1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Variation of a functional

  1. Feb 19, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider the functional ##S(a,b) = \int_0^∞ r(1-b)a' \, dr ## of two functions ##a(r)## and ##b(r)## (with ##a' = \frac{da}{dr}##). Find the ##a(r)## and ##b(r)## that extremize ##S##, with boundary conditions ##a(∞) = b(∞) = 1##.

    2. Relevant equations


    3. The attempt at a solution
    I know how to find ##b(r)##, my problem is ##a(r)##. This is what I've done,

    ##δS = \int_0^∞ r(1-η)a' \, dr = \int_0^∞ ra' \, dr + \int_0^∞ ra'η \, dr##
    where ##η## is the variation in ##b##.

    Can I say that since both terms should be ##0## so for the right term, since ##η## is arbitrary ##a' = 0## which implies the left term is also zero?
     
  2. jcsd
  3. Feb 19, 2016 #2

    Mark44

    Staff: Mentor

    I could be way off-base here, but isn't ##dS = \frac{\partial S}{\partial a} \cdot da + \frac{\partial S}{\partial b} \cdot db##?
    The two partial derivatives above are the partials of S(a, b), one with respect to a and the other with respect to b.
     
  4. Feb 19, 2016 #3

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    ##S## is a functional, not a function. In the expression, ##a## and ##b## are not numbers, they are functions. As such the derivatives should be functional derivatives and not partial ones. Still, the variation with respect to ##b## is not done correctly.
     
  5. Feb 19, 2016 #4
    Should it be ##δb(r) = (b(r + η(r)) - b(r)) = b'(r)η(r)##? So that ##δS = \int_0^∞ r(1-b'η)a' \, dr##?
     
  6. Feb 20, 2016 #5

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    No, I suggest you start from the definition of the functional derivative
     
  7. Feb 20, 2016 #6
    ##S(a,b) = \int_0^∞ f(b, a', r) \, dr = \int_0^∞ r(1-b)a' \, dr, \quad b = b(r) + αη(r)##

    ##\frac{dS}{dα} = \int_0^∞ \frac{∂f(b(r) + αη(r),~ a(r),~ r)}{∂α}~dr = \int_0^∞ r(-η)a'(r)~dr = -~ \int_0^∞ ra'(r)η(r)~dr = 0##

    This implies ##a'(r) = 0##
     
  8. Feb 20, 2016 #7

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Right, and therefore ...
     
  9. Feb 20, 2016 #8
    ##a = constant## but from the boundary condition, ##a(∞) = 1##, so ##a(r) = 1##. Thanks!!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Variation of a functional
Loading...