I just begin with the Calculus of variation. Here is the problem:
Consider the set of n Euler-Lagrange equations for a system with n degrees of freedom:
$\frac{\partial f}{\partial y_i}-\frac{d}{dx}\frac{\partial f}{\partial y'_i}=0$ for $i=1,2 \cdots, n$
Show that if $f(y_1,\cdots,y_n; y'_1 ,\cdots,y'_n;x)$ is replaced by $$g=f(y_1,\cdots,y_n; y'_1 ,\cdots,y'_n;x)+\frac{dF(y_1,\cdots,y_n)}{dx},$$ where F is an arbitrary differentiable function, the Euler-Lagrange equations remain the same.
Also, show that the functions $\{y_i\}$ that extremize the functional are the same as this new functional with g as the integrand.
Could anyone give me some idea? Thanks in advance!
The functional $S[f] := \int_a^b f \,\mathrm{d}x$ applied on $g = f+F'$, where the prime denotes the derivative with respect to $x$, leads to $$ S[g] = \int_a^b f + F' \,\mathrm{d}x = \int_a^b f \,\mathrm{d}x + \int_{F(a)}^{F(b)}\mathrm{d}F = S[f] + F(b)-F(a). $$ The additional term $F(b)-F(a)$ is a function of the boundary points $a$ and $b$ but does not depend on other quantities, so that its functional derivative will vanish $-$ in other words, $F(b)-F(a)$ is constant with respect to $x,y_i$ and $y_i'$ $-$, whence the same Euler-Lagrange equations for $S[f]$ and $S[g]$.