0
$\begingroup$

I am trying to solve this problem but i am stack , does anyone have any idea how to continue ?

The problem : Find differential function $f:\Bbb R^2 \rightarrow \Bbb R$ in area around $p=(1,2)$ such that $X(f)=0$ and $f(x,2x)=x $ for $x$ close to $1$ where $X=x_1\frac{\partial}{\partial x_1}+x_2\frac{\partial}{\partial x_2}$ the vector field in $\Bbb R^2$ ($id_{\Bbb R^2}=(x_1,x_2))$

What i have done so far : i changed chart in order to simplify X , ${\psi}^{-1}(t_1,t_2)={{\Phi}_{t_1}}^{X}((1,2+t_2))$ where ${{\Phi}_{t}}^{X}(x,y)=(xe^t,ye^t)$ the flow of $X$. Then $X=\frac{\partial}{\partial \psi_1}$ and now am trying to solve the $X(f)=0$ but i havent found anything :(

Any ideas on how can i approach this ?

$\endgroup$
  • 1
    $\begingroup$ I think this is impossible. The PDE $Xf=0$ implies $f(x,y)=g(x/y)$ for some function $g$. But then $x=f(x,2x)=g(1/2)$ cannot hold. $\endgroup$ – Spenser Jan 18 '18 at 15:50
  • $\begingroup$ how did you deduce that $f(x,y)=g(x/y)$? $\endgroup$ – dem0nakos Jan 18 '18 at 16:33
  • $\begingroup$ math.stackexchange.com/q/2287141/39285 $\endgroup$ – Spenser Jan 18 '18 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.