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Solving a differential equation in classical mechanics

  1. May 18, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m is subject to a force
    [tex]F(v)=-bv^2[/tex]
    The initial position is zero, and initial speed is [itex]v_o[/itex] Find x(t)

    3. The attempt at a solution
    for convenience's sake, define
    [tex]Q=-\frac{b}{m}[/tex]
    therefore,
    [tex]{\ddot{x}}=Q{\dot{x}}^2[/tex]

    Let
    [tex] u=\dot{x}[/tex]
    [tex]\dot{u}=Qu^2[/tex]

    then I just made a guess, not hard to see
    [tex]u=-\frac{1}{Qt}[/tex]
    therefore
    [tex]\dot{x}=\frac{-1}{Qt}[/tex]
    intergrate both side, and define x(1)=0
    [tex]\int_{x(1)}^{x'} dx=\int_{1}^{t'}\frac{-dt}{Qt}[/tex]

    therefore
    [tex]x'-x(1)=\frac{1}{Q}(In|1|-In|t'|)[/tex]
    implies
    [tex]e^{x}=t^\frac{m}{b}[/tex]

    so
    1.)is it the solution? or is there another better solution?
    actually, I really have a problem. As it is a not bounded function, somehow I think there might by another more reasonable answer, as it makes no sense when t is approaching to zero.

    2.) what physical meaning is it when t is approaching 0?

    thanks!
     
    Last edited: May 19, 2010
  2. jcsd
  3. May 18, 2010 #2
    i guess you got it wrong, i'm not good at mathematics, but i suppose you did something not right when you wrote that u = -(1/Qt), which is not true. Cause u as you wrote it is speed and here the problem you mentioned comes out - when t = 0, speed is infinity, but in problem statement it is said that at t = 0, speed has some defined value. I think you could do this problem first by finding velocity as function of time, and then coordinate as function of time - you will need to do integration twice. Just write:
    m(dv/dt) = -bv^2
    get (dt) on the ledt side, and v^2 on the left, and then integrate over reasonable interval (from say v0 to v, and t0 = 0 to t). Hope this will help you a bit ;]
     
  4. May 18, 2010 #3
    Are you sure the question reads "t approaching 0"? How can t approach 0 if t starts at 0?

    If you define define x(1)=0 as you said you were doing, simplifying your solution gives [tex]x}=-\ln{t}[/tex],

    The way you started is somewhat messy.
    Instead of lots of substitutions, let the acceleration be [tex]a=v\frac{dv}{dx}[/tex], then integrate and see what happens. :wink:

    Do you have an answer to work towards?
     
    Last edited: May 18, 2010
  5. May 18, 2010 #4

    vela

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    You solved the differential equation incorrectly. Instead of guessing, use separation of variables to find the complete solution.
     
  6. May 18, 2010 #5
    The answer I got when I worked it out is

    [tex]x(t)=\frac{m}{b}ln(1 + V_o \frac{b}{m}t)[/tex]

    I could be wrong though, but at least it's something to work towards.
     
  7. May 18, 2010 #6
    well I got the same answer as Maybe Memorie ;]
     
  8. May 19, 2010 #7
    Thank everybody !! Somehow I make a much more reasonable answer that is same as Maybe_Memorie =)

    [tex]
    v\frac{dv}{dx}=Qv^2[/tex]
    assuming v>0 for all t, and integrate both sides

    [tex]
    \int_{v_0}^{v'}\frac{dv}{v}=\int_{0}^{t'}Qdt[/tex]

    [tex]
    In\frac{v}{v_0}=Qx[/tex]

    implies
    [tex]
    e^{Qx}=\frac{v}{v_0}[/tex]

    [tex]
    v_0\int_{0}^{t'}dt=\int_{0}^{x'}\frac{dx}{e^{Qx}}[/tex]

    and after some calculation (just too lazy to type it here =P)

    [tex]
    x(t)=\frac{m}{b}ln(1 + V_o \frac{b}{m}t)[/tex]

    However, I have a question here for Math.
    Does the below definition still works even
    [tex]\int_{b}^{a}f(x)dx=F(a)-F(b)[/tex]
    F(x) is unbounded? From Apostol's book, the function must be bounded, such that its integral exists. But can the definite integral exist? Thanks! (sorry, I haven't read it to the section discussing about it)
     
    Last edited: May 19, 2010
  9. May 19, 2010 #8
    besides, is it why that my first approach false?
    [tex]
    u=-\frac{1}{Qt}
    [/tex]
    in which, I assume t>0, whereas, in fact, t can be and starts from 0.

    but what if I only consider t>0?
    will the equation works well for t>0 (obviously not, just doesn't make the physical sense, but why it fails? is it because [itex]
    u=-\frac{1}{Qt}
    [/itex] in which the fact that t is not zero is a false assumption, by the logic, a F-if can lead to a F-then, even if the logic is right here?)
     
  10. May 19, 2010 #9

    vela

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    Like I said yesterday, when you claimed u=-1/(Qt), you didn't solve the differential equation correctly. If you separate variables and integrate, you'll get the correct answer.
     
  11. May 20, 2010 #10
    Thanks, but why if I claimed u=-1/(Qt), I won't solve the differential equation correctly?
    Even if I check it back, which matches the conditions?
     
  12. May 20, 2010 #11
    Cause dx/dt is not equal to -1/(Qt), why do you think it is?
     
  13. May 20, 2010 #12

    vela

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    Have you actually solved the differential equation the way I've suggested? If you do that, it's pretty clear why your guess didn't work for this problem.
     
  14. May 21, 2010 #13
    um... actually, I have done the calculation at two other approaches that arrive the expected answer.
    however, sorry for my lousy way of asking questions
    I mean I just don't understand.

    [tex] \dot{x}=-\frac{1}{Qt}
    [/tex]

    [tex] \Rightarrow \ddot{x}=\frac{1}{Qt^2}
    [/tex]
    and
    [tex] Q\dot{x}^2=\frac{1}{Qt^2}
    [/tex]

    so why [itex] \dot{x}[/itex] can't be [itex]-\frac{1}{Qt}[/itex]?
     
  15. May 21, 2010 #14
    I compared the two dx/dt

    a.)the "guess work appraoch", but with a unreasonable answer:
    [tex]\dot{x}=\frac{b}{mt}
    [/tex]

    b.)the "separating the variables appraoch" but with a much more reasonable answer:
    [tex]\dot{x}=\frac{1}{\frac{1}{v_0}+\frac{b}{m}t}
    [/tex]

    And here is my opinion:

    For a.), during the calculation: I assumed: 1.) t will never be zero. So whenever it comes to t=0, contradictions appear.


    it is simply contradicted to the whole question WHEN t=0.
    1.)As it is a force function of speed, if given [itex]v_0=0[/itex], then it will NEVER move at all. However, the equation tells that there is such a [itex]\ddot{x}[/itex] independent of initial velocity.

    still lots of questions in my head... It takes some times to manage them at once
     
    Last edited: May 21, 2010
  16. May 21, 2010 #15

    vela

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    After you separate the variables, you have

    [tex]\frac{dv}{v^2} = -\frac{b}{m}dt[/tex]

    When you integrate that, what do you get for v(t)?
     
  17. May 21, 2010 #16

    vela

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    Note that the only difference between the two solutions is the 1/v0 term in the denominator. It has to be there so the solution can satisfy the initial condition. Your guessed solution satisfies the differential equation, but it's for an initial condition different than the one given in the problem.

    When you solve a differential equation, you don't get just one solution; you get a family of solutions because of the arbitrary constant of integration. Then you use the initial condition to determine what that constant is equal to -- which is what I was trying to get you to do -- thus choosing one solution out of the family of solutions.
     
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