My question is about the derivation of the prolongation formula from Olver's book:"Applications of Lie groups to differential equations" Page 109.
Considering a differential equation with independent variable(x) and one dependent variable(u): (x,u) $\subset$ $X \times U$
The coordinate in first jet space $M^{(1)}$ is (x,$u^{(1)}$) = (x,u,$u_j$).
Let u = f(x) is any function with $u_j = \frac{\partial u}{\partial x_j} $
First prolongation of a group action on M is given as: $pr^{1} g_\epsilon . (x,u^{(1)}) = (\tilde{x},\tilde{u}^{(1)})$
The dependent variable is unchanged here; I mean
$\tilde{u} = \tilde{f}_\epsilon(\tilde{x}) = f[\Xi^{-1}_\epsilon(\tilde{x})] = f[\Xi_{-\epsilon} (\tilde{x})] $
To find the infinitesimal generator of pr(1) go we must differentiate the formulas for the prolonged transformations with respect to e and set e = O. Thus $pr ^{1} v= \xi ^{i}(x)\frac{\partial}{\partial x^i}+ \phi ^{j}(x,u^{1})\frac{\partial}{\partial u^j},$ where $ \phi ^{j}(x,u^{1})= \frac{d}{ d\epsilon}_{\epsilon|=0} [\frac{\partial \Xi^k_{-\epsilon}}{\partial\tilde{x}^j} ] (\Xi_{\epsilon}(x)). u_{k}= - \frac{\partial \xi^k}{\partial x^j}(x). $
It seems simple, but I don't know how to calculate $ \phi ^{j}(x,u^{1}) $. Can anybody help me to find the answer? What kind of derivative is it?
Olver obtained two types of terms multiplying $u_{k}$, first:
$\frac{\partial}{\partial\tilde{x}^j} [\frac{d\Xi^k_{-\epsilon}}{d\epsilon} ] (\Xi_{\epsilon}(x)) |_{\epsilon = 0} = \frac{\partial}{\partial x^j} [\frac{d\Xi^k_{\epsilon}}{d\epsilon} ] |_{\epsilon = 0} = - \frac{\partial \xi^k}{\partial x^j}(x)$
and second,
$ \frac{\partial^{2} [\Xi^k_{-\epsilon} ]}{\partial\tilde{x}^j \partial\tilde{x}^l} ((\Xi_{-\epsilon}(x)) [\frac{d\Xi^k_{\epsilon}}{d\epsilon} ](x) |_{\epsilon = 0} =0.$
Shoud I add this two term? How can I obtained them? Thank you in advance!