TS-GCN: A novel tumor segmentation method integrating transformer and GCN

Haiyan Song; Cuihong Liu; Shengnan Li; Peixiao Zhang; Haiyan Song; Cuihong Liu; Shengnan Li; Peixiao Zhang

doi:10.3934/mbe.2023807

Mathematical Biosciences and Engineering

2023, Volume 20, Issue 10: 18173-18190. doi: 10.3934/mbe.2023807

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Research article Special Issues

TS-GCN: A novel tumor segmentation method integrating transformer and GCN

Haiyan Song ^{1
,
,},
Cuihong Liu ^{2,3
,
,},
Shengnan Li ^{1
,
,},
Peixiao Zhang ¹

1.
The Second Affiliated Hospital of Shandong University of Traditional Chinese Medicine, Jinan, China
2.
Affiliated Eye Hospital of Shandong University of Traditional Chinese Medicine, Jinan, China
3.
School of Nursing, Shandong University of Traditional Chinese Medicine, Jinan, China

Received: 15 May 2023 Revised: 29 May 2023 Accepted: 30 May 2023 Published: 21 September 2023

As one of the critical branches of medical image processing, the task of segmentation of breast cancer tumors is of great importance for planning surgical interventions, radiotherapy and chemotherapy. Breast cancer tumor segmentation faces several challenges, including the inherent complexity and heterogeneity of breast tissue, the presence of various imaging artifacts and noise in medical images, low contrast between the tumor region and healthy tissue, and inconsistent size of the tumor region. Furthermore, the existing segmentation methods may not fully capture the rich spatial and contextual information in small-sized regions in breast images, leading to suboptimal performance. In this paper, we propose a novel breast tumor segmentation method, called the transformer and graph convolutional neural (TS-GCN) network, for medical imaging analysis. Specifically, we designed a feature aggregation network to fuse the features extracted from the transformer, GCN and convolutional neural network (CNN) networks. The CNN extract network is designed for the image's local deep feature, and the transformer and GCN networks can better capture the spatial and context dependencies among pixels in images. By leveraging the strengths of three feature extraction networks, our method achieved superior segmentation performance on the BUSI dataset and dataset B. The TS-GCN showed the best performance on several indexes, with Acc of 0.9373, Dice of 0.9058, IoU of 0.7634, F1 score of 0.9338, and AUC of 0.9692, which outperforms other state-of-the-art methods. The research of this segmentation method provides a promising future for medical image analysis and diagnosis of other diseases.

Keywords:

Citation: Haiyan Song, Cuihong Liu, Shengnan Li, Peixiao Zhang. TS-GCN: A novel tumor segmentation method integrating transformer and GCN[J]. Mathematical Biosciences and Engineering, 2023, 20(10): 18173-18190. doi: 10.3934/mbe.2023807

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Abstract

1. Introduction

The 3D incompressible resistive Hall-Magnetohydrodynamics system (Hall-MHD in short) is the following system of PDEs for $(u, p, B)$ :

$\begin{align} &u_{t} +u\cdot \nabla u - B\cdot \nabla B+\nabla p-\mu\Delta u = 0, \end{align}$

(1a)

$\begin{align} &B_{t} + u\cdot \nabla B -B\cdot \nabla u +{\rm{curl}}\left(\left({\rm{curl}} \;B\right)\times B\right)- \nu\Delta B = 0, \end{align}$

(1b)

$\begin{align} &{\rm{div}}\; u = 0, \quad {\rm{div}}\; B = 0, \end{align}$

(1c)

where $u = (u_{1}, u_{2}, u_{3})$ is the plasma velocity field, $p$ is the pressure, and $B = (B_{1}, B_{2}, B_{3})$ is the magnetic field. $\mu$ and $\nu$ are the viscosity and the resistivity constants, respectively. The Hall-MHD is important in describing many physical phenomena [2,17,19,23,26,27,33]. In particular, the Hall MHD explains magnetic reconnection on the Sun which is very important role in acceleration plasma by converting magnetic energy into bulk kinetic energy.

The Hall-MHD recently has been studied intensively. The Hall-MHD can be derived from either two fluids model or kinetic models in a mathematically rigorous way [1]. Global weak solution, local classical solution, global solution for small data, and decay rates are established in [4,5,6]. There have been many follow-up results of these papers; see [7,8,12,13,14,15,16,18,29,30,31,32,34,35] and references therein.

1.1. The Hall equation with fractional Laplacian

We note that the Hall term ${\rm{curl}}\left(\left({\rm{curl}} \;B\right)\times B\right)$ is dominant in mathematical analysis of (1) and so we only consider the Hall equations ( $(u, p) = 0$ in (1)). Also motivated by [7], we consider the Hall equation with fractional Laplacian:

$\begin{eqnarray} B_{t}+{\rm{curl}} \left(({\rm{curl}} \;B)\times B\right)+\Lambda^{\beta}B = 0, \quad {\rm{div}}\; B = 0, \end{eqnarray}$

(2)

where we take $\nu = 1$ for simplicity. (2) is locally well-posed [7] when $\beta>1$ . But, we do not know whether (2) is locally well-posed when $\beta = 1$ :

$\begin{eqnarray} B_{t}+{\rm{curl}} \left(({\rm{curl}} \;B)\times B\right)+\Lambda B = 0, \quad {\rm{div}}\; B = 0. \end{eqnarray}$

(3)

However, we can show the existence of solutions globally in time if initial data is sufficiently small.

Theorem 1.1. Let $B_{0} \in H^{k}$ with $k>\frac{5}{2}$ and ${\rm{div}}\; B_{0} = 0$ . There exists a constant $\epsilon_{0}>0$ such that if $\|B_{0}\|_{H^{k}} \leq \epsilon_{0}$ , there exists a unique global-in-time solution of (3) satisfying

$\left\|B(t)\right\|^{2}_{H^{k}}+(1-C\epsilon_{0})\int^{t}_{0}\left\|\Lambda^{\frac{1}{2}}B(s)\right\|^{2}_{H^{k}}ds\leq \left\|B_{0}\right\|^{2}_{H^{k}} \quad {{for\; all \; t > 0 .}}$

Moreover, $B$ decays in time

$\begin{eqnarray} \left\|\Lambda^{l}B(t)\right\|_{L^{2}}\leq \frac{C_{0}}{(1+t)^{l}}, \quad 0 < l\leq k, \end{eqnarray}$

(4)

where $C_{0}$ depends on $\|B_{0}\|_{H^{k}}$ which is expressed in (27) explicitly.

Remark 1. The decay rate (4) is consistent with the decay rates of the linear part of (3).

Remark 2. After this work was completed, the referee pointed out that the same result is proved in [37,Theorem 1.1]. Compared to the proof in [37] where they use the Littlewood-Paley decomposition, we use the standard energy energy estimates and classical commutator estimates.

As one of a minimal modification of (3) to show the existence of unique local in time solutions, we now take a logarithmic correction of (3):

$\begin{eqnarray} B_{t}+{\rm{curl}} \left(({\rm{curl}} \;B)\times B\right)+\ln(2+\Lambda)\Lambda B = 0, \end{eqnarray}$

(5)

where the Fourier symbol of $\ln(2+\Lambda)\Lambda$ is $\ln(2+|\xi|)|\xi|$ .

Theorem 1.2. Let $B_{0} \in H^{k}$ with $k>\frac{5}{2}$ and ${\rm{div}}\; B_{0} = 0$ . There exists $T_\ast = T_{\ast}(\|B_{0}\|_{H^{k}})>0$ such that there exists a unique local-in-time solution of (5) satisfying

$\begin{eqnarray} \|B(t)\|_{H^{k}}\leq \ln \left(\frac{1}{e^{-\|B_{0}\|_{H^{k}}}-Ct}\right), \quad 0 < t < T^{\ast} = \frac{\exp\left(-\|B_{0}\|_{H^{k}}\right)}{C}. \end{eqnarray}$

(6)

1.2. 2D models

In this paper, we also deal with 2D models closely related to the $2\frac{1}{2}$ dimensional (3). If we take $B$ of the form

$\begin{eqnarray} B(t, x, y) = \left(-\psi_{y}(t, x, y), \psi_{x}(t, x, y), Z(t, x, y)\right), \end{eqnarray}$

(7)

we can rewrite (3) as

$\begin{align} & \psi_{t}+\Lambda\psi = [\psi, Z], \end{align}$

(8a)

$\begin{align} & Z_{t}+\Lambda Z = [\Delta \psi, \psi], \end{align}$

(8b)

where $[f, g] = \nabla f\cdot \nabla^{\perp}g = f_{x}g_{y}-f_{y}g_{x}$ . (7) is used to show a finite-time collapse to a current sheet [3,20,21,24] and is used in [10] to study regularity of stationary weak solutions.

1.2.1. Case 1

Although (8) is defined in 2D and has nice cancellation properties (18), the local well-posedness seems unreachable. But, suppose that we redistribute the power of the fractional Laplacians in (8) in such a way that (8b) has the full Laplacian and (8a) is inviscid:

$\begin{eqnarray} \psi_{t} = [\psi, Z], \quad Z_{t}-\Delta Z = [\Delta \psi, \psi]. \end{eqnarray}$

(9)

(9) has no direct link to (2), but we may interpret (9) as the $2\frac{1}{2}$ dimensional model of the Hall equations where only $B_{3}$ has the full Laplacian in (2). In this case, we can show that (9) is locally well-posed. Let

$\begin{eqnarray} \mathcal{E}(t) = \left\|\psi(t)\right\|^{2}_{H^{4}}+\left\|Z(t)\right\|^{2}_{H^{3}}, \quad \mathcal{E}_{0} = \left\|\psi_{0}\right\|^{2}_{H^{4}}+\left\|Z_{0}\right\|^{2}_{H^{3}}. \end{eqnarray}$

(10)

Theorem 1.3. There exists $T_{\ast} = T_{\ast}(\mathcal{E}_{0})>0$ such that there exists a unique solution of (9) satisfying

$\mathcal{E}(t)\leq \frac{\mathcal{E}_{01}}{1-Ct\mathcal{E}_{0}} \quad {{for \;all}}\; \ 0 < t\leq T_{\ast} < \frac{1}{C \mathcal{E}_{0}}.$

Moreover, we have the following blow-up criterion:

$\mathcal{E}(t)+ \int^{t}_{0}\left\|\nabla Z(s)\right\|^{2}_{H^{2}}ds < \infty\iff \int^{t}_{0}\left(\left\|\nabla^{2} Z(s)\right\|_{L^{\infty}} +\left\|\nabla^{2}\psi(s)\right\|^{2}_{L^{\infty}} \right)ds < \infty.$

Since there is no dissipative effect in the equation of $\psi$ in (9), we only have the local in tim result in Theorem 1.3. Among the possible conditions for the global existence, we find that adding a damping term to the equation of $\psi$ works. More precisely, we deal with the following

$\begin{eqnarray} \psi_{t}+\psi = [\psi, Z], \quad Z_{t}-\Delta Z = [\Delta \psi, \psi]. \end{eqnarray}$

(11)

In this case, we can show the existence of global in time solutions with small initial data having regularity one higher than the regularity in Theorem 1.3. Moreover, we can find decay rates of $\psi$ by using the structure of equation of $\psi$ which is a damped transport equation, and this is also the reason why the same method cannot be applied to $Z$ . Let

$\begin{split} &\mathcal{F}(t) = \left\|\psi(t)\right\|^{2}_{H^{5}}+\left\|Z(t)\right\|^{2}_{H^{4}}, \quad \mathcal{F}_{0} = \left\|\psi_{0}\right\|^{2}_{H^{5}}+\left\|Z_{0}\right\|^{2}_{H^{4}}, \\ & \mathcal{N}_{1}(t) = \left\|\nabla\psi(t)\right\|^{2}_{H^{4}}+\left\|\nabla Z(t)\right\|^{2}_{H^{4}}. \end{split}$

Theorem 1.4. There exists a constant $\epsilon_{0}>0$ such that if $\mathcal{F}_{0} \leq \epsilon_{0}$ , there exists a unique global-in-time solution of (11) satisfying

$\mathcal{F}(t)+(1-C \epsilon_{0})\int^{t}_{0}\mathcal{N}_{1}(s)ds\leq \mathcal{F}_{0} \quad {{for\; all\; t > 0 .}}$

Moreover, $\psi$ decays exponentially in time

$\left\|\psi(t)\right\|_{L^{2}}\leq \left\|\psi_{0}\right\|_{L^{2}} e^{-t}, \quad \left\|\Lambda^{k}\psi(t)\right\|_{L^{2}}\leq \mathcal{F}^{\frac{k-1}{8}}_{0}\left\|\nabla \psi_{0}\right\|^{\frac{5-k}{4}}_{L^{2}} e^{-\frac{(5-k)(1-C\epsilon_{0})}{4}t}$

with $1\leq k<5.$

1.2.2. Case 2

As another way to redistribute the derivatives in (8), we also deal with

$\begin{eqnarray} \psi_{t}-\Delta \psi = [\psi, Z], \quad Z_{t} = [\Delta \psi, \psi]. \end{eqnarray}$

(12)

Let $\mathcal{E}(t)$ and $\mathcal{E}_{0}$ be defined as before (10).

Theorem 1.5. There exists $T_{\ast}>0$ , which is depending on $\mathcal{E}_{0}$ , such that there exists a unique solution of (12) satisfying

$\mathcal{E}(t)\leq \frac{\mathcal{E}_{0}}{1-Ct\mathcal{E}_{0}} \quad {{for \;all}}\; \ 0 < t\leq T_{\ast} < \frac{1}{C \mathcal{E}_{0}}.$

Moreover, we have the following blow-up criterion

$\mathcal{E}(t)+\int^{t}_{0}\left\|\nabla \psi\right\|^{2}_{H^{4}}ds < \infty \iff \int^{t}_{0}\left\|\nabla^{2}\psi\right\|^{2}_{L^{\infty}}ds.$

We now add a damping term to the equation of $Z$ in (12):

$\begin{eqnarray} \psi_{t}-\Delta \psi = [\psi, Z], \quad Z_{t} + Z = [\Delta \psi, \psi]. \end{eqnarray}$

(13)

In this case, we can use the same regularity used in Theorem 1.5 because the dissipative effect in $\psi$ helps to control $\Delta \psi$ in the equation of $Z$ . Let $\mathcal{N}_{2}(t) = \left\|\nabla\psi(t)\right\|^{2}_{H^{5}}+\left\|Z(t)\right\|^{2}_{H^{3}}$ .

Theorem 1.6. There exists a constant $\epsilon_{0}>0$ such that if $\mathcal{E}_{0} \leq \epsilon_{0}$ , there exists a unique global-in-time solution of (13) satisfying

$\mathcal{E}(t)+(1-C \epsilon_{0})\int^{t}_{0}\mathcal{N}_{2}(s)ds\leq \mathcal{E}_{0} \quad {{for \;all \; t > 0 .}}$

Remark 3. Compared to Theorem 1.3, we only need one term in the blow-up criterion in Theorem 1.5 which is due to the dissipative effect in the equation of $\psi$ . Compared to Theorem 1.4, the proof of Theorem 1.6 is simpler, but we are not able to derive decay rates of $\psi$ and $Z$ .

2. Preliminaries

All constants will be denoted by $C$ and we follow the convention that such constants can vary from expression to expression and even between two occurrences within the same expression. And repeated indices are summed over.

The fractional Laplacian $\Lambda^{\beta} = (\sqrt{-\Delta})^{\beta}$ has the Fourier transform representation

$\widehat{\Lambda^{\beta} f}(\xi) = |\xi|^{\beta}\widehat{f}(\xi).$

For $s>0$ , $H^{s}$ is a energy space equipped with

$\|f\|_{H^{s}} = \|f\|_{L^{2}}+\|f\|_{\dot{H}^{s}}, \quad \|f\|_{\dot{H}^{s}} = \left\|\Lambda^{s}f\right\|_{L^{2}}.$

In the energy spaces, we have the following interpolations: for $s_{0}<s<s_{1}$

$\begin{eqnarray} \left\|f\right\|_{\dot{H}^{s}}\leq \left\|f\right\|^{\theta}_{\dot{H}^{s_{0}}}\left\|f\right\|^{1-\theta}_{\dot{H}^{s_{1}}}, \quad s = \theta s_{0}+(1-\theta)s_{1}. \end{eqnarray}$

(14)

2.1. Inequalities

We begin with two inequalities in 3D:

$\begin{align} & \left\|f\right\|_{L^{\infty}}\leq C\|f\|_{H^{s}}, \quad s > \frac{3}{2}, \end{align}$

(15a)

$\begin{align} & \left\|f\right\|_{L^{p}}\leq C\|f\|_{\dot{H}^{s}}, \quad \frac{1}{p} = \frac{1}{2}-\frac{s}{3}. \end{align}$

(15b)

We also provide the following inequalities in 2D

$\left\|f\right\|_{L^{4}}\leq C\left\|f\right\|^{\frac{1}{2}}_{L^{2}}\left\|\nabla f\right\|^{\frac{1}{2}}_{L^{2}}, \quad \left\|f\right\|_{L^{\infty}} \leq C\left\|f\right\|^{\frac{1}{2}}_{L^{2}} \left\|\Delta f\right\|^{\frac{1}{2}}_{L^{2}}$

which will be used repeatedly in the proof of Theorem 1.3, Theorem 1.4, Theorem 1.5, and Theorem 1.6. We also recall

$\left\|\nabla^{2}f\right\|_{L^{2}} = \left\|\Delta f\right\|_{L^{2}}$

which holds in any dimension.

We finally provide the Kato-Ponce commutator cstimate [22]

$\begin{eqnarray} \begin{split} \left\|\left[\Lambda^{k}, f\right]g\right\|_{L^{2}}& = \left\|\Lambda^{k}(fg)-f\Lambda^{k}g\right\|_{L^{2}}\\ &\leq C \left\|\nabla f\right\|_{L^{\infty}}\left\|\Lambda^{k-1}g\right\|_{L^{2}}+C \left\|g\right\|_{L^{\infty}}\left\|\Lambda^{k}f\right\|_{L^{2}} \end{split} \end{eqnarray}$

(16)

and the fractional Leibniz rule [11]: for $1\leq p<\infty$ and $p_{i}, q_{i} \ne 1$ ,

$\begin{eqnarray} \begin{split} &\left\|\Lambda^{s}(fg)\right\|_{L^{p}} \le C\left\|\Lambda^{s}f\right\|_{L^{p_{1}}} \|g\|_{L^{q_{1}}} +C\|f\|_{L^{p_{2}}} \left\|\Lambda^{s}g\right\|_{L^{q_{2}}}, \\ & \frac{1}{p} = \frac{1}{p_{1}}+\frac{1}{q_{1}} = \frac{1}{p_{2}}+\frac{1}{q_{2}}. \end{split} \end{eqnarray}$

(17)

2.2. Commutator

We recall the commutator $[f, g] = \nabla f\cdot \nabla^{\perp}g = f_{x}g_{y}-f_{y}g_{x}$ . Then, the commutator has the following properties:

$\begin{align} & \Delta [f, g] = [\Delta f, g] +[f, \Delta g] + 2[f_{x}, g_{x}] +2[f_{y}, g_{y}], \end{align}$

(18a)

$\begin{align} & \int f[f, g] = 0, \end{align}$

(18b)

$\begin{align} & \int f[g, h] = \int g [h, f]. \end{align}$

(18c)

3. Proof of Theorem 1.1 and Theorem 1.2

3.1. Proof of Theorem 1.1

We recall (3):

$\begin{eqnarray} B_{t}+{\rm{curl}} (\left({\rm{curl}} \;B)\times B\right)+\Lambda B = 0. \end{eqnarray}$

(19)

3.1.1. Approximation

We first approximate (19) by putting $\epsilon \Delta B$ to the right-hand side of (19):

$\begin{eqnarray} B_{t}+{\rm{curl}} (\left({\rm{curl}} \;B\right)\times B)+\Lambda B = \epsilon \Delta B. \end{eqnarray}$

(20)

We then mollify (20) as follows

$\begin{eqnarray} \begin{split} &\partial_{t} B^{(\epsilon)}+{\rm{curl}} \left(\mathcal{J}_{\epsilon}\left({\rm{curl}} \mathcal{J}_{\epsilon} B^{(\epsilon)}\right)\times \mathcal{J}_{\epsilon} B^{(\epsilon)}\right)+\Lambda \mathcal{J}^{2}_{\epsilon} B^{(\epsilon)} = \epsilon \mathcal{J}^{2}_{\epsilon}\Delta B^{(\epsilon)}, \\ & B^{(\epsilon)}_{0} = \mathcal{J}_{\epsilon} B_{0}, \end{split} \end{eqnarray}$

(21)

where $\mathcal{J}_{\epsilon}$ is the standard mollifier described in [25,Chapter 3.2]. Then, as proved in [4,Proposition 3.1], there exists a unique global-in-time solution $\left\{B^{(\epsilon)}\right\}$ of (21). Since the bounds in Section 3.1.2 are independent of $\epsilon>0$ , we can pass to the limit in a subsequence and show the existence of smooth solutions globally in time when $B_{0}\in H^{k}$ , $k>\frac{5}{2}$ , is sufficiently small as in [37,Section 3.2].

3.1.2. A priori estimates

We begin with the $L^{2}$ bound:

$\begin{eqnarray} \frac{1}{2}\frac{d}{dt}\left\|B\right\|^{2}_{L^{2}}+\left\|\Lambda^{\frac{1}{2}}B\right\|^{2}_{L^{2}} = 0. \end{eqnarray}$

(22)

We now take $\Lambda^{k}$ to (19) and take the inner product of the resulting equation with $\Lambda^{k}B$ . Then,

$\begin{split} &\frac{1}{2}\frac{d}{dt}\left\|\Lambda^{k}B\right\|^{2}_{L^{2}}+\left\|\Lambda^{\frac{1}{2}+k}B\right\|^{2}_{L^{2}} = -\int\Lambda^{k}{\rm{curl}} (\left({\rm{curl}} \;B)\times B\right) \cdot \Lambda^{k}B\\ & = \int \left(\left[\Lambda^{\frac{1}{2}+k}, B\right]\times {\rm{curl}} \;B\right)\cdot \Lambda^{k-\frac{1}{2}}{\rm{curl}} \;B\leq \left\|\left[\Lambda^{\frac{1}{2}+k}, B\right]\times {\rm{curl}} \;B\right\|_{L^{2}}\left\|\Lambda^{\frac{1}{2}+k}B\right\|_{L^{2}}. \end{split}$

By (16) and (15a) with $k>\frac{5}{2}$ ,

$\begin{eqnarray} \begin{split} \left\|\left[\Lambda^{\frac{1}{2}+k}, B\right]\times {\rm{curl}} \;B\right\|_{L^{2}}&\leq C \left\|\nabla B\right\|_{L^{\infty}}\left\|\Lambda^{k-\frac{1}{2}}{\rm{curl}} \;B\right\|_{L^{2}}\\ &\leq C \left\|B\right\|_{H^{k}}\left\|\Lambda^{\frac{1}{2}+k}B\right\|^{2}_{L^{2}}. \end{split} \end{eqnarray}$

(23)

So, we obtain

$\begin{eqnarray} \frac{d}{dt}\left\|\Lambda^{k}B\right\|^{2}_{L^{2}}+\left\|\Lambda^{\frac{1}{2}+k}B\right\|^{2}_{L^{2}}\leq C \left\|B\right\|_{H^{k}}\left\|\Lambda^{\frac{1}{2}+k}B\right\|^{2}_{L^{2}}. \end{eqnarray}$

(24)

By (22) and (24),

$\frac{d}{dt}\left\|B\right\|^{2}_{H^{k}}+\left\|\Lambda^{\frac{1}{2}}B\right\|^{2}_{H^{k}}\leq C \left\|B\right\|_{H^{k}}\left\|\Lambda^{\frac{1}{2}+k}B\right\|^{2}_{L^{2}}.$

If $\left\|B_{0}\right\|_{H^{k}} = \epsilon_{0}$ is sufficiently small, we can derive a uniform bound

$\begin{eqnarray} \left\|B(t)\right\|^{2}_{H^{k}}+(1-C\epsilon_{0})\int^{t}_{0}\left\|\Lambda^{\frac{1}{2}}B(s)\right\|^{2}_{H^{k}}ds\leq \left\|B_{0}\right\|^{2}_{H^{k}} \quad {{\rm{for}} \;{\rm{all}} \;t > 0 .} \end{eqnarray}$

(25)

3.1.3. Uniqueness

Let $B_{1}$ and $B_{2}$ be two solutions of (19). Then, $B = B_{1}-B_{2}$ satisfies

$\begin{eqnarray} B_{t}+\Lambda B+{\rm{curl}} \left(\left({\rm{curl}} \;B_{1}\right)\times B\right) - {\rm{curl}} \left(\left({\rm{curl}} \;B\right)\times B_{2}\right) = 0 \end{eqnarray}$

(26)

with $B_{0} = 0$ . We take the inner product of (26) with $B$ . By (17) with $k>\frac{5}{2}$ ,

$\begin{equation*} \begin{split} \frac{1}{2}\frac{d}{dt}\|B\|^{2}_{L^{2}}+\left\|\Lambda^{\frac{1}{2}} B\right\|^{2}_{L^{2}}& = -\int \left({\rm{curl}}\left(\left({\rm{curl}} \;B_{1}\right)\times B\right)\right) \cdot B\\ & = -\int \Lambda^{\frac{1}{2}}\left(\left(\left({\rm{curl}} \;B_{1}\right)\times B\right)\right) \cdot \Lambda^{-\frac{1}{2}}{\rm{curl}} \;B\\ &\leq C\left\|\nabla B_{1}\right\|_{L^{\infty}}\left\|\Lambda^{\frac{1}{2}} B\right\|^{2}_{L^{2}} +C\left\|\nabla \Lambda^{\frac{1}{2}}B_{1}\right\|_{L^{6}}\|B\|_{L^{3}}\left\|\Lambda^{\frac{1}{2}} B\right\|_{L^{2}}\\ &\leq C\left\|\nabla B_{1}\right\|_{L^{\infty}}\left\|\Lambda^{\frac{1}{2}} B\right\|^{2}_{L^{2}} +C\left\|\Lambda^{\frac{5}{2}}B_{1}\right\|_{L^{2}}\left\|\Lambda^{\frac{1}{2}} B\right\|^{2}_{L^{2}} \\ &\leq C\left\|B_{1}\right\|_{H^{k}} \left\|\Lambda^{\frac{1}{2}} B\right\|^{2}_{L^{2}}, \end{split} \end{equation*}$

where we use (15b) to control $L^{6}$ and $L^{3}$ terms. If $C\epsilon_{0}<1$ , (25) implies $B = 0$ in $L^{2}$ which gives the uniqueness of a solution.

3.1.4. Decay rates

By (14), it is enough to derive the decay rate with $k = l$ to show (4). Since

$\left\|\Lambda^{k}B\right\|^{\frac{2k+1}{k}}_{L^{2}}\leq \left\|B\right\|^{\frac{1}{k}}_{L^{2}}\left\|\Lambda^{\frac{1}{2}+k}B\right\|^{2}_{L^{2}} \leq \left\|B_{0}\right\|^{\frac{1}{k}}_{L^{2}}\left\|\Lambda^{\frac{1}{2}+k}B\right\|^{2}_{L^{2}}$

by (14) and (22), we create the following ODE from (24)

$\frac{d}{dt}\left\|\Lambda^{k}B\right\|^{2}_{L^{2}}+\frac{1-C\epsilon_{0}}{\left\|B_{0}\right\|^{\frac{1}{k}}_{L^{2}}}\left\|\Lambda^{k}B\right\|^{\frac{2k+1}{k}}_{L^{2}}\leq 0.$

By solving this ODE, we find the following decay rate

$\begin{eqnarray} \left\|\Lambda^{k}B(t)\right\|_{L^{2}}\leq \frac{\left((2k)^k \left\|B_{0}\right\|_{L^{2}}\left\|\Lambda^{k}B_{0}\right\|_{L^{2}}\right)}{\left(2k \left\|B_{0}\right\|^{\frac{1}{k}}_{L^{2}} +(1-C\epsilon_{0})\left\|\Lambda^{k}B_{0}\right\|^{\frac{1}{k}}_{L^{2}}t\right)^k}. \end{eqnarray}$

(27)

3.2. Proof of Theorem 1.2

We recall (5):

$B_{t}+{\rm{curl}} \left(({\rm{curl}} \;B)\times B\right)+\ln(2+\Lambda)\Lambda B = 0,$

The the uniqueness part of Theorem 1.2 is the same as that of Theorem 1.1 and we only derive a priori bounds. Let

$\left\|\sqrt{\ln(2+\Lambda)}\Lambda^{s}f\right\|^{2}_{L^{2}} = \int \left(\ln(2+|\xi|)\right)|\xi|^{2s}\left|\widehat{f}(\xi)\right|^{2}d\xi.$

We begin with the $L^{2}$ bound:

$\begin{eqnarray} \frac{1}{2}\frac{d}{dt}\left\|B\right\|^{2}_{L^{2}}+\left\|\sqrt{\ln(2+\Lambda)}\Lambda^{\frac{1}{2}}B\right\|^{2}_{L^{2}} = 0. \end{eqnarray}$

(28)

Following the computations in the proof of Theorem 1.1, we also have

$\begin{eqnarray} \frac{d}{dt}\left\|\Lambda^{k}B\right\|^{2}_{L^{2}}+\left\|\sqrt{\ln(2+\Lambda)}\Lambda^{\frac{1}{2}+k}B\right\|^{2}_{L^{2}}\leq C \left\|B\right\|_{H^{k}}\left\|\Lambda^{\frac{1}{2}+k}B\right\|^{2}_{L^{2}}. \end{eqnarray}$

(29)

For each $N\in \mathbb{N}$ , we have

$\begin{equation*} \begin{split} &\left\|\Lambda^{\frac{1}{2}+k}B\right\|^{2}_{L^{2}} = \int_{|\xi|\leq 2^{N}} |\xi|^{2k+1}\left|\widehat{B}(\xi)\right|^{2}d\xi+\int_{|\xi|\ge 2^{N}} |\xi|^{2k+1}\left|\widehat{B}(\xi)\right|^{2}d\xi\\ &\leq 2^{N}\int_{|\xi|\leq 2^{N}} |\xi|^{2k}\left|\widehat{B}(\xi)\right|^{2}d\xi+\frac{1}{\ln(2+2^{N})}\int_{|\xi|\ge 2^{N}} \ln(2+|\xi|)|\xi|^{2k+1}\left|\widehat{B}(\xi)\right|^{2}d\xi\\ &\leq 2^{N}\left\|\Lambda^{k}B\right\|^{2}_{L^{2}} +\frac{1}{\ln(2+2^{N})}\left\|\sqrt{\ln(2+\Lambda)}\Lambda^{\frac{1}{2}+k}B\right\|^{2}_{L^{2}}. \end{split} \end{equation*}$

So, (29) is replaced by

$\begin{split} &\frac{d}{dt}\left\|\Lambda^{k}B\right\|^{2}_{L^{2}}+\left\|\sqrt{\ln(2+\Lambda)}\Lambda^{\frac{1}{2}+k}B\right\|^{2}_{L^{2}} \\ &\leq C 2^{N}\left\|\Lambda^{k}B\right\|^{2}_{L^{2}}\|B\|_{H^{k}} + \frac{C\|B\|_{H^{k}}}{\ln(2+2^{N})}\left\|\sqrt{\ln(2+\Lambda)}\Lambda^{\frac{1}{2}+k}B\right\|^{2}_{L^{2}}. \end{split}$

We now choose $N>0$ such that

$\frac{1}{2}\ln(2+2^{N}) < C\left\|B\right\|_{H^{k}} < \ln(2+2^{N})$

and so $N \sim \|B\|_{H^{k}}$ . Then, (29) is reduced to

$\begin{eqnarray} \frac{d}{dt}\left\|\Lambda^{k}B\right\|^{2}_{L^{2}} \leq C \exp\left(\|B\|_{H^{k}}\right)\|B\|_{H^{k}} \left\|\Lambda^{k}B\right\|_{L^{2}}. \end{eqnarray}$

(30)

By (28) and (30), we obtain

$\frac{d}{dt}\left\|B\right\|^{2}_{H^{k}}\leq C \exp\left(\|B\|_{H^{k}}\right)\|B\|^{2}_{H^{k}}$

and so we have

$\frac{d}{dt}\left\|B\right\|_{H^{k}}\leq C \exp\left(\|B\|_{H^{k}}\right)\|B\|_{H^{k}} \leq C \exp\left(\|B\|_{H^{k}}\right).$

By solving this ODE, we can derive (6).

4. Proof of Theorem 1.3 and Theorem 1.4

4.1. Proof of Theorem 1.3

We recall (9):

$\begin{align} & \psi_{t} = [\psi, Z], \end{align}$

(31a)

$\begin{align} & Z_{t}-\Delta Z = [\Delta \psi, \psi]. \end{align}$

(31b)

4.1.1. Approximation

We first approximate (31a) by putting $\epsilon \Delta \psi$ to the right-hand side and mollify the resulting equations as (21). Then, we have

$\begin{eqnarray} \begin{split} &\partial_{t}\psi^{(\epsilon)} = \mathcal{J}_{\epsilon}[\mathcal{J}_{\epsilon}\psi^{(\epsilon)}, \mathcal{J}_{\epsilon}Z^{(\epsilon)}]+\epsilon \mathcal{J}^{2}_{\epsilon}\Delta \psi^{(\epsilon)}, \\ & \partial_{t}Z^{(\epsilon)}-\Delta \mathcal{J}^{2}_{\epsilon}Z^{(\epsilon)} = \mathcal{J}_{\epsilon}[\Delta \mathcal{J}_{\epsilon}\psi^{(\epsilon)}, \mathcal{J}_{\epsilon}\psi^{(\epsilon)}] \end{split} \end{eqnarray}$

(32)

with $\psi^{(\epsilon)}_{0} = \mathcal{J}_{\epsilon}\psi_{0}$ and $Z^{(\epsilon)}_{0} = \mathcal{J}_{\epsilon}Z_{0}$ . Since (32) is defined in $\mathbb{R}^{2}$ , the proof of the existence of a unique global-in-time solution of (32) is relatively easier than the one to (21). Moreover, the bounds in Section 4.1.2 are independent of $\epsilon>0$ and so we can pass to the limit in a subsequence and show the existence of smooth solutions locally in time when $\psi_{0}\in H^{4}$ and $Z_{0}\in H^{3}$ .

4.1.2. A priori estimates

We first note that

$\begin{eqnarray} \frac{1}{2}\frac{d}{dt}\left\|\psi\right\|^{2}_{L^{2}} = \int \psi[\psi, Z] = 0. \end{eqnarray}$

(33)

We next multiply (31a) by $-\Delta \psi$ , (31b) by $Z$ , and integrate over $\mathbb{R}^{2}$ . By (18c),

$\begin{eqnarray} \frac{1}{2}\frac{d}{dt} \left(\left\|\nabla \psi\right\|^{2}_{L^{2}}+\left\|Z\right\|^{2}_{L^{2}}\right) + \left\|\nabla Z\right\|^{2}_{L^{2}} = \int \left(-\Delta \psi[\psi, Z]+Z[\Delta \psi, \psi]\right) = 0. \end{eqnarray}$

(34)

We also multiply (31a) by $\Delta^{4} \psi$ , (31b) by $-\Delta^{3} Z$ and integrate over $\mathbb{R}^{2}$ . Then,

$\begin{eqnarray} \begin{split} \frac{1}{2}\frac{d}{dt} \left(\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}+\left\|\nabla\Delta Z\right\|^{2}_{L^{2}} \right) +\left\|\Delta^{2} Z\right\|^{2}_{L^{2}} & = \int \Delta^{4} \psi [\psi, Z] -\int\Delta^{3} Z[\Delta \psi, \psi]\\ & = \mathcal{R}. \end{split} \end{eqnarray}$

(35)

We now compute the right-hand side of (35). By (18a), (18b), and (18c),

$\begin{equation} \begin{split} \mathcal{R}& = 2\int\Delta^{2}\psi[\Delta \psi, \Delta Z]+4\int \Delta^{2} \psi[\psi_{x}, \Delta Z_{x}]+4\int \Delta^{2} \psi[\psi_{y}, \Delta Z_{y}]\\ & +4\int \Delta^{2} \psi[\Delta\psi_{x}, Z_{x}]+4\int \Delta^{2} \psi[\Delta \psi_{y}, Z_{y}]+4\int \Delta^{2} \psi[\psi_{xx}, Z_{xx}]\\ &+8\int \Delta^{2} \psi[\psi_{xy}, Z_{xy}]+4\int \Delta^{2} \psi[\psi_{yy}, Z_{yy}]-2\int\Delta^{2}Z[\Delta \psi_{x}, \psi_{x}]\\ &-2\int\Delta^{2}Z[\Delta \psi_{y}, \psi_{y}]. \end{split} \end{equation}$

(36)

So, we find that the number of derivatives acting on $(\psi, \psi, Z)$ are $(4, 4, 2)$ , $(3, 4, 3)$ , and $(4, 2, 4)$ up to multiplicative constants. Hence,

$\begin{equation*} \begin{split} & \frac{d}{dt} \left(\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}+\left\|\nabla\Delta Z\right\|^{2}_{L^{2}} \right) +\left\|\Delta^{2} Z\right\|^{2}_{L^{2}}\\ & \leq C\int \left|\nabla^{4}\psi\right| \left|\nabla^{4}\psi\right| \left|\nabla^{2} Z\right|+ C\int \left|\nabla^{3}\psi\right| \left|\nabla^{4}\psi\right| \left|\nabla^{3} Z\right|+ C\int \left|\nabla^{4}\psi\right| \left|\nabla^{2}\psi\right| \left|\nabla^{4} Z\right|\\ & \leq C\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}} \left\|\nabla^{2} Z\right\|_{L^{\infty}}+ C\left\|\nabla^{3}\psi\right\|_{L^{4}} \left\|\Delta^{2}\psi\right\|_{L^{2}} \left\|\nabla^{3} Z\right\|_{L^{4}}\\ &+ C\left\|\Delta^{2}\psi\right\|_{L^{2}} \left\|\nabla^{2}\psi\right\|_{L^{\infty}} \left\|\Delta^{2} Z\right\|_{L^{2}}\\ & \leq C\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}} \left\|\nabla^{2} Z\right\|_{L^{\infty}}+ C\left\|\Delta^{2}\psi\right\|^{\frac{3}{2}}_{L^{2}} \left\|\nabla\Delta\psi\right\|^{\frac{1}{2}}_{L^{2}} \left\|\Delta^{2} Z\right\|_{L^{2}}\\ &+C\left\|\Delta^{2}\psi\right\|_{L^{2}} \left\|\nabla^{2}\psi\right\|_{L^{\infty}} \left\|\Delta^{2} Z\right\|_{L^{2}}\\ &\leq C \mathcal{E}^{2}+\frac{1}{4}\left\|\Delta^{2} Z\right\|^{2}_{L^{2}}+\delta \left\|\nabla^{2} Z\right\|^{2}_{L^{\infty}} \leq C \mathcal{E}^{2}+\frac{1}{2}\left\|\Delta^{2} Z\right\|^{2}_{L^{2}}+\frac{1}{4}\left\|\nabla Z\right\|^{2}_{L^{2}}, \end{split} \end{equation*}$

where we use

$\begin{split} \left\|\nabla^{2} Z\right\|^{2}_{L^{\infty}}&\leq C \left\|\Delta Z\right\|_{L^{2}}\left\|\Delta^{2} Z\right\|_{L^{2}}\leq C \left\|\nabla Z\right\|^{\frac{2}{3}}_{L^{2}}\left\|\Delta^{2} Z\right\|^{\frac{4}{3}}_{L^{2}}\\&\leq C \left\|\nabla Z\right\|^{2}_{L^{2}}+C\left\|\Delta^{2} Z\right\|^{2}_{L^{2}} \end{split}$

with $\delta$ satisfying $4C\delta = 1$ . So, we have

$\begin{eqnarray} \frac{d}{dt} \left(\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}+\left\|\nabla\Delta Z\right\|^{2}_{L^{2}} \right) +\left\|\Delta^{2} Z\right\|^{2}_{L^{2}} \leq C \mathcal{E}^{2}+\frac{1}{2}\left\|\nabla Z\right\|^{2}_{L^{2}}. \end{eqnarray}$

(37)

By (33), (34), and (37), we derive $\mathcal{E}' \leq C \mathcal{E}^{2}$ from which we deduce

$\begin{eqnarray} \mathcal{E}(t)\leq \frac{\mathcal{E}_{0}}{1-Ct\mathcal{E}_{0}} \quad {{\rm{for}}\; {\rm{all}}} \;\ 0 < t\leq T_{\ast} < \frac{1}{C \mathcal{E}_{0}}. \end{eqnarray}$

(38)

4.1.3. Uniqueness

Let $(\psi_{1}, Z_{1})$ and $(\psi_{2}, Z_{2})$ be two solutions of (31) and let $\psi = \psi_{1}-\psi_{2}$ and $Z = Z_{1}-Z_{2}$ . Then, $(\psi, Z)$ satisfies the following equations:

$\psi_{t} = [\psi, Z_{1}]+[\psi_{2}, Z], \quad Z_{t}-\Delta Z = [\Delta\psi, \psi_{1}]+[\Delta\psi_{2}, \psi]$

with $\psi(0, x) = Z(0, x) = 0$ . For these equations, we have

$\begin{equation*} \begin{split} &\frac{1}{2}\frac{d}{dt}\left(\left\|\nabla \psi\right\|^{2}_{L^{2}}+ \|Z\|^{2}_{L^{2}}\right)+ \left\|\nabla Z\right\|^{2}_{L^{2}}\\ & = -\int \Delta \psi [\psi, Z_{1}]-\int \Delta \psi [\psi_{2}, Z]+\int Z [\Delta\psi, \psi_{1}]+\int Z[\Delta\psi_{2}, \psi]\\ & = \text{(I)+(II)+(III)+(IV)}. \end{split} \end{equation*}$

The first term is bounded using the definition of $[f, g]$ and ${\rm{div}}\; \nabla^{\perp}Z_{1} = 0$ :

$\text{(I)} = \int \left(\nabla^{\perp}Z_{1}\cdot \nabla \psi\right) \Delta \psi = -\int \left(\nabla^{\perp}\partial_{l}Z_{1}\cdot \nabla \psi\right) \partial_{l} \psi\leq C \left\|\nabla^{2}Z_{1}\right\|_{L^{\infty}}\left\|\nabla \psi\right\|^{2}_{L^{2}}.$

We next bound $\text{(II)+(III)}$ as

$\begin{split} \text{(II)+(III)}& = -\int Z[\Delta \psi, \psi]\leq C \left\|\nabla^{2}\psi\right\|_{L^{\infty}}\left\|\nabla \psi\right\|_{L^{2}}\left\|\nabla Z\right\|_{L^{2}} \\ &\leq C \left(\left\|\nabla^{2}\psi_{1}\right\|^{2}_{L^{\infty}}+\left\|\nabla^{2}\psi_{2}\right\|^{2}_{L^{\infty}}\right)\left\|\nabla \psi\right\|^{2}_{L^{2}}+\frac{1}{4}\left\|\nabla Z\right\|^{2}_{L^{2}}. \end{split}$

The last term is bounded as

$\text{(IV)}\leq C \left\|\nabla^{2}\psi_{2}\right\|_{L^{\infty}}\left\|\nabla \psi\right\|_{L^{2}}\left\|\nabla Z\right\|_{L^{2}} \leq C \left\|\nabla^{2}\psi_{2}\right\|^{2}_{L^{\infty}}\left\|\nabla \psi\right\|^{2}_{L^{2}}+\frac{1}{4}\left\|\nabla Z\right\|^{2}_{L^{2}}.$

So, we have

$\begin{eqnarray} \begin{split} &\frac{d}{dt}\left(\left\|\nabla \psi\right\|^{2}_{L^{2}}+ \|Z\|^{2}_{L^{2}}\right)\\ &\leq C\left( \left\|\nabla^{2}Z_{1}\right\|_{L^{\infty}}+\left\|\nabla^{2}\psi_{1}\right\|^{2}_{L^{\infty}}+\left\|\nabla^{2}\psi_{2}\right\|^{2}_{L^{\infty}}\right)\left(\left\|\nabla \psi\right\|^{2}_{L^{2}}+ \|Z\|^{2}_{L^{2}} \right). \end{split} \end{eqnarray}$

(39)

By (38), $\left\|\nabla^{2}\psi_{1}\right\|^{2}_{L^{\infty}}+\left\|\nabla^{2}\psi_{2}\right\|^{2}_{L^{\infty}}$ is integrable in time. Integrating (34) and (35) in time, we have

$\int^{t}_{0}\left(\left\|\nabla Z(s)\right\|^{2}_{L^{2}}+\left\|\Delta^{2} Z(s)\right\|^{2}_{L^{2}}\right)ds < \infty \quad {{\rm{for}}} \; 0 < t\leq \frac{T_{\ast}}{2}$

which gives the integrability of the first term in the parentheses on the right-hand side of (39). By repeating the same argument one more time, we have the uniqueness up to $T_{\ast}$ .

4.1.4. Blow-up criterion

Let

$\mathcal{B}(s) = \left\|\nabla^{2}Z(s)\right\|_{L^{\infty}} +\left\|\nabla^{2}\psi(s)\right\|^{2}_{L^{\infty}} .$

We first deal with

$\begin{split} &\frac{1}{2}\frac{d}{dt} \left(\left\|\Delta\psi\right\|^{2}_{L^{2}}+\left\|\nabla Z\right\|^{2}_{L^{2}} \right) +\left\| \Delta Z\right\|^{2}_{L^{2}} = \int \Delta^{2} \psi [\psi, Z] -\int\Delta Z[\Delta \psi, \psi]\\ & = 2\int \Delta \psi[\psi_{x}, Z_{x}]+2\int \Delta \psi[\psi_{y}, Z_{y}]\leq C \left\|\nabla^{2}Z\right\|_{L^{\infty}}\left\| \Delta \psi\right\|^{2}_{L^{2}} \end{split}$

and so we have

$\frac{d}{dt} \left(\left\|\Delta\psi\right\|^{2}_{L^{2}}+\left\|\nabla Z\right\|^{2}_{L^{2}} \right) +\left\| \Delta Z\right\|^{2}_{L^{2}} \leq C \left\|\nabla^{2}Z\right\|_{L^{\infty}}\left\| \Delta \psi\right\|^{2}_{L^{2}}.$

This implies

$\begin{eqnarray} \begin{split} &\left\|\Delta \psi(t)\right\|^{2}_{L^{}}+ \left\|\nabla Z(t)\right\|^{2}_{L^{2}}+\int^{t}_{0}\left\| \Delta Z(s)\right\|^{2}_{L^{2}} ds < \infty \\ &\iff \int^{t}_{0}\left\|\nabla^{2}Z(s)\right\|_{L^{\infty}}ds < \infty. \end{split} \end{eqnarray}$

(40)

We also deal with

$\begin{split} &\frac{1}{2}\frac{d}{dt} \left(\left\|\nabla \Delta\psi\right\|^{2}_{L^{2}}+\left\|\Delta Z\right\|^{2}_{L^{2}} \right) +\left\| \nabla \Delta Z\right\|^{2}_{L^{2}} = -\int \Delta^{3} \psi [\psi, Z] +\int\Delta^{2} Z[\Delta \psi, \psi]\\ & = -\int \Delta^{2}\psi [\Delta \psi, Z] -2\int \Delta^{2}\psi \left([\psi_{x}, Z_{x}]+[\psi_{y}, Z_{y}]\right) \\ &- 2\int \Delta\psi \left([\psi_{x}, \Delta Z_{x}]+[\psi_{y}, \Delta Z_{y}]\right) = \text{(I)+(II)+(III)}. \end{split}$

As in Section 4.1.3,

$\begin{eqnarray} \text{(I)} = \int \left(\nabla \nabla^{\perp}Z\cdot \nabla \Delta \psi\right)\cdot \nabla \Delta\psi \leq C\left\|\nabla^{2}Z\right\|_{L^{\infty}} \left\|\nabla\Delta \psi \right\|^{2}_{L^{2}}. \end{eqnarray}$

(41)

We next estimate $\text{(II)}+\text{(III)}$ :

$\begin{equation} \begin{split} &\text{(II)}+\text{(III)} = -4\int \Delta\psi \left([\Delta \psi_{y}, Z_{y}]+[\psi_{y}, \Delta Z_{y}]+[\psi_{xy}, Z_{xy}]+ [\psi_{yy}, Z_{yy}]\right)\\ &\leq C\int \left|\nabla^{2}Z\right| \left|\nabla^{3}\psi \right|^{2}+C\int \left|\nabla^{2}\psi \right| \left|\nabla^{3}\psi \right|\left|\nabla^{3}Z\right|\\ &\leq C\left\|\nabla^{2}Z\right\|_{L^{\infty}} \left\|\nabla\Delta \psi \right\|^{2}_{L^{2}}+C\left\|\nabla^{2}\psi\right\|^{2}_{L^{\infty}} \left\|\nabla \Delta \psi \right\|^{2}_{L^{2}}+\frac{1}{2}\left\|\nabla \Delta Z \right\|^{2}_{L^{2}}. \end{split} \end{equation}$

(42)

By (41) and (42), we have

$\frac{d}{dt}\left(\left\|\nabla \Delta \psi\right\|^{2}_{L^{2}}+\left\|\Delta Z\right\|^{2}_{L^{2}}\right) +\left\|\nabla \Delta Z\right\|^{2}_{L^{2}}\leq C\left(\left\|\nabla^{2} Z\right\|_{L^{\infty}} +\left\|\nabla^{2}\psi\right\|^{2}_{L^{\infty}} \right)\left\|\nabla\Delta \psi \right\|^{2}_{L^{2}}$

which implies

$\begin{eqnarray} \left\|\nabla \Delta \psi(t)\right\|^{2}_{L^{2}}+\left\|\Delta Z(t)\right\|^{2}_{L^{2}}+\int^{t}_{0} \left\|\nabla \Delta Z(s)\right\|^{2}_{L^{2}}ds < \infty \iff \int^{t}_{0}\mathcal{B}(s)ds < \infty. \end{eqnarray}$

(43)

We finally deal with

$\frac{1}{2}\frac{d}{dt} \left(\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}+\left\|\nabla\Delta Z\right\|^{2}_{L^{2}} \right) +\left\|\Delta^{2}Z\right\|^{2}_{L^{2}} = \int \Delta^{4} \psi [\psi, Z] -\int\Delta^{3} Z[\Delta \psi, \psi] = \mathcal{R}$

with the same $\mathcal{R}$ in (36). So, we have

$\begin{equation*} \begin{split} &\frac{1}{2} \frac{d}{dt} \left(\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}+\left\|\nabla\Delta Z\right\|^{2}_{L^{2}} \right) +\left\|\Delta^{2} Z\right\|^{2}_{L^{2}}\\ & \leq C\left\|\nabla^{2} Z\right\|_{L^{\infty}}\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}+ C\left\|\nabla^{2}\psi\right\|_{L^{\infty}}\left\|\Delta^{2} Z\right\|_{L^{2}} \left\|\Delta^{2}\psi\right\|_{L^{2}} \\ &+C\left\|\nabla\Delta Z\right\|_{L^{4}}\left\|\nabla\Delta \psi\right\|_{L^{4}}\left\|\Delta^{2}\psi\right\|_{L^{2}}\\ &\leq C\left(\left\|\nabla^{2}Z\right\|_{L^{\infty}} +\left\|\nabla^{2}\psi\right\|^{2}_{L^{\infty}} +\left\|\nabla \Delta Z\right\|^{\frac{3}{2}}_{L^{2}}\left\|\nabla \Delta \psi\right\|^{\frac{3}{2}}_{L^{2}}\right)\left\|\Delta^{2} \psi \right\|^{2}_{L^{2}}+\frac{1}{2}\left\|\Delta^{2} Z\right\|^{2}_{L^{2}} \end{split} \end{equation*}$

which gives

$\begin{eqnarray} \begin{split} &\frac{d}{dt} \left(\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}+\left\|\nabla\Delta Z\right\|^{2}_{L^{2}} \right) +\left\|\Delta^{2} Z\right\|^{2}_{L^{2}}\\ &\leq C\left(\mathcal{B}(s)+\left\|\nabla \Delta Z\right\|^{\frac{3}{2}}_{L^{2}}\left\|\nabla \Delta \psi\right\|^{\frac{3}{2}}_{L^{2}}\right)\left\|\Delta^{2} \psi \right\|^{2}_{L^{2}}. \end{split} \end{eqnarray}$

(44)

By (40) and (43), (44) implies

$\left\|\Delta^{2} \psi(t)\right\|^{2}_{L^{2}}+\left\|\nabla \Delta Z(t)\right\|^{2}_{L^{2}}+\int^{t}_{0} \left\|\Delta ^{2}Z(s)\right\|^{2}_{L^{2}}ds < \infty \iff \int^{t}_{0}\mathcal{B}(s)ds < \infty.$

4.2. Proof of Theorem 1.4

We recall (11):

$\psi_{t}+\psi = [\psi, Z], \quad Z_{t}-\Delta Z = [\Delta \psi, \psi]$

Since the uniqueness is already proved in Section 4.1.3 even without the damping term, we only focus on the a priori bounds and the decay rates.

4.2.1. A priori estimates

We first have

$\begin{equation} \begin{split} &\frac{1}{2}\frac{d}{dt}\left\|\psi\right\|^{2}_{L^{2}}+\left\|\psi\right\|^{2}_{L^{2}} = 0, \\ &\frac{1}{2}\frac{d}{dt} \left(\left\|\nabla \psi\right\|^{2}_{L^{2}}+\left\|Z\right\|^{2}_{L^{2}}\right) +\left\|\nabla \psi\right\|^{2}_{L^{2}}+ \left\|\nabla Z\right\|^{2}_{L^{2}} = 0. \end{split} \end{equation}$

(45)

We now consider the highest order part:

$\begin{split} &\frac{1}{2}\frac{d}{dt} \left(\left\|\nabla \Delta^{2}\psi\right\|^{2}_{L^{2}}+\left\|\Delta^{2} Z\right\|^{2}_{L^{2}} \right)+\left\|\nabla\Delta^{2}\psi\right\|^{2}_{L^{2}} +\left\|\nabla\Delta^{2} Z\right\|^{2}_{L^{2}} \\ & = -\int \Delta^{5} \psi [\psi, Z] +\int\Delta^{4} Z[\Delta \psi, \psi]. \end{split}$

We compute the right-hand side of this. By (18a), (18b), and (18c),

$\begin{equation} \begin{split} &-\int \Delta^{5} \psi [\psi, Z] +\int\Delta^{4} Z[\Delta \psi, \psi]\\ & = 2\int \Delta^{3}Z\left[\Delta \psi_{x}, \psi_{x}\right]+2\int \Delta^{3}Z\left[\Delta \psi_{y}, \psi_{y}\right]+2\int \Delta^{2}Z\left[\Delta^{2} \psi_{x}, \psi_{x}\right]\\ &+2\int \Delta^{2}Z\left[\Delta^{2} \psi_{y}, \psi_{y}\right] +2\int \Delta Z\left[\Delta^{2} \psi_{x}, \Delta\psi_{x}\right]+2\int \Delta Z\left[\Delta^{2} \psi_{y}, \Delta\psi_{y}\right] \\ &-\int \Delta^{3}\psi[\Delta\psi, \Delta Z]-2\int\Delta^{3}\psi[\Delta \psi_{x}, Z_{x}] -2\int\Delta^{3}\psi[\Delta \psi_{y}, Z_{y}] -2\int\Delta^{3}\psi[ \psi_{x}, \Delta Z_{x}] \\ &-2\int\Delta^{3}\psi[ \psi_{y}, \Delta Z_{y}] -2\int\Delta^{4}\psi[ \psi_{x}, Z_{x}] -2\int\Delta^{4}\psi[ \psi_{y}, Z_{y}] -\int \Delta^{3}\psi[\Delta^{2}\psi, Z]. \end{split} \end{equation}$

(46)

We now count the number of derivatives hitting on $(Z, \psi, \psi)$ using the integration by parts and (18b) and (18c) up to multiplicative constants. Except for the last integral, we have

$\begin{split} &(6, 2, 4) \mapsto (5, 2, 5), \ (5, 3, 4)\, \quad (4, 2, 6) \mapsto (5, 5, 2), \ (4, 3, 5)\\ &(2, 2, 8) \mapsto (3, 2, 7) \mapsto (4, 2, 6), \ (3, 3, 6) \mapsto (5, 5, 2), \ (4, 3, 5)\\ & (2, 4, 6) \mapsto (2, 5, 5), \ (3, 4, 5). \end{split}$

The last integral is

$\int \left(\nabla^{\perp}Z\cdot \nabla \Delta^{2}\psi\right)\Delta^{3}\psi = -\int \left(\nabla^{\perp}\partial_{l}Z\cdot \nabla \Delta^{2}\psi\right)\partial_{l}\Delta^{2}\psi$

and so this gives $(2, 5, 5)$ . So, the combinations of the numbers of derivatives taken on $(Z, \psi, \psi)$ are

$(2, 5, 5), \ (3, 4, 5), \ (4, 3, 5), \ (5, 2, 5), \ (5, 3, 4).$

The first and the fourth cases are bounded by

$\begin{split} &C\left\|\nabla^{2}Z\right\|_{L^{\infty}}\left\|\nabla \Delta^{2}\psi\right\|^{2}_{L^{2}}\leq C\left\|\nabla^{2}Z\right\|^{2}_{L^{\infty}}\left\|\nabla \Delta^{2}\psi\right\|^{2}_{L^{2}}+\frac{1}{6}\left\|\nabla \Delta^{2}\psi\right\|^{2}_{L^{2}}, \\ & C\left\|\nabla^{2}\psi\right\|_{L^{\infty}}\left\|\nabla \Delta^{2}Z\right\|^{2}_{L^{2}}\leq C\left\|\nabla^{2}\psi\right\|_{L^{\infty}}\left\|\nabla \Delta^{2}Z\right\|^{2}_{L^{2}}+\frac{1}{4}\left\|\nabla \Delta^{2}Z\right\|^{2}_{L^{2}}. \end{split}$

The second case is bounded by

$\begin{split} &C\left\|\nabla^{3}Z\right\|_{L^{4}}\left\|\nabla^{4}\psi\right\|_{L^{4}}\left\|\nabla\Delta^{2}\psi\right\|_{L^{2}}\leq C\left\|\Delta Z\right\|^{\frac{1}{2}}_{L^{2}}\left\|\nabla\Delta^{2}Z\right\|^{\frac{1}{2}}_{L^{2}} \left\|\Delta^{2}\psi\right\|^{\frac{1}{2}}_{L^{2}}\left\|\nabla\Delta^{2}\psi\right\|^{\frac{3}{2}}_{L^{2}}\\ & \leq C\left\|\Delta Z\right\|^{2}_{L^{2}}\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}\left\|\nabla\Delta^{2}Z\right\|^{2}_{L^{2}} +\frac{1}{6}\left\|\nabla\Delta^{2}\psi\right\|^{2}_{L^{2}}. \end{split}$

The third case is bounded by

$\begin{split} &C\left\|\nabla^{4}Z\right\|_{L^{4}}\left\|\nabla^{3}\psi\right\|_{L^{4}}\left\|\nabla\Delta^{2}\psi\right\|_{L^{2}}\leq C\left\|\Delta^{2}Z\right\|^{\frac{1}{2}}_{L^{2}}\left\|\nabla\Delta^{2}Z\right\|^{\frac{1}{2}}_{L^{2}} \left\|\Delta \psi\right\|^{\frac{1}{2}}_{L^{2}}\left\|\nabla\Delta^{2}\psi\right\|^{\frac{3}{2}}_{L^{2}}\\ & \leq C\left\|\Delta \psi\right\|^{2}_{L^{2}}\left\|\Delta^{2}Z\right\|^{2}_{L^{2}}\left\|\nabla\Delta^{2}Z\right\|^{2}_{L^{2}} +\frac{1}{6}\left\|\nabla\Delta^{2}\psi\right\|^{2}_{L^{2}}. \end{split}$

The last one is bounded by

$\begin{split} &C\left\|\nabla^{3}\psi\right\|_{L^{4}}\left\|\nabla^{4}\psi\right\|_{L^{4}}\left\|\nabla\Delta^{2}Z\right\|_{L^{2}}\leq C\left\|\nabla\Delta \psi\right\|^{\frac{1}{2}}_{L^{2}}\left\|\Delta^{2}\psi\right\|_{L^{2}}\left\|\nabla\Delta^{2}\psi\right\|^{\frac{1}{2}}_{L^{2}}\left\|\nabla\Delta^{2}Z\right\|_{L^{2}}\\ & \leq C\left\|\nabla \Delta \psi\right\|_{L^{2}}\left\|\nabla \Delta^{2}\psi\right\|_{L^{2}}\left\|\nabla\Delta^{2}Z\right\|_{L^{2}} \leq C\left\|\nabla \Delta \psi\right\|^{2}_{L^{2}}\left\|\nabla \Delta^{2}\psi\right\|^{2}_{L^{2}} +\frac{1}{4}\left\|\nabla\Delta^{2}Z\right\|^{2}_{L^{2}}. \end{split}$

So, we obtain

$\begin{equation} \begin{split} &\frac{d}{dt} \left(\left\|\nabla \Delta^{2}\psi\right\|^{2}_{L^{2}}+\left\|\Delta^{2} Z\right\|^{2}_{L^{2}} \right)+\left\|\nabla\Delta^{2}\psi\right\|^{2}_{L^{2}} +\left\|\nabla\Delta^{2} Z\right\|^{2}_{L^{2}}\\ & \leq C\left\|\nabla^{2}Z\right\|^{2}_{L^{\infty}}\left\|\nabla \Delta^{2}\psi\right\|^{2}_{L^{2}}+ C\left\|\nabla^{2}\psi\right\|^{2}_{L^{\infty}}\left\|\nabla \Delta^{2}Z\right\|^{2}_{L^{2}}+C\left\|\nabla \Delta \psi\right\|^{2}_{L^{2}}\left\|\nabla \Delta^{2}\psi\right\|^{2}_{L^{2}}\\ &+C\left\|\Delta Z\right\|^{2}_{L^{2}}\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}\left\|\nabla\Delta^{2}Z\right\|^{2}_{L^{2}}+C\left\|\Delta \psi\right\|^{2}_{L^{2}}\left\|\Delta^{2}Z\right\|^{2}_{L^{2}}\left\|\nabla\Delta^{2}Z\right\|^{2}_{L^{2}} \end{split} \end{equation}$

(47)

By (45) and (47),

$\mathcal{F}'(t)+\mathcal{N}_{1}(t)\leq C \left(\mathcal{F}(t)+\mathcal{F}^{2}(t)\right)\mathcal{N}_{1}(t).$

So, if $\mathcal{F}_{0} = \epsilon_{0}$ is sufficiently small, we obtain

$\label{A priori damping dd} \mathcal{F}(t)+(1-C \epsilon_{0})\int^{t}_{0}\mathcal{N}_{1}(s)ds\leq \mathcal{F}_{0} \quad {{\rm{for}} \;{\rm{all}} \;t > 0 .}$

4.2.2. Decay rates

From (45), $\left\|\psi(t)\right\|_{L^{2}}\leq \left\|\psi_{0}\right\|_{L^{2}} e^{-t}$ . Since

$\begin{split} \frac{1}{2}\frac{d}{dt}\left\|\nabla \psi\right\|^{2}_{L^{2}}+\left\|\nabla \psi\right\|^{2}_{L^{2}}& = -\int \Delta \psi[\psi, Z] = \int \left(\nabla^{\perp}Z\cdot \nabla \psi\right)\Delta \psi\\ & = -\int \left(\partial_{l}\nabla^{\perp}Z\cdot \nabla \psi\right)\partial_{l} \psi\\ &\leq \left\|\nabla^{2}Z\right\|_{L^{\infty}}\left\|\nabla \psi\right\|^{2}_{L^{2}} \leq C\epsilon_{0}\left\|\nabla \psi\right\|^{2}_{L^{2}}, \end{split}$

we have

$\left\|\nabla \psi(t)\right\|_{L^{2}}\leq \left\|\nabla \psi_{0}\right\|_{L^{2}} e^{-(1-C\epsilon_{0})t}.$

By using (14), we also obtain

$\left\|\Lambda^{k}\psi(t)\right\|_{L^{2}}\leq \mathcal{F}^{\frac{k-1}{8}}_{0}\left\|\nabla \psi_{0}\right\|^{\frac{5-k}{4}}_{L^{2}} e^{-\frac{(5-k)(1-C\epsilon_{0})}{4}t}, \quad 1\leq k < 5.$

5. Proof of Theorem 1.5 and Theorem 1.6

5.1. Proof of Theorem 1.5

We recall (12):

$\psi_{t}-\Delta \psi = [\psi, Z], \quad Z_{t} = [\Delta \psi, \psi].$

By applying the same approximation and mollification methods in Section 4.1.1, we can show the existence of smooth solutions locally in time when $\psi_{0}\in H^{4}$ and $Z_{0}\in H^{3}$ .

5.1.1. A priori estimates

We first have

$\begin{equation} \begin{split} &\frac{1}{2}\frac{d}{dt}\left\|\psi\right\|^{2}_{L^{2}}+\left\|\nabla \psi\right\|^{2}_{L^{2}} = 0, \\ & \frac{1}{2}\frac{d}{dt} \left(\left\|\nabla \psi\right\|^{2}_{L^{2}}+\left\|Z\right\|^{2}_{L^{2}}\right) + \left\|\Delta \psi\right\|^{2}_{L^{2}} = 0. \end{split} \end{equation}$

(48)

We next deal with

$\frac{1}{2}\frac{d}{dt} \left(\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}+\left\|\nabla\Delta Z\right\|^{2}_{L^{2}} \right) +\left\|\nabla \Delta^{2} \psi\right\|^{2}_{L^{2}} = \int \Delta^{4} \psi [\psi, Z] -\int\Delta^{3} Z[\Delta \psi, \psi] = \mathcal{R}$

with the same $\mathcal{R}$ in (36). In this case, we choose the the number of derivatives acting on $(\psi, \psi, Z)$ different from Section 4.1.2, which are given by $(3, 5, 2)$ , $(2, 5, 3)$ , and $(4, 4, 2)$ after several integration by parts. Hence, we have

and so we have the following bound

$\begin{eqnarray} \frac{d}{dt} \left(\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}+\left\|\nabla\Delta Z\right\|^{2}_{L^{2}} \right) +\left\|\nabla \Delta^{2} \psi\right\|^{2}_{L^{2}} \leq C \mathcal{E}^{2}. \end{eqnarray}$

(49)

By (48) and (49), we derive $\mathcal{E}' \leq C \mathcal{E}^{2}$ from which we deduce

$\begin{eqnarray} \mathcal{E}(t)\leq \frac{\mathcal{E}_{0}}{1-Ct\mathcal{E}_{0}} \quad {{\rm{for}}\; {\rm{all}}} \;\ 0 < t\leq T_{\ast} < \frac{1}{C \mathcal{E}_{0}}. \end{eqnarray}$

(50)

5.1.2. Uniqueness

Let $(\psi_{1}, Z_{1})$ and $(\psi_{2}, Z_{2})$ be two solutions and let $\psi = \psi_{1}-\psi_{2}$ and $Z = Z_{1}-Z_{2}$ . As in Section 4.1.3

$\begin{equation*} \begin{split} &\frac{1}{2}\frac{d}{dt}\left(\left\|\nabla \psi\right\|^{2}_{L^{2}}+ \|Z\|^{2}_{L^{2}}\right)+ \left\|\Delta \psi\right\|^{2}_{L^{2}}\\ & = -\int \Delta \psi [\psi, Z_{1}]-\int \Delta \psi [\psi_{2}, Z]+\int Z [\Delta\psi, \psi_{1}]+\int Z[\Delta\psi_{2}, \psi]\\ & = \text{(I)+(II)+(III)+(IV)}. \end{split} \end{equation*}$

The first term three terms are bounded as

$\begin{split} \text{(I)}&\leq \left\|\nabla Z_{1}\right\|_{L^{\infty}}\left\|\nabla \psi\right\|_{L^{2}}\left\|\Delta \psi\right\|_{L^{2}} \leq C\left\|\nabla Z_{1}\right\|^{2}_{L^{\infty}}\left\|\nabla \psi\right\|^{2}_{L^{2}}+\frac{1}{3}\left\|\Delta \psi\right\|^{2}_{L^{2}}, \\ \text{(II)+(III)}& = -\int Z[\Delta \psi, \psi]\leq C \left\|\nabla Z\right\|_{L^{\infty}}\left\|\nabla \psi\right\|_{L^{2}}\left\|\Delta \psi\right\|_{L^{2}} \\ &\leq C \left(\left\|\nabla Z_{1}\right\|^{2}_{L^{\infty}}+\left\|\nabla Z_{2}\right\|^{2}_{L^{\infty}}\right)\left\|\nabla \psi\right\|^{2}_{L^{2}}+\frac{1}{3}\left\|\Delta \psi\right\|^{2}_{L^{2}} \end{split}$

The last term is bounded as

$\begin{split} \text{(IV)}&\leq C \left\|\nabla^{3}\psi_{2}\right\|_{L^{4}}\left\|\nabla \psi\right\|_{L^{4}}\left\|Z\right\|_{L^{2}} \leq C \left\|\nabla^{3}\psi_{2}\right\|_{L^{4}}\left\|\nabla \psi\right\|^{\frac{1}{2}}_{L^{2}}\left\|\Delta \psi\right\|^{\frac{1}{2}}_{L^{2}}\left\|Z\right\|_{L^{2}} \\ & \leq C \left\|\nabla^{3}\psi_{2}\right\|^{\frac{4}{3}}_{L^{4}}\left\|\nabla \psi\right\|^{\frac{2}{3}}_{L^{2}}\left\|Z\right\|^{\frac{4}{3}}_{L^{2}} +\frac{1}{3}\left\|\Delta \psi\right\|^{2}_{L^{2}} \leq C \left\|\nabla^{3}\psi_{2}\right\|^{4}_{L^{4}}\left\|\nabla \psi\right\|^{2}_{L^{2}}\\ &+ C \left\|Z\right\|^{2}_{L^{2}} +\frac{1}{3}\left\|\Delta \psi\right\|^{2}_{L^{2}}\\ & \leq C \left\|\nabla\Delta \psi_{2}\right\|^{2}_{L^{2}}\left\|\Delta^{2}\psi_{2}\right\|^{2}_{L^{2}} \left\|\nabla \psi\right\|^{2}_{L^{2}}+ C \left\|Z\right\|^{2}_{L^{2}} +\frac{1}{3}\left\|\Delta \psi\right\|^{2}_{L^{2}}. \end{split}$

So, we have

$\begin{split} &\frac{d}{dt}\left(\left\|\nabla \psi\right\|^{2}_{L^{2}}+ \|Z\|^{2}_{L^{2}}\right)\\&\leq C\left(\left\|\nabla Z_{1}\right\|^{2}_{L^{\infty}}+\left\|\nabla Z_{2}\right\|^{2}_{L^{\infty}} +\left\|\nabla\Delta \psi_{2}\right\|^{2}_{L^{2}}\left\|\Delta^{2}\psi_{2}\right\|^{2}_{L^{2}}\right)\left(\left\|\nabla \psi\right\|^{2}_{L^{2}}+ \|Z\|^{2}_{L^{2}}\right). \end{split}$

By (50), the terms in the parentheses are integrable up to $\frac{T_{\ast}}{2}$ . By repeating the same argument one more time, we have the uniqueness up to $T_{\ast}$ .

5.1.3. Blow-up criterion

To derive the blow-up criterion, we first bound

$\begin{split} &\frac{1}{2}\frac{d}{dt} \left(\left\|\Delta\psi\right\|^{2}_{L^{2}}+\left\|\nabla Z\right\|^{2}_{L^{2}} \right) +\left\| \nabla \Delta \psi\right\|^{2}_{L^{2}} = \int \Delta^{2} \psi [\psi, Z] -\int\Delta Z[\Delta \psi, \psi]\\ & = 2\int \Delta \psi[\psi_{x}, Z_{x}]+2\int \Delta \psi[\psi_{y}, Z_{y}]\leq C \left\|\nabla^{2}\psi\right\|_{L^{\infty}}\left\|\nabla Z\right\|_{L^{2}}\left\|\nabla \Delta \psi\right\|_{L^{2}}\\ &\leq C \left\|\nabla^{2}\psi\right\|^{2}_{L^{\infty}}\left\|\nabla Z\right\|^{2}_{L^{2}}+\frac{1}{2}\left\|\nabla \Delta \psi\right\|^{2}_{L^{2}} \end{split}$

and so we have

$\frac{d}{dt} \left(\left\|\Delta\psi\right\|^{2}_{L^{2}}+\left\|\nabla Z\right\|^{2}_{L^{2}} \right) +\left\| \nabla \Delta \psi\right\|^{2}_{L^{2}} \leq C \left\|\nabla^{2}\psi\right\|^{2}_{L^{\infty}}\left\|\nabla Z\right\|^{2}_{L^{2}}.$

This implies

$\begin{eqnarray} \begin{split} &\left\|\Delta \psi(t)\right\|^{2}_{L^{}}+ \left\|\nabla Z(t)\right\|^{2}_{L^{2}}+\int^{t}_{0}\left\| \nabla \Delta \psi(s)\right\|^{2}_{L^{2}} ds < \infty\\& \iff \int^{t}_{0}\left\|\nabla^{2}\psi(s)\right\|^{2}_{L^{\infty}}ds < \infty \end{split} \end{eqnarray}$

(51)

We also take

$\begin{split} &\frac{1}{2}\frac{d}{dt} \left(\left\|\nabla \Delta\psi\right\|^{2}_{L^{2}}+\left\|\Delta Z\right\|^{2}_{L^{2}} \right) +\left\| \Delta^{2} \psi\right\|^{2}_{L^{2}} = -\int \Delta^{3} \psi [\psi, Z] +\int\Delta^{2} Z[\Delta \psi, \psi]\\ & = -\int \Delta^{2}\psi [\Delta \psi, Z] -2\int \Delta^{2}\psi \left([\psi_{x}, Z_{x}]+[\psi_{y}, Z_{y}]\right) \\&\qquad- 2\int \Delta\psi \left([\psi_{x}, \Delta Z_{x}]+[\psi_{y}, \Delta Z_{y}]\right)\\ & = \text{(I)+(II)+(III)}. \end{split}$

By using the computation in (41),

$\begin{split} \text{(I)}& = \int \left(\nabla \nabla^{\perp}Z\cdot \nabla \Delta \psi\right)\cdot \nabla \Delta\psi \leq C\left\|\nabla^{2}Z\right\|_{L^{2}} \left\|\nabla^{3}\psi \right\|^{2}_{L^{4}} \\ &\leq C\left\|\nabla^{2}Z\right\|^{2}_{L^{2}} \left\|\nabla\Delta \psi \right\|^{2}_{L^{2}}+\frac{1}{6}\left\|\Delta^{2}\psi \right\|^{2}_{L^{2}}. \end{split}$

We next estimate $\text{(II)}+\text{(III)}$ using (42):

$\begin{equation*} \begin{split} \text{(II)}+\text{(III)}&\leq C\int \left|\nabla^{2}Z\right| \left|\nabla^{3}\psi \right|^{2}+C\int \left|\nabla^{2}\psi \right| \left|\nabla^{4}\psi \right|\left|\nabla^{2}Z\right|\\ &\leq C\left\|\Delta Z\right\|^{2}_{L^{2}} \left\|\nabla\Delta \psi \right\|^{2}_{L^{2}}+C\left\|\nabla^{2}\psi\right\|^{2}_{L^{\infty}} \left\|\Delta Z \right\|^{2}_{L^{2}}+\frac{1}{3}\left\|\Delta^{2}\psi \right\|^{2}_{L^{2}} \end{split} \end{equation*}$

So, we have

$\begin{eqnarray} \begin{split} &\frac{d}{dt}\left(\left\|\nabla \Delta \psi\right\|^{2}_{L^{2}}+\left\|\Delta Z\right\|^{2}_{L^{2}}\right) +\left\|\Delta^{2} \psi\right\|^{2}_{L^{2}}\\&\leq C\left(\left\|\nabla\Delta \psi \right\|^{2}_{L^{2}}+\left\|\nabla^{2}\psi\right\|^{2}_{L^{\infty}}\right)\left\|\Delta Z\right\|^{2}_{L^{2}}. \end{split} \end{eqnarray}$

(52)

By (51), (52) implies

$\begin{eqnarray} \begin{split} &\left\|\nabla \Delta \psi(t)\right\|^{2}_{L^{2}}+\left\|\Delta Z(t)\right\|^{2}_{L^{2}}+\int \left\|\Delta^{2} \psi(s)\right\|^{2}_{L^{2}}ds < \infty \\ &\iff \int^{t}_{0}\left\|\nabla^{2}\psi(s)\right\|^{2}_{L^{\infty}}ds < \infty. \end{split} \end{eqnarray}$

(53)

We finally deal with

where we count the number of derivatives acting on $(\psi, \psi, Z)$ in (46) as $(3, 5, 2)$ , $(2, 5, 3)$ , and $(4, 4, 2)$ . Then, we obtain

$\begin{equation*} \begin{split} &\frac{1}{2} \frac{d}{dt} \left(\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}+\left\|\nabla\Delta Z\right\|^{2}_{L^{2}} \right) +\left\|\nabla\Delta^{2} \psi\right\|^{2}_{L^{2}}\\ & \leq C\left\|\Delta Z\right\|^{2}_{L^{2}}\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}\\&\qquad+ C\left\|\nabla^{2}\psi\right\|^{2}_{L^{\infty}}\left\|\nabla \Delta Z\right\|^{2}_{L^{2}} +C\left\|\Delta Z\right\|^{2}_{L^{4}}\left\|\nabla^{3}\psi\right\|^{2}_{L^{4}}+\frac{1}{2}\left\|\nabla \Delta^{2} \psi\right\|^{2}_{L^{2}}\\ &\leq C\left\|\Delta Z\right\|^{2}_{L^{2}}\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}+ C\left\|\nabla^{2}\psi\right\|^{2}_{L^{\infty}}\left\|\nabla \Delta Z\right\|^{2}_{L^{2}} +C\left\|\Delta Z\right\|^{2}_{L^{2}}\left\|\nabla\Delta Z\right\|^{2}_{L^{2}}\\ &+C\left\|\nabla \Delta \psi\right\|^{2}_{L^{2}}\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}+\frac{1}{2}\left\|\nabla \Delta^{2} \psi\right\|^{2}_{L^{2}} \end{split} \end{equation*}$

and so we have

$\begin{equation} \begin{split} &\frac{d}{dt} \left(\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}+\left\|\nabla\Delta Z\right\|^{2}_{L^{2}} \right) +\left\|\nabla\Delta^{2} \psi\right\|^{2}_{L^{2}}\\ &\leq C\left(\left\|\nabla^{2}\psi\right\|^{2}_{L^{\infty}}+\left\|\Delta Z\right\|^{2}_{L^{2}}\right)\left\|\nabla \Delta Z\right\|^{2}_{L^{2}} +C\left(\left\|\nabla \Delta \psi\right\|^{2}_{L^{2}}+\left\|\Delta Z\right\|^{2}_{L^{2}}\right)\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}} \end{split} \end{equation}$

(54)

By (51) and (53), (54) implies

$\label{BW 5}\begin{split} &\left\|\Delta^{2} \psi(t)\right\|^{2}_{L^{2}}+\left\|\nabla \Delta Z(t)\right\|^{2}_{L^{2}}+\int^{t}_{0} \left\|\nabla\Delta^{2} \psi(s)\right\|^{2}_{L^{2}}ds < \infty\\& \iff \int^{t}_{0}\left\|\nabla^{2}\psi(s)\right\|^{2}_{L^{\infty}}ds < \infty. \end{split}$

5.2. Proof of Theorem 1.6

We recall (13):

$\psi_{t}-\Delta \psi = [\psi, Z], \quad Z_{t}+Z = [\Delta \psi, \psi].$

Since the uniqueness is already proved in Section 5.1.2 even without the damping term, we only focus on the a priori bounds.

We first have

(55)

We also have

$\begin{split} \frac{1}{2}\frac{d}{dt} \left(\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}+\left\|\nabla\Delta Z\right\|^{2}_{L^{2}} \right) &+\left\|\nabla \Delta^{2} \psi\right\|^{2}_{L^{2}}+\left\|\nabla\Delta Z\right\|^{2}_{L^{2}} \\ & = \int \Delta^{4} \psi [\psi, Z] -\int\Delta^{3} Z[\Delta \psi, \psi] = \mathcal{R} \end{split}$

with the same $\mathcal{R}$ in (36). In this case, we also choose the the number of derivatives acting on $(\psi, \psi, Z)$ as $(3, 5, 2)$ , $(2, 5, 3)$ , and $(4, 4, 2)$ . Hence,

$\begin{equation*} \begin{split} & \frac{d}{dt} \left(\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}+\left\|\nabla\Delta Z\right\|^{2}_{L^{2}} \right) +\left\|\nabla\Delta^{2} \psi\right\|^{2}_{L^{2}} +\left\|\nabla\Delta Z\right\|^{2}_{L^{2}}\\ & \leq C \left\|\nabla^{2} Z\right\|_{L^{2}}\left\|\Delta^{2}\psi\right\|^{2}_{L^{4}} + C\left\|\nabla \Delta Z\right\|_{L^{2}} \left\|\nabla^{2}\psi\right\|_{L^{\infty}} \left\|\nabla \Delta^{2} \psi\right\|_{L^{2}}\\ &+ C\left\|\Delta Z\right\|_{L^{4}}\left\|\nabla^{3}\psi\right\|_{L^{4}} \left\|\nabla \Delta^{2}\psi\right\|_{L^{2}} \\ & \leq C \left\|\nabla Z\right\|^{\frac{1}{2}}_{L^{2}}\left\|\nabla\Delta Z\right\|^{\frac{1}{2}}_{L^{2}} \left\|\nabla \Delta\psi\right\|^{\frac{1}{2}}_{L^{2}}\left\|\nabla \Delta^{2}\psi\right\|^{\frac{3}{2}}_{L^{2}}+C\left\|\nabla \Delta Z\right\|_{L^{2}} \left\|\nabla^{2}\psi\right\|_{L^{\infty}} \left\|\nabla \Delta^{2} \psi\right\|_{L^{2}}\\ &+ C\left\|\Delta Z\right\|^{\frac{1}{2}}_{L^{2}}\left\|\nabla\Delta Z\right\|^{\frac{1}{2}}_{L^{2}} \left\|\Delta\psi\right\|^{\frac{1}{2}}_{L^{2}} \left\|\nabla \Delta^{2}\psi\right\|^{\frac{3}{2}}_{L^{2}}\\ & \leq C\left(\left\|\nabla Z\right\|^{2}_{L^{2}}\left\|\nabla \Delta\psi\right\|^{2}_{L^{2}}+\left\|\Delta Z\right\|^{2}_{L^{2}}\left\|\Delta\psi\right\|^{2}_{L^{2}}+\left\|\nabla^{2}\psi\right\|^{2}_{L^{\infty}}\right)\left\|\nabla\Delta Z\right\|^{2}_{L^{2}}\\&\qquad +\frac{1}{2} \left\|\nabla \Delta^{2} \psi\right\|_{L^{2}}. \end{split} \end{equation*}$

So, we obtain

$\begin{equation} \begin{split} &\frac{d}{dt} \left(\left\|\Delta^{2}\psi\right\|^{2}_{L^{2}}+\left\|\nabla\Delta Z\right\|^{2}_{L^{2}} \right) +\left\|\nabla\Delta^{2} \psi\right\|^{2}_{L^{2}} +\left\|\nabla\Delta Z\right\|^{2}_{L^{2}}\\ & \leq C\left(\left\|\nabla Z\right\|^{2}_{L^{2}}\left\|\nabla \Delta\psi\right\|^{2}_{L^{2}}+\left\|\Delta Z\right\|^{2}_{L^{2}}\left\|\Delta\psi\right\|^{2}_{L^{2}}+\left\|\nabla^{2}\psi\right\|^{2}_{L^{\infty}}\right)\left\|\nabla\Delta Z\right\|^{2}_{L^{2}} \end{split} \end{equation}$

(56)

By (55) and (56),

$\mathcal{E}'(t)+\mathcal{N}_{2}(t)\leq C \left(\mathcal{E}(t)+\mathcal{E}^{2}(t)\right)\mathcal{N}_{2}(t).$

So, if $\mathcal{E}_{0} = \epsilon_{0}$ is sufficiently small, we obtain

$\mathcal{E}(t)+(1-C \epsilon_{0})\int^{t}_{0}\mathcal{N}_{2}(s)ds\leq \mathcal{E}_{0} \quad {\rm{for}}\; {\rm{all}}\; t > 0 .$

Acknowledgments

H.B. was supported by NRF-2018R1D1A1B07049015. H. B. acknowledges the Referee for his/her valuable comments that highly improve the manuscript.

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沈阳化工大学材料科学与工程学院沈阳 110142

Mathematical Biosciences and Engineering

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Mathematical Biosciences and Engineering

TS-GCN: A novel tumor segmentation method integrating transformer and GCN

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Abstract

1. Introduction

1.1. The Hall equation with fractional Laplacian

1.2. 2D models

1.2.1. Case 1

1.2.2. Case 2

2. Preliminaries

2.1. Inequalities

2.2. Commutator

3. Proof of Theorem 1.1 and Theorem 1.2

3.1. Proof of Theorem 1.1

3.1.1. Approximation

3.1.2. A priori estimates

3.1.3. Uniqueness

3.1.4. Decay rates

3.2. Proof of Theorem 1.2

4. Proof of Theorem 1.3 and Theorem 1.4

4.1. Proof of Theorem 1.3

4.1.1. Approximation

4.1.2. A priori estimates

4.1.3. Uniqueness

4.1.4. Blow-up criterion

4.2. Proof of Theorem 1.4

4.2.1. A priori estimates

4.2.2. Decay rates

5. Proof of Theorem 1.5 and Theorem 1.6

5.1. Proof of Theorem 1.5

5.1.1. A priori estimates

5.1.2. Uniqueness

5.1.3. Blow-up criterion

5.2. Proof of Theorem 1.6

Acknowledgments

References

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Abstract

1. Introduction

1.1. The Hall equation with fractional Laplacian

1.2. 2D models

1.2.1. Case 1

1.2.2. Case 2

2. Preliminaries

2.1. Inequalities

2.2. Commutator

3. Proof of Theorem 1.1 and Theorem 1.2

3.1. Proof of Theorem 1.1

3.1.1. Approximation

3.1.2. A priori estimates

3.1.3. Uniqueness

3.1.4. Decay rates

3.2. Proof of Theorem 1.2

4. Proof of Theorem 1.3 and Theorem 1.4

4.1. Proof of Theorem 1.3

4.1.1. Approximation

4.1.2. A priori estimates

4.1.3. Uniqueness

4.1.4. Blow-up criterion

4.2. Proof of Theorem 1.4

4.2.1. A priori estimates

4.2.2. Decay rates

5. Proof of Theorem 1.5 and Theorem 1.6

5.1. Proof of Theorem 1.5

5.1.1. A priori estimates

5.1.2. Uniqueness

5.1.3. Blow-up criterion

5.2. Proof of Theorem 1.6

Acknowledgments