Stability of Blowup for a 1D Model of Axisymmetric 3D Euler Equation

Abstract

The question of the global regularity versus finite- time blowup in solutions of the 3D incompressible Euler equation is a major open problem of modern applied analysis. In this paper, we study a class of one-dimensional models of the axisymmetric hyperbolic boundary blow-up scenario for the 3D Euler equation proposed by Hou and Luo (Multiscale Model Simul 12:1722–1776, 2014) based on extensive numerical simulations. These models generalize the 1D Hou–Luo model suggested in Hou and Luo Luo and Hou (2014), for which finite-time blowup has been established in Choi et al. (arXiv preprint. arXiv:1407.4776, 2014). The main new aspects of this work are twofold. First, we establish finite-time blowup for a model that is a closer approximation of the three-dimensional case than the original Hou–Luo model, in the sense that it contains relevant lower-order terms in the Biot–Savart law that have been discarded in Hou and Luo Choi et al. (2014). Secondly, we show that the blow-up mechanism is quite robust, by considering a broader family of models with the same main term as in the Hou–Luo model. Such blow-up stability result may be useful in further work on understanding the 3D hyperbolic blow-up scenario.

Appendix: Real Line Case

One can also consider the model Eq. (5) and (6) with the law (9) for compactly supported data on $\mathbb {R}$. We only outline main ideas and changes involved, leaving all details to the interested reader. Without loss of generality, we assume the domain of the initial data is $[-1,1]$. In this case, similar argument like in Sect. 2 can show that the corresponding modified Hou–Luo kernel will be

$$\begin{aligned} F(x,y,a)=\frac{y}{x}\left[ \log \left( \frac{(x-y)^2}{(x+y)^2}\right) +\log \left( \frac{(x+y)^2+a}{(x-y)^2+a}\right) \right] , \end{aligned}$$

(43)

for $a>0$.

The analogue of Lemma 3.3 will be the following:

Lemma 5.1

(a) For any $a\ne 0$, there is a constant $C(a)> 0$ such that for any $0<x<y<1$, $F(x,y,a)\le -C(a)$.

(b) For any $0<y<x<\infty$, F(x, y, a) is increasing in x.

(c) For any $0<x,y<\infty$, $\frac{1}{y}(\partial _xF)(x,y,a)+\frac{1}{x}(\partial _x F)(y,x,a)$ is positive.

Proof

First it is easy to see that F(x, y, a) is non-positive. For part (a), one can follow the similar but easier argument as in the proof of part (a) of Lemma 3.3. Now let us prove part (b) and (c).

Proof of (b)

By direct computation

$$\begin{aligned} \frac{1}{y}\partial _xF(x,y,a)&=-\frac{1}{x^2}\left[ \log \left( \frac{(x-y)^2}{(x+y)^2}\right) +\log \left( \frac{(x+y)^2+a}{(x-y)^2+a}\right) \right] \\&\quad +\frac{1}{x}\left[ \frac{2(x-y)}{(x-y)^2}-\frac{2(x-y)}{(x-y)^2+a}-\frac{2(x+y)}{(x+y)^2}+\frac{2(x+y)}{(x+y)^2+a}\right] \\&=-\frac{1}{x^2}\left[ \log \left( \frac{(x-y)^2}{(x+y)^2}\right) +\log \left( \frac{(x+y)^2+a}{(x-y)^2+a}\right) \right] \\&\quad +\frac{1}{x}\left[ \frac{2a(x-y)}{(x-y)^2((x-y)^2+a)}-\frac{2a(x+y)}{(x+y)^2((x+y)^2+a)}\right] \\&=I+II. \end{aligned}$$

The term I, by the same argument as in the proof of the periodic analog, is positive. For the term II, we have

$$\begin{aligned} II=\frac{1}{x}(g(x-y)-g(x+y)), \end{aligned}$$

where $g(t)=\frac{2a}{t(t^2+a)}$. It is easy to see that for $t>0$, g(t) is decreasing in t, which means $II\ge 0$ whenever $0<y<x$.

Proof of (c)

First of all, let us call our target function G(x, y, a), which means

$$\begin{aligned} G(x,y,a)&=\frac{1}{y}(\partial _x F)(x,y,a)+\frac{1}{x}(\partial _x F)(y,x,a)\\&=-\left( \frac{1}{x^2}+\frac{1}{y^2}\right) \left[ \log \left( \frac{(x-y)^2}{(x+y)^2}\right) +\log \left( \frac{(x+y)^2+a}{(x-y)^2+a}\right) \right] \\&\quad +\left( \frac{1}{x}-\frac{1}{y}\right) \left( \frac{2a(x-y)}{(x-y)^2((x-y)^2+a)}\right) \\&\quad -\left( \frac{1}{y}+\frac{1}{x}\right) \left( \frac{2a(x+y)}{(x+y)^2((x+y)^2+a)}\right) \\&=-\left( \frac{1}{x^2}+\frac{1}{y^2}\right) \left[ \log \left( \frac{(x-y)^2}{(x+y)^2}\right) +\log \left( \frac{(x+y)^2+a}{(x-y)^2+a}\right) \right] \\&\quad -\frac{2a}{xy((x-y)^2+a)}-\frac{2a}{xy((x+y)^2+a)}. \end{aligned}$$

Now our aim is to prove the positivity of G(x, y, a). Notice that when $a=0$, $G(x,y,a)=0$, as a consequence, to prove the positivity of G(x, y, a), the only thing we need to show is this function is increasing in a for any x, y in the domain. On the other hand,

$$\begin{aligned} \partial _a G(x,y,a)&=-\left( \frac{1}{x^2}+\frac{1}{y^2}\right) \left( \frac{1}{(x+y)^2+a}-\frac{1}{(x-y)^2+a}\right) \\&\quad -\frac{2}{xy}\left[ \frac{(x-y)^2}{((x-y)^2+a)^2}+\frac{(x+y)^2}{((x+y)^2+a)^2}\right] . \end{aligned}$$

As a conclusion,

$$\begin{aligned}&((x-y)^2+a)^2((x+y)^2+a)^2\partial _a G(x,y,a)\\&\quad =\left( \frac{1}{x^2}+\frac{1}{y^2}\right) ((x+y)^2-(x-y)^2)((x+y)^2+a)((x-y)^2+a)\\&\quad \quad -\frac{2}{xy}\left[ (x-y)^2((x+y)^2+a)^2+(x+y)^2((x-y)^2+a)^2\right] \end{aligned}$$

It is easy to see this is a quadratic polynomial in a. Let us call the coefficient of the second-order term $A_2$ , then

$$\begin{aligned} A_2&=\left( \frac{1}{x^2}+\frac{1}{y^2}\right) ((x+y)^2-(x-y)^2)-\frac{2}{xy}[(x-y)^2+(x+y)^2]\\&=\left( \frac{1}{x^2}+\frac{1}{y^2}\right) \cdot 4xy-\frac{2}{xy}[2x^2+2y^2]\\&=\frac{4}{x^2y^2}((x^2+y^2)xy-xy(x^2+y^2))\\&=0. \end{aligned}$$

Similarly, for coefficient of the first-order term $A_1$, we have

$$\begin{aligned} A_1&=\left( \frac{1}{x^2}+\frac{1}{y^2}\right) (4xy)((x+y)^2+(x-y)^2)\\&\quad -\frac{2}{xy}[2(x-y)^2(x+y)^2+2(x+y)^2(x-y)^2]\\&=\frac{1}{x^2y^2}[(x^2+y^2)^2\cdot 8xy-8xy(x^2-y^2)^2]\\&\ge 0. \end{aligned}$$

Lastly, for the coefficient of the constant term $A_0$, we have

$$\begin{aligned} A_0&=\left( \frac{1}{x^2}+\frac{1}{y^2}\right) (4xy)(x+y)^2(x-y)^2\\&\quad -\frac{2}{xy}[(x-y)^2(x+y)^4+(x+y)^2(x-y)^4]\\&=\frac{(x+y)^2(x-y)^2}{x^2y^2}[(x^2+y^2)\cdot 4xy-2xy((x+y)^2+(x-y)^2)]\\&=0. \end{aligned}$$

In all, we have $\partial _a G(x,y,a)\ge 0$ for $x,y>0$. $\square$

From this lemma, one can do the same argument to get the blow-up result, which is the following theorem:

Theorem 5.2

There exist initial data such that solutions to (5) and (6), with velocity given by (16), and F(x, y, a) defined by (43), blow up in finite time.

In fact, we can prove the following type of initial data will lead to blowup:

• $\theta _{0x}, \omega _0$ smooth odd and are supported in $[-1,1]$.

• $\theta _{0x}, \omega _0 \ge 0$ on [0, 1].

• $\theta _0(0)=0$.

• $\Vert \theta _0\Vert _\infty \le M$.

And similarly, for general perturbation (analogue of theorem 4.1), we also have the similar blow-up result.

Assume the velocity u is given by the following choice of Biot–Savart Law

$$\begin{aligned} u(x) = \frac{1}{\pi } \int _{-1}^1 \left( \log |(x-y)]| + f(x,y)\right) \omega (y)\, {\text {d}}y, \end{aligned}$$

(44)

where f is a smooth function whose precise properties we will specify later. We view f as a perturbation, and we will show solutions to the system (5) and (6) can still blow up in finite time.

Theorem 5.3

Let $f\in C^2$ be supported on $[-1,1]$, such that $f(x,y)=f(-x,-y)$ for all y. Then there exist initial data $\omega _0$, $\theta _0$ such that solutions of (5) and (6), with velocity given by (44), blow up in finite time.

Again we can prove the following type of initial data will form finite-time singularity:

• $\theta _{0x}, \omega _0$ smooth odd and are supported in $[-1,1]$.

• $\theta _{0x}, \omega _0 \ge 0$ on [0, 1].

• $\theta _0(0)=0$.

• $\text{ supp }\, \omega _0 \subset [0,\epsilon ]$.

• $\Vert \theta _0\Vert _\infty \le M$.

We leave the proofs of these theorems as exercises for interested reader.