Results 1 to 14 of 14

Math Help - Weierstass's Excess Function - minimizing curve

  1. #1
    Junior Member BrooketheChook's Avatar
    Joined
    Sep 2009
    From
    Gold Coast
    Posts
    27

    Weierstass's Excess Function - minimizing curve

    Using the Weierstass condition, find the strongly minimizing curve and the value of J_{min} for the cases

    \displaystyle \int_1^2 (t^3 \dot{x}^2 + 3tx^2) \ dt

    where x(1) = 0, x(2) = 3.

    I've gotten to a point where I have tried to continue, but I can't seem to get anything that resembles a correct solution.

    Let f = t^3 \dot{x} ^2 + 3tx^2

    Using the Euler-Largrange eqn, I get

    6tx - 2t^3 \ddot{x} = 0 so \ddot{x} t^2 - 3x = 0 ...(1)

    which is a Euler's type DE which can be solved by changing variable t = e^s.

    Let x(s) = x(t) = x(e^s). By the chain rule and noting that \frac{dt}{ds} = e^s = t :

    \displaystyle \frac{dx}{ds} = \frac{dx}{dt} \frac{dt}{ds} = e^s \frac{dx}{dt} = t \frac{dx}{dt} ........(2)

    and

    \displaystyle \frac{d^2 x}{d^2 s} = \frac{d}{ds} \big( t \frac{dx}{dt} \big) = t \frac{dx}{dt} + t \frac{d}{ds} \frac{dx}{dt} = t \frac{dx}{dt} + t^2 \frac{d^2 x}{dt^2} ....(3)

    Subtracting (2) and (3) I get,

    \displaystyle \frac{d^2 x}{d^2 s} - \frac{dx}{ds} = t^2 \frac{d^2 x}{dt^2} = 3x

    implying \displaystyle \frac{d^2 x}{d^2 s} - \frac{dx}{ds} - 3x = 0

    which is equivalent to equation (1).

    The extremal is therefore

    \displaystyle x = x(t) = x(s) = Ae^{\frac{1}{2}(1 + \sqrt{13}) s} + Be^{\frac{-1}{2} (-1 + \sqrt{13})s}

    Now, when using x(1) = 0, x(2) = 3 I don't get a solution that appears to be correct.....and I don't know how to continue. Any help would be greatly appreciated.
    Thank-you.
    Last edited by BrooketheChook; October 11th 2010 at 05:41 PM. Reason: Typo in calculation just before eqn 1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Mar 2010
    Posts
    280
    Euler-Largrange equation is

    <br />
\displaystyle{\frac{\partial L}{\partial x} - \frac{d}{d t}\frac{\partial L}{\partial x \dot}=0}<br />

    <br />
\displaystyle{L=t^3 \dot{x}^2 + 3tx^2}<br />

    <br />
\displaystyle{\frac{\partial L}{\partial x}=6tx}<br />

    <br />
\displaystyle{\frac{\partial L}{\partial x \dot}=2t^3 x \dot}<br />

    Please calculate this

    <br />
\displaystyle{<br />
6tx-\frac{d}{d t}(2t^3 {x \dot})=0<br />
}<br />
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member BrooketheChook's Avatar
    Joined
    Sep 2009
    From
    Gold Coast
    Posts
    27
    Please calculate this
    \displaystyle{<br />
6tx-\frac{d}{d t}(2t^3 {x \dot})=0<br />
}
    Maybe I did this incorrectly.

    I get 6tx - 2t^3 \ddot{x} = 0 Maybe this line of calculation is wrong? In my first post there was a typo here which has now been fixed.

    \therefore 6x = 2t^2 \ddot{x}

    3x = \ddot{x} t^2

    so \ddot{x} t^2 - 3x = 0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    As zzzoak mentioned, you must compute

    \displaystyle \frac{d}{dt}(2t^{3}\dot{x}).

    Now, you're viewing x=x(t), which means that the derivative in question must be computed using the product rule (it's a total derivative, not a partial derivative!). I get

    2(3t^{2}\dot{x}+t^{3}\ddot{x})=6t^{2}\dot{x}+2t^{3  }\ddot{x}.

    Another puzzle: you're asked to use the Weierstrass condition. Is that the Weierstrass-Erdmann Corner Condition?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member BrooketheChook's Avatar
    Joined
    Sep 2009
    From
    Gold Coast
    Posts
    27
    I messed that up....twice.

    Now I have  t^2 \ddot{x} - 3t \dot{x} - 3x = 0

    There is a hint to this question that I forgot to mention earlier, which says "Look for solutions of the Euler-Lagrange equation of the form t^{\alpha}.

    That doesn't really make sense to me...how can I get the extremal from here?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    I agree with your DE now.

    Your hint is hinting at the fact that the DE you now have to solve is a Cauchy-Euler equation. To solve Cauchy-Euler equations, your ansatz looks like t^{\alpha}. So plug in x(t)=t^{\alpha} and solve for \alpha. What do you get?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member BrooketheChook's Avatar
    Joined
    Sep 2009
    From
    Gold Coast
    Posts
    27
    Assume x(t) = t^{\alpha}

    Differentiating

    \dot{x} = \alpha t ^{\alpha - 1}

    and

    \ddot{x} = \alpha(\alpha - 1) t^{\alpha - 2}

    Substituting back into the DE,

    t^2 (\alpha(\alpha - 1)t^{\alpha - 2} - 3t(\alpha t ^{\alpha - 1})  - 3t^{\alpha} = 0
    Last edited by BrooketheChook; October 12th 2010 at 09:24 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Looking good so far... What next?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member BrooketheChook's Avatar
    Joined
    Sep 2009
    From
    Gold Coast
    Posts
    27
    which I think (just quickly) is

     \alpha(\alpha - 1) t^{\alpha} - 3 \alpha t^{\alpha} - 3t^{\alpha}

    so

     (\alpha (\alpha - 1) - 3 \alpha - 3) t^{\alpha}

    So, either  x = 0 (trivial) or  x^{\alpha} = 0. Solving the quadratic yields

      \alpha =  2 \pm \ \sqrt{7}

    So general solution is

    x(t) = At^{2 + \sqrt{7}} + Bt^{2 -  \sqrt{7}}
    Follow Math Help Forum on Facebook and Google+

  10. #10
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Whoops. I think there was a sign error way back in post # 5. The double derivative and single derivative terms should have the same sign.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member BrooketheChook's Avatar
    Joined
    Sep 2009
    From
    Gold Coast
    Posts
    27
    Thanks for picking that up.....

    Okay, it will change the sign in a few posts....

    but the solving the quadratic yields x = -3, 1

    so the general solution is

    x(t) = At + \frac{B}{t^3}

    Then

    x(1) = 0 = A + B \Rightarrow \ B = - A

    x(2) = 3 = 2A - \frac{A}{8} \Rightarrow \ A = \frac{8}{5}

    B = - \frac{8}{5}

    Then x = x(t) = \frac{8t}{5} - \frac{8}{5t^3}

    Im not sure how to calculate J_{min} for this though....
    Follow Math Help Forum on Facebook and Google+

  12. #12
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    That all looks correct.

    Isn't

    \displaystyle J=\int_1^2 (t^3 \dot{x}^2 + 3tx^2) \ dt?

    Then, if you've found the minimizing function x(t), can't you just plug that into the expression for J and plug and chug?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Junior Member BrooketheChook's Avatar
    Joined
    Sep 2009
    From
    Gold Coast
    Posts
    27
    Then, if you've found the minimizing function  x(t), can't you just plug that into the expression for J and plug and chug?
    The way you worded this really made things clear.

    I get \frac{228}{5}
    Follow Math Help Forum on Facebook and Google+

  14. #14
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Looks good to me.

    Glad I was able to word things in a clear fashion. That's something I strive for constantly!

    Have a good one!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Minimizing a quadratic function w.r.t. a rotation matrix
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: October 3rd 2011, 06:02 AM
  2. Minimizing the expected value of a density function
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 16th 2010, 11:28 PM
  3. Replies: 1
    Last Post: October 1st 2010, 10:37 AM
  4. Replies: 4
    Last Post: January 6th 2010, 08:34 PM
  5. Transform curve of a function f(x)
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 21st 2009, 04:32 AM

/mathhelpforum @mathhelpforum