- #1
adriank
- 534
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I'm thinking about generalizations of Lagrangian mechanics to systems with infinitely many degrees of freedom, but what I've got uses some extremely sketchy math that still appears to give a correct result. I only consider conservative systems that do not explicitly depend on time.
Of course, if our system is described by finitely many coordinates q_i and a Lagrangian L = L(\mathbf{q}, \dot{\mathbf{q}}), then the motion obeys Lagrange's equation
\frac{\partial L}{\partial q_i} - \frac{d}{dt} \frac{\partial L}{\partial \dot q_i} = 0,
or equivalently,
\frac{\partial L}{\partial \mathbf{q}} - \frac{d}{dt} \frac{\partial L}{\partial \dot{\mathbf{q}}} = 0,
where \partial L / \partial \mathbf{q} is a directional derivative
\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},
and similarly for \partial L / \partial \dot{\mathbf{q}}.
Now say our system is described by some continuous parameter s, so we again have L = L(q, \dot q), but this time q is a function of both s and t. I make the bold claim that Lagrange's equation still holds in this sense:
\frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot q} = 0,
where the derivatives of the Lagrangian are some functional directional derivative:
\frac{\partial L}{\partial q}(f) = \left. \frac{d}{d\tau} L(q + \tau f, \dot q) \right\rvert_{\tau = 0},
where f is some function of s.
As an example, suppose we have a continuous spring, with an unextended length of 0, confined to the y axis. Its position is described by a function y(s, t), where s ranges from 0 to \ell; the parameter s perhaps represents length along the coil from one end. Its mass and spring constant per unit length are \mu and \kappa. I write y' to mean \partial y / \partial s. The potential energy and kinetic energy are
V = \int_0^\ell \mu g y(s) + \frac12 \kappa \left( \frac{\partial y(s)}{\partial s} \right)^2 \,ds, \quad<br /> T = \int_0^\ell \frac12 \mu \dot y(s)^2 \,ds,
so the Lagrangian is
L = \int_0^\ell \frac12 \mu \dot y(s)^2 - \mu gy(s) - \frac12 \kappa y'(s)^2 \,ds.
The directional derivatives with respect to y and y' of the Lagrangian satisfy
\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
\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
Thus, Lagrange's equation gives
\int_0^\ell \mu g f(s) + \kappa y'(s) f'(s) + \frac{d}{dt} \mu \dot y(s) f(s) \,ds = 0.
Integration by parts on the second term and simplifying gives
\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.
I claim that somehow the term on the left disappears. Since this equation must be true for any function f, the expression in brackets must be zero; we then get the equation of motion
\ddot y(s) = \frac{\kappa}{\mu} y''(s) - g.
It seems like a physically reasonable equation; solutions oscillate over time, and if we say fix y(\ell) = 0, then in an equilibrium solution y(s) = (\mu g/2\kappa) (s^2 - \ell^2) the tension \kappa y'(s) at any point is equal to the weight \mu gs 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.
Of course, if our system is described by finitely many coordinates q_i and a Lagrangian L = L(\mathbf{q}, \dot{\mathbf{q}}), then the motion obeys Lagrange's equation
\frac{\partial L}{\partial q_i} - \frac{d}{dt} \frac{\partial L}{\partial \dot q_i} = 0,
or equivalently,
\frac{\partial L}{\partial \mathbf{q}} - \frac{d}{dt} \frac{\partial L}{\partial \dot{\mathbf{q}}} = 0,
where \partial L / \partial \mathbf{q} is a directional derivative
\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},
and similarly for \partial L / \partial \dot{\mathbf{q}}.
Now say our system is described by some continuous parameter s, so we again have L = L(q, \dot q), but this time q is a function of both s and t. I make the bold claim that Lagrange's equation still holds in this sense:
\frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot q} = 0,
where the derivatives of the Lagrangian are some functional directional derivative:
\frac{\partial L}{\partial q}(f) = \left. \frac{d}{d\tau} L(q + \tau f, \dot q) \right\rvert_{\tau = 0},
where f is some function of s.
As an example, suppose we have a continuous spring, with an unextended length of 0, confined to the y axis. Its position is described by a function y(s, t), where s ranges from 0 to \ell; the parameter s perhaps represents length along the coil from one end. Its mass and spring constant per unit length are \mu and \kappa. I write y' to mean \partial y / \partial s. The potential energy and kinetic energy are
V = \int_0^\ell \mu g y(s) + \frac12 \kappa \left( \frac{\partial y(s)}{\partial s} \right)^2 \,ds, \quad<br /> T = \int_0^\ell \frac12 \mu \dot y(s)^2 \,ds,
so the Lagrangian is
L = \int_0^\ell \frac12 \mu \dot y(s)^2 - \mu gy(s) - \frac12 \kappa y'(s)^2 \,ds.
The directional derivatives with respect to y and y' of the Lagrangian satisfy
\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
\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
Thus, Lagrange's equation gives
\int_0^\ell \mu g f(s) + \kappa y'(s) f'(s) + \frac{d}{dt} \mu \dot y(s) f(s) \,ds = 0.
Integration by parts on the second term and simplifying gives
\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.
I claim that somehow the term on the left disappears. Since this equation must be true for any function f, the expression in brackets must be zero; we then get the equation of motion
\ddot y(s) = \frac{\kappa}{\mu} y''(s) - g.
It seems like a physically reasonable equation; solutions oscillate over time, and if we say fix y(\ell) = 0, then in an equilibrium solution y(s) = (\mu g/2\kappa) (s^2 - \ell^2) the tension \kappa y'(s) at any point is equal to the weight \mu gs 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.