# On Classical Solutions for A Kuramoto-Sinelshchikov-Velarde-Type Equation
## 2 Proof of Theorem 1, under the Assumptions (2) and (4)
\begin{split}\frac{d}{dt}\left\|\partial_{x}^{3}u(t,\cdot)\right\| _{L^{2}(\mathbb{R})}^{2}+\beta^{2}\left(2-4D_{4}\right)\left\|\partial_{x}^{5} u(t,\cdot)\right\|_{L^{2}(\mathbb{R})}^{2}\\ \leq\frac{C(T)}{D_{4}}+\frac{C(T)}{D_{4}}\left\|\partial_{x}^{3} u(t,\cdot)\right\|_{L^{2}(\mathbb{R})}^{2}+2\mu\left\|\partial_{x}^{4}u(t, \cdot)\right\|_{L^{2}(\mathbb{R})}^{2}.\end{split}  \tag{54}\]  
Observe that  
\[2\mu\left\|\partial_{x}^{4}u(t,\cdot)\right\|_{L^{2}(\mathbb{R})}^{2}=2\mu \int_{\mathbb{R}}\partial_{x}^{4}u\partial_{x}^{4}udx=-2\mu\int_{\mathbb{R}} \partial_{x}^{3}u\partial_{x}^{5}udx.\]  
Thus, by the Young inequality,  
\[ \begin{split} 2|\mu|\left\|\partial_{x}^{4}u(t,\cdot)\right\|_{L^{2}( \mathbb{R})}^{2}\leq& 2|\mu|\int_{\mathbb{R}}\partial_{x}^{3}u \partial_{x}^{5}udx\\ =& 2\int_{\mathbb{R}}\left|\frac{\mu\partial_{x}^{3}u}{ \beta\sqrt{D_{4}}}\right|\left|\beta\sqrt{D_{4}}\partial_{x}^{4}u\right|dx\\ \leq&\frac{\mu^{2}}{\beta^{2}D_{4}}\left\|\partial_{ x}^{3}u(t,\cdot)\right\|_{L^{2}(\mathbb{R})}^{2}+\beta^{2}D_{4}\left\|\partial_{x}^{5}u (t,\cdot)\right\|_{L^{2}(\mathbb{R})}^{2}.\end{split} \]  
It follows from (54) that  
\[ \begin{split}\frac{d}{dt}\left\|\partial_{x}^{3}u(t,\cdot)\right\| _{L^{2}(\mathbb{R})}^{2}&+\beta^{2}\left(2-5D_{4}\right)\left\| \partial_{x}^{5}u(t,\cdot)\right\|_{L^{2}(\mathbb{R})}^{2}\\ \leq&\frac{C(T)}{D_{4}}+\frac{C(T)}{D_{4}}\left\| \partial_{x}^{3}u(t,\cdot)\right\|_{L^{2}(\mathbb{R})}^{2}.\end{split} \]Choosing \(D_{4}=\frac{1}{5}\), we have that  
\[\frac{d}{dt}\left\|\partial_{x}^{3}u(t,\cdot)\right\|_{L^{2}(\mathbb{ R})}^{2}+\beta^{2}\left\|\partial_{x}^{5}u(t,\cdot)\right\|_{L^{2}(\mathbb{R})}^{2}\] \[\leq C(T)+C(T)\left\|\partial_{x}^{3}u(t,\cdot)\right\|_{L^{2}( \mathbb{R})}^{2}.\]  
Integrating on \((0,t)\), by (5) and (47), we obtain that  
\[\left\|\partial_{x}^{3}u(t,\cdot)\right\|_{L^{2}(\mathbb{R})}^{2}+ \beta^{2}\int_{0}^{t}\left\|\partial_{x}^{5}u(s,\cdot)\right\|_{L^{2}(\mathbb{ R})}^{2}ds\] \[\leq\left\|u_{0}\right\|_{H^{3}(\mathbb{R})}^{2}+C(T)t+C(T)\int_{ 0}^{t}\left\|\partial_{x}^{3}u(s,\cdot)\right\|_{L^{2}(\mathbb{R})}^{2}ds\leq C (T),\]  
which gives (51). Finally, we prove (52). Due to (49), (51) and the Holder inequality,  
\[(\partial_{x}^{2}u(t,x))^{2}= 2\int_{-\infty}^{x}\partial_{x}^{2}u\partial_{x}^{3}udy\leq\int _{\mathbb{R}}|\partial_{x}^{2}u||\partial_{x}^{3}u|dx\] \[\leq \left\|\partial_{x}^{2}u(t,\cdot)\right\|_{L^{2}(\mathbb{R})} \left\|\partial_{x}^{3}u(t,\cdot)\right\|_{L^{2}(\mathbb{R})}\leq C(T).\]  
Hence,  
\[\left\|\partial_{x}^{2}u\right\|_{L^{\infty}((0,T)\times\mathbb{R})}^{2}\leq C (T),\]  
which gives (52). Now, we prove Theorem 1. Proof of Theorem 1.: Let \(T>0\). Assuming (3) and (4), thanks to [PERSON] 6, 7, 8 and the Cauchy-Kovalevskaya Theorem [50], we have that \(u\) is solution of (1) and (14) holds. Instead, assuming (3) and (5), thanks to [PERSON] 6, 7, 8, 9 and the Cauchy-Kovalevskaya Theorem [50], we have that \(u\) is solution of (1) and (15) holds. We prove (16), under Assumptions (3) and (5). Let \(u_{1}\) and \(u_{2}\) be two solutions of (1), which verify (15) that is  
\[ \begin{cases}\partial_{t}u_{1}+2\gamma(\partial_{x}u_{1})^{2}+ \delta\partial_{x}^{3}u_{1}+\mu\partial_{x}^{2}u_{1}+\beta^{2}\partial_{x}^{ 4}u_{1}+\gamma u_{1}\partial_{x}^{2}u_{1}=0,&t>0,\,x\in\mathbb{R},\\ u_{1}(0,x)=u_{1,0}(x),&x\in\mathbb{R},\\ \end{cases} \] \[ \begin{cases}\partial_{t}u_{2}+2\gamma(\partial_{x}u_{2})^{2}+ \delta\partial_{x}^{3}u_{2}+\mu\partial_{x}^{2}u_{2}+\beta^{2}\partial_{x}^{ 4}u_{2}+\gamma u_{2}\partial_{x}^{2}u_{2}=0,&t>0,\,x\in\mathbb{R},\\ u_{1}(0,x)=u_{1,0}(x),&x\in\mathbb{R},\end{cases} \]  
Then, the function \(\omega\), defined in (36), is the solution of the following Cauchy problem:  
\[