Finding invariants of infinitesimal transformation I am studying "Ordinary Differential Equations" by James Morris Page and was stuck with verifying that a family of curves is indeed an invariant for a given transformation.
The finite transformation in $x-y$ plane is simply a uniform translation:
\begin{align}
x_1= x+t,~y_1= y+t
\end{align}
where $t$ is the parameter of transformation and obviously $t=0$ gives the identity transformation. The task is to verify that the family of circles
\begin{align}
\Omega (x,y)=(x-a)^2+(y-b)^2=r^2
\end{align}
where $a,b,r,$ are arbitrary, admits of the transformation above. That it does, is easy to verify by direct substitution:
\begin{align}
(x_1-t-a)^2+(y_1-t-b)^2 & =r^2 \\
(x_1-a')^2+(y_1-b')^2 & =r^2 \quad (a'\equiv a+t,b'\equiv b+t)
\end{align}
which therefore belongs to the same family of circles.
However there is another way of verifying the same, using infinitesimal transformation and that does not seem to give the correct answer. For some function $f(x,y)$ the infinitesimal transformation (corresponding to the finite transformation mentioned in the beginning) is represented by
\begin{align}
Uf\equiv \frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}
\end{align}
If $\Omega (x,y)$ is an invariant of the said transformation we must have
\begin{align}
U\Omega = \frac{\partial \Omega}{\partial x}+\frac{\partial \Omega}{\partial y}=g(\Omega)
\end{align}
for some function $g$. But
\begin{align}
U\Omega = 2(x-a)+2(y-b)\neq g(\Omega)
\end{align}
and hence one would conclude that the family of circles $\Omega (x,y)$ does not admit of the said transformation, which is incorrect. What did I do wrong? Did I mis-apply concepts somewhere? Thanks in advance for any help.
 A: The condition $U \Omega = g\circ \Omega$ applies when you look at level sets for a given function $\Omega$ and want to show that they are invariant under the (Lie-) group of transformations $\{\phi_t\}_{t\in {\Bbb R}}$ generated by $U$.
The given problem is more general. Here, you are looking at a differentiable family of functions $\Omega_{a,b}$ depending smoothly on parameters $a,b$. You want to know if $\phi_t$ maps a level set  $\Omega_{a,b}=r$ into the level set of some $\Omega_{a(t),b(t)}=r_t^2$, a possibly different function in the family and with a different level set-value. The more general condition reads:
   $$ \Omega_{a(t),b(t)} (\phi_t(x,y)) = G_t(\Omega_{a,b}(x,y)) $$
for some real valued functions $a,b,G$.
In order to proceed we assume that they are all smooth functions.
 Taking derivatives w.r.t. $t$, and setting $t=0$, you get:
 $$ U \Omega_{a,b} + \partial_a \Omega_{a,b} \dot{a} + \partial_b \Omega_{a,b} \dot{b} = \partial_t G_t(\Omega_{a,b}(x,y))_{|t=0} = g(\Omega_{a,b}(x,y)) $$
where $U\Omega = \partial_x \Omega v_x(x,y) + \partial_y \Omega v_y(x,y)$ and $\partial_t \phi_t(x,y)_{|t=0} = (v_x(x,y),v_y(x,y)$. 
In the present case we have $v_x=v_y=1$ so the differential condition reduces to
$$ 2(x-a) + 2(y-b)  + 2 (a-x) \dot{a} + 2 (b-y) \dot{b} = g(\Omega_{a,b}(x,y)), \; \; \forall x,y $$
which as  solution has $\dot{a}=\dot{b}=1$ and $g\equiv 0$, corresponding to the solution you described.
Update: In order to see better what is going it is of interest to look at a less trivial example. We take this time the multiplicative group $({\Bbb R}^*_+,\times)$, acting by dilation on the plane, centered at the origin:
$$ \phi_t(x,y)= (tx,ty) $$
Indeed it is a multiplicative action as $\phi_t \circ \phi_s=\phi_{ts}$, $t,s>0$.
Let us write  $\Omega_{a,b}(x,y) = \Omega(a,b; x,y) = (x-a)^2+(y-b)^2$. Then we claim that $\phi_t$, $t>0$ maps  any given level set $\Omega_{a,b}=r$ to another level set $\Omega_{a_t,b_t}=r_t$, but even better, $(a_t,b_t)=q_t(a,b)$ and $r_t=G_t(r)$ are both given by multiplicative actions. We have:
  $$ G_t \circ \Omega(a,b; x,y) = \Omega ( q_t(a,b); \phi_t(x,y)) $$
(which is equivalent to the statement about level sets).
To deduce $q_t$ and $G_t$ we look at the derivative at the neutral element, which is here $t=1$:
$$ U \Omega = \partial_x \Omega \; \dot{x}_{t=1} 
 + \partial_y \Omega \; \dot{y}_{t=1} =
  2(x-a)x + 2(y-b)y$$
$$ V\Omega = \partial_a \Omega \; \dot{a}_{t=1} 
 + \partial_b \Omega \; \dot{b}_{t=1} =
  2(a-x)\dot{a} + 2(b-y)\dot{b} $$
Note that $\dot{a}$ and $\dot{b}$ should not depend upon $x,y$ but may depend upon $a,b$ themselves. We want $U\Omega + V\Omega$ to be a function of $\Omega$ and looking not too hard at this you realize that setting $\dot{a}=a$ and $\dot{b}=b$ is on the right track as it yields:
 $$ U \Omega + V\Omega = 2(x-a)^2+2(y-b)^2 = 2 \Omega = \dot{G}_{t=1}(\Omega)$$
From here we may already deduce that our claim holds (at least in a neighborhood of $t=1$) since each of the equations for $a,b,G$ admits at least local solutions. We may also calculate these solutions:
 $$ \dot{a}_{t=1} =a , \; \; a_{t=1}=a \; \Rightarrow a_t=t a, \forall t>0 $$
(note that this is not the ODE for the exponential function; instead we have for small $h$: $a_{1+h}= a+ ah +o(h) = (1+h) a + o(h)$. As the action is multiplicative we deduce: $a_{(1+h)^n} = (1+h)^n a+ o(nh)$ from which taking suitable limits $a_{t}=ta$). Similarly you show that $G_t(\Omega)=t^2 \Omega$, $t>0$ is the unique solution for $G_t$. All in all we find something which we could have said already from the start:
 $$ \Omega(q_t(a,b);\phi_t(x,y)) = (tx-ta)^2+(ty-yb)^2=t^2 \Omega(a,b;x,y)=G_t(\Omega)$$
a somewhat complicated way to deduce that dilations maps circles to circles.
If you have the courage with calculations you may try to change the center for the dilation to some other point $(a_0,b_0)$. The result still holds but the transformation group for $a,b$  changes, whereas that for $G$ is the same.
A: This is the reference that I have in my book shelves:
Sophus Lie, Vorlesungen über Differentialgleichungen
mit bekannten Infinitesimalen Transformationen, bearbeitet und
herausgegeben von Dr. Georg Wilhelm Scheffers,Leipzig (1891). Availability :
Amazon ,
online (!) at GDZ.

Partial answer. A question that resembles the one as formulated by the OP is found in the (online) book at page 89:
$$
(x-c)^2+y^2 = r^2 \quad \mbox{for} \quad x_1 = x+t \quad \mbox{and} \quad y_1=y
$$
The author says that one must solve for the parameter $\,c\,$ in order to be able to proceed:
$$
\Omega(x,y) \equiv x \pm \sqrt{r^2-y^2} = c \\
\Omega(x_1,y_1) = x_1 \pm \sqrt{r^2-y_1^2} = x+t \pm \sqrt{r^2-y^2} = \Omega(x,y) + t
$$
Then, in our case, it follows that:
$$
U f\equiv \frac{\partial f}{\partial x} \quad \Longrightarrow \quad
U\Omega = \frac{\partial \Omega}{\partial x} = \frac{\partial\left(x+t \pm \sqrt{r^2-y^2}\right)}{\partial x} \equiv 1 
$$
So $\,U\Omega\,$ is trivialiter a function of $\,\Omega\,$ alone, namely a constant. 
But, at this moment, I have no idea how to proceed with the more complicated case $\;x_1= x+t\; , \;y_1= y+t$ .
