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Aug 3, 2018 at 3:50 comment added Ludwig So do you (or anybody) agree with my previous comment? It would be great to have a feedback...
Jul 27, 2018 at 21:24 comment added Ludwig In the last part of your proof, I don't understand why it should be $\langle v_+,v_-\rangle \ne 0$. Since $B$ is symmetric, $v_+$ and $v_-$ must be orthogonal, right? (Or perhaps I'm missing something...)
Jul 26, 2018 at 22:01 comment added Ludwig Okay thanks (just forgot a square factor in the diagonal in my calculations...) I'll check the remaining part of the proof asap
Jul 26, 2018 at 21:44 comment added David Hughes Sure, I have added the formula for $\dot{y}_1$ and $\dot{y}_2$.
Jul 26, 2018 at 21:43 history edited David Hughes CC BY-SA 4.0
Edited to clarify slope formula
Jul 26, 2018 at 17:47 comment added Ludwig Thanks for fixing this. Also, it is not clear to me why $m(t):=\dot{y}_2/\dot{y}_1=-\tanh(\sinh(\omega t)/\omega)$. Could you please expand this a little bit?
Jul 26, 2018 at 16:35 history edited David Hughes CC BY-SA 4.0
Edited per Ludwig's comment to correct the expression for $y$.
Jul 26, 2018 at 15:34 comment added David Hughes Yes, I made a mistake. However, this can be fixed by multiplying $y$ by $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, I will add this later.
Jul 26, 2018 at 3:31 comment added Ludwig Thanks for your answer! Could you please elaborate a little more on the derivation of the closed-form expression of the matrix exponential in your second formula? (Because, applying the expression in Theorem 2.2 of this paper, I get $$e^{t+\frac{1}{\omega}\sin(\omega t)}\begin{bmatrix} \cosh(\sin(\omega t)/\omega) & -\sinh(\sin(\omega t)/\omega) \\ -\sinh(\sin(\omega t)/\omega) & \cosh(\sin(\omega t)/\omega) \end{bmatrix},$$ which looks different from yours..)
Jul 26, 2018 at 2:06 history answered David Hughes CC BY-SA 4.0