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) & \text{for }x \in (0,L)\\
> u(0,t) &= g(t) & \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.


I [also posted this question on Mathematics Stack Exchange](https://math.stackexchange.com/q/3152541/228959).



Two PDE for one unknown?
A solution exists if and only if the following compatibility conditions are satisfied:
$$\partial\_xf\_1=\partial\_tf\_2,\quad g'(t)=g(t)f\_1(0,t),\quad u\_0'(x)=u\_0(x)f\_2(x,0).$$
For the existence, you can solve the Cauchy problem in $x$, then that in $t$, and verify that both solutions coincide.


