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I'm somewhat baffled: I have a problem in calculus of variations:

$$ \int_0^T \!(x-\dot x^2)dt,\qquad x(0)=0,\qquad x(T)=T^2-2. $$

Let $ F(t,x, \dot x) =x-\dot x^2. $

I calculate all the necessary derivatives: $$F_x=1 \qquad F_{\dot x} = -2\dot x \qquad \frac{d}{dt}F_{\dot x} = -2\ddot x$$

and write down an Euler-Lagrange equation: $1+2\ddot x=0$

And calculate my quotients: $$x=c_1t+c_2-\frac{t^2}{4} \qquad x(0)=0 \Longrightarrow c_2=0$$

And then I get stuck: I know that if I have $$ J = \int_{t_1}^{t_2} F(t,x,\dot x) dt $$and the right end follows a curve $x=\varphi(t)$, the transversality condition should hold: $$ F(t_2,x_2, \dot x_2) + [\dot \varphi(t_2)-\dot x_2]F_{\dot x}(t_2,x_2, \dot x_2) =0 $$ where $x_2=x(t_2)$.

But I can't wrap my mind about it: what is $\varphi$ in my case, and how would the transversality equation look like? Please help with any hints.

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  • $\begingroup$ Thanks, Sharkos, for adding a tag. $\endgroup$
    – Chiffa
    Commented May 26, 2013 at 23:50
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    $\begingroup$ Shouldn't you just use the condition $x(T) = T^2 -2$ to solve for $c_1$ and be done with it? $\endgroup$
    – Will Nelson
    Commented May 27, 2013 at 7:16
  • $\begingroup$ Well, I tried that. As you can see, I've found that $c_2=0$, so if I solve $x(T) = T^2 -2$ I'd have an equation about both $T$ and $c_1$, and I think I should only have an equation about one or the other. $\endgroup$
    – Chiffa
    Commented May 27, 2013 at 9:25
  • $\begingroup$ I guess you can take $T$ as parameter and solve for boundary condition such as $x(T)=c_1T-\frac {T^2}4=T^2-2$ $\endgroup$
    – AnilB
    Commented May 27, 2013 at 20:41
  • $\begingroup$ From the way the problem was presented, I assumed $T$ was a parameter. (Anil Baseski's point as well, I think.) I'd check the source of the problem to figure out whether $T$ is indeed a parameter. And even if it isn't, you could just solve the problem as if it is, then compute the target integral as a function of $T$, then optimize that function to choose $T$. $\endgroup$
    – Will Nelson
    Commented May 28, 2013 at 0:09

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