a)
Given L = t^2(dotx^2/2-x^6/6) the movement equations are obtained by doing
d/dt((partial L)/(partial dot x))-(partial L)/(partial x)=0 giving
ddot x = -(2dot x/t+x^5)
b)
The condition for invariance of int L dt is given by
(partial L)/(partial t) tau+(partial L)/(partial x) xi + (partial L)/(partial dot x)((d xi)/(dt)-dot x (d tau)/(dt))+L(d tau)/(dt)=0
Now expanding the total derivatives (d xi)/(dt), (d tau)/(dt) and substituting we arrive at
(partial L)/(partial t)tau+(partial L)/(partial x)xi+(partial L)/(partial dot x)((partial xi)/(partial t)+(partial xi)/(partial x)dot x-dot x(partial tau)/(partial t)-dot x^2 (partial tau)/(partial x))+L((partial tau)/(partial t)+(partial tau)/(partial x)dot x)=0
From L we have
(partial L)/(partial t) = t(dot x^2-1/3x^6)
(partial L)/(partial x) =-t^2x^5
(partial L)/(partial dot x) =t^2 dot x
Substituting and grouping the coefficients associated to the powers of dot x and equating to zero we have
{
((x tau)/3+t xi + t x/6 (partial tau)/(partial t)=0),
((partial xi)/(partial t)-x^6/6 (partial tau)/(partial x) = 0),
(t tau+t^2(partial xi)/(partial x) - t^2/2 (partial tau)/(partial t) = 0),((partial tau)/(partial x)=0):}
Now from the last equation, tau=tau(t) substituting into the second equation xi = xi(x) remaining the system of differential equations
{
(x/3 tau+t xi + t/6x (d tau)/(dt)=0),
(t tau+t^2 (d xi)/(dx)-t^2/2 (d tau)/(dt)=0):}
From inspection, the transformation is
tau=t and xi = -x/2 so
int L dt is invariant under the transformation group
t'=t+epsilon t
x'=x-1/2epsilon x
c) Suppose that L(t,x,dot x) is variationally invariant on t_i,t_f under the former transformation. Then
(partial L)/(partial dot x) xi+(L-dot x (partial L)/(partial dot x))tau=C^(te)
Substituting (partial L)/(partial dot x), L,xi, tau we get
(t^2x dot x)/2+(t^3 dot x^2)/2+(t^3x^6)/6 = C^(te)
