I would like to say that this is one of the best presentation I ever seen before. very clear and simple and staitforward presentation. It would be beter if you add the full derivation of the four k's slips of RK4. One more thing, please would you tell me which apps or software you are using to create your tutorial. thank you in advance and good luck.
Your videos are awesome! they have helped me a lot, thanks. I would like to know if you will talk about finite differences, i'm working with the quarter car model and i'm stuck with the equations. I'm making a real time numerical simulation of a suspension and your videos have been a great tool, thanks for all.
Thanks for the feedback, it's nice to hear. Finite difference is conceptually much the same thing as this - a slight re-working of the math. Where are you stuck?
That's nice. You can actually also read the documentation for ode45, the standard MATLAB ode solver. It has plenty of examples there including solving systems of ODEs or higher ODEs, which can guide you well and save time.
so Runge Kutta is derived from the terms in the Taylor series but what I don't understand is how we can say these K_i vectors could be equivalent to terms in the taylor expansion.
That's what makes the RK method so smart. The math to derive it is not that difficult, but quite tedious - especially for the RK4 case. I thought that all that math would clutter this video and I'd lose people. Here is a derivation for the RK2 case, which is somewhat simpler than RK4, if you are interested. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-X-_qCcYDbuY.html
Thank you for this video. May I ask what if the initial value is at a singular point. For example, dy/dx = f(y,x)/x but the initial value is at y(0) = 0. Can we still use Runge Kutta method? How can we solve this problem like that?
That's an excellent question! Not sure that I have an equally excellent answer. My initial instinct is to say that it not possible because of the singularity. However, I'm wondering if perhaps someone has figured out a way around this. I'm not aware of it, but if you find out, please share.
I don't think such a problem exists where you have a denominator x and the first time point starts at 0. Usually, the integration will start 1 or somewhere to the right of 0, so as not to be at the singularity. Plus it wouldn't make sense that your function has a value at that 0 since it literally approaches infinity there. So essentially you don't have to worry about such a hypothetical problem because I haven't seen such a problem in numerical methods problems I solved. The closest I've seen are boundary-value problems with a singularity at the right (or left) boundary and the derivative value is specified there. If 0.1 is the singularity, for example, it is approximated with 0.0999999.
