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I am trying to derive a Lie bracket, and then find the related integral curve at the point $(x_0,y_0)$.

The problem gives the vector fields $X = y \frac{\partial }{\partial x}$ ,$Y = \frac{x^2}{2} \frac{\partial }{\partial y}$. The goal is then to find $[X,Y] = XY-YX$, which is the Lie bracket (and another vector field) and then find the integral curve through $(x_0,y_0)$.

I've tried applying the definition of the Lie bracket as follows:

$[X,Y]= y\frac{\partial\frac{x^2}{2}}{\partial x}\frac{\partial }{\partial x}-\frac{x^2}{2}\frac{\partial y}{\partial y}\frac{\partial }{\partial y}$, from which you can simplify and get: $yx\frac{\partial }{\partial x} -\frac{x^2}{2}\frac{\partial }{\partial y}$.

I then proceed to get the integral curve, but don't quite get what is listed as a solution. I try listing this as a system of two ODE's (i.e $x'(t) = yx, y'(t) =\frac{x^2}{2}$, but once I finish solving, I don't get the result ($x(t)= \frac{2x_0}{x_0t+2}$, $y(t) = \frac{1}{4}y_0(x_0t+2)^2$. Can anyone see where I've may have made any mistakes?

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The error is when you compute the Lie bracket: $$[X,Y]=y\frac\partial{\partial x}\left(\frac{x^2}2\frac\partial{\partial y}\right)-\frac{x^2}2\frac\partial{\partial y}\left(y\frac\partial{\partial x}\right)=yx\frac\partial{\partial y}-\frac{x^2}2\frac\partial{\partial x}.$$

Notice that you have the vector fields wrong (you inverted $x$ and $y$).


Then you have to solve: $$\begin{cases}x'=-x^2/2\\y'=xy.\end{cases}$$ Integrating $x'=-x^2/2$ first yields $$x(t)=\frac{2x_0}{x_0t+2}$$ and substituting in $y'=xy$ and solving yields $$y=\frac{y_0}4(x_0t+2)^2.$$

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    $\begingroup$ Thanks for the explanation, I see now where I went wrong. However, I still don't see how the last step follows (the substitution step). After I solve for x(t), I go and plug it in to y' = xy and get a differential equation with logarithm for y. Can you explain how you got it? $\endgroup$ Feb 16 '16 at 2:23
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    $\begingroup$ Please disregard... I already got it :) I used the definition of y' = xy and redefined it in terms of the new x(t) and worked backwards. $\endgroup$ Feb 16 '16 at 3:16

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