UnkleRhaukus
  • UnkleRhaukus
\[a(x) = \frac{\mathrm d^2x}{\mathrm dt^2} = -4\pi^2x+x^2\] \[x_0 = \pi^2,\qquad v_0=\left.\frac{\mathrm dx}{\mathrm dt}\right|_0=0\] Numeric Integration (or any other method that works )
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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UnkleRhaukus
  • UnkleRhaukus
*second order*
Empty
  • Empty
Now having two initial conditions actually makes sense haha
Empty
  • Empty
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.

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UnkleRhaukus
  • UnkleRhaukus
the program i have wrote has a step size of 0.01, and goes t = [0:5]
UnkleRhaukus
  • UnkleRhaukus
doesn't look right , i doubt that the leap frog method is stable for this equation
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UnkleRhaukus
  • UnkleRhaukus
it should decay
UnkleRhaukus
  • UnkleRhaukus
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Empty
  • Empty
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.
Empty
  • Empty
Ohhh it should decay ok let me see.
UnkleRhaukus
  • UnkleRhaukus
i was expecting something like this dotted line here
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UnkleRhaukus
  • UnkleRhaukus
i think the leap frog method dosen't handle the sharp corner very well
Empty
  • Empty
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.
UnkleRhaukus
  • UnkleRhaukus
bad things happen
Empty
  • Empty
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.
UnkleRhaukus
  • UnkleRhaukus
and A = x_0 should be less than 4pi^2
Astrophysics
  • Astrophysics
\[x''= -w_0^2x+2 \beta x'\] I thought if b >0 then the force would be resisting for this at least
Astrophysics
  • Astrophysics
err that should be \[x''= -w_0^2x-2 \beta x'\]
Empty
  • Empty
@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
  • UnkleRhaukus
equilibrium is not at x = 0?
Empty
  • Empty
Woah my bad I don't know what I'm saying, I don't know why I said that part
Empty
  • Empty
I was thinking about the initial position being at \(\pi^2\) and for some reason I said that nonsense lol
UnkleRhaukus
  • UnkleRhaukus
is there some other ( stable ) simple second order numeric method, i can compare with ?
Empty
  • Empty
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.
UnkleRhaukus
  • UnkleRhaukus
RK4 won't work, it's a second order DE
Empty
  • Empty
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.
UnkleRhaukus
  • UnkleRhaukus
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UnkleRhaukus
  • UnkleRhaukus
the shape of the green leapfrog approximation is unconvincing,
anonymous
  • anonymous
$$x''=-4\pi^2 x+x^2\\x''=(x-2\pi^2)^2-4\pi^4\\y''=y^2-k$$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
IrishBoy123
  • IrishBoy123
no decay at dt = 0.00001
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