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ODE Problem

  1. Aug 14, 2011 #1
    1. The problem statement, all variables and given/known data
    Consider a mechanical system describe by the conservative 2nd-order ODE
    [itex]\frac{\partial^{2}x}{\partial t^{2}}=f(x)[/itex]
    (which could be non linear). If the potential energy is [itex]V(x)=-\int^{x}_{0} f(\xi) d \xi[/itex], show that the system satisfies conservation of energy [itex]\frac{1}{2}x^{2}+V(x)=E[/itex] (E is a constant).

    2. Relevant equations
    As above.

    3. The attempt at a solution
    I've missed almost everything we've done on ODEs, so I don't really have any idea how to being. Even knowing what to call the problem, or a link to some notes/worked examples/relevant textbook would be great. I think the start is to define:
    x_{1}(t) \\
    x_{2}(t) \\
    Then sub into the first equation:
    [itex]\frac{\partial^{2} x_{1}}{\partial t^{2}}=f_{1}(x_{1},x_{2})[/itex]
    [itex]\frac{\partial^{2} x_{2}}{\partial t^{2}}=f_{2}(x_{1},x_{2})[/itex]
    Now I think I should use the chain rule, and integrate equation 2, but I can't see how.
    Last edited: Aug 14, 2011
  2. jcsd
  3. Aug 14, 2011 #2


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    I think you mean [itex]\frac{1}{2}\dot{x}^2 + V(x) = E[/itex]. Just integrate the original equation with respect to x. You'll want to use the chain rule to evaluate the integral on the LHS.
  4. Aug 14, 2011 #3
    I just checked, it's definitely [itex]\frac{1}{2}x^{2}+V(x)=E[/itex]. I did have to fix the V(X) and the domain of the integration.

    You're saying do this?
    [itex]\int f(x) dx = \int \frac{\partial^{2}x}{\partial t^{2}} dx[/itex]
    [itex] \int \frac{\partial^{2}x}{\partial t^{2}} dx = \int \frac{\partial}{\partial t} ( \frac{\partial x}{\partial t} ) dx [/itex]
    The problem is I have no idea how to apply the chain rule to this case.
  5. Aug 14, 2011 #4


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    That can't be right. It would only hold if V(x)=-1/2 x2+V0. The first term is supposed to be the kinetic energy, so it needs to depend on v2, not x2.
    You want to use
    [tex]\frac{\partial}{\partial t} = \frac{\partial x}{\partial t} \frac{\partial}{\partial x}[/tex]
  6. Aug 14, 2011 #5
    Ah thanks, that should be great help.

    You can view the assignment here http://www.maths.utas.edu.au/People/Forbes/KYA314Ass2in2011.pdf [Broken]. Question 2. (a) is the one in question. I think I've typed it correctly though.

    Edit: I have confirmed it is not a typo.
    Last edited by a moderator: May 5, 2017
  7. Aug 14, 2011 #6


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    That's a typo for sure.
  8. Aug 15, 2011 #7


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    For problems like these, "conservation of energy" arises from using "quadrature" on the differential equation.

    That is, if x''= f(x), a function of x, only, we can let v= x' and then use the chain rule:
    [tex]\frac{d^2x}{dt^2}= \frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}= v\frac{dv}{dx}[/tex]
    so that our equation x''= f(x) becomes v v'= f(x) where the differentiation is now with respect to x:
    [tex]v\frac{dv}{dx}= f(x)[/tex]
    [tex]v dv= f(x)dx[/tex]
    [tex]\frac{1}{2}v^2= \int_0^x f(x)dx+ C[/tex]
    [tex]\frac{1}{2}v^2- \int_0^x f(x)dx= C[/tex]
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