Let $P(u,v)$ be a smooth real function defined on some open simply connected domain in $\mathbb{R}^2$, and satisfying the PDE $P_{uv}=P$.
Consider the following system of PDE's for a function $F(u,v)$:
$$ (1) \, \, \, F_{uu}+F=P \\ (2) \, \, \, F_{uv}=F$$
Does this system always have a local solution? That is, is it true that for any smooth $P$ satisfying $P_{uv}=P$, there is a solution $F$ to the system above?
Note that the equation $P_{uv}=P$ is hyperbolic, so has many solutions locally.
Edit:
Here I am verifying some of the details in Harry49's answer:
- First, let's check that $$F(u,v) = F_0(v-u) + \int_0^u P_v(u-\tau, v - \tau) \, \text d \tau$$ form a solution to the derived equation $F_{u} + F_{v} = P_v$.
Differentiating under the integral sign, we obtain
$F_u=-F'_0(v-u)+\int_0^u \frac{\partial}{\partial u}\big(P_v(u-\tau, v - \tau)\big) \,d \tau+\left.P_v(u-\tau, v - \tau)\right|_{\tau=u}=\\-F'_0(v-u)+ \int_0^u P_{vu}(u-\tau, v - \tau) \, \text d \tau+P_v(0, v - u) \tag{3}$
and similarly
$$ F_v=F'_0(v-u)+\int_0^u \frac{\partial}{\partial v}\big(P_v(u-\tau, v - \tau)\big) \,d \tau=\\F'_0(v-u)+\int_0^u P_{vv}(u-\tau, v - \tau) \, \text d \tau \tag{4}.$$
Summing $(3),(4)$, we obtain
$$ F_u+F_v=\int_0^u P_{vu}(u-\tau, v - \tau)+P_{vv}(u-\tau, v - \tau) \, \text d \tau+P_v(0, v - u). \tag{5}$$
Now, write $G=P_v$. Then, rewrite the last equation as $$ F_u+F_v=\int_0^u G_u(u-\tau, v - \tau)+G_v(u-\tau, v - \tau) \, \text d \tau+G(0, v - u). \tag{5'}$$ Now, write $\alpha(\tau)=(u-\tau, v - \tau)$; then $$\frac{d}{d\tau}G(\alpha(\tau))=\langle \big(G_u(\alpha(\tau)),G_v(\alpha(\tau))\big), \alpha'(\tau) \rangle=-\big(G_u(\alpha(\tau))+G_v(\alpha(\tau))\big),$$ so $(5')$ becomes $$ F_u+F_v=\int_0^u G_u(\alpha(\tau))+G_v(\alpha(\tau)) \, \text d \tau+G(0, v - u)=\\ -\int_0^u \frac{d}{d\tau}G(\alpha(\tau)) \, \text d \tau+G(\alpha(u))=\\ G(\alpha(0))-G(\alpha(u))+G(\alpha(u))=G(\alpha(0))=G(u,v)=P_v(u,v). \tag{5''} $$
- Now, let's see when this $F$ solves the original system of equations $(1),(2)$ (This would give us a condition on $F_0$):
As demonstrated above, $F$ satisfies $F_{u} + F_{v} = P_v$. Differentiating w.r.t $u$, we get $F_{uu} + F_{uv} = P_{uv}=P$. One can easily show that this last equation together with either $(1)$ or $(2)$ implies the third one. (In fact any two out of these three imply the remaining equation). Thus, it suffices to check when $F_{vu}=F$.
Differentiating equation $(4)$, we get $$ F_{vu}=-F''_0(v-u)+P_{vv}(0,v-u)+\int_0^u P_{vvu}(u-\tau, v - \tau) \, \text d \tau=\\ -F''_0(v-u)+P_{vv}(0,v-u)+\int_0^u P_{v}(u-\tau, v - \tau) \, \text d \tau, \tag{6}$$
where in the last equality we have used the fact that $P_{uv}=P$ and the equality of mixed derivatives, to deduce that $P_{vvu}=P_v$.
Using equation $(6)$ and the defining formula of $F$, we see that $F_{vu}=F$ if and only if $$ F_0(v-u)+F''_0(v-u)=P_{vv}(0,v-u).$$
Writing $H(t)=P_{vv}(0,t)$, we get the equivalent ODE $$ F_0(t)+F_0''(t)=H(t).$$
Earlier (unsuccessfull) attempts:
I tried fixing the variable $v$, and then treat $F_{uu}+F=P$ as an ODE in the $u$-variable, then get a family of solutions parametrized by $v$, and plugging them into $F_{uv}=F$.
Set $G:=F_u$; then $G_u+G_v=P$ and $G_v=F$...(not sure how to continue from here).
Comment: The system $(1),(2)$ implies $P_{uv}=P$. However, $P_{uv}=P$ plus either one of the equations $(1)$ and $(2)$ do not imply the other one.
Differentiating $(1)$ w.r.t. $v$ gives $$ F_{uuv} + F_{v} = P_{v} . $$ Differentiating $(2)$ w.r.t. $u$ gives $$ F_{uvu} = F_{u} , $$ so that the equality of mixed derivatives gives $F_{u} + F_{v} = P_v$. The method of characteristics provides solutions of the form $$ F(u,v) = F_0(v-u) + \int_0^u P_v(u-\tau, v - \tau) \, \text d \tau $$ where $F_0$ is an arbitrary function. The unknown $P$ solves a second-order linear hyperbolic PDE in its canonical form $P_{uv} = P$. Some solutions of the form $P(u,v) = A e^{ku + v/k}$ can be found by separation of variables.