At the moment I am working through the Book "Applications of Lie-Groups to Differential-Equations" by Peter Olver.
On page 120 he calculates the symmetrie groups of the heat equation. $$ u_{t} = u_{xx} \quad \text{with} \quad u = f(x,t), $$
with the notation
$$ \frac{\partial u}{\partial x} = u_x $$
He finds the infinitesimal generators of the symmetrie groups of the differential equation and then procceds to calculate the group action of the symmetrie groups. He finds 7 vectorfields (the infinitesimal generators) which are the basis of the Lie-Algebra of a local group of transformations (which is a Lie-Group). For example he finds
$$ v_1 = \partial_x $$
and the symmetrie group is
$$ G_1: (x,t,u) \rightarrow (\tilde{x},\tilde{t},\tilde{u}) = (x + \varepsilon,t,u). $$
This caluclation is clear for me, as he just solves the differential equation
$$ \frac{dx}{d\varepsilon} = 1, $$
which, as I understand it, is finding the flow generated by the vectorfield and that the flow is equal to a one-parameter group of transformations.
But then he proceeds to calculate the symmetrie groups of the vector fields
$$ v_2 = 2t\partial_x -xu\partial_u\\ v_3 = 4tx\partial_x +4t^2\partial_t - (x^2+2t)u\partial_u$$
which are given by
$$ G_2: \Bigl(x + 2t\varepsilon,t,u \cdot exp(-\varepsilon x - \varepsilon^2 t)\Bigr),\\ G_3: \Bigl(\frac{x}{1-\varepsilon t},\frac{t}{1-\varepsilon t}, u \cdot \sqrt{1-4\varepsilon t} \cdot exp\{\frac{-\varepsilon x^2}{1-4\varepsilon t}\} \Bigr). $$
I dont seem to understand how he calculated these group actions, especially the action on u.
Any help would be appreciated. Thank you!