Boundedness for a fully parabolic Keller-Segel model with sublinear segregation and superlinear aggregation

Research paper by Silvia Frassu, Giuseppe Viglialoro

Indexed on: 19 May '20Published on: 16 May '20Published in: arXiv - Mathematics - Analysis of PDEs


This work deals with a fully parabolic chemotaxis model with nonlinear production and chemoattractant. The problem is formulated on a bounded domain and, depending on a specific interplay between the coefficients associated to such production and chemoattractant, we establish that the related initial-boundary value problem has a unique classical solution which is uniformly bounded in time. To be precise, we study this zero-flux problem \begin{equation}\label{problem_abstract} \tag{$\Diamond$} \begin{cases} u_t= \Delta u - \nabla \cdot (f(u) \nabla v) & \text{ in } \Omega \times (0,T_{max}),\\ v_t=\Delta v-v+g(u) & \text{ in } \Omega \times (0,T_{max}),\\ \end{cases} \end{equation} where $\Omega$ is a bounded and smooth domain of $\mathbb{R}^n$, for $n\geq 2$, and $f(u)$ and $g(u)$ are reasonably regular functions generalizing, respectively, the prototypes $f(u)=u^\alpha$ and $g(u)=u^l$, with proper $\alpha, l>0$. After having shown that any sufficiently smooth $ u(x,0)=u_0(x)\geq 0, \, v(x,0)=v_0(x)\geq 0$ emanate a unique classical and nonnegative solution $(u,v)$ to problem \eqref{problem_abstract}, which is defined on $\Omega \times (0,T_{max})$ with $T_{max}$ denoting the maximum time of existence, we establish that for any $l\in (0,\frac{2}{n})$ and $\frac{2}{n}\leq \alpha<1+\frac{1}{n}-\frac{l}{2}$, $T_{max}=\infty$ and $u$ and $v$ are actually uniformly bounded in time. This paper is in line with the contribution by Horstmann and Winkler, moreover, extends the result by Liu and Tao. Indeed, in the first work it is proved that for $g(u)=u$ the value $\alpha=\frac{2}{n}$ represents the critical blow-up exponent to the model, whereas in the second, for $f(u)=u$, corresponding to $\alpha=1$, boundedness of solutions is shown under the assumption $0<l<\frac{2}{n}.$