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Extreme value of functional

  1. Mar 6, 2014 #1
    1. The problem statement, all variables and given/known data
    We have functional ##A(y)=\int_{-1}^{1}(4y+({y}')^2)dx## where ##y\in C^1(\mathbb{R})## and ##y(-1)=1## and ##y(1)=3##.
    a) Calculate ##A(y)## if graph for ##y## is line segment.
    b) Calculate the extreme value of ##A(y)## for that ##y##. That does it represent?


    2. Relevant equations



    3. The attempt at a solution

    a)

    Since ##y## is line segment I GUESS it can be written as ##y=kx+n##. The functional is therefore calcualted as:

    ##A(y)=\int_{-1}^{1}(4y+({y}')^2)dx=A(y)=\int_{-1}^{1}(4kx+4n+16k^2)dx=32k^2+8n##

    Because ##y(-1)=1## and ##y(1)=3##, ##k=1## and ##n=2##, so ##A(y)=48##

    b)

    We have a so called Euler–Lagrange equation. Lets say that ##L=4y+({y}')^2##.

    ##\frac{\partial L}{\partial y}-\frac{\mathrm{d} }{\mathrm{d} x}\frac{\partial L}{\partial {y}'}=0##

    ##4-2{y}''=0##

    ##y=x^2+Cx+D##

    For ##y(-1)=1## we get an equation saying that ##D=C## and from ##y(1)=3## another one saying ##C+D=2## therefore ##C=D=1##.

    So finally ##y(x)=x^2+x+1## and ##A(y)= 0##. ##y(x)## is quadratic function and extreme value of functional represents minimum, i guess?
     
  2. jcsd
  3. Mar 6, 2014 #2

    pasmith

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    Homework Helper

    Are you sure? [itex]4y + (y')^2 = 4x^2 + 4x + 4 + (2x + 1)^2 = 2(2x + 1)^2 + 3 > 0[/itex], so [itex]A(x^2 + x + 1) > 0[/itex].
     
  4. Mar 6, 2014 #3
    Ahhh, you are right... I made A mistake in my notes.. I used ##4y-({y}')^2##....

    Thank you!
     
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