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Extremals - calc of variations

  1. Dec 10, 2003 #1
    I am trying to find the extremal that minimizes [tex]\int_{0}^{1} \sqrt{y(1+y'^2)} dx[/tex]

    Because it is not explicitly a function of the free variable x, I can use the shortcut:

    constant=F-y'*(dF/dy') to solve for y(x)

    My problem is that after grinding through the algebra my y(x) is a function of itself, in other words I cannot isolate the variable I want to.

    If anybody can offer some tips on either another way to go about this
    problem or maybe argue that y(x) can be isolated it would be greatly appreciated.

    Thanks in advance for the help!

    (In case the formatting doesn't work, everything inside the integral is raised to the 1/2)
  2. jcsd
  3. Dec 10, 2003 #2


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    I'm not sure what you mean by "my y(x) is a function of itself"

    Assuming that by "F" you mean the integrand: F= y1/2(1+y'2)1/2 then I get dF/dy'= y1/2y'(1+y'2)-1/2 so

    F- y'dF/dy'= constant becomes
    y1/2(1+y'2)1/2-y'2y1/2(1+y'2)-1/2= const.

    That's a rather complicated differential equation for y. You might be able to simplify it by multiplying the equation by (1+y'2)-1/2, then squaring both sides.
  4. Dec 10, 2003 #3
    right, I think that is the same result that I get. Upon the simplification that you mentioned, and then solving the diff eq, I think it turns out to be:

    Kx - x*y^-1/2 + C = 2y^1/2 : c,k are constants

    and the y's cannont be combined and therefore y is a function of y.

    Hopefully I am just missing something here and there is an obvious solution.

    Thanks for the quick response.
  5. Dec 11, 2003 #4


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    I don't understand what you mean by "y's cannnot be combined and therefore y is a function of y." The fact that you (or I!) cannot solve an equation doesn't mean that there is no solution.

    If there is not a single solution for y as a function of x then there exist more than one solution to the differential equation. You might be able to determine which is correct for this problem when you apply the initial conditions to determine C.

    For this particular equation, Kx - x*y^-1/2 + C = 2y^1/2, you might try multiplying throug by y-1/2 to get
    (Kx+C)y1/2- xy- 2= 0. Now replace y by z2 so that the equation becomes (Kx+C)z- xz2- 2= 0 which is a quadratic equation. Use the quadratic formula to solve for z and then take the square root to get y.
  6. Dec 11, 2003 #5
    Point well taken, good idea. I'll let you know if it works out.
  7. Dec 12, 2003 #6
    Won't you get the same answer if you extremize

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