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Thread: Slightly more difficult extremal question

  1. #1
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    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 $\displaystyle x = x^*(t)$ 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.
    Last edited by OliviaB; Sep 12th 2010 at 04:20 AM. Reason: Forgot to add in hint
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  2. #2
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    My working continued...

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

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

    Then integration by parts in (A) yields

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

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

    we have

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

    Substituting this back into (A) yields

    $\displaystyle \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.........
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  3. #3
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    Working Continued

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

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

    Then we have

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

    and

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

    Then

    $\displaystyle \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 \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

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

    which I think is

    $\displaystyle \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 $\displaystyle x = x(t)$ with $\displaystyle x(0) =0 $ and $\displaystyle x(1) = x^*(1)$.
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