A Cahn–Hilliard equation with a proliferation term for biological and chemical applications

Abstract

In this paper, we are interested in the study of the solution to a generalization of the Cahn–Hilliard equation endowed with Neumann boundary conditions. This model has, in particular, applications in biology and in chemistry. We show that the solutions blow up in finite time or exist globally in time. We further prove that the relevant, from a biological and a chemical point of view, solutions converge to a constant as time goes to infinity. We finally give some numerical simulations which confirm the theoretical results.

Keywords

Cahn–Hilliard equation proliferation term blow up convergence to a steady state simulations

1. Introduction

The Cahn–Hilliard equation is very important in materials science. It describes important qualitative features of two phase systems, related with phase separation processes. In materials science, this pattern formation is referred to as the microstructures of the material and these microstructures are highly influential in determining many of the properties of the material, such as strength, hardness, and conductivity. The Cahn–Hilliard model is rather broad ranged in its evolutionary scope; it can serve as a good model for many systems during early times, it can give a reasonable qualitative description for these systems during intermediate times, and it can serve as a good model for even more systems at late times. Often, the late time evolution is so slow that the pattern formation becomes effectively frozen into the system over time scales of interest and, hence, it is the long time behavior of the system which is seen in practice. We refer the reader to, e.g., [2,3,6,12], and [21] for more details.

The Cahn–Hilliard equation also appears in the modeling of many other phenomena. These include population dynamics [9], bacterial films [17], thin films [23,25], image processing [4,11], and even the rings of Saturn [26].

In [7], the authors studied the following model proposed in [16]: $\frac{\partial u}{\partial t} + Δ^{2} u - Δ f (u) + g (u) = 0,$ (1.1) where $g (s) = s (s - 1)$ (1.2) and $f (s) = {(s - \frac{1}{2})}^{3} - (s - \frac{1}{2}) .$ (1.3) Equation (1.1) models, e.g., the clustering of malignant brain tumor cells in two space dimensions, see [16].

Furthermore, the authors in [27] proposed the following related equation: $\frac{\partial u}{\partial t} + M Δ (2 κ Δ u - f (u)) + u^{2} - P (1 - u) = 0$ (1.4) to model the formation of adsorbate islands on a proposed surface (in the modeling of a two-dimensional liquid–gas). Here, u is the distance from the critical coverage $u^{*} = u - \frac{1}{2}$ , M is the surface mobility which is supposed to be constant, $u^{2} - P (1 - u)$ is the nonlinear reaction rate, $P = \frac{κ_{ad} s}{κ_{des}} p$ is the reduced pressure of the gas phase, p being the pressure of the gas above the absorbed layer, and s is the stinking coefficient. Furthermore, $κ_{ad}$ and $κ_{des}$ are constants which represent the adsorption rate and the desorption one, respectively. Finally, F is the free energy of the corresponding homogeneous system, $f (s) = F^{'} (s) = a_{0} (\frac{T}{T_{c}} - 1) (s - \frac{1}{2}) + \frac{4}{3} \frac{T}{T_{c}} {(s - \frac{1}{2})}^{3},$ (1.5) where $a_{0} = 4 κ_{B} T_{c}$ and $κ = κ_{B} T_{c} ξ_{0}^{2}$ , $ξ_{0}$ is a phenomenological length related to the range of the interactions, and T and $T_{c}$ are the absolute temperature and a critical one, respectively.

Actually, in what follows, we will, for simplicity, set all physical constants equals to 1 and we will solve the problem in space dimension $N ⩽ 3$ , i.e., we consider in Section 2 the generalized Cahn–Hilliard equation $\frac{\partial u}{\partial t} + Δ^{2} u - Δ f (u) + g (u) = 0,$ (1.6) where $g (s) = α s^{2} + β s + γ, α > 0, β, γ \in R .$ (1.7) We note that, in two space dimensions, if $α = 1$ and $γ = - β$ ( $β \in R$ ), the equation models, e.g., the literal attractive interactions between adsorbed molecules which may induce a transition in the chemisorbed overlayer, see [27]. Furthermore, if $γ = 0$ and $β = - α$ ( $α > 0$ ), the equation models, e.g., the clustering of malignant brain tumor cells, see [16]. Finally, if $α = γ = 0$ and $β > 0$ , the equation (named Cahn–Hilliard–Oono) accounts for long-ranged (nonlocal) interactions in the phase separation process (see [22,28]; see also [18] for the study of the limit dynamics as β goes to zero, [5] for the study of the local existence with a logarithmic nonlinear term and [10] for the study of the global existence of the nonlocal equation with a transport term).

We further take, for simplicity, $f (s) = s^{3} - s$ (1.8) and we set $Θ = β^{2} - 4 α γ,$ (1.9) Θ being the discriminant of g (g is defined here as in (1.7)).

A third related model, proposed in [1], reads $\frac{\partial u}{\partial t} + Δ^{2} u - Δ f (u) + g (x, u) = 0,$ (1.10) where $g (x, s) = \frac{α}{2} (s + 1) - β {(1 - s)}^{2} {(1 + s)}^{2} + σ (x, t),$ (1.11) α and β being constants which represent the growth coefficient and the death one, respectively. Here, the term $σ (x, t)$ is used as an artificial growth term which allows us to manufacture a desired PDE solution, but it could also serve to model biologically relevant phenomena. Equation (1.10) has applications in biology and, more specifically, in tumor growth and wound healing. The uncontrolled tumor growth of abnormal cells often results in cancer. The morphological evolution of a growing solid tumor is the result of many factors, including cell–cell and cell–matrix adhesion, mechanical stress, cell motility, and the degree of heterogeneity of cell proliferation.

In that follows, we will, for simplicity, set $σ \equiv 0$ and we will solve the problem in space dimension $N ⩽ 3$ , i.e., we consider in Section 3 the generalized Cahn–Hilliard equation $\frac{\partial u}{\partial t} + Δ^{2} u - Δ f (u) + g (u) = 0,$ (1.12) where $g (s) = α s - β s^{q} {(s - 1)}^{q}, q ⩾ 2 .$ (1.13) We further take $f (s) = \sum_{i = 0}^{2 p + 1} a_{i} s^{i}, a_{2 p + 1} > 0, p ⩾ 1$ (1.14) (in particular, we will sometimes need compatibility conditions on p and q).

Problems (1.6) and (1.10), endowed with Dirichlet boundary conditions (and with more general nonlinear terms f and g), are considered in [19]; see also [14]. There, the well-posedness and the existence of the finite-dimensional global attractor are established. Furthermore, Eq. (1.6), when the nonlinear term g has odd degree and endowed with Neumann boundary conditions (and with more general nonlinear terms f), is considered in [14]. In that case, we also have the well-posedness and the existence of the finite-dimensional global attractor.

In this paper, Eqs (1.6) and (1.10) are endowed with Neumann boundary conditions, $\frac{\partial u}{\partial ν} = \frac{\partial Δ u}{\partial ν} = 0 on Γ,$ where Γ is the boundary of the domain Ω occupied by the system (we assume that it is a bounded and regular domain of $R^{N}$ ) and ν is the unit outer normal vector.

We set $v = u - ⟨ u ⟩,$ (1.15) where $⟨ \cdot ⟩ = \frac{1}{Vol (Ω)} \int_{Ω} \cdot d x .$ (1.16) Then, assuming that g is defined as in (1.7), with $α = - β = 1$ and $γ = 0$ , the authors of [7] proved that v is bounded and $⟨ u ⟩$ can blow up in finite time. Furthermore, when $⟨ u ⟩$ is bounded, the solution exists globally in time and is dissipative. They finally proved that, if $u (t) \in [0, 1]$ , $\forall t ⩾ 0$ , then u tends to 1 as time goes to infinity.

We note that Neumann boundary conditions are crucial here and we have global in time existence for the same problem with Dirichlet boundary conditions (see, e.g., [19]).

In Section 2, we take g as in (1.7) with $α > 0$ and $β, γ \in R$ . We prove that $⟨ u ⟩$ can blow up in finite time in general. Furthermore, when $⟨ u ⟩$ is bounded, the solution exists globally in time and is dissipative. We then prove that, if $Θ > 0$ and u is a solution such that $u (t) \in [\frac{- β - \sqrt{Θ}}{2 α}, \frac{- β + \sqrt{Θ}}{2 α}]$ or $u (t) \in [\frac{- β + \sqrt{Θ}}{2 α}, a]$ , $\forall t ⩾ 0$ , where $a > \frac{- β + \sqrt{Θ}}{2 α}$ is arbitrary, then u tends to $\frac{- β + \sqrt{Θ}}{2 α}$ as time goes to infinity. Finally, if $Θ = 0$ and u is a solution such that $u (t) \in [\frac{- β}{2 α}, a]$ , $\forall t ⩾ 0$ , where $a > \frac{- β}{2 α}$ , then u tends to $\frac{- β}{2 α}$ as time goes to infinity in some cases.

In Section 3, we take g as in (1.13). We prove that, if $α = 2 β$ , $β > 0$ (resp., $β < 0$ ) and $q = 2$ and u is a solution such that $u (t) \in [0, 2]$ , then u tends to 0 (resp., to 2) as time goes to infinity. Finally, we can again prove that $⟨ u ⟩$ can blow up in finite time in general. Note that, if $β > 0$ (resp., $β < 0$ ), the blow up solution tends to $+ \infty$ (resp., to $- \infty$ ).

We finally give some numerical simulations which confirm these results. They also show that, for g defined as in (1.7), even when the initial datum belongs to the relevant interval $[\frac{- β - \sqrt{Θ}}{2 α}, \frac{- β + \sqrt{Θ}}{2 α}]$ , we can have blow up in finite time.

Throughout the paper, the same letter c (and, sometimes, $c^{'}$ ) denotes constants which may vary from line to line.

2. A chemisorbed overlayer and tumor growth model

We consider the following initial and boundary value problem: $\begin{matrix} \frac{\partial u}{\partial t} + Δ^{2} u - Δ f (u) + g (u) = 0, & (2.1) \\ \frac{\partial u}{\partial ν} = \frac{\partial Δ u}{\partial ν} = 0 on Γ, & (2.2) \\ u |_{t = 0} = u_{0}, & (2.3) \end{matrix}$ where f and g are defined in (1.7) and (1.8), respectively.

Integrating (2.1) over Ω, we have $\frac{d}{d t} \int_{Ω} u d x + \int_{Ω} g (u) d x = 0,$ (2.4) hence $\frac{d}{d t} ⟨ u ⟩ + ⟨ g (u) ⟩ = 0 .$ (2.5)

We rewrite (2.5) in the form $\frac{d ⟨ u ⟩}{d t} + g (⟨ u ⟩) = g (⟨ u ⟩) - ⟨ g (u) ⟩ .$ Noting that $g (⟨ u ⟩) - ⟨ g (u) ⟩ = - α ⟨ v^{2} ⟩,$ where $v = u - ⟨ u ⟩$ , we obtain $\frac{d ⟨ u ⟩}{d t} + g (⟨ u ⟩) = - α ⟨ v^{2} ⟩ .$ (2.6)

We first have the following proposition.

Proposition 2.1.

For every $u_{0} \in L^{2} (Ω)$ , there exists $T_{0} = T_{0} (∥ u_{0} ∥) > 0$ and a unique solution u to ( 2.1 )–( 2.3 ) such that $u \in C ([0, T_{0}); L^{2} (Ω)) \cap L^{2} (0, T; H^{2} (Ω))$ , $\forall T < T_{0}$ .

Proof.

See [20] and [24]. □

Remark 2.2.

In the case when g has two distinct zeros, i.e., $Θ > 0$ (Θ is defined in (1.9)), we give below an example which proves that, even if we start in the physically relevant interval, namely $[\frac{- β - \sqrt{Θ}}{2 α}, \frac{- β + \sqrt{Θ}}{2 α}]$ , the solution can leave this interval. We consider the following equation: $\frac{\partial u}{\partial t} + \frac{\partial^{4} u}{\partial x^{4}} - \frac{\partial^{2}}{\partial x^{2}} ({(u - \frac{1}{2})}^{3} - (u - \frac{1}{2})) + α u^{2} + β u + γ = 0 .$ (2.7) We assume that $Θ > 0$ , take $u_{0} (x) = \frac{- β + \sqrt{Θ}}{2 α} - {(x - \frac{1}{2})}^{4}$ in a neighborhood of $\frac{1}{2}$ , and extend $u_{0}$ by a smooth function defined in $Ω = (0, 1)$ , with values in $[\frac{- β - \sqrt{Θ}}{2 α}, \frac{- β + \sqrt{Θ}}{2 α}]$ . We note that $u_{0}^{'} (\frac{1}{2}) = u_{0}^{″} (\frac{1}{2}) = 0$ and $u_{0}^{(4)} (\frac{1}{2}) = - 24$ . Furthermore, $\frac{\partial^{2}}{\partial x^{2}} ({(u_{0} - \frac{1}{2})}^{3} - (u_{0} - \frac{1}{2})) |_{x = \frac{1}{2}} = 0 .$ (2.8) We thus have $\frac{\partial u}{\partial t} (\frac{1}{2}, 0) = 24 > 0 .$ (2.9) Noting finally that $\begin{array}{rclr} u (\frac{1}{2}, t) & = & u (\frac{1}{2}, 0) + t \frac{\partial u}{\partial t} (\frac{1}{2}, 0) + o (t) \\ = & \frac{- β + \sqrt{Θ}}{2 α} + 24 t + o (t) \\ > & \frac{- β + \sqrt{Θ}}{2 α}, & (2.10) \end{array}$ for $t > 0$ small, we see that u cannot stay in $[\frac{- β - \sqrt{Θ}}{2 α}, \frac{- β + \sqrt{Θ}}{2 α}]$ . Similarly, if we take $u_{0} (x) = \frac{- β - \sqrt{Θ}}{2 α} + {(x - \frac{1}{2})}^{4}$ in a neighborhood of $\frac{1}{2}$ and extend it as above, we have $\frac{\partial u}{\partial t} (\frac{1}{2}, 0) = - 24 < 0,$ (2.11) hence $\begin{array}{rclr} u (\frac{1}{2}, t) & = & \frac{- β - \sqrt{Θ}}{2 α} - 24 t + o (t) \\ < & \frac{- β - \sqrt{Θ}}{2 α}, & (2.12) \end{array}$ for $t > 0$ small, and we see that u cannot stay in $[\frac{- β - \sqrt{Θ}}{2 α}, \frac{- β + \sqrt{Θ}}{2 α}]$ .

Lemma 2.3.

Let u be a solution to ( 2.1 )–( 2.3 ). Then, $⟨ u ⟩$ is bounded from above.

Proof.

If $β < 0$ , we note that $- 2 β s ⩽ α s^{2} + \frac{β^{2}}{α} .$ It thus follows from (2.6) that $\frac{d ⟨ u ⟩}{d t} - β ⟨ u ⟩ ⩽ \frac{β^{2}}{α} - γ,$ (2.13) which yields, owing to Gronwall’s lemma, $⟨ u (t) ⟩ ⩽ (⟨ u_{0} ⟩ - \frac{γ}{β}) e^{β t} - \frac{β}{α} + \frac{γ}{β}, t ⩾ 0,$ (2.14) and $⟨ u ⟩$ is bounded from above.

If $β = 0$ , we further note that $2 s ⩽ α s^{2} + \frac{1}{α} .$ (2.15) It thus follows from (2.6) that $\frac{d}{d t} ⟨ u ⟩ + 2 ⟨ u ⟩ ⩽ \frac{1}{α} - γ,$ (2.16) which yields, owing to Gronwall’s lemma, $⟨ u (t) ⟩ ⩽ (⟨ u_{0} ⟩ + \frac{γ}{2} - \frac{1}{2 α}) e^{- 2 t} + \frac{1}{2 α} - \frac{γ}{2}, t ⩾ 0,$ (2.17) and $⟨ u ⟩$ is bounded from above.

Finally, if $β > 0$ , we have, owing to (2.5), $\frac{d ⟨ u ⟩}{d t} + β ⟨ u ⟩ + γ ⩽ 0$ (2.18) and, also by Gronwall’s lemma, we obtain $⟨ u (t) ⟩ ⩽ (⟨ u_{0} ⟩ + \frac{γ}{β}) e^{- β t} - \frac{γ}{β}, t ⩾ 0,$ (2.19) and $⟨ u ⟩$ is bounded from above. □

Lemma 2.4.

Let u be a solution to ( 2.1 )–( 2.3 ). Then, if $β < 0$ and $⟨ u (T) ⟩ < - \frac{γ}{β}$ , for some $T > 0$ (and, in particular, if $⟨ u_{0} ⟩ < - \frac{γ}{β}$ ), there exists $t_{0} ⩾ T$ such that $⟨ u (t) ⟩ < 0, \forall t ⩾ t_{0},$ and, if the solution exists globally in time, $lim_{t \to + \infty} ⟨ u (t) ⟩ = - \infty .$

Proof.

It follows from (2.5) that $\frac{d ⟨ u ⟩}{d t} + β ⟨ u ⟩ + γ ⩽ 0,$ (2.20) which yields, owing to Gronwall’s lemma, $⟨ u (t) ⟩ ⩽ e^{- β (t - T)} (⟨ u (T) ⟩ + \frac{γ}{β}) - \frac{γ}{β} .$ (2.21) Therefore, if $⟨ u (T) ⟩ < - \frac{γ}{β},$ (2.22) then $\exists t_{0} ⩾ T$ such that $⟨ u (t) ⟩ < 0, \forall t ⩾ t_{0}$ (2.23) and, moreover, if the solution exits globally in time, $lim_{t \to + \infty} ⟨ u (t) ⟩ = - \infty . □$ (2.24)

Lemma 2.5.

Let u be a solution to ( 2.1 )–( 2.3 ). We have:

If $Θ < 0$ , then, for all $u_{0}$ , there exists $t_{0} ⩾ 0$ such that $⟨ u (t) ⟩ < 0, \forall t ⩾ t_{0},$ and, if the solution exists globally in time, $lim_{t \to + \infty} ⟨ u (t) ⟩ = - \infty .$

If $Θ = 0$ and $⟨ u_{0} ⟩ < 0$ , then $⟨ u (t) ⟩ < 0, \forall t ⩾ 0 .$

Proof.

It follows from (2.5) that ${\frac{d ⟨ u ⟩}{d t} + α ⟨ (u + \frac{β}{2 α})}^{2} ⟩ - \frac{Θ}{4 α} = 0 .$ (2.25) Setting $w = u + \frac{β}{2 α}$ , we rewrite (2.25) in the form $\frac{d ⟨ w ⟩}{d t} + α ⟨ w^{2} ⟩ - \frac{Θ}{4 α} = 0 .$ (2.26) It follows from (2.26) that $\frac{d ⟨ w ⟩}{d t} ⩽ \frac{Θ}{4 α},$ hence $⟨ w (t) ⟩ ⩽ ⟨ w_{0} ⟩ + \frac{Θ}{4 α} t,$ (2.27) which yields $⟨ u (t) ⟩ ⩽ ⟨ u_{0} ⟩ + \frac{Θ}{4 α} t .$ (2.28)

If $Θ < 0$ , then, $\forall u_{0}$ , $\exists t_{0}$ such that $⟨ u (t) ⟩ < 0, \forall t ⩾ t_{0},$ (2.29) and, if the solution exists globally in time, $lim_{t \to + \infty} ⟨ u (t) ⟩ = - \infty .$ (2.30)

Now, assuming that $Θ = 0$ , we have, owing to (2.28), $⟨ u (t) ⟩ ⩽ ⟨ u_{0} ⟩, \forall t ⩾ 0 .$ (2.31) Therefore, if $⟨ u_{0} ⟩ < 0,$ (2.32) then $⟨ u (t) ⟩ < 0, \forall t ⩾ 0 . □$ (2.33)

Proposition 2.6.

Let u be a solution to ( 2.1 )–( 2.3 ). Then, there exists a monotone increasing function Q such that ${∥ u (t) - ⟨ u (t) ⟩ ∥}^{2} ⩽ Q (∥ u_{0} - ⟨ u_{0} ⟩ ∥), \forall t ⩾ 0 .$

Proof.

Setting $v = u - ⟨ u ⟩$ , we rewrite problem (2.1)–(2.3) in the form $\begin{matrix} \frac{\partial v}{\partial t} + Δ^{2} v - Δ f (v + ⟨ u ⟩) + g (v + ⟨ u ⟩) - ⟨ g (v + ⟨ u ⟩) ⟩ = 0, & (2.34) \\ \frac{\partial v}{\partial ν} = \frac{\partial Δ v}{\partial ν} = 0 on Γ, & (2.35) \\ v |_{t = 0} = v_{0} = u_{0} - ⟨ u_{0} ⟩ . & (2.36) \end{matrix}$

We multiply (2.34) by ${(- Δ)}^{- 1} v$ and have $\begin{array}{l} \frac{1}{2} \frac{d}{d t} {∥ v ∥}_{- 1}^{2} + {∥ \nabla v ∥}^{2} + ((f (v + ⟨ u ⟩) - f (⟨ u ⟩), v)) \\ + ((g (v + ⟨ u ⟩) - g (⟨ u ⟩), {(- Δ)}^{- 1} v)) = 0 . & (2.37) \end{array}$ Here, $\begin{array}{l} ((f (v + ⟨ u ⟩) - f (⟨ u ⟩), v)) \\ = \int_{Ω} (v^{4} + 3 v^{3} ⟨ u ⟩ + 3 v^{2} {⟨ u ⟩}^{2}) d x - {∥ v ∥}^{2} \\ ⩾ \int_{Ω} (v^{4} + 3 v^{2} {⟨ u ⟩}^{2}) d x - 3 \int_{Ω} | v^{3} ⟨ u ⟩ | d x - {∥ v ∥}^{2} \\ ⩾ \frac{1}{8} \int_{Ω} (v^{4} + v^{2} {⟨ u ⟩}^{2}) d x - {∥ v ∥}^{2}, \end{array}$ owing to Young’s inequality (e.g., $3 a b ⩽ \frac{7}{8} a^{2} + \frac{18}{7} b^{2}$ , $a, b ⩾ 0$ ). Furthermore, $\begin{array}{rcl} | ((g (v + ⟨ u ⟩) - g (⟨ u ⟩), {(- Δ)}^{- 1} v)) | & = & | ((α v^{2} + 2 α v ⟨ u ⟩ + β v, {(- Δ)}^{- 1} v)) | \\ ⩽ & \frac{1}{16} \int_{Ω} (v^{4} + v^{2} {⟨ u ⟩}^{2}) d x + c {∥ v ∥}^{2}, \end{array}$ which yields $\frac{d}{d t} {∥ v ∥}_{- 1}^{2} + {∥ \nabla v ∥}^{2} + \frac{1}{8} \int_{Ω} (v^{4} + v^{2} {⟨ u ⟩}^{2}) d x ⩽ c {∥ v ∥}^{2} .$ (2.38) Therefore, $\frac{d}{d t} {∥ v ∥}_{- 1}^{2} + {∥ \nabla v ∥}^{2} + \frac{1}{16} \int_{Ω} (v^{4} + v^{2} {⟨ u ⟩}^{2}) d x ⩽ c,$ (2.39) hence $\frac{d}{d t} {∥ v ∥}_{- 1}^{2} + c {∥ v ∥}_{- 1}^{2} ⩽ c^{'}, c > 0 .$ (2.40) It thus follows from Gronwall’s lemma that ${∥ v ∥}_{- 1}^{2} ⩽ e^{- c t} {∥ v_{0} ∥}_{- 1}^{2} + c^{'}, c > 0, t ⩾ 0 .$ (2.41)

We set ${\dot{H}}^{- 1} (Ω) = {ϕ \in H^{- 1} (Ω), {⟨ ϕ, 1 ⟩}_{H^{- 1} (Ω), H^{1} (Ω)} = 0}$ .

Let B be a bounded subset of ${\dot{H}}^{- 1} (Ω)$ and $t_{0}$ be such that $v_{0} \in B$ and $t ⩾ t_{0}$ implies $v (t) \in B_{0}$ , where $B_{0} = {ϕ \in {\dot{H}}^{- 1} (Ω), {∥ ϕ ∥}_{- 1}^{2} ⩽ 2 c^{'}}$ , $c^{'}$ being the constant in (2.41). We then deduce from (2.39) that, for $t ⩾ t_{0}$ , $\int_{t}^{t + r} {∥ \nabla v ∥}^{2} d s ⩽ c (r), \int_{t}^{t + r} d s \int_{Ω} (v^{4} + v^{2} {⟨ u ⟩}^{2}) d x ⩽ c (r), r > 0 fixed .$

Finally, multiplying (2.34) by v, it is easy to obtain, noting that $\begin{array}{rcl} | ((g (v + ⟨ u ⟩) - g (⟨ u ⟩), v)) | & = & | ((α v^{2} + 2 α v ⟨ u ⟩ + β v, v)) | \\ ⩽ & c \int_{Ω} (α^{2} | v |^{4} + α^{2} v^{2} {| ⟨ u ⟩ |}^{2}) d x + c^{'} (| β | + 1) {∥ v ∥}^{2} \\ ⩽ & c (\int_{Ω} (v^{4} + v^{2} {⟨ u ⟩}^{2}) d x + {∥ v ∥}^{2}), \end{array}$ the inequality $\frac{d}{d t} {∥ v ∥}^{2} + {∥ Δ v ∥}^{2} ⩽ c (\int_{Ω} (v^{4} + v^{2} {⟨ u ⟩}^{2}) d x + {∥ \nabla v ∥}^{2} + 1) .$ (2.42) It thus follows from (2.42) and the uniform Gronwall’s lemma that ${∥ v (t) ∥}^{2} ⩽ c, t ⩾ t_{0} + r,$ (2.43) where r is fixed and the constant c is independent of $v_{0}$ and t, hence, owing to (2.43) and integrating (2.39) and (2.42) between 0 and $t_{0} + r$ , ${∥ v (t) ∥}^{2} ⩽ Q (∥ v_{0} ∥), t ⩾ 0,$ (2.44) where the function Q is monotone increasing. □

Theorem 2.7.

If $Θ > 0$ and $⟨ u (T) ⟩ < - \frac{β + \sqrt{Θ}}{2 α}$ , for some $T ⩾ 0$ (and, in particular, if $⟨ u_{0} ⟩ < - \frac{β + \sqrt{Θ}}{2 α}$ ), then the solution to ( 2.1 )–( 2.3 ) blows up in finite time. Furthermore, the blow up time $T^{+}$ satisfies $T^{+} ⩽ T + \frac{1}{2 α z_{1}} ln (\frac{z (T) - z_{1}}{z (T) + z_{1}}),$ where $z_{1} = \frac{\sqrt{Θ}}{2 α}$ and $z (T) = ⟨ u (T) ⟩ + \frac{β}{2 α}$ .

Proof.

We can rewrite (2.6) in the form $\frac{d}{d t} ⟨ u ⟩ + α {⟨ u ⟩}^{2} + β ⟨ u ⟩ + γ = - α ⟨ v^{2} ⟩,$ (2.45) hence the ODE $y^{'} + α y^{2} + β y + γ = - α ⟨ v^{2} ⟩,$ (2.46) where $y = ⟨ u ⟩$ . Thus, $y^{'} + α {(y + \frac{β}{2 α})}^{2} + \frac{4 α γ - β^{2}}{4 α} = - α ⟨ v^{2} ⟩ .$ (2.47) We finally set $z = y + \frac{β}{2 α}$ and have $z^{'} + α z^{2} - \frac{Θ}{4 α} = - α ⟨ v^{2} ⟩ .$ (2.48)

Since $Θ > 0$ , we can rewrite (2.48) in the form $z^{'} + α z^{2} - \frac{Θ}{4 α} = c (t),$ (2.49) where $c (t) = - α ⟨ v^{2} ⟩$ is nonpositive and uniformly bounded. Noting that the solution to the Riccati ODE $z^{'} + α z^{2} - \frac{Θ}{4 α} = 0$ reads $z (t) = z_{1} + \frac{1}{{\frac{z (T) + z_{1}}{2 z_{1} (z (T) - z_{1})} e}^{2 α z_{1} (t - T)} - \frac{1}{2 z_{1}}}, t ⩾ T,$ where $z_{1} = \frac{\sqrt{Θ}}{2 α}$ , we conclude in view of the comparison principle. Indeed, we note that it follows from (2.44) that $∥ v ∥$ is bounded and $φ \mapsto ∥ φ - ⟨ φ ⟩ ∥ + ⟨ φ ⟩$ is a norm in $L^{2} (Ω)$ which is equivalent to the usual one. □

As a consequence of Theorem 2.7, (2.14), (2.17), and (2.19), we have the following corollary.

Corollary 2.8.

We assume that $Θ > 0$ . Let u be a solution to ( 2.1 )–( 2.3 ). Then, either u blows up in finite time or u exists globally in time and, $\forall t ⩾ 0$ , $- \frac{β + \sqrt{Θ}}{2 α} ⩽ ⟨ u (t) ⟩ ⩽ \{\begin{matrix} | ⟨ u_{0} ⟩ - \frac{γ}{β} | - \frac{β}{α} + \frac{γ}{β} & if β < 0, \\ | ⟨ u_{0} ⟩ + \frac{γ}{β} | - \frac{γ}{β} & if β > 0, \\ | ⟨ u_{0} ⟩ + \frac{γ}{2} - \frac{1}{2 α} | - \frac{γ}{2} + \frac{1}{2 α} & if β = 0 . \end{matrix}$

Theorem 2.9.

If $Θ ⩽ 0$ and $⟨ u (T) ⟩ < - \frac{β}{2 α}$ , for some $T ⩾ 0$ (and, in particular, if $⟨ u_{0} ⟩ < - \frac{β}{2 α}$ ), then the solution to ( 2.1 )–( 2.3 ) blows up in finite time. Furthermore, the blow up time $T^{+}$ satisfies $T^{+} ⩽ T - \frac{1}{α z (T)} .$

Proof.

Since $Θ ⩽ 0$ , we can rewrite (2.48) in the form $z^{'} + α z^{2} = c (t),$ (2.50) where $c (t) = - α ⟨ v^{2} ⟩ + \frac{Θ}{4 α}$ is nonpositive and uniformly bounded. Noting that the solution to the Riccati ODE $z^{'} + α z^{2} = 0$ reads $z (t) = \frac{1}{α (t - T) + \frac{1}{z (T)}}, t ⩾ T,$ we conclude, owing to the comparison principle. □

As a consequence of Theorem 2.9, (2.14), (2.17), and (2.19), we have the following corollary.

Corollary 2.10.

We assume that $Θ = 0$ . Let u be a solution to ( 2.1 )–( 2.3 ). Then, either u blows up in finite time or u exists globally in time and, $\forall t ⩾ 0$ , $- \frac{β}{2 α} ⩽ ⟨ u (t) ⟩ ⩽ \{\begin{matrix} | ⟨ u_{0} ⟩ - \frac{γ}{β} | - \frac{β}{α} + \frac{γ}{β} & if β < 0, \\ | ⟨ u_{0} ⟩ + \frac{γ}{β} | - \frac{γ}{β} & if β > 0, \\ | ⟨ u_{0} ⟩ + \frac{γ}{2} - \frac{1}{2 α} | - \frac{γ}{2} + \frac{1}{2 α} & if β = 0 . \end{matrix}$

Note that, as a consequence of Lemma 2.4 and Lemma 2.5, we also have the following corollaries.

Corollary 2.11.

We assume that $β < 0$ . Let u be a solution to ( 2.1 )–( 2.3 ). If $⟨ u_{0} ⟩ < - \frac{γ}{β}$ , then u blows up in finite time.

Corollary 2.12.

We assume that $Θ < 0$ . Let u be a solution to ( 2.1 )–( 2.3 ). Then, for all $u_{0}$ , u blows up in finite time.

We finally deduce from (2.14), (2.17), (2.19), (2.43), Corollary 2.8, and Corollary 2.10 the following one.

Corollary 2.13.

Let u be a global in time solution to ( 2.1 )–( 2.3 ). Then, u is dissipative in $L^{2} (Ω)$ .

Remark 2.14.

It is not difficult to also prove the dissipativity in $H^{1} (Ω)$ and $H^{2} (Ω)$ (see also [19,20], and [24]).

Remark 2.15.

We assume that $Θ > 0$ . We saw in Remark 2.2 that u can become smaller than $\frac{- β - \sqrt{Θ}}{2 α}$ even if $u_{0} \in [\frac{- β - \sqrt{Θ}}{2 α}, \frac{- β + \sqrt{Θ}}{2 α}]$ . Similarly, we can prove that $⟨ u ⟩$ can become smaller than $\frac{- β - \sqrt{Θ}}{2 α}$ (and blow up) even if $⟨ u_{0} ⟩ \in [\frac{- β - \sqrt{Θ}}{2 α}, \frac{- β + \sqrt{Θ}}{2 α}]$ . Indeed, assume that $⟨ u_{0} ⟩ = \frac{- β - \sqrt{Θ}}{2 α} and u_{0} \neq \frac{- β - \sqrt{Θ}}{2 α} .$ Then, it follows from (2.6) that, at $t = 0$ , $\frac{d}{d t} ⟨ u ⟩ < 0,$ meaning that $⟨ u ⟩$ is smaller than $\frac{- β - \sqrt{Θ}}{2 α}$ , for $t > 0$ , and blows up in finite time. Of course, an interesting question is whether $⟨ u ⟩$ can become smaller than $\frac{- β - \sqrt{Θ}}{2 α}$ even if $⟨ u_{0} ⟩ \in] \frac{- β - \sqrt{Θ}}{2 α}, \frac{- β + \sqrt{Θ}}{2 α}]$ . As a partial answer in this direction (see also the numerical simulations below), let us consider the Riccati ODE $z^{'} + α z^{2} - \frac{Θ}{4 α} = c_{0} .$ (2.51) Then, when $- \frac{Θ}{4 α} < c_{0} < 0$ , the solution to this equation reads $z (t) = c_{1} + \frac{1}{\frac{z_{0} + c_{1}}{2 c_{1} (z_{0} - c_{1})} e^{2 α c_{1} t} - \frac{1}{2 c_{1}}}, c_{1}^{2} = \frac{c_{0}}{α} + \frac{Θ}{4 α^{2}} .$ Therefore, when $\frac{- \sqrt{Θ}}{2 α} ⩽ z_{0} < - c_{1}$ (i.e., $\frac{- β - \sqrt{Θ}}{2 α} ⩽ y_{0} < - c_{1} - \frac{β}{2 α}$ ), the solution blows up in finite time and becomes smaller than $\frac{- β - \sqrt{Θ}}{2 α}$ . Furthermore, when $c_{0} = - \frac{Θ}{4 α}$ , the solution to (2.51) reads $z (t) = \frac{1}{α t + \frac{1}{z_{0}}}$ and the same holds, for $\frac{- \sqrt{Θ}}{2 α} ⩽ z_{0} < 0$ (i.e., $\frac{- β - \sqrt{Θ}}{2 α} ⩽ y_{0} < - \frac{β}{2 α}$ ). Finally, when $c_{0} < - \frac{Θ}{4 α}$ , then $z^{'} ⩽ c_{0} + \frac{Θ}{4 α} (< 0),$ which yields that $z (t) ⩽ z_{0} + (c_{0} + \frac{Θ}{4 α}) t, t ⩾ 0 .$ Therefore, $y (t) ⩽ y_{0} + (c_{0} + \frac{Θ}{4 α}) t, t ⩾ 0,$ and the solution becomes smaller than $\frac{- β - \sqrt{Θ}}{2 α}$ , for every initial datum $y_{0}$ in $[\frac{- β - \sqrt{Θ}}{2 α}, \frac{- β + \sqrt{Θ}}{2 α}]$ .

Theorem 2.16.

Let $Θ > 0$ , and let u be a solution to ( 2.1 )–( 2.3 ) such that $u (t) \in [\frac{- β - \sqrt{Θ}}{2 α}, \frac{- β + \sqrt{Θ}}{2 α}]$ , $\forall t ⩾ 0$ . Then, u tends to $\frac{- β + \sqrt{Θ}}{2 α}$ in $H^{1} (Ω)$ as $t \to + \infty$ .

Proof.

We first note that, if $u_{0} \equiv \frac{- β - \sqrt{Θ}}{2 α}$ , then $u \equiv \frac{- β - \sqrt{Θ}}{2 α}$ . We thus assume that $u_{0} ≢ \frac{- β - \sqrt{Θ}}{2 α}$ .

Setting again $y = ⟨ u ⟩$ , we have $y^{'} = - ⟨ g (u) ⟩ .$ (2.52) Then, since $u (t) \in [\frac{- β - \sqrt{Θ}}{2 α}, \frac{- β + \sqrt{Θ}}{2 α}]$ , $\forall t ⩾ 0$ , we note that $g (u) ⩽ 0, \forall t ⩾ 0,$ hence $⟨ g (u) ⟩ ⩽ 0, \forall t ⩾ 0,$ and we can deduce from (2.52) that y is monotone increasing. Since it is bounded from above (by $\frac{- β + \sqrt{Θ}}{2 α}$ ), it follows that y tends to some limit $\bar{y}$ as $t \to + \infty$ .

Let now $t_{n}$ , $n \in N$ , be such that $y^{'} (t_{n})$ tends to 0 as $n \to + \infty$ (indeed, note that $y (t + 1) - y (t) = y^{'} (τ)$ , for some $τ \in (t, t + 1)$ ). Then, $⟨ u (t_{n}) ⟩$ , $∥ u (t_{n}) ∥$ , and ${∥ u (t_{n}) ∥}_{H^{1} (Ω)}$ (see Remark 2.14), and also $g (u (t_{n}))$ , $⟨ g (u (t_{n})) ⟩$ , and ${∥ g (u (t_{n})) ∥}_{L^{1} (Ω)}$ , are bounded, independently of n, so that, at least for a subsequence which we do not relabel, $u (t_{n}) \to \bar{u} a.e. and in L^{2} (Ω), y (t_{n}) \to ⟨ \bar{u} ⟩,$ as $n \to + \infty$ . Thus, passing to the limit $n \to + \infty$ in (2.52), we have $⟨ g (\bar{u}) ⟩ = 0$ and ${∥ g (\bar{u}) ∥}_{L^{1} (Ω)} = 0,$ hence $g (\bar{u}) = 0,$ which yields that $\bar{u}$ is constant and $\bar{u} \equiv \frac{- β + \sqrt{Θ}}{2 α}$ (indeed, $⟨ u ⟩$ is monotone increasing and cannot tend to $\frac{- β - \sqrt{Θ}}{2 α}$ ). Finally, we see that y tends to $\frac{- β + \sqrt{Θ}}{2 α}$ as $t \to + \infty$ .

Now, noting that the trajectory is asymptotically compact in $H^{1} (Ω)$ (see again Remark 2.14), the ω-limit set of $u_{0}$ is nonempty and compact in $H^{1} (Ω)$ . Then, if $\bar{u}$ belongs to this set, necessarily, $⟨ \bar{u} ⟩ = \frac{- β + \sqrt{Θ}}{2 α}$ , whence $\bar{u} \equiv \frac{- β + \sqrt{Θ}}{2 α}$ . □

We also have the following theorem. Theorem 2.17.

Let $Θ ⩾ 0$ , and let u be a solution to ( 2.1 )–( 2.3 ) such that $u (t) \in [\frac{- β + \sqrt{Θ}}{2 α}, a]$ , $\forall t ⩾ 0$ , where $a > \frac{- β + \sqrt{Θ}}{2 α}$ is arbitrary. Then, u tends to $\frac{- β + \sqrt{Θ}}{2 α}$ in $H^{1} (Ω)$ as $t \to + \infty$ .

Remark 2.18.

The proof of Theorem 2.17 is similar to the one of Theorem 2.16, but, now, $y = ⟨ u ⟩$ is monotone decreasing.

Remark 2.19.

The results in Theorem 2.16 and Theorem 2.17 also hold in the $H^{2}$ -norm (see Remark 2.14).

Remark 2.20.

We can more generally consider a nonlinear term f of the form $f (s) = \sum_{k = 0}^{2 p + 1} a_{k} s^{k}, a_{2 p + 1} > 0$ (see [7]).

3. A general tumor growth model

We consider in this section the following initial and boundary value problem: $\begin{matrix} \frac{\partial u}{\partial t} + Δ^{2} u - Δ f (u) + g (u) = 0, & (3.1) \\ \frac{\partial u}{\partial ν} = \frac{\partial Δ u}{\partial ν} = 0 on Γ, & (3.2) \\ u |_{t = 0} = u_{0}, & (3.3) \end{matrix}$ where the nonlinear term f is defined as $f (s) = \sum_{i = 0}^{2 p + 1} a_{i} s^{i}, a_{2 p + 1} > 0, p ⩾ 1,$ (3.4) and the proliferation term is defined as $g (s) = α s - β s^{q} {(s - 1)}^{q}, q > 1,$ (3.5) where $α, β \in R$ .

Remark 3.1.

Note that, here, we assume that the parameters α and β are constants but, in general biological applications, they can depend on the spatial variable and on time.

We set $L (s) = s^{q} {(s - 1)}^{q}, q > 1 .$ (3.6)

We note that it is easy to prove that there exist $c_{0}$ and $c_{1} ⩾ 0$ such that $f^{'} (s) ⩾ - c_{0}$ (3.7) and $\frac{1}{2} s^{2 q} - c_{1} ⩽ L (s) ⩽ \frac{3}{2} s^{2 q} + c_{1} .$ (3.8)

Integrating (3.1) over Ω, we have $\frac{d}{d t} ⟨ u ⟩ + ⟨ g (u) ⟩ = 0,$ (3.9) hence $\frac{d}{d t} ⟨ u ⟩ + α ⟨ u ⟩ - β ⟨ L (u) ⟩ = 0 .$ (3.10)

We first have the following proposition.

Proposition 3.2.

For every $u_{0} \in L^{2} (Ω)$ , there exists $T_{0} = T_{0} (∥ u_{0} ∥) > 0$ and a unique solution u to ( 3.1 )–( 3.3 ) such that $u \in C ([0, T_{0}); L^{2} (Ω)) \cap L^{2} (0, T; H^{2} (Ω))$ , $\forall T < T_{0}$ .

Proof.

See [20] and [24]. □

Lemma 3.3.

We assume that $α > 0$ . Let u be a solution to (3.1)–(3.3). We have:

If $β ⩾ 0$ , then $⟨ u ⟩$ is bounded from below.

If $β ⩽ 0$ , then $⟨ u ⟩$ is bounded from above.

Proof.

First, owing to (3.10), we have $\frac{d ⟨ u ⟩}{d t} + α ⟨ u ⟩ = β ⟨ L (u) ⟩ .$ (3.11) We assume that $β ⩾ 0$ . Then, owing to (3.8) and (3.11), $\frac{d ⟨ u ⟩}{d t} + α ⟨ u ⟩ ⩾ \frac{β}{2} ⟨ u^{2 q} ⟩ - β c_{1} ⩾ - β c_{1},$ (3.12) hence $\frac{d}{d t} (e^{α t} ⟨ u ⟩) ⩾ - β c_{1} e^{α t} .$ (3.13) It thus follows from (3.13) that $⟨ u (t) ⟩ ⩾ (⟨ u_{0} ⟩ + \frac{β c_{1}}{α}) e^{- α t} - \frac{β c_{1}}{α}, \forall t ⩾ 0,$ (3.14) i.e., $⟨ u ⟩$ is bounded from below.

Now, assuming that $β ⩽ 0$ , we have, owing to (3.8) and (3.11), $\frac{d ⟨ u ⟩}{d t} + α ⟨ u ⟩ ⩽ \frac{β}{2} ⟨ u^{2 q} ⟩ - β c_{1} ⩽ - β c_{1},$ (3.15) hence $\frac{d}{d t} (e^{α t} ⟨ u ⟩) ⩽ - β c_{1} e^{α t} .$ (3.16) It thus follows from (3.16) that $⟨ u (t) ⟩ ⩽ (⟨ u_{0} ⟩ + \frac{β c_{1}}{α}) e^{- α t} - \frac{β c_{1}}{α}, \forall t ⩾ 0,$ (3.17) i.e., $⟨ u ⟩$ is bounded from above. □

Lemma 3.4.

We assume that $α < 0$ . Let u be a solution to (3.1)–(3.3). We have:

If $β ⩾ 0$ and $⟨ u (T) ⟩ > - \frac{β c_{1}}{α}$ (and, in particular, if $⟨ u_{0} ⟩ > - \frac{β c_{1}}{α}$ ), then there exists $t_{0} ⩾ T$ such that $⟨ u (t) ⟩ > 0, \forall t ⩾ t_{0},$ and, if the solution exists globally in time, $lim_{t \to + \infty} ⟨ u (t) ⟩ = + \infty .$

If $β ⩽ 0$ and $⟨ u (T) ⟩ < - \frac{β c_{1}}{α}$ (and, in particular, if $⟨ u_{0} ⟩ < - \frac{β c_{1}}{α}$ ), then there exists $t_{0} ⩾ T$ such that $⟨ u (t) ⟩ < 0, \forall t ⩾ t_{0},$ and, if the solution exists globally in time, $lim_{t \to + \infty} ⟨ u (t) ⟩ = - \infty .$

Proof.

We assume that $β ⩾ 0$ . It thus follows from (3.8) and (3.10) that $\frac{d ⟨ u ⟩}{d t} + α ⟨ u ⟩ ⩾ - β c_{1},$ (3.18) which yields, owing to Gronwall’s lemma, $⟨ u (t) ⟩ ⩾ (⟨ u_{0} ⟩ + \frac{β c_{1}}{α}) e^{- α t} - \frac{β c_{1}}{α} .$ (3.19) Therefore, if $⟨ u_{0} ⟩ > - \frac{β c_{1}}{α},$ (3.20) then there exists $t_{0} ⩾ 0$ such that $⟨ u (t) ⟩ > 0, \forall t ⩾ t_{0},$ (3.21) and, moreover, if the solution exists globally in time, $lim_{t \to + \infty} ⟨ u (t) ⟩ = + \infty .$ (3.22)

Now, assuming that $β ⩽ 0$ , it also follows from (3.8) and (3.10) that $\frac{d ⟨ u ⟩}{d t} + α ⟨ u ⟩ ⩽ - β c_{1},$ (3.23) which yields, owing to Gronwall’s lemma, $⟨ u (t) ⟩ ⩽ (⟨ u_{0} ⟩ + \frac{β c_{1}}{α}) e^{- α t} - \frac{β c_{1}}{α} .$ (3.24) Therefore, if $⟨ u_{0} ⟩ < - \frac{β c_{1}}{α},$ (3.25) then there exists $t_{0} ⩾ 0$ such that $⟨ u (t) ⟩ < 0, \forall t ⩾ t_{0},$ (3.26) and, moreover, if the solution exists globally in time, $lim_{t \to + \infty} ⟨ u (t) ⟩ = - \infty . □$ (3.27)

Lemma 3.5.

We assume that $α = 0$ and q is even. Let u be a solution to (3.1)–(3.3). We have:

If $β ⩾ 0$ , then $⟨ u ⟩$ is bounded from below.

If $β ⩽ 0$ , then $⟨ u ⟩$ is bounded from above.

Proof.

We assume that $β ⩾ 0$ . We have, owing to (3.11), $\frac{d ⟨ u ⟩}{d t} ⩾ 0,$ (3.28) since q is even, hence $⟨ u (t) ⟩ ⩾ ⟨ u_{0} ⟩, \forall t ⩾ 0,$ (3.29) i.e., $⟨ u ⟩$ is bounded from below.

Now, assuming that $β ⩽ 0$ , we have, owing to (3.11), $\frac{d ⟨ u ⟩}{d t} ⩽ 0,$ (3.30) since q is even, hence, $⟨ u ⟩ ⩽ ⟨ u_{0} ⟩, \forall t ⩾ 0,$ (3.31) i.e., $⟨ u ⟩$ is bounded from above. □

Proposition 3.6.

We assume that $p = 2 q - 1$ . Let u be a solution to ( 3.1 )–( 3.3 ). Then, there exists a monotone increasing function Q such that ${∥ u (t) - ⟨ u (t) ⟩ ∥}^{2} ⩽ Q (∥ u_{0} - ⟨ u_{0} ⟩ ∥), \forall t ⩾ 0 .$

Proof.

Setting again $v = u - ⟨ u ⟩$ , we rewrite problem (3.1)–(3.3) in the form $\begin{matrix} \frac{\partial v}{\partial t} + Δ^{2} v - Δ f (v + ⟨ u ⟩) + g (v + ⟨ u ⟩) - ⟨ g (v + ⟨ u ⟩) ⟩ = 0, & (3.32) \\ \frac{\partial v}{\partial ν} = \frac{\partial Δ v}{\partial ν} = 0 on Γ, & (3.33) \\ v |_{t = 0} = v_{0} = u_{0} - ⟨ u_{0} ⟩ . & (3.34) \end{matrix}$

We multiply (3.32) by ${(- Δ)}^{- 1} v$ and have $\begin{array}{l} \frac{1}{2} \frac{d}{d t} {∥ v ∥}_{- 1}^{2} + {∥ \nabla v ∥}^{2} + ((f (v + ⟨ u ⟩) - f (⟨ u ⟩), v)) + α {∥ v ∥}_{- 1}^{2} \\ - β ((L (v + ⟨ u ⟩) - ⟨ L (v + ⟨ u ⟩) ⟩, {(- Δ)}^{- 1} v)) = 0 . & (3.35) \end{array}$ Here, $((f (v + ⟨ u ⟩) - f (⟨ u ⟩), v)) ⩾ ξ \int_{Ω} \sum_{k = 0}^{p} v^{2 p + 2 - 2 k} {⟨ u ⟩}^{2 k} d x,$ where $ξ > 0$ (see [7]). Furthermore, $\begin{array}{l} | β ((L (v + ⟨ u ⟩) - ⟨ L (v + ⟨ u ⟩) ⟩, {(- Δ)}^{- 1} v)) | \\ ⩽ c ∥ L (v + ⟨ u ⟩) - L (⟨ u ⟩) ∥ ∥ v ∥, \end{array}$ which yields $\begin{array}{l} \frac{d}{d t} {∥ v ∥}_{- 1}^{2} + 2 {∥ \nabla v ∥}^{2} + ξ \int_{Ω} \sum_{k = 0}^{p} v^{2 p + 2 - 2 k} {⟨ u ⟩}^{2 k} d x \\ ⩽ | α | {∥ v ∥}_{- 1}^{2} + c (ε) {∥ v ∥}^{2} + ε {∥ L (v + ⟨ u ⟩) - L (⟨ u ⟩) ∥}^{2} . & (3.36) \end{array}$ Actually, it suffices to treat the case $L (s) = s^{2 q}$ (see Remark 3.9) and we have $\begin{array}{rcl} {∥ L (v + ⟨ u ⟩) - L (⟨ u ⟩) ∥}^{2} & ⩽ & \int_{Ω} {({(v + ⟨ u ⟩)}^{2 q} - {⟨ u ⟩}^{2 q})}^{2} d x \\ ⩽ & \int_{Ω} {(\sum_{k = 0}^{2 q - 1} C_{2 q}^{k} v^{2 q - k} {⟨ u ⟩}^{k} d x)}^{2} d x \\ ⩽ & c \int_{Ω} \sum_{k = 0}^{2 q - 1} v^{4 q - 2 k} {⟨ u ⟩}^{2 k} d x + c^{'} \\ ⩽ & c \int_{Ω} \sum_{k = 0}^{p} v^{2 p + 2 - 2 k} {⟨ u ⟩}^{2 k} d x + c^{'}, \end{array}$ since $4 q = 2 p + 2$ . Therefore, $\frac{d}{d t} {∥ v ∥}_{- 1}^{2} + {∥ \nabla v ∥}^{2} + ξ \int_{Ω} \sum_{k = 0}^{p} v^{2 p + 2 - 2 k} {⟨ u ⟩}^{2 k} d x ⩽ c,$ (3.37) hence $\frac{d}{d t} {∥ v ∥}_{- 1}^{2} + c {∥ v ∥}_{- 1}^{2} ⩽ c^{'}, c > 0 .$ (3.38) It thus follows from Gronwall’s lemma that ${∥ v (t) ∥}_{- 1}^{2} ⩽ e^{- c t} {∥ v_{0} ∥}_{- 1}^{2} + c^{'}, c > 0, t ⩾ 0 .$ (3.39)

Let B be a bounded subset of ${\dot{H}}^{- 1} (Ω)$ and $t_{0}$ be such that $v_{0} \in B$ and $t ⩾ t_{0}$ implies $v (t) \in B_{0}$ , where $B_{0} = {ϕ \in {\dot{H}}^{- 1} (Ω), {∥ ϕ ∥}_{- 1}^{2} ⩽ 2 c^{'}}$ , $c^{'}$ being the constant in (3.39). We then deduce from (3.37) that, for $t ⩾ t_{0}$ , $\int_{t}^{t + r} {∥ \nabla v ∥}^{2} d s ⩽ c (r), \int_{t}^{t + r} d s \int_{Ω} \sum_{k = 0}^{p} v^{2 p + 2 - 2 k} {⟨ u ⟩}^{2 k} d x ⩽ c (r),$ where $r > 0$ is fixed.

Now, multiplying (3.32) by v, it is easy to obtain, noting that $\begin{array}{rcl} | ((L (v + ⟨ u ⟩) - L (⟨ u ⟩), v)) | & ⩽ & c ∥ L (v + ⟨ u ⟩) ∥ ∥ v ∥ \\ ⩽ & c (\int_{Ω} \sum_{k = 0}^{p} v^{2 p + 2 - 2 k} {⟨ u ⟩}^{2 k} d x + {∥ v ∥}^{2}), \end{array}$ the inequality $\frac{d}{d t} {∥ v ∥}^{2} + {∥ Δ v ∥}^{2} ⩽ c (\int_{Ω} \sum_{k = 0}^{p} v^{2 p + 2 - 2 k} {⟨ u ⟩}^{2 k} d x + {∥ v ∥}_{H^{1} (Ω)}^{2} + 1) .$ (3.40) It thus follows from (3.40) and Gronwall’s lemma that ${∥ v (t) ∥}^{2} ⩽ c, t ⩾ t_{0} + r,$ (3.41) where $0 < r < 1$ is fixed and the constant c is independent of $v_{0}$ and t, hence, owing to (3.41) and integrating (3.37) and (3.40) between 0 and $t_{0} + r$ (r is fixed in $(0, 1)$ ), ${∥ v (t) ∥}^{2} ⩽ Q (∥ v_{0} ∥), t ⩾ 0,$ (3.42) where the function Q is monotone increasing. □

Corollary 3.7.

Let u be a global in time solution to ( 3.1 )–( 3.3 ). Then, $v = u - ⟨ u ⟩$ is dissipative in $L^{2} (Ω)$ .

Remark 3.8.

It is not difficult to also prove the dissipativity of v in $H^{1} (Ω)$ and $H^{2} (Ω)$ (see also [19,20], and [24]).

Remark 3.9.

We treated in Proposition 3.6 the term $∥ L (v + ⟨ u ⟩) - L (⟨ u ⟩) ∥$ in the particular case $L (s) = s^{2 q}$ , but, in general, we are interested in the case when $L (s) = s^{q} {(s - 1)}^{q}$ . We have, more generally, $\begin{array}{rcl} L (v + ⟨ u ⟩) - L (⟨ u ⟩) & = & {({(v + ⟨ u ⟩)}^{2} - (v + ⟨ u ⟩))}^{q} - {{(⟨ u ⟩}^{2} - ⟨ u ⟩)}^{q} \\ = & \sum_{k = 0}^{q} C_{k}^{q} {(- 1)}^{k} {(v + ⟨ u ⟩)}^{2 q - 2 k} {(v + ⟨ u ⟩)}^{k} - \sum_{k = 0}^{q} C_{k}^{q} {(- 1)}^{k} {⟨ u ⟩}^{2 q - 2 k} {⟨ u ⟩}^{k} \\ = & \sum_{k = 0}^{q} C_{k}^{q} {(- 1)}^{k} ({(v + ⟨ u ⟩)}^{2 q - k} - {⟨ u ⟩}^{2 q - k}) \\ = & \sum_{k = 0}^{q} C_{k}^{q} {(- 1)}^{k} \sum_{j = 0}^{2 q - k - 1} C_{j}^{2 q - k} v^{2 q - k - j} {⟨ u ⟩}^{j}, \end{array}$ hence $\begin{array}{rcl} | L (v + ⟨ u ⟩) - L (⟨ u ⟩) | ⩽ c \sum_{k = 0}^{q} \sum_{j = 0}^{2 q - k - 1} | v |^{2 q - k - j} {| ⟨ u ⟩ |}^{j} ⩽ c \sum_{i = q - 1}^{2 q - 1} \sum_{j = 0}^{i} | v |^{i + 1 - j} {| ⟨ u ⟩ |}^{j}, \end{array}$ which yields $\begin{array}{rcl} {∥ L (v + ⟨ u ⟩) - L (⟨ u ⟩) ∥}^{2} ⩽ c \int_{Ω} \sum_{i = q - 1}^{2 q - 1} \sum_{j = 0}^{i} | v |^{2 i + 2 - 2 j} {| ⟨ u ⟩ |}^{2 j} ⩽ c \int_{Ω} \sum_{l = 0}^{2 q - 1} | v |^{4 q - 2 l} {| ⟨ u ⟩ |}^{2 l} + c^{'} . \end{array}$

We also have the following theorem. Theorem 3.10.

Let u be a solution to (3.1)–(3.2) (note that the same holds for the solution to (2.1)–(2.2)). Then, $v = u - ⟨ u ⟩$ tends to 0 in $H^{1} (Ω)$ as $t \to + \infty$ .

Proof.

We note that $⟨ v ⟩ = 0$ , $\forall t ⩾ 0$ . It thus follows from Proposition 3.6 that $∥ v (t) ∥$ and ${∥ v (t) ∥}_{H^{1} (Ω)}$ (see Remark 3.8) are bounded, so that, at least for a subsequence which we do not relabel, $v (t_{n}) \to \bar{v} in L^{2} (Ω), n \in N,$ hence $v (t_{n}) \to \bar{v} a.e. in Ω .$ Since $⟨ v ⟩ = 0$ , we have $⟨ \bar{v} ⟩ = 0,$ whence $\bar{v} \equiv 0 .$ Finally, as a consequence of Remark 2.14 and Remark 3.8, v tends to 0 in $H^{1} (Ω)$ as $t \to + \infty$ . □

Theorem 3.11.

We assume that $α = 0$ and q is even. Let u be a nonvanishing solution to ( 3.1 )–( 3.3 ) such that $u (t) \in [0, 1]$ , $\forall t ⩾ 0$ . Then:

If $β > 0$ , u tends to 1 in $H^{1} (Ω)$ as $t \to + \infty$ .

If $β < 0$ , u tends to 0 in $H^{1} (Ω)$ as $t \to + \infty$ .

Proof.

We first note that, if $u_{0} \equiv 0$ , then $u \equiv 0$ . We thus assume that $u_{0} ≢ 0$ .

Setting again $y = ⟨ u ⟩$ , we have $y^{'} + ⟨ g (u) ⟩ = 0,$ (3.43) hence $y^{'} = β ⟨ L (u) ⟩ .$ (3.44)

We note that $L (u) ⩾ 0, \forall t ⩾ 0,$ hence $⟨ L (u) ⟩ ⩾ 0, \forall t ⩾ 0,$ and, if $β > 0$ , we can deduce from (3.44) that y is monotone increasing. Since it is bounded from above (by 1), it follows that y tends to some limit $\bar{u}$ as $t \to + \infty$ .

Let now $t_{n}$ , $n \in N$ , be such that $y^{'} (t_{n})$ tends to 0 as $n \to + \infty$ (indeed, note that $y (t + 1) - y (t) = y^{'} (τ)$ , for some $τ \in (t, t + 1)$ ). Then, $⟨ u (t_{n}) ⟩$ , $∥ u (t_{n}) ∥$ , and ${∥ u (t_{n}) ∥}_{H^{1} (Ω)}$ (see Remark 3.8), and also $L (u (t_{n}))$ , $⟨ L (u (t_{n})) ⟩$ , and ${∥ L (u (t_{n})) ∥}_{L^{1} (Ω)}$ , are bounded, independently of n, so that, at least for a subsequence which we do not relabel, $u (t_{n}) \to \bar{u} a.e. and in L^{2} (Ω), y (t_{n}) \to ⟨ \bar{u} ⟩,$ as $n \to + \infty$ . Thus, passing to the limit $n \to + \infty$ in (3.44), we have $⟨ L (\bar{u}) ⟩ = 0$ and ${∥ L (\bar{u}) ∥}_{L^{1} (Ω)} = 0,$ hence $L (\bar{u}) = 0,$ which yields that $\bar{u}$ is constant and that $\bar{u} \equiv 1$ (indeed, $⟨ u ⟩$ is monotone increasing and cannot tend to 0). Finally, we see that y tends to 1 as $t \to + \infty$ .

Now, noting that the trajectory is asymptotically compact in $H^{1} (Ω)$ (see again Remark 3.8), the ω-limit set of $u_{0}$ is nonempty and compact in $H^{1} (Ω)$ . Then, since $\bar{u}$ belongs to this set, necessarily, $⟨ \bar{u} ⟩ = 1$ , whence $\bar{u} \equiv 1$ .

Similarly, if $β < 0$ , we can deduce from (3.44) that y is monotone decreasing. Since it is bounded from below (by 0), we can prove that u tends to 0. □

Theorem 3.12.

We assume that $α = 0$ and q is odd. Let u be a nonvanishing solution to ( 3.1 )–( 3.3 ) such that $u (t) \in [0, 1]$ , $\forall t ⩾ 0$ . Then:

If $β > 0$ , u tends to 0 in $H^{1} (Ω)$ as $t \to + \infty$ .

If $β < 0$ , u tends to 1 in $H^{1} (Ω)$ as $t \to + \infty$ .

Proof.

Since $u (t) \in [0, 1]$ , $\forall t ⩾ 0$ , we note that $L (u) ⩽ 0, \forall t ⩾ 0,$ hence $⟨ L (u) ⟩ ⩽ 0, \forall t ⩾ 0,$ and we deduce that (proceeding as the proof of the previous theorem), if $β > 0$ , u tends to 0 in $H^{1} (Ω)$ as $t \to + \infty$ and, if $β < 0$ , u tends to 1 in $H^{1} (Ω)$ as $t \to + \infty$ . □

3.1. The particular case

g (s) = 2 β s - β s^{2} {(s - 1)}^{2}

Theorem 3.13.
Let $β > 0$ , and let u be a solution to ( 3.1 )–( 3.3 ) such that $u (t) > 2$ , $\forall t ⩾ 0$ . Then, u blows up in finite time. Furthermore, the blow up time $T^{+}$ satisfies $T^{+} ⩽ T + \frac{1}{2 β} ln (\frac{y (T)}{y (T) - 2}),$ where $y (T) = ⟨ u (T) ⟩$ .
Proof.
We first note that $g (u) = - β u (u - 2) (u^{2} + 1),$ hence, we can rewrite (2.6) in the form $\frac{d}{d t} ⟨ u ⟩ = β ⟨ u (u - 2) (u^{2} + 1) ⟩ .$ (3.45) We set $ϑ (s) = s^{2} + 1$ and note that $ϑ (s) ⩾ 1, \forall s \in R .$ Therefore, $\frac{d}{d t} ⟨ u ⟩ ⩾ β ⟨ u (u - 2) ⟩,$ (3.46) since, by assumption, $u (u - 2) > 0$ . It thus follows that $\frac{d}{d t} ⟨ u ⟩ - β {⟨ u ⟩}^{2} + 2 β ⟨ u ⟩ ⩾ β ⟨ v^{2} ⟩,$ (3.47) hence the ODE $y^{'} - β y^{2} + 2 β y ⩾ β ⟨ v^{2} ⟩,$ (3.48) where $y = ⟨ u ⟩$ . Thus, $y^{'} - β y^{2} + 2 β y ⩾ c (t),$ (3.49) where $c (t) = β ⟨ v^{2} ⟩$ is nonnegative and uniformly bounded. Noting that the solution to the Riccati ODE $y^{'} - β y^{2} + 2 β y = 0$ reads $y (t) = \frac{2 y (T)}{e^{2 β (t - T)} (2 - y (T)) + y (T)}, t ⩾ T,$ we conclude, owing to the comparison principle. □
Theorem 3.14.
Let $β < 0$ , and let u be a solution to ( 3.1 )–( 3.3 ) such that $u (t) < 0$ , $\forall t ⩾ 0$ . Then, u blows up in finite time. Furthermore, the blow up time $T^{+}$ satisfies $T^{+} ⩽ T + \frac{1}{2 β} ln (\frac{y (T)}{y (T) - 2}),$ where $y (T) = ⟨ u (T) ⟩$ .
Remark 3.15.
The proof of Theorem 3.14 is similar to the one of Theorem 3.13, but, now, the solution tends to $- \infty$ .

We also have the following theorems. Theorem 3.16.
Let $β > 0$ , and let u be a solution to (3.1)–(3.3) such that $u (t) ⩽ 2$ , $\forall t ⩾ 0$ . Then, u tends to 0 in $H^{1} (Ω)$ as $t \to + \infty$ .
Theorem 3.17.
Let $β < 0$ , and let u be a solution to (3.1)–(3.3) such that $u (t) ⩾ 0$ , $\forall t ⩾ 0$ . Then, u tends to 0 in $H^{1} (Ω)$ as $t \to + \infty$ .
Remark 3.18.
The proof of Theorem 3.16 and Theorem 3.17 is similar to the one of Theorem 3.11; see also Theorem 2.16.

4. Numerical results

As far as the numerical simulations are concerned, we rewrite the problem in the form $\begin{matrix} \frac{\partial u}{\partial t} + Δ μ + g (u) = 0 in Ω, & (4.1) \\ μ = Δ u - f (u) in Ω, & (4.2) \\ \frac{\partial u}{\partial ν} = \frac{\partial μ}{\partial ν} = 0 on Γ, & (4.3) \\ u_{| t = 0} = u_{0}, & (4.4) \end{matrix}$ which has the advantage of splitting the fourth-order (in space) equation into a system of two second-order ones (see [8] and [13]). Consequently, we use a P1-finite element for the space discretization, together with an implicit Euler time discretization. The numerical simulations are performed with the software Freefem++ [15].

In the numerical results presented below, Ω is a $(0, 5) \times (0, 1)$ -rectangle. The triangulation is obtained by dividing Ω into $200 \times 40$ rectangles and by dividing each rectangle along the same diagonal. The time step is taken as $δ t = 0.01$ . We further take $f (s) = {(s - \frac{1}{2})}^{3} - (s - \frac{1}{2})$ .

The left pictures below represent the evolution of $u_{min} = {min}_{x \in Ω} u (x)$ (green curve), $u_{max} = {max}_{x \in Ω} u (x)$ (blue curve), and $⟨ u ⟩$ with respect to time (black curve), u being the numerical solution to (4.1), while the right pictures represent the evolution of $∥ u - ⟨ u ⟩ ∥$ with respect to t, for different choices of the initial datum $u_{0}$ .

4.1. The case when $g (s) = α s^{2} + β s + γ$

4.1.1. First case: $Θ > 0$

We show in Fig. 1 that, for $g (s) = \frac{1}{2} s^{2} - s - \frac{1}{2}$ with zeros $1 - \sqrt{2}$ and $1 + \sqrt{2}$ and $u_{0} = 1 - {sin}^{2} (x + y)$ , leading to $u_{0} \in [1 - \sqrt{2}, 1 + \sqrt{2}]$ and $⟨ u_{0} ⟩ = 0.422$ , the solution stays between $[1 - \sqrt{2}, 1 + \sqrt{2}]$ for all $t ⩾ 0$ and converges to $\bar{u} = 1 + \sqrt{2}$ .

In Fig. 2, we take $g (s) = s^{2} + s$ with zeros $- 1$ and 0. The solution remains greater than 0 and converges to 0. Here, $u_{0} = 1 - {sin}^{2} (x - y)$ , leading to $u_{0} ⩾ 0$ and $⟨ u_{0} ⟩ = 0.552$ .

Figure 3 corresponds to the function $g (s) = \frac{3}{4} s^{2} + 5 s + 3$ with zeros $- 6$ and $- \frac{2}{3}$ and to the initial datum $u_{0} = \frac{- 5}{(x^{6} + 1) (y^{3} + 1)} - 6$ , leading to $⟨ u_{0} ⟩ = - 6.892 < - 6$ . The solution blows up in finite time and $∥ u - ⟨ u ⟩ ∥$ approaches 0, in agreement with the theoretical results.

Fig. 1.

$u_{0} \in [1 - \sqrt{2}, 1 + \sqrt{2}]$ and $⟨ u_{0} ⟩ = 0.422$ ; the solution tends to $1 + \sqrt{2}$ . (The colors are visible in the online version of the article; https://dx-doi-org.web.bisu.edu.cn/10.3233/ASY-151306.)

Fig. 2.

$u_{0} ⩾ 0$ and $⟨ u_{0} ⟩ = 0.552$ ; the solution tends to 0. (The colors are visible in the online version of the article; https://dx-doi-org.web.bisu.edu.cn/10.3233/ASY-151306.)

Fig. 3.

$⟨ u_{0} ⟩ = - 0.250 < 1 - \frac{\sqrt{2}}{2}$ ; the solution blows up, while $∥ u - ⟨ u ⟩ ∥$ tends to 0. (The colors are visible in the online version of the article; https://dx-doi-org.web.bisu.edu.cn/10.3233/ASY-151306.)

Fig. 4.

$⟨ u_{0} ⟩ = 0.569$ ; the solution blows up, while $∥ u - ⟨ u ⟩ ∥$ tends to 0. (The colors are visible in the online version of the article; https://dx-doi-org.web.bisu.edu.cn/10.3233/ASY-151306.)

In Fig. 4, we take $u_{0} = \frac{1}{5 (x^{6} + 1) (y^{5} + \frac{1}{2})} + \frac{1}{2} \in [\frac{1}{2}, 1]$ , leading to $⟨ u_{0} ⟩ = 0.569$ , and $g (s) = s^{2} - \frac{3}{2} s + \frac{1}{2}$ with two zeros $\frac{1}{2}$ and 1. In that case, the solution blows up in finite time, while $∥ u - ⟨ u ⟩ ∥$ approaches 0.

4.1.2. Second case:

Θ = 0

In Fig. 5, the solution blows up in finite time. Here, we take $g (s) = \frac{1}{2} s^{2} - s + \frac{1}{2}$ with a double zero 1 and $u_{0} = \frac{16 (sin (x - y) + 1)}{x^{6} + 5}$ , hence $⟨ u_{0} ⟩ = 1.043 > 1$ .

Fig. 5.

$⟨ u_{0} ⟩ = 1.043 > 1$ ; the solution blows up, while $∥ u - ⟨ u ⟩ ∥$ tends to 0. (The colors are visible in the online version of the article; https://dx-doi-org.web.bisu.edu.cn/10.3233/ASY-151306.)

Fig. 6.

$⟨ u_{0} ⟩ = - 0.740$ ; the solution tends to $- 1$ . (The colors are visible in the online version of the article; https://dx-doi-org.web.bisu.edu.cn/10.3233/ASY-151306.)

Finally, we take $g (s) = \frac{1}{2} s^{2} + s + \frac{1}{2}$ with a double zero $- 1$ . Here, $u_{0} = \frac{- 4 cos (x - y)}{x^{6} + 1}$ and $⟨ u_{0} ⟩ = - 0.740$ . We emphasize that, at $t = 1.04$ , $u > - 1$ and the solution does not blow up and converges to $- 1$ as shown in Fig. 6.

4.1.3. Third case:

Θ < 0

Figure 7 corresponds to the case $4 α γ - β^{2} > 0$ ; more precisely, $g (s) = \frac{1}{2} s^{2} - s + 1$ . The solution blows up in finite time, in agreement with the theoretical results. Here, $u_{0} = \frac{8 cos (x - y)}{x^{6} + 1}$ , leading to $⟨ u_{0} ⟩ = 1.481$ .

Fig. 7.

$⟨ u_{0} ⟩ = 1.481$ ; the solution blows up, while $∥ u - ⟨ u ⟩ ∥$ tends to 0. (The colors are visible in the online version of the article; https://dx-doi-org.web.bisu.edu.cn/10.3233/ASY-151306.)

Fig. 8.

$u_{0} > 2$ ; the solution blows up, while $∥ u - ⟨ u ⟩ ∥$ tends to 0. (The colors are visible in the online version of the article; https://dx-doi-org.web.bisu.edu.cn/10.3233/ASY-151306.)

4.2. The case when

g (s) = 2 β s - β s^{2} {(s - 1)}^{2}

4.2.1. First case: $β > 0$

In Fig. 8, we take $g (s) = 0.03 s - 0.015 s^{2} {(s - 1)}^{2}$ . Here, $u_{0} = 2 + \frac{1}{2 (x^{6} + 0.1)} > 2$ and $⟨ u_{0} ⟩ = 2.476$ . The solution blows up in finite time and $∥ u - ⟨ u ⟩ ∥$ tends to 0, in agreement with the theoretical results.

Figure 9 corresponds to the function $g (s) = 2 s - s^{2} {(s - 1)}^{2}$ and to the initial datum $u_{0} = - \frac{0.1 cos (x - y)}{x^{6} + 1} + 2 \in [0, 2]$ , leading to $⟨ u_{0} ⟩ = 1.981$ . The solution remains between $[0, 2]$ and converges to $\bar{u} = 0$ , which confirms the theoretical results.

Fig. 9.

$u_{0} \in [0, 2]$ ; the solution tends to 0. (The colors are visible in the online version of the article; https://dx-doi-org.web.bisu.edu.cn/10.3233/ASY-151306.)

Fig. 10.

$u_{0} < 0$ ; the solution tends to 0. (The colors are visible in the online version of the article; https://dx-doi-org.web.bisu.edu.cn/10.3233/ASY-151306.)

In Fig. 10, we take $g (s) = 0.6 s - 0.3 s^{2} {(s - 1)}^{2}$ and the solution remains smaller than 0 and converges to $\bar{u} = 0$ , which confirms the theoretical results. Here, $u_{0} = - 1 - {cos}^{2} (x + y)$ , leading to $u_{0} < 0$ and $⟨ u_{0} ⟩ = - 1.422 < 0$ .

4.2.2. Second case:

β < 0

In Fig. 11, we take $u_{0} = - \frac{4 cos (x - y)}{(x^{6} + 1)}$ , leading to $⟨ u_{0} ⟩ = - 0.740$ , and $g (s) = - 0.04 s + 0.02 s^{2} {(s - 1)}^{2}$ . In that case, the solution blows up in finite time, while $∥ u - ⟨ u ⟩ ∥$ approaches 0.

Fig. 11.

$⟨ u_{0} ⟩ = - 0.740$ ; the solution blows up, while $∥ u - ⟨ u ⟩ ∥$ tends to 0. (The colors are visible in the online version of the article; https://dx-doi-org.web.bisu.edu.cn/10.3233/ASY-151306.)

Fig. 12.

$⟨ u_{0} ⟩ = 0.569$ ; the solution tends to 1. (The colors are visible in the online version of the article; https://dx-doi-org.web.bisu.edu.cn/10.3233/ASY-151306.)

Fig. 13.

(a) Initial datum at $t = 0$ . (b) Solution after $2300$ iterations. (c) Solution after $3500$ iterations. (d) Solution after $4500$ iterations. (The colors are visible in the online version of the article; https://dx-doi-org.web.bisu.edu.cn/10.3233/ASY-151306.)

Remark 4.1.

We come back to the proliferation term considered in Section 4.1. We assume that $Θ > 0$ . We saw in Remark 2.15 that $⟨ u ⟩$ can become smaller than $\frac{- β - \sqrt{Θ}}{2 α}$ , for $t > 0$ , and blow up in finite time (note that the partial answer given in Remark 2.15 is independent of the choice of potential f). An interesting question is whether the solutions u can blow up in finite time even if the initial datum $u_{0} \in [\frac{- β - \sqrt{Θ}}{2 α}, \frac{- β + \sqrt{Θ}}{2 α}]$ and the potential f has zeros in this interval, e.g., $f (s) = (s - \frac{- β - \sqrt{Θ}}{2 α}) (s + \frac{β}{2 α}) (s - \frac{- β + \sqrt{Θ}}{2 α}) .$ We give a partial answer via numerical simulations which seem to indicate that there is no blow up. We saw in Fig. 4 that the solution u blows up in finite time, even if $u_{0}$ is chosen between the zeros of g. We took $g (s) = s^{2} - \frac{3}{2} s + \frac{1}{2}$ , with zeros $\frac{1}{2}$ and 1, and $f (s) = {(s - \frac{1}{2})}^{3} - (s - \frac{1}{2})$ with three zeros $- \frac{1}{2}$ , $\frac{1}{2}$ , and $\frac{3}{2}$ . Furthermore, $u_{0} = \frac{1}{5 (x^{6} + 1) (y^{5} + \frac{1}{2})} + \frac{1}{2} \in [\frac{1}{2}, 1]$ , leading to $⟨ u_{0} ⟩ = 0.569$ . We run again the modified Cahn–Hilliard Eqs (4.1)–(4.4) with the same initial datum as in Fig. 4 and the same function g, but we now take $f (s) = (s - \frac{1}{2}) (s - \frac{3}{4}) (s - 1)$ . The solution u remains in $[\frac{1}{2}, 1]$ (no blow up) and converges to 1 as shown in Fig. 12.

Remark 4.2.

We come back to the proliferation term considered (1.11). We give real tumor growth results as in [1]. We rewrite the problem in the form $\begin{matrix} \frac{\partial u}{\partial t} + Δ μ + \frac{1}{ε} g (u) = 0 in Ω, & (4.5) \\ μ = ε Δ u - \frac{1}{ε} f (u) in Ω, & (4.6) \\ \frac{\partial u}{\partial ν} = \frac{\partial μ}{\partial ν} = 0 on Γ, & (4.7) \\ u_{| t = 0} = u_{0}, & (4.8) \end{matrix}$ where Ω is a circle of radius $1.2$ centered at $(0.5, - 0.5)$ and the triangulation is obtained by dividing Ω into $15, 676$ equal triangles. The time step is taken as $δ t = 10^{- 5}$ and the diffuse interface thickness $ε = 0.0125$ . We take $f (s) = s^{3} - s$ . Here, $u_{0} = - tanh (\frac{1}{\sqrt{2} ε} \sqrt{2 {{(x - 0.5)}^{2} + \frac{1}{4} (y + 0.5)}^{2}} - 0.1) \in [- 1, 1]$ , leading to $⟨ u_{0} ⟩ = - 0.225$ . Furthermore, $g (s) = 46 (u + 1) - 280 {{(u - 1)}^{2} (u + 1)}^{2}$ and the variation of the solution u is shown in Fig. 13. We note that, without mesh adaptation strategy, it seems too heavy in calculation time to reach a steady state in this example.

Footnotes

Acknowledgements

I would like to thank L. Cherfils and A. Miranville, my supervisors, for many stimulating discussions and useful comments on the subject of the paper.

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A Cahn–Hilliard equation with a proliferation term for biological and chemical applications

Abstract

Keywords

1. Introduction

4.1. The case when g ( s ) = α s 2 + β s + γ

4.1.1. First case: Θ > 0

4.2.1. First case: β > 0

Footnotes

Acknowledgements

References

4.1. The case when $g (s) = α s^{2} + β s + γ$

4.1.1. First case: $Θ > 0$

4.2.1. First case: $β > 0$