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.

  • $\begingroup$ $f_1dt + f_2dx$ should be exact ($=d\log u$)... $\endgroup$ Mar 18, 2019 at 16:22

1 Answer 1


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.


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