# Magnetic moment of the $\Xi_{b}(6227)$ as a molecular pentaquark state

Ulaş Özdem<EMAIL_ADDRESS>Health Services Vocational School of Higher
Education, Istanbul Aydin University, Sefakoy-Kucukcekmece, 34295 Istanbul,
Turkey

###### Abstract

In this study, considering that the $\Xi_{b}(6227)$ state is in molecular
structure, the magnetic moment of this state is extracted in the light-cone
QCD sum rules. The numerical result is obtained as $\mu_{\Xi_{b}}=0.12\pm
0.03~{}\mu_{N}$. The magnetic moment of this state contains important
information of its internal structure and shape deformations. Measurement of
the magnetic moment of the $\Xi_{b}(6227)$ state in future experimental
facilities can be very helpful in identification of the quantum numbers, as
well as comprehension of the inner structure of this state.

Magnetic moment, Exotic states, Light-cone QCD

##  I Introduction

In recent years, more and more heavy baryons have been observed by LHCb, Belle
and CDF collaborations and the spectra of the heavy baryon sectors have become
more and more abundance. Among the baryons, the singly bottom baryons are
especially fascinating, because the heavy quark symmetry is preserved well in
them, and their mass splitting is remarkably smaller than that in the case of
light and charmed baryons. Studies involving the spectroscopic parameters and
decay channels of singly heavy baryons enhance our understanding of the non-
perturbative regime of strong interaction, as well as our knowledge of their
nature and internal structure.

In 2018, the LHCb Collaboration was reported a bottom baryon, $\Xi_{b}(6227)$
($\Xi_{b}$ for short), in both the $\Lambda^{0}_{b}K^{-}$ and
$\Xi_{b}^{0}\pi^{-}$ final states with mass $m_{\Xi_{b}}=6226.9\pm
2.0(\mbox{stat})\pm 0.3(\mbox{syst})\pm 0.2~{}\mbox{MeV}$ and width
$\Gamma_{\Xi_{b}}=18.1\pm 5.4(\mbox{stat})\pm 1.8(\mbox{syst})$ [1]. However,
spin-parity of this state remains unknown. After the experimental discovery,
many studies have been carried out to understand the properties of this
particle. In some of these studies, this state is considered as the excited
state of the $\Xi_{b}$ baryon [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], while in
others it is considered as the pentaquark state [13, 14, 15, 16]. Even though
the studies of Refs. [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] seem to show that
this state is a conventional three-quark state, the $\Xi_{b}$ might still be a
$K\Sigma_{b}$ hadronic molecule state, because the mass gap between the
$\Xi^{*}_{b}$ and the ground $\Xi_{b}$ baryons, about 440 MeV, is large enough
to excite a light quark-anti-quark pair to form a molecular state. In Ref.
[13], mass of the $\Xi_{b}$ state was studied in molecular pentaquark state
with quantum numbers $J^{P}=1/2^{\pm}$ via QCD sum rules. They found that
their results support assigning $\Xi_{b}$ state to be a molecular state with
quantum number $J^{P}=1/2^{-}$. In Ref. [14], the radiative decays of the
$\Xi_{b}$ state have been obtained in the framework of effective Lagrangians
by assuming this state as a molecular state with the quantum number
$J^{P}=1/2^{-}$. In Ref. [15], the authors studied the strong decays of the
$\Xi_{b}$ baryon assuming that it is a molecular state with different spin-
party assignments. Their results showed that spin-parity $J^{P}=1/2^{-}$
assignment is preferred while spin-parity $J^{P}=1/2^{+}$ and
$J^{P}=3/2^{\pm}$ are disfavored. In Ref. [16], they studied mass of the
$\Xi_{b}$ state in the molecular picture via an extension of the local hidden
gauge approach in the Bethe–Salpeter equation. They found that the mass and
width of $\Xi_{b}$ state remarkably close to the experimental data, for both
the $J^{P}=1/2^{-}$ and $J^{P}=3/2^{-}$ sectors. However, the internal
structure of this particle is still not clearly elucidated. In addition to
spectroscopic properties such as mass and decay channels, different studies
are needed to understand the properties of this state.

The investigation on the electromagnetic properties may provide a way of
learning about the internal structure of the $\Xi_{b}$ state. Inspired by
this, in the present study, we consider the magnetic moments of the $\Xi_{b}$
with the spin-parity $J^{P}=1/2^{-}$, using light-cone QCD sum rule (LCSR)
method [17, 18, 19] and assuming that the $\Xi_{b}$ is a hadronic molecular
state. In the LCSR, the time-ordered product of the interpolating currents is
sandwiched between the vacuum and an on-shell state. The on-shell state in the
present work is the photon. As it is known, an appropriate correlation
function is written while performing calculations in LCSR. Then, this
correlation function is calculated in terms of both QCD degrees of freedom,
QCD side, and hadronic parameters, hadronic side. The results obtained with
these two different approaches are equalized with the dispersion relation by
choosing the same Lorentz structures. Finally, Borel transform and continuum
subtraction are performed to eliminate the contribution of excited and higher
state terms.

This article is structured as follows. Section II is set aside for the details
of the LCSR evaluations for the magnetic moment of the $\Xi_{b}$ state.
Section III is devoted to the numerical calculations of the magnetic moment
and discussion. Some parameters and formulas related to the calculations are
presented in Appendices A and B.

##  II Formalism

In this section, we explain how to compute the magnetic moment of $\Xi_{b}$
state in molecular picture. To derive the LCSR for the magnetic moment, we
start by considering the subsequent correlation function,

$\displaystyle\Pi(p,q)$ $\displaystyle=$ $\displaystyle i\int d^{4}xe^{ip\cdot
x}\langle
0|T\left\\{J^{\Xi_{b}}(x)\bar{J}^{\Xi_{b}}(0)\right\\}|0\rangle_{\gamma}\,,$
(1)

where $\gamma$ is the external electromagnetic field and the $J^{\Xi_{b}}(x)$
stands for interpolating current of the considered state and it is given as
follows:

$\displaystyle
J^{\Xi_{b}}(x)=[\bar{u}_{d}(x)i\gamma_{5}s_{d}(x)][\varepsilon^{abc}u^{T}_{a}(x)C\gamma_{\mu}d_{b}(x)\gamma^{\mu}\gamma_{5}b_{c}(x)],$
(2)

where $a,b,c~{}\mbox{and}~{}d$ are color indices. It should be noted that the
interpolating current couples to the five quark components of baryons. For a
conventional baryon, this coupling would be small, but for a pentaquark state,
this coupling can be large. In this work, we assume that $\Xi_{b}$ particle is
dominantly a molecular state and hence the coupling of the molecular
interpolating current to $\Xi_{b}$ is not negligible.

As we mentioned above, the correlation function in the LCSR method is
calculated in two different ways: hadronic and QCD sides. First, we focus on
computation of the hadronic side. For this aim, we insert a full set of
hadronic $\Xi_{b}$ state into Eq. (1) and carry out the integral over x to
obtain

$\displaystyle\Pi^{Had}(p,q)$ $\displaystyle=\frac{\langle 0\mid
J^{\Xi_{b}}(x)\mid{{\Xi_{b}}}(p,s)\rangle}{[p^{2}-m_{{\Xi_{b}}}^{2}]}\langle{{\Xi_{b}}}(p,s)\mid{{\Xi_{b}}}(p+q,s)\rangle_{\gamma}\frac{\langle{{\Xi_{b}}}(p+q,s)\mid\bar{J}^{\Xi_{b}}(0)\mid
0\rangle}{[(p+q)^{2}-m_{{{\Xi_{b}}}}^{2}]}+\cdots$ (3)

The matrix element, $\langle 0\mid J^{\Xi_{b}}(x)\mid{{\Xi_{b}}}(p,s)\rangle$
is written with respect to the residue and the polarization vector of the
$\Xi_{b}$ state as

$\displaystyle\langle 0\mid
J^{\Xi_{b}}(x)\mid{{\Xi_{b}}}(p,s)\rangle=\lambda_{\Xi_{b}}u(p,s),$ (4)

while the matrix element
$\langle{{\Xi_{b}}}(p,s)\mid{{\Xi_{b}}}(p+q,s)\rangle_{\gamma}$ is written
with respect to the form factors,

$\displaystyle\langle{{\Xi_{b}}}(p,s)\mid{{\Xi_{b}}}(p+q,s)\rangle_{\gamma}$
$\displaystyle=\varepsilon^{\mu}\,\bar{u}(p,s)\Big{[}\big{[}F_{1}(q^{2})+F_{2}(q^{2})\big{]}\gamma_{\mu}+F_{2}(q^{2})\frac{(2p+q)_{\mu}}{2m_{{\Xi_{b}}}}\Big{]}\,u(p+q,s).$
(5)

We insert Eq. (4) and Eq. (5) in Eq. (3), and summing over the polarization
vector of the $\Xi_{b}$ state, we obtain the following result for the hadronic
side of the correlation function:

$\displaystyle\Pi^{Had}(p,q)=$
$\displaystyle\lambda^{2}_{{\Xi_{b}}}\gamma_{5}\frac{\Big{(}p\\!\\!\\!/+m_{{\Xi_{b}}}\Big{)}}{[p^{2}-m_{{{\Xi_{b}}}}^{2}]}\varepsilon^{\mu}\Bigg{[}\big{[}F_{1}(q^{2})+F_{2}(q^{2})\big{]}\gamma_{\mu}+F_{2}(q^{2})\,\frac{(2p+q)_{\mu}}{2m_{{\Xi_{b}}}}\Bigg{]}\gamma_{5}\frac{\Big{(}p\\!\\!\\!/+q\\!\\!\\!/+m_{{\Xi_{b}}}\Big{)}}{[(p+q)^{2}-m_{{{\Xi_{b}}}}^{2}]}.$
(6)

The value of form factors $F_{1}(q^{2})$ and $F_{2}(q^{2})$ gives us the
magnetic form factor $F_{M}(q^{2})$ at different $q^{2}$ :

$\displaystyle F_{M}(q^{2})=F_{1}(q^{2})+F_{2}(q^{2}).$ (7)

At $q^{2}=0$, the magnetic moment $\mu_{{\Xi_{b}}}$ can be described with the
$F_{M}(q^{2}=0)$ as follows

$\displaystyle\mu_{{\Xi_{b}}}=\frac{e}{2\,m_{{\Xi_{b}}}}\,F_{M}(q^{2}=0).$ (8)

Now, we turn our attention to calculate the QCD side of the above-mentioned
correlation function. It is evaluated in deep Euclidean region where
$p^{2}\rightarrow\infty$ and $-(p+q)^{2}\rightarrow\infty$. Employing the
explicit expressions of the interpolating currents and contracting out the
light and heavy quark pairs using the Wick’s theorem, we obtain

$\displaystyle\Pi^{QCD}(p,q)$ $\displaystyle=$
$\displaystyle-i\,\varepsilon_{abc}\varepsilon_{a^{\prime}b^{\prime}c^{\prime}}\,\int
d^{4}x\,e^{ip\cdot x}\langle
0\mid\gamma^{\mu}\gamma_{5}S_{b}^{cc^{\prime}}(x)\gamma_{5}\gamma^{\nu}$ (9)
$\displaystyle\mbox{Tr}\Big{[}\gamma_{5}S_{u}^{dd^{\prime}}(-x)\gamma_{5}S_{s}^{dd^{\prime}}(x)\Big{]}\,\mbox{Tr}\Big{[}\gamma_{\mu}S_{d}^{bb^{\prime}}(x)\gamma_{\nu}\widetilde{S}_{u}^{aa^{\prime}}(x)\Big{]}\mid
0\rangle_{\gamma}\,,$

where $\widetilde{S}_{c(q)}^{ij}(x)=CS_{c(q)}^{ij\mathrm{T}}(x)C$ and ,
$S_{q}(x)$ and $S_{b}(x)$ are the light and b quark propagators, respectively.
Their explicit expressions in the x-space are presented as

$\displaystyle S_{q}(x)$ $\displaystyle=\frac{1}{2\pi
x^{2}}\Big{(}i\frac{x\\!\\!\\!/}{x^{2}}-\frac{m_{q}}{2}\Big{)}-\frac{\langle\bar{q}q\rangle}{12}\Big{(}1-i\frac{m_{q}x\\!\\!\\!/}{4}\Big{)}-\frac{\langle\bar{q}q\rangle}{192}m_{0}^{2}x^{2}\Big{(}1-i\frac{m_{q}x\\!\\!\\!/}{6}\Big{)}-\frac{ig_{s}}{32\pi^{2}x^{2}}~{}G^{\mu\nu}(x)\Big{[}\hbox
to0.0pt{/\hss}{x}\sigma_{\mu\nu}+\sigma_{\mu\nu}\hbox
to0.0pt{/\hss}{x}\Big{]},$ (10) $\displaystyle S_{b}(x)$
$\displaystyle=\frac{m_{b}^{2}}{4\pi^{2}}\Bigg{[}\frac{K_{1}\Big{(}m_{b}\sqrt{-x^{2}}\Big{)}}{\sqrt{-x^{2}}}+i\frac{{x\\!\\!\\!/}~{}K_{2}\Big{(}m_{b}\sqrt{-x^{2}}\Big{)}}{(\sqrt{-x^{2}})^{2}}\Bigg{]}-\frac{g_{s}m_{b}}{16\pi^{2}}\int_{0}^{1}dv\,G^{\mu\nu}(vx)\Bigg{[}(\sigma_{\mu\nu}{x\\!\\!\\!/}+{x\\!\\!\\!/}\sigma_{\mu\nu})\frac{K_{1}\Big{(}m_{b}\sqrt{-x^{2}}\Big{)}}{\sqrt{-x^{2}}}$
$\displaystyle+2\sigma_{\mu\nu}K_{0}\Big{(}m_{c}\sqrt{-x^{2}}\Big{)}\Bigg{]},$
(11)

where $K_{0}$, $K_{1}$ and $K_{2}$ are modified the second kind Bessel
functions and $G^{\mu\nu}$ is the gluon field strength tensor. The first term
of the light and heavy quark propagators corresponds to perturbative or free
part and the rest belong to the interacting parts. In the LCSR method, the
non-perturbative contribution shows up when a photon is emitted at long
distances. To take into account these contributions, it is required to expand
the light quark propagator near the $x^{2}\sim 0$. In this case, the matrix
elements of two and three-particle non-local operators show up between the
photon states and vacuum such as
$\langle\gamma(q)\left|\bar{q}(x)\Gamma_{i}q(0)\right|0\rangle$ and
$\langle\gamma(q)\left|\bar{q}(x)\Gamma_{i}G_{\mu\nu}q(0)\right|0\rangle$.
These matrix elements are parameterized in connection with the photon
distribution amplitudes (DAs), which were determined in Ref. [20]. The
explicit expressions of these terms are given in Appendix A. The QCD side of
the correlation function can be obtained associated with the quark-gluon
properties via the photon DAs, and after performing an integration over x, the
expression of the correlation function in the momentum representation can be
calculated straightforwardly.

As we explained in detail above, we have calculated the correlation function
in two different ways. The next step will be to match this differently
calculated correlation function with each other using certain approaches. To
do this, we will use the quark-hadron duality approach and we apply the double
Borel transform and continuum subtraction to suppress the contribution of
excited and higher states. Computations performed according to a scheme
briefly explained above yield

$\displaystyle\mu_{{\Xi_{b}}}\,\lambda^{2}_{{\Xi_{b}}}\,m_{{\Xi_{b}}}=e^{\frac{m^{2}_{{\Xi_{b}}}}{M^{2}}}\,\Delta^{QCD}.$
(12)

The explicit expression of $\Delta^{QCD}$ is presented Appendix B.

##  III Numerical analysis and discussions

The LCSR for magnetic moment of the ${\Xi_{b}}$ state contains many input
parameters that we need their numerical values. We use $m_{u}=m_{d}=0$,
$m_{s}=96^{+8}_{-4}\,\mbox{MeV}$, $m_{b}=4.18^{+0.03}_{-0.02}\,\mbox{GeV}$
[21], $m_{\Xi_{b}}=6226.9\pm 2.0(\mbox{stat})\pm 0.3(\mbox{syst})\pm
0.2~{}\mbox{MeV}$ [1], $\lambda_{\Xi_{b}}=(1.26^{+0.26}_{-0.27})\times
10^{-3}$ GeV6 [13], $f_{3\gamma}=-0.0039~{}\mbox{GeV}^{2}$ [20],
$\langle\bar{u}u\rangle=\langle\bar{d}d\rangle=(-0.24\pm
0.01)^{3}\,\mbox{GeV}^{3}$,
$\langle\bar{s}s\rangle=0.8\,\langle\bar{u}u\rangle$ $\,\mbox{GeV}^{3}$ [22],
$m_{0}^{2}=0.8\pm 0.1\,\mbox{GeV}^{2}$ [22], $\langle
g_{s}^{2}G^{2}\rangle=0.88~{}\mbox{GeV}^{4}$ [23] and $\chi=-2.85\pm
0.5~{}\mbox{GeV}^{-2}$ [24]. We also need the photon DAs and the input
parameters used in these amplitudes to proceed in the calculations. The
explicit expressions of the photon DAs and numerical values of input
parameters are given in Appendix A.

In addition to these parameters mentioned above, QCD sum rules contain two
extra parameters known as Borel mass parameter ($M^{2}$) and continuum
threshold ($s_{0}$). According to the principles of the method used, the
physical quantities to be calculated should not be very sensitive to the
variation of these parameters. In practice, it is necessary to apply some
physical constraints such as pole dominance and convergence of operator
product expansion to ensure this situation. Considering these constraints,
$50$ GeV2 $\leq s_{0}\leq 52$ GeV2 and $7$ GeV2 $\leq M^{2}\leq 9$ GeV2
working regions have been determined for these extra parameters. In Fig. 1, we
give the dependence of the magnetic moment of the $\Xi_{b}$ state on Borel
mass parameter $M^{2}$ , at three fixed values of the continuum threshold
$s_{0}$. It follows from this figure that the magnetic moment indicates a weak
dependence on $M^{2}$ in its working region.

Figure 1:  The dependence of magnetic moment of the $\Xi_{b}$ state on $M^{2}$
at three fixed values of $s_{0}$.

We have specified all the input parameters necessary to determine the
numerical value for the magnetic moment of the $\Xi_{b}$ state. As a result of
our detailed numerical calculations, where we consider the uncertainties in
the input parameters, uncertainties entering the photon DAs as well as
uncertainties because of the variations of Borel mass parameter $M^{2}$ and
continuum threshold $s_{0}$, the magnetic moment of $\Xi_{b}$ state is finally
found to have the value

$\displaystyle\mu_{\Xi_{b}}=0.80\pm 0.19~{}\frac{e}{2\,m_{\Xi_{b}}}=0.12\pm
0.03~{}\mu_{N}.$ (13)

The order of magnetic moment shows that it is accessible in the experiment.
Detailed analysis of the calculations shows that the value of the magnetic
moment is determined by the light quarks. Our analysis shows that the
experimental measurement of the magnetic moment of the $\Xi_{b}$ will be able
to tell whether it is a molecular state or a conventional baryon. If the
measured magnetic moment value is consistent with our prediction, we can say
that it is the molecular state; if not, we can say that it is the conventional
baryon state. With the increased luminosity in future runs and our prediction,
the facilities like PANDA, LHCb, BESIII, Belle II and so on might have the
capability to measure magnetic moment of $\Xi_{b}$ state.

Comparing the estimates obtained in this study with the results obtained using
different approaches will be a test for the consistency of our results. The
study of the magnetic moment of $\Xi_{b}$ with lattice QCD is strongly
recommended. In Refs. [25, 26], the mass and magnetic moment of
$\Lambda(1405)$ state have been calculated by considering it as the molecular
picture in the framework of lattice QCD. A similar analysis can be performed
using the Lattice QCD for the magnetic moment of the state $\Xi_{b}$. In
comparison with the predicted magnetic moment of lattice QCD, the results in
present study can ensure further information for the experimental search for
the $\Xi_{b}$, and, moreover, the experimental measurements for this magnetic
moment could be a important test for the molecule configuration of the
$\Xi_{b}$.

In summary, inspired by experimental observation of the $\Xi_{b}(6227)$ state,
the magnetic moment of the this state has been calculated in the framework of
the light-cone QCD sum rules employing the photon DAs and assuming that the
$\Xi_{b}(6227)$ is a hadronic molecular state. The acquired result may be
useful in exact determinations of the nature of this state. Checking our
prediction with different theoretical models and by future experiments can be
very helpful of the comprehension of the inner structure as well as the
geometric shape of the $\Xi_{b}(6227)$ state.

## Appendix A: Photon distribution amplitudes and wave functions

In this Appendix, we give the matrix elements
$\langle\gamma(q)\left|\bar{q}(x)\Gamma_{i}q(0)\right|0\rangle$ and
$\langle\gamma(q)\left|\bar{q}(x)\Gamma_{i}G_{\mu\nu}q(0)\right|0\rangle$ in
connection with the photon DAs and wave functions of different twists [20],

$\displaystyle\langle\gamma(q)|\bar{q}(x)\gamma_{\mu}q(0)|0\rangle=e_{q}f_{3\gamma}\left(\varepsilon_{\mu}-q_{\mu}\frac{\varepsilon
x}{qx}\right)\int_{0}^{1}due^{i\bar{u}qx}\psi^{v}(u)$
$\displaystyle\langle\gamma(q)|\bar{q}(x)\gamma_{\mu}\gamma_{5}q(0)|0\rangle=-\frac{1}{4}e_{q}f_{3\gamma}\epsilon_{\mu\nu\alpha\beta}\varepsilon^{\nu}q^{\alpha}x^{\beta}\int_{0}^{1}due^{i\bar{u}qx}\psi^{a}(u)$
$\displaystyle\langle\gamma(q)|\bar{q}(x)\sigma_{\mu\nu}q(0)|0\rangle=-ie_{q}\langle\bar{q}q\rangle(\varepsilon_{\mu}q_{\nu}-\varepsilon_{\nu}q_{\mu})\int_{0}^{1}due^{i\bar{u}qx}\left(\chi\varphi_{\gamma}(u)+\frac{x^{2}}{16}\mathbb{A}(u)\right)$
$\displaystyle-\frac{i}{2(qx)}e_{q}\bar{q}q\left[x_{\nu}\left(\varepsilon_{\mu}-q_{\mu}\frac{\varepsilon
x}{qx}\right)-x_{\mu}\left(\varepsilon_{\nu}-q_{\nu}\frac{\varepsilon
x}{qx}\right)\right]\int_{0}^{1}due^{i\bar{u}qx}h_{\gamma}(u)$
$\displaystyle\langle\gamma(q)|\bar{q}(x)g_{s}G_{\mu\nu}(vx)q(0)|0\rangle=-ie_{q}\langle\bar{q}q\rangle\left(\varepsilon_{\mu}q_{\nu}-\varepsilon_{\nu}q_{\mu}\right)\int{\cal
D}\alpha_{i}e^{i(\alpha_{\bar{q}}+v\alpha_{g})qx}{\cal S}(\alpha_{i})$
$\displaystyle\langle\gamma(q)|\bar{q}(x)g_{s}\tilde{G}_{\mu\nu}(vx)i\gamma_{5}q(0)|0\rangle=-ie_{q}\langle\bar{q}q\rangle\left(\varepsilon_{\mu}q_{\nu}-\varepsilon_{\nu}q_{\mu}\right)\int{\cal
D}\alpha_{i}e^{i(\alpha_{\bar{q}}+v\alpha_{g})qx}\tilde{\cal S}(\alpha_{i})$
$\displaystyle\langle\gamma(q)|\bar{q}(x)g_{s}\tilde{G}_{\mu\nu}(vx)\gamma_{\alpha}\gamma_{5}q(0)|0\rangle=e_{q}f_{3\gamma}q_{\alpha}(\varepsilon_{\mu}q_{\nu}-\varepsilon_{\nu}q_{\mu})\int{\cal
D}\alpha_{i}e^{i(\alpha_{\bar{q}}+v\alpha_{g})qx}{\cal A}(\alpha_{i})$
$\displaystyle\langle\gamma(q)|\bar{q}(x)g_{s}G_{\mu\nu}(vx)i\gamma_{\alpha}q(0)|0\rangle=e_{q}f_{3\gamma}q_{\alpha}(\varepsilon_{\mu}q_{\nu}-\varepsilon_{\nu}q_{\mu})\int{\cal
D}\alpha_{i}e^{i(\alpha_{\bar{q}}+v\alpha_{g})qx}{\cal V}(\alpha_{i})$
$\displaystyle\langle\gamma(q)|\bar{q}(x)\sigma_{\alpha\beta}g_{s}G_{\mu\nu}(vx)q(0)|0\rangle=e_{q}\langle\bar{q}q\rangle\left\\{\left[\left(\varepsilon_{\mu}-q_{\mu}\frac{\varepsilon
x}{qx}\right)\left(g_{\alpha\nu}-\frac{1}{qx}(q_{\alpha}x_{\nu}+q_{\nu}x_{\alpha})\right)\right.\right.q_{\beta}$
$\displaystyle-\left(\varepsilon_{\mu}-q_{\mu}\frac{\varepsilon
x}{qx}\right)\left(g_{\beta\nu}-\frac{1}{qx}(q_{\beta}x_{\nu}+q_{\nu}x_{\beta})\right)q_{\alpha}-\left(\varepsilon_{\nu}-q_{\nu}\frac{\varepsilon
x}{qx}\right)\left(g_{\alpha\mu}-\frac{1}{qx}(q_{\alpha}x_{\mu}+q_{\mu}x_{\alpha})\right)q_{\beta}$
$\displaystyle+\left.\left(\varepsilon_{\nu}-q_{\nu}\frac{\varepsilon
x}{q.x}\right)\left(g_{\beta\mu}-\frac{1}{qx}(q_{\beta}x_{\mu}+q_{\mu}x_{\beta})\right)q_{\alpha}\right]\int{\cal
D}\alpha_{i}e^{i(\alpha_{\bar{q}}+v\alpha_{g})qx}{\cal T}_{1}(\alpha_{i})$
$\displaystyle+\left[\left(\varepsilon_{\alpha}-q_{\alpha}\frac{\varepsilon
x}{qx}\right)\left(g_{\mu\beta}-\frac{1}{qx}(q_{\mu}x_{\beta}+q_{\beta}x_{\mu})\right)\right.q_{\nu}$
$\displaystyle-\left(\varepsilon_{\alpha}-q_{\alpha}\frac{\varepsilon
x}{qx}\right)\left(g_{\nu\beta}-\frac{1}{qx}(q_{\nu}x_{\beta}+q_{\beta}x_{\nu})\right)q_{\mu}$
$\displaystyle-\left(\varepsilon_{\beta}-q_{\beta}\frac{\varepsilon
x}{qx}\right)\left(g_{\mu\alpha}-\frac{1}{qx}(q_{\mu}x_{\alpha}+q_{\alpha}x_{\mu})\right)q_{\nu}$
$\displaystyle+\left.\left(\varepsilon_{\beta}-q_{\beta}\frac{\varepsilon
x}{qx}\right)\left(g_{\nu\alpha}-\frac{1}{qx}(q_{\nu}x_{\alpha}+q_{\alpha}x_{\nu})\right)q_{\mu}\right]\int{\cal
D}\alpha_{i}e^{i(\alpha_{\bar{q}}+v\alpha_{g})qx}{\cal T}_{2}(\alpha_{i})$
$\displaystyle+\frac{1}{qx}(q_{\mu}x_{\nu}-q_{\nu}x_{\mu})(\varepsilon_{\alpha}q_{\beta}-\varepsilon_{\beta}q_{\alpha})\int{\cal
D}\alpha_{i}e^{i(\alpha_{\bar{q}}+v\alpha_{g})qx}{\cal T}_{3}(\alpha_{i})$
$\displaystyle+\left.\frac{1}{qx}(q_{\alpha}x_{\beta}-q_{\beta}x_{\alpha})(\varepsilon_{\mu}q_{\nu}-\varepsilon_{\nu}q_{\mu})\int{\cal
D}\alpha_{i}e^{i(\alpha_{\bar{q}}+v\alpha_{g})qx}{\cal
T}_{4}(\alpha_{i})\right\\}~{},$

where $\varphi_{\gamma}(u)$ is the distribution amplitude of leading twist-2,
$\psi^{v}(u)$, $\psi^{a}(u)$, ${\cal A}(\alpha_{i})$ and ${\cal
V}(\alpha_{i})$, are the twist-3 amplitudes, and $h_{\gamma}(u)$,
$\mathbb{A}(u)$, ${\cal S}(\alpha_{i})$, ${\cal{\tilde{S}}}(\alpha_{i})$,
${\cal T}_{1}(\alpha_{i})$, ${\cal T}_{2}(\alpha_{i})$, ${\cal
T}_{3}(\alpha_{i})$ and ${\cal T}_{4}(\alpha_{i})$ are the twist-4 photon DAs.
The measure ${\cal D}\alpha_{i}$ is defined as

$\displaystyle\int{\cal
D}\alpha_{i}=\int_{0}^{1}d\alpha_{\bar{q}}\int_{0}^{1}d\alpha_{q}\int_{0}^{1}d\alpha_{g}\delta(1-\alpha_{\bar{q}}-\alpha_{q}-\alpha_{g})~{}.$

The expressions of the DAs entering into the above matrix elements are defined
as:

$\displaystyle\varphi_{\gamma}(u)$ $\displaystyle=$ $\displaystyle
6u\bar{u}\left(1+\varphi_{2}(\mu)C_{2}^{\frac{3}{2}}(u-\bar{u})\right),$
$\displaystyle\psi^{v}(u)$ $\displaystyle=$ $\displaystyle
3\left(3(2u-1)^{2}-1\right)+\frac{3}{64}\left(15w^{V}_{\gamma}-5w^{A}_{\gamma}\right)\left(3-30(2u-1)^{2}+35(2u-1)^{4}\right),$
$\displaystyle\psi^{a}(u)$ $\displaystyle=$
$\displaystyle\left(1-(2u-1)^{2}\right)\left(5(2u-1)^{2}-1\right)\frac{5}{2}\left(1+\frac{9}{16}w^{V}_{\gamma}-\frac{3}{16}w^{A}_{\gamma}\right),$
$\displaystyle h_{\gamma}(u)$ $\displaystyle=$
$\displaystyle-10\left(1+2\kappa^{+}\right)C_{2}^{\frac{1}{2}}(u-\bar{u}),$
$\displaystyle\mathbb{A}(u)$ $\displaystyle=$ $\displaystyle
40u^{2}\bar{u}^{2}\left(3\kappa-\kappa^{+}+1\right)+8(\zeta_{2}^{+}-3\zeta_{2})\left[u\bar{u}(2+13u\bar{u})\right.$
$\displaystyle+\left.2u^{3}(10-15u+6u^{2})\ln(u)+2\bar{u}^{3}(10-15\bar{u}+6\bar{u}^{2})\ln(\bar{u})\right],$
$\displaystyle{\cal A}(\alpha_{i})$ $\displaystyle=$ $\displaystyle
360\alpha_{q}\alpha_{\bar{q}}\alpha_{g}^{2}\left(1+w^{A}_{\gamma}\frac{1}{2}(7\alpha_{g}-3)\right),$
$\displaystyle{\cal V}(\alpha_{i})$ $\displaystyle=$ $\displaystyle
540w^{V}_{\gamma}(\alpha_{q}-\alpha_{\bar{q}})\alpha_{q}\alpha_{\bar{q}}\alpha_{g}^{2},$
$\displaystyle{\cal T}_{1}(\alpha_{i})$ $\displaystyle=$
$\displaystyle-120(3\zeta_{2}+\zeta_{2}^{+})(\alpha_{\bar{q}}-\alpha_{q})\alpha_{\bar{q}}\alpha_{q}\alpha_{g},$
$\displaystyle{\cal T}_{2}(\alpha_{i})$ $\displaystyle=$ $\displaystyle
30\alpha_{g}^{2}(\alpha_{\bar{q}}-\alpha_{q})\left((\kappa-\kappa^{+})+(\zeta_{1}-\zeta_{1}^{+})(1-2\alpha_{g})+\zeta_{2}(3-4\alpha_{g})\right),$
$\displaystyle{\cal T}_{3}(\alpha_{i})$ $\displaystyle=$
$\displaystyle-120(3\zeta_{2}-\zeta_{2}^{+})(\alpha_{\bar{q}}-\alpha_{q})\alpha_{\bar{q}}\alpha_{q}\alpha_{g},$
$\displaystyle{\cal T}_{4}(\alpha_{i})$ $\displaystyle=$ $\displaystyle
30\alpha_{g}^{2}(\alpha_{\bar{q}}-\alpha_{q})\left((\kappa+\kappa^{+})+(\zeta_{1}+\zeta_{1}^{+})(1-2\alpha_{g})+\zeta_{2}(3-4\alpha_{g})\right),$
$\displaystyle{\cal S}(\alpha_{i})$ $\displaystyle=$ $\displaystyle
30\alpha_{g}^{2}\\{(\kappa+\kappa^{+})(1-\alpha_{g})+(\zeta_{1}+\zeta_{1}^{+})(1-\alpha_{g})(1-2\alpha_{g})+\zeta_{2}[3(\alpha_{\bar{q}}-\alpha_{q})^{2}-\alpha_{g}(1-\alpha_{g})]\\},$
$\displaystyle\tilde{\cal S}(\alpha_{i})$ $\displaystyle=$
$\displaystyle-30\alpha_{g}^{2}\\{(\kappa-\kappa^{+})(1-\alpha_{g})+(\zeta_{1}-\zeta_{1}^{+})(1-\alpha_{g})(1-2\alpha_{g})+\zeta_{2}[3(\alpha_{\bar{q}}-\alpha_{q})^{2}-\alpha_{g}(1-\alpha_{g})]\\}.$

Numerical values of parameters used in distribution amplitudes are:
$\varphi_{2}(1~{}GeV)=0$, $w^{V}_{\gamma}=3.8\pm 1.8$, $w^{A}_{\gamma}=-2.1\pm
1.0$, $\kappa=0.2$, $\kappa^{+}=0$, $\zeta_{1}=0.4$, $\zeta_{2}=0.3$.

## Appendix B: The explicit expression of $\Delta^{QCD}$ function

In this Appendix, we give the explicit expression for the function
$\Delta^{QCD}$ acquired from the light-cone QCD sum rule in Sect. II. It is
obtained by choosing the $\varepsilon\\!\\!\\!/p\\!\\!\\!/$ Lorentz structure
as follows:

$\displaystyle\Delta^{QCD}$
$\displaystyle=\frac{1}{9830400\,\pi^{8}}\Big{[}-m_{b}^{13}\Big{(}(18e_{d}+e_{b}+18e_{u})\,I[-7,5]-15(e_{d}+e_{u})\,I[-6,4]\Big{)}-3m_{b}^{11}(25e_{d}+e_{b}+25e_{u})I[-6,5]\Big{]}$
$\displaystyle+\frac{m_{b}^{9}}{14155776\,\pi^{8}}\Big{[}P_{1}(12e_{d}-5e_{b}+12e_{u})+1152\,m_{s}\,\pi^{2}P_{2}(8e_{d}+e_{b}+8e_{u})-576\,m_{s}\,\pi^{2}P_{3}(8e_{d}+e_{b}+8e_{u})\Big{]}$
$\displaystyle\times
I[-5,3]+\frac{m_{b}^{9}\,(e_{d}+e_{u})}{655360\,\pi^{8}}\Big{[}5m_{b}^{2}I[-5,4]-8I[-5,5]\Big{]}+\Big{[}\frac{m_{b}^{9}\,P_{1}(e_{d}+e_{u})}{294912\,\pi^{8}}+\frac{m_{b}^{7}\,m_{s}}{24576\,\pi^{6}}\Big{(}3\,P_{2}\big{(}(3e_{d}+2e_{b}+3e_{u})$
$\displaystyle\times
m_{0}^{2}+4m_{b}^{2}(e_{d}+e_{u})\big{)}-P_{3}\,\big{(}m_{0}^{2}(3e_{d}+2e_{b}+3e_{u})+6\,m_{b}^{2}(e_{d}+e_{u})\big{)}\Big{)}+\frac{m_{b}^{7}\,P_{2}P_{3}}{1536\,\pi^{4}}(3e_{d}+2e_{b}+3e_{u})\Big{]}I[-4,2]$
$\displaystyle+\frac{m_{b}^{7}}{7077888\,\pi^{8}}\Big{[}P_{1}(33\
e_{d}-5e_{b}+33e_{u})+16m_{s}\pi^{2}P_{2}(23e_{d}+72e_{b}+648e_{u})-576\,m_{s}\pi^{2}P_{3}(9e_{d}+e_{b}+9e_{u})\Big{]}$
$\displaystyle\times
I[-4,3]+\frac{m_{b}^{7}}{983040\,\pi^{8}}\Big{[}-15\,m_{b}^{2}\,(e_{u}+e_{d})\,I[-4,4]-(9e_{d}+e_{b}-9e_{u})I[-4,5]\Big{]}+\frac{m_{b}^{5}}{147456\,\pi^{6}}\Big{[}-2m_{0}^{2}m_{b}^{2}\Big{(}P_{1}$
$\displaystyle\times(8e_{d}-e_{b}+8e_{u})+54\,m_{s}P_{2}(e_{d}+e_{u})\Big{)}+P_{3}\Big{(}m_{s}P_{1}(8e_{d}-e_{b}+8e_{u})+36m_{s}m_{0}^{2}m_{b}^{2}(e_{d}+e_{u})+576\pi^{2}P_{2}$
$\displaystyle\times\big{(}m_{0}^{2}(2e_{d}+e_{b}+2e_{u})-m_{b}^{2}(e_{d}+e_{u})\big{)}\Big{)}\Big{]}I[-3,1]+\frac{m_{b}^{5}}{2359296\,\pi^{8}}\Big{[}29e_{u}P_{1}m_{b}^{2}-864e_{b}m_{0}^{2}m_{s}\pi^{2}P_{2}-3456e_{u}P_{2}$
$\displaystyle\times\pi^{2}m_{s}(m_{0}^{2}+m_{b}^{2})+288\pi^{2}P_{3}\Big{(}m_{s}\big{(}m_{0}^{2}(e_{b}+4e_{u})+6e_{u}m_{b}^{2}\big{)}-16\pi^{2}P_{2}(e_{b}+4e_{u})\Big{)}+e_{d}\Big{(}29P_{1}m_{b}^{2}-576\pi^{2}$
$\displaystyle\times\Big{(}m_{b}^{2}m_{s}(6P_{2}-3P_{3})+m_{0}^{2}m_{s}(6P_{2}-2P_{3})+32\pi^{2}P_{2}P_{3}\Big{)}\Big{)}\Big{]}I[-3,2]+\frac{m_{b}^{5}\,(e_{d}+e_{u})}{131072\pi^{8}}\Big{[}P_{1}+64\pi^{2}m_{s}$
$\displaystyle\times(2P_{2}-P_{3})\Big{]}I[-3,3]+\frac{m_{b}^{5}}{655360\,\pi^{8}}\Big{[}10\,m_{b}^{2}\,(e_{d}+e_{u})\,I[-3,4]+(-2e_{d}+e_{b}-2e_{u})\,I[-3,5]\Big{]}+\frac{m_{b}^{3}}{147456\pi^{6}}\Big{[}$
$\displaystyle\times-2m_{s}P_{2}\Big{(}P_{1}(10e_{d}-e_{b}+10e_{u})+108m_{0}^{2}m_{b}^{2}(e_{d}+e_{u})\Big{)}+P_{3}\Big{(}m_{s}\big{(}P_{1}(10e_{d}-e_{b}+10e_{u})+72m_{0}^{2}m_{b}^{2}$
$\displaystyle\times(e_{d}+e_{u})\big{)}+576\pi^{2}P_{2}\big{(}m_{0}^{2}(e_{d}+e_{b}+e_{u})-2m_{b}^{2}(e_{d}+e_{u})\big{)}\Big{)}\Big{]}I[-2,1]+\frac{m_{b}^{3}}{7077888\,\pi^{8}}\Big{[}-9(e_{d}+e_{u})\Big{(}13P_{1}m_{b}^{2}$
$\displaystyle-288\pi^{2}(m_{b}^{2}m_{s}(4P_{2}-2P_{3})+m_{0}^{2}m_{s}(3P_{2}-P_{3})+16\pi^{2}P_{2}P_{3})\Big{)}I[-2,2]+\Big{(}(33e_{d}+5e_{b}+33e_{u})P_{1}$
$\displaystyle+1152m_{s}\pi^{2}P_{2}(e_{d}-e_{b}+e_{u})-576m_{s}\pi^{2}P_{3}(e_{d}-e_{b}+e_{u})\Big{)}I[-2,3]\Big{]}-\frac{m_{b}^{3}}{3276800\pi^{8}}\Big{[}25(e_{d}+e_{u})I[-2,4]+(e_{d}$
$\displaystyle-3e_{b}+e_{u})I[-2,5]\Big{]}-\frac{m_{b}}{147456\,\pi^{6}}\Big{[}2m_{s}P_{2}\Big{(}P_{1}(3\
e_{d}-e_{b}+3e_{u})+54m_{0}^{2}m_{b}^{2}(e_{d}+e_{u})\Big{)}+P_{3}\Big{(}-m_{s}(P_{1}$
$\displaystyle\times(3e_{d}-e_{d}+3e_{u})+36m_{0}^{2}m_{b}^{2}(e_{d}+e_{u}))+576\pi^{2}P_{2}(-e_{b}m_{0}^{2}+m_{b}^{2}(e_{d}+e_{u}))\Big{)}\Big{]}I[-1,1]$
$\displaystyle+\frac{m_{b}}{2359296\,\pi^{8}}\Big{[}288e_{b}m_{0}^{2}m_{s}\pi^{2}P_{2}+m_{b}^{2}(e_{d}+e_{u})\big{(}23P_{1}-1152m_{s}\pi^{2}P_{2}\big{)}+96\pi^{2}P_{3}\Big{(}6m_{b}^{2}m_{s}(e_{d}+e_{u})$
$\displaystyle+e_{b}(-m_{0}^{2}m_{s}+16\pi^{2}P_{2})\Big{)}\Big{]}I[-1,2]+\frac{m_{b}}{14155776\,\pi^{8}}\Big{[}(12e_{d}+12e_{u}+5e_{b})P_{1}+e_{b}576m_{s}\pi^{2}(-2P_{2}+P_{3})\Big{]}$
$\displaystyle\times
I[-1,3]+\frac{m_{b}}{9830400\,\pi^{8}}\Big{[}15m_{b}^{2}(e_{d}+e_{u})I[-1,4]+2e_{b}I[-1,5]\Big{]}+\frac{1}{884736\,m_{b}\,\pi^{6}}\Big{[}m_{s}P_{1}\Big{(}6(e_{d}+e_{u})(10m_{b}^{2}(2P_{2}$
$\displaystyle-
P_{3})+m_{0}^{2}(3P_{2}-P_{3}))+e_{b}m_{0}^{2}(-3P_{2}+P_{3})\Big{)}-16\Big{(}e_{b}(27m_{0}^{4}+P_{1})-3P_{2}P_{3}\pi^{2}(e_{d}+e_{u})(9m_{0}^{4}+2P_{1})\Big{)}\Big{]}I[0,0]$
$\displaystyle+\frac{1}{36864\,m_{b}\,\pi^{6}}\Big{[}m_{s}P_{1}(2P_{2}-P_{3})(e_{b}-8e_{d}-8e_{u})+144m_{0}^{2}P_{2}P_{3}\pi^{2}(4e_{b}+e_{d}+e_{u})\Big{]}I[0,1]$
$\displaystyle+\frac{(e_{d}+e_{u})}{2359296\,m_{b}\,\pi^{8}}\Big{[}-5m_{b}^{2}P_{1}I[0,2]+4\pi^{2}(11P_{1}+960m_{s}(2P_{2}-P_{3}))I[0,3]\Big{]}+\frac{2e_{b}-21(e_{d}+e_{u})}{614400\,m_{b}\,\pi^{8}}I[0,5]$
$\displaystyle-\frac{m_{b}}{147456\,\pi^{6}}\Big{[}(e_{b}-11e_{d}-11e_{u})m_{s}P_{1}(2P_{2}-P_{3})+576e_{b}m_{0}^{2}P_{2}P_{3}\pi^{2}\Big{]}I[1,0]$
$\displaystyle-\frac{m_{b}^{7}\,f_{3\gamma}}{393216\,\pi^{6}}\Big{[}\Big{(}(e_{d}+e_{u})I_{1}[\mathcal{A}]+(e_{d}-e_{u})I_{1}[\mathcal{V}]\Big{)}\Big{(}m_{b}^{2}I[-5,4]-3I[-4,4]\Big{)}\Big{]}+\frac{m_{b}^{5}\,f_{3\gamma}}{4718592\,\pi^{6}}\Big{[}\Big{(}((7e_{d}-4e_{s}-21e_{u})P_{1}$
$\displaystyle-576(e_{d}-3e_{u})m_{s}\pi^{2}(2P_{2}-P_{3}))I_{1}[\mathcal{A}]+(-3e_{d}P_{1}+11e_{u}P_{1}+576\pi^{2}(e_{d}-e_{u})m_{s}(2P_{2}-P_{3}))I_{1}[\mathcal{V}]\Big{)}I[-3,2]$
$\displaystyle-36\Big{(}(e_{d}+e_{u})I_{1}[\mathcal{A}]+(e_{d}-e_{u})I_{1}[\mathcal{V}]\Big{)}I[-3,4]\Big{]}+\frac{m_{b}^{3}\,f_{3\gamma}}{4718592\,\pi^{6}}\Big{[}\Big{(}192\pi^{2}(5e_{d}-3e_{u})\big{(}m_{0}^{2}m_{s}(3P_{2}-P_{3})+16P_{2}P_{3}\pi^{2}\big{)}$
$\displaystyle\times
I[-2,1]+\big{(}(-21e_{d}+4e_{s}+35e_{u})P_{1}+576\pi^{2}(3e_{d}-5e_{u})m_{s}(2P_{2}-P_{3})\big{)}I[-2,2]\Big{)}I_{1}[\mathcal{A}]+(192\pi^{2}(e_{d}-e_{u})\big{(}m_{0}^{2}m_{s}$
$\displaystyle\times(3P_{2}-P_{3})+16\pi^{2}P_{2}P_{3}\big{)}I[-2,1]-\big{(}(e_{d}+15e_{u})P_{1}+576\pi^{2}(e_{d}-e_{u})m_{s}(2P_{2}-P_{3})\big{)}I[-2,2])I_{1}[\mathcal{V}]\Big{]}$
$\displaystyle+\frac{f_{3\gamma}}{24576\,m_{b}\pi^{4}}\Big{(}m_{0}^{2}m_{s}(3P_{2}-P_{3})+16\pi^{2}P_{2}P_{3}\Big{)}\Big{(}(e_{d}+e_{u})I_{1}[\mathcal{A}]+(e_{d}-e_{u})I_{1}[\mathcal{V}]\Big{)}I[0,1],$
(14)

where $P_{1}=\langle g_{s}^{2}G^{2}\rangle$ is gluon condensate,
$P_{2}=\langle\bar{q}q\rangle$ stands for u/d quark condensate, and
$P_{3}=\langle\bar{s}s\rangle$ represents s-quark condensate. It should be
noted that in Eq. (14), for the sake of brevity, we only give expressions that
make substantial contributions to the numerical value of the magnetic moment
and do not present much higher-dimensional contributions, even though they are
considered in numerical calculations.

The functions $I[n,m]$ and $I_{1}[\mathcal{A}]$ are defined as:

$\displaystyle I[n,m]$
$\displaystyle=\int_{m_{b}^{2}}^{s_{0}}ds\int_{m_{b}^{2}}^{s}dl~{}e^{-s/M^{2}}~{}\frac{(s-l)^{m}}{l^{n}},$
$\displaystyle I_{1}[\mathcal{F}]$ $\displaystyle=\int
D_{\alpha_{i}}\int_{0}^{1}dv~{}\mathcal{F}(\alpha_{\bar{q}},\alpha_{q},\alpha_{g})\delta(\alpha_{\bar{q}}+v\alpha_{g}-u_{0}),$
(15)

where $\mathcal{F}$ stands for the corresponding photon DAs.

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The Nepali Math. Sc. Report
Vol. 36, No.1, 2019

# A priori estimates in terms of the maximum norm for the solution of the
Navier-Stokes equations with periodic initial data

Santosh Pathak Department of Mathematics and Statistics
University of New Mexico, Albuquerque, NM 87131, USA
<EMAIL_ADDRESS>
Received May 31, 2016

Abstract: In this paper, we consider the Cauchy problem for the incompressible
Navier-Stokes equations in $\mathbb{R}^{n}$ for $n\geq 3$ with smooth periodic
initial data and derive a priori estimtes of the maximum norm of all
derivatives of the solution in terms of the maximum norm of the initial data.
This paper is a special case of a paper by H-O Kreiss and J. Lorenz which also
generalizes the main result of their paper to higher dimension.
Key Words: Incompressible Navier-Stokes equation; Maximum norm estimates,
Periodic initial data
AMS (MOS) Subject Classification. Classification here.

##  1\. Introduction

We consider the Cauchy problem of the Navier-Stokes equations in
$\mathbb{R}^{n},n\geq 3$:

(1.1) $\displaystyle u_{t}+u\cdot\nabla u+\nabla p=\triangle u,\hskip
15.0pt\nabla\cdot u=0,$

with initial condition

(1.2) $\displaystyle u(x,0)=f(x),\hskip 15.0ptx\in\mathbb{R}^{n},$

where $u=u(x,t)=(u_{1}(x,t),\cdots u_{n}(x,t))$ and $p=p(x,t)$ stand for the
unknown velocity vector field of the fluid and its pressure, while
$f=f(x)=(f_{1}(x),\cdots f_{n}(x))$ is the given initial velocity vector
field. In what follows, we will use the same notations for the space of vector
valued and scalar functions for convenience in writing.

There is a large literature on the existence and uniqueness of solution of the
Navier-Stokes equations in $\mathbb{R}^{n}$. For given initial data, solutions
of (1.1) and (1.2) have been constructed in various function spaces. For
example, if $f\in L^{r}$ for some $r$ with $3\leq r<\infty$, then it is well
known that there is a unique classical solution in some maximum interval of
time $0\leq t<T_{f}$ where $0<T_{f}\leq\infty$. But for the uniqueness of the
pressure one requires $|p(x,t)|\to 0$ as $|x|\to\infty$. (See [6] and [11] for
$r=3$ and [1] for $3<r<\infty$.)

If $f\in L^{\infty}(\mathbb{R}^{n})$ then existence of a regular solution
follows from [2]. The solution is only unique if one puts some growth
restrictions on the pressure as $|x|\to\infty$. A simple example of non-
uniqueness is demonstrated in [7] where the velocity $u$ is bounded but
$|p(x,t)|\leq C|x|$. In addition, an estimate $|p(x,t)|\leq C(1+|x|^{\sigma})$
with $\sigma<1$ ( see [3] ) implies uniqueness. Also the assumption $p\in
L^{1}_{loc}(0,T;BMO)$ (see [5]) implies uniqueness.

In this paper we consider the initial function $f\in
C^{\infty}_{per}(\mathbb{R}^{n})$ which is the space of smooth $2\pi$ periodic
functions. Since $C^{\infty}_{per}(\mathbb{R}^{n})$ is a closed subspace of
the Banach space $L^{\infty}(\mathbb{R}^{n})$, the existence of a regular
solution of the Navier-Stokes equations (1.1) and (1.2) can be guaranteed by
[2]. In addition, in a paper by Giga and others [5] they consider $f\in
BUC(\mathbb{R}^{n})$ where $BUC(\mathbb{R}^{n})$ is the space of bounded
uniformly continuous functions. In the paper they construct a regular solution
of the Navier-Stokes equations in some maximum interval of time $0\leq
t<T_{f}$ where $T_{f}\leq\infty$. Clearly, our case is also a special case of
their paper where we put extra assumption of smooth periodic on their initial
function $f\in BUC(\mathbb{R}^{n})$. Moreover, for smooth periodic initial
data, the existence of smooth periodic solution is proved by H-O Kreiss and J.
Lorenz in their book [9] for $n=3$ where they use successive iteration using
the the vorticity formulation. On the other hand, Giga and others use
iteration on the integral equation of the transformed abstract ordinary
differential equations to construct a mild solution of the Navier-Stokes
equations and later prove such mild solution is indeed a regular solution
(local in time) of the Navier-Stokes equations (1.1) and (1.2) for $f\in
BUC(\mathbb{R}^{n})$ . Readers are referred to the paper by Giga and others
[5] for details on existence of the smooth periodic solution of the Navier-
Stokes equations of (1.1) and (1.2) for $f\in
C^{\infty}_{per}(\mathbb{R}^{n})$ with necessary alternations in their proofs
of case $f\in BUC(\mathbb{R}^{n})$.

The work in this paper reproves Theorem 4.1 of the Kreiss and Lorenz paper [8]
in periodic case assuming smooth periodic solution exists for some maximum
interval of time $0\leq t<T_{f}$. Since we are in a special case of their
paper, result of Theorem 4.1 must be true for smooth periodic initial data as
well, but what makes our work interesting and different is the approach taken
to handle the pressure term of the Navier-Stokes equations while deriving the
result of Theorem 4.1 of the Kreiss and Lorenz paper as I have adopted in my
first paper [10] Notice, pressure term of the Navier-Stokes equations can be
determined from the Poisson equation

(1.3) $\displaystyle\triangle p=-\nabla\cdot(u\cdot\nabla)u$

which is given by

(1.4) $\displaystyle p=\sum_{i,j}R_{i}R_{j}(u_{i}u_{j}),$

where $R_{i}=(-\triangle)^{-1/2}D_{i}$ is the $i$-th Riesz transform. Since
the Riesz transforms are not bounded in $L^{\infty}(\mathbb{R}^{n})$, the
pressure term $p\in L^{1}_{loc}(0,T;BMO)$ where $BMO$ is the space of
functions of bounded mean oscillation. Because of the non-local nature of the
pressure, the proof of Theorem 4.1 of the Kreiss and Lorenz paper is
complicated, however.

The main objective of this paper is to derive a priori estimates of the
maximum norm of the derivatives of $u$ in terms of the maximum norm of the
initial function, $u(x,0)=f(x)$, assuming the solution to exist and to be
$C^{\infty}_{per}(\mathbb{R}^{n}),n\geq 3$ for $0\leq t<T_{f}$. Before we
start formulating the problem, we introduce the following notations

$\displaystyle|f|_{\infty}=\mathop{\sup}_{x}{|f(x)|}\hskip
15.0pt\text{with}\hskip 15.0pt|f(x)|^{2}=\sum_{i}f_{i}^{2}(x),$

and $D^{\alpha}=D_{1}^{\alpha_{1}}\cdots
D_{n}^{\alpha_{n}},D_{i}=\partial/{\partial x_{i}}\hskip 5.0pt\text{for a
multiindex}\hskip 5.0pt\alpha=(\alpha_{1},\cdots,\alpha_{n})$. In what
follows, for any $j=0,1,\cdots$, if $|\alpha|=j$ then we will denote
$D^{\alpha}$ by $D^{j}$. We also set

$\displaystyle|\mathcal{D}^{j}u(t)|_{\infty}:=|\mathcal{D}^{j}u(\cdot,t)|_{\infty}=\mathop{\max}_{|\alpha|=j}|D^{\alpha}u(\cdot,t)|_{\infty}.$

Clearly, $|\mathcal{D}^{j}u(t)|_{\infty}$ measures all space derivatives of
order $j$ in maximum norm.

Proving the following theorem is the main goal of this paper whereas Kreiss
and Lorenz in their paper [8] prove the same theorem for $f\in
L^{\infty}(\mathbb{R}^{n})$ for $n=3$ from rather difficult approach while
dealing with the pressure term $p(x,t)$.

######  Theorem 1.1.

Consider the Cauchy problem for the Navier-Stokes equations (1.1), (1.2),
where $f\in C^{\infty}_{per}(\mathbb{R}^{n})$ for $n\geq 3$ with $\nabla\cdot
f=0$. There is a constant $c_{0}>0$ and for every $j=0,1,\cdots$ there is a
constant $K_{j}$ so that

(1.5) $\displaystyle t^{j/2}|\mathcal{D}^{j}u(t)|_{\infty}\leq
K_{j}|f|_{\infty}\hskip 15.0pt\text{for}\hskip
15.0pt0<t\leq\frac{c_{0}}{|f|_{\infty}^{2}}.$

The constants $c_{0}$ and $K_{j}$ are independent of $t$ and $f$.

For the purpose of proving Theorem 1.1, we start by transforming the momentum
equation (1.1) of the Navier-Stokes equations into the abstract ordinary
differential equation for $u$

(1.6) $\displaystyle u_{t}=\triangle u-\mathbb{P}(u\cdot\nabla)u$

by eliminating the pressure, where $\mathbb{P}$ is the Leray projector defined
by

$\displaystyle\mathbb{P}=(\mathbb{P}_{ij})_{1\leq i,j\leq n},\hskip
15.0pt\mathbb{P}_{ij}=\delta_{ij}+R_{i}R_{j};$

where $R_{i}$ is same as in (1.4) and $\delta_{ij}$ is the Kronecker delta
function. Note that the equation (1.6) is obtained from (1.1 ) by applying the
Leray projector with the properties $\mathbb{P}(\nabla
p)=0,\mathbb{P}(\triangle u)=\triangle u$, since $\nabla\cdot u=0$.

Since $\mathbb{P}(u\cdot\nabla u)=\sum_{i}D_{i}\mathbb{P}(u_{i}u)$, therefore
it is very appropriate to consider an analogous system of (1.6) as below:

(1.7) $\displaystyle u_{t}=\triangle u+D_{i}\mathbb{P}g(u)\hskip
15.0ptx\in\mathbb{R}^{n},\hskip 15.0ptt>0$

with initial condition

(1.8) $\displaystyle u(x,0)=f(x)\hskip 15.0pt\text{where}\hskip 15.0ptf\in
C^{\infty}_{per}(\mathbb{R}^{n}).$

Here $g:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is assumed to be quadratic in $u$.
The maximal interval of existence is again $0\leq t<T_{f}$. We would like to
prove the estimates of the maximum norm of the derivatives of the solution of
(1.7) and (1.8) in terms of the maximum norm of the initial data.

######  Theorem 1.2.

Under the above assumptions on $f$ and $g$ the solution of (1.7) and (1.8)
satisfies the following
(a) There is a constant $c_{0}>0$ with

(1.9) $\displaystyle T(f)>\frac{c_{0}}{|f|_{\infty}^{2}}$

and

(1.10) $\displaystyle|u(t)|_{\infty}\leq 2|f|_{\infty}\hskip
15.0pt\text{for}\hskip 15.0pt0\leq t\leq\frac{c_{0}}{|f|_{\infty}^{2}}.$

(b) For every $j=1,2,\cdots$, there is a constant $K_{j}>0$ with

(1.11) $\displaystyle t^{j/2}|\mathcal{D}^{j}u(.,t)|_{\infty}\leq
K_{j}|f|_{\infty}\hskip 15.0pt\text{for}\hskip
15.0pt0<t\leq\frac{c_{0}}{|f|_{\infty}^{2}}.$

The constant $c_{0}$ and $K_{j}$ are independent of $t$ and $f$.

In section 2, we will introduce some auxiliary results for the solution of the
heat equation and few other important estimates which are used later in
section 3 and 4. Proof of Theorem 1.2 will be provided in section 3. Then we
prove Theorem 1.1 in section 4. Finally, in section 5 we outline some remarks
on the use of the result obtained in Theorem 1.1.

##  2\. Some Auxiliary results

Let us consider $f\in C^{\infty}_{per}(\mathbb{R}^{n})$. The solution of

$\displaystyle u_{t}=\triangle u,\hskip 15.0ptu=f\hskip 15.0pt\text{at}\hskip
15.0ptt=0,$

is denoted by

$\displaystyle u(t):=u(\cdot,t)=e^{\triangle
t}f=\frac{1}{(2\pi)^{n}}\int_{\mathbb{T}^{n}}\theta(x-y,t)f(y)dy$

where

(2.1)
$\displaystyle\theta(x,t)=\sum_{k\in\mathbb{Z}^{n}}e^{-|k|^{2}t}e^{ik\cdot
x},\hskip 15.0ptt>0$

is the periodic heat kernel in $\mathbb{R}^{n}$. Using the Poisson summation
formula, (2.1) can be written as

(2.2)
$\displaystyle\theta(x,t)=\sum_{k\in\mathbb{Z}^{n}}\bigg{(}\frac{\pi}{t}\bigg{)}^{n}\exp\bigg{[}\frac{-|x+2\pi
k|^{2}}{4t}\bigg{]},\hskip 15.0ptt>0.$

With the use of (2.2), it is well known that

(2.3) $\displaystyle|e^{t\triangle}f|_{\infty}\leq|f|_{\infty},\hskip
15.0ptt\geq 0$

and

(2.4) $\displaystyle|\mathcal{D}^{j}e^{t\triangle}f|_{\infty}\leq
C_{j}t^{-j/2}|f|_{\infty}$

for some $C_{j}>0$ independent of $t$ and $f$.

######  Lemma 2.1.

Let $f\in C^{\infty}_{per}(\mathbb{R}^{n})$ then for any $j\geq 1$

(2.5) $\displaystyle|\mathcal{D}^{j}e^{\triangle t}\mathbb{P}f|_{\infty}\leq
C_{j}t^{-j/2}|f|_{\infty}\hskip 15.0pt\text{for}\hskip 15.0ptt>0$

for some constant $C_{j}>0$ independent of $t$ and $f$.

###### Proof.

Let us first denote $e^{\triangle t}f=\theta*f$ where $\theta(x,t)$ is given
by (2.2). Let us denote the Fourier coefficient of a function by
$\mathcal{F}$. For $\xi\in\mathbb{Z}^{n}$, notice
$\mathcal{F}({\theta(x,t)})(\xi)=Ce^{-t|\xi|^{2}},t>0$, where $C$ depends on
the normalizing constant in the definition of the Fourier coefficient. In the
proof of this lemma, we will allow the constant $C$ to change line to line as
per the need. Now, for any $t>0$, any choice of $k,l\in\\{1,2,\cdots,n\\}$;
and for any multiindex $\alpha$ such that $|\alpha|=j$, the operator
$D^{j}e^{\triangle t}\mathbb{P}_{kl}$ on the Fourier side is given by

$\displaystyle\mathcal{F}(D^{j}e^{\triangle t}\mathbb{P}_{kl}f_{l})(\xi)$
$\displaystyle=(-i\xi)^{\alpha}\mathcal{F}(e^{\triangle
t}\mathbb{P}_{kl}f_{l})(\xi)$
$\displaystyle=(-i\xi)^{\alpha}\mathcal{F}(\theta*\mathbb{P}_{kl}f_{l})(\xi)$
$\displaystyle=C(-i\xi)^{\alpha}\mathcal{F}(\theta(x,t))(\xi)\mathcal{F}(\mathbb{P}_{kl}f_{l})(\xi)$
$\displaystyle=C(-i\xi)^{\alpha}e^{-t|\xi|^{2}}\bigg{(}\delta_{kl}-\frac{\xi_{k}\xi_{l}}{|\xi|^{2}}\bigg{)}\mathcal{F}(f_{l})(\xi)$
$\displaystyle=C(-i\xi)^{\alpha}e^{-t|\xi|^{2}}\delta_{kl}\mathcal{F}(f_{l})(\xi)$
$\displaystyle-C(-i\xi)^{\alpha}\xi_{k}\xi_{l}\mathcal{F}(f_{l})(\xi)\int_{t}^{\infty}e^{-\tau|\xi|^{2}}d\tau.$

Using Fourier expansion we can write

$\displaystyle D^{j}(e^{\triangle t}\mathbb{P}_{kl}f_{l})(x)$
$\displaystyle=C\sum_{\xi\in\mathbb{Z}^{n}}(-i\xi)^{\alpha}\delta_{kl}e^{-t|\xi|^{2}}\mathcal{F}(f_{l})(\xi)e^{i\xi\cdot
x}$
$\displaystyle+C\sum_{\xi\in\mathbb{Z}^{n}}(-i\xi)^{\alpha}(i\xi_{k})(i\xi_{l})\mathcal{F}(f_{l})(\xi)e^{i\xi\cdot
x}\int_{t}^{\infty}e^{-\tau|\xi|^{2}}d\tau$
$\displaystyle=(-1)^{j}C\delta_{kl}D^{\alpha}\sum_{\xi\in\mathbb{Z}^{n}}e^{-t|\xi|^{2}}\mathcal{F}(f_{l})(\xi)e^{i\xi\cdot
x}$
$\displaystyle+C(-1)^{j}\int_{t}^{\infty}\sum_{\xi\in\mathbb{Z}^{n}}e^{-\tau|\xi|^{2}}(i\xi)^{\alpha}(i\xi_{k})(i\xi_{l})\mathcal{F}(f_{l})(\xi)e^{i\xi\cdot
x}d\tau$ $\displaystyle=(-1)^{j}C\delta_{kl}D^{\alpha}e^{\triangle
t}f_{l}+(-1)^{j}C\int_{t}^{\infty}D^{\alpha}D_{k}D_{l}e^{\triangle\tau}f_{l}d\tau$
$\displaystyle=I_{1}+I_{2}.$

From (2.4) we have $|I_{1}|_{\infty}\leq C_{j}t^{-j/2}|f_{l}|_{\infty}$. By
the use of (2.4) one more time we obtain

$\displaystyle|I_{2}|_{\infty}$ $\displaystyle\leq
C_{j}|f_{l}|_{\infty}\int_{t}^{\infty}\tau^{-(j+2)/2}d\tau$ $\displaystyle\leq
C_{j}t^{-j/2}|f_{l}|_{\infty}.$

Therefore

$\displaystyle|D^{j}e^{\triangle t}\mathbb{P}_{kl}f_{l}|_{\infty}$
$\displaystyle\leq|I_{1}|_{\infty}+|I_{2}|_{\infty}$ $\displaystyle\leq
C_{j}t^{-j/2}|f_{l}|_{\infty}.$

Hence Lemma 2.1 is proved. ∎

######  Corollary 2.2.

Let $g\in C^{\infty}_{per}(\mathbb{R}^{n}\times[0,T])$ for some $T>0$, then
the solution of

(2.6) $\displaystyle u_{t}=\triangle u+D_{i}\mathbb{P}g,\hskip 15.0ptu=0\hskip
15.0pt\text{at}\hskip 15.0ptt=0$

satisfies

(2.7) $\displaystyle|u(t)|_{\infty}\leq Ct^{1/2}\mathop{\max}_{0\leq s\leq
t}|g(s)|_{\infty}.$

###### Proof.

The solution of (2.6) is given by

$\displaystyle u(t)=\int_{0}^{t}e^{\triangle(t-s)}D_{i}\mathbb{P}g(u))(s)ds$

and

$\displaystyle|u(t)|_{\infty}\leq\int_{0}^{t}|e^{\triangle(t-s)}D_{i}\mathbb{P}g(u)(s)|_{\infty}ds.$

After commuting $D_{i}$ with the heat semi-group, we can use Lemma 2.1 to
obtain

$\displaystyle|u(t)|_{\infty}\leq\mathop{\max}_{0\leq s\leq
t}|g(s)|_{\infty}\int_{0}^{t}(t-s)^{-1/2}ds.$

Hence we obtain

$\displaystyle|u(t)|_{\infty}\leq Ct^{1/2}\mathop{\max}_{0\leq s\leq
t}|g(s)|_{\infty}.$

∎

##  3\. Estimates for $u_{t}=\triangle u+D_{i}\mathbb{P}g(u)$ : proof of
Theorem 1.2

In this section we consider the system $u_{t}=\triangle u+D_{i}\mathbb{P}g(u)$
with the initial condition $u=f$ at $t=0$ where $f\in
C^{\infty}_{per}(\mathbb{R}^{n})$. It is well-known that the solution is
smooth $2\pi$ periodic in a maximal interval $0\leq t<T_{f}$ where
$0<T_{f}\leq\infty$.

Let us consider $u$ is the solution of the inhomogeneous equation
$u_{t}=\triangle u+D_{i}\mathbb{P}(g(u(x,t)))$ and recall $g(u)$ is quadratic
in $u$. Thus, there is a constant $C_{g}$ such that we have the following:

(3.1) $\displaystyle|g(u)|\leq C_{g}|u|^{2},\hskip 15.0pt|g_{u}(u)|\leq
C_{g}|u|,\hskip 15.0pt\text{for all}\hskip 15.0ptu\in\mathbb{R}^{n}$

We first estimate the maximum norm of $u$.

######  Lemma 3.1.

Let $C_{g}$ denote the constant in (3.1) and let $C$ denote the constant in
(2.7); set $c_{0}=\frac{1}{16C^{2}C_{g}^{2}}$. Then we have
$T_{f}>c_{0}/{|f|_{\infty}^{2}}$ and

(3.2) $\displaystyle|u(t)|_{\infty}<2|f|_{\infty}\hskip 15.0pt\text{ for
}\hskip 15.0pt0\leq t<\frac{c_{0}}{|f|_{\infty}^{2}}.$

###### Proof.

Suppose (3.2) does not hold, then we can find the smallest time $t_{0}$ such
that $|u(t_{0})|_{\infty}=2|f|_{\infty}$. Since $t_{0}$ is the smallest time
so we have $t_{0}<c_{0}/|f|_{\infty}^{2}$. Now by (2.3) and (2.7) we have

$\displaystyle 2|f|_{\infty}$ $\displaystyle=|u(t_{0})|_{\infty}$
$\displaystyle\leq|f|_{\infty}+Ct_{0}^{1/2}\mathop{\max}_{0\leq s\leq
t_{0}}|g(s)|_{\infty}$
$\displaystyle\leq|f|_{\infty}+CC_{g}t_{0}^{1/2}\mathop{\max}_{0\leq s\leq
t_{0}}|u(s)|_{\infty}^{2}$
$\displaystyle\leq|f|_{\infty}+CC_{g}t_{0}^{1/2}4|f|_{\infty}^{2}.$

This gives

$\displaystyle 1\leq 4CC_{g}t_{0}^{1/2}|f|_{\infty},$

therefore $t_{0}\geq
1/{(16C^{2}C_{g}^{2}|f|_{\infty}^{2})}={c_{0}}/{|f|_{\infty}^{2}}$ which is a
contradiction. There (3.2) must hold. The estimate
$T_{f}>c_{0}/{|f|_{\infty}^{2}}$ is valid since $\limsup_{t\to
T_{f}}|u(t)|_{\infty}=\infty$ if $T_{f}$ is finite. ∎

Now we prove estimate (1.11) of Theorem 1.2 by induction on $j$. Let $j\geq
1$, and assume

(3.3) $\displaystyle t^{k/2}|\mathcal{D}^{k}u(t)|_{\infty}\leq
K_{k}|f|_{\infty},\hskip 15.0pt\text{for}\hskip 15.0pt0\leq
t\leq\frac{c_{0}}{|f|_{\infty}^{2}}\hskip 15.0pt\text{and}\hskip 15.0pt0\leq
k\leq j-1.$

Let us apply $D^{j}$ to the equation $u_{t}=\triangle u+D_{i}\mathbb{P}g(u)$
to obtain

$\displaystyle v_{t}=\triangle v+D^{j+1}\mathbb{P}g(u),\hskip
15.0ptv:=D^{j}u,$ $\displaystyle v(t)=D^{j}e^{\triangle
t}f+\int_{0}^{t}e^{\triangle(t-s)}D^{j+1}(\mathbb{P}g(u))(s)ds.$

Using (2.4) we get

(3.4) $\displaystyle t^{j/2}|v(t)|_{\infty}\leq
C|f|_{\infty}+t^{j/2}\biggl{|}\int_{0}^{t}e^{\triangle(t-s)}D^{j+1}(\mathbb{P}g(u))(s)ds\biggr{|}_{\infty}.$

We split the integral into

$\displaystyle\int_{0}^{t/2}+\int_{t/2}^{t}=:I_{1}+I_{2}$

and obtain

$\displaystyle|I_{1}(t)|$
$\displaystyle=\biggl{|}\int_{0}^{t/2}D^{j+1}e^{\triangle(t-s)}(\mathbb{P}g(u))(s)ds\biggr{|}_{\infty}$
$\displaystyle\leq\int_{0}^{t/2}|D^{j+1}e^{\triangle(t-s)}(\mathbb{P}g(u))(s)ds|_{\infty}ds.$

Using the inequality (2.5) in Lemma 2.1, we get

$\displaystyle|I_{1}(t)|_{\infty}$ $\displaystyle\leq
C\int_{0}^{t/2}(t-s)^{-(j+1)/2}|g(u(s))|_{\infty}ds$ $\displaystyle\leq
C|f|_{\infty}^{2}t^{(1-j)/2}.$

The integrand in $I_{2}$ has singularity at $s=t$. Therefore, we can move only
one derivative from $D^{j+1}\mathbb{P}g(u)$ to the heat semigroup.( If we move
two or more derivatives then the singularity becomes non-integrable.) Thus, we
have

$\displaystyle|I_{2}(t)|_{\infty}=\biggl{|}-\int_{t/2}^{t}De^{\triangle(t-s)}(D^{j}\mathbb{P}g(u))(s)ds\biggr{|}_{\infty}.$

Since the Leray projector commutes with any order derivatives, therefore

$\displaystyle|I_{2}(t)|_{\infty}=\bigg{|}-\int_{t/2}^{t}De^{\triangle(t-s)}(\mathbb{P}D^{j}g(u))(s)ds\bigg{|}_{\infty}.$

If we use Lemma 2.1 for $j=1$, we obtain

(3.5) $\displaystyle|I_{2}(t)|_{\infty}\leq
C\int_{t/2}^{t}(t-s)^{-1/2}|D^{j}g(u)(s)|_{\infty}ds.$

Since $g(u)$ is quadratic in $u$, therefore

$\displaystyle|D^{j}g(u)|_{\infty}\leq
C|u|_{\infty}|\mathcal{D}^{j}u|_{\infty}+\sum_{k=1}^{j-1}|\mathcal{D}^{k}u|_{\infty}|\mathcal{D}^{j-k}u|_{\infty}.$

By induction hypothesis (3.3) we obtain

(3.6)
$\displaystyle\sum_{k=1}^{j-1}|\mathcal{D}^{k}u(s)|_{\infty}|\mathcal{D}^{j-k}u(s)|_{\infty}\leq
Cs^{-j/2}|f|_{\infty}^{2}.$

Expression in (3.5) can be estimated as below:

$\displaystyle|I_{2}(t)|_{\infty}$ $\displaystyle\leq
C\int_{t/2}^{t}(t-s)^{-1/2}\bigg{(}C|u(s)|_{\infty}|\mathcal{D}^{j}u(s)|_{\infty}+\sum_{k=1}^{j-1}|\mathcal{D}^{k}u(s)|_{\infty}|\mathcal{D}^{j-k}u(s)|_{\infty}\bigg{)}ds$
$\displaystyle=J_{1}+J_{2}.$

Using (3.6), and since $\int_{t/2}^{t}(t-s)^{-1/2}s^{-j/2}ds=Ct^{(1-j)/2}$,
where $C$ is independent of $t$, we obtain $|J_{2}(t)|_{\infty}\leq
C|f|_{\infty}^{2}t^{(1-j)/2}$.
For $J_{1}$, we have

$\displaystyle|J_{1}(t)|_{\infty}$
$\displaystyle=C\int_{t/2}^{t}(t-s)^{-1/2}|u(s)|_{\infty}|\mathcal{D}^{j}u(s)|_{\infty}ds$
$\displaystyle\leq
C|f|_{\infty}\int_{t/2}^{t}(t-s)^{-1/2}s^{-j/2}s^{j/2}|\mathcal{D}^{j}u(s)|_{\infty}ds$
$\displaystyle\leq C|f|_{\infty}t^{(1-j)/2}\mathop{\max}_{0\leq s\leq
t}\\{s^{j/2}\mathcal{D}^{j}u(s)|_{\infty}\\}.$

We use these bounds to bound the integral in (3.4). We have $v=D^{j}u$. Then
maximizing the resulting estimate for $t^{j/2}|D^{j}u(t)|_{\infty}$ over all
derivatives $D^{j}$ of order $j$ and setting

$\displaystyle\phi(t):=t^{j/2}|\mathcal{D}^{j}u(t)|_{\infty}$

and from (3.4), we obtain the following estimate

$\displaystyle\phi(t)\leq
C|f|_{\infty}+Ct^{1/2}|f|_{\infty}^{2}+C|f|_{\infty}t^{1/2}\mathop{\max}_{0\leq
s\leq t}\phi(s)\hskip 15.0pt\text{for}\hskip 15.0pt0\leq
t\leq\frac{c_{0}}{|f|_{\infty}^{2}}.$

Since $t^{1/2}|f|_{\infty}\leq\sqrt{c_{0}}$ then $Ct^{1/2}|f|_{\infty}^{2}\leq
C\sqrt{c_{0}}|f|_{\infty}$. Therefore

(3.7) $\displaystyle\phi(t)\leq
C_{j}|f|_{\infty}+C_{j}|f|_{\infty}t^{1/2}\mathop{\max}_{0\leq s\leq
t}\phi(s)\hskip 15.0pt\text{for}\hskip 15.0pt0\leq t\leq
c_{0}/{|f|_{\infty}^{2}}.$

Let us fix $C_{j}$ so that the above estimate holds, and set

$\displaystyle c_{j}=\min\bigg{\\{}c_{0},\frac{1}{4C_{j}^{2}}\bigg{\\}}.$

First, let us prove the following

$\displaystyle\phi(t)<2C_{j}|f|_{\infty}\hskip 15.0pt\text{for}\hskip
15.0pt0\leq t<\frac{c_{j}}{|f|_{\infty}^{2}}.$

Suppose there is a smallest time $t_{0}$ such that
$0<t_{0}<c_{j}/{|f|_{\infty}^{2}}$ with $\phi(t_{0})=2C_{j}|f|_{\infty}$. Then
using (3.7) we obtain

$\displaystyle 2C_{j}|f|_{\infty}=\phi(t_{0})\leq
C_{j}|f|_{\infty}+2C_{j}^{2}|f|_{\infty}^{2}t_{0}^{1/2},$

thus

$\displaystyle 1\leq 2C_{j}|f|_{\infty}t_{0}^{1/2}\hskip
15.0pt\text{gives}\hskip 15.0ptt_{0}\geq c_{j}/{|f|_{\infty}^{2}}$

which contradicts the assertion. Therefore, we proved the estimate

(3.8) $\displaystyle t^{j/2}|\mathcal{D}^{j}u(t)|_{\infty}\leq
2C_{j}|f|_{\infty}\hskip 15.0pt\text{for}\hskip 15.0pt0\leq t\leq
c_{j}/{|f|_{\infty}^{2}}.$

If

(3.9) $\displaystyle
T_{j}:=\frac{c_{j}}{|f|_{\infty}^{2}}<t\leq\frac{c_{0}}{|f|_{\infty}^{2}}=:T_{0}$

then we start the corresponding estimate at $t-T_{j}$. Using Lemma 3.1, we
have $|u(t-T_{j})|_{\infty}\leq 2|f|_{\infty}$ and obtain

(3.10) $\displaystyle T_{j}^{j/2}|\mathcal{D}^{j}u(t)|_{\infty}\leq
4C_{j}|f|_{\infty}.$

Finally, for any $t$ satisfying (3.9)

$\displaystyle t^{j/2}\leq
T_{0}^{j/2}=\bigg{(}\frac{c_{0}}{c_{j}}\bigg{)}^{j/2}T_{j}^{j/2}$

and (3.10) yield

$\displaystyle t^{j/2}|\mathcal{D}^{j}u(t)|_{\infty}\leq
4C_{j}\bigg{(}\frac{c_{0}}{c_{j}}\bigg{)}^{j/2}|f|_{\infty}.$

This completes the proof of Theorem 1.2.

##  4\. Estimates For the Navier-Stokes Equations

Recall the transformed abstract ordinary differential equation (1.6)

(4.1) $\displaystyle u_{t}=\triangle u-\mathbb{P}(u\cdot\nabla u),\hskip
15.0pt\nabla\cdot u=0$

with

(4.2) $\displaystyle u(x,0)=f(x).$

Solution of (4.1) and (4.2) is given by

(4.3) $\displaystyle u(t)=e^{\triangle
t}f-\int_{0}^{t}e^{\triangle(t-s)}\mathbb{P}(u\cdot\nabla u)(s)ds.$

Using (4.3) with previous estimates (2.3), (2.4) and (2.5), we prove the
following lemma.

######  Lemma 4.1.

Set

(4.4) $\displaystyle
V(t)=|u(t)|_{\infty}+t^{1/2}|\mathcal{D}u(t)|_{\infty},\hskip 15.0pt0<t<T(f).$

There is a constant $C>0$, independent of $t$ and $f$, so that

(4.5) $\displaystyle V(t)\leq C|f|_{\infty}+Ct^{1/2}\mathop{\max}_{0\leq s\leq
t}{V^{2}(s)},\hskip 15.0pt0<t<T(f).$

###### Proof.

Using estimate (2.3) of the heat equation in (4.3), we obtain

$\displaystyle|u(t)|_{\infty}$
$\displaystyle\leq|f|_{\infty}+\bigg{|}\int_{0}^{t}e^{\triangle(t-s)}\mathbb{P}(u\cdot\nabla
u)(s)ds\bigg{|}_{\infty}.$

Apply identity $\mathbb{P}(u\cdot\nabla u)=\sum_{i}D_{i}\mathbb{P}(u_{i}u)$
with the fact, heat semi-group commutes with $D_{i}$, then use of inequality
(2.5) in Lemma 2.1 for $j=1$ to proceed

$\displaystyle|u(t)|_{\infty}$
$\displaystyle\leq|f|_{\infty}+C\int_{0}^{t}(t-s)^{-1/2}|u(s)|_{\infty}^{2}ds$
$\displaystyle=|f|_{\infty}+C\int_{0}^{t}(t-s)^{-1/2}s^{-1/2}s^{1/2}|u(s)|_{\infty}^{2}ds$
$\displaystyle\leq|f|_{\infty}+C\mathop{\max}_{0\leq s\leq
t}\\{s^{1/2}|u(s)|_{\infty}^{2}\\}\int_{0}^{t}(t-s)^{-1/2}s^{-1/2}ds.$

Since $\int_{0}^{t}(t-s)^{-1/2}s^{-1/2}ds=C>0$, which is independent of $t$,
we have the following estimate

$\displaystyle|u(t)|_{\infty}$
$\displaystyle\leq|f|_{\infty}+C\mathop{\max}_{0\leq s\leq
t}\\{s^{1/2}|u(s)|_{\infty}^{2}\\}$ (4.6) $\displaystyle|u(t)|_{\infty}$
$\displaystyle\leq|f|_{\infty}+Ct^{1/2}\mathop{\max}_{0\leq s\leq
t}{V^{2}(s)}.$

Apply $D_{i}$ to (4.1), and the Duhamel’s principle to obtain

(4.7) $\displaystyle v(t)=D_{i}e^{\triangle
t}f-\int_{0}^{t}e^{\triangle(t-s)}D_{i}\mathbb{P}(u\cdot\nabla)u(s)ds.$

We can estimate the integral in (4.7) using Lemma 2.1 for $j=1$ in the
following way:

$\displaystyle\bigg{|}\int_{0}^{t}D_{i}e^{\triangle(t-s)}\mathbb{P}(u\cdot\nabla
u)(s)ds\bigg{|}$
$\displaystyle\leq\int_{0}^{t}|D_{i}e^{\triangle(t-s)}\mathbb{P}(u\cdot\nabla
u)(s)|ds$ $\displaystyle\leq
C\int_{0}^{t}(t-s)^{-1/2}|u(s)|_{\infty}|\mathcal{D}u(s)|_{\infty}ds$
$\displaystyle=C\int_{0}^{t}(t-s)^{-1/2}s^{-1/2}s^{1/2}|u(s)|_{\infty}|\mathcal{D}u(s)|_{\infty}ds$
$\displaystyle\leq C\mathop{\max}_{0\leq s\leq
t}{\\{s^{1/2}|u(s)|_{\infty}|\mathcal{D}u(s)|_{\infty}\\}}\int_{0}^{t}(t-s)^{-1/2}s^{-1/2}ds$
$\displaystyle\leq C\mathop{\max}_{0\leq s\leq
t}{\\{|u(s)|_{\infty}^{2}+s|\mathcal{D}u(s)|_{\infty}^{2}\\}}.$

Therefore, using (2.4) $j=1$ in expression (4.7), we arrive at

$\displaystyle|v(t)|_{\infty}$ $\displaystyle\leq
Ct^{-1/2}|f|_{\infty}+C\mathop{\max}_{0\leq s\leq
t}{\\{|u(s)|_{\infty}^{2}+s|\mathcal{D}u(s)|_{\infty}^{2}\\}}$ (4.8)
$\displaystyle t^{1/2}|\mathcal{D}u(t)|_{\infty}$ $\displaystyle\leq
C|f|_{\infty}+Ct^{1/2}\mathop{\max}_{0\leq s\leq t}{V^{2}(t)}.$

Using (4.6) and (4.8), we have proved Lemma 4.1. ∎

######  Lemma 4.2.

Let $C>0$ denote the constant in estimate (4.5) and set

$\displaystyle c_{0}=\frac{1}{16C^{4}}.$

Then $T_{f}>c_{0}/{|f|_{\infty}^{2}}$ and

(4.9)
$\displaystyle|u(t)|_{\infty}+t^{1/2}|\mathcal{D}u(t)|_{\infty}<2C|f|_{\infty}\hskip
15.0pt\text{for}\hskip 15.0pt0\leq t<\frac{c_{0}}{|f|_{\infty}^{2}}.$

###### Proof.

We prove this lemma by contradiction after recalling the definition of $V(t)$
in (4.4). Suppose that (4.9) does not hold, then denote by $t_{0}$ the
smallest time with $V(t_{0})=2C|f|_{\infty}$. Use (4.5) to obtain

$\displaystyle 2C|f|_{\infty}$ $\displaystyle=V(t_{0})$ $\displaystyle\leq
C|f|_{\infty}+Ct_{0}^{1/2}4C^{2}|f|_{\infty}^{2},$

thus

$\displaystyle 1\leq 4C^{2}t_{0}^{1/2}|f|_{\infty}^{2},$

therefore $t_{0}\geq c_{0}/|f|_{\infty}^{2}$. This contradiction proves (4.9)
and $T_{f}>c_{0}/{|f|_{\infty}^{2}}$. ∎

Lemma 4.2 proves Theorem 1.1 for $j=0$ and $j=1$. By an induction argument as
in the proof of Theorem 1.2 one proves Theorem 1.1 for any $j=0,1,\cdots$

##  5\. Remarks

We can apply estimate (1.5) of Theorem 1.1 for

(5.1) $\displaystyle\frac{c_{0}}{2|f|_{\infty}^{2}}\leq
t\leq\frac{c_{0}}{|f|_{\infty}^{2}}$

and obtain

(5.2) $\displaystyle|\mathcal{D}^{j}u(t)|_{\infty}\leq
C_{j}|f|_{\infty}^{j+1}$

in interval (5.1). Starting the estimate at $t_{0}\in[0,T_{f})$ we have

(5.3) $\displaystyle|\mathcal{D}^{j}u(t_{0}+t)|_{\infty}\leq
C_{j}|u(t_{0})|_{\infty}^{j+1}$

for

(5.4) $\displaystyle\frac{c_{0}}{2|u(t_{0})|_{\infty}^{2}}\leq
t\leq\frac{c_{0}}{|u(t_{0})|_{\infty}^{2}}.$

Then, if $t_{1}$ is fixed with

(5.5) $\displaystyle\frac{c_{0}}{2|f|_{\infty}^{2}}\leq t_{1}<T_{f},$

we can maximize both sides of (5.3) over $0\leq t_{0}\leq t_{1}$ and obtain

(5.6)
$\displaystyle\max\bigg{\\{}|\mathcal{D}^{j}u(t)|_{\infty}:\frac{c_{0}}{2|f|_{\infty}^{2}}\leq
t\leq t_{1}+\tau\bigg{\\}}\leq C_{j}\max\\{|u(t)|_{\infty}^{j+1}:0\leq t\leq
t_{1}\\}$

with

$\displaystyle\tau=\frac{c_{0}}{|u(t_{1})|_{\infty}^{2}}$

Estimate (5.6) says, essentially, that the maximum of the $j$-th derivatives
of $u$ measured by $|\mathcal{D}^{j}u|_{\infty}$ , can be bounded in terms of
$|u|_{\infty}^{j+1}$. The positive value of $\tau$ on the left-hand side of
(5.6) shows that $|u|_{\infty}^{j+1}$ controls $|\mathcal{D}^{j}u|_{\infty}$
for some time into the future.
As is well known, if $(u,p)$ solves the Navier-Stokes equations and
$\lambda>0$ is any scaling parameter, then the functions
$u_{\lambda},p_{\lambda}$ defined by

$\displaystyle u_{\lambda}(x,t)=\lambda u(\lambda x,\lambda^{2}t),\hskip
15.0ptp_{\lambda}(x,t)=\lambda^{2}p(\lambda x,\lambda^{2}t)$

also solve the Navier-Stokes equations. Clearly,

$\displaystyle|u_{\lambda}(t)|_{\infty}=\lambda|u(\lambda^{2}t)|_{\infty},\hskip
15.0pt|\mathcal{D}^{j}u_{\lambda}(t)|_{\infty}=\lambda^{j+1}|\mathcal{D}^{j}u(\lambda^{2}t)|_{\infty}.$

Therefore, $|\mathcal{D}^{j}u|_{\infty}$ and $|u|_{\infty}^{j+1}$ both scale
like $\lambda^{j+1}$, which is, of course, consistent with the estimate (5.6).
We do not know under what assumptions $|u|_{\infty}^{j+1}$ can conversely be
estimated in terms of $|\mathcal{D}^{j}u|_{\infty}$.

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