# 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) & \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.

• $f_1dt + f_2dx$ should be exact ($=d\log u$)... – Francois Ziegler Mar 18 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.