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 one year ago
Ordinary differential equation I'm having trouble solving
Empty
 one year ago
Ordinary differential equation I'm having trouble solving

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Empty
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{\sin x(y'\sin x)'}{y} + c=k^2 \sin^2 x \] k is an arbitrary real number, c is (I think only negative) but it could possibly be positive. I am fairly certain it's going to be only one sign, if that helps. I can show the PDE I came from, but I don't want to clutter this so much.

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Real problem I have separated: \[\Psi(\theta, \phi)\] \(\phi\) is the azimuthal angle (goes \(2 \pi\) around equator) and \(\theta\) is the zenith angle, so it only goes \(\pi\). This will give us some boundary conditions I think. Here is the differential equation: \[\frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial \Psi}{\partial \theta} \right)+ \frac{1}{\sin^2 \theta}\frac{\partial^2 \Psi}{\partial \phi^2} = k^2 \Psi\] \[\Psi(\theta, \phi) =\Theta(\theta) \Phi(\phi) \] After plugging that in, the equation for \(\Phi\) is kind of trivial, just going to be an exponential, and I assume it will be a sine or cosine so this will make c a negative number to meet our boundary conditions. So here I replaced this for my equation: \[y=\Theta, \ x = \theta \] \[\frac{\sin x(y'\sin x)'}{y} c^2=k^2 \sin^2 x \] And here I'm going to go ahead and claim that c can be any real number now since it will always give a negative number in that spot now, but if my reasoning is off change it to positive if you like.

Empty
 one year ago
Best ResponseYou've already chosen the best response.2... Unrelated to the math, but if you want motivation, I'm trying to find a generalization of this idea of aromaticity in organic chemistry similar to the Huckel rule that a cyclic compound with conjugated systems has 4n+2 electrons in what are essentially porbitals. However I would like to try to derive for particles on a sphere to see if I can try to match the energy levels and stability of buckyballs, which is similar but I am curious how close "particle on a sphere" gets to modelling this. https://en.wikipedia.org/wiki/Particle_in_a_ring

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Hmm, related to you unrelated post, I've done something similar to particle in a ring, as I think it maybe related to this one question I have done regarding a bead along a hoop which rotates, there is a nice way to find the hamiltonian to it and seeing if the energy is conserved or not...probably not the place for this discussion, just thought I'd share it lol if it has to do with it in anyway.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I think it was called particle in a ring..so yeah lol

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Yeah sounds useful and interesting if you wanna explain it!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ok since it's been some time since I've done this mechanics, ok so this is what I believe the problem went like it's a particle moving on a ring while it's rotating with angular velocity w dw:1442817703513:dw so we can start of by finding the Lagrangian of this particle \[L = TU = \frac{ 1 }{ 2 }(\dot x^2+ \dot y^2+ \dot z ^2) mgz\] and we know the bead is constraint to the hoop...I don't know if this is related at all, still want me to continue?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1T = kinetic energy U = potential energy

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Yeah sure keep going this is interesting, this looks similar, except in my case there's no potential energy

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I can give you the general procedure of finding the hamiltonian as well xD, \[L = TU\] where L is the lagranian then you go on to find the linear momentum of the system \[P_i=\frac{ \partial L }{ \partial \dot q_i }\] then you find the total energy of the system from \[H = T+U = H(q,P)\] which you can get by finding \[\dot q_i = \frac{ \partial H }{ \partial P_i }~~~~~\text{and}~~~~~\dot P_i =  \frac{ \partial H }{ \partial q_i }\] This is the general procedure, and a bit over simplified.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1We need a better coordinate system so we have to use polar/ spherical :P

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Ok that sounds reasonable, like we could lock this particle in a spherical shell, like maybe a metal ball rolling around over the surface of a giant magnetic sphere haha

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Although that might, if you get technical, cause some problems if you really look at a magnetic field so don't take me too literally lol

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Lol we can do what ever, I think the difficult part is just finding the constraint really

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442818995523:dw the constraint will be \[z=r\sqrt{r^2\rho^2}\] then

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[L = \frac{ 1 }{ 2 }(\dot \rho ^2 + \rho ^2 \dot \theta ^2 + \dot z^2)mgz\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I think the rest should be ok if you follow the steps, it's setting it up that can be tricky, now there would be a lot of LaTeX xD

Empty
 one year ago
Best ResponseYou've already chosen the best response.2hmmm wait I thought \(\rho\) was going to be the radius, but it's not?

Empty
 one year ago
Best ResponseYou've already chosen the best response.2I'm confused how this constraint is found, what'd you do to find it? I don't really know what it's supposed to be, like constraining it to the circular tube path or describing potential energy of gravity?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Rho here is the distance of the particle from the z  axis not necessarily the radius, the constraint we get from the equation of the circle, so we look at where it's not moving and set up an equation so it would probably be better if I started of as such \[\rho^2 +(zr)^2 = r^2\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Note \[y^2 = \rho^2\]
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