I have two 2nd order systems.
System 1: $$\ddot{\theta}_{1}(t) + \sin{\theta}_{1}(t) = \cos{\theta}_{1}(t)f(t), \text{ with } t \in [0, T],$$ and $\theta_{1}(0)=\dot{\theta}_{1}(0)=0$ and $\theta_{1}(T)=a,\: \dot{\theta}_{1}(T) = 0$.
System 2: $$\ddot{\theta}_{2}(\mu) + \sin{\theta}_{2}(\mu) = \cos{\theta}_{2}(\mu)f(2T-\mu), \text{ with } \mu = t + T \in [T,2T],$$ and $\theta_{2}(T)= a, \: \dot{\theta}_{2}(T)=0,$ namely
- System 2 is identical to System 1.
- The initial condition of the System 2 is the terminal condition of the System 1.
- The input of the System 2 is the reversed version of the System 1's input.
Note that $f(t)$ is a real function, $T$ is a positive terminal time, and $a$ is a positive real number.
My purpose is: prove that $\theta_{2}(2T) = \dot{\theta}_{2}(2T) = 0$.
I very much appreciate any suggestion or hint!
In the second equation, set $s=2T-μ$, $y(s)=θ_2(μ)$, then the second equation reads as $$ \ddot y(s)+\sin(y(s))=\cos(y(s))f(s), ~~ s\in[0,T],\\~\\ y(T)=a,~\dot y(T)=0. $$ This is exactly identical to the first equation with the same boundary conditions at $T$, $y(T)=θ_1(T)$, $\dot y(T)=\dot θ_1(T)$. Since these are complete as initial conditions, there is only one solution to the resulting IVP, thus $y(s)=θ_1(s)$ for all $s\in[0,T]$, and so you get also the same values at $s=0$ which translate as values for $θ_2$ at $s=2T$.
Note that the original system for $θ_1$ is overdetermined, resp. that $f$ has to be chosen correctly so that all 3 homogeneous boundary conditions are satisfied simultaneously. With a general $f$ the problem for $θ_1$ is usually not solvable.