Citation: Blessing O. Emerenini, Stefanie Sonner, Hermann J. Eberl. Mathematical analysis of a quorum sensing induced biofilm dispersal model and numerical simulation of hollowing effects[J]. Mathematical Biosciences and Engineering, 2017, 14(3): 625-653. doi: 10.3934/mbe.2017036
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Yuhui Chen, Ronghua Pan, Leilei Tong .
The sharp time decay rate of the isentropic Navier-Stokes system in |
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In this paper, we are concerned with the sharp decay rates of solutions to the Cauchy problem for the isentropic Navier-Stokes equations:
{∂tρ+div(ρu)=0,(t,x)∈R+×R3,∂t(ρu)+div(ρu⊗u)+∇p(ρ)=divT,(t,x)∈R+×R3,lim|x|→∞ρ=ˉρ,lim|x|→∞u=0,t∈R+,(ρ,u)|t=0=(ρ0,u0),x∈R3, | (1.1) |
which governs the motion of a isentropic compressible viscous fluid. The unknown functions
Using the classical spectral method, the optimal time decay rate (upper bound) of the linearized equations of the isentropic Navier-Stokes equations are well known. One may then expect that the small solution of the nonlinear equations (1.1) have the same decay rate as the linear one. Our work is devoted to proving the sharp time decay rate (for both upper and lower bound) for the nonlinear system.
In the case of one space dimension, Zeng [24] and Liu-Zeng [15] offered a detailed analysis of the solution to a class of hyperbolic-parabolic system through point-wise estimate, including the isentropic Navier-Stokes system. For multi-dimensional Navier-Stokes equations (and/or Navier-Stokes-Fourier system), the
When additional external force is taken into account, the external force does affect the long time behavior of solutions. The upper bound of time decay rates were studied intensively, see for instance [1] and [2] on unbounded domain, [22], [23] on the convergence of the non-stationary flow to the corresponding steady flow when the initial date are small in
The main goal of current paper is to establish the sharp decay rate, on both upper and lower bounds, to the solutions of (1.1) using relatively simple energy method. We remark that similar results had been pursued by M. Schonbek [20], [21] for incompressible Navier-Stokes equations, and by Li, Matsumura-Zhang [13] for isentropic Navier-Stokes-Poisson system. Although they share the same spirit in obtaining the lower bound decay rates, the feature of the spectrum near zero exhibits quite different behaviors, leading to different analysis. For instance, we explored the elegant structure of the higher order nonlinear terms of Navier-Stokes, when choosing conservative variables: density and momentum. The conservative form of the sharp equations provided a natural derivative structure in these terms, leading to the possibility of a faster decay rate estimate. We will make a more detailed comparison later in this paper.
Define
{∂tn+divm=0,(t,x)∈R+×R3,∂tm+c2∇n−ˉμ△m−(ˉμ+ˉν)∇divm=F,(t,x)∈R+×R3,lim|x|→∞n=0,lim|x|→∞m=0,t∈R+,(n,m)|t=0=(ρ0−ˉρ,ρ0u0),x∈R3, | (1.2) |
where
F=−div{m⊗mn+ˉρ+ˉμ∇(nmn+ˉρ)}−∇{(ˉμ+ˉν)div(nmn+ˉρ)+(p(n+ˉρ)−p(ˉρ)−c2n)}. |
It is this structure of
Our aim is to obtain a clear picture of the large time behavior of
{∂t˜n+div˜m=0,(t,x)∈R+×R3,∂t˜m+c2∇˜n−ˉμ△˜m−(ˉμ+ˉν)∇div˜m=0,(t,x)∈R+×R3,lim|x|→∞˜n=0,lim|x|→∞˜m=0,t∈R+,(˜n,˜m)|t=0=(ρ0−ˉρ,ρ0u0),x∈R3, | (1.3) |
where
Notation. For
We now state our main result.
Theorem 1.1. Assume that
\begin{equation} \int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3}(n_0,m_0)dx\neq 0, \end{equation} | (1.4) |
then there is a unique global classical solution
\begin{eqnarray*} \begin{split} C^{-1}(1+t)^{-\frac34-\frac{k}2}&\leq\|{\nabla}^k \widetilde n(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \leq C(1+t)^{-\frac34-\frac{k}2}, \quad k = 0,1,2,3,\\ C^{-1}(1+t)^{-\frac34-\frac{k}2}&\leq\|{\nabla}^{k} \widetilde m(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \leq C(1+t)^{-\frac34-\frac{k}2}, \quad k = 0,1,2,3, \end{split} \end{eqnarray*} |
and the initial value problem (1.2) has a unique solution
\begin{eqnarray*} \begin{split} &\|{\nabla}^{k} ( n_h,m_h)(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \lesssim \delta_0^2(1+t)^{-\frac54-\frac{k}2},\quad k = 0,1,2,\\ &\|{\nabla}^{3} m_h(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \lesssim \delta_0^2(1+t)^{-\frac{11}4},\quad \|{\nabla}^{3} n_h(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \lesssim \delta_0(1+t)^{-\frac74}. \end{split} \end{eqnarray*} |
As a consequence, there exists a positive constant
\begin{eqnarray*} \begin{split} C_1^{-1}(1+t)^{-\frac34-\frac{k}2}\leq&\| {\nabla}^{k} n (t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \leq C_1(1+t)^{-\frac34-\frac{k}2},\quad k = 0,1, 2,\\ C_1^{-1}(1+t)^{-\frac34-\frac{k}2}\leq&\| {\nabla}^{k} m (t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \leq C_1(1+t)^{-\frac34-\frac{k}2},\quad k = 0,1, 2,3.\\ \end{split} \end{eqnarray*} |
Remark 1.1. We remark that this theorem is valid under the condition (1.4) which is important in the lower bound estimate to the linearized problem. When (1.4) fails, the decay rate of the linearized system (1.3) depends on the order of the degeneracy of moments. Assume
Remark 1.2. In [13], Li, Matsumura-Zhang proved the lower bound decay rate of the linearized isentropic Navier-Stokes-Poisson system, they only require
In what follows, we will set
\begin{equation} \left\{\begin{array}{l} \partial_t n + \bar \rho{\mathop{{\rm{div}}}\nolimits} u = -n{\mathop{{\rm{div}}}\nolimits} u - u\cdot {\nabla} n,\\ \partial_t u + \gamma\bar \rho{\nabla} n -\bar \mu△ u-(\bar \mu+\bar\nu){\nabla}{\mathop{{\rm{div}}}\nolimits} u\\ \qquad = -u\cdot {\nabla} u- \bar \mu f(n)△ u-(\bar \mu+\bar\nu)f(n){\nabla}{\mathop{{\rm{div}}}\nolimits} u-g(n){\nabla} n, \\ \lim\limits_{|x|\to\infty}n = 0, \quad\lim\limits_{|x|\to\infty}u = 0,\\ (n,u)\big|_{t = 0} = (\rho_0-\bar \rho,u_0), \end{array}\right. \end{equation} | (2.1) |
where
\begin{equation} f(n): = \frac n{n+\bar \rho}, \quad \quad g(n): = \frac{p'(n+\bar \rho)}{n+\bar \rho}-\frac{p'(\bar \rho)}{\bar \rho}. \end{equation} | (2.2) |
We assume that there exist a time of existence
\begin{equation} \|n(t)\|_{H^3}+\|u(t)\|_{H^3}\leq \delta, \end{equation} | (2.3) |
holds for any
\begin{eqnarray*} \frac{\bar \rho}2\leq n+\bar \rho\leq 2\bar \rho. \end{eqnarray*} |
Hence, we immediately have
\begin{equation} \left|f(n)\right|,\left|g(n)\right|\leq C|n|, \quad \big|{\nabla}^kf(n)\big|,\big|{\nabla}^kg(n)\big|\leq C \quad \forall k\in{\mathop{\mathbb N\kern 0pt}\nolimits}^+, \end{equation} | (2.4) |
where
Next, we begin with the energy estimates including
Theorem 2.1. Assume that
\begin{eqnarray*} \|n_0\|_{H^3}+\|u_0\|_{H^3} \leq \delta_0, \end{eqnarray*} |
then the problem (2.1) admits a unique global solution
\begin{eqnarray*} \|n(t)\|_{H^3}^2+\|u(t)\|_{H^3}^2+\int_0^t\left (\|{\nabla} n(\tau)\|_{H^2}^2+\|{\nabla} u(\tau)\|_{H^3}^2 \right)d\tau \leq C\left(\|n_0\|^2_{H^3}+\|u_0\|^2_{H^3}\right), \end{eqnarray*} |
where
The proof of this theorem is divided into several subsections.
For
\begin{equation} \begin{split} &\frac12 \frac{d}{d t}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3}\left (\gamma|n|^2+|u|^2\right)dx+ \int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3}\left(\bar\mu|{\nabla} u|^2+(\bar\mu+\bar\nu)|{\mathop{{\rm{div}}}\nolimits} u|^2\right)dx\\ = &\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3}\gamma\left(-n{\mathop{{\rm{div}}}\nolimits} u - u\cdot {\nabla} n\right)n-\big(u\cdot {\nabla} u+\bar\mu f(n)△ u\\ &\quad+(\bar \mu+\bar\nu)f(n){\nabla}{\mathop{{\rm{div}}}\nolimits} u+g(n){\nabla} n\big)\cdot u dx\\ \lesssim&\|n\|_{L^3}\|{\nabla} u\|_{L^2}\| n\|_{L^6}+\left(\|u\|_{L^3}\|{\nabla} u\|_{L^2}+\|n\|_{L^3}\|{\nabla} n\|_{L^2}\right)\| u\|_{L^6}\\ &\quad +\left(\|u\|_{L^\infty}\|{\nabla} n\|_{L^2}+\|n\|_{L^\infty}\|{\nabla} u\|_{L^2}\right)\|{\nabla} u\|_{L^2}\\ \lesssim&\left(\|n\|_{L^3}+\|u\|_{L^3}+\|n\|_{L^\infty}+\|u\|_{L^\infty}\right)\left(\|{\nabla} n\|_{L^2}^2+\|{\nabla} u\|^2_{L^2}\right). \end{split} \end{equation} | (2.5) |
Now for
\begin{equation} \begin{split} &\frac12 \frac{d}{dt}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} \left(\gamma|{\nabla} n|^2+|{\nabla} u|^2 \right)dx + \int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3}\left( \bar\mu|{\nabla}^2u|^2 +(\bar\mu+\bar\nu)|{\nabla} {\mathop{{\rm{div}}}\nolimits} u|^2\right)dx\\ \lesssim&\left(\|n\|_{L^\infty}+\|u\|_{L^\infty}+\|{\nabla} n\|_{L^\infty}+\|{\nabla} u\|_{L^\infty}\right)\left(\|{\nabla} n\|_{L^2}^2+\|{\nabla} u\|_{L^2}^2+\|{\nabla}^2 u\|_{L^2}^2\right). \end{split} \end{equation} | (2.6) |
For
\begin{equation} \begin{split} &\frac12 \frac{d}{dt}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} \left(\gamma|{\nabla}^2 n|^2+|{\nabla}^2 u|^2 \right)dx + \int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} \left(\bar\mu|{\nabla}^3 u|^2 +(\bar\mu+\bar\nu)|{\nabla}^2{\mathop{{\rm{div}}}\nolimits} u|^2\right)dx\\ \lesssim& \left(\|n\|_{L^\infty}+\|u\|_{L^\infty}+\|{\nabla} n\|_{L^\infty}+\|{\nabla} u\|_{L^\infty}\right)\left(\|{\nabla}^2 n\|_{L^2}^2+\|{\nabla}^2 u\|_{L^2}^2+\|{\nabla}^3 u\|_{L^2}^2\right).\\ \end{split} \end{equation} | (2.7) |
For
\begin{equation} \begin{split} &\frac12 \frac{d}{dt}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} \left(\gamma|{\nabla}^3 n|^2+|{\nabla}^3 u|^2 \right)dx + \int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3}\left(\bar\mu |{\nabla}^4 u|^2 +(\bar\mu+\bar\nu)|{\nabla}^3{\mathop{{\rm{div}}}\nolimits} u|^2\right)dx\\ \lesssim& \left(\|n\|_{L^\infty}+\|u\|_{L^\infty}+\|{\nabla} n\|_{L^\infty}+\|{\nabla} u\|_{L^\infty}\right)\left(\|{\nabla}^3 n\|_{L^2}^2+\|{\nabla}^3 u\|_{L^2}^2+\|{\nabla}^4 u\|_{L^2}^2\right)\\ &\quad+\|{\nabla} n\|_{L^3}\|{\nabla}^4 u\|_{L^2}\|{\nabla}^2 n\|_{L^6} +\|{\nabla} u\|_{L^3}\|{\nabla}^4 u\|_{L^2}\|{\nabla}^2 u\|_{L^6}\\ &\quad +\|{\nabla}^2 n\|_{L^3}\left(\|{\nabla}^3 n\|_{L^2}+\|{\nabla}^4 u\|_{L^2}\right)\|{\nabla}^2 u\|_{L^6}. \end{split} \end{equation} | (2.8) |
Summing up the above estimates, noting that
\begin{equation} \begin{split} \frac{d}{dt}\sum\limits_{0\leq k\leq3}\left(\gamma\|{\nabla}^k n\|^2_{L^2}+\|{\nabla}^k u\|^2_{L^2} \right) + C_1\sum\limits_{1\leq k\leq4}\|{\nabla}^{k} u\|^2_{L^2} \leq C_2\delta\sum\limits_{1\leq k\leq3}\|{\nabla}^{k} n\|^2_{L^2}. \end{split} \end{equation} | (2.9) |
For
\begin{equation} \begin{split} &\frac{d}{dt}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} u\cdot {\nabla} n dx +\gamma \bar \rho\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} |{\nabla} n|^2 dx\\ \lesssim &\|{\nabla} u\|_{L^2}^2+\|{\nabla} n\|_{L^2}\|{\nabla}^2 u\|_{L^2}+ (\|n\|_{L^\infty}+ \|u\|_{L^\infty})\left(\|{\nabla} n\|_{L^2}^2+\|{\nabla} u\|_{L^2}^2\right), \end{split} \end{equation} | (2.10) |
for
\begin{equation} \begin{split} &\frac{d}{dt}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} {\nabla} u\cdot {\nabla}^2 n dx +\gamma \bar \rho\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} |{\nabla}^2 n|^2 dx\\ \lesssim&\|{\nabla}^2 u\|_{L^2}^2+\|{\nabla}^2 n\|_{L^2}\|{\nabla}^3 u\|_{L^2}+ (\|(n, u)\|_{L^\infty}+\|({\nabla} n, {\nabla} u)\|_{L^\infty})\\ &\quad\times\left(\|{\nabla} n\|_{L^2}^2+\|{\nabla}^2 n\|_{L^2}^2+\|{\nabla}^2 u\|_{L^2}^2\right), \end{split} \end{equation} | (2.11) |
and for
\begin{equation} \begin{split} &\frac{d}{dt}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} {\nabla}^2 u\cdot {\nabla}^3 n dx +\gamma \bar \rho\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} |{\nabla}^3 n|^2 dx\\ \lesssim &\|{\nabla}^3 u\|_{L^2}^2+\|{\nabla}^3 n\|_{L^2}\|{\nabla}^4 u\|_{L^2}+ (\|(n, u)\|_{L^\infty}+\|({\nabla} n, {\nabla} u)\|_{L^\infty})\\ &\quad\times\left(\|{\nabla}^2 n\|_{L^2}^2+\|{\nabla}^2 u\|_{L^2}^2+\|{\nabla}^3 n\|_{L^2}^2+\|{\nabla}^3 u\|_{L^2}^2\right). \end{split} \end{equation} | (2.12) |
Plugging the above estimates, using the smallness of
\begin{equation} \begin{split} \frac{d}{dt}\sum\limits_{0\leq k\leq 2}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} {\nabla}^k u\cdot {\nabla}^{k+1} n dx +C_3\sum\limits_{1\leq k\leq 3}\|{\nabla}^{k} n\|_{L^2}^2 \leq C_4\sum\limits_{1\leq k\leq 4}\|{\nabla}^{k} u\|_{L^2}^2. \end{split} \end{equation} | (2.13) |
Proof of Theorem 2.1. Multiplying (2.13) by
\begin{equation} \begin{split} \frac{d}{dt}&\bigg\{\sum\limits_{0\leq k\leq3}\left(\gamma\|{\nabla}^k n\|^2_{L^2}+ \|{\nabla}^k u\|^2_{L^2} \right)+\frac{2C_2\delta}{C_3}\sum\limits_{0\leq k\leq 2}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} {\nabla}^k u\cdot {\nabla}^{k+1} n dx \bigg\}\\ &\qquad+ C_5\bigg\{\sum\limits_{1\leq k\leq 3}\|{\nabla}^{k} n\|_{L^2}^2+\sum\limits_{1\leq k\leq4}\|{\nabla}^{k} u\|^2_{L^2}\bigg\} \leq 0. \end{split} \end{equation} | (2.14) |
Next, we define
\begin{equation} \frac{d}{dt}\mathcal E(t)+\|{\nabla} n(t)\|_{H^2}^2+\|{\nabla} u(t)\|^2_{H^3} \leq 0. \end{equation} | (2.15) |
Observe that since
\begin{eqnarray*} C_6^{-1}\left(\| n(t)\|^2_{H^3}+\|u(t)\|^2_{H^3}\right) \leq\mathcal E(t)\leq C_6\left(\| n(t)\|^2_{H^3}+\|u(t)\|^2_{H^3}\right). \end{eqnarray*} |
Then integrating (2.15) directly in time, we get
\begin{eqnarray*} \begin{split} &\sup\limits_{0\leq t \leq T}\left (\|n(t)\|^2_{H^3}+\|u(t)\|^2_{H^3}\right)+C_6\int_0^T \left(\|{\nabla} n(\tau)\|_{H^2}^2+\|{\nabla} u(\tau)\|^2_{H^3}\right)d\tau\\ \leq & C_6^2\left(\|n_0\|^2_{H^3}+\|u_0\|^2_{H^3}\right). \end{split} \end{eqnarray*} |
Using a standard continuity argument along with classical local wellposedness theory, this closes the a priori assumption (2.3) if we assume
In this section, we consider the initial value problem for the linearized Navier-Stokes system
\begin{equation} \left\{\begin{array}{l} \partial_t \widetilde{n} + {\mathop{{\rm{div}}}\nolimits} \widetilde{m} = 0, \qquad (t,x)\in{\mathop{\mathbb R\kern 0pt}\nolimits}^+\times{\mathop{\mathbb R\kern 0pt}\nolimits}^3,\\ \partial_t \widetilde{m} + c^2 {\nabla} \widetilde{n} - \bar\mu △ \widetilde{m}-(\bar \mu+\bar\nu){\nabla}{\mathop{{\rm{div}}}\nolimits} \widetilde{m} = 0,\qquad (t,x)\in{\mathop{\mathbb R\kern 0pt}\nolimits}^+\times{\mathop{\mathbb R\kern 0pt}\nolimits}^3,\\ \lim\limits_{|x|\to\infty}\widetilde n = 0, \quad\lim\limits_{|x|\to\infty}\widetilde m = 0,\qquad t\in{\mathop{\mathbb R\kern 0pt}\nolimits}^+,\\ (\widetilde{n} ,\widetilde{m})\big|_{t = 0} = (\rho_0-\bar\rho, \rho_0u_0),\qquad x\in{\mathop{\mathbb R\kern 0pt}\nolimits}^3, \end{array}\right. \end{equation} | (3.1) |
where
In terms of the semigroup theory for evolutionary equations, the solution
\begin{eqnarray*} \widetilde U_t = B \widetilde U,\quad t \geq 0, \qquad \widetilde U(0) = \widetilde U_0, \end{eqnarray*} |
which gives rise to
\begin{eqnarray*} \widetilde U(t) = S(t)\widetilde U_0 = e^{tB}\widetilde U_0, \quad t \geq 0, \end{eqnarray*} |
where
B = {\left( \begin{matrix} 0 & -{\mathop{{\rm{div}}}\nolimits} \\ -c^2 \nabla & \bar\mu △angle+(\bar\mu+\bar\nu)\nabla{\mathop{{\rm{div}}}\nolimits} \end{matrix} \right).} |
What left is to analyze the differential operator
\begin{eqnarray*} \partial_t \widehat{\widetilde U}(t,\xi) = A(\xi)\widehat{\widetilde U}(t,\xi), \quad t \geq 0, \qquad \widehat{\widetilde U}(0,\xi) = \widehat{\widetilde U}_0(\xi), \end{eqnarray*} |
where
\begin{eqnarray*} A(\xi) = {\left( \begin{matrix} 0 & -i\xi^t \\ -c^2 i\xi & -\bar\mu |\xi|^2I_{3\times3}-(\bar\mu+\bar\nu)\xi\otimes\xi \end{matrix} \right).} \end{eqnarray*} |
The eigenvalues of the matrix
\begin{eqnarray*} \det(A(\xi)-\lambda I) = -(\lambda+\bar\mu|\xi|^2)^2(\lambda^2+(2\bar\mu+\bar\nu) |\xi|^2\lambda + c^2|\xi|^2) = 0, \end{eqnarray*} |
which implies
\begin{eqnarray*} \lambda_0 = -\bar\mu|\xi|^2 (\text{double}), \quad \lambda_1 = \lambda_1(|\xi|), \quad \lambda_2 = \lambda_2(|\xi|). \end{eqnarray*} |
The semigroup
\begin{eqnarray*} e^{tA} = e^{\lambda_0 t}P_0+e^{\lambda_1 t}P_1+e^{\lambda_2 t}P_2, \end{eqnarray*} |
where the project operators
\begin{eqnarray*} P_i = \prod\limits_{i\ne j}\frac{A(\xi)-\lambda_j I}{\lambda_i-\lambda_j}. \end{eqnarray*} |
By a direct computation, we can verify the exact expression for the Fourier transform
\begin{eqnarray*} \begin{split} \widehat G(t,\xi)& = e^{tA} = {\left( \begin{matrix} \frac{\lambda_1e^{\lambda_2 t}-\lambda_2e^{\lambda_1 t}}{\lambda_1-\lambda_2}& -\frac{i\xi^t(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2} \\ -\frac{c^2 i\xi(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2} & e^{-\lambda_0 t}(I-\frac{\xi \otimes \xi} {|\xi|^2})+\frac{\xi \otimes \xi} {|\xi|^2}\frac{\lambda_1e^{\lambda_1 t}-\lambda_2e^{\lambda_2 t}}{\lambda_1-\lambda_2} \end{matrix} \right)} = {\left( \begin{matrix} \widehat N \\ \widehat M \end{matrix} \right).} \end{split} \end{eqnarray*} |
Indeed, we can make the following decomposition for
\begin{eqnarray*} \begin{split} \widehat {\widetilde n} = \widehat N \cdot \widehat{\widetilde U}_0 = (\widehat {\mathcal N} +\widehat {\mathfrak N} )\cdot \widehat{\widetilde U}_0,\quad \widehat {\widetilde m} = \widehat M \cdot \widehat{\widetilde U}_0 = (\widehat {\mathcal M} +\widehat {\mathfrak M} )\cdot \widehat{\widetilde U}_0, \end{split} \end{eqnarray*} |
where
\begin{eqnarray*} \begin{split} &\widehat {\mathcal N} = \left( \frac{\lambda_1e^{\lambda_2 t}-\lambda_2e^{\lambda_1 t}}{\lambda_1-\lambda_2} \quad 0\right),\quad \widehat {\mathfrak N} = \left(0 \quad -\frac{i\xi^t(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2}\right),\\ &\widehat {\mathcal M} = \left(-\frac{c^2 i\xi(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2} \quad 0\right),\quad \widehat {\mathfrak M} = \left(0 \quad e^{-\lambda_0 t}(I-\frac{\xi \otimes \xi} {|\xi|^2})+\frac{\xi \otimes \xi} {|\xi|^2}\frac{\lambda_1e^{\lambda_1 t}-\lambda_2e^{\lambda_2 t}}{\lambda_1-\lambda_2}\right). \end{split} \end{eqnarray*} |
We further decompose the Fourier transform
Define
\begin{eqnarray*} \begin{split} \widehat {\mathcal N} = \widehat {\mathcal N}_1+\widehat {\mathcal N}_2,\quad \widehat {\mathfrak N} = \widehat {\mathfrak N}_1+\widehat {\mathfrak N}_2,\quad \widehat {\mathcal M} = \widehat {\mathcal M}_1+\widehat {\mathcal M}_2,\quad \widehat {\mathfrak M} = \widehat {\mathfrak M}_1+\widehat {\mathfrak M}_2, \end{split} \end{eqnarray*} |
where
\begin{eqnarray*} \begin{split} \chi(\xi) = { \left\{\begin{array}{l} 1,\quad |\xi|\leq R,\\ 0,\quad |\xi|\geq R+1. \end{array}\right.} \end{split} \end{eqnarray*} |
Then we have the following decomposition for
\begin{equation} \begin{split} &\widehat {\widetilde n} = \widehat N \cdot \widehat{\widetilde U}_0 = \widehat {N}_1 \cdot \widehat{\widetilde U}_0+\widehat {N}_2 \cdot \widehat{\widetilde U}_0 = (\widehat {\mathcal N}_1+\widehat {\mathfrak N}_1)\cdot \widehat{\widetilde U}_0 +(\widehat {\mathcal N}_2+\widehat {\mathfrak N}_2)\cdot \widehat{\widetilde U}_0,\\ &\widehat {\widetilde m} = \widehat M \cdot \widehat{\widetilde U}_0 = \widehat {M}_1 \cdot \widehat{U}_0+\widehat {M}_2 \cdot \widehat{\widetilde U}_0 = (\widehat {\mathcal M}_1+\widehat {\mathfrak M}_1)\cdot \widehat{\widetilde U}_0 +(\widehat {\mathcal M}_2+\widehat {\mathfrak M}_2)\cdot \widehat{\widetilde U}_0. \end{split} \end{equation} | (3.2) |
To derive the long time decay rate of solution, we need to use accurate approximation to the Fourier transform
\begin{equation} \begin{split} \lambda_1 = -\frac{2\bar\mu+\bar\nu}2 |\xi|^2 + \frac i 2\sqrt{4c^2|\xi|^2-(2\bar\mu+\bar\nu)^2|\xi|^4} = a+bi,\\ \lambda_2 = -\frac{2\bar\mu+\bar\nu}2 |\xi|^2 - \frac i 2 \sqrt{4c^2|\xi|^2-(2\bar\mu+\bar\nu)^2|\xi|^4} = a-bi, \end{split} \end{equation} | (3.3) |
and we have
\begin{eqnarray*} \begin{split} \frac{\lambda_1e^{\lambda_2 t}-\lambda_2e^{\lambda_1 t}}{\lambda_1-\lambda_2} & = e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t}\left[\cos(bt)+\frac12 (2\bar\mu+\bar\nu)|\xi|^2\frac{\sin(bt)}{b}\right]\\ &\sim O(1)e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t},\quad |\xi|\leq\eta, \end{split} \end{eqnarray*} |
\begin{eqnarray*} \begin{split} \frac{\lambda_1e^{\lambda_1 t}-\lambda_2e^{\lambda_2 t}}{\lambda_1-\lambda_2} & = e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t}\left[\cos(bt)-\frac12 (2\bar\mu+\bar\nu)|\xi|^2\frac{\sin(bt)}{b}\right]\\ & \sim O(1)e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t},\quad |\xi|\leq\eta, \end{split} \end{eqnarray*} |
\begin{eqnarray*} \begin{split} \frac{e^{\lambda_1 t}-e^{\lambda_2 t}}{\lambda_1-\lambda_2} = e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t}\frac{\sin(bt)}{b}\sim O(1)\frac 1{|\xi|}e^{-\frac12(2\bar\mu+\bar\nu)|\xi|^2t},\quad |\xi|\leq\eta, \end{split} \end{eqnarray*} |
where
\begin{eqnarray*} b = \frac12 \sqrt{4c^2|\xi|^2-(2\bar\mu+\bar\nu)^2|\xi|^4} \sim c|\xi|+O(|\xi|^3), \quad |\xi|\leq\eta. \end{eqnarray*} |
For the high frequency
\begin{equation} \begin{split} \lambda_1 = -\frac{2\bar\mu+\bar\nu} 2 |\xi|^2 - \frac12 \sqrt{(2\bar\mu+\bar\nu)^2|\xi|^4 - 4c^2|\xi|^2} = a-b,\\ \lambda_2 = -\frac{2\bar\mu+\bar\nu} 2 |\xi|^2 + \frac12 \sqrt{(2\bar\mu+\bar\nu)^2|\xi|^4 - 4c^2|\xi|^2} = a+b, \end{split} \end{equation} | (3.4) |
and we have
\begin{eqnarray*} \begin{split} \frac{\lambda_1e^{\lambda_2 t}-\lambda_2e^{\lambda_1 t}}{\lambda_1-\lambda_2} & = \frac12 e^{(a+b)t}[1+e^{-2bt}] - \frac{a}{2b} e^{(a+b)t}[1-e^{-2bt}] \sim O(1)e^{-R_0 t},\quad |\xi|\geq \eta, \end{split} \end{eqnarray*} |
\begin{eqnarray*} \begin{split} \frac{\lambda_1e^{\lambda_1 t}-\lambda_2e^{\lambda_2 t}}{\lambda_1-\lambda_2} & = \frac{a+b}{2b} e^{(a+b)t}[1-e^{-2bt}] +e^{(a-b)t} \sim O(1)e^{-R_0 t},\quad |\xi|\geq \eta, \end{split} \end{eqnarray*} |
\begin{eqnarray*} \begin{split} \frac{e^{\lambda_1 t}-e^{\lambda_2 t}}{\lambda_1-\lambda_2} = \frac{1}{2b}e^{(a+b)t}[1- e^{-2bt}] \sim O(1)\frac1{|\xi|^2}e^{-R_0 t},\quad |\xi|\geq \eta, \end{split} \end{eqnarray*} |
where
\begin{eqnarray*} b = \frac12 \sqrt{(2\bar\mu+\bar\nu)^2|\xi|^4 - 4c^2|\xi|^2} \sim \frac12 (2\bar\mu+\bar\nu) |\xi|^2-\frac{2c^2}{2\bar\mu+\bar\nu}+O(|\xi|^{-2}), \quad |\xi|\geq \eta. \end{eqnarray*} |
Here
In this section, we apply the spectral analysis to the semigroup for the linearized Navier-Stokes system. We will establish the
With the help of the formula for Green's function in Fourier space and the asymptotic analysis on its elements, we are able to establish the
Proposition 4.1. Let
\begin{eqnarray*} \|{\nabla}^{k} (\widetilde n,\widetilde m)(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \leq C(1+t)^{-\frac34-\frac{k}2}\big(\|U_0\|_{L^1({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}+\|{\nabla}^k U_0\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}\big), \end{eqnarray*} |
where
Proof. A straightforward computation together with the formula of the Green's function
\begin{eqnarray*} \begin{split} \widehat {\widetilde n}(t,\xi)& = \frac{\lambda_1e^{\lambda_2 t}-\lambda_2e^{\lambda_1 t}}{\lambda_1-\lambda_2}\widehat{n}_0-\frac{i\xi\cdot\widehat{m}_0(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2} \\ &\sim \left\{\begin{array}{l} O(1)e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t}(|\widehat{n}_0|+|\widehat{m}_0|),\quad |\xi|\leq\eta,\\ O(1)e^{-R_0 t}(|\widehat{n}_0|+|\widehat{m}_0|),\quad |\xi|\geq\eta, \end{array}\right.\\ \widehat {\widetilde m}(t,\xi)& = -\frac{c^2 i\xi(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2}\widehat{n}_0+e^{-\lambda_0 t}\widehat{m}_0+\left(\frac{\lambda_1e^{\lambda_1 t}-\lambda_2e^{\lambda_2 t}}{\lambda_1-\lambda_2}-e^{-\lambda_0 t}\right)\frac{\xi (\xi\cdot\widehat{m}_0)} {|\xi|^2} \\ &\sim \left\{\begin{array}{l} O(1)e^{- \bar\mu|\xi|^2t}(|\widehat{n}_0|+|\widehat{m}_0|),\quad |\xi|\leq\eta,\\ O(1)e^{-R_0 t}(|\widehat{n}_0|+|\widehat{m}_0|),\quad |\xi|\geq\eta, \end{array}\right. \end{split} \end{eqnarray*} |
here and below,
\begin{eqnarray*} \begin{split} &\quad\|(\widehat {\widetilde n},\widehat {\widetilde m})(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}^2\\ & = \int_{|\xi|\leq \eta}|(\widehat {\widetilde n},\widehat {\widetilde m})(t,\xi)|^2 d\xi+\int_{|\xi|\geq \eta}|(\widehat {\widetilde n},\widehat {\widetilde m})(t,\xi)|^2 d\xi\\ &\lesssim\int_{|\xi|\leq \eta}e^{-2\bar\mu|\xi|^2t}(|\widehat{n}_0|^2+|\widehat{m}_0|^2) d\xi +\int_{|\xi|\geq \eta}e^{-2R_0 t}(|\widehat{n}_0|^2+|\widehat{m}_0|^2)d\xi\\ &\lesssim(1+t)^{-\frac32}\|({n}_0,{m}_0)\|^2_{L^1({\mathop{\mathbb R\kern 0pt}\nolimits}^3) \cap L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}. \end{split} \end{eqnarray*} |
And the
\begin{eqnarray*} \begin{split} &\quad\|(\widehat {{\nabla}^{k}\widetilde n},\widehat {{\nabla}^{k}\widetilde m})(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}^2\\ & = \int_{|\xi|\leq \eta}|\xi|^{2k}|(\widehat {\widetilde n},\widehat {\widetilde m})(t,\xi)|^2 d\xi+\int_{|\xi|\geq \eta}|\xi|^{2k}|(\widehat {\widetilde n},\widehat {\widetilde m})(t,\xi)|^2 d\xi\\ &\lesssim\int_{|\xi|\leq \eta}e^{-2\bar\mu|\xi|^2t}|\xi|^{2k}(|\widehat{n}_0|^2+|\widehat{m}_0|^2) d\xi +\int_{|\xi|\geq \eta}e^{-2R_0 t}|\xi|^{2k}(|\widehat{n}_0|^2+|\widehat{m}_0|^2)d\xi\\ &\lesssim(1+t)^{-\frac32-k}\big(\|({n}_0,{m}_0)\|^2_{L^1({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}+\|({\nabla}^{k}{n}_0,{\nabla}^{k}{m}_0)\|^2_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}\big). \end{split} \end{eqnarray*} |
The proof of the Proposition 4.1 is completed.
It should be noted that the
Proposition 4.2. Let
\begin{eqnarray*} \begin{split} C^{-1}(1+t)^{-\frac34-\frac{k}2}&\leq \|{\nabla}^{k} \widetilde n(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \leq C(1+t)^{-\frac34-\frac{k}2},\\ C^{-1}(1+t)^{-\frac34-\frac{k}2}&\leq \|{\nabla}^{k} \widetilde m(t)\|_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \leq C(1+t)^{-\frac34-\frac{k}2}, \end{split} \end{eqnarray*} |
where
Proof. We only show the case of
\begin{eqnarray*} \begin{split} &\quad\widehat {\widetilde n}(t,\xi)\\& = \frac{\lambda_1e^{\lambda_2 t}-\lambda_2e^{\lambda_1 t}}{\lambda_1-\lambda_2}\widehat{n}_0-\frac{i\xi\cdot\widehat{m}_0(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2} \\ & = e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t}\left[\cos(b t)\widehat{n}_0-i\xi\cdot\widehat{m}_0\frac{\sin(b t)}{b}\right]+e^{-\frac12(2\bar\mu+\bar\nu)|\xi|^2t}\left[\frac12 (2\bar\mu+\bar\nu)|\xi|^2\frac{\sin(b t)}{b}\widehat{n}_0\right]\\ & = T_1+T_2, \quad \text{for}\quad|\xi|\leq \eta,\\ \end{split} \end{eqnarray*} |
\begin{eqnarray*} \begin{split} &\quad\widehat {\widetilde m}(t,\xi)\\ & = -\frac{c^2 i\xi(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2}\widehat{n}_0+e^{-\lambda_0 t}\widehat{m}_0+\left(\frac{\lambda_1e^{\lambda_1 t}-\lambda_2e^{\lambda_2 t}}{\lambda_1-\lambda_2}-e^{-\lambda_0 t}\right)\frac{\xi (\xi\cdot\widehat{m}_0)} {|\xi|^2} \\ & = \bigg[e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t}\left[\cos(b t)\frac{\xi (\xi\cdot\widehat{m}_0)} {|\xi|^2}-c^2i\xi\frac{\sin(b t)}{b}\widehat{n}_0\right]+e^{- \bar\mu|\xi|^2t}\left[\widehat{m}_0-\frac{\xi (\xi\cdot\widehat{m}_0)} {|\xi|^2}\right]\bigg]\\ &\quad\quad-e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t}\left[\frac12 (2\bar\mu+\bar\nu)|\xi|^2\frac{\sin(b t)}{b}\frac{\xi (\xi\cdot\widehat{m}_0)} {|\xi|^2}\right]\\ & = S_1+S_2, \quad \text{for}\quad|\xi|\leq \eta,\\ \end{split} \end{eqnarray*} |
here and below,
It is easy to check that
\begin{equation} \begin{split} \|\widehat {\widetilde n}(t,\xi)\|^2_{L^2}& = \int_{|\xi|\leq \eta}|\widehat {\widetilde n}(t,\xi)|^2 d\xi+\int_{|\xi|\geq \eta}|\widehat {\widetilde n}(t,\xi)|^2 d\xi\\ &\geq \int_{|\xi|\leq \eta}|T_1+T_2|^2 d\xi \geq \int_{|\xi|\leq \eta}\frac12|T_1|^2-|T_2|^2 d\xi. \end{split} \end{equation} | (4.1) |
We then calculate that
\begin{equation} \begin{split} \int_{|\xi|\leq \eta}|T_2|^2 d\xi &\lesssim \|\widehat{n}_0\|_{L^\infty}^2\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}|\xi|^4\left(\frac{\sin(b t)}{b}\right)^2 d\xi \\ &\lesssim\|\widehat{n}_0\|_{L^\infty}^2\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}|\xi|^2 d\xi\lesssim(1+t)^{-\frac52}\|{n}_0\|_{L^1}^2. \end{split} \end{equation} | (4.2) |
Since
\begin{eqnarray*} \begin{split} |\widehat{n}_0(\xi)|^2\geq \frac1C\left|\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3}n_0(x) d x\right|^2 \geq \frac{M_n^2}C,\quad \text{for}\quad |\xi|\leq\eta. \end{split} \end{eqnarray*} |
For
\begin{eqnarray*} \begin{split} \frac{\left|\xi\cdot\widehat{m}_0(\xi)\right|^2}{|\xi|^2} \geq \frac{\left|\xi\cdot M_m\right|^2}{C|\xi|^2},\quad \text{for}\quad |\xi|\leq\eta. \end{split} \end{eqnarray*} |
When
\begin{equation} \label{optimal3} \begin{split} &\quad\int_{|\xi|\leq \eta}|T_1|^2 d\xi\nonumber\\ &\geq \frac{M_n^2}C\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\cos^2(b t)d\xi+\frac1C\int_{|\xi|\leq \eta}\frac{\left|\xi\cdot M_m\right|^2}{b^2}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\sin^2(b t)d\xi \nonumber \end{split} \end{equation} |
\begin{equation} \begin{split}&\geq \frac{\min\{{M_n^2},\frac{M_m^2}{3c^2}\}}C\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\big(\cos^2(b t)+\sin^2(b t)\big)d\xi\\ &\geq C_1\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}d\xi\\ &\geq C^{-1}(1+t)^{-\frac32}. \end{split} \end{equation} | (4.3) |
If
\begin{eqnarray*} |\widehat{m}_0(\xi)|^2 < \epsilon,\quad \text{for}\quad |\xi|\leq\eta. \end{eqnarray*} |
We thus use the help of spherical coordinates and the change of variables
\begin{equation} \begin{split} &\quad\int_{|\xi|\leq \eta}|T_1|^2 d\xi\\ &\geq \frac{M_n^2}C\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\cos^2(b t)d\xi-\frac{\epsilon}{Cc^2}\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\sin^2(b t)d\xi\\ &\geq \frac{M_n^2}Ct^{-\frac32}\int_0^{\eta\sqrt t}e^{-(2\bar\mu+\bar\nu)r^2}\cos^2(cr\sqrt t)r^2dr-\frac{\epsilon}{Cc^2}t^{-\frac32}\int_0^{\eta\sqrt t}e^{-(2\bar\mu+\bar\nu)r^2}\sin^2(cr\sqrt t)r^2dr\\ &\geq \frac{M_n^2}Ct^{-\frac32}\sum\limits_{k = 0}^{[\frac{c\eta t}\pi]-1}\int_{\frac{k\pi}{c\sqrt t}}^{\frac{k\pi+\frac\pi4}{c\sqrt t}}e^{-(2\bar\mu+\bar\nu)r^2}\cos^2(cr\sqrt t)r^2dr-\frac{\epsilon}{Cc^2}(1+t)^{-\frac32}\\ &\geq \frac{M_n^2}{2C}t^{-\frac32}\sum\limits_{k = 0}^{[\frac{c\eta t}\pi]-1}\int_{\frac{k\pi}{c\sqrt t}}^{\frac{k\pi+\frac\pi4}{c\sqrt t}}e^{-(2\bar\mu+\bar\nu)r^2}r^2dr-\frac{\epsilon}{Cc^2}(1+t)^{-\frac32}\\ &\geq C_1^{-1}(1+t)^{-\frac32}-C_2^{-1}\epsilon(1+t)^{-\frac32}\\ .&\geq C^{-1}(1+t)^{-\frac32} \end{split} \end{equation} | (4.4) |
In the case of
\begin{equation} \begin{split} &\quad\int_{|\xi|\leq \eta}|T_1|^2 d\xi\\ &\geq -\frac\epsilon C\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\cos^2(b t)d\xi+\frac{M_m^2}{3Cc^2}\int_{|\xi|\leq \eta}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\sin^2(b t)d\xi\\ &\geq C^{-1}(1+t)^{-\frac32}. \end{split} \end{equation} | (4.5) |
Combining the above estimates (4.1), (4.2), (4.3), (4.4) and (4.5), we obtain the lower bound of the time decay rate for
\begin{eqnarray*} \| {\widetilde n}(t,x)\|^2_{L^2} = \|\widehat {\widetilde n}(t,\xi)\|^2_{L^2}\geq C^{-1}(1+t)^{-\frac32}. \end{eqnarray*} |
The lower bound of the time decay rate for
\begin{equation} \begin{split} \|\widehat {\widetilde m}(t,\xi)\|^2_{L^2}\geq \int_{|\xi|\leq \eta}\frac12|S_1|^2-|S_2|^2 d\xi, \end{split} \end{equation} | (4.6) |
then we find that
\begin{equation} \begin{split} \int_{|\xi|\leq \eta}|S_2|^2 d\xi \lesssim(1+t)^{-\frac52}\|{m}_0\|_{L^1}^2. \end{split} \end{equation} | (4.7) |
We then calculate that
\begin{eqnarray*} \begin{split} &\quad\int_{|\xi|\leq \eta}|S_1|^2 d\xi\\ &\geq \bigg\{\frac{c^4M_n^2}C\int_{|\xi|\leq \eta}\frac{|\xi|^2}{b^2}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\sin^2(b t)d\xi\\ &\qquad+\frac{1}C\int_{|\xi|\leq \eta}\frac{\left|\xi\cdot M_m\right|^2}{|\xi|^2}e^{-(2\bar\mu+\bar\nu)|\xi|^2t}\cos^2(bt)d\xi\bigg\}\\ &\qquad +\bigg\{\int_{|\xi|\leq \eta}e^{-\frac12(4\bar\mu+\bar\nu)|\xi|^2t}\cos(bt)\frac{\xi(\xi\cdot{\widehat m}_0)}{|\xi|^2}\left({\widehat m}_0-\frac{\xi(\xi\cdot{\widehat m}_0)}{|\xi|^2}\right)d\xi\bigg\}\\ & = J_1+J_2. \end{split} \end{eqnarray*} |
A direct computation gives rise to
\begin{equation} J_1\geq C^{-1}(1+t)^{-\frac32},\qquad J_2 = 0. \end{equation} | (4.8) |
Combining the above estimates (4.6), (4.7) and (4.8), we obtain the lower bound of the time decay rate for
\begin{eqnarray*} \| {\widetilde m}(t,x)\|^2_{L^2} = \|\widehat {\widetilde m}(t,\xi)\|^2_{L^2}\geq C^{-1}(1+t)^{-\frac32}. \end{eqnarray*} |
Then the proof of Proposition 4.2 is completed.
In this subsection, we establish the following
Proposition 4.3. Let
\begin{eqnarray*} \|{\nabla}^{k} (\widetilde n,\widetilde m)(t)\|_{L^p({\mathop{\mathbb R\kern 0pt}\nolimits}^3)} \leq C(1+t)^{-\frac32(1-\frac1p)-\frac{k}2}\big(\|U_0\|_{L^1({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}+\|{\nabla}^k U_0\|_{L^p({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}\big), \end{eqnarray*} |
where
To prove Proposition 4.3, the following two lemmas in [6] are helpful.
Lemma 4.1. Let
\begin{eqnarray*} \begin{split} |{\nabla}^\alpha _\xi \hat f(\xi)| \leq C'{ \left\{\begin{array}{l} |\xi|^{-|\alpha|+\sigma_1},\quad |\xi|\leq R, |\alpha| = n,\\ |\xi|^{-|\alpha|-\sigma_2},\quad |\xi|\geq R, |\alpha| = n-1,n,n+1, \end{array}\right.} \end{split} \end{eqnarray*} |
where
\begin{eqnarray*} f = m_1+m_2\delta, \end{eqnarray*} |
where
\begin{eqnarray*} m_2 = (2\pi)^{-\frac n 2}\lim\limits_{|\xi| \to \infty} \hat f(\xi), \end{eqnarray*} |
and
\begin{eqnarray*} \|f\ast g\|_{L^p} \leq C\| g\|_{L^p},\quad 1\leq p \leq \infty, \end{eqnarray*} |
where
Lemma 4.2. Let
\begin{eqnarray*} |{\nabla}_\xi ^\beta \hat f(\xi)|\leq C'|\xi|^{-|\beta|},\quad |\beta|\leq n+1. \end{eqnarray*} |
Then
\begin{eqnarray*} \|{\nabla}_x ^\alpha g(t,\cdot)\|_{L^p}\leq C(|\alpha|)t^{-\frac n 2(1-\frac 1 p)-\frac{|\alpha|}{2}}. \end{eqnarray*} |
In particular,
Now let us turn to the proof of Proposition 4.3.
Proof of Proposition 4.3. We first analyze above higher frequency terms denoted by
\begin{eqnarray*} \begin{split} \lambda_1 = -(2\bar\mu+\bar\nu) |\xi|^2+\frac{2c^2}{2\bar\mu+\bar\nu}+O(|\xi|^{-2}),\quad \lambda_2 = -\frac{2c^2}{2\bar\mu+\bar\nu}+O(|\xi|^{-2}),\quad |\xi|\geq \eta. \end{split} \end{eqnarray*} |
We shall prove that the higher frequency terms are
\begin{eqnarray*} \frac{\lambda_1e^{\lambda_2 t}-\lambda_2e^{\lambda_1 t}}{\lambda_1-\lambda_2} = e^{\lambda_2 t}+\frac{\lambda_2 e^{\lambda_2 t}}{\lambda_1-\lambda_2}-\frac{\lambda_2 e^{\lambda_1 t}}{\lambda_1-\lambda_2}. \end{eqnarray*} |
By a direct computation, it is easy to verify
\begin{eqnarray*} |{\nabla}_\xi^k \lambda_2|\lesssim|\xi|^{-2-k},\quad |\xi|\geq\eta, \end{eqnarray*} |
which gives rise to
\begin{eqnarray*} \begin{split} \bigg|{\nabla}_\xi^k \Big[(1-\chi(\cdot))e^{\lambda_2 t}\Big]\bigg|, \left|{\nabla}_\xi^k \Big[(1-\chi(\cdot))\frac{\lambda_2 e^{\lambda_2 t}}{\lambda_1-\lambda_2}\Big]\right|\lesssim{ \left\{\begin{array}{l} 0, \quad |\xi|\leq R,\\ e^{-c_1t}|\xi|^{-2-k},\quad |\xi|\geq R, \end{array}\right.} \end{split} \end{eqnarray*} |
here and below,
\begin{eqnarray*} (1-\chi(\cdot))\frac{\lambda_2 e^{\lambda_1 t}}{\lambda_1-\lambda_2} \sim e^{-\frac12(2\bar\mu+\bar\nu) |\xi|^2 t}\Big[(1-\chi(\cdot))\frac{e^{(-\lambda_2-\frac12(2\bar\mu+\bar\nu) |\xi|^2)t}}{\lambda_1-\lambda_2}\Big]. \end{eqnarray*} |
We can regard
\begin{equation} \|({\nabla}_x^k({\mathcal N}_2 \ast f),{\nabla}_x^k({\mathfrak N}_2 \ast f),{\nabla}_x^k( {\mathcal M}_2 \ast f),{\nabla}_x^k( {\mathfrak M}_2 \ast f))(t)\|_{L^p} \leq Ce^{-c_1t}\|{\nabla}_x^k f\|_{L^p}, \end{equation} | (4.9) |
for all integer
We also need to deal with the corresponding lower frequency terms denoted by
\begin{eqnarray*} \begin{split} &\frac{\lambda_1e^{\lambda_2 t}-\lambda_2e^{\lambda_1 t}}{\lambda_1-\lambda_2}, \frac{\lambda_1e^{\lambda_1 t}-\lambda_2e^{\lambda_2 t}}{\lambda_1-\lambda_2},\frac{|\xi|(e^{\lambda_1 t}-e^{\lambda_2 t})}{\lambda_1-\lambda_2}\sim O(1)e^{-\frac12 (2\bar\mu+\bar\nu)|\xi|^2t},\quad |\xi|\leq\eta, \end{split} \end{eqnarray*} |
which imply that for
\begin{eqnarray*} |\widehat{\mathcal N}_1|\sim O(1)e^{-c_2|\xi|^2t},\quad |\widehat{\mathfrak N}_1|\sim O(1)e^{-c_2|\xi|^2t},\\ |\widehat{\mathcal M}_1|\sim O(1)e^{-c_2|\xi|^2t},\quad |\widehat{\mathfrak M}_1|\sim O(1)e^{-c_2|\xi|^2t}, \end{eqnarray*} |
for some constants
\begin{equation} \begin{split} \|({\nabla}^k{\mathcal N}_1,{\nabla}^k{\mathfrak N}_1,{\nabla}^k {\mathcal M}_1,{\nabla}^k {\mathfrak M}_1)(t)\|_{L^p} \leq& C\left(\int_{|\xi|\leq \eta}\big||\xi|^k e^{-c_2|\xi|^2t}\big|^q d\xi\right)^{\frac 1q}\\ \leq& C(1+t)^{-\frac32(1-\frac1p)-\frac k2}. \end{split} \end{equation} | (4.10) |
Combining (4.9) and (4.10), we finally have for
\begin{eqnarray*} \begin{split} \|({\nabla}^k(N \ast f),{\nabla}^k(M \ast f))(t)\|_{L^p}& = \|({\nabla}^k((N_1+N_2) \ast f),{\nabla}^k((M_1+M_2) \ast f))(t)\|_{L^p}\\ &\leq C(1+t)^{-\frac32(1-\frac1p)-\frac k2}\|f\|_{L^1}+Ce^{-c_1t}\|{\nabla}^k f\|_{L^p}\\ &\leq C(1+t)^{-\frac32(1-\frac1p)-\frac k2}(\|f\|_{L^1}+\|{\nabla}^k f\|_{L^p}). \end{split} \end{eqnarray*} |
The proof of Proposition 4.3 is completed.
We are ready to prove Theorem 1.1 on the sharp time decay rate of the global solution to the initial value problem for the nonlinear Navier-Stokes system.
In what follows, we will set
\begin{equation} \left\{\begin{array}{l} \partial_t n_h + {\mathop{{\rm{div}}}\nolimits} m_h = 0,\qquad (t,x)\in{\mathop{\mathbb R\kern 0pt}\nolimits}^+\times{\mathop{\mathbb R\kern 0pt}\nolimits}^3,\\ \partial_t m_h + c^2 {\nabla} n_h- \bar\mu △ m_h- (\bar\mu+\bar\nu){\nabla}{\mathop{{\rm{div}}}\nolimits} m_h = F,\qquad (t,x)\in{\mathop{\mathbb R\kern 0pt}\nolimits}^+\times{\mathop{\mathbb R\kern 0pt}\nolimits}^3,\\ \lim\limits_{|x|\to\infty}n_h = 0, \quad\lim\limits_{|x|\to\infty} m_h = 0,\qquad t\in{\mathop{\mathbb R\kern 0pt}\nolimits}^+,\\ (n_h,m_h)\big|_{t = 0} = (0,0),\qquad x\in{\mathop{\mathbb R\kern 0pt}\nolimits}^3, \end{array}\right. \end{equation} | (5.1) |
where
\begin{eqnarray*} \begin{split} F = &-{\mathop{{\rm{div}}}\nolimits}\Big\{ \frac{(m_h+\widetilde m)\otimes (m_h+\widetilde m)}{n_h+\widetilde n+\bar\rho}+\bar\mu{\nabla}\big(\frac{(n_h+\widetilde n)(m_h+\widetilde m)}{n_h+\widetilde n+\bar\rho}\big)\Big\}\\ & - {\nabla}\Big\{(\bar\mu+\bar\nu){\mathop{{\rm{div}}}\nolimits}(\frac{(n_h+\widetilde n)(m_h+\widetilde m)}{n_h+\widetilde n+\bar\rho})+\big(p(n_h+\widetilde n+\bar\rho)-p(\bar\rho)-c^2(n_h+\widetilde n)\big)\Big\}. \end{split} \end{eqnarray*} |
Denote
\begin{eqnarray*} \partial_t U_h = BU_h+H,\quad t\geq0,\qquad U_h(0) = 0, \end{eqnarray*} |
where the nonlinear term
\begin{eqnarray*} U_h(t) = S(t)\ast U_{h}(0)+\int_0^t S(t-\tau)\ast H(\widetilde U, U_h)(\tau)d\tau, \end{eqnarray*} |
which
\begin{equation} n_h = N\ast U_{h}(0)+\int_0^t \mathfrak N (t-\tau)\ast H(\tau)d\tau, \end{equation} | (5.2) |
\begin{equation} m_h = M\ast U_{h}(0)+\int_0^t \mathfrak M (t-\tau)\ast H(\tau)d\tau. \end{equation} | (5.3) |
Furthermore, in view of the above definition for
\begin{eqnarray*} |\widehat{\mathfrak N}(\xi)|\sim O(1)e^{-c_3|\xi|^2t}, \quad |\widehat{\mathfrak M}(\xi)|\sim O(1)e^{-c_3|\xi|^2t}, \quad|\xi|\leq\eta, \end{eqnarray*} |
\begin{eqnarray*} |\widehat{\mathfrak N}(\xi)|\sim O(1)\frac1{|\xi|}e^{-R_0t}, \quad |\widehat{\mathfrak M}(\xi)|\sim O(1)\frac1{|\xi|^2}e^{-R_0t}+O(1)e^{-c_4|\xi|^2t}, \quad |\xi|\geq\eta. \end{eqnarray*} |
Thus, applying a similar argument as in the proof of Proposition 4.1, we have
\begin{equation} \|({\nabla}^k {\mathfrak N}\ast H, {\nabla}^k {\mathfrak M}\ast H)(t)\|_{L^2} \leq C(1+t)^{-\frac32(\frac1q-\frac12)-\frac12-\frac k 2}\big(\|Q\|_{L^q}+\|{\nabla}^{k+1} Q\|_{L^2}\big),\quad q = 1,2, \end{equation} | (5.4) |
\begin{equation} \|({\nabla}^k {\mathfrak N}\ast H, {\nabla}^k {\mathfrak M}\ast H)(t)\|_{L^2} \leq C(1+t)^{-\frac32(\frac1q-\frac12)-\frac12-\frac k 2}\big(\|Q\|_{L^q}+\|{\nabla}^{k} Q\|_{L^2}\big),\quad q = 1,2, \end{equation} | (5.5) |
\begin{equation} \|{\nabla}^k {\mathfrak M}\ast H(t)\|_{L^2} \leq C(1+t)^{-\frac32(\frac1q-\frac12)-\frac12-\frac k 2}\big(\|Q\|_{L^q}+\|{\nabla}^{k-1} Q\|_{L^2}\big),\quad q = 1,2, \end{equation} | (5.6) |
for any non-negative integer
\begin{equation} \begin{split} Q = &\Big|\frac{(m_h+\widetilde m)\otimes (m_h+\widetilde m)}{n_h+\widetilde n+\bar\rho}+\bar\mu{\nabla}\big(\frac{(n_h+\widetilde n)(m_h+\widetilde m)}{n_h+\widetilde n+\bar\rho}\big)\Big|\\ &+\Big|(\bar\mu+\bar\nu){\mathop{{\rm{div}}}\nolimits}(\frac{(n_h+\widetilde n)(m_h+\widetilde m)}{n_h+\widetilde n+\bar\rho})+\big(p(n_h+\widetilde n+\bar\rho)-p(\bar\rho)-c^2(n_h+\widetilde n)\big)\Big|. \end{split} \end{equation} | (5.7) |
For readers' convenience, we show how to estimate
\begin{eqnarray*} \begin{split} &\quad\|{\nabla}^k {\mathfrak M}\ast H(t)\|_{L^2}^2\\ &\lesssim\int_{|\xi|\leq \eta}e^{-2c_3|\xi|^2t}|\xi|^{2k}|\widehat H|^2d\xi +\int_{|\xi|\geq \eta}e^{-2R_0 t}|\xi|^{2k-4}|\widehat H|^2d\xi\\ &\quad+\int_{|\xi|\geq \eta}e^{-2c_4|\xi|^2 t}|\xi|^{2k}|\widehat H|^2d\xi\\ &\lesssim\int_{|\xi|\leq \eta}e^{-2c_3|\xi|^2t}|\xi|^{2k+2}|\widehat Q|^2d\xi +\int_{|\xi|\geq \eta}e^{-2R_0 t}|\xi|^{2k-2}|\widehat Q|^2d\xi\\ &\quad+\int_{|\xi|\geq \eta}e^{-2c_4|\xi|^2t}|\xi|^{2k+2}|\widehat Q|^2d\xi\\ &\lesssim(1+t)^{-3(\frac1q-\frac12)-1-k}\big(\|Q\|^2_{L^q({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}+\|{\nabla}^{\tilde k} Q\|^2_{L^2({\mathop{\mathbb R\kern 0pt}\nolimits}^3)}\big),\quad q = 1,2,\quad k-1\leq\tilde k\in{\mathop{\mathbb N\kern 0pt}\nolimits}^+. \end{split} \end{eqnarray*} |
In this subsection, we establish the faster decay rate for
We begin with following Lemma.
Lemma 5.1. Let
\begin{eqnarray*} \begin{split} \int_0^{\frac t 2}(1+t-\tau)^{-r_1}(1+\tau)^{-r_2} d\tau = &\int_0^{\frac t 2}(1+\frac t 2+\tau)^{-r_1}(1+\frac t 2-\tau)^{-r_2} d\tau\\ \lesssim&{ \left\{\begin{array}{l} (1+t)^{-r_1}, \quad \mathit{\text{for}} \quad r_2 > 1,\\ (1+t)^{-(r_1-\epsilon)},\quad \mathit{\text{for}} \quad r_2 = 1,\\ (1+t)^{-(r_1+r_2-1)},\quad \mathit{\text{for}} \quad r_2 < 1, \end{array}\right.} \end{split} \end{eqnarray*} |
and
\begin{eqnarray*} \begin{split} \int_{\frac t 2}^t(1+t-\tau)^{-r_1}(1+\tau)^{-r_2} d\tau = &\int_0^{\frac t 2}(1+t-\tau)^{-r_2}(1+\tau)^{-r_1} d\tau\\ \lesssim&{ \left\{\begin{array}{l} (1+t)^{-r_2}, \quad \mathit{\text{for}} \quad r_1 > 1,\\ (1+t)^{-(r_2-\epsilon)},\quad \mathit{\text{for}} \quad r_1 = 1,\\ (1+t)^{-(r_1+r_2-1)},\quad \mathit{\text{for}} \quad r_1 < 1, \end{array}\right.} \end{split} \end{eqnarray*} |
where
Proposition 5.1. Under the assumptions of Theorem 1.1, the solution
\begin{eqnarray*} \begin{split} &\|({\nabla} ^k n_h,{\nabla}^{k} m_h)\|_{L^2}\leq C\delta_0^2(1+t)^{-\frac54-\frac {k} 2},\\ &\|{\nabla} ^3 m_h\|_{L^2}\leq C\delta_0^2(1+t)^{-\frac{11}4},\quad \|{\nabla} ^3 n_h\|_{L^2}\leq C\delta_0(1+t)^{-\frac74}, \end{split} \end{eqnarray*} |
where
From (5.7), we deduce
\begin{eqnarray*} Q(\widetilde U, U_h) = Q_1+Q_2+Q_3+Q_4, \end{eqnarray*} |
which implies for a smooth solution
\begin{eqnarray*} \begin{split} &Q_1 = Q_1(\widetilde U, U_h)\sim O(1)\left(n_h^2+m_h\otimes m_h+\widetilde n^2+\widetilde m\otimes\widetilde m \right),\\ &Q_2 = Q_2(\widetilde U, U_h)\sim O(1)\left(\widetilde n n_h+\widetilde m\otimes m_h\right),\\ &Q_3 = Q_3(\widetilde U, U_h)\sim O(1)\left({\nabla}(n_h\cdot m_h)+{\nabla}(\widetilde n\cdot\widetilde m)\right),\\ &Q_4 = Q_4(\widetilde U, U_h)\sim O(1)\left({\nabla}(\widetilde n\cdot m_h)+{\nabla}( n_h\cdot\widetilde m) \right). \end{split} \end{eqnarray*} |
Define
\begin{equation} \begin{split} \Lambda(t) = :&\sup\limits_{0\leq s \leq t}\bigg\{\sum\limits_{k = 0}^2(1+s)^{\frac54+\frac k 2}{\delta_0}^{-\frac34}\|({\nabla} ^k n_h,{\nabla}^{k}m_h)(s)\|_{L^2}\\ &\quad+(1+s)^{\frac74}\|({\nabla}^3 n_h, {\nabla}^3 m_h)(s)\|_{L^2}\bigg \}. \end{split} \end{equation} | (5.8) |
Proposition 5.2. Under the assumptions of Theorem 1.1, if for some
\begin{eqnarray*} \begin{split} \Lambda(t)\leq C\delta_0^{\frac34},\quad t\in[0,T], \end{split} \end{eqnarray*} |
where
The proof of this Proposition 5.2 consists of following three steps.
Starting with (5.4), (5.5), (5.6) and (5.8), we have after a complicate but straightforward computation that
\begin{equation} \begin{split} \|(n_h, m_h)\|_{L^2}&\lesssim\int_0^t \|(\mathfrak N (t-\tau)\ast H(\tau), \mathfrak M (t-\tau)\ast H(\tau))\|_{L^2}d\tau\\ &\lesssim\int_0^{t} (1+t-\tau)^{-\frac54}\big(\|Q(\tau)\|_{L^1}+\| Q(\tau)\|_{L^2}\big)d\tau\\ &\lesssim\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right)\int_0^{t} (1+t-\tau)^{-\frac54}(1+\tau)^{-\frac32}d\tau\\ &\lesssim(1+t)^{-\frac54}\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right). \end{split} \end{equation} | (5.9) |
It is easy to verify that
\begin{eqnarray*} \begin{split} \|Q(t)\|_{L^1}\lesssim&\|Q_1\|_{L^1}+\|Q_2\|_{L^1}+\|Q_3\|_{L^1}+\|Q_4\|_{L^1}\\ \lesssim &\|(\widetilde n,\widetilde m)\|_{L^2}^2+\|( n_h, m_h)\|_{L^2}^2+ \|( n_h, m_h)\|_{L^2}\big(\|({\nabla}\widetilde n,{\nabla}\widetilde m)\|_{L^2}\\ &+\|({\nabla} n_h,{\nabla} m_h)\|_{L^2}\big)+\|(\widetilde n,\widetilde m)\|_{L^2}\left(\|({\nabla}\widetilde n,{\nabla}\widetilde m)\|_{L^2} +\| ({\nabla} n_h,{\nabla} m_h)\|_{L^2}\right)\\ \lesssim & (1+t)^{-\frac32}\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right). \end{split} \end{eqnarray*} |
Indeed, by virtue of Hölder's inequality and Gagliardo-Nirenberg's inequality, we obtain that
\begin{eqnarray*} \|u\|_{L^\infty}\lesssim \|{\nabla} u\|_{L^2}^{\frac12}\|{\nabla} ^2 u\|_{L^2}^{\frac12}, \end{eqnarray*} |
which implies that
\begin{eqnarray*} \begin{split} &\|Q(t)\|_{L^2}\\ \lesssim &\|(\widetilde n,\widetilde m)\|_{L^\infty}\big(\|(\widetilde n,\widetilde m)\|_{L^2}+\|({\nabla}\widetilde n,{\nabla}\widetilde m)\|_{L^2}+\|(n_h,m_h)\|_{L^2}\\ &+\|({\nabla} n_h,{\nabla} m_h)\|_{L^2}\big)+\|( n_h,m_h)\|_{L^\infty}\left(\|( n_h,m_h)\|_{L^2}+\|( {\nabla} n_h,{\nabla} m_h)\|_{L^2}\right)\\ &+\|({\nabla}\widetilde n,{\nabla}\widetilde m)\|_{L^\infty}\|(n_h,m_h)\|_{L^2}\\ \lesssim &(1+t)^{-\frac94}\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right). \end{split} \end{eqnarray*} |
Furthermore, exactly as in the estimate of the high order derivatives, we have
\begin{equation} \begin{split} &\|({\nabla} n_h, {\nabla} m_h)\|_{L^2}\\ \lesssim &\int_0^{\frac t 2} \|({\nabla}\mathfrak N , {\nabla}\mathfrak M)(t-\tau)\ast H(\tau)\|_{L^2}d\tau+\int_{\frac t 2}^t \|(\mathfrak N, \mathfrak M )(t-\tau)\ast {\nabla} H(\tau)\|_{L^2}d\tau\\ \lesssim &\int_0^{\frac t 2} (1+t-\tau)^{-\frac74}\big(\| Q(\tau)\|_{L^1}+\|{\nabla} Q(\tau)\|_{L^2}\big)d\tau +\int_{\frac t 2}^t (1+t-\tau)^{-\frac12}\|{\nabla} Q(\tau)\|_{L^2}d\tau\\ \lesssim&\left(\delta_0^2+\delta_0^{\frac98}\Lambda^2(t)\right)\Bigg(\int_0^{\frac t 2} (1+t-\tau)^{-\frac74}(1+\tau)^{-\frac32}d\tau+\int_{\frac t 2}^t (1+t-\tau)^{-\frac12}(1+\tau)^{-\frac{11}4}d\tau\Big)\\ \lesssim&(1+t)^{-\frac74}\left(\delta_0^2+\delta_0^{\frac98}\Lambda^2(t)\right), \end{split} \end{equation} | (5.10) |
Similarly, it holds that
\begin{eqnarray*} \begin{split} &\|{\nabla} Q(t)\|_{L^2}\nonumber\\ \lesssim &\|(\widetilde n,\widetilde m)\|_{L^\infty}\big(\|({\nabla}\widetilde n,{\nabla}\widetilde m)\|_{L^2}+\|({\nabla}^2\widetilde n,{\nabla}^2\widetilde m)\|_{L^2}+\|( {\nabla} n_h,{\nabla} m_h)\|_{L^2}\nonumber \end{split} \end{eqnarray*} |
\begin{eqnarray*} \begin{split}&\quad+\|( {\nabla}^2 n_h,{\nabla}^2 m_h)\|_{L^2}\big)+\|({\nabla}\widetilde n,{\nabla}\widetilde m)\|_{L^\infty}\big(\|({\nabla}\widetilde n,{\nabla}\widetilde m)\|_{L^2}+\|(n_h,m_h)\|_{L^2}\\ &\quad+\|( {\nabla} n_h,{\nabla} m_h)\|_{L^2}\big)+\|( n_h,m_h)\|_{L^\infty}\big(\|( {\nabla}^2\widetilde n,{\nabla}^2\widetilde m)\|_{L^2}+\|( {\nabla} n_h,{\nabla} m_h)\|_{L^2}\\ &\quad+\|({\nabla}^2 n_h,{\nabla}^2 m_h)\|_{L^2}\big)+\|( {\nabla} n_h,{\nabla} m_h)\|_{L^\infty}\|( {\nabla} n_h,{\nabla} m_h)\|_{L^2}\\ \lesssim & (1+t)^{-\frac{11}4}\left(\delta_0^2+\delta_0^{\frac98}\Lambda^2(t)\right). \end{split} \end{eqnarray*} |
Thus, we also get that
\begin{equation} \begin{split} &\|({\nabla}^2 n_h, {\nabla}^2 m_h)(t)\|_{L^2}\\ \lesssim& \int_0^{\frac t 2} \|({\nabla}^2 \mathfrak N, {\nabla}^2 \mathfrak M) (t-\tau)\ast H(\tau)\|_{L^2}d\tau\\ &\quad+\int_{\frac t 2}^t \|(\mathfrak N, \mathfrak M) (t-\tau)\ast {\nabla}^2 H(\tau)\|_{L^2}d\tau\\ \lesssim &\int_0^{\frac t 2}(1+t-\tau)^{-\frac94}\big(\|Q(\tau)\|_{L^1}+\|{\nabla}^2 Q(\tau)\|_{L^2}\big)d\tau\\ &\quad+\int_{\frac t 2}^t (1+t-\tau)^{-\frac12}\|{\nabla} ^2Q(\tau)\|_{L^2}d\tau\\ \lesssim& \left(\delta_0^2+\delta_0\Lambda(t)+\delta_0^{\frac34}\Lambda^2(t)\right)\bigg(\int_0^{\frac t 2} (1+t-\tau)^{-\frac94}(1+\tau)^{-\frac32}d\tau\\ &\quad+\int_{\frac t 2}^t (1+t-\tau)^{-\frac12}(1+\tau)^{-\frac{13}4}d\tau\bigg)\\ \lesssim&(1+t)^{-\frac94}\left(\delta_0^2+\delta_0\Lambda(t)+\delta_0^{\frac34}\Lambda^2(t)\right). \end{split} \end{equation} | (5.11) |
Finally, we have
\begin{eqnarray*} \begin{split} &\|{\nabla}^2 Q(t)\|_{L^2}\\ \lesssim &(\|(\widetilde n,\widetilde m)\|_{L^\infty}+\|(n_h,m_h)\|_{L^\infty})(\|({\nabla}^3\widetilde n,{\nabla}^3 \widetilde m)\|_{L^2}+\|({\nabla}^3n_h,{\nabla}^3 m_h)\|_{L^2})\\ &\quad+(\|({\nabla} \widetilde n,{\nabla}\widetilde m)\|_{L^\infty}+\|({\nabla} n_h,{\nabla} m_h)\|_{L^\infty}) (\|({\nabla}\widetilde n,{\nabla} \widetilde m)\|_{L^2}+\|({\nabla} n_h,{\nabla} m_h)\|_{L^2})\\ &\quad+(\|(\widetilde n,\widetilde m)\|_{L^\infty}+\|(n_h,m_h)\|_{L^\infty}+\|({\nabla} \widetilde n,{\nabla}\widetilde m)\|_{L^\infty}+\|({\nabla} n_h,{\nabla} m_h)\|_{L^\infty})\\ &\quad\times(\|({\nabla}^2\widetilde n,{\nabla}^2 \widetilde m)\|_{L^2}+\|({\nabla}^2 n_h,{\nabla}^2 m_h)\|_{L^2})\\ \lesssim&(1+t)^{-\frac{13}4}\left(\delta_0^2+\delta_0\Lambda(t)+\delta_0^{\frac34}\Lambda^2(t)\right). \end{split} \end{eqnarray*} |
In this subsection, we will close the a priori estimates and complete the proof of Proposition 5.2. For this purpose, we need to derive the time decay rate of higher order derivatives of
Lemma 5.2. Under the assumption of Theorem 1.1, one has
\begin{eqnarray*} \|{\nabla}^2 n(t)\|_{H^1}+\|{\nabla}^2 u(t)\|_{H^1}\lesssim (1+t)^{-\frac74}\left(\delta_0+\delta_0^{\frac34}\Lambda(t)\right). \end{eqnarray*} |
In particular, it holds that
\begin{eqnarray*} \|{\nabla}^3 (n_h, m_h)(t)\|_{L^2}\lesssim (1+t)^{-\frac74}\left(\delta_0+\delta_0^{\frac34}\Lambda(t)\right). \end{eqnarray*} |
Proof. First of all, in view of (2.12), recovering the dissipation estimate for
\begin{equation} \begin{split} &\frac{d}{dt}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} {\nabla}^2 u\cdot {\nabla}^3 n dx +C_1\|{\nabla}^3 n\|_{L^2}^2 dx\\ \leq &C_2\left(\|{\nabla}^3 u\|_{L^2}^2+\|{\nabla}^4 u\|_{L^2}^2\right)+C(1+t)^{-\frac{3}2}\left(\delta_0+\delta_0^{\frac38}\Lambda(t)\right)\\ &\quad\times\left(\|{\nabla}^2 n\|_{L^2}^2+\|{\nabla}^2 u\|_{L^2}^2+\|{\nabla}^3 u\|_{L^2}^2\right). \end{split} \end{equation} | (5.12) |
Summing up (2.7) and (2.8) in the energy estimate for
\begin{equation} \begin{split} &\frac{d}{dt}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} \left(\gamma |{\nabla}^2 n|^2+|{\nabla}^2 u|^2 + \gamma|{\nabla}^3 n|^2 +|{\nabla}^3 u|^2 \right)dx + C_3\left(\|{\nabla}^3 u|^2 _{L^2}+\|{\nabla}^4 u\|^2 _{L^2}\right) \\ \leq &C(1+t)^{-\frac{3}2}\left(\delta_0+\delta_0^{\frac38}\Lambda(t)\right)\left(\|{\nabla}^2 n\|_{L^2}^2+\|{\nabla}^2 u\|_{L^2}^2+\|{\nabla}^3 n\|_{L^2}^2\right). \end{split} \end{equation} | (5.13) |
Multiplying (5.12) by
\begin{eqnarray*} \begin{split} &\frac{d}{dt}\bigg\{\sum\limits_{2\leq k\leq3}\left(\gamma \|{\nabla}^k n\|^2_{L^2}+\|{\nabla}^k u\|^2_{L^2} \right)+\epsilon_1\frac{C_3}{C_2}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} {\nabla}^2 u\cdot {\nabla}^{3} n dx \bigg\}\\ &\quad+ C_4\Big(\|{\nabla}^{3} n\|_{L^2}^2+\sum\limits_{3\leq k\leq4}\|{\nabla}^{k} u\|^2_{L^2}\Big)\\ \leq &C(1+t)^{-\frac{3}2}\left(\delta_0+\delta_0^{\frac38}\Lambda(t)\right)\left(\|{\nabla}^2 n\|_{L^2}^2+\|{\nabla}^2 u\|_{L^2}^2\right). \end{split} \end{eqnarray*} |
Next, we define
\mathcal E_1(t) = \bigg\{\sum\limits_{2\leq k\leq3}\left(\gamma \|{\nabla}^k n\|^2_{L^2}+\|{\nabla}^k u\|^2_{L^2} \right)+\epsilon_1\frac{C_3}{C_2}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} {\nabla}^2 u\cdot {\nabla}^{3} n dx \bigg\}. |
Observe that since
\begin{eqnarray*} C_5^{-1}\left(\|{\nabla}^2 n(t)\|^2_{H^1}+\|{\nabla}^2 u(t)\|^2_{H^1}\right) \leq\mathcal E_1(t)\leq C_5\left(\|{\nabla}^2 n(t)\|^2_{H^1}+\|{\nabla}^2 u(t)\|^2_{H^1}\right). \end{eqnarray*} |
Then we arrive at
\begin{eqnarray*} \frac{d}{dt}\mathcal E_1(t)+C_4\Big(\|{\nabla}^{3} n(t)\|_{L^2}^2+\|{\nabla}^3 u(t)\|^2_{H^1}\Big) \leq C(1+t)^{-5}\left(\delta_0+\delta_0^{\frac38}\Lambda(t)\right)\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right). \end{eqnarray*} |
Denote
\begin{eqnarray*} \begin{split} &\frac{C_4}{3}\|{\nabla}^{3} (n, u)(x)\|_{L^2}^2 \geq\frac{C_4}{3}\int_{S(t)^c} |\xi|^6|(\widehat{n}, \widehat{u})(\xi)|^2d\xi\\ \geq&(1+\gamma)(1+t)^{-1}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} |\xi|^4|(\widehat{n}, \widehat{u})(\xi)|^2d\xi-(1+\gamma)(1+t)^{-1}\int_{S(t)} |\xi|^4|(\widehat{n}, \widehat{u})(\xi)|^2d\xi. \end{split} \end{eqnarray*} |
Hence we have
\begin{eqnarray*} \begin{split} &\frac{d}{dt}\mathcal E_1(t)+(1+t)^{-1}\mathcal E_1(t)+\|{\nabla}^{3} n\|_{L^2}^2+\|{\nabla}^3 u\|^2_{H^1}\\ \lesssim&(1+t)^{-5}\left(\delta_0+\delta_0^{\frac38}\Lambda(t)\right)\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right)+(1+t)^{-1}\int_{S(t)} |\xi|^4|(\widehat{n}, \widehat{u})(\xi)|^2d\xi\\ &\quad+(1+t)^{-1}\int_{{\mathop{\mathbb R\kern 0pt}\nolimits}^3} {\nabla}^2 u\cdot {\nabla}^{3} n dx. \end{split} \end{eqnarray*} |
Multiplying the above equation by
\begin{eqnarray*} \begin{split} &\frac{d}{dt}\Big\{(1+t)^5\mathcal E_1(t)\Big\}+(1+t)^5\Big(\|{\nabla}^{3} n\|_{L^2}^2+\|{\nabla}^3 u\|^2_{H^1}\Big) \lesssim(1+t)^{\frac12}\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right). \end{split} \end{eqnarray*} |
Integrating it with respect to time from
\begin{eqnarray*} \begin{split} &(1+t)^5\mathcal E_1(t)+\int_0^T(1+t)^5\Big(\|{\nabla}^{3} n\|_{L^2}^2+\|{\nabla}^3 u\|^2_{H^1}\Big)dt\\ \lesssim& \mathcal E_1(0)+(1+t)^{\frac32}\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right), \end{split} \end{eqnarray*} |
which implies that
\begin{eqnarray*} \|{\nabla}^3 n\|^2_{L^2}+\|{\nabla}^3 u\|^2_{L^2}\lesssim\mathcal E_1(t)\lesssim (1+t)^{-5}\delta_0^2+(1+t)^{-\frac72}\left(\delta_0^2+\delta_0^{\frac32}\Lambda^2(t)\right). \end{eqnarray*} |
Finally, we have
\begin{eqnarray*} \|{\nabla}^3 n_h\|_{L^2}+\|{\nabla}^3 m_h\|_{L^2}\lesssim (1+t)^{-\frac74}\left(\delta_0+\delta_0^{\frac34}\Lambda(t)\right). \end{eqnarray*} |
This completes the proof of this Lemma.
In this subsection, we first combine the above a priori estimates of (5.8), (5.9), (5.10), (5.11) and Lemma 5.2 together to give the proof of the Proposition 5.2. In deed, for any
\begin{equation} \Lambda(t)\leq C\left(\delta_0+\delta_0^{\frac14}\Lambda(t)+\Lambda^2(t)\right) \leq C\delta_0^{\frac34}. \end{equation} | (5.14) |
With the help of standard continuity argument, Proposition 5.2 and the smallness of
\begin{eqnarray*} \begin{split} &\|({\nabla}^k n_h, {\nabla}^k m_h)\|_{L^2}\lesssim \delta_0^2(1+t)^{-\frac54-\frac k2},\quad k = 0,1,\\ &\|{\nabla}^2 (n_h, m_h)\|_{L^2}\lesssim\delta_0^{\frac74}(1+t)^{-\frac94},\quad \|{\nabla}^3 (n_h, m_h)\|_{L^2}\lesssim\delta_0(1+t)^{-\frac74}. \end{split} \end{eqnarray*} |
Consequently, for any
\begin{equation} \Lambda(t)\leq C\delta_0. \end{equation} | (5.15) |
From (5.11) and (5.15), thus we also get that
\begin{eqnarray*} \|{\nabla}^2 (n_h, m_h)\|_{L^2}\lesssim\delta_0^2(1+t)^{-\frac94}. \end{eqnarray*} |
For
\begin{eqnarray*} \begin{split} &\|{\nabla}^3 m_h(t)\|_{L^2} \\\lesssim &\int_0^{\frac t 2}(1+t-\tau)^{-\frac{11}4}\big(\|Q(\tau)\|_{L^1}+\|{\nabla}^2 Q(\tau)\|_{L^2}\big)d\tau\\ &\quad+\int_{\frac t 2}^t (1+t-\tau)^{-\frac12}\|{\nabla} ^2Q(\tau)\|_{L^2}d\tau\\ \lesssim& \delta_0^2\bigg(\int_0^{\frac t 2} (1+t-\tau)^{-\frac{11}4}(1+\tau)^{-\frac32}d\tau+\int_{\frac t 2}^t (1+t-\tau)^{-\frac12}(1+\tau)^{-\frac{13}4}d\tau\bigg)\\ \lesssim&\delta_0^2(1+t)^{-\frac{11}4}. \end{split} \end{eqnarray*} |
Hence, we finish the proof of the Proposition 5.1. Theorem 1.1 follows.
Y. Chen is partially supported by the China Postdoctoral Science Foundation under grant 2019M663198, Guangdong Basic and Applied Basic Research Foundation under grant 2019A1515110733, NNSF of China under grants 11801586, 11971496 and China Scholarship Council. The research of R. Pan is partially supported by National Science Foundation under grants DMS-1516415 and DMS-1813603, and by National Natural Science Foundation of China under grant 11628103. L. Tong's research is partially supported by China Scholarship Council.
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