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

Analytical solution to mx"(t)+b(x'(t))x'(t)+k(p)x(t)=0

  1. Aug 13, 2015 #1
    Hello guys!

    I have following differential equation mx"(t)+b(x'(t))x'(t)+k(p)x(t)=0. As can be seen, "attenuation term" is dependent of velocity x'(t).

    Also stiffness term k(p) is dependent of term p, which is p=k(p)x(t)/A. In this equation A is constant and k(p) means, of course, same term as in our differential equation. And there we have x(t) again... It looks very hard, I know.. :)

    This whole mx"(t)+b(x'(t))x'(t)+k(p)x(t)=0, p=k(p)x(t)/A describes pressurizing of liquid in cylinder as function of displacement.

    I have no idea can this kind of equation be even solved analytically. I tried to solve it numerically in Matlab and got good results, thats not a problem. I would be however interested to see what kind of analytic solution there might be.

    So, my question is that is there some analytic solution?

    If there is no analytic solution, would there be analytic solution to even mx"(t)+b(x'(t))x'(t)+kx(t)=0? In this, we have deleted nasty dependence of p term in k(p)-->k

    Thanks already for your answers and follow-up questions,

    Best regards,

    -DRmushroom
     
  2. jcsd
  3. Aug 13, 2015 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    With an unspecified b, this won't have an analytic solution, otherwise a very general class of differential functions would have one.
     
  4. Aug 13, 2015 #3
    Hello mfb,

    I checked and b(x') could be approximated with for example b(x')=100*(x')^(1/4), when x'>0. It is not exactly that, but quite close with few percent max. error. And anyway b is not very dominant term in this case anyway. So in this case we have solution? That's Interesting.. how it looks like and how could this be derived?

    -DRmushroom
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook