Self-similar blowup in low regularity

This lecture covers mainly the singularity works On the effect of advection and vortex stretching and Finite time singularity formation for $C^{1,\alpha}$ solutions to the incompressible Euler equations in $\R^3$ by Elgindi. In the former work Elgindi investigates a blowup mechanism for a simplified 1D model for Euler equation, and then he applies this mechanism to 3D Euler case in the latter one. Conclusively, he constructs a self-similar type blowup vorticity formulated as
\begin{equation*}
\omega(x,t)=\frac{1}{1-(1+\mu)t}F\br{\frac{|x|^\alpha}{(1-(1+\mu)t)^{1+\lambda}} }
\end{equation*}for the 3D incompressible Euler flow with initial data $\omega_0\in C^\alpha$($\alpha$ small enough), where $F$ is a bounded profile. The basic idea is to investigate a fundamental model(which preserves the singularity but is easy enough to solve out the explicit solution), and analyze the singularity under a perturbation by the advection term $u\cdot \nabla \omega$ up to some bounded quantities. Some further discussions and open problems will also be discussed in related topics.

1 Introduction

We recall that the incompressible Euler flow is governed by the following equations
\begin{equation*}
\begin{cases}
\partial_t u+u\cdot \nabla u +\nabla p=0,\\
\nabla \cdot u=0,
\end{cases}
\end{equation*}here $u(x,t)$ is the velocity and $p$ is the pressure. The classic open problem is whether a $C^\infty$-initial data $u_0$ can develop a global smooth formulation. A well-known Beale-Kato-Majda criterion tells us that a classic solution loses its regularity as $t\to T$ iff the vorticity $\omega\coloneqq \nabla \times u$ satisfies
\begin{equation*}
\lim_{t\to T} \norm{\omega}_{L^1(0,t; L^\infty )}=+\infty.
\end{equation*}Consequently, it is natural to consider the vorticity equation\begin{equation*}
\partial_t \omega+ u\cdot \nabla \omega=\omega \cdot \nabla u,
\end{equation*}and investigate the blowup mechanism of $\omega$. The term $u\cdot \nabla \omega$ is a transport term and cannot cause any growth of the vorticity since $u$ is divergence-free. The term $\omega \cdot \nabla u$ is called the vortex stretching term and can lead to amplification of the vorticity, which is the main difficulty and causes the singularity. It is conjectured that when the transport term is weaker than the vortex stretching term, finite time singularities will occur. As Biot-Savart law gives out the representation of the velocity $u$ by the vorticity $\omega$:
$$u=-\nabla \times (\Delta^{-1} \omega )=-K* \omega, $$where $\nabla K$ is a Calderon-Zygmund kernel, we can view $\nabla u=-\nabla K* \omega\sim \omega$ with particular singularity. Omitting this singularity and the advection term, it is enlightening is reduce the vorticity equation into the following ODE (with only time variable $t$):
$$\partial_t \omega=\omega^2, $$which has explicit solution $\omega=\frac{1}{1-t}$ if equipped with unit data $w_0=1$. Based on this observation, we will consider a backward self-similar type formulation in our further discussion.

The main content of this lecture is organized as follows: In Section 2, we will consider a simplified 1D model which reveals the significant singularity of vortex stretching, and investigate the self-similar type blowup initiated above under high regular($H^k, k\ge 3$) case and low regular($C^\alpha, 0\sl \alpha \sl1$) case respectively. As the blowup is caused by $\omega \cdot \nabla u$, we take the system without advection as the fundamental model (Section 2.1) and view the advection as a perturbation under some small assumption, whose convergence can be obtained by some Hardy type estimate(Section 2.2). The idea will be applied to 3D Euler flow in Section 3. Particularly, the main extra difficulties is that the selection of fundamental model(Section 3.1) and the analysis of perturbation part will be more complex, for which we will calculate some important elliptic(Section 3.3) and coercive estimate (Section 3.2), and obtain the existence via a propri estimates and the classic compactness argument(Section 3.4-3.5). The appendix will gives some related facts.

2 Simple 1D system

As we reveals in the introduction, the vortex stretching behaves as a Calderon-Zygmund singularity integral of $\omega$. The only C-Z operator in 1D case is exactly the Hilbert transform(up to a constant scaling), so it is proper to consider the following 1D model
\begin{equation}
\label{origin 1D model}
\partial_t \omega + u\partial_x \omega = \omega H\omega .
\end{equation}Since $\nabla u\sim -H\omega$, we assume
\begin{equation*}
u=\Lambda^{-1} \omega \coloneqq - \int_0^x H\omega (y)dy.
\end{equation*}For the convenience of the discussion, we'd like to dilate the blowup time to the unit and add a smallness¹ ahead of the advection, i.e. the following system is considered:
\begin{equation}
\label{1D system}
\begin{cases}
\partial_t \omega +a u\partial_x \omega =2\omega H\omega,\\
u=\Lambda^{-1}\omega.
\end{cases}
\end{equation}When $a=0$, the advection term $u\partial_x \omega$ is eliminated and we will view
\begin{equation}
\label{1D fundamental}
\partial_t \omega =2\omega H\omega
\end{equation} as the fundamental model. It can be solved explicitly using the method of analytical signal(see Appendix \ref{Hilbert transform}). Particularly for regular case($H^k,k\ge 3$), we will establish the only self-similar ansatz
\begin{equation*}
\omega (x,t)=\frac 1{1-t}F_0 \br{\frac{x}{1-t} }
\end{equation*} as long as oddness of the initial data is required. Based on this fundamental formulation, we'd like to consider the ansatz of $\omega$ for perturbed system \eqref{1D system} formulated as
$$\omega (x,t)=\frac 1{1-t}F\br{\frac{x}{(1-t)^{1+\lambda}} }, $$where $F, \lambda$ depends on $a$ analytically, that is,
\begin{equation*}
F(a)= F_0 +\sum_{n=1}^\infty a^n F_n(a), \lambda=\sum_{n=1}^\infty a^n \lambda^n(a).
\end{equation*}The existence of blowup solution will be obtained once we prove the convergence of $F$ in $H^k$ norm. Similarly, the low regularity case($C^\alpha, 0<\alpha<1$) can be processed if we consider the following variable transform \begin{equation*} f(z)= \tilde f(z^\alpha)= \tilde f(w). \end{equation*}Some extra estimates is need for transformed operators, and we can find the convergence holds if the perturbation and regular index satisfies the following balance $$a\alpha \le c, $$where $c$ is a given constant. This implies the origin 1D model \eqref{origin 1D model} admits this blowup mechanism if we require $\alpha$ small enough, which is a motivation for the investigation of $C^\alpha$-blowup for 3D Euler system in later sections. In Section 2.1, we will give the explicit solution of the fundamental model \eqref{1D fundamental} and some initiative examples for our later self-similar type formation. Then we obtain the self-similar solution under $H^k$ and $C^\alpha$ case respectively. In Section 2.2, we derive the transport operator $L$ for the remained part $F_n$ and investigate the solvability of $L$. Some Hardy type estimates lead to the convergence of the remained part finally.

2.1 Fundamental model

The investigation of the fundamental model \eqref{1D fundamental} origins from the work by Constantin-Lax-Majda. They calculated out the explicit solution by analytical method and the Tricomi identity of Hilbert transform.

It is clear from \eqref{CLM Solution} that the solution $\omega(x,t)$ turns singular at $(x,t)$ iff $\br{1+tH\omega_0 (x)}^2 +t^2 \omega_0^2 (x)=0$. Consequently, the singular set
\begin{equation*}
S=\set{z|z=\br{x,-\frac{1}{H\omega_0 (x) } } \text{ such that } \omega_0 (x)=0 \text{ and } H\omega_0 (x)\sl0 }.
\end{equation*}

We choose $\omega_0(x)=\sin x$ so that $H\omega_0(x)=-\cos x$ by following calculation:²
\begin{equation*}
\begin{split}
H(\sin x )=& -\frac 1\pi \int \frac{\sin y }{y-x }dy=-\frac 1\pi \int \frac{\sin (x+z) }{z} d z\\
= &- \frac 1{\pi}\br{\sin x \int \frac {\cos z }{z}dz+\cos x \int \frac{\sin z}{z}dz }\\
=&- \frac 1\pi \cdot \pi \cos x =-\cos x.
\end{split}
\end{equation*}Then it comes the vorticity and velocity
\begin{equation}
\label{Solution Formula}
\omega (x,t)= \frac{\sin x}{1+t^2 -2t\cos x },
\end{equation} We can plot the graph of $\omega (x,t)$ as following(at time $t=0,0.7,0.9$ repsectively):

In this case, the singular point occurs at $x=0$ and the blow-up time $T^*=1$. Particularly,
\begin{equation*}
\omega (x,1)=\frac{\sin x}{2(1-\cos x)}= \frac{x+O(x) }{2(1-1+\frac 1 2 x^2 +O(x^2) ) }=\frac 1x+O\br{\frac 1x }.
\end{equation*} This indicates $\omega$ develops a local singularity like $x^{-1}$ near $\pi /2$ as $t\to 1$, then we can compute out respectively
\begin{equation}
\label{UB}
\begin{split}
\int_{-\pi}^\pi \abs{\omega (x,t) }^p dx \to +\infty \text{ as } t\to T^*,\forall p\in [1,\infty], \end{split}
\end{equation} Particularly, the above implies $\norm{\omega (t)}_{L^\infty}$ blows up. Indeed, we can show \eqref{UB} holds for any $\omega_0$ as long as the singular points in $P_x S$ are simple zeros of $\omega_0$.

The second example as well as the model $\partial_t \omega=\omega^2$ mentioned in the introduction motivates the backward self-similar kind of formulation. Indeed, we can show the initial data in the above example is the only possible one up to constant scaling which develops a such finite-time singularity if we assume oddness.

The proof of these two case are quite similar. Indeed, the proof of low regular case covers the high regular one if we assume $\alpha=1$. So we'd like to omit the proof of Theorem 5 and prove Theorem 6 directly.

The explicit self-similar solutions of fundamental model provide the basic singularity observation of 1D system \eqref{1D system}. In the following section, we will view the advection term $a u\partial_x \omega$ as a perturbation and investigate the stability of such finite-time singularity.

2.2 Perturbation stability

The discussion will be conveyed in $H^k$-case and $C^\alpha-$case respectively. The basic idea is to figure out the explicit form of remained part $F_n$ expressed by the inverse of transport operator $L^{-1}$, where $\lambda_n$ is determined by the solvability of $L$. With these expressions, we can give out the estimates of $F_n$ and $\lambda_n$ under correspondent norm and determine the convergence of $F$ and $\lambda$.

2.2.1 High regular ($H^k,k\ge 3$) case

Now it is convenient to consider the 1D model \eqref{1D system} with $|a|\sl a_0 $ for some small $a_0>0$. As we investigated before, if $a=0$, then
\begin{equation*}
\omega (x,t)=\frac 1{1-t}F_0 \br{\frac x{1-t} }, F_0=\frac{z}{1+z^2}
\end{equation*}is the only one of the above self-similar type solution assuming oddness. Then for the perturbed case $|a|\sl a_0$, we consider the solution in the following formation:
\begin{equation}
\label{ansatz}
\omega (x,t)=\frac{1}{1-t} F\br{\frac{x}{(1-t)^{1+\lambda}} }.
\end{equation}Here $F,\lambda$ depends on $a$, and particularly $F(0)=F_0$ and $\lambda(0)=0$ according to the fundamental model. So
it is reasonable to seek for a solution in the expansion form
\begin{equation}
\label{expansion formation}
F(a)=F_0 +\sum_{n=1}^\infty a^n F_n, \lambda(a) =\sum_{n=1}a^n \lambda_n.
\end{equation}Our goal to is find proper $a$ and $\lambda_n$ such that $F$ and $\lambda$ converges in $H^k$, which
establish the existence immediately. Now we substitute \eqref{expansion formation} into the system \eqref{1D system}. Then it is transformed into the following profile equation
\begin{equation}
\label{profile}
F+(1+\lambda)D_zF-a\Lambda^{-1}(F) \partial_z F+2F HF =0.
\end{equation} (Indeed, set $z=x/(1-t)^{1+\lambda }$, then we can compute out that
\begin{equation*}
\begin{split}
\partial_t \omega (x,t)=& \frac{1}{(1-t)^2}F +\frac{1}{1-t}(F )' \cdot \frac{(1+\lambda )x }{(1-t)^{2+\lambda } }=(1-t)^{-2}\br{F +(1+\lambda )z(F )' }(z),\\
\omega H\omega(x,t)= & \frac 1{1-t}F \br{\frac{x}{(1-t)^{1+\lambda }} } \int \frac{\frac 1{1-t}F\br{\frac y{(1-t)^{1+\lambda }} } }{\frac{x}{(1-t)^{1+\lambda }}-\frac{y}{(1-t)^{1+\lambda }} }d\frac{y}{(1-t)^{1+\lambda }}\\
=&\frac{1}{(1-t)^2}F (z)\int \frac{F (w)}{z-w} dw=\frac{1}{(1-t)^2}F HF (z),\\
-u\partial_x\omega (x,t)=& \Lambda^{-1} (\omega )\partial_x\omega (x,t)=\int_0^x H\omega(y,t)dy\cdot \partial_x \omega (x,t)\\
=& \int_0^z \frac{1}{1-t}HF(1-t )^{1+\lambda }dw \cdot \frac1{1-t}F'(z)\frac{1}{(1-t)^{1+\lambda }}\\
=&\frac{1}{(1-t)^2} \int_0^z HF(w )dw F'(z)=-\frac{1}{(1-t)^2}\Lambda^{-1}(F)F'(z).
\end{split}
\end{equation*}Combining the above together, we can finally get \eqref{profile}.) Moreover, take the expansion form \eqref{expansion formation} into the profile equation \eqref{profile}, then it comes
\begin{equation*}
\begin{split}
\br{F_0+\sum_{n=0} a^n F_n }+\br{\br{1+\sum_{n=0}^\infty a^n \lambda _n }z-a\Lambda^{-1}\br{F_0+\sum_{n=0}^\infty a^n F_n } }\\
\times\br{F_0+\sum_{n=0}^\infty a^n F_n }'+2\br{F_0+\sum_{n=0}F_n }H\br{F_0+\sum_{n=0}F_n } =0
\end{split}
\end{equation*}Sort the terms by the power of $a$, then
\begin{equation*}
\begin{split}
\sum_{n=1}^\infty a^n \left(F_n +\br{\sum_{j=0}^n \lambda_j (F_{n-j} )' }z-\sum_{j=0}^{n-1}\Lambda^{-1}(F _j)(F_{n-j-1} )'\right. \\
\left.+2\sum_{j=0}^n F_j HF_{n-j} \right) +F_0+zF_0'+2F_0HF_0=0.
\end{split}
\end{equation*}(For convenience we denote $F_0 \coloneqq F_0, \lambda_0 =1$ and squeeze them into the series, then
\begin{equation*}
\begin{split}
z\br{\sum_{n=0}^\infty a^n \lambda _n }\br{\sum_{n=0}^\infty a^n F_n }'= & z\sum_{n=0}^\infty a^n \br{\sum_{j=0}^n \lambda_j (F_{n-j} )' },\\
a\Lambda^{-1}\br{\sum_{n=0}^\infty a^nF_n }\br{\sum_{n=0}^\infty a^n F_n }'=& \sum_{n=0}^\infty a^{n+1}\br{\sum_{j=0}^n \Lambda^{-1}(F_j)(F _{n-k})' }\\
=&\sum_{n=0}^\infty a^{n}\br{\sum_{j=0}^{n-1} \Lambda^{-1}(F_j)(F _{n-1-k})' }\\
2\br{\sum_{n=0}^\infty a^nF_n }H\br{\sum_{n=0}^\infty a^nF_n }=&2\sum_{n=0}^\infty a^n \br{\sum_{j=0}^n F_j H F_{n-j} }.\\
\end{split}
\end{equation*}Combining together, we finally get the equation.) Extract the term related to $F_n$, then
\begin{equation}
\label{LF}
\begin{split}
L(F_n )\coloneqq & F_n +z(F_n )'+2F_0 HF_n +2HF_0 F_n \\
=&\sum_{j=0}^{n-1} \Lambda^{-1} (F_j )(F_{n-1-j} )'-2\sum_{j=1}^{n-1} F_j HF_{n-j} -\br{\sum_{j=1}^n \lambda_j (F_{n-j} )' }z\\
\eqqcolon & G_n(\lambda_1,\dots,\lambda_n; F_1,\dots,F_{n-1} )-\lambda_n (F_0 )'z\\
\end{split}
\end{equation}Consequently, it is necessary to figure out the solvability(i.e. invertibility) of $L$ under certain conditions. Here we set $ F_0={z}/\br{1+z^2}$ and $Lf=g$ as $g\in C^1_0$ is odd³, then after Hilbert transform on the both side,
\begin{equation}
\label{matrix}
\begin{cases}
f+zf'-\frac{2}{1+z^2}f+2\frac{z}{1+z^2} Hf=g,\\
Hf+z(Hf)'-\frac{2}{1+z^2}Hf- \frac{2z}{1+z^2}f=Hg.
\end{cases}
\end{equation} (Here we use the Tricomi identity and the fact $f\to 0$ as $|x|\to 0$:
\begin{equation*}
\begin{split}
H(zf')=\int_{-\infty}^{+\infty} \frac{(z-w)f(z-w)}{\pi w}dw& =z\int_{-\infty}^{+\infty} \frac{f'(z-w)}{\pi w}\ dw+\frac 1 \pi \int_{-\infty}^{+\infty} f'(v)dv=zHf',\\
H\br{-\frac{2}{1+z^2}f+ \frac{2z}{1+z^2}Hf}=& -\frac{2}{1+z^2}H f-\frac{2z}{1+z^2}f- H\br{\frac{2z}{1+z^2}Hf }\\
&- \frac{2z}{1+z^2}f-\frac{2}{1+z^2} Hf+H\br{\frac{2}{1+z^2} f }\\
\Longrightarrow H\br{-\frac{2}{1+z^2}f+ \frac{2z}{1+z^2}Hf}=&-\frac{2}{1+z^2}H f-\frac{2z}{1+z^2}f.
\end{split}
\end{equation*}Proof of basic Hilbert transform and Tricomi identity see in Appendix \ref{Hilbert transform}.) A necessary condition of solvability comes immediately if we take $\lim_{z\to 0+} z^{-1}\cdot$ in \eqref{matrix}-1 and evaluate $z=0$ in \eqref{matrix}-2:
\begin{equation}
\label{relation}
2Hf(0)=g'(0)\text{ and } Hf(0)=-Hg(0)
\end{equation}since $f(0)=g(0)=0$ by the oddness, which implies eventually:
\begin{equation}
\label{Initial Condition}
g'(0)+2Hg(0)=0.
\end{equation}Now we set the following quantities:
\begin{equation*}
U\coloneqq f+iHf, G\coloneqq g+iHg
\end{equation*}and denote $\hat d\coloneqq d-d(0)$⁴. First rewrite \eqref{matrix} into the following form:
\begin{equation}
\label{rewrite}
\begin{cases}
f'+\frac{z^2-1}{z(z^2+1)}f+\frac{2}{1+z^2}\br{Hf-Hf(0) }=\frac{g}{z}-\frac{1}{1+z^2}g'(0),\\
(Hf)'-\frac{2}{z(1+z^2)}f+\frac{z^2-1}{z(1+z^2)}\br{Hf-Hf(0) }=\frac{Hg-Hg(0)}{z}+\frac{2z}{1+z^2}Hg(0)
\end{cases}
\end{equation}using the necessary relations \eqref{relation}. Denote the right side of two equations as $g_1,g_2$ and $\hat G=g_1+ig_2$ respectively.
Then we can derive the ODE system for $U-U(0)$ after summing \eqref{rewrite}-1 and $i\times \eqref{rewrite}$-2:
\begin{equation*}
(U-U(0))'+\frac{z-i}{z(z+i)}(U -U(0))=\hat G, \hat G=g_1+ig_2.
\end{equation*}A direct calculation gives out the formation of $\hat U$:
\begin{equation*}
\hat U=\frac{-z}{(z+i)^2}\br{c+\int_0^z \frac{-(w+i)^2}{w}\tilde G (w) dw}.
\end{equation*} (Multiplying both side by the integrating factor $-\frac{(z+i)^2}{z} $ results in
$$-\frac{(z+i)^2}{z}u'-\frac{(z+i)(z-i)}{z^2}u=-\frac{(z+i)^2}{z}\tilde G$$and thus
$$\frac d{dz} \br{-\frac{(z+i)^2}{z} u }=-\frac{(z+i)^2}{z}\tilde G. $$Integrating both side, we get the formation finally.)
Extracting the real part of the formation, we finally get the expression of $f$ in matrix form:⁵
\begin{equation*}
\begin{pmatrix}
f \\
Hf-Hf(0)
\end{pmatrix}=\frac{z}{(z^2+1)^2}
\begin{pmatrix}
1-z^2 & -2z\\
2z & 1-z^2
\end{pmatrix}
\br{
\begin{pmatrix}
c_1\\
c_2
\end{pmatrix}+
\int_0^z
\begin{pmatrix}
\frac{1-w^2}{w} & 2\\
-2 & \frac{1-w^2}{w}
\end{pmatrix}
\begin{pmatrix}
g_1 (w) \\ g_2 (w)
\end{pmatrix} dw
}.
\end{equation*}We can observe that for $|z|$ small, the integral term is of order $Q(z)$, and $\begin{pmatrix}
1-z^2 & -2z\\
2z & 1-z^2
\end{pmatrix}\to I$. So we have precisely $c_1=f'(0),c_2=Hf'(0)$. Then evenness of $Hf$ implies $c_2=0$,
and thus we have:
\begin{equation*}
\begin{split}
f(z)=&\frac{ z(1-z^2)}{(z^2+1)^2}c_1+\frac{z(1-z^2)}{(z^2+1)^2} \int_0^z \br{\frac{1-w^2}{w}g_1(w)+2g_2(w) }dw\\
&+ \frac{-2z^2}{(z^2+1)^2} \int_0^z \br{-2g_1(w)+\frac{1-w^2}{w}g _2(w) }dw.
\end{split}
\end{equation*}This clearly implies that $g\in H^k\Longrightarrow f\in H^k$ for any $k\ge 3$.
Particularly, if we require $f'(0)=0$, then the solution $L^{-1}g$ is uniquely determined by
\begin{equation}
\label{Formation}
\begin{split}
L^{-1}g(z)=&\frac{z(z^2-1)}{(z^2+1)^2} \int_0^z \br{\frac{w^2-1}{w}g_1(w)-2g_2(w) }dw\\
&+ \frac{2z^2}{(z^2+1)^2} \int_0^z \br{2g_1(w)+\frac{w^2-1}{w}g _2(w) }dw.
\end{split}
\end{equation}It remains to show that the formation is indeed the solution of $Lf=g$. (Omit ) We finally conclude the above results as the following invertibility.

Now by the formation \eqref{Formation} and definition of $g_1,g_2$, we can write
\begin{equation*}
L^{-1}=T^{1,1}-T^{1,-1}-2S^{1,0}+2S^{2,0}+T^{2,1}-T^{2,-1},
\end{equation*}where $T^{l,\sigma},S^{l,\sigma}, l\in \{1,2\},\sigma\in \{1,0,-1\}$ is defined by
\begin{equation*}
\begin{split}
T^{1,\sigma}(g)=& \frac{z(1-z^2)}{(z^2+1)^2} \int_0^z s^\sigma\br{\frac{g(s)}{s}-\frac{1}{1+s^2}g'(0) }ds,\\
T^{2,\sigma}(g)=& \frac{2z^2}{(z^2+1)^2} \int_0^z s^\sigma\br{\frac{g(s)}{s}-\frac{1}{1+s^2}g'(0) }ds,\\
S^{1,\sigma}(g)=& \frac{z(1-z^2)}{(z^2+1)^2} \int_0^z s^\sigma\br{\frac{Hg(s)-Hg(0)}{s}+\frac{2s}{1+s^2}Hg(0) }ds,\\
S^{2,\sigma}(g)=& \frac{2z^2}{(z^2+1)^2} \int_0^z s^\sigma\br{\frac{Hg(s)-Hg(0)}{s}+\frac{2s}{1+s^2}Hg(0) }ds.
\end{split}
\end{equation*}As we will set $g=G_n(\lambda_1 ,\dots,\lambda_{n-1} ;F_1 ,\dots,F_{n-1} )-\lambda_n (F _0)'z$, it is necessary to obtain following $H^{s}$ estimates for $T^{l,\sigma}, S^{l,\sigma}$:

The estimates will be proved in the end of the section. Now we will apply them to derive the perturbation results.

With these estimates, now we are able to establish the convergence of the series \eqref{Formation} under certain choice of $\lambda_n $. Indeed, according to Lemma 7, we substitute the invertible initial condition \eqref{Initial Condition} into the equation \eqref{LF}, which indicates
\begin{equation*}
\br{G_n -\lambda_n \frac{z(1-z^2)}{(1+z^2)^2} }'(0)+ 2H\br{G_n -\lambda_n \frac{z(1-z^2)}{(1+z^2)^2} }(0)=0\Longrightarrow \lambda_n =G_n'(0)+2HG_n (0) .
\end{equation*} (Here
$$\br{\frac{z(1-z^2) }{(1+z^2)^2}}'(0)=\frac{(1-3z^2)(1+z^2)^2-4(1+z^2)z^2(1-z^2)}{(1+z^2)^4 }\Big|_{z=0}=1, $$ and
\begin{equation*}
\begin{split}
H\br{\frac{z(1-z^2) }{(1+z^2)^2}}(0)=& \frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{(z^2-1) }{(1+z^2)^2}dz=\frac{z}{\pi(1+z^2)}\Big|_{z=-\infty}^{z=+\infty}=0.
\end{split}
\end{equation*} ) Moreover, we can compute out that
\begin{equation*}
\begin{split}
G_n'(0)=& \sum_{j=0}^{n-1} HF_j (0)(F _{n-1-j})'(0)-2\sum_{j=1 }^{n-1} (F_j )'(0)HF_{n-j} (0)-\sum_{j=1}^{n-1}\lambda_j (F_{n-j}^{a})'(0),\\
H G_n (0)=& \sum_{j=1}^{n-1} H\br{\Lambda^{-1}(F_j ) F_{n-1-j}' }(0)- \sum_{j=1}^{n-1} (H(F_j )H( F_{n-j} )- F_j F_{n-j} )(0).
\end{split}
\end{equation*} ( Here we use the Tricomi identity:
\begin{equation*}
\begin{split}
2\sum_{j=1}^{n-1} H (HF_j F_{n-j} )(0)=& \sum_{j=1}^{n-1} H (HF_j F_{n-j}+HF_{n-j} F_j )(0)\\
=& \sum_{j=1}^{n-1} H (H(F_j F_{n-j})- H( H F_j H F_{n-j}) )(0)\\
=& \sum_{j=1}^{n-1} (H(F_j)H( F_{n-j})- F_j F_{n-j} )(0).
\end{split}
\end{equation*} and the fact $H(zf')(0)=0$ for $f$ is odd .) Denote $\norm{F_{j} }_{H^3}=\mu_j$, then applying the above results and Sobolev embeddings, we finally get the recurrence estimate for $|\lambda_n |, \mu_n$:
\begin{equation*}
\begin{split}
|\lambda_n |\le & \norm{HF_0 }_{L^\infty}\norm{F_{n-1}'}_{L^\infty}+\sum_{j=1 }^{n-1}C( \norm{F_j' }_{L^\infty}\norm{HF_{n-1-j}}_{L^\infty}\\
&+\norm{\Lambda^{-1}(F_j) }_{L^\infty}\norm{F_{n-1-j}'}_{L^\infty}+\norm{F_j}_{L^\infty}\norm{F_{n-j}}_{L^\infty}+ |\lambda_j | \norm{F_{n-j}'}_{L^\infty} )\\
\le & C\mu_0 \mu_{n-1-j}+C\sum_{j=1}^{n-1}\br{\mu_j \mu_{j}+\mu_j\mu_{n-j}+|\lambda_j |\mu_{n-j} },\\
\end{split}
\end{equation*} and by the $H^3$-estimates \eqref{H3 estimates},
\begin{equation*}
\begin{split}
\mu_{n}= &\norm{F_j}_{H^3}\le |\lambda_n| \norm{L^{-1} (zF_0') }_{H^3}+ \norm{L^{-1}(\Lambda^{-1}(F_0)F_{n-1}')}_{H^3}\\
&+\sum_{j=1}^{n-1}\norm{L^{-1}(\Lambda^{-1} (F_j)F_{n-1-j}' ) +2L^{-1}(F_j HF_{n-j})+|\lambda_j | L^{-1}(zF_{n-j}') }_{H^3} \\
\le &C|\lambda_n| \mu_0+ C\mu_0\mu_{n-1}+C\sum_{j-1}^n\br{\mu_j \mu_{n-1-j}+\mu_j \mu_{n-j}+|\lambda_j|\mu_{n-j} }.
\end{split}
\end{equation*} Consequently, there exists come absolute constant $C_0>0$ such that $\zeta_n \coloneqq \mu_n +C_0(|\lambda_n |+\mu_{n-1} ) $ for $n\ge 1$ satisfies the inequality:
\begin{equation*}
\zeta_n \le C\sum_{j=1}^{n-1} \zeta_j \zeta_{n-j} \Longrightarrow \zeta_j \le C^n \zeta_0^n,
\end{equation*} where $ = \frac{1}{n}\binom{2(n-1) }{n-1} $ is the well-known Catalan number⁶ for $n\ge 1$, growing asymptotically as
\begin{equation*}
C_n \sim \frac{4^n}{(n-1)^\frac 32 \sqrt \pi}.
\end{equation*} Consequently, we see there exists $r>0$ such that
$$\zeta_n \le r^n\Longrightarrow \mu_n \le r^n, |\lambda_n |\le r^n.$$ Now if we set $a_0=r^{-1}$, then for any $|a|\sl a_0$, we see $\norm{F }_{H^3}$ and $\lambda $ converges as series. Choose $\omega _0=F $, then we finally obtains:

Though the solution $w $ constructed above blows up in $L^\infty_{t,x}([0,1]\times\R )$ sense, some weaker uniform boundedness is still preserved.

2.2.2 Low regular ($C^\alpha,0\sl\alpha\sl 1$) case

Inspired by the fundamental model, it is considerable to discuss the perturbed system \eqref{1D system} in variable $w=z^\alpha$ for $z\in\ \R^+$ for $C^{\alpha}$ data. So we choose the following ansatz:
\begin{equation*}
z=\frac{x}{(1-t)^{\frac{1+\lambda^{\alpha}}{\alpha} } }, \omega^{\alpha} (t,x)=\frac{1}{1-t}F^{\alpha}\br{z },
\end{equation*}which corresponds to the equation
\begin{equation}
\label{alpha profile equation}
F^\alpha +\frac{1+\lambda^\alpha}{\alpha} z\partial_z F^\alpha -a\Lambda^{-1}(F^\alpha )\partial_z F^\alpha+2F^\alpha HF^\alpha=0.
\end{equation}Based on the analysis on the fundamental model, it is reasonable to seek a solution as
\begin{equation*}
F^\alpha (a)=F^\alpha_0+\sum_{n=1}^\infty a^n F_n^\alpha, \lambda^\alpha (a)=\sum_{n=1}^\infty a^n \lambda_n^\alpha.
\end{equation*}Substitute them into the profile equation \eqref{alpha profile equation}, then
\begin{equation*}
\begin{split}
L^\alpha (F_n^\alpha)\coloneqq & F_n^\alpha +\frac 1 \alpha z\partial_z F_n^\alpha+2H(F_0^\alpha ) F_n^\alpha+2F_0^\alpha HF_n^\alpha\\
=& \sum_{j=0}^{n-1} \Lambda^{-1}(F_j^\alpha)\partial_z F_{n-1-j}^\alpha-\frac 1\alpha \sum_{j=1}^{n-1}\lambda_j^\alpha z\partial_z F_{n-j}^\alpha-2\sum_{j=1}^{n-1}F_j^\alpha HF_{n-j}^\alpha -\frac 1\alpha \lambda_n^\alpha z \partial_z F_{n-1}^\alpha\\
\coloneqq & G_n^\alpha (F_{n-1}^\alpha,\dots,\cdot ,F_1^\alpha;\lambda_{n-1}^\alpha,\dots,\lambda_1^\alpha )-\frac 1\alpha \lambda_n z\partial_z F_{n-1}^\alpha.
\end{split}
\end{equation*}Then the existence is reduced to the solvability of $L^\alpha$. Before the further investigation, we observe from the fundamental model that the involved functions are in better form if viewed as the function of $z^\alpha$. So it is proper to set
$$f(z)= f(w^\frac 1\alpha)=\tilde f(w).$$

Then the above is written as
\begin{equation*}
\tilde L^\alpha (\tilde F_n^\alpha )(w)\coloneqq \widetilde{L^\alpha (F^\alpha_n) }(w)= \tilde F^\alpha_n (w)+w\partial_w \tilde F^\alpha_n (w)+2 \widetilde{ H (F_0^\alpha) }(w) \tilde F^\alpha_n (w)+2\tilde F^\alpha_0(w) \tilde H^\alpha \tilde F_n^\alpha (w).
\end{equation*}We consider $\alpha\in \set{1/l}_{l\in \N}$ for simplicity, as the main concern is the results when $\alpha\to 0$. Setting $\alpha=1/l$, now we take the ansatz where particularly $\lambda^{\alpha}(0)=0$. The idea is similar with one in the $H^k$ case.

3 Blowup of 3D Euler flow

The blowup mechanism we introduce here is analogue to the simplified 1D model. Particularly, we will establish the discussion under the axisymmetric flow without swirl. The point is that this model can reduce the Cartesian coordinates $(x_1,x_2,x_3)$ into radial variable $z$ and angular variable $\theta$. The radial part is correspondent to 1D case, where we can handle the angular part under the separation assumption($f(z,\theta)=f^1(z)f^2(\theta)$). Recall the classic vorticity-stream formulation of Euler equations:⁷
\begin{equation*}
\begin{cases}
\frac 12 \partial_t \omega+ u\cdot \nabla \omega = \omega \cdot \nabla u,\\
-\Delta \psi=\omega,\\
u=-\nabla \times \psi.
\end{cases}
\end{equation*}They can be obtained immediately by the Biot-Savart law we mentioned before. As for the axisymmetric flow without flow
\begin{equation*}
u=u^r(r,x_3,t) e_r+u^3 (r,x_3,t) e_3.
\end{equation*}The above system is transformed as following model:
\begin{equation*}
\begin{cases}
\frac 12 \partial_t \omega +u^r \partial_r \omega +u^3\partial_3 \omega =\frac{u^r}{r} \omega,\\
\partial_r^2 \psi +\frac 1r \partial_r \psi-\frac 1{r^2}\psi +\partial_3 \psi=-\omega,\\
(u^r,u^3)=(\partial_3 \psi, -\partial_r \psi-\frac 1r\psi ).
\end{cases}
\end{equation*}Here $\omega, \psi$ correspond to $\varphi-$component of the original system. The next spherical transform is significant: as we concern about $C^\alpha$-data case, it is inspired by $1$D discussion to consider the following variable transform:
\begin{equation*}
\rho =\sqrt{ r^2+x_3^2}, \tan \theta=\frac {x_3}{r}, R=\rho^\alpha,
\end{equation*}and homogeneity⁸ $\omega(r,x_3,t)=\Omega(R,\theta,t), \psi(r,x_3,t)=\rho^2 \Psi (R, \theta,t)$. Then the spherical form is
\begin{equation}
\label{spherical form}
\begin{cases}
\frac 12 \partial_t \Omega +U(\Psi )\partial_\theta \Omega+V(\Psi)\alpha D_R \Omega=R(\Psi)\Omega,\\
L(\Psi )=-\Omega,
\end{cases}
\end{equation}where $D_R=R\partial_R$ and the linear operators involved are defined as
\begin{equation*}
\begin{split}
&U\coloneqq-3\Id-\alpha D_R,V\coloneqq\partial_\theta-\tan \theta,\\
&R\coloneqq\frac 1{\cos \theta}\br{2\sin \theta +\alpha \sin \theta D_z+\cos \theta \partial_\theta },\\
& L\coloneqq L_R+L_\theta\coloneqq (\alpha^2 D_R^2 +\alpha (5+\alpha)D_R) +(\partial_\theta +\partial_\theta(\tan \theta \cdot )-6\Id ).
\end{split}
\end{equation*}Still, $R(\Psi )\Omega$ is the vortex stretching which causes the finite-time singularity, and $U(\Psi)\partial_\theta \Omega+V(\Psi)\alpha D_R\Omega$ is the advection regularizing the vorticity. The main trouble comes if we view the advection as a $\alpha-$perturbation, as $\frac 12 \partial_t \Omega=R(\Psi)\Omega $ is not practical to figure out a explicit solution as the 1D case, for which we cannot analyze the singularity precisely. Our pillar observation, which arises in Section 3.1 and 3.3, is that
$$R(\Psi)\Omega=\frac 1{2\alpha}\Omega L_{12}(\Omega)+\text{low order terms about }\alpha,$$with
$$L_{12}(f)(R)=\int_R^\infty \int_0^\frac \pi 2\frac{f(R',\theta')K(\theta')}{R'} d\theta'dR', K(\theta)=3\sin \theta \cos^2 \theta. $$Consequently, it is enlightening to consider
\begin{equation*}
\frac 12 \partial_t\Omega=\frac 1{2\alpha} \Omega L_{12} (\Omega)
\end{equation*}as the fundamental model, and view the remain part and advection as perturbations. We can solve this model explicitly with following self-similar ansatz in Section 3.1:
$$\Omega(R,\theta,t) =\frac{1}{1-t}F_* (\frac{R}{1-t},\theta ),F_*(z,\theta)=\alpha \frac{\Gamma(\theta)}{c^*}\frac{2z}{(1+z)^2}. $$Based on this result and analysis on 1D model, the linearization of the perturbation can be done via following self-similar type transform for 3D Euler flow:
\begin{equation*}
\Omega (R,\theta,t)=\frac{1}{1-(1+\mu)t}F(z,\theta), \Psi(R,\theta,t)=\frac{1}{1-(1+\mu)t}\Phi(z,\theta), z=\frac{R}{(1+(1+\mu)t)^{1+\lambda}},
\end{equation*}where $\mu,\lambda$ are some constants undetermined but related to $\alpha$.
Under this transform, we obtain the following profile equations from the spherical form \eqref{spherical form} of Euler flow:
\begin{equation}
\begin{cases}
(1+\mu)F+(1+\mu)(1+\lambda)D_z F+2U(\Phi)\partial_\theta F+2 V(\Phi )\alpha D_z F=2R(\Phi)F,\\
L \Phi= L_z \Phi +L_\theta\Phi=-F.
\end{cases}
\end{equation} Still, we express the profile $F$ as a perturbation of $F_*:$
$$F=F_*+g.$$With suitable choice of constant $\lambda,\mu$ and \eqref{SS FM} holds, we can calculate out the equation for the remained part $g$:
\begin{equation}
\Li_{\Gamma}^T g=\P(\Ni_*+\Ni+\Ni_0),
\end{equation}where $\Li_{\Gamma}^T$ is the transport operator, and $\P(\Ni_*+\Ni+\Ni_0)$ related to $F_*$ and $g$ (we will deduce them later in Section 3.2).

The significant difference from 1D model arises: in simplified 1D case we can view $g$ depends on perturbation analytically, i.e. $g=\sum_{n=1}^\infty a^n F_n$, and $F_n $ can be expressed precisely by the solvability of transport operator, from which we can convey a direct estimate on $g$. However, it does not work for 3D case. Indeed, the full transport operator $\Li_\Gamma^T$ for 3D Euler flow is too complex to give a explicit inverse. However, the idea will work if returning to the traditional (but more universal) routine: we will obtain a propri estimates of $g$:
$$\norm{g}^2 \le C\br{\alpha^2 \norm{g}^2 +\alpha^\frac 12 \norm{g }^2+\alpha^{-\frac{3}{2}}\norm{g}^3+\alpha^{-\frac{5}{2}}\norm{g}^4 }, $$ and establish the existence via a compactness argument(See Section 3.5). The a propri estimates is based on two calculations: The coercivity of the transport operator:
\begin{equation*}
\br{\Li_\Gamma^T g,g }\ge C\norm{g}^2,
\end{equation*} and the boundedness of bilinear form:
\begin{equation*}
\br{\P(\Ni_*+\Ni+\Ni_0),g }\le C\br{\alpha^2 \norm{g}^2 +\alpha^\frac 12 \norm{g }^2+\alpha^{-\frac{3}{2}}\norm{g}^3+\alpha^{-\frac{5}{2}}\norm{g}^4 }.
\end{equation*}We will give detailed calculations in Section 3.2 and Section 3.4 respectively. All above estimates will be processed under new-introduced Sobolev type weighted spaces $\Hi^k_{z,\theta}$ and $\Wi^{k,\infty}_{z,\theta}$, defined by
\begin{equation*}
\begin{split}
\norm{f}_{\Hi^k}^2& \coloneqq \sum_{i\le k } \norm{D^i_z f w_z v_\theta}_{L^2}^2+\sum_{i+j\le k, j>0} \norm{D_z^iD_\theta^j f w_z w_\theta}_{L^2}^2,\\
\norm{f}_{W^{l,\infty}}& \coloneqq \sum_{i\le l}\norm{\tilde D_z^i f }_{L^\infty }+\sum_{i+j \le l,j\ge 0 } \norm{\tilde D^i_zD^j_\theta f \frac{\sin^{-\frac \alpha 5 } 2\theta }{\alpha+\sin 2\theta} }_{L^\infty},\\
D_z & =z\partial_z , \tilde D_z=(z+1)\partial_z, D_\theta =\sin (2\theta)\partial_\theta,\\
w_z(z)& =\frac{(1+z)^2}{z^2}, w_\theta=(\sin \theta \cos^2 \theta)^{-\frac \gamma 2}, v_\theta =(\sin \theta \cos^2\theta )^{-\frac {\eta}{2}},
\end{split}
\end{equation*}where the weights $w_z,w_\theta,v_\theta$ are selected to ensure the coercivity of $\Li_\Gamma^T$ and boundedness of the elliptic estimate. Some basic properties of these spaces will be listed in Appendix.

3.1 Fundamental model

3.2 Transport coercivity

3.3 Elliptic estimate

3.4 A propri estimate

3.5 Compactness argument

Appendix

Weighted Sobolev spaces

1. Indeed, we will show $a=2$ is admissible if we require low regularity in $C^\alpha$-case. ↩

2. See Math Stack Exchange for the calculation of $\int \sin y/y dy$ by Cauchy theorem. ↩

3. This will implies $f\in C_0^1$ is odd as $L$ preserves the parity. ↩

4. Particularly, this implies $\hat U=f+i(Hf-Hf(0)),\hat G=g+i(Hg-Hg(0)).$ ↩

5. Notice that $(a+ib)(c+id)= \begin{pmatrix}
   1 & i
   \end{pmatrix}
   \begin{pmatrix}
   a & -b\\
   b & a
   \end{pmatrix}
   \begin{pmatrix}
   c\\ d
   \end{pmatrix}. $ ↩

6. See pedia for details. ↩

7. Here we dilate the time variable to ensure blowup near unit time $t=1$, and add negative sign ahead of stream equation to simplified the elliptic estimate. ↩

8. to ensure no extra radial variable occurring in the equations. ↩