For an arbitrary $G(x,t)$, does $f_t=2G_xf+Gf_x$, $f(x,0)=0$ have a unique solution for $f$? 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 $(x,t)\in\mathbb R^2$?

EDIT: I asked this question on MathOverflow, and I got a correct answer. It turns out it is not necessarily true that $f(x,t)=0$ for all $(x,t)$.
 A: Regards @DarrenOng . I would like to contribute. The PDE may be viewed as a 1D conservation law with a source function :
$$  f_{t} + F(x,t)_{x} = S(x,t)   $$ 
with
$$ F(x,t) = G(x,t)f(x,t), \:\:\: \text{and} \:\: S(x,t) = f(x,t) G_{x}(x,t)$$
$F$ here is the flux function, and $S$ is the source function. Now the conservation law (the PDE, which is in local form) is derived from the integral form (or can be called the global conservation law) :
$$ \frac{d}{dt} \left( \int_{a}^{b} f(x,t) dx \right) = F(a,t) - F(b,t) + \int_{a}^{b} S(x,t) dx $$
Continuing this global form, we get :
\begin{align*} 
\frac{d}{dt} \left( \int_{a}^{b} f(x,t) dx \right) &= - \int_{a}^{b} F_{x}(x,t) dx + \int_{a}^{b} S(x,t) dx \\ &= - \int_{a}^{b} -(G(x,t)f(x,t))_{x} dx + \int_{a}^{b} f(x,t) G_{x} dx 
\end{align*}
The total value : $\int f(x,t) dx$ will change iff the right-hand side of the global form is not zero. Since $f(x,0) = 0$, we have
\begin{align*} 
\frac{d}{dt} \left( \int_{a}^{b} f(x,t) dx \right) |_{t=0}  &= 0 
\end{align*}

Notice that 
  \begin{align*}  \frac{d}{dt} \left(\int_{a}^{b} f(x,t) dx \right) &=  \left( \int_{a}^{b} f_{t}(x,t)dx\right) \end{align*}  so \begin{align*}  \frac{d}{dt} \left( \int_{a}^{b} f(x,t) dx \right) |_{t=0} &= 0 \\ \left( \int_{a}^{b} f_{t}(x,0)dx \right) &= 0 \\ f_{t}(x,0) &= 0 \end{align*}
  The last one is obtained because the integral applies for any value of $a$ and $b$. Using this result, and write $f(x,t)$ as : 
  $$ f(x, t) = f(x,0) + f_{t}(x,0) t +
 \frac{f_{tt}(x,0) t^{2}}{2!} +.... $$
  we get $ f(x,t) = 0 $, for any $t>0$.
Nonrigorous proof : $f_{t}(x,0)=0$, $f(x,0)=0$, are enough to ensure that $f(x,t)=0$ for all $t$. This is because : if $f(x,0)=0$,
  then $f_{x}(x,0)=0$ and $f_{t}(x,0) = 0$. Now since $f_{t}(x,0)$
  (which means there is no change), then what is the value of $f$
  after $t=0$, say $f(x,0+\delta t)$? of course it will be the same as
  before, $f(x,\delta t)=f(x,0)=0$ because there is no change. Now we
  will have again $f_{t}(x,\delta t) = 2G_{x}f(x,\delta t) + f_{x}(x,\delta t) G = 0$, which means there is no change again.
  This will continue to happen. 
This makes sense. But for me its difficult to show the rigorous
  proof...

My conclusion is $f(x,t)=0$ for $t \ge 0 $. Thanks.
