# Transversality condition equation

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.

• Thanks, Sharkos, for adding a tag. – Chiffa May 26 '13 at 23:50
• Shouldn't you just use the condition $x(T) = T^2 -2$ to solve for $c_1$ and be done with it? – Will Nelson May 27 '13 at 7:16
• 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. – Chiffa May 27 '13 at 9:25
• I guess you can take $T$ as parameter and solve for boundary condition such as $x(T)=c_1T-\frac {T^2}4=T^2-2$ – AnilB May 27 '13 at 20:41
• 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 $T$. – Will Nelson May 28 '13 at 0:09