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Calculus of Variations

  1. Oct 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Optimize the following cost integral

    [tex] x(1)^2 + \displaystyle \int_0^1 (x^2 + \dot{x}^2) dx [/tex]

    subject to x(0) =1, x(1) is free

    2. Relevant equations

    Now our prof showed us a method of doing this. In general, if we want to minimize

    [tex] f(b,x(b)) + \displaystyle \int_a^b L(t,x,\dot{x}) dx [/tex]
    where x(b) is free, then we can change the problem to minimizing
    [tex] \displaystyle \int_a^b L(t,x,\dot{x}) + \frac{\partial f}{\partial t} + \sum_i \frac{\partial f}{\partial x_i} \dot{x}_i dx [/tex]

    3. The attempt at a solution

    Now we he goes through the example above, he changes the Lagrangian to

    [tex] \displaystyle \int_0^1 \left[ 2x\dot{x} + (x^2 + \dot{x}^2) \right] dx [/tex]

    My problem is that I don't see where [itex] 2x\dot{x} [/itex] comes from. The only way this conforms to the above equation is if f has the form of the original Lagrangian. At least in this case, I figure that [itex] f(t,x(t)) = x(t) [/itex] in which case

    [tex] \displaystyle \frac{\partial f}{\partial t} + \sum_i \frac{\partial f}{\partial x_i} \dot{x}_i = \dot{x} + \dot{x} = 2\dot{x} [/tex]

    which varies from what he got by the factor of x
  2. jcsd
  3. Oct 9, 2008 #2
    If nothing else, can someone please confirm that if [itex] f(t,x(t)) = x(t) [/itex] then

    [tex] \displaystyle \frac{\partial f}{\partial t} + \sum_i \frac{\partial f}{\partial x_i} \dot{x}_i = \dot{x} + \dot{x} = 2\dot{x} [/tex]
  4. Oct 9, 2008 #3
    Nevermind, everything has been figured out. I knew it looked weird that I was getting a coefficient of 2. It turns out I didn't notice the fact that x(1) was actually x(1)², so no worries.
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