3.4.10.1 Example 1 on how to find Lie group \(\left ( x,y\right ) \) given Lie infinitesimal xi and eta

Given\(\ \xi =1,\eta =2x\) find Lie group \(\bar {x},\bar {y}\).  Since

\[ \xi \left ( x,y\right ) =\left . \frac {\partial \bar {x}}{\partial \epsilon }\right \vert _{\epsilon =0}\]

Then

\begin{align} \frac {d\bar {x}}{d\epsilon } & =\xi \left ( \bar {x},\bar {y}\right ) \nonumber \\ & =1 \tag {1}\end{align}

Similarly, since

\[ \eta \left ( x,y\right ) =\left . \frac {\partial \bar {y}}{\partial \epsilon }\right \vert _{\epsilon =0}\]

Then

\begin{align} \frac {d\bar {y}}{d\epsilon } & =\eta \left ( \bar {x},\bar {y}\right ) \nonumber \\ & =2\bar {y} \tag {2}\end{align}

Where in both odes (1,2) we have the condition that at \(\epsilon =0\) then \(\bar {x}=x,\bar {y}=y\). Starting with (1), solving it gives

\[ \bar {x}=\epsilon +c_{1}\left ( x,y\right ) \]

Where \(c_{1}\left ( x,y\right ) \) is arbitrary function which acts like constant of integration since \(\bar {x}\left ( x,y\right ) \) is function of two variables. At \(\epsilon =0\) then \(c_{1}\left ( x,y\right ) =x\). Hence the above is

\begin{equation} \bar {x}=\epsilon +x \tag {3}\end{equation}

And from (2), solving give

\[ \bar {y}=2\bar {x}\epsilon +c_{2}\left ( x,y\right ) \]

But at \(\epsilon =0\) \(,\bar {y}=y,\bar {x}=x\) then the above gives \(c_{2}=y\). Hence the above becomes

\[ \bar {y}=2\bar {x}\epsilon +y \]

But \(\bar {x}=\epsilon +x\) from (3), hence  the above becomes

\begin{align*} \bar {y} & =2\left ( \epsilon +x\right ) \epsilon +y\\ & =2\epsilon ^{2}+2\epsilon x+y \end{align*}

Therefore Lie group is

\begin{align*} \bar {x} & =\epsilon +x\\ \bar {y} & =2\epsilon ^{2}+2\epsilon x+y \end{align*}