13.8.2  Euler-Lagrange equations

The Euler-Lagrange equations for a Lagrangian function f(x,y,y′) are differential equations which must be satisfied by extrema of the functional F(y).

The euler_lagrange command finds the Euler-Lagrange equations for a Lagrangian f.

• euler_lagrange takes one mandatory argument and two optional arguments:
  • expr, an expression involving an independent variable, a dependent variable, and the dependent variable derivative.
  • Optionally, indvar, the independent variable (by default x).
  • Optionally, depvar, the dependent variable (by default y).

    If a function y∈ℝⁿ is required (by default n=1), you can enter y=(y₁,y₂,…,y_n) as a vector [y₁,y₂,…,y_n]. In that case, y′=(y₁′,y₂′,…,y_n′).

  An alternate way to specify the independent and dependent variables is by replacing both optional arguments by either, for example, y(x) or [y₁(x),y₂(x),…,y_n(x)].
• euler_lagrange(expr ⟨,indvar,depvar ⟩) returns the system of differential Euler-Lagrange equations.

  For n=1, a single equation is returned: ∂ f/∂ y=d/dx ∂ f/∂ y′ .

  In general, n equations are returned: ∂ f/∂ y_k=d/dx ∂ f/∂ y_k′, k=1,2,…,n.

The degrees of these differential equations are kept as low as possible. If, for example, ∂ f/∂ y=0, the equation ∂ f/∂ y′=K is returned, where K∈ℝ is an arbitrary constant. Similarly, using the Hamiltonian

the Euler-Lagrange equation is simplified in the case n=1 and ∂ f/∂ t=0 to:

since it can be shown that d/dx H(y,y′,x)=0. Therefore the Euler-Lagrange equations, which are generally of order two in y, are returned in a simpler form of order one in the aforementioned cases. If n=1 and ∂ f/∂ x=0, then both equations are returned, each of them being sufficient to determine y (one of the returned equations is usually simpler than the other).

Examples

Minimize the functional F for 0<a<b and f(x,y,y′)=x² y′(x)²+y(x)².

This can be solved by assuming y(x)=x^r for some r∈ℝ.

The same pair of solutions is also returned by the kovacicsols command (see Section 13.4.3):

You can conclude that y=C₁ x^−√5+1/2+C₂ x^√5+1/2. The values of C₁ and C₂ are determined from the boundary conditions. Finally, to prove that f is convex:

Therefore, y minimizes F on [a,b].

Find the function y in {y∈ C¹[1/2,1]:y(1/2)=−√3/2, y(1)=0} which minimizes the functional

To obtain the corresponding Euler-Lagrange equation:

The sought solution is the function of the above form which satisfies the boundary conditions.

If the integrand in F(y) is convex, then y₀(x)=−√1−x² is a minimizer for F. Indeed:

You can similarly find the minimizer for

where y∈ C¹[0,π] and y(0)=y(π)=0.

The above function is the sought minimizer as the integrand 2 sin(x) y(x)+y′(x)² is convex:

Minimize the functional F(y)=∫₀¹(y′(x)⁴−4 y(x)) dx on C¹[0,1] with boundary conditions y(0)=1 and y(1)=2.

First, solve the associated Euler-Lagrange equation:

Next, check if the integrand in F(y) is convex:

Hence the minimizer is y₀(x)=−3/4 (1.52832425067−x)^4/3+2.32032831141, 0≤ x≤ 1.

Find Euler-Lagrange equations for a bivariate functional F:

where K₀,K₁∈ℝ are arbitrary constants (note that these symbols are generated automatically).

It can be proven that if f is convex (as a function of three independent variables, see Section 13.7.8), then a solution y to the Euler-Lagrange equations minimizes the functional F.