Two PDE for one unknown? Let $x \in (0,L)$, $t \in (0,T)$, and let $f_1 = f_1(x,t) \in \mathbb{R}$, $f_2 = f_2(x,t) \in \mathbb{R}$, $u^0 = u^0(x) \in \mathbb{R}$ and $g= g(t) \in \mathbb{R}$ be continuous functions.
My question is:

Can we find a function $u = u(x,t) \in \mathbb{R}$ that satisfies
  \begin{equation} 
\partial_t u(x,t) = u(x,t) f_1(x,t) \qquad \text{in } (0,L)\times(0,T)
\end{equation}
  and
  \begin{equation} 
\partial_x u(x,t) = u(x,t) f_2(x,t) \qquad \text{in } (0,L)\times(0,T)
\end{equation}
  with additional initial and boundary conditions:
  \begin{align*}
u(x,0) &= u^0(x) \qquad \text{for }x \in (0,L)\\
u(0,t) &= g(t) \qquad \text{for }t \in (0,T).
\end{align*}

(Here $\partial_t$ and $\partial_x$ denote the partial derivative with respect to time and space respectively.)
I had though about choosing $u$ as the solution of the transport equation
\begin{align*}
\begin{cases}
\partial_t u + \partial_x u = (f_1+f_2)u & \text{in }(0,L)\times (0,T)\\
u(x,0) = u^0(x) & \text{for }x \in (0,L)\\
u(0,t) = g(t) & \text{for }t \in (0,T).
\end{cases}
\end{align*}
However, I do not know if some supplementary assumption may allow to have $u$ satisfying both equations $\partial_t u = u f_1$ and $\partial_x u = u f_2$ separately. 
Any suggestion, reference (e.g. where a function has to satisfy two separate equations as for here), explanation of why it is/is not possible, would be welcome. Thank you.
 A: $$\partial_t u(x,t) = u(x,t) f_1(x,t) \quad\implies\quad \frac{\partial\ln|u|}{\partial t}=f_1(x,t) \tag 1$$
$$\partial_x u(x,t) = u(x,t) f_2(x,t) \quad\implies\quad \frac{\partial\ln|u|}{\partial x}=f_2(x,t) \tag2$$ 
$$\frac{\partial^2\ln|u|}{\partial x \partial t}= \frac{\partial f_1}{\partial x} = \frac{\partial f_2}{\partial t}$$
First case : $\quad \frac{\partial f_1}{\partial x} \neq \frac{\partial f_2}{\partial t}\quad$ your problem has no solution.
Second case : $\quad\frac{\partial f_1}{\partial x} = \frac{\partial f_2}{\partial t}\quad$  your problem is likely to have a solution.
From the integration of Eq.$(1)$ and with condition $u(x,0)=u^0(x)$ :
$$u(x,t)=u^0(x)\exp\left(\int_0^t f_1(x,\tau)d\tau\right)$$
From the integration of Eq.$(2)$ and with condition $u(0,t)=g(t)$ :
$$u(x,t)= g(t)\exp\left(\int_0^x f_2(\chi,t)d\chi\right)$$ 
This supposes $\quad u^0(x)=u^0(0)\:\exp\left(\int_0^x f_2(\chi,t)d\chi\right)\quad$ and $\quad g(t)=g(0)\:\exp\left(\int_0^t f_1(x,\tau)d\tau\right)$.
$$u(x,t)=u(0,0)\:\exp\left(\int_0^t f_1(x,\tau)d\tau + \int_0^x f_2(\chi,t)d\chi \right)$$
If $\quad u^0(x)\neq u^0(0)\:\exp\left(\int_0^x f_2(\chi,\tau)d\chi\right)\quad$ or $\quad g(t)\neq g(0)\:\exp\left(\int_0^t f_1(x,\tau)d\tau\right)\quad$ the problem has no solution.
