# Well-posedness for the Non-integrable Periodic Fifth Order KdV in Bourgain
Spaces

Ryan McConnell

###### Abstract.

We study well-posedness for a non-integrable generalization of the fifth order
KdV, the second member in the KdV heirarchy. In particular, we use
differentiation-by-parts to establish well-posedness for $s>35/64$ in low
modulation restricted norm spaces, as well as non-linear smoothing of order
$\varepsilon<\min(2(s-35/64),1)$. As corollaries, we obtain unconditional
well-posedness for the non-integrable fifth order KdV for $s>1$ and global
well-posedness for the integrable fifth order KdV for $s\geq 1$. We also show
local well-posedness for the non-integrable fifth order KdV for $s>1/2$,
contingent upon the conjectured $L^{8}$ Strichartz estimate. As an application
of the nonlinear smoothing we obtain non-trivial upper bounds on the upper
Minkowski dimension of the solution to the non-integrable fifth order KdV.

The author was partially supported by NSF grant DMS-2154031.

###### Contents

  1. 1 Introduction
  2. 2 Notation & Background
  3. 3 Symbol Analysis
  4. 4 Reductions
    1. 4.1 First Differentiation-by-Parts
    2. 4.2 Second Differentiation-by-Parts
  5. 5 Duhamel Estimates & Proof of Theorem 1.1
    1. 5.1 Main Terms in the Duhamel Estimate
    2. 5.2 Proof of Theorem 1.3
  6. 6 Smoothing
    1. 6.1 Auxilliary Smoothing Lemmas
    2. 6.2 Smoothing for the Main terms
  7. 7 Unconditional Well-posedness
  8. 8 Acknowledgements
  9. 9 Appendix: Toy Equation Reduction Justification
    1. 9.1 Resonance Analysis
  10. 10 Appendix: Bounds on the Dimension of solutions to (1)

##  1\. Introduction

We study the local well-posedness of the non-integrable fifth order KdV given
by

(1)
$\displaystyle\begin{cases}u_{t}-\partial_{x}^{5}u+\alpha\partial_{x}(u^{3})+\beta\partial_{x}(\partial_{x}u)^{2}+\gamma\partial_{x}(u\partial_{x}^{2}u)=0\\\
u(x,0)=u_{0}\in H^{s}(\mathbb{T}),\end{cases}$

for $\alpha,\beta,\gamma\in\mathbb{R}$, and real initial data $u_{0}$. Under
certain assumptions on the coefficients (1) is completely integrable and has
been extensively studied, with well-posedness established in
$L^{2}(\mathbb{T})$ by [20]. Additionally, it has recently been showed that
the integrable (1) posed on $\mathbb{R}$ is actually well-posed to $s=-1$, in
complete analogy with the KdV, [4].

In order to attack (1), one must develop tools that don’t rely on inverse
scattering. In this direction, Kwon [27] introduced the modified energy method
which enabled them to prove well-posedness for the non-integrable fifth order
equation in $H^{s}(\mathbb{R})$ for $s>5/2$, which was generalized by Kenig
and Pilod for higher order equations in the KdV heirarchy, [24]. Well-
posedness on $\mathbb{R}$ was then pushed to the $H^{2}$ level by Kenig and
Pilod in [23]. It’s also worth mentioning that [21] proves local well-
posedness in $H^{s}(\mathbb{R})\cap\dot{H}^{a}(\mathbb{R})$111Under the norm
$\|\langle\xi\rangle^{s-a}\langle\xi\rangle^{a}\widehat{f}\|_{L^{2}_{x}(\mathbb{R})}$.
for $s\geq-1/4$ and a certain range of $a$, and [28] proves well-posedness in
$H^{3/4}(\mathbb{R})$ for the the modified fifth order KdV.

Similar efforts have taken place on $\mathbb{T}$, where the absence of
hallmark dispersive effects causes great difficulty. Indeed, the most recent
results for the periodic domain are restricted to considering $\gamma=2\beta$
as in [26], where well-posedness is proved for the nearly integrable fifth
order equation at the $s=2$ level using the modified energy method and short-
time $X^{s,b}$ spaces. Similarly, [22] proves unconditional well-posedness in
$H^{s}$ for $s\geq 3/2$, and [36] proved well-posedness using the modified
energy method of Kwon for certain polynomials in $\partial_{x}^{3}$,
$\partial_{x}^{2}$ and $\partial_{x}$ for $s\geq 13$.

In particular, [36] begins to explore what they call “Parabolic Resonances”,
which are terms of the form

$in^{2\ell}\widehat{u}(n)\sum_{0=n_{1}+\cdots+n_{k}}m(n_{1},\cdots,n_{k})\prod_{j=1}^{k}\widehat{u}(n_{j}),$

where $m(-n_{1},\cdots,-n_{k})=-m(n_{1},\cdots,n_{k})$. The importance of the
oddness condition on $m$ is that for real initial data $u_{0}$, the quantity

(2)
$in^{2\ell}\sum_{0=n_{1}+\cdots+n_{j}}m(n_{1},\cdots,n_{k})\prod_{j=1}^{k}\widehat{u}(n_{j}),$

is purely real. This then becomes another term in the linear group that
smooths the equation in one direction of time and induces ill-posedness in the
other. In this language, the result of Tsugawa is that polynomials in
$\partial_{x}^{3}$, $\partial_{x}^{2}$ and $\partial_{x}$ for which the
induced $\ell=1$ parabolic resonances decay are well-posed for $s\geq 13$ in
both directions of time, whereas polynomials for which (2) does not decay are
well-posed in only one direction of time, and ill-posed in the other.

Much like [24], we will instead work with the non-integrable equation

(3)
$\displaystyle\begin{cases}u_{t}-\partial_{x}^{5}u+2\partial_{x}(u\partial_{x}^{2}u)=0\\\
u(x,0)=u_{0}\in H^{s}(\mathbb{T}),\end{cases}$

which not only conserves the mean, but also contains the worst portion of the
non-linearity in (1). Indeed, there is a $High\times Low$ interaction in which
the high term will have to experience $3$ derivatives, which also demonstrates
why the fifth order equation is hard to work with. In particular, the standard
smoothing associated to the $X^{s,1/2}$ space will result in savings of

$\sqrt{|(n_{1}+n_{2})^{5}-n_{1}^{5}-n_{2}^{5}|}=\sqrt{|5nn_{1}n_{2}(n_{1}^{2}+n_{1}n_{2}+n_{2}^{2})|}\sim|n_{1}|^{2}\sqrt{|n_{2}|},$

when $n_{1}\sim n\gg n_{2}$. As this is a whole derivative off of ameliorating
the derivative losses in the non-linearity, we are forced to work a little
harder.

The benefit of working with this reduction is twofold: 1) every result
obtained for this equation extends to the full equation (1) (see the Section
9), and 2) dramatically simplified equations. In particular, we show the
following theorem, with no assumption on the integrability of the equation.

######  Theorem 1.1.

Let $s>35/64$ and $u_{0}\in H^{s}(\mathbb{T})$. Then for $0\leq t\leq
T=T(\|u_{0}\|_{H^{s}_{x}})$ there is a unique solution $\tilde{u}\in
Y^{s}_{T}$ (defined in Section 2) to (29) with a continuous data-to-solution
map from $H^{s}$ to $Y^{s}_{T}$. Furthermore, the following bound holds:

$\|\tilde{u}\|_{Y^{s}_{T}}\lesssim\|u_{0}\|_{H^{s}_{x}}.$

In particular, since the solution to (29) is related to the solution of (3) by
a bi-continuous map on $H^{s}$, we find that (3) is well-posed in $H^{s}$ in
the sense of ([35], Definition 3.4)222This is a well-posedness result in the
sense of gauge conjugation. See, e.g. [5, 6, 30].

The lower bound of $35/64$ is not optimal, but we can say more. In that
direction, we define ($P_{a}$) to be the claim that the $L^{8}$ Strichartz
estimate holds with $a+$ losses.

######  Definition 1.2 (Property $(P_{a})$).

Property $P_{a}$ holds for $a\geq 0$ if for all $f\in H^{a+}(\mathbb{T})$ we
have the estimate

($P_{a}$) $\|W_{t}f\|_{L^{8}_{x,t}(\mathbb{T}^{2})}\lesssim\|f\|_{H^{a+}}.$

It follows by (4) that ($P_{a}$) holds with $s\geq 3/32$. As it’s conjectured
that the $L^{12}$ Strichartz estimate should only produce $\varepsilon$
losses, we see that $a=0$ is optimal. With this convention, we find the
following enhanced version of Theorem 1.1.

######  Theorem 1.3.

Let $a\geq 0$, assume ($P_{a}$) holds, $s>\frac{1+a}{2}$, and $u_{0}\in
H^{s}(\mathbb{T})$. Then for $0\leq t\leq T=T(\|u_{0}\|_{H^{s}_{x}})$ there is
a unique solution $\tilde{u}\in Y^{s}_{T}$ (defined in Section 2) to (29) with
a continuous data-to-solution map from $H^{s}$ to $Y^{s}_{T}$. Furthermore,
the following bound holds:

$\|\tilde{u}\|_{Y^{s}_{T}}\lesssim\|u_{0}\|_{H^{s}_{x}}.$

In particular, since the solution to $\eqref{Equation: Tilde u differential
equation before smoothing}$ is related to the solution of (3) by a bi-
continuous map on $H^{s}$, we find that (3) is well-posed in $H^{s}$ in the
sense of ([35], Definition 3.4).

Theorem 1.1 is proved through a differentiation-by-parts procedure inspired by
the work of Shatah [33] and Babin-Ilyin-Titi [1], with the added intricacy of
attempting to perform the fixed point argument in an $X^{s,b}$ type space and
not simply $H^{s}$. In particular, through two differentiation-by-parts
applications and the extra smoothing that the $X^{s,b}$ norms are known for,
we are able to obtain the above result. The well-posedness itself is proved
using a contraction argument in the Bourgain space $Y^{s}$ on the integral
equation (30) (which is obtain through the differentiation-by-parts process).
While Bourgain spaces and differentiation-by-parts have been used together in
the past (e.g. [10]), this is the first occurrence, to the author’s knowledge,
in which well-posedness itself has been proved using a contraction argument,
differentiation-by-parts, and $X^{s,b}$.

Additionally, since it’s possible, in principle, to work in
$C^{0}_{t}L^{2}_{x}$, the main new ingredient is fully embracing the non-
integrability of the system. That is, the work of, say, [13, 22] seek to only
remove resonances that correspond to conserved quantities of the equation,
whereas work on the KdV such as [2, 6, 30] begin to explore the removal of
time dependent (spatially independent) factors. We fully embrace this
perspective, turning the problem of solving the rather complicated non-
integrable (3), into simply applying an, albeit messy, rather straightforward
procedure.

In Section 9 we demonstrate that no parabolic resonances are created when
considering the full equation, (1). Because of this, we immediately establish
the following corollary.

######  Corollary 1.4.

Let $a\geq 0$ be such that ($P_{a}$) holds, $s>\frac{1+a}{2}$, and
$\alpha,\beta,\gamma\in\mathbb{R}$. Then (1) is well-posed in the sense of
([35], Definition 3.4). In particular, (1) is well-posed for $s>35/64.$

Additionally, as a consequence of the the integrable fifth order KdV sharing
the same conservation laws as the KdV, we find the following global well-
posedness result as a corollary.

######  Corollary 1.5.

Let $s\geq 1$, $\alpha,\beta,\gamma\in\mathbb{R}$, $\gamma=2\beta$, and
$u_{0}\in H^{s}(\mathbb{T})$. Then the solution, $u\in C^{0}_{t}H^{s}_{x}$, of
(1) emanating from $u_{0}$ extends globally.

In addition to proving well-posedness, we continue the study of the regularity
of the integral term in the Duhamel representation of the solution $u$
emanating from $u_{0}$. Specifically, we establish the following bound on the
integral term in the Duhamel representation.

######  Theorem 1.6.

Let $a\geq 0$ be such that ($P_{a}$) holds, $s>\frac{1+a}{2}$,
$\varepsilon<\min(2s-1-a,1)$, $u_{0}\in H^{s}(\mathbb{T})$, and $u$ be the
solution to (3) emanating from $u_{0}$ for $T=T(\|u_{0}\|_{H^{s}})$. Then
there is an invertible transformation333This is the same transformation used
for gauge well-posedness. $u\mapsto\tilde{u}$ with
$\|\tilde{u}\|_{C^{0}_{t}H^{s}_{x}}=\|u\|_{C^{0}_{t}H^{s}_{x}}$,
$u(x,0)=\tilde{u}(x,0)$ so that

$\|\tilde{u}-e^{t\partial_{x}^{5}}u_{0}\|_{C^{0}_{t}([0,T])H^{s+\varepsilon}_{x}(\mathbb{T})}\lesssim
C(\|u_{0}\|_{H^{s}_{x}(\mathbb{T})}).$

This phenomenon, nonlinear smoothing, was fundamental to Bourgain’s high-low
method, [3], and is of interest in periodic problems due to the lack of
hallmark dispersive traits. Indeed, through the use of differentiation-by-
parts, Erdogan and Tzirakis were able to establish nonlinear smoothing for the
periodic KdV (see: [11, 12]) as well as for the fractional cubic NLS and the
quintic NLS on the real line with Gurel, [10]. Oh and Stefanov, [30],
established smoothing for the generalized KdV using a resonant decomposition
similar to Proposition 3.2, as well as a normal form transformation.
Differentation-by-parts was also recently used to show smoothing for the
periodic dNLS, [19], solving a once intractable problem. For information about
real line smoothing see, e.g. [8, 10, 25].

In Section 9 we demonstrate that the high-frequency approximation to the
single resonances of Equation (1) experiences the same cancellation effects
that were necessary to establish smoothing for Equation (3). That is, we show
that there are no parabolic resonances with $k=0$, which requires a rather
delicate cancellation that does not depend on the value of the coefficients
$\alpha,\beta,\gamma$. In particular, we immediately find the following
corollary.

######  Corollary 1.7.

Let $a\geq 0$ be such that ($P_{a}$) holds, $s>\frac{1+a}{2}$,
$\alpha,\beta,\gamma\in\mathbb{R}$, $\varepsilon<\min(2s-1-a,1)$, $u_{0}\in
H^{s}(\mathbb{T})$, and $u$ be the solution to (1) emanating from $u_{0}$ for
$T=T(\|u_{0}\|_{H^{s}})$. Then there is an invertible transformation
$u\mapsto\tilde{u}$ with
$\|\tilde{u}\|_{C^{0}_{t}H^{s}_{x}}=\|u\|_{C^{0}_{t}H^{s}_{x}}$,
$u(x,0)=\tilde{u}(x,0)$ so that

$\|\tilde{u}-e^{t\partial_{x}^{5}}u_{0}\|_{C^{0}_{t}([0,T])H^{s+\varepsilon}_{x}(\mathbb{T})}\lesssim
C(\|u_{0}\|_{H^{s}_{x}(\mathbb{T})}).$

As a corollary of the equivalent equation constructed to prove Theorem 1.6, we
are able to establish unconditional well-posedness. By unconditional well-
posedness, we mean that for every real $u_{0}\in H^{s}(\mathbb{T})$ there is a
solution $u\in C^{0}_{t}H^{s}_{x}$ that is unique in $C^{0}_{t}H^{s}_{x}$– see
[14, 37] for the KdV, [29] for the modified KdV, and [22] for the fifth order
KdV for $s\geq 3/2$. This differs from the well-posedness result above in that
we make no appeal to the auxiliary space $Y^{s}$. Specifically, we find the
following corollary to hold true.

######  Corollary 1.8.

Let $s>1$ and $u_{0}\in H^{s}(\mathbb{T})$. Then (3) is unconditionally well-
posed in $H^{s}$.

It then follows by the work in Appendix 9 that Corollary 1.8 extends to the
full equation. In particular, we find the following corollary.

######  Corollary 1.9.

Let $s>1$, $\alpha,\beta,\gamma\in\mathbb{R}$, and $u_{0}\in
H^{s}(\mathbb{T})$. Then (1) is unconditionally well-posed in $H^{s}$.

It’s important to note that the decay of the $\ell=0$ Parabolic Resonances is
not important for the conclusions of Corollaries 1.8 or Corollary 1.9. All
that is required for these to hold is the decay of the $\ell=1$ parabolic
resonances, showing that unconditional well-posedness is, in a sense, easier
to obtain than nonlinear smoothing.

######  Remark 1.10.

It’s possible that a finer analysis, as in [29], could obtain the endpoint
$s=1$.

The outline of the paper is as follows. In Section 2 we define the slightly
unusual Bourgain space we will be using. In Section 4 we set out to perform
differentiation-by-parts in order to reduce the equation to one that is easier
to use. In particular, we construct $\sim$ and prove Lemma 4.1. Section 5 will
be comprised of all of the estimates required to close the contraction and
prove Theorem 1.3, and hence Theorem 1.1. In Section 6 we perform another
differentiation-by-parts and prove the remaining estimates required to prove
Theorem 1.6. Lastly, we piggyback off of the equation derived in Section 6 in
order to establish unconditional well-posedness in Section 7. In the
appendices we justify the reduction to (3), and demonstrate that the same
cancellation properties extend to that equation.

##  2\. Notation & Background

Let $n$ be the dual Fourier variable to $x\in\mathbb{T}$ and $\tau$ be the
dual Fourier variable to $t\in\mathbb{R}$. We denote the Fourier transform of
a function $u$ on the torus $\mathbb{T}$ by $\widehat{u}(n)$. When $u$ is a
spacetime function on $\mathbb{T}\times\mathbb{R}$, we denote both the space-
time Fourier transform and the spacial Fourier transform as $\widehat{u}$, as
there will be little confusion in context.

We write $A\lesssim_{\epsilon}B$ when there is a constant $0<C(\epsilon)$ such
that $|A|\leq C|B|$; $A\gg_{\epsilon}B$ to be the negation of
$A\lesssim_{\epsilon}B$; and $A\sim_{\epsilon}B$ if, in addition,
$B\lesssim_{\epsilon}A$. We will also write $g=O(f)$ to denote
$|g|\lesssim|f|$ and $a+$ (resp. $a-$) to denote $a+\epsilon$ (resp.
$a-\epsilon$) for all $\epsilon>0$, with implicit constants depending on
$\epsilon$.

Define $\langle h\rangle:=(1+|h|^{2})^{1/2}$ and $J^{s}$ to be the Fourier
multiplier given by $\langle n\rangle^{s}$. Let $W_{t}=e^{t\partial_{x}^{5}}$
be the propagator for the fifth order linear group, and for clarity we ignore
factors of $2\pi$. We then define the Bourgain space $X^{s,b}$ first
introduced in the seminal paper [2], by

$\|u\|_{X^{s,b}}=\|\langle
n\rangle^{s}\langle\tau+n^{5}\rangle^{b}\widehat{u}\|_{L^{2}_{\tau}\ell^{2}_{n}}=\|\langle
n\rangle^{s}\langle\tau\rangle^{b}\widehat{u}(n,\tau-n^{5}\rangle)\|_{\ell^{2}_{n}L^{2}_{\tau}}=\|W_{-t}u\|_{H^{s}_{x}H^{b}_{t}},$

which measures, in a sense, how much $u$ deviates from the free solution
$W_{t}u_{0}$. Given $T>0$ we define the restricted Bourgain space as the space
of equivalence classes of functions endowed with the norm

$\|u\|_{X^{s,b}_{T}}:=\inf\\{\|v\|_{X^{s,b}}\,|\,u|_{[0,T]}=v|_{[0,T]}\\},$

with dual space $X^{-s,-b}.$ Because of this natural pairing and the fact that
our norms are defined in terms of the absolute value of the Fourier transform,
we will often assume that all Fourier transforms are non-negative and invoke
duality, where it will be natural to define the hyper-plane by

$\Gamma_{p}:=\\{\tau-\tau_{1}-\cdots-\tau_{p}=0,\,\,n-n_{1}-\cdots-
n_{p}=0\\},$

with obvious inherited measure denoted by $d\Gamma$.

Due to the non-negativity of the Fourier transforms, we note that if
$p(n,n_{1},\cdots,n_{k})$ is a symbol associated to a multilinear operator of
the form

$P(u_{1},\cdots,u_{k})(t,x)=\mathcal{F}_{n}^{-1}\bigg{(}\sum_{n=n_{1}+\cdots
n_{k}}p(n,n_{1},\cdots,n_{k})\prod_{i=1}^{k}\widehat{u}(n_{i},t)\bigg{)}(x),$

and $|p|\leq cq$ for some symbol $q(n,n_{1},\cdots,n_{k})$ with associated
multilinear operator $Q$, then

$\|P(u_{1},\cdots,u_{k})\|_{X^{s,b}_{T}}\lesssim\|Q(u_{1},\cdots,u_{k})\|_{X^{s,b}_{T}}.$

Since we have mean conservation and the transformation $u\mapsto
u-\int_{\mathbb{T}}u\,dx$ simply adds another lower order term to the linear
group, we will assume that $u_{0}$ is mean-zero and that all indexed variables
$n_{j}$ are summed over $\mathbb{Z}\setminus\\{0\\}.$ We will often need to
deal with tuples $(n_{1},\cdots,n_{k})$ for some $k>1$ and
$n_{i}\in\mathbb{Z}\setminus\\{0\\}$, where it will be convenient to denote
the decreasing rearrangement of $\\{|n_{i}|\\}_{i=1}^{k}$ by $n_{i}^{*}$. That
is, $n_{i}^{*}$ denotes the $i$’th largest term, in absolute value, among
$n_{1},\cdots,n_{k}.$

Due to the the method we employ, we will need to handle boundary terms in
$X^{s,b}$ for $b>0$ of the form

$\left\|\mathcal{F}_{x}^{-1}\left(\sum_{n=n_{1}+n_{2}}\frac{\widehat{u}(n_{1})\widehat{u}(n_{2})}{n_{1}n_{2}}\right)\right\|_{X^{s,b}}.$

Because of this, we will need to choose $b$ so that for $n_{1}\gg n$

$\left|(n_{1}+n_{2})n_{1}n_{2}((n_{1}+n_{2})^{2}+n_{1}^{2}+n_{3}^{2})\right|^{b}\lesssim
n_{1},$

which necessitates the choice $b=1/4$. This space, however, fails to control
the $C^{0}_{t}H^{s}_{x}$ norm. To remedy this issue, we define the norms

$\|\cdot\|_{Y^{s}}:=\|\cdot\|_{X^{s,1/4}}+\|\langle
n\rangle^{s}\mathcal{F}_{x,t}(\cdot)\|_{\ell^{2}_{n}\ell^{1}_{\tau}}$

for the initial data, and

$\|\cdot\|_{Z^{s}}:=\|\cdot\|_{X^{s,-3/4}}+\left\|\frac{\langle
n\rangle^{s}}{\langle\tau+n^{5}\rangle}\mathcal{F}_{x,t}(\cdot)\right\|_{\ell^{2}_{n}\ell^{1}_{\tau}}$

for the nonlinearity. We analogously define the restricted $Y^{s}_{T}$ space.

We now record some standard facts (See e.g. [16]):

######  Lemma 2.1.

For any $\chi\in\mathcal{S}(\mathbb{R})$ and $f\in
C^{\infty}_{x}(\mathbb{T})$,

$\displaystyle\|\chi(t)e^{it\partial_{x}^{2}}f\|_{Y^{s}}$
$\displaystyle\lesssim\|f\|_{H^{s}}$
$\displaystyle\left\|\chi(t)\int_{0}^{t}e^{i(t-s)\partial_{x}^{2}}F(s)\,ds\right\|_{Y^{s}}$
$\displaystyle\lesssim\|F\|_{Z^{s}}.$

Additionally, for $\eta\in\mathcal{S}(\mathbb{R})$ positive, supported on
$[-2,2]$, and $1$ on $[-1,1]$, Sobolev embedding implies

$\|\eta(t/T)u\|_{L^{q}_{t}L^{p}_{x}}\lesssim
T^{\frac{1}{q}}\|u\|_{Y^{s}_{T}}\qquad\qquad
s>\frac{1}{2}-\frac{1}{p}\qquad\qquad p\geq 2.$

Similarly, we have the well-known relationship between the modulation variable
and the time, $T,$ given by

$\|u\|_{X^{s,b}_{T}}\lesssim
T^{b^{\prime}-b}\|u\|_{X^{s,b^{\prime}}_{T}}\qquad\qquad-\frac{1}{2}<b\leq
b^{\prime}<\frac{1}{2}.$

We will also need the following Strichartz estimates.

######  Lemma 2.2.

For $u$ on $\mathbb{T}\times\mathbb{R}$ and $\eta\in\mathcal{S}(\mathbb{R})$
defined previously, we have

$\|\eta(t/T)u\|_{L^{4}_{x,t}(\mathbb{T}\times\mathbb{R})}\lesssim\|u\|_{X^{0,\frac{3}{10}}_{T}},$
$\|\eta(t/T)u\|_{L^{6}_{x,t}(\mathbb{T}\times\mathbb{R})}\lesssim\|u\|_{X^{0+,\frac{1}{2}+}_{T}}.$

and

$\|\eta(t/T)u\|_{L^{30}_{x,t}(\mathbb{T}\times\mathbb{R})}\lesssim\|u\|_{X^{3/10+,\frac{1}{2}+}_{T}}.$

###### Proof.

The proof of the first estimate is exactly the same as Bourgain’s in [2] since
we again have a factorization of

$(n_{1}+n_{2})^{5}-n_{1}^{5}-n_{2}^{5}=5/2(n_{1}+n_{2})n_{1}n_{2}(n^{2}+n_{1}^{2}+(n_{1}+n_{2})^{2}).$

The second estimate follows from [17] Theorem 1.4, and the last from [18]
Theorem 3.1. ∎

Interpolation between the $L^{4}$ estimate with the trivial $L^{2}$ estimate
gives,

$\|\eta(t/T)u\|_{L^{24/7}_{x,t}(\mathbb{T}\times\mathbb{R})}\lesssim\|u\|_{X^{0,1/4}_{T}},$

whereas interpolation between the $L^{6}$ and the $L^{30}$ estimates above
gives

(4)
$\|\eta(t/T)u\|_{L^{8}_{x,t}(\mathbb{T}\times\mathbb{R})}\lesssim\|u\|_{X^{3/32+,1/2+}_{T}}.$

######  Remark 2.3.

Since we’re above the Sobolev embedding level, the standard game one plays
with Strichartz estimates loses its potency. However, the above estimates will
still be useful in going below $s=1$.

##  3\. Symbol Analysis

Before we proceed with the paper, we need a more number theoretic result that
will enable us to handle large multilinear estimates. In order to state the
proposition we need to allow for an abuse of notation that will be more
convenient. We let $n_{1}^{*}+n_{2}^{*}=0$ denote the case when the sum of the
two largest (in magnitude) frequencies is zero. That is, if

$|n_{1}|\geq|n_{2}|\geq\cdots\geq|n_{p}|,$

then $n_{1}^{*}+n_{2}^{*}=0$ denotes the situation $n_{1}+n_{2}=0$. For the
remainder of this section we fix $n,n_{1},\cdots,n_{p}$ so that

$n=n_{1}+\cdots+n_{p}.$

With that in mind, we define

$\Phi_{p}(n_{1},\cdots,n_{p}):=\bigg{(}\sum_{i=1}^{p}n_{i}\bigg{)}-\sum_{i=1}^{p}n_{i}^{5}=n^{5}-\sum_{i=1}^{p}n_{i}^{5},$

for which we find the following propositions.

######  Proposition 3.1.

Let $|n_{1}|\geq|n_{2}|\geq|n_{3}|$. Then at least one of the following must
be true

  * I)

“Resonance”- $n_{i}=n$ for some $1\leq i\leq 3$,

  * II)

$|n_{3}|\gtrsim|n_{1}|$,

  * III)

$|\Phi_{3}(n_{1},n_{2},n_{3})|=|n^{5}-n_{1}^{5}-\cdots n_{3}^{5}|\gtrsim
n_{1}^{4}$.

###### Proof.

Note that we have the factorization

(5)
$n^{5}-n_{1}^{5}-n_{2}^{5}-n_{3}^{5}=\frac{5}{2}(n_{1}+n_{2})(n_{1}+n_{3})(n_{2}+n_{3})(n^{2}+n_{1}^{2}+n_{2}^{2}+n_{3}^{2}).$

If we first assume that there is no resonance, then we find that the above is
non-zero. Assuming now that $|n_{3}|\ll|n_{1}|$, we find that
$|(n_{1}+n_{3})(n_{1}+n_{2})(n_{2}+n_{3})|\gtrsim n_{1}^{2}$, so that

$|(n_{1}+n_{2})(n_{1}+n_{3})(n_{2}+n_{3})(n^{2}+n_{1}^{2}+n_{2}^{2}+n_{3}^{2})|\gtrsim
n_{1}^{4},$

as desired. ∎

Piggybacking off of Proposition 3.1 we establish the following more general
decomposition lemma.

######  Proposition 3.2.

Let $|n_{1}|\geq|n_{2}|\geq\cdots\geq|n_{p}|$, and $p\geq 4$. Denote
$\tilde{n}_{i}=\sum_{j=i}^{p}n_{j}$, so we have that $n=\tilde{n}_{1}$ and
$n_{p}=\tilde{n}_{p}$. Then at least one of the following must be true

  * I)

“Resonance”- $n_{i}=n$ for some $1\leq i\leq p$,

  * II)

$n_{1}^{*}+n_{2}^{*}=0$,

  * III)

$|n_{3}|\gtrsim|n_{1}|$,

  * IV)

$|\Phi_{p}(n_{1},\cdots,n_{p})|=|n^{5}-n_{1}^{5}-\cdots n_{p}^{5}|\gtrsim
n_{1}^{4}$,

  * V)

$n_{3}^{4}|n_{4}|\gtrsim n_{1}^{4}$.

######  Remark 3.3.

The third condition is also contained in the last condition by mean-zero
preservation, so we will often drop it and handle it with the case
$n_{3}^{4}n_{4}\gtrsim n_{1}^{4}$. Additionally, we don’t care about the
specifics of the decomposition of the Fourier space into these cases– any such
will work.

###### Proof.

For $p\geq 4$ we may write

$\displaystyle n^{5}-n_{1}^{5}-\cdots n_{p}^{5}$
$\displaystyle=(n^{5}-n_{1}^{5}-n_{2}^{5}-\tilde{n}_{3}^{5})+(\tilde{n}_{3}^{5}-n_{3}^{5}-\tilde{n}_{4}^{5})+\cdots+(\tilde{n}_{p-1}^{5}-n_{p-1}^{5}-n_{p}^{5})$
$\displaystyle=\Phi_{3}(n_{1},n_{2},\tilde{n}_{3})+\sum_{j=3}^{p-1}\Phi_{2}(n_{j},\tilde{n}_{j+1}).$

Hence, if there is no resonance, $n_{1}+n_{2}\neq 0$, and $|n_{3}|\ll|n_{1}|$,
then by (5) the first term satisfies

$\displaystyle|\Phi_{3}(n_{1},n_{2},\tilde{n}_{3})|$
$\displaystyle=5/2|(n_{1}+n_{2})(n_{1}+\tilde{n}_{3})(n_{2}+\tilde{n}_{3})|(n^{2}+n_{1}^{2}+n_{2}^{2}+\tilde{n}_{3}^{2})\gtrsim
n_{1}^{4}.$

Indeed, the first term is non-zero by assumption, and the lack of resonance
ensures that $n_{1}+\tilde{n}_{3},n_{2}+\tilde{n_{3}}\neq 0$. Additionally, if
$|n_{1}|\gg|n_{2}|$, then the first two factors are, in magnitude,
$\gtrsim|n_{1}|$ each, and if $|n_{2}|\sim|n_{1}|$, then the middle factors
must each be $\gtrsim|n_{1}|$ in magnitude.

It follows that if $|\Phi_{p}(n_{1},\cdots,n_{p})|\ll n_{1}^{4}$, then we must
have

$\big{|}\sum_{j=3}^{p-1}\Phi_{2}(n_{j},\tilde{n}_{j+1})\big{|}\gtrsim
n_{1}^{4}.$

However, the remaining terms must satisfy

$|\Phi_{2}(n_{j},\tilde{n}_{j+1})|=|\tilde{n}_{j}n_{j}\tilde{n}_{j+1}|(\tilde{n}_{j}^{2}+n_{j}^{2}+\tilde{n}_{j+1}^{2})\lesssim
n_{3}^{4}|n_{4}|,$

so that

$n_{1}^{4}\lesssim\big{|}\sum_{j=3}^{p-1}\Phi_{2}(n_{j},\tilde{n}_{j+1})\big{|}\lesssim
n_{3}^{4}|n_{4}|,$

which is the final case. ∎

######  Remark 3.4.

For brevity, we denote the property that
$|\Phi_{p}(n_{1},\cdots,n_{p})|\gtrsim(n_{1}^{*})^{4}$ by
$\mathcal{P}_{p}(n_{1},\cdots,n_{p})$. Similarly, we denote the property that
$|\Phi_{p}(n_{1},\cdots,n_{p})|\ll(n_{1}^{*})^{4}$ by
$\mathcal{P}_{p}^{c}(n_{1},\cdots,n_{p})$.

##  4\. Reductions

Before continuing with the reductions, we warn the reader: in the interest of
readability we have, by and large, chosen not to track what the explicit
constants are. This pattern is only broken when explicit cancellation is
sought, but in most other situations we state results modulo constants.

This section is organized so that each application of differentiation-by-parts
is separated into a subsection. The overarching goal of this section is the
following lemma.

######  Lemma 4.1.

Let $a\geq 0$ be such that ($P_{a}$) holds, $s>\frac{1+a}{2}$ and $u_{0}\in
H^{s}(\mathbb{T})$. Then there is an invertible transformation
$u\mapsto\tilde{u}$ with
$\|\tilde{u}\|_{C^{0}_{t}H^{s}_{x}}=\|u\|_{C^{0}_{t}H^{s}_{x}}$ and
$u(x,0)=\tilde{u}(x,0)$, such that if $u$ satisfies (6), then $\tilde{u}$ is a
fixed point of (30).

As mentioned, the proof of the above lemma follows from a differentiation-by-
parts process. This fundamentally is the observation that

$\widehat{u}_{t}=\sum_{\Omega\neq
0}e^{it\Omega}m(n_{1},n_{2})\widehat{u}(n_{1})\widehat{u}(n_{2})\\\
=\sum_{\Omega\neq
0}\frac{\partial_{t}e^{it\Omega}m(n_{1},n_{2})}{i\Omega}\widehat{u}(n_{1})\widehat{u}(n_{2})=\partial_{t}\sum_{\Omega\neq
0}\frac{e^{it\Omega}m(n_{1},n_{2})}{i\Omega}\widehat{u}(n_{1})\widehat{u}(n_{2})\\\
-\sum_{\Omega\neq
0}\frac{e^{it\Omega}m(n_{1},n_{2})}{i\Omega}\partial_{t}\widehat{u}(n_{1})\widehat{u}(n_{2})-\sum_{\Omega\neq
0}\frac{e^{it\Omega}m(n_{1},n_{2})}{i\Omega}\widehat{u}(n_{1})\partial_{t}\widehat{u}(n_{2}),$

where $m$ is some time-independent symbol, such as a polynomial in $n_{1}$ and
$n_{2}$. In our situation we then have a formula for $\partial_{t}\widehat{u}$
from the differential equation, and we may substitute this in and hope to use
$\Omega$ to mitigate losses from $m$.

The general strategy will then be to separate portions of the sum over which
$\Omega$ is possibly very small – Resonant or nearly Resonant portions – from
the portions where $\Omega$ is large. The portions over which it is small will
dictate the well-posedness level, and hence will be of the most interest.

Now, by the work in Appendix 9 we see that the existence of a solution to the
the above equation is equivalent to a solution to the fifth order KdV, we
simply resign ourselves to proving Lemma 4.1 and the associated bounds on it,
and iteratively constructing the transformation $\sim$.

###  4.1. First Differentiation-by-Parts

We symmetrize the nonlinearity and rewrite (3) on the Fourier side as

(6)
$\displaystyle\begin{cases}\widehat{u}_{t}+in^{5}\widehat{u}-in\sum_{n_{1}+n_{2}=n}(n_{1}^{2}+n_{2}^{2})\widehat{u}(n_{1})\widehat{u}(n_{2})=0\\\
\widehat{u}(n,0)=\widehat{u_{0}}(n).\end{cases}$

We now perform the change of variables

(7) $\widehat{v}(n)=\widehat{W_{-t}u}(n)=e^{-itn^{5}}\widehat{u}(n),$

so that we obtain

(8)
$\displaystyle\begin{cases}\widehat{v}_{t}(n)=in\sum_{n_{1}+n_{2}=n}(n_{1}^{2}+n_{2}^{2})e^{-it\Phi_{2}(n,n_{1},n_{2})}\widehat{v}(n_{1})\widehat{v}(n_{2})\\\
\widehat{v}(n,0)=\widehat{u_{0}}(n),\end{cases}$

for

(9)
$\Phi_{2}(n_{1},n_{2})=(n_{1}+n_{2})^{5}-n_{1}^{5}-n_{2}^{5}=\frac{5}{2}nn_{1}n_{2}(n^{2}+n_{1}^{2}+n_{2}^{2}),$

when $n=n_{1}+n_{2}$.

Notice that if either frequency is equal to $n$, then conservation of mean
forces the term to vanish. Hence we may assume that the phase doesn’t vanish
and differentiation-by-parts yields

$\displaystyle\partial_{t}\widehat{v}(n)$
$\displaystyle=-\partial_{t}\sum_{n=n_{1}+n_{2}}\frac{n(n_{1}^{2}+n_{2}^{2})}{\Phi_{2}(n,n_{1},n_{2})}e^{-it\Phi_{2}(n_{1},n_{2})}\widehat{v}(n_{1})\widehat{v}(n_{2})$
$\displaystyle\qquad+2\sum_{n=n_{1}+n_{2}}\frac{n(n_{1}^{2}+n_{2}^{2})}{\Phi_{2}(n_{1},n_{2})}e^{-it\Phi_{2}(n,n_{1},n_{2})}\widehat{v}(n_{1})\partial_{t}\widehat{v}(n_{2})$
(10)
$\displaystyle=-\partial_{t}\sum_{n=n_{1}+n_{2}}\frac{n(n_{1}^{2}+n_{2}^{2})}{\Phi_{2}(n_{1},n_{2})}e^{-it\Phi_{2}(n,n_{1},n_{2})}\widehat{v}(n_{1})\widehat{v}(n_{2})$
$\displaystyle\qquad+2i\sum_{\begin{subarray}{c}n=n_{1}+n_{2}+n_{3}\\\
n_{2}+n_{3}\neq 0\\\ n,n_{1},n_{2},n_{3}\neq
0\end{subarray}}\frac{n(n_{1}^{2}+(n_{2}+n_{3})^{2})(n_{2}+n_{3})(n_{2}^{2}+n_{3}^{2})}{\Phi_{2}(n_{1},n_{2}+n_{3})}$
$\displaystyle\qquad\qquad\qquad\times
e^{-it\Phi_{3}(n_{1},n_{2},n_{3})}\widehat{v}(n_{1})\widehat{v}(n_{2})\widehat{v}(n_{3}),$

where

(11)
$\Phi_{3}(n_{1},n_{2},n_{3})=\big{(}\sum_{j=1}^{3}n_{j}\big{)}^{5}-n_{1}^{5}-n_{2}^{5}-n_{3}^{5}\\\
=\frac{5}{2}(n_{1}+n_{2})(n_{2}+n_{3})(n_{1}+n_{3})(n^{2}+n_{1}^{2}+n_{2}^{2}+n_{3}^{2}).$

While we had no resonances in the nonlinearity of (6), this new trilinear term
has resonance. Specifically, if $n_{1}+n_{2}=0$ or $n_{1}+n_{3}=0$ then we
won’t be able to use modulation considerations– that is, $n=n_{2}$ or
$n=n_{3}$. If $n_{2}=n$, then by (9) we have $n_{1}=-n_{3}$ and the fraction
becomes

$\displaystyle\frac{4i}{5}\frac{n(n_{3}^{2}+(n+n_{3})^{2})(n+n_{3})(n^{2}+n_{3}^{2})}{nn_{3}(n+n_{3})(n^{2}+n_{3}^{2}+(n+n_{3})^{2})}=\frac{4i}{5}\frac{(n_{3}^{2}+(n+n_{3})^{2})(n^{2}+n_{3}^{2})}{n_{3}(n^{2}+n_{3}^{2}+(n+n_{3})^{2})}$
(12)
$\displaystyle\qquad\quad=\frac{4i}{5}\left\\{\frac{n^{2}+n_{3}^{2}}{n_{3}}-\frac{n^{2}}{2n_{3}}+\frac{1}{2}n-\frac{n^{2}n_{3}+nn_{3}^{2}}{n^{2}+(n+n_{3})^{2}+n_{3}^{2}}\right\\}.$

Incorporating the cases when $n_{3}=n$ and $n_{2}=n_{3}=n$ we find that the
resonant sum will then be

$\displaystyle\frac{8i}{5}\widehat{v}(n)\sum_{n_{3}}$
$\displaystyle\left\\{\frac{n^{2}+n_{3}^{2}}{n_{3}}-\frac{n^{2}}{2n_{3}}+\frac{1}{2}n-\frac{n^{2}n_{3}+nn_{3}^{2}}{n^{2}+(n+n_{3})^{2}+n_{3}^{2}}\right\\}\widehat{v}(n_{3})\widehat{v}(-n_{3})$
$\displaystyle\qquad\qquad-\frac{4i}{3}n\widehat{v}(n)\widehat{v}(n)\widehat{v}(-n).$

Since $\widehat{v}(n_{3})\widehat{v}(-n_{3})$ is invariant under the sign of
$n_{3}$, we see that the resonant sum weighted by the first two summands of
(12) will vanish. It follows that the full resonant portion of (10) reduces to

(13) $\displaystyle\frac{4i}{5}n\widehat{v}(n)$
$\displaystyle\sum_{n_{3}}\widehat{v}(n_{3})\widehat{v}(-n_{3})$
$\displaystyle\qquad-\frac{8i}{5}\widehat{v}(n)\sum_{n_{3}}\frac{n^{2}n_{3}+nn_{3}^{2}}{n^{2}+(n+n_{3})^{2}+n_{3}^{2}}\widehat{v}(n_{3})\widehat{v}(-n_{3})$
$\displaystyle\qquad-\frac{4i}{3}n\widehat{v}(n)\widehat{v}(n)\widehat{v}(-n).$

The second and third portions of the above are suitable to estimate, whereas
the first we must remove.

Letting,

$\mathcal{K}(v)(t)=\frac{4i}{5}\int_{\mathbb{T}}v^{2}(x,t)\,dx=\frac{4i}{5}n\sum_{n_{3}}\widehat{v}(n_{3})\widehat{v}(-n_{3}),$

we define

(14)
$\widehat{v}(n)=e^{n\int_{0}^{t}\mathcal{K}(v)\,ds}\widehat{w}(n)=e^{n\int_{0}^{t}\mathcal{K}(w)\,ds}\widehat{w}(n).$

######  Remark 4.2.

As $v$ is real, the above quantity, $\mathcal{K}$, is technically an imaginary
multiple of its $L^{2}_{x}$ norm. We write it in this fashion because it is
not technically conserved under the flow of (3).

It follows that

$\partial_{t}\widehat{v}(n)=n\mathcal{K}(v)\widehat{v}(n)+e^{n\int_{0}^{t}\mathcal{K}(v)\,ds}\partial_{t}\widehat{w}(n).$

Using (14) and the display below it, we find $\widehat{w}$ satisfies (looking
at (6))

(15)
$\displaystyle\begin{cases}\widehat{w}_{t}(n)=in\sum_{n_{1}+n_{2}=n}(n_{1}^{2}+n_{2}^{2})e^{-it\Phi_{2}(n_{1},n_{2})}\widehat{w}(n_{1})\widehat{w}(n_{2})\\\
\qquad\qquad-n\mathcal{K}(w)\widehat{w}(n)\\\
\widehat{w}(n,0)=\widehat{u_{0}}(n).\end{cases}$

Performing the differentiation-by-parts procedure again, we see

(16) $\displaystyle\partial_{t}w$
$\displaystyle=-\partial_{t}\sum_{n=n_{1}+n_{2}}\frac{n(n_{1}^{2}+n_{2}^{2})}{\Phi_{2}(n_{1},n_{2})}e^{-it\Phi_{2}(n_{1},n_{2})}\widehat{w}(n_{1})\widehat{w}(n_{2})$
(17) $\displaystyle\qquad+2i\sum_{\begin{subarray}{c}n=n_{1}+n_{2}+n_{3}\\\
\star\end{subarray}}\frac{n(n_{1}^{2}+(n_{2}+n_{3})^{2})(n_{2}+n_{3})(n_{2}^{2}+n_{3}^{2})}{\Phi_{2}(n_{1},n_{2}+n_{3})}$
$\displaystyle\qquad\qquad\qquad\times
e^{-it\Phi_{3}(n_{1},n_{2},n_{3})}\widehat{w}(n_{1})\widehat{w}(n_{2})\widehat{w}(n_{3})$
$\displaystyle\qquad-\mathcal{K}(w)\sum_{\begin{subarray}{c}n=n_{1}+n_{2}\end{subarray}}\frac{(n_{1}+n_{2})(n_{1}^{2}+n_{2}^{2})}{\Phi_{2}(n_{1},n_{2})}e^{it\Phi_{2}(n_{1},n_{2})}\widehat{w}(n_{1})\widehat{w}(n_{2})$
$\displaystyle\qquad-\frac{8i}{5}\widehat{w}(n)\sum_{n_{3}}\frac{n^{2}n_{3}+nn_{3}^{2}}{n^{2}+(n+n_{3})^{2}+n_{3}^{2}}\widehat{w}(n_{3})\widehat{w}(-n_{3})-\frac{4i}{3}n\widehat{w}(n)\widehat{w}(n)\widehat{w}(-n),$

where $\star$ denotes the non-resonance restriction– the complement of case 1
of Proposition 3.2. Note the cancellation of the single resonance in the
above, as opposed to (10).

###  4.2. Second Differentiation-by-Parts

In this section we perform the second differentiation-by-parts and obtain the
equation we will work with. Specifically, notice that when
$|n_{3}|\gg\max(|n_{1}|,|n_{2}|)$, then the trilinear term, (17), has,
morally, two derivatives. Since we can only recover one derivative from
modulation considerations, we are forced to perform another differentiation-
by-parts procedure for this term. There’s nothing wrong, in principle, with
performing the differentiation-by-parts on the full trilinear term, but with
the aid of Proposition 3.1 we can do better.

We now recall Remark 3.4 and define the symbols

$\displaystyle m_{D_{1}}$
$\displaystyle=\frac{n((n_{1}+n_{2})^{2}+(n_{3}+n_{4})^{2})(n_{3}+n_{4})(n_{3}^{2}+n_{4}^{2})(n_{1}+n_{2})(n_{1}^{2}+n_{2}^{2})}{\Phi_{2}(n_{1}+n_{2},n_{3}+n_{4})\Phi_{3}(n_{1}+n_{2},n_{3},n_{4})}\chi_{\mathcal{P}_{3}(n_{1}+n_{2},n_{3},n_{4})}$
$\displaystyle m_{D_{2}}$
$\displaystyle=\frac{n(n_{1}^{2}+(n_{2}+n_{3}+n_{4})^{2})(n_{2}+n_{3}+n_{4})(n_{2}^{2}+(n_{3}+n_{4})^{2})(n_{3}+n_{4})(n_{3}^{2}+n_{4}^{2})}{\Phi_{2}(n_{1},n_{2}+n_{3}+n_{4})\Phi_{3}(n_{1},n_{2},n_{3}+n_{4})}$
$\displaystyle\qquad\qquad\times\chi_{\mathcal{P}_{3}(n_{1},n_{2},n_{3}+n_{4})}$
$\displaystyle m_{A_{i}}$ $\displaystyle=m_{D_{1}}\chi_{n=n_{i}}\qquad i=1,2$
$\displaystyle m_{B_{i}}$ $\displaystyle=m_{D_{2}}\chi_{n=n_{i}}\qquad i=3,4,$

and let

$\displaystyle\widehat{B_{1}}(\phi)(n)$
$\displaystyle=\sum_{\begin{subarray}{c}n=n_{1}+n_{2}\\\
\star\end{subarray}}\frac{n(n_{1}^{2}+n_{2}^{2})}{\Phi_{2}(n_{1},n_{2})}\widehat{\phi}(n_{1})\widehat{\phi}(n_{2})$
$\displaystyle\widehat{B_{2}}(\phi)(n)$
$\displaystyle=\sum_{\begin{subarray}{c}n=n_{1}+n_{2}+n_{3}\\\
\mathcal{P}_{3}(n_{1},n_{2},n_{3})\\\
\star\end{subarray}}\frac{n(n_{1}^{2}+(n_{2}+n_{3})^{2})(n_{2}+n_{3})(n_{2}^{2}+n_{3}^{2})}{\Phi_{2}(n_{1},n_{2}+n_{3})\Phi_{3}(n_{1},n_{2},n_{3})}\widehat{\phi}(n_{1})\widehat{\phi}(n_{2})\widehat{\phi}(n_{3})$
$\displaystyle\widehat{D_{1}}(\phi)(n)$
$\displaystyle=\sum_{\begin{subarray}{c}n=n_{1}+n_{2}+n_{3}+n_{4}\\\
\star\end{subarray}}m_{D_{1}}\widehat{\phi}(n_{1})\widehat{\phi}(n_{2})\widehat{\phi}(n_{3})\widehat{\phi}(n_{4})$
$\displaystyle\widehat{D_{2}}(\phi)(n)$
$\displaystyle=\sum_{\begin{subarray}{c}n=n_{1}+n_{2}+n_{3}+n_{4}\\\
\star\end{subarray}}m_{D_{2}}\widehat{\phi}(n_{1})\widehat{\phi}(n_{2})\widehat{\phi}(n_{3})\widehat{\phi}(n_{4})$
$\displaystyle\widehat{\mathcal{NR}_{1}}(\phi)(n)$
$\displaystyle=\sum_{\begin{subarray}{c}n=n_{1}+n_{2}+n_{3}\\\
\mathcal{P}_{3}^{c}(n_{1},n_{2},n_{3})\\\
\star\end{subarray}}\frac{n(n_{1}^{2}+(n_{2}+n_{3})^{2})(n_{2}+n_{3})(n_{2}^{2}+n_{3}^{2})}{\Phi_{2}(n_{1},n_{2}+n_{3})}\widehat{\phi}(n_{1})\widehat{\phi}(n_{2})\widehat{\phi}(n_{3}),$

where $\star$ again corresponds to the lack of resonance. That is, we write

$\eqref{Equation: Second DBP trlinear
term}=-2\partial_{t}\widehat{W_{-t}}\widehat{B}_{2}(\widehat{W_{t}}w)+\widehat{W_{-t}}\big{\\{}2i\widehat{\mathcal{NR}}_{1}(\widehat{W_{t}}w)\\\
+4i(\widehat{D}_{1}(\widehat{W_{t}}w)+\widehat{\mathcal{RE}}_{3}(\widehat{W}_{t}w)+2\widehat{D}_{2}(\widehat{W_{t}}w)+2\widehat{\mathcal{RE}}_{4}(\widehat{W}_{t}w))\\\
-2\widehat{R}_{2}(\widehat{W_{t}}w)\big{\\}}+\text{mult. resonances},$

where $R_{2}$ is defined below. Notice that $m_{D_{1}}$ corresponds to
$\partial_{t}$ landing on the $n_{1}$ term, while $m_{D_{2}}$ corresponds to
$\partial_{t}$ landing on $n_{2}$ or $n_{3}$. $m_{A_{i}}$ and $m_{B_{i}}$ then
correspond to the symbols of the additional resonant terms. Notice that we
have separated out the region $\Phi_{3}\ll(n_{1}^{*})^{4}$, and only
differentiated-by-parts the non-resonant complement of this region.

The prior terms will be the main terms of study, but we will also need to
define the additional cast of characters:

(18) $\displaystyle\widehat{R_{1}}(\phi)(n)$
$\displaystyle=\mathcal{K}(\phi)\sum_{\begin{subarray}{c}n=n_{1}+n_{2}\\\
\star\end{subarray}}m_{R}^{1}(n_{1}+n_{2})\widehat{\phi}(n_{1})\widehat{\phi}(n_{2})$
(19) $\displaystyle\widehat{R_{2}}(\phi)(n)$
$\displaystyle=\mathcal{K}(\phi)\sum_{\begin{subarray}{c}n=n_{1}+n_{2}+n_{3}\\\
\star\end{subarray}}m_{R}^{2}(n_{1}+n_{2}+n_{3})\widehat{\phi}(n_{1})\widehat{\phi}(n_{2})\widehat{\phi}(n_{3})$
(20) $\displaystyle\widehat{\mathcal{RE}_{1}}(\phi)(n)$
$\displaystyle=\frac{8i}{5}\widehat{\phi}(n)\sum_{n_{3}}\frac{n^{2}n_{3}+nn_{3}^{2}}{n^{2}+(n+n_{3})^{2}+n_{3}^{2}}\phi(n_{3})\widehat{\phi}(-n_{3})$
(21) $\displaystyle\widehat{\mathcal{RE}_{2}}(\phi)(n)$
$\displaystyle=\frac{4i}{3}n\widehat{\phi}(n)\widehat{\phi}(n)\widehat{\phi}(-n)+\mbox{Second
Order Resonances}$ (22) $\displaystyle\widehat{\mathcal{RE}_{3}}(\phi)(n)$
$\displaystyle=\widehat{\phi}(n)\sum_{\begin{subarray}{c}0=n_{2}+n_{3}+n_{4}\\\
n_{3},n_{4}\neq n\\\
\mathcal{P}_{3}(n+n_{2},n_{3},n_{4})\end{subarray}}m_{A_{1}}\widehat{\phi}(n_{2})\widehat{\phi}(n_{3})\widehat{\phi}(n_{4})$
(23) $\displaystyle\widehat{\mathcal{RE}_{4}}(\phi)(n)$
$\displaystyle=\widehat{\phi}(n)\sum_{\begin{subarray}{c}0=n_{1}+n_{2}+n_{4}\\\
n_{1},n_{2}\neq n\\\
\mathcal{P}_{3}(n,n_{1},n_{2},n+n_{4})\end{subarray}}m_{B_{3}}\widehat{\phi}(n_{1})\widehat{\phi}(n_{2})\widehat{\phi}(n_{4}),$

with symbols $m_{R}^{i}$, $i=1,2$ defined by

$\displaystyle m_{R}^{1}$
$\displaystyle=\frac{n(n_{1}^{2}+n_{2}^{2})}{\Phi_{2}(n_{1},n_{2})}$
$\displaystyle m_{R}^{2}$
$\displaystyle=\frac{n(n_{1}^{2}+(n_{2}+n_{3})^{2})(n_{2}+n_{3})(n_{2}^{2}+n_{3}^{2})}{\Phi_{2}(n_{1},n_{2}+n_{3})\Phi_{3}(n_{1},n_{2},n_{3})}\chi_{\mathcal{P}_{3}(n_{1},n_{2},n_{3})}.$

Note that these symbols are simply the symbols that arise (up to symmetry)
after performing the differentiation-by-parts procedure and substituting the
correction terms into the temporally differentiated term.

######  Remark 4.3.

The

Second Order Resonances

term in $\mathcal{RE}_{2}$ contains the double resonances coming from
$n_{1}=n_{2}=n$ in $D_{1}$ and $n_{3}=n_{4}=n$ in $D_{2}$. By the next two
lemmas we find that the symbols satisfy

$m_{D_{1}}(n,n,n_{3},n_{4}),m_{D_{2}}(n_{1},n_{2},n,n)=O(n).$

It follows that we will handle them in nearly the exact the same way as the
first term in $\mathcal{RE}_{2}$.

It’s important to notice that the above terms lack the complex exponential
terms that have phases given by $\Phi_{\ell}$. We choose the above
representations for ease of representation of the Duhamel formula, (30), that
will come. It is recommended that a reader, on a first read through of this
section, first compares the terms above to the final form of the equation,
(29), before moving on to the following lemmas.

We now perform analysis on several of the symbols that appear above which will
be useful later.

######  Lemma 4.4.

With $D_{1}$, $D_{2},$ $m_{D_{1}}$, and $m_{D_{2}}$ defined as above, we have:

(24)
$m_{D_{1}}=O\left(\frac{\max(n_{1}^{2},n_{2}^{2})}{\max\\{|n_{1}+n_{2}|,|n_{3}|,|n_{4}|\\}^{2}}\right),$

and

(25)
$m_{D_{2}}=O\left(\frac{\max(n_{3}^{2},n_{4}^{2})}{n_{1}\max\\{|n_{1}|,|n_{2}|,|n_{3}+n_{4}|\\}}\right).$

###### Proof.

This follows immediately from the summation restriction, the definition of the
symbols, and the factorization of $\Phi_{2}$. ∎

While $D_{1}$ and $D_{2}$ will correspond to the worst terms integrated in the
Duhamel formula, we still have access to Proposition 3.2 that will enable us
to handle the $D_{1}$ and $D_{2}$. This leaves us with having to understand
the resonances that we removed from these two terms, which occur when:

  * A)

In $D_{1}$: $n_{3},n_{4}\neq n$ and

    * $A_{1}$)

$n_{2}+n_{3}+n_{4}=0$,

    * $A_{2}$)

$n_{1}+n_{3}+n_{4}=0$,

  * B)

In $D_{2}$: $n_{1},n_{2}\neq n$ and

    * $B_{3})$

$n_{1}+n_{2}+n_{4}=0$,

    * $B_{4})$

$n_{1}+n_{2}+n_{3}=0$.

Note that Cases $A_{i}$ $i=1,2$ correspond to $D_{1}$ with symbol $m_{A_{i}}$
and similarly for cases $B_{i}$ and $m_{B_{i}}$.

######  Corollary 4.5.

Suppose that $n_{1},n_{2},n_{3},n_{4}$ are fixed, $n=n_{1}+n_{2}+n_{3}+n_{4}$,
and $|n|\gg n_{2}^{*}$.

  * I)

If $n=n_{i}$ for $i\in\\{1,2\\}$, then we have

(26) $m_{A_{i}}=O\left(\frac{n_{2}^{*}}{n}\right).$

  * II)

If $n=n_{i}$ for $i\in\\{3,4\\}$ and $j\in\\{3,4\\}\setminus\\{i\\}$, then we
have

(27)
$m_{B_{i}}+\frac{1}{25}\frac{n}{n_{1}n_{j}}+\frac{3n_{2}+6n_{j}}{25n_{1}n_{j}}=O\left(\frac{\max(|n_{2}|^{2},|n_{j}|^{2})}{n}\right).$

###### Proof.

Before beginning, we would like to comment on the $n_{i}^{*}$ notation in the
setting of the above lemma. In the context of case $I$ with $i=1$ we have that
$n_{i}^{*}$ is the $i^{\prime}$th largest entry of
$(|n|,|n_{2}|,|n_{3}|,|n_{4}|)$. And, as $n_{2}+n_{3}+n_{4}=0$, we always have
that $|n_{j_{1}}|\sim|n_{j_{2}}|$ where $j_{1}\neq j_{2}$ and
$|n_{j_{1}}|=\max(|n_{2}|,|n_{3}|,|n_{4}|).$

The first portion follows almost immediately from Lemma 4.4. Instead, we find
that

$|m_{A_{i}}|\lesssim\bigg{|}\frac{(n_{1}^{2}+n_{2}^{2})(n_{3}^{2}+n_{4}^{2})}{\max\\{|n_{1}+n_{2}|,|n_{3}|,|n_{4}|\\}^{4}}\bigg{|}\lesssim\bigg{|}\frac{n^{2}(n_{2}^{*})^{2}}{n^{4}}\bigg{|}\lesssim\frac{n_{2}^{*}}{|n|},$

as desired.

For the second portion, we assume by symmetry that $n_{3}=n$ and note that the
expansion of the numerator of $m_{B_{i}}$ is

$n^{9}+(3n_{2}+6n_{4})n^{8}+l.o.t.,$

where $l.o.t.$ denotes terms with $n$ degree strictly less than $8$ and total
degree equal to $9$. Similarly, we have for for $n\gg n_{2}^{*}$ and the
factorizations for $\Phi_{2}$ and $\Phi_{3}$ that

$\displaystyle\Phi_{2}(n_{1},n_{2}+n+n_{4})$
$\displaystyle\Phi_{3}(n_{1},n_{2},n+n_{4})$
$\displaystyle=25n^{8}n_{1}(n_{1}+n_{2})+O(n^{7}n_{4}^{2}n_{1}+n^{7}n_{4}n_{1}^{2})$
$\displaystyle=-25n^{8}n_{4}n_{1}+O(n^{7}n_{4}^{2}n_{1}+n^{7}n_{4}n_{1}^{2})$

that we have the desired expansion by crude big-$O$ manipulation.

∎

######  Remark 4.6.

By symmetry we have that

$\displaystyle\sum_{n_{1}+n_{2}+n_{4}=0}\frac{1}{n_{1}n_{4}}\widehat{u}(n_{1})\widehat{u}(n_{2})\widehat{u}(n_{4})=\sum_{\begin{subarray}{c}n_{1}+n_{2}+n_{4}=0\\\
|n|\gg\max(|n_{1}|,|n_{2}|,|n_{4}|)\end{subarray}}\frac{1}{n_{1}n_{4}}\widehat{u}(n_{1})\widehat{u}(n_{2})\widehat{u}(n_{4})=0.$

Additionally, we have by construction that

$\frac{3n_{2}+6n_{4}}{n_{1}n_{4}}=\frac{3(n_{2}+n_{4})}{n_{1}n_{4}}+\frac{3}{n_{1}}=\frac{-3}{n_{4}}+\frac{3}{n_{1}},$

and hence by symmetry that

$\sum_{n_{1}+n_{2}+n_{4}=0}\frac{3n_{2}+6n_{4}}{n_{1}n_{4}}\widehat{u}(n_{1})\widehat{u}(n_{2})\widehat{u}(n_{4})=\sum_{\begin{subarray}{c}n_{1}+n_{2}+n_{4}=0\\\
|n|\gg\max(|n_{1}|,|n_{2}|,|n_{4}|)\end{subarray}}\frac{3n_{2}+6n_{4}}{n_{1}n_{4}}\widehat{u}(n_{1})\widehat{u}(n_{2})\widehat{u}(n_{4})=0,$

for real $u$.

######  Remark 4.7.

It’s important to observe that the second cancellation structure explored in
Remark 4.6 is unimportant for proving local well-posedness. The use of the
above is purely in obtaining a better asymptotic bound in order to establish
smoothing.

Similarly, the cancellation of the first term is unimportant for the well-
posedness of the system. Because there is only one power of $n$ associated
with it we could have removed it in exactly the same way that we removed
$\mathcal{K}$.

We now undo the interaction representation. It follows that if

(28)
$\widehat{\tilde{u}}(n)=e^{n\int_{0}^{t}\mathcal{K}(\tilde{u})\,ds}\widehat{u}(n)=e^{n\int_{0}^{t}\mathcal{K}(v)\,ds}\widehat{u}(n),$

then $\tilde{u}$ satisfies (modulo constants)

(29) $\displaystyle\partial_{t}$ $\displaystyle
W_{-t}\tilde{u}=\partial_{t}W_{-t}B_{1}(\tilde{u})+\partial_{t}W_{-t}B_{2}(\tilde{u})+W_{-t}D_{1}(\tilde{u})+W_{-t}D_{2}(\tilde{u})$
$\displaystyle\qquad+W_{-t}(R_{1}(\tilde{u})+R_{2}(\tilde{u})+\mathcal{NR}_{1}(\tilde{u}))$
$\displaystyle\qquad+W_{-t}(\mathcal{RE}_{1}(\tilde{u})+\mathcal{RE}_{2}(\tilde{u})+\mathcal{RE}_{3}(\tilde{u})+\mathcal{RE}_{4}(\tilde{u})).$

with

$\tilde{u}(x,0)=u_{0}(x).$

As the the transformation mapping $u$ to $\tilde{u}$ is bi-continuous on
$H^{s}$ for $s>1/2$, we conclude the proof of Lemma 4.1.

Similarly, by invoking Duhamel we must have, modulo constants, that
$\tilde{u}$ is a fixed point of $\Gamma$ defined by

(30) $\displaystyle\Gamma[\phi]$
$\displaystyle:=B_{1}(\phi)(x,t)+B_{2}(\phi)(x,t)-W_{t}B_{1}(u_{0})(x)-W_{t}B_{2}(u_{0})(x)+W_{t}u_{0}(x)$
$\displaystyle+\int_{0}^{t}W_{t-s}\big{\\{}(D_{1}+D_{2}+\mathcal{NR}_{1})(\phi)(x,s)+(R_{1}+R_{2})(\phi)(x,s)$
$\displaystyle\quad+(\mathcal{RE}_{1}+\mathcal{RE}_{2}+\mathcal{RE}_{3}+\mathcal{RE}_{4})(\phi)(x,s)\big{\\}}\,ds.$

In the next few sections we will estimate the portions of $\Gamma$ in order to
establish that $\Gamma$ is a contraction in $Y^{s}$.

##  5\. Duhamel Estimates & Proof of Theorem 1.1

Space considerations necessitate the introduction of an extra parameter,
$\varepsilon.$ This parameter is a measure of extra smoothness and will help
shorten the subsequent sections and reduce repetition of the arguments. It is
recommended that the reader ignore the $\varepsilon$ factors on a first read
through of local well-posedness, and return to them when they reach the
nonlinear smoothing section.

We begin this section with a general estimate that will reduce repetitive
arguments when we have modulation considerations.

######  Proposition 5.1.

Let $\varphi_{i}\in Y^{1/2+}_{T}$ for $i\in\\{1,\cdots,m\\}$ and denote
$\mathcal{F}_{x}(\varphi_{i})(n)=\widehat{\varphi}_{n}$. Suppose that
$\varphi_{i}$ satisfy
$supp\,\mathcal{F}_{x}(\varphi_{i})\subset[-N_{i},N_{i}]$ for $N_{i}$ dyadic
with $N_{1}\geq\cdots\geq N_{m}$. Then there is $\theta>0$ so that

$\displaystyle\left\|P_{N}\mathcal{F}_{x}^{-1}\left(\sum_{\begin{subarray}{c}n=n_{1}+\cdots+n_{m}\\\
\mathcal{P}_{m}(n_{1},\cdots,n_{m})\end{subarray}}\prod_{i=1}^{m}\widehat{\varphi}_{n_{i}}\right)\right\|_{Z^{0}_{T}}\lesssim\frac{T^{\theta}N^{\frac{1}{2}-\frac{1}{p}}}{N_{1}}\|\varphi_{1}\|_{Y^{0}}\|\varphi_{2}\|_{Y^{1/p}}\prod_{i=2}^{m}\|\varphi_{i}\|_{Y^{1/2+}},$

for any $p\geq 2$.

Additionally, the above bound holds under the space-time constraint that
either $\langle\tau+n^{5}\rangle\gtrsim N_{1}^{4}$, or
$\langle\tau_{i}+n^{5}_{i}\rangle\gtrsim N_{1}^{4}$ for some
$i\in\\{1,\cdots,m\\},$ without the summation constraint
$\mathcal{P}_{m}(n_{1},\cdots,n_{m})$.

###### Proof.

The proof of the above proposition is standard, but included for completeness.
Before beginning we note that we may assume that the space-time Fourier
transforms of $\varphi_{i}$ are all positive, and that $n_{i}^{*}=|n_{i}|\sim
N_{i}$.

Note that the assumption $\mathcal{P}_{m}(n_{1},\cdots,n_{m})$ implies that

$\max(\langle\tau+n^{5}\rangle,\max_{j}(\langle\tau_{j}+n_{j}^{5}\rangle))\gtrsim
N_{1}^{4},$

and hence we may consider both claims in the proposition at once. We now
handle the two cases

  * I)

$\langle\tau+n^{5}\rangle\gtrsim n_{1}^{4}$,

  * II)

$\langle\tau_{j}+n_{j}^{5}\rangle\gtrsim n_{1}^{4}$.

Assuming that we have $I$, we find

$\bigg{\|}P_{N}\mathcal{F}_{x}^{-1}\bigg{(}\sum_{\begin{subarray}{c}n=n_{1}+\cdots+n_{m}\\\
\mathcal{P}_{m}(n_{1},\cdots,n_{m})\end{subarray}}\prod_{i=1}^{m}\widehat{\varphi}_{n_{i}}\bigg{)}\bigg{\|}_{X^{0,-3/4}_{T}}\lesssim
T^{\theta}\bigg{\|}P_{N}\mathcal{F}_{x}^{-1}\bigg{(}\sum_{\begin{subarray}{c}n=n_{1}+\cdots+n_{m}\\\
\mathcal{P}_{m}(n_{1},\cdots,n_{m})\end{subarray}}\prod_{i=1}^{m}\widehat{\varphi}_{n_{i}}\bigg{)}\bigg{\|}_{X^{0,-1/2+}}\\\
\lesssim\frac{T^{\theta}}{N_{1}^{2-}}\bigg{\|}\chi_{N}(n)\sum_{\begin{subarray}{c}n=n_{1}+\cdots+n_{m}\\\
\mathcal{P}_{m}(n_{1},\cdots,n_{m})\end{subarray}}\prod_{i=1}^{m}\widehat{\varphi}_{n_{i}}\bigg{\|}_{X^{0,0}}\lesssim\frac{T^{\theta}}{N_{1}^{2-}}\|\varphi_{1}\|_{X^{0,0}}\prod_{i=2}^{m}\|\varphi_{i}\|_{L^{\infty}_{t}L^{\infty}_{x}},$

so that Sobolev embedding concludes the proof. The second portion of the norm
follows from the above estimate by Cuachy-Schwarz in $\tau$:

$\big{\|}\tfrac{1}{\langle\tau+n^{5}\rangle}\mathcal{F}_{x,t}(\cdot)\big{\|}_{\ell^{2}_{n}L^{1}_{\tau}}\lesssim\|\cdot\|_{X^{0,-1/2+}}.$

We now assume case $II$, in which case we first take a minor loss, which will
make later work easier:

(31)
$\displaystyle\bigg{\|}P_{N}\mathcal{F}_{x}^{-1}\bigg{(}\sum_{\begin{subarray}{c}n=n_{1}+\cdots+n_{m}\\\
\mathcal{P}_{m}(n_{1},\cdots,n_{m})\end{subarray}}\prod_{i=1}^{m}\widehat{\varphi}_{n_{i}}\bigg{)}\bigg{\|}_{X^{0,-3/4}_{T}}$
$\displaystyle\qquad\qquad\lesssim
T^{\theta}\bigg{\|}P_{N}\mathcal{F}_{x}^{-1}\bigg{(}\sum_{\begin{subarray}{c}n=n_{1}+\cdots+n_{m}\\\
\mathcal{P}_{m}(n_{1},\cdots,n_{m})\end{subarray}}\prod_{i=1}^{m}\widehat{\varphi}_{n_{i}}\bigg{)}\bigg{\|}_{X^{0,-(1/2-)}}.$

and invoke duality for $w\in X^{0,1/2-}$

$\displaystyle\bigg{\|}P_{N}\mathcal{F}_{x}^{-1}\bigg{(}\sum_{\begin{subarray}{c}n=n_{1}+\cdots+n_{m}\\\
\mathcal{P}_{m}(n_{1},\cdots,n_{m})\end{subarray}}\prod_{i=1}^{m}\widehat{\varphi}_{n_{i}}\bigg{)}\bigg{\|}_{X^{0,-(1/2-)}}$
$\displaystyle\qquad\qquad=\int_{\tau=\tau_{1}+\cdots\tau_{m}}\sum_{\begin{subarray}{c}n=n_{1}+\cdots+n_{m}\\\
\mathcal{P}_{m}(n_{1},\cdots,n_{m})\end{subarray}}\widehat{w}(n,\tau)\prod_{i=1}^{m}\mathcal{F}_{x,t}\left(\varphi_{i}\right)(n_{i},\tau_{i})\,d\Gamma$
(32)
$\displaystyle\qquad\qquad\lesssim\frac{1}{N_{1}}\int_{\tau=\tau_{1}+\cdots\tau_{m}}\sum_{\begin{subarray}{c}n=n_{1}+\cdots+n_{m}\\\
\mathcal{P}_{m}(n_{1},\cdots,n_{m})\end{subarray}}\widehat{w}(n,\tau)\langle\tau_{j}+n_{j}^{5}\rangle^{1/4}\prod_{i=1}^{m}\mathcal{F}_{x,t}\left(\varphi_{i}\right)(n_{i},\tau_{i})\,d\Gamma.$

If $j=1$, then we find

$\eqref{Equation: Modulation proposition xsb to
show}\lesssim\frac{1}{N_{1}}\|w\|_{L^{\infty-}_{t}L^{p}_{x}}\left\|\mathcal{F}^{-1}_{n_{1},\tau_{1}}\left(\langle\tau_{1}+n_{1}^{5}\rangle^{1/4}\mathcal{F}_{x,t}(\varphi_{1})\right)\right\|_{L^{2}_{t}L^{2}_{x}}\\\
\times\|\varphi_{2}\|_{L^{2(m-1)+}_{t}L_{x}^{\frac{2p}{p-2}}}\prod_{i=3}^{m}\|\varphi_{i}\|_{L^{2(m-1)+}_{t}L^{\infty}_{x}},$

where Sobolev embedding and modulation completes the estimate. When $j>1$ then
a similar estimate holds

$\displaystyle\eqref{Equation: Modulation proposition xsb to
show}\lesssim\frac{1}{N_{1}}\|w\|_{L^{\infty-}_{t}L^{p}_{x}}\|\varphi_{1}\|_{L^{2+}_{t}L^{2}_{x}}\left\|\mathcal{F}^{-1}_{n_{j},\tau_{j}}\left(\langle\tau_{j}+n_{j}^{5}\rangle^{1/4}\mathcal{F}_{x,t}(\varphi_{j})\right)\right\|_{L^{2}_{t}L^{\frac{2p}{p-2}}_{x}}\prod_{\begin{subarray}{c}i=2\\\
i\neq j\end{subarray}}^{m}\|\varphi_{i}\|_{L^{\infty}_{t}L^{\infty}_{x}},$

where again Sobolev embedding and modulation will complete the claim. The case
that $j\neq 2$ is exactly the same as above, except one would use
$L^{2}_{t}L^{\infty}_{x}$ on $\varphi_{j}$ and
$L^{\infty}_{t}L^{\frac{2p}{p-2}}$ on $\varphi_{2}$.

The second portion of the norm in case $II$ follows from the same Cuachy-
Schwarz argument:

$\big{\|}\tfrac{1}{\langle\tau+n^{5}\rangle}\mathcal{F}_{x,t}(\cdot)\big{\|}_{\ell^{2}_{n}L^{1}_{\tau}}\lesssim\|\cdot\|_{X^{0,-(1/2-)}},$

and the prior estimates via the reduction done on (31). ∎

######  Lemma 5.2 (Propagator Terms).

Let $s>1/2$ and $0\leq\varepsilon<1$. Then

$\|B_{1}(\tilde{u})+B_{2}(\tilde{u})\|_{C^{0}_{t}H^{s+\varepsilon}_{x}}\lesssim\|\tilde{u}\|_{C^{0}_{t}H^{s}_{x}}^{2}+\|\tilde{u}\|_{C^{0}_{t}H^{s}_{x}}^{3}.$

###### Proof.

We first apply Plancherel’s and estimate the $B_{1}$ term:

$\|B_{1}(\tilde{u})\|_{C^{0}_{t}H^{s+\varepsilon}_{x}}\sim\bigg{\|}\underset{\begin{subarray}{c}n=n_{1}+n_{2}\\\
\star\end{subarray}}{\int\sum}\frac{n(n_{1}^{2}+n_{2}^{2})\langle
n\rangle^{s+\varepsilon}}{\Phi_{2}(n_{1},n_{2})}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\bigg{\|}_{C^{0}_{t}\ell^{2}_{n}}\\\
\lesssim\bigg{\|}\underset{n=n_{1}+n_{2}}{\int\sum}\frac{\langle
n\rangle^{s+\varepsilon}}{n_{1}n_{2}}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\bigg{\|}_{C^{0}_{t}\ell^{2}_{n}}\lesssim\|\tilde{u}\|_{C^{0}_{t}H^{s}}\|J_{x}^{-1}\tilde{u}\|_{C^{0}_{t}L^{\infty}_{x}}\lesssim\|\tilde{u}\|_{C^{0}_{t}H^{s}}^{2},$

for $0\leq\varepsilon\leq 1$.

As for $B_{2}$, note that the symbol is defined at the beginning of Section
4.2 and by assumption we have that $\mathcal{P}_{3}(n_{1},n_{2},n_{3})$ holds.
Thus $|\Phi_{3}(n_{1},n_{2},n_{3})|\gtrsim(n_{1}^{*})^{4}$ and the symbol is
$O(1/n),$ so

$\|B_{2}(\tilde{u})\|_{C^{0}_{t}H^{s+\varepsilon}}\lesssim\|\tilde{u}\|_{C^{0}_{t}H^{s}}^{3},$

by Sobolev embedding in exactly the same way we found the prior estimate. ∎

######  Lemma 5.3 (Boundary Terms).

Let $s>1/2$. Then

$\|B_{1}(\tilde{u})(t)+B_{2}(\tilde{u})(t)\|_{Y^{s}_{T}}\lesssim\|\tilde{u}\|_{Y^{s}_{T}}^{2}+\|\tilde{u}\|_{Y^{s}_{T}}^{3}.$

###### Proof.

We first estimate the $X^{s,1/4}$ portion of $B_{1}$. If, on the one hand,
there exists an $1\leq i\leq 2$ with
$\langle\tau_{i}+n_{i}^{5}\rangle\gtrsim\langle\tau+n^{5}\rangle$, then we
have

$\displaystyle\|B_{1}(\tilde{u})\|_{X^{s,1/4}}$
$\displaystyle\lesssim\bigg{\|}\underset{\begin{subarray}{c}n=n_{1}+n_{2}\\\
\tau=\tau_{1}+\tau_{2}\end{subarray}}{\int\sum}\frac{n(n_{1}^{2}+n_{2}^{2})\langle
n\rangle^{s}\langle\tau_{1}+n_{1}^{5}\rangle^{1/4}}{\Phi_{2}(n_{1},n_{2})}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\bigg{\|}_{\ell^{2}_{t}L^{2}_{\tau}}$
$\displaystyle\lesssim\bigg{\|}\underset{\begin{subarray}{c}n=n_{1}+n_{2}\\\
\tau=\tau_{1}+\tau_{2}\end{subarray}}{\int\sum}\frac{\langle
n\rangle^{s}\langle\tau_{1}+n_{1}^{5}\rangle^{1/4}}{n_{1}n_{2}}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\bigg{\|}_{\ell^{2}_{t}L^{2}_{\tau}}.$

At worst we have that $|n_{1}|\gtrsim|n_{2}|$, which results in

$\left\|\mathcal{F}^{-1}_{n,\tau}\left(\langle
n\rangle^{s-1}\langle\tau+n^{5}\rangle^{1/4}\widehat{\tilde{u}}\right)\right\|_{L^{2}_{x,t}}\|J_{x}^{-1}\tilde{u}\|_{L^{\infty}_{x,t}}\lesssim\|\tilde{u}\|_{X^{s,1/4}}\|\tilde{u}\|_{Y^{s}}.$

This holds for $B_{2}$ as well, since the symbol will be

$O\bigg{(}\frac{1}{n_{1}(n_{1}+n_{2})(n_{2}+n_{3})(n_{1}+n_{3})}\bigg{)}.$

As $n_{1}(n_{1}+n_{2})(n_{2}+n_{3})(n_{1}+n_{3})\gtrsim n_{1}^{*}n_{3}^{*}$,
we may carry out the same argument as before.

We may now assume that

$\langle\tau_{i}+n_{i}^{5}\rangle\ll\langle\tau+n^{5}\rangle\sim|nn_{1}n_{2}|(n_{1}^{2}+n_{2}^{2}+n^{2})\sim|\Phi_{2}(n_{1},n_{2})|,$

for $i=1,2$. It follows then, that for $|n_{1}|\gtrsim|n_{2}|$

$\|B_{1}(\tilde{u})\|_{X^{s,1/4}}\lesssim\bigg{\|}\underset{\begin{subarray}{c}n=n_{1}+n_{2}\\\
\tau=\tau_{1}+\tau_{2}\end{subarray}}{\int\sum}\frac{n(n_{1}^{2}+n_{2}^{2})\langle
n\rangle^{s}}{\Phi_{2}(n_{1},n_{2})^{3/4}}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\bigg{\|}_{\ell^{2}_{t}L^{2}_{\tau}}\\\
\lesssim\bigg{\|}\underset{\begin{subarray}{c}n=n_{1}+n_{2}\\\
\tau=\tau_{1}+\tau_{2}\end{subarray}}{\int\sum}\frac{\langle
n\rangle^{s}}{n_{2}^{3/4}}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\bigg{\|}_{\ell^{2}_{t}L^{2}_{\tau}}\lesssim\|\tilde{u}\|_{X^{s,0}_{T}}\|J_{x}^{-3/4}\tilde{u}\|_{L^{\infty}_{x,t}}\lesssim\|\tilde{u}\|_{Y^{s}_{T}}^{2}.$

For $B_{2}$, the fact that $\mathcal{P}_{3}(n_{1},n_{2},n_{3})$ holds implies
that the symbol multiplied by $\Phi_{3}^{1/4}$ will be

$O\bigg{(}\frac{(n_{2}^{2}+n_{3}^{2})}{n_{1}n_{1}^{*}}\bigg{)}=O(1),$

from which it follows that

$\|B_{2}(\tilde{u})\|_{X^{s,1/4}_{T}}\lesssim\|\tilde{u}\|_{X^{s,0}_{T}}\|\tilde{u}\|_{Y^{s}_{T}}^{2}.$

The $\ell^{2}_{n}L^{1}_{\tau}$ portion of the norms is fairly straight
forward. For $B_{1}$ we assume $|n_{1}|\gtrsim|n_{2}|$ so that the convolution
structure together with the symbol of $B_{1}$ being $O(1)$ gives

$\displaystyle\|\langle
n\rangle^{s}B_{1}(\tilde{u})\|_{\ell^{2}_{n}L^{1}_{\tau}}$
$\displaystyle\lesssim\left\|\|J^{s}\mathcal{F}_{t}(\tilde{u})\|_{L^{1}_{\tau}}\|\mathcal{F}_{t}(\tilde{u})\|_{L^{1}_{\tau}}\right\|_{L^{2}_{x}}\lesssim\|\langle
n_{1}\rangle^{s}\widehat{\tilde{u}}\|_{\ell^{2}_{n}L^{1}_{\tau}}\|\mathcal{F}_{t}(\tilde{u})\|_{L^{\infty}_{x}L^{1}_{\tau}}\lesssim\|\tilde{u}\|_{Y^{s}_{T}}^{2}.$

Since the second portion for $B_{2}$ follows in the exact same manner, we
conclude the proof. ∎

######  Lemma 5.4 (Correction Term Penalties).

Let $s>1/2$ and $0\leq\varepsilon<1$. Then there is a $\theta>0$ so that

$\|R_{1}(\tilde{u})+R_{2}(\tilde{u})\|_{Z^{s+\varepsilon}_{T}}\lesssim
T^{\theta}\left(\|\tilde{u}\|_{Y^{s}_{T}}^{4}+\|\tilde{u}\|_{Y^{s}_{T}}^{5}\right).$

###### Proof.

Recall that these terms are defined on (18) and (19). We begin with the
$X^{s,-3/4}$ portion, where the symbols associated to these are (modulo
$\mathcal{K}$ terms)

(33) $\displaystyle|m_{R}^{1}(n_{1}+n_{2})|$
$\displaystyle=\bigg{|}\frac{n(n_{1}+n_{2})(n_{1}^{2}+n_{2}^{2})}{\Phi_{2}(n_{1},n_{2})}\bigg{|}\lesssim
1,$ (34) $\displaystyle|m_{R}^{2}(n_{1}+n_{2}+n_{3})|$
$\displaystyle=\bigg{|}\frac{n(n_{1}^{2}+(n_{2}+n_{3})^{2})(n_{2}+n_{3})(n_{2}^{2}+n_{3}^{2})(n_{1}+n_{2}+n_{3})}{\Phi_{2}(n_{1},n_{2}+n_{3})\Phi_{3}(n_{1},n_{2},n_{3})}\bigg{|}$
$\displaystyle\qquad\lesssim\bigg{|}\frac{n_{1}+n_{2}+n_{3}}{n_{1}(n_{1}+n_{2})(n_{2}+n_{3})(n_{1}+n_{3})}\bigg{|}\lesssim
1.$

Furthermore, we note that the second carries the restriction
$|\Phi_{3}(n,n_{1},n_{2},n_{3})|\gtrsim(n_{1}^{*})^{4}$. Since we may use
modulation considerations for all the terms present, we focus instead on the
trilinear term $R_{2}$. We first turn the trilinear operator into a 5-linear
operator by rewriting it as

$\displaystyle\widehat{R_{2}}(\tilde{u})(n)$
$\displaystyle=\frac{4i}{5}\sum_{\begin{subarray}{c}n=n_{1}+n_{2}+n_{3}+n_{4}+n_{5}\\\
0=n_{4}+n_{5}\\\ n\neq
n_{1},\,n_{2},\,n_{3}\end{subarray}}\frac{m_{R}^{2}(n_{1}+n_{2})}{n_{4}n_{5}}\prod_{j=1}^{5}\widehat{\tilde{u}}(n_{j}),$

and note that $\mathcal{P}_{5}(n_{1},n_{2},n_{3},n_{4},-n_{4})$ holds true, as

$|\Phi_{5}(n_{1},n_{2},n_{3},n_{4},-n_{4})|=|\Phi_{3}(n_{1},n_{2},n_{3})|\gtrsim(n_{1}^{*})^{4}.$

We then perform a Littlewood-Paley decomposition and invoke Proposition 5.1
under the assumption that $N\sim N_{1}$ and $p=2$ to obtain

$\displaystyle\|R_{2}(\tilde{u}))\|_{Z^{s+\varepsilon}_{T}}$
$\displaystyle\lesssim T^{\theta}\sum_{N\sim N_{1}\geq N_{2}\geq
N_{3}}\frac{N^{s+\varepsilon}}{N_{1}}N_{2}^{1/2+}N_{3}^{1/2+}\prod_{i=1}^{3}\|P_{N_{i}}(\eta(t/T)\tilde{u})\|_{Y^{0}}\lesssim
T^{\theta}\|\tilde{u}\|_{Y^{s}_{T}}^{5},$

for $s>1/2$ and $0\leq\varepsilon<1$. Since the case that $N_{1}\sim N_{2}\gg
N$ follows similarly, we conclude the proof of the lemma. ∎

The following lemma concerns the resonant terms from our analysis. The main
tool in our analysis will be Corollary 4.5. These estimate are slightly
involved due to the summation restrictions, but the key tool is still the
high-frequency approximation from the aforementioned corollary.

######  Lemma 5.5 (Resonant Terms).

Let $s>1/2$ and $\varepsilon<\min(2s-1,1)$. Then there is a $\theta>0$ so that

$\|\mathcal{RE}_{1}(\tilde{u})+\mathcal{RE}_{2}(\tilde{u})+\mathcal{RE}_{3}(\tilde{u})+\mathcal{RE}_{4}(\tilde{u})\|_{Z^{s+\varepsilon}}\lesssim
T^{\theta}\left(\|\tilde{u}\|_{Y^{s}_{T}}^{3}+\|\tilde{u}\|_{Y^{s}_{T}}^{4}\right).$

###### Proof.

We first note that these terms are defined on (20), (21), (22), and (23). We
will proceed by bounding the $X^{s,b}$ portion of the norm for every term
under consideration.

Consider first the $\mathcal{RE}_{1}$ term. We may split the sum into when
$n_{3}>0$ and when $n_{3}<0$. We get

$\displaystyle\widehat{\mathcal{RE}_{1}}(\tilde{u})$
$\displaystyle\sim\widehat{\tilde{u}}(n)\sum_{n_{3}>0}\left(\frac{n^{2}n_{3}}{n^{2}+(n+n_{3})^{2}+n_{3}^{2}}-\frac{n^{2}n_{3}}{n^{2}+(n-n_{3})^{2}+n_{3}^{2}}\right)\widehat{\tilde{u}}(n_{3})\widehat{\tilde{u}}(-n_{3})$
$\displaystyle\qquad+\widehat{\tilde{u}}(n)\sum_{n_{3}}\frac{nn_{3}^{2}}{n^{2}+(n+n_{3})^{2}+n_{3}^{2}}\widehat{\tilde{u}}(n_{3})\widehat{\tilde{u}}(-n_{3})$
(35)
$\displaystyle=-\widehat{\tilde{u}}(n)\sum_{n_{3}>0}\frac{4n^{3}n_{3}^{2}}{(n^{2}+(n+n_{3})^{2}+n_{3}^{2})(n^{2}+(n-n_{3})^{2}+n_{3}^{2})}\widehat{\tilde{u}}(n_{3})\widehat{\tilde{u}}(-n_{3})$
(36)
$\displaystyle\qquad+\widehat{\tilde{u}}(n)\sum_{n_{3}}\frac{nn_{3}^{2}}{n^{2}+(n+n_{3})^{2}+n_{3}^{2}}\widehat{\tilde{u}}(n_{3})\widehat{\tilde{u}}(-n_{3}).$

If $|n|\gg|n_{3}|$ then we find that both symbols are $O(n_{3}^{2}/n)$,
whereas if $|n|\lesssim|n_{3}|$ they are $O(n)=O(n_{3})$. Since the latter is
easy to handle, we assume that $|n|\gg|n_{3}|$ and write

$\frac{n_{3}^{2}\langle n\rangle^{s+\varepsilon}}{n}\lesssim\langle
n\rangle^{s+\varepsilon}\langle n_{3}\rangle^{2s}\langle
n_{3}\rangle^{2(1-s)}\langle n\rangle^{-1}\lesssim\langle
n\rangle^{s+\varepsilon}\langle n_{3}\rangle^{2s}\langle
n\rangle^{\max(2(1-s),0)-1}.$

Hence, for $s>1/2$ both of these sums satisfy

$\|\eqref{Equation: RE1 first sum}+\eqref{Equation: RE1 second
equation}\|_{X^{s+\varepsilon,-3/4}}\lesssim\|\eqref{Equation: RE1 first
sum}+\eqref{Equation: RE1 second
equation}\|_{X^{s+\varepsilon,0}}\lesssim\|\tilde{u}\|_{X^{s+\varepsilon-\min(2(s-1/2),1),0}}\|\tilde{u}\|_{L^{\infty}_{t}H^{s}_{x}}^{2}.$

Note that this also handles the second portion of the norm by Cauchy-Schwarz,
as in the proof of Proposition 5.1.

We now handle $\mathcal{RE}_{2}$ and the other multiple resonances from Remark
4.3. We will handle just a prototypical 4-linear term that appears from double
resonances in $D_{2}$. By lemma 4.4 and the fact that the trilinear term in
$\mathcal{RE}_{2}$ has symbol $O(n)$, we may assume that the symbol on the
$4$-linear term under consideration is $O(n)$. It then suffices by Cauchy-
Schwarz to bound

$\left\|\langle
n\rangle^{s+\varepsilon}n\widehat{\tilde{u}}(n)\widehat{\tilde{u}}(n)\sum_{-n=n_{1}+n_{2}}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\right\|_{L^{2}_{t}\ell^{2}_{n}}.$

We then bound this by

$\displaystyle\bigg{\|}\langle
n\rangle^{1/2+\varepsilon/2}\widehat{\tilde{u}}(n)$ $\displaystyle\langle
n\rangle^{1/2+\varepsilon/2}\widehat{\tilde{u}}(n)\langle
n\rangle^{s}\sum_{-n=n_{1}+n_{2}}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\bigg{\|}_{L^{2}_{t}\ell^{2}_{n}}$
$\displaystyle\lesssim\bigg{\|}\|\langle
n\rangle^{1/2+\varepsilon/2}\widehat{\tilde{u}}(n)\|_{\ell^{2}_{n}}^{2}\big{\|}\langle
n\rangle^{s}\sum_{-n=n_{1}+n_{2}}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\big{\|}_{\ell^{2}_{n}}\bigg{\|}_{L^{2}_{t}}$
$\displaystyle\lesssim\bigg{\|}\|\langle
n\rangle^{1/2+\varepsilon/2}\widehat{\tilde{u}}(n)\|_{\ell^{2}_{n}}^{2}\|\langle
n\rangle^{s}\widehat{\tilde{u}}(n)\|_{\ell^{2}_{n}}\|\widehat{\tilde{u}}(n)\|_{\ell^{1}_{n}}\bigg{\|}_{L^{2}_{t}}$
$\displaystyle\lesssim\bigg{\|}\|\langle
n\rangle^{1/2+\varepsilon/2}\widehat{\tilde{u}}(n)\|_{\ell^{2}_{n}}^{2}\|\langle
n\rangle^{s}\widehat{\tilde{u}}(n)\|_{\ell^{2}_{n}}\|n^{1/2+}\widehat{\tilde{u}}(n)\|_{\ell^{2}_{n}}\bigg{\|}_{L^{2}_{t}}$
$\displaystyle\lesssim\|\tilde{u}\|_{X^{s,0}_{T}}\|\tilde{u}\|_{Y^{1/2+\varepsilon/2}_{T}}^{2}\|\tilde{u}\|_{Y^{1/2+}_{T}}.$

This is acceptable so long as $0\leq\varepsilon<2(s-1/2).$

As for $\mathcal{RE}_{3}$ and $\mathcal{RE}_{4}$, we restrict ourselves to
simply bounding the harder term, $\mathcal{RE}_{4}$. We may assume that $i=3$
and decompose the symbol of the operator into

(37) $\displaystyle m_{B_{3}}=m_{B_{3}}\chi_{|n|\gg
n_{2}^{*}}+m_{B_{3}}\chi_{|n|\lesssim n_{2}^{*}},$

and define the symbol (with vanishing associated multilinear operator, by
Remark 4.6)

$\Psi=\left(\frac{n}{25n_{1}n_{4}}+\frac{3n_{2}+6n_{4}}{25n_{1}n_{4}}\right),$

and the splitting

(38) $\displaystyle\Psi$
$\displaystyle=\Psi\chi_{\mathcal{P}_{3}(n_{1},n_{2},n+n_{4})}\chi_{|n|\gg
n_{2}^{*}}$ (39)
$\displaystyle+\Psi\chi_{\mathcal{P}_{3}^{c}(n_{1},n_{2},n+n_{4})}\chi_{|n|\gg
n_{2}^{*}}$ (40) $\displaystyle+\Psi\chi_{|n|\lesssim n_{2}^{*}}.$

We handle first the (40) term. Note that, after Cauchy-Schwarz in $\tau$, the
quartilinear operator with symbol (40) reduces to bounding

(41) $\displaystyle\bigg{\|}\langle
n\rangle^{s+\varepsilon}\tilde{u}(n)\sum_{\begin{subarray}{c}0=n_{1}+n_{2}+n_{4}\\\
|n|\lesssim
n_{2}^{*}\end{subarray}}\frac{n_{2}^{*}}{n_{1}n_{4}}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\widehat{\tilde{u}}(n_{4})\bigg{\|}_{L^{2}_{t}\ell^{2}_{n}}.$

By the algebraic relationship we may assume that $n_{1}\sim n_{2}\gtrsim n$,
so that

$\displaystyle\eqref{Equation: Approximation with large
frequencies}\lesssim\bigg{\|}\langle
n\rangle^{s}\tilde{u}(n)\sum_{0=n_{1}+n_{2}+n_{4}}\langle
n_{1}n_{2}\rangle^{\varepsilon/2}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\widehat{\tilde{u}}(n_{4})\bigg{\|}_{L^{2}_{t}\ell^{2}_{n}}\lesssim\|\tilde{u}\|_{X^{s,0}_{T}}\left\|(J_{x}^{\frac{\varepsilon}{2}}\tilde{u})^{2}\tilde{u}\right\|_{L^{\infty}_{t}L^{1}_{x}}.$

Sobolev Embedding and Cauchy-Schwarz on the last term completes the estimate,
which will be good so long as $0\leq\varepsilon\leq 2s-1.$

As for (39), we find from the assumption that

$\Phi_{3}(n_{1},n_{2},n+n_{4})\ll\max\\{|n_{1}|,|n_{2}|,|n+n_{4}|\\}^{4}$

and Proposition 3.1 that one of the following must hold:

  * I)

$n\in\\{n_{1},n_{2},n+n_{4}\\}$, or

  * II)

$n_{1}\sim n_{2}\sim n+n_{4}$.

If $I$ holds, then by the assumption of mean-zero $\tilde{u}$ we must have
that either $n_{1}=n$ or $n_{2}=n$. If $II$ holds then we find that $n_{1}\sim
n_{2}\sim n+n_{4}\sim n$. In both cases we then have, by the algebraic
relationship $0=n_{1}+n_{2}+n_{4}$, that two of $n_{1},n_{2},n_{4}$ are
$\gtrsim n$, where the implied constant might be different from the one in the
relationship $n\gg n_{2}^{*}$. It then follows that we may conclude this case
as in the bound for (41), which holds for $0\leq\varepsilon<\min(2s-1,1)$.

The last easy bound to show is the bound for $\mathcal{RE}_{4}$ with symbol
given by $m_{B_{3}}\chi_{n\lesssim n_{2}^{*}}$. In this case we find by (25)
that

$m_{B_{3}}\chi_{|n|\lesssim
n_{2}^{*}}\lesssim\max\\{|n_{1}|,|n_{2}|,|n_{4}|\\}.$

By symmetry and Cauchy-Schwarz we find that it suffices to bound

(42) $\bigg{\|}\langle
n\rangle^{s+\varepsilon}\widehat{\tilde{u}}(n)\sum_{\begin{subarray}{c}0=n_{1}+n_{2}+n_{4}\\\
|n_{1}|\gtrsim
n\end{subarray}}n_{1}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\widehat{\tilde{u}}(n_{4})\bigg{\|}_{L^{2}_{\tau}\ell^{2}_{n}}.$

Since the algebraic relation forces there to be two large frequencies within
the sum, we find

$\eqref{Equation: RE4 multiple big n2
big}\lesssim\|\tilde{u}\|_{X^{s,0}_{T}}\left\|(J_{x}^{\frac{1+\varepsilon}{2}}\tilde{u})^{2}\tilde{u}\right\|_{L^{\infty}_{t}L^{1}_{x}}\lesssim\|\tilde{u}\|_{X^{s,0}_{T}}\|\tilde{u}\|_{Y^{s}_{T}}^{3},$

by taking the $\sup$ in $n$ of the interior sum, for $0\leq\varepsilon<2s-1.$

The last bound we need to show is (38) grouped with the first term of (37),
which reduces to bounding

(43) $\bigg{\|}\langle
n\rangle^{s+\varepsilon}\widehat{\tilde{u}}(n)\sum_{\begin{subarray}{c}0=n_{1}+n_{2}+n_{4}\\\
|n|\gg|n_{1}|,|n_{2}|,|n_{4}|\end{subarray}}\frac{(n_{2}^{*})^{2}}{n}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\widehat{\tilde{u}}(n_{4})\bigg{\|}_{L^{2}_{\tau}\ell^{2}_{n}},$

by Corollary 4.5. Without loss of generality we assume $|n_{1}|\sim|n_{2}|$,
so that for $0\leq\varepsilon<\min(2(s-1/2),1)$ we find

$\frac{(n_{1})^{2}}{n^{1-\varepsilon}}\lesssim(n_{1}n_{2})^{\min(s,1)},$

which yields

$\|\tilde{u}\|_{X^{s,0}_{T}}\|\tilde{u}\|_{Y^{\min(s,1)}_{T}}^{2}\|\tilde{u}\|_{Y^{1/2+}_{T}},$

by the same method as in (42).

In every instance the second portion of the norm follows from the first
portion via Cauchy-Schwarz, and we similarly find a factor of $T$ by
modulation considerations for every bound, completing the proof. ∎

###  5.1. Main Terms in the Duhamel Estimate

Before we begin with the main Duhamel terms, we would like to note that we
could have consolidated the two conditions $n_{1}^{*}+n_{2}^{*}=0$ and
$n_{3}\gtrsim n_{1}$ into $n_{3}\gtrsim n$ in Proposition 3.2 However, the
condition $n_{1}^{*}+n_{2}^{*}=0$ provides better bounds, as is evident by
Corollary 5.6. Moreover, from a well-posedness perspective, a bound relating
$n_{3}$ to $n$ is not nearly as good as a bound relating $n_{3}$ with $n_{1}$.
In particular, the latter bound gives more information about $high\times
high\times low\cdots$ interactions.

######  Corollary 5.6.

Suppose that $n_{1}^{*}+n_{2}^{*}=0$, in the language of Proposition 3.2. Then

$m_{D_{1}}=O(1),\qquad\qquad m_{D_{2}}=O(n_{2}^{*}).$

###### Proof.

Note that by the external derivatives in the nonlinearity of (3) we have

  * I)

for $m_{D_{1}}$ we cannot have $n_{1}+n_{2}=0$, and

  * II)

for $m_{D_{2}}$ we cannot have $n_{3}+n_{4}=0$.

It follows that the denominators in Lemma 4.4 are large. The result then
follows from the bounds (24) and (25). ∎

We now perform a decomposition motivated by the next section, and disjointly
decompose $\mathbb{Z}^{k}$ into $\mathcal{A}_{k}\sqcup\mathcal{B}_{k}$, where

$\mathcal{A}_{k}:=\left\\{(n_{1},\cdots,n_{k})|\,\,\mathcal{P}_{k}(n_{1},\cdots,n_{k})\right\\}=\left\\{(n_{1},\cdots,n_{k})|\,\,|\Phi_{k}(n_{1},\cdots,n_{k})|\gtrsim(n_{1}^{*})^{4}\right\\}.$

Using this, we split $D_{i}$ into

$\displaystyle D_{1}$ $\displaystyle=\mathcal{NR}_{2}+D_{\mathcal{A},1}$
$\displaystyle D_{2}$ $\displaystyle=\mathcal{NR}_{3}+D_{\mathcal{A},2},$

where $D_{\mathcal{A},i}$ are defined by the symbols
$\chi_{\mathcal{A}_{4}}m_{D_{i}}$, and $\mathcal{NR}_{i}$ by
$\chi_{\mathcal{B}_{4}}m_{D_{i}}$.

######  Lemma 5.7 ($\mathcal{NR}$ Terms444This should be read as nearly-
resonant, as opposed to non-resonant.).

Let $a\geq 0$ be such that ($P_{a}$) holds, $s>\frac{1+a}{2}$, and

$0<\varepsilon<\min\left(2s-1-a,1\right).$

Then

$\|\mathcal{NR}_{1}(\tilde{u})+\mathcal{NR}_{2}(\tilde{u})+\mathcal{NR}_{3}(\tilde{u})\|_{Z_{T}^{s+\varepsilon}}\lesssim_{\varepsilon}T^{\theta}\|\tilde{u}\|_{X_{T}^{s,1/4}}\left(\|\tilde{u}\|_{Y_{T}^{s}}^{2}+\|\tilde{u}\|_{Y_{T}^{s}}^{3}\right).$

###### Proof.

Since $\mathcal{NR}_{1}$ has three large similar frequencies, we will instead
show the bound for the harder terms. Note that we have no resonance by the
definition of $D_{1}$ and $D_{2}$, and by the definition of $\mathcal{B}_{4}$,
Proposition 3.2, and Remark 3.3 we have the following cases:

  * I)

$n_{1}^{*}+n_{2}^{*}=0$,

  * II)

$(n_{3}^{*})^{4}n_{4}^{*}\gtrsim(n_{1}^{*})^{4}$,

where the symbols under consideration are
$O\left(\frac{(n_{1}^{*})^{2}}{n_{3}^{*}n_{4}^{*}}\right)$.

Case I, $n_{1}^{*}+n_{2}^{*}=0$: By Corollary 5.6 it suffices (for
$\mathcal{NR}_{3}$, which is strictly harder than $\mathcal{NR}_{2}$) to
estimate

(44)
$\displaystyle\bigg{\|}\mathcal{F}_{x}^{-1}\bigg{(}\sum_{\begin{subarray}{c}n=n_{1}+n_{2}+n_{3}+n_{4}\\\
n_{1}+n_{2}=0\end{subarray}}n_{2}^{*}\prod_{i=1}^{4}\widehat{\tilde{u}}(n_{i})\bigg{)}\bigg{\|}_{X^{s+\varepsilon,-3/4}}.$

We find, by taking the supremum in $n$ of the inner sum and Cauchy-Schwarz:

$\eqref{Equation: Main WP estimate 3 large first term}\lesssim\bigg{\|}\langle
n\rangle^{s}\sum_{n=n_{3}+n_{4}}\widehat{\tilde{u}}(n_{3})\widehat{\tilde{u}}(n_{4})\sum_{n_{1}}|n_{1}|^{1+\varepsilon}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(-n_{1})\bigg{\|}_{L^{2}_{t}\ell^{2}_{n}}\lesssim\|\tilde{u}\|_{C^{0}_{t}H_{x}^{\frac{1+\varepsilon}{2}}}^{2}\|\tilde{u}^{2}\|_{X^{s,0}_{T}}\\\
\lesssim\|\tilde{u}\|_{Y_{T}^{\frac{1+\varepsilon}{2}}}^{2}\|\tilde{u}\|_{Y^{s}_{T}}\|\tilde{u}\|_{X^{1/2+,0}_{T}}\lesssim
T^{\theta}\|\tilde{u}\|_{X^{s,1/4}_{T}}\|\tilde{u}\|_{Y^{s}_{T}}^{3},$

which holds for $0\leq\varepsilon<2s-1.$

The other portion of the norm is handled in exactly the same way, since we may
reduce to the $X^{s,0}$ norm, which is the starting portion of the above
display.

Case II, $(n_{3}^{*})^{4}n_{4}^{*}\gtrsim(n_{1}^{*})^{4}$: By Lemma 4.4 and
the assumption of Case II, the symbols are (on this region of
$\\{n\\}\times\mathbb{N}^{4}$)

(45)
$O\left(\frac{(n_{1}^{*})^{2}}{n_{3}^{*}n_{4}^{*}}\right)=O\left(\frac{n_{3}^{*}}{\sqrt{n_{4}^{*}}}\right)=O\left(\sqrt{\frac{n_{2}^{*}n_{3}^{*}}{n_{4}^{*}}}\right).$

If $\langle\tau+n^{5}\rangle\gtrsim(n_{1}^{*})^{4}$, then by the reduction in
(45) we find

$O\left(\sqrt{\frac{n_{2}^{*}n_{3}^{*}}{n_{4}^{*}\langle\tau+n^{5}\rangle^{\frac{1}{2}-}}}\right)=O\left(\sqrt{\frac{(n_{2}^{*}n_{3}^{*})^{0+}}{n_{4}^{*}}}\right).$

Writing

(46) $\langle
n\rangle^{\varepsilon}\lesssim(n_{2}^{*}n_{3}^{*})^{\varepsilon/2}(n_{4}^{*})^{\varepsilon/4},$

and invoking Parseval’s, it then suffices to estimate

$\|J_{x}^{s}\tilde{u}(J_{x}^{\varepsilon/2+}\tilde{u})^{2}J_{x}^{\varepsilon/4-1/2}\tilde{u}\|_{X^{0,-(1/2-)}}.$

Sobolev Embedding then easily provides

$\|\mathcal{NR}_{2}(\tilde{u})+\mathcal{NR}_{3}(\tilde{u})\|_{X^{s+\varepsilon,-(1/2-)}}\lesssim\|(J_{x}^{s}\tilde{u})(J_{x}^{\varepsilon/2+}\tilde{u})^{2}(J_{x}^{\varepsilon/4-1/2}\tilde{u})\|_{L^{2}_{x,t}}\\\
\lesssim\|\tilde{u}\|_{X^{s,0}_{T}}\|\tilde{u}\|_{Y_{T}^{1/2+\varepsilon/2+}}^{2}\|\tilde{u}\|_{Y_{T}^{\varepsilon/4+}},$

which holds for $0\leq\varepsilon<2s-1.$ A factor of $T^{1/4}$ arises from the
space in the modulation of the first term.

It follows that we may assume that

(47) $\langle\tau+n^{5}\rangle\ll(n_{1}^{*})^{4},$

and hence

(48)
$\langle\tau+n^{5}\rangle^{0+}|n|^{a}\lesssim(n_{2}^{*}n_{3}^{*})^{a/2+}(n_{4}^{*})^{a/4+}.$

We then invoke duality for $w\in X^{-s-\varepsilon,1/2+}$, the above
reduction, (48), (45), and (46) to find

$\displaystyle\|\mathcal{NR}_{2}(\tilde{u})+\mathcal{NR}_{3}(\tilde{u})\|_{X^{s+\varepsilon,-(1/2-)}}$
$\displaystyle\lesssim\|\mathcal{NR}_{2}(\tilde{u})+\mathcal{NR}_{3}(\tilde{u})\|_{X^{s+\varepsilon+,-(1/2+)}}$
$\displaystyle\lesssim\sup_{\|w\|_{X^{-s-\varepsilon,1/2-}=1}}\|J_{x}^{-s-\varepsilon-(a+)}w\|_{L^{8+}_{x,t}}\|J_{x}^{s}\tilde{u}\|_{L^{24/7-}_{x,t}}$
$\displaystyle\qquad\times\|J_{x}^{1/2+a/2+\varepsilon/2+}\tilde{u}\|_{L^{24/7}_{x,t}}^{2}\|J_{x}^{\varepsilon/4+a/4-1/2}\tilde{u}\|_{L^{\infty}_{x,t}}$
$\displaystyle\lesssim\sup_{\|w\|_{X^{-s-\varepsilon,1/2-}=1}}\|w\|_{X^{-s-\varepsilon,1/2-}}\|\tilde{u}\|_{X^{s,1/4-}_{T}}\|\tilde{u}\|_{Y^{\frac{1+a+\varepsilon}{2}+}_{T}}^{3},$

for $s\geq(1+a+\varepsilon)/2$ and $0<\varepsilon<2(s-(1+a)/2)$. A (small)
factor of $T$ arrises in this estimate through fudging the implicit space in
the $-$ notation. Note that (48) was necessary because of the loss of $0+$
temporal derivatives and $a$ spatial derivatives following from the $L^{8}$
estimate.

Since the other potion of the norm is bounded by the same quantity via Cauchy-
Schwarz, we are done with the proof of this lemma. ∎

The following terms, as well as the boundary terms in $Y^{s}$, are the only
terms that we are unable to (in this section) prove smoothing for, which is
why they were separated from the $D_{1}$ and $D_{2}$ terms. While we are
unable to prove smoothing in this section, we are still able to establish
local well-posedness.

######  Lemma 5.8 (Main Terms).

If $s>1/2$, then there is $\theta>0$ so that

$\|D_{\mathcal{A},1}+D_{\mathcal{A},2}\|_{Z^{s}_{T}}\lesssim
T^{\theta}\|\tilde{u}\|_{Y^{s}_{T}}^{4}.$

###### Proof.

Recall that by Lemma 4.4 we have the (slightly non-optimal)

$\displaystyle m_{D_{1}},m_{D_{2}}$
$\displaystyle=O\left(\frac{(n_{1}^{*})^{2}}{\max(n_{3}^{*},|n|)}\right),$

so that it suffices to estimate

(49)
$\bigg{\|}\mathcal{F}_{n,\tau}^{-1}\bigg{(}\underset{\begin{subarray}{c}n=n_{1}+n_{2}+n_{3}+n_{4}\\\
\tau=\tau_{1}+\tau_{2}+\tau_{3}+\tau_{4}\\\
\mathcal{P}_{4}(n_{1},n_{2},n_{3},n_{4})\end{subarray}}{\int\sum}\tfrac{(n_{1}^{*})^{2}\langle
n\rangle^{s}}{\max(n_{3}^{*},|n|)}\prod_{j=1}^{4}\widehat{\tilde{u}}(n_{j},\tau_{j})\,d\Gamma_{4}\bigg{)}(x,t)\bigg{\|}_{Z^{s}}$

Notice that after the symbol reduction we may assume that
$|n_{1}|\geq|n_{2}|\geq|n_{3}|\geq|n_{4}|$, so that $|n_{1}|=n_{1}^{*}$.
Moreover, by construction, $D_{\mathcal{A},1}$ and $D_{\mathcal{A},2}$ are
non-resonant quartilinear operators supported where
$\mathcal{P}_{4}(n_{1},n_{2},n_{3},n_{4})$ holds. It follows that we must
necessarily have

$\max(\langle\tau+n^{5}\rangle,\langle\tau_{1}+n_{1}^{5}\rangle,\cdots,\langle\tau_{4}+n_{4}^{5}\rangle)\gtrsim
n_{1}^{4}.$

We now perform another Littlewood-Paley decomposition, invoke Proposition 5.1
with $p=2$, and assume $N_{1}\sim N$ to obtain

$\displaystyle\eqref{Equation: 5.8 to show}$ $\displaystyle\lesssim
T^{\theta}\sum_{N\sim N_{1}\geq\cdots\geq
N_{4}}\frac{N_{1}^{2}N^{s}}{N_{1}\max(N_{3},N)}N_{2}^{1/2+}N_{3}^{1/2+}N_{4}^{1/2+}\prod_{i=1}^{4}\|P_{N_{i}}(\tilde{u})\|_{Y^{0}_{T}}$
$\displaystyle\lesssim T^{\theta}\|\tilde{u}\|_{Y^{0}_{T}}^{4}.$

When $N_{1}\sim N_{2}\gg N$, we again use Proposition 5.1 for $p\geq 2$ to
find

$\displaystyle\eqref{Equation: 5.8 to show}$ $\displaystyle\lesssim
T^{\theta}\sum_{\begin{subarray}{c}N_{1}\sim N_{2}\geq\cdots\geq N_{4}\\\
N_{1}\sim N_{2}\gg
N\end{subarray}}\frac{N_{1}^{2}N^{s+\frac{1}{2}-\frac{1}{p}}}{N_{1}\max(N_{3},N)}N_{2}^{1/p}N_{3}^{1/2+}N_{4}^{1/2+}\prod_{i=1}^{4}\|P_{N_{i}}(\tilde{u})\|_{Y^{0}_{T}}.$

When $1/2<s<1$ we may choose $\frac{1}{p}=s-\frac{1}{2}$, so that we find

$\displaystyle\frac{N_{1}^{2}N^{s+\frac{1}{2}-\frac{1}{p}}}{N_{1}\max(N_{3},N)}N_{2}^{1/p}N_{3}^{1/2+}N_{4}^{1/2+}$
$\displaystyle=\frac{N_{1}}{\max(N_{3},N)}N_{2}^{s-1/2}N_{3}^{1/2+}N_{4}^{1/2+}$
$\displaystyle\sim\frac{N_{1}^{s}}{\max(N_{3},N)}N_{2}^{1/2}N_{3}^{1/2+}N_{4}^{1/2+},$

which allows us to conclude the proof when $1/2<s<1$. When $s\geq 1$ we simply
let $p=2$, and the above proofs go through. We note that the second case here
actually is smoother, but since the $N_{1}\sim N$ case is not, there is no
real gain here. ∎

As noted, the bound used for $m_{D_{1}}$ in Lemma 5.8 is non-optimal. In fact,
we have the bound

$m_{D_{1}}=O\left(\frac{(n_{1}^{*})^{2}}{n\max(n_{3}^{*},|n|)}\right)$

by Lemma 4.4. We then have the following corollary.

######  Corollary 5.9.

If $s>1/2$ and $0\leq\varepsilon\leq 1$ then there is $\theta>0$ so that

$\|D_{\mathcal{A},1}(\tilde{u})\|_{Z^{s+\varepsilon}}\lesssim
T^{\theta}\|\tilde{u}\|^{4}_{Y^{s}_{T}}.$

###### Proof.

We find by the above comment that

$\langle
n\rangle^{\varepsilon}m_{D_{1}}=O\left(\frac{(n_{1}^{*})^{2}}{\max(n_{3}^{*},|n|)}\right),$

for $0\leq\varepsilon\leq 1$. The proof then follows by the proof of Lemma
5.8, as $\langle n\rangle^{\varepsilon}m_{D_{1}}=O(m_{D_{2}}).$ ∎

###  5.2. Proof of Theorem 1.3

###### Proof.

Consider $\Gamma$ as defined at (30). $\Gamma:Y^{s}\mapsto Y^{s}$ being a
contraction on small enough balls for $\|u_{0}\|_{H^{s}}\ll\delta$ follows
from standard arguments and Lemmas 5.2, 5.3, 5.4, 5.5, 5.7, and 5.8 with
$\varepsilon=0$.

Since the fifth order KdV has the same scaling as the KdV, we may easily scale
in order to obtain arbitrary data well-posedness. For more details, see [7].
Alternatively, one could define a different operator by splitting the
frequency space into $n>N$ and $n\leq N$, as in [14]. The benefit of this
second approach is that it doesn’t require the use of scaling, and will be
demonstrated in the proof of Corollary 1.8.

As noted, $P_{a}$ holds with $a=3/32$, so we find Theorem 1.1. Since we may
obtain $u$ from $\tilde{u}$, we have the well-posedness of $\eqref{Equation:
Toy Fifth Order}$ and hence well-posedness of (1) by Appendix 9. ∎

##  6\. Smoothing

Recall that our equation is given by a fixed point of $\Gamma$, and is, up to
constants, given by (30). Lemma 5.2 shows that $B_{1}(u_{0})$ and
$B_{2}(u_{0})$ are actually fairly smooth, while we can handle the other $B$
terms in $C^{0}_{t}H^{s+\varepsilon}_{x}$. In fact, every term but $D_{2}$ is
a full derivative smoother.

The first issue we need to tackle is that we will need to differentiate a
portion of $D_{2}$ again in order to place the term that will eventually have
smoothing into $C^{0}_{t}H^{s+\varepsilon}_{x}$. In particular, Lemma 5.8 is
the best we may obtain without any more work.

In light of Proposition 3.2, we may again return to the decomposition of
$\mathbb{Z}^{k}$ into $\mathcal{A}_{k}\sqcup\mathcal{B}_{k}$, where
$\mathcal{A}_{k}=\left\\{(n_{1},\cdots,n_{k})|\,\,\mathcal{P}_{k}(n_{1},\cdots,n_{k})\right\\}.$
Using this, we again split $D_{2}$ into
$D_{2}=\mathcal{NR}_{3}+D_{\mathcal{A},2},$ where $D_{\mathcal{A},2}$ is
defined by the symbol $\chi_{\mathcal{A}_{4}}m_{D_{2}}$. We then
differentiate-by-parts $D_{\mathcal{A},2}$ and write (thanks to (15))

(50) $\displaystyle
W_{-t}\widehat{D}_{\mathcal{A},2}(\tilde{u})=-\partial_{t}W_{-t}\widehat{B_{3}}(\tilde{u})-W_{-t}\widehat{C}(\tilde{u})+l.o.t.,$

where

$\displaystyle\widehat{B_{3}}(\tilde{u})$
$\displaystyle=\sum_{\begin{subarray}{c}n=n_{1}+\cdots+n_{4}\\\
\mathcal{A}_{4}\end{subarray}}\frac{m_{D_{2}}}{\Phi_{4}(n_{1},\cdots,n_{4})}\prod_{j=1}^{4}\widehat{\tilde{u}}(n_{j})$
$\displaystyle\widehat{C}(\tilde{u})$
$\displaystyle=\sum_{\begin{subarray}{c}n=n_{1}+\cdots+n_{5}\end{subarray}}(m_{\mathcal{A}}^{1}+\cdots+m_{\mathcal{A}}^{4})\prod_{j=1}^{5}\widehat{\tilde{u}}(n_{j})$
$\displaystyle m_{\mathcal{A}}^{\ell}$
$\displaystyle=\frac{m_{D_{2}}(n_{\ell}+n_{\ell+1})(n_{\ell}^{2}+n_{\ell+1}^{2})}{\Phi_{4}(n_{1},\cdots,n_{\ell}+n_{\ell+1},\cdots,n_{5})}\chi_{\mathcal{P}_{4}(n_{1},\cdots,n_{\ell}+n_{\ell+1},\cdots,n_{5})},$

for $1\leq\ell\leq 4$. Notice that, by the definition of $\mathcal{A}_{4}$ and
Lemma 4.4, we find that

(51)
$m_{\mathcal{A}}^{\ell}=O\left(\frac{n_{\ell}^{2}+n_{\ell+1}^{2}}{|n|\max(|n_{1}|,\cdots,|n_{\ell}+n_{\ell+1}|,\cdots,|n_{5}|)}\right),$

on the domain of the sum in $\widehat{C}$.

We will need to handle the resonant terms that appear from $C$ by performing
the following decomposition.

######  Lemma 6.1.

Let $1\leq\ell\leq 4$, $1\leq j\leq 2$, and $m_{\mathcal{A},R_{j}}^{\ell}$
denote resonance in $C$ of the form $n_{\ell+j-1}=n.$ Then

$\displaystyle m^{1}_{\mathcal{A},R_{j}},m^{2}_{\mathcal{A},R_{j}}$
$\displaystyle=O\left(\frac{n_{2}^{*}}{|n|}\right),$ $\displaystyle
m^{3}_{\mathcal{A},R_{j}},m^{4}_{\mathcal{A},R_{j}}$
$\displaystyle=O\left(1+\frac{n_{2}^{*}}{n}\right).$

In particular, for $\ell\in\\{3,4\\}$ we let

$H_{\ell,j}=125n^{12}\left(\sum_{\begin{subarray}{c}r=2\\\
r\neq\ell+j-1\end{subarray}}^{5}n_{r}\right)\left(\sum_{\begin{subarray}{c}r=3\\\
r\neq\ell+j-1\end{subarray}}^{5}n_{r}\right)(n_{\ell+j-1}+n_{\ell+j}-n).$

Then, off the zero-set of $H_{\ell,j}$ and for $|n|\gg(n_{2}^{*})^{5/4}$, we
have

$\displaystyle m^{\ell}_{\mathcal{A},R_{j}}$
$\displaystyle-\frac{1}{H_{\ell,j}}=O_{|n|\gg(n_{2}^{*})^{5/4}}\left(\frac{\max(n_{1},\cdots,n_{\ell+j-2},n_{\ell+j},\cdots,n_{5})}{n}\right).$

Furthermore, on the zero-set of $H_{\ell,j}$ we must have that the resonant
multilinear operator with symbol $m^{\ell}_{\mathcal{A},R_{j}}$ satisfies
$|n|\lesssim(n_{2}^{*})^{5/4}.$

###### Proof.

For $\ell=1,2$ we need a little more than (51) gives us. Specializing to
$\ell=2$ and $j=1$ we find by (25) that these symbols satisfy

$m^{2}_{\mathcal{A},R_{1}}=O\left(\frac{(n^{2}+n_{3}^{2})(n_{4}^{2}+n_{5}^{2})}{\max(|n_{1}|,|n+n_{3}|,|n_{4}|,|n_{5}|)^{4}}\right)=O\left(\frac{n_{2}^{*}}{n}\right),$

with a slightly better bound holding for $m^{1}_{\mathcal{A},R_{j}}$. The
bound when $\ell=3,4$ follows immediately from (51) and considering when
$n\gg\max_{j\neq\ell+k-1}(|n_{j}|)$ and
$n\lesssim\max_{j\neq\ell+k-1}(|n_{j}|)$.

When $\ell=3,4$ we now have $4$ terms giving an $n^{3}$ factor, which leaves
us a main term that we must handle. We now assume that $\ell=3$ and $j=1$, as
the other configurations are exactly the same. We then have that the
denominator expands as

$\displaystyle\Phi_{2}(n_{1},n_{2}+n+n_{4}+n_{5})$
$\displaystyle\Phi_{3}(n_{1},n_{2},n+n_{4}+n_{5})\Phi_{4}(n_{1},n_{2},n+n_{4},n_{5})$
$\displaystyle=125n^{12}(n_{2}+n_{4}+n_{5})(n_{4}+n_{5})n_{4}+O((n_{2}^{*})^{12}+n^{11}n_{2}^{*})$
$\displaystyle=-125n^{12}n_{1}(n_{4}+n_{5})n_{4}+O((n_{2}^{*})^{12}+n^{11}n_{2}^{*}).$

Similarly, the numerator is $n^{12}+O(n^{11}n_{2}^{*}).$ Taking

$H_{3,1}=-125n^{12}n_{1}(n_{4}+n_{5})n_{4}$

immediately gives the bound claimed.

In order to show the bound on the zero-set, we note that the mean-zero
assumption implies that

$n_{2}+n_{4}+n_{5},n_{4}\neq 0,$

so that we only have to worry about when $n_{4}+n_{5}=0$.

We notice that $m^{3}_{\mathcal{A},R_{1}}$ comes with the restriction that

$|\Phi_{3}(n_{1},n_{2},n+n_{4}+n_{5})|\gtrsim\max(n_{1},n_{2},n+n_{4}+n_{5})^{4},$

and hence if $n_{4}+n_{5}=0$, then

$|\Phi_{3}(n_{1},n_{2},n)|\gtrsim n^{4}.$

From this it easily follows that $(n_{1}^{*})^{5}\gtrsim n^{4}$, and the
algebraic relationship among the remaining frequencies forces
$(n_{2}^{*})^{5/4}\gtrsim|n|$. ∎

When summing over a symmetric subset we find a property analogous to the one
discussed in Remark 4.6. In particular, we make the following remark.

######  Remark 6.2.

Much like Remark 4.6, we have a vanishing property in the above resonance. In
particular,

$f:=\widehat{\tilde{u}}(n)\sum_{\begin{subarray}{c}0=n_{1}+n_{2}+n_{4}+n_{5}\\\
|n|\gg(n_{2}^{*})^{5/4}\end{subarray}}m^{3}_{\mathcal{A},R_{1}}\frac{1}{H_{3,1}}\prod_{\begin{subarray}{c}r=1\\\
r\neq 3\end{subarray}}^{5}\widehat{\tilde{u}}(n_{r})=0,$

where the same result holds for the other configurations of $3\leq\ell\leq 4$
and $1\leq j\leq 2$. Indeed, recall that, up to a constant,
$H_{3,1}=cn^{12}n_{1}(n_{4}+n_{5})n_{4}.$ Letting $j=n_{4}+n_{5}$ we may
rewrite $f$ as

$\displaystyle\sum_{j\neq
0}\frac{c}{j}\widehat{\tilde{u}}(n)\sum_{\begin{subarray}{c}-j=n_{1}+n_{2}\\\
|n|\gg\max(|n_{1}|,|n_{2}|)^{5/4}\end{subarray}}\frac{1}{n_{1}}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\sum_{\begin{subarray}{c}j=n_{4}+n_{5}\\\
|n|\gg\max(|n_{4}|,|n_{5}|)^{5/4}\end{subarray}}\frac{1}{n_{4}}\widehat{\tilde{u}}(n_{4})\widehat{\tilde{u}}(n_{5}).$

It then follows that the $j$ and $-j$ terms in the above sum cancel, so that
$f=0$.

We now define the operator $\mathcal{RE}_{5}(\phi)$ to be

(52)
$\widehat{\mathcal{RE}_{5}}(\phi)=\widehat{\phi}(n)\sum_{\begin{subarray}{c}0=n_{1}+\cdots+n_{\ell-1}+n_{\ell+1}+\cdots+n_{5}\\\
1\leq\ell\leq
4\end{subarray}}m^{\ell}_{\mathcal{A},R_{j}}\prod_{\begin{subarray}{c}k=1\\\
k\neq\ell+j-1\end{subarray}}^{4}\widehat{\phi}(n_{k}),$

which, by symmetry, will handle all of the the new single resonances coming
from $C$. After removal of $\mathcal{RE}_{5}$ and placing the double
resonances of $C$ into $\mathcal{RE}_{2}$, we denote $C_{*}$ to be the non-
resonant part of $C$.

######  Remark 6.3.

Bear in mind, much like in the definition of $\mathcal{RE}_{2}$, we have
multiple resonances popping up in the above, and we overload the definition of
$\mathcal{RE}_{2}$ to contain them as well. The higher order resonances
arising from $m_{A_{i}}$ and $m_{B_{i}}$ are already smoother, as they will
have 3 high frequencies ($\gtrsim|n|$). Similarly, the bound (51) with the
added information that the denominator must be $\gtrsim n^{2}$ gives
$m^{\ell}_{\mathcal{A},R_{j}}=O(1)$, and the fact that there are three
frequencies $\gtrsim|n|$ gives $2s$ smoothing (beginning at the local
wellposedness level). With this in mind, we see that it really is sufficient
to prove smoothing for the cubic double resonance term of $\mathcal{RE}_{2}$.

Because of the property discussed in Remark 6.2, we find that we don’t have to
perform another removal. Our final equation then reads (up to constants)

(53) $\displaystyle\partial_{t}$
$\displaystyle(W_{-t}\tilde{u})=\partial_{t}W_{-t}B_{1}(\tilde{u})+\partial_{t}W_{-t}B_{2}(\tilde{u})+\partial_{t}W_{-t}B_{3}(\tilde{u})$
$\displaystyle+W_{-t}(\mathcal{NR}_{1}(\tilde{u})+\mathcal{NR}_{2}(\tilde{u})+\mathcal{NR}_{3}(\tilde{u})+C_{*}(\tilde{u}))$
$\displaystyle+W_{-t}(R_{1}(\tilde{u})+R_{2}(\tilde{u})+R_{3}(\tilde{u}))$
$\displaystyle+W_{-t}(\mathcal{RE}_{1}(\tilde{u})+\mathcal{RE}_{2}(\tilde{u})+\mathcal{RE}_{3}(\tilde{u})+\mathcal{RE}_{4}(\tilde{u})+\mathcal{RE}_{5}(\tilde{u})),$

where $R_{3}$ is the remainder term resulting from the substitution of

(54)
$\mathcal{F}^{-1}_{x}\left(in\mathcal{K}(\tilde{u})\widehat{\tilde{u}}(n)\right)$

into the various components of $B_{3}$. In the interest if readability, we
note that the $B$ terms can be understood as Boundary terms, the
$\mathcal{NR}$ terms as Nearly Resonant, the $R$ terms as Remainder terms,
$\mathcal{RE}$ as the Resonant terms, and $C_{*}$ as the non-resonant portion
of $C$.

###  6.1. Auxilliary Smoothing Lemmas

Because of the extra work we did in the prior section, we only have to prove
smoothing for a small subset of the terms present in Equation (53). In
particular, we only555We have the additional higher order resonances from $C$
that were added to $\mathcal{RE}_{2}$, but, as mentioned, they follow in the
exact same way as the estimate for $\mathcal{RE}_{2}$. See Lemma 5.5 and
Remark 6.3. have to deal with $B_{3}$, $\mathcal{RE}_{5}$, $R_{3}$, and
$C_{*}$.

######  Lemma 6.4 (Propagator Terms).

Let $s>1/2$ and $0<\varepsilon<1$. Then

$\|B_{3}(\tilde{u})\|_{C^{0}_{t}H^{s+\varepsilon}_{x}}\lesssim\|\tilde{u}\|_{C^{0}_{t}H^{s}_{x}}^{4}.$

###### Proof.

The proof is exactly the same as in Lemma 5.2, as the symbol of $B_{3}$ is
$O(1/n_{1}^{*})$. ∎

######  Lemma 6.5 (Correction Term Penalties).

Let $s>1/2$. Then for $0<\varepsilon<1$,

$\|R_{3}(\tilde{u})\|_{Z^{s+\varepsilon}_{T}}\lesssim\|\tilde{u}\|_{Y^{s}_{T}}^{6}.$

###### Proof.

Recall the definition of $R_{3}$ given on (54). In particular, it is a
quartilinear operator with symbol given by

(55)
$\bigg{|}\frac{m_{D_{2}}\chi_{\mathcal{P}(n_{1},\cdots,n_{4})}\sum_{j=1}^{4}n_{j}\mathcal{K}(\tilde{u})}{\Phi_{4}(n_{1},\cdots,n_{4})}\bigg{|}\lesssim\frac{n_{1}^{*}|\mathcal{K}(\tilde{u})m_{D_{2}}|\chi_{\mathcal{P}(n_{1},\cdots,n_{4})}}{|\Phi_{4}(n_{1},\cdots,n_{4})|}\lesssim\frac{|\mathcal{K}(\tilde{u})|}{n_{1}^{*}n_{3}^{*}},$

where the final inequality followed from Lemma 4.4 and the restriction to
$\mathcal{P}(n_{1},\cdots,n_{4})$. In other words, we have found (after
Cauchy-Schwarz in $\tau$) that it is sufficient to estimate

$\|R_{3}(\tilde{u})\|_{Z^{s+\varepsilon}}\lesssim\|R_{3}\|_{X^{s+\varepsilon,0}}\lesssim\bigg{\|}\sum_{\begin{subarray}{c}n=n_{1}+\cdots+n_{4}\\\
\mathcal{A}_{4}\end{subarray}}\frac{\langle
n\rangle^{s+\varepsilon}|\mathcal{K}(\tilde{u})|}{n_{1}^{*}n_{3}^{*}}\prod_{j=1}^{4}\widehat{\tilde{u}}(n_{j})\big{)}\bigg{\|}_{L^{2}_{t}\ell^{2}_{x}}.$

Thus, by turning it into a 6-linear operator as in Lemma 5.4 and cancelling
the $\langle n\rangle^{\varepsilon}$ factor with the $n_{1}^{*}$ factor, we
find that the result follows immediately from

$\bigg{\|}\sum_{\begin{subarray}{c}n=n_{1}+\cdots+n_{4}\\\
\mathcal{A}_{4}\end{subarray}}\frac{\langle
n\rangle^{s+\varepsilon}|\mathcal{K}(\tilde{u})|}{n_{1}^{*}n_{3}^{*}}\prod_{j=1}^{4}\widehat{\tilde{u}}(n_{j})\big{)}\bigg{\|}_{L^{2}_{t}\ell^{2}_{x}}\lesssim\|\tilde{u}^{6}\|_{X^{s,0}}\lesssim\|\tilde{u}\|_{L^{2}_{t}H^{s}_{x}}\|\tilde{u}\|_{Y^{1/2+}_{T}}^{6},$

where we have invoked Sobolev Embedding and the $Y^{s}\hookrightarrow
C^{0}_{t}H^{s}_{x}$ embedding. This similarly handles the other portion of the
norm as demonstrated in the prior lemmas.

It’s interesting to note that we did not have to use modulation
considerations, as in the proof of Lemma 5.4. ∎

######  Lemma 6.6 (Resonant Terms).

Let $s>1/2$ and $0<\varepsilon<\min(2(s-1/2),1)$. Then there is a $\theta>0$
so that

$\displaystyle\left\|\mathcal{RE}_{3}(\tilde{u})\right\|_{Z^{s+\varepsilon}_{T}}\lesssim
T^{\theta}\|\tilde{u}\|_{Y^{s}_{T}}^{5}.$

###### Proof.

Recall that this term is defined on (52) as

$\mathcal{RE}_{3}(\tilde{u})=\widehat{\tilde{u}}(n)\sum_{\begin{subarray}{c}0=n_{1}+\cdots+n_{\ell-1}+n_{\ell+1}+\cdots+n_{5}\\\
1\leq\ell\leq
4\end{subarray}}m^{\ell}_{\mathcal{A},R_{j}}\prod_{\begin{subarray}{c}k=1\\\
k\neq\ell+j-1\end{subarray}}^{4}\widehat{\tilde{u}}(n_{k}).$

Additionally, this lemma almost follows from Lemma 5.5 and Remarks 6.2 and
6.3.

We ignore the emergence of double resonances, as they are bounded in exactly
the same way as $\mathcal{RE}_{2}$. Additionally, we see by Lemma 6.1 that the
cases $\ell=1,2$ follow in exactly the same way as the proof for
$\mathcal{RE}_{3}$. We now assume that resonance is occurring on $\ell=3$ with
$j=1$.

We now, in order to introduce the summation restrictions, split
$m^{3}_{\mathcal{A},R_{1}}$ into

(56)
$m^{3}_{\mathcal{A},R_{1}}=m^{3}_{\mathcal{A},R_{1}}\chi_{|n|\gg(n_{2}^{*})^{5/4}}+m^{3}_{\mathcal{A},R_{1}}\chi_{|n|\lesssim(n_{2}^{*})^{5/4}},$

and

(57) $\displaystyle\frac{1}{H_{3,1}}$
$\displaystyle=\frac{1}{H_{3,1}}\chi_{|n|\gg(n_{2}^{*})^{5/4}}\chi_{\mathcal{P}_{3}(n_{1},n_{2},n+n_{4}+n_{5})}\chi_{\mathcal{P}_{4}(n_{1},n_{2},n+n_{4},n_{5})}$
(58)
$\displaystyle+\frac{1}{H_{3,1}}\chi_{n\gg(n_{2}^{*})^{5/4}}\chi_{\mathcal{P}_{3}(n_{1},n_{2},n+n_{4}+n_{5})}\chi_{\mathcal{P}_{4}^{c}(n_{1},n_{2},n+n_{4},n_{5})}$
(59)
$\displaystyle+\frac{1}{H_{3,1}}\chi_{n\gg(n_{2}^{*})^{5/4}}\chi_{\mathcal{P}_{3}^{c}(n_{1},n_{2},n+n_{4}+n_{5})}\chi_{\mathcal{P}_{4}(n_{1},n_{2},n+n_{4},n_{5})}$
(60)
$\displaystyle+\frac{1}{H_{3,1}}\chi_{n\gg(n_{2}^{*})^{5/4}}\chi_{\mathcal{P}_{3}^{c}(n_{1},n_{2},n+n_{4}+n_{5})}\chi_{\mathcal{P}_{4}^{c}(n_{1},n_{2},n+n_{4},n_{5})}$
(61) $\displaystyle+\frac{1}{H_{3,1}}\chi_{n\lesssim(n_{2}^{*})^{5/4}},$

where we are assuming that we are off the zero set of $H_{3,1}$. We note that
by Lemma 6.1 the zero set of $H_{3,1}$ is contained in the set defined by
$n\lesssim(n_{2}^{*})^{5/4}$, and hence to handle $m_{\mathcal{A},R_{1}}^{3}$
when $H_{3,1}=0$ it is sufficient to handle the second term of (56).

We handle (58), (59), and (60) with (61). In order to do this, we handle only
the reduction for the situation that $\Phi_{4}(n,n_{1},n_{2},n+n_{4},n_{5})\ll
n^{4}$, as the other is similar after the observation that $n_{4}+n_{5}=0$ is
not in the domain of $H_{3,1}$. In this situation we must have, by Proposition
3.2 and Remark 3.3, that one of the following must be true:

  * I)

$n\in\\{n_{1},n_{2},n+n_{4},n_{5}\\}$, or

  * II)

$n_{1}\sim n_{2}\sim n+n_{3}$, or

  * III)

$n_{1}^{*}+n_{2}^{*}=0$, or

  * IV)

$(n_{3}^{*})^{4}n_{4}^{*}\gtrsim(n_{1}^{*})^{4}$.

First assume that $I$ occurs, in which case we find that $n+n_{4}\neq n$ by
the mean-zero assumption on $\tilde{u}$. This leaves us with $n_{1},n_{2}$ or
$n_{5}$ being equal to $n$. This is ruled out by the assumption that
$n\gg(n_{2}^{*})^{5/4}$. This similarly rules out $II$ and $III$. All that
remains is $IV$, which may hold with a constant differing from the one in the
characteristic function. Regardless, though, we may handle this with
(61)666The only difference in the situation that $\Phi_{3}\ll n^{4}$ is the
observation that $0\neq n_{2}+n_{4}$ as it is not in the domain of $H_{3,1}$.
In particular, the final claim of Lemma 6.1 allows us to handle this with
(61)..

We now handle (61) with the second portion of (56), as
$m^{3}_{\mathcal{A},R_{1}}=O(1)$. By the algebraic relationship among the
remaining frequencies and (51) we find

$\eqref{Equation: H symbol split
5}+m^{3}_{\mathcal{A},R_{1}}\chi_{n\lesssim(n_{2}^{*})^{5/4}}=O_{n\lesssim(n_{2}^{*})^{5/4}}\left(1+\frac{n^{2}+(n_{2}^{*})^{2}}{n\max(n_{1},n_{2},n+n_{4},n_{5})}\right)=O_{n\lesssim(n_{2}^{*})^{5/4}}\left(1+\frac{n_{2}^{*}}{n}\right).$

The $O\left(\frac{n_{2}^{*}}{n}\right)$ provides smoothing after killing the
$\langle n\rangle^{\varepsilon}$ factor using the denominator, and then
estimating in the same way we estimated (42). This leaves us with estimating
the $O(1)$ term.

By using the fact that $n\lesssim(n_{2}^{*})^{5/4}$ and the algebraic
relationship among the remaining frequencies, we find $\langle
n\rangle^{\varepsilon}\lesssim(n_{2}^{*}n_{3}^{*})^{5/8\varepsilon}.$ We then
have by Cauchy-Schwarz in $\tau$ that

$\displaystyle\bigg{\|}\langle n\rangle^{s+\varepsilon}\widehat{\tilde{u}}(n)$
$\displaystyle\sum_{\begin{subarray}{c}0=n_{1}+n_{2}+n_{4}+n_{5}\\\
n\lesssim(n_{2}^{*})^{5/4}\end{subarray}}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\widehat{\tilde{u}}(n_{4})\widehat{\tilde{u}}(n_{5})\bigg{\|}_{L^{2}_{t}\ell^{2}_{n}}$
$\displaystyle\lesssim\bigg{\|}\langle
n\rangle^{s}\widehat{\tilde{u}}(n)\sum_{\begin{subarray}{c}0=n_{1}+n_{2}+n_{4}+n_{5}\\\
n\lesssim(n_{2}^{*})^{5/4}\end{subarray}}\langle
n_{1}\rangle^{5/8\varepsilon}\widehat{\tilde{u}}(n_{1})\langle
n_{2}\rangle^{5/8\varepsilon}\widehat{\tilde{u}}(n_{2})\widehat{\tilde{u}}(n_{4})\widehat{\tilde{u}}(n_{5})\bigg{\|}_{L^{2}_{t}\ell^{2}_{n}}$
$\displaystyle\lesssim\|\tilde{u}\|_{X^{s,0}_{T}}\|(J_{x}^{5/8\varepsilon}\tilde{u})^{2}\tilde{u}\|_{L^{\infty}_{t}L^{1}_{x}}$
$\displaystyle\lesssim\|\tilde{u}\|_{X^{s,0}_{T}}\|\tilde{u}\|_{Y^{s}_{T}},$

for $0\leq\varepsilon<8s/5.$

All that remains is the first portion of (56) with (57). By Lemma 6.1 we see
that $H_{3,1}$ is non-zero and the symbol after adding these two terms
together is

$m^{3}_{\mathcal{A},R_{1}}\chi_{n\gg(n_{2}^{*})^{5/4}}-\frac{1}{H_{3,1}}\chi_{n\gg(n_{2}^{*})^{5/4}}\chi_{\Phi_{3}(n,n_{1},n_{2},n+n_{4}+n_{5})\gtrsim
n^{4}}\chi_{\Phi_{4}(n,n_{1},n_{2},n+n_{4},n_{5})\gtrsim n^{4}}\\\
=O\left(\frac{\max(|n_{1}|,|n_{2}|,|n_{4}|,|n_{5}|)}{n}\right)=O\left(\frac{n_{2}^{*}}{n}\right).$

This then follows in a similar (and easier) manner than (43).

As the second portion of the norm follows in the same manner in all of the
above cases, we’re done after noting that we always have space in the
modulation for a factor of $T$. ∎

###  6.2. Smoothing for the Main terms

We now proceed to prove smoothing for the main term. The main tools for this
will be (51) and Lemma 6.1.

######  Lemma 6.7.

Let $a\geq 0$ be such that ($P_{a}$) holds, $s>\frac{1+a}{2}$, and
$0<\varepsilon<\min(2s-1-a,1)$. Then there is a $\theta>0$ so that

$\|C_{*}(\tilde{u})\|_{Z^{s+\varepsilon}}\lesssim
T^{\theta}\|\tilde{u}\|_{Y^{s}_{T}}^{5}.$

###### Proof.

Recall that this term is given by

$\widehat{C}(\tilde{u})=\sum_{\begin{subarray}{c}n=n_{1}+\cdots+n_{5}\end{subarray}}(m_{\mathcal{A}}^{1}+\cdots+m_{\mathcal{A}}^{4})\prod_{j=1}^{5}\widehat{\tilde{u}}(n_{j}),$

where, by (51), we have

$\displaystyle
m_{\mathcal{A}}^{\ell}=O\left(\frac{n_{\ell}^{2}+n_{\ell+1}^{2}}{n\max(n_{1},\cdots,n_{\ell}+n_{\ell+1},\cdots,n_{5})}\right).$

We then either have a $high\times high\mapsto low$ interaction leaving us with

(62) $\displaystyle
m_{\mathcal{A}}^{\ell}=O\left(\frac{n_{1}^{*}n_{2}^{*}}{n\max(n_{1},\cdots,n_{\ell}+n_{\ell+1},\cdots,n_{5})}\right),$

or we have $high\times low\mapsto high$ or $high\times high\mapsto high$,
giving

(63) $\displaystyle m_{\mathcal{A}}^{\ell}=O\left(\frac{n_{1}^{*}}{n}\right).$

By Proposition 3.2 and the fact that we have already handled resonances, one
of the following must be true

  * I)

$n^{5}-n_{1}^{5}-n_{3}^{5}-n_{4}^{5}-n_{5}^{5}\gtrsim(n_{1}^{*})^{4}$,

  * II)

$n_{1}^{*}+n_{2}^{*}=0$,

  * III)

$(n_{3}^{*})^{4}n_{4}^{*}\gtrsim(n_{1}^{*})^{4}$.

Case $I$,
$n^{5}-n_{1}^{5}-n_{3}^{5}-n_{4}^{5}-n_{5}^{5}\gtrsim(n_{1}^{*})^{4}$: We must
have that

$\max(\langle\tau+n^{5}\rangle,\langle\tau_{1}+n_{1}^{5}\rangle,\cdots,\langle\tau_{5}+n_{5}^{5}\rangle)\gtrsim(n_{1}^{*})^{4}.$

Moreover, we assume for the proof of Case I that
$|n_{1}|\geq|n_{2}|\geq\cdots\geq|n_{5}|$, and $\ell=1$.

We first perform a Littlewood-Paley decomposition as in Proposition 3.2. If we
have $N_{1}\sim N$, then we have by Proposition 5.1 that

$\displaystyle\|C_{*}(\tilde{u})\|_{Z^{s+\varepsilon}}$ $\displaystyle\lesssim
T^{\theta}\sum_{N\sim N_{1}\geq N_{2}\cdots
N_{5}}N^{\varepsilon-1}\|P_{N_{1}}(\tilde{u})\|_{Y^{s}}\prod_{i=2}\|P_{N_{i}}(\tilde{u})\|_{Y^{1/2+}}$
$\displaystyle\lesssim T^{\theta}\|\tilde{u}\|_{Y^{s}_{T}}^{5},$

for $0\leq\varepsilon<1.$ If we have $N_{1}\sim N_{2}\gg N$ then we find

$\displaystyle\|C_{*}(\tilde{u})\|_{Z^{s+\varepsilon}}$ $\displaystyle\lesssim
T^{\theta}\sum_{\begin{subarray}{c}N_{1}\sim N_{2}\geq\cdots\geq N_{5}\\\
N_{1}\sim N_{2}\gg
N\end{subarray}}\frac{N_{1}^{2}N^{s+\varepsilon+\frac{1}{2}-\frac{1}{p}}}{N_{1}N^{2}}N_{2}^{1/p}N_{3}^{1/2+}N_{4}^{1/2+}\prod_{i=1}^{5}\|P_{N_{i}}(\tilde{u})\|_{Y^{0}}.$

When $1/2<s<1$ we may again choose $\frac{1}{p}=s-\frac{1}{2}$ and
$0\leq\varepsilon\leq 1$, so that we find

$\displaystyle\frac{N_{1}^{2}N^{s+\varepsilon+\frac{1}{2}-\frac{1}{p}}}{N_{1}N^{2}}N_{2}^{1/p}N_{3}^{1/2+}N_{4}^{1/2+}N_{5}^{1/2+}$
$\displaystyle=N_{1}N_{2}^{s-1/2}N_{3}^{1/2+}N_{4}^{1/2+}N_{5}^{1/2+}$
$\displaystyle\sim N_{1}^{s}N_{2}^{1/2}N_{3}^{1/2+}N_{4}^{1/2+}.$

Cauchy-Schwarz then finishes both of these cases. When $s\geq 1$ the proof
follows by the above estimates and taking $p=2.$

Case II, $n_{1}^{*}+n_{2}^{*}=0$: In this case we will assume that
$n_{2}^{*}\gg n_{3}^{*}$, as otherwise we can handle it in Case III. Since we
cannot have $n_{\ell}+n_{\ell+1}=0$, we see that (just as in the $High\times
High\mapsto Low$ situation):

$\max(n_{1},\cdots,n_{\ell}+n_{\ell+1},\cdots,n_{5})\gtrsim n_{1}^{*},$

and hence $m_{\mathcal{A}}^{\ell}=O(n_{1}^{*}/n)=O(n_{2}^{*}/n)$, as we’re in
case II.

For $0<\varepsilon\leq 1$ and $s>1/2$ it follows by taking the supremum in $t$
and invoking the algebra property of $H^{s}$ that

$\displaystyle\|C_{*}(\tilde{u})\|_{X^{s+\varepsilon,0}}$
$\displaystyle\lesssim\left\|\sum_{n_{1}+n_{2}=0}|n_{1}|\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\right\|_{C^{0}_{t}}\left\|\langle
n\rangle^{s+\varepsilon-1}\sum_{n_{1}+n_{2}+n_{3}=n}\widehat{\tilde{u}}(n_{1})\widehat{\tilde{u}}(n_{2})\widehat{\tilde{u}}(n_{3})\right\|_{L^{2}_{t}\ell^{2}_{n}}$
$\displaystyle\lesssim\|\tilde{u}\|_{C^{0}_{t}H^{1/2}_{x}}^{2}\|\tilde{u}\|_{Y^{s}_{T}}^{3}\lesssim\|\tilde{u}\|_{Y^{s}_{T}}^{5}.$

Case III, $(n_{3}^{*})^{4}n_{4}\gtrsim(n_{1}^{*})^{4}$: Note first that
$\max(n_{1},\cdots,n_{\ell}+n_{\ell+1},\cdots,n_{5})\gtrsim n_{3}^{*}$, and
hence if $n\gtrsim n_{4}^{*}$ then $m^{\ell}_{\mathcal{A}}$ satisfies (45),
and the proof of Lemma 5.7 holds with an extra application of
$Y^{1/2+}\hookrightarrow L^{\infty}_{x,t}$. We now assume that $n_{4}^{*}\gg
n$, in which case

$\langle
n\rangle^{s+\varepsilon}m_{\mathcal{A}}^{\ell}=O\left(\frac{|n|^{s+\varepsilon}(n_{1}^{*})^{s}(n_{3}^{*})^{2}\sqrt{n_{4}^{*}}}{nn_{3}^{*}}\right)=O\left(\frac{(n_{1}^{*})^{s-(1/2+)}(n_{2}^{*}n_{3}^{*}n_{4}^{*})^{\frac{1}{2}+\frac{\varepsilon}{3}+}}{|n|^{1/2+}}\right).$

It then follows by duality with $w\in X^{-s-\varepsilon,1/2-}$ that we have

$\displaystyle\|C_{*}(\tilde{u})\|_{X^{s+\varepsilon,-(1/2-)}}$
$\displaystyle\lesssim\|J_{x}^{-s-\varepsilon-(1/2+)}w\|_{L^{\infty}_{x}L^{\infty-}_{t}}\|J_{x}^{s-(1/2+)}\tilde{u}\|_{L^{\infty}_{x,t}}$
$\displaystyle\qquad\times\|J_{x}^{1/2+\varepsilon/3+}\tilde{u}\|_{L^{24/7-}_{x,t}}^{3}\|\tilde{u}\|_{L^{\infty}_{x,t}}$
$\displaystyle\lesssim\|w\|_{X^{-s-\varepsilon,1/2-}}\|\tilde{u}\|_{Y^{s}_{T}}\|\tilde{u}\|_{X^{s,1/4-}_{T}}\|\tilde{u}\|_{Y^{\frac{1}{2}+}_{T}}^{3},$

for

$0\leq\varepsilon<3\left(s-\frac{1}{2}\right).$

We find a small factor of $T$ from the space in the $L^{{24/7-}}_{x,t}$
estimates. As the other portion of the norm follows in exactly the same
fashion after Cauchy-Schwarz in $\tau$ we conclude the proof.

Note that the case when $n\gtrsim n_{4}^{*}$ requires the hypothesis on the
$L^{8}$ estimate, as in Lemma 5.7. Additionally, the other portion of the $Z$
norm is handled by the above estimates as in the previous lemmas. ∎

###### Proof of Smoothing.

Upon integration we find the equivalent Duhamel formula as

(64) $\displaystyle\tilde{u}-$
$\displaystyle\\{B_{1}(\tilde{u})+B_{2}(\tilde{u})+B_{3}(\tilde{u})\\}+W_{t}\\{B_{1}(u_{0})+B_{2}(u_{0})+B_{3}(u_{0})-u_{0}\\}$
$\displaystyle=\int_{0}^{t}W_{t-s}\left(C_{*}(\tilde{u})+smooth\right)\,ds.$

For $s>\frac{1+a}{2},$ $0\leq\varepsilon<\min(2s-1-a,1)$, and
$t<T=T(\|u_{0}\|_{H^{s}})$ we apply the lemmas in this section to obtain

$\displaystyle\left\|\mbox{LHS of }\eqref{Equation: Duhamel formula for
smoothing}\right\|_{C^{0}_{t}H^{s+\varepsilon}_{x}}\lesssim\|C_{*}(\tilde{u})+smooth\|_{Z^{s+\varepsilon}_{T}}\lesssim
C(\|u\|_{Y^{s}})\lesssim C(\|u_{0}\|_{H^{s}_{x}}).$

An application of Lemma 6.4 and the local well-posedness bound handles all of
the terms remaining on the left hand side of (64), giving the desired result

$\|\tilde{u}-W_{t}u_{0}\|_{C^{0}_{t}H^{s+\varepsilon}_{x}}\lesssim
C(\|u_{0}\|_{H^{s}_{x}}),$

by the triangle inequality. ∎

##  7\. Unconditional Well-posedness

For this section we will not need to rely on showing cancellation in the
resonant terms with symbols that are $O(1)$, so our equation under
consideration (again, modulo constants) is

(65) $\displaystyle\partial_{t}W_{-t}$
$\displaystyle\tilde{u}=\partial_{t}W_{-t}B_{1}(\tilde{u})+\partial_{t}W_{-t}B_{2}(\tilde{u})+\partial_{t}W_{-t}B_{3}(\tilde{u})$
$\displaystyle\qquad+W_{-t}(\mathcal{NR}_{1}(\tilde{u})+\mathcal{NR}_{2}(\tilde{u})+\mathcal{NR}_{3}(\tilde{u})+C(\tilde{u}))$
$\displaystyle\qquad+W_{-t}(R_{1}(\tilde{u})+R_{2}(\tilde{u})+R_{3}(\tilde{u}))$
$\displaystyle\qquad+W_{-t}(\mathcal{RE}_{1}(\tilde{u})+\mathcal{RE}_{2}(\tilde{u})+\mathcal{RE}_{3}(\tilde{u})+\mathcal{RE}_{4}(\tilde{u})).$

Since the only meaningful differences between estimates for $Y^{s}$ and
$C^{0}_{t}H^{s}_{x}$ are applications of Strichartz estimates and modulation
considerations, we will only show the proofs for the $\mathcal{NR}_{i}$,
$R_{i}$, and $C(\tilde{u})$ terms.

######  Lemma 7.1.

Let $s>1$. Then following bounds hold

$\displaystyle\|(\mathcal{NR}_{1}+\mathcal{NR}_{2}+\mathcal{NR}_{3})(\tilde{u})\|_{C^{0}_{t}H^{s}_{x}}$
$\displaystyle\lesssim\|\tilde{u}\|_{C^{0}_{t}H^{s}_{x}}^{3}+\|\tilde{u}\|_{C^{0}_{t}H^{s}_{x}}^{4}$
$\displaystyle\|(R_{1}+R_{2}+R_{3})(\tilde{u})\|_{C^{0}_{t}H^{s}_{x}}$
$\displaystyle\lesssim\|\tilde{u}\|_{C^{0}_{t}H^{s}_{x}}^{4}+\|\tilde{u}\|_{C^{0}_{t}H^{s}_{x}}^{6}$
$\displaystyle\|C(\tilde{u})\|_{C^{0}_{t}H^{s}_{x}}$
$\displaystyle\lesssim\|\tilde{u}\|_{C^{0}_{t}H^{s}_{x}}^{5}.$

###### Proof.

Beginning with the first term, we recall that we lack resonance and that the
situation $n_{1}^{*}+n_{2}^{*}=0$ was already handled in Lemma 5.7 (without
appeal to Strichartz estimates or modulation considerations). It thus suffices
to consider the case that $(n_{3}^{*})^{4}n_{4}^{*}\gtrsim(n_{1}^{*})^{4}$.

As $\mathcal{NR}_{1}$ carries three large similar frequencies, we have by
Lemma 4.4 that the symbol for all terms is $O((n_{1}^{*})^{2}/n_{3}^{*})$. It
follows by the case restriction that all of the symbols satisfy

$O\left(\frac{(n_{1}^{*})^{2}}{n_{3}^{*}}\right)=O\left(\sqrt{n_{2}^{*}n_{3}^{*}n_{4}^{*}}\right).$

Restricting ourselves to the harder $\mathcal{NR}_{2}$ and $\mathcal{NR}_{3}$,
we find by Hölders and Sobolev embedding that

$\displaystyle\|\mathcal{NR}_{2}(\tilde{u})+\mathcal{NR}_{3}(\tilde{u})\|_{C^{0}_{t}H^{s}_{x}}$
$\displaystyle\lesssim\|(J_{x}^{s}\tilde{u})(J_{x}^{1/2}\tilde{u})^{3}\|_{C^{0}_{t}L^{2}_{x}}\lesssim\|\tilde{u}\|_{C^{0}_{t}H^{s}_{x}}^{4},$

for $s>1$.

The second bound follows from the fact that, by (33), (34), and (55) all of
their symbols are $O(1)$. Restricting ourselves to $R_{3}$, and find by the
symbol being $O(1)$ that

$\displaystyle\|R_{3}\|_{C^{0}_{t}H^{s}_{x}}\lesssim\|\tilde{u}^{4}\|_{C^{0}_{t}H^{s}_{x}}\lesssim\|\tilde{u}\|_{C^{0}_{t}H^{s}_{x}}^{4},$

for $s>1/2$.

As for the last bound, we note that (51) gives

$\displaystyle
m_{\mathcal{A}}^{\ell}=O\left(\frac{n_{\ell}^{2}+n_{\ell+1}^{2}}{|n|\max(n_{1},\cdots,n_{\ell}+n_{\ell+1},\cdots,n_{5})}\right)=O\left(\frac{(n_{1}^{*})^{2}}{n^{2}}\right).$

If $n_{1}^{*}\gg n_{2}^{*}$, then the symbol is $O(1)$ and the result is
immediate as in the $R_{j}$ estimate. We now assume that $n_{1}^{*}\sim
n_{2}^{*}\gg|n|$, so that we obtain by Cauchy-Schwarz and Sobolev embedding:

$\|C(\tilde{u})\|_{C^{0}_{t}H^{s}_{x}}\lesssim\|J_{x}^{s-2}((J_{x}\tilde{u})^{2}u^{3})\|_{C^{0}_{t}L^{2}_{x}}\\\
\lesssim\|(J_{x}^{s}\tilde{u})^{2}\tilde{u}^{3}\|_{C^{0}_{t}L^{1}_{x}}\lesssim\|\tilde{u}\|_{C^{0}_{t}H^{s}_{x}}^{2}\|\tilde{u}\|_{C^{0}_{t}H^{1/2+}_{x}}^{3},$

for $s\geq 1$. ∎

###### Proof of Corollary 1.8.

Modulo constants, we define the operators $\Gamma_{1}$ and $\Gamma_{2}$ by

$\displaystyle\Gamma_{1}[v]$
$\displaystyle=W_{t}\\{B_{1}(u_{0})+B_{2}(u_{0})+B_{3}(u_{0})-u_{0}\\}-\\{B_{1}(v)+B_{2}(v)+B_{3}(v)\\}$
$\displaystyle\qquad+\int_{0}^{t}W_{t-s}\left(C(v)+\mathcal{NR}_{1}(v)+\mathcal{NR}_{2}(v)+\mathcal{NR}_{3}(v)+l.o.t\right)\,ds,$

and

$\displaystyle\Gamma_{2}[v]$
$\displaystyle=W_{t}u_{0}+\int_{0}^{t}W_{t-s}\left(\partial_{x}\left((\partial_{x}^{2}v)v+v(\partial_{x}^{2}v)\right)+\mathcal{K}(v)\partial_{x}v\right)\,ds.$

These are the operators associated to the Duhamel representation of the
equivalent formulation of the problem after and before the differentiation-by-
parts procedure, respectively. Combining these two, we obtain the operator

$\Gamma^{N}[v]=\begin{cases}\widehat{\Gamma_{1}[v]}(n)&|n|\geq N\\\
\widehat{\Gamma_{2}[v]}(n)&|n|<N,\end{cases}$

which acts as a workaround for the lack of any time factor $T$ on the $B$
terms.

Since $B_{1}$, $B_{2},$ and $B_{3}$ are all smoother, we find

$\displaystyle\|\widehat{\Gamma_{1}(v)}\|_{C^{0}_{t}H^{s}_{x}}$
$\displaystyle\lesssim
T(\|v\|_{C^{0}_{t}H^{s}_{x}}^{3}+\|v\|_{C^{0}_{t}H^{s}_{x}}^{5})$
$\displaystyle+N^{-\alpha}(\|v\|_{C^{0}_{t}H^{s}_{x}}^{2}+\|u_{0}\|_{H^{s}_{x}}^{2}+\|v\|_{C^{0}_{t}H^{s}_{x}}^{5}+\|u_{0}\|_{H^{s}_{x}}^{5})+\|u_{0}\|_{H^{s}_{x}},$

for some $\alpha>0$. We similarly find

$\displaystyle\|\widehat{\Gamma_{2}(v)}\|_{C^{0}_{t}H^{s}_{x}}$
$\displaystyle\lesssim
T(N^{3}\|v\|_{C^{0}_{t}H^{s}_{x}}^{3}+N\|v\|_{C^{0}_{t}H^{s}_{x}}^{3})+\|u_{0}\|_{H^{s}_{x}}.$

Since similar bounds hold for the difference, we see that we may select
$N=N(\|u_{0}\|_{H^{s}_{x}})$ large enough to make

$N^{-\alpha}(\|u_{0}\|_{H^{s}_{x}}^{2}+\|u_{0}\|_{H^{s}_{x}}^{5})\ll\|u_{0}\|_{H^{s}_{x}},$

and then $T$ small enough to make both

$\displaystyle T(\|u_{0}\|_{H^{s}_{x}}^{3}+\|u_{0}\|_{H^{s}_{x}}^{5})$
$\displaystyle\ll\|u_{0}\|_{H^{s}_{x}}$ $\displaystyle
T(N^{3}\|u_{0}\|_{H^{s}_{x}}^{2}+N\|u_{0}\|_{H^{s}_{x}}^{3})$
$\displaystyle\ll\|u_{0}\|_{H^{s}_{x}}.$

This enables us to perform a contraction argument in $C^{0}_{t}H^{s}_{x}$ for
$s>1$, and hence standard arguments allow us to conclude unconditional well-
posedness. ∎

##  8\. Acknowledgements

The author would like to thank Professor Burak Erdoğan for not only
recommending this problem, but for his support and guidance. Similarly, the
author would like to thank Professor Nikolaos Tzirakis for several fruitful
conversations over the vaguest of questions.

##  9\. Appendix: Toy Equation Reduction Justification

Here we will justify the reduction to (3). Consider the full Equation (1),
given again here in the more symmetric form on the Fourier side as

$\displaystyle\begin{cases}\partial_{t}\widehat{u}-ik^{5}\widehat{u}=\alpha_{1}in\sum_{n=n_{1}+n_{2}+n_{3}}\widehat{u}_{1}\widehat{u}_{2}\widehat{u}_{3}\\\
\qquad\qquad+\alpha_{2}in\sum_{n=n_{1}+n_{2}}n_{1}n_{2}\widehat{u}_{1}\widehat{u}_{2}+\alpha_{3}in\sum_{n=n_{1}+n_{2}}(n_{1}^{2}+n_{2}^{2})\widehat{u}_{1}\widehat{u}_{2}\\\
u(x,0)=u_{0}\in H^{s}(\mathbb{T}).\end{cases}$

First note that the cubic term is given as

$\displaystyle\alpha_{1}in\sum_{n=n_{1}+n_{2}+n_{3}}\widehat{u}(n_{1})\widehat{u}(n_{2})\widehat{u}(n_{3})$
$\displaystyle=\alpha_{1}in\sum_{\begin{subarray}{c}n=n_{1}+n_{2}+n_{3}\\\
\star\end{subarray}}\widehat{u}(n_{1})\widehat{u}(n_{2})\widehat{u}(n_{3})$
$\displaystyle\qquad+3\alpha_{1}in\widehat{u}(n)\sum_{\begin{subarray}{c}0=n_{2}+n_{3}\end{subarray}}\widehat{u}(n_{2})\widehat{u}(n_{3})$
$\displaystyle\qquad-2\alpha_{1}in\widehat{u}(n)\widehat{u}(n)\widehat{u}(-n)$
(66)
$\displaystyle=\alpha_{1}in\sum_{\begin{subarray}{c}n=n_{1}+n_{2}+n_{3}\\\
\star\end{subarray}}\widehat{u}(n_{1})\widehat{u}(n_{2})\widehat{u}(n_{3})$
$\displaystyle\qquad+3\alpha_{1}in\widehat{u}(n)\sum_{\begin{subarray}{c}n_{2}\end{subarray}}|\widehat{u}|^{2}(n_{2})$
(67) $\displaystyle\qquad-2\alpha_{1}in\widehat{u}(n)|\widehat{u}|^{2}(n),$

under the assumption of real $u$. By using a change of variables as before we
may remove

$3\alpha_{1}in\widehat{u}(n)\sum_{\begin{subarray}{c}n_{2}\end{subarray}}|\widehat{u}|^{2}(n_{2}).$

With only one derivative appearing in the cubic expressions (66) and (67), we
see that these term estimatable in $Y^{s}$ (resp. $C^{0}_{t}H^{s}_{x}$) for
$s>1/2$ (resp. $s>1$), so this term will not need to be differentiated-by-
parts.

We then switch to interaction variables and perform differentiate-by-parts for
each remaining term. Specifically, we let $v=W_{-t}u.$ We will often ignore
subscripts on the summations and, when obvious, ignore the entries in $\Phi$.
Moreover, we ignore terms that ought to appear due to the decomposition
Proposition 3.2, such as when $\Phi_{k}\ll(n_{1}^{*})^{4}$. For simplicity we
write $v_{j}$ for $\widehat{v}(n_{j})$ and define the notation

$\mathcal{N}_{k}=e^{-it\Phi_{k}}\prod_{j=1}^{k}v_{j}.$

For multilinear operators $\mathcal{M},\mathcal{O},$ and $\mathcal{P}$ of the
form

$\widehat{\mathcal{Q}}=\sum_{n=n_{1}+\cdots+n_{k}}q(n,n_{1},\cdots,n_{k})\prod_{j=1}^{k}v_{j}$

with symbols $m,o,$ and $p$ respectively, we write

$\mathcal{M}=\mathcal{O}+O\left(\mathcal{P}\right),\mbox{ if }m-o=O(p).$

For multilinear operators of the form

$\widehat{v}(n)\sum_{0=n_{2}+\cdots+n_{k}}q(n,n_{2},\cdots,n_{k})\prod_{j=2}^{k}v_{j}$

we write $n_{2}^{*}$ to denote the largest frequency, in magnitude, among
$n_{2},\cdots,n_{k}$ (This is inherited from the fact that such an operator
arises from setting $n_{1}=n$). Additionally, we will frequently perform large
frequency single resonance approximations, in that we will assume that
$n_{j}=n$ for some $j$ and that $n^{4}\gg(n_{2}^{*})^{5}$ (We note that the
latter restriction is only used for one resonance as in Lemma 6.6). In order
to efficiently work with these large frequency approximations, we adopt the
notation

$\mathcal{M}=\mathcal{O}+O_{A}\left(\mathcal{P}\right),\mbox{ if }m-o=O(p)$

for $(n,n_{1},\cdots,n_{k})\in A$, where $\mathcal{M},$ $\mathcal{O}$, and
$\mathcal{P}$ are $k$-linear operators of the form above and
$A\subset\mathbb{Z}^{k+1}$. Specifically for $k\in\mathbb{N}$ and $1\leq i\leq
k$, we define the set

$\displaystyle\mathcal{C}_{i}$
$\displaystyle=\\{(n,n_{1},\cdots,n_{k})\,:\,n=n_{i}\gg n_{2}^{*}\\},$

which will be used in the above context.

For what is to come, we focus solely on the resonance interactions, for
otherwise the estimates are strictly easier than those performed in the body
of this manusript. We also make the observation that resonance, by assumption,
can only occur in the variables introduced by the most recent temporal
differentiation. Indeed, we always remove prior resonances in order to assume
that $\Phi_{k}\gtrsim(n_{1}^{*})^{4}$.

The last comment we make before continuing with the calculation is that we
drop references to $\Phi_{k}(n_{1},\cdots
n_{k})\gtrsim\max(n_{1},\cdots,n_{k})^{4}$. Inded, all resonances technically
carry a chain of such assumptions as in the proof of Lemma 6.6 and they are
handled in the same manner.

Cubic Term: We find

$\displaystyle\alpha_{1}in\sum_{\begin{subarray}{c}n=n_{1}+n_{2}+n_{3}\\\
\star\end{subarray}}e^{-it\Phi_{3}(n,n_{1},n_{2},n_{3})}v_{1}v_{2}v_{3}$
$\displaystyle=-\alpha_{1}\partial_{t}\left(\sum\tfrac{n}{\Phi_{3}(n,n_{1},n_{2},n_{3})}\mathcal{N}_{3}\right)$
(68)
$\displaystyle+3\alpha_{1}^{2}\sum\tfrac{in(n_{1}+n_{2}+n_{3})}{\Phi_{3}(n,n_{1}+n_{2}+n_{3},n_{4},n_{5})}\mathcal{N}_{5}$
(69)
$\displaystyle+3\alpha_{1}\alpha_{2}\sum\tfrac{in(n_{1}+n_{2})n_{1}n_{2}}{\Phi_{3}(n,n_{1}+n_{2},n_{3},n_{4})}\mathcal{N}_{4}$
(70)
$\displaystyle+3\alpha_{1}\alpha_{3}\sum\tfrac{in(n_{1}+n_{2})(n_{1}^{2}+n_{2}^{2})}{\Phi_{3}(n,n_{1}+n_{2},n_{3},n_{4})}\mathcal{N}_{4},$

where we find that the symbols of (68) and (69) are $O(1/n^{2})$ and
$O(\tfrac{n_{1}^{*}n_{2}^{*}}{n\max(|n_{3}^{*}|,|n|)})$, respectively. It
follows that their resonances have symbols that are $O(n_{2}^{*}/n)$, and
hence we may estimate these terms in the same manner as (43) and Lemma 6.7.

As the symbol of (70) is
$O(\tfrac{(n_{1}^{*})^{2}}{\max(|n_{3}^{*}|,|n|)^{2}})$, we find that the only
non-smooth777That is, a resonance with symbol worse than $O(1)$. In
particular, the term above (70) has $\mathcal{C}_{i}$ approximation that has a
symbol that is $O(n_{2}^{*}/n)$. resonance here is the single resonance in
(70). We find easily that the $n\gg n_{2}^{*}$ approximation to the single
resonant symbol is

(71) $\eqref{Equation: Appendix A1A3
term}=3\alpha_{1}\alpha_{3}\widehat{v}(n)\sum_{0=n_{2}+n_{3}+n_{4}}\frac{i}{5n_{2}}v_{2}v_{3}v_{3}+O_{\mathcal{C}_{1}}\bigg{(}v(n)\sum_{0=n_{2}+n_{3}+n_{4}}\frac{n_{2}^{*}}{nn_{3}^{*}}v_{2}v_{3}v_{4}\bigg{)},$

where the double resonance has symbol that is $O(1)$. It follows that we may
handle all but the single resonance using the same methods as in Lemmas 5.5
and 6.7.

First Quadratic Term:

$\displaystyle\alpha_{2}in\sum_{\begin{subarray}{c}n=n_{1}+n_{2}\\\
\star\end{subarray}}ie^{-it\Phi_{2}(n,n_{1},n_{2})}n_{1}v_{1}n_{2}v_{2}$
$\displaystyle=-\alpha_{2}\partial_{t}\left(\sum\tfrac{nn_{1}n_{2}}{\Phi_{2}(n,n_{1},n_{2})}\mathcal{N}_{2}\right)$
(72)
$\displaystyle+2\alpha_{2}\alpha_{1}\sum\tfrac{in(n_{1}+n_{2}+n_{3})^{2}n_{4}}{\Phi_{2}(n,n_{1}+n_{2}+n_{3},n_{4})}\mathcal{N}_{4}$
(73)
$\displaystyle+2\alpha_{2}^{2}\sum\tfrac{in(n_{1}+n_{2})^{2}n_{1}n_{2}n_{3}}{\Phi_{2}(n,n_{1}+n_{2},n_{3})}\mathcal{N}_{3}$
(74)
$\displaystyle+2\alpha_{2}\alpha_{3}\sum\tfrac{in(n_{1}+n_{2})^{2}(n_{1}^{2}+n_{2}^{2})n_{3}}{\Phi_{2}(n,n_{1}+n_{2},n_{3})}\mathcal{N}_{3},$

where the symbol of (72) is $O(n_{2}^{*}/n)$. It follows that we may estimate
(72) in the same manner as before. However, the other two terms have symbols
that are $O(n_{1}^{*}n_{2}^{*}/\max(|n|,|n_{3}|))$ and
$O((n_{1}^{*})^{2}/\max(|n|,|n_{3}|))$, respectively. The non-resonant portion
of these can be estimated as in Lemma 6.7, but the resonant portion will
require more care. The first of these terms has large frequency single
resonance approximation

$\displaystyle\eqref{Equation: Appendix A2A2 term}$
$\displaystyle=\frac{2\alpha_{2}^{2}}{5}\widehat{v}(n)\sum_{0=n_{2}+n_{3}}in_{3}v_{2}v_{3}+O_{\mathcal{C}_{1}}\bigg{(}\widehat{v}(n)\sum_{0=n_{2}+n_{3}}\frac{n_{2}^{*}n_{3}}{n}v_{2}v_{3}\bigg{)}$
(75)
$\displaystyle=O_{\mathcal{C}_{1}}\bigg{(}\widehat{v}(n)\sum_{0=n_{2}+n_{3}}\frac{n_{2}^{*}n_{3}}{n}v_{2}v_{3}\bigg{)},$

by realness and parity considerations.

Similarly, we find

$\displaystyle\eqref{Equation: Appendix A2A3 term}$
$\displaystyle=\frac{2\alpha_{2}\alpha_{3}in}{5}\widehat{v}(n)\sum_{0=n_{2}+n_{3}}\frac{in_{3}}{n_{2}}v_{2}v_{3}+\frac{4\alpha_{2}\alpha_{3}i}{5}\sum_{0=n_{2}+n_{3}}n_{3}v_{2}v_{3}$
$\displaystyle\qquad+O_{\mathcal{C}_{1}}\bigg{(}\widehat{v}(n)\sum_{0=n_{2}+n_{3}}\frac{n_{2}^{2}}{n}v_{2}v_{3}\bigg{)}$
(76)
$\displaystyle=\frac{2\alpha_{2}\alpha_{3}in}{5}\widehat{v}(n)\sum_{0=n_{2}+n_{3}}\frac{n_{3}}{n_{2}}v_{2}v_{3}+O_{\mathcal{C}_{1}}\bigg{(}\widehat{v}(n)\sum_{0=n_{2}+n_{3}}\frac{n_{2}^{2}}{n}v_{2}v_{3}\bigg{)},$

by similar considerations as in the prior display.

Second Quadratic Term:

$\displaystyle\alpha_{3}in\sum_{\begin{subarray}{c}n=n_{1}+n_{2}\\\
\star\end{subarray}}e^{-it\Phi_{2}(n,n_{1},n_{2})}v_{1}v_{2}$
$\displaystyle=-\alpha_{3}\partial_{t}\left(\sum\tfrac{n(n_{1}^{2}+n_{2}^{2})}{\Phi_{2}(n,n_{1},n_{2})}\mathcal{N}_{2}\right)$
(77)
$\displaystyle+2\alpha_{3}\alpha_{1}\sum\tfrac{in(n_{1}+n_{2}+n_{3})((n_{1}+n_{2}+n_{3})^{2}+n_{4}^{2})}{\Phi_{2}(n,n_{1}+n_{2}+n_{3},n_{4})}\mathcal{N}_{4}$
(78)
$\displaystyle+2\alpha_{3}\alpha_{2}\sum\tfrac{in(n_{1}+n_{2})((n_{1}+n_{2})^{2}+n_{3}^{2})n_{1}n_{2}}{\Phi_{2}(n,n_{1}+n_{2},n_{3})}\mathcal{N}_{3}$
(79)
$\displaystyle+2\alpha_{3}^{2}\sum\tfrac{in(n_{1}+n_{2})((n_{1}+n_{2})^{2}+n_{3}^{2})(n_{1}^{2}+n_{2}^{2})}{\Phi_{2}(n,n_{1}+n_{2},n_{3})}\mathcal{N}_{3},$

where we find the above terms to have symbols that are $O(1/n_{4}^{*})$,
$O(n_{1}^{*}n_{2}^{*}/n_{3}^{*})$, and $O((n_{1}^{*})^{2}/n_{3}^{*})$,
respectively. When non-resonant, we note that the proof of Lemma 5.7 handles
the all of these (the first two explicitly, and the last is easier by
Proposition 3.1 by noting that we’ll further differentiate-by-parts the last
term, so that we only must worry about the case that $n_{3}^{*}\sim
n_{1}^{*}$).

We now demonstrate their large frequency resonant approximations in order.

(80) $\displaystyle\eqref{Equation: Appendix A3A1
term}=\frac{2\alpha_{3}\alpha_{1}}{5}\sum_{0=n_{2}+n_{3}+n_{4}}\frac{i}{5(n_{2}+n_{3})}v_{2}v_{3}v_{4}+O_{\mathcal{C}_{1}}\bigg{(}\widehat{v}(n)\sum_{0=n_{2}+n_{3}+n_{4}}\frac{n_{2}^{*}}{n}v_{2}v_{3}v_{4}\bigg{)}.$

The second term will have a non-trivial $n$ term:

$\displaystyle\eqref{Equation: Appendix A3A2 term}$
$\displaystyle=\frac{2\alpha_{3}\alpha_{2}in}{5}\widehat{v}(n)\sum_{0=n_{2}+n_{3}}v_{2}v_{3}+\frac{8\alpha_{3}\alpha_{2}}{5}\sum_{0=n_{2}+n_{3}}n_{2}v_{2}v_{3}$
$\displaystyle\qquad+O_{\mathcal{C}_{1}}\bigg{(}\widehat{v}(n)\sum_{0=n_{2}+n_{3}}\frac{n_{2}^{2}}{n}v_{2}v_{3}\bigg{)}$
$\displaystyle=\frac{2\alpha_{3}\alpha_{2}in}{5}\widehat{v}(n)\sum_{0=n_{2}+n_{3}}v_{2}v_{3}+O_{\mathcal{C}_{1}}\bigg{(}\widehat{v}(n)\sum_{0=n_{2}+n_{3}}\frac{n_{2}^{2}}{n}v_{2}v_{3}\bigg{)},$

again by parity considerations.

Finally, we have

(81) $\displaystyle\eqref{Equation: Appendix A3A3 term}$
$\displaystyle=\frac{2\alpha_{3}^{2}in^{2}}{5}\widehat{v}(n)\sum_{0=n_{2}+n_{3}}\frac{1}{n_{2}}v_{2}v_{3}+\frac{6\alpha_{3}^{2}in}{5}\sum_{0=n_{2}+n_{3}}v_{2}v_{3}$
$\displaystyle\qquad+\frac{2\alpha_{3}^{2}}{5}\sum_{0=n_{2}+n_{3}}in_{2}v_{2}v_{3}+O_{\mathcal{C}_{1}}\bigg{(}\widehat{v}(n)\sum_{0=n_{2}+n_{3}}\frac{n_{2}^{2}}{n}v_{2}v_{3}\bigg{)}$
(82)
$\displaystyle=\frac{6\alpha_{3}^{2}in}{5}\sum_{0=n_{2}+n_{3}}v_{2}v_{3}+O_{\mathcal{C}_{1}}\bigg{(}\widehat{v}(n)\sum_{0=n_{2}+n_{3}}\frac{n_{2}^{2}}{n}v_{2}v_{3}\bigg{)},$

D.B.P. of Non-resonant (74) Term:

We now assume that it is non-resonant and perform the differentiation-by-parts
procedure to it to find

$\displaystyle\eqref{Equation: Appendix A2A3 term}$
$\displaystyle=2\alpha_{2}\alpha_{3}\partial_{t}\left(\sum_{\begin{subarray}{c}n=n_{1}+n_{2}+n_{3}\\\
\star\end{subarray}}\tfrac{n(n_{1}+n_{2})^{2}(n_{1}^{2}+n_{2}^{2})n_{3}}{\Phi_{2}(n,n_{1}+n_{2},n_{3})\Phi_{3}}\mathcal{N}_{3}\right)$
(83)
$\displaystyle-2\alpha_{2}\alpha_{3}\alpha_{3}\sum\tfrac{in(n_{1}+n_{2}+n_{3})^{2}(n_{1}+n_{2})((n_{1}+n_{2})^{2}+n_{3}^{2})(n_{1}^{2}+n_{2}^{2})n_{4}}{\Phi_{2}(n,n_{1}+n_{2}+n_{3},n_{4})\Phi_{3}(n,n_{1}+n_{2},n_{3},n_{4})}\mathcal{N}_{4}$
$\displaystyle+l.o.t.,$

where $l.o.t.$ denotes the substitution of the cubic and first quadratic into
any of the terms, as well as the second quadratic into the $n_{3}$ term. We do
not include these because they are smoother already. Indeed, note that we find
(on the region we differentiated-by-parts) by the factorization of $\Phi_{2}$
and the bound on $\Phi_{3}$ that

$\frac{n(n_{1}+n_{2})^{2}(n_{1}^{2}+n_{2}^{2})n_{3}}{\Phi_{2}(n,n_{1}+n_{2},n_{3})\Phi_{3}}=O\left(\frac{|n_{1}|+|n_{2}|}{\max(|n_{1}+n_{2}|,|n_{3}|)\max(|n_{1}|,|n_{2}|,|n_{3}|)^{3}}\right).$

It follows that if we substitute the cubic term or the first quadratic term
into $n_{1}$ or $n_{2}$ then the symbol becomes at worst

$\displaystyle
O\big{(}\tfrac{(|n_{1}+n_{2}|+|n_{3}|)nn_{1}n_{2}}{\max(|n_{1}+n_{2}+n_{3}|,|n_{3}|)\max(|n_{1}+n_{2}|,|n_{3}|,|n_{4}|)^{3}}\big{)}$
$\displaystyle=O\big{(}\tfrac{n_{1}n_{2}}{\max(|n_{1}+n_{2}+n_{3}|,|n_{3}|)\max(|n_{1}+n_{2}|,|n_{3}|,|n_{4}|)}\big{)}$
$\displaystyle=O\big{(}\tfrac{n_{1}n_{2}}{nn_{3}^{*}}\big{)}.$

As the high frequency resonant approximation is
$O_{\mathcal{C}_{1}}(n_{2}^{*}/n)$, we may estimate these as in Lemmas 5.5 and
6.7. Similarly, if we substitute the second cubic into the $n_{3}$ term, then
we find that the symbol is

$\displaystyle O\big{(}$
$\displaystyle\tfrac{(|n_{1}|+|n_{2}|)(n_{3}+n_{4})(n_{3}^{2}+n_{4}^{2})}{\max(|n_{1}+n_{2}|,|n_{3}+n_{4}|)\max(|n_{1}|,|n_{2}|,|n_{3}+n_{4}|)^{3}}\big{)}=O\big{(}\tfrac{n_{1}^{*}n_{2}^{*}}{nn_{3}^{*}}\big{)}.$

As the resonant approximation is again seen to have symbol
$O_{\mathcal{C}_{i}}(n_{2}^{*}/n)$, we find that we may estimate these using
the same methods as in the prior two cited lemmas.

As for the remaining term, we find the following large frequency resonant
approximation:

(84) $\displaystyle\eqref{Equation: Apppendix A2A3A3 term}$
$\displaystyle=-\frac{4\alpha_{2}\alpha_{3}^{2}i}{25}\widehat{v}(n)\sum_{0=n_{2}+n_{3}+n_{4}}\frac{n_{4}}{n_{2}(n_{2}+n_{3})}v_{2}v_{3}v_{4}+O_{\mathcal{C}_{1}}\bigg{(}\widehat{v}(n)\sum_{0=n_{2}+n_{3}+n_{4}}\frac{n_{2}^{*}}{n}v_{2}v_{3}v_{4}\bigg{)}$

D.B.P. of Non-resonant (78) Term:

We obtain

$\displaystyle\eqref{Equation: Appendix A3A2 term}$
$\displaystyle=2\alpha_{3}\alpha_{2}\partial_{t}\left(\sum\tfrac{n(n_{1}+n_{2})((n_{1}+n_{2})^{2}+n_{3}^{2})n_{1}n_{2}}{\Phi_{2}(n,n_{1}+n_{2},n_{3})\Phi_{3}(n,n_{1},n_{2},n_{3})}\mathcal{N}_{3}\right)$
(85)
$\displaystyle-4\alpha_{3}\alpha_{2}\alpha_{3}\sum\tfrac{in(n_{1}+n_{2}+n_{3})(n_{1}+n_{2})^{2}((n_{1}+n_{2}+n_{3})^{2}+n_{4}^{2})(n_{1}^{2}+n_{2}^{2})n_{3}}{\Phi_{2}(n,n_{1}+n_{2}+n_{3},n_{3})\Phi_{3}(n,n_{1}+n_{2},n_{3},n_{4})}\mathcal{N}_{4}$
$\displaystyle+l.o.t.,$

where again the $l.o.t.$ are generated by substitution of the cubic term
anywhere, and substitution of any term into the $n_{3}$ variable. Indeed, we
find by the factorization of $\Phi_{2}$ and the assumed region of the sum that

$O\left(\frac{n(n_{1}+n_{2})((n_{1}+n_{2})^{2}+n_{3}^{2})n_{1}n_{2}}{\Phi_{2}(n,n_{1}+n_{2},n_{3})\Phi_{3}(n,n_{1},n_{2},n_{3})}\right)=O\left(\frac{n_{1}}{n_{3}\max(|n_{1}|,|n_{2}|,|n_{3}|)^{3}}\right).$

It follows that the substitution of the cubic into any term will be easily
boundable, and that the substitution into $n_{3}$ of either quadratic term
will have the exact same bounds as in the (74) case.