Results 1 to 3 of 3

Math Help - Slightly more difficult extremal question

  1. #1
    Newbie
    Joined
    Aug 2010
    From
    Queenstown
    Posts
    16

    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.
    Last edited by OliviaB; September 12th 2010 at 05:20 AM. Reason: Forgot to add in hint
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Aug 2010
    From
    Queenstown
    Posts
    16
    My working continued...

    (b) Multiplying \frac{d}{dt} (\dot{x}^*(t)) = 3x^*(t)<br />
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.........
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2010
    From
    Queenstown
    Posts
    16
    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).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Slightly difficult word problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: December 20th 2011, 03:13 AM
  2. slightly unusual IVP
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: October 29th 2009, 07:54 AM
  3. IVP, slightly confused
    Posted in the Calculus Forum
    Replies: 5
    Last Post: August 25th 2009, 11:35 AM
  4. Replies: 8
    Last Post: June 6th 2009, 08:26 PM
  5. Slightly different area between curve
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 29th 2008, 09:48 AM

Search Tags


/mathhelpforum @mathhelpforum