Singularities and unsteady separation for the inviscid two-dimensional Prandtl system

Abstract

We consider the inviscid unsteady Prandtl system in two dimensions, motivated by the fact that it should model to leading order separation and singularity formation for the original viscous system. We give a sharp expression for the maximal time of existence of regular solutions, showing that singularities only happen at the boundary or on the set of zero vorticity, and that they correspond to boundary layer separation. We then exhibit new Lagrangian formulae for backward self-similar profiles, and study them also with a different approach that was initiated by Elliott–Smith–Cowley and Cassel–Smith–Walker. One particular profile is at the heart of the so-called Van-Dommelen and Shen singularity, and we prove its generic appearance (that is, for an open and dense set of blow-up solutions) for any prescribed Eulerian outer flow. We comment on the connection between these results and the full viscous Prandtl system. This paper combines ideas for transport equations, such as Lagrangian coordinates and incompressibility, and for singularity formation, such as self-similarity and renormalisation, in a novel manner, and designs a new way to study singularities for quasilinear transport equations.

Appendices

Parametrisation and volume preservation

Lemma A.1

Let $\Omega \subset {\mathbb {R}}^2$ be open, and $f=(f_1,f_2)\in C^1(\Omega ,{\mathbb {R}}^2)$ be such that $\nabla f_1\ne 0$ on $\Omega$, and that for each q in the range of $f_1$, the level set $\Gamma [q]=\{z\in \Omega , \ f_1(z)=q\}$ is diffeomorphic to ${\mathbb {R}}$. Then

• (i) Let $\gamma :{\mathcal {O}}\rightarrow \Omega$ be a diffeomorphism between an open set ${\mathcal {O}}\subset {\mathbb {R}}^2$ and $\Omega$, such that for each $q \in {\mathbb {R}}$, $\gamma (q,\cdot )$ is an arclength parametrisation of $\Gamma [q]$. Then f preserves volume if and only if for all q in the range of $f_1$, and any two $z=\gamma (q,s)$ and $z'=\gamma (q,s')$ it holds that

  $$\begin{aligned} |f_2(z)-f_2(z')|=\left| \int _{s}^{s'} \frac{\mathrm{d}{{\tilde{s}}}}{|\nabla f_1(\gamma (q,{{\tilde{s}}}))|}\right| . \end{aligned}$$

  (A.1)

• (ii) If $\partial _{z_1}f_1\ne 0$ on $\Omega$, let ${{\tilde{z}}}_1:{\mathcal {O}}\rightarrow \Omega$ be the only diffeomorphism between an open set ${\mathcal {O}}\subset {\mathbb {R}}^2$ and $\Omega$, such that for each $q\in {\mathbb {R}}$, ${{\tilde{z}}}_1(q,\cdot )$ is the parametrisation of $\Gamma [q]$ by the second component, i.e. $\Gamma [q]=\{({{\tilde{z}}}_1(q,{{\tilde{z}}}_2),{{\tilde{z}}}_2), \ (q,{{\tilde{z}}}_2)\in {\mathcal {O}} \}$, then f preserves volume if and only if, for all q in the range of $f_1$, and any two $z=({{\tilde{z}}}_1(q,z_2),z_2)$ and $z'=({{\tilde{z}}}_1(q,z_2'),z_2')$,

  $$\begin{aligned} |f_2(z)-f_2(z')| =\left| \int _{z_2}^{z_2'} \frac{\mathrm{d}\tilde{z_2}}{\partial _{z_1} f_1({{\tilde{z}}}_1(q,{{\tilde{z}}}_2),{{\tilde{z}}}_2)}\right| . \end{aligned}$$

  (A.2)

Proof

Proof of (i). We denote by $\gamma ^{-1}=( q,s):\Omega \rightarrow {\mathcal {O}}$ the inverse of $\gamma$, and note that $q=f_1$. At a point $z\in \Omega$, considering the orthonormal vectors $v_1=\nabla f_1(z) |\nabla f_1(z)|^{-1}$ and $v_2= \nabla ^{\perp }f_1(z) |\nabla f_1(z)|^{-1}$ where $(e_1,e_2)^{\perp }=(-e_2,e_1)$ we get the identities $\nabla q.v_1=|\nabla f_1(z)|$, $\nabla q.v_2=0$ and $\nabla s. v_2=1$. This implies $|\text {Det}\left( J \gamma ^{-1} \right) |=|\nabla f_1(z)|.$ Consider now the mapping $\phi :(q,s)\mapsto (q,f_2(\gamma (q,s))$. Since $\frac{\partial q}{\partial s}_{|_q}=0$ and $\frac{\partial q}{\partial q}_{|s}=1$ we have $|\text {Det}\left( J\phi \right) |=|\frac{\partial f_2}{\partial s}_{|q}|.$ Since $f=\phi \circ \gamma ^{-1}$, we get from the two previous computations that f preserves volume, that is, $|\text {Det}Jf|=1$, if and only if

$$ |\frac{\partial f_2}{\partial s}_{|q}|=\frac{1}{|\nabla f_1(z)|}. $$

This is equivalent to (i) upon integrating, and noticing that these quantities cannot change sign.

Proof of (ii). From the assumptions on f, there always exists a parametrisation by arclength of the level sets $\gamma :{\mathcal {O}}' \rightarrow \Omega$ satisfying the properties of (i). We change variables ${{\tilde{z}}}_1^{-1}\circ \gamma :(q,s)\mapsto (q,{{\tilde{z}}}_2)$. Differentiating the identity $f_1({{\tilde{z}}}_1(q,{{\tilde{z}}}_2),{{\tilde{z}}}_2)=q$ we find $\frac{\partial {{\tilde{z}}}_1}{\partial _{{{\tilde{z}}}_2}}_{|_{q}}=-\partial _{z_2}f_1/\partial _{z_1}f_1$. Hence

$$ |\frac{\partial s}{\partial {{\tilde{z}}}_2}_{|_q}|=\sqrt{1+\left( \frac{\partial _{z_2}f_1}{\partial _{z_1}f_1} \right) ^2}=\frac{|\nabla f_1|}{|\partial _{z_1}f_1|}. $$

Changing variables $s \mapsto {{\tilde{z}}}_2$ in (A.1), then yields that f preserves volume if and only if (A.2) holds true. $\quad \square$

Computing the normal component of the characteristics

proof of Lemma 5.4

At several moments in the proof, we will use the following: the function $\Psi _1$ enjoys

$$\begin{aligned}&c<\left| \frac{X\partial _X\Psi _1(X)}{\Psi _1(X)} \right| \leqq \frac{1}{c}, \ \ \ \ |\Psi _1(X)|\lesssim |X|^{\frac{1}{3}}, \nonumber \\&\Psi _1(X+X')=\Psi _1(X)+O \left( |X'|^{\frac{1}{3}}\right) \end{aligned}$$

(B.1)

uniformly for $X,X'\in {\mathbb {R}}$. From the first bound above, given a small quantity O(|z|) that has size |z|, as an application of the implicit function theorem, it is true that

$$\begin{aligned} \Psi _1\left( X(1+O(|z|))\right) =\Psi _1\left( X\right) (1+O(|z|)), \end{aligned}$$

(B.2)

where the “new” O(|z|) in the right hand side enjoys the same differentiability properties and similar bounds, as the one in the left hand side. To rearrange the O()’s in what follows, we shall simplify $O(|z_1|)+O(|z_2|)=O(|z_2|)$ in zones where $|z_1|\lesssim |z_2|$, and $O(|z|^\alpha )=O(|z|^\beta )$ if $0< \beta \leqq \alpha$ as all quantities in the O()’s will be small.

To compute the vertical component, we will use the following result. Assume that (a, b) and $(a',b')$ belong to the same level set curve $\Gamma [x]=\{x({{\tilde{a}}},{\tilde{b}})=x \}$. Assume that, when ordering points on $\Gamma$ with their distance to the boundary, $(a',b')$ is after (a, b), and that either $b\leqq b'$ or $a\leqq a'$. Assume moreover that $\Gamma _{(a,b)}^{(a',b')}$, the part of $\Gamma$ between (a, b) and $(a',b')$, can be either parametrised with the variable ${\tilde{b}}$ as ${{\tilde{a}}}={{\tilde{a}}}({\mathcal {X}},{\tilde{b}})$ for $b\leqq {\tilde{b}}\leqq b'$ or with the variable ${{\tilde{a}}}$ as ${\tilde{b}}={\tilde{b}}({\mathcal {X}},{{\tilde{a}}})$ for $a\leqq {{\tilde{a}}}\leqq a'$. Then by applying Lemma A.1 and (5.5) one obtains the following identities:

$$\begin{aligned} \int _{\Gamma _{(a,b)}^{(a',b')}} \frac{\mathrm{d}s}{|\nabla x|}=\frac{1}{k_6{\mathsf {t}}^{\frac{1}{4}}} \int _{b}^{b'} \frac{\mathrm{d}{\tilde{b}}}{|\partial _a {\mathcal {X}}|} \ \ \text {or} \ \ \int _{\Gamma _{(a,b)}^{(a',b')}} \frac{\mathrm{d}s}{|\nabla x|}=\frac{1}{k_6{\mathsf {t}}^{\frac{1}{4}}} \int _{a}^{a'} \frac{\mathrm{d}{\tilde{b}}}{|\partial _b {\mathcal {X}}|}. \end{aligned}$$

(B.3)

Step 1 The normal component for left part of the sides, the core, and the bottom, of the self-similar zone. We first derive rough estimates that will be used in the next steps. Fix $(X,Y)\in Z_0^c$ such that either $-\epsilon \leqq x(a,b)-x^*\leqq K{\mathsf {t}}^{3/2}$, or, $K{\mathsf {t}}^{3/2}\leqq x-x^*\leqq \epsilon$ and $b<0$ and $N_2 |{\mathcal {X}}|\leqq |b|^2$. Note that in the second case, from (5.11), one has necessarily $|a|\approx |b|^{2/3}\gg 1$. Let $\Gamma$ denote the part of the curve $\Gamma [x]$ which joins the boundary of the upper half plane and (X, Y). We decompose it in two parts:

$$\begin{aligned} \Gamma _1:= \Gamma \cap Z_0, \ \ \Gamma _2=\Gamma \cap Z_0^c, \ \ {\mathcal {Y}}= k_6{\mathsf {t}}^{\frac{1}{4}}\left( \int _{\Gamma _1} \frac{\mathrm{d}s}{|\nabla x|}+\int _{\Gamma _2} \frac{\mathrm{d}s}{|\nabla x|} \right) . \end{aligned}$$

(B.4)

The integral in $Z_0$ is at distance one to $(X_0,Y_0)$, and everything then remains regular:

$$\begin{aligned} \int _{\Gamma _1} \frac{\mathrm{d}s}{|\nabla x|}=O(1), \ \ \partial _{{\mathcal {X}}}\left( \int _{\Gamma _1} \frac{\mathrm{d}s}{|\nabla x|}\right) =O\left( {\mathsf {t}}^{\frac{3}{2}}\right) . \end{aligned}$$

(B.5)

In $Z_0^c$, from Lemma 5.3 the curve $\Gamma [x]$ lies in $Z_1$, so it can be parametrised with the variable ${\tilde{b}}$. Also, from (B.3), (5.20) and, as $|{\tilde{b}}|{\mathsf {t}}^{3/4}\ll 1$ for $-\delta {\mathsf {t}}^{-3/4}\leqq {\tilde{b}}\leqq b$,

$$\begin{aligned} k_6{\mathsf {t}}^{\frac{1}{4}}\int _{\Gamma _2} \frac{\mathrm{d}s}{|\nabla x|}= & {} \int _{-\frac{\delta }{{\mathsf {t}}^{-\frac{3}{4}}}}^b \frac{\mathrm{d}{\tilde{b}}}{\partial _a{\mathcal {X}}} \ =\ \int _{-\frac{\delta }{{\mathsf {t}}^{-\frac{3}{4}}}}^b \frac{1+O\left( {\mathsf {t}}^{\frac{1}{12}}+|{\tilde{b}}|^{\frac{1}{3}}{\mathsf {t}}^{\frac{1}{4}}+|{\mathcal {X}}|^{\frac{1}{6}}{\mathsf {t}}^{\frac{1}{4}}\right) }{1+3\Psi _1^2( p^* ({\mathcal {X}}-{\tilde{b}}^2) )}\mathrm{d}{\tilde{b}}\nonumber \\= & {} \int _{-\infty }^b \frac{1+O\left( {\mathsf {t}}^{\frac{1}{12}}+|{\tilde{b}}|^{\frac{1}{4}}{\mathsf {t}}^{\frac{3}{16}}+|{\mathcal {X}}|^{\frac{1}{6}}{\mathsf {t}}^{\frac{1}{4}}\right) }{1+3\Psi _1^2( p^* ({\mathcal {X}}-{\tilde{b}}^2) )}\mathrm{d}{\tilde{b}}+O\left( \int _{-\infty }^{-\frac{\delta }{{\mathsf {t}}^{-\frac{3}{4}}}} \frac{\mathrm{d}{\tilde{b}}}{|{\tilde{b}}|^{\frac{4}{3}}} \right) \nonumber \\= & {} \int _{-\infty }^b \frac{1+O\left( {\mathsf {t}}^{\frac{1}{12}}+|{\tilde{b}}|^{\frac{1}{4}}{\mathsf {t}}^{\frac{3}{16}}+|{\mathcal {X}}|^{\frac{1}{6}}{\mathsf {t}}^{\frac{1}{4}} \right) }{1+3\Psi _1^2( p^* ({\mathcal {X}}-{\tilde{b}}^2) )}\mathrm{d}{\tilde{b}}+O\left( {\mathsf {t}}^{\frac{1}{4}} \right) , \end{aligned}$$

(B.6)

where we used the fact that $|{\mathcal {X}}|\leqq \epsilon {\mathsf {t}}^{-3/2}$, and that for ${\tilde{b}}\leqq -\delta {\mathsf {t}}^{-3/4}$, ${\tilde{b}}^2 \gg {\mathcal {X}}$ if $\epsilon$ is small enough, implying $\Psi _1^2(p^*({\mathcal {X}}-{\tilde{b}}^2))\approx |{\tilde{b}}|^{4/3}$. Hence, injecting (B.5) and (B.6) in (B.4),

$$\begin{aligned} {\mathcal {Y}}(a,b)= \int _{-\infty }^b \frac{1+O\left( {\mathsf {t}}^{\frac{1}{12}}+|{\tilde{b}}|^{\frac{1}{4}}{\mathsf {t}}^{\frac{3}{16}}+|{\mathcal {X}}|^{\frac{1}{6}}{\mathsf {t}}^{\frac{1}{4}} \right) }{1+3\Psi _1^2( p^* ({\mathcal {X}}-{\tilde{b}}^2) )}\mathrm{d}{\tilde{b}}+O\left( {\mathsf {t}}^{\frac{1}{4}}\right) . \end{aligned}$$

(B.7)

We now study the derivative of ${\mathcal {Y}}$. As ${\mathcal {X}}({{\tilde{a}}},{\tilde{b}})={\mathcal {X}}$ is inverted through ${{\tilde{a}}}={{\tilde{a}}}({\mathcal {X}},{\tilde{b}})$ one has

$$\begin{aligned} \frac{\partial }{\partial {\mathcal {X}}}_{|{\tilde{b}}}\left( {{\tilde{a}}}({\mathcal {X}},{\tilde{b}}) \right)= & {} \frac{1}{\partial _a {\mathcal {X}}({{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}})}, \ \ \frac{\partial }{\partial {\mathcal {X}}}_{|{\tilde{b}}} \left( \frac{1}{\partial _a {\mathcal {X}}({{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}})} \right) \\= & {} -\frac{\partial _{aa}{\mathcal {X}}({{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}}) }{(\partial _a {\mathcal {X}}({{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}}))^3} \end{aligned}$$

One deduces from (B.3), (5.20), (5.21) that

$$\begin{aligned} \frac{\partial }{\partial {\mathcal {X}}}_{|b} \left( k_6{\mathsf {t}}^{\frac{1}{4}}\int _{\Gamma _2} \frac{\mathrm{d}s}{|\nabla x|}\right)= & {} \frac{\partial }{\partial {\mathcal {X}}}_{|b} \left( \int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^b \frac{\mathrm{d}{\tilde{b}}}{\partial _a{\mathcal {X}}({{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}})}\right) \\= & {} -\int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^b \frac{\partial _{aa}{\mathcal {X}}({{\tilde{a}}},{\tilde{b}})\mathrm{d}{\tilde{b}}}{(\partial _a {\mathcal {X}}({{\tilde{a}}},{\tilde{b}}))^3}\\= & {} 6p^*\int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^b \frac{\Psi _1\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2 \right) \right) }{\left( 1+3\Psi _1^2\left( p^*({\mathcal {X}}-{\tilde{b}}^2) \right) \right) ^3} \\&\left( 1+O\left( {\mathsf {t}}^{\frac{1}{12}}+|{\mathcal {X}}|^{\frac{1}{6}}{\mathsf {t}}^{\frac{1}{4}}+|{\tilde{b}}|^{\frac{1}{3}}{\mathsf {t}}^{\frac{1}{4}} \right) \right) \mathrm{d}{\tilde{b}}\\&+\int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^b \frac{O\left( {\mathsf {t}}^{\frac{1}{12}} \right) }{\left( 1+3\Psi _1^2\left( p^*({\mathcal {X}}-{\tilde{b}}^2) \right) \right) ^3} \mathrm{d}{\tilde{b}}. \end{aligned}$$

Note that in the integrals above, for ${\tilde{b}}\leqq -\delta {\mathsf {t}}^{-3/4}$ there holds $\Psi _1(p^*({\mathcal {X}}-{\tilde{b}}^2)) > rsim {\tilde{b}}^{2/3}$. Hence, integrating from infinity instead of $-\delta {\mathsf {t}}^{-3/4}$ produces an error which is $O({\mathsf {t}}^{7/4})$ and

$$\begin{aligned} \frac{\partial }{\partial {\mathcal {X}}}_{|b} \left( k_6{\mathsf {t}}^{\frac{1}{4}}\int _{\Gamma _2} \frac{\mathrm{d}s}{|\nabla x|}\right)&= 6p^*\int _{-\infty }^b \frac{\Psi _1\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2 \right) \right) }{\left( 1+3\Psi _1^2\left( p^*({\mathcal {X}}-{\tilde{b}}^2) \right) \right) ^3}\\&\quad \left( 1+O\left( {\mathsf {t}}^{\frac{1}{12}}+|{\mathcal {X}}|^{\frac{1}{6}}{\mathsf {t}}^{\frac{1}{4}}+|{\tilde{b}}|^{\frac{1}{4}}{\mathsf {t}}^{\frac{3}{16}} \right) \right) \mathrm{d}{\tilde{b}}\\&\quad +\int _{-\infty }^b \frac{O\left( {\mathsf {t}}^{\frac{1}{12}} \right) }{\left( 1+3\Psi _1^2\left( p^*({\mathcal {X}}-{\tilde{b}}^2) \right) \right) ^3} \mathrm{d}{\tilde{b}}+O\left( {\mathsf {t}}^{\frac{7}{4}} \right) \end{aligned}$$

Therefore, injecting (5.20), (B.5) and the above identity in (B.4), we get

$$\begin{aligned} \partial _{a}{\mathcal {Y}}(a,b) \,= & {} \, \partial _{a}{\mathcal {X}}\left( k_6{\mathsf {t}}^{\frac{1}{4}}\frac{\partial }{\partial {\mathcal {X}}} \int _{\Gamma _1} \frac{\mathrm{d}s}{|\nabla x|} +\frac{\partial }{\partial {\mathcal {X}}}_{|b} \left( k_6{\mathsf {t}}^{\frac{1}{4}}\int _{\Gamma _2} \frac{\mathrm{d}s}{|\nabla x|}\right) \right) \nonumber \\= & {} \partial _{a}{\mathcal {X}}\Biggl ( 6 p^*\int _{-\infty }^b \frac{\Psi _1\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2 \right) \right) }{\left( 1+3\Psi _1^2\left( p^*({\mathcal {X}}-{\tilde{b}}^2) \right) \right) ^3}\nonumber \\&\left( 1+O\left( {\mathsf {t}}^{\frac{1}{12}}+|{\mathcal {X}}|^{\frac{1}{6}}{\mathsf {t}}^{\frac{1}{4}}+|{\tilde{b}}|^{\frac{1}{4}}{\mathsf {t}}^{\frac{3}{16}} \right) \right) \mathrm{d}{\tilde{b}}\nonumber \\&+\int _{-\infty }^b \frac{O\left( {\mathsf {t}}^{\frac{1}{12}} \right) }{\left( 1+3\Psi _1^2\left( p^*({\mathcal {X}}-{\tilde{b}}^2) \right) \right) ^3} \mathrm{d}{\tilde{b}}+O\left( {\mathsf {t}}^{\frac{7}{4}} \right) \Biggr ) \nonumber \\= & {} 6p^* \left( 1+3\Psi _1^2\left( p^*\left( {\mathcal {X}}-b^2 \right) \right) \right) \nonumber \\&\Biggl ( \int _{-\infty }^b \frac{O\left( {\mathsf {t}}^{\frac{1}{12}} \right) }{\left( 1+3\Psi _1^2\left( p^*({\mathcal {X}}-{\tilde{b}}^2) \right) \right) ^3} \mathrm{d}{\tilde{b}}+O({\mathsf {t}}^{\frac{7}{4}}) \nonumber \\&+ \int _{-\infty }^{b} \frac{\Psi _1\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2 \right) \right) }{\left( 1+3\Psi _1^2\left( p^*({\mathcal {X}}-{\tilde{b}}^2) \right) \right) ^3} \nonumber \\&\quad \left( 1+O\left( {\mathsf {t}}^{\frac{1}{12}}+|{\mathcal {X}}|^{\frac{1}{6}}{\mathsf {t}}^{\frac{1}{4}}+|{\tilde{b}}|^{\frac{1}{4}}{\mathsf {t}}^{\frac{3}{16}}+{\mathsf {t}}^{\frac{1}{4}}|b|^{\frac{1}{3}} \right) \right) \mathrm{d}{\tilde{b}}\Biggr ). \end{aligned}$$

(B.8)

The estimates (5.27) and (5.28), and (1.17), prove (5.27) and (5.28) for the first case.

Step 2 The normal component for the sides of the self-similar zone. We now prove the estimates (5.30) and (5.31) in the second case described by (5.29). The parametrisation of the curve $\Gamma [x]$ has to be done more carefully. In the case of the left side, i.e. $x<0$, we are in the case considered in Step 1, which has already been covered. In the case of the right side, i.e. $x>0$, we are in case (ii) of Lemma 5.3, and then inside $Z_0^c$, the curve $\Gamma$ can be decomposed in five curves, $\Gamma _i$ for $i=1,...,5$ that enjoy the properties described there. We use different variables to parametrise the curves $\Gamma _i$, applying Lemma 5.3. On $\Gamma _1$ we use the variable ${\tilde{b}}$, on $\Gamma _2$ ${{\tilde{a}}}$, on $\Gamma _3$ ${\tilde{b}}$, on $\Gamma _4$ ${{\tilde{a}}}$ and on $\Gamma _5$ ${\tilde{b}}$. Without loss of generality for the argument, we assume that $({{\overline{a}}},{{\overline{b}}})$ is located on $\Gamma _2$. Indeed, treating the case of three or more different parametrisation can be done the very same way. We consider a point (a, b) close to $({{\overline{a}}},{{\overline{b}}})$, which still belongs to $\Gamma _2$ (up to changing slightly the constants in the definition of $Z_1$ and $Z_2$). We denote by $\Gamma _0$ the part of the curve outside $Z_0^c$. Hence, from (B.3),

$$\begin{aligned} {\mathcal {Y}}=k_6{\mathsf {t}}^{\frac{1}{4}} \int _{\Gamma _0} \frac{\mathrm{d}s}{|\nabla x|}+\int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^{b_1} \frac{\mathrm{d}{\tilde{b}}}{\partial _a {\mathcal {X}}}+\int _{a_1}^{a} \frac{\mathrm{d}{{\tilde{a}}}}{\partial _b {\mathcal {X}}}, \end{aligned}$$

(B.9)

where we recall that $(a_1,b_1)$ is the endpoint of $\Gamma _1$ and the starting point of $\Gamma _2$, defined in Lemma 5.3. The integral over $\Gamma _0$ is at distance one to $(X_0,Y_0)$ and hence, in a zone where everything remains regular,

$$\begin{aligned} \int _{\Gamma _0} \frac{\mathrm{d}s}{|\nabla x|}=O(1), \ \ \partial _{{\mathcal {X}}}\left( \int _{\Gamma _0} \frac{\mathrm{d}s}{|\nabla x|}\right) =O\left( {\mathsf {t}}^{\frac{3}{2}}\right) . \end{aligned}$$

(B.10)

We now consider the second integral, corresponding to the part $\Gamma _1$ of the curve joining the points $(a_{in},-\delta {\mathsf {t}}^{-3/4})$ and $(a_1,b_1)$. Since this part is in $Z_1$, one obtains from (5.20), injecting (5.67), that

$$\begin{aligned} \int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^{b_1} \frac{\mathrm{d}{\tilde{b}}}{\partial _a {\mathcal {X}}}= & {} \int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^{b_1} \frac{\mathrm{d}{\tilde{b}}}{1+3\Psi _1^2(p^*({\mathcal {X}}-{\tilde{b}}^2))} \nonumber \\&\left( 1+O\left( {\mathsf {t}}^{\frac{1}{12}}|{\mathcal {X}}|^{\frac{1}{18}}+ {\mathsf {t}}^{\frac{1}{12}}|{\tilde{b}}|^{\frac{1}{9}} \right) \right) \end{aligned}$$

(B.11)

We turn to the third integral, corresponding to the part $\Gamma _2$ of the curve joining the points $(a_{1},b_1)$ and (a, b). There, since this part is in $Z_2$, one obtains from (5.23) that

$$ \int _{a_1}^{a} \frac{\mathrm{d}{{\tilde{a}}}}{\partial _b {\mathcal {X}}} = \int _{a_1}^a \frac{\mathrm{d}{{\tilde{a}}}}{2\sqrt{{\mathcal {X}}-{{\tilde{a}}}-p^{*2}{{\tilde{a}}}^3}} \left( 1+O\left( |{\mathcal {X}}|^{\frac{1}{6}}{\mathsf {t}}^{\frac{1}{4}}\right) \right) . $$

In $Z_2$, one has from (5.10) that $|{{\tilde{a}}}|\ll |{\tilde{b}}|^{2/3}$ and $|{\tilde{b}}|\gg 1$, so that from (5.11) one has

$$\begin{aligned} {\mathcal {X}}\approx |{\tilde{b}}|^2\gg |{{\tilde{a}}}|^3 \ \ \text {and} \ \ \sqrt{{\mathcal {X}}-{{\tilde{a}}}-p^{*2}{{\tilde{a}}}^3}\approx {\mathcal {X}}^{1/2} \end{aligned}$$

(B.12)

uniformly for ${{\tilde{a}}}$ in $Z_2$, as well as $|a_1|,|a|\ll {\mathcal {X}}^{1/3}$. Using these bounds, one infers from the above identity that

$$\begin{aligned} \int _{a_1}^{a} \frac{\mathrm{d}{{\tilde{a}}}}{\partial _b {\mathcal {X}}}= & {} \int _{a_1}^a \frac{\mathrm{d}{{\tilde{a}}}}{2\sqrt{{\mathcal {X}}-{{\tilde{a}}}-p^{*2}{{\tilde{a}}}^3}} +O\left( \frac{|a-a_1|}{|{\mathcal {X}}|^{\frac{1}{2}}} |{\mathcal {X}}|^{\frac{1}{6}}{\mathsf {t}}^{\frac{1}{4}}\right) \\= & {} \int _{a_1}^a \frac{\mathrm{d}{{\tilde{a}}}}{2\sqrt{{\mathcal {X}}-{{\tilde{a}}}-p^{*2}{{\tilde{a}}}^3}} +O\left( |{\mathcal {X}}|^{-\frac{1}{6}+\frac{1}{6}}{\mathsf {t}}^{\frac{1}{4}}\right) . \end{aligned}$$

Also, from the identity (5.15), using the above bound (B.12),

$$\begin{aligned}&\left| a+\frac{1}{p^*}\Psi _1\left( p^*\left( {\mathcal {X}}-b^2\right) \right) \right| +\left| a+\frac{1}{p^*}\Psi _1\left( p^*\left( {\mathcal {X}}-b^2\right) \right) \right| \lesssim |{\mathcal {X}}|^{\frac{1}{3} +\frac{1}{18}}{\mathsf {t}}^{\frac{1}{12}}, \nonumber \\&\left| \int _{a}^{-\frac{1}{p^*}\Psi _1\left( p^*\left( {\mathcal {X}}-b^2\right) \right) } \frac{\mathrm{d}{{\tilde{a}}}}{\sqrt{{\mathcal {X}}-{{\tilde{a}}}-p^{*2}{{\tilde{a}}}^3}} \right| \nonumber \\&\quad \lesssim \frac{\left| a+\frac{1}{p^*}\Psi _1\left( p^*\left( {\mathcal {X}}-b^2\right) \right) \right| }{|{\mathcal {X}}|^{\frac{1}{2}}}\lesssim |{\mathcal {X}}|^{-\frac{1}{6}+\frac{1}{18}}{\mathsf {t}}^{\frac{1}{12}}, \nonumber \\&\left| \int _{a_1}^{-\frac{1}{p^*}\Psi _1\left( p^*\left( {\mathcal {X}}-4b_1^2\right) \right) } \frac{\mathrm{d}{{\tilde{a}}}}{\sqrt{{\mathcal {X}}-{{\tilde{a}}}-p^{*2}{{\tilde{a}}}^3}} \right| \nonumber \\&\quad \lesssim \frac{\left| a_1+\frac{1}{p^*}\Psi _1\left( p^*\left( {\mathcal {X}}-4b_1^2\right) \right) \right| }{|{\mathcal {X}}|^{\frac{1}{2}}}\lesssim |{\mathcal {X}}|^{-\frac{1}{6}+\frac{1}{18}}{\mathsf {t}}^{\frac{1}{12}}. \end{aligned}$$

(B.13)

Therefore

$$ \int _{a_1}^{a} \frac{\mathrm{d}{{\tilde{a}}}}{\partial _b {\mathcal {X}}} =\int _{-\frac{1}{p^*}\Psi _1\left( p^*\left( {\mathcal {X}}-4b_1^2\right) \right) }^{-\frac{1}{p^*}\Psi _1\left( p^*\left( {\mathcal {X}}-b^2\right) \right) } \frac{\mathrm{d}{{\tilde{a}}}}{2\sqrt{{\mathcal {X}}-{{\tilde{a}}}-p^{*2}{{\tilde{a}}}^3}} +O\left( |{\mathcal {X}}|^{-\frac{1}{6}+\frac{1}{18}}{\mathsf {t}}^{\frac{1}{12}} \right) . $$

We now change variables in the above integral, taking ${\tilde{b}}=-\sqrt{{\mathcal {X}}-{{\tilde{a}}}-p^{*2}{{\tilde{a}}}^3}$. The left and right endpoints of the integral are precisely $b_1$ and b, and this produces

$$\begin{aligned} {{\tilde{a}}}= & {} -\frac{1}{p^*} \Psi _1\left( p^*({\mathcal {X}}-{\tilde{b}}^2)\right) , \ \ \mathrm{d}{\tilde{b}}=\frac{1}{2} \frac{1+3p^{*2}{{\tilde{a}}}^2}{\sqrt{{\mathcal {X}}-{{\tilde{a}}}-p^{*2}{{\tilde{a}}}^3}}d{{\tilde{a}}}, \\ \int _{a_1}^{a} \frac{\mathrm{d}{{\tilde{a}}}}{\partial _b {\mathcal {X}}}= & {} \int _{b_1}^{b} \frac{\mathrm{d}{\tilde{b}}}{1+3\Psi _1^2\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2\right) \right) }+O\left( |{\mathcal {X}}|^{-\frac{1}{6}+\frac{1}{18}}{\mathsf {t}}^{\frac{1}{12}} \right) . \end{aligned}$$

We inject the identities (B.10), (B.11) and the above identity in the expression (B.9), giving the following expression for ${\mathcal {Y}}$:

$$\begin{aligned} {\mathcal {Y}}= & {} \int _{-\frac{\delta }{{\mathsf {t}}^{\frac{3}{4}}}}^{b} \frac{\mathrm{d}{\tilde{b}}\left( 1+O\left( |{\mathcal {X}}|^{\frac{1}{18}}{\mathsf {t}}^{\frac{1}{12}}+ |{\tilde{b}}|^{\frac{1}{9}}{\mathsf {t}}^{\frac{1}{12}}\right) \right) }{1+3\Psi _1^2 \left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2\right) \right) }+O\left( |{\mathcal {X}}|^{-\frac{1}{6}+\frac{1}{18}}{\mathsf {t}}^{\frac{1}{12}} \right) \nonumber \\= & {} {\mathcal {Y}}^\Theta (a,b) \left( 1+O\left( {\mathsf {t}}^{\frac{1}{12}}|{\mathcal {X}}|^{\frac{1}{18}}\right) \right) . \end{aligned}$$

(B.14)

Here we used (5.36). This shows the first desired bound in (5.30). We now consider the derivatives of ${\mathcal {Y}}$. The point $(a_1,b_1)$ changes as ${\mathcal {X}}$ changes, but the identity ${\mathcal {X}}(a_1,b_1)={\mathcal {X}}(a,b)$ ensures that

$$\begin{aligned} \partial _{\mathcal {X}}a_1\partial _a{\mathcal {X}}(a_1,b_1)+\partial _{\mathcal {X}}b_1 \partial _b {\mathcal {X}}(a_1,b_1)=1. \end{aligned}$$

(B.15)

Hence, differentiating the sum of the two leading order integrals in (B.9) one obtains

$$\begin{aligned}&{\mathsf {t}}^{\frac{1}{4}}k_6\partial _a \left( \int _{\Gamma _1} \frac{\mathrm{d}s}{|\nabla x|}+\int _{\Gamma _2} \frac{\mathrm{d}s}{|\nabla x|}\right) \nonumber \\&\quad = \partial _a \left( \int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^{b_1(a,b)} \frac{\mathrm{d}{\tilde{b}}}{\partial _a {\mathcal {X}}({{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}})}+\int _{a_1(a,b)}^a -\frac{\mathrm{d}{{\tilde{a}}}}{\partial _b {\mathcal {X}}({{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}))} \right) \nonumber \\&\quad = \partial _a {\mathcal {X}}(a,b) \nonumber \\&\qquad \Biggl (\int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^{b_1} \partial _{\mathcal {X}}\left( \frac{1}{\partial _a {\mathcal {X}}({{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}})}\right) \mathrm{d}{\tilde{b}}+\int _{a_1}^{a} \partial _{\mathcal {X}}\left( -\frac{1}{\partial _b {\mathcal {X}}({{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}) )}\right) d{{\tilde{a}}}\nonumber \\&\qquad \qquad +\frac{\partial _{\mathcal {X}}b_1}{\partial _a{\mathcal {X}}(a_1,b_1)}+\frac{\partial _{\mathcal {X}}a_1}{\partial _b{\mathcal {X}}(a_1,b_1)} \Biggr )-\frac{1}{\partial _b {\mathcal {X}}(a,b)} \nonumber \\&\quad = \partial _a {\mathcal {X}}(a,b) \nonumber \\&\qquad \Biggl (\int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^{b_1} \frac{-\partial _{aa}{\mathcal {X}}({{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}})\mathrm{d}{\tilde{b}}}{\left( \partial _a {\mathcal {X}}({{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}})\right) ^3} +\int _{a_1}^{a} \frac{\partial _{bb}{\mathcal {X}}({{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}) )d{{\tilde{a}}}}{\left( \partial _b {\mathcal {X}}({{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}) )\right) ^3} \nonumber \\&\qquad \quad + \frac{1}{\partial _a {\mathcal {X}}(a_1,b_1) \partial _b {\mathcal {X}}(a_1,b_1)}\Biggr ) \nonumber \\&\qquad -\frac{1}{\partial _b {\mathcal {X}}(a,b)}. \end{aligned}$$

(B.16)

The first integral is located in $Z_1$ with $|a|\gg 1$, hence from (5.20) and (5.21),

$$\begin{aligned} \begin{aligned}&\int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^{b_1} \frac{-\partial _{aa}{\mathcal {X}}\left( {{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}}\right) }{\left( \partial _a {\mathcal {X}}\left( {{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}}\right) \right) ^3}\mathrm{d}{\tilde{b}}\\&\quad = 6p^* \int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^{b_1} \frac{\Psi _1\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2\right) \right) \left( 1+O\left( {\mathsf {t}}^{\frac{1}{12}}|{\mathcal {X}}|^{\frac{1}{18}} +{\mathsf {t}}^{\frac{1}{12}}|{\tilde{b}}|^{\frac{1}{9}}\right) \right) }{\left( 1 +3\Psi _1^2\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2\right) \right) \right) ^3}\mathrm{d}{\tilde{b}}. \end{aligned} \end{aligned}$$

(B.17)

The second integral is located in $Z_2$, hence from (5.23) and (5.24),

$$ \int _{a_1}^{a} \frac{\partial _{bb}{\mathcal {X}}\left( {{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}) \right) }{\left( \partial _b {\mathcal {X}}\left( {{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}) \right) \right) ^3}d{{\tilde{a}}}= -\frac{1}{4} \int _{a_1}^{a} \frac{1+O\left( |{\mathcal {X}}|^{\frac{1}{6}}{\mathsf {t}}^{\frac{1}{4}}\right) }{\left( {\mathcal {X}}-{{\tilde{a}}}-p^{*2}{{\tilde{a}}}^3 \right) ^{\frac{3}{2}}}d{{\tilde{a}}}. $$

From (B.12), (B.13),

$$\begin{aligned}&\int _{a_1}^{a} \frac{\partial _{bb}{\mathcal {X}}\left( {{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}) \right) }{\left( \partial _b {\mathcal {X}}\left( {{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}) \right) \right) ^3}d{{\tilde{a}}}\\&\quad = -\frac{1}{4} \int _{-\frac{1}{p^*} \Psi _1\left( p^*\left( {\mathcal {X}}-4b_1^2\right) \right) }^{-\frac{1}{p^*} \Psi _1\left( p^*\left( {\mathcal {X}}-b^2\right) \right) } \frac{\mathrm{d}{{\tilde{a}}}}{\left( {\mathcal {X}}-{{\tilde{a}}}-p^{*2}{{\tilde{a}}}^3 \right) ^{\frac{3}{2}}} +O\left( |a-a_1|\frac{|{\mathcal {X}}|^{\frac{1}{6}}{\mathsf {t}}^{\frac{1}{4}}}{|{\mathcal {X}}|^{\frac{3}{2}}} \right) \\&\qquad +O\left( \left| \frac{\left| a+\frac{1}{p^*} \Psi _1 \left( p^*\left( {\mathcal {X}}-b^2 \right) \right) \right| +\left| a_1+\frac{1}{p^*} \Psi _1 \left( p^*\left( {\mathcal {X}}-4b_1^2 \right) \right) \right| }{|{\mathcal {X}}|^{\frac{3}{2}}} \right| \right) \\&\quad = -\frac{1}{4} \int _{-\frac{1}{p^*} \Psi _1\left( p^*\left( {\mathcal {X}}-4b_1^2\right) \right) }^{-\frac{1}{p} \Psi _1\left( p^*\left( {\mathcal {X}}-b^2\right) \right) } \frac{\mathrm{d}{{\tilde{a}}}}{\left( {\mathcal {X}}-{{\tilde{a}}}-p^{*2}{{\tilde{a}}}^3 \right) ^{\frac{3}{2}}}+O\left( \left| {\mathcal {X}}\right| ^{-\frac{7}{6}+\frac{1}{18}}{\mathsf {t}}^{\frac{1}{12}} \right) . \end{aligned}$$

We now change variables, taking ${\tilde{b}}=-\sqrt{{\mathcal {X}}-{{\tilde{a}}}-p^{*2}{{\tilde{a}}}^3}$. Note that this change of variables ensures ${\mathcal {X}}^\Theta ({{\tilde{a}}},{\tilde{b}})={{\tilde{a}}}+p^{*2}{{\tilde{a}}}^3+{\tilde{b}}^2=Cte={\mathcal {X}}$, and

$$ -\frac{1}{4} \frac{1}{\left( {\mathcal {X}}-{{\tilde{a}}}-p^{*2}{{\tilde{a}}}^3 \right) ^{\frac{3}{2}}}=\frac{\partial _{bb}{\mathcal {X}}^\Theta }{(\partial _b {\mathcal {X}}^\Theta )^3}. $$

There holds in this case a general formula when integrating on the curve $\{{\mathcal {X}}(a,b)=Cte\}$, obtained by performing a change of variables and an integration by parts (note the signs $\partial _a {\mathcal {X}}^\Theta >0$ and $\partial _b {\mathcal {X}}^\Theta <0$ in the present case):

$$\begin{aligned} \int _{a_1}^{a_2} \frac{\partial _{bb}{\mathcal {X}}^\Theta }{(\partial _{b}{\mathcal {X}}^\Theta )^3}\mathrm{d}a&= - \int _{b_1}^{b_2} \frac{\partial _{bb}{\mathcal {X}}^\Theta \mathrm{d}b}{(\partial _{b}{\mathcal {X}}^\Theta )^2\partial _a {\mathcal {X}}^\Theta } \\&= - \int _{b_1}^{b_2} \left( \frac{\mathrm{d}}{\mathrm{d}b} \partial _b {\mathcal {X}}^\Theta +\frac{\partial _b {\mathcal {X}}^\Theta }{\partial _a {\mathcal {X}}^\Theta }\partial _{ba} {\mathcal {X}}^\Theta \right) \frac{\mathrm{d}b}{(\partial _{b} {\mathcal {X}}^\Theta )^2\partial _a {\mathcal {X}}^\Theta } \\&= -\int _{b_1}^{b_2} \frac{\mathrm{d}}{\mathrm{d}b} (\partial _b {\mathcal {X}}^\Theta )\frac{1}{(\partial _{b} {\mathcal {X}}^\Theta )^2\partial _a {\mathcal {X}}^\Theta } \mathrm{d}b- \int _{b_1}^{b_2} \frac{\partial _{ab} {\mathcal {X}}^\Theta }{\partial _b {\mathcal {X}}^\Theta (\partial _a {\mathcal {X}}^\Theta )^2} \\&= -\int _{b_1}^{b_2} \frac{\mathrm{d}}{\mathrm{d}b} \left( \partial _b {\mathcal {X}}^\Theta \frac{1}{(\partial _{b} {\mathcal {X}}^\Theta )^2\partial _a {\mathcal {X}}^\Theta }\right) \mathrm{d}a\\&\quad +\int _{b_1}^{b_2} \partial _b {\mathcal {X}}^\Theta \frac{\mathrm{d}}{\mathrm{d}b}\left( \frac{1}{(\partial _{b} {\mathcal {X}}^\Theta )^2\partial _a {\mathcal {X}}^\Theta }\right) \mathrm{d}b \\&\quad -\int _{b_1}^{b_2} \frac{\partial _{ab} {\mathcal {X}}^\Theta \mathrm{d}b}{\partial _b {\mathcal {X}}^\Theta (\partial _a {\mathcal {X}}^\Theta )^2} \\&= - \frac{1}{\partial _a {\mathcal {X}}^\Theta (a_2,b_2)\partial _b {\mathcal {X}}^\Theta (a_2,b_2)}+\frac{1}{\partial _a {\mathcal {X}}^\Theta (a_1,b_1)\partial _b {\mathcal {X}}^\Theta (a_1,b_1)} \\&\quad -2\int _{b_1}^{b_2} \left( \partial _{bb} {\mathcal {X}}^\Theta -\frac{\partial _b {\mathcal {X}}^\Theta }{\partial _a {\mathcal {X}}^\Theta }\partial _{ab} {\mathcal {X}}^\Theta \right) \frac{1}{(\partial _{b} {\mathcal {X}}^\Theta )^2\partial _a {\mathcal {X}}^\Theta } \mathrm{d}b\\&\quad -\int _{b_1}^{b_2} \left( \partial _{ab} {\mathcal {X}}^\Theta -\frac{\partial _b {\mathcal {X}}^\Theta }{\partial _{a} {\mathcal {X}}^\Theta }\partial _{aa} {\mathcal {X}}^\Theta \right) \frac{1}{\partial _b {\mathcal {X}}^\Theta (\partial _a {\mathcal {X}}^\Theta )^2}\mathrm{d}b \\&\quad -\int _{b_1}^{b_2} \frac{\partial _{ab} {\mathcal {X}}^\Theta }{\partial _b {\mathcal {X}}^\Theta (\partial _a {\mathcal {X}}^\Theta )^2} \\&= -\frac{1}{\partial _a {\mathcal {X}}^\Theta (a_2,b_2)\partial _b {\mathcal {X}}^\Theta (a_2,b_2)}+\frac{1}{\partial _a {\mathcal {X}}^\Theta (a_1,b_1)\partial _b {\mathcal {X}}^\Theta (a_1,b_1)} \\&\quad +2\int _{a_1}^{a_2} \frac{\partial _{bb} {\mathcal {X}}^\Theta }{(\partial _{b} {\mathcal {X}}^\Theta )^3} \mathrm{d}a+\int _{b_1}^{b_2} \frac{\partial _{aa} {\mathcal {X}}^\Theta }{(\partial _a {\mathcal {X}}^\Theta )^3}\mathrm{d}b, \end{aligned}$$

from which one deduces the change of parametrisation identity

$$\begin{aligned} \int _{a_1}^{a_2} \frac{\partial _{bb} {\mathcal {X}}^\Theta }{(\partial _{b} {\mathcal {X}}^\Theta )^3}\mathrm{d}a= & {} \frac{1}{\partial _a {\mathcal {X}}^\Theta (a_2,b_2)\partial _b {\mathcal {X}}^\Theta (a_2,b_2)}-\frac{1}{\partial _a {\mathcal {X}}^\Theta (a_1,b_1)\partial _b {\mathcal {X}}^\Theta (a_1,b_1)} \\&-\int _{b_1}^{b_2} \frac{\partial _{aa} {\mathcal {X}}^\Theta }{(\partial _a {\mathcal {X}}^\Theta )^3}\mathrm{d}b. \end{aligned}$$

Applied to our case, this produces, noticing that the endpoint, are $(a_1,b_1)$ and (a, b),

$$\begin{aligned}&-\frac{1}{4} \int _{-\frac{1}{p^*} \Psi _1\left( p^*\left( {\mathcal {X}}-4b_1^2\right) \right) }^{-\frac{1}{p^*} \Psi _1\left( p^*\left( {\mathcal {X}}-b^2\right) \right) } \frac{1}{\left( {\mathcal {X}}-{{\tilde{a}}}-p^{*2}{{\tilde{a}}}^3 \right) ^{\frac{3}{2}}}d{{\tilde{a}}}\\&\quad = \frac{1}{\partial _a{\mathcal {X}}^\Theta (a,b)\partial _b {\mathcal {X}}^\Theta (a,b)}-\frac{1}{\partial _a {\mathcal {X}}^\Theta (a_1,b_1)\partial _b {\mathcal {X}}^\Theta (a_1,b_1)}-\int _{b_1}^b \frac{\partial _{aa}{\mathcal {X}}^\Theta }{(\partial _a {\mathcal {X}}^\Theta )^3}\mathrm{d}{\tilde{b}}\\&\quad = \frac{1}{(1+3p^{*2}a^2)2b}-\frac{1}{(1+3p^{*2}a_1^2)2b_1} \\&\qquad +6p^* \int _{b_1}^b \frac{\Psi _1\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2\right) \right) }{\left( 1+3\Psi _1^2\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2 \right) \right) \right) ^3}\mathrm{d}{\tilde{b}}. \end{aligned}$$

Note that, since $(a_1,b_1)$ belong to both $Z_1$ and $Z_2$, from (B.12), (5.23) and (5.20),

$$\begin{aligned} \frac{1}{\partial _a {\mathcal {X}}(a_1,b_1)\partial _b{\mathcal {X}}(a_1,b_1)}= & {} \frac{1+O\left( |{\mathcal {X}}|^{\frac{1}{6}}{\mathsf {t}}^{\frac{1}{4}}\right) }{\left( 1+3p^{*2}a_1^2\right) 2b_1} \\= & {} \frac{1}{\left( 1+3p^{*2}a_1^2\right) 2b_1}+O\left( |{\mathcal {X}}|^{-\frac{7}{6}+\frac{1}{6}}{\mathsf {t}}^{\frac{1}{4}} \right) , \end{aligned}$$

so that

$$\begin{aligned} \int _{a_1}^{a} \frac{\partial _{bb}{\mathcal {X}}\left( {{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}) \right) }{\left( \partial _b {\mathcal {X}}\left( {{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}) \right) \right) ^3}d{{\tilde{a}}}= & {} \frac{1}{(1+3p^{*2}a^2)2b}-\frac{1}{\partial _a {\mathcal {X}}(a_1,b_1)\partial _b{\mathcal {X}}(a_1,b_1)} \\&+6p^* \int _{b_1}^b \frac{\Psi _1\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2\right) \right) }{\left( 1+3\Psi _1^2\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2 \right) \right) \right) ^3}\mathrm{d}{\tilde{b}}\\&+O\left( |{\mathcal {X}}|^{-\frac{7}{6}+\frac{1}{18}}{\mathsf {t}}^{\frac{1}{12}} \right) . \end{aligned}$$

From the above identity and (B.17) one concludes that

$$\begin{aligned}&\int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^{b_1} \frac{-\partial _{aa}{\mathcal {X}}\left( {{\tilde{a}}}({\mathcal {X}},{\tilde{b}}), {\tilde{b}}\right) }{\left( \partial _a {\mathcal {X}}\left( {{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}}\right) \right) ^3}\mathrm{d}{\tilde{b}}+\int _{a_1}^{a} \frac{\partial _{bb}{\mathcal {X}}\left( {{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}) \right) }{\left( \partial _b {\mathcal {X}}\left( {{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}) \right) \right) ^3}d{{\tilde{a}}}\\&\qquad +\frac{1}{\partial _a {\mathcal {X}}(a_1,b_1) \partial _b {\mathcal {X}}(a_1,b_1)}\\&\quad = 6p^*\int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^{b} \frac{\Psi _1\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2\right) \right) \left( 1+O\left( {\mathsf {t}}^{\frac{1}{12}}|{\mathcal {X}}|^{\frac{1}{18}}+{\mathsf {t}}^{\frac{1}{12}}|{\tilde{b}}|^{\frac{1}{9}}\right) \right) }{\left( 1+3\Psi _1^2\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2 \right) \right) \right) ^3}\mathrm{d}{\tilde{b}}\\&\qquad + \frac{1}{(1+3p^{*2}a^2)2b}+O\left( |{\mathcal {X}}|^{-\frac{7}{6}+\frac{1}{18}}{\mathsf {t}}^{\frac{1}{12}}\right) \end{aligned}$$

From (5.37), and using the fact that for ${\tilde{b}}\leqq -\delta {\mathsf {t}}^{3/4}$, one has $|\Psi _1\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2\right) \right) |\approx |{\tilde{b}}|^{2/3}$, and

$$\begin{aligned}&\int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^{b} \frac{\Psi _1\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2\right) \right) \left( 1+O\left( {\mathsf {t}}^{\frac{1}{12}}|{\mathcal {X}}|^{\frac{1}{18}}+{\mathsf {t}}^{\frac{1}{12}}|{\tilde{b}}|^{\frac{1}{9}}\right) \right) }{\left( 1+3\Psi _1^2\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2 \right) \right) \right) ^3}\mathrm{d}{\tilde{b}}\\&\quad = \int _{-\infty }^{b} \frac{\Psi _1\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2\right) \right) }{\left( 1+3\Psi _1^2\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2 \right) \right) \right) ^3}\mathrm{d}{\tilde{b}}+O\left( |{\mathcal {X}}|^{-\frac{7}{6}+\frac{1}{18}}{\mathsf {t}}^{\frac{1}{12}} \right) . \end{aligned}$$

Therefore

$$\begin{aligned}&\int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^{b_1} \frac{-\partial _{aa}{\mathcal {X}}\left( {{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}}\right) }{\left( \partial _a {\mathcal {X}}\left( {{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}}\right) \right) ^3}\mathrm{d}{\tilde{b}}+\int _{a_1}^{a} \frac{\partial _{bb}{\mathcal {X}}\left( {{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}) \right) }{\left( \partial _b {\mathcal {X}}\left( {{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}) \right) \right) ^3}d{{\tilde{a}}}\nonumber \\&\qquad +\frac{1}{\partial _a {\mathcal {X}}(a_1,b_1) \partial _b {\mathcal {X}}(a_1,b_1)} \nonumber \\&\quad = \int _{-\infty }^{b} \frac{\Psi _1\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2\right) \right) }{\left( 1+3\Psi _1^2\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2 \right) \right) \right) ^3}\mathrm{d}{\tilde{b}}+ \frac{1}{(1+3p^{*2}a^2)2b} \nonumber \\&\qquad +O\left( |{\mathcal {X}}|^{-\frac{7}{6}+\frac{1}{18}}{\mathsf {t}}^{\frac{1}{12}}\right) \end{aligned}$$

(B.18)

Since (a, b) is in $Z_2$, from (B.12), (5.23) and (5.20),

$$ \frac{\partial _a {\mathcal {X}}(a,b)}{(1+3p^{*2}a^2)2b}= \frac{1}{\partial _b {\mathcal {X}}(a,b)}+O\left( |{\mathcal {X}}|^{-\frac{1}{2}+\frac{1}{6}}{\mathsf {t}}^{\frac{1}{4}}\right) . $$

Injecting the two identities above, (B.17) and $|\partial _a{\mathcal {X}}|\lesssim |{\mathcal {X}}|^{2/3}$ in the identity (B.16) gives

$$\begin{aligned}&{\mathsf {t}}^{\frac{1}{4}}k_6\partial _a \left( \int _{\Gamma _1} \frac{\mathrm{d}s}{|\nabla x|}+\int _{\Gamma _2} \frac{\mathrm{d}s}{|\nabla x|}\right) \\&\quad = 6 p^*\partial _a {\mathcal {X}}(a,b) \int _{-\infty }^{b} \frac{\Psi _1\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2\right) \right) \mathrm{d}{\tilde{b}}}{\left( 1+3\Psi _1^2\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2 \right) \right) \right) ^3} \\&\qquad +O\left( |{\mathcal {X}}|^{-\frac{1}{2} +\frac{1}{18}}{\mathsf {t}}^{\frac{1}{12}} \right) \end{aligned}$$

From this identity, (B.9), (B.10) and (5.68), we have proved that

$$ \partial _a {\mathcal {Y}}=\partial _a {\mathcal {Y}}^\Theta +O\left( |{\mathcal {X}}|^{-\frac{1}{2} +\frac{1}{18}}{\mathsf {t}}^{\frac{1}{12}} \right) , $$

which is the second identity in (5.30) that we had to show. We now turn to the partial derivative with respect to b. From (B.9) and (B.15), and then injecting (B.18), (5.68), (5.69) and (5.67),

$$\begin{aligned}&{\mathsf {t}}^{\frac{1}{4}}k_6 \partial _b \left( \int _{\Gamma _1} \frac{\mathrm{d}s}{|\nabla x|}+\int _{\Gamma _2} \frac{\mathrm{d}s}{|\nabla x|}\right) \\&\quad = \partial _b \left( \int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^{b_1(a,b)} \frac{\mathrm{d}{\tilde{b}}}{\partial _a {\mathcal {X}}({{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}})}+\int _{a_1(a,b)}^a -\frac{\mathrm{d}{{\tilde{a}}}}{\partial _b {\mathcal {X}}({{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}))} \right) \\&\quad = \partial _b {\mathcal {X}}(a,b) \\&\qquad \left( \int _{-\frac{\delta }{{\mathsf {t}}^{3/4}}}^{b_1} \frac{-\partial _{aa}{\mathcal {X}}({{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}})\mathrm{d}{\tilde{b}}}{\left( \partial _a {\mathcal {X}}({{\tilde{a}}}({\mathcal {X}},{\tilde{b}}),{\tilde{b}})\right) ^3}+\int _{a_1}^{a} \frac{\partial _{bb}{\mathcal {X}}({{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}))d{{\tilde{a}}}}{\left( \partial _b {\mathcal {X}}({{\tilde{a}}},{\tilde{b}}({\mathcal {X}},{{\tilde{a}}}))\right) ^3} \right. \\&\qquad \qquad \left. +\frac{1}{\partial _a {\mathcal {X}}(a_1,b_1) \partial _b {\mathcal {X}}(a_1,b_1)} \right) \\&\quad = \partial _b {\mathcal {X}}(a,b) \\&\qquad \left( \int _{-\infty }^{b} \frac{\Psi _1\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2\right) \right) }{\left( 1+3\Psi _1^2\left( p^*\left( {\mathcal {X}}-{\tilde{b}}^2 \right) \right) \right) ^3}\mathrm{d}{\tilde{b}}+ \frac{1}{(1+3p^{*2}a^2)2b} \right. \\&\qquad \qquad \left. +O\left( |{\mathcal {X}}|^{-\frac{7}{6}+\frac{1}{18}}{\mathsf {t}}^{\frac{1}{12}}\right) \right) \\&\quad = \partial _b {\mathcal {Y}}^\Theta +O\left( |{\mathcal {X}}|^{-\frac{2}{3} +\frac{1}{18}} {\mathsf {t}}^{\frac{1}{12}} \right) , \end{aligned}$$

which was the last estimate (5.31) we had to show. We claim that the computations we performed for this right side of the self-similar zone can be adapted in a straightforward way in the case where one has to consider more parts of the curve $\Gamma$ inside $Z_0^c$ to parametrise; the integral over $\Gamma _3$, $\Gamma _4$ and $\Gamma _5$ can be treated the very same way, leading to the same result. $\quad \square$