A regularity criterion for the Cahn-Hilliard-Boussinesq system with zero viscosity

Abstract

This paper studies a coupled Cahn-Hilliard-Boussinesq system with zero viscosity. We prove a regularity criterion in terms of vorticity in the homogeneous Besov space ${\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^0$.

MSC:35Q30, 76D03, 76D05, 76D07.

1 Introduction

In this paper, we study the following Cahn-Hilliard-Boussinesq system with zero viscosity [1]:

$${\partial }_{t}u+(u\cdot \mathrm{\nabla })u+\mathrm{\nabla }\pi =\mu \mathrm{\nabla }\varphi +\theta e_3,$$

(1.1)

$$divu=0,$$

(1.2)

$${\partial }_t\theta +u\cdot \mathrm{\nabla }\theta =\mathrm{\Delta }\theta ,$$

(1.3)

$${\partial }_t\varphi +u\cdot \mathrm{\nabla }\varphi =\mathrm{\Delta }\mu ,$$

(1.4)

$$-\mathrm{\Delta }\varphi +f^{\prime }(\varphi )=\mu ,$$

(1.5)

$$(u,\theta ,\varphi )(x,0)=(u_0,{\theta }_0,{\varphi }_0)(x),x\in {\mathbb{R}}^3,$$

(1.6)

with u the fluid velocity field, θ the temperature, ϕ the order parameter, and π the pressure, are the unknowns. $e_3:={(0,0,1)}^t$. μ is the chemical potential. $f(\varphi ):=\frac{1}{4}{({\varphi }^2-1)}^2$ is the double well potential.

When $\theta =\varphi \equiv 0$, (1.1) and (1.2) are the well-known Euler system; Kozono, Ogawa and Taniuchi [2] proved the following regularity criterion:

$$\omega :=curlu\in L^1(0,T;{\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^0).$$

(1.7)

Here ${\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^0$ denotes the homogeneous Besov space.

When $\varphi =0$, (1.1), (1.2), and (1.3) are the well-known Boussinesq system with zero viscosity; Fan and Zhou [3] also showed the regularity criterion (1.7).

When $u=0$, (1.4) and (1.5) are the well-known Cahn-Hilliard system.

It is easy to show that the problem (1.1)-(1.6) has a unique local smooth solution, thus we omit the details here. However, the global regularity is still open. The aim of this paper is to study the blow up criterion. We will prove the following.

Theorem 1.1 Let $T>0$ and $u_0\in H^3$, ${\theta }_0\in H^2$, ${\varphi }_0\in H^3$ with $div{u}_0=0$ in ${\mathbb{R}}^3$. Suppose that $(u,\theta ,\varphi )$ is a local smooth solution to the problem (1.1)-(1.6). Then $(u,\theta ,\varphi )$ is smooth up to time T provided that (1.7) is satisfied.

We will use the following logarithmic Sobolev inequality [2]:

$${\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}\le C(1+{\parallel curlu\parallel }_{{\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^0}log(e+{\parallel {\mathrm{\Lambda }}^{3}u\parallel }_{L^2})),$$

(1.8)

and the bilinear product and commutator estimates due to Kato and Ponce [4]:

$${\parallel {\mathrm{\Lambda }}^s(fg)\parallel }_{L^p}\le C({\parallel {\mathrm{\Lambda }}^{s}f\parallel }_{L^{p_1}}{\parallel g\parallel }_{L^{q_1}}+{\parallel f\parallel }_{L^{p_2}}{\parallel {\mathrm{\Lambda }}^{s}g\parallel }_{L^{q_2}}),$$

(1.9)

$${\parallel {\mathrm{\Lambda }}^s(fg)-f{\mathrm{\Lambda }}^{s}g\parallel }_{L^p}\le C({\parallel \mathrm{\nabla }f\parallel }_{L^{p_1}}{\parallel {\mathrm{\Lambda }}^{s-1}g\parallel }_{L^{q_1}}+{\parallel {\mathrm{\Lambda }}^{s}f\parallel }_{L^{p_2}}{\parallel g\parallel }_{L^{q_2}}),$$

(1.10)

with $s>0$, $\mathrm{\Lambda }:={(-\mathrm{\Delta })}^{1/2}$ and $\frac{1}{p}=\frac{1}{p_1}+\frac{1}{q_1}=\frac{1}{p_2}+\frac{1}{q_2}$.

2 Proof of Theorem 1.1

First, it follows from (1.2), (1.3), and the maximum principle that

$${\parallel \theta \parallel }_{L^{\mathrm{\infty }}(0,T;L^{\mathrm{\infty }})}\le C.$$

(2.1)

Testing (1.3) by θ, using (1.2), we see that

$$\frac{1}{2}\frac{d}{dt}\int {\theta }^{2}dx+\int {|\mathrm{\nabla }\theta |}^{2}dx=0,$$

whence

$${\parallel \theta \parallel }_{L^{\mathrm{\infty }}(0,T;L^2)}+{\parallel \theta \parallel }_{L^2(0,T;H^1)}\le C.$$

(2.2)

Testing (1.4) by ϕ, using (1.2) and (1.5), we find that

$$\begin{array}{r}\frac{1}{2}\frac{d}{dt}\int {\varphi }^{2}dx+\int {|\mathrm{\Delta }\varphi |}^{2}dx\\ =\int f^{\prime }(\varphi )\mathrm{\Delta }\varphi dx=\int ({\varphi }^3-\varphi )\mathrm{\Delta }\varphi dx\\ =-\int 3{\varphi }^2{|\mathrm{\nabla }\varphi |}^{2}dx-\int \varphi \mathrm{\Delta }\varphi dx\\ \le -\int \varphi \mathrm{\Delta }\varphi dx\le {\parallel \varphi \parallel }_{L^2}{\parallel \mathrm{\Delta }\varphi \parallel }_{L^2}\\ \le \frac{1}{2}{\parallel \mathrm{\Delta }\varphi \parallel }_{L^2}^2+\frac{1}{2}{\parallel \varphi \parallel }_{L^2}^2,\end{array}$$

which gives

$${\parallel \varphi \parallel }_{L^{\mathrm{\infty }}(0,T;L^2)}+{\parallel \varphi \parallel }_{L^2(0,T;H^2)}\le C.$$

(2.3)

Testing (1.1) and (1.4) by u and μ, respectively, using (1.2), (1.5), and (2.2), summing up the result, we deduce that

$$\begin{array}{r}\frac{d}{dt}\int \frac{1}{2}{|\mathrm{\nabla }\varphi |}^2+f(\varphi )+\frac{1}{2}{u}^{2}dx+\int {|\mathrm{\nabla }\mu |}^{2}dx\\ =\int \theta e_{3}udx\le {\parallel \theta \parallel }_{L^2}{\parallel u\parallel }_{L^2}\le C{\parallel u\parallel }_{L^2},\end{array}$$

which gives

$${\parallel \varphi \parallel }_{L^{\mathrm{\infty }}(0,T;H^1)}\le C,$$

(2.4)

$${\parallel u\parallel }_{L^{\mathrm{\infty }}(0,T;L^2)}\le C,$$

(2.5)

$${\parallel \mathrm{\nabla }\mu \parallel }_{L^2(0,T;L^2)}\le C.$$

(2.6)

In the following calculations, we will use the following Gagliardo-Nirenberg inequality:

$${\parallel \varphi \parallel }_{L^{\mathrm{\infty }}}^2\le C{\parallel \mathrm{\nabla }\varphi \parallel }_{L^2}{\parallel \mathrm{\Delta }\varphi \parallel }_{L^2}.$$

(2.7)

It follows from (2.6), (1.5), (2.3), (2.4), and (2.7) that

$$\begin{array}{rcl}{\int }_0^T\int {|\mathrm{\nabla }\mathrm{\Delta }\varphi |}^{2}dxdt& =& {\int }_0^T\int {\left|\mathrm{\nabla }(f^{\prime }(\varphi )-\mu )\right|}^{2}dxdt\\ \le & C{\int }_0^T\int {|\mathrm{\nabla }\mu |}^{2}dxdt+C{\int }_0^T\int {|\mathrm{\nabla }{f}^{\prime }(\varphi )|}^{2}dxdt\\ \le & C+C{\int }_0^T\int {\left|\mathrm{\nabla }({\varphi }^3-\varphi )\right|}^{2}dxdt\\ \le & C+C{\int }_0^T\int {\varphi }^4{|\mathrm{\nabla }\varphi |}^{2}dxdt\\ \le & C+C{\parallel \mathrm{\nabla }\varphi \parallel }_{L^{\mathrm{\infty }}(0,T;L^2)}^2{\int }_0^T{\parallel \varphi \parallel }_{L^{\mathrm{\infty }}}^{4}dt\\ \le & C+C{\int }_0^T{\parallel \varphi \parallel }_{L^{\mathrm{\infty }}}^{4}dt\\ \le & C+C{\int }_0^T{\parallel \mathrm{\nabla }\varphi \parallel }_{L^2}^2{\parallel \mathrm{\Delta }\varphi \parallel }_{L^2}^{2}dt\\ \le & C+C\underset{t}{sup}{\parallel \mathrm{\nabla }\varphi (t)\parallel }_{L^2}^2{\int }_0^T{\parallel \mathrm{\Delta }\varphi \parallel }_{L^2}^{2}dt\le C,\end{array}$$

(2.8)

which yields

$${\parallel \mathrm{\nabla }\varphi \parallel }_{L^2(0,T;L^{\mathrm{\infty }})}\le C.$$

(2.9)

Applying ${\mathrm{\Lambda }}^3$ to (1.1), testing by ${\mathrm{\Lambda }}^{3}u$, using (1.2), (1.9), (1.10), and noting that

$$\mathrm{\Delta }\varphi \mathrm{\nabla }\varphi =\sum _j{\partial }_j({\partial }_j\varphi \mathrm{\nabla }\varphi )-\frac{1}{2}\mathrm{\nabla }{|\mathrm{\nabla }\varphi |}^2,$$

we derive

$$\begin{array}{rcl}\frac{1}{2}\frac{d}{dt}\int {\left|{\mathrm{\Lambda }}^{3}u\right|}^{2}dx& =& -\int ({\mathrm{\Lambda }}^3(u\cdot \mathrm{\nabla }u)-u\mathrm{\nabla }{\mathrm{\Lambda }}^{3}u){\mathrm{\Lambda }}^{3}udx\\ -\int \sum _j{\mathrm{\Lambda }}^3{\partial }_j({\partial }_j\varphi \mathrm{\nabla }\varphi )\cdot {\mathrm{\Lambda }}^{3}udx+\int {\mathrm{\Lambda }}^3\theta e_3\cdot {\mathrm{\Lambda }}^{3}udx\\ \le & C{\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}{\parallel {\mathrm{\Lambda }}^{3}u\parallel }_{L^2}^2+C{\parallel \mathrm{\nabla }\varphi \parallel }_{L^{\mathrm{\infty }}}{\parallel {\mathrm{\Lambda }}^5\varphi \parallel }_{L^2}{\parallel {\mathrm{\Lambda }}^{3}u\parallel }_{L^2}\\ +C{\parallel {\mathrm{\Lambda }}^3\theta \parallel }_{L^2}{\parallel {\mathrm{\Lambda }}^{3}u\parallel }_{L^2}\\ \le & C{\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}{\parallel {\mathrm{\Lambda }}^{3}u\parallel }_{L^2}^2+C{\parallel \mathrm{\nabla }\varphi \parallel }_{L^{\mathrm{\infty }}}^2{\parallel {\mathrm{\Lambda }}^{3}u\parallel }_{L^2}^2+C{\parallel {\mathrm{\Lambda }}^{3}u\parallel }_{L^2}^2\\ +ϵ{\parallel {\mathrm{\Lambda }}^5\varphi \parallel }_{L^2}^2+ϵ{\parallel {\mathrm{\Lambda }}^3\theta \parallel }_{L^2}^2\end{array}$$

(2.10)

for any $0<ϵ<1$.

Taking Δ to (1.3), testing by Δθ, using (1.2), (1.10), and (2.1), we obtain

$$\begin{array}{r}\frac{1}{2}\frac{d}{dt}\int {|\mathrm{\Delta }\theta |}^{2}dx+\int {|\mathrm{\nabla }\mathrm{\Delta }\theta |}^{2}dx\\ =-\int (\mathrm{\Delta }(u\cdot \mathrm{\nabla }\theta )-u\mathrm{\nabla }\mathrm{\Delta }\theta )\mathrm{\Delta }\theta dx\\ \le C{\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}{\parallel \mathrm{\Delta }\theta \parallel }_{L^2}^2+C{\parallel \mathrm{\nabla }\theta \parallel }_{L^4}{\parallel \mathrm{\Delta }u\parallel }_{L^4}{\parallel \mathrm{\Delta }\theta \parallel }_{L^2}\\ \le C{\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}{\parallel \mathrm{\Delta }\theta \parallel }_{L^2}^2+C{\parallel \theta \parallel }_{L^{\mathrm{\infty }}}^{1/2}{\parallel \mathrm{\Delta }\theta \parallel }_{L^2}^{1/2}\cdot {\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}^{1/2}{\parallel \mathrm{\nabla }\mathrm{\Delta }u\parallel }_{L^2}^{1/2}\cdot {\parallel \mathrm{\Delta }\theta \parallel }_{L^2}\\ \le C{\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}{\parallel \mathrm{\Delta }\theta \parallel }_{L^2}^2+C({\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}{\parallel \mathrm{\Delta }\theta \parallel }_{L^2}+{\parallel \mathrm{\nabla }\mathrm{\Delta }u\parallel }_{L^2}){\parallel \mathrm{\Delta }\theta \parallel }_{L^2}\\ \le C{\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}{\parallel \mathrm{\Delta }\theta \parallel }_{L^2}^2+C{\parallel \mathrm{\Delta }\theta \parallel }_{L^2}^2+C{\parallel \mathrm{\nabla }\mathrm{\Delta }u\parallel }_{L^2}^2.\end{array}$$

(2.11)

Here we have used the Gagliardo-Nirenberg inequalities:

$$\begin{array}{c}{\parallel \mathrm{\nabla }\theta \parallel }_{L^4}^2\le C{\parallel \theta \parallel }_{L^{\mathrm{\infty }}}{\parallel \mathrm{\Delta }\theta \parallel }_{L^2},\hfill \\ {\parallel \mathrm{\Delta }u\parallel }_{L^4}^2\le C{\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}{\parallel \mathrm{\nabla }\mathrm{\Delta }u\parallel }_{L^2}.\hfill \end{array}$$

Taking ∇Δ to (1.4), testing by $\mathrm{\nabla }\mathrm{\Delta }\varphi$, using (1.5), (1.2), and (1.10), we have

$$\begin{array}{r}\frac{1}{2}\frac{d}{dt}\int {|\mathrm{\nabla }\mathrm{\Delta }\varphi |}^{2}dx+\int {\left|{\mathrm{\Lambda }}^5\varphi \right|}^{2}dx\\ =-\sum _i\int (\mathrm{\nabla }\mathrm{\Delta }(u_i{\partial }_i\varphi )-u_i\mathrm{\nabla }\mathrm{\Delta }{\partial }_i\varphi )\mathrm{\nabla }\mathrm{\Delta }\varphi dx+\int \mathrm{\nabla }\mathrm{\Delta }\cdot \mathrm{\Delta }{f}^{\prime }(\varphi )\cdot \mathrm{\nabla }\mathrm{\Delta }\varphi dx\\ \le C{\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}{\parallel \mathrm{\nabla }\mathrm{\Delta }\varphi \parallel }_{L^2}^2+C{\parallel \mathrm{\nabla }\varphi \parallel }_{L^{\mathrm{\infty }}}{\parallel {\mathrm{\Lambda }}^{3}u\parallel }_{L^2}{\parallel \mathrm{\nabla }\mathrm{\Delta }\varphi \parallel }_{L^2}\\ +\int \mathrm{\nabla }\mathrm{\Delta }{f}^{\prime }(\varphi )\cdot \mathrm{\nabla }{\mathrm{\Delta }}^2\varphi dx.\end{array}$$

(2.12)

Using (2.4), we get

$$\begin{array}{r}\int \mathrm{\nabla }\mathrm{\Delta }{f}^{\prime }(\varphi )\cdot \mathrm{\nabla }{\mathrm{\Delta }}^2\varphi dx\\ \le C\int (|\mathrm{\nabla }\mathrm{\Delta }\varphi |+\left|{\varphi }^2\right||\mathrm{\nabla }\mathrm{\Delta }\varphi |+|\varphi ||\mathrm{\nabla }\varphi |\left|{\mathrm{\nabla }}^2\varphi \right|+{|\mathrm{\nabla }\varphi |}^3)\left|\mathrm{\nabla }{\mathrm{\Delta }}^2\varphi \right|dx\\ \le C({\parallel \mathrm{\nabla }\mathrm{\Delta }\varphi \parallel }_{L^2}+{\parallel \varphi \parallel }_{L^{\mathrm{\infty }}}^2{\parallel \mathrm{\nabla }\mathrm{\Delta }\varphi \parallel }_{L^2}\\ +{\parallel \varphi \parallel }_{L^{\mathrm{\infty }}}{\parallel \mathrm{\nabla }\varphi \parallel }_{L^2}{\parallel {\mathrm{\nabla }}^2\varphi \parallel }_{L^{\mathrm{\infty }}}+{\parallel \mathrm{\nabla }\varphi \parallel }_{L^{\mathrm{\infty }}}^2{\parallel \mathrm{\nabla }\varphi \parallel }_{L^2}){\parallel {\mathrm{\Lambda }}^5\varphi \parallel }_{L^2}\\ \le C{\parallel \mathrm{\nabla }\mathrm{\Delta }\varphi \parallel }_{L^2}^2+C{\parallel \varphi \parallel }_{L^{\mathrm{\infty }}}^4{\parallel \mathrm{\nabla }\mathrm{\Delta }\varphi \parallel }_{L^2}^2+C{\parallel \varphi \parallel }_{L^{\mathrm{\infty }}}{\parallel {\mathrm{\nabla }}^2\varphi \parallel }_{L^{\mathrm{\infty }}}{\parallel {\mathrm{\Lambda }}^5\varphi \parallel }_{L^2}\\ +C{\parallel \mathrm{\nabla }\varphi \parallel }_{L^{\mathrm{\infty }}}^2{\parallel {\mathrm{\Lambda }}^5\varphi \parallel }_{L^2}+\frac{1}{4}{\parallel {\mathrm{\Lambda }}^5\varphi \parallel }_{L^2}^2.\end{array}$$

(2.13)

Now we use the following Gagliardo-Nirenberg inequalities:

$$\begin{array}{c}{\parallel {\mathrm{\nabla }}^2\varphi \parallel }_{L^{\mathrm{\infty }}}\le C{\parallel \mathrm{\nabla }\mathrm{\Delta }\varphi \parallel }_{L^2}^{3/4}{\parallel {\mathrm{\Lambda }}^5\varphi \parallel }_{L^2}^{1/4},\hfill \\ {\parallel \mathrm{\nabla }\varphi \parallel }_{L^{\mathrm{\infty }}}^2\le C{\parallel \mathrm{\Delta }\varphi \parallel }_{L^2}{\parallel {\mathrm{\Lambda }}^3\varphi \parallel }_{L^2}.\hfill \end{array}$$

We obtain

$$\begin{array}{r}\int \mathrm{\nabla }\mathrm{\Delta }{f}^{\prime }(\varphi )\cdot \mathrm{\nabla }{\mathrm{\Delta }}^2\varphi dx\\ \le C{\parallel \mathrm{\nabla }\mathrm{\Delta }\varphi \parallel }_{L^2}^2+C{\parallel \varphi \parallel }_{L^{\mathrm{\infty }}}^4{\parallel \mathrm{\nabla }\mathrm{\Delta }\varphi \parallel }_{L^2}^2\\ +C{\parallel \varphi \parallel }_{L^{\mathrm{\infty }}}^{8/3}{\parallel \mathrm{\nabla }\mathrm{\Delta }\varphi \parallel }_{L^2}^2+C{\parallel \mathrm{\Delta }\varphi \parallel }_{L^2}^2{\parallel {\mathrm{\Lambda }}^3\varphi \parallel }_{L^2}^2+\frac{1}{2}{\parallel {\mathrm{\Lambda }}^5\varphi \parallel }_{L^2}.\end{array}$$

(2.14)

Combining (2.10), (2.11), (2.12), and (2.14), taking ϵ small enough, using (1.8), (2.5), Gronwall’s inequality and

$${\parallel \varphi \parallel }_{L^4(0,T;L^{\mathrm{\infty }})}\le C,$$

we conclude that

$$\begin{array}{c}{\parallel u\parallel }_{L^{\mathrm{\infty }}(0,T;H^3)}\le C,\hfill \\ {\parallel \theta \parallel }_{L^{\mathrm{\infty }}(0,T;H^2)}+{\parallel \theta \parallel }_{L^2(0,T;H^3)}\le C,\hfill \\ {\parallel \varphi \parallel }_{L^{\mathrm{\infty }}(0,T;H^3)}+{\parallel \varphi \parallel }_{L^2(0,T;H^5)}\le C.\hfill \end{array}$$

This completes the proof.