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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!

Pragabhava
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user43378
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1 Answers1

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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 $F$.

Martijn
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