We first consider the corresponding problem with the linear equation
$$\begin{aligned} \left \{ \textstyle\begin{array}{@{}l} u_{t}+L_{p}u_{t}+u_{x}+M_{\alpha}u=0,\quad x\in\mathbb {T}, t\in(0,T),\\ u(0,x)=u_{0}(x),\quad x\in\mathbb{T},\\ \int_{\mathbb{T}}u(t,x)\,dx=0. \end{array}\displaystyle \right . \end{aligned}$$
(3.1)
If \(u(t)\in L^{2}(\mathbb{T})\), \(\forall t>0\), then the kth Fourier coefficient \(\widehat{u}_{k}(t)\) of \(u(t,x)\) in x satisfies
$$\begin{aligned} \bigl(1+|k|^{p+1}\bigr)\widehat{u}_{k}'(t)+ \bigl(ik+|k|^{2\alpha}\bigr)\widehat {u}_{k}(t)=0, \quad k\in \mathbb{Z}, \end{aligned}$$
that is,
$$\widehat{u}_{k}(t)=e^{-\frac{ik+|k|^{2\alpha}}{1+|k|^{p+1}}t}\widehat{u}_{k}(0). $$
Therefore, we have
$$\begin{aligned} |u|^{2}_{\frac{p+1}{2}}=\sum _{k\in\mathbb{Z}\setminus\{0\} }|k|^{p+1}\bigl|\widehat{u}_{k}(t)\bigr|^{2}= \sum_{k\in\mathbb{Z}\setminus\{0\} }|k|^{p+1}e^{-\frac{2|k|^{2\alpha}}{1+|k|^{p+1}}t}\bigl| \widehat{u}_{k}(0)\bigr|^{2}. \end{aligned}$$
(3.2)
Theorem 3.1
If
\(u_{0}(x)\in\dot{H}^{\frac {p+1}{2}}(\mathbb{T})\), then the unique solution
\(u(t,x)\)
of the periodic initial value problem (3.1) satisfies
$$\begin{aligned} \bigl|u(t)\bigr|_{\frac{p+1}{2}}^{2}\leq|u_{0}|_{\frac{p+1}{2}}^{2},\quad 0< \alpha \leq\frac{p+1}{2}. \end{aligned}$$
(3.3)
Furthermore, we have
$$\begin{aligned}& \bigl|u(t)\bigr|_{\alpha}^{2}\leq \frac{1}{et}|u_{0}|_{\frac{p+1}{2}}^{2},\quad 0< \alpha< \frac{p+1}{2}, \forall t>0, \end{aligned}$$
(3.4)
$$\begin{aligned}& \bigl|u(t)\bigr|_{\alpha}^{2}\leq e^{-t}|u_{0}|_{\frac{p+1}{2}}^{2},\quad \alpha =\frac{p+1}{2}, \forall t>0. \end{aligned}$$
(3.5)
Proof
Equation (3.2) implies that
$$\begin{aligned} |u|^{2}_{\frac{p+1}{2}} =&\sum _{k\in\mathbb{Z}\setminus\{0\} }|k|^{p+1}e^{-\frac{2|k|^{2\alpha}}{1+|k|^{p+1}}t}\bigl|\widehat {u}_{k}(0)\bigr|^{2} \\ \leq&\sum_{k\in\mathbb{Z}\setminus\{0\}}|k|^{p+1}\bigl|\widehat {u}_{k}(0)\bigr|^{2} \\ =&|u_{0}|_{\frac{p+1}{2}}^{2},\quad 0< \alpha\leq \frac{p+1}{2}. \end{aligned}$$
(3.6)
On the other hand, for \(0<\alpha<\frac{p+1}{2}\) we have
$$\begin{aligned} |u|^{2}_{\alpha} =&\sum _{k\in\mathbb{Z}\setminus\{0\}}|k|^{2\alpha }e^{-\frac{2|k|^{2\alpha}}{1+|k|^{p+1}}t}\bigl|\widehat {u}_{k}(0)\bigr|^{2} \\ =&\sum_{k\in\mathbb{Z}\setminus\{0\}}|k|^{p+1}\frac {1+|k|^{p+1}}{|k|^{p+1}} \frac{|k|^{2\alpha}}{1+|k|^{p+1}}e^{-\frac {2|k|^{2\alpha}}{1+|k|^{p+1}}t}\bigl|\widehat{u}_{k}(0)\bigr|^{2} \\ \leq&\sum_{k\in\mathbb{Z}\setminus\{0\}}|k|^{p+1} \varphi_{\alpha }(k)e^{-\varphi_{\alpha}(k)t}\bigl|\widehat{u}_{k}(0)\bigr|^{2}, \end{aligned}$$
(3.7)
where
$$\begin{aligned} \varphi_{\alpha}(k)=\frac{2|k|^{2\alpha}}{1+|k|^{p+1}},\quad 0< \alpha< \frac {p+1}{2}, k\in\mathbb{Z}\setminus\{0\}. \end{aligned}$$
(3.8)
Since the function \(xe^{-xt}\) is uniformly bounded by \(\frac{1}{et}\), we have
$$\begin{aligned} |u|^{2}_{\alpha} \leq&\frac{1}{et}\sum _{k\in\mathbb{Z}\setminus\{0\} }|k|^{p+1}\bigl|\widehat{u}_{k}(0)\bigr|^{2}= \frac{1}{et}|u_{0}|_{\frac{p+1}{2}}^{2}. \end{aligned}$$
(3.9)
For \(\alpha=\frac{p+1}{2}\), we have
$$\begin{aligned} |u|^{2}_{\frac{p+1}{2}} =&\sum _{k\in\mathbb{Z}\setminus\{0\} }|k|^{p+1}e^{-\frac{2|k|^{p+1}}{1+|k|^{p+1}}t}\bigl|\widehat {u}_{k}(0)\bigr|^{2} \\ \leq&\sum_{k\in\mathbb{Z}\setminus\{0\}}|k|^{p+1}e^{-t}\bigl| \widehat {u}_{k}(0)\bigr|^{2} \\ =&e^{-t}\sum_{k\in\mathbb{Z}\setminus\{0\}}|k|^{p+1}\bigl| \widehat {u}_{k}(0)\bigr|^{2} \\ =&e^{-t}|u_{0}|_{\frac{p+1}{2}}^{2}. \end{aligned}$$
(3.10)
The proof is complete. □
We now deal with the nonlinear equation (1.2), that is, \(u_{t}+L_{p}u_{t}+u_{x}+uu_{x}+M_{\alpha}u=0\). We can find similar kind of decreasing properties but less explicit than in the linear case.
Theorem 3.2
If
\(u_{0}(x)\in\dot{H}^{\frac {p+1}{2}}(\mathbb{T})\), then the unique solution
\(u(t,x)\)
of the periodic initial value problem (1.2) satisfies
$$\begin{aligned} \lim_{t\rightarrow+\infty}\bigl|u(t)\bigr|_{\frac{p+1}{2}}^{2}=0. \end{aligned}$$
(3.11)
Proof
Since \(\int_{\mathbb{T}}uu_{x}\,dx=0\), \(\int_{\mathbb {T}}u^{2}u_{x}\,dx=0\), and
$$\begin{aligned}& \begin{aligned}[b] \int_{\mathbb{T}}u(u_{t}+L_{p}u_{t})\,dx&= \pi\frac{d}{dt} \biggl(\sum_{k\in \mathbb{Z}\setminus\{0\}}\bigl| \widehat{u}_{k}(t)\bigr|^{2}+\sum_{k\in\mathbb {Z}\setminus\{0\}}|k|^{p+1}\bigl| \widehat{u}_{k}(t)\bigr|^{2} \biggr) \\ &=\pi\frac{d}{dt}\sum_{k\in\mathbb{Z}\setminus\{0\} } \bigl(1+|k|^{p+1}\bigr)\bigl|\widehat{u}_{k}(t)\bigr|^{2}, \end{aligned} \\& \int_{\mathbb{T}}uM_{\alpha}u\,dx=2\pi\sum _{k\in\mathbb{Z}\setminus\{0\} }|k|^{2\alpha}\bigl|\widehat{u}_{k}(t)\bigr|^{2}=2 \pi\bigl|u(t)\bigr|_{\alpha}^{2}. \end{aligned}$$
The equation and zero mean condition in (1.2) imply that
$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\sum _{k\in\mathbb{Z}\setminus\{0\} }\bigl(1+|k|^{p+1}\bigr)\bigl|\widehat{u}_{k}(t)\bigr|^{2}+\bigl|u(t)\bigr|_{\alpha}^{2}=0, \end{aligned}$$
(3.12)
hence
$$\begin{aligned} \frac{d}{dt}\sum_{k\in\mathbb{Z}\setminus\{0\}} \bigl(1+|k|^{p+1}\bigr)\bigl|\widehat {u}_{k}(t)\bigr|^{2}=-2\bigl|u(t)\bigr|_{\alpha}^{2} \leq0. \end{aligned}$$
(3.13)
It implies that \(\sum_{k\in\mathbb{Z}\setminus\{0\} }(1+|k|^{p+1})|\widehat{u}_{k}(t)|^{2}\) is decreasing in t, so we have
$$\begin{aligned} \sum_{k\in\mathbb{Z}\setminus\{0\}}\bigl(1+|k|^{p+1} \bigr)\bigl|\widehat {u}_{k}(t)\bigr|^{2}\leq\sum _{k\in\mathbb{Z}\setminus\{0\} }\bigl(1+|k|^{p+1}\bigr)\bigl|\widehat{u}_{k}(0)\bigr|^{2},\quad \forall t\geq0. \end{aligned}$$
(3.14)
Therefore,
$$\begin{aligned} \bigl|u(t)\bigr|^{2}_{\frac{p+1}{2}} =&\sum _{k\in\mathbb{Z}\setminus\{0\} }|k|^{p+1}\bigl|\widehat{u}_{k}(t)\bigr|^{2} \\ \leq&\sum_{k\in\mathbb{Z}\setminus\{0\}}\bigl(1+|k|^{p+1}\bigr)\bigl| \widehat {u}_{k}(0)\bigr|^{2} \\ \leq& 2|u_{0}|^{2}_{\frac{p+1}{2}},\quad \forall t\geq0. \end{aligned}$$
(3.15)
Equation (3.15) and \(u_{0}(x)\in\dot{H}^{\frac{p+1}{2}}\) lead to \(u(t)\in\dot{H}^{\frac{p+1}{2}}\) and then \(u(t)\in\dot{H}^{\alpha }\) for \(\frac{1}{2}<\alpha\leq\frac{p+1}{2}\).
On the other hand, (3.13) implies that \(\sum_{k\in\mathbb {Z}\setminus\{0\}}(1+|k|^{p+1})|\widehat{u}_{k}(t)|^{2}\) is decreasing in t and bounded below by zero, then the limit
$$\lim_{t\rightarrow+\infty}\sum_{k\in\mathbb{Z}\setminus\{0\} } \bigl(1+|k|^{p+1}\bigr)\bigl|\widehat{u}_{k}(t)\bigr|^{2} $$
exists, we denote it by A.
Denote
$$\lim_{t\rightarrow+\infty}\bigl|u(t)\bigr|_{\alpha}^{2}=B. $$
If \(B>0\), for large enough t, we have \(|u(t)|_{\alpha}^{2}>B/2\), then there is a constant \(T>0\) such that
$$\begin{aligned} \frac{d}{dt}\sum_{k\in\mathbb{Z}\setminus\{0\}}\bigl(1+|k|^{p+1} \bigr)\bigl|\widehat {u}_{k}(t)\bigr|^{2}< -B, \quad t>T, \end{aligned}$$
hence
$$\begin{aligned} \sum_{k\in\mathbb{Z}\setminus\{0\}}\bigl(1+|k|^{p+1}\bigr)\bigl|\widehat {u}_{k}(t)\bigr|^{2}< \sum_{k\in\mathbb{Z}\setminus\{0\} } \bigl(1+|k|^{p+1}\bigr)\bigl|\widehat{u}_{k}(T)\bigr|^{2}-B(t-T), \end{aligned}$$
and then
$$\begin{aligned} \lim_{t\rightarrow+\infty}\sum_{k\in\mathbb{Z}\setminus\{0\} } \bigl(1+|k|^{p+1}\bigr)\bigl|\widehat{u}_{k}(t)\bigr|^{2}\leq- \infty. \end{aligned}$$
This contradiction leads to \(\lim_{t\rightarrow+\infty}|u(t)|_{\alpha }^{2}=B=0\), that is,
$$\begin{aligned} \lim_{t\rightarrow+\infty}\sum_{k\in\mathbb{Z}\setminus\{0\} }|k|^{2\alpha}\bigl| \widehat{u}_{k}(t)\bigr|^{2}=0, \end{aligned}$$
then \(\forall k\in\mathbb{Z}\), \(\lim_{t\rightarrow+\infty}|\widehat {u}_{k}(t)|^{2}=0\), therefore we have
$$\begin{aligned} \lim_{t\rightarrow+\infty}\sum_{k\in\mathbb{Z}\setminus\{0\} } \bigl(1+|k|^{p+1}\bigr)\bigl|\widehat{u}_{k}(t)\bigr|^{2}=A=0 \end{aligned}$$
and
$$\begin{aligned} \lim_{t\rightarrow+\infty}\sum_{k\in\mathbb{Z}\setminus\{0\} }|k|^{p+1}\bigl| \widehat{u}_{k}(t)\bigr|^{2}=0, \quad\textit{i.e.} \lim _{t\rightarrow+\infty}\bigl|u(t)\bigr|_{\frac{p+1}{2}}^{2}=0. \end{aligned}$$
The proof is complete. □
Remark 3.1
This paper gives the local well-posedness for the subcritical index \(\alpha>\frac{1}{2}\). The interesting case would be to consider the supercritical case \(0<\alpha<\frac{1}{2}\) and the critical case \(\alpha=\frac{1}{2}\). In the supercritical case \(0<\alpha<\frac{1}{2}\), there will be less dissipation, so the dispersive part comes to play a principal role. In the cases \(0<\alpha <\frac{1}{2}\) and \(\alpha=\frac{1}{2}\), \(\dot{H}^{\alpha}(\mathbb{T})\) is not an algebra, we must find another way to establish the estimates on the nonlinear term. We will consider the supercritical case in future work.