# On the Well-Posedness of A High Order Convective Cahn-Hilliard Type Equations
## 2 A Priori Estimates  
In this section, we prove some a priori estimates on \(u\). We denote with \(C_{0}\) the constants which depend only on the initial data, and with \(C(T)\), the constants which depend also on \(T\). We begin by proving the following lemma. Lemma 1: For each \(t>0\), we have that (9). Proof: Integrating (1) on \(\mathbb{R}\), we have that  
\[\int_{\mathbb{R}}\partial_{t}u(t,x)dx=\frac{d}{dt}\int_{\mathbb{R}}u(t,x)dx=0 \tag{11}\]  
(9) follows from (3) and (11). Remark 1: In light of (9), we can consider the following function:  
\[P(t,x)=\int_{-\infty}^{x}u(t,y)dy. \tag{12}\]  
Moreover, again by (9), we have that  
\[P(t,-\infty)=P(t,\infty)=0. \tag{13}\]  
Lemma 2: Let \(T>0\). There exists a constant \(C(T)>0\), such that  
\[\left\|P(t,\cdot)\right\|_{L^{2}(\mathbb{R})}^{2} +2\beta^{2}e^{t}\int_{0}^{t}e^{-s}\left\|\partial_{x}^{2}u(s, \cdot)\right\|_{L^{2}(\mathbb{R})}^{2}ds \tag{14}\] \[+2\delta^{2}e^{t}\int_{0}^{t}e^{-s}\left\|u(s,\cdot)\partial_{x} u(s,\cdot)\right\|_{L^{2}(\mathbb{R})}^{2}ds\] \[\leq C(T)+C(T)\int_{0}^{t}\left\|u(s,\cdot)\right\|_{L^{4}(\mathbb{R} )}^{4}ds+C(T)\int_{0}^{t}\left\|\partial_{x}u(s,\cdot)\right\|_{L^{2}(\mathbb{ R})}^{2}ds,\]_for every \(0\leq t\leq T\)._  
Proof. : Let \(0\leq t\leq T\). We begin by observing that  
\[\partial_{x}u^{3}=3u^{2}\partial_{x}u. \tag{15}\]  
Integrating (1) on \((-\infty,\,x)\), we have that  
\[\int_{-\infty}^{x}\partial_{t}udy+\kappa u^{2}-\beta^{2}\partial_{x}^{5}u+ \alpha\partial_{x}^{3}u+\delta^{2}\partial_{x}^{3}\left(u^{3}\right)=0. \tag{16}\]  
Differentiating (12) with respect to \(t\), we obtain that  
\[\partial_{t}P(t,x)=\frac{d}{dt}\int_{-\infty}^{x}u(t,y)dy=\int_{-\infty}^{x} \partial_{t}u(t,y)dy. \tag{17}\]  
From (16) and (17)  
\[\partial_{t}P+\kappa u^{2}-\beta^{2}\partial_{x}^{5}u+\alpha\partial_{x}^{3}u +\delta^{2}\partial_{x}^{3}\left(u^{3}\right)=0. \tag{18}\]  
Observe that by (13) and (18),  
\[ \begin{split}-2\beta^{2}\int_{\mathbb{R}}P\partial_{x}^{5}udx=& 2\beta^{2}\int_{\mathbb{R}}\partial_{x}P\partial_{x}^{4}udx= \beta^{2}\int_{\mathbb{R}}u\partial_{x}^{4}udx\\ =&-2\beta^{2}\int_{\mathbb{R}}\partial_{x}u\partial_ {x}^{3}udx=2\beta^{2}\left\|\partial_{x}^{2}u(t,\cdot)\right\|_{L^{2}(\mathbb{ R})}^{2},\\ 2\alpha\int_{\mathbb{R}}P\partial_{x}^{3}udx=&-2 \alpha\int_{\mathbb{R}}\partial_{x}P\partial_{x}^{2}udx=-2\alpha\int_{ \mathbb{R}}u\partial_{x}^{2}udx=2\alpha\left\|\partial_{x}u(t,\cdot)\right\|_ {L^{2}(\mathbb{R})}^{2}.\end{split}  \tag{19}\]  
Moreover, by (15) and (16),  
\[ \begin{split} 2\delta^{2}\int_{\mathbb{R}}P\partial_{x}^{3} \left(u^{3}\right)dx=&-2\delta^{2}\int_{\mathbb{R}}\partial_{x}P \partial_{x}^{2}\left(u^{3}\right)dx=-2\delta^{2}\int_{\mathbb{R}}u\partial_{x }^{2}\left(u^{3}\right)dx\\ =& 2\delta^{2}\int_{\mathbb{R}}\partial_{x}u\partial_ {x}\left(u^{3}\right)dx=6\delta^{2}\left\|u(t,\cdot)\partial_{x}u(t,\cdot) \right\|_{L^{2}(\mathbb{R})}^{2}.\end{split}  \tag{20}\]  
Therefore, multiplying (18) by \(2P\), an integration on \(\mathbb{R}\), (19) and (20) give  
\[ \begin{split}\frac{d}{dt}\left\|P(t,\cdot)\right\|_{L^{2}( \mathbb{R})}^{2}&+2\beta^{2}\left\|\partial_{x}^{2}u(t,\cdot) \right\|_{L^{2}(\mathbb{R})}^{2}+6\delta^{2}\left\|u(t,\cdot)\partial_{x}u(t, \cdot)\right\|_{L^{2}(\mathbb{R})}^{2}\\ =&-2\kappa\int_{\mathbb{R}}Pu^{2}dx-2\alpha\left\| \partial_{x}u(t,\cdot)\right\|_{L^{2}(\mathbb{R})}^{2}.\end{split}  \tag{21}\]  
Due to the Young inequality,  
\[2|\kappa|\int_{\mathbb{R}}|P|u^{2}dx\leq\left\|P(t,\cdot)\right\|_{L^{2}( \mathbb{R})}^{2}+\kappa^{2}\left\|u(t,\cdot)\right\|_{L^{4}(\mathbb{R})}^{4}.\]  
It follows from (21) that  
\[