How can I prove this problem?

Question

For the equation $\ddot{x}+\mu \dot{x}+V^{\prime}(x)=0$ where $V(x)=-\left(1-x^{2}\right)^{2}$, prove that the periodic solutions are trivial.

My attempt:

Rewrite the system to $$ \begin{array}{l}{\dot{x}=v} \\ {\dot{v}=-\mu v-V^{\prime}(x)}\end{array} $$

Suppose on the contrary that $x(t)$ is a non-trivial periodic solution i.e. $x(t)=x(T+t)$ for some $T>0$ and $x(t)\ne x(s+t)$ for all $0<s<t$. Multiplying the equation for $\dot{v}$ by $v$ and integrating it from $t_{0}$ to $T+t_{0}$, we can get $$ \int_{t_{0}}^{T+t_{0}}v(t)\dot{v}(t)dt=\int_{t_{0}}^{T+t_{0}}v(t)\left(-\mu v(t)-V^{\prime}(x(t))\right)dt=\int_{t_{0}}^{T+t_{0}}-\mu v^{2}(t)dt-\int_{t_{0}}^{T+t_{0}}v(t)V^{\prime}(x(t))dt $$ For the left-hand side, we can express it as the kinetic energy $$ \int_{t_{0}}^{T+t_{0}}v(t)\dot{v}(t)dt=\left.\frac{v^{2}(t)}{2}\right|_{t_{0}}^{T+t_{0}} $$ For the right-hand side, we have two terms. Let us focus on the latter one firstly. By $\displaystyle dx(t)=v(t)dt$, we can get $$ \int_{t_{0}}^{T+t_{0}}v(t)V^{\prime}(x(t))dt=\int_{t_{0}}^{T+t_{0}}V^{\prime}(x(t))d(x(t))=\int_{x(t_{0})}^{x(T+t_{0})}V^{\prime}(\phi)d\phi=0 $$ Then we can get $$ \left.\frac{v^{2}(t)}{2}\right|_{t_{0}}^{T+t_{0}}=\int_{t_{0}}^{T+t_{0}}-\mu v^{2}(t)dt $$ But what's the next?

Answer 1

You are already done: Since you assumed the solution to be periodic also $v(t)=v(T+t)$. Thus, your last equation simplifies to $$ 0=-\int_{t_0}^{T+t_0}\mu v^2(t) dt $$ which implies for a sufficiently smooth solution $v=0$ for all $t$. Hence, $x$ is constant.

Answer 2

Just note that for any time points $t_{0}$, we have $$ v(t_{0})=\dot{x}(t_{0})=\lim _{h \rightarrow 0} \frac{x(t_{0}+h)-x(t_{0})}{h}=\lim _{h \rightarrow 0} \frac{x(T+t_{0}+h)-x(T+t_{0})}{h}=\dot{x}(T+t_{0})=v(T+t_{0}) $$ which implies $$ \int_{t_{0}}^{T+t_{0}}v(t)\dot{v}(t)dt=\frac{v^{2}(T+t_{0})}{2}-\frac{v^{2}(t_{0})}{2}=0=\int_{t_{0}}^{T+t_{0}} -\mu v^{2}(t) d t + 0 $$

Then the contradiction works.