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.

  • 1
    $\begingroup$ Are these really the equations you want to consider? It would look more nice and symmetric if the derivatives of $x,y$ with respect to $u$ involved derivatives of $f$ with respect to $u$ just as the derivatives of $x,y$ with respect to $v$ involve derivatives of $f$ with respect to $v$. In that case it seems really promising to complexify the equation by defining $z \equiv x + i y, w \equiv u + i v$. $\endgroup$
    – David
    Jul 17, 2021 at 22:06

1 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*}

  • $\begingroup$ thanks for the analysis. I would like to observe the following. If we take $x(u,v)-x_{0}=\int^{u}dt\:f(t,v)\cos(t-v)$ then it satisfies the 1st eqtn. Now, $\partial x(u,v)/\partial v=\int^{u}dt\:\partial\left(f(t,v)\cos(t-v)\right)/\partial v=\int^{u}dt\:\partial\Big(\partial f(t,v)/\partial v\sin(t-v)\Big)/\partial t=\partial f(u,v)/\partial v\sin(u-v)$, then both equations are satisfied (similarly for $y(u,v)$) so by construction the solution should be just $x(u,v)-x_{0}=\int^{u}dt\:f(t,v)\cos(t-v)$, isn't it ? (By $\int^{u}$ I mean integration without evaluating a lower limit). $\endgroup$ Jul 20, 2021 at 16:30
  • $\begingroup$ I don't know what you mean by integration without evaluating a lower limit. That's not a well-defined function. $\endgroup$
    – Deane
    Jul 20, 2021 at 17:39
  • $\begingroup$ I mean $\int^t dt' \:\partial f(t',s)/\partial t'\equiv f(t,s)$. (This notation is, by the way, a common practice in physics) $\endgroup$ Jul 20, 2021 at 18:09
  • $\begingroup$ This appears to be the indefinite integral, which means that the right side should be written $f(t,s) + C$, since adding any constant to $f$ also gives an antiderivative of $\partial f(t',s)/\partial t'$. The value of this constant is important in the formulas for the solution. $\endgroup$
    – Deane
    Jul 20, 2021 at 19:16
  • $\begingroup$ exactly, so I'm setting this value of $C$ to zero in my previous comment. My reasoning is this: the solution is unique up to an additive constante, so if I manage to find $one$ solution this is $the$ solution. Now, by the Leibniz rule the solution in my previous comment seems to satisfy the system, but I'm trying to check if this is the same that you kindly sketched. $\endgroup$ Jul 20, 2021 at 19:19

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