image 1

I'm trying to derive the error bound for the Ralston Runge-Kutta method (2nd order). Image 1 is taken from Ralston's book "A First Course in Numerical Analysis". In it he expands a step in y where h is the corresponding step in x using the Taylor series expansion. His purpose is to find Coefficients for the Runge Kutta method that produce the minimum error bound.

My problem is the operator D that he defines. What exactly is its function and why and how does it allow him to express the Taylor series in a different form?

In equation (5.6-10) it looks to me as if the RHS is a number and the LHS is still a function?

f(x,y) = y'(x) is the ODE he is examining.

This is his original paper on the subject


The operator $D_f$ is also called the 'flow-derivative'. Denote the exact solution by $y(x)$ and let $\phi(x,y)$ be some smooth function of $x,y$. Then the derivative of $\phi$ along the flow is given by: $$ \frac{d}{dx} \phi(x,y(x)) = \partial_x \phi + (y'(x)\cdot \partial_y) \phi = (\partial_x + f(x,y) \partial_y) \phi = D_f\phi $$ In particular, $$\frac{d^2}{dx^2} y(x) = \frac{d}{dx} f(x,y(x)) = D_f f(x,y(x))$$ The text uses the operator $D= \partial_x + f_n(x_n,y_n) \partial_y= \partial_x + f_n \partial_y$ (here $f_n=f_n(x_n,y_n)$ is a constant). So one has extra cross-terms when squaring the 'non-fixed' flow derivative and evaluating at $(x_n,y_n)$: $$ (\partial_x + f \partial_y)^2 f_{|n}= (\partial_x + f_n \partial_y) (\partial_x + f \partial_y)f_{|n} = D (\partial_x + f \partial_y)f_{|n}= (Df) (\partial_y f)_{|n} + D^2 f_{|n} $$

You may write down a Taylor-expansion to order 4 for $y(x+h)$ in terms of flow derivatives of $f$. The idea in the RK method is that this 4th order Taylor expansion should coincide with the 4th order Taylor-expansion of the 'RK-method'. This then yields a certain number of equations to determine the choices of the coefficients in the method. I personally think it becomes more readable if you consider the origin $(0,0)$ as initial evaluation point.

  • $\begingroup$ Thank you very much for your help. So this flow-derivative is essentially finding the derivative with respect to x at a point n? So when you square the operator its like finding the second derivative with respect to x at a point n? That said, how exactly does he derive equation 5.6-10? $\endgroup$ – MHilton Mar 11 '17 at 16:04
  • $\begingroup$ I have added a calculation for this. In fact in the book the author fixes the point where $f$ is evaluated in the $D$ operator (perhaps it is easier for his calculations? but I wouldn't a priori do so). Then for the 'exact' flow derivative you get extra terms compared to his $D$ $\endgroup$ – H. H. Rugh Mar 11 '17 at 16:30
  • $\begingroup$ In the calculation you wrote out, the right hand term seems simple enough to understand. What I don't understand is the left hand term. How does the fixed fn come about? $\endgroup$ – MHilton Mar 11 '17 at 22:25
  • $\begingroup$ I have regrouped the terms in a perhaps more clear way? $\endgroup$ – H. H. Rugh Mar 11 '17 at 22:47
  • $\begingroup$ Thank you very much. I understand now! First rate help $\endgroup$ – MHilton Mar 13 '17 at 2:24

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