5

The example I'm doing gives an equation $$L(y, y') = \frac{y'}{y}$$

then $$\frac{\partial L}{\partial y} = -\frac{y'}{y^2}$$ and $$\frac{\partial L}{\partial y'} = \frac{1}{y}$$ ... $$\frac{d}{dx}\frac{\partial L}{\partial y'} = \frac{d}{dx}\frac{1}{y} = -\frac{y'}{y^2}$$

substituting into the Euler Lagrange $$\frac{\partial L}{\partial y} = \frac{d}{dx} \frac{\partial L}{\partial y'}$$ then yields $$-\frac{y'}{y^2} = -\frac{y'}{y^2}$$

Does this mean that any differentiable function $y$ is a solution to the Euler-Lagrange equation for that Lagrangian??

Rawb
  • 443

1 Answers1

9

The reason is that $$ \mathcal{L}=\int_a^b L\big(y(t),y'(t)\big)\,dt =\int_a^b\frac{y'(t)}{y(t)}\,dt=\big[\ln y(t)\big]_a^b=\ln y(b)-\ln y(a) $$ is independent of the path, and hence, all paths are stationary.

Einar Rødland
  • 9,707
  • very good... thanks a lot – Rawb Sep 06 '12 at 15:21
  • 1
    I am not completely understanding how this is "path independent"... this will give you a different answer if $y(t) = t^2$ versus $y(t) = t$. What do you mean by "path independent" @EinarRodland? – makansij Jan 03 '18 at 04:17
  • @Sother It depends only on the end-points, $y(a)$ and $y(b)$, not on the path between them. In the Lagrange formalism, you fix the end-points and find the path that minimises $\mathcal{L}$, but in this case all paths are minimal. – Einar Rødland Jan 04 '18 at 08:48
  • 1
    I see. So, it "path independent" CAN still depend on the function $y$, but it just doesn't depend on what happens BETWEEN $a$ and $b$? – makansij Jan 20 '18 at 16:57
  • @Sother: That's right. – Einar Rødland Jan 21 '18 at 22:13