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 $(x,t)$.

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.

• Thanks @Arief But could you explain why $\frac{d}{dt} \left( \int_{a}^{b} f(x,t) dx \right) |_{t=0} = 0$ is enough? Just because the time derivative is zero at $t=0$ doesn't imply that $\int_{a}^{b} f(x,t) dx \ =0$ for all $t$, right? – Darren Ong Jun 2 '17 at 4:26
• @DarrenOng I edited the answer. I presume that this is due to the cancellation in the source term : $S = f(x,t) G_{x}$ because $f(x,0)=0$. If the source does not have $f(x,t)$ as factor, then it is not necessarily that $f(x,t)= 0$. Since at first $f(x,0)=0$, then $f$ will increase or decrease only if there is a source. But the source depends on $f(x,t)$, so at each time the source will cancel out with 0. – Arief Anbiya Jun 2 '17 at 15:22
• thanks for the edits @Arief! The answer looks clearer now. But I think you are assuming that $f(x,t)$ is analytic in $t$, when it is actually just smooth in $t$ (for example, let's say $f(x,t)=\exp(-1/t^2)$), then $f$ is smooth but not analytic). If $f(x,t)$ is not analytic, it does not necessarily equal to its Taylor series, so I don't think we can conclude that $f(x,t)=0$ for any $t>0$. – Darren Ong Jun 3 '17 at 12:41
• @DarrenOng I added another explanation. But $f = exp(-1/t^{2})$ is not a solution, because $f(x,0)$ must be 0. – Arief Anbiya Jun 3 '17 at 17:16
• Thanks for the update @Arief! I am afraid that the assertion "$f_{t}(x,0)=0$, $f(x,0)=0$, are enough to ensure that $f(x,t)=0$ for all $t$" cannot be true. Take for example the function $f$ defined by $f(x,t)=exp(-1/t^2)$ for $t\neq 0$, and $f(x,0)=0$. This function is smooth everywhere, including at $t=0$. However, note that even though $\dfrac{\partial}{\partial t^n} f(x,0)=0$ for every positive integer $n$, it is still not true that $f(x,t)=0$ for all $t>0$. – Darren Ong Jun 5 '17 at 1:19