Perturbation at Blow-Up Time of Self-Similar Solutions for the Modified Korteweg–de Vries Equation

Abstract

We prove a first stability result of self-similar blow-up for the modified Korteweg–de Vries equation on the line. More precisely, given a self-similar solution and a sufficiently small regular profile, there is a unique global solution which behaves at $t=0$ as the sum of the self-similar solution and the smooth perturbation.

Appendix A: Proof of Estimate (2.1)

Proof of Lemma 9

As we are to prove a trilinear estimate, we can rescale and we will assume, without loss of generality, that

$$\begin{aligned} \Vert f\Vert _{\mathscr {E}(t)} = \Vert g\Vert _{\mathscr {E}(t)} = \Vert h \Vert _{\mathscr {E}(t)} =1, \end{aligned}$$

so that

$$\begin{aligned} \Vert \tilde{f} \Vert _{L^\infty } \leqq 1 \quad \text {and} \quad \Vert \partial _{\xi }\tilde{f} \Vert _{L^2} \leqq t^{1/6}, \end{aligned}$$

(A.1)

and the same for ${\tilde{g}}$ and ${\tilde{h}}$.

As in Lemma 11, given a domain $\mathscr {D} \subset \mathbb {R}^2$, we write

$$\begin{aligned} J(\mathscr {D}) = \iint _{\mathscr {D}} e^{it\Phi }\tilde{f}(\xi _1)\tilde{g}(\xi _2)\xi _3\tilde{h}(\xi _3)\textrm{d}\xi _1 \textrm{d}\xi _2. \end{aligned}$$

We will only do the proof when the integral is restricted to the domain

$$\begin{aligned} \mathscr {A}:= \{ (\xi _1,\xi _2) \in \mathbb {R}^2: |\xi _3|\geqq |\xi _2|\geqq |\xi _1| \}, \end{aligned}$$

where $\xi _3$ is the largest frequency: this is actually the worst-case scenario, and it is clear from the computations below how to adapt the estimates to the others domains.

• 1) For small frequencies, we have a crude bound: let $\mathscr {A}_0 = \{ (\xi _1,\xi _2) \in \mathscr {A}: |\xi _3| \leqq t^{-1/3} \}$. Then

  $$\begin{aligned} |J(\mathscr {A}_0)| \leqq \iint _{\mathscr {A}_0} |\xi _3| \textrm{d}\xi _1 \textrm{d}\xi _2 \Vert {\tilde{f}}\Vert _{L^\infty } \Vert {\tilde{g}} \Vert _{L^\infty } \Vert {\tilde{h}} \Vert _{L^\infty } \leqq \frac{1}{t}. \end{aligned}$$

• 2) We now consider the case when $|\xi _2|$ is significantly smaller than $|\xi _3|$ (and so $|\xi _1|$ as well):

  $$\begin{aligned} \mathscr {A}_1 = \left\{ (\xi _1,\xi _2) \in \mathscr {A} {\setminus } \mathscr {A}_0: |\xi _3| - |\xi _2| \geqq \frac{|\xi _3|}{10} \right\} . \end{aligned}$$

  Notice that, on $\mathscr {A}_1$,

  $$\begin{aligned} |\partial _{\xi _1} \Phi | = 3(\xi _3^2 - \xi _1^2) \gtrsim |\xi _3|^2 \text{ and } |\partial _{\xi _1}^2 \Phi | \lesssim |\xi _3,| \end{aligned}$$

  which allows us to perform an IBP in $\xi _1$ using that $e^{i t \Phi } = \frac{1}{it \partial _{1} \Phi } \partial _{\xi _1} (e^{it \Phi })$:

  $$\begin{aligned} J(\mathscr {A}_1)&= \iint _{\mathscr {A}_1} e^{it\Phi } \partial _{\xi _1} \left( \frac{\xi _3}{it \partial _{1} \Phi } \tilde{f}(\xi _1)\tilde{g}(\xi _2) \tilde{h}(\xi _3) \right) \textrm{d}\xi _1 \textrm{d}\xi _2 \nonumber \\ {}&\qquad + \int _{\partial \mathscr {A}_1} \frac{\xi _3 e^{it\Phi }}{it \partial _{1} \Phi } \tilde{f}(\xi _1)\tilde{g}(\xi _2) \tilde{h}(\xi _3) \textrm{d}\sigma (\xi _1,\xi _2). \end{aligned}$$

  (A.2)

Unfortunately, one can check that a direct bound on these integrals (after distributing the derivative $\partial _{\xi _1}$ on all factors of the first integral) lead to at least a logarithmic divergence.

For the terms in (A.2) where the derivative in $\xi _1$ does not fall on $\tilde{h}(\xi _3)$, we perform a second IBP, this time in $\xi _2$, writing this time $e^{i t \Phi } = \frac{1}{it \partial _{2} \Phi } \partial _{\xi _2} (e^{it \Phi })$. Notice that on $\mathscr {A}_1$, $|\partial _{2} \Phi | \gtrsim |\xi _3|^2$ and $|\nabla ^2 \Phi | \lesssim |\xi _3|$, $|\nabla ^3 \Phi | \lesssim 1$. For example, when the derivative in (A.2) falls on the phase $\partial _{1} \Phi$:

$$\begin{aligned} J_1&: =\left| \iint _{\mathscr {A}_1} e^{it\Phi } \left( \frac{\xi _3 \partial _{1 1}^2 \Phi }{it (\partial _{1} \Phi )^2} \tilde{f}(\xi _1)\tilde{g}(\xi _2) \tilde{h}(\xi _3) \right) \textrm{d}\xi _1 \textrm{d}\xi _2 \right| \nonumber \\ {}&\leqq \left| \iint _{\mathscr {A}_1} e^{it\Phi } \partial _{\xi _2} \left( \frac{\xi _3 \partial _{1 1}^2 \Phi }{t^2 (\partial _{1} \Phi )^2 \partial _{2} \Phi } \tilde{f}(\xi _1)\tilde{g}(\xi _2) \tilde{h}(\xi _3) \right) \textrm{d}\xi _1 \textrm{d}\xi _2 \right| \\ {}&\qquad + \left| \int _{\partial \mathscr {A}_1} \frac{e^{it\Phi } \xi _3 \partial _{1 1}^2 \Phi }{t^2 (\partial _{1} \Phi )^2 \partial _{2} \Phi } \tilde{f}(\xi _1)\tilde{g}(\xi _2) \tilde{h}(\xi _3) \textrm{d}\sigma (\xi _1,\xi _2) \right| \\ {}&\lesssim \frac{1}{t^2} \iint _{\mathscr {A}_1} \frac{1}{|\xi _3|^5} \textrm{d}\xi _1 \textrm{d}\xi _2 \Vert {\tilde{f}} \Vert _{L^\infty } \Vert {\tilde{g}} \Vert _{L^\infty } \Vert {\tilde{h}} \Vert _{L^\infty } \\ {}&\qquad + \frac{1}{t^2}\iint _{\mathscr {A}_1} \Vert {\tilde{f}} \Vert _{L^\infty } (|\partial _\xi {\tilde{g}}(\xi _2)| \Vert {\tilde{h}} \Vert _{L^\infty } +\Vert {\tilde{g}} \Vert _{L^\infty } |\partial _\xi {\tilde{h}}(\xi _3)| ) \frac{\textrm{d}\xi _1 \textrm{d}\xi _2}{|\xi _3|^4} \\ {}&\qquad + \frac{1}{t^2} \int _{\partial \mathscr {A}_1} \frac{\textrm{d}\sigma (\xi _1,\xi _2)}{|\xi _3|^4} \Vert {\tilde{f}} \Vert _{L^\infty } \Vert {\tilde{g}} \Vert _{L^\infty } \Vert {\tilde{h}} \Vert _{L^\infty }. \end{aligned}$$

Now with (A.1) in mind, and with the change of variable in $\xi _2$ defined by $\xi _2' = \xi _3 =\xi -\xi _1-\xi _2$ and $\mathscr {B}_1 = \{(\xi _1,\xi _2'): (\xi _1, \xi -\xi _1-\xi _2') \in \mathscr {A}_1 \}$ we estimate

$$\begin{aligned} \iint _{\mathscr {A}_1} \frac{\textrm{d}\xi _1 \textrm{d}\xi _2}{|\xi _3|^5}&= \iint _{\mathscr {B}_1} \frac{\textrm{d}\xi _1 \textrm{d}\xi _2'}{|\xi _2'|^5} \leqq \int _{|\xi _2'| \geqq t^{-1/3}} \left( \int _{|\xi _1| \leqq |\xi _2'|} \frac{\textrm{d}\xi _1 }{|\xi _2'|^5} \right) \textrm{d}\xi _2' \\&= 2\int _{|\xi _2'| \geqq t^{-1/3}} \frac{\textrm{d}\xi _2'}{|\xi _2'|^4} \lesssim t. \end{aligned}$$

Similarly, using the Cauchy-Schwarz inequality,

$$\begin{aligned} \iint _{\mathscr {A}_1} |\partial _\xi {\tilde{g}}(\xi _2)| \frac{\textrm{d}\xi _1 \textrm{d}\xi _2}{|\xi |^4}&\leqq \int _{|\xi _1| \geqq t^{-1/3}} \Vert \partial _\xi {\tilde{g}} \Vert _{L^2}\left( \int _{|\xi _2| \geqq |\xi _1|} \frac{\textrm{d}\xi _2}{|\xi _2|^8} \right) ^{1/2} \textrm{d}\xi _1 \\ {}&\leqq \int _{|\xi _1| \geqq t^{-1/3}} \frac{\textrm{d}\xi _1}{|\xi _1|^{7/2}} \Vert \partial _\xi {\tilde{g}} \Vert _{L^2} \lesssim t^{5/2 \cdot 1/3 + 1/6} \lesssim t, \end{aligned}$$

which allows to take care of the two terms the second last line. As for the boundary term,

$$\begin{aligned} \int _{\partial \mathscr {A}_1} \frac{\textrm{d}\sigma (\xi _1,\xi _2)}{|\xi _3|^4} = \int _{\partial \mathscr {B}_1} \frac{\textrm{d}\sigma (\xi _1,\xi _2')}{|\xi _2'|^4} \lesssim t. \end{aligned}$$

We hence obtained

$$\begin{aligned} J_1 \lesssim \frac{1}{t}. \end{aligned}$$

As mentioned, similar computations can be done for all terms, except for the term with a derivative on ${\tilde{h}}$:

$$\begin{aligned} J_1'&:= \iint _{\mathscr {A}_1} \left( \frac{e^{it\Phi } \xi _3}{it \partial _{1} \Phi } \tilde{f}(\xi _1)\tilde{g}(\xi _2) \partial _\xi \tilde{h}(\xi _3) \right) \textrm{d}\xi _1 \textrm{d}\xi _2 \end{aligned}$$

Here, we first perform the change of variable in $\xi _1$ defined by $\xi _1':= \xi - \xi _1 - \xi _2 = \xi _3$ so that $\xi _3':= \xi - \xi _1' - \xi _2 = \xi _1$, with $\mathscr {B}_1' = \{(\xi _1',\xi _2): (\xi - \xi _1'-\xi _2, \xi _2) \in \mathscr {A}_1 \}$:

$$\begin{aligned} J_1'&= \iint _{\mathscr {B}_1'} \left( \frac{e^{it\Phi } \xi _2'}{it \partial _{1} \Phi (\xi _3', \xi _2)} \tilde{f}(\xi _3')\tilde{g}(\xi _2) \partial _\xi \tilde{h}(\xi _1') \right) \textrm{d}\xi _1' \textrm{d}\xi _2. \end{aligned}$$

Then we are in a position to perform once again an IBP in $\xi _2$ (because there will be no derivative falling on $\partial _\xi {\tilde{h}}$). On $\mathscr {B}_1'$,

$$\begin{aligned} \partial _{1} \Phi (\xi _3', \xi _2) = 3|\xi _1'^2 - \xi _3'| \gtrsim |\xi _1'|^2, \quad |\partial _{2} \Phi (\xi _3',\xi _2)| = 3|\xi _1'^2 - \xi _2^2| \gtrsim |\xi _1'|^2, \end{aligned}$$

$|\xi _1'| \geqq t^{-1/3}$ and $|\xi _1'| \geqq |\xi _2| \geqq |\xi _3'|$. These estimates allow us to argue as for $J_1$ and derive

$$\begin{aligned} |J(\mathscr {A}_1)| \lesssim \frac{1}{t}. \end{aligned}$$

• 3) We then consider the case when only $\xi _1$ is significantly smaller:

  $$\begin{aligned} \mathscr {A}_2 = \left\{ (\xi _1,\xi _2) \in \mathscr {A} {\setminus } \mathscr {A}_0: | |\xi _3| - |\xi _2| | \leqq \frac{|\xi _3|}{10} \text { and } |\xi _3| - |\xi _1| \geqq \frac{|\xi _3|}{10} \right\} . \end{aligned}$$

  We start as in 2), first performing an IBP in $\xi _1$:

  $$\begin{aligned} J(\mathscr {A}_2)&= \iint _{\mathscr {A}_2} e^{it\Phi } \partial _{\xi _1} \left( \frac{\xi _3}{it \partial _{1} \Phi } \tilde{f}(\xi _1)\tilde{g}(\xi _2) \tilde{h}(\xi _3) \right) \textrm{d}\xi _1 \textrm{d}\xi _2 \nonumber \\ {}&\qquad + \int _{\partial \mathscr {A}_2} \frac{\xi _3 e^{it\Phi }}{it \partial _{1} \Phi } \tilde{f}(\xi _1)\tilde{g}(\xi _2) \tilde{h}(\xi _3) \textrm{d}\sigma (\xi _1,\xi _2). \end{aligned}$$

  (A.3)

  Except on the term where the derivative $\partial _{\xi _1}$ falls on ${\tilde{h}}$ (see step 3.3 below), we perform on (A.3) an IBP in $\xi _2$, using the slighlty different identity

  $$\begin{aligned} e^{i t \Phi } = \frac{1}{1 + it \xi _2 \partial _2 \Phi } \partial _{\xi _2} (\xi _2 e^{i t \Phi }). \end{aligned}$$

  For example, if in (A.3) the $\partial _{\xi _1}$ derivative fell on ${\tilde{f}}$, we are to bound

  $$\begin{aligned} J_2&: = \left| \iint _{\mathscr {A}_2} e^{it\Phi } \left( \frac{\xi _3}{it \partial _{1} \Phi } \partial _\xi \tilde{f}(\xi _1) \tilde{g}(\xi _2) \tilde{h}(\xi _3) \right) \textrm{d}\xi _1 \textrm{d}\xi _2 \right| \\ {}&\leqq \left| \iint _{\mathscr {A}_2} e^{it\Phi } \partial _{\xi _2} \left( \frac{\xi _2 \xi _3}{(1 + it \xi _2 \partial _2 \Phi )(it \partial _{1} \Phi )} \partial _\xi \tilde{f}(\xi _1) \tilde{g}(\xi _2) \tilde{h}(\xi _3) \right) \textrm{d}\xi _1 \textrm{d}\xi _2 \right| \\ {}&\qquad + \left| \iint _{\partial \mathscr {A}_2} e^{it\Phi } \frac{\xi _2 \xi _3}{(1 + it \xi _2 \partial _2 \Phi )(it \partial _{1} \Phi )} \partial _\xi \tilde{f}(\xi _1) \tilde{g}(\xi _2) \tilde{h}(\xi _3) \textrm{d}\sigma (\xi _1,\xi _2) \right| \end{aligned}$$

We now split into two cases.

3.1) Let $\mathscr {B}_2 = \{ (\xi _1, \xi _2) \in \mathscr {A}_2: |\xi _3 - \xi _2| \leqq \frac{|\xi _3|}{100} \}$.

On $\mathscr {B}_2$, $|\partial _1 \Phi | =3|\xi _3^2 - \xi _1^2| \gtrsim |\xi _3|^2$ and

$$\begin{aligned} |\partial _2 \Phi | =3|\xi _3^2 - \xi _2^2| \gtrsim |\xi _3| |\xi _3-\xi _2| = |\xi _3| |\xi -\xi _1 - 2 \xi _2|. \end{aligned}$$

Also using that $|\xi _2| \gtrsim |\xi _3| \geqq |\xi _2|$ and $|\nabla ^2 \Phi | \lesssim |\xi _3|$, we infer

$$\begin{aligned} J_2&\lesssim \iint _{\mathscr {A}_1} \frac{|\xi _2| | \partial _\xi {\tilde{f}}(\xi _1)|}{(1+t|\xi _2|^2 |\xi _3-\xi _2|) t |\xi _2|^2} \textrm{d}\xi _1 \textrm{d}\xi _2 \Vert {\tilde{g}} \Vert _{L^\infty } \Vert {\tilde{h}}\Vert _{L^\infty } \end{aligned}$$

(A.4)

$$\begin{aligned}&\qquad + \iint _{\mathscr {A}_1} \frac{|\xi _2|^3 | \partial _\xi {\tilde{f}}(\xi _1)|}{(1+t|\xi _2|^2 |\xi _3-\xi _2|)^2 t |\xi _2|^2} \textrm{d}\xi _1 \textrm{d}\xi _2 \Vert {\tilde{g}} \Vert _{L^\infty } \Vert {\tilde{h}}\Vert _{L^\infty } \nonumber \\ {}&\qquad + \iint _{\mathscr {A}_1} \frac{|\xi _2|^2 | \partial _\xi {\tilde{f}}(\xi _1)| ( | \partial _\xi {\tilde{g}}(\xi _2)| \Vert {\tilde{h}}\Vert _{L^\infty } + \Vert {\tilde{g}}\Vert _{L^\infty } | \partial _\xi {\tilde{h}}(\xi _3)| )}{(1+t|\xi _2|^2 ||\xi _3-\xi _2||) t |\xi _2|^2} \textrm{d}\xi _1 \textrm{d}\xi _2 \nonumber \\ {}&\qquad + \iint _{\partial \mathscr {A}_1} \frac{|\xi _2|^2 | \partial _\xi {\tilde{f}}(\xi _1)|}{(1+t|\xi _2|^2 ||\xi _3-\xi _2||) t |\xi _2|^2} \textrm{d}\sigma (\xi _1,\xi _2) \Vert {\tilde{g}} \Vert _{L^\infty } \Vert {\tilde{h}}\Vert _{L^\infty } \end{aligned}$$

(A.5)

Then for fixed $\xi _2$, we can bound

$$\begin{aligned} \int _{\xi _1} \frac{ | \partial _\xi {\tilde{f}}(\xi _1)|}{(1+t|\xi _2|^2 |\xi _3-\xi _2|)} \textrm{d}\xi _1&\leqq \left( \int \frac{\textrm{d}\xi _1}{(1+t|\xi _2|^2 |\xi -\xi _1-2\xi _2|)^2} \right) ^{1/2} \Vert \partial _\xi {\tilde{f}} \Vert _{L^2} \\&\lesssim \frac{1}{t^{1/2} |\xi _2|}\Vert \partial _\xi {\tilde{f}} \Vert _{L^2} \end{aligned}$$

so that for the term in (A.4)

$$\begin{aligned} \iint _{\mathscr {A}_1} \frac{|\xi _2| | \partial _\xi {\tilde{f}}(\xi _1)|}{(1+t|\xi _2|^2 |\xi _3-\xi _2|) t |\xi _2|^2} \textrm{d}\xi _1 \textrm{d}\xi _2&\lesssim \frac{1}{t^{3/2}} \int _{|\xi _2| \geqq t^{-1/3}} \frac{\textrm{d}\xi _2}{|\xi _2|^{2}} \Vert \partial _\xi {\tilde{f}} \Vert _{L^2} \\&\lesssim t^{-3/2+ 1/3+1/6} \leqq \frac{1}{t}. \end{aligned}$$

Similarly, for (A.4)

$$\begin{aligned}&\iint _{\mathscr {A}_1} \frac{|\xi _2|^2 | \partial _\xi {\tilde{f}}(\xi _1)| | \partial _\xi {\tilde{g}}(\xi _2)|}{(1+t|\xi _2|^2 |\xi _3-\xi _2|) t |\xi _2|^2} \textrm{d}\xi _1 \textrm{d}\xi _2\\ \lesssim \ {}&\frac{1}{t} \int _{|\xi _2| \geqq t^{-1/3}} | \partial _\xi {\tilde{g}}(\xi _2)| \left( \int _{\xi _1} \frac{ | \partial _\xi {\tilde{f}}(\xi _1)| \textrm{d}\xi _1}{(1+t|\xi _2|^2 |\xi -\xi _1-2\xi _2|) } \right) \textrm{d}\xi _2 \\ \lesssim \ {}&\frac{1}{t^{3/2}} \int _{|\xi _2| \geqq t^{-1/3}}|\partial _\xi {\tilde{g}}(\xi _2)| \frac{\textrm{d}\xi _2}{|\xi _2|} \Vert \partial _\xi {\tilde{f}} \Vert _{L^2} \lesssim t^{-3/2+1/3 \cdot 1/2} \Vert \partial _\xi {\tilde{g}} \Vert _{L^2} \Vert \partial _\xi {\tilde{f}} \Vert _{L^2} \lesssim \frac{1}{t}. \end{aligned}$$

One can bound the other terms in $J_2$ in the same fashion.

3.2) Let $\mathscr {C}_2 = \{ (\xi _1, \xi _2) \in \mathscr {A}_2: |\xi _3 + \xi _2| \leqq \frac{|\xi _3|}{100} \}$. Then on $\mathscr {C}_2$, $|\partial _1 \Phi | =3|\xi _3^2 - \xi _1^2| \gtrsim |\xi _3|^2$ and

$$\begin{aligned} |\partial _2 \Phi | =3|\xi _3^2 - \xi _2^2| \gtrsim |\xi _3| |\xi _3+\xi _2| = |\xi _3| |\xi -\xi _1|. \end{aligned}$$

We still have $|\xi _2| \gtrsim |\xi _3| \geqq |\xi _2| \gtrsim |\xi |, |\xi _1|$ and $|\nabla ^2 \Phi | \lesssim |\xi _3|$.

We can then proceed as in 3.1), integrating first in $\xi _1$ for $\xi _2$ fixed.

Summing up, as $\mathscr {A}_2 = \mathscr {B}_2 \cup \mathscr {C}_2$, we get that

$$\begin{aligned} J_2 \lesssim \frac{1}{t}. \end{aligned}$$

3.3) To complete the bound on $J(\mathscr {A}_2)$, it remains to consider the term in (A.3) when the derivative $\partial _{\xi _1}$ of the (first) IBP falls on ${\tilde{h}}$:

$$\begin{aligned} J_2': = \left| \iint _{\mathscr {A}_2} e^{it\Phi } \left( \frac{\xi _3}{it \partial _{1} \Phi } \tilde{f}(\xi _1) \tilde{g}(\xi _2) \partial _\xi \tilde{h}(\xi _3) \right) \textrm{d}\xi _1 \textrm{d}\xi _2 \right| \\ \end{aligned}$$

As in case 2), we perform the change of variable in $\xi _1$ defined by $\xi _1':= \xi - \xi _1 - \xi _2 = \xi _3$ so that $\xi _3':= \xi - \xi _1' - \xi _2 = \xi _1$, with $\mathscr {B}_2' = \{(\xi _1',\xi _2): (\xi - \xi _1'-\xi _2, \xi _2) \in \mathscr {A}_2 \}$:

$$\begin{aligned} J_2'&= \iint _{\mathscr {B}_2'} \left( \frac{e^{it\Phi } \xi _2'}{it \partial _{1} \Phi (\xi _3', \xi _2)} \tilde{f}(\xi _3')\tilde{g}(\xi _2) \partial _\xi \tilde{h}(\xi _1') \right) \textrm{d}\xi _1' \textrm{d}\xi _2, \end{aligned}$$

and then we are in a position for an IBP in $\xi _2$. We conclude by doing the same computations as in 3.1) and 3.2), and we obtain, as desired,

$$\begin{aligned} J(\mathscr {A}_2) \lesssim \frac{1}{t}. \end{aligned}$$

4) We are now in the case when $|\xi _1|, |\xi _2|, |\xi _3|$ are similar, which corresponds to the stationary points. There are two cases: either they all have same signs (corresponding to $(\xi /3,\xi /3,\xi /3)$), or they don’t (corresponding to $(\xi ,\xi ,-\xi )$ and its two symmetric). We thus consider

$$\begin{aligned} \mathscr {A}_3 := \left\{ (\xi _1,\xi _2) \in \mathscr {A} : |\xi _3-\xi _2|, |\xi _3-\xi _1| \leqq \frac{|\xi _3|}{10} \right\} \\ \mathscr {A}_4 : = \left\{ (\xi _1,\xi _2) \in \mathscr {A}: |\xi _3 + \xi _2|, |\xi _3+\xi _1| \leqq \frac{|\xi _3|}{10} \right\} . \end{aligned}$$

The other two stationary points can be treated as for $\mathscr {A}_4$.

In these stationary regions, we must exploit as much as possible the phase oscillations. To that end, we adapt our reference frame to the directions of maximal oscillation (see (A.6)). This analysis was crucial to derive the refined estimates in [18] and [7]. The computations below follow closely the arguments therein.

As $\xi _1$, $\xi _2$, $\xi _3$ are all of the order of $\xi$, it is convenient to rescale and denote $q_i = \xi _i/\xi$ so that $q_1+q_2+q_3 =1$, and the phase

$$\begin{aligned} \Phi (\xi _1,\xi _2) = \xi ^3 Q(q_1,q_2) = 3 \xi ^3 (1-q_1)(1-q_2)(1-q_3). \end{aligned}$$

By symmetry, we can also assume without loss of generality that $\xi >0$, and we denote the rescaling factor

$$\begin{aligned} \tau := t \xi ^3 \geqq 1 \quad \text {on } \mathscr {A}_3 \cup \mathscr {A}_4. \end{aligned}$$

($|\tau | \leqq 1$ corresponds to $\mathscr {A}_0$). Denote the nonlinear function without the $\xi _3$ factor

$$\begin{aligned} F:= {\tilde{f}}(\xi q_1) {\tilde{g}}(\xi q_2) {\tilde{h}}(\xi q_3). \end{aligned}$$

We now introduce the change of variable

$$\begin{aligned} 1-q_1=\lambda - \mu ,\quad 1-q_2= \lambda + \mu ,\quad 1-q_3=2(1-\lambda ), \end{aligned}$$

so that

$$\begin{aligned} Q(q_1,q_2) = 6(1-\lambda ) (\lambda -\mu )(\lambda +\mu ). \end{aligned}$$

The stationary point correspond to $(\lambda ,\mu ) = (2/3,0)$ and (0, 0), respectively. We integrate by parts using the relation

$$\begin{aligned} e^{i\tau Q}=\frac{1}{1+12i\tau \mu ^2(1-\lambda )}\partial _ \mu ( \mu e^{i\tau Q}) \end{aligned}$$

so that, with $\mathscr {D}' = \{ (\lambda , \mu ): (1-(\lambda -\mu ), 1-(\lambda +\mu ) \in \mathscr {D} \}$, where $\mathscr {D}$ is an integration domain and

$$\begin{aligned} A = 1+12i\tau \mu ^2(1-\lambda ),\quad \partial _\mu A = 24 i \tau \mu (1-\lambda ), \end{aligned}$$

and depending whether the derivative falls on the phase or not,

$$\begin{aligned} | J(\mathscr {D})|&= 2\xi ^3 \iint _{\mathscr {D}} e^{i \tau Q} F (2\lambda -1) \textrm{d}\lambda \textrm{d}\mu = - 2\xi ^3 \iint _{\mathscr {D}} \frac{1}{A} \partial _\mu ( \mu e^{i \tau Q}) F (2\lambda -1) \textrm{d}\lambda \textrm{d}\mu \\ {}&= 2\xi ^3 \iint _{\mathscr {D}'} e^{i\tau Q} F \frac{\mu \partial _\mu A (2\lambda -1)}{A^2} \textrm{d}\lambda \textrm{d}\mu + 2\xi ^3 \iint _{\mathscr {D}'} e^{i\tau Q}\partial _{1} F \frac{ \mu (2\lambda -1)}{A}\textrm{d}\lambda \textrm{d}\mu \\ {}&\qquad - 2\xi ^3 \iint _{\mathscr {D}'} e^{i\tau Q}\partial _{2} F \frac{ \mu (2\lambda -1)}{A}\textrm{d}\lambda \textrm{d}\mu \\ {}&=: 2\xi ^3 (J_\sharp (\mathscr {D}') + J_\flat (\mathscr {D}') + J_\natural (\mathscr {D}')). \end{aligned}$$

(Observe that there is no derivative on ${\tilde{h}}$ in this computation.) By symmetry between $\partial _1 F$ and $\partial _2 F$, it suffices to bound $J_\sharp (\mathscr {D}')$ and $J_\flat (\mathscr {D}')$.

5) Here we bound $J_\sharp$. We do yet another change of variable of $\mu$ defined by $\nu = \mu \sqrt{1-\lambda }$ so that Q has separate variables:

$$\begin{aligned} Q = 6 (1-\lambda ) \lambda ^2 - 6\nu ^2, \quad \partial _\lambda Q = 6 \lambda (2-3\lambda ). \end{aligned}$$

(A.6)

We perform an integration by parts using, for fixed $\lambda _0 \in \{ 0, 2/3 \}$

$$\begin{aligned} e^{i\tau Q}=\frac{1}{1 + 6i\tau (\lambda -\lambda _0) \lambda (2-3\lambda )}\partial _\lambda ((\lambda -\lambda _0)e^{i\tau Q}). \end{aligned}$$

Denoting

$$\begin{aligned} B = 1 + 6 i \tau (\lambda -\lambda _0) \lambda (2-3\lambda ), \quad \partial _\lambda B = 6i\tau (\lambda (2-3\lambda ) + 2 (\lambda -\lambda _0)(1-3\lambda )), \end{aligned}$$

we have

$$\begin{aligned} J_\sharp (\mathscr {D}')&= \iint _{\mathscr {D}'} e^{i\tau Q}F \frac{24i\tau (1-\lambda ) \mu ^2}{A^2}\textrm{d}\lambda \textrm{d}\mu \\ {}&= \iint _{\mathscr {D}'} \partial _{\lambda } (\lambda - \lambda _0) e^{i\tau Q}) F \frac{24i\tau \nu ^2}{A^2B} \frac{2\lambda -1 }{\sqrt{1-\lambda }} \textrm{d}\lambda \textrm{d}\nu \\ {}&= - \iint _{\mathscr {D}'} e^{i\tau Q}\left[ - \partial _1 F - \partial _2 F + 2 \partial _3 F \right] \times \frac{24i\tau (\lambda -\lambda _0) \nu ^2 }{A^2 B} \frac{2\lambda -1}{\sqrt{1-\lambda }} d \lambda \textrm{d}\nu \\ {}&\quad + \iint _{\mathscr {D}'} e^{i\tau Q} F \frac{24i\tau (\lambda - \lambda _0) \nu ^2}{A^2 B} \left( 2+ \frac{1}{2(1-\lambda ) } + \frac{\partial _\lambda B}{B}\right) \frac{2\lambda -1}{\sqrt{1-\lambda }} \textrm{d}\lambda \textrm{d}\nu \\ {}&\quad + \int _{\partial \mathscr {D}'} (\lambda - \lambda _0) e^{i\tau Q} F \frac{24i\tau \nu ^2}{A^2B}\frac{2\lambda -1}{\sqrt{1-\lambda }} \textrm{d}\sigma (\tilde{\lambda },\nu ). \end{aligned}$$

5.1) On $\mathscr {A}_3'$, we choose $\lambda _0 = 2/3$. Then we have $|\lambda - 2/3| \leqq 1/10$ and $|\nu | \leqq 1/10$ so that

$$\begin{aligned} \lambda , 1-\lambda , 1-3\lambda , 2\lambda -1 \end{aligned}$$

can be bounded (above and below) by uniform constants. In particular, letting $\tilde{\lambda }:= \lambda - 2/3$,

$$\begin{aligned} |A| \gtrsim 1+ \tau \nu ^2, \quad |B| \gtrsim 1+ \tau \tilde{\lambda }^2, \end{aligned}$$

and we can bound

$$\begin{aligned} |J_\sharp (\mathscr {A}_3')|&\lesssim \iint _{\mathscr {A}_3^\sharp } \frac{\tau \nu ^2 |\tilde{\lambda }|}{(1+\tau \nu ^2)^2(1+ \tau \tilde{\lambda }^2)}(|\partial _1 F| + |\partial _2 F| + |\partial _3 F|)\textrm{d}\tilde{\lambda }\textrm{d}\nu \nonumber \\&\quad + \iint _{\mathscr {A}_3^\sharp } \frac{\tau \nu ^2 |\tilde{\lambda }|}{(1+\tau \tilde{\lambda }^2) (1+\tau \nu ^2)^2} \left( 1 + \frac{\tau | \tilde{\lambda }|}{1 + \tau \tilde{\lambda }^2} \right) |F| \textrm{d}\tilde{\lambda }\textrm{d}\nu \nonumber \\&\quad + \int _{\partial \mathscr {A}_3^\sharp } \frac{\tau \nu ^2 \tilde{|}\lambda |}{(1+\tau \tilde{\lambda }^2) (1+\tau \nu ^2)^2} |F| \textrm{d}\sigma (\tilde{\lambda },\nu ). \end{aligned}$$

(A.7)

First consider the first term. Recall that $\Vert F \Vert _{L^\infty } \lesssim \Vert {\tilde{f}} \Vert _{L^\infty } \Vert {\tilde{g}} \Vert _{L^\infty } \Vert {\tilde{h}} \Vert _{L^\infty } \lesssim 1$ and

$$\begin{aligned} \left( \int |\partial _1 F|^2 \textrm{d}\tilde{\lambda }\right) ^{1/2} \lesssim |\xi |^{1/2} \Vert \partial _\xi {\tilde{f}} \Vert _{L^2} \Vert {\tilde{g}} \Vert _{L^\infty } \Vert {\tilde{h}} \Vert _{L^\infty } \lesssim \tau ^{1/6}, \end{aligned}$$

and similarly for $\partial _2 F$ and $\partial _3 F$. Keeping in mind that $\tilde{\lambda }\in [-1,1]$ uniformly on $\mathscr {A}_3^\sharp = \{ (\tilde{\lambda },\nu ): (\lambda ,\mu ) \in \mathscr {A}_3' \}$, we bound

$$\begin{aligned}&\int _{-1}^1 \frac{|\tilde{\lambda }|}{1+ \tau \tilde{\lambda }^2} |\partial _j F| \textrm{d}\tilde{\lambda }\lesssim \Vert \partial _j F \Vert _{L^2} \left( \int _{-1}^1 \frac{\tilde{\lambda }^2 \textrm{d}\tilde{\lambda }}{(1+\tau \tilde{\lambda }^2)^2} \right) ^{1/2} \lesssim \tau ^{1/6-3/4},\quad j=1,2,3. \end{aligned}$$

Hence the first term in (A.7) is bounded by

$$\begin{aligned}&\tau ^{1/6-3/4} \int \frac{\tau \nu ^2}{(1+\tau \nu ^2)^2} \textrm{d}\nu \lesssim \tau ^{1/6-3/4- 1/2} \lesssim \tau ^{-13/12} \end{aligned}$$

(we used that $\tau \geqq 1$).

Similarly, for the second term, observe that

$$\begin{aligned} \int _{-1}^1 \frac{|\tilde{\lambda }|}{1+ \tau \tilde{\lambda }^2} \left( 1 + \frac{\tau |\tilde{\lambda }|}{1 + \tau \tilde{\lambda }^2} \right) \textrm{d}\tilde{\lambda }\lesssim \tau ^{-1} \ln (1+\tau ) + \tau ^{-1/2} \lesssim \tau ^{-1/2}, \end{aligned}$$

so that the second term is bounded by

$$\begin{aligned} \tau ^{-1/2} \int \frac{\tau \nu ^2}{(1+\tau \nu ^2)^2} \textrm{d}\nu \lesssim \tau ^{-1}. \end{aligned}$$

The boundary term can be bounded accordingly by

$$\begin{aligned} \int _{-1}^1 \frac{\tau |\rho ^3| \textrm{d}\rho }{(1+ \tau \rho ^2)^3} \lesssim \tau ^{-1}. \end{aligned}$$

Hence, as $\tau \geqq 1$, we obtain

$$\begin{aligned} |J_\sharp (\mathscr {A}_3')| \lesssim \tau ^{-1}. \end{aligned}$$

5.2) On $\mathscr {A}_4'$, we choose $\lambda _0 = 0$. Then we have $|\lambda | \leqq 1/10$ and $|\nu | \leqq 1/10$ so that

$$\begin{aligned} 1-\lambda , 1-3\lambda , 2 - 3\lambda \end{aligned}$$

can be bounded (above and below) by uniform constants and

$$\begin{aligned} |A| \gtrsim 1+ \tau \nu ^2, \quad |B| \gtrsim 1+ \tau \lambda ^2. \end{aligned}$$

Hence, we can bound (with $\mathscr {A}_4^\sharp = \{ (\lambda , \nu ): (\lambda ,\mu ) \in \mathscr {A}_4' \}$)

$$\begin{aligned} |J_\sharp (\mathscr {A}_4')|&\lesssim \iint _{\mathscr {A}_4^\sharp } \frac{\tau \nu ^2 |\lambda |}{(1+(\tau \nu ^2)^2)(1+ \tau \lambda ^2)}(|\partial _1 F| + |\partial _2 F| + |\partial _3 F|)d \lambda \textrm{d}\nu \\ {}&\quad + \iint _{\mathscr {A}_4^\sharp } \frac{\tau \nu ^2 |\lambda |}{(1+\tau \lambda ^2) (1+\tau \nu ^2)^2} \left( 1 + \frac{| \lambda |}{1 + \tau \lambda ^2} \right) |F| d \lambda \textrm{d}\nu \\ {}&\quad + \int _{\partial \mathscr {A}_4^\sharp } \frac{\tau \nu ^2 |\lambda |}{(1+\tau \lambda ^2) (1+\tau \nu ^2)^2} \textrm{d}\sigma ( \lambda ,\nu ). \end{aligned}$$

The exact same computations as in 5.1) give the bound

$$\begin{aligned} |J_\sharp (\mathscr {A}_4')| \lesssim \tau ^{-1}. \end{aligned}$$

6) Here we bound $J_\flat$. We consider the change of variables

$$\begin{aligned} \mu =\frac{3\zeta +\chi -2}{2},\quad \lambda =\frac{2-\zeta +\chi }{2}. \end{aligned}$$

One may obtain this transformation by going back to the $\xi$ variables, switching $q_1$ with $q_3$ and then redoing the $\lambda ,\mu$ transformation. In this way, $q_1$ depends on a single variable $\zeta$, and more precisely

$$\begin{aligned} q_1 = 2 \zeta , \quad 1- q_2 = \zeta +\chi , \quad 1-q_3 = \zeta - \chi . \end{aligned}$$

This permits the integration by parts in $\chi$ without the introduction of second-order derivatives in f. In this coordinate system, the stationary points are

$$\begin{aligned} (\chi ,\zeta )=(0,2/3) \text { and } (-1,1). \end{aligned}$$

Also

$$\begin{aligned} Q = 6 ( \zeta +\chi ) (\zeta - \chi ) (1-\zeta ), \quad \partial _\chi Q = 12 \chi (\zeta -1). \end{aligned}$$

6.1) On $\mathscr {A}_3'$, we use the relation

$$\begin{aligned} e^{i\tau Q}=\frac{1}{1+12i\tau \chi ^2(1-\zeta )}\partial _\chi (\chi e^{i\tau Q}). \end{aligned}$$

Define

$$\begin{aligned} A{} & {} =1+4i\tau \mu ^2(1-\lambda )=1+i\tau (3\zeta +\chi -2)^2(\zeta -\chi )/2, \quad \\ C{} & {} = 1+12i\tau \chi ^2(1-\zeta ). \end{aligned}$$

We now integrate by parts:

$$\begin{aligned} J_\flat (\mathscr {A}_3')&= \iint _{\mathscr {A}_3'} e^{i\tau Q}\frac{ \mu \partial _1 F}{A}\textrm{d}\lambda \textrm{d}\mu = \iint _{\mathscr {A}_3^\flat } \frac{1}{C} \partial _{\chi } (\chi e^{i\tau Q}) \frac{\mu }{2A} \partial _1 F \textrm{d}\zeta \textrm{d}\chi \nonumber \\ {}&= - \iint _{\mathscr {A}_3^\flat } \chi e^{i\tau Q} \left[ - \partial _{12} F+ \partial _{13} F \right] \frac{\mu }{2AC} \textrm{d}\chi \textrm{d}\zeta \nonumber \\ {}&\qquad - \iint _{\mathscr {A}_3^\flat } \chi e^{i\tau Q} \frac{\partial _{1} F}{2AC} \left( 1 - \mu \left( \frac{\partial _\chi A }{A} + \frac{\partial _\chi C}{C} \right) \right) \textrm{d}\chi \textrm{d}\zeta \nonumber \\ {}&\qquad - \iint _{\partial \mathscr {A}_3^\flat } e^{i\tau Q} \frac{\chi \mu }{2AC} \partial _1 F \textrm{d}\sigma (\zeta ,\chi ). \end{aligned}$$

On $\mathscr {A}_3^\flat$, $1-\zeta$, $\zeta -\chi$ are bounded above and below and

$$\begin{aligned} A \gtrsim 1 + \tau \mu ^2, \quad C \gtrsim 1 + \tau \chi ^2, \quad |\partial _\chi A| \lesssim \tau |\mu |, \quad |\partial _{\chi } C| \lesssim \tau |\chi |. \end{aligned}$$

Therefore, after changing again the variable $\zeta$ back for $\mu$ (and $\mathscr {A}_3^\flat = \{ (\chi , \mu ): (\lambda ,\mu ) \in \mathscr {A}_3' \}$),

$$\begin{aligned}&| J_\flat (\mathscr {A}_3')| \lesssim \iint _{\mathscr {A}_3^\flat } \frac{|\chi \mu |}{(1+\tau \mu ^2)(1+\tau \chi ^2)} | \partial _{12} F| + | \partial _{13} F|) \textrm{d}\chi \textrm{d}\mu \end{aligned}$$

(A.8)

$$\begin{aligned}&\quad + \iint _{\mathscr {A}_3^\flat } \frac{|\chi |}{(1+\tau \mu ^2)(1+\tau \chi ^2)} \left( 1 +\frac{|\tau | \mu ^2}{1+\tau \mu ^2} + \frac{\tau |\mu \chi |}{1+\tau \chi ^2} \right) |\partial _1 F| \textrm{d}\chi \textrm{d}\mu \end{aligned}$$

(A.9)

$$\begin{aligned}&\quad + \iint _{\partial \mathscr {A}_3^\flat } \frac{|\chi \mu |}{(1+\tau \mu ^2)(1+\tau \chi ^2)} | \partial _{1} F| \textrm{d}\sigma (\chi ,\mu ). \end{aligned}$$

(A.10)

By the Cauchy-Schwarz inequality,

$$\begin{aligned} \left( \iint | \partial _{12} F|^2 \textrm{d}\chi \textrm{d}\mu \right) ^{1/2} \lesssim |\xi | \Vert \partial _\xi {\tilde{f}} \Vert _{L^2} \Vert \partial _\xi {\tilde{g}} \Vert _{L^2} \Vert {\tilde{h}} \Vert _{L^\infty }&\lesssim \tau ^{1/3}, \\ \left( \int | \partial _{1} F|^2 \textrm{d}\chi \right) ^{1/2} \lesssim |\xi |^{1/2} \Vert \partial _\xi {\tilde{f}} \Vert _{L^2} \Vert {\tilde{g}} \Vert _{L^\infty } \Vert {\tilde{h}} \Vert _{L^\infty }&\lesssim \tau ^{1/6}, \quad \text {so that} \\ \Vert \partial _1 F \Vert _{L^\infty _\mu L^2_\chi } , \Vert \partial _1 F \Vert _{L^\infty _\chi L^2_\mu }, \Vert \partial _1 F \Vert _{L^2(\partial \mathscr {A}_3^\flat )}&\lesssim \tau ^{1/6}. \end{aligned}$$

and the first bound also holds with $\partial _{13} F$. Hence the integral in (A.8) is bounded by

$$\begin{aligned}&\left( \iint \frac{|\mu |^2}{(1+\tau \mu ^2)^2} \frac{|\chi |^2}{(1+\tau \chi ^2)^2} \textrm{d}\mu \textrm{d}\chi \right) ^{1/2} \left( \iint (|\partial _{12} F| + |\partial _{13} F|)^2 \textrm{d}\chi \textrm{d}\mu \right) ^{1/2} \\&\lesssim \tau ^{-3/2+1/3} \lesssim \tau ^{-7/6}. \end{aligned}$$

The first two terms of the integral in (A.9) can be controlled as

$$\begin{aligned} \int \frac{\textrm{d}\mu }{1+\tau \mu ^2} \left( \int \frac{\chi ^2 \textrm{d}\chi }{1+(\tau \chi ^2)^2} \right) ^{1/2} \Vert \partial _1 F \Vert _{L^\infty _\mu L^2_\chi } \lesssim \tau ^{-1/2- 3/4+ 1/6} \lesssim \tau ^{-13/12}, \end{aligned}$$

while the third term in (A.9) is bounded by

$$\begin{aligned} \int \frac{\tau \chi ^2 \textrm{d}\chi }{(1+\tau \chi ^2)^2} \left( \int \frac{|\mu |^2 \textrm{d}\mu }{(1+\tau \mu ^2)^2} \right) ^{1/2} \Vert \partial _1 F \Vert _{L^\infty _\chi L^2_\mu } \lesssim \tau ^{-1/2-3/4+ 1/6} \lesssim \tau ^{-13/12}. \end{aligned}$$

Finally, the boundary term (A.10) can be bounded by

$$\begin{aligned} \Vert \partial _1 F \Vert _{L^2(\partial \mathscr {A}_3^\flat )} \left( \int \left( \frac{\rho ^2}{(1+\tau \rho ^2)^2} \right) ^2 \textrm{d}\rho \right) ^{1/2} \lesssim \tau ^{1/6-5/4} \lesssim \tau ^{-13/12} \end{aligned}$$

and we obtain

$$\begin{aligned} J_\flat (\mathscr {A}_3') \lesssim \tau ^{-13/12}. \end{aligned}$$

6.2) On $\mathscr {A}_4'$, we use the relation

$$\begin{aligned} e^{i\tau Q}=\frac{1}{1+12i\tau \chi (\chi +1)(1-\zeta )}\partial _\chi ((\chi +1) e^{i\tau Q}). \end{aligned}$$

Let

$$\begin{aligned} A{} & {} =1+4i\tau \mu ^2(1-\lambda )=1+i\tau (3\zeta +\chi -2)^2(\zeta -\chi )/2, \quad \\ D{} & {} = 1+12i\tau \chi (\chi +1)(1-\zeta ). \end{aligned}$$

The integration by parts yields

$$\begin{aligned} J_\flat (\mathscr {A}_4')&= \iint _{\mathscr {A}_4'} e^{i\tau Q}\frac{ \mu \partial _1 F}{A}\textrm{d}\lambda \textrm{d}\mu = \iint _{\mathscr {A}_4^\flat } \frac{1}{D} \partial _{\chi } ((\chi +1) e^{i\tau Q}) \frac{\mu }{2A} \partial _1 F \textrm{d}\zeta \textrm{d}\chi \\ {}&= - \iint _{\mathscr {A}_4^\flat } (\chi +1) e^{i\tau Q} \left[ - \partial _{12} F+ \partial _{13} F \right] \frac{\mu }{2AD} \textrm{d}\chi \textrm{d}\zeta \\ {}&\qquad - \iint _{\mathscr {A}_4^\flat } (\chi +1) e^{i\tau Q} \frac{\partial _{1} F}{2AD} \left( 1 - \mu \left( \frac{\partial _\chi A}{A} + \frac{\partial _\chi D}{D} \right) \right) \textrm{d}\chi \textrm{d}\zeta \\ {}&\qquad - \iint _{\partial \mathscr {A}_4^\flat } (\chi +1) e^{i\tau Q} \frac{\chi \mu }{2AD} \partial _1 F \textrm{d}\sigma (\zeta ,\chi ). \end{aligned}$$

On $\mathscr {A}_4^\flat$, $1-\zeta$, $\zeta -\chi$ are bounded above and below. Moreover, denoting $\tilde{\chi }= \chi +1$, $\tilde{\zeta }= 1-\zeta$, there holds

$$\begin{aligned} A \gtrsim 1 + \tau \mu ^2, \quad D \gtrsim 1 + \tau |\tilde{\chi }\tilde{\zeta }|, \quad |\partial _\chi A| \lesssim \tau |\mu |, \quad |\partial _{\chi } D| \lesssim \tau |\tilde{\zeta }|. \end{aligned}$$

After we perform once again the change of variable $(\mu ,\chi ) \rightarrow (\tilde{\zeta }, \tilde{\chi })$ with $\tilde{\zeta }= 1-\zeta = \frac{2\mu -\tilde{\chi }}{3}$), we find

$$\begin{aligned}&\ |J_\flat (\mathscr {A}_4')| \lesssim \iint _{\mathscr {A}_4^\flat } \frac{|\tilde{\chi }\mu |}{(1+\tau \mu ^2)(1+ \tau |\tilde{\chi }\tilde{\zeta }|)} (|\partial _{12} F|+ |\partial _{13} F) \textrm{d}\tilde{\chi }\textrm{d}\mu \end{aligned}$$

(A.11)

$$\begin{aligned}&\quad + \iint _{\mathscr {A}_4^\flat } \frac{|\tilde{\chi }|}{(1+\tau \mu ^2)(1+ \tau |\tilde{\chi }\tilde{\zeta }|)} \left( 1 + \frac{\tau \mu ^2}{1+\tau \mu ^2} + \frac{\tau |\mu \tilde{\zeta }|}{1+\tau |\tilde{\chi }\tilde{\zeta }|} \right) |\partial _{1} F| \textrm{d}\tilde{\chi }\textrm{d}\mu \end{aligned}$$

(A.12)

$$\begin{aligned}&\quad + \int _{\partial \mathscr {A}_4^\flat } \frac{|\tilde{\chi }\mu |}{(1+\tau \mu ^2)(1+ \tau |\tilde{\chi }\tilde{\zeta }|)} |\partial _1 F| \textrm{d}\sigma (\tilde{\chi },\mu ). \end{aligned}$$

(A.13)

As in 6.1),

$$\begin{aligned} \left( \iint (|\partial _{12} F|+ |\partial _{13} F)^2 \textrm{d}\tilde{\chi }\textrm{d}\mu \right) ^{1/2}&\lesssim \tau ^{1/3}, \\ \Vert \partial _1 F \Vert _{L^\infty _{\tilde{\chi }} L^2_\mu }, \Vert \partial _1 F \Vert _{L^\infty _\mu L^2_{\tilde{\chi }}}, \Vert \partial _1 F \Vert _{L^2(\mathscr {A}_4^\flat )}&\lesssim \tau ^{1/6}. \end{aligned}$$

Hence, rescaling $\check{\chi }= \sqrt{\tau } \tilde{\chi }$, $\check{\mu }= \sqrt{\tau } \mu$ and using

$$\begin{aligned} \int \frac{\check{\chi }^2 \textrm{d}\check{\chi }}{(1+|\check{\chi }(2\check{\mu }- \check{\chi })|)^2}{} & {} \lesssim \int _{|2 \check{\mu }- \check{\chi }| \leqq 1/|\check{\mu }|} \check{\chi }^2 \textrm{d}\check{\chi }+ \int _{ |2 \check{\mu }- \check{\chi }| \geqq 1/|\check{\mu }|} \frac{\textrm{d}\check{\chi }}{(2\check{\mu }- \check{\chi })^2}\nonumber \\{} & {} \lesssim |\check{\mu }|, \end{aligned}$$

(A.14)

we can bound the integral in (A.11) by

$$\begin{aligned}&( \Vert \partial _{12} F \Vert _{L^2} + \Vert \partial _{13} F \Vert _{L^2} ) |\tau |^{-3/2} \left( \iint _{| \check{\mu }|, |\check{\chi }| \leqq 10 \sqrt{\tau }} \left( \frac{|\check{\chi }\check{\mu }|}{(1+ \check{\mu }^2)(1+|\check{\chi }(2 \check{\mu }- \check{\chi })|)} \right) ^2 d \check{\chi }\textrm{d}\check{\mu }\right) ^{1/2}\\ {}&\lesssim \tau ^{1/3-3/2}\left( \int _{|\check{\mu }|\leqq 10\sqrt{\tau }} \frac{|\check{\mu }|^3}{(1+\check{\mu }^2)^2}\textrm{d}\check{\mu }\right) ^{1/2} \lesssim \tau ^{-7/6} \ln (1+\tau ) \lesssim \tau ^{-1}. \end{aligned}$$

We now consider the integral in (A.12). For the first two terms, we do Cauchy-Schwarz in $\tilde{\chi }$ and rescale as before $\check{\chi }= \sqrt{\tau } \tilde{\chi }$ and $\check{\mu }= \sqrt{\tau } \mu$, so as to get the bound

$$\begin{aligned}&\Vert \partial _1 F \Vert _{L^\infty _{\mu } L^2_{\tilde{\chi }}} \int _{|\mu | \leqq 1} \left( \int _{|\tilde{\chi }| \leqq 1} \frac{\tilde{\chi }^2d \tilde{\chi }}{(1+|\tau \tilde{\chi }\tilde{\zeta }|)^2} \right) ^{1/2} \frac{\textrm{d}\mu }{(1+\tau \mu ^2)} \\ {}&\lesssim \tau ^{1/6-1/2-3/4} \int _{|\check{\mu }| \leqq 10 \sqrt{\tau }} \left( \int _{|\check{\chi }| \leqq 10 \sqrt{\tau }} \frac{\check{\chi }^2 d \check{\chi }}{(1+| \check{\chi }\check{\zeta }|)^2} \right) ^{1/2} \frac{\textrm{d}\check{\mu }}{(1+\check{\mu }^2)} \lesssim \tau ^{-13/12}, \end{aligned}$$

where we used once again (A.14).

For the third term of the integral in (A.12), we do Cauchy-Schwarz in $\tilde{\chi }$ and rescale again $\check{\chi }= \sqrt{\tau } \tilde{\chi }$, $\check{\mu }= \sqrt{\tau } \mu$, and get the bound

$$\begin{aligned}&\Vert \partial _1 F \Vert _{L^\infty _{\mu } L^2_{\tilde{\chi }}} \int _{|\mu | \leqq 1} \left( \int _{|\tilde{\chi }| \leqq 1} \frac{d \tilde{\chi }}{(1+|\tau \tilde{\chi }\tilde{\zeta }|)^2} \right) ^{1/2} \frac{\mu \textrm{d}\mu }{(1+\tau \mu ^2)} \\ {}&\lesssim \tau ^{1/6- 1-1/4} \int _{|\check{\mu }| \leqq 10 \sqrt{\tau }} \left( \int _{|\check{\chi }| \leqq 10 \sqrt{\tau }} \frac{d \check{\chi }}{(1+| \check{\chi }\check{\zeta }|)^2} \right) ^{1/2} \frac{|\check{\mu }|\textrm{d}\check{\mu }}{(1+\check{\mu }^2)} \lesssim \tau ^{-13/12}. \end{aligned}$$

Indeed, we show that the integral $\lesssim 1$ as follows. We can assume $| \check{\mu }| \geqq 10$. Then, for fixed $\check{\mu }$,

$$\begin{aligned} \int _{10 \leqq |\check{\chi }|} \frac{d \check{\chi }}{(1+| \check{\chi }(2\check{\mu }- \check{\chi })|)^2}&\lesssim \int _{|2 \check{\mu }- \check{\chi }| \leqq 1/|\check{\mu }|} \textrm{d}\check{\chi }\\&\qquad + \int _{1/|\check{\mu }| \leqq |2 \check{\mu }- \check{\chi }| \leqq |\check{\mu }|/100 } \frac{\textrm{d}\check{\chi }}{ \check{\mu }^2 (2\check{\mu }- \check{\chi })^2} \\ {}&\qquad + \int _{|2 \check{\mu }- \check{\chi }| \geqq |\check{\mu }|/100 } \frac{\textrm{d}\check{\chi }}{(2\check{\mu }- \check{\chi })^2} \lesssim \frac{1}{|\check{\mu }|}. \end{aligned}$$

Hence

$$\begin{aligned} \int _{|\check{\mu }| \leqq 10 \sqrt{\tau }} \left( \int _{|\check{\chi }| \leqq 10 \sqrt{\tau }} \frac{d \check{\chi }}{(1+| \check{\chi }\check{\zeta }|)^2} \right) ^{1/2} \frac{\check{\mu }\textrm{d}\check{\mu }}{(1+\check{\mu }^2)} \lesssim \int \frac{|\check{\mu }|^{1/2} \textrm{d}\check{\mu }}{(1+\check{\mu }^2)} \lesssim 1. \end{aligned}$$

Finally, we consider the boundary term (A.13): there, $\tilde{\chi }$, $\mu$ and $\tilde{\zeta }$ are proportional; by Cauchy-Schwarz inequality, this term is bounded up to a constant by

$$\begin{aligned} \Vert \partial _1 F \Vert _{L^2(\mathscr {A}_4^\flat )} \left( \int _{[-1,1]} \frac{\rho ^4 \textrm{d}\rho }{(1+\tau \rho ^2)^4} \right) ^{1/2} \lesssim \tau ^{1/6-5/4} \lesssim \tau ^{-13/12}. \end{aligned}$$

7) In summary, (recalling $\tau \geqq 1$), we proved in 5) and 6) that

$$\begin{aligned} |J_\sharp (\mathscr {A}_3)|+ |J_\flat (\mathscr {A}_3)| + |J_\sharp (\mathscr {A}_4)| + |J_\flat (\mathscr {A}_4)| \lesssim \tau ^{-1}. \end{aligned}$$

Hence $|J(\mathscr {A}_3)| + |J(\mathscr {A}_4)| \lesssim \tau ^{-1} \xi ^3 \lesssim 1/t$. Summing up with the bounds in 1)-3), we infer that $|J(\mathscr {A})| \lesssim 1/t$, and the same estimate holds for $J(\mathbb {R}^2)$, as claimed. $\square$