Critical mass threshold for the 2D Patlak-Keller-Segel-Navier-Stokes system
June 01, 2026
Abstract
In this paper, we investigate critical mass threshold for the Patlak-Keller-Segel-Navier-Stokes system on the two-dimensional whole space and obtain global existence of strong solutions if the initial mass is less than or equal to \(8\pi\), regardless of the initial norm of the velocity. One new observation is that the local mass of the density function rearrangement satisfies a good inequality that is independent of velocity; and then an improved maximum principle is applied by choosing a nice auxiliary function.
Keywords: Patlak-Keller-Segel-Navier-Stokes; critical mass threshold; global existence; maximum principle; blow-up
2010 Mathematics Subject Classification: 92C17, 35Q30, 35A01, 76D05, 35B44.
1 Introduction↩︎
In this paper, we investigate the critical mass threshold of the following two-dimensional parabolic-elliptic Patlak-Keller-Segel (PKS) system coupled with Navier-Stokes (NS) equations in \(\mathbb{R}^{2}\times(0,T)\): \[\label{ksns} \left\{ \begin{array}{lr} \partial_tn+u\cdot\nabla n=\triangle n-\nabla\cdot(n\nabla c), \\ \triangle c+n=0, \\ \partial_tu+u\cdot\nabla u+\nabla \pi=\triangle u+n\nabla \phi, \\ \nabla\cdot u=0, \end{array} \right.\tag{1}\] along with initial conditions \[(n,u)\big|_{t=0}=(n_{\rm in},v_{\rm in}),\] where \(n\) represents the cell density, \(c\) denotes the chemoattractant density, and \(u\) denotes the velocity of fluid. In addition, \(\pi\) is the pressure and \(\phi\) is the given potential function.
When \(u=0\), \(\pi=0\) and \(\phi=0\), the system (1 ) is reduced to the following classical parabolic-elliptic PKS system: \[\label{eq:pks} \left\{ \begin{array}{lr} \partial_tn=\triangle n-\nabla\cdot(n\nabla c), \\ \triangle c+n=0. \end{array} \right.\tag{2}\] The system, jointly developed by Patlak [1], Keller and Segel [2], serves as a classical mathematical framework for modeling aggregation behavior driven by both random motion and chemotaxis. There has been much progress in the study of the well-posedness of this system (2 ) in \(\mathbb{R}^d\), and we will briefly list some of it.
It is well known that solutions to the one-dimensional PKS system are globally well-posed. However, in spatial dimensions \(d \ge 2\), solutions to the classical PKS system (2 ) may develop singularities in finite time. In the two-dimensional setting, provided the integrability condition \((1 + |x|^2 + |\ln n|)n \in L^\infty([0, T), L^1(\mathbb{R}^2))\) holds, the solution satisfies the following free energy inequality: \[F[n(0)] \ge F[n(t)] + \int_{0}^{t} \int_{\mathbb{R}^2} n(x, s)|\nabla \ln n(x, s) - \nabla c(x, s)|^2 dx ds,\] for a.e. \(t \in (0, T)\), where the free energy functional \(F(n)\) is defined by \[F[n] = \int_{\mathbb{R}^2} n \ln n dx - \frac{1}{2} \int_{\mathbb{R}^2} n c dx, \quad c(x, s) = -\frac{1}{2\pi} \int_{\mathbb{R}^2} \ln |x - y|n(y,s)dy.\] This free energy functional was originally introduced in the context of chemotaxis models by Nagai, Senba, and Yoshida [3]. Blanchet, Dolbeault, and Perthame [4] established the free energy inequality and identified a critical mass threshold \(M_c = 8\pi\), where they proved that free energy solutions of 2 exist globally for initial data satisfying \(n_{in} \in L^1_+(\mathbb{R}^2, (1+|x|^2)dx)\) and \(n_{in} \log n_{in} \in L^1(\mathbb{R}^2)\), provided the total mass \(M := \|n_{in}\|_{L^1} < 8\pi\). These results were previously announced in [5] and their proof relies on the free energy inequality combined with the logarithmic Hardy-Littlewood-Sobolev inequality. Another alternative proof, based on ideas from [6], can be found in [7]. Global existence for the critical mass case, \(M = 8\pi\), was established by Blanchet, Carrillo, and Masmoudi [8] (see also [9] for the radially symmetric case). Conversely, in the super-critical regime \(M > 8\pi\), solutions are known to blow up in finite time. This is a consequence of the virial identity: \[\label{eq:blow-up} \frac{d}{dt} \int_{\mathbb{R}^2} |x|^2 n(x, t)dx = 4M \left( 1 - \frac{M}{8\pi} \right);\tag{3}\] see, for example, [10] or [4]. In the aforementioned works, the finite second moment assumption \(|x|^2 n_{in} \in L^1(\mathbb{R}^2)\) plays a crucial role. Recently, Wei [11] established global well-posedness for \(M \leq 8\pi\) by assuming only \(n_{in} \in L^1(\mathbb{R}^2)\), thereby removing the extra moment constraints. The similar results also hold for the parabolic-parabolic form of (2 ) (\(\triangle c\) is replaced by \(\triangle c-\partial_t c\)), which was obtained by Hosono in [12] with the help of a reconstructed Lyapunov functional. For further results on this topic, we refer the reader to [13]–[16] and references therein. Finally, in spatial dimensions \(d \ge 3\), finite-time blow-up may occur for arbitrarily small initial mass (see [17]–[21] and related works).
It’s interesting that whether Wei’s result can be extended to the PKS system coupling with Navier-Stokes equations 1 . Gong and He in [22] considered the case of \(\phi\equiv c\) and proved that classical solutions exist for any finite time by assuming that the initial mass \(M<8\pi\) and \(n_{in}(1+|x|^2)\in L^1(\mathbb{R}^2)\). Recently, Lai, Wei and Zhou in [23] improved this result and obtained the critical mass threshold \(M\leq 8\pi\) when \(\phi\equiv c\), that is the solutions are global if the initial mass \(\int_{\mathbb{R}^2}n_{in}dx\leq 8\pi\) along with \(n_{in}\log(n_{in}), n_{in}\log(1+|x|)\in L^1(\mathbb{R}^2)\) by using a good cancellation of \(n\nabla c.\) It’s still an open question whether the same conclusion holds for general the potential function \(\phi.\) Here we answer this question.
Assume that \(\phi\in L^\infty\left([0,\infty); \dot{W}^{1,\infty}(\mathbb{R}^{2})\right)\), and we say that \[0\leq n \in C_w([0, T), L^1(\mathbb{R}^2))\bigcap L^\infty_{loc}((0, T),
L^\infty(\mathbb{R}^2))\] along with \(u\in L^\infty([0, T), H^1(\mathbb{R}^2))\) is a strong solution to (1 ) with the initial data \(0\leq n_{in}\in
L^1(\mathbb{R}^{2})\) and \(u_{in}\in W^{1,2}(\mathbb{R}^{2})\) if
(i) for all test functions \(\psi \in \mathcal{D}(\mathbb{R}^2)\), there holds \[\begin{align}
\frac{d}{dt} \int_{\mathbb{R}^2} \psi(x)n(x, t)dx = & \int_{\mathbb{R}^2} \Delta\psi(x)n(x, t)dx + \int_{\mathbb{R}^2}u\cdot \nabla\psi(x)n(x, t)dx \\
& - \frac{1}{4\pi} \int_{\mathbb{R}^2} \int_{\mathbb{R}^2} [\nabla\psi(x) - \nabla\psi(y)] \cdot \frac{x - y}{|x - y|^2} n(x, t)n(y,t)dx dy;
\end{align}\] (ii) \(\nabla \cdot u = 0\) is true in the sense of distributions;
(iii) for all divergence-free test functions \(\varphi \in C_\sigma^\infty(\mathbb{R}^2; \mathbb{R}^2)\) (the space of smooth, compactly supported, solenoidal vector fields), the following equality holds: \[\begin{align}
\frac{d}{dt} \int_{\mathbb{R}^2} u(x, t) \cdot \varphi(x) \, dx = & \, \int_{\mathbb{R}^2} u(x, t) \cdot \Delta \varphi(x) \, dx \\
& + \int_{\mathbb{R}^2} (u(x, t) \otimes u(x, t)) : \nabla \varphi(x) \, dx+\int_{\mathbb{R}^2} n\nabla\phi \cdot \varphi(x) \, dx,
\end{align}\] in the sense of distributions on \((0, T)\). Here \(C_w([0, T), L^1(\mathbb{R}^2))\) is a subspace of \(L^\infty([0, T),
L^1(\mathbb{R}^2))\) such that \(t \mapsto \int_{\mathbb{R}^2} \psi(x) n(x,t)dx\) is continuous for any \(\psi \in \mathcal{D}(\mathbb{R}^2)\).
our main result is stated as follows.
Theorem 1. Assume that \(\phi\in L^\infty\left([0,\infty); \dot{W}^{1,\infty}(\mathbb{R}^{2})\right)\), \(0\leq n_{in}\in L^1\cap L^{\infty}(\mathbb{R}^{2})\) and \(u_{in}\in W^{1,2}(\mathbb{R}^{2})\). Then there exists a global strong solution \((n, u, \pi)\) to the Keller-Segel-Navier-Stokes equations 1 with the initial data \((n_{in},u_{in})\) provided that the initial mass \(\int_{\mathbb{R}^{2}} n_{in}dx\leq 8\pi.\)
Remark 2. Theorem 1 generalized the result on the Keller-Segel system in [11] to the coupled Navier-Stokes system 1 , which is sharp since the solution will blow up due to the inequality 3 .
Remark 3. Theorem 1 also improved and generalized the main theorem of Lai, Wei and Zhou in [23]. On the one hand, we removed the redundant initial conditions of \(n_{in}\log(n_{in}), n_{in}\log(1+|x|)\in L^1(\mathbb{R}^2)\), so it is not really necessary for the initial value to satisfy this condition; on the other hand, we removed the restriction on the gravitational potential function \(\phi\), which can be any function belonging to \(L^\infty_t\dot{W}^{1,\infty}\). When \(\phi\equiv c\) as in [23], according to the following Theorem 5 and 1 one can get \(\nabla c\in L^\infty\) and then the conclusion follows.
Remark 4. Our new observation is based on the local mass of the density function rearrangement satisfies a good inequality that is independent of velocity, which is shown in Proposition 6, \[\begin{align} \frac{\partial k}{\partial t}-4\pi s\frac{\partial^2k}{\partial s^2}-k\frac{\partial k}{\partial s}\leq 0, \end{align}\] where \[\begin{align} k(s,t)=\int_0^sn^*(\tau,t)d\tau. \end{align}\] Then we proved an improved maximum principle and applied it by choosing a nice auxiliary function.
In detail, Theorem 1 follows from the following a priori estimates of the two dimensional Keller-Segel equation with the drift term \(u.\)
Theorem 5. Assume that \(0\leq n_{in}\in L^1\cap L^\infty(\mathbb{R}^{2})\), \(u\in L^\infty\left([0,T]; W^{1,2}(\mathbb{R}^{2})\right)\). Then there exists a global strong solution \(n\) to the Keller-Segel equation in \(\mathbb{R}^{2}\times [0,T]\) \[\label{eq:pks-u} \left\{ \begin{array}{lr} \partial_tn+u\cdot \nabla n=\triangle n-\nabla\cdot(n\nabla c), \\ \triangle c+n=0. \end{array} \right.\qquad{(1)}\] with the initial data \(n_{in}\) provided that the initial mass \(\int_{\mathbb{R}^{2}} n_{in}dx=M\leq 8\pi.\) Specifically, the following estimates hold uniformly:
\(M<8\pi\): \[\begin{align} \label{eq:Mless8pi}\|n(t)\|_{L^\infty}\leq \frac{8\pi}{8\pi-M}\|n_{in}\|_{L^\infty},\quad 0<t<T; \end{align}\qquad{(2)}\]
\(M=8\pi\): \[\begin{align} \label{eq:M618pi}\|n(t)\|_{L^\infty}\leq 8 \|n_{in}\|_{L^\infty} \left(1 + 4T \|n_{in}\|_{L^\infty}\right),\quad 0<t<T. \end{align}\qquad{(3)}\]
The rest of the paper is organized as follows. In Section 2 we finish the proof of Theorem 1 under the assumption of Theorem 5. In section 3, the improved maximum principle is applied to prove the uniform boundedness of the density function by choosing a suitable auxiliary function and we complete the proof of Theorem 5. In Section 4 we show the local mass of the density function rearrangement satisfies a good inequality that is independent of velocity. In Section 5 we prove an improved maximum principle. At last, the last section is devoted to proving the local wellposed result of the system 1 .
2 Proof of Theorem 1↩︎
In this section, assuming that Theorem 5 holds, we complete the proof of Theorem 1.
Proof of Theorem 1. By the local well-posed result of Theorem 9, there exists a local strong solution to the system 1 . Standard bootstrapping arguments allow us to upgrade this mild solution to a strong solution for \(t > 0\). Let \(T\) be a possible blow-up time. We have \(n\geq 0\) due to the maximum principle and \(\|n(t)\|_{L^1}=\int_{\mathbb{R}^2}n(x,t)dx=\|n_0\|_{L^1}\) by the first equation of 1 , thus \(n\in L^\infty(0,T;L^1(\mathbb{R}^2))\). As \(n_{in}\in L^1(\mathbb{R}^2)\bigcap L^\infty(\mathbb{R}^2)\) and \(u\in L^\infty(0,T';H^1(\mathbb{R}^2))\) for all \(0<T'<T\). If \(T<\infty\), then by Theorem 5 we have \(\|n(t)\|_{L^\infty}\leq C(1+T')\) for all \(0<t<T'<T\).
Hence \(n\in L^\infty(0,T;L^1\cap L^\infty(\mathbb{R}^2))\). Let \(\omega=\partial_yu_1-\partial_xu_2=\nabla^T\cdot u\) be the vorticity of the velocity \(u\), which satisfies the following equation due to 1 : \[\begin{align} \partial_t \omega+u\cdot\nabla \omega-\triangle \omega=\nabla^T\cdot(n\nabla\phi) \end{align}\] which implies that \[\begin{align} \frac{d}{dt}\|\omega\|_{L^2(\mathbb{R}^2)}^2\leq \|n\nabla\phi\|_{L^2(\mathbb{R}^2)}^2\leq C, \quad 0<t<T \end{align}\] due to \(n\in L^\infty(0,T;L^1\cap L^\infty(\mathbb{R}^2))\). Similarly by the third equation of 1 we have \(\frac{d}{dt}\|u\|_{L^2(\mathbb{R}^2)}\leq \|n\nabla\phi\|_{L^2(\mathbb{R}^2)}\leq C\). Thus \(u\in L^\infty(0,T;H^1(\mathbb{R}^2))\), which is a contradiction by Remark 10. Thus we must have \(T=+\infty\). ◻
3 Proof of Theorem 5↩︎
In this section, we will state the local mass of the density function rearrangement \(k\) satisfies a good inequality that is independent of velocity, and establish a maximum principle. By using the two important properties, our main objective is to complete the proof of Theorem 5.
Let \(T\) be the possible blow-up time, and \(s\geq 0.\) Define \[\begin{align} \mu(\sigma,t)=|\{x\in \mathbb{R}^2; n(x,t)>\sigma\}| \end{align}\] and the rearrangement function of \(n(x,t)\) is \[\begin{align} n^*(s,t)=\inf_{\sigma>0}\{\sigma>0; \mu(\sigma,t)\leq s\}. \end{align}\] Then it follows that \[\begin{align} \label{eq:s-n} s=n^*(\mu(s,t), t ). \end{align}\tag{4}\] Let \[\begin{align} \label{eq:k-def} k(s,t)=\int_0^sn^*(\tau,t)d\tau. \end{align}\tag{5}\]
Motivated by Lemma 4 of Diaz and Nagai [24] or Lemma 4.3 in [25], we prove The following lemma, which plays an important role in the proof of the main theorem.
Proposition 6. Let \((n,u)\) be a local strong solution of (1 ) in \([0,T)\) as stated in Theorem 9 and and \(k\) is defined in 5 . Then there holds \[\begin{align} \label{eq:mono} \frac{\partial k}{\partial t}-4\pi s\frac{\partial^2k}{\partial s^2}-k\frac{\partial k}{\partial s}\leq 0 \end{align}\qquad{(4)}\] for a.e. \((s,t)\in (0,\infty)\times (2\tau,T)\) with \(\tau>0\). Moreover, \[\begin{align} \label{eq:mono-b} k|_{s=0}=0, \frac{\partial k}{\partial s}|_{s=\infty}=0. \end{align}\qquad{(5)}\]
Next we aim to prove a maximum principle on the inequality of ?? , which is important for the arguments of Theorem 5.
Let \(\Omega\) be a domain in the two dimensional space, \(Q_T=\Omega\times (0,T)\), \(0<\Lambda\leq |\Omega|\), \(\Omega^*=(0,\Lambda)\), and \(Q_T^*=\Omega^*\times (0,T)\). \(C_1(t)\) is a positive bounded function locally on \([0,T)\). Motivated by Prop A.1 in [24] and Lemma 4.3 in [25] we prove the following slightly different maximum principle.
Proposition 7. Let \(f,g\) be functions on \(\overline{Q_T^*}\) satisfying the following:
(i) \(f,g\in L_{loc}^\infty(Q_T^*)\cap H_{loc}^1(0,T; L^2(\Omega^*))\cap L_{loc}^2(0,T; W^{2,2}(\delta,\Lambda))\) with \(\delta>0\);
(ii) \[\begin{align}
\left| f(s,t)\right|+\left| \frac{\partial f}{\partial s}(s,t)\right|\leq C_1(t), \left| \frac{\partial g}{\partial s}(s,t)\right|\leq C_1(t), 0<s<\Lambda, 0<t<T;
\end{align}\] (iii) \[\begin{align}
\frac{\partial f}{\partial t}-4\pi s\frac{\partial^2f}{\partial s^2}-f\frac{\partial f}{\partial s}\leq \frac{\partial g}{\partial t}-4\pi s\frac{\partial^2g}{\partial s^2}-g\frac{\partial g}{\partial s}
\end{align}\] a.e. in \(\overline{Q_T^*}\);
(iv) \(0=f(0,t)\leq g(0,t)\) for any \(t\in (0,T)\);
(v) \(\frac{\partial f}{\partial s}(\Lambda, t) \leq \frac{\partial g}{\partial s}(\Lambda, t)\) or \(f(\Lambda, t) \leq g(\Lambda, t)\) with \(\Lambda<\infty\), for any \(t\in (0,T)\);
(vi) \(f(s,0)\leq g(s,0)\) on \(\Omega^*\);
(vii) \(g(s,t)\geq 0\) on \(\overline{Q_T^*}\).
Then we have \(f\leq g\) on \(\overline{Q_T^*}\).
Proof of Theorem 5. Note that it follows from Proposition 6 that \[\partial_t k - 4\pi s \partial_s^2 k - k \partial_s k \le 0, \quad k|_{s=0}=0, \quad k|_{s=\infty}=M,\] and \[\partial_s k|_{s=\infty}=n^*(\infty,t)=0,\quad \partial_s k \ge 0.\]
Define an auxiliary function as follows: \[\begin{align} \label{eq:auxiliary-g} g(s):=\frac{8\pi q s}{1 + q s}\geq 0, \end{align}\tag{6}\] which implies that \[\begin{align} - 4\pi s \partial_s^2 g - g \partial_s g = 0. \end{align}\] Then there exists \(q > 0\) such that \(k(s,0) \le g(s)\) for any \(s \ge 0\). In fact, by choosing \(s_0>0\) and \(q\) large enough we have \[\begin{align} \label{eq:ks1} \frac{k(s,0)}{s} \le \|n_{in} \|_{L^\infty(\mathbb{R}^2)}\leq \frac{8\pi q }{1 + qs },\quad 0<s\leq s_0, \end{align}\tag{7}\] and \[\begin{align} \label{eq:ks2} {k(s,0)} \le M \leq \frac{8\pi qs }{1 + qs },\quad s\geq s_0, \end{align}\tag{8}\] where the sharp value \((s_0, q_0)\) is the intersection point of the curve \(L_1: q=\frac{\|n_{in} \|_{L^\infty} }{8\pi-\|n_{in} \|_{L^\infty}s}\) and \(L_2: q=\frac{M}{(8\pi-M)s}\) shown as follows. More precisely if \(s_0=\frac{M}{\|n_{in} \|_{L^\infty}}\), \(q=q_0=\frac{\|n_{in} \|_{L^\infty} }{8\pi-M}\), then we have \(\frac{8\pi q_0 }{1 + q_0s_0}=\|n_{in} \|_{L^\infty(\mathbb{R}^2)}\), \(\frac{8\pi q_0s_0}{1 + q_0s_0}=M\), which implies 7 , 8 .
Figure 1: 
.
Hence take \(q\geq q_0=\frac{\|n_{in}\|_{L^\infty}}{8\pi - M}\) and we have \[\begin{align} \label{eq:ks0} k(s,0) \le g(s), \quad 0\leq s\leq s_0= \frac{M}{\|n_{in}\|_{L^\infty}}. \end{align}\tag{9}\] Moreover, by 9 we have \[\begin{align} {k(s,t)} \le M \leq g(s), \quad s\geq s_0. \end{align}\] Then by Proposition 7 in the domain \((0,s_0)\times(0,T)\) we have \[k(s,t) \le \frac{8\pi q s}{1 + q s} ,\] which implies \[\begin{align} \label{eq:bound-Mless8pi}\|n (\cdot,t)\|_{L^\infty(\mathbb{R}^2)}=n^*(0,t)=\partial_s k|_{s=0}= \lim_{s\rightarrow 0} \frac{k(s,t)}{s}\le 8\pi q= \frac{8\pi\|n_{in}\|_{L^\infty}}{8\pi - M}, \end{align}\tag{10}\] for any \(t<T\) and specially \(q\) is independent of \(t.\)
Take an auxiliary function \[\begin{align} \phi(s,t) = (2T - t)^{-2} e^{-\frac{s}{4\pi(2T - t)}}, \end{align}\]which satisfies \[\begin{align} \label{eq:phi-eq} \partial_t \phi + \partial_s^2 (4\pi s \phi) = 0. \end{align}\tag{11}\] Furthermore, there holds \[\begin{align} \partial_t \int_0^\infty k \phi ds &= \int_0^\infty (\partial_t k \phi + k \partial_t \phi)ds \leq \int_0^\infty [(4\pi s \partial_s^2 k + k \partial_s k) \phi + k \partial_t \phi] ds\\ &= \int_0^\infty \left[ k \partial_s^2 (4\pi s \phi) - \frac{k^2}{2} \partial_s \phi + k \partial_t \phi \right]ds \\ &= -\frac{1}{2} \int_0^\infty k^2 \partial_s \phi ds= \frac{1}{8\pi(2T - t)} \int_0^\infty k^2 \phi ds\le \frac{M}{8\pi(2T - t)} \int_0^\infty k \phi ds, \end{align}\] which implies \[\partial_t \int_0^\infty (2T - t) k \phi ds\le 0\] due to \(M = 8\pi\). Let \[\begin{align} R(t)=\int_0^\infty (2T - t) k \phi ds \end{align}\]then \[\frac{d}{dt} R(t)\leq0 ,\quad R(t) \leq R(0) ,\quad \forall\;0\le t\le T.\] Using \(\frac{k(s,0)}{s} \le \|n_{in} \|_{L^\infty(\mathbb{R}^2)}\) and \(k(s,0)\leq M=8\pi\) we have \[\begin{align} R(0) &= \int_0^\infty (2T) k(s,0) \phi(s,0) ds= (2T)^{-1} \int_0^\infty e^{-\frac{s}{8\pi T}} k(s,0) \, ds \\ &\leq (2T)^{-1}\int_0^{\infty}e^{-\frac{s}{8\pi T}}\min\{\|n_{in} \|_{L^\infty}s,8\pi\} ds\\ &= (2T)^{-1}\int_0^{\infty}8\pi Te^{-\frac{s}{8\pi T}}\partial_s\min\{\|n_{in} \|_{L^\infty}s,8\pi\} ds\\ &=4\pi\|n_{in} \|_{L^\infty}\int_0^{{8\pi}/\|n_{in} \|_{L^\infty}}e^{-\frac{s}{8\pi T}}ds=4\pi\|n_{in} \|_{L^\infty}8\pi T(1-e^{-\frac{1}{\|n_{in} \|_{L^\infty} T}}) \end{align}\] Consequently, we have \[\begin{align} &R(0)\geq R(t) =(2T - t)^{-1} \int_{0}^\infty e^{-\frac{s}{4\pi(2T - t)}} k(s, t) \, ds\\ \ge& (2T - t)^{-1} \int_{s_0}^\infty e^{-\frac{s}{4\pi(2T - t)}} k(s_0, t) \, ds = 4\pi e^{-\frac{s_0}{4\pi(2T - t)}} k(s_0, t)\ge4\pi e^{-\frac{s_0}{4\pi T }} k(s_0, t), \end{align}\] which implies \[{ k(s_0, t) \le\frac{R(0)}{4\pi} e^{\frac{s_0}{4\pi T}} \le \|n_{in} \|_{L^\infty}8\pi T(1-e^{-\frac{1}{\|n_{in} \|_{L^\infty} T}})e^{\frac{s_0}{4\pi T}},} \quad \forall~ 0 \le t \le T.\] Let \(a=\frac{1}{\|n_{in} \|_{L^\infty} T}\), \(b=\frac{s_0}{4\pi T}\) then we have \[\label{b1} { k(s_0, t) \le 8\pi (1-e^{-a})e^{b}/a,} \quad \forall~ 0 \le t \le T,\quad b>0,\quad s_0=4\pi Tb.\tag{12}\] Now we claim that\[\begin{align} \label{b2} &\frac{1-e^{-a}}{a}e^{\frac{a}{a+4}}\le\frac{4}{a+4},\quad\forall~ a>0. \end{align}\tag{13}\] In fact let \(F(a)=\frac{1-e^{-a}}{a}e^{\frac{a}{a+4}}\frac{a+4}{4}\) then for \(a>0\) we have \(F(a)>0\) and\[\begin{align} &\frac{F'(a)}{F(a)}=\frac{1}{e^a-1}-\frac{1}{a}+\frac{4}{(a+4)^2}+\frac{1}{a+4}=\frac{1}{e^a-1}-\frac{4^2}{a(a+4)^2}<0, \end{align}\]as \(e^a-1>a+a^2/2+a^3/6>a(1+a/4)^2\). Thus \(F\) is decreasing for \(a>0\) and\[\begin{align} &{F(a)}\leq \lim_{a\to0+}F(a)=\lim_{a\to0+}\frac{1-e^{-a}}{a}=1,\quad \forall~ a>0, \end{align}\]which implies 13 . Let \(b=\frac{a}{a+4}\), \(s_0=4\pi Tb\), \(q=\frac{4}{as_0}=\frac{1}{\pi Tab}=\frac{a+4}{\pi Ta^2}\), \(g(s)=\frac{8\pi qs}{1+qs}\), then \(- 4\pi s \partial_s^2 g - g \partial_s g = 0\), by 12 and 13 we have\[\label{b3} { k(s_0, t) \le 8\pi \frac{1-e^{-a}}{a}e^{\frac{a}{a+4}}\le8\pi \frac{4}{a+4}=\frac{8\pi qs_0}{1+qs_0}=g(s_0),\quad \forall~ 0 \le t \le T.}\tag{14}\] Moreover, for \(0<s<s_0\) we have \[\begin{align} \frac{k(s,0)}{s}\leq \|n_{in} \|_{L^\infty}\leq 8 \|n_{in} \|_{L^\infty} =\frac{8\pi q}{1+qs_0}\leq \frac{8\pi q}{1+qs}. \end{align}\] Then by Proposition 7 in the domain \((0,s_0)\times(0,T)\) we have \[k(s,t) \le g(s)=\frac{8\pi q s}{1 + q s} ,\] which implies (recall that \(a=\frac{1}{\|n_{in} \|_{L^\infty} T}\)) \[\begin{align} \label{eq3}&\|n (\cdot,t)\|_{L^\infty(\mathbb{R}^2)}=n^*(0,t)=\partial_s k|_{s=0}=\lim_{s\rightarrow 0} \frac{k(s,t)}{s}\le 8\pi q= \frac{8(a+4)}{Ta^2}\\&=8\|n_{in} \|_{L^\infty}(1+4T\|n_{in} \|_{L^\infty}),\nonumber \end{align}\tag{15}\] for any \(t<T\). 0one can choose \[\begin{align} \label{eq:s0m39} s_0<\min\left\{ \dfrac{8\pi}{\|n_{in}\|_{L^\infty}}, 4\pi T\left[\ln\left(1+\frac{32\pi^2}{R(0)}\right)-\ln 2\right]\right\},\\ M'=\frac{1}{2}\left[ \frac{R(0)}{4\pi} e^{\frac{s_0}{4\pi T}} +8\pi\right]<8\pi.\nonumber \end{align}\tag{16}\] Then as Step I there exists \(q\geq \frac{M'}{(8\pi-M')s_0}\) such that [eq:1] holds. Due to the monotonicity of \(k(s,0)\), we have \[\begin{align} k(s,0) \le k(s_0,0)\leq \frac{8\pi q s_0}{1 + q s_0}=g(s_0),\quad \forall ~0 \le s \le s_0, \end{align}\] and \[\begin{align} k(s,0)\leq g(s),\quad {\rm if }~ q\geq \frac{\|n_{in} \|_{L^\infty} }{8\pi-\|n_{in} \|_{L^\infty}s_0}. \end{align}\] Hence, we take \[\begin{align} \label{eq:bound-q2} q=\frac{M'}{(8\pi-M')s_0}. \end{align}\tag{17}\] Moreover, note that \[\begin{align} k(0,t)=0. \end{align}\] Thus by Proposition 7 again we have \[k(s,t) \le \frac{8\pi q s}{1 + q s} \quad \forall~ 0 \le s \le s_0, \;0< t < T,\] which implies \[\|n (\cdot,t)\|_{L^\infty(\mathbb{R}^2)} \le 8\pi q\leq \frac{8\pi M'}{(8\pi-M')s_0}, \quad \forall~ t<T\] due to the similar arguments as Step I. Specially \(q\) is independent of \(t.\) Hence \(n\) can be extended beyond time \(T\).
We divide the time evolution into a short-time local regime (\(t \le \delta\)) and a long-time global regime (\(t > \delta\)), where we set: \[\delta = \frac{1}{10 \|n_{in}\|_{L^\infty}}.\] For very small times \(t \le \delta\), the local maximum principle applied directly to the cell density equation yields the standard differential inequality: \[\frac{d}{dt} \|n(t)\|_{L^\infty} \le \|n(t)\|_{L^\infty}^2.\] Solving this yields: \[\|n(t)\|_{L^\infty} \le \frac{\|n_{in}\|_{L^\infty}}{1 - t \|n_{in}\|_{L^\infty}}.\] Since \(t \le \delta = \frac{1}{10 \|n_{in}\|_{L^\infty}}\), we have \(t \|n_{in}\|_{L^\infty} \le 0.1\). Using the inequality \(\frac{1}{1-x} \le 1 + 1.2 x\) for \(0 \le x \le 0.1\), we obtain: \[\|n(t)\|_{L^\infty} \le \|n_{in}\|_{L^\infty} \left(1 + 1.2 t \|n_{in}\|_{L^\infty}\right), \quad \forall t \in [0, \delta].\]
For \(t > \delta\), as in Step II we define \[\begin{align} R_t(\tau) &= \int_0^\infty (2t - \tau) k \phi ds= (2t - \tau)^{-1} \int_0^\infty e^{-\frac{s}{4\pi(2t - \tau)}} k(s,\tau) \, ds \end{align}\] and choose \(s_0\) such that \(s_0 < 4\pi t \ln\left(1/\theta_t\right)\), where: \[\theta_t = \frac{R_t(0)}{32\pi^2} = \frac{1}{8\pi} \int_0^\infty n_{in}^*(s) e^{-\frac{s}{8\pi t}} \, ds < 1.\] By the decay estimate of the auxiliary functional \(R_t(\tau)\), we have: \[k(s_0, \tau) \le M' := 4\pi \left(1 + \theta_t e^{\frac{s_0}{4\pi t}}\right) < 8\pi, \quad \forall \tau \in (0, t].\] Thus, to ensure \(k(s_0, \tau) \le g(s_0)\), it suffices to require \(g(s_0) \ge M'\), which yields: \[\frac{8\pi q s_0}{1 + q s_0} \ge M' \iff q \ge q_1 := \frac{1}{s_0} \frac{1 + \theta_t e^{\frac{s_0}{4\pi t}}}{1 - \theta_t e^{\frac{s_0}{4\pi t}}}.\]
Since \(k(s, 0) \le s \|n_{in}\|_{L^\infty}\), we require \(s \|n_{in}\|_{L^\infty} \le \frac{8\pi q s}{1 + q s}\) on \(s \in [0, s_0]\). This holds if: \[q \ge q_2 := \frac{\|n_{in}\|_{L^\infty}}{8\pi - s_0 \|n_{in}\|_{L^\infty}},\] which is valid provided we choose \(s_0 < \frac{8\pi}{\|n_{in}\|_{L^\infty}}\).
Choosing \(q = \max\{q_1, q_2\}\), Proposition 2 guarantees \(k(s, t) \le g(s)\) on \([0, s_0]\). Dividing by \(s\) and taking the limit \(s \to 0\) yields: \[\|n(t)\|_{L^\infty} = \lim_{s \to 0} \frac{k(s, t)}{s} \le \lim_{s \to 0} \frac{g(s)}{s} = 8\pi q = 8\pi \max\{q_1, q_2\}.\] Setting \(s_0 = \min \left\{ \frac{4\pi}{\|n_{in}\|_{L^\infty}}, \, 2\pi t \ln\left(\frac{1}{\theta_t}\right) \right\}\), we obtain: \[\|n(t)\|_{L^\infty} \le \max \left\{ 2 \|n_{in}\|_{L^\infty}, \, \frac{4}{t \ln\left(\frac{1}{\theta_t}\right)} \right\} \frac{1 + \sqrt{\theta_t}}{1 - \sqrt{\theta_t}}.\]
To make the bound fully explicit, we estimate the supremum of \(\theta_t\). By the Hardy-Littlewood rearrangement inequality, the functional \(\int_0^\infty n_{in}^*(s) e^{-\frac{s}{8\pi t}} \, ds\) under the constraints \(0 \le n_{in}^* \le \|n_{in}\|_{L^\infty}\) and \(\int_0^\infty n_{in}^* \, ds = 8\pi\) is maximized by the step function: \[n_{in}^*(s) = \begin{cases} \|n_{in}\|_{L^\infty} & 0 \le s \le \frac{8\pi}{\|n_{in}\|_{L^\infty}} \\ 0 & s > \frac{8\pi}{\|n_{in}\|_{L^\infty}} \end{cases}\] This gives: \[R_t(0) \le 4\pi \int_0^{\frac{8\pi}{\|n_{in}\|_{L^\infty}}} \|n_{in}\|_{L^\infty} e^{-\frac{s}{8\pi t}} \, ds = 32\pi^2 t \|n_{in}\|_{L^\infty} \left(1 - e^{-\frac{1}{t \|n_{in}\|_{L^\infty}}}\right).\] Setting \(z = \frac{1}{t \|n_{in}\|_{L^\infty}} > 0\), we get: \[\theta_t \le \frac{1 - e^{-z}}{z}.\] We apply the globally valid inequality \(e^{-z} \ge 1 - \frac{2z}{2+z}\) for all \(z \ge 0\), which yields: \[\theta_t \le 1 - \frac{z}{2+z} \implies 1 - \theta_t \ge \frac{z}{2+z}.\] Using \(1 - \sqrt{\theta_t} \ge \frac{1-\theta_t}{2}\), we obtain: \[1 - \sqrt{\theta_t} \ge \frac{z}{2(2+z)}.\] Thus, the concentration factor is bounded by: \[\frac{1 + \sqrt{\theta_t}}{1 - \sqrt{\theta_t}} \le \frac{2}{1 - \sqrt{\theta_t}} \le \frac{4(2+z)}{z} = 8 + 8 t \|n_{in}\|_{L^\infty}.\]
Next, since \(\ln(1/\theta_t) \ge 1 - \theta_t \ge \frac{z}{2+z}\), we estimate: \[\frac{4}{t \ln(1/\theta_t)} \le \frac{4(2+z)}{t z} = 8 \|n_{in}\|_{L^\infty} + \frac{4}{t}.\] Since we are in the regime \(t > \delta\), we have \(t \|n_{in}\|_{L^\infty} > 0.1\), which implies \(\frac{4}{t} < 40 \|n_{in}\|_{L^\infty}\). Hence, the maximum term is bounded by: \[\max \left\{ 2 \|n_{in}\|_{L^\infty}, \, \frac{4}{t \ln(1/\theta_t)} \right\} \le 8 \|n_{in}\|_{L^\infty} + 40 \|n_{in}\|_{L^\infty} = 48 \|n_{in}\|_{L^\infty}.\] Multiplying these bounds together yields the long-time linear estimate: \[\|n(t)\|_{L^\infty} \le 48 \|n_{in}\|_{L^\infty} \left(8 + 8 t \|n_{in}\|_{L^\infty}\right) = 384 \|n_{in}\|_{L^\infty} \left(1 + t \|n_{in}\|_{L^\infty}\right), \quad \forall t > \delta.\]
Combining the short-time estimate and the long-time estimate, we conclude that for all \(t > 0\): \[\|n(t)\|_{L^\infty} \le 384 \|n_{in}\|_{L^\infty} \left(1 + t \|n_{in}\|_{L^\infty}\right).\] This completes the proof. ◻
4 Monotonicity property and proof of Proposition 6↩︎
In this section, we are committed to proving Proposition 6.
Before the proof of Proposition 6, we give the following useful lemma.
Lemma 8. There hold \[\begin{align} \label{eq:mr} \int_{\{x\in \mathbb{R}^2; n(x,t)>s\}}\partial_t n(x,t) dx=\frac{\partial k}{\partial t}(\mu(s,t),t ), \end{align}\qquad{(6)}\] and \[\begin{align} \label{eq:mr-cor} k(\mu(s,t),t)=\int_{\{x\in \mathbb{R}^2; n(x,t)>s\}} n(x,t) dx. \end{align}\qquad{(7)}\]
Proof. It follows from a theorem of Mossino and Rakotoson ((2.12), [26]). Here we give a complete proof.
Proof of ?? . First, by the classical theory of decreasing rearrangements, the function \(n^*(\sigma, t)\) is equimeasurable with \(n(x,t)\). This implies that the integral of \(n(x,t)\) over any of its super-level sets is exactly equal to the integral of \(n^*(\sigma, t)\) over the corresponding interval \([0, \mu(s,t)]\), where \(\mu(s,t)\) is the Lebesgue measure of the super-level set \(\{x \in \mathbb{R}^2 \mid n(x,t) > s\}\). Therefore, we have \[\int_{\{n(x,t)>s\}} n(x,t) \,dx = \int_0^{\mu(s,t)} n^*(\tau,t) \,d\tau.\] Recalling the definition of \(k(s,t)\) in 5 , the right-hand side is exactly \(k(\mu(s,t), t)\). This proves ?? .
Proof of ?? . We introduce an auxiliary function \(G(t)\) representing the excess mass above the threshold level \(s\): \[\label{eq:G95def} G(t) = \int_{\{n(x,t)>s\}} \big( n(x,t) - s \big) \,dx.\tag{18}\] First, differentiating 18 directly with respect to \(t\) in the physical space yields: \[\label{eq:dt95G951} \frac{d}{dt} G(t) = \int_{\{n(x,t)>s\}} \partial_t n(x,t) \,dx.\tag{19}\] Notice that the boundary terms arising from the differentiation of the domain of integration vanish, since \(n(x,t) = s\) on the boundary \(\partial \{n(x,t)>s\}\).
Second, we rewrite \(G(t)\) using the rearrangement space identity ?? : \[\begin{align} G(t) &= \int_{\{n(x,t)>s\}} n(x,t) \,dx - \int_{\{n(x,t)>s\}} s \,dx \\ &= k(\mu(s,t), t) - s \mu(s,t). \end{align}\] Now, differentiating this expression with respect to \(t\) using the chain rule, we obtain: \[\label{eq:dt95G952} \frac{d}{dt} G(t) = \frac{\partial k}{\partial \mu}(\mu(s,t), t) \frac{\partial \mu}{\partial t} + \frac{\partial k}{\partial t}(\mu(s,t), t) - s \frac{\partial \mu}{\partial t}.\tag{20}\] By the Fundamental Theorem of Calculus applied to definition 5 , we know that \(\frac{\partial k}{\partial \mu} = n^*(\mu, t)\). Substituting \(\mu = \mu(s,t)\) and utilizing the critical identity 4 , which states \(n^*(\mu(s,t), t) = s\), we have \[\frac{\partial k}{\partial \mu}(\mu(s,t), t) = s.\] Plugging this back into 20 , the terms involving \(\frac{\partial \mu}{\partial t}\) perfectly cancel out: \[\begin{align} \frac{d}{dt} G(t) &= s \frac{\partial \mu}{\partial t} + \frac{\partial k}{\partial t}(\mu(s,t), t) - s \frac{\partial \mu}{\partial t} \\ &= \frac{\partial k}{\partial t}(\mu(s,t), t), \end{align}\] which and 19 implies ?? . ◻
Proof of Proposition 6. The proof is divided into three steps.
Step I: The energy estimate. For \(t\in (\tau,T)\), \(h>0\) and \(\rho\in (0, n^*(0,t))\), let
\[\begin{align} \label{eq:test-1}\,\, \Phi(s)=\left\{
\begin{aligned}
0,\quad &s\leq \rho,\\
s-\rho,\quad & \rho<s\leq \rho+h,\\
h,\quad & s>\rho+h.
\end{aligned}
\right.
\end{align}\tag{21}\] Multiplying \(\Phi(n)\) on both sides of \(\eqref{ksns}_1\), we have \[\begin{align}
\label{eq:ks-energy}
\int_{\mathbb{R}^2}\frac{\partial}{\partial t}n \Phi(n)dx&=& -\int_{\mathbb{R}^2}\nabla n \cdot \nabla \Phi(n)dx+\int_{\mathbb{R}^2}n \nabla c \cdot \nabla \Phi(n)dx\nonumber\\
&&+ \int_{\mathbb{R}^2} n u\cdot \nabla \Phi(n)dx=:I_1+{\color{red}I_2}+I_3.
\end{align}\tag{22}\] For the first term \(I_1\) of the right hand of 22 , by 21 we get \[\begin{align}
\label{eq:laplace32n}
&&\lim_{h\searrow 0}\frac{1}{h}\int_{\mathbb{R}^2}\nabla n \cdot \nabla \Phi(n)dx\nonumber\\
&=& \lim_{h\searrow 0}\frac{1}{h}\left(\int_{n>\rho}|\nabla n|^2 dx-\int_{n>\rho+h}|\nabla n|^2 dx\right)\nonumber\\
&=&-\frac{\partial}{\partial \rho}\int_{n>\rho}|\nabla n|^2 dx.
\end{align}\tag{23}\] For the second term \(I_2\) of the right hand of 22 , let \[\begin{align}
\label{eq:test-2}\,\, \Psi(s)=\left\{
\begin{aligned}
0,\quad &s\leq \rho,\\
\frac{1}{2}(s^2-\rho^2),\quad & \rho<s\leq \rho+h,\\
h(\rho+\frac{h}{2}),\quad & s>\rho+h.
\end{aligned}
\right.
\end{align}\tag{24}\] Then 21 , 24 and integration by parts imply \[\begin{align}
\int_{\mathbb{R}^2}n \nabla c \cdot \nabla \Phi(n)dx=\int_{\mathbb{R}^2} \nabla c \cdot \nabla \Psi(n)dx=\int_{\mathbb{R}^2}n \Psi(n)dx.
\end{align}\] Consequently, using 24 again we have \[\begin{align}
\label{eq:n32nablac}
&&\lim_{h\searrow 0}\frac{1}{h}\int_{\mathbb{R}^2}n \nabla c \cdot \nabla \Phi(n)dx\nonumber\\
&=& \lim_{h\searrow 0}\frac{1}{2h}\int_{\rho<n\leq \rho+h} n(n^2-\rho^2)dx+\int_{n> \rho+h}n (\rho+\frac{h}{2})dx\nonumber\\
&=&\rho \int_{n> \rho}ndx\nonumber\\
&= &n^*(\mu(\rho,t),t)\int_{n> \rho}n dx\nonumber\\
&=& \frac{\partial k}{\partial s}(\mu(\rho,t),t) k(\mu(\rho,t),t),
\end{align}\tag{25}\] where we used 4 and ?? .
For the term \(I_3\), there holds \[\begin{align}
\label{eq:u32term}
\int_{\mathbb{R}^2} n u\cdot \nabla \Phi(n)dx=\int_{\mathbb{R}^2} u \cdot \nabla \Psi(n)dx=0.
\end{align}\tag{26}\] Moreover, by ?? we have \[\begin{align}
\label{eq:partial32t32n}
\lim_{h\searrow 0}\frac{1}{h}\int_{\mathbb{R}^2}\frac{\partial}{\partial t}n \Phi(n)dx=\int_{n>\rho}\partial_t ndx=\frac{\partial k}{\partial t}(\mu(\rho,t),t).
\end{align}\tag{27}\]
Consequently, recalling 22 and concluding the above computations such as 23 , 25 , 26 and 27 , we have \[\begin{align} \label{eq:energy-k} \frac{\partial k}{\partial t}(\mu(\rho,t),t)-\frac{\partial}{\partial \rho}\int_{n>\rho}|\nabla n|^2 dx= \frac{\partial k}{\partial s}(\mu(\rho,t),t) k(\mu(\rho,t),t). \end{align}\tag{28}\]
Step II: The lower bound of the diffusion term. Let \[\begin{align} E(\rho):= \int_{n>\rho}|\nabla n|^2 dx, \end{align}\] and we claim that \[\begin{align} \label{eq:bound32of32mu} -\frac{\partial}{\partial \rho} E(\rho)\geq 4\pi \mu\left( -\frac{\partial\mu}{\partial\rho}\right)^{-1}. \end{align}\tag{29}\] Proof of 29 . For any integrable function \(f\), by Co-area formula (for example Theorem 3.11 [27]) we have \[\begin{align} \int_{\mathbb{R}^2} f(x) |\nabla n| \, dx = \int_{-\infty}^{\infty} \left( \int_{\Gamma_s} f(x) \, d\mathcal{H}^1 \right) ds, \end{align}\] where \(\mathcal{H}^1\) denotes the 1-dimensional Hausdorff measure (arc length) and the level set \(\Gamma_s=\{x; n(x,t)=s\}\). Specifically, for integrals over the super-level set \(n > \rho\) we get \[\begin{align} \label{eq:co-area} \int_{n > \rho} g(x) \, dx = \int_{\rho}^{\infty} \left( \int_{\Gamma_s} \frac{g(x)}{|\nabla n|} \, d\mathcal{H}^1 \right) ds. \end{align}\tag{30}\] Taking \(g(x) = 1\) in 30 , then we have \[\begin{align} \mu(\rho) = \int_{\rho}^{\infty} \left( \int_{\Gamma_s} \frac{1}{|\nabla n|} \, d\mathcal{H}^1 \right) ds, \end{align}\] which yields that \[\begin{align} \label{eq:derivative32of32mu} -\frac{\partial \mu}{\partial \rho} = \int_{\Gamma_\rho} \frac{1}{|\nabla n|} \, d\mathcal{H}^1. \end{align}\tag{31}\] Moreover, choosing \(g(x) = |\nabla n|^2\) in 30 we have \[\begin{align} E(\rho) = \int_{\rho}^{\infty} \left( \int_{\Gamma_s} |\nabla n| \, d\mathcal{H}^1 \right) ds, \end{align}\] which implies that \[\begin{align} \label{eq:dE} -\frac{\partial E}{\partial \rho} = -\frac{\partial}{\partial \rho}\int_{n>\rho}|\nabla n|^2 \, dx = \int_{\Gamma_\rho} |\nabla n| \, d\mathcal{H}^1. \end{align}\tag{32}\]
The classical isoperimetric inequality in \(\mathbb{R}^2\) states that for a domain with area \(A\) and perimeter \(L\), \(4\pi A \leq L^2\) (see, for example, [28]). In our context, \(A = \mu(\rho)\) and perimeter \(= L(\rho)\). Therefore: \[\label{eq:iso} 4\pi \mu(\rho) \leq L(\rho)^2,\tag{33}\] where \(L(\rho) = \mathcal{H}^1(\Gamma_\rho)\) be the length (perimeter) of the level set \(\Gamma_\rho\). Then \[\begin{align} L(\rho) = \int_{\Gamma_\rho} 1 \, d\mathcal{H}^1 = \int_{\Gamma_\rho} \sqrt{|\nabla n|} \cdot \frac{1}{\sqrt{|\nabla n|}} \, d\mathcal{H}^1. \end{align}\] Applying the Cauchy-Schwarz inequality, (31 ) and (32 ), we obtain that \[\begin{align} \label{eq:CS} L(\rho)^2 \leq \left( -\frac{\partial E}{\partial \rho} \right) \cdot \left( -\frac{\partial \mu}{\partial \rho} \right), \end{align}\tag{34}\] which and 33 implies the claim 29 .
Step III: Conclusions. By 28 and 29 we have \[\begin{align} 4\pi \mu \left( -\frac{\partial\mu}{\partial\rho}\right)^{-1}\leq -\frac{\partial k}{\partial t}(\mu(\rho,t),t)+\frac{\partial k}{\partial s}(\mu(\rho,t),t) k(\mu(\rho,t),t). \end{align}\] By 4 and 5 , this implies \[\begin{align} -\frac{\partial n^*}{\partial s}(s,t)=-\frac{\partial^2 k}{\partial s^2}(s,t)\leq \frac{1}{4\pi s}\left[-\frac{\partial k}{\partial t}(s,t)+\frac{\partial k}{\partial s}(s,t) k(s,t)\right] \end{align}\] The proof of ?? is complete and ?? follows from 5 and the property of \(n^*\). ◻
5 An improved maximum principle and proof of Proposition 7↩︎
In this section, we are committed to proving Proposition 7.
Proof of Proposition 7. Let \(w(s, t) = f(s, t) - g(s, t)\). Next we prove that \(w(s, t) \le 0\) almost everywhere in \(Q_T^*\). Subtracting the inequality for \(g\) from the inequality for \(f\) in condition (iii), we obtain the differential inequality for \(w\): \[\label{eq:diff} \frac{\partial w}{\partial t} - 4\pi s \frac{\partial^2 w}{\partial s^2} - \left( f \frac{\partial f}{\partial s} - g \frac{\partial g}{\partial s} \right) \le 0.\tag{35}\] Define the positive part of the difference as \(w_+(s, t) = \max\{0, w(s, t)\}\). Assume that \(w_+ \in H^1(0, \Lambda)\) for a.e. \(t\) without loss of generality, otherwise we can use an approximation method. Multiplying the inequality by \(w_+\) and integrating over the spatial domain \(\Omega^* = (0, \Lambda)\), we get: \[\label{eq:integral} \frac{1}{2} \frac{d}{dt} \int_0^{\Lambda} (w_+)^2 \, ds - 4\pi \int_0^{\Lambda} s \frac{\partial^2 w}{\partial s^2} w_+ \, ds - \int_0^{\Lambda} \left( f \frac{\partial f}{\partial s} - g \frac{\partial g}{\partial s} \right)w_+ \, ds \le 0.\tag{36}\]
Step I. The Diffusion term. We perform integration by parts on the second term. Note that \(w_+\) is non-zero only where \(w > 0\), so \(\partial_s w = \partial_s w_+\) on the support of \(w_+\). \[\begin{align} - \int_0^{\Lambda} s \frac{\partial^2 w}{\partial s^2} w_+ \, ds &= - \left[ s \frac{\partial w}{\partial s} w_+ \right]_0^{\Lambda} + \int_0^{\Lambda} \frac{\partial}{\partial s} (s w_+) \frac{\partial w}{\partial s} \, ds \\ &= - \Lambda \frac{\partial w}{\partial s}(\Lambda, t) w_+(\Lambda, t) + \int_0^{\Lambda} \left( w_+ \frac{\partial w_+}{\partial s} + s \left(\frac{\partial w_+}{\partial s}\right)^2 \right) \, ds. \end{align}\] From one condition of (v), \(\frac{\partial f}{\partial s}(\Lambda, t) \le \frac{\partial g}{\partial s}(\Lambda, t)\), which implies \(\frac{\partial w}{\partial s}(\Lambda, t) \le 0\). Since \(w_+ \ge 0\), the boundary term \(- \Lambda \frac{\partial w}{\partial s} w_+\) is non-negative. The other condition of (v) \(f(\Lambda, t) \leq g(\Lambda, t)\) implies \(- \Lambda \frac{\partial w}{\partial s} w_+=0\) if \(\Lambda<\infty\). Also, \[\begin{align} \int_0^{\Lambda} w_+ \frac{\partial w_+}{\partial s} \, ds = \frac{1}{2} (w_+(\Lambda, t))^2 \ge 0 \end{align}\] due to the condition (iv). Thus: \[\begin{align} \label{eq:32bound32of32diffusion} - 4\pi \int_0^{\Lambda} s \frac{\partial^2 w}{\partial s^2} w_+ \, ds \ge 4\pi \int_0^{\Lambda} s \left(\frac{\partial w_+}{\partial s}\right)^2 \, ds. \end{align}\tag{37}\]
Step II. The convection term. Note that the third term in (36 ) satisfies \[\begin{align} \int_0^{\Lambda} w_+\left(f\frac{\partial}{\partial s}f-g\frac{\partial}{\partial s}g\right) \, ds &=& \int_0^{\Lambda} w_+^2\frac{\partial}{\partial s} f+g\frac{\partial w_+}{\partial s} w_+\, ds\\ &\leq &C_1(t) \int_0^{\Lambda} w_+^2ds+\int_0^{\Lambda} \left| f\frac{\partial w_+}{\partial s}\right| w_+\, ds \end{align}\] where we used the conditions (ii), (vii) and \(g\leq f\) when \(w_+\) does not vanish. Using conditions (ii) and (iv) we have \[\begin{align} |f|\leq \min\{C_1(t), C_1(t)s\}\leq C_1(t)\sqrt{s}, \end{align}\] which implies \[\begin{align} \int_0^{\Lambda} \left| f\frac{\partial w_+}{\partial s}\right| w_+\, ds\leq \int_0^{\Lambda} s (\partial_s w_+)^2+ C_1(t)^2 \int_0^{\Lambda} w_+^2ds. \end{align}\] Hence, we have \[\begin{align} \label{eq:third32term} \int_0^{\Lambda} w_+\left(f\frac{\partial}{\partial s}f-g\frac{\partial}{\partial s}g\right) \, ds &\leq &\left(C_1(t) + C_1(t)^2 \right)\int_0^{\Lambda} w_+^2ds+\int_0^{\Lambda} s(\partial_s w_+)^2 ds\nonumber\\ \end{align}\tag{38}\]
Combining the estimates 37 and 38 into (36 ), we have \[\frac{1}{2} \frac{d}{dt} \|w_+\|_{L^2}^2 + 4\pi \int_0^{\Lambda} s (\partial_s w_+)^2 \le 2\pi \int_0^{\Lambda} s (\partial_s w_+)^2 + \left(C_1(t) + C_1(t)^2 \right) \int_0^{\Lambda} (w_+)^2.\] Absorbing the gradient term: \[\frac{d}{dt} \|w_+(t)\|_{L^2}^2 \le 2C \|w_+(t)\|_{L^2}^2.\] From the condition (vi), \(w(s, 0) \le 0\), so \(w_+(s, 0) = 0\), implying \(\|w_+(0)\|_{L^2}^2 = 0\). By Gronwall’s inequality, \(\|w_+(t)\|_{L^2}^2 = 0\) for all \(t \in [0, T]\). Therefore, \(w_+(s, t) = 0\) a.e., which implies \(f(s, t) \le g(s, t)\) on \(Q_T^*\). ◻
6 Appendix: Local well-posed result of the system 1↩︎
In this section, we establish the local existence and uniqueness of mild solutions to the system (1.1) using the Contraction Mapping Principle.
Theorem 9. Assume that \(\phi\in L^\infty\left([0,\infty); \dot{W}^{1,\infty}(\mathbb{R}^{2})\right)\), \(0\leq n_{in}\in L^1{\cap L^{\infty}}(\mathbb{R}^{2})\) and \(u_{in}\in W^{1,2}(\mathbb{R}^{2})\). Then there exists a local strong solution \((n, u, \pi)\) to the Keller-Segel-Navier-Stokes equations 1 with the initial data \((n_{in},u_{in})\) .
Proof. Let \(P\) denote the Leray projection operator onto the divergence-free vector fields in \(L^2(\mathbb{R}^2)\). The system (1.1) can be rewritten as the following integral equations: \[\label{eq:mild95form} \left\{ \begin{align} n(t) &= e^{t\Delta} n_{in} - \int_0^t e^{(t-s)\Delta} \nabla \cdot (n u + n \nabla c)(s) \, ds, \\ u(t) &= e^{t\Delta} u_{in} - \int_0^t e^{(t-s)\Delta} P \left( u \cdot \nabla u + n \nabla \phi \right)(s) \, ds, \end{align} \right.\tag{39}\] where \(c(t) = (-\Delta)^{-1} n(t) = -\frac{1}{2\pi} \ln|\cdot| * n(t)\).
Step I: Function Spaces. We seek solutions in a Banach space that captures the critical scaling of the Keller-Segel system while maintaining the regularity required for the Navier-Stokes equations. For \(T > 0\), define the space \(\mathcal{X}_T\) as: \[\mathcal{X}_T := \left\{ (n, u) \;\middle|\; \begin{align} &n \in L^\infty(0, T; L^1{\cap L^{\infty}}(\mathbb{R}^2)),\\&u \in L^\infty(0, T; H^1(\mathbb{R}^2)), \end{align} \right\}\] where the norm on \(\mathcal{X}_T\) is defined by \[\|(n, u)\|_{\mathcal{X}_T} := \sup_{0<t<T} \|n(t)\|_{L^1\cap L^{\infty}} + \sup_{0<t<T} \|u(t)\|_{H^1}.\] Define a closed ball in \(\mathcal{X}_T\) centered at the origin with radius \(R\) with \(T\leq T_1\): \[B_{R}(T) := \left\{ (n, u) \in \mathcal{X}_T \;\middle|\; \|(n, u)\|_{\mathcal{X}_T} \le R \right\},\] where \(R\) is chosen sufficiently large, e.g., \(R > 2(\|n_{in}\|_{L^1\cap L^{\infty}} + \|u_{in}\|_{H^1})\).
Step II: Linear estimates and contraction mapping. Define the map
\(\Phi(n, u) = (\Phi_1(n, u), \Phi_2(n, u))\) corresponding to the right-hand side of 39 . We aim to show that for sufficiently small \(T\), \(\Phi\) maps \(B_{R}(T)\) into itself and is a contraction.
II.1: Estimates for the Cell Density \(n\). Recall the \(L^p\)-\(L^q\) smoothing estimates for the heat semigroup \(e^{t\Delta}\) in \(\mathbb{R}^2\) (see, for example, Proposition A.16 in [29]): \[\begin{align} \label{eq:semigroup} \|\nabla^k e^{t\Delta} f\|_{L^q} \le C t^{\frac{1}{q}-\frac{1}{p} - \frac{k}{2}} \|f\|_{L^p}, \quad 1 \le p \le q \le \infty. \end{align}\tag{40}\] By the Hardy-Littlewood-Sobolev (HLS) inequality, since \(n \in L^1\cap L^{\infty}\subset L^{4/3}\), \[\begin{align} \label{eq:HLS} \|\nabla c\|_{L^{4}}= \|\nabla (-\Delta)^{-1} n\|_{L^{4}} \leq C \|n\|_{L^{4/3}}\leq C\|n\|_{L^1\cap L^{\infty}}. \end{align}\tag{41}\] For \(\Phi_1\), by 39 and 40 we estimate the \(L^{4/3}\) norm with the time weight \(t^{1/4}\): \[\begin{align} &\|\Phi_1(n, u)(t)\|_{L^1\cap L^{\infty}} \le \|e^{t\Delta} n_{in}\|_{L^1\cap L^{\infty}} + \int_0^t \left\| e^{(t-s)\Delta} \nabla \cdot (nu + n\nabla c)(s) \right\|_{L^1\cap L^{\infty}} ds \\ \le& \| n_{in}\|_{L^1\cap L^{\infty}} + C \int_0^t ((t-s)^{-1/2}+(t-s)^{-3/4}) \left( \|nu\|_{L^1\cap L^4} + \|n\nabla c\|_{L^1\cap L^4} \right) ds. \end{align}\] Using Hölder’s inequality, 41 and Gagliardo-Nirenberg interpolation inequality we have \[\|n \nabla c\|_{L^1\cap L^4} \le \|n\|_{L^1\cap L^{\infty}} \|\nabla c\|_{L^4} \le C\|n\|_{L^1\cap L^{\infty}}^2,\] \[\|n u\|_{L^1\cap L^4} \le C \|n\|_{L^1\cap L^{\infty}} \|u\|_{L^4} \le C \|n\|_{L^1\cap L^{\infty}}\|u\|_{H^1}.\] Then, \[\begin{align} \|\Phi_1(t)\|_{L^1\cap L^{\infty}} &\le \| n_{in}\|_{L^1\cap L^{\infty}}+ C \int_0^t ((t-s)^{-1/2}+(t-s)^{-3/4}) \|(n,u)\|_{\mathcal{X}_T}^2 \, ds \\ &\leq \| n_{in}\|_{L^1\cap L^{\infty}}+ C (t^{1/2}+t^{1/4}) \|(n,u)\|_{\mathcal{X}_T}^2. \end{align}\] Thus, for \(t \in (0, T)\), we have \[\sup_{0<t<T} \|\Phi_1(t)\|_{L^1\cap L^{\infty}} \le \| n_{in}\|_{L^1\cap L^{\infty}}+ C(T^{1/2}+ T^{1/4})R^2\le \| n_{in}\|_{L^1\cap L^{\infty}}+R/4,\] if \(T\) is sufficiently small.
II.2: Estimates for the Velocity \(u\). For \(\Phi_2\), we need to bound the \(H^1\) norm. \[\|\Phi_2(t)\|_{H^1} \le \|e^{t\Delta} u_{in}\|_{H^1} + \int_0^t \|e^{(t-s)\Delta} P \nabla \cdot (u \otimes u)\|_{H^1} ds + \int_0^t \|e^{(t-s)\Delta} P (n \nabla \phi)\|_{H^1} ds.\] For the convection term, using 40 again we have \[\begin{align} \int_0^t \|\nabla e^{(t-s)\Delta} (u \otimes u)\|_{L^2} ds &\le C \int_0^t (t-s)^{-1/2} \|u\|_{L^4}^2 ds \\ &\le C T^{1/2} \sup_t \|u(t)\|_{H^1}^2, \end{align}\] and \[\begin{align} \int_0^t \|\nabla^2 e^{(t-s)\Delta} (u \otimes u)\|_{L^2} ds &\le C \int_0^t (t-s)^{-2/3} \|u\nabla u\|_{L^\frac{3}{2}} ds \\ &\le C T^{1/3} \sup_t \|u(t)\|_{H^1}^2, \end{align}\]
For the coupling term \(n \nabla \phi\), since \(\phi \in \dot{W}^{1,\infty}\): \[\begin{align} \int_0^t \|e^{(t-s)\Delta} (n \nabla \phi)\|_{H^1} ds &\leq \int_0^t \| e^{(t-s)\Delta} (n \nabla \phi)\|_{L^2} ds+\int_0^t \|\nabla e^{(t-s)\Delta} (n \nabla \phi)\|_{L^2} ds \\ &\le C \int_0^t (1+(t-s)^{-1/2}) \|n\|_{L^2} \|\nabla \phi\|_{L^\infty} ds\\ &\le C(t +t^{1/2})\sup_t\|n\|_{L^1\cap L^{\infty}} \|\nabla \phi\|_{L^\infty}. \end{align}\] Now we have \[\sup_{0<t<T} \|\Phi_2(t)\|_{H^1} \le \| u_{in}\|_{H^1}+ C(T+ T^{1/3})(R^2+R\|\nabla \phi\|_{L^{\infty}((0,T),L^\infty)})\le \| u_{in}\|_{H^1}+ R/4,\] and \[\|\Phi(n,u)\|_{\mathcal{X}_T}\le \| n_{in}\|_{L^1\cap L^{\infty}}+\| u_{in}\|_{H^1}+R/2<R,\] if \(T\) is sufficiently small.
Step III: Conclusion. Choose \(T\in(0,1)\) small enough. Then \(\Phi\) maps \(B_{R}(T)\) into itself. Similar estimates on the difference \((\Phi(n_1, u_1) - \Phi(n_2, u_2))\) show that \(\Phi\) is a contraction map. Thus, by the Banach Fixed Point Theorem, there exists a unique mild solution to (1 ) on \([0, T)\). Moreover, \(n\geq 0\) due to the maximum principle. Standard bootstrapping arguments allow us to upgrade this mild solution to a strong solution for \(t > 0\). ◻
Remark 10. In the proof, \(T\in(0,1)\) depends only on \(\| n_{in}\|_{L^1\cap L^{\infty}}\), \(\| u_{in}\|_{H^1}\) and \(\|\nabla \phi\|_{L^{\infty}((0,1),L^\infty)}\). Thus if the solution blows up in finite time \(T^*\) then
\(n \not\in L^\infty(0, T^*; L^1{\cap L^{\infty}}(\mathbb{R}^2))\) or \(u \not\in L^\infty(0, T^*; H^1(\mathbb{R}^2))\).
W. Wang was supported by National Key R&D Program of China (No. 2023YFA1009200) and NSFC under grant 12471219. Z. Zhang is supported by NSF of China under Grant No. 12288101.
Data Availability Statement: Data sharing is not applicable to this article as no data sets were generated or analysed during the current study.
Conflict of Interest: The authors declare that they have no conflict of interest.