Optimizing Cost Integrals with Free Endpoint: Calculus of Variations

[Kreizhn]

Homework Statement

Optimize the following cost integral

x(1)^2 + \displaystyle \int_0^1 (x^2 + \dot{x}^2) dx

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

Homework Equations

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

f(b,x(b)) + \displaystyle \int_a^b L(t,x,\dot{x}) dx
where x(b) is free, then we can change the problem to minimizing
\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

The Attempt at a Solution

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

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

My problem is that I don't see where 2x\dot{x} 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 f(t,x(t)) = x(t) in which case

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

which varies from what he got by the factor of x

 

[Kreizhn]

If nothing else, can someone please confirm that if f(t,x(t)) = x(t) then

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

 

[Kreizhn]

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.