A Note on $L^{\infty}$-Bound and Uniqueness to a Degenerate Keller-Segel Model

Abstract

In this note we establish the uniform $L^{\infty}$-bound for the weak solutions to a degenerate Keller-Segel equation with the diffusion exponent $\frac {2n}{n+2}< m<2-\frac{2}{n}$ under a sharp condition on the initial data for the global existence. As a consequence, the uniqueness of the weak solutions is also proved.

Appendix

For the convenience of the reader, this section will outline proofs of Lemmas 3.1 and 3.2.

Proof of Lemma 3.1

From (1.2), we know that

$$\begin{aligned} |\nabla c(x)-\nabla c(y)| =& \bigg|-\frac{1}{n\alpha(n)} \int _{\mathbb{R}^{n}} \biggl(\frac{x-z}{|x-z|^{n}}-\frac {y-z}{|y-z|^{n}} \biggr) \rho dz \bigg| \\ \leq& \frac{1}{n\alpha(n)}\int_{\mathbb{R}^{n}} \bigg|\frac {x-z}{|x-z|^{n}}- \frac{y-z}{|y-z|^{n}} \bigg| \rho dz. \end{aligned}$$

(4.1)

Let $r=|x-y|$. If $r \ge1$, using the boundedness of $\|\rho\|_{L^{1}}$ and $\|\rho\|_{L^{\infty}}$, we know (3.1) is true. Next we need only to consider the case $0< r<1$. By dividing the region in (4.1), we have

$$\begin{aligned} &|\nabla c(x)-\nabla c(y)| \\ &\quad \leq \frac{1}{n\alpha(n)} \biggl(\int_{|x-z|\geq2r} \bigg| \frac {x-z}{|x-z|^{n}}-\frac{y-z}{|y-z|^{n}} \bigg| \rho dz +\int_{|x-z|< 2r} \bigg|\frac{x-z}{|x-z|^{n}}-\frac{y-z}{|y-z|^{n}}\bigg| \rho dz \biggr)\\ &\quad=:I_{1}+I_{2}. \end{aligned}$$

For $I_{2}$, from the boundedness of $\|\rho\|_{L^{\infty}}$, we have

$$\begin{aligned} I_{2} =& \frac{1}{n\alpha(n)}\int_{|x-z|< 2r} \bigg| \frac {x-z}{|x-z|^{n}}-\frac{y-z}{|y-z|^{n}} \bigg| \rho dz \\ \leq&C \|\rho\|_{L^{\infty}} \biggl(\int_{|x-z|< 2r} \frac {1}{|x-z|^{n-1}} dz+\int_{|y-z|< 3r}\frac{1}{|y-z|^{n-1}} dz \biggr) \\ \leq& C\|\rho\|_{L^{1}\cap L^{\infty}} r. \end{aligned}$$

For $I_{1}$, we have directly

$$\begin{aligned} I_{1} =& \frac{1}{n\alpha(n)}\int_{|x-z|\geq2r}\bigg| \frac {x-z}{|x-z|^{n}}-\frac{y-z}{|y-z|^{n}}\bigg| \rho dz \\ \leq& C \int_{|x-z|\geq2r}\max \biggl\{ \frac{1}{|x-z|^{n}}, \frac {1}{|y-z|^{n}} \biggr\} |x-y| \rho dz \\ \leq& C|x-y| \biggl(\int_{2r\leq|x-z|< 2 }\max \biggl\{ \frac {1}{|x-z|^{n}},\frac{1}{|y-z|^{n}} \biggr\} \rho dz \\ &{}+\int_{|x-z|\geq2 }\max \biggl\{ \frac{1}{|x-z|^{n}}, \frac {1}{|y-z|^{n}} \biggr\} \rho dz \biggr) \\ \leq& C r \biggl(\|\rho\|_{L^{\infty}} \biggl(\int_{2r\leq|x-z|< 2 } \frac{1}{|x-z|^{n}} dz +\int_{r\leq|y-z|< 3 }\frac{1}{|y-z|^{n}} dz \biggr)+M_{0} \biggr) \\ \leq& C\|\rho\|_{L^{1}\cap L^{\infty}} r \biggl(\int_{2r}^{2} \frac {1}{s^{n}}s^{n-1} ds +\int_{r}^{3} \frac{1}{s^{n}} s^{n-1} ds \biggr) + \bar{C}\|\rho\|_{L^{1}\cap L^{\infty}} r \\ \leq& C\|\rho\|_{L^{1}\cap L^{\infty}} r(1-\log r). \end{aligned}$$

In summary, the relation (3.1) holds for any $r>0$, and a simple computation gives (3.2). □

Proof of Lemma 3.2

Let $\rho_{s}:=\tau_{s\sharp} \rho$ be the interpolation along the optimal transport map $\tau_{1}$ between $\rho$ and $\bar{\rho}$, where $\tau_{s}=(1-s)I+s\tau_{1}$. Taking $\psi\in C_{c}^{\infty}(\mathbb {R}^{n})$, we obtain

$$\begin{aligned} &\frac{d}{ds}\int_{\mathbb{R}^{n}}\psi(x) \rho_{s}(x) dx \\ &\quad=\frac{d}{ds}\int_{\mathbb{R}^{n}}\psi\bigl((1-s)x+s \tau_{1}(x)\bigr) \rho (x) dx \\ &\quad=\int_{\mathbb{R}^{n}}\nabla\psi\bigl((1-s)x+s\tau_{1}(x) \bigr)\cdot\bigl(\tau _{1}(x)-x\bigr) \rho(x) dx \\ &\quad\leq \biggl(\int_{\mathbb{R}^{n}}|\nabla\psi\bigl((1-s)x+s \tau_{1}(x)\bigr)\big|^{2} \rho(x) dx \biggr)^{1/2} \biggl(\int_{\mathbb{R}^{n}}|\tau_{1}(x)-x|^{2} \rho(x) dx \biggr)^{1/2} \\ &\quad\leq \biggl(\int_{\mathbb{R}^{n}}|\nabla\psi(x)|^{2} \rho_{s}(x) dx \biggr)^{1/2}W_{2}(\rho, \bar{\rho}). \end{aligned}$$

(4.2)

Integrating (4.2) respect to $s$ from 0 to 1, we get

$$\begin{aligned} \int_{\mathbb{R}^{n}} \psi(x) \bigl(\bar{\rho}(x)-\rho \bigr) dx =& \int_{0}^{1} \biggl(\int _{\mathbb{R}^{n}}|\nabla\psi(x)|^{2} \rho_{s}(x) dx \biggr)^{1/2} ds W_{2}(\bar{\rho},\rho) \end{aligned}$$

(4.3)

$$\begin{aligned} \leq& \Bigl(\sup_{s\in[0,1]}\|\rho_{s} \|_{L^{\infty}} \Bigr)^{1/2} \biggl(\int_{\mathbb{R}^{n}}| \nabla\psi(x)|^{2} dx \biggr)^{1/2}W_{2}(\rho, \bar{\rho}). \end{aligned}$$

(4.4)

On the other hand, taking $\psi(x)$ as a test function in the second equation of (1.1), we have

$$\begin{aligned} \int_{\mathbb{R}^{n}}\psi(x) (\bar{\rho}-\rho) dx =-\int _{\mathbb{R}^{n}}\psi(x) \Delta(\bar{c}- c) dx=\int_{\mathbb {R}^{n}} \nabla\psi(x) \cdot(\nabla\bar{c}-\nabla c) dx. \end{aligned}$$

(4.5)

From (4.3) and (4.5), we can deduce

$$\int_{\mathbb{R}^{n}}\nabla\psi(x) \cdot(\nabla\bar{c}-\nabla c) dx\leq \Bigl(\sup_{s\in[0,1]}\|\rho_{s}\|_{L^{\infty}} \Bigr)^{1/2}\|\nabla\psi(x)\|_{L^{2}}W_{2}(\rho, \bar{\rho}). $$

From [14, Corollary 2.7]), one has $\|\rho_{s}\|_{L^{\infty}}\leq \max\{\|\bar{\rho}\|_{L^{\infty}}, \|\rho\|_{L^{\infty}}\}$. Thus

$$\|\nabla\bar{c}-\nabla c\|_{L^{2}}\leq \bigl(\max\{\|\bar{\rho}\| _{L^{\infty}}, \|\rho\|_{L^{\infty}}\} \bigr)^{1/2}W_{2}( \rho, \bar{\rho}). $$

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