Thread: Slightly more difficult extremal question

1. Slightly more difficult extremal question

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

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

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

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

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

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

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

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

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

My Working (haven't made much progress)

(a) Applying the Euler-Lagrange equation

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

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

and

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

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

2. My working continued...

(b) Multiplying $\frac{d}{dt} (\dot{x}^*(t)) = 3x^*(t)
$
with $x^*(t) - x(t)$, integrating in $t$ yields

$\displaystyle \int_0^1 \frac{d}{dt} [\dot{x}^*(t)] [x^*(t) - x(t)] \ dt = \int_0^1 3x^* (t) [x^*(t) - x(t)] \ dt$.........(A)

Then integration by parts in (A) yields

$\displaystyle \dot{x}^*(1) [x^*(1) - x(1)] - \dot{x}^*(0) [x^*(0) - x(0)] - \int_0^1 \dot{x}^*(t) \frac{d}{dt} [x^*(t) - x(t)] \ dt$

Using that $x^*(0) - x(0) = 0$ and $x^*(1) - x(1) = 0$

we have

$\displaystyle = - \int_0^1 \dot{x}^*(t) [x^*(t) - x(t)] \ dt$.

Substituting this back into (A) yields

$\displaystyle \int_0^1 \dot{x}^*(t) [x^*(t) - x(t)] \ dt + \int_0^1 3x^* (t) [x^*(t) - x(t)] \ dt = 0$

I'm not sure how to continue here.........

3. Working Continued

(c) We use the inequality $ab \leq \frac{1}{2}(a^2 + b^2)$ for any two real numbers $a, b \in \mathbb{R}$ in

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

Then we have

$\dot{x}^*(t) \dot{x}(t) \leq \ \frac{1}{2} (|\dot{x}^*(t)|^2 + |\dot{x}(t)|^2)$

and

$x^*(t)x(t) \leq \ \frac{1}{2} (|x^*(t)|^2 + |x(t)|^2)$

Then

$\displaystyle \int_0^1 \dot{x}^*(t) \dot{x}(t) \ dt = \frac{1}{2} \int_0^1 \ (|\dot{x}^*(t)|^2 + |\dot{x}(t)|^2) dt$

and

$\displaystyle \int_0^1 \ x^*(t)x(t) \ dt \leq \frac{1}{2} \ \int_0^1 (|x^*(t)|^2 + |x(t)|^2) \ dt$

which implies

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

which I think is

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

for all curves $x = x(t)$ with $x(0) =0$ and $x(1) = x^*(1)$.