Hello everyone,

I have another question involving extremals. There are 3 parts to it, and trickier than my last post.

(a) Find the extremal $\displaystyle x = x^*(t)$ for

$\displaystyle \displaystyle \int_0^1 (\dot{x}^2 + 3x^2) \ dt \ \text{with} \ x(0) = 0, x(1) = 1$

i.e. satisfying $\displaystyle \frac{d}{dt} (\dot{x}^*(t)) = 3x^*(t)$

(b) For the extremal $\displaystyle x^*(t)$ show that

$\displaystyle \displaystyle \int_0^1 [[\dot{x}^*(t)]^2 + 3[x^*(t)]^2] \ dt = \int_0^1 [\dot{x}^*(t) \dot{x} t + 3x^*(t)x(t)] dt$

for all $\displaystyle x = x(t) \in \ C^2 \ \text{with} \ x(0) = 0, x(1) = 1$

Hint: $\displaystyle x^*(0) - x(0) = 0$ and $\displaystyle x^*(1) - x(1) = 0$

(c) Prove that $\displaystyle x = x^*(t)$ is a minimizing curve.

My Working(haven't made much progress)

(a) Applying the Euler-Lagrange equation

$\displaystyle \ddot{x} = 3x$ which has solution $\displaystyle x = \frac{3xt^2}{2} + At + B$

Now$\displaystyle x(0) = 0 \ \text{so} \ B = 0$

and

$\displaystyle x(1) = \frac{3x}{2} + A = 1$ so $\displaystyle A = 1 - \frac{3x}{2}$

which is really as far as I've gotten so far. What's throwing me off is the $\displaystyle x$ in the value of $\displaystyle A$. I'm not sure how to handle this.