Finding the zeroes of a function

Question

How do I find the zeroes of the function $S$, below?

I want to find the zeroes of $$S = 4\psi_2$$ Where $$\dot{\psi}_1=2\psi_2$$ $$\dot{\psi}_2=-2\psi_1$$

and I have that(from below) $$\begin{pmatrix}\psi_1 \\ \psi_2\end{pmatrix} = Ae^{2it}\begin{pmatrix}i \\ 1\end{pmatrix}+Be^{-2it}\begin{pmatrix}-i \\ 1\end{pmatrix}$$ I have two equilibrium points that are centre nodes at $(2,0)$ and $(-2,0)$

So I suppose this is equivalent to finding the zeroes of the function:

$$S = 4(Ae^{2it}+Be^{-2it})$$

And eigenvector: $\underset{\sim}{v}_1=\begin{pmatrix}i \\ 1\end{pmatrix}$ for $\lambda=2i$

And eigenvector: $\underset{\sim}{v}_2=\begin{pmatrix}-i \\1\end{pmatrix}$ for $\lambda = -2i$

Answer 1

When you have \begin{equation} \dot{\psi}_1 = 2\psi_2 \end{equation} and \begin{equation} \dot{\psi}_2 = -2\psi_1, \end{equation} you can re-write this as \begin{equation} \left(\begin{array}{c} \dot{\psi}_1 \\ \dot{\psi}_2 \end{array}\right) = \left(\begin{array}{rc} 0 & 2 \\ -2 & 0 \end{array}\right)\left(\begin{array}{c} \psi_1 \\ \psi_2 \end{array}\right). \end{equation} Call the matrix above $A$. $A$ takes the general form of a matrix $M$ defined as \begin{equation} M = \left(\begin{array}{rc} \sigma & \omega \\ - \omega & \sigma \end{array}\right), \end{equation} and the exponential for such matrices takes the form \begin{equation} \exp(Mt) = e^{\sigma t}\left(\begin{array}{rc} \cos(\omega t) & \sin(\omega t) \\ - \sin(\omega t) & \cos(\omega t) \end{array}\right). \end{equation} Using the values of $\sigma = 0$ and $\omega = 2$ we have in $A$ above, we get \begin{equation} \exp(At) = \left(\begin{array}{rc} \cos(2 t) & \sin(2 t) \\ - \sin(2 t) & \cos(2 t) \end{array}\right). \end{equation} Then solving for $\psi_1$ and $\psi_2$ gives \begin{equation} \left(\begin{array}{c} \psi_1(t) \\ \psi_2(t) \end{array}\right) = \left(\begin{array}{rc} \cos(2 t) & \sin(2 t) \\ - \sin(2 t) & \cos(2 t) \end{array}\right)\left(\begin{array}{c} \psi_1(t_0) \\ \psi_2(t_0) \end{array}\right). \end{equation}

From this we get that \begin{equation} S = -4\sin(2t)\psi_1(t_0) + 4\cos(2t)\psi_2(t_0). \end{equation} You can plug in any initial condition, $\big(\psi_1(t_0), \psi_2(t_0)\big)^T$ of interest and solve for the zeros of $S$ from there.