Finite Differences - ODEs

Mar 01, 2020

(7.1)

$y(x + h) = y(x) + \frac{h}{2}(f(x, y(x)) + f[x + h, y(x) + hf(x, y(x))]) \\ y(x + h) = y(x) + \frac{h}{2}(f(x, y(x)) + \frac{\partial y(x)}{\partial x} + \frac{\partial y(x)}{\partial h} + \frac{\partial hf(x, y(x))}{\partial x} + \frac{\partial hf(x, y(x))}{\partial h}) \\ y(x + h) = y(x) + \frac{h}{2}(f(x, y(x)) + f(x, y(x)) + 0 + h\frac{\partial f(x, y(x))}{\partial x} + f(x, y(x))) \\ y(x + h) = y(x) + \frac{3h}{2}f(x, y(x)) + \frac{h^2}{2} \frac{\partial f(x, y(x))}{\partial x}$

From Taylor exapnsion around $x$ we know:

$y(x + h) = y(x) + hf(x, y(x)) + \frac{h^2}{2} \frac{\partial f(x, y(x))}{\partial x} + \mathcal{O}(h^3)$

So the approximation error is:

$\frac{h}{2}f(x, y(x)) + \mathcal{O}(h^3)$

(7.2) & (7.3)

See Jupyter Notebook here

(7.4)

For the platform, we define $z = A cos(\omega t)$ . Our initial conditions are $z(0) = A$ . For the pendulum, we solve numerically using Runge Kutta, where the step size is the actual time passed between frames (multiplied by some configurable factor). The step is similar to (7.2) except the second derivative is expressed as:

$\ddot{\theta} = -\frac{g + \ddot{z}}{l} sin(\theta) = -\frac{g - A \omega^2 cos(\omega t)}{l} sin(\theta)$

Here is a live animation generated by $A = 20, \omega = \pi, l = 200, g = 9.8, \theta(0) = \frac{\pi}{8}, \dot{\theta}(0) = 0$ :

Alternatively, to play around with different parameters, open this JSFiddle