3.1 A priori estimate on the velocity field
In this subsection, we will prove the estimate of solution to problem (2.17)–(2.19) on the velocity field u.
Lemma 3.1
It holds that for any \(t\in [0,T]\)
$$\begin{aligned} &\frac{d}{dt} \Vert u \Vert _{X_{\tau ,\alpha}} +\sum _{m\geq 0}\tau ^{m}M_{m} \frac{ \Vert \theta _{\alpha}\partial ^{m}_{x}\partial _{y}u \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}- \frac{\alpha}{2\langle t\rangle} \Vert u \Vert _{X_{\tau ,\alpha}}- \frac{C}{\langle t\rangle} \Vert b \Vert _{X_{\tau ,\alpha}} \\ &\quad \leq \dot{\tau}(t) \Vert u \Vert _{Y_{\tau ,\alpha}}+ \frac{C}{\tau (t)^{\frac{1}{2}}} \bigl(\langle t\rangle ^{-1/4}\bigl( \Vert u \Vert _{X_{ \tau ,\alpha}}+ \Vert b \Vert _{X_{\tau ,\alpha}}\bigr)+\langle t \rangle ^{1/4}\bigl( \Vert u \Vert _{D_{\tau ,\alpha}}+ \Vert b \Vert _{D_{\tau ,\alpha}}\bigr) \bigr) \\ &\quad \quad{}\times \bigl( \Vert u \Vert _{Y_{\tau ,\alpha}}+ \Vert b \Vert _{Y_{\tau ,\alpha}}\bigr)+ \Vert w \Vert _{X_{\tau ,\alpha}} - \frac{1}{4}\sum_{m\geq 0} \frac{ \Vert \theta _{\alpha}\partial _{x}^{m}\partial _{y}w \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}\tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}w \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}. \end{aligned}$$
Proof
For \(m\geq 0\), applying the operator \(\partial ^{m}_{x}\) on (2.17)1 and multiplying the resulting equation by \(\theta ^{2}_{\alpha}\partial _{x}^{m}u\), we derive that
$$\begin{aligned} &\int _{\mathbb{R}_{+}^{2}}\partial ^{m}_{x} \bigl( \partial _{t} u+\bigl(u^{s}+ \tilde{u}\bigr)\partial _{x}u+\tilde{v}\partial _{y}u-2\partial _{y}w- \partial ^{2}_{y}u-(b+1)\partial _{x}b \\ &\quad {}-g\partial _{y}b-2\partial _{y}^{2}u^{s}b+ \tilde{v}\partial _{y}^{2}u^{s} \psi \bigr) \theta ^{2}_{\alpha}\partial _{x}^{m}u\, dx\, dy=0. \end{aligned}$$
(3.1)
We now deal with each term in (3.1) as follows. For the first term, integrating it by parts with respect to time t, we have
$$\begin{aligned} \int _{\mathbb{R}_{+}^{2}}\partial ^{m}_{x}\partial _{t} u\theta ^{2}_{ \alpha}\partial _{x}^{m}u\, dx\, dy &= \frac{1}{2} \int _{\mathbb{R}_{+}^{2}} \partial _{t}\bigl(\partial ^{m}_{x}u\bigr)^{2}\theta ^{2}_{\alpha }\,dx\, dy \\ &=\frac{1}{2}\frac{d}{dt} \int _{\mathbb{R}_{+}^{2}}\bigl(\partial ^{m}_{x}u \bigr)^{2} \theta ^{2}_{\alpha }\,dx\, dy - \int _{\mathbb{R}_{+}^{2}}\bigl(\partial ^{m}_{x}u \bigr)^{2} \theta _{\alpha}\frac{d}{dt}\theta _{\alpha }\,dx\, dy \\ &=\frac{1}{2}\frac{d}{dt} \bigl\Vert \theta _{\alpha}\partial ^{m}_{x}u \bigr\Vert ^{2}_{L^{2}( \mathbb{R}_{+}^{2})}+\frac{\alpha}{4\langle t\rangle} \bigl\Vert \theta _{ \alpha }z\partial ^{m}_{x}u \bigr\Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}. \end{aligned}$$
Therefore, we obtain
$$\begin{aligned} &\frac{\int _{\mathbb{R}_{+}^{2}}\partial ^{m}_{x}\partial _{t} u\theta ^{2}_{\alpha}\partial _{x}^{m}u\, dx\, dy}{ \Vert \theta _{\alpha}\partial ^{m}_{x}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}= \frac{1}{2}\frac{d}{dt} \bigl\Vert \theta _{\alpha}\partial ^{m}_{x}u \bigr\Vert _{L^{2}( \mathbb{R}_{+}^{2})}+\frac{\alpha}{4\langle t\rangle} \frac{ \Vert \theta _{\alpha }z\partial ^{m}_{x}u \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial ^{m}_{x}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}. \end{aligned}$$
(3.2)
For the fifth term, integrating it by parts over \(\mathbb{R}_{+}^{2}\), we obtain
$$\begin{aligned} - \int _{\mathbb{R}_{+}^{2}}\partial ^{2}_{y}\partial ^{m}_{x} u \theta ^{2}_{\alpha} \partial _{x}^{m}u\, dx\, dy & = \bigl\Vert \theta _{\alpha} \partial ^{m}_{x}\partial _{y}u \bigr\Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}+ \int _{\mathbb{R}_{+}^{2}}\partial _{y}\partial ^{m}_{x} u\partial _{y}\bigl( \theta ^{2}_{\alpha}\bigr)\partial _{x}^{m}u\, dx\, dy \\ & = \bigl\Vert \theta _{\alpha}\partial ^{m}_{x} \partial _{y}u \bigr\Vert ^{2}_{L^{2}( \mathbb{R}_{+}^{2})}- \frac{1}{2} \int _{\mathbb{R}_{+}^{2}}\bigl(\partial ^{m}_{x} u \bigr)^{2}\partial ^{2}_{y}\bigl(\theta ^{2}_{\alpha}\bigr)\,dx\, dy \\ & = \bigl\Vert \theta _{\alpha}\partial ^{m}_{x} \partial _{y}u \bigr\Vert ^{2}_{L^{2}( \mathbb{R}_{+}^{2})}- \frac{\alpha}{2{\langle t\rangle}} \bigl\Vert \theta _{ \alpha}\partial ^{m}_{x}u \bigr\Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})} \\ &\quad {}- \frac{\alpha ^{2}}{2\langle t\rangle} \bigl\Vert \theta _{\alpha }z\partial ^{m}_{x}u \bigr\Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}, \end{aligned}$$
where we have used the fact that \(\partial ^{2}_{y}(\theta ^{2}_{\alpha})= \frac{\alpha}{{\langle t\rangle}}\theta ^{2}_{\alpha}+ \frac{\alpha ^{2}}{\langle t\rangle}z^{2}\theta ^{2}_{\alpha}\) and the boundary condition \(\partial _{x}^{m}u|_{y=0}=0\).
Thus,
$$\begin{aligned} &\frac{-\int _{\mathbb{R}_{+}^{2}}\partial ^{2}_{y}\partial ^{m}_{x} u\theta ^{2}_{\alpha}\partial _{x}^{m}u\, dx\, dy}{ \Vert \theta _{\alpha}\partial ^{m}_{x}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad = \frac{ \Vert \theta _{\alpha}\partial ^{m}_{x}\partial _{y}u \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial ^{m}_{x}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}- \frac{\alpha}{2{\langle t\rangle}} \bigl\Vert \theta _{\alpha}\partial ^{m}_{x}u \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} -\frac{\alpha ^{2}}{2\langle t\rangle} \frac{ \Vert \theta _{\alpha }z\partial ^{m}_{x}u \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial ^{m}_{x}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}. \end{aligned}$$
(3.3)
Next, we establish the estimates of the remainder terms in (3.1),
$$\begin{aligned} R_{1}&\overset{\bigtriangleup}{=} \int _{\mathbb{R}_{+}^{2}}\partial ^{m}_{x} \bigl( \bigl(\tilde{u}+u^{s}\bigr)\partial _{x}u\bigr)\theta ^{2}_{\alpha}\partial _{x}^{m}u\, dx\, dy= \sum_{i=0}^{m}\binom{m}{i} \int _{\mathbb{R}_{+}^{2}}\partial ^{m-i}_{x} \tilde{u}\partial ^{i+1}_{x}u\theta ^{2}_{\alpha} \partial _{x}^{m}u\, dx\, dy \\ & \leq \sum_{i=0}^{[m/2]}\binom{m}{i} \bigl\Vert \partial ^{m-i}_{x} \tilde{u} \bigr\Vert _{L^{2}_{x}L_{y}^{\infty}} \bigl\Vert \theta _{\alpha}\partial ^{i+1}_{x}u \bigr\Vert _{L^{\infty}_{x}L_{y}^{2}} \bigl\Vert \theta _{\alpha}\partial _{x}^{m}u \bigr\Vert _{L^{2}( \mathbb{R}_{+}^{2})} \\ &\quad{}+\sum_{i=[m/2]+1}^{m}\binom{m}{i} \bigl\Vert \partial ^{m-i}_{x} u \bigr\Vert _{L_{xy}^{\infty}} \bigl\Vert \theta _{\alpha}\partial ^{i+1}_{x}u \bigr\Vert _{L^{2}( \mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha}\partial _{x}^{m}u \bigr\Vert _{L^{2}( \mathbb{R}_{+}^{2})}. \end{aligned}$$
(3.4)
For \(0\leq i\leq [m/2]\), by using (2.9)1 and Lemma 2.2 (Agmon inequality in x), we deduce
$$\begin{aligned} \bigl\Vert \partial ^{m-i}_{x} \tilde{u} \bigr\Vert _{L^{2}_{x}L_{y}^{\infty}} & = \bigl\Vert \partial ^{m-i}_{x} \bigl(u+\partial _{y}u^{s}\psi \bigr) \bigr\Vert _{L^{2}_{x}L_{y}^{ \infty}} \leq \bigl\Vert \partial ^{m-i}_{x}u \bigr\Vert _{L^{2}_{x}L_{y}^{\infty}}+ \bigl\Vert \partial _{y}u^{s} \partial ^{m-i}_{x}\psi \bigr\Vert _{L^{2}_{x}L_{y}^{\infty}} \\ & \leq C \bigl\Vert \theta _{\alpha }\partial ^{m-i}_{x} u \bigr\Vert ^{\frac{1}{2}}_{L^{2}( \mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha }\partial _{y}\partial ^{m-i}_{x} u \bigr\Vert ^{\frac{1}{2}}_{L^{2}(\mathbb{R}_{+}^{2})}+C\langle t\rangle ^{- \frac{1}{4}} \bigl\Vert \theta _{\alpha}\partial ^{m-i}_{x} b \bigr\Vert _{L^{2}( \mathbb{R}_{+}^{2})} \end{aligned}$$
(3.5)
and
$$\begin{aligned} \bigl\Vert \theta _{\alpha}\partial ^{i+1}_{x}u \bigr\Vert _{L^{\infty}_{x}L_{y}^{2}} \leq C \bigl\Vert \theta _{\alpha} \partial ^{i+1}_{x}u \bigr\Vert ^{\frac{1}{2}}_{L^{2}( \mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha}\partial ^{i+2}_{x}u \bigr\Vert ^{ \frac{1}{2}}_{L^{2}(\mathbb{R}_{+}^{2})}. \end{aligned}$$
(3.6)
For \([m/2]\leq i\leq m\), by using (2.9)2, we arrive at
$$\begin{aligned} &\bigl\Vert \partial ^{m-i}_{x} \tilde{u} \bigr\Vert _{L_{xy}^{\infty}} \\ &\quad \leq C \bigl\Vert \theta _{\alpha }\partial ^{m-i}_{x} u \bigr\Vert ^{\frac{1}{4}}_{L^{2}( \mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha }\partial ^{m-i+1}_{x} u \bigr\Vert ^{ \frac{1}{4}}_{L^{2}(\mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha }\partial _{y} \partial ^{m-i}_{x} u \bigr\Vert ^{\frac{1}{4}}_{L^{2}(\mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha }\partial ^{m-i+1}_{x}\partial _{y} u \bigr\Vert ^{\frac{1}{4}}_{L^{2}( \mathbb{R}_{+}^{2})} \\ &\quad \quad{}+C\langle t\rangle ^{-\frac{1}{4}} \bigl\Vert \theta _{\alpha}\partial ^{m-i}_{x} b \bigr\Vert ^{\frac{1}{2}}_{L^{2}(\mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha} \partial ^{m-i+1}_{x} b \bigr\Vert ^{\frac{1}{2}}_{L^{2}(\mathbb{R}_{+}^{2})}. \end{aligned}$$
(3.7)
From now on, we apply \(X_{i}\), \(Y_{i}\), \(D_{i}\) to stand for the seminorms of function u, \(\bar{X}_{i}\), \(\bar{Y}_{i}\), \(\bar{D}_{i}\) to stand for the seminorms of function b, \(\tilde{X}_{i}\), \(\tilde{Y}_{i}\), and \(\tilde{D}_{i}\) denote the seminorms of function w.
Therefore,
$$\begin{aligned} &\sum_{m\geq 0} \frac{ \vert R_{1} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha} \partial _{x}^{m}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \leq \frac{C}{\tau (t)^{\frac{1}{2}}} \Biggl(\sum_{m\geq 0} \sum _{i=0}^{[m/2]}\binom{m}{i} \bigl(X^{\frac{1}{2}}_{m-i}D^{ \frac{1}{2}}_{m-i}+ \langle t\rangle ^{-\frac{1}{4}}\bar{X}^{ \frac{1}{2}}_{m-i} \bigr)Y^{\frac{1}{2}}_{i+1}Y^{\frac{1}{2}}_{i+2} \\ &\quad \quad{}+\sum_{m\geq 0}\sum _{i=[m/2]+1}^{m} \binom{m}{i} \bigl(X^{\frac{1}{4}}_{m-i}X^{\frac{1}{4}}_{m-i+1}D^{ \frac{1}{4}}_{m-i}D^{\frac{1}{4}}_{m-i+1}+ +\langle t\rangle ^{- \frac{1}{4}}\bar{X}^{\frac{1}{2}}_{m-i} \bar{X}^{\frac{1}{2}}_{m-i+1} \bigr)Y_{i+1} \Biggr). \end{aligned}$$
(3.8)
Similarly, we also have
$$\begin{aligned} R_{2}&\overset{\bigtriangleup}{=} \int _{\mathbb{R}_{+}^{2}}\partial ^{m}_{x} ( \tilde{v}\partial _{y}u)\theta ^{2}_{\alpha} \partial _{x}^{m}u\, dx\, dy= \sum _{i=0}^{m}\binom{m}{i} \int _{\mathbb{R}_{+}^{2}}\partial ^{m-i}_{x} \tilde{v}\partial ^{i}_{x}\partial _{y}u \theta ^{2}_{\alpha}\partial _{x}^{m}u\, dx\, dy \\ & \leq \sum_{i=0}^{[m/2]}\binom{m}{i} \bigl\Vert \partial ^{m-i}_{x} \tilde{v} \bigr\Vert _{L^{2}_{x}L_{y}^{\infty}} \bigl\Vert \theta _{\alpha}\partial ^{i}_{x} \partial _{y}u \bigr\Vert _{L^{\infty}_{x}L_{y}^{2}} \bigl\Vert \theta _{\alpha}\partial _{x}^{m}u \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} \\ &\quad{}+\sum_{i=[m/2]+1}^{m}\binom{m}{i} \bigl\Vert \partial ^{m-i}_{x} \tilde{v} \bigr\Vert _{L_{xy}^{\infty}} \bigl\Vert \theta _{\alpha}\partial ^{i}_{x} \partial _{y}u \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha} \partial _{x}^{m}u \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})}. \end{aligned}$$
(3.9)
For \(0\leq i\leq [m/2]\), by using (2.9)1 and Lemma 2.2 (Agmon inequality in x), we derive
$$\begin{aligned} \bigl\Vert \partial ^{m-i}_{x} \tilde{v} \bigr\Vert _{L^{2}_{x}L_{y}^{\infty}}& \leq \langle t\rangle ^{\frac{1}{4}} \bigl\Vert \theta _{\alpha }\partial _{y} \partial ^{m-i}_{x} \tilde{v} \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} \\ & \leq \langle t\rangle ^{\frac{1}{4}} \bigl\Vert \theta _{\alpha} \partial ^{m-i+1}_{x} \tilde{u} \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} \\ & \leq \langle t\rangle ^{\frac{1}{4}} \bigl( \bigl\Vert \theta _{\alpha} \partial ^{m-i+1}_{x} u \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})}+ \bigl\Vert \partial _{y}u^{s} \partial ^{m-i+1}_{x}\psi \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} \bigr) \\ & \leq \langle t\rangle ^{\frac{1}{4}} \bigl( \bigl\Vert \theta _{\alpha} \partial ^{m-i+1}_{x} u \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})}+\langle t \rangle ^{\frac{1}{2}} \bigl\Vert \partial ^{m-i+1}_{x}\psi \bigr\Vert _{L^{2}( \mathbb{R}_{+}^{2})} \bigr) \\ & \leq \langle t\rangle ^{\frac{1}{4}} \bigl( \bigl\Vert \theta _{\alpha} \partial ^{m-i+1}_{x} u \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})}+ \bigl\Vert \partial ^{m-i+1}_{x}b \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} \bigr) \end{aligned}$$
(3.10)
and
$$\begin{aligned} \bigl\Vert \theta _{\alpha}\partial ^{i}_{x} \partial _{y}u \bigr\Vert _{L^{\infty}_{x}L_{y}^{2}} \leq \bigl\Vert \theta _{\alpha}\partial ^{i}_{x}\partial _{y}u \bigr\Vert ^{\frac{1}{2}}_{L^{2}( \mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha}\partial ^{i+1}_{x}\partial _{y}u \bigr\Vert ^{\frac{1}{2}}_{L^{2}(\mathbb{R}_{+}^{2})}. \end{aligned}$$
(3.11)
For \([m/2]\leq i\leq m\), by using (2.9)2 and Lemma 2.3, we discover
$$\begin{aligned} \bigl\Vert \partial ^{m-i}_{x} \tilde{v} \bigr\Vert _{L^{\infty}_{x}L_{y}^{\infty}} & \leq \langle t\rangle ^{\frac{1}{4}} \bigl( \bigl\Vert \theta _{\alpha} \partial ^{m-i+1}_{x} u \bigr\Vert _{L^{2}_{y}L^{\infty}_{x}}+ \bigl\Vert \partial ^{m-i+1}_{x}b \bigr\Vert _{L_{y}^{2}L^{\infty}_{x}} \bigr) \\ & \leq \langle t\rangle ^{\frac{1}{4}} \bigl( \bigl\Vert \theta _{\alpha} \partial ^{m-i+1}_{x} u \bigr\Vert ^{\frac{1}{2}}_{L^{2}(\mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha}\partial ^{m-i+2}_{x} u \bigr\Vert ^{\frac{1}{2}}_{L^{2}( \mathbb{R}_{+}^{2})} \\ &\quad{}+ \bigl\Vert \partial ^{m-i+1}_{x}b \bigr\Vert ^{\frac{1}{2}}_{L^{2}(\mathbb{R}_{+}^{2})} \bigl\Vert \partial ^{m-i+2}_{x}b \bigr\Vert ^{\frac{1}{2}}_{L^{2}(\mathbb{R}_{+}^{2})} \bigr). \end{aligned}$$
(3.12)
Hence,
$$\begin{aligned} &\sum_{m\geq 0} \frac{ \vert R_{2} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \leq \frac{C}{\tau (t)^{\frac{1}{2}}} \Biggl(\sum_{m \geq 0} \sum_{i=0}^{[m/2]}\binom{m}{i} \bigl(\langle t \rangle ^{ \frac{1}{4}}Y_{m-i+1}+\langle t\rangle ^{\frac{1}{4}} \bar{Y}_{m-i+1} \bigr)D^{\frac{1}{2}}_{i}D^{\frac{1}{2}}_{i+1} \\ &\quad \quad{}+\sum_{m\geq 0}\sum _{i=[m/2]+1}^{m} \binom{m}{i} \bigl(\langle t\rangle ^{\frac{1}{4}}Y^{\frac{1}{2}}_{m-i+1}Y^{ \frac{1}{2}}_{m-i+2}+ \langle t\rangle ^{\frac{1}{4}}\bar{Y}^{ \frac{1}{2}}_{m-i+1} \bar{Y}^{\frac{1}{2}}_{m-i+2} \bigr)D_{i} \Biggr). \end{aligned}$$
(3.13)
For the following term, we apply the Hölder inequality, leading to
$$\begin{aligned} R_{3}\overset{\bigtriangleup}{=} \int _{\mathbb{R}_{+}^{2}}\partial ^{m}_{x} \partial _{y}w\theta ^{2}_{\alpha}\partial _{x}^{m}u\, dx\, dy\leq \bigl\Vert \theta _{\alpha} \partial _{x}^{m}\partial _{y}w \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha}\partial _{x}^{m}u \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})}. \end{aligned}$$
Therefore,
$$\begin{aligned} &\sum_{m\geq 0} \frac{ \vert R_{3} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \leq \frac{1}{4}\sum_{m\geq 0} \frac{ \Vert \theta _{\alpha}\partial _{x}^{m}\partial _{y}w \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}\tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}w \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} +\sum_{m\geq 0} \bigl\Vert \theta _{\alpha}\partial _{x}^{m}w \bigr\Vert _{L^{2}( \mathbb{R}_{+}^{2})}\tau ^{m}M_{m}. \end{aligned}$$
(3.14)
Also, \(b=\tilde{b}\) such that
$$\begin{aligned} R_{4}&\overset{\bigtriangleup}{=} \int _{\mathbb{R}_{+}^{2}}\partial ^{m}_{x} \bigl((b+1)\partial _{x}b\bigr)\theta ^{2}_{\alpha} \partial _{x}^{m}u\, dx\, dy=\sum_{i=0}^{m} \binom{m}{i} \int _{\mathbb{R}_{+}^{2}}\partial ^{m-i}_{x} b \partial ^{i+1}_{x}b\theta ^{2}_{\alpha} \partial _{x}^{m}u\, dx\, dy \\ & \leq \sum_{i=0}^{[m/2]}\binom{m}{i} \bigl\Vert \partial ^{m-i}_{x} b \bigr\Vert _{L^{2}_{x}L_{y}^{ \infty}} \bigl\Vert \theta _{\alpha}\partial ^{i+1}_{x}b \bigr\Vert _{L^{\infty}_{x}L_{y}^{2}} \bigl\Vert \theta _{\alpha}\partial _{x}^{m}u \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} \\ &\quad{}+\sum_{i=[m/2]+1}^{m}\binom{m}{i} \bigl\Vert \partial ^{m-i}_{x} b \bigr\Vert _{L_{xy}^{\infty}} \bigl\Vert \theta _{\alpha}\partial ^{i+1}_{x}b \bigr\Vert _{L^{2}( \mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha}\partial _{x}^{m}u \bigr\Vert _{L^{2}( \mathbb{R}_{+}^{2})}. \end{aligned}$$
(3.15)
For \(0\leq i\leq [m/2]\), by using (2.9)1 and Lemma 2.2 (Agmon inequality in x), we conclude
$$\begin{aligned} \bigl\Vert \partial ^{m-i}_{x} b \bigr\Vert _{L^{2}_{x}L_{y}^{\infty}}\leq \bigl\Vert \theta _{ \alpha }\partial ^{m-i}_{x} b \bigr\Vert ^{\frac{1}{2}}_{L^{2}(\mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha }\partial _{y}\partial ^{m-i}_{x} b \bigr\Vert ^{\frac{1}{2}}_{L^{2}( \mathbb{R}_{+}^{2})} \end{aligned}$$
(3.16)
and
$$\begin{aligned} \bigl\Vert \theta _{\alpha}\partial ^{i+1}_{x}b \bigr\Vert _{L^{\infty}_{x}L_{y}^{2}} \leq \bigl\Vert \theta _{\alpha}\partial ^{i+1}_{x}b \bigr\Vert ^{\frac{1}{2}}_{L^{2}( \mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha}\partial ^{i+2}_{x}b \bigr\Vert ^{ \frac{1}{2}}_{L^{2}(\mathbb{R}_{+}^{2})}. \end{aligned}$$
(3.17)
For \([m/2]\leq i\leq m\), by using (2.9)2, we arrive at
$$\begin{aligned} \bigl\Vert \partial ^{m-i}_{x} b \bigr\Vert _{L_{xy}^{\infty}} &\leq \bigl\Vert \theta _{\alpha } \partial ^{m-i}_{x} b \bigr\Vert ^{\frac{1}{4}}_{L^{2}(\mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha }\partial ^{m-i+1}_{x} b \bigr\Vert ^{\frac{1}{4}}_{L^{2}( \mathbb{R}_{+}^{2})} \\ &\quad{}\times \bigl\Vert \theta _{\alpha }\partial _{y} \partial ^{m-i}_{x} b \bigr\Vert ^{\frac{1}{4}}_{L^{2}(\mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha } \partial ^{m-i+1}_{x} \partial _{y} b \bigr\Vert ^{\frac{1}{4}}_{L^{2}( \mathbb{R}_{+}^{2})}. \end{aligned}$$
(3.18)
Therefore,
$$\begin{aligned} \sum_{m\geq 0} \frac{ \vert R_{4} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} & \leq \frac{C}{\tau (t)^{\frac{1}{2}}} \Biggl(\sum_{m\geq 0} \sum _{i=0}^{[m/2]}\binom{m}{i}\bar{X}^{\frac{1}{2}}_{m-i} \bar{D}^{\frac{1}{2}}_{m-i}\bar{Y}^{\frac{1}{2}}_{i+1} \bar{Y}^{ \frac{1}{2}}_{i+2} \\ &\quad{} + \sum_{m\geq 0}\sum _{i=[m/2]+1}^{m} \binom{m}{i}\bar{X}^{\frac{1}{4}}_{m-i} \bar{X}^{\frac{1}{4}}_{m-i+1} \bar{D}^{\frac{1}{4}}_{m-i} \bar{D}^{\frac{1}{4}}_{m-i+1}\bar{Y}_{i+1} \Biggr). \end{aligned}$$
(3.19)
As a consequence, we obtain
$$\begin{aligned} \sum_{m\geq 0} \frac{ \vert R_{5} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} &\leq \frac{C}{\tau (t)^{\frac{1}{2}}} \Biggl(\sum_{m\geq 0} \sum _{i=0}^{[m/2]}\binom{m}{i}\langle t\rangle ^{\frac{1}{4}} \bar{Y}_{m-i+1}\bar{D}^{\frac{1}{2}}_{i} \bar{D}^{\frac{1}{2}}_{i+1} \\ &\quad{}+\sum_{m\geq 0}\sum _{i=[m/2]+1}^{m} \binom{m}{i}\langle t\rangle ^{\frac{1}{4}}\bar{Y}^{\frac{1}{2}}_{m-i+1} \bar{Y}^{\frac{1}{2}}_{m-i+2} \bar{D}_{i} \Biggr). \end{aligned}$$
(3.20)
Similar to [30], we have
$$\begin{aligned} \biggl\vert \int _{\mathbb{R}_{+}^{2}}\partial _{y}^{2}u^{s} \partial _{x}^{m}b \theta ^{2}_{\alpha} \partial _{x}^{m}u\, dx\, dy \biggr\vert \leq C\langle t \rangle ^{-1} \bigl\Vert \theta _{\alpha}\partial _{x}^{m}u \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha}\partial _{x}^{m}b \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})}, \end{aligned}$$
(3.21)
which implies
$$\begin{aligned} \sum_{m\geq 0} \frac{ \vert R_{6} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}b \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \leq C\langle t \rangle ^{-1}\sum_{m\geq 0} \bigl\Vert \theta _{\alpha} \partial _{x}^{m}b \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})}\tau ^{m}M_{m}, \end{aligned}$$
(3.22)
where we used the fact that \(\|\partial _{y}^{2}u^{s}\|_{L^{\infty}_{y}}\leq C\langle t\rangle ^{-1}\).
Analogously, we have
$$\begin{aligned} R_{7}&\overset{\bigtriangleup}{=} \biggl\vert \int _{\mathbb{R}_{+}^{2}} \partial _{y}^{2}u^{s} \partial _{x}^{m}(\tilde{v}\psi ) \theta ^{2}_{ \alpha}\partial _{x}^{m}u\, dx\, dy \biggr\vert \\ & \leq \sum_{i=0}^{[m/2]}\binom{m}{i} \bigl\Vert \partial ^{m-i}_{x} \tilde{v} \bigr\Vert _{L^{2}_{x}L_{y}^{\infty}} \bigl\Vert \theta _{\alpha}\partial _{y}^{2}u^{s} \bigr\Vert _{L_{y}^{2}} \bigl\Vert \partial ^{i}_{x} \psi \bigr\Vert _{L_{xy}^{\infty}} \bigl\Vert \theta _{ \alpha}\partial _{x}^{m}u \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} \\ &\quad{}+\sum_{i=[m/2]+1}^{m}\binom{m}{i} \bigl\Vert \partial ^{m-i}_{x} \tilde{v} \bigr\Vert _{L_{xy}^{\infty}} \bigl\Vert \theta _{\alpha}\partial _{y}^{2}u^{s} \bigr\Vert _{L_{y}^{2}} \bigl\Vert \partial ^{i}_{x} \psi \bigr\Vert _{L^{2}_{x}L_{y}^{\infty}} \bigl\Vert \theta _{\alpha}\partial _{x}^{m}u \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})}. \end{aligned}$$
(3.23)
For \(0\leq i\leq [m/2]\), by using (2.9)1 and Lemma 2.2 (Agmon inequality in x), leads to
$$\begin{aligned} \bigl\Vert \partial ^{m-i}_{x} \tilde{v} \bigr\Vert _{L^{2}_{x}L_{y}^{\infty}} \leq \langle t\rangle ^{\frac{1}{4}} \bigl( \bigl\Vert \theta _{\alpha}\partial ^{m-i+1}_{x} u \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})}+ \bigl\Vert \partial ^{m-i+1}_{x}b \bigr\Vert _{L^{2}( \mathbb{R}_{+}^{2})} \bigr) \end{aligned}$$
(3.24)
and
$$ \bigl\Vert \theta _{\alpha}\partial _{y}^{2}u^{s} \bigr\Vert _{L_{y}^{2}}\leq C\langle t \rangle ^{-\frac{3}{4}}. $$
By using (2.9)2 and Lemma 2.3, we obtain
$$\begin{aligned} \bigl\Vert \partial ^{i}_{x} \psi \bigr\Vert _{L_{xy}^{\infty}}\leq C\langle t\rangle ^{ \frac{1}{4}} \bigl\Vert \theta _{\alpha }\partial ^{i}_{x} b \bigr\Vert ^{\frac{1}{2}}_{L^{2}( \mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha }\partial ^{i+1}_{x} b \bigr\Vert ^{ \frac{1}{2}}_{L^{2}(\mathbb{R}_{+}^{2})}. \end{aligned}$$
(3.25)
For \([m/2]\leq i\leq m\), by using (2.9)2 and Lemma 2.3, we find
$$\begin{aligned} \bigl\Vert \partial ^{m-i}_{x} \tilde{v} \bigr\Vert _{L^{\infty}_{x}L_{y}^{\infty}} &\leq C\langle t\rangle ^{\frac{1}{4}} \bigl( \bigl\Vert \theta _{\alpha}\partial ^{m-i+1}_{x} u \bigr\Vert ^{\frac{1}{2}}_{L^{2}(\mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha} \partial ^{m-i+2}_{x} u \bigr\Vert ^{\frac{1}{2}}_{L^{2}(\mathbb{R}_{+}^{2})} \\ &\quad{}+ \bigl\Vert \partial ^{m-i+1}_{x}b \bigr\Vert ^{\frac{1}{2}}_{L^{2}(\mathbb{R}_{+}^{2})} \bigl\Vert \partial ^{m-i+2}_{x}b \bigr\Vert ^{\frac{1}{2}}_{L^{2}(\mathbb{R}_{+}^{2})} \bigr) \end{aligned}$$
(3.26)
and
$$\begin{aligned} \bigl\Vert \partial ^{i}_{x} \psi \bigr\Vert _{L^{2}L_{y}^{\infty}}\leq C\langle t \rangle ^{\frac{1}{4}} \bigl\Vert \theta _{\alpha }\partial ^{i}_{x} b \bigr\Vert _{L^{2}( \mathbb{R}_{+}^{2})} . \end{aligned}$$
(3.27)
Accordingly,
$$\begin{aligned} &\sum_{m\geq 0} \frac{ \vert R_{7} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \leq \frac{C}{\tau (t)^{\frac{1}{2}}} \Biggl(\sum_{m\geq 0} \sum _{i=0}^{[m/2]}\binom{m}{i}\langle t\rangle ^{-\frac{1}{4}}({Y}_{m-i+1}+ \bar{Y}_{m-i+1}) \bar{X}^{\frac{1}{2}}_{i}\bar{X}^{\frac{1}{2}}_{i+1} \\ &\quad \quad{}+\sum_{m\geq 0}\sum _{i=[m/2]+1}^{m} \binom{m}{i}\langle t\rangle ^{-\frac{1}{4}}\bigl({Y}^{\frac{1}{2}}_{m-i+1}{Y}^{ \frac{1}{2}}_{m-i+2}+ \bar{Y}^{\frac{1}{2}}_{m-i+1}\bar{Y}^{ \frac{1}{2}}_{m-i+2} \bigr)\bar{X}_{i} \Biggr). \end{aligned}$$
(3.28)
Collecting the previous estimates (3.8), (3.13), (3.14), (3.19), (3.22), and (3.28) and summing up \(m\geq 0\) gives
$$\begin{aligned} &\frac{d}{dt} \Vert u \Vert _{X_{\tau ,\alpha}}+\sum _{m\geq 0} \frac{\alpha (1-2\alpha )}{4\langle t\rangle}\tau ^{m}M_{m} \frac{ \Vert \theta _{\alpha }z\partial ^{m}_{x}u \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} +\sum_{m\geq 0}\tau ^{m}M_{m} \frac{ \Vert \theta _{\alpha}\partial ^{m}_{x}\partial _{y}u \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \quad{}-\frac{\alpha}{2\langle t\rangle} \Vert u \Vert _{X_{\tau ,\alpha}}- \frac{C}{\langle t\rangle} \Vert b \Vert _{X_{\tau ,\alpha}} \\ &\quad \leq \dot{\tau}(t) \Vert u \Vert _{Y_{\tau ,\alpha}}+ \frac{C}{\tau (t)^{\frac{1}{2}}} \bigl(\langle t\rangle ^{-1/4}\bigl( \Vert u \Vert _{X_{ \tau ,\alpha}}+ \Vert b \Vert _{X_{\tau ,\alpha}}\bigr)+\langle t \rangle ^{1/4}\bigl( \Vert u \Vert _{D_{\tau ,\alpha}}+ \Vert b \Vert _{D_{\tau ,\alpha}}\bigr) \bigr) \\ &\quad \quad{}\times \bigl( \Vert u \Vert _{Y_{\tau ,\alpha}}+ \Vert b \Vert _{Y_{\tau ,\alpha}}\bigr)+ \Vert w \Vert _{X_{\tau ,\alpha}}- \frac{1}{4}\sum_{m\geq 0} \frac{ \Vert \theta _{\alpha}\partial _{x}^{m}\partial _{y}w \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})} \tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}w \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}, \end{aligned}$$
(3.29)
where we have used the fact that for any positive sequences \(\{a_{j}\}_{j\geq 0}\) and \(\{b_{j}\}_{j\geq 0}\),
$$\begin{aligned} \sum_{m\geq 0}\sum_{j\geq 0}^{m}a_{j}b_{m-j} \leq \sum_{j\geq 0}a_{j}\sum _{j\geq 0}b_{j}. \end{aligned}$$
Choosing suitable \(\alpha \leq \frac{1}{2}\) in (3.29) gives
$$\begin{aligned} &\frac{d}{dt} \Vert u \Vert _{X_{\tau ,\alpha}} +\sum _{m\geq 0}\tau ^{m}M_{m} \frac{ \Vert \theta _{\alpha}\partial ^{m}_{x}\partial _{y}u \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}- \frac{\alpha}{2\langle t\rangle} \Vert u \Vert _{X_{\tau ,\alpha}}- \frac{C}{\langle t\rangle} \Vert b \Vert _{X_{\tau ,\alpha}} \\ &\quad \leq \dot{\tau}(t) \Vert u \Vert _{Y_{\tau ,\alpha}}+ \frac{C}{\tau (t)^{\frac{1}{2}}} \bigl(\langle t\rangle ^{-1/4}\bigl( \Vert u \Vert _{X_{ \tau ,\alpha}}+ \Vert b \Vert _{X_{\tau ,\alpha}}\bigr)+\langle t \rangle ^{1/4}\bigl( \Vert u \Vert _{D_{\tau ,\alpha}}+ \Vert b \Vert _{D_{\tau ,\alpha}}\bigr) \bigr) \\ &\quad \quad{}\times \bigl( \Vert u \Vert _{Y_{\tau ,\alpha}}+ \Vert b \Vert _{Y_{\tau ,\alpha}}\bigr)+ \Vert w \Vert _{X_{\tau ,\alpha}} - \frac{1}{4}\sum_{m\geq 0} \frac{ \Vert \theta _{\alpha}\partial _{x}^{m}\partial _{y}w \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}\tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}w \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}. \end{aligned}$$
(3.30)
The proof is then complete. □
3.2 A priori estimate on the magnetic field
In this subsection, we will derive the estimate of solution to problem (2.17)–(2.19) on the magneto field b.
Lemma 3.2
It holds that for any \(t\in [0,T]\)
$$\begin{aligned} &\frac{d}{dt} \Vert b \Vert _{X_{\tau ,\alpha}} +\sum _{m\geq 0}\tau ^{m}M_{m} \frac{ \Vert \theta _{\alpha}\partial ^{m}_{x}\partial _{y}b \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}b \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} -\frac{\alpha}{2\langle t\rangle} \Vert b \Vert _{X_{\tau ,\alpha}} \\ &\quad \leq \dot{\tau}(t) \Vert b \Vert _{Y_{\tau ,\alpha}}+ \frac{C}{\tau (t)^{\frac{1}{2}}} \bigl(\langle t\rangle ^{-1/4}\bigl( \Vert u \Vert _{X_{ \tau ,\alpha}}+ \Vert b \Vert _{X_{\tau ,\alpha}}\bigr)+\langle t \rangle ^{1/4}\bigl( \Vert u \Vert _{D_{\tau ,\alpha}}+ \Vert b \Vert _{D_{\tau ,\alpha}}\bigr) \bigr) \\ &\quad \quad{}\times \bigl( \Vert u \Vert _{Y_{\tau ,\alpha}}+ \Vert b \Vert _{Y_{\tau ,\alpha}}\bigr)+ \Vert w \Vert _{X_{\tau ,\alpha}}. \end{aligned}$$
Proof
For \(m\geq 0\), applying the operator \(\partial ^{m}_{x}\) on (2.17)2 and multiplying the resulting equation by \(\theta ^{2}_{\alpha}\partial _{x}^{m}b\), we derive that
$$\begin{aligned} &\int _{\mathbb{R}_{+}^{2}}\partial ^{m}_{x} \bigl( \partial _{t} b+\bigl(u^{s}+ \tilde{u}\bigr)\partial _{x}b+\tilde{v}\partial _{y}b-\partial ^{2}_{y}b-(b+1) \partial _{x}u \\ &\quad {}-g\partial _{y}u-g\partial _{y}^{2}u^{s} \psi \bigr) \theta ^{2}_{\alpha}\partial _{x}^{m}b\, dx\, dy=0. \end{aligned}$$
(3.31)
Similar to the estimates of (3.1), we now deal with each term in (3.31) as follows. For the first term, we have
$$\begin{aligned} \frac{\int _{\mathbb{R}_{+}^{2}}\partial ^{m}_{x}\partial _{t} b\theta ^{2}_{\alpha}\partial _{x}^{m}b\, dx\, dy}{ \Vert \theta _{\alpha}\partial ^{m}_{x}b \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}= \frac{1}{2}\frac{d}{dt} \bigl\Vert \theta _{\alpha}\partial ^{m}_{x}b \bigr\Vert _{L^{2}( \mathbb{R}_{+}^{2})}+\frac{\alpha}{4\langle t\rangle} \frac{ \Vert \theta _{\alpha }z\partial ^{m}_{x}b \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial ^{m}_{x}b \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}. \end{aligned}$$
(3.32)
For the third term,
$$\begin{aligned} &\frac{-\int _{\mathbb{R}_{+}^{2}}\partial ^{2}_{y}\partial ^{m}_{x} b\theta ^{2}_{\alpha}\partial _{x}^{m}b\, dx\, dy}{ \Vert \theta _{\alpha}\partial ^{m}_{x}b \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad = \frac{ \Vert \theta _{\alpha}\partial ^{m}_{x}\partial _{y}b \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial ^{m}_{x}b \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} -\frac{\alpha}{2{\langle t\rangle}} \bigl\Vert \theta _{\alpha}\partial ^{m}_{x}b \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} -\frac{\alpha ^{2}}{2\langle t\rangle} \frac{ \Vert \theta _{\alpha }z\partial ^{m}_{x}b \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial ^{m}_{x}b \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}, \end{aligned}$$
(3.33)
where we have used the fact that \(\partial ^{2}_{y}(\theta ^{2}_{\alpha})= \frac{\alpha}{{\langle t\rangle}}\theta ^{2}_{\alpha}+ \frac{\alpha ^{2}}{\langle t\rangle}z^{2}\theta ^{2}_{\alpha}\) and boundary condition \(\partial _{x}^{m}b|_{y=0}=0\).
Indeed, similar to Sect. 3.1, the nonlinear terms of (3.31) can be estimated as below
$$\begin{aligned}& \begin{aligned}[b] &\sum_{m\geq 0} \frac{ \vert R_{8} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}b \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \leq \frac{C}{\tau (t)^{\frac{1}{2}}} \Biggl(\sum_{m\geq 0} \sum _{i=0}^{[m/2]}\binom{m}{i} \bigl(X^{\frac{1}{2}}_{m-i}D^{ \frac{1}{2}}_{m-i}+ \langle t\rangle ^{-1/4}\bar{X}_{m-i} \bigr) \bar{Y}^{ \frac{1}{2}}_{i+1}\bar{Y}^{\frac{1}{2}}_{i+2} \\ &\quad \quad{}+\sum_{m\geq 0}\sum _{i=[m/2]+1}^{m} \binom{m}{i} \bigl(X^{\frac{1}{4}}_{m-i}X^{\frac{1}{4}}_{m-i+1}D^{ \frac{1}{4}}_{m-i}D^{\frac{1}{4}}_{m-i+1}+ \langle t\rangle ^{-1/4} \bar{X}^{\frac{1}{2}}_{m-i} \bar{X}^{\frac{1}{2}}_{m-i+1} \bigr)\bar{Y}_{i+1} \Biggr), \end{aligned} \end{aligned}$$
(3.34)
$$\begin{aligned}& \begin{aligned}[b] &\sum_{m\geq 0} \frac{ \vert R_{9} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha} \partial _{x}^{m}b \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \leq \frac{C}{\tau (t)^{\frac{1}{2}}} \Biggl(\sum_{m\geq 0} \sum _{i=0}^{[m/2]}\binom{m}{i} \bigl(\langle t \rangle ^{ \frac{1}{4}}Y_{m-i+1}+\langle t\rangle ^{\frac{1}{4}} \bar{Y}_{m-i+1} \bigr)\bar{D}^{\frac{1}{2}}_{i} \bar{D}^{\frac{1}{2}}_{i+1} \\ &\quad \quad{}+\sum_{m\geq 0}\sum _{i=[m/2]+1}^{m} \binom{m}{i} \bigl(\langle t\rangle ^{\frac{1}{4}}Y^{\frac{1}{2}}_{m-i+1}Y^{ \frac{1}{2}}_{m-i+2}+ \langle t\rangle ^{\frac{1}{4}}\bar{Y}^{ \frac{1}{2}}_{m-i+1} \bar{Y}^{\frac{1}{2}}_{m-i+2} \bigr)\bar{D}_{i} \Biggr), \end{aligned} \end{aligned}$$
(3.35)
$$\begin{aligned}& \begin{aligned}[b] &\sum_{m\geq 0} \frac{ \vert R_{10} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}b \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \leq \frac{C}{\tau (t)^{\frac{1}{2}}} \Biggl(\sum_{m\geq 0} \sum _{i=0}^{[m/2]}\binom{m}{i}\bar{X}^{\frac{1}{2}}_{m-i} \bar{D}^{\frac{1}{2}}_{m-i}{Y}^{\frac{1}{2}}_{i+1}{Y}^{\frac{1}{2}}_{i+2} \\ &\quad \quad{}+\sum_{m\geq 0}\sum _{i=[m/2]+1}^{m} \binom{m}{i}\bar{X}^{\frac{1}{4}}_{m-i} \bar{X}^{\frac{1}{4}}_{m-i+1} \bar{D}^{\frac{1}{4}}_{m-i} \bar{D}^{\frac{1}{4}}_{m-i+1}{Y}_{i+1} \Biggr), \end{aligned} \end{aligned}$$
(3.36)
$$\begin{aligned}& \begin{aligned} &\sum_{m\geq 0} \frac{ \vert R_{11} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}b \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \leq \frac{C}{\tau (t)^{\frac{1}{2}}} \Biggl(\sum_{m\geq 0} \sum _{i=0}^{[m/2]}\binom{m}{i}\langle t\rangle ^{\frac{1}{4}} \bar{Y}_{m-i+1}{D}^{\frac{1}{2}}_{i}{D}^{\frac{1}{2}}_{i+1} \\ &\quad \quad{}+\sum_{m\geq 0}\sum _{i=[m/2]+1}^{m} \binom{m}{i}\langle t\rangle ^{\frac{1}{4}}\bar{Y}^{\frac{1}{2}}_{m-i+1} \bar{Y}^{\frac{1}{2}}_{m-i+2}{D}_{i} \Biggr) \end{aligned} \end{aligned}$$
(3.37)
and
$$\begin{aligned} \begin{aligned}[b] &\sum_{m\geq 0} \frac{ \vert R_{12} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}b \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \leq \frac{C}{\tau (t)^{\frac{1}{2}}} \Biggl(\sum_{m\geq 0} \sum _{i=0}^{[m/2]}\binom{m}{i}\langle t\rangle ^{-\frac{1}{4}} \bar{Y}_{m-i+1}\bar{X}^{\frac{1}{2}}_{i} \bar{X}^{\frac{1}{2}}_{i+1} \\ &\quad \quad{}+\sum_{m\geq 0}\sum _{i=[m/2]+1}^{m} \binom{m}{i}\langle t\rangle ^{-\frac{1}{4}}\bar{Y}^{\frac{1}{2}}_{m-i+1} \bar{Y}^{\frac{1}{2}}_{m-i+2} \bar{X}_{i} \Biggr). \end{aligned} \end{aligned}$$
(3.38)
Combining the above estimates (3.31)–(3.38) and summing over \(m\geq 0\) yields
$$\begin{aligned} &\frac{d}{dt} \Vert b \Vert _{X_{\tau ,\alpha}}+\sum _{m\geq 0} \frac{\alpha (1-2\alpha )}{4\langle t\rangle}\tau ^{m}M_{m} \frac{ \Vert \theta _{\alpha }z\partial ^{m}_{x}b \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}b \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \quad{}+\sum_{m\geq 0}\tau ^{m}M_{m} \frac{ \Vert \theta _{\alpha}\partial ^{m}_{x}\partial _{y}b \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}b \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}- \frac{\alpha}{2\langle t\rangle} \Vert b \Vert _{X_{\tau ,\alpha}} \\ &\quad \leq \dot{\tau}(t) \Vert b \Vert _{Y_{\tau ,\alpha}}+ \frac{C}{\tau (t)^{\frac{1}{2}}} \bigl(\langle t\rangle ^{-1/4}\bigl( \Vert u \Vert _{X_{ \tau ,\alpha}}+ \Vert b \Vert _{X_{\tau ,\alpha}}\bigr)+\langle t \rangle ^{1/4}\bigl( \Vert u \Vert _{D_{\tau ,\alpha}}+ \Vert b \Vert _{D_{\tau ,\alpha}}\bigr) \bigr) \\ &\quad \quad{}\times \bigl( \Vert u \Vert _{Y_{\tau ,\alpha}}+ \Vert b \Vert _{Y_{\tau ,\alpha}}\bigr)+ \Vert w \Vert _{X_{\tau ,\alpha}}. \end{aligned}$$
(3.39)
Analogously, by choosing \(\alpha \leq \frac{1}{2}\), we conclude
$$\begin{aligned} &\frac{d}{dt} \Vert b \Vert _{X_{\tau ,\alpha}} +\sum _{m\geq 0}\tau ^{m}M_{m} \frac{ \Vert \theta _{\alpha}\partial ^{m}_{x}\partial _{y}b \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}b \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} -\frac{\alpha}{2\langle t\rangle} \Vert b \Vert _{X_{\tau ,\alpha}} \\ &\quad \leq \dot{\tau}(t) \Vert b \Vert _{Y_{\tau ,\alpha}}+ \frac{C}{\tau (t)^{\frac{1}{2}}} \bigl(\langle t\rangle ^{-1/4}\bigl( \Vert u \Vert _{X_{ \tau ,\alpha}}+ \Vert b \Vert _{X_{\tau ,\alpha}}\bigr)+\langle t \rangle ^{1/4}\bigl( \Vert u \Vert _{D_{\tau ,\alpha}}+ \Vert b \Vert _{D_{\tau ,\alpha}}\bigr) \bigr) \\ &\quad \quad{}\times \bigl( \Vert u \Vert _{Y_{\tau ,\alpha}}+ \Vert b \Vert _{Y_{\tau ,\alpha}}\bigr)+ \Vert w \Vert _{X_{\tau ,\alpha}}. \end{aligned}$$
(3.40)
The proof is hence complete. □
3.3 A priori estimate on the microrotational velocity
In this subsection, we will establish the estimate of solution to problem (2.17)–(2.19) on the microrotational velocity w.
Lemma 3.3
It holds that for any \(t\in [0,T]\)
$$\begin{aligned} &\frac{d}{dt} \Vert w \Vert _{X_{\tau ,\alpha}} +\sum _{m\geq 0}\tau ^{m}M_{m} \frac{ \Vert \theta _{\alpha}\partial ^{m}_{x}\partial _{y}w \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}w \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \leq \dot{\tau}(t) \Vert w \Vert _{Y_{\tau ,\alpha}}+ \frac{C}{\tau (t)^{\frac{1}{2}}} \bigl(\langle t\rangle ^{-1/4}\bigl( \Vert u \Vert _{X_{ \tau ,\alpha}}+ \Vert b \Vert _{X_{\tau ,\alpha}}\bigr)+\langle t \rangle ^{1/4}\bigl( \Vert u \Vert _{D_{\tau ,\alpha}}+ \Vert w \Vert _{D_{\tau ,\alpha}}\bigr) \bigr) \\ &\quad \quad{}\times \bigl( \Vert u \Vert _{Y_{\tau ,\alpha}}+ \Vert b \Vert _{Y_{\tau ,\alpha}}+ \Vert w \Vert _{Y_{\tau ,\alpha}}\bigr)+ \Vert u \Vert _{X_{\tau ,\alpha}}+\langle t\rangle ^{-1/2}\bigl( \Vert u \Vert _{X_{\tau ,\alpha}}+ \Vert w \Vert _{X_{\tau ,\alpha}}\bigr) \\ &\quad \quad{}-\frac{1}{4}\sum_{m\geq 0} \frac{ \Vert \theta _{\alpha}\partial _{x}^{m}\partial _{y}u \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}\tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}. \end{aligned}$$
Proof
For \(m\geq 0\), applying the operator \(\partial ^{m}_{x}\) on (2.17)3 and multiplying the resulting equation by \(\theta ^{2}_{\alpha}\partial _{x}^{m}w\), we derive that
$$\begin{aligned} &\int _{\mathbb{R}_{+}^{2}}\partial ^{m}_{x} \bigl( \partial _{t} w+\bigl(u^{s}+ \tilde{u}\bigr)\partial _{x}w+\tilde{v}\partial _{y}w+2\partial _{y}\bigl(u+u^{s}\bigr) \\ &\quad{}+2\partial _{y}u^{s}b+2\partial _{y}^{2}u^{s}\psi -\partial ^{2}_{y}w \bigr) \theta ^{2}_{\alpha} \partial _{x}^{m}w\, dx\, dy=0. \end{aligned}$$
(3.41)
Similar to the estimates of (3.1), we now deal with each term in (3.41) as follows. For the first term, we have
$$\begin{aligned} \frac{\int _{\mathbb{R}_{+}^{2}}\partial ^{m}_{x}\partial _{t} w\theta ^{2}_{\alpha}\partial _{x}^{m}w\, dx\, dy}{ \Vert \theta _{\alpha}\partial ^{m}_{x}w \Vert _{L^{2}(\mathbb{R}^{+})}}= \frac{1}{2}\frac{d}{dt} \bigl\Vert \theta _{\alpha}\partial ^{m}_{x}w \bigr\Vert _{L^{2}( \mathbb{R}^{+})}+\frac{\alpha}{4\langle t\rangle} \frac{ \Vert \theta _{\alpha }z\partial ^{m}_{x}w \Vert ^{2}_{L^{2}(\mathbb{R}^{+})}}{ \Vert \theta _{\alpha}\partial ^{m}_{x}w \Vert _{L^{2}(\mathbb{R}^{+})}}. \end{aligned}$$
(3.42)
For the last term,
$$\begin{aligned} &\frac{-\int _{\mathbb{R}_{+}^{2}}\partial ^{2}_{y}\partial ^{m}_{x} w\theta ^{2}_{\alpha}\partial _{x}^{m}w\, dx\, dy}{ \Vert \theta _{\alpha}\partial ^{m}_{x}w \Vert _{L^{2}(\mathbb{R}^{+})}} \\ &\quad = \frac{ \Vert \theta _{\alpha}\partial ^{m}_{x}\partial _{y}w \Vert ^{2}_{L^{2}(\mathbb{R}^{+})}}{ \Vert \theta _{\alpha}\partial ^{m}_{x}w \Vert _{L^{2}(\mathbb{R}^{+})}}- \frac{\alpha}{2{\langle t\rangle}} \bigl\Vert \theta _{\alpha}\partial ^{m}_{x}w \bigr\Vert _{L^{2}(\mathbb{R}^{+})} -\frac{\alpha ^{2}}{2\langle t\rangle} \frac{ \Vert \theta _{\alpha }z\partial ^{m}_{x}w \Vert ^{2}_{L^{2}(\mathbb{R}^{+})}}{ \Vert \theta _{\alpha}\partial ^{m}_{x}w \Vert _{L^{2}(\mathbb{R}^{+})}}, \end{aligned}$$
(3.43)
where we have used the fact that \(\partial ^{2}_{y}(\theta ^{2}_{\alpha})= \frac{\alpha}{{\langle t\rangle}}\theta ^{2}_{\alpha}+ \frac{\alpha ^{2}}{\langle t\rangle}z^{2}\theta ^{2}_{\alpha}\) and boundary condition \(\partial _{x}^{m}w|_{y=0}=0\).
Similar to Sects. 3.1 and 3.2 we have
$$\begin{aligned}& \begin{aligned}[b] &\sum_{m\geq 0} \frac{ \vert R_{12} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}w \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \leq \frac{C}{\tau (t)^{\frac{1}{2}}} \Biggl(\sum_{m\geq 0} \sum _{i=0}^{[m/2]}\binom{m}{i} \bigl(X^{\frac{1}{2}}_{m-i}D^{ \frac{1}{2}}_{m-i}+ \langle t\rangle ^{-1/4}\bar{X}_{m-i} \bigr) \tilde{Y}^{\frac{1}{2}}_{i+1}\tilde{Y}^{\frac{1}{2}}_{i+2} \\ &\quad \quad{}+\sum_{m\geq 0}\sum _{i=[m/2]+1}^{m} \binom{m}{i} \bigl(X^{\frac{1}{4}}_{m-i}X^{\frac{1}{4}}_{m-i+1}D^{ \frac{1}{4}}_{m-i}D^{\frac{1}{4}}_{m-i+1}+ \langle t\rangle ^{-1/4} \bar{X}^{\frac{1}{2}}_{m-i} \bar{X}^{\frac{1}{2}}_{m-i+1} \bigr) \tilde{Y}_{i+1} \Biggr), \end{aligned} \end{aligned}$$
(3.44)
$$\begin{aligned}& \begin{aligned}[b] &\sum_{m\geq 0} \frac{ \vert R_{13} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}w \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \leq \frac{C}{\tau (t)^{\frac{1}{2}}} \Biggl(\sum_{m\geq 0} \sum _{i=0}^{[m/2]}\binom{m}{i} \bigl(\langle t \rangle ^{ \frac{1}{4}}Y_{m-i+1}+\langle t\rangle ^{\frac{1}{4}} \bar{Y}_{m-i+1} \bigr)\tilde{D}^{\frac{1}{2}}_{i} \tilde{D}^{\frac{1}{2}}_{i+1} \\ &\quad \quad{}+\sum_{m\geq 0}\sum _{i=[m/2]+1}^{m} \binom{m}{i} \bigl(\langle t\rangle ^{\frac{1}{4}}Y^{\frac{1}{2}}_{m-i+1}Y^{ \frac{1}{2}}_{m-i+2}+ \langle t\rangle ^{\frac{1}{4}}\bar{Y}^{ \frac{1}{2}}_{m-i+1} \bar{Y}^{\frac{1}{2}}_{m-i+2} \bigr)\tilde{D}_{i} \Biggr), \end{aligned} \end{aligned}$$
(3.45)
also,
$$\begin{aligned} &\sum_{m\geq 0} \frac{ \vert R_{14} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}w \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \leq \frac{1}{4}\sum_{m\geq 0} \frac{ \Vert \theta _{\alpha}\partial _{x}^{m}\partial _{y}u \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}\tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} +\sum_{m\geq 0} \bigl\Vert \theta _{\alpha}\partial _{x}^{m}u \bigr\Vert _{L^{2}( \mathbb{R}_{+}^{2})}\tau ^{m}M_{m} \end{aligned}$$
(3.46)
and
$$\begin{aligned} \biggl\vert \int _{\mathbb{R}_{+}^{2}}\partial _{y}u^{s}\partial _{x}^{m}w \theta ^{2}_{\alpha} \partial _{x}^{m}w\, dx\, dy \biggr\vert \leq C\langle t \rangle ^{-1/2} \bigl\Vert \theta _{\alpha}\partial _{x}^{m}w \bigr\Vert ^{2}_{L^{2}( \mathbb{R}_{+}^{2})}, \end{aligned}$$
(3.47)
imply
$$\begin{aligned} \sum_{m\geq 0} \frac{ \vert R_{15} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}w \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \leq C\langle t \rangle ^{-1/2}\sum_{m\geq 0} \tilde{X}_{m}, \end{aligned}$$
(3.48)
where we have used the fact that \(\|\partial _{y}u^{s}\|_{L^{\infty}_{y}}\leq C\langle t\rangle ^{-1/2}\).
Finally, we deal with the remainder term in (3.41) as follows
$$\begin{aligned} \vert R_{16} \vert & \leq 2 \bigl\Vert \partial ^{2}_{y}u^{s} \bigr\Vert _{L^{\infty}_{y}} \bigl\Vert \theta _{ \alpha}\partial _{x}^{m} \psi \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} \bigl\Vert \theta _{ \alpha}\partial _{x}^{m}w \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} \\ & \leq 2\langle t\rangle ^{-1} \bigl\Vert \theta _{\alpha}\partial _{x}^{m} \psi \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha}\partial _{x}^{m}w \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} \\ & \leq 2\langle t\rangle ^{-1/2} \bigl\Vert \theta _{\alpha}\partial _{x}^{m}b \bigr\Vert _{L^{2}(\mathbb{R}_{+}^{2})} \bigl\Vert \theta _{\alpha}\partial _{x}^{m}w \bigr\Vert _{L^{2}( \mathbb{R}_{+}^{2})}, \end{aligned}$$
implies
$$\begin{aligned} \sum_{m\geq 0} \frac{ \vert R_{16} \vert \tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}w \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \leq C\langle t \rangle ^{-1/2}\sum_{m\geq 0} \bar{X}_{m}, \end{aligned}$$
(3.49)
where we have used the fact that \(\|\partial ^{2}_{y}u^{s}\|_{L^{\infty}_{y}}\leq C\langle t\rangle ^{-1}\).
Combining the estimates (3.41)–(3.49) and summing over \(m\geq 0\) yields
$$\begin{aligned} &\frac{d}{dt} \Vert w \Vert _{X_{\tau ,\alpha}}+\sum _{m\geq 0} \frac{\alpha (1-2\alpha )}{4\langle t\rangle}\tau ^{m}M_{m} \frac{ \Vert \theta _{\alpha }z\partial ^{m}_{x}w \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}w \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} +\sum_{m\geq 0}\tau ^{m}M_{m} \frac{ \Vert \theta _{\alpha}\partial ^{m}_{x}\partial _{y}w \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}w \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \leq \dot{\tau}(t) \Vert w \Vert _{Y_{\tau ,\alpha}}+ \frac{C}{\tau (t)^{\frac{1}{2}}} \bigl(\langle t\rangle ^{-1/4}\bigl( \Vert u \Vert _{X_{ \tau ,\alpha}}+ \Vert b \Vert _{X_{\tau ,\alpha}}\bigr)+\langle t \rangle ^{1/4}\bigl( \Vert u \Vert _{D_{\tau ,\alpha}}+ \Vert w \Vert _{D_{\tau ,\alpha}}\bigr) \bigr) \\ &\quad \quad{}\times \bigl( \Vert u \Vert _{Y_{\tau ,\alpha}}+ \Vert b \Vert _{Y_{\tau ,\alpha}}+ \Vert w \Vert _{Y_{\tau ,\alpha}}\bigr)+ \Vert u \Vert _{X_{\tau ,\alpha}}+\langle t\rangle ^{-1/2}\bigl( \Vert u \Vert _{X_{\tau ,\alpha}}+ \Vert w \Vert _{X_{\tau ,\alpha}}\bigr) \\ &\quad \quad{}-\frac{1}{4}\sum_{m\geq 0} \frac{ \Vert \theta _{\alpha}\partial _{x}^{m}\partial _{y}u \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}\tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}. \end{aligned}$$
Analogously, by choosing \(\alpha \leq \frac{1}{2}\), we derive
$$\begin{aligned} &\frac{d}{dt} \Vert w \Vert _{X_{\tau ,\alpha}} +\sum _{m\geq 0}\tau ^{m}M_{m} \frac{ \Vert \theta _{\alpha}\partial ^{m}_{x}\partial _{y}w \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}w \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \leq \dot{\tau}(t) \Vert w \Vert _{Y_{\tau ,\alpha}}+ \frac{C}{\tau (t)^{\frac{1}{2}}} \bigl(\langle t\rangle ^{-1/4}\bigl( \Vert u \Vert _{X_{ \tau ,\alpha}}+ \Vert b \Vert _{X_{\tau ,\alpha}}\bigr)+\langle t \rangle ^{1/4}\bigl( \Vert u \Vert _{D_{\tau ,\alpha}}+ \Vert w \Vert _{D_{\tau ,\alpha}}\bigr) \bigr) \\ &\quad \quad{}\times \bigl( \Vert u \Vert _{Y_{\tau ,\alpha}}+ \Vert b \Vert _{Y_{\tau ,\alpha}}+ \Vert w \Vert _{Y_{\tau ,\alpha}}\bigr)+ \Vert u \Vert _{X_{\tau ,\alpha}}+\langle t\rangle ^{-1/2}\bigl( \Vert u \Vert _{X_{\tau ,\alpha}}+ \Vert w \Vert _{X_{\tau ,\alpha}}\bigr) \\ &\quad \quad{}-\frac{1}{4}\sum_{m\geq 0} \frac{ \Vert \theta _{\alpha}\partial _{x}^{m}\partial _{y}u \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}\tau ^{m}M_{m}}{ \Vert \theta _{\alpha}\partial _{x}^{m}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}. \end{aligned}$$
(3.50)
The proof is now complete. □
Up to now, we have completed all the estimates of solution \((u,b,w)\) to problem (2.17)–(2.19).
Combining the estimates (3.30), (3.40), and (3.50), we conclude
$$\begin{aligned} &\frac{d}{dt} \bigl\Vert (u,b,w) \bigr\Vert _{X_{\tau ,\alpha}} + \frac{3}{4}\sum_{m\geq 0}\tau ^{m}M_{m} \frac{ \Vert \theta _{\alpha}\partial ^{m}_{x}\partial _{y}u \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} +\frac{3}{4} \sum_{m\geq 0}\tau ^{m}M_{m} \frac{ \Vert \theta _{\alpha}\partial ^{m}_{x}\partial _{y}b \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}b \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \\ &\quad \quad{}+\frac{3}{4}\sum_{m\geq 0}\tau ^{m}M_{m} \frac{ \Vert \theta _{\alpha}\partial ^{m}_{x}\partial _{y}w \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}w \Vert _{L^{2}(\mathbb{R}_{+}^{2})}}- \frac{\alpha}{2\langle t\rangle} \bigl\Vert (u,b,w) \bigr\Vert _{X_{\tau ,\alpha}} \\ &\quad \leq \dot{\tau}(t) \bigl\Vert (u,b,w) \bigr\Vert _{Y_{\tau ,\alpha}}+ \frac{C}{\tau (t)^{\frac{1}{2}}} \bigl(\langle t\rangle ^{-1/4} \bigl\Vert (u,b,w) \bigr\Vert _{X_{\tau ,\alpha}}+\langle t\rangle ^{1/4} \bigl\Vert (u,b,w) \bigr\Vert _{D_{\tau , \alpha}} \bigr) \\ &\quad \quad{}\times \bigl\Vert (u,b,w) \bigr\Vert _{Y_{\tau ,\alpha}}+C \bigl\Vert (u,b,w) \bigr\Vert _{X_{\tau ,\alpha}}. \end{aligned}$$
(3.51)
Using Lemma 2.3, we discover
$$\begin{aligned}& \sum_{m\geq 0} \frac{ \Vert \theta _{\alpha}\partial _{y}\partial _{x}^{m}u \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}u \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \tau ^{m}M_{m}\geq \frac{\alpha ^{\frac{1}{2}}\beta}{2\langle t\rangle ^{\frac{1}{2}}} \Vert u \Vert _{D_{\tau ,\alpha}}+\frac{\alpha (1-\beta )}{\langle t\rangle} \Vert u \Vert _{X_{ \tau ,\alpha}}, \end{aligned}$$
(3.52)
$$\begin{aligned}& \sum_{m\geq 0} \frac{ \Vert \theta _{\alpha}\partial _{y}\partial _{x}^{m}b \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2})}}{ \Vert \theta _{\alpha}\partial _{x}^{m}b \Vert _{L^{2}(\mathbb{R}_{+}^{2}) }}\tau ^{m}M_{m}\geq \frac{\alpha ^{\frac{1}{2}}\beta}{2\langle t\rangle ^{\frac{1}{2}}} \Vert b \Vert _{D_{\tau ,\alpha}}+\frac{\alpha (1-\beta )}{\langle t\rangle} \Vert b \Vert _{X_{ \tau ,\alpha}} \end{aligned}$$
(3.53)
and
$$\begin{aligned} \sum_{m\geq 0} \frac{ \Vert \theta _{\alpha}\partial _{y}\partial _{x}^{m}w \Vert ^{2}_{L^{2}(\mathbb{R}_{+}^{2}) }}{ \Vert \theta _{\alpha}\partial _{x}^{m}w \Vert _{L^{2}(\mathbb{R}_{+}^{2})}} \tau ^{m}M_{m}\geq \frac{\alpha ^{\frac{1}{2}}\beta}{2\langle t\rangle ^{\frac{1}{2}}} \Vert w \Vert _{D_{\tau ,\alpha}}+\frac{\alpha (1-\beta )}{\langle t\rangle} \Vert w \Vert _{X_{ \tau ,\alpha}}, \end{aligned}$$
(3.54)
for \(\beta \in (0,1/3)\).
Substituting (3.52)–(3.54) into (3.51), leads to
$$\begin{aligned} &\frac{d}{dt} \bigl\Vert (u,b,w) \bigr\Vert _{X_{\tau ,\alpha}}+ \frac{\alpha (1-3\beta )}{4}\frac{1}{\langle t\rangle} \bigl\Vert (u,b,w) \bigr\Vert _{X_{ \tau ,\alpha}}+ \frac{3\alpha ^{\frac{1}{2}}\beta}{8\langle t\rangle ^{\frac{1}{2}}} \bigl\Vert (u,b,w) \bigr\Vert _{D_{\tau ,\alpha}} \\ &\quad \leq \biggl(\dot{\tau}(t)+\frac{C}{\tau (t)^{\frac{1}{2}}} \bigl( \langle t\rangle ^{-1/4} \bigl\Vert (u,b,w) \bigr\Vert _{X_{\tau ,\alpha}}+\langle t \rangle ^{1/4} \bigl\Vert (u,b,w) \bigr\Vert _{D_{\tau ,\alpha}} \bigr) \biggr) \bigl\Vert (u,b,w) \bigr\Vert _{Y_{ \tau ,\alpha}} \\ &\quad \quad{}+C \bigl\Vert (u,b,w) \bigr\Vert _{X_{\tau ,\alpha}}. \end{aligned}$$
(3.55)
We will choose a suitable function \(\tau (t)\) such that the following ordinary differential equation holds
$$\begin{aligned} \frac{d}{dt}\bigl({\tau}(t)\bigr)^{\frac{3}{2}}+ \frac{3C}{2} \bigl(\langle t \rangle ^{-1/4} \bigl\Vert (u,b,w) \bigr\Vert _{X_{\tau ,\alpha}}+\langle t\rangle ^{1/4} \bigl\Vert (u,b,w) \bigr\Vert _{D_{\tau ,\alpha}} \bigr)=0. \end{aligned}$$
(3.56)
Hence, from (3.55) this implies
$$\begin{aligned} &\frac{d}{dt} \bigl\Vert (u,b,w) \bigr\Vert _{X_{\tau ,\alpha}}+ \frac{\alpha (1-3\beta )}{4}\frac{1}{\langle t\rangle} \bigl\Vert (u,b,w) \bigr\Vert _{X_{ \tau ,\alpha}}+ \frac{3\alpha ^{\frac{1}{2}}\beta}{8\langle t\rangle ^{\frac{1}{2}}} \bigl\Vert (u,b,w) \bigr\Vert _{D_{\tau ,\alpha}} \\ &\quad \leq C \bigl\Vert (u,b,w) \bigr\Vert _{X_{\tau ,\alpha}}. \end{aligned}$$
(3.57)
Using the classical bootstrap argument [25], we first assume that there exists a \(T_{\ast}>0\) such that
$$\begin{aligned} &\bigl\Vert (u,b,w) \bigr\Vert _{X_{\tau ,\alpha}}+ \int _{0}^{t} \biggl( \frac{\alpha (1-3\beta )}{4\langle s\rangle} \bigl\Vert (u,b,w) \bigr\Vert _{X_{\tau , \alpha}}+ \frac{3\alpha ^{\frac{1}{2}}\beta}{8\langle s\rangle ^{\frac{1}{2}}} \bigl\Vert (u,b,w) \bigr\Vert _{D_{\tau ,\alpha}} \biggr)\,ds \\ &\quad \leq 2 \bigl\Vert (u_{0},b_{0},w_{0}) \bigr\Vert _{X_{\tau _{0},\alpha}}, \end{aligned}$$
(3.58)
for \(t\in [0,T_{\ast}]\).
From (3.56), this implies
$$\begin{aligned} &{\tau}(t)^{\frac{3}{2}}-{\tau}(0)^{\frac{3}{2}} \\ &\quad =-\frac{3C}{2} \int _{0}^{t} \bigl(\langle s\rangle ^{-1/4} \bigl\Vert (u,b,w) \bigr\Vert _{X_{ \tau ,\alpha}}+\langle s \rangle ^{1/4} \bigl\Vert (u,b,w) \bigr\Vert _{D_{\tau ,\alpha}} \bigr)\,ds \\ &\quad =-\frac{3C}{2} \int _{0}^{t} \biggl(\frac{4}{\alpha (1-3\beta )} \langle t\rangle ^{3/4}\frac{\alpha (1-3\beta )}{4\langle s\rangle} \bigl\Vert (u,b,w) \bigr\Vert _{X_{\tau ,\alpha}} \\ &\quad \quad {}+\frac{8}{3\alpha ^{\frac{1}{2}}\beta}\langle s \rangle ^{3/4} \frac{3\alpha ^{\frac{1}{2}}\beta}{8\langle t\rangle ^{\frac{1}{2}}} \bigl\Vert (u,b,w) \bigr\Vert _{D_{\tau ,\alpha}} \biggr)\,ds \\ &\quad \geq -3C \bigl\Vert (u_{0},b_{0},w_{0}) \bigr\Vert _{X_{\tau _{0},\alpha}} \biggl( \frac{4}{\alpha (1-3\beta )}+ \frac{8}{3\alpha ^{\frac{1}{2}}\beta} \biggr)\langle t\rangle ^{3/4} \\ &\quad =-C_{2}\langle t\rangle ^{4/3}, \end{aligned}$$
(3.59)
where \(C_{2}=12C\|(u_{0},b_{0},w_{0})\|_{X_{\tau _{0},\alpha}} ( \frac{1}{\alpha (1-3\beta )}+\frac{2}{3\alpha ^{\frac{1}{2}}\beta} )\) and \(T=\min \{ (\frac{7}{8C_{2}}\tau _{0}^{\frac{3}{2}} )^{ \frac{3}{4}}-1,T_{1}\}\) with \(T_{1}\) to be determined later.
Therefore, taking positive \(T_{1}=\frac{1}{4C}\leq T_{\ast}\), for \(\forall t\in [0,T_{1}]\), from (3.57) gives
$$\begin{aligned} &\bigl\Vert (u,b,w) \bigr\Vert _{X_{\tau ,\alpha}}+ \int _{0}^{t} \biggl( \frac{\alpha (1-3\beta )}{4\langle s\rangle} \bigl\Vert (u,b,w) \bigr\Vert _{X_{\tau , \alpha}}+ \frac{3\alpha ^{\frac{1}{2}}\beta}{8\langle s\rangle ^{\frac{1}{2}}} \bigl\Vert (u,b,w) \bigr\Vert _{D_{\tau ,\alpha}} \biggr)\,ds \\ &\quad \leq C \int _{0}^{t} \bigl\Vert (u,b,w) \bigr\Vert _{X_{\tau ,\alpha}}\,ds + \bigl\Vert (u_{0},b_{0},w_{0}) \bigr\Vert _{X_{\tau _{0},\alpha}} \\ &\quad \leq \frac{3}{2} \bigl\Vert (u_{0},b_{0},w_{0}) \bigr\Vert _{X_{\tau _{0},\alpha}}. \end{aligned}$$
(3.60)
The proof of Theorem 2.1 is thus finished.