Multiplying throughout by y ′ = d y / d x, There is a simple first integral of this equation. An Important First Integral of the Euler-Lagrange Equation This general result is called the Euler-Lagrange equation. ∂ f y, y ′ ∂ y − d d x ∂ f y, y ′ ∂ y ′ = 0. One point in the interval, and deduce that Since this is true for any infinitesimal variation, we canĬhoose a variation which is only nonzero near Δ J y = ∫ x 1 x 2 ∂ f y, y ′ ∂ y − d d x ∂ f y, y ′ ∂ y ′ δ y x d x = 0. Then integrate the second term by parts, remembering δ y = 0 at the endpoints, to get To make further progress, we write δ y ′ = δ d y / d x = d / d x δ y, Δ J y = ∫ x 1 x 2 ∂ f y, y ′ ∂ y δ y x + ∂ f y, y ′ ∂ y ′ δ y ′ x d x = 0. Then under any infinitesimal variation δ y x (equal to zero at the fixed endpoints) J y = ∫ x 1 x 2 f y, y ′ d x y ′ = d y / d x. To emphasize the generality of the method, we’ll just write General Method for the Minimization Problem Of course, this also means y ′ x → y ′ x + δ y ′ x where δ y ′ = δ d y / d x = d / d x δ y. That is, we want δ J = 0 to first order, if we make a change y x → y x + δ y x. To minimize the total area, so this is a different problem!) The soap film is not constrained in that way, it can stretch or contract (You might be thinking at this point: isn’t this identical to the catenary equation?Ĭhain has an additional requirement: it has a fixed length. Subject to given values of y at the two ends. J y x = 2 π ∫ x 1 x 2 y d s = 2 π ∫ x 1 x 2 y 1 + y ′ 2 d x The total area is given by integrating, adding Sequence of rings or collars, of radius y , and therefore area 2 π y d s. We need to find the function y x that minimizes the total area ( d s is measured along the curve of the Taking the axis of rotational symmetry to be the x -axis, and the radius y x, Propagates, its path traces out as “world sheet” and the string dynamics isĭetermined by that sheet having minimal area.) This problem is also closely related to string Neglecting gravity, which is a small effect). The minimum possible total area compatible with the fixed boundaries (and This problem is very similar to the catenary: surface tension will pull the soap film to A Soap Film Between Two Horizontal Rings: the Euler-Lagrange Equation Infinitesimal change in the curve of its position, subject to fixed endpoints,Īs a warm up, we’ll consider a simpler-butĬlosely related-problem. So, we’re looking for the configuration where the potentialĮnergy doesn’t change to first order for any Links, that is, or equivalently a continuous rope). The position of every point on the chain (in the limit of infinitely small Variables-it’s a function of a function, it depends on Hanging change isn’t just a function of a variable, or even of a number of The difference here is that the potential energy of the Infrequent in the problems we’re likely to encounter.) Is nonzero- what about x 3 near the origin? But such situations are ( Nitpicking footnote: Actually this assumes the second order term This method of solving the problem is called the calculus of variations: in ordinaryĬalculus, we make an infinitesimal change in a variable, and compute theĬorresponding change in a function, and if it’s zero to leading order in the Or internal friction, it will settle down again in the catenary configuration.įormally speaking, there will be no change in that potentialĮnergy to leading order if we make an infinitesimal change in the curve, y x → y x + δ y x (subject of course to keeping the length the Other words, if we nudge the chain somewhere, and its motion is damped by air Number of variables, and our problem is to minimize it with respect to arbitrary Here the potential energy is a function of a function, equivalent to an infinite Minimize a function with respect to a single variable, or several variables. J y x = 2 π ∫ x 1 x 2 y d s = 2 π ∫ x 1 x 2 y 1 + y ′ 2 d x, y ′ = d y / d x where we have taken the rope density and gravity both equal to unity for mathematical Minimizes the gravitational potential energy The catenary is the curved configuration y = y x of a uniform inextensible rope with two fixedĮndpoints at rest in a constant gravitational field. Technique developed can be successfully applied of a vast array of Here we’ll find howĪnalyzing that leads to a differential equation for the curve, and how the Second-order changes in the gravitational potential energy. Now going to look at a completely different approach: the equilibriumĬonfiguration is an energy minimum, so small deviations from it can only make Length of chain, the tension at the two ends and its weight, balanced. Shape of chain hanging between two places, by finding how the forces on a We’ve seen how Whewell solved the problem of the equilibrium
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