# On the Well-Posedness of A High Order Convective Cahn-Hilliard Type Equations
## 3 Proof of Theorem 1  
This section devoted to the proof of Theorem 1. Proof of Theorem 1.: Fix \(T>0\). Thanks to Lemmas 3, 4, 5, 6, (7) and the Cauchy-Kovalevskaya Theorem [67], we have that \(u\) is solution of (1) and (8) holds. In particular, by Lemma 1, we get (9). Moreover, by Lemmas 2, 3, 4, 5, 6, 8 and Remark 1, we have that  
\[P\in H^{1}((0,T)\times\mathbb{R})\cap L^{\infty}(0,T;H^{4}(\mathbb{R})).\]  
We prove (10). Let \(u_{1}\) and \(u_{2}\) be two solutions of (1), which verify (8), which is  
\[ \begin{cases}\partial_{t}u_{1}+\kappa\partial_{x}u_{1}^{2}-\beta^{2}\partial_ {x}^{6}u_{1}+\alpha\partial_{x}^{4}u_{1}+\delta^{2}\partial_{x}^{4}\left(u_{1 }^{3}\right)=0,&t>0,\,x\in\mathbb{R},\\ u_{1}(0,x)=u_{1,\,0}(x),&x\in\mathbb{R},\\ \end{cases} \] \[ \begin{cases}\partial_{t}u_{2}+\kappa\partial_{x}u_{2}^{2}-\beta^{2} \partial_{x}^{6}u_{2}+\alpha\partial_{x}^{4}u_{2}+\delta^{2}\partial_{x}^{4} \left(u_{2}^{3}\right)=0,&t>0,\,x\in\mathbb{R},\\ u_{2}(0,x)=u_{2,\,0}(x),&x\in\mathbb{R},\end{cases} \]  
Then, the function  
\[\omega=u_{1}-u_{2} \tag{65}\]  
is the solution of the following Cauchy problem:  
\[ \begin{cases}\partial_{t}\omega+\kappa\partial_{x}\left(u_{1}^{2}-u_{2}^{2} \right)-\beta^{2}\partial_{x}^{6}\omega+\alpha\partial_{x}^{4}\omega+\delta^{ 2}\partial_{x}^{4}\left(u_{1}^{3}-u_{2}^{3}\right)=0,&t>0,\,x\in\mathbb{R},\\ \omega(0,x)=u_{1,0}(x)-u_{2,\,0}(x),&x\in\mathbb{R}.\end{cases}  \tag{66}\]  
Since  
\[2\int_{\mathbb{R}}\omega\partial_{t}\omega dx= \frac{d}{dt}\left\|\omega(t,\cdot)\right\|_{L^{2}(\mathbb{R})}^{ 2},\] \[2\kappa\int_{\mathbb{R}}\partial_{x}\left(u_{1}^{2}-u_{2}^{2} \right)\omega dx= -2\kappa\int_{\mathbb{R}}(u_{1}^{2}-u_{2}^{2})\partial_{x}\omega dx,\] \[-2\beta^{2}\int_{\mathbb{R}}\omega\partial_{x}^{6}\omega dx= 2\beta^{2}\left\|\partial_{x}^{3}\omega(t,\cdot)\right\|_{L^{2}( \mathbb{R})}^{2},\] \[2\alpha\int_{\mathbb{R}}\omega\partial_{x}^{4}\omega dx= -2\kappa\int_{\mathbb{R}}\partial_{x}\omega\partial_{x}^{3} \omega dx,\] \[2\delta^{2}\int_{\mathbb{R}}\omega\partial_{x}^{4}\left(u_{1}^{ 3}-u_{2}^{3}\right)dx= -2\delta^{2}\int_{\mathbb{R}}\partial_{x}^{3}\omega\partial_{x} \left(u_{1}^{3}-u_{2}^{3}\right)dx,\]  
\[\leq C(T)\int_{\mathbb{R}}|\partial_{x}\omega||\partial_{x}^{3} \omega|dx=\int_{\mathbb{R}}\left|\frac{\mathcal{C}(T)\partial_{x}\omega}{\beta} \right|\left|\partial\partial_{x}^{3}u\right|\ dx\] \[\leq C(T)\left\|\partial_{x}\omega(t,\cdot)\right\|_{L^{2}( \mathbb{R})}^{2}+\frac{\beta^{2}}{2}\left\|\partial_{x}^{3}\omega(t,\cdot) \right\|_{L^{2}(\mathbb{R})}^{2}.\]  
It follows from (69) that  
\[ \begin{split}\frac{d}{dt}\left\|\omega(t,\cdot)\right\|_{L^{2}( \mathbb{R})}^{2}+\frac{\beta^{2}}{2}\left\|\partial_{x}^{3}\omega(t,\cdot) \right\|_{L^{2}(\mathbb{R})}^{2}\\ \leq& C(T)\left\|\omega(t,\cdot)\right\|_{L^{2}( \mathbb{R})}^{2}+C(T)\left\|\partial_{x}\omega(t,\cdot)\right\|_{L^{2}( \mathbb{R})}^{2}.\end{split}  \tag{72}\]  
Observe that  
\[C(T)\left\|\partial_{x}\omega(t,\cdot)\right\|_{L^{2}(\mathbb{R})}^{2}=C(T) \int_{\mathbb{R}}\partial_{x}\omega\partial_{x}\omega dx=-C(T)\int_{\mathbb{R }}\omega\partial_{x}^{2}\omega dx.\]  
Therefore, by the Young inequality,  
\[ \begin{split} C(T)\left\|\partial_{x}\omega(t,\cdot)\right\|_{L^ {2}(\mathbb{R})}^{2}\leq& C(T)\int_{\mathbb{R}}\left|\frac{ \omega}{\sqrt{D_{5}}}\right|\left|\sqrt{D_{5}}\partial_{x}^{2}\omega\right| dx\\ \leq&\frac{C(T)}{2D_{5}}\left\|\omega(t,\cdot) \right\|_{L^{2}(\mathbb{R})}^{2}+\frac{C(T)D_{5}}{2}\left\|\partial_{x}^{2} \omega(t,\cdot)\right\|_{L^{2}(\mathbb{R})}^{2},\end{split}  \tag{73}\]  
where \(D_{5}\) is a positive constant, which will specified later. Observe again that  
\[ \begin{split}\frac{C(T)D_{5}}{2}\left\|\partial_{x}^{2}\omega(t,\cdot)\right\|_{L^{2}(\mathbb{R})}^{2}=\frac{C(T)D_{5}}{2}\int_{\mathbb{R}} \partial_{x}^{2}\omega\partial_{x}^{2}\omega dx=-\frac{C(T)D_{5}}{2}\int_{ \mathbb{R}}\partial_{x}\omega\partial_{x}^{3}\omega dx.\end{split}  \tag{74}\]  
Consequently, by the Young inequality,  
\[