Euler-Lagrange Differential Equation

A fundamental equation of Calculus of Variations which states that if $J$ is defined by an Integral of the form

$$J = \int f(x,y,\dot y)\,dx,$$ (1)

where

$$\dot y \equiv {dy\over dt},$$ (2)

then $J$ has a Stationary Value if the Euler-Lagrange differential equation

$${\partial f\over \partial y} - {d\over dt}\left({\partial f\over \partial \dot y}\right)= 0$$ (3)

is satisfied. If time Derivative Notation is replaced instead by space variable notation, the equation becomes

$${\partial f\over \partial y}-{d\over dx} {\partial f\over \partial y_x}=0.$$ (4)

In many physical problems, $f_x$ (the Partial Derivative of $f$ with respect to $x$) turns out to be 0, in which case a manipulation of the Euler-Lagrange differential equation reduces to the greatly simplified and partially integrated form known as the Beltrami Identity,

$$f-y_x{\partial f\over\partial y_x}=C.$$ (5)

For three independent variables (Arfken 1985, pp. 924-944), the equation generalizes to

$${\partial f\over \partial u}-{\partial \over \partial x} {\p... ...{\partial \over \partial z} {\partial f\over \partial u_z}=0.$$ (6)

Problems in the Calculus of Variations often can be solved by solution of the appropriate Euler-Lagrange equation.

To derive the Euler-Lagrange differential equation, examine

$$\delta J \equiv \delta \int L(q, \dot q, t)\,dt = \int \left({{\partial L\over \partial q} \delta q+{\partial L\over \partial \dot q}\delta \dot q}\right)\,dt$$

$$= \int \left[{{\partial L\over \partial q}\,\delta q+{\partial L\over \partial \dot q} {d(\delta q)\over dt}}\right]\,dt,$$ (7)

since $\delta\dot q=d(\delta q)/dt$. Now, integrate the second term by Parts using

$$u = {\partial L\over \partial \dot q} \qquad\qquad dv=d(\delta q)$$ (8)

$$du = {d\over dt}\left({\partial L\over \partial \dot q}\right)\,dt \qquad v=\delta q,$$ (9)

so

$$\int {\partial L\over \partial \dot q} {d(\delta q)\over dt... ...r dt} {\partial L\over \partial \dot q}\,dt}\right)\,\delta q.$$ (10)

Combining (7) and (10) then gives

$$\delta J=\left[{{\partial L\over \partial \dot q} \delta q}\... ...er dt}{\partial L\over \partial \dot q}}\right)\,\delta q\,dt.$$ (11)

But we are varying the path only, not the endpoints, so $\delta q(t_1)=\delta q(t_2)=0$ and (11) becomes

$$\delta J = \int_{t_1}^{t_2} \left({{\partial L\over \partial... ...er dt}{\partial L\over \partial \dot q}}\right)\,\delta q\,dt.$$ (12)

We are finding the Stationary Values such that $\delta J=0$. These must vanish for any small change $\delta q$, which gives from (12),

$${\partial L\over \partial q}-{d\over dt}\left({\partial L\over \partial \dot q}\right)=0.$$ (13)

This is the Euler-Lagrange differential equation.

The variation in $J$ can also be written in terms of the parameter $\kappa$ as

$$\delta J = \int [f(x,y+\kappa v,\dot y+\kappa\dot v)-f(x,y,\dot y)]\,dt$$

$$= \kappa I_1+{\textstyle{1\over 2}}\kappa^2 I_2+{\textstyle{1\over 6}}\kappa^3 I_3+{\textstyle{1\over 24}}\kappa^4 I_4+\ldots,$$ (14)

where

$$v = \delta y$$ (15)

$$\dot v = \delta\dot y$$ (16)

and the first, second, etc., variations are

$$I_1 = \int (vf_y+\dot v f_{\dot y})\,dt$$ (17)

$$I_2 = \int (v^2f_{yy}+2v\dot v f_{y\dot y}+{\dot v}^2f_{\dot y\dot y})\,dt$$ (18)

$$I_3 = \int (v^3f_{yyy}+3v^2\dot vf_{yy\dot y}+3v{\dot v}^2f_{y\dot y\dot y}+{\dot v}^3f_{\dot y\dot y\dot y})\,dt$$ (19)

$$I_4 = \int (v^4f_{yyyy}+4v^3{\dot v}f_{yyy\dot y}+6v^2{\dot v}^2f_{yy\d... ...dot v}^3f_{y\dot y\dot y\dot y} + {\dot v}^4 f_{\dot y\dot y\dot y\dot y})\,dt.$$ (20)

The second variation can be re-expressed using

$${d\over dt}(v^2\lambda)=v^2\dot\lambda+2v\dot v\lambda,$$ (21)

so

$$I_2+[v^2\lambda]_2^1=\int_1^2 [v^2(f_{yy}+\dot\lambda)+2v\dot v(f_{y\dot y}+\lambda)+{\dot v}^2f_{\dot y\dot y}]\,dt.$$ (22)

But

$$[v^2\lambda]_2^1=0.$$ (23)

Now choose $\lambda$ such that

$$f_{\dot y\dot y}(f_{yy}+\dot\lambda)=(f_{y\dot y}+\lambda)^2$$ (24)

and $z$ such that

$$f_{y\dot y}+\lambda=-{f_{\dot y\dot y}\over z}{dz\over dt}$$ (25)

so that $z$ satisfies

$$f_{\dot y\dot y}\ddot z+\dot f_{\dot y\dot y}\dot z-(f_{yy}-\dot f_{y\dot y})z=0.$$ (26)

It then follows that

$$I_2=\int f_{\dot y\dot y}\left({\dot v+{f_{y\dot y}+\lambda\... ..._{\dot y\dot y}\left({\dot v-{v\over z}{dz\over dt}}\right)^2.$$ (27)

See also Beltrami Identity, Brachistochrone Problem, Calculus of Variations, Euler-Lagrange Derivative

References

Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, 1985.

Forsyth, A. R. Calculus of Variations. New York: Dover, pp. 17-20 and 29, 1960.

Morse, P. M. and Feshbach, H. ``The Variational Integral and the Euler Equations.'' §3.1 in Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 276-280, 1953.