Given domain $[0,T]$, can we have a 2nd order ODE describing $[0,T/2]$...?

Question

If we have a whole time domain from $[0,T]$ can we have a second order ODE describing $[0,T/2]$ and then a first order ODE describing $[T/2,0]$ ensuring that the solutions of the ODE match up at $0$ and $T/2?$

My initial thought was probably not, as this would mean two states mapping into one state on the boundary, but I am lost on this one.

Answer 1

The solution to the 2nd order ODE on the first time interval may, as you note, be incompatible with the first order ODE. That is, the pair $ (f(T/2^-), f'(T/2^-)) $ may not satisfy the first order ODE. However, if you're willing to allow for $ f'(x) $ to have a discontinuity at $T/2$ then this is fine.

Answer 2

If I understand your question correctly, there are many such situations.

For a given smooth function $f$ and order $n$, consider the following $n$^th order ODE in terms of an unknown function $\psi$: \begin{gather*} 0=f^{(n)}(x)\psi(x)-f(x)\psi^{(n)}(x) \\ f^{(k)}(0)=\psi^{(k)}(0) \end{gather*} This ODE is linear; since $f$ is smooth, it is well-posed. And so the only solution to this ODE is $f$.

So pick an function $f$ on $[0,T]$. There is at least one first-order ODE that $f$ satisfies; we impose that on $[T/2,T]$. Likewise, there is at least one second-order ODE that $f$ satisfies; we impose that on $[0,T/2]$. Now $f$ certainly matches with itself at $T/2$.