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Calculus of variations: Euler-Lagrange

  1. Apr 25, 2012 #1
    This is from a past paper (from a lecturer I don't particularly understand)
    1. The problem statement, all variables and given/known data
    a) {4 marks} Find the Euler-Lagrange equations governing extrema of [itex] I [/itex] subject to [itex] J=\text{constant} [/itex], where[tex]I=\int_{t_1}^{t_2}\text{d}t \frac{1}{2}(x\dot{y}-y\dot{x})=\int f(t,x,y,\dot{x},\dot{y})[/tex]
    and[tex]J=\int_{t_1}^{t_2}\text{d}t (\dot{x}^2+\dot{y}^2)=\int g(t,x,y,\dot{x},\dot{y})[/tex]
    b) {8 marks} show that for the problem in part a) the extremal curves satisfy [itex](x-\alpha)\dot{x}+(y-\beta)\dot{y}=0[/itex] where [itex]\alpha[/itex] and [itex]\beta[/itex] are constants.
    2. Relevant equations
    From an earlier part of the question I have two Euler-Lagrange equations (one differentiating w.r.t. [itex]y[/itex] aswell)[tex]\frac{\partial (f-\lambda g)}{\partial x}-\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial (f-\lambda g)}{\partial \dot{x}}=0[/tex]
    and I think I can write, due to no dependence on [itex]t[/itex] (another one with [itex]y[/itex] again)[tex](f-\lambda g) - \dot{x}\frac{\partial (f-\lambda g)}{\partial \dot{x}}=\mathrm{constant}[/tex]
    3. The attempt at a solution
    For part a) I'm not particularly sure what I am being asked for, or if the equation above is the answer. for part b) I have tried subbing into the equations above and can get out linear equations for [itex]x(t) \text{ and } y(t)[/itex] and get a few dead ends, I'm not really sure what approach to use (a definite answer to part a) would probably help).
     
  2. jcsd
  3. Apr 26, 2012 #2

    vela

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    For part (a), you want to take the specific f and g you've been given and substitute them into the general Euler-Lagrange equation you cited as a relevant equation.
     
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