RK derivation

  • #1
bangthatdrum
11
0
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?
 

Answers and Replies

  • #2
bangthatdrum
11
0


Please ignore this.
 
  • #3
D H
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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:

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


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

[tex]\frac{dg(t,y(t))}{dt} = \frac{\partial g(t,y)}{\partial t} + \frac{\partial g(t,y)}{\partial y}\frac{dy}{dt}[/tex]

With this, the second time derivative of y is

[tex]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}[/tex]

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

[tex]y''(t) = \frac{\partial f(t,y)}{\partial t} + \frac{\partial f(t,y)}{\partial y}f(t,y)[/tex]
 
  • #5
bangthatdrum
11
0


Thank you D H. This is wonderful.
 

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