# Inconsistency? Lagrangian with its Euler–Lagrange equation as condition

Consider the action

$$A_{1} = \int{L(q, \dot{q})}{dt}\tag{1}$$

and the corresponding Euler–Lagrange equation

$$\frac{\partial{L}}{\partial{q}} - \frac{d}{dt}\left(\frac{\partial{L}}{\partial{\dot{q}}}\right)=0.\tag{2}$$

This equation is a general condition that $$L$$ fulfill. Therefore this condition you can add to the original action as Lagrange multiplier (this change has no, in principle, affect for Euler–Lagrange equation).

$$A_{2} = \int{\left[L(q, \dot{q}) + \lambda\left(\frac{\partial{L}}{\partial{q}} - \frac{d}{dt}\left(\frac{\partial{L}}{\partial{\dot{q}}}\right)\right)\right]}dt\tag{3}$$

Where $$\lambda \in R$$ and so the last term is

$$\frac{d}{dt}\left(\lambda\frac{\partial{L}}{\partial{\dot{q}}}\right)\tag{4}$$

This term is in the form of total derivative and so we can dropped it from the lagrangian (generates the same equation). We're getting that expression

$$A_{2} = \int{\left(L(q, \dot{q}) + \lambda\frac{\partial{L}}{\partial{q}} \right)}dt\tag{5}$$

But this lagrangian generally generates a different equation than the original lagrangian $$L$$.

I can't figure out where I made a mistake.

• Commented May 27, 2019 at 21:26
• Why assuming $\lambda=\rm const.$? Lagrange multipliers for functionals are in fact functions. I would refer you to chapter 17 of "Mathematical Methods for Physicists: A Comprehensive Guide" by Arfken, Weber and Harris. Commented May 27, 2019 at 21:50

## 1 Answer

Here we assume that $$\lambda$$ is not a function of time, i.e. there is only one time-averaged constraint. Then OP's derivation has the following shortcomings:

• Firstly, the new time-averaged constrained term in eq. (3) subtly changes the EOM for $$q$$.

Example. Consider for simplicity the static model $$L(q)~=~\frac{1}{2}q^2+\frac{1}{3}q^3$$. Then the stationary points for the action (1) are $$q\approx 0$$ and $$q\approx -1$$, while the constraint yields the time-average $$\langle q\rangle \approx - \frac{1}{2}$$. The EOM for $$q$$ becomes $$q^2+q\approx\lambda (2q+1)$$. See also this related Phys.SE post.

• Secondly, removing the boundary term (4) alters the EL equation for $$\lambda$$. More generally, boundary terms do matter if they don't vanish/aren't fixed by the pertinent boundary conditions of the theory.