Runge-Kutta Method - Need help with the calculus

In summary, when deriving the Runge-Kutta Method to solve a first-order ODE y' = f(x), we use Taylor expansion and differentiate the function multiple times. The third derivative involves a term that is related to a rooted tree, which can be understood by referencing the Butcher group. This notation may seem confusing at first, but it can be broken down into terms that correspond to the different parts of the rooted tree.
  • #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.
 
Physics news on Phys.org
  • #2
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
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.
 

Attachments

  • rk_trees.png
    rk_trees.png
    17.3 KB · Views: 478
  • #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)).
 
  • #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 ?
 

Related to Runge-Kutta Method - Need help with the calculus

1. What is the Runge-Kutta Method?

The Runge-Kutta Method is a numerical method used to solve ordinary differential equations. It is a popular method because it is accurate and efficient for solving a wide range of differential equations.

2. How does the Runge-Kutta Method work?

The Runge-Kutta Method works by breaking down a differential equation into smaller steps and using these steps to approximate the solution. It involves calculating intermediate values at each step and using these values to improve the accuracy of the approximation.

3. What are the advantages of using the Runge-Kutta Method?

Some advantages of using the Runge-Kutta Method include its flexibility in handling a wide range of differential equations, its accuracy in approximating solutions, and its ability to handle stiff equations (equations with rapidly changing solutions).

4. Are there any limitations to the Runge-Kutta Method?

One limitation of the Runge-Kutta Method is that it may not be suitable for solving equations with discontinuities or singularities. It also requires knowledge of the initial conditions of the differential equation.

5. How can I implement the Runge-Kutta Method?

The Runge-Kutta Method can be implemented using computer software or by hand. Many programming languages have built-in functions for solving differential equations using the Runge-Kutta Method. If solving by hand, it involves a series of calculations based on the specific form of the differential equation.

Similar threads

Replies
65
Views
2K
  • Differential Equations
Replies
3
Views
1K
  • Differential Equations
Replies
3
Views
652
Replies
1
Views
2K
  • Differential Equations
Replies
5
Views
811
  • Differential Equations
Replies
6
Views
2K
  • Differential Equations
Replies
3
Views
1K
  • Differential Equations
Replies
2
Views
1K
  • Differential Equations
Replies
2
Views
2K
Replies
18
Views
12K
Back
Top