Results 1 to 14 of 14

Thread: 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 $\displaystyle J_{min}$ for the cases

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

    where $\displaystyle x(1) = 0, x(2) = 3$.

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

    Using the Euler-Largrange eqn, I get

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

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

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

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

    and

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

    Subtracting (2) and (3) I get,

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

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

    which is equivalent to equation (1).

    The extremal is therefore

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

    Now, when using $\displaystyle x(1) = 0, x(2) = 3$ 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; Oct 11th 2010 at 04:41 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

    $\displaystyle
    \displaystyle{\frac{\partial L}{\partial x} - \frac{d}{d t}\frac{\partial L}{\partial x \dot}=0}
    $

    $\displaystyle
    \displaystyle{L=t^3 \dot{x}^2 + 3tx^2}
    $

    $\displaystyle
    \displaystyle{\frac{\partial L}{\partial x}=6tx}
    $

    $\displaystyle
    \displaystyle{\frac{\partial L}{\partial x \dot}=2t^3 x \dot}
    $

    Please calculate this

    $\displaystyle
    \displaystyle{
    6tx-\frac{d}{d t}(2t^3 {x \dot})=0
    }
    $
    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 \displaystyle{
    6tx-\frac{d}{d t}(2t^3 {x \dot})=0
    } $
    Maybe I did this incorrectly.

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

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

    $\displaystyle 3x = \ddot{x} t^2$

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

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    Awards
    2
    As zzzoak mentioned, you must compute

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

    Now, you're viewing $\displaystyle x=x(t),$ 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

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

    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 $\displaystyle 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 $\displaystyle t^{\alpha}$.

    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
    7
    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 $\displaystyle t^{\alpha}.$ So plug in $\displaystyle x(t)=t^{\alpha}$ and solve for $\displaystyle \alpha.$ 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 $\displaystyle x(t) = t^{\alpha}$

    Differentiating

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

    and

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

    Substituting back into the DE,

    $\displaystyle t^2 (\alpha(\alpha - 1)t^{\alpha - 2} - 3t(\alpha t ^{\alpha - 1}) - 3t^{\alpha} = 0$
    Last edited by BrooketheChook; Oct 12th 2010 at 08:24 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    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

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

    so

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

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

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

    So general solution is

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

  10. #10
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    7
    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 $\displaystyle x = -3, 1$

    so the general solution is

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

    Then

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

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

    $\displaystyle B = - \frac{8}{5}$

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

    Im not sure how to calculate $\displaystyle J_{min}$ 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
    7
    Awards
    2
    That all looks correct.

    Isn't

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

    Then, if you've found the minimizing function $\displaystyle x(t),$ can't you just plug that into the expression for $\displaystyle J$ 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 $\displaystyle x(t), $ can't you just plug that into the expression for $\displaystyle J $ and plug and chug?
    The way you worded this really made things clear.

    I get $\displaystyle \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
    7
    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: Oct 3rd 2011, 05:02 AM
  2. Minimizing the expected value of a density function
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Nov 16th 2010, 10:28 PM
  3. Replies: 1
    Last Post: Oct 1st 2010, 09:37 AM
  4. Replies: 4
    Last Post: Jan 6th 2010, 07:34 PM
  5. Transform curve of a function f(x)
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Nov 21st 2009, 03:32 AM

/mathhelpforum @mathhelpforum