Is action principle trivial?

Question

Given a function $f(t)$, is it possible to construct Lagrangian $\mathcal{L}$ such that $\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial\mathcal{L}}{\partial \dot{f}}=\frac{\partial\mathcal{L}}{\partial f}$, when solved with appropriate boundary conditions, gives $f(t)$.

Answer 1

Yes, just take $\mathcal{L}(t, x, \dot x) = (x - f(t))^2$, since:

• The action $S[x] = \int \mathcal{L}(t, x(t),\dot x(t)) dt = \int (x(t) - f(t))^2 dt $ is minimized for $x(t) = f(t)$, since it is non-negative for any $x(t)$, and zero only for $x(t) = f(t)$.
• Euler-Lagrange gives $$ 0 = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot x} = \frac{\partial \mathcal{L}}{\partial x} = 2 (x(t) - f(t))$$