爲什麼我的龍格庫塔實現振盪到0？

Question

我在Haskell中實現了Runge-Kutta。但是，如果在位置爲（0，0，-10）且速度爲（0,0,0）的物體上運行，它會非常快速地來回振盪，直到 達到（0,0,0 ）。

什麼問題？

（注意：我基於this resource實施和*|和|*運營由矢量和矢量由一個標量，分別乘以一個標量

integrate :: PhysicalObject a => a -> Second -> Second -> (Position3, Velocity3) 
integrate object t dt = (p, v) 
    where p = (position object) + (dxdt |* dt) 
     v = (velocity object) + (dvdt |* dt) 
     dxdt = (1.0/6.0) *| (dx1 + (2 *| (dx2 + dx3)) + dx4) 
     dvdt = (1.0/6.0) *| (dv1 + (2 *| (dv2 + dv3)) + dv4) 
     (dx1, dv1) = (velocity object, acceleration (position object, velocity object) t) 
     (dx2, dv2) = evaluate (position object, velocity object) t (0.5*dt) (dx1, dv1) 
     (dx3, dv3) = evaluate (position object, velocity object) t (0.5*dt) (dx2, dv2) 
     (dx4, dv4) = evaluate (position object, velocity object) t dt (dx3, dv3) 

evaluate :: (Vec3 Double, Vec3 Double) -> Second -> Second -> (Vec3 Double, Vec3 Double) -> (Vec3 Double, Vec3 Double) 
evaluate (ix, iv) t dt (dx, dv) = (odx, odv) 
    where odx = sv 
     odv = acceleration (sx, sv) (t + dt) 
     sx = ix + (dx |* dt) 
     sv = iv + (dv |* dt) 

acceleration :: (Vec3 Double, Vec3 Double) -> Second -> Vec3 Double 
acceleration (sx, sv) t = (-k *| sx) - (b *| sv) 
    where k = 10 
     b = 1 

Answer 1

的acceleration函數建模的彈簧，在

這種情況下，我已經把它改爲：

acceleration :: (Vec3 Double, Vec3 Double) -> Second -> Vec3 Double 
acceleration (sx, sv) t = sv