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Let $x in (0,L)$, $t in (0,T)$, and let $f_1 = f_1(x,t) in mathbbR$, $f_2 = f_2(x,t) in mathbbR$, $u^0 = u^0(x) in mathbbR$ and $g= g(t) in mathbbR$ be continuous functions.

My question is:

Can we find a function $u = u(x,t) in mathbbR$ that satisfies
beginequation
partial_t u(x,t) = u(x,t) f_1(x,t) qquad textin (0,L)times(0,T)
endequation
and
beginequation
partial_x u(x,t) = u(x,t) f_2(x,t) qquad textin (0,L)times(0,T)
endequation
with additional initial and boundary conditions:
beginalign*
u(x,0) &= u^0(x) qquad textfor x in (0,L)\
u(0,t) &= g(t) qquad textfor t in (0,T).
endalign*

(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
beginalign*
begincases
partial_t u + partial_x u = (f_1+f_2)u & textin (0,L)times (0,T)\
u(x,0) = u^0(x) & textfor x in (0,L)\
u(0,t) = g(t) & textfor t in (0,T).
endcases
endalign*
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.

$$partial_t u(x,t) = u(x,t) f_1(x,t) quadimpliesquad fracpartiallnpartial t=f_1(x,t) tag 1$$

$$partial_x u(x,t) = u(x,t) f_2(x,t) quadimpliesquad fracpartiallnpartial x=f_2(x,t) tag2$$

$$fracupartial x partial t= fracpartial f_1partial x = fracpartial f_2partial t$$

First case : $quad fracpartial f_1partial x neq fracpartial f_2partial tquad$ your problem has no solution.

Second case : $quadfracpartial f_1partial x = fracpartial f_2partial tquad$ your problem is likely to have a solution.

From the integration of Eq.$(1)$ and with condition $u(x,0)=u^0(x)$ :
$$u(x,t)=u^0(x)expleft(int_0^t f_1(x,tau)dtauright)$$
From the integration of Eq.$(2)$ and with condition $u(0,t)=g(t)$ :
$$u(x,t)= g(t)expleft(int_0^x f_2(chi,t)dchiright)$$

This supposes $quad u^0(x)=u^0(0):expleft(int_0^x f_2(chi,t)dchiright)quad$ and $quad g(t)=g(0):expleft(int_0^t f_1(x,tau)dtauright)$.

$$u(x,t)=u(0,0):expleft(int_0^t f_1(x,tau)dtau + int_0^x f_2(chi,t)dchi right)$$

If $quad u^0(x)neq u^0(0):expleft(int_0^x f_2(chi,tau)dchiright)quad$ or $quad g(t)neq g(0):expleft(int_0^t f_1(x,tau)dtauright)quad$ the problem has no solution.