Let $C$ be a positive constant. Consider the following system of differential equation with inial value \begin{eqnarray} z(t)+\frac{\sqrt{2}}{2} u'(t)-1=0 \\ 2u''(t)+C \sin(2u(t))=0 \end{eqnarray} with $z(0)=z_0$ and $u(0)=u_0\neq 0$. Would you help me to prove that $z(t)$ that satisfying the system is periodic.
Here's my effort:
First, we can simplify $2u''(t)+C \sin(2u(t))=0$ by linearization to become $u''(t)+C u(t)=0$. Hence, $u(t)=A_1\cos(\sqrt{C}t)+A_2\sin(\sqrt{C}t)$. Thus, $z(t)=-\frac{\sqrt{2}}{2} u'(t)+1=\frac{\sqrt{2C}}{2} (A_1\sin(\sqrt{C}t)-A_2\cos(\sqrt{C}t))+1$. So, $z(t)$ is periodic.
Thanks for your help.