Proof. We present the (well-known) argument for completeness. For any \(\varphi \in C_c^\infty(\overline{\mathbb{H}})\) with \(\partial_y\varphi|_{y=0}=0\), we have \[\label{dualeq1} \begin{align} \varphi(t_0,x_0,y_0)=&\int_{\mathbb{H}}(\partial_t+y\partial_x-\partial_y^2)\Gamma(z;w) \varphi(t,x,y) dz\\ =& \int_{\mathbb{H}} \Gamma(z;w)(-\partial_t-y\partial_x-\partial_y^2)\varphi(t,x,y)dz. \end{align}\tag{21}\] Define for \((t,x,y)\in\mathbb{H}\), \[\label{dualeq2} h(t,x,y):=(-\partial_t-y\partial_x-\partial_y^2)\varphi(t,x,y).\tag{22}\] Define \(\Gamma^\ast(t_0,x_0,y_0;t,x,y)\) as the fundamental solution to \(-\partial_{t_0}-y_0\partial_{x_0}-\partial_{y_0}^2\) on \(\mathbb{H}\) with zero Neumann boundary condition, i.e, \[\label{dualeq3} \begin{align} (-\partial_{t_0}-y_0\partial_{x_0}-\partial_{y_0}^2)\Gamma^\ast(t_0,x_0,y_0;t,x,y)=&\,\delta(t_0-t,x_0-x,y_0-y),\\ \partial_{y_0}\Gamma^\ast(t_0,x_0,y_0;t,x,y)|_{y_0=0}=&\,0. \end{align}\tag{23}\] It follows from the identity 22 that \[\label{dualeq3461} \varphi(t_0,x_0,y_0)=\int_{\mathbb{H}} \Gamma^\ast(t_0,x_0,y_0;t,x,y)h(t,x,y)\,dtdxdy.\tag{24}\] Comparing 21 and 24 , we conclude that \[\label{dualeq4} \Gamma(t,x,y;t_0,x_0,y_0)=\Gamma^\ast(t_0,x_0,y_0;t,x,y).\tag{25}\] The desired conclusions follow from 25 and 23 . ◻

Proof. By transition invariance (see Proposition 1), we can assume without loss of generality that \(z=(t,x,y)\) and \(w=(0,0,y_{0})\).

We can write \[\label{decom} \begin{align} \Gamma(z;w)&=y_{0}^{-4}\Gamma(D(y_{0})z; 0,0,1)\\ &=y_{0}^{-4}\big(\widetilde{\Gamma}(D(y_{0})z; 0,0,1)+(\Gamma-\widetilde{\Gamma})(D(y_{0})z; 0,0,1)\big)\\ &=\widetilde{\Gamma}(z,w)+y_{0}^{-4}(\Gamma-\widetilde{\Gamma})(D(y_{0})z; 0,0,1). \end{align}\tag{27}\]

Let \[V(t,x,y)=(\Gamma-\widetilde{\Gamma})(t,x,y;0,0,1).\] By 19 and 26 , \(V\) satisfies the equation \[\label{Vequa} \begin{cases} \partial_{t}V+y\partial_{x}V-\partial_{y}^{2}V=0,\quad t>0,x\in\mathbb{R},y>0;\\ \partial_{y}V|_{y=0}=\displaystyle\frac{\sqrt{3}}{2\pi t^{2}}\Big(\frac{1}{t}-\frac{3x}{t^{2}}\Big)\exp\left\{-\frac{1}{4t}-\frac{3}{t^{3}}(x-\frac{t}{2})^{2}\right\}=:g(t,x),\\ V|_{t=0}=0. \end{cases}\tag{28}\]

Step 1: smoothness of \(V\). We first prove that the solution \(V(t,x,y)\) in 28 are all its derivatives are uniformly bounded for \(t\ge0, x\in\mathbb{R}, y\ge0\). Decompose \(V(t,x,y)\) as \[V(t,x,y)=\mathcal{V}(t,x,y)+g(t,x)\sqrt{1+t}\cdot\chi\Big(\frac{y}{\sqrt{1+t}}\Big),\] where \(\chi(z)\in C_{c}^{\infty}([0,1))\) and \(\chi'(0)=1\). By the expression of \(g(t,x)\) in 28 and simple calculations, we have \[\left|\nabla_{(t,x,y)}^{k}\Big(g(t,x)\sqrt{1+t}\cdot\chi\Big(\frac{y}{\sqrt{1+t}}\Big)\Big)\right|\lesssim_{k} t^{-5/2}e^{-\frac{1}{8t}},\;\forall k\geq 0.\] Thus we only need to show \(\mathcal{V}\) and its derivatives are bounded.

Thanks to 28 and \(g|_{t=0}\equiv 0\), we have \(\mathcal{V}(t,x,y)\) satisfies the following equation \[\label{Vequ} \begin{cases} \partial_{t}\mathcal{V}+y\partial_{x}\mathcal{V}-\partial_{y}^{2}\mathcal{V}=\mathcal{G}(t,x,y),\quad t>0,x\in\mathbb{R},y>0;\\ \partial_{y}\mathcal{V}|_{y=0}=0,\mathcal{V}|_{t=0}=0,\\ \end{cases}\tag{29}\] where \[\begin{align} \mathcal{G}(t,x,y)=&\,\frac{1}{2}g(t,x)\frac{y}{1+t}\chi'\Big(\frac{y}{\sqrt{1+t}}\Big)+g(t,x)(1+t)^{-\frac{1}{2}}\chi''\Big(\frac{y}{\sqrt{1+t}}\Big)\\ &-\Big[g_{t}(t,x)\sqrt{1+t}+\frac{1}{2}g(t,x)(1+t)^{-\frac{1}{2}}+yg_{x}(t,x)\sqrt{1+t}\Big]\chi\Big(\frac{y}{\sqrt{1+t}}\Big). \end{align}\] By the standard estimate for the transport–diffusion equation, we have \[\label{boundnessofv} |\mathcal{V}(t,x,y)|\leq \int_{0}^{t}\|\mathcal{G}(s)\|_{L^{\infty}_{x,y}}ds\lesssim \min\{t,1\}.\tag{30}\] We can also obtain the bounds for the higher–order derivatives of \(V\) by the same approach, by taking \(t,x\) derivatives in 29 and using 30 to control the \(t,x\) derivatives, and then using equation 29 to control the \(y\) derivatives.

Step 2: Proof of ?? . By the translation invariance of \(\Gamma\), we can assume \(w=(0,0,y_{0})\) for some \(y_{0}>0\); Furthermore by the dilation invariance of \(\Gamma\), we have \[\Gamma(z;w)=\|z-w\|^{-4}\Gamma(D(\|z-w\|z, D(\|z-w\|)w)).\] Hence, we only need to show that \[\label{goal} \sup_{\|z-w\|=1}|\Gamma(z;w)|\lesssim 1,\;\text{uniformly in}\; y_{0}\in[0,10].\tag{31}\]

We consider two separate cases.

Case 1: \(10\geq y_{0}>\frac{1}{100}\). 27 implies that \[\Gamma(t,x,y;0,0,y_{0})=\widetilde{\Gamma}(t,x,y;0,0,y_{0})+y_{0}^{-4}V\left(D(y_{0})z\right).\]

By 26 and 30 , we obtain that \[\label{wholespacebd} \sup_{\|z-w\|=1, y_{0}\leq 10}|\widetilde{\Gamma}(z,w)|\lesssim 1.\tag{32}\]

Hence, 31 follows from 30 and 32 .

Case 2: \(y_{0}\leq \frac{1}{100}\). This case is more complicated since the source is close to the boundary and the bounds on \(V\) from the last step are not sufficient for our purposes. Notice that if \(y_{0}\leq \frac{1}{100}\) and \(\|z-(0,0,y_{0})\|=1\), then at least one of the following three cases must happen: (i) \((t,x,y)\in\Omega_{1}:=\left\{\frac{1}{2}\leq y\leq 1,\;0\leq t\leq 1,\;|x|\leq 1\right\}\); (ii) \((t,x,y)\in\Omega_{2}:=\left\{\frac{1}{2}\leq t \leq 1,\;0\leq y\leq 1,\;|x|\leq 1\right\}\); (iii) \((t,x,y)\in\Omega_{3}:=\left\{\frac{1}{2}\leq|x|\leq 1,\;0\leq y\leq 1, 0\leq t\leq 1\right\}\).

We distinguish three subcases depending on (i) - (iii).

Subcase 2.1: \((t,x,y)\in\Omega_{1}.\) By the non-negativity and conservation of mass, we have \[\label{l1conser} \int_{\mathbb{R}\times\mathbb{R}^{+}}\Gamma(z;w)\,dxdy\equiv 1,\quad \text{for}\,\, t>0,y_{0}>0,\tag{33}\] Hence, in this case, the desired bound 31 follows from the uniform bound 33 and the interior estimate of hypoelliptic operator (Proposition 3.2 in [51]) since \(y\ge1/2\).

Subcase 2.2: \((t,x,y)\in\Omega_{2}\). Taking Fourier transform in the \(x\)–variable, and denoting \[\widehat{\Gamma}(t,\xi,y;0,0,y_{0})=\int_{-\infty}^{\infty}\Gamma(t,x,y;0,0,y_{0})e^{-2\pi i x\xi}dx,\] \(\widehat{\Gamma}\) satisfies the equation for \(y\ge0, t\ge0\), \[\label{ftx} \begin{cases} \partial_{t}\widehat{\Gamma}+2\pi \mathrm{i} y\xi\widehat{\Gamma}-\partial_{y}^{2}\hat{\Gamma}=0,\\ \widehat{\Gamma}|_{t=0}=\delta(y-y_{0}),\;\partial_{y}\hat{\Gamma}|_{y=0}=0. \end{cases}\tag{34}\] By 33 , we have \[\label{l1conserft} \sup_{t>0,\,\xi\in\mathbb{R}}\int_{0}^{\infty}|\widehat{\Gamma}(t,\xi,y;0,0,y_0)|\,dy\lesssim 1.\tag{35}\]

Multiplying 34 by the complex conjugate of \(\widehat{\Gamma}\), integrating in \(y\), and taking the real part, we obtain the following energy inequality: \[\label{diffineq} \frac{1}{2}\frac{d}{dt}\|\widehat{\Gamma}(t,\xi,y;w)\|^2_{L^{2}_{y}}+\|\partial_{y}\widehat{\Gamma}(t,\xi,y;w)\|^2_{L^{2}_{y}}\leq 0.\tag{36}\] By Nash’s inequality (see [55]) \[\int_{0}^{\infty}|F(y)|^{2}dy\lesssim \left(\int_{0}^{\infty}|\partial_{y}F|^{2}dy\right)^{\frac{1}{3}}\left(\int_{0}^{\infty}|F(y)|dy\right)^{\frac{4}{3}},\] and 35 , we have \[\label{nash} \int_{0}^{\infty}|\widehat{\Gamma}(t,\xi,y;w)|^{2}dy\lesssim \left(\int_{0}^{\infty}|\partial_{y}\widehat{\Gamma}(t,\xi,y;w)|^{2}dy\right)^{\frac{1}{3}}.\tag{37}\]

Combining 36 and 37 , we conclude that there exists a constant \(C\) such that for all \(\xi\in\mathbb{R}\) and \(y_{0}>0\), \[\label{ftl2bd} \|\widehat{\Gamma}(t,\xi,y;w)\|_{L^{2}_{y}}\leq Ct^{-\frac{1}{4}}.\tag{38}\]

By 38 and enhanced dissipation estimates in Proposition 20, we have for \(t\ge 1/3,\, y_0>0\) and some \(c>0\), \[\label{gerveyesti1} \|\widehat{\Gamma}(t,\xi,y;w)\|_{L^{2}_{y}}\lesssim e^{-c|\xi|^{\frac{2}{3}}t}.\tag{39}\] By 39 and the standard energy estimate of the heat equation, denoting \(D:=\{t\ge1/2, y\ge0\}\), we also have \[\label{gerveyesti2} \|\partial_{t}\widehat{\Gamma}(t,\xi,y;w)(1+y)^{-1}\|_{L^2(D)}+\|\partial_{y}^{2}\widehat{\Gamma}(t,\xi,y;w)(1+y)^{-1}\|_{L^2(D)}\lesssim e^{-c|\xi|^{\frac{2}{3}}}.\tag{40}\] The desired bound 31 in this case then follows from 39 , 40 , Fourier inversion and Sobolev inequality.

Subcase 2.3: \((t,x,y)\in\Omega_{3}\). This is perhaps the most complicated case since the bound 38 is not very effective for small \(t>0\). To overcome this difficulty, let \(\chi(x)\) be a smooth cut off function in the \(x\) variable such that \[\mathrm{supp}\,\chi\subset\{x:1/10\leq |x|\leq 3/2\},\;\chi(x)\equiv 1\;\text{in}\;\{x: 1/5\leq |x|\leq 5/4\}.\] Let \(\Gamma_{1}(t,x,y;0,0,y_{0})=\Gamma(t,x,y;0,0,y_{0})\chi(x)\). Then \(\Gamma_{1}\) satisfies the following equation \[\label{gamma1eq} \begin{cases} \partial_{t}\Gamma_{1}+y\partial_{x}\Gamma_{1}-\partial_{y}^{2}\Gamma_{1}=y\chi'(x)\Gamma,\\ \Gamma_{1}|_{t=0}=0,\;\partial_{y}\Gamma_{1}|_{y=0}=0. \end{cases}\tag{41}\] Taking Fourier transform in \(x\), we get that \[\label{gamma1fteq} \begin{cases} \partial_{t}\widehat{\Gamma}_{1}+2\pi\mathrm{i}y\xi\widehat{\Gamma}_{1}-\partial_{y}^{2}\widehat{\Gamma}_{1}=y\,\widehat{\chi'\Gamma},\\ \widehat{\Gamma}_{1}|_{t=0}=0,\;\partial_{y}\widehat{\Gamma}_{1}|_{y=0}=0. \end{cases}\tag{42}\] Let \[\label{eqphi} \phi(y)=\frac{1}{1+y}-\frac{1}{2(1+y)^{2}}.\tag{43}\] Clearly, \(\phi'(0)=0\) and \(\phi(y)\sim (1+y)^{-1}\).

Multiplying 42 with the complex conjugate of \(\widehat{\Gamma}_{1}\phi^{2}(y)\), integrating and taking the real part, we get the following energy identity \[\label{ftenergyid} \begin{align} &\frac{1}{2}\frac{d}{dt}\left(\int_{0}^{\infty}|\widehat{\Gamma}_{1}(t,\xi,y;w)|^{2}\phi^{2}(y)dy\right)+\int_{0}^{\infty}|\partial_{y}\widehat{\Gamma}_{1}(t,y,\xi;w)|^{2}\phi^{2}(y)dy\\ &+2\mathrm{Re}\left(\int_{0}^{\infty}\partial_{y}\widehat{\Gamma}_{1}(t,\xi,y;w)\cdot\overline{\widehat{\Gamma}_{1}}(t,\xi,y;w)\phi(y)\phi'(y)dy\right)\\ &=\mathrm{Re}\left(\int_{0}^{\infty}y\,\widehat{\chi'\Gamma}(t,\xi,y;w)\overline{\widehat{\Gamma}_{1}}(t,\xi,y;w)\phi^{2}(y)dy\right). \end{align}\tag{44}\] Notice from 38 that \[\begin{align} &\left|\int_{0}^{\infty}y\,\widehat{\chi'\Gamma}(t,\xi,y;w)\overline{\widehat{\Gamma}_{1}}(t,\xi,y;w)\phi^{2}(y)dy\right|\\ &\lesssim \|\widehat{\Gamma}_{1}(t,\xi,y;w)\phi(y)\|_{L^{2}_{y}}\|\widehat{\Gamma}(t,\xi,y;w)\|_{L^{2}_{y}}\lesssim t^{-\frac{1}{4}} \|\widehat{\Gamma}_{1}(t,\xi,y;w)\phi(y)\|_{L^{2}_{y}}. \end{align}\] Therefore, by 44 and Gronwall’s inequality, for any \(T>0\) we have \[\label{ftenergybd} \sup_{t\in[0,T]}\left(\int_{0}^{\infty}|\widehat{\Gamma}_{1}(t,\xi,y;w)|^{2}\phi^{2}(y)\,dy\right)+\int_{0}^{T}\int_{0}^{\infty}|\partial_{y}\widehat{\Gamma}_{1}(t,\xi,y;w)|^{2}\phi^{2}(y)\,dydt\lesssim_T1.\tag{45}\]

Set \(\Gamma^{*}=\widehat{\Gamma}_{1}(t,\xi,y;w)\phi(y)\). Then we have \[\label{gammastareq} \begin{cases} \partial_{t}\Gamma^{*}+2\pi \mathrm{i}y\xi\Gamma^{*}-\partial_{y}^{2}\Gamma^{*}=y\phi(y)\widehat{\chi'\Gamma}-2\partial_{y}\widehat{\Gamma}_{1}\phi'(y)-\widehat{\Gamma}_{1}\phi''(y),\\ \Gamma^{*}|_{t=0}=0,\;\partial_{y}\Gamma^{*}|_{y=0}=0. \end{cases}\tag{46}\] If \(|\xi|\geq 1\), by Duhamel’s principle, and the enhanced dissipation estimate (Proposition 20), we get that \[\label{eq:que1} \|\Gamma^{*}(t,\xi,y;w)\|_{L^{2}_{y}}\lesssim \int_{0}^{t}e^{-c|\xi|^{\frac{2}{3}}(t-s)}\left\|y\phi(y)\widehat{\chi'\Gamma}-2\partial_{y}\widehat{\Gamma}_{1}\phi'(y)-\widehat{\Gamma}_{1}\phi''(y)\right\|_{L^{2}_{y}}ds,\tag{47}\] for some \(c>0\).

Using the bounds 38 and 45 , together with the pointwise inequality \(|\phi'|+|\phi''|\lesssim|\phi|\), we obtain that \[\label{gammaonebd} \int_{0}^{T}\int_{0}^{\infty}|\Gamma_{1}(t,x,y;w)|^{2}\phi^{2}(y)\,dydt\lesssim_{T} (1+|\xi|)^{-\frac{4}{3}}.\tag{48}\]

We next set \(\Gamma_{2}=\Gamma(t,x,y;0,0,y_{0})\chi_{2}(x)\), where \(\chi_{2}\) is a standard smooth cutoff function with \[\mathrm{supp}\,\chi_{2}\subset\{x:1/5\leq|x|\leq 5/4\},\;\chi_{2}(x)\equiv 1\;\text{in}\;\{1/4\leq|x|\leq 6/5\}.\] Then \(\Gamma_{2}\) satisfies \[\label{gamma2eq} \begin{cases} \partial_{t}\Gamma_{2}+y\partial_{x}\Gamma_{2}-\partial_{y}^{2}\Gamma_{2}=y\chi_{2}'(x)\Gamma_{1},\\ \Gamma_{2}|_{t=0},\;\partial_{y}\Gamma_{2}|_{y=0}=0, \end{cases}\tag{49}\] and correspondingly, \[\label{gamma2fteq} \begin{cases} \partial_{t}\widehat{\Gamma}_{2}+2\pi\mathrm{i}y\xi\,\widehat{\Gamma}_{2}-\partial_{y}^{2}\widehat{\Gamma}_{2}=y\,\widehat{\chi_{2}'\Gamma_{1}},\\ \widehat{\Gamma}_{2}|_{t=0}=0,\;\partial_{y}\Gamma_{2}|_{y=0}=0. \end{cases}\tag{50}\] By the same argument as in the derivation of 48 and using 48 , we obtain \[\label{gammatwobd} \int_{0}^{T}\int_{0}^{\infty}|\widehat{\Gamma}_{2}(t,y,\xi;w)|^{2}\phi^{4}(y)\,dydt\lesssim _{T}(1+|\xi|)^{-\frac{8}{3}}.\tag{51}\]

By induction, for any \(k\in\mathbb{N}^{+}\), we obtain that \[\label{gammakbd} \int_{0}^{T}\int_{0}^{\infty}|\widehat{\Gamma}_{k}(t,y,\xi;w)|^{2}\phi^{2k}(y)\,dydt\lesssim _{k,T}(1+|\xi|)^{-\frac{4k}{3}},\tag{52}\] where \(\Gamma_{k}=\Gamma(t,x,y;0,0,y_{0})\chi_{k}(x)\), with \[\mathrm{supp}\,\chi_{k}\subset\{x:\chi_{k-1}(x)=1\},\;\text{and}\;\chi_{k}(x)\equiv 1\;\text{in}\;\{2/5\leq |x|\leq 11/10\}\;\text{for all}\;k\in\mathbb{N}^{+}.\] As a consequence, we have \[\label{gammaphybd} \int_{0}^{T}\int_{0}^{\infty}|(\partial_{x}^{k}\Gamma)(t,x,y;0,0,y_{0})|^{2}(1+|y|)^{-8k}dydt\lesssim_{k,T}1,\;\forall k\in\mathbb{N},\;\frac{1}{2}\leq|x|\leq 1.\tag{53}\]

Finally, for fixed \(\frac{1}{2}\leq |x|\leq 1\), \(\Gamma=\Gamma(t,x,y;0,0,y_{0})\) satisfies the following one dimensional inhomogeneous heat equation \[\begin{cases} \partial_{t}\Gamma-\partial_{y}^{2}\Gamma=-y\partial_{x}\Gamma,\;t> 0, y>0,\\ \Gamma|_{t=0}=0,\;\partial_{y}\Gamma|_{y=0}=0. \end{cases}\] By 53 and properties of the heat equation, for any \(\frac{1}{2}\leq|x|\leq 1\), we have \[\label{gammaphysbd} \int_{0}^{T}\int_{0}^{1}\left(|\Gamma|^{2}+|\partial_{t}\Gamma|^{2}+|\partial_{y}^{2}\Gamma|^{2}\right)dydt\lesssim_{T} 1.\tag{54}\] Therefore, the desired bound 31 in this case follows from 53 , 54 and Sobolev inequalities. This completes the proof of Proposition 2. ◻

Proof. The proof is similar to Proposition 2. For completeness we outline the proof below.

By the argument in Proposition 2, we need to show that for \(y_{0}\leq 10\|z-w\|\), \[\label{highgoal} \sup_{\|z-w\|=1}\left|\partial_{t}^{\alpha_{1}}\partial_{t_{0}}^{\alpha_{2}}\partial_{x}^{\beta_{1}}\partial_{x_{0}}^{\beta_{2}}\partial_{y}^{\gamma_{1}}\partial_{y_{0}}^{\gamma_{2}}\Gamma(z;w)\right|\lesssim 1.\tag{55}\] By Proposition 1, we only need to consider the case \(\alpha_{2}=\beta_{2}=0\). Since \(\mathcal{L}_{0}\) commutes with \(\partial_{t}\) and \(\partial_{x}\), we can obtain 55 by similar arguments as in Proposition 2 when \(\gamma_{1}=\gamma_{2}=0\).

Next, notice that \[(\partial_{t}+y\partial_{x}-\partial_{y}^{2})\Gamma(z;w)=(\partial_{t_{0}}+y_{0}\partial_{x_{0}}+\partial_{y_{0}}^{2})\Gamma(z;w)=0,\quad z\neq w,\] Hence 55 holds when \(\gamma_{1}\) and \(\gamma_{2}\) are even.

When one of \(\gamma_{i} (i=1,2)\) is odd, we can interpolate in \(y\)–direction and apply 55 when \(\gamma_{i} (i=1,2)\) are both even. Finally, we can interpolate again to handle the case when \(\gamma_{i} (i=1,2)\) are both odd. The proof of Proposition 3 is then complete. ◻

Proof. By Proposition 2 and Proposition 3 we have \[\begin{align} &0\leq \Gamma(z;w)\leq \widetilde{\Gamma}(z;w)+C_{1}\|z-w\|^{-4},\\ &|(\partial_{y},\partial_{y_{0}})\Gamma(z;w)|\leq |(\partial_{y},\partial_{y_{0}})\widetilde{\Gamma}(z;w)+C_{2}\|z-w\|^{-5}. \end{align}\]

Recall ?? , we have \[\|\cdot\|^{-4}\in L^{\frac{3}{2},\infty},\;\|\cdot\|^{-5}\in L^{\frac{6}{5},\infty},\] where \(L^{p,q}\) is the classical Lorentz space with respect to the Lebesgue measure.

Therefore, proposition 4 follows from the potential estimate in the whole space (see e.g. [49], Corollary 2.2), as well as Young’s inequality. ◻

Proof. By translation invariance, we can assume that \(t_{0}=x_{0}=0\). We first note that \(\Gamma(z;w)\) is smooth away from the initial time by Proposition 2 and Proposition 3, and \(\Gamma(z;w)\) is non–negative as proven previously. Next we will exclude the following four cases one by one: \[\begin{align} &\mathrm{Case}\, (\alpha): \quad\Gamma(\bar{t},\bar{x},\bar{y};0,0,\bar{y}_{0})=0,\;\text{for some}\;\bar{t}>0,\bar{x}>0,\bar{y}>0,\bar{y}_{0}>0,\\ &\mathrm{Case}\, (\beta):\quad \Gamma(\bar{t},\bar{x},\bar{y};0,0,0)=0,\;\text{for some}\;\bar{t}>0,\bar{x}>0,\bar{y}>0,\\ &\mathrm{Case}\, (\gamma): \quad\Gamma(\bar{t},\bar{x},0;0,0,y_{0})=0,\;\text{for some}\;\bar{t}>0,\bar{x}>0,\bar{y}_{0}>0,\\ &\mathrm{Case}\, (\theta): \quad \Gamma(\bar{t},\bar{x},0;0,0,0)=0,\;\text{for some}\;\bar{t}>0,\bar{x}>0. \end{align}\]

We divide the proof in several steps.

Step 1: excluding case (\(\alpha\)). By the dilation invariance of \(\Gamma(\cdot,\cdot)\) (Proposition 1), we can assume that \(\bar{t}=1\). Our first important observation is that there exists a universal constant \(M>0\) such that \[\label{eq:po1} \Gamma(t,x,xt^{-1};0,0,xt^{-1})>0,\tag{56}\] for all \(0<t\leq 1\) and \(x\geq Mt^{\frac{3}{2}}\).

To see 56 , we recall the decomposition ?? . By 26 , we have \[\begin{align} \Gamma(t,x,xt^{-1};0,0,xt^{-1})&\geq\widetilde{\Gamma}(t,x,xt^{-1};0,0,xt^{-1})-\|V\|_{L^{\infty}}\cdot x^{-4}t^{4}\\ &=\frac{\sqrt{3}}{2\pi t^{2}}-\|V\|_{L^{\infty}}\cdot x^{-4}t^{4} \end{align}\] Hence 56 holds for all \(0<t\leq 1\) and \(x\geq Mt^{\frac{3}{2}}\).

Applying the strong maximal principle due to Nirenberg [46] in the \(y\)–direction as well as \(y_{0}\)–direction (by considering the dual equation of \(\Gamma(\cdot,w)\) for \(w\)–variable), we have \[\label{positivedomain} \Gamma(t,x,y;0,0,y_{0})>0,\;\forall\;0<t\leq 1,\;x>Mt^{\frac{3}{2}},\;y>0,\;y_{0}>0.\tag{57}\]

Next, define the dyadic time sequence \[t_{k}=2^{-k},\;\;k\geq 0.\] By 57 , taking an integer \(N\gg1\), we have \[\Gamma(t_{N},x,y;0,0,y_{0})>0,\;\forall\;x>M\cdot 2^{-\frac{3}{2}N},\;y>0,\;y_{0}>0.\] By the reproducing formula and the translation invariance in time of the fundamental solution, for any \(x>2M\cdot 2^{-\frac{3}{2}N}\), and \(y>0,y_{0}>0\), we have \[\label{eqreproduce} \begin{align} \Gamma(t_{N-1},x,y;0,0,y_{0})&=\int_{0}^\infty\int_{0}^x\Gamma(t_{N},x-\xi,y;0,0,\eta)\Gamma(t_{N},\xi,\eta;0,0,y_{0})\,d\xi d\eta\\ &\geq \int_{0}^\infty\int_{ 2^{-3N/2}M}^{x- 2^{-3N/2}M}\Gamma(t_{N},x-\xi,y;0,0,\eta)\Gamma(t_{N},\xi,\eta;0,0,y_{0})\,d\xi d\eta\\ &>0. \end{align}\tag{58}\]

Repeating the iteration procedure for \(N\) times, we have \[\Gamma(1,x,y;0,0,y_{0})>0,\;\forall x>M\cdot 2^{-\frac{1}{2}N},\;y,y_{0}>0,\] contradiction with case (\(\alpha\)), provided that \(N\gg 1\) is chosen sufficiently large.

Step 2: excluding case (\(\beta\)). Assume case (\(\beta\)) happens, then by the reproductive formula of the fundamental solution, we obtain that for all \(0<\tau<t\), \[0=\Gamma(\bar{t},\bar{x},\bar{y};0,0,0)=\int_{0<\xi<\bar{x}}\int_{\eta>0}\Gamma(t-\tau,\bar{x}-\xi,\bar{y};0,0,\eta)\Gamma(\tau,\xi,\eta;0,0,0)\,d\xi d\eta.\] Since the case (\(\alpha\)) is already excluded, we have for all \(0<\tau<\bar{t},0<x<\bar{x}\) and \(y>0\) that \(\Gamma(t,x,y;0,0,0)=0\).

By the dilation invariance of \(\Gamma\), it follows that \[\Gamma(t,x,y;0,0,0)\equiv 0,\;\forall t>0,x>0,y>0,\] which contradicts the fact that \[\int_{0}^{\infty}\int_{0}^{\infty}\Gamma(t,x,y;0,0,0)\,dxdy=\int_{-\infty}^{\infty}\int_{0}^{\infty}\Gamma(t,x,y;0,0,0)\,dxdy\equiv 1,\;\forall t>0.\]

Step 3: excluding case (\(\gamma\)). Notice that \[\partial_{y}\Gamma|_{y=0}\equiv 0.\] Case \(\gamma\) can then be ruled out using the classical Hopf’s boundary point lemma [56] (see also [57], [58]).

Step 4: excluding case (\(\theta\)). It follows from Step 2 that \[\Gamma(\bar{t},\bar{x},0;0,0,0)=\int_{0<\xi<\bar{x}}\int_{\eta>0}\Gamma(\frac{\bar{t}}{2},x-\xi,0;0,0,\eta)\Gamma(\frac{\bar{t}}{2},\xi,\eta;0,0,0)d\xi d\eta>0.\] Combining Step 1 – Step 4, the proof of Proposition 5 is then complete. ◻

Proof. If \((z,w)\) satisfies the assumptions, then by dilation invariance (see Proposition 1), \[\Gamma(z;w)\geq r^{-4}\min_{(Z,W)\in\mathcal{Q}}\Gamma(Z;W),\] where \(\mathcal{Q}\) is the compact subset defined by \[\begin{align} \mathcal{Q}=\Big\{Z=(t,x,y),\;W=(t_{0},x_{0},y_{0}):\,\, &\|Z\|\leq M, \|W\|\leq M, \\ &t\geq-\alpha,t_{0}\leq-\beta,\gamma\leq x\leq \delta, |x_{0}|\leq \kappa, 0\leq y,y_{0}\leq \iota \Big\}. \end{align}\] By Proposition 5, \(\Gamma(z;w)\) is positive in \(\mathcal{Q}\). Hence the (positive) minimal value of \(\Gamma(z;w)\) in \(\mathcal{Q}\) can be attained by the regularity of \(\Gamma\) stated in Proposition 2 and Proposition 3. The proof of Corollary 1 then follows. ◻

Proof. We remark that he proof of Lemma 4 is similar to [49] (the interior estimate case). For completeness, we give a self–contained proof below.

Step 1: energy estimates. For \(0<r_{2}<r_{1}\leq 1\), choose a smooth cut–off function \(\chi(t,x,y)=\chi_{1}(t)\chi_{2}(x)\chi_{3}(y)\) such that \[\begin{align} &\mathrm{supp}(\chi_{1})\subset (-r_{1}^{2},0],\;\chi_{1}(t)\equiv 1\;\text{for}\;t\in[-r_{2}^{2},0],\\ &\mathrm{supp}(\chi_{2})\subset (-r_{1}^{3},r_{1}^{3}),\;\chi_{2}(x)\equiv 1\;\text{for}\;x\in[-r_{2}^{3},r_{2}^{3}],\\ &\mathrm{supp}(\chi_{3})\subset [0,r_{1}),\;\chi_{3}(y)\equiv 1\;\text{for}\;y\in[0,r_{2}]. \end{align}\] Multiplying ?? by \(w\chi^{2}\), and integrating on \(\mathcal{B}_{r_{1}}^{+}\), for any \(t\in(-r_{1}^{2},0]\) we obtain that \[\begin{align} &\iint_{[-r_{1}^{3},r_{1}^{3}]\times[0,r_{1}]}|w|^{2}(0,x,y)\chi^{2}dxdy+\iiint_{\mathcal{B}_{r_{1}}^{+}}a(t,x,y)|\partial_{y}w|^{2}\chi^{2}dtdxdy\\ &\leq 2\iiint_{\mathcal{B}_{r_{1}}^{+}}|w|^{2}\chi(\partial_{t}\chi+y\partial_{x}\chi)\,dtdxdy-2\iiint_{\mathcal{B}_{r_{1}}^{+}}a(t,x,y)w\partial_{y}w\cdot \chi\partial_{y}\chi \,dtdxdy. \end{align}\] Direct calculations show that \[\label{caccioppoli} \iiint_{\mathcal{B}_{r_{2}}^{+}}|\partial_{y}w|^{2}\chi^{2}dtdxdy\lesssim(r_{1}-r_{2})^{-3}\iiint_{\mathcal{B}_{r_{1}}^{+}}|w|^{2}dtdxdy.\tag{60}\]

Step 2: Sobolev–type inequality. Recall the fundamental solution \(\Gamma(\cdot,\cdot)\) for the operator \(\mathcal{L}_{0}=\partial_{t}+y\partial_{x}-\partial_{y}^{2}\) with the Neumann boundary condition (see Section 2). For almost every \((t,x,y)\in \mathcal{B}_{r_{2}}^{+}\), we have \[\begin{align} w(t,x,y)=(w\chi)(t,x,y)&=\iiint_{\mathcal{B}_{r_{1}}^{+}}\Gamma(t,x,y;\tau,\xi,\eta)\left(\partial_{\tau}+\eta\partial_{\xi}-\partial_{\eta}^{2}\right)(w\chi)\,d\tau d\xi d\eta\\ &=\iiint_{\mathcal{B}_{r_{1}}^{+}}\Gamma(t,x,y;\tau,\xi,\eta)\left(\partial_{\tau}+\eta\partial_{\xi}\right)(w\chi)\,d\tau d\xi d\eta\\ &\quad+\iiint_{\mathcal{B}_{r_{1}}^{+}}\partial_{\eta}\Gamma(t,x,y;\tau,\xi,\eta)\partial_{\eta}(w\chi)\,d\tau d\xi d\eta:=\mathcal{T}+\mathcal{D}. \end{align}\] Moreover, we have by integration by parts, \[\begin{align} &\mathcal{T}\leq \iiint_{\mathcal{B}_{r_{1}}^{+}}\Gamma(t,x,y;\tau,\xi,\eta)w(\tau,\xi,\eta)\left(\partial_{\tau}+\eta\partial_{\xi}\right)\chi\,d\tau d\xi d\eta\\ &\quad -\iiint_{\mathcal{B}_{r_{1}}^{+}}a(\tau,\xi,\eta)\partial_{\eta}\Gamma(t,x,y;\tau,\xi,\eta)\cdot \partial_{\eta}w(\tau,\xi,\eta)\cdot\chi(\tau,\xi,\eta)d\tau d\xi d\eta\\ &\quad-\iiint_{\mathcal{B}_{r_{1}}^{+}}a(\tau,\xi,\eta)\Gamma(t,x,y;\tau,\xi,\eta)\cdot \partial_{\eta}w(\tau,\xi,\eta)\cdot\partial_{\eta}\chi(\tau,\xi,\eta)d\tau d\xi d\eta=:I_1+I_2+I_3, \end{align}\] and \[\begin{align} \mathcal{D}&=\iiint_{\mathcal{B}_{r_{1}}^{+}}\partial_{\eta}\Gamma(t,x,y;\tau,\xi,\eta)\partial_{\eta}w(\tau,\xi,\eta)\cdot \chi(\tau,\xi,\eta)d\tau d\xi d\eta\\ &\qquad+\iiint_{\mathcal{B}_{r_{1}}^{+}}\partial_{\eta}\Gamma(t,x,y;\tau,\xi,\eta)w(\tau,\xi,\eta)\cdot \partial_{\eta}\chi(\tau,\xi,\eta)d\tau d\xi d\eta=:I_4+I_5. \end{align}\] In above, we have used the boundary condition of \(w\) and the assumption that \(w\) is the subsolution of ?? .

By Proposition 4, we have the following bounds \[\label{eq:hIs} \begin{align} &\|I_{1}\|_{L^{6}(\mathbb{H})}\lesssim (r_{1}-r_{2})^{-3}\|w\|_{L^{2}(\mathcal{B}_{r_{1}}^{+})},\quad \|I_{2}\|_{L^{3}(\mathbb{H})}\lesssim \|\partial_{y}w\|_{L^{2}(\mathcal{B}_{r_{1}}^{+})},\\ &\|I_{3}\|_{L^{6}(\mathbb{H})}\lesssim (r_{1}-r_{2})^{-1}\|\partial_{y}w\|_{L^{2}(\mathcal{B}_{r_{1}}^{+})},\quad \|I_{4}\|_{L^{3}(\mathbb{H})}\lesssim \|\partial_{y}w\|_{L^{2}(\mathcal{B}_{r_{1}}^{+})},\\ &\|I_{5}\|_{L^{3}(\mathbb{H})}\lesssim (r_{1}-r_{2})^{-1}\|w\|_{L^{2}(\mathcal{B}_{r_{1}}^{+})}. \end{align}\tag{61}\] It follows from 61 that \[\|w\|_{L^{3}(\mathcal{B}_{r_{2}}^{+})}\leq\|w\chi\|_{L^{3}(\mathbb{H})}\lesssim (r_{1}-r_{2})^{-3}\big(\|w\|_{L^{2}(\mathcal{B}_{r_{1}}^{+})}+\|\partial_{y}w\|_{L^{2}(\mathcal{B}_{r_{1}}^{+})}\big).\]

Step 3: A reversed Hölder’s inequality. For fixed \(0<r_{2}<r_{1}\leq 1\), by Step 1 and Step 2, we obtain that \[\label{reverseholder} \begin{align} \|w\|_{L^{3}(\mathcal{B}_{r_{2}}^{+})}&\lesssim (r_{1}-r_{2})^{-3}\Big(\|w\|_{L^{2}(\mathcal{B}_{\frac{r_{1}+r_{2}}{2}}^{+})}+\|\partial_{y}w\|_{L^{2}(\mathcal{B}_{\frac{r_{1}+r_{2}}{2}}^{+})}\Big)\lesssim (r_{1}-r_{2})^{-\frac{9}{2}}\|w\|_{L^{2}(\mathcal{B}_{r_{1}}^{+})}. \end{align}\tag{62}\]

Step 4: Moser’s Iteration. Set \[r^{(n)}=r_{2}+2^{-n}(r_{1}-r_{2}),\;\sigma_{n}=\left(3/2\right)^{n},\;w_{n}=w^{\sigma_{n}}.\] One can check that \(w_{n}\) are also nonnegative subsolutions of ?? . By 62 , we have \[\|w_{n}\|_{L^{3}(\mathcal{B}_{r^{(n+1)}}^{+})}\lesssim (r^{(n)}-r^{(n+1)})^{-\frac{9}{2}}\|w_{n}\|_{L^{2}(\mathcal{B}_{r^{(n)}}^{+})}.\] Equivalently, \[\|w\|_{L^{2\sigma_{n+1}}(\mathcal{B}_{r^{(n+1)}}^{+})}\lesssim (r^{(n)}-r^{(n+1)})^{-\frac{9}{2\sigma_{n}}}\|w\|_{L^{2\sigma_{n}}(\mathcal{B}_{r^{(n)}}^{+})}.\] Sending \(n\to\infty\), we get \[\begin{align} \sup_{\mathcal{B}_{r_{2}}^{+}}w&\lesssim\left(\prod_{k=0}^{\infty}(r^{(k)}-r^{(k+1)})^{-\frac{9}{2\sigma_{k}}}\right)\|w\|_{L^{2}(\mathcal{B}_{r_{1}}^{+})}\lesssim (r_{1}-r_{2})^{-l}\|w\|_{L^{2}(\mathcal{B}_{r_{1}}^{+})}, \end{align}\] for some \(l>0\). The proof of Lemma 4 is now complete. ◻

Proof of Lemma 6. One can check that \[\label{subsoleq} \partial_{t}(w\psi-H)+y\partial_{x}(w\psi-H)-\partial_{y}^{2}(w\psi-H)\leq \partial_{y}\left((a-1)\partial_{y}w\right)\psi-2\partial_{y}w\partial_{y}\psi.\tag{63}\] Then ?? follows from testing 63 by \((w\psi-H)^{+}\), and integrating over \(t\in[-\frac{1}{4},0], x\in\mathbb{R}, y\in\mathbb{R}^{+}\). For more details we refer to Lemma 3.1 in [50]. ◻

Proof. Let \(\psi(t,x,y)\) be a cut–off function in \(\mathcal{B}^+_{\frac{1}{2}}(0,0,0)\) such that

• \(\psi\) is compactly supported in \(\mathcal{B}_{\frac{1}{2}}^{+}(0,0,0)\), and \(\psi|_{\mathcal{B}_{\theta^{2}}^{+}(0,\frac{99}{100}\theta^{3},0)}\equiv 1\),

• \(\psi(t,x,y)\equiv 0\) if \(t\leq-\theta^{2}\),

• \(|\partial_{y}\psi,\partial_{y}^{2}\psi|\lesssim 1\), \(\partial_{t}\psi+y\partial_{x}\psi\) is non–negative, and

  \[\partial_{t}\psi+y\partial_{x}\psi\gtrsim \frac{1}{\theta^{2}},\;\forall t\in (-\frac{1}{100}\theta^{2},-\frac{1}{200}\theta^{2}),\;|x|<\theta^{3}, 0\leq y<\theta.\]

As a concrete example, we can choose \[\psi(t,x,y)=\chi\left(\frac{x^{2}}{\theta}-\frac{t}{\theta^{2}}\right)\chi(y),\;\mathrm{supp}(\chi)\subset [0,\frac{1}{10}), \,\chi|_{[0, \frac{1}{1000}]}\equiv 1,\,-\chi'\gtrsim 1 \,{\rm for}\,y\in[1/500,1/5],\] with \(\theta\in(0,1)\) sufficiently small.

Without loss of generality, we can assume that \(\sup_{\mathcal{B}_{1}^{+}}w=1\). By superimposition principle, we can decompose \[H(t,x,y)=\psi(t,x,y)-P(t,x,y)+E(t,x,y),\] where \(P\) satisfies \[\begin{cases} \partial_{t}P+y\partial_{x}P-\partial_{y}^{2}P=(\partial_{t}\psi+y\partial_{x}\psi)(1-w),t>-\frac{1}{4},\;x\in\mathbb{R},\;y\in\mathbb{R}^{+},\\ P|_{t=-\frac{1}{4}}=0,\;\,\partial_{y}P|_{y=0}, \end{cases}\] and \(E\) satisfies \[\begin{cases} \partial_{t}E+y\partial_{x}E-\partial_{y}^{2}E=(1-w)\partial_{y}^{2}\psi, t>-\frac{1}{4},\;x\in\mathbb{R},\;y\in\mathbb{R}^{+},\\ E|_{t=-\frac{1}{4}}=0,\;\partial_{y}E|_{y=0}. \end{cases}\]

We first consider the \(P\) term. Write \[P(t,x,y)=\int_{-\frac{1}{4}}^{t}\int_{-\infty}^{\infty}\int_{0}^{\infty}\Gamma(t,x,y;\tau,\xi,\eta)(\partial_{\tau}\psi+\eta\partial_{\xi}\psi)(1-w) (\tau,\xi,\eta)\,d\eta d\xi d\tau.\] By the properties of \(\psi\), denoting \(z=(t,x,y), \,w=(\tau,\xi,\eta)\), for each \((t,x,y)\in\mathcal{B}_{\theta^{2}}^{+}(0,\frac{99}{100}\theta^{3},0)\), we have \[P(t,x,y)\gtrsim \int_{-\frac{1}{100}\theta^{2}}^{-\frac{1}{200}\theta^{2}}\iint_{\left\{w(\tau,\xi,\eta)=0,\;|\xi|<\frac{49}{50}\theta^{3},\;0<\eta<\theta\right\}}\Gamma(z;w)\theta^{-2}d\tau d\xi d\eta.\] By Corollary 1, and ?? , for each \((t,x,y)\in\mathcal{B}_{\theta^{2}}^{+}(0,\frac{99}{100}\theta^{3},0)\), we have \[P(t,x,y)\gtrsim \int_{-\frac{1}{100}\theta^{2}}^{-\frac{1}{200}\theta^{2}}\iint_{\left\{w(\tau,\xi,\eta)=0,\;|\xi|<\frac{49}{50}\theta^{3},\;0<\eta<\theta\right\}}\theta^{-6}d\tau d\xi d\eta\geq c_{0}>0.\]

We now turn to the bounds on \(E\). By the property of fundamental solution ?? , \[\begin{align} E&\leq 2\int_{-\theta^{2}}^{0}\iint_{\mathbb{R}\times\mathbb{R}^{+}}\Gamma(t,x,y;\tau,\xi,\eta)|\partial_{\eta}^{2}\psi|d\tau d\xi d\eta\\ &\lesssim \int_{-\theta^{2}}^{0}\iint_{\mathbb{R}\times\mathbb{R}^{+}}\Gamma(t,x,y;\tau,\xi,\eta)d\tau d\xi d\eta\leq c_{1}\theta^{2}. \end{align}\] Hence, for each \((t,x,y)\in \mathcal{B}_{\theta^{2}}^{+}(0,\frac{99}{100}\theta^{3},0)\), we have \[H(t,x,y)\leq 1-c_{0}+c_{1}\theta^{2}\leq \lambda_{0}<1,\] provided that \(\theta\in(0,1)\) is chosen sufficiently small. ◻

Proof. The proof is essentially the same as that for Proposition 4.2 in [4] and we present the details here for the reader’s convenience.

For \(h\in(0,\frac{1}{2})\) and \(r\in(0,\theta)\), define \[\begin{align} &C_{r}:=\left\{(x,y):|x|<r^{3},\;0<y<r\right\},\\ &C_{r,h}(t):=\left\{(x,y)\in C_{r},\;w(t,x,y)\geq h\right\},\\ &\mu_{r}(t):=\left|\left\{(x,y)\in C_{r},\;w(t,x,y)\geq 1\right\}\right|. \end{align}\] We introduce the auxiliary function \[\widetilde{w}(t,x,y):=\log^{+}\frac{1}{w+h^{\frac{101}{100}}}.\] One can check that \[\label{tildew} \begin{cases} \partial_{t}\widetilde{w}+y\partial_{x}\widetilde{w}-\partial_{y}(a\partial_{y}\widetilde{w})+a(\partial_{y}\widetilde{w})^{2}\leq 0,\\ \partial_{y}\widetilde{w}|_{y=0}=0,\\ 0\leq\widetilde{w}\leq\log h^{-\frac{101}{100}}. \end{cases}\tag{64}\] Let \(\chi(y)\) be smooth the cut–off function such that \[\chi(y)\equiv 1,\;0\leq y\leq\beta\theta,\quad \chi(y)\equiv 0,\;y\geq \theta,\quad |\chi'(y)|\leq \frac{2}{(1-\beta)\theta}.\]

Multiplying 64 with \(\chi^{2}(y)\) and integrating in \((\tau,t)\times C_{\theta}\) where \(-\theta^{2}<\tau<t<1\), we obtain the inequality \[\label{massesti} \begin{align} &\int_{C_{\theta}}\chi^{2}(y)\widetilde{w}(t,x,y)dxdy+\frac{c}{2}\int_{\tau}^{t}\int_{C_{\theta}}\chi^{2}|\partial_{y}\widetilde{w}|^{2}dsdxdy\\ &\leq\int_{C_{\theta}}\chi^{2}(y)\widetilde{w}(\tau,x,y)dxdy+\frac{16}{c\beta^{4}(1-\beta)^{2}}|C_{\beta\theta}|-\int_{\tau}^{t}\int_{0}^{\theta}\chi^{2}y\widetilde{w}\big|_{x=-\theta^{3}}^{x=\theta^{3}}\,dyds. \end{align}\tag{65}\] We first estimate the last term on the right hand side. Thanks to \(0\leq\widetilde{w}\leq\log h^{-\frac{101}{100}}\), \[-\int_{\tau}^{t}\int_{0}^{\theta}\chi^{2}y\widetilde{w}\big|_{x=-\theta^{3}}^{x=\theta^{3}}dyds\leq\big(\log h^{-\frac{101}{100}}\big)(t-\tau)\int_{0}^{\theta}ydy\leq \frac{1}{4}\beta^{-4}\big(\log h^{-\frac{101}{100}}\big)|C_{\beta\theta}|.\]

On the other hand, notice that \[\begin{align} \int_{-\theta^{2}}^{-\frac{1}{100}\theta^{2}}\mu_\theta(t)dt&=\int_{-\theta^{2}}^{0}\mu_\theta(t)dt-\int_{-\frac{1}{100}\theta^{2}}^{0}\mu_\theta(t)dt\\ &\geq \frac{1}{2}|\mathcal{B}_{\theta}^{+}|-\frac{1}{100}\theta^{2}|C_{\theta}|=\frac{49}{100}\theta^{2}|C_{\theta}|. \end{align}\] By mean value’s theorem, there exists \(\tau^{\ast}\in(-\theta^{2},-\frac{1}{100}\theta^{2})\), such that \[\mu_\theta(\tau^{\ast})\geq \frac{49}{99}|C_{\theta}|.\] As a consequence, \[\int_{C_{\theta}}\widetilde{w}(\tau,x,y)dxdy=\int_{C_{\theta}\cap\{w<1\}}\widetilde{w}(\tau,x,y)dxdy\leq\frac{50}{99}\big(\log h^{-\frac{101}{100}}\big)|C_{\theta}|.\] By virtue of the mass estimate 65 , we have \[\begin{align} &\big(\log\frac{1}{h+h^{\frac{101}{100}}}\big)|C_{\beta\theta}\setminus C_{\beta,h}(t)|\leq \int_{C_{\beta\theta}}\widetilde{w}(t,x,y)dxdy\leq\int_{C_{\theta}}\chi^{2}\widetilde{w}dxdy\\ &\leq\Big(\frac{50}{99}\beta^{-4}+\frac{1}{4}\beta^{-4}+\frac{16}{c\beta^{4}(1-\beta)^{2}\log h^{-\frac{101}{100}}}\Big)\big(\log h^{-\frac{101}{100}}\big)|C_{\beta\theta}|. \end{align}\] Note that \[\lim_{h\to 0+}\Big(\frac{50}{99}\beta^{-4}+\frac{1}{4}\beta^{-4}+\frac{16}{c\beta^{4}(1-\beta)^{2}\log h^{-\frac{101}{100}}}\Big)\cdot\frac{\log h^{-\frac{101}{100}}}{\log(h+h^{\frac{101}{100}})^{-1}}<\frac{99}{100},\] hence the conclusion holds by choosing \(h\in(0,\frac{1}{2})\) sufficiently small. ◻

Proof of Lemma 5. Define \[V(t,x,y)=\log^{+}\left(\frac{h}{w+h^{2}}\right),\] where \(0<h<h_{0}(c,\theta)\) will be determined later. By Lemma 8, we have \[\label{slidevanishing} \Big|\Big\{V(t,x,y)=0,\;|x|<\frac{49}{50}\theta^{3},\;0<y<\theta\Big\}\Big|\geq\frac{1}{100}\theta^{4},\;\text{for a.e.}\;-\frac{1}{100}\theta^{2}<t<0.\tag{66}\] Notice that \[\label{logeq} \begin{cases} \partial_{t}V+y\partial_{x}V-\partial_{y}(a(t,x,y)\partial_{y}V)+a(t,x,y)(\partial_{y}V)^{2}\leq0,\;\text{in}\; \mathcal{B}_{1}^{+},\\ \partial_{y}V|_{y=0}=0. \end{cases}\tag{67}\] Thus \(V\) is a nonnegative subsolution of ?? . By Lemma 6 and Lemma 7, for a suitable \(0<\lambda_{0}<1\), we have \[\label{Vestimat} \iiint_{\mathcal{B}_{\theta^{2}}^{+}(0,\frac{99}{100}\theta^{3},0)}\big|(V-\lambda_{0}\log\frac{1}{h})^{+}\big|^{2}dtdxdy\lesssim \iiint_{\mathcal{B}_{\frac{1}{2}}^{+}}|\partial_{y}V|^{2}dtdxdy.\tag{68}\] To estimate the right hand side of 68 , let \(\chi(\cdot)\) be a standard cut–off function supported in \((-1,1)\) with \(\chi|_{[-\frac{1}{2},\frac{1}{2}]}\equiv 1\). Testing 67 by \(\chi(t)\chi(x)\chi^{2}(y)\), we have for some \(c_0\in(0,1)\), \[\begin{align} &c\iiint_{\mathcal{B}_{1}^{+}}|\partial_{y}V|^{2}\chi(t)\chi(x)\chi^{2}(y)\,dtdxdy\leq \iiint_{\mathcal{B}_{1}^{+}}a(t,x,y)|\partial_{y}V|^{2}\chi(t)\chi(x)\chi^{2}(y)dtdxdy\\ &\leq\left|\iiint_{\mathcal{B}_{1}^{+}}V(t,x,y)\left(\chi'(t)\chi(x)\chi^{2}(y)+y\chi(t)\chi'(x)\chi^{2}(y)\right)dtdxdy\right|\\ &\quad\quad\quad\;+2\iiint_{\mathcal{B}_{1}^{+}}a(t,x,y)\partial_{y}V(t,x,y)\chi(t)\chi(x)\chi(y)\chi'(y)dtdxdy\\ &\leq \left|\iiint_{\mathcal{B}_{1}^{+}}V(t,x,y)\left(\chi'(t)\chi(x)\chi^{2}(y)+y\chi(t)\chi'(x)\chi^{2}(y)\right)dtdxdy\right|\\ &\quad\quad\quad\;+\frac{c}{2}\iiint_{\mathcal{B}_{1}^{+}}|\partial_{y}V|^{2}\chi(t)\chi(x)\chi^{2}(y)+c^{-3}\iiint_{\mathcal{B}_{1}^{+}}\chi(t)\chi(x)|\chi'(y)|^{2}dtdxdy. \end{align}\] Hence, \[\label{vybound} \iiint_{\mathcal{B}_{1}^{+}}|\partial_{y}V|^{2}\chi(t)\chi(x)\chi^{2}(y)\leq 50\log\left(\frac{1}{h}\right),\tag{69}\] provided \(h\in(0,1)\) small enough.

Combining 68 and 69 , and applying Lemma 4 (with a suitable rescaling), we can conclude that there exists a positive constant \(C(\theta)\) such that \[\label{eq:hol3} \sup_{\mathcal{B}_{\frac{\theta^{2}}{2}}^{+}(0,\frac{99}{100}\theta^{3},0)}V(t,x,y)\leq \lambda_{0}\log\frac{1}{h}+C(\theta)\left(\log\frac{1}{h}\right)^{\frac{1}{2}},\tag{70}\] for any \(h\leq h(\theta)\). 70 implies that \[\inf_{\mathcal{B}_{\frac{\theta^{2}}{2}}^{+}(0,\frac{99}{100}\theta^{3},0)}w(t,x,y)\geq h^{*},\] for some \(h^{*}=h^{*}(\theta)\). This completes the proof of Proposition 5. ◻

Proof. For \(0<r\leq 1\), \(x_0\in \mathbb{R}\), we denote \[\label{shiftball} \mathcal{B}^S_r(0,x_0,0):= \mathcal{B}^+_r(0,x_0-\frac{99}{100}\theta^3,0),\quad \mathcal{B}^S_r:=\mathcal{B}^S_r(0,0,0).\tag{71}\] Define \[M=\sup_{\mathcal{B}_{\theta}^{S}}w,\;m=\inf_{\mathcal{B}_{\theta}^{S}}w,\;w^{+}=2\frac{w-m}{M-m}+\frac{h^{\ast}}{2},\;w^{-}=2\frac{M-w}{M-m}+\frac{h^{\ast}}{2},\] where \(\theta>0, h^{\ast}>0\) are same as in Lemma 5.

Case 1: At least one of \(w^{+}\) and \(w^{-}\) achieves a negative value at some point in \(\mathcal{B}_{1}^{S}\). Assume without loss of generality that \(w^{+}\) is negative at some point in \(\mathcal{B}_{1}^{S}\), it follows that \[\inf_{\mathcal{B}_{1}^{S}}w\leq m-\frac{h^{\ast}}{4}(M-m).\] Hence for sufficiently small \(\theta>0\), \[\mathrm{osc}_{\mathcal{B}_{2}^{+}}w\geq\mathrm{osc}_{\mathcal{B}_{1}^{S}}w\geq \big(1+\frac{h^{\ast}}{4}\big)\cdot\mathrm{osc}_{\mathcal{B}_{\theta}^{S}}w\geq \big(1+\frac{h^{\ast}}{4}\big)\cdot\mathrm{osc}_{\mathcal{B}_{\theta^2}^{+}}w.\]

Case 2: Both \(w^{+}\) and \(w^{-}\) are non–negative in \(\mathcal{B}_{1}^{S}\). Clearly, at least one of the two satisfies ?? . Without loss of generality, assume that \(w^{+}\) satisfies ?? . Then Proposition 5 implies that \[w\geq \frac{h^{\ast}}{4}(M-m)+m,\;\text{in}\;\mathcal{B}_{\frac{\theta^{2}}{2}}^{S}\big(0,\frac{99}{100}\theta^{3},0\big)=\mathcal{B}_{\frac{\theta^{2}}{2}}^{+}.\] It follows that \[\label{oscesti1} \mathrm{osc}_{\mathcal{B}_{\frac{\theta^{2}}{2}}^{+}}w\leq M-\big(\frac{h^{\ast}}{4}(M-m)+m\big)=\big(1-\frac{h}{2}\big)(M-m)\leq\big(1-\frac{h^{\ast}}{4}\big)\mathrm{osc}_{\mathcal{B}_{2}^{+}}w.\tag{72}\] Hence the conclusion of Corollary 2 follows from 72 . ◻

Proof. Taking Fourier transform with respect to the \(x\)–variable, we get \[\label{fteq} \begin{cases} \partial_{t}\widehat{w}(t,\xi,y)+i2\pi y\xi\widehat{w}(t,\xi,y)-\partial_{y}^{2}\widehat{w}(t,\xi,y)=\widehat{g\,}(t,\xi,y),\\ \partial_{y}\widehat{w}|_{y=0}=0. \end{cases}\tag{97}\]

We need to prove that for any \(\xi\in\mathbb{R}\), \[\label{fourieresti} \|\partial_{y}^{2}\widehat{w}\|_{L^{2}_{t,y}}\lesssim \|\widehat{g\,}\|_{L^{2}_{t,y}}.\tag{98}\] Lemma 11 follows from 98 and the Plancherel’s identity.

Notice that 98 is obvious when \(\xi=0\) by the standard energy estimates for the one dimensional heat equation.

We assume without loss of generality \(\xi>0\), and apply the enhanced dissipation estimate (Proposition 20) and Duhamel’s principle. It follows that for some \(\delta_0>0\), \[\|\widehat{w}(t,\xi,\cdot)\|_{L^{2}_{y}}\lesssim \int_{-\infty}^{t}e^{-\delta_0\xi^{\frac{2}{3}}(t-s)}\|\widehat{g\,}(s,\xi,\cdot)\|_{L^{2}_{y}}ds.\] By Young’s inequality, we conclude that \[\label{enhancebound} \|\widehat{w}(\cdot,\xi,\cdot)\|_{L^{2}_{t,y}}\lesssim \xi^{-\frac{2}{3}}\|\widehat{g\,}(\cdot,\xi,\cdot)\|_{L^{2}_{t,y}}.\tag{99}\]

Next we turn to the \(H^{2}_{y}\) estimate. Testing 97 by \(-\overline{\partial_{y}^{2}\widehat{w}}\), integraing by parts and taking the real part, we have (with \(D:=\mathbb{R}\times[0,\infty)\)) \[\iint_{D} |\partial_{y}^{2}\widehat{w}|^{2}dtdy\lesssim\xi\iint_D |\widehat{w}||\partial_{y}\widehat{w}|\,dtdy+\iint_D |\widehat{g\,}|^{2}dtdy.\] Thanks to \(\partial_{y}\widehat{w}|_{y=0}=0\), applying the interpolation inequality in \(y\)–direction, and combining with the enhanced dissipation estimate 99 , we obtain that \[\label{enhancedl2added} \begin{align} \iint_D |\partial_{y}^{2}\widehat{w}|^{2}dtdy&\lesssim \xi \|\widehat{w}\|_{L^{2}_{t,y}}\|\partial_{y}\widehat{w}\|_{L^{2}_{t,y}}+\|\widehat{g\,}\|_{L^{2}_{t,y}}^{2}\lesssim \xi \|\widehat{w}\|_{L^{2}_{t,y}}^{\frac{3}{2}}\|\partial_{y}^{2}\widehat{w}\|_{L^{2}_{t,y}}^{\frac{1}{2}}+\|\widehat{g\,}\|_{L^{2}_{t,y}}^{2}\\ &\leq \frac{1}{2}\|\partial_{y}^{2}\widehat{w}\|_{L^{2}_{t,y}}^{2}+C\big(\xi^{\frac{4}{3}}\|\widehat{w}\|_{L^{2}_{t,y}}^{2}+\|\widehat{g\,}\|_{L^{2}_{t,y}}^{2}\big)\leq\frac{1}{2}\|\partial_{y}^{2}\widehat{w}\|_{L^{2}_{t,y}}^{2}+C\|\widehat{g\,}\|_{L^{2}_{t,y}}^{2}. \end{align}\tag{100}\] The proof of Lemma 11 follows from 100 by integrating in \(\xi\). ◻

Proof of Lemma 10. Let \(p'\) be the dual exponent of \(p\), that is \(\frac{1}{p}+\frac{1}{p'}=1\). By duality, we need to prove that, for any \(\phi(t,x,y)\in C_{0}^{\infty}(\mathbb{H})\), \[\label{dualgoal} \left|\int_{\mathbb{H}}(\partial_{y}^{2}w)(t,x,y)\phi(t,x,y)\,dxdydt\right|\lesssim \|g\|_{L^{p}(\mathbb{H})}\|\phi\|_{L^{p'}(\mathbb{H})},\tag{101}\] where with \(z=(t,x,y), \,w=(\tau,\xi,\eta)\), \[\label{expression} w(z) = \int_{-\infty}^{t}\int_{-\infty}^{\infty}\int_{0}^{\infty}\Gamma(z;w)g(w)\,d\eta d\xi d\tau.\tag{102}\]

Due to the singularity of the fundamental solution near the source, we can not take \(\partial_{y}^{2}\) into the integrand in 102 directly. To overcome this technical difficulty, for any \(\epsilon>0\), we decompose \[\label{singudecom} \begin{align} w(z)&=\int_{-\infty}^{t}\int_{-\infty}^{\infty}\int_{0}^{\infty}\chi\left(\frac{t-\tau}{\epsilon}\right)\Gamma(z;w)g(w)\, d\eta d\xi d\tau\\ &\quad+\int_{-\infty}^{t}\int_{-\infty}^{\infty}\int_{0}^{\infty}\left[1-\chi\left(\frac{t-\tau}{\epsilon}\right)\right]\Gamma(z;w)g(w)\,d\eta d\xi d\tau\\ &=:w_{1}(z)+w_{2}(z). \end{align}\tag{103}\] Where \(\chi(\cdot)\) is a standard smooth cut–off function on the line (supported in the far field), satisfying \(\chi(\sigma)= 0\,\,{\rm for}\,\, \sigma\leq 5\,\,{\rm and}\,\, \chi(\sigma)= 1\,\,{\rm for}\,\, \sigma\geq 10.\)

Hence, \[\begin{align} \left|\int_{\mathbb{H}}(\partial_{y}^{2}w)(t,x,y)\phi(t,x,y)\,dxdydt\right|&\leq \left|\int_{\mathbb{H}}(\partial_{y}^{2}w_{1})(t,x,y)\phi(t,x,y)\,dxdydt\right|\\ &\quad+\left|\int_{\mathbb{H}}(\partial_{y}^{2}w_{2})(t,x,y)\phi(t,x,y)\,dxdydt\right|\\ &=:A_{g,\phi}+B_{g,\phi}. \end{align}\]

Estimates of \(B_{g,\phi}\). For the term \(B_{g,\phi}\), integrating by parts and applying Proposition 3 and Schur’s test (Lemma 21), we get that \[\label{bgphibd} B_{g,\phi}\leq \|\partial_{y}w_{2}\|_{L^{p}}\|\partial_{y}\phi\|_{L^{p'}}\lesssim \epsilon^{\frac{1}{2}}\|g\|_{L^{p}}\|\partial_{y}\phi\|_{L^{p'}}.\tag{104}\]

Estimates of \(A_{g,\phi}\). Recall that the fundamental solution \(\Gamma(z;w)=\Gamma(t,x,y;\tau,\xi,\eta)\) satisfies different estimates for \(\eta\gtrsim \|z-w\|\) and \(\eta\lesssim \|z-w\|\). To estimate \(A_{g,\phi}\), using ?? we decompose \(w_{1}(z)\) into three parts, \[\label{decomw1} \begin{align} w_{1}(z)=&\int_{-\infty}^{t}\int_{-\infty}^{\infty}\int_{0}^{\infty}\chi\left(\frac{t-\tau}{\epsilon}\right)\left[1-\chi\left(\frac{\eta}{\|z-w\|}\right)\right]\Gamma(z;w)g(w) \,d\eta d\xi d\tau\\ &+\int_{-\infty}^{t}\int_{-\infty}^{\infty}\int_{0}^{\infty}\chi\left(\frac{t-\tau}{\epsilon}\right)\chi\left(\frac{\eta}{\|z-w\|}\right)\widetilde{\Gamma}(z,w)g(w)\,d\eta d\xi d\tau\\ &+\int_{-\infty}^{t}\int_{-\infty}^{\infty}\int_{0}^{\infty}\chi\left(\frac{t-\tau}{\epsilon}\right)\chi\left(\frac{\eta}{\|z-w\|}\right)\eta^{-4}V\Big(\frac{t-\tau}{\eta^{2}},\frac{x-\xi}{\eta^{3}},\frac{y}{\eta}\Big)g(w)\,d\eta d\xi d\tau\\ =&w_{11}(z)+w_{12}(z)+w_{13}(z). \end{align}\tag{105}\]

Notice from the definition that \(w_{12}(z)\) can be extended \((t,x,y)\in\mathbb{R}^{3}\) from \(\mathbb{H}\). By direct computation, \(w_{12}(z)\) satisfies the equation \[\label{w12eq} \begin{align} &\partial_{t}w_{12}+y\partial_{x}w_{12}-\partial_{y}^{2}w_{12}=g_{12}(z),\;(t,x,y)\in\mathbb{R}^{3},\\ &g_{12}(z)=\int_{-\infty}^{t}\int_{-\infty}^{\infty}\int_{0}^{\infty}\mathcal{K}_{12}(z,w)g(w)dw. \end{align}\tag{106}\] In the above we have used the notation that \[\label{kernelk12} \begin{align} \mathcal{K}_{12}=&\frac{1}{\epsilon}\chi'\big(\frac{t-\tau}{\epsilon}\big)\chi\big(\frac{\eta}{\|z-w\|}\big)\widetilde{\Gamma}(z,w)-\chi\big(\frac{t-\tau}{\epsilon}\big)\chi'\big(\frac{\eta}{\|z-w\|}\big)\widetilde{\Gamma}(z,w)\frac{\eta (\partial_{t}+y\partial_{x})\|z-w\|}{\|z-w\|^{2}}\\ &+2\chi\big(\frac{t-\tau}{\epsilon}\big)\chi'\big(\frac{\eta}{\|z-w\|}\big)\partial_{y}\widetilde{\Gamma}(z,w)\frac{\eta}{\|z-w\|^{2}}\partial_{y}\|z-w\|\\ &-\chi\big(\frac{t-\tau}{\epsilon}\big)\widetilde{\Gamma}(z,w)\bigg[\chi''\big(\frac{\eta}{\|z-w\|}\big)\eta^{2}\|z-w\|^{-4}(\partial_{y}\|z-w\|)^{2}\\ &+2\chi'\big(\frac{\eta}{\|z-w\|}\big)\eta\|z-w\|^{-3}(\partial_{y}\|z-w\|)^{2}-2\chi'\big(\frac{\eta}{\|z-w\|}\big)\eta\|z-w\|^{-2}(\partial_{y}^{2}\|z-w\|)\bigg]. \end{align}\tag{107}\] By the \(W^{2,p}_{y}\) estimate for the operator \(\mathcal{L}_{0}=\partial_{t}+y\partial_{x}-\partial_{y}^{2}\) in the whole space [53], we have \[\label{wholespacew2p} \|\partial_{y}^{2}w_{12}(z)\|_{L^{p}}\lesssim_{p} \|g_{12}(z)\|_{L^{p}}\lesssim \|g\|_{L^{p}}.\tag{108}\] We have employed Schur’s test in the last inequality, as well as the explicit expression of \(\widetilde{\Gamma}(z,w)\).

Next, we need to estimate \(w_{11}(z)\) and \(w_{13}(z)\). Let \[\begin{align} &T_{11}(g)(z):=\partial_{y}^{2}w_{11}(z)=\int_{-\infty}^{t}\int_{-\infty}^{\infty}\int_{0}^{\infty}\mathcal{K}_{11}(z,w)g(w)\,dw,\\ &\mathcal{K}_{11}(z,w)=\chi\big(\frac{t-\tau}{\epsilon}\big)\partial_{y}^{2}\left\{\left[1-\chi\left(\frac{\eta}{\|z-w\|}\right)\right]\Gamma(z;w)\right\}. \end{align}\] We need to deduce that \(T_{11}\) is bounded on \(L^{p}(\mathbb{H})\).

First we show that \(T_{11}\) is bounded on \(L^{2}(\mathbb{H})\). To this end, notice that \(w_{11}(z)\) satisfies the following equation \[\label{w11eq} \begin{cases} \partial_{t}w_{11}+y\partial_{x}w_{11}-\partial_{y}^{2}w_{11}=:\widetilde{g}(t,x,y),\\ \partial_{y}w_{11}|_{y=0}=0, \end{cases}\tag{109}\] where \(\widetilde{g}\) is given by \[\widetilde{g}(t,x,y)=\int_{-\infty}^{t}\int_{-\infty}^{\infty}\int_{0}^{\infty}\widetilde{\mathcal{K}}_{11}(z,w)g(w)dw,\] with \[\begin{align} \widetilde{\mathcal{K}}_{11}(z,w)=&-2\chi\big(\frac{t-\tau}{\epsilon}\big)\chi'\big(\frac{\eta}{\|z-w\|}\big)\partial_{y}\Gamma(z;w)\frac{\eta}{\|z-w\|^{2}}\partial_{y}\|z-w\|\\ &+\frac{1}{\epsilon}\chi'\big(\frac{t-\tau}{\epsilon}\big)\left[1-\chi\big(\frac{\eta}{\|z-w\|}\big)\right]\Gamma(z;w)\\ &+\chi\big(\frac{t-\tau}{\epsilon}\big)\chi'\big(\frac{\eta}{\|z-w\|}\big)\Gamma(z;w)\frac{\eta (\partial_{t}+y\partial_{x})\|z-w\|}{\|z-w\|^{2}}\\ &+\chi\big(\frac{t-\tau}{\epsilon}\big)\Gamma(z;w)\bigg[\chi''\big(\frac{\eta}{\|z-w\|}\big)\eta^{2}\|z-w\|^{-4}(\partial_{y}\|z-w\|)^{2}\\ &\quad\quad+2\chi'\big(\frac{\eta}{\|z-w\|}\big)\eta\|z-w\|^{-3}(\partial_{y}\|z-w\|)^{2}-\eta\|z-w\|^{-2}(\partial_{y}^{2}\|z-w\|)\\ &\quad\quad-2\chi'\big(\frac{\eta}{\|z-w\|}\big)\eta\|z-w\|^{-2}(\partial_{y}^{2}\|z-w\|)\bigg]. \end{align}\] By Lemma 11, equation 109 , and Schur’s test, we have \[\|T_{11}(g)\|_{L^{2}(\mathbb{H})}\lesssim \|\widetilde{g}\|_{L^{2}(\mathbb{H})}\lesssim \|g\|_{L^{2}(\mathbb{H})},\] which means \(T_{11}\) is continuous from \(L^{2}(\mathbb{H})\) to \(L^{2}(\mathbb{H})\)

On the other hand, by Proposition 2 and Proposition 3, we have the following pointwise bound and continuity bounds of the Kernel \(\mathcal{K}_{11}(z,w)\), which means the Kernel satisfies Hörmander’s condition of the non–convolution type of SIO in the homogeneous metric space \(\mathbb{H}\) equipped with the metric \(d(z,w)=\|z-w\|\) and the Lebesgue measure. \[\label{bdkernel} \begin{align} &|\mathcal{K}_{11}(z,w)|\lesssim \frac{1}{\|z-w\|^{6}},\;z\neq w,\\ &|\mathcal{K}_{11}(z,w)-\mathcal{K}_{11}(z_{0},w)|\lesssim \frac{\|z_{0}-z\|}{\|z_{0}-w\|^{7}}, \end{align}\tag{110}\] provided \(\|z-z_{0}\|\leq c\|w-z_{0}\|\); and \[\begin{align} &|\mathcal{K}_{11}(z,w)-\mathcal{K}(z,w_{0})|\lesssim \frac{\|w_{0}-w\|}{\|z-w_{0}\|^{7}}, \end{align}\] provided \(\|w-w_{0}\|\leq c\|z-w_{0}\|,\) for some \(0<c<1\).

Notice that \(T_{11}\) is bounded from \(L^{2}(\mathbb{H})\) to \(L^{2}(\mathbb{H})\). Hence, by 110 and Theorem 4.7 in [66] (see also [67], [68], [69], [70]), we have \(T_{11}\) is bounded from \(L^{p}(\mathcal{B}_{1}^{+})\) to \(L^{p}(\mathcal{B}_{1}^{+})\) for any \(p\in(1,\infty)\).

Finally, we estimate \(w_{13}(z)\). As before, define \[\begin{align} &T_{13}(g)(z):=\partial_{y}^{2}w_{13}(z)=\int_{-\infty}^{t}\int_{-\infty}^{\infty}\int_{0}^{\infty}\mathcal{K}_{13}(z,w)g(w)\,dw,\\ &\mathcal{K}_{13}(z,w)=\chi\left(\frac{t-\tau}{\epsilon}\right)\eta^{-4}\partial_{y}^{2}\left[\chi\left(\frac{\eta}{\|z-w\|}\right)V\left(\frac{t}{\eta^{2}},\frac{x}{\eta^{3}},\frac{y}{\eta}\right)\right]. \end{align}\] First, notice that \[\|T_{13}\|_{L^{2}\to L^{2}}\leq\|T_{1}\|_{L^{2}\to L^{2}}+\|T_{11}\|_{L^{2}\to L^{2}}+\|T_{12}\|_{L^{2}\to L^{2}},\quad T_{1}(g)(z)=\partial_{y}^{2}w_{1}(z).\]

Similarly to the \(L^{2}\) boundedness of \(T_{11}\), we can deduce that \(T_{1}\) is bounded on \(L^{2}(\mathbb{H})\) by employing Lemma 11 again. The \(L^{2}\) boundedness of \(T_{12}\) follows from 108 .

Thus, \(T_{13}\) is bounded on \(L^{2}(\mathbb{H})\). Furthermore, by directly calculation, the kernel \(\mathcal{K}_{13}\) also satisfies 110 . Therefore, \(T_{13}\) is bounded on \(L^{p}(\mathbb{H})\) for any \(p\in(1,\infty)\),

In summary, we conclude that the operator \(T_{1}\) is bounded on \(L^{p}(\mathbb{H})\) for any \(p\in(1,\infty)\). By Hölder’s inequality, we have \[\label{agphibound} A_{g,\phi}\leq \|T_{1}g\|_{L^{p}(\mathbb{H})}\|\phi\|_{L^{p'}(\mathbb{H})}\lesssim \|g\|_{L^{p}(\mathbb{H})}\|\phi\|_{L^{p'}(\mathbb{H})}.\tag{111}\] Combining with 104 , we obtain 101 by the arbitrary of \(\epsilon>0\). Therefore we complete the proof of Proposition 10. ◻

Proof. ?? follows from standard energy estimates, by choosing a smooth cut–off function \(\eta(y)\) such that \(\eta(y)|_{[0,y_{0}]}\equiv 1\) and \(\eta(y)\) is supported in \([0,1)\) and then testing 113 against \(w(t,x,y)\eta^{2}(y)\), using integrating by parts arguments. ◻

Proof. Fix a smooth cut–off function \(\eta(y)\) such that \(\eta(y)|_{[0,y_{0}]}\equiv 1\) and \(\eta(y)\) is supported in \([0,1)\). Let \(W=w\eta^{2}\), it is easy to check that \(W\) satisfies the following equation in \(\mathbb{H}^-\): \[\label{mjherpwv} \begin{align} &\partial_{t}W+y\partial_{x}W-\partial_{y}^{2}W=\eta^{2}f+\eta^{2}\partial_{y}\left((a-1)\partial_{y}w\right)-4\eta\eta'\partial_{y}w-\left(2\eta\eta''+2(\eta')^{2}\right)w,\\ &\partial_{y}W|_{y=0}=0. \end{align}\tag{115}\] By the superposition principle, \(W=W_{1}+W_{2}\), where on \(\mathbb{H}^-\), \(W_1\) is the solution to \[\label{W1eq} \begin{align} &\partial_{t}W_{1}+y\partial_{x}W_{1}-\partial_{y}^{2}W_1=\eta^{2}f-4\eta\eta'\partial_{y}w-\left(2\eta\eta''+2(\eta')^{2}\right)w-2\eta\eta'(a-1)\partial_{y}w,\\ &\partial_{y}W_{1}|_{y=0}=0,\\ \end{align}\tag{116}\] and \(W_2\) is the solution to \[\label{W2eq} \begin{align} &\partial_{t}W_{2}+y\partial_{x}W_{2}-\partial_{y}^{2}W_{2}=\partial_{y}\left((a(t,x,y)-1)\eta^{2}\partial_{y}w\right)\\ &\partial_{y}W_{2}|_{y=0}=0. \end{align}\tag{117}\] It suffices to prove that for \(h\in\{W_1, W_2\}\), \[\label{high2} \|(Id-S_{10})h\|_{\widetilde{H}^{\frac{1}{3}}(\mathcal{S}_{y_{0}})}\lesssim_{c,y_{0}} \|w\|_{L^{2}(\mathcal{S}_{1})}+\|\partial_{y}w\|_{L^{2}(\mathcal{S}_{1})}+\|f\|_{L^{2}(\mathcal{S}_{1})}.\tag{118}\]

Step 1: Proof of 118 for \(h=W_{1}\). By Lemma 11 and 99 , we have \[\label{estiw1} \begin{align} &\left\|\partial_{y}^{2}W_{1}\right\|_{L^{2}(\mathbb{H}^-)}+\left\||D_{x}|^{\frac{2}{3}}W_{1}\right\|_{L^{2}(\mathbb{H}^-)}\\ &\lesssim\left\|\eta^{2}f(t,x,y)-4\eta(y)\eta'(y)\partial_{y}w-\left(2\eta\eta''+2(\eta')^{2}\right)w-2\eta(y)\eta'(y)(a(t,x,y)-1)\partial_{y}w\right\|_{L^{2}(\mathbb{H}^-)}\\ &\lesssim_{c,y_{0}} \|w\|_{L^{2}(\mathcal{S}_{1})}+\|\partial_{y}w\|_{L^{2}(\mathcal{S}_{1})}+\|f\|_{L^{2}(\mathcal{S}_{1})}. \end{align}\tag{119}\] We then complete the proof of 118 in the case \(h=W_1\).

Step 2: Proof of 118 for \(h=W_{2}\). The analysis of \(W_{2}\) is more difficult due to the extra derivative in the force term.

Write \[\label{specificF} F(t,x,y)=\left(a(t,x,y)-1\right)\partial_{y}w\cdot\eta^{2}.\tag{120}\] Then \(W_{2}\) satisfies the following linear system the force term in a divergence form \[\label{toysystem} \begin{cases} \partial_{t}W_{2}+y\partial_{x}W_2-\partial_{y}^{2}W_{2}=\partial_{y}F,\\ \partial_{y}W_{2}|_{y=0}=0. \end{cases}\tag{121}\] We claim that for any \(0<|\delta|<1\), we have \[\label{differentialquo} \begin{align} &\left\|W_{2}(T,X+\delta,Y)-W_{2}(T,X,Y)\right\|_{L^{2}(\mathbb{H}^-)}\lesssim |\delta|^{\frac{1}{3}}\|F\|_{L^{2}(\mathbb{H}^-)}. \end{align}\tag{122}\] For 122 we do not need \(F\) to be given in the specific form 120 . In fact, we only need suitable decay conditions at infinity so that \(W_2\) admits the integral formula 126 below.

We assume 122 momentarily and complete the proof of 118 for \(h=W_2\). Recall \(\Delta_j\) be the standard Littlewood-Paley projection in \(x\), to the frequency \(|\xi|\sim 2^j\) (for more details, kindly see Appendix 12). As a consequence of 122 , we obtain the following estimate for the linear system 121 for all \(j\geq 10\), \[\label{besovesti} \sup_{j\geq 10}2^{\frac{1}{3}j}\|\Delta_{j}W_{2}\|_{L^{2}((-\infty,0]\times\mathbb{R}\times[0,10])}\lesssim\|F\|_{L^{2}(\mathbb{H}^-)}.\tag{123}\] Since \[\Delta_{j}=\Delta_{j}\widetilde{\Delta}_{j},\qquad\widetilde{\Delta}_{j}=\sum_{|l|\leq 2}\Delta_{j+l},\] acting \(\widetilde{\Delta}_{j}\) on 121 , and applying the linear estimate 123 once again, we obtain \[\label{refinebesovesti} 2^{\frac{1}{3}j}\|\Delta_{j}W_{2}\|_{L^{2}(\mathbb{R}^{2}\times[0,10])}\lesssim\sum_{|l|\leq 2}\|\Delta_{j+l}F\|_{L^{2}(\mathbb{H})}.\tag{124}\] Therefore, \[\label{zerow2} \sum_{j\geq 10}2^{\frac{2}{3}j}\|\Delta_{j}W_{2}\|_{L^{2}((-\infty,0]\times\mathbb{R}\times[0,10])}^{2}\lesssim\sum_{j}\|\Delta_{j}F\|_{L^{2}(\mathbb{H})}^{2}\sim \|F\|_{L^{2}(\mathbb{H})}^{2}.\tag{125}\] 118 then follows from 119 and 125 .

To complete the proof, it remains to prove the bounds 122 .

In view of 117 , for \((T,X,Y)\in\mathbb{H}^-\), we can write \[\label{Wexpress} W_{2}(T,X,Y)=-\int_{\mathbb{H}^-}(\partial_{y}\Gamma)(T,X,Y;t,x,y)F(t,x,y)dtdxdy,\tag{126}\] where \(\Gamma\) is the fundamental solution of the operator \(\mathcal{L}_{0}\) on the upper half space with zero Neumann boundary conditions from section 2.

We have \[\label{decomw2} \begin{align} &|W_{2}(T,X+\delta,Y)-W_{2}(T,X,Y)|\\ &\leq \int_{\mathbb{H}^-}|(\partial_{y}\Gamma)(T,X+\delta,Y;t,x,y)-(\partial_{y}\Gamma)(T,X,Y;t,x,y)|\cdot |F(t,x,y)|\,dtdxdy\\ &\leq W_{21}+W_{22}+W_{23}+W_{24}. \end{align}\tag{127}\] In the above, for \(j\in\{1,2,3,4\}\), \[\label{decomw23939} W_{2j}:=\int_{D_{2j}}|(\partial_{y}\Gamma)(T,X+\delta,Y;t,x,y)-(\partial_{y}\Gamma)(T,X,Y;t,x,y)|\cdot |F(t,x,y)|\,dtdxdy,\tag{128}\] \[D_{21}:= \big\{(t,x,y)\in\mathbb{H}^-:\,\|(T-t,X-x,Y-y)\|\geq 5\delta^{\frac{1}{3}}, \|(T-t,X-x-(T-t)Y,Y-y)\|\geq 5\delta^{\frac{1}{3}}\big\},\] \[D_{22}:= \big\{(t,x,y)\in\mathbb{H}^-:\,\|(T-t,X-x,Y-y)\|\leq 20\delta^{\frac{1}{3}}, \|(T-t,X-x-(T-t)Y,Y-y)\|\leq 20\delta^{\frac{1}{3}}\big\},\] \[D_{23}:= \big\{(t,x,y)\in\mathbb{H}^-:\,\|(T-t,X-x,Y-y)\|\ge 20\delta^{\frac{1}{3}}, \|(T-t,X-x-(T-t)Y,Y-y)\|\leq 5\delta^{\frac{1}{3}}\big\},\] \[D_{24}:= \big\{(t,x,y)\in\mathbb{H}^-:\,\|(T-t,X-x,Y-y)\|\leq 5\delta^{\frac{1}{3}}, \|(T-t,X-x-(T-t)Y,Y-y)\|\ge 20\delta^{\frac{1}{3}}\big\}.\]

We need to show that for \(j\in\{1,2,3,4\}\), \[\label{high3} \left\|W_{2j}\right\|_{L^{2}(\mathbb{H}^-)}\lesssim |\delta|^{\frac{1}{3}}\|F\|_{L^{2}(\mathbb{H}^-)},\tag{129}\] We consider \(W_{2j}\) case by case.

Case (I): the bound 129 for \(W_{21}\). By the mean value theorem and Proposition 3, for suitable \(\theta\in(0,1)\), we have \[\begin{align} &|(\partial_{y}\Gamma)(T,X+\delta,Y;t,x,y)-(\partial_{y}\Gamma)(T,X,Y;t,x,y)|\lesssim \delta|(\partial_{x}\partial_{y}\Gamma)(T,X+\theta\delta,Y;t,x,y)|\\ &\lesssim \delta\left(\|(T-t,X+\theta\delta-x,Y-y)\|^{-8}+\|(T-t,X+\theta\delta-x-(T-t)Y,Y-y)\|^{-8}\right)\\ &\lesssim \delta\left(\|(T-t,X-x,Y-y)\|^{-8}+\|(T-t,X-x-(T-t)Y,Y-y)\|^{-8}\right), \end{align}\] provided that \[\|(T-t,x-x,Y-y)\|\geq 5\delta^{\frac{1}{3}},\quad \|(T-t,X-x-(T-t)Y,Y-y)\|\geq 5\delta^{\frac{1}{3}}.\] Therefore, by Young’s inequality, \[\|W_{21}\|_{L^{2}_{(T,X,Y)}(\mathbb{H}^-)}\lesssim\delta\left(\int_{\|z\|\geq 5\delta^{\frac{1}{3}}}\|z\|^{-8}dtdxdy\right)\times\|F\|_{L^{2}(\mathbb{H}^-)}\lesssim\delta^{\frac{1}{3}}\|F\|_{L^{2}(\mathbb{H}^-)}.\]

Case (II): the bound 129 for \(W_{22}\). By ?? , 26 , and Proposition 3, we have \[\begin{align} &|(\partial_{y}\Gamma)(T,X+\delta,Y;t,x,y)-(\partial_{y}\Gamma)(T,X,Y;t,x,y)|\\ &\lesssim \|(T-t,X+\delta-x,Y-y)\|^{-5}+\|(T-t,X-x,Y-y)\|^{-5}\\ &\quad+\|(T-t,X+\delta-x-(T-t)Y,Y-y)\|^{-5}+\|(T-t,X-x-(T-t)Y,Y-y)\|^{-5}. \end{align}\] Hence, applying Young’s inequality again, we have \[\|W_{22}\|_{L^{2}_{(T,X,Y)}(\mathbb{H}^-)}\lesssim\Big(\int_{\|z\|\leq 25\delta^{\frac{1}{3}}}\|z\|^{-5}dtdxdy\Big)\times\|F\|_{L^{2}(\mathbb{H}^-)}\lesssim\delta^{\frac{1}{3}}\|F\|_{L^{2}(\mathbb{H}^-)}.\]

Case (III): the bound 129 for \(W_{23}\). In \(D_{23}\), we use the decomposition ?? . Set \[\label{high4} p:= \Big(\frac{T-t}{y^{2}},\frac{X+\delta-x}{y^{3}},\frac{Y}{y}\Big), \qquad q:= \Big(\frac{T-t}{y^{2}},\frac{X-x}{y^{3}},\frac{Y}{y}\Big).\tag{130}\]

By the Green decomposition ?? , we get that \[\begin{align} W_{23}&\lesssim\int_{D_{23}}|(\partial_{y}\widetilde{\Gamma})(T,X+\delta,Y;t,x,y)-(\partial_{y}\widetilde{\Gamma})(T,X,Y;t,x,y)|\cdot |F(t,x,y)|\,dtdxdy\\ &+\int_{D_{23}}y^{-5}\left|V(p)-V(q)\right|\cdot |F(t,x,y)|\,dtdxdy\\ &+\int_{D_{23}}\frac{|T-t|}{y^{7}}\left|(\partial_{1}V)(p)-(\partial_{1}V)(q)\right|\cdot |F(t,x,y)|\,dtdxdy\\ &+\int_{D_{23}}\frac{|X-x|}{y^{8}}\left|(\partial_{2}V)(p)-(\partial_{2}V)(q)\right|\cdot |F(t,x,y)|\,dtdxdy\\ &+\int_{D_{23}}\frac{Y}{y^{6}}\left|(\partial_{3}V)(p)-(\partial_{3}V)(q)\right|\cdot |F(t,x,y)|\,dtdxdy\\ &=:W_{23,1}+W_{23,2}+W_{23,3}+W_{23,4}+W_{23,5}. \end{align}\]

By 26 , we have \[\begin{align} &|(\partial_{y}\widetilde{\Gamma})(T,X+\delta,Y;t,x,y)-(\partial_{y}\widetilde{\Gamma})(T,X,Y;t,x,y)|\\ &\lesssim\|(T-t,X-x-(T-t)Y,Y-y)\|^{-5}+\|(T-t,X+\delta-x-(T-t)Y,Y-y)\|^{-5}. \end{align}\] Therefore, \(W_{23,1}\) can be estimated via Young’s inequality \[\|W_{23,1}\|_{L^{2}_{(T,X,Y)}(\mathbb{H}^-)}\lesssim\left(\int_{\|z\|\leq 10\delta^{\frac{1}{3}}}\|z\|^{-5}dtdxdy\right)\times\|F\|_{L^{2}(\mathbb{H}^-)}\lesssim\delta^{\frac{1}{3}}\|F\|_{L^{2}(\mathbb{H}^-)}.\] To estimate \(W_{23,2}\), by mean–value theorem we have for some \(\theta\in(0,1)\), \[\begin{align} &\left|V\Big(\frac{T-t}{y^{2}},\frac{X+\delta-x}{y^{3}},\frac{Y}{y}\Big)-V\Big(\frac{T-t}{y^{2}},\frac{X-x}{y^{3}},\frac{Y}{y}\Big)\right|\\ &= y^{-3}\delta\left|(\partial_{2}V)\Big(\frac{T-t}{y^{2}},\frac{X+\theta\delta-x}{y^{3}},\frac{Y}{y}\Big)\right|. \end{align}\] Notice that in \(D_{23}\), we have \[y\gtrsim \|(T-t,X-x,Y-y)\|.\] Hence, by Young’s inequality and the fact that \(|\partial_{2}V|\lesssim 1\), we have \[\|W_{23,2}\|_{L^{2}_{(T,X,Y)}(\mathbb{H}^-)}\lesssim\delta\left(\int_{\|z\|\geq 20\delta^{\frac{1}{3}}}\|z\|^{-8}dtdxdy\right)\times\|F\|_{L^{2}(\mathbb{H}^-)}\lesssim\delta^{\frac{1}{3}}\|F\|_{L^{2}(\mathbb{H}^-)}.\] The remaining terms \(W_{23,3},W_{23,4},W_{23,5}\) can be estimated in the same fashion as \(W_{22}\). Thus we obtain that \[\|W_{23}\|_{L^{2}_{(T,X,Y)}(\mathbb{H}^-)}\lesssim\delta^{\frac{1}{3}}\|F\|_{L^{2}(\mathbb{H}^-)}.\] Case (IV): the bound 129 for \(W_{24}\). The treatment of \(W_{24}\) is similar to that of \(W_{23}\), and we will be somewhat brief. As before, we decompose \[\label{w24} \begin{align} W_{24}&\lesssim\int_{D_{24}}|(\partial_{y}\widetilde{\Gamma})(T,X+\delta,Y;t,x,y)-(\partial_{y}\widetilde{\Gamma})(T,X,Y;t,x,y)|\cdot |F(t,x,y)|\,dtdxdy\\ &+\int_{D_{24}}y^{-5}\left|V(p)-V(q)\right|\cdot |F(t,x,y)|\,dtdxdy\\ &+\int_{D_{24}}\frac{|T-t|}{y^{7}}\left|(\partial_{1}V)(p)-(\partial_{1}V)(q)\right|\cdot |F(t,x,y)|\,dtdxdy\\ &+\int_{D_{24}}\frac{|X-x|}{y^{8}}\left|(\partial_{2}V)(p)-(\partial_{2}V)(q)\right|\cdot |F(t,x,y)|\,dtdxdy\\ &+\int_{D_{24}}\frac{Y}{y^{6}}\left|(\partial_{3}V)(p)-(\partial_{3}V)(q)\right|\cdot |F(t,x,y)|\,dtdxdy\\ &=:W_{24,1}+W_{24,2}+W_{24,3}+W_{24,4}+W_{24,5}. \end{align}\tag{131}\]

In view of 26 , as well as Young’s inequality, denoting \(r:=(T-t,X-x-(T-t)Y,Y-y)\) for brevity of notations, we have \[\begin{align} &\|W_{24,1}\|_{L^{2}(\mathbb{H}^-)}\lesssim \delta\Big\|\int_{\|r\|\geq10\delta^{\frac{1}{3}}}\|r\|^{-8}\cdot |F(t,x,y)|\,dtdxdy\Big\|_{L^{2}(\mathbb{H}^-)}\lesssim \delta^{\frac{1}{3}}\|F\|_{L^{2}(\mathbb{H}^-)}. \end{align}\] Notice that in \(D_{24}\), we have \[y\gtrsim \|(T-t,X-x,Y-y)\|.\] Hence denoting \(s:=(T-t,X-x,Y-y)\) for brevity of notations, we have \[\begin{align} &\|W_{24,2}\|_{L^{2}(\mathbb{H}^-)}\lesssim \Big\|\int_{\|s\|\leq5\delta^{\frac{1}{3}}}\|s\|^{-5}\cdot |F(t,x,y)|\,dtdxdy\Big\|_{L^{2}(\mathbb{H}^-)}\lesssim \delta^{\frac{1}{3}}\|F\|_{L^{2}(\mathbb{H}^-)}. \end{align}\] The remaining terms in 131 can be handled in a similar way as \(W_{24,2}\). The proof of 122 is now complete. ◻

Proof of Lemma 14. Let \(\Delta_{j}\) be the standard Littlewood–Paley projection operator in \(x\). By 113 , we have for \((t,x,y)\in\mathcal{S}_1\), \[\label{Vjeq} \begin{cases} \partial_{t}\Delta_{j}w+y\partial_{x}\Delta_{j}w-\partial_{y}\Delta_{j}\left(a(t,x,y)\partial_{y}w\right)=\Delta_{j}f,\\ \partial_{y}\Delta_{j}V|_{y=0}=0. \end{cases}\tag{132}\]

Fix \(K\gg1\) to be determined below and a standard smooth cutoff function \(\eta(y)\in C_0^\infty([0,y_1))\) with \(\eta\equiv1\) on \([0,y_0]\). Testing 132 against \(2^{2jks}\Delta_{j}w\,\eta^{2}(y)\) with \(j\geq K,\) and integrating by parts, we get \[\label{energyide} \begin{align} I_{j}:&=\int_{\mathcal{S}_{1}}2^{2jks}\Delta_{j}(a\partial_{y}w)\partial_{y}\Delta_{j}w\,\eta^{2}(y)\,dtdxdy\\ &\leq\int_{\mathcal{S}_{1}}2^{2jks}\Delta_{j}f\Delta_{j}w\,\eta^{2}(y)\,dtdxdy -2\int_{\mathcal{S}_{1}}2^{2jks}\Delta_{j}(a\partial_{y}w)\Delta_{j}w\,\eta(y)\eta'(y)\,dtdxdy\\ &=:J_{j,1}-2J_{j,2}. \end{align}\tag{133}\] By Bony’s paraproduct decomposition 259 , we have \[\label{paraproduct} \begin{align} \Delta_{j}(a\partial_{y}w)&=\sum_{|l|\leq 5}\Delta_{j}\left(S_{j+l-1}a\,\Delta_{j+l}\partial_{y}w\right)+\sum_{|l|\leq 5}\Delta_{j}\left(S_{j+l-1}\partial_{y}w\,\Delta_{j+l}a\right)\\ &\quad +\sum_{l> j-5,\,|q-l|\leq 1}\Delta_{j}\left(\Delta_{l}a\,\Delta_{q}\partial_{y}w\right) \end{align}\tag{134}\] Hence, the commutator term \([\Delta_{j},a]\partial_{y}w\) can be expressed as \[\label{commutator} \begin{align} \Delta_{j}(a\partial_{y}w)-a\Delta_{j}\partial_{y}w&=(S_{j-6}-\mathrm{Id})a\,\Delta_{j}\partial_{y}w+\sum_{|l|\leq 5}[\Delta_{j}, S_{j-6}a]\Delta_{j+l}\partial_{y}w\\ &\quad+\sum_{|l|\leq 5}\Delta_{j}\left(S_{j+l-1}\partial_{y}w\,\Delta_{j+l}a\right)+\sum_{l> j-5,|q-l|\leq 1}\Delta_{j}\left(\Delta_{l}a\,\Delta_{q}\partial_{y}w\right)\\ &\quad+\sum_{-5<l\leq5}\sum_{q=-6}^{l-2}\Delta_{j}\left(\Delta_{j+q}a\,\Delta_{j+l}\partial_{y}w\right). \end{align}\tag{135}\] Therefore, by 135 , \[\label{ijformula} \begin{align} I_{j}=I_{j,1}+I_{j,2}+I_{j,3}+I_{j,4}+I_{j,5}+I_{j,6}, \end{align}\tag{136}\] where the terms \(I_{j,l},\,l\in\mathbb{Z}\cap[1,6]\) are defined as \[\begin{align} &I_{j,1}:=\int_{\mathcal{S}_{1}}2^{2jks}a(t,x,y)|\partial_{y}\Delta_{j}w|^{2}\eta^{2}(y)\,dtdxdy,\\ &I_{j,2}:=\int_{\mathcal{S}_{1}}2^{2jks}\big[(S_{j-6}-\mathrm{Id})a\,\Delta_{j}\partial_{y}w\big]\partial_{y}\Delta_{j}w\,\eta^{2}(y)\,dtdxdy,\\ &I_{j,3}:=\int_{\mathcal{S}_{1}}2^{2jks}\Big(\sum_{|l|\leq 5}[\Delta_{j}, S_{j-6}a]\Delta_{j+l}\partial_{y}w\Big)\partial_{y}\Delta_{j}w\,\eta^{2}(y)\,dtdxdy,\\ &I_{j,4}:=\int_{\mathcal{S}_{1}}2^{2jks}\Big(\sum_{|l|\leq 5}\Delta_{j}(S_{j+l-1}\partial_{y}w\Delta_{j+l}a)\Big)\partial_{y}\Delta_{j}w\,\eta^{2}(y)\,dtdxdy,\\ &I_{j,5}:=\int_{\mathcal{S}_{1}}2^{2jks}\Big(\sum_{-5<l\leq5}\sum_{q=-6}^{l-2}\Delta_{j}\left(\Delta_{j+q}a\Delta_{j+l}\partial_{y}w\right)\Big)\partial_{y}\Delta_{j}w\,\eta^{2}(y)\,dtdxdy,\\ &I_{j,6}:=\int_{\mathcal{S}_{1}}2^{2jks}\Big(\sum_{l> j-5,|q-l|\leq 1}\Delta_{j}\left(\Delta_{l}a\,\Delta_{q}\partial_{y}w\right)\Big)\partial_{y}\Delta_{j}w\,\eta^{2}(y)\,dtdxdy. \end{align}\]

We bound the terms \(I_{j,l}, \,l\in \mathbb{Z}\cap[1,6]\) as follows.

(I): Estimate of \(I_{j,1}\). By the property of \(a(t,x,y)\), we have \[I_{j,1}\geq c\int_{\mathcal{S}_{1}}2^{2jks}|\partial_{y}\Delta_{j}w|^{2}\eta^{2}dtdxdy.\]

(II): Estimate of \(I_{j,2}\). Since \(j\geq K\), by Bernstein’s inequality, \[\begin{align} |I_{j,2}|&\leq \Big(\sum_{q\geq K-6}\|\Delta_{q}a\|_{L^{\infty}}\Big)\int_{\mathcal{S}_{1}}2^{2jks}|\partial_{y}\Delta_{j}w|^{2}\eta^{2}dtdxdy\\ &\lesssim 2^{-\alpha K}\|a\|_{C^{\alpha}(\mathcal{S}_{1})}\int_{\mathcal{S}_{1}}2^{2jks}|\partial_{y}\Delta_{j}w|^{2}\eta^{2}dtdxdy\leq \frac{1}{10}I_{j,1}, \end{align}\] if \(K\gg1\) is chosen sufficiently large, depending on \([a]_{C^{\alpha}(\mathcal{S}_{1})}\).

(III): Estimate of \(I_{j,3}\). Recall the standard commutator estimate (see e.g. Lemma 2.97 in [71]), \[\label{standardcomm} \|[\Delta_{j},b]f\|_{L^{2}(\mathbb{R})}\lesssim 2^{-j}\|\nabla_x b\|_{L^{\infty}(\mathbb{R})}\|f\|_{L^{2}(\mathbb{R})}.\tag{137}\] Hence, by 137 and Bernstein’s inequality, for \(y\in(0,1)\) and \(D:=(-\infty,0]\times\mathbb{R}\), \[\begin{align} &\left\|[\Delta_{j},S_{j-6}a]\Delta_{j+l}\partial_{y}w\right\|_{L^{2}_{t,x}(D)}\lesssim 2^{-j}\|\nabla_x S_{j-6}a\|_{L^{\infty}}\left\|\Delta_{j+l}\partial_{y}w\right\|_{L^{2}_{t,x}(D)}\\ &\leq 2^{-j}\Big(\sum_{q<j-6}\|\nabla_x\Delta_{q}a\|_{L^{\infty}}\Big)\left\|\Delta_{j+l}\partial_{y}w\right\|_{L^{2}_{t,x}(D)} \lesssim 2^{-j\alpha}\|a\|_{C^{\alpha}(D)}\left\|\Delta_{j+l}\partial_{y}w\right\|_{L^{2}_{t,x}(D)}. \end{align}\] By Cauchy–Schwarz inequality, \[|I_{j,3}|\leq \frac{1}{10}I_{j,1}+C2^{-2j(\alpha-s)}\|a\|_{C^{\alpha}}^{2}\Big(\sum_{|l|\leq 5}\int_{\mathcal{S}_{1}}2^{2j(k-1)s}|\partial_{y}\Delta_{j+l}w|^{2}\eta^{2}dtdxdy\Big).\]

(IV): Estimate of \(I_{j,4}\). By Cauchy–Schwarz inequality, we have \[\begin{align} |I_{j,4}|&\lesssim \sum_{|l|\leq5}\left(2^{2jks}\|\partial_{y}\Delta_{j}w\cdot \eta(y)\|_{L^{2}(\mathcal{S}_{1})}\cdot\|\partial_{y}w\|_{L^{\infty}(\mathcal{S}_{1})}\cdot\|\Delta_{j+l}a\|_{L^{2}(\mathcal{S}_{1})}\right)\\ &\leq \frac{1}{10}I_{j,1}+C\|\partial_{y}w\|_{L^{\infty}(\mathcal{S}_{1})}^{2}\sum_{|l|\leq5}2^{2jks}\|\Delta_{j+l}a\|_{L^{2}(\mathcal{S}_{1})}^{2}. \end{align}\]

(V): Estimate of \(I_{j,5}\). By Bernstein’s inequality, \[\begin{align} |I_{j,5}|&\lesssim 2^{-j\alpha}[a]_{C^{\alpha}}\Big(\int_{\mathcal{S}_{1}}2^{2jks}|\partial_{y}\Delta_{j}w|^{2}\eta^{2}\Big)^{\frac{1}{2}}\Big(\sum_{-5<l\leq 5}\int_{\mathcal{S}_{1}}2^{2jks}|\partial_{y}\Delta_{j+l}w|^{2}\eta^{2}dtdxdy\Big)^{\frac{1}{2}}\\ &\leq \frac{1}{10}I_{j,1}+C[a]_{C^{\alpha}}^{2}2^{-2j(\alpha-s)}\Big(\sum_{-5<l\leq 5}\int_{\mathcal{S}_{1}}2^{2j(k-1)s}|\partial_{y}\Delta_{j+l}w|^{2}\eta^{2}dtdxdy\Big). \end{align}\]

(VI): Estimate of \(I_{j,6}\). For \(I_{j,6}\), we rewrite this term as \[I_{j,6}=\sum_{l> j-5,|q-l|\leq 1}\int_{\mathcal{S}_{1}}2^{(j-l)ks}\left(\Delta_{j}\left(2^{lks}\Delta_{l}a\cdot\Delta_{q}\partial_{y}w\right)\right)2^{jks}\partial_{y}\Delta_{j}w\eta^{2}(y)dtdxdy.\] By Cauchy–Schwarz inequality, \[\begin{align} I_{j,6}&\leq\frac{1}{10}I_{j,1}+C\|\partial_{y}w\|_{L^{\infty}}^{2}\Big(\sum_{l>j-5}2^{(j-l)ks}\cdot 2^{lks}\|\Delta_{l}a\|_{L^{2}}\Big)^{2}\\ &=\frac{1}{10}I_{j,1}+C\|\partial_{y}w\|_{L^{\infty}}^{2}\|(Id-\Delta_{-1})a\|_{\widetilde{H}^{ks}(\mathcal{S}_{1})}^{2}\cdot c_{j}, \end{align}\] where \(c_{j}\) is a non–negative sequence with \(\sum_{j\geq 10}c_{j}\lesssim 1\). In the above we have applied Young’s inequality: \[\sum_{j\geq 10}\Big(\sum_{l>j-5}2^{(j-l)ks}\cdot 2^{lks}\|\Delta_{l}a\|_{L^{2}}\Big)^{2}\lesssim \sum_{l\geq 5}2^{2lks}\|\Delta_{l}a\|_{L^{2}}^{2}.\]

We now turns to estimate \(J_{j,1}\) and \(J_{j,2}\) in 133 . We can bound \(J_{j,1}\) straightforwardly as \[|J_{j,1}|\leq \left(\int_{\mathcal{S}_{1}}2^{2jks}|\Delta_{j}f|^{2}\eta^{2}(y)dtdxdy\right)^{\frac{1}{2}}\left(\int_{\mathcal{S}_{1}}2^{2jks}|\Delta_{j}w|^{2}\eta^{2}(y)dtdxdy\right)^{\frac{1}{2}}.\]

It remains to estimate \(J_{j,2}\). We write \[\begin{align} J_{j,2}=&\int_{\mathcal{S}_{1}}2^{2jks}\Delta_{j}(a\partial_{y}w)\Delta_{j}w\,\eta(y)\eta'(y)dtdxdy\\ =&\int_{\mathcal{S}_{1}}2^{2jks}a(t,x,y)\partial_{y}\Delta_{j}w\Delta_{j}w\,\eta(y)\eta'(y)dtdxdy\\ &+\int_{\mathcal{S}_{1}}2^{2jks}[\Delta_{j},a(t,x,y)]\partial_{y}w\Delta_{j}w\,\eta(y)\eta'(y)dtdxdy, \end{align}\] which, by@eq:commutator , can be further decomposed as \[J_{j,2} =:J_{j,2,1}+\cdots+J_{j,2,6},\] where the terms \(J_{j,2,l},l\in\mathbb{Z}\cap[1,3]\) are given by \[\begin{align} J_{j,2,1}=&\int_{\mathcal{S}_{1}}2^{2jks}a(t,x,y)\partial_{y}\Delta_{j}w\Delta_{j}w\,\eta(y)\eta'(y)\,dtdxdy,\\ J_{j,2,2}=&\int_{\mathcal{S}_{1}}2^{2jks}(S_{j-6}-Id)a\,\Delta_{j}\partial_{y}w\cdot\Delta_{j}w\,\eta(y)\eta'(y)\,dtdxdy,\\ J_{j,2,3}=&\sum_{|l|\leq 5}\int_{\mathcal{S}_{1}}2^{2jks}[\Delta_{j},S_{j-6}a]\Delta_{j+l}\partial_{y}w\cdot\Delta_{j}w\,\eta(y)\eta'(y)\,dtdxdy, \end{align}\] and the terms \(J_{j,2,l},l\in\mathbb{Z}\cap[4,6]\) are given by \[\begin{align} J_{j,2,4}=&\sum_{|l|\leq 5}\int_{\mathcal{S}_{1}}2^{2jks}\Delta_{j}(S_{j+l-1}\partial_{y}w\Delta_{j+l}a)\cdot\Delta_{j}w\,\eta(y)\eta'(y)\,dtdxdy,\\ J_{j,2,5}=&\sum_{-5<l\leq 5}\sum_{q=-6}^{l-2}\int_{\mathcal{S}_{1}}2^{2jks}\Delta_{j}(\Delta_{j+q}a\Delta_{j+l}\partial_{y}w)\cdot \Delta_{j}w\,\eta(y)\eta'(y)\,dtdxdy,\\ J_{j,2,6}=&\sum_{l>j-5,|q-l|\leq1}\int_{\mathcal{S}_{1}}2^{2jks}\Delta_{j}(\Delta_{l}a\Delta_{q}\partial_{y}w)\cdot\Delta_{j}w\,\eta(y)\eta'(y)\,dtdxdy. \end{align}\] For \(J_{j,2,1}\), integrating by parts, we get \[|J_{j,2,1}|\lesssim_{y_{0}}(\|a\|_{L^{\infty}}+\|\partial_{y}a\|_{L^{\infty}})\cdot \int_{\mathcal{S}_{1}}2^{2jks}|\Delta_{j}w|^{2}dtdxdy.\] The remaining term \(J_{j,2,2}\) to \(J_{j,2,6}\) can be estimated in the same fashion as \(I_{j,l}\) terms. Indeed, we have \[\begin{align} &|J_{j,2,2}|\leq\frac{1}{10}I_{j,1}+C_{y_{0}}\|2^{jks}\Delta_{j}w\|_{L^{2}(\mathcal{S}_{1})}^{2},\\ &|J_{j,2,3}|\lesssim_{y_{0}}\|2^{jks}\Delta_{j}w\|_{L^{2}(\mathcal{S}_{y_1})}^{2}+\|a\|_{C^{\alpha}}^{2}2^{-2j(\alpha-s)}\Big(\sum_{|l|\leq 5}\|2^{j(k-1)s}\partial_{y}\Delta_{j+l}w\|_{L^{2}(\mathcal{S}_{y_1})}^{2}\Big), \end{align}\] as well as \[\begin{align} &|J_{j,2,4}|\lesssim_{y_{0}}\|2^{jks}\Delta_{j}w\|_{L^{2}(\mathcal{S}_{1})}^{2}+\|\partial_{y}w\|_{L^{\infty}}^{2}\Big(\sum_{|l|\leq 5}\|2^{jks}\Delta_{j+l}a\|_{L^{2}(\mathcal{S}_{1})}^{2}\Big),\\ &|J_{j,2,5}|\lesssim_{y_{0}}\|2^{jks}\Delta_{j}w\|_{L^{2}(\mathcal{S}_{1})}^{2}+[a]_{C^{\alpha}}^{2}2^{-2j(\alpha-s)}\Big(\sum_{-5<l\leq 5}\|2^{j(k-1)s}\|\partial_{y}\Delta_{j+l}w\|_{L^{2}(\mathcal{S}_{1})}^{2}\Big)\\ &|J_{j,2,6}|\lesssim_{y_{0}}\|2^{jks}\Delta_{j}w\|_{L^{2}(\mathcal{S}_{1})}^{2}+\|\partial_{y}w\|_{L^{\infty}}^{2}\|(Id-\Delta_{-1})a\|_{\widetilde{H}^{ks}(\mathcal{S}_{1})}^{2}\cdot c_{j}. \end{align}\]

Combining above all estimates, and summing in \(j\), we complete the proof of Lemma 14. ◻

Proof of Lemma 15. The proof of Lemma 15 is somewhat similar to Lemma 13, and we will be somewhat brief. We decompose \(W\) into two parts, \(W_1\) and \(W_2\), where \(W_{1}\) satisfies 116 and \(W_{2}\) satisfies 117 .

\(\bullet\) Bounds on \(W_{1}\). By Lemma 11 and 99 , we have \[\label{higherw1} \begin{align} &\Big\|\partial_{y}^{2}|D_{x}|^{ks}W_{1}\Big\|_{L^{2}(\mathbb{H}^-)}^{2}+\Big\||D_{x}|^{\frac{2}{3}}|D_{x}|^{ks}W_{1}\Big\|_{L^{2}(\mathbb{H}^-)}^{2}\\ &\lesssim \left\||D_{x}|^{ks}\left(\eta^{2}f-4\eta(y)\eta'(y)\partial_{y}w-\left(2\eta\eta''+2(\eta')^{2}\right)w-2\eta(y)\eta'(y)(a-1)\partial_{y}w\right)\right\|_{L^{2}(\mathbb{H}^-)}^{2}\\ &\lesssim \left\|\eta^{2}f(t,x,y)-4\eta(y)\eta'(y)\partial_{y}w-\left(2\eta\eta''+2(\eta')^{2}\right)w-2\eta(y)\eta'(y)(a-1)\partial_{y}w\right\|_{L^{2}(\mathbb{H}^-)}^{2}\\ &\quad+\sum_{j\geq 10}2^{2jks}\left\|\Delta_{j}\left(\eta^{2}f-4\eta(y)\eta'(y)\partial_{y}w-\left(2\eta\eta''+2(\eta')^{2}\right)w-2\eta(y)\eta'(y)(a-1)\partial_{y}w\right)\right\|_{L^{2}(\mathbb{H}^-)}\\ &\lesssim C_{y_{0}}\big(\|w\|_{\widetilde{H}^{ks}(\mathcal{S}_{y_1})},\|\partial_{y}w\|_{\widetilde{H}^{ks}(\mathcal{S}_{y_1})},\|f\|_{\widetilde{H}^{ks}(\mathcal{S}_{y_1})}\big). \end{align}\tag{138}\] In the last inequality, we used Bony’s paraproduct decomposition 259 .

\(\bullet\) Estimation of \(W_{2}\). Write \[F(t,x,y)=\left(a(t,x,y)-1\right)\partial_{y}w\cdot \eta^{2}.\] By 124 , we have for \(j\geq 10\), \[\label{higherw2} 2^{(2ks+\frac{2}{3})j}\|\Delta_{j}W_{1}\|_{L^{2}(\mathbb{R}^{-}\times\mathbb{R}\times[0,10])}^{2}\lesssim\sum_{|l|\leq2}2^{2jks}\|\Delta_{j+l}F\|_{L^{2}(\mathbb{H}^-)}^{2}.\tag{139}\] Applying Bony para–product decomposition 259 , we have \[\sum_{j\geq 8}2^{2jks}\|\Delta_{j}F\|_{L^{2}(\mathbb{H}^-)}^{2}\lesssim\|\partial_{y}w\|_{L^{\infty}\cap\widetilde{H}^{ks}(\mathcal{S}_{y_1})}^{2}\times \Big(\|a\|_{C^{\alpha}(\mathcal{S}_{1})}+\|(Id-\Delta_{-1})a\|_{\widetilde{H}^{ks}(\mathcal{S}_{1})}\Big)^{2}.\] Hence, the conclusion of Lemma 15 follows from 138 , and summation 139 for \(j\). ◻

Proof (Sketch). By Hölder estimates in section 3 and Proposition 9, we have \[w\in C^{\alpha}_{\rm loc}(\mathcal{B}_{1}^{+}),\;\text{for\;some}\;\alpha>0,\] and \[\partial_{y}^{2}w\in L^{p}_{\rm loc}(\mathcal{B}_{1}^{+}),\;\text{for\;all}\;2\leq p<\infty,\] Applying the anisotropic interpolation (see Lemma 23), we get that \[\partial_{y}w\in L^{\infty}_{\rm loc}(\mathcal{B}_{1}^{+}).\] To bootstrap to higher regularity, let us recall the enhanced dissipation estimate ( see Lemma 11) and 99 for a suitable smooth cutoff function \(\varphi\in C_c^\infty(\mathcal{B}_1^+)\), \[|D_{x}|^{\frac{2}{3}}(\varphi w)\in L^{2}_{\rm loc}(\mathbb{H}^{-}).\] Therefore, applying Proposition 13 inductively (again with suitable localizations), we obtain that \[\partial_{x}^{k}w\in L^{2}_{\rm loc}(\mathcal{B}_{1}^{+}),\quad \forall k\geq 1.\] Finally, the regularity of \(w\) in \(t\) and \(y\) follows from standard parabolic theory. This completes the outline of the proof. ◻

Proof of Theorem 1. Let \(u(t,x,y)\) satisfies the assumptions of Theorem 1, recall that we can represent \(u\) via \[y=\int_{0}^{u(t,x,y)}\frac{d\eta}{w(t,x,\eta)},\] where \[w(t,x,\eta):=\partial_{y}u(t,x,y),\quad \text{for}\;u(t,x,y)=\eta.\] By direct calculation, we obtain that \[(\partial_{t},\partial_{x})u(t,x,y)=w(t,x,u(t,x,y))\cdot\int_{0}^{u(t,x,y)}\frac{(\partial_{t},\partial_{x})w}{w^{2}}d\eta.\] We can also represent the higher–order derivatives of \(u\) in terms of \(w\) via Faà Di Bruno’s formula. Moreover, for any \(M\in\mathbb{N}\) and \(\widetilde{D}\Subset D_{T,X}\), we have \[\label{pfmainbasedesti} \|u\|_{C^{M}(\widetilde{D})}\lesssim_{M,m_{\widetilde{D}},M_{\widetilde{D}},\|w\|_{C^{M}(D^{\ast})}} 1,\tag{152}\] where \(D^{\ast}\) is the image of \(\widetilde{D}\) under Crocco transform ?? .

Now assume \(D^{\star}\) is another compact subset of \(D_{T,X}\) such that \(\widetilde{D}\Subset D^{\star}\Subset D_{T,X}\), and \(D^{\#}\) is the image of \(D^{\star}\) under Crocco transform ?? . Then by 152 and 141 , we have \[\label{basedesti2} \|u\|_{C^{M}(\widetilde{D})}\lesssim_{M,\mathrm{dist}(D^{\ast}, D^{\#}),m_{D^{\star}},M_{D^{\star}}} 1.\tag{153}\] We conclude the proof of Theorem 1 using 153 and the fact that \[m_{D^{\star}}\cdot\mathrm{dist}(\widetilde{D},D^{\star})\leq \mathrm{dist}(D^{\ast},D^{\#})\leq M_{D^{\star}}\cdot\mathrm{dist}(\widetilde{D},D^{\star}).\] ◻

Proof. Assume that \(w_{1},w_{2}\) are two weak solutions satisfying the assumptions of Proposition 16. For \(i=1,2\), define \(v_{i}=w_{i}^{-1}\), and let \(v=v_{1}-v_{2}\), \(w=w_{1}-w_{2}\). Then in \(R_{T,X}\) we have the equation \[\label{reciequ} \partial_{t}v+y\partial_{x}v+\partial_{y}^{2}w=0.\tag{156}\] Choosing \(m=2k+10\), and then testing 156 by \({\rm sign}(v)(1-y)^{m}\) and integrating on \(y\in[0,1]\), we get that \[\label{uniqueesti} \begin{align} &\partial_{t}\left(\int_{0}^{1}|v|(1-y)^{m}dy\right)+\partial_{x}\left(\int_{0}^{1}|v|y(1-y)^{m}dy\right)-\int_{0}^{1}\partial_{y}^{2}w\cdot {\rm sign}(w)(1-y)^{m}dy=0. \end{align}\tag{157}\] The above formula can be justified rigorously by replacing \({\rm sign}(v)\) with \(\phi(v/\epsilon)\) and taking \(\epsilon\to0+\), where the smooth function \(\phi(x)\) is odd, equal to \(-1\) for \(x\leq-1\) and equal to \(1\) for \(x\ge1\). By direct calculations, we obtain \[\begin{align} -\int_{0}^{1}\partial_{y}^{2}w\cdot {\rm sign}(w)(1-y)^{m}dy&\ge-m(m-1)\int_{0}^{1}|w|(1-y)^{m-2}dy. \end{align}\] For \(t\ge0\), let \[F(t):=\iint_{[0,X]\times[0,1]}|v|(1-y)^{m}dxdy.\] By 157 and the identity \(w=\frac{1}{v_1}-\frac{1}{v_2}=-\frac{v_1-v_2}{v_1v_2}=-(v_1-v_2)w_1w_2\), we have \[\label{gia1} \begin{align} F'(t)&\leq m(m-1)\int_{0}^{X}\int_{0}^{1}|w|(1-y)^{m-2}dxdy\\ &=m(m-1)\int_{0}^{X}\int_{0}^{1}|v|(1-y)^{m}{\boxed{\frac{w_{1}\cdot w_{2}}{(1-y)^{2}}}}dxdy. \end{align}\tag{158}\] In view of ?? , the boxed term in the integrand is not necessarily bounded, and as a consequence we can not apply Grönwall’s inequality directly. To overcome this difficulty, we split the integral on the right-hand side of 158 in \(y\) into two parts, one over \((0,y_0)\) and the other over \((y_0,1)\) with \(1/2\leq y_0\leq1\), to obtain that \[\begin{align} F'(t)&\leq m(m-1)\int_{0}^{X}\int_{0}^{1}|w|(1-y)^{m-2}dxdy\\ &=m(m-1)\int_{0}^{X}\int_{0}^{y_{0}}|v|(1-y)^{m}\frac{w_{1}\cdot w_{2}}{(1-y)^{2}}dxdy+m(m-1)\int_{0}^{X}\int_{y_{0}}^{1}|w|(1-y)^{m-2}dxdy\\ &\leq C_{1}\log\left(\frac{1}{1-y_{0}}\right)F(t)+C_{2}(1-y_{0})^{m-1}. \end{align}\] By Grönwall’s inequality, we get \[\sup_{t\in[0,T^{*}]}F(t)\leq C_{2}\sup_{t\in[0,T^\ast]}\left[t(1-y_{0})^{m-1}\exp\left\{C_{1}t\log\left(\frac{1}{1-y_{0}}\right)\right\}\right]\leq C_{2}T^{*}(1-y_{0})^{m-1-C_{1}T^{*}}.\] By letting \(y_{0}\to 1\), we obtain that \[F(t)\equiv 0,\;0\leq t\leq T^{*},\;T^{*}:=\frac{m-1}{2C_{1}}.\] Therefore, \(F(t)\) vanishes identically by dividing the temporal interval and repeating the above argument, which completes the proof of Proposition 16. ◻

Proof of Proposition 17. We divide the proof of Proposition 17 into six steps:

Step 1: Approximation. Fix \(\epsilon>0\). For any \(T>0\) and \(X>0\), let \(w^{\epsilon}\in C^\infty(\overline{R}_{T,X})\) be the unique smooth solution to the following approximation system \[\label{approsys} \begin{cases} \partial_{t}w^{\epsilon}+(y+\epsilon)\partial_{x}w^{\epsilon}-(w^{\epsilon}+\epsilon)^{2}\partial_{y}^{2}w^{\epsilon}=0,\;(t,x,y)\in(0,T]\times(0,X]\times(0,1),\\ w^{\epsilon}|_{t=0}=w_{0}^{\epsilon},\;w^{\epsilon}|_{x=0}=w_{1}^{\epsilon},\\ w^{\epsilon}|_{y=1}=\partial_{y}w^{\epsilon}|_{y=0}=0, \end{cases}\tag{161}\] where \(w_{0}^{\epsilon},w_{1}^{\epsilon}\) are suitable mollification of \(w_{0},w_{1}\) respectively, satisfying the compatibility conditions corresponding to 161 and certain uniform–in–\(\epsilon\) bounds, which are presented in Appendix 10 (see (WP1),(WP2) and (WP3) in Appendix 10.3). We refer to Appendix 11 that the classical solution \(w^\epsilon\) of 161 exists globally and is unique.

Step 2: Pointwise Estimates. We first derive an uniform–in–\(\epsilon\) upper bound of \(w^{\epsilon}\). More precisely, we shall prove that for any \(T>0,\, X>0\) and \((t,x,y)\in R_{T,X}\), \[\label{boundone} w^{\epsilon}(t,x,y)\lesssim_{T,X} (1-y)\log^{\frac{1}{2}}\Big(\frac{10}{1-y}\Big).\tag{162}\] Let \[h(t,x,y):=w^{\epsilon}(t,x,y)-C(1-y)\log^{\frac{1}{2}}\Big(\frac{10}{1-y}\Big),\] Consider the operator \[\mathcal{L}_{\epsilon}=\partial_{t}+(y+\epsilon)\partial_{x}-(w^{\epsilon}+\epsilon)^{2}\partial_{y}^{2}.\] By direct calculations, we get that \[\label{signoperator} \mathcal{L}_{\epsilon}[h]=-C\frac{(w^{\epsilon}+\epsilon)^{2}}{1-y}\Big(\frac{1}{2}\log^{-\frac{1}{2}}\Big(\frac{10}{1-y}\Big)+\frac{1}{4}\log^{-\frac{3}{2}}\Big(\frac{10}{1-y}\Big)\Big)<0.\tag{163}\] Taking \(C>1\) sufficiently large (depending on \(T\) and \(X\)), we also have \[\label{signboundary} \begin{align} &h|_{t=0}\leq 0,\quad \forall (x,y)\in[0,X]\times[0,1],\\ &h|_{x=0}\leq 0, \quad\forall (t,y)\in[0,T]\times[0,1],\\ &h|_{y=1}=0,\quad \partial_{y}h|_{y=0}>0, \quad \forall (t,x)\in[0,T]\times[0,1]. \end{align}\tag{164}\] Hence \(h\leq 0\) for any \((t,x,y)\in[0,T]\times[0,X]\times[0,1]\) by 163 –164 , and the maximum principle.

Next we turn to the lower bound of \(w^{\epsilon}\). Note that \(w^{\epsilon}\) is non–negative by maximal principle. We claim that that for \((t,x,y)\in R_{T,X}\), \[\label{boundtwo} w^{\epsilon}(t,x,y) \gtrsim_{T,X} (1-y)^{m}.\tag{165}\] To this end, we define for some \(c>0\) and \(M>1\) to be determined below, \[g(t,x,y):=w^{\epsilon}(t,x,y)-ce^{-Mt}e^{2my}(1-y)^{m}.\] It suffices to prove that \(g(t,x,y)\) is non–negative in \([0,T]\times [0,X]\times [0,1]\). By direct calculation, \[\mathcal{L}_{\epsilon}[g]=ce^{-Mt}e^{2my}(1-y)^{m-2}\left[M(1-y)^{2}+(w^{\epsilon}+\epsilon)^{2}\left(4m^{2}(1-y)^{2}-4m^{2}(1-y)+m^{2}-m\right)\right].\] Hence, \[\mathcal{L}_{\epsilon}[g]>0,\;\text{in}\; (0,T]\times (0,X]\times (0,1),\] by taking \(M>1\) large enough, depending on \(T>0, X>0\).

Notice that by taking \(c>0\) small (depending on \(T>0, X>0\)), we have \[\label{signcondi} \begin{align} &g|_{t=0}\geq 0,\;\forall (x,y)\in[0,X]\times[0,1],\\ &g|_{x=0}\geq 0,\;\forall (t,y)\in[0,T]\times[0,1]\\ &g|_{y=1}=0,\quad \partial_{y}g|_{y=0}=-cme^{-Mt}<0. \end{align}\tag{166}\] Hence, \(g\) is non–negative by the maximal principle.

Combining 162 and 165 , we obtain the uniform-in-\(\epsilon\) pointwise bound for \((t,x,y)\in R_{T,X}\), \[\label{ptwe1} (1-y)^{m}\lesssim_{T,X} w^\epsilon(t,x,y)\lesssim_{T,X}(1-y)\log^{\frac{1}{2}}\left(\frac{10}{1-y}\right).\tag{167}\]

Step 3: Energy Estimates. Testing 161 by \[\frac{1}{w^{\epsilon}+\epsilon}(1+\epsilon-y)^{-1}\log^{-3}\left(\frac{10}{1+\epsilon-y}\right),\] and integrating by parts, we get the following energy identity \[\label{energyidee} \begin{align} &\iint_{[0,X]\times [0,1]}(\log (w^{\epsilon}+\epsilon))(1+\epsilon-y)^{-1}\log^{-3}\left(\frac{10}{1+\epsilon-y}\right)dxdy\bigg|^{t=T}_{t=0}\\ &+\iint_{[0,T]\times [0,1]}(\log (w^{\epsilon}+\epsilon)) (y+\epsilon)(1+\epsilon-y)^{-1}\log^{-3}\left(\frac{10}{1+\epsilon-y}\right)dtdy\bigg|^{x=X}_{x=0}\\ &+\iiint_{[0,T]\times[0,X]\times[0,1]}|\partial_{y}w^{\epsilon}|^{2}(1+\epsilon-y)^{-1}\log^{-3}\left(\frac{10}{1+\epsilon-y}\right)dtdxdy\\ &+\iiint_{[0,T]\times[0,X]\times[0,1]}(w^{\epsilon}+\epsilon)\partial_{y}w^{\epsilon}(1+\epsilon-y)^{-2}\log^{-3}\left(\frac{10}{1+\epsilon-y}\right)dtdxdy\\ &-3\iiint_{[0,T]\times[0,X]\times[0,1]}(w^{\epsilon}+\epsilon)\partial_{y}w^{\epsilon}(1+\epsilon-y)^{-2}\log^{-4}\left(\frac{10}{1+\epsilon-y}\right)dtdxdy\\ &-\log^{-3}\left(\frac{10}{\epsilon}\right)\cdot\iint_{[0,T]\times[0,X]}\partial_{y}w^{\epsilon}dtdx\bigg|_{y=1}=0. \end{align}\tag{168}\] The last term in 168 is non–negative, since \(w^{\epsilon}|_{y=1}=0\) and \(w^{\epsilon}\) is non–negative, which implies that \(\partial_{y}w^{\epsilon}|_{y=1}\leq 0\). Therefore, by 162 , 165 and 168 , we have \[\label{boundthree} \iiint_{[0,T]\times[0,X]\times[0,1]}|\partial_{y}w^{\epsilon}|^{2}(1+\epsilon-y)^{-1}\log^{-3}\left(\frac{10}{1+\epsilon-y}\right)dtdxdy\lesssim_{T,X} 1.\tag{169}\]

Step 4: Tangential derivative estimates. We also need to estimate \(\partial_{t}w^{\epsilon}\) and \(\partial_{x}w^{\epsilon}\). Let \(v^{\epsilon}=(w^{\epsilon}+\epsilon)^{-1}\), then \(v^{\epsilon}(t,x,y)\) satisfies the following equation for \((t,x,y)\in R_{T,X}\), \[\partial_{t}v^{\epsilon}+(y+\epsilon)\partial_{x}v^{\epsilon}+\partial_{y}^{2}w^{\epsilon}=0.\] Denote the tangential gradient by \(\nabla_{\text{tan}}=(\partial_{t},\partial_{x})\), then we have the following equation for the vector field \(\nabla_{\text{tan}}v^{\epsilon}\) in \(\Omega\): \[\label{recieq} \partial_{t}\nabla_{\text{tan}}v^{\epsilon}+(y+\epsilon)\partial_{x}\nabla_{\text{tan}}v^{\epsilon}-\partial_{y}^{2}\left(\frac{\nabla_{\text{tan}}v^{\epsilon}}{(v^{\epsilon})^{2}}\right)=0.\tag{170}\] Denoting \(\mathrm{sgn}(\nabla_{\text{tan}}v^{\epsilon}):=\left(\mathrm{sgn}(\partial_{t}v^{\epsilon}),\mathrm{sgn}(\partial_{x}v^{\epsilon})\right)\) and testing 170 against \((\nabla_{\text{tan}}v^{\epsilon})\cdot(1-y),\) and integrating with respect to \(y\in [0,1]\), we obtain that \[\label{bvesti} \begin{align} &\partial_{t}\int_{0}^{1}|\nabla_{\text{tan}}v^{\epsilon}|(1-y)dy+\partial_{x}\int_{0}^{1}|\nabla_{\text{tan}}v^{\epsilon}|(y+\epsilon)(1-y)dy\\ &-\int_{0}^{1}\partial_{y}^{2}\left(\frac{\nabla_{\text{tan}}v^{\epsilon}}{(v^{\epsilon})^{2}}\right)\mathrm{sgn}(\nabla_{\text{tan}}v^{\epsilon})(1-y)dy:=I_1+I_2+I_3=0. \end{align}\tag{171}\] By direct calculations, \[\label{bvesti3} \begin{align} I_3&=-\int_{0}^{1}\partial_{y}^{2}\nabla_{\text{tan}}w^{\epsilon} \cdot \mathrm{sgn}(\nabla_{\text{tan}}w^{\epsilon})(1-y)dy\\ &\geq -\int_{0}^{1}\partial_{y}\nabla_{\text{tan}}w^{\epsilon}\cdot {\rm sgn}(\nabla_{\text{tan}}w^{\epsilon})dy\\ &=-\int_{0}^{1}\partial_{y}|\nabla_{\text{tan}}w^{\epsilon}|dy=|\nabla_{\text{tan}}w^{\epsilon}(t,x,0)|, \end{align}\tag{172}\] As discussed just below equation 157 , the above calculations can be justified rigorously through replacing the \({\rm sign}\) function by \(\phi(\cdot/\delta)\) where \(\phi\) is smooth, odd, equal to \(-1\) for \(x\leq-1\) and equal to \(1\) for \(x\ge1\) with \(\phi'\ge0\), and taking \(\delta\to0+\).

Hence by 171 , we get that \[\label{boundfour} \begin{align} &\sup_{t\in[0,T]}\int_{0}^{X}\int_{0}^{1}|\nabla_{\text{tan}}v^{\epsilon}|(1-y)dxdy\\ &\leq\int_{0}^{X}\int_{0}^{1}|\nabla_{\text{tan}}v^{\epsilon}|(1-y)dxdy\bigg|_{t=0}+\int_{0}^{T}\int_{0}^{1}|\nabla_{\text{tan}}v^{\epsilon}|(y+\epsilon)(1-y)dxdy\bigg|_{x=0}\lesssim_{T,X} 1. \end{align}\tag{173}\]

Using the equation 161 and the bound 173 , we conclude that \[\label{boundfive} \sup_{t\in[0,T]}\int_{0}^{X}\int_{0}^{1}|\nabla_{\text{tan}}v^{\epsilon}|(1-y)dxdy+\sup_{t\in[0,T]}\int_{0,X}\int_{0}^{1}|\partial_{y}^{2}w^{\epsilon}|(1-y)dxdy\lesssim_{T,X}1.\tag{174}\]

Step 5: Higher regularity estimates. In this step, we establish the uniform–in–\(\epsilon\) ( up–to–boundary \(\{y=0\}\)) smoothness of \(w^{\epsilon}\). For any fixed \(1>\delta>\epsilon\) and \(t_{0}>\delta, x_{0}>\delta\), we first prove the uniform–in–\(\epsilon\) regularity of \(w^{\epsilon}\) in the ball \[\left\{(t,x,y):-\delta^{2}<t-t_{0}\leq0, |x-x_{0}|<\delta^{3}, 0\leq y<\delta\right\}.\] By the translation invariance of the equation 161 with respect to \((t,x)\), we can assume that \(t_{0}=x_{0}=0\). Define \[\widetilde{w}^{\epsilon}(t,x,y)=w^{\epsilon}\left(\frac{\delta^{2}}{4}t,\frac{\delta^{3}}{8}x+\frac{\epsilon\delta^{2}}{4}t,\frac{\delta}{2}y\right).\] We have \[\begin{cases} \partial_{t}\widetilde{w}^{\epsilon}+y\partial_{x}\widetilde{w}^{\epsilon}-(\widetilde{w}^{\epsilon}+\epsilon)^{2}\partial_{y}^{2}\widetilde{w}^{\epsilon}=0,\\ \partial_{y}\widetilde{w}^{\epsilon}|_{y=0}=0,\\ \end{cases} \quad \text{in}\;\left\{-1<t\leq0,|x|<1,0\leq y<1\right\}.\] By the same procedure as in the derivation of 142 , we obtain that \[\|\widetilde{w}^{\epsilon}\|_{C^{k}(-\frac{1}{4}\leq t\leq0,|x|\leq \frac{1}{8},y\leq \frac{1}{2})}\leq C(k,\tilde{m},\tilde{M}),\quad \forall k\geq 0,\] where \[\tilde{m}=\inf_{-1<t<0,|x|<1,|y|<1}\widetilde{w}^{\epsilon}(t,x,y),\quad \tilde{M}=\sup_{-1<t<0,|x|<1,|y|<1}\widetilde{w}^{\epsilon}(t,x,y).\] Therefore, \[\label{wepsilonboundary} \|w^{\epsilon}\|_{C^{k}(-\frac{\delta^{2}}{100}\leq t-t_{0}\leq 0,|x-x_{0}|\leq \frac{\delta^{3}}{100},0\leq y\leq \frac{\delta}{100})}\leq C(k,\delta,m,M),\tag{175}\] where \[m=\inf_{-\delta^{2}<t-t_{0}<0,|x-x_{0}|<\delta^{3},|y|<\delta}\widetilde{w}^{\epsilon}(t,x,y),\quad M=\sup_{-\delta^{2}<t-t_{0}<0,|x-x_{0}|<\delta^{3},|y|<\delta}\widetilde{w}^{\epsilon}(t,x,y).\]

Next, for any fixed \(1>\kappa>\epsilon\) and \(t_{0}>\kappa,x_{0}>\kappa,\kappa<y_{0}<1-\kappa\), we prove the uniform–in–\(\epsilon\) (interior) regularity of \(w^{\epsilon}\) in the ball \[\left\{(t,x,y):-\kappa^{2}<t-t_{0}\leq0,|x-x_{0}|<\kappa^{3},|y-y_{0}|<\kappa\right\}.\] By translation invariance with respect to \((t,x)\)–variable, we can assume \(t_{0}=x_{0}=0\). Define \[\label{rescale1} w_{1}^{\epsilon}(t,x,y)=w^{\epsilon}\left(\frac{\kappa^{2}}{4}t,\frac{\kappa^{3}}{8}x+\frac{\epsilon\kappa^{2}}{4}t,\frac{\kappa}{2}(y+y_{0})\right).\tag{176}\] Then \(w_{1}^{\epsilon}(t,x,y)\) satisfies \[\partial_{t}w_{1}^{\epsilon}+(y+y_{0})\partial_{x}w_{1}^{\epsilon}-(w_{1}^{\epsilon}+\epsilon)^{2}\partial_{y}^{2}w^{\epsilon}=0,\;\text{in}\;\{-1<t\leq 0,|x|<1,|y|<1\}.\] Next, define \[\label{rescale2} w_{2}^{\epsilon}(t,x,y)=w_{1}^{\epsilon}(t,x+y_{0}t,y),\tag{177}\] then \(w_{2}^{\epsilon}\) satisfies \[\partial_{t}w_{2}^{\epsilon}+y\partial_{x}w_{2}^{\epsilon}-(w_{2}^{\epsilon}+\epsilon)^{2}\partial_{y}^{2}w_{2}^{\epsilon}=0,\;\text{in}\;\{-\frac{1}{2}<t\leq 0,|x|<\frac{1}{2},|y|<1\}.\] By the same procedure as in the derivation of 151 , we get that \[\|w_{2}^{\epsilon}\|_{C^{k}(-\frac{1}{4}\leq t\leq 0,|x|\leq \frac{1}{8},|y|\leq \frac{1}{2})}\leq C(k,\kappa,\bar{m},\bar{M}),\] where \[\bar{m}=\inf_{-\frac{1}{2}<t<0,|x|<\frac{1}{2},|y|<1}w_{2}^{\epsilon}(t,x,y),\quad \bar{M}=\sup_{-\frac{1}{2}<t<0,|x|<\frac{1}{2},|y|<1}w_{2}^{\epsilon}(t,x,y).\] Therefore by 176 and 177 , \[\label{wepsiloninter} \|w^{\epsilon}\|_{C^{k}(-\frac{\kappa^{2}}{100}\leq t-t_{0}\leq0,|x-x_{0}|\leq\frac{\kappa^{3}}{100},|y-y_{0}|\leq\frac{\kappa}{100})}\leq C(k,\kappa,m,M),\tag{178}\] where \[m=\inf_{-\kappa^{2}<t-t_{0}<0,|x-x_{0}|<\kappa,|y-y_{0}|<\kappa}w^{\epsilon}(t,x,y),\quad M=\inf_{-\kappa^{2}<t-t_{0}<0,|x-x_{0}|<\kappa,|y-y_{0}|<\kappa}w^{\epsilon}(t,x,y).\] Combining 175 , 178 , as well as the uniform–in–\(\epsilon\) upper bound 162 , and the lower bound 165 , we conclude that for any \(\delta>100\epsilon\), and any \(k\geq 0\), \[\label{wepsilonlocalsmooth} \|w^{\epsilon}\|_{C^{k}(\delta\leq t\leq \delta^{-1},\delta\leq x\leq \delta^{-1},0\leq y\leq 1-\delta)}\lesssim_{k,\delta}1.\tag{179}\]

Step 6: Passing to the limit. By the interior smoothness estimate 179 , there exist a subsequence of \(w^{\epsilon}\) (denoted by \(w^{\epsilon_{n}}\)), and a function \[w(t,x,y)\in C^{\infty}\left((0,+\infty)\times(0,+\infty)\times [0,1)\right),\] such that \(w^{\epsilon_{n}}\) and its derivatives uniformly convergent to \(w\) in any compact subset of \((0,+\infty)\times(0,+\infty)\times [0,1)\). Furthermore, by the estimates 167 , 169 , 173 and 179 on \(w^{\epsilon}\), we conclude that \(w\in\mathcal{X}\), where the functional space \(\mathcal{X}\) is defined in 159 . By the weak compactness, there also exists a subsequence of \(w^{\epsilon}\) (denoted by \(w^{\epsilon}\) as well), such that for any \(T>0,X>0\), \[\partial_{t}w^{\epsilon_{n}}, \partial_{x}w^{\epsilon_{n}} \rightharpoonup \partial_{t}w, \partial_{x}w,\] in the sense of distributions in \((0,T)\times(0,X)\times (0,1)\).

Finally, we need to show that \(w\) satisfies 154 as well as the required boundary conditions. Since \(w^{\epsilon_{n}}\) and its derivatives convergent uniformly to \(w\) in any compact subset of \((0,+\infty)\times(0,+\infty)\times [0,1)\), we have \[\begin{cases} \partial_{t}w+y\partial_{x}w-w^{2}\partial_{y}^{2}=0,\;(t,x,y)\in(0,+\infty)\times(0,+\infty)\times (0,1),\\ w|_{y=1}=0,\quad \partial_{y}w|_{y=0}=0. \end{cases}\] Therefore, to verify that \(w\) satisfies all boundary conditions in 154 , it remains to check \[w|_{t=0}=w_{0}(x,y),\;w|_{x=0}=w_{1}(x,y).\] Here \(w|_{t=0}\) and \(w|_{x=0}\) are defined as the boundary traces of \(W^{1,1}\) functions.

For any \(\phi(x,y)\in C_{c}^{\infty}\left((0,+\infty)\times(0,1)\right)\), and \(\eta(t)\in C_{c}^{\infty}([0,+\infty))\) with \(\eta(0)=1\). For convenience denote \(S:=(0,\infty)\times(0,\infty)\times(0,1)\) and \(\partial S:=(0,\infty)\times(0,1)\). Notice that \[\begin{align} &\iiint_{S} \partial_{t}w(t,x,y)\phi(x,y)\eta(t)dtdxdy\\ &=\lim_{n\to\infty}\iiint_{S} \partial_{t}w^{\epsilon_{n}}(t,x,y)\phi(x,y)\eta(t)dtdxdy\\ &=-\lim_{n\to\infty}\bigg(\iiint_{S} w^{\epsilon_{n}}(t,x,y)\phi(x,y)\eta'(t)dtdxdy+\iint_{\partial S} w^{\epsilon_{n}}(0,x,y)\phi(x,y)dxdy\bigg)\\ &=-\iiint_{S} w(t,x,y)\phi(x,y)\eta'(t)dtdxdy-\iint_{\partial S} w_{0}(x,y)\phi(x,y)dxdy\\ &=\iiint_{S} \partial_{t}w(t,x,y)\phi(x,y)\eta(t)dtdxdy+\iint_{\partial S} \left[w(0,x,y)-w_{0}(x,y)\right]\phi(x,y)dxdy. \end{align}\] Hence \(w|_{t=0}=w_{0}(x,y)\). Similarly we can check that \(w|_{x=0}=w_{1}(t,y)\). Therefore, we finish the proof of Proposition 17. ◻

Proof of Theorem 3. By the continuity and (strict) monotonicity in \(y\) of \(u_{0}\) and \(u_{1}\), we can define \(w_{0}(\tau,\xi,\eta)\) and \(w_{1}(\tau,\xi,\eta)\) using (for \(i\in\{0,1\}\)) \[w_{i}(\tau,\xi,\eta)=(\partial_{y}u_{i})(\tau,\xi,y_{i}),\;\text{where}\;\;u_{i}(\tau,\xi,y_{i})=\eta.\] Under assumptions 5 , 6 , and ?? , one can check that \(w_{0}(\tau,\xi,\eta)\) and \(w_{1}(\tau,\xi,\eta)\) satisfy the assumptions in Proposition 17. Indeed, the bounds ?? follow from 5 , and the bounds ?? follow from 6 , ?? , and the fact that \[\int_{0}^{\infty}(1-u_{i})dy=\int_{0}^{\infty}\frac{1-\eta}{w_{i}}d\eta.\]

By Proposition 16 and Proposition 17, for each positive integer \(N\), the transformed Prandtl system admits a unique weak solution \(w^{N}(\tau,\xi,\eta)\) in \([0,N]\times [0,N] \times [0,1]\) with the corresponding initial and boundary conditions. Also, \(w^{N}\in \mathcal{X}_{N,N}^{m}\) for all \(N\in\mathbb{Z}\cap[1,\infty)\) (recall the definition of space \(\mathcal{X}_{T, N}^m\) in 159 ). Moreover, we have the up–to–boundary regularity \[w^{N}\in C^{\infty}\left((0,N]\times(0,N]\times[0,1)\right).\] It follows from Proposition 16 that for any \(M\geq N \geq 1\), \(w^{M}\equiv w^{N}\) in \([0,N]\times[0,N]\times[0,1]\). Then for any \((t,x,y)\in[0,+\infty)\times[0,+\infty)\times[0,1]\), we can define \[w(t,x,y):=\lim_{N\to\infty}w^{N}(t,x,y).\] Clearly, \(w\equiv w^{N}\) in \([0,N]\times [0,N]\times[0,1]\). Furthermore, one can check that \[\begin{cases} w\in\mathcal{X}_{T,X}^{m},\;\forall\, T,X>0,\; w\in C^{\infty}\left((0,\infty)\times(0,\infty)\times[0,1)\right),\\ w|_{t=0}=w_{0},\;w|_{x=0}=w_{1},\;\partial_{y}w|_{y=0}=0, \end{cases}\] and that \(w(t,x,y)\) satisfies the transformed Prandtl equation 154 in \((0,\infty)\times(0,\infty)\times[0,1)\).

To get back to the physical variables, we define the function \(u(t,x,y)>0\) through the equation \[\label{invercroccoweak} y=\displaystyle \int_{0}^{u(t,x,y)}\frac{d\eta}{w(t,x,\eta)}.\\\tag{180}\] Clearly, \(u(t,x,y)\) is strictly monotone in \(y\) direction, with \[u|_{y=0}\equiv 0,\quad \lim_{y\to\infty}u(t,x,y)\equiv 1\] in view of the pointwise bound \((1-\eta)^{m}\lesssim w\lesssim (1-\eta)\log^{\frac{1}{2}}\left(\frac{10}{1-\eta}\right)\). We note that since \(w\) is smooth in \((0,\infty)\times(0,\infty)\times[0,1)\), we can differentiate 180 up to any order. By direct calculations, we get that \[\begin{cases} \partial_{y}u(t,x,y)=w(t,x,u(t,x,y)),\\ (\partial_{t},\partial_{x})u(t,x,y)=w(t,x,u(t,x,y))\displaystyle \int_{0}^{u(t,x,y)}\frac{(\partial_{\tau},\partial_{\xi})w}{w^{2}}d\eta. \end{cases}\] Hence, for any \(T,X>0\), \(u(t,x,y)\) satisfies ?? and ?? thanks to 159 and standard interpolations. The global \(W^{1,1}\) bounds in ?? follows from the identity \[\int_{0}^{\infty}|(\partial_{t},\partial_{x}) u|dy=\int_{0}^{1}\frac{|(\partial_{t},\partial_{x})w|}{w^{2}}\cdot(1-\eta)d\eta.\] Furthermore, \(u(t,x,y)\in C^{\infty}\left((0,\infty)\times(0,\infty)\times[0,\infty)\right)\) follows from 180 and the fact \[w\in C^{\infty}\left((0,\infty)\times(0,\infty)\times[0,1)\right).\] By 180 , we can check that \(u|_{y=0} \equiv 0\), as well as \[\label{physicaltrace} u|_{t=0} = u_{0},\qquad u|_{x=0} = u_{1}.\tag{181}\]

Define \(v\) using the second equation in 3 , we can then check that \((u,v)\) satisfies the dynamical Prandtl equation 3 .

Finally, the uniqueness statement in Theorem 3 follows from Proposition 16 and the fact that \[w_{1}(t,x,\eta) \equiv w_{2}(t,x,y)\Rightarrow \int_{0}^{u_{1}(t,x,y)}\frac{d\eta}{w_{1}(t,x,\eta)}=\int_{0}^{u_{2}(t,x,y)}\frac{d\eta}{w_{2}(t,x,\eta)}\Rightarrow u_{1}(t,x,y)\equiv u_{2}(t,x,y).\] Therefore, we complete the proof of Theorem 3. ◻

Proof. The proof is based on the maximal principle, and follows the same argument as that in the second step of the proof of Proposition 17. ◻

Proof. Step 1: \(H^{1}\) Estimates. Differentiate 182 with respect to \(\tau\). Then \(w_{\tau}\) satisfies the following equation \[\label{wteq} \begin{cases} \partial_{\tau}w_{\tau}+\eta\partial_{\xi}w_{\tau}-2w\partial_{\eta}^{2}ww_{\tau}-w^{2}\partial_{\eta}^{2}w_{\tau}=0,\\ w_{\tau}|_{\eta=1} = \partial_{\eta}w_{\tau}|_{\eta=0}=0. \end{cases}\tag{189}\] Testing 189 by \(w_{\tau}w^{-2}\), and integrating in \([0,T]\times [0,X]\times [0,1]\), we obtain the following energy inequality: \[\begin{align} \label{h1energyid} &\sup_{\tau\in[0,T]}\iint_{[0,X]\times [0,1]}\frac{|w_\tau|^{2}}{w^{2}}d\xi d\eta+\sup_{\xi\in[0,X]}\iint_{[0,T]\times [0,1]}\frac{\eta|w_{\tau}|^{2}}{w^{2}}d\tau d\eta+\iiint_{[0,T]\times [0,X]\times [0,1]}|\partial_{\eta}w_{\tau}|^{2}d\tau d\xi d\eta\nonumber\\ &\lesssim \iint_{[0,X]\times [0,1]}\frac{|w_\tau(0,\xi,\eta)|^{2}}{w^{2}(0,\xi,\eta)}d\xi d\eta+\iint_{[0,T]\times [0,1]}\frac{\eta|w_{\tau}(\tau,0,\eta)|^{2}}{w^{2}(\tau,0,\eta)}d\tau d\eta\nonumber\\ &\quad+\iiint_{[0,T]\times[0,X]\times[0,1]}\frac{|w_{\tau}|^{2}}{w^{2}}\frac{|w_{\tau}+\eta w_{\xi}|}{w}d\tau d\xi d\eta+\iiint_{[0,T]\times[0,X]\times[0,1]}\frac{|w_{\tau}|^{2}}{w^{2}}|w\partial_{\eta}^{2}w|d\tau d\xi d\eta\nonumber\\ &\lesssim \iint_{[0,X]\times [0,1]}\left|\frac{w_\tau(0,\xi,\eta)}{w_{0}(\xi,\eta)}\right|^{2}d\xi d\eta+\iint_{[0,T]\times [0,1]}\eta\cdot\left|\frac{w_{\tau}(\tau,0,\eta)}{w_{1}(\tau,\eta)}\right|^{2}d\tau d\eta\nonumber\\ &\quad+\left(\left\|\frac{w_{\tau}+\eta w_{\xi}}{w}\right\|_{L^{\infty}_{\tau,\xi,\eta}}\right)\times\iiint_{[0,T]\times[0,X]\times[0,1]}\frac{|w_{\tau}|^{2}}{w^{2}}d\tau d\xi d\eta\nonumber\\ &\lesssim \mathcal{N}_{0}+\mathcal{N}_{1}\iiint_{[0,T]\times[0,X]\times[0,1]}\frac{|w_{\tau}|^{2}}{w^{2}}d\tau d\xi d\eta \end{align}\tag{190}\] Similarly, differentiating 182 with respect to \(\xi\) and testing against \(w_{\xi}w^{-2}\), we can also obtain \[\begin{align} &\sup_{\tau\in[0,T]}\iint_{[0,X]\times [0,1]}\frac{|w_\xi|^{2}}{w^{2}}d\xi d\eta+\sup_{\xi\in[0,X]}\iint_{[0,T]\times [0,1]}\frac{\eta|w_{\xi}|^{2}}{w^{2}}d\tau d\eta\\ &+\iiint_{[0,T]\times [0,X]\times [0,1]}|\partial_{\eta}w_{\xi}|^{2}d\tau d\xi d\eta\lesssim \mathcal{N}_{0}+\mathcal{N}_{1}\iiint_{[0,T]\times[0,X]\times[0,1]}\frac{|w_{\xi}|^{2}}{w^{2}}d\tau d\xi d\eta. \end{align}\]

Step 2: \(H^{2}\) Estimates. Next we estimate \(w_{\tau\tau},w_{\tau\xi}\) and \(w_{\xi\xi}\). To this end, we differentiate 189 with respect to \(\tau\), then \(w_{\tau\tau}\) satisfies the following equation \[\label{wtteq} \begin{cases} \partial_{\tau}w_{\tau\tau}+\eta\partial_{\xi}w_{\tau\tau}-2w\partial_{\eta}^{2}w\cdot w_{\tau\tau}-2\partial_{\eta}^{2}w(w_{\tau})^{2}-4ww_{\tau}\partial_{\eta}^{2}w_{\tau}-w^{2}\partial_{\eta}^{2}w_{\tau\tau}=0,\\ w_{\tau\tau}|_{\eta=1}=\partial_{\eta}w_{\tau\tau}|_{\eta=0}=0. \end{cases}\tag{191}\] Testing 191 by \(w_{\tau\tau}w^{-2}\) and then integrating in \([0,T]\times[0,X]\times[0,1]\), we get the following energy inequality \[\label{h2energyid} \begin{align} &\sup_{\tau\in[0,T]}\iint_{[0,X]\times [0,1]}\frac{|w_{\tau\tau}|^{2}}{w^{2}}d\xi d\eta+\sup_{\xi\in[0,X]}\iint_{[0,T]\times [0,1]}\frac{\eta|w_{\tau\tau}|^{2}}{w^{2}}d\tau d\eta\\ &\quad\quad+\iiint_{[0,T]\times [0,X]\times [0,1]}|\partial_{\eta}w_{\tau\tau}|^{2}d\tau d\xi d\eta\lesssim\mathcal{N}_{0}+\mathcal{I}_{1}+\mathcal{I}_{2}+\mathcal{I}_{3}. \end{align}\tag{192}\] Where \[\mathcal{I}_{1}:= \iiint_{[0,T]\times[0,X]\times[0,1]}\frac{|w_{\tau\tau}|^{2}\cdot|w_{\eta\eta}|}{w}d\tau d\xi d\eta,\] \[\mathcal{I}_{2}:=\iiint_{[0,T]\times[0,X]\times[0,1]}\frac{|w_{\tau}|^{2}\cdot|w_{\eta\eta}w_{\tau\tau}|}{w^{2}}d\tau d\xi d\eta,\] and \[\mathcal{I}_{3}:=\iiint_{[0,T]\times[0,X]\times[0,1]}\frac{|w_{\tau}w_{\tau\tau}\partial_{\eta}^{2}w_{\tau}|}{w}d\tau d\xi d\eta.\] Estimate of \(\mathcal{I}_{1}\). By 182 , \[\mathcal{I}_{1}\lesssim \mathcal{N}_{1}\iiint_{[0,T]\times[0,X]\times[0,1]}\frac{|w_{\tau\tau}|^{2}}{w^{2}}d\tau d\xi d\eta.\] Estimate of \(\mathcal{I}_{2}\). By Cauchy–Schwarz inequality, we have \[\begin{align} \mathcal{I}_{2}&\lesssim \left\|\frac{w_{\tau}}{w}\right\|_{L^{\infty}_{\tau,\xi,\eta}}\times\left\|w\partial_{\eta}^{2}w\right\|_{L^{\infty}_{\tau,\xi,\eta}}\times\left\|\frac{w_{\tau}}{w}\right\|_{L^{2}_{\tau,\xi,\eta}}\times \left\|\frac{w_{\tau\tau}}{w}\right\|_{L^{2}_{\tau,\xi,\eta}}\\ &\lesssim (\mathcal{N}_{1})^{2} \left\|\frac{w_{\tau}}{w}\right\|_{L^{2}_{\tau,\xi,\eta}}\times \left\|\frac{w_{\tau\tau}}{w}\right\|_{L^{2}_{\tau,\xi,\eta}}. \end{align}\] Estimate of \(\mathcal{I}_{3}\). The estimate of \(\mathcal{I}_{3}\) is a bit complicated since more derivatives are involved. Thanks to 182 and 189 , we have \[\begin{align} \partial_{\eta}^{2}w_{\tau}&=\frac{w_{\tau\tau}+\eta w_{\xi\tau}-2w\partial_{\eta}^{2}w\cdot w_{\tau}}{w^{2}}=\frac{w_{\tau\tau}+\eta w_{\xi\tau}}{w^{2}}-2\frac{w_{\tau}^{2}+\eta w_{\tau}w_{\xi}}{w^{3}}. \end{align}\] Hence, \[\begin{align} \mathcal{I}_{3}&\lesssim\mathcal{N}_{1}\left(\iiint_{[0,T]\times[0,X]\times[0,1]}\frac{|w_{\tau\tau}|^{2}}{w^{2}}d\tau d\xi d\eta+\iiint_{[0,T]\times[0,X]\times[0,1]}\frac{|w_{\xi\tau}|^{2}}{w^{2}}d\tau d\xi d\eta\right)\\ &\quad\quad+(\mathcal{N}_{1})^{2}\left\|\frac{w_{\tau}+\eta w_{\xi}}{w}\right\|_{L^{2}_{\tau,\xi,\eta}}\times \left\|\frac{w_{\tau\tau}}{w}\right\|_{L^{2}_{\tau,\xi,\eta}}. \end{align}\] Therefore, the conclusion of Lemma 17 follows from 190 , 192 , and bounds on \(\mathcal{I}_1, \mathcal{I}_2\), \(\mathcal{I}_3\). ◻

Proof. We only give a detailed proof in the case of arbitrarily large \(T>0\), and the proof of the case when \(X>0\) is arbitrary follows from a similar (indeed, simpler) argument. Since we only need to obtain a local solution in \(\xi\), we assume \(X\leq 1\) in the following calculations. The proof is divided into several steps.

Step 1. Energy bounds. By the bootstrap assumption ?? and Lemma 17, we have \[\label{2thenergy} \begin{align} \|w\|_{E_{2}}^2:=&\sup_{\tau\in[0,T]}\sum_{1\leq\alpha+\beta\leq 2}\int_{0}^{X}\int_{0}^{1}\bigg|\frac{\partial_{\tau}^{\alpha}\partial_{\xi}^{\beta}w}{w}\bigg|^{2}d\xi d\eta+\sup_{\xi\in[0,X]}\sum_{1\leq\alpha+\beta\leq 2}\int_{0}^{T}\int_{0}^{1}\eta\bigg|\frac{\partial_{\tau}^{\alpha}\partial_{\xi}^{\beta}w}{w}\bigg|^{2}d\tau d\eta\\ &+\sum_{1\leq \alpha+\beta\leq 2}\int_{0}^{T}\int_{0}^{X}\int_{0}^{1}\big|\partial_{\tau}^{\alpha}\partial_{\xi}^{\beta}\partial_{\eta}w\big|^{2}d\tau d\eta d\xi\\ \lesssim& \mathcal{N}_{0}+M^{2}\sum_{1\leq\alpha+\beta\leq 2}\int_{0}^{T}\int_{0}^{X}\int_{0}^{1}\bigg|\frac{\partial_{\tau}^{\alpha}\partial_{\xi}^{\beta}w}{w}\bigg|^{2}d\tau d\xi d\eta. \end{align}\tag{193}\] By Lemma 16, for any \(0<\iota\ll 1\), \[\begin{align} &\int_{0}^{T}\int_{0}^{X}\int_{0}^{1}\bigg|\frac{\partial_{\tau}^{\alpha}\partial_{\xi}^{\beta}w}{w}\bigg|^{2}d\tau d\xi d\eta\\ &=\int_{0}^{T}\int_{0}^{X}\int_{0}^{\iota}\bigg|\frac{\partial_{\tau}^{\alpha}\partial_{\xi}^{\beta}w}{w}\bigg|^{2}d\tau d\xi d\eta+\int_{0}^{T}\int_{0}^{X}\int_{\iota}^{1}\bigg|\frac{\partial_{\tau}^{\alpha}\partial_{\xi}^{\beta}w}{w}\bigg|^{2}d\tau d\xi d\eta\\ &\lesssim_{T} \iota\int_{0}^{T}\int_{0}^{X}\|\partial_{\tau}^{\alpha}\partial_{\xi}^{\beta}w\|_{L^{\infty}_{\eta}[0,1]}^{2}d\tau d\xi+\frac{X}{\iota}\sup_{\xi\in[0,X]}\int_{0}^{T}\int_{0}^{1}\eta\bigg|\frac{\partial_{\tau}^{\alpha}\partial_{\xi}^{\beta}w}{w}\bigg|^{2}d\tau d\eta\\ &\lesssim_{T} \iota\int_{0}^{T}\int_{0}^{X}\int_{0}^{1}|\partial_{\tau}^{\alpha}\partial_{\xi}^{\beta}\partial_{\eta}w|^{2}d\tau d\xi d\eta+\frac{X}{\iota}\sup_{\xi\in[0,X]}\int_{0}^{T}\int_{0}^{1}\eta\bigg|\frac{\partial_{\tau}^{\alpha}\partial_{\xi}^{\beta}w}{w}\bigg|^{2}d\tau d\eta. \end{align}\] Therefore, choosing \(\iota=\sqrt{X}\) and setting \[\label{requireone} X\ll_{T} M^{-4},\tag{194}\] the second term in the RHS of 193 can then be absorbed by the LHS of 193 . As a consequence, \[\label{2thenergyineq} \|w\|_{E_{2}}^2\lesssim \mathcal{N}_{0},\tag{195}\] provided that 194 holds.

Step 2. The maximal principle. For \(f\in\{w_{\tau},w_{\xi}\}\), one can check that \(f\) satisfies the following equation \[\begin{cases} \partial_{\tau}f+\eta\partial_{\xi}f-w^{2}\partial_{\eta}^{2}f-2w\partial_{\eta}^{2}w\cdot f=0,\\ f|_{\eta=1}=\partial_{\eta}f|_{\eta=0}=0. \end{cases}\] Define the function \(F\) and the differential operator \(\mathcal{L}\) as \[F(\tau,\xi,\eta):=f(\tau,\xi,\eta)-Me^{100M\xi}w,\qquad \mathcal{L}:=\partial_{\tau}+\eta\partial_{\xi}-w^{2}\partial_{\eta}^{2}-2w\partial_{\eta}^{2}w.\] By direct calculations we have \[\mathcal{L}[F]=-Me^{100M\xi}w\left(100\eta M-w\partial_{\eta}^{2}w\right).\] Therefore it follows from 182 and ?? that \[\mathcal{L}[F](\tau,\xi,\eta)<0,\quad \forall \tau\in[0,T], \xi\in[0,X], \eta\in[1/2,1).\] By 195 , Lemma 16 and Lemma 20 below (which are consequences of general anisotropic Sobolev type inequalities), where the lower bounds on \(w\) and the upper bound on \(\partial_\tau w, \partial_\xi w\) are independent of \(X\) when \(0<X\leq1\), we have \[\label{requiretwo} F|_{\eta=\frac{1}{2}}<0,\,\,F|_{\tau=0}<0,\,\, F|_{\xi=0}<0,\quad \text{provided\,\,that}\,\, M\gg \sqrt{\mathcal{N}_{0}}\,\,.\tag{196}\]

Therefore by the maximum principle, we can conclude that \[\label{oneside0461} F\leq 0,\;\forall (\tau,\xi,\eta)\in[0,T]\times[0,X]\times[1/2,1].\tag{197}\]

By applying the same argument for \(\widetilde{F}=f+Me^{100M\xi}w\), we obtain that \(\widetilde{F}\ge0\) on \([0,T]\times[0,X]\times[1/2,1]\). Combining this with 197 , we conclude that for \(M\gg\sqrt{\mathcal{N}_{0}}\), \[\label{farfieldbd} |w_{\tau},w_{\xi}|\leq M e^{100M\xi}w,\quad\forall (\tau,\xi,\eta)\in[0,T]\times[0,X]\times[1/2,1].\tag{198}\]

Therefore, the improved bound ?? holds in \([0,T]\times[0,X]\times[1/2,1]\) follows by taking \[\label{requirefour} X\leq\frac{M}{200}.\tag{199}\]

Step 3. Closing the bootstrap. To fully close the bootstrap, it remains to derive the improved bound ?? in \([0,T]\times[0,X]\times[0,\frac{1}{2}]\). To this end, by 195 and Lemma 20 below, we obtain for any \(1<p<\infty\) that \[\sup_{\xi\in[0,X],\tau\in[0,T]}\Big[\|w_{\tau}\|_{L^{p}(0,1/2)}+\|w_{\xi}\|_{L^{p}(0,1/2)}\Big]\lesssim_{p,T}\sqrt{\mathcal{N}_{0}}.\] Recall the standard interpolation inequality, \[\|f\|_{L^{\infty}(0,\frac{1}{2})}\lesssim\|f\|_{L^{p}(0,1/2)}^{\frac{\alpha}{\alpha+\frac{1}{p}}}\|f\|_{C^{\alpha}(0,1/2)}^{\frac{\frac{1}{p}}{\alpha+\frac{1}{p}}}.\] By Lemma 19 below and taking \(p\gg1\) and \(0<\alpha\ll1\), we then get that \[\label{nearfieldbd} \sup_{\xi\in[0,X], \tau\in[0,T]}\Big[\|w_{\tau}\|_{L^{\infty}(0,1/2)}+\|w_{\xi}\|_{L^{\infty}(0,1/2)}\Big]\lesssim_{T} X^{-\frac{1}{10}}\sqrt{\mathcal{N}_{0}}.\tag{200}\] By 200 and Lemma 16, we have \[\sup_{\tau\in[0,T],\xi\in[0,X],\eta\in[0,\frac{1}{2}]}\left|\frac{w_{\tau},w_{\xi}}{w}\right|\lesssim_{T}X^{-\frac{1}{10}}\sqrt{\mathcal{N}_{0}}.\] If we set \[\label{requirefive} X^{-\frac{1}{10}}\sqrt{\mathcal{N}_{0}}\ll_{T} M,\tag{201}\] then the improved bound ?? holds in \([0,T]\times[0,X]\times[0,\frac{1}{2}]\).

Combining all constraints of \(X\) and \(M\) in 194 , 196 , 199 and 201 , we can choose \(M\gg 1\) and \(X\ll1\) depending on \(T\) and \(w_{0},w_{1}\), such that ?? holds. Indeed, we can firstly fix \(X=c_1(T)M^{-4}\) with a sufficiently small \(c_1(T)>0\) to satisfy 194 and 199 , and then fix \(M=C_2(T)\langle\mathcal{N}_{0}\rangle^{5/6}\) with a sufficiently large \(C_2(T)>1\) to satisfy 196 and 201 . The higher–order energy bounds ?? follows from the same argument as in the first step. The proof of Lemma 18 is complete. ◻

Proof. For notational brevity, we assume \(T=1\). By the definition of \(\|\cdot\|_{E_{2}}\), we only need to prove that there exists \(\alpha>0\) such that \[\label{embedtwo} \begin{align} &\|f(t,x,y)\|_{C^{\alpha}([0,1]\times[0,X]\times[0,1])}\\ &\lesssim X^{-2}\bigg[\sup_{t\in[0,1]}\left(\|f\|_{L^{2}_{x,y}}+\|\partial_{t}f\|_{L^{2}_{x,y}}+\|\partial_{x}f\|_{L^{2}_{x,y}}\right)+\big\||\partial_{y}f|+|\partial_{t}\partial_{y}f|+|\partial_{x}\partial_{y}f|\big\|_{L^{2}_{t,x,y}}\bigg]. \end{align}\tag{202}\] In our applications, we take \(f\in\{\partial_\tau w, \partial_\xi w\}\). By standard rescaling argument, we only need to show that \[\label{embedone} \begin{align} &\|f(t,x,y)\|_{C^{\alpha}([0,1]\times[0,1]\times[0,1])}\\ &\lesssim \bigg[\sup_{t\in[0,1]}\left(\|f\|_{L^{2}_{x,y}}+\|\partial_{t}f\|_{L^{2}_{x,y}}+\|\partial_{x}f\|_{L^{2}_{x,y}}\right)+\big\||\partial_{y}f|+|\partial_{t}\partial_{y}f|+|\partial_{x}\partial_{y}f|\big\|_{L^{2}_{t,x,y}}\bigg]. \end{align}\tag{203}\]

Proof of 203 . We need to show that there exists \(\alpha>0\) such that \[\label{holderint} \begin{align} &|f(w)-f(\widetilde{w})|\\ &\lesssim |w-\widetilde{w}|^{\alpha}\times\left[\sup_{t\in[0,1]}\left(\|f\|_{L^{2}_{x,y}}+\|\partial_{t}f\|_{L^{2}_{x,y}}+\|\partial_{x}f\|_{L^{2}_{x,y}}\right)+\big\||\partial_{y}f|+|\partial_{t}\partial_{y}f|+|\partial_{x}\partial_{y}f|\big\|_{L^{2}_{t,x,y}}\right], \end{align}\tag{204}\] for \(w, \widetilde{w}\in [0,1]\times[0,1]\times[0,1]\).

Choosing a rectangle \(Q\) such that \(w, \widetilde{w}\in Q\), and the length of the side parallel to \(t,x,y\)-axis is \(\kappa,\kappa ,\kappa^{\frac{3}{4}}\), respectively. By Hölder’s inequality and Sobolev inequality, we have \[\label{controlonoscil} \begin{align} &\left|f(w)-\frac{1}{\kappa^{\frac{11}{4}}}\iiint_{Q}f(z)dz\right| \lesssim \frac{1}{\kappa^{\frac{11}{4}}}\iiint_{Q}|f(w)-f(z)|dz\\ &\quad \lesssim \frac{1}{\kappa^{\frac{11}{4}}}\iiint_{Q}\left|\int_{0}^{1}\frac{d}{ds}f\left(w_{1}+s(z_{1}-w_{1}),w_{2}+s(z_{2}-w_{2}),w_{3}+s^{\frac{3}{4}}(z_{3}-w_{3})\right)ds\right|dz\\ &\quad \lesssim \kappa^{\frac{1}{8}}\|\partial_{t}f,\partial_{x}f\|_{L^{\infty}_{t}L^{2}_{x,y}}+\kappa^{\frac{1}{8}}\|f_{y}\|_{L^{2}_{y}L^{14}_{t,x}}\\ &\quad \lesssim \kappa^{\frac{1}{8}}\left(\left\|\partial_{t}f,\partial_{x}f\right\|_{L^{\infty}_{t}L^{2}_{x,y}}+\left\|f\right\|_{L^{\infty}_{t}L^{2}_{x,y}}+\left\|f_{y}\right\|_{L^{2}_{y}H^1_{t,x}}\right). \end{align}\tag{205}\]

We can see that 205 implies 204 by the triangle inequality, hence we complete the proof. ◻

Proof. By the definition of \(\|w\|_{E_{2}}\), using 182 , Sobolev inequalities and Lemma 16, we only need to prove the following general functional inequalities \[\label{lemma8467goal1} \Big(\sup_{t\in[0,T]}\|f\|_{L^{p}[0,\frac{1}{2}]}\Big)^{2}\lesssim_{p,T} \int_{0}^{T}\int_{0}^{\frac{1}{2}}y\left(|f|^{2}+|f_{t}|^{2}+|f_{yy}|^{2}\right)dtdy,\tag{206}\]

\[\label{lemma8467goal2} \Big(\sup_{t\in[0,T],\eta\in[\frac{1}{4},\frac{3}{4}]}|f(t,y)|\Big)^{2}\lesssim_{T} \int_{0}^{T}\int_{\frac{1}{4}}^{\frac{3}{4}}\left(|f|^{2}+|f_{t}|^{2}+|f_{yy}|^{2}\right)dtdy.\tag{207}\] We only focus on 206 since the proof of 207 follows from a similar but slightly simpler argument. For notational conciseness we assume that \(T=1\). We can extend \(f(t,y)\) to \(\mathbb{T}\times \mathbb{R}^{+}\) satisfying \[\int_{\mathbb{T}}\int_{\mathbb{R}^+}y\left(|f|^{2}+|f_{t}|^{2}+|f_{yy}|^{2}\right)dtdy\lesssim\int_{0}^{T}\int_{0}^{\frac{1}{2}}y\left(|f|^{2}+|f_{t}|^{2}+|f_{yy}|^{2}\right)dtdy.\] We can expand the extended \(f\) using Fourier series in \(t\) as \[f(t,y)=\sum_{k\in\mathbb{Z}}\widehat{f}_{k}(y)e^{ik\cdot t}.\] By Plancheral’s identity, we only need to show that \[\label{ineqfourier} \|f\|_{L^{\infty}_{t}L^{p}_{y}(\mathbb{T}\times\mathbb{R}^+)}^{2}\lesssim_{p}\sum_{k\in\mathbb{Z}}\int_{0}^{\infty}y\left(\langle k\rangle^{2}\big|\widehat{f}_{k}(y)\big|^{2}+\big|\partial_{y}^{2}\widehat{f}_{k}(y)\big|^{2}\right)dy.\tag{208}\] Using the interpolation inequality 261 and a standard rescaling argument, we have the following estimates \[\label{ineqf4460} \begin{align} \|f\|_{L^{\infty}_{t}L^{p}_{y}(\mathbb{T}\times\mathbb{R}^+)}&\leq \sum_{k\in\mathbb{Z}}\Big\|\widehat{f}_{k}(y)\Big\|_{L^{p}_{y}}\lesssim\sum_{k\in\mathbb{Z}}\Big\|\sqrt{y}\widehat{f}_{k}(y)\Big\|_{L^{2}_{y}}^{\frac{p+1}{2p}}\Big\|\sqrt{y}\partial_{y}^{2}\widehat{f}_{k}(y)\Big\|_{L^{2}_{y}}^{\frac{p-1}{2p}}\\ &=\sum_{k\in\mathbb{Z}}\langle k\rangle^{-\frac{p+1}{2p}}\Big\|\langle k\rangle\sqrt{y}\widehat{f}_{k}(y)\Big\|_{L^{2}_{y}}^{\frac{p+1}{2p}}\Big\|\sqrt{y}\partial_{y}^{2}\widehat{f}_{k}(y)\Big\|_{L^{2}_{y}}^{\frac{p-1}{2p}}. \end{align}\tag{209}\] Then 208 follows from 209 and Hölder’s inequality, which completes the proof of Lemma 20. ◻

Proof of Theorem 4. Similarly to the proof of Proposition 17, the proof is contained with three main parts: (i) Approximation; (ii) Uniform bounds; (iii) Passing to limit.

Step 1: Approximation. Same to 161 , let \(w^{\epsilon}\) be the global solution of \[\label{approsyss} \begin{cases} \partial_{\tau}w^{\epsilon}+(\eta+\epsilon)\partial_{\xi}w^{\epsilon}-(w^{\epsilon}+\epsilon)^{2}\partial_{\eta}^{2}w^{\epsilon}=0,\;(\tau,\xi,\eta)\in \mathbb{R}^{+}\times\mathbb{R}^{+}\times (0,1),\\ w^{\epsilon}|_{\tau=0}=w_{0}^{\epsilon},\;w^{\epsilon}|_{\xi=0}=w_{1}^{\epsilon},\\ w^{\epsilon}|_{\eta=1}=\partial_{\eta}w^{\epsilon}|_{\eta=0}=0, \end{cases}\tag{210}\] where \(w_{0}^{\epsilon},w_{1}^{\epsilon}\) are suitable mollification of \(w_{0},w_{1}\) respectively, satisfying the compatibility conditions corresponding to 161 and certain uniform–in–\(\epsilon\) bounds, which are presented in Appendix 10 (see (CP1),(CP2) and (CP3) in Appendix 10.3). We refer to Appendix 11 that the classical solution \(w^\epsilon\) of 161 exists globally and is unique.

Step 2: Uniform bounds. Similar to the second step of Proposition 17, for any fixed \(T>0\), we have \[\label{approxiaboun} (1-\eta)^{m}\lesssim_{T}w^{\epsilon}(\tau,\xi,\eta)\lesssim_{T} (1-\eta)\log^{\frac{1}{2}}\left(\frac{10}{1-\eta}\right) \;\forall \tau\in[0,T],\xi\in[0,1],\eta\in [0,1].\tag{211}\] Furthermore, by the same argument of Lemma 17 and Lemma 18, for above fixed \(T>0\), there exists \(M>0\) and \(0<X\leq 1\) depends both on \(T\) and the given data \((w_{0},w_{1})\), but independent with \(\epsilon\), such that \[\label{approxiabound} \begin{align} &\sup_{\tau\in[0,T],\xi\in[0,X],\eta\in[0,1]}\left(\left|\frac{\partial_{\tau}w^{\epsilon}}{w^{\epsilon}+\epsilon}\right|+\left|\frac{\partial_{\xi}w^{\epsilon}}{w^{\epsilon}+\epsilon}\right|\right)+\sum_{\tau\in[0,T]}\sum_{1\leq\alpha+\beta\leq k}\int_{0}^{X}\int_{0}^{1}\left|\frac{\partial_{\tau}^{\alpha}\partial_{\xi}^{\beta}w^{\epsilon}}{w^{\epsilon}+\epsilon}\right|^{2}d\xi d\eta\\ &+\sum_{\xi\in[0,X]}\sum_{1\leq\alpha+\beta\leq k}\int_{0}^{T}\int_{0}^{1}\eta\left|\frac{\partial_{\tau}^{\alpha}\partial_{\xi}^{\beta}w^{\epsilon}}{w^{\epsilon}+\epsilon}\right|^{2}d\tau d\eta+\sum_{\alpha+\beta\leq k}\int_{0}^{T}\int_{0}^{X}\int_{0}^{1}|\partial_{\tau}^{\alpha}\partial_{\xi}^{\beta}\partial_{\eta}w^{\epsilon}|^{2}d\tau d\eta d\xi\leq M. \end{align}\tag{212}\]

Step 3: Passing to Limits. By Lemma 19, we have \[\sup_{\epsilon\in(0,1]}\left(\|w^{\epsilon}\|_{C^{\alpha}([0,T]\times[0,X]\times[0,1])}+\|\partial_{\tau}w^{\epsilon},\partial_{\xi}w^{\epsilon}\|_{C^{\alpha}([0,T]\times[0,X]\times[0,1])}\right)\lesssim M,\] for some generic constant \(\alpha>0\). Thanks to Arzela–Ascoli Theorem, there exists a subsequence \(w^{\epsilon_{n}}\) and a function \(w(\tau,\xi,\eta)\in C^{\alpha}([0,T]\times[0,X]\times[0,1])\) such that \[\label{convergenceunif} \begin{cases} w^{\epsilon_{n}}\rightarrow w,\;\text{uniformly\;in}\;[0,T]\times[0,X]\times[0,1],\\ \partial_{\tau}w^{\epsilon_{n}}\rightarrow \partial_{\tau}w,\;\text{uniformly\;in}\;[0,T]\times[0,X]\times[0,1],\\ \partial_{\xi}w^{\epsilon_{n}}\rightarrow \partial_{\xi}w,\;\text{uniformly\;in}\;[0,T]\times[0,X]\times[0,1],\\ (w^{\epsilon_{n}}+\epsilon_{n})^{2}\partial_{\eta}^{2}w^{\epsilon_{n}}\rightarrow w^{2}\partial_{\eta}^{2}w,\;\text{uniformly\;in}\;[0,T]\times[0,X]\times [0,1]. \end{cases}\tag{213}\] By 210 , we obtain \(w\) satisfies the original equation 182 . The desired bounds in Theorem 4 follow from 211 , 212 , 213 , and the lower semi–continuity of norms. The uniqueness follows from Proposition 16. Therefore, we finish the proof of Theorem 4. ◻

Proof of Theorem 5. The proof follows from the same argument as in the proof of Theorem [pfg1], see section 7.2. ◻

Proof of Theorem 2. Theorem 2 essentially follows from the combination of Theorem 3 on the global well-posedness of weak solutions and Theorem 5 on the local existence of classical solutions. We first note that the uniqueness statement in Theorem 2 follows from Proposition 16. Therefore, in the following we focus on the existence part.

Assume the initial data \(u_{0}\) and the in–flow data \(u_{1}\) satisfy Assumption 1. By Theorem 3, there exists a unique global weak solution \(u(t,x,y)\) satisfying 3 , with the properties stated in Theorem 3. In particular, \[\label{pfmainsmooth} u\in C^{\infty}((0,\infty)\times(0,\infty)\times[0,\infty)).\tag{214}\]

On the other hand, for arbitrary \(T,X>0\), by Theorem 5, there exist \(X_{T}>0\) and \(T_{X}>0\), such that Prandtl equation 3 admits a local–in–\(x\) classical solution \(u^{(1)}(t,x,y)\) in \([0,T]\times[0,X_{T}]\times[0,\infty)\) and a local–in–\(t\) classical solution \(u^{(2)}(t,x,y)\) in \([0,T_{X}]\times[0,X]\times[0,\infty)\).

By uniqueness of weak solutions, we have \[\label{weaktoclassical} \begin{align} u(t,x,y)\equiv u^{(1)}(t,x,y)\quad \text{in}\;[0,T]\times[0,X_{T}]\times[0,\infty),\\ u(t,x,y)\equiv u^{(2)}(t,x,y)\quad \text{in}\;[0,T_{X}]\times[0,X]\times[0,\infty). \end{align}\tag{215}\]

Therefore, the continuous differentiability of \(u(t,x,y)\) and the Hölder continuity of the associated derivatives \(\{\partial_{t}u,\partial_{x}u,\partial_{y}u,\partial_{y}^{2}u\}\) for \((t,x,y)\in[0,\infty)\times[0,\infty)\times[0,\infty)\) follow from 214 , 215 , and Theorem 5.

The proof of Theorem 2 is now complete. ◻

Proof. Without loss of generality we can assume that \(X=1\). We construct the solution of 233 in the following steps:

Step 1: Iterated system. We use the following iteration scheme for \(n\in \mathbb{Z}\cap[1,\infty)\), \[\label{iterappendix} \begin{cases} \partial_{t}w^{(n)}+(y+1)\partial_{x}w^{(n)}-(w^{(n-1)}+1)^{2}\partial_{y}^{2}w^{(n)}=0,\\ w^{(n)}|_{t=0}=w_{0},\;w^{(n)}|_{x=0}=w_{1},\\ w^{(n)}|_{y=1}=\partial_{y}w^{(n)}|_{y=0}=0. \end{cases}\tag{237}\] In the above \(w^{(0)}(t,x,y)\) is a suitable Whitney–type boundary extension of \(w_{0}\) and \(w_{1}\). Indeed, the compatibility conditions on \((w_{0},w_{1})\) ensures the following extension problem (\(g_k, h_k\) are computed through \((w_{0},w_{1})\) and the equation 233 ) \[\partial_{t}^{k}w^{(0)}|_{t=0}=g_{k}(x,y),\quad \partial_{x}^{k}w^{(0)}|_{t=0}=h_{k}(t,y),\] can be solved for some sufficiently regular extension function \(w^{(0)}\) on \([0,\infty)\times[0,\infty)\times[0,1]\), see Whitney [72] and Lions [73].

Step 2: Uniform–in–\(n\) energy bounds. By standard energy estimates and the Sobolev inequalities, we obtain the key energy inequality \[\label{standardenergyappendix} \sup_{t\in[0,\delta]}\mathcal{E}_{s,w^{(n)}}(t)+\mathcal{D}_{s,w^{(n)}}(\delta)\leq C(w_{0},w_{1})+\int_{0}^{\delta}\mathcal{P}_{s}\left(\mathcal{E}_{s,w^{(n-1)}}(\tau)\right)\cdot\mathcal{E}_{s,w^{(n)}}(\tau)\,d\tau,\tag{238}\] for any \(\delta>0\). Here \(\mathcal{P}_{s}\) is a polynomial depending on \(s\). We briefly present the key steps of deriving 238 :

• 238 is derived by standard energy estimates, through differentiating 233 to derive the equation of \(\partial_{x}^{k}w\) (respectively \(\partial_{y}^{k}w\)) and testing against \(\partial_{x}^{k}w\) (respectively \(\partial_{y}^{k}w\cdot\chi^{2}_{k}(y)\)).

• The terms that derivatives fall on cut–off functions \(\chi_{k}\) are bounded by the lower–level diffusion terms.

• Using the inequality \[\label{embedtohs} \|w(t)\|_{H^{s}_{x,y}}\lesssim \mathcal{E}_{s,w}(t)\tag{239}\] to control the lower–order derivatives in cubic (or higher power) terms appearing in the energy estimates;239 follows from 234 , 236 , and standard elliptic estimates 251 .

• In the energy estimates for \(\partial_{x}^{k}w\), the boundary conditions on \(y=0\) and \(y=1\) are identical as those for \(w\), and therefore do not contribute to the energy estimates; however, in the energy estimates for \(\partial_{y}^{k}w\), the boundary conditions become nontrivial and requires a careful treatment. The key observation is that, at each level of the energy estimates thanks to our choice of the cutoff functions \(\chi_k\), the boundary terms can be expressed through lower–order terms which are then bounded by the energy norm and diffusion terms. We will illustrate this point in detail when going through the energy estimates in subsection 11.2 to avoid repetitive arguments.

Now choosing \(K>1\) and \(0<\delta_{1}<1\) such that \[\label{choosepara} \sup_{t\in[0,1]}\mathcal{E}_{s,w^{(0)}}(t)\leq K,\quad C(w_{0},w_{1})+\delta_{1}\cdot\mathcal{P}_{s}(K)\cdot K\leq K,\tag{240}\] then by induction we obtain from 238 that \[\sup_{n\geq 1}\left(\sup_{t\in[0,\delta_{1}]}\mathcal{E}_{s,w^{(n)}}(t)+\mathcal{D}_{s,w^{(n)}}(\delta_{1})\right)\leq K.\]

Step 3. Convergence of the iteration sequence. For \(n\geq 2\), define \(\overline{w}^{(n)}=w^{(n)}-w^{(n-1)}\). It is clear that \(\overline{w}^{(n)}\) satisfies \[\begin{cases} \partial_{t}\overline{w}^{(n)}+y\partial_{x}\overline{w}^{(n)}-(w^{(n-1)}+1)^{2}\partial_{y}^{2}\overline{w}^{(n)}-(w^{(n-1)}+w^{(n-2)}+2)\partial_{y}^{2}w^{(n-1)}\cdot\overline{w}^{(n-1)}=0,\\ \overline{w}^{(n)}|_{t=0}=\overline{w}^{(n)}|_{x=0}=0,\\ \overline{w}^{(n)}|_{y=1}=\partial_{y}\overline{w}^{(n)}|_{y=0}=0. \end{cases}\] By standard energy estimates over \([0,1]\times[0,1]\), we obtain that \[\frac{d}{dt}\|\overline{w}^{(n)}(t)\|_{L^{2}_{x,y}}^{2}\lesssim (1+\|w^{(n-1)}(t)\|_{W^{2,\infty}_{y}}^{2})\|\overline{w}^{(n)}(t)\|_{L^{2}_{x,y}}^{2}+\|\overline{w}^{(n-1)}(t)\|_{L^{2}_{x,y}}^{2}.\] By Grönwall’s inequalities and Sobolev inequalities, there exists \(0<\delta\leq \delta_{1}\) such that \[\sup_{t\in[0,\delta]}\|\overline{w}^{(n)}\|_{L^{2}_{x,y}}\leq \frac{1}{2}\sup_{t\in[0,\delta]}\|\overline{w}^{(n-1)}\|_{L^{2}_{x,y}}.\] Therefore, the sequence \(w^{(n)}\) strongly converges to \(w(t,x,y)\) in \(C^{2}([0,\delta]\times[0,1]\times[0,1])\) with \[\label{solutionspace} \left(\sup_{t\in[0,\delta]}\mathcal{E}_{s,w}(t)+\mathcal{D}_{s,w}(\delta)\right)\leq K.\tag{241}\] Letting \(n\to\infty\) in 237 , we conclude the existence part. The uniqueness follows from standard maximal principle argument, which completes the proof of Proposition 18. ◻

Proof of Proposition [bdwy].. We can assume that \(0<\delta\ll 1\). Following analogous arguments as in Proposition 17, we have \[|\partial_{y}w(t,x,y)|\lesssim_{\delta} 1,\quad\text{provided\,\,that\,\,}\;t\geq \delta,\,x\geq \delta,\,0\leq y\leq 1-\delta.\]

It remains to consider the region \(1-\delta< y\leq1\). For fixed \((t_{0},x_{0},y_{0})\) with \(t_{0}\geq \delta, x_{0}\geq \delta\) and \(1-\delta<y_{0}<1\), we define \[V(t,x,y)=w\left(t_{0}+r_{0}^{2}t,x_{0}+r_{0}^{3}x+(y_{0}+1)r_{0}^{2}t,y_{0}+r_{0}y\right),\quad r_{0}=1-y_{0}.\] By direct computation, we see that \(V\) satisfies for \(|t|<1,\,|x|<1,\,|y|<1\), \[\partial_{t}V+y\partial_{x}V-(V+1)^{2}\partial_{y}^{2}V=0,\] and \[\|V\|_{L^{\infty}_{\{|t|<1,|x|<1,|y|<1\}}}\lesssim r_{0}\ll 1.\] In the above we used the following fact: \[\label{vanishrate} 0\leq w(t,x,y)\leq A\cdot (1-y),\quad A:=\left(\sup_{x,y}\frac{w_{0}}{1-y}+\sup_{t,y}\frac{w_{1}}{1-y}\right),\tag{256}\] which can be demonstrated by standard maximal principle argument.

By the same argument in the proof of Lemma 10 and the \(W^{2,p}\) estimate in the whole space [53], we have for all \(p\in(1,\infty)\), \[\label{w2pinter} \|(\partial_{y}V,\partial_{y}^{2}V)\|_{L^{p}_{\{|t|<\frac{1}{2},|x|<\frac{1}{2},|y|<\frac{1}{2}\}}}\lesssim \|V\|_{L^{\infty}_{\{|t|<1,|x|<1,|y|<1\}}}\lesssim_{p} r_{0}.\tag{257}\]

Let \(\chi(t,x,y)\) be the cut–off function supported in \(\{|t|<\frac{1}{2},|x|<\frac{1}{2},|y|<\frac{1}{2}\}\) and \[\chi(t,x,y)\equiv 1,\;|t|<\frac{1}{4},|x|<\frac{1}{4},|y|<\frac{1}{4}.\]

Then we have \[V(t,x,y)\chi(t,x,y)=\int_{|\tau|<\frac{1}{2},|\xi|<\frac{1}{2},|\eta|<\frac{1}{2}}\widetilde{\Gamma}(t,x,y;\tau,\xi,\eta)\mathcal{W}(\tau,\xi,\eta)\,d\tau d\xi d\eta,\] where \(\widetilde{\Gamma}(t,x,y;\tau,\xi,\eta)\) is the fundamental solution for the operator \(\mathcal{L}_{0}=\partial_{t}+y\partial_{x}-\partial_{y}^{2}\) in the whole space (see 26 ), and \[\mathcal{W}(\tau,\xi,\eta)=(V^{2}\partial_{\eta}^{2}V+2V\partial_{y}^{2}V)\cdot\chi+V(\partial_{\tau}\chi+\eta\partial_{\xi}\chi-\partial_{\eta}^{2}\chi)-2\partial_{\eta}V\partial_{\eta}\chi.\]

By 26 , 257 and Hölder’s inequality, we obtain that \[|\partial_{y}V(0,0,0)|\lesssim r_{0}.\] Therefore, \[|\partial_{y}w(t_{0},x_{0},y_{0})|\leq r_{0}^{-1}|\partial_{y}V(0,0,0)|\lesssim 1.\] The proof of Proposition [bdwy] is then complete. ◻

Proof. Proof of 260 : Clearly, \[\label{obviousbd} \int_{1}^{\infty}|f|^{2}dy\leq \int_{1}^{\infty}y|f|^{2}dy.\tag{263}\] By the mean–value theorem, there exists \(z\in [1,2]\) such that \[|f(z)|^{2}\lesssim \int_{1}^{2}y|f|^{2}dy.\] Hence, for any \(y\in [0,1]\), \[\label{ptcontrol} |f(y)|\lesssim \int_{0}^{2}|f'(y)|dy +|f(z)|\lesssim \left(\int_{0}^{2}|f'(y)|^{2}dy+\int_{1}^{2}y|f(y)|^{2}dy\right)^{\frac{1}{2}}.\tag{264}\] 260 follows from 263 and 264 .

Proof of 261 : We have \[\|f\|_{L^{\infty}(1,\infty)}\lesssim \|f\|_{L^{2}(1,\infty)}+\|f'\|_{L^{2}(1,\infty)}.\] Integrating by parts, we get that \[\begin{align} \int_{1}^{\infty}|f'(y)|^{2}dy\leq \|f\|_{L^{2}(1,\infty)}\|f''\|_{L^{2}(1,\infty)}+|f(1)|\cdot|f'(1)|. \end{align}\] By mean–value theorem, there exists \(y_{1}\in[\frac{1}{2},1],y_{2}\in[\frac{3}{2},2]\), such that \[|f(y_{1})|^{2}+|f(y_{2})|^{2}\lesssim \int_{\frac{1}{2}}^{2}y|f(y)|^{2}dy.\] Applying mean–value theorem again, we can conclude that there exists \(y_{3}\in[y_{1},y_{2}]\), such that \[|f'(y_{3})|^{2}\lesssim \int_{\frac{1}{2}}^{2}y|f(y)|^{2}dy.\] Thus, by Newton–Leibniz and Cauchy–Schwarz, \[\label{standard1d} |f'(1)|^{2}\lesssim \int_{\frac{1}{2}}^{2}y|f(y)|^{2}dy+ \int_{\frac{1}{2}}^{2}y|f''(y)|^{2}dy.\tag{265}\] Therefore, for \(p\geq 2\), \[\|f\|_{L^{p}(1,\infty)}\leq \|f\|_{L^{2}(1,\infty)}^{\frac{2}{p}}\|f\|_{L^{\infty}(1,\infty)}^{\frac{p-2}{p}}\lesssim \|\sqrt{y}f(y)\|_{L^{2}(\mathbb{R}^{+})}+\|\sqrt{y}f''(y)\|_{L^{2}(\mathbb{R}^{+})}.\] For \(1<p<2\), by Hölder’s inequality, \[\|f\|_{L^{p}(1,\infty)}\lesssim \|\sqrt{y}f\|_{L^{2}(1,\infty)}.\]

Next, we focus on the interval \(y\leq 1\). Integrating by parts, and applying 265 , we have \[\begin{align} \int_{0}^{1}|f'(y)|^{2}dy&=|f'(1)|^{2}-2\int_{0}^{1}yf'(y)f''(y)dy.\\ &\lesssim \int_{\frac{1}{2}}^{2}y|f(y)|^{2}dy+ \int_{\frac{1}{2}}^{2}y|f''(y)|^{2}dy+\left(\int_{0}^{1}|f'(y)|^{2}dy\right)^{\frac{1}{2}}\left(\int_{0}^{1}y^{2}|f''(y)|^{2}dy\right)^{\frac{1}{2}}. \end{align}\] Therefore, \[\|f\|_{L^{\infty}([0,1])}\leq |f(1)|+\int_{0}^{1}|f'(y)|dy\lesssim \|\sqrt{y}f(y)\|_{L^{2}(\mathbb{R}^{+})}+\|\sqrt{y}f''(y)\|_{L^{2}(\mathbb{R}^{+})}.\]

Proof of 262 : Notice that 262 is obvious if \(f''\equiv 0\), hence we may assume that \(f''\not\equiv 0\).

For \(p=\infty\), by the same method of 265 , we can obtain \[\label{standard1dd} \|f'(y)\|_{L^{\infty}(\mathbb{R}^{+})}\lesssim \|f(y)\|_{L^{\infty}(\mathbb{R}^{+})}+\|f''(y)\|_{L^{\infty}(\mathbb{R}^{+})}.\tag{266}\] Then 262 follows from standard rescaling argument.

For \(1\leq p<\infty\), by the same method of 265 , we can also obtain for each \(k\in\mathbb{Z}\cap[0,\infty)\), \[\int_{k}^{k+1}|f'(y)|^{p}dy\lesssim \int_{k}^{k+1}|f(y)|^{p}dy+\int_{k}^{k+1}|f''(y)|^{p}dy.\] Hence \[\label{standardoned} \|f'\|_{L^{p}(\mathbb{R}^{+})}\lesssim \left(\|f\|_{L^{p}(\mathbb{R}^{+})}^{p}+\|f''\|_{L^{p}(\mathbb{R}^{+})}^{p}\right)^{\frac{1}{p}}\leq \|f\|_{L^{p}(\mathbb{R}^{+})}+\|f''\|_{L^{p}(\mathbb{R}^{+})}.\tag{267}\] 262 follows from 267 and a standard rescaling argument. ◻

Proof. Let \(\Delta_{j}\) be the Littlewood–Paley projection operator. We shall use the interpolation inequality \[\label{newinterpo5} \|g'(y)\|_{L^{\infty}(\mathbb{R})}\lesssim \|g(y)\|_{L^{\infty}(\mathbb{R})}^{\frac{p-1}{2p-1}}\cdot\|g''(y)\|_{L^{p}(\mathbb{R})}^{\frac{p}{2p-1}}.\tag{268}\] The proof of 268 follows from the same argument as that for 262 .

In view of 268 and Bernstein’s inequality, we get that \[\begin{align} \|\partial_{y}f\|_{L^{\infty}(\mathbb{R}^{3})}&\leq \sum_{j\ge-1}\|\partial_{y}\Delta_{j}f\|_{L^{\infty}_{t,x}L^\infty_y(\mathbb{R}^{3})}\lesssim \sum_{j\ge-1}\|\Delta_{j}f\|_{L^{\infty}_{t,x,y}(\mathbb{R}^3)}^{\frac{p-1}{2p-1}}\|\partial_{y}^{2}\Delta_{j}f\|_{L^{\infty}_{t,x}L^{p}_{y}(\mathbb{R}^3)}^{\frac{p}{2p-1}}\\ &\lesssim \sum_{j\ge-1}\left(2^{-j\alpha}\|f\|_{C^{\alpha}(\mathbb{R}^3)}\right)^{\frac{p-1}{2p-1}}\left(2^{\frac{2j}{p}}\|\partial_{y}^{2}f\|_{L^{p}(\mathbb{R}^3)}\right)^{\frac{p}{2p-1}}\\ &\lesssim_{p,\alpha} \|f\|_{C^{\alpha}(\mathbb{R}^{3})}+\|\partial_{y}^{2}f\|_{L^{p}(\mathbb{R}^{3})}, \end{align}\] provided that \(p>\frac{2}{\alpha}+1\) (so that \(-\alpha\frac{p-1}{2p-1}+\frac{2}{2p-1}<0\)). ◻

Proof. Let \(s=\xi t\), and \(v(t,y)=V(s,y)\) then \(V\) satisfies the following parabolic equation \[\label{smallvis} \begin{cases} \partial_{s}V+iyV-\xi^{-1}\partial_{y}^{2}V=0,\\ V|_{s=0}=v_{0}(y),\;\partial_{y}V|_{y=0}=0. \end{cases}\tag{270}\] To establish ?? , we only need to show that (with \(\nu=\xi^{-1}\)) \[\label{goalenhanced} \|V\|_{L^{2}_{y}}\lesssim e^{-\nu^{\frac{1}{3}}s}\|v_{0}\|_{L^{2}_{y}}.\tag{271}\] By the Gearhart–Prüss type theorem proved in [75] (see also [76], [77], [78]), we only need to show that \[\label{opbound} \left\|(i\lambda+iy-\nu\partial_{y}^{2})^{-1}\right\|_{L^{2}_{y}(\mathbb{R}^+)\to L^{2}_{y}(\mathbb{R}^+)}\lesssim \nu^{-\frac{1}{3}},\tag{272}\] for each \(\lambda\in\mathbb{R}\). Equivalently, we need to prove that assuming \(f\) satisfies the following elliptic equation for \(y\ge0\), \[\begin{cases}\label{ode} i(y+\lambda)f-\nu\partial_{y}^{2}f=g,\\ \partial_{y}f|_{y=0}=0, \end{cases}\tag{273}\] then \[\|f\|_{L^{2}(\mathbb{R}^+)}\lesssim \nu^{-\frac{1}{3}}\|g\|_{L^{2}(\mathbb{R}^+)},\] holds for any \(\lambda\in\mathbb{R}\). By rescaling, we only need to consider \(\nu=1\). By linearity, we assume that \(\|g\|_{L^{2}(\mathbb{R}^+)}=1\).

Case 1. \(\lambda\geq 0\). First, testing \(\bar{f}\) in 273 , integraing by parts, and taking the real part, we get \[\label{H1control} \int_{0}^{\infty}|\partial_{y}f|^{2}dy\leq \|f\|_{L^{2}(\mathbb{R}^+)}\tag{274}\] Next, testing \(\bar{f}\) in 273 again, integrating by parts, and taking the imaginary part, we obtain that \[\label{L1control} \int_{0}^{\infty}(y+\lambda)|f|^2dy\leq \|f\|_{L^{2}(\mathbb{R}^+)}\tag{275}\] Hence we have \(\|f\|_{L^{2}(\mathbb{R}^+)}\lesssim 1\) by 274 , 275 , and Lemma 22.

Case 2. \(\lambda<0\). This case is more complicated as \(y+\lambda\) changes sign on \(\mathbb{R}^+\). To overcome this difficulty, define the following cut–off function \[\Theta(y):= \begin{cases} 1,\quad y\geq -\lambda+\delta,\\ 0,\quad -\lambda-\frac{\delta}{2}\leq y \leq -\lambda+\frac{\delta}{2},\\ -1,\quad y\leq -\lambda-\delta, \end{cases} \quad |\Theta'(y)|\leq \frac{4}{\delta},\] where the parameter \(\delta>0\) will be chosen later.

First, testing \(\bar{f}\) in 273 , we obtain 274 as above. Then testing 273 by \(\Theta(y)\bar{f}\), and taking imaginary part, by 274 we have \[\begin{align} \int_{|y+\lambda|\geq \delta}|y+\lambda||f|^{2}dy&\leq \|f\|_{L^{2}}+4\|\partial_{y}f\|_{L^{2}}\|f\|_{L^{2}}\delta^{-1}\leq\|f\|_{L^{2}}+4\delta^{-1}\|f\|_{L^{2}}^{\frac{3}{2}}. \end{align}\] Applying Lemma 22 we get that \[\label{L2control} \begin{align} \int_{0}^{\infty}|f|^{2}dy&\lesssim \int_{0}^{\infty}|\partial_{y}f|^{2}dy+\int_{0}^{\infty}|y+\lambda|\cdot|f|^{2}dy\\ &\lesssim\delta\int_{0}^{\infty}|f|^{2}dy+\int_{|y+\lambda|\geq \delta, \,y\in\mathbb{R}^+}|y+\lambda||f|^{2}dy+\int_{0}^{\infty}|\partial_{y}f|^{2}dy\\ &\lesssim \|f\|_{L^{2}}+4\delta^{-1}\|f\|_{L^{2}}^{\frac{3}{2}}+2\delta\|f\|_{L^{2}}^{2}. \end{align}\tag{276}\] Choosing \(\delta>0\) sufficiently small, by 276 and Hölder inequality we obtain that \(\|f\|_{L^{2}}\lesssim 1\). The proof of Proposition 20 is complete. ◻