I asked the following question at Math Stackexchange a while ago [here](https://math.stackexchange.com/questions/2212682/for-an-arbitrary-gx-t-does-f-t-2g-xfgf-x-fx-0-0-have-a-unique-solut) but did not get a correct answer. 



> 
> Let $f(x,t)$ and $G(x,t)$ be smooth functions from 
>  $\mathbb R^2\to\mathbb R$.
> 
> 
> The PDE $$\dfrac{\partial}{\partial t}f(x,t)=2f(x,t)
>  \dfrac{\partial}{\partial x}G(x,t)+G(x,t)\dfrac{\partial}{\partial x}f(x,t)$$ 
>  applies on all of
>  $\mathbb R^2$. Furhermore, let us impose the condition $$f(x,0)=0,
> \forall x\in \mathbb R$$
> 
> 
> Is it necessarily true that $f(x,t)=0$ for all $(x,t)\in\mathbb R^2$?
> 
> 
> 


I will comment that it is easy to show this is true if $f$ is assumed to be analytic, but it seems rather difficult if I don't have this assumption. Through some basic PDE tricks (method of characteristics, etc) it is possible to show a local version of this result: that for a fixed $x$, there is a small $\epsilon(x)$ such that $f(x,t)=0$ for all $0<t<\epsilon(x)$, but this is not good enough for my purposes.



For an arbitrary $G(x,t)$, does $f_t=2G_xf+Gf_x$, $f(x,0)=0$ have a unique solution for $f$?
The answer is "No, it is not necessarily true that $f(x,t)=0$ for all $(x,t)\in\mathbb{R}^2$".


Here is a counterexample: Let $G(x,t) = x^2$ and let $h:\mathbb{R}\to\mathbb{R}$ be *any* smooth function that vanishes to infinite order at $0\in\mathbb{R}$. Now define $f:\mathbb{R}^2\to\mathbb{R}$ by requiring that $f(x,t) = 0$ when $xt\le1$ (which includes the 'initial line' $t=0$ as well as the line $x=0$), while
$$
f(x,t) = \frac{1}{x^4}\,h\left(\frac{xt-1}{x}\right)
$$
when $xt\ge1$. Then $f$ is smooth and satisfies the equation and initial conditions, but $f$ need not vanish in the regions where $xt > 1$.


