Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Minimal Function

  1. Nov 23, 2005 #1
    The function of arc-length [tex]c[/tex] that minimises [tex]\int^b_{a}{y dx}[/tex] is the catenary, but why? I have tried using calculus of variations, [tex]\frac{\partial f}{\partial y} - \frac{d}{dx}(\frac{\partial f}{\partial \dot{y}}) + \lambda (\frac{\partial g}{\partial y}) = 0[/tex], however I don't know which constraint function to use.
     
    Last edited: Nov 24, 2005
  2. jcsd
  3. Nov 24, 2005 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Surely that is the wrong integral to look at (clearly the constant function y(x)=0 for all x minimizes the integral) (edit: and equally clearly I've jsut said something stupid, haven't I? That integral has no minimum: assumin y(a)=y(b)=0, and why not, that a=0, b=pi then -ksin(x) can be made arbitrarily small by making k arbitrarily large).

    Try instead minimizing

    [tex] \int y \sqrt{1+y'^2}dx[/tex]

    with boundary conditions y(a)=y(b)
     
    Last edited: Nov 24, 2005
  4. Nov 24, 2005 #3
    Why is the arc-length multiplied by [tex]y[/tex]?
     
  5. Nov 24, 2005 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Long answer with hand waving justification: Well, assuming a uniformly dense, unifrom cross section, the contribution of an infinitesimal bit of chain, dx, is the length of the chain times a constant (cross section times density) times the (gravitational) potential which is, ignoring constants

    [tex] y \sqrt{1+y'^2} dx[/tex]

    adding/integrating all those gives the answer.

    Short answer for those who understand mechanics: y is the gravitational potential.
     
  6. Nov 24, 2005 #5
    [tex]y[/tex] is the minimal function, [tex]\sqrt{1 + \dot{y}^2}dx[/tex] is the arc-length. What is [tex]y \sqrt{1 + \dot{y}^2}dx[/tex]?
    Why don't I need a constraint equation?
     
    Last edited: Nov 24, 2005
  7. Nov 24, 2005 #6

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    it is the contribution to the gravitational potential of a small part of the chain (potential times the infinitesimal length of the chain)

    suppose i raise a mass of m kgs a distance of d metres, what is the increase in its gravitational potential energy? in your idea it is just d and independent of m, which is (sadly) not true.




    you do not want to minimize y, you want to minimize the potential energy *of the chain*, that is the integral of y.arclength.dx

    http://mathworld.wolfram.com/CalculusofVariations.html

    what do you mean by constraint equation? boundary conditions? perhaps the method you've been taught differs from the one I know.
     
  8. Nov 24, 2005 #7
    [tex]\frac{\partial f}{\partial y} - \frac{d}{dx}(\frac{\partial f}{\partial \dot{y}}) + \lambda (\frac{\partial g}{\partial y}) = 0[/tex]

    The constaint equation is [tex]g(y,\dot{y}, x) = 0[/tex]
     
  9. Nov 24, 2005 #8

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Then you are doing something that differs from the method as I know it. It wouldn't hurt you to write more words, eg, try explaining the method you've been taught. I already understood that g from your first post was 'the constraint' equation, but had no idea what that meant. Seaching many sites on calc of variations doesn't reveal what the constraint function is either
     
  10. Nov 24, 2005 #9
    I originally thought that the way to tackle the problem would be to set [tex]\int^b_{a}{\sqrt{1 + \dot{y}^2}dx} = c[/tex] as a constraint equation, and minimise [tex]\int^b_{a}{ydx}[/tex].

    Can you explain geometrically [tex]\int^b_{a}{y \sqrt{1 + \dot{y}^2}dx[/tex]?
     
  11. Nov 24, 2005 #10

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    I thought I already did? length of an infinitesimal part of the chain (which is thus proportional to the mass of the infinitesimal part of the chain) times the potential that that infinitesimal mass contributes added up (integrated) over the whole length. i think that's the third time i've explained it. hopefully it is correct: i've taken this from the only source I could with enough detail in. (google is your friend.)

    It appears you're trying to solve this by a combination of calculus of variation and lagrangian multipliers.
     
  12. Nov 24, 2005 #11
    How is [tex]y[/tex] the gravitational potential?
     
  13. Nov 24, 2005 #12

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    it's the 'height' of the chain, relative to the 'ground' or reference level y=0. an object of mass m and distance d above the 'ground' has gravitational potential energy m*d*g relative to the 'ground' .


    (Don't conflate the potential of the field with the potential energy of something in the field, something I hoped i'd managed not to do), though reading back my last post i think i failed to do so, hence the probable confusion)

    g is just a constant, and the mass is a constant times the arclength, plus the lagrange equation is homogenous (minmizing L(x,y,y') is the same as minmizing kL for any constant k) so we can divide by constants hence we need to minimize y.arclength over the arc.
     
    Last edited: Nov 24, 2005
  14. Nov 24, 2005 #13
    [tex]f = y\sqrt{1 + \dot{y}^2}[/tex]
    [tex]\frac{\partial{f}}{\partial{y}} = \sqrt{1 + \dot{y}^2}[/tex] [tex]\frac{\partial{f}}{\partial{\dot{y}}} = \frac{y\dot{y}}{\sqrt{1 + \dot{y}^2}}[/tex]
    [tex]\frac{d}{dx}(\frac{y\dot{y}}{\sqrt{1 + \dot{y}^2}}) = \frac{\dot{y}^2 + y\ddot{y} - \frac{y\dot{y}^2\ddot{y}}{\sqrt{1 + \dot{y}^2}}}{1 + \dot{y}^2} = \frac{(\dot{y}^2 + y\ddot{y})\sqrt{1 + \dot{y}^2} - y\dot{y}^2\ddot{y}}{\sqrt{1 + \dot{y}^2}^3}[/tex]
    [tex]\sqrt{1 + \dot{y}^2} - \frac{(\dot{y}^2 + y\ddot{y})\sqrt{1 + \dot{y}^2} - y\dot{y}^2\ddot{y}}{\sqrt{1 + \dot{y}^2}^3} = 0[/tex]
     
    Last edited: Nov 24, 2005
  15. Nov 24, 2005 #14

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

  16. Nov 24, 2005 #15
    Thats interesting, why not if m is just a constant ?
     
  17. Nov 24, 2005 #16

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Then you need to do another integral/differential equation if m varies with time, or position or something. See for instance the standard mechanics 101 exercise of the work done to launch a rocket with burning fuel making the mass change.
     
    Last edited: Nov 24, 2005
  18. Nov 25, 2005 #17
    [tex]\frac{\partial f}{\partial x} = \dot{y} \sqrt{1 + \dot{y}^2} + \frac{y \dot{y}}{\sqrt{1 + \dot{y}^2}} = \dot{y} \frac{1 + \dot{y}^2 + y}{\sqrt{1 + \dot{y}^2}}[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?