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Calculus of Variations: Functional is product of 2 integrals

  1. Oct 2, 2016 #1
    1. The problem statement, all variables and given/known data
    Minimize the functional: ∫01 dx y'2⋅ ∫01 dx(y(x)+1) with y(0)=0, y(1)=a


    2. Relevant equations
    (1) δI=∫ dx [∂f/∂y δy +∂f/∂y' δy']
    (2) δy'=d/dx(δy)
    (3) ∫ dx ∂f/∂y' δy' = δy ∂f/∂y' |01 - ∫ dx d/dx(∂f/∂y') δy
    where the first term goes to zero since there is no variation at the endpoints



    3. The attempt at a solution
    Using (1) to get
    δI= ∫dx 2y'δy' ∫dx(y+1) + ∫dx y'2 ∫dx δy
    Using (2) and (3) to get
    δI= ∫dx y'2 ∫dx δy - ∫dx d/dx(2y')δy ∫dx(y+1) = 0
    I then tried various ways of manipulating this equation to something that I could minimize trying to isolate δy but was unable to do so since it was always within an integral. I tried to simplify d/dx(2 y') to 2y'' to see if that helped but could not find any benefit to doing so. I also thought about using an integral by parts again but couldn't see where to do so. Using the values at the end points seems like something I should be doing but I don't know how I would. Any advice on how to proceed?
     
  2. jcsd
  3. Oct 3, 2016 #2

    stevendaryl

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    You're on the right track. You have:

    [itex]\int_0^1 dx (y')^2 \int_0^1 \delta y dx - \int_0^1 dx 2y'' \delta y\int_0^1 (y+1) dx = 0 [/itex]

    (Yes, it is better to write [itex]\frac{d}{dx} 2 y'[/itex] as [itex]2 y''[/itex].)

    At this point, let's just name two integrals:
    [itex]I_1 = \int_0^1 dx (y')^2[/itex]

    [itex]I_2 = \int_0^1 dx (y+1)[/itex]

    So we have:
    [itex]I_1 \int_0^1 \delta y dx - I_2 \int_0^1 dx 2y'' \delta y = 0[/itex]

    Now, the important thing to recognize is that [itex]I_1[/itex] and [itex]I_2[/itex] are not functions of [itex]x[/itex] (since [itex]x[/itex] has been integrated out). So you can treat them as if they were constants, and you can combine the two integrals:

    [itex]\int_0^1 \delta y (I_1 - I_2 2y'') dx = 0[/itex]
     
  4. Oct 3, 2016 #3
    Thank you! Yes, the integrals becoming constants after being evaluated was what I was failing to think about.
     
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