Incorrect documentation of variational derivative in Euler-Lagrange equations

⚠️ Found by a local LLM, not a human. This was flagged by an automated documentation-mistake finder (qwen3:14b) and passed 10/10 independent verification runs before filing, but it has not been reviewed by a person. Please check it's a genuine error before acting on it.

Summary

The documentation in Physlib/ClassicalMechanics/EulerLagrange.lean (around line 63) is incorrect.

Current text

- The variational derivative of `L t (q' t) (deriv q' t))` for a lagrangian `L` is equal to the `eulerLagrangeOp`.

Why this is wrong

The docstring for euler_lagrange_varGradient claims the variational derivative of L t (q' t) (deriv q' t) equals eulerLagrangeOp, but this is incorrect. The theorem actually proves that the variational derivative of the action functional — the integral of the Lagrangian over time — equals the Euler-Lagrange operator.

The confusion arises because the docstring refers to the Lagrangian itself rather than its time integral, which is the correct object whose variational derivative yields the operator. The theorem's proof and statement in the code correctly handle the action functional, not the pointwise Lagrangian value.

The correction aligns the documentation with the mathematical content of euler_lagrange_varGradient, which applies to the integral of L over time, ensuring consistency between the declaration and its formal proof.

Suggested correction

- The variational derivative of the action `∫ t, L t (q' t) (deriv q' t)` for a Lagrangian `L` is equal to the `eulerLagrangeOp`.

Open Physlib/ClassicalMechanics/EulerLagrange.lean around lines 63–64 and inspect the docstring for euler_lagrange_varGradient alongside its theorem statement and proof. Confirm that the variational derivative applies to the time-integrated action, then update the documentation so it describes that action rather than the pointwise Lagrangian.