The question is. Determine the solutions of the Euler-Lagrange equation when


Which is fairly straightforward. At this point i would note the Beltrami equation is a simplification of the EL eqn.

Via EL $$\frac{\partial f}{\partial y}-\frac{d}{dx}[\frac{\partial f}{\partial y'}] = 1$$ This implies $1 = 0$ and so we have no solutions.

But via the Beltrami equation we have $$f-\frac{\partial f}{\partial y'}y'=y=0$$ which does have a solution (unless I'm being thick in which case please say so)

Any explanation would be great. (Side note i apologise for formatting. I'm using mobile app and mathjax doesn't always render on preview for me).

  • $\begingroup$ $$\frac{\partial f}{\partial y} =1,\quad \frac{\partial f}{\partial y'} =0 $$ $\endgroup$ – Hyperkähler Jan 31 '18 at 15:23
  • $\begingroup$ did you mean this here martin-ueding.de/articles/euler-lagrange-equation-derivation $\endgroup$ – Dr. Sonnhard Graubner Jan 31 '18 at 15:24
  • $\begingroup$ Yes. Thats essentially the same dervivation im using. For 2 points $x_1,x_2$ we consider all paths from $x_1$ to $x_2$. And then look for the infinium of that set. So in essence we're looking for stationary points of an integral. So we have $K = { y \in C^1((x_1,x_2), \mathbb{R}) | y(x_1)=y_1, y(x_2)=y_2}$ with $F:K \rightarrow \mathbb{R}$. With $\int_{x_1}^{x_2} f(x,y,y')dx$ $\endgroup$ – Vaas Jan 31 '18 at 15:35
  • $\begingroup$ Id also like to apologise again. I get no mathjax in comments on mobile $\endgroup$ – Vaas Jan 31 '18 at 15:36
  1. In physics terminology, OP's Lagrangian $f$ doesn't have a kinetic term, i.e. the Lagrangian (and the EL equation) are of zeroth order in derivatives. Hence the variational problem does not need boundary conditions (BCs), and in fact OP doesn't mention any BCs.

  2. The Lagrangian is just (minus) a linear potential, which doesn't have a stationary point. In fact, the system tends to $y\to-\infty$ trying to maximize this potential.

  3. The Beltrami identity produces constant energy solutions, but they are not stable.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.