6.22.2 Euler-Lagrange equation(s): euler_lagrange

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. The function f can be given in one of two ways. For the first way:

• euler_lagrange takes one mandatory argument and two optional arguments:
  • expr, an expression involving an independent variable, a dependent variable, and the dependent variable prime.
  • 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.
  If n=1, a single equation is returned:

  ∂ f ∂ y

  =

  d dx

  ∂ f ∂ y′

  . (1)

  In general, n equations are returned:

  ∂ f ∂ yk

  =

  d dx

  ∂ f ∂ yk′

  ,  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 ‍(1) and ‍(2) are returned, each of them being sufficient to determine y (one of the returned equations is usually simpler than the other).

Example.
Input:

Output

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

It can be proven that if f is convex (as a function of three independent variables, see Section ‍6.21.8), then a solution y to the Euler-Lagrange equations minimizes the functional F.

Example.
Minimize the functional F for 0<a<b and

Input:

Output:

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

Output:

Note that a pair of independent solutions is also returned by the kovacicsols command (see Section ‍6.57.3):
Input:

Output:

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:
Input:

Output:

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

Example.
Find the function y in

which minimizes the functional

To obtain the corresponding Euler-Lagrange equation:
Input:

Output:

Input:

Output:

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

Output:

Input:

Output:

To prove that y₀(x)=−√1−x² is indeed a minimizer for F, you need to show that the integrand in F(y) is convex.
Input:

Output:

You can similarly find the minimizer for

where y∈ C¹[0,π] and y(0)=y(π)=0.
Input:

Output:

Input:

Output:

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

Output:

Example.
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:
Input:

Output:

Input:

Output:

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

Output:

Hence the minimizer is