Introduction

We are interested in the 2D Prandtl equations

$$\begin{aligned}&\partial _t U^P + U^P \partial _x U^P + V^P \partial _y U^P - \partial _y^2 U^P = \partial _t U^E +U^E \partial _x U^E,\nonumber \\&\quad \quad \partial _x U^P + \partial _y V^P = 0, \end{aligned}$$
(1)

set in the domain \(\Omega = {\mathbb {T}}\times \mathbb {R}_+\), completed with boundary conditions

$$\begin{aligned} U^P\vert _{y=0} = V^P\vert _{y=0} = 0, \quad \lim _{y \rightarrow +\infty } U^P = U^E. \end{aligned}$$
(2)

These equations are a degenerate Navier-Stokes model, introduced by Prandtl in 1904 to describe the boundary layer, which is the region of high velocity gradients that forms near solid boundaries in incompressible flows at high Reynolds number. They can be derived from the Navier-Stokes equation under the formal asymptotics

$$\begin{aligned} (u^\nu , v^\nu )(t,x,z)\approx & {} (U^P(t,x,z/\sqrt{\nu }), \sqrt{\nu } V^P(t,x,z/\sqrt{\nu })), \nonumber \\ (U^P,V^P)= & {} (U^P,V^P)(t,x,y), \end{aligned}$$
(3)

where \(\nu \) is the inverse Reynolds number, and \((u^\nu , v^\nu )\) is the Navier-Stokes solution. This asymptotics is supposed to apply to the flow in the boundary layer region: the typical scale \(\sqrt{\nu }\) of the boundary layer in this model is inspired by the heat part of the Navier-Stokes equation. Away from the boundary, one rather expects an inviscid asymptotics of the type

$$\begin{aligned} (u^\nu , v^\nu )(t,x,z) = (u^E, v^E)(t,x,z), \end{aligned}$$

where \((u^E, v^E)\) is the solution of the Euler equation. In order to match the two asymptotic expansions, one must impose the condition

$$\begin{aligned} \lim _{y \rightarrow +\infty } U^P(t,x,y) = U^E(t,x) := u^E(t,x,0), \end{aligned}$$

which yields the boundary condition for \(y \rightarrow \infty \) in (2). The other two boundary conditions at \(y = 0\) express the usual no-slip condition at the boundary. We refer to [6] for a more detailed derivation. Let us stress that the pressure in the Prandtl model is independent of y: its value is given by the pressure in the Euler flow at \(z=0\). This explains the right-hand side of (1), which depends only on tx, and is coherent with the third boundary condition in (2).

The Prandtl system (1)–(2) is very classical, as it appears in most textbooks on fluid dynamics. Still, it is well-known from physicists that its range of applications is narrow, due to underlying instabilities. Among those instabilities, one can mention the phenomenon of separation, which is related to the development of a reverse flow in the boundary layer [4, 8]. Another example is the so-called Tollmien-Schlichting wave, that is typical of viscous flows at high but finite Reynolds number [5, 15]. Of course, such instability mechanisms create difficulties at the PDE level, making the mathematical analysis of boundary layer theory an interesting topic. The two main problems that one needs to address are the well-posedness of the reduced model (1), and the validity of the asymptotics (3). We shall focus on the former in the present paper. About the validity of boundary layer expansions in the unsteady setting, there are many possible references, among which [11, 14, 16, 26, 31, 32]. About the steady setting, see the recent works [10, 17, 19].

To analyse the well-posedness of the Prandtl model is uneasy, even at the level of local in time smooth solutions. The key difference with Navier-Stokes is that there is no time evolution for the vertical velocity, which is recovered only through the divergence-free condition. Hence, the term \(v \partial _y u\) can be seen as a first order nonlinear operator in x. Moreover, this operator is not skew-symmetric in \(H^s\). As the diffusion in (1) is only transverse, this prevents the derivation of standard Sobolev estimates. The first rigorous study of the Prandtl equations goes back to Oleinik [29], who tackled the case of data \(U\vert _{t=0}\) that are monotonic in y. She established local well-posedness of the system using the so-called Crocco transform, a tricky change of variables and unknowns. Let us stress that such monotonicity assumption excludes the phenomenon of reverse flow and therefore prevents boundary layer separation. More recently, the local well-posedness result of Oleinik was revisited using the standard Eulerian form of the equations, see [1, 21, 27] for the local theory in Sobolev spaces.

The analysis of non-monotonic data is much more recent, and has experienced some strong impetus over the last years. Surprisingly, it was shown in [9] that the Prandtl system is ill-posed in the Sobolev setting (cf. [13, 18, 24] for improvements). Specifically, paper [9] centers on the linearization of (1)–(2) around shear flows, given by \((U,V) = (U_s(y),0)\). The linearized system reads

$$\begin{aligned} \begin{aligned} \partial _t u + U_s \partial _x u + U'_s v - \partial ^2_y u = 0, \\ \partial _x u + \partial _y v = 0, \\ u\vert _{y=0} = v\vert _{y=0} = 0, \quad \lim _{y \rightarrow +\infty } u = 0. \end{aligned} \end{aligned}$$
(4)

In the case where \(U_s\) has one non-degenerate critical point, one can show that (4) has unstable solutions of the form \(u(t,x,y) = \mathrm {e}^{\mathrm {i}k x} \mathrm {e}^{\sigma _k t} u_k(y)\) for k arbitrarily large and \(\mathfrak {R}\sigma _k \sim \lambda \sqrt{k}\). Such high frequency instability forbids the construction of Sobolev solutions. To obtain positive results, one must start from initial data \(u_\mathrm {in}\) that are strongly localized in Fourier, typically for which \(|\hat{u}_0(k,y)| \lesssim \mathrm {e}^{-\delta |k|^\gamma }\) for some positive \(\delta > 0\), \(\gamma \le 1\). Such localization condition corresponds to Gevrey regularity in x (Gevrey class \(1/\gamma \)). The first result in this direction is due to Sammartino and Caflisch [30], who established existence of local in time solutions in the analytic setting (\(\gamma = 1\)). See also the nice paper [22]. Note that the requirement for analyticity is natural in view of standard estimates. For instance, at the level of the linearized equations (4), one gets directly by testing against u that

$$\begin{aligned} \partial _t \Vert \hat{u}(t,k,\cdot )\Vert _{L^2_y} \le C |k| \, \Vert \hat{u}(t,k,\cdot )\Vert _{L^2_y} \end{aligned}$$

so that \(\Vert \hat{u}(t,k,\cdot )\Vert _{L^2_y} \le \mathrm {e}^{C|k|t} \Vert \hat{u}_0(k, \cdot )\Vert _{L^2_y}\). Hence, if \(\Vert \hat{u} _0(k,\cdot )\Vert _{L^2_y} \lesssim \mathrm {e}^{-\delta |k| }\), a uniform control will be provided as long as \(t \le \delta /C\).

To relax the analyticity condition is much harder. In the special case where \(u_\mathrm {in}\) has for each value of x a single non-degenerate critical point in y, the first author and N. Masmoudi proved the local well-posedness of system (1)–(2) for data that are in Gevrey class 7 / 4 with respect to x [12]. Well-posedness was extended to Gevrey class 2 in article [23], for data that are small perturbations of a shear flow with a single non-degenerate critical point. Note that this exponent (corresponding to \(\gamma = 1/2\)) is optimal in view of the instability mechanism of [9].

All the recent results mentioned above rely heavily on the structure of the initial data: monotonicity for the Sobolev setting, single non-degenerate critical points for the Gevrey setting. It is therefore natural to ask about the optimal regularity under which local well-posedness of the Prandtl equations holds, without additional structural assumption. This is the problem that we solve in the present paper: we establish the short-time well-posedness of the Prandtl equations for general data with Gevrey 2 regularity in x and Sobolev regularity in y. We recall once more that such regularity framework is the best possible. Indeed, from the results of [9], high frequency modes k in x may experience exponential growth with rate \(\sqrt{k}\). This means that to hope for short time stability, the amplitude of these modes should be \(O(\mathrm {e}^{-C\sqrt{k}})\), which is the Fourier translation of a Gevrey 2 requirement.

Result

Let \(\gamma \ge 1\), \(\tau > 0\), \(r \in \mathbb {R}\). For functions \(f = f(x)\) of one variable, we define the Gevrey norm

$$\begin{aligned} |f|_{\gamma ,\tau , r}^2 = \sum _{j \in \mathbb {N}} \left( \frac{\tau ^{j+1} (j{+}1)^r}{(j!)^{\gamma }} \right) ^2 \Vert f^{(j)} \Vert ^2_{L^2({\mathbb {T}})} \end{aligned}$$
(5)

and for functions \(f = f(x,y)\) of two variables, the norm

$$\begin{aligned} \Vert f \Vert _{\gamma ,\tau , r}^2 = \sum _{j \in \mathbb {N}} \left( \frac{\tau ^{j+1} (j{+}1)^r}{(j!)^{\gamma }} \right) ^2 \Vert \partial _x^j f \Vert ^2_j, \end{aligned}$$
(6)

where \(\Vert \cdot \Vert _j\), \(j \ge 0\), denotes a family of weighted \(L^2\) norms. Namely,

$$\begin{aligned} \Vert f \Vert _j^2 = \int _{{\mathbb {T}}\times \mathbb {R}^+} |f(x,y)|^2 \rho _j(y)\, \mathrm {d}x\, \mathrm {d}y, \end{aligned}$$
(7)

where \(\rho _j\), \(j \ge 0\), is the family of weights given by

$$\begin{aligned} \rho _0(y) = (1+y)^{2m}, \quad \rho _j(y) = \frac{\rho _{j-1}(y)}{\left( 1+\frac{y}{j^\alpha }\right) ^2} = \rho _0(y) \prod _{k=1}^j \left( 1 + \frac{y}{k^{\alpha }}\right) ^{-2}, \quad j \ge 1, \end{aligned}$$

for fixed constants \(\alpha \ge 0\) and \(m \ge 0\) chosen later (m large enough and \(\alpha \) matching the constraints found from the estimates). The need for this family of weights will be clarified later. Let us note that locally in y, this family of norms is comparable to more classical families such as

$$\begin{aligned} ||| f |||_{\gamma ,\tau , r}^2 = \sum _{j \in \mathbb {N}} \left( \frac{\tau ^{j+1} (j{+}1)^r}{(j!)^{\gamma }} \right) ^2 \Vert \partial _x^j f \Vert ^2_{L^2}. \end{aligned}$$
(8)

For instance, for functions f which are zero for \(|y|\ge M\), one has

$$\begin{aligned} \Vert f \Vert _{\gamma , \tau ,r} \le C_M ||| f |||_{\gamma ,\tau ,r}, \quad ||| f |||_{\gamma ,\tau ,r} \le C_{M,\tau '} \Vert f \Vert _{\gamma ,\tau ',r} \hbox { for any}\ \tau ' > \tau . \end{aligned}$$

The only difference is when y goes to infinity, where the family of weights \(\rho _j\) puts less constraints on the decay of the derivatives compared to a fixed weight \(\rho _0\) for derivatives of any order.

With these spaces, we can now state our main result.

Theorem 1

There exists m and \(\alpha \) such that: for all \(0< \tau _1 < \tau _0\), \(r \in \mathbb {R}\),  for all \(T_0 > 0\),   for all \(U^E\) satisfying

$$\begin{aligned}&\sup _{[0,T_0]} \, |\partial _t U^E|_{2,\tau _0,r} + |U^E|_{2,\tau _0,r}<+\infty ,\\&\sup _{[0,T_0]} \max _{l=0,\dots ,3} \Vert \partial _t^l(\partial _t + U^E\partial _x) U^E \Vert _{H^{6-2l}({\mathbb {T}})} < +\infty \end{aligned}$$

for all \(U^P_\mathrm {in}\) satisfying

$$\begin{aligned}&\Vert U^P_\mathrm {in}-U^E\vert _{t=0} \Vert _{2,\tau _0,r}< +\infty , \quad \Vert (1+y)\partial _y U^P_\mathrm {in}\Vert _{2,\tau _0,r}< +\infty , \\&\quad \Vert (1+y)^{m+6}\partial _y U^P_\mathrm {in}\Vert _{H^6({\mathbb {T}}\times \mathbb {R}_+)} < +\infty \end{aligned}$$

and under usual compatibility conditions (see the last remark below), there exists \(0 < T \le T_0\) and a unique solution \(U^P\) of (1)–(2) over (0, T) with initial data \(U^P_\mathrm {in}\) that satisfies

$$\begin{aligned}&\sup _{t \in [0,T]} \Vert U^P(t)-U^E(t) \Vert ^2_{2,\tau _1,r} + \sup _{t \in [0,T]} \Vert (1+y)\partial _y U^P(t) \Vert ^2_{2,\tau _1,r} \\&\quad + \int _0^T \Vert (1+y)\partial ^2_y U^P(t) \Vert ^2_{2,\tau _1,r} \, \mathrm {d}t < +\infty \end{aligned}$$

Remark

  • The main novelty of the theorem is that we reach the optimal Gevrey regularity although no structural assumption is made on the data: no monotonicity, or hypothesis on the number and order of the critical points is needed. Only Gevrey regularity of the data and natural compatiblity conditions are required.

  • Our method of proof, explained below, is inspired by the hyperbolic part of the Prandtl equations. It is based on both a tricky change of unknown and appropriate choice of test function. This method would also allow to recover the Sobolev well-posedness of the hyperbolic version of the Prandtl system by means of energy methods. As far as we know, the well-posedness of this inviscid Prandtl model had been only established in \(C^k\) spaces using the method of characteristics: see [20] for more. This part will be detailed elsewhere. In the case of the usual Prandtl equations studied here, our methodology has to be slightly modified to handle in an optimal way the diffusion term. Still, commutators are responsible for the loss of Sobolev regularity: only Gevrey 2 smoothness in x can be established.

  • There is a loss on the Gevrey radius \(\tau \) of the solutions through time, going from \(\tau _0\) to \(\tau _1\). This loss, which appears technical in the paper, is actually unavoidable. This is due to the instabilities described in [9]: exponential growth of perturbations at rate \(\sqrt{k}\) causes a decay of the Gevrey radius linearly with time.

  • Besides the regularity requirements mentioned in Theorem 1, the initial data must satisfy compatibility conditions. It is typical of parabolic problems in domains with boundaries, cf. [28, Chapter 3] for a general discussion. Here, the value of \(U^P_\mathrm {in}\) and of some of its derivatives at \(y=0\) cannot be arbitrary: they must be related to \(U^E\) accordingly to the equation and to the amount of regularity asked for u (with respect to the y-variable). Let us note that locally near \(y=0\), most of our estimates only involve \(U^P-U^E\) in \(L^2_t H^2_y\) (not mentioning the Gevrey regularity in x). Such estimates could be carried with the single compatibility condition \(U^P_\mathrm {in}\vert _{y=0}=0\). Still, the low norm \(\Vert (U^P, V^P) \Vert _{low}\) introduced in (19) involves more y-derivatives: its control through Lemma 15 implies therefore a few more compatibility conditions. For the sake of brevity, we do not provide their explicit expressions, and refer to [33, Proposition 2.3] for a detailed discussion on a variation of the Prandtl equations.

Outline of the strategy As mentioned earlier, our analysis of the Prandtl equations relies on the identification of new controlled quantities because the usual unknown u and kinetic energy do not give enough information. To help to identify the relevant quantities, it is a good idea to start from the study of the linearized system (4). After Fourier transform in x and Laplace transform in time, we are left with the ODE

$$\begin{aligned} (\lambda + \mathrm {i}k U_s) \partial _y \Psi - \mathrm {i}k U'_s \Psi - \partial ^3_y \Psi = u_\mathrm {in}\end{aligned}$$
(9)

where \(\Psi \) corresponds to the Fourier-Laplace transform of the stream function. At high frequencies k, a natural idea (although not legitimate in the end) is to neglect the diffusion term. We are then left with the first order ODE

$$\begin{aligned} (\lambda + \mathrm {i}k U_s) \partial _y \Psi - \mathrm {i}k U'_s \Psi = u_\mathrm {in}. \end{aligned}$$
(10)

We note that the standard estimate (based on taking \(\partial _y \overline{\Psi }\) as a test function) yields a control of the type

$$\begin{aligned} \mathfrak {R}\lambda \, \Vert \partial _y \Psi \Vert _{L^2}^2&\lesssim k \Vert \partial _y \Psi \Vert _{L^2} \Vert U'_s \Psi \Vert _{L^2} + \Vert u_\mathrm {in}\Vert _{L^2} \Vert \partial _y \Psi \Vert _{L^2} \\&\lesssim k \Vert \partial _y \Psi \Vert _{L^2}^2 + \Vert u_\mathrm {in}\Vert _{L^2} \Vert \partial _y \Psi \Vert _{L^2} \end{aligned}$$

where the last line comes from the Hardy inequality (as soon as \(|U'_s(y)| = \mathcal {O}(y^{-1})\) at infinity). Such bound ensures the solvability of the resolvent equation (10) only for \(\lambda \sim k\). This in turn yields a semigroup bound of the type \(\mathrm {e}^{C kt}\), only compatible with stability in the analytic setting.

To reach stability in lower regularity, an important point is to notice that the homogeneous equation has \(\Psi _s = (\lambda + i k U_s)\) as a special solution. With the integrating factor method in mind, it is then natural to set \(\Psi = (\lambda + i k U_s) \psi \). The first order equation (10) becomes

$$\begin{aligned} (\lambda + i k U_s)^2 \partial _y \psi = u_\mathrm {in}\end{aligned}$$

which is much better than the original formulation. Indeed, we can test the equation against \(\phi = \frac{1}{\lambda + i k U_s} \partial _y \overline{\psi }\) to obtain a control of \(\partial _y \psi \) in terms of \(u_\mathrm {in}\), and from there a control of \(\Psi \) for any \(\lambda > 0\).

Back to the full resolvent equation (9) we find for the same unknown \(\psi \)

$$\begin{aligned} (\lambda + \mathrm {i}k U_s)^2 \partial _y \psi - (\lambda + \mathrm {i}k U_s) \partial _y^3 \psi = u_\mathrm {in}+ [\lambda +\mathrm {i}k U_s,\partial _y^3] \psi . \end{aligned}$$

Testing again against \(\phi = \frac{1}{\lambda + i k U_s} \partial _y \overline{\psi }\), the LHS allows the control

$$\begin{aligned} \mathfrak {R}\lambda \, \Vert \partial _y \psi \Vert _{L^2}^2 + \Vert \partial _y^2 \psi \Vert _{L^2}^2. \end{aligned}$$

In the commutator at the RHS, the worst error term is \(3\mathrm {i}k \partial _y U_s \partial _y^2 \psi \), which is bounded as

$$\begin{aligned} C \frac{k}{|\mathfrak {R}\lambda |} \Vert \partial _y \psi \Vert _{L^2} \Vert \partial _y^2 \psi \Vert _{L^2} \le \frac{1}{2} \Vert \partial _y^2 \psi \Vert _{L^2}^2 + \frac{C^2k^2}{|\mathfrak {R}\lambda |^2} \Vert \partial _y \psi \Vert _{L^2}^2. \end{aligned}$$

We see that under the constraint \(\mathfrak {R}\lambda \sim k^{2/3}\), the estimate can be closed, and this can be shown to imply short time stability for data with Gevrey regularity 3 / 2. This estimate around a shear flow is detailed as Lemma 4.1 in [3].

In order to reach the optimal Gevrey exponent 2, we need to get rid of the commutator term containing \(\partial _y^2 \psi \), which comes with a worse control than \(\partial _y \psi \). To do so, we change a bit our new unknown \(\psi \): we now define \(\psi \) through the relation

$$\begin{aligned} \Psi = (\lambda +\mathrm {i}k U_s - \partial _y^2) \psi \end{aligned}$$
(11)

including the diffusion term. Hence, (9) becomes

$$\begin{aligned} (\lambda + \mathrm {i}k U_s - \partial _y^2)^2 \partial _y \psi + (\lambda + \mathrm {i}k U_s - \partial _y^2) (\mathrm {i}k U_s' \psi ) - \mathrm {i}k U_s (\lambda + \mathrm {i}k U_s - \partial _y^2) \psi = u_\mathrm {in}. \end{aligned}$$

Testing this time against the solution \(\phi \) of \((\lambda +\mathrm {i}k U_s - \partial _y^2) \phi = \partial _y \overline{\psi }\) (again with the diffusion term), the LHS yields the same control, but the error term is now

$$\begin{aligned} \mathfrak {R}\int [\lambda +\mathrm {i}k U_s - \partial _y^2, \mathrm {i}k U_s'] \psi \; \phi . \end{aligned}$$

From the definition of \(\phi \) it can be shown that \(\Vert \phi \Vert \lesssim \lambda ^{-1} \Vert \partial _y \psi \Vert \) so that the error can be bounded by

$$\begin{aligned} \frac{k}{|\mathfrak {R}\lambda |} \Vert \partial _y \psi \Vert _{L^2}^2. \end{aligned}$$

The estimate can now be closed for \(\mathfrak {R}\lambda \sim k^{1/2}\) yielding Gevrey regularity 2.

Obviously, such approach is no longer applicable as such to the nonlinear system (1)–(2): we not only lose the linearity of the equations, but the coefficients are no longer of shear flow type. They notably depend on t and x, which forbids an easy use of Fourier or Laplace transform. Rather than turning to the characterization of Gevrey spaces in the Fourier variable k, we consider norms based on the x-variable, see (5) and (6). Roughly, the idea is to work with time dependent norms, that is with the quantities

$$\begin{aligned} \Vert (U^P - U^E)(t) \Vert _{\gamma ,\tau (t),r}, \quad \tau (t) = \tau _0 \mathrm {e}^{-\beta t}. \end{aligned}$$

By differentiating j-times the first line of the Prandtl system, we can derive an equation on

$$\begin{aligned} u_j(t) \, := \, \frac{\tau (t)^{j+1} (j{+}1)^r}{(j!)^{\gamma }} \partial _x^j \left( U^P(t) - U^E(t) \right) \end{aligned}$$

that can be written as

$$\begin{aligned}&(\partial _t + \beta (j+1) ) u_j + U^P \partial _x u_j + V^P \partial _y u_j + v_j \partial _y U^P -\partial _y^2 u_j = F_j,\nonumber \\&\quad v_j = - \int _0^y \partial _x u_j. \end{aligned}$$
(12)

Roughly, inspired by the shear flow case, the idea will be to introduce as a new unknown the solution \(\psi _j = \int _0^y H_j\) of

$$\begin{aligned} (\partial _t + \beta (j{+}1)+ U^P \partial _x - \partial _y^2) \psi _j = \int _0^y u_j\, \mathrm {d}z. \end{aligned}$$

which is reminiscent of the Fourier relation (11). The test function \(\phi _j\) should then solve the reverse equation

$$\begin{aligned} (-\partial _t + \beta (j{+}1)- U^P \partial _x - \partial _y^2) \phi _j = \partial _y \psi _j \end{aligned}$$

and be solved backward in time. Performing the same estimate as in the shear flow case, we expect to find an inequality of the type

$$\begin{aligned} \beta (j{+}1)\Vert \partial _y \psi _j \Vert ^2 + \Vert \partial _y^2 \psi _j \Vert ^2 \lesssim \frac{1}{\beta ^3 (j{+}1)^3} \Vert F_j \Vert ^2 + \frac{1}{\beta ^3 (j{+}1)^3} \Vert \partial _x \partial _y \psi _j \Vert ^2 \end{aligned}$$

By exploiting a relation of the form \(\Vert \partial _x \partial _y \psi _j\Vert \sim j^\gamma \Vert \partial _y \psi _{j+1}\Vert \) (that needs to be shown!) and using that \(\gamma \le 2\), we will then be able to sum over j and establish for large enough \(\beta \) a control of \(\sum _j \Vert \partial _y \psi _j\Vert ^2\) in terms of \(\sum _j \Vert F_j \Vert ^2\).

In fact, in implementing this strategy, several refinements are necessary, and the relations satisfied by \(\psi _j = \int _0^y H_j\) or \(\phi _j\) need to be slightly modified. Particularly problematic is the term \(V^P \partial _y\) because \(V^P \sim -\partial _x U^E y\) increases linearly with y: this prevents from closing an energy estimate with a fixed weigth \(\rho = \rho (y)\). This difficulty appears in various places in the literature on the Prandtl equations. This is for instance the reason why article [12] is limited to the special case \(U^E = 0\) and decaying initial data. One can also mention [22], where this difficulty is overcome by a clever change of variables, which is reminiscent of the method of characteristics and allows to remove the bad part of the convection term from the momentum equation. Energy estimates can then be established in these new coordinates \(x',y'\), and yield some local well-posedness result, with solutions that are analytic in \(x'\) and \(L^2\) in \(y'\). The disadvantage of this approach is that the regularity of the solution in the original variables x and y is no longer clear at positive times. Here, we stick to the eulerian variables, but overcome the difficulty by introducing the family of weights \(\rho _j\), \(j \ge 0\). These weights allow to trade a power of y against a derivative in x, which is appropriate to the commutator terms. Moreover, they put very little conditions on the derivatives of the solution, so that they provide a very general framework for well-posedness. Note that the specific expression of \(\rho _j\) is important: it could not be for instance replaced by the more natural guess \((1+y)^{2(m-j)}\), as commutators with the diffusion term would not be under control. Note also that the strategy used in [27], where Sobolev well-posedness is established under monotonicity assumptions by increasing the weight with the number of y-derivatives, does not extend to the Gevrey framework in variable x.

The plan of the paper is as follows. In the next section, we first collect several properties of the weight \(\rho _j\). We then write the equations satisfied by the x-derivatives of the Prandtl solution in a form analogue to (12). This means that we put most of the nonlinear terms at the right-hand side, and consider those equations as linear. We finish the section by introducing the adapted quantities \(H_j\) and \(\phi _j\). The main section is Sect. 4: a priori Gevrey estimates for the linear equations are performed, that provide a control of the \(u_j\)’s in terms of the nonlinear terms \(F_j's\). Note that such estimates are obtained under a condition of the form \(\beta > C (1+ \Vert (U^P, V^P) \Vert _{low})^2\), where \(\Vert (U^P, V^P)\Vert _{low}\) is a low regularity norm of the solution. The treatment of the nonlinearity \(F_j\) is then handled in Sect. 5. The last step in the derivation of a priori estimates is to recover the control of the low regularity norm \(\Vert (U^P,V^P)\Vert _{low}\), see Sect. 6. Finally, issues regarding the construction and uniqueness of solutions are discussed in Sect. 7.

Preliminaries

The explicit form of the weights \(\rho _j\) is only needed in the Sect. 5. In the other parts, we just need a sufficient control of the logarithmic derivative (Lemma 2), a bound for antiderivatives (Lemma 3) and relate \(\rho _j\) to \(\rho _{j+1}\) (Lemma 4).

Lemma 2

Let \(m \ge 0\) and \(\alpha \ge 0\). There exists a constant \(C_l\) such that for all \(y \in \mathbb {R}^+\), \(j \in \mathbb {N}\)

$$\begin{aligned} \left| \frac{\partial _y \rho _j(y)}{\rho _j(y)} \right| \le {\left\{ \begin{array}{ll} C_l (j{+}1)^{1-\alpha } &{} \text {if } \alpha < 1 \\ C_l \log (j{+}1)&{} \text {if } \alpha = 1 \\ C_l &{} \text {if } \alpha > 1 \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} (1+y) \left| \frac{\partial _y \rho _j(y)}{\rho _j(y)} \right| \le C_l \, (j{+}1). \end{aligned}$$

Proof

Given the explicit form of \(\rho _j\), we can compute the logarithmic derivative of \(\rho \) directly as

$$\begin{aligned} \frac{\partial _y \rho _j}{\rho _j} = \partial _y \log \rho _j= & {} \partial _y \log \rho _0 - 2 \sum _{k=1}^{j} \partial _y \log \left( 1 + \frac{y}{k^\alpha }\right) \\= & {} \frac{2m}{1+y} - 2 \sum _{k=1}^{j} \frac{1}{k^\alpha \left( 1+\frac{y}{k^\alpha }\right) }. \end{aligned}$$

From this expression the result follows directly. \(\square \)

Lemma 3

For \(m > \frac{1}{2}\) introduce the constant

$$\begin{aligned} C_m = \sqrt{\frac{1}{2m-1}}. \end{aligned}$$

Then for all \(\alpha \ge 0\), \(j \in \mathbb {N}\) and all \(f = f(y)\),

$$\begin{aligned} \sup _{y \ge 0} \left( \frac{\rho _j(y)}{\rho _0(y)}\right) ^{1/2} \int _0^y |f(z)|\, \mathrm {d}z \le C_m \, \Vert f \Vert _{L^2(\rho _j)}. \end{aligned}$$

More generally, for \(0 \le n \le j\) with \(n < m - \frac{1}{2}\) one has

$$\begin{aligned} \sup _{y \ge 0} \left( \frac{\rho _j(y)}{\rho _n(y)}\right) ^{1/2} \int _0^y |f(z)|\, \mathrm {d}z \le C_{m-n} \, \Vert f \Vert _{L^2(\rho _j)}. \end{aligned}$$

Eventually, for all \(A = A(x,y)\) and \(B = B(x,y)\), the following inequality holds:

$$\begin{aligned} \Vert A \int _0^y B(z)\, \mathrm {d}z \Vert _j \le C_m \Vert A \Vert _{L^\infty _xL^2_y(\rho _0)} \Vert B \Vert _{j}. \end{aligned}$$

Proof

Note that \(\rho _j/\rho _n\) for \(j \ge n\) is non-increasing. Hence

$$\begin{aligned} \left( \frac{\rho _j(y)}{\rho _n(y)}\right) ^{1/2} \int _0^y |f(z)|\, \mathrm {d}z\le & {} \int _0^y \left( \frac{\rho _j(z)}{\rho _n(z)}\right) ^{1/2} |f(z)| \, \mathrm {d}z \\\le & {} \Vert f \Vert _{L^2(\rho _j)} \left( \int _0^y \frac{1}{\rho _n(z)} \, \mathrm {d}z \right) ^{1/2}, \end{aligned}$$

where we used the Cauchy-Schwarz inequality in the second inequality.

As \(\alpha \ge 0\) we find directly that

$$\begin{aligned} \frac{1}{\rho _n(y)} \le (1+y)^{-2m} \prod _{k=1}^j \left( 1 + \frac{y}{k^{\alpha }}\right) ^{2} \le \frac{1}{(1+y)^{2m-2n}} \end{aligned}$$

whose integral over \(y \in \mathbb {R}^+\) gives \(C_{m-n}^2\). This proves the first and second bounds. The remaining estimate with A and B follows directly. \(\square \)

The weights are decaying so that \(\rho _j \le \rho _k\) for \(j \ge k\). As \(\alpha \ge 0\), we have for \(j \in \mathbb {N}\) that \((1+y)^2 \rho _{j+1} \le (j{+}1)^{2\alpha } \rho _j\) and \(\rho _{j+1} \ge \frac{\rho _j}{(1+y)^2}\). This shows:

Lemma 4

Let \(\alpha \ge 0\). For \(j\in \mathbb {N}\), \(A = A(x,y)\) and \(B = B(x,y)\) it holds that for

$$\begin{aligned} \Vert A \Vert _{j+1} \le \Vert A \Vert _j,\qquad \Vert (1+y) A \Vert _{j+1} \le (j{+}1)^{\alpha } \Vert A \Vert _j \end{aligned}$$

and

$$\begin{aligned} \left\| \frac{A}{(1+y)} \right\| _j \le \Vert A \Vert _{j+1}, \qquad \Vert A \int _0^y B(z)\, \mathrm {d}z \Vert _j \le C_m \Vert (1+y) A \Vert _{L^\infty _xL^2_y(\rho _0)} \Vert B \Vert _{j+1}. \end{aligned}$$

Let us insist again that most parts of the proof would work with constant weight \(\rho \) instead of \(\rho _j\). The dependency on j will be only needed to treat the commutator terms coming from \(V^P \partial _y U^P\). The difficulty is that \(V^P\) grows like y as soon as \(U^E\) is non-constant. Here the crucial property that we will use is that we can control \(\Vert (1+y) A \Vert _{j+1}\) by \((j{+}1)^{\alpha } \Vert A \Vert _j\).

The Prandtl equations are given for \((U^P,V^P)\) with inhomogeneous boundary conditions at \(y \rightarrow \infty \). In order to work with homogeneous boundary conditions at zero and infinity, we introduce

$$\begin{aligned} U^e(t,x,y) = (1-\mathrm {e}^{-y}) U^E(t,x), \quad V^e(t,x,y) = - (y + \mathrm {e}^{-y}-1)\partial _x U^E(t,x) \end{aligned}$$

and set \(u = U^P - U^e\), \(v = -\int _0^y \partial _x u = V^P - V^e\). Then,

$$\begin{aligned} \partial _t u + (u \partial _x + v \partial _y) u + (U^e \partial _x + V^e \partial _y) u + (u \partial _x + v \partial _y) U^e - \partial ^2_y u = f^e \end{aligned}$$
(13)

where

$$\begin{aligned} f^e = \partial _t U^E + U^E \partial _x U^E - \partial _t U^e - U^e \partial _x U^e - V^e \partial _y U^e + \partial ^2_y U^e. \end{aligned}$$
(14)

In the new variables (uv) the boundary conditions are

$$\begin{aligned} u = v = 0 \quad \text { at } y=0,\qquad \text { and } \lim _{y\rightarrow \infty } u =0. \end{aligned}$$
(15)

The condition at \(y\rightarrow \infty \) will be encoded in the functional space of u.

To prove Theorem 1, the point is to obtain good estimates for Gevrey norms of u of type (6) for time-dependent radius \(\tau = \tau (t)\). More precisely, we give ourselves parameters \(m,\alpha ,\gamma ,r\), to be fixed later, as well as the time-dependent radius \(\tau (t) = \tau _0 \mathrm {e}^{-\beta t}\), with \(\beta > 0\) to be fixed later. Then, for any function \(f = f(t,x)\) or \(f=f(t,x,y)\) and \(j \in \mathbb {N}\) we set

$$\begin{aligned} f_j(t,\cdot ) := M_j\, \partial _x^j f(t,\cdot ) \quad \text {with} \quad M_j := \frac{\tau (t)^{j+1} (j{+}1)^r}{(j!)^{\gamma }}. \end{aligned}$$

Taking j derivatives in x of (13) and multiplying by \(M_j\) yields

$$\begin{aligned}&\Big (\partial _t + \beta (j{+}1)+ U^P \partial _x + (j{+}1)\partial _x U^P + V^P \partial _y - \partial _y^2\Big ) u_j \nonumber \\&\quad +\,\partial _y U^P v_j + j \partial _{xy} U^P \partial _x^{-1} v_j = F_j \end{aligned}$$
(16)

where \(F_j\) collects all terms with less than j derivatives in x as well as the weighted derivative of the forcing \(f^e\). It is given by

$$\begin{aligned} \begin{aligned} F_j&= f_j^e + M_j \left[ u \partial _x, \partial _x^j\right] u + M_j \, \partial _x u\, \partial _x^j u \\&\quad + M_j \left[ \partial _y u, \partial _x^j\right] v + M_j j\, \partial _{xy} u\, \partial _x^{j-1} v + M_j v\, \partial _x^j \partial _y u \\&\quad + M_j \left[ U^e \partial _x, \partial _x^j\right] u + M_j j\, \partial _x U^e\, \partial _x^j u \\&\quad + M_j \left[ V^e \partial _y, \partial _x^j\right] u \\&\quad + M_j \left[ \partial _x U^e, \partial _x^j\right] u \\&\quad + M_j \left[ \partial _y U^e, \partial _x^j \right] v + M_j j\, \partial _{xy} U^e\, \partial _x^{j-1} v. \end{aligned} \end{aligned}$$

We now introduce our crucial auxiliary functions \(H_j(t,x,y)\) defined by

$$\begin{aligned} \begin{aligned}&\Big (\partial _t + \beta (j{+}1)+ U^P \partial _x + (j{+}1)\partial _x U^p + V^P \partial _y - \partial _y^2\Big ) \int _0^y H_j \, \mathrm {d}z = \int _0^y u_j \, \mathrm {d}z, \\&\quad H_j|_{t=0} = 0, \qquad \partial _y H_j|_{y=0} = 0, \qquad H_j|_{y\rightarrow \infty } = 0. \end{aligned} \end{aligned}$$
(17)

For the existence of \(H_j\), one can consider (17) as a convection-diffusion equation for \(A_j = \int _0^y H_j\, \mathrm {d}z\), with boundary conditions \(A_j|_{y=0} = \partial _y A_j|_{y\rightarrow \infty } = 0\), which has a solution by the classical theory of parabolic PDEs. The PDE (17) itself then implies that \(\partial _y^2 A_j|_{y=0}=0\) so that taking \(H_j = \partial _y A_j\) gives the required solution.

We further introduce the corresponding test functions \(\phi _j\) by

$$\begin{aligned} \begin{aligned}&\left( -\partial _t + \beta (j{+}1)- U^P \partial _x + j \partial _x U^p - V^P \partial _y - \partial _y V^P - V^P \frac{\partial _y \rho _j}{\rho _j} \right. \\&\left. \quad - \left( \partial _y + \frac{\partial _y \rho _j}{\rho _j}\right) ^2 \right) \phi _j = H_j, \\&\quad \phi _j|_{t=T} = 0, \qquad \phi _j|_{y=0} = 0, \qquad \phi _j|_{y\rightarrow \infty } = 0. \end{aligned} \end{aligned}$$
(18)

Note here that the operator acting on \(\phi _j\) is the formal adjoint operator of the operator acting on \(\int _0^y H_j \, \mathrm {d}z\) in (17), with respect to the \(L^2(\rho _j)\) scalar product, denoted \(\left\langle \,,\right\rangle _j\). This is a backward heat equation solved backward in time for \(t \in [0,T]\).

Testing (18) against \(\phi _j\) in \(\Vert \cdot \Vert _j\) and integrating over [tT] yields

$$\begin{aligned} \begin{aligned}&\frac{1}{2} \Vert \phi _j(t) \Vert _j^2 + \beta (j{+}1)\int _t^T \Vert \phi _j(s) \Vert _j^2\, \mathrm {d}s + \left( j{+}\frac{1}{2}\right) \int _t^T \left\langle \partial _x U^P \phi _j,\phi _j\right\rangle _j\, \mathrm {d}s \\&\qquad - \frac{1}{2} \int _t^T \left\langle \left( \partial _y V^P + V^P \frac{\partial _y \rho _j}{\rho _j}\right) \phi _j,\phi _j\right\rangle _j \, \mathrm {d}s + \int _t^T \Vert \partial _y \phi _j(s) \Vert _j^2 \, \mathrm {d}s \\&\qquad + \int _t^T \left\langle \frac{\partial _y \rho _j}{\rho _j} \phi _j,\partial _y \phi _j\right\rangle _j\, \mathrm {d}s \\&\quad = \int _t^T \left\langle H_j,\phi _j\right\rangle _j\, \mathrm {d}s. \end{aligned} \end{aligned}$$

Hence we find

$$\begin{aligned} \begin{aligned}&\frac{1}{2} \Vert \phi _j(t) \Vert _j^2 + \frac{3\beta (j{+}1)}{4} \int _t^T \Vert \phi _j(s) \Vert _j^2\, \mathrm {d}s + \frac{1}{2} \int _t^T \Vert \partial _y \phi _j(s) \Vert _j^2 \, \mathrm {d}s \\&\quad \le \frac{1}{\beta (j{+}1)} \int _t^T \Vert H_j(s)\Vert _j^2\, \mathrm {d}s + \left( \left( j{+}\frac{1}{2}\right) \Vert \partial _x U^P \Vert _{\infty } + \frac{1}{2} \left\| \partial _y V^P {+} V^P \frac{\partial _y \rho _j}{\rho _j} \right\| _{\infty } \right. \\&\left. \qquad + \frac{1}{2} \left\| \frac{\partial _y \rho _j}{\rho _j} \right\| _{\infty }^2 \right) \int _t^T \Vert \phi _j(s) \Vert _j^2\, \mathrm {d}s. \end{aligned} \end{aligned}$$

By Lemma 2, under the condition \(\alpha \ge \frac{1}{2}\), we get the following control:

Lemma 5

Fix \(m\ge 0\) and \(\alpha \ge \frac{1}{2}\). Then there exist a constant \(\mathcal {C} = \mathcal {C}(m,\alpha )\) such that for all \(j \in \mathbb {N}\) it holds that

$$\begin{aligned}&\Vert \phi _j(t) \Vert _j^2 + \beta (j{+}1)\int _t^T \Vert \phi _j(s) \Vert _j^2\, \mathrm {d}s + \int _t^T \Vert \partial _y \phi _j(s) \Vert _j^2 \, \mathrm {d}s \\&\quad \le \frac{2}{\beta (j{+}1)} \int _t^T \Vert H_j(s)\Vert _j^2\, \mathrm {d}s \end{aligned}$$

if

$$\begin{aligned} \beta \ge \mathcal {C} \left( 1 + \Vert \partial _x U^P \Vert _{\infty } + \Vert \partial _y V^P \Vert _{\infty } + \left\| \frac{V^P}{1+y} \right\| _{\infty } \right) . \end{aligned}$$

Note that for \(\alpha < \frac{1}{2}\), the term with \(\Vert \frac{\partial _y \rho _j}{\rho _j} \Vert _{\infty }^2\) could not have been absorbed. This a priori estimate also ensure the existence of \(\phi _j\) as solution of (18). A similar estimate holds for \(H_j\) which ensures the existence of \(H_j\) as solution of (17).

Linear Estimates

In this section we analyse the linearised equation (16) and obtain an estimate for the solution in terms of the \(F_j\) containing the forcing and lower-order terms. For this, we shall first analyse (16) for a fixed j. We will obtain a control of \(H_j\) in terms of the forcing \(F_j\) and an error term \(\partial _x H_j\), which will be shown to be approximately \((j{+}1)^{\gamma } H_{j+1}\). By summing over j, we will find the following control.

Lemma 6

Fix \(m > \frac{1}{2}, \frac{1}{2} \le \alpha \le \frac{1}{2} + \gamma , 1 \le \gamma \le 2, r \in \mathbb {R}\). Then there exists a constant \(\mathcal {C} = \mathcal {C}(m,\alpha ,\gamma ,r)\) such that for all \(\tau _1\), \(\beta \) and T such that

$$\begin{aligned} \beta \ge \mathcal {C} (1 + \Vert (U^P,V^P) \Vert _{low})\, \left( 1 + \frac{1}{\tau _1} + \Vert (U^P,V^P) \Vert _{low}\right) \text { and } \tau (T) \ge \tau _1 \end{aligned}$$

the \(H_j\)’s defined by (17) for solutions \(u_j\)’s of (16) satisfy

$$\begin{aligned} \begin{aligned}&\sum _{j=0}^{\infty } \beta ^2 (j{+}1)^{2\gamma } \left[ \int _0^T \Vert H_j(t) \Vert _j^2 \, \mathrm {d}t + \frac{1}{\beta (j{+}1)} \Vert H_j(T) \Vert _j^2 + \frac{1}{\beta (j{+}1)} \int _0^T \Vert \partial _y H_j \Vert _j^2 \, \mathrm {d}t \right] \\&\quad \le 16 \sum _{j=0}^{\infty } \left[ \frac{(j{+}1)^{2\gamma -4}}{\beta ^2} \int _0^T \Vert F_j(t) \Vert _j^2\, \mathrm {d}t + \frac{(j{+}1)^{2\gamma -3}}{\beta } \Vert u_{\mathrm {in},j} \Vert _j^2 \right] . \end{aligned} \end{aligned}$$

Here we use a low-order control of \(U^P\) and \(V^P\) in order to control the commutator error terms. From the required bounds, we define the low-order norm as

$$\begin{aligned} \begin{aligned}&\Vert (U^P,V^P) \Vert _{low} = \sup _{t\in [0,T]} \max \left( \max _{0\le k \le 3}\Vert \partial _x^k U^P \Vert _{\infty }, \Vert \partial _x \partial ^2_y U^P \Vert _\infty , \right. \\&\left. \quad \Vert (1+y) \partial _y U^P \Vert _{\infty }, \Vert (1+y) \partial _y^2 U^P \Vert _{\infty },\right. \\&\left. \quad \Vert (1+y) \partial _y U^P \Vert _{L^\infty _xL^2_y(\rho _0)}, \Vert \partial _{xy} U^P \Vert _{L^\infty _xL^2_y(\rho _0)}, \Vert \partial _{xxy} U^P \Vert _{L^\infty _xL^2_y(\rho _0)}, \right. \\&\left. \quad \Vert (1+y)^2 \partial _y^2 U^P \Vert _{L^\infty _xL^2_y(\rho _0)}, \Vert (1+y) \partial _{x} \partial _y^2 U^P \Vert _{L^\infty _xL^2_y(\rho _0)},\quad \max _{0\le k \le 2}\left\| \frac{\partial _x^k V^P}{1+y} \right\| _{\infty } \right) . \end{aligned} \end{aligned}$$
(19)

Although a main ingredient of our proof, the unknown \(H_j\) is less natural than the usual \(u_j\), notably for the future treatment of the nonlinearity, which involves \(u_j\) and \(\omega _j = \partial _y u_j\). This is why we shall we relate the control of \(H_j\) to \(u_j\) and show:

Proposition 7

Fix \(m > \frac{1}{2}, \frac{1}{2} \le \alpha \le \frac{1}{2} + \gamma , 1 \le \gamma \le 2, r \in \mathbb {R}\). Then there exist constants \(C = C(m,\alpha ,\gamma ,r)\) and \(\mathcal {C} = \mathcal {C}(m,\alpha ,\gamma ,r)\) such that for all \(\tau _1\), \(\beta \) and T such that

$$\begin{aligned} \beta \ge \mathcal {C} (1 + \Vert (U^P,V^P) \Vert _{low})\, \left( 1 + \frac{1}{\tau _1} + \Vert (U^P,V^P) \Vert _{low}\right) \text { and } \tau (T) \ge \tau _1 \end{aligned}$$

the solution u of (16) satisfies

$$\begin{aligned} \begin{aligned}&\int _0^T \Vert u \Vert _{\gamma ,\tau ,r}^2\, \mathrm {d}t + \sup _{t\in [0,T]} \frac{1}{\beta } \Vert u \Vert _{\gamma ,\tau ,r-\frac{\gamma }{2}}^2 + \int _0^T \frac{1}{\beta } \Vert (1+y) \omega \Vert _{\gamma ,\tau ,r+1-\gamma }^2 \, \mathrm {d}t\\&\qquad + \sup _{t\in [0,T]} \frac{1}{\beta ^2} \Vert (1+y) \omega \Vert _{\gamma ,\tau ,r+\frac{1}{2} - \gamma }^2 + \frac{1}{\beta ^2} \int _0^T \Vert (1+y) \partial _y \omega \Vert _{\gamma ,\tau ,r+\frac{1}{2} -\gamma }^2\, \mathrm {d}t\\&\quad \le C \left[ \frac{1}{\beta ^2} \sum _{j=0}^{\infty } \int _0^T \frac{1}{(j{+}1)^{4-2\gamma }} \Vert F_j \Vert _j^2\, \mathrm {d}t \right. \\&\left. \qquad + \frac{1}{\beta ^2} \sum _{j=0}^{\infty } \int _0^T \frac{1}{(j{+}1)^{2\gamma -1}} \Vert (1+y) F_j \Vert _j^2 \, \mathrm {d}t \right] \\&\qquad + \frac{C}{\beta ^2} \sum _{j=0}^{\infty } \int _0^T \frac{1}{(j{+}1)^{2\gamma -1}} \Vert F_j|_{y=0} \Vert _{L^2_x}^2 \, \mathrm {d}t \\&\qquad + C \left[ \frac{1}{\beta } \Vert u_{\mathrm {in}} \Vert _{\gamma ,\tau _0,r+\gamma -\frac{3}{2}}^2 + \frac{1}{\beta ^2} \Vert (1+y) \omega _{\mathrm {in}} \Vert _{\gamma ,\tau _0,r+\frac{1}{2} - \gamma }^2 \right] . \end{aligned} \end{aligned}$$

For \(\gamma \ge 5/4\) this is

$$\begin{aligned} \begin{aligned}&\int _0^T \Vert u \Vert _{\gamma ,\tau ,r}^2\, \mathrm {d}t + \sup _{t\in [0,T]} \frac{1}{\beta } \Vert u \Vert _{\gamma ,\tau ,r-\frac{\gamma }{2}}^2 + \int _0^T \frac{1}{\beta } \Vert (1+y) \omega \Vert _{\gamma ,\tau ,r+1-\gamma }^2 \, \mathrm {d}t\\&\qquad + \sup _{t\in [0,T]} \frac{1}{\beta ^2} \Vert (1+y) \omega \Vert _{\gamma ,\tau ,r+\frac{1}{2} - \gamma }^2 + \frac{1}{\beta ^2} \int _0^T \Vert (1+y) \partial _y \omega \Vert _{\gamma ,\tau ,r+\frac{1}{2} -\gamma }^2\, \mathrm {d}t\\&\quad \le C \left[ \frac{1}{\beta ^2} \sum _{j=0}^{\infty } \int _0^T \frac{1}{(j{+}1)^{4-2\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{2\gamma - \frac{5}{2}}}\right) F_j \right\| _j^2\, \mathrm {d}t \right] \\&\qquad + \frac{C}{\beta ^2} \sum _{j=0}^{\infty } \int _0^T \frac{1}{(j{+}1)^{2\gamma -1}} \Vert F_j|_{y=0} \Vert _{L^2_x}^2 \, \mathrm {d}t \\&\qquad + C \left[ \frac{1}{\beta } \Vert u_{\mathrm {in}} \Vert _{\gamma ,\tau _0,r+\gamma -\frac{3}{2}}^2 + \frac{1}{\beta ^2} \Vert (1+y) \omega _{\mathrm {in}} \Vert _{\gamma ,\tau _0,r+\frac{1}{2} - \gamma }^2 \right] . \end{aligned} \end{aligned}$$

Estimate for \(H_j\)

We focus first on Lemma 6. The idea is to use the solution \(\phi _j\) of (18) as a test function in (16). Taking the weighted scalar product and integrating over [0, T], we find for the first term in (16):

$$\begin{aligned} \begin{aligned}&\int _0^T \left\langle \left( \partial _t + \beta (j{+}1)+ U^P \partial _x + (j{+}1)\partial _x U^p + V^P \partial _y - \partial _y^2\right) u_j,\phi _j\right\rangle _j \, \mathrm {d}t \\&\quad = - \left\langle u_{\mathrm {in},j},\phi _j(0)\right\rangle _j\\&\qquad + \int _0^T \left\langle u_j,\left( -\partial _t + \beta (j{+}1)- U^P \partial _x + j \partial _x U^p - V^P \partial _y - \partial _y V^P - V^P \frac{\partial _y \rho _j}{\rho _j} \right. \right. \\&\left. \left. \qquad - \left( \partial _y + \frac{\partial _y \rho _j}{\rho _j}\right) ^2 \right) \phi _j\right\rangle _j\, \mathrm {d}t \\&\quad = - \left\langle u_{\mathrm {in},j},\phi _j(0)\right\rangle _j + \int _0^T \left\langle u_j,H_j\right\rangle _j\, \mathrm {d}t. \end{aligned} \end{aligned}$$

Note that there is no boundary term as \(u_j\) and \(\phi _j\) vanish at the boundaries. Differentiating (17), we can replace \(u_j\) in the last integral and find

$$\begin{aligned} \begin{aligned}&\int _0^T \left\langle u_j,H_j\right\rangle _j\, \mathrm {d}t \\&\quad = \int _0^T \left\langle \Big (\partial _t + \beta (j{+}1)+ U^P \partial _x + (j{+}1)\partial _x U^p + V^P \partial _y - \partial _y^2\Big ) H_j,H_j\right\rangle _j\, \mathrm {d}t \\&\qquad + \int _0^t \left\langle (\partial _y U^P \partial _x + j \partial _{xy} U^P) \int _0^y H_j \, \mathrm {d}z,H_j\right\rangle _j\, \mathrm {d}t \\&\qquad + \int _0^t \left\langle \partial _{xy} U^P \int _0^y H_j\, \mathrm {d}z,H_j\right\rangle _j \, \mathrm {d}t + \int _0^T \left\langle \partial _y V^P H_j,H_j\right\rangle _j \, \mathrm {d}t \\&\quad = \frac{1}{2} \Vert H_j(T) \Vert _j^2 + \beta (j{+}1)\int _0^T \Vert H_j(t) \Vert _j^2\, \mathrm {d}t + \int _0^T \Vert \partial _y H_j(t) \Vert _j^2\, \mathrm {d}t \\&\qquad + \int _0^t \left\langle (\partial _y U^P \partial _x + j \partial _{xy} U^P) \int _0^y H_j \, \mathrm {d}z,H_j\right\rangle _j\, \mathrm {d}t \\&\qquad + \int _0^t \left\langle \partial _{xy} U^P \int _0^y H_j\, \mathrm {d}z,H_j\right\rangle _j \, \mathrm {d}t + \int _0^T \left\langle \partial _y V^P H_j,H_j\right\rangle _j \, \mathrm {d}t \\&\qquad + \left( j{+}\frac{1}{2}\right) \int _0^T \left\langle \partial _x U^P H_j,H_j\right\rangle _j\, \mathrm {d}t - \frac{1}{2} \int _0^T \left\langle \left( \partial _y V^P + V^P \frac{\partial _y \rho _j}{\rho _j}\right) H_j,H_j\right\rangle _j \, \mathrm {d}t \\&\qquad + \int _0^T \left\langle \frac{\partial _y \rho _j}{\rho _j} H_j,H_j\right\rangle _j \, \mathrm {d}t. \end{aligned} \end{aligned}$$

By the boundary values of \(H_j\) there are again no boundary terms from partial integration in \(y\). In the last expression, the first line contains the good controlled terms, the second line will cancel the leading contribution from the bad terms \(\partial _y U^P v_j + j \partial _{xy} U^P \partial _x^{-1} v_j\) (see below), while the last two lines collect the error terms.

Next, we compute the contribution from the terms with \(v_j\) using \(v_j = -\partial _x \int _0^y u_j \, \mathrm {d}z\):

$$\begin{aligned} \begin{aligned}&\int _0^T \left\langle \partial _y U^P v_j,\phi _j\right\rangle _j\,\mathrm {d}t \\&\quad = - \int _0^T \left\langle \partial _y U^P \partial _x \left[ \Big (\partial _t + \beta (j{+}1)+ U^P \partial _x + (j{+}1)\partial _x U^p + V^P \partial _y - \partial _y^2\Big ) \right. \right. \\&\left. \left. \qquad \int _0^y H_j\, \mathrm {d}z \right] ,\phi _j\right\rangle _j\,\mathrm {d}t \\&\quad = - \int _0^T \left\langle \partial _y U^P \Big (\partial _t + \beta (j{+}1)+ U^P \partial _x + (j{+}1)\partial _x U^p + V^P \partial _y - \partial _y^2\Big ) \right. \\&\left. \qquad \partial _x \int _0^y H_j\, \mathrm {d}z ,\phi _j\right\rangle _j\,\mathrm {d}t \\&\qquad - \int _0^T \left\langle \partial _y U^P (\partial _x U^P \partial _x + (j{+}1)\partial _x^2 U^P + \partial _x V^P \partial _y) \int _0^y H_j\, \mathrm {d}z,\phi _j\right\rangle _j\,\mathrm {d}t \\&\quad = - \int _0^T \left\langle \Big (\partial _t + \beta (j{+}1)+ U^P \partial _x + (j{+}1)\partial _x U^p + V^P \partial _y - \partial _y^2\Big )\right. \\&\left. \qquad \left[ \partial _y U^P \partial _x \int _0^y H_j\, \mathrm {d}z\right] ,\phi _j\right\rangle _j\,\mathrm {d}t \\&\qquad + \int _0^T \left\langle \Big ( (\partial _t \!+\! U^P \partial _x \!+\! V^P \partial _y) \partial _y U^P \!-\! 2 \partial _y^2 U^P \partial _y - \partial _y^3 U^P \Big ) \partial _x \int _0^y H_j\, \mathrm {d}z,\phi _j\right\rangle _j\,\mathrm {d}t \\&\qquad - \int _0^T \left\langle \partial _y U^P (\partial _x U^P \partial _x + (j{+}1)\partial _x^2 U^P + \partial _x V^P \partial _y) \int _0^y H_j\, \mathrm {d}z,\phi _j\right\rangle _j\,\mathrm {d}t \\&\quad = - \int _0^T \left\langle \partial _y U^P \partial _x \int _0^y H_j\, \mathrm {d}z,H_j\right\rangle _j\,\mathrm {d}t \\&\qquad + \int _0^T \left\langle \Big ( (\partial _t \!+\! U^P \partial _x \!+\! V^P \partial _y) \partial _y U^P - 2 \partial _y^2 U^P \partial _y \!-\! \partial _y^3 U^P \Big ) \partial _x \int _0^y H_j\, \mathrm {d}z,\phi _j\right\rangle _j\,\mathrm {d}t \\&\qquad - \int _0^T \left\langle \partial _y U^P (\partial _x U^P \partial _x + (j{+}1)\partial _x^2 U^P + \partial _x V^P \partial _y) \int _0^y H_j\, \mathrm {d}z,\phi _j\right\rangle _j\,\mathrm {d}t \\ \end{aligned} \end{aligned}$$

and

$$\begin{aligned}&\int _0^T\left\langle j \partial _{xy} U^P \partial _x^{-1} v_j,\phi _j\right\rangle _j\,\mathrm {d}t \\&\quad = -j \int _0^T\left\langle \partial _{xy} U^P \Big (\partial _t + \beta (j{+}1)+ U^P \partial _x + (j{+}1)\partial _x U^p + V^P \partial _y - \partial _y^2\Big ) \right. \\&\left. \qquad \int _0^y H_j \, \mathrm {d}z,\phi _j\right\rangle _j\,\mathrm {d}t \\&\quad = -j \int _0^T\left\langle \partial _{xy} U^P \int _0^y H_j\, \mathrm {d}z,H_j\right\rangle _j\,\mathrm {d}t \\&\qquad + j\int _0^T\left\langle \Big ( (\partial _t + U^P \partial _x + V^P \partial _y) \partial _{xy} U^P - 2 \partial _x \partial _y^2 U^P \partial _y - \partial _x \partial _y^3 U^P \Big ) \right. \\&\left. \qquad \int _0^y H_j\, \mathrm {d}z,\phi _j\right\rangle _j\,\mathrm {d}t. \end{aligned}$$

In both cases the leading order term cancels. Hence collecting the terms we arrive at

$$\begin{aligned} \begin{aligned}&\frac{1}{2} \Vert H_j(T) \Vert _j^2 + \beta (j{+}1)\int _0^T \Vert H_j(t) \Vert _j^2\, \mathrm {d}t + \int _0^T \Vert \partial _y H_j(t) \Vert _j^2\, \mathrm {d}t \\&\quad \le \int _0^T \left\langle F_j,\phi _j\right\rangle \, \mathrm {d}t + \left\langle u_{\mathrm {in},j},\phi _j(0)\right\rangle _j + \int _0^T \sum _{i=1}^{5} E_i\, \mathrm {d}t \end{aligned} \end{aligned}$$

where \(E_1,\dots ,E_5\) collect the lower-order error terms as

$$\begin{aligned} E_1&= - \left\langle \partial _{xy} U^P \int _0^y H_j\, \mathrm {d}z,H_j\right\rangle _j - \left\langle \partial _y V^P H_j,H_j\right\rangle _j, \\ E_2&= - \left( j{+}\frac{1}{2}\right) \left\langle \partial _x U^P H_j,H_j\right\rangle _j + \frac{1}{2} \left\langle \left( \partial _y V^P + V^P \frac{\partial _y \rho _j}{\rho _j}\right) H_j,H_j\right\rangle _j\\&\quad - \left\langle \frac{\partial _y \rho _j}{\rho _j} H_j,H_j\right\rangle _j, \\ E_3&= - \left\langle \Big ( (\partial _t + U^P \partial _x + V^P \partial _y) \partial _y U^P - 2 \partial _y^2 U^P \partial _y - \partial _y^3 U^P \Big ) \partial _x \int _0^y H_j\, \mathrm {d}z,\phi _j\right\rangle _j, \\ E_4&= \left\langle \partial _y U^P (\partial _x U^P \partial _x + (j{+}1)\partial _x^2 U^P + \partial _x V^P \partial _y) \int _0^y H_j\, \mathrm {d}z,\phi _j\right\rangle _j, \\ E_5&= -j \left\langle \Big ( (\partial _t \!+\! U^P \partial _x \!+\! V^P \partial _y) \partial _{xy} U^P \!- 2 \partial _x \partial _y^2 U^P \partial _y - \partial _x \partial _y^3 U^P \Big ) \int _0^y H_j\, \mathrm {d}z,\phi _j\right\rangle _j. \end{aligned}$$

Here \(E_3\) and \(E_4\) contain the worst terms, as they involve x-derivatives of \(H_j\). They are responsible for the Gevrey regularity requirement.

Assume \(m \ge 0\), \(\alpha \ge \frac{1}{2}\) and \(\beta \) large enough so that Lemma 5 applies. We can then estimate the forcing terms as

$$\begin{aligned} \int _0^T \left\langle F_j,\phi _j\right\rangle _j \, \mathrm {d}t \le \frac{2}{\beta ^3 (j{+}1)^3} \int _0^T \Vert F_j(t) \Vert _j^2 \, \mathrm {d}t + \frac{\beta (j{+}1)}{4} \int _0^T \Vert H_j(t) \Vert _j^2\, \mathrm {d}t \end{aligned}$$

and

$$\begin{aligned} \left\langle u_{\mathrm {in},j},\phi _j(0)\right\rangle _j \le \frac{2}{\beta ^2 (j{+}1)^2} \Vert u_{\mathrm {in},j} \Vert _j^2 + \frac{\beta (j{+}1)}{4} \int _0^T \Vert H_j(t) \Vert _j^2\, \mathrm {d}t. \end{aligned}$$

Absorbing the terms with \(H_j\) we therefore find

$$\begin{aligned} \begin{aligned}&\Vert H_j(T) \Vert _j^2 + \beta (j{+}1)\int _0^T \Vert H_j(t) \Vert _j^2\, \mathrm {d}t + 2 \int _0^T \Vert \partial _y H_j(t) \Vert _j^2\, \mathrm {d}t \\&\quad \le \frac{4}{\beta ^3 (j{+}1)^3} \int _0^T \Vert F_j(t) \Vert _j^2 \, \mathrm {d}t + \frac{4}{\beta ^2 (j{+}1)^2} \Vert u_{\mathrm {in},j} \Vert _j^2 + 2 \int _0^T \sum _{i=1}^{5} E_i\, \mathrm {d}t. \end{aligned} \end{aligned}$$

We now estimate the error terms, where we repeatedly use Lemma 3. For \(E_1\) we find

$$\begin{aligned} E_1 \le \left[ C_m \Vert \partial _{xy} U^P \Vert _{L^\infty _xL^2_y(\rho _0)} + \Vert \partial _y V^P \Vert _{\infty } \right] \Vert H_j \Vert _j^2. \end{aligned}$$

For \(E_2\) we also use Lemma 2 and assume \(\alpha \ge \frac{1}{2}\)

$$\begin{aligned} E_2 \le \left[ \left( j{+}\frac{1}{2}\right) \Vert \partial _x U^P \Vert _{\infty } + \frac{1}{2} \Vert \partial _y V^P \Vert _{\infty } + C_l (j{+}1)\left( 1+ \left\| \frac{V^P}{1+y} \right\| _{\infty }\right) \right] \Vert H_j \Vert _j^2. \end{aligned}$$

In the term \(E_3\) we have terms with \(\partial _x H_j\), which we want to estimate in \(\Vert \cdot \Vert _{j+1}\) as they will be later controlled by \(H_{j+1}\). Using Lemma 4 we find

$$\begin{aligned} \begin{aligned} E_3&\le C_m \Vert (1{+}y) (\partial _t + U^P \partial _x + V^P \partial _y - \partial _y^2) \partial _{y} U^P \Vert _{L^\infty _xL^2_y(\rho _0)} \Vert \partial _x H_j \Vert _{j+1} \Vert \phi _j \Vert _j \\&+ 2 \Vert (1{+}y) \partial _y^2 U^P \Vert _{\infty } \Vert \partial _x H_j \Vert _{j+1} \Vert \phi _j \Vert _j \\&\le 2 \Vert (1{+}y) \partial _y^2 U^P \Vert _{\infty } \Vert \partial _x H_j \Vert _{j+1} \Vert \phi _j \Vert _j \end{aligned} \end{aligned}$$

where we used the identity

$$\begin{aligned} (\partial _t + U^P \partial _x + V^P \partial _y) \partial _y U^P - \partial ^2_y \partial _y U^P = 0. \end{aligned}$$
(20)

Similarly, we find for \(E_4\) that

$$\begin{aligned} \begin{aligned} E_4&\le C_m \Vert (1{+}y) \partial _y U^P \partial _x U^P \Vert _{L^\infty _xL^2_y(\rho _0)} \Vert \partial _x H_j \Vert _{j+1} \Vert \phi _j \Vert _j \\&+ (j{+}1)C_m \Vert \partial _y U^P \partial _x^2 U^P \Vert _{L^\infty _xL^2_y(\rho _0)} \Vert H_j \Vert _j \Vert \phi _j \Vert _j \\&+ \Vert \partial _y U^P \partial _x V^P \Vert _{\infty } \Vert H_j \Vert _j \Vert \phi _j \Vert \end{aligned} \end{aligned}$$

And finally for \(E_5\) we find

$$\begin{aligned} \begin{aligned} E_5&\le \bigl ( j C_m \Vert (\partial _t + U^P \partial _x + V^P \partial _y - \partial _y^2) \partial _{xy} U^P \Vert _{L^\infty _xL^2_y(\rho _0)} + 2j \\&\quad \Vert \partial _x \partial _y^2 U^P \Vert _{\infty } \bigr ) \Vert H_j \Vert _j \Vert \phi _j \Vert _j \\&\le \bigl ( j C_m \Vert (\partial _x U^P \partial _x + \partial _x V^P \partial _y) \partial _{y} U^P \Vert _{L^\infty _xL^2_y(\rho _0)} \\&\quad + 2j \Vert \partial _x \partial _y^2 U^P\Vert _{\infty } \bigr ) \Vert H_j \Vert _j \Vert \phi _j \Vert _j \end{aligned} \end{aligned}$$

where we took again advantage of (20).

We collect the various factors in constants \(D_1,D_2,D_3\) defined as folllows:

$$\begin{aligned} \begin{aligned} D_1&= 4 \Big ( \Vert (1{+}y) \partial _y^2 U^P \Vert _{\infty } + C_m \Vert (1{+}y) \partial _y U^P \partial _x U^P \Vert _{L^\infty _xL^2_y(\rho _0)} \Big ) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} D_2&= 2 \Big ((j{+}1)C_m \Vert \partial _y U^P \partial _x^2 U^P \Vert _{L^\infty _xL^2_y(\rho _0)} + \Vert \partial _y U^P \partial _x V^P \Vert _{\infty } \\&\quad + j C_m \Vert (\partial _x U^P \partial _x + \partial _x V^P \partial _y) \partial _{y} U^P \Vert _{L^\infty _xL^2_y(\rho _0)} + 2j \Vert \partial _x \partial _y^2 U^p \Vert _{\infty }\Big ) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} D_3= & {} 2 \left( C_m \Vert \partial _{xy} U^P \Vert _{L^\infty _xL^2_y(\rho _0)} + \Vert \partial _y V^P \Vert _{\infty } + \left( j{+}\frac{1}{2}\right) \Vert \partial _x U^P \Vert _{\infty } \right. \\&\left. + \frac{1}{2} \Vert \partial _y V^P \Vert _{\infty } + C_l (j{+}1)\left( 1+ \left\| \frac{V^P}{1+y} \right\| _{\infty }\right) \right) . \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} 2 \int _0^T \sum _{i=1}^{5} E_i\, \mathrm {d}t&\le D_1 \int _0^T \Vert \partial _x H_j \Vert _{j+1} \Vert \phi _j \Vert _j\, \mathrm {d}t + D_2 \int _0^T \Vert H_j \Vert _j \Vert \phi _j \Vert _j\, \mathrm {d}t\\&\quad + D_3 \int _0^T \Vert H_j \Vert _j^2 \, \mathrm {d}t \\&\le \frac{1}{4} \int _0^T \beta ^3 (j{+}1)^3 \Vert \phi _j \Vert _j^2\, \mathrm {d}t + \frac{2D_1^2}{\beta ^3(j{+}1)^3} \int _0^T \Vert \partial _x H_j \Vert _{j+1}^2 \, \mathrm {d}t \\&\quad + \left( \frac{2D_2^2}{\beta ^3(j{+}1)^3}+D_3\right) \int _0^T \Vert H_j \Vert _j^2 \, \mathrm {d}t \end{aligned} \end{aligned}$$

With Lemma 5 the \(\phi \) integral can be estimated as

$$\begin{aligned} \frac{1}{4} \int _0^T \beta ^3 (j{+}1)^3 \Vert \phi _j \Vert _j^2\, \mathrm {d}t \le \frac{1}{2} \beta (j{+}1)\int _0^T \Vert H_j(t) \Vert _j^2\, \mathrm {d}t \end{aligned}$$

and thus can be absorbed in the LHS.

Here \(\Vert (U^P,V^P)\Vert _{low}\) has been designed such that we can find numerical constants \(c_1,c_2,c_3\) such that

$$\begin{aligned} D_1&\le c_1 (1 + \Vert (U^P,V^P) \Vert _{low})^2, \\ D_2&\le c_2 (j{+}1)\, (1 + \Vert (U^P,V^P) \Vert _{low})^2, \\ D_3&\le c_3 (j{+}1)\, (\Vert (U^P,V^P) \Vert _{low}). \end{aligned}$$

Combining all the estimates we arrive at the following lemma.

Lemma 8

Assume \(\alpha \ge \frac{1}{2}\) and \(m > \frac{1}{2}\). Then there exist a constant \(\mathcal {C} = \mathcal {C}(m,\alpha )\) such that for

$$\begin{aligned} \beta \ge \mathcal {C} (1 + \Vert (U^P,V^P) \Vert _{low}) \end{aligned}$$

and \(j \in \mathbb {N}\) the \(H_j\) defined by (17) for a solution \(u_j\) of (13) satisfy

$$\begin{aligned} \begin{aligned}&2 \Vert H_j(T) \Vert _j^2 + \beta (j{+}1)\int _0^T \Vert H_j(t) \Vert _j^2\, \mathrm {d}t + 4 \int _0^T \Vert \partial _y H_j(t) \Vert _j^2\, \mathrm {d}t \\&\quad \le \frac{8}{\beta ^3 (j{+}1)^3} \int _0^T \Vert F_j(t) \Vert _j^2 \, \mathrm {d}t + \frac{8}{\beta ^2 (j{+}1)^2} \Vert u_{\mathrm {in},j} \Vert _j^2 \\&\qquad + \frac{4c_1^2(1 + \Vert (U^P,V^P) \Vert _{low})^4}{\beta ^3 (j{+}1)^3} \int _0^T \Vert \partial _x H_j \Vert _{j+1}^2 \, \mathrm {d}t. \end{aligned} \end{aligned}$$

Proof

Use the previous estimates. Note that the condition on \(\beta \) also implies that the hypothesis of Lemma 5 is satisfied by choosing \(\mathcal {C}\) large enough. \(\square \)

Relating \(\partial _x H_j\) with \(H_{j+1}\)

To conclude the proof of Lemma 6, that will be achieved by summation of the previous estimate over j, we need first to control \(\partial _x H_j\) by \(H_{j+1}\).

Lemma 9

Let \(m > \frac{1}{2}\) and \(\alpha \ge \frac{1}{2}\). Then there exist constants \(\mathcal {C} = \mathcal {C}(m,\alpha )\) and \(C = C(m,\alpha ,r)\) such that for all \(\tau _1\), \(\beta \) and T with

$$\begin{aligned} \beta \ge \mathcal {C} \left( 1 + \Vert (U^P,V^P) \Vert _{low} \right) ^2, \quad \tau (T) \ge \tau _1, \end{aligned}$$

it holds that

$$\begin{aligned} \begin{aligned} \int _0^T \Vert \partial _x H_j \Vert _{j+1}^2 \mathrm {d}t&\le C \frac{(j{+}1)^{2\gamma }}{\tau _1^2} \int _0^T \Vert H_{j+1}\Vert ^2_{j+1} \, \mathrm {d}t \\&\quad + C \frac{(j{+}1)^{2\alpha -2}}{\beta } \int _0^T \Vert \partial _y H_j\Vert _{j}^2 \, \mathrm {d}t + \frac{C}{\beta } \int _0^T \Vert H_j \Vert _{j}^2 \, \mathrm {d}t. \end{aligned} \end{aligned}$$

Proof

From the definition of \(u_j\), it holds that \(\partial _x u_j(t) = \left( \frac{j+2}{j+1}\right) ^r \frac{(j+1)^\gamma }{\tau (t)} u_{j+1}(t)\). Hence we anticipate that

$$\begin{aligned} \partial _x H_j(t) \approx \left( \frac{j{+}2}{j{+}1}\right) ^r \frac{(j{+}1)^\gamma }{\tau (t)} H_{j+1}(t). \end{aligned}$$

Therefore we estimate the difference

$$\begin{aligned} \Delta _j := \partial _x H_j - \left( \frac{j{+}2}{j{+}1}\right) ^r \frac{(j{+}1)^\gamma }{\tau (t)} H_{j+1}. \end{aligned}$$

From equation (17) (used with indices j and \(j{+}1\)), we find that

$$\begin{aligned}&\Big (\partial _t + \beta (j{+}1)+ U^P \partial _x + (j{+}2) \partial _x U^P + V^P \partial _y - \partial _y^2\Big ) \int _0^y \Delta _j \, \mathrm {d}z\nonumber \\&\quad = - \left[ (j{+}1)\partial _{xx} U^P + \partial _x V^P \partial _y \right] \int _0^y H_j\, \mathrm {d}z. \end{aligned}$$
(21)

We stress that \(\int _0^y \Delta _j\, \mathrm {d}z\) does not converge to zero at infinity, so that one can not perform \(L^2\) estimates on this quantity. However, we can notice by 3 that

$$\begin{aligned} \begin{aligned} \Vert \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2} \int _0^y \Delta _j\, \mathrm {d}z\Vert _{L^2}&\le \Vert \bigl (\frac{\rho _{1}}{\rho _0}\bigr )^{1/2} \Vert _{L^2_y} \, \Vert \bigl (\frac{\rho _{j+1}}{\rho _1}\bigr )^{1/2} \int _0^y \Delta _j \,\mathrm {d}z \Vert _{L^2_x L^\infty _y} \\&\le C_{m-1} \Vert \Delta _j \Vert _{j+1} \\&< +\infty . \end{aligned} \end{aligned}$$
(22)

The square integrable quantity \(\delta _j = \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2} \int _0^y \Delta _j\, \mathrm {d}z\) satisfies the equation

$$\begin{aligned} \begin{aligned}&\Big (\partial _t + \beta (j{+}1)+ U^P \partial _x + (j{+}2) \partial _x U^P + V^P \partial _y - \partial _y^2\Big ) \delta _j \\&\quad = - (j{+}1)\partial _{xx} U^P \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2} \int _0^y H_j\,\mathrm {d}z - \partial _x V^P \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2} H_j \\&\qquad + V^P \partial _y \Bigl ( \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2}\Bigr ) \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{-1/2} \delta _j - 2 \partial _y \Bigl ( \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2} \Bigr ) \Delta _j\\&\qquad - \partial ^2_y \Bigl ( \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2} \Bigr ) \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{-1/2} \delta _j. \end{aligned} \end{aligned}$$
(23)

As in (22), we obtain

$$\begin{aligned} \Vert (j{+}1)\partial _{xx} U^P \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2} \int _0^y H_j \, \mathrm {d}z\Vert _{L^2} \le C_{m-1} (j{+}1)\Vert \partial _{xx} U^P \Vert _{\infty } \Vert H_j \Vert _{j+1}.\qquad \end{aligned}$$
(24)

We also get

$$\begin{aligned} \Vert \partial _x V^P \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2} H_j \Vert _{L^2}\le \Vert \frac{1}{1+y}\partial _x V^P \Vert _{\infty } \Vert H_j \Vert _{j+1}. \end{aligned}$$

By Lemma 2, we find

$$\begin{aligned} \Vert V^P \partial _y \Bigl ( \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2}\Bigr ) \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{-1/2} \delta _j \Vert _{L^2} \le \Vert \frac{1}{1+y}V^P \Vert _{\infty } C_l (j+1) \Vert \delta _j \Vert _{L^2}. \end{aligned}$$

Using again Lemma 2 and the identity

$$\begin{aligned} \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2} \Delta _j = \partial _y \delta _j - \partial _y \Bigl ( \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2} \Bigr ) \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{-1/2} \delta _j \end{aligned}$$

and defining

$$\begin{aligned} A_{j,\alpha } = \max ((j{+}1)^{1-\alpha }, \log (j{+}1), 1) \end{aligned}$$
(25)

we obtain

$$\begin{aligned} \Vert 2 \partial _y \Bigl ( \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2} \Bigr ) \Delta _j \Vert _{L^2} \le 2 C_l A_{j,\alpha } \Vert \partial _y \delta _j \Vert _{L^2} + 2 C_l^2 A_{j,\alpha }^2 \Vert \delta _j \Vert _{L^2}. \end{aligned}$$

Eventually,

$$\begin{aligned}&\Vert \partial ^2_y \Bigl ( \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2} \Bigr ) \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{-1/2} \delta _j \Vert _{L^2} \\&\quad = \Vert \partial _y \Bigl ( \sum _{k=1}^{j+1}\frac{1}{k^\alpha (1+\frac{y}{k^\alpha })} \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2} \Bigr ) \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{-1/2} \delta _j \Vert _{L^2} \le C A_{j,\alpha }^2 \Vert \delta _j \Vert _{L^2} \end{aligned}$$

for some constant \(C = C(\alpha )\). The previous bounds combined with an energy estimate yield that for \(\mathcal {C}\) large enough (we remind that \(\alpha \ge \frac{1}{2}\)):

$$\begin{aligned}&\Vert \delta _j(T)\Vert _{L^2}^2 + \beta (j+1) \int _0^T \Vert \delta _j \Vert _{L^2}^2\, \mathrm {d}t \nonumber \\&\quad + \int _0^T \Vert \partial _y \delta _j \Vert _{L^2}^2\, \mathrm {d}t \le (j{+}1)\int _0^T \Vert H_j \Vert _{j+1}^2 \, \mathrm {d}t. \end{aligned}$$
(26)

We can then take the x-derivative of equation (23) and proceed as above. For \(\mathcal {C}\) large enough, we get

$$\begin{aligned}&\Vert \partial _x \delta _j(T)\Vert _{L^2}^2 + \beta (j{+}1)\int _0^T \Vert \partial _x \delta _j \Vert _{L^2}^2\,\mathrm {d}t + \int _0^T \Vert \partial _x \partial _y \delta _j \Vert _{L^2}^2 \,\mathrm {d}t\\&\quad \le (j{+}1)\int _0^T (\Vert \partial _x H_j \Vert _{j+1}^2 +\Vert H_j \Vert _{j+1}^2)\,\mathrm {d}t \\&\qquad + \int _0^T \left( 2 \Vert \partial _x V^P \partial _y \Bigl ( \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2}\Bigr ) \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{-1/2} \delta _j \Vert _{L^2} \right. \\&\left. \qquad +\,2 (j{+}2) \Vert \partial ^2_x U^P \delta _j \Vert _{L^2} + 2 \Vert \partial _x V^P \partial _y \delta _j \Vert _{L^2} \right) \Vert \partial _x \delta _j \Vert _{L^2}\,\mathrm {d}t \end{aligned}$$

We then use that

$$\begin{aligned} \Vert \partial _x V^P \partial _y \Bigl ( \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{1/2}\Bigr ) \bigl (\frac{\rho _{j+1}}{\rho _0}\bigr )^{-1/2} \delta _j \Vert _{L^2} \le \left\| \frac{\partial _x V^P}{1+y} \right\| _{\infty } C_l (j{+}1)\Vert \delta _j \Vert _{L^2}, \end{aligned}$$

and

$$\begin{aligned} \Vert \partial _x V^P \partial _y \delta _j \Vert _{L^2} \le \left\| \frac{\partial _x V^P}{1+y} \right\| _{\infty } C_l (j{+}1)\Vert \delta _j \Vert _{L^2} + \left\| \frac{\partial _x V^P}{1+y} \right\| _{\infty } \Vert \Delta _j \Vert _{j+1} \end{aligned}$$

and the bound (26) to end up with

$$\begin{aligned} \begin{aligned}&\Vert \partial _x \delta _j(T)\Vert _{L^2}^2 + \frac{\beta (j{+}1)}{2} \int _0^T \Vert \partial _x \delta _j \Vert _{L^2}^2\, \mathrm {d}t + \int _0^T \Vert \partial _x \partial _y \delta _j \Vert _{L^2}^2 \, \mathrm {d}t\\&\quad \le 2(j{+}1)\int _0^T ( \Vert \partial _x H_j \Vert _{j+1}^2 + \Vert H_j \Vert _{j+1}^2 + \Vert \Delta _j \Vert _{j+1}^2) \, \mathrm {d}t. \end{aligned} \end{aligned}$$
(27)

To estimate directly \(\Delta _j\), we differentiate the Eq. (21) with respect to y, which gives

$$\begin{aligned}&\Big (\partial _t + \beta (j{+}1)+ U^P \partial _x + (j{+}2) \partial _x U^p + V^P \partial _y + \partial _y V^P - \partial _y^2\Big ) \Delta _j \\&\quad = -(j{+}1)\partial _{xx} U^P H_j - (j{+}1)\partial _{xxy} U^P \int _0^y H_j - \partial _{xy} V^P H_j - \partial _x V^P \partial _y H_j \\&\qquad - \partial _y U^P \partial _x \int _0^y \Delta _j - (j{+}2) \partial _{xy} U^P \int _0^y \Delta _j. \end{aligned}$$

We take the \(\left\langle \,,\right\rangle _{j+1}\) scalar product with \(\Delta _j\):

$$\begin{aligned} \begin{aligned}&\left( \frac{1}{2} \partial _t + \beta (j{+}1)\right) \left\| \Delta _j \right\| _{j+1}^2 \\&\qquad - \Bigl [ \left( j{+}2\right) \Vert \partial _x U^P \Vert _{\infty } + \frac{1}{2} \Vert \partial _y V^P \Vert _{\infty } + \frac{1}{2} \Vert V^P \frac{\partial _y \rho _{j+1}}{\rho _{j+1}} \Vert _{\infty } \Bigr ] \left\| \Delta _j \right\| _{j+1}^2 \\&\qquad + \Vert \partial _y \Delta _j \Vert _{j+1}^2 - \left\langle \partial _y \Delta _j,\frac{\partial _y \rho _{j+1}}{\rho _{j+1}} \Delta _j\right\rangle _{j+1} \\&\quad \le (j{+}1)\bigl (\Vert \partial _{xx} U^P \Vert _{\infty } + C_m \Vert \partial _{xxy} U^P \Vert _{L^\infty _xL^2_y(\rho _0)} + \Vert \partial _{xy} V^P \Vert _{\infty }\bigr ) \Vert H_j \Vert _{j+1} \left\| \Delta _j \right\| _{j+1} \\&\qquad + \left\| \frac{\partial _x V^P}{1+y} \right\| _{\infty } \Vert (1{+}y) \partial _y H_j \Vert _{j+1} \left\| \Delta _j \right\| _{j+1} \\&\qquad + \left( \Vert \partial _{y} U^P \sqrt{\rho _0} \Vert _{\infty } \Vert \partial _x \delta _j \Vert _{L^2} + (j{+}2) \Vert \partial _{xy} U^P \sqrt{\rho _0}\Vert _{\infty } \Vert \delta _j \Vert _{L^2} \right) \left\| \Delta _j \right\| _{j+1} . \end{aligned} \end{aligned}$$

By the 1d Sobolev imbedding theorem, we find that for a constant \(C = C(m)\) it holds that

$$\begin{aligned} \Vert \partial _{y} U^P \sqrt{\rho _0} \Vert _{\infty } \le C \Vert (U^P,V^P)\Vert _{low} \text { and } \Vert \partial _{xy} U^P \sqrt{\rho _0}\Vert _{\infty } \le C \Vert (U^P,V^P)\Vert _{low}. \end{aligned}$$

Combining these last two inequalities with (26), (27) and the inequality

$$\begin{aligned} \Vert (1+y)\partial _y H_j\Vert _{j+1} \le (j{+}1)^\alpha \Vert \partial _y H_j \Vert _j, \end{aligned}$$

and taking \(\mathcal {C}\) large enough, we obtain

$$\begin{aligned} \begin{aligned}&\Vert \Delta _j(T) \Vert _{j+1}^2 + \beta (j{+}1)\int _0^T \Vert \Delta _j\Vert _{j+1}^2 \, \mathrm {d}t + \int _0^T \Vert \partial _y \Delta _j\Vert _{j+1}^2 \, \mathrm {d}t\\&\quad \le (j{+}1)\int _0^T \Vert H_j\Vert _{j+1}^2 \, \mathrm {d}t + (j{+}1)^{2\alpha -1} \int _0^T \Vert \partial _y H_j\Vert _{j}^2\, \mathrm {d}t \\&\qquad + \frac{1}{\beta (j{+}1)} \int _0^T \Vert \partial _x H_j \Vert _{j+1}^2 \, \mathrm {d}t. \end{aligned} \end{aligned}$$
(28)

Lemma 9 follows straightforwardly. \(\square \)

Combining Lemmas 8 and 9, we will now prove Lemma 6.

Proof of Lemma 6

We choose \(\mathcal {C}\) such that Lemmas 8 and 9 apply. We multiply the inequality in Lemma 8 by \(\beta (j{+}1)^{2\gamma -1}\) and sum over j to get

$$\begin{aligned} \begin{aligned}&\sum _{j=0}^\infty \beta ^2 (j{+}1)^{2\gamma } \left[ \int _0^T \Vert H_j(t) \Vert _j^2\, \mathrm {d}t + \frac{1}{\beta (j{+}1)} \Vert H_j(T) \Vert _j^2 \right. \\&\left. \qquad + \frac{1}{\beta (j{+}1)} \int _0^T \Vert \partial _y H_j(t) \Vert _j^2\, \mathrm {d}t \right] \\&\quad \le 8 \sum _{j=0}^{\infty } \left[ \frac{(j{+}1)^{2\gamma -4}}{\beta ^2} \int _0^T \Vert F_j(t) \Vert _j^2\, \mathrm {d}t + \frac{(j{+}1)^{2\gamma -3}}{\beta } \Vert u_{\mathrm {in},j} \Vert _j^2 \right] \\&\qquad + \sum _{j=0}^{\infty } \frac{4c_1^2(1 + \Vert (U^P,V^P) \Vert _{low})^4}{\beta ^2} (j{+}1)^{2\gamma -4} \int _0^T \Vert \partial _x H_j \Vert _{j+1}^2 \, \mathrm {d}t. \end{aligned} \end{aligned}$$

Taking \(\mathcal {C}\) large enough, we can then find by Lemma 9 a constant \(C = C(m,\alpha ,r)\) such that

$$\begin{aligned}&\sum _{j=0}^{\infty } (j{+}1)^{2\gamma -4} \int _0^T \Vert \partial _x H_j \Vert _{j+1}^2\, \mathrm {d}t \\&\quad \le C \left( 1+\frac{1}{\tau ^2}\right) \sum _{j=0}^\infty (j{+}1)^{4\gamma -4} \int _0^T \Vert H_j \Vert _j^2\, \mathrm {d}t\\&\qquad + \frac{C}{\beta } \sum _{j=0}^\infty (j{+}1)^{2(\gamma +\alpha )-6} \int _0^T \Vert \partial _y H_j \Vert _j^2\, \mathrm {d}t \\&\quad \le C \left( 1+\frac{1}{\tau ^2}\right) \sum _{j=0}^\infty (j{+}1)^{4\gamma -4} \left[ \int _0^T \Vert H_j \Vert _j^2 \, \mathrm {d}t\right. \\&\left. \qquad + \frac{1}{\beta (j{+}1)} \int _0^T \Vert \partial _y H_j \Vert _j^2\, \mathrm {d}t \right] . \end{aligned}$$

We have used here that \(\alpha \le \gamma +\frac{1}{2}\). Hence, the last term at the right-hand side can be absorbed if

$$\begin{aligned} \frac{1}{2} \beta ^2 (j{+}1)^{2\gamma } \ge \frac{4Cc_1^2(1 + \Vert (U^P,V^P) \Vert _{low})^4}{\beta ^2} \left( 1+\frac{1}{\tau ^2}\right) (j{+}1)^{4\gamma -4}, \end{aligned}$$

which can be ensured by a suitable large \(\mathcal {C}\) if \(\gamma \le 2\). \(\square \)

Control of \(u_j\) and \(\omega _j\)

We now relate the estimates on \(H_j\) to \(u_j\) and start with an estimate for the \(L^2\) norm.

Lemma 10

Let \(m > \frac{1}{2}\) and \(\alpha \ge \frac{1}{2}\). Then there exists a constant \(\mathcal {C} = \mathcal {C}(m,\alpha )\) such that for

$$\begin{aligned} \beta \ge \mathcal {C} \left( 1 + \Vert (U^P,V^P) \Vert _{low} \right) \end{aligned}$$

and for any \(\epsilon _1, \epsilon _2, \epsilon _3, \epsilon _4 > 0\) it holds that

$$\begin{aligned} \begin{aligned}&\frac{1}{2} \int _0^T \Vert u_j \Vert _j^2\, \mathrm {d}t - \frac{\epsilon _1}{(j{+}1)^{2\gamma }} \int _0^T \Vert \partial _x u_j \Vert _{j+1}^2\, \mathrm {d}t - \frac{\epsilon _2}{\beta (j{+}1)^{\gamma }} \int _0^T \Vert \partial _y u_j \Vert _j^2\, \mathrm {d}t\\&\quad - \frac{\epsilon _3}{4\beta (j{+}1)^\gamma } \Vert u_j (T) \Vert _j^2 - \frac{\epsilon _4}{\beta ^2 (j{+}1)^{2\gamma }} \int _0^T \Vert \partial _y^2 u_j(t) \Vert _j^2\, \mathrm {d}t\\&\le \frac{\beta (j{+}1)^\gamma }{\epsilon _3} \Vert H_j(T) \Vert _j^2 \\&\quad + \left[ 16 \beta ^2 (j{+}1)^2 + \frac{(j{+}1)^{2\gamma }}{\epsilon _1} C_m^2 \Vert (1{+}y) \partial _y U^P \Vert _{L^\infty _{t,x}L^2_y(\rho _0)}^2 + \frac{\beta ^2(j{+}1)^{2\gamma }}{4\epsilon _4} \right] \\&\qquad \int _0^T \Vert H_j \Vert _j^2 \, \mathrm {d}t \\&\quad + \left[ \frac{\beta (j{+}1)^{\gamma }}{4\epsilon _2} + 16 C_l^2 A_{j,\alpha }^2 \right] \int _0^T \Vert \partial _y H_j \Vert _j^2 \, \mathrm {d}t + \int _0^T \Vert H_j \Vert _j \Vert F_j \Vert _j\, \mathrm {d}t \end{aligned} \end{aligned}$$

where \(u_j\) is satisfying (16), \(A_{j,\alpha }\) is defined in (25) and \(H_j\) is defined by (17).

Proof

Using the definition (17) of \(H_j\) we find

$$\begin{aligned} \begin{aligned} \int _0^T \Vert u_j(t) \Vert _j^2 \, \mathrm {d}t =&\int _0^T \left\langle \Big (\partial _t + \beta (j{+}1)+ U^P \partial _x + (j{+}1)\partial _x U^p\right. \\&\left. + V^P \partial _y + \partial _y V^P - \partial _y^2\Big ) H_j,u_j\right\rangle _j\, \mathrm {d}t \\&\,+ \int _0^T \left\langle (\partial _y U^P \partial _x + (j{+}1)\partial _{xy} U^P) \int _0^y H_j\, \mathrm {d}z,u_j\right\rangle _j\, \mathrm {d}t. \end{aligned} \end{aligned}$$
(29)

By the evolution equation (16) for \(u_j\), the first term can be written (from the partial integration in \(y\) there is no boundary term as \(u|_{y=0} = 0\))

$$\begin{aligned} \begin{aligned}&\int _0^T \left\langle \Big (\partial _t + \beta (j{+}1)+ U^P \partial _x + (j{+}1)\partial _x U^p + V^P \partial _y + \partial _y V^P - \partial _y^2\Big ) H_j,u_j\right\rangle _j\, \mathrm {d}t \\&= \left\langle H_j(T),u_j(T)\right\rangle _j \\ {}&\quad + \int _0^T \left\langle H_j, \left( -\partial _t + \beta (j{+}1)- U^P\partial _x + j \partial _x U^P - V^P \partial _y - V^P \frac{\partial _y \rho _j}{\rho _j}\right) u_j\right\rangle \mathrm {d}t\\&\quad + \int _0^T \left\langle \partial _y H_j,\left( \partial _y + \frac{\partial _y \rho _j}{\rho _j}\right) u_j\right\rangle _j\mathrm{d}t \\&= \left\langle H_j(T),u_j(T)\right\rangle _j + \int _0^T \left\langle H_j, \left( 2\beta (j{+}1)+ (2j{+}1) \partial _x U^P - V^P \frac{\partial _y \rho _j}{\rho _j}\right) u_j\right\rangle _j\, \mathrm {d}t \\&\quad + \int _0^T \left\langle H_j,\partial _y U^P v_j + j \partial _{xy} U^P \partial _x^{-1} v_j\right\rangle _j\, \mathrm {d}t - \int _0^T \left\langle H_j,F_j\right\rangle _j\, \mathrm {d}t \\&\quad + \int _0^T \left\langle \partial _y H_j,\left( \partial _y + \frac{\partial _y \rho _j}{\rho _j}\right) u_j\right\rangle _j\, \mathrm {d}t - \int _0^T \left\langle H_j,\partial _y^2 u_j\right\rangle _j\, \mathrm {d}t. \end{aligned} \end{aligned}$$

The terms can now be bounded using Lemma 2:

$$\begin{aligned} \begin{aligned}&\left\langle H_j, \left( 2\beta (j{+}1)+ (2j{+}1) \partial _x U^P - V^P \frac{\partial _y \rho _j}{\rho _j}\right) u_j\right\rangle _j \\&\quad \le \left\| 2\beta (j{+}1)+ (2j{+}1) \partial _x U^P - V^P \frac{\partial _y \rho _j}{\rho _j} \right\| _{\infty } \Vert H_j \Vert _j \Vert u_j \Vert _j \\&\quad \le (j{+}1)\left[ 2\beta + 2 \Vert \partial _x U^P \Vert _{\infty } + C_l \left\| \frac{V^P}{1+y} \right\| _{\infty } \right] \Vert H_j \Vert _j \Vert u_j \Vert _j. \end{aligned} \end{aligned}$$

Recalling that \(v_j = -\partial _x \int _0^y u_j \, \mathrm {d}z\) we find

$$\begin{aligned} \begin{aligned}&\left\langle H_j,\partial _y U^P v_j + j \partial _{xy} U^P \partial _x^{-1} v_j\right\rangle _j\, \mathrm {d}t \\&\quad \le C_m \Vert (1{+}y) \partial _y U^P \Vert _{L^\infty _xL^2_y(\rho _0)} \Vert H_j \Vert _{j} \Vert \partial _x u_j \Vert _{j+1} + j \Vert \partial _{xy} U^P \Vert _{L^\infty _xL^2_y(\rho _0)} \Vert H_j \Vert _j \Vert u_j \Vert _j. \end{aligned} \end{aligned}$$

For the forcing terms we find

$$\begin{aligned} - \left\langle H_j,F_j\right\rangle _j \le \Vert H_j \Vert _j \Vert F_j \Vert _j. \end{aligned}$$

The diffusion terms give

$$\begin{aligned} \begin{aligned}&\left\langle \partial _y H_j,(\partial _y + \frac{\partial _y \rho _j}{\rho _j}) u_j\right\rangle _j - \left\langle H_j,\partial _y^2 u_j\right\rangle _j \\&\quad \le \Vert \partial _y H_j \Vert _j\, \Vert \partial _y u_j \Vert _j + C_l A_{j,\alpha } \Vert \partial _y H_j \Vert _j\, \Vert u_j \Vert _j + \Vert H_j \Vert _j \Vert \partial _y^2 u_j \Vert _j. \end{aligned} \end{aligned}$$

The integrand in the second integral in (29) can be estimated as

$$\begin{aligned} \begin{aligned}&\left\langle (\partial _y U^P \partial _x + (j{+}1)\partial _{xy} U^P) \int _0^y H_j\, \mathrm {d}z,u_j\right\rangle _j\\&\quad \le C_m \Vert (1{+}y) \partial _y U^P \Vert _{L^\infty _xL^2_y(\rho _0)} \Vert H_j \Vert _j \Vert \partial _x u_j \Vert _{j+1} + j C_m \Vert \partial _{xy} U^P \Vert _{L^\infty _xL^2_y(\rho _0)} \Vert H \Vert _j \Vert u \Vert _j. \end{aligned} \end{aligned}$$

Collecting the terms we find by choosing \(\mathcal {C}\) large enough that

$$\begin{aligned} \begin{aligned} \int _0^T \Vert u_j \Vert ^2\, \mathrm {d}t&\le \left\langle H_j(T),u_j(T)\right\rangle _j + 4 \beta (j{+}1)\int _0^T \Vert H_j \Vert _j \Vert u \Vert _j \, \mathrm {d}t\\&\quad + 2C_m \Vert (1{+}y) \partial _y U^P \Vert _{L^\infty _xL^2_y(\rho _0)} \int _0^T \Vert H_j \Vert _j \Vert \partial _x u_j \Vert _{j+1}\, \mathrm {d}t \\&\quad + \int _0^T \Big ( \Vert \partial _y H_j \Vert _j\, \Vert \partial _y u_j \Vert _j + C_l A_{j,\alpha } \Vert \partial _y H_j \Vert _j\, \Vert u_j \Vert _j \\&\quad + \Vert H_j \Vert _j \Vert \partial _y^2 u_j \Vert _j \Big )\, \mathrm {d}t \\&\quad + \int _0^T \Vert H_j \Vert _j \Vert F_j \Vert _j \, \mathrm {d}t. \end{aligned} \end{aligned}$$

Splitting the products gives the claimed estimate. \(\square \)

The missing terms can be estimated by the evolution of \(u_j\) and \(\omega _j = \partial _y u_j\). For \(u_j\) we find:

Lemma 11

Let \(m > \frac{1}{2}\) and \(\alpha \ge \frac{1}{2}\). Then there exists a constant \(\mathcal {C} = \mathcal {C}(m,\alpha )\) such that for

$$\begin{aligned} \beta \ge \mathcal {C} \left( 1 + \Vert (U^P,V^P) \Vert _{low} \right) \end{aligned}$$

the solution \(u_j\) of (16) satisfies

$$\begin{aligned} \begin{aligned}&\frac{1}{2} \Vert u_j(T) \Vert _j^2 + \frac{1}{2} \int _0^T \Vert \partial _y u_j \Vert ^2\, \mathrm {d}t\\&\qquad -4 \beta (j{+}1)^\gamma \int _0^T \Vert u_j \Vert _j^2\, \mathrm {d}t - \frac{C_m^2 \Vert (1{+}y) \partial _y U^P \Vert _{L^\infty _xL^2_y(\rho _0)}^2}{\beta (j{+}1)^\gamma } \int _0^T \Vert \partial _x u_{j} \Vert _{j+1}^2 \, \mathrm {d}t\\&\quad \le \frac{1}{2} \Vert u_{\mathrm {in},j} \Vert _j^2 + \frac{1}{\beta (j{+}1)^\gamma } \int _0^T \Vert F_j \Vert _j^2\, \mathrm {d}t. \end{aligned} \end{aligned}$$

Proof

By (16) we find

$$\begin{aligned} \left\langle \partial _t u_j,u_j\right\rangle _j= & {} \left\langle \Big (-\beta (j{+}1)- U^P \partial _x - (j{+}1)\partial _x U^p - V^P \partial _y + \partial _y^2\Big )u_j,u_j\right\rangle _j\\&\quad - \left\langle \partial _y U^P v_j + j \partial _{xy} U^P \partial _x^{-1} v_j,u_j\right\rangle _j + \left\langle F_j,u_j\right\rangle _j \\\le & {} - \frac{1}{2} \Vert \partial _y u_j \Vert _j^2 + 4 \beta (j{+}1)^\gamma \Vert u_j \Vert _j^2 \\&\quad + \frac{C_m^2 \Vert (1{+}y) \partial _y U^P \Vert _{L^\infty _xL^2_y(\rho _0)}^2}{\beta (j{+}1)^\gamma } \Vert \partial _x u_{j+1} \Vert _{j+1}^2 \\&\quad + \frac{1}{\beta (j{+}1)^\gamma } \Vert F_j \Vert _j^2, \end{aligned}$$

where there is no boundary term from the partial integration in \(y\) as \(u_j\) vanishes at the boundary and we used in the inequality that \(\mathcal {C}\) can be chosen large enough. Integrating this over [0, T] gives the claimed result. \(\square \)

By differentiating (16) in y and find

$$\begin{aligned}&\Big (\partial _t + \beta (j{+}1)+ U^P \partial _x + (j{+}1)\partial _x U^p + V^P \partial _y + \partial _y V^P - \partial _y^2\Big ) \omega _j \nonumber \\&\quad + \partial _{yy} U^P v_j + j \partial _{xyy} U^P \partial _x^{-1} v_j + \partial _{xy} U^P u_j = \partial _y F_j. \end{aligned}$$
(30)

This immediately yields the following control for \(\omega _j\).

Lemma 12

Let \(m > \frac{1}{2}\) and \(\alpha \ge \frac{1}{2}\). Then there exists a constant \(\mathcal {C} = \mathcal {C}(m,\alpha )\) such that for

$$\begin{aligned} \beta \ge \mathcal {C} \left( 1 + \Vert (U^P,V^P) \Vert _{low} \right) \end{aligned}$$

the vorticity \(\omega _j = \partial _y u_j\) satisfies

$$\begin{aligned} \begin{aligned}&\Vert (1+y) \omega _j(T) \Vert _j^2 + \int _0^T \beta (j{+}1)\Vert (1+y) \omega _j \Vert _j^2\, \mathrm {d}t + \int _0^T \Vert (1+y) \partial _y \omega _j \Vert _j^2\, \mathrm {d}t\\&\quad \le \frac{4C_m^2 \Vert (1+y)^2 \partial _{yy} U^P \Vert _{L^\infty _{t,x}L^2_y(\rho _0)}^2}{\beta (j{+}1)} \int _0^T \Vert \partial _x u_{j} \Vert _{j+1}^2\, \mathrm {d}t \\&\qquad + \frac{4C_m^2 (j{+}1)\Vert (1+y) \partial _{xyy} U^P \Vert _{L^\infty _{t,x}L^2_y(\rho _0)}^2}{\beta } \int _0^T \Vert u_j \Vert _j^2\, \mathrm {d}t\\&\qquad + \Vert (1+y) \omega _{\mathrm {in},j} \Vert _j^2 + 4 \int _0^T \Vert (1+y) F_j \Vert _j^2\, \mathrm {d}t + 4 \int _0^T \Vert F_j|_{y=0} \Vert _{L^2_x}^2\, \mathrm {d}t. \end{aligned} \end{aligned}$$

Proof

Integrate (30) against \((1+y)^2\omega _j\) in \(\Vert \cdot \Vert _j\). This yields

$$\begin{aligned}&\frac{1}{2} \partial _t \Vert (1+y) \omega _j \Vert _j^2 + \beta (j{+}1)\Vert (1+y) \omega _j \Vert _j^2 + \Vert (1+y) \partial _y \omega _j \Vert _j^2 \\&\quad \le j \Vert \partial _x U^P \Vert _{\infty } \Vert (1+y) \omega _j \Vert _j^2 + \left\| V^P \left( \frac{\partial _y \rho _j}{\rho _j} + \frac{\partial _y(1+y)^2}{(1+y)^2}\right) \right\| _{\infty } \Vert (1+y) \omega _j \Vert _j^2 \\&\qquad +\, \Vert \omega _j|_{y=0} \Vert _{L^2_x} \Vert \partial _y \omega _j|_{y=0} \Vert _{L^2_x}\\&\qquad +\, \Vert (1+y) \partial _y \omega _j \Vert _j \left\| (1+y) \left( \frac{\partial _y \rho _j}{\rho _j} + \frac{\partial _y(1+y)^2}{(1+y)^2}\right) \omega _j \right\| _j \\&\qquad +\, C_m \Vert (1+y)^2 \partial _{yy} U^P \Vert _{L^\infty _xL^2_y(\rho _0)} \Vert \partial _x u_{j} \Vert _{j+1} \Vert (1+y) \omega _j \Vert _j\\&\qquad +\, j C_m \Vert (1+y) \partial _{xyy} U^P \Vert _{L^\infty _xL^2_y(\rho _0)} \Vert u_{j} \Vert _{j} \Vert (1+y) \omega _j \Vert _j\\&\qquad +\, \Vert (1+y) F_j \Vert _j\, \left\| (1+y) \left( \partial _y + \frac{\partial _y \rho _j}{\rho _j} + \frac{\partial _y (1+y)^2}{(1+y)^2}\right) \omega _j \right\| _j, \end{aligned}$$

where we find a boundary term from the diffusion and there is no boundary term from \(V^P\partial _y\) because \(V^P|_{y=0}=0\).

From (16) we find \(\partial _y \omega _j|_{y=0} = F_j|_{y=0}\). For \(\omega _j|_{y=0}\) write

$$\begin{aligned} |\omega _j(y=0)| \le \int _0^1 \left[ \omega _j \sqrt{\rho _j} + \int _0^y |(\omega _j \sqrt{\rho _j})'|\, \mathrm {d}z \right] \, \mathrm {d}y \end{aligned}$$

to get

$$\begin{aligned} \Vert \omega _j|_{y=0} \Vert _{L^2_x}^2 \le 2 \left( 1+ \left\| \frac{\partial _y \rho _j}{\rho _j} \right\| \right) ^2 \Vert \omega _j \Vert _j^2 + 2 \Vert \partial _y \omega _j \Vert _j^2. \end{aligned}$$

By choosing \(\mathcal {C}\) large enough and using that \(\alpha \ge \frac{1}{2}\), the result follows after integration over time. \(\square \)

Combining the results, we can conclude this section.

Proof of Proposition 7

Adding the control of Lemma 11 with a factor \(\epsilon _3 (j{+}1)^{-\gamma } / \beta \) and Lemma 12 with a factor \((j{+}1)^{1-2\gamma } / \beta ^2\) to the inequality of Lemma 10 yields

$$\begin{aligned}&\left( \frac{1}{2} - 4 \epsilon _3 - \frac{4C_m^2 \Vert (1+y) \partial _{xyy} U^P \Vert _{L^\infty _{t,x}L^2_y(\rho _0)}^2}{(j{+}1)^{2\gamma -2}\beta ^3} \right) \int _0^T \Vert u_j \Vert _j^2 \, \mathrm {d}t\\&\qquad +\, \frac{\epsilon _3}{4\beta (j{+}1)^\gamma } \Vert u_j(T) \Vert _j^2 + \left( \frac{\epsilon _3}{2\beta (j{+}1)^\gamma } - \frac{\epsilon _2}{\beta (j{+}1)^\gamma }\right) \int _0^T \Vert \partial _y u_j \Vert _j^2\, \mathrm {d}t \\&\qquad - \frac{\epsilon _4}{\beta ^2 (j{+}1)^{2\gamma }} \int _0^T \Vert \partial _y^2 u_j(t) \Vert _j^2\, \mathrm {d}t \\&\qquad - \left( \frac{\epsilon _1}{(j{+}1)^{2\gamma }} + \frac{\epsilon _3 C_m \Vert (1{+}y) \partial _y U^P \Vert _{L^\infty _{t,x}L^2_y(\rho _0)}^2}{\beta ^2 (j{+}1)^{2\gamma }} x\right. \\&\left. \qquad + \frac{4C_m^2 \Vert (1+y) \partial _{xyy} U^P \Vert _{L^\infty _{t,x}L^2_y(\rho _0)}^2}{(j{+}1)^{2\gamma }\beta ^3} \right) \int _0^T \Vert \partial _x u_j \Vert _{j+1}^2\, \mathrm {d}t \\&\, \qquad +\frac{(j{+}1)^{1-2\gamma }}{\beta ^2} \Vert (1+y) \omega _j(T) \Vert _j^2 + \int _0^T \frac{(j{+}1)^{2-2\gamma }}{\beta } \Vert (1+y) \omega _j \Vert _j^2\, \mathrm {d}t\\&\qquad + \int _0^T \frac{(j{+}1)^{1-2\gamma }}{\beta ^2} \Vert (1+y) \partial _y \omega _j \Vert _j^2\, \mathrm {d}t\\&\quad \le \frac{\beta (j{+}1)^\gamma }{\epsilon _3} \Vert H_j(T) \Vert _j^2 \\&\qquad + \left[ 16 \beta ^2 (j{+}1)^2 + \frac{(j{+}1)^{2\gamma }}{\epsilon _1} C_m^2 \Vert (1{+}y) \partial _y U^P \Vert _{L^\infty _t L^\infty _xL^2_y(\rho _0)}^2 + \frac{\beta ^2(j{+}1)^{2\gamma }}{4\epsilon _4} \right] \\&\quad \qquad \int _0^T \Vert H_j \Vert _j^2 \, \mathrm {d}t \\&\qquad + \left[ \frac{\beta (j{+}1)^{\gamma }}{4\epsilon _2} + 16 C_l^2 A_{j,\alpha }^2 \right] \int _0^T \Vert \partial _y H_j \Vert _j^2 \, \mathrm {d}t + \int _0^T \Vert H_j \Vert _j \Vert F_j \Vert _j\, \mathrm {d}t \\&\qquad + \frac{\epsilon _3}{2\beta (j{+}1)^\gamma } \Vert u_{\mathrm {in},j} \Vert _j^2 + \frac{\epsilon _3}{\beta ^2 (j{+}1)^{2\gamma }} \int _0^T \Vert F_j \Vert _j^2\, \mathrm {d}t \\&\qquad + \frac{(j{+}1)^{1-2\gamma }}{\beta ^2} \Vert \omega _{\mathrm {in},j} \Vert _j^2 + \frac{4(j{+}1)^{1-2\gamma }}{\beta ^2} \int _0^T \Vert (1+y) F_j \Vert _j^2\, \mathrm {d}t \\&\qquad + \frac{4(j{+}1)^{1-2\gamma }}{\beta ^2}\int _0^T \Vert F_j|_{y=0} \Vert _{L^2_x}^2\, \mathrm {d}t. \end{aligned}$$

Using that \(\partial _x u_j(t) = \left( \frac{j+2}{j+1}\right) ^r \frac{(j+1)^\gamma }{\tau (t)} u_{j+1}(t)\), we can sum over j and choose \(\epsilon _1,\epsilon _2,\epsilon _3,\epsilon _4\) appropriately to arrive for \(m> \frac{1}{2}, \alpha \ge \frac{1}{2},\gamma \ge 1, \tau _1 > 0, r \in \mathbb {R}\) at the control

$$\begin{aligned} \begin{aligned}&\sum _{j=0}^{\infty } \left\{ \int _0^T \Vert u_j \Vert _j^2 \, \mathrm {d}t + \frac{1}{\beta (j{+}1)^\gamma } \Vert u_j(T) \Vert _j^2 + \frac{(j{+}1)^{2-2\gamma }}{\beta } \int _0^T \Vert (1+y) \omega _j \Vert _j^2\, \mathrm {d}t \right\} \\&\qquad + \sum _{j=0}^{\infty } \frac{(j{+}1)^{1-2\gamma }}{\beta ^2} \left\{ \Vert (1+y) \omega _j(T) \Vert _j^2 + \int _0^T \Vert (1+y) \partial _y\omega _j(t) \Vert _j^2 \, \mathrm {d}t \right\} \\&\quad \le C \sum _{j=0}^{\infty } \left\{ \beta (j{+}1)^\gamma \Vert H_j(T) \Vert _j^2 \right. \\&\left. \qquad + \beta ^2 (j{+}1)^{2\gamma } \int _0^T \Vert H_j \Vert _j^2 \, \mathrm {d}t + \beta (j{+}1)^\gamma \int _0^T \Vert \partial _y H_j \Vert _j^2\, \mathrm {d}t \right\} \\&\qquad + C \sum _{j=0}^{\infty } \left\{ \frac{1}{\beta (j{+}1)^\gamma } \Vert u_{\mathrm {in},j} \Vert _j^2 + \frac{(j{+}1)^{1-2\gamma }}{\beta } \Vert (1+y) \omega _{\mathrm {in},j} \Vert _j^2 \right\} \\&\qquad + C \sum _{j=0}^{\infty } \left\{ \int _0^T \frac{1}{\beta ^2(j{+}1)^{2\gamma }} \Vert F_j \Vert _j^2\, \mathrm {d}t + \frac{(j{+}1)^{1-2\gamma }}{\beta ^2} \int _0^T \Vert (1+y) F_j \Vert _j^2\, \mathrm {d}t\right. \\&\left. \qquad + \frac{(j{+}1)^{1-2\gamma }}{\beta ^2} \int _0^T \Vert F_j|_{y=0} \Vert _{L^2_x}^2\, \mathrm {d}t. \right\} \end{aligned} \end{aligned}$$
(31)

if

$$\begin{aligned} \beta \ge \mathcal {C} (1 + \Vert (U^P,V^P) \Vert _{low})\, \left( 1 + \frac{1}{\tau _1} + \Vert (U^P,V^P) \Vert _{low}\right) \text { and } \tau (T) \ge \tau _1 \end{aligned}$$

where C and \(\mathcal {C}\) are constant only depending on \(m,\alpha , \gamma ,r\) (and not \(\tau _1\)).

Controlling H by Lemma 6 then yields the result for a fixed time \(T\). Applying this estimate for all T in \([0,T^*]\) then shows the claimed estimate.

For \(\gamma \ge 5/4\) we find that \((j{+}1)^{2\gamma -4} \ge (j{+}1)^{1-2\gamma }\) so that

$$\begin{aligned} (j{+}1)^{2\gamma -4} \Vert F_j \Vert _j^2 + (j{+}1)^{1-2\gamma } \Vert (1+y)F_j \Vert _j^2 \le 2 (j{+}1)^{2\gamma -4} \Vert \left( 1+\frac{y}{(j{+}1)^{2\gamma - \frac{5}{2}}}\right) F_j \Vert _j^2, \end{aligned}$$

which proves the expression in this case. \(\square \)

Nonlinear Estimates

In order to close the estimate, we have to estimate \(F_j\).

Proposition 13

Fix the parameters \(m,\alpha ,\gamma ,r\) and an additional parameter R such that

$$\begin{aligned} \begin{aligned}&\gamma \in \left[ \frac{3}{2},2\right] , \quad \alpha \le \gamma - 1, \quad m \ge \frac{2\gamma -1}{\alpha } + 1, \\&r> 2\gamma , \quad R > 2\gamma + 1, \quad R \ge r + 3\gamma - 2. \end{aligned} \end{aligned}$$
(32)

Then there exists a constant \(C = C(m,\alpha ,\gamma ,r)\) such that for \(\beta \), \(\tau _1\) and T with \(\tau (T) \ge \tau _1\),

$$\begin{aligned} \begin{aligned}&\sum _{j=0}^{\infty } \frac{1}{(j{+}1)^{4-2\gamma }} \int _0^T \left\| \left( 1+\frac{y}{(j{+}1)^{2\gamma - \frac{5}{2}}} \right) F_j \right\| _j^2 \, \mathrm {d}t\\&\quad \le 2 \int _0^T \Vert (1+y) f^e_j \Vert _{\gamma ,\tau ,r-2+\gamma }^2\,\mathrm{d}t \\&\qquad + \frac{C \beta }{\tau _1^4} \left[ \sup _{[0,T]} \left( \Vert u \Vert _{\gamma ,\tau ,r-\frac{\gamma }{2}}^2 + \frac{\Vert (1+y) \omega \Vert _{\gamma ,\tau ,r+\frac{1}{2}-\gamma }^2}{\beta } + | U^E |_{\gamma ,\tau ,R}^2 \right) \right] \\&\quad \qquad \int _0^T \left[ \Vert u \Vert _{\gamma ,\tau ,r}^2 + \frac{\Vert (1+y) \omega \Vert _{\gamma ,\tau ,r+1-\gamma }^2}{\beta } \right] \, \mathrm {d}t. \end{aligned} \end{aligned}$$

We restrict to the case of \(\gamma \ge 3/2\) because we need \(\alpha \ge \gamma -1\) in order to control the terms \(\partial _x^k u \partial _x^{l-k+1} u\) in \(F_j\). Combined with the earlier requirement that \(\alpha \ge 1/2\) this yields \(\gamma \ge 3/2\).

Proof

Write \(F_j = f_j^e + \sum _{i=1}^6 F_j^i\) with

$$\begin{aligned} F_j^1&= M_j \left[ u \partial _x, \partial _x^j\right] u + M_j (j{+}1)\, \partial _x u\, \partial _x^j u, \\ F_j^2&= M_j \left[ \partial _y u, \partial _x^j\right] v + M_j j\, \partial _{xy} u\, \partial _x^{j-1} v + M_j v\, \partial _x^j \partial _y u, \\ F_j^3&= M_j \left[ U^e \partial _x, \partial _x^j\right] u + M_j j\, \partial _x U^e\, \partial _x^j u, \\ F_j^4&= M_j \left[ V^e \partial _y, \partial _x^j\right] u, \\ F_j^5&= M_j \left[ \partial _x U^e, \partial _x^j\right] u, \\ F_j^6&= M_j \left[ \partial _y U^e, \partial _x^j \right] v + M_j j\, \partial _{xy} U^e\, \partial _x^{j-1} v. \end{aligned}$$

As \(\gamma \ge 3/2\) and \(\alpha \le \gamma -1\), we have \(2\gamma - \frac{5}{2} \ge \alpha \), so that

$$\begin{aligned} \left\| \left( 1+\frac{y}{(j{+}1)^{2\gamma - \frac{5}{2}}} \right) F_j \right\| _j \le \left\| \left( 1+\frac{y}{(j{+}1)^\alpha }\right) F_j \right\| _j \end{aligned}$$

so that it suffices to bound the right-hand side.

Analysis of \(F_j^1\). We write

$$\begin{aligned} F^1_j= & {} \sum _{l=2}^{\left\lfloor \frac{j+1}{2} \right\rfloor } \left( {\begin{array}{c}j\\ l\end{array}}\right) \frac{M_j}{M_l M_{j-l+1}} u_l u_{j-l+1} \, \\&\quad + \, \sum _{l=\left\lfloor \frac{j+1}{2} \right\rfloor +1}^{j-1} \left( {\begin{array}{c}j\\ l\end{array}}\right) \frac{M_j}{M_l M_{j-l+1}} u_l u_{j-l+1} \, =: \, F^1_{j,low} + F^1_{j,high}. \end{aligned}$$

For \(F_{j,low}^1\), we notice that for \(l \le \left\lfloor \frac{j+1}{2} \right\rfloor \) there exist a constant \(C = C(r)\) with

$$\begin{aligned} \left( {\begin{array}{c}j\\ l\end{array}}\right) \frac{M_j}{M_l M_{j-l+1}} \le \frac{C}{\tau _1} \left( {\begin{array}{c}j\\ l\end{array}}\right) ^{1-\gamma } \frac{(j{+}1)^{\gamma }}{(l{+}1)^r}. \end{aligned}$$

This shows

$$\begin{aligned} \frac{1}{(j{+}1)^{2-\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F^1_{j,low} \right\| _j\le & {} \frac{C}{\tau _1} \sum _{l=2}^{\left\lfloor \frac{j+1}{2} \right\rfloor } \left( {\begin{array}{c}j\\ l\end{array}}\right) ^{1-\gamma } \frac{(j{+}1)^{2\gamma -2}}{(l{+}1)^r} \Vert u_l u_{j-l+1} \Vert _{j-1} \\\le & {} \frac{C}{\tau _1} \sum _{l=2}^{\left\lfloor \frac{j+1}{2} \right\rfloor } \left( {\begin{array}{c}j\\ l\end{array}}\right) ^{1-\gamma } \frac{(j{+}1)^{2\gamma -2}}{(l{+}1)^r} \\&\left\| \left( \frac{\rho _{j-1}}{\rho _{j-l+1}}\right) ^{1/2} u_l \right\| _{L^\infty _{x,y}} \Vert u_{j-l+1} \Vert _{j-l+1}. \end{aligned}$$

Note that for an absolute constant \(C_a\),

$$\begin{aligned} \left( {\begin{array}{c}j\\ l\end{array}}\right) ^{1-\gamma } (j+1)^{2\gamma -2} \le C_a \quad \hbox { for all}\ 2 \le l \le \left\lfloor \frac{j+1}{2} \right\rfloor . \end{aligned}$$
(33)

From the 1d Sobolev embedding and Lemma 3, we find that for \(n \le \min (m-1,l)\):

$$\begin{aligned} \left\| \left( \frac{\rho _{j-1}}{\rho _{j-l+1}}\right) ^{1/2} u_l \right\| _{L^\infty _{x,y}}&\le C_A \left\| \left( \frac{\rho _{j-1}}{\rho _{j-l+1}}\right) ^{1/2} \partial _x u_l \right\| _{L^2_xL^\infty _y} \\&\le C_A \, C_{1}\, \sup _y \left( \frac{\rho _{j-1} \rho _{n}}{\rho _{l} \rho _{j-l+1}}\right) ^{1/2} \Vert \partial _x \partial _y u_l \Vert _{l} \\&\le \frac{C}{\tau _1} \sup _y\left( \frac{\rho _{j-1} \rho _{n}}{\rho _{l} \rho _{j-l+1}}\right) ^{1/2} (l{+}1)^\gamma \, \Vert (1+y) \omega _{l+1} \Vert _{l+1} \end{aligned}$$

where \(C_A\) is an absolute constant, C is a constant depending on mr. Note that we used here Lemma 4 to bound \(\Vert \omega _{l+1}\Vert _l\) by \(\Vert (1+y)\omega _{l+1}\Vert _{l+1}\).The factor with the \(\rho \) is explicit:

$$\begin{aligned} \left( \frac{\rho _{j-1} \rho _{n}}{\rho _{l} \rho _{j-l+1}}\right) ^{1/2} = \frac{\prod _{k=1}^{l} \left( 1+\frac{y}{k^\alpha }\right) }{\prod _{k=j-l+2}^{j-1} \left( 1+\frac{y}{k^\alpha }\right) \prod _{k=1}^n \left( 1+\frac{y}{k^\alpha }\right) }. \end{aligned}$$

For \(l \le m-1\), we take \(n=l\) and find that

$$\begin{aligned} \left( \frac{\rho _{j-1} \rho _{n}}{\rho _{l} \rho _{j-l+1}}\right) ^{1/2} \le 1. \end{aligned}$$

For \(l > m-1\), we take \(n=m-1\) and find that

$$\begin{aligned} \begin{aligned} \left( \frac{\rho _{j-1} \rho _{n}}{\rho _{l} \rho _{j-l+1}}\right) ^{1/2}&\le \frac{\prod _{k=1}^{l} \left( 1+\frac{y}{k^\alpha }\right) }{\prod _{k=j-l+2}^{j-m+2} \left( 1+\frac{y}{k^\alpha }\right) \prod _{k=1}^{m-1} \left( 1+\frac{y}{k^\alpha }\right) } \\&\le \left( \frac{(j-l+2)\cdots (j-m+2)}{m\cdots l} \right) ^\alpha \\&\le C \left( {\begin{array}{c}j\\ l\end{array}}\right) ^\alpha (j{+}1)^{-\alpha (m-1)} \end{aligned} \end{aligned}$$

for a constant \(C = C(m,\alpha )\) and using that \(l \le \left\lfloor \frac{j+1}{2} \right\rfloor \).

Hence we find for a constant \(C = C(m,\alpha ,r)\) that

$$\begin{aligned}&\frac{1}{(j{+}1)^{2-\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F^1_{j,low} \right\| _j \\&\quad \le \frac{C}{\tau _1^2} \sum _{l=2}^{\left\lfloor \frac{j+1}{2} \right\rfloor } (l{+}1)^{\gamma -r} \Vert (1+y) \omega _{l+1} \Vert _{l+1} \Vert u_{j-l+1} \Vert _{j-l+1} \end{aligned}$$

using that \(1-\gamma +\alpha \le 0\) and \(2\gamma -2 \le \alpha (m-1)\).

The discrete Young’s convolution inequality implies for all \(t \in [0,T]\) that

$$\begin{aligned}&\sum _{j=0}^{\infty } \frac{1}{(j{+}1)^{4-2\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F^1_{j,low} \right\| _j^2\\&\quad \le \frac{\mathcal {C}}{\tau _1^4} \left( \sum _{l=0}^\infty (l{+}1)^{\gamma -r} \Vert (1+y) \omega _l \Vert _l \right) ^2 \sum _{j=0}^{\infty } \Vert u_j \Vert _j^2 \\&\quad \le \frac{\mathcal {C}}{\tau _1^4} \left( \sum _{l=0}^{\infty } (l{+}1)^{4\gamma -1-2r} \right) \left( \sum _{l=0}^{\infty } (l{+}1)^{1-2\gamma } \Vert (1+y) \omega _l \Vert _l^2\right) \left( \sum _{j=0}^{\infty } \Vert u_j \Vert _j^2 \right) . \end{aligned}$$

As \(4\gamma - 1 - 2r < -1\), the first integral is finite. Hence we arrive at the required estimate

$$\begin{aligned} \sum _{j=0}^{\infty } \frac{1}{(j{+}1)^{4-2\gamma }} \int _0^T \Vert F^1_{j,low} \Vert _j^2\, \mathrm {d}t \le \frac{C}{\tau _1^2} \sup _{t\in [0,T]} \Vert (1+y) \omega \Vert _{\gamma ,\tau ,r+1-\gamma }^2 \int _0^T \Vert u \Vert _{\gamma ,\tau ,r}^2\, \mathrm {d}t \end{aligned}$$

with a constant \(C=C(m,\alpha ,\gamma ,r)\).

For the treatment of \(F^1_{j,high}\) swap the roles of \(u_l\) and \(u_{j-l+1}\) so that

$$\begin{aligned} F_{j,high}^1 = \sum _{l=2}^{j- \left\lfloor \frac{j+1}{2} \right\rfloor } \left( {\begin{array}{c}j\\ l-1\end{array}}\right) \frac{M_j}{M_l M_{j-l+1}} u_l u_{j-l+1}. \end{aligned}$$

In the given range \(l=2,\dots ,j-\left\lfloor \frac{j+1}{2} \right\rfloor \) we find

$$\begin{aligned} \left( {\begin{array}{c}j\\ l-1\end{array}}\right) \le \left( {\begin{array}{c}j\\ l\end{array}}\right) \end{aligned}$$

so that it can be bounded as \(F_{j,low}^1\).

Analysis of \(F_j^2\). We write

$$\begin{aligned} \begin{aligned} F_j^2&= - \sum _{l=2}^{\left\lfloor \frac{j+1}{2} \right\rfloor } \left( {\begin{array}{c}j\\ l\end{array}}\right) \frac{M_j}{M_l M_{j-l+1}} \partial _y u_l\, \partial _x^{-1} v_{j-l+1}\\&\quad - \sum _{l=\left\lfloor \frac{j+1}{2} \right\rfloor +1}^{j-1} \left( {\begin{array}{c}j\\ l\end{array}}\right) \frac{M_j}{M_l M_{j-l+1}} \partial _y u_l\, \partial _x^{-1} v_{j-l+1} =: F_{j,low}^2 + F_{j,high}^2 \end{aligned} \end{aligned}$$

and note that it vanishes unless \(j \ge 3\).

By \(v_{j-l+1} = -\partial _x \int _0^y u_{j-l+1}\, \mathrm {d}z\) we find for \(n \le \min (m-1,j-l+1)\) using the 1d Sobolev inequality and Lemma 3 that

$$\begin{aligned} \begin{aligned}&\left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) \partial _y u_l \partial _x^{-1} v_{j-l+1} \right\| _j\\&\quad \le \left\| \partial _y u_l \partial _x^{-1} v_{j-l+1} \right\| _{j-1}\\&\quad \le C_{m-n} \left\| \left( \frac{\rho _{j-1} \rho _{n}}{\rho _{l} \rho _{j-l+1}}\right) ^{1/2} \partial _y u_l \right\| _{L^\infty _xL^2_y(\rho _{l})} \Vert u_{j-l+1} \Vert _{j-l+1} \\&\quad \le \frac{\mathcal {C}}{\tau _1} \sup _y\left( \frac{\rho _{j-1} \rho _{n}}{\rho _{l} \rho _{j-l+1}}\right) ^{1/2} (l{+}1)^\gamma \, \Vert (1+y) \omega _{l+1} \Vert _{l+1} \Vert u_{j-l+1} \Vert _{j-l+1} \end{aligned} \end{aligned}$$

for a constant \(\mathcal {C} = \mathcal {C}(m,r)\).

In the range \(l=2,\dots ,\left\lfloor \frac{j+1}{2} \right\rfloor \) for \(F_{j,low}^2\) we find that \(j-l+1 \ge \frac{j+1}{2}\) and as we can assume that \(j \ge 3\) we can always ensure that this is at least 2.

For \(\frac{j+1}{2} \le m-1\), we can take \(n=2\) and find a constant \(C = C(m,r)\) such that

$$\begin{aligned} \sup _y\left( \frac{\rho _{j-1} \rho _{n}}{\rho _{l} \rho _{j-l+1}}\right) ^{1/2} \le C \end{aligned}$$

and otherwise we can taken \(n=m-1\) and find the same control as for \(F_{j,low}^1\) as

$$\begin{aligned}&\frac{1}{(j{+}1)^{2-\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F^2_{j,low} \right\| _j \\&\quad \le \frac{C}{\tau _1^2} \sum _{l=2}^{\left\lfloor \frac{j+1}{2} \right\rfloor } (l{+}1)^{\gamma -r} \Vert (1+y) \omega _{l+1} \Vert _{l+1} \Vert u_{j-l+1} \Vert _{j-l+1} \end{aligned}$$

and we can conclude as for \(F_{j,low}^1\).

For \(F_{j,high}^2\) we find

$$\begin{aligned} F_{j,high}^2 = - \sum _{l=2}^{j-\left\lfloor \frac{j+1}{2} \right\rfloor } \left( {\begin{array}{c}j\\ l-1\end{array}}\right) \frac{M_j}{M_l M_{j-l+1}} \partial _x^{-1} v_l\, \partial _y u_{j-l+1}. \end{aligned}$$

For \(n = \min (m-1,l+1)\) we find

$$\begin{aligned} \begin{aligned}&\left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) \partial _x^{-1} v_l\, \partial _y u_{j-l+1} \right\| _{j}\\&\quad \le \left\| \left( \frac{\rho _{j-1}}{\rho _{j-l}}\right) ^{1/2} \partial _x^{-1} v_l \right\| _{L^\infty _{x,y}} \Vert (1+y) \omega _{j-l+1} \Vert _{j-l+1} \\&\quad \le \frac{C}{\tau _1} \sup _y\left( \frac{\rho _{j-1} \rho _{n}}{\rho _{l+1} \rho _{j-l}}\right) ^{1/2} (l{+}1)^{\gamma } \Vert u_{l+1} \Vert _{l+1} \Vert (1+y) \omega _{j-l+1} \Vert _{j-l+1}. \end{aligned} \end{aligned}$$

For \(l+1<m-1\) we can find a constant \(C = C(m)\) such that

$$\begin{aligned} \left( {\begin{array}{c}j\\ l-1\end{array}}\right) \le \left( {\begin{array}{c}j\\ l\end{array}}\right) (j{+}1)^{-1}. \end{aligned}$$

Using the stronger assumption \(2\gamma -1 \le \alpha (m-1)\), we can then conclude as in the treatment of \(F_{j,low}^1\) that

$$\begin{aligned}&\frac{1}{(j{+}1)^{2-\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F^2_{j,high} \right\| _j \\&\quad \le \frac{C}{\tau _1^2} \sum _{l=2}^{j-\left\lfloor \frac{j+1}{2} \right\rfloor } (l{+}1)^{\gamma -r} \Vert u_{l+1} \Vert _{l+1}\, (j{+}1)^{-1} \Vert (1+y) \omega _{j-l+1} \Vert _{j-l+1}. \end{aligned}$$

Hence we find

$$\begin{aligned}&\sum _{j=0}^{\infty } \frac{1}{(j{+}1)^{4-2\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F^2_{j,high} \right\| _j^2 \\&\quad \le \frac{C}{\tau _1^4} \left( \sum _{l=0}^\infty (l{+}1)^{\gamma -r} \Vert u_l \Vert _l \right) ^2 \sum _{j=0}^\infty (j{+}1)^{-2} \Vert (1+y) \omega _j \Vert _j^2 \\&\quad \le \frac{C}{\tau _1^4} \left( \sum _{l=0}^{\infty } (l{+}1)^{3\gamma -2r}\right) \left( \sum _{l=0}^{\infty } (l{+}1)^{-\gamma } \Vert u_l \Vert _l^2\right) \left( \sum _{j=0}^{\infty } (j{+}1)^{-2} \Vert (1+y) \omega _j \Vert _j^2 \right) . \end{aligned}$$

As \(3\gamma -2r < -1\), this gives the required estimate

$$\begin{aligned} \sum _{j=0}^{\infty } \frac{1}{(j{+}1)^{4-2\gamma }} \int _0^T \Vert F^2_{j,high} \Vert _j^2\, \mathrm {d}t \le \frac{C}{\tau _1^4} \sup _{t\in [0,T]} \Vert u \Vert _{\gamma ,\tau ,r-\frac{\gamma }{2}}^2 \int _0^T \Vert (1+y) \omega \Vert _{\gamma ,\tau ,r-1}^2\, \mathrm {d}t \end{aligned}$$

with a constant \(C=C(m,\alpha ,\gamma ,r)\). As \(r-1 \le r+1-\gamma \) this is the required control.

Analysis of \(F_{j}^3\) and \(F_{j}^5\). We write

$$\begin{aligned} \begin{aligned} F_j^3 + F_j^5&= - \sum _{l=2}^{j} \left( {\begin{array}{c}j\\ l\end{array}}\right) \frac{M_j}{M_l M_{j-l+1}} U_l^e u_{j-l+1} - \sum _{l=1}^{j} \left( {\begin{array}{c}j\\ l\end{array}}\right) \frac{M_j}{M_{l+1} M_{j-l}} U_{l+1}^e u_{j-l} \\&= - \sum _{l=2}^{\left\lfloor \frac{j+1}{2} \right\rfloor } \left[ \left( {\begin{array}{c}j\\ l\end{array}}\right) + \left( {\begin{array}{c}j\\ l-1\end{array}}\right) \right] \frac{M_j}{M_l M_{j-l+1}} U_l^e u_{j-l+1} \\&\quad + \sum _{l=\left\lfloor \frac{j+1}{2} \right\rfloor +1}^{j+1} \left[ \left( {\begin{array}{c}j\\ l\end{array}}\right) + \left( {\begin{array}{c}j\\ l-1\end{array}}\right) \right] \frac{M_j}{M_l M_{j-l+1}} U_l^e u_{j-l+1}\\&=: F_{j,low}^{3,5} + F_{j,high}^{3,5} \end{aligned} \end{aligned}$$

with the convention that

$$\begin{aligned} \left( {\begin{array}{c}j\\ j+1\end{array}}\right) = 0. \end{aligned}$$

Using the definition of \(U^e\) and the 1d Sobolev embedding theorem we find

$$\begin{aligned} \Vert U_l^e \Vert _{L^\infty _{x,y}} \le \Vert U^E_l \Vert _{L^\infty _x} \le \frac{C_s (l{+}1)^\gamma }{\tau _1} \Vert U^E_{l+1} \Vert . \end{aligned}$$

As \(l\ge 2\), this implies

$$\begin{aligned} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) U_l^e u_{j-l+1} \right\| _j \le \frac{C_s(l{+}1)^\gamma }{\tau _1} \Vert U^E_{l+1} \Vert \, \Vert u_{j-l+1} \Vert _{j-l+1}. \end{aligned}$$

For \(l=2,\dots ,\left\lfloor \frac{j+1}{2} \right\rfloor \) we find for a constant \(C = C(\gamma ,r)\)

$$\begin{aligned} \left[ \left( {\begin{array}{c}j\\ l\end{array}}\right) + \left( {\begin{array}{c}j\\ l-1\end{array}}\right) \right] \frac{M_j}{M_l M_{j-l+1}} \le \frac{C}{\tau _1} \left( {\begin{array}{c}j\\ l\end{array}}\right) ^{1-\gamma } \frac{(j{+}1)^\gamma }{(l{+}1)^r} \end{aligned}$$

so that as \(l \ge 2\)

$$\begin{aligned} \frac{1}{(j{+}1)^{2-\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F_{j,low}^{3,5} \right\| _j \le \frac{C}{\tau _1^2} \sum _{l=2}^{\left\lfloor \frac{j+1}{2} \right\rfloor } (l{+}1)^{\gamma -r} \Vert U^E_{l+1} \Vert \, \Vert u_{j-l+1} \Vert _{j-l+1}. \end{aligned}$$

Hence we find

$$\begin{aligned}&\sum _{j=0}^{\infty } \frac{1}{(j{+}1)^{4-2\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F_{j,low}^{3,5} \right\| _j^2\\&\quad \le \frac{C}{\tau _1^4} \left( \sum _{l=0}^\infty (l{+}1)^{\gamma -r} \Vert U^E_l \Vert \right) ^2 \sum _{j=0}^\infty \Vert u_j \Vert _j^2 \\&\quad \le \frac{C}{\tau _1^4} \left( \sum _{l=0}^{\infty } (l{+}1)^{2\gamma -2R} \right) \left( \sum _{l=0}^{\infty } (l{+}1)^{2R-2r} \Vert U^E_l \Vert ^2\right) \left( \sum _{j=0}^{\infty } \Vert u_j \Vert _j^2 \right) . \end{aligned}$$

As \(2\gamma -R < -1\) this gives the bound

$$\begin{aligned}&\int _0^T \sum _{j=0}^{\infty } \frac{1}{(j{+}1)^{4-2\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F_{j,low}^{3,5} \right\| _j^2 \, \mathrm {d}t \\&\quad \le \frac{C}{\tau _1^4} \sup _{t\in [0,T]} |U^E|_{\gamma ,\tau ,R}^2 \int _0^T \Vert u \Vert _{\gamma ,\tau ,r}^2\, \mathrm {d}t. \end{aligned}$$

For \(l=\left\lfloor \frac{j+1}{2} \right\rfloor +1,\dots ,j\) we find

$$\begin{aligned} \left[ \left( {\begin{array}{c}j\\ l\end{array}}\right) + \left( {\begin{array}{c}j\\ l-1\end{array}}\right) \right] \frac{M_j}{M_l M_{j-l+1}} \le \frac{C}{\tau _1} \left( {\begin{array}{c}j\\ l-1\end{array}}\right) ^{1-\gamma } \frac{(l{+}1)^\gamma }{(j{-}l{+}1)^r} \end{aligned}$$

so that

$$\begin{aligned}&\frac{1}{(j{+}1)^{2-\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F_{j,high}^{3,5} \right\| _j \\&\quad \le \frac{C}{\tau _1^2} \sum _{l=\left\lfloor \frac{j+1}{2} \right\rfloor +1}^j (l{+}1)^{3\gamma -2} \Vert U^E_{l+1} \Vert \, (j{-}l{+}1)^{-r} \Vert u_{j-l+1} \Vert _{j-l+1}. \end{aligned}$$

Hence we find

$$\begin{aligned}&\sum _{j=0}^{\infty } \frac{1}{(j{+}1)^{4-2\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F_{j,high}^{3,5} \right\| _j^2\\&\quad \le \frac{\mathcal {C}}{\tau _1^4} \left( \sum _{j=0}^\infty (j{+}1)^{-r} \Vert u_l \Vert _l \right) ^2 \sum _{l=0}^\infty (l{+}1)^{6\gamma -4} \Vert U^E_{l+1} \Vert \\&\quad \le \frac{\mathcal {C}}{\tau _1^4} \left( \sum _{j=0}^{\infty } (j{+}1)^{-2r} \right) \left( \sum _{j=0}^{\infty } \Vert u_l \Vert _l^2\right) \left( \sum _{l=0}^{\infty } (l{+}1)^{6\gamma -4} \Vert U^E_{l+1} \Vert ^2 \right) . \end{aligned}$$

As \(r > \frac{1}{2}\) this gives the bound

$$\begin{aligned}&\int _0^T \sum _{j=0}^{\infty } \frac{1}{(j{+}1)^{4-2\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F_{j,high}^{3,5} \right\| _j^2 \, \mathrm {d}t \\&\quad \le \frac{\mathcal {C}}{\tau _1^4} \sup _{t\in [0,T]} |U^E|_{\gamma ,\tau ,r+3\gamma -2}^2 \int _0^T \Vert u \Vert _{\gamma ,\tau ,r}^2\, \mathrm {d}t, \end{aligned}$$

which is the required bound as \(R \ge r+3\gamma -2\).

Analysis of \(F_j^4\). This term is creating trouble with the integrability in y as \(V^e \sim y\) and is the reason for most technical difficulties.

We write

$$\begin{aligned} \begin{aligned} F_{j}^4&= - \sum _{l=1}^{\left\lfloor \frac{j+1}{2} \right\rfloor } \left( {\begin{array}{c}j\\ l\end{array}}\right) \frac{M_j}{M_{l+1}M_{j-l}} \partial _x^{-1} V^e_{l+1} \partial _y u_{j-l}\\&\quad \,-\, \sum _{l=\left\lfloor \frac{j+1}{2} \right\rfloor +1}^j \left( {\begin{array}{c}j\\ l\end{array}}\right) \frac{M_j}{M_{l+1}M_{j-l}} \partial _x^{-1} V^e_{l+1} \partial _y u_{j-l} \\&=: F_{j,low}^4 + F_{j,high}^4. \end{aligned} \end{aligned}$$

As \(l \ge 1\) we find

$$\begin{aligned} \begin{aligned}&\left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) \partial _x^{-1} V^e_{l+1} \partial _y u_{j-l} \right\| _{j} \\&\quad \le \left\| \frac{\partial _x^{-1} V^e_{l+1}}{1+y} \right\| _{L^\infty _{x,y}} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) (1+y) \omega _{j-l} \right\| _j \\&\quad \le \frac{C}{\tau } (l{+}1)^{\gamma } \Vert U^E_{l+2} \Vert \, \Vert (1+y) \omega _{j-l} \Vert _{j-l} \end{aligned} \end{aligned}$$

where \(C=C(r)\) is constant. In the last line we used the 1d Sobolev inequality and that

$$\begin{aligned} \sqrt{\frac{\rho _j}{\rho _{j-l}}} \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) \le C. \end{aligned}$$

For \(l=1,\dots ,\left\lfloor \frac{j+1}{2} \right\rfloor \) we find

$$\begin{aligned} \left( {\begin{array}{c}j\\ l\end{array}}\right) \frac{M_j}{M_{l+1}M_{j-l}} \le \frac{C}{\tau _1} \left( {\begin{array}{c}j\\ l\end{array}}\right) ^{1-\gamma } (l{+}1)^{\gamma -r} \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned}&\quad \frac{1}{(j{+}1)^{2-\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F_{j,low}^4 \right\| _j \\&\quad \le \frac{C}{\tau _1^2} \sum _{l=1}^{\left\lfloor \frac{j+1}{2} \right\rfloor } \left( {\begin{array}{c}j\\ l\end{array}}\right) ^{1-\gamma } (l{+}1)^{2\gamma -r} (j{+}1)^{\gamma -2} \Vert U^E_{l+2} \Vert \, \Vert (1+y) \omega _{j-l} \Vert _{j-l} \\&\quad \le \frac{C}{\tau _1^2} \sum _{l=1}^{\left\lfloor \frac{j+1}{2} \right\rfloor } (l{+}1)^{2\gamma -r} \Vert U^E_{l+2} \Vert \, (j{+}1)^{-1} \Vert (1+y) \omega _{j-l} \Vert _{j-l}. \end{aligned} \end{aligned}$$

Hence we find

$$\begin{aligned} \begin{aligned}&\sum _{j=0}^{\infty } \frac{1}{(j{+}1)^{4-2\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F_{j,low}^4 \right\| _j^2 \\&\quad \le \frac{\mathcal {C}}{\tau _1^4} \left( \sum _{l=0}^{\infty } (l{+}1)^{2\gamma -r} \Vert U^E_{l+2}\Vert \right) ^2 \left( \sum _{j=0}^{\infty } (j{+}1)^{-2} \Vert (1+y) \omega _j \Vert _j^2 \right) \\&\quad \le \frac{\mathcal {C}}{\tau _1^4} | U^E|_{\gamma ,\tau ,R}^2 \left( \sum _{j=0}^{\infty } (j{+}1)^{-2} \Vert (1+y) \omega _j \Vert _j^2 \right) \\ \end{aligned} \end{aligned}$$

as \(4\gamma -2R < -1\). This gives the bound

$$\begin{aligned}&\int _0^T \sum _{j=0}^{\infty } \frac{1}{(j{+}1)^{4-2\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F_{j,low}^4 \right\| _j^2 \, \mathrm {d}t \\&\quad \le \frac{\mathcal {C}}{\tau _1^4} \sup _{t\in [0,T]} |U^E|_{\gamma ,\tau ,R}^2 \int _0^T \Vert (1+y) \omega \Vert _{\gamma ,\tau ,r-1}^2\, \mathrm {d}t, \end{aligned}$$

which is the required bound as \(-1 \le 1-\gamma \).

For \(F_{j,high}^4\) we find

$$\begin{aligned}&\frac{1}{(j{+}1)^{2-\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F_{j,high}^4 \right\| _j \\&\quad \le \frac{C}{\tau ^2} \sum _{l=\left\lfloor \frac{j+1}{2} \right\rfloor +1}^{j} \frac{(l{+}1)^{3\gamma -2}}{(j{-}l{+}1)^{r}} \Vert U^E_{l+1} \Vert \, \Vert (1+y) \omega _{j-l} \Vert _{j-l}. \end{aligned}$$

As \(-1+\gamma -r < - \frac{1}{2}\) this gives the bound

$$\begin{aligned}&\int _0^T \sum _{j=0}^{\infty } \frac{1}{(j{+}1)^{4-2\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F_{j,high}^4 \right\| _j^2 \, \mathrm {d}t \\&\quad \le \frac{\mathcal {C}}{\tau _1^4} \sup _{t\in [0,T]} |U^E|_{\gamma ,\tau ,r+3\gamma -2}^2 \int _0^T \Vert (1+y)\omega \Vert _{\gamma ,\tau ,r+1-\gamma }^2\, \mathrm {d}t. \end{aligned}$$

As \(R \ge r+3\gamma -2\) this is the required result.

Analysis of \(F_j^6\). We write

$$\begin{aligned} F_j^6= & {} -\sum _{l=2}^{\left\lfloor \frac{j+1}{2} \right\rfloor } \left( {\begin{array}{c}j\\ l\end{array}}\right) \frac{M_j}{M_l M_{j-l+1}} \partial _y U^e_l \partial _x^{-1} v_{j-l+1}\\&\quad -\sum _{l=\left\lfloor \frac{j+1}{2} \right\rfloor +1}^{j} \left( {\begin{array}{c}j\\ l\end{array}}\right) \frac{M_j}{M_l M_{j-l+1}} \partial _y U^e_l \partial _x^{-1} v_{j-l+1} =: F_{j,low}^6 + F_{j,high}^6. \end{aligned}$$

As \(\partial _y U^e\) is exponentially decaying, we find

$$\begin{aligned} \begin{aligned} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) \partial _y U^e_l \partial _x^{-1} v_{j-l+1} \right\| _j&\le C \Vert U^E_l \Vert _{L^\infty _x} \Vert u_{j-l+1} \Vert _{j-l+1}\\&\le \frac{C(l{+}1)^{\gamma }}{\tau } \Vert U^E_{l+1} \Vert \, \Vert u_{j-l+1} \Vert _{j-l+1}. \end{aligned} \end{aligned}$$

For \(F_{j,low}^6\) we find (using that \(\left( {\begin{array}{c}j\\ l\end{array}}\right) \frac{M_j}{M_l M_{j-l+1}}\le C (l+1)^{-r}\) for \(l=2,\ldots , \left\lfloor \frac{j+1}{2} \right\rfloor \)):

$$\begin{aligned} \frac{1}{(j{+}1)^{2-\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F_{j,low}^6 \right\| _j \le \frac{C}{\tau _1^2} \sum _{l=2}^{\left\lfloor \frac{j+1}{2} \right\rfloor +1} (l{+}1)^{\gamma -r} \Vert U^E_{l+1} \Vert \, \Vert u_{j-l+1} \Vert _{j-l+1}. \end{aligned}$$

As \(\gamma -R < - \frac{1}{2}\) this gives the control

$$\begin{aligned}&\int _0^T \sum _{j=0}^{\infty } \frac{1}{(j{+}1)^{4-2\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F_{j,low}^6 \right\| _j^2 \, \mathrm {d}t \\&\quad \le \frac{\mathcal {C}}{\tau _1^4} \sup _{t\in [0,T]} |U^E|_{\gamma ,\tau ,R}^2 \int _0^T \Vert u \Vert _{\gamma ,\tau ,r}^2\, \mathrm {d}t. \end{aligned}$$

For \(F_{j,high}^6\) we find

$$\begin{aligned}&\frac{1}{(j{+}1)^{2-\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F_{j,high}^6 \right\| _j \\&\quad \le \frac{C}{\tau _1^2} \sum _{l=\left\lfloor \frac{j+1}{2} \right\rfloor +1}^{j} (l{+}1)^{2\gamma -2} \Vert U^E_{l+1} \Vert \, (j{+}1)^{\gamma -r} \Vert u_{j-l+1} \Vert _{j-l+1}. \end{aligned}$$

As \(\gamma -r < - \frac{1}{2}\) this gives the control

$$\begin{aligned}&\int _0^T \sum _{j=0}^{\infty } \frac{1}{(j{+}1)^{4-2\gamma }} \left\| \left( 1+\frac{y}{(j{+}1)^{\alpha }} \right) F_{j,high}^6 \right\| _j^2 \, \mathrm {d}t \\&\quad \le \frac{\mathcal {C}}{\tau _1^4} \sup _{t\in [0,T]} |U^E|_{\gamma ,\tau ,r+1-\gamma }^2 \int _0^T \Vert u \Vert _{\gamma ,\tau ,r}^2\, \mathrm {d}t, \end{aligned}$$

which is the required control as \(R \ge r+1-\gamma \). \(\square \)

As a direct consequence of Proposition 7 and Proposition 13, we can state the following corollary, where we use that \(F_j|_{y=0} = f^e_j|_{y=0}\) as \(u\) and \(v\) vanish at \(y=0\).

Corollary 14

Fix the parameters \(m, \alpha , \gamma , r, R\) as in (32) and \(\alpha \ge 1/2\). There exists \(\mathcal {C}\) and \(\mathbf {C}\) such that for all \(\beta , \tau _1, T\) with

$$\begin{aligned} \beta \ge \mathcal {C} (1 + \Vert (U^P,V^P) \Vert _{low})\, \left( 1 + \frac{1}{\tau _1} + \Vert (U^P,V^P) \Vert _{low}\right) , \quad \text { and } \tau (T) \ge \tau _1 \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned} {\left| \left| \left| u \right| \right| \right| }^2&\le \mathbf {C} \left[ \frac{1}{\beta } \Vert u_{\mathrm {in}} \Vert _{\gamma ,\tau _0,r+\gamma -\frac{3}{2}}^2 + \frac{1}{\beta ^2} \Vert (1+y) \omega _{\mathrm {in}} \Vert _{\gamma ,\tau _0,r+\frac{1}{2} - \gamma }^2\right] \\&+ \mathbf {C} \left[ + \frac{1}{\beta ^2} \int _0^T \Vert f^e_j|_{y=0} \Vert _{\gamma ,\tau ,r-2+\gamma }^2\, \mathrm {d}t \right. \\&\left. + \frac{1}{\beta ^2} \int _0^T \Vert (1+y) f^e_j \Vert _{\gamma ,\tau ,r-2+\gamma }^2 \, \mathrm {d}t\right] \, \\&+ \frac{\mathbf {C}}{\tau _1^4} \left( \frac{1}{\beta }|U^E|_{\gamma ,\tau ,R}^2 + {\left| \left| \left| u \right| \right| \right| }^2 \right) {\left| \left| \left| u \right| \right| \right| }^2 \end{aligned} \end{aligned}$$
(34)

where

$$\begin{aligned} \begin{aligned} {\left| \left| \left| u \right| \right| \right| }^2 =&\int _0^T \Vert u \Vert _{\gamma ,\tau ,r}^2\, \mathrm {d}t + \sup _{t\in [0,T]} \frac{1}{\beta } \Vert u \Vert _{\gamma ,\tau ,r-\frac{\gamma }{2}}^2 + \frac{1}{\beta } \int _0^T \ \Vert (1+y) \omega \Vert _{\gamma ,\tau ,r+1-\gamma }^2\, \mathrm {d}t\\&\quad + \sup _{t\in [0,T]} \frac{1}{\beta ^2} \Vert (1+y) \omega \Vert _{\gamma ,\tau ,r+\frac{1}{2} - \gamma }^2 + \frac{1}{\beta ^2} \int _0^T \Vert (1+y) \partial _y \omega \Vert _{\gamma ,\tau ,r+\frac{1}{2} -\gamma }^2\, \mathrm {d}t \end{aligned}\nonumber \\ \end{aligned}$$
(35)

Control of the Low Norm and Final A Priori Estimate

Corollary 14, which shows an a priori bound on the Gevrey norm of u, was derived under a lower bound on \(\beta \) involving \(\Vert (U^P,V^P) \Vert _{low}\). The last step is to see how this low norm relates to \({\left| \left| \left| u \right| \right| \right| }\). A convenient approach is to establish an additional estimate on a weighted Sobolev norm, namely

$$\begin{aligned} \Vert f \Vert ^2_{\mathcal {H}^s} = \sum _{|\bar{\alpha }|\le s} \int _{{\mathbb {T}}\times \mathbb {R}_+} |\partial ^{\bar{\alpha }} f|^2 (1+y)^{2\bar{\alpha }_2} \rho _0(y) \,\mathrm {d}x\, \mathrm {d}y, \end{aligned}$$

where the summation variable is the multiindex \(\bar{\alpha } = (\bar{\alpha }_1,\bar{\alpha }_2)\). In this setting, we can state the following estimate.

Lemma 15

Let \(s \ge 3\) be an even integer, \(m\ge s+2\), \(\alpha \ge 0\), \(r \in \mathbb {R}\), \(\gamma \ge 1\), and define \({\left| \left| \left| u \right| \right| \right| }\) as in (35). Then, there exists C depending on \(s,m,\alpha ,\gamma ,r\) such that

$$\begin{aligned} \begin{aligned} \frac{\mathrm {d}}{\mathrm {d}t}\Vert \omega ^P \Vert ^2_{\mathcal {H}^s} + \Vert \partial _y \omega ^P \Vert ^2_{\mathcal {H}^s}&\le C \Vert \omega ^P \Vert _{\mathcal {H}^s}^s + C \left( 1+ \Vert U^E \Vert _{H^{s+1}({\mathbb {T}})} + {\left| \left| \left| u \right| \right| \right| }\,\right) \, \Vert \omega ^P \Vert _{\mathcal {H}^s}^2 \\&\quad + \sum _{l=0}^{\frac{s}{2}} \Vert \partial _t^l (\partial _t + U^E \partial _x U^E) \Vert ^2_{H^{s-2l}}. \end{aligned} \end{aligned}$$
(36)

where \(\omega ^P = \partial _y U^P\).

Proof

A similar estimate was established in [27, Proposition 5.6], so that we will only explain the main steps. The starting point is the advection-diffusion equation on the vorticity

$$\begin{aligned} (\partial _t + U^P \partial _x + V^P \partial _y) \omega ^P - \partial ^2_y \omega ^P = 0. \end{aligned}$$
(37)

One applies \(\partial ^\alpha \) to the equation, test it against \((1+y)^{2\alpha _2} \rho _0 \partial ^\alpha \omega ^P\), and sum over \(|\alpha | \le s\). Then,

$$\begin{aligned}&\frac{1}{2} \partial _t \Vert \omega ^P \Vert _{\mathcal {H}^s}^2 + \Vert \partial _y \omega ^P \Vert ^2_{\mathcal {H}^s} \le \sum _{|\alpha | \le s} \int V^P \partial _y ((1+y)^{2\alpha _2} \rho _0) |\partial ^\alpha \omega ^P|^2 \\&\quad -\sum _{|\alpha | \le s}\int [ \partial ^\alpha , (U^P \partial _x + V^P \partial _y) ] \omega ^P \, \partial ^\alpha \omega ^P (1+y)^{2\alpha _2} \rho _0 \\&\quad - \sum _{|\alpha | \le s} \int _{\Omega } \partial _y ((1+y)^{2\alpha _2} \rho _0) \partial _y \partial ^\alpha \omega ^P \, \partial ^\alpha \omega ^P \\&\quad - \sum _{|\alpha | \le s} \int _{\{y=0\}} \partial _y \partial ^\alpha \omega ^P \, \partial ^\alpha \omega ^P. \end{aligned}$$

Using the equation on \(U^P\), one can obtain recursively boundary conditions for the odd derivatives \(\partial _y^{2k+1} \omega ^P\), starting from the Neumann condition

$$\begin{aligned} \partial _y \omega ^P\vert _{y=0} = - \partial _t U^E - U^E \partial _x U^E. \end{aligned}$$

More precisely, the boundary data \(\partial _y^{2k+1} \omega ^P\vert _{y=0}\) can be expressed in terms of the data \(U^E\) and of products of mixed derivatives \(\partial _x^{\rho _1} \partial _y^{\rho _2}\omega ^P\vert _{y=0}\) with \(\rho _2\le 2k-2\). We refer to [27, Lemma 5.9] for the expressions of these boundary conditions. This allows to establish the following bound, cf equations (5.20)-(5.22) in [27]:

$$\begin{aligned}&-\sum _{|\alpha | \le s} \int _{\{y=0\}} \partial _y \partial ^\alpha \omega ^P \, \partial ^\alpha \omega ^P \\&\quad \le C_s \Vert \omega ^P \Vert _{\mathcal {H}^s}^s + C_s \sum _{l=0}^{\frac{s}{2}} \Vert \partial _t^l (\partial _t + U^E \partial _x U^E) \Vert _{H^{s-2l}}^2 + \frac{1}{4} \Vert \partial _y \omega ^P \Vert _{\mathcal {H}^s}^2. \end{aligned}$$

The diffusion term does not raise any difficulty: we find

$$\begin{aligned} - \sum _{|\alpha | \le s} \int _{\Omega } \partial _y ((1+y)^{2\alpha _2} \rho _0) \partial _y \partial ^\alpha \omega ^P \, \partial ^\alpha \omega ^P \le C \Vert \omega ^P\Vert _{\mathcal {H}^s}^2 + \frac{1}{4} \Vert \partial _y\omega ^P \Vert _{\mathcal {H}^s}^2. \end{aligned}$$

where C depends on s and m. Also, through standard estimates, we find

$$\begin{aligned} \sum _{|\alpha | \le s} \int V^P \partial _y ((1+y)^{2\alpha _2} \rho _0) |\partial ^\alpha \omega ^P|^2 \le C \Vert \omega ^P\Vert _{\mathcal {H}^s}^3 \end{aligned}$$

and

$$\begin{aligned} \sum _{|\alpha | \le s}\int [\partial ^\alpha , U^P \partial _x ] \omega ^P \, \partial ^\alpha \omega ^P (1+y)^{2\alpha _2} \rho _0 \le C \Vert \omega ^P\Vert _{\mathcal {H}^s}^3. \end{aligned}$$

The other part of the commutator is slightly more delicate. First, one can show that

$$\begin{aligned} \sum _{|\alpha | \le s, \alpha _1 \ne s}\int [\partial ^\alpha , V^P \partial _y ] \omega ^P \, \partial ^\alpha \omega ^P (1+y)^{2\alpha _2} \rho _0 \le C \Vert \omega ^P\Vert _{\mathcal {H}^s}^3. \end{aligned}$$

Note that the weight (that grows with the number of y-derivatives) allows to compensate for the linear growth in y of \(V^P\). The success of this trick comes from the fact that we are interested here in Sobolev estimates (contrary to the former Gevrey estimates). When \(\alpha _1 = s\), namely \(\alpha = (s,0)\), one can show similarly that

$$\begin{aligned}&\int [\partial ^\alpha , V^P \partial _y ] \omega ^P \, \partial ^\alpha \omega ^P \rho _0 - \int \partial ^\alpha V^P \partial _y \omega _P \, \partial ^\alpha \omega ^P \rho _0 \, \le \, C \Vert \omega ^P\Vert _{\mathcal {H}^s}^3. \end{aligned}$$

However, the term where the s derivatives with respect to x apply to \(V^P\) cannot be handled with usual manipulations. It is the well-known loss of x-derivative peculiar to the Prandtl equation: in particular, one cannot control \(\Vert (1+y)^{-1}\partial ^s_x V^P \Vert _{L^2(\rho _0)}\) by \(\Vert \omega ^P\Vert _{\mathcal {H}^s}\). This is where \({\left| \left| \left| u \right| \right| \right| }\) is involved. We find that

$$\begin{aligned} \int \partial ^\alpha V^P \partial _y \omega _P \, \partial ^\alpha \omega ^P \rho _0&\le \Vert (1+y)^{-1} \partial _x^s V^P\Vert _{\infty }\, \Vert (1+y) \partial _y \omega ^P\Vert _{L^2_xL^2(\rho _0)}\, \Vert \omega ^P\Vert _{L^2_xL^2(\rho _0)} \\&\le C (\Vert U^E\Vert _{H^{s+1}} +{\left| \left| \left| u \right| \right| \right| }) \Vert \omega ^P\Vert _{\mathcal {H}^s}^2 \end{aligned}$$

using that \(\Vert (1+y)^{-1} \partial _x^s V^P\Vert _{\infty } \le C \Vert \partial _x^{s+1}u^P\Vert _\infty \le C {\left| \left| \left| u \right| \right| \right| }\) as soon as \(m \ge s+2\). Putting together the previous estimates yields the result. \(\square \)

We conclude this section with

Proposition 16

Let us fix \(s=6\), \(m \ge s+2\), \(\alpha \), \(\gamma \), r, R as in (32) and \(\alpha \ge 1/2\). Further fix \(\tau _1 >0\). Let

$$\begin{aligned} M_\mathrm {in}= 2 \max (\mathbf {C},1) \left( \Vert u_{\mathrm {in}} \Vert _{\gamma ,\tau _0,r+\gamma -\frac{3}{2}}^2 + \Vert (1+y) \omega _{\mathrm {in}} \Vert _{\gamma ,\tau _0,r+\frac{1}{2} - \gamma }^2 + \Vert \omega ^P\vert _{t=0}\Vert _{\mathcal {H}^s}^2\right) \end{aligned}$$

where \(\mathbf {C}\) is the constant appearing in Corollary 14. There exists \(\beta _*\) and \(T_*\) depending on \(\tau _1\), \(M_\mathrm {in}\), on \(\Vert \omega ^P\vert _{t=0}\Vert _{\mathcal {H}^s}\), on \(\sup _{[0,T_0]} |U^E|_{\gamma ,\tau _0,R}^2\) and on various Sobolev norms of \(U^E\), such that, for all \(\beta > \beta _*\) and for all \(T \le T_*\) with \(\tau (T) \ge \tau _1\): if \({\left| \left| \left| u \right| \right| \right| }^2 \le \frac{2M_\mathrm {in}}{\beta }\), then \({\left| \left| \left| u \right| \right| \right| }^2 \le \frac{3M_\mathrm {in}}{2\beta }\).

Proof

Let \(\beta , T\) such that \({\left| \left| \left| u \right| \right| \right| }^2 \le \frac{2M_\mathrm {in}}{\beta } \le 2 M_\mathrm {in}\) (assuming \(\beta \ge 1\)). We first apply Lemma 15, which yields

$$\begin{aligned} \begin{aligned} \frac{\mathrm {d}}{\mathrm {d}t}\Vert \omega ^P \Vert ^2_{\mathcal {H}^s} + \Vert \partial _y \omega ^P \Vert ^2_{\mathcal {H}^s}&\le C \Vert \omega ^P \Vert _{\mathcal {H}^s}^s + C (1+ \Vert U^E \Vert _{H^{s+1}({\mathbb {T}})} + \sqrt{2M_\mathrm {in}}) \Vert \omega ^P \Vert _{\mathcal {H}^s}^2 \\&+ \sum _{l=0}^{\frac{s}{2}} \Vert \partial _t^l (\partial _t + U^E \partial _x U^E) \Vert ^2_{H^{s-2l}}. \end{aligned} \end{aligned}$$
(38)

Integrating this differential inequality shows

$$\begin{aligned} \sup _{t \in [0,T]} \Vert \omega ^P(t) \Vert _{\mathcal {H}^s} \le 2 \Vert \omega ^P\vert _{t=0} \Vert _{\mathcal {H}^s} \end{aligned}$$
(39)

for \(T \le T_1\), where \(T_1\) depends on \(M_\mathrm {in}\), \(\sup _{t \in [0,T_0]} \Vert U^e(t) \Vert _{H^{s+1}({\mathbb {T}})}\), \(\int _0^{T_0} \Vert \partial _t^l (\partial _t + U^E \partial _x U^E) \Vert ^2_{H^{s-2l}}\, \mathrm {d}t\) and on \(\Vert \omega ^P\vert _{t=0} \Vert _{\mathcal {H}^s}\).

Standard Sobolev imbeddings imply that

$$\begin{aligned} \max _{0\le k \le 3}\Vert \partial _x^k U^P \Vert _{\infty } + \max _{0\le k \le 2}\left\| \frac{\partial _x^k V^P}{1+y} \right\| _{\infty } \le C \left( \max _{0\le k \le 3} \Vert \partial _x^k U^E\Vert _{\infty } + {\left| \left| \left| u \right| \right| \right| }\right) . \end{aligned}$$

As regards the other terms defining \(\Vert (U^P,V^P)\Vert _{low}\), cf (19), they all involve \(\omega ^P\) and are controlled by \(\Vert \omega ^P\Vert _{\mathcal {H}^s}\) as soon as \(s \ge 5\). Hence, it follows from (39) that

$$\begin{aligned} \Vert (U^P,V^P)\Vert _{low} \le K \end{aligned}$$

for \(T \le T_1\) and for some K depending on \(M_\mathrm {in}\), \(\Vert \omega ^P\vert _{t=0} \Vert _{\mathcal {H}^s}\) and various norms of \(U^E\). If we now choose

$$\begin{aligned} \beta _* \ge \mathcal {C} (1+K)\, \left( 1 + \frac{1}{\tau _1} +K\right) , \quad \text { and } \tau (T) \ge \tau _1 \end{aligned}$$

where \(\mathcal {C}\) is the constant appearing in Corollary 14, we obtain for \(\beta \ge \beta _*\):

$$\begin{aligned} {\left| \left| \left| u \right| \right| \right| }^2 \le \frac{M_\mathrm {in}}{2\beta } + \frac{\mathbf {C}}{\beta ^2} \int _0^T \Vert (1+y) f^e_j \Vert _{\gamma ,\tau ,r-2+\gamma }^2 \,\mathrm {d}t + \frac{\mathbf {C}}{\beta \tau _1^4} \left( |U^E|_{\gamma ,\tau ,R}^2 + 2M_\mathrm {in}\right) {\left| \left| \left| u \right| \right| \right| }^2. \end{aligned}$$

Taking \(\beta _*\) large enough so that

$$\begin{aligned} \frac{\mathbf {C}}{\beta _* \tau _1^4} \sup _{t \in [0,T_0]} \left( |U^E|_{\gamma ,\tau _0,R}^2 + 2M_\mathrm {in}\right) \le \frac{1}{2} \end{aligned}$$

we get

$$\begin{aligned} {\left| \left| \left| u \right| \right| \right| }^2 \le M_\mathrm {in}+ \frac{2 \mathbf {C}}{\beta ^2} \int _0^T \Vert (1+y) f^e_j \Vert _{\gamma ,\tau ,r-2+\gamma }^2 \, \mathrm {d}t + \frac{2 \mathbf {C}}{\beta ^2} \int _0^T \Vert f^e_j|_{y=0} \Vert _{\gamma ,\tau ,r-2+\gamma }^2 \, \mathrm {d}t \end{aligned}$$

If we take \(T_* \le T_1\) such that \(2 \mathbf {C} \int _0^{T_*} \Vert (1+y) f^e_j \Vert _{\gamma ,\tau ,r-2+\gamma }^2\, \mathrm {d}t \le \frac{1}{2} M_\mathrm {in}\), the result follows. \(\square \)

Existence and Uniqueness

On the basis of the previous a priori estimates, we now complete the proof of Theorem 1: we construct a unique solution of (1)–(2) with data \(U^P_\mathrm {in}\). This obviously amounts to constructing a unique solution of (13)–(15) with data \(u_\mathrm {in}:= U^P_\mathrm {in}- U^e\vert _{t=0}\).

We fix \(s=6\), \(\gamma =2\). We take \(m \ge s+2\) and \(\displaystyle \alpha \ge \tfrac{1}{2}\) that satisfy the inequalities in the first line of (32). Let \(0< \tau _1 < \tau _0\), \(r \in \mathbb {R}\), \(T_0 > 0\), and \(\displaystyle U^E\), \(U^P_\mathrm {in}= u_\mathrm {in}+ U^e\vert _{t=0}\) satisfying the assumptions of the theorem. Let now \((\tau '_0, \tau '_1)\) with \(\displaystyle 0< \tau _1< \tau '_1< \tau '_0 < \tau _0\). Let \(r'\) and \(R'\) as in the second line of (32). As \(\tau _0 > \tau '_0\), we have

$$\begin{aligned} \begin{aligned}&\Vert u_{\mathrm {in}} \Vert _{\gamma ,\tau '_0,r'+\gamma -\frac{3}{2}}^2 + \Vert (1+y) \omega _{\mathrm {in}} \Vert _{\gamma ,\tau '_0,r+\frac{1}{2} - \gamma }^2 \\&\quad \le C \left( \Vert u_{\mathrm {in}} \Vert _{\gamma ,\tau _0,r}^2 + \Vert (1+y) \omega _{\mathrm {in}} \Vert _{\gamma ,\tau _0,r}^2 \right) < +\infty \end{aligned} \end{aligned}$$

while

$$\begin{aligned} \Vert \omega ^P\vert _{t=0}\Vert _{\mathcal {H}^s}^2 \le C ( \sup _{[0,T_0]} |U^E|_{2,\tau _0,r} + \Vert (1+y)^{m+6} \omega _\mathrm {in}\Vert _{H^6({\mathbb {T}}\times \mathbb {R}_+)}) < +\infty \end{aligned}$$

and

$$\begin{aligned} \sup _{[0,T_0]} |U^E|_{2,\tau '_0,R'} \le C \sup _{[0,T_0]} |U^E|_{2,\tau _0,r} < +\infty \end{aligned}$$

for a constant C possibly depending on \(\tau _0, \tau _0', r, r', R'\).

The idea is then to apply Proposition 16 to a solution of an approximate system, for which well-posedness is granted. Inspired by [27], we consider the regularized equation

$$\begin{aligned} \partial _t u + (u \partial _x + v \partial _y) u + (U^e_\epsilon \partial _x + V^e_\epsilon \partial _y) u + (u \partial _x + v \partial _y) U^e_\epsilon - \epsilon \partial _x^2 u - \partial ^2_y u = f^e_\epsilon ,\nonumber \\ \end{aligned}$$
(40)

adding a tangential diffusion \(- \epsilon \partial _x^2 u\). The modified vector field \((U^e_\epsilon , V^{e}_\epsilon )\) takes the form

$$\begin{aligned} U^e_\epsilon = \partial _y (\mathrm{e}^{-\epsilon y} (y + \mathrm{e}^{-y} -1)) U^E_\epsilon , \quad V^{e}_\epsilon = - \mathrm{e}^{-\epsilon y} (y + \mathrm{e}^{-y} -1) \partial _x U^E_\epsilon \end{aligned}$$

where \(U^E_\epsilon \) is an analytic approximation of \(U^E\), converging to \(U^E\) in the norm \(| \ |_{2,\tau _0, r}\) as \(\epsilon \rightarrow 0\). Note that \((U^e_\epsilon , V^e_\epsilon )\) is still divergence-free, but has now fast decay in y, so that all difficulties generated by the linear growth of \(V^e\) vanish. Accordingly, the right-hand side \(f^e\) is modified into \(f^e_\epsilon \) replacing \(U^E\) by \(U^E_\epsilon \), resp. \((U^e, V^e)\) by \((U^e_\epsilon , V^e_\epsilon )\) in (14). Similarly, one regularizes the initial data to obtain some \(u_{\mathrm {in},\epsilon }\) real analytic in xy, with fast decay at infinity in y (and obeying suitable compatibility conditions).

One can show that system (40) is well-posed following classical methods for fully parabolic equations. For instance, for \(T_{\epsilon ,max}\) small enough, one can prove the existence of a Sobolev solution \(u_\epsilon \) on \((0, T_{\epsilon ,max})\) through a fixed point argument applied to

$$\begin{aligned} \mathcal {T}_\epsilon u(t) = \mathrm{e}^{t (\epsilon \partial _x^2 + \partial ^2_y)} u_{\mathrm {in},\epsilon } + \int _0^t \mathrm{e}^{(t-s) (\epsilon \partial _x^2 + \partial ^2_y)} F_\epsilon [u](s)\, \mathrm{d}s \end{aligned}$$

with \(F_\epsilon [u] = f^e_\epsilon - (u \partial _x + v \partial _y) u - (U^e_\epsilon \partial _x + V^e_\epsilon \partial _y) u - (u \partial _x + v \partial _y) U^e_\epsilon \). Moreover, \(u^\epsilon \) remains (real) analytic in (xy) as long as the Sobolev norm of \(u_\epsilon \) does not blow up, that is on \((0,T_{\epsilon ,max})\). This property, related to the analytic regularization of the heat kernel is well-known, even in the more difficult context of the Navier-Stokes equation: see [2, 7, 25] and references therein.

We now claim that all a priori estimates obtained for a solution u of (13) can be established for \(u_\epsilon \) solution of (40), uniformly in \(\epsilon \). For this, one just needs to adapt the definitions of the auxiliary quantities \(H_j\) and \(\phi _j\): we rather consider

$$\begin{aligned}&\Big (\partial _t \!+\! \beta (j{+}1)\!+\! U^P \partial _x + (j{+}1)\partial _x U^p \!+\! V^P \partial _y - \epsilon \partial _x^2 - \partial _y^2\Big ) \int _0^y H_j \, \mathrm {d}z = \int _0^y u_j \, \mathrm {d}z, \nonumber \\&H_j|_{t=0} = 0, \qquad \partial _y H_j|_{y=0} = 0, \qquad H_j|_{y\rightarrow \infty } = 0. \end{aligned}$$
(41)

and

$$\begin{aligned} \begin{aligned}&\left( -\partial _t + \beta (j{+}1)- U^P \partial _x + j \partial _x U^p - V^P \partial _y - \partial _y V^P - V^P \frac{\partial _y \rho _j}{\rho _j} - \epsilon \partial _x^2 \right. \\&\left. \quad - \left( \partial _y + \frac{\partial _y \rho _j}{\rho _j}\right) ^2 \right) \phi _j = H_j, \\&\phi _j|_{t=T} = 0, \qquad \phi _j|_{y=0} = 0, \qquad \phi _j|_{y\rightarrow \infty } = 0. \end{aligned} \end{aligned}$$
(42)

The additional good terms coming from \(-\epsilon \partial ^2_x\) allow to control the extra commutator terms that it generates. Hence, we can apply Proposition 16 with \(\tau '_0, \tau '_1\), \(r'\) and \(R'\) instead of \(\tau _0, \tau _1, r\), and R. Let \(\beta _*\) and \(T_*\) given by the proposition (note that they are independent of \(\epsilon \)). We then introduce

$$\begin{aligned} T_{\epsilon , *} = \sup \{ T \le T_{\epsilon , max}, {\left| \left| \left| u \right| \right| \right| }^2 \le 2 M_\mathrm {in}/\beta \} \end{aligned}$$

where \(\beta > \beta _*\) is fixed, and \({\left| \left| \left| u \right| \right| \right| }\) is defined in (35). Note that \({\left| \left| \left| u \right| \right| \right| }\) implicitly depends on T. By continuity in time of \(u^\epsilon \), one has \(T_{\epsilon , *} > 0\). But from Proposition 16, one deduces easily that for any \(T \le T_*\), \(\, T_{\epsilon , max} \ge T_{\epsilon , *} \ge T\).

From there, by standard compactness arguments, one obtains a solution to the Prandtl system over [0, T], with the regularity properties stated in the theorem. It remains to show uniqueness. For this, we take two solutions \(u^1\) and \(u^2\) up to time T. The difference \(u^d\) then satisfies (from (13))

$$\begin{aligned}&\partial _t u^d + u^1 \partial _x u^d + u^d \partial _x u^2 + v^d \partial _y u^1 + v^2 \partial _y u^d \\&\quad + (U^e \partial _x +\, V^e \partial _y) u^d + (u^d \partial _x + v^d \partial _y) U^e - \partial ^2_y u^d = f^{d,e}. \end{aligned}$$

We then find for \(u^d_j\) that

$$\begin{aligned}&\Big (\partial _t + \beta (j{+}1)+ U^{1,P} \partial _x + (j{+}1)\partial _x U^{1,P} + V^{2,P} \partial _y - \partial _y^2\Big ) u^d_j \\&\quad +\, \partial _y U^{1,P} v^d_j + j \partial _{xy} U^{1,P} \partial _x^{-1} v^d_j = F^d_j + \partial _x u^d u_j^d, \end{aligned}$$

where again \(F^d_j\) consists of \(f_j^{d,e}\) and mixed terms with less than j derivatives on \(u^1\), \(u^2\) or \(u^d\). Comparing with (16), we see that the only difference is the replacement of \((U^P, V^P)\) by \((U^{1,P}, V^{2,P}\)). Let us stress that the latter field not being divergence-free is not an issue: none of the a priori estimates carried in Sect. 4 and Sect. 5 were using the fact that \((U^P, V^P)\) was divergence-free. One can therefore obtain a similar Gevrey bound on \(u^d\), under a lower bound on \(\beta \) (involving the low norms of \((U^{1,P}, V^{1,P})\) and \((U^{2,P}, V^{2,P})\)). This provides a stability estimate which shows uniqueness. These considerations now finish the proof of our main result Theorem 1.