## extremals no. 2

(a) Find the extremal $\displaystyle x = x^*(t)$ for the following:
$\displaystyle \int^{1}_{0} (4\dot{x}^2 + x^2) dt, x(0) = 0, x(1) = 1$
i.e. satisfying
(1) $\displaystyle \frac{d}{dt}(4\dot{x}^*(t)) = x^*(t)$

(b) For the extremal $\displaystyle x^*(t)$, show that
(2) $\displaystyle \int^{1}_{0}[4[\dot{x}^*(t)]^2 + [x^*(t)]^2] dt = \int^{1}_{0}[4\dot{x}^*(t)\dot{x}(t) + x^*(t)x(t)]dt$
for all $\displaystyle x = x(t) \in C^2$ with $\displaystyle x(0) = 0, x(1) = 1$.

(Hint: Multiplying equation (1) with $\displaystyle x^*(t) - x(t)$, integrating in $\displaystyle t$ yields

(3) $\displaystyle \int^{1}_{0}\frac{d}{dt}[4\dot{x}^*(t)][x^*(t) - x(t)] = \int^1_0 x^*(t)[x^*(t) - x(t)] dt$
and then integrating by parts in (3) with $\displaystyle x^*(0) - x(0) = 0$ and $\displaystyle x^*(1) - x(1) = 0$.

(c) prove that $\displaystyle x = x^*(t)$ is a minimizing curve.
(Hint: use inequality $\displaystyle ab \leq \frac{1}{2}(a^2 + b^2)$ for any two real number $\displaystyle a, b \in \Re$. Then you use the identity (2).)