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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.

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$z$ is not always periodic. $z$ is given by $z(t)= 1-\frac{1}{\sqrt{2}} u'(t)$.

The system governing $u$ is that of a simple, undamped pendulum. It can have many different sorts of solution depending on $C, u(0), u'(0)$.

One can choose initial conditions so that the pendulum trajectory asymptotically approaches vertical. In this case, $u'$ is not periodic (it has limit $0$), and hence neither is $z$.

Addendum: Here is an explicit demonstration of the latter point.

Let $y_1(t) = 2u(t), y_2(t) = 2 u'(t)$. Then we have the system $\dot{y} = ( y_2 , -C \sin y_1)^T$. Let $V(y) = \frac{1}{2C} y_2^2 - \cos y_1$, and notice that if $y$ is a solution to the differential equation, then $\dot{V}(y(t)) = 0$, hence $V$ is constant on a trajectory.

Choose the initial condition $y_0 = (\frac{\pi}{2}, \sqrt{2C})^T$. Notice that $V(y_0) = 1$, hence $V(y(t)) = 1$ for all $t \geq 0$. Notice that $(\pi, 0)$ is an equilibrium point of the system. We have $\dot{y}_1(0) >0$, $\dot{y}_2(0) <0$, and $\frac{1}{2C} y_2(t)^2 = 1+ \cos y_1(t)$ for all $t \geq 0$.

Now I claim that $y_2(t) >0$ for all $t \geq 0$. If not, then let $t'$ be the first time that $y_2(t') = 0$ (hence $y_1$ is non-decreasing on $[0,t']$). This implies that $\cos y_1(t') = -1$, from which we get $y_1(t') = \pi$. However, since $(\pi, 0)$ is an equilibrium point, this would contradict uniqueness of solution, hence we have $y_2(t) >0$ for all $t \geq 0$. It follows from this that $y_1$ is increasing for all $t \geq 0$, and by the same reasoning, we see that $y_1(t) < \pi$ for all $t \geq 0$, and hence we have $\dot{y}_2(t) < 0$, from which it follows that $y_2$ is decreasing and $y_2(t) >0$ for all $t \geq 0$. It follows that $y$ is bounded and not periodic. Hence $z(t) = 1 - \frac{1}{2 \sqrt{2}} y_2(t)$ is not periodic.

As an aside, a small amount of extra work shows that $\lim_{t \to \infty} y(t) = (\pi, 0)^T$.

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The typo is in the last $z(t)$. the first $z(t)$ is ok. No, we restrict $u(0)\neq 0$. The main point is: Is it true if the linearization system of differential equation have a periodic solution then the original nonlinear have a periodic solution too – beginner Dec 6 '12 at 5:50
More explicitly: Is This argument correct? since $u(t)=A_1\cos(\sqrt{C}t)+A_2\sin(\sqrt{C}t)$ is a periodic solution of$u''(t)+C u(t)=0$, then solution of $2u''(t)+C \sin(2u(t))=0$ is also periodic. – beginner Dec 6 '12 at 5:55
OK, I will delete my answer. – copper.hat Dec 6 '12 at 5:57
In short, is there any theorem that if the solution of the linearization is periodic, hence the solution of the original nonlinear is periodic too. – beginner Dec 6 '12 at 5:58
Two things: First, there are simple $\mathbb{R}^2$ systems whose linearizations have periodic solutions, but the original systems do not have periodic solutions. Second, the pendulum equation has an unbounded solution (corresponding to swinging a pendulum really hard so it keeps swinging around), and the corresponding $z$ will also be unbounded. – copper.hat Dec 6 '12 at 6:00

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