Understanding the RK Derivation Intuitively

[bangthatdrum]

Im looking at the RK derivation and as part of that I know it is the case that:

y'' = partial f / partial x + partial f / partial y * y'

But at an intuitive level I cannnot understand why. How does the second derivative equal the sum of the two partial derivatives times the first?

 

[bangthatdrum]

Please ignore this.

 

[D H]

Numerical initial value techniques such as Runge-Kutta address the problem of approximating y(t) where the value of y(t) is known at some initial point t₀ and where time derivative of y(t) is a function of the variable of interest and time:

$$\aligned y'(t) &\equiv \frac{dy(t)}{dt} = f(t,y(t)) \\[4pt] y(t_0) &= y_0 \endaligned$$


Suppose some function g(t,y) is a differentiable function of t and y, by which I mean that the partial derivatives of g with respect to t and to y exist at all points where g is defined. Now suppose that y is a function of t. The total derivative of g with respect to t is given by

$$\frac{dg(t,y(t))}{dt} = \frac{\partial g(t,y)}{\partial t} + \frac{\partial g(t,y)}{\partial y}\frac{dy}{dt}$$

With this, the second time derivative of y is

$$y''(t) = \frac{dy'}{dt} = \frac{df(t,y(t))}{dt} = \frac{\partial f(t,y)}{\partial t} + \frac{\partial f(t,y)}{\partial y}\frac{dy}{dt}$$

But we already know dy/dt: It is f(t,y). Thus

$$y''(t) = \frac{\partial f(t,y)}{\partial t} + \frac{\partial f(t,y)}{\partial y}f(t,y)$$

 

[D H]

Please ignore this.

Too late.

 

[bangthatdrum]

Thank you D H. This is wonderful.