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Runge-Kutta Method - Need help with the calculus

  1. Jul 21, 2014 #1
    When deriving the Runge-Kutta Method to solve y'=f(x) we need to use Taylor expansion. Hence we need to differentiate the function many times.

    y'(x)=f(y(x))

    y''(x)=f'(y(x))y'(x) = f'(y(x))f(y(x))

    y''' = f''(y(x))(f(y(x)),y'(x)) + f'(y(x))f'(y(x))y'(x)

    I can understand the second derivative but not the third derivative especially the term
    f''(y(x))(f(y(x)),y'(x))

    Can someone please explain the meaning of the notation used here.
     
  2. jcsd
  3. Jul 22, 2014 #2

    HallsofIvy

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    Are you sure you have copied that correctly? The parentheses with the comma is NOT standard operation. If have no idea what "(f(y(x)), y'(x))" could mean.
     
  4. Jul 22, 2014 #3
    I got the expression from Prof. J.L. Butcher note, an authoritative person in Runge-Kutta method. That term is related to a rooted tree.

    Just google rooted tree Runge-Kutta for detail.
     

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  5. Jul 22, 2014 #4
    I've never heard of the Butcher group, but I just read the wiki page and it sounds really powerful.

    You can derive everything by hand the old-fashioned way:
    don't write the dependencies for clarity
    [itex]y(x) = y[/itex]
    [itex]f(y(x)) = f[/itex]

    the first order ODE you gave is:
    [itex]y'=f[/itex]

    The chain rule gives the second derivative:
    [itex]y''=(f)' = f' \cdot y' = f' \cdot f[/itex]

    The product rule and chain rule give the third derivative:
    [itex]y'''=(f'\cdot f)' = (f')' \cdot f + f' \cdot (f)' = f'' \cdot f^2 + (f')^2 \cdot f[/itex]

    Now compare this with the rooted tree with 3 nodes, figure 2 in http://en.wikipedia.org/wiki/Butcher_group
    The first tree corresponds to the first term, as this is f''(f,f) in multivariate form. The second tree corresponds to the second term, as this is f'(f'(f)).
     
  6. Jul 23, 2014 #5
    That explain partly. But why can't we also write the first term as f''(f(f)) and the second term as (f',f')f ?
     
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