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Math Help - Optimization: Lagrangian with two functions

  1. #1
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    Optimization: Lagrangian with two functions

    Find y_1(x), y_2(x) so that J is minimized.

    J(y_1,y_2)=\int_0^\frac{\pi}{4}(4y_1^2+y_2^2+y_1'y  _2')dx

    y_1(0)=1, y_2(0)=0

    y_1(\frac{\pi}{4})=0, y_2(\frac{\pi}{4})=1


    My work so far


    I set up the Euler-Lagrange equation for each of the functions, giving me a set of second order differential equations:

    y_1''=2y_2

    y_2''=8y_1

    I tried putting the equations in matrix form and using Wolfram to calculate the eigenvalues and eigenvectors. But solving a 4x4 system with complex eigenvalues seems to be beyond the scope of my calculus of variations course. I suspect there is a simpler way to do it, something that can be done by hand.

    According to my textbook, a first integral of a Lagrangian of several functions, not explicitly dependent on x, is given by:

    L-\sum_{i=0}^{n}y_i'L_{y_{i}^{'}} = C

    Using that property I obtained another equation:

    4y_1^2+y_2^2-y_{1}^{'}y_{2}^{'}=C

    I'm not sure what to do with this though. So basically I'm stuck.

    Any help is appreciated!
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  2. #2
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    Re: Optimization: Lagrangian with two functions

    Quote Originally Posted by TwoPlusTwo View Post
    y_1''=2y_2

    y_2''=8y_1
    I haven't checked the boundary conditions, but from the first equation we get:
    y_2 = \frac{1}{2}y_1''

    So
    y_2'' = \frac{1}{2} y_1^{( \text{iv} )}

    Thus the second equation reads:
    y_2'' = \frac{1}{2} y_1^{( \text{iv} )} = 8y_1

    Solve that and let's see what you get.

    -Dan
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  3. #3
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    Re: Optimization: Lagrangian with two functions

    Quote Originally Posted by topsquark View Post
    I haven't checked the boundary conditions, but from the first equation we get:
    y_2 = \frac{1}{2}y_1''

    So
    y_2'' = \frac{1}{2} y_1^{( \text{iv} )}

    Thus the second equation reads:
    y_2'' = \frac{1}{2} y_1^{( \text{iv} )} = 8y_1

    Solve that and let's see what you get.

    -Dan
    Great, thanks! That was just what I needed.

    I first solved for y_1, then differentiated twice (and divided by two) to get y_2. Using the boundary conditions, I obtained the four constants from a set of four equations. The final solution was:

    y_{1}(x)=c_{1}e^{2x}+c_{2}e^{-2x}+c_{3}\sin(2x)+c_{4}\cos(2x)

    y_{2}(x)=2c_{1}e^{2x}+2c_{2}e^{-2x}-2c_{3}\sin(2x)-2c_{4}\cos(2x),

    where

    c_1=0.1403
    c_2=0.3596
    c_3=0.25
    c_4=0.5

    I still used Wolfram to solve the first system, but the solution was much shorter and easier to work with
    Last edited by TwoPlusTwo; September 30th 2013 at 02:33 AM.
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