Boundedness of Solutions of Nonautonomous Degenerate Logistic Equations

Abstract

In this work we analyze the boundedness properties of the solutions of a nonautonomous parabolic degenerate logistic equation in a bounded domain. The equation is degenerate in the sense that the logistic nonlinearity vanishes in a moving region, K(t), inside the domain. The boundedness character of the solutions depends not only on, roughly speaking, the first eigenvalue of the Laplace operator in K(t) but also on the way this moving set evolves inside the domain and in particular on the speed at which it moves.

1 Introduction

In this work we study the behavior of the solutions of a nonautonomous evolution problem of the type

$$\begin{aligned} \left\{ \begin{array}{lr} u_t-\Delta u = \lambda u - n(t,x) |u|^{\rho -1}u, &{} x\in \Omega , \; t>t_0,\\ u=0, &{} x\in \partial \Omega , \; t>t_0,\\ u(t_0,x)=u_0(x)\ge 0, &{} x\in \Omega , \end{array}\right. \end{aligned}$$

(1.1)

where $\Omega$ is a bounded and smooth domain in $\mathbb {R}^N$, for some positive integer N, $\rho >1, t_0,\lambda \in \mathbb {R}$, $n\in L^\infty (\mathbb {R}\times \Omega )$ with $n\ge 0$ and $0\le u_0\in L^q(\Omega )$ where $1<q\le \infty$ will be chosen appropriately large enough. We consider Dirichlet boundary conditions but the analysis could be developed for other boundary conditions.

The set where the function $n(t,\cdot )$ vanishes is denoted by K(t), that is,

$$\begin{aligned} K(t):=\left\{ x\in \overline{\Omega }: n(t,x)=0 \right\} , \; \; t\in \mathbb {R}, \end{aligned}$$

(1.2)

that we assume is compact for every $t\in \mathbb {R}$, probably empty for some times or even the entire domain for other intervals of time.

Notice that we are not assuming any special structure of the function n(t, x). We are not restricting ourselves to the periodic case, that is $n(t+T,x)=n(t,x)$ for certain $T>0$ or even to quasiperiodic or almost periodic cases. These are relevant and very interesting cases, but we are aiming to cover general nonatuonomous situation, including these ones.

From a population dynamics interpretation, see [7], Eq. (1.1) represents the evolution of the density of certain population, u(t, x), , in certain habitat given by the domain $\Omega$. The species diffuses inside the domain and the fact that the boundary condition are homogeneous Dirichlet represents that the exterior of the habitat is very hostile for the species and no individual of the species survives outside the habitat. The term $\lambda u$ represent the natural growth rate of the species, that we assume to be constant and the term $-n(t,x) |u|^{\rho -1}u$ represents a logistic term that opposes to the natural growth of the population. If for instance $n(t,x)\ge \beta >0$ throughout the domain, we have that if u is large then the term $\lambda u-n(t,x)|u|^{\rho -1}u<0$ and the population decreases so that we do not expect large values of u.

The case where $n(t,x)\equiv n(x)$, so that the equation becomes autonomous, that is

$$\begin{aligned} \left\{ \begin{array}{lr} u_t-\Delta u = \lambda u - n(x) |u|^{\rho -1}u, &{} x\in \Omega , \; t>t_0,\\ u=0, &{} x\in \partial \Omega , \; t>t_0,\\ u(t_0,x)=u_0(x)\ge 0, &{} x\in \Omega , \end{array}\right. \end{aligned}$$

(1.3)

has been addressed in the literature when $n(x)\ge \beta >0$ throughout the domain, see for instance [7, 21], and also when $n(\cdot )$ vanishes at some region $K_0$ of the domain, that is,

$$\begin{aligned} K_0=\{ x\in {\bar{\Omega }}: \, n(x)=0\}, \end{aligned}$$

see [5, 19] and references therein. Following again the population dynamics interpretation, the set $K_0$ represents a gifted location where the species is able to grow whereas $\Omega \setminus K_0$ represents a location where the species encounters hardships such as predators, harsh terrain, low resources or intraspecies competition. We refer to $K_0$ as a “sanctuary” for the species. Then, by means of the diffusion, part of the growth attained in $K_0$ is migrated outside $K_0$ where the hard conditions threaten the growth of the species. It is natural to see that the size of $K_0$ and the strength of the intrinsic growth rate $\lambda$ will play a key role determining whether the species keeps growing forever or remains bounded.

Also, in many places in the literature it is considered the case where the function n(x) is regular (at least continuous) and $K_0$ is the closure of a regular open set $\Omega _0$, that is $K_0 = {\bar{\Omega }}_0$, see for instance [19] and references therein. With these conditions, the asymptotic behavior of solutions of (1.1) has been studied. Denoting by $\lambda _1^\Omega$ the first eigenvalue of the Laplace operator with homogeneous Dirichlet boundary conditions on $\Omega$, it is known that if $\lambda < \lambda _1^\Omega$ then 0 is the unique equilibrium which is globally asymptotically stable and therefore all solutions approach 0 when $t\rightarrow \infty$. If $\lambda > \lambda _1^\Omega$ then 0 becomes unstable, and solutions tend to grow away from it. As a matter of fact, if $\lambda _1^\Omega< \lambda < \lambda _1^{\Omega _0}$ then solutions tend to the unique globally asymptotically stable positive equilibrium $u_{eq}$. On the other hand, if $\lambda > \lambda _1^{\Omega _0}$ then solutions have been shown to grow and become unbounded as $t\rightarrow \infty$ only on $K_0$ but they remain bounded outside $K_0$, which will be referred to as “grow up”. We refer to [19] for a rather complete study of this case. Hence, these results can be summarised as follows:

$$\begin{aligned} \left\{ \begin{array}{ll} \text{ If } \lambda< \lambda _1^{\Omega } &{}\Rightarrow u\rightarrow 0,\\ \text{ If } \lambda _1^\Omega< \lambda< \lambda _1^{\Omega _0} &{}\Rightarrow u\rightarrow u_{eq},\\ \text{ If } \lambda _1^{\Omega _0} < \lambda &{}\Rightarrow u(x)\rightarrow \infty ,\quad x\in \Omega _0. \end{array}\right. \end{aligned}$$

(1.4)

An extension of (1.4) to the case where $K_0$ is only assumed to be compact without any further regularity assumptions was studied in [5]. In [5], the role of $\lambda _1^{\Omega _0}$ is substituted by $\lambda _0(K_0)$, the characteristic value of the set $K_0$, which is defined as follows: consider $\{\Omega _\delta \}_{0<\delta \le \delta _0}$ a nested decreasing family of smooth open sets satisfyng $K_0=\bigcap _{0<\delta \le \delta _0} \Omega _\delta$ and define

$$\begin{aligned} \lambda _0(K_0):= \lim \limits _{\delta \rightarrow 0^+}\lambda _1^{\Omega _\delta }. \end{aligned}$$

Notice that $\lambda _0(K_0)$ can be $+\infty$ if $K_0$ is “small”, for example, when $K_0$ is a point in $\Omega$. If $K_0$ is regular, that is, $K_0 = {\bar{\Omega }}_0$ for some regular open set $\Omega _0$, then it is possible to show that $\lambda _0(K_0) = \lambda _1^{\Omega _0}$. In [5] authors show that if $\lambda <\lambda _0(K_0)$ then all solutions are bounded and if $\lambda \ge \lambda _0(K_0)$ solutions grow up without bound as time goes to $+\infty$.

We summarize this results in Fig. 1.

Fig. 1

A representation of the cases shown in (1.4)

Therefore, for the autonomous case, the key factor that decides whether the solutions are bounded or unbounded is the relation between $\lambda$ and $\lambda _0(K_0)$: if $\lambda < \lambda _0(K_0)$ solutions remain bounded whereas if $\lambda \ge \lambda _0(K_0)$ solutions become unbounded as $t\rightarrow \infty$. Moreover, in general terms observe that the smaller $K_0$ is, the larger $\lambda _0(K_0)$ is. Therefore, smaller $K_0$ imply that it is more likely to have $\lambda < \lambda _0(K_0)$ and thus, boundedness. And viceversa, the larger the set $K_0$, the smaller $\lambda _0(K_0)$ is and the more likely that solutions will grow.

In the nonautonomous case, that is problem (1.1), the situation is not so simple and, as we will see in this work, the relation between $\lambda$ and $\lambda _0(K(t))$ will not be enough to decide the boundedness or unboundedness of the solutions. Note that $\lambda _0(K(t))$ is time dependent and we may not have $\lambda <\lambda _0(K(t))$ for all t or $\lambda >\lambda _0(K(t))$ for all t. Moreover, as we will see below, even in the case $\lambda >\lambda _0(K(t))$ for all t we may not have grow–up and if $\lambda < \lambda _0(K(t))$ may not have boundedness. Therefore, the characterization of boundedness or unboundedness of solutions for the nonautonomous case is not so straightforward as in the case of autonomous problems. As a matter of fact, many interesting questions appear in the nonautonomous case. Does the geometry of the sets K(t) determine the boundedness character of the solutions? Is the velocity at which the sets K(t) evolve important to decide the boundedness? We will see that both the geometry and velocity of the sets K(t) are important factors to take into consideration to decide the boundeness character of the solutions.

To conclude this introduction, we would like to mention that the results of the present article are a necessary first step to analyze the asymptotic behavior of the solutions of these nonautonomous degenerate logistic equations. In the cases where boundedness is obtained we should question ourselves about the existence of attractors, their fine structure and behavior. In the case of unboundedness, we should try to understand how and where the solution becomes unbounded as $t\rightarrow +\infty$. Most likely, the set where the solution becomes unbounded (the “grow-up set”) will be a moving set. These issues will be treated in future publications. With respect to boundedness, let us mention that there are several interesting works in the literature addressing the general problem of the asymptotic behavior of solutions for nonautonomous problems, analyzing the existence of different kind of asymptotic sets which respond to different concepts of attraction like pullback, forwards, uniform attractors and which describe the behavior of the systems for large times. Without being exhaustive we would like to mention the monograph [8] and references therein for a nice and rather complete reference in this respect. We also mention the works [24, 25] where the authors construct complete trajectores for both autonomous and nonoautonomous parabolic evolution equations, which bound the asymptotic dynamics of the evolution processes. The works [17, 18, 23] are also relevant in this respect.

Moreover, there are some other interesting works that deal with the periodic case, that is $n(t+T,x)=n(t,x)$ for all $t\in \mathbb {R}$, $x\in \Omega$ and obtain results on existence of periodic solutions under different hypotheses. In particular, if K(t) is static, that is $K(t)\equiv K_0$, existence results of periodic solutions (and therefore, boundedness) were already proven in [1]. Very recently in [2], authors generalise results in [1] and characterize the existence of periodic solutions without imposing K(t) being static. In particular, authors prove a very interesting result where they give an if and only if characterization for the existence of periodic solutions in terms of the behavior of certain nontrivial periodic parabolic eigenvalue problems. This interesting technique is specific for periodic situations and do not apply directly to our most general nonautonomous scenario. See also [10, 11] for some related results on periodic parabolic problems.

With respect to unboundedness, we should try to understand how and where the solution becomes unbounded as $t\rightarrow +\infty$. Most likely, the set where the solution becomes unbounded (the “grow-up set”) will be a moving set but to identify it and study the asympotic properties of these systems requires further analysis. These issues will be treated in future publications.

We describe now the contents of the paper. In Sect. 2 we include some preliminary results like local and global existence of mild solutions and regularity of solutions. We also recall comparison results with respect to initial data, nonlinearity and domain.

In Sect. 3 we include an important result that says that the boundedness or unboundedness character of the solution evaluated at a particular point $x\in \Omega$ does not depend on initial time or the initial data $u_0$ (as long as $u_0\ge 0$ and $u_0\not \equiv 0$). Therefore if for one particular nontrivial initial data $u_0$ and initial time $\tau _0$ its solution is asymptotically bounded at x, that is $\limsup _{t\rightarrow +\infty } u(t,x)<+\infty$, then for any other nontrivial initial data and initial time we also have that the limsup is finite. This result is based on comparison results, regularity and the parabolic Hopf Lemma.

In Sect. 4 we prove a first result that applies to many situations since we do not impose many restrictions on the configuration of the family of sets K(t). Roughly speaking, the idea is the following: if we assume a nondegeneracy condition of the function n(t, x) (see Assmption (N)) and consider the upper and lower limit of the family of sets K(t), that is $\overline{\bigcup \limits _{t\ge \tau _0}K(t)}$, $\bigcap \limits _{t\ge \tau _0}K(t)$, then if $\lambda <\lambda _0(\overline{\bigcup \limits _{t\ge \tau _0}K(t)})$ we have boundedness of solutions, while if $\lambda >\lambda _0(\bigcap \limits _{t\ge \tau _0}K(t))$ we have unboundeness of solutions, see Proposition 4.3 and Corollary 4.4. We also include in this section some relevant examples that allow us to clarify the meaning of these sets.

In Sect. 5 we address situations where the rather general condition of the previous section does not apply and in particular we have $\lambda _0(\overline{\bigcup \limits _{t\ge \tau _0}K(t)})<\lambda <\lambda _0(\bigcap \limits _{t\ge \tau _0}K(t))$.

A first important case covered in this section says, roughly speaking, that if $n(t,x)\ge \nu _0>0$ (and therefore $K(t)=\emptyset$) for a sequence of time intervals $I_1$, $I_2$, etc. with the condition that the length of each of them is bounded below by a quantity $\eta >0$ and the distance between two consecutive intervals is no larger than a quantity $\Xi$, then, independently of the value of $\lambda$, all solutions are bounded, see Proposition 5.1. That is, the effective presence of the logistic term in the entire domain and at the intervals of time $I_1$, $I_2$, etc. which do not degenerate and they are not too far away from the next, guarantees the boundedness of solutions.

Next, we show that, roughly speaking and oversimplifying the conditions, if $\lambda <\lambda _0(K(t))$ for all t large enough then we have boundedness of solutions. The precise statement is contained in Theorem 5.7. A very important example where this theorem applies is contained in Remark 5.9 where a fix set $K_0$ is considered and we assume that K(t) consists in moving (translation by a curve $\gamma (t)$ and a rotation) the set $K_0$ inside $\Omega$, see Fig. 2.

Fig. 2

Domain $\Omega$ and a continuously moving K(t)

In this section we also consider the case where the set K(t) allows jumps. A good example of this situation is when we consider the existence of a sequence $t_0<t_1<t_2<\ldots$, satisfying certain conditions, such that $K(t)=K_0$ for $t\in (t_{2k},t_{2k+1})$ and $K(t)=K_1$ for $t\in (t_{2k+1},t_{2k+2})$. Hence, the set K(t) jumps from $K_0$ to $K_1$ and back to $K_0$ at certain instants of time. We show that if $K_0\cap K_1=\emptyset$ then independently of the value of $\lambda$ all solutions are bounded, see Theorem 5.11.

Finally, in Sect. 6 we analyze conditions under which solutions become unbounded. In view of the results of Theorem 5.7 where boundedness is established for, roughly speaking, $\lambda <\lambda _0(K(t))$ for all t, it seems natural to ask whether for $\lambda >\lambda _0(K(t))$ for all t, we have unboundedness of solutions. This is actually true if the equation is autonomous, see [5] for instance. But notice that Theorem 5.11 and the example where K(t) jumps between two sets $K_0$ and $K_1$ with $K_0\cap K_1=\emptyset$ and $\lambda$ arbitrary (in particular we could consider $\lambda >\lambda _0(K_0), \lambda _0(K_1)$), means that this is not true in a general setting. As a matter of fact, we will be able to show that if, roughly speaking, K(t) moves “slowly” and $\lambda >\lambda _0(K(t))$ for all t, then we have unboundedness of solutions. The precise statement of this result is Theorem 6.1. We also analyze a situation where the condition $\lambda >\lambda _0(K(t))$ for all t can be weakened, see Proposition 6.5.

2 Preliminaries

Equation (1.1) is a particular case of a more general class of parabolic nonautonomous equations:

$$\begin{aligned} \left\{ \begin{array}{lr} u_t- \Delta u = f(t,x,u), &{} x\in \Omega , \; t>t_0,\\ u=0, &{} x\in \partial \Omega , \; t>t_0,\\ u(t_0,x)=u_0(x), &{} x\in \Omega , \end{array}\right. \end{aligned}$$

(2.1)

where $f:\mathbb {R}\times \Omega \times \mathbb {R}\rightarrow \mathbb {R}$ satisfies

$$\begin{aligned} \begin{array}{l} \left| f(t,x,u) - f(t,x,v)\right| \le C \left| u - v \right| \left( \left| u\right| ^{\rho -1} + \left| v\right| ^{\rho -1} + 1\right) , \; \text{ for } u,v\in \mathbb {R}, \,\, t\ge t_0, \end{array}\nonumber \\ \end{aligned}$$

(2.2)

for certain constant C uniformly for $t\ge t_0$ on bounded sets and $x\in \Omega$. Observe that if $n\in L^\infty (\mathbb {R}\times \Omega )$ we easily show that $f(t,x,u)=\lambda u-n(t,x)|u|^{\rho -1}u$ satisfies (2.2) with $C=C(\lambda ,\rho ,\Vert n\Vert _{L^\infty })$.

2.1 Local Existence

For problems of the type (2.1) with $u_0\in L^q(\Omega )$ and for $1<q<\infty$ large enough we have existence and uniqueness of local mild solutions, that is, solutions of the variations of constants formula associated to (2.2). As a matter of fact, following the general techniques to obtain well posedness by [16] and in particular the results from [3] we can show that if $q>\displaystyle \max \left\{ \frac{N}{2}\rho , \rho \right\}$ then problem (2.1) has a unique $u\in C([t_0,\tau _0),L^q(\Omega ))$, local mild solution for some $\tau _0>t_0$. Moreover, for any $0<\eta <1$ $u\in C\left( (t_0,\tau _0); L^\infty (\Omega )\right) \cap C\left( (t_0,\tau _0); C^{1,\eta }({\bar{\Omega }})\right)$.

We refer to [20] for more details on this.

2.2 Positivity, Comparison and Global Existence of Solutions

Comparison results for mild solutions as studied in [4] can be applied to (2.1) and in particular to (1.1) with $f(t,x,u)=\lambda u - n(t,x)|u|^{\rho -1}u$. Let us denote by $u(t,x,t_0,u_0,f,\Omega )$ the unique solution of (2.1) where we make explicit the dependence of the solution in the initial time $t_0$ and initial data $u_0$, the nonlinearity f and the domain $\Omega$. When we want to stress the dependence of the solution on just the initial data and nonlinearity, we may write $u(t,x, u_0,f)$.

With the techniques and results from [4] it can be easily shown the following

Lemma 2.1

Let $f_i(t,x,u)=\lambda u - n_i(t,x) |u|^{\rho -1} u$ with $n_i\in L^\infty (\mathbb {R}\times \Omega )$, $n_1(t,x)\ge n_2(t,x)\ge 0$ a.e. $t\in (t_0,+\infty )$, a.e. $x\in {\bar{\Omega }}$. Then, for as long as solutions exist, we have that:

$$\begin{aligned} 0\le \left| u(t,x,u_0,f_1)\right| \le u(t,x,|u_0|,f_1) \le u(t,x,|u_0|,f_2), \quad x\in {\bar{\Omega }}. \end{aligned}$$

(2.3)

Moreover, if ${\tilde{\Omega }}\subset \Omega$ then for any initial condition $u_0\ge 0$ defined in $\Omega$ if we denote by $\tilde{u}_0= {u_0}_{|{\tilde{\Omega }}}$, we have

$$\begin{aligned} 0\le u(t,x,{\tilde{u}}_0, f,{\tilde{\Omega }})\le u(t,x, u_0,f, \Omega ),\quad x\in {\tilde{\Omega }}, \end{aligned}$$

as long as both solutions exist.

From the previous Lemma 2.1 and taking $n_2(t,x)\equiv 0$, we have

$$\begin{aligned} 0\le |u(t,x)| \le U(t,x), \end{aligned}$$

(2.4)

as long as the solutions exist and a.e. $x\in \Omega$, where we have denoted by U the solution of the linear problem

$$\begin{aligned} \left\{ \begin{array}{lr} U_t - \Delta U = \lambda U, &{} x\in \Omega ,\; t\in (t_0,\infty ),\\ U=0, &{} x\in \partial \Omega , \; t\in (t_0,\infty ),\\ U(t_0)=|u_0|, &{} x\in \Omega . \end{array}\right. \end{aligned}$$

(2.5)

This solution can be expressed also as the linear semigroup $U(t,\cdot )=e^{\lambda (t-t_0)} e^{\Delta (t-t_0)}|u_0|$, which is globally defined and uniformly bounded in compact sets of $t>t_0$.

Corollary 2.2

(Global Existence) With the assumptions above, we have that $u(t,x,t_0,u_0)$ is globally defined and bounded in compact sets of $t\in (t_0,+\infty )$. In particular, for $0\le u_0\in L^q(\Omega )$ with $q>\displaystyle \max \left\{ \frac{N}{2}\rho , \rho \right\}$, u verifies

$$\begin{aligned} 0\le u(t,x,t_0,u_0) \le e^{\lambda t} e^{\Delta (t-t_0)} u_0(x), \end{aligned}$$

which implies that

$$\begin{aligned} \left\{ \begin{array}{l} \Vert u(t,\cdot ,u_0)\Vert _{L^\infty (\Omega )} \le M e^{(\lambda -\lambda _1^\Omega )(t-t_0)}\Vert u_0\Vert _{L^\infty (\Omega )},\quad t>t_0,\\ \\ \Vert u(t,\cdot ,u_0)\Vert _{L^\infty (\Omega )} \le M e^{(\lambda -\lambda _1^\Omega )(t-t_0)}((t-t_0)^{-\frac{N}{2q}}+1) \Vert u_0\Vert _{L^q(\Omega )},\quad t>t_0, \end{array}\right. \end{aligned}$$

(2.6)

where the constant M is defined so that

$$\begin{aligned} \left\{ \begin{array}{l} \Vert e^{\Delta t}u_0\Vert _{L^\infty (\Omega )} \le M e^{-t \lambda _1^\Omega }\Vert u_0\Vert _{L^\infty (\Omega )},\quad t>0, \\ \\ \Vert e^{\Delta t}u_0\Vert _{L^\infty (\Omega )} \le M e^{-t \lambda _1^\Omega }(t^{-\frac{N}{2q}}+1) \Vert u_0\Vert _{L^q(\Omega )},\quad t>0. \end{array} \right. \end{aligned}$$

(2.7)

Remark 2.3

Observe that the constant M above depends only on $\lambda _1^\Omega$ and $|\Omega |$, see [9].

3 Initial Data Independence

We prove now that the boundedness or unboundedness character of the solutions is independent of initial data and initial time. This was shown for the autonomous case in [5]. This means that if for some particular initial data $u_0$ and initial time $t_0$ the solution is bounded at certain point $x\in \Omega$, then for any other initial data and initial time, the solution will also be bounded at this point.

Let us start with the following auxiliary lemma.

Lemma 3.1

Let $u(t,x,u_0)$ be the solution of (1.1) and let $\lambda \in \mathbb {R}$, $u_0\ge 0$ and $\alpha >0$. We have:

i) If $\alpha >1$, then $\alpha u(t,x,u_0)\ge u(t,x,\alpha u_0).$

ii) If $\alpha <1$, then $\alpha u(t,x,u_0)\le u(t,x, \alpha u_0).$

Proof

The function $v(t,x)=\alpha u(t,x, u_0)$ satisfies the problem

$$\begin{aligned} \left\{ \begin{array}{lr} v_{t}-\Delta v=\lambda v - \frac{n(t,x)}{\alpha ^{\rho -1}} v^\rho ,&{} x\in \Omega , t>0,\\ v(t,x)=0, &{} x\in \partial \Omega , t>0,\\ v(0,x)=\alpha u_0,&{} x\in \Omega . \end{array} \right. \end{aligned}$$

(3.1)

By comparison results, see Lemma 2.1, since $\frac{n(t,x)}{\alpha ^{\rho -1}}\le n(t,x)$ if $\alpha >1$ and $\frac{n(t,x)}{\alpha ^{\rho -1}}\ge n(t,x)$ if $\alpha <1$, we have

(i) if $\alpha >1$, then v is a supersolution of $u(t,x,\alpha u_0)$, so $\alpha u(t,x,u_0) \ge u(t,x,\alpha u_0)$.

(ii) if $\alpha <1$, then v is a subsolution of $u(t,x,\alpha u_0)$, so $\alpha u(t,x,u_0)\le u(t,x,\alpha u_0)$.

This proves the result. $\square$

Before stating and proving the main result of this section, let us state a Hopf parabolic Lemma for our Eq. (1.1) and a technical result. We we will prove both of them below, after the main result.

Lemma 3.2

(Hopf parabolic Lemma) Let $\Omega \subset \mathbb {R}^N$ satisfy the interior sphere condition, $n\in L^\infty (\mathbb {R}\times \Omega )$ and let $u(t,x,t_0,u_0)$ be the solution of (1.1) with $u_0\in L^\infty (\Omega )$, $u_0\ge 0$ and $u_0\not \equiv 0$. Then we have that

$$\begin{aligned} \frac{\partial u}{\partial \textbf{n}} (t,x,u_0) < 0, \; \; \; \forall t>t_0, x\in \partial \Omega , \end{aligned}$$

where $\textbf{n}$ is the outward normal vector of $\partial \Omega$.

Recall that a domain $\Omega$ satisfies the interior sphere condition if for each $x\in \partial \Omega$ there exists $z\in \Omega$ such that if $a=|x-z|$ then $B(z,a)\subset \Omega$ and ${\bar{B}}(z,a) \cap {\bar{\Omega }}=\{ x\}$, see [15].

Lemma 3.3

Let $\Omega$ be as above and let $\varphi ,\phi \in C^1({\bar{\Omega }})$ with $\varphi (x),\phi (x)>0$ for $x\in \Omega$ and $\varphi (x)=\phi (x)=0$ for $x\in \partial \Omega$. Then if $\frac{\partial \varphi }{\partial n}(x)<0$ for each $x\in \partial \Omega$, there exists $\alpha <1$ such that

$$\begin{aligned} \alpha \phi (x)\le \varphi (x), \hbox { for all }x\in {\bar{\Omega }}. \end{aligned}$$

Now, we can prove,

Proposition 3.4

The solution of (1.1) is bounded or unbounded independently of initial data and initial time.

Proof

Let $u_0,v_0$ be two non–trivial, non–negative initial data, let $\tau _0\ge t_0$ and let us denote by $u(t,x;t_0,u_0)$ be the solution of (1.1) starting at $t_0$ with initial condition $u_0$.

First, we want to show that $u(t,\cdot ; \tau _0,u_0)$ is bounded for $t>\tau _0$ iff $u(t,\cdot ;\tau _0,v_0)$ is bounded for $t>\tau _0$. We let the process evolve a small amount of time $\delta >0$ and by the regularity stated in Sect. 2.1, $u(\tau _0+\delta ,\cdot ;\tau _0, u_0),u(\tau _0+\delta ,\cdot ;\tau _0, v_0)\in C^1({\bar{\Omega }})$. Moreover, these solutions vanish in $\partial \Omega$, both are strictly positive inside $\Omega$ and by the parabolic Hopf Lemma (see Lemma 3.2) they also satisfy

$$\begin{aligned} \frac{\partial u(\tau _0+\delta ,\cdot ;\tau _0, u_0)}{\partial \textbf{n}}<0,\qquad \frac{\partial u(\tau _0+\delta ,\cdot ;\tau _0, v_0)}{\partial \textbf{n}}<0. \end{aligned}$$

Applying now Lemma 3.3 to these two functions, $x\rightarrow u(\tau _0+\delta ,x;\tau _0, u_0)$ and $x\rightarrow u(\tau _0+\delta ,x;\tau _0, v_0)$ both playing the role of $\varphi$ and $\phi$, we have the existence of $\alpha ,{\tilde{\alpha }}<1$ such that

$$\begin{aligned} \alpha u(\tau _0+\delta ,\cdot ;\tau _0,u_0)\le u(\tau _0+\delta ,\cdot ;\tau _0,v_0), \quad {\tilde{\alpha }} u(\tau _0+\delta ,\cdot ;\tau _0,v_0)\le u(\tau _0+\delta ,\cdot ;\tau _0,u_0). \end{aligned}$$

Therefore, denoting $\beta =1/{\tilde{\alpha }}$, we get that

$$\begin{aligned} \alpha u(\tau _0+\delta ,\cdot ;\tau _0,u_0)\le u(\tau _0+\delta ,\cdot ;\tau _0,v_0) \le \beta u(\tau _0+\delta ,\cdot ;\tau _0,u_0). \end{aligned}$$

Hence, taking into account that $u(t,\cdot ;\tau _0,v_0)=u(t,\cdot ;\tau _0+\delta ,u(\tau _0+\delta ,\cdot ;\tau _0,v_0))$, for $t>\tau _0+\delta$ we have

$$\begin{aligned} u(t,\cdot ;\tau _0+\delta ,\alpha u(\tau _0+\delta ,\cdot ;\tau _0,u_0)) \le u(t,\cdot ;\tau _0,v_0) \le u(t,\cdot ;\tau _0+\delta ,\beta u(\tau _0+\delta ,\cdot ;\tau _0,u_0)). \end{aligned}$$

But from Lemma 3.1 for $\alpha<1<\beta$ and $t> t_0 + \delta$

$$\begin{aligned}\begin{array}{c} \alpha u(t,\cdot ;\tau _0,u_0) = \alpha u(t,\cdot ;\tau _0+\delta ,u(\tau _0+\delta ,\cdot ;\tau _0,u_0)) \le u(t,\cdot ;\tau _0+\delta ,\alpha u(\tau _0+\delta ,\cdot ;\tau _0,u_0)), \\ u(t,\cdot ;\tau _0+\delta ,\beta u(\tau _0+\delta ,\cdot ;\tau _0,u_0)) \le \beta u(t,\cdot ;\tau _0+\delta ,u(\tau _0+\delta ,\cdot ;\tau _0,u_0))= \beta u(t,\cdot ;\tau _0,u_0). \end{array}\end{aligned}$$

Hence, we arrive at

$$\begin{aligned} \alpha u(t,\cdot ;\tau _0,u_0)\le u(t,\cdot ;\tau _0,v_0) \le \beta u(t,\cdot ;\tau _0,u_0), \quad t>\tau _0+\delta . \end{aligned}$$

Thus, one solution is bounded if and only if the other is bounded.

Now let us take two different initial times $\tau _1,\tau _2\ge t_0$ and two initial data $u_0, v_0$ as before. Without any loss of generality, we may assume $\tau _1<\tau _2$. Let the process evolve from $\tau _1$ to $\tau _2$ with initial data $u_0$ and denote by ${\tilde{u}}_0=u(\tau _2,\cdot ;\tau _1,u_0)$. From the proof above, we know that

$$\begin{aligned} u(t,\cdot ;\tau _2,{\tilde{u}}_0)=u(t,\cdot ;\tau _1,u_0) \quad \text{ is } \text{ bounded } \text{ for } t\ge \tau _2 \quad \Leftrightarrow \quad u(t,\cdot ;\tau _2,v_0) \text{ is } \text{ bounded }. \end{aligned}$$

This shows the result. $\square$

For the sake of completeness we include a short proof of Lemma 3.2 and 3.3.

Proof

(Hopf parabolic Lemma) Denoting by U the solution of the linear problem (2.5), by Lemma 2.1 we have that for $t\ge t_0$, $U(t) \in L^\infty (\Omega )$ and $\Vert U(t)\Vert _\infty \le e^{\lambda (t-t_0)} \Vert u_0\Vert _\infty$. Moreover,

$$\begin{aligned} 0\le u(t,x; u_0) \le U(t,x;u_0), \; \text{ for } t>t_0, x\in {\bar{\Omega }}. \end{aligned}$$

Then,

$$\begin{aligned} \lambda u - n(t,x) |u|^{\rho -1} u \ge \lambda u - \Vert n\Vert _\infty \left( e^{\lambda (t-t_0)} \Vert u_0\Vert _\infty \right) ^{\rho -1} u \ge \lambda u - N_0 u, \quad t\in [t_0,t_1], \end{aligned}$$

where

$$\begin{aligned} N_0=\displaystyle \sup _{t\in [t_0,t_1]}\Vert n\Vert _\infty \left( \Vert u_0\Vert _\infty e^{\lambda (t-t_0)}\right) ^{\rho -1} =\Vert n\Vert _\infty \left( \Vert u_0\Vert _\infty e^{|\lambda | (t_1-t_0)} \right) ^{\rho -1}. \end{aligned}$$

This implies that the solution v of

$$\begin{aligned} \left\{ \begin{array}{lr} v_t-\Delta v-(\lambda -N_0)v=0 &{} t\in (t_0,t_1),\; x\in \Omega ,\\ v=0, &{} t\in (t_0,t_1),\; x\in \partial \Omega ,\\ v(t_0)=u_0 &{} x\in \Omega , \end{array}\right. \end{aligned}$$

(3.2)

satisfies, using Lemma 2.1, $u(t,x)\ge v(t,x)$. Moreover, multiplying $e^{-(\lambda -N_0)(t-t_0)}$ to both sides of equation for v, we arrive at $v(t,x)=e^{(\lambda -N_0)(t-t_0)}w(t,x)$ where

$$\begin{aligned} \left\{ \begin{array}{lr} w_t-\Delta w=0 &{} (t,x)\in (t_0,t_1)\times \Omega ,\\ w=0, &{} x\in \partial \Omega ,\\ w(t_0)=u_0 &{} x\in \Omega . \end{array}\right. \end{aligned}$$

(3.3)

Therefore,

$$\begin{aligned} u(t,x)\ge v(t,x) = e^{(\lambda - N_0)t} w(t,x), \; \text{ for } \text{ all } (t,x) \in (t_0,t_1)\times \Omega . \end{aligned}$$

(3.4)

Since $u(t,x)=v(t,x)=w(t,x)=0$ at the boundary and by the regularity of $u(t,\cdot )$ we have

$$\begin{aligned} \frac{\partial u}{\partial \textbf{n}}(t,x)\le \frac{\partial v}{\partial \textbf{n}}(t,x)=e^{(\lambda - N_0)t}\frac{\partial w}{\partial \textbf{n}}(t,x), \; \text{ for } \text{ all } (t,x) \in (t_0,t_1)\times \partial \Omega .\end{aligned}$$

(3.5)

Applying the Hopf parabolic Lemma to w which is a classical solution, see [22, Chapter 3 Theorem 7], [15, section 6.4.2], we have $\frac{\partial w}{\partial \textbf{n}} (t,x,u_0) < 0$ for all $t>t_0$, $x\in \partial \Omega$. This statement together with (3.5) imply the result. $\square$

Proof

(Lemma 3.3) Let us denote by $A>0$ a constant such that $|\nabla \varphi (x)|,|\nabla \phi (x)|, \varphi (x),\phi (x)\le A$ for all $x\in {\bar{\Omega }}$. Let us argue by contradiction. If the result is not true, then there exists $\alpha _n\rightarrow 0$ and $x_n\in {\bar{\Omega }}$ such that $\alpha _n\phi (x_n)>\varphi (x_n)$. This implies that $x_n\in \Omega$, $\varphi (x_n)\le A\alpha _n\rightarrow 0$ and therefore $d(x_n,\partial \Omega )\rightarrow 0$. Taking a subsequence if necessary we obtain the existence of $x^*\in \partial \Omega$ such that $x_n\rightarrow x^*$. Since $\frac{\partial \varphi }{\partial n}(x)<0$, by continuity of $x\rightarrow \nabla \varphi (x)$ in ${\bar{\Omega }}$ and $x\rightarrow \textbf{n}(x)$ in $\partial \Omega$, we get that there exists $\delta , a>0$ such that

$$\begin{aligned} \nabla \varphi (x)\cdot \textbf{n}(y)\le -a, \quad \forall x\in B(x*,\delta )\cap {\bar{\Omega }},\quad \forall y\in B(x*,\delta )\cap \partial \Omega . \end{aligned}$$

(3.6)

Since $x_n\rightarrow x^*$ we have that there exists $n_0\in \mathbb {N}$ such that $|x_n-x^*|<\delta /2$ for $n\ge n_0$. If we take now $\rho _n$ such that $B(x_n,\rho _n)\subset \Omega$ and $\bar{B}(x_n,\rho _n)\cap \partial \Omega \ni z_n^*$ then we have $\rho _n<\delta /2$ and $|z_n^*-x^*|\le |z_n^*-x_n|+|x_n-x^*|<\delta /2+\delta /2=\delta$.

Using (3.6) and the mean value theorem, we have:

$$\begin{aligned} \frac{\varphi (x_n)-\varphi (z_n^*)}{|x_n-z_n^*|}=\nabla \varphi (\xi _n)\cdot \frac{x_n-z_n^*}{|x_n-z_n^*|}=-\nabla \varphi (\xi _n)\textbf{n}(z_n^*)\ge a>0,\quad \forall n\ge n_0, \end{aligned}$$

where $\xi _n$ lies in the segment joining $x_n$ and $z_n^*$. Observe that we have used that the unit exterior normal vector at $z_n^*$ is given $-\frac{x_n-z_n^*}{|x_n-z_n^*|}$.

On the other hand,

$$\begin{aligned} \frac{\varphi (x_n)-\varphi (z_n^*)}{|x_n-z_n^*|}=\frac{\varphi (x_n)}{|x_n-z_n^*|}\le \alpha _n\frac{\phi (x_n)}{|x_n-z_n^*|}=\alpha _n\frac{\phi (x_n)-\phi (z_n^*)}{|x_n-z_n^*|}\le \alpha _nA\rightarrow 0, \end{aligned}$$

which is in contradiction with the inequality just obtained above. $\square$

4 A General Result

We start giving in this section a rather general result, which can be applied in many situations with not many geometric restrictions on the sets K(t), where boundedness or unboundedness is proved. These results are obtained via comparison techniques with certain autonomous reference problems involving, roughly speaking, the upper and lower limit of the sets K(t).

Moreover, we will often assume the following condition:

Assumption

(N) There exists a strictly increasing continuous function $\nu :\mathbb {R}^+\rightarrow \mathbb {R}^+$ with $\nu (0)=0$ and

$$\begin{aligned} n(t,x)\ge \nu (d(x,K(t))),\; \text{ for } \text{ all } x\in \overline{\Omega }, t\in \mathbb {R}. \end{aligned}$$

Notice that K(t) may be empty for some values of t. For the values of t when K(t) is empty, we will assume that there exists $\nu _0>0$ such that $n(t,x)\ge \nu _0$ for all $x\in \overline{\Omega }$.

Let us define the following functions and sets which will be important in the analysis below. For a fixed $\tau _0\ge t_0$,

$$\begin{aligned} \begin{array}{l} {\underline{n}}_{\tau _0}(x):=\inf \limits _{t\ge \tau _0} n(t,x), \; \text{ for } x\in \overline{\Omega }, \qquad K^{\sup }_{\tau _0}:=\overline{\{ x\in \overline{\Omega }: {\underline{n}}_{\tau _0}(x)=0\}}, \\ {\overline{n}}_{\tau _0}(x):=\sup \limits _{t\ge \tau _0} n(t,x), \; \text{ for } x\in \overline{\Omega }, \qquad K^{\inf }_{\tau _0}:=\{ x\in \overline{\Omega }: {\overline{n}}_{\tau _0}(x)=0\}. \end{array} \end{aligned}$$

(4.1)

We can show,

Lemma 4.1

With the notations above, $K^{\sup }_{\tau _0}$ is decreasing in $\tau _0$, $K^{\inf }_{\tau _0}$ is increasing in $\tau _0$ and

$$\begin{aligned} \overline{\bigcup \limits _{t\ge \tau _0}K(t)}\subset K^{\sup }_{\tau _0}, \qquad \bigcap \limits _{t\ge \tau _0}K(t)=K^{\inf }_{\tau _0}. \end{aligned}$$

Moreover if (N) holds, then $\overline{\bigcup \limits _{t\ge \tau _0}K(t)}= K^{\sup }_{\tau _0}$.

Proof

That $K^{\sup }_{\tau _0}$ is decreasing in $\tau _0$, $K^{\inf }_{\tau _0}$ is increasing in $\tau _0$ follows directly from the definition in (4.1).

Let us see all inclusions. Let $x\in \bigcup \limits _{t\ge \tau _0} K(t)$, then $x\in K(t)$ for some $t\ge \tau _0$, which implies $n(t,x)=0$ and therefore ${\underline{n}}_{\tau _0}(x)=0$. Hence,

$$\begin{aligned} \bigcup \limits _{t\ge \tau _0} K(t)\subset \{ x\in \Omega : {\underline{n}}_{\tau _0}(x)=0\} \Rightarrow \overline{\bigcup \limits _{t\ge \tau _0} K(t)}\subset \overline{\{ x\in \Omega : {\underline{n}}_{\tau _0}(x)=0\}}=K^{\sup }_{\tau _0}. \end{aligned}$$

Now, let us take $x\in K^{\inf }_{\tau _0}$, then $\sup \limits _{t\ge \tau _0} n(t,x)=0$. Therefore

$$\begin{aligned} 0\le n(t,x) \le \sup \limits _{t\ge \tau _0} n(t,x)=0, \;\; \forall t\ge \tau _0\;\; \Rightarrow n(t,x)=0,\;\; \forall t\ge \tau _0. \end{aligned}$$

Hence, $x\in K(t)$ for every $t\ge \tau _0$ and thus, $x\in \bigcap \limits _{t\ge \tau _0} K(t)$ and therefore, $K^{\inf }_{\tau _0}\subset \bigcap \limits _{t\ge \tau _0}K(t)$.

Now, let $x\in \bigcap \limits _{t\ge \tau _0} K(t)$, then $x\in K(t)$ for every $t\ge \tau _0$. Therefore, ${\overline{n}}_{\tau _0}(x)=0$ which implies that $x\in K^{\inf }_{\tau _0}$ and, thus, $\bigcap \limits _{t\ge \tau _0}K(t)\subset K^{\inf }_{\tau _0}$.

Finally, let us assume (N) holds and $x\in \{ x\in \Omega : {\underline{n}}_{\tau _0}(x)=0\}$ for some $\tau _0\ge t_0$. Then, ${\underline{n}}_{\tau _0}(x)=\inf \limits _{t\ge \tau _0} n(t,x)=0$. Therefore, either

• $\exists t\ge \tau _0$ with $n(t,x)=0$. This implies $x\in K(t)$ and therefore $x\in \bigcup \limits _{t\ge \tau _0} K(t)$ or,

• $n(t,x)>0$ for every $t\ge \tau _0$ and there exists a sequence $(t_i)_{i\in \mathbb {N}}$ such that $\lim \limits _{i\rightarrow \infty } n(t_i,x)=0$, which implies that

  $$\begin{aligned} 0\le \lim \limits _{i\rightarrow \infty } \nu (d(x,K(t_i))) \le \lim \limits _{i\rightarrow \infty } n(t_i,x)=0. \end{aligned}$$

  Hence, $\lim \limits _{i\rightarrow \infty } d(x,K(t_i)) = 0$ and, therefore, $x\in \overline{\bigcup \limits _{t\ge \tau _0} K(t)}$.

Thus, $K^{\sup }_{\tau _0}\subset \overline{\bigcup \limits _{t\ge \tau _0} K(t)}$. This concludes the proof of the result. $\square$

Remark 4.2

Assumption (N) is avoiding that the function n(t, x) degenerates outside K(t), that is, if $x(t)\in \Omega$ is at a positive distance from K(t), that is, $d(x(t),K(t))\ge d_0>0$, then $n(t,x(t))\ge \nu (d_0)>0$ for all t. This is important for Lemma 4.1 and also for many results in this work. Notice that we can construct the function

$$\begin{aligned} n(t,x)=\left\{ \begin{array}{lr} 0,&{} t\ge t_0,x\in K_0, \\ \frac{1}{t-t_0+1}, &{} t\ge t_0, \; x\in \overline{\Omega }\setminus K_0, \end{array}\right. \end{aligned}$$

where $K(t)\equiv K_0$ compact but ${\underline{n}}_{\tau _0}(x)=0$ for all $\tau _0\ge t_0$ and all $x\in \overline{\Omega }$. Therefore $K^{\sup }_{\tau _0}={\bar{\Omega }}$ but $\overline{\bigcup \limits _{t\ge \tau _0} K(t)}=K_0 \subsetneq {\bar{\Omega }}$, and therefore the last inequality in Lemma 4.1 does not hold true.

We can show now

Proposition 4.3

Let us consider problem (1.1) and let us assume (N) holds. We have,

• If $\lambda <\lambda _0(K^{\sup }_{\tau _0})$ for some $\tau _0\ge t_0$, then all solutions of (1.1) are bounded.

• If $\lambda \ge \lambda _0(K^{\inf }_{\tau _0})$ for some $\tau _0\ge t_0$, then all solutions of (1.1) are unbounded.

Proof

Let ${\tau _0}\ge t_0$ and define

$$\begin{aligned}{\underline{n}}_{\tau _0}^*:=\left\{ \begin{array}{lr} 0, &{} x\in K^{\sup }_{\tau _0},\\ {\underline{n}}_{\tau _0}(x), &{}x\in \overline{\Omega }\setminus K^{\sup }_{\tau _0}. \end{array}\right. \end{aligned}$$

Let ${\underline{v}}$ and ${\overline{v}}$ be the respective solutions of the autonomous problems

$$\begin{aligned}{} & {} \left\{ \begin{array}{ll} {\underline{v}}_t-\Delta {\underline{v}} = \lambda {\underline{v}} - {\underline{n}}_{\tau _0}^* |{\underline{v}}|^{\rho -1}{\underline{v}}, &{} x\in \Omega , \; t>\tau _0,\\ {\underline{v}}=0, &{} x\in \partial \Omega , \; t>\tau _0,\\ {\underline{v}}(\tau _0,x)=u_0(x)\ge 0, &{} x\in \Omega , \end{array}\right. \\{} & {} \left\{ \begin{array}{ll} {\overline{v}}_t-\Delta {\overline{v}} = \lambda {\overline{v}} - {\underline{n}}_{\tau _0} |{\overline{v}}|^{\rho -1}{\overline{v}}, &{} x\in \Omega , \; t>\tau _0,\\ {\overline{v}}=0, &{} x\in \partial \Omega , \; t>\tau _0,\\ {\overline{v}}(\tau _0,x)=u_0(x)\ge 0, &{} x\in \Omega . \end{array}\right. \end{aligned}$$

Then, since

$$\begin{aligned} {\overline{n}}_{\tau _0}(x)\ge n(t,x)\ge {\underline{n}}_{\tau _0}(x)\ge {\underline{n}}_{\tau _0}^*(x), \; t\ge \tau _0, \; x\in \overline{\Omega }, \end{aligned}$$

by comparison, see Corollary 2.1, we have that

$$\begin{aligned} {\overline{v}}(t,x) \le u(t,x) \le {\underline{v}}(t,x), \; t\ge \tau _0, \text{ a.e. } x\in \Omega . \end{aligned}$$

Now,

• If $\lambda <\lambda _0(K^{\sup }_{\tau _0})$ then ${\underline{v}}$ is bounded due to [5]. Thus u is a bounded solution.

• If $\lambda \ge \lambda _0(K^{\inf }_{\tau _0})$, then ${\overline{v}}$ is unbounded due to [5]. Thus u is an unbounded solution.

$\square$

If we define $\lambda _0^{-}=\lim _{\tau _0\rightarrow +\infty }\lambda _0(K^{\sup }_{\tau _0})\in \mathbb {R}^+\cup \{+\infty \}$ and $\lambda _0^{+}=\lim _{\tau _0\rightarrow +\infty }\lambda _0(K^{\inf }_{\tau _0})\in \mathbb {R}^+\cup \{+\infty \}$, which both limits exist since the sequences $\lambda _0(K^{\sup }_{\tau _0})$ and $\lambda _0(K^{\inf }_{\tau _0})$ are monotone decreasing and increasing respectively, we can obtain

Corollary 4.4

In the assumptions of Proposition 4.3, we have,

• If $\lambda <\lambda _0^{-}$ all solutions of (1.1) are bounded.

• If $\lambda >\lambda _0^{+}$ all solutions of (1.1) are unbounded.

Remark 4.5

It is natural to define $K^{\sup }=\cap _{\tau _0\ge t_0}K^{\sup }_{\tau _0}$ and $K^{\inf }=\overline{\cup _{\tau _0>t_0}K^{\inf }_{\tau _0}}$ and these two sets are like the upper and lower limits of the one parameter family of sets K(t). We would expect that $\lambda _0^-=\lambda _0(K^{\sup })$ and $\lambda _0^+=\lambda _0(K^{\inf })$ but this is not true in general although it is true in many situations, as we will see in the examples below.

The results above express the importance of the family of sets $K^{\sup }_{\tau _0}, K^{\inf }_{\tau _0}$ and the numbers $\lambda _0^{-}$ and $\lambda _0^{+}$ in order to obtain boundedness or unboundedness of solutions in a rather general situation. Observe that the proposition and the corollary do not say anything if $\lambda \in (\lambda _0^{-},\lambda _0^{+})$.

Fig. 3

Impact of the time-dependency of K(t) on the variety of scenarios

Let us see some examples of sets $K^{\sup }_{\tau _0}$ and $K^{\inf }_{\tau _0}$.

Example 4.6

(A Ball of Changing Radius) Let $x_0\in \Omega$ and $r_0>0$ such that $\overline{B(x_0,r_0)}\subset \Omega$ and let

$$\begin{aligned} n(t,x)=1-\chi _{K(t)}(x), \quad \text{ for } t > 0,\; x\in \overline{\Omega }, \end{aligned}$$

where

$$\begin{aligned} K(t)=\overline{B\left( x_0,r(t)\right) }, \; t>0, \end{aligned}$$

where r(t) is specified according to different cases.

Case 1.:

Consider $r(t)=r_0(1-\frac{1}{t+1})$. Then, $K^{\inf }_{\tau _0}=\overline{B(x_0,r(\tau _0))}$, $K^{\sup }_{\tau _0}= \overline{B(x_0,r_0)}$. Moreover, $\lambda _0(K^{\inf }_{\tau _0})=\lambda _0(\overline{B(x_0,r_0)})(1-\frac{1}{\tau _0+1})^{-2}\rightarrow \lambda _0(\overline{B(x_0,r_0)})$ and hence, $\lambda _0^{+}=\lambda _0(\overline{B(x_0,r_0)})$. Also, since $K^{\sup }_{\tau _0}$ is independent of $\tau _0$, $\lambda _0^{-}=\lambda _0(K_{\tau _0}^{\sup })$. Therefore, if $\lambda < \lambda _0(\overline{B(x_0,r)})$ we have boundedness of solutions and if $\lambda > \lambda _0 (\overline{B(x_0,r)})$ we have unboundedness of solutions.

Case 2.:

If $r(t)=\frac{r_0}{t+1}$, that K(t) is a continuously shrinking ball, then, $K^{\sup }_{\tau _0}=B(x_0,\frac{r_0}{\tau _0+1})$ and $\lambda _0^{-}=\lim _{\tau _0\rightarrow +\infty } \lambda _0(B(x_0,r_0))(\tau _0+1)^2=+\infty$, $K^{\inf }_{\tau _0}=\{x_0\}$ and $\lambda _0^{+}=+\infty$. Hence solutions are bounded for each $\lambda \in \mathbb {R}$.

Case 3.:

Finally, if $r(t)=r_0(1+|\sin (\omega t)|)$ for some $0\ne \omega \in \mathbb {R}$ then, $K^{\inf }_{\tau _0}=\overline{B(x_0,r_0)}$ and $K^{\sup }_{\tau _0}=\overline{B(x_0,2r_0)}$. Hence if $\lambda < \lambda _0(\overline{B(x_0,2r_0)})$ we have boundedness of solutions and if $\lambda > \lambda _0(\overline{B(x_0,r_0)})$ we have unboundedness of solutions.

Observe that in this last case Proposition 4.3 and Corollary 4.4 do not provide any information about the behavior of solutions when $\lambda \in (\lambda _0(\overline{B(x_0,2r_0)}), \lambda _0(\overline{B(x_0,r_0)}))$.

We can conclude that the nonautonomous problem presents a broader spectrum of cases than its autonomous version. Also $\lambda$ and $\lambda _0(K(t))$ are no longer the only elements involved in determining asymptotic behavior of the solutions as, for example, $\lambda _0^{-}$ and $\lambda _0^{+}$ arise naturally in the non–autonomous setting.

Let us study other configurations of K(t).

Example 4.7

(Rotating sector) Consider an open domain $\Omega \subset \mathbb {R}^2$ with $\overline{ B(0,r_0)}\subset \Omega$ and let the following set expressed in polar coordinates

$$\begin{aligned} {\tilde{K}}=\{ (r,\theta ): 0\le r\le r_0,\, \theta _0\le \theta \le \theta _1\}. \end{aligned}$$

Define K(t) as the set ${\tilde{K}}$, rotating clockwise at certain angular speed $\omega >0$, see Fig. 4, that is

$$\begin{aligned} K(t)=\{(r,\theta ): 0\le r\le r_0, \, \theta _0-\omega t\le \theta \le \theta _1-\omega t\}. \end{aligned}$$

In this particular case, $K^{\sup }_{\tau _0}$ is the entire circle $\overline{B(0,r_0)}$ as every point x in the circle verifies that n(t, x) vanishes for some $t>\tau _0$ for any $\tau _0>0$.

Fig. 4

Domain $\Omega$ with a rotating sector

Furthermore, for any $\tau _0>0$, $K^{\inf }_{\tau _0}=\{ 0\}$, the center of the circle, as it is the only point where n vanishes for every $t>t_0$. Thus, in this example we have that $\lambda _0(K(t))=\lambda _0({\tilde{K}})$ is constant and

$$\begin{aligned} \lambda _0(\overline{B(0,r_0)})=\lambda _0^{-}<\lambda _0(K(t))<\lambda _0^{+}=\infty , \; \text{ for } \text{ any } t>0. \end{aligned}$$

Now, although from the previous proposition we can show that the solutions are bounded for $\lambda < \lambda _0(\overline{B(0,r_0)})$, we cannot decide the behavior for $\lambda >\lambda _0(\overline{B(0,r_0)})$ since we cannot apply Proposition 4.3 to arrive to a conclusion. In fact, as seen later on, whether the solution is bounded or not will also depend on the speed at which K(t) rotates.

Example 4.8

(Jumping Sets) Let us consider problem (1.1) and n(t, x) such that K(t) “jumps” from a compact set $K_0$ to another compact set $K_1$ periodically:

$$\begin{aligned} K(t)=\left\{ \begin{array}{lr} K_0, &{} t\in (n t_0, n t_0+T_1], n\in \mathbb {N},\\ K_1, &{} t\in (n t_0+T_1, (n+1)t_0], n\in \mathbb {N},\\ \end{array}\right. \end{aligned}$$

where $0<T_1<t_0$ and $K_0,K_1$ are compact disjoint sets as in Fig. 5. This is an example of special significance as, following the autonomous logic, it is reasonable to expect the existence of unbounded solutions for $\lambda >\lambda _0(K(t))$ for all $t>t_0$. However, as shown later on, for every $\lambda \in \mathbb {R}$ the solutions remain bounded. The relationship between a jumping K(t) and boundedness will be explored in the following Sect. 5.

Fig. 5

Domain $\Omega$ with “jumping” subdomains $K_0$ and $K_1$

From the previous two examples, it is clear that the relation between $\lambda$ and $\lambda _0(K(t))$ although important, does not completely assess the asymptotic behavior of solutions of (1.1): the geometry of K(t) and how K(t) moves also have to be factored in. In fact, we will present several examples of boundedness and unboundedness linked to the geometry of K(t): not only its size, but also taking into consideration how K(t) moves.

5 Boundedness

The main objective of this section is to obtain boundedness results that go beyond Proposition 4.3 and Corollary 4.4 in the previous section. In these results, we have established boundedness for $\lambda < \lambda _0^{-}$ and unboundedness for $\lambda >\lambda _0^{+}$. This section is devoted to explore under which conditions we can obtain boundedness for $\lambda \in (\lambda _0^{-},\lambda _0^{+})$.

Let us start with the particular but important situation where $K(t)=\emptyset$ for some sequence of intervals of times. We can show:

Proposition 5.1

Let us assume there exists a sequence of times $t_0<t_1<t_2\ldots <t_i\rightarrow +\infty$ and constants $\eta$, $\Xi >0$ such that

$$\begin{aligned} \left\{ \begin{array}{lr} \left| t_{2i-1}-t_{2i-2}\right| \le \Xi , &{} i=1,2,\ldots ,\\ \left| t_{2i}-t_{2i-1}\right| > \eta , &{} i=1,2,\ldots . \end{array}\right. \end{aligned}$$

(5.1)

Assume also that there exists a $\nu _0>0$ with $n(t,x)\ge \nu _0$ for $t\in (t_{2i-1},t_{2i})$ for all $i\in \mathbb {N}$. Then, for all $\lambda \in \mathbb {R}$, all solutions of (1.1) are bounded. Moreover, there exist $\tau >0$ and $\Lambda >0$ independent of $u_0$ such that

$$\begin{aligned} \left| u(t,x; t_0,u_0)\right| \le \Lambda , \quad \forall t\ge \tau , \forall x\in \Omega . \end{aligned}$$

Proof

Since the boundedness character of solutions is independent of initial conditions due to Proposition 3.4 we may take a particular nontrivial initial conditon $u_0\in L^\infty (\Omega )$. First, we estimate the growth of u in the interval $[t_{2i-2},t_{2i-1})$. In this interval, the only information we have of the function n is that $n(t,x)\ge 0$ but it may vanish in part or all of the domain. We can always bound the growth of u with the linear equation (that is, with $n\equiv 0$). With the notation of Corollary 2.2, we have

$$\begin{aligned} \Vert u(t,\cdot ; t_{2i-1}, u(t_{2i-2},\cdot ;t_0,u_0))\Vert _\infty\le & {} Me^{(\lambda -\lambda _1^\Omega )(t-t_{2i-2})}\Vert u(t_{2i-2},\cdot ; t_0,u_0)\Vert _\infty , \;\\{} & {} t\in [t_{2i-2},t_{2i-1}), i\in \mathbb {N}. \end{aligned}$$

Moreover, since $|t_{2i-1}-t_{2i-2}|\le \Xi$, we have

$$\begin{aligned} \Vert u(t,\cdot ; u(t_{2i-2},\cdot ;u_0))\Vert _\infty \le Me^{\left| \lambda -\lambda _1^\Omega \right| \Xi }\Vert u(t_{2i-2},\cdot ; u_0)\Vert _\infty , \; t\in [t_{2i-2},t_{2i-1}), i\in \mathbb {N}. \end{aligned}$$

Due to the fact that in $[t_{2i-1},t_{2i})$ the nonlinear term is applied to all the domain, we can obtain a supersolution for the time intervals $[t_{2i-1},t_{2i})$ for $i\in \mathbb {N}$ by using the following ODE problem:

$$\begin{aligned} \left\{ \begin{array}{lr} W(t)_t=\lambda W(t) - \nu _0 W(t)^\rho , &{} t\in (0,t_{2i}-t_{2i-1}),\\ W(0)=\Vert u(t_{2i-1},\cdot )\Vert _{L^\infty (\Omega )},&{} t=0, \end{array} \right. \end{aligned}$$

(5.2)

that is, $W(t-t_{2i-1})\ge u(t,x)$ for $t\in (t_{2i-1},t_{2i}]$ for every $i\in \mathbb {N}$ and a.e. $x\in \Omega$. As a matter of fact the function W(t) is a solution of

$$\begin{aligned}\left\{ \begin{array}{lr} W_t(t)-\Delta W(t) =\lambda W(t) - \nu _0 W(t)^\rho \ge \lambda W(t) - n(t,x) W(t)^\rho , &{} t\in (0,t_{2i}-t_{2i-1}),\; x\in \Omega ,\\ W(t,x)\ge 0,&{} t\in (0,t_{2i}-t_{2i-1}),\; x\in \partial \Omega ,\\ W(0)=\Vert u(t_{2i-1},\cdot )\Vert _{L^\infty (\Omega )},&{} t=0,\; x\in \Omega , \end{array}\right. \end{aligned}$$

which shows that W is a supersolution of u. By direct inspection the solution W of (5.2) is:

$$\begin{aligned} W(t)=\left[ \frac{\nu _0}{\lambda }\left( 1-e^{-\lambda (\rho -1)t} \right) + e^{-\lambda (\rho -1)t} W(0)^{1-\rho }\right] ^{-\frac{1}{\rho -1}}, \end{aligned}$$

which is bounded above by:

$$\begin{aligned} W_\infty (t)=\left[ \frac{\nu _0}{\lambda }\left( 1-e^{-\lambda (\rho -1)t} \right) \right] ^{-\frac{1}{\rho -1}}. \end{aligned}$$

(5.3)

The function $W_\infty (t - t_{2i-1})$ is specially important not only because it bounds u(t, x) for $t\in [t_{2i-1},t_{2i})$ for every $i\in \mathbb {N}$ and every $x\in \Omega$ but it also reaches, for a fixed amount of time $(t_{2i}-t_{2i-1})$, the same threshold $W_\infty (t_{2i}-t_{2i-1})$ wherever the solution u may start at time $t_{2i-1}$. Now, since $|t_{2i}-t_{2i-1}|> \eta$ for every $i\in \mathbb {N}$, we have that $W_\infty (\eta )$ bounds $u(t_{2i})$ for $i\in \mathbb {N}$ and a.e. $x\in \Omega$, see Fig. 6. In particular,

$$\begin{aligned} \left| u(t_2,x)\right| \le W_\infty (\eta ),\quad \text{ a.e. }\; x\in \Omega . \end{aligned}$$

Fig. 6

$W_\infty$ bounds u for $t\in (t_{2i-1},t_{2i}]$

Thus, if we define the function (see Fig. 7),

$$\begin{aligned} {\overline{u}}(t):=\left\{ \begin{array}{lr} M e^{\left| \lambda -\lambda _1^\Omega \right| (t-t_{2i-2})}W_\infty (\eta ), &{} t\in (t_{2i-2},t_{2i-1}+\eta /2],\; i=2,3,\ldots \\ W_\infty (\eta /2), &{} t\in (t_{2i-1}+\eta /2,t_{2i}],\; i=2,3,\ldots . \end{array}\right. \end{aligned}$$

Fig. 7

Construction of the upper bound for u

then, we have

$$\begin{aligned} \left| u(t,x;u_0) \right| \le {\bar{u}}(t), \quad t\ge {t_2},\; x\in \Omega \end{aligned}$$

and in particular

$$\begin{aligned} \left| u(t,x;u_0) \right| \le \Lambda := \max \limits _{s\in { [t_2,t_4]}} {\bar{u}}(s), \quad \forall t\ge { t_2}. \end{aligned}$$

Choosing $\tau =t_2$, we conclude the proof of the proposition. $\square$

Remark 5.2

Observe that Proposition 5.1 applies to all $\lambda \in \mathbb {R}$. In this case, $\lambda _0^+ = \infty$ since $K^{\inf }_{\tau _0}=\emptyset$.

5.1 Estimates of the Solution Locally in Space and Time

The effect of having the logistic term acting over the whole domain $\Omega$ during certain intervals of time, allows us to obtain global bounds in space with the use of the function $W_\infty (t)$ (see Proposition 5.1). In this subsection, which is inspired by the results from [14], we will obtain bounds of the solutions which are local in space and time under certain conditions on the vanishing logistic term.

Let us start with a partial result that gives us estimates of the solution locally in space but globally in time. In order to accomplish this, we will need to consider the following singular elliptic Dirichlet problem

$$\begin{aligned} \left\{ \begin{array}{lr} -\Delta z = \lambda z - \beta z^\rho , &{} \text{ in }\;B(0,a),\\ z=\infty , &{} \text{ on }\;\partial B(0,a). \end{array}\right. \end{aligned}$$

(5.4)

Problem (5.4) has been studied in numerous articles, see [6, 12, 14]. It is known that it has a unique positive radial solution denoted by $z_a(x)$ and this solution is used to construct a supersolution in any ball where the logistic term n(t, x) is strictly positive. In particular, let us assume that $\overline{B(x_0,a)} \subset \Omega$ and $n(t,x)\ge \beta > 0$ for all $x\in B(x_0,a)$ and $t\ge t_0$. Then if $u_0(x) \in L^\infty (\Omega )$ is smaller than $z_a(0)$, we have that

$$\begin{aligned} 0 \le u_0(x) \le z_a(0)=\inf \limits _{y\in B(0,a)}z_a(y)\le z_a(y),\quad \forall x\in B(x_0,a),\; \forall y\in B(0,a). \end{aligned}$$

Hence, $z_a(x-x_0)$ is a supersolution of u(t, x), that is,

$$\begin{aligned} 0 \le u(t,x) \le z_a(x-x_0),\quad x\in B(x_0,a),\; t\ge t_0, \end{aligned}$$

and therefore, if $C=\sup \limits _{|y|\le b} z_a(y)$ for $0<b<a$ then

$$\begin{aligned} 0 \le u(t,x) \le C,\quad x\in B(x_0,b),\; t\ge t_0. \end{aligned}$$

Observe that $C=C(\lambda ,\beta ,\rho ,a,b)$.

Now we can show the following result:

Lemma 5.3

Let us consider problem (1.1) with n(t, x) satisfying (N) and let $K\subset \overline{\Omega }$ be a compact set with $K(t)\subset K$ for every $t\ge t_0$ (for instance $K=K^{\sup }_{t_0}$).

For each $\delta >0$ small enough there exists $\Lambda =\Lambda (\Omega ,\lambda ,\rho ,n,u_0,\delta )>0$ satisfying

$$\begin{aligned} 0\le u(t,x)\le \Lambda ,\; \text{ for } t>t_0, x\in \Omega \setminus \Omega _\delta , \end{aligned}$$

(5.5)

where $\Omega _\delta =\{ x\in \Omega : d(x,K)<\delta \}$ is an open neighbourhood of K.

Proof

Let us consider $\Omega _{\delta /2}$ and define

$$\begin{aligned}n_{\delta /2}(x):=\left\{ \begin{array}{ll} 0,&{} x\in \overline{\Omega _{\delta /2}},\\ \nu (\delta /2),&{} x\in \Omega \setminus \overline{\Omega _{\delta /2}}, \end{array}\right. \end{aligned}$$

where $\nu (\cdot )$ is the function in (N). It is clear that $n(t,x)\ge n_{\delta /2}(x),$ $\forall x\in \overline{\Omega _{\delta /2}}$, $t\ge t_0$. Moreover, from (N) we have $n(t,x)\ge \nu (d(x,K(t)))$ for every $x\in \Omega \setminus \overline{\Omega _{\delta /2}}$ But since $d(x,K(t))\ge \delta /2$ for $x\in \Omega {\setminus }\overline{\Omega _{\delta /2}}$ and $\nu$ is increasing, then we have $n(t,x)\ge \nu (\delta /2)>0$. Hence, $n(t,x)\ge n_{\delta /2}(x)$ $\forall x\in \overline{\Omega }$ and $\forall t\ge t_0$ and by comparison $u(t,x;u_0)\le {\hat{u}}(t,x,;u_0)$ where ${\hat{u}}$ is the solution of (1.3) with $n(x)=n_{\delta /2}(x)$.

But estimates (5.5) were shown in [5] for ${\hat{u}}$ (since ${\hat{u}}$ solves an autonomous problem) as it has been shown above. $\square$

Note that the bounds obtained in the previous Lemma 5.3 apply only to points outside $K^{\sup }$. But in many situations we may have points $x_0\in \Omega$ with $\overline{B(x_0,a)}\subset \Omega \setminus K(t)$ for $t\in (t_1,t_2)$ while $x_0\in K(t)$ for other times. For instance, consider K(t) is a continuously moving ball within a ring for which $K^{\sup }$ is the ring, see Fig. 8. If $x\in K^{\sup }$ we have times when $x\in K(t)$, but we have other times when $x\in \Omega \setminus K(t)$. It is very relevant to study the behavior of the solution in these points.

Fig. 8

A continuously moving K(t) along the ring $\bigcup \limits _{t\ge t_0}K(s)$

We have the following:

Proposition 5.4

Let u be the solution of (1.1) for $0\le u_0\in L^\infty (\Omega )$. Let $x_0\in \Omega$ and $a>0$ be such that $B(x_0,a)\subset \Omega$ and let n satisfy

$$\begin{aligned} n(t,x)\ge \beta >0,\; t\in (t_1,t_2),\; x\in B(x_0,a), \end{aligned}$$

for some $\beta >0$ and $t_1<t_2$. Then, for any $\tau _0\in (0,t_2-t_1)$ and any $r\in (0,a)$ there exists $\Lambda =\Lambda (\lambda ,\beta ,\rho ,\tau _0,r,a)$ such that

$$\begin{aligned} 0\le u(t,x) \le \Lambda ,\quad \text{ for } t\in [t_1+\tau _0,t_2],\; x\in B(x_0,r). \end{aligned}$$

Proof

Let us consider $U^*(t,x)=z_{a,x_0}(x)+W_\infty (t-t_1)$ for $t\in (t_1,t_2)$ and $x\in B(x_0,a)$ where $z_{a,x_0}(x)=z_a(x-x_0)$ and $z_a$ is the unique positive solution of (5.4) and $W_\infty$ is defined in (5.3). Then, since $z_a,W_\infty >0$ and $\rho >1$, we have

$$\begin{aligned}\begin{array}{rl} U^*_t-\Delta U^* &{} = {W_\infty }_t-\Delta z_{a,x_0}=\lambda W_\infty - \beta W_\infty ^\rho +\lambda z_{a,x_0} - \beta z_{a,x_0}^\rho \\ \\ \; &{} = \lambda U^* - \beta \left( W_\infty ^\rho + z_{a,x_0}^\rho \right) \\ &{}= \lambda U^* -\beta (W_\infty + z_{a,x_0})^\rho \left( \left( \frac{ W_\infty }{W_\infty + z_{a,x_0}}\right) ^\rho +\left( \frac{ z_{a,x_0}}{W_\infty + z_{a,x_0}}\right) ^\rho \right) \\ \\ \; &{} \ge \lambda U^* -\beta (W_\infty + z_{a,x_0})^\rho \left( \left( \frac{ W_\infty }{W_\infty + z_{a,x_0}}\right) +\left( \frac{ z_{a,x_0}}{W_\infty + z_{a,x_0}}\right) \right) = \lambda U^* -\beta (U^*)^\rho . \end{array}\end{aligned}$$

Moreover $U^*(t,x)=\infty$ for $t\in (t_1,t_2)$, $x\in \partial B(x_0,a)$ and for $t=t_1$, $x\in B(x_0,a)$. Therefore, $U^*$ is a supersolution for u in $(t_1,t_2)\times B(x_0,a)$ for any initial condition $u_0\in L^\infty (\Omega )$. Notice that for any $\tau _0\in (0,t_2-t_1)$ and any $r\in (0,a)$ if $t\in [t_1+\tau _0,t_2),\; x\in B\left( x_0,r\right)$ then

$$\begin{aligned} 0\le u(t,x) \le U^*(t,x) \le \Vert W_\infty (\tau _0)\Vert _\infty +\Vert z_{a,x_0}(\cdot )\Vert _{L^\infty \left( B\left( 0,r\right) \right) }<\infty . \end{aligned}$$

Which concludes the proof, choosing $\Lambda =\Vert W_\infty (\tau _0)\Vert _\infty +\Vert z_{a,x_0}(\cdot )\Vert _{L^\infty \left( B\left( 0,r\right) \right) }$. $\square$

Remark 5.5

Other bounds for $U^*$ when $x\rightarrow \partial B(x_0,a)$ or, $t\rightarrow t_2$ for $\beta =\beta (t)$ continuous, $\beta (t)>0$ for every $t\in (t_1,t_2)$ and $\beta (t_2)=0$ can be seen in [13].

With Proposition 5.4 it is not difficult to obtain bounds not only in balls but in other subsets at a positive distance from the sets K(t).

Corollary 5.6

Let D be a subset of $\Omega$, $\tau _0>0$, and $\delta >0$. If $d\left( D,\bigcup \limits _{t\in [t_1,t_2]} K(t)\right)>\delta >0$ for any $t_1,t_2\in \mathbb {R}$ with $t_2-t_1>\tau >0$, then, there exists $\Lambda =\Lambda (\lambda ,\rho ,\tau , \nu ,\delta )$ such that

$$\begin{aligned} 0\le u(t,x) \le \Lambda ,\quad \forall t\in [t_1+\tau ,t_2],\; \forall x\in D. \end{aligned}$$

Proof

Let us take the open set $\mathcal {U}$

$$\begin{aligned} \mathcal {U}:=\bigcup \limits _{z\in D}B\left( z,\frac{\delta }{2}\right) \supset D. \end{aligned}$$

Since D is at least a distance $\delta$ away from K(t) for $t\in [t_1,t_2]$ we know that $\mathcal {U}$ is at least a distance $\delta /2$ away from K(t) for the same time interval. Therefore,

$$\begin{aligned} n(t,x)\ge \nu (d(x,K(t))\ge \nu \left( \frac{\delta }{2}\right) >0,\quad \forall t\in [t_1,t_2], \; x\in \mathcal {U}. \end{aligned}$$

Moreover, applying Proposition 5.4 for every ball $B(z,\delta /2)$ with $z\in D$ and using the continuity of u in t we obtain that there exists a constant $\Lambda =\Lambda (\lambda ,\rho ,\tau , \nu (\delta /2))$ such that

$$\begin{aligned} 0\le u(t,x) \le \Lambda ,\quad \forall t\in [t_1+\tau ,t_2],\; \forall x\in B\left( z,\frac{\delta }{4}\right) ,\; \forall z\in D, \end{aligned}$$

which concludes the proof.

5.2 Boundedness for $\lambda < \lambda _0(K(t))$, for all $t \ge t_0$

In this subsection we obtain one of the main results of this section which consists of a boundedness result for the case where, roughly speaking, $\lambda < \lambda _0(K(t))$ $\forall t\ge t_0$. As a matter of fact, we will show the following important result:

Theorem 5.7

Assume the function $n(\cdot )$ satisfies (N). Assume also that there exist $\tau _0>0$, $\delta >0$ and a family of smooth bounded open sets $\left\{ \Omega _\delta (t)\right\} _{t\ge t_0}$ with $\Omega _\delta (t)\subset \Omega$ such that for each $t\ge t_0+\tau _0$

$$\begin{aligned} \bigcup \limits _{s\in [t-\tau _0,t+\tau _0]} K(s) \subset \Omega _\delta (t), \qquad \text{ and } \qquad d\left( \bigcup \limits _{s\in [t-\tau _0,t+\tau _0]} K(s), \,\, \partial \Omega _\delta (t)\right) \ge \delta . \end{aligned}$$

(5.6)

Then if $\lambda <\lambda _1\equiv \inf \limits _{t\ge t_0}\lambda _1(\Omega _\delta (t))$ we have that all solutions of (1.1) are bounded. Moreover, the solutions are uniformly asymptotically bounded in $L^\infty$.

Remark 5.8

Condition (5.6) implies that K(t) is at a distance at least $\delta >0$ from $\partial \Omega$. See Corollary 5.10 below to extend this result to the case where $K(t)\cap \partial \Omega \ne \emptyset$.

Proof

Let us define $t_1 = t_0+\tau _0$, and in general, $t_{i+1} = t_i +\tau _0$ for $i\ge 1$, $\lambda _1:=\inf \limits _{t\ge t_0} \lambda _1(\Omega _\delta (t))$ and let $D_i$ be a $\delta /2$–neighbourhood of $\Omega \setminus \Omega _\delta (t_i)$, that is, $D_i=\{ x\in \Omega : d(x,\Omega {\setminus }\Omega _\delta (t_i))< \delta /2\}$ for $i\ge 1$, see Fig. 9. Now, from (5.6) we have that for any $x\in D_i$ and $i\ge 1$

$$\begin{aligned} d\left( x,\bigcup \limits _{s\in [t_i-\tau _0,t_i+\tau _0]}K(s)\right) \ge \delta /2. \end{aligned}$$

Fig. 9

$\Omega _\delta (t)$ contains $\displaystyle \bigcup _{s\in [t-\tau _0,t+\tau _0]} K(s)$ and it lies at a distance $\delta$ from $\partial \Omega _\delta (t)$

Applying Corollary 5.6 to $D_i$ for $i\ge 1$ we obtain that there exists a bound $\Lambda =\Lambda (\lambda ,\rho ,\tau _0,\nu ,\delta )$ such that $u(t,x)\le \Lambda$ for $t\in [t_i,t_i+\tau _0]$ and $x\in D_{i}$. In particular, since $\Omega {\setminus } \Omega _\delta (t_i) \subset D_{i}$ and $\partial \Omega _\delta (t_i) \subset D_{i}$, constant $\Lambda$ is independent of i and satisfies

$$\begin{aligned} u(t,x) \le \Lambda , \qquad \text{ for } t\in [t_i,t_{i+1}],\; \text{ a.e. } x\in \Omega \setminus \Omega _\delta (t_i),\; i\ge 1. \end{aligned}$$

(5.7)

In particular we have $u(t,x) \le \Lambda$, for $t\in [t_i,t_{i+1}]$, a.e. $x\in \partial \Omega _\delta (t_i)$ and any $i\ge 1$.

Let us prove that we have

$$\begin{aligned} \Vert u(t_{i+1},\cdot )\Vert _{L^2(\Omega _\delta (t_i))}\le \frac{\lambda |\Omega |^{1/2}}{\lambda -\lambda _1^{\Omega }}\Lambda +e^{(\lambda -\lambda _1)\tau _0}\Vert u(t_{i},\cdot )\Vert _{L^2(\Omega _\delta (t_{i}))}, \quad i=1,2,\ldots \end{aligned}$$

(5.8)

To do so, let us consider the problem

$$\begin{aligned} \left\{ \begin{array}{lr} v_t-\Delta v = \lambda v, &{} x\in \Omega _\delta (t_i), \; t\in (t_i,t_{i+1}),\\ v = \Lambda , &{} x\in \partial \Omega _\delta (t_i), \; t\in (t_i,t_{i+1}),\\ v(t_i,x)= u(t_i,x), &{} x\in \Omega _\delta (t_i). \end{array}\right. \end{aligned}$$

(5.9)

By comparison, we have

$$\begin{aligned} v(t,x)\ge u(t,x), \qquad t\in [t_i,t_{i+1}], \text{ a.e. } x\in \Omega _\delta (t_i). \end{aligned}$$

Fig. 10

Neighbourhoods $\Omega _\delta (t_1)$ and $\Omega _\delta (t_2)$ (dotted lines)

Now, v can be split as $v=v_1+v_2$ where

$$\begin{aligned} \left\{ \begin{array}{lr} -\Delta v_1 = \lambda v_1, &{} x\in \Omega _\delta (t_i),\\ v_1= \Lambda , &{} x\in \partial \Omega _\delta (t_i), \end{array}\right. \end{aligned}$$

(5.10)

$$\begin{aligned} \left\{ \begin{array}{lr} v_{2_t} -\Delta v_2= \lambda v_2, &{} x\in \Omega _\delta (t_i), \; t\in (t_i,t_{i+1}),\\ v_2=0, &{} x\in \partial \Omega _\delta (t_i), \; t\in (t_i,t_{i+1}),\\ v_2(t_i,x)= u(t_i,x)-v_1(x), &{} x\in \Omega _\delta (t_i). \end{array}\right. \end{aligned}$$

(5.11)

Observe that since $\lambda <\lambda _1\le \lambda _1(\Omega _\delta (t_1))$, $v_1$ is well defined. Moreover, standard estimates show that

$$\begin{aligned} \Vert v_1\Vert _{L^2(\Omega _{\delta }(t_i))}\le \frac{\lambda |\Omega |^{1/2}}{\lambda -\lambda _1(\Omega )}\Lambda . \end{aligned}$$

Since, by the maximum principle, $v_1(x)\ge 0$ we have that standard $L^2$-estimates applied to (5.11) show

$$\begin{aligned} \Vert v_2(t_{i+1},\cdot )\Vert _{L^2(\Omega _\delta (t_i))}\le e^{(\lambda -\lambda _1)\tau _0}\Vert v_2(t_{i},\cdot )\Vert _{L^2(\Omega _\delta (t_i))}. \end{aligned}$$

Hence, since $0\le u(t_{i+1},x)\le v_1(x)+v_2(t_{i+1},x)$ from the two previous estimates we obtain (5.8).

But observe now that if $x\in \Omega _\delta (t_{i+1}){\setminus } \Omega _\delta (t_{i})$ we have $u(t_{i+1},x)\le \Lambda$. Hence

$$\begin{aligned} \Vert u(t_{i+1},\cdot )\Vert _{L^2(\Omega _\delta (t_{i+1}))}^2\le & {} \Vert u(t_{i+1},\cdot )\Vert _{L^2(\Omega _\delta (t_{i+1})\cap \Omega _\delta (t_{i}) )}^2+\Lambda ^2|\Omega _\delta (t_{i+1})\setminus \Omega _\delta (t_{i})|\\\le & {} \Vert u(t_{i+1},\cdot )\Vert _{L^2(\Omega _\delta (t_{i}))}^2+\Lambda ^2|\Omega | \end{aligned}$$

and therefore, plugging in (5.8) in this last expression we get

$$\begin{aligned} \Vert u(t_{i+1},\cdot )\Vert _{L^2(\Omega _\delta (t_{i+1}))}\le \frac{2\lambda -\lambda _1}{\lambda -\lambda _1}|\Omega |^{1/2} \Lambda +e^{(\lambda -\lambda _1)\tau _0}\Vert u(t_{i},\cdot )\Vert _{L^2(\Omega _\delta (t_{i}))}, \quad i=1,2,\ldots \nonumber \\ \end{aligned}$$

(5.12)

If we denote by $z_i=\Vert u(t_{i},\cdot )\Vert _{L^2(\Omega _\delta (t_{i}))}$, $A=\displaystyle \frac{2\lambda -\lambda _1}{\lambda -\lambda _1}|\Omega |^{1/2} \Lambda$ and $\rho =e^{(\lambda -\lambda _1)\tau _0}$, we have $z_{i+1}\le A+\rho z_i$, $i=1,2,\ldots$ and iterating this inequality we obtain

$$\begin{aligned} \Vert u(t_{i+1},\cdot )\Vert _{L^2(\Omega _\delta (t_{i+1}))}\le & {} A(1+\rho +\ldots +\rho ^{i-1})+\rho ^{i}\Vert u(t_{1},\cdot )\Vert _{L^2(\Omega _\delta (t_{1}))}\\\le & {} \frac{A}{1-\rho }+\rho ^i\Vert u(t_{1},\cdot )\Vert _{L^2(\Omega _\delta (t_{1}))}. \end{aligned}$$

Now since $t_1=t_0+\tau _0$, standard $L^2$-estimates in $\Omega$ show that

$$\begin{aligned} \Vert u(t_{1},\cdot )\Vert _{L^2(\Omega _\delta (t_{1}))}\le \Vert u(t_{0}+\tau _0,\cdot )\Vert _{L^2(\Omega )}\le e^{(\lambda -\lambda _1^\Omega )\tau _0}\Vert u_0\Vert _{L^2(\Omega )}, \end{aligned}$$

and choosing $i_0\ge 1$ large enough so that $\rho ^{i_0}e^{(\lambda -\lambda _1^\Omega )\tau _0}\Vert u_0\Vert _{L^2(\Omega )}\le 1$, we get

$$\begin{aligned} \Vert u(t_{i+1},\cdot )\Vert _{L^2(\Omega _\delta (t_{i+1}))}\le \frac{A}{1-\rho }+1,\qquad i\ge i_0. \end{aligned}$$

But, since $u(t_{i+1},x)\le \Lambda$ for $x\in \Omega {\setminus }\Omega _\delta (t_{i+1})$ and taking into account the definition of A and $\rho$, we have that if we define

$$\begin{aligned} R=\Lambda |\Omega |^{1/2}+ \frac{2\lambda -\lambda _1}{\lambda -\lambda _1}|\Omega |^{1/2} \Lambda \frac{1}{1-e^{(\lambda -\lambda _1)\tau _0}}+1, \end{aligned}$$

then

$$\begin{aligned} \Vert u(t_{i+1},\cdot )\Vert _{L^2(\Omega )}\le R,\qquad i\ge i_0. \end{aligned}$$

Using now $L^2-L^\infty$-estimates, we get

$$\begin{aligned} \Vert u(t,\cdot )\Vert _{L^\infty (\Omega )}\le CR, \qquad t\in [t_{i+2},t_{i+3}], \quad i\ge i_0. \end{aligned}$$

This shows the result. $\square$

Remark 5.9

Conditions (5.6) can be easily met, for example, whenever $K(t)=\gamma (t)+R(t) K_0$, where $\gamma (t)$ is a $\mathcal {C}^1$ curve in $\Omega$ with $\sup _{t\ge t_0}|\gamma '(t)|<\infty$, R(t) is a rigid motion for all $t\ge t_0$ and is continuous with respect to $t\ge t_0$ with the condition that $\gamma (t)+R(t) K_0\subset \Omega$ $\forall t\ge t_0$, and $K_0$ is a compact subset. If $\lambda <\lambda _0(K_0)$ and $K_0$ can be seen as the intersection of a family of nested open neighbourhoods of $K_0$, then there exists an open neighbourhood $\Omega _\delta$ such that $\lambda < \lambda _1(\Omega _\delta )$ and $d(\partial \Omega _\delta , K_0)>2\delta$ for some $\delta >0$.

Since K(t) moves continuously, for any positive distance to the boundary of $\Omega _\delta$, let us consider $\delta$, there exists a time $\tau$ such that

$$\begin{aligned} \bigcup \limits _{[t-\tau ,t+\tau ]} K(s) \subset \Omega _\delta (t), \qquad \text{ and } \qquad d\left( \bigcup \limits _{[t-\tau ,t+\tau ]} K(s), \partial \Omega _\delta (t)\right) \ge \delta . \end{aligned}$$

Finally, since there is a maximum speed at which it travels, there is a minimum time $\tau _0$ which verifies all the above.

Finally, as we mentioned above, we can consider the case where K(t) touches $\partial \Omega$.

Corollary 5.10

Let $R>0$ such that $\overline{\Omega }\subset B(0,R)$. Let us assume n satisfies (N) and there exist $\tau _0>0,\delta >0$ and a family of smooth bounded open sets $\{\Omega _\delta (t)\}_{t\ge t_0}$ with $\Omega _\delta (t) \subset B(0,R)$ such that for each $t\ge t_0+\tau _0$, (5.6) in Theorem 5.7 is satisfied.

Then, if $\lambda <\inf \limits _{t\ge t_0}\lambda _1(\Omega _\delta (t))$ and $\tau _0>\omega$, we have that all solutions of (1.1) are uniformly asymptotically bounded in $L^\infty$.

Proof

If we define

$$\begin{aligned} {\tilde{n}}(t,x):= \left\{ \begin{array}{lr} n(t,x), &{} t\ge t_0, \; x\in \overline{\Omega },\\ \nu (2R), &{} t\ge t_0, \; x\in B(0,R)\setminus \overline{\Omega }, \end{array}\right. \end{aligned}$$

we have ${\tilde{n}}$ satiesfies condition (N) also with the same function $\nu$. Moreover, by comparison, the solution $u(t,t_0, u_0, {\tilde{n}}, B(0,R))$ is a supersolution of $u(t,t_0, u_0, n, \Omega )$. But we can apply Theorem 5.7 to $\tilde{u}$, obtaining boundedness for ${\tilde{u}}$ and therefore for u.

5.3 Boundedness for a Jumping K(t)

This final boundedness result characterizes boundedness with a discontinuity in K(t) independently of the size of $\lambda$. It is an interesting result since it does not consider the linear growth occurring in the sets K(t) and its relationship with $\lambda$: it only focuses on the fact that the sets K(t) jump at certain instants of time.

As a prototype of this situation, see Example 4.8 and Fig. 11 below.

Fig. 11

A jumping K(t) between compact subsets $K_0$ and $K_1$

That is, by “jumps” or discontinuity in K(t) we refer to a sequence of times $(t_j)_{j\in \mathbb {N}}$ with $t_1<t_2<t_3< \ldots \rightarrow \infty$ where the moving set K(t) jumps at each $t_j$. As a matter of fact we have:

Theorem 5.11

Let u be the solution of (1.1), $\lambda \in \mathbb {R}$ and let n satisfying (N). Moreover, let us assume there exists an unbounded increasing sequence $(t_j)_{j\in \mathbb {N}}$ and $\tau _0,\delta ,\Xi >0$ such that

$$\begin{aligned} \left\{ \begin{array}{l} 2\tau _0\le \inf \limits _{j\in \mathbb {N}} |t_{j+1}-t_j|\le \sup \limits _{j\in \mathbb {N}} |t_{j+1}-t_j| =\Xi < \infty \;, \\ \\ \qquad d\left( \bigcup \limits _{(t_j-\tau _0,t_j)} K(s),\bigcup \limits _{(t_j,t_j+\tau _0)} K(s)\right) > \delta . \\ \end{array} \right. \end{aligned}$$

(5.13)

Then, all solutions of (1.1) are bounded.

Proof

In this proof we will use that K(t) is bounded away from the boundary of $\Omega$. Corollary 5.12 below, extends this result even when $K(t)\cap \partial \Omega \ne \emptyset$.

Let us start showing that we have a constant $\Lambda ^*>0$ such that for any initial condition we can obtain the bounds:

$$\begin{aligned} \Vert u(t_j+\tau _0,\cdot )\Vert _{L^\infty (\Omega )}\le \Lambda ^*,\quad \hbox { for }j=1,2\ldots \end{aligned}$$

(5.14)

Let us consider for each $j\in \mathbb {N}$ the set

$$\begin{aligned} \Omega ^+_{\delta }(t_j):= \left\{ x\in \Omega : d\left( x, \bigcup \limits _{(t_j,t_j+\tau _0)} K(s)\right) <\frac{\delta }{3} \right\} \supset \bigcup \limits _{(t_j,t_j+\tau _0)} K(s), \end{aligned}$$

which satisfies

$$\begin{aligned} d\left( \partial \Omega ^+_{\delta }(t_j), \bigcup \limits _{(t_j,t_j+\tau _0)} K(s)\right) = \delta /3. \end{aligned}$$

(5.15)

Notice that from (5.13) and (5.15) we have

$$\begin{aligned} d\left( \Omega ^+_{\delta }(t_j), \bigcup \limits _{(t_j-\tau _0,t_j)} K(s)\right) > \delta /3. \end{aligned}$$

(5.16)

Fig. 12

Neighbourhood $\Omega _\delta ^+(t_j)$ stays at least at a distance $\delta /3$ from $\bigcup \limits _{s\in (t_j-\tau _0,t_j)} K(s)$ and its boundary stays at least at the same distance from $\bigcup \limits _{s\in (t_j,t_j+\tau _0)} K(s)$

Let $j\in \mathbb {N}$, since (5.16) holds, operating as in Theorem 5.7 and applying Corollary 5.6 to the interval $(t_j-\tau _0,t_j)$ and the subset $\Omega ^+_{\delta }(t_j)$ there exists a constant $\Lambda _1=\Lambda _1(\lambda ,\rho ,\tau _0,\nu ,\delta )$ such that

$$\begin{aligned} u(t_j,\cdot ) \le \Lambda _1, \qquad \text{ for } \text{ a.e. } x\in \Omega _{\delta }^+(t_j). \end{aligned}$$

Applying again Corollary 5.6 to the interval $(t_j-\tau _0,t_j+\tau _0)$ and the subset $\partial \Omega ^+_{\delta }(t_j)$, there exists a constant $\Lambda _1^*=\Lambda _1^*(\lambda ,\rho ,\tau _0,\nu ,\delta )$ such that

$$\begin{aligned} u(t,\cdot ) \le \Lambda _1^*, \qquad \forall t\in [t_j,t_j+\tau _0],\; x\in \partial \Omega _{\delta }^+(t_j). \end{aligned}$$

Observe that $\Lambda _1$ and $\Lambda _1^*$ do not depend on the initial condition nor the time $t_j$. If we take $\Lambda = \max \{ \Lambda _1, \Lambda _1^*\}$, which do not depend on the initial condition nor the time $t_j$ and we consider problem

$$\begin{aligned} \left\{ \begin{array}{lr} v_t-\Delta v = \lambda v, &{} x\in \Omega _{\delta }^+(t_j), \; t\in (t_j,t_j+\tau _0),\\ v = \Lambda , &{} x\in \partial \Omega _{\delta }^+(t_j), \; t\in (t_j,t_j+\tau _0),\\ v(t_j,x)= \Lambda , &{} x\in \Omega _{\delta }^+(t_j), \end{array}\right. \end{aligned}$$

(5.17)

then, by comparison we have that $u(t,\cdot )\le v(t,\cdot )$ for $t\in (t_j,t_j+\tau _0)$ and a.e. $x\in \Omega _{\delta }^+(t_j)$.

Defining $w(t,x)=e^{-\lambda (t-t_j)}(v(t,x)-\Lambda )$, we have that w satisfies

$$\begin{aligned} \left\{ \begin{array}{lr} w_t-\Delta w = \lambda \Lambda e^{-\lambda (t-t_j)}, &{} x\in \Omega _{\delta }^+(t_j), \; t\in (t_j,t_j+\tau _0),\\ w = 0, &{} x\in \partial \Omega _{\delta }^+(t_j), \; t\in (t_j,t_j+\tau _0),\\ w(t_j,x)= 0, &{} x\in \Omega _{\delta }^+(t_j). \end{array}\right. \end{aligned}$$

(5.18)

But,

$$\begin{aligned} w(t,x)\le \int _{t_j}^t\lambda \Lambda e^{-\lambda (s-t_j)}ds=\Lambda (1-e^{-\lambda (t-t_j)}), \end{aligned}$$

which implies that

$$\begin{aligned} v(t,x)\le \Lambda +e^{\lambda (t-t_j)}\Lambda (1-e^{-\lambda (t-t_j)})=\Lambda e^{\lambda (t-t_j)}. \end{aligned}$$

Therefore, choosing $\Lambda ^*=\Lambda e^{\lambda \tau _0}$ and taking into account that $u(t,x)\le v(t,x)$, we prove the claim (5.14).

To show that the solution is bounded, we observe that by comparison

$$\begin{aligned} u(t,x)\le U(t-(t_j+\tau _0),x), \qquad t\in [t_j+\tau _0,t_{j+1}+\tau _0],\quad j=1,2,\ldots \end{aligned}$$

where U(t, x) is the solution of the linear problem

$$\begin{aligned} \left\{ \begin{array}{lr} U_t - \Delta U = \lambda U, &{} x\in \Omega ,\; t\in (0,\infty ),\\ U=0, &{} x\in \partial \Omega , \; t\in (0,\infty ),\\ U(0)=\Lambda ^*, &{} x\in \Omega , \end{array}\right. \end{aligned}$$

(5.19)

which obviously satisfies $U(t,x)\le e^{\lambda t}\Lambda ^*$ by the maximum principle and therefore

$$\begin{aligned} \Vert u(t,\cdot )\Vert _{L^\infty (\Omega )}\le \Lambda ^*e^{|\lambda | \Xi }, \qquad t\ge t_1, \end{aligned}$$

which shows the result. $\square$

Also, with a similar proof as Corollary 5.10, we can also show

Corollary 5.12

Let $R>0$ such that $\overline{\Omega }\subset B(0,R)$. Then, assuming the same conditions as Theorem 5.11, we have that all solutions of (1.1) are bounded.

6 Unboundedness

This section is devoted to obtaining conditions under which solutions grow up, that is, under which solutions become unbounded as $t\rightarrow \infty$.

Observe that we have proven before, see Theorem 5.7, that if, roughly speaking, $\lambda <\lambda _0(K(t))$ for all $t\ge t_0$, then we have boundedness of solutions. The equivalent situation for unboundedness would be to show that if $\lambda > \lambda _0(K(t))$ $\textrm{for}\,\,\textrm{all}\,\,t\ge t_0$, the growth given by the linear part is strong enough to produce solutions which are unbounded as $t\rightarrow \infty$. But the previous section has also provided us with a clear counterexample to this hypothetical result. Actually we may have $K_0, K_1\ne \emptyset$ and $\lambda >\lambda _0(K_0)$, $\lambda >\lambda _0(K_1)$ but solutions remain bounded as $t\rightarrow \infty$ as detailed in Theorem 5.11 and seen in Example 4.8. This again proves that the size of $\lambda _0(K(t))$ is insufficient to determine the boundedness or unboundedness of solutions. Actually, the geometry and the speed at which the sets K(t) move are also important parameters.

The assumptions in the following result can be interpreted as first of all having K(t) large enough to allow the solution to grow inside these set. Moreover, the sets K(t) move slowly enough so that the growth inside K(t) is transported together with K(t), allowing the function u to grow without bounds inside the slowly moving set K(t).

We can state now:

Theorem 6.1

Let us assume that n satisfies (N). Let us also assume that there exists a sequence of times $(t_i)_{i\ge 0}$ with $t_0<t_1<\ldots<t_i<\ldots$ and a family of nonempty, smooth open subsets

$$\begin{aligned} E_i\subset \bigcap \limits _{s\in [t_i,t_{i+1}]} K(s), \end{aligned}$$

(6.1)

with rigid motions $T_i:\mathbb {R}^N\rightarrow \mathbb {R}^N$ such that $E_i:= T_i(E_0)$, and such that there exists $(x_i)_{i\ge 0}$ and $r>0$ such that $B(x_i,2r)\subset E_i\cap E_{i+1}$ for $i=1,2,\ldots$, see Fig. 13. Assume also that $\lambda >\lambda _1(E_0)$.

Then, there exists $\tau >0$ such that if $t_{i+1}-t_i\ge \tau$ $for\, all\, i=0,1,2,\ldots$ all solutions are unbounded.

Observe that having the sets $E_i$ satisfying $\lambda >\lambda _0(E_i)$ is saying that the sets $E_i$ are large enough and therefore K(t) are large enough from (6.1). Also, intuitively, having $t_{i+1}-t_i\ge \tau$ with property (6.1) means that the velocity at which the sets K(t) move is slow enough.

Fig. 13

The sets $E_i$ and $B(x_i,2r)$

The key point to prove this result is to find subsolutions which are unbounded as $t\rightarrow \infty$. Since the nonlinear logistic term is applied outside K(t), unbounded growth will only be possible within K(t). Therefore, we will consider the problem restricted to the sets $E_i$ which will be linear problems. Under certain conditions, the solutions of these linear problems will grow without bounds, providing us with the key factor to get the unboundedness for the nonlinear problem.

In this respect, we have the following interesting auxiliary result on the growth of the linear problem and its relation with the first eigenfunctions.

Let us denote by $\phi ^A_n$ the n–th eigenfunction of the laplacian with homogeneous Dirichlet boundary conditions on the domain A normalized in the $L^2$–norm and by $\lambda _n^A$ its eigenvalue. Assume also that $\phi _1^A$ is chosen to be positive in A.

Lemma 6.2

Let us consider the following linear problem

$$\begin{aligned} \left\{ \begin{array}{ll} v_t-\Delta v = \lambda v, &{} t>0,\; x\in E, \\ v = 0, &{} t> 0,\; x\in \partial E,\\ 0\le v(0,x)= v_0(x)\in L^\infty (E), &{} x\in E. \end{array}\right. \end{aligned}$$

(6.2)

where E is a smooth bounded open connected domain.

Let $D\subset E$ be a smooth open subset with ${\overline{D}}\subset E$ and let $\lambda >\lambda ^E_{1}$. Then, if $\gamma >1$ is given, $v_0$ is not identically 0 and $\tau >0$ is given by:

$$\begin{aligned} \tau= & {} \max \left\{ \frac{N\lambda _2^E}{2e(\lambda _2^E-\lambda _1^E)}\left( \frac{2C_\infty \Vert v_0\Vert _{L^2(E)}}{\left\langle v_0, \phi ^E_1 \right\rangle _{L^2(E)} \inf \limits _{D}{\phi ^E_1}}\right) ^{\frac{2}{N}},\frac{1}{\lambda -\lambda _1^E}\log \left( \frac{2\gamma \max \limits _{D}\phi _1^D}{\left\langle v_0, \phi ^E_1 \right\rangle _{L^2(E)} \inf \limits _{D}{\phi ^E_1} }\right) \right\} ,\nonumber \\ \end{aligned}$$

(6.3)

where $C_\infty =C_\infty (E,N)$ is the constant of the embedding $H^N(E)\hookrightarrow L^\infty (E)$, then

$$\begin{aligned} v(t,x;v_0) \ge \gamma \phi ^{D}_1(x), \quad t\ge \tau ,\; x\in D. \end{aligned}$$

Proof

Let us define

$$\begin{aligned} I =\inf \limits _{D}{\phi ^E_1}>0,\qquad \mu =\frac{I}{\displaystyle 2\max \limits _{D} \phi _1^D},\qquad \alpha _j=\left\langle v_0, \phi ^E_j \right\rangle _{L^2(E)},\quad j\in \mathbb {N}. \end{aligned}$$

We have for $x\in D$

$$\begin{aligned} v(t,x)= & {} \sum _{j=1}^{\infty }e^{\left( \lambda -\lambda ^E_j\right) t} \langle v_0,\phi ^E_j\rangle _{L^2(E)}\phi ^E_j(x)\\= & {} e^{\left( \lambda -\lambda ^{E}_1\right) t}\alpha _1 \phi ^E_1(x)+\sum _{i=2}^{\infty }e^{\left( \lambda -\lambda ^E_j\right) t} \alpha _j\phi ^E_j(x)\\\ge & {} e^{\left( \lambda -\lambda ^E_1\right) t}\alpha _1(\mu \phi ^D_1+I/2)+\sum _{i=2}^{\infty }e^{\left( \lambda -\lambda ^E_j\right) t} \alpha _j \phi ^E_j(x) \end{aligned}$$

where we have used that

$$\begin{aligned} \phi _1^E(x)=\frac{\phi _1^E(x)}{2}+\frac{\phi _1^E(x)}{2}\ge \frac{I}{2}+\frac{I}{2}\ge \frac{I\phi _1^D}{2\max \limits _{D} \phi _1^D}+\frac{I}{2}=\mu \phi _1^D+\frac{I}{2}. \end{aligned}$$

Now, let us observe that

$$\begin{aligned} v(t,x)\ge e^{(\lambda -\lambda ^E_1)t}\left( \alpha _1\mu \phi ^D_1(x)+\alpha _1I/2 + J(x)\right) , \end{aligned}$$

where

$$\begin{aligned} J(x)=\sum _{j=2}^{\infty }e^{\left( \lambda ^E_1-\lambda ^E_j\right) t} \alpha _j\phi ^E_j(x). \end{aligned}$$

We want to arrive at an $L^\infty$–estimate for J. If N is the dimension of the space and if $k>N/2$ we have the Sobolev inclusion $H^k(\Omega )\hookrightarrow L^\infty (\Omega )$. Thus, denoting by $C_{\infty }=C_{\infty }(N,E)$ the constant of the embedding $H^{N}(E)\hookrightarrow L^\infty (E)$, we have

$$\begin{aligned} \left\| J \right\| ^2_{L^\infty (E)}\le & {} C_{\infty }^2\left\| \sum _{j=2}^{\infty }e^{\left( \lambda ^E_1-\lambda ^E_j\right) t} \alpha _j\phi ^E_j\right\| ^2_{H^{N}(E)} = C_{\infty }^2\sum _{j=2}^{\infty }e^{-2\left( \lambda ^E_j-\lambda ^E_1\right) t} \alpha _j^2 (\lambda ^E_j)^{N} \\= & {} C_{\infty }^2 \sum _{j=2}^{\infty }e^{-2\left( \lambda ^E_j-\lambda ^E_1\right) t} \left( (\lambda ^E_j-\lambda ^E_1)t\right) ^{N}\alpha _j^2 \frac{(\lambda ^E_j)^{N}}{\left( (\lambda ^E_j-\lambda ^E_1)t\right) ^{N}} \\\le & {} C_{\infty }^2 \left( \frac{N}{2e}\right) ^N \sum _{j=2}^{\infty } \alpha _j^2 \left( \frac{\lambda ^E_j}{(\lambda ^E_j-\lambda ^E_1)t}\right) ^N \\\le & {} C_{\infty }^2\left( \frac{N}{2e}\right) ^N \sup _{j\ge 2}\left( \frac{\lambda ^E_j}{(\lambda ^E_j-\lambda ^E_1)t}\right) ^N \Vert v_0\Vert ^2_{L^2(E)}, \end{aligned}$$

where we are bounding

$$\begin{aligned} e^{-2\left( \lambda ^E_j-\lambda ^E_1\right) t} \left( (\lambda ^E_j-\lambda ^E_1)t\right) ^{N}\le \sup _{a\ge 0}\{e^{-2a}a^N\}=\left( \frac{N}{2e}\right) ^N,\quad \forall j\ge 2, \end{aligned}$$

and using

$$\begin{aligned} \sum _{j=2}^\infty \alpha _j^2\le \sum _{j=1}^\infty \alpha _j^2=\Vert v_0\Vert ^2_{L^2(E)}. \end{aligned}$$

Moreover, since $\lambda _j^E\ge \lambda _2^E>\lambda _1^E$ for $j\ge 2$, we can easily check that $\frac{\lambda ^E_j}{\lambda ^E_j-\lambda ^E_1}$ is decreasing in $j\ge 2$ and therefore $\frac{\lambda ^E_j}{\lambda ^E_j-\lambda ^E_1}\le \frac{\lambda ^E_2}{\lambda ^E_2-\lambda ^E_1}$. Hence,

$$\begin{aligned} \left\| J \right\| ^2_{L^\infty (E)}\le & {} C_{\infty }^2\left( \frac{N}{2e}\right) ^{N}\left( \frac{\lambda ^E_2}{(\lambda _2^E-\lambda _1^E)t }\right) ^{N}\Vert v_0\Vert _{L^2(E)}^2. \end{aligned}$$

Therefore, if we take

$$\begin{aligned} t\ge \frac{N\lambda _2^E}{2e(\lambda _2^E-\lambda _1^E)} \left( \frac{C_\infty 2\Vert v_0\Vert _{L^2(E)}}{\alpha _1 I}\right) ^{2/N}, \end{aligned}$$

we have $\Vert J\Vert ^2_{L^\infty (E)}\le \alpha _1I/2$ and therefore,

$$\begin{aligned} v(t,x) \ge e^{(\lambda -\lambda ^E_1)t}\left( \alpha _1\mu \phi ^{D}_1(x) +\alpha _1I/2 -\Vert J\Vert _{L^\infty (E)}\right) \ge e^{(\lambda -\lambda ^E_1)t}\alpha _1\mu \phi ^{D}_1(x). \end{aligned}$$

Moreover, if $\gamma >1$ is given and we choose $t\ge \frac{1}{\lambda -\lambda _1^E}\log \left( \frac{\gamma }{\alpha _1\mu }\right)$, then $e^{(\lambda -\lambda ^E_1)t}\alpha _1\mu \ge \gamma$.

Hence, if we take

$$\begin{aligned} \tau =\max \left\{ \frac{N\lambda _2^E}{2e(\lambda _2^E-\lambda _1^E)} \left( \frac{C_\infty 2\Vert v_0\Vert _{L^2(E)}}{\alpha _1 I}\right) ^{2/N}, \frac{1}{\lambda -\lambda _1^E}\log \left( \frac{\gamma }{\alpha _1\mu }\right) \right\} >0, \end{aligned}$$

we have $v(t,x)\ge \gamma \phi _1^D(x)$ for all $t\ge \tau$ and $x\in D$. This concludes the proof. $\square$

Let us consider now the following corollary, which will be important in the proof of the main result.

Corollary 6.3

In the setting of the previous proposition, if we assume that $B(a,2r)\subset E$ and take $v_0=\phi _1^{B(a,r)}$, then there exists $\rho >0$, independent of a, such that

$$\begin{aligned} \Vert v(t,x,v_0)\Vert _{L^\infty (E)}\ge \rho , \quad t\ge 0. \end{aligned}$$

Proof

Let us fix $a_0\in E$ such that $B(a_0,2r)\subset E$. Let us also fix $\gamma >1$. Applying Proposition 6.2, with $D=B(a_0,r)\subset \subset E$, we have that the value of $\tau$ is independent of a. Hence, we have that

$$\begin{aligned} v(t,x,\phi _1^{B(a,r)})\ge \gamma \phi _1^{B(a_0,r)}(x), \quad x\in B(a,r),\quad t\ge \tau . \end{aligned}$$

This implies that for all $a\in E$ with $B(a,2r)\subset E$, we have

$$\begin{aligned} \Vert v(t,\cdot , \phi _1^{B(a,r)})\Vert _{L^\infty (E)}\ge \gamma \Vert \phi _1^{B(a_0,r)}\Vert _{L^\infty (E)}=\gamma \Vert \phi _1^{B(0,r)}\Vert _{L^\infty (E)}, \quad \forall t\ge \tau . \end{aligned}$$

Now, by comparison with the problem posed in B(a, r) and using that $\phi _1^{B(a,r)}$ is the first eigenfunction in B(a, r), we have

$$\begin{aligned} v(t,x,\phi _1^{B(a,r)})\ge e^{(\lambda -\lambda _1^{B(a,r)})t}\phi _1^{B(a,r)}(x), \quad t\ge 0,\quad x\in B(a,r). \end{aligned}$$

But this implies that

$$\begin{aligned} \Vert v(t,\cdot ,\phi _1^{B(a,r)})\Vert _{L^\infty (E)}\ge e^{-|\lambda -\lambda _1^{B(0,r)}|\tau }\Vert \phi _1^{B(0,r)}\Vert _{L^\infty (E)}, \qquad t\in [0,\tau ]. \end{aligned}$$

Therefore, choosing

$$\begin{aligned} \rho =\min \{ \gamma \Vert \phi _1^{B(0,r)}\Vert _{L^\infty (E)}, e^{-|\lambda -\lambda _1^{B(0,r)}|\tau }\Vert \phi _1^{B(0,r)}\Vert _{L^\infty (E)}\}, \end{aligned}$$

we prove the result. $\square$

We are in a position now to give a proof of the main result, Theorem 6.1.

Proof

(of Theorem 6.1). First let us state some immediate remarks derived from the assumptions. Since $B(x_i,2r)\subset E_i\cap E_{i+1}$ then $d(x_i,\partial E_i)\ge 2r$ and $d(x_i,\partial E_{i+1})\ge 2r$. Moreover, since $T_i:\mathbb {R}^N\rightarrow \mathbb {R}^N$ are rigid motions and $\lambda > \lambda _0(E_0)$ then $\lambda > \lambda _0(E_i)$ for any $i\ge 0$, and if we define ${\tilde{x}}_i=T_i^{-1}(x_i)$ then we have $d(\tilde{x}_i,\partial E_0)\ge 2r$ for all $i=0,1,2,\ldots$.

To simplify, let us denote by $\psi$ the first eigenfunction of $-\Delta$ in B(0, r) with Dirichlet boundary conditions (which is explicitly known), that is $\psi =\phi _1^{B(0,r)}$. Also, denote by $\psi _i:= \psi (x-x_i)$ the first eigenfunction of $-\Delta$ in $B(x_i,r)$ with Dirichlet homogeneous boundary conditions and extended by zero outside $B(x_i,r)$ for every $i\ge 0$.

We are going to apply Proposition 6.2 for $i=1,2,\ldots$, with $E=E_i$, $D=B(x_i,r)$ and with initial condition $v_0=\psi _{i-1}$, which has support in $B(x_{i-1},r)\subset E_i$. The value of $\tau$ obtained in the proposition, see (6.3), is given for each i by:

$$\begin{aligned} \tau _i=\max \left\{ \frac{N\lambda _2^{E_i}}{2e(\lambda _2^{E_i}-\lambda _1^{E_i})}\left( \frac{2C_{\infty }\Vert \psi _{i-1}\Vert _{L^2(E_i)}}{\left\langle \psi _{i-1}, \phi ^{E_i}_1 \right\rangle _{L^2(E_i)} \inf \limits _{B(x_i,r)}{\phi ^{E_i}_1}}\right) ^{2/N}, \right. \\ \left. \frac{1}{\lambda -\lambda _1^{E_i}} \log \left( \frac{2\gamma \max \limits _{B(x_i,r)}\phi _1^{B(x_i,r)}}{\left\langle \psi _{i-1}, \phi ^{E_i}_1 \right\rangle _{L^2(E_i)}\inf \limits _{B(x_i,r)}{\phi ^{E_i}_1}} \right) \right\} >0. \end{aligned}$$

Observe that since all sets $E_i$ are rigid transformations of the same set $E_0$, then $\lambda _k^{E_i}=\lambda _k^{E_0}$, for all $k=1,2,..$, and the embedding constant $C_\infty$ is independent of i. Moreover since $B(x_i,r)\subset E_i^r\equiv \{x\in E_i: d(x,\partial E_i)>r\}$ and the maps $T_i$ are rigid motions, then $\inf \limits _{B(x_i,r)}\phi _1^{E_i}\ge \inf \limits _{E_i^r}\phi _1^{E_i}=\inf \limits _{E_0^r}\phi _1^{E_0}$, which is independent of i. Moreover, $\Vert \psi _{i-1}\Vert _{L^2(E_i)}=1$ and $\max \limits _{B(x_i,r)}\phi _1^{B(x_i,r)}=\max \limits _{B(0,r)}\phi _1^{B(0,r)}>0$ which are also independent of i. Also,

$$\begin{aligned} \left\langle \psi _{i-1}, \phi ^{E_i}_1 \right\rangle _{L^2(E_i)}=\int _{E_i}\psi _{i-1}\phi _1^{E_i}\ge \left( \int _{B(x_{i-1},r)}\psi _{i-1}\right) \inf \limits _{B(x_{i-1},r)}\phi _{1}^{E_i} \\ \ge \left( \int _{B(0,r)}\psi \right) \inf \limits _{E_i^r}\phi _{1}^{E_i}=\left( \int _{B(0,r)}\psi \right) \inf \limits _{E_i^0}\phi _{1}^{E_0}>0, \end{aligned}$$

which is also independent of i. Therefore, if we define

$$\begin{aligned} \tau= & {} \max \left\{ \frac{N\lambda _2^{E_0}}{2e(\lambda _2^{E_0}-\lambda _1^{E_0})}\left( \frac{2C_\infty }{\int _{B(0,r)}\psi (\inf \limits _{E_0^r}{\phi ^{E_0}_1})^2}\right) ^{2/N}, \frac{1}{\lambda -\lambda _1^{E_0}} \log \left( \frac{2\gamma \max \limits _{B(0,r)}\psi }{\int _{B(0,r)}\psi (\inf \limits _{E_0^r}{\phi ^{E_0}_1})^2} \right) \right\} \\> & {} 0, \end{aligned}$$

we can apply the previous proposition for intervals of time greater than or equal to $\tau$.

Hence, if for $i=1,2\ldots$, we have $t_{i}-t_{i-1}\ge \tau$ and we consider problems

$$\begin{aligned} \left\{ \begin{array}{ll} v_{i_t}-\Delta v_i = \lambda v_i, &{} x\in E_{i}, \; t\in (t_{i-1},t_{i}],\\ v_i(t,x) = 0, &{} x\in \partial E_{i}, \; t\in (t_{i-1},t_{i}],\\ v_i(t_{i-1},x)=\gamma ^{i-1}\psi _{i-1}, &{} x\in E_{i}, \end{array}\right. \end{aligned}$$

(6.4)

then we have that $v_i(t_i,x)\ge \gamma ^i\psi _{i}(x)$ in $B(x_{i+1},r)$. If we extend the functions $v_i(t,\cdot )$ by 0 to all $\Omega$ and define the function v(t, x) as $v(t,x)=v_i(t,x)$ $t\in [t_{i-1},t_i)$, then by comparison it is not difficult to see that

$$\begin{aligned} v(t,x)\le u(t,x,t_0,\psi _0(\cdot )),\quad t\ge t_0. \end{aligned}$$

Indeed, if $t\in [t_0,t_1]$ we have that $v_1(t,x)$ is a subsolution of u(t, x) and therefore $u(t_1,x)\ge v_1(t_1,x)\ge \gamma \psi _1(x)$. Observing that $\gamma \psi _1(x)$ is the initial condition of $v_2$, we get again that in $[t_1,t_2]$ the function $v_2(t,x)$ is a subsolution of u(t, x). Therefore $u(t_2,x)\ge v_2(t_2,x)\ge \gamma ^2 \psi _2(x)$. By induction it is not difficult to see that in $[t_{i-1},t_i]$ we have that $v_i(t,x)$ is a subsolution of u(t, x) and therefore $u(t_i,x)\ge \gamma ^i\psi _i(x)$. This means that

$$\begin{aligned} u(t,x)\ge v(t,x),\qquad t\ge t_0, \quad x\in \Omega , \end{aligned}$$

and in particular

$$\begin{aligned} \Vert u(t_i,x)\Vert _{L^\infty (\Omega )}\ge \gamma ^i\Vert \psi _i\Vert _{L^\infty (\Omega )}=\gamma ^i \Vert \psi \Vert _{L^\infty (B(0,r))}\rightarrow \infty , \hbox { as } i\rightarrow \infty . \end{aligned}$$

Moreover, with Corollary 6.3 we can actually see that $\Vert v_i(t,x)\Vert _{L^\infty (\Omega )}\ge \gamma ^{i-1}\rho$ for each $t\in [t_{i-1},t_i]$. Hence, we get

$$\begin{aligned} \lim _{t\rightarrow +\infty } \Vert u(t,\cdot )\Vert _{L^\infty (\Omega )}=+\infty . \end{aligned}$$

This concludes the proof of the theorem. $\square$

Remark 6.4

Observe that if we have the situation where $E_i\equiv E_0$ for all $i=1,2,\ldots$, condition (6.1) is read as $E_0\subset K(t)$ for all $t\ge t_0$ and since we are assuming $\lambda >\lambda _0(E_0)$ we can show that the solutions are unbounded. For this, notice that $n(t,x)\le {\tilde{n}}(x)$ where ${\tilde{n}}(x)=\Vert n\Vert _{L^\infty }\chi _{\Omega \setminus E_0}(x)$. Via comparison with the autonomous problems with ${\tilde{n}}$, we show that the solutions are all unbounded.

In the line of the above remark, in the next result we show that the assumption that $E_0\subset K(t)$ for all t can be weakened. For this, let us consider problem (1.1) where n satisfy (N) and K(t) satisfies

$$\begin{aligned} \left\{ \begin{array}{lr} K(t)\supset K_1, &{} t\in (T_{0,i},T_{1,i}], x\in \Omega ,\\ K(t)\supset K_0, &{} t\in (T_{1,i},T_{0,i+1}], x\in \Omega , \end{array}\right. \end{aligned}$$

(6.5)

for some unbounded time sequences $t_0\le T_{0,i}\le T_{1,i}\le T_{0,i+1} \rightarrow \infty$ as $i\rightarrow \infty$ and for some sets $K_0=\bar{\Omega _0}$, $K_1=\bar{\Omega _1}$ with $\Omega _0\subset \Omega _1$, smooth open sets.

We can show the following

Proposition 6.5

With the notation above, if we assume

$$\begin{aligned} \lambda _0(K_1)<\lambda<\lambda _0(K_0)<\infty . \end{aligned}$$

(6.6)

and that there exist $\tau$ and $\eta$ such that for the time sequences $t_0\le T_{0,i}< T_{1,i} <T_{0,i+1}\rightarrow \infty$ satisfy

$$\begin{aligned} \left\{ \begin{array}{lr} \left| T_{1,i}-T_{0,i}\right| > \tau , &{} i\in \mathbb {N},\\ \left| T_{0,i+1}-T_{1,i}\right| \le \eta , &{} i\in \mathbb {N}, \end{array}\right. \end{aligned}$$

(6.7)

then, if $\tau$ is sufficiently large, the solution u(t, x) of (1.1) is unbounded.

Proof

We want to exhibit a subsolution of u(t, x) that grows up as time goes to infinity. Without any loss of generaliy, we can assume that

$$\begin{aligned} \lambda \in (\lambda ^{\Omega _1}_1, \lambda ^{\Omega _1}_2), \end{aligned}$$

(6.8)

since, if greater than or equal to $\lambda ^{\Omega _1}_2$, we can take a smaller $\lambda$ which verifies (6.8) and it would result in a subsolution of the original solution. Let us consider the following family of problems for each $i\ge 1$:

$$\begin{aligned} \left\{ \begin{array}{lll} U_{i_t}(t,x)-\Delta U_i(t,x) = \lambda U_i(t,x),\ \ {} &{}t\in (T_{0,i},T_{1,i}],&{} x\in \Omega _1,\\ U_i(t,x) = 0,&{}t\in (T_{0,1},T_{1,i}],&{} x\in \partial \Omega _1,\\ U_i(T_{0,i},x) = \phi _1^{\Omega _0}, &{} &{} x\in \Omega _1, \end{array}\right. \end{aligned}$$

(6.9)

$$\begin{aligned} \left\{ \begin{array}{lll} W_{i_t}(t,x)-\Delta W_i(t,x) = \lambda W_i(t,x),\ \ {} &{}t\in (T_{1,i},T_{0,i+1}],&{} x\in \Omega _0,\\ W_i(t,x) = 0,&{}t\in (T_{1,i},T_{0,i+1}],&{} x\in \partial \Omega _0,\\ W_i(T_{1,i},x)=\phi _1^{\Omega _0}, &{} &{} x\in \Omega _0. \end{array}\right. \end{aligned}$$

(6.10)

Observe that $U_i$ is posed in the large set $\Omega _1$ and therefore the solution there tends to grow, while $W_i$ is posed in the smaller set $\Omega _0$ where the solution tends to decay. As a matter of fact,

$$\begin{aligned} W_i(t)=e^{(\lambda -\lambda _1^{\Omega _0})(t-T_{1,i})}\phi _1^{\Omega _0}, \end{aligned}$$

and if we define

$$\begin{aligned} \alpha = e^{(\lambda -\lambda _1^{\Omega _0})\eta }\in (0,1), \end{aligned}$$

we have

$$\begin{aligned} W_i(t)\ge \alpha \phi _1^{\Omega _0}, \qquad t\in [T_{1,i},T_{0,i+1}]. \end{aligned}$$

(6.11)

Applying now Proposition 6.2 to $U_i$, with $E=\Omega _1$, $D=\Omega _0$, and $U_{i,0}=\phi _1^{\Omega _0}$, we have that if $\gamma >1$ and define

$$\begin{aligned} \tau= & {} \max \{ \frac{N\lambda _2^{\Omega _1}}{2e(\lambda _2^{\Omega _1}-\lambda _1^{\Omega _1})}\left( \frac{2C_\infty \Vert \phi _1^{\Omega _0}\Vert _{L^2(\Omega _1)}}{\left\langle \phi _1^{\Omega _0}, \phi ^{\Omega _1}_1 \right\rangle _{L^2(\Omega _1)} \inf \limits _{\Omega _0}{\phi ^{\Omega _1}_1}}\right) ^{2/N}, \\{} & {} \frac{1}{\lambda -\lambda _1^{\Omega _1}} \log \left( \frac{2\frac{\gamma }{\alpha }\max \limits _{\Omega _0}\phi _1^{\Omega _0}}{\left\langle \phi _1^{\Omega _0}, \phi ^{\Omega _1}_1 \right\rangle _{L^2(\Omega _1)} \inf \limits _{\Omega _0}{\phi ^{\Omega _1}_1} }\right) \}, \end{aligned}$$

which does not depend on i. Then, choosing $T_{1,i}-T_{0,i}\ge \tau$, we have

$$\begin{aligned} U_i(t)\ge \frac{\gamma }{\alpha } \phi _1^{\Omega _0}, \quad t\in [T_{0,i}+\tau , T_{1,i}]. \end{aligned}$$

(6.12)

Also, notice that comparing problem (6.9)) with the same problem but posed in $\Omega _0\subset \Omega _1$, we have that

$$\begin{aligned} U_i(t)\ge e^{(\lambda -\lambda _1^{\Omega _0})t}\phi _1^{\Omega _0}\ge e^{(\lambda -\lambda _1^{\Omega _0})\tau }\phi _1^{\Omega _0}, \quad t\in [T_{0,i}, T_{0,i}+\tau ]. \end{aligned}$$

(6.13)

Putting together (6.12) and (6.13), we get

$$\begin{aligned} U_i(t)\ge \rho \phi _1^{\Omega _0},\qquad t\in [T_{0,i},T_{1,i}] \end{aligned}$$

(6.14)

where $\rho =\min \{\frac{\gamma }{\alpha },e^{(\lambda -\lambda _1^{\Omega _0})\tau }\}$. Let us construct the function Z(t, x) where for every $i\in \mathbb {N}$:

$$\begin{aligned}Z(t,\cdot )=\left\{ \begin{array}{lr} \gamma ^{i-1} U_i(t,\cdot ), &{} t\in (T_{0,i},T_{1,i}],\\ \frac{\gamma ^{i}}{\alpha }W_i(t,\cdot ), &{} t\in (T_{1,i},T_{0,i+1}],\\ \end{array}\right. \end{aligned}$$

where we assume $U_i, W_i$ are extended by 0 to the whole domain $\Omega$. With the analysis above, we can conclude that Z(t, x) is a subsolution of u(t, x) and with (6.11) and (6.14) we get that $\lim _{t\rightarrow +\infty }\Vert Z(t,\cdot )\Vert _{L^\infty }=+\infty$, which proves the result. $\square$