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Solving higher order ODE as system of first order

  1. Aug 11, 2014 #1

    Maylis

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    For this problem, I am stuck on the actual system. I don't see what substitution I can make, and the fact that ##u(v)## is a piece-wise function is tripping me up. How the heck do I approach this?? This doesn't look like a standard problem at all.
     

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  3. Aug 12, 2014 #2

    HallsofIvy

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    Actually, that's very much a standard problem- except perhaps that the function on the right, u, is much simpler than usual- it only has two values. No, you do not let "[itex]v=\theta[/itex]" First, it isn't the independent variable you want to change to reduce the order and second, setting one variable equal to a new variable is just changing its name- not really changing anything.

    Instead, let the new variable be equal to the derivative of [itex]\theta[/itex]: [itex]\phi= d\theta/dt[/itex] so that the second derivative of [itex]\theta[/itex] is the first derivative of [itex]\phi[/itex]- [itex]d^2\theta/dt^2= d\phi/dt[/itex]. Now what does [itex]d^2\theta/dt^2= v(\theta)[/itex] become?
     
  4. Aug 12, 2014 #3

    Maylis

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    okay,
    Are these two my new system of two first order equations? They don't have to be derivatives with respect to the same variable?
    [itex] \phi = \frac{d{\theta}}{dt} [/itex]
    [itex] \frac{d^2{\theta}}{dt^2} = \frac{d{\phi}}{dt}[/itex]

    Therefore,
    [itex]\frac{d^2{\theta}}{dt^2} = u(\int \phi{dt})[/itex] ???

    How do I specify the new initial conditions
     
    Last edited: Aug 12, 2014
  5. Aug 12, 2014 #4

    Zondrina

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    Specifically:

    ##\frac{d \phi}{dt} = u(\int \phi(t) dt)##

    Really there are two first order equations here. Do you see them?
     
  6. Aug 12, 2014 #5

    Maylis

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    I'm afraid I don't see them
     
  7. Aug 12, 2014 #6

    Zondrina

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    You are given a specified piecewise function. It only has two values.
     
  8. Aug 12, 2014 #7

    Maylis

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    So does that mean the two equations are

    [itex] \frac {d{\phi}}{dt} = -1 [/itex] if ##u(\int \phi{dt}) \ge 0##
    [itex] \frac {d{\phi}}{dt} = 1 [/itex] if ##u(\int \phi{dt}) \lt 0##.

    and if so, now how do I determine the new conditions?
     
    Last edited: Aug 12, 2014
  9. Aug 12, 2014 #8

    HallsofIvy

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    No, you put "[itex]\phi[/itex]" in the wrong place!
    [itex]\frac{d\phi}{dt}= u(\theta)[/itex]

    Your initial conditions were originally [itex]\theta(5)= \pi/2[/itex] and [itex]\theta'(5)= \pi/10[/tex]
    so in terms of [itex]\theta[/itex] and [itex]\phi[/tex] would be [itex]\theta(5)= \pi/2[/itex], [itex]\phi(5)= \pi/10[/itex].
     
  10. Aug 12, 2014 #9

    Maylis

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    HallsOfIvy,

    So does that mean my two equation system is

    ##\frac {d{\phi}}{dt} = -1## if ##\theta \ge 0##

    ##\frac {d{\phi}}{dt} = 1## if ##\theta \lt 0## ??

    with initial condition ##\phi(5) = \pi/10## and ##\dot{\theta}(5) = \pi/2##?? But then only one condition is in terms of ##\phi##, and the other in terms of ##\dot{\theta}##, is there a way to make the second one in terms of ##\phi##?
     
    Last edited: Aug 12, 2014
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