Question

\[a(x) = \frac{\mathrm d^2x}{\mathrm dt^2} = -4\pi^2x+x^2\]

\[x_0 = \pi^2,\qquad v_0=\left.\frac{\mathrm dx}{\mathrm dt}\right|_0=0\]

Numeric Integration (or any other method that works )

Answer 1

*second order*

Answer 2

Now having two initial conditions actually makes sense haha

Answer 3

So 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.

Answer 4

the program i have wrote has a step size of 0.01, and goes t = [0:5]

Answer 5

doesn't look right , i doubt that the leap frog method is stable for this equation

Answer 6

it should decay

Answer 7

Well 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.

Answer 8

Ohhh it should decay ok let me see.

Answer 9

i was expecting something like this dotted line here

Answer 10

i think the leap frog method dosen't handle the sharp corner very well

Answer 11

I 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.

Answer 12

bad things happen

Answer 13

Hahaha 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.

Answer 14

and A = x_0 should be less than 4pi^2

Answer 15

\[x''= -w_0^2x+2 \beta x'\] I thought if b >0 then the force would be resisting for this at least

Answer 16

err that should be \[x''= -w_0^2x-2 \beta x'\]

Answer 17

@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.

Answer 18

equilibrium is not at x = 0?

Answer 19

Woah my bad I don't know what I'm saying, I don't know why I said that part

Answer 20

I was thinking about the initial position being at \(\pi^2\) and for some reason I said that nonsense lol

Answer 21

is there some other ( stable ) simple second order numeric method, i can compare with ?

Answer 22

Hmm I think the only other options I know of is midpoint method and runge-kutta, but not very well. I can't really say how accurate they'll be.

Answer 23

RK4 won't work, it's a second order DE

Answer 24

I 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.

Answer 25

the shape of the green leapfrog approximation is unconvincing,

Answer 26

$$x''=-4\pi^2 x+x^2\\x''=(x-2\pi^2)^2-4\pi^4\\y''=y^2-k$$now consider $$2y'y''=2y^2y'-ky'\\\frac{d}{dt}(y')^2=\frac{d}{dt}\left(\frac23y^3-ky+C\right)\\(y')^2=\frac23y^3-ky+C$$which looks like a Weierstrass function

Answer 27

no decay at dt = 0.00001