erwachsenc6

2022-10-17

How to find the zeroes of the function $S$, below?
want to find the zeroes of $S=4{\psi }_{2}$
Where
${\stackrel{˙}{\psi }}_{1}=2{\psi }_{2}$
${\stackrel{˙}{\psi }}_{2}=-2{\psi }_{1}$

Do you have a similar question?

wlanauee

Expert

When you have
${\stackrel{˙}{\psi }}_{1}=2{\psi }_{2}$
and
${\stackrel{˙}{\psi }}_{2}=-2{\psi }_{1},$
you can re-write this as
$\left(\begin{array}{c}{\stackrel{˙}{\psi }}_{1}\\ {\stackrel{˙}{\psi }}_{2}\end{array}\right)=\left(\begin{array}{rc}0& 2\\ -2& 0\end{array}\right)\left(\begin{array}{c}{\psi }_{1}\\ {\psi }_{2}\end{array}\right).$
Call the matrix above $A$. $A$ takes the general form of a matrix $M$ defined as
$M=\left(\begin{array}{rc}\sigma & \omega \\ -\omega & \sigma \end{array}\right),$
and the exponential for such matrices takes the form
$\mathrm{exp}\left(Mt\right)={e}^{\sigma t}\left(\begin{array}{rc}\mathrm{cos}\left(\omega t\right)& \mathrm{sin}\left(\omega t\right)\\ -\mathrm{sin}\left(\omega t\right)& \mathrm{cos}\left(\omega t\right)\end{array}\right).$
Using the values of $\sigma =0$ and $\omega =2$ we have in $A$ above, we get
$\mathrm{exp}\left(At\right)=\left(\begin{array}{rc}\mathrm{cos}\left(2t\right)& \mathrm{sin}\left(2t\right)\\ -\mathrm{sin}\left(2t\right)& \mathrm{cos}\left(2t\right)\end{array}\right).$
Then solving for ${\psi }_{1}$ and ${\psi }_{2}$ gives
$\left(\begin{array}{c}{\psi }_{1}\left(t\right)\\ {\psi }_{2}\left(t\right)\end{array}\right)=\left(\begin{array}{rc}\mathrm{cos}\left(2t\right)& \mathrm{sin}\left(2t\right)\\ -\mathrm{sin}\left(2t\right)& \mathrm{cos}\left(2t\right)\end{array}\right)\left(\begin{array}{c}{\psi }_{1}\left({t}_{0}\right)\\ {\psi }_{2}\left({t}_{0}\right)\end{array}\right).$
From this we get that
$S=-4\mathrm{sin}\left(2t\right){\psi }_{1}\left({t}_{0}\right)+4\mathrm{cos}\left(2t\right){\psi }_{2}\left({t}_{0}\right).$
You can plug in any initial condition, $\left({\psi }_{1}\left({t}_{0}\right),{\psi }_{2}\left({t}_{0}\right){\right)}^{T}$ of interest and solve for the zeros of $S$ from there.

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