3
$\begingroup$

I am looking for a list of ODEs to use as examples in the teaching of a numerical methods course for engineers.

I am looking for first and second order examples - the more applied (to engineering) the better, and the harder (or impossible) to solve analytically the better.

Here I list some the examples that I do I know about (in no particular order - many have easily found exact solutions.):

  1. Object under variable gravity and drag with a variable drag coefficient.
  2. Oscillators: undamped, subject to and not subject to external forces.
  3. Temperature in a rod: insulated (transient) and uninsulated.
  4. Pendulum: Real and Approximate
  5. Van der Pol Oscillator
  6. Newton Cooling
  7. Euler-Bernoulli Beam Equations
  8. Charge on a capacitor with and without inductance.
  9. Height of a liquid.
  10. Salt concentrations.
  11. Coupled temperatures.

Thank you.

Improve this question
$\endgroup$
8
  • 2
    $\begingroup$ See matheducators.stackexchange.com/q/8577/77 and the answers there. $\endgroup$ – Joel Reyes Noche Sep 16 '19 at 13:26
  • 1
    $\begingroup$ @JoelReyesNoche I have all of those except for the hanging cable. I can use that thank you. $\endgroup$ – JP McCarthy Sep 16 '19 at 13:44
  • $\begingroup$ I would add soft and hard springs (usually modeled by a cubic nonlinearity). $\endgroup$ – Dan Fox Sep 16 '19 at 15:51
  • $\begingroup$ @DanFox this is the Duffing equation? Thank you. $\endgroup$ – JP McCarthy Sep 16 '19 at 16:34
  • 1
    $\begingroup$ @JPMcCarthy: I have no idea about the names people attach to these things. I mean the following: linearize a one-dimensional mechanical system with potential (e.g. simple pendulum) around an equilibrium of the potential.Then one obtains the harmonic oscillator. If instead of linearizing, one includes higher order terms, and one supposes, as is often physically reasonable, that the potential is odd, then the next lower order term is cubic. Such a system models a spring with a nonlinear response, and is called soft or hard depending on the sign of the cubic term. $\endgroup$ – Dan Fox Sep 22 '19 at 9:16
3
$\begingroup$

I would consider to add two items to the list, both from a systems slant:

  1. Predator prey relations. The behavior can be graphically investigated, but the actual solution function is not analytically soluble. Any solid ODE book will cover this. E.g. Speigel's

https://www.amazon.com/Applied-Differential-Equations-Murray-Spiegel/dp/0130400971#customerReviews

  1. Xenon transients in a nuclear reactor. Long term equilibria can be calculated (limited of time as infinity), but the function of time is not analytically soluble. The standard problem is response to a power change (assume a homogenous reactor). In actuality, very large civilian reactors can have strange oscillations in the geometry and time dimension. But I would avoid this as it is a PDE and too complicated.

See: http://mafija.fmf.uni-lj.si/seminar/files/2016_2017/Seminar_-_Xenon_oscillations_rev3.pdf

[Edit adding derivation]

Xe is a reactor "poison" (absorbs neutrons in competition with uranium). Iodine is a precursor of Xe (look at the periodic table.) For each of them, the $\frac{\mathrm{d}}{\mathrm{d}t}$Concentration = production rate - loss rate. Xe has two PR and two LR. I (iodine) has one of each.

$$\frac{\mathrm{d}I}{\mathrm{d}t}=PR-LR=aPWR-bI$$

Addition is correlated to power as a certain percent of the fissions generate this fission product. Loss rate is just exponential decline.

$$\frac{\mathrm{d}}{\mathrm{d}t}Xe=PR-LR=PR_1+PR_2-LR_1-LR_2$$

$$\frac{\mathrm{d}}{\mathrm{d}t}Xe=cPWR+bI-dXe-ePWR$$

Xe has an extra production term as it is produced directly from fission and also from Iodine decay. It has an extra loss term as it goes away both by decay and by "burnout" (the poisoning reaction).

Note, where I write PWR, it really should be the local neutron flux. Which is related to overall power but not perfectly linearly (there is leakage at the reactor border for instance).

Note, that the Xe situation while it has the trappings of mechE (power reactor) and physics (radioactive decay), it is really a chem E problem in sheep's clothing. As you are looking at concentration. Many feed/bleed problems with salt tanks and the like live in ODE world. And I would consider to included one or two. Have a lot of application in chemistry, environment, etc. I also find concentration somewhat intuitive (more so than bending beams), but maybe that's me.

Comments:

I'm not an expert on this topic, but FWIW, your (2) is the very standard core of analytical ODE course. Unless the forcing function is quite unusual, the problem is soluble.

I believe your (3) is typically a PDE, not ODE, problem. Perhaps some of the others as well. However there area lot of good PLATE (not rod) heat transfer problems that are ODEs. (I seem to recall them as analytically soluble, but there may be some that are not.)

Improve this answer
$\endgroup$
9
  • $\begingroup$ Thank you: I was going to include predator-prey, and will definitely do so now. The uninsulated rod steady state is an ODE. I do the insulated transient PDE with them also. I do some PDE with them also but I think your second example might be too complicated. But thank you. $\endgroup$ – JP McCarthy Sep 16 '19 at 13:18
  • $\begingroup$ Sounds good. The xenon one looks hairy, but it's actually quite pretty imnsho. Will elaborate in answer. I engrained this thing in my head, without memorizing equations per se, so that I can still recall it 30 years later. $\endgroup$ – guest Sep 16 '19 at 13:28
  • $\begingroup$ Great stuff. If it is accessible with fairly straightforward finite differences it could be a nice example to show them if not examine. $\endgroup$ – JP McCarthy Sep 16 '19 at 13:42
  • 1
    $\begingroup$ Yeah, you actually may want to keep the engineering overly simple. Just teach them the methods. If there are too many constants and lambdas and physical relations going on, it may slow down just learning the approximation method. And they can learn heat transfer in heat transfer class. $\endgroup$ – guest Sep 16 '19 at 13:46
  • 1
    $\begingroup$ Kewl. Good luck, man. $\endgroup$ – guest Sep 16 '19 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.