Euler-Lagrange Differential Equation
The Euler-Lagrange differential equation is the fundamental equation of calculus of variations. It states that if J is defined by an integral of the form
| [画像: J=intf(t,y,y^.)dt, ] |
(1)
|
where
| [画像: y^.=(dy)/(dt), ] |
(2)
|
then J has a stationary value if the Euler-Lagrange differential equation
is satisfied.
If time-derivative notation y^. is replaced instead by space-derivative notation y_x, the equation becomes
The Euler-Lagrange differential equation is implemented as EulerEquations [f, u[x], x] in the Wolfram Language package VariationalMethods` .
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,
For three independent variables (Arfken 1985, pp. 924-944), the equation generalizes to
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
since deltaq^.=d(deltaq)/dt. Now, integrate the second term by parts using
so
| int(partialL)/(partialq^.)(d(deltaq))/(dt)dt=int(partialL)/(partialq^.)d(deltaq)=[(partialL)/(partialq^.)deltaq]_(t_1)^(t_2)-int_(t_1)^(t_2)(d/(dt)(partialL)/(partialq^.)dt)deltaq. |
(13)
|
Combining (◇) and (◇) then gives
But we are varying the path only, not the endpoints, so deltaq(t_1)=deltaq(t_2)=0 and (14) becomes
We are finding the stationary values such that deltaJ=0. These must vanish for any small change deltaq, which gives from (15),
This is the Euler-Lagrange differential equation.
The variation in J can also be written in terms of the parameter kappa as
where
and the first, second, etc., variations are
The second variation can be re-expressed using
so
| [画像: I_2+[v^2lambda]_2^1=int_1^2[v^2(f_(yy)+lambda^.)+2vv^.(f_(yy^.)+lambda)+v^.^2f_(y^.y^.)]dt. ] |
(26)
|
But
| [v^2lambda]_2^1=0. |
(27)
|
Now choose lambda such that
| f_(y^.y^.)(f_(yy)+lambda^.)=(f_(yy^.)+lambda)^2 |
(28)
|
and z such that
so that z satisfies
| f_(y^.y^.)z^..+f^._(y^.y^.)z^.-(f_(yy)-f^._(yy^.))z=0. |
(30)
|
It then follows that
See also
Beltrami Identity, Brachistochrone Problem, Calculus of Variations, Euler-Lagrange Derivative, Functional Derivative, VariationExplore with Wolfram|Alpha
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.Goldstein, H. Classical Mechanics, 2nd ed. Reading, MA: Addison-Wesley, p. 44, 1980.Lanczos, C. The Variational Principles of Mechanics, 4th ed. New York: Dover, pp. 53 and 61, 1986.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.Referenced on Wolfram|Alpha
Euler-Lagrange Differential EquationCite this as:
Weisstein, Eric W. "Euler-Lagrange Differential Equation." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/Euler-LagrangeDifferentialEquation.html