This exam question is asked every year, but I am struggling to understand the difference between numerical methods even though I can solve all the exercises. Thanks a lot in advance for your help!

QUESTION: Consider the autonomous, scalar differential equation $y'=f(y)$ and the numerical method defined by $$y_{n+1}=y_n+\frac{h}{2}(y_n'+y_{n+1}')+\frac{h^2}{12}(y_n''-y_{n+1}'').$$

State whether the method is:

  1. Explicit
  2. A Runge-Kutta method
  3. A Taylor series method
  4. A One-step method

Note: in a previous part of this exercise I wrote the differential equation as $$y_{n+1}=y_n+\frac{h}{2}\left[f(y_n)+f(y_{n+1})\right]+\frac{h^2}{12}\left[f'(y_n)-f'(y_{n+1})\right]$$

ATTEMPT: What observations do you make to say what type this method is?

  1. I don't know. How can I tell if a method is implicit or explicit?
  2. I believe it is, doesn't it satisfy the formula for an $s$-stage Runge-Kutta method? $$y_{n+1}=y_n+\sum_{i=1}^s b_i f(t_n+c_ih,Y_i),$$ where $$Y_i=y_n+h\sum_{j=1}^s a_{ij}f(t_n+c_jh,Y_j),\quad i=1,2,\dots,s.$$ I am not 100% sure though, so I would really appreciate it if you could give me an explanation or show me how it satisfies it.
  3. I believe it is not a Taylor series method. If it was, it would be $\frac{h^2}{2}$ instead of $\frac{h^2}{12}$.
  4. I believe it does not satisfy the formula for a $k$-step method, which is $$\sum_{j=0}^k \alpha_jy_{n+j}=h\sum_{j=0}^k\beta_j y_{n+j}',\quad y_{n+j}'=f(t_n+jh,y_{n+1}).$$
  • $\begingroup$ Just so you know: RK methods are explicit, and 1-step methods can be (Euler's Method). $\endgroup$ – FundThmCalculus Aug 17 '15 at 19:40
  • $\begingroup$ @FundThmCalculus RK method may be implicit $\endgroup$ – uranix Aug 17 '15 at 19:54
  • $\begingroup$ True, I was thinking of most of the common RK methods, such as those implemented in MATLAB. $\endgroup$ – FundThmCalculus Aug 18 '15 at 11:49

First, note that $y''(x_n) \neq f'(y(x_n))$. $$ y''(x) = (y'(x))' = (f(y(x)))' = f'(y(x)) y'(x) = f'(y(x)) f(y(x))\\ y''_n = f'(y_n) f(y_n) $$

  • It is implicit, since you cannot solve it for $y_{n+1}$ (note $f(y_{n+1})$ and $f'(y_{n+1})$ on the right).
  • It is not an RK since RK works only with ODEs of first order and does not use derivatives of $f'$, which are needed for $y''$.
  • I don't know exactly what a Taylor method stands for, but I suggest that Taylor method is explicit, thus not the case. I suggest Taylor method is based on Taylor series, i.e. $$ y_{n+1} = y_n + h y'_n + \frac{h^2}{2} y''_n = y_n + hf(y_n) + \frac{h^2}{2}f'(y_n)f(y_n). $$ The other possible Taylor method may be obtained combining expansions around $x = x_n$ and $x = x_{n+1}$: $$ y_{n+1/2} = y_n + \frac{h}{2}y'_n + \frac{h^2}{4} y''_n + O(h^3)\\ y_{n+1/2} = y_{n+1} - \frac{h}{2}y'_{n+1} + \frac{h^2}{4} y''_{n+1} + O(h^3) $$ Eliminating $y_{n+1/2}$ gives $$ y_{n+1} = y_{n} + \frac{h}{2}y'_n + \frac{h^2}{4} y''_n + \frac{h}{2}y'_{n+1} - \frac{h^2}{4} y''_{n+1} + O(h^3) = y_n + \frac{h}{2}(y'_n + y'_{n+1}) + \frac{h^2}{4}(y''_n - y''_{n+1}) + O(h^3) $$ Close, but not it.
  • It is a one step method, since the only necessary value to compute $y_{n+1}$ is $y_n$, no $y_{n-1}, y_{n-2}, \dots $ is needed.

The method looks like the Euler–Maclaurin formula was applied to the following integral representation of the ODE: $$ y(x_{n+1}) = y(x_{n}) + \int_{x_n}^{x_{n+1}} y'(\xi)d\xi \approx \\ \approx y(x_{n}) + \frac{h}{2} (y'(x_n) + y'(x_{n+1})) + \frac{h^2}{12} (y''(x_n) - y''(x_{n+1})). $$


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