Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am now reading the book Calculus of Variations written by Jost and I have a problem in the proof of Noether's theorem:

Theorem 1.5.1. Let $F\in C^2([a, b]\times \mathbb R^d \times \mathbb R^d, \mathbb R)$ and a one-parameter family of maps $$h_s:\mathbb R^d\rightarrow \mathbb R^d$$ be of class $C^2\bigl((-\varepsilon_0, \varepsilon_0)\times \mathbb R^d, \mathbb R\bigr)$ for some $\varepsilon_0>0$ with $$h_0(z)=z \quad \forall z\in \mathbb R^d$$ satisfying $$\int_a^b F\Big(t,h_s\bigl(u(t)\bigr),\frac {d}{dt}h_s\bigl(u(t)\bigr)\Big) dt=\int_a^bF\Big(t,u(t),\dot{u}(t)\Big)dt$$ for all $s\in(-\varepsilon_0, \varepsilon_0)$ and all $u\in C^2([a, b], \mathbb R^d)$.

Then, for any solution $u(t)$ of the Euler-Lagrange equation for $$I(u)=\int_a^bF\Big(t,u(t),\dot{u}(t)\Big)dt,$$ $$F_p\bigl(t,u(t),\dot{u}(t)\bigr) \frac{d}{ds}h_s\bigl(u(t)\bigr)\Bigl|_{s=0}$$ is a constant $\forall t\in [a, b]$.

The proof begins with saying the invariance of the integral gives $\forall t_0\in [a, b]$, $$\frac {d}{ds} \int_a^{t_0} F\Big(t,h_s\bigl(u(t)\bigr),\frac {d}{dt}h_s\bigl(u(t)\bigr)\Big) dt\Bigl|_{s=0}=0.$$However, this is where I find my difficulties. I can understand it when $t_0=b$, but I cannot see the reason why otherwise. I do think it is possible since we require the integral to be unchanged $\forall u\in C^2$. I have tried changing the variable $t$ so that the case where $t_0\in [a,b]$ is arbitrary is reduced to $t_0=b$, but I have yet to make any progress.

So is there any hint that anyone can give me? It would be of great help if there are any and thanks in advance!

share|cite|improve this question

The above proof seems to be false, or at best incomplete.

In the original publication by E. Noether, it is assumed that the integral is invariant for arbitrary integration domains. In this case, this is equivalent to the assumption that the integrand is invariant, i.e. $$F\left(t, h_s(z), h_{s*}(p)\right)=F(t, z, p).$$ This is also the assumption made by, for instance, Arnold, Mathematical Methods of Classical Mechanics.

The proof above also implies that this is the case: suppose that $$\int_a^{t_0} F\Big(t,h_s\bigl(u(t)\bigr),\frac {d}{dt}h_s\bigl(u(t)\bigr)\Big) dt=\int_a^{t_0}F\Big(t,u(t),\dot{u}(t)\Big)dt$$ for all $t_0$. By additivity of the integral, this implies $$\int_{t_1}^{t_0} F\Big(t,h_s\bigl(u(t)\bigr),\frac {d}{dt}h_s\bigl(u(t)\bigr)\Big) dt=\int_{t_1}^{t_0}F\Big(t,u(t),\dot{u}(t)\Big)dt$$ for all $a\le t_1<t_0\le b$, but this implies that $$F\left(t, h_s(u(t)), \frac{d}{dt}h_s(u(t))\right)=F(t, u(t), \dot{u}(t))$$ for all $t\in [a, b]$ by the continuity assumptions. As $u$ is arbitrary, this is the same assumption as made by Noether and Arnold.

The remaining question is whether invariance of the integral over $[a, b]$ implies invariance of the integral over all subintervals. In case $F$ does not depend on $t$ explicitly, this can be done by scaling the variable $t$. Even though the assumption that invariance holds for any $u\in C^2$ is very strong, it seems unlikely that this is the case for general $F$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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