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Proof of Lagrange multipliers method in calculus of variations.

  1. Aug 21, 2012 #1
    I have been reading a little about calculus of variations. I understand the basic method and it's proof. I also understand Lagrange multipliers with regular functions, ie since you are moving orthogonal to one gradient due to the constraint, unless you are also moving orthogonal to the other gradient you have some component of movement along the gradient and it will change the value of the function, so the gradients have to be multiples of each other.

    But when reading about the use of Lagrange multipliers in calculus of variations with an integral constraint, I haven't found an explanation which I could understand. If anyone has a good link or an explanation I would be glad.

    For example, when I read this one:

    https://netfiles.uiuc.edu/liberzon/www/teaching/cvoc/node38.html [Broken]

    Why are the Lagrange multipliers introduced? Both of the integrals have to be zero, when n is such a function that preserves the constraint, but why does the Euler-Lagrange equation of one have to be a multiple of the other? Wouldn't they have to be both zero, or what is going on?
     
    Last edited by a moderator: May 6, 2017
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  3. Aug 21, 2012 #2

    HallsofIvy

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    This depends upon the fact that if g(x,y)= constant is a curve in the xy-plane (or g(x,y,z)= constant) then [itex]\nabla g[/itex] is perpendicular to the curve (or plane).

    Lets take a simple example: Find the point on the curve y= f(x) closest to [itex](x_0, y_0)[/itex]. The function to be minimized is the distance function [itex]D= \sqrt{(x- x_0)^2+ (y- y_0)^2}[/itex]. (Yes, it would be simpler to minimize "distance squared" and give the same answer but I want to be as straight forward as possible.)

    It's not difficult to see that the "gradient" vector of the distance function just points directly from (x, y) to [itex](x_0, y_0)[/itex]. With no constraint, we could just move in the direction of the gradient vector, eventually getting to [itex](x_0, y_0)[/itex], the point that obviously makes the distance minimum (in fact, 0).

    But if we are restricted to stay on curve y= f(x), we cannot move directly toward the point. What we can do is move along the curve- and an obvious strategy is to look at the projection of the gradient onto the curve. If the gradient is "to the right", that is, if its projection onto the curve points to the right, move to the right. If the gradient is "to the left", move to the left. We can keep doing that, and getting closer to [itex](x_0, y_0)[/itex], as long as the gradient has such a projection- as long as the gradient is not perpendicular to the constraint curve. And we can write y= f(x) as g(x,y)= y- f(x)= 0: a "level curve". Finally, we know that the gradient of such a function is perpendicular to any level curve: that is, [itex]\nabla g[/itex] is perpendicular to g(x,y)= 0 at every point on the curve.

    Putting those together, [itex]\nabla D[/itex] must be perpendicular to g(x,y)= y- f(x)= 0 and [itex]\nabla g[/itex] is always perpendicular to that curve. That means the two vectors [itex]\nabla D[/itex] and [itex]\nabla g[/itex] must be parallel: [itex]\nabla D= \lambda\nabla g[/itex] where [itex]\lambda[/itex] is some real number.

    For maximizing functions, h, other than just "distance", remember that [itex]\nabla h[/itex] always points in the direction of fastest increase of h so the same applies to that situation. If [itex]\nabla h[/itex] were not perpendicular to the curve g(x,y)= y- f(x)= 0, we could move "left" or "right" on the curve, depending on the direction of the projection of [itex]\nabla h[/itex] on the curve. We cannot do that when there is no projection- when [itex]\nabla h[/itex] is perpendicular to the curve which is the same as saying it is parallel to [itex]\nabla g[/itex]: [itex]\nabla h= \lambda\nabla f[/itex].
     
  4. Aug 21, 2012 #3
    That makes perfect sense, but how does the idea work when for example I have a constraint that a line segment has to be of a particular length and then I have to find a line segment of that given length that maximizes a particular integral? Then I don't have a line of the constraint, the constraint could be any line, as long as it is of given length. How does the idea work with functionals in calculus of variations?
     
  5. Aug 22, 2012 #4
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