Solving overdetermined, well posed, linear system of PDEs

Question

Let $f=f(u,v)$ be a (given) solution of the following PDE,

$$ \begin{equation} \frac{\partial^2 f}{\partial u\partial v}=f,\label{1}\tag{$*$} \end{equation} $$

and consider the overdetermined system(s) of PDEs

$$ \begin{cases} \dfrac{\partial x}{\partial u}=f\cos\left(u-v\right)\\ \\ \dfrac{\partial x}{\partial v}=\dfrac{\partial f}{\partial v}\sin\left(u-v\right)\\ \\ \dfrac{\partial y}{\partial u}=f\sin\left(u-v\right)\\ \\ \dfrac{\partial y}{\partial v}=-\dfrac{\partial f}{\partial v}\cos\left(u-v\right). \end{cases} $$

Since $f$ solves \eqref{1} it is guaranteed that $$ \frac{\partial^2 x}{\partial u\partial v}=\frac{\partial^2 x}{\partial v\partial u} $$ and similarly for $y$, so the system is well posed. The question is, is it possible to write down a general solution for $x(u,v)$ and $y(u,v)$ in terms of integrals of $f$ ?

If I try, for instance, to integrate the first equation to get $$ x=\int \mathrm{d}u\;f\cos(u-v)+g(v), $$ with $g(v)$ some unknown function of $v$, and then plug in the second equation for $x$, I cannot solve for $g(v)$, so I need to try some other ansatz.

Answer 1

You can do this as follows: Start with any values for $(x,y)$ at $(0,0)$, say $$ (x,y)(0,0) = (x_0,y_0). $$ Integrate the first and third equations along the line $v = 0$: \begin{align*} x(u,0) &= x_0 + \int_{s=0}^{s=u} f(s,0)\cos(s)\,ds\\ y(u,0) &= y_0 + \int_{s=0}^{s=u} f(s,0)\sin(s)\,ds \end{align*} Next, for each $u$, integrate the second and fourth equations along the line $(u,\cdot)$ to get your solution: \begin{align*} x(u,v) &= x(u,0) + \int_{t=0}^{t=v} \frac{\partial f}{\partial v}(u,t)\sin(u-t)\,dt\\\ y(u,v) &= y(u,0) + \int_{t=0}^{t=v} \frac{\partial f}{\partial u}(u,t)\cos(u-t)\,dt. \end{align*}