Mixing solutions for the Muskat problem

Abstract

We prove the existence of mixing solutions of the incompressible porous media equation for all Muskat type $H^5$ initial data in the fully unstable regime. The proof combines convex integration, contour dynamics and a basic calculus for non smooth semiclassical type pseudodifferential operators which is developed.

Appendices

Appendix

In this “Appendix” we will prove lemma 4.8, lemma 4.9 and the required estimates for the velocity u and for the coefficient of the transport term a. Throughout the whole section, there are integrals which are interpreted in the principal value sense, both at 0 and $\infty$. Since, this is standard and harmless in our context, we will not make it explicit.

1.1 Lower order terms. Proof of lemma 4.8

We can write

$$\begin{aligned} {\mathcal {M}} u =-\frac{1}{4\pi }\int _{-1}^1\int _{-1}^1 \int _{{\mathbb {R}}} k_{\theta }(x,y)\partial _x \theta dy d\lambda 'd\lambda . \end{aligned}$$

The main part of the proof of the lemma will be showing that

$$\begin{aligned}&\left| \left| {\mathcal {D}}^{-1}\left( \partial ^5_x {\mathcal {M}} u+\frac{1}{4\pi }\int _{-1}^1\int _{-1}^1 \int _{{\mathbb {R}}} k_{\theta }(\cdot ,y)\partial ^6_x \theta dy d\lambda 'd\lambda \right) \right| \right| _{L^2}\nonumber \\&\le C\left( ||f||_{H^4},||{\mathcal {D}}^{-1}\partial ^5_x f||_{L^2}\right) . \end{aligned}$$

(A.1)

In order to accomplish this, we will need to compute the derivatives of the function

$$\begin{aligned} k_\theta (x,y)=\frac{y}{y^2+\theta ^2}, \end{aligned}$$

where

$$\begin{aligned} \theta =\Delta f(x,x-y)+\varepsilon (x)\lambda -\varepsilon (x-y)\lambda '=\Delta f+\Delta \varepsilon \lambda ' +\varepsilon (x)(\lambda -\lambda '). \end{aligned}$$

In addition we introduce

$$\begin{aligned} h= f(x)+\lambda \varepsilon (x)\quad \text {and} \quad h'=f(x)+\lambda '\varepsilon (x),\gamma =\varepsilon (x)(\lambda -\lambda '). \end{aligned}$$

Thus,

$$\begin{aligned} \theta =\Delta h' +\varepsilon (x)(\lambda -\lambda ')=\Delta h'+\gamma , \end{aligned}$$

where we remark that $\gamma$ depends on x and on t although we will not make this dependence explicit. We recall that c(x, t) is as in the statement of theorem 4.1. Since $\varepsilon (x,t)=c(x,t)t$, with $0<{\overline{c}}<c(x,t)$ and $||c(x,t)||_{C^5}\le C$,

$$\begin{aligned} \Vert \gamma \Vert _{C^5} \le C t|\lambda -\lambda '|. \end{aligned}$$

(A.2)

A big part of our proof will be based on comparing $\theta$ with its linearized version

$$\begin{aligned} \theta _{lin}=\partial _x h'(x) y+\gamma . \end{aligned}$$

Remark A.1

Notice that in order to obtain selfadjoint pseudodifferential operators we could deal as well with

$$\begin{aligned} \theta _{lin}^W=\partial _x h'(\frac{x+y}{2}) y+\gamma , \end{aligned}$$

Here the W in the exponent stands for the Weyl quantization. We do not pursue this issue here.

To make the notation even more compact we will write $\Delta f=\Delta f(x,x-y)$ and $\partial ^k_x f(x)=\partial ^k_x f$. Then we have

$$\begin{aligned}&\partial _x k_\theta (x,y)=-2\frac{y\theta \partial _x\theta }{(y^2+\theta ^2)^2}\equiv -2k_\theta ^{11}(x,y), \\&\partial ^2_x k_\theta (x,y)=-2\frac{y((\partial _x\theta )^2+(\theta )\partial _x^2\theta )}{(y^2+(\theta )^2)^2} +8\frac{y(\left( \theta )\partial _x\theta \right) ^2}{(y^2+(\theta )^2)^3}\\&\equiv c_{21} k_{\theta }^{21}(x,y)+c_{22} k_\theta ^{22}(x,y), \end{aligned}$$

$$\begin{aligned} \partial _x^3 k_\theta (x,y)&=-2y \frac{3\partial _x\theta \partial ^2_x\theta +(\theta )\partial ^3_x\theta }{(y^2+(\theta )^2)^2}+24 y \frac{(\theta )\partial _x\theta ((\partial _x\theta )^2+(\theta )\partial ^2_x\theta )}{(y^2+(\theta )^2)^3}\nonumber \\&\quad -48y\frac{(\theta )^3(\partial _x\theta )^3 }{(y^2+(\theta )^2)^4}\nonumber \\&\equiv c_{31}k_\theta ^{31}(x,y)+c_{32}k_\theta ^{32}(x,y)+c_{33}k_\theta ^{33}(x,y), \end{aligned}$$

(A.3)

$$\begin{aligned} \partial ^4_x k_\theta (x,y)&=384y\frac{(\theta )^4(\partial _x\theta )^4}{(y^2+(\theta )^2)^5}-288y\frac{(\theta )^2(\partial _x\theta )^2((\partial _x \theta )^2+(\theta ) \partial _{x}^2\theta )}{(y^2+(\theta )^2)^4}\\&\quad +y\frac{24(\partial _x\theta )^4+144(\theta )(\partial _x\theta )^2\partial ^2_x\theta +24(\theta )^2(\partial ^2_x\theta )^2}{(y^2+(\theta )^2)^3}\\&\quad +y\frac{32(\theta )^2\partial _x\theta \partial ^3_x \theta }{(y^2+(\theta )^2)^3}\\&\quad -y\frac{6(\partial _x^2\theta )^2+8\partial _x\theta \partial ^3_x\theta -2(\theta )\partial ^4_x\theta )}{(y^2+(\theta )^2)^2}\\&\equiv c_{41}k_\theta ^{41}(x,y)+c_{42}k_\theta ^{42}(x,y)+c_{43}k_\theta ^{43}(x,y)+c_{44}k_\theta ^{44}(x,y), \\ \partial ^5_x k_\theta (x,y)&=-3840y\frac{(\theta )^5(\partial _x\theta )^5}{(y^2+(\theta )^2)^6}+3840y\frac{(\theta )^3(\partial _x\theta )^5+(\theta )^4(\partial _x \theta )^3\partial ^2_x\theta }{(y^2+(\theta )^2)^5}\\&\quad +y\frac{240 (\partial _x\theta )^3\partial ^2_x\theta +360(\theta )\partial _x \theta (\partial _x^2\theta )^2+240(\theta )(\partial _x\theta )^2\partial _x^3\theta }{(y^2+(\theta )^2)^4}\\&\quad +y\frac{80(\theta )^2\partial ^2_x\theta \partial ^3_x\theta +40(\theta )^2\partial _x\theta \partial ^4_x\theta }{(y^2+(\theta )^2)^3}\\&\quad -y\frac{10\partial _x\theta \partial ^4_x \theta +2(\theta )\partial ^5_x \theta }{(y^2+(\theta )^2)^2}\equiv c_{51}k_\theta ^{51}(x,y)\\&\quad +c_{52}k_\theta ^{52}(x,y)+c_{53}k_\theta ^{53}(x,y)\\&\quad +c_{54}k_\theta ^{54}(x,y)+c_{55}k_\theta ^{55}(x,y), \end{aligned}$$

where we notice that the numbers $c_{ij}$ $i,j=1,2,3,4,5$ are harmless coefficients. Then, by applying Minkowski inequality, we need to bound

$$\begin{aligned}&\sum _{j=1}^5\left| \left| {\mathcal {D}}^{-1}\int _{{\mathbb {R}}}\partial ^j_xk_{\theta }(\cdot ,y)\partial ^{5-j}_x\partial _x\theta dy\right| \right| _{L^2}\\&\quad =\sum _{j=1}^5\sum _{i=1}^j\left| \left| {\mathcal {D}}^{-1}\int _{{\mathbb {R}}}k^{ji}_{\theta }(\cdot ,y)\partial ^{5-j}_x\partial _x\theta dy\right| \right| _{L^2} \end{aligned}$$

independently of t, $\lambda$ and $\lambda '$. The highest order terms in this sum are given by

$$\begin{aligned} {\mathcal {D}}^{-1}\int _{{\mathbb {R}}}k^{11}_{\theta }(x,y)\partial ^5_x\theta dy \quad \text {and}\quad {\mathcal {D}}^{-1}\int _{{\mathbb {R}}}k^{55}_{\theta }(x,y)\partial _x\theta dy. \end{aligned}$$

(A.4)

Since there are 5 derivatives of the function $\theta$ in both terms we have to use the operator $D^{-1}$. Since $D^{-1}$ is bounded in $L^2$ it holds that

$$\begin{aligned}&\sum _{j=2}^5\sum _{i=1}^{\min (j,\,4)}\left| \left| {\mathcal {D}}^{-1}\int _{{\mathbb {R}}}k^{ji}_{\theta }(\cdot ,y)\partial ^{5-j}_x\partial _x\theta dy\right| \right| _{L^2}\nonumber \\&\le \sum _{j=2}^5\sum _{i=1}^{\min (j,\,4)}\left| \left| \int _{{\mathbb {R}}}k^{ji}_{\theta }(\cdot ,y)\partial ^{5-j}_x\partial _x\theta dy\right| \right| _{L^2}, \end{aligned}$$

(A.5)

which make the computation easy. We first deal with the sum (A.5) and finally with (A.4), which are somewhat more delicate.

We will use the following convection. We will write:

• 1. $\precsim$ meaning “is bounded in absolute value by”.

• 2. $f(x)\sim g(x)$ if $||f-g||_{L^2}\le \langle A \rangle$.

• 3. We will denote by $k^{ji}(x)$ the integral

  $$\begin{aligned} \int _{{\mathbb {R}}}k^{ji}_\theta (x,y)\partial ^{6-j}_x\theta dy. \end{aligned}$$

• 4. $C_{k+\alpha }$ will be a constant depending on $||f||_{C^{k+\alpha }}$, with k an integer and $0\le \alpha <\frac{1}{2}$. $C_{k+\alpha }(x)$ will be a function whose $L^\infty -$norm is bounded by a constant depending on $||f||_{C^{k+\alpha }}$.

• 5. Given an integral $\int _{{\mathbb {R}}}f(x,y)dy$ we will estimate separately, $\int _{|y|>1}f(x,y)dy$ its $in-$part and $\int _{|y|<1}f(x,y)dy$ its $out-$part. Several terms $k^{ji}(x)$, with i and j integers, will arise in the computations above. In these terms there always will be an integration of the form

  $$\begin{aligned} k^i_j(x)=\int _{{\mathbb {R}}}\cdots dy. \end{aligned}$$

  We will call $k^{ji}_{\,in}(x)$ and $k^{ji}{j\, out}(x)$ to its $in-$part and to its $out-$part respectively.

• 6. For any f, we will write $\Delta _{t|\lambda -\lambda '|} f \equiv \Delta f (x, x-|t(\lambda -\lambda ')|y).$

• 7. We always assume that $\varepsilon <1$.

• 8. In every integral we take a principal value.

1.1.1 Preliminary lemmas

The proof is rather long and will be armed by the lemmas below. They could be ordered as follows.

• i) Estimates of pointwise of the kernel. Lemmas A.2, A.3 and A.4 estimate operators with non singular terms.

• ii) Comparison between the kernels depending $\theta$ and those depending on $\theta _{lin}.$ Lemmas A.5 and A.6.

• iii) Lemmas on kernels depending on $\theta ,\theta _{lin}$.

The proof of iii) either use i) to show that the kernels are not singular or rely on properties of the Hilbert transform.

The first two lemmas are pointwise properties of the functions involved. The proofs follows from the mean value theorem.

Lemma A.2

There exists a constant $1<C_A<\infty$ depending only on the $L^\infty$-norm of the $\partial _xh'(x)$ such that

$$\begin{aligned}&\frac{1}{(y\pm \sigma _h(x) A_h(x)c(x))^2+c(x)^2\sigma (x)^2}\le {\mathcal {C}}_A(y)\\&\quad \equiv \left\{ \begin{array}{cc}C_A \quad |y|\le C_{A} \\ \frac{C_A}{y^2} \quad |y|>C_A\end{array}\qquad \forall x\in {\mathbb {R}}\right. \end{aligned}$$

Here $A_h(x)=\partial _xh'(x)$ and $\sigma _h(x)=\frac{1}{1+A_h(x)^2}$.

Proof

Along this proof $C_A$ denotes a constant bigger that 1 and depending only in $||A_h||_{L^\infty }$. Firstly we notice that

$$\begin{aligned}&\frac{1}{(y\pm \sigma _h(x) A_h(x)c(x))^2+c^2\sigma ^2}\\&\quad \le \frac{1}{(y\pm \sigma _h(x) A_h(x)c(x))^2+\left( \inf _{x\in {\mathbb {R}}}c(x)\sigma _h(x)\right) ^2}, \end{aligned}$$

where

$$\begin{aligned} \inf _{x\in {\mathbb {R}}} \sigma _h(x)=\frac{1}{1+||A_h||_{L^\infty }^2}\equiv \sigma _{inf}. \end{aligned}$$

Fixed x, the function $\frac{1}{(y\pm \sigma (x)A(x)c(x))^2+(\sigma _{inf})^2}$ is a translation of the function $\frac{1}{y^2+(\sigma _{\inf })^2}$. This is bounded by $\frac{1}{(\sigma _{inf})^2}$ and decay like $\frac{C_A}{y^2}$. But $-C_{A}\le \sigma _h(x)A_h(x)c(x)\le C_{A}$. Then the conclusion of the lemma follows easily. $\square$

The following lemma will allow us to show that numerators of various kernels are in fact bounded for |y| sufficiently small.

Lemma A.3

There exists a constant $c_A$ which depends on $||f||_{C^1} +||\varepsilon ||_{C^1}$ and on ${\overline{c}}$ such that for $|y|<c_A$ the following inequality holds:

$$\begin{aligned} |c(x,t)|-\left| \frac{\Delta _{t|\lambda -\lambda '|} h }{t|\lambda -\lambda '|y}y\right|&\ge \frac{{\overline{c}}}{2}\\ |c(x,t)|-|\partial _x h'(x) y|&\ge \frac{{\overline{c}}}{2}. \end{aligned}$$

Proof

By the mean value theorem,

$$\begin{aligned} \left| \frac{\Delta f(x,x-t|\lambda -\lambda '| y)+\Delta \varepsilon (x,x-t|\lambda | y)}{t|\lambda -\lambda '|y}\right| \le ||f||_{C^1} +||\varepsilon ||_{C^1} \end{aligned}$$

thus the claim follows. $\square$

For the reiterative use we state that $L^1$ kernels gives us good bounds.

Lemma A.4

Consider a kernel $J_\lambda (x,y)$ satisfying $|J_\lambda (x,y)|\le j(y)\in L^1({\mathbb {R}})$, for all $x\in {\mathbb {R}}$ and $\lambda \in [-1,1]$. Then, the the integral

$$\begin{aligned} I_\lambda (x)=\int _{{\mathbb {R}}}J_\lambda (x,y)f(x-\lambda y)dy \end{aligned}$$

is bounded in $L^2$ as $||I_\lambda ||_{L^2}\le C(||j||_{L^1})||f||_{L^2}$.

Proof

Again the proof is straightforward by using Minkowski inequality. $\square$

The next two lemmas allow us to compare $\theta$ with $\theta _{lin}=\partial _x h' y +\gamma$ in various expressions. We start with a pointwise bound.

Lemma A.5

The following bound holds for every $y \in {\mathbb {R}}$.

$$\begin{aligned}&\frac{1}{\left( y^2+\theta ^2\right) ^a}-\frac{1}{\left( y^2+\theta _{lin}^2\right) ^a}\\&\quad \precsim C_2\sum _{l=-1}^{2(a-1)}\frac{|y|^{2a-l}|\gamma |^{l+1}}{\left( y^2+\theta )^2\right) ^a\left( y^2+\theta _{lin}^2\right) ^a} \end{aligned}$$

for $a\ge 2$.

Proof

We just write that

$$\begin{aligned}&\frac{1}{\left( y^2+\left( \theta \right) ^2\right) ^a}-\frac{1}{\left( y^2+\theta _{lin}^2\right) ^a}\\&\quad = \frac{\left( y^2+\theta _{lin}^2\right) ^a-\left( y^2+\left( \theta \right) ^2\right) ^a}{\left( y^2+\left( \theta \right) ^2\right) ^a\left( y^2+\theta _{lin}^2\right) ^a} \end{aligned}$$

and, since, $c^a-b^a= (c-b)\sum _{l=1}^{a} c^{a-l}b^{l-1}$ for $c,b\in {\mathbb {R}}$, we have that

$$\begin{aligned}&\left( y^2+\theta _{lin}^2\right) ^a-\left( y^2+\left( \theta \right) ^2\right) ^a\\&\quad =\left( \theta _{lin}^2-\left( \theta \right) ^2\right) \sum _{l=1}^a \left( y^2+\theta _{lin}^2\right) ^{a-l}\left( y^2+\theta ^2\right) ^{l-1}. \end{aligned}$$

Next we introduce the expansions

$$\begin{aligned}&\left( \theta _{lin}^2-\left( \theta \right) ^2\right) =\left( \partial _x h' y-\Delta h'\right) \left( \partial _x h' y +\Delta h' +2\gamma \right) \\&\quad \precsim C_2 |y|^2\left( |y|+|\gamma |\right) , \end{aligned}$$

Here, we have used that since $h \in H^4$ , we have uniform $L^\infty$ bound of $\partial _x h',\partial ^2_x h'$ and thus its uniform Lipschitz continuity. Next, since

$$\begin{aligned}&\left( y^2+\theta _{lin}^2\right) ^{a-l}=\sum _{i=0}^{a-l}c(i,a-l)y^{2(a-l-i)}(\theta _{lin})^{2i}\\&\quad =\sum _{i=0,n=0}^{a-l,2i}c(i,a-l)c(n,2i)y^{2(a-l)-n}(\partial _x h')^{2i-n}\gamma ^n\\&\quad \precsim C_1 \sum _{i=0,n=0}^{a-l,2i}|y|^{2(a-l)-n}|\gamma |^n \end{aligned}$$

and

$$\begin{aligned}&\left( y^2+\left( \theta \right) ^2\right) ^{l-1}\\&\quad \precsim C_1 \sum _{j=0,m=0}^{l-1,2i}|y|^{2(l-1)-m}|\gamma |^m, \end{aligned}$$

it follows that,

$$\begin{aligned}&\left( y^2+\theta _{lin}^2\right) ^a-\left( y^2+\left( \theta \right) ^2\right) ^a\\&\quad \precsim C_2 |y|^2(|y|+|\gamma |)\sum _{i=0,n=0}^{a-l,2i}\sum _{j=0,m=0}^{l-1,2i}|y|^{2(a-1)-(n+m)}|\gamma |^{n+m}\\&\quad \precsim C_2 |y|^2(|y|+|\gamma |)|y|^{2(a-1)}\sum _{l=0}^{2(a-1)}|y|^{-l}|\gamma |^l\\&\quad =C_2 |y|^{2a}(|y|+|\gamma |)\sum _{l=0}^{2(a-1)}|y|^{-l}|\gamma |^l\\&\quad =C_2 y^{2a}\left( \sum _{l=0}^{2(a-1)}|y|^{-l+1}|\gamma |^l +\sum _{l=0}^{2(a-1)}|y|^{-l}|\gamma |^{l+l}\right) \\&\quad \precsim C_2 |y|^{2a}\sum _{l=-1}^{2(a-1)}|y|^{-l}|\gamma |^{l+1}. \end{aligned}$$

From this last inequality is easy to achieve the conclusion of the lemma. $\square$

Next we show that we can also compare operators depending on $\theta$ by those depending on its linearization $\theta _{lin}$.

Lemma A.6

Let $a=2,3,4$ or 5 and define

$$\begin{aligned}&k[g](x)=\int _{|y|<1}\left( \sum _{i=0}^{2a-1}|\gamma |^i|y|^{2a-1-i}\right) \\&\quad \times \left( \frac{1}{\left( y^2+\theta ^2\right) ^a}-\frac{1}{\left( y^2+\theta _{lin}^2\right) ^a}\right) g(x-y)dy. \end{aligned}$$

Then,

$$\begin{aligned} \Vert k[g]\Vert _{L^2} \le \langle A \rangle ||g||_{L^2}, \Vert k[g]\Vert _{L^\infty } \le \langle A \rangle ||g||_{L^\infty }, \Vert k[1]\Vert _{L^\infty } \le \langle A \rangle , \end{aligned}$$

where $g\in L^2$ in the first estimate and $g\in L^\infty$ in the second one.

Proof

By lemma A.5 we have that

$$\begin{aligned} k(x)\precsim&C_2\int _{|y|<1}\sum _{i=0}^{2a-1}|\gamma |^i|y|^{2a-1-i}\\&\quad \times \sum _{l=-1}^{2(a-1)}\frac{|y|^{2a-l}|\gamma |^{l+1}}{\left( y^2+\left( \theta \right) ^2\right) ^a\left( y^2+\theta _{lin}^2\right) ^a}|g(x-y)|dy\\&=C_2\sum _{l=-1,\, i=0}^{2(a-1),\, 2a-1}\int _{|y|<1}\frac{|y|^{4a-1-(i+l)}|\gamma |^{i+l+1}}{\left( y^2+\left( \theta \right) ^2\right) ^a\left( y^2+\theta _{lin}^2\right) ^a}|g(x-y)|dy. \end{aligned}$$

By the upper bound on $\gamma$ we need to estimate

$$\begin{aligned} k_{l,\,i}(x)=\int _{|y|<1}\frac{|y|^{4a-1-(i+l)}|t|\lambda -\lambda '||^{i+l+1}}{\left( y^2+\left( \theta \right) ^2\right) ^a\left( y^2+\theta _{lin}^2\right) ^a}|g(x-y)|dy. \end{aligned}$$

After the change of variable $y'=\frac{y}{t|\lambda -\lambda '|}$ we have that

$$\begin{aligned}&k_{l,\,i}(x)=t|\lambda -\lambda '|\int _{|y|<\frac{1}{t|\lambda -\lambda '|}}\\&\quad \quad \frac{|y|^{4a-1-l-i} |g(x-t|\lambda -\lambda '| y)|}{\left( y^2+\left( \frac{\Delta _{t|\lambda -\lambda '|} h'}{t|\lambda -\lambda '|y}y+c(x)\text {sign}(t(\lambda -\lambda '))\right) ^2\right) ^a\left( y^2+\left( \partial _xf y+c(x)\text {sign}(t(\lambda -\lambda '))\right) ^2\right) ^a} dy\\&\quad =\int _{|y|<c_A} \cdots \,dy + \int _{c_A<|y|<\frac{1}{t|\lambda -\lambda '|}} \cdots \, dy. \end{aligned}$$

The integrand in $k_{l,\,i}(x)$ is bounded in $|y|<c_A$ by lemma A.3 for every $-1\le l\le 2(a-1)$ and $0\le i \le 2a-1$. In $|y|>c_A$, the integrand is bounded by $C |y|^{-1-i-l}$ for every $-1\le l\le 2(a-1)$ and $0\le i \le 2a-1$. Then Minkowski inequality yields

$$\begin{aligned} ||k_{l,\,i}||_{L^2}\le C_1 ||g||_{L^2}\left( 1+|\gamma |\int _{c_A\le |y|\le \frac{1}{t|\lambda -\lambda '|}}|y|^{-1-i-l}dy\right) \le C_1||g||_{L^2} \end{aligned}$$

for every $-1\le l\le 2(a-1)$ and $0\le i \le 2a-1$. The $L^\infty$ bound simply follows by extracting $\Vert g\Vert _\infty$ as a constant. $\square$

The next lemma gives $L^2$ and $L^\infty$ bounds for the various operators. It turns out that after a change of variable, lemma A.3 shows that in fact the kernels are not singular near cero, whereas away from cero direct $L^\infty$ bounds are available for the kernel. This yields direct proofs for $L^\infty$ bounds and a further use of Minkowski inequality yields $L^2$ bound. We state separately the action of the operator on 1 for later use.

Lemma A.7

Let $g\in L^2$.

• a) Let $a\ge 2$ and k[g](x) be given by

  $$\begin{aligned}&k[g](x)=\sum _{i=0}^{2a}\int _{|y|<1}\frac{|y|^{2a-i}|\gamma |^i}{\left( y^2+\theta ^2\right) ^a}|g(x-y)|dy. \end{aligned}$$

  Then

  $$\begin{aligned} \Vert k[g]\Vert _{L^2} \le \langle A \rangle ||g||_{L^2}, \quad \Vert k[1]\Vert _{L^\infty } \le \langle A \rangle . \end{aligned}$$

• b) Let $a=2,\ldots ,10$ and k[g] be given by

  $$\begin{aligned}&k[g](x)=\sum _{i=1}^{2a-1}\int _{|y|<1}\frac{|y|^{2a-1-i} |\gamma |^i}{\left( y^2+\theta ^2\right) ^a}g(x-y)dy. \end{aligned}$$

  Then

  $$\begin{aligned} \Vert k[g]\Vert _{L^2} \le \langle A \rangle ||g||_{L^2}, \quad \Vert k[1]\Vert _{L^\infty } \le \langle A \rangle . \end{aligned}$$

Proof

We prove first the $L^2$ bound for a). We notice that, by the upper bound on $\gamma$,it is enough to estimate

$$\begin{aligned} k_i(x)=\int _{|y|<1}\frac{|y|^{2a-i}|t|\lambda -\lambda '||^i}{\left( y^2+\theta \right) ^a}|g(x-y)|dy. \end{aligned}$$

After the change of variables $y'=\frac{y}{t|\lambda -\lambda '|}$ we have that

$$\begin{aligned} k_i(x)&=\int _{|y|<\frac{1}{t|\lambda -\lambda '|}}t|\lambda -\lambda '|\frac{|y|^{2a-i}}{\left( y^2+\left( \frac{\Delta _{t|\lambda -\lambda '} h' }{t|\lambda -\lambda '| y}y+c(x,t)\text {sign}(\lambda -\lambda ')\right) ^2\right) ^a}|g(x-t|\lambda -\lambda '|y)|dy\\&=\int _{|y|<c_A}\cdots \,dy +\int _{c_A<|y|<\frac{1}{t|\lambda -\lambda '|}}\cdots \,dy. \end{aligned}$$

For every $i=0,\ldots ,2a$, in the region $|y|<c_A$ we can apply lemma A.3, obtaining that the kernel is uniformly bounded. In the region $c_A<|y|<\frac{1}{t|\lambda -\lambda '|}$ we can estimate

$$\begin{aligned}&t|\lambda -\lambda '|\frac{|y|^{2a-i}}{\left( y^2+\left( \frac{\Delta _{t|\lambda -\lambda '|} h}{t|\lambda -\lambda '| y}y+c(x,t)\text {sign}(\lambda -\lambda ')\right) ^2\right) ^a}\\&\quad \le C_A t|\lambda -\lambda '| |y|^{-i}\le C_A t|\lambda -\lambda '|, \end{aligned}$$

for every $i=0,\ldots ,2a$. Then we can apply Minkowski inequality to prove the lemma. Indeed,

$$\begin{aligned}&C_At|\lambda -\lambda '|\left| \left| \int _{c_A<|y|<\frac{1}{t|\lambda -\lambda '|}}|g(x-t|\lambda -\lambda '|y)|dy\right| \right| _{L^2}\\&\quad \le C_At|\lambda -\lambda '|\int _{c_A<|y|<\frac{1}{t|\lambda -\lambda '|}}|\left| \left| g(\cdot -t|\lambda -\lambda '|y)\right| \right| _{L^2}dy\\&\quad \le C_At|\lambda -\lambda '|||g||_{L^2}\int _{c_A<|y|<\frac{1}{t|\lambda -\lambda '|}}dy \le C_A ||g||_{L^2}. \end{aligned}$$

The $L^\infty$ bound follows in the same way. The case b) is dealt with by the same change of variables. $\square$

In the next lemmas the kernel scales as $y^{-1}$ and thus the estimates are more delicate. For the outer integral we show that the kernel can be decomposed as the sum of $c(x)\frac{1}{y}$ and a function which decays as $|y|^{-2}$. Thus, the Hilbert transform controls the first and the second is not singular.

Lemma A.8

Let $g\in L^2$, $a\ge 2$ and

$$\begin{aligned}&k(x)=\int _{|y|>1}\frac{y^{2a-1}}{\left( y^2+\theta _{lin}^2\right) ^a}g(x-y)dy. \end{aligned}$$

Then,

$$\begin{aligned} \Vert k[g]\Vert _{L^2} \le \langle A \rangle ||g||_{L^2}, \quad \Vert k[1]\Vert _{L^\infty } \le \langle A \rangle . \end{aligned}$$

Proof

Direct computation shows hat

$$\begin{aligned} \frac{y^{2a-1}}{\left( y^2+\theta _{lin}^2\right) ^a}-\frac{1}{(1+(\partial _x h')^2)^a}\frac{1}{y}=j(x,y), \end{aligned}$$

where $j(x,y)=\frac{(1+(\partial _x h')^2)^a y^{2a}-(y^2+(\partial _x h' y+\gamma )^2)^a}{\left( y^2+\theta _{lin}^2\right) ^a(1+(\partial _x h')^2)^2y}.$ Since $|j(x,y)| \le C|y|^{-2}$ the claim for $L^2$ follows from the $L^2$ boundedness of the truncated Hilbert transform and lemma A.4. In order to estimate k[1] notice that $\int _{|y|>1} \frac{1}{y} dy =0$. $\square$

The proof of the next lemma is more subtle as it uses an explicit computation of the Hilbert transform of our kernel.

Lemma A.9

Let $g\in L^2$ and $a=2,3,4$ or 5. Then, the integral

$$\begin{aligned} I[g](x)=\int _{{\mathbb {R}}}\frac{y^{2a-1}}{\left( y^2+\theta _{lin}^2\right) ^a}g(x-y)dy, \end{aligned}$$

satisfies

$$\begin{aligned} \Vert I[g]\Vert _{L^2} \le \langle A \rangle ||g||_{L^2},\quad \Vert I[1]\Vert _{L^\infty } \le \langle A \rangle . \end{aligned}$$

Proof

In analogy with other estimates, we denote $\sigma _h=(1+\partial _xh'(x)^2)^{-1}$ and $A_h=\partial _x h'(x)$. Therefore

$$\begin{aligned}&y^2+(\partial _x h'y+\gamma )^2=\sigma _h^{-1}((y+A_h\sigma _h\gamma )^2+\sigma _h^2\gamma ^2). \end{aligned}$$

(A.6)

and

$$\begin{aligned} I(x)&= C_1(x)\int _{{\mathbb {R}}}\frac{y^{2a-1}}{((y+A_h\sigma _h\gamma )^2+\sigma _h^2\gamma ^2)^a}g(x-y)dy\nonumber \\&=C_1(x)\int _{{\mathbb {R}}}\frac{(y+A_h\sigma _h\gamma )^{2a-1}}{((y+A_h\sigma _h \gamma )^2+\sigma _h^2\gamma ^2)^a}g(x-y)dy\nonumber \\&\quad +C_1(x)\int _{{\mathbb {R}}}\frac{y^{2a-1}-(y+A_h\sigma _h\gamma )^{2a-1}}{((y+A_h\sigma _h \gamma )^2+\sigma _h^2\gamma ^2)^a}g(x-y)dy\nonumber \\&\equiv I_1(x)+I_2(x). \end{aligned}$$

(A.7)

Now we use the identity (for fixed $\sigma _h$ and $\gamma$)

$$\begin{aligned} H\left[ \frac{x^{2a-1}}{(x^2+\sigma _h^2\gamma ^2)^{2a-1}}\right] (x)=-\frac{\sigma _h |\gamma |}{(x^2+\sigma _h^2\gamma ^2)^a}\sum _{l=0}^{a-1}\alpha _{al}\left( \sigma _h|\gamma |\right) ^{2(a-1-l)}x^{2l} \end{aligned}$$

where the $\alpha _{al}$’s are harmless coefficients. Then

$$\begin{aligned}&I_1(x)=\int _{{\mathbb {R}}}\frac{\sigma _h |\gamma |}{((y+A_h \sigma _h\gamma )^2+\sigma _h^2\gamma ^2)^a}\\&\quad \sum _{l=0}^{a-1}\alpha _{al}\left( \sigma _h|\gamma |\right) ^{2(a-1-l)}(y+A_h \sigma _h \gamma )^{2l}Hg(x-y)dy, \end{aligned}$$

and after the usual change of variables

$$\begin{aligned}&I_1(x)\precsim C_1\sum _{i=0}^{a-1}\int _{{\mathbb {R}}}\frac{(y+c A_h\sigma _h t\text {sign}((\lambda -\lambda '))^{2i}}{((y+c\sigma _h A_h\text {sign}(\lambda -\lambda '))^2+c^2\sigma _h^2)^a}H\ g(x-t|\lambda -\lambda '| y)dy\\&\quad \precsim C_1\sum _{i=0}^{a-1}\int _{{\mathbb {R}}}(1+|y|^{2i}){\mathcal {C}}_A(y)^a Hg(x-t|\lambda -\lambda '|y)dy, \end{aligned}$$

where we have applied lemma A.2. Then $I_1(x)$ is bounded in $L^2$ thanks to lemma A.4. To bound $I_2(x)$ we notice that we can write this term in the following way:

$$\begin{aligned} I_2(x)=\int _{{\mathbb {R}}}\frac{y^{2a-1}-(y+c\sigma _h A_h \text {sign}(\lambda -\lambda '))^{2a-1}}{((y-c\sigma _h A_h\text {sign}(\lambda -\lambda '))^2+c^2\sigma _h^2)^a} g(x-t|\lambda -\lambda '| y)dy. \end{aligned}$$

Since the numerator is a polynomial in y of order $2(a-1)$ we can apply again lemmas A.2 and A.4 to obtain a suitable estimate in $L^2$.

In order to estimate I[1] simply replace $g(x-y)$ by 1 and notice that the analogous term to $I_1(x)$ in (A.7) is equal to zero in this case.

$\square$

1.1.2 Estimation of the terms in the sum (A.5)

We need to estimate the terms $k_\theta ^{ji}\partial _x^{5-j}\theta$ which can be further decomposed into a sum of products of various derivatives of $\theta$ divided by $(y^2+\theta ^2)^a$ for $a=2,3,4$. It will be important that since we assume that h and hence $\theta$ belongs to $H^4$ we can assume that $\partial _x^2\theta$ is uniformly Lipschitz. Thus if in the product of derivatives $\partial _x^{j_1} \theta \partial _x^{j_2} \theta \cdots \partial _x^{j_n} \partial _x \theta$ there is only one $j_i \ge 3$ we can interpreted as an operator acting on $\partial ^{j_i}_x \theta$ and we could put our hands on the kernel, linearizing it using the Lipschitz continuity. We illustrate this in the first section with $j=2$ and leave the rest to the reader. Unfortunately, there are some terms with $j=3$ with e.g $(\partial ^3_x \theta )^2$ appears. Those have to be dealt with independently. We do it by means of another pseudodifferential operator.

1.1. Terms in (A.5) with $j=2$, $i=1,2$.

1.1.1. Let us estimate $k^{21}(x)$. We split it into two terms $k^{21}(x)=k^{21}_{1}(x)+k^{21}_2(x)$, with

$$\begin{aligned} k^{21}_1(x)=-\int _{{\mathbb {R}}}k^{21}_\theta (x,y)\partial ^4_x h'(x-y) dy \quad \text {and}&k^{21}_2(x)=\partial ^4_x h\int _{{\mathbb {R}}} k^{21}_\theta (x,y)dy. \end{aligned}$$

To bound $k^{21}_1$ we proceed as follows,

$$\begin{aligned} k^{21}_1(x)=-\int _{{\mathbb {R}}}y\frac{(\partial _x\theta )^2+\theta \partial ^2_x\theta }{(y^2+\theta ^2)^2}\partial ^4_x h'(x-y) dy. \end{aligned}$$

Since $y\frac{(\partial _x\theta )^2+\theta \partial ^2_x\theta }{(y^2+\theta ^2)^2}\precsim C_2|y|^{-3}$, $k^{21}_{1\,out}$ is bounded by lemma A.4. In order to bound the inner part we further split it into two terms, one with only terms depending on $h'$ and the other where $\gamma$ and its derivatives appear. Namely, $k^{21}_{1\,in}=k^{21}_{11\,in}+k^{21}_{11\,in}$, with

$$\begin{aligned} k^{21}_{11\,in}&=\int _{|y|<1} y \frac{(\partial _x\gamma )^2+2\Delta \partial _x h'\partial _x\gamma }{(y^2+\theta ^2)^2}\partial ^4_x h'(x-y)dy\nonumber \\&+\int _{|y|<1} y \frac{\gamma (\Delta \partial ^2_x h'+ \partial ^2_x\gamma )+\Delta h' \partial _x^2\gamma }{(y^2+\theta ^2)^2}\partial ^4_x h'(x-y)dy \end{aligned}$$

(A.8)

and

$$\begin{aligned} k^{21}_{12\,in}=\int _{|y|<1} y \frac{(\partial _x\Delta h')^2+\Delta h'\partial ^2_x\Delta h'}{(y^2+\theta ^2)^2}\partial ^4_x h'(x-y)dy. \end{aligned}$$

(A.9)

To estimate $k^{21}_{11\, in}$ we use the regularity of $h'$ and mean value theorem to bound its integrand by

$$\begin{aligned} C_3\frac{t^2|\lambda -\lambda '|^2|y|+|y|^2t|\lambda -\lambda '|}{(y^2+\theta ^2)^2}|\partial ^4h(x-y)| \end{aligned}$$

and after that we apply lemma A.7.

To bound $k^{21}_{12}(x)$ we write

$$\begin{aligned} k^{21}_{12}(x)&=-\int _{{\mathbb {R}}}\left( \frac{y\left( (\partial _x\Delta h')^2+\Delta \partial ^2_x\Delta h'\right) }{\left( y^2+\theta ^2\right) ^2}-\frac{y^3\left( (\partial ^2_xh')^2+\partial _x h'\partial ^3_x h\right) }{\left( y^2+\theta _{lin}^2\right) ^2}\right) \nonumber \\&\qquad \partial _x^4h'(x-y)dy\nonumber \\&\quad -\left( (\partial ^2_xh')^2+\partial _xh'\partial ^3_xh'\right) \int _{{\mathbb {R}}}\frac{y^3}{\left( y^2+\theta _{lin}^2\right) ^2}\partial ^4_xh'(x-y)dy\nonumber \\&\equiv k^{21}_{121}(x)+k^{21}_{122}(x). \end{aligned}$$

(A.10)

To bound $k^{21}_{121\, out}(x)$ we notice that $\frac{(\partial _x\Delta h')^2+\Delta \partial ^2_x\Delta h'}{\left( y^2+\theta ^2\right) ^2}\precsim C_2 |y|^{-3}$ and we can directly apply A.8 with $a=2$ to the second term other part. To bound $k^{21}_{121\,in}(x)$ we split it into two new terms,

$$\begin{aligned}&k^{21}_{121\, in}(x)=-\int _{|y|<1}\left( \frac{y((\partial _x\Delta h') ^2+\Delta h' \partial _x^2\Delta h')-y^3((\partial ^2_xh')^2+\partial _xh'\partial ^3_xh')}{\left( y^2+\theta ^2\right) ^2}\right) \\&\qquad \quad \partial ^4_x h'(x-y)dy\\&\qquad -((\partial ^2_xh')^2+\partial _xh'\partial ^3_xh')\int _{{\mathbb {R}}}y^3\left( \frac{1}{\left( y^2+\theta ^2\right) ^2}-\frac{1}{\left( y^2+\theta _{lin}^2\right) ^2}\right) \partial ^4_x h'(x-y)dy\\&\quad \sim -\int _{|y|<1}\left( \frac{y((\partial _x\Delta h')^2+\Delta h' \partial _x^2\Delta h')-y^3((\partial ^2_xh')^2+\partial _xh'\partial ^3_xh')}{\left( y^2+\theta ^2\right) ^2}\right) \\&\qquad \quad \partial ^4_x h'(x-y)dy, \end{aligned}$$

where we have applied lemma A.6, with $a=2$. Therefore, applying lemma A.7 , with $a=2$ we check that

$$\begin{aligned}&k^{21}_{121\,in}(x)\precsim \int _{|y|<1}\frac{C_3 |y|^4+C_{3+\alpha }|y|^{3+\alpha }}{\left( y^2+\theta ^2\right) ^2}|\partial ^4_x h'(x-y)|dy\\&\quad \sim \int _{|y|<1}\frac{C_{3+\alpha }|y|^{3+\alpha }}{\left( y^2+\theta ^2\right) ^2}|\partial ^4_x h'(x-y)|dy\precsim \int _{|y|<1}|y|^{-1+\alpha }|\partial ^4_x h'(x-y)|dy. \end{aligned}$$

Thus, we can apply lemma A.4 to finish the estimate of $k^{21}_{121\, in}(x)$.

To bound $k^{21}_{122}(x)$ we apply lemma A.9 with $a=2$. This finishes the bound for $k^{21}_1(x)$.

To bound $k^{21}_2(x)$ in $L^2$ is enough to bound ${\overline{k}}^{21}_2(x)\equiv \int _{{\mathbb {R}}}k^{21}_\lambda (x,y)dy$ in $L^\infty$. The outer part is estimated as we did for $k^{21}_1$. For the inner part we split ${\overline{k}}^{21}_{2\, in}(x)\equiv {\overline{k}}^{21}_{21\, in}(x)+{\overline{k}}^{21}_{22\,in}$, with ${\overline{k}}^{21}_{21\,in}$ as $k^{22}_{21\,in}(x)$ in (A.8) and ${\overline{k}}^{21}_{22\, in}(x)$ as $k^{21}_{12\, in}(x)$ in (A.9) but replacing $\partial ^4_xh'(x-y)$ by 1. ${\overline{k}}^{21}_{21\,in}(x)$ is bounded by applying lemma A.7. To bound ${\overline{k}}^{21}_{22}(x)$ we split this term into ${\overline{k}}^{21}_{221}(x)+{\overline{k}}^{21}_{221}$ (analogous to $k^{21}_{121}(x)$ and $k^{21}_{122}(x)$ in (A.10)). ${\overline{k}}^{21}_{221\, out}(x)$ is bounded by using lemmas A.4 and A.8. For ${\overline{k}}^{21}_{221\,in}(x)$ we do the analogous splitting than for $k^{21}_{121\,in}$ and we apply the same argument together with lemma A.7. To bound ${\overline{k}}^{21}_{222}(x)$ we use lemma A.9. This finishes the estimate of $k^{21}(x)$ in $L^\infty$ and completes the proof of the estimate of $k^{21}(x)$ in $L^2$.

To bound $k^{21}_{12}(x)$ we write

$$\begin{aligned} k^{21}_{12}(x)&=-\int _{{\mathbb {R}}}\left( \frac{y\left( (\partial _x\Delta h')^2+\Delta \partial ^2_x\Delta h'\right) }{\left( y^2+\theta ^2\right) ^2}-\frac{y^3\left( (\partial ^2_xh')^2+\partial _x h'\partial ^3_x h\right) }{\left( y^2+\theta _{lin}^2\right) ^2}\right) \nonumber \\&\qquad \partial _x^4h'(x-y)dy\nonumber \\&\quad -\left( (\partial ^2_xh')^2+\partial _xh'\partial ^3_xh'\right) \int _{{\mathbb {R}}}\frac{y^3}{\left( y^2+\theta _{lin}^2\right) ^2}\partial ^4_xh'(x-y)dy\nonumber \\&\equiv k^{21}_{121}(x)+k^{21}_{122}(x). \end{aligned}$$

(A.11)

To bound $k^{21}_{121\, out}(x)$ we notice that $\frac{(\partial _x\Delta h')^2+\Delta \partial ^2_x\Delta h'}{\left( y^2+\theta ^2\right) ^2}\precsim C_2 |y|^{-3}$ and we can directly apply A.8 with $a=2$ to the second term. To bound $k^{21}_{121\,in}(x)$ we split it into two new terms,

$$\begin{aligned}&k^{21}_{121\, in}(x)=-\int _{|y|<1}\left( \frac{y((\partial _x\Delta h') ^2+\Delta h' \partial _x^2\Delta h')-y^3((\partial ^2_xh')^2+\partial _xh'\partial ^3_xh')}{\left( y^2+\theta ^2\right) ^2}\right) \\&\quad \quad \partial ^4_x h'(x-y)dy\\&\qquad -((\partial ^2_xh')^2+\partial _xh'\partial ^3_xh')\int _{{\mathbb {R}}}y^3\left( \frac{1}{\left( y^2+\theta ^2\right) ^2}-\frac{1}{\left( y^2+\theta _{lin}^2\right) ^2}\right) \partial ^4_x h'(x-y)dy\\&\quad \sim -\int _{|y|<1}\left( \frac{y((\partial _x\Delta h')^2+\Delta h' \partial _x^2\Delta h')-y^3((\partial ^2_xh')^2+\partial _xh'\partial ^3_xh')}{\left( y^2+\theta ^2\right) ^2}\right) \\&\quad \quad \partial ^4_x h'(x-y)dy, \end{aligned}$$

where we have applied lemma A.6, with $a=2$. Therefore, applying lemma A.7 , with $a=2$ we check that

$$\begin{aligned}&k^{21}_{121\,in}(x)\precsim \int _{|y|<1}\frac{C_3 |y|^4+C_{3+\alpha }|y|^{3+\alpha }}{\left( y^2+\theta ^2\right) ^2}|\partial ^4_x h'(x-y)|dy\\&\quad \sim \int _{|y|<1}\frac{C_{3+\alpha }|y|^{3+\alpha }}{\left( y^2+\theta ^2\right) ^2}|\partial ^4_x h'(x-y)|dy\precsim \int _{|y|<1}|y|^{-1+\alpha }|\partial ^4_x h'(x-y)|dy. \end{aligned}$$

Thus, we can apply lemma A.4 to finish the estimate of $k^{21}_{121\, in}(x)$.

To bound $k^{21}_{122}(x)$ we apply lemma A.9 with $a=2$. This finishes the bound for $k^{21}_1(x)$.

To bound $k^{21}_2(x)$ in $L^2$, it is enough to bound ${\overline{k}}^{21}_2(x)\equiv \int _{{\mathbb {R}}}k^{21}_\lambda (x,y)dy$ in $L^\infty$. The outer part is estimated as we did for $k^{21}_1$. For the inner part we split ${\overline{k}}^{21}_{2\, in}(x)\equiv {\overline{k}}^{21}_{21\, in}(x)+{\overline{k}}^{21}_{22\,in}$, with ${\overline{k}}^{21}_{21\,in}$ as $k^{22}_{21\,in}(x)$ in (A.8) and ${\overline{k}}^{21}_{22\, in}(x)$ as $k^{21}_{12\, in}(x)$ in (A.9) but replacing $\partial ^4_xh'(x-y)$ by 1. ${\overline{k}}^{21}_{21\,in}(x)$ is bounded by applying lemma A.7. To bound ${\overline{k}}^{21}_{22}(x)$ we split this term into ${\overline{k}}^{21}_{221}(x)+{\overline{k}}^{21}_{221}$ (analogous to $k^{21}_{121}(x)$ and $k^{21}_{122}(x)$ in (A.11)). ${\overline{k}}^{21}_{221\, out}(x)$ is bounded by using lemmas A.4 and A.8. For ${\overline{k}}^{21}_{221\,in}(x)$ we do an analogous splitting to that for $k^{21}_{121\,in}$ and we apply the same argument together with lemma A.7. To bound ${\overline{k}}^{21}_{222}(x)$ we use lemma A.9. This finishes the estimate of $k^{21}(x)$ in $L^\infty$ and completes the proof of the estimate of $k^{21}(x)$ in $L^2$.

1.1.2. Let us estimate $k^{22}(x)$. We split into two terms

$$\begin{aligned} k^{22}(x)=k^{22}_1(x)+k^{22}_2(x),\end{aligned}$$

(A.12)

with

$$\begin{aligned} k^{22}_1(x)=-\int _{{\mathbb {R}}}k^{22}_\theta (x,y)\partial ^4_x h'(x-y)dy,\quad {\text {and}}&k^{22}_2=\partial ^4_x h \int _{{\mathbb {R}}}k^{22}_\theta (x,y)dy. \end{aligned}$$

We split $k^{22}_1(x)$ in four terms, $k^{22}_1(x)=k^{22}_{11}(x)+k^{22}_{12}(x)+k^{22}_{13}+k^{22}_{14}$, with

$$\begin{aligned}&k^{22}_{11}(x)=-\int _{{\mathbb {R}}}y\frac{(\Delta \partial _x h')^2(\gamma ^2+2\gamma \Delta h') }{(y^2+\theta ^2)^3}\partial ^4_x h'(x-y)dy \end{aligned}$$

(A.13)

$$\begin{aligned}&k^{22}_{12}(x)=-\int _{{\mathbb {R}}}y\frac{(\Delta \partial _x h')^2(\Delta h')^2}{(y^2+\theta ^2)^3}\partial _x^4h'(x-y)dy.\end{aligned}$$

(A.14)

$$\begin{aligned}&k^{22}_{13}(x)=-\int _{{\mathbb {R}}}y\frac{(2\partial _x \gamma \Delta \partial _x h'+\partial _x \gamma ^2)(\Delta h')^2 }{(y^2+\theta ^2)^3}\partial ^4_x h'(x-y)dy\end{aligned}$$

(A.15)

$$\begin{aligned}&k^{22}_{14}(x)=-\int _{{\mathbb {R}}}y\frac{((\partial _x \gamma )^2+2\partial _x \gamma \Delta \partial _x h')(\gamma ^2+2\gamma \Delta h') }{(y^2+\theta ^2)^3}\partial ^4_x h'(x-y)dy. \end{aligned}$$

(A.16)

The function $k^{22}_{11}(x)$ can be bounded in $L^2$ as follows. The integrand of $k^{22}_{11}(x) \precsim C_1|y|^{-5}|\partial ^4_x h'(x-y)|$. Then $k^{22}_{11\,out}(x)$ is estimated by lemma A.4. Also the integrand of $k^{22}_{11}(x)\precsim C_2\frac{y^3(\gamma ^2+2|\gamma |y)}{(y^3+(\theta )^2)^3}|\partial ^4_x h'(x-y)|$ and we can apply lemma A.7 with $a=3$ to bound $k^{22}_{11\, in}(x)$. Similarly, we can deal with $k^{22}_{13}$ and $k^{22}_{14}$ as we can obtain the correct estimates in powers of |y| and $|\gamma |$ to apply Lemma A.7.

To bound $k^{22}_{12}(x)$ we write

$$\begin{aligned} k^{22}_{12}(x)&=-\int _{{\mathbb {R}}}\left( \frac{y(\Delta \partial _x h')^ 2 (\Delta h')^2}{\left( y^2+\theta ^2\right) ^3}-\frac{(\partial ^2_x h')^2(\partial _x h')^2 y^5}{\left( y^2+\theta _{lin}^2\right) ^3}\right) \partial ^4_x h'(x-y) dy\nonumber \\&\qquad -(\partial _x h')^2(\partial ^2_x h')^2 \int _{{\mathbb {R}}}\frac{y^5}{\left( y^2+\theta _{lin}^2\right) ^3}\partial _x^4h'(x-y)\nonumber \\&\equiv k^{22}_{121}(x)+k^{22}_{122}(x). \end{aligned}$$

(A.17)

To bound $k^{22}_{121,\, out}(x)$ we notice that $\frac{y(\Delta \partial _x h')^2(\Delta h')^2}{\left( y^2+(\theta )^2\right) ^3}\precsim C_1 |y|^{-5}$ and we can apply A.8 to the other part with $a=3$. To bound $k^{22}_{121,in}(x)$ we split in two terms,

$$\begin{aligned} k^{22}_{121,\,in}&=\int _{|y|<1}\left( \frac{y(\Delta h')^2 (\Delta \partial _x h')^2 -y^5(\partial _x h)^2(\partial ^2_x h)^2}{\left( y^2+(\theta )^2\right) ^3}\right) \partial ^4_x h'(x-y)dy\\&\quad +(\partial _x h')^2(\partial ^2_x h')^2\int _{|y|<1}y^5\left( \frac{1}{\left( y^2+\theta ^2\right) ^3}-\frac{1}{\left( y^2+\theta _{lin}^2\right) ^3}\right) \partial ^4_x h'(x-y)dy, \end{aligned}$$

in such a way that we can apply lemma A.7 with $a=3$, since $|y(\Delta h')^2 \Delta (\partial _x h')^2 -y^5(\partial _x h')^2(\partial ^2_x h')^2|\precsim C_3 y^6$, and lemma A.6 with $a=3$.

To bound $k^{22}_{122}(x)$ we apply lemma A.9 with $a=3$. This finishes the bound for $k^{22}_1(x)$.The term $k^{22}_2(x)$is estimated in a similar way to $k^{21}_2$. We have completed the proof of the estimate of $k^{22}(x)$ in $L^2$.

1.2. Terms in (A.5) with $j=3$ and $i=1$.

Let us bound $k^{31}(x)$. Unfortunately the proof of the estimation of $k^{31}(x)$ does not follow the same steps than the rest of the functions $k^{i}_{j}(x)$. Indeed we need to do something different and use the boundedness of pseudodifferential operators used in the body of the text.

We split into two terms $k^{31}(x)=k^{31u}(x)+k^{31d}_2(x)$ with

$$\begin{aligned}&k^{31u}(x)=\int _{{\mathbb {R}}}y\frac{\partial _x \theta \partial ^2_x \theta }{\left( y^2+\theta ^2\right) ^2}\partial ^3_x \Delta h' dy\\&k^{31d}(x)=\int _{{\mathbb {R}}}y\frac{(\theta )\partial ^3_x \theta }{\left( y^2+\theta ^2\right) ^2} \partial ^3_x \Delta h' dy. \end{aligned}$$

The proof for $k^{31u}(x)$ will be similar to the above.

For $k^{31d}(x)$ we need new estimates since as $h \in H^4$ we can not bound $\partial ^3_x \theta$ by |y| uniformly.

Let us bound $k^{31d}(x)$ first. We split this function in four parts,

$$\begin{aligned}&k^{31d}_1(x)= \int _{{\mathbb {R}}}y\frac{\gamma \partial ^3_x h'}{\left( y^2+\theta ^2\right) ^2} \partial ^3_x \Delta h' dy,\quad \text {and} \end{aligned}$$

(A.18)

$$\begin{aligned}&k^{31d}_{2}=\int _{{\mathbb {R}}}y\frac{ \Delta h' \partial ^3_x \Delta h'}{\left( y^2+\theta ^2\right) ^2}\partial ^3_x \Delta h' dy \quad \text {and} \end{aligned}$$

(A.19)

$$\begin{aligned}&k^{31d}_{3}=\int _{{\mathbb {R}}}y\frac{ \Delta h' \partial ^3_x \gamma }{\left( y^2+\theta ^2\right) ^2}\partial ^3_x \Delta h' dy \end{aligned}$$

(A.20)

$$\begin{aligned}&k^{31d}_4(x)= \int _{{\mathbb {R}}}y\frac{\gamma \partial ^3_x \gamma }{\left( y^2+\theta ^2\right) ^2} \partial ^3_x \Delta h' dy.\quad \text {and} \end{aligned}$$

(A.21)

The last two terms are easily bounded by Lemma A.7 for the in part and by Lemma A.4 for the outer part.

To bound $k^{31d}_{2}(x)$ we use that $\Delta (\partial _x^3h )^2=2\partial ^3_x h \Delta \partial ^3_x h - \Delta \left( (\partial ^3_x h)^2\right)$, and then, we need to show that the integral

$$\begin{aligned} N(x)=\int _{{\mathbb {R}}}\frac{y\Delta h'}{\left( y^2+\theta ^2\right) ^2}\Delta gdy \end{aligned}$$

(A.22)

is in $L^2$ with either $g=\partial ^3_x h'$ or $(\partial ^3_x h')^2$. Notice that, in both cases, we can allow in our estimates that $||g||_{H^1}$ appears. We will split N(x) in two terms $N_1(x)$ and $N_2(x)$ with

$$\begin{aligned}&N_1(x)=\int _{{\mathbb {R}}}y\left( \frac{\Delta h'}{\left( y^2+\theta ^2\right) ^2}-\frac{\partial _x h' y}{\left( y^2+\theta _{lin}^2\right) ^2}\right) \Delta g dy,\end{aligned}$$

(A.23)

$$\begin{aligned}&N_2(x)=\partial _x h'\int _{{\mathbb {R}}}\frac{y^2}{\left( y^2+\theta _{lin}^2\right) ^2}\Delta g dy . \end{aligned}$$

(A.24)

That is $N_1$ compares with the linearized version and $N_2$ treats with the linearized kernel. We can not deal directly with $N_1$ with our previous lemmas by replacing $\Delta h'$ by y as the denominator is too singular. However we can add an subtract a term

$$\begin{aligned} \int \frac{\frac{1}{2}\partial ^2_x h' y^2}{\left( y^2+\theta _{lin}^2\right) ^2} \Delta g dy. \end{aligned}$$

Then several terms appear but since $h \in H^4$ we have that

$$\begin{aligned} \Delta h'-\partial _x h'y-\partial _x^2 h'(x) y^2 \le |y|^3. \end{aligned}$$

(A.25)

After splitting in various terms, all of them can be deal with using our lemmas  A.6, A.7, A.9 and a small modification.

In order to deal with $N_2(x)$ we need to introduce another new idea. We first use (A.6) and treat the two terms in $\Delta g$ separately. proceed as follows

$$\begin{aligned} N_2(x)=&\partial _x h'\gamma \sigma _h^2 \left( g(x)\int _{{\mathbb {R}}}\frac{y^2}{((y+\sigma _h A_h \gamma )^2+\sigma ^2\gamma ^2)^2}dy\right. \\&\left. \qquad \qquad -\int _{{\mathbb {R}}}\frac{y^2}{((y+\sigma _h A_h \gamma )^2+\sigma _h^2\gamma ^2)^2}g(x-y)dy\right) \end{aligned}$$

Now we can compute that

$$\begin{aligned} \int _{{\mathbb {R}}}\frac{y^2}{((y+\sigma _h A_h \gamma )^2+\sigma _h^2\gamma ^2)^2}dy=\frac{\pi }{2\sigma _h^2|\gamma |} \end{aligned}$$

For the convolution term, the following Fourier transform computation

$$\begin{aligned} \begin{aligned}&{\mathcal {F}}\left[ \frac{y^2}{\left( (y+\sigma A \lambda )^2+\sigma ^2\lambda ^2\right) ^2}\right] (\xi )\\&=\frac{\pi }{2\sigma ^2|\lambda |}e^{2\pi i A\sigma \xi \lambda }e^{-2\pi \sigma |\lambda ||\xi |}\left( 1+2\pi \sigma ^2(A^2-1)+4\pi iA\sigma ^2\xi \lambda \right) \end{aligned} \end{aligned}$$

(A.26)

yields that

$$\begin{aligned}&\int _{{\mathbb {R}}}\frac{y^2}{((y+\sigma _h A_h \gamma )^2+\sigma _h^2\gamma ^2)^2}g(x-y)dy\\&= \frac{\pi }{2\sigma _h^2|\gamma |}\int _{{\mathbb {R}}}e^{2\pi i\xi x}{\hat{g}}(\xi )e^{2\pi i \xi A_h\sigma _h \gamma }e^{-2\pi \sigma _h |\gamma ||\xi |}\\&\qquad \qquad \qquad \times (1+2\pi \sigma _h^2|\gamma ||\xi |(A^2-1)+4\pi i A\sigma _h^2 \xi \gamma )d\xi , \end{aligned}$$

so that

$$\begin{aligned} N_2(x)=\partial _x h'\frac{\pi }{2}\text {Op}(p)(\Lambda g) \end{aligned}$$

where the symbol is given by,

$$\begin{aligned} \begin{aligned} p(x,\xi )=&\frac{1}{|\xi ||\gamma |}\left( 1-e^{2\pi i A_h\sigma _h \xi \gamma }e^{-2\pi \sigma _h |\xi ||\gamma |}\right) \\&-\sigma _h^2\pi ^2 e^{2\pi i A_h\sigma _h \xi \gamma }e^{-2\pi \sigma _h |\xi ||\gamma |}((A_h^2-1)+2i \text {sign}(\xi )\text {sign}(\gamma ))\end{aligned} \end{aligned}$$

Therefore by applying lemma 5.1 we obtain that the $L^2-$norm of $N_2(x)$ is bounded by $\langle A \rangle (||\Lambda g||_{L^2}+||\partial _xg||_{L^2})$.

This concludes the proof of the estimate of the $L^2-$norm of $k^{31d}_2(x)$.

To deal with $k_1^{31d}$, we use again that $(\Delta \partial _x^3 h')^2=2\partial ^3_x h' \Delta \partial ^3_x h' - \Delta \left( (\partial ^3_x h')^2\right)$. Thus, it suffices to bound the integral

$$\begin{aligned} M(x)=\int _{{\mathbb {R}}}\frac{y\gamma }{(y^2+(\theta )^2)^2} \Delta g(x-y) dy \end{aligned}$$

(A.27)

in terms of the $H^1$-norm of g. The proof is analogous since at the only delicate point it holds that

$$\begin{aligned} \int _{\mathbb {R}}\frac{y\gamma }{\left( y^2+\theta _{lin}^2\right) ^2}dy = -\frac{A\pi }{2\sigma ^2 |\gamma |}, \end{aligned}$$

1.3.The rest of the terms in (A.5).

The estimation of the rest of the terms in (A.5) follow the same steps than the estimation for either $k^{21}(x)$ or $k^{22}(x)$.

1.1.3 Estimation of the terms in (A.4)

We will show how to estimate is $k^{11}(x)$ in (A.4), since $k^{55}(x)$ is analogous. Here we recall that we are concerned with $||D^{-1} k^{11}||_{L^2}$. In order to bound this norm we will proceed as follows

$$\begin{aligned} k^{11}(x)=\int _{{\mathbb {R}}}k^{11}_\theta (x,y)\partial ^5_x \theta dy=\int _{{\mathbb {R}}}k^{11}_\theta (x,y)D\Theta dy, \end{aligned}$$

where $\Theta \equiv =D^{-1}\partial ^5_x\theta$ (we clarify that the operator $D=(1+t\partial _x)$ acts on x rather than y). Then we would like to estimate $||D^{-1} k^{11}||_{L^2}\le \langle A \rangle ||\Theta ||_{L^2}.$ In order to do it we notice that

$$\begin{aligned} \partial _x \int _{{\mathbb {R}}} k^{11}_\theta (x,y) \Theta dy =\int _{{\mathbb {R}}}\partial _x k^{11}_\theta (x,y) \Theta dy +\int _{{\mathbb {R}}}k^{11}_\theta (x,y)\partial _x \Theta dy, \end{aligned}$$

so that

$$\begin{aligned}&D^{-1}k^{11}(x)= \int _{{\mathbb {R}}} k^{11}_\theta (x,y) \Theta dy-tD^{-1}\int _{{\mathbb {R}}}\partial _x k^{11}_\theta (x,y) \Theta dy\\&\equiv S_1(x)+tD^{-1}S_2(x). \end{aligned}$$

Happily, the proof of the estimation for $S_1$ and $S_2$ follow the same steps that the estimation of $k^{22}(x)$ in (A.12). Thus we have proven (A.1). That is

$$\begin{aligned} \partial ^5_x{\mathcal {M}} u =-\frac{1}{4\pi }\int _{-1}^1\int _{-1}^1 \int _{{\mathbb {R}}} k_{\theta }(\cdot ,y)\partial ^6_x \theta dy d\lambda 'd\lambda + l.o.t. \end{aligned}$$

In order to finish the proof of lemma 4.8 notice that

$$\begin{aligned} k_\theta \partial _x^6\theta = k_\theta \partial ^6_x \Delta f(x,x-y)+ k_\theta \left( \partial _x^6\varepsilon (x)\lambda -\partial _x^6\varepsilon (x-y)\lambda '\right) \end{aligned}$$

and, by assumptions on c(x, t), $\partial _x^6c(\cdot ,t)\in L^2$ uniformly in t. We then have that

$$\begin{aligned} \partial ^5_x{\mathcal {M}} u =-\frac{1}{4\pi }\int _{-1}^1\int _{-1}^1 \int _{{\mathbb {R}}} k_{\theta }(\cdot ,y)\partial ^6_x \Delta f(x,x-y) dy d\lambda 'd\lambda + l.o.t. \end{aligned}$$

Lemma 4.8 is proved.

1.2 Proof of Lemma 4.9

In this section we will prove Lemma 4.9. We will use the same convection as in the previous section.

Since the transport term in lemma 4.9 arises in an obvious way, t he main issue is to linearize $\theta$ to $\theta _{lin}$ in the integral

$$\begin{aligned} \int _{-1}^1\int _{-1}^1\int _{{\mathbb {R}}}\frac{y}{y^2+\theta ^2}\partial ^6_x f(x-y)dyd\lambda \lambda '. \end{aligned}$$

This is the content of the following $L^2$ estimate.

Lemma A.10

Let $f\in H^6$, c as in theorem 4.1 and

$$\begin{aligned}&F(x)=D^{-1}\frac{1}{4}\int _{-1}^1\int _{-1}^1\int _{{\mathbb {R}}} \left( \frac{y}{y^2+\theta ^2}-\frac{y}{y^2+\theta _{lin}^2}\right) \partial ^6_x f(x-y)dyd\lambda \lambda '. \end{aligned}$$

Then,

$$\begin{aligned} \Vert F\Vert _{L^2}\le \langle A \rangle ||D^{-1}\partial ^5_x f||_{L^2}. \end{aligned}$$

After applying Minkowski inequality, to obtain the estimate, it is enough to show that

$$\begin{aligned} D^{-1}\int _{{\mathbb {R}}} \left( \frac{y}{y^2+\theta ^2}-\frac{y}{y^2+\theta _{lin}^2}\right) \partial ^6_x f(x-y)dy \end{aligned}$$

is in $L^2$ with $L^2-$norm bounded by $\langle A \rangle \left( ||D^{-1}\partial ^5_x f||_{L^2}+1\right)$ uniformly in t (for small t), $\lambda$ and $\lambda '$.

Let us call $g(x)=D^{-1}\partial ^5_x f(x)$. Then we proceed as when investigating the commutators $[D^{-1},\text {Op}(p)]$ but this time working directly with the kernel

$$\begin{aligned} k(x,y)=\left( \frac{y}{y^2+\theta ^2}-\frac{y}{y^2+\theta _{lin}^2}\right) . \end{aligned}$$

Then we have to estimate the function

$$\begin{aligned} P(x)=D^{-1}\int _{{\mathbb {R}}} k(x,y)D \partial _x g(x-y)dy. \end{aligned}$$

By direct application of the definition of D, it holds that $\int k(x,y) D\partial _xg(x-y)dy=D\int k(x,y)\partial _x g(x-y)dy-t\int \partial _xk(x,y)\partial _x g(x-y)dy$ we have

$$\begin{aligned} P(x)=&\int _{{\mathbb {R}}} k(x,y) \partial _x g(x-y)dy-tD^{-1} \int _{{\mathbb {R}}} \partial _xk(x,y) \partial _x g(x-y)dy\\ \equiv&M(x)+tD^{-1}L(x), \end{aligned}$$

We need to estimate both M(x) and L(x) in $L^2$.

To bound M(x) we first integrate by parts, recalling that $\partial _x g(x-y)=-\partial _y g(x-y)$,

$$\begin{aligned}&M(x)=\int _{{\mathbb {R}}} \partial _y\left( \frac{y}{y^2+\theta ^2}-\frac{y}{y^2+\theta _{lin}^2}\right) g(x-y)dy\nonumber \\&=\int _{{\mathbb {R}}} \left( \frac{1}{y^2+\theta ^2}-\frac{1}{y^2+\theta _{lin}^2}\right) g(x-y)dy\nonumber \\&+2\int _{{\mathbb {R}}}y\left( \frac{y+\theta _{lin}\partial _x h'}{\left( y^2+\theta _{lin}^2\right) ^2}-\frac{y+\theta \partial _x h'(x-y)}{\left( y^2+\theta ^2\right) ^2}\right) g(x-y) dy\nonumber \\&\equiv M_1(x)+2M_2(x). \end{aligned}$$

(A.28)

As a matter of fact both $M_1, M_2$ can be estimated by means our previous lemmas. The outer integrals can be estimated by brute force as the kernels decay fast enough. The inner ones, by using the first and second Taylor polynomial of $h'$, can be split into terms with the appropriate powers of y in order to apply Lemmas A.6, A.9 as before.

It remains to bound L(x). This function is given by

$$\begin{aligned} L(x)&=-2\int _{{\mathbb {R}}}y\left( \frac{\theta \partial _x\theta }{\left( y^2+\theta ^2\right) ^2}-\frac{\theta _{lin}(\partial ^2_x h'y+\partial _x\gamma )}{\left( y^2+\theta _{lin}^2\right) ^2}\right) \partial _x g(x-y)dy\\&=-2\int _{{\mathbb {R}}}y\left( \frac{(\Delta h'\Delta \partial _x h+\gamma \Delta \partial _x h')}{\left( y^2+\theta ^2\right) ^2}-\frac{\partial ^2_x h'\partial _x h y^2+\gamma \partial ^2_x h' y }{\left( y^2+\theta _{lin}^2\right) ^2}\right) \partial _x g(x-y)dy\\&\quad -2\int _{{\mathbb {R}}}y\left( \frac{\partial _x\gamma \Delta h'}{\left( y^2+\theta ^2\right) ^2}-\frac{\partial _x \gamma \partial _xh' y }{\left( y^2+\theta _{lin}^2\right) ^2}\right) \partial _x g(x-y)dy\\&\quad -2\int _{{\mathbb {R}}}y\left( \frac{\gamma \partial _x\gamma }{\left( y^2+\theta ^2\right) ^2}-\frac{\gamma \partial _x\gamma }{\left( y^2+\theta _{lin}^2\right) ^2}\right) \partial _x g(x-y)dy \\&\quad \equiv S(x)+ {\tilde{S}}(x)+{\overline{S}}(x). \end{aligned}$$

Firstly, we will carefully bound S(x) since the numerators of the terms ${\tilde{S}}$ and ${\overline{S}}$ have the same behaviour.

We repeat the trick of observing that $\partial _x g(x-y)=-\partial _y g(x-y)$ to integrate by parts and obtain that

$$\begin{aligned} S(x)&=-2\int _{{\mathbb {R}}}\left( \frac{(\Delta h'+\gamma )\partial _x\Delta h'}{\left( y^2+\theta ^2\right) ^2}-\frac{\theta _{lin}\partial ^2_x h'y}{\left( y^2+\theta _{lin}^2\right) ^2}\right) g(x-y)dy\\&\quad -2\int _{{\mathbb {R}}}y\left( \frac{\partial _x h(x-y)\partial _x\Delta h'+(\Delta h'+\gamma )\partial ^2_x h'(x-y)}{\left( y^2+\theta ^2\right) ^2}\right. \\&\qquad \left. -\frac{\partial _xh' \partial ^2_x h'y+(\theta _{lin})\partial ^2_x h'}{\left( y^2+\theta _{lin}^2\right) ^2}\right) g(x-y)dy\\&\quad +8\int _{{\mathbb {R}}}y\left( \frac{\theta \partial _x\Delta h'(y^2+\theta ^2)(y+(\Delta h'+\gamma ))\partial ^2_x h'(x-y)}{\left( y^2+\theta ^2\right) ^2} \right. \\&\qquad \left. -\frac{\theta _{lin}\partial _x^2h' y (y^2+\theta _{lin}^2)(y+\theta _{lin})\partial ^2_xh'}{\left( y^2+\theta _{lin}^2\right) ^2}\right) g(x-y)dy\\&\equiv -2S_1(x)-2S_2(x)+8S_3(x). \end{aligned}$$

To bound $S_1(x)$ we split it in the following way

$$\begin{aligned}&S_1(x)=\int _{{\mathbb {R}}}\frac{\theta \partial _x\Delta h'-\theta _{lin}\partial ^2_x h' y}{\left( y^2+\theta ^2\right) ^2}g(x-y)dy\nonumber \\&\qquad +\int _{{\mathbb {R}}}(\theta _{lin})\partial ^2_x h' y \left( \frac{1}{\left( y^2+\theta ^2\right) ^2}-\frac{1}{\left( y^2+\theta _{lin}^2\right) ^2}\right) g(x-y)dy\nonumber \\&\equiv S_{11}(x)+S_{12}(x). \end{aligned}$$

(A.29)

To bound $S_{11}(x)$ we split into two terms

$$\begin{aligned} S_{11}(x)&=\int _{{\mathbb {R}}}\frac{\Delta h' \partial _x\Delta h' -\partial _x h'\partial ^2_xh' y^2}{\left( y^2+\theta ^2\right) ^2}g(x-y)dy\\&\quad +\int _{{\mathbb {R}}} \frac{\gamma (\partial _x\Delta h' -\partial ^2_x h'y)}{\left( y^2+\theta ^2\right) ^2}g(x-y)dy\\&\quad \sim \int _{{\mathbb {R}}}\frac{\Delta h' \partial _x\Delta h' -\partial _x h'\partial ^2_xh' y^2}{\left( y^2+\theta ^2\right) ^2}g(x-y)dy. \end{aligned}$$

The last line follows from the bound $(\partial _x\Delta h'-\partial ^2_x h' y)\precsim C_2(|y|+1)$ and lemma A.7.

The next term is more complicated as in principle the numerator scales as $y^2$ which is not enough to apply our lemmas. As before we Taylor $\Delta _h$ up to second order and differentiate to obtain $\Delta h' \partial _x \Delta h'=pa_x h' \partial ^2_x h' y^2+G(\partial ^j_x h')|y|^3+ C|y|^\alpha$. Being explicit,

$$\begin{aligned} G=\partial _x h' \partial _x^3 h'+(\partial _x^2 h')^2+\partial ^2_x h' \partial ^3_x h') \end{aligned}$$

which is uniformly bounded in x since $h' \in H^4$. Since terms of the type

$$\begin{aligned} \int _{R} \frac{G(h')}{(y^2 +\theta _{lin}^2)^2} g(x-y) \end{aligned}$$

are estimated like our terms $k^{ji}(x)$ we can subtract them freely. Hence

$$\begin{aligned} |S_{11in} (x)| \le \int \frac{|y|^{3+\alpha }}{(y^2+\theta ^2)} dy + G(h')(x) \int y^3 \left( \frac{1}{(y^2 +\theta _{lin}^2)^2} -\frac{1}{(y^2 +\theta ^2)^2}\right) . \end{aligned}$$

$S_{12}(x)$ is easy by now. It can be estimated as $M_2(x)$ in (A.28). This finishes the estimation of $S_1(x)$.

To bound $S_2(x)$ we split it into two terms

$$\begin{aligned} S_2(x)&=\int _{{\mathbb {R}}}y\left( \frac{\partial _xh'(x-y)\partial _x\Delta h' +\theta \partial ^2_xh'(x-y)-\partial _x h' \partial ^2_x h' y -(\partial _xh' y +\gamma )\partial ^2_x h'}{\left( y^2+\theta ^2\right) ^2}\right) g(x-y)dy\nonumber \\&\qquad + \int _{{\mathbb {R}}}y\left( \partial _x h' \partial ^2_x h'y+(\partial _xh' y +\gamma )\partial ^2_x h'\right) \left( \frac{1}{\left( y^2+\theta ^2\right) ^2}-\frac{1}{\left( y^2+\theta _{lin}^2\right) }\right) g(x-y)dy\nonumber \\&\quad \equiv S_{21}(x)+S_{22}(x). \end{aligned}$$

(A.30)

The term $S_{22}$ has the correct behaviour in powers of y and $\gamma$ to deal with them $S_{21}(x)$ we add freely a term

$$\begin{aligned} \frac{1}{2} \partial _x^3 h'(x) \int \frac{y^3}{(y^2+\theta _{lin}^2)} dy \end{aligned}$$

and proceed exactly as with $S_{11}$.

Finally, it remains to bound $S_3(x).$ Since the computation are longer but no new idea is needed we skip the details.

Then we have achieved the conclusion of lemma 4.9.

1.3 Estimates for the velocity

Lemma A.11

Let $\mathbf{u }$ be like in expression (4.17) with f and $\varepsilon =ct$ in theorem 4.1. Then $\mathbf{u }\in L^\infty ({\mathbb {R}}^2)$ and

$$\begin{aligned} ||\mathbf{u }(\cdot ,t)||_{L^\infty ({\mathbb {R}}^2)}\le P(||f||_{H^4}) \end{aligned}$$

for some smooth function P.

Proof

In this proof C stands for a constant that may depend on $||f||_{H^4}$ and on the regularity of c(x, t). The velocity in (4.17) reads

$$\begin{aligned} \mathbf{u }(\mathbf{x })&=\frac{1}{\pi }P.V.\int _{{\mathbb {R}}}\frac{1}{2}\int _{-1}^1 \frac{x_1-y}{|(x_1-y, x_2-f(y)-\varepsilon (y)\lambda ')|^2} (1,\partial _x f(y)\\&\quad +\partial _x\varepsilon (y)\lambda ')d\lambda ' dy. \end{aligned}$$

And evaluating in at $(x,f(x)+\lambda ), (x,\lambda ) \in {\mathbb {R}}^2$, we have that

$$\begin{aligned}&\mathbf{u }(x, \lambda +f(x))=\frac{1}{\pi }P.V.\int _{{\mathbb {R}}}\frac{1}{2}\int _{-1}^1 \frac{x-y}{|(x-y, \Delta f(x,y)+(\lambda -\varepsilon (y)\lambda '))|^2}\\&\quad (1,\partial _x, f(y)+\partial _x\varepsilon (y)\lambda ')d\lambda 'dx' \end{aligned}$$

with $\Delta f(x,y)=f(x)-f(y)$. Next we check that the integral

$$\begin{aligned} I(x)&=P.V.\int _{{\mathbb {R}}}\frac{x-y}{(x-y)^2+(\Delta f(x,y)+\lambda -\varepsilon (y)\lambda ')^2}dy\nonumber \\&=P.V.\int _{{\mathbb {R}}}\frac{y}{(y^2+(\Delta f(x,x-y)+\lambda -\varepsilon (x-y)\lambda ')^2}dy \end{aligned}$$

(A.31)

belongs to $L^\infty (dx)$ uniformly in $\lambda \in {\mathbb {R}}$ and $\lambda '\in [-1,1]$. In order to do it we split (A.31) into two parts

$$\begin{aligned} I_1(x)&=P.V.\int _{{\mathbb {R}}}\frac{y}{y^2+(\Delta f(x,x-y)+\lambda -\varepsilon (x-y)\lambda ')^2}\\&\quad -\frac{y}{y^2+((\partial _xf(x)+\partial _x\varepsilon (x)\lambda ')y+\lambda -\varepsilon (x)\lambda ')^2}dy\\ I_2(x)&=P.V.\int _{{\mathbb {R}}}\frac{y}{y^2+((\partial _x f(x)+\partial _x\varepsilon (x)\lambda ')y+\lambda -\varepsilon (x)\lambda ')^2}dy. \end{aligned}$$

We will denote

$$\begin{aligned} A_\lambda '=\partial _xf(x)+\partial _x\varepsilon (x)\lambda ',&\sigma _{\lambda '}=\frac{1}{1+A_{\lambda '}^2},&\gamma =\lambda +\varepsilon (x)\lambda '. \end{aligned}$$

Thus

$$\begin{aligned} I_{2}(x)&=\sigma _{\lambda '}P.V.\int _{{\mathbb {R}}}\frac{y}{(y+\sigma _{\lambda '} A_\lambda '\gamma )^2+\gamma ^2\sigma _{\lambda '}^2}dy\\&=\sigma _{\lambda '}P.V.\int _{{\mathbb {R}}}\frac{y+\sigma _{\lambda '}A_{\lambda '}\gamma }{(y+\sigma _{\lambda '}A_{\lambda '}\gamma )^2+\gamma ^2\sigma _{\lambda '}^2}dy\\&\quad -\sigma _{\lambda '}\int _{{\mathbb {R}}}\frac{\sigma _{\lambda '}A_{\lambda '}\gamma }{(y+\sigma _{\lambda '}A_{\lambda '}\gamma )^2+\gamma ^2\sigma _{\lambda '}^2}dy. \end{aligned}$$

The first integral on the right hand side of the previous equation is equal to zero. The second one is a bounded integral for every value of $\gamma \in {\mathbb {R}}$ and $\lambda '\in [-1,1]$.

In order to bound $I_1(x)$ we split it into two terms

$$\begin{aligned} I_{11}(x)&=\int _{|y|<1}\frac{y}{y^2+(\Delta f(x,x-y)+\lambda -\varepsilon (x-y)\lambda ')^2}\\&\quad -\frac{y}{y^2+(A_{\lambda '}y+\gamma )^2}dy\\ I_{12}(x)&=P.V.\int _{|y|>1}\frac{y}{y^2+(\Delta f(x,x-y)+\lambda -\varepsilon (x-y)\lambda ')^2}\\&\quad -\frac{y}{y^2+(A_{\lambda '}y+\gamma )^2}dy. \end{aligned}$$

To bound $I_{12}(x)$ we consider $I_{121}(x)$ and $I_{122}(x)$ with

$$\begin{aligned} I_{121}(x)&=P.V.\int _{|y|>1}\frac{y}{y^2+(\Delta f(x,x-y)+\lambda -\varepsilon (x-y)\lambda ')^2}dy\\&=P.V.\int _{|y|>1}\left( \frac{y}{y^2+(\Delta f(x,x-y)+\lambda -\varepsilon (x-y)\lambda ')^2}-\frac{y}{y^2+\lambda ^2}\right) dy \end{aligned}$$

and

$$\begin{aligned} I_{122}(x)&=P.V.\int _{|y|>1}\frac{y}{y^2+(A_{\lambda '}y+\gamma )^2}dy\\&=P.V.\int _{|y|>1}\left( \frac{y}{y^2+(A_{\lambda '}y+\gamma )^2}-\frac{y}{(1+A_{\lambda '}^2)y^2+\gamma ^2}\right) dy. \end{aligned}$$

Then

$$\begin{aligned} |I_{121}(x)|\le C\int _{|y|>1}\frac{(1+|\lambda |)}{|y|(y^2+\lambda ^2)}dy\le C, \end{aligned}$$

and

$$\begin{aligned} |I_{122}(x)|\le C \int _{|y|>1}\frac{|\gamma |}{(y^2+\gamma ^2)}dy\le C. \end{aligned}$$

To bound $I_{11}(x)$ we notice that

$$\begin{aligned}&(A_{\lambda '}y+\gamma )^2-(\Delta f(x,x-y)+\lambda -\varepsilon (x-y)\lambda ')^2\\&\quad =(A_{\lambda '}y +\gamma -\Delta f(x,x-y)-\lambda +\varepsilon (x-y)\lambda ')(A_{\lambda '}y+\Delta f(x,x-y)\\&\qquad +\gamma +\lambda -\varepsilon (x-y)\lambda '). \end{aligned}$$

In addition,

$$\begin{aligned}&A_{\lambda '}y +\gamma -\Delta f(x,x-y)-\lambda +\varepsilon (x-y)\lambda '\\&\quad =\partial _xf(x)y-\Delta f(x,x-y)+(-\varepsilon (x)+\varepsilon (x-y)+\partial _x\varepsilon (x)y)\lambda ' \end{aligned}$$

Then

$$\begin{aligned}&\left| (A_{\lambda '}y+\gamma )^2-(\Delta f(x,x-y)+\lambda -\varepsilon (x-y)\lambda ')^2\right| \le C y^2, \end{aligned}$$

and

$$\begin{aligned}&\left| A_{\lambda '}y+\Delta f(x,x-y)+\gamma +\lambda -\varepsilon (x-y)\lambda '\right| \\&\quad \le C(|y|+|2\lambda -(\varepsilon (x)+\varepsilon (x-y))\lambda ')\\&\quad \le C(|y|+2|(\lambda -\varepsilon (x)\lambda '|+|\varepsilon (x)-\varepsilon (x-y)||\lambda '|\le C(|y|+|\gamma |). \end{aligned}$$

$$\begin{aligned} |I_{11}|\le C\int _{|y|<1}\frac{|y|^3(|y|+|\gamma |)}{(y^2+(A_{\lambda '}y+\gamma )^2)(y^2+(\Delta f (x,x-y)+\lambda -\varepsilon (x-y)\lambda ')^2)}dy. \end{aligned}$$

(A.32)

Since we can bound

$$\begin{aligned} \frac{|y|^4}{(y^2+(A_{\lambda '}y+\gamma )^2)(y^2+(\Delta f (x,x-y)+\lambda -\varepsilon (x-y)\lambda ')^2)}\le C, \end{aligned}$$

the first integral in (A.32) is easy to bound. In addition for $|\gamma |>4 (||f||_{L^\infty }+||\varepsilon ||_{L^\infty })$ we have that

$$\begin{aligned} (\Delta f+\lambda -\varepsilon (x-y)\lambda ')^2=(\Delta f+(\varepsilon (x)-\varepsilon (x-y))\lambda '+\gamma )^2\ge \frac{\gamma ^2}{4}, \end{aligned}$$

so that, in this range

$$\begin{aligned}&\frac{|y|^3|\gamma |}{(y^2+(A_{\lambda '}y+\gamma )^2)(y^2+(\Delta f (x,x-y)+\lambda -\varepsilon (x-y)\lambda ')^2)}\\&\quad \le \frac{|y||\gamma |}{y^2+\frac{1}{4}\gamma ^2}, \end{aligned}$$

and we can estimate the second integral. In the range $|\gamma |\le 4||A_{\lambda '}||_{L^\infty }$ we can apply lemma A.7. This concludes the proof of the bound of I(x).

The bound

$$\begin{aligned}&J(x)=P.V.\int _{{\mathbb {R}}}\frac{y}{(y^2+(\Delta f(x,x-y)+\lambda +\varepsilon (x-y)\lambda ')^2}\\&\quad (\partial _x f(x-y)+\partial _x\varepsilon (x-y)\lambda ')dy. \end{aligned}$$

follows similar steps.

Then we have achieved the conclusion of lemma A.11. $\square$

Lemma A.12

Let f and $\varepsilon =c t$ be as in theorem 4.1. Then the velocity $u_c^\sharp (x,\lambda )$ satisfies

$$\begin{aligned} |u_c^\sharp (x,\lambda )-u_c^\sharp (x,\lambda ')|\le Ct, \end{aligned}$$

where C depends on $||f||_{H^4}$ but it does not depend on either x, $\lambda$ or $\lambda '$.

Proof

Recall that

$$\begin{aligned} u_c^\sharp (x,\lambda )=\int _{-1}^1 k_\theta (x,y) \partial _x \theta dy d \lambda '. \end{aligned}$$

Then it is enough to prove that the function

$$\begin{aligned} h(x,\lambda )=\int _{{\mathbb {R}}}\frac{y}{y^2+\theta ^2}\partial _x \theta dy \end{aligned}$$

satisfies $||\partial _\lambda h||_{L^\infty (dx)}\le |||f|||t$ for every $\lambda$. In addition, by Sobolev’s embedding we reduce the problem to prove that $||\partial _\lambda h||_{H^1(dx)}\le |||f||| t$. We notice that

$$\begin{aligned} \partial _\lambda h(x,\lambda )= \varepsilon (x)\int _{{\mathbb {R}}}\frac{2y\theta }{\left( y^2+\theta ^2\right) ^2}\partial _x \theta . \end{aligned}$$

The $L^2-norm$ of this function can be bounded in the same way we bounded $k^{31d}_2(x)$ in (A.18) and N(x) in (A.22). By taking a derivative with respect to x we have

$$\begin{aligned} \partial _x\partial _\lambda h(x,\lambda )&= \varepsilon _x \int _{{\mathbb {R}}}\frac{2y\theta }{\left( y^2+\theta ^2\right) ^2}\partial _x \theta + \varepsilon \big ( \int _{{\mathbb {R}}}\frac{2y\theta }{\left( y^2+\theta ^2\right) ^2}\partial ^2_x \theta \\&\quad + \int _{{\mathbb {R}}}\frac{2y}{\left( y^2+\theta ^2\right) ^2}(\partial _x \theta )^2-2\int _{{\mathbb {R}}}\frac{2y\theta ^2 }{\left( y^2+\theta ^2\right) ^3}\partial _x \theta \big ). \end{aligned}$$

The first two terms can be bounded exactly as we bound N(x) in (A.22). The third can be bounded in the same way that $k^{31d}_2(x)$ in (A.18) and N(x) in (A.22). The last term can be bounded by using a similar strategy, though a different pseudodifferential operator arises.

$\square$

1.4 Estimates on the coefficient a(x, t)

The function a(x, t) is given by the expression

$$\begin{aligned} a(x,t)=P.V.\int _{{\mathbb {R}}}K(x,y)dy \end{aligned}$$

where the principal value is taken at 0 and at the infinity. We need to prove the next lemma

Lemma A.13

Let f and $\varepsilon =c t$ be as in theorem 4.1. The following estimate holds:

$$\begin{aligned} ||\partial _x a||_{H^2}\le |||f|||. \end{aligned}$$

Proof

We will use the same convection than in “Appendix A.1”. Recall that we can express

$$\begin{aligned} a(x,t)=\int _{-1}^1\int _{-1}^1 \frac{y}{y_2+\theta ^2} dyd\lambda d\lambda ' \end{aligned}$$

and thus, if we set

$$\begin{aligned} \int k_\theta ^{ji}(x,y) dy={\bar{k}}^{ij}, \end{aligned}$$

we need to show that for $j=1,2,3,4$ and every i.

$$\begin{aligned} {\bar{k}}^{ij} \in L^2 \end{aligned}$$

The estimation in fact are often easier than in Sect. A.1.2. All the outer integrals are automatic since the terms $\partial _{j-i} \Delta \theta$ do not appear in the numerators. The inner integrals can also be dealt with.

We sketch the case with $j=3$ which is the most singular and leave the rest to the energetic reader. From the fourth terms corresponding to (A.18) we can directly bound the inner integrals of ${\bar{k}}_1^{31},{\bar{k}}_3^{31},{\bar{k}}_4^{31}$ by means of lemma A.4 whereas ${\bar{k}}_2^{31}$ is equal to N in (A.22), with $g= \partial _x \theta ^3$ directly. The other two terms are easily bounded. Namely we have the bounds,

$$\begin{aligned}&|{\bar{k}}^{32}_{in}(x) |=\int _{|y|\le 1} \frac{y \theta \partial _x \theta (\partial _x \theta )^2+\theta \partial _x^2 \theta }{(y^2+\theta ^2)^3}dy \le \frac{y^5+y \gamma ^4}{(y^2+\theta ^2)^3}\\&|{\bar{k}}_{in}^{33}(x)|=\int \frac{y \theta ^3+\partial _x \theta ^3}{(y^2+\theta ^2)^4}dy \le C \int \frac{|y|^7+ y \gamma ^6}{{(y^2+\theta ^2)^4}} dy. \end{aligned}$$

For $a=3,4$, the terms with $y|\gamma |^{2a-2}$ in the numerator are directly bounded by lemma A.7, whereas the term with $|y|^{2a-1}$ is bounded in the usual two steps. Firstly, we use Lemma A.6 to replace $\theta$ by $\theta _{lin}$ and then obtain the estimate with lemma A.9. The rest of the derivatives are bounded in the same way.

$\square$

Symbols and estimates

1.1 Fourier transform of $K^{c,c'}_{A}(x)$

Lemma B.1

Let $K^{c',c}_{A}$ the function given by the expression

$$\begin{aligned} K^{c,c'}_A(x)=\frac{1}{4\pi }\int _{-1}^{1}\int _{-1}^1 \frac{y}{y^2+(Ay+c't\lambda ' y +ct(\lambda -\lambda '))^2}d\lambda d\lambda '. \end{aligned}$$

Then

$$\begin{aligned} \widehat{K^{c,c'}_{A}}(\xi )&=\frac{-i\text {sign}(\xi )}{4\cdot 2\pi c t |\xi |}\int _{-1}^{1} \left( 2-e^{2\pi \sigma _{\lambda '}ct|\xi |(-1-\lambda ')(A_{\lambda '}i\text {sign}(\xi )+1)}\right. \\&\quad \left. -e^{2\pi \sigma _{\lambda '}ct|\xi |(1-\lambda ')(A_{\lambda '}i\text {sign}(\xi )-1)}\right) , \end{aligned}$$

and

$$\begin{aligned} {\widehat{K}}^{c,0}_A(\xi )= & {} \frac{-i\text {sign}(\xi )}{2\pi c t|\xi |}\left( 1+\frac{1}{4\pi c t |\xi |}\left( e^{-4\pi \sigma c t |\xi |}\left( \cos (4\pi \sigma A c t |\xi |)\right. \right. \right. \nonumber \\&\quad \left. \left. \left. -A\sin (4\pi \sigma A c t |\xi |)\right) -1\right) \right) . \end{aligned}$$

(B.1)

where

$$\begin{aligned} A_{\lambda '}=A+c't\lambda ',&\sigma _{\lambda '}=\frac{1}{1+A_{\lambda '}^2}. \end{aligned}$$

Proof

We first notice that, if we call

$$\begin{aligned} \gamma =ct|\lambda -\lambda '|, \end{aligned}$$

we can write

$$\begin{aligned}&y^2+(Ay+c'\lambda 'ty+ct(\lambda -\lambda '))^2= (1+(A+c'\lambda 't)^2)y^2+c^2t^2(\lambda -\lambda ')^2\\&\qquad +2(A+c'\lambda 't')yct(\lambda -\lambda ')\\&\quad =\frac{1}{\sigma _{\lambda '}}\left( y^2+2\sigma _{\lambda '}A_{\lambda '}\gamma \text {sign}(\lambda -\lambda ')y+\sigma _{\lambda '}\gamma ^2\right) \\&\quad =\frac{1}{\sigma _{\lambda '}}\left( (y+\sigma _{\lambda '}A_{\lambda '}\gamma \text {sign}(\lambda -\lambda '))^2+\sigma _{\lambda '}\gamma ^2 -\sigma _{\lambda '}^2A_{\lambda '}^2\gamma ^2\right) \\&\quad =\frac{1}{\sigma _{\lambda '}}\left( (y+\sigma _{\lambda '}A_{\lambda '}\gamma \text {sign}(\lambda -\lambda '))^2+\sigma _{\lambda '}^2\gamma ^2\right) . \end{aligned}$$

And then

$$\begin{aligned} K^{c,c'}_A(x)=\frac{1}{4\pi }\int _{-1}^1\int _{-1}^1\frac{\sigma _{\lambda '}y}{(y+\mu )^2+\nu ^2}d\lambda d\lambda ' \end{aligned}$$

with $\mu =\sigma _{\lambda '}A_{\lambda '}\gamma \text {sign}(\lambda -\lambda ')$ and $\nu =\sigma _{\lambda '}\gamma .$

With these notations Fourier transforms are easier as they resemble the relation between Poisson and Abel kernels. We can compute that

$$\begin{aligned}&{\mathcal {F}}\left[ \frac{y}{(y+\mu )^2+\nu ^2}\right] (\xi )= {\mathcal {F}}\left[ \frac{y+\mu }{(y+\mu )^2+\nu ^2}\right] (\xi )\\&\qquad -\mu {\mathcal {F}}\left[ \frac{1}{(y+\mu )^2+\nu ^2}\right] (\xi )\\&\quad =e^{2\pi i\xi \mu }\left( {\mathcal {F}}\left[ \frac{y}{y^2+\nu ^2}\right] (\xi )-\mu {\mathcal {F}}\left[ \frac{1}{y^2+\nu ^2}\right] (\xi )\right) \\&\quad =e^{2\pi i\xi \mu }e^{-2\pi \nu |\xi |}\pi \left( -i\text {sign}(\xi )-\frac{\mu }{\nu }\right) . \end{aligned}$$

Therefore

$$\begin{aligned}&{\widehat{K}}^{c,c'}_{A}(\xi )=\frac{1}{4}\int _{-1}^{1}\int _{-1}^1\sigma _{\lambda '}e^{2\pi i \sigma _{\lambda '}A_{\lambda '}ct(\lambda -\lambda ')\xi }e^{-2\pi \sigma _{\lambda '}ct|\lambda -\lambda '||\xi |}\nonumber \\&\quad \left( -i\text {sign}(\xi )-A_{\lambda '}\text {sign}(\lambda -\lambda ')\right) d\lambda d\lambda '. \end{aligned}$$

(B.2)

For visualization set $ct\sigma _\lambda ' 2 \pi i |\xi |=a$ in the next estimates

$$\begin{aligned}&\int _{-1}^1 e^{a A_{\lambda '}i \text {sign}(\xi ) (\lambda -\lambda ')}e^{- a|\lambda -\lambda '| } \left( -i\text {sign}(\xi )-A_{\lambda '} \text {sign}(\lambda -\lambda ')\right) d\lambda \\&\quad =\int _{-1}^{\lambda '}e^{ a (1+i A_{\lambda '} \text {sign}(\xi )) (\lambda -\lambda ')} \left( -i\text {sign}(\xi )+A_{\lambda '}\right) d \lambda \\&\qquad +\int _{\lambda '}^{1}e^{a(-1+iA_{\lambda '} \text {sign}(\xi ) )(\lambda -\lambda ')} \left( -i\text {sign}(\xi )-A_{\lambda '}\right) d\lambda ,\\&\quad =\frac{-i\text {sign}(\xi )+A_{\lambda '}}{a(1+iA_{\lambda '}\text {sign}(\xi ))} \left( 1-e^{a(-1-\lambda ')(A_{\lambda '}i\text {sign}(\xi )+1)}\right) \\&\qquad +\frac{-i\text {sign}(\xi )-A_{\lambda '}}{a(iA_{\lambda '}\text {sign}(\xi )-1)} \left( e^{a (1-\lambda ')(A_{\lambda '}i\text {sign}(\xi )-1)}-1\right) \\&\quad =-i\text {sign}(\xi )\frac{1}{a}\left( 2-e^{a(-1-\lambda ')(A_{\lambda '}i\text {sign}(\xi )+1)} -e^{a(1-\lambda ')(A_{\lambda '}i\text {sign}(\xi )-1)}\right) . \end{aligned}$$

Thus

$$\begin{aligned}&{\widehat{K}}^{c,c'}_{A}(\xi )=\frac{-i\text {sign}(\xi )}{4\cdot 2\pi c t |\xi |}\int _{-1}^{1} \left( 2-e^{2\pi \sigma _{\lambda '}ct|\xi |(-1-\lambda ')(A_{\lambda '}i\text {sign}(\xi )+1)}\right. \\&\quad \left. -e^{2\pi \sigma _{\lambda '}|\xi |(1-\lambda ')(A_{\lambda '}i\text {sign}(\xi )-1)}\right) d \lambda '. \end{aligned}$$

This proves the first identity of lemma B.1.

By taking $c'=0$ we have that

$$\begin{aligned}&{\widehat{K}}^{c,0}_{A}(\xi )=\frac{-i\text {sign}(\xi )}{4\cdot 2\pi c t |\xi |}\int _{-1}^{1} \left( 2-e^{2\pi \sigma ct|\xi |(-1-\lambda ')(A i\text {sign}(\xi )+1)} \right. \nonumber \\&\quad \left. -e^{2\pi \sigma |\xi |(1-\lambda ')(A i\text {sign}(\xi )-1)}\right) d\lambda ', \end{aligned}$$

(B.3)

where $\sigma =\frac{1}{1+A^2}$. Integrating in $\lambda '$ yields

$$\begin{aligned} I&=\int _{-1}^1 e^{-2\pi \sigma c t (1+\lambda ')|\xi |(1+iA\text {sign}(\xi ))}d\lambda '\\&=\frac{1}{2\pi \sigma c t|\xi | (1+iA\text {sign}(\xi ))}\left( 1-e^{-4\pi \sigma c t (1+iA\text {sign}(\xi ))|\xi |}\right) \end{aligned}$$

and

$$\begin{aligned} II&=\int _{-1}^1e^{2\pi \sigma c t (-1+\lambda ')|\xi |(1-iA\text {sign}(\xi ))}d\lambda '\\&=\frac{1}{2\pi \sigma c t|\xi | (1-iA\text {sign}(\xi ))}\left( 1-e^{-4\pi \sigma c t (1-iA\text {sign}(\xi ))|\xi |}\right) . \end{aligned}$$

However

$$\begin{aligned} I+II=\frac{2}{2\pi \sigma ct|\xi |} \text {Re}\left( I\right) . \end{aligned}$$

Now if we notice that

$$\begin{aligned}&\text {Re}\left( \frac{1}{1+iA\text {sign}(\xi )}\right) =\sigma , \quad \text {Re}\left( \frac{e^{-4\pi \sigma c t i A \xi }}{1+iA\text {sign}(\xi )}\right) \\&\quad =\sigma \left( \cos \left( 4\pi \sigma A c t |\xi |\right) -A\sin \left( 4\pi \sigma c t A |\xi |\right) \right) , \end{aligned}$$

equality (B.1) follows. $\square$

1.2 Estimations of the various symbols

In this section we will use the notation $t\xi =\tau$. In the following estimates:

• 1. A is function in $H^3$.

• 2. The function c is as in the statement of theorem 4.1.

• 3. We can consider that the time $t<<1.$

To alleviate the notation we introduce the following auxiliar function

$$\begin{aligned} h(x,\tau )= & {} \frac{1}{c\tau }\left\{ 1+\frac{1}{4\pi c\tau }\left( e^{-4\pi c\tau \sigma }\left( \cos (4\pi c\tau \sigma A)-A\sin (4\pi c\tau \sigma A)\right) -1\right) \right\} \nonumber \\= & {} \frac{1}{c\tau }(1+h_2(\tau )) \end{aligned}$$

(B.4)

with

$$\begin{aligned} h_2(x,\tau )=\frac{1}{4\pi c\tau }\left( -1+e^{-4\pi \sigma c \tau }\left( \cos (4\pi \sigma A c\tau )-A\sin (4\pi \sigma A c\tau )\right) \right) \end{aligned}$$

and $\sigma =\frac{1}{1+A^2}.$

We emphasis that h and $h_2$ depend on x just through A and c.

Notice that (B.1) implies that

$$\begin{aligned} 2\pi i \text {sign}(\xi ){\hat{K}}^{c,0}_{A}(\xi )=h(x,\tau ). \end{aligned}$$

We will omit that h depends on A as well. We study the regularity of the function h in detail.

Lemma B.2

The following identities holds:

$$\begin{aligned} h_2(x,\tau )= & {} -\int _{0}^1e^{-4\pi \sigma c\tau \tau _1}\cos (4\pi \sigma A c \tau \tau _1)d\tau _1, \end{aligned}$$

(B.5)

$$\begin{aligned} h(x,\tau )= & {} 4\pi \sigma \int _{0}^1\int _{0}^1 e^{-4\pi \sigma c \tau \tau _1\tau _2}\tau _1\left( \cos (4\pi \sigma A c \tau \tau _1\tau _2)\right. \nonumber \\&\left. +A\sin (4\pi \sigma A c \tau \tau _1\tau _2)\right) d\tau _2d\tau _1. \end{aligned}$$

(B.6)

The following estimate not only gives us how the symbol p grows but it also implies lemma 4.11, the key in showing that $p_+$ is positive for small times.

Lemma B.3

Let $\varphi ,h$ defined as above. The following estimates hold:

$$\begin{aligned} \begin{aligned}&|h(x,\tau ) |\le \frac{\frac{1}{c}}{1+\tau }+\frac{2|A|+5+8\pi }{1+(\tau )^2}, \\&|\partial _A h(x,\tau )|+|\partial _c h(x,\tau )| \le \frac{\langle A \rangle }{1+\tau }, \end{aligned} \end{aligned}$$

(B.7)

and

$$\begin{aligned} \left( \varphi (\tau )-h(x,\tau )\right) \ge \frac{1-\frac{1}{c}}{1+\tau }+\frac{B-(2|A|+5+8\pi )}{1+\tau ^2}. \end{aligned}$$

(B.8)

Proof

We start with $\tau =t|\xi |\ge 1$. Since $c \ge 1$ we have that

$$\begin{aligned}&\frac{1}{c\tau }\left( 1+\frac{1}{4\pi c \tau }\left( e^{-4\pi \sigma c\tau }\left( \cos (4\pi \sigma A c \tau )-A\sin (4\pi \sigma A c \tau )\right) -1\right) \right) \\&\quad \le \frac{1}{c\tau }+\frac{|A|+2}{\tau ^2}. \end{aligned}$$

But since for $\tau \ge 1$, $\frac{1}{\tau ^2}\le \frac{2}{1+\tau ^2}$ and $\frac{1}{\tau }=\frac{1}{1+\tau }+\frac{1}{\tau }-\frac{1}{1+\tau }=\frac{1}{1+\tau }+ \frac{1}{\tau +\tau ^2}\le \frac{1}{1+\tau }+\frac{1}{1+\tau ^2}$. Then

$$\begin{aligned}&\frac{1}{c\tau }\left( 1+\frac{1}{4\pi c \tau }\left( e^{-4\pi \sigma c\tau }\left( \cos (4\pi \sigma A c \tau )-A\sin (4\pi \sigma A c \tau )\right) -1\right) \right) \\&\quad \le \frac{\frac{1}{c}}{1+\tau }+\frac{2|A|+5}{1+\tau ^2}. \end{aligned}$$

For $\tau \le 1$ we use the expression (B.2) to get uniform bounds on h and its derivatives.

The derivatives of h for $\tau \ge 1$ are controlled by

$$\begin{aligned}&|\partial _c h| \le C\left( \frac{1}{c^2 \tau }+\frac{1}{c^3\tau ^2}(1+|A|)+ \frac{1}{c^2 \tau ^2} \tau (1+|A|) \right. \\&\quad \left. + \tau (|A|+|A|^2) \le \frac{\langle A \rangle }{\tau }\right) ,\\&|\partial _A h| \le C\left( \frac{1}{c^2 \tau ^2} \tau ( |A|+|A|^2)\right) . \end{aligned}$$

If we combine (B.7) with the definition of $\varphi$, (B.8) follows as well. $\square$

Lemma B.4

Let $k=1,2,3$

$$\begin{aligned}&|\partial _\tau h(x,\tau )|+|\partial _{\tau ,A} h|+ |\partial _{\tau ,c} h| \le \langle A \rangle \frac{1}{1+\tau ^2}, \\&\partial ^k_x h(x,\tau ) \le \langle A \rangle \left( \frac{ |\partial _x c|}{1+\tau }+\frac{|\partial ^k_x c|+|\partial ^k_x A|}{1+\tau ^2}\right) ,\\&\partial ^k_x \partial _\tau h(x,\tau ) \le \langle A \rangle \left( \frac{ |\partial _x c|}{1+\tau ^2}+\frac{|\partial ^k_x c|+|\partial ^k_x A|}{1+|\tau |^3}\right) , \end{aligned}$$

where the constant C depends on $|\partial ^{i}_x c|+ |\partial _x^{i} A |$ for $i<k$.

Proof

For $\tau <1$ we use lemma B.2 and the result follows. For $\tau >1$, from the expression

$$\begin{aligned} h=\frac{1}{c\tau }-\frac{1}{4\pi c^2\tau ^2}+\frac{1}{4\pi c^2\tau ^2}e^{-4\pi \sigma \tau }\left( \cos (4\pi c\sigma A\tau )-A\sin (4\pi c\sigma A\tau )\right) , \end{aligned}$$

(B.9)

we see that $|\partial _\tau h|\le \langle A \rangle (1+\tau ^2)^{-1}$. In order to bound the derivatives on $\sigma$ and A of $\partial _\tau h$ we see that the two first terms in (B.9) do not cause any difficulty. The third one in (B.9) brings down a factor $\tau$ but since $c>1$ and $\sigma \ge \frac{1}{1+||A||_{L^\infty }}$ we still have exponential decay. In order to bound the derivatives with respect to x of h and $\partial _\tau h$ a similar argument applies.

$\square$

We recall that

$$\begin{aligned} p&=2\pi i \xi {\hat{K}}^{c,0}_{A}, \quad p_b=2\pi i \xi \left( p_{main}-p\right) ,\\ p_{\text {good}}&=\frac{1}{|\xi |}p-\varphi (t|\xi |), \quad \text {and} \quad p_+=-(1+|\xi |) p_{\text {good}}. \end{aligned}$$

Lemma B.5

Given $t>0$, the symbols $p_b, t\partial _x p_{main}, p_{good} \in {\mathcal {S}}_{1,1}$ with the following estimates.

• i) $\Vert p_b\Vert _{1,1}+\Vert p_{good}\Vert _{1,1} \le \langle A \rangle$

• ii) $\Vert tp_{main}\Vert _{1,1}+\Vert t \partial _{x} p_{main}\Vert _{1,1}+\Vert \partial _\xi p_{main} \Vert _{1,0} \le \langle A \rangle$

Proof

We start with the $L^\infty$ estimates (no x derivatives). The estimation of $p_b$ itself is the most subtle so we address it first.

• 1. Estimation of the $L^\infty$ norm of $p_b$.

  The fundamental theorem of calculus tell us that,

  $$\begin{aligned} p-p_{main}&=\frac{-i\text {sign}(\xi )}{4\cdot 2\pi c t |\xi |}\int _{-1}^1\int _{0}^1\frac{d}{ds} e^{-2\pi \sigma _{s\lambda }ct|\xi |(1+\lambda )(1+iA_{s\lambda }sign(\xi ))} dsd\lambda \\&\quad +\frac{-i\text {sign}(\xi )}{4\cdot 2\pi c t |\xi |}\int _{-1}^1\int _{0}^1\frac{d}{ds} e^{-2\pi \sigma _{s\lambda }ct|\xi |(1-\lambda )(1-iA_{s\lambda }sign(\xi ))} ds d\lambda \\&\equiv T_1+T_2. \end{aligned}$$

  Now we use the chain rule and that $\partial _s A_{s\lambda }=c't\lambda$ and $\partial _s \sigma _{s\lambda }=-2\sigma _{s\lambda }^2A_{s\lambda }c't\lambda$ to obtain that

  $$\begin{aligned}&2\pi i \xi T_1\nonumber \\&\quad =\frac{1}{4ct}\int _{-1}^1 \int _{0}^1 2\sigma _{s\lambda }^2A_{s\lambda }c't\lambda 2\pi c t |\xi |(1+\lambda )(1+iA_{s\lambda }\text {sign}(\xi ))\nonumber \\&\quad \qquad e^{-2\pi \sigma _{s\lambda } c t|\xi |(1+\lambda )(1+iA_{s\lambda }\text {sign}{\xi }) } ds d\lambda \nonumber \\&\qquad +\frac{1}{4ct}\int _{-1}^1\int _{0}^1 c't\lambda i\text {sign}(\xi ) (-2\pi \sigma _{s\lambda } c t|\xi |(1+\lambda ))\nonumber \\&\quad \qquad e^{-2\pi \sigma _{s\lambda } c t|\xi |(1+\lambda )(1+iA_{s\lambda })\text {sign}(\xi )}dsd\lambda \end{aligned}$$

  (B.10)
  $$\begin{aligned}&\quad \equiv T_{11}+T_{12}. \end{aligned}$$

  (B.11)

  Now observe that the dangerous t in the denominator cancels out in both $T_{11}, T_{12}$. Thus, we are entitled to take modulus and obtain the elementary bound:

  $$\begin{aligned} |2\pi i \xi T_1|\le C\int _{-1}^1\int _{0}^1 2\pi \sigma _{s\lambda }c t |\xi |(1+\lambda ) e^{-2\pi \sigma _{s\lambda }ct|\xi |(1+\lambda )}dsd\lambda \le \langle A \rangle . \end{aligned}$$

  The estimation involving $T_2$ is exactly analogous and thus $\Vert p_b\Vert _{L^\infty } \le \langle A \rangle$.

• 2. Estimation of the $L^\infty$-norm of $\partial _\xi (2\pi i \xi ( {\hat{K}}^{c,0}_A(\xi )-{\hat{K}}^{c,c'}_A(\xi )))$. We recall that

  $$\begin{aligned}&2\pi i \xi ({\hat{K}}^{c,0}_A(\xi )-{\hat{K}}^{c,c'}_A(\xi ))\\&\quad =\frac{1}{4 c t}\int _{-1}^{1} \left( e^{2\pi \sigma _{\lambda '}ct|\xi |(-1-\lambda ')(A_{\lambda '}i\text {sign}(\xi )+1)}\right. \\&\qquad \left. +e^{2\pi \sigma _{\lambda '}|\xi |(1-\lambda ')(A_{\lambda '}i\text {sign}(\xi )-1)}\right) d\lambda '\\&\qquad +\frac{1}{4 c t }\int _{-1}^{1} \left( e^{2\pi \sigma ct|\xi |(-1-\lambda ')(A i\text {sign}(\xi )+1)} +e^{2\pi \sigma (1-\lambda ')(A i\text {sign}(\xi )-1)}\right) d\lambda '\\&\quad \equiv U_1+U_2. \end{aligned}$$

  In this case we can bound $U_1$ and $U_2$ separately and it is enough to estimate $U_1$. In addition, we can split $U_1= U_{11}+U_{12}$ with $U_{11}=\frac{1}{4ct}\int _{-1}^1 e^{2\pi \sigma _{\lambda '}ct|\xi |(-1-\lambda ')(A_{\lambda '}i\text {sign}(\xi )+1)}d\lambda '$ and it easy to see that the estimation of $U_{11}$ and $U_{12}$ follows similar steps. We have that

  $$\begin{aligned}&\partial _\xi U_{11} =\frac{1}{4c t}\int _{-1}^1 (-2\pi \sigma _{\lambda '}c t(1+\lambda ')(\text {sign}(\xi )+iA_{\lambda '})\nonumber \\&\qquad e^{2\pi \sigma _{\lambda '}ct|\xi |(-1-\lambda ')(A_{\lambda '}i\text {sign}(\xi )+1)}d\lambda '\nonumber \\&\quad =\frac{1}{4}\int _{-1}^1 (-2\pi \sigma _{\lambda '}(1+\lambda ')(\text {sign}(\xi )+iA_{\lambda '})e^{2\pi \sigma _{\lambda '}ct|\xi |(-1-\lambda ')(A_{\lambda '}i\text {sign}(\xi )+1)}d\lambda ', \end{aligned}$$

  (B.12)

  and then $|\partial _\xi U_{11}|\le \langle A \rangle$.

• 3. The $L^\infty$ estimate for $p_{good}$ follows directly from lemma B.3 and the definition of $\varphi$.

• 4. To estimate $p_{main}$ we notice that $p_{main}=p_{b}+p$. We have already proved that $p_{b}$ is bounded and recall that $p=\xi h(x,\tau )$. Thus $|p(x,\xi )|= \frac{1}{t} \tau h(x,\tau )$ and thus the estimate (B.7) implies that $\Vert tp_{main} \Vert _{L^\infty } \le \langle A \rangle$. Similarly, when estimating $\partial _\xi p_{main}$, by our uniform estimate on $p_b$ we are reduce to estimate $\partial _\xi p$. Now notice that by the definition of p and the chain rule, $\Vert \partial _\xi p\Vert _{L^\infty } =\frac{t}{t}\Vert \partial _\tau (\tau h)\Vert \le \langle A \rangle$ where the last bound follows from the lemma B.4

• 5. Finally, we deal with the derivatives respect to x. Firstly, observe that $tp(x,\tau )=\tau h(x,\tau )$ and thus the estimates of the derivatives in x follow directly from those of h, which are explicitly bounded in lemma B.4. Hence we obtain that

  $$\begin{aligned} \Vert t p \Vert _{1,1}+ \Vert t \partial _x p \Vert _{1,1}+ \Vert \partial _\xi p_b \Vert _{1,0} \le \langle A \rangle . \end{aligned}$$

  (B.13)

  Next we look at the x derivatives of $p_{b},\partial _\xi p_{b}$. We have done all the work as in the expressions, of $T_{11},T_{12},U_1,U_2$ the only difficulty occurs when after the use of chain rule we differentiate A and c. Thus we obtain that

  $$\begin{aligned} \Vert p_b\Vert _{1,1}+ \Vert \partial _x p_b\Vert _{1,1} \le \langle A \rangle . \end{aligned}$$

  (B.14)

  Since, $\frac{p}{\xi }=h(x,\tau )$ and $\varphi$ is explicit lemma B.3, lemma B.4 imply readily the bounds for $\Vert p_{good}\Vert _{1,1} \le \langle A \rangle$ and thus the claim i) follows. Claim ii), which deals with $p_{main}$, is an straightforward consequence of (B.13) and (B.14).

$\square$

The following lemma is cumbersome as considering $p_+^\frac{1}{2}$ instead of $p_+$ is less innocent than it seems. In fact here is the only place where the existence of the constant $c_\infty$ is required.

Lemma B.6

Let $2|A|+5+8\pi <\frac{B}{2}$. The symbols $p^\frac{1}{2}_+$ and $q=\partial _\xi p^\frac{1}{2}_+$ satisfy that for $0<\varepsilon <1$ it holds that,

$$\begin{aligned}&\Vert t^{\frac{1}{2}} p_+^\frac{1}{2} \Vert _{1,1} \le \langle A \rangle ,\quad ||\partial _\xi p_+^\frac{1}{2}||_{L^\infty ({\mathbb {R}}^2)}\le \langle A \rangle , \quad \\&\Vert \partial _x \partial _\xi p_+^\frac{1}{2}\Vert _{L^\infty ({\mathbb {R}}^2)} \le \langle A \rangle , \quad \sup _{\xi \in {\mathbb {R}}}\left( \Vert \partial _x q \Vert _{H^2}+ \Vert \partial _{x} q \Vert _{{\dot{H}}^{-\varepsilon }}\right) \le \langle A \rangle ,\\&||t^\frac{1}{2}\partial ^2_x p_{+}^\frac{1}{2}||_{L^\infty ({\mathbb {R}}^2)}\le \langle A \rangle . \end{aligned}$$

Proof

We give the proof in the case $c\ge 1+\kappa$, and will explain at the end of the proof the modifications for the case $c=1$. We explain first the $L^\infty$ bounds then $\partial _xq \in {\dot{H}}^{-\varepsilon }$ and finally how to control the x derivatives of both symbols.

• 1. Since,

  $$\begin{aligned} tp_{+}(x,\xi )=(t+\tau )\left( \varphi (\tau )-h(x,\tau )\right) \end{aligned}$$

  (B.15)

  it follows from (B.8) that

  $$\begin{aligned} |p_+^\frac{1}{2}|\le \frac{\langle A \rangle }{\sqrt{t}}.\end{aligned}$$

  (B.16)

  Now we deal with $q=\partial _\xi p_+^\frac{1}{2}(x,\xi )$. By product rule for derivatives,

  $$\begin{aligned} q(x,\xi )&=\frac{1}{2}\frac{sign(\xi )}{(1+|\xi |)^\frac{1}{2}}\left( \varphi (\tau )-h(x,\tau )\right) ^\frac{1}{2}\nonumber \\&\quad +(1+|\xi |)^\frac{1}{2}\frac{t\text {sign}(\xi )}{2}\left( \varphi (\tau )-h(x,\tau )\right) ^{-\frac{1}{2}} \times \partial _{\tau }\left( \varphi (\tau )-h_2(\tau )\right) . \end{aligned}$$

  (B.17)

  The first term is innocent since $(\varphi -h)$ is bounded. For the second we notice, that

  $$\begin{aligned} t(1+|\xi |)^\frac{1}{2} \le t^\frac{1}{2}(1+|\tau |)^\frac{1}{2}, (\varphi -h)^{-\frac{1}{2}} \le C (1+\tau )^{\frac{1}{2}}, \end{aligned}$$

  (B.18)

  where the second inequality comes from lemma B.3. In addition, lemma B.4 implies that $|\partial _\tau (\varphi -h)| \le \frac{\langle A \rangle }{1+\tau ^2}$. Thus, the desired

  $$\begin{aligned} |q(x,\xi )|\le \langle A \rangle . \end{aligned}$$

  follows.

• 2. That $\partial _xq\in {\dot{H}}^{-\varepsilon }$ follows from the existence of a constant $q_\infty$ such that $q-q_\infty \in L^2$. Since the only x dependence of q is through A, c mwe will declare $q_\infty (t,\xi )=q(0,c_\infty , t,\xi )$. Now by plugging into (B.17) the bounds from lemmas B.3 amd B.4, it follows that $|\nabla _{A,c}| q\le \langle A \rangle$. Then, the mean value theorem applied to q as a function of A, c, yields that for every $x \in {\mathbb {R}}$,

  $$\begin{aligned} |q(x)-q_\infty | \le \langle A \rangle (|A(x)-0|+ |c(x)-c_\infty |) \end{aligned}$$

  and since A is $H^3$ and $\int |c-c_\infty |^2 dx<C$ by assumption, the result follows. The proof to bound $||p^\frac{1}{2}_+||_{{\dot{H}}^{-\varepsilon }}$ follows similar steps.

• 3. Now we compute $\partial ^k_x p^\frac{1}{2}_+(x,\xi )$ and $\partial ^k_{x} q$. By chain and product rule $\partial ^k_x p^\frac{1}{2}_+(x,\xi )$ it is a sum of terms of the type $(1+|\xi |)^\frac{1}{2} (\varphi -h)^{\frac{1}{2}-i} \Pi _{\sum \alpha _i=k} \partial ^{\alpha _i} _x (\varphi - h)$ for $i=1,\ldots ,k$. By (B.18) and lemma B.4 the most singular term is

  $$\begin{aligned} (1+|\xi |)^\frac{1}{2} (\varphi -h)^{\frac{1}{2}-k} |\partial _x h|^k\le & {} t^{-\frac{1}{2}} \langle A \rangle |c_x|+|A_x| \frac{(1+\tau )^k}{ (1+\tau )^k} \\\le & {} t^{-\frac{1}{2}} \langle A \rangle \left( |c_x|+|A_x| \right) . \end{aligned}$$

  We move to q. We need to show that $\partial _{x} q \in L^\infty$ for $p_+^\frac{1}{2} \in {\mathcal {S}}_{1,1}$ and that $\partial ^k_x q \in L^2$ for $k=1,2,3$. By lemma B.2 we only need to give the details of the case $\tau >1$. Notice that when we differentiate in (B.17) the derivatives of the first term still remain innocent as $\partial _x^k\left( \varphi (\tau )-h(x,\tau )\right) ^\frac{1}{2}$ has been shown to be bounded by $\sup _{1 \le i \le k}|\partial _x^i c|+|\partial _x^i A|$. For the second term, again product rule combined with lemma B.3 and B.4 implies that the most singular term is

  $$\begin{aligned}&t(1+|\xi |)^\frac{1}{2} (\varphi -h(x,\tau )^{-\frac{1}{2}-k} |\partial _x h|^k \partial _{\tau }\left( \varphi (\tau )-h(x,\tau )\right) \\&\quad \le \langle A \rangle \left( |\partial _x c|+ |\partial _x A| \right) \frac{ (1+\tau )^{k+1} }{(1+\tau )^k(1+\tau ^2)}. \end{aligned}$$

  Notice that the terms $\partial ^{\alpha _i} _x (\varphi - h)$ and $\partial ^{\alpha _i} _x \partial _\tau (\varphi - h)$ will be less singular respect to $\tau$ and will be controlled by $\langle A \rangle (|\partial ^{\alpha _i}_x c|+ |\partial ^{\alpha _i}_x A|)$.

Finally, we mention that in the case $c=1$, $\partial _x c=0$, thus their derivatives are bounded by powers of $\frac{1}{1+\tau ^2}$ and lemma B.3 implies that $(\varphi -h)^{-1} \le \frac{C}{B} (1+\tau ^2)$. The terms appearing in our various derivatives compensate exactly in the same way. $\square$