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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,
\end{displaymath} (1)

\dot y \equiv {dy\over dt},
\end{displaymath} (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
\end{displaymath} (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.
\end{displaymath} (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.
\end{displaymath} (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.
\end{displaymath} (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

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

since $\delta\dot q=d(\delta q)/dt$. Now, integrate the second term by Parts using
$\displaystyle u$ $\textstyle =$ $\displaystyle {\partial L\over \partial \dot q} \qquad\qquad dv=d(\delta q)$ (8)
$\displaystyle du$ $\textstyle =$ $\displaystyle {d\over dt}\left({\partial L\over \partial \dot q}\right)\,dt \qquad v=\delta q,$ (9)


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

Combining (7) and (10) then gives
\delta J=\left[{{\partial L\over \partial \dot q} \delta q}\... dt}{\partial L\over \partial \dot q}}\right)\,\delta q\,dt.
\end{displaymath} (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... dt}{\partial L\over \partial \dot q}}\right)\,\delta q\,dt.
\end{displaymath} (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.
\end{displaymath} (13)

This is the Euler-Lagrange differential equation.

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

$\displaystyle \delta J$ $\textstyle =$ $\displaystyle \int [f(x,y+\kappa v,\dot y+\kappa\dot v)-f(x,y,\dot y)]\,dt$  
  $\textstyle =$ $\displaystyle \kappa I_1+{\textstyle{1\over 2}}\kappa^2 I_2+{\textstyle{1\over 6}}\kappa^3 I_3+{\textstyle{1\over 24}}\kappa^4 I_4+\ldots,$ (14)

$\displaystyle v$ $\textstyle =$ $\displaystyle \delta y$ (15)
$\displaystyle \dot v$ $\textstyle =$ $\displaystyle \delta\dot y$ (16)

and the first, second, etc., variations are

$\displaystyle I_1$ $\textstyle =$ $\displaystyle \int (vf_y+\dot v f_{\dot y})\,dt$ (17)
$\displaystyle I_2$ $\textstyle =$ $\displaystyle \int (v^2f_{yy}+2v\dot v f_{y\dot y}+{\dot v}^2f_{\dot y\dot y})\,dt$ (18)
$\displaystyle I_3$ $\textstyle =$ $\displaystyle \int (v^3f_{yyy}+3v^2\dot vf_{yy\dot y}+3v{\dot v}^2f_{y\dot y\dot y}+{\dot v}^3f_{\dot y\dot y\dot y})\,dt$ (19)
$\displaystyle I_4$ $\textstyle =$ $\displaystyle \int (v^4f_{yyyy}+4v^3{\dot v}f_{yyy\dot y}+6v^2{\dot v}^2f_{yy\d... v}^3f_{y\dot y\dot y\dot y} + {\dot v}^4 f_{\dot y\dot y\dot y\dot y})\,dt.$ (20)

The second variation can be re-expressed using
{d\over dt}(v^2\lambda)=v^2\dot\lambda+2v\dot v\lambda,
\end{displaymath} (21)

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.
\end{displaymath} (22)

\end{displaymath} (23)

Now choose $\lambda$ such that
f_{\dot y\dot y}(f_{yy}+\dot\lambda)=(f_{y\dot y}+\lambda)^2
\end{displaymath} (24)

and $z$ such that
f_{y\dot y}+\lambda=-{f_{\dot y\dot y}\over z}{dz\over dt}
\end{displaymath} (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.
\end{displaymath} (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.
\end{displaymath} (27)

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


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.

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© 1996-9 Eric W. Weisstein