Euler–Lagrange equation

In calculus of variations, the Euler–Lagrange equation, or Lagrange's equation, is a differential equation whose solutions are the functions for which a given functional is stationary. It was developed by Swiss mathematician Leonhard Euler and Italian mathematician Joseph Louis Lagrange in the 1750s.

Because a differentiable functional is stationary at its local maxima and minima, the Euler–Lagrange equation is useful for solving optimization problems in which, given some functional, one seeks the function minimizing (or maximizing) it. This is analogous to Fermat's theorem in calculus, stating that where a differentiable function attains its local extrema, its derivative is zero.

In Lagrangian mechanics, because of Hamilton's principle of stationary action, the evolution of a physical system is described by the solutions to the Euler–Lagrange equation for the action of the system. In classical mechanics, it is equivalent to Newton's laws of motion, but it has the advantage that it takes the same form in any system of generalized coordinates, and it is better suited to generalizations (see, for example, the "Field theory" section below).

1 History
2 Statement
3 Examples
- 3.1 Classical mechanics
  - 3.1.1 Basic method
  - 3.1.2 Particle in a conservative force field
- 3.2 Field theory
4 Variations for several functions, several variables, and higher derivatives
5 Notes
6 References
7 See also

History

The Euler–Lagrange equation was developed in the 1750s by Euler and Lagrange in connection with their studies of the tautochrone problem. This is the problem of determining a curve on which a weighted particle will fall to a fixed point in a fixed amount of time, independent of the starting point.

Lagrange solved this problem in 1755 and sent the solution to Euler. The two further developed Lagrange's method and applied it to mechanics, which led to the formulation of Lagrangian mechanics. Their correspondence ultimately led to the calculus of variations, a term coined by Euler himself in 1766.^[1]

Statement

The Euler–Lagrange equation is an equation satisfied by a function, q, of a real argument, t, which is a stationary point of the functional

$\displaystyle S(q) = \int_a^b L(t,q(t),q'(t))\, \mathrm{d}t$

where:

q is the function to be found:

$\begin{align} q \colon [a, b] \subset \mathbb{R} & \to X \\ t & \mapsto x = q(t) \end{align}$

such that q is differentiable, q(a) = x_a, and q(b) = x_b;

q′ is the derivative of q:

$\begin{align} q' \colon [a, b] & \to T_{q(t)}X \\ t & \mapsto v = q'(t) \end{align}$

TX being the tangent bundle of X (the space of possible values of derivatives of functions with values in X);

L is a real-valued function with continuous first partial derivatives:

$\begin{align} L \colon [a, b] \times X \times TX & \to \mathbb{R} \\ (t, x, v) & \mapsto L(t, x, v). \end{align}$

The Euler–Lagrange equation, then, is given by

$L_x(t,q(t),q'(t))-\frac{\mathrm{d}}{\mathrm{d}t}L_v(t,q(t),q'(t)) = 0.$

where L_x and L_v denote the partial derivatives of L with respect to the second and third arguments, respectively.

If the dimension of the space X is greater than 1, this is a system of differential equations, one for each component:

$\frac{\partial L(t,q(t),q'(t))}{\partial x_i}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L(t,q(t),q'(t))}{\partial v_i} = 0 \quad \text{for } i = 1, \dots, n.$

Derivation of one-dimensional Euler-Lagrange equation

The derivation of the one-dimensional Euler–Lagrange equation is one of the classic proofs in mathematics. It relies on the fundamental lemma of calculus of variations.

We wish to find a function $f$ which satisfies the boundary conditions $f(a) = A$ , $f(b) = B$ , and which extremizes the cost functional

$J = \int_a^b F(x,f(x),f'(x))\, dx. \,\!$

We assume that F has continuous first partial derivatives. A weaker assumption can be used, but the proof becomes more difficult.

If f extremizes the cost functional subject to the boundary conditions, then any slight perturbation of f that preserves the boundary values must either increase J (if f is a minimizer) or decrease J (if f is a maximizer).

Let g_ε(x) = f(x) + εη(x) be such a perturbation of f, where η(x) is a differentiable function satisfying η(a) = η(b) = 0. Then define

$J_\varepsilon(\varepsilon) = \int_a^b F(x,g_\varepsilon(x), g_\varepsilon'(x) )\, dx. \,\!$

We now wish to calculate the total derivative of J with respect to ε or the first variation of J.

$\frac{\mathrm{d} J_\varepsilon}{\mathrm{d} \varepsilon} = \frac{\mathrm d}{\mathrm d\varepsilon}\int_a^b F(x,g_\varepsilon(x), g_\varepsilon'(x) )\, dx = \int_a^b \frac{\mathrm{d}F}{\mathrm{d}\epsilon}(x,g_\varepsilon(x), g_\varepsilon'(x) )\, dx$

It follows from the total derivative that

$\frac{\mathrm d}{\mathrm d\varepsilon} = \frac{\partial}{\partial \varepsilon} + \frac{\mathrm d g_i}{\mathrm d\varepsilon}\frac{\partial}{\partial g_i}$

$\frac{\mathrm d F}{\mathrm d\varepsilon}=\frac{\mathrm d x}{\mathrm d\varepsilon}\frac{\partial F}{\partial x}+\frac{\mathrm d g_\varepsilon}{\mathrm d\varepsilon}\frac{\partial F}{\partial g_\epsilon}+\frac{\mathrm d g'_\varepsilon}{\mathrm d\varepsilon}\frac{\partial F}{\partial g'_\epsilon}$

$\frac{\mathrm{d}F}{\mathrm{d}\varepsilon} = \eta(x) \frac{\partial F}{\partial g_\varepsilon} + \eta'(x) \frac{\partial F}{\partial g_\varepsilon'}.$

$\frac{\mathrm{d} J_\varepsilon}{\mathrm{d} \varepsilon} = \int_a^b \left[\eta(x) \frac{\partial F}{\partial g_\varepsilon} + \eta'(x) \frac{\partial F}{\partial g_\varepsilon'} \, \right]\,dx.$

When ε = 0 we have g_ε = f and since f is an extreme value it follows that $\frac{\mathrm d J_\varepsilon}{\mathrm d\varepsilon}(0) = 0$ , i.e.

$\frac{\mathrm d J_\varepsilon}{\mathrm d\varepsilon}(0) = \int_a^b \left[ \eta(x) \frac{\partial F}{\partial f} + \eta'(x) \frac{\partial F}{\partial f'} \,\right]\,dx = 0.$

The next crucial step is to use integration by parts on the second term, yielding

$0 = \int_a^b \left[ \frac{\partial F}{\partial f} - \frac{d}{dx} \frac{\partial F}{\partial f'} \right] \eta(x)\,dx + \left[ \eta(x) \frac{\partial F}{\partial f'} \right]_a^b.$

Using the boundary conditions on η, we get that

$0 = \int_a^b \left[ \frac{\partial F}{\partial f} - \frac{d}{dx} \frac{\partial F}{\partial f'} \right] \eta(x)\,dx. \,\!$

Applying the fundamental lemma of calculus of variations now yields the Euler–Lagrange equation

$0 = \frac{\partial F}{\partial f} - \frac{d}{dx} \frac{\partial F}{\partial f'}.$

Alternate derivation of one-dimensional Euler-Lagrange equation

Given a functional

$J = \int^b_aF(t, y(t), y'(t))dt$

on $C^1([a, b])$ with the boundary conditions $y(a) = A$ and $y(b) = B$ , we proceed by approximating the extremal curve by a polygonal line with $n$ segments and passing to the limit as the number of segments grows arbitrarily large.

Divide the interval $[a, b]$ into $n + 1$ equal segments with endpoints $t_0 = a, t_1, t_2, \ldots, t_n, t_{n + 1} = b$ and let $\Delta t = t_k - t_{k - 1}$ . Rather than a smooth function $y(t)$ we consider the polygonal line with vertices $(t_0, y_0),\ldots,(t_{n + 1}, y_{n + 1})$ , where $y_0 = A$ and $y_{n + 1} = B$ . Accordingly, our functional becomes a real function of $n$ variables given by

$J(y_1, \ldots, y_n) = \sum^n_{k = 0}F\left(t_k, y_k, \frac{y_{k + 1} - y_k}{\Delta t}\right)\Delta t.$

Extremals of this new functional defined on the discrete points $t_0,\ldots,t_{n + 1}$ correspond to points where

$\frac{\partial J(y_1,\ldots,y_n)}{\partial y_m} = 0.$

Evaluating this partial derivative gives that

$\frac{\partial J}{\partial y_m} = F_y\left(t_m, y_m, \frac{y_{m + 1} - y_m}{\Delta t}\right)\Delta t + F_{y'}\left(t_{m - 1}, y_{m - 1}, \frac{y_m - y_{m - 1}}{\Delta t}\right) - F_{y'}\left(t_m, y_m, \frac{y_{m + 1} - y_m}{\Delta t}\right).$

Dividing the above equation by $\Delta t$ gives

$\frac{\partial J}{\partial y_m \Delta t} = F_y\left(t_m, y_m, \frac{y_{m + 1} - y_m}{\Delta t}\right) - \frac{1}{\Delta t}\left[F_{y'}\left(t_m, y_m, \frac{y_{m + 1} - y_m}{\Delta t}\right) - F_{y'}\left(t_{m - 1}, y_{m - 1}, \frac{y_m - y_{m - 1}}{\Delta t}\right)\right],$

and taking the limit as $\Delta t \to 0$ of the right-hand side of this expression yields

$\frac{\delta J}{\delta y} = F_y - \frac{d}{dt}F_{y'}.$

The term $\frac{\delta J}{\delta y}$ denotes the variational derivative of the functional $J$ , and a necessary condition for a differentiable functional to have an extremum on some function is that its variational derivative at that function vanishes.

Examples

A standard example is finding the real-valued function on the interval [a, b], such that f(a) = c and f(b) = d, the length of whose graph is as short as possible. The length of the graph of f is:

$\ell (f) = \int_{a}^{b} \sqrt{1+f'(x)^2}\,\mathrm{d}x,$

the integrand function being L(x, y, y′) = √1 + y′ ² evaluated at (x, y, y′) = (x, f(x), f′(x)).

The partial derivatives of L are:

$\frac{\partial L(x, y, y')}{\partial y'} = \frac{y'}{\sqrt{1 + y'^2}} \quad \text{and} \quad \frac{\partial L(x, y, y')}{\partial y} = 0.$

By substituting these into the Euler–Lagrange equation, we obtain

$\frac{\mathrm{d}}{\mathrm{d}x} \frac{f'(x)}{\sqrt{1 + f'(x)^2}} = 0 \Rightarrow \frac{f'(x)}{\sqrt{1 + f'(x)^2}} = \text{constant} \Rightarrow f'(x) = \text{constant:}$

that is, the function must have constant first derivative, and thus its graph is a straight line.

Classical mechanics

Basic method

To find the equations of motions for a given system, one only has to follow these steps:

From the kinetic energy $T$ , and the potential energy $V$ , compute the Lagrangian $L = T - V$ .
Compute $\frac{\partial L}{\partial q}$ .
Compute $\frac{\partial L}{\partial \dot{q}}$ and from it, $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$ . It is important that $\dot{q}$ be treated as a complete variable in its own right, and not as a derivative.
Equate $\frac{\partial L}{\partial q} = \frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$ . This is the Euler–Lagrange equation.
Solve the differential equation obtained in the preceding step. At this point, $\dot{q}$ is treated "normally". Note that the above might be a system of equations and not simply one equation.

Particle in a conservative force field

The motion of a single particle in a conservative force field (for example, the gravitational force) can be determined by requiring the action to be stationary, by Hamilton's principle. The action for this system is

$S = \int_{t_0}^{t_1} L(t, \mathbf{x}(t), \mathbf{\dot{x}}(t))\,\mathrm{d}t$

where x(t) is the position of the particle at time t. The dot above is Newton's notation for the time derivative: thus ẋ(t) is the particle velocity, v(t). In the equation above, L is the Lagrangian (the kinetic energy minus the potential energy):

$L(t, \mathbf{x}, \mathbf{v}) = \frac{1}{2}m \sum_{i=1} ^{3} v_i^2 - U(\mathbf{x}),$

where:

m is the mass of the particle (assumed to be constant in classical physics);
v_i is the i-th component of the vector v in a Cartesian coordinate system (the same notation will be used for other vectors);
U is the potential of the conservative force.

In this case, the Lagrangian does not vary with its first argument t. (By Noether's theorem, such symmetries of the system correspond to conservation laws. In particular, the invariance of the Lagrangian with respect to time implies the conservation of energy.)

By partial differentiation of the above Lagrangian, we find:

$\frac{\partial L(t,\mathbf{x},\mathbf{v})}{\partial x_i} = -\frac{\partial U(\mathbf{x})}{\partial x_i} = F_i (\mathbf{x})\quad \text{and} \quad \frac{\partial L(t,\mathbf{x},\mathbf{v})}{\partial v_i} = m v_i = p_i,$

where the force is F = −∇U (the negative gradient of the potential, by definition of conservative force), and p is the momentum. By substituting these into the Euler–Lagrange equation, we obtain a system of second-order differential equations for the coordinates on the particle's trajectory,

$F_i(\mathbf{x}(t)) = \frac{\mathrm d}{\mathrm{d}t} m \dot{x}_i(t) = m \ddot{x}_i(t),$

which can be solved on the interval [t₀, t₁], given the boundary values x_i(t₀) and x_i(t₁). In vector notation, this system reads

$\mathbf{F}(\mathbf{x}(t)) = m\mathbf{\ddot x}(t)$

or, using the momentum,

$\mathbf{F} = \frac {\mathrm{d}\mathbf{p}} {\mathrm{d}t}$

which is Newton's second law.

Field theory

Field theories, both classical field theory and quantum field theory, deal with continuous coordinates, and like classical mechanics, has its own Euler–Lagrange equation of motion for a field,

$\partial_\mu \left( \frac{\partial \mathcal{L}}{\partial ( \partial_\mu \psi )} \right) - \frac{\partial \mathcal{L}}{\partial \psi} = 0. \,$

where

$\psi \,$ is the field, and
$\partial\,$ is a vector differential operator:

$\partial_\mu = \left(\frac{1}{c} \frac{\partial}{\partial t}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right). \,$

Note: Not all classical fields are assumed commuting/bosonic variables, (like the Dirac field, the Weyl field, the Rarita-Schwinger field) are fermionic and so, when trying to get the field equations from the Lagrangian density, one must choose whether to use the right or the left derivative of the Lagrangian density (which is a boson) with respect to the fields and their first space-time derivatives which are fermionic/anticommuting objects.

There are several examples of applying the Euler–Lagrange equation to various Lagrangians:

Dirac equation;
Proca equation;
electromagnetic tensor;
Korteweg–de Vries equation;
quantum electrodynamics.

Variations for several functions, several variables, and higher derivatives

Single function of single variable with higher derivatives

The stationary values of the functional

$I[f] = \int_{x_0}^{x_1} \mathcal{L}(x, f, f', f'', \dots, f^{(n)})~\mathrm{d}x ~;~~ f'�:= \cfrac{\mathrm{d}f}{\mathrm{d}x}, ~f''�:= \cfrac{\mathrm{d}^2f}{\mathrm{d}x^2}, ~ f^{(n)}�:= \cfrac{\mathrm{d}^nf}{\mathrm{d}x^n}$

can be obtained from the Euler-Lagrange equation^[2]

$\cfrac{\partial \mathcal{L}}{\partial f} - \cfrac{\mathrm{d}}{\mathrm{d} x}\left(\cfrac{\partial \mathcal{L}}{\partial f'}\right) + \cfrac{\mathrm{d}^2}{\mathrm{d} x^2}\left(\cfrac{\partial \mathcal{L}}{\partial f''}\right) - \dots + (-1)^n \cfrac{\mathrm{d}^n}{\mathrm{d} x^n}\left(\cfrac{\partial \mathcal{L}}{\partial f^{(n)}}\right) = 0$

Several functions of one variable

If the problem involves finding several functions ( $f_1, f_2, \dots, f_n$ ) of a single independent variable ( $x$ ) that define an extremum of the functional

$I[f_1,f_2, \dots, f_n] = \int_{x_0}^{x_1} \mathcal{L}(x, f_1, f_2, \dots, f_n, f_1', f_2', \dots, f_n')~\mathrm{d}x ~;~~ f_i'�:= \cfrac{\mathrm{d}f_i}{\mathrm{d}x}$

then the corresponding Euler-Lagrange equations are^[2]

$\begin{align} \cfrac{\partial \mathcal{L}}{\partial f_1} - \cfrac{\partial}{\partial x}\left(\cfrac{\partial \mathcal{L}}{\partial f_1'}\right) - \sum_{i=1}^n \cfrac{\partial}{\partial f_i}\left(\cfrac{\partial \mathcal{L}}{\partial f_1'}\right)~\cfrac{\mathrm{d}f_i}{\mathrm{d}x} - \sum_{i=1}^n \cfrac{\partial}{\partial f_i'}\left(\cfrac{\partial \mathcal{L}}{\partial f_1'}\right)~\cfrac{\mathrm{d}^2f_i}{\mathrm{d}x^2} & = 0 \\ \cfrac{\partial \mathcal{L}}{\partial f_2} - \cfrac{\partial}{\partial x}\left(\cfrac{\partial \mathcal{L}}{\partial f_2'}\right) - \sum_{i=1}^n \cfrac{\partial}{\partial f_i}\left(\cfrac{\partial \mathcal{L}}{\partial f_2'}\right)~\cfrac{\mathrm{d}f_i}{\mathrm{d}x} - \sum_{i=1}^n \cfrac{\partial}{\partial f_i'}\left(\cfrac{\partial \mathcal{L}}{\partial f_2'}\right)~\cfrac{\mathrm{d}^2f_i}{\mathrm{d}x^2} & = 0 \\ \dots \dots \dots \dots \dots & \\ \cfrac{\partial \mathcal{L}}{\partial f_n} - \cfrac{\partial}{\partial x}\left(\cfrac{\partial \mathcal{L}}{\partial f_n'}\right) - \sum_{i=1}^n \cfrac{\partial}{\partial f_i}\left(\cfrac{\partial \mathcal{L}}{\partial f_n'}\right)~\cfrac{\mathrm{d}f_i}{\mathrm{d}x} - \sum_{i=1}^n \cfrac{\partial}{\partial f_i'}\left(\cfrac{\partial \mathcal{L}}{\partial f_n'}\right)~\cfrac{\mathrm{d}^2f_i}{\mathrm{d}x^2} & = 0 \end{align}$

Single function of several variables

A multi-dimensional generalization comes from considering a function on n variables. If Ω is some surface, then

$I[f] = \int_{\Omega} \mathcal{L}(x_1, \dots , x_n, f, f_{x_1}, \dots , f_{x_n})\, \mathrm{d}\mathbf{x}\,\! ~;~~ f_{x_i}�:= \cfrac{\partial f}{\partial x_i}$

is extremized only if f satisfies the partial differential equation

$\frac{\partial \mathcal{L}}{\partial f} - \sum_{i=1}^{n} \frac{\partial}{\partial x_i} \frac{\partial \mathcal{L}}{\partial f_{x_i}} = 0. \,\!$

When n = 2 and $\mathcal{L}$ is the energy functional, this leads to the soap-film minimal surface problem.

Several functions of several variables

If there are several unknown functions to be determined and several variables such that

$I[f_1,f_2,\dots,f_m] = \int_{\Omega} \mathcal{L}(x_1, \dots , x_n, f_1, \dots, f_m, f_{1,1}, \dots , f_{1,n}, \dots, f_{m,1}, \dots, f_{m,n}) \, \mathrm{d}\mathbf{x}\,\! ~;~~ f_{j,i}�:= \cfrac{\partial f_j}{\partial x_i}$

the system of Euler-Lagrange equations is^[2]

$\begin{align} \frac{\partial \mathcal{L}}{\partial f_1} - \sum_{i=1}^{n} \frac{\partial}{\partial x_i} \frac{\partial \mathcal{L}}{\partial f_{1,i}} & = 0 \\ \frac{\partial \mathcal{L}}{\partial f_2} - \sum_{i=1}^{n} \frac{\partial}{\partial x_i} \frac{\partial \mathcal{L}}{\partial f_{2,i}} & = 0 \\ \dots \quad \dots \qquad \dots & \\ \frac{\partial \mathcal{L}}{\partial f_m} - \sum_{i=1}^{n} \frac{\partial}{\partial x_i} \frac{\partial \mathcal{L}}{\partial f_{m,i}} & = 0 \end{align}$

Single function of two variables with higher derivatives

If there is a single unknown function to be determined that is dependent on two variables and their higher derivatives such that

$\begin{align} I[f] & = \int_{\Omega} \mathcal{L}(x_1, x_2, f, f_{,1}, f_{,2}, f_{,11}, f_{,12}, f_{,22}, \dots f_{,11\dots 1},\dots, f_{,1\dots 2})\, \mathrm{d}\mathbf{x}\,\! \\ & \qquad \quad f_{,i}�:= \cfrac{\partial f}{\partial x_i} ~;~~ f_{,ij}�:= \cfrac{\partial^2 f}{\partial x_i\partial x_j} ~;~~ f_{,1\dots n}�:= \cfrac{\partial^{n} f}{\partial x_1\dots\partial x_n} \end{align}$

the Euler-Lagrange equation is^[2]

$\begin{align} \frac{\partial \mathcal{L}}{\partial f} & - \frac{\partial}{\partial x_1}\left(\frac{\partial \mathcal{L}}{\partial f_{,1}}\right) - \frac{\partial}{\partial x_2}\left(\frac{\partial \mathcal{L}}{\partial f_{,2}}\right) + \frac{\partial^2}{\partial x_1^2}\left(\frac{\partial \mathcal{L}}{\partial f_{,11}}\right) + \frac{\partial^2}{\partial x_1\partial x_2}\left(\frac{\partial \mathcal{L}}{\partial f_{,12}}\right) + \frac{\partial^2}{\partial x_2^2}\left(\frac{\partial \mathcal{L}}{\partial f_{,22}}\right) \\ & + \dots + (-1)^n \frac{\partial^n}{\partial x_2^n}\left(\frac{\partial \mathcal{L}}{\partial f_{,22\dots 2}}\right) = 0 \end{align}$

Notes

↑ A short biography of Lagrange
↑ ^2.0 ^2.1 ^2.2 ^2.3 Courant, R. and Hilbert, D., 1953, Methods of Mathematical Physics: Vol I, Interscience Publishers, New York.

References

Weisstein, Eric W., "Euler-Lagrange Differential Equation" from MathWorld.
Calculus of Variations on PlanetMath
Izrail Moiseevish Gelfand (1963). Calculus of Variations. Dover. ISBN 0-486-41448-5.
Calculus of Variations at Example Problems.com (Provides examples of problems from the calculus of variations that involve the Euler–Lagrange equations.)