I have written a project which I must 'defend' so to speak. In this project, I must compare THE Runge-Kutta method (4th order) with Euler to explore the advantages and disadvantages. I'm sorry for any incorrect mathematical terms, I'm translating them the best I can. The level is final year high-school maths.
Context and understanding
As far as I am able to understand, forward Euler's local truncation error can be found by looking into Taylor's series:
Let y'(x) = f(x,y(x))
A point on the actual function y(x0) = y0 is known.
then y(x0+h) ≈ y(x0) + h * f(x0,y0) where h is a small step size, and f(x0,y0) is the derivative in the given first point (x0,y0)
The error is the next term in Taylor's series: 1/2*h2*y''(ξ)
for a suitable ξ ∈[x, x+h]. I understand that 'series' means infinite terms, but the ξ is in there to cut it short, so that this equation is true:
y(x0+h) = y(x) + h * y'(x) + 1/2*h2*y''(ξ)
Hence the 'suitable' ξ.
As we can see, the local truncation error in forward Euler must be proportional to h squared: y(x0+h) - y(x) + h * y'(x) + = 1/2*h2*y''(ξ) = error
So Local Truncation Error: LTE = O(h2) (Not that I really understand 'big-O' notation)
I also choose an interval in which I solve the differential equation, which entails that I also look at the global error. This means that the number of steps is inverse proportional to the step size (O(h-1)). Apparently, you write this out with 'big-O' notation like so: O(h-1) * O(h2) = O(h)
And so, with the help of taylor's series, you can figure out both the local and global error.
The actual question: RK4
Is it possible to figure out the error of RK4 in the same way? Is there a connection between taylor's series and RK4 which will provide the result that the local error of RK4 is proportional to h to the fifth power (O(h5), and that the global error is proportional to the fourth power (O(h4))?
I've done an 'experiment' where I calculate the error of each method by numerically solving a differential equation that is also analytically solvable. Here, it is clear the RK4 indeed has a global error proportional to the step size of h to the fourth power.
This is the RK4 I speak of:
In my project, I tossed in this equation:
where delta x is h
and wrote that the error first appears in the fifth term (besides y(x)) and that therefore must be proportional to h to the fifth power. Im going to want to know if that's wrong
Thanks for reading
Context and understanding
As far as I am able to understand, forward Euler's local truncation error can be found by looking into Taylor's series:
Let y'(x) = f(x,y(x))
A point on the actual function y(x0) = y0 is known.
then y(x0+h) ≈ y(x0) + h * f(x0,y0) where h is a small step size, and f(x0,y0) is the derivative in the given first point (x0,y0)
The error is the next term in Taylor's series: 1/2*h2*y''(ξ)
for a suitable ξ ∈[x, x+h]. I understand that 'series' means infinite terms, but the ξ is in there to cut it short, so that this equation is true:
y(x0+h) = y(x) + h * y'(x) + 1/2*h2*y''(ξ)
Hence the 'suitable' ξ.
As we can see, the local truncation error in forward Euler must be proportional to h squared: y(x0+h) - y(x) + h * y'(x) + = 1/2*h2*y''(ξ) = error
So Local Truncation Error: LTE = O(h2) (Not that I really understand 'big-O' notation)
I also choose an interval in which I solve the differential equation, which entails that I also look at the global error. This means that the number of steps is inverse proportional to the step size (O(h-1)). Apparently, you write this out with 'big-O' notation like so: O(h-1) * O(h2) = O(h)
And so, with the help of taylor's series, you can figure out both the local and global error.
The actual question: RK4
Is it possible to figure out the error of RK4 in the same way? Is there a connection between taylor's series and RK4 which will provide the result that the local error of RK4 is proportional to h to the fifth power (O(h5), and that the global error is proportional to the fourth power (O(h4))?
I've done an 'experiment' where I calculate the error of each method by numerically solving a differential equation that is also analytically solvable. Here, it is clear the RK4 indeed has a global error proportional to the step size of h to the fourth power.
This is the RK4 I speak of:
In my project, I tossed in this equation:
where delta x is hand wrote that the error first appears in the fifth term (besides y(x)) and that therefore must be proportional to h to the fifth power. Im going to want to know if that's wrong
Thanks for reading