Blow-up criteria for a two-fluid model of the truncated Euler equations

Abstract

In this paper, we study a two-fluid model of the truncated Euler equations with partial viscosity. We obtain new blow-up criteria for a solution of the system in terms of vorticity in the homogeneous Besov space ${\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^{-1}$.

1 Introduction and main results

In this paper, we are concerned with the regularity of the following two-fluid model of the truncated Euler equations with partial viscosity:

$$\{\begin{array}{l}{u}_t-\mathrm{\Delta }u+u\cdot \mathrm{\nabla }u+\mathrm{\nabla }p=\theta e_3,(x;t)\in {\mathbb{R}}^3\times (0;1),\\ {\theta }_t+u\cdot \mathrm{\nabla }\theta =|\mathrm{\nabla }u{|}^2,\\ \mathrm{\nabla }\cdot u=0,\\ u(x,0)=u_0,\theta (x,0)={\theta }_0,\end{array}$$

(1.1)

where $u=(u^1(x,t),u^2(x,t),u^3(x,t))$ denotes the fluid velocity vector field, $p=p(x,t)$ is the scalar pressure, $e_3={(0,0,1)}^T$, while $u_0$ and ${\theta }_0$ are given initial velocity and initial temperature, respectively, with $\mathrm{\nabla }\cdot u_0=0$. For more detailed background of the system, we refer the readers to [1] and the references therein. When $\theta =0$, the system reduces to the well-known Navier-Stokes equations.

It is well known that the question of global existence or finite time blow-up of smooth solutions for the 3D incompressible Navier-Stokes equations is still unsolved. This challenging problem has attracted significant attention. Therefore, it is interesting to study the blow-up criterion of the solutions for system (1.1). Recently, Sun and Fan [2] first proved the following blow-up criterion for system (1.1):

$$\begin{array}{c}curlu\in L^1(0,T;{\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^0\left({\mathbb{R}}^3\right)),\hfill \\ u\in L^2(0,T;{\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^0\left({\mathbb{R}}^3\right)).\hfill \end{array}$$

Our purpose in this paper is to establish blow-up criteria of a strong solution for the two-fluid model of the truncated Euler equations in terms of vorticity in the homogeneous Besov space ${\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^{-1}$. The main difficulty is without viscosity in the second equation for system (1.1).

Now we state our main results as follows.

Theorem 1.1 Let $(u_0,{\theta }_0)\in H^2({\mathbb{R}}^3)$, $(u(\cdot ,t),\theta (\cdot ,t))$ be a strong solution to system (1.1) with the initial data $(u_0,{\theta }_0)$ for $0\le t<T$. If the solution u satisfies the following condition:

$$curlu\in L^2(0,T;{\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^{-1}),$$

(1.2)

then the solution $(u,\theta )$ can be extended smoothly beyond $t=T$.

2 Preliminaries and lemmas

Throughout this paper, we introduce some function spaces, notations and important inequalities.

Let $e^{t\mathrm{△}}$ denote the heat semigroup defined by

$$e^{t\mathrm{△}}f=K_t\ast f,K_t={(4\pi t)}^{-\frac{3}{2}}exp(-\frac{|x{|}^2}{4t})$$

for $t>0$ and $x\in {\mathbb{R}}^3$, where ∗ denotes the convolution of functions defined on ${\mathbb{R}}^3$.

We now recall the definition of the homogeneous Besov space with negative indices ${\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^{-\alpha }$ on ${\mathbb{R}}^3$ and the homogeneous Sobolev space ${\dot{H}}_q^{\alpha }$ of exponent $\alpha >0$. It is known (p.192 of Ref. [3]) that $f\in {\mathcal{S}}^{\prime }({\mathbb{R}}^3)$ belongs to ${\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^{-\alpha }$ if and only if $e^{t\mathrm{\Delta }}f\in L^{\mathrm{\infty }}$ for all $t>0$ and $t^{\frac{\alpha }{2}}{\parallel e^{t\mathrm{\Delta }}f\parallel }_{\mathrm{\infty }}\in L^{\mathrm{\infty }}(0,\mathrm{\infty };L^{\mathrm{\infty }})$. The norm of ${\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^{-\alpha }$ is defined, up to equivalence, by

$${\parallel f\parallel }_{{\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^{-\alpha }}=\underset{t>0}{sup}\left(t^{\frac{\alpha }{2}}{\parallel e^{t\mathrm{\Delta }}f\parallel }_{\mathrm{\infty }}\right).$$

We introduce now the homogeneous Sobolev space ${\dot{H}}_q^{\alpha }({\mathbb{R}}^3)$, which is defined as the set of functions $f\in L^r({\mathbb{R}}^3)$, $\frac{1}{r}=\frac{1}{q}-\frac{\alpha }{3}$ such that ${(-\mathrm{\Delta })}^{\frac{\alpha }{2}}f\in L^q({\mathbb{R}}^3)$. This space is endowed with the norm

$${\parallel f\parallel }_{{\dot{H}}_q^{\alpha }}={\parallel {(-\mathrm{\Delta })}^{\frac{\alpha }{2}}f\parallel }_{L^q},$$

and when $q=2$, we just let ${\dot{H}}_2^{\alpha }({\mathbb{R}}^3)={\dot{H}}^{\alpha }({\mathbb{R}}^3)$.

Lemma 2.1 [4]

Let $1<p<q<\mathrm{\infty }$ and $s=\alpha (\frac{q}{p}-1)>0$. Then there exists a constant depending only on α, p and q such that the estimate

$${\parallel f\parallel }_{L^q}\le C{\parallel {(-\mathrm{\Delta })}^{\frac{s}{2}}f\parallel }_{L^p}^{\frac{p}{q}}{\parallel f\parallel }_{{\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^{-\alpha }}^{1-\frac{p}{q}}$$

(2.1)

holds for all $f\in {\dot{H}}_p^{\alpha }({\mathbb{R}}^3)\cap {\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^{-\alpha }({\mathbb{R}}^3)$.

In particular, for $s=1$, $p=2$ and $q=4$, we get $\alpha =1$ and

$${\parallel f\parallel }_{L^4}\le C{\parallel f\parallel }_{{\dot{H}}^1}^{\frac{1}{2}}{\parallel f\parallel }_{{\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^{-1}}^{\frac{1}{2}}.$$

(2.2)

Lemma 2.2 [5]

For any function $f\in W^{1,s}({\mathbb{R}}^3)$ ($s\ge 1$), and $r\ge 1$, we have

$${\parallel f\parallel }_{L^{\gamma }}\le C{\parallel f\parallel }_{L^2}^{1-\alpha }{\parallel \mathrm{\nabla }f\parallel }_{L^2}^{\alpha },$$

(2.3)

where

$$\alpha =\frac{\frac{1}{r}-\frac{1}{\gamma }}{\frac{1}{3}-\frac{1}{s}-\frac{1}{r}}$$

and C is a constant independent of f.

3 Proof of the main results

In this section, we prove Theorem 1.1.

Proof of Theorem 1.1 First, we multiply the both sides of the first equation of (1.1) by u and the second equation of (1.1) by θ, respectively; after integration by parts over ${\mathbb{R}}^3$, we get

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

(3.1)

$$\frac{1}{2}\frac{d}{dt}\int {\theta }^{2}dx=\int \theta |\mathrm{\nabla }u{|}^{2}dx.$$

(3.2)

Next, we multiply the both sides of the first equation of (1.1) by $(-\mathrm{\Delta }u)$; after integration by parts over ${\mathbb{R}}^3$, we get

$$\frac{1}{2}\frac{d}{dt}\int |\mathrm{\nabla }u{|}^{2}dx+\int |\mathrm{\Delta }u{|}^{2}dx\le \int u\cdot \mathrm{\nabla }u\mathrm{\Delta }udx+\int |\mathrm{\Delta }u|\theta dx.$$

(3.3)

Using the Young inequality and adding (3.2) to (3.3), we obtain

$$\begin{array}{r}\frac{d}{dt}(\int |\mathrm{\nabla }u{|}^{2}dx+\int {\theta }^{2}dx)+\int |\mathrm{\Delta }u{|}^{2}dx\\ \le \int u\cdot \mathrm{\nabla }u\mathrm{\Delta }udx+\int \theta |\mathrm{\nabla }u{|}^{2}dx+\int {\theta }^{2}dx.\end{array}$$

(3.4)

It is easy to see that

$$\begin{array}{rl}\int u\cdot \mathrm{\nabla }u\mathrm{\Delta }udx& =-\sum _{i,k}\int {\partial }_{k}u\cdot {\partial }_{i}u\cdot {\partial }_{k}udx\\ \le {\parallel \mathrm{\nabla }u\parallel }_{L^2}{\parallel \mathrm{\nabla }u\mathrm{\nabla }u\parallel }_{L^2}\\ \le {\parallel \mathrm{\nabla }u\parallel }_{L^2}{\parallel \mathrm{\nabla }u\parallel }_{L^4}^2\\ \le {\parallel \mathrm{\nabla }u\parallel }_{L^2}{\parallel \mathrm{\nabla }u\parallel }_{{\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^{-1}}{\parallel \mathrm{\Delta }u\parallel }_{L^2}\\ \le C{\parallel \mathrm{\nabla }u\parallel }_{L^2}^2{\parallel \mathrm{\nabla }\times u\parallel }_{{\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^{-1}}^2+\frac{1}{2}{\parallel \mathrm{\Delta }u\parallel }_{L^2}^2,\end{array}$$

(3.5)

where we use the Hölder inequality, the Young inequality and (2.2).

Similarly,

$$\begin{array}{rl}\int \theta |\mathrm{\nabla }u{|}^{2}dx& \le {\parallel \theta \parallel }_{L^2}{\parallel \mathrm{\nabla }u\mathrm{\nabla }u\parallel }_{L^2}\\ \le {\parallel \theta \parallel }_{L^2}{\parallel \mathrm{\nabla }u\parallel }_{L^4}^2\\ \le {\parallel \theta \parallel }_{L^2}{\parallel \mathrm{\nabla }u\parallel }_{{\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^{-1}}{\parallel \mathrm{\Delta }u\parallel }_{L^2}\\ \le C{\parallel \theta \parallel }_{L^2}^2{\parallel \mathrm{\nabla }\times u\parallel }_{{\dot{B}}_{\mathrm{\infty },\mathrm{\infty }}^{-1}}^2+\frac{1}{2}{\parallel \mathrm{\Delta }u\parallel }_{L^2}^2.\end{array}$$

(3.6)

Putting (3.5) and (3.6) into (3.4) and applying the Gronwall inequality yields

$$u\in L^{\mathrm{\infty }}(0,T;H^1)\cap L^2(0,T;H^2),\theta \in L^{\mathrm{\infty }}(0,T;L^2).$$

Multiplying the both sides of the second equation of (1.1) by $|\theta {|}^{p-2}\theta$ ($2<p<\mathrm{\infty }$), after integration by parts over ${\mathbb{R}}^3$, we deduce that

$$\frac{1}{p}\frac{d}{dt}\int {\theta }^{p}dx\le \int |\theta {|}^{p-1}|\mathrm{\nabla }u{|}^{2}dx\le {\parallel \theta \parallel }_{L^p}^{p-1}{\parallel \mathrm{\nabla }u\parallel }_{L^{2p}}^2\le {\parallel \theta \parallel }_{L^p}^{p-1}{\parallel u\parallel }_{H^2}^2,$$

which implies that

$$\theta \in L^{\mathrm{\infty }}(0,T;L^p)(2<p<\mathrm{\infty }).$$

(3.7)

Taking curl on the both sides of the first equation of (1.1), we obtain the following vorticity formulation for the vorticity field $\omega =curlu$:

$${\partial }_t\omega +u\cdot \mathrm{\nabla }\omega -\mathrm{\Delta }\omega =curl(\theta e_3)+\omega \cdot \mathrm{\nabla }u,$$

(3.8)

and then multiplying the both sides of the vorticity equation by $|\omega {|}^{p-2}\omega$ ($p>2$), after integration by parts over ${\mathbb{R}}^3$, we infer that

$$\begin{array}{rl}\frac{d}{dt}\int v^{2}dx+\int |\mathrm{\nabla }v{|}^{2}dx& \le C\int |\theta |\mathrm{\nabla }||\omega {|}^{p-2}\omega |dx+C\int \mathrm{\nabla }u|\omega {|}^{p}dx\\ \le C\int |\theta ||v{|}^{\frac{p-2}{p}}|\mathrm{\nabla }v|dx+C\int \mathrm{\nabla }u|\omega {|}^{\frac{p}{2}}|\omega {|}^{\frac{p}{2}}dx\\ \le C{\parallel \theta \parallel }_{L^{\frac{4p}{p+2}}}{\parallel v\parallel }_{L^4}^{1-\frac{2}{p}}{\parallel \mathrm{\nabla }v\parallel }_{L^2}+C{\parallel \mathrm{\nabla }u\parallel }_{L^2}{\parallel v\parallel }_{L^4}^2\\ \le C{\parallel v\parallel }_{L^4}^{1-\frac{2}{p}}{\parallel \mathrm{\nabla }v\parallel }_{L^2}+C{\parallel v\parallel }_{L^2}^{\frac{1}{2}}{\parallel \mathrm{\nabla }v\parallel }_{L^2}^{\frac{3}{2}}\\ \le C{\parallel v\parallel }_{L^2}^{\frac{1}{2}-\frac{1}{p}}{\parallel \mathrm{\nabla }v\parallel }_{L^2}^{\frac{3}{2}-\frac{1}{p}}+C{\parallel v\parallel }_{L^2}^{\frac{1}{2}}{\parallel \mathrm{\nabla }v\parallel }_{L^2}^{\frac{3}{2}}\\ \le C{\parallel v\parallel }_{L^2}^2+\frac{1}{2}{\parallel \mathrm{\nabla }v\parallel }_{L^2}^2,\end{array}$$

(3.9)

where $v:=|\omega {|}^{\frac{p}{2}}$. Using the Gronwall inequality, we obtain

$$\omega \in L^{\mathrm{\infty }}(0,T;L^p)(p>2),$$

which implies that

$$\mathrm{\nabla }u\in L^{\mathrm{\infty }}(0,T;L^p)(p>2).$$

(3.10)

On the other hand, since u is a solution of the Stokes system

$$u_t-\mathrm{\Delta }u+\mathrm{\nabla }p=-u\cdot \mathrm{\nabla }u+\theta e_3\in L^{\mathrm{\infty }}(0,T;L^p),$$

it follows from the $W^{2,p}$-theory of the Stokes system that

$$u\in L^p(0,T;W^{2,p})(p>2).$$

(3.11)

So,

$$\mathrm{\nabla }u\in L^p(0,T;L^{\mathrm{\infty }})(p>2).$$

(3.12)

Differentiating the second equation of (1.1) with respect to $x_i$ ($1\le i\le 3$) and multiplying the resulting equation by ${\partial }_i\theta$, and then integrating by parts over ${\mathbb{R}}^3$, we get

$$\begin{array}{rl}\frac{1}{2}\frac{d}{dt}\int |\mathrm{\nabla }\theta {|}^{2}dx& \le C\int |\mathrm{\nabla }u||\mathrm{\nabla }\theta {|}^{2}dx+C\int |\mathrm{\nabla }u||\mathrm{\Delta }u||\mathrm{\nabla }\theta |dx\\ \le C{\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}{\parallel \mathrm{\nabla }\theta \parallel }_{L^2}^2+C{\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}{\parallel \mathrm{\Delta }u\parallel }_{L^2}{\parallel \mathrm{\nabla }\theta \parallel }_{L^2}.\end{array}$$

By (3.12) and the Gronwall inequality, we have

$$\theta \in L^{\mathrm{\infty }}(0,T;H^1).$$

(3.13)

We multiply the both sides of the equation of (3.8) by $(-\mathrm{\Delta }\omega )$; after integration by parts over ${\mathbb{R}}^3$, we get

$$\begin{array}{rl}\frac{1}{2}\frac{d}{dt}\int |\mathrm{\nabla }\omega {|}^{2}dx+\int |\mathrm{\Delta }\omega {|}^{2}dx& \le \int |u||\mathrm{\nabla }\omega ||\mathrm{\Delta }\omega |dx+\int |\mathrm{\Delta }\omega ||\mathrm{\nabla }\theta |dx\\ \le {\parallel u\parallel }_{L^4}{\parallel \mathrm{\Delta }\omega \parallel }_{L^2}{\parallel \mathrm{\nabla }\omega \parallel }_{L^4}+{\parallel \mathrm{\nabla }\theta \parallel }_{L^2}{\parallel \mathrm{\Delta }\omega \parallel }_{L^2}\\ \le C{\parallel \mathrm{\Delta }\omega \parallel }_{L^2}^{\frac{1}{2}}{\parallel \mathrm{\nabla }\omega \parallel }_{L^2}^{\frac{3}{2}}+C{\parallel \mathrm{\Delta }\omega \parallel }_{L^2}\\ \le C{\parallel \mathrm{\nabla }\omega \parallel }_{L^2}^2+\frac{1}{2}{\parallel \mathrm{\Delta }\omega \parallel }_{L^2}^2+C,\end{array}$$

the Gronwall inequality yields

$$u\in L^{\mathrm{\infty }}(0,T;H^2)\cap L^2(0,T;H^3).$$

(3.14)

In the following calculations, we use the following commutator estimate and bilinear estimate [6]:

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

(3.15)

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

(3.16)

with $s>0$, ${\mathrm{\Lambda }}^s={(-\mathrm{\Delta })}^{\frac{s}{2}}$ and $\frac{1}{p}=\frac{1}{p_1}+\frac{1}{q_1}=\frac{1}{p_2}+\frac{1}{q_2}$. Taking the operation ${\mathrm{\Lambda }}^2$ on both sides of the second equation of (1.1), then multiplying them by ${\mathrm{\Lambda }}^2\theta$, and integrating by parts over ${\mathbb{R}}^3$, we have

$$\begin{array}{rl}\frac{1}{2}\frac{d}{dt}\int |{\mathrm{\Lambda }}^2\theta {|}^{2}dx\le & |\int ({\mathrm{\Lambda }}^2(u\cdot \mathrm{\nabla }\theta )-u\mathrm{\nabla }{\mathrm{\Lambda }}^2\theta )dx|+\int {\mathrm{\Lambda }}^2(|\mathrm{\nabla }u{|}^2){\mathrm{\Lambda }}^2\theta dx\\ \le & C({\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}{\parallel {\mathrm{\Lambda }}^2\theta \parallel }_{L^2}+{\parallel {\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}{\parallel {\mathrm{\Lambda }}^2\theta \parallel }_{L^2}u\parallel }_{L^3}{\parallel \mathrm{\nabla }\theta \parallel }_{L^6}){\parallel {\mathrm{\Lambda }}^2\theta \parallel }_{L^2}\\ +C{\parallel \mathrm{\nabla }u\parallel }_{L^{\mathrm{\infty }}}{\parallel {\mathrm{\Lambda }}^{3}u\parallel }_{L^2}{\parallel {\mathrm{\Lambda }}^2\theta \parallel }_{L^2}.\end{array}$$

Combining (3.12) with (3.14) and using the Gronwall inequality, we obtain

$$\theta \in L^{\mathrm{\infty }}(0,T;H^2).$$

This completes the proof. □