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 ($=d\log u$)... $\endgroup$ – Francois Ziegler Mar 18 '19 at 16:22

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.