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I Calculus of variations question

  1. Aug 16, 2017 #1
    Hey, I have a theorem I cannot prove.

    We have a function [itex]x^*[/itex] that maximizes or minimizes the integral:
    [tex]\int^{t_1}_{t_0} F(t,x(t),\dot{x}(t))dt[/tex]

    Our end point conditions are:
    [tex]x(t_0) = x_0, x(t_1) \geq x_1[/tex]

    I am told that [itex]x^*[/itex] has to satisfy the Euler equation. That I can fully understand since [itex]x^*(t_1)[/itex] can be equal to [itex]x_1[/itex]. However, then it gives me the transversality condition:
    [tex]\left(\frac{\partial F}{\partial \dot{x}}\right)_{t=t_1} \leq 0 \text{ ( = 0 if $x^*(t_1) > x_1$)}[/tex]

    I can understand the statement in the parentheses. However, I do not understand why [itex]\left(\frac{\partial F}{\partial \dot{x}}\right)_{t=t_1}[/itex] must be less than or equal to zero if [itex]x^*(t_1) = x_1[/itex]. Why can it not be more than zero?
  2. jcsd
  3. Aug 16, 2017 #2


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    The original variation of the functional (call the functional ##\mathcal F##) is given by
    \delta \mathcal F = \int_{t_0}^{t_1} \left( \frac{\partial F}{\partial x} \delta x + \frac{\partial F}{\partial \dot x} \delta \dot x\right) dt.
    Integration by parts of the second term now leads to
    \delta \mathcal F = \int_{t_0}^{t_1} \left( \frac{\partial F}{\partial x} - \frac{d}{dt} \frac{\partial F}{\partial \dot x} \right) \delta x \, dt
    + \left[\frac{\partial F}{\partial \dot x} \delta x\right]_{t=t_0}^{t_1}.
    Since ##\delta x## is arbitrary, the Euler-Lagrange equation has to hold if ##x## is a stationary function of the functional. This, together with ##\delta x(t_0) = 0## leads to
    \delta \mathcal F = \left. \frac{\partial F}{\partial \dot x}\right|_{t=t_1} \delta x(t_1).
    It follows that to have a local stationary function, you must have
    \left. \frac{\partial F}{\partial \dot x}\right|_{t=t_1} = 0.
    However, there is also the option of having a function at the boundary ##x(t_1) = x_1##. If you are at the boundary, then only variations ##\delta x(t_1) \geq 0## are allowed and therefore ##\delta \mathcal F## can only be positive if
    \left. \frac{\partial F}{\partial \dot x}\right|_{t=t_1} > 0
    implying that the value of ##\mathcal F## would increase with the allowed variation. Your condition
    \left. \frac{\partial F}{\partial \dot x}\right|_{t=t_1} \leq 0
    is therefore equivalent to requiring that ##\delta \mathcal F \leq 0##, i.e., that ##x^*## maximises the functional (at least locally). For a minimisation, you would get the opposite inequality.
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