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Of course, if our system is described by finitely many coordinates [itex]q_i[/itex] and a Lagrangian [itex]L = L(\mathbf{q}, \dot{\mathbf{q}})[/itex], then the motion obeys Lagrange's equation

[tex]\frac{\partial L}{\partial q_i} - \frac{d}{dt} \frac{\partial L}{\partial \dot q_i} = 0,[/tex]

or equivalently,

[tex]\frac{\partial L}{\partial \mathbf{q}} - \frac{d}{dt} \frac{\partial L}{\partial \dot{\mathbf{q}}} = 0,[/tex]

where [itex]\partial L / \partial \mathbf{q}[/itex] is a directional derivative

[tex]\frac{\partial L}{\partial \mathbf{q}}(\mathbf{v}) = \left. \frac{d}{d\tau} L(\mathbf{q} + \tau \vect{v}, \dot{\mathbf{q}}) \right\rvert_{\tau = 0},[/tex]

and similarly for [itex]\partial L / \partial \dot{\mathbf{q}}[/itex].

Now say our system is described by some continuous parameter [itex]s[/itex], so we again have [itex]L = L(q, \dot q)[/itex], but this time [itex]q[/itex] is a function of both [itex]s[/itex] and [itex]t[/itex]. I make the bold claim that Lagrange's equation still holds in this sense:

[tex]\frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot q} = 0,[/tex]

where the derivatives of the Lagrangian are some functional directional derivative:

[tex]\frac{\partial L}{\partial q}(f) = \left. \frac{d}{d\tau} L(q + \tau f, \dot q) \right\rvert_{\tau = 0},[/tex]

where [itex]f[/itex] is some function of [itex]s[/itex].

As an example, suppose we have a continuous spring, with an unextended length of 0, confined to the [itex]y[/itex] axis. Its position is described by a function [itex]y(s, t)[/itex], where [itex]s[/itex] ranges from [itex]0[/itex] to [itex]\ell[/itex]; the parameter [itex]s[/itex] perhaps represents length along the coil from one end. Its mass and spring constant per unit length are [itex]\mu[/itex] and [itex]\kappa[/itex]. I write [itex]y'[/itex] to mean [itex]\partial y / \partial s[/itex]. The potential energy and kinetic energy are

[tex]V = \int_0^\ell \mu g y(s) + \frac12 \kappa \left( \frac{\partial y(s)}{\partial s} \right)^2 \,ds, \quad

T = \int_0^\ell \frac12 \mu \dot y(s)^2 \,ds,[/tex]

so the Lagrangian is

[tex]L = \int_0^\ell \frac12 \mu \dot y(s)^2 - \mu gy(s) - \frac12 \kappa y'(s)^2 \,ds.[/tex]

The directional derivatives with respect to [itex]y[/itex] and [itex]y'[/itex] of the Lagrangian satisfy

[tex]\frac{\partial L}{\partial y}(f) = \left. \frac{d}{d\tau} \int_0^\ell \frac12 \mu \dot y(s)^2 - \mu g (y(s) + \tau f(s)) - \frac12 \kappa (y'(s) + \tau f'(s))^2 \,ds \right\rvert_{\tau = 0} = -\int_0^\ell \mu g f(s) + \kappa y'(s) f'(s) \,ds[/tex]

[tex]\frac{\partial L}{\partial \dot y}(f) = \left. \frac{d}{d\tau} \int_0^\ell \frac12 \mu (\dot y(s) + \tau f(s))^2 - \mu gy(s) - \frac12 \kappa y'(s)^2 \right\rvert_{\tau = 0} = \int_0^\ell \mu \dot y(s) f(s) \,ds[/tex]

Thus, Lagrange's equation gives

[tex]\int_0^\ell \mu g f(s) + \kappa y'(s) f'(s) + \frac{d}{dt} \mu \dot y(s) f(s) \,ds = 0.[/tex]

Integration by parts on the second term and simplifying gives

[tex]\kappa (y'(\ell) f(\ell) - y'(0) f(0)) - \int_0^\ell \left[ \mu g - \kappa y''(s) + \mu \ddot y(s) \right] f(s) \,ds = 0.[/tex]

I claim that somehow the term on the left disappears. Since this equation must be true for any function [itex]f[/itex], the expression in brackets must be zero; we then get the equation of motion

[tex]\ddot y(s) = \frac{\kappa}{\mu} y''(s) - g.[/tex]

It seems like a physically reasonable equation; solutions oscillate over time, and if we say fix [itex]y(\ell) = 0[/itex], then in an equilibrium solution [itex]y(s) = (\mu g/2\kappa) (s^2 - \ell^2)[/itex] the tension [itex]\kappa y'(s)[/itex] at any point is equal to the weight [itex]\mu gs[/itex] of the spring below that point.

Has anyone seen anything that looks somewhat like this? I looked around and couldn't really find anything that looked like this.