{5s/(3 + s)}(0,T;W^{1,5s/(3 + s)}(\Omega )), \\ \partial _t u_n \text{ is } \text{ bounded } \text{ in } L^{5s/(3 + s)}(0,T;(W^{1,5s/(4s - 3)}(\Omega ))'), \end{array} \end{aligned}$$

for $s \ge 2$ we have

$$\begin{aligned} \begin{array}{c} u_n \text{ is } \text{ bounded } \text{ in } L^2(0,T;H^1(\Omega )), \\ \partial _t u_n \text{ is } \text{ bounded } \text{ in } L^2(0,T;(H^1(\Omega ))'), \end{array} \end{aligned}$$

and, for $s \ge 1$, we have

$$\begin{aligned} \begin{array}{c} u_n \text{ is } \text{ bounded } \text{ in } L^{\infty }(0,T;L^s(\Omega )) \cap L^{5s/3}(Q), \\ v_n \text{ is } \text{ bounded } \text{ in } L^{\infty }(Q) \cap L^4(0,T;W^{1,4}(\Omega )) \cap L^2(0,T;H^2(\Omega )), \\ \partial _t v_n \text{ is } \text{ bounded } \text{ in } L^{5/3}(Q). \end{array} \end{aligned}$$

In view of these n-uniform bounds we follow the reasoning of Subsects. 3.4.2 and 3.4.3 and conclude that, up to a subsequence, there is $(\overline{u}, \overline{v},\overline{f})$ such that, if $s \in [1,2)$, we have

$$\begin{aligned} \begin{array}{c} u_n \longrightarrow \overline{u} \text{ weakly } \text{ in } L^{5s/3}(Q) \cap L^{5s/(3+s)}(0,T;W^{1,5s/(3+s)}(\Omega )), \\ u_n \longrightarrow \overline{u} \text{ strongly } \text{ in } L^p(Q), \ \forall p \in [1,5s/3) \\ \text{ and } \partial _t u_n \rightarrow \partial _t \overline{u} \text{ weakly } \text{ in } L^{5s/(3 + s)}(0,T;(W^{1,5s/(4s-3)}(\Omega ))'), \end{array} \end{aligned}$$

(4.5)

for $s \ge 2$ we have

$$\begin{aligned} \begin{array}{c} u_n \longrightarrow \overline{u} \text{ weakly } \text{ in } L^{5s/3}(Q) \cap L^2(0,T;H^1(\Omega )), \\ u_n \longrightarrow \overline{u} \text{ strongly } \text{ in } L^p(Q), \ \forall p \in [1,5s/3) \\ \text{ and } \partial _t u_n \rightarrow \partial _t \overline{u} \text{ weakly } \text{ in } L^2(0,T;(H^1(\Omega ))'), \end{array} \end{aligned}$$

(4.6)

and, for $s \ge 1$,

$$\begin{aligned} \begin{array}{c} v_n \rightarrow \overline{v} \text{ weakly* } \text{ in } L^{\infty }(Q) \cap L^{\infty }(0,T;H^1(\Omega )), \\ v_n \rightarrow \overline{v} \text{ weakly } \text{ in } L^4(0,T;W^{1,4}(\Omega )) \cap L^2(0,T;H^2(\Omega )), \\ \text{ and } \partial _t v_n \rightarrow \partial _t \overline{v} \text{ weakly } \text{ in } L^{5/3}(Q) \end{array} \end{aligned}$$

(4.7)

and

$$\begin{aligned} f_n \rightarrow \overline{f} \text{ weakly } \text{ in } L^q(Q). \end{aligned}$$

(4.8)

With these convergences we can pass to the limit as $n \rightarrow \infty$ in (4.1) and conclude that $(\overline{u},\overline{v})$ is a weak solution of (1.2) with control $\overline{f}$.

Now we are going to prove that $(\overline{u},\overline{v})$ satisfies the energy inequality (4.2) and therefore $(\overline{u},\overline{v},\overline{f}) \in S_{ad}^M$. Let $\overline{z} = \sqrt{\overline{v} + \alpha ^2}$, if we continue following the ideas of Subsects. 3.4.2 and (3.4.3) we conclude the convergences

$$\begin{aligned} \begin{array}{c} (u_n + 1)^{s/2} \longrightarrow (\overline{u} + 1)^{s/2} \text{ weakly* } \text{ in } L^{\infty }(0,T;L^2(\Omega )), \\ (u_n + 1)^{s/2} \longrightarrow (\overline{u} + 1)^{s/2} \text{ weakly } \text{ in } L^2(0,T;H^1(\Omega )), \\ u_n^{s/2} \nabla z_n \longrightarrow \overline{u}^{s/2} \nabla \overline{z} \text{ weakly } \text{ in } L^2(Q), \\ \dfrac{\nabla z_n}{\sqrt{z_n}} \longrightarrow \dfrac{\nabla \overline{z}}{\sqrt{\overline{z}}} \text{ weakly } \text{ in } L^4(Q), \\ D^2 z_n \longrightarrow D^2 \overline{z} \text{ weakly } \text{ in } L^2(Q), \\ \nabla z_n \longrightarrow \nabla \overline{z} \text{ strongly } \text{ in } L^2(Q), \end{array} \end{aligned}$$

(4.9)

$$\begin{aligned} E(u_n,z_n)(t) \rightarrow E(\overline{u},\overline{z})(t) \ a.e. \ t \in (0,T), \end{aligned}$$

(4.10)

and the weak continuity regularity

$$\begin{aligned} \overline{v} \in C_w([0,T];H^1(\Omega )) \text{ and } \overline{u} \in C_w([0,T];L^s(\Omega )), \text{ for } s \ge 1. \end{aligned}$$

(4.11)

Therefore, using (4.9), (4.10) and (4.11) we are able to pass to the limit in the energy inequality (4.2) and conclude that $(\overline{u},\overline{v},\overline{f}) \in S_{ad}^M$.

Finally, we prove that the infimum is attained in $(\overline{u},\overline{v},\overline{f})$. Since $(\overline{u},\overline{v},\overline{f}) \in S_{ad}^M$, we have $J_{inf} \le J(\overline{u},\overline{v},\overline{f})$. On the other hand, considering again Lemma 2.2, the functional J is weakly lower semicontinuous and then

$$\begin{aligned} J(\overline{u},\overline{v},\overline{f}) \le \liminf _{n \rightarrow \infty }{J(u^n,v^n,f^n)} = J_{inf}. \end{aligned}$$

Therefore we conclude that there is at least one $(\overline{u},\overline{v},\overline{f}) \in S_{ad}^M$ such that $J(\overline{u},\overline{v},\overline{f}) = J_{inf}$, as we wanted to prove.

4.2 Proof of Theorem 1.7

Since the functional J in (1.8) is nonnegative,

$$\begin{aligned} J_{inf}^E: = \displaystyle {\inf _{(u,v,f) \in S_{ad}^E}} J(u,v,f) \ge 0 \end{aligned}$$

is well defined and there is a sequence $\{ (u_n,v_n,f_n) \} \subset S_{ad}^E$ such that

$$\begin{aligned} \lim _{n \rightarrow \infty }{J(u_n,v_n,f_n)} = J_{inf}^E. \end{aligned}$$

(4.12)

Since $(u_n,v_n,f_n) \in S_{ad}^E$, in particular it satisfies the system (4.1) and the energy inequality

$$\begin{aligned} \begin{array}{l} E(u_n,z_n)(t_2) + \beta \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{| {\nabla [u_n + 1]^{s/2}} | _{}^2 dx} \ dt + \dfrac{1}{4} \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{u_n^s | {\nabla z_n} | _{}^2 dx} \ dt \\ \quad + \beta \Bigg ( \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{| {D^2 z_n} | _{}^2 \ dx} \ dt + \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{\frac{| {\nabla z_n} | _{}^4}{z^2} \ dx} \ dt \Bigg ) \\ \quad \le E(u_n,z_n)(t_1) + {\mathcal {K}}\big (\Vert {f_n} \Vert _{L^q(Q)},\Vert {v_0} \Vert _{W^{2-2/q,q}(\Omega )}\big ). \end{array} \end{aligned}$$

(4.13)

Following the proof of Theorem 1.5, in Subsect. 4.1, but this time using (4.13), we conclude that there is a continuous and increasing function of $\Vert {f_n} \Vert _{L^q(Q)}$, let us denote it by $C(\Vert {f_n} \Vert _{L^q(Q)}) > 0$, such that

$$\begin{aligned} \begin{array}{c} \Vert {u_n} \Vert _{L^{5s/3}(Q)}, \Vert {\nabla u_n} \Vert _{L^{5s/(3+s)}(Q)} \le C(\Vert {f_n} \Vert _{L^q(Q)}), \\ \Vert {\partial _t u_n} \Vert _{L^{5s/(3 + s)}(0,T;W^{1,5s/(4s-3)}(\Omega )')} \le C(\Vert {f_n} \Vert _{L^q(Q)}), \end{array} \end{aligned}$$

for $s \in [1,2)$,

$$\begin{aligned} \Vert {u_n} \Vert _{L^{5s/3}(Q)}, \Vert {\nabla u_n} \Vert _{L^2(Q)}, \Vert {\partial _t u_n} \Vert _{L^2(0,T;H^1(\Omega )')} \le C(\Vert {f_n} \Vert _{L^q(Q)}), \end{aligned}$$

for $s \ge 2$ and

$$\begin{aligned} \begin{array}{c} \Vert {v_n} \Vert _{L^{\infty }(Q)}, \Vert {\nabla v_n} \Vert _{L^4(Q)} \le C(\Vert {f_n} \Vert _{L^q(Q)}) \\ \Vert {\nabla v_n} \Vert _{L^\infty (0,T;L^2(\Omega ))\cap L^2(0,T;H^1(\Omega ))}, \Vert {\partial _t v_n} \Vert _{L^{5/3}(Q)} \le C(\Vert {f_n} \Vert _{L^q(Q)}) \end{array} \end{aligned}$$

for $s \ge 1$. From the definition of the functional J and (4.12) we also conclude that

$$\begin{aligned} f_n \text{ is } \text{ bounded } \text{ in } L^q(Q). \end{aligned}$$

Analogously to Subsect. 4.1, from the latter and (4.13) we prove that there is $(\overline{u},\overline{v},\overline{f})\in X_u\times X_v\times L^q(Q)$ such that, up to a subsequence, we have the convergences (4.5), (4.6), (4.7) and (4.8). These convergences allow us to conclude that $(\overline{u},\overline{v})$ is a weak solution of (1.2) with control $\overline{f}$. Because of the weakly lower semicontinuity of the norm (Lemma 2.2)

$$\begin{aligned} J(\overline{u},\overline{v},\overline{f}) \le \liminf J(u_n,v_n,f_n) = J_{inf}^E. \end{aligned}$$

(4.14)

However, we are not able to prove the $(\overline{u},\overline{v},\overline{f}) \in S_{ad}^E$ and then we can not guarantee that $J_{inf}^E = J(\overline{u},\overline{v},\overline{f})$. In fact, following the ideas of Subsect. 4.1, we are able to take the $\liminf$ in (4.13) and, using that the map $r\in {\mathbb {R}}_+\mapsto {\mathcal {K}}(r, \Vert {v_0} \Vert _{W^{2-2/q,q}(\Omega )})$ is continuous and therefore

$$\begin{aligned} \liminf _{n \rightarrow \infty }{{\mathcal {K}}\big (\Vert {f_n} \Vert _{L^q(Q)}, \Vert {v_0} \Vert _{W^{2-2/q,q}(\Omega )}\big )} \le {\mathcal {K}}\big (\liminf _{n \rightarrow \infty }{\Vert {f_n} \Vert _{L^q(Q)}}, \Vert {v_0} \Vert _{W^{2-2/q,q}(\Omega )}\big ), \end{aligned}$$

we obtain

$$\begin{aligned} \begin{array}{l} E(\overline{u},\overline{z})(t_2) + \beta \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{| {\nabla [\overline{u} + 1]^{s/2}} | _{}^2 \ dx} \ dt + \dfrac{1}{4} \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{\overline{u}^s | {\nabla \overline{z}} | _{}^2 \ dx} \ dt \\ + \beta \Bigg ( \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{| {D^2 \overline{z}} | _{}^2 \ dx} \ dt + \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{\frac{| {\nabla \overline{z}} | _{}^4}{z^2} \ dx} \ dt \Bigg ) \\ \le E(\overline{u},\overline{z})(t_1) + {\mathcal {K}}\big (\displaystyle {\liminf _{n \rightarrow \infty }} \Vert {f_n} \Vert _{L^q(Q)},\Vert {v_0} \Vert _{W^{2-2/q,q}(\Omega )}\big ). \end{array} \end{aligned}$$

(4.15)

Since by Lemma 2.2 we have $\Vert {\overline{f}} \Vert _{L^q(Q)} \le \displaystyle {\liminf _{n \rightarrow \infty }} \Vert {f_n} \Vert _{L^q(Q)}$, it is not clear that $(\overline{u},\overline{z})$ satisfies (1.6), that is, we can not guarantee that $(\overline{u},\overline{v},\overline{f}) \in S_{ad}^E$.

On the other hand, we can prove that for $M^{q} \ge \frac{q}{\gamma _f} J_{inf}^E$ we have $(\overline{u},\overline{v},\overline{f}) \in S_{ad}^M$. Indeed, because of the convergence (4.8), the weakly lower semicontinuity of the norm (see Lemma 2.2), the inequality $\Vert {f} \Vert _{L^q(Q)}^q \le \frac{q}{\gamma _f} J(u,v,f)$, and (4.12), one has

$$\begin{aligned} \Vert {\overline{f}} \Vert _{L^q(Q)}^q \le \liminf _{n \rightarrow \infty } \Vert {f_n} \Vert _{L^q(Q)}^q \le \frac{q}{\gamma _f} \liminf _{n \rightarrow \infty } J(u_n,v_n,f_n) = \frac{q}{\gamma _f} J_{inf}^E. \end{aligned}$$

(4.16)

Then, taking $M^q \ge \frac{q}{\gamma _f} J_{inf}^E$, we have $\Vert {\overline{f}} \Vert _{L^q(Q)} \le M$. Moreover, from (4.15) and (4.16) we also have that $(\overline{u},\overline{z})$ satisfies

$$\begin{aligned} \begin{array}{l} E(\overline{u},\overline{z})(t_2) + \beta \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{| {\nabla [\overline{u} + 1]^{s/2}} | _{}^2 dx} \, dt + \dfrac{1}{4} \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{\overline{u}^s | {\nabla \overline{z}} | _{}^2 dx} \, dt \\ \quad + \beta \Bigg ( \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{| {D^2 \overline{z}} | _{}^2 dx} \, dt + \displaystyle {\int _{t_1}^{t_2} \int _{\Omega }}{\frac{| {\nabla \overline{z}} | _{}^4}{z^2} dx} \, dt \Bigg ) \\ \quad \le E(\overline{u},\overline{z})(t_1) + {\mathcal {K}}\big (M,\Vert {v_0} \Vert _{W^{2-2/q,q}(\Omega )}\big ), \end{array} \end{aligned}$$

which allows us to conclude that $(\overline{u},\overline{v},\overline{f}) \in S_{ad}^M$. Therefore, using (4.14),

$$\begin{aligned} \min _{(u,v,f) \in S_{ad}^M} J(u,v,f) \le J\big (\overline{u},\overline{v},\overline{f}\big ) \le \inf _{(u,v,f) \in S_{ad}^E} J(u,v,f), \end{aligned}$$

as we wanted to prove.