Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Stuck on second order non-linear ODE

  1. Feb 11, 2010 #1
    Out of boredom* I derived some simplified equations to describe the dynamics of a pneumatic gun. These equations are extremely simplified, but they should provide some insight into the system that my numerical models can't. This system had one non-linear equation that could be rewritten as the following after some manipulation.

    [itex]\ddot{z} (C_2 z - C_5 t^2 + x_0) = C_4 + C_1 \tau (1 - e^{-t/\tau})[/itex]

    [itex]z = \int_0^t \int_0^t P(t) dt dt[/itex]

    z is pressure in the barrel integrated twice.

    I am certain the t^2 term can be dropped as the projectile is launched in less than 30 ms for the slowest launchers I have made. Edit: Okay, maybe not as the x_0 term is actually reasonably comparable to the t^2 term.

    A few different methods to obtain an approximate solution were attempted. I attempted to obtain a perturbed solution and nearly had success. Unfortunately, one term in each iteration of the series grew exponentially larger, indicating that this solution does not converge. I could provide the work for this if someone would like to check it. I have little experience with perturbation theory aside from what Farlow's PDE book says.

    Linearization of the non-linear term with an average does not seem to preserve important physical characteristics of the solution (i.e. what is pressure grows asymptotically when it should grow to a peak, decrease, and then oscillate) and the resulting solution is fairly messy.

    Replacing z in the non-linear term with the Taylor expansion about t=0 yields another t^2 term, which as I've noted, isn't helpful (or too accurate).

    I looked at book of exact solutions for ODEs (about 6200 I believe!) without any luck in even finding a particular solution. I've also tried to add small terms to fit the equation to a form from the book, but due to some restrictions, this wouldn't work.

    I've tried a number of guess solutions, guess change of variables, and even tried integrating the equation directly (after some appropriate change of variables) to see if I could drop any terms integration by parts leaves for the non-linear part (the answer to that is obviously no). I tried a number of other things I can't remember now too.

    I figure I could guess a form of the second derivative in the non-linear equation and substitute it in to get a linear equation with variable coefficients, but that would make things a little messier. I'm going to plug in an approximate solution for z that I found unacceptable earlier to see if this would result in something reasonable. Maybe a few iterations of this would return a good solution?

    Any ideas on how to get a reasonably correct approximate solution?

    * This is what happens when I unexpectedly get a week off due to snow.
    Last edited: Feb 12, 2010
  2. jcsd
  3. Feb 16, 2010 #2
    gato_ posted something here that seems to have been deleted... I know because I got a notification email. Thanks for posting that. Why was it deleted? Was it incorrect? Unfortunately, [itex]t/\tau[/itex] will vary between 0 and something on the order of 5. After that the assumptions I made regarding flow through the valve might not apply. (The flow is choked in the valve and the barrel pressure is assumed to always be lower than the gas chamber pressure... this is reasonable for most pneumatic guns but a contrived situation where flow will go backwards can occur.)

    At the moment I'm looking into finding an exact solution to a similar problem, maybe not one with all the terms I have above, and perturbing the difference between the similar problem and the full problem.

    By dropping some inconvenient terms, changing variables to [itex]- t/\tau = ln \xi[/itex] and [itex]y = C_2 z + x_0[/itex], and defining some new terms you could get an equation like [itex]y (\xi y_{\xi \xi} + y_{\xi}) = K_1 - K_2 \xi[/itex]. To me this one seems more likely to be solved. Any ideas?

    I did find a solution to one similar problem but unfortunately, it was implicit. I considered making an explicit approximation, but I don't know much about that.
  4. Feb 16, 2010 #3
    Actualy, I did a loong post, just before discovering an error at the beggining, lol.
    What I did:
    1) I change to [tex]u=z-C_{5}/C_{2} t^2+x_{0}/C_{2}[/tex] so that [tex]P=\ddot{z}=\ddot{u}+2C_{5}/C_{2}[/tex]. Next I rescale everything (I assume all the constants are possitive. Change signs if otherwise):
    [tex]u\rightarrow C_{4}C_{5}^{3}/2 u, t\rightarrow C_{5}\sqrt(C_{2}C_{4})/2 t , \tau\rightarrow C_{5}\sqrt(C_{2}C_{4})/2 \tau , P\rightarrow 2C_{5}/C_{2} P[/tex]
    the equation now looks like:
    [tex]u(\ddot{u}+1)=1+\mu\tau (1-e^{-t/\tau}) , P=\ddot{u}+1[/tex]
    This has, up to my knowledge, no magical variable change that transforms it into a solvable equation. so we must rely on asymptotics...
  5. Feb 17, 2010 #4
    Brettel, to go on I need more information, mainly about the coefficients and their signs. Assuming everything positive constantly throws oscillating solutions, and I am not sure that is what you are looking for. Can you tell me a bit about the derivation of the euqation?
  6. Feb 17, 2010 #5
    I am not looking specifically for oscillating solutions. That is merely what I would expect in a certain time range for the physical problem (that I have seen in my numerical model). If I had an exact solution I would expect this behavior, but only for a certain time range.

    Thinking about how this should respond physically for [itex]t>>\tau[/itex], the projectile would probably go towards the left due to a non-physical assumption I made about the direction of the friction force. (Edit: Ignore this. It is wrong.)

    I'm reasonably certain all coefficients are positive (I'll check). I'll post the derivation later tonight. It is derived mass conservation, the ideal gas law, and Newtonian mechanics. Notably not used are the conservation of momentum and energy. The fluid's temperature is not affected by energy transfer and there are no dynamic pressure terms. These assumptions are reasonable for low velocities.
    Last edited: Feb 17, 2010
  7. Feb 17, 2010 #6
    Then it contains oscilations. When [tex]t>>\tau[/tex], the equation reads
    for which [tex]u=1+\mu\tau[/tex] is a solution. Actually, u oscilates around that value. If [tex]u=1+\mu\tau+v[/tex], then:
    Which represents, for small v, oscilations with slightly variable frequency. If v is not that small (making u approach 0), then u just bounces at 0. At that point, the pressure becomes infinite ([tex]P=\ddot{v}=-v/(1+\mu\tau+v)[/tex]).
    Last edited: Feb 17, 2010
  8. Feb 18, 2010 #7
    Thanks for your help gato. I think I might use an asymptotic approximation for some part of the analysis. I had not considered that approach earlier.

    Unfortunately I am busier than I had anticipated. I'll return to this and post the derivation sometime in the future as I think it might help find a better approximate solution. Expect a bump in a while...
  9. Mar 9, 2010 #8
    Through an approximation that should be good for up to t = 2 * tau I could derive an equation of the following form:

    [itex]y'' y = K \xi[/itex]

    [itex]y = y(\xi)[/itex]

    Any ideas on how to solve this one? A general solution would be best but particulars might be helpful too. I'm confident there is a general solution as it's fairly simple. I'm going to take a look at the book of exact solutions I looked at before, but I thought I'd post this here in case someone could point out a solution as I have limited time at the moment.

    Edit: Seems this is an Emden–Fowler equation... I'm investigating when I should be studying for a heat transfer exam. Though, I know that subject well enough.
    Last edited: Mar 9, 2010
  10. Mar 11, 2010 #9
    I'm sticking to the other one. [tex]u(\ddot{u}+1)=1+\mu\tau[/tex] has a first integral,


    It is then integrable, but no explicit expression for u is at hand. The phase space and solutions, however, are easy to draw (first figure). Pressure is well defined for small oscilations around the equilibrium, but not for large ones. Maybe that speaks about effects you neglected for the derivation

    Attached Files:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook