Introduction to Calculus of Variations

15. Introduction to Calculus of Variations#

Calculus of variation deals with variations - i.e. “small changes” - of functions and functionals.

The meaning of the term functional may vary on the subfield of interest. In the field of calculus of variation, a functional can be defined as a function of function, i.e. a function whose argument is another function.

15.1. Lagrange equations#

Given the functional $S$ , with arguments a function $q (t)$ and the independent variable $t$ ,

S [q (t), t] = \int_{t = t_{0}}^{t_{1}} L (\dot{q} (t), q (t), t) d t

its variation w.r.t. the function $q (t)$ reads

δ S [q (t), t] = lim_{ε \to 0} \frac{1}{ε} (S [q (t) + ε w (t), t] - S [q (t), t])

where the function $w (t)$ is arbitrary, among those satisfying the constraint of the problems: as an example here, if the function $q (t)$ has prescribed values $q^{*}$ for some values of the independent variable, $t^{*}$ , the variation $w (t)$ of the function $q (t)$ is zero there, $w (t^{*})$ so that the variated function $q (t) + ε w (t)$ satisfies the constraint as well, i.e. $q (t^{*}) + ε w (t^{*}) = q^{*}$ .

Variation involves only small changes of function arguments, since these ones are the elements that can be effectively changed, while the independent variable is not.

Direct computation of the variation gives

\begin{array}{r} \begin{aligned} δ S [q (t), t] & = lim_{ε \to 0} \frac{1}{ε} (S [q (t) + ε w (t), t] - S [q (t), t]) = \\ = lim_{ε \to 0} \frac{1}{ε} (\int_{t = t_{0}}^{t_{1}} L (\dot{q} (t) + ε w (t), q (t) + ε w (t), t) - \int_{t = t_{0}}^{t_{1}} L (\dot{q} (t), q (t), t) d t) = \\ = lim_{ε \to 0} \frac{1}{ε} \int_{t = t_{0}}^{t_{1}} (L (\dot{q} (t) + ε w (t), q (t) + ε w (t), t) - L (\dot{q} (t), q (t), t)) d t = \\ = lim_{ε \to 0} \frac{1}{ε} \int_{t = t_{0}}^{t_{1}} {L (\dot{q} (t), q (t), t) + ε [\frac{\partial L}{\partial \dot{q}} \dot{w} (t) + \frac{\partial L}{\partial q} w (t)] + o (ε) - L (\dot{q} (t), q (t), t)} d t = \\ = lim_{ε \to 0} \frac{1}{ε} \int_{t = t_{0}}^{t_{1}} {ε [\frac{\partial L}{\partial \dot{q}} \dot{w} (t) + \frac{\partial L}{\partial q} w (t)] + o (ε)} d t = \\ = \int_{t = t_{0}}^{t_{1}} {\frac{\partial L}{\partial \dot{q}} \dot{w} (t) + \frac{\partial L}{\partial q} w (t)} d t = \\ = {[w (t) \frac{\partial L}{\partial \dot{q}}] |}_{t = t_{0}}^{t_{1}} + \int_{t = t_{0}}^{t_{1}} {- \frac{d}{d t} (\frac{\partial L}{\partial \dot{q}}) + \frac{\partial L}{\partial q}} w (t) d t . \end{aligned} \end{array}

The solution depends on the boundary conditions at the extreme points $t_{0}$ , $t_{1}$ . If the value of the function $q (t)$ is prescribed in $t_{0}$ and $t_{1}$ , $q (t_{0}) = q_{0}$ , $q (t_{1}) = q_{1}$ , then its variation is zero, $w (t_{0}) = w (t_{1}) = 0$ , for the reason that has been discussed above. The variation of the functional with prescribed boundary values of the argument function thus reads

δ S [q (t), t] = \int_{t = t_{0}}^{t_{1}} {- \frac{d}{d t} (\frac{\partial L}{\partial \dot{q}}) + \frac{\partial L}{\partial q}} δ q (t) d t,

having called $w (t) =: δ q (t)$ to stress that is the variation of function $q (t)$ . This notation - it’s just notation, it has no special properties - could be useful if the functional depends on several arguments.

Stationary conditions, $δ S = 0$ . Stationary condition of the functional $S$ implies that $δ S = 0$ for all the possible variations of the argument function, $\forall δ q (t)$ . This condition implies that the integrand is identically zero, i.e. Lagrange equations,

\frac{d}{d t} (\frac{\partial L}{\partial \dot{q}}) - \frac{\partial L}{\partial q} = 0,

15.1.1. Euler-Beltrami equation#

If the Lagrangian function $L$ is not an explicit function of the independent variable, $L (q^{'} (x), q (x))$ , Euler-Berltrami equation follows from the derivative of the Lagrangian,

\begin{array}{r} \begin{aligned} \frac{d L}{d x} & = \frac{\partial L}{\partial q^{'}} q^{″} (x) + \frac{\partial L}{\partial q} q^{'} = \\ = \frac{\partial L}{\partial q^{'}} q^{″} + \frac{d}{d x} (\frac{\partial L}{\partial q^{'}}) q^{'} = \\ = \frac{d}{d x} (\frac{\partial L}{\partial q^{'}} q^{'}), \end{aligned} \end{array}

and thus

\frac{d}{d x} [L - q^{'} \frac{\partial L}{\partial q^{'}}] = 0 \to L - q^{'} \frac{\partial L}{\partial q^{'}} = C const.

Note 1. While Lagrange equations are a set of $N$ equations if the functional depends on $N$ argument functions $q_{k} (t)$ , $k = 1 : N$ , Euler-Beltrami equation is an equation only. Indeed for multiple argument functions

\begin{array}{r} \begin{aligned} \frac{d L}{d x} & = \frac{\partial L}{\partial q_{k}^{'}} q_{k}^{″} (x) + \frac{\partial L}{\partial q_{k}} q_{k}^{'} = \\ = \frac{\partial L}{\partial q_{k}^{'}} q_{k}^{″} + \frac{d}{d x} (\frac{\partial L}{\partial q_{k}^{'}}) q_{k}^{'} = \frac{d}{d x} (\frac{\partial L}{\partial q_{k}^{'}} q_{k}^{'}), \end{aligned} \end{array}

where Einstein’s summation notation of repeated index is used. Euler-Beltrami thus reads

L (q_{l}^{'} (x), q_{l} (x)) - q_{k}^{'} (x) \frac{\partial L}{\partial q_{k}^{'}} (q_{l}^{'} (x), q_{l} (x)) = C .

Note 2. If the Lagrangian function is an explicit function of the independent variable $x$ , $L (q^{'} (x), q (x), x)$ , it’s not hard to realize that the derivative of the Lagrangian function, along with the use of the Lagrange equation, gives

\frac{d}{d x} [L - q^{'} \frac{\partial L}{\partial q^{'}}] = \frac{\partial L}{\partial x} .

Example 15.1 (Euler-Beltrami with $L (q^{'} (x), q (x), x)$ , Hamiltonian, energy and E.Noether)

Euler-Beltrami equation shows that if $L (q^{'} (x), q (x))$ , thus $L - q^{'} \partial_{q^{'}} L$ is constant, (or an integral of motion in dynamics). In analytical mechanics (Lagrange mechanics, Hamiltonian mechanics), Lagrangian and Hamiltonian functions of a system read

\begin{array}{r} \begin{aligned} L ({\dot{q}}_{k} (t), q_{k} (t), t) & = T (\dot{q}, q, t) + U (q, t) \\ H (p, q, t) & := p_{k} {\dot{q}}_{k} - L = {\dot{q}}_{k} \frac{\partial L}{\partial {\dot{q}}^{k}} - L, \end{aligned} \end{array}

having used the common definition of the generalized momenta $p_{k} := \frac{\partial L}{\partial {\dot{q}}^{k}}$ . It should be immediate to realize that the Hamiltonian is just the quantity appearing in Euler-Beltrami equation (or in its “modified version” if $\partial_{t} L \neq 0$ ), and thus

\frac{d H}{d t} = \frac{\partial L}{\partial t} .

In mechanics, if $\partial_{t} L = 0$ , the Hamiltonian is a constant of motion. In this case, it can be prove that the Hamiltonian is equal to the eneergy of the system.

Classical examples.

Example 15.2 (Brachistochrone)

Find the trajectory…

Elementary length: $d s = v d t$
Energy: $E (y) = \frac{1}{2} m v^{2} - m g y + C$ . Setting $E = 0$ at starting point, from rest, at $y_{0} = 0$ , it implies $C = 0$ ; thus $v = \sqrt{2 g y}$
$x (s), y (s)$ ,
$d s = \sqrt{d x^{2} + d y^{2}} = \sqrt{1 + y^{' 2} (x)} d x$

T = \int_{t_{0}}^{t_{1}} d t = \int_{s_{0}}^{s_{1}} \frac{d s}{v} = \int_{x_{0}}^{x_{1}} \frac{\sqrt{1 + y^{' 2} (x)}}{\sqrt{2 g y (x)}} d x

The Lagrangian doesn’t explicitly depend on $x$ , thus Euler-Beltrami equation can be used. Partial derivative of the Lagrangian function w.r.t. $q^{'}$ reads

\frac{\partial L}{\partial y^{'}} = . . . = \frac{1}{\sqrt{2 g y} \sqrt{1 + y^{' 2}}} y^{'},

and thus Euler-Beltrami equation reads

C = L - q^{'} \frac{\partial L}{\partial y^{'}} = \frac{1 + y^{' 2} - y^{' 2}}{\sqrt{2 g y} \sqrt{1 + y^{' 2}}} = \frac{1}{\sqrt{2 g y} \sqrt{1 + y^{' 2}}}

Squaring $2 g C^{2} = \frac{1}{y (1 + y^{' 2})}$ , it’s possible to write

y (x) = \frac{1}{2 g C^{2} (1 + y^{' 2} (x))},

Making the substitution $y (x)^{'} = \dots$

Example 15.3 (Catenary)

Example 15.4 (Isoperimetric problem)