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UnkleRhaukus
 one year ago
\[a(x) = \frac{\mathrm d^2x}{\mathrm dt^2} = 4\pi^2x+x^2\]
\[x_0 = \pi^2,\qquad v_0=\left.\frac{\mathrm dx}{\mathrm dt}\right_0=0\]
Numeric Integration (or any other method that works )
UnkleRhaukus
 one year ago
\[a(x) = \frac{\mathrm d^2x}{\mathrm dt^2} = 4\pi^2x+x^2\] \[x_0 = \pi^2,\qquad v_0=\left.\frac{\mathrm dx}{\mathrm dt}\right_0=0\] Numeric Integration (or any other method that works )

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Empty
 one year ago
Best ResponseYou've already chosen the best response.3Now having two initial conditions actually makes sense haha

Empty
 one year ago
Best ResponseYou've already chosen the best response.3So what are you looking for exactly, a program or a specific result? Without a step size for the time and without an ending time there's not a lot I can really do for you. I can help walk you through the leapfrog method though if you like.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2the program i have wrote has a step size of 0.01, and goes t = [0:5]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2doesn't look right , i doubt that the leap frog method is stable for this equation

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Well it's not terribly bad for an approximation, I guess it's all a matter of what percent error you're willing to tolerate. You can make a smaller step size and it should make the error smaller.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Ohhh it should decay ok let me see.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2i was expecting something like this dotted line here

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2i think the leap frog method dosen't handle the sharp corner very well

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I was looking through your code for bugs and couldn't find anything, just out of curiosity what happens when you change b=1 to something else like b=10 or b=100 or something ridiculous.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2bad things happen

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Hahaha I guess I just wanted to make sure it was still working. I think you're probably right about leapfrogging over those narrow curves, I don't think it's able to quite turn fast enough so instead of decaying it's rotated slightly off and is instead decreasing your frequency which is sorta what it looks like.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2and A = x_0 should be less than 4pi^2

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[x''= w_0^2x+2 \beta x'\] I thought if b >0 then the force would be resisting for this at least

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1err that should be \[x''= w_0^2x2 \beta x'\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.3@Astrophysics Think of it like this, an undamped spring will have this equation: x''=x That means the acceleration is larger the further away you are from equilibrium (x=0). This is what pushes it back. So if you have some other factor getting bigger, then you're damping the spring force to come back. Of course here it's not quite the same situation, equilibrium is not at x=0, it's shifted up so that we don't end up with a greater restoring force than normal on the other side of the swing.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2equilibrium is not at x = 0?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Woah my bad I don't know what I'm saying, I don't know why I said that part

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I was thinking about the initial position being at \(\pi^2\) and for some reason I said that nonsense lol

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2is there some other ( stable ) simple second order numeric method, i can compare with ?

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Hmm I think the only other options I know of is midpoint method and rungekutta, but not very well. I can't really say how accurate they'll be.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2RK4 won't work, it's a second order DE

Empty
 one year ago
Best ResponseYou've already chosen the best response.3I don't know, what happens when you make the step size smaller on your current program? Does it make the wavelength shorter and dampen it? Try d=0.001 or d=0.00005 maybe and see what happens.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2the shape of the green leapfrog approximation is unconvincing,

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$x''=4\pi^2 x+x^2\\x''=(x2\pi^2)^24\pi^4\\y''=y^2k$$now consider $$2y'y''=2y^2y'ky'\\\frac{d}{dt}(y')^2=\frac{d}{dt}\left(\frac23y^3ky+C\right)\\(y')^2=\frac23y^3ky+C$$which looks like a Weierstrass function

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0no decay at dt = 0.00001
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