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

Second order Autonomous Differential Equations

  1. Aug 20, 2006 #1
    Hi All,
    I have a differntial equation that I came up with on a little engineering problem posted here https://www.physicsforums.com/showthread.php?t=129247 that I can't solve. It is d^2R/dt^2=W^2*R where R is radial position and W is angular velocity and t is time. I think it is an autonomous diff eq. but don't know, its been a while since math class. Any ideas?? Thanks
     
  2. jcsd
  3. Aug 21, 2006 #2
    W is a constant?

    if so we can solve it relatively easy.

    Introduce v = dR/dt. Then the differential equation is
    v dv/dR = W^2 R.

    Integrating once gives
    v^2 - v0^2 = W^2 R^2 - W^2 R0^2
    Where i have assumed v(t=0) = v0 and R(t=0) = R0.

    a quick arrangement
    v = +/- sqrt( v0^2 - W^2 R0^2 + W^2 R^2 )

    and thus dR/dt = +/- sqrt( v0^2 - W^2 R0^2 + W^2 R^2 )

    This is a seperable first order ODE

    define a such that W^2 a^2 = v0^2 - W^2 R0^2

    Then
    dR/dt = +/- sqrt(W^2 a^2 + W^2 R^2)

    dR/sqrt(a^2 + R^2) = +/- W dt

    The right hand side is +/- Wt.
    To integrate the left hand side Put R = a*sinh(x) so that
    dR = a cosh(x) dx then
    sqrt( a^2 + R^2 ) = sqrt(a^2 + a^2 sinh^2(x)) = a sqrt(1 + sinh(x)^2)
    = a sqrt(cosh^2(x)) = a*cosh(x).

    This dR/sqrt(a^2 + R^2) -> dx
    The integral is thus

    arcsinh(R/a) - arcsinh(R0 / a) = +/- Wt
    and therefore

    R = a sinh( +/- Wt + arcsinh(R0 / a))
    and a = sqrt(V0^2/W^2 - R0^2).

    Plugging this in and checking shows us that the
    - sign gives the v(0) = - v0

    Therefore

    R = a sinh( Wt + arcsinh(R0 / a))
     
    Last edited: Aug 21, 2006
  4. Aug 21, 2006 #3
    just shows to go you, when your tool is a hammer....
    i was thinking about the trick v =dR/dt so that was the
    first way i did this problem.

    ergh! the easier (more standard way) assume
    R = e^kt then the differential equation gives
    k^2 = W^2 so that

    R = A exp(Wt) + B exp(-Wt)

    introducing the initial values you get the same
    answer as above but easier.
     
  5. Aug 21, 2006 #4
    qbert,
    Thanks for the answers. I know exactly what you mean about going about problems with the wrong tool, still it is at least good to know you can grind out answers in more than one way even if your first choice isn't the easiest.
     
  6. Aug 23, 2006 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    "autonomous" equations are those in which the independent variable does not appear explicitly. Second order autonomous equations are those of the form [itex]\frac{d^2y}{dx^2}= f(y,y')[/itex] and, as qbert said, letting v= y' is a standard method (its' called "quadrature"). If v= y', then
    [tex]\frac{d^2y}{dx^2}= \frac{dv}{dt}= \frac{dv}{dy}\frac{dy}{dt}= v\frac{dv}{dy}[/tex]
    so that the second order equation reduces to a first order equation
    [tex]v\frac{dv}{dy}= f(y,v)[/itex]. After you have solved for v, integrate to find y.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Second order Autonomous Differential Equations
Loading...