Runge-Kutta Method - Need help with the calculus

  • #1
matematikawan
338
0
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.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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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.
 
  • #3
matematikawan
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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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  • #4
bigfooted
Gold Member
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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)).
 
  • #5
matematikawan
338
0
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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