Finding Singular points

ash4sigh

hi,

given the system ml[tex]^{2}[/tex][tex]\theta''[/tex]+b[tex]\theta'[/tex]+mglsin([tex]\theta[/tex])

how do I find the singular points??

or any system for that matter - trying the isocline method just not working!! tedious..

HallsofIvy

First that isn't a system, it is a single equation (actually what you wrote isn't even an equation but I assume that was supposed to be "= 0").

Start by writing it as a system of equations: let [itex]\omega= \theta'[/itex] so that [itex]\theta"= \omega'[/itex] and your one equation becomes two first order equations:
[itex]ml^2\omega'+ b\omega+ mglsin(\theta)= 0[/itex] and [itex]\theta'= \omega[/itex] or
[itex]ml^2\omega'= -b\omega- mglsin(\theta)[/itex] and [itex]\theta'= \omega[/itex].

Now "singular points" (or "equilibrium points), points that are single point solutions to the system, are those [itex](\theta, \omega)[/itex] points where the right hand sides of those equations are 0. (I'm very surprised you didn't know that.)

In other words, you must solve the pair of equations [itex]b\omega+ mgl sin(\theta)= 0[/itex] and [itex]\omega= 0[/itex]. And that, obviously, reduces to solving [itex]\sin(\theta)= 0[/itex].

ash4sigh

it is a pendulum system - not sure where the second theta came from in the first term though..

so it'll be [itex](\theta= 0+k\Pi,w=0)[/itex] where k is an integer

thank you very much - I have a million and one questions to ask

the hard part with this one is that I am trying to see it from a phase plane perspective - and visualising where the isoclines converge when you can only draw a couple by hand is tough for a newbie..

thank you for your time, I will be sure to be back.