Proof. We square the left-hand side of (3.15) and localize the integral based on z via

(3.16) $$ \begin{align} \int f^2 \,\mathrm{d} y = \int f^2 \chi(\frac{z}{\gamma}) \,\mathrm{d} y + \int f^2 (1 - \chi(\frac{z}{\gamma})) \,\mathrm{d} y. \end{align} $$

For the localized component, we integrate by parts in y via

(3.17) $$ \begin{align} \int f^2 \chi(\frac{z}{\gamma}) \,\mathrm{d} y = \int \partial_y (y) f^2 \chi(\frac{z}{\gamma}) \,\mathrm{d} y = - \int 2 y f f_y \chi(\frac{z}{\gamma}) \,\mathrm{d} y - \frac{1}{\gamma} \int \frac{y}{\sqrt{x}} f^2 \chi'(\frac{z}{\gamma}) \,\mathrm{d} y. \end{align} $$

We estimate each of these terms via

(3.18) $$ \begin{align} \Big| \int y f f_y \chi(\frac{z}{\gamma})\,\mathrm{d} y \Big| &\lesssim \| f \|_{L^2_y} \| \sqrt{x} \sqrt{\bar{u}^0_p} \sqrt{\gamma} f_y \|_{L^2_y} \le \delta \| f \|_{L^2_y}^2 + C_\delta \gamma x \| \sqrt{\bar{u}^0_p} f_y \|_{L^2_y}^2, \end{align} $$

(3.19) $$ \begin{align} \Big| \frac{1}{\gamma} \int \frac{y}{\sqrt{x}} f^2 \chi'(\frac{z}{\gamma}) \,\mathrm{d} y \Big| &= \Big| \frac{1}{\gamma} \int z^2 \frac{1}{z} f^2 \chi'(\frac{z}{\gamma}) \,\mathrm{d} y \Big| \lesssim \frac{1}{\gamma^2} \| \bar{u}^0_p f \|_{L^2_y}^2, \end{align} $$

where above, we have used (2.12). For the far-field term, we estimate again by invoking (2.12) via

(3.20) $$ \begin{align} |\int f^2 (1 - \chi(\frac{z}{\gamma})) \,\mathrm{d} y| = |\int \frac{1}{|\bar{u}^0_p|^2} |\bar{u}^0_p|^2 f^2 (1 - \chi(\frac{z}{\gamma})) \,\mathrm{d} y| \lesssim \frac{1}{\gamma^2} \| \bar{u}^0_p f \|_{L^2_y}^2. \end{align} $$

We have thus obtained

(3.21) $$ \begin{align} \| f \|_{L^2_y}^2 \le \delta \| f \|_{L^2_y}^2 + C_\delta \gamma x \| \sqrt{\bar{u}^0_p} f_y \|_{L^2_y}^2 + \frac{C}{\gamma^2} \| \bar{u}^0_p f \|_{L^2_y}^2, \end{align} $$

and the desired result follows from taking $\delta $ small relative to universal constants and absorbing to the left-hand side.

Proof. Estimate (3.22) follows immediately upon taking $f = U^{(k)}_y \phi _{k+1}\langle x \rangle ^k$ , $\sqrt {\varepsilon } U^{(k)}_x \phi _{k+1}\langle x \rangle ^k$ , or $\varepsilon V^{(k)}_x \phi _{k+1}\langle x \rangle ^k$ in (3.15). Estimates (3.23) and (3.24) follow upon choosing $f = U^{(k)}_x \langle x \rangle ^{k + \frac 1 2}$ and $f = \sqrt {\varepsilon }V^{(k)}_x \langle x \rangle ^{k + \frac 1 2}$ , respectively.

Proof. The U and V bounds follow upon choosing $f = U \langle x \rangle ^{-\frac 12 - \frac {1}{200}}$ in (3.15), whereas the q bound follows from the standard Hardy inequality valid as $q|_{x = 0} = 0$ : $\| \sqrt {\varepsilon }q \langle x \rangle ^{-\frac 32 - \frac {1}{200}} \| \lesssim \| \sqrt {\varepsilon } q_x \langle x \rangle ^{-\frac 12 - \frac {1}{200}} \| = \| \sqrt {\varepsilon } V \langle x \rangle ^{-\frac 12 - \frac {1}{200}} \|$ .

Proof. We compute the quantity on the left-hand side of above via

(3.27) $$ \begin{align} \nonumber \int \langle x \rangle^{-3.01} \bar{u}^2 f^2 \,\mathrm{d} x & = - \int \frac{\partial_x}{2.01} \langle x \rangle^{-2.01} \bar{u}^2 f^2 \,\mathrm{d} x \\ \nonumber & = \frac{2}{2.01} \int \langle x \rangle^{-2.01} \bar{u}^2 f f_x \,\mathrm{d} x + \frac{2}{2.01} \int \langle x \rangle^{-2.01} \bar{u} \bar{u}_x f^2 \,\mathrm{d} x \\ & \le \frac{1}{2.01} \int \langle x \rangle^{-3.01} \bar{u}^2 f^2 \,\mathrm{d} x + \frac{1}{2.01} \int \langle x \rangle^{-1.01} \bar{u}^2 f_x^2 \,\mathrm{d} x + \frac{2}{2.01} \int \langle x \rangle^{-2.01} \bar{u} \bar{u}_x f^2 \,\mathrm{d} x, \end{align} $$

which, upon bringing the first term on the right-hand side to the left, gives the inequality (3.26) with the precise constants.

Proof. We begin with (3.28). For this, we first freeze x and integrate from $y = \infty $ to obtain

(3.30) $$ \begin{align} \nonumber \varepsilon^{\frac 1 2}|V^{(j)}|^2 \langle x \rangle^{2j} \phi_{j+1}^2 &= 2 \varepsilon^{\frac 1 2} |\int_y^\infty V^{(j)} V^{(j)}_y\langle x \rangle^{2j} \phi_{j+1}^2 \,\mathrm{d} y' | \lesssim \| \varepsilon^{\frac 1 2} V^{(j)} \langle x \rangle^{j - \frac 1 2} \phi_{j+1} \|_{L^2_y} \| U^{(j)}_x \langle x \rangle^{j+ \frac 1 2} \phi_{j+1} \|_{L^2_y} \\ & = \| \varepsilon^{\frac 1 2} V^{(j-1)}_x \langle x \rangle^{(j-1)+ \frac 1 2} \phi_{j+1} \|_{L^2_y} \| U^{(j)}_x \langle x \rangle^{j + \frac 1 2} \phi_{j+1} \|_{L^2_y}. \end{align} $$

We now take $L^2_x$ and appeal to (3.23)–(3.24).

Similarly, we compute

(3.31) $$ \begin{align} \nonumber \varepsilon^{\frac 1 2}\bar{u}_\ast^2 |V^{(j)}|^2 \langle x \rangle^{2j} \phi_{j+1}^2 &\le 2 \varepsilon^{\frac 1 2} \bar{u}_\ast^2 |\int_y^\infty V^{(j)} V^{(j)}_y\langle x \rangle^{2j} \phi_{j+1}^2 \,\mathrm{d} y' | \lesssim \varepsilon^{\frac 1 2} \int_y^\infty \bar{u}_\ast^2| V^{(j-1)}_x ||U^{(j)}_x|\langle x \rangle^{2j} \phi_{j+1}^2 \,\mathrm{d} y' \\ &\lesssim \varepsilon^{\frac 1 2} \| \bar{u}_\ast V^{(j-1)}_x \langle x \rangle^{j - \frac 1 2} \phi_{j+1}\| \| \bar{u}_\ast U^{(j)}_x \langle x \rangle^{j + \frac 1 2} \phi_{j+1} \| \end{align} $$

where above we have used that $\bar {u}_\ast (y') \ge \bar {u}_\ast (y) \ge 0$ when $y' \ge y$ to bring the $\bar {u}_\ast ^2$ factor inside the integral. We now use (2.12) to conclude.

Proof. To prove these estimates, we simply use (2.23) to express $(u, v)$ in terms of $(U, V)$ . We do this now, starting with (3.32). Differentiating (2.23) $k+1$ times in x, we obtain

$$ \begin{align*} \| u^{(k)}_x x^{k+ \frac 1 2} \phi_{k+1} \| \le I_1 + I_2, \end{align*} $$

where

$$ \begin{align*} I_1 & := \sum_{l = 0}^{k} \binom{k+1}{l} ( \| \partial_x^l \bar{u} \partial_x^{k - l} U_x x^{k+\frac 1 2} \phi_{k+1} \| + \| \partial_x^{l} \bar{u}_y \partial_x^{k-l}V x^{k + \frac 1 2} \phi_{k+1} \| ), \\ I_2 & := \| \partial_x^{k+1} \bar{u} U x^{k+ \frac 1 2} \phi_{k+1} \| + \| \partial_x^{k+1} \bar{u}_y q x^{k+ \frac 1 2} \phi_{k+1} \| \end{align*} $$

We first estimate $I_1$ as follows

(3.36) $$ \begin{align} \nonumber I_1 & \le \sum_{l = 0}^{k} \binom{k+1}{l} ( \| \partial_x^l \bar{u} \partial_x^{k - l} U_x x^{k+\frac 1 2} \phi_{k+1} \| + \| \partial_x^{l} \bar{u}_y \partial_x^{k-l}V x^{k + \frac 1 2} \phi_{k+1} \| ) \\ \nonumber &\lesssim \sum_{l = 0}^{k} \| \frac{\partial_x^l \bar{u}}{ \bar{u}} x^l \|_\infty \| \bar{u} \partial_x^{k-l} U_x \langle x \rangle^{k - l + \frac 1 2} \phi_{k+1} \| + \| \partial_x^l \bar{u}_y y x^l \|_\infty \| \partial_x^{k-l} V_y x^{k-l + \frac 1 2} \phi_{k+1} \| \\ \nonumber & \le \sum_{l = 0}^{k}C_\delta \| \bar{u} \partial_x^{k-l} U_x \langle x \rangle^{k - l + \frac 1 2} \phi_{k+1} \| + C_\delta \| \bar{u} \partial_x^{k-l} V_y x^{k-l + \frac 1 2} \phi_{k+1} \| +\delta \| \sqrt{\bar{u}} \partial_x^{k+1-l} U_y x^{k+1 -l} \phi_{k+1} \| \\ & \le \sum_{l = 0}^{k} ( C_\delta\| U, V \|_{\mathcal{X}_{\le k-l + \frac 1 2}} + \delta \| U, V \|_{\mathcal{X}_{\le k+1-l}} ), \end{align} $$

where we have invoked (3.23).

We turn to $I_2$ , which requires a different estimate. We recall the splitting of the background profile into Euler and Prandtl components, (2.1). For the Eulerian piece, we estimate as follows

$$ \begin{align*} I_{2,E} & := \| \partial_x^{k+1} \bar{u}_E U x^{k+ \frac 1 2} \phi_{k+1} \| + \| \partial_x^{k+1} \partial_y\bar{u}_E q x^{k+ \frac 1 2} \phi_{k+1} \| \\ &\lesssim (\| x^{k + \frac32} \partial_x^{k+1}\bar{u}_E \|_{L^\infty} + \| x^{k + \frac32} (y \partial_y) \partial_x^{k+1}\bar{u}_E \|_{L^\infty} ) \| U \langle x \rangle^{-1} \| \lesssim \| U, V \|_{X_0}. \end{align*} $$

For the Prandtl component, we estimate as follows

$$ \begin{align*} I_{2,P} & := \| \partial_x^{k+1} \bar{u}_p U x^{k+ \frac 1 2} \phi_{k+1} \| + \| \partial_x^{k+1} \partial_y\bar{u}_p q x^{k+ \frac 1 2} \phi_{k+1} \| \\ &\lesssim \| \partial_x^{k+1} \bar{u}_p (U - U(x, 0)) x^{k+ \frac 1 2} \phi_{k+1} \| + \| \partial_x^{k+1} \bar{u}_p U(x, 0) x^{k+ \frac 1 2} \phi_{k+1} \| \\ &\quad + \| \partial_x^{k+1} \partial_y\bar{u}_p (q - y U(x, 0)) x^{k+ \frac 1 2} \phi_{k+1}\| + \| \partial_x^{k+1} \partial_y\bar{u}_p y U(x, 0) x^{k+ \frac 1 2} \phi_{k+1} \| \\ &\lesssim \| \partial_x^{k+1} \bar{u}_p y x^{k + \frac12} \|_{L^\infty} \| \frac{U - U(x, 0)}{y} \| + \| \partial_x^{k+1} \bar{u}_p x^{k + \frac34} \|_{L^\infty_x L^2_y} \| U(x, 0) \langle x \rangle^{-\frac14} \|_{L^2_x} \\ &\quad + \| \partial_x^{k+1} \partial_y \bar{u}_p y^2 x^{k + \frac12} \|_{L^\infty} \| \frac{q - y U(x, 0)}{y^2} \| + \| \partial_x^{k+1} y \partial_y \bar{u}_p x^{k + \frac34} \|_{L^\infty_x L^2_y} \| U(x, 0) \langle x \rangle^{-\frac14} \|_{L^2_x} \\ &\lesssim ( \| \partial_x^{k+1} \bar{u}_p y x^{k + \frac12} \|_{L^\infty} + \| \partial_x^{k+1} \partial_y \bar{u}_p y^2 x^{k + \frac12} \|_{L^\infty}) \| U_y \| \\ &\quad +( \| \partial_x^{k+1} \bar{u}_p x^{k + \frac34} \|_{L^\infty_x L^2_y} +\| \partial_x^{k+1} y\partial_y \bar{u}_p x^{k + \frac34} \|_{L^\infty_x L^2_y} ) \| U(x, 0) \langle x \rangle^{-\frac14} \|_{L^2_x} \\ &\lesssim \| U, V \|_{Y_{\frac12}} + \| U, V \|_{X_0}, \end{align*} $$

where we have used the inequality (3.22). This establishes the bound of the first quantity in (3.32). The analogous proof works for the second quantity in (3.32).

For the first quantity in (3.33), we obtain the identity

(3.37) $$ \begin{align} u_y^{(m)} = \sum_{l =0}^m \binom{m}{l} (\partial_x^l \bar{u} \partial_x^{m-l} U_y + 2 \partial_x^l \bar{u}_y \partial_x^{m-l} U + \partial_x^l \bar{u}_{yy} \partial_x^{m-l} q). \end{align} $$

We now estimate

(3.38) $$ \begin{align} \nonumber \| u_y^{(m)} x^m \phi_m \| &\lesssim\sum_{l = 0}^m \Big\| \frac{\partial_x^l \bar{u}}{\bar{u}} \langle x \rangle^l \Big\|_\infty \| \sqrt{\bar{u}} \partial_x^{m-l} U_y \langle x \rangle^{m-l} \phi_{m-l} \| \\ \nonumber &\quad + \sum_{l = 0}^{m-1} \| \partial_x^l \bar{u}_y \langle x \rangle^{l + \frac 1 2}\|_\infty \| \partial_x^{m-1-l} U_x \langle x \rangle^{m-1-l +\frac 1 2} \phi_{m-1-l} \| \\ \nonumber &\quad + \| \partial_x^m \bar{u}_y y x^m \|_\infty \Big\| \frac{U - U(x, 0)}{y} \Big\| + \| \partial_x^m \bar{u}_y \langle x \rangle^{m+\frac 1 4} \|_{L^\infty_x L^2_y} \| U(x, 0) \langle x \rangle^{-\frac 1 4} \|_{L^2_x} \\ \nonumber &\quad + \sum_{l = 0}^{m-1} \| \partial_x^l \bar{u}_{yy} \langle x \rangle^{l + \frac 1 2} y \|_\infty \| \partial_x^{m-1-l} \frac{q_x}{y} \langle x \rangle^{m-1-l +\frac 1 2} \phi_{m-1-l} \| \\ &\quad + \| \partial_x^m \bar{u}_{yy} y^2 x^m \|_\infty \| \frac{q - y U(x, 0)}{\langle y \rangle^2} \| + \| \partial_x^m y \bar{u}_{yy} \langle x \rangle^{m+\frac 1 4} \|_{L^\infty_x L^2_y} \| U(x, 0) \langle x \rangle^{-\frac 1 4} \|_{L^2_x}, \end{align} $$

where we use above that $m \ge 1$ in (3.33).

We now arrive at the mixed norm estimates in (3.34). An immediate computation gives

(3.39) $$ \begin{align} \nonumber \| \frac{1}{\bar{u}} \partial_x^j v \langle x \rangle^j \phi_j \|_{L^2_x L^\infty_y} &\lesssim\sum_{l = 0}^{j-1} \Big\| \frac{\partial_x^l \bar{u}}{ \bar{u}} \langle x \rangle^l \Big\|_\infty \| \partial_x^{j-l} V \langle x \rangle^{j-l} \phi_{j} \|_{L^2_x L^\infty_y} \\ \nonumber &\quad + \sum_{l = 0}^{j-2} \Big\| \frac{\partial_x^{l+1} \bar{u}}{ \bar{u}} \langle x \rangle^{l+1} \Big\|_\infty \| \partial_x^{j-l} q \langle x \rangle^{j-l-1} \phi_{j} \|_{L^2_x L^\infty_y} \\ \nonumber &\quad + \| \partial_x^j \bar{u} \langle x \rangle^{j - \frac 1 2} y \|_\infty \Big\| \frac{V}{y} \langle x \rangle^{\frac 1 2} \phi_j\Big\|_{L^2_x L^\infty_y} + \| \partial_x^{j+1} \bar{u} y \langle x \rangle^{j + \frac 1 2} \|_\infty \Big\| \frac{q}{y} \langle x \rangle^{- \frac 1 2} \phi_j \Big\|_{L^2_x L^\infty_y} \\ &\lesssim \sum_{l = 0}^{j-1} \| U, V \|_{\mathcal{X}_{\le l+1}} + \Big\| \frac{V}{y} \langle x \rangle^{\frac 1 2} \phi_j \Big\|_{L^2_x L^\infty_y} +\Big\| \frac{q}{y} \langle x \rangle^{- \frac 1 2} \phi_j \Big\|_{L^2_x L^\infty_y}, \end{align} $$

where we have invoked (3.28). To conclude, we need to estimate the final two terms appearing above. First, we have by using $V|_{y = 0} = 0$ ,

(3.40) $$ \begin{align} \Big\| \frac{V}{y} \langle x \rangle^{\frac 1 2} \phi_j \Big\|_{L^2_x L^\infty_y} \lesssim \| U_x \langle x \rangle^{\frac 1 2} \phi_j \|_{L^2_x L^\infty_y} \lesssim \| U, V \|_{\mathcal{X}_{\le 1.5}}, \end{align} $$

where above we have used the estimate

$$ \begin{align*} \nonumber \langle x \rangle U_x^2 & = |\int_y^\infty \langle x \rangle U^{(1)} U^{(1)}_{y} \,\mathrm{d} y'| \lesssim \| U_x \langle x \rangle^{\frac 1 2} \|_{L^2_y} \| U_{xy} \langle x \rangle \|_{L^2_y} \\ \nonumber &\lesssim (\| \bar{u} U_x \langle x \rangle^{\frac 1 2} \|_{L^2_y} + \| \sqrt{\bar{u}} U_{xy} \langle x \rangle \|_{L^2_y} )(\| \sqrt{\bar{u}} U^{(1)}_y \langle x \rangle \|_{L^2_y} + \| \sqrt{\bar{u}} U^{(1)}_{yy} \langle x \rangle^{\frac 3 2} \|_{L^2_y} ), \end{align*} $$

which, upon taking supremum in y and subsequently integrating in x, yields $\| U_x \langle x \rangle ^{\frac 1 2}\phi _j \|_{L^2_x L^\infty _y}^2 \lesssim \| U, V \|_{\mathcal {X}_{\le 1.5}}^2.$ An analogous estimate applies to the third term from (3.39). The estimate (3.35) works in a nearly identical manner, invoking (3.29) instead of (3.28).

Proof. We first establish the U decay via

(3.42) $$ \begin{align} \nonumber U_x^2 \langle x \rangle^{\frac 5 2} \phi_{12}^2 &= \Big| - \int_y^\infty 2 U_x U_{xy} \langle x \rangle^{\frac 5 2} \phi_{12}^2 \Big|\\ \nonumber &\le \Big| \int_y^\infty \int_x^\infty 2 U_{xx} U_{xy} \langle x \rangle^{\frac 5 2} \phi_{12}^2 \Big| + \Big| 2 \int_y^\infty \int_x^\infty U_x U_{xxy} \langle x \rangle^{\frac 5 2} \phi_{12}^2\Big| \\ \nonumber &\quad + 5 \Big| \int_y^\infty \int_x^\infty U_x U_{xy} \langle x \rangle^{\frac 3 2}\phi_{12}^2 \Big| + \Big| \int_y^\infty \int_x^\infty 4 U_x U_{xy} \phi_{12} \phi_{12}' \Big| \\ \nonumber &\lesssim \| U_{xx} \langle x \rangle^{\frac 3 2} \phi_{12} \| \| U_{xy} \langle x \rangle \phi_{12} \| + \| U_x \langle x \rangle^{\frac 1 2} \phi_{12} \| \| U_{yxx} \langle x \rangle^2 \phi_{12} \| \\ &\quad + \| U_x \langle x \rangle^{\frac 1 2} \phi_{12}\| \| U_{xy} \langle x \rangle \phi_{12} \| + \| U_x \phi_{11} \| \| U_{xy} \phi_{11} \|. \end{align} $$

We now perform the same calculation for the V decay in (3.41). We begin with $V_x$ , via

(3.43) $$ \begin{align} \nonumber \varepsilon V_x^2 \langle x \rangle^{3} \phi_{12}^2 &= \Big| \int_y^\infty 2 \varepsilon V_x V_{xy} \langle x \rangle^3 \phi_{12}^2 \,\mathrm{d} y' \Big| \\ \nonumber &\le \int_y^\infty \int_x^\infty 2 \varepsilon | V_{xx} | | V_{xy} | \langle x \rangle^{3} \phi_{12}^2 \,\mathrm{d} y' \,\mathrm{d} x' + \int_y^\infty \int_x^\infty 2 \varepsilon | V_{x} | | V_{xxy} | \langle x \rangle^{3} \phi_{12}^2 \,\mathrm{d} y' \,\mathrm{d} x' \\ \nonumber &\quad + \int_y^\infty \int_x^\infty 6 \varepsilon | V_{x} | | V_{xy} | \langle x \rangle^{2} \phi_{12}^2 \,\mathrm{d} y' \,\mathrm{d} x' + |\int_y^\infty \int_x^\infty 4\varepsilon |V_x| |V_{xy}| \langle x \rangle^{3} \phi_{12} \phi_{12}' \,\mathrm{d} y' \,\mathrm{d} x' \\ \nonumber &\lesssim \sqrt{\varepsilon} \| \sqrt{\varepsilon} V^{(1)}_x \langle x \rangle^{\frac 3 2} \phi_{12}\| \| U^{(1)}_x \langle x \rangle^{\frac 3 2} \phi_{12} \| + \sqrt{\varepsilon} \| \sqrt{\varepsilon} V_x \langle x \rangle^{\frac 1 2} \phi_{12} \| \| U^{(2)}_x \langle x \rangle^{2.5} \phi_{12} \| \\ &\quad + \sqrt{\varepsilon} \| \sqrt{\varepsilon} V_x \langle x \rangle^{\frac 1 2} \phi_{12} \| \| U^{(1)}_x \langle x \rangle^{\frac 3 2} \phi_{12} \| + \sqrt{\varepsilon} \| \sqrt{\varepsilon} V_x \langle x \rangle^{\frac 1 2} \phi_{11} \| \| U^{(1)}_x \langle x \rangle^{\frac 3 2} \phi_{11} \|. \end{align} $$

From here the result follows upon invoking (3.23)–(3.24). The same computation can be done for higher derivatives, and for $U, V$ themselves we use the estimate

(3.44) $$ \begin{align} U \phi_{12} = \phi_{12} \int_x^\infty U_x \lesssim \|U, V \|_{\mathcal{X}} \int_x^\infty \langle x' \rangle^{- \frac 5 4} \,\mathrm{d} x' \lesssim \langle x \rangle^{- \frac 1 4} \| U, V \|_{\mathcal{X}}, \end{align} $$

and similarly for V, we integrate

(3.45) $$ \begin{align} V \phi_{12} = \phi_{12} \int_x^\infty V_x \lesssim \varepsilon^{- \frac 1 2} \|U, V \|_{\mathcal{X}} \int_x^\infty \langle x' \rangle^{- \frac 3 2} \,\mathrm{d} x' \lesssim \langle x \rangle^{- \frac 1 2} \varepsilon^{- \frac 1 2} \| U, V \|_{\mathcal{X}}. \end{align} $$

This concludes the proof.

Proof. We note that standard Sobolev embeddings gives $\| u^{(k)} \psi _{12} \|_\infty \lesssim \| u^{(k)} \psi _{12} \|_{H^2} \lesssim \varepsilon ^{-M_1} \| U, V \|_{\mathcal {X}}$ , and similarly for the remaining quantities in (3.47)–(3.49). Next, we appeal to (2.23) to obtain

$$ \begin{align*} \nonumber \| u^{(k)} x^{k + \frac 1 4} \phi_{12} \|_{\infty} & \le \sum_{l = 0}^k \binom{k}{l} (\| \partial_x^l \bar{u} \partial_x^{k-l} U x^{k + \frac 1 4} \phi_{12} \|_\infty + \| \partial_x^l \bar{u}_y \partial_x^{k-l} q x^{k + \frac 1 4} \phi_{12} \|_\infty) \\ \nonumber &\lesssim \sum_{l = 0}^k \| \partial_x^l \bar{u} x^l \|_\infty \| \partial_x^{k-l} U x^{k-l + \frac 1 4} \phi_{12} \|_\infty + \| \partial_x^l \bar{u}_y y x^{l} \|_\infty \| \partial_x^{k-l} \frac{q}{y} x^{k-l + \frac 1 4} \phi_{12} \|_\infty \\ \nonumber &\lesssim\sum_{l = 0}^k \| \partial_x^{k-l} U x^{k-l + \frac 1 4} \phi_{12} \|_\infty \lesssim \| U, V \|_{\mathcal{X}}, \end{align*} $$

where we have invoked estimate (3.41) as well as the bound (applied to $\partial _x^{k-l} q$ ):

$$ \begin{align*} \| \frac{f}{y} \|_{L^\infty} = \| \frac{1}{y} \int_0^y f_y \|_{L^\infty} \lesssim \| \frac{1}{y} y \| f_y \|_{L^\infty} \|_{L^\infty} \lesssim \| f_y \|_{L^\infty}, \end{align*} $$

for a generic function $f(y)$ satisfying $f(0) = 0$ . To conclude, by using that $x < 400$ is bounded on the set $\{\psi _{12} = 1 \} \cap \{ \phi _{12} < 1 \}$ , we have

(3.51) $$ \begin{align} \nonumber \| u^{(k)} x^{k + \frac 1 4} \psi_{12} \|_\infty & \le \| u^{(k)} x^{k + \frac 1 4} \psi_{12} (1 - \phi_{12}) \|_\infty + \| u^{(k)} x^{k + \frac 1 4} \psi_{12} \phi_{12} \|_\infty \\ &\lesssim \varepsilon^{-M_1} \| U, V \|_{\mathcal{X}} + \| u^{(k)} x^{k + \frac 1 4} \phi_{12} \|_\infty \lesssim \varepsilon^{-M_1} \| U, V \|_{\mathcal{X}}. \end{align} $$

The remaining estimates in (3.47)–(3.50) work in largely the same manner.

Proof. We apply the multiplier

(4.5) $$ \begin{align} \mathbf{M}_{X_0} := [U g^2, \varepsilon V g^2 - \varepsilon q \partial_x (g^2)] = [U g^2, \varepsilon V g^2 + \varepsilon \frac{1}{100} q \langle x \rangle^{-1 - \frac{1}{100}}] \end{align} $$

to the system (2.26)–(2.28). This produces the long identity

$$ \begin{align*} \nonumber \text{Pressure}_{X_0} + \text{Transport}_{X_0} + \text{Error}_{X_0} + \text{Diffusion}_{X_0} = \mathcal{T}_{X_0} + \mathcal{F}_{X_0}. \end{align*} $$

We now provide the definition of each of these terms appearing on the left-hand side.

$$ \begin{align*} \text{Pressure}_{X_0} & := \int P_x U g^2 \,\mathrm{d} y \,\mathrm{d} x+ \int \frac{P_y}{\varepsilon} (\varepsilon V g^2 + \varepsilon \frac{1}{100} q \langle x \rangle^{-1 - \frac{1}{100}} ) \,\mathrm{d} y \,\mathrm{d} x, \\ \text{Transport}_{X_0} & := \int \mathcal{T}_1[U] U g(x)^2 \,\mathrm{d} y \,\mathrm{d} x + \int \mathcal{T}_2[V] (\varepsilon V g^2 + \varepsilon (.01) q \langle x \rangle^{-1.01} ) \,\mathrm{d} y \,\mathrm{d} x \\ \text{Diffusion}_{X_0} & := \int \Big( - \partial_y^2 u + \bar{u}^0_{pyyy} q \Big)U g^2 \,\mathrm{d} y \,\mathrm{d} x- \int \varepsilon u_{xx} U g(x)^2 \,\mathrm{d} y \,\mathrm{d} x\\ &\quad - \int v_{yy} (\varepsilon V g^2 + \varepsilon \frac{1}{100} q \langle x \rangle^{-1 - \frac{1}{100}}) \,\mathrm{d} y \,\mathrm{d} x- \int \varepsilon^2 v_{xx} V g^2 \,\mathrm{d} y \,\mathrm{d} x\\ \text{Error}_{X_0} & := \int 2 \zeta U^2 g^2 \,\mathrm{d} y \,\mathrm{d} x + \int \zeta_y q U g^2 \,\mathrm{d} y \,\mathrm{d} x+ \int \varepsilon \alpha UV g^2 \,\mathrm{d} y \,\mathrm{d} x \\ &\quad + \int \varepsilon \alpha_y q V g^2 \,\mathrm{d} y \,\mathrm{d} x + \int \varepsilon \zeta V^2 g^2 \,\mathrm{d} y \,\mathrm{d} x + \frac{1}{100} \int \varepsilon \alpha U q \langle x \rangle^{-1 - \frac{1}{100}} \,\mathrm{d} y \,\mathrm{d} x \\ &\quad + \frac{1}{100} \int \varepsilon \alpha_y q^2 \langle x \rangle^{-1 - \frac{1}{100}} \,\mathrm{d} y \,\mathrm{d} x + \frac{1}{100} \int \varepsilon \zeta V q \langle x \rangle^{-1 - \frac{1}{100}} \,\mathrm{d} y \,\mathrm{d} x. \end{align*} $$

We observe that $\text {div}_\varepsilon (\mathbf{M}_{X_0}) = 0$ , and moreover that the normal component vanishes at $y = 0, y = \infty $ . Therefore, the $\text {Pressure}_{X_0}$ term vanishes via

$$ \begin{align*} \nonumber \int P_x U g^2 + \int \frac{P_y}{\varepsilon} (\varepsilon V g^2 + \varepsilon \frac{1}{100} q \langle x \rangle^{-1 - \frac{1}{100}} ) = - \int P \text{div}_\varepsilon (\mathbf{M}_{X_0})= 0. \end{align*} $$

The strategy of the proof will now be to establish the following bounds

(4.6) $$ \begin{align} |\text{Error}_{X_0}| &\lesssim \sqrt{\varepsilon} \| U, V \|_{X_0}^2, \qquad \text{Transport}_{X_0} + \text{Diffusion}_{X_0} \ge c \| U, V \|_{X_0}^2 - C ( \sqrt{\varepsilon} + \delta_\ast) \| U, V \|_{X_0}^2, \end{align} $$

for two constants $c, C> 0$ which are independent of $\varepsilon $ and $\delta _\ast $ . This will then give the desired inequality, (4.2) for small enough values of $\varepsilon , \delta _\ast $ .

Over the course of the many calculations appearing in this lemma, several subtle cancellations need to be obtained between the $\text {Transport}_{X_0}$ and the $\text {Diffusion}_{X_0}$ term (which is why we keep them together in (4.6)). Due to the interaction between the terms in $\mathcal {T}_1[U]$ and the terms in $\text {Diffusion}_{X_0}$ , we treat these first (Steps 1 and 2), and postpone the treatment of the $\mathcal {T}_2[V]$ component of $\text {Transport}_{X_0}$ until Step 3 below. We treat the $\text {Error}_{X_0}$ contributions in Step 4 below.

Step 1: $\mathcal {T}_1[U]$ Terms: We now arrive at the transport terms from $\mathcal {T}_1$ , defined in (2.24),

(4.7) $$ \begin{align} \nonumber \int \mathcal{T}_1[U] U g(x)^2 \,\mathrm{d} y \,\mathrm{d} x & = \int (\bar{u}^2 U_x + \bar{u} \bar{v} U_y + 2 \bar{u}^0_{pyy} U) U g^2 \,\mathrm{d} y \,\mathrm{d} x \\ \nonumber & = \frac{1}{200}\int \bar{u}^2 U^2 \langle x \rangle^{-1-\frac{1}{100}} - \int \bar{u} \bar{u}_x U^2 g^2 - \frac 12 \int \partial_y (\bar{u} \bar{v}) U^2 g^2 + \int 2 \bar{u}^0_{pyy} U^2 g^2 \\\nonumber & = \frac{1}{200} \int \bar{u}^2 U^2 \langle x \rangle^{-1 - \frac{1}{100}} - \frac 1 2 \int (\bar{u} \bar{u}_x + \bar{v} \bar{u}_y) U^2 g^2 + \int 2 \bar{u}^0_{pyy} U^2 g^2 \\ & = \frac{1}{200} \int \bar{u}^2 U^2 \langle x \rangle^{-1 - \frac{1}{100}} + \frac 3 2 \int \bar{u}^0_{pyy} U^2 g^2 - \frac 1 2 \int \zeta U^2 g^2 =: \sum_{i = 1}^3 T^0_i, \end{align} $$

where we have invoked (2.14). The term $T^0_1$ is a positive contribution towards the $X_0$ norm. The term $T^0_2$ will be cancelled out below, see (4.10), and so we do not need to estimate it now. The third term from (4.7), $T^0_3$ , will be estimated via

(4.8) $$ \begin{align} \Big| \int \zeta U^2 g^2 \Big| &\lesssim \sqrt{\varepsilon} \int U^2 \langle x \rangle^{-(1+\frac{1}{50})} \lesssim \sqrt{\varepsilon} \| U, V \|_{X_0}^2, \end{align} $$

where we have invoked the pointwise estimates (2.15), as well as the Hardy-type inequality (3.25).

Step 2: Diffusive Terms We would now like to treat the diffusive terms from (2.26)–(2.27). We group the term $\bar {u}^0_{pyyy}q$ from (2.26) with this treatment for the purpose of achieving a cancellation. More precisely, we will begin by treating the following quantity, for which several integrations by parts produce

(4.9) $$ \begin{align} \nonumber \int \Big( - \partial_y^2 u + \bar{u}^0_{pyyy} q \Big)U g^2 & = \int u_y U_y g^2 + \int_{y = 0} u_y U g^2 \,\mathrm{d} x - \frac 1 2 \int \bar{u}^0_{pyyyy} q^2 g^2 \\ \nonumber & = \int \bar{u} U_y^2 g^2- 2\int \bar{u}_{yy} U^2 g^2 + \frac 1 2 \int \partial_y^4 (\bar{u} - \bar{u}^0_p) q^2 g^2+ \int_{y = 0} \bar{u}_y U^2 g^2 \,\mathrm{d} x \\ & = D^0_1 + D^0_2 + D^0_3 + D^0_4. \end{align} $$

We have used that $u_y|_{y = 0} = ( \bar {u} U_y + 2 \bar {u}_y U + \bar {u}_{yy}q)|_{y = 0} = 2 \bar {u}_y U|_{y = 0}$ . Both $D^0_1, D^0_4$ are positive contributions, thanks to (2.13). We now note that the main contribution from the $D^0_2$ term cancels the contribution $T^0_2$ , and generates a positive damping term of

(4.10) $$ \begin{align} \nonumber T^0_2 + D^0_2 & = \frac 3 2 \int \bar{u}^0_{pyy} U^2 g^2 - 2 \int \bar{u}_{yy} U^2 g^2 = - \frac 1 2 \int \bar{u}^0_{pyy} U^2 g^2 - 2 \int (\bar{u}_{yy} - \bar{u}^0_{pyy}) U^2 g^2 \\& = - \frac 1 2 \int \partial_{yy} \bar{u}_\ast U^2 g^2 - \frac 1 2 \int \partial_y^2 (\bar{u}^0_{p} - \bar{u}_\ast) U^2 g^2 - 2 \int (\bar{u}_{yy} - \bar{u}^0_{pyy}) U^2 g^2 \\\nonumber &=: D^0_{2,1} + D^0_{2,2} + D^0_{2,3}. \end{align} $$

The first term on the right-hand side above, due to $\bar {u}_\ast $ , is a positive contribution due to (2.5). For the term $D^0_{2,2,}$ , we estimate via

$$ \begin{align*} | \frac 1 2 \int \partial_y^2 (\bar{u}^0_{p} - \bar{u}_\ast) U^2 g^2| \lesssim \delta_\ast \int \langle x \rangle^{- \frac 5 4+ \sigma_\ast} U^2 \lesssim \delta_\ast \|U, V \|_{X_0}^2, \end{align*} $$

where we have appealed to estimate (1.36), as well as (3.25).

The remaining contribution, $D^0_{2,2}$ , we estimate by invoking the pointwise estimate $|\bar {u}_{yy} - \bar {u}^0_{pyy}| \lesssim \sqrt {\varepsilon } \langle x \rangle ^{-1-\frac {1}{50}}$ due to (2.11). The third term from (4.9), $D^0_3$ , is estimated by

(4.11) $$ \begin{align} |\int \partial_y^4 (\bar{u} - \bar{u}^0_p) q^2 g^2| \lesssim \sqrt{\varepsilon} \| \frac{q}{y} \langle x \rangle^{- \frac 1 2 - \frac{1}{100}} \|^2 \lesssim \sqrt{\varepsilon} \| U \langle x \rangle^{- \frac 1 2 - \frac{1}{100}} \|^2 \lesssim \sqrt{\varepsilon} \| U, V \|_{X_0}^2, \end{align} $$

where we have invoked estimate (2.11) and subsequently the standard Hardy inequality in y, as $q|_{y = 0} = 0$ , as well as (3.25).

The next diffusive term is

(4.12) $$ \begin{align} - \int \varepsilon u_{xx} U g(x)^2 = \int \varepsilon u_x U_x g(x)^2 + 2 \int \varepsilon u_x U g(x) g'(x). \end{align} $$

upon using that $U|_{x = 0} = 0$ . The $g'$ term above is easily controlled by $\sqrt {\varepsilon } \| U \langle x \rangle ^{- \frac 1 2 - \frac {1}{200}} \|^2 + \sqrt {\varepsilon } \| \sqrt {\varepsilon } \sqrt {\bar {u}} U_x g \|^2 + \sqrt {\varepsilon } \| \sqrt {\varepsilon }V \langle x \rangle ^{- \frac 1 2 - \frac {1}{200}} \|^2$ upon consulting the definition of g to compute $g'$ , and definition (2.23) to expand $u_x$ in terms of $(U, V, q)$ .

We now address the first term on the right-hand side of (4.12), which yields,

(4.13) $$ \begin{align} \nonumber \int \varepsilon u_x U_x g^2 & = \int \varepsilon \partial_x (\bar{u} U + \bar{u}_y q) U_x g^2 \\ \nonumber & = \int \varepsilon \bar{u} U_x^2 g^2- \frac 1 2 \int \varepsilon \bar{u}_{xx} U^2g^2 + \frac 1 2 \int \varepsilon \bar{u}_{xxyy} q^2 g^2 + \int \varepsilon \bar{u}_{xy} UV g^2 \\ &\quad - \frac 1 2\int \varepsilon \bar{u}_{yy} V^2 g^2 + \int \varepsilon \bar{u}_{xyy} q^2 gg' - \varepsilon \int \bar{u}_x g g' U^2 = \sum_{i = 1}^7 D^1_i. \end{align} $$

We observe that $D^1_1$ is a positive contribution. We estimate $D^1_7$ via

(4.14) $$ \begin{align} |\varepsilon \int \bar{u}_x gg' U^2| \lesssim \varepsilon \| U \langle x \rangle^{-1} \|^2 \lesssim \varepsilon \| U, V \|_{X_0}^2, \end{align} $$

and similarly for $D^1_6$ , where we have invoked the pointwise decay estimate on $\bar {u}_x$ in (2.6). The remaining terms, $D^1_2,...,D^1_5$ will be placed into the term $\mathcal {J}_1$ defined below in (4.21).

We now arrive at the $v_{yy}$ diffusive term from (2.27), which reads after integrating by parts in y,

(4.15) $$ \begin{align} - \int v_{yy} (\varepsilon V g^2 + \varepsilon \frac{1}{100} q \langle x \rangle^{-1 - \frac{1}{100}}) = \int \varepsilon v_y V_y g^2 - \frac{1}{100} \int \varepsilon v U_y \langle x \rangle^{- 1 - \frac{1}{100}}. \end{align} $$

The second contribution on the right-hand side above is easily estimated by appealing to (2.23) and estimate (2.7), which generates

$$ \begin{align*} \nonumber |\int \varepsilon (\bar{u} V - \bar{u}_x q) U_y \langle x \rangle^{- 1- \frac{1}{100}}| &\lesssim \sqrt{\varepsilon} \Big( \| \sqrt{\varepsilon} V \langle x \rangle^{- \frac 1 2 - \frac{1}{200}} \| + \sqrt{\varepsilon} \| \bar{u}_{x} y \|_\infty \| U \langle x \rangle^{- \frac 1 2 - \frac{1}{200}} \| \Big)\| \sqrt{\bar{u}} U_y \| \\ \nonumber &\lesssim \sqrt{\varepsilon} \|U, V \|_{X_0} \|U, V \|_{X_0}, \end{align*} $$

where we have invoked the Hardy type inequality, (3.15) to estimate the $U, V$ terms in parenthesis above, and the definition of the $\|U, V \|_{X_0}$ norm which contains the term $\| \sqrt {\bar {u}} U_y \|$ .

For the first contribution on the right-hand side of (4.15), we have the following identity

(4.16) $$ \begin{align} \nonumber \int \varepsilon v_y V_y g^2& = \int \varepsilon \bar{u} V_y^2 g^2 - \frac 1 2 \int \varepsilon \bar{u}_{yy} V^2 g^2 + \frac 1 2 \int \varepsilon \bar{u}_{xxyy} q^2 g^2+ \int \varepsilon \bar{u}_{xy} UV g^2 \\ &\quad- \frac 1 2 \int \varepsilon \bar{u}_{xx} U^2 g^2 + \int \varepsilon \bar{u}_{xyy} q^2 gg' - \int \varepsilon \bar{u}_x U^2 gg' = \sum_{i = 1}^7 D^2_i. \end{align} $$

We observe that $D^2_1$ is a positive contribution, whereas $D^2_6, D^2_7$ are estimated identically to $D^1_6, D^1_7$ . The remaining terms, $D^2_2,...,D^2_5$ will be placed into the term $\mathcal {J}_1$ , defined below in (4.21).

We now arrive at the final diffusive term, for which we first integrate by parts using the boundary condition $V|_{x = 0} = q|_{x = 0} = 0$ ,

(4.17) $$ \begin{align} &\int - \varepsilon^2 v_{xx} V g^2 - 2 \int \varepsilon^2 v_{xx} q gg' = \int \varepsilon^2 v_x V_x g^2 + \int 2 \varepsilon^2 v_x q (gg')'. \end{align} $$

The second term above, which contains a $g'$ factor, is easily controlled by a factor of $\sqrt {\varepsilon } \| U, V \|_{X_0}^2$ by again appealing to the definition of g and (2.23). For the first term on the right-hand side of (4.17), we have

(4.18) $$ \begin{align} \nonumber \int \varepsilon^2 v_x V_x g^2& = \int \varepsilon^2 \bar{u} V_x^2 g^2 - 2 \int \varepsilon^2 \bar{u}_{xx} V^2 g^2+ \frac 1 2\int \varepsilon^2 \bar{u}_{xxxx} q^2 g^2 \\ &\quad + \int \varepsilon^2 \bar{u}_{xx} q^2 (gg')' + \int 2 \varepsilon^2 \bar{u}_{xxx} q^2 gg' - \int 2 \varepsilon^2 \bar{u}_{xx} V^2 g^2 = \sum_{i = 1}^6 D^3_i. \end{align} $$

The terms with $g'$ above are easily controlled by a factor of $\sqrt {\varepsilon } \| U, V \|_{X_0}^2$ by again appealing to the definition of g and estimate (2.6).

We now expand the damping terms, which are terms $D^1_5$ and $D^2_2$ ,

(4.19) $$ \begin{align} D^1_5 + D^2_2 & = - \int \varepsilon \bar{u}_{yy} V^2 g^2 = - \int \varepsilon \bar{u}^0_{pyy} V^2 g^2 - \int \varepsilon^2 (\bar{u}_{yy} - \bar{u}^0_{pyy}) V^2 g^2. \end{align} $$

We estimate the latter term above via an appeal to (2.11), which gives

(4.20) $$ \begin{align} \Big| \int \varepsilon^2 (\bar{u}_{yy} - \bar{u}^0_{pyy}) V^2 g^2 \Big| \lesssim \varepsilon \| \sqrt{\varepsilon} V \langle x \rangle^{- \frac 1 2 - \frac{1}{100}} \|^2 \lesssim \varepsilon \| U, V \|_{X_0}^2. \end{align} $$

We now consolidate the remaining terms from (4.13), (4.16), (4.18). Specifically, we obtain

$$ \begin{align*} D^1_2 + D^1_3 + D^1_4 + D^2_3 + D^2_4 + D^2_5 + D^3_2 + D^3_3 + D^3_6 = \mathcal{J}_1, \end{align*} $$

where we have defined

(4.21) $$ \begin{align} \mathcal{J}_1 := - \int \varepsilon \bar{u}_{xx} U^2 g^2 - 2 \int \varepsilon^2 \bar{u}_{xx} V^2 g^2+ \int 2 \varepsilon \bar{u}_{xy} UV g^2+ \int \Big( \varepsilon \bar{u}_{xxyy} + \frac 1 2 \varepsilon^2 \bar{u}_{xxxx} \Big) q^2 g^2. \end{align} $$

To estimate these contributions, we simply use the fact that $\| \bar {u}_{xx} \langle x \rangle ^2 \|_\infty \lesssim 1$ , $\| \bar {u}_{xy} \langle x \rangle ^{\frac 3 2} \|_\infty \lesssim 1$ , $\| y^2 (\bar {u}_{xxyy} + \bar {u}^P_{xxxx}) \langle x \rangle ^2 \|_\infty \lesssim 1$ , and $\| \bar {u}_{E xxxx} \|_{L^\infty _y} \lesssim \sqrt {\varepsilon } \langle x \rangle ^{-\frac 9 2}$ according to the estimates (2.6)–(2.7). Upon invoking these pointwise bounds, we estimate

$$ \begin{align*} |\mathcal{J}_1| &\lesssim \varepsilon \|U \langle x \rangle^{- 1} \|^2 + \varepsilon \|\sqrt{\varepsilon}V \langle x \rangle^{- 1} \|^2 + \sqrt{\varepsilon} \|U \langle x \rangle^{-\frac34} \| \| \sqrt{\varepsilon} V \langle x \rangle^{-\frac34} \| + \varepsilon \| \frac{q}{y} \langle x \rangle^{-1} \|^2 + \varepsilon \|\sqrt{\varepsilon} q \langle x \rangle^{-2} \|^2 \\ &\lesssim \sqrt{\varepsilon} \|U, V \|_{X_0}^2, \end{align*} $$

upon invoking the Hardy-type inequality (3.25).

Step 3: $\mathcal {T}_2[V]$ Terms We now treat the terms arising from $\mathcal {T}_2[V]$ , which has been defined in (2.25). Specifically, the result of applying the multiplier (4.5) is

(4.22) $$ \begin{align} \nonumber &\int \mathcal{T}_2[V] (\varepsilon V g^2 + \varepsilon (.01) q \langle x \rangle^{-1.01} ) \\\nonumber & \quad = \int \varepsilon \mathcal{T}_2[V] V g^2 + \varepsilon (.01) \int \bar{u}^2 V_x q \langle x \rangle^{-1.01} + \varepsilon (.01) \int (\bar{u} \bar{v} V_y + \bar{u}^0_{pyy} V) q \langle x \rangle^{-1.01} \\& \quad =: \tilde{T}^1_1 + \tilde{T}^1_2 + \tilde{T}^1_3. \end{align} $$

For the first term on the right-hand side of (4.22), $\tilde {T}^1_1$ , we have the following identity

(4.23) $$ \begin{align} \nonumber \tilde{T}^1_1 = \int \varepsilon \mathcal{T}_2[V] V g^2 & = \int \varepsilon \bar{u}^2 V_x V g^2 + \int \varepsilon \bar{u} \bar{v} V_y V g^2 + \int \varepsilon \bar{u}^0_{pyy} V^2 g^2 \\ & = \frac 1 2 \int \varepsilon \bar{u}^0_{pyy} V^2 g^2 - \frac 1 2 \int \varepsilon \zeta V^2 g^2 + \frac{1}{200} \int \varepsilon \bar{u}^2 V^2 \langle x \rangle^{-1 - \frac{1}{100}} = \sum_{i = 1}^3 T^1_i, \end{align} $$

where we have invoked (2.14). The term $T^1_2$ is estimated in an analogous manner to (4.8), whereas $T^1_3$ is a positive contribution to the $X_0$ norm. On the other hand, $T^1_1$ combines with $D^1_5 + D^2_2$ to produce a damping term of the form $- \frac 12 \int \bar {u}^0_{pyy} \varepsilon V^2 g^2$ .

We now need to address the contribution of $\tilde {T}^1_2$ . Integrating by parts, we get

(4.24) $$ \begin{align} \nonumber \tilde{T}^1_2 & = .01 \int \varepsilon \bar{u}^2 V^2 \langle x \rangle^{-1.01} -2 (.01) \int \varepsilon \bar{u} \bar{u}_x V q \langle x \rangle^{-1.01} + (.01)(1.01) \int \varepsilon \bar{u}^2 V q \langle x \rangle^{-2.01} \\ \nonumber & = .01 \int \varepsilon \bar{u}^2 V^2 \langle x \rangle^{-1.01} -2 (.01) \int \varepsilon \bar{u} \bar{u}_x V q \langle x \rangle^{-1.01} \\ &\quad - \frac 1 2 (.01)(1.01) (2.01) \int \varepsilon \bar{u}^2 q^2 \langle x \rangle^{-3.01} - \frac 1 2 (.01)(1.01) (2.01) \int \varepsilon \bar{u} \bar{u}_x q^2 \langle x \rangle^{-2.01} \\ \nonumber & = \tilde{T}^1_{2,1} + \tilde{T}^1_{2,2} + \tilde{T}^1_{2,3} + \tilde{T}^1_{2,4}. \end{align} $$

Of these, the third term, $\tilde {T}^1_{2,3}$ is very dangerous due to a lack of decay in z for the coefficient. To treat it, we combine $\tilde {T}^1_{2,1}, \tilde {T}^1_{2,3}$ and $T^1_3$ to obtain the expression

(4.25) $$ \begin{align} \nonumber \tilde{T}^1_{2,1} + \tilde{T}^1_{2,3} + T^1_3 & = \frac{3}{2}(.01) \int \varepsilon \bar{u}^2 V^2 \langle x \rangle^{-1.01} - \frac{(.01)(1.01)(2.01)}{2} \int \varepsilon \bar{u}^2 q^2 \langle x \rangle^{-3.01} \\ \nonumber & \ge \Big(\frac{3}{2}(.01) - \frac{(.01)(1.01)(2.01)}{2} \frac{1}{1.01} \Big) \int \varepsilon \bar{u}^2 V^2 \langle x \rangle^{-1.01}\\ \nonumber &\quad - \frac{(.01)(1.01)(2.01)}{2} \frac{2}{1.01} \int \langle x \rangle^{-2.01} \bar{u} \bar{u}_x q^2 \\ & \ge \frac{.01}{2} \int \varepsilon \bar{u}^2 V^2 \langle x \rangle^{-1.01} - (.01)(2.01) \int \varepsilon \langle x \rangle^{-2.01} \bar{u} \bar{u}_x q^2, \end{align} $$

where we have used the precise constants appearing in (3.26).

We now estimate the error term from (4.25) by now splitting $\bar {u} = \bar {u}_p + \bar {u}_E$ , according to (2.1). First, we have

(4.26) $$ \begin{align} |\int \varepsilon \langle x \rangle^{-2.01} \bar{u} \bar{u}_{px} q^2| \lesssim \| \bar{u}_{px} y^2 \|_\infty \varepsilon \| \frac{q}{y} \langle x \rangle^{-1.01} \|^2 \lesssim \varepsilon \| U \langle x \rangle^{-1.01} \|^2 \lesssim \varepsilon \| U, V \|_{X_0}^2, \end{align} $$

where we have used estimate (2.9). For the $\bar {u}_E$ component, we may use the small amplitude and importantly the enhanced decay in x from (2.8) to obtain

(4.27) $$ \begin{align} |\int \varepsilon \langle x \rangle^{-2.01} \bar{u} \bar{u}_{Ex} q^2| \lesssim \varepsilon^{\frac 3 2} | \int \langle x \rangle^{- 3.51} q^2| \lesssim \varepsilon^{\frac 1 2} \| \sqrt{\varepsilon}V \langle x \rangle^{- \frac 3 4} \|^2 \lesssim \sqrt{\varepsilon} \| U, V \|_{X_0}^2. \end{align} $$

The error terms $\tilde {T}^1_{2,2}$ and $\tilde {T}^1_{2,4}$ are estimated in a nearly identical manner.

We now address the third terms from (4.22), $\tilde {T}^1_3$ , which upon integration by parts in y gives

(4.28) $$ \begin{align} \nonumber &\varepsilon (.01) |\int (\bar{u} \bar{v} V_y + \bar{u}^0_{pyy} V) q \langle x \rangle^{-1.01}| \lesssim \varepsilon | \int (\bar{u}^0_{pyy} - (\bar{u} \bar{v})_y) V q \langle x \rangle^{-1.01}| + \varepsilon | \int \bar{u} \bar{v} V U \langle x \rangle^{-1.01}| \\ &\lesssim \sqrt{\varepsilon} \Big( \| (\bar{u}^0_{pyy} - (\bar{u} \bar{v})_y) y \langle x \rangle^{\frac 1 2} \|_\infty + \| \bar{v} \langle x \rangle^{\frac 1 2} \|_\infty \Big) \| U \langle x \rangle^{- \frac 3 4} \| \| \sqrt{\varepsilon} V \langle x \rangle^{- \frac 3 4} \| \lesssim \sqrt{\varepsilon} \|U, V \|_{X_0}^2, \end{align} $$

where we have appealed to the estimates (2.6) as well as (3.25).

Step 4: $\text {Error}_{X_0}$ Terms The first term in $\text {Error}_{X_0}$ is identical to (4.8), which has already been bounded above by a factor of $\sqrt {\varepsilon }\| U, V \|_{X_0}^2$ . We next move to the second term in $\text {Error}_{X_0}$ :

(4.29) $$ \begin{align} \Big| \int \zeta_y q U g^2 \Big| \lesssim \sqrt{\varepsilon} \| \frac{q}{y} \langle x \rangle^{- \frac 1 2 - \frac{1}{100}} \| \| U \langle x \rangle^{- \frac 1 2 - \frac{1}{100}} \| \lesssim \sqrt{\varepsilon} \| U \langle x \rangle^{- \frac 1 2 - \frac{1}{100}} \|^2 \lesssim \sqrt{\varepsilon} \|U, V \|_{X_0}^2, \end{align} $$

where we have invoked the pointwise estimates (2.15) on the quantity $|y \partial _y \zeta |$ , and used the standard Hardy inequality in y, admissible as $q|_{y = 0} = 0$ . We next treat the third and fourth terms from $\text {Error}_{X_0}$ , which we estimate via

(4.30) $$ \begin{align} \Big| \int \varepsilon \alpha UV g^2\Big| \lesssim \varepsilon \int \langle x \rangle^{- \frac 3 2}|UV| \lesssim \sqrt{\varepsilon} \| U \langle x \rangle^{- \frac 3 4} \| \| \sqrt{\varepsilon} V \langle x \rangle^{- \frac 3 4} \| \lesssim \sqrt{\varepsilon} \|U, V \|_{X_0}^2, \end{align} $$

(4.31) $$ \begin{align} \Big| \int \varepsilon \alpha_y qV g^2\Big| \lesssim \varepsilon \int \langle x \rangle^{- \frac 3 2}|\frac{q}{\langle y \rangle}|V| \lesssim \sqrt{\varepsilon} \| \frac{q}{y} \langle x \rangle^{- \frac 3 4} \| \| \sqrt{\varepsilon} V \langle x \rangle^{- \frac 3 4} \| \lesssim \sqrt{\varepsilon} \|U, V \|_{X_0}^2. \end{align} $$

Above we have relied on the coefficient estimate in (2.16). The fifth term in $\text {Error}_{X_0}$ is treated analogously to the first term. For the remaining terms, we estimate as follows

$$ \begin{align*} |\text{Error}_{X_0}^{(6)}| &\lesssim \sqrt{\varepsilon} \| \alpha \langle x \rangle^{\frac32} \|_\infty \| U \langle x \rangle^{-\frac34} \| \| \sqrt{\varepsilon}q \langle x \rangle^{-\frac74} \| \lesssim \sqrt{\varepsilon} \|U, V \|_{X_0}^2, \\ |\text{Error}_{X_0}^{(7)}| &\lesssim \sqrt{\varepsilon} \| \alpha_y y \langle x \rangle^{\frac32} \|_\infty \| \frac{q}{y} \langle x \rangle^{-\frac34} \| \| \sqrt{\varepsilon}q \langle x \rangle^{-\frac74} \| \lesssim \sqrt{\varepsilon} \|U, V \|_{X_0}^2\\ |\text{Error}_{X_0}^{(8)}| &\lesssim \sqrt{\varepsilon} \| \frac{\zeta}{\sqrt{\varepsilon}}\langle x \rangle^{1 + \frac{1}{50}} \|_\infty \| \sqrt{\varepsilon}V \langle x \rangle^{-\frac12 - \frac{1}{100}} \| \| \sqrt{\varepsilon} q \langle x \rangle^{-\frac32 - \frac{1}{100}} \| \lesssim \sqrt{\varepsilon} \|U, V \|_{X_0}^2, \end{align*} $$

where we have repeatedly appealed to the coefficient estimates (2.16) as well as the Hardy-type inequality (3.25).

Step 5: Conclusion The bounds in Step 4 establish that $|\text {Error}_{X_0}| \lesssim \sqrt {\varepsilon } \|U, V \|_{X_0}^2$ . We now consolidate the $\text {Transport}_{X_0} + \text {Diffusion}_{X_0}$ bounds. Recall the definition of the $X_0$ norm, (3.2). The first three quantities in (3.2) are controlled by $D^0_1$ , (4.9), $D^1_1$ , (4.13), $D^2_1$ , (4.16), and $D^3_1$ , (4.18). The fourth and fifth terms appearing in $X_0$ are damping terms, which are found in (4.10) and (4.19). The boundary trace term appearing in $X_0$ is controlled by $D^0_4$ appearing in (4.9). Finally, the $U, \sqrt {\varepsilon }V$ terms appearing in $X_0$ are controlled by the lower bound in (4.25) as well as $T^0_1$ in (4.7). Therefore, all components of the $X_0$ norm have been controlled from above by $\text {Transport}_{X_0} + \text {Diffusion}_{X_0}$ . All of the (many) remaining contributions from $\text {Transport}_{X_0} + \text {Diffusion}_{X_0}$ have been shown to be bounded above by $\sqrt {\varepsilon } \|U, V \|_{X_0}^2$ or by $\delta _\ast \|U, V \|_{X_0}^2$ . This then establishes the inequalities in (4.6), and therefore concludes the proof of Lemma 4.1.

Proof. We apply the weighted in x vector-field

(4.35) $$ \begin{align} \mathbf{M}_{X_{\frac 1 2}} := [U_x x \phi_1(x)^2 , \varepsilon V_x x \phi_1(x)^2 + \varepsilon V \phi_1(x)^2 + 2\varepsilon V x \phi_1 \phi_1'] \end{align} $$

as a multiplier to (2.26)–(2.28). We first of all notice that $\text {div}_\varepsilon (\mathbf{M}_{X_{\frac 1 2}}) = 0$ , and thus,

$$ \begin{align*} \nonumber \int P_x U_x x \phi_1^2 + \int \frac{P_y}{\varepsilon} (\varepsilon V_x x \phi_1^2 + \varepsilon V \phi_1^2 + 2 \varepsilon V x \phi_1 \phi_1') = - \int P \text{div}_\varepsilon(\mathbf{M}_{X_{\frac 1 2}}) = 0, \end{align*} $$

where we use that $V|_{y = 0} = V_x|_{y = 0} = 0$ and $\phi _1$ to eliminate any contributions from $\{x = 0\}$ .

Step 1: $\mathcal {T}_1[U]$ terms: We address the terms from $\mathcal {T}_1$ which produces the identity

(4.36) $$ \begin{align} \nonumber \int \mathcal{T}_1[U] U_x x \phi_1^2 & = \int \bar{u}^2 U_x^2 x \phi_1^2+ \int \bar{u} \bar{v} U_y U_x x \phi_1^2- \int \bar{u}_{yy} U^2 \phi_1^2- \int \bar{u}_{xyy} x U^2 \phi_1^2 \\ &\quad - 2\int \bar{u}_{yy} U^2 x \phi_1 \phi_1' = :\sum_{i = 1}^5 T^{(2)}_i. \end{align} $$

First, we observe $T^{(2)}_1$ is a positive contribution. We estimate $T^{(2)}_2$ via

(4.37) $$ \begin{align} &|\int \bar{v} \bar{u} U_y U_x x \phi_1^2 | \lesssim \| \frac{\bar{v}}{\bar{u}} x^{\frac 1 2} \|_\infty \| \sqrt{\bar{u}} U_y \| \| \bar{u} U_x x^{\frac 1 2} \phi_1 \| \le \delta \| \bar{u} U_x x^{\frac 1 2} \phi_1 \|^2 + C_\delta \| \sqrt{\bar{u}} U_y \|^2, \end{align} $$

where above we have invoked estimate (2.7) for $\bar {v}$ .

For $T^{(2)}_3$ , we need to split $U = U(x, 0) + (U - U(x, 0))$ , and subsequently estimate via

(4.38) $$ \begin{align} \nonumber |\int \bar{u}_{yy} U^2 \phi_1^2 | &\lesssim \int |\bar{u}_{yy}| (U - U(x, 0))^2 \phi_1^2 + \int |\bar{u}_{yy}| U(x, 0)^2 \phi_1^2 \\ \nonumber &\lesssim \| y^2 \bar{u}_{yy} \|_\infty \Big\| \frac{U - U(x, 0)}{y} \phi_1 \Big\|^2 +\Big( \sup_x \int |\bar{u}_{yy} | x^{\frac 1 2} \,\mathrm{d} y \Big) \| U x^{-\frac 1 4} \|_{L^2(x = 0)}^2 \\ &\lesssim \| U_y \phi_1\|^2 + \| U, V \|_{X_0}^2 \le C_\delta \| U, V \|_{X_0}^2 + \delta \| U, V \|_{Y_{\frac 1 2}}^2, \end{align} $$

where above, we used Hardy inequality in y, which is admissible as $(U - U(x, 0))|_{y = 0} = 0$ , as well as the inequality (3.22). $T^{(2)}_4$ works in an analogous manner, and so we omit it. For $T^{(2)}_5$ , we note that the support of $\phi _1'$ is bounded in x, and so this term can trivially be controlled by $\| U, V \|_{X_0}^2$ .

Step 2: $\mathcal {T}_2[V]$ terms: We now address the contributions from $\mathcal {T}_2[V]$ . For this, we first note that, examining the multiplier (4.35), the contribution from $\varepsilon V$ has already been treated in Lemma 4.1, and we therefore just need to estimate the contribution from the principal term, $\varepsilon V_x x$ . More precisely, we have already established the following estimate,

(4.39) $$ \begin{align} |\int \mathcal{T}_2[V] \varepsilon (V \phi_1^2 + 2 V x \phi_1 \phi_1' )| \lesssim \|U, V \|_{X_0}^2. \end{align} $$

We now estimate the contribution of the principal term, $\varepsilon V_x x$ . For this, recall the definition (2.25),

$$ \begin{align*} \nonumber \int \varepsilon \mathcal{T}_2[V] V_x x \phi_1^2 & = \int \varepsilon \Big( \bar{u}^2 V_x + \bar{u} \bar{v} V_y + \bar{u}_{yy} V \Big) V_x x \phi_1^2 \\ \nonumber & = \int \varepsilon \bar{u}^2 V_x^2 x \phi_1^2 + \int \varepsilon \bar{u} \bar{v} V_y V_x x \phi_1^2 - \frac 1 2 \int \varepsilon \partial_x (x \bar{u}_{yy}) V^2 \phi_1^2 - \int \varepsilon x \bar{u}_{yy} V^2 \phi_1 \phi_1' \\ \nonumber & = T^{(3)}_1 + ... + T^{(3)}_4. \end{align*} $$

We observe that $T^{(3)}_1$ is a positive contribution. The integrand in the term $T^{(3)}_4$ has a bounded support of x and so can immediately be controlled by $\|U, V \|_{X_0}^2$ . We may estimate $T^{(3)}_2$ , and $T^{(3)}_3$ via

(4.40) $$ \begin{align} &|\int \varepsilon \bar{u} \bar{v} V_y V_x x \phi_1^2| \lesssim \sqrt{\varepsilon} \| \frac{ \bar{v}}{\bar{u}} x^{\frac 1 2} \|_\infty \| \bar{u} U_x x^{\frac 1 2} \phi_1 \| \sqrt{\varepsilon} \bar{u} V_x x^{\frac 1 2}\phi_1 \| \lesssim \sqrt{\varepsilon} \| U, V \|_{X_{\frac 1 2}}^2, \end{align} $$

(4.41) $$ \begin{align} &|\int \varepsilon \partial_x (x \bar{u}_{yy}) V^2 \phi_1^2| \lesssim \| y^2 \partial_x (x \bar{u}_{yy}) \|_\infty \varepsilon \Big\| \frac{V}{y} \phi_1\Big\|^2 \lesssim \varepsilon \| V_y \phi_1\|^2 \le C_\delta \| U, V \|_{X_0}^2 + \delta \| U, V \|_{Y_{\frac 1 2}}^2,\end{align} $$

where we have invoked the Hardy inequality (3.15).

Step 3: Diffusive Terms We now address the main diffusive term, which is the contribution of $- u_{yy}$ in (2.26). We, again, group the term $\bar {u}^0_{pyyy}q$ from (2.26) with this term. More precisely, after a long series of integrations by parts, we have the following identity

(4.42) $$ \begin{align} \nonumber \int \Big(- u_{yy} &\quad + \bar{u}^0_{pyyy} q \Big) U_x x\phi_1^2 \\ \nonumber & = - \int 2 x \bar{u}_y U_x U_y\phi_1^2 + \frac 3 2 \int \partial_x (x \bar{u}_{yy}) U^2 \phi_1^2- \frac 1 2 \int (\bar{u} + x \bar{u}_x ) U_y^2\phi_1^2 \\ \nonumber &\quad - \int_{y = 0} \partial_x (x \bar{u}_y) U^2 \phi_1^2 \,\mathrm{d} x -\int (\bar{u}_{yyy} - \bar{u}^0_{pyyy}) qU_x x \phi_1^2 + \int x \bar{u}_{yy} U^2 \phi_1 \phi_1' \\ &\quad -2 \int_{y = 0} U^2 x \bar{u}_y \phi_1 \phi_1' - \int \bar{u} x U_y^2 \phi_1 \phi_1' - \int 4 \bar{u}_y UU_y x \phi_1 \phi_1' = \sum_{i = 1}^{9} D^{(4)}_i. \end{align} $$

For the first term above, $D^{(4)}_1$ , we localize in z using the cutoff function $\chi (\cdot )$ (see (1.62)), via

(4.43) $$ \begin{align} D_1^{(4)} = - \int 2 x \bar{u}_y U_x U_y\phi_1^2 (1 - \chi(z)) - \int 2 x \bar{u}_y U_x U_y\phi_1^2 \chi(z). \end{align} $$

The far-field component is controlled easily via

(4.44) $$ \begin{align} |\int x \bar{u}_y U_x U_y (1- \chi(z)) \phi_1^2| \lesssim \| \bar{u}_y x^{\frac 1 2} \|_{\infty} \| \bar{u} U_x x^{\frac 1 2} \phi_1 \| \| \sqrt{\bar{u}} U_y \| \le \delta \| U, V \|_{X_{\frac 1 2}}^2 + C_\delta \| U, V \|_{X_0}^2, \end{align} $$

where we have used $\bar {u} \gtrsim 1$ when $z \gtrsim 1$ , according to (2.12).

The localized piece requires the use of higher order norms, and we estimate it via

(4.45) $$ \begin{align} \nonumber |\int x \bar{u}_y U_x U_y \chi(z) \phi_1^2| &\lesssim \| \sqrt{x} \bar{u}_y \|_\infty \| U_x x^{\frac 1 2} \chi(z) \phi_1 \| \| U_y \phi_1 \| \\ \nonumber &\lesssim ( \| \sqrt{\bar{u}} U_{xy} x\phi_1 \| + \| \bar{u} U_x \sqrt{x} \phi_1\| ) ( \delta \| U, V \|_{Y_{\frac 1 2}} + C_\delta \| U, V \|_{X_0} ) \\ &\lesssim \delta \| U, V \|_{X_1}^2 + \delta \| U, V \|_{X_{\frac 1 2}}^2 + \delta \| U, V \|_{Y_{\frac 1 2}}^2 + C_\delta \| U, V \|_{X_0}^2, \end{align} $$

where above, we have appealed to both (3.22) as well as (3.23). For $D^{(4)}_2$ , we estimate in the same manner as (4.38), whereas $D^{(4)}_3$ can easily be controlled upon using $\| \bar {u} + x \partial _x \bar {u} \|_\infty \lesssim 1$ , according to (2.6). The term $D^{(4)}_4$ is immediately controlled by $\|U, V \|_{X_0}^2$ . The term $D^{(4)}_5$ we estimate via

(4.46) $$ \begin{align} \nonumber |\int (\bar{u}_{yyy} - \bar{u}^0_{pyyy}) q U_x x \phi_1^2| &\lesssim \| (\bar{u}_{yyy} - \bar{u}^0_{pyyy}) y x^{1.01} \|_\infty \| U \langle x \rangle^{-1.01} \| \| U_x \langle x \rangle^{\frac 1 2} \phi_1 \| \\ &\lesssim \sqrt{\varepsilon} (\|U, V \|_{X_0}^2 + \| U, V \|_{X_{\frac 1 2}}^2 + \| U, V \|_{X_1}^2), \end{align} $$

where we have invoked (2.11). Finally, for the remaining four terms from (4.42), $D^{(4)}_k$ , $k = 6,7,8,9$ , due to the presence of $\phi _1'$ , the x weights are all bounded, and these terms can thus be easily controlled by $\|U, V \|_{X_0}^2$ .

We now move to the contribution of the tangential diffusive term, $-\varepsilon u_{xx}$ , which produces the following identity

(4.47) $$ \begin{align} \nonumber - \int \varepsilon u_{xx} U_x x \phi_1^2 & = \int \varepsilon u_x U_{xx} x \phi_1^2 + \int \varepsilon u_x U_x \phi_1^2 + 2 \int \varepsilon u_x U_x x \phi_1 \phi_1' \\ \nonumber & = \frac 1 2 \int \varepsilon \bar{u} U_x^2 \phi_1^2 - \frac 3 2 \int \varepsilon x \bar{u}_x U_x^2 \phi_1^2 + \int \varepsilon (\bar{u}_{xx} + \frac{1}{2} \bar{u}_{xxx}x) U^2 \phi_1^2 + \int \varepsilon \bar{u}_{xyy} q V_x \phi_1^2 \\ \nonumber &\quad + \int \varepsilon \bar{u}_{xy} U V_x x \phi_1^2 - \int \varepsilon \bar{u}_x U V_y \phi_1^2 - \int \varepsilon \bar{u}_{xy} q V_y \phi_1^2 + \frac{\varepsilon}{2} \int \bar{u}_{xyy} x V^2 \phi_1^2 \\ &\quad + \int \varepsilon \bar{u}_y V_y V_x x \phi_1^2 + E_{loc}^{(1)} =: \sum_{i = 1}^{9} D^{(5)}_i + E^{(1)}_{loc}, \end{align} $$

where $E_{loc}^{(1)}$ contains localized terms and satisfies $|E_{loc}^{(1)}| \lesssim \|U, V \|_{X_0}^2$ . We now estimate each of the remaining terms in (4.47). $D^{(5)}_1$ and $D^{(5)}_2$ are controlled by the right-hand side of (4.32) upon invoking (3.22) and upon using $\| \bar {u} + x \bar {u}_x \|_\infty \lesssim 1$ . We estimate $D^{(5)}_3$ by noting that $|\bar {u}_{xx}| + |x \bar {u}_{xxx}| \lesssim \langle x \rangle ^{-2}$ , after which it is easily controlled by $\| U, V \|_{X_0}$ .

For the fourth term, we estimate via localizing in z using the cut-off $\chi $ , defined in (1.62), via

$$ \begin{align*} D^{(5)}_4 = \int \varepsilon \bar{u}_{xyy} q V_x \phi_1^2 \chi(z) + \int \varepsilon \bar{u}_{xyy} q V_x \phi_1^2 (1- \chi(z)) =: D^{(5)}_{4, near} + D^{(5)}_{4,\mathit{far}} \end{align*} $$

First for the far-field component, we have

$$ \begin{align*} \nonumber | D^{(5)}_{4,\mathit{far}} | \lesssim \sqrt{\varepsilon} \| \bar{u}_{xyy} y x^{\frac 3 2} \|_\infty \| \sqrt{\varepsilon} \bar{u} V_x x^{\frac 1 2} \phi_1\| \| \frac{q}{y} \langle x \rangle^{-1} \| \lesssim \sqrt{\varepsilon} \| U, V \|_{X_{\frac 1 2}} \| U \langle x \rangle^{-1} \|, \end{align*} $$

which is an admissible contribution according to Hardy type inequality (3.23). For the localized component, we have

(4.48) $$ \begin{align} \nonumber | D^{(5)}_{4, near}| &= | \int \varepsilon \bar{u}_{xyy} \frac{q}{y} \frac{y}{\sqrt{x}} V_x x^{\frac 32} \phi_1^2 \chi(z)| \lesssim \int \varepsilon |\bar{u}_{xyy}| |\frac{q}{y}| \bar{u} V_x x^{\frac 32} \phi_1^2| \\ &\lesssim \sqrt{\varepsilon} \| U \langle x \rangle^{-1} \| \| \sqrt{\varepsilon} \bar{u} V_x \langle x \rangle^{\frac 1 2} \phi_1 \| \lesssim \sqrt{\varepsilon} \| U, V \|_{X_0}^2 + \sqrt{\varepsilon} \| U, V \|_{X_{\frac 12}}^2, \end{align} $$

where we have used the pointwise decay estimate $|\bar {u}_{xyy} \langle x \rangle ^2| \lesssim 1$ , according to (2.6). The fifth term, $D^{(5)}_5$ , follows by a nearly identical calculation.

For the sixth term from (4.47), $D^{(5)}_6$ , it is convenient to integrate by parts in x, which produces

$$ \begin{align*} \nonumber |\int \varepsilon \bar{u}_{xy} U V_x x \phi_1^2| & = | \int \varepsilon \bar{u}_{xxy} U V x \phi_1^2 + \int \frac{\varepsilon}{2} \bar{u}_{xyy} V^2 x \phi_1^2 + \int \varepsilon \bar{u}_{xy} UV \phi_1^2 - \int \varepsilon \bar{u}_{xy} UV x 2 \phi_1 \phi_1' | \\ \nonumber &\lesssim \sqrt{\varepsilon} (\| \bar{u}_{xxy} x^{2+2\sigma} \|_\infty + \| \bar{u}_{xy} x^{1+2\sigma} \|_\infty ) \| U \langle x \rangle^{- \frac 1 2 - \sigma} \| \| \sqrt{\varepsilon} V \langle x \rangle^{-\frac 1 2 - \sigma} \| \\ \nonumber &\quad + \| \bar{u}_{xyy} y^2 x \|_\infty \Big\| \sqrt{\varepsilon} \frac{V}{y} \phi_1 \Big\|^2 \\ \nonumber &\lesssim \sqrt{\varepsilon} ( \| U, V \|_{X_0} + \| U, V \|_{X_{\frac 1 2}} ) + C_\delta \| U, V \|_{X_0} + \delta \| U, V\|_{Y_{\frac 1 2}}^2. \end{align*} $$

We estimate $D^{(5)}_7$ via

(4.49) $$ \begin{align} |\int \varepsilon \bar{u}_{xy} q V_y \phi_1^2| \lesssim \sqrt{\varepsilon} \| y x \bar{u}_{xy} \|_\infty \| \sqrt{\varepsilon} U_x \phi_1 \| \| \frac{q}{y} \langle x \rangle^{-1} \| \lesssim \sqrt{\varepsilon} \| \sqrt{\varepsilon}U_x \phi_1\| \| U \langle x \rangle^{-1} \|, \end{align} $$

which is an admissible contribution according to (3.23).

We estimate $D^{(5)}_8$ via

(4.50) $$ \begin{align} |\int \varepsilon x \bar{u}_{xyy} V^2 \phi_1^2| \le \| x \bar{u}_{xyy} y^2 \|_\infty \Big\| \sqrt{\varepsilon} \frac{V}{y} \phi_1 \Big\|^2 \lesssim \| \sqrt{\varepsilon} V_y \phi_1\|^2, \end{align} $$

upon which we invoke (3.22).

Finally, we estimate $D^{(5)}_9$ via

(4.51) $$ \begin{align} \nonumber |\int \varepsilon \bar{u}_y V_y V_x x \phi_1^2| &\lesssim \| \bar{u}_y x^{\frac 1 2} \|_\infty \| \sqrt{\varepsilon} U_x \phi_1 \| \| \sqrt{\varepsilon} V_x x^{\frac 1 2}\phi_1 \| \\ \nonumber & \le (C_{\delta_1} \| U, V \|_{X_0} + \delta_1 \| U, V \|_{Y_{\frac 1 2}} )( C_{\delta_2} \| U, V \|_{X_{\frac 1 2}} + \delta_2 \| U, V \|_{X_1} ) \\ & \le \delta \| U, V \|_{Y_{\frac 1 2}}^2 + \delta \| U, V \|_{X_1}^2 + C_\delta \| U, V \|_{X_0}^2 + \delta \| U, V \|_{X_{\frac 1 2}}^2. \end{align} $$

We move to the third diffusive term, by which we mean

(4.52) $$ \begin{align} - \varepsilon \int v_{yy} (V_x x \phi_1^2 + V \phi_1^2 + 2 V x \phi_1 \phi_1') = - \int \varepsilon v_{yy} V_x x \phi_1^2 + \int \varepsilon v_y V_y \phi_1^2 + \int 2 \varepsilon v_y V_y x \phi_1 \phi_1'. \end{align} $$

We easily estimate the final two terms above via

(4.53) $$ \begin{align} |\int \varepsilon v_y V_y \phi_1^2 + \int 2 \varepsilon v_y V_y x \phi_1 \phi_1' | \lesssim \|U, V \|_{X_0}^2. \end{align} $$

We thus deal with the principal contribution, which gives the following integrations by parts identity

(4.54) $$ \begin{align} \nonumber - \varepsilon \int v_{yy} V_x x \phi_1^2 & = - \int \frac{\varepsilon}{2} \partial_x (x \bar{u}) V_y^2 \phi_1^2 - \int \varepsilon \bar{u}_y V_y V_x x\phi_1^2 + \frac 1 2 \int \varepsilon \partial_x (x \bar{u}_{yy}) V^2 \phi_1^2\\ \nonumber &\quad + \int \varepsilon \bar{u}_{xyy} q V_x x \phi_1^2 + \int \varepsilon \bar{u}_{xy} U V_x x \phi_1^2+ \int \varepsilon \bar{u}_x U_y V_x x \phi_1^2 \\ &\quad - \int \varepsilon x \bar{u} V_y^2 \phi_1 \phi_1' + \frac 1 2 \int \varepsilon x \bar{u}_{yy} V^2 \phi_1 \phi_1'. \end{align} $$

These terms are largely identical to those in (4.47). The only slightly different term is the sixth term of (4.54), which is estimated as

(4.55) $$ \begin{align} |\varepsilon \int \bar{u}_x U_y V_x x| \lesssim \| \frac{\bar{u}_x}{\bar{u}} x \|_\infty \| \sqrt{\bar{u}} U_y \| \| \varepsilon \sqrt{\bar{u}} V_x \| \lesssim \| U, V \|_{X_0}^2. \end{align} $$

We now move to the fourth and final diffusive term, by which we mean

(4.56) $$ \begin{align} - \varepsilon^2 \int v_{xx} V_x x \phi_1^2 - \int \varepsilon^2 v_{xx} V \phi_1^2 -2 \int \varepsilon^2 v_{xx} V x \phi_1 \phi_1', \end{align} $$

An integration by parts in x demonstrates that the final two terms above are estimated above by $\|U, V \|_{X_0}^2$ . The first term above gives

(4.57) $$ \begin{align} \nonumber - \varepsilon^2 \int v_{xx} V_x x \phi_1^2 & = \int \varepsilon^2 v_x V_{xx} x \phi_1^2 + \int \varepsilon^2 v_x V_x \phi_1^2 + 2\int \varepsilon^2 v_x V_x x \phi_1 \phi_1' \\ \nonumber & = \int \varepsilon^2 \partial_x (\bar{u} V - \bar{u}_x q) V_{xx} x \phi_1^2+ \int \varepsilon^2 \partial_x (\bar{u} V - \bar{u}_x q) V_x \phi_1^2 + 2 \int \varepsilon^2 v_x V_x x \phi_1 \phi_1' \\ & = \tilde{D}^{(6)}_1 + \tilde{D}^{(6)}_2 + \tilde{D}^{(6)}_3. \end{align} $$

The term $\tilde {D}^{(6)}_3$ is easily controlled by a factor of $\|U, V \|_{X_0}^2$ . A few integrations by parts produces for the first term, $\tilde {D}^{(6)}_1$ , the following identity

(4.58) $$ \begin{align} \nonumber \tilde{D}^{(6)}_1 & = - \frac{\varepsilon^2}{2} \int (x \bar{u})_x V_x^2 \phi_1^2- 2 \varepsilon^2 \int x \bar{u}_x V_x^2 \phi_1^2 + \varepsilon^2 \int (x \bar{u}_x)_{xx} V^2 \phi_1^2 \\ \nonumber &\quad + \int \varepsilon^2 V_x q \partial_x (x \bar{u}_{xx}) \phi_1^2+ \varepsilon^2 \int \partial_x (x \bar{u}_{xx}) V^2 \phi_1^2 - 4 \int \varepsilon^2 \bar{u}_x x V V_x \phi_1 \phi_1' \\ &\quad + 2 \int \varepsilon^2 \bar{u}_{xx} q V_x x \phi_1 \phi_1' =: \sum_{i = 1}^{7} D^{(6)}_i. \end{align} $$

We now proceed to estimate all the terms above. The first term of (4.58), $D^{(6)}_1$ , we estimate via

(4.59) $$ \begin{align} |\frac{\varepsilon^2}{2} \int (x \bar{u})_x V_x^2| \lesssim \| \partial_x (x \bar{u}) \|_\infty \varepsilon^2 \| V_x \|^2 \lesssim \| U, V \|_{X_0}^2. \end{align} $$

$D^{(6)}_2$ and $D^{(6)}_3$ are estimated in an analogous manner. For $D^{(6)}_4$ and $D^{(6)}_6$ , we invoke the Hardy type inequality (3.24) coupled with the estimate $\| \partial _x^2(x \bar {u}_x) x^2\|_\infty \lesssim 1$ . We estimate $D^{(6)}_5$ via

(4.60) $$ \begin{align} \nonumber |\int \varepsilon^2 V_x q \partial_x (x \bar{u}_{xx})| &\lesssim \varepsilon \Big\| \frac{\partial_x (x \bar{u}_{xx})}{\bar{u}} x y \Big\|_\infty \| \| \frac{q}{y} x^{-1} \| \| \varepsilon \sqrt{\bar{u}} V_x \| \lesssim \varepsilon \| U \langle x \rangle^{-1} \| \| \varepsilon \sqrt{\bar{u}} V_x \| \\ &\lesssim \varepsilon ( \| U, V \|_{X_0} + \| U, V \|_{X_{\frac 1 2}} ) \| U, V \|_{X_0}, \end{align} $$

where we have invoked the Hardy-type inequality (3.23).

For the second term from (4.57), $\tilde {D}^{(6)}_2$ , we expand and integrate by parts to generate

(4.61) $$ \begin{align} \nonumber |\int \varepsilon^2 \partial_x (\bar{u} V - \bar{u}_x q) V_x | & =| \int \varepsilon^2 \bar{u} V_x^2 - \int 2 \varepsilon^2 \bar{u}_{xx} V^2 + \frac 1 2 \int \varepsilon^2 \bar{u}_{xxxx} q^2 | \\ \nonumber &\lesssim \| \sqrt{\bar{u}} \varepsilon V_x \|^2 + \varepsilon \| \bar{u}_{xx} x^2 \|_\infty \| \sqrt{\varepsilon} V \langle x \rangle^{-1} \|^2 + \varepsilon^2 \| \bar{u}_{xxxx} y^2 x^2 \| \frac{q}{y} \langle x \rangle^{-1} \|^2 \\ &\lesssim \| U, V \|_{X_0}^2 + \varepsilon (\| U, V \|_{X_0}^2 + \| U, V \|_{X_{\frac 1 2}}^2), \end{align} $$

where we have invoked (3.23)–(3.24).

Step 4: Error Terms We now move to the remaining error terms, the first of which is the $\zeta U$ term from (2.26). For this, we estimate via

(4.62) $$ \begin{align} \nonumber \Big| \int \zeta U U_x x \phi_1^2 \Big| &\lesssim \sqrt{\varepsilon} \int \langle x \rangle^{- (1 + \frac{1}{50})} |U| |U_x| x \phi_1^2 \lesssim \sqrt{\varepsilon} \| U \langle x \rangle^{- \frac 1 2 - \frac{1}{50}} \| \| U_x x^{\frac 1 2} \phi_1 \| \\ \nonumber &\lesssim \sqrt{\varepsilon} ( \| \bar{u} U \langle x \rangle^{- \frac 1 2 - \frac{1}{200}} \| + \| \sqrt{\bar{u}} U_y \| ) ( \| \bar{u} U_x x^{\frac 1 2} \phi_1 \| + \| \sqrt{\bar{u}} U_{xy} x \phi_1 \| ) \\ &\lesssim \sqrt{\varepsilon} \| U, V \|_{X_0} (\| U, V\|_{X_{\frac 1 2}} + \| U, V \|_{X_1} ), \end{align} $$

where we have invoked estimate (2.15) for pointwise decay of $\zeta $ . The $\zeta _y q$ term from (2.26) and $\zeta V$ term from (2.27) is estimated in an identical manner.

We now address the remaining error terms in equation (2.27). The first of these is the term

$$ \begin{align*} \Big| \int \varepsilon \alpha U (V_x x + V ) \phi_1^2 \Big| \lesssim \sqrt{\varepsilon} \| U \langle x \rangle^{-\frac 3 4} \| ( \| \sqrt{\varepsilon} \bar{u} V_x x^{\frac 1 2} \phi_1 \| + \| \sqrt{\varepsilon} \bar{u} V \langle x \rangle^{- \frac 3 4} \| ) \lesssim \sqrt{\varepsilon} \| U, V \|_{X_0} \| U, V \|_{X_{\frac 1 2}}, \end{align*} $$

where we have invoked the pointwise decay estimate on $\alpha $ from (2.16). The estimate on the $\alpha _y q$ term follows in an identical manner. This concludes the proof of Lemma 4.2.

Proof. For this proof, it is convenient to work in the vorticity formulation, (2.30). We apply the multiplier $U_y x \phi _1(x)^2$ to (2.30).

Step 1: $\mathcal {T}_1$ Terms We first note that since $\mathcal {T}_1[U](x, 0) = 0$ , we may integrate by parts in y to view the product in the velocity form, and subsequently integrate by parts several times in y and x to produce

(4.65) $$ \begin{align} \nonumber \int \partial_y \mathcal{T}_1[U] U_y x \phi_1^2& = \int 2 \bar{u} \bar{u}_y U_x U_y x \phi_1^2 - \frac 1 2 \int \bar{u}^2 U_y^2 \phi_1^2 - \int \bar{u} \bar{u}_x U_y^2 x \phi_1^2 + \frac 1 2 \int (\bar{u} \bar{v})_y U_y^2 x \phi_1^2 \\ &\quad + \int 2 \bar{u}^0_{pyy} U_y^2 x - \int \partial_y^4 \bar{u}^0_p U^2 x - \int \bar{u}^2 U_y^2 x \phi_1 \phi_1' =: \sum_{i = 1}^7 A^{(1)}_i. \end{align} $$

Note that above, we used the integration by parts identity

(4.66) $$ \begin{align} - \int 2 \bar{u}^0_{pyy} U U_{yy} x \phi_1^2 & = \int 2 \bar{u}^0_{pyy} U_y^2 x \phi_1^2 - \int \bar{u}^0_{pyyyy} U^2 x \phi_1^2, \end{align} $$

which is available due to the condition that $\bar {u}^0_{pyy}|_{y = 0} = 0$ and $\bar {u}^0_{pyyy}|_{y = 0} = 0$ .

We now estimate each of the terms in (4.65), starting with $A^{(1)}_1$ , which is controlled by

(4.67) $$ \begin{align} \nonumber |\int 2\bar{u} \bar{u}_y U_x U_y x \phi_1^2| &\lesssim \| \bar{u}_y x^{\frac 1 2} \|_\infty \| \bar{u} U_x x^{\frac 1 2} \phi_1\| \| U_y \phi_1 \| \le C_{\delta_1} \| U_y \phi_1 \|^2 + \delta_1 \| U, V \|_{X_{\frac 1 2}}^2 \\ \nonumber & \le C_{\delta_1} C_{\delta_2} \| \sqrt{\bar{u}} U_y \|^2 + C_{\delta_1} \delta_2 \| U, V \|_{Y_{\frac 1 2}}^2 + \delta_1 \| U, V \|_{X_{\frac 1 2}}^2 \\ & \le C_\delta \| U, V \|_{X_0}^2 + \delta \| U, V \|_{Y_{\frac 1 2}}^2 + \delta \| U, V \|_{X_{\frac 1 2}}^2, \end{align} $$

where we have invoked (3.22).

For $A^{(1)}_2$ , $A^{(1)}_3$ , $A^{(1)}_4$ , and $A^{(1)}_5$ , we appeal to the coefficient estimate

(4.68) $$ \begin{align} \|\frac{1}{2} \bar{u}^2\|_\infty + \| \bar{u} \bar{u}_x x \|_\infty + \frac 1 2 \| x \partial_y (\bar{u} \bar{v}) \|_\infty + \| 2 \bar{u}_{yy} x \|_\infty \lesssim 1, \end{align} $$

to control these terms by $C_\delta \| U, V \|_{X_0}^2 + \delta \| U, V \|_{Y_{\frac 1 2}}^2$ .

We estimate $A^{(1)}_6$ via

(4.69) $$ \begin{align} \nonumber |\int \partial_y^4 \bar{u}^0_p U^2 x \phi_1^2| &\lesssim |\int \partial_y^4 \bar{u}^0_p \mathring{U}^2 x \phi_1^2| + |\int \partial_y^4 \bar{u}^0_p U(x, 0)^2 x \phi_1^2| \\ \nonumber &\lesssim \| \partial_y^4 \bar{u}^0_p x y^2 \|_\infty \| U_y \phi_1 \|^2 + \| \partial_y^4 \bar{u}^0_p x^{\frac 32} \|_{L^\infty_x L^1_y} \| U(x, 0) x^{-\frac 1 4} \|_{x = 0}^2 \\ & \le C_\delta \| U, V \|_{X_0}^2 + \delta \| U, V \|_{Y_{\frac 1 2}}^2. \end{align} $$

The final term in (4.65), $A^{(1)}_7$ , is localized in x, and is clearly bounded above by a factor of $\| U, V \|_{X_0}^2$ .

Step 2: $\mathcal {T}_2$ Terms: We now estimate the contributions from $\mathcal {T}_2$ via first integrating by parts in x to produce

(4.70) $$ \begin{align} - \varepsilon \int \partial_x \mathcal{T}_2[V] U_y x \phi_1^2 = \int \varepsilon \mathcal{T}_2[V] U_{xy} x \phi_1^2 + 2\int \varepsilon \mathcal{T}_2[V] U_y \phi_1 \phi_1'. \end{align} $$

We now appeal to the definition of $\mathcal {T}_2[V]$ in (2.25) to produce the identity

(4.71) $$ \begin{align} \nonumber \int \varepsilon \mathcal{T}_2[V] U_{xy} x \phi_1^2 & = \int 2 \varepsilon \bar{u} \bar{u}_y V_x V_y x \phi_1^2 - \int \frac 1 2 \varepsilon \bar{u}^2 V_y^2 \phi_1^2 - \int \varepsilon \bar{u} \bar{u}_x V_y^2 x \phi_1^2 \\ \nonumber &\quad + \frac 1 2 \int \varepsilon (\bar{u} \bar{v})_y V_y^2 x \phi_1^2 + \int \varepsilon \bar{u}_{yy} V_y^2 x \phi_1^2 - \int \frac{\varepsilon}{2} \bar{u}_{yyyy} V^2 x \phi_1^2 - \int \varepsilon \bar{u}^2 V_y^2 x \phi_1 \phi_1' \\ & = A^{(2)}_1 + ... + A^{(2)}_7. \end{align} $$

For the first term, $A^{(2)}_1$ , we estimate via

(4.72) $$ \begin{align} \nonumber | \int \varepsilon \bar{u} \bar{u}_y V_x V_y x \phi_1^2 | &\lesssim\| \bar{u}_y x^{\frac 1 2} \|_\infty \| \sqrt{\varepsilon} V_y \phi_1 \| \| \sqrt{\varepsilon} \bar{u} V_x x^{\frac 1 2} \phi_1 \| \\ & \le C_\delta \| U, V \|_{X_0}^2 + \delta \| U, V \|_{Y_{\frac 1 2}}^2 + \delta \| U, V \|_{X_{\frac 1 2}}^2. \end{align} $$

For $A^{(2)}_2, A^{(2)}_3, A^{(2)}_4, A^{(2)}_5$ , we estimate using (4.68). We estimate $A^{(2)}_6$ via

$$ \begin{align*} \nonumber |\int \varepsilon \bar{u}_{yyyy} V^2 x| \lesssim \| \bar{u}_{yyyy} xy^2 \|_\infty \| \sqrt{\varepsilon} V_y \|^2 \le C_\delta \| U, V \|_{X_0}^2 + \delta \| U, V \|_{Y_{\frac 1 2}}^2. \end{align*} $$

The final term, $A^{(2)}_7$ , can be controlled by $\|U, V \|_{X_0}^2$ upon invoking the bounded support of $\phi _1'$ .

Step 3: Diffusive Terms: We now compute (in the vorticity form) via a long series of integrations by parts the following identity

(4.73) $$ \begin{align} \nonumber - \int \partial_y^3 u U_y x \phi_1^2& = \int \bar{u} U_{yy}^2 x \phi_1^2 - \frac 9 2 \int \bar{u}_{yy} U_y^2 x \phi_1^2+ \int 3 \bar{u}_{yyyy} U^2 x \phi_1^2- \int \frac 1 2 \partial_y^6 \bar{u} q^2 x \phi_1^2 \\ &\quad + \frac 3 2 \int_{y = 0} \bar{u}_y U_y^2 x \phi_1^2 \,\mathrm{d} x + \frac 3 2 \int_{y = 0} \bar{u}_{yy} UU_y x \phi_1^2 = \sum_{i = 1}^6 B^{(1)}_i. \end{align} $$

We first notice that $B^{(1)}_1$ and $B^{(1)}_5$ are positive contributions. $B^{(1)}_2$ is easily estimated by $\| U, V \|_{X_0}^2$ upon using $\|\frac {\bar {u}_{yy}}{\bar {u}}x \|_\infty \lesssim 1$ . We estimate $B^{(1)}_3$ by

(4.74) $$ \begin{align} \nonumber |\int 3 \bar{u}_{yyyy} U^2 x \phi_1^2| &\lesssim |\int \bar{u}_{yyyy} (U - U(x, 0))^2 x \phi_1| + |\int \bar{u}_{yyyy} U(x, 0)^2 x| \\ \nonumber &\lesssim \| \bar{u}_{yyyy} x y^2 \|_\infty \Big\| \frac{U - U(x, 0)}{y} \phi_1 \Big\|^2 + \| \bar{u}_{yyyy} x^{\frac 3 2} \|_{L^\infty_x L^1_y} \| U(x, 0) x^{-\frac 1 4} \|_{y = 0}^2 \\ &\lesssim \| U_y \|^2 + \| U(x, 0) x^{-\frac 1 4} \|_{y = 0}^2 \le C_\delta \|U, V \|_{X_0}^2 + \delta \|U, V \|_{Y_{\frac 1 2}}^2. \end{align} $$

The term $B^{(1)}_4$ is estimated in an entirely analogous manner. The term $B^{(1)}_6$ is estimated by

(4.75) $$ \begin{align} |\int_{y = 0} \bar{u}_{yy} UU_y x \phi_1^2| \lesssim \varepsilon^{\frac 1 2} \| U \langle x \rangle^{- \frac 1 4} \|_{y = 0} \| U_y \phi_1 \langle x \rangle^{\frac 1 4} \|_{y = 0} \lesssim \varepsilon^{\frac 1 2} \| U, V\|_{X_0} \|U, V \|_{Y_{\frac 1 2}}, \end{align} $$

upon invoking the bound $|\bar {u}_{yy}(x, 0)| \lesssim \varepsilon ^{\frac 1 2} \langle x \rangle ^{-1}$ due to (2.6) coupled with the fact that $\bar {u}^0_{pyy}(x, 0) = 0$ .

In addition to this term, we need to estimate the term $\bar {u}^0_{p yyy}q$ from (2.26), we do so via

$$ \begin{align*} \nonumber \int \partial_y (\bar{u}^0_{pyyy} q) U_y x = \int \bar{u}^0_{p yyyy} q U_y x + \int \bar{u}^0_{p yyy} UU_y x = \frac 1 2 \int \partial_y^6 \bar{u}^0_p q^2 x - \frac 1 2 \int \partial_y^4 \bar{u}^0_p U^2 x, \end{align*} $$

which we estimate in an identical manner to (4.74).

The next diffusive term is

(4.76) $$ \begin{align} - 2 \int \varepsilon u_{xxy} U_y x \phi_1^2& = \int 2 \varepsilon \partial_{xy} (\bar{u} U + \bar{u}_y q) U_{xy} x \phi_1^2 + \int 2 \varepsilon \partial_x (\bar{u} U + \bar{u}_y q) U_y \phi_1^2\\ \nonumber &\quad + \int 4 \varepsilon u_{xy} U_y x \phi_1 \phi_1' =: \sum_{i = 1}^{5} \tilde{B}^{(2)}_i. \end{align} $$

Due to the localization in x of $\phi _1'$ that the term $\tilde {B}^{(2)}_5$ above is estimated by

(4.77) $$ \begin{align} |\int \varepsilon u_{xy} U_y x \phi_1 \phi_1'| \lesssim \| \varepsilon u_{xy} \phi_1' \| \| U_y \phi_1 \| \lesssim \|U, V \|_E (C_\delta \|U, V \|_{X_0} + \delta \|U, V \|_{Y_{\frac 1 2}}). \end{align} $$

Due to the length of the forthcoming expressions, we handle each of the remaining four terms in (4.76), $\tilde {B}^{(2)}_k$ , $k = 1,2,3,4$ , individually. First, integration by parts yields for $\tilde {B}^{(2)}_1$ ,

(4.78) $$ \begin{align} \nonumber \int 2 \varepsilon \partial_{xy}(\bar{u} U) U_{xy} x \phi_1^2 & = \int 2 \varepsilon \bar{u} U_{xy}^2 x\phi_1^2 - \int 4 \varepsilon \bar{u}_{xxy} UU_y x \phi_1^2 - \int 4 \varepsilon \bar{u}_{xy} U_x U_y x \phi_1^2- \int 4 \varepsilon \bar{u}_{xy} U U_y \phi_1^2\\ \nonumber &\quad - \int \varepsilon \partial_x (x \bar{u}_x) U_y^2 \phi_1^2 - \int 4 \varepsilon \bar{u}_{yy} U_x^2 x \phi_1^2- \int_{y = 0} 2 \varepsilon \bar{u}_y U_x(x, 0)^2 x \phi_1^2 \,\mathrm{d} x \\ &\quad + \int \varepsilon \bar{u}_{yyyy} V^2 x \phi_1^2+ \int 2 \varepsilon \bar{u}_{xyy} q U_{xy}x \phi_1^2 + E_{loc}^{(2)} =: \sum_{i = 1}^{9} B^{(2)}_i + E_{loc}^{(2)}, \end{align} $$

where $E_{loc}^{(2)}$ are localized contributions that can be controlled by a large factor of $\| U, V \|_{X_0}^2 + \|U, V \|_E^2$ .

The first term, $B^{(2)}_1$ , is a positive contribution. The terms $B^{(2)}_2$ and $B^{(2)}_4$ are estimated via

(4.79) $$ \begin{align} \nonumber | \int 4 \varepsilon \bar{u}_{xxy} UU_y x \phi_1^2| + |\int 4 \varepsilon \bar{u}_{xy} UU_y \phi_1^2| &\lesssim \varepsilon \Big( \| \bar{u}_{xxy} x^2 \|_\infty + \| \bar{u}_{xy} x \|_\infty \Big) \| U \langle x \rangle^{-1} \phi_1 \| \| U_y \phi_1 \| \\ &\lesssim \varepsilon (\| U, V \|_{X_0}^2 + \| U, V \|_{Y_{\frac 1 2}}^2 + \| U, V \|_{X_{\frac 1 2}}^2), \end{align} $$

where we have appealed to (3.22) and (3.23).

The terms $B^{(2)}_3$ , $B^{(2)}_5$ , $B^{(2)}_6$ , and $B^{(2)}_8$ are estimated via

$$ \begin{align*} \nonumber &|\int 4 \varepsilon \bar{u}_{xy} U_x U_y x \phi_1^2| \lesssim \sqrt{\varepsilon} \| \bar{u}_{xy} x \|_\infty \| \sqrt{\varepsilon} U_x \phi_1 \| \| U_y \phi_1 \| \lesssim \sqrt{\varepsilon} \Big( \| U, V \|_{X_0}^2 + \| U, V \|_{Y_{\frac 1 2}}^2 \Big), \\ \nonumber &|\int \varepsilon \partial_x (x \bar{u}_x) U_y^2 \phi_1^2| \lesssim \| \partial_x (x \bar{u}_x) \|_\infty \varepsilon \| U_y \phi_1 \|^2 \lesssim \varepsilon \Big( \| U, V \|_{X_0}^2 + \| U, V \|_{Y_{\frac 1 2}}^2 \Big) \\\nonumber &|\int 4 \varepsilon \bar{u}_{yy} U_x^2 x \phi_1^2 | \lesssim \| \bar{u}_{yy} x \|_\infty \| \sqrt{\varepsilon} U_x \phi_1 \|^2 \le C_\delta \| U, V \|_{X_0}^2 + \delta \| U, V \|_{Y_{\frac 1 2}}^2, \\ \nonumber &|\int \varepsilon \bar{u}_{yyyy} V^2 x \phi_1^2| \lesssim \varepsilon \| \bar{u}_{yyyy} x y^2 \|_\infty \Big\| \frac{V}{y} \phi_1 \Big\|^2 \le C_\delta \| U, V \|_{X_0}^2 + \delta \| U, V \|_{Y_{\frac 1 2}}^2, \end{align*} $$

and $B^{(2)}_9$ is estimated via

(4.80) $$ \begin{align} \nonumber | \int\varepsilon \bar{u}_{xyy} q U_{xy} x \phi_1^2| &\lesssim\sqrt{\varepsilon} \| \bar{u}_{xyy} x^{\frac 3 2} y \|_\infty \Big\| \frac{q}{y} \langle x \rangle^{-1} \phi_1 \Big\| \| \sqrt{\varepsilon} \sqrt{\bar{u}} U_{xy} x^{\frac 1 2} \phi_1 \| \\ \nonumber &\lesssim \sqrt{\varepsilon} \| U \langle x \rangle^{-1} \| \| \sqrt{\varepsilon} \sqrt{\bar{u}} U_{xy} x^{\frac 1 2} \phi_1 \| \\ &\lesssim \sqrt{\varepsilon} ( \| U, V \|_{X_0} + \| U, V \|_{X_{\frac 1 2}} ) \| U, V \|_{Y_{\frac 1 2}}. \end{align} $$

The term $B^{(2)}_7$ requires us to use the $X_1$ norm, albeit with a prefactor of $\varepsilon $ and with a weaker weight in x:

(4.81) $$ \begin{align} |\int_{y = 0} \varepsilon \bar{u}_y U_x(x, 0)^2 x \phi_1^2 \,\mathrm{d} x| \lesssim \varepsilon \| \sqrt{\bar{u}_y} U_x(x, 0) x \phi_1 \|_{x = 0}^2 \lesssim \varepsilon \| U, V \|_{X_1}^2, \end{align} $$

where we use that the choice of $\phi _1$ is the same as that of $\| \cdot \|_{X_1}$ . This concludes treatment of $\tilde {B}^{(2)}_1$ .

The second term from (4.76), $\tilde {B}^{(2)}_2$ , gives

(4.82) $$ \begin{align} \nonumber \int 2 \varepsilon \partial_{xy}( \bar{u}_y q) U_{xy} x \phi_1^2 & = \int 2 \varepsilon \bar{u}_{xyy} q U_{xy} x \phi_1^2 - \int 2 \varepsilon \bar{u}_{xy} U_x U_y x \phi_1^2 - \int 2 \varepsilon \bar{u}_{xy} UU_y \phi_1^2 \\ \nonumber &\quad - \int 2 \varepsilon \bar{u}_{xxy} x UU_y \phi_1^2 - \int 2 \varepsilon \bar{u}_{yy} V_y^2 x \phi_1^2 + \int \varepsilon \bar{u}_{yyyy} V^2 x \phi_1^2\\ &\quad - \int_{y = 0} \varepsilon \bar{u}_y U_x(x, 0)^2 x \phi_1^2 \,\mathrm{d} x - 4 \int \varepsilon \bar{u}_{xy} U U_y x \phi_1 \phi_1' = : \sum_{i = 1}^8 J_i. \end{align} $$

The final term above, $J_8$ , is localized in x contribution, can easily be controlled by $\| U, V \|_{X_0}^2$ . $J_1$ is treated in the same manner as (4.80). $J_2, J_3$ , and $J_4$ are estimated by

$$ \begin{align*} \nonumber &|\int \varepsilon \bar{u}_{xy} U_x U_y x \phi_1^2| \lesssim \sqrt{\varepsilon} \| \bar{u}_{xy} x \|_\infty \| \sqrt{\varepsilon} U_x \phi_1 \| \| U_y \phi_1 \| \lesssim \sqrt{\varepsilon} (\| U, V \|_{X_0}^2 + \| U, V \|_{Y_{\frac 1 2}}^2), \\ \nonumber &|\int \varepsilon (x \bar{u}_{xy})_x UU_y \phi_1| \lesssim \varepsilon \| \bar{u}_{xy} x \|_\infty \| U \langle x \rangle^{-1} \phi_1 \| \| U_y \phi_1 \| \lesssim \varepsilon (\| U, V \|_{X_0}^2 + \| U, V \|_{X_{\frac 1 2}}^2+ \| U, V \|_{Y_{\frac 1 2}}^2). \end{align*} $$

Terms $J_5$ , $J_6$ , $J_7$ are identical to $B^{(2)}_6$ , $B^{(2)}_8$ , and $B^{(2)}_7$ . This concludes treatment of $\tilde {B}^{(2)}_2$ .

The third and fourth terms from (4.76), $\tilde {B}^{(2)}_3$ and $\tilde {B}^{(2)}_4$ together give the identity

(4.83) $$ \begin{align} \int 2\varepsilon \partial_x (\bar{u} U + \bar{u}_y q) U_y \phi_1^2 = \int 2 \varepsilon \bar{u} U_x U_y \phi_1^2+ \int 2 \varepsilon \bar{u}_{xy} q U_y \phi_1^2+ \int 2 \varepsilon \bar{u}_{yy} UV \phi_1^2 + \int 2 \varepsilon \bar{u}_y U^2 \phi_1 \phi_1'. \end{align} $$

The first and final terms from (4.83) can easily be estimated by $\sqrt {\varepsilon } \| U, V \|_{X_0}^2$ , while the second term (4.83)

$$ \begin{align*} \nonumber |\int 2 \varepsilon \bar{u}_{xy} q U_y \phi_1^2| \lesssim \varepsilon \| \bar{u}_{xy} y x \|_\infty \| \frac{q}{y} \langle x \rangle^{-1} \phi_1 \| \| U_y \phi_1 \| \lesssim \varepsilon \| U \langle x \rangle^{-1} \| \| U_y \phi_1 \| \end{align*} $$

and the third term from (4.83) can be estimated via

(4.84) $$ \begin{align} |\int 2 \varepsilon \bar{u}_{yy} UV \phi_1^2| &\lesssim \varepsilon | \int \bar{u}_{yy} \mathring{U} V \phi_1^2| + \varepsilon |\int \bar{u}_{yy} U(x, 0) V \phi_1^2| \\ \nonumber &\lesssim \sqrt{\varepsilon} \| \bar{u}_{yy} y^2 \|_\infty \| U_y \phi_1 \| \| \sqrt{\varepsilon} V_y \phi_1 \| + \sqrt{\varepsilon} \| \bar{u}_{yy} y x^{- \frac 1 4} \|_{L^\infty_x L^1_y} \| U(x, 0) x^{\frac 1 4} \|_{L^2(x = 0)} \| \sqrt{\varepsilon} V_y \phi_1 \|. \end{align} $$

We now arrive at the final diffusive term, which we integrate by parts in x via

(4.85) $$ \begin{align} \nonumber \int \varepsilon^2 v_{xxx} U_y x \phi_1^2 & = - \varepsilon^2 \int v_{xx} U_{xy} x \phi_1^2- \int \varepsilon^2 v_{xx} U_y \phi_1^2 - 2 \varepsilon^2 \int v_{xx} U_y x \phi_1 \phi_1' \\ \nonumber & = \int \varepsilon^2 v_{xx} V_{yy} x \phi_1^2 - \int \varepsilon^2 v_x V_{yy} \phi_1^2 + 2 \int \varepsilon^2 v_x U_y \phi_1 \phi_1' - 2 \varepsilon^2 \int v_{xx} U_y x \phi_1 \phi_1' \\ & = \tilde{P}_1 + \tilde{P}_2 + \tilde{P}_3 + \tilde{P}_4. \end{align} $$

Again, the terms with $\phi _1'$ above, $\tilde {P}_3, \tilde {P}_4$ , are easily estimated above by a factor of $\| U, V \|_{X_0}^2 + \|U, V \|_E^2$ due to the localization in x, in an analogous manner to (4.77).

For the second term on the right-hand side of (4.85), $\tilde {P}_2$ , we produce the following identity

(4.86) $$ \begin{align} \nonumber - \int \varepsilon^2 v_x V_{yy} \phi_1^2& = \int \varepsilon^2 \bar{u}_y V_x V_y \phi_1^2- \frac 3 2 \int \varepsilon^2 \bar{u}_x V_y^2\phi_1^2 - \int \varepsilon^2 \bar{u}_{xyy} V^2 \phi_1^2- \frac{\varepsilon^2}{2} \int \bar{u}_{xxx} U^2 \phi_1^2\\ \nonumber &\quad + \int \varepsilon^2 \bar{u}_{xxy} UV \phi_1^2 + \frac{\varepsilon^2}{2} \int \bar{u}_{xxxyy} q^2 \phi_1^2- \int \varepsilon^2 \bar{u} V_y^2 \phi_1 \phi_1' + \int \varepsilon^2 \bar{u}_{xxyy} q^2 \phi_1 \phi_1' \\ &\quad - \int \varepsilon^2 \bar{u}_{xx} U^2 \phi_1 \phi_1' = \sum_{i = 1}^{9} P^{(1)}_i. \end{align} $$

Again, the terms with a $\phi _1'$ , $P^{(1)}_7, P^{(1)}_8, P^{(1)}_9$ , are easily controlled by a factor of $\| U, V \|_{X_0}^2$ .

We now proceed to estimate each of the remaining terms above via

$$ \begin{align*} &|\int \varepsilon^2 \bar{u}_y V_x V_y \phi_1^2| \lesssim \sqrt{\varepsilon} \| \varepsilon V_x \phi_1\| \| \sqrt{\varepsilon} V_y \phi_1 \| \lesssim \sqrt{\varepsilon} (\| U, V \|_{X_0}^2 + \| U, V \|_{Y_{\frac 1 2}}^2), \\ &|\int \varepsilon^2 \bar{u}_x V_y^2 \phi_1^2| \lesssim \varepsilon \| \sqrt{\varepsilon} V_y \phi_1 \|^2 \lesssim \varepsilon (\| U, V \|_{X_0}^2 + \| U, V \|_{Y_{\frac 1 2}}^2), \\ &|\int \varepsilon^2 \bar{u}_{xyy} V^2 \phi_1^2| \lesssim \varepsilon \| \bar{u}_{xyy} y^2 \|_\infty \Big\| \sqrt{\varepsilon} \frac{V}{y} \phi_1 \Big\|^2 \lesssim \varepsilon \| \sqrt{\varepsilon} V_y \phi_1\|^2 \lesssim \varepsilon (\| U, V \|_{X_0}^2 + \| U, V \|_{Y_{\frac 1 2}}^2), \\ &|\int \varepsilon^2 \bar{u}_{xxx} U^2 \phi_1^2| \lesssim \varepsilon^2 \| \bar{u}_{xxx} x^2 \|_\infty \| U \langle x \rangle^{-1} \|^2 \lesssim \varepsilon^2 (\| U, V \|_{X_0}^2 + \| U, V \|_{X_{\frac 1 2}}^2), \\ &|\int \varepsilon^2 \bar{u}_{xxy} UV \phi_1^2| \lesssim \varepsilon^{\frac 32} \| U \langle x \rangle^{-1} \| \| \sqrt{\varepsilon} V_y \phi_1\| \lesssim \varepsilon^{\frac 3 2} (\| U, V \|_{X_0}^2 + \| U, V \|_{X_{\frac 1 2}}^2 + \| U, V \|_{Y_{\frac 1 2}}^2), \\ &| \frac{\varepsilon^2}{2} \int \bar{u}_{xxxyy} q^2 \phi_1^2| \lesssim \varepsilon^2 \| \bar{u}_{xxxyy} x^2 y^2 \|_\infty \| U \langle x \rangle^{-1} \|^2 \lesssim \varepsilon^2 (\| U, V \|_{X_0}^2 + \| U, V \|_{X_{\frac 1 2}}^2). \end{align*} $$

This concludes the treatment of $\tilde {P}_2$ .

We now treat $\tilde {P}_1$ . We further integrate by parts using that $v = \bar {u} V - \bar {u}_x q$ , which produces the following identity

(4.87) $$ \begin{align} \tilde{P}_1 = \int \varepsilon^2 v_{xx} V_{yy} x \phi_1^2& = \int \varepsilon^2 \partial_{xx} (\bar{u} V - \bar{u}_x q) V_{yy} x \phi_1^2\\ \nonumber & = \int \varepsilon^2 (\bar{u} V_{xx} + 3 \bar{u}_x V_x + 3 \bar{u}_{xx} V - \bar{u}_{xxx} q) V_{yy} x \phi_1^2 =: \tilde{P}_{1,1} + \tilde{P}_{1,2} + \tilde{P}_{1,3} + \tilde{P}_{1,4}. \end{align} $$

For $\tilde {P}_{1,1}$ , we integrate by parts several times in x and y to produce the identity

(4.88) $$ \begin{align} \nonumber \int \varepsilon^2 \bar{u} V_{xx} V_{yy} x \phi_1^2 & = \int \varepsilon^2 \bar{u} V_{xy}^2 x \phi_1^2 - \varepsilon^2 \frac 1 2 \int \partial_x (x \bar{u}_x) V_y^2 \phi_1^2 - \frac 1 2 \int \varepsilon^2 \bar{u}_x V_y^2 \phi_1^2 \\ \nonumber &\quad + \int \varepsilon^2 \bar{u}_y V_x V_y \phi_1^2+ \int \varepsilon^2 \bar{u}_{xy} V_x V_y x\phi_1^2 - \int \frac{\varepsilon^2}{2} \bar{u}_{yy} V_x^2 x \phi_1^2 \\ \nonumber &\quad - \int \varepsilon^2 \bar{u} V_y^2 \phi_1 \phi_1' - \int \varepsilon^2 x \bar{u}_x V_y^2 \phi_1 \phi_1' + 2\int \varepsilon^2 \bar{u} V_{xy} V_y x \phi_1 \phi_1' \\ &\quad + 2 \int \varepsilon^2 \bar{u}_y V_x V_y x \phi_1 \phi_1' =: \sum_{i = 1}^{10} H^{(1)}_i. \end{align} $$

All of the terms with $\phi _1'$ can again be controlled by a factor of $\| U, V \|_{X_0}^2 + \|U, V \|_E^2$ . The first term, $H^{(1)}_1$ , is a positive contribution. $H^{(1)}_2$ and $H^{(1)}_3$ are easily estimated above by $ \varepsilon (\| U, V \|_{X_0}^2 + \| U, V \|_{Y_{\frac 1 2}}^2)$ , while the $H^{(1)}_4, H^{(1)}_5$ and $H^{(1)}_6$ are estimated via

$$ \begin{align*} &| \int \varepsilon^2 \partial_x (x \bar{u}_y) V_x V_y\phi_1^2 | \lesssim \sqrt{\varepsilon} \| \partial_x (x \bar{u}_y) \|_\infty \| \varepsilon V_x \| \| \sqrt{\varepsilon} V_y \| \lesssim \sqrt{\varepsilon} (\| U, V \|_{X_0}^2 + \| U, V \|_{Y_{\frac 1 2}}^2), \\ &|\int \varepsilon^2 \bar{u}_{yy} V_x^2 x\phi_1^2| \lesssim \| \bar{u}_{yy} x \|_\infty \| \varepsilon V_x \|^2 \le \delta \| U, V \|_{Y_{\frac 1 2}}^2 + \| U, V \|_{X_0}^2. \end{align*} $$

This concludes the treatment of $\tilde {P}_{1,1}$ .

The terms $\tilde {P}_{1,k}$ , $k = 2, 3, 4$ , are equivalent to

(4.89) $$ \begin{align} \nonumber &\varepsilon^2 \int (3 \bar{u}_x V_x + 3 \bar{u}_{xx} V - \bar{u}_{xxx} q) V_{yy} x \phi_1^2\\ \nonumber & = - \int 3 \varepsilon^2 \bar{u}_{xy} V_x V_y x \phi_1^2+ \frac 3 2 \int \varepsilon^2 \partial_x (x\bar{u}_{x}) V_y^2 x \phi_1^2 - \int 3 \varepsilon^2 \bar{u}_{xx} V_y^2 x \phi_1^2 \\ &\quad + \frac 3 2 \int \varepsilon^2 \bar{u}_{xxyy} V^2 x \phi_1^2 + \int \varepsilon^2 \bar{u}_{xxxy} q V_y x + \int \varepsilon^2 \bar{u}_{xxx} U V_y x \phi_1^2. \end{align} $$

We estimate each of these contributions in a nearly identical fashion to the terms from (4.88), and so omit repeating these details.

Step 4: Error Terms We estimate the terms on the right-hand side of (2.30), starting with

(4.90) $$ \begin{align} \int \partial_y(\zeta U) U_y x \phi_1^2 & = \int \zeta U_y^2 x \phi_1^2 - \frac 1 2 \int \partial_y^2 \zeta U^2 x - \frac 1 2 \int_{y = 0} \partial_y \zeta U^2 x \phi_1^2. \end{align} $$

The first term above is estimated via

$$ \begin{align*} \nonumber |\int \zeta U_y^2 x \phi_1^2| \lesssim \sqrt{\varepsilon} \| U_y \phi_1 \|^2 \lesssim \sqrt{\varepsilon} (\| U, V \|_{X_0}^2 + \| U, V \|_{Y_{\frac 1 2}}^2), \end{align*} $$

where we have appealed to the estimate (2.15) as well as the Hardy type inequality (3.22).

For the second and third terms from (4.90), we estimate via

(4.91) $$ \begin{align} |\int \partial_y^2 \zeta U^2 x| + |\int_{y = 0} \partial_y \zeta U^2 x \phi_1^2| \lesssim \sqrt{\varepsilon} \| U \langle x \rangle^{- \frac 1 2 - \frac{1}{100}} \|^2 + \sqrt{\varepsilon} \| U \langle x \rangle^{- \frac 1 2} \|_{y = 0}^2 \lesssim \sqrt{\varepsilon} \| U, V \|_{X_0}^2, \end{align} $$

where we have appealed to (2.15).

The $(\zeta _y q)_y$ and $(\zeta V)_x$ terms on the right-hand side of (2.30) are estimated in a completely analogous manner. We now estimate the term

(4.92) $$ \begin{align} \nonumber | \int \varepsilon (\alpha U)_x U_y x \phi_1^2 | & \le |\int \varepsilon \alpha U_x U_y x \phi_1^2| + |\int \varepsilon \alpha_x U U_y x \phi_1^2| \\ \nonumber &\lesssim \sqrt{\varepsilon} \| \sqrt{\varepsilon} U_x \phi_1 \| \| U_y \phi_1 \| + \varepsilon \| U \langle x \rangle^{-1} \| \| U_y \phi_1 \| \\ &\lesssim \sqrt{\varepsilon} (\| U, V \|_{X_0}^2 + \| U, V \|_{Y_{\frac 1 2}}^2) + \varepsilon \| U, V \|_{X_0}^2, \end{align} $$

where we have appealed to estimate (2.16) to estimate the coefficient $\alpha $ . The remaining term with $(\alpha _y q)_x$ is estimated in a completely analogous manner. This concludes the proof of Lemma 4.3.

Proof. We start with the estimation of $\mathcal {C}_1^n$ , defined in (2.34), which we do via

(4.94) $$ \begin{align} \nonumber \| \mathcal{C}_1^n \langle x \rangle^{n + \frac 1 2} \phi_n \| &\lesssim \sum_{k = 0}^{n-1} ( \| \partial_x^{n-k} \zeta \langle x \rangle^{(n-k) + 1.01} \|_\infty \| U^{(k)} \langle x \rangle^{k - \frac 1 2 - .01} \| \\ \nonumber &\quad + \| \partial_x^{n-k} \partial_y \zeta \langle x \rangle^{(n-k) + 1.01} y \|_\infty \| \frac{q^{(k)}}{y} \langle x \rangle^{k - \frac 1 2 - .01} \| \\ \nonumber &\quad + \| \frac{\partial_x^{n-k}(\bar{u}^2)}{\bar{u}} \langle x \rangle^{n-k} \|_{\infty} \| \bar{u} U^{(k)}_x \langle x \rangle^{k + \frac12} \phi_n \| \\ \nonumber &\quad + \| \frac{\partial_x^{n-k}(\bar{u} \bar{v})}{\sqrt{\bar{u}}} x^{n-k + \frac12} \|_{\infty} \| \sqrt{\bar{u}} U^{(k)}_y x^k \phi_n \| )\\ &\quad + \sum_{k = 1}^{n-1} \| \frac{\partial_x^{n-k} \bar{u}^0_{pyy}}{\bar{u}} \langle x \rangle^{n-k + 1} \|_\infty \| \bar{u} U^{(k-1)}_x \langle x \rangle^{k - \frac12} \phi_n \| + \| \partial_x^{n} \bar{u}^0_{pyy} U \langle x \rangle^{n + \frac12} \phi_n \| \end{align} $$

(4.95) $$ \begin{align} &\lesssim\|U, V \|_{\mathcal{X}_{\le n-\frac12}}, \end{align} $$

where we have appealed to estimate (2.15) for the coefficient of $\zeta $ . For the final term on the right-hand side of (4.94), we have proceeded in an identical manner to estimate (4.38).

We now address the terms in $\mathcal {C}_2^{n}$ via

(4.96) $$ \begin{align} \nonumber \| \sqrt{\varepsilon} \mathcal{C}_2^n \langle x \rangle^{n + \frac 1 2} \phi_n \| &\lesssim \sum_{k = 0}^{n-1} \sqrt{\varepsilon} \| \partial_x^{n-k} \alpha \langle x \rangle^{(n-k) + \frac 3 2} \|_\infty \| U^{(k)} \langle x \rangle^{k-1} \phi_n \| \\ \nonumber &\quad + \sqrt{\varepsilon} \| \partial_x^{n-k} \alpha_y \langle x \rangle^{(n-k) + \frac 3 2} y \|_\infty \| \frac{ q^{(k)} }{y} \langle x \rangle^{k-1} \phi_n \| \\ \nonumber &\quad + \| \partial_x^{n-k} \zeta \langle x \rangle^{(n-k) + 1.01} \|_\infty \| \sqrt{\varepsilon} V^{(k)} \langle x \rangle^{k - \frac 1 2 - .01} \| \\ \nonumber &\quad + \| \frac{\partial_x^{n-k}(\bar{u}^2)}{\bar{u}} \langle x \rangle^{n-k} \|_\infty \| \sqrt{\varepsilon} \bar{u} V^{(k)}_x \langle x \rangle^{k + \frac12} \phi_n \| \\ \nonumber &\quad + \| \frac{\partial_x^{n-k}(\bar{u} \bar{v})}{\sqrt{\bar{u}}} \langle x \rangle^{n-k + \frac12} \|_\infty \| \sqrt{\varepsilon} \sqrt{\bar{u}} V^{(k)}_y \langle x \rangle^{k} \phi_n \| \\ &\quad + \| \partial_x^{n-k} \bar{u}^0_{pyy} \langle x \rangle^{n - k + \frac12}y \|_\infty \| \sqrt{\varepsilon} V^{(k)}_y \langle x \rangle^{k} \phi_n \| \lesssim \|U, V \|_{\mathcal{X}_{\le n-\frac12}}, \end{align} $$

where we have appealed to estimate (2.16) to estimate the coefficient $\alpha $ . The higher order y derivative works in an identical manner.

Proof. We apply the multiplier

(4.100) $$ \begin{align} [U^{(n)} \langle x \rangle^{2n} \phi_{n}^2, \varepsilon V^{(n)} \langle x \rangle^{2n} \phi_n^2 + 2n \varepsilon V^{(n-1)} \langle x \rangle^{2n-1} \phi_n^2+ 2 \varepsilon V^{(n-1)} \langle x \rangle^{2n} \phi_n \phi_n'] \end{align} $$

to the system (2.31)–(2.33). The interaction of the multipliers (4.100) with the left-hand side of (2.31)–(2.32) is essentially identical to that of Lemma 4.1. As such, we treat the new commutators arising from the $\mathcal {C}_1^n, \mathcal {C}_2^n$ terms, defined in (2.34)–(2.35). First, we have

$$ \begin{align*} \nonumber |\int \mathcal{C}_1^{n} U^{(n)} \langle x \rangle^{2n} \phi_n^2| \lesssim \| \mathcal{C}_1^{n} \langle x \rangle^{n + \frac 1 2} \phi_n \| \| U^{(n)} \langle x \rangle^{n - \frac 1 2} \phi_n \| \lesssim \|U, V \|_{\mathcal{X}_{\le n - \frac12}}^2. \end{align*} $$

Next,

$$ \begin{align*} \nonumber |\int \varepsilon \mathcal{C}_2^{n} V^{(n)} \langle x \rangle^{2n} \phi_n^2| \lesssim \| \sqrt{\varepsilon} \mathcal{C}_2^{n} \langle x \rangle^{n + \frac 1 2} \phi_n \| \| \sqrt{\varepsilon} V^{(n)} \langle x \rangle^{n - \frac 1 2} \phi_n \| \lesssim \|U, V \|_{\mathcal{X}_{\le n - \frac12}}^2. \end{align*} $$

The identical estimate works as well for the middle term from the multiplier in (4.100), whereas the final term with $\phi _n'$ is localized in x, lower order, and trivially bounded by $\|U, V\|_{\mathcal {X}_{\le n - \frac 1 2}}^2$ .

Proof. We apply the multiplier

(4.104) $$ \begin{align} [U^{(n)}_x \langle x \rangle^{1+2n} \phi_{n+1}^2, \varepsilon V^{(n)}_x \langle x \rangle^{1+2n} \phi_{n+1}^2 + \varepsilon (1 + 2n) V^{(n)} \langle x \rangle^{2n} \phi_{n+1}^2 + 2\varepsilon V^{(n)} \langle x\rangle^{1+2n} \phi_{n+1} \phi_{n+1}'] \end{align} $$

to the system (2.31)–(2.33). Again, the interaction of these multipliers with the left-hand side of (2.31)–(2.33) is nearly identical to that of Lemma 4.2, and so we proceed to treat the commutators arising from $\mathcal {C}_1^{n}, \mathcal {C}_2^{n}$ . We also may clearly estimate the contribution of the $\phi _{n+1}'$ term by a factor of $\|U, V \|_{\mathcal {X}_{\le n}}$ . We have

(4.105) $$ \begin{align} \nonumber |\int \mathcal{C}_1^{n} U^{(n)}_x \langle x \rangle^{1+2n} \phi_{n+1}^2| &\lesssim \| \mathcal{C}_1^{n} \langle x \rangle^{n + \frac 1 2} \phi_{n+1} \| \| U_x^{(n)} \langle x \rangle^{n + \frac 1 2} \phi_{n+1} \| \\ &\lesssim \|U \|_{\mathcal{X}_{\le n-\frac12}} ( \|U \|_{X_{n+\frac 1 2}} + \|U, V \|_{X_{n+1}} ) \end{align} $$

and similarly

(4.106) $$ \begin{align} \nonumber &|\int \mathcal{C}_2^{n} (\varepsilon V^{(n)}_x \langle x \rangle^{1+2n} \phi_{n+1}^2 + \varepsilon (1 + 2n) V^{(n)} \langle x \rangle^{2n} \phi_{n+1}^2 )| \\ \nonumber &\lesssim \| \sqrt{\varepsilon} C_2^{n} \langle x \rangle^{n + \frac 12} \phi_{n+1} \| ( \| \sqrt{\varepsilon} V^{(n)}_x \langle x \rangle^{n + \frac 1 2} \phi_{n+1} \| + \| \sqrt{\varepsilon} V^{(n)} \langle x \rangle^{n - \frac 12} \phi_{n+1} \| ) \\ &\lesssim \|U \|_{\mathcal{X}_{\le n-\frac12}} ( \|U \|_{X_{n+\frac 1 2}} + \|U, V \|_{X_{n+1}} ), \end{align} $$

where above we have invoked estimate (4.93). From here, the conclusion of the lemma follows from an application of Young’s inequality for products.

Proof. We again only need to estimate the commutator terms, which are

(4.110) $$ \begin{align} \nonumber |\int ( \partial_y \mathcal{C}_1^n - \varepsilon \partial_x \mathcal{C}_2^n) U^{(n)}_y\langle x \rangle^{1 + 2n} \phi_{n+1}^2| &\lesssim\| ( \partial_y \mathcal{C}_1^n - \varepsilon \partial_x \mathcal{C}_2^n) \langle x \rangle^{n + 1} \phi_{n+1} \| U^{(n)}_y \langle x \rangle^{n } \phi_{n+1} \| \\ &\lesssim \|U, V \|_{\mathcal{X}_{\le n }} \|U, V \|_{\mathcal{X}_{\le n + \frac 1 2}}, \end{align} $$

with the help again of estimate (4.93). The conclusion again follows from an application of Young’s inequality for products.

Proof. We use the formulas (5.7) to write

(5.23) $$ \begin{align} \mu_s Q = \partial_x^{11} \psi = \partial_x^{11} (\bar{u} q) = \bar{u} \partial_x^{11} q + \sum_{k = 1}^{11} \binom{11}{k} \partial_x^k \bar{u} \partial_x^{11-k} q. \end{align} $$

We divide through both sides by $\mu _s$ , multiply by $\sqrt {\varepsilon } x^{9.5} \phi _{10}$ , and compute the $L^2$ norm, which gives

(5.24) $$ \begin{align} \nonumber \| \sqrt{\varepsilon} Q x^{9.5} \phi_{10} \| &\lesssim \| \sqrt{\varepsilon} V^{(9)}_x x^{9.5} \phi_{10} \| + \sum_{k = 1}^9 \| \frac{\partial_x^k \bar{u}}{\bar{u}} x^k \|_\infty \| \sqrt{\varepsilon} V^{(9-k)}_x x^{9-k + \frac 1 2} \phi_{11} \| \\ \nonumber &\quad + \| \partial_x^{10} \bar{u}_p x^9 y \|_\infty \| \sqrt{\varepsilon} \frac{V}{y} x^{\frac 1 2} \phi_{11} \| + \| \partial_x^{10} \bar{u}_E x^{10.5} \|_\infty \| \sqrt{\varepsilon} V \langle x \rangle^{-1} \| \\ \nonumber &\quad + \| \partial_x^{11} \bar{u}_p x^{10.5}y \|_\infty \| \sqrt{\varepsilon} \frac{q}{y} \langle x \rangle^{-1} \| + \| \partial_x^{11} \bar{u}_E x^{11.5} \|_\infty \| \sqrt{\varepsilon} q \langle x \rangle^{-2} \| \\ &\lesssim \|U, V \|_{\mathcal{X}_{\le 10}}, \end{align} $$

where we need to treat the lower order terms corresponding to $k = 9, 10$ in the sum (5.23) differently, in order to avoid using the critical Hardy inequality. We have invoked estimates (3.24), (2.8), (2.9).

We now address the second inequality, (5.20). We divide (5.23) by $\mu _s$ , multiply by $\sqrt {\varepsilon } x^{10} \phi _{11}$ , and compute the $L^2_x L^\infty _y$ norm, which gives

(5.25) $$ \begin{align} \nonumber \| \sqrt{\varepsilon} Q x^{10} \phi_{11} \|_{L^2_x L^\infty_y} &\lesssim \| V^{(10)} x^{10} \phi_{11} \|_{L^2_x L^\infty_y} + \sum_{k = 1}^9 \| \frac{\partial_x^{k} \bar{u}}{\bar{u}} x^{k} \|_{\infty} \| V^{(10-k)} x^{10-k} \phi_k \|_{L^2_x L^\infty_y} \\ \nonumber &\quad + \| \partial_x^{10} \bar{u}_p x^{9.5} y \|_{\infty} \| \frac{V}{y} x^{\frac 1 2} \|_{L^2_x L^\infty_y} + \| \partial_x^{10} \bar{u}_E x^{10.5} \|_\infty \| V \langle x \rangle^{- \frac 1 2} \|_{L^2_x L^\infty_y} \\ &\quad + \| \partial_x^{11} \bar{u}_p x^{10.5} y \|_{\infty} \| \frac{q}{y} x^{-\frac 1 2} \|_{L^2_x L^\infty_y} + \| \partial_x^{11} \bar{u}_E x^{11.5} \|_\infty \| q \langle x \rangle^{- \frac 3 2} \|_{L^2_x L^\infty_y} \\ \nonumber &\lesssim \|U, V \|_{\mathcal{X}}, \end{align} $$

where we have again invoked estimates (2.8), (2.9) for the $\bar {u}$ terms, the mixed norm estimate (3.28), as well as the following Sobolev interpolation estimates

(5.26) $$ \begin{align} &\| \frac{V}{y} x^{\frac 1 2} \phi_{11} \|_{L^2_x L^\infty_y} \lesssim \| V_y x^{\frac 1 2} \phi_{11} \|_{L^2_x L^\infty_y} \lesssim \| V_y x^{\frac 1 2} \phi_{11} \|^{\frac 1 2} \| U^{(1)}_y x^{\frac 1 2} \phi_{11} \|^{\frac 1 2} \lesssim \|U, V \|_{\mathcal{X}_{\le 1.5}}, \end{align} $$

(5.27) $$ \begin{align} &\| \sqrt{\varepsilon} V \langle x \rangle^{- \frac 1 2} \|_{L^2_x L^\infty_y} \lesssim \| \sqrt{\varepsilon} V \langle x \rangle^{-1} \phi_{11} \|^{\frac 1 2} \| \sqrt{\varepsilon} U_x \phi_{11} \| \lesssim \| U, V \|_{\mathcal{X}_{\le 1}}, \end{align} $$

and the analogous estimates for q instead of V for the final two terms from (5.25).

Dividing through by $\mu _s$ and differentiating in y yields

(5.28) $$ \begin{align} \tilde{U} = \frac{\bar{u}}{\mu_s} U^{(11)} - \partial_y ( \frac{\bar{u}}{\mu_s} ) V^{(10)} + \sum_{k = 1}^{11} \binom{11}{k} \partial_y ( \frac{\partial_x^k \bar{u}}{\mu_s} ) \partial_x^{11-k} q + \sum_{k = 1}^{11} \binom{11}{k} \frac{\partial_x^k \bar{u}}{\mu_s} U^{(11-k)}, \end{align} $$

and similarly, dividing through by $\mu _s$ and differentiating in x yields

(5.29) $$ \begin{align} \tilde{V} = \frac{\bar{u}}{\mu_s} V^{(11)} + \partial_x (\frac{\bar{u}}{\mu_s}) V^{(10)} + \sum_{k = 1}^{11} \binom{11}{k} \frac{\partial_x^k \bar{u}}{\mu_s} V^{(11-k)} - \sum_{k = 1}^{11} \binom{11}{k} \partial_x (\frac{\partial_x^k \bar{u}}{\mu_s}) \partial_x^{11-k} q. \end{align} $$

We first establish the following auxiliary estimate, which will be needed in forthcoming calculations due to the second term from (5.28).

$$ \begin{align*} \nonumber \partial_y (\frac{\bar{u}}{\mu_s}) = \partial_y (\frac{\mu_s - \varepsilon^{\frac{N_2}{2}} u }{\mu_s}) = - \varepsilon^{\frac{N_2}{2}} \partial_y (\frac{u}{\mu_s}) = - \varepsilon^{\frac{N_2}{2}} \partial_y ( \frac{u}{\bar{u}} \frac{\bar{u}}{\mu_s} ) = - \varepsilon^{\frac{N_2}{2}} \frac{\bar{u}}{\mu_s} \partial_y (\frac{u}{\bar{u}}) - \varepsilon^{\frac{N_2}{2}} \frac{u}{\bar{u}} \partial_y (\frac{\bar{u}}{\mu_s}), \end{align*} $$

which, rearranging for the quantity on the left-hand side, yields the identity

(5.30) $$ \begin{align} \partial_y (\frac{\bar{u}}{\mu_s}) = - \frac{\varepsilon^{\frac{N_2}{2}}}{1 + \varepsilon^{\frac{N_2}{2}} \frac{u}{\bar{u}} } \frac{\bar{u}}{\mu_s} \partial_y (\frac{u}{\bar{u}}),\\[-24pt]\nonumber \end{align} $$

from which we estimate

(5.31) $$ \begin{align} \| \partial_y (\frac{\bar{u}}{\mu_s}) x^{\frac 12} \psi_{12} \|_{L^\infty_x L^2_y} &\lesssim \varepsilon^{\frac{N_2}{2}} \| \frac{\bar{u}}{\mu_s} \|_{L^\infty} \frac{1}{1 - \varepsilon^{\frac{N_2}{2}} \| \frac{u}{\bar{u}} \|_{L^\infty} } \| \partial_y (\frac{u}{\bar{u}}) x^{\frac 1 2} \psi_{12} \|_{L^\infty_x L^2_y} \lesssim \varepsilon^{\frac{N_2}{2}-M_1} \|U,V \|_{\mathcal{X}}, \end{align} $$

where we have invoked estimates (3.49) and (3.50).

From these formulas, we provide the estimate (5.22) via

(5.32) $$ \begin{align} \nonumber \| \mu_s \tilde{U} x^{10.5} \phi_{11} \| & \lesssim \| \bar{u} U^{(10)}_x x^{10.5} \phi_{11} \| + \| \partial_y ( \frac{\bar{u}}{\mu_s}) x^{\frac 1 2} \psi_{12} \|_{L^\infty_x L^2_y} \| V^{(10)} x^{10} \phi_{11} \|_{L^2_x L^\infty_y} \\ \nonumber &\quad + \sum_{k = 1}^{10} \| \frac{\bar{u}}{\mu_s} \|_{\infty} \| \partial_y (\frac{\partial_x^{k} \bar{u}}{\bar{u}}) y \langle x \rangle^k \|_{\infty} \| \frac{\partial_x^{11-k} q}{y} \langle x \rangle^{11-k - \frac 1 2} \| \\ \nonumber &\quad + \| \frac{\bar{u}}{\mu_s} \|_\infty \| \partial_y (\frac{\partial_x^{11} \bar{u}_p}{\bar{u}}) y^2 \langle x \rangle^{10.5} \|_\infty \| \frac{q - y U(x, 0)}{\langle y \rangle^2} \| + \| \frac{\bar{u}}{\mu_s} \|_\infty \| \partial_y (\frac{\partial_x^{11} \bar{u}_E}{\bar{u}}) y \langle x \rangle^{11.5} \|_\infty \| U \langle x \rangle^{-1} \| \\ \nonumber &\quad + \sum_{k = 1}^{10} \| \frac{\bar{u}}{\mu_s} \|_\infty \| \frac{\partial_x^k \bar{u}}{\bar{u}} x^k \|_\infty \| U^{(11-k)} \langle x \rangle^{(11-k- \frac 1 2)} \| + \| \frac{\bar{u}}{\mu_s} \|_\infty \| \frac{\partial_x^{11} \bar{u}_p}{\bar{u}} y \langle x \rangle^{10.5} \|_\infty \| \frac{U - U(x, 0)}{\langle y \rangle} \| \\ \nonumber &\quad + \| \frac{\bar{u}}{\mu_s} \|_\infty \| \frac{\partial_x^{11} \bar{u}_p}{\bar{u}} \langle x \rangle^{11-\frac 1 4} \|_{L^\infty_x L^2_y} \| U(x, 0) \langle x \rangle^{- \frac 1 4} \|_{L^2_x} + \| \frac{\partial_x^{11} u_E}{\bar{u}} \langle x \rangle^{11.5} \|_\infty \| U \langle x \rangle^{-1} \| \\ & \lesssim \| U, V \|_{\mathcal{X}_{\le 10.5}} + \varepsilon^{\frac{N_2}{2} - M_1} \| U, V \|_{\mathcal{X}}, \end{align} $$

where we have invoked estimates (2.6), (2.8), (2.9), (5.31), and (3.28). An essentially identical proof applies also to the $\tilde {V}$ quantity from (5.21), so we omit repeating these details. This establishes estimate (5.21), with (5.22) following similarly, upon using the Hardy-type inequality (3.23).

Proof. Dividing through equation (5.23) by $\bar {u}$ and computing $\partial _y^2$ gives

(5.34) $$ \begin{align} \nonumber U^{(11)}_y & = \frac{\mu_s}{\bar{u}} \tilde{U}_y + 2\partial_y (\frac{\mu_s}{\bar{u}}) \tilde{U} + \partial_y^2 (\frac{\mu_s}{\bar{u}}) Q - \sum_{k = 1}^{11} \binom{11}{k} \partial_y^2 (\frac{\partial_x^k \bar{u}}{\bar{u}}) \partial_x^{11-k} q \\ &\quad - 2 \sum_{k = 1}^{11} \binom{11}{k} \partial_y^2 (\frac{\partial_x^k \bar{u}}{\bar{u}} ) \partial_x^{11-k} U - \sum_{k = 1}^{11} \binom{11}{k} \frac{\partial_x^k \bar{u}}{\bar{u}} \partial_x^{11-k} U_y \end{align} $$

From here, we obtain the estimate

(5.35) $$ \begin{align} \nonumber \| \sqrt{\bar{u}} U^{(11)}_y x^{11} \phi_{11} \| &\lesssim \| \sqrt{ \frac{\mu_s}{\bar{u}} } \|_\infty \| \sqrt{\mu_s} \tilde{U}_y x^{11} \phi_{11} \| + \| \sqrt{\bar{u}} \partial_y (\frac{\mu_s}{\bar{u}}) x^{\frac 1 2} \|_\infty \| \tilde{U} x^{10.5} \phi_{11} \| \\ \nonumber &\quad + \varepsilon^{\frac{N_2}{2}} \| \bar{u} \partial_y^2 (\frac{u}{\bar{u}}) x \psi_{12} \|_{L^\infty_x L^2_y} \| \frac{ Q }{\sqrt{\bar{u}}} x^{10} \phi_{11} \|_{L^2_x L^\infty_y} \\ \nonumber &\quad + \sum_{k = 1}^{11} \| \partial_y^2(\frac{\partial_x^k \bar{u}}{\bar{u}}) y x^{k + \frac 1 2} \|_\infty \| \frac{\partial_x^{11-k} q}{y} x^{11-k - \frac 1 2} \phi_{11} \| \\ \nonumber &\quad + \sum_{k = 1}^{11} \| \frac{\partial_x^k \bar{u}}{\bar{u}} x^k \|_\infty \| U^{(11-k)}_y x^{11-k} \phi_{11} \| \\ &\lesssim \| \tilde{U}, \tilde{V} \|_{\Theta_{11}} + \| U, V \|_{\mathcal{X}_{\le 10.5}} + \varepsilon^{\frac{N_2}{2} - M_1} \| U, V \|_{\mathcal{X}} \| U, V \|_{\mathcal{X}_{\le 10.5}}, \end{align} $$

where we have invoked (3.50), (3.49), (5.20), and (3.22).

An essentially identical calculation applies to the remaining terms from the $\| U, V \|_{X_{11}}$ norm, and also an essentially identical computation enables us to prove the second inequality in (5.33). We note, however, that to compare the quantities $ \| \mu _{sy} \tilde {U} x^{11} \phi _{11} \|_{y = 0}$ and $\| \bar {u}_y U^{(11)} x^{11} \phi _{11} \|_{y = 0}$ , we also need to demonstrate boundedness of the coefficients $|\frac {\bar {u}_y}{\mu _{sy}}| \phi _{11}$ and $|\frac {\mu _{sy}}{\bar {u}_y}| \phi _{11}$ . For this purpose, we estimate

(5.36) $$ \begin{align} \nonumber |\mu_{sy}(x, 0) - \bar{u}^0_{py}(x, 0)| \phi_{11} & \le \sum_{i = 1}^{N_1} \varepsilon^{\frac i 2} (\sqrt{\varepsilon} \| u^i_{EY} \|_{L^\infty_y} + \| u^i_{py} \|_{L^\infty_y} ) + \varepsilon^{\frac{N_2}{2}} \| u_y \psi_{12} \|_{L^\infty_y} \\ &\lesssim \sum_{i = 1}^{N_1} \varepsilon^{\frac i 2} (\sqrt{\varepsilon} \langle x \rangle^{- \frac 3 2} + \langle x \rangle^{- \frac 3 4 + \sigma_\ast}) + \varepsilon^{\frac{N_2}{2}- M_1} \langle x \rangle^{- \frac 3 4} \| U, V \|_{\mathcal{X}} \lesssim \varepsilon^{\frac 1 2} \langle x \rangle^{- \frac 3 4 + \sigma_\ast}, \end{align} $$

where we have invoked estimates (1.41), (1.38), (3.47), the identity that $\phi _{11} = \psi _{12} \phi _{11}$ , the fact that $\frac {N_2}{2} - M_1>> 0$ , and finally the assumption that $\|U, V \|_{\mathcal {X}} \le 1$ .

Proof. We begin with the first estimate, (5.37). For this, we consider

(5.39) $$ \begin{align} \nonumber \bar{u} \tilde{W}^2 \langle x \rangle^{21.5} & \le \langle x \rangle^{21.5} |\int_y^\infty 2 \bar{u} \tilde{W} \tilde{W}_y(y') \,\mathrm{d} y' | + \langle x \rangle^{21.5} |\int_y^\infty 2 \tilde{W}^2 \bar{u}_y \,\mathrm{d} y' | \\ &\lesssim \| \tilde{W} \langle x \rangle^{10.5} \|_{L^2_y} \| \sqrt{\bar{u}} \tilde{W}_y \langle x \rangle^{11} \|_{L^2_y} + \| \bar{u}_y \langle x \rangle^{\frac 1 2} \|_\infty \| \tilde{W} \langle x \rangle^{10.5} \|_{L^2_y}^2. \end{align} $$

Multiplying both sides by $\phi _{11}^2$ and placing both sides in $L^2_x$ gives estimate (5.37).

We turn now to (5.38). We split

(5.40) $$ \begin{align} \| \tilde{W} \langle x \rangle^{10.5} \phi_{11} \|_{L^2_x L^4_y}^2 \le \| \tilde{W} \langle x \rangle^{10.5} \phi_{11} \chi(z) \|_{L^2_x L^4_y}^2 + \| \tilde{W} \langle x \rangle^{10.5} \phi_{11} (1 - \chi(z)) \|_{L^2_x L^4_y}^2 \end{align} $$

We first consider the z-localized contribution above. To compute the $L^4_y$ , we raise to the fourth power and integrate by parts via

(5.41) $$ \begin{align} \nonumber \| \tilde{W} \langle x \rangle^{10.5} \chi(z) \|_{ L^4_y}^4 & = \int \tilde{W}^4 \chi(z)^4 \langle x \rangle^{42} \,\mathrm{d} y = \int \partial_y (y) \tilde{W}^4 \chi(z)^4 \langle x \rangle^{42} \,\mathrm{d} y \\ & = - \int 4 y \tilde{W}^3 \tilde{W}_y \chi(z)^4 \langle x \rangle^{42} - \int 4 y \tilde{W}^4 \frac{1}{\sqrt{x}} \chi^3 \chi'(z) \langle x \rangle^{42} \,\mathrm{d} y, \end{align} $$

from which we deduce that (using that $\sqrt {a + b} \le \sqrt {a} + \sqrt {b}$ for $a, b \ge 0$ )

(5.42) $$ \begin{align} \| \tilde{W} \langle x \rangle^{10.5} \chi(z) \|_{ L^4_y}^2 \le |\int 4 y \tilde{W}^3 \tilde{W}_y \chi(z)^4 \langle x \rangle^{42} |^{\frac12} + | \int 4 y \tilde{W}^4 \frac{1}{\sqrt{x}} \chi(z)^3 \chi'(z) \langle x \rangle^{42} \,\mathrm{d} y|^{\frac12}. \end{align} $$

We will first handle the first integral above. Using that $z \le 1$ on the support of $\chi (z)$ , we can estimate via

(5.43) $$ \begin{align} \nonumber |\int 4y \tilde{W}^3 \tilde{W}_y \chi(z) \langle x \rangle^{42}|^{\frac12} &\lesssim (\int |\tilde{W}|^3 \bar{u} |\tilde{W}_y| \langle x \rangle^{42.5} \chi^4 \,\mathrm{d} y)^{\frac12} \\ \nonumber &\lesssim (\| \sqrt{\bar{u}} \tilde{W} \langle x \rangle^{10.5} \|_{L^\infty_y} \| \tilde{W} \langle x \rangle^{10.5} \chi \|_{L^4_y}^2 \| \sqrt{\bar{u}} \tilde{W}_y \langle x \rangle^{11} \|_{L^2_y})^{\frac12} \\ & \le \frac{1}{100} \| \tilde{W} \langle x \rangle^{10.5}\chi \|_{L^4_y}^2 + C \| \sqrt{\bar{u}} \tilde{W} \langle x \rangle^{10.5}\|_{L^\infty_y} \| \sqrt{\bar{u}} \tilde{W}_y \langle x \rangle^{11} \|_{L^2_y} \end{align} $$

The term with $1/100$ prefactor gets absorbed to the left-hand side of (5.42). For the remaining term on the right-hand side above, we multiply by $\phi _{11}^2$ , integrate over x, and appeal to (5.37) to give the desired estimate.

For the far-field integral from (5.41), we estimate it by first noting that $1 \lesssim |y/\sqrt {x}| \lesssim 1$ on the support of $\chi '$ . Thus,

(5.44) $$ \begin{align} \nonumber |\int \tilde{W}^4 \langle x \rangle^{42} z \chi'(z) \,\mathrm{d} y|^{\frac12} &\lesssim ( \| \bar{u} \tilde{W} \langle x \rangle^{10.5} \|_{L^\infty_y}^2 \| \bar{u} \tilde{W} \langle x \rangle^{10.5} \|_{L^2_y}^2)^{\frac12} \\ \nonumber &\lesssim \| \bar{u} \tilde{W} \langle x \rangle^{10.5} \|_{L^2_y}^{\frac32} \| \sqrt{\bar{u}} \tilde{W}_y \langle x \rangle^{10.5} \|_{L^2_y}^{\frac12} \\ &\lesssim \| \bar{u} \tilde{W} \langle x \rangle^{10.5} \|_{L^2_y}^{2} + \| \sqrt{\bar{u}} \tilde{W}_y \langle x \rangle^{10.5} \|_{L^2_y}^2. \end{align} $$

Multiplying by $\phi _{11}^2$ , integrating over x, and appealing to (5.37) gives the desired estimate. The $1 - \chi (z)$ contribution from (5.40) is treated similarly to the $\chi '$ term.

Proof. We begin first with $\mathcal {R}_1^{(1)}$ , defined in (5.4). For the first term in (5.4), we assume $1 \le j \le 6$ , in which case we bound